Alan Jeffrey University of Newcastle-upon-Tyne

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To Lisl and our family

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C O N T E N T S

Preface

PART ONE

CHAPTER

1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9

1.10 1.11 1.12 1.13 1.14

xv

REVIEW MATERIAL

1

Review of Prerequisites

3

Real Numbers, Mathematical Induction, and Mathematical Conventions 4 Complex Numbers 10 The Complex Plane 15 Modulus and Argument Representation of Complex Numbers 18 Roots of Complex Numbers 22 Partial Fractions 27 Fundamentals of Determinants 31 Continuity in One or More Variables 35 Differentiability of Functions of One or More Variables 38 Tangent Line and Tangent Plane Approximations to Functions 40 Integrals 41 Taylor and Maclaurin Theorems 43 Cylindrical and Spherical Polar Coordinates and Change of Variables in Partial Differentiation 46 Inverse Functions and the Inverse Function Theorem 49

vii

PART TWO

CHAPTER

2 2.1 2.2 2.3 2.4

CHAPTER

Vectors and Vector Spaces

55

2.5 2.6 2.7

3

Matrices and Systems of Linear Equations

3.5 3.6 3.7 3.8 3.9 3.10

4 4.1 4.2 4.3 4.4 4.5

viii

53

Vectors, Geometry, and Algebra 56 The Dot Product (Scalar Product) 70 The Cross Product (Vector Product) 77 Linear Dependence and Independence of Vectors and Triple Products 82 n -Vectors and the Vector Space R n 88 Linear Independence, Basis, and Dimension 95 Gram–Schmidt Orthogonalization Process 101

3.1 3.2 3.3 3.4

CHAPTER

VECTORS AND MATRICES

105

Matrices 106 Some Problems That Give Rise to Matrices 120 Determinants 133 Elementary Row Operations, Elementary Matrices, and Their Connection with Matrix Multiplication 143 The Echelon and Row-Reduced Echelon Forms of a Matrix 147 Row and Column Spaces and Rank 152 The Solution of hom*ogeneous Systems of Linear Equations 155 The Solution of Nonhom*ogeneous Systems of Linear Equations 158 The Inverse Matrix 163 Derivative of a Matrix 171

Eigenvalues, Eigenvectors, and Diagonalization Characteristic Polynomial, Eigenvalues, and Eigenvectors 178 Diagonalization of Matrices 196 Special Matrices with Complex Elements Quadratic Forms 210 The Matrix Exponential 215

205

177

PART THREE

CHAPTER

5 5.1 5.2

5.3 5.4 5.5 5.6 5.7 5.8 5.9 5.10

CHAPTER

6 6.1 6.2 6.3 6.4 6.5 6.6 6.7

6.8 6.9 6.10 6.11 6.12

ORDINARY DIFFERENTIAL EQUATIONS

225

First Order Differential Equations

227

Background to Ordinary Differential Equations Some Problems Leading to Ordinary Differential Equations 233 Direction Fields 240 Separable Equations 242 hom*ogeneous Equations 247 Exact Equations 250 Linear First Order Equations 253 The Bernoulli Equation 259 The Riccati Equation 262 Existence and Uniqueness of Solutions 264

228

Second and Higher Order Linear Differential Equations and Systems

269

hom*ogeneous Linear Constant Coefﬁcient Second Order Equations 270 Oscillatory Solutions 280 hom*ogeneous Linear Higher Order Constant Coefﬁcient Equations 291 Undetermined Coefﬁcients: Particular Integrals 302 Cauchy–Euler Equation 309 Variation of Parameters and the Green’s Function 311 Finding a Second Linearly Independent Solution from a Known Solution: The Reduction of Order Method 321 Reduction to the Standard Form u + f (x)u = 0 324 Systems of Ordinary Differential Equations: An Introduction 326 A Matrix Approach to Linear Systems of Differential Equations 333 Nonhom*ogeneous Systems 338 Autonomous Systems of Equations 351 ix

CHAPTER

7 7.1 7.2 7.3 7.4

CHAPTER

8 8.1 8.2

8.3 8.4 8.5 8.6 8.7 8.8 8.9 8.10 8.11

PART FOUR

CHAPTER

9 9.1 9.2 9.3 9.4 9.5 9.6

x

The Laplace Transform Laplace Transform: Fundamental Ideas 379 Operational Properties of the Laplace Transform 390 Systems of Equations and Applications of the Laplace Transform 415 The Transfer Function, Control Systems, and Time Lags

379

437

Series Solutions of Differential Equations, Special Functions, and Sturm–Liouville Equations A First Approach to Power Series Solutions of Differential Equations 443 A General Approach to Power Series Solutions of hom*ogeneous Equations 447 Singular Points of Linear Differential Equations 461 The Frobenius Method 463 The Gamma Function Revisited 480 Bessel Function of the First Kind Jn(x) 485 Bessel Functions of the Second Kind Yν (x) 495 Modiﬁed Bessel Functions I ν (x) and K ν (x) 501 A Critical Bending Problem: Is There a Tallest Flagpole? Sturm–Liouville Problems, Eigenfunctions, and Orthogonality 509 Eigenfunction Expansions and Completeness 526

443

504

FOURIER SERIES, INTEGRALS, AND THE FOURIER TRANSFORM

543

Fourier Series

545

Introduction to Fourier Series 545 Convergence of Fourier Series and Their Integration and Differentiation 559 Fourier Sine and Cosine Series on 0 ≤ x ≤ L 568 Other Forms of Fourier Series 572 Frequency and Amplitude Spectra of a Function 577 Double Fourier Series 581

CHAPTER

10 10.1 10.2 10.3

PART FIVE

CHAPTER

11 11.1 11.2 11.3 11.4 11.5 11.6

CHAPTER

12 12.1 12.2 12.3 12.4

PART SIX

CHAPTER

13 13.1 13.2 13.3 13.4

Fourier Integrals and the Fourier Transform The Fourier Integral 589 The Fourier Transform 595 Fourier Cosine and Sine Transforms

589

611

VECTOR CALCULUS

623

Vector Differential Calculus

625

Scalar and Vector Fields, Limits, Continuity, and Differentiability 626 Integration of Scalar and Vector Functions of a Single Real Variable 636 Directional Derivatives and the Gradient Operator Conservative Fields and Potential Functions 650 Divergence and Curl of a Vector 659 Orthogonal Curvilinear Coordinates 665

644

Vector Integral Calculus Background to Vector Integral Theorems 678 Integral Theorems 680 Transport Theorems 697 Fluid Mechanics Applications of Transport Theorems

677

704

COMPLEX ANALYSIS

709

Analytic Functions

711

Complex Functions and Mappings 711 Limits, Derivatives, and Analytic Functions 717 Harmonic Functions and Laplace’s Equation 730 Elementary Functions, Inverse Functions, and Branches 735

xi

CHAPTER

14 14.1 14.2 14.3 14.4

CHAPTER

15 15.1 15.2 15.3 15.4 15.5

CHAPTER

16 16.1

CHAPTER

17 17.1 17.2

PART SEVEN

CHAPTER

18 18.1 18.2 18.3 18.4

xii

Complex Integration

745

Complex Integrals 745 Contours, the Cauchy–Goursat Theorem, and Contour Integrals 755 The Cauchy Integral Formulas 769 Some Properties of Analytic Functions 775

Laurent Series, Residues, and Contour Integration Complex Power Series and Taylor Series 791 Uniform Convergence 811 Laurent Series and the Classiﬁcation of Singularities 816 Residues and the Residue Theorem 830 Evaluation of Real Integrals by Means of Residues

791

839

The Laplace Inversion Integral The Inversion Integral for the Laplace Transform

Conformal Mapping and Applications to Boundary Value Problems

863 863

877

Conformal Mapping 877 Conformal Mapping and Boundary Value Problems 904

PARTIAL DIFFERENTIAL EQUATIONS

925

Partial Differential Equations

927

What Is a Partial Differential Equation? 927 The Method of Characteristics 934 Wave Propagation and First Order PDEs 942 Generalizing Solutions: Conservation Laws and Shocks 951

18.5 18.6

18.7 18.8 18.9 18.10 18.11 18.12

PART EIGHT

CHAPTER

19 19.1 19.2 19.3 19.4 19.5 19.6 19.7

The Three Fundamental Types of Linear Second Order PDE 956 Classiﬁcation and Reduction to Standard Form of a Second Order Constant Coefﬁcient Partial Differential Equation for u(x, y) 964 Boundary Conditions and Initial Conditions 975 Waves and the One-Dimensional Wave Equation 978 The D’Alembert Solution of the Wave Equation and Applications 981 Separation of Variables 988 Some General Results for the Heat and Laplace Equation 1025 An Introduction to Laplace and Fourier Transform Methods for PDEs 1030

NUMERICAL MATHEMATICS

1043

Numerical Mathematics

1045

Decimal Places and Signiﬁcant Figures 1046 Roots of Nonlinear Functions 1047 Interpolation and Extrapolation 1058 Numerical Integration 1065 Numerical Solution of Linear Systems of Equations 1077 Eigenvalues and Eigenvectors 1090 Numerical Solution of Differential Equations 1095

Answers 1109 References 1143 Index 1147

xiii

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P R E F A C E

T

his book has evolved from lectures on engineering mathematics given regularly over many years to students at all levels in the United States, England, and elsewhere. It covers the more advanced aspects of engineering mathematics that are common to all ﬁrst engineering degrees, and it differs from texts with similar names by the emphasis it places on certain topics, the systematic development of the underlying theory before making applications, and the inclusion of new material. Its special features are as follows.

Prerequisites

T

he opening chapter, which reviews mathematical prerequisites, serves two purposes. The ﬁrst is to refresh ideas from previous courses and to provide basic self-contained reference material. The second is to remove from the main body of the text certain elementary material that by tradition is usually reviewed when ﬁrst used in the text, thereby allowing the development of more advanced ideas to proceed without interruption.

Worked Examples

T

he numerous worked examples that follow the introduction of each new idea serve in the earlier chapters to illustrate applications that require relatively little background knowledge. The ability to formulate physical problems in mathematical terms is an essential part of all mathematics applications. Although this is not a text on mathematical modeling, where more complicated physical applications are considered, the essential background is ﬁrst developed to the point at which the physical nature of the problem becomes clear. Some examples, such as the ones involving the determination of the forces acting in the struts of a framed structure, the damping of vibrations caused by a generator and the vibrational modes of clamped membranes, illustrate important mathematical ideas in the context of practical applications. Other examples occur without speciﬁc applications and their purpose is to reinforce new mathematical ideas and techniques as they arise. A different type of example is the one that seeks to determine the height of the tallest ﬂagpole, where the height limitation is due to the phenomenon of xv

buckling. Although the model used does not give an accurate answer, it provides a typical example of how a mathematical model is constructed. It also illustrates the reasoning used to select a physical solution from a scenario in which other purely mathematical solutions are possible. In addition, the example demonstrates how the choice of a unique physically meaningful solution from a set of mathematically possible ones can sometimes depend on physical considerations that did not enter into the formulation of the original problem.

Exercise Sets

T

he need for engineering students to have a sound understanding of mathematics is recognized by the systematic development of the underlying theory and the provision of many carefully selected fully worked examples, coupled with their reinforcement through the provision of large sets of exercises at the ends of sections. These sets, to which answers to odd-numbered exercises are listed at the end of the book, contain many routine exercises intended to provide practice when dealing with the various special cases that can arise, and also more challenging exercises, each of which is starred, that extend the subject matter of the text in different ways. Although many of these exercises can be solved quickly by using standard computer algebra packages, the author believes the fundamental mathematical ideas involved are only properly understood once a signiﬁcant number of exercises have ﬁrst been solved by hand. Computer algebra can then be used with advantage to conﬁrm the results, as is required in various exercise sets. Where computer algebra is either required or can be used to advantage, the exercise numbers are in blue. A comparison of computer-based solutions with those obtained by hand not only conﬁrms the correctness of hand calculations, but also serves to illustrate how the method of solution often determines its form, and that transforming one form of solution to another is sometimes difﬁcult. It is the author’s belief that only when fundamental ideas are fully understood is it safe to make routine use of computer algebra, or to use a numerical package to solve more complicated problems where the manipulation involved is prohibitive, or where a numerical result may be the only form of solution that is possible.

New Material

T

ypical of some of the new material to be found in the book is the matrix exponential and its application to the solution of linear systems of ordinary differential equations, and the use of the Green’s function. The introductory discussion of the development of discontinuous solutions of ﬁrst order quasilinear equations, which are essential in the study of supersonic gas ﬂow and in various other physical applications, is also new and is not to be found elsewhere. The account of the Laplace transform contains more detail than usual. While the Laplace transform is applied to standard engineering problems, including

xvi

control theory, various nonstandard problems are also considered, such as the solution of a boundary value problem for the equation that describes the bending of a beam and the derivation of the Laplace transform of a function from its differential equation. The chapter on vector integral calculus ﬁrst derives and then applies two fundamental vector transport theorems that are not found in similar texts, but which are of considerable importance in many branches of engineering.

Series Solutions of Differential Equations

U

nderstanding the derivation of series solutions of ordinary differential equations is often difﬁcult for students. This is recognized by the provision of detailed examples, followed by carefully chosen sets of exercises. The worked examples illustrate all of the special cases that can arise. The chapter then builds on this by deriving the most important properties of Legendre polynomials and Bessel functions, which are essential when solving partial differential equations involving cylindrical and spherical polar coordinates.

Complex Analysis

B

ecause of its importance in so many different applications, the chapters on complex analysis contain more topics than are found in similar texts. In particular, the inclusion of an account of the inversion integral for the Laplace transform makes it possible to introduce transform methods for the solution of problems involving ordinary and partial differential equations for which tables of transform pairs are inadequate. To avoid unnecessary complication, and to restrict the material to a reasonable length, some topics are not developed with full mathematical rigor, though where this occurs the arguments used will sufﬁce for all practical purposes. If required, the account of complex analysis is sufﬁciently detailed for it to serve as a basis for a single subject course.

Conformal Mapping and Boundary Value Problems

S

ufﬁcient information is provided about conformal transformations for them to be used to provide geometrical insight into the solution of some fundamental two-dimensional boundary value problems for the Laplace equation. Physical applications are made to steady-state temperature distributions, electrostatic problems, and ﬂuid mechanics. The conformal mapping chapter also provides a quite different approach to the solution of certain two-dimensional boundary value problems that in the subsequent chapter on partial differential equations are solved by the very different method of separation of variables.

xvii

Partial Differential Equations

A

n understanding of partial differential equations is essential in all branches of engineering, but accounts in engineering mathematics texts often fall short of what is required. This is because of their tendency to focus on the three standard types of linear second order partial differential equations, and their solution by means of separation of variables, to the virtual exclusion of ﬁrst order equations and the systems from which these fundamental linear second order equations are derived. Often very little is said about the types of boundary and initial conditions that are appropriate for the different types of partial differential equations. Mention is seldom if ever made of the important part played by nonlinearity in ﬁrst order equations and the way it inﬂuences the properties of their solutions. The account given here approaches these matters by starting with ﬁrst order linear and quasilinear equations, where the way initial and boundary conditions and nonlinearity inﬂuence solutions is easily understood. The discussion of the effects of nonlinearity is introduced at a comparatively early stage in the study of partial differential equations because of its importance in subjects like ﬂuid mechanics and chemical engineering. The account of nonlinearity also includes a brief discussion of shock wave solutions that are of fundamental importance in both supersonic gas ﬂow and elsewhere. Linear and nonlinear wave propagation is examined in some detail because of its considerable practical importance; in addition, the way integral transform methods can be used to solve linear partial differential equations is described. From a rigorous mathematical point of view, the solution of a partial differential equation by the method of separation of variables only yields a formal solution, which only becomes a rigorous solution once the completeness of any set of eigenfunctions that arises has been established. To develop the subject in this manner would take the text far beyond the level for which it is intended and so the completeness of any set of eigenfunctions that occurs will always be assumed. This assumption can be fully justiﬁed when applying separation of variables to the applications considered here and also in virtually all other practical cases.

Technology Projects

T

o encourage the use of technology and computer algebra and to broaden the range of problems that can be considered, technology-based projects have been added wherever appropriate; in addition, standard sets of exercises of a theoretical nature have been included at the ends of sections. These projects are not linked to a particular computer algebra package: Some projects illustrating standard results are intended to make use of simple computer skills while others provide insight into more advanced and physically important theoretical questions. Typical of the projects designed to introduce new ideas are those at the end of the chapter on partial differential equations, which offer a brief introduction to the special nonlinear wave solutions called solitons.

xviii

Numerical Mathematics

A

lthough an understanding of basic numerical mathematics is essential for all engineering students, in a book such as this it is impossible to provide a systematic account of this important discipline. The aim of this chapter is to provide a general idea of how to approach and deal with some of the most important and frequently encountered numerical operations, using only basic numerical techniques, and thereafter to encourage the use of standard numerical packages. The routines available in numerical packages are sophisticated, highly optimized and efﬁcient, but the general ideas that are involved are easily understood once the material in the chapter has been assimilated. The accounts that are given here purposely avoid going into great detail as this can be found in the quoted references. However, the chapter does indicate when it is best to use certain types of routine and those circ*mstances where routines might be inappropriate. The details of references to literature contained in square brackets at the ends of sections are listed at the back of the book with suggestions for additional reading. An instructor’s Solutions Manual that gives outline solutions for the technology projects is also available.

Acknowledgments

I

wish to express my sincere thanks to the reviewers and accuracy readers, those cited below and many who remain anonymous, whose critical comments and suggestions were so valuable, and also to my many students whose questions when studying the material in this book have contributed so fundamentally to its development. Particular thanks go to: Chun Liu, Pennsylvania State University William F. Moss, Clemson University Donald Hartig, California Polytechnic State University at San Luis Obispo Howard A. Stone, Harvard University Donald Estep, Georgia Institute of Technology Preetham B. Kumar, California State University at Sacramento Anthony L. Peressini, University of Illinois at Urbana-Champaign Eutiquio C. Young, Florida State University Colin H. Marks, University of Maryland Ronald Jodoin, Rochester Institute of Technology Edgar Pechlaner, Simon Fraser University Ronald B. Guenther, Oregon State University Mattias Kawski, Arizona State University L. F. Shampine, Southern Methodist University In conclusion, I also wish to thank my editor, Barbara Holland, for her invaluable help and advice on presentation; Julie Bolduc, senior production editor, for her patience and guidance; Mike Sugarman, for his comments during the early stages of writing; and, ﬁnally, Chuck Glaser, for encouraging me to write the book in the ﬁrst place.

xix

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PART

ONE

REVIEW MATERIAL

Chapter

1

Review of Prerequisites

1

This Page Intentionally Left Blank

1

C H A P T E R

Review of Prerequisites

E

very account of advanced engineering mathematics must rely on earlier mathematics courses to provide the necessary background. The essentials are a ﬁrst course in calculus and some knowledge of elementary algebraic concepts and techniques. The purpose of the present chapter is to review the most important of these ideas that have already been encountered, and to provide for convenient reference results and techniques that can be consulted later, thereby avoiding the need to interrupt the development of subsequent chapters by the inclusion of review material prior to its use. Some basic mathematical conventions are reviewed in Section 1.1, together with the method of proof by mathematical induction that will be required in later chapters. The essential algebraic operations involving complex numbers are summarized in Section 1.2, the complex plane is introduced in Section 1.3, the modulus and argument representation of complex numbers is reviewed in Section 1.4, and roots of complex numbers are considered in Section 1.5. Some of this material is required throughout the book, though its main use will be in Part 5 when developing the theory of analytic functions. The use of partial fractions is reviewed in Section 1.6 because of the part they play in Chapter 7 in developing the Laplace transform. As the most basic properties of determinants are often required, the expansion of determinants is summarized in Section 1.7, though a somewhat fuller account of determinants is to be found later in Section 3.3 of Chapter 3. The related concepts of limit, continuity, and differentiability of functions of one or more independent variables are fundamental to the calculus, and to the use that will be made of them throughout the book, so these ideas are reviewed in Sections 1.8 and 1.9. Tangent line and tangent plane approximations are illustrated in Section 1.10, and improper integrals that play an essential role in the Laplace and Fourier transforms, and also in complex analysis, are discussed in Section 1.11. The importance of Taylor series expansions of functions involving one or more independent variables is recognized by their inclusion in Section 1.12. A brief mention is also made of the two most frequently used tests for the convergence of series, and of the differentiation and integration of power series that is used in Chapter 8 when considering series solutions of linear ordinary differential equations. These topics are considered again in Part 5 when the theory of analytic functions is developed. The solution of many problems involving partial differential equations can be simpliﬁed by a convenient choice of coordinate system, so Section 1.13 reviews the theorem for the

3

4

Chapter 1

Review of Prerequisites change of variable in partial differentiation, and describes the cylindrical polar and spherical polar coordinate systems that are the two that occur most frequently in practical problems. Because of its fundamental importance, the implicit function theorem is stated without proof in Section 1.14, though it is not usually mentioned in ﬁrst calculus courses.

1.1

Real Numbers, Mathematical Induction, and Mathematical Conventions

N

umbers are fundamental to all mathematics, and real numbers are a subset of complex numbers. A real number can be classiﬁed as being an integer, a rational number, or an irrational number. From the set of positive and negative integers, and zero, the set of positive integers 1, 2, 3, . . . is called the set of natural numbers. The rational numbers are those that can be expressed in the √ form m/n, where m and n are integers with n = 0. Irrational numbers such as π , 2, and sin 2 are numbers that cannot be expressed in rational form, so, for example, for no √ integers m and n is it true that 2 is equal to m/n. Practical calculations can only be performed using rational numbers, so all irrational numbers that arise must be approximated arbitrarily closely by rational numbers. Collectively, the sets of integers and rational and irrational numbers form what is called the set of all real numbers, and this set is denoted by R. When it is necessary to indicate that an arbitrary number a is a real number a shorthand notation is adopted involving the symbol ∈, and we will write a ∈ R. The symbol ∈ is to be read “belongs to” or, more formally, as “is an element of the set.” If a is not a member of set R, the symbol ∈ is negated by writing ∈, / and we will write a ∈ / R where, of course, the symbol ∈ / is to be read as “does not belong to,” or “is not an element of the set.” As real numbers can be identiﬁed in a unique manner with points on a line, the set of all real numbers R is often called the real line. The set of all complex numbers C to which R belongs will be introduced later. One of the most important properties of real numbers that distinguishes them from other complex numbers is that they can be arranged in numerical order. This fundamental property is expressed by saying that the real numbers possess the order property. This simply means that if x, y ∈ R, with x = y, then either x < y or

x > y,

where the symbol < is to be read “is less than” and the symbol > is to be read “is greater than.” When the foregoing results are expressed differently, though equivalently, if x, y ∈ R, with x = y, then either x − y < 0

absolute value

or

x − y > 0.

It is the order property that enables the graph of a real function f of a real variable x to be constructed. This follows because once length scales have been chosen for the axes together with a common origin, a real number can be made to correspond to a unique point on an axis. The graph of f follows by plotting all possible points (x, f (x)) in the plane, with x measured along one axis and f (x) along the other axis. The absolute value |x| of a real number x is deﬁned by the formula x if x ≥ 0 |x| = −x if x < 0.

Section 1.1

Real Numbers, Mathematical Induction, and Mathematical Conventions

5

This form of deﬁnition is in reality a concise way of expressing two separate statements. One statement is obtained by reading |x| with the top condition on the right and the other by reading it with the bottom condition on the right. The absolute value of a real number provides a measure of its magnitude without regard to its sign so, for example, |3| = 3, |−7.41| = 7.41, and |0| = 0. Sometimes the form of a general mathematical result that only depends on an arbitrary natural number n can be found by experiment or by conjecture, and then the problem that remains is how to prove that the result is either true or false for all n. A typical example is the proposition that the product (1 − 1/4)(1 − 1/9)(1 − 1/16) . . . [1 − 1/(n + 1)2 ] = (n + 2)/(2n + 2),

mathematical induction

for n = 1, 2, . . . .

This assertion is easily checked for any speciﬁc positive integer n, but this does not amount to a proof that the result is true for all natural numbers. A powerful method by which such propositions can often be shown to be either true or false involves using a form of argument called mathematical induction. This type of proof depends for its success on the order property of numbers and the fact that if n is a natural number, then so also is n + 1. The steps involved in an inductive proof can be summarized as follows.

Proof by Mathematical Induction Let P(n) be a proposition depending on a positive integer n. STEP 1 STEP 2 STEP 3 STEP 4

Show, if possible, that P(n) is true for some positive integer n0 . Show, if possible, that if P(n) is true for an arbitrary integer n = k ≥ n0 , then the proposition P(k + 1) follows from proposition P(k). If Step 2 is true, the fact that P(n0 ) is true implies that P(n0 + 1) is true, and then that P(n0 + 2) is true, and hence that P(n) is true for all n ≥ n0 . If no number n = n0 can be found for which Step 1 is true, or if in Step 2 it can be shown that P(k) does not imply P(k + 1), the proposition P(n) is false.

The example that follows is typical of the situation where an inductive proof is used. It arises when determining the nth term in the Maclaurin series for sin ax that involves ﬁnding the nth derivative of sin ax. A result such as this may be found intuitively by inspection of the ﬁrst few derivatives, though this does not amount to a formal proof that the result is true for all natural numbers n. EXAMPLE 1.1

Prove by mathematical induction that dn /dx n [sin ax] = a n sin(ax + nπ/2),

for n = 1, 2, . . . .

Solution The proposition P(n) is that dn /dx n [sin ax] = a n sin(ax + nπ/2), STEP 1

for n = 1, 2, . . . .

Differentiation gives d/dx[sin ax] = a cos ax,

6

Chapter 1

Review of Prerequisites

but setting n = 1 in P(n) leads to the result d/dx[sin ax] = a sin(ax + π/2) = a cos ax, showing that proposition P(n) is true for n = 1 (so in this case n0 = 1). STEP 2 Assuming P(k) to be true for k > 1, differentiation gives d/dx{dk/dx k[sin ax]} = d/dx[a k sin(ax + kπ/2)], so dk+1 /dx k+1 [sin ax] = a k+1 cos(ax + kπ/2). However, replacing k by k + 1 in P(k) gives dk+1 /dx k+1 [sin ax] = a k+1 sin[ax + (k + 1)π/2] = a k+1 sin[(ax + kπ/2) + π/2] = a k+1 cos(ax + kπ/2), showing, as required, that proposition P(k) implies proposition P(k + 1), so Step 2 is true. STEP 3 As P(n) is true for n = 1, and P(k) implies P(k + 1), it follows that the result is true for n = 1, 2, . . . and the proof is complete. The binomial theorem ﬁnds applications throughout mathematics at all levels, so we quote it ﬁrst when the exponent n is a positive integer, and then in its more general form when the exponent α involved is any real number. Binomial theorem when n is a positive integer If a, b are real numbers and n is a positive integer, then n(n − 1) n−2 2 a b 2! n(n − 1)(n − 2) n−3 3 a b + · · · + bn , + 3!

(a + b)n = a n + na n−1 b +

binomial coefﬁcient

or more concisely in terms of the binomial coefﬁcient n n! , = r (n − r )!r ! we have (a + b)n =

n n n−r r a b, r r =0

where m! is the factorial function deﬁned as m! = 1 · 2 · 3 · · · m with m > 0 an integer, and 0! is deﬁned as 0! = 1. It follows at once that n n = = 1. 0 n

Section 1.1

Real Numbers, Mathematical Induction, and Mathematical Conventions

7

The binomial theorem involving the expression (a + b)α , where a and b are real numbers with |b/a| < 1 and α is an arbitrary real number takes the following form. General form of the binomial theorem when α is an arbitrary real number If a and b are real numbers such that |b/a| < 1 and α is an arbitrary real number, then

α

α

α b α(α − 1) b 2 (a + b) = a =a 1+ + 1! a 2! a α(α − 1)(α − 2) b 3 + + ··· . 3! a α

b 1+ a

α

The series on the right only terminates after a ﬁnite number of terms if α is a positive integer, in which case the result reduces to the one just given. If α is a negative integer, or a nonintegral real number, the expression on the right becomes an inﬁnite series that diverges if |b/a| > 1.

EXAMPLE 1.2

Expand (3 + x)−1/2 by the binomial theorem, stating for what values of x the series converges. Solution Setting b/a = 13 x in the general form of the binomial theorem gives 1 −1/2 1 2 1 1 5 3 −1/2 −1/2 x + ··· . 1+ x =3 = √ 1− x+ x − (3 + x) 3 6 24 432 3 The series only converges if | 13 x| < 1, and so it is convergent provided |x| < 3.

Some standard mathematical conventions Use of combinations of the ± and ∓ signs The occurrence of two or more of the symbols ± and ∓ in an expression is to be taken to imply two separate results, the ﬁrst obtained by taking the upper signs and the second by taking the lower signs. Thus, the expression a ± b sin θ ∓ c cos θ is an abbreviation for the two separate expressions a + b sin θ − c cos θ

and a − b sin θ + c cos θ.

Multi-statements

multi-statement

When a function is deﬁned sectionally on n different intervals of the real line, instead of formulating n separate deﬁnitions these are usually simpliﬁed by being combined into what can be considered to be a single multi-statement. The following example is typical of a multi-statement: ⎧ ⎨sin x, x < π 0, π ≤ x ≤ 3π/2 f (x) = ⎩ −1, x > 3π/2.

8

Chapter 1

Review of Prerequisites

It is, in fact, three statements. The ﬁrst is obtained by reading f (x) in conjunction with the top line on the right, the second by reading it in conjunction with the second line on the right, and the third by reading it in conjunction with the third line on the right. An example of a multi-statement has already been encountered in the deﬁnition of the absolute value |x| of a number x. Frequent use of multi-statements will be made in Chapter 9 on Fourier series, and elsewhere. Polynomials polynomials

A polynomial is an expression of the form P(x) = a0 x n + a1 x n−1 + · · · + an−1 x + an . The integer n is called the degree of the polynomial, and the numbers ai are called its coefﬁcients. The fundamental theorem of algebra that is proved in Chapter 14 asserts that P(x) = 0 has n roots that may be either real or complex, though some of them may be repeated. (a0 = 0 is assumed.) Notation for ordinary and partial derivatives If f (x) is an n times differentiable function then f (n) (x) will, on occasion, be used to signify dn f/dx n , so that f (n) (x) =

sufﬁx notation for partial derivatives

dn f . dx n

If f (x, y) is a suitably differentiable function of x and y, a concise notation used to signify partial differentiation involves using sufﬁxes, so that ∂f ∂ , fyx = ( fy )x = fx = ∂x ∂x

∂f ∂y

=

∂2 f ∂2 f , fyy = ,..., ∂ y∂ x ∂ y2

with similar results when f is a function of more than two independent variables. Inverse trigonometric functions The periodicity of the real variable trigonometric sine, cosine, and tangent functions means that the corresponding general inverse trigonometric functions are many √ valued. So, for example, if y = sin x and we ask for what values of x is y = 1/ 2, we ﬁnd this is true for x = π/4 ± 2nπ and x = 3π/4 ± 2nπ for n = 0, 1, 2, . . . . To overcome this ambiguity, we introduce the single valued inverses, denoted respectively by x = Arcsin y, x = Arccos y, and x = Arctan y by restricting the domain and range of the sine, cosine, and tangent functions to one where they are either strictly increasing or strictly decreasing functions, because then one value of x corresponds to one value of y and, conversely, one value of y corresponds to one value of x. In the case of the function y = sin x, by restricting the argument x to the interval −π/2 ≤ x ≤ π/2 the function becomes a strictly increasing function of x. The corresponding single valued inverse function is denoted by x = Arcsin y, where y is a number in the domain of deﬁnition [−1, 1] of the Arcsine function and x is a number in its range [−π/2, π/2]. Similarly, when considering the function y = cos x, the argument is restricted to 0 ≤ x ≤ π to make cos x a strictly decreasing function of x. The corresponding single valued inverse function is denoted by x = Arccos y, where y is a number in the domain of deﬁnition [−1, 1] of the Arccosine function and x is a number in its range [0, π]. Finally, in the case of the function y = tan x, restricting

Section 1.1

Real Numbers, Mathematical Induction, and Mathematical Conventions

9

the argument to the interval −π/2 < x < π/2 makes the tangent function a strictly increasing function of x. The corresponding single valued inverse function is denoted by x = Arctan y where y is a number in the domain of deﬁnition (−∞, ∞) of the Arctangent function and x is a number in its range (−π/2, π/2). As the inverse trigonometric functions are important in their own right, the variables x and y in the preceding deﬁnitions are interchanged to allow consideration of the inverse functions y = Arcsin x, y = Arccos x, and y = Arctan x, so that now x is the independent variable and y is the dependent variable. With this interchange of variables the expression y = arcsin x will be used to refer to any single valued inverse function with the same domain of deﬁnition as Arcsin x, but with a different range. Similar deﬁnitions apply to the functions y = arccos x and y = arctan x. Double summations An expression involving a double summation like ∞ ∞ amn sin mx sin ny, m=1 n=1

double summation

means sum the terms amn sin mx sin ny over all possible values of m and n, so that ∞ ∞

amn sin mx sin ny = a11 sin x sin y + a12 sin x sin 2y

m=1 n=1

+ a21 sin 2x sin y + a22 sin 2x sin 2y + · · · . A more concise notation also in use involves writing the double summation as ∞

amn sin mx sin nx.

m=1,n=1

The signum function signum function

The signum function, usually written sign(x), and sometimes sgn(x), is deﬁned as 1 if x > 0 sign(x) = −1 if x < 0. We have, for example, sign(cos x) = 1 for 0 < x < π/2, and sign(cos x) = −1 for π/2 < x < π or, equivalently, 1, 0 < x < 12 π sign(cos x) = −1, 12 π < x < π. Products Let {uk}nk=1 be a sequence of numbers or functions u1 , u2 , . . . ; then the product of the n members of this sequence is denoted by nk=1 uk, so that n uk = u1 u2 · · · un . k=1

inﬁnite product

When the sequence is inﬁnite, lim

n→∞

n k=1

uk =

∞ k=1

uk

10

Chapter 1

Review of Prerequisites

is called an inﬁnite product involving the sequence {uk}. Typical examples of inﬁnite products are ∞ ∞ 1 1 x2 sin x 1− 2 = 1− 2 2 = and . k 2 k π x k=2 k=1 More background information and examples can be found in the appropriate sections in any of references [1.1], [1.2], and [1.5]. Logarithmic functions the functions ln and Log

The notation ln x is used to denote the natural logarithm of a real number x, that is, the logarithm of x to the base e, and in some books this is written loge x. In this book logarithms to the base 10 are not used, and when working with functions of a complex variable the notation Log z, with z = r eiθ means Log z = ln r + iθ .

EXERCISES 1.1 √ √ 1. Prove √that if a > 0, b > 0, then a/ b + b/ a ≥ √ a + b. Prove Exercises 2 through 6 by mathematical induction.

n−1 2. k=0 (a + kd) = (n/2)[2a + (n − 1)d] (sum of an arithmetic series).

n−1 k n 3. r = (1 − r )/(1 − r ) (r = 1) k=0 (sum of a geometric series).

n 2 k = (1/6)n(n + 1)(2n + 1) (sum of squares). 4. k=1 5. dn /dx n [cos ax] = a n cos(ax + nπ/2), with n a natural number. 6. dn /dx n [ln(1 + x)] = (−1)n+1 (n − 1)!/(1 + x)n , with n a natural number.

1.2

7. Use the binomial theorem to expand (3 + 2x)4 . 8. Use the binomial theorem and multiplication to expand (1 − x 2 )(2 + 3x)3 . In Exercises 9 through 12 ﬁnd the ﬁrst four terms of the binomial expansion of the function and state conditions for the convergence of the series. 9. 10. 11. 12.

(3 + 2x)−2 . (2 − x 2 )1/3 . (4 + 2x 2 )−1/2 . (1 − 3x 2 )3/4 .

Complex Numbers Mathematical operations can lead to numbers that do not belong to the real number system R introduced in Section 1.1. In the simplest case this occurs when ﬁnding the roots of the quadratic equation ax 2 + bx + c = 0

with a, b, c ∈ R, a = 0

by means of the quadratic formula x= discriminant of a quadratic

−b ±

√ b2 − 4ac . 2a

The discriminant of the equation is b2 − 4ac, and if b2 − 4ac < 0 the formula involves the square root of a negative real number; so, if the formula is to have meaning, numbers must be allowed that lie outside the real number system. The inadequacy of the real number system when considering different mathematical operations can be illustrated in other ways by asking, for example, how to ﬁnd the three roots that are expected of a third degree algebraic equation as

Section 1.2

Complex Numbers

11

simple as x 3 − 1 = 0, where only the real root 1 can be found using y = x 3 − 1, or by seeking to give meaning to ln(−1), both of which questions will arise later. Difﬁculties such as these can all be overcome if the real number system is extended by introducing the imaginary unit i deﬁned as i 2 = −1, so expressions like (−k2 ) where k a positive real number may be written (−1) (k2 ) = ±ik. Notice that as the real number k only scales the imaginary unit i, it is immaterial whether the result is written as ik or as ki. The extension to the real number system that is required to resolve problems of the type just illustrated involves the introduction of complex numbers, denoted collectively by C, in which the general complex number, usually denoted by z, has the form z = α + iβ,

real and imaginary part notation

with α, β real numbers.

The real number α is called the real part of the complex number z, and the real number β is called its imaginary part. When these need to be identiﬁed separately, we write Re{z} = α

and

Im{z} = β,

so if z = 3 − 7i, Re{z} = 3 and Im{z} = −7. If Im{z} = β = 0 the complex number z reduces to a real number, and if Re{z} = α = 0 it becomes a purely imaginary number, so, for example, z = 5i is a purely imaginary number. When a complex number z is considered as a variable it is usual to write it as z = x + i y, where x and y are now real variables. If it is necessary to indicate that z is a general complex number we write z ∈ C. When solving the quadratic equation az2 + bz + c = 0 with a, b, and c real numbers and a discriminant b2 − 4ac < 0, by setting 4ac − b2 = k2 in the quadratic formula, with k > 0, the two roots z1 and z2 are given by the complex numbers z1 = −(b/2a) + i(k/2a)

and

z2 = −(b/2a) − i(k/2a).

Algebraic rules for complex numbers Let the complex numbers z1 and z2 be deﬁned as z1 = a + ib and

z2 = c + id,

with a, b, c, and d arbitrary real numbers. Then the following rules govern the arithmetic manipulation of complex numbers. Equality of complex numbers The complex numbers z1 and z2 are equal, written z1 = z2 if, and only if, Re{z1 } = Re{z2 } and Im{z1 } = Im{z2 }. So a + ib = c + id if, and only if, a = c and b = d.

12

Chapter 1

Review of Prerequisites

EXAMPLE 1.3

(a) 3 − 9i = 3 + bi if, and only if, b = −9. (b) If u = −2 + 5i, v = 3 + 5i, w = a + 5i, then u = w if, and only if, a = −2 but u = v, and v = w if, and only if, a = 3. Zero complex number The zero complex number, also called the null complex number, is the number 0 + 0i that, for simplicity, is usually written as an ordinary zero 0.

EXAMPLE 1.4

If a + ib = 0, then a = 0 and b = 0. Addition and subtraction of complex numbers The addition (sum) and subtraction (difference) of the complex numbers z1 and z2 is deﬁned as z1 + z2 = Re{z1 } + Re{z2 } + i[Im{z1 } + Im{z2 }] and z1 − z2 = Re{z1 } − Re{z2 } + i[Im{z1 } − Im{z2 }]. So, if z1 = a + ib and z2 = c + id, then z1 + z2 = (a + ib) + (c + id) = (a + c) + i(b + d), and z1 − z2 = (a + ib) − (c + id) = (a − c) + i(b − d).

EXAMPLE 1.5

If z1 = 3 + 7i and z2 = 3 + 2i, then the sum z1 + z2 = (3 + 3) + (7 + 2)i = 6 + 9i, and the difference z1 − z2 = (3 − 3) + (7 − 2)i = 5i. Multiplication of complex numbers The multiplication (product) of the two complex numbers z1 = a + ib and z2 = c + id is deﬁned by the rule z1 z2 = (a + ib)(c + id) = (ac − bd) + i(ad + bc). An immediate consequence of this deﬁnition is that if k is a real number, then kz1 = k(a + ib) = ka + ikb. This operation involving multiplication of a complex

Section 1.2

Complex Numbers

13

number by a real number is called scaling a complex number. Thus, if z1 = 3 + 7i and z2 = 3 + 2i, then 2z1 − 3z2 = (6 + 14i) − (9 + 6i) = −3 + 8i. In particular, if z = a + ib, then −z = (−1)z = −a − ib. This is as would be expected, because it leads to the result z − z = 0. In practice, instead of using this formal deﬁnition of multiplication, it is more convenient to perform multiplication of complex numbers by multiplying the bracketed quantities in the usual algebraic manner, replacing every product i 2 by −1, and then combining separately the real and imaginary terms to arrive at the required product.

EXAMPLE 1.6

(a) 5i(−4 + 3i) = −15 − 20i. (b) (3 − 2i)(−1 + 4i)(1 + i) = (−3 + 12i + 2i − 8i 2 )(1 + i) = [(−3 + 8) + (12 + 2)i](1 + i) = (5 + 14i)(1 + i) = 5 + 14i + 5i + 14i 2 = (5 − 14) + (5 + 14)i = −9 + 19i.

Complex conjugate If z = a + ib, then the complex conjugate of z, usually denoted by z and read “z bar,” is deﬁned as z = a − ib. It follows directly that (z) = z and

zz = a 2 + b2 .

In words, the complex conjugate operation has the property that taking the complex conjugate of a complex conjugate returns the original complex number, whereas the product of a complex number and its complex conjugate always yields a real number. If z = a + ib, then adding and subtracting z and z gives the useful results z + z = 2Re{z} = 2a

and

z − z = 2i Im{z} = 2ib.

These can be written in the equivalent form Re{z} = a =

1 (z + z) 2

and

Im{z} = b =

1 (z − z). 2i

Quotient (division) of complex numbers Let z1 = a + ib and z2 = c + id. Then the quotient z1 /z2 is deﬁned as z1 (ac + bd) + i(bc − ad) = , z2 = 0. z2 c2 + d2

14

Chapter 1

Review of Prerequisites

In practice, division of complex numbers is not carried out using this deﬁnition. Instead, the quotient is written in the form z1 z1 z2 = , z2 z2 z2 where the denominator is now seen to be a real number. The quotient is then found by multiplying out and simplifying the numerator in the usual manner and dividing the real and imaginary parts of the numerator by the real number z2 z2 . EXAMPLE 1.7

Find z1 /z2 given that z1 = (3 + 2i) and z2 = 1 + 3i. Solution (3 + 2i)(1 − 3i) 3 − 9i + 2i − 6i 2 9 7i 3 + 2i = = = − . 1 + 3i (1 + 3i)(1 − 3i) 10 10 10 Modulus of a complex number The modulus of the complex number z = a + ib denoted by |z|, and also called its magnitude, is deﬁned as |z| = (a 2 + b2 )1/2 = (zz)1/2 .

It follows directly from the deﬁnitions of the modulus and division that |z| = |z| = (a 2 + b2 )1/2 , and z1 /z2 = z1 z2 /|z2 |2 . EXAMPLE 1.8

If z = 3 + 7i, then |z| = |3 + 7i| = (32 + 72 )1/2 =

√ 58.

It is seen that the foregoing rules for the arithmetic manipulation of complex numbers reduce to the ordinary arithmetic rules for the algebraic manipulation of real numbers when all the complex numbers involved are real numbers. Complex numbers are the most general numbers that need to be used in mathematics, and they contain the real numbers as a special case. There is, however, a fundamental difference between real and complex numbers to which attention will be drawn after their common properties have been listed. Properties shared by real and complex numbers Let z, u, and w be arbitrary real or complex numbers. Then the following properties are true: z + u = u + z. This means that the order in which complex numbers are added does not affect their sum. 2. zu = uz. This means that the order in which complex numbers are multiplied does not affect their product. 1.

Section 1.3

The Complex Plane

3.

(z + u) + w = z + (u + w). This means that the order in which brackets are inserted into a sum of ﬁnitely many complex numbers does not affect the sum.

4.

z(uw) = (zu)w. This means that the terms in a product of complex numbers may be grouped and multiplied in any order without affecting the resulting product.

5.

z(u + w) = zu + zw. This means that the product of z and a sum of complex numbers equals the sum of the products of z and the individual complex numbers involved in the sum.

6.

z + 0 = 0 + z = z. This result means that the addition of zero to any complex number leaves it unchanged.

7.

z · 1 = 1 · z = z. This result means that multiplication of any complex number by unity leaves the complex number unchanged.

15

Despite the properties common to real and complex numbers just listed, there remains a fundamental difference because, unlike real numbers, complex numbers have no natural order. So if z1 and z2 are any complex numbers, a statement such as z1 < z2 has no meaning.

EXERCISES 1.2 Find the roots of the equations in Exercises 1 through 6. 1. z2 + z + 1 = 0. 2. 2z2 + 5z + 4 = 0. 3. z2 + z + 6 = 0.

4. 3z2 + 2z + 1 = 0. 5. 3z2 + 3z + 1 = 0. 6. 2z2 − 2z + 3 = 0.

7. Given that z = 1 is a root, ﬁnd the other two roots of 2z3 − z2 + 3z − 4 = 0. 8. Given that z = −2 is a root, ﬁnd the other two roots of 4z3 + 11z2 + 10z + 8 = 0.

1.3

9. Given u = 4 − 2i, v = 3 − 4i, w = −5i and a + ib = (u + iv)w, ﬁnd a and b. 10. Given u = −4 + 3i, v = 2 + 4i, and a + ib = uv2 , ﬁnd a and b. 11. Given u = 2 + 3i, v = 1 − 2i, w = −3 − 6i, ﬁnd |u + v|, u + 2v, u − 3v + 2w, uv, uvw, |u/v|, v/w. 12. Given u = 1 + 3i, v = 2 − i, w = −3 + 4i, ﬁnd uv/w, uw/v and |v|w/u.

The Complex Plane

cartesian representation of z

Complex numbers can be represented geometrically either as points, or as directed line segments (vectors), in the complex plane. The complex plane is also called the z-plane because of the representation of complex numbers in the form z = x + i y. Both of these representations are accomplished by using rectangular cartesian coordinates and plotting the complex number z = a + ib as the point (a, b) in the plane, so the x-coordinate of z is a = Re{z} and its y-coordinate is b = Im{z}. Because of this geometrical representation, a complex number written in the form z = a + ib is said to be expressed in cartesian form. To acknowledge the Swiss amateur mathematician Jean-Robert Argand, who introduced the concept of the complex plane in 1806, and who by profession was a bookkeeper, this representation is also called the Argand diagram.

16

Chapter 1

Review of Prerequisites

Imaginary axis

Imaginary axis

4 3

4 z = 3i

3 z = 2 + 2i

2

z = 3i z = 2 + 2i

2 1

1

z=4

z=4 0

1

2

3

5 Real axis

1

2

3

4

5 Real axis

−1

−1 −2

4

z = 2 − 2i (a)

−2

z = 2 − 2i (b)

FIGURE 1.1 (a) Complex numbers as points. (b) Complex numbers as vectors.

triangle and parallelogram laws

For obvious reasons, the x-axis is called the real axis and the y-axis the imaginary axis. Purely real numbers are represented by points on the real axis and purely imaginary ones by points on the imaginary axis. Examples of the representation of typical points in the complex plane are given in Fig. 1.1a, where the numbers 4, 3i, 2 + 2i, and 2 − 2i are plotted as points. These same complex numbers are shown again in Fig. 1.1b as directed line segments drawn from the origin (vectors). The arrow shows the sense along the line, that is, the direction from the origin to the tip of the vector representing the complex number. It can be seen from both ﬁgures that, when represented in the complex plane, a complex number and its complex conjugate (in this case 2 + 2i and 2 − 2i) lie symmetrically above and below the real axis. Another way of expressing this result is by saying that a complex number and its complex conjugate appear as reﬂections of each other in the real axis, which acts like a mirror. The addition and subtraction of two complex numbers have convenient geometrical interpretations that follow from the deﬁnitions given in Section 1.2. When complex numbers are added, their respective real and imaginary parts are added, whereas when they are subtracted, their respective real and imaginary parts are subtracted. This leads at once to the triangle law for addition illustrated in Fig. 1.2a, in which the directed line segment (vector) representing z2 is translated without rotation or change of scale, to bring its base (the end opposite to the arrow) into coincidence with the tip of the directed line element representing z1 (the end at which the arrow is located). The sum z1 + z2 of the two complex numbers is then represented by the directed line segment from the base of the line segment representing z1 to the tip of the newly positioned line segment representing z2 . The name triangle law comes from the triangle that is constructed in the complex plane during this geometrical process of addition. Notice that an immediate consequence of this law is that addition is commutative, because both z1 + z2 and z2 + z1 are seen to lead to the same directed line segment in the complex plane. For this reason the addition of complex numbers is also said to obey the parallelogram law for addition, because the commutative property generates the parallelogram shown in Fig. 1.2a.

Section 1.3

The Complex Plane

17

Imaginary axis

Imaginary axis b+d d d

z1

z2

+

z2

b −c

b

z1 a−c

−z2 z1

z2

z1 − zc

−z2 a

2

Real axis

b−d −d

a a + c Real axis

c (a)

(b)

FIGURE 1.2 Addition and subtraction of complex numbers using the triangle/parallelogram law.

The geometrical interpretation of the subtraction of z2 from z1 follows similarly by adding to z1 the directed line segment −z2 that is obtained by reversing of the sense (arrow) along z2 , as shown in Fig. 1.2b. It is an elementary fact from Euclidean geometry that the sum of the lengths of the two sides |u| and |v| of the triangle in Fig. 1.3 is greater than or equal to the length of the hypotenuse |u + v|, so from geometrical considerations we can write |u + v| ≤ |u| + |v|. triangle inequality

This result involving the moduli of the complex numbers u and v is called the triangle inequality for complex numbers, and it has many applications. An algebraic proof of the triangle inequality proceeds as follows: |u + v|2 = (u + v)(u + v) = uu + vu + uv + vv = |u|2 + |v|2 + (uv + uv) ≤ |u2 | + |v2 | + 2|uv| = (|u| + |v|)2 . The required result now follows from taking the positive square root. A similar argument, the proof of which is left as an exercise, can be used to show that u| − |v ≤ |u + v|, so when combined with the triangle inequality we have u| − |v ≤ |u + v| ≤ |u| + |v|.

Imaginary axis

+ ⎢u

v⎥

⎢v⎥ ⎢u⎥

0 FIGURE 1.3 The triangle inequality.

Real axis

18

Chapter 1

Review of Prerequisites

EXERCISES 1.3 In Exercises 1 through 8 use the parallelogram law to form the sum and difference of the given complex numbers and then verify the results by direct addition and subtraction. 1. 2. 3. 4.

u = 2 + 3i, v = 1 − 2i. u = 4 + 7i, v = −2 − 3i. u = −3, v = −3 − 4i. u = 4 + 3i, v = 3 + 4i.

1.4

5. 6. 7. 8.

u = 3 + 6i, v = −4 + 2i. u = −3 + 2i, v = 6i. u = −4 + 2i, v = −4 − 10i. u = 4 + 7i, v = −3 + 5i.

In Exercises 9 through 11 use the parallelogram law to verify the triangle inequality |u + v| ≤ |u| + |v| for the given complex numbers u and v. 9. u = −4 + 2i, v = 3 + 5i. 10. u = 2 + 5i, v = 3 − 2i. 11. u = −3 + 5i, v = 2 + 6i.

Modulus and Argument Representation of Complex Numbers

polar representation of z

When representing z = x + i y in the complex plane by a point P with coordinates (x, y), a natural alternative to the cartesian representation is to give the polar coordinates (r, θ) of P. This polar representation of z is shown in Fig. 1.4, where OP = r = |z| = (x 2 + y2 )1/2

and

tan θ = y/x.

(1)

The radial distance OP is the modulus of z, so r = |z|, and the angle θ measured counterclockwise from the positive real axis is called the argument of z. Because of this, a complex number expressed in terms of the polar coordinates (r, θ ) is said to be in modulus–argument form. The argument θ is indeterminate up to a multiple of 2π, because the polar coordinates (r, θ ), and (r, θ + 2kπ ), with k = ±1, ±2, . . . , identify the same point P. By convention, the the angle θ is called the principal value of the argument of z when it lies in the interval −π < θ ≤ π . To distinguish the principal value of the argument from all of its other values, we write Arg z = θ,

when −π < θ ≤ π.

(2)

The values of the argument of z that differ from this value of θ by a multiple of 2π are denoted by arg z, so that arg z = θ + 2kπ,

with k = ± 1, ±2, . . . .

Imaginary axis P(r, θ)

y = r sin θ

r= θ O

⎢z⎥

z x = r cos θ

Real axis

FIGURE 1.4 The complex plane and the (r, θ ) representation of z.

(3)

Section 1.4

Modulus and Argument Representation of Complex Numbers

19

The signiﬁcance of the multivalued nature of arg z will become apparent later when the roots of complex numbers are determined. The connection between the cartesian coordinates (x, y) and the polar coordinates (r, θ ) of the point P corresponding to z = x + i y is easily seen to be given by x = r cos θ

modulus–argument representation of z

and

y = r sin θ.

This leads immediately to the representation of z = x + i y in the alternative modulus–argument form z = r (cos θ + i sin θ ).

(4)

A routine calculation using elementary trigonometric identities shows that (cos θ + i sin θ )2 = (cos 2θ + i sin 2θ). An inductive argument using the above result as its ﬁrst step then establishes the following simple but important theorem. THEOREM 1.1

De Moivre’s theorem (cos θ + i sin θ)n = (cos nθ + i sin nθ),

EXAMPLE 1.9

for n a natural number.

Use de Moivre’s theorem to express cos 4θ and sin 4θ in terms of powers of cos θ and sin θ . Solution The result is obtained by ﬁrst setting n = 4 in de Moivre’s theorem and expanding (cos θ + i sin θ )4 to obtain cos4 θ + 4i cos3 θ sin θ − 6 cos2 θ sin2 θ − 4i cos θ sin3 θ + sin4 θ = cos 4θ + i sin 4θ. Equating the respective real and imaginary parts on either side of this identity gives the required results cos 4θ = cos4 θ − 6 cos2 θ sin2 θ + sin4 θ and sin 4θ = 4 cos3 θ sin θ − 4 cos θ sin3 θ. As the complex number z = cos θ + i sin θ has unit modulus, it follows that all numbers of this form lie on the unit circle (a circle of radius 1) centered on the origin, as shown in Fig. 1.5. Using (5), we see that if z = r (cos θ + i sin θ), then zn = r n (cos nθ + i sin nθ), for n a natural number. θ

(5)

The relationship between e , sin θ, and cos θ can be seen from the following well-known series expansions of the functions eθ = sin θ =

∞ θn n=0 ∞

n!

=1+θ +

(−1)n

n=0

θ2 θ3 θ4 θ5 θ6 + + + + + ···; 2! 3! 4! 5! 6!

θ3 θ5 θ7 θ 2n+1 =θ− + − + ···; (2n + 1)! 3! 5! 7!

∞ θ2 θ4 θ6 θ 2n cos θ = =1− + − + ···. (−1)n (2n)! 2! 4! 6! n=0

20

Chapter 1

Review of Prerequisites

Imaginary axis

i

z = cos θ + i sin θ

y = sin θ

−1

θ x = cos θ

1

Real axis

−i FIGURE 1.5 Point z = cos θ + i sin θ on the unit circle centered on the origin.

Euler formula

By making a formal power series expansion of the function eiθ , simplifying powers of i, grouping together the real and imaginary terms, and using the series representations for cos θ and sin θ, we arrive at what is called the real variable form of the Euler formula eiθ = cos θ + i sin θ,

for any real θ .

(6)

This immediately implies that if z = r eiθ , then zα = r α eiαθ ,

for any real α.

(7)

When θ is restricted to the interval −π < θ ≤ π , formula (6) leads to the useful results 1 = ei0 ,

i = eiπ/2 ,

−1 = eiπ ,

−i = e−iπ/2

and, in particular, to 1 = e2kπi

for k = 0, ±1, ±2, . . . .

The Euler form for complex numbers makes their multiplication and division very simple. To see this we set z1 = r1 eiα and z2 = r2 eiβ and then use the results z1 z2 = r1r2 ei(α+β)

and

z1 /z2 = r1 /r2 ei(α−β) .

(8)

These show that when complex numbers are multiplied, their moduli are multiplied and their arguments are added, whereas when complex numbers are divided, their moduli are divided and their arguments are subtracted. EXAMPLE 1.10

Find uv, u/v, and u25 given that u = 1 + i, v =

√

3 − i.

√ √ √ , v = 3 −√ i = 2e−iπ/6 , so uv = 2√ 2eiπ/12 , u/v = Solution u = 1 + i = 2eiπ/4 √ √ i5π/12 2)e while u25 = √ ( 2eiπ/4 )25 = ( √2)25 (eiπ/4 )25 = 4096 2(ei(6+1/4)π ) = (1/ √ i6π iπ/4 4096 2(e )(e ) = 4096 2(eiπ/4 ) = 4096 2(1 + i). To ﬁnd the principal value of the argument of a given complex number z, namely Arg z, use should be made of the signs of x = Re{z}, and y = Im{z} together

Section 1.4

Modulus and Argument Representation of Complex Numbers

21

with the results listed below, all of which follow by inspection of Fig. 1.5. Signs of x and y x < 0, y < 0 x > 0, y < 0 x > 0, y > 0 x < 0, y > 0 EXAMPLE 1.11

Arg z = θ −π < θ < −π/2 −π/2 < θ < 0 0 < θ < π/2 π/2 < θ < π

Find r = |z|, Arg z, arg z, and the modulus–argument form of the following values of z. √ √ √ (a) −2 3 − 2i (b) −1 + i 3 (c) 1 + i (d) 2 − i2 3. √ Solution (a) r = {(−2 3)2 + (−2)2 }1/2 = 4, Argz = θ = −5π/6 and arg z = −5π/6 + 2kπ , k = ± 1, ±2, . . . , z = 4(cos(−5π/6) + i sin(−5π/6)). √ (b) r = {(−1)2 + ( 3)2 }1/2 = 2, Arg z = θ = 2π/3 and arg z = 2π/3 + 2kπ, k = ±1, ±2, . . . , z = 2(cos(2π/3) + i sin(2π/3)). √ (c) r = {(1)2 + (1)2 }1/2 = 2, Arg z = θ = π/4 and arg z = π/4 + 2kπ, √ k = ±1, ±2, . . . , z = 2(cos(π/4) + i sin(π/4)). √ (d) r = {(2)2 + (−2 3)2 }1/2 = 4, Arg z = θ = −π/3 and arg z = −π/3 + 2kπ, k = ±1, ±2, . . . , z = 4(cos(−π/3) + i sin(−π/3)).

EXERCISES 1.4 1. Expand (cos θ + i sin θ)2 and then use trigonometric identities to show that (cos θ + i sin θ )2 = (cos 2θ + i sin 2θ). 2. Give an inductive proof of de Moivre’s theorem (cos θ + i sin θ)n = (cos nθ + i sin nθ ), for n a natural number. 3. Use de Moivre’s theorem to express cos 5θ and sin 5θ in terms of powers of cos θ and sin θ . 4. Use de Moivre’s theorem to express cos 6θ and sin 6θ in terms of powers of cos θ and sin θ. 5. Show by expanding (cos α + i sin α)(cos β + i sin β) and using trigonometric identities that (cos α + i sin α)(cos β + i sin β) = cos(α + β) + i sin(α + β). 6. Show by expanding (cos α + i sin α)/(cos β + i sin β) and using trigonometric identities that (cos α + i sin α)/(cos β + i sin β) = cos(α − β) + i sin(α − β).

7. If z = cos θ + i sin θ = eiθ , show that when n is a natural number, 1 1 1 1 n n cos(nθ) = and sin(nθ) = z + n z − n . 2 z 2i z Use these results to express cos3 θ sin3 θ in terms of multiple angles of θ. Hint: z¯ = 1/z. 8. Use the method of Exercise 7 to express sin6 θ in terms of multiple angles of θ. 9. By expanding (z + 1/z)4 , grouping terms, and using the method of Exercise 7, show that cos4 θ = (1/8)(3 + 4 cos 2θ + cos 4θ). 10. By expanding (z − 1/z)5 , grouping terms, and using the method of Exercise 7, show that sin5 θ = (1/16)(sin 5θ − 5 sin 3θ + 10 sin θ). 11. Use the method of Exercise 7 to show that cos3 θ + sin3 θ = (1/4)(cos 3θ + 3 cos θ − sin3 θ + 3 sin θ).

22

Chapter 1

Review of Prerequisites

In Exercises 12 through 15 express the functions of u, v, and w in modulus-argument form. √ 12. uv, u/v, and v5 , given that u = 2 − 2i and v = 3 + i3 3. √ 13. uv, u/v, and u7 , given that u = −1 − i 3, v = −4 + 4i. √ 14. uv, u/v, and v6 , given that u = 2 − 2i, v = 2 − i2 3. 15. uvw, √ uw/v, and w 3 /u4 , given that u = 2 − 2i, v = 3 − i3 3 and w = 1 + i. √ 16. Express [(−8 + i8 3)/(−1 − i)]2 in modulus–argument form. √ 17. Find in modulus–argument form [(1 + i 3)3 / (−1 + i)2 ]3 . 18. Use the factorization (1 − zn+1 ) = (1 − z)(1 + z + z2 + · · · + zn )

1.5

with z = eiθ = exp(iθ) to show that n

exp(ikθ) =

k=1

exp(inθ) − 1 . 1 − exp(−iθ)

19. Use the ﬁnal result of Exercise 18 to show that n

exp(ikθ) =

k=1

exp[i(n + 1/2)θ] − exp(iθ/2) , exp(iθ/2) − exp(−iθ/2)

and then use the result to deduce the Lagrange identity 1 + cos θ + cos 2θ + · · · + cos nθ sin[(n + 1/2)θ] = 1/2 + , for 0 < θ < 2π. 2 sin(θ/2)

(z = 1)

Roots of Complex Numbers It is often necessary to ﬁnd the n values of z1/n when n is a positive integer and z is an arbitrary complex number. This process is called ﬁnding the nth roots of z. To determine these roots we start by setting w = z1/n ,

which is equivalent to w n = z.

Then, after deﬁning w and z in modulus–argument form as w = ρeiφ

and

z = reiθ ,

(9)

we substitute for w and z in w n = z to obtain ρ n einφ = r eiθ . It is at this stage, in order to ﬁnd all n roots, that use must be made of the manyvalued nature of the argument of a complex number by recognizing that 1 = e2kπi for k = 0, ±1, ±2, . . . . Using this result we now multiply the right-hand side of the foregoing result by by e2kπi (that is, by 1) to obtain ρ n einφ = reiθ e2kπi = rei(θ +2kπ) . Equality of complex numbers in modulus–argument form means the equality of their moduli and, correspondingly, the equality of their arguments, so applying this to the last result we have ρn = r

and

nφ = θ + 2kπ,

ρ = r 1/n

and

φ = (θ + 2kπ)/n.

showing that

Here r 1/n is simply the nth positive root of r : ρ =

√ n r.

Section 1.5

Roots of Complex Numbers

23

Imaginary axis w2

w1

(θ + 4π)/ n

2π/n

(θ

+

)/n 2π

w0

θ/n 0

Real axis wn−1

FIGURE 1.6 Location of the roots of z1/n .

nth roots of a complex number z

Finally, when we substitute these results into the expression for w, we see that the n values of the roots denoted by w0 , w1 , . . . , wn−1 are given by wk = r 1/n {cos[(θ + 2kπ)/n] + i sin[(θ + 2kπ )/n]}, for k = 0, 1, . . . , n − 1.

(10)

Notice that it is only necessary to allow k to run through the successive integers 0, 1, . . . , n − 1, because the period of the sine and cosine functions is 2π, so allowing k to increase beyond the value n − 1 will simply repeat this same set of roots. An identical argument shows that allowing kto run through successive negative integers can again only generate the same n roots w0 , w1 , . . . , wn−1 . Examination of the arguments of the roots shows them to be spaced uniformly around a circle of radius r 1/n centered on the origin. The angle between the radial lines drawn from the origin to each successive root is 2π/n, with the radial line from the origin to the ﬁrst root w0 making an angle θ/n to the positive real axis, as shown in Fig. 1.6. This means that if the location on the circle of any one root is known, then the locations of the others follow immediately. Writing unity in the form 1 = ei0 shows its modulus to be r = 1 and the principal value of its argument to be θ = 0. Substitution in formula (10) then shows the n roots of 11/n , called the nth roots of unity, to be w0 = 1,

w1 = eiπ/n ,

w2 = ei2π/n , . . . , wn−1 = ei(n−1)π/n .

(11)

By way of example, the ﬁfth roots of unity are located around the unit circle as shown in Fig. 1.7. If we set ω = w1 , it follows that the nth roots of unity can be written in the form 1, ω, ω2 , . . . , ωn−1 . As ωn = 1 and ωn − 1 = (ω − 1)(1 + ω + ω2 + · · · + ωn−1 ) = 0, as ω1 = 1 we see that the the nth roots of unity satisfy 1 + ω + ω2 + · · · + ωn−1 = 0.

(12)

24

Chapter 1

Review of Prerequisites Imaginary axis i

w1

w2

2π /5 2π /5 w0

2π/5

1

Real axis

2π /5 2π /5

w3

w4 FIGURE 1.7 The ﬁfth roots of unity.

This result remains true if ω is replaced by any one of the other nth roots of unity, with the exception of 1 itself. EXAMPLE 1.12

Find w = (1 + i)1/3 . √ √ Solution Setting z = 1 + i = 2eiπ/4 shows that r = |z| = 2 and θ = π/4. Substituting these results into formula (1) gives wk = 21/6 {cos[(1/12)(1 + 8k)π ] + i sin[(1/12)(1 + 8k)π ]},

for k = 0, 1, 2.

The square root of a complex number ζ = α + iβ is often required, so we now derive a useful formula for its two roots in terms of |ζ |, α and the sign of β. To obtain the result we consider the equation z2 = ζ,

where ζ = α + iβ,

and let Arg ζ = θ . Then we may write z2 = |ζ |eiθ , and taking the square root of this result we ﬁnd the two square roots z− and z+ are given by z± = ±|ζ |1/2 eiθ/2 = ±|ζ |1/2 {cos(θ/2) + i sin(θ/2)}. Now cos θ = α/|ζ |, but cos2 (θ/2) = (1/2)(1 + cos θ ),

and

sin2 (θ/2) = (1/2)(1 − cos θ),

Section 1.5

Roots of Complex Numbers

25

so cos2 (θ/2) = (1/2)(1 + α/|ζ |),

and

sin2 (θ/2) = (1/2)(1 − α/|ζ |).

As −π < θ ≤ π, it follows that in this interval cos(θ/2) is nonnegative, so taking the square root of cos2 (θ/2) we obtain 1/2 |ζ | + α cos(θ/2) = . 2|ζ | However, the function sin(θ/2) is negative in the interval −π < θ < 0 and positive in the interval 0 < θ < π, and so has the same sign as β. Thus, the square root of sin2 (θ/2) can be written in the form 1/2 |ζ | − α sin(θ/2) = sign (β) . 2|ζ | Using these expressions for cos(θ/2) and sin(θ/2) in the square roots z± brings us to the following useful rule. Rule for ﬁnding the square root of a complex number Let z2 = ζ , with ζ = α + iβ. Then the square roots z+ and z− of ζ are given by

|ζ | + α 1/2 |ζ | − α 1/2 z+ = + i sign (β) 2 2 1/2 |ζ | + α |ζ | − α 1/2 z− = − − i sign (β) . 2 2

EXAMPLE 1.13

Find the square roots of (a) ζ = 1 + i and (b) ζ = 1 − i. √ Solution (a) ζ = 1 + i so |ζ | = 2, α = 1 and sign(β) = 1, so the square roots of ζ = 1 + i are ⎧ √ √ 1/2 1/2 ⎫ ⎨ ⎬ 2+1 2−1 z± = ± +i . ⎩ ⎭ 2 2 √ (b) ζ = 1 − i, so |ζ | = 2, α = 1 and sign(β) = −1, from which it follows that the square roots of ζ = 1 − i are ⎧ √ √ 1/2 1/2 ⎫ ⎨ ⎬ 2+1 2−1 −i . z± = ± ⎩ ⎭ 2 2 The theorem that follows provides information about the roots of polynomials with real coefﬁcients that proves to be useful in a variety of ways.

26

Chapter 1

Review of Prerequisites

THEOREM 1.2

Roots of a polynomial with real coefﬁcients

Let

P(z) = zn + a1 zn−1 + a2 zn−2 + · · · an−1 z + an be a polynomial of degree n in which all the coefﬁcients a1 , a2 , . . . , an are real. Then either all the n roots of P(z) = 0 are real, that is, the n zeros of P(z) are all real, or any that are complex must occur in complex conjugate pairs. Proof The proof uses the following simple properties of the complex conjugate operation. 1. If a is real, then a = a. This result follows directly from the deﬁnition of the complex conjugate operation. 2. If b and c are any two complex numbers, then b + c = b + c. This result also follows directly from the deﬁnition of the complex conjugate operation. 3. If b and c are any two complex numbers, then bc = bc and br = (b)r . We now proceed to the proof. Taking the complex conjugate of P(z) = 0 gives zn + a1 zn−1 + a2 zn−2 + · · · + an−1 z + an = 0, but the ar are all real so ar zn−r = ar zn−r = ar zn−r = ar (z)n−r , allowing the preceding equation to be rewritten as (z)n + a1 (z)n−1 + a2 (z)n−2 + · · · + an−1 z + an = 0. This result is simply P(z) = 0, showing that if z is a complex root of P(z), then so also is z. Equivalently, z and z are both zeros of P(z). If, however, z is a real root, then z = z and the result remains true, so the ﬁrst part of the theorem is proved. The second part follows from the fact that if z = α + iβ is a root, then so also is z = α − iβ, and so (z − α − iβ) and (z − α + iβ) are factors of P(z). The product of these factors must also be a factor of P(z), but (z − α − iβ)(z − α + iβ) = z2 − 2αz + α 2 + β 2 , and the expression on the right is a quadratic in z with real coefﬁcients, so the ﬁnal result of the theorem is established. EXAMPLE 1.14

Find the roots of z3 − z2 − z − 2 = 0, given that z = 2 is a root. Solution If z = 2 is a root of P(z) = 0, then z − 2 is a factor of P(z), so dividing P(z) by z − 2 we obtain z2 + z + 1. The remaining two roots of P(z) = 0 are the 2 (−1 ± roots √ we ﬁnd that z =√ √ of z + z + 1 = 0. Solving this quadratic equation i 3)/2, so the three roots of the equation are 2, (−1 + i 3)/2, and (−1 − i 3)/2.

For more background information and examples on complex numbers, the complex plane and roots of complex numbers, see Chapter 1 of reference [6.1], Sections 1.1 to 1.5 of reference [6.4], and Chapter 1 of reference [6.6].

Section 1.6

Partial Fractions

27

EXERCISES 1.5 In Exercises 1 through 8 ﬁnd the square roots of the given complex number by using result (10), and then conﬁrm the result by using the formula for ﬁnding the square root of a complex mumber. 1. 2. 3. 4.

−1 + i. 3 + 2i. i. −1 + 4i.

5. 6. 7. 8.

2 − 3i. −2 − i. 4 − 3i. −5 + i.

In Exercises 9 through 14 ﬁnd the roots of the given complex number. √ 12. (−1 − i)1/3 . 9. (1 + i 3)1/3 . 1/4 13. (−i)1/3 . 10. i . 1/4 14. (4 + 4i)1/4 . 11. (−1) . 15. Find the roots of z3 + z(i − 1) = 0. 16. Find the roots of z3 + i z/(1 + i) = 0.

1.6

17. Use result (12) to show that 1 + cos(2π/n) + cos(4π/n) + · · · + cos[(2(n − 1)π/n)] = 0 and sin(2π/n) + sin(4π/n) + · · · + sin[(2(n − 1)π/n)] = 0. 18. Use Theorem 1.1 and the representation z = r eiθ to prove that if a and b are any two arbitrary complex numbers, then ab = ab and (a r ) = (a)r . 19. Given z = 1 is a zero of the polynomial P(z) = z3 − 5z2 + 17z − 13, ﬁnd its other two zeros and verify that they are complex conjugates. 20. Given that z = −2 is a zero of the polynomial P(z) = z5 + 2z4 − 4z − 8, ﬁnd its other four zeros and verify that they occur in complex conjugate pairs. 21. Find the two zeros of the quadratic P(z) = z2 − 1 + i, and explain why they do not occur as a complex conjugate pair.

Partial Fractions Let N(x) and D(x) be two polynomials. Then a rational function of x is any function of the form N(x)/D(x). The method of partial fractions involves the decomposition of rational functions into an equivalent sum of simpler terms of the type P2 P1 , ,... ax + b (ax + b)2

and

Q2 x + R2 Q1 x + R1 , ,..., 2 2 Ax + Bx + C (Ax + Bx + C)2

where the coefﬁcients are all real together with, possibly, a polynomial in x. The steps in the reduction of a rational function to its partial fraction representation are as follows: STEP 1 Factorize D(x) into a product of linear factors and quadratic factors with real coefﬁcients with complex roots, called irreducible factors. This is the hardest step, and real quadratic factors will only arise when D(x) = 0 has pairs of complex conjugate roots (see Theorem 1.2). Use the result to express D(x) in the form s D(x) = (a1 x + b1 )r1 . . . (am x + bm)rm A1 x 2 + B1 x + C1 1 s . . . Ak x 2 + Bk x + Ck k , where ri is the number of times the linear factor (ai x + bi ) occurs in the factorization of D(x), called its multiplicity, and s j is the corresponding multiplicity of the quadratic factor (Aj x 2 + Bj x + C j ).

28

Chapter 1

Review of Prerequisites

STEP 2 Suppose ﬁrst that the degree n of the numerator is less than the degree d of the denominator. Then, to every different linear factor (ax + b) with multiplicity r , include in the partial fraction expansion the terms P1 P2 Pr + + ··· + , (ax + b) (ax + b)2 (ax + b)r

partial fraction undetermined coefﬁcients

where the constant coefﬁcients Pi are unknown at this stage, and so are called undetermined coefﬁcients. STEP 3 To every quadratic factor ( Ax 2 + Bx + C)s with multiplicity s include in the partial fraction expansion the terms Q1 x + R1 Q2 x + R2 Qs x + Rs + + ··· + , 2 2 2 2 (Ax + Bx + C) (Ax + Bx + C) (Ax + Bx + C)s where the Q j and Rj for j = 1, 2, . . . , s are undetermined coefﬁcients. STEP 4 Take as the partial fraction representation of N(x)/D(x) the sum of all the terms in Steps 2 and 3. STEP 5 Multiply the expression N(x)/D(x) = Partial fraction representation in Step 4 by D(x), and determine the unknown coefﬁcients by equating the coefﬁcients of corresponding powers of x on either side of this expression to make it an identity (that is, true for all x). STEP 6 Substitute the values of the coefﬁcients determined in Step 5 into the expression in Step 4 to obtain the required partial fraction representation. STEP 7 If n ≥ d, use long division to divide the denominator into the numerator to obtain the sum of a polynomial of degree n − d of the form T0 + T1 x + T2 x 2 + · · · + Tn−d x n−d , together with a remainder term in the form of a rational function R(x) of the type just considered. Find the partial fraction representation of the rational function R(x) using Steps 1 to 6. The required partial fraction representation is then the sum of the polynomial found by long division and the partial fraction representation of R(x).

EXAMPLE 1.15

Find the partial fraction representations of (a) F(x) =

x2 (x + 1)(x − 2)(x + 3)

and

(b) F(x) =

2x 3 − 4x 2 + 3x + 1 . (x − 1)2

Solution (a) All terms in the denominator are linear factors, so by Step 1 the appropriate form of partial fraction representation is A B C x2 = + + . (x + 1)(x − 2)(x + 3) x+1 x−2 x+3 Cross multiplying, we obtain x 2 = A(x − 2)(x + 3) + B(x + 1)(x + 3) + C(x + 1)(x − 2).

Section 1.6

Partial Fractions

29

Setting x = −1 makes the terms in B and C vanish and gives A = −1/6. Setting x = 2 makes the terms in A and C vanish and gives B = 4/15, whereas setting x = −3 makes the terms in A and B vanish and gives C = 9/10, so x2 −1 4 9 = + + . (x + 1)(x − 2)(x + 3) 6(x + 1) 15(x − 2) 10(x + 3) (b) The degree of the numerator exceeds that of the denominator, so from Step 7 it is necessary to start by dividing the denominator into the numerator longhand to obtain 2x 3 − 4x 2 + x + 3 3−x = 2x + . 2 (x − 1) (x − 1)2 We now seek a partial fraction representation of (3 − x)/(x − 1)2 by using Step 1 and writing 3−x B A + = . 2 (x − 1) x − 1 (x − 1)2 When we multiply by (x − 1)2 , this becomes 3 − x = A(x − 1) + B. Equating the constant terms gives 3 = −A+ B, whereas equating the coefﬁcients of x gives −1 = Aso that B = 2. Thus, the required partial fraction representation is 2x 3 − 4x 2 + x + 3 2 1 + = 2x + . (x − 1)2 1 − x (x − 1)2 An examination of the way the undetermined coefﬁcients were obtained in (a) earlier, where the degree of the numerator is less than that of the denominator and linear factors occur in the denominator, leads to a simple rule for ﬁnding the undetermined coefﬁcients called the “cover-up rule.” The cover-up rule Let a partial fraction decomposition be required for a rational function N(x)/D(x) in which the degree of the numerator N(x) is less than that of the denominator D(x) and, when factored, let D(x) contain some linear factors (factors of degree 1). Let (x − α) be a linear factor of D(x). Then the unknown coefﬁcient K in the term K/(x − α) in the partial fraction decomposition of N(x)/D(x) is obtained by “covering up” (ignoring) all of the other terms in the partial fraction expansion, multiplying the remaining expression N(x)/D(x) = K/(x − α) by (x − α), and then determining K by setting x = α in the result. To illustrate the use of this rule we use it in case (a) given earlier to ﬁnd Afrom the representation x2 A B C = + + . (x + 1)(x − 2)(x + 3) x+1 x−2 x+3

30

Chapter 1

Review of Prerequisites

We “cover up” (ignore) the terms involving B and C, multiply through by (x + 1), and ﬁnd A from the result x2 =A (x − 2)(x + 3)

completing the square

by setting x = −1, when we obtain A = −1/6. The undetermined coefﬁcients B and C follow in similar fashion. Once a partial fraction representation of a function has been obtained, it is often necessary to express any quadratic x 2 + px + q that occurs in a denominator in the form (x + A)2 + B, where A and B may be either positive or negative real numbers. This is called completing the square, and it is used, for example, when integrating rational functions and when ﬁnding inverse Laplace transforms. To ﬁnd A and B we set x 2 + px + q = (x + A)2 + B = x 2 + 2Ax + A2 + B, and to make this an identity we now equate the coefﬁcients of corresponding powers of x on either side of this expression: (coefﬁcients of x 2 ) (coefﬁcients of x) (constant terms)

1 = 1 (this tells us nothing) p = 2A q = A2 + B.

Consequently A = (1/2) p and B = q − (1/4) p2 , and so the result obtained by completing the square is x 2 + px + q = [x + (1/2) p]2 + q − (1/4) p2 . If the more general quadratic ax 2 + bx + c occurs, all that is necessary to reduce it to this same form is to write it as ax 2 + bx + c = a[x 2 + (b/a)x + c/a], and then to complete the square using p = b/a and q = c/a. EXAMPLE 1.16

Complete the square in the following expressions: (a) x 2 + x + 1. (b) x 2 + 4x. (c) 3x 2 + 2x + 1. Solution (a) p = 1, q = 1, so A = 1/2, B = 3/4, and hence x 2 + x + 1 = (x + 1/2)2 + 3/4. (b) p = 4, q = 0, so A = 2, B = −4, and hence x 2 + 4x = (x + 2)2 − 4. (c) 3x 2 + 2x + 1 = 3[x 2 + (2/3)x + 1/3] and so p = 2/3, q = 1/3, from which it follows that A = 1/3 and B = 2/9, so 3x 2 + 2x + 1 = 3{(x + 1/3)2 + 2/9}. Further information and examples of partial fractions can be found in any one of references [1.1] to [1.7].

Section 1.7

Fundamentals of Determinants

31

EXERCISES 1.6 Express the rational functions in Exercises 1 through 8 in terms of partial fractions using the method of Section 1.6, and verify the results by using computer algebra to determine the partial fractions. 1. 2. 3. 4. 5.

(3x + 4)/(2x 2 + 5x + 2). (x 2 + 3x + 5)/(2x 2 + 5x + 3). (3x − 7)/(2x 2 + 9x + 10). (x 2 + 3x + 2)/(x 2 + 2x − 3). (x 3 + x 2 + x + 1)/[(x + 2)2 (x 2 + 1)].

1.7

6. (x 2 − 1)/(x 2 + x + 1). 7. (x 3 + x 2 + x + 1)/{(x + 2)2 (x + 1)}. 8. (x 2 + 4)/(x 3 + 3x 2 + 3x + 1). Complete the square in Exercises 9 through 14. 9. x 2 + 4x + 5. 10. x 2 + 6x + 7. 11. 2x 2 + 3x − 6.

12. 4x 2 − 4x − 3. 13. 2 − 2x + 9x 2 . 14. 2 + 2x − x 2 .

Fundamentals of Determinants A determinant of order n is a single number associated with an array A of n2 numbers arranged in n rows and n columns. If the number in the ith row and jth column of a determinant is ai j , the determinant of A, denoted by det A and sometimes by |A|, is written a11 a12 . . . a1n a a22 . . . a2n (13) det A = |A| = 21 . . . . . . . . . . . . . an1 an2 . . . ann It is customary to refer to the entries ai j in a determinant as its elements. Notice the use of vertical bars enclosing the array A in the notation |A| for the determinant of A, as opposed to the use of the square brackets in [A] that will be used later to denote the matrix associated with an array A of quantities in which the number of rows need not be equal to the number of columns. The value of a ﬁrst order determinant det A with the single element a11 is deﬁned as a11 so that det[a11 ] = a11 or, in terms of the alternative notation for a determinant, |a11 | = a11 . This use of the notation |.| to signify a determinant should not be confused with the notation used to signify the absolute value of a number. The second order determinant associated with an array of elements containing two rows and two columns is deﬁned as a11 a12 = a11 a22 − a12 a21 , det A = (14) a21 a22 so, for example, using the alternative notation for a determinant we have 9 3 −7 −4 = 9(−4) − (−7)3 = −15. Notice that interchanging two rows or columns of a determinant changes its sign. We now introduce the terms minor and cofactor that are used in connection with determinants of all orders, and to do so we consider the third order determinant a11 a12 a13 (15) det A = a21 a22 a23 . a31 a32 a33

32

Chapter 1

Review of Prerequisites

minors and cofactors

The minor Mi j associated with ai j , the element in the ith row and jth column of det A, is deﬁned as the second order determinant obtained from det A by deleting the elements (numbers) in its ith row and jth column. The cofactor Ci j of an element in the ith row and jth column of the det A in (15) is deﬁned as the signed minor using the rule Ci j = (−1)i+ j Mi j .

(16)

With these ideas in mind, the determinant det A in (15) is deﬁned as det A =

3

a1 j (−1)1+ j det M1 j

j=1

= a11 M11 − a12 M12 + a13 M13 . If we introduce the cofactors Ci j , this last result can be written det A = a11 C11 + a12 C12 + a13 C13 ,

(17)

and more concisely as det A =

3

a1 j C1 j .

(18)

j=1

Result (18), or equivalently (17), will be taken as the deﬁnition of a third order determinant. EXAMPLE 1.17

Evaluate the determinant

1 2 −2

3 −3 1 0 . 1 1

Solution

The minor M11 = 11 01 = (1)(1) − (0)(1) = 1, so the cofactor C11 = (−1)(1+1) M11 = 1. The minor M12 = −22 01 = (2)(1) − (0)(−2) = 2, so the cofactor C12 = (−1)(1+2) M12 = −2. The minor M13 = −22 11 = (2)(1) − (1)(−2) = 4, so the cofactor C13 = (−1)(1+3) M13 = 4.

Using (17) we have 1 3 −3 2 1 0 = (1)C11 + (3)C12 + (−3)C13 = (1)(1) + (3)(−2) + (−3)(4) = −17. −2 1 1

When expanded, (17) becomes det A = a11 a22 a33 − a11 a32 a23 − a12 a21 a33 + a12 a31 a23 + a13 a21 a32 − a13 a31 a22 ,

Section 1.7

Fundamentals of Determinants

33

and after regrouping these terms in the form det A = −a21 a12 a33 + a21 a32 a13 + a22 a11 a33 − a22 a31 a13 − a23 a11 a32 + a23 a31 a12 , we ﬁnd that det A = a21 C21 + a22 C22 + a23 C23 . Proceeding in this manner, we can easily show that det A may be obtained by forming the sum of the products of the elements of A and their cofactors in any row or column of det A. These results can be expressed symbolically as follows. Expanding in terms of the elements of the ith row:

det A = ai1 Ci1 + ai2 Ci2 + ai3 Ci3 =

3

ai j Ci j .

(19)

j=1

Laplace expansion theorem

Expanding in terms of the elements of the jth column:

det A = a1 j C1 j + a2 j C2 j + a3 j C3 j =

3

ai j Ci j .

(20)

i=1

Results (19) and (20) are the form taken by the Laplace expansion theorem when applied to a third order determinant. The extension of the theorem to determinants of any order will be made later in Chapter 3, Section 3.3. EXAMPLE 1.18

Expand the following determinant (a) in terms of elements of its ﬁrst row, and (b) in terms of elements of its third column: 1 2 4 |A| = 1 0 2 . 1 2 1 Solution (a) Expanding in terms of the elements of the ﬁrst row requires the three cofactors C11 = M11 , C12 = −M12 , and C13 = M13 , where 0 2 1 2 1 0 = 2, = −4, M12 = = −1, M13 = M11 = 2 1 1 1 1 2 so C11 = (−1)(1+1) (−4) = −4, C12 = (−1)(1+2) (−1) = 1, C13 = (−1)(1+3) (2) = 2, and so |A| = (1)(−4) + (2)(1) + (4)(2) = 6. (b) Expanding in terms of the elements of the third column requires the three cofactors C13 = M13 , C23 = −M23 , and C33 = M33 , where 1 0 1 2 1 2 = −2, = 2, M23 = = 0, M33 = M13 = 1 2 1 2 1 0 so C13 = (−1)(1+3) (2) = 2, C23 = 0, C33 = (−1)(3+3) (−2) = −2 and so |A| = (4)(2) + (2)(0) + (1)(−2) = 6.

34

Chapter 1

Review of Prerequisites

Two especially simple third order determinants are of the form a11 0 a11 a12 a13 0 det A = 0 a22 a23 and det A = a21 a22 0 . a31 a32 a33 0 0 a33 The ﬁrst of these determinants has only zero elements below the diagonal line drawn from its top left element to its bottom right one, and the second determinant has only zero elements above this line. This diagonal line in every determinant is called the leading diagonal. The value of each of the preceding determinants is easily seen to be given by the product a11 a22 a33 of the terms on its leading diagonal. Simpler still in form is the third order determinant a11 0 0 det A = 0 a22 0 = a11 a22 a33 , 0 0 a33 whose value a11 a22 a33 is again the product of the elements on the leading diagonal. For another approach to the elementary properties of determinants, see Appendix A16 of reference [1.2], and Chapter 2 of reference [2.1].

EXERCISES 1.7 Evaluate the determinants in Exercises 1 through 6 (a) in terms of elements of the ﬁrst row and (b) in terms of elements of the second column. −1 3 6 1 5 7 4. 2 1 4 . 1. 1 −1 1 . −1 3 1 1 2 1 1 0 −6 2 1 −1 3 . 5. 2 1 2. 2 6 −1 . 4 3 21 5 1 −1 1 5 −1 5 2 4 6. 2 1 −3 . 3. 1 2 1 . −4 1 3 1 5 1 7. On occasion the elements of a matrix may be functions, in which case the determinant may be a function. Evaluate the functional determinant 1 0 0 0 sin x − cos x . 0 cos x sin x 8. Determine the values of λ that make the following determinant vanish: 3 − λ 2 2 2 2−λ 0 . 2 0 4 − λ Hint: This is a polynomial in λ of degree 3. 9. A matrix is said to be transposed if its ﬁrst row is written as its ﬁrst column, its second row is written as its second

column . . . , and its last row is written as its last column. If the determinant is |A|, the determinant of AT , the transpose matrix A, is denoted by |AT |. Write out the expansion of |A| using (17) and reorder the terms to show that |A| = |AT |. 10. Use elimination to solve the system of linear equations a11 x1 + a12 x2 = b1 a21 x1 + a22 x2 = b2 for x1 and x2 , in which not both b1 and b2 are zero, and show that the solution can be written in the form x1 = D1 /|A|

and

x2 = D2 /|A|,

provided |A| = 0,

where |A| is the determinant of the matrix of coefﬁcients of the system b1 a12 a11 b1 a11 a12 . , D1 = , and D2 = |A| = a21 a22 b2 a22 a21 b2 Notice that D1 is obtained from |A| by replacing its ﬁrst column by b1 and b2 , whereas D2 is obtained from |A| by replacing its second column by b1 and b2 . This is Cramer’s rule for a system of two simultaneous equations. Use this method to ﬁnd the solution of x1 + 5x2 = 3 7x1 − 3x2 = −1.

Section 1.8

Continuity in One or More Variables

where |A| is the determinant of the matrix of coefﬁcients and Di is the determinant obtained from |A| by replacing its ith column by b1 , b2 , and b3 for i = 1, 2, 3. This is Cramer’s rule for a system of three simultaneous equations, and the method generalizes to a system of n linear equations in n unknowns. Use this method to ﬁnd the solution of

11. Repeat the calculation in Exercise 10 using the system of equations a11 x1 + a12 x2 + a13 x3 = b1 a21 x1 + a22 x2 + a23 x3 = b2 a31 x1 + a32 x2 + a33 x3 = b3 , in which not all of b1 , b2 , and b3 are zero, and show that provided |A| = 0, x1 = D1 /|A|,

1.8

x2 = D2 /|A|,

and

35

x1 + 2x2 − x3 = 2 x1 − 3x2 − 2x3 = −1 2x1 + x2 + 2x3 = 1.

x3 = D3 /|A|,

Continuity in One or More Variables If the function y = f (x) is deﬁned in the interval a ≤ x ≤ b, the interval is called the domain of deﬁnition of the function. The function f is said to have a limit at a point c in a ≤ x ≤ b, written limx→c f (x) = L, if for every arbitrarily small number ε > 0 there exists a number δ > 0 such that | f (x) − L| < ε

when |x − c| < δ.

(21)

This technical deﬁnition means that as x either increases toward c and becomes arbitrarily close to it, or decreases toward c and becomes arbitrarily close to it, so f (x) approaches arbitrarily close to the value L. Notice that it is not necessary for f (x) to be deﬁned at x = c, or, if it is, that f (c) assumes the value L. If f (x) has a limit L as x → c and in addition f (c) = L, so that lim f (x) = f (c) = L,

x→c

continuity from the right

then the function f is said to be continuous at c. It must be emphasized that in this deﬁnition of continuity the limiting operation x → c must be true as x tends to c from both the left and right. It is convenient to say that x approaches c from the left when it increases toward c and, correspondingly, to say that x approaches c from the right when it decreases toward it. The function f is continuous from the right at x = c if lim f (x) = f (c),

x→c+

continuity from the left

lim f (x) = f (c),

(24)

where now x → c− means that x increases toward c, causing x to tend to c from the left. The relationship among deﬁnitions (22), (23), and (24) is that f is continuous at the point c if lim f (x) = lim+ f (x) = f (c).

x→c−

continuous function

(23)

where the notation x → c+ means that x decreases toward c, causing x to tend to c from the right. Similarly, f is continuous from the left at x = c if x→c−

continuity at x = c

(22)

x→c

(25)

When expressed in words, this says that f is continuous at x = c if the limits of f as x tends to c from both the left and right exist and, furthermore, the limits equal the functional value f (c). A function f that is continuous at all points of a ≤ x ≤ b is said to be a continuous function on that interval. Graphically, a continuous function on a ≤ x ≤ b is a function whose graph is unbroken but not necessarily smooth. A function f is said

36

Chapter 1

Review of Prerequisites

Discontinuous Continuous from the left at x = d

y

Continuous at x = c

y

Discontinuous at x = c

k2

f (c)

k1 y = f(x )

y = f (x ) Continuous from the right 0

a

c

d

b x

a

c

(a)

b

x

(b)

FIGURE 1.8 (a) A continuous function for a < x < b. (b) A discontinuous function.

smooth function

continuous and piecewise smooth function discontinuous function

to be smooth over an interval if at each point of the graph the tangent lines to the left and right of the point are the same. Figure 1.8a shows the graph of a continuous function that is smooth over the intervals a ≤ x < c and c < x < b but has different tangent lines to the immediate left and right of x = c where the function is not smooth. A function such as this is said to be continuous and piecewise smooth over the interval a ≤ x ≤ b. A function f is said to be discontinuous at a point c if it is not continuous there. For a jump discontinuity we have lim f (x) = k1

x→c−

piecewise continuity

and

lim f (x) = k2 ,

x→c+

but k1 = k2 .

(26)

A function f is said to have a removable discontinuity at a point c if k1 = k2 in (26), but f (c) = k1 , as at the point c2 in Fig. 1.9. An example of a discontinuous function is shown in Fig. 1.8b where a jump discontinuity occurs at x = c. A function f is said to be piecewise continuous on an interval a ≤ x ≤ b if it is continuous on a ﬁnite number of adjacent subintervals, but discontinuous at the end points of the subintervals, as shown in Fig. 1.9. The notion of continuity of a function of several variables is best illustrated by considering a function f (x, y) of the two independent variables x and y. The function f deﬁned in some region of the (x, y)-plane D, say, is said to be continuous y Discontinuous at c1

Discontinuous at c3

y = f (x ) Discontinuous at c2

a

c1

c2

FIGURE 1.9 A piecewise continuous function.

c3

b

x

Section 1.8

Continuity in One or More Variables

37

at the point (a, b) in D if continuity of f(x, y)

lim

x→a,y→b

discontinuity of f(x, y)

f (x, y) = f (a, b),

(27)

and to be discontinuous otherwise. In this deﬁnition of continuity, it is important to recognize that a general point P at (x, y) is allowed to tend to the point (a, b) in D along any path in the (x, y)-plane that lies in D. Expressed differently, f will only be continuous at (a, b) if the limit in (27) is independent of the way in which the point (x, y) approaches the point (a, b). When this is true for all points in D, the function f is said to be continuous in D. The function f is, for instance, discontinuous at (a, b) if lim

x→a,y→b

f (x, y) = k,

but f (a, b) = k.

Sufﬁcient for showing that a function f is discontinuous at a point (a, b) is by demonstrating that two different limiting values of f are obtained if the point P at (x, y) is allowed to tend to (a, b) along two different straight-line paths. This approach can be used to show that the function xy f (x, y) = 2 x + a 2 y2 has no limit at the origin. If we allow the point P at (x, y) to tend to the origin along the straight line y = kx, with k an arbitrary constant, the function f becomes f (x, kx) =

k , 1 + a 2 k2

and it is seen from this that f is constant along each such line. However, the value of f on each line, and hence at the origin, depends on k, so f has no limit at the origin and so is discontinuous at that point, though f is deﬁned and continuous at all other points of the (x, y)-plane. An example of a function f (x, y) that is continuous everywhere except at points along a curve in the (x, y)-plane is shown in Fig. 1.10.

z = f (x, y )

us

o tinu con Dis long Γ a

z

y 0

Γ x

D

FIGURE 1.10 A function f (x, y) continuous everywhere except at points on .

38

Chapter 1

Review of Prerequisites

The extension of these deﬁnitions to functions of n variables is immediate and so will not be discussed. Discussions on continuity and its consequences can be found in any one of references [1.1] to [1.7].

1.9

Differentiability of Functions of One or More Variables The function f (x) deﬁned in a ≤ x ≤ b is said to be differentiable with the derivative f (c) at a point c inside the interval if the following limit exists: lim

h→0

differentiability of f(x)

left- and right-hand derivatives of f(x)

f (c + h) − f (c) = f (c). h

(28)

Here, as in the deﬁnition of continuity, for f to be differentiable at point c the limit must remain unchanged as h tends to zero through both positive and negative values. The function f is said to be differentiable in the interval a ≤ x ≤ b if it is differentiable at every point in the interval. When f is differentiable at a point c with derivative f (c), the number f (c) is the gradient, or slope, of the tangent line to the graph at the point (c, f (c)). A function with a continuous derivative throughout an interval is said to be a smooth function over the interval. The function f will be said to be nondifferentiable at any point c where the limit in (28) does not exist. Even when a function f is nondifferentiable at a point, it is possible that a special form of derivative can still be deﬁned to the left and right of the point if the requirement that the limit in (28) exists as h → 0 through both positive and negative values is relaxed. The function f has a right-hand derivative at a if the limit lim+

h→0

f (a + h) − f (a) h

(29)

exists, and a left-hand derivative at b if the limit lim

h→0−

ﬁrst order partial derivatives of f(x, y)

f (b + h) − f (b) . h

(30)

exists. When c is a speciﬁc point, f (c) is a number, but when x is a variable, f (x) becomes a function. Left- and right-hand derivatives are illustrated in Fig. 1.11. An important consequence of differentiability is that differentiability implies continuity, but the converse is not true. The ﬁrst order partial derivative with respect to x of the function f (x, y) of the two independent variables x and y at the point (a, b) is the number deﬁned by lim

h→0

f (a + h, b) − f (a, b) , h

(31)

Section 1.9

Differentiability of Functions of One or More Variables

39

left-hand derivative equal to slope of line

y

f (b) f (c) y = f (x)

right-hand derivative equal to slope of line right-hand derivative equal to slope of line

left-hand derivative equal to slope of line f (a)

a

c

b

x

FIGURE 1.11 Left- and right-hand derivatives as tangent lines.

provided the limit exists. The value of this partial derivative is denoted either by ∂ f/∂ x at (a, b), or by fx (a, b). The corresponding partial derivative at a general point (x, y) is the function fx (x, y). Similarly, the ﬁrst order partial derivative with respect to y of the function f (x, y) at the point (a, b) is the number deﬁned by the limit lim

k→0

second order partial derivatives of f(x, y)

f (a, b + k) − f (a, b) , k

(32)

provided the limit exists. The value of this partial derivative is denoted either by ∂ f/∂ y at (a, b), or by fy (a, b). At a general point (x, y) this partial derivative becomes the function fy (x, y). Higher order partial derivatives are deﬁned in a similar fashion leading, for example, to the second order partial derivatives ∂ 2 f/∂ x 2 = ∂/∂ x(∂ f/∂ x), ∂ 2 f/∂ y2 = ∂/∂ y(∂ f/∂ y), ∂ 2 f/∂ x∂ y = ∂/∂ y(∂ f/∂ x),

and

∂ 2 f/∂ y∂ x = ∂/∂ x(∂ f/∂ y).

A more compact notation for these same derivatives is fxx , fyy , fxy , and fyx , so that, for example fyx = ∂ 2 f/∂ y∂ x and fyy = ∂ 2 f/∂ y2 . mixed partial derivatives

THEOREM 1.3

The derivatives fxy and fyx are called mixed partial derivatives, and their relationship forms the statement of the next theorem, the proof of which can be found in any one of references [1.1] to [1.7]. Equality of mixed partial derivatives Let f, fx , fxy , and fyx all be deﬁned and continuous at a point (a, b) in a region. Then fxy (a, b) = fyx (a, b).

40

Chapter 1

Review of Prerequisites

total differential

This result, given conditions for the equality of mixed partial derivatives, is an important one, and use will be made of it on numerous occasions as, for example, in Chapter 18 when second order partial differential equations are considered. If z = f (x, y), the total differential dz of f is deﬁned as dz = (∂ f/∂ x) dx + (∂ f/∂ y) dy,

(33)

where dz, dx, and dy are differentials. Here, a differential means a small quantity, and the differential dz is determined by (33) when the differentials dx and dy are speciﬁed. When ∂ f/∂ x and ∂ f/∂ y are evaluated at a speciﬁc point (a, b), result (33) provides a linear approximation to f (x, y) near to the point (a, b). Although ﬁnite, the limits of the quotients of the differentials dz ÷ dx and dy ÷ dx as the differential dx → 0 are such that they become the values of the derivatives dz/dx and dy/dx, respectively, at a point (x, y) where ∂ f/∂ x and ∂ f/∂ y are evaluated.

1.10

Tangent Line and Tangent Plane Approximations to Functions

tangent line approximation

Let y = f (x) be deﬁned in the interval a ≤ x ≤ b and be differentiable throughout it. Then a tangent line (linear) approximation to f near a point x0 in the interval is given by yT = f (x0 ) + (x − x0 ) f (x0 ).

(34)

This linear expression approximates the function f close to x0 by the tangent to the graph of y = f (x) at the point (x0 , f (x0 )). This simple approximation has many uses; one will be in the Euler and modiﬁed Euler methods for solving initial value problems for ordinary differential equations developed in Chapter 19. EXAMPLE 1.19

Find a tangent line approximation to y = 1 + x 2 + sin x near the point x = α. Solution Setting x0 = α and substituting into (34) gives y ≈ 1 + α 2 + sin α + (x − α)(2α + cos α) for x close to α.

tangent plane approximation

Let the function z = f (x, y) be deﬁned in a region Dof the (x, y)-plane where it possesses continuous ﬁrst order partial derivatives ∂ f/∂ x and ∂ f/∂ y. Then a tangent plane (linear) approximation to f near any point (x0 , y0 ) in D is given by zT = f (x0 , y0 ) + (x − x0 ) fx (x0 , y0 ) + (y − y0 ) fy (x0 , y0 ).

(35)

This linear expression approximates the function f close to the point (x0 , y0 ) by a plane that is tangent to the surface z = f (x, y) at the point (x0 , y0 , f (x0 , y0 )). The tangent plane approximation in (35) is an immediate extension to functions of two variables of the tangent line approximation in (34), to which it simpliﬁes when only one independent variable is involved.

Section 1.11

Integrals

41

Both of these approximations are derived from the appropriate Taylor series expansions of functions discussed in Section 1.12 by retaining only the linear terms. EXAMPLE 1.20

Find the tangent plane approximation to the function z = x 2 − 3y2 near the point (1, 2). Solution Setting x0 = 1, y0 = 2 and substituting into (35) gives z ≈ −11 + 2(x − 1) − 12(y − 2) for (x, y) close to (1, 2).

1.11

Integrals

indeﬁnite and deﬁnite integrals

A differentiable function F(x) is called an antiderivative of the function f (x) on some interval if at each point of the interval dF/dx = f (x). If F(x) is any antiderivative of f (x), the indeﬁnite integral of f (x), written f (x) dx, is f (x) dx = F(x) + c, where c is an arbitrary constant called the constant of integration. The function f (x) is called the integrand of the integral. Thus, an indeﬁnite integral is a function, and an antiderivative and an indeﬁnite integral can only differ by an arbitrary additive constant. b The expression a f (x) dx, called a deﬁnite integral, is a number and may be interpreted geometrically as the area between the graph of f (x) and the lines x = a and x = b, for b > a, with areas above the x-axis counted as positive and those below it as negative. The relationship between deﬁnite integrals that are numbers and indeﬁnite integrals that are functions is given in the next theorem, included in which is also the mean value theorem for integrals. See the references at the end of the chapter for proofs and further information.

THEOREM 1.4

Fundamental theorem of integral calculus and the mean value theorem for integrals If F (x) is continuous in the interval a ≤ x ≤ b, throughout which F (x) = f (x), then

b

f (x) dx = F(b) − F(a).

a

Another result is

b

f (x) dx = (b − a) f (ξ ),

a

if f is differentiable, where the number ξ , although unknown, lies in the interval a < ξ < b. In this form the result is called the mean value theorem for integrals.

42

Chapter 1

Review of Prerequisites

An improper integral is a deﬁnite integral in which one or more of the following cases arises: (a) the integrand becomes inﬁnite inside or at the end of the interval of integration, or (b) one (or both) of the limits of integration is inﬁnite.

Types of Improper Integrals Case (a)

convergence and divergence of improper integrals

If the integrand of an integral becomes inﬁnite at a point c inside the interval of integration a ≤ x ≤ b as shown in Fig. 1.12a, the improper integral is said to exist if the limits in (36) exist. When the improper integral exists it is said to converge to the (ﬁnite) value of the following limit:

b

f (x) dx = lim

h→0 a

a

Cauchy principal value

c−h

f (x) dx + lim

b

k→0 c+k

f (x) dx.

(36)

In this deﬁnition h > 0 and k > 0 are allowed to tend to zero independently of each other. If, when the limit is taken, the integral is either inﬁnite or indeterminate, the integral is said to diverge. Some integrals of this type diverge when h and k are allowed to tend to zero independently of each other, but converge when the limit is taken with h = k, in which case the result of the limit is called the Cauchy principal value of the integral. Integrals of this type arise frequently when certain types of deﬁnite integral are evaluated in the complex plane by means of contour integration (see Chapter 15, Section 15.5).

Case (b) If a limit of integration in a deﬁnite integral is inﬁnite, say the upper limit as shown in Fig. 1.12b, then, when it exists, the improper integral is said to converge to the value of the limit

∞

f (x) dx = lim

R→∞ a

a

y

R

f (x) dx,

(37)

y

y = f (x)

y = f(x)

a

c (a)

b

x

x

a (b)

FIGURE 1.12 (a) f (x) is inﬁnite inside the interval of integration. (b) The interval of integration is inﬁnite in length.

Section 1.12

Taylor and Maclaurin Theorems

43

and the integral is divergent if the limit is either inﬁnite or indeterminate. If both limits are inﬁnite, the improper integral is said to converge to the value of the limit

∞

f (x) dx =

−∞

R

lim

f (x) dx

R→∞,S→∞ −S

(38)

when it exists, and the integral is said to be divergent if the limit is either inﬁnite or indeterminate. In (38) R and S are allowed to tend to inﬁnity independently of each other. Integrals of this type also have Cauchy principal values if the foregoing process leads to divergence, but the integrals are convergent when the limit is taken with R = S. Integrals of this type also occur when certain real integrals are evaluated by means of contour integration (see Chapter 15, Section 15.5). Elementary examples of convergent improper integrals of the types shown in (36) to (38) are

1 x p − x− p dx = − π cot pπ, x − 1 p 0 ∞ exp(−x) sin xdx = 1/2 and 1

∞

−∞

THEOREM 1.5

( p2 < 1), dx = π. 1 + x2

Differentiation under the integral sign — Leibniz’ rule If ξ (t), η(t), dξ/dt, dη/dt, f (x, t), and ∂ f/∂t are continuous for t0 ≤ t ≤ t1 and for x in the interval of integration, then

d dt

η(t) ξ (t)

f (x, t) dx =

η(t)

ξ (t)

dη dξ ∂ f (x, t) dx + f (η(t), t) − f (ξ (t), t) . ∂t dt dt

This theorem is used, for example, in Chapter 18 when discussing discontinuous solutions of a class of partial differential equations called conservation laws. Extensions of the theorem to functions of more variables are developed in Chapter 12, Section 12.3, where certain vector integral theorems are developed, and applications of the results of that section to ﬂuid mechanics are to be found in Chapter 12, Section 12.4. An application of Theorem 1.5 that is easily checked by direct calculation is d dt

t2

2t

(x 2 + t) dx =

t2

dx + (t 4 + t) · 2t − (4t 2 + t) · 2 = 2t 5 − 5t 2 − 4t.

2t

A proof of Leibniz’ rule can be found, for example, in Chapter 12 of reference [1.6].

1.12

Taylor and Maclaurin Theorems THEOREM 1.6

Taylor’s theorem for a function of one variable Let a function f (x) have derivatives of all orders in the interval a < x < b. Then for each positive integer n and

44

Chapter 1

Review of Prerequisites

each x0 in the interval f (x) = f (x0 ) + (x − x0 ) f (1) (x0 ) + +

(x − x0 )2 (2) f (x0 ) + · · · 2!

(x − x0 )n (n) f (x0 ) + Rn+1 (x), n!

where f (r ) (x) = dr f/dxr , and the remainder term Rn+1 (x) is given by Rn+1 (x) =

(x − x0 )n+1 (n+1) f (ξ ), (n + 1)!

for some ξ between x0 and x.

Taylor polynomial

Maclaurin’s theorem

Taylor’s theorem becomes the Taylor series for f (x) when n is allowed to become inﬁnite, and if the remainder term is neglected in Taylor’s theorem the result is called the Taylor polynomial approximation to f (x) of degree n. The Taylor polynomial of degree 1 is simply the tangent line approximation to f at x0 given in (34). Taylor’s theorem reduces to Maclaurin’s theorem if x0 = 0, and if we allow n to become inﬁnite in Maclaurin’s theorem, it becomes the Maclaurin series for f (x). A special case of Theorem 1.6 arises when Taylor’s theorem is terminated with the term R1 (x), corresponding to n = 0, because the result can be written f (x) − f (x0 ) = f (ξ ), x − x0

mean value theorem

(39)

with ξ between x0 and x, and in this form it is called the mean value theorem for derivatives (see the last result of Theorem 1.4). A Taylor series is an example of an inﬁnite series called a power series, the general form of which is ∞

an (x − x0 )n = a0 + a1 (x − x0 ) + a2 (x − x0 )2 + · · · .

(40)

n=0

In (40) the quantity x is a variable, the numbers ai are the coefﬁcients of the power series, the constant x0 is called the center of the series, or the point about which the series is expanded, and unless otherwise stated, x, x0 , and the ai are real numbers, so the power series is a function of x. A power series is said to converge for a given value of x if the sum of the inﬁnite series for this value of x is ﬁnite. If the sum is inﬁnite, or is not deﬁned, the power series will be said to diverge for that value of x. Power series converge in an interval x0 − R < x < x0 + R, where the number R is called the radius of convergence of the series. Expressions for R are derived in Section 15.1. The interval x0 − R < x < x0 + R is called the interval of convergence of the power series. A power series converges for all x inside the interval of convergence and diverges for all x outside it, and the series may, or may not, converge at the end points of the interval. The convergence properties of power series are shown diagramatically in Fig. 1.13, and results (40) and combining expressions for R with

Section 1.12

Taylor and Maclaurin Theorems

Interval of Convergence

Divergence x0 − R

x0

45

Divergence x0 + R

FIGURE 1.13 Interval of convergence of a power series with center x0 .

(40) gives the following theorem (see the references at the end of the chapter for real variable proofs of the following results and for more information). THEOREM 1.7

Ratio test and nth root test for the convergence of power series The power series ∞

an (x − x0 )n = a0 + a1 (x − x0 ) + a2 (x − x0 )2 + · · ·

n=0

radius and interval of convergence

converges in the interval of convergence x0 − R < x < x0 + R, where the radius of convergence R is determined by either of the formulas (a) R = 1/ lim |an+1 /an | n→∞

or

(b) R = 1/ lim |an |1/n . n→∞

The power series will diverge outside the interval of convergence, and its behavior at the ends of the interval of convergence must be determined separately. A simple result on the convergence of a series that is often useful is the alternating series test. An alternating series is so named because the signs of successive terms of the series alternate in sign. THEOREM 1.8

The alternating series test for convergence The alternating series converges if an > 0 and an+1 < an for all n and limn→∞ an = 0.

∞

n+1 an n=1 (−1)

The following theorem on the differentiation and integration of power series is often needed, and it is a real variable form of a result proved later in Chapter 15 when complex power series are studied. THEOREM 1.9

Differentiation and integration of power series Let a power series have an interval of convergence x0 − R < x < x0 + R. Then the series may be differentiated and integrated term by term, and in each case the resulting series will have the same interval of convergence as the original series. In addition, within an interval of convergence common to any two power series, the series may be scaled by a constant and added or subtracted term by term and the resulting power series will have the same common interval of convergence. The simplest form of Taylor’s theorem for a function of two variables that ﬁnds many applications is given in the next theorem.

THEOREM 1.10

Taylor’s theorem for a function of two variables Let f (x, y) be deﬁned for a < x < b and c < y < d and have continuous partial derivatives up to and including

46

Chapter 1

Review of Prerequisites

those of order 2. Then for x0 and y0 any points such that a < x0 < b and c < y0 < d, f (x, y) = f (x0 , y0 ) + (x − x0 ) fx (x0 , y0 ) + (y − y0 ) fy (x0 , y0 ) 1 (x − x0 )2 fxx (x0 + ξ, y0 + η) + 2(x − x0 )(y − y0 ) + 2! × fxy (x0 + ξ, y0 + η)(y − y0 )2 fyy (x0 + ξ, y0 + η) , where the numbers ξ and η are unknown, but ξ lies between x0 and x and η lies between y0 and y. The group of second order partial derivatives in Theorem 1.10 forms the remainder term, and when these derivatives are ignored, the result reduces to the tangent plane approximation to f (x, y) at the point (x0 , y0 ) given in (35). More information on Taylor’s theorem and series can be found, for example, in reference [1.2].

1.13

Cylindrical and Spherical Polar Coordinates and Change of Variables in Partial Differentiation Mathematical problems formulated using a particular coordinate system, such as cartesian coordinates, often need to be reexpressed in terms of a different coordinate system in order to simplify the task of ﬁnding a solution. When partial derivatives occur in the formulation of problems, it becomes necessary to know how they transform when a different coordinate system is used. The fundamental theorem governing the transformation of partial derivatives under a change of variables takes the following form (see the references at the end of the chapter for the proof of Theorem 1.11 and for more examples of its use).

THEOREM 1.11

Change of variables in partial differentiation Let f (x1 , x2 , . . . , xn ) be a differentiable function with respect to the n independent variables x1 , x2 , . . . , xn , and let the n new independent variables u1 , u2 , . . . , un be determined in terms of x1 , x2 , . . . , xn by x1 = X1 (u1 , u2 , . . . , un ),

x2 = X2 (u1 , u2 , . . . , un ), . . . ,

xn = Xn (u1 , u2 , . . . , un ),

where X1 , X2 , . . . , Xn are differentiable functions of their arguments. Then, if as a result of the change of variables the function f (x1 , x2 , . . . , xn ) becomes the function F(X1 , X2 , . . . , Xn ), and using chain rules we have ∂F ∂ f ∂ X1 ∂ f ∂ X2 ∂ f ∂ Xn = + +···+ ∂u1 ∂ x1 ∂u1 ∂ x2 ∂u1 ∂ xn ∂u1 ∂F ∂ f ∂ X1 ∂ f ∂ X2 ∂ f ∂ Xn = + +···+ ∂u2 ∂ x1 ∂u2 ∂ x2 ∂u2 ∂ xn ∂u2 .................................... ∂ f ∂ X1 ∂ f ∂ X2 ∂ f ∂ Xn ∂F = + +···+ . ∂un ∂ x1 ∂un ∂ x2 ∂un ∂ xn ∂un

(41)

Section 1.13

Cylindrical and Spherical Polar Coordinates and Change of Variables in Partial Differentiation

47

To ﬁnd higher order partial derivatives it is necessary to express the relationships between the operations of differentiation in the two coordinate systems, rather than between the actual derivatives themseves. This can be accomplished by rewriting the results of Theorem 1.11 in the form of partial differential operators as follows: ∂ ∂ X1 ∂ ∂ X2 ∂ ∂ Xn ∂ ≡ + + ··· + ∂u1 ∂u1 ∂ x1 ∂u1 ∂ x2 ∂u1 ∂ xn ∂ ∂ X1 ∂ ∂ X2 ∂ ∂ Xn ∂ ≡ + + ··· + ∂u2 ∂u2 ∂ x1 ∂u2 ∂ x2 ∂u2 ∂ xn ....................................

(42)

∂ ∂ X1 ∂ ∂ X2 ∂ ∂ Xn ∂ ≡ + + ··· + . ∂un ∂un ∂ x1 ∂un ∂ x2 ∂un ∂ xn When expressed in this form the relationships between the partial differentiation operations ∂/∂ x1 , ∂/∂ x2 , . . . , ∂/∂ xn and ∂/∂u1 , ∂/∂u2 , . . . , ∂/∂un become clear. This interpretation is needed when ﬁnding higher order partial derivatives such as ∂ 2 F/∂u2 ∂u1 , because ∂F ∂ X1 ∂ ∂F ∂ ∂ X2 ∂ ∂ Xn ∂ ∂2 F = . = + + ··· + ∂u2 ∂u1 ∂u1 ∂u2 ∂u1 ∂ x1 ∂u1 ∂ x2 ∂u1 ∂ xn ∂u2 An important combination of partial derivatives that occurs throughout physics and engineering is called the Laplacian of a function. When a twice differentiable function f (x, y, z) of the cartesian coordinates x, y, and z is involved, the Laplacian of f , denoted by f and sometimes by ∇ 2 f , read “del squared f ,” takes the form f = ∇2 f =

∂2 f ∂2 f ∂2 f + 2 + 2. 2 ∂x ∂y ∂z

(43)

Cylindrical Polar Coordinates (r, θ, z) The cylindrical polar coordinate system (r, θ, z) is illustrated in Fig. 1.14, and its relationship to cartesian coordinates is given by x = r cos θ,

y = r sin θ,

z = z,

with 0 ≤ θ < 2π and r ≥ 0.

(44)

Spherical Polar Coordinates (r, φ, θ) The spherical polar coordinate system (r, φ, θ ) shown in Fig. 1.15 is related to cartesian coordinates by x = r sin θ cos φ,

y = r sin θ sin φ, z = r cos θ, with 0 ≤ θ ≤ π, 0 ≤ φ < 2π.

(45)

The derivation of the formulas for the change of variables in functions of several variables can be found in any one of references [1.1] to [1.7], where cylindrical and

48

Chapter 1

Review of Prerequisites

z

z

P (r, θ, z)

P (r, φ, θ) θ

z

r 0

0 θ

r

y

x

y

P

Q

P y

x

φ y

x

z

x

FIGURE 1.14 Cylindrical polar coordinates (r, θ, z).

FIGURE 1.15 Spherical polar coordinates (r, φ, θ ).

spherical polar coordinates are also discussed. Information on general orthogonal coordinate systems can be found in references [G.3] and [2.3].

EXERCISES 1.13 1. By making the change of variables x = r cos θ, y = r sin θ, z = z, in the function f (x, y, z), when it becomes the function F(r, θ, z), show that in cylindrical polar coordinates ∂F ∂f ∂f = cos θ + sin θ , ∂r ∂x ∂y ∂f ∂f ∂F ∂f ∂F = −r sin θ + r cos θ , = . ∂θ ∂x ∂y ∂z ∂z 2. Use the results of Exercise 1 to show that in cylindrical polar coordinates the Laplacian ∂ f ∂ f ∂ f + + 2 becomes ∂ x2 ∂ y2 ∂z 2 ∂ F 1 ∂2 F 1 ∂F ∂2 F F = + 2 + + , 2 2 ∂r r ∂r r ∂θ ∂z2 and hence that an equivalent form of F is ∂F 1 ∂ ∂F ∂ ∂F 1 ∂ r + + r . F = r ∂r ∂r r ∂θ ∂θ ∂z ∂z f =

2

2

2

3. By making the change of variable x = r sin θ cos φ, y = r sin θ sin φ, z = r cos θ in the function f (x, y, z), when it

becomes F(r, φ, θ ), show that in spherical polar coordinates ∂F ∂f ∂f ∂f = sin θ cos φ + sin φ sin θ + cos φ ∂r ∂x ∂y ∂z ∂f ∂f ∂f ∂F = r cos φ cos θ + r cos φ sin θ − r sin φ ∂φ ∂x ∂y ∂z ∂f ∂f ∂F = −r sin φ sin θ + r sin φ cos θ . ∂z ∂x ∂y 4. Use the results of Exercise 3 to show that in spherical polar coordinates the Laplacian f =

∂2 f ∂2 f ∂2 f + + 2 2 2 ∂x ∂y ∂z

becomes 2 ∂ F 1 1 ∂ 2∂F r + F = 2 r ∂r ∂r r 2 sin2 θ ∂φ 2 1 ∂F ∂ + 2 sin θ . r sin θ ∂θ ∂θ

Section 1.14

1.14

Inverse Functions and the Inverse Function Theorem

49

Inverse Functions and the Inverse Function Theorem In mathematics and its applications it is often necessary to ﬁnd the inverse of a function y = f (x) so x can be expressed in the form x = g(y), and when this can be done the function g is called the inverse of f and is such that y = f (g(y)). When f is an arbitrary function its inverse is often denoted by f −1 , and this superscript notation is also used to denote the inverse of trigonometric functions so if, for example, y = sin x, the inverse sine function is written sin−1 , so that x = sin−1 y. However, the notation y = arcsin y is also used with the understanding that the notations arcsin and sin−1 are equivalent. A trivial example of a function whose inverse can be found unambiguously is y = ax + b, because provided a = 0 we can write x = (y − b)/a for all x and y. This is not the case, however, when trigonometric functions are involved, because the function y = sin x will give a unique value of y for any given x, but given y there are inﬁnitely many values of x for which y = sin x. This and similar inverse trigonometric functions are considered in elementary calculus courses. There the multivalued nature of the inverse sine function is resolved by restricting it to make y lie in a speciﬁc interval chosen so that one y corresponds to one x and, conversely, one x corresponds to one y. This situation is described by saying that the relationship between x and y is one-to-one. Speciﬁcally, in the case of the sine function, this is accomplished by requiring that if x = sin y, the inverse function y = Arcsin x is restricted so its principal value lies in the interval −π/2 ≤ Arcsinx ≤ π/2, where the domain of deﬁnition of the inverse function is −1 ≤ x ≤ 1. A different possibility that arises frequently is when x and y are related by an equation of the form f (x, y) = 0 from which it is impossible to extract either x as a function of y, or y as a function of x in terms of known functions. A typical example of this type is f (x, y) = x 2 − 2y2 − sin xy. To make matters precise, if x and y are related by an equation f (x, y) = 0, then if a function y = g(x) exists such that f (x, g(x)) = 0, the function y = g(x) is said to be deﬁned implicitly by f (x, y) = 0. Although it is often not possible to ﬁnd the function g(x), it is still necessary to know when, in a neighborhood of a point (x0 , y0 ), given a value of x, a unique value of y can be found, sometimes only numerically. The implicit function theorem that follows is seldom mentioned in ﬁrst calculus courses because its proof involves certain technicalities, but it is quoted here in the simplest possible form because of its fundamental importance and the fact that is it frequently used by implication.

THEOREM 1.12

The implicit function theorem Let f (x, y) and fy (x, y) be continuous in a region D of the (x, y)-plane and let (x0 , y0 ) be a point inside D, where f (x0 , y0 ) = 0 and fy (x0 , y0 ) = 0. Then (i) There is a rectangle R inside D containing (x0 , y0 ) at all points of which there can be found a unique y such that f (x, y) = 0. (ii) If the value of y is denoted by g(x), then y0 = g(x0 ), with f (x, g(x)) = 0, and g(x) is continuous inside R.

50

Chapter 1

Review of Prerequisites

(iii) If, in addition, fx (x, y) is continuous in D then g(x) is differentiable in R and g (x) = − ffxy (x,g(x)) . (x,g(x)) In general terms, the implicit function theorem gives conditions that ensure the existence of an inverse function that is continuous and smooth enough to be differentiable. The theorem has a more general form involving functions f (x1 , x2 , . . . , xn ) of n variables, though this will not be given here. The interested reader can ﬁnd accounts of the implicit function theorem and some of its generalizations in references [1.4], [1.6], and [5.1].

CHAPTER 1

TECHNOLOGY PROJECTS Project 1 Linear Difference Equations and the Fibonacci Sequence In Italy in 1202, Leonardo of Pisa, also known as Fibonacci, posed the following question. Let a newly born pair of rabbits produce two offspring each month, with breeding starting when they are 2 months old. Assuming that the pair of offspring start breeding in the same fashion when 2 months old, and that the process continues thereafter in a similar manner with no deaths, how many pairs of rabbits will there be after n months? If un , is the number of pairs of rabbits after n months, the production of rabbits can be represented by the linear difference equation, or recurrence relation, un+2 = un+1 + un , where the sequence of numbers ur with r = 1, 2, . . . is generated by setting u1 = 1 and u2 = 1, since this represents the initial pair of rabbits that began the breeding process. A simple calculation using this difference equation shows that the sequence of numbers generated in this manner that represents the number of pairs of rabbits present each month is 1, 1, 2, 3, 5, 8, . . . , and this is called the Fibonacci sequence. This sequence is found to occur in the study of regular solids, in numerical analysis, and elsewhere in mathematics. A linear difference equation of the form un+2 = aun+1 + bun , with a and b real numbers, can be solved by substituting un = Aλn into the difference equation and ﬁnding the two roots λ1 and λ2 of the resulting quadratic equation in λ. When λ1 = λ2 , the general solution is un = A1 λn1 + Aλn2 , and when λ1 = λ2 = λ , say, the general solution is un = (A1 + nA2 )μn . The arbitrary constants A1 and A2 are found by requiring un to satisfy some given conditions of the form u1 = α and u2 = β,

where the numbers α and β specify the way the sequence starts (the initial conditions). Use this method to show that the solution un for the Fibonacci sequence is √ n √ n 1 1 5 1+ 5 un = √ , 2 2 5

(

) (

)

for n = 1, 2, . . . . Make use of computer algebra to generate the ﬁrst 30 terms of the Fibonacci sequence directly from the difference equation, and verify that the results are in agreement wïth the preceding formula. Use computer √algebra to show that limn→∞ (un /un 1 ) = 12 ( 5 + 1). This number is called the golden mean, and in art and architecture it represents the ratio of the sides of a rectangle that is considered to have the most pleasing appearance. Project 2 Erratic Behavior of a Sequence Generated by a Difference Equation 1. Not all difference equations generate sequences of numbers that evolve steadily as happens with the Fibonacci sequence. Use computer algebra to generate the ﬁrst 20 terms of the sequence produced by the difference equation un+2 = 2un+1

5un

with u1 = 1, u2 =

3,

and observe its erratic behavior. Use the method of Project 1 to determine the analytical solution, and by means of computer algebra conﬁrm that the two results are in agreement. Examine the analytical solution and explain why the behavior of the sequence of terms is so erratic. 2. Construct a difference equation of your own in which the roots λ1 and λ2 are equal. Find the analytical solution and use computer algebra to determine the ﬁrst 20 terms of the sequence. Verify that these terms are in agreement with the ones generated directly from the difference equation.

51

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PART

TWO

VECTORS AND MATRICES

2 Chapter 3 Chapter

Chapter

4

Vectors and Vector Spaces Matrices and System of Linear Equations Eigenvalues, Eigenvectors, and Diagonalization

53

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2

C H A P T E R

Vectors and Vector Spaces

E

ngineers, scientists, and physicists need to work with systems involving physical quantities that, unlike the density of a solid, cannot be characterized by a single number. This chapter is about the algebra of important and useful quantities called vectors that arise naturally when studying physical systems, and are deﬁned by an ordered group of three numbers (a, b, c). Vectors are of fundamental importance and they play an essential role when the laws governing engineering and physics are expressed in mathematical terms. A scalar quantity is one that is completely described when its magnitude is known, such as pressure, temperature, and area. A vector is a quantity that is completely speciﬁed when both its magnitude and direction are given, such as force, velocity, and momentum. A vector can be described geometrically as a directed straight line segment, with its length proportional to the magnitude of the vector, the line representing the vector parallel to the line of action of the vector, and an arrow on the line showing the direction along the line, or the sense, in which the vector acts. This geometrical interpretation of a vector is valuable in many ways, as it can be used to add and subtract vectors and to multiply them by a scalar, since this merely involves changing their magnitude and sense, while leaving the line to which they are parallel unchanged. However, to perform more general algebraic operations on vectors some other form of representation is required. The one that is used most frequently involves describing a vector in terms of what are called its components along a set of three mutually orthogonal axes, which are usually taken to be the axes O{x, y, z} in the cartesian coordinate system. Here, by the component of a vector along a given line l , we mean the length of the perpendicular projection of the vector onto the line l . We will see later that this cartesian representation of a vector identiﬁes it completely in terms of three components and enables algebraic operations to be performed on it. In particular, it allows the introduction of the scalar product, or dot product, of two vectors that results in a scalar, and a vector product, or cross product, of two vectors that leads to a vector. Finally, vectors and their algebra will be generalized to n space dimensions, leading to the concept of a vector space and to some related ideas.

55

56

Chapter 2

2.1

Vectors and Vector Spaces

Vectors, Geometry, and Algebra

M

scalar

vector

directed straight line segment

translation

any quantities are completely described once their magnitude is known. A typical example of a physical quantity of this type is provided by the temperature at a given point in a room that is determined by the number specifying its value measured on a temperature scale, such as degrees F or degrees C. A quantity such as this is called a scalar quantity, and different examples of mathematical and physical scalar quantities are real numbers, length, area, volume, mass, speed, pressure, chemical concentration, electrical resistance, electric potential, and energy. Other physical quantities are only fully speciﬁed when both their magnitude and direction are given. Quantities like this are called vector quantities, and a typical example of a vector quantity arises when specifying the instantaneous motion of a ﬂuid particle in a river. In this case both the particle speed and its direction must be given if the description of its motion is to be complete. Speed in a given direction is called velocity, and velocity is a vector quantity. Some other examples of vector quantities are force, acceleration, momentum, the heat ﬂow vector at a point in a block of metal, the earth’s magnetic ﬁeld at a given location, and a mathematical quantity called the gradient of a scalar function of position that will be deﬁned later. By deﬁnition, the magnitude of a vector quantity is a nonnegative number (a scalar) that measures its size without regard to its direction, so, for example, the magnitude of a velocity is a speed. A convenient geometrical representation of a vector is provided by a straight line segment drawn in space parallel to the required direction, with an arrowhead indicating the sense in which the vector acts along the line segment, and the length of the line segment proportional to the magnitude of the vector. This is called a directed straight line segment, and by deﬁnition all directed straight line segments that are parallel to one another and have the same sense and length are regarded as equal. Expressed differently, moving a directed straight line segment parallel to itself so that its length remains the same and its arrow still points in the same direction leaves the vector it represents unchanged. A shift of a directed straight line segment of this type is called a translation of the vector it represents. For this reason the terms directed straight line segment and vector can be used interchangeably. Some examples of vectors that are equal through translation are shown in Fig. 2.1. It must be emphasized that geometrical representations of vectors as directed straight line segments in space are deﬁned without reference to a speciﬁc coordinate system. This purely geometrical interpretation of vectors ﬁnds many applications, though a different form of representation is necessary if an effective vector algebra is to be developed for use with the calculus. An analytical representation of vectors that allows a vector algebra to be constructed with this purpose in mind can be based on a general coordinate system. However, throughout this chapter only rectangular cartesian coordinates will be used because they provide a simple and natural way of representing vectors.

FIGURE 2.1 Equal geometrical vectors.

Section 2.1

Vectors, Geometry, and Algebra

57

z

y

x FIGURE 2.2 A right-handed rectangular cartesian coordinate system.

right-handed system

In rectangular cartesian coordinates the x-, y-, and z-axes are all mutually orthogonal (perpendicular), and the positive sense along the axes is taken to be in the direction of increasing x, y, and z. The orientation of the axes will always be such that the positive direction along the z-axis is the one in which a right-handed screw (such as a corkscrew) aligned with the z-axis will advance when rotated from the positive x-axis to the positive y-axis, as shown in Fig. 2.2. A system of axes with this property is called a right-handed system. The end of a vector toward which the arrow points will be called the tip of the vector, and the other end its base. Because a vector is invariant under a translation, there is no loss of generality in taking its base to be located at the origin O of the coordinate system, and its tip at a point P with the coordinates (a1 , a2 , a3 ), say, as shown in Fig. 2.3. An application of the Pythagoras theorem to the triangle OPP

z

a3

OP

=

2 + (a 1

1 /2 2) 2 + a3 a2

P (a1, a2, a3)

a2

O

y

OP

'= (a

1

a1

2

+a 2 2)

1/ 2

P

x FIGURE 2.3 The vector from O to P and its components a1 , a2 , and a3 in the x-, y-, z-coordinate system.

58

Chapter 2

Vectors and Vector Spaces

magnitude, unit vector, and components

ordered number triple

norm and modulus

shows the length of the line from O to P to be (a12 + a22 + a32 )1/2 . This length is proportional to the magnitude of the vector it represents, and as the base of the vector is at O, the sense of the vector is from O to P. For convenience, the constant of proportionality will be taken to be 1, so a directed straight line segment of unit length will represent a vector of magnitude 1 and so will be called a unit vector. Using this convention, the vector represented by the line from O to P in Fig. 2.3 has magnitude (a12 + a22 + a32 )1/2 . The three numbers a1 , a2 , and a3 , in this order, that deﬁne the vector from O to P are called its components in the x, y, and z directions, respectively. A set of three numbers a1 , a2 , and a3 in a given order, written (a1 , a2 , a3 ), is called an ordered number triple. As the coordinates (a1 , a2 , a3 ) of point P in Fig. 2.3 completely deﬁne the vector from O to P, this ordered number triple may be taken as the deﬁnition of the vector itself. In general, changing the order of the numbers in an ordered number triple changes the vector it deﬁnes. Sometimes it is necessary to consider a vector whose base does not coincide with the origin. Suppose that when this occurs the base C is at the point (c1 , c2 , c3 ) and the tip D is at the point (d1 , d2 , d3 ). Then Fig. 2.4 shows the components of this vector in the x, y, and z directions to be d1 − c1 , d2 − c2 , and d3 − c3 . These components determine both the magnitude and direction of the vector. The vector is described by the ordered number triple (d1 − c1 , d2 − c2 , d3 − c3 ), and the length of CD that is equal to the magnitude of the vector is [(d1 − c1 )2 + (d2 − c2 )2 + (d3 − c3 )2 ]1/2 . For convenience, it is usual to represent a vector by a single boldface character such as a, and its magnitude (length) by a , called the norm of a. It is necessary to say here that in applications of vectors to mechanics, and in some purely geometrical applications of vectors, the norm of vector r is often called its modulus and written |r|. When this convention is used, because |r| is a scalar it is usual to denote it by the corresponding ordinary italic letter r , so that r = |r|. If the base and tip of a vector need to be identiﬁed by letters, a vector such as the one from C to D in Fig. 2.4 is written CD, with underlining used to indicate that a vector is involved, and the ordering of the letters is such that the ﬁrst shows the

z

d3

D

c3

C 0

c2

d2 y

c1 d1

C' D

x FIGURE 2.4 Vector directed from point C at (c1 , c2 , c3 ) to point D at (d1 , d2 , d3 ).

Section 2.1

Vectors, Geometry, and Algebra

59

base and the second the tip of the vector. Thus, CD and DC are vectors of equal magnitude but opposite sense, and when these vectors are represented by arrows, the arrows are parallel and of equal length, but point in opposite directions. EXAMPLE 2.1

If, in Fig. 2.4, C is the point (−3, 4, 9) and D the point (2, 5, 7), the vector CD has components 2 − (−3) = 5, 5 − 4 = 1, and 7 − 9 = −2, and so is represented by the ordered number triple (5, 1, −2), whereas vector DC has components −5, −1, and 2 and is represented by the ordered number triple (−5, −1, 2). Having illustrated the concepts of scalars and vectors using some familiar examples, we now develop the algebra of vectors in rather more general terms. Vectors A vector quantity a is an ordered number triple (a1 , a2 , a3 ) in which a1 , a2 , and a3 are real numbers, and we shall write a = (a1 , a2 , a3 ). The numbers a1 , a2 , and a3 , in this order, are called the ﬁrst, second, and third components of vector a or, equivalently, its x-, y-, and z-components.

Null vector The null (zero) vector, written 0, has neither magnitude nor direction and is the ordered number triple 0 = (0, 0, 0).

Equality of vectors Two vectors a = (a1 , a2 , a3 ) and b = (b1 , b2 , b3 ) are equal, written a = b, if, and only if, a1 = b1 , a2 = b2 , and a3 = b3 . EXAMPLE 2.2

If a = (a1 , −5, 6), b = (3, b2 , b3 ) and c = (3, −5, 1), then a = b if a1 = 3, b2 = −5 and b3 = 6, and b = c if b2 = −5 and b3 = 1, but a = c for any choice of a1 because 6 = 1. Norm of a vector The norm of vector a = (a1 , a2 , a3 ), denoted by a , is the non-negative real number 1/2 a = a12 + a22 + a32 , and in geometrical terms a is the length of vector a. The norm of the null vector 0 is 0 = 0. For example, if a is in m/sec, “length” of a is in m/sec.

EXAMPLE 2.3

If a = (1, −3, 2), then a = [12 + (−3)2 + 22 ]1/2 =

√

14, as illustrated in Fig. 2.5.

60

Chapter 2

Vectors and Vector Spaces z

2 A

OA =⎥ ⎢a ⎥

a

⎢

−3 0

y

1

A

x FIGURE 2.5 Vector a and its norm a .

The sum of two vectors If a = (a1 , a2 , a3 ) and b = (b1 , b2 , b3 ) have the same dimensions, say, both are m/sec, their sum, written a + b, is deﬁned as the ordered number triple (vector) obtained by adding corresponding components of a and b to give a + b = (a1 + b1 , a2 + b2 , a3 + b3 ).

EXAMPLE 2.4

If a = (1, 2, −5) and b = (−2, 2, 4), then a + b = (1 + (−2), 2 + 2, −5 + 4) = (−1, 4, −1). Multiplying a vector by a scalar Let a = (a1 , a2 , a3 ) and λ be an arbitrary real number. Then the product λa is deﬁned as the vector λa = (λa1 , λa2 , λa3 ).

EXAMPLE 2.5

Let a = (2, −3, 5), b = (−1, 2, 4). Then 2a = (4, −6, 10), 4b = (−4, 8, 16), and 2a + 4b = (4 + (−4), −6 + 8, 10 + 16) = (0, 2, 26). This deﬁnition of the product of a vector and a scalar, called scaling a vector, shows that when vector a is multiplied by a scalar λ, the norm of a is multiplied by |λ|, because 1/2 λa = λ2 a12 + λ2 a22 + λ2 a32 = |λ| · a . It also follows from the deﬁnition that the sense of vector a is reversed when it is multiplied by −1, though its norm is left unaltered. The deﬁnition of the difference

Section 2.1

Vectors, Geometry, and Algebra

61

y

y

a2 + b2 b2

a+

b a2

b

b

a2 a

a 0

b1

a1

x

a1 a1 + b1 x

FIGURE 2.6 The vector sum a + b.

of two vectors is seen to be contained in the deﬁnition of their sum, because a − b = a + (−b). In particular, when a = b, we ﬁnd that that a − a = 0, showing that −a is the additive inverse of a. The geometrical interpretations of the sum a + b, the difference a − b, and the scaled vector λa in terms of their components are shown in Figs. 2.6 to 2.8, though to simplify the diagrams only the two-dimensional cases are illustrated. This involves no loss of generality, because it is always possible to choose the (x, y)-plane to coincide with the plane containing the vectors a and b.

Vector Addition by the Triangle Rule

triangle rule for addition

Consideration of Fig. 2.6 shows that the addition of vector b to vector a is obtained geometrically by translating vector b until its base is located at the tip of vector a, and then the vector representing the sum a + b has its base at the base of vector a and its tip at the tip of the repositioned vector b. Because of the triangle involving vectors a, b, and a + b, this geometrical interpretation of a vector sum is called the triangle rule for vector addition. The triangle rule also applies to the difference of two vectors, as may be seen by considering Fig. 2.7, because after obtaining −b from b by reversing its sense, the difference a − b can be written as the vector sum a + (−b), where −b is added to vector a by means of the triangle rule. The algebraic results discussed so far concerning the addition and scaling of vectors, together with some of their consequences, are combined to form the following theorem.

y

y b2 a2

a2

b

−b1 −b

a a1 − b1

a 0

b1

a1 x

−b2 FIGURE 2.7 The vector difference a − b.

0 a2 − b2

a−b

−b a1

x

62

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Vectors and Vector Spaces

y

y 2a2

a2

1/2 a

2a 1/2 a2

a a1

y

y

x

x

2a1

(k = 2)

−2a1 1/2 a1

x

x

(k = 1/2)

−2a −2a2 (k = −2)

FIGURE 2.8 The vector ka for different values of k. THEOREM 2.1

Addition and scaling of vectors Let P, Q, and R be arbitrary vectors and let α and β be arbitrary real numbers. Then: 1.

P+Q=Q+P

2.

P+0=0+P=P

3.

(P + Q) + R = P + (Q + R)

4.

α(P + Q) = αP + αQ

5.

(αβ)P = α(βP) = β(αP)

6.

(α + β)P = αP + βP

7.

αP = |α| · P

(vector addition is commutative); (0 is the identity element in vector addition); (vector addition is associative); (multiplication by a scalar is distributive over vector addition); (multiplication of a vector by a product of scalars is associative); (multiplication of a vector by a sum of scalars is distributive); (scaling P by α scales the norm of P by |α|).

Proof The results of this theorem are all immediate consequences of the above deﬁnitions so as the proofs of results 1 to 6 are all very similar, and result 7 has already been established, we only prove result 4. Let P = ( p1 , p2 , p3 ) and Q = (q1 , q2 , q3 ); then α(P + Q) = α( p1 + q1 , p2 + q2 , p3 + q3 ) = α[( p1 , p2 , p3 ) + (q1 , q2 , q3 )] = α( p1 , p2 , p3 ) + α(q1 , q2 , q3 ) = αP + αQ, as was to be shown.

The Representation of Vectors in Terms of the Unit Vectors i, j, and k The components of a vector, together with vector addition, can be used to describe vectors in a very convenient way. The idea is simple, and it involves using the standard convention that i, j, and k are vectors of unit length that point in the positive sense along the x-, y-, and z-axes, respectively. Vectors such as i, j, and k that have a unit norm (length) are called unit vectors, so i = j = k = 1.

Section 2.1

Vectors, Geometry, and Algebra

63

z a3 k

j

a=

i

a 1i

j + a2

A(a1, a2, a3)

k + a3

a3k

a2 y

a1i a1 a2 j x FIGURE 2.9 Vector a in terms of the unit vectors i, j, and k.

An arbitrary vector a can be represented by an “arrow,” with its base at the origin and its tip at the point A with cartesian coordinates (a1 , a2 , a3 ) where, of course, a1 , a2 , and a3 are also the components of a. Consequently, scaling the unit vectors i, j, and k by the respective x, y, and z components a1 , a2 , and a3 of a, followed by vector addition of these three vectors, shows that a can be written a = a1 i + a2 j + a3 k,

(1)

as can be seen from Fig. 2.9. The representation of vector a in terms of the unit vectors i, j, and k in (1), and the ordered triple notation, are equivalent, so a = a1 i + a2 j + a3 k = (a1 , a2 , a3 ). position vector

(2)

In some applications a vector deﬁnes a point in space, so vectors of this type are called position vectors. The symbol r is normally used for a position vector, so if point P with coordinates (x, y, z) is a general point in space, as in Fig. 2.10, its

z

k

j

r=

i

xi

j+ +y

P (x, y, z)

zk

zk

0 y xi yj

1/

OP = ⎥⎢r⎥⎢ = (x2 + y2 + z2) 2

x FIGURE 2.10 Position vector of a general point P in space.

64

Chapter 2

Vectors and Vector Spaces

position vector relative to the origin is r = xi + yj + zk,

(3)

r = (x 2 + y2 + z2 )1/2 .

(4)

and its norm (length) is

EXAMPLE 2.6

(a) Find the distance of point P from the origin given that its position vector is r = 2i + 4j − 3k. (b) If a general point P in space has position vector r = xi + yj + zk, describe the surface deﬁned by r = 3 and ﬁnd its cartesian equation. Solution (a) As r is the position vector of P relative to√the origin, the distance of point P from the origin is r = [22 + 42 + (−3)2 ]1/2 = 29. (b) As r = 3 (constant), it follows that the required surface is one for which every point lies at a distance 3 from the origin, so the surface must be a sphere of radius 3 centered on the origin. As r = xi + yj + zk is the general position vector of a point on this sphere, the result r = 3 is equivalent to (x 2 + y2 + z2 )1/2 = 3, so the cartesian equation of the sphere is x 2 + y2 + z2 = 9. Because of the equivalence of the ordered number triple notation and the representation of vectors in terms of the unit vectors i, j, and k given in (2), both systems obey the same rules governing the addition and scaling of vectors in terms of their components. Thus, the following rules apply to the combination of any two vectors a = a1 i + a2 j + a3 k, b = b1 i + b2 j + b3 k expressed in terms of i, j, and k, and an arbitrary real number λ. The sum a + b is given by a + b = (a1 + b1 )i + (a2 + b2 )j + (a3 + b3 )k.

(5)

The product λ a is given by λ a = λa1 i + λa2 j + λa3 k.

(6)

The norm of scaled vector λ a is given by λ a = |λ| · a 1/2 = |λ| a12 + a22 + a32 . EXAMPLE 2.7

(7)

If a = 5i + j − 3k and b = 2i − 2j − 7k, ﬁnd (a) a + b, (b) a − b, (c) 2a + b, and (d) |−2a|. Solution (a)

a + b = (5i + j − 3k) + (2i − 2j − 7k) = (5 + 2)i + (1 − 2)j + (−3 − 7)k = 7i − j − 10k.

(b)

a − b = (5i + j − 3k) − (2i − 2j − 7k) = (5 − 2)i + (1 − (−2))j + (−3 − (−7))k = 3i + 3j + 4k.

Section 2.1

Vectors, Geometry, and Algebra

65

2a + b = 2(5i + j − 3k) + (2i − 2j − 7k) = (10i + 2j − 6k) + (2i − 2j − 7k) = (10 + 2)i + (2 + (−2))j + (−6 + (−7))k = 12i − 13k. √ 1/2 |−2a| = [(−10)2 + (−2)2 + 62 ] = 2 35

(c)

(d) or, equivalently,

√ |−2a| = |−2| · a = 2 a = 2[52 + 12 + (−3)2 ]1/2 = 2 35.

Finding a Unit Vector in the Direction of an Arbitrary Vector It is often necessary to ﬁnd a unit vector in the direction of an arbitrary vector a = a1 i + a2 j + a3 k. This is accomplished by dividing a by its norm a , because the vector a/ a has the same sense as a and its norm is 1. It is convenient to use a symbol related to an arbitrary vector a to indicate the unit vector in its direction, so from ˆ read “a hat.” So if a = a1 i + a2 j + a3 k, now on such a vector will be denoted by a, 1/2 aˆ = a/ a = (a1 i + a2 j + a3 k)/ a12 + a22 + a32 1/2 = (a1 /a)i + (a2 /a)j + (a3 /a)k, with a = a12 + a22 + a32 .

(8)

As the symbols i, j, and k are used exclusively for the unit vectors in the x-, y-, and ˆ and k. ˆ z-directions, it is not necessary to write ˆi, j, ˆ and a can be put in the useful form The relationship between a, a, ˆ a = a a,

(9)

showing that a general vector a can always be written as the unit vector aˆ scaled by a . Unless otherwise stated, a = 0. EXAMPLE 2.8

Find a unit vector in the direction of a = 3i + 2j + 5k. √ Solution As a = (32 + 22 + 52 )1/2 = 38, it follows that √ √ √ aˆ = a/ a = (3/ 38)i + (2/ 38)j + (5/ 38)k.

EXAMPLE 2.9

It is known from experiments in mechanics that forces are vector quantities and so combine according to the laws of vector algebra. Use this fact to ﬁnd the sum and difference of a force of 9 units in the direction of 2i + j − 2k and a force of 10 units in the direction of 4i − 3j, and determine the magnitudes of these forces. Solution We will use the convention that a unit vector represents a force of 1 unit. Let F be the force of 9 units. Then as 2i + j − 2k = [22 + 12 + (−2)2 ]1/2 = 3, the unit vector in the direction of F is Fˆ = (1/3)(2i + j − 2k) = (2/3)i + (1/3)j − (2/3)k, so F = 9Fˆ = 6i + 3j − 6k units.

66

Chapter 2

Vectors and Vector Spaces

Similarly, let G be the force of 10 units. Then as 4i − 3j = 5, the unit vector in the direction of G is ˆ = (1/5)(4i − 3j) = (4/5)i − (3/5)j, G ˆ = 8i − 6j units. so G = 10G Combining these results shows that F + G = 14i − 3j − 6k units, and F − G = −2i + 9j − 6k units, from which it follows that the magnitudes of the forces are given by √ F + G = 241 units and F − G = 11 units. Equality of vectors expressed in terms of unit vectors As the difference of two equal and opposite vectors is the null vector 0, this shows that if a = b, where a = a1 i + a2 j + a3 k, and b = b1 i + b2 j + b3 k, then the respective components of vectors a and b must be equal, leading to the result that (10) a = b if, and only if, a1 = b1 , a2 = b2 , and a3 = b3 .

Simple Geometrical Applications of Vectors Although our use of vectors will be mainly in connection with the calculus, the following simple geometrical applications are helpful because they illustrate basic vector arguments and properties. Although we have seen how an arbitrary vector can be expressed in terms of unit vectors associated with a cartesian coordinate system, it must be remembered that the fundamental concept of a vector and its algebra is independent of a coordinate system. Because of this, it is often possible to use the rules governing elementary vector algebra given in Theorem 2.1 to establish equations in a purely vectorial manner, without the need to appeal to any coordinate system. Once a general vector equation has been established, the representation of the vectors involved in terms of their components and the unit vectors i, j, and k can be used to convert the vector equation into the equivalent cartesian equations. The purely vectorial approach to geometrical problems is well illustrated by ﬁnding the vector AB in terms of the position vectors of points A and B, and then using the result to ﬁnd the position vector of the mid-point of AB. After this, the purely vectorial derivation of a geometrical result followed by its interpretation in cartesian form will be illustrated by ﬁnding the equation of a straight line in three space dimensions.

Vector AB in terms of the position vectors of A and B Let a and b be the position vectors of points A and B relative to an origin O, as shown in Fig. 2.11. An application of the triangle rule for the addition of vectors gives OA + AB = OB, but OA = a and OB = b, so a + AB = b,

Section 2.1

Vectors, Geometry, and Algebra

67

B

AB A b a

O FIGURE 2.11 Vectors a, b, and AB.

giving AB = b − a.

(11)

When expressed in words, this simple but useful result asserts that vector AB is obtained by subtracting the position vector a of point A from the position vector b of point B. EXAMPLE 2.10

Find the position vector of the mid-point of AB if point A has position vector a and point B has position vector b relative to an origin O. Solution Let point C, with position vector c relative to origin O, be the mid-point of AB, as shown in Fig. 2.12. By the triangle rule, OA + AC = OC, but OA = a, and from (11) AC = (1/2)(b − a), so OC = a + (1/2)(b − a), so the required result is c = OC = (1/2)(b + a).

B C

AC

A

c

b

a

O FIGURE 2.12 C is the mid-point of AB.

68

Chapter 2

Vectors and Vector Spaces

b A

a

λb AP =

L

P

r

O FIGURE 2.13 The straight line L.

The vector and cartesian equations of a straight line Let line L be a straight line through point A with position vector a relative to an origin O, and let the line be parallel to a vector b. If P is an arbitrary point on line L with position vector r relative to O, an application of the triangle rule for vector addition to the vectors shown in Fig. 2.13 gives r = OA + AP.

vector equation of straight line

But OA = a, and as AP is parallel to b, a number λ can always be found such that AP = λb, so the vector equation of line L becomes r = a + λb.

(12)

Notice that result (12) determines all points P on L if λ is taken to be a number in the interval −∞ < λ < ∞. The cartesian equations of line L follow by setting a = a1 i + a2 j + a3 k, b = b1 i + b2 j + b3 k, and r = xi + yj + zk in result (12), and then using the deﬁnition of equality of vectors given in (10) to obtain the corresponding three scalar cartesian equations. Proceeding in this way we ﬁnd that xi + yj + zk = a1 i + a2 j + a3 k + λ(b1 i + b2 j + b3 k), cartesian and standard form of straight line

so equating corresponding components of i, j, and k on each side of this equation brings us to the required cartesian equations for L in the form x1 = a1 + λb1 ,

x2 = a2 + λb2 ,

x3 = a3 + λb3 .

(13)

An equivalent form of these equations is obtained by solving each equation for λ and equating the results to get x − a1 y − a2 z − a3 = = = λ. b1 b2 b3

(14)

Section 2.1

Vectors, Geometry, and Algebra

69

This is the standard form (also called the canonical form) of the cartesian equations of a straight line. It is important to notice that when written in standard form the coefﬁcients of x, y, and z are all unity. Once the equation of a straight line is written in standard form, equating each numerator to zero determines the components (a1 , a2 , a3 ) of a position vector of a point on the line, while the denominators in the order (b1 , b2 , b3 ) determine the components of a vector parallel to the line. EXAMPLE 2.11

A straight line L is given in the form 3−y z+ 1 2x − 3 = = . 4 2 3 Find the position vector of a point on L and a vector parallel to L. Solution When the equation is written in standard form it becomes x − 3/2 y−3 z+ 1 = = = λ. 2 −2 3 Comparing these equations with (14) shows that (a1 , a2 , a3 ) = (3/2, 3, −1) and b = (b1 , b2 , b3 ) = (2, −2, 3). So the position vector of a point on the line is a = (3/2)i + 3j − k, and a vector parallel to the line is b = 2i − 2j + 3k. Neither of these results is unique, because μb is also parallel to the line for any scalar μ = 0, and any other point on L would sufﬁce. For example, the vector 14i − 14j + 21k is also parallel to the line, while setting λ = 2 leads to the result (a1 , a2 , a3 ) = (11/2, −1, 5), corresponding to a different point on the same line, this time with position vector a = (11/2)i − j + 5k.

Summary

This section has introduced vectors both as geometrical quantities that can be represented by directed line segments and, using a right-handed system of cartesian axes, as ordered number triples. Deﬁnitions of the scaling, addition, and subtraction of vectors have been given, and a general vector has been deﬁned in terms of the set of three unit vectors i, j, and k that lie along the orthogonal cartesian axes O{x, y, z}. Finally, the vector and cartesian equations of a straight line in space have been derived, and the standard form of the cartesian equations has been introduced from which a vector parallel to the line may be found by inspection.

EXERCISES 2.1 1. Prove Results 1, 3, and 6 of Theorem 2.1. 2. Given that a = 2i + 3j − k, b = i − j + 2k, and c = 3i + 4j + k, ﬁnd (a) a + 2b − c, (b) a vector d such that a + b + c + d = 0, and (c) a vector d such that a − b + c + 3d = 0. 3. Given a = i + 2j + 3k, b = 2i − 2j + k, ﬁnd (a) a vector c such that 2a + b + 2c = i + k, (b) a vector c such that 3a − 2b + c = i + j − 2k. 4. Given that a = 3i + 2j − 3k, b = 2i − j + 5k, and c = 2i + 5j + 2k, ﬁnd (a) 2a + 3b − 3c, (b) a vector d such that a + 3b − 2c + 3d = 0, and (c) a vector d such that 2a − 3d = b + 4c.

5. Given that Aand B have the respective position vectors 2i + 3j − k and i + 2j + 4k, ﬁnd the vector AB and a unit vector in the direction of AB. 6. Given that A and B have the respective position vectors 3i − j + 4k and 2i + j + k, ﬁnd the vector AB and the position vector c of the mid-point of AB. 7. Given that Aand B have the respective position vectors a and b, ﬁnd the position vector of a point P on the line AB located between A and B such that (length AP)/(length PB) = m/n,

where m, n > 0

are any two real numbers.

70

Chapter 2

Vectors and Vector Spaces

8. Find the position vector r of a point P on the straight line joining point Aat (1, 2, 1) and point B at (3, −1, 2) and between A and B such that

13. A straight line L is given in the form 2x + 1 3y + 2 2 − 4z = = . 3 4 −1

(length AP)/(length PB) = 3/2. 9. It is known from Euclidean geometry that the medians of a triangle (lines drawn from a vertex to the mid-point of the opposite side) all meet at a single point P, and that P is two-thirds of the distance along each median from the vertex through which it passes. If the vertices A, B, and C of a triangle have the respective position vectors a, b, and c, show that the position vector of P is (1/3)(a + b + c). 10. Forces of 1, 2, and 3 units act through the origin along, and in the positive directions of, the respective x-, y-, and z-axes. Find the vector sum S of these forces, the magnitude S of the sum of the vectors, and a unit vector in the direction of S. 11. Forces of 2, 1, and 4 units act through the origin along, and in the positive directions of, the respective x-, y-, and z-axes. Find the vector sum S of these forces, the magnitude S of the sum of the vectors, and a unit vector in the direction of S. 12. A straight line L is given in the form 3x − 1 2y + 3 2 − 3z = = . 4 2 1 Find the position vectors of two different points on L and a unit vector parallel to L.

2.2

14.

15.

16.

17.

18.

Find position vectors of two different points on L and a unit vector parallel to L. Given that a straight line L1 passes through the points (−2, 3, 1) and (1, 4, 6), ﬁnd (a) the position vector of a point on the line and a vector parallel to it, and (b) a straight line L2 parallel to L1 that passes through the point (1, 2, 1). Given that a straight line L1 passes through the points (3, 2, 4) and (2, 1, 6), ﬁnd (a) the position vector of a point on the line and a vector parallel to it, and (b) a straight line L2 parallel to L1 that passes through the point (−2, 1, 2). A straight line has the vector equation r = a + λb, where a = 3j + 2k, and b = 2i + j + 2k. Find the cartesian equations of the line and the coordinates of three points that lie on it. A straight line passes through the point (3, 2, −3) parallel to the vector 2i + 3j − 3k. Find the cartesian equations of the line and the coordinates of three points that lie on it. In mechanics, if a point A moves with velocity vA and point B moves with velocity vB , the velocity vR of A relative to B (the relative velocity of A with respect to B) is deﬁned as vR = vA − vB . Power boat A moves northeast at 20 knots and power boat B moves southeast at 30 knots. Find the velocity of boat Arelative to boat B, and a unit vector in the direction of the relative velocity.

The Dot Product (Scalar Product) A product of two vectors a and b can be formed in such a way that the result is a scalar. The result is written a · b and called the dot product of a and b. The names scalar product and inner product are also used in place of the term dot product.

Dot Product dot or scalar product

Let a and b be any two vectors that after a translation to bring their bases into coincidence are inclined to one another at an angle θ , as shown in Fig. 2.14, where 0 ≤ θ ≤ π. Then the dot product of a and b is deﬁned as the number a · b = a · b cos θ. This geometrical deﬁnition of the dot product has many uses, but when working with vectors a and b that are expressed in terms of their components in the i, j, and

Section 2.2

The Dot Product (Scalar Product)

71

a

θ b FIGURE 2.14 Vectors a and b inclined at an angle θ.

k directions, a more convenient form is needed. An equivalent deﬁnition that is easier to use is given later in (23). properties of the dot product

Properties of the dot product The following results, in which a and b are any two vectors and λ and μ are any two scalars, are all immediate consequences of the deﬁnition of the dot product. The dot product is commutative a · b = b · a and

λa · μb = μa · λb = λμa · b

(15)

The dot product is distributive and linear a · (b + c) = a · b + a · c

and

a · (λb + μc) = λa · b + μa · c.

(16)

The angle between two vectors The angle θ between vectors a and b is given by cos θ =

a·b , a · b

with 0 ≤ θ ≤ π.

(17)

Parallel vectors (θ = 0) If vectors a and b are parallel, then a · b = a · b

and, in particular,

a · a = a 2 .

(18)

Orthogonal vectors (θ = π/2) If vectors a and b are orthogonal, then a · b = 0.

(19)

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Product of unit vectors ˆ are unit vectors, then If aˆ and b ˆ = cos θ, aˆ · b

with 0 ≤ θ ≤ π.

(20)

An immediate consequence of properties (15), (19), and (20) is that i · i = j · j = k · k = 1,

(21)

i · j = j · i = i · k = k · i = j · k = k · j = 0.

(22)

and

We now use results (21) and (22) to arrive at a simple expression for the dot product in terms of the components of a and b. To arrive at the result we set a = a1 i + a2 j + a3 k, b = b1 i + b2 j + b3 k and form the dot product a · b = (a1 i + a2 j + a3 k) · (b1 i + b2 j + b3 k).

dot product in terms of components

Expanding this product using (15) and (16) and making use of results (21) and (22) brings us to the following alternative deﬁnition of the dot product expressed in terms of the components of a and b: a · b = a1 b1 + a2 b2 + a3 b3 .

(23)

Using (23) in (17) produces the following useful expression that can be used to ﬁnd the angle θ between a and b: a1 b1 + a2 b2 + a3 b3 cos θ = 1/2 2 1/2 where 0 ≤ θ ≤ π. 2 2 a1 + a2 + a32 b1 + b22 + b32 EXAMPLE 2.12

(24)

Find a · b and the angle between the vectors a and b, given that a = i + 2j + 3k and b = 2i − j − 2k. √ Solution a = 14, b = 3, and a · b = 1 · 2 + 2 · (−1) + 3 · (−2) = −6. Using these results in (24) gives √ √ cos θ = −6/(3 14) = −2/ 14, so as 0 ≤ θ ≤ π we see that θ = 2.1347 radians, or θ = 122.3◦ .

projecting a vector onto a line

The projection of a vector onto the line of another vector The projection of vector a onto the line of vector b is a scalar, and it is the signed length of the geometrical projection of vector a onto a line parallel to b, with the sign positive for 0 ≤ θ < π/2 and negative for π/2 < θ ≤ π . This is illustrated in Fig. 2.15, from which it is seen that the signed length of the projection of a onto the line of vector b is ON, where ON = a cos θ.

Section 2.2 OA = ⎥⎢a⎥⎢

a

A

The Dot Product (Scalar Product)

A

a

OA = ⎥⎢a⎥⎢ a

a θ b

73

O

N

N

b

^

b

θ O

^

b

π/2 < θ ≤ π

0 ≤ θ < π/2

FIGURE 2.15 The projection of vector a onto the line of vector b.

ˆ is the unit vector along b, then as a = a a , ˆ = cos θ , the projection ˆ If b and aˆ · b ON = a cos θ can be written as the dot product ˆ =a·b ˆ = ON = a aˆ · b

EXAMPLE 2.13

a·b b

(25)

Find the strength of the magnetic ﬁeld vector H = 5i + 3j + 7k in the direction of 2i − j + 2k, where a unit vector represents one unit of magnetic ﬂux. Solution We are required to ﬁnd the projection of vector H in the direction of ˆ = (1/3)(2i − j + 2k), the vector 2i − j + 2k. Setting b = 2i − j + 2k, b = 3, so b so the strength of the vector H in the direction of b is ˆ = (1/3)(5i + 3j + 7k) · (2i − j + 2k) = 7. H·b Direction cosines and direction ratios If a = a1 i + a2 j + a3 k is an arbitrary vector, the unit vector aˆ in the direction of a is aˆ = (a1 i + a2 j + a3 k)/ a 1/2 = (a1 i + a2 j + a3 k)/ a12 + a22 + a32 . (26) Taking the dot product of a with i, j, and k, and setting l = a1 / (a12 + a22 + a32 )1/2 , m = a2 /(a12 + a22 + a32 )1/2 , and n = a3 /(a12 + a22 + a32 )1/2 gives ˆ l = i · a,

ˆ m = j · a,

and

ˆ n = k · a,

so we may write aˆ = li + mj + nk.

(27)

The dot product aˆ · aˆ = l 2 + m2 + n2 = (a12 + a22 + a32 )/ a 2 , but a 2 = a12 + a22 + a32 , so l 2 + m2 + n2 = 1.

(28)

The number l is the cosine of the angle β1 between a and the x-axis, the number m is the cosine of the angle β2 between a and the y-axis, and the number n is the cosine of the angle β3 between a and the z-axis, as shown in

74

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z

β3 O

a⎥ ⎢ = ⎥⎢ OA a

A

β2

β1

y

x FIGURE 2.16 The angles β1 , β2 , and β3 .

direction cosines

Fig. 2.16. The numbers (l, m, n) are called the direction cosines of a, because they determine the direction of the unit vector aˆ that is parallel to a. Notice that when any two of the three direction cosines l, m, and n of a vector a are given, the third is related to them by l 2 + m2 + n2 = 1. Because of result (27) it is always possible to write a = a (li + mj + nk),

direction ratios

EXAMPLE 2.14

(29)

where l, m, and n are the direction cosines of a. As the components a1 , a2 , and a3 of a are proportional to the direction cosines, they are called the direction ratios of a. Find the direction cosines and direction ratios of a = 3i + j − 2k. √ √ √ Solution√ As a = 14, the direction cosines are l = 3/ 14, m = 1/ 14, and n = −2/ 14. The direction ratios of√a are 3,√1, and −2, or any √ nonnegative multiple of these three numbers such as 15/ 14, 5/ 14, and −10/ 14. The triangle inequality The following result will be needed in the proof of the triangle inequality that is to follow. The absolute value of a · b = a · b cos θ is |a · b| = a · b |cos θ|, but | cos θ | ≤ 1, so using this in the above result we obtain the Cauchy–Schwarz inequality, |a · b| ≤ a · b .

(30)

Section 2.2

THEOREM 2.2

The Dot Product (Scalar Product)

75

The triangle inequality If a and b are any two vectors, then a + b ≤ a + b . Proof

From (18) we have a + b 2 = (a + b) · (a + b) = a · a + 2a · b + b · b = a 2 + 2a · b + b 2 ,

but a · b ≤ |a · b|, so from the Cauchy–Schwarz inequality (30) a + b 2 ≤ a 2 + 2 a · b + b 2 = ( a + b )2 . Taking the positive square root of this last result, we obtain the triangle inequality a + b ≤ a + b . The triangle inequality will be generalized in Section 2.5, but in its present form it is the vector equivalent of the Euclidean theorem that “the sum of the lengths of any two sides of a triangle is greater than or equal to the length of the third side,” and it is from this theorem that the inequality derives its name.

Equation of a Plane

vector equation of a plane

When working with the vector calculus it is sometimes necessary to consider a plane that is locally tangent to a point on a surface in space so it will be useful to derive the general equation of a plane in both its vector and cartesian forms. A plane can be deﬁned by specifying a ﬁxed point belonging to the plane and a vector n that is perpendicular to the plane. This follows because if n is perpendicular at a point on the plane, it must be perpendicular at every point on the plane. Any vector n that is perpendicular to a plane is called a normal to the plane. Clearly a normal to a plane is not unique, because a plane has two sides, so if a normal n is directed away from one side of the plane, the vector −n is a normal directed away from the other side. Both n and −n can be scaled by any nonzero number and still remain normals; consequently, if n is a normal to a plane, so also are all vectors of the form λn, with λ = 0 any real number. Let a ﬁxed point Aon plane with normal n have position vector a relative to an origin O, and let P be a general point on plane with position vector r relative to O. Then, as may be seen from Fig. 2.17, the vector r − a lies in the plane, and so is perpendicular (normal) to n. Forming the dot product of n and r − a, and using (19), shows that the vector equation of plane is n · (r − a) = 0,

(31)

n · r = n · a.

(32)

or, equivalently,

cartesian equation of a plane

The cartesian form of this equation follows by considering a general point with coordinates (x, y, z) on plane , setting r = xi + yj + zk, a = a1 i + a2 j + a3 k,

76

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Vectors and Vector Spaces n π A

r−a AP =

P

a ^ n.a = ^ n.r

^ n

r

O FIGURE 2.17 Plane with normal n passing through point A.

and n = n1 i + n2 j + n3 k, and then substituting into (32) to get (n1 i + n2 j + n3 k) · (xi + yj + zk) = (n1 i + n2 j + n3 k) · (a1 i + a2 j + a3 k). Taking the dot products and using results (21) and (22) show the cartesian equation of plane to be n1 x + n2 y + n3 z = n1 a1 + n2 a2 + n3 a3 = d, a constant. EXAMPLE 2.15

(33)

Find the cartesian equation of the plane through the point (2, 5, 3) with normal 3i + 2j − 7k. Solution Here n1 = 3, n2 = 2, n3 = −7 and a1 = 2, a2 = 5, and a3 = 3, so substituting into (33) shows the plane has the equation 3x + 2y − 7z = −5.

Summary

This section has introduced the dot or scalar product of two vectors in geometrical terms and, more conveniently for calculations, in terms of the components of the two vectors involved. The applications given include the important operation of projecting a vector onto the line of another vector and the derivation of the vector equation and cartesian equation of a plane.

EXERCISES 2.2 1. Find the dot products of the following pairs of vectors: (a) i − j + 3k, 2i + 3j + k. (b) 2i − j + 4k, −i + 2 j + 2k. (c) i + j − 3k, 2i + j + k. 2. Find the dot products of the following pairs of vectors: (a) i − 2 j + 4k, i + 2 j + 3k. (b) 3i + j + 2k, 4i − 3j + k. (c) 5i − 3j + 3k, 2i − 3j + 5k. 3. Find which of the following pairs of vectors are orthogonal: (a) 3i + 2 j − 6k, −9i − 6j + 18k. (b) 3i − j + 7k, 3i + 2 j + k.

(c) 2i + j + k, i + j − k. (d) i + j − 3k, 2i + j + k. 4. Find which, if any, of the following pairs of vectors are orthogonal: (a) 2i + j + k, 8i + 2 j + 2k. (b) i + 2 j + 3k, 2i − 2 j − 3k. (c) i + 2 j + 4k, 2i + j + 3k. (d) i + j, 2 j + 3k. 5. Given that a = 2i + 3j − 2k, b = i + 3j + k and c = 3i + j − k, ﬁnd (a) (a + b) · c. (b) (2b − 3c) · a. (c) a · a. (d) c · (a − 2b).

Section 2.3 6. Given that a = 3i + 2 j − 3k, b = 2i + j + 2k, and c = 5i + 2 j − 2k, ﬁnd (a) b · (b + (a · c)c). (b) (a + 2b) · (2b − 3c). (c) (c · c)b − (a · a)c. 7. Find the angle between the following pairs of vectors: (a) i + j + k, 2i + j − k. (b) 2i − j + 3k, 2i + j + 3k. (c) 3i − j + k, i − 2 j + 3k. (d) i − 2 j + k, 4i − 8j + 16k. 8. Given a = 2i − 3j − 3k, b = i + j + 2k, and c = 3i − 2 j − k, ﬁnd the angles between the following pairs of vectors: (a) a + b, b − 2c. (b) 2a − c, a + b − c. (c) b + 3c, a − 2c. 9. Find the component of the force F = 4i + 3j + 2k in the direction of the vector i + j + k. 10. Find the component of the force F = 2i + 5j − 3k in the direction of the vector 2i + j − 2k. 11. Given that a = i + 2 j + 2k and b = 2i − 3j + k, ﬁnd (a) the projection of a onto the line of b, and (b) the projection of b onto the line of a. 12. Given that a = 3i + 6j + 9k and b = i + 2 j + 3k, (a) ﬁnd the projection of a onto the line of b and (b) compare the magnitude of a with the result found in (a) and comment on the result. 13. Find the direction cosines and corresponding angles for the following vectors: (a) i + j + k. (b) i − 2 j + 2k. (c) 4i − 2 j + 3k. 14. Find the direction cosines and corresponding angles for the following vectors: (a) i − j − k. (b) 2i + 2 j − 5k. (c) −4j − k. 15. Verify the triangle inequality for vectors a = i + 2 j + 3k and b = 2i + j + 7k. 16. Verify the triangle inequality for vectors a = 2i − j − 2k and 3i + 2 j + 3k. 17. Find the equation of the plane with normal 2i − 3j + k that contains the point (1, 0, 1). 18. Find the equation of the plane with normal i − 2 j + 2k that contains the point (2, −3, 4). 19. Given that a plane passes through the point (2, 3, −5), and the vector 2i + k is normal to the plane, ﬁnd the cartesian form of its equation.

2.3

The Cross Product

77

20. The equation of a plane is 3x + 2y − 5z = 4. Find a vector that is normal to the plane, and the position vector of a point on the plane. 21. Explain why if the vector equation of plane in (32) is divided by n to bring it into the form r · n = a · n, the number |a · n| is the perpendicular distance of origin O from the plane. Explain also why if a · n > 0 the plane lies to the side of O toward which n is directed, as in Fig. 2.15, but that if a · n < 0 it lies on the opposite side of O toward which −n is directed. 22. Use the result of Exercise 21 to ﬁnd the perpendicular distance of the plane 2x − 4y − 5z = 5 from the origin. 23. The angle between two planes is deﬁned as the angle between their normals. Find the angle between the two planes x + 3y + 2z = 4 and 2x − 5y + z = 2. 24. Find the angle between the two planes 3x + 2y − 2z = 4 and 2x + y + 2z = 1. 25. Let a and b be two arbitrary skew (nonparallel) vectors, and set a = ab + ap , where ab is parallel to b and ap is perpendicular to b and lies in the plane of a and b. Find ab and ap in terms of a and b. 26. The law of cosines for a triangle with sides of length a, b, and c, in which the angle opposite the side of length c is C, takes the form c2 = a 2 + b2 − 2ab cos C. Prove this by taking vectors a, b, and c such that c = a − b and considering the dot product c · c = (a − b) · (a − b). 27. The work units W done by a constant force F when moving its point of application along a straight line L parallel to a vector a are deﬁned as the product of the component of F in the direction of a and the distance d moved along line L. Express W in terms of F, a, and d. 28. If a and b are arbitrary vectors and λ and μ are any two scalars, prove that λa + μb 2 ≤ λ2 a 2 + 2λμa · b + μ2 b 2 . 29. Verify the result of Exercise 28 by setting λ = 2, μ = −3, a = 3i + j − 4k, and b = 2i + 3j + k.

The Cross Product A product of two vectors a and b can be deﬁned in such a way that the result is a vector. The result is written a × b and called the cross product of a and b. The name vector product is also used in place of the term cross product. Before deﬁning the cross product we ﬁrst formulate what is called the right-hand rule. Given any two skew vectors a and b, the right-hand rule is used to determine

78

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the sense of a third vector c that is required to be normal to the plane containing vectors a and b. right-hand rule

The Right-Hand Rule Let a and b be two arbitrary skew vectors with the same base point, with c a vector normal to the plane containing them. If the ﬁngers of the right hand are curled in such a way that they point from vector a to vector b through the angle θ between them, with 0 < θ < π , then when the thumb is extended away from the palm it will point in the direction of vector c. When applying the right-hand rule, the order of the vectors is important. If vectors a, b, and c obey the right-hand rule, they will always be written in the order a, b, c, with the understanding that c is normal to the plane of a and b, with its sense determined by the right-hand rule. Figure 2.18 illustrates the right-hand rule. An important special case of the right-hand rule has already been encountered in connection with the unit vectors i, j, and k that obey the rule, and because the vectors are mutually orthogonal the vectors j, k, i and k, i, j also obey the right-hand rule.

geometrical deﬁnition of a cross product

The cross product (a geometrical interpretation) Let a and b be two arbitrary vectors, with nˆ a unit vector normal to the plane of ˆ in this order, obey the right-hand rule. Then a and b chosen so that a, b, and n, the cross product of vectors a and b, written a × b, is deﬁned as the vector ˆ a × b = a . b sin θ n.

(34)

This geometrical deﬁnition of the cross product is useful in many situations, but when the vectors a and b are speciﬁed in terms of their cartesian components a different form of the deﬁnition will be needed. The cross product can be interpreted as a vector area, in the sense that it can ˆ where S = OA· BN = a · b sin θ is the geometrical area be written a × b = Sn,

c

θ

b a

FIGURE 2.18 The right-hand rule.

Section 2.3

The Cross Product

79

B ^ n

S

b

A

θ

a

O FIGURE 2.19 The cross product interpreted as the vector area of a parallelogram.

of the parallelogram in Fig. 2.19, and the unit vector nˆ is normal to the area. This shows that the geometrical area S of the vector parallelogram with sides a and b is simply the modulus of the cross product a × b, so S = a × b . properties of the cross product

Properties of the cross product The following results are consequences of the deﬁnition of the cross product. The cross product is anticommutative a × b = −b × a

(35)

a × (b + c) = a × b + a × c.

(36)

The cross product is associative

Parallel vectors (θ = 0) If vectors a and b are parallel, then a × b = 0.

(37)

Orthogonal vectors (θ = π/2) If vectors a and b are orthogonal, then ˆ a × b = a . b n.

(38)

ˆ a × b = sin θ n.

(39)

Product of unit vectors If a and b are unit vectors, then

An immediate consequence of properties (34), (35), and (37) is that i × i = j × j = k × k = 0,

(40)

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Vectors and Vector Spaces

and i × j = k,

j × i = −k,

j × k = i,

k × j = −i,

k × i = j, i × k = −j. (41)

Only results (35) and (36) require some comment, as the other results are obvious. The change of sign in (35) that makes the cross product anticommutative occurs because when the vectors a and b are interchanged, the right-hand rule causes the direction of nˆ to be reversed. Result (36) can be proved in several ways, but we shall postpone its proof until a different expression for the cross product has been derived. To obtain a more convenient expression for the cross product that can be used when a and b are known in terms of their components, we proceed as follows. Let a = a1 i + a2 j + a3 k and b = b1 i + b2 j + b3 k, and consider the cross product a × b = (a1 i + a2 j + a3 k) × (b1 i + b2 j + b3 k). Expanding this expression term by term is justiﬁed because of the associative property given in (36), and it leads to the result a × b = a1 b1 i × i + a1 b2 i × j + a1 b3 i × k + a2 b1 j × i + a2 b2 j × j + a2 b3 j × k + a3 b1 k × i + a3 b2 k × j + a3 b3 k × k.

cross product in terms of components

Results (40) cause three terms on the right-hand side to vanish, and results (41) allow the remaining six terms to be collected into three groups as follows to give a × b = (a2 b3 − a3 b2 )i − (a1 b3 − a3 b1 )j + (a1 b2 − a2 b1 )k.

(42)

This alternative expression for the cross product in terms of the cartesian components of vectors a and b can be further simpliﬁed by making formal use of the third-order determinant, i j k a × b = a1 a2 a3 , b1 b2 b3 because a formal expansion in terms of elements of the ﬁrst row generates result (42). We take this result as an alternative but equivalent deﬁnition of the cross product. practical deﬁnition of a cross product using a determinant

The cross product (cartesian component form) Let a = a1 i + a2 j + a3 k and b = b1 i + b2 j + b3 k. Then i a × b = a1 b1

j a2 b2

k a3 , b3

(43)

When expressing a × b as the determinant in (43), purely formal use was made of the method of expansion of a determinant in terms of the elements of its ﬁrst row, because (43) is not a determinant in the ordinary sense as its elements are a mixture of vectors and numbers.

Section 2.3

EXAMPLE 2.16

The Cross Product

81

Given that a = 3i − 2 j − k and b = i + 4j + 2k, ﬁnd a × b and a unit vector nˆ normal to the plane containing a and b such that a, b, and n, in this order, obey the right-hand rule. Solution Substitution into expression (43) gives i j k a × b = 3 −2 −1 1 4 2 = [(−2) · 2 − 4 · (−1)]i − [3 · 2 − 1 · (−1)] j + [3 · 4 − 1 · (−2)]k = −7j + 14k. The required unit vector nˆ is simply the unit vector in the direction of a × b, so √ nˆ = (a × b)/ a × b = (−7j + 14k)/(7 5). √ √ = (−1/ 5)j + (2/ 5)k. We now return to the proof of the associative property stated in (35) and establish it by means of result (43). Setting a = a1 i + a2 j + a3 k, b = b1 i + b2 j + b3 k, and c = c1 i + c2 j + c3 k, we have i j k a2 a3 . a × (b + c) = a1 (b1 + c1 ) (b2 + c2 ) (b3 + c3 ) Expanding the determinant in terms of elements of its ﬁrst row and grouping terms gives a × (b + c) = (a2 b3 − a3 b2 )i − (a1 b3 − a3 b1 )j + (a1 b2 − a2 b1 )k + (a2 c3 − a3 c2 )i − (a1 c3 − a3 c1 )j + (a1 c2 − a2 c1 )k = a × b + a × c, and the result is proved.

Summary

This section ﬁrst introduced the vector or cross product of two vectors in geometrical terms and then used the result to show that the vector product is anticommutative, in the sense that a × b = −b × a. Important results involving the vector product are given in terms of the components of the two vectors that are involved. Finally, the vector product was expressed in a form that is most convenient for calculations by writing it in determinantal form, the rows of which contain the unit vectors i, j, and k and the components of the respective vectors.

EXERCISES 2.3 In Exercises 1 through 6 use (43) to ﬁnd a × b. 1. 2. 3. 4. 5.

For a = 2i − j − 4k, b = 3i − j − k. For a = −3i + 2 j + 4k, b = 2i + j − 2k. For a = 7i + 6k, b = 3j + k. For a = 3i + 7j + 2k, b = i − j + k. For a = 2i + j + k, b = 2i − j + k.

6. For a = 3i − 2 j + 6k, b = 2i + j + 3k. In Exercises 7 through 10 verify the equivalence of the definitions of the cross product in (34) and (43) by ﬁrst using ˆ and then (43) to calculate a × b, and hence a × b and n, calculating a and b directly, using result (17) to ﬁnd cos θ and hence sin θ, and using the results to ﬁnd a × b from (34).

82 7. 8. 9. 10.

Chapter 2

Vectors and Vector Spaces

For a = i + j + 3k and b = 3i + 2 j + k. For a = i + j + k and b = 4i + 2 j + 2k. For a = 2i + j − 3k and b = 5i − 2k. For a = −2i − 3j + k and b = 3i + j + 2k.

21. 22. 23. 24.

In Exercises 11 through 14, verify by direct calculation that (b + c) × a = −a × (b + c). 11. 12. 13. 14.

a = 3j + 2k, b = i − 4j + k, and c = 5i − 2 j + 3k. a = −i + 5j + 2k, b = 4i + k, and c = −2i − 4j + 3k. a = i + k, b = 3i − j − 2k, and c = 3i + j + k. a = 5i + j + k, b = 2i − j − k, and c = 4i + 2 j + 3k.

In Exercises 15 through 18 ﬁnd a unit vector normal to a plane containing the given vectors. 3i + j + k and i + 2 j + k. 2i − j + 2k and 2i + 3j + k. i + j + k and 2i + 3j − k. 2i + 2 j − k and 3i + j + 4k. Find a unit vector normal to a plane containing vectors a + b and a + c, given that a = i + 2 j + k, b = 2i + j − 2k, and c = 3i + 2 j + 4k. 20. Given that a = 3i + j + k, b = 2i − j + 2k, and c = i + j + k, ﬁnd (a) a vector normal to the plane containing the vectors a + (a · b)b and c and, (b) explain why the normal to a plane containing the vectors a and b and the normal to a plane containing the vectors (a · b)a and (b · c)b are parallel. 15. 16. 17. 18. 19.

In Exercises 21 through 24, ﬁnd the cartesian equation of the plane that passes through the given points.

2.4

25. 26. 27. 28. 29.

(1, 3, 2), (2, 0, −4), and (1, 6, 11). (1, 4, 3), (2, 0, 1), and (3, 4, −6). (1, 2, 3), (2, −4, 1), and (3, 6, −1). (1, 0, 1), (2, 5, 7), and (2, 3, 9). Three points with position vectors a, b, and c will be collinear (lie on a line) if the parallelogram with adjacent sides a − b and a − c has zero geometrical area. Use this result in Exercises 25 through 28 to determine which sets of points are collinear. (2, 2, 3), (6, 1, 5), (−2, 4, 3). (1, 2, 4), (7, 0, 8), (−8, 5, −2). (2, 3, 3), (3, 7, 5), (0, −5, −1). (1, 3, 2), (4, 2, 1), (1, 0, 2). A vector N normal to the plane containing the skew vectors a and b can be found as follows. N is normal to a and b, so a · N = 0 and b · N = 0. If a component of N is assigned an arbitrary nonzero value c, say, the other two components can be found from these two equations as multiples of c, and N will then be determined as a multiple of c. A suitable choice of c will make N a ˆ Apply this method to vectors a and b in unit normal N. ˆ Compare the result with Exercise 7 to ﬁnd a vector N. the unit vector nˆ = (a × b)/ a × b ˆ found from (43). Explain why although both nˆ and N are normal to the plane containing a and b they may have opposite senses.

Linear Dependence and Independence of Vectors and Triple Products The dot and cross products can be combined to provide a simple test that determines whether or not an arbitrary set of three vectors possesses a property of fundamental importance to the algebra of vectors. First, however, some introductory remarks are necessary. Given a set of n vectors a1 , a2 , . . . , an , and a set of n constants c1 , c2 , . . . , cn , the sum c1 a1 + c2 a2 + · · · + cn an

linear combination of vectors

basis

is called a linear combination of the vectors. Linear combinations of the vectors i, j, and k were used in Section 2.1 to express every vector in three-dimensional space as a linear combination of these three vectors. A triad of vectors such as i, j, and k with the property that all vectors in three-dimensional space can be represented as linear combinations of these three vectors is said to form a basis for the space.

Section 2.4

Linear Dependence and Independence of Vectors and Triple Products

83

z

a3

a2

a1

y

x FIGURE 2.20 Nonorthogonal triad forming a basis in three-dimensional space.

It is a fundamental property of three-dimensional space that a basis for the space comprises a set of three vectors a1 , a2 , and a3 , with the property that the linear combination c1 a1 + c2 a2 + c3 a3 = 0

linear independence and linear dependence

(44)

is only true when c1 = c2 = c3 = 0. Vectors a1 , a2 , and a3 satisfying this condition are said to be linearly independent vectors, and a vector d of the form d = c1 a1 + c2 a2 + c3 a3 , where not all of c1 , c2 , and c3 are zero, is said to be linearly dependent on the vectors a1 , a2 , and a3 . The vectors i, j, and k that form a basis for three-dimensional space are linearly independent vectors, but the position vector r = 2i − 3j + 5k is linearly dependent on vectors i, j, and k. Clearly, vectors i, j, and k do not form the only basis for three-dimensional space, because any triad of linearly independent vectors a1 , a2 , and a3 will serve equally well, as, for example, the nonorthogonal set of vectors shown in Fig. 2.20. The dot and cross products will now be combined to develop a test for linear dependence and independence based on the elementary geometrical idea of the volume of the parallelepiped shown in Fig. 2.21, three edges a, b, and c of which meet at the origin. z y

n^ V

c b θ 0

a

FIGURE 2.21 Volume V of a parallelepiped.

x

84

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Vectors and Vector Spaces

The volume V of a parallelepiped is a nonnegative number given by the product of the area of its base and its height. Suppose vectors a and b are chosen to form two sides of the base of the parallelepiped. Then the vector area of this base has already been interpreted as a × b. The vertical height of the parallelepiped is the projection of vector c in the direction of the unit vector nˆ normal to the base, and ˆ it follows that so is given by nˆ · c. Consequently, as a × b = a · b sin θ n, V = |(a × b) · c|.

(45)

The absolute value of the right-hand side of (45) has been taken because a volume must be a nonnegative quantity, whereas the dot product (a × b) · c may be of either sign. If vectors a, b, and c form a basis for three-dimensional space, vector c cannot be linearly dependent on vectors a and b, and so the parallelepiped in Fig. 2.21 with these vectors as its sides must have a nonzero volume. If, however, vectors a, b, and c are coplanar (all lie in the same plane), and so cannot form a basis for the space, the volume of the parallelepiped will be zero. These simple geometrical observations lead to the following test for the linear independence of three vectors in three-dimensional space. THEOREM 2.3 a test for linear independence

scalar triple product

Test for linear independence of vectors in three-dimensional space Let a, b, and c be any three vectors. Then the vectors are linearly independent if (a × b)· c = 0, and they are linearly dependent if (a × b) · c = 0. A product of the type (a × b) · c is called a scalar triple product. The name arises because the result is a scalar. It is also called a mixed triple product since both · and × appear. Three vectors are involved in this dot (scalar) product, one of which is the vector a × b and the other is the vector c. Scalar triple products are easily evaluated, because taking the dot product of a × b in the form given in (42) with c = c1 i + c2 j + c3 k gives (a × b) · c = (a2 b3 − a3 b2 )c1 − (a1 b3 − a3 b1 )c2 + (a1 b2 − a2 b1 )c3 . The right-hand side of this expression is simply the value of a determinant with successive rows given by the components of a, b, and c, so we have arrived at the following convenient formula for the scalar triple product.

scalar triple product as a determinant

Scalar triple product Let a = a1 i + a2 j + a3 k, b = b1 i + b2 j + b3 k, and c = c1 i + c2 j + c3 k. Then a1 (a × b) · c = b1 c1

a2 b2 c2

a3 b3 . c3

(46)

Interchanging any two rows in a matrix changes the sign but not the value of its determinant. Two such switches in (46) leave the value unchanged, so the dot

Section 2.4

Linear Dependence and Independence of Vectors and Triple Products

85

product is commutative and so we arrive at the useful result (a × b) · c = a · (b × c).

(47)

So, in a scalar triple product the dot and cross may be interchanged without altering the result. EXAMPLE 2.17

Given the two sets of vectors (a) a = i + 2j − 5k, b = i + j + 2k, c = i + 4j − 19k and (b) a = 2i + j + k, b = 3i + 4k, c = i + j + k, ﬁnd if the vectors are linearly independent or linearly dependent. Solution We apply Theorem 2.3 to each set, using result (46) to evaluate the scalar triple products. 1 2 −5 2 = 0, (a) (a × b) · c = 1 1 1 4 −19 so the set of three vectors in (a) is linearly dependent. In fact this can be seen from the fact that c = 3a − 2b. 2 1 1 (b) (a × b) · c = 3 0 4 = −4 = 0, 1 1 1 so the set of three vectors in (b) is linearly independent. Although not required, the volume V of the parallelepiped formed by these three vectors is V = |(a × b) · c| = | − 4| = 4. Another notation for the scalar triple product of vectors a, b, and c is [a, b, c], so [a, b, c] = (a × b) · c, or, in terms of a determinant,

a1 [a, b, c] = b1 c1

alternative forms of a scalar triple product

a2 b2 c2

a3 b3 . c3

(48)

(49)

Using this deﬁnition of [a, b, c] with the row interchange property of determinants (see Section 1.7) shows that [a, b, c] = [b, c, a] = [c, a, b],

(50)

because two row interchanges are needed to arrive at [b, c, a] from [a, b, c], leaving the sign of the determinant unchanged, whereas two more are required to arrive at [c, a, b] from [b, c, a], again leaving the sign of the determinant unchanged. The order of the vectors in results (46), or in the equivalent notation of (48), is easily remembered when the results are abbreviated to a b b c c a

c a b

86

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In this pattern, row two follows from row one when the ﬁrst letter is moved to the end position, and row three follows from row two by means of the same process. The effect of applying this process to the third row is simply to regenerate the ﬁrst row. Rearrangements of this kind are called cyclic permutations of the three vectors. Again making use of the row interchange property of determinants (see Section 1.7), it follows that [a, b, c] = −[a, c, b], because this time only one row interchange is needed to produce the result on the right from the one on the left, so that a sign change is involved. A different product involving the three vectors a, b, and c that this time generates another vector is of the form a × (b × c), and products of this type are called vector triple products since the results are vectors. In these products it is essential to include the brackets because, in general, a × (b × c) = (a × b) × c. The most important results concerning vector triple products are given in the following theorem. THEOREM 2.4

Vector triple products (a)

vector triple product

If a, b, and c are any three vectors, then

a × (b × c) = (a · c)b − (a · b)c

and (b)

(a × b) × c = (a · c)b − (b · c)a.

Proof The proof of the results in Theorem 2.4 both follow in similar fashion, so we only prove result (a) and leave the proof of result (b) as an exercise. We write the cross product a × (b × c) in the form of the determinant in (43), with the components of a in the second row and those of b × c (obtained from (42)) in the third row when we ﬁnd that i j k . a1 a2 a3 a × (b × c) = (b2 c3 − b3 c2 ) (b3 c1 − b1 c3 ) (b1 c2 − b2 c1 ) Expanding this determinant in terms of the elements of its ﬁrst row and grouping terms gives a × (b × c) = [(a2 c2 + a3 c3 )b1 − (a2 b2 + a3 b3 )c1 ]i + [(a1 c1 + a3 c3 )b2 − (a1 b1 + a3 b3 )c2 ] j + [(a1 c1 + a2 c2 )b3 − (a1 b1 + a2 b2 )c3 ]k. As it stands, this result is not yet in the form that is required, but adding and subtracting a1 b1 c1 to the coefﬁcient of i, a2 b2 c2 to the coefﬁcient of j, and a3 b3 c3 to the coefﬁcient of k followed by grouping terms give a × (b × c) = (a · c)b − (a · b)c, and the result is established.

Section 2.4

EXAMPLE 2.18

Linear Dependence and Independence of Vectors and Triple Products

87

Find a × (b × c) and (a × b) × c, given that a = 3i + j − 4k, b = 2i + j + 3k, and c = i + 5j − k. Solution a · b = −5, a · c = 12, and b · c = 4, so a × (b × c) = (a · c)b − (a · b)c = 12b + 5c = 29i + 37j + 31k, and (a × b) × c = (a · c)b − (b · c)a = 12b − 4a = 12i + 8j + 52k. Accounts of geometrical vectors can be found, for example, in references [2.1], [2.3], [2.6], and [1.6].

Summary

This section introduced the two fundamental concepts of linear dependence and independence of vectors. It then showed how the scalar triple product involving three vectors, that gives rise to a scalar quantity, provides a simple test for the linear dependence or independence of the vectors involved. A simple and convenient way of calculating a scalar triple product was shown to be in terms of a determinant with the elements in its rows formed by the components of the three vectors involved in the product. Finally a vector triple product was deﬁned that gives rise to a vector quantity, and it was shown that to avoid ambiguity it is necessary to bracket a pair of vectors in such a product. A rule for the expansion of a vector triple product was derived and shown to involve a linear combination of two of the vectors multiplied by scalar products so that, for example, a × (b × c) = (a · c)b − (a · b)c.

EXERCISES 2.4 In Exercises 1 through 4 use the vectors a, b, and c to ﬁnd (a) the scalar triple product a · (b × c), and (b) the volume of the parallelepiped determined by these three vectors directed away from a corner. 1. 2. 3. 4.

a = 2i − j − 3k, b = 3i − 2k, c = i + j − 4k. a = i − j + 2k, b = i + j + 3k, c = 2i − j + 3k. a = −i − j + k, b = 2i + 2 j + 3k, c = −4i + j + 3k. a = 5i + 3k, b = 2i − j, c = −2i + 3j − 2k.

In Exercises 5 through 10 ﬁnd which sets of vectors are coplanar. 5. 6. 7. 8. 9. 10.

i + 3j + 2k, 2i + j + 4k, 4i + 7j + 8k. 2i + j + 4k, i + 2 j + k, 4i + 3j + 6k. 2i + k, i + 4j + 2k, 3i + 12 j + 7k. i + j + k, 2i + j + 2k, 4i + 3j + k. 2i + j − k, 3i + j + 2k, 5i + j + 8k. 2i + j − k, i + 2 j + 2k, 5i + 4j + k.

In Exercises 11 through 15 use computer algebra to verify that [a, b, c] = [c, a, b] = −[a, c, b]. 11. a = i + j + k, b = 2i + j − k, c = 3i − j + k. 12. a = i − j − k, b = −5i + 2 j − 3k, c = 2i + 3j − 2k.

13. a = −3i − 4j + k, b = 9i + 12 j − 3k, c = i + 2j + k. 14. a = 3i + 4k, b = i + 5k, c = 2 j + k. 15. Prove that if a, b, c, and d are any four vectors, and λ, μ are arbitrary scalars [λa + μb, c, d] = λ[a, c, d] + μ[b, c, d]. Use computer algebra with vectors a, b, c, d from Exercise 12 with d = 4c − 2 j + 6k, and scalars λ, μ of your choice, to verify this result. In Exercises 16 through 20 ﬁnd (a) the cartesian equation of the plane containing the given points, and (b) a unit vector normal to the plane. 16. 17. 18. 19. 20. 21.

(1, 2, 1), (3, 1, −2), (2, 1, 4). (2, 0, 3), (0, 1, 0), (2, 4, 5). (−1, 2, −3), (2, 4, 1), (3, 0, 1). (1, 2, 5), (−2, 1, 0), (0, 2, 0). Prove result (b) of Theorem 2.4. Show that a × (b × c) + b × (c × a) + c × (a × b) = 0.

22. The law of sines for a triangle with angles A, B, and C opposite sides with the respective lengths a, b, and c

88

Chapter 2

Vectors and Vector Spaces

takes the form b c a = = . sin A sin B sin C Prove this by considering a vector triangle with sides a, b, and c, where c = a + b, and taking the cross product of c = a + b ﬁrst with a, then with b, and ﬁnally with c. In Exercises 23 through 26 use the fact that four points with position vectors p, q, r, and s will be coplanar if the vectors p − q, p − r, and p − s are coplanar to ﬁnd which sets of points all lie in a plane. 23. 24. 25. 26. 27.

(1, 1, −1), (−3, 1, 1), (−1, 2, −1), (1, 0, 0). (1, 2, −1), (2, 1, 1), (0, 1, 2), (1, 1, 1). (0, −4, 0), (2, 3, 1), (3, −4, −2), (4, −2, −2). (1, 2, 3), (1, 0, 1), (2, 1, 2), (4, 1, 0). The volume of a tetrahedron is one-third of the product of the area of its base and its vertical height. Show the volume V of the tetrahedron in Fig. 2.22, in which three edges formed by the vectors a, b, and c are directed away from a vertex, is given by V = (1/6)|a · (b × c)|

28. Let a, b, c, and d be vectors and λ, μ, ν be scalars satisfying the equation λ(b × c) + μ(c × a) + ν(a × b) + d = 0. Show that if a, b, and c are linearly independent, then λ = −(a · d)/[a · (b × c)], ν = −(c · d)/[a · (b × c)].

2.5

μ = −(b · d)/[a · (b × c)],

a

b

c FIGURE 2.22 Tetrahedron.

29. Let a, b, c, and d be vectors and λ, μ, ν be scalars satisfying the equation λa + μb + νc + d = 0. By taking the scalar products of this equation ﬁrst with b × c, then with a × c, and ﬁnally with a × b, show that if a, b, and c are linearly independent, then λ = −d · (b × c)/[a · (b × c)], μ = −d · (c × a)/[a · (b × c)], ν = −d · (a × b)/[a · (b × c)]. 30. Show that a = i + 2 j + k, b = 2i − j − k, and c = 4i + 3j + ik are linearly independent vectors, and use them with a vector d of your choice to verify the results of Exercises 28 and 29. 31. Prove the Lagrange identity (a × b) · (c × d) = (a · c)(b · d) − (a · d)(b · c).

n-Vectors and the Vector Space R n

n-tuples

There are many occasions when it is convenient to generalize a vector and its associated algebra to spaces of more than three dimensions. A typical situation occurs in mechanics, where it is sometimes necessary to consider both the position and the momentum of a particle as functions of time. This leads to the study of a 6-vector, three components of which specify the particle position and three its momentum vector at a time t. Sets of n numbers (x1 , x2 , . . . , xn ) in a given order, that can be thought of either as n-vectors or as the coordinates of a point in n-dimensional space are called ordered n-tuples of real numbers or, simply, n-tuples.

n-Vectors and the Vector Space R n

Section 2.5

n-vector

89

An n-vector If n ≥ 2 is an integer, and x1 , x2 , . . . , xn are real numbers, an n-vector is an ordered n-tuple (x1 , x2 , . . . , xn ).

components and dimension

norm in R n

The numbers x1 , x2 , . . . , xn are called the components of the n-vector, xi is the ith component of the vector, and n is called the dimension of the space to which the n-vector belongs. For any given n, the set of all vectors with n real components is called a real n-space or, simply, an n-space, and it is denoted by the symbol Rn . A corresponding space exists when the n numbers x1 , x2 , . . . , xn are allowed to be complex numbers, leading to a complex n-space denoted by C n . In this notation R3 is the three-dimensional space used in previous sections. In R3 the length of a vector was taken as the deﬁnition of its norm, so if r = x1 i + x2 j + x3 k, then r = (x12 + x22 + x32 )1/2 . A generalization of this norm to R n leads to the following deﬁnition. The norm in Rn The norm of the n-vector (x1 , x2 , . . . , xn ), denoted by (x1 , x2 , . . . , xn ) is (x1 , x2 , . . . , xn ) =

x12 + x22 + · · · + xn2 1/2 n 2 = xi .

(51)

i=1

The laws for the equality, addition, and scaling of vectors in R3 in terms of the components of the vector generalize to R n as follows. Equality of n-vectors

algebraic rules for equality, addition, and scaling using components

Let (x1 , x2 , . . . , xn ) and (y1 , y2 , . . . , yn ) be two n-vectors. Then the vectors will be equal, written (x1 , x2 , . . . , xn ) = (y1 , y2 , . . . , yn ), if, and only if, corresponding components are equal, so that x1 = y1 , x2 = y2 , . . . , xn = yn . (52)

Addition of n-vectors Let (x1 , x2 , . . . , xn ) and (y1 , y2 , . . . , yn ) be any two n-vectors. Then the sum of these vectors, written (x1 , x2 , . . . , xn ) + (y1 , y2 , . . . , yn ), is deﬁned as the vector whose ith component is the sum of the corresponding ith components of the vectors for i = 1, 2, . . . , n, so that (x1 , x2 , . . . , xn ) + (y1 , y2 , . . . , yn ) = (x1 + y1 , x2 + y2 , . . . , xn + yn ). (53)

90

Chapter 2

Vectors and Vector Spaces

Scaling an n-vector Let (x1 , x2 , . . . , xn ) be an arbitrary n-vector and λ be any scalar. Then the result of scaling the vector by λ, written λ(x1 , x2 , . . . , xn ), is deﬁned as the vector whose ith component is λ times the ith component of the original vector, for i = 1, 2, . . . , n, so that λ(x1 , x2 , . . . , xn ) = (λx1 , λx2 , . . . , λxn ).

(54)

The null (zero) vector in Rn is the vector 0 in which every component is zero, so that 0 = (0, 0, . . . , 0).

(55)

As with vectors in R3 , so also with n-vectors in Rn , it is convenient to use a single boldface symbol for a vector and the corresponding italic symbols with sufﬁxes when it is necessary to specify the components. So we will write x = (x1 , x2 , . . . , xn )

and

y = (y1 , y2 , . . . , yn ).

The reasoning that led to the interpretation of Theorem 2.1 on the algebraic rules for the addition and scaling of vectors in R3 leads also the following theorem for n-vectors. THEOREM 2.5

Algebraic rules for the addition and scaling of n-vectors in R n Let x, y, and z be arbitrary n-vectors, and let λ and μ be arbitrary real numbers. Then: (i) (ii) (iii) (iv) (v) (vi) (vii)

x + y = y + x; x + 0 = 0 + x = x; (x + y) + z = x + (y + z); λ(x + y) = λx + λy; (λμ)x = λ(μx) = μ(λx); (λ + μ)x = λx + μx; λx = |λ| x .

Because of this similarity between vectors in R3 and in Rn , the space Rn is called a real vector space, though because the symbol R indicates real numbers this is usually abbreviated a vector space. Analogously, when the elements of the n-vectors are allowed to be complex, the resulting space is called the complex vector space C n . So far there would seem to be little difference between vectors in R3 and Rn , but major differences do exist, and they are best appreciated when geometrical analogies are sought for vector operations in Rn . dot product of n-vectors

The dot product of n-vectors Let x = (x1 , x2 , . . . , xn ) and y = (y1 , y2 , . . . , yn ) be any two n-vectors. Then the dot product of these two vectors, written x · y and also called their inner

Section 2.5

n-Vectors and the Vector Space R n

91

product, is deﬁned as the sum of the products of corresponding components, so that (x1 , x2 , . . . , xn ) · (y1 , y2 , . . . , yn ) = x1 y1 + x2 y2 + . . . + xn yn .

(56)

The following properties of this dot product are strictly analogous to those of the dot product in R3 and can be deduced directly from (56). THEOREM 2.6

Properties of the dot product in R n Let x, y, and z be any three n-vectors and λ be any scalar. Then: (i) (ii) (iii) (iv) (v) (vi)

x · y = y · x; x · (y + z) = x · y + x · z; (λx) · y = x · (λy) = λ(x · y); x · x = x 2 ; x · 0 = 0; x 2 = 0 if, and only if, x = 0.

The existence of a dot product in Rn allows the Cauchy–Schwarz and triangle inequalities to be generalized, both of which play a fundamental role in the study of vector spaces. Various forms of proof of these inequalities are possible, but the one given here has been chosen because it makes full use of the properties of the dot product listed in Theorem 2.6. THEOREM 2.7

The Cauchy–Schwarz and triangle inequalities Let x = (x1 , x2 , . . . , xn ) and y = (y1 , y2 , . . . , yn ) be any two n-vectors. Then

generalized inequalities for n-vectors

(a)

|x · y| ≤ x · y

(Cauchy–Schwarz inequality),

and (b)

x + y ≤ x + y

(triangle inequality).

Proof We start by proving the Cauchy–Schwarz inequality in (a). The inequality is certainly true if x · y = 0, so we need only consider the case x · y = 0. Let x and y be any two n-vectors, and λ be a scalar. Then, using properties (ii) to (iv) of Theorem 2.6, x + λy 2 = (x + λy) · (x + λy), = x 2 + λx · y + λy · x + λ2 y 2 . However, by result (1) of Theorem 2.6, y · x = x · y, so x + λy 2 = x 2 + 2λx · y + λ2 y 2 . We now set λ = − x 2 /(x · y) to obtain x + λy 2 = − x 2 + ( x 4 y 2 )/|x · y|2 , where we have used the fact that (x · y)2 = |x · y|2 . As x + λy 2 is nonnegative, this result is equivalent to − x 2 + ( x 4 · y 2 )/|x · y|2 ≥ 0.

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Vectors and Vector Spaces

Cancelling the nonnegative number x 2 , which leaves the inequality sign unchanged; rearranging the terms; and taking the square root of the remaining nonnegative result on each side of the inequality yields the Cauchy–Schwarz inequality |x · y| ≤ x · y . To prove the triangle inequality (b) we set λ = 1 and start from the result x + y 2 = x 2 + 2x · y + y 2 . As x · y may be either positive or negative, x · y ≤ |x · y|, so making use of the Cauchy–Schwarz inequality shows that x + y 2 ≤ x 2 + 2 x · y + y 2 = ( x + y )2 . The triangle inequality follows from taking the square root of each side of this inequality, which is permitted because both are nonnegative numbers. The dot product in R3 allowed the angle between vectors to be determined and, more importantly, it provided a test for the orthogonality of vectors. These same geometrical ideas can be introduced into the vector space Rn if the Cauchy–Schwarz inequality is written in the form − x · y ≤ x · y ≤ x · y . After division by the nonnegative number x · y , this becomes −1 ≤

x·y ≤ 1. x · y

This enables the angle θ between the two n-vectors x and y to be deﬁned by the result x·y . x · y

cos θ =

orthogonality of n-vectors unit n-vector

On account of this result, two n-vectors x and y in Rn will be said to be orthogonal when x · y = 0. By analogy with R3 we will call x = (x1 , x2 , . . . , xn ) a unit n-vector if x = 1. If we deﬁne the unit n-vectors e1 , e2 , . . . , en as e1 = (1, 0, 0, 0, . . . , 0), e2 = (0, 1, 0, 0, . . . , 0), . . . , en = (0, 0, 0, 0, . . . , 1), we see that

ei · e j =

1 0

for i = j for i = j,

showing that the ei are mutually orthogonal unit n-vectors in Rn . As a result of this the vectors e1 , e2 , . . . , en play the same role in Rn as the vectors i, j, and k in R3 . This allows the vector x = (x1 , x2 , . . . , xn ) to be written as x = x1 e1 + x2 e2 + · · · + xn en , where xi is the ith component of x.

Section 2.5

n-Vectors and the Vector Space R n

93

Now suppose that for n > 3, we set u1 = (1, 0, 0, 0, . . . , 0),

subspaces

u2 = (0, 1, 0, 0, . . . , 0),

u3 = (0, 0, 1, 0, . . . , 0),

and all other ui identically zero, so that ui = (0, 0, 0, 0, . . . , 0) for i = 4, 5, . . . , n. Then it is not difﬁcult to see that u1 , u2 , and u3 behave like the unit vectors i, j, and k, so that, in some sense the vector space R3 is embedded in the vector space Rn with vectors in both spaces obeying the same algebraic rules for addition and scaling. This is recognized by saying that R3 is a subspace of Rn . Subspace of Rn A subset S of vectors in the vector space Rn is called a subspace of Rn if S is itself a vector space that obeys the rules for the addition and the scaling of vectors in Rn .

EXAMPLE 2.19

Find the condition that the set S of vectors of the form (x, mx + c, 0), for any m and all real x forms a subspace of the vector space R3 , and give a geometrical interpretation of the result. Solution The set S can only contain the null vector (0, 0, 0) if c = 0, so if c = 0 the vectors in S cannot form a subspace of R3 . Now let c = 0, so that S contains the null vector. The vector addition law holds, because if (x, mx, 0) and (x , mx , 0) are vectors in S, the sum (x, mx, 0) + (x , mx , 0) = (x + x , m(x + x ), 0) is also a vector in S. The scaling λ(x, mx, 0) = (λx, mλx, 0) also generates a vector in S, so the scaling law for vectors also holds, showing that S is a subspace of R3 provided c = 0. If the three components of vectors in S are regarded as the x-, y-, and z-components of a vector in R3 , the vectors can be interpreted as points on the straight line y = mx passing through the origin and lying in the plane z = 0. This subspace is a one-dimensional vector space embedded in the three-dimensional vector space R3 .

EXAMPLE 2.20

Test the following subsets of Rn to determine if they form a subspace. (a) S is the set of vectors (x1 , x1 + 1, . . . , xn ) with all the xi real numbers. (b) S is the set of vectors (x1 , x2 , . . . , xn ) with x1 + x2 + · · · + xn = 0 and all the xi are real numbers. Solution (a) The set S does not contain the null vector and so cannot form a subspace of Rn . This result is sufﬁcient to show that S is not a subspace, but to see what properties of a subspace the set S possesses we consider both the summation and scaling of vectors in S. If (x1 , x1 + 1, . . . , xn ) and (x1 , x1 + 1, . . . , xn ) are two vectors in S, their sum (x1 , x1 + 1, . . . , xn ) + (x1 , x1 + 1, . . . , xn ) = (x1 + x1 , x1 + x1 + 2, . . . , xn + xn ) is not a vector in S, so the summation law is not satisﬁed.

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The scaling condition for vectors is not satisﬁed, because if λ is an arbitrary scalar, λ(x1 , x1 + 1, . . . , xn ) = (λx1 , λx1 + λ, . . . , λxn ) = (a, a + 1, . . .),

(λn1 = a)

showing that scaling generates another a vector in S. We have proved that the vectors in S do not form a subspace of Rn . (b) The set S does contain the null vector, because x1 = x2 = · · · = xn = 0 satisﬁes the constraint condition x1 + x2 + · · · + xn = 0. Both the summation law and the scaling law for vectors are easily seen to be satisﬁed, so this set S does form a subspace of Rn . EXAMPLE 2.21

Let C(a, b) be the space of all real functions of a single real variable x that are continuous for a < x < b, and let S(a, b) be the set of all functions belonging to C(a, b) that have a derivative at every point of the interval a < x < b. Show that S(a, b) forms a subspace of C(a, b). Solution In this case a vector in the space is simply any real function of a single real variable x that is continuous in the interval a < x < b. The null vector corresponds to the continuous function that is identically zero in the stated interval, so as the derivative of this function is also zero, it follows that the set S(a, b) must also contain the null vector. The sum of continuous functions in a < x < b is a continuous function, and the sum of differentiable functions in this same interval is a differentiable function, so the summation law for vectors is satisﬁed. Similarly, scaling continuous functions and differentiable functions does not affect either their continuity or their differentiability, so the scaling law for vectors is also satisﬁed. Thus, S(a, b) forms a subspace of C(a, b). Think of the dimension of these spores as inﬁnite; norm and inner product are easy to deﬁne.

Summary

This section generalized the concept of a three-dimensional vector to a vector with n components in R n . It was shown that the magnitude of a vector in three space dimensions generalizes to the norm of a vector in R n and that in terms of components, the equality, addition, and scaling of vectors in R n follow the same pattern as with three space dimensions. The dot product was generalized and two fundamental inequalities for vectors in R n were derived. The concept of orthogonality of vectors was generalized and the notion of a subspace of R n was introduced.

EXERCISES 2.5 In Exercises 1 through 8 ﬁnd the sum of the given pairs of vectors, their norms, and their dot product. 1. 2. 3. 4. 5. 6. 7. 8.

(2, 1, 0, 2, 2), (1, −1, 2, 2, 4). (3, −1, −1, 2, −4), (1, 2, 0, 0, 3). (2, 1, −1, 2, 1), (−2, −1, 1, −2, −1). (3, −2, 1, 1, 2, 0, 1), (1, −1, 1, −1, 1, 0, 1). (3, 0, 1, 0), (0, 2, 0, 4). (1, −1, 2, 2, 0, 1), (2, −2, 1, 1, 1, 0). (−1, 2, −4, 0, 1), (2, −1, 1, 0, 2). (3, 1, 2, 4, 1, 1, 1), (1, 2, 3, −1, −2, 1, 3).

In Exercises 9 through 12 ﬁnd the angle between the given pairs of n-vectors and the unit n-vector associated with each vector. 9. 10. 11. 12.

(3, 1, 2, 1), (1, −1, 2, 2). (4, 1, 0, 2), (2, −1, 2, 1). (2, −2, −2, 4), (1, −1, −1, 2). (2, 1, −1, 1), (1, −2, 2, 2).

In Exercises 13 through 18 determine if the set of vectors S forms a subspace of the given vector space. Give reasons why S either is or is not a subspace.

Section 2.6 13. S is the set of vectors of the form (x1 , x2 , . . . , xn ) in Rn , with the xi real numbers and x2 = x14 . 14. S is the set of vectors of the form (x1 , x2 , . . . , xn ) in Rn , with the xi real numbers and x1 + 2x2 + 3x3 + · · · + nxn = 0. 15. S is the set of vectors of the form (x1 , x2 , . . . , xn ) in Rn , with the xi real numbers and x1 + x2 + x3 + · · · + xn = 2. 16. S is the set of vectors of the form (x1 , x2 , . . . , x6 ) in R 6 , with the xi real numbers and x1 = 0 or x6 = 0. 17. S is the set of vectors of the form (x1 , x2 , . . . , x6 ) in R 6 , with the xi real numbers and x1 − x2 + x3 · · · + x6 = 0. 18. S is the set of vectors of the form (x1 , x2 , . . . , x5 ) in R5 , with the xi real numbers and x2 < x3 . In Exercises 19 to 23 determine if the given set S is a subspace of the space C[0, 1] of all real valued functions that are continuous on the interval 0 ≤ x ≤ 1. Give reasons why either S is a subspace, or it is not. 19. S is the set of all polynomials of degree two. 20. S is the set of all polynomial functions. 21. S is the set of all continuous functions such that f (0) = f (1) = 0. 22. S is the set of all continuous functions such that f (0) = 0 and f (1) = 2. 23. S is the set of all continuous once differentiable functions such that f (0) = 0 and f (x) > 0. 24. Prove that the set S of all vectors lying in any plane in R3 that passes through the origin forms a subspace of R3 . 25. Explain why the set S of all vectors lying in any plane in R3 that does not pass through the origin does not form a subspace of R3 .

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Linear Independence, Basis, and Dimension

95

26. Consider the polynomial P(λ) deﬁned as P(λ) = x + λy 2 , where x and y are vectors in Rn . Show, provided not both x and y are null vectors, that the graph of P(λ) as a function of λ is nonnegative, so P(λ) = 0 cannot have real roots. Use this result to prove the Cauchy–Schwarz inequality |x · y| ≤ x · y . 27. Let x and y be vectors in Rn and λ be a scalar. Prove that x + λy 2 + x − λy 2 = 2( x 2 + λ2 y 2 ). 28. If x and y are orthogonal vectors in Rn , prove that the Pythagoras theorem takes the form x + y 2 = x 2 + y 2 . 29. What conditions on the components of vectors x and y in the Cauchy–Schwarz inequality cause it to become an equality, so that 1/2 1/2 n n n 2 2 xi yi = xi yi ? i=1

i=1

i=1

30. Modify the method of proof used in Theorem 2.7 to prove the complex form of the Cauchy–Schwarz inequality 1/2 1/2 n n n 2 2 x y≤ |xi | + |yi | , i=1 i i i=1 i=1 where the xi and yi are complex numbers.

Linear Independence, Basis, and Dimension The concept of the linear independence of a set of vectors in R3 introduced in Section 2.4 generalizes to Rn and involves a linear combination of n-vectors. Linear combination of n-vectors Let x1 , x2 , . . . , xm be a set of n-vectors in Rn . Then a linear combination of the n-vectors is a sum of the form c1 x1 + c2 x2 + · · · + cmxm, where c1 , c2 , . . . , cm are nonzero scalars.

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An example of a linear combination of vectors in R5 is provided by the vector sum (m = 3, n = 5) 2x1 + x2 + 3x3 , where x1 = (1, 2, 3, 0, 4), x2 = (2, 1, 4, 1, −3), and x3 = (6, 0, 2, 2, −1). The vector in R5 formed by this linear combination is 2x1 + x2 + 3x3 = 2(1, 2, 3, 0, 4) + (2, 1, 4, 1, −3) + 3(6, 0, 2, 2, −1), = (22, 5, 16, 7, 2). A linear combination of n-vectors is the most general way of combining nvectors, and the deﬁnition of a linear combination of vectors contains within it the deﬁnition of the scaling of a single n-vector as a special case. This can be seen by setting m = 1, because this reduces the linear combination to the single scaled n-vector c1 x1 . linear dependence and independence of n-vectors

Linear dependence of n-vectors Let x1 , x2 , . . . , xm be a set of n-vectors in Rn . Then the set is said to be linearly dependent if, and only if, one of the n-vectors can be expressed as a linear combination of the remaining n-vectors. An example of linear dependence in R 4 is provided by the vectors x1 = (1, 0, 2, 5), x2 = (2, 1, 2, 1), x3 = (3, 2, 1, 0), and x4 = (−1, −1, −1, 7), because x4 = 2x1 − 3x2 + x3 . Linear independence of n-vectors Let x1 , x2 , . . . , xm be a set of n-vectors in Rn . Then the set is said to be linearly independent if, and only if, the n-vectors are not linearly dependent. A simple example of a set of linearly independent vectors in R 4 is provided by the vectors e1 = (1, 0, 0, 0), e2 = (0, 1, 0, 0), and e3 = (0, 0, 1, 0). The linear independence of these 4-vectors can be seen from the fact that for no choice of c1 and c2 can the vector c1 e1 + c2 e2 be made equal to e3 . To make effective use of the concept of linear independence, and to understand the notion of the basis and dimension of a vector space, it is necessary to have a test for linear independence. Such a test is provided by the following theorem.

THEOREM 2.8

Linear dependence and independence Let S be a set of non-zero n-vectors x1 , x2 , . . . , xm, with m ≥ 2. Then: (a) Set S is linearly dependent if the vector equation c1 x1 + c2 x2 + · · · + cmxm = 0 is true for some set of scalars (constants) c1 , c2 , . . . , cm that are not all zero;

Section 2.6

Linear Independence, Basis, and Dimension

97

(b) Set S is linearly independent if the vector equation c1 x1 + c2 x2 + · · · + cmxm = 0 is only true when c1 = c2 = · · · = cm = 0. Proof To establish result (a) it is necessary to show that the conditions of the deﬁnition of linear dependence are satisﬁed. First, if the set S of n-vectors is linearly dependent, scalars d1 , d2 , . . . , dm exist such that d1 x1 + d2 x2 + · · · + dmxm = 0. There is no loss of generality in assuming that d1 = 0, because if this is not the case a renumbering of the vectors can always make this possible. Consequently, x1 = (−d2 /d1 )x2 + (−d3 /d1 )x3 + · · · + (−dm/d1 )xm, which shows, as claimed, that the set S is linearly dependent, because x1 is linearly dependent on x2 , x3 , . . . , xm. A similar argument applies to show that xr is linearly dependent on the remaining n-vectors in S provided dr = 0, for r = 2, 3, . . . , m. Conversely, if one of the n-vectors in set S, say x1 , is linearly dependent on the remaining n-vectors in the set, scalars d1 , d2 , . . . , dm can be found such that x1 = d2 x2 + · · · + dmxm, so that x1 − d2 x2 − · · · − dmxm = 0. This result is of the form given in deﬁnition of linear dependence with c1 = 1, c2 = −d2 , . . . , cm = −dm, not all of which constants are zero, so again the set of n-vectors in S is seen to be linearly dependent. To establish result (b), suppose, if possible, that the set S of vectors is linearly independent, but that some scalars d1 , d2 , . . . , dm that are not all zero can be found such that d1 x1 + d2 x2 + · · · + dmxm = 0. Then if d1 = 0, say, is one of these scalars, it follows that x1 = (−d2 /d1 )x2 + (−d3 /d1 )x3 + · · · + (−dm/d1 )xm, which is impossible because this shows that, contrary to the hypothesis, x1 is linearly dependent on the remaining n-vectors in S. So we must have c1 = c2 = · · · = cm = 0. A systematic and efﬁcient computational method for the application of Theorem 2.8 to vectors in Rn will be developed in the next chapter for the three separate cases that arise, (a) m < n, (b) m = n, and (c) m > n. However, when n and m are small, a straightforward approach is possible, as illustrated in the next example. EXAMPLE 2.22

Test the following sets of vectors in R4 for linear dependence or independence. (a)

x1 = (2, 1, 1, 0),

x2 = (0, 2, 0, 1),

x3 = (1, 1, 0, 2),

x4 = (0, 2, 1, 1).

(b)

x1 = (4, 0, 2), x2 = (2, 2, 0), x3 = (1, 1, 0), x4 = (5, 1, 2).

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Solution In both (a) and (b) it is necessary to consider the vector equation c1 x1 + · · · + cmxm = 0. If the equation is only satisﬁed when c1 = c2 = · · · = cm = 0, the set of vectors will be linearly independent, whereas if a solution can be found in which not all of the constants c1 , c2 , c3 , c4 vanish, the set of vectors will be linearly dependent. (a) Substituting for x1 , x2 , x3 , x4 in the preceding equation and equating corresponding components show the coefﬁcients ci must satisfy the following equations 2c1 + c3 = 0 c1 + 2c2 + c3 + 2c4 = 0 c1 + c 4 = 0 c2 + 2c3 + c4 = 0. The third equation shows that c4 = −c1 , so the equations can be rewritten as 2c1 + c3 = 0 −c1 + 2c2 + c3 = 0 c2 − c1 + 2c3 = 0. Adding twice the third equation to the ﬁrst equation shows that c3 = 0, so c1 = 0, and it then follows that c2 = c3 = c4 = 0. This has established the linear independence of the set of vectors in (a). (b) Proceeding in the same manner with the set of vectors in (b) leads to the following equations for the coefﬁcients ci : 4c1 + 2c2 + c3 + 5c4 = 0 2c2 + c3 + c4 = 0 2c1 + 2c4 = 0. The third equation shows that c4 = −c1 , so using this result in the ﬁrst two equations reduces the ﬁrst one to −c1 + 2c2 + c3 = 0 and the second to −c1 + 2c2 + c3 = 0. There is only one equation connecting c1 , c2 , and c3 , and hence also c4 . This means that if c2 and c3 are given arbitrary values, not both of which are zero, the constants c1 and c4 will be determined in terms of them. Thus, a set of constants c1 , c2 , c3 , c4 that are not all zero can be found that satisfy the vector equation, showing that the set of vectors in (b) is linearly dependent. This set of constants is not unique, but this does affect the conclusion that the set of vectors is linearly dependent, because to establish linear dependence it is sufﬁcient that at least one such set of constants can be found.

Section 2.6

Linear Independence, Basis, and Dimension

99

Example 2.22 has shown one way in which Theorem 2.8 can be implemented for vectors in Rn , but it also illustrates the need for a systematic approach to the solution of the system of equations for the coefﬁcients when n is large. A trivial case of Theorem 2.8 arises when the set of vectors S contains the null vector 0, because then the set of vectors in S is always linearly dependent. This can be seen by assuming that x1 = 0, because then the vector equation in the theorem becomes c1 0 + c2 x2 + · · · + cmxm = 0. This vector equation is satisﬁed if c1 = 0 (arbitrary) and c2 = c3 = · · · = cm = 0, so, as not all of the coefﬁcients are zero, the set of vectors must be linearly dependent. We conclude this introduction to the vector space Rn by deﬁning the span, a basis, and the dimension of a vector space. span of a vector space

Span of a vector space Let the set of non-zero vectors x1 , x2 , . . . , xm belonging to a vector space V have the property that every vector in V can be expressed as a linear combination of these vectors. Then the vectors x1 , x2 , . . . , xm are said to span the vector space V.

EXAMPLE 2.23

All vectors v in the (x, y)-plane are spanned by the vectors i and j, because any vector v = (v1 , v2 ) can always be written v = v1 i + v2 j. This is an example of vectors spanning the space R2 .

EXAMPLE 2.24

The vector space Rn is spanned by the unit n-vectors e1 = (1, 0, 0, 0, . . . , 0), e2 = (0, 1, 0, 0, . . . , 0), . . . , en = (0, 0, 0, 0, . . . , 1).

EXAMPLE 2.25

The subspace R3 of the vector space R5 is spanned by the unit vectors e1 = (1, 0, 0, 0, 0), e2 = (0, 1, 0, 0, 0), e3 = (0, 0, 1, 0, 0), because all vectors v = (v1 , v2 , v3 ) in R3 can be written in the form of the linear combination v = v1 e1 + v1 e2 + v3 e3 .

basis of a vector space in R n

Basis of a vector space Let x1 , x2 , . . . , xn be vectors in Rn . Then the vectors are said to form a basis for the vector space Rn if: (i) The vectors x1 , x2 , . . . , xn are linearly independent. (ii) Every vector in Rn can be expressed as a linear combination of the vectors x1 , x2 , . . . , xn .

dimension of a vector space

Dimension of a vector space The dimension of a vector space is the number of vectors in its basis.

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EXAMPLE 2.26

A basis for the space of ordinary vectors in three dimensions is provided by the vectors i, j, and k, so the dimension of the space is 3.

EXAMPLE 2.27

A basis for Rn is provided by the n vectors e1 = (1, 0, 0, 0, . . . , 0), e2 = (0, 1, 0, 0, . . . , 0), . . . , en = (0, 0, 0, 0, . . . , 1), so its dimension is n.

EXAMPLE 2.28

It was shown in Example 2.20 (b) that the set S of vectors (x1 , x2 , . . . , xn ) with x1 + x2 + · · · + xn = 0 forms a subspace of Rn . The dimension of Rn is n, but the constraint condition x1 + x2 + · · · + xn = 0 implies that only n − 1 of the components x1 , x2 , . . . , xn can be speciﬁed independently, because the constraint itself determines the value of the remaining component. This in turn implies that the basis for the subspace S can only contain n − 1 linearly independent vectors, so S must have dimension n − 1. More information on linear vector spaces can be found in references [2.1] and [2.5] to [2.12].

Summary

In this section the concepts of linear dependence and independence were generalized to vectors in R n , and the span of a vector space was deﬁned as a set of vectors in R n with the property that every vector in R n can be expressed as a linear combination of these vectors. Naturally in R n , as in R 3 , a set of vectors spanning the space is not unique. The smallest set of n vectors spanning a vector space is said to form a basis for the vector space, and the dimension of a vector space is the number of vectors in its basis. This corresponds to the fact that the unit vectors i, j, and k form a basis for the ordinary three-dimensional space R 3 , because every vector in this space can be represented as a linear combination of i, j, and k.

EXERCISES 2.6 In Exercises 1 through 12 determine if the set of m vectors in three-dimensional space is linearly independent by solving for the scalars c1 , c2 . . . cm in Theorem 2.8. Where appropriate, verify the result by using Theorem 2.3. a = i + 2 j + k, b = i − j + k, c = 2i + k. a = 3i − j + k, b = i + 3k, c = 5i − j + 7k. a = 2i − j + k, b = 3i + j − k, c = 8i + j + 7k. a = 3i + 2k, b = i + j + 2k, c = 11i + 2 j − 2k. a = 4i − j + 3k, b = i + 4j − 2k, c = 3i − j − k. a = i + j − k, b = i − j + k, c = −i + j + k. a = i + 2 j + k, b = i + 3j − k, c = 3i + 10j − 5k. a = 2i + 3j + k, b = i − 3j + 2k, c = i + 15j − 4k. a = 3i − j + 2k, b = i + j + k (m = 2). a = i + j + k, b = i + 2 j + k, c = i + 3j + k, d = i − 4j + k (m = 4). 11. a = i − j + 3k, b = 2i − j + 2k, c = i + k, d = 3i + j + k (m = 4).

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

12. a = i + j, b = j + k, c = i − k. In Exercises 13 through 16, determine if the set of vectors in R 4 is linearly independent by using the method of Example 2.22. 13. 14. 15. 16.

(1, 3, −1, 0), (1, 2, 0, 1), (0, 1, 0, −1), (1, 1, 0, 1). (1, −2, 1, 2), (4, −1, 0, 2), (2, 1, −1, 1), (1, 0, 0, −1). (2, 1, 0, 1), (1, 0, 1, 1), (4, 1, 2, −1), (1, 0, 1, −1). (1, 2, 1, 1), (1, −2, 0, −1), (1, 1, 1, 2), (1, −1, 0, 0).

In Exercises 17 through 20, ﬁnd a basis and the dimension of the given subspace S. 17. The subspace S of vectors in R5 of the form (x1 , x2 , x3 , x4 , x5 ) with x1 = x2 . 18. The subspace S of vectors in R 4 of the form (x1 , x2 , x3 , x4 ) with x1 = 2x2 . 19. The subspace S of vectors in R5 of the form (x1 , x2 , x3 , x4 , x5 ) with x1 = x2 = 2x3 .

Section 2.7 20. The subspace S of vectors in R 6 of the form (x1 , x2 , x3 , x4 , x5 , x6 ) with x1 = 2x2 and x3 = −x4 . 21. Let u = cos2 x and v = sin2 x form a basis for a vector space V. Find which of the following can be represented in terms of u and v, and so lie in V.

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Gram–Schmidt Orthogonalization Process

101

(a) 2. (b) sin 2x. (c) 0. (d) cos 2x. (e) 2 + 3x. (f) 3 − 4 cos 2x. 22. Given that r ≤ n, prove that any subset S of r vectors selected from a set of n linearly independent vectors is linearly independent.

Gram–Schmidt Orthogonalization Process A set of vectors forming a basis for a vector space is not unique, and having obtained a basis by some means, it is often useful to replace it by an equivalent set of orthogonal vectors. The Gram–Schmidt orthogonalization process accomplishes this by means of a sequence of simple steps that have a convenient geometrical interpretation. We now develop the Gram–Schmidt orthogonalization process for geometrical vectors in R3 , though in Section 4.2 the method will be extended to vectors in Rn to enable orthogonal matrices to be constructed from a set of eigenvectors associated with a symmetric matrix. Let us now show how any basis for R3 , comprising three nonorthogonal linearly independent vectors a1 , a2 , and a3 , can be used to construct an equivalent basis involving three linearly independent orthogonal vectors u1 , u2 , and u3 . It is essential that the vectors a1 , a2 , and a3 be linearly independent, because if not, the vectors u1 , u2 , and u3 generated by the Gram–Schmidt orthogonalization process will be linearly dependent and so cannot form a basis for R3 . The derivation of the method starts by setting u1 = a1 , where the choice of a1 instead of a2 or a3 is arbitrary. The component of a2 in the direction of u1 is u1 · a2 , so the vector component of a2 in this direction is (u1 · a2 )u1 =

(u1 · a2 )u1 , u1 2

and this always exists because u1 2 > 0. Subtracting this vector from a2 gives a vector u2 that is normal to u1 , so u2 = a2 −

(u1 · a2 )u1 . u1 2

Similarly, to ﬁnd a vector normal to both u1 and u2 involving a3 , it is necessary to subtract from a3 the components of vector a3 in the direction of u1 and also in the direction of u2 , so that u3 = a3 −

(u1 · a3 )u1 (u2 · a3 )u2 − , u1 2 u2 2

and this also always exists, because u1 2 > 0 and u2 2 > 0. If an orthonormal basis is required, it is necessary to normalize the vectors u1 , u2 , and u3 by dividing each by its norm.

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Rule for the Gram–Schmidt orthogonalization process in R3 A set of nonorthogonal linearly independent vectors a1 , a2 , and a3 that form a basis in R3 can be used to generate an equivalent orthogonal basis involving the vectors, u1 , u2 , and u3 by setting

u1 = a1 ,

u2 = a2 −

u3 = a3 −

(u1 · a2 )u1 , u1 2

and

(u1 · a3 )u1 (u2 · a3 )u2 − . u1 2 u2 2

As already remarked, the choice of a1 as the vector with which to start the orthogonalization process was arbitrary, and the process could equally well have been started by setting u1 = a2 or u1 = a3 . Using a different vector will produce a different set of orthogonal vectors u1 , u2 , and u3 , but any basis for R3 is equivalent to any other basis, so unless there is a practical reason for starting with a particular vector, the choice is immaterial. EXAMPLE 2.29

Given the nonorthogonal basis a1 = i − j − k, a2 = i + j + k, and a3 = −i + 2k, use the Gram–Schmidt orthogonalization process to ﬁnd an equivalent orthogonal basis, and then ﬁnd the corresponding orthonormal basis. Solution Using the preceding rule we start with u1 = i − j − k, and to ﬁnd u2 we need to use the results u1 · a2 = −1 and u1 2 = 3, so that u2 = i + j + k − (−1/3)(i − j − k) = (4/3)i + (2/3)j + (2/3)k. To ﬁnd u3 we need to use the results u1 · a3 = −3, u1 2 = 3, u2 · a3 = 0, and u2 2 = 24/9, so that u3 = −i + 2k − (−3/3)(i − j − k) = −j + k. So the required equivalent orthogonal basis is u1 = i − j − k,

u2 = (4/3)i + (2/3)j + 2/3k,

and

u3 = −j + k.

The corresponding orthonormal basis obtained by dividing each of these vectors by its norm (modulus) is √ uˆ 1 = (1/ 3)u1 ,

uˆ 2 = (1/2) (3/2)u2

and

√ uˆ 3 = (1/ 2)u3 .

Other accounts of the Gram–Schmidt orthogonalization process are to be found in references [2.1] and [2.7] to [2.12].

Section 2.7

Summary

Gram–Schmidt Orthogonalization Process

103

In this section it is shown how in R 3 the Gram–Schmidt orthogonalization process converts any three nonorthogonal linearly independent vectors a1 , a2 , and a3 into three orthogonal vectors u1 , u2 , and u3 . If necessary, the vectors u1 , u2 , and u3 can then be normalized in the usual manner to form an orthogonal set of unit vectors.

EXERCISES 2.7 In Exercises 1 through 6, use the given nonorthogonal basis for vectors in R3 to ﬁnd an equivalent orthogonal basis by means of the Gram–Schmidt orthogonalization process. 1. 2. 3. 4. 5.

a1 a1 a1 a1 a1

= i + 2 j + k, a2 = i − j, a3 = 2 j − k. = j + 3k, a2 = i + j − k, a3 = i + 2k. = 2i + j, a2 = 2 j + k, a3 = k. = i + 3k, a2 = i − j + k, a3 = 2i + j. = −i + k, a2 = 2 j + k, a3 = i + j + k.

6. a1 = i + k, a2 = −j + k, a3 = i + j + 2k. In Exercises 7 and 8, ﬁnd two different but equivalent sets of orthogonal vectors by arranging the same three nonorthogonal vectors in the orders indicated. 7. (a) a1 = 3j − k, a2 = i + j, a3 = i + 2k. (b) a1 = i + j, a2 = 3j − k, a3 = i + 2k. 8. (a) a1 = j − k, a2 = i + k, a3 = −i − j + k. (b) a1 = −i − j + k, a2 = i + k, a3 = j − k.

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C H A P T E R

3

Matrices and Systems of Linear Equations

M

any types of problems that arise in engineering and physics give rise to linear algebraic simultaneous equations. A typical engineering example involves the determination of the forces acting in the struts of a pin-jointed structure like a truss that forms the side of a bridge supporting a load. The determination of the forces in a strut is important in order to know when it is in compression or tension, and to ensure that no truss exceeds its safe load. The analysis of the forces in structures of this type gives rise to a set of linear simultaneous equations that relate the forces in the struts and the external load. It is necessary to know when systems of linear equations are consistent so a solution exists, when they are inconsistent so there is no solution, and whether when a solution exists it is unique or nonunique in the sense that it involves a number of arbitrary parameters. In practical problems all of these mathematical possibilities have physical meaning, and in the case of a truss, the inability to determine the forces acting in a particular strut indicates that it is redundant and so can be removed without compromising the integrity of the structure. A more complicated though very similar situation occurs when linearly vibrating systems are coupled together, as may happen when an active vibration damper is attached to a spring-mounted motor. However, in this case it is a system of simultaneous linear ordinary differential equations determining the amplitudes of the vibrations of the motor and vibration damper that are coupled together. The analysis of this problem, which will be considered later, also gives rise to a linear system of simultaneous algebraic equations. Linear ordinary differential equations are also coupled together when working with linear control systems involving feedback. When such systems are solved by means of the Laplace transform to be described later, linear algebraic systems again arise and the nature of the zeros of the determinant of a certain quantity then determines the stability of the control system. Linear systems of simultaneous algebraic equations also play an essential role in computer graphics, where at the simplest level they are used to transform images by translating, rotating, and stretching them by differing amounts in different directions. Although each equation in a system of linear algebraic equations can be considered separately, such can be discovered about the properties of the physical problem that gave rise to the equations if the system of equations can be studied as a whole. This can be accomplished by using the algebra of matrices that provides a way of analyzing systems

105

106

Chapter 3

Matrices and Systems of Linear Equations as a single entity, and it is the purpose of this chapter to introduce and develop this aspect of what is called linear algebra. After deﬁning the notion of a matrix, this chapter develops the fundamental matrix operations of equality, addition, scaling, transposition, and multiplication. Various applications of matrices are given, and the brief review of determinants given in Chapter 1 is developed in greater detail, prior to its use when considering the solution of systems of linear algebraic equations. The concept of elementary row operations is introduced and used to reduce systems of linear algebraic equations to a form that shows whether or not a unique solution exists. When a solution does exist, which is either unique or determined in terms of some of the remaining variables, this reduction enables the solution to be found immediately. The inverse of an n × n matrix is deﬁned and shown only to exist when the determinant of the matrix is nonvanishing, and, ﬁnally, the derivative of a matrix whose elements are functions of a variable is introduced and some of its most important properties are derived.

3.1

Matrices

M

atrices arise naturally in many different ways, one of the most common being in the study of systems of linear equations such as a11 x1 + a12 x2 + · · · + a1n xn = b1 a21 x1 + a22 x2 + · · · + a2n xn = b2 · · · · · · · · · am1 x1 + am2 x2 + · · · + amn xn = bm.

(1)

In system (1) the numbers ai j are the coefﬁcients of the equations, the numbers bi are the nonhom*ogeneous terms, and the number of equations m may equal, exceed, or be less than n, the number of unknowns x1 , x2 , . . . , xn . System (1) is said to be hom*ogeneous when b1 = b2 = · · · = bm = 0, and to be nonhom*ogeneous when at least one of the bi is nonvanishing. The algebraic properties of the system are determined by the array of coefﬁcients ai j , the nonhom*ogeneous terms bi and the numbers m and n. From now on, the array of coefﬁcients and the nonhom*ogeneous terms on the right will be denoted by the single symbols A and b, respectively, where ⎡

a11 ⎢ a21 A=⎢ ⎣ am1

a12 a22 . . . am2

. . . .

. . . .

⎤ . a1n . a2n ⎥ ⎥ ⎦ . . amn

⎡

and

⎤ b1 ⎢ b2 ⎥ ⎥ b=⎢ ⎣ · ⎦. bm

(2)

The array of mn coefﬁcients ai j in m rows and n columns that form A is an example of an m × n matrix, where m × n is read “m by n.” The array b is an example of an m × 1 matrix, and it is called an m element column vector. We will use the convention that an array such as A, with two or more rows and two or more columns, will be denoted by a boldface capital letter. An array with a single row, or a column such as b, will be denoted by a boldface lowercase letter. Each entry in a matrix is called an element of the matrix, and entries may be numbers, functions, or even matrices themselves. The sufﬁxes associated with an element show its position in the matrix, because the ﬁrst sufﬁx is the row number

Section 3.1

Matrices

107

and the second is the column number. Because of this convention, the element a35 in a matrix belongs to the third row and the ﬁfth column of the matrix. So, for example, if A is a 3 × 2 matrix and its general element ai j = i + 3 j, then as i may only take the values 1, 2, and 3, and j the values 1 and 2, it follows that ⎡ ⎤ 4 7 A = ⎣5 8⎦ . 6 9 In a column vector c with elements c11 , c21 , c31 , . . . , cm1 , as only a single column is involved, it is usual to vary the sufﬁx convention by omitting the second sufﬁx and instead numbering the elements sequentially as c1 , c2 , c3 , . . . , cm, so that ⎡ ⎤ c1 ⎢ c2 ⎥ ⎢ ⎥ ⎢ .. ⎥ ⎥ c=⎢ ⎢ . ⎥. ⎢.⎥ ⎣ .. ⎦ cm Later it will be necessary to introduce row vectors, and in an s element row vector r with elements r11 , r12 , r13 , . . . , r1s , the notation is again simpliﬁed, this time by omitting the ﬁrst sufﬁx and numbering the elements sequentially as r1 , r2 , . . . , rs , so r = [r1 , r2 , . . . , rs ].

(3)

In general, row and column vectors will be denoted by boldface lowercase letters such as a, b, c, and x, and matrices such as the coefﬁcient matrix in (2) will be denoted by boldface capital letters such as A, B, P, and Q. A different convention that is also used to denote a matrix involves enclosing the array between curved brackets instead of the square ones used here. Thus, 1 5 9 1 5 9 and (4) −3 2 4 −3 2 4 denote the same 2 × 3 matrix. A matrix should never be enclosed between two vertical rules in order to avoid confusion with the determinant notation because 3 −4 3 −4 = 26 is a determinant. is a matrix, but 5 2 5 2 Deﬁnition of a matrix An m × n matrix is an array of mn entries, called elements, arranged in m rows and n columns. If a matrix is denoted by A, then the element in its ith row and jth column is denoted by ai j and ⎡

a11 ⎢ a21 A = [ai j ] = ⎢ ⎣ . am1

a12 a22 . . . am2

. . . . . . . . . . . . .

⎤ a1n a2n ⎥ ⎥. . ⎦ amn

108

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Matrices and Systems of Linear Equations

some typical matrices

The following are typical examples of matrices: A 1 × 1 matrix:

[3]; a single element may be regarded as a matrix. ⎡ ⎤ 1 3 5 0 A 3 × 4 matrix: ⎣2 −1 4 3⎦ ; a matrix with real numbers as elements. 7 2 1 6 1+i 1−i A 2 × 2 matrix: ; a matrix with complex numbers as 3 + 4i 2 − 3i elements. cosθ sin θ A 2 × 2 matrix: ; a matrix with functions as elements. −sinθ cos θ A 1 × 3 matrix: [2, −5, 7]; a three-element row vector. 11 A 2 × 1 matrix: ; a two-element column vector. 9

A square matrix is a matrix in which the number of rows m equals the number of columns n. A typical square matrix is the 3 × 3 matrix ⎡ ⎤ 2 0 5 ⎣1 −3 4⎦ . 3 1 7 Deﬁnition of the equality of matrices Let A = [ai j ] be an m × n matrix and B = [bi j ] be a p × q matrix. Then matrices A and B will be equal, written A = B, if, and only if: (a) A and B have the same number of rows, and the same number of columns, so that m = p and n = q, and (b) ai j = bi j , for each i and j. Equality of matrices means that if A and B are equal, then each is an identical copy of the other. EXAMPLE 3.1

2 3 a If A = , b 6 1

2 B= −3

3 6

9 , 1

⎡

and

2 C = ⎣−3 0

3 6 0

⎤ 9 1⎦ , 0

then A = B if and only if a = 9 and b = −3, but A = C and B = C. Deﬁnition of matrix addition The addition of matrices A and B is only deﬁned if the matrices each have the same number of rows and the same number of columns. Let A = [ai j ] and B = [bi j ] be m × n matrices. Then the the m × n matrix formed by adding A and B, called the sum of A and B and written A + B, is the matrix whose element in the ith row and jth column is ai j + bi j , for each i and j, so that A + B = [ai j + bi j ]. Matrices that can be added are said to be conformable for addition.

Section 3.1

Matrices

109

It is an immediate consequence of this deﬁnition that A + B = B + A, so matrix addition is commutative. Deﬁnition of the transpose of a matrix Let A = [ai j ] be an m × n matrix. Then the transpose of A, denoted by AT (and sometimes by A ), is the matrix obtained from A by interchanging rows and columns to produce the n × m matrix AT = [ai j ]T = [a ji ].

The deﬁnition of the transpose of a matrix means that the ﬁrst row of A becomes the ﬁrst column of AT , the second row of A becomes the second column of AT , . . . . , and, ﬁnally, the mth row of A becomes the mth column of AT . In particular, if A is a row vector, then its transpose is a column vector, and conversely. EXAMPLE 3.2

2 If A = 1

6 0

3 4

⎡

2 then AT = ⎣6 3

⎤ ⎡ ⎤ 1 7 0⎦ , and if A = [7, 3, 2] then AT = ⎣3⎦ . 4 2

Deﬁnition of scaling a matrix by a number Let A = [ai j ] be an m × n matrix and λ be a scalar (real or complex). Then if A is scaled by λ, written λA, every element of A is multiplied by λ to yield the m × n matrix λA = [λai j ].

EXAMPLE 3.3

If λ = 2

and

A=

2 1

−6 4

7 , 15

and if λ = −1, then

λA = (−1)A = −A =

difference (subtraction) of matrices

then λA = 2A =

−2 −1

4 2

−12 8

14 , 30

6 −7 . −4 −15

Taken together, the deﬁnitions of the addition and scaling of matrices show that if the matrices A and B are conformable for addition, then the subtraction of matrix B from A, called their difference and written A − B, is to be interpreted as A − B = A + (−1)B.

EXAMPLE 3.4

If A =

2 5 8 1 −4 5

and

B=

2 2

4 −4

5 , 1

then A − B =

0 −1

1 0

3 . 4

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Chapter 3

Matrices and Systems of Linear Equations

negative of a matrix

The null or zero matrix 0 is deﬁned as any matrix in which every element is zero. The introduction of the null matrix makes it appropriate to call −A the negative of A, because A − A = A + (−1)A = 0. When working with the null matrix the number of its rows and columns is never stated, because these are always taken to be whatever is appropriate for the equation that is involved. Deﬁnition of the product of a row and a column vector Let a = [a1 , a2 , . . . , ar ] be an r-element row vector, and b = [b1 , b2 , . . . , br ]T be an r -element column vector. Then the product ab, in this order, is the number deﬁned as ab = a1 b1 + a2 b2 · · · + ar br .

Notice that this product is only deﬁned when the number of elements in the row vector A equals the number of elements in the column vector B. EXAMPLE 3.5

Find the product ab given that a = [1, 4, −3, 10] and b = [2, 1, 4, −2]T . Solution ab = [1, 4, −3, 10]

⎡

⎤ 2 ⎢ 1⎥ ⎢ ⎥ ⎣ 4⎦ −2

= (1) · (2) + (4) · (1) + (−3) · (4) + (10) · (−2) = −26. Deﬁnition of the product of matrices Let A = [ai j ] be an m × n matrix in which the r th row is the row vector ar , and let B = [bi j ] be a p × q matrix in which the sth column is the column vector bs . The matrix product AB, in this order, is only deﬁned if the number of columns in A equals the number of rows in B, so that n = p. The product is then an m × q matrix with the element in its r th row and sth column deﬁned as ar bs . Thus, if cr s = ar bs , as cr s = ar 1 b1s + ar 2 b2s + · · · + ar n bns , AB = [cr s ] = [ar 1 b1s + ar 2 b2s + · · · + ar n bns ], for 1 ≤ r ≤ m and 1 ≤ s ≤ q, or, equivalently, ⎡

a1 b1 a1 b2 a1 b3 ⎢ a2 b1 a2 b2 a2 b3 AB = ⎢ . . . . . . . . . . ⎣ amb1 amb2 amb3

⎤ . . . a1 bq . . . a2 bq ⎥ ⎥. . . . . . ⎦ . . . ambq

Section 3.1

Matrices

111

When a matrix product AB is deﬁned, the matrices are said to be conformable for matrix multiplication in the given order. in general, matrix multiplication is noncommutative

EXAMPLE 3.6

It is important to notice that when the product AB is deﬁned, the product BA may or may not be deﬁned, and even when BA is deﬁned, in general AB = BA. This situation is recognized by saying that, in general, matrix multiplication is noncommutative. Provided matrices A and B are conformable for multiplication, the above rule for ﬁnding their product AB, in this order, is best remembered by saying that the element in the ith row and jth column of AB is the product of the ith row of A and the jth column of B. Form the matrix products AB and BA given that ⎡ 4 1 4 −3 A= and B = ⎣2 2 5 4 0

⎤ 1 6⎦ . 3

Solution Let us calculate the matrix product AB. The ﬁrst and second row vectors of A are a1 = [1, 4, −3] and a2 = [2, 5, 4], and the ﬁrst and second column vectors of B are b1 = [4, 2, 0]T and b2 = [1, 6, 3]T . As A is a 2 × 3 matrix and B is a 3 × 2 matrix, the product AB is conformable for multiplication and yields a 2 × 2 matrix (1 · 4 + 4 · 2 + (−3) · 0) (1 · 1 + 4 · 6 + (−3) · 3) ab a b AB = 1 1 1 2 = a2 b1 a2 b2 (2 · 4 + 5 · 2 + 4 · 0) (2 · 1 + 5 · 6 + 4 · 3) 12 16 = . 18 44 The product BA is also conformable for multiplication and yields a 3 × 3 matrix, where now we must use the row vectors of B that with an obvious change of notation are b1 = [4, 1], b2 = [2, 6], b3 = [0, 3], and the column vectors of A that are a1 = [1, 2]T , a2 = [4, 5]T , and a3 = [−3, 4]T , so that ⎡ ⎤ ⎡ ⎤ b1 a1 b1 a2 b1 a3 (4 · 1 + 1 · 2) (4 · 4 + 1 · 5) (4 · (−3) + 1 · 4) BA = ⎣b2 a1 b2 a2 b2 a3 ⎦ = ⎣(2 · 1 + 6 · 2) (2 · 4 + 6 · 5) (2 · (−3) + 6 · 4)⎦ (0 · 1 + 3 · 2) (0 · 4 + 3 · 5) (0 · (−3) + 3 · 4) b3 a1 b3 a2 b3 a3 ⎡ ⎤ 6 21 −8 = ⎣14 38 18⎦ . 6 15 12 This is an example of two matrices A and B that can be combined to form the products AB and BA, but AB = BA. EXAMPLE 3.7

Write the system of simultaneous equations (1) in matrix form. Solution Using the matrices A and b in (2) and setting x = [x1 , x2 , . . . , xn ]T allows the system of equations (1) to be written Ax = b. Here, as is usual, to save space the transpose operation has been used to display the elements of column vector x in the more convenient form x = [x1 , x2 , . . . , xn ]T .

112

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The deﬁnitions of matrix multiplication and addition lead almost immediately to the results of the following theorem, so the proof is left as an exercise. THEOREM 3.1 some important properties of matrices

THEOREM 3.2

Associative and distributive properties of matrices Let A, B, and C be matrices that are conformable for the operations that follow, and let λ be a scalar. Then: (i)

If AB and BA are both deﬁned, in general AB = BA;

(ii)

A(BC) = (AB)C = ABC;

(iii)

(λA)B = A(λB) = λAB;

(iv)

A(B + C) = AB + AC;

(v)

(A + B)C = AC + BC.

Transposition of a product If matrices A and B are conformable to form the product AB, then (AB)T = BT AT . Proof The products (AB)T and BT AT are both deﬁned, and each is an m × q matrix. Introduce the notation [M]i j to denote the element of M in row i and column j. Then from the transpose operation and the rule for matrix multiplication, for all permissible i, j, [AB]i,T j = [AB] j,i = (product of jth row of A with ith column of B) =

n

a jkbki .

k=1

Similarly, [BT AT ]i, j = (product of ith row of BT with jth column of AT ) = (product of ith column of B with jth row of A) =

n

a jkbki .

k=1

So [AB]iTj = [BT AT ]i j for all permissible i, j, showing that (AB)T = BT AT .

raising a matrix to a power

It is an immediate consequence of Theorem 3.1(ii) that if A is a square matrix and m and n are positive integers, n A $ ·A·A %&· . . . · A' = A

and

Am · An = Am+n .

n times

A useful result from the deﬁnition of addition is (A + B)T = AT + BT , while from Theorem 3.2 (ABC)T = CT BT AT .

Section 3.1

pre- and postmultiplication of matrices

Matrices

113

As the order in which a sequence of permissible matrix multiplications is performed inﬂuences the product, it is necessary to introduce a form of words that makes the order unambiguous. This is accomplished by saying that if matrix A multiplies matrix B from the left, as in AB, then B is premultiplied by A, while if A multiplies B from the right, as in BA, then B is postmultiplied by A. Equivalently, in the product AB, we can say that A is postmultiplied by B, or that B is premultiplied by A.

Important Differences Between Ordinary Algebraic Equations and Matrix Equations (i) The algebraic equation ab = 0, in which a and b are numbers, not both of which are zero, implies that either a = 0 or b = 0. However, if the matrix product AB is deﬁned and is such that AB = 0, then it does not necessarily follow that either A = 0 or B = 0. (ii) The algebraic equation ab = ac in which a, b, and c are numbers, with a = 0, allows cancellation of the factor a leading to the conclusion that b = c. However, if the matrix products AB and AC are deﬁned and are such that AB = AC, this does not necessarily imply that B = C, so that cancellation of matrix factors is not permissible. The validity of these two statements can be seen by considering the following simple examples. EXAMPLE 3.8

Consider matrices A and B given by 1 4 A= and 3 12

B=

4 −8 . −1 2

Then AB = 0, but neither A nor B is a null matrix. EXAMPLE 3.9

Consider the matrices A, B, and C given by 1 −1 2 4 6 A= , B= , 2 −2 2 3 4 Then

0 AB = AC = 0

and

1 2

3 C= 3

6 5

8 . 6

2 , 4

but B = C.

leading diagonal and trace of a matrix

In a square n × n matrix A = [ai j ], the elements on a line extending from top left to bottom right is called the leading diagonal of A, and it contains the n elements a11 , a22 , . . . , ann . So the leading diagonal of the 2 × 2 matrix A in Example 3.8 contains the elements 1 and 12, and the leading diagonal of the 2 × 2 matrix B contains the elements 4 and 2. Symbolically, the leading diagonal of the n × n matrix A = [ai j ] shown below comprises the n elements in the shaded diagonal strip, though these

114

Chapter 3

Matrices and Systems of Linear Equations

n elements do not form an n element vector.

⎡

a11 ⎢a21 ⎢ ⎢a31 A=⎢ ⎢ · ⎢ ⎣ · an1

a12 a22 a32 · · an2

a13 a23 a33 · · an3

· · · · · ·

· · · · · ·

· · · · · ·

⎤ a1n a2n ⎥ ⎥ a3n ⎥ ⎥. · ⎥ ⎥ · ⎦ ann

The trace of a square matrix A, written tr(A), is the sum of the terms on its leading diagonal, so for the foregoing matrix tr(A) = a11 + a22 + · · · + ann . Square matrices in which all elements away from the leading diagonal are zero, but not every element on the leading diagonal is zero, are called diagonal matrices. Of the class of diagonal matrices, the most important are the unit matrices, also called identity matrices, in which every element on the leading diagonal is the number 1. These n × n matrices are usually all denoted by the symbol I, with the value of n being understood to be appropriate to the context in which they arise. If, however, the value of n needs to be indicated, the symbol I can be replaced by In . It is easily seen from the deﬁnition of matrix multiplication that for any m × n matrix A it follows that ImA = AIn or, more simply, IA = AI = A, and that when A is an n × n matrix, IA = AI = A.

identity or unit matrix

When working with matrices, the unit matrix I plays the part of the unit real number, and it is because of this that I is called either the unit or the identity matrix. An example of a 4 × 4 diagonal matrix is ⎡ ⎤ 3 0 0 0 ⎢0 2 0 0⎥ ⎥ D=⎢ ⎣0 0 0 0⎦ , with the trace given by tr(D) = 3 + 2 + 0 + 1 = 6. 0 0 0 1 The 3 × 3 unit matrix is the diagonal matrix ⎡ ⎤ 1 0 0 I = I3 = ⎣0 1 0⎦ , and its trace is tr(I) = 1 + 1 + 1 = 3. 0 0 1 Various special square n × n matrices occur sufﬁciently frequently for them to be given names, and some of the most important of these are the following:

some special matrices

Upper triangular matrices are matrices in which all elements below the leading diagonal are zero. A typical example of a 4 × 4 upper triangular matrix is ⎡ ⎤ 1 3 −1 0 ⎢0 2 −6 1⎥ ⎥ U=⎢ ⎣0 0 −3 2⎦ . 0 0 0 4

Section 3.1

Matrices

115

Lower triangular matrices are matrices in which all elements above the leading diagonal are zero. A typical example of a 4 × 4 lower triangular matrix is ⎡ ⎤ 2 0 0 0 ⎢ 1 0 0 0⎥ ⎥ L=⎢ ⎣ 3 −2 5 0⎦ . −2 4 7 3 Symmetric matrices A = [ai j ] are matrices in which ai j = a ji for all i and j. If A is symmetric, then A = AT . A typical example of a symmetric matrix is ⎡ ⎤ 1 5 −3 2⎦ . M=⎣ 5 4 −3 2 7 Skew-symmetric matrices A = [a ji ] are matrices in which ai j = −a ji for all i and j. From the deﬁnition of an n × n skew-symmetric matrix we have aii = −aii for i = 1, 2, . . . , n, so the elements on the leading diagonal must all be zero. An equivalent deﬁnition of a skew-symmetric matrix A is that AT = −A. A typical example of a skew-symmetric matrix is ⎡ ⎤ 0 3 −5 6 ⎢−3 0 2 −4⎥ ⎥. S=⎢ ⎣ 5 −2 0 −1⎦ −6 4 1 0 An orthogonal matrix Q is a matrix such that QQT = QT Q = I. A typical orthogonal matrix is ⎡ 1 1 ⎤ −√ √ ⎢ 2 2⎥ ⎥. Q=⎢ ⎣ 1 1 ⎦ √ √ 2 2 More special than the preceding real valued matrices are matrices A = [ai j ] in which the elements ai j are complex numbers. We will write A to denote the matrix obtained from A by replacing each of its elements ai j by its complex conjugate a i j , so that A = [a i j ]. Then matrix A is said to be Hermitian if T

A = A. A typical Hermitian matrix is

A=

7 1 + 4i

1 − 4i . 3

The matrix A is said to be skew-Hermitian if T

A = −A. A typical skew-Hermitian matrix is 3i A= −5 + 2i

5 + 2i . 0

116

Chapter 3

Matrices and Systems of Linear Equations

block matrices

More will be said later about some of these special square matrices and the ways in which they arise. Finally, we mention that every m × n matrix A can be represented differently as a block matrix, in which each element is itself a matrix. This is accomplished by partitioning the matrix A into submatrices by considering horizontal and vertical lines to be drawn through A between some of its rows and columns, and then identifying each group of elements so deﬁned as a submatrix of A. Clearly there is more than one way in which a matrix can be partitioned. As an example of matrix partitioning, let us consider the 3 × 3 matrix ⎤ ⎡ 3 −1 2 2 0⎦ . A = ⎣1 2 1 0 One way in which this matrix can be partitioned is as follows: ⎡ ⎤ 3 −1 2 ⎢ ⎥ A = ⎣1 2 0 ⎦. 2

1

This can now be written in block matrix form as A11 A12 , A= A21 A22 where the submatrices are A11 = [3 −1],

A12 = [2],

A21 =

1 2

2 , 1

and

A22 =

0 . 0

The addition and scaling of block matrices follow the same rules as those for ordinary matrices, but care must be exercised when multiplying block matrices. To see how multiplication of block matrices can be performed, let us consider the product of matrix A above and the 3 × 4 matrix ⎡ ⎤ 1 2 2 1 ⎥ ⎢ B = ⎣3 1 1 0 ⎦ , 2 3 0 2 which are conformable for the product AB that is itself a 3 × 4 matrix. If B is partitioned as indicated by the dashed lines, it can be written as B11 B12 B= , B21 B22 where the submatrices are 1 2 B11 = , B12 = 3 1

2 1

1 , 0

B21 = [2],

and

B22 = [3, 0, 2].

Consideration of the deﬁnition of the product of matrices shows that we may now write the matrix product AB in the condensed form A11 B11 + A12 B21 A11 B12 + A12 B22 , AB = A21 B11 + A22 B21 A21 B12 + A22 B22

Section 3.1

Matrices

117

where the partitioned matrices have been multiplied as though their elements were ordinary elements. This result follows because of correct partitioning, as each product of submatrices is conformable for multiplication and all of the matrix sums are conformable for addition. In this illustration, routine calculations show that A11 B11 + A12 B21 = [4], A11 B12 + A12 B22 = [11, 5, 7], 7 4 A21 B11 + A22 B21 = , and A21 B12 + A22 B22 = 5 5

4 5

1 , 2

so ⎡

[4] AB = ⎣ 7 5

⎤ ⎡ [11, 5, 7] 4 ⎦ ⎣ 7 = 4 4 1 5 5 5 2

11 4 5

⎤ 5 7 4 1⎦. 5 2

This result is easily conﬁrmed by direct matrix multiplication. The calculation of a matrix product AB using partitioned matrices applies in general, provided the partitioning of A and B is performed in such a way that the products of all the submatrices involved are deﬁned. Matrix partitioning has various uses, one of which arises in machine computation when a very large ﬁxed matrix A needs to be multiplied by a sequence of very large matrices P, Q, R, . . . . If it happens that A can be partitioned in such a way that some of its submatrices are null matrices, the computational time involved can be drastically reduced, because the product of a submatrix and a null matrix is a null matrix, and so need not be computed. The economy follows from the fact that in machine computation multiplications occupy most of the time, so any reduction in their number produces a signiﬁcant reduction in the time taken to evaluate a matrix product, and the result is even more signiﬁcant when the same partitioned matrix with null blocks is involved in a sequence of calculations. Block matrices are also of signiﬁcance when describing complex oscillation problems governed by a large system of simultaneous ordinary differential equations. Their importance arises from the fact that the matrix of coefﬁcients of the equations often contains many null submatrices, and when this happens the structure of the nonnull blocks provides useful information about the fundamental modes of oscillation that are possible, and also about their interconnections. For other accounts of elementary matrices see the appropriate chapters in references [2.1], [2.5], and [2.7] to [2.12].

Summary

This section deﬁned m × n matrices, and the special cases of column and row vectors, and it introduced the fundamental algebraic operations of equality, addition, scaling, transposition, and multiplication of matrices. It was shown that, in general, matrix multiplication is not commutative, so that even when both of the products AB and BA are deﬁned, it is usually the case that AB = BA. Pre- and postmultiplication of matrices was deﬁned, and some important special types of matrices were introduced, such as the unit matrix I. It was also shown how a matrix A can be subdivided into blocks, and that a matrix operation performed on A can be interpreted in terms of matrix operations performed on block matrices obtained by subdivision of A.

118

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EXERCISES 3.1 In Exercises 1 through 4 ﬁnd the values of the constants a, b, and c in order that A = B. 2 a 1 c −a 1 4 . , B= 1. A = 2 b −1 2 3 a ⎤ ⎤ ⎡ ⎡ 1 4 3 1 4 3 2. A = ⎣a 2 4⎦ , B = ⎣2 2 4⎦. 9 1 c b 1 0 ⎤ ⎤ ⎡ 2 ⎡ 2 a a 1 a a 1 1 2⎦ , B = ⎣ 3 1 2⎦. 3. A = ⎣ b 1+a 2+c 6 2 4 6 ⎡ ⎤ ⎡ ⎤ 1 3+a 2 1 −1 c a 5 ⎦, B = ⎣4 a 5 ⎦. 4. A = ⎣1 + b b2 b2 1 a2 1 a2 In Exercises 5 through 8 ﬁnd A + B and A − B. ⎤ ⎤ ⎡ ⎡ 2 0 1 −2 1 4 3 6 1⎦. 1 0 2⎦ , B = ⎣1 1 −3 5. A = ⎣2 0 1 1 0 1 −1 0 1 ⎤ ⎤ ⎡ ⎡ 1 7 6 2 −1 6 6. A = ⎣ 0 2 4⎦ , B = ⎣1 −2 3⎦. −1 0 1 2 1 2 ⎤ ⎤ ⎡ ⎡ 0 2 3 1 2 4 ⎢3 −1 1⎥ ⎢3 1 0⎥ ⎥. ⎥ ⎢ 7. A = ⎢ ⎣1 1 0⎦ , B = ⎣0 1 1⎦ 1 3 2 2 2 4 ⎤ ⎤ ⎡ ⎡ 1 0 0 0 1 4 3 6 ⎢ 3 1 0 0⎥ ⎢0 2 1 4⎥ ⎥ ⎥ ⎢ 8. A = ⎢ ⎣0 0 3 1⎦ , B = ⎣1 2 4 0⎦. 1 1 1 3 0 0 0 2 In Exercises 9 through 12 form the sum λA + μB. ⎤ ⎡ 1 4 2 9. λ = 1, μ = 3, A = ⎣2 1 4⎦, 3 2 2 ⎤ ⎡ 2 3 −1 4⎦. B = ⎣1 2 1 0 3 1 4 1 10. λ = −1, μ = 2, A = , 2 4 0 2 1 1 . B= 0 2 4

⎡

11. λ = 4,

12. λ = 3,

μ = −2,

μ = −3,

4 A = ⎣2 1 ⎡ 6 B = ⎣2 1 ⎡ 3 A = ⎣2 3 ⎡ 3 B = ⎣4 2

3 1 2 1 4 1 1 2 6 2 2 1

⎤ 1 1⎦, 1 ⎤ 0 2⎦. 2 ⎤ 4 1⎦, 2 ⎤ 1 3⎦. 1

In Exercises 13 through 16 ﬁnd the product AB. 13. A = [1, 14. A = [2, 15. A = [1, B = [2, 16. A = [1, B = [−1,

4, 3, 4, 2, 3, 2,

−2, 3], B = [2, 1, −1, 2]T . 1, 4], B = [3, 1, 1, 3]T . 3, 7, 5], −1, −1, 3]T . −1, 2, 0], 13, 4, 1]T .

In Exercises 17 through 22 ﬁnd the product AB and, when it exists, the product BA. 2 1 3 1 4 . , B= 17. A = 1 4 1 2 0 18. A = [1, 4, 6, −7], B = [2, 3, −2, 3]T . ⎤ ⎤ ⎡ ⎡ 1 0 0 3 1 4 19. A = ⎣ 0 1 0 ⎦ , B = ⎣ 2 1 −5 ⎦ . 0 0 1 7 2 0 ⎤ ⎤ ⎡ ⎡ 2 0 0 9 −1 4 6 −2 ⎦ . 20. A = ⎣ 0 −3 0 ⎦ , B = ⎣ 1 0 0 5 2 2 3 ⎤ ⎡ ⎤ ⎡ 2 3 1 5 2 3 ⎢4 1 2⎥ ⎥, B = ⎣2 0 4⎦. ⎢ 21. A = ⎣ 2 2 6⎦ 1 4 7 1 5 2 ⎤ ⎤ ⎡ ⎡ 3 1 1 2 1 0 ⎢ 4 ⎢2 1 1 4⎥ 2⎥ ⎥ ⎥ ⎢ 22. A = ⎢ ⎣ 1 0 2 1 ⎦ , B = ⎣ 6 −2 ⎦ . −1 4 1 1 2 1 23. Given ⎤ ⎤ ⎡ ⎡ 2 5 −3 4 2 1 4⎦ and B = ⎣2 5 6⎦ , A=⎣ 5 1 −3 4 6 1 6 3 show that (AB)T = BA.

Section 3.1 In Exercises 24 through 28 write the given systems of equations in the matrix form Ax = b, where A is the coefﬁcient matrix, x is the vector of unknowns, and b is the nonhom*ogeneous vector term. 26. 5x + 3y − 6z = 14 6x − 5y + 11z = 20 x − 4y + 3z = 2 9x − 3y + 2z = 35.

24. 3x + 5y − 6z = 7 x − 7y + 4z = −3 2x + 4y − 5z = 4. 25. 4u + 5v − w + 7z = 25 3u + 2v + 3z = 6 v + 6w − 7z = 0.

27. 3x + 4y − 2z = λx 2x − 7y + 6z = λy 8x + 3y + 5z = λz.

28. 2x + 3y + 6z = λ(3x + 2y + 3z) 3x − 4y + 2z = λ(x − 5y + 2z) 4x + 9y + 2z = λ(x − 2y + 4z). 29. If

⎡

1 A = ⎣1 0

⎤ 6 0⎦ , 3

⎤ ⎡ 2 0 1 2 3⎦ , B = ⎣4 0 −1 1 ⎡ ⎤ x1 x2 x3 X = ⎣ y1 y2 y3 ⎦ , z1 z2 z3

3 2 1

and

solve for X given that 3X + A = A B − X + 3B. ⎡

2 A = ⎣1 3

1 2 0

⎤ ⎤ ⎡ 1 4 1 4 1⎦ , B = ⎣2 1 2⎦ , 1 1 2 2 ⎡ ⎤ x1 x2 x3 X = ⎣ y1 y2 y3 ⎦ , z1 z2 z3

solve for X given that 2ABT + X − 2I = 3X + 4B − 2A. 31. Given that

⎡ 3 A = ⎣2 2

2 2 0

⎤ 2 0⎦ , 4

show that A3 − 9A2 + 18A = 0. 32. Given that

⎡

0 A = ⎣0 2

1 0 1

⎤ 0 1⎦ , −2

show that A3 + 2A2 − A − 2I = 0.

and

119

33. Prove the second result in Theorem 3.1 that A(BC) = (AB)C = ABC. 34. Prove the third result in Theorem 3.1 that (λA)B = A(λB) = λAB. 35. Prove the fourth result in Theorem 3.1 that A(B + C) = AB + AC. In Exercises 36 through 39 verify that (AB)T = BT AT . ⎤ ⎤ ⎡ ⎡ 2 1 3 3 1 4 36. A = ⎣2 1 2⎦ , B = ⎣1 2 5⎦ . 0 2 1 4 2 3 ⎤ ⎡ ⎤ ⎡ 1 4 3 2 1 4 3 ⎢ 2 1 5⎥ ⎥ 2 1⎦ , B = ⎢ 37. A = ⎣1 6 ⎣−1 3 2⎦ . 1 1 −2 4 1 7 3 ⎤ ⎤ ⎡ ⎡ 3 1 −5 1 4 2 4⎦ . 38. A = ⎣7 3 −1⎦ , B = ⎣1 3 2 0 8 0 2 5 ⎤ ⎡ ⎤ ⎡ 1 2 1 1 4 6 2 ⎢−2 1 4⎥ ⎥ 39. A = ⎣2 1 4 1⎦ , B = ⎢ ⎣ 2 2 5⎦ . 3 0 0 2 1 1 1 40. Verify that (ABC)T = CT BT AT given that −2 3 −2 1 5 , and C = , B= A= 5 4 5 3 1

T

30. If

Matrices

41. Prove that if D is the n × n diagonal matrix ⎡ ⎤ k1 0 0 · · · 0 ⎢ 0 k2 0 · · · 0 ⎥ ⎢ ⎥ ⎥ D=⎢ ⎢ 0 0 k3 · · · 0 ⎥ , then ⎣. . . . . . . . . . . . ⎦ 0 0 0 · · · kn ⎡ m ⎤ k1 0 0 · · · 0 ⎢ 0 k2m 0 · · · 0 ⎥ ⎢ ⎥ 0 k3m · · · 0 ⎥ Dm = ⎢ ⎢0 ⎥, ⎣. . . . . . . . . . . . . ⎦ 0 0 0 · · · knm where m is a positive integer. 42. Find A2 , A3 , and A4 , given that ⎤ ⎡ 1 2 7 6⎦ . A = ⎣2 5 1 0 −1 43. Find A2 , A4 , and A6 , given that √ 1/2 −( 3)/2 . A= √ 1/2 ( 3)/2

3 . 7

120

Chapter 3

Matrices and Systems of Linear Equations

44. Use the matrix A in Exercise 42 to ﬁnd A3 , A5 , and A7 . 45. A square matrix A such that A2 = A is said to be idempotent. Find the three idempotent matrices of the form 1 p . A= q r 46. A square matrix A such that for some positive integer n has the property that An−1 = 0, but An = 0 is said to be nilpotent of index n (n ≥ 2). Show that the matrix ⎤ ⎡ 0 0 0 0 0⎦ A = ⎣4 1 −1 0 is nilpotent and ﬁnd its index. 47. A quadratic form in the variables x1 , x2 , x3 , . . . , xn is an expression of the form ax12 + bx1 x2 + cx22 + dx1 x3 + · · · + f xn−1 xn + gxn2 in which some of the coefﬁcients a, b, c, d, . . . , f, g may be zero. Explain why xT Ax is a quadratic form and ﬁnd the quadratic form for which ⎤ ⎡ ⎡ ⎤ x1 3 4 0 3 ⎢x2 ⎥ ⎢4 2 2 6⎥ ⎥ ⎢ ⎥ A=⎢ ⎣0 2 5 1⎦ and x = ⎣x3 ⎦ . 3 6 1 7 x4 48. Find the quadratic form xT Ax when ⎤ ⎡ ⎤ ⎡ x1 4 1 3 6 ⎢x2 ⎥ ⎢2 3 5 4⎥ ⎥ ⎢ ⎥ A=⎢ ⎣1 4 1 2⎦ and x = ⎣x3 ⎦ . 2 0 4 1 x4 49. Explain why the matrix A in the general expression for

3.2

a quadratic form xT Ax can always be written as a symmetric matrix. In Exercises 50 through 52 ﬁnd the symmetric matrix A for the given quadratic form when written xT Ax, with x = [x, y, z]T . 50. 51. 52. 53.

x 2 + 3xy − 4y2 + 4xz + 6yz − z2 . 2x 2 + 4xy + 6y2 + 7xz − 9z2 . 7x 2 + 7xy − 5y2 + 4xz + 2yz − 9z2 . A square matrix P is called a stochastic matrix if all its elements are nonnegative and the sum of the elements in each row is 1. Thus, the matrix ⎡ ⎤ p11 p12 · · · p1n ⎢ p21 p22 · · · p2n ⎥ ⎥ P=⎢ ⎣. . . . . . . . . . . ⎦ pn1 pn2 · · · pnn will be a stochastic matrix if pi j ≥ 0 for 0 ≤ i ≤ n, 0 ≤ j ≤ n, and n

pi j = 1

for i = 1, 2, . . . , n.

j=1

Let the n element column vector E = [1, 1, 1, . . . , 1]T . By considering the matrix product PE, and using mathematical induction, prove that Pm is a stochastic matrix for all positive integral values of m. 54. Construct a 3 × 3 stochastic matrix P. Find P2 and P3 , and by showing that all elements of these matrices are nonnegative and that all their row-sums are 1, verify the result of Exercise 53 that each of these matrices is a stochastic matrix.

Some Problems That Give Rise to Matrices (a) Electric Circuits with Resistors and Applied Voltages A simple electric circuit involving ﬁve resistors and three applied voltages is shown in Fig. 3.1. The directions of the currents i 1 , i 2 , and i 3 ﬂowing in each branch of the circuit are shown by arrows. The currents themselves can be determined by an application of Ohm’s law and the Kirchhoff laws that can be stated as follows: (a) Voltage = current × resistance (Ohm’s law); (b) The algebraic sum of the potential drops around each closed circuit is zero (Kirchhoff’s second law); (c) The current entering each junction must equal the algebraic sum of the currents leaving it (Kirchhoff’s ﬁrst law).

equations and matrices for electric circuits

An application of these laws to the circuit in Fig. 3.1, where the potentials are in volts, the resistances are in ohms, and the currents are in amps, leads to the following

Section 3.2

Some Problems That Give Rise to Matrices

8

121

12

i1 10

8

i2

4

i3

6

4

6

FIGURE 3.1 An electric circuit with resistors and applied voltages.

set of simultaneous equations: 8 = 12i 1 + 10(i 1 − i 2 ) + 8(i 1 − i 3 ) 4 = 10(i 2 − i 1 ) + 6(i 2 − i 3 ) 6 = 8(i 3 − i 1 ) + 6(i 3 − i 2 ) + 4i 3 . After collecting terms this system can be written as the matrix equation Ax = b, with ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 8 30 −10 −8 i1 16 −6⎦ , x = ⎣i 2 ⎦ , b = ⎣4⎦ . A = ⎣−10 6 −8 −6 18 i3 The directions assumed for the currents ir for r = 1, 2, 3 are shown by the arrows in Fig. 3.1, but if after the system of equations is solved, the value of the current is found to be negative, the direction of its arrow must be reversed. The circuit in Fig. 3.1 is simple, so in this example the currents can be found by routine elimination between the three equations. When many coupled circuits are involved a matrix approach is useful, and it then becomes necessary to develop a method for solving for x the matrix equation Ax = b, the elements of which are the required currents. If the number of equations is small, x can be found by making use of the matrix A−1 , inverse to A, that will be introduced later, though the most computationally efﬁcient approach is to use one of the numerical methods for solving systems of linear simultaneous equations described in Chapter 19.

(b) Combinatorial Problems: Graph Theory Matrices play an important role in combinatorial problems of many different types and, in particular, in graph theory. The purpose of the brief account offered here will be to illustrate a particular application of matrices, and no attempt will be made to discuss their subsequent use in the solution of the associated problems. Combinatorial problems involve dealing with the possible arrangements of situations of various different kinds, and computing the number and properties of such arrangements. The arrangements may be of very diverse types, involving at one extreme the ordering of matches that are to take place in a tennis tournament,

122

Chapter 3

Matrices and Systems of Linear Equations

4

5

3

1 2 FIGURE 3.2 The graph representing routes.

graphs, vertices, edges, and adjacency matrix

and at the other extreme ﬁnding an optimum route for a delivery truck or for the most efﬁcient routing of work through a machine shop. The ideas involved are most easily illustrated by means of examples, the ﬁrst of which involves the delivery from a storage depot of a consumable product to a group of supermarkets in a large city where it is important that daily deliveries be made as rapidly as possible. One possibility involves a delivery truck making a delivery to each supermarket in turn and returning to the storage depot between each delivery before setting out on the next delivery. An alternative is to travel between supermarkets after each delivery without returning to the storage depot. The question that then arises is which approach to routing is the best, and how it is to be determined. A typical situation is illustrated in Fig. 3.2, in which supermarkets numbered 1 to 5 are involved, with circles representing supermarkets and lines and arcs representing the routes. The representation in Fig. 3.2 is called a graph, and it is to be regarded as a set of points represented by the circles called vertices of the graph, and edges of the graph represented by the lines and arcs. In Fig. 3.2 the vertices are the circles 1, 2, . . . , 5 and the seven edges are the lines and arcs connecting the vertices. A special type of matrix associated with such a graph is an adjacency matrix, that is, a matrix whose only entries are 0 or 1. The rules for the entries in an adjacency matrix A = [ai j ] are that 1, if vertices i and j are joined by an edge ai j = 0, otherwise. The adjacency matrix for the graph in Fig. 3.2 is seen to be the symmetric matrix ⎤ ⎡ 0 1 0 1 1 ⎢1 0 1 0 0⎥ ⎥ ⎢ ⎥ A=⎢ ⎢0 1 0 1 1⎥ . ⎣1 0 1 0 1⎦ 1 0 1 1 0 It is to be expected that an adjacency matrix is symmetric, because if i is adjacent to j, then j is adjacent to i. Although we shall not attempt to do so here, the interconnection properties of the problem represented by the graph in Fig. 3.2 can be analyzed in terms of its adjacency matrix A. The optimum routing problem can then be resolved once the traveling times along roads (lines or arcs) are known. Sometimes it happens that the edges in a graph represent connections that only operate in one direction, so then arrows must be added to the graph to indicate these

Section 3.2

FIGURE 3.3 A typical digraph.

digraph

Some Problems That Give Rise to Matrices

123

FIGURE 3.4 The Konigsberg ¨ bridge problem.

directions. A graph of this type is called a digraph (directed graph). The rules for the entries in the adjacency matrix A = [ai j ] of a digraph are that 1, if vertices i and j are joined by an edge with an arrow from i to j ai j = 0, otherwise. A typical digraph is shown in Fig. 3.3, and it has the associated adjacency matrix ⎡ ⎤ 0 1 0 1 ⎢0 0 1 0⎥ ⎥ A=⎢ ⎣1 0 0 0⎦ . 0 0 1 0

K¨ onigsberg bridge problem

The adjacency matrix A characterizes all the possible interconnections between the four vertices and, as with the previous example, an analysis of the properties of any situation capable of representation in terms of this digraph can be performed using the matrix A. Problems of this type can arise in transportation problems in cities with one-way streets, and in chemical processes where a ﬂuid is piped to different parts of a plant through an interconnecting network of pipes through which ﬂuid may only ﬂow in a given direction. Before closing this brief introduction to graph theory, mention should be made of a problem of historical signiﬁcance, since it represented the start of graph theory ¨ as it is known today. The problem is called the Konigsberg bridge problem, and it was solved by Euler (1707–1783). During the early 18th century the Prussian town of Konigsberg ¨ was established on two adjacent islands in the middle of the river Pregel. The islands were linked to the land on either side of the river, and to one another, by seven bridges, as shown in Fig. 3.4a. It was suggested to Euler that he should resolve the conjecture that it ought to be possible to walk through the town, starting and ending at the same place, while crossing each of the seven bridges only once.

124

Chapter 3

Matrices and Systems of Linear Equations

Euler replaced the picture in Fig. 3.4a by the graph in Fig. 3.4b, though it was not until much later that the term graph in the sense used here was introduced. In Fig. 3.4b the vertices S and Q represent the two islands and, using the same lettering, P and R represent the riverbanks. The number of edges incident on each vertex represents the number of bridges connected to the corresponding land mass. Euler introduced the concept of a connected graph, in which each pair of vertices is linked by a set of edges, and also what is now called an eulerian circuit, comprising a path through all vertices that starts and ends at the same vertex and uses every edge only once. He called the number of edges incident upon a vertex the degree of the vertex, and by using these ideas he was able to prove the impossibility of the conjecture. The arguments involved are not difﬁcult, but their details would be out of place here. Many more practical problems are capable of solution by graph theory, which itself belongs to the branch of mathematics called combinatorics. In elementary accounts, graph theory and related combinatorial issues are usually called discrete mathematics. More information about combinatorics and its connection with matrices can be found in References [2.2] and [2.13].

(c) Translations, Rotations, and Scaling of Graphs: Computer Graphics matrices and computer graphics

The simplest operations in computer graphics involve copying a picture to a different location, rotating a picture about a ﬁxed point, and scaling a picture, where the scaling can be different in the horizontal and vertical directions. These operations are called, respectively, a translation, a rotation, and a scaling of the picture. Operations of this nature can all be represented in terms of matrices, and they involve what are called linear transformations of the original picture.

Translation A translation of a two-dimensional picture involves copying it to a different location without either rotating it or changing its horizontal and vertical scales. Figure 3.5 shows the original cartesian axes O(x, y) and the shifted axes O (x , y ), where the respective axes remain parallel to their original directions and the origin O is located at the point (h, k) relative to the O(x, y) axes. The relationship between the two sets of coordinates is given by x = x + h

y

k

O

and

y = y + k.

y′

O′

x′

h

FIGURE 3.5 A translation.

x

Section 3.2

Some Problems That Give Rise to Matrices

125

y y P

x φ

Q θ x

O FIGURE 3.6 A rotation through an angle θ.

If x = [x, y]T , x = [x , y ]T , and b = [h, k]T , the coordinate transformation can be written in matrix form as x = x + b, where matrix b represents the translation.

Rotation A rotation of the coordinate axes through an angle θ is shown in Fig. 3.6, where P(x, y) is an arbitrary point. The coordinates of P in the (x, y) reference frame and the (x , y ) reference frame are related as x = OR = OP cos(φ + θ ) = OP cos φ cos θ − OP sin φ sin θ = OQ cos θ − PQ sin θ = x cos θ − y sin θ, and y = P R = OP sin(φ + θ) = OP sin φ cos θ + OP cos φ sin θ = PQ cos θ + OQ sin θ = y cos θ + x sin θ, so x = x cos θ − y sin θ

and

y = y cos θ + x sin θ.

Deﬁning the matrices x, x , and R as x x x= , x = , and y y

R=

cos θ sin θ

allows the coordinate transformation to be written as x = Rx .

Scaling If S is a matrix of the form S=

kx 0

0 , ky

−sin θ cos θ

126

Chapter 3

Matrices and Systems of Linear Equations

where kx and ky are positive constants, and x = Sx, it follows that x = kx x

and

y = ky y ,

showing that x is obtained by scaling x by kx , while y is obtained by scaling y by ky . This form of scaling is represented by premultiplication of x by S, and if, for example, 4 0 S= , 0 3 the effect of this transformation on a circle of radius a will be to map it into an ellipse with semimajor axis of length 4a parallel to the x-axis and a semiminor axis of length 3a parallel to the y-axis.

Composite transformations By combining the preceding matrix operations to form a composite transformation, it is possible to carry out several transformations simultaneously. As an example, the effect of a rotation R followed by a translation b when performed on a vector x are seen to be described by the matrix equation x = Rx + b, the effect of which is shown in Fig. 3.7. If a scaling S is performed before the rotation and translation, the effect on a vector x is described by the matrix equation x = RSx + b. This is illustrated in Fig. 3.8b, which shows the effect when a transformation of this type is performed on the circle of radius a centered on the origin shown in Fig. 3.8a, with h cos π/3 −sin π/3 3 0 b= , R= , and S = . k sin π/3 cos π/3 0 2 It is seen that the circle has ﬁrst been scaled to become an ellipse with semiaxes 3a and 2a, after which the ellipse has been rotated through an angle π/3, and ﬁnally its center has been translated to the point (h, k).

y

P(x, y) y′ φ

k

O

x′ θ

O′

h

FIGURE 3.7 A rotation and a translation.

x

Section 3.2

Some Problems That Give Rise to Matrices

127

y

x′

3a y

y′

x′

2a

y′

a

π /3

k

π/3

−2a x

h

x

−3a (a)

(b)

FIGURE 3.8 The composite transformation x = RSx + b.

It is essential to remember that the order in which transformations are performed will, in general, inﬂuence the result. This is easily seen by considering the two transformations x = RSx + b and x = SRx + b. If the ﬁrst of these is performed on the circle in Fig. 3.8a, it produces Fig. 3.8b, but when the second is performed on the same circle, it ﬁrst converts it into an ellipse with its major axis horizontal, and then translates the center of the ellipse to the point (h, k). In this case the effect of the rotation cannot be seen, because the circle is symmetric with respect to rotations. A relationship of the form x = F(x ) can be interpreted geometrically in two distinct ways which are equally valid: 1. As the change in the way we describe the location of a point P. Then the relationship is called a transformation of coordinates (Figs. 3.5, 3.6, 3.7). 2. As a mapping of a point P from one location to a new one.

(d) Matrix Analysis of Framed Structures A framed structure is a network of straight struts joined at their ends to form a rigid three-dimensional structure. A typical framed structure is the steel work for a large building before the walls and ﬂoors have been added. A simple example of a framed structure, called a truss, is a plane construction in which the struts are joined together to form triangles, as in the side section of the small bridge shown in Fig. 3.9. For safety, to ensure that no strut fails when the bridge carries the largest permitted load, it is necessary to determine the force experienced by each strut in the truss when the bridge supports its maximum load in several different positions. Typically, the largest load could be due to a heavy truck crossing the bridge. The analysis of trusses is usually simpliﬁed by making the following assumptions:

r The structure is in the vertical plane; r The weight of each strut can be neglected; r Struts are rigid and so remain straight;

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Matrices and Systems of Linear Equations

F4

B

B

F1

D

A

A

C

E

FIGURE 3.9 A typical truss found in a side section of a bridge.

π /3

F3

F5

π /3

F2 L /2

L /2

D

π /3 C

F7

F6

π /3

L

R1

E R2

3m FIGURE 3.10 A truss supporting a concentrated load.

r Each joint is considered to be hinged, so the only forces acting at a joint are along the struts meeting at the joint if forces are applied at joints only.

r There are no redundant struts, so that removing a strut will cause the truss to collapse. We now write down the simultaneous equations that must be solved to ﬁnd the forces acting in the seven struts of length L that form the truss shown in Fig. 3.10, when a concentrated load 3m is located at point C midway between A and E. This load could be considered to be a heavily laden truck standing in the center of the bridge. To determine the reactions at the support points A and E, we use the fact that for equilibrium the turning moments about these two points must be zero. The turning moment of the load 3m about the point A must be cancelled by the turning moment of the reaction R2 at E, so 3m(L) = R2 (2L), showing that R2 = 3m/2. Similarly, the turning moment of the load 3m about the point E must be cancelled by the turning moment of the reaction R1 at A, so 3m(L) = R1 (2L), showing that R1 = 3m/2. The directions in which the forces F1 to F7 are assumed to act are shown by arrows, and if later a force is found to be negative, the direction of the associated arrows must be reversed. For equilibrium the sum of the vertical components of all forces acting at each joint must be zero, as must be the sum of the horizontal components of all forces acting at each joint. The equations representing the balance of forces at each joint are as follows, where when resolving the forces acting at joint C, the effect of the load 3m which acts vertically downwards must be taken into account: equations and matrices for a framed structure

Joint A(vertical) Joint A(horizontal) Joint B(vertical) Joint B(horizontal) Joint C(vertical) Joint C(horizontal) Joint D(vertical)

F1 sin π/3 − 3m/2 = 0 F1 cos π/3 + F2 = 0 F1 sin π/3 + F3 sin π/3 = 0 F1 cos π/3 − F3 cos π/3 − F4 = 0 F3 sin π/3 + F5 sin π/3 + 3m = 0 F2 + F3 cos π/3 − F5 cos π/3 − F6 = 0 F5 sin π/3 + F7 sin π/3 = 0

Section 3.2

Some Problems That Give Rise to Matrices

129

F4 + F5 cos π/3 − F7 cos π/3 = 0 F7 sin π/3 − 3m/2 = 0 F6 + F7 cos π/3 = 0.

Joint D(horizontal) Joint E(vertical) Joint E(horizontal)

After substituting for sin π/3 and cos π/3, these equations can be written in the matrix form Ax = b, where ⎤ ⎡1√ 3 0 0 0 0 0 0 2 ⎢ 1 ⎥ ⎤ ⎡ 1 0 0 0 0 0 ⎥ ⎢ 2 3m/2 ⎡ ⎤ ⎥ ⎢ √ √ F1 ⎢ 0 ⎥ ⎢1 3 0 1 3 0 0 0 0 ⎥ ⎥ ⎢ ⎥ ⎢2 2 ⎢ ⎥ ⎢ 0 ⎥ ⎥ ⎢ 1 F ⎢ ⎥ 1 2 ⎥ ⎢ ⎥ ⎢ 0 − −1 0 0 0 ⎢ ⎥ ⎢ 0 ⎥ ⎥ ⎢ 2 ⎢ F3 ⎥ √2 √ ⎥ ⎢ ⎥ ⎢ 1 1 ⎢ ⎥ 0 2 3 0 2 3 0 0 ⎥ ⎢ −3m ⎥ ⎢ 0 ⎢ ⎥ ⎥. ⎢ ⎥ ⎢ , x = ⎢ F4 ⎥ , b = ⎢ A=⎢ 1 0 ⎥ ⎢ ⎥ 1 0 − 12 −1 0 ⎥ ⎥ ⎢ ⎥ ⎢ 0 2 ⎢ F5 ⎥ √ √ ⎥ ⎢ 0 ⎥ ⎢ ⎢ ⎥ 1 1 ⎥ ⎢ ⎥ ⎢ 0 3 0 3 0 0 0 ⎢ ⎥ 2 2 ⎢ 0 ⎥ ⎥ ⎢ ⎣ F6 ⎦ ⎥ ⎢ ⎢ 0 1 1 ⎥ 0 −2 ⎥ 0 0 1 ⎢ ⎣3m/2⎦ 2 F7 ⎥ ⎢ √ ⎢ 0 0 0 0 0 0 0 12 3⎥ ⎦ ⎣ 0

1

1 2

These are 10 equations for the 7 unknown forces F1 to F7 , so unless 3 of the equations represented in Ax = b are combinations of the remaining 7 equations, we cannot expect there to be a solution. When the rank of a matrix is introduced in Section 3.6, we will see how systems of this type can be checked for consistency and, when appropriate, simpliﬁed and solved. In this case the equations are sufﬁciently simple that they can be solved sequentially, without the use of matrices. The solution is seen to be √ √ √ √ F1 = m 3, F2 = −m/( 3/2), F3 = −m 3, F4 = m 3, √ √ √ F5 = −m 3, F6 = −m( 3/2), F7 = m 3.

k1

m1

x1

The signs show that the arrows in Fig. 3.10 associated with forces F2 , F3 , F5 , and F6 should be reversed, so these struts are in tension, while the others are in compression. Notice that matrix A is determined by the geometry of the truss, and so does not change when forces are applied to more than one of the joints on the truss (bridge). This means that after the 10 equations have been reduced to seven, the same modiﬁed matrix A can be use to ﬁnd the forces in the struts for any form of concentrated loading. Had a more complicated struss been involved, many more equations would have been involved, so that a matrix approach becomes necessary. This approach also identiﬁes any redundant struts in a structure, because the force in a redundant strut is indeterminate.

(e) A Compound Mass–Spring System

k2 x2 m2 FIGURE 3.11 A compound mass–spring system.

Matrices can have variables as elements, and an analysis of the compound mass– spring system shown in Fig. 3.11 shows one way in which this can arise. Figure 3.11 represents a mass m1 suspended from a rigid support by a spring of negligible mass with spring constant k1 , and a mass m2 suspended from mass m1 by a spring of negligible mass with spring constant k2 . The vertical displacement of m1 from its

130

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Matrices and Systems of Linear Equations

equations of motion of a coupled mass–spring system

equilibrium position is x1 , and the vertical displacement of m2 from its equilibrium position is x2 . Each spring is assumed to be linearly elastic, so the restoring force exerted by a spring is equal to the product of the displacement from its equilibrium position and the spring constant. The product of the mass m1 and its acceleration is m1 d2 x1 /dt 2 , and the restoring force due to spring k1 is k1 x1 , while the restoring force due to spring k2 is k2 (x1 − x2 ), so the equation of motion of m1 is d2 x1 = −k1 x1 − k2 (x1 − x2 ). dt 2 Similarly, the equation of motion of m2 is m1

d2 x2 = −k2 (x2 − x1 ), dt 2 where the negative signs are necessary because the springs act to restore the masses to their original positions. This system can be written as the matrix differential equation x¨ + Ax = 0, by deﬁning A and x as ⎡ 2 ⎤ ⎡ (k + k ) k2 ⎤ d x1 1 2 − ⎥ ⎢ ⎥ ⎢ m1 m1 ⎥ x1 ⎢ dt 2 ⎥ ¨ A=⎢ , and x = , x = ⎢ ⎥. ⎣ x2 ⎣ d2 x2 ⎦ k2 ⎦ k2 − m2 m2 dt 2 The solution of this system will not be considered here as ordinary differential equations and systems of the type derived here are discussed in detail in Chapter 6, where matrix methods are also developed. Chapter 7 develops Laplace transform methods for the solution of differential equations and systems. It will sufﬁce to mention here that the dynamical behavior of the compound mass–spring system in Fig. 3.11 is completely characterized by matrix A. m2

(f) Stochastic Processes Certain problems arise that are not of a deterministic nature, so that both the formulation of the problem and its outcome must be expressed in terms of probabilities. The probability p that a certain event occurs is a number in the interval 0 ≤ p ≤ 1. An event with probability p = 0 is one that never occurs, and an event with probability p = 1 is one that is certain to occur. So, for example, when tossing a coin N times and recording each outcome as an H (head) or a T (tail), if the number of heads is NH and the number of tails is NT , so that N = NH + NT , the numbers NH /N and NT /N will be approximations to the respective probabilities that a head or a tail occurs when the coin is tossed. If the coin is unbiased, it is reasonable to expect that as N increases both NH /N and NT /N will approach the value 1/2. This will mean, of course, that the chances of either a head or a tail occurring on each occasion are equal. The example we now outline is called a stochastic process and is illustrated by considering a process that evolves with time and is such that at any given moment it may be in precisely one of N different situations, usually called states, where N is ﬁnite. We shall denote the N states in which the process may ﬁnd itself at any given time tm by S1 , S2 , . . . , SN , with m = 0, 1, 2, . . . , and tm−1 < tm, it being assumed that the outcome at each time depends on probabilities, and so is not deterministic.

Section 3.2

Some Problems That Give Rise to Matrices

131

To formulate the problem we assume that what are called the conditional probabilities pki (also called transition probabilities) that determine the probability with which the process will be in state S j at time tm are all known, given that it was in state Sk at time tm−1 , and that these probabilities are the same from t1 to t2 as from tm−1 to tm for m = 0, 1, 2, . . . . This last assumption means that the probability with which the transition from state Sk to S j occurs is independent of the time at which the process was in state Sk. The conditional probabilities can be arranged as the N × N matrix P = [ p jk], so as probabilities are involved, all the p jk are nonnegative, and as each stage must have an outcome, the sum of the elements in every row of matrix P must equal 1. A matrix P with these properties, namely that 0 ≤ p jk ≤ 1,

0 ≤ j ≤ N,

0 ≤ k ≤ N,

and

N

p jk = 1,

j=1

stochastic matrix and a Markov process

is called a stochastic matrix (see Exercise 53, Section 3.1). Processes like these, whose condition at any subsequent instant does not depend on how the process arrived at its present state, are called Markov processes. Simple but typical examples of such processes involving only two states are gambling wins and losses, the reliability of machines that may either be operational or under repair, shells ﬁred from a gun that either hit or miss the target and errors that introduce an incorrect digit 1 or 0 when transferring binary coded information. To develop the argument a little further, let us now consider a process that can be in one of two states, and that the matrix P describing its transitions is given by 2/3 1/3 P= . 1/4 3/4 Now suppose that initially the probability distribution is given by the row matrix E(0) = [ p, q], where, of course, p + q = 1. Then if E(m) denotes the probability distribution of the states at time tm, it follows that E(1) = E(0)P, but as P is independent of the time we conclude that after m transitions the general result must be E(m) = E(0)Pm, so in this case E(m) = [ p, q]

m 2/3 1/3 . 1/4 3/4

Direct calculation shows that E(3) = [0.470 p + 0.398q, 0.530 p + 0.602q], E(6) = [0.432 p + 0.426q, 0.568 p + 0.574q], and E(10) = [0.429 p + 0.429q, 0.571 p + 0.571q], so it is reasonable to ask if E(m) tends to a limiting vector as m → ∞ and, if so, what this is? As this problem is simple, an analytical answer is possible, though it involves using a diagonalizing matrix P which will be discussed later.

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Matrices and Systems of Linear Equations

We will see later that P can be written as ADB, where D is a diagonal matrix and AB = I. In this case 1 4 1 0 3/7 4/7 A= , D= , and B = , 1 −3 0 5/12 1/7 −1/7 so

1 P= 1

4 1 0 3/7 −3 0 5/12 1/7

4/7 . −1/7

In what follows we will need to make repeated use of the fact that 3/7 4/7 1 4 1 0 BA = = = I. 1/7 −1/7 1 −3 0 1 Using this last property we ﬁnd that 1 4 1 0 3/7 4/7 1 P2 = 1 −3 0 5/12 1/7 −1/7 1 2 1 4 1 0 3/7 4/7 = . 1 −3 0 5/12 1/7 −1/7

4 −3

1 0 3/7 0 5/12 1/7

4/7 −1/7

However, when a diagonal matrix is raised to a power, each of its elements is raised to that same power (see Problem 41, Section 3.1), so 0 3/7 4/7 1 4 1 2 P = 1 −3 0 (5/12)2 1/7 −1/7 and, in general, Pm = Thus,

1 1

4 −3

1 0

⎡

0 (5/12)m

3 + 4(5/12)m ⎢ 7 Pm = ⎢ ⎣ 3 − 3(5/12)m 7

3/7 1/7

4/7 . −1/7

⎤ 4 − 4(5/12)m ⎥ 7 ⎥, 4 + 3(5/12)m ⎦ 7

showing that as m → ∞, so lim E(m)Pm = [3( p + q)/7, 4( p + q)/7] = [3/7, 4/7],

m→∞

and we have found the limiting state of the system. Stochastic processes also occur that involve more than two states. The problem of determining the probability with which such processes will be in a given state, and when a limiting state exists, the limiting values of the probabilities involved, is of considerable practical importance. An introduction to stochastic process can be found in reference [2.4].

Summary

This section has introduced some of the many areas in which matrices play an essential role. These range from electric circuits needing the application of Kirchhoff’s laws, through routing problems involving the concepts of directed graphs and adjacency matrices, to the classical K¨onigsberg bridge problem, computer graphic operations performed by linear transformations, the matrix analysis of forces in a framed structure, the oscillations of a coupled mass–spring system, and stochastic processes.

Section 3.3

Determinants

133

EXERCISES 3.2 1. State which of the following matrices is a stochastic matrix, giving a reason when this is not the case. ⎤ ⎡ ⎤ ⎡ 0.5 0.2 0.3 0.5 0.3 0.2 (c) ⎣0.7 0.3 0.2⎦ . (a) ⎣0.25 0 0.75⎦. 0.4 0.2 0.4 0.5 0.5 0 ⎤ ⎤ ⎡ ⎡ 0.3 0.1 0.6 1.2 0 −0.2 0.2⎦ . (d) ⎣0.8 0 0.2⎦ . (b) ⎣ 0 0.8 0 1 0 0.6 0.3 0.1

4.

2. Given the stochastic matrix 3/4 P= 1/2

5.

1/4 1/2

4

3

1 2 FIGURE 3.13

4 5

and the initial probability distribution E(0) = [ p, q], with p, q ≥ 0 and p + q = 1, the probability distribution of the two states at time tm is given by

3

E(m) = E(0)Pm. Find E(2), E(4), and E(6), together with their values when p = 1/4, q = 3/4. In Exercises 3 through 6 ﬁnd the adjacency matrices for the given graphs and digraphs.

FIGURE 3.14

6. 4

3.

5 1

5 6 6

3

4 2

3

2

FIGURE 3.15

FIGURE 3.12

3.3

1

Determinants Every square matrix A with numbers as elements has associated with it a single unique number called the determinant of A, which is written detA. If A is an n × n matrix, the determinant of A is indicated by displaying the elements ai j of A between two vertical bars as follows:

notation for a determinant

a11 a detA = 21 · · · an1

a12 a22 ··· an2

· · · a1n · · · a2n . · · · · · · · · · ann

(5)

The number n is called the order of determinant A, and in (5) the vertical bars are used to distinguish detA, that is a number, from matrix A that is an n × n array of numbers.

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A general deﬁnition of the value of detA in terms of its elements ai j will be given later, so for the moment we deﬁne only the value of ﬁrst and second order determinants (see Section 1.7). If A only contains a single element a11 so A = [a11 ] then, by deﬁnition, detA = a11 , and if A is the 2 × 2 matrix a12 a , A = 11 a21 a22 then, by deﬁnition,

a det A = 11 a21

a12 = a11 a22 − a21 a12 . a22

(6)

Notice that in (6) the numerical value of detA is obtained by forming the product of the two terms a11 and a22 on the leading diagonal, and subtracting from it the product of the two terms a21 and a12 on the cross diagonal. This process, called expanding the determinant, is easily remembered by representing the method by which the determinant is expanded as a12 = a11 a22 − a21 a12 , @ a21 @ R a22 a11

where the product involving the downward arrow generates the ﬁrst pair of terms on the right and the product involving the upward arrow indicates that the product of the associated pair of terms is to be subtracted. EXAMPLE 3.10

Find detA given 3 −1 (a) detA = 2 6

and

1 + i (b) detA = −3i

i . 2

Solution (a) Using (5) we have 3 −1 = 3 · 6 − 2 · (−1) = 20. detA = 2 6 (b) Again using (5) we have 1 + i detA = −3i

i = (1 + i) · 2 − (−3i) · i = −1 + 2i. 2

To provide some motivation for the introduction of determinants, we solve by elimination the two linear simultaneous algebraic equations a11 x1 + a12 x2 = b1 a21 x1 + a22 x2 = b2 .

(7)

To eliminate x2 we multiply the ﬁrst equation by a22 and the second equation by a12 , and then subtract the results to obtain (a11 a22 − a21 a12 )x1 = a22 b1 − a12 b2 . This shows that when a11 a22 − a21 a12 = 0, x1 =

a22 b1 − a12 b2 . a11 a22 − a21 a12

Section 3.3

Determinants

135

This result can be expressed in terms of detA as x1 = (a22 b1 − a12 b2 )/detA.

(8)

Similarly, when x1 is eliminated from equations (7) we ﬁnd that x2 = (a11 b2 − a21 b1 )/detA. Cramer’s rule for a system of two equations

(9)

Examination of (8) and (9) shows that their numerators can be written in terms of determinants that are closely related to detA, because x1 = where D = detA,

D1 D

b D1 = 1 b2

and a12 , a22

x2 =

and

D2 , D a D2 = 11 a21

(10) b1 . b2

(11)

The form of solution of equations (7) in terms of the determinants in (10) and (11) is called Cramer’s rule. The rule itself says that xi = Di /D for i = 1, 2, where determinant D1 is obtained from D = detA by replacing the ﬁrst column of A by the nonhom*ogeneous terms b1 and b2 on the right of equations (7), and determinant D2 is obtained from D by replacing the second column of A by these same two terms. EXAMPLE 3.11

Use Cramer’s rule to solve the equations 3x1 + 5x2 = 4 2x1 − 4x2 = 1. Solution The three determinants required by Cramer’s are 4 3 3 5 5 = −21, D2 = = −22, D1 = D = detA = 1 −4 2 2 −4

4 = −5, 1

so x1 = D1 /D = 21/22 and x2 = D2 /D = 5/22.

minors and cofactors

This example shows how determinants enter naturally into the solution of a system of equations. As determinants of order n > 2 occur in the study of differential equations, analytical geometry, throughout linear algebra, and elsewhere, it is necessary to generalize the deﬁnition of a determinant of order 2 given in (6) to determinants of any order n. With this objective in mind, we ﬁrst deﬁne the minors and cofactors of a determinant of order n. The minor Mi j associated with the element ai j in the ith row and jth column of the nth order determinant in (5) is the determinant of order n − 1 formed from detA by deleting the elements in the ith row and jth column. As each element of detA has an associated minor, a determinant of order n has n2 minors. By way of example, the minor M3 j of the nth order determinant in (5) is the determinant of order n − 1 a11 a12 · · · a1 j−1 a1 j+1 · · · a1n a21 a22 · · · a2 j−1 a2 j+1 · · · a2n (12) M3 j = a41 a42 · · · a4 j−1 a4 j+1 · · · a4n . · · · · · · · · · · · · · · · · · · · · · an1 an2 · · · anj−1 anj+1 · · · ann

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Matrices and Systems of Linear Equations

The cofactor Ci j associated with the element ai j in determinant (5) is deﬁned in terms of the minor Mi j as Ci j = (−1)i+ j Mi j

for i, j = 1, 2, . . . , n,

(13)

so an nth order determinant has n2 cofactors. EXAMPLE 3.12

Find the minors and cofactors of

2 detA = 1

−3 . 4

Solution Inspection shows that M11 = 4, M12 = 1, M21 = −3, and M22 = 2. Using deﬁnition (12), the cofactors are seen to be C11 = (−1)1+1 M11 = 4,

C12 = (−1)1+2 M12 = −1,

C21 = (−1)2+1 M21 = 3,

and C22 = (−1)2+2 M22 = 2. Recognizing that the cofactors of the second order determinant a a12 are C11 = a22 , C12 = −a21 , C21 = −a12 , and C22 = a11 , detA = 11 a21 a22 expanding a second order determinant in terms of rows or columns

we see from the deﬁnition detA = a11 a22 − a21 a12 that detA can be expressed in terms of these cofactors in four different ways: detA = a11 C11 + a12 C12 , using elements and cofactors from the ﬁrst row of A; detA = a21 C21 + a22 C22 , using elements and cofactors from the second row of A; detA = a11 C11 + a21 C21 , using elements and cofactors from the ﬁrst column of A; detA = a12 C12 + a22 C22 , using elements and cofactors from the second column of A. This has proved by direct calculation that the value of the general second order determinant detA is given by the sum of the products of the elements and their associated cofactors in any row or column of the determinant. When the deﬁnition of a determinant is extended to the case n > 2 it will be seen that this same property remains true. There are various ways of deﬁning an nth order determinant, and from among these we have chosen to use one that involves a recursive process. More will be said about this recursive process, and how it can be used to evaluate the determinant, once the deﬁnition has been formulated. Deﬁnition of a determinant of order n The nth order determinant detA in which the element ai j has the associated cofactor Ci j for i, j = 1, 2, . . . , n is deﬁned as a11 a detA = 21 · · · an1

a12 a22 ··· an2

· · · a1n n · · · a2n = a1 j C1 j . · · · · · · j=1 · · · ann

(14)

Section 3.3

Determinants

137

Recalling the different ways in which a second order determinant can be evaluated, we see that the expansion of detA in (14) is in terms of the elements and cofactors of the ﬁrst row, so for conciseness this expansion is said to be in terms of the elements of the ﬁrst row. The recursive process enters this deﬁnition through the fact that each cofactor C1 j is a determinant of order n − 1, as can be seen from (12), so each cofactor in turn can be expanded in terms of determinants of order n − 2, and the process continued until determinants of order 2 are obtained that can then be calculated using (6). EXAMPLE 3.13

Expand

1 detA = 2 1

4 −1 0 3 . 2 1

Solution To expand this third order determinant using (14), we must ﬁnd the cofactors of the elements of the ﬁrst row, so to do this we ﬁrst ﬁnd the minors and then use (13) to ﬁnd the cofactors, as a result of which we ﬁnd that 0 3 = −6, so C11 = (−1)1+1 (−6) = −6 M11 = 2 1 2 3 = −1, so C12 = (−1)1+2 (−1) = 1 M12 = 1 1 2 0 = 4, so C13 = (−1)1+3 (4) = 4. M13 = 1 2 As the elements of the ﬁrst row are a11 = 1, a12 = 4, and a13 = −1, we ﬁnd from (12) that detA = (1)C11 + (4)C12 + (−1)C13 = (1)(−6) + (4)(1) + (−1)(4) = −6. The determinant associated with either an upper or a lower triangular matrix A of any order is easily expanded, because repeated application of (12) shows that it reduces to the product of the terms on the leading diagonal, so the expansion of the nth order upper triangular determinant with elements a11 , a22 , . . . , ann on its leading diagonal a11 a12 · · · a1n 0 a22 · · · a2n = a11 a22 . . . ann , detA = (15) 0 · · · · · · 0 0 0 0 ann and a corresponding result is true for a lower triangular matrix. Deﬁnition (14) can be used to prove that nth order determinants, like second order determinants, have the property that their value is given by the sum of the products of the elements and their cofactors in any row or column. This result, together with a generalization concerning the vanishing of the sum of the products of the elements in any row (or column) and the corresponding cofactors in a different row (or column), forms the next theorem. The details of the proof can be found in linear algebra texts, for example, [2.1], [2.5], [2.7], [2.9], but the method used has no other application in what is to follow, so the proof will be omitted.

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Matrices and Systems of Linear Equations

THEOREM 3.3 Laplace expansion theorem

Laplace expansion theorem and an extension Let A be an n × n matrix with elements ai j . Then, (i) detA can be expanded in terms of elements of its ith row and the cofactors Ci j of the ith row as detA = ai1 Ci1 + ai2 Ci2 + · · · + ain Cin =

n

ai j Ci j

j=1

for any ﬁxed i with 1 ≤ i ≤ n. (ii) detA can be expanded in terms of elements of its jth column and the cofactors Ci j of the jth column as detA = a1 j C1 j + a2 j C2 j + · · · + anj Cnj =

n

ai j Ci j

j=1

for any ﬁxed j with 1 ≤ j ≤ n. (iii) The sum of the products of the elements of the ith row with the corresponding cofactors of the kth row is zero when i = k. (iv) The sum of the products of the elements in the jth column with the corresponding cofactors of the kth column is zero when j = k. Results (i) and (ii) are often used to advantage when a row or column contains many zeros, because if the determinant is expanded in terms of the elements of that row or column, the cofactors associated with each zero element need not be calculated. Results (iii) and (iv) simply say that the sum of the products of the elements in any row (or column) with the corresponding cofactors in a different row (or column) is zero. PIERRE SIMON LAPLACE (1749–1827) A French mathematician of remarkable ability who made contributions to analysis, differential equations, probability, and celestial mechanics. He used mathematics as a tool with which to investigate physical phenomena, and made fundamental contributions to hydrodynamics, the propagation of sound, surface tension in liquids, and many other topics. His many contributions had a wide-ranging effect on the development of mathematics.

EXAMPLE 3.14

Verify Theorem 3.3(i) by expanding the determinant in Example 3.13 in terms of the elements of its second row. Use the determinant to check the result of Theorem 3.3(iii). Solution The second row contains a zero element in its mid position, so the cofactor C22 associated with the zero element need not be calculated. The necessary cofactors in the second row that follow from the minors are 4 −1 = 6 so C21 = (−1)2+1 (6) = −6 M21 = 2 1 1 4 = −2 so C23 = (−1)2+3 (−2) = 2. M23 = 1 2

Section 3.3

Determinants

139

As a21 = 2 and a23 = 3, it follows from Theorem 3.3(i) that when detA is expanded in terms of elements of its second row, detA = (2)(−6) + (3)(2) = −6, conﬁrming the result obtained in Example 3.13. As a particular case of Theorem 3.3(iii), let us show that the sum of the products of the cofactors in the ﬁrst row of detA and the corresponding elements in the third row is zero. In Example 3.13 it was found that C11 = −6, C12 = 1, and C13 = 4, so as the elements of the third row are a31 = 1, a32 = 2, and a33 = 1, we have a31 C11 + a32 C12 + a33 C13 = (−6)(1) + (2)(1) + (1)(4) = 0, conﬁrming the result of Theorem 3.3(iii) when the elements of row 3 and the cofactors of row 1 are used. Determinants have a number of special properties that can be used to simplify their expansion, though their main uses are found elsewhere in mathematics, where determinants often characterize some important theoretical feature of a problem. The most important and useful of these properties are contained in the next theorem. THEOREM 3.4 basic properties of determinants

Properties of determinants A determinant detA has the following properties: (i) If any row or column of a determinant detA only contains zero elements, then detA = 0. (ii) If A is a square matrix with the transpose AT , then detA = detAT . (iii) If each element of a row or column of a square matrix A is multiplied by a constant k, then the value of the determinant is kdetA. (iv) If two rows (or columns) of a square matrix are interchanged, the sign of the determinant is changed. (v) If any two rows or columns of a square matrix A are proportional, then detA = 0. (vi) Let the square matrix A be such that each element ai j of the ith row (or the (1) (2) jth column) can be written as ai j = ai j + ai j . Then if A1 is the matrix derived from (1) A by replacing its ith row (or jth column) by the elements ai j and A2 is the matrix (2) derived from A by replacing its ith row (or jth column) by the elements ai j , detA = detA1 + detA2 . (vii) The addition of a multiple of a row (or column) of a determinant to another row (or column) of the determinant leaves the value of the determinant unchanged. (viii) Let A and B be two n × n matrix, then det(AB) = detA detB. Proof (i) The result follows by expanding the determinant in terms of the row or column that only contains zero elements.

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(ii) The result follows from the fact that expanding detA in terms of the elements of its ﬁrst row is the same as expanding detAT in terms of the elements of its ﬁrst column. (iii) The result follows by expanding the determinant in terms of the row or column in which each element has been multiplied by the constant k, because k appears as a factor in each term, so the result becomes kdetA. (iv) The proof is by induction, starting with a second order determinant for which the result can be seen to be true from deﬁnition (6). To proceed with an inductive proof we assume the results to be true for a determinant of order r − 1, and show it must be true for a determinant of order r. Expand a row of a determinant of order r in terms of the elements of a row (or column) that has not been interchanged. Then, by hypothesis, as the cofactors are determinants of order r − 1, their signs will all be reversed. This establishes that if the hypothesis is true for a determinant of order r − 1 it must also be true for a determinant of order r . As the result is true for r = 2, it follows by induction that it is true for all integers r > 2, and the result is proved. (v) If the value of the determinant is detA, and one row is k times another, then from (ii) by removing the factor k from the row the value of the determinant will be kdetA1 , where A1 is now a determinant with two identical rows. From (ii), interchanging two rows changes the sign of the determinant, but the rows are identical, leaving the determinant invariant, so detA1 = 0. A similar argument shows the result to be true when two columns are proportional, so the result is proved. (vi) The result is proved directly by expanding the determinant in terms of the elements of the ith row (or the jth column). (vii) Let the square matrix B be obtained from A by adding k times the ith row (or a column) to the jth row (or column). Then from (iii) and (vi), detB = detA + kdetC, where C is obtained from A by replacing the ith row (or column) by the jth row (or column). As detC has two identical rows (or columns), it follows from (v) that detC = 0, so detB = detA and the result is proved. (viii) A proof of this result will be given later after the introduction of elementary row operation matrices. Cramer’s rule, which was ﬁrst encountered when seeking the solution of the two equations in (7), can be extended to a system of n equations in a very straightforward manner, and it takes the following form. Cramer’s rule Cramer’s rule for a system of n equations in n unknowns

The solution of the system of n equations in the n unknowns x1 , x2 , . . . , xn a11 x1 + a12 x2 + · a21 x1 + a22 x2 + · · · an1 x1 + an2 x2 + ·

· · · ·

· + a1n xn = b1 · + a2n xn = b2 · · · · + ann xn = bn

Section 3.3

Determinants

141

is given by xi = detAi /detA

for i = 1, 2, . . . , n,

where detA is the determinant of the coefﬁcient matrix with elements ai j , and detAi is the determinant obtained from the coefﬁcient matrix by replacing its ith column by the column containing the number b1 , b2 , . . . , bn . The justiﬁcation for Cramer’s rule in this more general form will be postponed until after the introduction of inverse matrices, when a simple proof can be given. Cramer’s rule is mainly of theoretical importance and, in general, it should not be used to solve equations when n > 3. This is because the number of multiplications required to evaluate a determinant of order n is (n − 1)n!, so to solve for n unknowns (n + 1) determinants must be evaluated leading to a total of (n2 − 1)n! multiplications, and this calculation becomes excessive when n > 3. An efﬁcient way of solving large systems by means of elimination is given in Chapter 19. EXAMPLE 3.15

Use Cramer’s rule to solve x1 − 2x2 + x3 = 1 2x1 + x2 − 2x3 = 3 −x1 + 3x2 + 4x3 = −2. Solution The determinants involved are 1 −2 1 1 1 −2 = 29, detA1 = 3 detA = 2 −1 −2 3 4 1 1 1 3 −2 = 1, detA2 = 2 −1 −2 4

1 detA3 = 2 −1

−2 1 1 −2 = 37 3 4 −2 1 1 3 = −6, 3 −2

so x1 = 37/29, x2 = 1/29, and x3 = −6/29. A purely algebraic approach to the study of determinants and their properties is to be found in reference [2.8], and many examples of their applications are given in references [2.11] and [2.12].

Summary

This section has extended to an nth order determinant the basic notion of a second order determinant that was reviewed in Chapter 1, and then established its most important properties. The Laplace expansion formulas that were established are of theoretical importance, but it will be seen later that the practical evaluation of a determinant is most easily performed by reducing the n × n matrix associated with a determinant to its echelon form.

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EXERCISES 3.3 In Exercises 1 through 4 ﬁnd detA. 2 1 −1 3. 1. detA = 0 4 3 2 −2 −1 2 1 2. detA = 1 3 2. −4 1 2 2 4 −3 1 0. 3. detA = −2 5 −2 4 4 0 0 4. detA = −2 cos x − sin x . 5 sin x cos x 5. Given that

−3 1 4 detA = 2 −1 5 = 87, 4 2 5

conﬁrm by direct calculation that (a) interchanging the ﬁrst and last rows changes the sign of detA and (b) interchanging the second and third columns changes the sign of detA. 6. Given that 2 1 3 detA = 5 −2 2 = −24, −1 1 3 conﬁrm by direct calculation that (a) adding twice row two to row three leaves detA unchanged and (b) subtracting three times column three from column one leaves detA unchanged. Establish the results in Exercises 7 through 12 without a direct expansion of the determinant by using the properties listed in Theorem 3.4. 1 + a a a 1+b b = (1 + a + b + c). 7. b c c 1+c 1 a b+ c 8. 1 b c + a = 0. 1 c a + b 2 a b2 c2 9. a b c = (a − b)(a − c)(b − c). 1 1 1

2 x + a2 ab ac 2 2 x +b bc = x 4 (x 2 + a 2 + b2 + c2 ). 10. ab 2 ac cb x + c2 1 a b 11. a 1 b = (a + b + 1)(a − 1)(b − 1). a b 1 k 1 12. 1 1

1 k 1 1

1 1 k 1

1 1 = (k + 3)(k − 1)3 . 1 k

In Exercises 13 and 14 use Cramer’s rule to solve the system of equations. 13. 2x1 − 3x2 + x3 = 4 x1 + 2x2 − 2x3 = 1 3x1 + x2 − 2x3 = −2. 14. 3x1 + x2 + 2x3 = 5 2x1 − 4x2 + 3x3 = −3 x1 + 2x2 + 4x3 = 2. 15. Let P(λ) be given by 3 − λ P(λ) = 2 4

0 1 2−λ 2 , 2 1 − λ

where λ is a parameter. Expand the determinant to ﬁnd the form of the polynomial P(λ) and use the result to ﬁnd for what values of λ the determinant vanishes. 16. Let P(λ) be given by 4 − λ 1 P(λ) = −1

0 1 −λ 1 , −2 2 − λ

where λ is a parameter. Expand the determinant to ﬁnd the form of the polynomial P(λ) and use the result to ﬁnd for what values of λ the determinant vanishes. 17. Given that ⎡ −3 A=⎣ 1 1

⎤ 0 4 2 −1⎦ 0 1

⎡

and

1 2 B = ⎣2 3 3 1

⎤ 3 1⎦ , 2

calculate det(AB), detA, detB, and hence verify that det(AB) = detAdetB.

Section 3.4

3.4

Elementary Row Operations, Elementary Matrices, and Their Connection with Matrix Multiplication

143

Elementary Row Operations, Elementary Matrices, and Their Connection with Matrix Multiplication To motivate what is to follow we will examine the processes involved when solving by elimination the system of linear equations a11 x1 + a12 x2 + · · · + a1n xn = b1 a21 x1 + a22 x2 + · · · + a2n xn = b2 · · · · · · · · · am1 x1 + am2 x2 + · · · + amn xn = bm,

(16)

though later more will need to be said about the details of this important problem, and how it is inﬂuenced by the number of equations m and the number of unknowns n.

Elementary Row Operations the three basic types of elementary row operation

The three types of elementary row operations used when solving equations (16) by elimination are: TYPE I The interchange of two equations TYPE II The scaling of an equation by a nonzero constant TYPE III The addition of a scalar multiple of an equation to another equation In matrix notation the system of equations (16) becomes Ax = b,

(17)

where A = [ai j ] is an m × n matrix, x = [x1 , x2 , . . . , xn ]T , and b = [b1 , b2 , . . . , bm]T . The three elementary row operations of types I to III that can be performed on the equations in (16) can be interpreted as the corresponding operations performed on the rows of the matrices A and b. This is equivalent to performing these same operations on the rows of the new matrix denoted by (A, b), deﬁned as ⎡

a11 a12 . . . ⎢ a21 a22 . . . ⎢ (A, b) = ⎢ · · · · · · ⎣ am1

the augmented matrix

am2

a1n a2n

. . . amn

⎤ b1 b2 ⎥ ⎥ ⎥, ⎦

(18)

bm

that has m rows and n + 1 columns and is obtained by inserting the column vector b containing the nonhom*ogeneous terms on the right of matrix A. When considering the system of linear equations in (16), matrix (A, b) is called the augmented matrix associated with the system. The separation of the last column in (18) by a vertical dashed line is to indicate partitioning of the matrix to show that the elements of the last column are not elements of the coefﬁcient matrix A.

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We are now in a position to introduce a notation for the three elementary row operations that are necessary when using an elimination process to ﬁnd the solution of a system of equations in matrix form (ordinary or augmented). Elementary row operations The three elementary row operations that may be performed on a matrix are: (i) The interchange of the ith and jth rows, which will be denoted by R{i → j, j → i}. (ii) The replacement of each element in the ith row by its product with a nonzero constant α, which will be denoted by R{(α)i → i}. (iii) The replacement of each element of the jth row by the sum of β times the corresponding element in the ith row and the element in the jth row, which will be denoted by R{(β)i + j → j}. EXAMPLE 3.16

To illustrate the elementary row operations, we consider the matrix ⎡ ⎤ 1 6 4 −3 2 7 4⎦ . A = ⎣2 0 1 5 2 8 2 3 An example of an elementary row operation of type (i) performed on A is provided by R{1 → 3, 3 → 1}. This requires rows 1 and 3 to be interchanged to give the new matrix ⎡ ⎤ 5 2 8 2 3 7 4⎦ . R{1 → 3, 3 → 1}A = ⎣2 0 1 1 6 4 −3 2 An example of an elementary row operation of type (ii) performed on A is provided by R{(−3)1 → 1}. This requires each element in row 1 to be multiplied by −3 to give the new matrix ⎡ ⎤ −3 −18 −12 9 −6 0 1 7 4⎦ . R{(−3)1 → 1}A = ⎣ 2 5 2 8 2 3 An example of an elementary row operation of type (iii) performed on A is provided by R{(4)1 + 2 → 2}, which requires the elements of row 1 to be multiplied by 4 and then added to the corresponding elements of row 2 to give the new matrix ⎡ ⎤ 1 6 4 −3 2 R{(4)1 + 2 → 2}A = ⎣6 24 17 −5 12⎦ . 5 2 8 2 3 A sequence of elementary row operations performed on the augmented matrix (A, b) will lead to a different augmented matrix (A , b ). However, as this is equivalent to performing the corresponding sequence of operations on the actual equations in (16), although (A, b) and (A , b ) will look different, the interpretation of (A , b ) in terms of the solution of the system of equations in (16) will, of course, be the same as that of (A, b). It will be seen later that the purpose of carrying out these operations on a matrix is to simplify it while leaving its essential

Section 3.4

Elementary Row Operations, Elementary Matrices, and Their Connection with Matrix Multiplication

145

algebraic structure unaltered, e.g., without changing the solution x1 , . . . , xn of the corresponding system of equations. The deﬁnition that now follows is a consequence of the equivalence, in terms of equations (16), of matrix (A, b) and any matrix (A , b ) that can be derived from it by means of a sequence of elementary row operations, though the deﬁnition applies to matrices in general, and not only to augmented matrices. Row equivalence of matrices Two m × n matrices will be said to be row equivalent if one can be obtained from the other by means of a sequence of elementary row operations. Row equivalence between matrices A and B is denoted by writing A ∼ B. The row equivalence of matrices has the useful properties listed in the following theorem. THEOREM 3.5

Reﬂexive, symmetric, and transitive properties of row equivalence (i) Every m × n matrix A is row equivalent to itself (reﬂexive property). (ii) Let A and B be m × n matrices. Then if A is row equivalent to B, B is row equivalent to A (symmetric property). (iii) Let A, B, and C be m × n matrices. Then if matrix A is row equivalent to B and B is row equivalent to C, A is row equivalent to C (transitive property). Proof (i) The property is self-evident. (ii) To establish this property we must show the three elementary row operations involved are reversible. In the case of elementary row operations of type (i) the result follows from the fact that if an application of the operation R{i → j, j → i} to matrix A yields a new matrix B, an application of the operation R{ j → i, i → j} to matrix B generates the original matrix A. Similarly, in the case of elementary row operations of type (ii), if an application of the operation R{(α)i → i} to matrix A yields a new matrix B, an application of the operation R{(1/α)i → i} to matrix B reproduces the original matrix A. Finally we consider the case of elementary row operations of type (iii). If an application of the operation R{(β)i + j → j} to matrix A yields a new matrix B, an application of the operation R{(−β)i + j → j} to B returns the original matrix A. Taken together these results establish property (ii). (iii) Using property (ii) in (iii) establishes the row equivalence ﬁrst of A and B, and then of B and C, and hence of A and C, so property (iii) is proved. Let us now deﬁne what are called elementary matrices and examine the effect they have when used to premultiply a matrix. Elementary matrices An n × n elementary matrix is any matrix that is obtained from an n × n unit matrix I by performing a single elementary row operation.

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The following concise notation will be used to identify the elementary matrices that correspond to each of the three elementary row operations. the three basic types of elementary matrix

EXAMPLE 3.17

TYPE I

Ei j will denote the elementary matrix obtained from the unit matrix I by interchanging its ith and jth rows. TYPE II Ei (c) will denote the matrix obtained from the unit matrix I by multiplying its ith row by the nonzero scalar c. TYPE III Ei j (c) will denote the matrix obtained from the unit matrix I by adding c times its ith row to its jth row. Let I be the 3 × 3 unit matrix. Then ⎡ ⎤ ⎡ 1 0 0 1 0 I = ⎣0 1 0⎦ , E23 = ⎣0 0 0 0 1 0 1

⎤ ⎡ 0 1 1⎦ , E3 (4) = ⎣0 0 0 ⎡ ⎤ 1 0 0 E13 (5) = ⎣0 1 0⎦ . 5 0 1

0 1 0

⎤ 0 0⎦ , 4

and

Determinants of Elementary Matrices It follows directly from the deﬁnitions of elementary matrices that: (a) The determinant of an elementary matrix of Type I is −1, because two rows of a unit matrix have been interchanged so, in terms of Ei j , we have det(Ei j ) = −1. (b) The determinant of an elementary matrix of Type II in which a row is multiplied by a nonzero constant c is c, because a row of a unit matrix has been multiplied by c so, in terms of Ei (c), we have det(Ei (c)) = c. (c) The determinant of an elementary matrix of Type III in which c times one row has been added to another row is 1, because the addition of a multiple of a row of a unit matrix to another row leaves its value unchanged so, in terms of Ei j (c), we have det(Ei j (c)) = 1. The next theorem shows that premultiplication of a matrix A by an elementary matrix E that is conformable for multiplication performs on A the same elementary row operation that was used to generate E from I. THEOREM 3.6

Row operations performed by elementary matrices Let E be an m × m elementary matrix produced by performing an elementary row operation on the unit matrix I, and let A be an m × n matrix. Then the matrix product EA is the matrix that is obtained when the row operation that generated E from I is performed on A. Proof The proof of the theorem follows directly from the deﬁnition of a matrix product and the fact that, with the exception of the ith element in the ith row of I, which is 1, all the other elements in that row are zero. So if E is the elementary matrix obtained from I by replacing the element 1 in its ith row by α, the result of the matrix product EA will be that the elements in the ith row of A will be multiplied by α. As the form of argument used to establish the effect on A of premultiplication by P to form PA can also be employed when the other two elementary row operations are used to generate an elementary matrix E, the details will be left as an exercise.

Section 3.5

EXAMPLE 3.18

The Echelon and Row-Reduced Echelon Forms of a Matrix

Let A be the matrix

⎡

2 A = ⎣1 6

4 3 1

147

⎤ 5 7⎦ . 2

If we use the notation for elementary matrices, and introduce the elementary matrix E23 from Example 3.17 obtained by interchanging the last two rows of I3 , a routine calculation shows that ⎡ ⎤⎡ ⎤ ⎡ ⎤ 1 0 0 2 4 5 2 4 5 E23 A = ⎣0 0 1⎦ ⎣1 3 7⎦ = ⎣6 1 2⎦ , 0 1 0 6 1 2 1 3 7 so the product E23 A has indeed interchanged the last two rows of A. Similarly, again using the elementary matrices in Example 3.17, it is easily checked that E3 (4)A multiplies the elements in the third row of A by 4, while E13 (5)A adds ﬁve times the ﬁrst row of A to the last row. The main use of Theorem 3.6 is to be found in the theory of matrix algebra, and in the justiﬁcation it provides for various practical methods that are used when working with matrices. This is because when solving purely numerical problems the necessary row operations need only be performed on the rows of the augmented matrix instead of on the system of equations itself. Typical uses of the theorem will occur later after a discussion of the linear independence of equations, the deﬁnition of what is called the rank of a matrix, and the introduction of the inverse of an n × n matrix A. In this last case, the results of the theorem will be used to provide an elementary method by which what is called the inverse matrix of an n × n matrix can be obtained when n is small.

Summary

3.5

This section introduced the three types of elementary row operations that are used when manipulating matrices together with the corresponding three types of elementary matrix that can be used to perform elementary row operations.

The Echelon and Row-Reduced Echelon Forms of a Matrix We now use the row equivalence of matrices to reduce a matrix A to one of two slightly different but related standard forms called, respectively, its echelon form and its row-reduced echelon form. It is helpful to introduce these two new concepts by considering the solution of the system of m equations in n unknowns introduced in (16) and written in an equivalent but more condensed form as (A, b), where ⎡

a11 ⎢ a21 ⎢ (A, b) = ⎣ am1

a12 a22 · · am2

. . . a1n . . . a2n · · · · . . . amn

⎤ b1 b2 ⎥ ⎥, ⎦ bm

because this is equivalent to the full matrix equation Ax = b.

(19)

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Echelon and row-reduced echelon forms of a matrix echelon and row-reduced echelon forms

A matrix A is said to be in echelon form if: (i) The ﬁrst nonzero element in each row, called its leading entry, is 1; (ii) In any two successive rows i and i + 1 that do not consist entirely of zeros the leading element in the (i + 1)th row lies to the right of the leading element in ith row; (iii) Any rows that consist entirely of zeros lie at the bottom of the matrix. Matrix A is said to be in row-reduced echelon form if, in addition to conditions (i) to (iii), it is also true that (iv) In a column that contains the leading entry of a row, all the other elements are zero. In summary, this deﬁnition means that a matrix A is in echelon form if the ﬁrst nonzero entry in any row is a 1, the entry appears to the right of the ﬁrst nonzero entry in the row above, and all rows of zeros lie at the bottom of the matrix. Furthermore, matrix A is in row-reduced echelon form if, in addition to these conditions, the ﬁrst nonzero entry in any row is the only nonzero entry in the column containing that entry.

EXAMPLE 3.19

The following matrices are in echelon form: ⎡ 1 ⎡ ⎤ ⎢0 1 0 5 7 ⎢ ⎣0 0 1 0⎦ and ⎢0 ⎢ ⎣0 0 0 0 0 0 The matrices ⎡ 0 1 0 ⎢0 0 1 ⎢ ⎣0 0 0 0 0 0

2 1 0 0

0 0 1 0

5 3 1 0

⎤ 0 2⎥ ⎥, 0⎦ 0

⎡

1 ⎣0 0

0 1 0

0 0 1

1 0 0 0 0

9 2 1

1 1 0 0 0

1 2 1 0 0

⎤ 2 3⎦ , 0

1 0 5 1 0

⎤ 1 1⎥ ⎥ 2⎥ ⎥. 3⎦ 0 ⎡

and

1 ⎣0 0

0 1 0

0 0 1

⎤ 5 2⎦ 1

are in row-reduced echelon form. Rules for the reduction of a matrix to echelon form rules for ﬁnding the echelon form

The reduction of the m × n matrix to its echelon form is accomplished by means of the following steps: 1. Find the row whose ﬁrst nonzero element is furthest to the left and, if necessary, move it into row 1; if there is more than one such row, choose the row whose ﬁrst nonzero element has the largest absolute value. 2. Scale row 1 to make its leading entry 1. 3. Subtract multiples of row 1 from the m − 1 rows below it to reduce to zero all entries that lie below the leading entry in the ﬁrst column. 4. In the m − 1 rows below row 1, ﬁnd the row whose ﬁrst nonzero entry is furthest to the left and, if necessary, move it into row 2; if there is more

Section 3.5

5. 6. 7.

8.

The Echelon and Row-Reduced Echelon Forms of a Matrix

149

than one such row, choose the row whose ﬁrst nonzero entry has the largest absolute value. Scale row 2 to make its leading entry 1. Subtract multiples of row 2 from the m − 2 rows below it to reduce to zero all entries in the column below the leading entry in row 2. Continue this process until either the ﬁrst nonzero entry in the mth row is 1, or a stage is reached at which all subsequent rows consist entirely of zeros. The matrix is then in its echelon form.

Remark The selection in Step 1, and the steps corresponding to Step 4, of a row whose ﬁrst nonzero entry has the largest magnitude is made to reduce computational errors, and is not necessary mathematically. This criterion is introduced to ensure that the elimination procedure does not use an unnecessary scaling of a nonzero entry of small absolute magnitude to reduce to zero an entry of large absolute magnitude.

rules for ﬁnding the row-reduced echelon form

Rules for the reduction of a matrix to row-reduced echelon form 1. Proceed as in the reduction of a matrix to echelon form, but when steps equivalent to Step 6 are reached, in addition to subtracting multiples of the row containing a leading entry 1 from the rows below to reduce to zero all elements in the column below the leading entry, this same process must be repeated to reduce to zero all elements in the column above the leading entry. 2. An equivalent approach is ﬁrst to reduce the matrix to echelon form and then, starting with row 2 and working downwards, to subtract multiples of successive rows from the rows above to generate columns with leading entries to ones with the single nonzero entry 1. Each of these methods reduces a matrix to its row-reduced echelon form.

The row equivalence of a matrix with either its echelon or its row-reduced echelon form means that the different-looking systems of equations represented by these three matrices all have identical solution sets. The simpliﬁed structure of the row echelon and row-reduced echelon forms of the original augmented matrix makes the solution of the associated system of equations particularly easy, as can be seen from the following examples. EXAMPLE 3.20

Reduce the following matrix to its echelon and its row-reduced echelon form: ⎡

0 ⎢2 ⎢ ⎣1 1

1 4 2 3

2 8 4 6

0 2 2 1

⎤ 3 4⎥ ⎥. 2⎦ 5

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Matrices and Systems of Linear Equations

⎡

0 ⎢2 Solution ⎢ ⎣1 1

1 4 2 3

2 8 4 6

0 2 2 1 ⎡ 1 ⎢ ∼ ⎢0 divide row 1 by 2 ⎣1 1

⎤ ⎡ ⎤ 3 2 4 8 2 4 ∼ ⎢ ⎥ 4⎥ ⎥ switch rows ⎢0 1 2 0 3⎥ 2⎦ 2 and 1 ⎣1 2 4 2 2⎦ 5 1 3 6 1 5 ⎤ ⎡ 2 4 1 2 1 ∼ ⎢0 1 2 0 3⎥ ⎥ subtract row 1 ⎢ 2 4 2 2⎦ from rows 3 and 4 ⎣0 3 6 1 5 0

⎡ 1 ∼ ⎢0 subtract row 2 ⎢ ⎣0 from row 4 0

2 1 0 0

4 2 0 0

2 1 0 1

4 2 0 2

1 0 1 0

⎤ 2 3⎥ ⎥ 0⎦ 3

⎤ 2 3⎥ ⎥ 0⎦ 0

1 0 1 0

and the matrix is now in echelon form. Having already obtained the echelon form of the matrix, we now use it to obtain the row-reduced echelon form. We already have ⎡ ⎤ ⎡ ⎤ 0 1 2 0 3 1 2 4 1 2 ∼ ⎢2 4 8 2 4⎥ ⎢0 1 2 0 3⎥ ⎢ ⎥ ⎢ ⎥subtract twice row 2 ⎣1 2 4 2 2⎦ ∼ ⎣0 0 0 1 0⎦ from row 1 1 3 6 1 5 0 0 0 0 0 ⎤ ⎡ ⎡ ⎤ 1 0 0 0 −4 1 0 0 1 −4 ∼ ⎢ ⎢0 1 2 0 3⎥ 3⎥ ⎥, ⎢ ⎥ subtract row 3 ⎢0 1 2 0 ⎦ ⎣ ⎣0 0 0 1 0 0 0 1 0⎦ 0 from row 1 0 0 0 0 0 0 0 0 0 0 and the matrix is now in its row-reduced echelon form. EXAMPLE 3.21

Solve the system of equations x2 + 2x3 = 3 2x1 + 4x2 + 8x3 + 2x4 = 4 x1 + 2x2 + 4x3 + 2x4 = 2 x1 + 3x2 + 6x3 + x4 = 5. Solution The augmented matrix (A, b) for this system is the matrix in Example 3.20 that was shown to be equivalent to the row-reduced echelon form ⎤ ⎡ 1 0 0 0 −4 ⎢ ⎥ ⎢0 1 2 0 3⎥ ⎢ ⎥. ⎢0 0 0 1 0⎥ ⎦ ⎣ 0 0 0 0 0 If we recall that the ﬁrst four columns of this matrix contain the coefﬁcients of x1 , x2 , x3 , and x4 , while the last column contains the nonhom*ogeneous terms, the matrix implies the much simpler system of equations x4 = 0,

x2 + 2x3 = 3,

and

x1 = −4.

Section 3.5

The Echelon and Row-Reduced Echelon Forms of a Matrix

151

As there are only three equations connecting four unknowns, it follows that in the second equation either x2 or x3 can be assigned arbitrarily, so if we choose to set x3 = k (an arbitrary number), the solution set of the system in terms of the parameter k becomes x1 = −4,

x2 = 3 − 2k,

x3 = k,

and

x4 = 0.

The same solution could have been obtained from the echelon form of the matrix ⎡ ⎤ 1 2 4 1 2 ⎢ ⎥ ⎢0 1 2 0 3⎥ ⎢ ⎥ ⎢0 0 0 1 0⎥ , ⎣ ⎦ 0 0 0 0 0 because this implies the system of equations x1 + 2x2 + 4x3 + x4 = 2,

x2 + 2x3 = 3,

and

x4 = 0.

Starting from the last equation we ﬁnd x4 = 0, and setting x3 = k in the middle equation gives, as before, x2 = 3 − 2k. Finally, substituting x2 , x3 , and x4 in the ﬁrst equation gives x1 = −4. This process of arriving at a solution of a system of equations whose coefﬁcient matrix is in upper triangular form is called back substitution. It should be noticed that the system of equations would have had no solution if the row-reduced echelon form had been ⎤ ⎡ 1 0 0 0 −4 ⎢ ⎥ ⎢0 1 2 0 3⎥ ⎥. ⎢ ⎢0 0 0 1 0⎥ ⎦ ⎣ 5 0 0 0 0

back substitution

This is because although the equations corresponding to the ﬁrst three rows of this matrix would have been the same as before, the fourth row implies 0 = 5, which is impossible. This corresponds to a system of equations where one equation contradicts the others, so that no solution is possible.

Summary

This section deﬁned two related types of fundamental matrix that can be obtained from a general matrix by means of elementary row operations. The ﬁrst was a reduction to echelon form and the second, derived from the ﬁrst form, was a reduction to row-reduced echelon form. Each of the reduced forms retains the essential properties of the original matrix, while simplifying the task of solving the associated system of linear algebraic equations.

EXERCISES 3.5 Let P, Q, and R be the matrices ⎡

3 P = ⎣0 0

0 1 0

⎤

0 0⎦ , 1

⎡

0 Q = ⎣0 1

0 1 0

⎤ 1 0⎦ , 0

⎡ 1 R = ⎣0 0

2 1 0

⎤

0 0⎦ . 1

In Exercises 1 through 4 verify by direct calculation that (a) premultiplication by P multiplies row 1 by 3; (b) premultiplication by Q interchanges rows 1 and 3; and

(c) premultiplication by R adds twice row 2 to row 1. ⎤ ⎤ ⎡ ⎡ 4 0 1 2 1 1 3. ⎣2 0 3⎦. 1. ⎣1 3 0⎦. 1 2 5 1 2 4 ⎡ ⎤ ⎡ ⎤ 9 1 3 1 −1 2 1 3⎦. 4. ⎣2 4 7⎦. 2. ⎣2 1 2 2 3 0 7

152

Chapter 3

Matrices and Systems of Linear Equations

In Exercises 5 and 6 write down the required elementary matrices. 5. When I is the 3 × 3 unit matrix, write down E12 , E2 (3), and E12 (6). 6. When I is the 4 × 4 unit matrix, write down E41 , E4 (3), and E23 (4). In Exercises 7 through 12, reduce the given matrices to their row-reduced echelon form. ⎤ ⎡ ⎤ ⎡ 3 2 1 1 0 3 4 1 ⎢ 2 5 1 2⎥ 7. ⎣3 1 2 2⎦. ⎥ ⎢ ⎥ 1 5 2 1 10. ⎢ ⎢3 1 1 3⎥. ⎣ 0 1 3 4⎦ ⎤ ⎡ 4 1 3 1 3 2 1 3 1 8. ⎣2 1 1 2 0⎦. ⎤ ⎡ 2 2 4 1 4 3 2 1 1 0 ⎢1 1 3 2 1⎥ ⎤ ⎡ ⎥ 11. ⎢ 4 −2 2 3 1 ⎣3 2 5 1 4⎦. ⎦ ⎣ 0 0 3 2 . 9. 2 1 0 3 1 2 4 1 2 5 1 ⎤ ⎡ 3 2 3 2 12. ⎣3 7 1 −1⎦. 5 1 1 3

3.6

In Exercises 13 through 18, reduce the given augmented matrices to their row-reduced echelon form and, where appropriate, use the result to solve the related system of equations in terms of an appropriate number of the unknowns x1 , x2 , . . . . ⎤ ⎤ ⎡ ⎡ 2 3 1 0 2 1 0 2 1 ⎥ ⎥ ⎢ ⎢ ⎥ ⎢1 3 1 4 2⎥ 13. ⎢ ⎥ ⎢ ⎣1 3 1 4⎦. 16. ⎢ ⎥. ⎢2 1 2 3 1⎥ 6 9 4 8 ⎦ ⎣ ⎤ ⎡ 4 7 4 11 7 2 1 1 0 ⎥ ⎢ ⎤ ⎡ ⎥ 14. ⎢ 1 0 1 0 2 0 ⎣2 3 1 4⎦. ⎥ ⎢ ⎢2 2 6 0 6 0⎥ 4 9 4 8 ⎥ ⎢ 17. ⎢ ⎥. ⎡ ⎤ ⎥ ⎢ 1 0 1 1 6 0 0 2 1 1 1 ⎦ ⎣ ⎢ ⎥ ⎢ ⎥ 2 3 2 7 0 8 15. ⎣1 3 1 2 1⎦. ⎡ ⎤ 3 9 4 3 0 3 0 6 0 6 ⎢ ⎥ ⎥ 18. ⎢ ⎣1 1 5 1 9 ⎦. 2 0 4 2 10

Row and Column Spaces and Rank The reduction of an m × n matrix A to either its echelon or its row-reduced echelon form will produce a row of zeros whenever the row is a linear combination of some (or all) of the rows above it. So if an echelon form contains r ≤ m nonzero rows, it follows that these r rows are linearly independent, and hence that the remaining m − r rows are linearly dependent on the ﬁrst r rows. The number r is called the row rank of matrix A. This means that if the r nonzero rows of an echelon form u1 , u2 , . . . , ur are regarded as n element row vectors belonging to a vector space Rn , the r vectors will span a subspace of Rn . Consequently, as these vectors form a basis for this subspace, every vector in it can be expressed as a linear combination of the form a1 u1 + a2 u2 + · · · + ar ur ,

row and column ranks and spaces

where the a1 , a2 , . . . , ar are scalar constants. This subspace of Rn is called the row space of matrix A. It should be remembered that the vectors forming a basis for a space are not unique, and that any basis can be transformed to any other one by means of suitable linear combinations of the vectors involved. So although the r nonzero rows of the echelon form of A and those of its row-reduced echelon form look different, they are equivalent, and each forms a basis for the row space of A. Just as there may be linear dependence between the rows of A, so also may there be linear dependence between its columns. If s of the n columns of an m × n matrix A are linearly independent, the number s is called the column rank of matrix A. When the s nonzero columns v1 , v2 , . . . , vs are regarded as m element column vectors belonging to a vector space Rm, these vectors will span a subspace of Rm.

Section 3.6

Row and Column Spaces and Rank

153

Consequently, as these vectors form a basis for this subspace, every vector in it can be expressed as a linear combination of the form b1 v1 + b2 v2 + · · · + bs vs , where the b1 , b2 , . . . , bs are scalar constants. This subspace of Rm is called the column space of matrix A. The connection between the row and column ranks of a matrix is provided by the following theorem. THEOREM 3.7 equality of the rank of a matrix and its transpose

The equality of the row and column ranks Let A be any matrix. Then the row rank and column rank of A are equal. Proof Let an m × n matrix A have row rank r . Then in its row-reduced echelon form it must contain r columns v1 , v2 , . . . , vr , in each of which only the single nonzero entry 1 appears. Call these columns the leading columns of the row-reduced echelon form, and let them be arranged so that in the ith column v, the entry 1 appears in the ith row. The row-reduced echelon form of A will comprise the leading columns arranged in numerical order with, possibly, columns between the ith and the (i + 1)th leading columns in which zero elements lie below the ith row but nonzero elements may occur above it. Furthermore, there may be columns to the right of column vr in which zero elements lie below the r th row but nonzero elements may lie above it. By subtracting suitable multiples of the leading columns from any columns that lie between them or to the right of vr , it is possible to reduce all entries in such columns to zero. Consequently, at the end of this process, the only remaining nonzero columns will be the r linearly independent leading columns v1 , v2 , . . . , vr . This establishes the equality of the row and column ranks. Rank The rank of matrix A, denoted by rank (A), is the value common to the row and column ranks of A.

THEOREM 3.8

Rank of A and AT Let A be any matrix. Then rank (A) = rank (AT ). Proof The columns of A are the rows of AT , so the column rank of A is the row rank of AT . However, by Theorem 3.7 these two ranks are equal, so the result is proved.

EXAMPLE 3.22

Let

⎡

1 ⎢2 A=⎢ ⎣1 1

0 1 0 0

3 7 3 3

0 0 2 0

4 10 6 4

⎤ 0 1⎥ ⎥. 4⎦ 0

154

Chapter 3

Matrices and Systems of Linear Equations

Then the row-reduced echelon form of A is B (B ∼ A) ⎡ ⎤ 1 0 3 0 4 0 ⎢0 1 1 0 2 1⎥ ⎥ B=⎢ ⎣0 0 0 1 1 2⎦, 0 0 0 0 0 0 showing that the number of leading columns is 3, so the row rank of A is 3, and hence its rank is 3. Three row vectors spanning a subspace of R6 , and so forming a basis for this subspace, are the three nonzero row vectors in this 4 × 6 matrix, u1 = [1, 0, 3, 0, 4, 0],

u2 = [0, 1, 1, 0, 2, 1],

The row-reduced echelon form of AT is ⎡ 1 0 ⎢0 1 ⎢ ⎢0 0 ⎢ ⎢0 0 ⎢ ⎣0 0 0 0

and

u3 = [0, 0, 0, 1, 1, 2].

⎤ 1 0⎥ ⎥ 0⎥ ⎥, 0⎥ ⎥ 0⎦ 0

0 0 1 0 0 0

showing that the number of leading columns is 3, conﬁrming as would be expected that the column rank of A (the row rank of AT ) is 3. The three row vectors of AT spanning a subspace of R4 , and so forming a basis for this subspace, are the three nonzero rows in this 6 × 4 matrix, namely, [1, 0, 0, 1],

[0, 1, 0, 0],

and

[0, 0, 1, 0].

The three linearly independent column vectors of A are obtained by transposing these vectors to obtain ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 0 0 ⎢0⎥ ⎢1⎥ ⎢0⎥ ⎢ ⎥ ⎢ ⎥ ⎥ v1 = ⎢ ⎣0⎦ , v2 = ⎣0⎦ , and v3 = ⎣1⎦ . 1 0 0

Summary

This section introduced the important algebraic concepts of the rank of a matrix, and of the row and column spaces of a matrix. The equality of the row and column ranks of a matrix was then proved. It will be seen later that the rank of a matrix plays a fundamental role when we seek a solution of a linear algebraic system of equations.

EXERCISES 3.6 In Exercises 1 through 14 ﬁnd the row-reduced echelon form of the given matrix, its rank, a basis for its row space, and a basis for its column space. ⎤ ⎡ ⎡ ⎤ 1 3 2 1 1 3 1 0 1 1 ⎢2 0 2 1⎥ 1. ⎣2 2 1 0 0 1⎦ . ⎥ 2. ⎢ ⎣1 0 4 5⎦ . 0 2 1 4 1 3 0 1 2 4

⎡ 3 ⎢4 3. ⎢ ⎣2 3 2 4. 1

0 2 1 0 0 2 0 0

⎤ 6 0 11 3⎥ ⎥. 4 0⎦ 6 3

3 2

0 0

1 1

0 4

2 1

⎡

4 . 2

1 5. ⎣2 3 ⎡ 3 6. ⎣1 8

2 3 2 2 2 8

⎤ 3 1⎦ . 1 ⎤ 4 2 ⎦. 12

Section 3.7 ⎡

1 ⎢3 7. ⎢ ⎣2 0 ⎡ 2 ⎢1 ⎢ 8. ⎢ ⎢1 ⎣3 2

3.7

3 0 3 3

⎤ 4 4⎥ ⎥. 1⎦ 5

1 1 2 3 3

3 0 1 4 1

⎤ 1 3⎥ ⎥ 0⎥ ⎥. 1⎦ 3

⎡ 1 9. ⎣2 3

2 1 3

1 0 1

4 5 1 2 5 7

⎡ 2 10. ⎣0 2

4 2 6

0 1 1

10 3 13

The Solution of hom*ogeneous Systems of Linear Equations ⎤ 7 1⎦ . 8 ⎤ 8 1⎦ . 9

⎡

0 −1

⎢ 0 ⎢0 11. ⎢ ⎢0 0 ⎣ 1 0 ⎡ 1 0 ⎢ 1 −1 ⎢ 12. ⎢ ⎢2 5 ⎣ 1 3

4

3

⎤

⎥ 1 2⎥ ⎥. 0 −1 ⎥ ⎦ 0 0 ⎤ 0 0 ⎥ 0 0⎥ ⎥. −1 0 ⎥ ⎦ 2 1

⎡

1

⎤

7

2

5

7⎥ ⎦.

3

5

3

1

1

2

3

3

4

4 5

7

⎢ 13. ⎣ 0 0 ⎡ 1 ⎢2 ⎢ ⎢ 1 14. ⎢ ⎢ ⎢3 ⎣

4

155

⎤

1⎥ ⎥ ⎥ 2⎥. ⎥ 3⎥ ⎦ 5

The Solution of hom*ogeneous Systems of Linear Equations Having now introduced the echelon and row-reduced echelon forms of an m × n matrix A, we are in a position to discuss the nature of the solution set of the system of linear equations

hom*ogeneous and nonhom*ogeneous systems of equations

a11 x1 + a12 x2 + · · · + a1n xn = b1 a21 x1 + a22 x2 + · · · + a2n xn = b2 · · · · · · · · · am1 x1 + am2 x2 + · · · + amn xn = bm,

(20)

which will be nonhom*ogeneous when at least one of the terms bi on the right is nonzero, and hom*ogeneous when b1 = b2 = · · · = bm = 0. In this section we will only consider hom*ogeneous systems. Rather than working with the full system of hom*ogeneous equations corresponding to bi = 0, i = 1, 2, . . . , m in (20), it is more convenient to work with its coefﬁcient matrix ⎡

a11 ⎢ a21 A=⎢ ⎣ am1

a12 a22 . . . am2

. . . .

. . . .

⎤ . a1n . a2n ⎥ ⎥, ⎦ . . amn

(21)

which contains all the information about the system. The coefﬁcients in the ﬁrst column of A are multipliers of x1 , those in the second column are multipliers of x2 , . . . , and those in the nth column are multipliers of xn . Denote by AE either the echelon or the row-reduced echelon form of the coefﬁcient matrix A. Then, as elementary row operations performed on a coefﬁcient matrix are equivalent in all respects to performing the same operations on the corresponding full system of equations, the solution set of the matrix equation Ax = 0

(22)

will be the same as the solution set of an echelon form of the hom*ogeneous equations AE x = 0.

(23)

156

Chapter 3

Matrices and Systems of Linear Equations

trivial solution

It is obvious that x = 0, corresponding to x = [0, 0, . . . , 0]T , is always a solution of (22) and, of course of (23), and it is called the trivial solution of the hom*ogeneous system of equations. To discover when nontrivial solutions exist it is necessary to work with the equivalent echelon form of the equations given in (23). If rank(A) = r , the ﬁrst r rows of AE will be nonzero rows, and the last m − r rows will be zero rows. As there are m rows in A, we must consider the three separate cases (a) m < n, (b) m = n, and (c) m > n. Case (a): m < n. In this case there are more variables than equations. As rank(A) = r , and there are m equations, it follows that r = rank(A) ≤ m. The system in (22) will thus contain only r linearly independent equations corresponding to the ﬁrst r rows of AE . So working with system (23), we see that r of the variables x1 , x2 , . . . , xn will be determined in terms of the remaining m − r variables regarded as parameters (see Example 3.23). Case (b): m = n. In this case the number of variables equals the number of equations. If rank(A) = r < n we have the same situation as in Case (a), and the variables x1 , x2 , . . . , xn will be determined by the system of equations in (23) in terms of the remaining m − r variables regarded as parameters. However, if r = n, only the trivial solution x = 0 is possible, because in this case AE becomes the unit matrix In , from which it follows directly that x = 0. Case (c): m > n. In this case the number of equations exceeds the number of variables and r = rank(A) ≤ n. This is essentially the same situation as in Case (b), because if r = rank(A) < n, the variables x1 , x2 , . . . , xn will be determined by the system of equations in (22) in terms of the remaining m − r variables regarded as parameters, while if rank(A) = n only the trivial solution x = 0 is possible. The practical determination of solution sets to hom*ogeneous systems of linear equations is illustrated in the next example.

EXAMPLE 3.23

Find the solution sets of the hom*ogeneous systems of linear equations with coefﬁcient matrices given by: ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 2 3 6 1 1 2 1 7 0 1 3 2 ⎢1 4 2 2⎥ ⎥ (a) A = ⎣3 6 4 24 3⎦, (b) A = ⎣2 1 0⎦, (c) A = ⎢ ⎣4 11 10 5⎦, 1 4 4 12 3 1 2 1 1 0 1 1 ⎡ ⎤ 1 4 1 2 ⎡ ⎤ ⎢1 3 0 1⎥ 1 2 3 1 4 3 ⎢ ⎥ ⎥ ⎣ ⎦ (d) A = ⎢ ⎢2 1 1 1⎥ , (e) A = 0 1 3 0 1 5 ⎣4 9 3 5⎦ 3 1 2 3 1 4 5 5 2 3 Solution (a) The row-reduced echelon form of the matrix is ⎡ ⎤ 1 0 0 8 3 AE = ⎣0 1 0 −2 −3⎦ , 0 0 1 3 3

Section 3.7

The Solution of hom*ogeneous Systems of Linear Equations

157

showing that rank(A) = 3. This corresponds to the following three equations between the ﬁve variables x1 , x2 , x3 , x4 , and x5 : x1 + 8x4 + 3x5 = 0,

x2 − 2x4 − 3x5 = 0,

x3 + 3x4 + 3x5 = 0.

and

Letting x4 = α and x5 = β be arbitrary numbers (parameters) allows the solution set to be written x1 = −8α − 3β,

x2 = 2α + 3β,

x3 = −3α − 3β,

x4 = α,

x5 = β.

(b) The row-reduced echelon form of the matrix is ⎡ ⎤ 1 0 0 AE = ⎣0 1 0⎦ , 0 0 1 showing that rank(A) = 3. This corresponds to the trivial solution x1 = x2 = x3 = 0. (c) The row-reduced echelon form of the matrix is ⎡ ⎤ 1 0 0 20/13 ⎢0 1 0 5/13 ⎥ ⎥ AE = ⎢ ⎣0 0 1 −7/13 ⎦ , 0 0 0 0 showing that rank(A) = 3. This corresponds to the solution set x1 + (20/13)x4 = 0, x2 + (5/13)x4 = 0, and x3 − (7/13)x4 = 0. Setting x4 = k, an arbitrary number (a parameter), shows the solution set to be given by x1 = −(20/13)k,

x2 = −(5/13)k,

x3 = (7/13)k,

and

x4 = k.

(d) The row-reduced echelon form of the matrix is ⎡ ⎤ 1 0 0 0 ⎢0 1 0 1/3⎥ ⎢ ⎥ ⎥ AE = ⎢ ⎢0 0 1 2/3⎥ , ⎣0 0 0 0 ⎦ 0 0 0 0 showing that rank(A) = 3. This corresponds to the following three equations for the four variables x1 , x2 , x3 , and x4 : x1 = 0,

x2 + (1/3)x4 = 0,

and

x3 + (2/3)x4 = 0.

Setting x4 = k, an arbitrary number (a parameter), shows the solution set to be given by x1 = 0,

x2 = −k/3 = 0,

x3 = −2k/3,

and

x4 = k.

(e) The row-reduced echelon form of the matrix is ⎡ ⎤ 1 0 0 1 −1/4 1/2 AE = ⎣0 1 0 0 13/4 −5/2⎦ , 0 0 1 0 −3/4 5/2 showing that rank(A) = 3. This corresponds to the following three equations for the six variables x1 to x6 : x1 + x4 − (1/4)x5 + (1/2)x6 = 0,

x2 + (13/4)x5 − (5/2)x6 = 0 x3 − (3/4)x5 + (5/2)x6 = 0.

158

Chapter 3

Matrices and Systems of Linear Equations

Setting x4 = α, x5 = β, and x6 = γ , where α, β, and γ are arbitrary numbers (parameters), shows the solution set to be given by x1 = −α + (1/4)β − (1/2)γ , x2 = −(13/4)β + (5/2)γ , x4 = α, x5 = β, and x6 = γ .

Summary

x3 = (3/4)β − (5/2)γ

This section made use of the rank of a matrix to determine when a nontrivial solution of a linear system of hom*ogeneous linear algebraic equations exists and, when it does, its precise form.

EXERCISES 3.7 In Exercises 1 through 10, use the given form of the matrix A to ﬁnd the solution set of the associated hom*ogeneous linear system of equations Ax = 0. ⎤ ⎡ ⎤ ⎡ 1 2 4 1 1 3 2 1 1 ⎢0 3 1 3⎥ 1. ⎣1 1 0 1 2⎦. ⎥ 3. ⎢ ⎣1 4 1 3⎦. 0 1 2 1 3 ⎤ ⎡ 2 6 5 4 1 2 0 1 1 ⎤ ⎡ ⎢0 3 1 0 1⎥ 1 2 1 0 ⎥ 2. ⎢ ⎣2 0 2 0 1⎦. ⎢2 1 0 1⎥ ⎥ 4. ⎢ 1 0 3 1 1 ⎣0 3 5 1⎦. 1 0 1 5

3.8

⎡

1 ⎢2 ⎢ 5. ⎢ ⎢1 ⎣3 2 ⎡ 2 ⎢1 ⎢ 6. ⎢ ⎢0 ⎣1 0 ⎡ 1 ⎢0 ⎢ 7. ⎣ 1 2

3 1 0 1 3

⎡

⎤ 4 3⎥ ⎥ 2⎥ ⎥. 1⎦ 1

1 2 1 3 4

1 3 4 1 1

⎤ 3 0⎥ ⎥ 2⎥ ⎥. 2⎦ 1

5 1 2 3

2 4 1 0

2 1 0 1

1 0 0 1

3 1 2 0

⎤ 2 1⎥ ⎥. 0⎦ 2

1 ⎢2 ⎢ 8. ⎣ 5 2 ⎡ 1 9. ⎣2 0 ⎡ 1 ⎢2 10. ⎢ ⎣0 1

4 1 6 1

1 3 7 0

⎤ 0 1⎥ ⎥. 2⎦ 1

1 3 1

5 1 0

0 2 1

3 5 1 0

2 1 2 3

1 0 0 1

⎤ 0 1 1 3⎦. 3 0 ⎤ 1 2⎥ ⎥. 3⎦ 2

The Solution of Nonhom*ogeneous Systems of Linear Equations We now turn our attention to the solution of the nonhom*ogeneous system of equations in (20) that may be written in the matrix form Ax = b,

(24)

where A is an m × n matrix and b is an m × 1 nonzero column vector. In many respects the arguments we now use parallel the ones used when seeking the form of the solution set for a hom*ogeneous system, but there are important differences. This time, rather than working with the matrix A, we must work with the augmented matrix (A, b) and use elementary row operations to transform it into either an echelon or a row-reduced echelon form that will be denoted by (A, b)E . When this is done, system (24) and the echelon form corresponding to (A, b)E will, of course, each have the same solution set. It is important to recognize that rank(A) is not necessarily equal to rank (A, b)E , so that in general rank(A) ≤ rank((A, b)E ). The signiﬁcance of this observation will become clear when we seek solutions of systems like (24).

Section 3.8

The Solution of Nonhom*ogeneous Systems of Linear Equations

159

Case (a): m < n. In this case there are more variables than equations, and it must follow that rank((A, b)E ) ≤ m. If rank(A) = rank((A, b)E ) = r , it follows that r of the equations in (24) are linearly independent and m − r are linear combinations of these r equations. This means that the ﬁrst r rows of (A, b)E are linearly independent while the last m − r rows are rows of zeros. Thus, r of the variables x1 to xn will be determined by the equations corresponding to these r nonzero rows, in terms of the remaining m − r variables as parameters. It can happen, however, that rank(A) = r < rank((A, b)E ), and then the situation is different, because one or more of the rows following the r th row will have zeros in its ﬁrst n entries and nonzero numbers for their last entries. When interpreted as equations, these will imply contradictions, because they will assert expressions such as 0 = c with c = 0 that are impossible. Thus, no solution will exist if rank(A) = rank((A, b)E ). Case (b): m = n. In this case the number of variables equals the number of equations, and it must follow that rank((A, b)E ) ≤ n. The situation now parallels that of Case (a), because if rank(A) = rank((A, b)E ) = r < m, then r of the equations in (24) will be linearly independent, while m − r will be linear combinations of these r equations. So, as before, the ﬁrst r rows of (A, b)E will be linearly independent while the last m − r rows will be rows of zeros. Thus, r of the variables x1 to xn will be determined by the equations corresponding to these r nonzero rows in terms of the remaining m − r variables as parameters. In the case r = n, the solution will be unique, because then AE = I. Finally, if rank(A) = rank((A, b)E ), it follows, as in Case (a), that no solution will exist. Case (c): m > n. In this case there are more equations than variables, and it must follow that rank((A, b)E ) ≤ n. If rank(A) = rank((A, b)E ) = r , it follows, as in Case (b), that r of the equations in (24) are linearly independent while m − r are linear combinations of these r equations. Thus, again, the ﬁrst r rows of (A, b)E will be linearly independent while the last m − r rows will be rows of zeros. Consequently, r of the variables x1 to xn will be determined by the equations corresponding to these r nonzero rows in terms of the remaining m − r variables as parameters. If rank(A) = rank((A, b)E ), then as before no solution will exist. These considerations bring us to the deﬁnition of consistent and inconsistent systems of nonhom*ogeneous equations, with consistent systems having solutions, sometimes in terms of parameters, and inconsistent systems have no solution.

consistent and inconsistent systems

Consistent and inconsistent nonhom*ogeneous systems The nonhom*ogeneous system Ax = b is said to be consistent when it has a solution; otherwise, it is said to be inconsistent.

As with hom*ogeneous systems, the practical determination of solution sets of nonhom*ogeneous systems of linear equations will be illustrated by means of examples.

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EXAMPLE 3.24

Find the solution sets for each of the following augmented matrices (A, b), where the matrices A are those given in Example 3.23. ⎡ ⎤ ⎡ ⎤ 1 2 1 7 0 1 1 3 2 2 ⎥ ⎢ ⎥ ⎢ 1⎦ (a) (A, b) = ⎣3 6 4 24 3 0⎦ (b) (A, b) = ⎣2 1 0 1 2 1 −3 1 4 4 12 3 3 ⎤ ⎡ 1 4 1 2 2 ⎡ ⎤ 2 3 6 1 2 ⎢1 3 0 1 0⎥ ⎥ ⎢ ⎥ ⎢ ⎢1 4 ⎥ ⎢ 2 2 3⎥ ⎢ ⎥ ⎢ (c) (A, b) = ⎢ (d) (A, b) = ⎢2 1 1 1 3⎥ ⎥ ⎥ ⎥ ⎢ ⎣4 11 10 5 1⎦ ⎣4 9 3 5 7⎦ 1 0 1 1 2 5 5 2 3 0 ⎡ ⎤ 1 2 3 1 4 3 −2 ⎢ ⎥ 0⎦ . (e) (A, b) = ⎣0 1 3 0 1 5 3 1 2 3 1 4 1 Solution (a) In this case, ⎡ 1 0 0 ⎢ (A, b)E = ⎣0 1 0 0 0 1

8

3

−2

−3

3

3

−7

⎤

⎥ 11/2⎦ . −3

As rank(A, b)E = 3, and the rank of matrix A is the rank of the matrix formed by deleting the last column of (A, b)E , it follows that rank(A) = 3. So rank(A, b)E = rank(A), showing the equations to be consistent, so they have a solution. If we remember that the ﬁrst column contains the coefﬁcients of x1 , the second column the coefﬁcients of x2 , . . . , and the ﬁfth column the coefﬁcients of x5 , while the last column contains the nonhom*ogeneous terms, we can see that the matrix (A, b)E is equivalent to the three equations x1 + 8x4 + 3x5 = −7,

x2 − 2x4 − 3x5 = 11/2,

x3 + 3x4 + 3x5 = −3.

So, if we set x4 = α and x5 = β, with α and β arbitrary numbers (parameters), the solution set becomes x1 = −8α − 3β − 7, x2 = 2α + 3β + 11/2, x4 = α and x5 = β.

x3 = −3α − 3β − 3,

(b) In this case, ⎡

1

⎢ (A, b)E = ⎣0 0

1

1

⎤ 9 ⎥ −17⎦ . 22

Here A is a 3 × 3 matrix and rank(A) = rank((A, b)E ) = 3, so the equations are consistent and the solution is unique. The solution set is seen to be x1 = 9,

x2 = −17,

and

x3 = 22.

Section 3.8

The Solution of Nonhom*ogeneous Systems of Linear Equations

161

(c) In this case, ⎡

1 0 0 ⎢0 1 0 ⎢ (A, b)E = ⎢ ⎢0 0 0 ⎣0 0 0 0 0 0

20/13 5/13 −7/13 0 0

⎤ 0 0⎥ ⎥ 0⎥ ⎥. 1⎦ 0

This system has no solution because the equations are inconsistent. This follows from the fact that rank(A) = 3, as can be seen from the ﬁrst four columns, while the ﬁve columns show that rank((A, b)E ) = 4, so that rank(A) = rank((A, b)E ). The inconsistency can be seen from the contradiction contained in the last row, which asserts that 0 = 1. (d) In this case ⎡ ⎤ 1 0 0 0 0 ⎢0 1 0 1/3 0⎥ ⎢ ⎥ ⎥ (A, b)E = ⎢ ⎢0 0 1 2/3 0⎥ . ⎣0 0 0 0 1⎦ 0 0 0 0 0 This system also has no solution because the equations are inconsistent. This follows from the fact that rank(A) = 3 and rank((A, b)E ) = 4, so that rank(A) = rank((A, b)E ). The inconsistency can again be seen from the contradiction in the last row, which again asserts that 0 = 1. (e) In this case ⎡

1

⎢ (A, b)E = ⎣0 0

1 −1/4

1

13/4

−5/2

1

0 −3/4

5/2

1/2

5/8

⎤

⎥ −21/8⎦ , 7/8

showing that rank(A) = rank((A, b)E ) = 3, so the equations are consistent. Reasoning as in (a) and setting x4 = α, x5 = β, and x6 = γ , with α, β, and γ arbitrary numbers (parameters), shows the solution set to be given by x1 = −α + (1/4)β − (1/2)γ + 5/8, x2 = −(13/4)β + (5/2)γ − 21/8, x3 = (3/4)β − (5/2)γ + 7/8, x4 = α, x5 = β, x6 = γ .

general solution of a nonhom*ogeneous system

A comparison of the corresponding solution sets in Examples 3.23 and 3.24 shows that whenever the nonhom*ogeneous system has a solution, it comprises the sum of the solution set of the corresponding hom*ogeneous system, containing arbitrary parameters, and numerical constants contributed by the nonhom*ogeneous terms. This is no coincidence, because it is a fundamental property of nonhom*ogeneous linear systems of equations. The combination of solutions comprising the sum of a solution of the hom*ogeneous system Ax = 0 containing arbitrary constants, and a particular ﬁxed solution of the nonhom*ogeneous system Ax = b that is free from arbitrary constants, is called the general solution of a nonhom*ogeneous system. The result is important, so it will be recorded as a theorem.

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THEOREM 3.9

General solution of a nonhom*ogeneous system The nonhom*ogeneous system of equations Ax = b for which rank(A) = rank ((A, b)E ) has a general solution of the form x = xH + x P , where xH is the general solution of the associated hom*ogeneous system AxH = 0 and xP is a particular (ﬁxed) solution of the nonhom*ogeneous system AxP = b. Proof Let x be any solution of the nonhom*ogeneous system Ax = b, and let xP be a solution of the nonhom*ogeneous system AxP = b that contains no arbitrary constants (a ﬁxed solution). Then, as the equations are linear, A(x − xP ) = Ax − AxP = b − b = 0, showing that the difference xD = x − xP is itself a solution of the hom*ogeneous system. Consequently, all solutions of the nonhom*ogeneous system are contained in the solution set of the hom*ogeneous system to which xD belongs, and the theorem is proved.

Summary

This section used the rank of a matrix to determine when a solution of a linear system of nonhom*ogeneous equations exists and to determine its precise form. If the ranks of a matrix and an augmented matrix are equal, it was shown that a solution exists, furthermore, if there are n equations and the rank r < n, then r unknowns can be expressed in terms of arbitrary values assigned to the remaining n − r unknowns. The system was shown to have a unique solution when r = n, and no solution if the ranks of the matrix and the augmented matrix are different.

EXERCISES 3.8 In Exercises 1 through 10 write down a system of equations with an appropriate number of unknowns x1 , x2 , . . . corresponding to the augmented matrix. Find the solution set when the equations are consistent, and state when the equations are inconsistent. ⎤ ⎡ ⎡ ⎤ 1 −2 1 3 11 1 3 1 1 0 ⎥ ⎢ ⎢ ⎥ ⎢1 1 3 2 1 ⎥ ⎢0 3 −2 1 11⎥ ⎢ ⎥ ⎢ ⎥ 3. ⎢ ⎥. ⎢ ⎥ ⎢1 1 0 3 1 ⎥ 1. ⎢2 1 0 4 23⎥ . ⎦ ⎣ ⎢ ⎥ ⎢ ⎥ 2 −1 2 21⎦ 2 0 2 1 0 ⎣3 1 −1 ⎡ 2 1 3 ⎢ 2. ⎢ ⎣0 1 4 3 0 0

3 1 1 2

2 4 ⎤ 1 ⎥ 1⎥ ⎦. 1

⎡

1 ⎢ 4. ⎢ ⎣2 5

4

2

3

3

1

4

8

5

4

⎤

⎥ 2⎥ ⎦. 8

⎤ ⎡ 1 −1 2 −1 −4 ⎥ ⎢ 3 1 2 12⎥ ⎢2 ⎥ ⎢ ⎥ ⎢ 2 −2 3 15⎥ . 5. ⎢1 ⎢ ⎥ ⎢ ⎥ 1 −1 1 11⎦ ⎣3 1 ⎡ 1 ⎢ ⎢2 ⎢ ⎢ 6. ⎢0 ⎢ ⎢ ⎣2 ⎡

1 −1 2

3

1

2 ⎤

1

1

2

1

6

7

⎥ 3⎥ ⎥ ⎥ 3⎥ . ⎥ ⎥ 5⎦

1 −2

1

⎤

1 2 3 0 1 ⎢ ⎥ ⎢0 1 0 2 1 ⎥ ⎢ ⎥ 7. ⎢ ⎥. ⎢2 1 3 1 0 ⎥ ⎣ ⎦ 1 4 1 5 2

⎡

2 1 0 0 3 1

⎤

⎢ ⎥ ⎥ 8. ⎢ ⎣1 2 1 1 3 0⎦ . 0 1 2 5 1 2

3 ⎡

1

⎢ ⎢1 ⎢ 9. ⎢ ⎢2 ⎣ 0 ⎡

1

2

1

1

2

1

1

3

5

4

⎤

⎥ 0⎥ ⎥ ⎥. 4⎥ ⎦ 1

3 1 1 2 1

⎤

⎢ ⎥ ⎥ 10. ⎢ ⎣1 −2 1 3 1 0⎦. 2 0 1 0 3 0

Section 3.9

3.9

The Inverse Matrix

163

The Inverse Matrix

multiplicative inverse matrix

The operation of division is not deﬁned for matrices. However, we will see that n × n matrices A for which detA = 0 have associated with them an n × n matrix B, called its multiplicative inverse, with the property that AB = BA = I. The purpose of this section will be to develop ways of ﬁnding the multiplicative inverse of a matrix, which for simplicity is usually called the inverse matrix, but ﬁrst we give a formal deﬁnition of the inverse of a matrix. The inverse of a matrix Let A and B be two n × n matrices. Then matrix A is said to be invertible and to have an associated inverse matrix B if AB = BA = I. Interchanging the order of A and B in this deﬁnition shows that if B is the inverse of A, then A must be the inverse of B. To see that not all n × n matrices have inverses, it will be sufﬁcient to try to ﬁnd a matrix B such that the product AB = I, where 1 2 a b A= and B = . 1 2 c d The product AB is

AB =

1 1

2 2

a c

b a + 2c = d a + 2c

b + 2d , b + 2d

so if this product is to equal the 2 × 2 unit matrix I, it is necessary that a + 2c b + 2d 1 0 = . a + 2c b + 2d 0 1 Equating corresponding elements in the ﬁrst columns shows that this can only hold if a + 2c = 1 and a + 2c = 0, while equating corresponding elements in the second columns shows that b + 2d = 0 and b + 2d = 1, which is impossible, so matrix A has no inverse. In this case detA = 0, and we will see later why the nonvanishing of detA is necessary if A is to have an inverse. Nonsingular and singular matrices singular and nonsingular n × n matrices

EXAMPLE 3.25

An n × n matrix is said to be nonsingular when its inverse exists, and to be singular when it has no inverse. We have already seen that the matrix A=

1 1

2 , 2

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for which detA = 0, has no inverse and so is singular. However, in the case of matrix A that follows, a simple matrix multiplication conﬁrms that it has associated with it an inverse B, where ⎡ ⎤ ⎡ ⎤ 1 0 1 2 1 −2 1 −1⎦ , A = ⎣−1 2 0⎦ and B = ⎣ 1 0 1 1 −1 −1 2 because AB = BA = I. Furthermore, detA = 0, so A is nonsingular, as is B, and each is the inverse of the other. Before proceeding further it is necessary to establish that, when it exists, the inverse matrix is unique. THEOREM 3.10

Uniqueness of the inverse matrix A nonsingular matrix A has a unique inverse. Proof Suppose, if possible, that the nonsingular n × n matrix A has the two different inverses B and C. Then as AC = I, we have B = BI = B(AC) = (BA)C = IC = C, showing that B = C, so the inverse matrix is unique. It is convenient to denote the inverse of a nonsingular n × n matrix A by the symbol A−1 . This is suggested by the exponentation notation (raising to a power), because if for the moment we write A = A1 , then AA−1 = A1 A−1 = I, showing that exponents may be combined in the usual way, with the understanding that A1 A−1 = A(1−1) = A0 = I.

THEOREM 3.11 basic properties of the inverse matrix

Basic properties of inverse matrices (i) (ii) (iii) (iv)

The unit matrix I is its own inverse, so I = I−1 . If A is nonsingular, so also is A−1 , and (A−1 )−1 = A. If A is nonsingular, so also is AT , and (A−1 )T = (AT )−1 . If A and B are nonsingular n × n matrices, so is AB, and (AB)−1 = B−1 A−1 .

(v) If A is nonsingular, then (A−1 )m = (Am)−1 for m = 1, 2, . . . . Proof We prove only (i) and (iv), and leave the proofs of (ii), (iii), and (v) as exercises. The proof of (i) is almost immediate, because I2 = I, showing that I = I−1 . To prove (iv) we premultiply B−1 A−1 by AB to obtain ABB−1 A−1 = AIA−1 = AA−1 = I, which shows that (AB)−1 is B−1 A−1 , so the proof is complete. A simple method of ﬁnding the inverse of an n × n matrix is by means of elementary row operations, but to justify the method we ﬁrst need the following theorem.

Section 3.9

THEOREM 3.12

The Inverse Matrix

165

Elementary row operation matrices are nonsingular Every n × n matrix E that represents an elementary row operation is nonsingular. Proof Every n × n matrix E that represents an elementary row operation is derived from the unit matrix I by means of one of the three operations deﬁned at the start of Section 3.4. So, as rank(I) = n and E and I are row similar, it follows that rank(E) = n, and so E is also nonsingular.

ﬁnding an inverse matrix using elementary row operations

We can now describe an elementary way of ﬁnding an inverse matrix by means of elementary row transformations. Let A be a nonsingular n × n matrix, and let E1 , E2 , . . . , Em represent a sequence of elementary row operations of Types I, II, and III that reduces A to I, so that EmEm−1 . . . E2 E1 A = I. Then postmultiplying this result by A−1 gives EmEm−1 . . . E2 E1 I = A−1 , so A−1 is given by A−1 = EmEm−1 · · · E2 E1 I, where the product of the ﬁrst m matrices on the right is nonsingular because of Theorem 3.11. Expressed in words, this result states that when a sequence of elementary row operations is used to reduce a nonsingular matrix A to the unit matrix I, performing the same sequence of elementary row operations on I, in the same order, will generate the inverse matrix A−1 . If matrix A is singular, this will be indicated by the generation of either a complete row or a complete column of zeros before I is reached. If A is nonsingular, it is reducible to the unit matrix I, and clearly detA = 0. However, if A is singular, the attempt to reduce it to I will generate either a row or a column of zeros, so that then detA = 0. The vanishing or nonvanishing of detA provides a simple and convenient test for the singularity or nonsingularity of A whenever n is small, say n ≤ 3, because only then is it a simple matter to calculate detA. The practical way in which to implement this result is not to use the matrices Ei to reduce A to I, but to perform the operations directly on the rows of the partitioned matrix (A, I), because when A in the left half of the partitioned matrix has been reduced to I, the matrix I in the right half will have been transformed into A−1 .

EXAMPLE 3.26

Use elementary row operations to ﬁnd A−1 given that ⎡

1 A = ⎣ −1 0

0 2 1

⎤ 1 0⎦. 1

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Solution We form the augmented matrix (A, I) and proceed as described earlier. ⎡ ⎤ ⎤ ⎡ 1 0 1 1 0 0 1 0 1 1 0 0 ∼ ⎥ ⎢ ⎥ ⎢ (A, I) = ⎣−1 2 0 0 1 0⎦ add row 1 ⎣0 2 1 1 1 0⎦ to row 2 0 1 1 0 0 1 0 1 1 0 0 1 ⎡ 1 0 1 ∼ subtract row 3 ⎢ ⎣0 1 0 from row 2 0 1 1

1

1

1

⎤ ⎡ 1 0 1 0 ∼ ⎥ subtract row 2 ⎢ −1⎦ ⎣0 1 0 from row 3 1 0 0 1

⎡ 1 ∼ subtract row 3 ⎢ ⎣0 from row 1 0

2

1

1

1

1

1

−1

−1

−2

1

1

1

−1

−1

⎤

⎥ −1⎦ 2

⎤

⎥ −1⎦ . 2

The 3 × 3 matrix on the left of this row-equivalent partitioned matrix is now the unit matrix I, so the required inverse matrix is the one to the right of the partition, namely, ⎡ ⎤ 2 1 −2 1 −1⎦ . A−1 = ⎣ 1 −1 −1 2 Once A−1 has been obtained, it is always advisable to check the result by verifying that AA−1 = I. Before proceeding further we will use elementary matrices to provide the promised proof of Theorem 3.4(viii).

the proof that det(AB) = detA detB

Proof that det(AB) = detA detB Let E1 be a row matrix of Type I. Then if A is a nonsingular matrix, det(EI A) = −detA, because only a row interchange is involved. However, det(EI ) = −1, so det(EI A) = detEI detA. Similar arguments show this to be true for elementary row operation matrices of the other two types, so if E is an elementary row operation of any type, then det(EA) = detEdetA. If detA = 0, premultiplication by a sequence of elementary row operation matrices E1 , E2 , . . . , Er will reduce A to I, so performing them on I in the reverse order allows us to write A = E1 E2 . . . Er I = E1 E2 . . . Er . A repetition of the result det(EA) = detEdetA shows that detA = detE1 detE2 . . . detEr . If B is conformable for multiplication with A, using the preceding result we have det(AB) = det(E1 E2 . . . Er B) = detE1 detE2 . . . detEr detB, but detE1 detE2 . . . detEn = detA,

and so

det(AB) = detAdetB.

Section 3.9

The Inverse Matrix

167

To complete the proof we must show this result remains true if A is singular, in which case detA = 0. When detA = 0, the attempt to reduce it to the unit matrix I by elementary row operation matrices will fail because at one stage it will produce a determinant in which a row will contain only zero elements. Consequently, a determinant detEm, say, will be zero, which is impossible, so det(AB) = 0. However, if detA = 0, then detAdetB = 0, so that once again det(AB) = detAdetB, and the result is proved. EXAMPLE 3.27

Use (a) elementary row operations and (b) the determinant test to show matrix A is singular, given that ⎡ ⎤ 1 1 0 A = ⎣1 0 1⎦ . 4 3 1 Solution (a) Using elementary row operations on the augmented matrix gives ⎡ ⎤ ⎡ ⎤ 1 1 0 1 0 0 1 1 0 1 0 0 ∼ ⎢ ⎥ ⎥ ⎢ (A, I) = ⎣1 0 1 0 1 0⎦ subtract row 1 ⎣0 −1 1 −1 1 0⎦ from row 2 4 3 1 0 0 1 0 0 1 4 3 1 ⎡

1

1

∼ subtract 4 times ⎢ ⎣0 −1 1 row 1 from row 3 0 −1 1 ⎡ 1 1 0 ∼ subtract row 2 ⎢ ⎣0 −1 1 from row 3 0 0 0

−1

1

⎥ 0⎦

−4

1

1

−1

1

−3

−1

⎤

1

⎤

⎥ 0⎦ .

1

The reduction is terminated at this stage by the appearance of a row of zeros on the matrix to the left of the partition, showing that A cannot be reduced to I, and hence that A is singular. (b) Applying the determinant test to A, we ﬁnd that detA = 0, showing that A is singular. Although in this case this is by far the quickest way to establish the singularity of A, this would not have been so had the order of detA been much greater than 3. This is because when n > 3, the effort involved in performing the elementary row operations in an attempt to reduce A to I is considerably less than the effort involved when calculating detA. The following very different way of ﬁnding the inverse of an n × n matrix A is mainly of theoretical importance, though it is a practical method when n is small. The method is based on the properties of the sum of products of elements and cofactors of a determinant. Let A = [ai j ] be an n × n matrix, C = [Ci j ] be the associated n × n matrix of cofactors and form the matrix product ⎡ ⎤⎡ ⎤ a11 a12 . . . a1n C11 C21 . . . Cn1 ⎢a21 a22 . . . a2n ⎥ ⎢C12 C22 . . . Cn2 ⎥ ⎥⎢ ⎥ ACT = ⎢ ⎣ . . . . . . . . . ⎦⎣ . . . . . . . . . ⎦. an1 an2 . . . ann C1n C2n . . . Cnn

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If we write B = ACT , with B = [bi j ], it follows from the rule for matrix multiplication that bi j = ai1 C1 j + ai2 C2 j + · · · + ain Cnj . Thus, bi j is seen to be the sum of the product of the elements of the ith row of A and the corresponding cofactors of the elements of the jth row of A. It then follows from the Laplace expansion theorem for determinants that bi j = detA, for i = j = 1, 2, . . . , n and bi j = 0, for i = j. Using these results in the matrix product, we ﬁnd that ⎡ ⎤ detA 0 0 . . . 0 ⎢ 0 ⎥ detA 0 ⎢ ⎥ ⎥ 0 0 detA . . . 0 ACT = ⎢ ⎢ ⎥ ⎣ ⎦ . . . . . . . . . . . . 0 0 0 . . . detA = detA I. Consequently, provided detA = 0, it follows that (1/detA)ACT = I. Writing this as A{(1/detA)CT } = I shows that A−1 = (1/detA)CT . adjoint matrix

The matrix CT , called the adjoint of A and written adjA, is the transpose of the matrix of cofactors of A. So the formula for the inverse of A becomes A−1 = (1/detA)adjA.

(25)

We have arrived at the following deﬁnition and theorem. Adjoint matrix If A is an n × n matrix, and C is the associated matrix of cofactors, the transpose CT of the matrix of cofactors is called the adjoint of A and is written adjA.

THEOREM 3.13 formal deﬁnition of an inverse matrix

The inverse matrix in terms of the adjoint of A Let A be a nonsingular n × n matrix. Then the inverse of A is given by A−1 = (1/detA)adjA.

Section 3.9

EXAMPLE 3.28

Use Theorem 3.13 to ﬁnd A−1 , given that ⎡ 1 3 A = ⎣2 1 1 0 Solution The matrix of cofactors ⎡ ⎤ 1 −1 −1 1 3⎦ , C = ⎣−3 3 −1 −5

The Inverse Matrix

169

⎤ 0 1⎦ . 1 ⎡

1 so CT = ⎣−1 −1

⎤ −3 3 1 −1⎦ . 3 −5

Expanding detA in terms of the elements of its ﬁrst row (we already have its associated cofactors in the ﬁrst row of C) gives detA = 1 · 1 + (−1) · 3 + 1 · 0 = −2, so from Theorem 3.13, ⎤ ⎡ 1 3 − 32 −2 2 ⎥ ⎢ 1 1⎥ 1 A−1 = (−1/2)CT = ⎢ − 2 2 2 ⎦. ⎣ 5 3 1 −2 2 2 Although the result of Theorem 3.13 is of considerable theoretical importance, unless n is small, the task of evaluating the determinants involved makes it impractical for the determination of inverse matrices. In general, for large n, an inverse matrix is found by means of a computer using elementary row operations to reduce A to I.

General Proof of Cramer’s Rule proof of Cramer’s rule for a system of n equations

In conclusion, we will use Theorem 3.13 to arrive at a simple proof of Cramer’s rule for the system of equations a11 x1 + a12 x2 + · · · + a1n xn = b1 a21 x1 + a22 x2 + · · · + a2n xn = b2 · · · · · · · · · an1 x1 + an2 x2 + · · · + ann xn = bn . If we write the system as Ax = b, then, provided detA = 0, the solution can be written x = A−1 b = (1/detA)(adjA)b = (1/detA)CT b, where CT is the transpose of the matrix of cofactors of A. If x = (x1 , x2 , . . . , xn )T and b = (b1 , b2 , . . . , bn )T , the ith element of x is given by xi = (1/detA)(C1i b1 + C2i b2 + · · · + Cni bn )

for i = 1, 2, . . . , n.

This is simply the expansion of detAi in terms of the elements of its ith column, where Ai is the matrix obtained from A by replacing the elements of the ith column by the elements of b. This has established that xi = detAi /detA, and the proof is complete.

for i = 1, 2, . . . , n,

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Chapter 3

Matrices and Systems of Linear Equations

More information about the material in Sections 3.4 to 3.9 is to be found in the appropriate chapters of references [2.1], [2.5], and [2.7] to [2.12]. GABRIEL CRAMER (1704–1752): A Swiss mathematician who made many contributions to algebra and geometry. The result called Cramer’s rule was, in fact, ﬁrst formulated by Maclaurin around 1729 and published posthumously in his Treatise on Algebra (1748). The form of the rule attributed to Cramer appeared in his book Traite des courbes algebraiques (1750), which became a standard reference work during the remainder of the century. The work was so well written and so often quoted that after his death Cramer was, on occasions, considered to be the originator of the rule.

Summary

Division by matrices is not deﬁned, but the introduction of a multiplicative inverse A−1 of a nonsingular n × n matrix A, called the inverse of A, enables certain operations that in some sense are similar to matrix division to be performed. This section gave the formal deﬁnition of the inverse of a matrix and established its most important algebraic properties. The inverse matrix was used to prove Cramer’s rule for a general system of n nonhom*ogeneous linear algebraic equations when the determinant of the coefﬁcient matrix is nonsingular.

EXERCISES 3.9 In Exercises 1 through 8, construct a suitable augmented matrix and ﬁnd the inverse of the given matrix using elementary row operations. ⎤ ⎡ ⎤ ⎡ 2 3 1 1 3 7 5. ⎣1 2 0⎦ . 1. ⎣2 1 −1⎦ . 2 4 1 2 1 5 ⎤ ⎡ ⎤ ⎡ −4 1 0 3 0 1 2. ⎣ 1 −3 1⎦ . 6. ⎣1 −1 1⎦ . 2 1 4 0 4 5 ⎤ ⎡ ⎤ ⎡ 1 1 3 1 2 0 1 ⎢1 3. ⎣5 2 1⎦ . 0 −3 4⎥ ⎥. 7. ⎢ 1 6 2 ⎣0 1 2 5⎦ ⎤ ⎡ 2 −1 2 2 2 −6 1 ⎤ ⎡ ⎦ ⎣ 3 4 . 4. 1 0 1 2 3 ⎢2 2 −4 2⎥ 0 −2 1 ⎥. 8. ⎢ ⎣1 3 0 1⎦ 3 1 1 0 9. Given that ⎡ 3 −1 4 A = ⎣1 2 1

⎤ 1 0⎦ −3

and

⎡ 1 B = ⎣2 3

verify that (AB)−1 = B−1 A−1 .

⎤ −3 1 0 5⎦ , 1 2

10. Given that ⎡ 4 1 A = ⎣3 1 3 2

⎤ 2 0⎦, verify that (A−1 )T = (AT )−1 and 1

(A−1 )2 = (A2 )−1 . In Exercises 11 through 16, use Theorem 3.13 to ﬁnd the inverse of the given matrix, and check the result by showing that AA−1 = I. ⎤ ⎡ ⎤ ⎡ −3 2 6 2 4 −5 7⎦ . 1⎦ . 14. ⎣ 2 −1 11. ⎣2 7 5 4 −2 1 3 4 ⎤ ⎡ ⎤ ⎡ 2 0 1 2 3 −7 8 ⎢3 1 3 4⎥ 4 3⎦ . 12. ⎣1 ⎥. 15. ⎢ ⎣ 1 0 −2 3⎦ 0 −5 1 ⎤ ⎡ 1 −2 2 7 9 2 1 ⎤ ⎡ 0 1 −4 1 13. ⎣1 4 10⎦ . ⎢3 7 5 2⎥ 3 1 2 ⎥. 16. ⎢ ⎣1 −2 6 0⎦ 0 1 3 1 In the following two exercises, use the determinant test to show the given matrix is singular, and then verify this by using elementary row operations applied to a suitable augmented matrix, as in Example 3.27. Compare the effort involved in each case. ⎤ ⎤ ⎡ ⎡ 0 2 1 0 1 3 0 1 ⎢ 1 1 3 0⎥ ⎢1 1 2 1⎥ ⎥. ⎥ 17. ⎢ 18. ⎢ ⎣2 1 4 2 ⎦ . ⎣1 1 2 5⎦ 4 3 10 2 0 −1 1 2

Section 3.10

3.10

Derivative of a Matrix

171

Derivative of a Matrix When the elements of matrix A are differentiable functions of a single variable, say t, so that A = A[ai j (t)], calculus can be performed on matrices, so it becomes necessary to deﬁne the derivative of a matrix. An illustration of the need for this was given in Section 3.2(e), where the matrix differential equation x¨ + Ax = 0 was obtained as the system of second order differential equations determining the motion of a compound mass–spring system. Derivative of a matrix

fundamental deﬁnition of dA/dt

Let the m × n matrix A have elements ai j (t) that are differentiable functions of the variable t. Then the ﬁrst order derivative of A with respect to t, written dA/dt, is deﬁned as dA/dt = [d(ai j )/dt], and its nth order derivative with respect to t is deﬁned recursively as dn A/dt n = d/dt[dn−1 A/dt n−1 ],

for n = 1, 2, . . . ,

with the convention that d0 (ai j )/dt0 = ai j , so that d0 A/dt 0 = A. The derivative of a constant matrix is the null (zero) matrix 0.

EXAMPLE 3.29

Find dA/dt and d2 A/dt 2 given that 2 t te t 3t cosh t (a) A = . , (b) A = cos 3t 2t + 1 et sin 2t Solution (a) By deﬁnition, dA/dt = (b) dA/dt =

THEOREM 3.14 derivative of a sum, a product, and an inverse matrix

2t 2

3 et t

et + te −3 sin 3t

2 0 0 et 2et + tet 2 2 d A/dt = . −9 cos 3t

sinh t 2 cos 2t and

and

d2 A/dt 2 =

cosh t . −4 sin 2t

Derivative of the sum of two matrices Let A(t) and B(t) be an m × n matrices, each with differentiable elements. Then d/dt{A + B} = dA/dt + dB/dt. Proof The result follows immediately from the deﬁnition of the sum of two matrices.

172

Chapter 3

Matrices and Systems of Linear Equations

THEOREM 3.15

Derivative of a matrix product Let A(t) be an m × n matrix and B(t) be an n × q matrix, each with differentiable elements. Then, if the m × q matrix C(t) = A(t) B(t), dC/dt = {dA/dt}B + A{dB/dt}. Proof It follows from the deﬁnition of the matrix product of two matrices A and B that are conformable for multiplication that cr s = ar 1 b1s + ar 2 b2s + · · · + ar n bns , so each term in cr s is a product of two differentiable functions. Differentiating cr s establishes the theorem in which the order of the matrix products must be as shown.

THEOREM 3.16

Derivative of an inverse matrix Let A(t) be an n × n nonsingular matrix with differentiable elements. Then dA−1/dt = −A−1 {dA/dt}A−1 . Proof As A is nonsingular, its inverse A−1 exists and AA−1 = I. Differentiating the matrix product AA−1 = I gives {dA/dt}A−1 + AdA−1 /dt = 0. Premultiplication by A−1 followed by a rearrangement establishes the theorem.

EXAMPLE 3.30

Find dA−1 /dt given that

A=

Solution We have dA/dt =

−sin t cos t

−sin t . cos t

cos t sin t

−cos t −sin t

and

so from Theorem 3.16 −1

−1

−1

dA /dt = −A {dA/dt}A

A−1 =

cos t −sin t

−sin t = −cos t

sin t , cos t

cos t . −sin t

In this case the result is easily checked by direct differentiation of A−1 . Applications of the derivative of a matrix are to be found in reference [2.11] and, for example, in connection with systems of ordinary differential equations in reference [3.15].

Summary

Matrices can occur with functions as their elements as, for example, when a matrix describes a rotation through an angle θ about the origin of a cartesian coordinate system O{x, y}, or when a column vector contains the unknown functions u1 (t), u2 (t), . . . , un (t) that form the solution set of a system of linear differential equations with independent variable t. Because of this, it is necessary to understand how to differentiate a matrix with respect to an independent variable that is present in functions forming its elements. This section addressed this matter by ﬁrst deﬁning the fundamental operation of differentiation

Section 3.10

Derivative of a Matrix

173

of a matrix, and then establishing the way in which it is to be applied to the sum and product of two matrices and to the inverse matrix.

EXERCISES 3.10 In Exercises 1 through 4, ﬁnd dC/dt and d2 C/dt 2 . 3 t t t sin t 1. C = A + B, where A = 2 and t cos t sin 2t 1 2t 2 cosh t . B= t 3 cos t 2t e 1 tan t and 2. C = A − B, where A = t sin t cos 3t 2 2t sinh t . B= t t sin t t + 2 2t t 3 3. C = A − 2B, where A = and 3 3t e2t 2t t t3 e . B= 2 1 t sinh t (t + 1)2 t t 2 4. C = A + 3B, where A = and 2t 1 ln t t sin t 4 t . B= t t cosh t In Exercises 5 and 6, use Theorem 3.15 to ﬁnd dC/dt, where C = AB, and check the result by direct differentiation of C.

−cos 3t sin t cosh t cos t 6. A = sinh t sin t 5. A =

sin t cos t

and

1 + 2t 2

B=

and

B=

ln(2t) t

2 sin t . cos t t . cos t

In Exercises 7 and 8 ﬁnd dA−1 /dt by means of Theorem 3.16 and then verify the result by direct differentiation of A−1 . ⎤ ⎡ cos t sin t 0 ⎥ ⎢ 7. A = ⎣ −sin t cos t 0 ⎦ . 2 t 1 t t 2 2t . −t 3t 9. Find an expression for

8. A =

d2 {A−1 }/dt 2 in terms of A−1 , dA/dt, and d2 A/dt 2 . Apply the result to cos t −sin t A= sin t cos t and verify it by direct differentiation of A−1 .

CHAPTER 3

TECHNOLOGY PROJECTS Project 1 Simpliﬁcation of det C When C = [ci j + di j ] The purpose of this project is to provide practice with the computer algebra of determinants and to extend the result of Theorem 3.4(vi) to the case when each element of a determinant is the sum of two numbers. 1. Let a1 , a2 , a3 , b1 , b2 , b3 be arbitrary 3 ⫻ 1 element column vectors. Then, by repeated application of Theorem 3.4(vi), extend its result to the case when C ⫽ [a1 ⫹ b1 , a2 ⫹ b2 , a3 ⫹ b3 ] by expressing det C as a sum of 3 ⫻ 3 determinants with columns formed from a1 , a2 , a3 , b1 , b2 , and b3 . 2. Deﬁne an arbitrary matrix C of the form C ⫽ [a1 ⫹ b1 , a2 ⫹ b2 , a3 ⫹ b3 ], and with the aid of a computer algebra determinant package ﬁnd det C by using the result of Step 1. Conﬁrm the result by applying the computer algebra package directly to ﬁnd det C. Project 2 The Row-Reduced Echelon Form of a Matrix and Its Rank The purpose of this project is to provide practice with elementary row operations performed by means of computer algebra. It involves reducing a matrix step by step, using the rules given in Section 3.5, to its rowreduced echelon form, from which its rank can then be determined by inspection. 1. Let A be the matrix ⎤ ⎡ 0 1 3 2 4 2 ⎢ 1 2 1 ⫺3 1 1⎥ ⎥ ⎢ ⎢ 0 1 2 0 1⎥ A = ⎢⫺4 ⎥. ⎣ 0 ⫺3 ⫺4 5 0 ⫺3⎦ 2 1 ⫺2 ⫺1 2 ⫺1 Using computer algebra, apply sequentially the steps in the rule in Section 3.5 to reduce A to 174

its row-reduced echelon form, and hence ﬁnd rank (A). 2. Conﬁrm the result obtained in Step 1 by using a computer algebra package to ﬁnd directly the row-reduced echelon form of A. Take note that in some computer algebra packages the rowreduced echelon form of a matrix A is called the Gauss–Jordan form of A. Project 3 A Theorem on the Rank of a Matrix Product ABC The purpose of this project is to provide practice with matrix multiplication and the reduction of matrices to their row-reduced echelon forms using computer algebra. 1. If A, B, and C are arbitrary rectangular matrices, it can be shown that when the matrix product ABC exists, then Rank(AB) ⫹ Rank(BC) ⱕ Rank(B) ⫹ Rank(ABC). 2. Deﬁne three arbitrary rectangular matrices A, B, and C for which the product ABC is deﬁned. Using computer algebra matrix multiplication and computer algebra row-reduction to echelon form, ﬁnd the ranks of AB, BC, B, and ABC, and hence conﬁrm the inequality in Step 1 for this particular case. Project 4 Consistency of Augmented Coefficient Matrices, Solution by Back Substitution and Cramer's Rule The purpose of this project is to use computer algebra to determine the consistency of two 6 ⫻ 7 augmented coefﬁcient matrices. The solution for the corresponding consistent set of linear equations is then found after the reduction of its augmented coefﬁcient matrix to row-reduced echelon form followed by back

Section 3.10

substitution. Finally, the solution is checked using Cramer's rule, which, despite the large determinants involved, becomes feasible when computer algebra is used. 1. Use computer algebra to determine which of the augmented coefﬁcient matrices A and B is consistent, given that ⎡ ⎤ 1 4 7 3 0 2 4 ⎢3 1 0 2 3 4 1⎥ ⎢ ⎥ ⎢1 2 1 1 4 3 2⎥ ⎢ ⎥ and A ⎢ 1 6 3⎥ ⎢2 4 0 0 ⎥ ⎣0 1 2 1 2 1 0⎦ 2 5 2 1 1 7 5 ⎡ ⎤ 4 1 3 0 1 4 2 ⎢1 1 1 3 2 1 1⎥ ⎢ ⎥ ⎢0 ⎥ 1 1 2 2 1 3 ⎢ ⎥. B ⎢ ⎥ 4 0 1 1 2 3 4 ⎢ ⎥ ⎣1 1 3 2 4 2 1⎦ 0 4 3 3 1 2 0 2. In the case of the consistent set of equations, using the reduction of the coefﬁcient matrix to its row-reduced echelon form, ﬁnd the solution by back substitution. 3. Using computer algebra, apply Cramer's rule to the consistent set of equations to ﬁnd the solution, and so conﬁrm the result found in step 2. Project 5 A One-Way Traffic Flow Problem The diagram shows the pattern of one way trafﬁc ﬂow at six road intersections at the corners of two city blocks. The arrows show the directions of trafﬁc ﬂow, and the associated numbers are the trafﬁc ﬂow rates in vehicles per hour at peak trafﬁc time.

Derivative of a Matrix

160

480 A

B

500

180

x1

x2

x7

F

C

880

110

x6

x3

D

980

175

x5

x4

760

E

150 700

By equating the ﬂow rate of trafﬁc into an intersection to the ﬂow rate out of it (no parking is allowed), ﬁnd equations relating the trafﬁc ﬂow rates x1 , x2 , . . . , x7 along each of the roads. Explain why with the given peak ﬂow rates it is impossible to close road DE, and comment on the effect on trafﬁc ﬂow if road CD is closed for repairs. Project 6 Forces in Bridge Struts Use matrix methods to ﬁnd the forces in the pinjointed framed bridge section shown in Fig. 3.10, given that a concentrated load m acts vertically downwards at joint B. Give a simple example of a pin-jointed framed structure that contains a redundant strut, and prove its redundancy by attempting to determine the forces acting in the strut when the structure is loaded.

175

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4

C H A P T E R

Eigenvalues, Eigenvectors, and Diagonalization

I

n engineering and physics, problems involving n linear algebraic equations in n independent variables with a constant coefﬁcient matrix A often arise where a solution vector x is required to be proportional to Ax. Setting the constant of proportionality equal to λ, this means that x must be a solution of the equation Ax = λx or, equivalently, of the equation (A − λI)x = 0. The numbers λi for which nonzero solutions xi exist are called the eigenvalues of matrix A, and the corresponding vectors xi are called the eigenvectors of A. Eigenvalues and eigenvectors arise, for example, when studying vibrational problems, where the eigenvalues represent fundamental frequencies of vibration and the eigenvectors characterize the corresponding fundamental modes of vibration. They also occur in many other ways; in mechanics, for example, the eigenvalues can represent the principal stresses in a solid body, in which case the eigenvectors then describe the corresponding principal axes of stress caused by the body being subjected to external forces. Also in mechanics, the moment of inertia of a solid body about lines through its center of gravity can be represented by an ellipsoid, with the length of a line drawn from its center to the surface of the ellipsoid proportional to the moment of inertia of the body about an axis through the center of gravity of the body drawn parallel to the line. In this case the eigenvalues represent the principal moments of inertia of the body about the principal axes of inertia, that are then determined by the eigenvectors. More precisely, if A is an n × n matrix, the polynomial Pn (λ) of degree n in the scalar λ deﬁned as Pn (λ) = det (A − λI) is called the characteristic polynomial of A. The roots of the equation Pn (λ) = 0 are called the eigenvalues of matrix A, and the column vectors x1 , x2 , . . . , xn satisfying the matrix equation (A − λi I)xi = 0 are called the eigenvectors of matrix A. This chapter explains how eigenvalues and eigenvectors are determined and establishes important properties of eigenvectors. The eigenvectors of an n × n matrix A with n linearly independent eigenvectors are then used to simplify the structure of A by means of a process called diagonalization. An important application of diagonalization will arise later when considering the solution of linear systems of ordinary differential equations that arise from the study of mechanical, electrical, and chemical reaction problems. Diagonalization is also an important tool when working with partial differential equations, different types of which describe the temperature distribution in a metal, electromagnetic wave propagation, and diffusion processes, to name a few examples.

177

178

Chapter 4

Eigenvalues, Eigenvectors, and Diagonalization After a brief discussion of some special n × n matrices with complex elements, real quadratic forms are deﬁned and the properties of eigenvectors are used to reduce a general quadratic form to a sum of squares. This is a process that ﬁnds many different applications, one of which occurs later when classifying the partial differential equations of engineering and physics in order to know the type of auxiliary conditions that must be imposed in order for them to give rise to physically meaningful solutions. The chapter ends with the introduction of the matrix exponential e A , where A is a real n × n matrix, and it is shown how this enters into the solution of a linear ﬁrst order matrix differential equation of the form dx/dt = Ax.

4.1

Characteristic Polynomial, Eigenvalues, and Eigenvectors

T

hroughout this chapter we will be considering the solutions of the hom*ogeneous system of algebraic equations Ax = λx,

(1)

where A[ai j ] is an n × n matrix, x is an n element column vector with elements x1 , x2 , . . . , xn , and λ is a scalar. For A given we wish to ﬁnd x and λ. Introducing the n × n unit matrix by I allows (1) to be written (A − λI)x = 0,

(2)

showing that x is a solution of a hom*ogeneous system of equations with the coefﬁcient matrix A − λI. It was seen in Chapter 3 that nontrivial solutions x of (2) are only possible if one or more rows of the coefﬁcient matrix A − λI are linearly dependent on its remaining rows. This means that nontrivial solutions x will exist if rank(A − λI) < n, but this, in turn, is equivalent to the more convenient condition det(A − λI) = 0. This is a polynomial equation for λ. Let Pn (λ) be the polynomial of degree n in λ deﬁned by the determinant a11 − λ a12 a13 a14 · · · · · a1n a21 a22 − λ a23 a24 · · · · · a2n a32 a33 − λ a34 · · · · · a3n . (3) Pn (λ) = a31 . . . . . . . . . . . . . . . . . . . an1 an2 an3 an4 . . . . ann − λ Inspection of the determinant deﬁning Pn (λ) shows the coefﬁcient of λn is (−1)n , so the polynomial is of the form Pn (λ) = (−1)n [λn + c1 λn−1 + c2 λn−2 − · · · + cn−1 λ + c0 ]. characteristic polynomial, equation, and eigenvalue

(4)

The polynomial Pn (λ) is called the characteristic polynomial of A and the associated polynomial equation Pn (λ) = 0 is the characteristic equation of A. As the characteristic equation of A is of degree n in λ, it will have n roots, some of which may be repeated. The roots of Pn (λ) = 0, or equivalently the zeros of Pn (λ), are called the eigenvalues of A or, sometimes, the characteristic values of A.

Section 4.1

Characteristic Polynomial, Eigenvalues, and Eigenvectors

179

Eigenvalues (characteristic values) of A The eigenvalues of an n × n matrix A are the n zeros of the polynomial P(λ) = det(A − λI), or, equivalently, the n roots of the nth degree polynomial equation det(A − λI) = 0.

spectrum and spectral radius

eigenvectors and eigenvalues

In general, a matrix with complex coefﬁcients will have complex eigenvalues, though even when the coefﬁcients of A are all real it is still possible for complex eigenvalues to arise. This is because then the characteristic equation will have real coefﬁcients, so if complex roots occur they must do so in complex conjugate pairs. If an eigenvalue λ∗ is repeated r times, corresponding to the presence of a factor (λ − λ∗ )r in the characteristic polynomial Pn (λ), the number r is called the algebraic multiplicity of the eigenvalue λ∗ . The set of all eigenvalues λ1 , λ2 , . . . ,λn of A is called the spectrum of A, and the number R = max{|λ1 |, |λ2 |, . . . , |λn |}, equal to the largest of the moduli of the eigenvalues, is called the spectral radius of A. The name comes from the fact that when the spectrum of A is plotted as points in the complex plane, they all lie inside or on a circle of radius R centered on the origin. An eigenvector of an n × n matrix A, corresponding to an eigenvalue λ = λi , is a nonzero n-element column vector xi that satisﬁes the matrix equation Axi = λi xi or, equivalently, that is a solution of the hom*ogeneous system of n algebraic equations (A − λi I)xi = 0.

(5)

Eigenvectors of A The eigenvector xi of the n × n matrix A, corresponding to the eigenvalue λ = λi , is a solution of the hom*ogeneous equation (A − λi I)xi = 0. It is important to recognize that because system (5) is hom*ogeneous, the elements of an eigenvector can only be determined as multiples of one of its nonzero elements as a parameter. This means that if for some choice of the parameter x is an eigenvalue, then kx will also be an eigenvalue for any k = 0. The next theorem is fundamental to the use of eigenvectors and shows that when an n × n matrix A has n distinct (different) eigenvalues, its n eigenvectors form a basis for the vector space associated with the matrix A. THEOREM 4.1 eigenvectors are linearly independent

Linear independence of eigenvectors The eigenvectors x1 , x2 , . . . , xm, corresponding to m distinct eigenvalues λ1 , λ2 , . . . , λm, of an n × n matrix A, are linearly independent. Furthermore, if m = n, the set of eigenvectors x1 , x2 , . . . , xn forms a basis for the n-dimensional vector space associated with A. Proof The proof will be by induction, starting with two vectors, and it uses the fact that Axi = λi xi for i = 1, 2, . . . , m.

180

Chapter 4

Eigenvalues, Eigenvectors, and Diagonalization

Let x1 and x2 correspond to distinct eigenvalues λ1 and λ2 , and let constants k1 and k2 be such that k1 x1 + k2 x2 = 0. Then A(k1 x1 + k2 x2 ) = 0, but Axi = λi xi , so this is equivalent to k1 λ1 x1 + k2 λ2 x2 = 0. Subtracting λ2 times the ﬁrst equation from the last result gives (λ1 − λ2 )k1 x1 = 0. By hypothesis, λ1 = λ2 , so as x1 = 0 it follows that k1 = 0. Using this result in k1 x1 + k2 x2 = 0 shows that k2 = 0, so we have established the linear independence of x1 and x2 . To proceed with an inductive proof we now assume that linear independence has been proved for the ﬁrst r − 1 vectors, and show that the r th vector must also be linearly independent. To accomplish this we consider the equation k1 x1 + k2 x2 + · · · + kr xr = 0. Premultiplying this equation by A and reasoning as before, we arrive at the result k1 λ1 x1 + k2 λ2 x2 + · · · + kr λr xr = 0. Subtracting λr times the ﬁrst equation from the last one gives (λ1 − λr )k1 x1 + (λ2 − λr )k2 x2 + · · · + (λr −1 − λr )kr −1 xr −1 = 0. By the inductive hypothesis x1 , x2 , . . . , xr −1 are linearly independent, so as xr = 0, (λ1 − λr )k1 = (λ2 − λr )k2 = · · · = (λr −1 − λr )kr −1 = 0. The eigenvalues are distinct, so the last result can only be true if k1 = k2 = · · · = kr −1 = 0. Thus kr = 0, and so the vector xr is linearly independent of the vectors x1 , x2 , . . . , xr −1 . It has been shown that x1 and x2 are linearly independent, so by induction we conclude that the set of vectors xi is linearly independent for i = 1, 2, . . . , m. A matrix A can have no more than n linearly independent eigenvectors, so when m = n the set of eigenvectors x1 , x2 , . . . , xn spans the n-dimensional vector space associated with matrix A and forms a basis for this space. The proof is complete.

algebraic and geometric multiplicity

It can happen that an eigenvalue with algebraic multiplicity r > 1 only has s different eigenvectors associated with it, where s < r , and when this occurs the number s is called the geometric multiplicity of the eigenvalue. The set of all eigenvectors associated with an eigenvalue with geometric multiplicity s together with the null vector 0 forms what is called the eigenspace associated with the eigenvalue. When one or more eigenvalues has a geometric multiplicity that is less than its algebraic multiplicity, it follows directly that the vector space associated with A must have dimension less than n.

Section 4.1

EXAMPLE 4.1

Characteristic Polynomial, Eigenvalues, and Eigenvectors

Find the characteristic polynomial, the matrix ⎡ 2 A = ⎣3 3

181

eigenvalues, and the eigenvectors of the 1 2 1

⎤ −1 −3⎦ . −2

Solution The characteristic polynomial P3 (λ) is given by 2 − λ 1 −1 3 2−λ −3 , P3 (λ) = 3 1 −2 − λ and after expanding the determinant we ﬁnd that P3 (λ) = −λ3 + 2λ2 + λ − 2. The characteristic equation P3 (λ) = 0 is λ3 − 2λ2 − λ + 2 = 0, and inspection shows it has the roots 2, 1, and −1. So the eigenvalues of A are λ1 = 2, λ2 = 1, and λ3 = −1, and as these roots are all distinct (there are no repeated roots), each has an algebraic and geometric multiplicity of 1 (each is a single root). The set of numbers −1, 1, 2 forms the spectrum of matrix A. As the spectral radius R of a matrix is deﬁned as the largest of the moduli of the eigenvalues, we see that R = 2. To ﬁnd the eigenvectors xi of A corresponding to the eigenvalues λ = λi , for i = 1, 2, 3, it will be necessary to solve the hom*ogeneous system of algebraic equations (A − λi I)xi = 0

for i = 1, 2, 3,

where xi = [x1 , x2 , x3 ]T .

Case λ1 = 2 The system of equations to be solved is ⎡ ⎤⎡ ⎤ ⎡ ⎤ 0 2−2 1 −1 x1 ⎣ 3 2−2 −3⎦ ⎣x2 ⎦ = ⎣0⎦ , 0 3 1 −2 − 2 x3 and this matrix equation is equivalent to the set of three linear algebraic equations x2 − x3 = 0,

3x1 − 3x3 = 0,

and

3x1 + x2 − 4x3 = 0.

The ﬁrst two equations are equivalent, so only one of the ﬁrst two equations and the third equation are linearly independent. Solving the last two equations for x1 and x2 in terms of x3 , we ﬁnd that x1 = x2 = x3 , so setting x3 = k1 where k1 is an arbitrary real number (a parameter) shows that the eigenvector x1 corresponding to the eigenvalue λ1 = 2 is given by ⎡ ⎤ ⎡ ⎤ k1 1 x1 = ⎣k1 ⎦ = k1 ⎣1⎦. k1 1

182

Chapter 4

Eigenvalues, Eigenvectors, and Diagonalization

As k1 is an arbitrary parameter, for convenience we set k1 = 1 and as a result obtain the eigenvector ⎡ ⎤ 1 x1 = ⎣1⎦ . 1

Case λ2 = 1 This time the system of equations to be solved to ﬁnd the eigenvector x2 is ⎡ ⎤⎡ ⎤ ⎡ ⎤ 2−1 1 −1 x1 0 ⎣ 3 2−1 −3⎦ ⎣x2 ⎦ = ⎣0⎦, 3 1 −2 − 1 0 x3 and this is equivalent to the three linear algebraic equations x1 + x2 − x3 = 0,

3x1 + x2 − 3x3 = 0,

and

3x1 + x2 − 3x3 = 0.

The last two equations are identical, so we must solve for x1 , x2 , and x3 using the ﬁrst two equations. It is easily seen from these two equations that x2 = 0 and x1 = x3 , so setting x1 = k2 , where k2 is an arbitrary real number (a parameter), gives ⎡ ⎤ 1 x2 = k2 ⎣0⎦. 1 Making the arbitrary choice k2 = 1 shows that the eigenvector x2 corresponding to λ2 = 1 is ⎡ ⎤ 1 x2 = ⎣0⎦ . 1

Case λ3 = −1 Setting λ = λ3 , and proceeding as before, shows that the elements of the eigenvector x3 must satisfy the three equations 3x1 + x2 − x3 = 0,

3x1 + 3x2 − 3x3 = 0,

and

3x1 + x2 − x3 = 0,

with the solution x1 = 0, x2 = x3 = k3 , where k3 is an arbitrary real number (a parameter). Making the arbitrary choice k3 = 1 allows the eigenvector x3 to be written as ⎡ ⎤ 0 x3 = ⎣1⎦ . 1 We have shown that matrix A has the three distinct eigenvalues λ1 = 2, λ2 = 1, and λ3 = −1, corresponding to which there are the three eigenvectors ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 1 0 x1 = ⎣1⎦ , x2 = ⎣0⎦ , and x3 = ⎣1⎦ . 1 1 1 These three eigenvectors form a basis for the three-dimensional vector space associated with A.

Section 4.1

Characteristic Polynomial, Eigenvalues, and Eigenvectors

183

As the eigenvectors x of matrix A satisfy the hom*ogeneous equation (2), they can be multiplied by an arbitrary nonzero number K, which is either positive or negative, and still remain an eigenvector. This property is used to scale the eigenvectors of A to produce what are called normalized eigenvectors. This scaling is used in numerical calculations involving the iteration of eigenvectors, because without normalization the elements of x may either grow or diminish in absolute value after each stage of the calculation, leading to a progressive loss of accuracy. Normalization of eigenvectors a frequently used way of normalizing eigenvectors

Various normalizations are in use. The most common one for eigenvectors with real elements involves scaling the eigenvector so that the square root of the sum of the squares of its elements is 1. So, for example, if ⎡ ⎤ a ⎣ x = b⎦ , c

the normalizing factor

K=

1 (a 2 + b2 + c2 )1/2

(6)

and the normalized eigenvector xˆ becomes ⎡ ⎤ a/(a 2 + b2 + c2 )1/2 xˆ = ⎣b/(a 2 + b2 + c2 )1/2 ⎦ . c/(a 2 + b2 + c2 )1/2

(7)

When the eigenvectors in Example 4.1 are normalized in this way, they become ⎡ √ ⎤ ⎡ √ ⎤ ⎡ ⎤ 0√ 1/√3 1/ 2 xˆ 1 = ⎣1/√3 ⎦ , xˆ 2 = ⎣ 0√ ⎦ , and xˆ 3 = ⎣1/√2⎦ . 1/ 2 1/ 2 1/ 3 EXAMPLE 4.2

Find the characteristic polynomial, eigenvalues, and eigenvectors of the matrix ⎡ ⎤ 0 0 1 1 ⎢−1 2 0 1⎥ ⎥. A=⎢ ⎣−1 0 2 1⎦ 1 0 −1 0 Solution The determinant deﬁning the characteristic polynomial is ⎡ ⎤ −λ 0 1 1 ⎢ −1 2 − λ 0 1⎥ ⎥, P4 (λ) = ⎢ ⎣ −1 0 2−λ 1⎦ 1 0 −1 −λ and after the determinant is expanded the characteristic equation P4 (λ) = 0 is found to be P4 (λ) = λ(λ3 − 4λ2 + 5λ − 2) = 0. Clearly, λ = 0 is a root of P4 (λ) = 0, and inspection shows the other three roots to be 1, 1, and 2. So the eigenvalues of A are λ1 = 0, λ2 = 1, λ3 = 1, and λ4 = 2. In this

184

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Eigenvalues, Eigenvectors, and Diagonalization

case λ2 = λ3 = 1, so the eigenvalue 1 has algebraic multiplicity 2, and the remaining two eigenvalues each have an algebraic multiplicity of 1. To ﬁnd the eigenvectors corresponding to these eigenvalues we proceed as in Example 4.1.

Case λ1 = 0 Setting λ = λ1 = 0 in (A − λI)x = 0 leads to the four equations x3 + x4 = 0,

−x1 + 2x2 + x4 = 0,

−x1 + 2x3 + x4 = 0, and

x1 − x3 = 0.

Proceeding as before we ﬁnd that x1 = x2 = x3 = −x4 , so solving for x1 , x2 , and x3 in terms of x4 , and setting x4 = 1 (an arbitrary choice), shows the eigenvector x1 to be ⎡ ⎤ −1 ⎢−1⎥ ⎥ x1 = ⎢ ⎣−1⎦ . 1

Case λ2 = λ3 = 1 The eigenvalue 1 has algebraic multiplicity 2, so we must attempt to ﬁnd two different eigenvectors that correspond to the single eigenvalue λ = 1. Setting λ = 1 in (A − λI)x = 0 leads to the four equations −x1 + x3 + x4 = 0,

−x1 + x2 + x4 = 0,

−x1 + x3 + x4 = 0,

x1 − x3 − x4 = 0.

The ﬁrst, third, and fourth equations are identical, so x1 , x2 , x3 , and x4 must be determined from the two equations −x1 + x3 + x4 = 0

and

−x1 + x2 + x4 = 0.

As there are four unknown quantities x1 , x2 , x3 , and x4 , and only two equations relating them, it will only be possible to solve for two of these quantities in terms of the remaining two. The equations show that x2 = x3 and x4 = x1 − x3 , so choosing to solve for x3 and x4 in terms of x1 and x2 by setting x1 = α and x2 = β, with α and β arbitrary constants, shows that the eigenvectors x2 and x3 are both of the form ⎡ ⎤ α ⎢ β ⎥ ⎥ x2,3 = ⎢ ⎣ β ⎦. α−β It is possible to obtain two different eigenvectors from this last result by choosing two different pairs of values for the arbitrary parameters α and β. We will deﬁne x2 by setting α = 1 and β = 1, and x3 by setting α = 1 and β = 0, and as a result we ﬁnd that ⎡ ⎤ ⎡ ⎤ 1 1 ⎢1⎥ ⎢0⎥ ⎥ ⎢ ⎥ x2 = ⎢ ⎣1⎦ and x3 = ⎣0⎦ . 0 1 Had other choices of the parameters α and β been made, two different eigenvectors would have been produced.

Section 4.1

Characteristic Polynomial, Eigenvalues, and Eigenvectors

185

Case λ4 = 2 Setting λ = λ4 = 2 in (A − λI)x = 0 leads to the four equations −2x1 + x3 + x4 = 0,

−x1 + x4 = 0,

−x1 + x4 = 0,

x1 − x3 − 2x4 = 0.

These equations have the solution x1 = x3 = x4 = 0, with no condition being imposed on x2 . For simplicity we choose to set x2 = 1 to obtain ⎡ ⎤ 0 ⎢1⎥ ⎥ x4 = ⎢ ⎣0⎦ . 0 In this example, the eigenvalue 1 has algebraic multiplicity 2, and two different eigenvectors can be associated with it, so the geometric multiplicity of the eigenvalue is also 2. The four eigenvectors x1 , x2 , x3 , and x4 form a basis for the four-dimensional vector space associated with matrix A. Had different values been used for α and β, the basis vectors for this vector space would have been different, though the vector space itself would have remained the same because linear combinations of basis vectors will produce an equivalent set of basis vectors. The spectrum of A is the set of numbers 0, 1, 2, and the spectral radius of A is seen to be R = 2. EXAMPLE 4.3

Show that the matrix

⎡

1 A = ⎣0 0

1 1 0

⎤ 0 0⎦ 0

has three eigenvalues, but only two linearly independent eigenvectors. Solution The characteristic polynomial 1 − λ 0 P3 (λ) = 0

1 0 1−λ 0 , 0 −λ

and after expanding the determinant the characteristic equation P3 (λ) = 0 becomes P3 (λ) = −λ(1 − λ)2 = 0. The eigenvalue λ1 = 0 occurs with algebraic multiplicity 1 and the eigenvalue λ2 = λ3 = 1 occurs with algebraic multiplicity 2. The equations determining the eigenvector x1 , corresponding to the eigenvalue λ = λ1 = 0, are x1 + x2 = 0

and

x2 = 0,

so x1 = x2 = 0 and x3 is arbitrary. Setting x3 = 1 gives ⎡ ⎤ 0 x1 = ⎣0⎦ . 1 The equations determining x2 and x3 , corresponding to λ = λ2 = λ3 = 1, are x1 = k(arbitrary)

and

x2 = x3 = 0,

186

Chapter 4

Eigenvalues, Eigenvectors, and Diagonalization

so setting k = 1, we ﬁnd that the eigenvalue λ2 = λ3 = 1 with algebraic multiplicity 2 only has associated with it the single eigenvector ⎡ ⎤ 1 x2,3 = ⎣0⎦ . 0 So the algebraic multiplicity of the eigenvalue λ = 1 is 2, but its geometric multiplicity is 1. The spectrum of A is the set of numbers 0, 1, so the spectral radius of A is R = 1. The eigenvalues of a diagonal matrix can be found immediately, and the corresponding eigenvectors take on a particularly simple form. Let D be the n × n diagonal matrix ⎡ ⎤ a1 0 0 · · · · 0 ⎢ 0 a2 0 · · · · 0 ⎥ ⎢ ⎥ ⎥ D=⎢ ⎢ . . . . . . . . . . ⎥, ⎣ . . . . . . . . . . ⎦ 0 0 0 · · · · an with entries a1 , a2 , . . . , an on its leading diagonal, not all of which are zero, and zeros elsewhere. Then it is easily seen that the eigenvalues of D are λ1 = a1 , λ2 = a2 , . . . , λn = an . The eigenvector xi corresponding to the eigenvalue λi = ai becomes an n-element column vector in which only the ith element is nonzero. It is not difﬁcult to show that this result remains true whatever the algebraic multiplicity of an eigenvalue, so every diagonal n × n matrix has n eigenvectors of this form. For convenience, the ith element in xi is usually taken to be 1 so, for example, the matrix ⎡ ⎤ 3 0 0 A = ⎣0 −5 0⎦ 0 0 4 has eigenvalues λ1 = 3, λ2 = −5, and λ3 = 4 and eigenvectors ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 0 0 x1 = ⎣0⎦ , x2 = ⎣1⎦ , and x3 = ⎣0⎦ . 0 0 1 Similarly, the diagonal matrix

⎡

−2 A=⎣ 0 0

0 4 0

⎤ 0 0⎦ 4

has an eigenvalue λ1 = −2 with multiplicity 1 and a double eigenvalue λ2 = λ3 = 4 with multiplicity 2, but the matrix still has the three distinct eigenvectors ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 0 0 x1 = ⎣0⎦ , x2 = ⎣1⎦ , and x3 = ⎣0⎦ . 0 0 1 When the degree of the characteristic equation of a matrix exceeds 2, its roots must usually be found by means of a numerical technique. In such circ*mstances the next theorem provides a simple and useful check for the values of the eigenvalues that have been computed.

Section 4.1

THEOREM 4.2 a check on the sum of the eigenvectors

Characteristic Polynomial, Eigenvalues, and Eigenvectors

187

The sum of eigenvalues Let the n × n matrix A[ai j ] have the n eigenvalues λ1 , λ2 , . . . , λn , which may be either real or complex. Then λ1 + λ2 + · · · + λn = (−1)n−1 (a11 + a22 + · · · + ann ) = (−1)n−1 tr(A). Proof As the multiplication of a column of a matrix by a number k is equivalent to multiplication of its determinant by k, we can write Pn (λ) = det(A − λI) = (−1)n det(λI − A). Expanding the determinant on the right in terms of the elements of the ﬁrst column and separating out the factors that can give rise to the terms in λn and λn−1 , we arrive at the result Pn (λ) = (−1)n {(λ − a11 )(λ − a22 ) · · · (λ − ann ) + Qn−2 (λ)}, where Qn−2 (λ) is a polynomial in λ of degree n − 2. Identifying the coefﬁcients of λn and λn−1 in the expression for Pn (λ) shows that Pn (λ) = (−1)n {λn − (a11 + a22 + · · · + ann )λn−1 + · · · + constant + Qn−2 (λ)}. An equivalent expression for Pn (λ) can be obtained by expanding it in terms of its factors (λ − λ1 ), (λ − λ2 ), . . . , (λ − λn ) to obtain Pn (λ) = (−1)n (λ − λ1 )(λ − λ2 ) · · · (λ − λn ) = (−1)n {λn − (λ1 + λ2 + · · · + λn )λn−1 + · · · + constant}. The statement of the theorem then follows by comparing the coefﬁcients of λn−1 in the two different expressions for Pn (λ), where it will be recalled that the trace of an n × n matrix A[ai j ], written tr(A), is the sum of the elements on its leading diagonal, so that tr(A) = a11 + a22 + · · · + ann .

EXAMPLE 4.4

Use Theorem 4.2 to check the eigenvalues of the matrices in Examples 4.1 and 4.2. Solution In Example 4.1, λ1 = 2, λ2 = 1, and λ3 = −1, so λ1 + λ2 + λ3 = 2, and tr(A) = 2 + 2 − 2 = 2, so the result of Theorem 4.2 is veriﬁed. Similarly, in Example 4.2, λ1 = 0, λ2 = 1, λ3 = 1, and λ4 = 2, so λ1 + λ2 + λ3 + λ4 = 4, and tr(A) = 0 + 2 + 2 + 0 = 4, showing that the result of Theorem 4.2 is again veriﬁed.

EXAMPLE 4.5

Find the characteristic polynomial, eigenvalues, and eigenvectors of ⎡ ⎤ −1 − 2i −1 − i 2 + 2i −i 4i ⎦ , A = ⎣ −4i −1 − 3i −1 − i 2 + 3i and use Theorem 4.2 to check the eigenvalues. Solution This matrix has complex elements. Expanding det(A − λI) = 0 shows that the characteristic polynomial P3 (λ) is P3 (λ) = λ3 − λ2 + λ − 1.

188

Chapter 4

Eigenvalues, Eigenvectors, and Diagonalization

Inspection shows the eigenvalues determined by P3 (λ) = 0 to be λ1 = 1, λ2 = i, and λ3 = −i. Finding the eigenvectors, as in Example 4.1, gives ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 0 1 (λ1 = 1) x1 = ⎣0⎦ , (λ2 = i) x2 = ⎣ 1 ⎦ , and (λ3 = −i) x3 = ⎣1⎦ . 1 1/2 1 In this example, although the matrix A has complex elements, the characteristic polynomial has real coefﬁcients, and one of its zeros (an eigenvalue) is real and its other two zeros (eigenvalues) are complex conjugates. The test in Theorem 4.2 is satisﬁed because tr(A) = λ1 + λ2 + λ3 = tr(A) = 1. Complex eigenvalues arise in numerous applications of matrices, and when this happens it is often useful to have qualitative information about a region in the complex plane that contains all of the eigenvalues, without the necessity of computing their actual values. This form of approach is particularly useful when the coefﬁcients of a polynomial are not speciﬁc, and all that is known is that they lie within given intervals or, if complex, that the modulus of each is bounded by a given number. Another need for this type of information occurs when working with systems of linear differential equations, because it will be seen in Chapter 6 that the roots of a characteristic polynomial equation determine the form of the general solution of a hom*ogeneous system. Roots of the form α + iβ will be seen to lead to real solutions of the form eαt sin βt and eαt cos βt, and these solutions will only remain bounded (stable) as t → +∞ if the real part of every root is negative. This means that the qualitative knowledge that all of the roots lie to the left of the imaginary axis will be sufﬁcient to ensure that the solution remains ﬁnite (is stable) as t → +∞. The theorem that follows is the simplest of many similar results that are available, all of which provide information about regions in the complex plane where all of the zeros of a characteristic polynomial are located. Two other results are to be found in the exercise set at the end of this section; the one called the Routh– Hurwitz stability criterion is particularly useful when working with systems of linear differential equations. Although the theorem to be proved in this section identiﬁes a region less precisely than many similar theorems, it has been included to illustrate how such regions can be found, and also because the derivation of the result is elementary. The proof only uses the basic properties of complex numbers extending as far as the triangle inequality. THEOREM 4.3 ﬁnding a region that contains all the eigenvalues

The Gerschgorin circle theorem Let A[ai j ] be an n × n matrix, and deﬁne the circles C1 , C2 , . . . , Cn in the complex plane such that circle Cr has its center at arr and the radius ρr =

n

|ar j | = |ar 1 | + |ar 2 | + · · · + |ar,r −1 | + |ar,r +1 | + · · · + |ar n |.

j=1, j=r

Then each of the eigenvalues of A lies in at least one of these circles.

Section 4.1

Proof

Characteristic Polynomial, Eigenvalues, and Eigenvectors

189

The r th equation of Ax = λx is ar 1 x1 + · · · + ar,r −1 xr −1 + (arr − λ)xr + ar,r +1 xr +1 + · · · + ar n xn = 0.

Solving for (arr − λ), taking the modulus of the result, and making repeated use of the triangle inequality |a + b| ≤ |a| + |b|, where a and b are arbitrary complex numbers, leads to the inequality |λ − arr |

n. . . . an

195

Then, r > 0 for r = 1, 2, . . . , n, if and only if every zero of Pn (λ) has a negative real part. 28. (a) Numerical computation shows that the matrix ⎡ −2 A=⎣ 2 0

1 3 4

⎤ 5 1⎦ 2

has the eigenvalues 5.7238, −1.3619 + 1.9328i, and −1.3619 − 1.9328i. Apply the Routh–Hurwitz stability criterion to conﬁrm that not every zero of the characteristic polynomial has a negative real part. (b) Numerical computation shows that the matrix ⎤ ⎡ −2 −2 −3 0⎦ A = ⎣ 3 −1 −4 0 −3 has the eigenvalues −5.4873, −0.2563 − 1.4564i, and −0.2563 + 1.4564i. Apply the Routh–Hurwitz stability criterion to conﬁrm that every zero of the characteristic polynomial has a negative real part. An n × n matrix A is said to be similar to an n × n matrix B if there exists a nonsingular n × n matrix M such that B = M−1 AM. The relationship between A and B is said to constitute a similarity transformation between the two matrices. 29. If A and B are similar, show that detA = detB, and by substituting B = M−1 AM in detB and expanding the result, show that similar matrices have the same eigenvalues. 30. Verify the result of Exercise 29 by direct calculation by using ⎤ ⎡ 3 1 −1 1 0 −1⎦ and M = ⎣1 A = ⎣4 4 −2 1 2 ⎡

4 0 1

⎤ 1 1⎦ 0

to show that both A and B have the eigenvalues −1, 2, and 3. 31. Let the n × n elementary matrix E be obtained from the unit matrix I by interchanging its ith and jth rows (columns). By considering the product EQ, where Q is an n × n orthogonal matrix, prove that an orthogonal matrix remains orthogonal when its rows (columns) are interchanged.

196

Chapter 4

4.2

Eigenvalues, Eigenvectors, and Diagonalization

Diagonalization of Matrices

diagonal matrix

Our purpose in this section will be to examine the possibility of diagonalizing an n × n matrix A. The reason for this is to try to simplify the structure of A so that, in some ways, it reﬂects the simple properties of a diagonal matrix. Diagonalization ﬁnds many applications, some of which will be discussed later. Let D be the general n × n diagonal matrix ⎡ ⎤ λ1 0 0 . . . . 0 ⎢ 0 λ2 0 . . . . 0 ⎥ ⎢ ⎥ ⎥ D=⎢ (13) ⎢ . . . . . . . . . ⎥. ⎣ . . . . . . . . . ⎦ 0 0 0 . . . . λn Then, as already seen in Section 4.1, the eigenvalues of D are the entries λ1 , λ2 , . . . , λn on its leading diagonal, and the corresponding n linearly independent eigenvectors can be taken to be ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 0 0 ⎢0⎥ ⎢1⎥ ⎢0⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢0⎥ ⎢0⎥ ⎢0⎥ ⎥ ⎢ ⎥ ⎢ ⎥ (14) x1 = ⎢ ⎢ · ⎥ , x2 = ⎢ · ⎥ , . . . , xn = ⎢ · ⎥ . ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣·⎦ ⎣·⎦ ⎣·⎦ 0 0 1 The rule for matrix multiplication shows that ⎡ m λ1 0 0 . . . ⎢ 0 λm 0 . . . 2 ⎢ . . . . . . . Dm = ⎢ ⎢ ⎣ . . . . . . . 0 0 0 . . .

⎤ . 0 . 0⎥ ⎥ . . ⎥ ⎥, . . ⎦ . λm n

(15)

for any positive integer m, so Dm is easily computed and will have the same set of m m eigenvectors as D, though its eigenvalues will be λm 1 , λ2 , . . . , λn . In addition to these properties, it is obvious that detD = λ1 · λ2 · · · λn , so D will be nonsingular provided no entry on its leading diagonal is zero. As a result, when D is nonsingular, the rule for matrix multiplication shows that DD−1 = I, where ⎡ ⎤ 1/λ1 0 0 . . . . 0 ⎢ 0 0 ⎥ 1/λ2 0 . . . . ⎢ ⎥ −1 ⎢ ⎥. . . . . . . . . . . . (16) D =⎢ ⎥ ⎣ ⎦ . . . . . . . . . . . 0 0 0 0 1/λn We now state and prove the fundamental theorem on the diagonalization of n × n matrices. THEOREM 4.6 how to diagonalize a matrix

Diagonalization of an n × n matrix Let the n × n matrix A have n eigenvalues λ1 , λ2 , . . . , λn , not all of which need be distinct, and let there be n corresponding distinct eigenvectors x1 , x2 , . . . , xn , so that Axi = λi xi ,

i = 1, 2, . . . , n.

Section 4.2

Diagonalization of Matrices

197

Deﬁne the matrix P to be the n × n matrix in which the ith column is the eigenvector xi , with i = 1, 2, . . . , n, so that in partitioned form P = [x1 x2 · · · xn ], and let D be the diagonal matrix ⎡

λ1 0 0 . ⎢ 0 λ2 0 . ⎢ D=⎢ ⎢ . . . . . . ⎣ . . . . . . 0 0 0 .

. . . . .

. . . . .

⎤ . 0 . 0⎥ ⎥ . . ⎥ ⎥, . . ⎦ . λn

where the eigenvalue λi is in the ith position in the ith row. Then P−1 AP = D. Proof Consider the product B = AP. Then, by expressing P in partitioned form, we can write B as B = [Ax1

Ax2

...

Axn ].

Using the fact that Axi = λi xi allows this to be rewritten as B = [λ1 x1

λ2 x2

...

λn xn ] = PD,

showing that PD = AP. As the columns of P are linearly independent, P is nonsingular, so P−1 exists and we can premultiply by P−1 to obtain D = P−1 AP, and the theorem is proved.

General Remarks About Diagonalization (i) An n × n matrix can be diagonalized provided it possesses n linearly independent eigenvectors. (ii) A symmetric matrix can always be diagonalized. (iii) The diagonalizing matrix for a real n × n matrix A may contain complex elements. This is because although the characteristic polynomial of A has real coefﬁcients, its zeros either will be real or will occur in complex conjugate pairs. (iv) A diagonalizing matrix is not unique, because its form depends on the order in which the eigenvectors of A are used to form its columns. A useful consequence of the diagonalized form of a matrix is that it enables it to be raised to a positive integral power with the minimum of effort. This property will be used later when the matrix exponential is introduced. To see the ease with which an n × n matrix can be raised to a power when it is diagonalizable, we start by writing A in the form A = PDP−1 . We then have A2 = (PDP−1 )(PDP−1 ) = PDP−1 PDP−1 = PDDP−1 = PD2 P−1 ,

198

Chapter 4

Eigenvalues, Eigenvectors, and Diagonalization

so that, in general, Am = PDmP−1 ,

for m = 1, 2, . . . .

As evaluating Dm simply involves raising each entry on its leading diagonal to the power m, the evaluation of Am only involves three matrix multiplications. This last result was used without justiﬁcation in Section 3.2(f) when a stochastic matrix was raised to the power m (do not confuse the stochastic matrix P in that section with the orthogonalizing matrix P just deﬁned). EXAMPLE 4.9

Diagonalize the matrix

⎡

2 A = ⎣3 3

1 2 1

⎤ −1 −3⎦ , −2

and use the result to ﬁnd A5 . Solution Matrix A was examined in Example 4.1 and shown to have the eigenvalues λ1 = 2, λ2 = 1, and λ3 = −1, and the corresponding eigenvectors ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 1 0 x1 = ⎣1⎦ , x2 = ⎣0⎦ , and x3 = ⎣1⎦ . 1 1 1 Theorem 4.5 shows that a diagonalizing matrix P is given by ⎡ ⎤ 1 1 0 P = ⎣1 0 1⎦ , 1 1 1 and a routine calculation shows that ⎡ P−1

1 1 = ⎣ 0 −1 −1 0

⎤ −1 1⎦ . 1

Before ﬁnding A5 , and although it is unnecessary for what is to follow, it is instructive to check that when the matrix P−1 AP is formed, the eigenvalues appearing in the diagonal matrix D do so in the order in which the corresponding eigenvectors of A have been used to form the columns of P. This is seen to be so in this case because ⎤ ⎡ 2 0 0 0⎦. D = P−1 AP = ⎣0 1 0 0 −1 Returning to the calculation of A5 and using the expressions for P, P−1 , and D in A5 = PD5 P−1 gives ⎡ ⎤⎡ 5 ⎤⎡ ⎤ ⎡ ⎤ 2 0 0 1 1 0 1 1 0 32 31 −31 A5 = ⎣1 0 1⎦ ⎣ 0 15 0 ⎦ ⎣1 0 1⎦ = ⎣33 32 −33⎦. 1 1 1 1 0 1 33 31 −32 0 0 (−1)5 Had the eigenvectors been arranged in a different order when constructing P, a different but equivalent diagonal matrix would have been obtained. For example,

Section 4.2

if P had been written

1 P = ⎣0 1

1 1 1

⎤ 0 1⎦ , 1

⎡ 1 D = ⎣0 0

0 2 0

⎤ 0 0⎦, −1

⎡

D would have become

Diagonalization of Matrices

199

though after P−1 was found and A5 = PD5 P−1 was computed, the matrix A5 would, of course, remain the same. EXAMPLE 4.10

Diagonalize the matrix ⎡

0 ⎢−1 A=⎢ ⎣−1 1

0 2 0 0

⎤ 1 1 0 1⎥ ⎥. 2 1⎦ −1 0

Solution Matrix A was considered in Example 4.2, which showed that it had the eigenvalues λ1 = 0, λ2 = 1, λ3 = 1, and λ4 = 2, and that although the eigenvalue 1 occurred with algebraic multiplicity 2, the matrix still had the four linearly independent eigenvectors ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ −1 1 1 ⎢−1⎥ ⎢1⎥ ⎢0⎥ ⎥ ⎢ ⎥ ⎢ ⎥ (λ1 = 0) x1 = ⎢ ⎣−1⎦ , (λ2 = 1) x2 = ⎣1⎦ , (λ3 = 1) x3 = ⎣0⎦ , 0 1 1 and

(λ4 = 2)

⎡ ⎤ 0 ⎢1⎥ ⎥ x1 = ⎢ ⎣0⎦ . 0

Using these eigenvectors to form P gives ⎡ −1 1 ⎢−1 1 P=⎢ ⎣−1 1 1 0 from which it follows that

⎡

P−1

−1 ⎢−1 =⎢ ⎣ 1 0

0 0 0 1

1 0 0 1

⎤ 0 1⎥ ⎥, 0⎦ 0

⎤ 1 1 2 1⎥ ⎥. −1 0⎦ −1 0

200

Chapter 4

Eigenvalues, Eigenvectors, and Diagonalization

Because of the ordering of the eigenvectors, the diagonal matrix D will be ⎤ ⎡ 0 0 0 0 ⎢0 1 0 0⎥ ⎥ D=⎢ ⎣0 0 1 0⎦ , 0 0 0 2 where P−1 AP = D. We saw in Theorem 4.4 that a real symmetric n × n matrix A with distinct eigenvalues has a set of n mutually orthogonal linearly independent eigenvectors. It follows at once that if when constructing the diagonalizing matrix for A the normalized eigenvectors of A are used to form the columns of P, the resulting diagonalizing matrix will be an orthogonal matrix. This is often advantageous, because the properties of orthogonal matrices can simplify subsequent calculations that may arise. However, if an eigenvalue is repeated, the corresponding eigenvectors will not, in general, be orthogonal to the other eigenvectors, so although there will still be a set of n linearly independent eigenvectors, the set will no longer form an orthogonal set. Because of the frequency with which symmetric matrices arise in applications, and the fact that symmetric matrices with repeated eigenvalues are not unusual, it is reasonable to ask if it is possible for symmetric matrices always to be diagonalized by an orthogonal matrix and, if so, how this can be achieved. The answer to the question about the possibility of diagonalization by an orthogonal matrix is in the afﬁrmative. The method of arriving at an orthonormal set of vectors to be used when constructing P involves using a generalization of the Gram–Schmidt orthogonalization process introduced in Section 2.7 in the context of geometrical vectors in R3 . As an n element matrix vector is simply a vector in a vector space, an extension of the Gram–Schmidt orthogonalization process to include n-element matrix vectors can be used to construct an orthonormal set of n vectors from any set of n linearly independent eigenvectors that are always associated with an n × n symmetric matrix A. The required generalization of the orthogonalization process that leads to an orthonormal system is an immediate extension of the one derived in Section 2.7, so the details of its derivation will be omitted. Rule for the Gram–Schmidt orthogonalization process for matrix vectors orthogonalization of a set of linearly independent vectors

Let x1 , x2 , . . . , xn be a set of n element linearly independent nonorthogonal matrix column vectors. Then an equivalent orthonormal set of vectors p1 , p2 , . . . , pn can be constructed from the vectors x1 , x2 , . . . , xn , via an intermediate set of orthogonal nonnormalized vectors v2 , v2 , . . . , vn . The steps involved in the determination of the vectors p1 , p2 , . . . , pn are as follows: p1 v2 p2 vr pr

= = = = =

x1 / x1 , x2 − (p1 · x2 )p1 , v2 / v2 , xr − {(p1 · xr )p1 + (p2 · xr )p2 + · · · + (pr −1 · xr )pr −1 } vr / vr , for r = 2, 3, . . . , n.

Section 4.2

Diagonalization of Matrices

201

When the Gram–Schmidt orthogonalization process is applied to the eigenvectors of a real symmetric matrix A with repeated eigenvalues, the diagonalizing matrix P is constructed by using the vectors p1 , p2 , . . . , pn , obtained from the preceding scheme after starting with any linearly independent set of eigenvectors x1 , x2 , . . . , xn of A. Then, in partitioned form, P = [p1

p2

...

pn ]

and, as before, D = P−1 AP, where D is again a diagonal matrix with its diagonal elements equal to the eigenvalues of A arranged in the same order as the corresponding columns of P. This time, however, entries on the leading diagonal will be repeated as many times as the multiplicity of the eigenvalues concerned. EXAMPLE 4.11

Use the Gram–Schmidt orthogonalization process to construct an orthonormal set of vectors from the vectors ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 1 1 x1 = ⎣1⎦ , x2 = ⎣ 0⎦ , and x3 = ⎣2⎦ . 1 −1 0 Solution In this case the Gram–Schmidt orthogonalization process involves the three vectors x1 , x2 , and x3 , so a set of orthonormal vectors p1 , p2 , and p3 is given by the scheme p1 = x1 / x1 v2 = x2 − (p1 · x2 )p1 p2 = v2 / v2 v3 = x3 − {(p1 · x3 )p1 + (p2 · x3 )p2 } p3 = v3 / v3 . A series of straightforward calculations gives √ ⎤ ⎡ √ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ 1/√3 1 1 1/ 2 0√ ⎦ , p1 = ⎣1/√3⎦ , and v2 = ⎣ 0⎦ − 0 p1 = ⎣ 0⎦ , so p2 = ⎣ −1 −1 −1/ 2 1/ 3 and, ﬁnally, √ ⎤ ⎡ ⎡ ⎤ ⎡ √ ⎤ ⎡ ⎤ 1 1/ 2 −1/2 √ 1/√3 √ 0√ ⎦ = ⎣ 1 ⎦ , v3 = ⎣2⎦ − 3 ⎣1/√3⎦ − 1/ 2 ⎣ 0 −1/2 −1/ 2 1/ 3 so

√ ⎤ −1/ 6 ⎢ ⎥ p3 = ⎣ (2/3)⎦ . √ −1/ 6 ⎡

202

Chapter 4

Eigenvalues, Eigenvectors, and Diagonalization

EXAMPLE 4.12

Construct an orthogonal diagonalizing matrix for the symmetric matrix ⎡ ⎤ 4 0 0 A = ⎣0 1 2⎦ . 0 2 1 Solution This has the distinct eigenvalues λ1 = −1, λ2 = 3, and λ1 = 4, so the corresponding eigenvectors x1 , x2 , and x3 are orthogonal. Simple calculations show that ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 0 0 1 x1 = ⎣−1⎦ , x2 = ⎣1⎦ , and x3 = ⎣0⎦ . 1 1 0 The normalized eigenvectors are ⎡ ⎡ ⎤ ⎤ 0√ 0√ xˆ 1 = ⎣−1/√2⎦ , xˆ 2 = ⎣1/√2⎦ , and 1/ 2 1/ 2

⎡ ⎤ 1 xˆ 3 = ⎣0⎦ , 0

so the diagonalizing matrix P and the corresponding diagonal matrix D are ⎤ ⎡ ⎤ ⎡ 0√ 0√ 1 −1 0 0 P = ⎣−1/√2 1/√2 0⎦ and D = ⎣ 0 3 0⎦ . 0 0 4 1/ 2 1/ 2 0 EXAMPLE 4.13

Construct an orthogonal diagonalizing matrix for the real symmetric matrix ⎡ ⎤ −1 2 4 2 −2⎦ . A=⎣ 2 4 −2 −1 Solution This has the eigenvalues λ1 = −6, λ2 = 3, and λ3 = 3, so as the eigenvalue 3 has multiplicity 2, the corresponding set of eigenvectors x1 , x2 , and x3 will not be orthogonal. The eigenvectors x1 , x2 , and x3 are easily shown to be ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ −2 1 0 x1 = ⎣ 1⎦ , x2 = ⎣2⎦ , and x3 = ⎣−2⎦ . 2 0 1 Applying the Gram–Schmidt orthogonalization process to vectors x1 , x2 , and x3 , as in Example 4.11, after some straightforward calculations we arrive at the orthonormal set √ ⎤ ⎡ ⎤ ⎡ √ ⎤ ⎡ 4/(3√5) −2/3 1/√5 p1 = ⎣ 1/3⎦ , p2 = ⎣2/ 5⎦ , and p3 = ⎣−2/(3 5)⎦ . √ 2/3 0 5/3 In this case an orthogonal diagonalizing matrix is √ √ ⎤ ⎡ −2/3 1/√5 4/(3√5) P = ⎣ 1/3 2/ 5 −2/(3 5)⎦ , √ 2/3 0 5/3

Section 4.2

and the corresponding diagonal matrix is ⎡ −6 0 D=⎣ 0 3 0 0

Diagonalization of Matrices

203

⎤ 0 0⎦ . 3

To close this section we state the important Cayley–Hamilton theorem, which is true for all square matrices, though before considering the theorem we ﬁrst deﬁne a matrix polynomial. A matrix polynomial involving an n × n matrix A is an expression of the form Am + b1 Am−1 + b2 Am−2 + · · · + bm−1 A + bmI, in which m is an integer and b1 , b2 , . . . , bm are real or complex numbers. THEOREM 4.7 a matrix satisﬁes its own characteristic equation

The Cayley–Hamilton theorem Let Pn (λ) be the characteristic polynomial of an arbitrary n × n square matrix A. Then A satisﬁes its own characteristic equation, and so is a solution of the matrix polynomial equation Pn (A) = 0. Proof For simplicity, we only prove the theorem for real symmetric matrices, though it is true for every n × n matrix. If A is a real n × n symmetric matrix, then from Theorem 4.6 we may write A = PDP−1 . Let the characteristic polynomial of A be Pn (λ) = (−1)n {λn + c1 λn−1 + · · · + cn−1 λ + cn }. Then replacing λ by A converts Pn (λ) to the matrix polynomial Pn (A) = (−1)n {An + c1 An−1 + · · · + cn−1 A + cn I}, but Ar = PDr P−1 , so Pn (A) = (−1)n {P{Dn + c1 Dn−1 + · · · + cn−1 D + cn In }P−1 }. The ith row of the matrix polynomial Dn + c1 Dn−1 + · · · + cn−1 D + cn I is simply λin + c1 λin−1 + · · · + cn−1 λi + cn , but this is Pn (λi ), and it must vanish for i = 1, 2, . . . , n because λi is an eigenvalue of A. Thus, Dn + c1 Dn−1 + · · · + cn−1 D + cn I = 0, showing that Pn (A) = P{0}P−1 = 0, and the result is proved.

EXAMPLE 4.14

Verify the Cayley–Hamilton theorem for the matrix 2 1 A= . 5 2 Solution The characteristic polynomial is P2 (λ) = λ2 − 4λ − 1, and 9 4 9 4 2 1 1 0 0 , so P2 (A) = −4 − = A2 = 20 9 20 9 5 2 0 1 0

0 . 0

Finding A−1 from the Cayley–Hamilton theorem If the n × n matrix A is nonsingular, the following interesting result can be obtained directly from the Cayley–Hamilton theorem. Let the characteristic

204

Chapter 4

Eigenvalues, Eigenvectors, and Diagonalization

polynomial of A be Pn (λ) = (−1)n {λn + c1 λn−1 + · · · + cn−1 λ + cn }, so from Theorem 4.7 An + c1 An−1 + · · · + cn−1 A + cn I = 0. The matrix A−1 exists because by hypothesis A is nonsingular, so premultiplication of the preceding equation by A−1 , followed by a rearrangement of terms, allows A−1 to be expressed in terms of powers of A through the result A−1 = (−1/cn ){An−1 + c1 An−2 + · · · + cn−1 I}.

EXAMPLE 4.15

(17)

Use the result of equation (17) to ﬁnd A−1 for the nonsingular matrix 2 1 A= . 5 2 Solution Matrix A was considered in Example 4.14, where it was found that the characteristic polynomial P2 (λ) = λ2 − 4λ − 1, so in terms of (17) we see that c1 = −4 and c2 = −1. Thus, ( 2 1 1 0 −2 1 A−1 = −1/(−1) −4 = . 5 2 0 1 5 −2

Summary

This section has described how an n × n matrix can be diagonalized when it possesses n linearly independent eigenvectors. The diagonalization was shown not to be unique, since its form depends on the order in which the eigenvectors are used to construct the diagonalizing matrix P. Sometimes, when a linearly independent set of n vectors has been obtained, it is desirable to replace it by an equivalent set of n orthogonal or orthonormal vectors. The section closed by showing how this can be accomplished by means of the Gram–Schmidt orthogonalization procedure.

EXERCISES 4.2 In Exercises 1 through 12, ﬁnd a diagonalizing matrix P for the given matrix, in each case using the fact that the zeros of the characteristic polynomial are small integers that can be found by trial and error. ⎡

⎤

−2 −3 −1 2 1⎦ . 1. ⎣ 1 3 3 2 ⎤ ⎡ 3 1 4 2. ⎣−4 −2 −4⎦ . −1 −1 2 ⎤ ⎡ 3 1 −2 3. ⎣6 2 −6⎦. 4 1 −3

⎡

−6 4. ⎣ 2 7 ⎡ −1 5. ⎣ 2 2 ⎡ 14 6. ⎣ −8 −26

−10 3 10 2 −1 −2 2 −3 −4

⎤ −4 2⎦ . 5 ⎤ −2 2⎦ . 3 ⎤ 8 −4⎦. −15

⎡

5 7. ⎣ 2 −2 ⎡ 12 8. ⎣ −6 −22 ⎡ 2 9. ⎣ 1 −2

−2 1 2

⎤ 2 2⎦ . 1

⎤ 4 6 −2 −3⎦ . −8 −11 ⎤ 0 0 −1 2⎦. 0 1

⎤ 12 −4 8 2 −4⎦ . 10. ⎣ −6 −20 8 −14 ⎤ ⎡ −6 2 −4 0 −4⎦ . 11. ⎣−4 4 −2 2 ⎤ ⎡ −7 0 −6 3⎦. 12. ⎣ 3 −1 9 0 8 ⎡

In Exercises 13 through 16 use the Gram–Schmidt orthogonalization process with the given set of vectors to ﬁnd (a) an equivalent set of orthogonal vectors and (b) an orthonormal set.

Section 4.3 ⎡ ⎤ 1 13. ⎣1⎦ , 1 ⎡ ⎤ 2 14. ⎣1⎦ , 1

⎡ ⎤ ⎡ ⎤ 0 0 ⎣1⎦ , ⎣0⎦ . 1 1 ⎡ ⎤ ⎡ ⎤ 1 0 ⎣−1⎦ , ⎣2⎦ . 1 1

⎤ −1 15. ⎣ 1⎦ , 0 ⎡ ⎤ −1 16. ⎣ 2⎦ , 0 ⎡

⎤ 2 ⎣ 1⎦ , −1 ⎡ ⎤ 1 ⎣ 1⎦ , −1 ⎡

⎤ 1 ⎣−2⎦ . 2 ⎡ ⎤ 1 ⎣−1⎦ . 1 ⎡

In Exercises 17 through 22 ﬁnd an orthogonal diagonalizing matrix P for the given symmetric matrix. ⎤ ⎤ ⎡ ⎡ 4 1 0 3 0 0 19. ⎣1 4 0⎦ . 17. ⎣0 3 1⎦ . 0 0 3 0 1 3 ⎤ ⎤ ⎡ ⎡ 2 1 1 5 1 0 20. ⎣1 2 1⎦ . 18. ⎣1 5 0⎦ . 1 1 2 0 0 2

4.3

⎡

4 21. ⎣2 0

Special Matrices with Complex Elements 2 4 0

⎤ 0 0⎦ . 2

⎡ 4 1 22. ⎣1 4 1 1

205

⎤ 1 1⎦ . 4

23. Verify by direct calculation that the matrix in Exercise 1 satisﬁes the Cayley–Hamilton theorem. 24. Verify by direct calculation that the matrix in Exercise 7 satisﬁes the Cayley–Hamilton theorem. In Exercises 25 through 28 use (17) to ﬁnd A−1 and check the result by showing that AA−1 = I. ⎤ ⎡ 2 1 0 2 3 . 25. A = 1 2⎦ . −1 4 27. A = ⎣−2 0 −1 −2 5 1 ⎤ ⎡ . 26. A = 1 0 2 3 −2 28. A = ⎣3 1 0⎦ . 0 2 4

Special Matrices with Complex Elements In the previous section it was seen that one way in which matrices with complex elements can occur is when the eigenvectors of an arbitrary n × n matrix are used to construct a diagonalizing matrix. This is not the only reason for considering n × n matrices with complex elements, because the following three special types of matrices arise naturally in applications of mathematics to physics and engineering, and elsewhere. Hermitian, skew-Hermitian, and unitary matrices Let A = [ai j ] be an n × n matrix with possibly complex elements. Then: T

A is called an Hermitian matrix if A = A, so that a kj = a jk; T

A is called a skew-Hermitian matrix if A = −A, so that a kj = −a jk; T

U is called a unitary matrix if U = U−1 . The basic properties of these three types of matrices follow almost directly from their deﬁnitions.

Basic Properties of Hermitian, Skew-Hermitian, and Unitary Matrices 1. The elements on the leading diagonal of an Hermitian matrix are real, because a ii = aii , and this is only possible if aii is real. 2. The elements on the leading diagonal of a skew-Hermitian matrix are either purely imaginary or 0. This follows from the fact that a ii = −aii , so the real part of aii must equal its negative, and this is only possible if aii is purely imaginary or 0.

206

Chapter 4

Eigenvalues, Eigenvectors, and Diagonalization

3. If the elements of an Hermitian matrix are real, then the matrix is a real T symmetric matrix, because then A = AT , and the deﬁnition of an Hermitian matrix reduces to the deﬁnition of a real symmetric matrix. 4. If the elements of a skew-Hermitian matrix are real, then the matrix is a skewsymmetric matrix, because then the deﬁnition of a skew-Hermitian matrix reduces to the deﬁnition of a skew-symmetric matrix. 5. Any n × n matrix A of the form A = B + iC, where B is a real symmetric matrix and C is a real skew-symmetric matrix, is an Hermitian matrix. This follows directly from Properties 3 and 4. 6. Any n × n matrix A can be written in the form A = B + C, where B is Hermitian and C is a skew-Hermitian. To see this we write T T T A = (1/2)(A + A ) + (1/2)(A − A ), and then set B = (1/2)(A + A ) and T T T C = (1/2)(A − A ). Then B = (1/2)(AT + A) = (1/2)(A + A ) = B and T CT = (1/2)(AT − A) = −(1/2)(A − A ) = −C, showing that B is Hermitian and C is skew-Hermitian. T 7. A real unitary matrix is an orthogonal matrix, because in that case A = AT , causing the deﬁnition of a unitary matrix to reduce to the deﬁnition of an orthogonal matrix. 8. The determinant of a unitary matrix is ±1. This result is established in essentially the same way as the result of Theorem 4.4(i), so the argument will not be repeated. EXAMPLE 4.16

The following are examples of Hermitian, skew-Hermitian, and unitary matrices. Hermitian matrix : ⎡

3 A = ⎣ 2 − 5i −7 − 3i

2 + 5i 0 1+i

⎤ −7 + 3i 1 −i ⎦. 4

−3 − 2i −2i −5

⎤ −6 − 4i 5 ⎦. 0

Skew-Hermitian matrix: ⎡

4i B = ⎣3 − 2i 6 − 4i Unitary matrix: ⎡1 + i

⎢ 2 ⎢ U=⎢ ⎢1 + i ⎣ 2 0

−1 + i 2 1−i 2 0

⎤ 0

⎥ ⎥ ⎥. 0⎥ ⎦ 1

It can be seen from Properties 3, 4, and 7 that Hermitian, skew-Hermitian, and unitary matrices are, respectively, generalizations of symmetric, skew-symmetric, and orthogonal real-valued matrices. Accordingly, it is to be expected that some of the properties exhibited by these real-valued matrices are shared by their complex generalizations, and this is indeed the case as we now show.

Section 4.3

THEOREM 4.8

Special Matrices with Complex Elements

207

Eigenvalues of Hermitian, skew-Hermitian, and unitary matrices (i) The eigenvalues of an Hermitian matrix are real. (ii) The eigenvalues of a skew-Hermitian matrix are either purely imaginary or 0. (iii) The eigenvalues λ of a unitary matrix are all such that |λ| = 1. Proof (i) Apart for the need to introduce the complex conjugate operation, the proof is essentially the same as that of Theorem 4.4 for symmetric matrices, and so it is omitted. (ii) Let x be the eigenvector of A corresponding to the eigenvalue λ, so Ax = λx. Then xT Ax = λxT x, from which we have λ = xT Ax/xT x, but xT x = x1 x 1 + x2 x 2 + · · · + xn x n is real. However, A = −AT , so xT Ax = −xT Ax, so we can write λ = xT Ax/xT x = −xT Ax/xT x. The product xT x is real, so this last result shows that the complex number λ equals the negative of its complex conjugate, and this is only possible if λ is purely imaginary or 0, so the proof is complete. (iii) Apart from the need to introduce the complex conjugate operation, the proof is essentially that of Theorem 4.5(iii), so it will be omitted. The location of the eigenvalues of these complex matrices and of their corresponding real forms are illustrated in Fig. 4.3.

Imaginary axis Skew-Hermitian and skew-symmetric matrix eigenvalues are located on the imaginary axis

−1

i

Unit circle

−i

Unitary and orthogonal matrix eigenvalues are located on the unit circle

1

Real axis

Hermitian and symmetric matrix eigenvalues are located on the real axis

FIGURE 4.3 The location of the eigenvalues of Hermitian, skew-Hermitian, and unitary matrices in the complex plane.

208

Chapter 4

Eigenvalues, Eigenvectors, and Diagonalization

If the deﬁnitions of an inner product and a norm are generalized, the concept of orthogonality can be extended to include vectors with complex elements. These generalizations have many applications, but they will only be used here to prove the orthogonality of the rows and columns of unitary matrices. As the norm of a vector is essentially its length and so must be nonnegative, the previous deﬁnition of a norm in terms of an inner product must be modiﬁed in such a way that the inner product and norm of a complex vector coincide with those for a real vector when purely real vectors are considered. This is achieved by introducing the complex conjugate operation into the deﬁnition of an inner product. Inner product of complex vectors Let w = [w1 , w2 , . . . , wn ]T and z = [z1 , z2 , . . . , zn ]T be two column vectors with complex elements. Then the inner product of the column vectors w and z, again denoted by w · z, is deﬁned as w · z = wT z, so that w · z = w 1 z1 + w 2 z2 + · · · + w n zn .

(18)

Norm of complex vectors The norm of a vector z, again denoted by z , is deﬁned as the nonnegative number z = (z · z)1/2 = (zT z)1/2 = (z1 z1 + z2 z2 + · · · + zn zn )1/2 = (|z1 |2 + |z2 |2 + · · · + |zn |2 )1/2 .

(19)

It can be seen from the preceding deﬁnition that the inner product of two arbitrary complex vectors is a complex number. However, the deﬁnition of the norm of a complex vector z is a real nonnegative number, as would be expected. EXAMPLE 4.17

If w = [1 + 2i, 3 − i, i]T and z = [2 + i, 1 − i, 1 + 3i]T , ﬁnd w · z and z . Solution w · z = (1 + 2i)(2 + i) + (3 − i)(1 − i) + i(1 + 3i) = 11 − 6i, and z = [|2 + i|2 + |1 − i|2 + |1 + 3i|2 ]1/2 = 171/2 . We are now in a position to generalize the concept of an orthonormal system of real vectors to a system of complex vectors that will be called a unitary system if the vectors satisfy the following conditions. A unitary system A set of complex vectors z1 , z2 , . . . , zn is said to form a unitary system if zi · z j =

ziT z j

=

0 1

if i = j if i = j.

(20)

Section 4.3

THEOREM 4.9

Special Matrices with Complex Elements

209

The eigenvectors of a unitary matrix The rows and columns of a unitary matrix each form a unitary system of vectors. T

T

Proof By deﬁnition the n × n matrix U is unitary if U = U−1 , so that U U = I. The element in the ith row and jth column of I is the inner product xi · x j = xiT x j , where xi and x j are the ith and jth columns of U. Consequently, xiT x j

=

0 1

if i = j if i = j,

showing that the columns of U form a unitary system. The rows also form a uniT T tary system, because taking the transpose of U U we ﬁnd that (U U)T = UT U = IT = I.

Summary

Matrices with complex elements arise in a variety of different applications, and from among these matrices, the most important are Hermitian, skew-Hermitian, and unitary matrices. Hermitian and skew-Hermitian matrices are the complex analogues of real symmetric and skew-symmetric matrices, respectively, and unitary matrices are the complex analogue of real orthogonal matrices. This section derived and illustrated by means of examples the most important properties of these matrices, and then introduced the inner product and norm of matrices with complex elements.

EXERCISES 4.3 In Exercises 1 through 4 write the given matrix as the sum of an Hermitian and a skew-Hermitian matrix. ⎤ ⎡ 1+i 3 + i 3 + 2i 2 4 + i ⎦. 1. ⎣−1 + 3i −3 − 2i 2 + 3i 4 + 2i ⎤ ⎡ 0 3 + i 1 + 2i 2 ⎦. 2. ⎣1 − 5i 1 + i 1 + 4i −2i 3 ⎤ ⎡ 4 − 2i 1 + i 2 + 2i 4 ⎦. 3. ⎣−1 − 3i 1 + 2i 0 2 0 ⎤ ⎡ 3 + i 4 − i 5 + 2i 2 ⎦. 4. ⎣2 + i 1 + 2i −1 2i 4−i In Exercises 5 through 8 ﬁnd the eigenvalues of the Hermitian matrices and hence conﬁrm the result of Theorem 4.8(a) that they are real. 3 2 − 3i 1 2−i . 7. . 5. 2 + 3i 1 2+i 2 2 2 + 2i −4 2 − 2i . 6. 8. . 1 − 2i 3 2 + 2i 3

In Exercises 9 through 12 ﬁnd the eigenvalues of the skewHermitian matrices and hence conﬁrm the result of Theorem 4.8(b) that they are purely imaginary. 0 3 + 2i i 3+i . 11. . 9. −3 + 2i 0 −3 + i 2i 4i 2 + 3i 3i 2−i . 12. . 10. −2 + 3i i −2 − i 0 13. Show the following matrix is unitary: √ √ 1/√ 2 −i/√ 2 . i/ 2 1/ 2 In Exercises 14 and 15 show the matrices are unitary, ﬁnd their eigenvalues and eigenvectors, and conﬁrm that the eigenvalues all lie on the unit circle. √ √ (i − 1)/√2 (1 − i)/√2 . (i − 1)/ 2 (i − 1)/ 2 √ √ (1 + i)/√2 −(1 + i)/√ 2 . 15. (1 + i)/ 2 (1 + i)/ 2 14.

210

Chapter 4

4.4

Eigenvalues, Eigenvectors, and Diagonalization

Quadratic Forms A hom*ogeneous polynomial P(x) of degree two of the form P(x) ≡ a11 x12 + a22 x22 + · · · + ann xn2 + 2a12 x1 x2 + 2a13 x1 x3 + · · · + 2an−1,n xn−1 xn ,

real quadratic form

(21)

in which the coefﬁcients ai j and the variables in x(x1 , x2 , . . . , xn ) are real numbers, is called a real quadratic form in the variables x1 , x2 , . . . , xn . The term hom*ogeneous of degree two or, more precisely, algebraically hom*ogeneous of degree two, means that each term in P is quadratic in the sense that it involves a product of precisely two of the variables x1 , x2 , . . . , xn . The terms involving the products xi x j with i = j are called the mixed product or cross-product terms. Real quadratic forms A real quadratic form P(x) is a hom*ogeneous polynomial in the real variables x1 , x2 , . . . , xn of the form shown in (21). If A is a real symmetric n × n matrix and x is an n-element column vector deﬁned as ⎡ ⎤ ⎡ x1 a11 ⎢ x2 ⎥ ⎢a12 ⎢ ⎥ ⎢ ⎥ ⎢ x=⎢ ⎢ · ⎥ and A = ⎢ . . ⎣·⎦ ⎣. . xn a1n

a12 . . a22 . . . . . . . . . . . . a2n . .

. . . . .

⎤ a1n a2n ⎥ ⎥ . .⎥ ⎥, . .⎦ ann

(22)

then P(x) can be written in the matrix form P(x) ≡ xT Ax.

(23)

There is no loss of generality in requiring A to be a symmetric matrix, because if the coefﬁcient of a cross-product term xi x j equals bi j , this can always be rewritten as bi j = 2ai j allowing the terms ai j to be positioned symmetrically about the leading diagonal, as shown in the matrix A in (22). Exercise 30 at the end of this section shows how the deﬁnition of a real quadratic form can be extended to any real n × n matrix. EXAMPLE 4.18

Express the quadratic form P(x) ≡ 3x12 − 2x22 + 4x32 + x1 x2 + 3x1 x3 − 2x2 x3 as the matrix product P(x) = xT Ax. Solution By deﬁning x and A as ⎡ ⎤ ⎡ x1 3 1/2 x = ⎣x2 ⎦ , A = ⎣1/2 −2 3/2 −1 x3 we can write P(x) = xT Ax.

⎤ 3/2 −1 ⎦ , 4

Section 4.4

Quadratic Forms

211

Quadratic forms arise in various ways; for example, in mechanics a quadratic form can describe the ellipsoid of inertia of a solid body, the angular momentum of a solid body rotating about an axis, and the kinetic energy of a system of moving particles. Other areas in which quadratic forms occur include the geometry of conics in two space dimensions and of quadrics in three space dimensions, optimization problems, crystallography, and in the classiﬁcation of partial differential equations (see Chapter 18). We now give a general deﬁnition of a quadratic form that allows both the matrix A and the vector x to contain complex elements. General quadratic forms quadratic form and vectors with complex elements

Let the elements of an n × n matrix A = [ai j ] and an n-element column vector z be complex numbers. Then a quadratic form P(z) involving the variables z1 , z2 , . . . , zn of vector z is an expression of the form n

P(z) = zT Az =

ai j zi z j .

(24)

i=1, j=1

This deﬁnition is seen to include real quadratic forms, because when the elements of A and z are real, result (24) reduces to the real quadratic form deﬁned in (23). The structure of a quadratic form becomes clearer if a change of variables is made that removes the mixed product terms, leaving only the squared terms. This is called the reduction of the quadratic form to its standard form, also known as its canonical form. The next theorem shows how such a simpliﬁcation can be achieved. THEOREM 4.10 how to reduce a quadratic form to a sum of squares

Reduction of a quadratic form Let the n × n real symmetric matrix A have the eigenvalues λ1 , λ2 , . . . , λn , and let Q be an orthogonal matrix that diagonalizes A, so that QT AQ = D, where D is a diagonal matrix with the eigenvalues of A as the elements on its leading diagonal. Then the change of variable x = Qy, involving the column vectors x = [x1 , x2 , . . . , xn ]T and y = [y1 , y2 , . . . , yn ]T , transforms the real quadratic form P(x) ≡ xT Ax into the standard form P(x) ≡

n

ai j xi x j = λ1 y12 + λ2 y22 + · · · + λn yn2 .

i=1, j=1

Proof The proof uses the fact that because Q is an orthogonal matrix, QT AQ = D. Substituting x = Qy into the real quadratic form xT Ax gives P(x) ≡ xT Ax = (Qy)T AQy = yT QT AQy = yT Dy = λ1 y12 + λ2 y22 + · · · + λn yn2 . It follows immediately from Theorem 4.10 that the standard form of P(x) is determined once the eigenvalues of A are known and, when needed, the transformation of coordinates between x and y is given by x = Qy or, equivalently, by y = QT x.

212

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Eigenvalues, Eigenvectors, and Diagonalization

quadratic forms and principal axes

EXAMPLE 4.19

The next example provides a geometrical interpretation of Theorem 4.10 in the context of rigid body mechanics. In order to understand its implications it is necessary to know that if an origin O is taken at an arbitrary point inside a solid body, and an orthogonal set of axes O{x1 , x2 , x3 } is located at O, nine moments and products of inertia of the body can be deﬁned relative to these axes and displayed in the form of a 3 × 3 inertia matrix. The moment of inertia of the body about any line passing through the origin O is proportional to the length of the segment of the line that lies between O and the point where it intersects a three-dimensional surface deﬁned by a quadratic form determined by the inertia matrix. When the surface determined by the inertia matrix is scaled so the length of the line from O to its point of intersection with the surface equals the reciprocal of the moment of inertia about that line, the surface is called the ellipsoid of inertia. If the orientation of the O{x1 , x2 , x3 } axes is chosen arbitrarily, the resulting quadratic form will be complicated by the presence of mixed product terms, but a suitable rotation of the axes can always remove these terms and lead to the most convenient orientation of the new system of axes O{y1 , y2 , y3 }. In the geometry of both conics and quadrics, and also in mechanics, new axes obtained in this way that lead to the elimination of mixed product terms are called the principal axes, and it is because of this that Theorem 4.10 is often known as the principal axes theorem. The ellipsoid of inertia of a solid body is given by P(x) ≡ 4x12 + 4x22 + x32 − 2x1 x2 . Find its standard form in terms of a new orthogonal set of axes O{y1 , y2 , y3 }, and ﬁnd the linear transformation that connects the two sets of coordinates. Solution The quadratic form P(x) can be written as xT Ax by deﬁning ⎡ ⎤ ⎡ ⎤ x1 4 −1 0 4 0⎦ . x = ⎣x2 ⎦ and A = ⎣−1 0 0 1 x3 The eigenvalues of A are λ1 = 1, λ2 = 5, and λ3 = 3, so the standard form of P(x) is P(x) ≡ y12 + 5y22 + 3y32 . The eigenvalues and corresponding normalized eigenvectors of A are √ ⎤ ⎡ ⎤ ⎡ ⎡ √ ⎤ 0 −1/√2 1/√2 λ1 = 1, xˆ 1 = ⎣0⎦ , λ2 = 5, xˆ 2 = ⎣ 1/ 2⎦ , λ3 = 3, xˆ 3 = ⎣1/ 2⎦ , 1 0 0 so the orthogonal diagonalizing matrix for A is √ √ ⎤ ⎡ 0 −1/√2 1/√2 Q = ⎣0 1/ 2 1/ 2⎦ , 1 0 0 and the change of variables between x and y determined by x = Qy becomes √ √ x1 = (−y2 + y3 )/ 2, x2 = (y2 + y3 )/ 2, x3 = y1 . The equation P(x) = constant is seen to be an ellipsoid for which O{y1 , y2 , y3 } are the principal axes.

Section 4.4

EXAMPLE 4.20

Quadratic Forms

213

Reduce the quadratic part of the following expression to its standard form involving the principal axes O{y1 , y2 }, and hence ﬁnd the form taken by the complete expression in terms of y1 and y2 : x12 + 4x1 x2 + 4x22 + x1 − 2x2 . Solution The quadratic part of the expression is x12 + 4x1 x2 + 4x22 , and this can be expressed in the form xT Ax by setting 1 2 x1 and A = . x= x2 2 4 The eigenvalues and eigenvectors of A are 1 −2 λ1 = 5, x1 = and λ2 = 0, x2 = , 2 1 so the orthogonal diagonalizing matrix is √ √ 5 1/√5 −2/√5 and D = Q= 0 2/ 5 1/ 5

0 . 0

Making the variable change x = Qy shows the standard form of the quadratic terms to be 5y12 . The variables x1 and x2 are related to y1 and y2 by the expressions √ √ √ √ √ x1 = y1 / 5 − 2y2 / 5 and x2 = 2y1 / 5 + y2 / 5, so x1 − 2x2 = −(3y1 + 4y2 )/ 5. In terms of the principal axes involving the coordinates y1 and y2 , the complete expression x12 + 4x1 x2 + 4x22 + x1 − 2x2 reduces to √ x12 + 4x1 x2 + 4x22 + x1 − 2x2 = 5y12 − (3y1 + 4y2 )/ 5. Quadratic forms P(x) are classiﬁed according to the behavior of the sign of P(x) when x is allowed to take all possible values. In terms of vector spaces, this amounts to saying that if the vector x in P(x) is an n vector, then x ∈ Rn . how to classify quadratic forms

Classiﬁcation of quadratic forms Let P(x) be a quadratic form. Then: 1. P(x) is said to be positive deﬁnite if P(x) > 0 for all x = 0 in Rn , with P(x) = 0 if, and only if, x = 0. P(x) is said to be negative deﬁnite if in this deﬁnition the inequality sign > is replaced by 3. 0⎦ 0

Substituting into (32) and adding the scaled matrices gives ⎡ ⎤ 1 2t t + 3t 2 t − (3/2)t 2 + t 3 ⎢0 1 3t −2t + (3/2)t 2 ⎥ ⎥. eAt = ⎢ ⎦ ⎣0 0 1 t 0 0 0 1 Setting t = 1 in this result, we ﬁnd that ⎡ 1 2 4 ⎢ 0 1 3 eA = ⎢ ⎣0 0 1 0 0 0

⎤ 1/2 −1/2⎥ ⎥. 1⎦ 1

Differentiation of the terms in the matrix eAt gives ⎡ ⎤ 0 2 1 + 6t 1 − 3t + 3t 2 ⎢0 0 3 −2 + 3t ⎥ ⎥, d[eAt ]/dt = ⎢ ⎣0 0 ⎦ 0 1 0 0 0 0 and as this is equal to AeAt , it conﬁrms the result d[eAt ]/dt = AeAt .

Section 4.5

The Matrix Exponential

219

It was possible to sum the inﬁnite series of matrices in Example 4.22 because only a diagonal matrix was involved, so its powers could be determined immediately. The situation was different in Example 4.23 because An = 0 for n > 3 so that only a ﬁnite sum of matrices was involved. Matrices such as those in Example 4.23, which vanish when raised to a ﬁnite power, are called nilpotent matrices. If A is neither diagonal nor nilpotent, but is diagonalizable, in order to determine Am it is ﬁrst necessary to ﬁnd the diagonalizing matrix P for A. Then, if D is the diagonalized form of A, so that D = P−1 AP, it follows that A = PDP−1 and A2 = (PDP−1 )(PDP−1 ) = PD2 P−1 ,

A3 = AA2 = (PDP−1 )(PD2 P−1 ) = PD3 P−1 ,

so that in general, Am = PDmP−1 . Using this result in the matrix exponential gives eAt = I + (PDP−1 )t + PD2 P−1

t2 + ···, 2!

and writing I = PP−1 reduces this to ( t2 t3 eAt = P In + Dt + D2 + D3 + · · · P−1 . 2! 3!

(36)

The form of eA follows directly from this by setting t = 1. EXAMPLE 4.24

Determine eAt given that −2 A= 6

−3 , 7

and use the result to ﬁnd eA . Solution The eigenvalues and eigenvectors of A are −1 1 λ1 = 1, x1 = and λ2 = 4, x2 = , 1 −2 so the diagonalizing matrix −1 1 −2 P= and P−1 = 1 −2 −1

−1 1 , while D = −1 0

0 . 4

Substituting these matrices into (36) gives 1 0 t3 1 0 t2 1 0 1 0 + + · · · P−1 eAt = P + t+ 0 1 0 4 0 43 3! 0 42 2! t 0 e (et − e4t ) (2et − e4t ) −1 . P =P = 0 e4t (2e4t − 2et ) (2e4t − et )

220

Chapter 4

Eigenvalues, Eigenvectors, and Diagonalization

Finally, setting t = 1 we ﬁnd that (2e − e4 ) A e = (2e4 − 2e)

(e − e4 ) . (2e4 − e)

So far, the properties of the matrix exponential have closely paralleled those of the ordinary exponential, but there are signiﬁcant differences, one of the most important being that in general, even when A + B is deﬁned, eA eB = e(A+B) . To determine under what conditions the equality is true, we consider the matrix exponentials eAt eBt and e(A+B)t and require their derivatives to be equal when t = 0. Differentiating each expression once with respect to t gives d[eAt eBt ]/dt = AeAt eBt + eAt BeBt and d e(A+Bt) /dt = (A + B)e(A+B)t , and these are seen to be equal when t = 0. Next, computing d2 [eAt eBt ]/dt 2 and d2 [e(A+B)t ]/dt 2 , we obtain d2 [eAt eBt ]/dt 2 = A2 eAt eBt + 2AeAt BeBt + eAt B2 eBt and d2 e(A+B)t /dt 2 = (A + B)2 e(A+B)t = (A2 + AB + BA + B2 )e(A+B)t . Setting t = 0 shows that these two expressions are only equal if AB = BA; that is, the matrices A and B must commute, and the same condition applies when all higher order derivatives are considered. This has established the fundamental result that when does e A e B = e(A+B)

eA eB = e(A+B)

if, and only if, AB = BA.

(37)

Replacing B by −A in (37) gives eA e−A = e0 = I,

(38)

from which we see, as would be expected, that e−A is the inverse of eA , and also that as e−A is nonsingular it always exists. This parallels the real variable situation, because e−x exists for all ﬁnite x. Having arrived at a satisfactory deﬁnition of eAt and determined its derivatives, we are now in a position to deﬁne the antiderivative eAt dt as the matrix obtained by integrating each element of eAt with respect to t, it being understood that when this is done an arbitrary constant n × n matrix must always be added to the result representing the arbitrary additative constant of integration that arises when each term of eAt is integrated. EXAMPLE 4.25

Find

eAt dt given that A is the matrix in Example 4.21.

Solution It was shown in Example 4.21 that if ⎡ 3t ⎡ ⎤ e 3 0 0 A = ⎣0 −2 0⎦ then eAt = ⎣ 0 0 0 4 0

e−2t 0

⎤ 0 0 ⎦, e4t

Section 4.5

so that

The Matrix Exponential

221

⎡ 3t ⎤ e /3 + c1 0 0 ⎦ 0 −e−2t/2 + c2 0 eAt dt = ⎣ 4t 0 0 e /4 + c3 ⎡ 3t ⎤ ⎡ ⎤ e /3 0 0 c1 0 0 −e−2t/2 0 ⎦ + ⎣ 0 c2 0 ⎦ , =⎣ 0 0 0 c3 0 0 e4t/4

where c1 , c2 , and c3 are arbitrary constants. Applications of the matrix exponential to ordinary differential equations are to be found in reference [3.15].

Summary

The matrix exponential e At arises as the natural extension of the exponential function when solving a system of linear ﬁrst order constant coefﬁcient differential equations in the matrix form dx/dt = Ax. This section has described how e At can be calculated in simple cases and shown that e A e B = e A+B if, and only if, AB = BA. A different way of ﬁnding e At using the Laplace transform is given later in Section 7.3(b).

EXERCISES 4.5 1. Given that

⎡ 0 ⎢0 A=⎢ ⎣0 0

3 0 0 0

1 2 0 0

⎤ 0 1⎥ ⎥, 3⎦ 0

show that it is nilpotent and ﬁnd the smallest power for which An = 0. 2. Given that ⎤ ⎡ 0 1 2 2 ⎢0 0 3 1⎥ ⎥ A=⎢ ⎣0 0 0 1⎦ , 0 0 0 0 ﬁnd eAt . 3. Given that

0 A= 0

2 and 0

0 B= 3

0 , 0

show that A and B do not commute, and by ﬁnding eAt , eBt , and e(A+B)t , verify that eAt eBt = e(A+B)t .

In Exercises 4 through 9, ﬁnd eAt . −2 0 1 7. A = . 4. A = 2 1 0 ⎡ 3 m 0 . 5. A = 0 n 8. A = ⎣6 2 0 −c ⎡ . 6. A = 0 c 0 9. A = ⎣2 2

2 . 1

⎤ −2 2 −4 6⎦. −1 3 ⎤ 1 −2 −1 2⎦. −2 4

10. By considering the deﬁnition of eAt show, provided the square matrices A and B commute, that AeBt = eBt A. 11. By considering the deﬁnition of eAt show that e−At dt = −A−1 e−At + C = e−At A−1 + C, where C is an arbitrary constant matrix that is conformable for addition with A. 12. Show that if the square matrices A and B commute, then the binomial theorem takes the form n n k n−k (A + B)n = AB . k k=0

222

Chapter 4

Eigenvalues, Eigenvectors, and Diagonalization

CHAPTER 4

TECHNOLOGY PROJECTS Project 1 Verifying and Using the Cayley–Hamilton Theorem The purpose of this project is to verify the Cayley--Hamilton theorem in a particular case by constructing an arbitrary 6 × 6 non-singular matrix A and, after ﬁnding its characteristic polynomial, showing by direct calculation that A satisﬁes its own characteristic matrix polynomial equation. The matrix polynomial equation is then to be used to compute the inverse matrix A−1 , after which the inverse is to be checked by showing that the product AA−1 = I. The project then explores the way in which this approach fails when A is singular.

1. Construct an arbitrary 6 × 6 matrix A and check that det A = 0 to ensure that it has an inverse A−1 . 2. Find the characteristic polynomial for matrix A. 3. Show by direct calculation that A satisﬁes its own characteristic matrix polynomial equation. 4. Use the characteristic matrix polynomial equation to ﬁnd A−1 , and check its correctness by showing that the product AA−1 = I. 5. Replace the last row of A by the entries in the row above to form a matrix B that is singular, and ﬁnd the characteristic polynomial for B. 6. Try to use the characteristic matrix polynomial equation for B to ﬁnd B−1 , and comment on the way in which this approach fails. Project 2 Diagonalization of a Matrix This project involves the diagonalization of a 5 × 5 matrix A when two of its ﬁve eigenvalues are equal, but there are ﬁve linearly independent eigenvectors. 1. Find a diagonalizing matrix for ⎤ ⎡ 13 31 30 51 −40 62 64 104 −88⎥ ⎢ 32 ⎥ ⎢ 80⎥ . A = ⎢−28 −56 −58 −88 ⎣−17 −33 −34 −55 48⎦ −13 −25 −26 −37 38 222

2. Diagonalize the matrix B = 12 A, and comment on the relationship between the diagonalizing matrices for A and B. Project 3 Orthogonal Vectors Computed by the Gram–Schmidt Method The purpose of this project is to develop a computer algebra procedure that generalizes the Gram-Schmidt process to n-dimensional vectors. The extension is almost immediate and follows from the fact that in the case of three-dimensional vectors one of them, say a1 , was taken as the ﬁrst vector u1 of an orthogonal basis, the second vector u2 was derived from a2 by subtracting from it the projection of u1 onto a2 , and, ﬁnally, the third vector u3 was obtained from a3 by subtracting from it both the projection of u1 onto a3 and the projection of u2 onto a3 . Starting with a set of n linearly independent vectors {a1 , a2 , . . . , an }, an orthogonal basis {u1 , u2 , . . . , un } for this space is obtained by extending the preceding method by setting u1 = a 1 u2 = a2 −

a2 .u1 u1 u1 .u1

u3 = a3 −

a3 .u1 a3 .u2 u1 − u2 u1 .u1 u2 .u2

.. . u n = an −

an .u1 an .u2 an .un−1 u1 − u2 − · · · − un−1 . u1 .u1 u2 .u2 un−1 .un−1

Write a computer algebra procedure that reproduces these results step by step for four-dimensional vectors. Check the procedure by applying it to the set of linearly independent vectors a1 = [−1, −1, 1, 2], a2 = [1, 0, 1, −2], a3 = [0, 1, −1, −1], and a4 = [2, −1, 1, 1], and showing that the corresponding set of orthogonal basis vectors is u1 = [−1, −1, 1, 2], 2 u2 = [ 37 , − 47 , 11 , − 67 ], u3 = [− 11 , 3 , 3 , − 13 ], and 7 26 13 26 2 4 2 2 u4 = [ 7 , 7 , 7 , 7 ].

Section 4.5

Deﬁne two other sets of linearly independent vectors and, after applying your procedure, verify that the resulting sets of vectors {u1 , u2 , u3 , u4 } are orthogonal.

The purpose of this project is to ﬁnd a transformation that reduces a given quadratic form in four variables to a sum of squares. 1. Given the quadratic form x22 − x12 − 2x1 x2 − 2x1 x3 + 2x1 x4 − 2x3 x4 , ﬁnd a transformation that reduces it to a sum of squares. 2. Find the simpliﬁed quadratic form produced by the transformation in Step 1. Project 5 The Hubble Space Telescope and Quadratic Forms When the Hubble space telescope in orbit around the earth is required to photograph a particular nebula it has to be rotated until it is pointing in the correct direction. As it is a rigid body, the kinetic energy W required to rotate it at an angular velocity ω about a suitable axis is given by W = 12 Iω2 , where I is the moment of inertia of the telescope about the axis of rotation. Because the telescope has an irregular shape, the moment of inertia I will depend on the axis of rotation, and a convenient way of representing the value of I about all possible axes through a given point in the telescope is by means of what is called the ellipsoid of inertia. The ellipsoid of inertia for a given rigid body of mass m relative to a ﬁxed point in the body is a threedimensional plot of the moment of inertia relative to all possible axes of rotation passing through the point. It is shown in texts on mechanics that this plot is an ellipsoidal surface, with the property that the length of the straight line drawn from the center of the ellipsoid to its surface is inversely proportional to the radius of gyration k of the body about that line, where I = mk2 . Given that an ellipsoid of inertia has the form 16x 2 − 4xy + 37y2 − 12xz + 18yz + 11z2 = 12,

223

use matrix methods to ﬁnd a linear transformation from the variables x, y, and z to new variables X, Y, and Z that reduces the expression to one of the form X2 Y2 Z2 + + = 1. a2 b2 c2

Project 4 Reduction of a Quadratic Form to Standard Form

The Matrix Exponential

Hence ﬁnd the radii of gyration 1/a, 1/b, and 1/c about the principal axes of the ellipsoid that form its three mutually orthogonal axes about which there is symmetry. Project 6 Dynamical Systems and Logging Operations Discrete dynamical systems are used to model situations in engineering, control theory, physics, ecology, and elsewhere that can be considered to evolve stage by stage, with each stage dependent on the previous one. For example, a logging operation to supply a saw mill in a speciﬁc area of forest, with tree replanting and the availability of a limited supply of logs from outside the area, can be described by a simple dynamical system that models the way the output of cut timber is inﬂuenced by the competition between the felling of trees, the importing of a limited amount of logs, and the regeneration of the forest. In the simplest case the long-term behavior of a dynamical system can be represented mathematically by the matrix equation xk+1 = Axk,

for k = 0, 1, 2, . . . ,

where A is an n × n matrix, and xk is an n element column vector whose elements describe the physical characteristics of the system at the kth stage. In a logging operation n = 2, and xk = [Tk, Rk]T , where Tk is the amount of timber remaining after k years and Rk is the amount of replanted timber that has matured after k years. In general, let A be diagonalizable with the real eigenvalues λ1 , λ2 , . . . , λn , and let the corresponding linearly independent eigenvectors be u1 , u2 , . . . , un . Then, if x0 describes the initial state of the system, since the eigenvectors form a basis for the system we may set x0 = c1 u1 + c2 u2 + . . . + cn un . Use the representation of x0 to ﬁnd a general expression for xk in terms of the eigenvalues, and comment on the approximate form taken by xk as k becomes large. 223

224

Chapter 4

Eigenvalues, Eigenvectors, and Diagonalization

Given that A interpret the meaning of the coefﬁcients of A in the context of a logging operation. Starting with . generate the ﬁrst 15 vectors xk, com-

224

pare the results with the approximation found earlier, and comment on the result in terms of a logging operation. Suggest a physical dynamical system where A is a 3 × 3 matrix. Deﬁne a suitable numerical matrix A and initial vector x0 , generate the ﬁrst 15 vectors xk, and interpret the results in terms of the model.

PART

THREE

ORDINARY DIFFERENTIAL EQUATIONS

Chapter

5

Chapter

6

7 Chapter 8 Chapter

First Order Differential Equations Second and Higher Order Linear Differential Equations and Systems The Laplace Transform Series Solutions of Differential Equations, Special Functions, and Sturm–Liouville Equations 225

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C H A P T E R

5

First Order Differential Equations

D

ifferential equations are fundamental to the study of engineering and physics, and this chapter marks the start of our discussion of this important topic. Typically, in an electrical problem, the dependent variable i (t) in an ordinary differential equation might be the current ﬂowing in a circuit at time t, in which case the independent variable would be the time. In all such examples, the nature of i (t) depends on the current ﬂow at the start, and the speciﬁcation of information of this type is called an initial condition for the differential equation. Similarly, in chemical engineering, a dependent variable m(t) might be the amount of a chemical produced by a reaction at time t. Here also the independent variable would be the time t, and to determine m(t) in any particular case it would be necessary to specify the amount of m(t) present at the start, that for convenience is usually taken to be when t = 0. Many physical problems are capable of description in terms of a single ﬁrst order ordinary differential equation, while other more complicated problems involve coupled ﬁrst order differential equations, that after the elimination of all but one of the independent variables, can be replaced by a single higher order equation for the remaining dependent variable. This happens, for example, when determining the current in an R-L-C electrical circuit. Thus ﬁrst order ordinary differential equations can be considered as the building blocks in the study of higher order equations, and their properties are particularly important and easy to obtain when the equations are linear. The study and properties of the specially simple class of equations called constant coefﬁcient equations is very important, as it forms the foundation of the study of higher order constant coefﬁcient equations that will be developed later and have many and varied applications. Motivation for the study of ordinary differential equations in general is provided by considering a number of typical problems that give rise to different types of differential equation. The ﬁrst application involves the determination of orthogonal trajectories. A typical example of orthogonal trajectories arises in steady state two-dimensional temperature distributions, where one family of trajectories corresponds to the lines along which the temperature is constant, while the other family corresponds to lines along which heat ﬂows. Other examples considered are the radioactive decay of a substance, the logistic equation and its connection with population growth, damped oscillations, the shape of a suspended power line, and the bending of beams. The chapter starts by deﬁning an mth order ordinary differential equation, of which a ﬁrst order equation is a special case. Various important terms are deﬁned, and the physical

227

228

Chapter 5

First Order Differential Equations signiﬁcance of initial and boundary conditions for differential equations are introduced and explained. The geometrical interpretation of the derivative dy/dx as the slope of a curve is used in Section 5.3 to develop the concept of the direction ﬁeld associated with the ﬁrst order equation dy/dx = f (x, y). This concept is particularly useful as it leads to a geometrical picture showing the qualitative behavior of all solutions of the differential equation. It will be seen later that the idea underlying a direction ﬁeld forms the basis of the simple Euler method for the numerical solution of an initial value problem. First order equations are considered, separable equations are deﬁned and solved, and some other special types of equation are introduced that arise in applications, of which the most important is the general linear ﬁrst order differential equation. Its solution is found by using what is called an integrating factor. The ﬁrst order linear differential equation is important, because the structure of its solution is typical of linear differential equations of all orders. Another special ﬁrst order equation that is considered is the Bernoulli equation. The Bernoulli equation is an important type of nonlinear equation with many applications, and in a sense it stands on the border between linear and nonlinear ﬁrst order differential equations. An application of the Bernoulli equation is outlined in the text, and another more detailed one is to be found in the Exercise set at the end of Section 5.8. The chapter ends by considering the important and practical questions concerning the existence and uniqueness of solutions of dy/dx = f (x, y).

5.1

Background to Ordinary Differential Equations

A

n ordinary differential equation (ODE) is an equation that relates a function y(x) to some of its derivatives y(r ) (x) = dr y/dxr . It is usual to call x the independent variable and y the dependent variable, and to write the most general ordinary differential equation as * ) F x, y, y(1) , y(2) , . . . , y(n) = 0. (1)

nth order linear variable coefﬁcient equation

The number n in (1) is called the order of the ordinary differential equation, and it is the order of the highest derivative of y that occurs in the equation. A class of ODEs of particular importance in engineering and science, because of their frequency of occurrence and the extensive analytical methods that are available for their solution, are the linear ordinary differential equations. The most general nth order linear differential equation can be written

a0 (x)

dn y dn−1 y dy + an (x)y = f (x), + a (x) + · · · + an−1 (x) 1 dx n dx n−1 dx

(2)

with a0 (x) = 0 and we will consider it to be deﬁned over some interval a ≤ x ≤ b. The functions a0 (x), a1 (x), . . . , an (x), called the coefﬁcients of the equation, are known functions, and the known function f (x) is called the nonhom*ogeneous term. The name forcing function is also sometimes given to f (x), because in applications it represents the inﬂuence of an external input that drives a physical system represented by the differential equation. Equation (2) is called hom*ogeneous if f (x) ≡ 0.

Section 5.1

nth order linear constant coefﬁcient equation

EXAMPLE 5.1

229

It will be seen later that the solution of the nonhom*ogeneous equation (2) is related in a fundamental manner to the solution of its associated hom*ogeneous equation. When one or more of the coefﬁcients of (2) depend on x, it is called a variable coefﬁcient equation. Simpler than variable coefﬁcient linear equations, but still of considerable importance, are the linear equations in which the coefﬁcients are the constants a0 , a1 , . . . , an , so that (2) becomes

a0

nonlinear equation and degree

Background to Ordinary Differential Equations

dn y dn−1 y dy + a + · · · + an−1 + an y = f (x) 1 n n−1 dx dx dx

for a ≤ x ≤ b.

(3)

Equations of this type are called constant coefﬁcient linear equations. If the interval a ≤ x ≤ b on which equations (2) and (3) are deﬁned is not speciﬁed, it is to be understood to be the largest one for which the equations have meaning. Sometimes, in the case of (2), this interval is determined by the variable coefﬁcients ar (x), whereas in applications it is often determined by the nature of the problem that restricts x to a speciﬁc interval. An ordinary differential equation that is not linear is said to be nonlinear. Nonlinearity arises in ordinary differential equations because of the occurrence of a nonlinear function of the dependent variable y that sometimes occurs in the form of a power or a radical. The terms hom*ogeneous and nonhom*ogeneous have no meaning for nonlinear equations. A term that is also in use, mainly as an indication of the complexity to be expected of a solution, is the degree of an equation. The degree is the greatest power to which the highest order derivative in the differential equation is raised after the radicals have been cleared from expressions involving the dependent variable y. (a) The ODE dy + 2xy = sin x dx is a linear variable coefﬁcient nonhom*ogeneous ﬁrst order equation. (b) The ODE (1 − x 2 )

d2 y dy + 6y = 0, − 2x dx 2 dx

with −1 < x < 1,

is a linear variable coefﬁcient hom*ogeneous second order equation. (c) The ODE d2 y dy + by = sin ωx, +a dx 2 dx

with ω = constant,

is a linear constant coefﬁcient nonhom*ogeneous second order equation. (d) The ODE d2 θ + k sin θ = 0, dt 2

with k = constant

is a nonlinear second order equation because θ occurs nonlinearly in the function sin θ .

230

Chapter 5

First Order Differential Equations

(e) The ODE k

d2 y = f (x)[1 + (dy/dx)2 ]3/2 , dx 2

with k > 0 a constant

is a nonlinear second order equation of degree 2 involving a power and a radical.

general and particular solutions, and integral curves

singular solution

EXAMPLE 5.2

A solution of an ordinary differential equation is a function y = (x) that, when substituted into the equation, makes it identically zero over the interval on which the equation is deﬁned. A solution of an nth order equation that contains n arbitrary constants is called the general solution of the equation. If the arbitrary constants in the general solution are assigned speciﬁc values, the result is called a particular solution of the equation. For obvious reasons the solution of an ordinary differential equation is also called an integral curve. A solution that cannot be obtained from the general solution for any choice of its arbitrary constants is called a singular solution. In the case of linear equations all possible solutions of the equation can be obtained from the general solution, so linear equations have no singular solutions. Nonlinear equations possess a more complicated structure that often allows the existence of one or more singular solutions. (a) The general solution of the linear constant coefﬁcient nonhom*ogeneous equation d2 y − 4y = x dx 2 is y = Ae2x + Be−2x − x/4, where A and B are arbitrary constants. This is easily checked, because substituting for y in the equation leads to the identity x ≡ x. (b) The nonlinear equation 2 dy + y2 = 1 dx has the general solution y = sin(x + A). However, y = ±1 are also seen to be solutions, though as these cannot be obtained from the general solution for any choice of A, they are singular solutions. The linear equation (2) is often written in the more compact form L[y] = f (x),

linear operator

(4)

where L is the linear operator L[·] ≡ a0 (x)

dn dn−1 d + an (x), + a (x) + · · · + an−1 (x) 1 dx n dx n−1 dx

(5)

with coefﬁcients that may or may not be functions of x. Only when L[·] acts on an n times differentiable function does it produce a function.

Section 5.1

Background to Ordinary Differential Equations

231

Equation (2) is called linear because if y1 and y2 are any two solutions of the hom*ogeneous form of the equation L[y] = 0, the linear combination y = C1 y1 + C2 y2 where C1 and C2 are constants is also a solution. In terms of the differential operator L[·] this property becomes L[C1 y1 + C2 y2 ] = C1 L[y1 ] + C2 L[y2 ], and it follows directly from the linearity of the differentiation operation, because dm dm y1 dm y2 (y1 + y2 ) = + , m m dx dx dx m for m = 0, 1, . . . , n, with d0 y/dx 0 ≡ y. If y1 (x), y2 (x), . . . , ym(x) are solutions of the nth order hom*ogeneous equation L[y] = 0, with m ≤ n and C1 , C2 , . . . , Cm arbitrary constants, the linear combination y(x) = C1 y1 (x) + C2 y(x) + · · · + Cm ym(x)

linear superposition

is called a linear superposition of the m solutions, and it is also a solution of the hom*ogeneous equation. Later we will deﬁne the linear independence of a set of functions over an interval and show that the hom*ogeneous form of (2) has precisely n linearly independent solutions y1 (x), y2 (x), . . . , yn (x), and that its general solution is yc (x) = C1 y1 (x) + C2 y(x) + · · · + Cn yn (x),

complementary solution, particular integral, and complete solution

where C1 , C2 , . . . , Cn are arbitrary constants. This general solution of the hom*ogeneous form of equation (2) is called the complementary function or the complementary solution of (2). A function yp (x) that is a solution of the nonhom*ogeneous equation (2) but contains no arbitrary constants is called a particular integral of (2). The complete solution y(x) of equation (2) is y(x) = yc (x) + yp (x).

boundary and initial conditions

(6)

(7)

In applications of ordinary differential equations the values of the arbitrary constants in speciﬁc problems are obtained by choosing them so the solution satisﬁes auxiliary conditions that identify a particular problem. Auxiliary conditions speciﬁed at a single point x = a, say, are called initial conditions, because x often represents the time so that conditions of this type describe how the solution starts. An initial value problem (i.v.p.) involves ﬁnding a solution of a differential equation that satisﬁes prescribed initial conditions. A different type of problem arises when the auxiliary conditions are speciﬁed at two different points x = a and x = b, say. Conditions of this type are called boundary conditions, because in such problems x usually represents a space variable, and the solution is required to be determined between two boundaries located at x = a and x = b where boundary conditions are prescribed. A boundary value problem (b.v.p.) involves ﬁnding a solution of a differential equation that satisﬁes prescribed boundary conditions.

232

Chapter 5

First Order Differential Equations

EXAMPLE 5.3

(a) The linear nonhom*ogeneous ordinary differential equation d2 y +y=x dx 2 has the general solution y = A cos x + B sin x + x. This equation together with the initial conditions y(0) = 0, y (0) = 0 speciﬁed at the point x = 0 constitutes an initial value problem for y. Choosing A and B to satisfy these initial conditions shows the unique solution of this i.v.p. to be y = x − sin x for x ≥ 0. (b) The linear hom*ogeneous ordinary differential equation d2 y +y=0 dx 2 has the general solution y = A cos x + B sin x. This equation together with the conditions y(0) = 0, y (π/3) = 3 speciﬁed at the two different points x = 0 and x = π/3 constitutes a boundary value problem for y. Choosing Aand B to satisfy these conditions shows that this b.v.p. has the unique solution y = 6 sin x for 0 < x < π/3. (c) Consider the linear hom*ogeneous ordinary differential equation d2 y −y=0 dx 2

unique and nonunique solutions

Summary

deﬁned for x ≥ 0,

which is easily seen to have the general solution y = Ae x + Be−x . Imposing the boundary conditions y(0) = 1 and y(+∞) = 0 constitutes a boundary value problem for y in which one condition is at x = 0 and the other is at plus inﬁnity. The condition at inﬁnity can only be satisﬁed if A = 0, so matching the solution y = Be−x to the condition y(0) = 1 shows that this b.v.p. has the unique (only) solution y = e−x . (d) It is possible for a boundary value problem to have a unique solution as in (b), more than one solution, or no solution at all. More will be said about this later, but for the moment we give a simple example that shows why a boundary value problem may have many solutions or no solution. The general solution of (b) is y = A cos x + B sin x, so if the boundary conditions y(0) = 0 and y(π ) = 0 are imposed we ﬁnd that A = 0 and B is indeterminate, so it may be assigned any value. In this case a solution certainly exists, as it is given by y = B sin x, but B is arbitrary, so there is more than one solution. When more than one solution can be found that satisﬁes the auxiliary conditions, the solution is said to be nonunique. If, in this example, the boundary conditions are replaced by y(0) = 0 and y(π ) = 1, no choice of constants A and B can make the general solution satisfy the boundary conditions, so in this case there is no solution.

This section introduced the concept of an nth order ordinary differential equation, and the initial and boundary conditions that such equations are often required to satisfy. Emphasis was placed on linear equations and, in particular, on the structure of the solution of a linear ﬁrst order equation, because the structure of the solution of this fundamental type of equation is shared by the solutions of all higher order linear equations.

Section 5.2

Some Problems Leading to Ordinary Differential Equations

233

EXERCISES 5.1 In Exercises 1 through 10, determine the order and degree of the equation and classify it as hom*ogeneous linear, nonhom*ogeneous linear, or nonlinear. 1. 2. 3. 4.

y + 3y + 4y − y = 0. y + 4y + y = x sin x. y + x(y )2 = cosh x. (y )3/2 + xy = [(1 + x)y ].

5.2

5. 6. 7. 8. 9. 10.

y + 3y + 2y = x 2 sin y. √ y(4) + x 2 y = 3 + x 3 . y + 3xy = 1 + x 2 . y + y = tan(y ). (2 + x 2 )y + x(1 − y2 ) = 0. y /y + sin x = 3.

Some Problems Leading to Ordinary Differential Equations Before we develop methods for the solution of ordinary differential equations, it will be helpful to examine some simple geometrical and physical problems that lead to ODEs. There are many such problems, so we only consider some representative examples.

(a) A Geometrical Problem: Orthogonal Trajectories The equation F(x, y, c) = 0,

orthogonal trajectory

isotherms, heat ﬂow, streamlines, equipotentials, and ﬂux lines

where the real variable c is a parameter, deﬁnes a one-parameter family of curves in the (x, y)-plane. This means that assigning a speciﬁc value to c determines a particular curve in the (x, y)-plane, and a different value of c will determine a different curve. It often happens that the equation F(x, y, c) = 0 deﬁnes y implicitly in terms of x, so that the equation cannot be solved explicitly as y = f (x, c). A curve that intersects every member of a one-parameter family of curves orthogonally (at right angles) is called an orthogonal trajectory of the family. A geometrical problem that often occurs is how to ﬁnd a family of curves that form orthogonal trajectories to a given family. When some applications of conformal mapping to two-dimensional physical problems are considered in Chapter 17, it will be seen that orthogonal trajectories arise in the study of steady state heat conduction, ﬂuid dynamics, and electromagnetic theory. In heat conduction (see Chapter 18), one family of curves represents lines of constant temperature called isotherms, and their orthogonal trajectories then represent heat ﬂow lines. In two-dimensional ﬂuid dynamics, orthogonal trajectories express the relationship between the curves followed by ﬂuid particles called streamlines, and the associated equipotential lines along which a function called the ﬂuid potential is constant. In two-dimensional electromagnetic theory an analogous situation arises where one family of curves describes lines of constant electric potential, again called equipotential lines, and the family of orthogonal trajectories that describes what are then called ﬂux lines.

234

Chapter 5

First Order Differential Equations

family 1

family 2

π/2

FIGURE 5.1 Two typical families of orthogonal trajectories.

Two typical families of orthogonal trajectories are illustrated in Fig. 5.1, and if these curves are related to steady state heat ﬂow, family 1 could represent the isotherms and family 2 the heat ﬂow lines. Two speciﬁc examples of families of orthogonal trajectories are shown in Fig. 5.2, where in case (a) the curves are given by x 2 + y2 = c2

and

y = kx

(with c and k real).

The ﬁrst equation describes a family of concentric circles centered on the origin, and the second family that forms their orthogonal trajectories comprises all the straight lines that pass through the origin. In case (b) the curves are given by x 2 − y2 = c

and

xy = k

(with c and k real),

where the two families of curves are families of mutually orthogonal rectangular hyperbolas.

y y

x

0 (a) FIGURE 5.2 Speciﬁc examples of orthogonal trajectories.

x (b)

Section 5.2

Some Problems Leading to Ordinary Differential Equations

235

In general the equation F(x, y, c) = 0,

(8)

with c a parameter, describes a family of curves. To ﬁnd their orthogonal trajectories we ﬁrst need to obtain the differential equation for the family of curves determined by (8). This can be done by differentiating (8) with respect to x and then eliminating c between (8) and the equation with dy/dx to arrive at a differential equation of the form dy = f (x, y). dx

(9)

If the family of curves described by this differential equation is to be orthogonal to another family, the products of the gradients of every pair of intersecting curves must equal −1. So the gradient dy/dx of the family of curves that are mutually orthogonal to those of (9) must be such that 1 dy =− . dx f (x, y)

(10)

This is the differential equation of the required family of orthogonal trajectories. In general (10) can often be solved by the method of separation of variables that will be discussed later.

(b) Chemical Reaction Rates and Radioactive Decay In many circ*mstances, for a limited period of time, the rate of reaction of a chemical process can be considered to be proportional only to the amount Q of the chemical that is present at a given time t. The differential equation governing such a process then has the form dQ = kQ, dt

(11)

where k ≥ 0 is a constant of proportionality. This is a hom*ogeneous linear ﬁrst order differential equation. An analogous situation applies to the radioactive decay of an isotope for which the decay takes place at a rate proportional to the amount of radioactive isotope that is present at any given instant of time. The equation governing the amount Q of the isotope as a function of time t is also of the form shown in (11), but instead of the amount growing as in the previous case, it is decreasing, so as in this case the constant of proportionality is usually denoted by a positive number λ, the equation for radioactive decay takes the form dQ = −λQ. dt

(12)

236

Chapter 5

First Order Differential Equations

It is not difﬁcult to see by inspection that the general solution of (12) is Q = Q0 e−λt ,

half-life

where Q0 is the amount of the isotope present at the start when t = 0. The so-called half-life Th of an isotope is the time taken for half of it to decay away, so setting Q = (1/2)Q0 in the above result shows the half-life to be given by Th = (1/λ) ln 2.

(c) The Logistic Equation: Population Growth In the study of phenomena involving the rate of increase of a quantity of interest, it often happens that the rate is inﬂuenced both by the amount of the quantity that is present at any given instant of time and by the limitation of a resource that is necessary to enable an increase to occur. Such a situation arises in a population of animals that compete for limited food resources, leading to the so-called predator– prey situations where an animal (the predator) feeds on another species (the prey) with the effect that overfeeding leads to starvation. This in turn leads to a reduction in the number of predators that in turn can lead to a recovery of the food stock. Similar situations arise in manufacturing when there is competition for scarce resources, and in a variety of similar situations. To model the situation we let P represent the amount of the quantity of interest present at a given time t, and M represent the amount of resources available at the start. Then a simple model for this process is provided by the differential equation dP = kP(M − P), dt

logistic equation

(13)

in which k is a constant of proportionality. When constructing this equation the assumption has been made that the rate of increase dP/dt is proportional to both the amount P that is present at time t and to the amount M − P that remains. Equation (13) is called the logistic equation, and it is nonlinear because of the presence of the term −kP2 on the right, though it is easily integrated by the method of separation of variables to be described later.

(d) A Differential Equation that Models Damped Oscillations

damping

Mechanical and electrical systems, and control systems in general, can exhibit oscillatory behavior that after an initial disturbance slowly decays to zero. The process producing the decay is a dissipative one that removes energy from the system, and it is called damping. To see the prototype equation that exhibits this phenomenon we need only consider the following very simple mechanical model. A mass M rests on a rough horizontal surface and is attached by a spring of negligible mass to a ﬁxed point. The mass–spring system is caused to oscillate along the line of the spring by being displaced from its equilibrium position by a small amount and then released. Figure 5.3a shows the system in its equilibrium conﬁguration, and Fig. 5.3b shows it when the mass has been displaced through a distance x from its rest position.

Section 5.2

Some Problems Leading to Ordinary Differential Equations

m

237

m

x

L

L (a)

(b)

FIGURE 5.3 Mass–spring system.

If t is the time, the acceleration of the mass is d2 x/dt 2 , so the force acting due to the motion is Md2 x/dt 2 . The forces opposing the motion are the spring force, assumed to be proportional to the displacement x from the equilibrium position, and the frictional force, assumed to be proportional to the velocity dx/dt of the mass M. If the spring constant of proportionality is p and the frictional constant of proportionality is k, the two opposing forces are kdx/dt due to friction and px due to the spring. Equating the forces acting along the line of the spring and taking account of the fact that the spring and frictional forces oppose the force due to the acceleration shows the equation of motion to be the hom*ogeneous second order linear equation M

dx d2 x − px, = −k 2 dt dt

or d2 x dx + bx = 0, +a dt 2 dt

(14)

where a = k/M and b = p/M. If an external force Mf (t) is applied to the spring, the equation governing the damped oscillations becomes the linear nonhom*ogeneous second order equation dx d2 x + bx = f (t). +a dt 2 dt An equation of the same form as (14) governs the oscillation of the charge q in the R–L–C electric circuit shown in Fig. 5.4. The open circuit is shown in Fig. 5.4a with the plates of the capacitor C carrying initial charges Q and −Q, while Fig. 5.4b shows the circuit when the switch S has been closed, causing a current i to ﬂow due to a charge is q at time t. The respective potential drops in the direction of the arrow across the resistor R, the inductance L, and the capacitor C are V = i R, where i = dq/dt, Ldi/dt, Q

C

−Q

q

L

S

R (a) FIGURE 5.4 An R–L–C circuit.

S

−q

C

i

R (b)

L

238

Chapter 5

First Order Differential Equations

FIGURE 5.5 Suspended cable.

and q/C. Applying Kirchhoff’s law, which requires the sum of the potential drops around the circuit to be zero, gives

L

di q + Ri + = 0. dt C

Eliminating i by using the result i = dq/dt leads to the following hom*ogeneous linear second order equation for q:

LC

d2 q dq + q = 0. + RC 2 dt dt

This ODE is of the same form as (14) with a = R/L and b = 1/LC.

(e) The Shape of a Suspended Power Line: The Catenary An analysis of the forces acting on a power line attached to pylons as shown in Fig. 5.5, or on the suspension cable of a cable car, shows the shape of the cable to be determined by the solution y(x) of the nonlinear differential equation d2 y = a 1 + (dy/dx)2 . dx 2 The shape taken by the cable is called a catenary, after the Latin word catena, meaning chain. Although this equation will not be solved here, it is not difﬁcult to show that its solution is a hyperbolic cosine curve.

(f) Bending of Beams An analysis of the forces and moments acting on a horizontal beam of uniform construction made from a material with Young’s modulus E and supported at its two end points, with the moment of inertia of its cross-section about the central horizontal axis of the beam equal to I, leads to the following equation for the vertical

Section 5.2

Some Problems Leading to Ordinary Differential Equations

239

w(x) load undeflected shape

y

x

y y = y(x)

deflected shape

FIGURE 5.6 Deﬂection of a loaded beam.

deﬂection y caused by the weight of the beam and any loads it is supporting: EI d2 y/dx 2 = M(x). [1 + (dy/dx)2 ]3/2

(15)

Here M(x) is the bending moment that acts to one side of a point x in the beam. If a b distributed load of line density w(x) acts along the beam creating a load a w(x)dx on the segment from x = a to x = b, as represented in Fig. 5.6, it can be shown that M(x) and w(x) are related by the result d2 M = −w(x). dx 2

(16)

Using this result in (15) shows that the deﬂection y(x) is determined by the solution of the nonlinear fourth order equation + , d2 EId2 y/dx 2 = w(x), dx 2 [1 + (dy/dx)2 ]3/2 ﬂexural rigidity

(17)

in which the product EI is called the ﬂexural rigidity of the beam. If the bending is small and the term (dy/dx)2 can be neglected, (17) simpliﬁes to the linear fourth order constant coefﬁcient equation d4 y w(x) , = dx 4 EI which can be solved by direct integration. Many applications of ordinary differential equations to physical problems are to be found in reference [3.6].

Summary

This section has provided mathematical and physical examples of problems that give rise to ordinary differential equations, some with initial conditions and others with boundary conditions. The logistic equation was seen to be nonlinear and ﬁrst order, whereas others such as the equation governing radioactive decay and the equation describing damped

240

Chapter 5

First Order Differential Equations oscillations were seen to be linear and of ﬁrst and second order, respectively. The beam equation is nonlinear, though when the bending is small it was seen to reduce to a simple linear fourth order equation that could be solved by direct integration.

EXERCISES 5.2 1. Derive the differential equation that describes the families of circles that are tangent to both the x- and y-axes. 2. Derive the differential equation satisﬁed by all curves such that the magnitude of the area under the curve between any two ordinates at x = a and x = b is proportional to the magnitude of the arc length of the curve from x = a to x = b. Verify that the catenary y(x) = k cosh (x/k − K) is such a curve, with k and K parameters.

5.3

3.* A launch travels along the y-axis a constant speed U, starting from the origin, and a police launch starting from a point a > 0 on the x-axis pursues it at a constant speed V > U. If t is the time measured from the start of the pursuit, write down the differential equation that describes the pursuit path. At all times the police launch steers toward the ﬁrst launch.

Direction Fields In certain applications of mathematics it is necessary to know the qualitative behavior of solutions of a general ﬁrst order equation dy = f (x, y) dx

global properties

(18)

over the entire (x, y)-plane, when either no analytical solution is available or, if one exists, it is too complicated to be useful. General properties of solutions of (18) that are known throughout the (x, y)-plane are called global properties. A typical global property might be that the solutions are known to be bounded for all x. A numerical solution of (18) can always be obtained for any given initial condition (see Chapter 19), but it is impracticable to obtain such solutions for a large enough set of initial conditions simply to enable general the behavior of solutions all over the (x, y)-plane to be understood. A convenient answer to this problem involves constructing a graphical representation of what is called the direction ﬁeld of (18) at a conveniently chosen mesh of points covering a region R of interest in the (x, y)-plane. The idea involved is simple and starts by dividing the interval a ≤ x ≤ b into m subintervals of equal length x = (b − a)/m, and the interval c ≤ y ≤ d into n subintervals of equal length y = (d − c)/n. The mesh of points to be used to cover R are then located at the points (xr , ys ), where xr = a + r x and ys = c + sy with r = 0, 1, . . . , m and s = 0, 1, . . . , n. Once the mesh has been chosen, the function f (x, y) is evaluated at each of the points (xr , ys ). It follows directly that the number f (xr , ys ) associated with the point (xr , ys ) is the gradient (slope) of the integral curve (solution curve) that passes through that point. Accordingly, the next step is to construct through each point (xr , ys ), a small straight line segment making an angle θr s = Arctan f (xr , ys ) with the x-axis, as in Fig. 5.7a.

Section 5.3

Direction Fields

241

y

tangent to integral curve at (xs , ys) gives the slope of the direction field at the point

ys

θrs = Arctan f (xr, ys) θrs xr

x

(a) 4 2 0 y

−2 −4 −6 −8 −4

−2

0 x

2

4

(b) FIGURE 5.7 (a) The construction of a direction ﬁeld vector at the point (xr , ys ). (b) The direction ﬁeld and integral curves for dy/dx = cos(x + y).

direction ﬁeld

By the nature of their construction, each line segment that is drawn in this manner is tangent to the integral curve that passes through the point through which the segment is drawn. An examination of the pattern of the line segments indicates the overall pattern of behavior of all of the integral curves passing through region R. The assignment of a gradient f (x, y) to each point of R is said to deﬁne the direction ﬁeld of the ODE in (18) over R, and the method just described is its geometrical interpretation at a ﬁnite number of points of R. The graphical interpretation of a direction ﬁeld can be used to obtain an approximation to the integral curve that passes through an initial point (x0 , y0 ) in R. This is accomplished by starting with the line segment through the point (x0 , y0 ) and then joining up successive line segments as they intersect one another. As the construction of a direction ﬁeld over a large region involves many calculations, it is usual to construct them with the aid of a computer. The direction ﬁeld for the nonlinear ﬁrst order equation dy = cos(x + y) dx over the region −4 ≤ x ≤ 4 and −8 ≤ y ≤ 5 is shown in Fig. 5.7b, to which have been added some integral curves to show their relationship to the direction ﬁeld.

242

Chapter 5

First Order Differential Equations

Summary

The concept of a direction ﬁeld of a ﬁrst order differential equation dy/dx = f (x, y) was introduced in this section. It is a graphical representation of the slope (gradient) of solution curves of the differential equation where they pass through a rectangular mesh of points inside a region of the (x, y)-plane where the solution of the differential equation is of interest. It involves plotting at each mesh point (xi , yi ) a short segment of the tangent to the solution curve with slope f (xi , yi ) that passes through that point, to which is added an arrow showing the direction in which the solution is changing as x increases. A direction ﬁeld provides a geometrical representation of the global nature of the solution inside the region of interest, and tracing successive line segments from one to another, starting from any mesh point, provides a rough picture of the solution curve that originates from the initial condition represented by that mesh point.

EXERCISES 5.3 In each of the following exercises, with the aid of a computer algebra package: (a) Construct the direction ﬁeld for the given equation at a suitable number of mesh points, (b) use the results of (a) to sketch some representative integral curves, and (c) compare an approximate integral curve through a chosen initial point (x0 , y0 ) with the exact solution found by requiring the given general solution to pass through that point.

5.4

1. 2. 3. 4. 5.

dy/dx dy/dx dy/dx dy/dx dy/dx

= y + 2x; y = Ce x − 2 − 2x. = y + 2 cos x; y = Ce x − cos x + sin x. = 2x − y; y = Ce−x − 2 + 2x. = x(1 + y/2); y = C exp(x 2 /4) − 2. = y + x 2 ; y = Ce x − 2 − 2x − x 2 .

Separable Equations Sometimes the function f (x, y) in the ﬁrst order differential equation dy = f (x, y) dx

(19)

can be written as the product of a function F(x) depending only on x and a function G(y) depending only on y, so that f (x, y) = F(x)G(y), allowing (19) to be written dy = F(x)G(y). dx two forms of a separable equation

(20)

When (19) can be expressed in this simple form, its variables x and y are said to be separable, and the equation itself to be of variables separable type. If we use differential notation, (20) becomes 1 dy = F(x)dx, G(y)

(21)

so provided G(y) = 0, equation (21) can be solved by routine integration of the left side with respect to y and of the right side with respect to x. Thus, in principle, the solution of a ﬁrst order differential equation in which the variables are separable can always be found, though in practice the integrals involved may be difﬁcult or sometimes impossible to evaluate analytically.

Section 5.4

Separable Equations

243

Separable ﬁrst order equations The differential equation dy = f (x, y) dx is said to be separable if it can be written in the form dy = F(x)G(y), dx or, in differential form, 1 dy = F(x)dx. G(y)

EXAMPLE 5.4 examples of separable equations

Solve the logistic equation dP = kP(M − P) dt given in equation (13) of Section 5.2(c), assuming k > 0 and 0 ≤ P ≤ M. Find the solution of the initial value problem in which P = P0 when t = 0, and draw some typical integral curves. Solution The equation is separable and can be written in the differential form dP = kdt. P(M − P) If we write the left-hand side in partial fraction form, the equation becomes dP dP + = Mkdt, P (M − P) and after integration we ﬁnd that P = Mkt + C, ln M − P where C is an arbitrary constant of integration. As the solution for P must lie in the interval 0 ≤ P ≤ M, this result simpliﬁes to P=

MA , A+ exp(−Mkt)

where A is an arbitrary constant. The arbitrary constant A is related to C by A = eC , but as C is arbitrary, the constant A is also arbitrary, so for simplicity we denote the arbitrary constant in this last result by A without mentioning how it is related to C. In general, arithmetic is not usually performed on arbitrary constants, so after algebraic manipulations, either constants are renamed or the same symbol is used for a related constant.

244

Chapter 5

First Order Differential Equations kM = 4

P/M 1

kM = 3 kM = 2

0.8

kM = 1

0.6 0.4 0.2 −2

−1

1

2

t

FIGURE 5.8 Integral curves for the logistic equation.

To solve the initial value problem we must ﬁnd A such that P = P0 when t = 0, from which it is easily seen that A = P0 /(M − P0 ). The required particular solution is thus MP0 . P= P0 + (M − P0 ) exp(−Mkt) Representative integral curves of P(t)/M obtained from this expression using P0 /M = 1/4 and kM = 1, 2, 3, and 4 are shown in Fig. 5.8 for −2 ≤ t ≤ 2.

EXAMPLE 5.5

Solve the initial value problem for the equation expressed in differential form x 2 y2 dx − (1 + x 2 )dy = 0,

given that y(0) = 1.

Solution The equation is separable because it can be written dy x2 dx. = y2 (1 + x 2 ) Integration gives

dy = y2

x2 dx, (1 + x 2 )

and after the integrations have been performed this becomes −1/y = x − Arctan x + C, where C is an arbitrary constant of integration. This general solution will satisfy the initial condition y(0) = 1 if C = −1, so the required solution is seen to be y = 1/(Arctan x − x + 1). EXAMPLE 5.6

Derive the differential equation that determines the orthogonal trajectories of the one parameter family of curves y = Cxe x , and solve it to ﬁnd the equation of these trajectories. Solution The differential equation describing the family of curves y = Cxe x is found by ﬁrst calculating y (x), and then using the original equation to eliminate C

Section 5.4

Separable Equations

245

from the result. We have y (x) = Cex (1 + x), but from the original equation C = y/xe x , so eliminating C between these two results shows that the required differential is y (x) = y(1 + x)/x. The product of the gradient y (x) of curves belonging to this family and the gradient of the family of orthogonal trajectories must equal −1 (see Section 5.2(a)), so the differential equation of the orthogonal trajectories is the separable equation dy x =− . dx y(1 + x) After separation of the variables and integration, this becomes x dx, ydy = − 1+x so that y2 = ln(1 + x)2 − 2x + C. EXAMPLE 5.7

A circular metal radiator pipe has inner radius R1 and outer radius R2 (R2 > R1 ). When operating under steady conditions the radial temperature distribution T(r ) in the metal wall of the pipe is known to be a solution of the ordinary differential equation (see the heat equation in cylindrical polar coordinates in Section 18.5) r

d2 T dT = 0. + 2 dr dr

(i) Find the radial temperature distribution in the pipe wall when the inner surface is maintained at a constant temperature T1 and the outer surface is maintained at a constant temperature T2 . (ii) Find the radial temperature distribution in the pipe wall when the inner surface is maintained at a constant temperature T1 and heat is lost by radiation from the outer surface according to Newton’s law of cooling that requires the heat ﬂux across the outer surface to be proportional to the difference in temperature between the surface and the surrounding air at a temperature T2 . Solution (i) Setting u = dT/dr the equation becomes the separable equation r

du +u=0 dr

and so

du dr =− , u r

from which it follows that ln u = − ln r + ln A, where for convenience the arbitrary integration constant has been written ln A. Thus ur = A, so after substituting for u and again separating variables we have dT A = . dr r

246

Chapter 5

First Order Differential Equations

A ﬁnal integration gives the general solution T(r ) = A ln r + B, where B is another arbitrary integration constant. Matching the arbitrary constants A and B to the required conditions T(R1 ) = T1 and T(R2 ) = T2 then gives the required solution T(r ) =

T1 ln(R2 /r ) + T2 ln(r/R1 ) . ln(R2 /R1 )

(ii) The heat ﬂux across the surface r = R2 is proportional to dT/dr at r = R2 , and this in turn is proportional to the temperature difference T(R2 ) − T2 , so the required boundary condition on the outer surface of the pipe is of the form dT = −h[T(R2 ) − T2 ], dr r =R2 where the negative sign is necessary because heat is being lost across the surface r = R2 , and h is a constant depending on the metal in the pipe and the heat transfer condition at its surface. The general solution is still T(r ) = A ln r + b, but now the arbitrary constants A and B must be matched to the condition T(R1 ) = T1 on the inside wall of the pipe, and to the above condition derived from Newton’s law of cooling. When this is done the temperature distribution in the pipe is found to be T(r ) = T1 +

Summary

r hR2 (T2 − T1 ) . ln 1 + hR2 ln(R2 /R1 ) R1

This section introduced the important class of separable differential equations dy/dx = F (x)G(y), so called because when written in the form dy/G(y) = F (x)dx the variables are separated by the = sign; they can be integrated immediately provided antiderivatives (indeﬁnite integrals) of 1/G(y) and F (x) can be found. This method was used to integrate the nonlinear logistic equation and to obtain the equation of some orthogonal trajectories.

EXERCISES 5.4 In Exercises 1 through 4 solve the given differential equation by hand and conﬁrm the result by using computer algebra. 1. 2. 3. 4.

2yy = x(1 − 2y) with y(1) = 1. 2x 2 y2 y + y4 = 4 with y(1) = 3. √ (x 2 − 4)y = x(1 − 2y) with y( 5) = 1. √ √ 2 (1 + x 2 )y = (1 − y2 ) with y(1) = 1.

In Exercises 5 through 14 ﬁnd the general solution of the given differential equation. √ √ 5. (1 + x 2 )y − 3x (y2 − 1) = 0. 6. e−3x y + x sin 2y = 0. 7. 2(1 + x)(1 + y)y + (y + 2)2 = 0.

8. 9. 10. 11. 12. 13. 14.

2(x − 1)y + (x 2 − 2x + 3) cos2 y = 0. (1 + 3y2 )y + 2y ln |1 + x| = 0. 2(1 − cos x)y + 3 sin y = 0. (1 + x 2 )yy − x(y2 + y + 1) = 0. √ (x 2 + 9)y2 y − (4 − y2 ) = 0. y ctg x + 2y = 4. (x + 1)y2 y = x(y2 + 4).

In Exercises 15 through 17 derive and then solve the differential equation that determines the orthogonal trajectories to the given one parameter family of curves. 15. y = b + k(x − a) with a and b constants and k a parameter.

Section 5.5 16. x 2 − 4y2 + y = c with c a parameter. 17. y = Cx 2 e2x with C a parameter. 18. A snowball of radius 2 inches is brought into a warm room at a constant temperature above freezing point, and it is found that after 6 hours it has melted to a radius of 1.5 inches. Assuming the melting occurs at a rate proportional to the surface area, write down the differential equation determining the radius as a function of time t in hours, and ﬁnd the general expression for the radius as a function of time. Comment on any deﬁciency exhibited by this mathematical model. 19. A simple model called Malthus’ law for the change in a bacterial population N(t) as a function of time t involves assuming the rate of change is proportional to the population present at time t. Write down the differential equation governing N(t) if the constant of proportionality is λ > 0, and ﬁnd an expression for N(t) given that initially N(0) = N0 . Find λ if N(t1 ) = N1 when t = t1 and N(t2 ) = N2 when t = t2 , with N1 > N2 and t2 > t1 . Give a reason why this model is unrealistic when t is large. 20. When a beam of light enters a parallel slab of transparent material at right angles to its plane surface, its intensity I decreases at a rate proportional to the intensity I(x) at a perpendicular distance x into the material. Given a slab of material where the intensity at a distance h into the slab is 40% of the initial intensity, write down the differential equation for I(x). Solve the equation for I(x) and ﬁnd the distance at which the intensity is 10% of its initial value. 21. The dating of a fossilized bone is based on the amount of radioactive isotope carbon-14 present in the bone.

5.5

hom*ogeneous Equations

247

The method uses the fact that the isotope is produced in the atmosphere at a steady rate by bombardment of nitrogen by cosmic radiation when it is absorbed into the living bone. The process stops when the bone is dead, after which the C-14 present in the bone decays exponentially. Assuming the half-life of C-14 is 5600 years, and a bone is found to contain 1/500th of the original amount of C-14 that was present originally, determine its age. This approach is called radioactive carbon dating. 22. A cylindrical tank of cross-sectional area A standing in a vertical position is ﬁlled with water to a depth h. At time t = 0 a circular hole of radius a in the bottom of the tank is opened and water is allowed to drain away under gravity. It is known from Torricelli’s law that the speed of ﬂow of the water through the hole √ when the water in the tank has depth x is equal to 2gx, this being the speed attained by a particle falling freely from rest under gravity through a distance x, where g is the acceleration due to gravity. Write down the differential equation determining the water height x(t) in the tank when t > 0, and solve the equation for x(t). If water is added to the tank at a rate V(t), write down the modiﬁed equation governing the water height. If V(t) = V0 is constant, and the ﬂow into and out of the tank reaches equilibrium, ﬁnd the equilibrium height of the water √ in the tank. Remark: √ In applications the expression 2gx is replaced by k 2gx, with 0 < k < 1 a constant. The factor k allows for the contraction of the jet after leaving the hole. In the case of water k ≈ 0.6.

hom*ogeneous Equations

hom*ogeneous equation of degree n

EXAMPLE 5.8

A function f (x, y) is said to be algebraically hom*ogeneous of degree n, or simply hom*ogeneous of degree n, if f (t x, t y) = t n f (x, y) for some real number n and all t > 0, for (x, y) = (0, 0). (a) If f (x, y) = x 2 + 3xy + 4y2 , then f (t x, t y) = t 2 (x 2 + 3xy + 4y2 ) = t 2 f (x, y), so f (x, y) is hom*ogeneous of degree 2. (b) If f (x, y) = ln |y| − ln |x| for (x, y) = (0, 0), then f (x, y) = ln |y/x|, so f (t x, t y) = f (x, y), showing that f (x, y) is hom*ogeneous of degree 0. (c) If f (x, y) =

x 3/2 + x 1/2 y + 3y3/2 , then f (t x, t y) = t 0 f (x, y), 2x 3/2 − xy1/2

showing that f (x, y) is hom*ogeneous of degree 0.

248

Chapter 5

First Order Differential Equations

(d) If f (x, y) = x 2 + 4y2 + sin(x/y), then f (t x, t y) = t 2 (x 2 + 4y2 ) + sin(x/y), so f (x, y) is not hom*ogeneous, because although both the ﬁrst group of terms and the last term are hom*ogeneous functions of x and y, they are not both hom*ogeneous of the same degree. (e) If f (x, y) = tan(xy + 1), then f (t x, t y) = tan(t 2 xy + 1), so f (x, y) is not hom*ogeneous. hom*ogeneous differential equations The ﬁrst order ODE in differential form P(x, y)dx + Q(x, y)dy = 0 is called hom*ogeneous if P and Q are hom*ogeneous functions of the same degree or, equivalently, if when written in the form dy = f (x, y), dx

the function f (x, y) can be written as

f (x, y) = g(y/x).

The substitution y = ux will reduce either form of the hom*ogeneous equation to an equation involving the independent variable x and the new dependent variable u in which the variables are separable. As with most separable equations the solution can be complicated, and it is often the case that y is determined implicitly in terms of x. EXAMPLE 5.9

Solve (y2 + 2xy)dx − x 2 dy = 0. Solution Both terms in the differential equation are hom*ogeneous of degree 2, so the equation itself is hom*ogeneous. Differentiating the substitution y = ux gives du dy =u+x , dx dx

or

dy = udx + xdu.

After substituting for y and dy in the differential equation and cancelling x 2 , we obtain the variables separable equation u(u + 1)dx = xdu,

or

dx du = . u(u + 1) x

This has the general solution u=

Cx , 1 − Cx

but

y = ux

and so y =

Cx 2 , 1 − Cx

where C is an arbitrary constant. In this case the general solution is simple and y is determined explicitly in terms of x.

Section 5.5

EXAMPLE 5.10

hom*ogeneous Equations

249

Solve y2 dy = . dx xy − x 2 Solution The equation is hom*ogeneous because it can be written dy (y/x)2 = . dx (y/x) − 1 Making the substitution y = ux, and again using the result dy/dx = u + xdu/dx, reduces this to the separable equation u2 1 du dx = , or 1 − . u+x du = dx u−1 u x Integration gives u − ln |u| = ln |x| + ln |C|, where C is an arbitrary integration constant. Finally, substituting u = y/x and simplifying the result we arrive at the following implicit solution for y: y = Ce y/x . An equation of the form ax + by + c dy = dx px + qy + r

near-hom*ogeneous

is called near-hom*ogeneous, because it can be transformed into a hom*ogeneous equation by means of a variable change that shifts the origin to the point of intersection of the two lines ax + by + c = 0

EXAMPLE 5.11

and

px + qy + r = 0.

Solve the initial value problem y+1 dy = dx x + 2y

with y(2) = 0.

Solution The equation is near-hom*ogeneous and the lines y + 1 = 0 and x + 2y = 0 intersect at the point x = 2 and y = −1, so we make the variable change x = X + 2 and y = Y − 1, as a result of which the equation becomes the hom*ogeneous equation dY Y = . dX X + 2Y Solving this as in Example 5.9 by setting Y = uX leads to the equation 1 + 2u dX − du = , 2 2u X with the solution 1/u = 2 ln |CuX |,

250

Chapter 5

First Order Differential Equations

where C is an arbitrary integration constant. If we set u = Y/X, this becomes X = 2Y ln |CY|, where C is an arbitrary constant. Returning to the original variables by substituting X = x − 2, Y = y + 1, we arrive at the required general solution x = 2 + 2(y + 1) ln |C(y + 1)|. Although this is an implicit solution for y, if we regard y as the independent variable and x as the dependent variable, solution curves (integral curves) are easily graphed. Substituting the initial condition y = 0 when x = 2 in the general solution shows that C = 1, so the solution of the initial value problem is x = 2 + 2(y + 1) ln |y + 1|.

Summary

This section introduced the special type of ﬁrst order ordinary differential equation known as an algebraically hom*ogeneous equation. This name is frequently shortened to the term hom*ogeneous equation, though this must not be confused with the sense in which the term hom*ogeneous is used in Section 5.1. After showing how such equations can be solved, it was shown how a simple linear change of variables changes a near-hom*ogeneous equation to a hom*ogeneous equation that can then be solved.

EXERCISES 5.5 In Exercises 1 through 14 ﬁnd by hand calculation the general solution of the given hom*ogeneous or nearhom*ogeneous equations and conﬁrm the result by using computer algebra. 1. 2. 3. 4. 5.

y y y y y

= y/(2x + y). = (2xy + y2 )/(3x 2 ). = (2x 2 + y2 )/xy. = (2xy + y2 )/x 2 . = (x − y)/(x + 2y).

5.6

6. 7. 8. 9. 10. 11. 12. 13. 14.

y y y y y y y y y

= (x + 4y)/x. = (2x + y cos2 (y/x))/(x cos2 (y/x)). = 3y2 /(1 + x 2 ). = (x + y sin2 (y/x))/(x sin2 (y/x)). = 3x exp(x + 2y)/y. = (y + 2)/(x + y + 2). = (y + 1)/(x + 2y + 2). = (x + y + 1)/(x − y + 1). = (x − y + 1)/(x + y).

Exact Equations The so-called exact equations have a simple structure, and they arise in many important applications as, for example, in the study of thermodynamics. After deﬁnition of an exact equation, a test for exactness will be derived and the general solution of such an equation will be found. Exact equations deﬁnition of an exact equation

The ﬁrst order ODE M(x, y)dx + N(x, y)dy = 0

Section 5.6

Exact Equations

251

is said to be exact if a function F(x, y) exists such that the total differential d[F(x, y)] = M(x, y)dx + N(x, y)dy.

It follows directly that if M(x, y)dx + N(x, y)dy = 0

(22)

is exact, then the total differential d[F(x, y)] = 0, so the general solution of (22) must be F(x, y) = constant. EXAMPLE 5.12

(23)

The total differential of F(x, y) = 3x 3 + 2xy2 + 4y3 + 2x is d[F(x, y)] = (∂ F/∂ x)dx + (∂ F/∂ y)dy = (9x 2 + 2y2 + 2)dx + (4xy + 12y2 )dy, so the exact differential equation (9x 2 + 2y2 + 2)dx + (4xy + 12y2 )dy = 0 has the general solution 3x 3 + 2xy2 + 4y3 + 2x = constant. Three questions now arise: (i) Is there a test for exactness? (ii) If an equation is exact, is it possible to ﬁnd its general solution? (iii) If an equation is not exact, is it possible to modify it to make it exact? There are satisfactory answers to the ﬁrst two questions, and a less satisfactory answer to the third question. We deal with the last question ﬁrst. It can be shown that an equation of the form (21) that is not exact can always be made exact if it is multiplied by a suitable factor μ(x, y), called an integrating factor, though there is no general method by which such an integrating factor can be found. Fortunately, however, an integrating factor can always be found for a variable coefﬁcient linear ﬁrst order ODE, and in the next section the integrating factor will be derived for such an ODE and then used to ﬁnd its general solution. We now turn our attention to the ﬁrst question. If F(x, y) = constant is a solution of the exact differential equation M(x, y)dx + N(x, y)dy = 0,

(24)

then M(x, y) = ∂ F/∂ x and N(x, y) = ∂ F/∂ y. So, provided the derivatives ∂ F/ ∂ x, ∂ F/∂ y, ∂ 2 F/∂ x∂ y, and ∂ 2 F/∂ y∂ x are deﬁned and continuous in the region within which the differential equation is deﬁned, the mixed derivatives will be equal so that ∂ 2 F/∂ x∂ y = ∂ 2 F/∂ y∂ x. This last result is equivalent to requiring that ∂ M/∂ y = ∂ N/∂ x in order that (24) is exact, so this provides the required test for exactness.

252

Chapter 5

First Order Differential Equations

THEOREM 5.1

a simple test for exactness

EXAMPLE 5.13

Test for exactness The differential equation M(x, y)dx + N(x, y)dy = 0 is exact if and only if ∂ M/∂ y = ∂ N/∂ x. Test for exactness the differential equations (a) {sin(xy + 1) + xy cos(xy + 1)}dx + x 2 cos(xy + 1)dy = 0. (b) (2x + sin y)dx + (2x cos y + y)dy = 0. Solution In case (a) M(x, y) = sin(xy + 1) + xy cos(xy + 1) and N(x, y) = x 2 cos(xy + 1), and ∂ M/∂ y = ∂ N/∂ x, so the equation is exact. In case (b) M(x, y) = 2x + sin y and N(x, y) = 2x cos y + y but ∂ M/∂ y = ∂ N/∂ x, so the equation is not exact. Having established a test for exactness, it remains for us to determine how the general solution of an exact equation can be found. The starting point is the fact that if F(x, y) = constant is a solution of the exact equation M(x, y)dx + N(x, y)dy = 0, then ∂ F/∂ x = M(x, y) and ∂ F/∂ y = N(x, y). Two expressions for F(x, y) can be obtained from these results by integrating M with respect to x while regarding y as a constant, and integrating N with respect to y while regarding x as a constant, because this reverses the process of partial differentiation by which M and N were obtained. However, after integrating M it will be necessary to add not only an arbitrary constant, but also an arbitrary function f (y) of y, because this will behave like a constant when F is differentiated partially with respect to x to obtain M. Similarly, after integrating N it will be necessary to add not only an arbitrary constant, but also an arbitrary function g(x) of x, because this will behave like a constant when F is differentiated partially with respect to y to obtain N. These two expressions for F will look different but must, of course, be identical. The arbitrary function f (y) can be found by identifying it with any function only of y that occurs in the expression for F obtained by integrating N, while the arbitrary function g(x) can be found by identifying it with any function only of x that occurs in the expression for F found by integrating M, where, of course, the true constants introduced after each integration must be identical.

EXAMPLE 5.14

Show the following equation is exact and ﬁnd its general solution: {3x 2 + 2y + 2 cosh(2x + 3y)}dx + {2x + 2y + 3 cosh(2x + 3y)}dy = 0. Solution In this equation M(x, y) = 3x 2 + 2y + 2 cosh(2x + 3y), and N(x, y) = 2x + 2y + 3 cosh(2x + 3y), so as My = Nx = 2 + 6 sinh(2x + 3y) the equation is exact: F(x, y) = M(x, y)dx = {3x 2 + 2y + 2 cosh(2x + 3y)}dx = x 3 + 2xy + sinh(2x + 3y) + f (y) + C,

Section 5.7

and

F(x, y) =

Linear First Order Equations

253

N(x, y)dy =

{2x + 2y + 3 cosh(2x + 3y)}dy

= 2xy + y2 + sinh(2x + 3y) + g(x) + D. For these two expressions to be identical, we must set f (y) ≡ y2 , g(x) ≡ x 3 , and D = C, so F(x, y) is seen to be F(x, y) = x 3 + 2xy + y2 + sinh(2x + 3y) + C, and so the general solution is x 3 + 2xy + y2 + sinh(2x + 3y) = C, where as C is an arbitrary constant we have chosen to write C rather than −C on the right of the solution.

Summary

This section introduced the class of ﬁrst order ordinary differential equations known as exact equations that arise in many different applications. It was then shown how the equality of mixed derivatives yields a simple test for exactness.

EXERCISES 5.6 In Exercises 1 through 8 test the equation for exactness, and when an equation is exact, ﬁnd its general solution. 1. (a) {sin(3y) + 4x 2 y}dx + {3x cos(3y) + y + 2x 3 }dy = 0; (b) {4x 3 + 3y2 + cos x}dx + {6xy + 2}dy = 0. 2. (a) {(2x + 3y2 )−1/2 + 4y3 + 2x}dx + {3y/(2x + 3y2 ) + 12xy2 }dy = 0; (b) {cos(x + 3y2 ) + 4xy3 }dx + {6y cos(x + 3y2 ) + 3x 2 y2 + 2y}dy = 0. 3. (a) {sin x + x cos x + cosh(x + 2y)}dx + {3y2 + 2cosh(x + 2y)}dy = 0; (b) {6x(2x 2 + y2 )1/2 + x 2 }dx + 2y(2x 2 + y2 )1/2 dy = 0. 4. (a) {6x/(3x 2 + y) + 4xy3 }dx + {1/(3x 2 + y) + 6x 2 y2 + 3y2 }dy = 0; (b) {sin(xy) + xy cos(xy) + y2 sin(xy)}dx + {x 2 cos(xy) + cos(xy) − xy sin(xy)}dy = 0.

5.7

5. (a)

3x 2

+

dx + 2 x 3 + y2

y x 3 + y2

, + 6y dy = 0;

(b) {y/x + 2xsinh(y2 )}dx + {ln x + 2x 2 y cosh(y2 )} dy = 0. 6. (a) {4xy + 1/x}dx + {2x 2 − 1/y}dy = 0; (b) {6xy − 2/(x 2 y)}dx + {3x 2 − 2/(xy2 )}dy = 0. 7. (a) {2xy + 6/x}dx + {x 2 + 4/y}dy = 0; (b) {2x/(2x + 3y2 ) − 2x 2 /(2x + 3y2 )2 + 2}dx − 6x 2 y/ (2x + 3y2 )2 dy = 0. √ 8. (a) {(5/2)x 3/2 + 14y3 }dx + {(3/2) y + 42xy2 }dy = 0; 2 (b) (y/x ) cos(y/x)dx + {(1/x) cos(y/x) + 6y exp(y2 )} dy = 0.

Linear First Order Equations The standard form of the linear ﬁrst order differential equation is standard form of linear ﬁrst order equation

dy + P(x)y = Q(x), dx

(25)

254

Chapter 5

First Order Differential Equations

where P(x) and Q(x) are known functions. An initial value problem (i.v.p) for a linear ﬁrst order ODE involves the speciﬁcation of an initial condition y(x0 ) = y0 ,

(26)

where this last condition means that y = y0 when x = x0 . Thus, the solution of the initial value problem will evolve away from the point (x0 , y0 ) in the (x, y)-plane as x increases from x0 . To ﬁnd the general solution of (25) we multiply the equation by a function μ(x), still to be determined, to obtain μ

dy + μP(x)y = μQ(x), dx

(27)

and seek a choice for μ that allows the left-hand side of (26) to be written as d(μy)/dx. With this choice of μ, equation (27) becomes d(μy) = μQ(x), dx

(28)

so integrating with respect to x and dividing by μ shows the general solution of (25) to be y(x) =

integrating factor

1 C + μ(x) μ(x)

μ(x)Q(x)dx,

(29)

where C is an arbitrary integration constant. Notice that it is essential to include the arbitrary integration constant immediately after the integration μ(x)Q(x)dx has been performed, and before dividing by μ(x); otherwise, the form of the general solution will be incorrect. To make use of (29) it is necessary to determine the function μ(x) called the integrating factor for the linear ﬁrst order ODE in (24). By deﬁnition dy d(μy) =μ + μP(x)y, dx dx so after expanding the left-hand side this becomes μ

dy dμ dy +y =μ + μP(x)y. dx dx dx

Cancelling the terms μdy/dx and dividing by y gives the following variables separable equation for the integrating factor μ(x): dμ = μP(x). dx This has the solution μ(x) = A exp

( P(x)dx ,

Section 5.7

ﬁnding the integrating factor

Linear First Order Equations

255

where A is an arbitrary integration constant. As μ multiplies the entire equation (27), the choice of A is immaterial, so for simplicity we will always set A = 1 and take the integrating factor to be μ(x) = exp

( P(x)dx .

(30)

Inserting (30) into (29) shows the general solution of (25) to be ( ( ( exp y(x) = C exp − P(x)dx + exp − P(x)dx P(x)dx Q(x)dx. (31)

complementary function, particular integral, and general solution

If an initial value problem is involved in which the solution of (25) is required subject to the initial condition y(x0 ) = y0 , the value of the arbitrary constant C in (31) must be chosen accordingly. The form of the general solution in (31) is mainly of importance for theoretical reasons, because it shows that the general solution is the sum of a complementary function ( yc (x) = C exp − P(x)dx

(32)

that contains the arbitrary constant belonging to the general solution of (25), and a particular integral ( ( yp (x) = exp − P(x)dx exp P(x)dx Q(x)dx

(33)

that contains no arbitrary constant and is determined by the nonhom*ogeneous term Q(x). Substitution of yc (x) into the hom*ogeneous form of (25) given by dy + P(x)y = 0 dx shows that yc (x) is its general solution. The general solution of the nonhom*ogeneous equation (25) is now seen to be the sum of the general solution of the hom*ogeneous form of the equation, and a particular integral determined by the nonhom*ogeneous term. It will be shown later that this is the pattern of the general solution for all linear nonhom*ogeneous differential equations, no matter what their order. Rather than trying to remember the form of general solution given in (31), it is better to obtain the solution by starting from the integrating factor μ(x) in (30) and integrating result (28), while not forgetting to include the arbitrary constant immediately after the integration before dividing by μ(x). For convenience, the steps in the determination of the general solution of (25) can be listed as follows.

256

Chapter 5

First Order Differential Equations

steps used when solving a linear ﬁrst order equation

Rule for solving linear ﬁrst order equations STEP 1

If the equation is not in standard form and is written a(x)

dy + b(x)y = c(x), dx

divide by a(x) to bring it to the standard form dy + P(x)y = Q(x), dx

STEP 2

with P(x) = b(x)/a(x) and Q(x) = c(x)/a(x) Find the integrating factor μ(x) = exp

STEP 3

( P(x)dx .

Rewrite the original differential equation in the form d(μy) = μQ(x). dx

STEP 4

Integrate the equation in Step 3 to obtain μ(x)y(x) =

STEP 5 STEP 6

EXAMPLE 5.15

μ(x)Q(x)dx + C.

Divide the result of Step 4 by μ(x) to obtain the required general solution of the linear ﬁrst order differential equation in Step 1. If an initial condition y(x0 ) = y0 is given, the required solution of the i.v.p. is obtained by choosing the arbitrary constant C in the general solution found in Step 5 so that y = y0 when x = x0 .

Solve the initial value problem cos x

dy + y = sin x, subject to the initial condition y(0) = 2. dx

Solution We follow the steps in the above rule. STEP 1

When written in standard form the equation becomes 1 dy + y = tan x, dx cos x

so P(x) = 1/ cos x and Q(x) = tan x.

Section 5.7

STEP 2

μ(x) = exp

dx cos x

(

= sec x + tan x =

= exp{ln |sec x + tan x|} 1 + sin x . cos x

The original differential equation can now be written d dx

STEP 4

257

The integrating factor

STEP 3

Linear First Order Equations

1 + sin x 1 + sin x y(x) = tan x. cos x cos x

Integrating the result of Step 3 gives

1 + sin x 1 + sin x y(x) = tan xdx + C cos x cos x = sec x tan xdx + tan2 xdx + C = sec x + tan x − x + C =

1 + sin x − x + C. cos x

STEP 5 Dividing the result of Step 4 by the integrating factor μ(x) = (1 + sin x)/ cos x shows that the required general solution is y(x) =

C cos x x cos x +1− , 1 + sin x 1 + sin x

for x such that 1 + sin x = 0. The complementary function is seen to be yc (x) =

C cos x , 1 + sin x

and the particular integral is yp (x) = 1 −

x cos x . 1 + sin x

STEP 6 The initial condition requires that y = 2 when x = 0, and the general solution is seen to satisfy this condition if C = 1, so the solution of the i.v.p. is y(x) = 1 + EXAMPLE 5.16

(1 − x) cos x . 1 + sin x

An R–L circuit contains an inductor and resistor in series, and a current is made to ﬂow through them by applying a voltage across the ends of the circuit. If the inductance varies linearly with time in such a way that L(t) = L0 (1 + kt), ﬁnd the current i(t) ﬂowing in the circuit when t > 0, given that a constant voltage V0 is applied at time t = 0 when i(t) = 0.

258

Chapter 5

First Order Differential Equations

Solution The voltage change due to a current i(t) ﬂowing through the inductance is d(L(t)i)/dt, and from Ohm’s law the corresponding voltage change across the resistance R is Ri, so as the sum of the these voltage changes must equal the imposed constant voltage V0 , the differential equation determining the current becomes d (L(t)i) + Ri = V0 for t > 0. dt Substituting for L(t) and rearranging terms we arrive at the following linear ﬁrst order variable coefﬁcient nonhom*ogeneous equation for i(t) V0 di kL0 + R i= + , dt L0 (1 + kt) L0 (1 + kt) subject to the initial condition i(0) = 0. V0 kL0 + R and Q(t) = , so In the notation of this section P(t) = L (1 + kt) L (1 + kt) 0 0 the integrating factor in Step 2 becomes ( μ(t) = exp P(t)dt = (1 + kt)[kL0 +R]/kL0 . Using μ(t) and Q(t) in Step 4 and applying the initial condition i(0) = 0 then shows that the current i(t) at a time t > 0 is determined by ) * kL0 +R V0 i(t) = . 1 − (1 + kt) kL0 kL0 + R

Summary

The study of the linear ﬁrst order differential equation considered in this section is important in its own right, and it also provides the key to understanding the nature of the solution of linear higher order differential equations. It was shown how, after an equation is written in standard form, it can be solved by means of an integrating factor that can be found directly from the coefﬁcient of y in the equation.

EXERCISES 5.7 In Exercises 1 through 10 ﬁnd the general solution for the linear ﬁrst order differential equation, and check your result by using computer algebra. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

dy/dx + 2y = 1. dy/dx + (1/x)y = x. (x + 1)dy/dx + y = 2x(x + 1). x 2 dy/dx + xy = x 2 sin x. x 2 dy/dx − 2xy = 1 + x. sin x dy/dx − y cos x = 2 sin2 x. x dy/dx + 2y = x 2 . (x + 3)dy/dx − 2y = x + 3. sin x dy/dx − y = 2 sin x. sin x dy/dx + y = sin x.

In Exercises 11 through 16 solve the initial value problem for the linear ﬁrst order differential equation, and check your result by using a computer algebra package.

x dy/dx − y = x 2 cos x, with y(π/2) = π . x 2 dy/dx + 2xy = 2 + x, with y(1) = 1. x dy/dx − 2y = 2 + x, with y(1) = 0. x dy/dx + 2y = 2x 4 , with y(1) = 1. sin x dy/dx + y cos x = 2 sin2 x, with y(π/2) = 0. 2 dy/dx + y = x 2 , with y(0) = 1. A 25-liter gas cylinder contains 80% oxygen and 20% helium. If helium is added at a rate of 0.2 liters a second, and the mixture is drawn off at the same rate, how long will it be before the cylinder contains 80% helium? 18. If in Exercise 17 the volume of the gas cylinder is 20 liters and initially it contains 90% oxygen and 10% helium, and the rate of supply of helium is q liters a second, what must be the value of q if the cylinder is to become 80% full of helium in 1 minute?

11. 12. 13. 14. 15. 16. 17.

Section 5.8

259

write down the differential equation for ν(t), and hence ﬁnd the value of k if the motion starts with ν(0) = ν0 , and at time t = 1/k its velocity is ν(1/k) = 14 ν0 .

19. A particle of unit mass moves horizontally in a resisting medium with velocity ν(t) at time t with a resistance opposing the motion given by kν(t), with k > 0. If the particle is also subject to an additional resisting force kt,

5.8

The Bernoulli Equation

The Bernoulli Equation The Bernoulli equation is a nonlinear ﬁrst order differential equation with the standard form dy + P(x)y = Q(x)yn , dx

standard form of the Bernoulli equation

(n = 1).

(34)

The substitution u = y1−n

(35)

reduces (34) to the linear ﬁrst order ODE du 1 + P(x)u = Q(x), (1 − n) dx

wavefront and acceleration wave

(36)

and this can be solved by the method described in Section 5.7. Once the general solution u(x) of (36) has been found, the general solution y(x) of (34) follows by returning to the original dependent variable by making the substitution u = y1−n . When using the general solution in (36) it is important to write the Bernoulli equation in standard form before identifying P(x), Q(x), and n. However, if the form of the equation corresponding to (36) is derived directly, starting from the substitution u = y1−n , there is no need for the equation to be in standard form. The Bernoulli equation occurs in various applications of mathematics that involve some form of nonlinearity. It occurs, for example, in solid and ﬂuid mechanics, where it is found to describe an important characteristic of special types of wave that propagate through space as time increases. To appreciate how this ODE enters into these problems, we consider a simple application to solid mechanics involving a long bar made of a composite material or a polymer whose properties are such that the extension caused by a force does not obey Hooke’s law, and so is not proportional to the force. Materials of this type are said to be nonlinearly elastic. If such a bar receives a blow at one end a disturbance will propagate along it at a ﬁnite speed, so that at any instant of time there will be a region in the bar through which the disturbance has passed, and a region ahead of the disturbance through which it has still to pass. When the blow is not large, the propagating boundary between these two regions is called a wavefront and t the function representing the displacement at position x at any given time t will be continuous along the bar, though its derivative with respect to x will be discontinuous across the wavefront. The propagating jump in the derivative of the displacement with respect to x at the wavefront as a function of time is called an acceleration wave, and we will denote it

260

Chapter 5

First Order Differential Equations

by a(t). For many nonlinear materials the magnitude a(t) of the acceleration wave obeys a Bernoulli equation of the form da + μ(t)a = β(t)a 2 . dt

(37)

It was shown by P. J. Chen (Selected Topics in Wave Propagation, Noordhoff, Leyden, 1976, p. 29) that μ(t) depends on the material properties of the medium through which the disturbance propagates and also the geometry involved, which in a one-dimensional case may be plane, cylindrically, or spherically symmetric, but that the function β(t) depends only on the material properties of the medium. This same equation governs the behavior of acceleration waves in three space dimensions and time. Because of the effects of nonlinearity, in many materials it is possible for the acceleration wave to strengthen as it propagates to the point at which the continuity of the displacement function breaks down and what is called a shock wave forms. When this occurs, the speed of propagation of disturbances and other physical quantities become discontinuous across the shock wave, and this in turn can lead to the fracture of the material. Once the material properties of such a medium are speciﬁed together with the nature of the initial disturbance, the Bernoulli equation in (37) can be used to determine whether or not a shock wave will form and, if it does, the point along the bar where this occurs. EXAMPLE 5.17 examples of the Bernoulli equation

Solve the Bernoulli equation da + a = ta 2 , dt and ﬁnd a condition that determines when the solution becomes unbounded. Solution The equation is in standard form with P(t) = 1, Q(t) = t, and n = 2. Making the substitution u = 1/a corresponding to (35) and substituting into (36) leads to the linear ﬁrst order equation du − u = −t. dt Solving this by the method described in Section 5.7 gives u(t) = Cet + 1 + t, so transforming back to the variable a(t), we ﬁnd that a(t) = 1/(Cet + 1 + t). The solution a(t) of the Bernoulli equation will become unbounded at t = tc if tc is a solution of the equation C exp(tc ) + 1 + tc = 0. This result shows that an acceleration wave starting at time t = 0 will decay instead of evolving into a shock wave if C > 0, because then the equation for tc has no positive solution, whereas a shock wave will always form if C < 0. Had a(t) represented the magnitude of an acceleration wave, the development of an inﬁnite gradient in the displacement corresponding to a(tc ) = ∞ would indicate shock formation.

Section 5.8

EXAMPLE 5.18

The Bernoulli Equation

261

Find the general solution of dy − 2y = xy1/2 . dx Solution In terms of the standard form of the Bernoulli equation given in (34), P(x) = −2, Q(x) = x, and n = 1/2. However, rather than substituting into equation (36) to obtain a linear differential equation for u(x), we will derive it directly starting from the substitution u = y1/2 , and differentiating it to ﬁnd du/dx in terms of dy/dx. We have 1 1 dy du dy = y−1/2 = , dx 2 dx 2u dx

so

dy du = 2u . dx dx

Substituting for y and dy/dx in the Bernoulli equation and cancelling a factor 2u gives the following linear equation (compare it with (36) after substituting for P(x), Q(x) and n): du 1 − u = x. dx 2 The method of Section 5.6 shows this equation to have the general solution u(x) = Ce x − (1/2)(1 + x), so as u = y1/2 , the required general solution of the Bernoulli equation is y(x) = [Ce x − (1/2)(1 + x)]2 . JACOB BERNOULLI (1654–1705) A Swiss mathematician born in Basel where he was professor of mathematics until his death. He was a member of one of the most distinguished families of mathematicians in all of the history of mathematics. His most important contributions were to the theory of probability and the calculus and theory of elasticity. Other members of the family contributed to many different parts of mathematics including hydrodynamics and the calculus of variations.

Summary

In a sense, the Bernoulli equation, which is a nonlinear ﬁrst order differential equation, stands on the boundary between linear and nonlinear ﬁrst order differential equations, so for this and other reasons it is important in applications. It arises in different applications, many of which themselves arise from problems bordering on linear and nonlinear regimes. This section showed how a straightforward change of variable transforms a Bernoulli equation into a linear ﬁrst order differential equation that can then be solved by the method of Section 5.6.

EXERCISES 5.8 In Exercises 1 through 8 ﬁnd the general solution of the Bernoulli equation. 1. dy/dx + 2y = 2xy1/2 . 2. dy/dx + y = 3y2 . 3. dy/dx − y = 2xy3/2 .

4. 5. 6. 7. 8.

x dy/dx + y = xy2 . dy/dx + 2y sin x = 2y2 sin x. x dy/dx + y = 2xy1/2 . x dy/dx − 2y = xy3/2 . dy/dx + 4xy = xy3 .

262

Chapter 5

First Order Differential Equations

9. A model for the variation of a ﬁnite amount of stock n(t) in a warehouse as a function of the time t caused by the supply of fresh stock and its removal by demand is dn = (a − bn)n with the constants a, b > 0, dt where n(0) = n0 . Find n(t) and discuss the nature of the change in the stock level as a function of time according as n0 is less than a/b, equal to a/b, or greater than a/b. 10.* This exercise concerns water in a canal of variable depth with the x-axis taken along the canal in the equilibrium surface of the water, and the y-axis vertically downwards. Let the equilibrium depth of water in a channel be h(x), and the cross-sectional area of water in the canal be a slowly varying function W(x). When a water wave advances along the channel into water at rest there will be a change of acceleration across the advancing line (wavefront) that separates the disturbed water from the undisturbed water. Such an advancing disturbance is called an acceleration

5.9

wave. If the change in acceleration across the wavefront at point x along the channel is a(x), it can be shown that the strength a(x) of the acceleration wave obeys the Bernoulli equation da 3h W 3a 2 + + a+ = 0. dx 4h 2W 2h If the initial condition for a(x) is a(0) = a0 , then a wave of elevation wave is one for which a0 < 0, and a wave of depression is one for which a0 > 0. In this approximation the wave will break, due to the water surface becoming vertical at the wavefront if, after propagating a critical distance xc along the channel, the strength of the acceleration a(xc ) = ∞. (i) Find a(x) in terms of a0 = a(0), h0 = h(0) and W0 = W(0). (ii) Discuss the breaking and non-breaking of waves of elevation and depression. (iii) If the water shelves to zero at x = l, so that h(l) = 0, ﬁnd a condition that ensures the wave breaks before x = l.

The Riccati Equation The Riccati equation is an important nonlinear equation with the standard form

standard form of the Riccati equation

dy + P(x)y + R(x)y2 = Q(x). dx

(38)

Its signiﬁcance derives from the fact that it stands at the boundary between linear and nonlinear equations, and it occurs in various applications of mathematics that involve nonlinear problems. The Riccati equation reduces to a linear ﬁrst order equation when R(x) ≡ 0, and to a Bernoulli equation when Q(x) ≡ 0. Obtaining the general solution of a Riccati equation is difﬁcult, but the task is simpliﬁed if a particular solution is known, or can be found by inspection. If a particular solution is y1 (x) is known, then

substitutions that simplify the Riccati equation

(i) The substitution y = y1 + 1/u reduces the equation to a linear ﬁrst order equation. (ii) The substitution y = y1 + u reduces the equation to a Bernoulli equation. (iii) The general substitution

y=

1 dz R(x)z dx

Section 5.9

The Riccati Equation

263

reduces the Riccati equation to the linear hom*ogeneous second order ODE ( d2 z R (x) dz + P(x) − − R(x)Q(x)z = 0 dx 2 R(x) dx discussed in Chapters 6 and 8. Substitution (i) is often the most convenient one to use, as will be seen from the next example. EXAMPLE 5.19

Find the general solution of the Riccati equation dy + x 2 y − xy2 = 1. dx Solution Inspection shows that y1 (x) = x is a particular solution, so we make the substitution y = x + 1/u, from which it follows that 1 du dy =1− 2 , dx u dx and after substitution for y and dy/dx in the Riccati equation it reduces to the linear ODE du + x 2 u = −x. dx Solving this by the method of Section 5.6 gives u(x) = C exp(−x /3) − exp(−x /3) 3

3

x exp(x 3 /3)dx,

where the integral in the last term cannot be expressed in terms of elementary functions. Transforming back to the variable y(x) shows the general solution of the Riccati equation to be y(x) = x +

exp(x 3 /3) . C − x exp(x 3 /3)dx

It is not unusual for solutions of ODEs to give rise to functions such as x exp(x 3 /3)dx that have no representation in terms of known functions, because not all functions have antiderivatives that are expressible in terms of elementary functions.

JACOPO FRANCESCO (COUNT) RICCATI (1676–1754) An Italian mathematician whose main contributions to mathematics were in the ﬁeld of differential equations, though he also contributed to geometry and the study of acoustics.

Additional information relevant to the material in Sections 5.4 to 5.9 is to be found in the appropriate chapters of any one of references [3.3] to [3.5], [3.15], [3.16], and [3.19]. A sophisticated and extremely enlightening discussion of ordinary differential equations is to be found in reference [3.1] that considers not only ﬁrst order equations, but also higher order equations and systems.

264

Chapter 5

First Order Differential Equations

Summary

This section introduced the Riccati equation, of which the Bernoulli equation is a special case. Solving the Riccati equation is difﬁcult, but some substitutions were given that simplify this task when one solution of the Riccati equation is already known, possibly by inspection.

EXERCISES 5.9 1. Show that the substitution y = y1 + 1/u reduces the Riccati equation in (38) to a linear ﬁrst order equation. 2. Show that the substitution y = y1 + u reduces the Riccati equation in (38) to a Bernoulli equation. In Exercises 3 through 6 verify that y1 (x) is a solution of the Riccati equation and use it to ﬁnd the general solution of the equation. 3. dy/dx + 2x 2 y − 2xy2 = 1, with y1 (x) = x. 4. dy/dx + 2y2 − y = 1, with y1 (x) ≡ 1.

5.10

5. dy/dx − 2y2 + 3y = 1, with y1 (x) ≡ 1. 6. dy/dx − 3x 2 y + 3xy2 = 1, with y1 (x) = x. 7. Verify that the substitution y=

1 dz R(x)z dx

reduces the Riccati equation (38) to the linear hom*ogeneous second order ODE ( R (x) dz d2 z − R(x)Q(x)z = 0. + P(x) − dx 2 R(x) dx

Existence and Uniqueness of Solutions

existence and uniqueness

The questions of whether a solution to an initial value problem for a ﬁrst order differential equation can be found and, when a solution does exist, whether it is the only solution are of fundamental importance in the theory of differential equations, and also in their applications. Establishing that a solution to an initial value problem can be found is called the existence problem, while ensuring that when a solution exists it is the only one is called the uniqueness problem. To show that the questions of existence and uniqueness arise even with very simple initial value problems we examine the following two examples. Let us consider the initial value problem dy 4 = y1/4 , dx 3

with y(0) = −1,

involving a variables separable equation. Integration shows the general solution to be y3 = (x + C)4 , from which it can be seen that y is essentially nonnegative. Clearly there can be no solution to this equation such that y = −1 when x = 0, so this is an example of an initial value problem that has no solution. Had the initial condition been y(0) = 1 the unique solution would have been y3 = (x + 1)4 . In fact this equation has a solution for any initial condition in which y(x) is positive, but no solution when it is negative. This is hardly surprising, because had we examined the function y1/4 carefully before proceeding with the integration we would have seen that it is a complex number whenever y is negative. Sometimes,

Section 5.10

Existence and Uniqueness of Solutions

265

as here, an inspection of the initial condition and the equation can show in advance whether or not the condition is appropriate, but more frequently constraints on an initial condition that allow a solution to the differential equation to exist only emerge when the form of the solution is known. To illustrate nonuniqueness, we need only consider the differential equation dy = 3y2/3 , subject to the initial condition y(0) = 0. dx The equation is variables separable, and integration shows it has the solution y = x 3 , but this is not the only solution because it also has the singular, though somewhat uninteresting, solution y = 0. However, these are not the only two solutions, because for any a > 0 the function 0, x < a y(x) = (x − a)3 , x ≥ a is continuous, has a continuous ﬁrst derivative, and satisﬁes both the differential equation and the initial condition, showing that it also is a solution. As a > 0 is arbitrary, we see that y(x) is a one-parameter family of solutions, so clearly this initial value problem does not have a unique solution. The following theorem on existence and uniqueness is stated without proof (see, for example, references [3.1],[3.3],[3.4],[3.10] and [3.12]). It is important to appreciate that though the conditions in the theorem are sufﬁcient to ensure existence and uniqueness, they are not necessary conditions, as examples can be constructed that fail to satisfy the conditions of the theorem, but nevertheless have a unique solution. THEOREM 5.2 conditions that deﬁnitely ensure existence and uniqueness

Existence and uniqueness of solutions Let f (x, y) be a continuous and bounded function of x and y in a rectangular region R of the (x, y)-plane that contains a given point (x0 , y0 ). Then for some suitably small positive number h the initial value problem dy = f (x, y), dx

with

y(x0 ) = y0

has at least one solution within the open interval x0 − h < x < x0 + h. If, in addition, ∂ f/∂ y is continuous and bounded in R, the solution is unique in an open interval centered on x0 that may lie within the interval x0 − h < x < x0 + h. Let us apply this theorem to the initial value problem dy = 3y2/3 , dx

with y(0) = 0,

that we have just shown does not have a unique solution. The function f (x, y) = 3y2/3 is continuous in any neighborhood of the origin where the initial condition is given, but ∂ f/∂ y = 2y−1/3 is unbounded at the origin. So the ﬁrst condition of Theorem 5.2 is satisﬁed but the second is not, showing that although this initial value problem has a solution, it is not unique.

266

Chapter 5

First Order Differential Equations

Summary

This section described what is meant by the existence of a solution of a differential equation, and the uniqueness of a solution that is usually expected in applications to physical problems. A theorem, stated without proof, was given that guarantees both the existence and uniqueness of a solution. However, the conditions of the theorem are more restrictive than necessary, so equations can be found that while not satisfying the conditions of the theorem nevertheless have a solution, and it is unique.

EXERCISES 5.10 In Exercises 1 through 6, ﬁnd any points at which the imposition of initial conditions will not lead to a unique solution. 1. dy/dx = (1 − x)1/2 . 2. dy/dx = xy + 1.

3. 4. 5. 6.

dy/dx dy/dx dy/dx dy/dx

= x 2 + y2 . = (x 2 + y2 − 1)−1/2 . = −y/x. = x ln|1 − y2 |.

Section 5.10

Existence and Uniqueness of Solutions

267

CHAPTER 5

TECHNOLOGY PROJECTS Project 1

(⫺6, 4). Superimpose the integral curves on the direction ﬁeld and compare them with the arrows in the direction ﬁeld. 3. Repeat Steps 1 and 2, but this time using the nonlinear ODE

Solution of First Order Linear Differential Equation The purpose of this project is to use computer algebra to solve a ﬁrst order equation step by step from ﬁrst principles, and then to obtain the same result by means of a computer software ODE solver.

1. Given the linear ﬁrst order differential equation

y ⫹ (3x sin x)y ⫽ 2x sin x, 2

Project 3

2

use computer integration to ﬁnd the general solution by reproducing the steps in the rule for the solution by means of an integrating factor given in Section 5.6, and check the result by substitution into the differential equation. 2. Use a computer ODE solver to ﬁnd the general solution and conﬁrm that it is the same as the result obtained in step 1. Project 2 Direction Fields and Integral Curves The purpose of the following project is to gain insight into the relationship between direction ﬁelds and integral curves by using a computer package to plot the direction ﬁelds for two nonlinear ﬁrst order differential equations, and then to add to the direction ﬁeld plots some typical integral curves obtained by using a standard numerical ODE solver package.

1. Construct the direction ﬁeld for the nonlinear ODE 1 1 y ⫽ sin x cos x ⫹ y for ⫺6 ⱕ x ⱕ 6, 2 2 ⫺6 ⱕ y ⱕ 6.

( ) (

y ⫽ x sin(y ⫺ 1)/(3 ⫹ cos x) for ⫺6 ⱕ x ⱕ 6, ⫺6 ⱕ y ⱕ 6.

)

2. Use a standard ODE numerical solver package to ﬁnd the solutions (the integral curves) through the points (⫺6, ⫺4), (⫺6, ⫺2), (⫺6, 2),

Direction Fields and Isoclines An isocline is a curve in the direction ﬁeld of the differential equation y ⫽ f (x, y) at each point of which the slope of the direction ﬁeld has the same constant value. This means that wherever a solution curve of the equation intersects an isocline, its tangent will have the same slope. The isoclines of the differential equation y ⫽ f (x, y) are the curves k ⫽ f (x, y), where k is the slope (gradient) of all solution curves at the points where they intersect the isocline. In general an isocline is not a solution curve and, depending on the function f (x, y), there may be no isoclines for some values of the constant k. The purpose of this project is to construct the direction ﬁeld for an ODE, and to superimpose on it some representative isoclines and solution curves to illustrate their interrelationship.

1. Use computer algebra to construct the direction ﬁeld for the ordinary differential equation y ⫽ x 2 ⫺ y ⫺ 1

for ⫺2 ⱕ x ⱕ 2, ⫺2 ⱕ y ⱕ 2,

and superimpose on the direction ﬁeld the isoclines corresponding to k ⫽ ⫺1, 0, 1, 2. Verify that all arrows intersecting an isocline are parallel. 2. Use a standard ODE numerical solver package to ﬁnd the solutions through the points (⫺2, ⫺1.5), (⫺2, ⫺0.5), (⫺2, 0.5), (⫺2, 1.5). Superimpose the solution curves on the isoclines found in Step 1 and conﬁrm that the tangents to solution curves where they intersect an isocline are all parallel.

267

This Page Intentionally Left Blank

6

C H A P T E R

Second and Higher Order Linear Differential Equations and Systems

L

inear second order differential equations with constant coefﬁcients are the simplest of the higher order differential equations, and they have many applications. They are of the general form y + Ay + B y = F (x) with A and B constants and F (x), called the nonhom*ogeneous term, a known function of x. The equation is called nonhom*ogeneous when F (x) is not identically zero; otherwise, it is called hom*ogeneous. All general solutions are shown to be the sum of two quite different parts, one being a solution of the hom*ogeneous equation called the complementary function that contains the expected two arbitrary constants of integration, and the other a special solution called a particular integral that depends only on F (x) and contains no arbitrary constants. Methods are developed for the solution of hom*ogeneous and nonhom*ogeneous second order equations and for the solution of associated initial value problems. Particular attention is paid to the second order equations that describe oscillatory phenomena, because equations of this type arise in practical problems involving oscillations in electrical circuits, in the description of many types of mechanical vibration, and elsewhere. It is shown that in stable oscillatory motions the particular integral describes the start-up of an initial value problem, after which it decays, leaving only the complementary function that describes the long-term behavior known as the steady state solution. The methods of solution for second order equations developed in this chapter include the simplest one, called the method of undetermined coefﬁcients; the powerful method of variation of parameters; and a related method involving a function called the Green’s function that is independent of the nonhom*ogeneous term F (x). Various useful special cases of second order equations are considered, after which higher order linear differential equations and ﬁrst order systems are introduced and solved, the solutions of which have the same general structure as the second order equations. Matrix methods are introduced for the description and solution of ﬁrst order systems of equations. The chapter concludes with a discussion of linear autonomous systems of equations, followed by a brief introduction to nonlinear autonomous systems that arise in many practical problems and can lead to oscillatory solutions of a nonlinear nature. The general behavior of solutions of both types of autonomous system is described in an interesting and useful geometrical manner involving what are called trajectories in the phase plane.

269

270

Chapter 6

6.1

Second and Higher Order Linear Differential Equations and Systems

hom*ogeneous Linear Constant Coefﬁcient Second Order Equations

linear constant coefﬁcient second order equation

T

he simplest general higher order hom*ogeneous differential equation that occurs in applications is the linear constant coefﬁcient second order equation d2 y dy + By = 0. +A 2 dx dx

(1)

Equations like this were derived in Section 5.2(d), where they were shown to describe the motion of a mass–spring system subject to frictional resistance, and also the variation of charge in an R–L–C electric circuit. The equation also describes the pendulum-like motion of a load suspended from a crane that is set in motion when the crane rotates to a new position and soon stops. The motion can be modeled as shown in Fig. 6.1, where is the length of the crane cable, m is the load, F is the resisting frictional force exerted by the air due to motion, and θ is the angular deﬂection of the cable from the vertical. The angular momentum of the load about a line through the support point of the cable at O normal to the plane of motion is m 2 (dθ/dt), so the rate of change of angular momentum about O is m 2 (d2 θ/dt 2 ). The moments acting to restore the load to its equilibrium position at Q are due to the air resistance F opposing the motion and the turning moment of the gravitational force mg about O. If the air resistance acting on the load is proportional to the speed of the load, and the constant of proportionality is μ, the resisting frictional force is F = μ(dθ/dt), so the restoring moment exerted by F about O is F = μ2 (dθ/dt). The turning moment exerted by the gravitational force mg about O is mgsin θ, so equating the rate of change of angular momentum to the sum of the two restoring moments gives

θ

l

mg

FIGURE 6.1 A deﬂected load supported by a crane cable.

Section 6.1

hom*ogeneous Linear Constant Coefﬁcient Second Order Equations

271

the equation of motion m 2

d2 θ dθ − mgsin θ. = −μ2 dt 2 dt

The negative signs on the right are necessary because the restoring moments act in the opposite sense to that of the rate of change of angular momentum. When the angle of swing is small sin θ can be approximated by θ , and the equation of motion simpliﬁes to g d2 θ μ dθ + θ = 0. + dt 2 m dt Because of its many applications we start our discussion of higher order equations by examining the properties and general solution of equation (1). Let y1 (x) and y2 (x) be any two solutions of (1). Then because each function satisﬁes the differential equation, it follows that d2 y1 dy1 d2 y2 dy2 + By + By2 = 0. + A = 0 and +A 1 dx 2 dx dx 2 dx Now consider the linear combination of the two solutions y(x) = c1 y1 (x) + c2 y2 (x),

(2)

(3)

where c1 and c2 are arbitrary constants. Substituting (3) into (1) and grouping terms gives d2 [c1 y1 + c2 y2 ] d[c1 y1 + c2 y2 ] + B[c1 y1 + c2 y2 ] +A dx 2 dx 2 2 d y1 d y2 dy1 dy2 + By1 + c2 + By2 = 0, = c1 +A +A dx 2 dx dx 2 dx

linear superposition, dependence, and independence

because each of the bracketed groups of terms vanishes on account of (2). This has shown that y(x) = c1 y1 (x) + c2 y2 (x) is also a solution of (1). This last result is described by saying equation (1) allows the linear superposition of solutions and it means that the sum of solutions is again a solution. Later we will see that linear superposition of solutions is a fundamental property of all hom*ogeneous linear equations, including those with variable coefﬁcients. Two functions y1 (x) and y2 (x) are said to be linearly independent over an interval a ≤ x ≤ b if the equation c1 y1 (x) + c2 y2 (x) = 0

(4)

is only true for all x in the interval if c1 = c2 = 0. The functions are said to be linearly dependent if (4) is true for some nonvanishing constants c1 and c2 . When the functions are linearly dependent, provided c1 = 0, equation (4) can be written y1 (x) = −

c2 y2 (x), c1

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with a corresponding result y2 (x) = −

c1 y1 (x), c2

if c2 = 0, showing that in each case the linear dependence of the functions means they are proportional. We have established the following simple test.

simple test for linear independence

EXAMPLE 6.1

Test for linear independence of y1 (x) and y2 (x) over a ≤ x ≤ b The two functions y1 (x) and y2 (x) will be linearly independent over a ≤ x ≤ b if they are not proportional over the interval; otherwise, they will be linearly dependent. Apply the test for linear independence to the following pairs of functions. (a) e x and e2x are linearly independent for all x because e2x /e x = e x is deﬁned for all x and e x is not a constant. (b) ln x 2 and ln x 3 are linearly dependent for x > 0, because ln x 2 = 2 ln x and ln x 3 = 3 ln x, so ln x 2 /ln x 3 = 2/3 is a constant, and the logarithmic function is deﬁned for x > 0. (c) sinh 2x and sinh x cosh x are linearly dependent for all x because sinh 2x = 2 sinh x cosh x.

general solution

The notion of the linear independence of functions is of special signiﬁcance when the functions are solutions of hom*ogeneous differential equations. This is because it will be seen later that all particular solutions of such differential equations can be represented in the form of suitable linear combinations of as many linearly independent solutions as the equation allows. In fact, the number of linearly independent solutions is equal to the order of the differential equation, so the second order differential equation (1) has two linearly independent solutions. So, if y1 (x) and y2 (x) are linearly independent solutions of (1), and c1 and c2 are arbitrary constants, the general solution of (1) from which all particular solutions can be obtained can be written y(x) = c1 y1 (x) + c2 y2 (x).

(5)

The justiﬁcation of this assertion will be postponed until the nature of the linearly independent solutions of (1) has been established. EXAMPLE 6.2

Direct substitution of the functions y1 (x) = sin 2x and y2 (x) = cos 2x into the second order differential equation y + 4y = 0 conﬁrms that they are solutions. The functions are linearly independent for all x because they are not proportional, so y(x) = c1 cos 2x + c2 sin 2x is the general solution of the differential equation.

Section 6.1

hom*ogeneous Linear Constant Coefﬁcient Second Order Equations

273

We will now ﬁnd the general solution of (1), and when doing so use will be made of the fact that if y(x) = ceλx , with c and λ constants, then d[ceλx ] dy = = cλeλx dx dx

and

d2 y d2 [ceλx ] = = cλ2 eλx . dx 2 dx 2

Substituting these results into (1) leads to the equation (λ2 + Aλ + B)eλx = 0. However, the factor eλx is nonvanishing for all x, so after its cancellation this equation is seen to be equivalent to the quadratic equation for λ λ2 + Aλ + B = 0.

(6)

When the quadratic equation (6) has two distinct (different) roots λ1 and λ2 , the functions y1 (x) = exp(λ1 x) and y2 (x) = exp(λ2 x) will be linearly independent for all x, because y1 (x)/y2 (x) = exp[(λ1 − λ2 )x] is not constant. Thus, then exp(λ1 x) and exp(λ2 x) are linearly independent solutions of (1), so the general solution is y(x) = c1 exp(λ1 x) + c2 exp(λ2 x),

(7)

where c1 and c2 are arbitrary constants. It is now necessary to introduce the type of initial conditions that are appropriate for (1). As (1) is a second order differential equation, it relates y(x), y (x), and y (x), so it follows that suitable initial conditions will be the speciﬁcation of y(x) and y (x) at some point x = a. Then the value of y (a) cannot be assigned arbitrarily, because the differential equation itself will determine its value in terms of y(a) and y (a). The solution of (1) satisfying these initial conditions can be found from the general solution (7) by determining c1 and c2 from the two equations: Initial condition on y(x) initial conditions

y(a) = c1 exp(λ1 a) + c2 exp(λ2 a), Initial condition on y(x) y (a) = λ1 c1 exp(λ1 a) + λ2 c2 exp(λ2 a).

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

(8)

When we considered systems of linear algebraic equations in Chapter 3, it was shown that equations (8) will determine c1 and c2 uniquely if the determinant of the coefﬁcients of c1 and c2 is nonvanishing. Thus, the speciﬁcation of y(a) and y (a) will be appropriate as initial conditions if exp(λ1 a) = λ1 exp(λ1 a)

exp(λ2 a) = 0. λ2 exp(λ2 a)

(9)

274

Chapter 6

Second and Higher Order Linear Differential Equations and Systems

Expanding the determinant gives = (λ2 − λ1 ) exp[(λ1 + λ2 )a]. However, by hypothesis λ1 = λ2 , while exp[(λ1 + λ2 )a] never vanishes, so = 0. The particular solution satisfying the initial conditions follows by using the values of c1 and c2 found from (8) in the general solution (7). EXAMPLE 6.3

Find the solution of the initial value problem y + 4y = 0,

if

y(π/4) = 1

and

y (π/4) = 1.

Solution In Example 6.2 direct substitution has already been used to show that cos 2x and sin 2x are linearly independent solutions of the differential equation, so its general solution is y(x) = c1 cos 2x + c2 sin 2x, from which it follows by differentiation that y (x) = −2c1 sin 2x + 2c2 cos 2x. Imposing the initial condition on y(x) at x = π/4 leads to the following equation that must be satisﬁed by c1 and c2 : 1 = c1 cos π/2 + c2 sin π/2. Similarly, imposing the initial condition on y (x) at x = π/4 leads to the second condition that must be satisﬁed by c1 and c2 : 1 = −2c1 sin π/2 + 2c2 cos π/2. These equations have the solution c1 = −1/2 and c2 = 1, so the particular solution satisfying the initial conditions y(π/4) = 1 and y (π/4) = 1 is y(x) = sin 2x −

1 cos 2x. 2

The quadratic equation determining the permissible values of λ in the exponential solutions y1 (x) = exp(λ1 x) and y2 (x) = exp(λ2 x) of differential equation (1), namely, λ2 + Aλ + B = 0, characteristic equation

(10)

is called the characteristic equation of the differential equation. Its two roots, √ √ −A+ A2 − 4B −A− A2 − 4B and λ2 = , (11) λ1 = 2 2 are the values of λ to be used in the general solution (7). When the roots λ1 and λ2 are real and distinct, the functions y1 (x) = exp(λ1 x)

and

y2 (x) = exp(λ2 x)

(12)

are said to form a basis for the solution space of (1). This means that the solution of every initial value problem for (1) can be obtained from the linear combination y(x) = c1 exp(λ1 x) + c2 exp(λ2 x) by assigning suitable values to c1 and c2 . A comparison of differential equation (1) and its characteristic equation (10) shows the characteristic equation can be written down immediately from the differential equation by simply replacing y by 1, dy/dx by λ and d2 y/dx 2 by λ2 . It is

Section 6.1

hom*ogeneous Linear Constant Coefﬁcient Second Order Equations

275

usual to use this method when obtaining the characteristic equation, as it avoids the unnecessary intermediate steps involved when substituting y(x) = exp(λx). Three different cases must now be considered, according to whether (i) λ1 and λ2 are real and distinct (λ1 = λ2 ), (ii) λ1 and λ2 are complex conjugates, or (iii) the possibility, excluded so far, that λ1 and λ2 are real and equal, so λ1 = λ2 = μ, say.

Case (I) (Real and Distinct Roots) how a solution depends on the roots

This case corresponds to the condition A2 − 4B > 0, with λ1 =

−A+

√

A2 − 4B 2

and

λ2 =

−A−

√

A2 − 4B . 2

(13)

No more need be said about this case because it has already been established that the functions exp(λ1 x) and exp(λ2 x) form a basis for the solution space of (1), which thus has the general solution y(x) = c1 exp(λ1 x) + c2 exp(λ2 x).

Case (II) (Complex Conjugate Roots) This case corresponds to the condition A2 − 4B < 0. A real solution y(x) corresponding to complex conjugate roots λ1 and λ2 is only possible if the arbitrary constants c1 and c2 are themselves complex conjugates. A routine calculation shows that if λ1 = α + iβ and λ2 = α − iβ, with α = −(1/2)A,

β = (1/2)(4B − A2 )1/2 ,

(14)

the two corresponding linearly independent solutions are y1 (x) = eαx cos βx

e

αx

and

y2 (x) = eαx sin βx.

(15)

A basis for the solution space of (1) is formed by the functions eαx cos βx and sin βx, corresponding to a general solution of the form y1 (x) = eαx [c1 cos βx + c2 sin βx].

(16)

The calculation required to establish the form of this result is left as an exercise.

Case (III) (Equal Real Roots) This case corresponds to the condition A2 − 4B = 0, with μ = λ1 = λ2 = −(1/2)A.

(17)

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Chapter 6

Second and Higher Order Linear Differential Equations and Systems

In this case only the one exponential solution y1 (x) = eμx

(18)

can be found. However, substitution of the function y2 (x) = xeμx

(19)

into the differential equation shows that it is also a solution. The functions y1 (x) and y2 (x) are linearly independent because y2 (x)/y1 (x) = x is not a constant, so in this case a basis for the solution space of (1) is formed by the functions eμx and xeμx , with the corresponding general solution y(x) = (c1 + c2 x)eμx .

(20)

Summary of the forms of solution of y + Ay + By = 0 summary of types of solution

Characteristic equation: λ2 + Aλ + B = 0 Case (I)

A2 − 4B > 0. The general solution is

y(x) = c1 exp(λ1 x) + c2 exp(λ2 x), √ −A+ A2 − 4B λ1 = and 2

with λ2 =

−A−

√

A2 − 4B . 2

Case (II) A2 − 4B < 0. The general solution is y1 (x) = eαx [c1 cos βx + c2 sin βx], α = −(1/2)A

and

with

β = (1/2)(4B − A2 )1/2 .

Case (III) A2 = 4B. The general solution is y(x) = (c1 + c2 x)eμx ,

EXAMPLE 6.4

with

μ = −(1/2)A.

Find the general solution and hence solve the stated initial value problem for (i) y + y − 2y = 0, with y(0) = 1 and y (0) = 2; (ii) y + 2y + 4y = 0, with y(0) = 2 and y (0) = 1; (iii) y + 4y + 4y = 0, with y(0) = 3 and y (0) = 1. Solution (i) The characteristic equation is λ2 + λ − 2 = 0, with the roots λ1 = 1, λ2 = −2, so this is Case (I). The general solution is y(x) = c1 e x + c2 e−2x .

Section 6.1

hom*ogeneous Linear Constant Coefﬁcient Second Order Equations

277

The initial condition y(0) = 1 is satisﬁed if 1 = c1 + c 2 ,

while the initial condition y (0) = 2 is satisﬁed if 2 = c1 − 2c2 . These equations have the solution c1 = 4/3 and c2 = −1/3, so the solution of the initial value problem is y(x) = (4/3)e x − (1/3)e−2x . (ii) The characteristic equation is λ2 + 2λ + 4 = 0, with A2 − 4B = −12, so this is Case (II) with α = −1 and β = solution is √ √ y(x) = e−x [c1 cos(x 3) + c2 sin(x 3)].

√

3. The general

The initial condition y(0) = 2 is satisﬁed if 2 = c1 , while the initial condition y (0) = 1 is satisﬁed if √ 1 = −2 + c2 3. √ Solving these equations gives c1 = 2 and c2 = 3, so the solution of the initial value problem is √ √ √ y(x) = e−x [ 3 sin(x 3) + 2 cos(x 3)]. (iii) The characteristic equation is λ2 + 4λ + 4 = 0, with A2 − 4B = 0, so this is Case (III) with μ = −2. The general solution is y(x) = (c1 + c2 x)e−2x . Using the initial condition y(0) = 3 shows that 3 = c1 , whereas the initial condition y (0) = 1 will be satisﬁed if 1 = −6 + c2 . Solving these equations gives c1 = 3 and c2 = 7, so the solution of the initial value problem is y(x) = (3 + 7x)e−2x . We now formulate the fundamental existence and uniqueness theorem for the hom*ogeneous linear second order constant coefﬁcient differential equation (1). This is a special case of a more general theorem that will be quoted later. THEOREM 6.1 existence and uniqueness of solutions

Existence and uniqueness of solutions of hom*ogeneous second order constant coefﬁcient equations Let differential equation (1) have two linearly independent solutions y1 (x) and y2 (x). Then, for any x = x0 and numbers μ1 and μ2 , a unique solution of (1) exists satisfying the initial conditions y(x0 ) = μ0 ,

y(1) (x0 ) = μ1 .

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Second and Higher Order Linear Differential Equations and Systems

Proof The existence of the solutions y1 (x) and y2 (x) was established when the cases (I), (II), and (III) were examined. The nonvanishing of the determinant in (9) showed c1 and c2 to be uniquely determined by the given initial conditions when the roots are real and distinct, so the solution of the initial value problem is also unique. An examination of the form of the determinant in cases (II) and (III) establishes the uniqueness of the solution in the remaining two cases, though the details are left as an exercise. two-point boundary conditions

A different type of problem that can arise with second order equations occurs when the solution is required to satisfy a condition at two distinct points x = a and x = b, instead of satisfying two initial conditions. Problems of this type are called two-point boundary value problems, because the points a and b can be regarded as boundaries between which the solution is required, and at which it must satisfy given boundary conditions. Problems of this type occur in the study of the bending of beams that are supported in different ways at each end, and elsewhere (see Section 8.10). Typical two-point boundary value problems involve either the speciﬁcation of y(x) at x = a and at x = b, or the speciﬁcation of y(x) at one boundary and y (x) at the other one. The most general two point boundary value problem involves ﬁnding a solution in the interval a < x < b such that y + Ay + By = 0, subject to the boundary condition at x = a αy(a) + βy (a) = μ, and the boundary condition at x = b γ y(b) + δy (b) = K, where α, β, γ , δ, μ, and K are known constants.

EXAMPLE 6.5

Solve the two-point boundary value problem y + 2y + 17y = 0,

with y(0) = 1 and y (π/4) = 0.

Solution The characteristic equation is λ2 + 2λ + 17 = 0 with the complex roots λ1 = −1 + 4i and λ2 = −1 − 4i, so the general solution is y(x) = e−x [c1 cos 4x + c2 sin 4x]. At the boundary x = 0 the general solution reduces to 1 = c1 , whereas at the boundary x = π/4 it reduces to 0 = −e−π/4 + 4c2 e−π/4 , showing that c2 = 1/4. So the solution of the two-point boundary value problem is 1 y(x) = e−x cos 4x + sin 4x , for 0 < x < π/4. 4

Summary

This section introduced the hom*ogeneous linear second order constant coefﬁcient equation and explained the importance of the linear independence of solutions. It showed how

Section 6.1

hom*ogeneous Linear Constant Coefﬁcient Second Order Equations

279

for this second order equation the general solution can be expressed as a linear combination of the two linearly independent solutions that can always be found. The form of the two linearly independent solutions was shown to depend on the relationship between the roots of the characteristic equation. A fundamental existence and uniqueness theorem was given and the nature of a simple two-point boundary value problem was explained.

EXERCISES 6.1 In Exercises 1 through 4 test the given pairs of functions for linear independence or dependence over the stated intervals. 1. (a) (b) (c) 2. (a) (b) (c) 3. (a) (b) (c) 4. (a) (b) (c)

sinh2 x, cosh2 x, for all x. x + ln |x|, x + 2 ln |x|, for |x| > 0. 1 + x, x + x 2 , for all x. sin x, cos x, for all x. sin x cos x, sin 2x, for all x. e2x , xe2x , for all x. |x|x 2 , x 3 , for −1 < x < 1. sin x, tan x, for −π/4 ≤ x ≤ π/4. x|x|, x 2 , for x ≥ 0. sin x, | sin x|, for π ≤ x ≤ 2π. x 3 − 2x + 4, −4x 3 + 8x − 16, for all x. x + 2|x|, x − 2|x| for all x.

Find the general solution of the differential equations in Exercises 5 through 20. 5. y + 3y − 4y = 0.

6. y + 2y + y = 0.

7. y − 2y + 2y = 0.

8. y + 2y + 2y = 0.

9. y + 2y − 3y = 0.

10. y + 5y + 4y = 0.

11. y + 6y + 9y = 0.

12. y − 2y + 4y = 0.

13. y − 4y + 5y = 0.

14. y + 3y + 3y = 0.

15. y + 6y + 25y = 0.

16. y − 4y + 20y = 0.

17. y + 5y + 4y = 0.

18. y + 4y + 5y = 0.

19. y − 3y + 3y = 0.

20. y + y + y = 0.

Solve initial value problems in Exercises 21 through 28 using the method of this section, and conﬁrm the solutions for even numbered problems by using computer algebra. 21. 22. 23. 24. 25. 26. 27. 28.

y + 5y + 6y = 0, with y + 4y + 5y = 0, with y + 2y + 2y = 0, with y + 6y + 8y = 0, with y − 5y + 6y = 0, with y − 3y + 3y = 0, with y − 3y − 4y = 0, with y − 2y + 3y = 0, with

y(0) = 1, y (0) = 2. y(0) = 1, y (0) = 3. y(0) = 3, y (0) = 1. y(0) = 1, y (0) = 0. y(0) = 2, y (0) = 1. y(0) = 0, y (0) = 2. y(0) = −1, y (0) = 2. y(0) = 1, y (0) = 0.

Solve the boundary value problems in Exercises 29 through 36 using the method of this section, and conﬁrm the solutions for even-numbered problems by using computer algebra. 29. 30. 31. 32. 33. 34. 35. 36.

y + 4y + 3y = 0, with y(0) = 1, y (1) = 0. y + 4y + 4y = 0, with y(0) = 2, y (1) = 0. y + 6y + 9y = 0, with y(−1) = 1, y (1) = 0. y + 4y + 5y = 0, with y(−π/2) = 1, y (π/2) = 0. y + 2y + 26y = 0, with y(0) = 1, y (π/4) = 0. y + 2y + 26y = 0, with y(0) = 0, y (π/4) = 2. y + 5y + 6y = 0, with y(0) = 0, y (1) = 1. y + 2y − 3y = 0, with y(0) = 1, y (1) = 1.

Theorem 6.1 ensures the existence and uniqueness of solutions of initial value problems for the differential equation in (1), but does not apply to two-point boundary value problems that may have no solution, a unique solution or inﬁnitely many solutions. In Exercises 37 and 38 use the general solution of y + y = 0 to ﬁnd if a solution exists and is unique, exists but is nonunique, or does not exist for each set of boundary conditions. √ 37. (a) y(0) = 0, y(π) = 0. (c) y (0) = 1, y(π/4) = 2. (b) y(0) = 1, y(2π ) = 2. 38. (a) y(0) = 1, y(π/2) = 1. (c) y (0) = 0, y (π) = 0. (b) y(0) = 0, y (π ) = 0. 39. For what values of λ will the following two-point boundary value problem have inﬁnitely many solutions, and what is the form of these solutions: y + λ2 y = 0,

with y(0) = 0, y(π) = 0.

40. A particle moves in a straight line in such a way that its distance x from the origin at time t obeys the differential equation x + x + x = 0. Assuming it starts from the origin with speed 30 ft/sec, what will be its distance√from the origin, its speed, and its acceleration after π/ 3 seconds? 41. The angular displacement θ of a damped simple pendulum obeys the equation θ + 2μθ + (μ2 + p2 )θ = 0,

280

Chapter 6

Second and Higher Order Linear Differential Equations and Systems

with 4 p2 > μ2 . Find the angular displacement θ (t), and the time t and angular displacement when it ﬁrst comes to rest, given that it starts with θ = 0 and dθ/dt = α. 42. The top of a vibration damper oscillates in a straight line in such a way that its position x from the origin at time t obeys the differential equation x + 2x + 4x = 0. Given that it starts from the origin with speed U, ﬁnd its position as a function of U and t and the distance from the origin when it ﬁrst comes to rest.

6.2

43. The free oscillations of all physical systems giving rise to oscillatory solutions obey an equation of the form x + 2μx + (μ2 + p2 )x = 0 with p2 > 0. Given that x(0) = 0, and (dx/dt)t=0 = Ap, solve the equation and show that x(t) = Aexp(−μt) sin pt. Use this result to prove that the ratio of the magnitude of successive extrema of x(t) forms a geometric series with common ratio r = exp[−μπ/( p)]. The number μπ/( p) is called the logarithmic decrement of the oscillations.

Oscillatory Solutions The nonhom*ogeneous constant coefﬁcient second order equation a0

forcing function and damping

d2 y dy + a2 y = f (t), + a1 2 dt dt

(21)

in which t can be regarded as the time and f (t) as an external input to the system, is the simplest mathematical model capable of representing the oscillatory behavior of a physical system. It was shown in Section 5.2(d) that one way this equation can arise is when describing the motion of a mass–spring system in which a mass moves on a rough horizontal surface, with the motion resisted by a frictional force proportional to the speed. Friction dissipates energy, so the motion will decay to zero as time increases unless it is sustained by some external input of energy in the form of a forcing function represented in (21) by the nonhom*ogeneous term f (t). The dissipation of energy due to friction, or to a friction-like effect in other applications, is called damping, and in the R–L–C circuit considered in Section 5.2(d), where the charge q on the capacitor was shown to satisfy a hom*ogeneous form of equation (21) with a0 = LC, a1 = RC, and a3 = 1, the damping was due to the dissipative (friction-like) term a1 = RC. Another way in which equation (21) can arise is when a cylindrical mass with moment of inertia I about its axis of symmetry is mounted on a ﬂexible shaft that can be twisted about its axis, with the resistance to torsion (twisting) proportional to the angle of twist θ , and damping proportional to the angular velocity dθ/dt about the shaft. This occurs, for example, in a torsional pendulum and also in heavy rotating machinery when a heavy ﬂywheel is attached to a shaft. The equation governing the torsional oscillations θ(t) as a function of the time t becomes dθ d2 θ + μθ = f (t), +k dt 2 dt where k and μ are constants and, as before, f (t) is a forcing function. A comparison of the second order constant coefﬁcient differential equations that govern mechanical, electrical, and torsional oscillations leads to Table 6.1, which relates analogous physical quantities in each of the different systems. Many other physical situations can be represented by this same constant coefﬁcient second order differential equation with varying degrees of approximation. It does, for example, provide a simple model that describes the effect of a ﬂuctuating vertical lift at the center of a ﬂexible suspension bridge caused by gusts of wind. If I

Section 6.2

Oscillatory Solutions

281

TABLE 6.1 A Comparison of Second Order Constant Coefﬁcient Differential Equations

Mechanical System

Electrical System with Elements in Series

Torsional System

Mass Damping constant Spring constant Applied force

Inductance Resistance Reciprocal of capacitance Applied voltage

Moment of inertia Torsional damping constant Shaft torsional constant Applied torque

this effect is sustained, and the gusts come at the natural frequency of the bridge, the amplitude of the oscillations can become dangerously large. On November 7, 1940, in the state of Washington, this effect caused the failure of the Tacoma Narrows Bridge over Puget Sound. Powerful gusting winds at around the natural frequency of this excessively ﬂexible bridge induced and then sustained vertical oscillations of the bridge that reached an amplitude of 28 feet before the bridge snapped and fell. When analyzing the oscillatory nature of solutions of (21), and looking at the effect of resonance that occurs if the natural frequency of oscillation of the system coincides with the frequency of a periodic forcing function, it is helpful to have in mind a mathematical model of a simple but typical mechanical system. The mechanical system we will consider here is shown in Fig. 6.2, and it involves a piece of heavy machinery that vibrates vertically and is mounted on a spring and damper system to reduce the transmission of the vibrations to the foundations of the building. The damper is usually a piston that moves in a viscous ﬂuid, with the resisting force considered to be proportional to the speed of the piston.

F(t)

M

Vibrating platform

y(t)

Damper

FIGURE 6.2 A vibrating machine mounted on a spring and damper system.

282

Chapter 6

Second and Higher Order Linear Differential Equations and Systems

If the mass of the machine is M, the displacement of the machine from the ﬂoor at time t is y(t), the spring constant is k, the constant of proportionality for ˜ the damper is μ, and the force exerted by the vibrating machine is F(t), the rate of change of momentum d/dt{M(dy/dt)} must be equated to the frictional resistance ˜ −μdy/dt, the restoring force of the spring −ky, and the external force F(t). So this system, with the displacement y(t) as its one degree of freedom, is seen to satisfy the differential equation M

d2 y dy ˜ = −μ − ky + F(t). dt 2 dt

For convenience this will be written in the standard form d2 y dy + by = F(t), +a 2 dt dt

(22)

˜ with a = μ/M, b = k/M, and F(t) = F(t)/M. Differential equation (22) is nonhom*ogeneous, so its solution will be more complicated than the solution of the hom*ogeneous equation considered in the previous section. The equation is linear, so as in Section 5.6 we will represent its general solution as the sum y(t) = yc (t) + yp (t),

(23)

with yc (t) the general solution of the hom*ogeneous form of equation (22) d2 yc dyc + byc = 0, +a dt 2 dt

(24)

and yp (t) a particular solution of d2 yp dyp + byp = F(t) +a (25) 2 dt dt that contains no arbitrary constants. The justiﬁcation for writing the general solution of (22) as y(t) = yc (t) + yp (t) follows if we notice that (22) can be written d2 [yc + yp ] d[yc + yp ] + b[yc + yp ] = F(t) +a 2 dt dt or, equivalently, as 2 dyp d2 yp d yc dyc + by + byp = F(t), + a +a + c dt 2 dt dt 2 dt

complementary function and particular integral

where the group of terms in the bracket vanishes because of (24), while yp (t) satisﬁes the remainder of the equation because of (25). As in Section 5.6, the solution yc (t) will be called the complementary function, and the solution yp (t) will be called a particular integral. It is important to recognize that the two arbitrary constants associated with the general solution of (22) occur in the complementary function yc (t), whereas the particular integral yp (t) contains no arbitrary constants.

Section 6.2

Oscillatory Solutions

283

We now introduce the notation a = 2ζ and b = 2 , when the characteristic equation of (22) becomes λ2 + 2ζ λ + 2 = 0,

(26)

with the roots λ1 = −ζ + (ζ 2 − 2 )1/2

and

λ2 = −ζ − (ζ 2 − 2 )1/2 .

(27)

The solution yc (t) of (22) will correspond to one of the Cases (I), (II), or (III) of Section 6.1, but before determining its form in each of these cases we further simplify the notation by setting k2 = ζ 2 − 2 , so that λ1 = −ζ + k

and

λ2 = −ζ − k.

(28)

Case (I): k 2 > 0 (ζ 2 > Ω2 ) The complementary function yc (t) is nonoscillatory and given by yc (t) = exp(−ζ t){C1 exp(kt) + C2 exp(−kt)}.

(29)

Case (II): k 2 < 0 (ζ 2 < Ω2 ) If we set k2 = −ω02 the complementary function is seen to be oscillatory and given by yc (t) = exp(−ζ t){C1 cos ω0 t + C2 sin ω0 t}.

(30)

Case (III): k 2 = 0 (Ω2 = ζ 2 ) The complementary function is nonoscillatory and given by yc (t) = {C1 + C2 t} exp(−ζ t).

critical damping and overdamping

transient and steady state solutions

(31)

Cases (I) and (III) exhibit no oscillatory behavior. Case (I) is said to be overdamped and Case (III) to be critically damped, because it marks the boundary between the overdamped behavior of Case (I) and the oscillatory behavior of Case (II). The parameter ζ entering into the exponential factor exp(−ζ t) that is present in all three cases is called the damping exponent and, provided ζ > 0, the factor exp(−ζ t) will cause all three complementary functions to decay to zero as time increases. This property of a complementary function with ζ > 0 has led to its being called the transient solution of the differential equation. Accordingly, after a suitable lapse of time, only the particular integral yp (t) will remain, and this property is recognized by calling yp (t) the steady state solution, with the understanding that it is the time-dependent solution that remains after the transient solution has become vanishingly small. Typical transient solution behavior in the critically damped case is shown in Fig. 6.3 for different initial conditions, some of which can cause an initial increase in the amplitude of yc (t) before it decays to zero. The behavior in the overdamped case is similar to that in the critically damped case.

284

Chapter 6

Second and Higher Order Linear Differential Equations and Systems yc(t)/yc(0) 1.5

1

0.5

1

2

3

4 t

FIGURE 6.3 yc (t) in the critically damped case for different initial conditions.

It is now necessary to determine the form of the particular integral yp (t), and to do so the function F(t) must be speciﬁed. A vibration is periodic in nature, so we shall model it by a nonhom*ogeneous term of the form F(t) = Acos ωt,

amplitude and angular frequency of a vibration

(32)

in which the amplitude A will be considered to be ﬁxed and the angular frequency ω will be regarded as a parameter. The angular frequency ω is expressed in terms of radians/unit time and corresponds to a time period of oscillation of T = 2π/ω time units, while the frequency of the vibration 1/T = ω/2π measures the number of cycles (vibrations) occurring in one time unit. If the unit of time T is 1 sec, the frequency is measured in cycles/sec, called hertz (Hz), so 20 Hz is 20 cycles/sec. Setting F(t) = Acos ωt in (22) shows that the differential equation to be considered is d2 y dy + 2 y = Acos ωt. + 2ζ dt 2 dt

(33)

A systematic approach to the determination of particular integrals will be developed in the next section, but here we will proceed from ﬁrst principles. As equation (33) has constant coefﬁcients, and its nonhom*ogeneous term is Acos ωt, the way this nonhom*ogeneous term can be obtained by differentiating a particular integral yp (t) is if the particular integral is of the form yp (t) = Csin ωt + Dcos ωt,

(34)

unless ζ = 0 and = ω (then try yp = t(Csin ωt + Dcos ωt). Substituting (34) into (33) and collecting terms gives [(2 − ω2 )C − 2ζ ωD]sin ωt + [(2 − ω2 )D + 2ζ ωC]cos ωt = Acos ωt. This must be true for all t, but this will only be possible if the respective coefﬁcients of sin ωt and cos ωt on either side of the equation are identical, so (2 − ω2 )C − 2ζ ωD = 0

and

(2 − ω2 )D + 2ζ ωC = A.

Solving for C and D gives C=

(2

2Aωζ − ω2 )2 + 4ζ 2 ω2

and

D=

(2

A(2 − ω2 ) , − ω2 )2 + 4ζ 2 ω2

(35)

Section 6.2

Oscillatory Solutions

285

so the required particular integral is yp (t) =

2Aωζ A(2 − ω2 ) sin ωt + cos ωt. (2 − ω2 )2 + 4ζ 2 ω2 (2 − ω2 )2 + 4ζ 2 ω2

(36)

A better understanding of the nature of this particular integral can be obtained if it is rewritten. To accomplish this we return to (34) and write it as C D 2 2 1/2 yp (t) = (C + D ) sin ωt + 2 cos ωt , (37) (C 2 + D2 )1/2 (C + D2 )1/2 and then deﬁne an angle φ by the requirement that sin φ =

(C 2

C , + D2 )1/2

and

cos φ =

(C 2

D , + D2 )1/2

(38)

or by the equivalent expression tan φ = C/D =

2ζ ω . − ω2

(39)

2

The trigonometric identity cos(ωt − φ) = cos ωtcos φ + sin ωtsin φ then allows yp (t) to be expressed in the more convenient form yp (t) =

[(2

−

A cos (ωt − φ). + 4ζ 2 ω2 ]1/2

ω 2 )2

(40)

Using this form for yp (t) gives the simpler expression for the general solution y(t) = yc (t) +

phase angle and phase lag

[(2

−

ω2 )

A sin (ωt − φ), + 4ζ 2 ω2 ]1/2

(41)

where yc (t) is one of the Cases (I), (II), or (III), depending on the sign of ζ 2 − 2 . The angle φ, which by convention is required to lie in the interval 0 < φ < π, is called the phase angle of the solution, and often the phase lag, because it represents the delay with which the steady-state solution (the output from the system) lags behind the input to the system determined by F(t). We have seen that provided ζ > 0, the transient solution yc (t) decays to zero as t increases, leaving only the steady state solution yp (t). The steady state solution (41), illustrated in Fig. 6.4, is a sinusoid with the same angular frequency as the function F(t) = Asin (ωt) that forces the oscillations, but with an amplitude that depends on both Aand the angular forcing frequency ω. The effect of the phase lag φ is seen to shift the origin.

yp(t)/P(ω) yP(t) = P(ω)cos(ωt − φ)

1 0

t

φ/ω −1

T = 2π/ω

FIGURE 6.4 The steady state solution yp (t)/P(ω).

286

Chapter 6

Second and Higher Order Linear Differential Equations and Systems

instability, ampliﬁcation factor, and resonance

If ζ < 0, the general solution y(t) will increase without bound as time increases, and in physical problems this behavior is called instability. In effect, when ζ < 0, energy is fed into the system as time increases, instead of it being removed by friction. The amplitude of the steady state solution is P(ω) =

[(2

−

A , + 4ζ 2 ω2 ]1/2

(42)

ω 2 )2

and P(ω)/A = [(2 − ω2 ) + 4ζ 2 ω2 ]−1/2 is called the ampliﬁcation factor, because it is the ratio of the amplitude of the solution (response) to the amplitude of the forcing function (input). The ampliﬁcation factor attains its maximum value Pmax when ω = ωc , with ωc2 = 2 − 2ζ 2 , in which case Pmax =

2ς (2

A . − ζ 2 )1/2

(43)

The angular frequency ωc is called the resonant angular frequency of the system that is said to experience resonance at the the frequency ωc . It is to avoid exciting resonance that troops marching across a ﬂexible suspension bridge are told to break step. Conversely, it is for this same reason that when one pushes a swing, successive pushes need to be synchronized with each oscillation if the amplitude of the motion is to be built up. If ζ = 0, result (42) shows that resonance occurs when ω = , leading to an inﬁnite ampliﬁcation factor. The critical role played by damping in limiting the amplitude of oscillations can be seen from (43). Figures 6.5a and 6.5b show the variation of the scaled ampliﬁcation factor 2 P(ω)/Aand the phase angle φ as functions of ω/ for a range of values of ζ / . Care must always be exercised when ﬁnding the phase angle φ, because the phase is required to lie in the interval 0 < φ < π, though the usual domain of deﬁnition of the inverse tangent function is (−π/2, π/2). The most extreme effect of resonance occurs when there is no damping (ζ = 0), though this can never happen in physical problems because there are always some dissipative effects. In the absence of damping, the natural angular frequency of oscillations is , and equation (42) shows that when the vibrations are forced by a φ Ω 2P(ω)/A

ζ/Ω = 0.1

5 4

ζ/Ω = 0.2

1 0

ζ/Ω = 0.1 ζ/Ω = 0.3 ζ/Ω = 0.6 ζ/Ω = 1.1

π/2

3 2

ζ/Ω = 0

π

ζ/Ω = 0.5 ζ/Ω = 0.8 0.5

1

1.5

2

ω/Ω

0.5

(a) FIGURE 6.5 (a) Amplitude as a function of ω/ . (b) Phase angle as a function of ω/ .

1 (b)

1.5

2 ω/Ω

Section 6.2 ΩP(ω)/A 5

Oscillatory Solutions

287

φ π

4 3 π/2

2 1 0

0.5

2 ω/Ω

1.5

1

0.5

(a)

1

1.5

2 ω/Ω

(b)

FIGURE 6.6 (a) Variation of amplitude. (b) Variation of phase.

sinusoidal input of angular frequency ω, the amplitude of the steady state solution is P(ω) =

A . |2 − ω2 |1/2

This shows that P(ω) becomes inﬁnite when the exciting angular frequency ω equals the natural angular frequency . The variation of P(ω)/A as ω/ varies is shown in Fig. 6.6a, while the corresponding variation of the phase is shown in Fig. 6.6b for the limiting case ω → . To understand how the solution becomes unstable when ω = , it is necessary to consider the solution of d2 y + 2 y = Asin t, dt 2

with y(0) = 0, (dy/dt) = 0, t = 0.

We ﬁnd that y(t) =

beats

A (sin t − t cos t), 22

and the variation of y(t) is shown in Fig. 6.7, from which it can be seen that when the damping is zero, forcing at the resonant angular frequency causes the amplitude of the oscillations to grow linearly with time. An interesting and important property of oscillatory solutions under conditions that allow dissipation to be ignored is to be found in the occurrence of beats in the

y(t)

FIGURE 6.7 Linear growth of amplitude with time when ζ = 0.

t

288

Chapter 6

Second and Higher Order Linear Differential Equations and Systems

steady state solution. Consider a solution of the form y(t) =

|2

A [cos ωt − cos t]. − ω2 |1/2

Subtracting the trigonometric identities cos(C − D) = cos Ccos D + sin Csin Dand cos(C + D) = cos Ccos D − sin Csin D, and then setting C = ( + ω)/2 and D = ( − ω)/2, gives ( + ω)t ( − ω)t cos ωt − cos t = 2sin sin , 2 2 so the solution becomes y(t) =

( + ω)t ( − ω)t 2A sin sin . |2 − ω2 |1/2 2 2

This result can be written ( + ω)t , y(t) = E(t) sin 2

with E(t) =

2A ( − ω)t , sin |2 − ω2 |1/2 2

showing that when ω is close to , the solution is in the form of a component with the “high angular frequency” ( + ω)/2, modulated by an amplitude 2A ( − ω)t E(t) = 2 ; sin | − ω2 |1/2 2 with the “low angular frequency” ( − ω)/2. This solution is seen to be in the form of “pulses” at the higher angular frequency ( + ω)/2 modulated by the lower angular frequency ( − ω)/2. A typical physical example of beats can be experienced when listening to two sound waves with similar frequencies 1 and 2 that interact. Then, provided the amplitudes are similar, the sound at the higher frequency is heard as pulses that arrive at the lower frequency. Figure 6.8 shows a typical situation where beats occur, and when listening to such interacting sound waves the high frequency would be heard as a slow throbbing sound. EXAMPLE 6.6

Solve the initial value problem 4

dy d2 y + 37y = 12cos t, +4 dt 2 dt

with y(0) = 1, y (0) = −2.

1/

(Ω2 − ω2) 2y(t)/2A

FIGURE 6.8 The phenomenon of beats produced when frequencies ω and are close.

t

Section 6.2

Oscillatory Solutions

289

Solution The characteristic equation is 4λ2 + 4λ + 37 = 0, an example showing the makeup of a typical solution

with the roots λ1 = −(1/2) + 3i and λ2 = −(1/2) − 3i, so the complementary function is yc (t) = exp[−t/2](C1 sin 3t + C2 cos 3t). When written in the standard form the differential equation becomes d2 y dy 37 + y = 3cos t. + dt 2 dt 4 Comparison with (33) shows that ζ = 1/2, 2 = 37/4, A = 3, and ω = 1, so ω0 = (2 − ζ 2 )1/2 = 3. Substituting these results into equations (35) gives C = 48/1105 and D = 396/1105, so the general solution is y(t) = exp[−t/2](C1 sin 3t + C2 cos 3t) +

396 48 sin t + cos t. 1105 1105

Imposing the initial condition y(0) = 1 on y(t) gives 1 = C2 + (396/1105),

so C2 = 709/1105.

Similarly, imposing the initial condition y (0) = −2 on y(t) gives −2 = 48/1105 − (1/2)C2 + 3C1 ,

so C1 = −1269/2210.

Finally, substituting the values of C1 and C2 into the general solution shows that the solution of the initial value problem is y(t) =

1 1 exp(−t/2)(1418cos 3t − 1269sin 3t) + (48sin t + 396cos t). 2210 1105

The steady state solution is yp (t) =

12 1 (48sin t + 396cos t) = √ cos (t − φ), 1105 1105

where the phase lag φ = arctan C/D = arctan(48/396) = 0.1206 radians, and the transient solution is yc (t) =

1 exp(−t/2)(1418cos 3t − 1269sin 3t). 2210

On the following page, the transient solution yp (t) is shown in Fig. 6.9a, the steady state solution yc (t) in Fig. 6.9b, and the complete solution y(t) of the initial value problem in Fig. 6.9c.

Summary

This section showed that the solution of a nonhom*ogeneous constant coefﬁcient equation where the independent variable is the time comprises two parts: one called the transient solution, which describes the startup of the solution, and another called the steady state solution, which describes the nature of the time-dependent solution that remains when the transient solution has decayed sufﬁciently to become negligible. The important case involving a sinusoidal forcing function was examined in detail, and the terms amplitude, frequency, and phase angle of the solution were explained, together with the important effect of resonance.

290

Chapter 6

Second and Higher Order Linear Differential Equations and Systems

yc(t)

yp(t)

0.5

0.5

2

4

6

8

10

t

2

0.5

4

8

6

10

12

14

t

0.5 (a)

(b) y(t) 1 0.5 0

5

10

15

20

t

0.5

(c) FIGURE 6.9 (a) The transient solution. (b) The steady state solution. (c) The complete solution.

EXERCISES 6.2 In Exercises 1 through 7 solve the initial value problem using the methods of this section, and identify the steady state and transient solutions. Conﬁrm the results for the even numbered problems by computer algebra and plot their solutions for some interval 0 < t < T.

1. y + 2y + 5y = 2 sin x, with y(0) = 1, y (0) = 0. 2. y + 2y + 5y = 3 sin x, with y(0) = 0, y (0) = 0. 3. y + 2y + y = sin x, with y(0) = 1, y (0) = 0. 4. y + 2y + y = sin 2x, with y(0) = 2, y (0) = 0. 5. y + 3y + 2y = sin 3x, with y(0) = 0, y (0) = 1. 6. y + 2y + 5y = sin x, with y(0) = 0, y (0) = 1. 7. y + 5y + 6y = Asin x, with y(0) = 3, y (0) = 1. 8. Use the argument in Section 6.2 when establishing the results in (35) to show that if the forcing function on the right of (33) is replaced by A sin ωt, and the particular integral is written yp (t) = Csin ωt + Dcos ωt, the constants C and D are given by C=

(2

A(2 − ω2 ) − ω2 )2 + 4ζ 2 ω2

and

D=

2ζ ω A , (2 − ω2 )2 + 4ζ 2 ω2

and that the phase angle φ is such that tan φ = −

2ζ ω . 2 − ω2

In Exercises 9 through 14 use the results of Exercise 8 when solving the initial value problem. Find the phase angle and identify the steady state and transient solutions. 9. 10. 11. 12. 13. 14. 15.

y + 5y + 6y = 2 cos x, with y(0) = 1, y (0) = 1. y + 7y + 6y = 2 cos 3x, with y(0) = 2, y (0) = 1. y + 6y + 9y = 2 cos 3x, with y(0) = 2, y (0) = 2. y + 2y + 2y = cos 4x, with y(0) = 0, y (0) = 2. y + 6y + 8y = 3 cos 2x, with y(0) = 4, y (0) = 1. y + 2y + 5y = 3 cos 3x, with y(0) = 2, y (0) = 3. The fall of a loaded parachute is determined by the differential equation m

d2 y dy + mg = 0, + kg dt 2 dt

where m is the weight of the payload in pounds, k is the drag coefﬁcient of the parachute, y(t) is its height above the ground at time t in seconds, and g is the acceleration due to gravity. Taking g = 32 ft/sec2 , k = 10 lb/ft/sec, and the initial speed of fall at time t = 0 when the parachute opens 2000 ft above the ground

Section 6.3

hom*ogeneous Linear Higher Order Constant Coefﬁcient Equations

to be dy/dt = −32 ft/sec (remember y(t) is measured upward but the speed is downward), ﬁnd y(t) and the speed of fall at time t as functions of m. Use the result to ﬁnd the largest payload M in pounds if the speed of fall on landing is not to exceed 24 ft/sec. Plot y(t) for m = M and estimate the time of descent in this case. 16. Stokes’ law F = 6πaηu determines the drag F on a sphere of radius a moving slowly through a ﬂuid with viscosity η at a speed u. Let the density of the sphere be ρ1 and the ﬂuid density be ρ2 (ρ1 > ρ2 ). Find the equation of motion of the sphere in terms of the distance x(t) from its point of release, if it falls from rest in the ﬂuid at time t = 0. Solve the equation of motion to ﬁnd x(t), and hence the speed of fall, as functions of time. Suggest how this result could be used to determine the viscosity of oil in an experiment involving the release from rest of a ball bearing that is allowed to fall vertically through oil contained in a long glass cylinder. 17. A spherical container of radius a and density ρ1 is released from rest on the sea bed at a depth h below the surface and allowed to ﬂoat slowly upward in still water of density ρ2 , where ρ2 − ρ1 is small and ρ2 > ρ1 . Assuming that Stokes’ law in Exercise 16 applies, and the

6.3

291

viscosity of the water is η, ﬁnd the distance x(t) of the container from the sea bed as a function of time, and use it to write down the equation determining the time T when the container reaches the surface. Estimate this time, and suggest how a more accurate value of T could be obtained. 18. As ω → 1, from either above or below, so the solution x(t) of x + x = sin ωt subject to the initial conditions x(0) = x (0) = 0 tends to the divergent resonance solution illustrated in Fig. 6.7. Use computer algebra to plot the solution for ω = 0.85, 0.95, 0.99, 1.0, 1.05, and 1.1 to illustrate how the amplitude of the oscillations tends to a linear growth as ω → 1. Show that for ω = 1, x = 12 (sin t − t cos t). 19. Typically, beats occur when two slowly varying oscillations with equal amplitudes and almost equal frequencies are superimposed. Use computer algebra to plot x(t) = cos ω1 t + cos ω2 t, with suitable values of ω1 and ω2 and a sufﬁciently long time interval 0 ≤ t ≤ T, to show a clear pattern of the beats. Find the equation determining the high-frequency oscillation and the equations forming the envelope of the high-frequency component.

hom*ogeneous Linear Higher Order Constant Coefﬁcient Equations A Typical Example Leading to a Fourth Order System Linear nth order constant coefﬁcient differential equations often arise as a result of the elimination of all but one of the unknowns in a system of simultaneous lower order differential equations. To see how this can happen, consider the longitudinal motion of three equal particles of mass m, coupled together by four identical springs each of unstrained length l with spring constant k, with the left and right ends of the system clamped, as illustrated in Fig. 6.10. Now suppose that the system oscillates in the direction of the springs, with y1 , y2 , and y3 the displacements of the masses from their equilibrium positions, as shown in Fig. 6.10. Equating the rate of change of momentum d/dt{m(dy1 /dt)} of the mass with coordinate y1 to the sum of the restoring force k(y2 − y1 ) due to the second spring and the force k(y3 − y1 ) due to the third spring shows that the equation of motion of the ﬁrst mass is d2 y1 = k(l − y1 ) + k(y2 − y1 − l) = k(y2 − 2y1 ). dt 2 Similar arguments applied to the second and third masses in this system with three degrees of freedom (the coordinates y1 , y2 , and y3 ) gives the other two coupled equations of motion, m

m

d2 y2 = k(l + y1 − y2 ) + k(y3 − y2 − l) = k(y1 + y3 − 2y2 ) dt 2

292

Chapter 6

Second and Higher Order Linear Differential Equations and Systems

k

k m l

k m

l

k m

l

l

Equilibrium position

m

m

m

y1(t) y2(t) y3(t) Disturbed state FIGURE 6.10 A three-mass–spring system with its ends clamped.

and m

d2 y3 = k(l + y2 − y3 ) + k(4l − y3 − l) = k(4l + y2 − 2y3 ). dt 2

Eliminating any two of the three unknowns y1 , y2 , and y3 from these three equations of motion leads to a hom*ogeneous sixth order constant coefﬁcient differential equation for the remaining unknown. Initial conditions for the system are the values yi (0) and yi (0) for i = 1, 2, and 3. More complicated systems of this type are used to study one-dimensional waves in various types of periodic structure ranging from chains of low-pass electrical ﬁlters to the vibration of molecules in crystal lattices. A different example that gives rise to a fourth order differential equation is the modeling of a two degree of freedom vibration damper for a motor generator of mass M. Unless damped, the vertical vibrations due to the periodic motion of the pistons are passed to the foundations of the building and can cause unacceptable vibrations throughout the building. One way of dealing with this problem is not only to mount the motor generator on a spring and damper system, but also to spring mount a smaller mass m on top of the motor generator, as in Fig. 6.11, and to adjust the two spring constants and the mass m so that the vertical oscillations of M are minimized and passed instead to the smaller mass m mounted on the motor generator. Let the mass M be connected to the foundation by a spring with spring constant K, and let the spring constant of the spring supporting mass m be k. To make the model more realistic, suppose that in addition there is a viscous damper ﬁtted between the mass M and the foundation that exerts a resistance proportional to the speed of its displacement with constant of proportionality μ, and let the displacements of the masses M and m from their equilibrium positions be x and y, respectively. Suppose also that the vibrational force acting on M due to the operation of the motor generator is F(t).

Section 6.3

hom*ogeneous Linear Higher Order Constant Coefﬁcient Equations

293

F(t)

m

y(t) m

k

M k

x(t)

M

K K

Equilibrium position

Disturbed state

FIGURE 6.11 A two degree of freedom vibration system with a viscous damper.

The equation of motion of the mass M obtained by equating its rate of change of momentum to the combined restoring forces of the two springs, the viscous damper, and the vibrational force F(t) is dx d2 x + F(t), = −k(x − y) − Kx − μ 2 dt dt and the equation of motion of the mass m obtained by equating its rate of change of momentum to the restoring force exerted by the top spring is M

d2 y = −k(y − x). dt 2 Eliminating y between these two equations gives the fourth order constant coefﬁcient equation for x d4 x 1 d3 x d2 x dx d2 F + γ δx = + α 3 + (β + γ + γ δ/β) 2 + αβ γ F(t) + 2 , dt 4 dt dt dt M dt m

where α = μ/M, β = k/m, γ = k/M, and δ = K/m. Similarly, eliminating x between the two equations gives d4 y γ d3 y d2 y dy + γ δy = F(t). + α 3 + (β + γ + γ δ/β) 2 + αβ 4 dt dt dt dt M When F(t) is a periodic force with frequency ω, and the constants k, K, and m are adjusted to take account of resonance in the spring and damper mounting, the system can be tuned so that the displacement x(t) is reduced almost to zero, and the vibration is transferred instead to the mass m mounted on top of the motor generator.

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General hom*ogeneous Higher Order Constant Coefﬁcient Equations The hom*ogeneous linear constant coefﬁcient nth order equation dn y dn−1 y dy + an y = 0 + a + · · · + an−1 1 n n−1 dx dx dx

(44)

has properties that are similar to those of second order equations. If y1 (x), y2 (x), . . . , yr (x) are any r solutions of (44), the linearity of the equation means that the linear combination of functions y(x) = c1 y1 (x) + c2 y2 (x) + · · · + cr yr (x),

linear superposition in higher order systems

with c1 , c2 , . . . , cr arbitrary constants, is also a solution. This linear superposition property of solutions of the hom*ogeneous equation is an extension of the same property encountered in Section 6.1 when considering hom*ogeneous constant coefﬁcient second order equations. The proof of this property follows by substituting y(x) into the left-hand side of (44), using the linearity of the differentiation operation ds ds y1 ds y2 ds yr (c y + c y + · · · + c y ) = c + c + · · · + c , 1 1 2 2 r r 1 2 r dx s dx s dx s dx s for s = 0, 1, . . . , n, where d0 y/dx 0 ≡ y and grouping terms, to obtain r expressions of the form n d yi dn−1 yi dyi + a ci + a + · · · + a y 1 n−1 n i . dx n dx n−1 dx Each of these expressions vanishes, because the yi (x) are solutions of the hom*ogeneous equation, so the result of substituting y(x) into the left side of (44) is to reduce it to zero, showing that y(x) = c1 y1 (x) + c2 y2 (x) + · · · + cr yr (x)

basis, solution space, and general solutions

is a solution. It will be shown later that the hom*ogeneous equation (44) has n linearly independent solutions y1 (x), . . . , yn (x), and that these form a basis for its solution space. This means that every particular solution of (44) can be written as y(x) = c1 y1 (x) + c2 y2 (x) + · · · + cn yn (x),

(45)

for some choice of constants c1 , c2 , . . . , cn . It is because of this property that (45) is called the general solution of (44). A more general test for linear independence than the one in Section 6.1 is needed to ensure that the n solutions y1 (x), y2 (x), . . . , yn (x) of (44) form a basis for the solution space. To obtain this test we must ﬁrst extend the earlier deﬁnition of linear independence in a natural way to a set of functions g1 (x), g2 (x), . . . , gn (x) deﬁned over an interval a ≤ x ≤ b. The set of functions will be said to be linearly

Section 6.3

hom*ogeneous Linear Higher Order Constant Coefﬁcient Equations

295

independent over the interval if for all x in the interval, k1 g1 (x) + k2 g2 (x) + · · · + kn gn (x) = 0 linear independence and dependence

(46)

is only true if k1 = k2 = · · · = kn = 0; otherwise, the set of functions will be said to be linearly dependent. As the test will be needed later for solutions of linear differential equations more general than (44), it will be derived for the variable coefﬁcient differential equation

a0 (x)

dn y dn−1 y dy + an (x)y = 0, + a (x) + · · · + an−1 (x) 1 n n−1 dx dx dx

(47)

where the coefﬁcients ai (x) are continuous functions of x for a ≤ x ≤ b. The test will also apply to solutions of (44), because a constant is a special case of a continuous function. The derivation starts from the fact that if the functions y1 (x), y2 (x), . . . , yn (x) are solutions of the nth order equation (47) with continuous coefﬁcients over an interval a ≤ x ≤ b, then they will be everywhere continuous and differentiable at least n − 1 times over this same interval. By deﬁnition, the functions will be linearly independent over the interval a ≤ x ≤ b if the equation c1 y1 (x) + c2 y2 (x) + · · · + cn (x)yn (x) = 0

(48)

is only true if c1 = c2 = · · · = cn = 0 for all x in the interval. Differentiating the equation n − 1 times gives c1 y1 (x) + c2 y2 (x) + . . . + cn (x)yn (x) = 0 (1)

(1)

(1)

c1 y1 (x) + c2 y2 (x) + · · · + cn yn (x) = 0 · · · · · · · · · · · · · · · · · · · · · (n−1) (n−1) (n−1) (x) + c2 y2 (x) + · · · + cn yn (x) = 0. c1 y1

Wronskian determinant

(49)

This hom*ogeneous system of equations can only have the null solution c1 = c2 = · · · = cn = 0 that is necessary to ensure the linear independence of the functions y1 (x), y2 (x), . . . , yn (x) if the determinant W of the coefﬁcients is nonvanishing, for a ≤ x ≤ b. This shows that the required condition for linear independence is W = 0, for a ≤ x ≤ b, where y1 (x) y2 (x) (1) (1) y2 (x) W = y1 (x) · · · · · · · · · (n−1) (n−1) y (x) y2 (x) 1

yn (x) (1) ... yn (x) . · · · · · · (n−1) . . . yn (x) ...

(50)

The determinant W is called the Wronskian of the set of functions y1 (x), y2 (x), . . . , yn (x), and it is named after the Polish mathematician who introduced the condition. We have proved the following theorem concerning the linear independence of solutions of hom*ogeneous linear differential equations with continuous coefﬁcients.

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JOZEF MARIA WRONSKI (1778–1853) A Polish philosopher and mathematician now remembered only because of his introduction of the functional determinant called the Wronskian.

THEOREM 6.2

the Wronskian test for linear independence

EXAMPLE 6.7

Wronskian test for linear independence Let y1 (x), y2 (x), . . . , yn (x) be n − 1 times differentiable solutions of a hom*ogeneous linear nth order differential equation with continuous coefﬁcients that is deﬁned over an interval a ≤ x ≤ b. Then a necessary and sufﬁcient condition for the functions to be linearly independent solutions of the differential equation is that their Wronskian W is nonvanishing over this interval. The solutions will be linearly dependent over the interval if W vanishes identically. (a) The set of continuous functions cosh x, sinh x, 1 is linearly independent, because the Wronskian cosh x sinh x 1 W = sinh x cosh x 0 = sinh2 x − cosh2 x = −1, for all x. cosh x sinh x 0 (b) The set of continuous functions 1, x, x 2 , (1 + x)2 is linearly dependent because the Wronskian 1 x x 2 (1 + x)2 0 x 2x 2 + 2x = 0 for all x. W= 2 0 1 2 0 0 0 0 This result is obvious without appeal to Theorem 6.2, because setting y1 = 1, y2 = x, y3 = x 2 , and y4 = (1 + x)2 , we have y4 = y1 + 2y2 + y3 , showing that y4 is a linear combination of y1 , y2 , and y3 .

initial value problem and initial conditions

THEOREM 6.3

It should be understood that when Theorem 6.2 is used as a general test for the linear independence of an arbitrary set of functions u1 , u2 , . . . un deﬁned over an interval I, the vanishing of their Wronskian is a necessary condition for their linear independence over the interval, but it is not a sufﬁcient condition if any of the functions involved are discontinuous within the interval. It is the requirement in Theorem 6.2 that the functions be solutions of a hom*ogeneous linear differential equation with continuous coefﬁcients that ensures that the vanishing of the Wronskian is both a necessary and sufﬁcient condition for their linear independence, though the details of the proof of this are omitted. An initial value problem for the nth order linear differential equations (44) and (47) at a point x = x0 involves specifying the initial conditions y(x0 ) = k0 , y(1) (x0 ) = k1 , . . . , y(n−1) (x0 ) = kn for y(x), and its ﬁrst n − 1 derivatives at the point x0 , where the constants k1 , k2 , . . . , kn can be speciﬁed arbitrarily. The derivative y(n) (x0 ) cannot be speciﬁed as an initial condition, because it is determined by the differential equation itself once the stated initial conditions have been given. The following is the fundamental existence and uniqueness theorem for linear higher order differential equations. Existence and uniqueness of solutions Let the coefﬁcients of the hom*ogeneous differential equation (47) be continuous functions over an interval a < x < b that

Section 6.3

existence and uniqueness of solutions

hom*ogeneous Linear Higher Order Constant Coefﬁcient Equations

297

contains the point x0 and a0 (x) = 0 in (a, b). Then a unique solution exists on this interval that satisﬁes the initial conditions y(x0 ) = k0 ,

y(1) (x0 ) = k1 , . . . , y(n−1) (x0 ) = kn .

Proof A proof of the existence of solutions of initial value problems for linear higher order variable coefﬁcient differential equations is beyond the level of this ﬁrst account, and so will be omitted. However, the existence and uniqueness of solutions of initial value problems for constant coefﬁcient equations will follow from the subsequent work in which the form of the general solution will be found and its constants matched so that it satisﬁes the initial conditions. It remains for us to establish the uniqueness of the initial value problem for linear higher order variable coefﬁcient equations with continuous coefﬁcients. Let us consider equation (47), and write its general solution y(x) = c1 y1 (x) + c2 y2 (x) + · · · + cn yn (x). Differentiating this result n − 1 times, and after each differentiation substituting the initial conditions, leads to the following system of simultaneous equations: c1 y1 (x0 ) + c2 y2 (x0 ) + · · · + cn (x)yn (x0 ) = k0 (1)

(1)

c1 y1 (x0 ) + c2 y2 (x0 ) + · · · + cn yn(1) (x0 ) = k1 · · · · · · · · · · · · · · · · · · · · · (n−1) (n−1) c1 y1 (x0 ) + c2 y2 (x0 ) + · · · + cn yn(n−1) (x0 ) = kn−1 . This nonhom*ogeneous system of linear equations will have a unique solution for the constant coefﬁcients c1 , c2 , . . . , cn provided the determinant of the coefﬁcients does not vanish. The determinant is simply the Wronskian W(x0 ), and by hypothesis the n solutions are linearly independent, so W(x0 ) = 0 for a ≤ x ≤ b. Consequently, the coefﬁcients c1 , c2 , . . . , cn are uniquely determined and, when substituted into the general solution, lead to a unique solution of the initial value problem. To solve the hom*ogeneous constant coefﬁcient equation dn y dn−1 y dy + an y = 0, + a + · · · + an−1 1 n n−1 dx dx dx

(51)

we proceed as with a second order equation and seek solutions of the form y(x) = ceλx , with c and λ constants. Substituting y(x) into (51) leads to the result (λn + a1 λn−1 + a2 λn−2 + · · · + an )eλx = 0, characteristic equation for higher order equations

after which cancellation of the nonvanishing factor eλx shows λ may be any of the roots of the characteristic equation λn + a1 λn−1 + a2 λn−2 + · · · + an = 0.

(52)

This polynomial of degree n has n roots λ1 , λ2 , . . . , λn that either will all be real or, if some are complex, will occur in complex conjugate pairs. To each root λi there will correspond a solution yi (x), and the linearly independent solutions y1 (x), y2 (x), . . . , yn (x) form a basis for the solution space. An arbitrary linear combination of the n basis functions forms the complementary function for (51).

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Rules for constructing the complementary function of an nth order constant coefﬁcient differential equation how to construct the complementary function

The differential equation dn y dn−1 y dy + an y = 0 + a + · · · + an−1 1 dx n dx n−1 dx with real coefﬁcients a1 , a2 , . . . , an has the characteristic equation λn + a1 λn−1 + a2 λn−2 + · · · + an = 0, with the n roots λ1 , λ2 , . . . , λn . 1. To a single real root λ = α there corresponds the single solution eαx , with A an arbitrary constant. 2. Substitution shows that to a real root λ = α with multiplicity r (repeated r times) there correspond the r linearly independent solutions eαx , xeαx , . . . , xr −1 eαx . 3. To a pair of complex conjugate roots λ = α ± iβ there correspond the two solutions eαx cos βx

and

eαx sin βx.

4. To a pair of complex conjugate roots λ = α ± iβ repeated s times, there correspond the 2s solutions eαx cos βx, eαx sin βx, eαx x cos βx, eαx x sin βx, . . . . . . , eαx x s−2 cos βx, eαx x s−2 sin βx, eαx x s−1 cos βx, eαx x s−1 sin βx. 5. The general solution of the differential equation is an arbitrary linear combination of all solutions produced by the preceding rules. To see why the functions in Rules 2 and 4 are solutions of the differential equation, we consider a typical case in which the differential equation has a real root λ = μ with multiplicity 2. Removing the factor (λ − μ)2 from the characteristic polynomial allows it to be written λn + a1 λn−1 + a2 λn−2 + · · · + an = (λ − μ)2 Q(λ), where Q(λ) is a polynomial of degree n − 2 in λ that does not vanish when λ = μ. Differentiating this result with respect to λ gives nλn−1 + (n − 1)a1 λn−2 + · · · + an−1 = 2(λ − μ)Q(λ) + (λ − μ)2 Q (λ), and setting λ = μ reduces this to nμn−1 + (n − 1)a1 μn−2 + · · · + an−1 = 0.

Section 6.3

hom*ogeneous Linear Higher Order Constant Coefﬁcient Equations

299

As the multiplicity of the root is 2, and eμx is known to be a solution, it is necessary to show that xeμx is also a solution. This will follow if when xeμx is substituted into the differential equation the result becomes an identity. Setting y(x) = xeμx and differentiating m times gives y(m) = mμm−1 eμx + m μx μ xe . Substituting this into the left-hand side of the differential equation leads to the result (nμn−1 + (n − 1)a1 μn−2 + · · · + an−1 )eμx + (μn + a1 μn−1 + · · · + an−1 μ + an )xeμx , but this is zero because we have shown that the coefﬁcient of eμx is zero, and the coefﬁcient of xeμx vanishes because μ is a root of the characteristic equation. Thus, xeμx satisﬁes the differential equation identically and so is a solution. The functions eμx and xeμx are linearly independent because they are not proportional. The same form of argument can be extended to the case when λ = μ is a real root of arbitrary multiplicity, whereas the linear independence of the solutions follows from Theorem 6.2. A similar argument can be used when a pair of complex conjugate roots occurs with arbitrary multiplicity, though the details of these extensions are left as exercises. EXAMPLE 6.8

some typical examples

Find the general solution of (i) y − 2y − 5y + 6y = 0; (ii) y + 2y + 4y = 0; (iii) y(iv) + y − 2y = 0. Solution (i) The characteristic equation is λ3 − 2λ2 − 5λ + 6 = 0. Inspection shows that λ = 1 is a root, so dividing the characteristic equation by the factor (λ − 1) shows that the other two roots are the solutions of λ2 − λ − 6 = 0, which are λ = −2 and λ = 3. Thus, from Rule 1 the general solution is y(x) = C1 e x + C2 e −2x + C3 e 3x . (ii) The characteristic equation is λ3 + 2λ2 + 4λ = 0

λ(λ2 + 2λ + 4) = 0, √ from which we see that λ = 0, or λ = −1 ± i 3. Combining Rules 1 and 3 shows the general solution to be √ √ y(x) = C1 + e−x (C2 cos(x 3) + C3 sin(x 3)). or

(iii) The characteristic equation is λ4 + λ2 − 2 = 0. This is a biquadratic equation, so if we set m = λ2 , this becomes m2 + m − √ 2 = 0, 2, and with the solutions m = −2 and m = 1. Thus, λ can take the values 1, −1, i √ −i 2. Combining Rules 1 and 3 shows the general solution to be √ √ y(x) = C1 e x + C2 e−x + C3 cos(x 2) + C4 sin(x 2).

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EXAMPLE 6.9

Find the general solution of a hom*ogeneous equation with the characteristic equation λ3 (λ + 4)2 (λ2 + 2λ + 5)2 = 0. Solution In this equation the real root λ = 0 occurs with multiplicity 3, the real root λ = −4 occurs with multiplicity 2, and the pair of complex conjugate roots (λ = −1 + 2i) and (λ = −1 − 2i) occur with multiplicity 2. The terms to be included in the general solution corresponding to the repeated root λ = 0 follow by setting λ = 0 and r = 3 in Rule 2 to obtain D1 + D2x + D3 x 2 . Similarly, the terms to be included corresponding to the repeated root λ = −4 follow by setting α = −4 and r = 2 in Rule 2 to obtain K1 e−4x + K2 xe−4x , where K1 and K2 are arbitrary constants. Finally, the terms to be included because of the repeated complex conjugate roots follow by setting α = −1, β = 2, and s = 2 in Rule 4 to obtain e−x {E1 cos 2x + F1 sin 2x + E2 xcos 2x + F2 xsin 2x}. Collecting terms shows that the general solution is y(x) = D1 + D2x + D3 x 2 + K1 e−4x + K2 xe−4x +e−x {E1 cos (2x) + F1 sin (2x) + E2 xcos (2x) + F2 xsin (2x)}. This general solution contains nine arbitrary constants, as would be expected because the characteristic polynomial is of degree 9.

EXAMPLE 6.10

Solve the initial value problem y − 2y − 5y + 6y = 0,

with y(0) = 1, y (0) = y (0) = 0.

Solution The general solution was shown in Example 6.8 (i) to be y(x) = C1 e x + C2 e−2x + C3 e3x . The initial conditions require that (y(0) = 1) (y (0) = 0) (y (0) = 0)

1 = C1 + C2 + C3 0 = C1 − 2C2 + 3C3 0 = C1 + 4C2 + 9C3 .

The solution of this system of equations is C1 = 1, C2 = 1/5, C3 = −1/5, so the solution of the initial value problem is 1 1 y(x) = e x + e−2x − e3x . 5 5 EXAMPLE 6.11

Solve the initial value problem y + 2y + 4y = 0,

with y(0) = 0, y (0) = 1, y (0) = 0.

Solution The general solution was found in Example 6.8 (ii) to be √ √ y(x) = C1 + e−x (C2 cos (x 3) + C3 sin (x 3)).

Section 6.3

hom*ogeneous Linear Higher Order Constant Coefﬁcient Equations

301

The initial conditions require that (y(0) = 0) C1 + C2 = 0

√ 1 = −C2 + C3 3 √ (y (0) = 0) 0 = C2 + C3 3. √ . . These equations have the solution C1 = 1 2, C2 = −1 2, and C3 = 3/6, so the solution is √ √ √ 3 −x 1 1 y(x) = + e sin x 3 − e−x cos x 3. 2 6 2 (y (0) = 1)

Summary

This section extended the discussion of linear second order constant coefﬁcient equations to higher order equations, and showed how the characteristic equation again determines the nature of the solutions that enter into the complementary function. The concept of linearly independent functions was extended, and it was shown that the set of linearly independent functions associated with a higher order equation forms a basis for its solution space. The Wronskian was deﬁned and shown to provide a test for the linear independence of a set of solutions of a higher order equation. Rules were given for construction of the complementary function of an nth order constant coefﬁcient equation, and then applied to some typical examples.

EXERCISES 6.3 1. Use the Wronskian test to prove the linear independence of the functions e x , xe x , x 2 e x for |x| < ∞. 2. Use the Wronskian test to prove the linear independence of the functions sin x, e x sin x, e x cos x. 3. Test the following functions for linear independence: 3, −x, x 2 , (1 + 2x)2 . 4. Test the following functions for linear independence: 1, ln x, ln x 1/2 , e x for x = 0. In Exercises 5 through 12 show that the given functions form a basis for the associated differential equation. Write down the general solution, state the interval in which it is deﬁned, and, where required, solve the given initial value problem. 5. xy − y − 4x 3 y = 0; cosh x 2 and sinh x 2 . 6. xy − y + 4x 3 y = 0; sin x 2 and cos x 2 . 7. y + 3y + 9y − 13y = 0; e x , e−2x cos 3x, e−2x sin 3x. Solve the initial value problem for which y(0) = 1, y (0) = 0, and y (0) = 0. 8. x 3 y − x 2 y + 2xy − 2y = 0; x, x 2 , x ln |x|. Solve the initial value problem for which y(1) = 1, y (1) = 1, and y (1) = 0. 9. (8x 2 + 1)y − 16xy + 16y = 0; 2x, 8x 2 − 1.

10. y − 16xy + (64x 2 − 8)y = 0; exp(4x 2 ), 2x exp(4x 2 ). 11. [4 − 2x cot(x/2)]y − xy + y = 0; x/2, sin(x/2). 12. 3x 3 y + xy − y = 0; 3x, 3x exp[1/(3x)]. In Exercises 13 through 18 solve the initial value problems using the ﬁve stated rules for the construction of the complementary function and, when available, use computer algebra to check the results. 13. y + y − 4y = 0, with y(0) = 1, y (0) = 1, y (0) = 0. 14. y + 3y − 4y = 0, with y(1) = −1, y (1) = 0, y (1) = 1. 15. y + 3y + 7y + 5y = 0, with y(0) = 1, y (0) = 0, y (0) = 0. 16. y − 2y + 5y + 26y = 0, with y(0) = 0, y (0) = 1, y (0) = 1. 17. y(iv) − y − 2y = 0, with y(0) = 1, y (0) = 0, y (0) = 0, y (0) = 0. 18. y(iv) − y − 6y = 0, with y(0) = 0, y (0) = 1, y (0) = 0, y (0) = 0. 19.* A gyrostatic pendulum is a pendulum bob (mass) suspended by a light inextensible string from a ﬁxed point, with the bob allowed to swing around its equilibrium position. If the displacement of the bob from its equilibrium position is small, the x and y coordinates of the bob as a function of time t can be shown to satisfy the

302

Chapter 6

Second and Higher Order Linear Differential Equations and Systems deﬁned on some interval I. Then the Abel formula for the Wronskian is

coupled differential equations dy d2 x + c2 x = 0 +a 2 dt dt

and

d2 y dx + c2 y = 0, −a 2 dt dt

with a > 0. Find the general solution for x(t) and y(t). By examination of the constants in the general solution identify two situation in which the motion of the bob will be in a circle (a circular pendulum), in each case commenting on the angular velocity of the bob. 20.* The discharge of capacitor in the primary circuit of an induction coil with a closed secondary circuit is oscillatory and governed by the equations dx dy 1 L +M + xdt = f (t) dt dt C dy dx +N = 0, M dt dt

and

where L, M, N, and C are all positive constants and f (t) is a forcing function. Find the differential equation satisﬁed by the discharge x(t), and show that when LN − M 2 is small and positive the complementary function for the discharge x(t) exhibits rapid oscillations.

W(y1 (x), y2 (x)) = W(y1 (x0 ), y2 (x0 )) x a1 (t) × exp − dt , x0 a0 (t) where x0 is any point in the interval I. Verify this result for the differential equation x 2 y − 2xy − 4y = 0, given that two linearly independent solutions over any interval that does not contain the origin are l/x and x 4 . Conclude that the choice of the point x0 entering into the constant factor W(y1 (x0 ), y2 (x0 )) is immaterial. 22.* Complete the details of the following outline proof of the Abel formula. Show that the derivative of the Wronskian of the functions in Exercise 21 can be written W(y1 (x), y2 (x)) = y1 (x)y2 − y2 (x)y1 (x). Use the fact that y1 (x) and y2 (x) are solutions of the differential equation to show that W = −

and by integrating over the interval x0 ≤ t ≤ x derive the result

Background material 21.* Let y1 (x) and y2 (x) be two linearly independent solutions of the differential equation a0 (x)y + a1 (x)y + a2 (x)y = 0,

6.4

a1 (x) W, a0 (x)

W(y1 (x), y2 (x)) = W(y1 (x0 ), y2 (x0 )) x a1 (t) × exp − dt . x0 a0 (t)

Undetermined Coefﬁcients: Particular Integrals Like the nonhom*ogeneous second order constant coefﬁcient differential equation considered in Section 6.2, a particular integral yp (x) of the nonhom*ogeneous linear higher order constant coefﬁcient differential equation dn y dn−1 y dy + an y = f (x) + a1 n−1 + · · · + an−1 n dx dx dx is a solution of the equation that does not contain arbitrary constants, so dn yp dn−1 yp dyp + an yp = f (x). + a + · · · + an−1 1 n n−1 dx dx dx

(53)

Section 6.4

particular integral, complementary function, and undetermined coefﬁcients

Undetermined Coefﬁcients: Particular Integrals

303

The complementary function yc (x) associated with (53) is the general solution of the hom*ogeneous form of the equation dn yc dn−1 yc dyc + an yc = 0, + a + · · · + an−1 1 n n−1 dx dx dx considered in Section 6.3. It follows from the deﬁnitions of yc (x) and yp (x) and the linearity of the equation that the general solution y(x) of (53) can be written y(x) = yc (x) + yp (x).

(54)

A particular integral of (53) can be found by the method of undetermined coefﬁcients whenever the nonhom*ogeneous term f (x) is a linear combination of elementary functions such as polynomials, exponentials, and sine or cosine functions. The method depends for its success on recognizing the general form of a function that when substituted into the left-hand side of (53) yields the general form of the nonhom*ogeneous term f (x) on the right-hand side. Undetermined coefﬁcients are involved because although the general form of a particular integral yp (x) can be guessed from the function f (x), any multiplicative constants (the undetermined coefﬁcients) involved will not be known. Their values are found by substituting the possible form for yp (x) into the left-hand side of (53) and equating the undetermined coefﬁcients of terms on the left of the equation to the known coefﬁcients of corresponding terms in f (x) on the right. The approach is illustrated in the following example. EXAMPLE 6.12

Find the general solution of y + 5y + 6y = 4e−x + 5sin x. Solution The general solution is y(x) = yc (x) + yp (x), where yc (x) is the complementary function satisfying the hom*ogeneous form of the equation yc + 5yc + 6yc = 0, and yp (x) is a particular integral that corresponds to the nonhom*ogeneous term 4e−x + 5sin x. The characteristic equation is λ2 + 5λ + 6 = 0, with the roots λ1 = −2 and λ2 = −3 corresponding to the linearly independent solutions e−2x and e−3x , so the complementary function is yc (x) = C1 e−2x + C2 e−3x , where C1 and C2 are arbitrary constants.

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Chapter 6

Second and Higher Order Linear Differential Equations and Systems

To ﬁnd a particular integral, we notice ﬁrst that neither the term e−x nor the term sin x is contained in the complementary function. This means that the only form of particular integral yp (x) that can produce the nonhom*ogeneous term 4e−x + 5sin x is yp (x) = Ae−x + Bsin x + Ccos x, undetermined coefﬁcients

where A, B, and C are the undetermined coefﬁcients that must be found. Substituting this expression for yp (x) into the differential equation leads to the result (Ae−x − Bsin x − Ccos x) + 5(−Ae−x + Bcos x − Csin x) + 6(Ae−x + Bsin x + Ccos x) = 4e−x + 5sin x. When we collect terms involving e−x , sin x, and cos x this becomes 2Ae−x + 5(B − C)sin x + 5(B + C)cos x = 4e−x + 5sin x. If yp (x) is a particular integral, this expression must be an identity (true for all x), but this is only possible if the coefﬁcients of corresponding functions of x on either side of the equation are identical. Equating corresponding coefﬁcients gives (coefﬁcients of e−x )

2A = 4,

(coefﬁcient of sin x)

5(B − C) = 5

(coefﬁcient of cos x)

5(B + C) = 0.

so A = 2

Solving the last two equations for B and C gives B = 1/2, C = −1/2, so the particular integral is yp (x) = 2e−x + (1/2)sin x − (1/2)cos x. Substituting yc (x) and yp (x) into y(x) = yc (x) + yp (x) shows that the general solution is y(x) = C1 e−2x + C2 e−3x + 2e−x + (1/2)sin x − (1/2)cos x. A complication arises if a term in the nonhom*ogeneous term f (x) is contained in the complementary function, as illustrated in the next example. EXAMPLE 6.13

Find a particular integral of the equation y + y − 12y = e3x . Solution This equation has the complementary function yc (x) = C1 e3x + C2 e−4x , so e3x is contained in both the nonhom*ogeneous term and the complementary function. An attempt to ﬁnd a particular integral of the form yp (x) = Ae3x will fail, because e3x is a solution of the hom*ogeneous form of the equation, so its substitution into the left-hand side of the differential equation will lead to the contradiction 0 = e3x . To overcome this difﬁculty we need to seek a more general particular integral that, when substituted into the differential equation, produces a multiple of e3x whose scale factor can be equated to the coefﬁcient of the nonhom*ogeneous

Section 6.4

Undetermined Coefﬁcients: Particular Integrals

305

term and other terms that cancel. As exponentials are involved, a natural choice is yp (x) = Axe3x . Differentiation of yp (x) gives yp (x) = Ae3x + 3Axe3x

and

yp (x) = 6Ae3x + 9Axe3x .

Substituting these results into the differential equation gives 6Ae3x + 9Axe3x + Ae3x + 3Axe3x − 12Axe3x = e3x , so after cancellation of the terms in Axe3x this reduces to 7Ae3x = e3x , showing that A = 1/7. So the required particular integral is yp (x) =

1 3x xe . 7

Table 6.2 lists the form of particular integral that correspond to the most common nonhom*ogeneous terms. Each of its entries can be constructed by using arguments similar to the one just given. When the nonhom*ogeneous term is a linear combination of terms in the table, the form of yp (x) is found by adding the forms of the corresponding particular integrals. EXAMPLE 6.14

Find the general solution of y − 5y + 6y = x 2 + sin x.

some typical examples

Solution

The characteristic equation is λ3 − 5λ2 + 6λ = 0,

or

λ(λ2 − 5λ + 6) = 0,

with the roots λ1 = 0, λ2 = 2, and λ3 = 3, so the complementary function is yc (x) = C1 + C2 e2x + C3 e3x . The function x 2 on the right-hand side is not contained in the complementary function, but there is no undifferentiated term involving y(x) in the equation, so from Step 2(b) in Table 6.2 the appropriate form of particular integral corresponding to this term is Ax + Bx 2 + Cx 3 . The function sin x is not contained in the complementary function, so the form of particular integral appropriate to this term is seen from Step 4(a) to be Dsin x + Ecos x. Combining these two forms shows that the general form of yp (x) is yp (x) = Ax + Bx 2 + Cx 3 + Dsin x + Ecos x. Substituting yp (x) into the differential equation gives (6C − Dcos x + Esin x) − 5(2B + 6Cx − Dsin x − Ecos x) + 6(A+ 2Bx + 3Cx 2 + Dcos x − Esin x) = x 2 + sin x.

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TABLE 6.2 Particular Integrals by the Method of Undetermined Coefﬁcients The method applies to the linear constant coefﬁcient differential equation how to ﬁnd a particular integral using undetermined coefﬁcients

dn y dn−1 y dy + a1 n−1 + · · · + an−1 + an y = f (x), n dx dx dx which has the characteristic equation λn + a1 λn−1 + · · · + an−1 λ + an = 0, with the roots λ1 , λ2 , . . . , λn , and the complementary function yc (x) = C1 y1 (x) + C2 y2 (x) + · · · + Cn yn (x), where y1 (x), y2 (x), . . . , yn (x) are the linearly independent solutions of the hom*ogeneous equation appropriate to the nature of the roots. 1.

f (x) = constant.

(λ = 0)

Include in yp (x) the constant term K. 2.

f (x) = a0 + a1 x + a2 x 2 + · · · + am x m. (a) If the left-hand side of the differential equation contains an undifferentiated term y(x), include in yp (x) the polynomial A0 x m + A1 x m−1 + · · · + Am. (b) If the left-hand side of the differential equation contains no undifferentiated function of y(x), and the lowest order derivative is ds y/dx s , include in yp (x) the polynomial A0 x m+s + A1 x m+s−1 + · · · + Am x s .

3. f (x) = Peax . (a) If eax is not contained in the complementary function, include in yp (x) the term Beax . (b) If the complementary function contains the terms eax , xeax , . . . , x meax , include in yp (x) the term Bx m+1 eax . 4. f (x) contains terms in cos px and/or sin px. (a) If cos px and/or sin px are not contained in the complementary function, include in yp (x) the terms Pcos px + Q sin px. (b) If the complementary function contains the terms x cos px and/or x sin px, include in yp (x) terms of the form x 2 (Pcos px + Q sin px). (continued )

Section 6.4

Undetermined Coefﬁcients: Particular Integrals

307

TABLE 6.2 (continued ) (c) If the complementary function contains the terms x 2 cos px and/or x 2 sin px, include in yp (x) terms of the form x 3 (Pcos px + Qsin px). 5. f (x) contains terms in e px cos qx and/or e px sin qx. (a) If e px cos qx and/or e px sin qx are not contained in the complementary function, include in yp (x) terms of the form e px (Rcos qx + Ssin qx). (b) If the complementary function contains xe px cos qx and/or xe px sin qx, include in yp (x) terms of the form x 2 e px (Rcos qx + Ssin qx). 6. The required particular integral yp (x) is the sum of all the terms produced by identifying each term belonging to f (x) with one of the types of term listed above. 7. The values of the undetermined coefﬁcients K, A0 , A1 , . . ., Am, B, P, Q, R, and S are found by substituting yp (x) into the differential equation, equating the coefﬁcients of corresponding functions on either side of the equation to make the result an identity, and then solving the resulting simultaneous equations for the undetermined coefﬁcients.

Equating coefﬁcients of corresponding functions on each side of this expression to make it an identity, we have (constant terms)

6C − 10B + 6A = 0,

(terms in x)

−30C + 12B = 0,

(terms in x 2 )

18C = 1,

(terms in sin x)

5D − 5E = 1,

(terms in cos x)

5D + 5E = 0.

Solving these simultaneous equations gives A = 19/108, B = 5/36, C = 1/18, D = 1/10, and E = −1/10, so the particular integral is yp (x) =

5 1 19 1 1 x + x2 + x3 + sin x − cos x. 108 36 18 10 10

Combining this with the complementary function shows the general solution to be y(x) = C1 + C2 e2x + C3 e3x +

5 1 19 1 1 x + x2 + x3 + sin x − cos x. 108 36 18 10 10

The existence and uniqueness of solutions of initial value problems for nonhom*ogeneous linear differential equations are guaranteed by the following theorem, which is a direct extension of Theorem 6.3.

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Second and Higher Order Linear Differential Equations and Systems

THEOREM 6.4 more on existence and uniqueness: this time for nonhom*ogeneous equations

Existence and uniqueness of solutions of nonhom*ogeneous linear equations Let the coefﬁcients and nonhom*ogeneous term of differential equation (53) be continuous functions over an interval a < x < b that contains the point x0 . Then a unique solution exists on this interval that satisﬁes the initial conditions y(x0 ) = k0 ,

y(1) (x0 ) = k1 , . . . , y(n−1) (x0 ) = kn−1 .

Proof As before, the proof of the existence of solutions of variable coefﬁcient equations will be omitted, while the existence of solutions of constant coefﬁcient equations has already been established. This only leaves the proof of uniqueness that follows along the same lines as those of Theorem 6.3, with y(x) replaced by y(x) = c1 y1 (x) + c2 y2 (x) + · · · + cn (x) + yp (x), and the system of equations determining c1 , c2 , . . . , cn replaced by c1 y1 (x0 ) + c2 y2 (x0 ) + · · · + cn (x)yn (x0 ) = k0 − yp (x0 ) c1 y1 (x0 ) + c2 y2 (x0 ) + · · · + cn yn(1) (x0 ) = k1 − yp (x0 ) . . . . . . . . . . . (1)

(1)

(n−1)

c1 y1

(n−1)

(x0 ) + c2 y2

(x0 ) + · · · + cn yn(n−1) (x0 ) = kn−1 − y(n−1) (x0 ). p

The constants c1 , c2 , . . . , cn are uniquely determined by this system because, as with Theorem 6.3, the determinant of the coefﬁcients is the Wronskian and so is nonvanishing for x = x0 . EXAMPLE 6.15

Solve the initial value problem y + 4y + 3y = e−x ,

with y(0) = 2,

y (0) = 1.

Solution The characteristic equation is λ2 + 4λ + 3 = 0, with the roots λ1 = −1 and λ2 = −3, so the complementary function is yc (x) = C1 e−x + C2 e−3x . The nonhom*ogeneous term e−x is contained in the complementary function, so by Step 3(b) in Table 6.2 we must seek a particular integral of the form yp (x) = Axe−x . Substituting the expression for yp (x) into the differential equation gives (−2Ae−x + Axe−x ) + 4(Ae−x − Axe−x ) + 3Axe−x = e−x ,

or

2Ae−x = e−x ,

showing that A = 1/2. So, in this case, the particular integral is yp (x) = (1/2)xe−x and the general solution is y(x) = C1 e−x + C2 e−3x + (1/2)xe−x . The initial condition y(0) = 2 will be satisﬁed if 2 = C1 + C2 ,

and the initial condition y (0) = 1 will be satisﬁed if 1/2 = −C1 − 3C2 ,

Section 6.5

Cauchy–Euler Equation

309

so C1 = 13/4 and C2 = −5/4. Substituting these values for C1 and C2 in the general solution gives the solution of the initial value problem 13 1 5 + x e−x − e−3x . y(x) = 4 2 4

Summary

The determination of particular integrals for nonhom*ogeneous equations is important, and the method of undetermined coefﬁcients that was described in this section is the simplest method by which they can be found. The method is only applicable to nonhom*ogeneous terms formed by a sum of polynomials, exponentials, trigonometric functions, and certain of their combinations. It depends for its success on recognizing the general form of function that, when substituted into the left of the differential equation, produces terms of the type found in the nonhom*ogeneous term on the right. The method involves substituting a linear combination of such terms with arbitrary constant multipliers (the undetermined coefﬁcients) into the left of the equation and matching the constants so the terms that result are identical to the terms on the right.

EXERCISES 6.4 17. 18. 19. 20.

Find the general solutions of the following differential equations. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16.

y + 2y − 3y = 4 + x + 4e2x . y + 4y + 4y = 2 − sin 3x. y + 2y + y = 5 + x 2 e x . y − 4y + 4y = 3x 2 + 2e3x . y + 4y + 4y = sin x − 2 cos x. y + 4y + 5y = sin x. y + 2y + 2y = 1 + x + e−x . y + 5y + 6y = 3 sin x + 5x + x 2 . y + 2y + 2y = 2 − 4x 2 . y + 2y + 2y = sin x. y − 7y + 12y = x + e2x + e3x . y + 4y + 5y = 3 + 2e−2x . y + 2y − 8y = 3xcos 4x. y + 2y − 15y = 3 + 2xsin x. y + 9y = 2 cos 3x + sin 3x. y − 4y = 3e2x + 4e−2x .

6.5

y + 3y + 2y = x 2 + 3e−2x . y + y + 3y − 5y = 4e−x . y + 4y + 5y = e−2x sin x. y + 4y + 5y = x 2 − e−2x cos x.

In Exercises 21 through 28 solve the initial value problems. Where the characteristic equation is of degree 3, at least one root is an integer and can be found by inspection. 21. y + 6y + 13y = e−3x cos x, with y(0) = 2, y (0) = 1. 22. y − 4y + 5y = e2x cos x, with y(0) = 0, y (0) = 2. 23. y + 9y = 7 + 2sin 3x − 4cos 3x, with y(0) = −1, y (0) = 1. 24. y + 4y + 5y = x + sin x, with y(0) = − 1, y (0) = 0. 25. y − 2y + 5y = 1 + e−x , with y(0) = 2, y (0) = 1. 26. y + 4y + 5y = 2 + e−2x sin x, with y(0) = 0, y (0) = 0. 27. y + y − 2y = 3 + 2 cos x, with y(0) = 0, y (0) = 1, y (0) = −1. 28. y + y − y − y = 2 + e−x , with y(0) = 1, y (0) = 1, y (0) = 0.

Cauchy–Euler Equation

Cauchy–Euler equation

One of the simplest linear variable coefﬁcient differential equations is the hom*ogeneous second order Cauchy–Euler equation, whose standard form is

x2

d2 y dy + a2 y = 0. + a1 x 2 dx dx

(55)

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The solution of this hom*ogeneous equation can be reduced to a simple algebraic problem by seeking a solution of the form y(x) = Ax m,

(56)

where A is an arbitrary constant, and the permissible values of m are to be determined. Differentiating y(x) to obtain dy = mAx m−1 dx

and

d2 y = m(m − 1)Ax m−2 dx 2

(57)

and substituting these expressions into the Cauchy–Euler equation gives the following quadratic equation for m: m(m − 1) + a1 m + a2 = 0.

(58)

When this equation has two distinct real roots m = α and m = β, the general solution of (55) is y(x) = C1 x α + C2 x β ,

(59)

but if the two roots are real and equal with m = μ, the general solution of (55) is y(x) = C1 x μ + C2 x μ ln |x|,

(60)

where C1 and C2 are arbitrary real constants. If the equation for m has the complex conjugate roots m = α ± iβ, substitution conﬁrms that the general solution of (55) is y(x) = C1 x α cos(β ln |x|) + C2 x α sin(β ln |x|).

(61)

The second solution x μ ln |x| in (60) can be obtained from the method of Section 6.7 by using the known solution y1 (x) = x μ to ﬁnd a second linearly independent solution y2 (x). The form of solution (61) follows from writing the general solution as y(x) = Aexp(α + iβ) + B exp(α − iβ), with A an arbitrary complex constant and B its complex conjugate so that y(x) is real. EXAMPLE 6.16

Find the general solution of x2

d2 y dy + 2y = 0 + 3x 2 dx dx

for x = 0.

Solution The equation for m is m(m − 1) + 3m + 2 = 0, with the roots m = −1 ± i. The general solution is thus y(x) = C1 x −1 cos (ln |x|) + C2 x −1 sin (ln |x|).

Section 6.6

Summary

Variation of Parameters and the Green’s Function

311

The Cauchy–Euler equation is the simplest linear variable coefﬁcient equation for which a closed form analytical solution can be found. The solution is obtained by recognizing that it must be of the form y(x) = Ax m and ﬁnding the permissible values of m.

EXERCISES 6.5 Find the general solutions of the following Cauchy–Euler equations. 1. 2. 3. 4. 5. 11.

x 2 y + 3xy − 3y = 0. 6. x 2 y + 2xy + 4y = 0. x 2 y + 3xy + 5y = 0. 7. x 2 y + 6xy + 4y = 0. 2 x y + 5xy + 9y = 0. 8. x 2 y + xy + 4y = 0. 2 x y − 3xy − 5y = 0. 9. x 2 y + 4xy + 4y = 0. 2 x y + 3xy − 8y = 0. 10. x 2 y + 3xy + 6y = 0. With the change of variable x = et , we ﬁnd using the chain rule that dy 1 dy d2 y 1 d2 y dy = and . = − dx x dt dx 2 x 2 dt 2 dt

12. Use the substitution y(x) = Ax m to solve the third order Cauchy–Euler equation x 3 y − 3x 2 y + 6xy − 6y = 0. 13. Use the substitution of Exercise 11 to solve the Cauchy– Euler equation in Exercise 12. 14. Express dy/dx, d2 y/dx 2 , and d3 y/dx 3 in terms of dy/dt, d2 y/dt 2 , and d3 y/dt 3 if ax + b = et . Use the substitution to show that the general solution of (2x + 3)3 y + 3(2x + 3)y − 6y = 0 is

Use these results to show that this change of variable transforms a Cauchy–Euler equation into a constant coefﬁcient equation, and solve Exercise 3 by this method.

6.6

y(x) = C1 (2x + 3) + C2 (2x + 3)1/2 + C3 (2x + 3)3/2 for x > 0.

Variation of Parameters and the Green’s Function Variation of Parameters The method of variation of parameters, perhaps more properly called variation of constants, is a powerful method used to ﬁnd a particular integral of a linear differential equation once its complementary function is known. In what follows the method will be developed for a general linear second order variable coefﬁcient differential equation, though it is easily extended to include linear variable coefﬁcient differential equations of any order. As linear constant coefﬁcient equations are a special case of variable coefﬁcient equations, the method enables particular integrals to be found for all linear equations. The method also has the advantage that no special cases arise due to the nonhom*ogeneous term being included in the complementary function. Consider the general linear second order differential equation d2 y dy + b(x)y = f (x), + a(x) 2 dx dx

idea underlying the method of variation of parameters

(62)

deﬁned on some interval α ≤ x ≤ β over which a(x), b(x), and f (x) are deﬁned and continuous. Let y1 (x) and y2 (x) be two known linearly independent solutions

312

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Second and Higher Order Linear Differential Equations and Systems

of the hom*ogeneous form of (62), so the complementary function is yc (x) = C1 y1 (x) + C2 y(x).

(63)

The idea underlying the method of variation of parameters, and from which it derives its name, is to replace the constants C1 and C2 by the unknown functions u1 (x) and u2 (x), and then to seek a particular integral of the form yp (x) = u1 (x)y1 (x) + u2 (x)y2 (x).

(64)

Two equations are needed in order to determine u1 (x) and u2 (x), and the ﬁrst of these is obtained as follows. Differentiation of (64) gives yp (x) = u1 (x)y1 (x) + u2 (x)y2 (x) + u1 (x)y1 (x) + u2 (x)y2 (x), so by requiring u1 (x) and u2 (x) to be such that the last two terms vanish, we have yp (x) = u1 (x)y1 (x) + u2 (x)y2 (x),

(65)

u1 (x)y1 (x) + u2 (x)y2 (x) = 0.

(66)

subject to the condition

Equation (66) is the ﬁrst condition to be imposed on u1 (x) and u2 (x), and a second condition is obtained as follows. Differentiating (65) gives yp (x) = u1 (x)y1 (x) + u2 (x)y2 (x) + u1 (x)y1 (x) + u2 (x)y2 (x),

(67)

so substituting (64), (65), and (67) into (62), followed by grouping terms, gives u1 [y1 + a(x)y1 + b(x)y1 ] + u2 [y2 + a(x)y2 + b(x)y2 ] + + u1 y1 + u2 y2 = f (x).

(68)

As y1 (x) and y2 (x) are both solutions of differential equation (62) with f (x) = 0, the expressions multiplying u1 (x) and u2 (x) both vanish identically, reducing (68) to the second condition on u1 (x) and u2 (x), u1 y1 + u2 y2 = f (x).

(69)

The functions u1 (x) and u2 (x) can now be found by solving equations (66) and (69). Solving these for u1 (x) and u2 (x) gives u1 (x) =

−y2 (x) f (x) W(x)

and

u2 (x) =

y1 (x) f (x) , W(x)

(70)

where y W(x) = 1 y1

y2 = y1 y2 − y1 y2 y2

is the Wronskian of y1 (x) and y2 (x) and so is never zero.

(71)

Section 6.6

Variation of Parameters and the Green’s Function

313

After integration, results (70) become u1 (x) = −

the general solution

y2 (x) f (x) dx W(x)

and

u2 (x) =

y1 (x) f (x) dx. W(x)

(72)

Finally, combining (64) and (72), we ﬁnd that y(x) = −y1 (x)

y2 (x) f (x) dx + y2 (x) W(x)

y1 (x) f (x) dx. W(x)

(73)

This result represents the general solution of (62), because each indeﬁnite integral has associated with it an additive arbitrary constant, and if these are −C1 and C2 , say, they include in y(x) the complementary function yc (x) = C1 y1 (x) + C2 y2 (x). When these constants are set equal to zero result (73) reduces to the particular integral yp (x). Rule for the method of variation of parameters 1. Write the differential equation in the standard form

how to apply the method of variation of parameters

d2 y dy + b(x)y = f (x). + a(x) dx 2 dx 2. Find two linearly independent solutions y1 (x) and y2 (x) of the hom*ogeneous form of the differential equation and construct the equations u1 (x)y1 (x) + u2 (x)y2 (x) = 0

and

u1 y1 + u2 y2 = f (x).

3. Solve the equations in Step 2 for u1 (x) and u2 (x) and integrate to ﬁnd u1 (x) and u2 (x), each with an arbitrary additive constant of integration. 4. The general solution of the differential equation is then given by y(x) = u1 (x)y1 (x) + u2 (x)y2 (x). Or, alternatively, after ﬁnding y1 (x) and y2 (x): 5. Substitute into y(x) = −y1 (x)

y2 (x) f (x) dx + y2 (x) W(x)

y1 (x) f (x) dx, W(x)

where y W(x) = 1 y1

y2 = y1 y2 − y1 y2 . y2

6. The result of Step 5 becomes the particular integral yp (x) if the arbitrary integration constants are set equal to zero.

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Second and Higher Order Linear Differential Equations and Systems

The example that follows shows how the method of variation of parameters deals automatically with the presence of a nonhom*ogeneous term in the complementary function of a constant coefﬁcient equation. EXAMPLE 6.17 a simple example that could also be solved by undetermined coefﬁcients

Find the general solution of the second order differential equation y + 2y + y = xe−x . Solution The characteristic equation is λ2 + 2λ + 1 = 0, with the repeated root λ = −1. Thus, the complementary function is yc (x) = C1 e−x + C2 xe−x . Two linearly independent solutions are thus y1 (x) = e−x

and

y2 (x) = xe−x ,

while the nonhom*ogeneous term is f (x) = xe−x . The Wronskian y1 y2 = e−x (e−x − xe−x ) + e−x xe−x = e−2x , W(x) = y1 y2 so substituting in (73) shows that the particular integral is 1 −x 2 −x yp (x) = −e x dx + xe xdx = x 3 e−x . 6 The general solution is 1 y(x) = C1 e−x + C2 xe−x + x 3 e−x . 6 This result could, of course, have been found by the method of undetermined coefﬁcients. The next example shows how the method of variation of parameters determines a particular integral for a constant coefﬁcient equation whose particular integral could not have been found by using undetermined coefﬁcients. EXAMPLE 6.18

Find the general solution of the differential equation y + y = csc x

an example that could not be solved by undetermined coefﬁcients

in any interval in which x = nπ , for n = 1, 2, . . . . Solution It follows at once that the complementary function is yc (x) = C1 cos x + C2 sin x, so two linearly independent solutions are y1 (x) = cos x

and

y2 (x) = sin x.

The Wronskian W(x) = y1 y2 − y1 y2 = cos2 x + sin2 x = 1, and f (x) = 1/ sin x, so substituting into (73) shows that the particular integral is yp (x) = − cos x dx + sin x cot xdx.

Section 6.6

As

Variation of Parameters and the Green’s Function

315

cot x dx = ln |sin x|, yp (x) = −x cos x + sin x ln |sin x|,

and the general solution is y(x) = C1 cos x + C2 sin x − xcos x + sin x ln |sin x|, in any interval in which x = nπ , for n = 1, 2, . . . , because ln |sin nπ| = ∞. Although this is a constant coefﬁcient equation, it is unlikely that its particular integral could have been found by the method of undetermined coefﬁcients. The last example shows how the method of variation of parameters determines a particular integral for a linear second order variable coefﬁcient equation. EXAMPLE 6.19

Find the general solution of the second order variable coefﬁcient equation x 2 y − 3xy + 4y = ln x

application to a variable coefﬁcient equation

(x > 0).

Solution This is a Cauchy–Euler equation, and the method of Section 6.5 shows that its complementary function is yc (x) = C1 x 2 + C2 x 2 ln x,

x > 0,

for

so two linearly independent solutions are y1 (x) = x 2

and

y2 (x) = x 2 ln x

for

x > 0.

A routine calculation shows the Wronskian W(x) = x 3 . Before identifying f (x) the equation must be written in the standard form with the coefﬁcient of y equal to 1. Dividing the differential equation by x 2 to bring it into the standard form shows that f (x) = (ln x)/x 2 . Substitution into (73) then gives ln x (ln x)2 2 yp (x) = −x 2 dx + x ln x dx. x3 x3 Integration by parts shows that (ln x)2 1 (ln x)2 1 ln x 1 = − − − 2 3 2 2 x 2 x 2 x 4x

and

1 ln x 1 ln x dx = − − 2, 3 2 x 2 x 4x

so using these results in the expression for yp (x) gives yp (x) =

1 1 + ln x 4 4

(x > 0).

The general solution is thus y(x) = C1 x 2 + C2 x 2 ln x +

1 1 + ln x 4 4

(x > 0).

Although the complementary function of a Cauchy–Euler equation is easily determined, a particular integral is usually sufﬁciently complicated that its general form cannot be guessed and so must be found by the method of variation of parameters.

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Finally, we remark that an application of the method of variation of parameters to the equation what happens if an integral has no known antiderivative

y + y = (1 + x 2 )1/2 gives a particular integral in the form yp (x) = − cos x (sin x)(1 + x 2 )1/2 dx + sin x (cos x)(1 + x 2 )1/2 dx. Neither of the two integrals involved can be evaluated in terms of known functions, so if an analytical solution is needed it must be obtained in series form. The Maclaurin series for the functions (sin x)(1 + x 2 )1/2 and (cos x)(1 + x 2 )1/2 are 1 (sin x)(1 + x 2 )1/2 = x + x 3 − 3 1 4 2 1/2 (cos x)(1 + x ) = x − x − 3

1 5 x + · · · and 5 13 6 x + ···. 90

Integrating these results and substituting in the expression for yp (x) gives 1 2 1 1 1 13 7 yp (x) = −(cos x) x + x 4 − x 6 + · · · + sin x x − x 5 + x + ··· . 2 12 30 15 630 Let y(x) satisfy the differential equation d2 y dy + b(x)y = f (x), + a(x) dx 2 dx

(74)

deﬁned on an interval α ≤ x ≤ β, and let a be any point inside this interval. Then the general solution of (74) given in (73) can be put into a convenient form for solving the initial value problem for (74) when the initial conditions are y(a) = 0 and y (a) = 0. We start from the general solution in (73) y2 (x) f (x) y1 (x) f (x) dx + y2 (x) dx. (75) y(x) = −y1 (x) W(x) W(x) f (x) Next, we rewrite the indeﬁnite integral y2 (x) dx as the deﬁnite integral with a W(x) x y2 (t) f (t) variable upper limit a W(t) dt and an arbitrary ﬁxed lower limit x = a. In this result, the additive arbitrary integration constant associated with the indeﬁnite integral has been replaced by the arbitrary constant a in the lower integration limit. The implications of the lower limit will become apparent when an initial value problem is considered. A corresponding result holds for the second indeﬁnite integral in (75). Using these results, taking the functions y1 (x) and y2 (x) under the respective integral signs as they are not involved in the integrations, and combining the integrals allows the general solution y(x) to be written in the form y(x) = a

x

y1 (t)y2 (x) − y1 (x)y2 (t) f (t)dt. W(t)

(76)

Section 6.6

Variation of Parameters and the Green’s Function

317

Setting x = a in this result shows that y(a) = 0. Differentiation of (76) with respect to x using Leibniz’s rule d dx

q(x) p(x)

dq dp g(x, t)dt = g(x, q) − g(x, p) + dx dx

q(x) p(x)

∂ g(x, t)dt ∂x

gives y1 (x)y2 (x) − y1 (x)y2 (x) y (x) = f (x) + W(x)

variation of parameters and initial value problems

x

a

y1 (t)y2 (x) − y1 (x)y2 (t) dt. W(t)

The ﬁrst term on the right vanishes, and setting x = a causes the integral to vanish, so we have shown that y (a) = 0. Consequently, the integral y(x) =

x

a

y1 (t)y2 (x) − y1 (x)y2 (t) f (t)dt W(t)

solves the initial value problem d2 y dy + b(x)y = f (x), + a(x) 2 dx dx EXAMPLE 6.20

with

y(a) = y (a) = 0.

Use result (76) to solve the initial value problem y + 4y = 1 + cos 2x,

with y(0) = y (0) = 0.

Solution Two linearly independent solutions of the hom*ogeneous equation are y1 (x) = sin 2x and y2 (x) = cos 2x, so W(t) = −2(sin2 2t + cos2 2t) = −2. Substituting into (76) with f (t) = 1 + cos 2t gives x 1 (sin 2xcos 2t − sin 2tcos 2x)(1 + cos 2t)dt, y(x) = 2 0

and so y(x) = 14 (1 − cos 2x + xsin 2x).

The Green’s Function An important result that can be derived from the general solution of (74) when expressed in the form given in (75) is obtained by considering a boundary value problem for the equation written in the standard form d2 y dy + b(x)y = f (x), + a(x) dx 2 dx

(77)

and deﬁned over the interval a ≤ x ≤ b. Evaluating the ﬁrst integral in (75) over the interval b ≤ t ≤ x, changing the sign by reversing the limits of integration, and then evaluating the second integral

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over the interval a ≤ t ≤ x gives y(x) = y2 (x) a

x

y1 (t) f (t)dt + y1 (x) W(t)

b x

y2 (t) f (t)dt. W(t)

(78)

As y2 (x) is not involved in the ﬁrst integral, and y1 (x) is not involved in the second integral, they may be taken under the respective integral signs so that (78) becomes y(x) = a

x

y1 (t)y2 (x) f (t)dt + W(t)

b x

y1 (x)y2 (t) f (t)dt. W(t)

(79)

This can be written y(x) =

b

G(x, t) f (t)dt,

(80)

a

the Green’s function

where the function G(x, t) is called the Green’s function for differential equation (77) deﬁned over the interval a ≤ x ≤ b and is deﬁned as ⎧ y1 (t)y2 (x) ⎪ ⎪ ⎨ W(t) , G(x, t) = ⎪ y1 (x)y2 (t) ⎪ ⎩ , W(t)

a≤t ≤x (81) x ≤ t ≤ b.

Inspection of (81) shows G(x, t) to be a continuous function of x for a ≤ x ≤ b. Differentiation of G(x, t) with respect to x gives ⎧ y1 (t)y2 (x) ⎪ ⎪ ⎨ W(t) , Gx (x, t) = ⎪ y (x)y2 (t) ⎪ ⎩ 1 , W(t)

a≤t ≤x (82) x ≤ t ≤ b.

Examination of (82) shows that as t increases across t = x, the function Gx (x, t) is discontinuous and experiences the jump Gx (x, x+ ) − Gx (x, x− ) =

y1 (x)y2 (x) − y1 (x)y2 (x) W(x) =− = −1, W(x) W(x)

where x+ is the limit at t decreases to x and x− is the limit as t increases to x. Now let y1 (x) and y2 (x) be two linearly independent solutions of the hom*ogeneous differential equation, with y1 (x) such that at x = a it satisﬁes the hom*ogeneous boundary condition k1 y1 (a) + K1 y1 (a) = 0, and y2 (x) such that at x = b it satisﬁes the hom*ogeneous boundary condition k2 y2 (b) + K2 y2 (b) = 0. Then G(x, t) is seen to satisfy these same hom*ogeneous boundary conditions, and differentiation of (80) with respect to x, again using Leibniz’s rule, shows that

Section 6.6

Variation of Parameters and the Green’s Function

319

the solution y(x) also satisﬁes these hom*ogeneous boundary conditions. Combining results shows that ⎧ y1 (t)y2 (x) ⎪ ⎪ , a≤t ≤x ⎪ b ⎨ W(t) y(x) = G(x, t) f (t)dt with G(x, t) = ⎪ y1 (x)y2 (t) a ⎪ ⎪ , x≤t ≤b ⎩ W(t) (83) is the solution of the boundary value problem for the nonhom*ogeneous linear second order equation d2 y dy + b(x)y = f (x), + a(x) 2 dx dx subject to the hom*ogeneous boundary conditions k1 y(a) + K1 y (a) = 0

with

k2 y(b) + K2 y (b) = 0.

When using this approach, unless the Green’s function itself is required, it is usually more convenient to obtain the solution directly from result (78). The advantage of the Green’s function is that it characterizes all the essential features of the differential equation without reference to the nonhom*ogeneous term f (x), so that once it is known (80) solves the hom*ogeneous boundary value problem for any function f (x). Properties of the Green’s function deﬁned over the interval a ≤ x ≤ b Consider the boundary value problem d2 y dy + b(x)y = 0, + a(x) dx 2 dx fundamental properties of the Green’s function

subject to the boundary conditions k1 y(a) + K1 y (a) = 0

and

k2 y(b) + K2 y (b) = 0

The Green’s function in (81) has the following properties: 1. The piecewise deﬁned Green’s function G(x, t) satisﬁes the differential equation in the respective intervals a ≤ x < t and t < x ≤ b. 2. G(x, t) is a continuous function of x for a ≤ x ≤ b. 3. G(x, t) satisﬁes the hom*ogeneous boundary conditions. 4. The function Gx (x, t) is continuous for a ≤ x < t and t < x ≤ b, but it is discontinuous across x = t where it experiences the jump Gx (x, x+ ) − Gx (x, x− ) = −1.

320

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EXAMPLE 6.21

Find the Green’s function for the differential equation x 2 y − 2xy + 2y = 3x 2 and use it to solve the boundary value problem when y(1) = 0 and y (2) = 0. Solution The hom*ogeneous form of the equation is a Cauchy–Euler equation, and the method of Section 6.5 shows that it has the two linearly independent solutions y1 (x) = x and y2 (x) = x 2 , so the general solution is y(x) = ax + bx 2 . For the solution y1 (x) we must use the form of this solution that satisﬁes the left boundary condition y(1) = 0, and this is easily seen to be y1 (x) = x − x 2 . For the linearly independent solution y2 (x) we must use the form of solution y(x) = ax + bx 2 that satisﬁes the right boundary condition y (2) = 0. As y (x) = a + 2bx, the condition y (2) = 0 shows that y2 (x) = 4x − x 2 . Using these results the Wronskian becomes W(t) = 3t 2 . The Green’s function for this differential equation deﬁned by (81) is ⎧ (t − t 2 )(4x − x 2 ) ⎪ ⎪ , 1≤t <x ⎨ 3t 2 G(x, t) = 2 2 ⎪ ⎪ ⎩ (4t − t )(x − x ) , x < t ≤ 2. 3t 2 To ﬁnd the function f (x) we must write the equation in the standard form where the coefﬁcient of y is 1, so y −

2 2 y + 2 y = 3, x x

showing that f (x) = 3. It now follows from (78), or from (80), that x 2 (t − t 2 ) (4t − t 2 )3 2 3dt + (x − x ) dt, y(x) = (4x − x 2 ) 3t 2 3t 2 1 x and so y(x) = x 2 (3 ln x − 2 − 4 ln 2) + 2x(1 + 2 ln 2). It is easily checked that this is the required solution, because y(1) = 0, y (2) = 0, and y(x) satisﬁes the differential equation. More information and examples relating to the material in Sections 6.1 to 6.6 can be found in any one of the references [3.3], [3.4], [3.15], and [3.16].

Summary

This section described the powerful method of variation of parameters that enables the general solution of a linear nonhom*ogeneous equation to be found from the linearly independent solutions (the basis functions) that enter into its complementary function. It takes automatic account of nonhom*ogeneous terms that contain one or more basis functions, and it enables particular integrals, and hence general solutions, to be found where the method of undetermined coefﬁcients fails. It was shown how the general solution obtained by the method of variation of parameters can be rewritten in terms of a Green’s function that characterizes all of the essential features of the differential equation without reference to the nonhom*ogeneous term. Knowledge of the Green’s function enables a hom*ogeneous boundary value problem to be solved for any given nonhom*ogeneous term on the right of the equation.

Section 6.7

The Reduction of Order Method

321

EXERCISES 6.6 In Exercises 1 through 13 ﬁnd the general solution. 1. 2. 3. 4. 5. 6. 7.

y + y − 2y = xe x . y − 5y + 6y = x 2 e3x . y + 5y + 6y = x 2 e−2x . y + 4y + 4y = xsin x. y − 2y + y = 2e x /x. y + 4y + 5y = e−2x sin x. y + 4y + 5y = xe−2x cos x.

y − 4y + 4y = e2x /x. y + 16y = x 2 e x . y + 16y = sec x. y + 3y + 2y = 3/(1 + e x ). 12. y + y = tan x. 13. y + y = sec2 x. 8. 9. 10. 11.

In Exercises 14 through 18 verify that the functions y1 (x) and y2 (x) are linearly independent solutions of hom*ogeneous form of the stated differential equation, and use them to ﬁnd a particular integral and a general solution of the given equation. 14. x 2 y − 4xy + 6y = 2x + ln x, where y1 (x) = x 2 and y2 (x) = x 3 . √ 15. x 2 y + 3xy − 3y = x, where y1 (x) = x and −3 y2 (x) = x . 16. x 2 y + 3xy − 8y = 2 ln x, where y1 (x) = x 2 and y2 (x) = x −4 . 17. (1 − x 2 )y − xy + 4y = x, where y1 (x) = 2x 2 − 1 and y2 (x) = x(x 2 − 1)1/2 . 18. (1 − x 2 )y − 2y = 1, where y1 (x) = 1 and y2 (x) = x + 2 ln(x − 1).

6.7

In Exercises 19 through 22 use result (76) to solve the stated initial value problem. 19. x 2 y − 3xy + 3y = 2x 2 ln x, with y(1) = 0 and y (1) = 0. 20. y + 5y + 6y = xe−2x , with y(1) = 0 and y (1) = 0. 21. y + y = 2 sec2 x, with y(0) = 0 and y (0) = 0. 22. y + 4y + 5y = x, with y(0) = 0 and y (0) = 0. In Exercises 23 through 26 ﬁnd the Green’s function for the given differential equation, subject to the associated hom*ogeneous boundary conditions. 23. 24. 25. 26.

y = f (x), with y(0) = 0 and y(1) = 0. y = f (x), with y(0) = 0 and y (1) = 0. y + λ2 y = f (x), with y(0) = 0 and y(1) = 0. y + λ2 y = f (x), with y(0) = 0 and y (1) = 0.

In Exercises 27 through 30 solve the given boundary value problem by means of a suitable Green’s function. 27. 28. 29. 30.

x 2 y + xy − y = x 2 e−x , with y(1) = 0 and y(2) = 0. x 2 y + 2xy − 2y = x 3 , with y(1) = 0 and y(2) = 0. x 2 y − 3xy + 3y = x 2 ln x, with y (1) = 0 and y(2) = 0. x 2 y − 3xy = x 2 , with y(1) = 0 and y(2) = 0.

Finding a Second Linearly Independent Solution from a Known Solution: The Reduction of Order Method

reduction of order method

In working with hom*ogeneous linear second order variable coefﬁcient equations, it can happen that one solution y1 (x) is known and it is necessary to ﬁnd a second linearly independent solution y2 (x). The method we now describe, called the reduction of order method, involves seeking a second solution of the form y2 (x) = u(x)y1 (x),

(84)

where the function u(x) is to be determined. Provided u(x) is not constant, the solutions y1 (x) and y2 (x) will be linearly independent, because y1 (x) and y2 (x) will not be proportional. The method will be developed using the hom*ogeneous second order variable coefﬁcient equation in the standard form d2 y dy + a(x) + b(x)y = 0. 2 dx dx

(85)

322

Chapter 6

Second and Higher Order Linear Differential Equations and Systems

Differentiating (84) gives du dy2 dy1 = y1 + u , dx dx dx

and

d2 y2 d2 y1 d2 u du dy1 + u = y + 2 . 1 dx 2 dx 2 dx dx dx 2

(86)

Substituting (84) and (86) into (85) and grouping terms gives y1 u + (2y1 + ay1 )u + (y1 + ay1 + by1 )u = 0.

(87)

As y1 (x) is a solution of (85), the factor y1 + ay1 + by1 multiplying u is zero, causing the equation to be reduced to 2y1 du d2 u . = − + a(x) dx 2 y1 dx

(88)

The substitution v = du/dx reduces (88) to the ﬁrst order variables separable equation 2y1 dv =− + a(x) v, dx y1

(89)

and it is from this reduction of order of the differential equation that the method derives its name. Separating variables and integrating (89) we ﬁnd that 2y1 dv =− + a(x) dx + ln C, v y1 or 2y1 + a(x) dx, ln(v/C) = − y1 so exp{− a(x)dx} v(x) = C . y12

(90)

As v = du/dx, integration of (90) gives exp{− a(x)dx} u(x) = C dx + D, y12 where D is another arbitrary constant. The arbitrary constant D can be set equal to zero, because when u(x) is substituted in (84) the constant D will simply scale the solution y1 (x). Furthermore, as any constant C that scales u(x) will scale each term in the differential equation, its value is immaterial, so for convenience we set C = 1. Thus, the expression for u(x) is given by u(x) =

exp{− a(x)dx} dx. y12

(91)

Section 6.7

The Reduction of Order Method

323

Using this expression for u(x) in (84) shows that the second linearly independent solution is y2 (x) = y1 (x)

exp{− a(x)dx} dx. y12

(92)

Thus, in terms of y1 (x), the general solution of (85) can be written y(x) = C1 y1 (x) + C2 y1 (x)

exp{− a(x)dx} dx, y12

(93)

where C1 and C2 are arbitrary constants. EXAMPLE 6.22

Given that y1 (x) = e−3x is a solution of y + 6y + 9y = 0, ﬁnd a second linearly independent solution, and hence ﬁnd the general solution. Solution The equation is in standard form with a(x) = 6 and y1 (x) = e−3x , so exp{− 6 dx} dx = dx = x, u(x) = exp(−6x) showing that y2 (x) = xe−3x . This result is to be expected, because the linear constant coefﬁcient equation corresponds to case (III) with μ = −3. The general solution is thus y(x) = (C1 + C2 x)e−3x .

EXAMPLE 6.23

Given that y1 (x) = x 2 is a solution of x 2 y − 3xy + 4y = 0 for x > 0, ﬁnd a second linearly independent solution, and hence ﬁnd the general solution. Solution Writing the equation in standard form (85) shows that a(x) = −3/x, so exp{− {−3/x)dx} exp{ln x 3 } dx = dx u(x) = 4 x x4 dx = ln x, = x from which it follows that the second linearly independent solution is y2 (x) = x 2 ln x

for x > 0.

The general solution is y(x) = x 2 (C1 + C2 ln x). The reduction of order method can lead to an expression for u(x) that cannot be integrated analytically. In such cases, in order to ﬁnd an analytical approximation to y2 (x), the integrand in (92) must be expanded in powers of x and integrated term by term. This approach will be used in Chapter 8 in connection with series solutions of second order variable coefﬁcient linear differential equations. See references [3.3] and [3.4].

324

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Second and Higher Order Linear Differential Equations and Systems

Summary

It is often the case that one solution of a linear second order variable-coefﬁcient hom*ogeneous variable-coefﬁcient equation can be found, often by inspection, though a second linearly independent solution cannot be found in similar fashion. This section showed how a known solution can be used to ﬁnd a second linearly independent solution. It was shown that the second linearly independent solution of the original second order equation is determined in terms of a ﬁrst order equation, and it is this feature that has caused this approach to be called the reduction of order method.

EXERCISES 6.7 In the following exercises, verify that y1 (x) is a solution of the given differential equation and use it to ﬁnd a second linearly independent solution. 1. 2. 3. 4. 5.

y − 5y − 14y = 0 with y1 (x) = e7x . y + 4y = 0, with y1 (x) = sin 2x. y + 4y + 5y = 0, with y1 (x) = e−x cos x. x 2 y + 3xy + y = 0, with y1 (x) = 1/x. x 2 y − xy + y = 0, with y1 (x) = x.

6.8

x 2 y + xy + y = 0, with y1 (x) = cos(ln x). xy + 2y + xy = 0, with y1 (x) = sin x/x. √ x 2 y + xy + (x 2 − 1/4)y = 0, with y1 (x) = sin x/ x. x 2 (ln x − 1)y − xy + y = 0, with y1 (x) = x. (1 − x cot x)y − xy + y = 0, with y1 (x) = x. (Hint: When ﬁnding −a(x)dx, make the substitution u = sin x − xcos, and in the ﬁnal integral make the substitution v = sin x/x.)

6. 7. 8. 9. 10.

Reduction to the Standard Form u + f (x)u = 0 When studying the properties of second order variable coefﬁcient equations it is sometimes advantageous to reduce the equation y + a(x)y + b(x)y = 0

the standard form of a linear variable coefﬁcient equation

(94)

to the standard form for a second order equation u + f (x)u = 0,

(95)

from which the ﬁrst derivative term u is missing. This reduction has many uses, one of which occurs in Section 8.6 when we derive the analytical form of Bessel functions of fractional order. To accomplish the reduction we seek a solution of (94) of the form y(x) = u(x)v(x),

(96)

and then try to choose v(x) so the ﬁrst derivative term in u vanishes. Differentiation of y = uv gives y = uv + u v and y = u v + 2u v + uv , so substitution into equation (94) gives u v + (2v + av)u + (v + av + bv)u = 0.

(97)

This result shows that the ﬁrst derivative term u will vanish if v(x) is such that 2v + av = 0,

(98)

Reduction to the Standard Form u + f (x)u = 0

Section 6.8

325

which has the solution

1 v(x) = exp − 2

a(x)dx .

(99)

From (98) we have v = −(1/2)av and v = −(1/2)(a v + av ), so eliminating v and v from (97) gives

1 1 2 u + − a (x) − a (x) + b(x) u = 0. 2 4

(100)

Because of its importance, we record this result in the form of a theorem. THEOREM 6.5

Reduction to the standard form u + f (x)u = 0 The substitution y(x) = u(x)v(x)

how to perform the reduction

, with 1 v(x) = exp − a(x)dx , 2

reduces the differential equation y + a(x)y + b(x)y = 0 to u + f (x)u = 0, where 1 1 f (x) = − a (x) − a 2 (x) + b(x). 2 4

EXAMPLE 6.24

Reduce the equation 4x 2 y + 4xy + (16x 2 − 1)y = 0 to standard form and hence ﬁnd the general solution. Solution Dividing the differential equation by 4x 2 to reduce it to the form given in (94) shows that a(x) = 1/x and b(x) = 4 − 1/(4x 2 ). Applying the result of Theorem 6.5 then shows that 1 (1/x)dx = x −1/2 and f (x) = 4. v(x) = exp − 2 The equation for u(x) is thus u + 4u = 0 with the general solution u(x) = C1 cos 2x + C2 sin 2x,

326

Chapter 6

Second and Higher Order Linear Differential Equations and Systems

but y(x) = u(x)v(x) = x −1/2 u(x), so the general solution is / / 1 1 cos 2x + C2 sin 2x. y(x) = C1 x x See references [3.3] and [3.4].

Summary

The study of the properties of some hom*ogeneous linear variable coefﬁcient equations of the form y + a(x)y + b(x)y = 0 is simpliﬁed if a change of variable can be found that reduces them to an equivalent form u + f (x)u = 0. This section showed how such a change of variable can be found, and used it to solve a variable coefﬁcient equation for which the two linearly independent functions entering into its general solution are by no means obvious.

EXERCISES 6.8 Reduce the equations in Exercises 1 and 2 to standard form, but do not attempt to ﬁnd their general solutions.

In Exercises 3 through 7 reduce the equation to standard form and hence ﬁnd its general solution.

1. x 2 y − xy + 9xy = 0. 2. x 2 y + xy + (x 2 − 9)y = 0.

3. y − 2y + y = 0. 4. y + 4y + 3y = 0. 5. y − 4y + 5y = 0.

6.9

6. x 2 y + xy + (36x 2 − 1) y = 0. 7. xy + 2y + xy = 0.

Systems of Ordinary Differential Equations: An Introduction Physical problems that give rise to ordinary differential equations often do so in the form of coupled systems of ﬁrst order linear differential equations, or systems of second order equations that are more easily treated if reduced to a ﬁrst order system. A very simple example of this type was encountered in Section 5.2(d), where two ﬁrst order equations were derived that linked the current i and the charge q ﬂowing in an R–L–C circuit at time t. In that case it was convenient to eliminate the current i to obtain a simple second order equation for the current q that could be solved by the methods of Section 6.1 and 6.2. Another example is the three-loop electric circuit shown in Fig. 6.12. In the circuit H is an inductance; C1 and C2 are capacitances; R1 , R2 , and R3 are resistors; V0 is an applied voltage; i 1 , i 2 , and i 3 are circulating currents; and q2 and q3 are the charges on the respective capacitances C1 and C2 . H

i1

q2

R1

q3

C1

i2

R2

C2

i3

V0

FIGURE 6.12 A three-loop electric circuit with an applied voltage.

R3

Section 6.9

an electrical problem leading to a ﬁrst order system

Systems of Ordinary Differential Equations: An Introduction

327

Applying Kirchhoff’s laws (see Section 5.2(d)) to each loop when the switch is closed leads to the three coupled equations di 1 + R1 (i 1 − i 2 ) = V0 dt R2 i 2 + R1 (i 2 − i 1 ) + q2 C1 = 0 R3 i 3 + R2 (i 3 − i 2 ) + q3 C2 = 0.

H

Using the results i 2 = dq2 /dt and i 3 = dq3 /dt reduces these equations to the coupled system of ﬁrst order equations dq2 di 1 + R1 i 1 − R1 = V0 dt dt dq2 − R1 i 1 + q2 C1 = 0 (R1 + R2 ) dt dq2 dq3 (R2 + R3 ) − R2 + q3 C2 = 0 dt dt H

for i 1 , q2 , and q3 . When these are solved the currents i 2 and i 3 follow from i 2 = dq2 /dt and i 3 = dq3 /dt. An example of a different kind is provided by the two degree of freedom vibration system with a damper in Fig. 6.11 that was shown to lead to the two coupled second order equations M

dx d2 x + F(t) = −k(x − y) − Kx − μ dt 2 dt

and m

d2 y = −k(y − x). dt 2

Instead of eliminating ﬁrst y and then x to obtain two fourth order differential equations for x and y, respectively, a different approach is to reduce these two equations to a system of four ﬁrst order equations by introducing ﬁrst order derivatives of x and y as new variables. To do this we set w = dx/dt and z = dy/dt, and as a result obtain the simultaneous system of four ﬁrst order equations dx =w dt dy =z dt dw M + (k + K)x − ky + μw = F(t) dt dz m + ky − kx = 0. dt

a general hom*ogeneous ﬁrst order system

This reduction of a higher order differential equation, or a coupled system of differential equations, to a ﬁrst order system is often useful. In Chapter 19 this approach is used when seeking the numerical solution of higher order differential equations by means of the Runge–Kutta method. This method provides accurate numerical solutions of ﬁrst order differential equations that may be either linear or nonlinear, and it can be adapted to solve higher order differential equations by reducing them to a coupled system of ﬁrst order equations.

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Second and Higher Order Linear Differential Equations and Systems

A general system of n ﬁrst order linear variable coefﬁcient differential equations involving the n dependent variables x1 (t), x2 (t), . . . , xn (t) that are functions of the independent variable t (in applications t is often the time), the variable coefﬁcients ai j (t), and the nonhom*ogeneous terms f1 (t), f2 (t), . . . , fn (t) has the form x1 (t) = a11 (t)x1 (t) + a12 (t)x2 (t) + · · · + a1n (t)xn (t) + f1 (t) x2 (t) = a21 (t)x1 (t) + a22 (t)x2 (t) + · · · + a2n (t)xn (t) + f2 (t) . . . . . . . . . . . xn (t) = an1 (t)x1 (t) + an2 x2 (t) + · · · + ann (t)xn (t) + fn (t).

(101)

System (101) is said to be hom*ogeneous when all the functions fi (t) are zero, and to be nonhom*ogeneous when at least one of them is nonzero. It is a linear system because it is linear in the functions x1 (t), x2 (t), . . . , xn (t) and their derivatives, and it is a variable coefﬁcient system whenever at least one of the coefﬁcients ai j (t) is a function of t; otherwise, it becomes a constant coefﬁcient system. An initial value problem for system (101) involves seeking a solution of (101) such that at t = t0 the variables x1 (t), x2 (t), . . . , xn (t) satisfy the initial conditions x1 (t0 ) = k1 , x2 (t0 ) = k2 , . . . , xn (t0 ) = kn ,

(102)

where k1 , k2 , . . . , kn are given constants. Matrix notation allows system (101) to be written in the concise form x (t) = A(t)x(t) + b(t), matrix notation for systems

(103)

or more simply as x = Ax + b, where a prime again indicates differentiation with respect to t, and the matrices in (103) are deﬁned as ⎡

x1 (t)

⎤

⎡

⎢ x2 (t) ⎥ ⎥ ⎢ ⎢ . ⎥ .. ⎥ , x(t) = ⎢ ⎥ ⎢ ⎢ . ⎥ ⎣ .. ⎦ ⎡

⎢ a (t) ⎢ 21 A(t) = ⎢ ⎢ ··· ⎣ ···

⎤

⎢ x (t) ⎥ ⎢ 2 ⎥ ⎢ . ⎥ . ⎥ x (t) = ⎢ ⎢ . ⎥, ⎢ . ⎥ ⎣ .. ⎦ xn (t)

xn (t) a11 (t)

x1 (t)

a12 (t)

··· ···

a22 (t) · · · ··· ··· ··· ··· an1 (t) an2 (t) · · ·

a1n (t)

⎤

· · · a2n (t) ⎥ ⎥ ··· ··· ⎥ ⎥, ··· ··· ⎦ · · · ann (t)

⎡

f1 (t)

⎤

⎢ f2 (t) ⎥ ⎥ ⎢ ⎢ . ⎥ .. ⎥ . b(t) = ⎢ ⎢ ⎥ ⎢ . ⎥ ⎣ .. ⎦ fn (t) (104)

The n × 1 vector x(t) is called the solution vector, the n × n matrix A(t) is called the coefﬁcient matrix, and the n × 1 vector b(t) is called the nonhom*ogeneous term of the system.

Section 6.9

Systems of Ordinary Differential Equations: An Introduction

329

System (103) becomes an initial value problem for the solution x(t) when at t = t0 the vector x(t) is required to satisfy the initial condition ⎤ k1 ⎢ k2 ⎥ ⎢ ⎥ ⎢ .. ⎥ ⎥ x(t0 ) = ⎢ ⎢ . ⎥, ⎢ . ⎥ ⎣ .. ⎦ kn ⎡

(105)

where x(t0 ) is the initial vector and k1 , k2 , . . . , kn are given constants. EXAMPLE 6.25

Express in matrix form the initial value problem x1 = 2x1 − x2 + 4 − t 2 x2 = −x1 + 2x2 + 1, with x1 (0) = 1

and

x2 (0) = 0.

Solution The system of equations can be written x (t) = A x(t) + b(t) where

x1 x(t) = , x2

2 A= −1

−1 , 2

and the initial vector is x(0) =

4 − t2 , b(t) = 1

and

1 . 0

As A is a constant matrix and b(t) = 0, this is a constant coefﬁcient nonhom*ogeneous system. solution by elimination: a ﬁrst approach

EXAMPLE 6.26

In what follows, our main objective will be to develop matrix methods for the solution of initial value problems for systems of ﬁrst order linear constant coefﬁcient differential equations. Before developing a matrix approach, we ﬁrst describe a simple way of solving system (102) when no more than three equations are involved. The method is straightforward and does not use matrix algebra, but it is often useful, and the examples that are solved show that systems can have oscillatory solutions even when no oscillatory term is present in the nonhom*ogeneous term. The approach used is called solution by elimination, because it involves eliminating all but one of the dependent variables in order to arrive at a single higher order equation for the remaining variable, say x1 (t). Once x1 (t) has been found, it is used in the system of equations to determine sequentially the remaining variables x2 (t), x3 (t), . . . , xn (t). The method will be illustrated by means of examples. Solve by elimination the initial value problem of Example 6.25. Solution The equations involved are x1 = 2x1 − x2 + 4 − t 2 x2 = −x1 + 2x2 + 1.

330

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The method will be to eliminate the dependent variable x2 between the two equations to obtain a single second order equation for x1 . After solving for x1 , the dependent variable x2 will be found by substituting for x1 in the ﬁrst equation. Thus, the solution of this system of two ﬁrst order equations will involve the solution of a single second order equation, and it will be through this equation that the two arbitrary constants expected to occur in the general solution of the system will enter. Differentiation of the ﬁrst equation belonging to the system gives x1 = 2x1 − x2 − 2t, and after substituting for x2 from the second equation in the system, this becomes x1 = 2x1 + x1 − 2x2 − 1 − 2t. Solving the ﬁrst equation belonging to the system for x2 gives x2 = 2x1 + 4 − t 2 − x1 , so using this result to eliminate x2 from the second order equation for x1 shows that x1 satisﬁes the equation x1 − 4x1 + 3x1 = 2t 2 − 2t − 9. Solving this equation by any method, say by the method of undetermined coefﬁcients, gives x1 (t) = C1 e3t + C2 et −

2 53 10 + t + t 2, 27 9 3

where C1 and C2 are arbitrary constants of integration. It now remains for us to ﬁnd x2 , and this is accomplished by substituting for x1 in the ﬁrst equation, which can be written in the form x2 = 2x1 + 4 − t 2 − x1 . As a result we ﬁnd that x2 (t) = −C1 e3t + C2 et −

1 28 8 + t + t 2, 27 9 3

so the general solution of the nonhom*ogeneous system is x1 (t) = C1 e3t + C2 et −

2 53 10 + t + t 2, 27 9 3

and x2 (t) = −C1 e3t + C2 et −

1 28 8 + t + t 2. 27 9 3

To solve the initial value problem, C1 and C2 must be chosen such that x1 (0) = 1 and x2 (0) = 0. Setting t = 0 in the general solution and using these initial conditions, we ﬁnd that C1 and C2 must satisfy the equations 1 = C1 + C2 −

53 27

and

0 = −C1 + C2 −

28 , 27

with the solution C1 = 26/27 and C2 = 2. Thus, the required solution of the initial value problem is x1 (t) =

26 3t 53 10 2 e + 2et − + t + t 2, 27 27 9 3

Section 6.9

Systems of Ordinary Differential Equations: An Introduction

331

and x2 (t) = −

1 26 3t 28 8 e + 2et − + t + t 2. 27 27 9 3

Unlike ﬁrst order linear differential equations whose complementary function can only contain an exponential function, systems of such equations can give rise to periodic solutions even when these do not occur in the nonhom*ogeneous term. This is illustrated by the next example. EXAMPLE 6.27

Solve by elimination the system of differential equations x1 + 2x1 − x2 = 1 + e−t ,

x2 + x1 + 2x2 = 3,

subject to the initial conditions x1 (0) = 5/2 and x2 (0) = −1/2. Solution Proceeding as in the previous example by differentiating the ﬁrst equation with respect to t and substituting for x2 from the second equation gives x1 + 2x1 + x1 + 2x2 − 3 = −e−t . Substituting for x2 from the ﬁrst equation belonging to the system then shows that x1 must satisfy the second order differential equation x1 + 4x1 + 5x1 = 5 + e−t , with the general solution x1 (t) = C1 e−2t cos t + C2 e−2t sin t + 1 + (1/2)e−t . Finally, solving the ﬁrst equation belonging to the system for x2 and substituting for x1 , we have x2 (t) = −C1 e−2t sin t + C2 e−2t cos t + 1 − (1/2)e−t . Thus, the general solution of the system is x1 (t) = C1 e−2t cos t + C2 e−2t sin t + 1 + (1/2)e−t and x2 (t) = −C1 e−2t sin t + C2 e−2t cos t + 1 − (1/2)e−t . To satisfy the initial conditions, the arbitrary constants C1 and C2 must be chosen such that x1 (0) = 5/2 and x2 (0) = −1/2. Inserting these conditions into the preceding general solution leads to the equations 5/2 = C1 + 3/2

and

− 1/2 = C2 + 1/2,

so that C1 = 1 and C2 = −1.

The solution of the initial value problem is then given by 1 x1 (t) = e−2t (cos t − sin t) + 1 + e−t 2 and 1 x2 (t) = −e−2t (sin t + cos t) + 1 − e−t . 2 This example illustrates the way in which oscillatory terms can enter into the solution through a higher order equation satisﬁed by one of the dependent variables, although they may not be present in the nonhom*ogeneous terms.

332

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Second and Higher Order Linear Differential Equations and Systems

As a ﬁnal example of the elimination method, we consider a hom*ogeneous system of three equations to show how this simple method becomes more difﬁcult when the number of equations is greater than two, and also to demonstrate how care must then be taken with the determination of the arbitrary constants of integration. EXAMPLE 6.28

Find the general solution of the system of equations x1 = x2 + x3 ,

x2 = x1 + x3 ,

and

x3 = x1 + x2 .

Solution Differentiating the ﬁrst equation with respect to t and substituting for x2 and x3 from the second and third equations and using the ﬁrst equation shows that x1 satisﬁes the second order equation x1 − x1 − 2x1 = 0, with the solution x1 (t) = C1 e−t + C2 e2t , where C1 and C2 are arbitrary constants of integration. Substituting for x1 (t) in the second equation belonging to the system, differentiating the result with respect to t, and then substituting for x3 from the third equation belonging to the system shows that x2 satisﬁes the nonhom*ogeneous second order equation x2 − x2 = 3C2 e2t , with the solution x2 (t) = C2 e2t + C3 e−t + C4 et . how to resolve the problem of the arbitrary constants

It now appears that an anomalous situation has arisen, because when seeking a solution of a system of three equations, four arbitrary integration constants have appeared. This apparent inconsistency will be resolved shortly, so for the moment we continue working with this form of solution for x2 (t). Subtracting the ﬁrst two equations belonging to the system gives x1 − x2 = x2 − x1 . After substituting for x1 (t) and x2 (t) in this equation and cancelling terms, this is seen to reduce to −C4 et = C4 et . As et = 0 for any t, it follows that C4 = 0, and the apparent inconsistency has been resolved because now only the three arbitrary constants C1 , C2 , and C3 appear in the general solutions for x1 (t) and x2 (t). In fact, no further integration is required to determine x3 (t), because substituting x1 (t) and x2 (t) into the ﬁrst equation belonging to the system and solving for x3 (t) gives x3 (t) = −(C1 + C3 )e−t + C2 e2t . Thus, the general solution of the system is given by x1 (t) = C1 e−t + C2 e2t x2 (t) = C2 e2t + C3 e−t x3 (t) = −(C1 + C3 )e−t + C2 e2t .

Section 6.10

Summary

A Matrix Approach to Linear Systems of Differential Equations

333

This section has shown how a system of ﬁrst order equations can arise from a typical electrical problem. A matrix notation for systems was introduced, and an elementary method for solving small systems of equations using elimination was described that avoided the use of matrices. This method was seen to lead to more arbitrary constants in the general solution than the number of equations involved, but a simple argument resolved this difﬁculty.

EXERCISES 6.9 Solve Exercises 1 through 6 by elimination. 1. 2x1 = x1 − x2 , 2x2 = 3x1 + 5x2 . 2. x1 = −10x1 − 18x2 , x2 = 6x1 + 11x2 . 3. x1 = 2x1 − 12x2 , 2x2 = 3x1 − 8x2 , with x1 (0) = 0 and x2 (0) = 1. 4. x1 = 3x2 + t, x2 = 2x1 + x2 − 3, with x1 (0) = 1 and x2 (0) = 1.

6.10

5. x1 = 2x2 + 4x3 + 3e−t , x2 = x1 + x2 − 2x3 + 1, x3 = −2x1 + 5x3 , with x1 (0) = 1, x2 (0) = 0, and x3 (0) = 0. 6. x1 = − 2x1 + 2x2 + 2x3 + 3et , x2 = −x1 − x2 − 2x3 + 1, x3 = x1 + 2x2 + 3x3 − 3, with x1 (0) = 1, x2 (0) = 1, and x3 (0) = 0.

A Matrix Approach to Linear Systems of Differential Equations We will now consider some general properties of the variable coefﬁcient system x (t) = A(t)x(t) + b(t),

a solution in matrix form

(106)

where the matrices x(t), A(t), and b(t) are as deﬁned in (103). A solution of system (106) is a vector x(t) with elements x1 (t), x2 (t), . . . , xn (t) that when substituted in system (106) satisﬁes it identically. Thus, a solution of the initial value problem in Example 6.26 is the vector ⎤ ⎡ 2 2 53 10 26 3t t ⎢ 27 e + 2e − 27 + 9 t + 3 t ⎥ x (t) ⎥. x(t) = 1 =⎢ ⎦ ⎣ 26 x2 (t) 8 1 28 + t + t2 − e3t + 2et − 27 27 9 3

Structure of Solutions of hom*ogeneous Systems (a) Linear superposition of solutions The properties of linear hom*ogeneous systems of differential equations are similar to those of a single linear higher order hom*ogeneous differential equation. A most important property that is common to both is that a linear superposition of solutions of a linear hom*ogeneous system of variable-coefﬁcient ﬁrst order differential equations is itself a solution of the hom*ogeneous system. This result is easily proved. Let Ψ1 (t), Ψ2 (t), . . . , Ψm(t) be any m solutions of the linear hom*ogeneous system x (t) = A(t)x(t), and taking C1 , C2 , . . . , Cm to be

334

Chapter 6

Second and Higher Order Linear Differential Equations and Systems

any set of m arbitrary constants form the vector Ψ(t) = C1 Ψ1 (t) + C2 Ψ2 (t) + · · · + CmΨm(t). Then Ψ(t) = (C1 Ψ1 + C 2 Ψ2 + · · · + CmΨm) = C1 Ψ1 + C2 Ψ2 + · · · + CmΨm, so the system Ψ(t) = A(t)Ψ(t) becomes C1 Ψ1 + C2 Ψ2 + · · · + CmΨm = A(C1 Ψ1 + C2 Ψ2 + · · · + CmΨm) = C1 AΨ1 + C2 AΨ2 + · · · + CmAΨm. Consequently, as Ψi (t) = A(t)Ψi (t), we have shown that Ψ(t) is also a solution of the hom*ogeneous system, and the result is proved.

(b) Existence and uniqueness We now state without proof the fundamental theorem on the existence and uniqueness of the solution to the initial value problem for a system of linear variable coefﬁcient ﬁrst order differential equations. (See, for example, references [3.4] and [3.5].) THEOREM 6.5

Existence and uniqueness of solutions of linear systems Let the vector x(t) with the n elements xi (t) (i = 1, 2, . . . , n) be the solution of the nonhom*ogeneous variable coefﬁcient system of ﬁrst order linear differential equations x (t) = A(t)x(t) + b(t), where the functions ai j (t) (i, j = 1, 2, . . . , n) forming the elements of A(t) and the elements fi (t) (i = 1, 2, . . . , n) forming the elements of the vector b(t) are continuous functions in some interval a < t < b. Furthermore, let the elements of x(t) satisfy the initial conditions xi (t0 ) = ki (i = 1, 2, . . . , n), where the ki are given constants and t0 is any point such that a < t0 < b. Then the solution of the initial value problem exists and is unique for all t such that a < t < b.

(c) Fundamental matrix and a test for linear independence of solutions As with single higher order linear differential equations, the general solution of a hom*ogeneous system will be constructed by the forming a linear combination of all possible linearly independent solutions of the system. For this reason it is necessary to know how many linearly independent solutions belong to a given hom*ogeneous system, and how to test the linear independence of a set of solutions. The answers to these two fundamental questions are provided by the next two theorems, the results of which should be remembered. As the proofs of these theorems may be omitted at a ﬁrst reading, they are given at the end of this section. THEOREM 6.6

Linearly independent solutions of a hom*ogeneous system Let the elements ai j (t) (i, j = 1, 2, . . . , n) of the n × n matrix A(t) be continuous in the interval a < t < b. Then the linear hom*ogeneous system x = A(t)x

Section 6.10

A Matrix Approach to Linear Systems of Differential Equations

335

possesses n linearly independent solutions Ψ1 (t), Ψ2 (t), . . . , Ψn (t), and every solution of the system is expressible as a linear combination of the form Ψ(t) = C1 Ψ1 (t) + C2 Ψ2 (t) + · · · + Cn Ψn (t) for some choice of the constants C1 , C2 , . . . , Cn .

a fundamental matrix

An n × n matrix Φ(t) whose columns are any n linearly independent solution vectors of the hom*ogeneous system x = A(t)x is called a fundamental matrix for the system, and Theorem 6.6 shows that the general solution of the system can always be written in the form x(t) = Φ(t)C, where C is an n-element column vector with arbitrary constant elements C1 , C2 , . . . , Cn . Clearly, a fundamental matrix is not unique, because any of its columns may be replaced by a linear combination of its columns and the result will remain a fundamental matrix. This follows because if the columns of a determinant are replaced by linear combinations of its columns, the value of the determinant is unaltered, so if initially the determinant was nonsingular, it will remain nonsingular.

THEOREM 6.7 a determinant test for linear independence of solution vectors

Determinant test for the linear independence of solution vectors Let the column (m) (m) (m) vectors Ψm(t) (m = 1, 2, . . . , n), whose elements 1 (t), 2 (t), . . . , n (t), be n solutions of the hom*ogeneous system x = A(t)x, in which the elements ai j (t) (i, j = 1, 2, . . . , n) of the n × n matrix A(t) are continuous functions for a < t < b. Then the n vectors Ψm(t) (m = 1, 2, . . . , n) are linearly independent solutions for a < t < b if, for some t0 in the interval, the determinant (1) (t0 ) 1 (1) 2 (t0 ) (t0 ) = . . . (1) (t ) n 0

(2)

1 (t0 ) (2)

2 (t0 ) .. . (2)

n (t0 )

(n) · · · 1 (t0 ) (n) · · · 2 (t0 ) = 0, .. .. . . (n) · · · n (t0 )

and the vectors Ψm(t) (m = 1, 2, . . . , n) form a basis for solutions of the system. Furthermore, if (t0 ) = 0, then (t) = 0, for all t in a < t < b. EXAMPLE 6.29

Find a set of linearly independent solution vectors for the system x1 = x2 + x3 ,

x2 = x1 + x3

and construct a fundamental matrix.

and

x3 = x1 + x2 ,

336

Chapter 6

Second and Higher Order Linear Differential Equations and Systems

Solution In Example 6.28 the solution of this system was shown to be x1 (t) = C1 e−t + C2 e2t x2 (t) = C2 e2t + C3 e−t x3 (t) = −(C1 + C3 )e−t + C2 e2t . Writing this solution in the form x(t) = Φ(t)C determined by Theorem 6.7, we obtain ⎤⎡ ⎤ ⎡ ⎤ ⎡ −t e e2t 0 C1 x1 (t) ⎣ x2 (t) ⎦ = ⎣ 0 e2t e−t ⎦ ⎣ C2 ⎦ . C3 x3 (t) −e−t e2t −e−t Thus, a fundamental matrix for the system, that is, a matrix whose columns are linearly independent solution vectors of the system, can be taken to be ⎤ ⎡ −t e e2t 0 e2t e−t ⎦ , (t) = ⎣ 0 −e−t e2t −e−t provided the solution vectors corresponding to the columns of this matrix are linearly independent. The test for this is provided by Theorem 6.7, and as it is easily shown that det Φ(t) = −3. So it follows from Theorem 6.7 that the three column vectors ⎡ −t ⎤ ⎡ 2t ⎤ ⎡ ⎤ e e 0 1 (t) = ⎣ 0 ⎦ , 2 (t) = ⎣ e2t ⎦ , and 3 (t) = ⎣ e−t ⎦ −e−t −e−t e2t are, indeed, linearly independent solution vectors.

Proofs of Theorems 6.6 and 6.7 Proof of Theorem 6.6 Consider any set of n linearly independent column vectors v1 , v2 , . . . , vn , each with constant elements, and for some t0 in a < t0 < b use them as initial conditions in the set of initial value problems x = A(t)x

with

x(t0 ) = vm,

for m = 1, 2, . . . , n.

By the existence and uniqueness theorem, each of these initial value problems has a unique solution Ψm(t) deﬁned on a < t < b. To establish the linear independence of these solutions on a < t < b, we suppose, if possible, that constants C1 , C2 , . . . , Cn can be found such that C1 Ψ1 (t) + C2 Ψ2 (t) + · · · + Cn Ψn (t) = 0 for every t in the interval. Setting t = t0 , this result becomes C1 v1 + C2 v2 + · · · + Cn vn = 0, but as the vm are linearly independent, this can only be true if C1 = C2 = · · · = Cn = 0, so we have proved that the solutions Ψm(t) (m = 1, 2, . . . , n) are linearly independent over the interval. We must now show that for some constants C1 , C2 , . . . , Cn , not all of which are zero, every solution of the system x = A(t)x can be written Ψ(t) = C1 Ψ1 (t) + C2 Ψ2 (t) + · · · + Cn Ψn (t), and in particular this result must be true when t = t0 .

Section 6.10

A Matrix Approach to Linear Systems of Differential Equations

Deﬁne a matrix Φ(t) whose columns are Ψ1 (t), Ψ2 (t), . . . , Ψn (t), where the elements (m) n (t), for m = 1, 2, . . . , n, so ⎡ (1) (2) 1 (t) 1 (t) ⎢ (1) ⎢2 (t) 2(2) (t) ⎢ Φ(t) = ⎢ . .. ⎢ . . ⎣ . (1)

n (t)

337

the n linearly independent vectors (m) (m) of Ψm(t) are 1 (t), 2 (t), . . . ,

(2)

n (t)

(n)

· · · 1 (t)

⎤

⎥ (n) · · · 2 (t)⎥ ⎥ . .. .. ⎥ ⎥ . . ⎦ (n)

· · · n (t)

Now set t = t0 and consider the matrix equation Φ(t0 )C = Ψ(t0 ), where C is a column vector with the n elements C1 , C2 , . . . , Cn . Expanding the expression on the left and grouping terms shows that Φ(t0 )C = C1 Ψ1 (t0 ) + C2 Ψ2 (t0 ) + · · · + Cn Ψn (t0 ), and so C1 Ψ1 (t0 ) + C2 Ψ2 (t0 ) + · · · + Cn Ψn (t0 ) = Ψ(t0 ). The existence of a unique set of constants C1 , C2 , . . . , Cn , not all of which are zero, follows from the fact that detΦ(t0 ) = 0, because of the linear independence of its columns. As Ψ(t) and C1 Ψ1 (t) + C2 Ψ2 (t) + · · · + Cn Ψn (t) are both solutions of the same initial value problem x = A(t)x

with

x(t0 ) = Ψ(t0 ),

the existence and uniqueness theorem shows that Ψ(t) = C1 Ψ1 (t) + C2 Ψ2 (t)1 + · · · + Cn Ψn (t) for all t such that a < t < b, and the theorem is proved. Proof of Theorem 6.7 The proof is in two parts. First we show that if the vectors are linearly independent, then detΦ(t) = 0 for all t in the interval. Then we assume the converse, namely that Φ(t) is a fundamental matrix, and show this implies detΦ(t) = 0 for all t in the interval. The fact that every solution of the system can be expressed as a linear combination of the n linearly independent solutions will then follow from Theorem 6.6. If Φ(t) is a matrix whose columns are solution vectors and det Φ(t) = 0, then the vectors are linearly independent. To show this, suppose constants C1 , C2 , . . . , Cn can be found such that C1 Ψ1 (t) + C2 Ψ2 (t) + · · · + Cn Ψn (t) = 0 for all t in the interval a < t < b. Then for any t0 in the interval, setting t = t0 the equation can be written Φ(t0 )C = 0, where C is a column matrix with elements C1 , C2 , . . . , Cn . As det Φ(t0 ) = 0, the only solution of this hom*ogeneous system of algebraic equations is C1 = C2 = · · · = Cn = 0, so the column vectors must be linearly independent for all t in the interval. We must now consider the converse situation and suppose that Φ(t) is a fundamental matrix. Then, if Ψ(t) is a solution of the system, from the deﬁnition of

338

Chapter 6

Second and Higher Order Linear Differential Equations and Systems

a fundamental solution a unique constant vector C can always be found such that Ψ(t) = Φ(t)C for all t in the interval. To ﬁnd C we need only set t = t0 in this last result, because as det Φ(t0 ) = 0 the hom*ogeneous system of algebraic equations must have a unique solution. The result is true for each t0 in the interval, and so it follows that det Φ(t) = 0 over the interval a < t < b. As the set of n vectors Ψm(t) (m = 1, 2, . . . , n) is linearly independent, it follows from Theorem 6.6 that every solution of the system is expressible as a linear combination of these vectors, so they form a basis for solutions of the system. For more information about the material in Sections 6.9 and 6.10 see, for example, references [3.3], [3.4], and [3.16].

Summary

The linear superposition of matrix vector solutions was shown to be permissible, and the concept of a fundamental matrix was introduced, the columns of which contained n linearly independent solution vectors of a linear system of n ﬁrst order equations. The fundamental matrix had the property that the general solution of the system could be expressed in terms of its product with a column vector containing n arbitrary constants. A determinant test was then developed that established when a set of n solution vectors was suitable to form the columns of a fundamental matrix—that is, to form a basis for the solution set of the system.

EXERCISES 6.10 In Exercises 1 through 6, verify by substitution that the functions x1 (t) and x2 (t) are solutions of the given system of equations. By writing the solution in matrix form, ﬁnd a fundamental matrix for the system and verify that its columns are linearly independent. 1. x1 = x1 + x2 , x2 = −x1 + x2 ; x1 (t) = et (C1 cos t + C2 sin t), x2 (t) = et (C2 cos t − C1 sin t). 2. x1 = 2x1 + x2 , x2 = −2x1 ; x1 (t) = et (C1 cos t + C2 sin t), x2 (t) = (C2 − C1 )et cos t − (C1 + C2 )et sin t.

6.11

3. x1 = x1 − 2x2 , x2 = x1 − x2 ; x1 (t) = C1 cos t + C2 sin t, x2 (t) = (1/2)(C1 − C2 ) cos t + (1/2)(C1 + C2 ) sin t. 4. x1 = −3x1 − x2 , x2 = 3x1 + x2 ; x1 (t) = C1 + C2 e−2t , x2 (t) = −3C1 − C2 e−2t . 5. 2x1 = 2x1 − x2 , x2 = x1 + 2x2 ; x1 (t) = C1 e3t/2 cos t/2 + C2 e3t/2 sin t/2, x2 (t) = −(C1 + C2 )e3t/2 cos t/2 + (C1 − C2 )e3t/2 sin t/2. 6. 2x1 = −x1 + x2 , x2 = x1 − x2 ; x1 (t) = C1 + C2 e−3t/2 , x2 (t) = C1 − 2C2 e−3t/2 .

Nonhom*ogeneous Systems A nonhom*ogeneous variable coefﬁcient system of ﬁrst order linear differential equations can be written x = A(t)x + b(t).

(107)

Its general solution can be expressed as the sum of the general solution of the associated hom*ogeneous system x = Ax that will contain the arbitrary constants, and a particular solution free from arbitrary constants that can be taken to be any solution of the nonhom*ogeneous equation x = Ax + b. This result is recorded and proved in the next theorem.

Section 6.11

Nonhom*ogeneous Systems

339

The Structure of the Solution THEOREM 6.8 nonhom*ogeneous system and the structure of the solution

Structure of the solution of x = A(t)x + b(t) Let Φ(t) be a fundamental matrix for the hom*ogeneous linear ﬁrst order system x = A(t)x, and let P(t) be any solution of the nonhom*ogeneous system x = A(t)x + b(t). Then the general solution of the nonhom*ogeneous system is x(t) = Φ(t)C + P(t), with C an n-element column matrix with arbitrary constants C1 , C2 , . . . , Cn as elements. Proof The result is almost immediate and follows by substitution. Setting x = Φ(t)C + P(t), we have x = Φ (t)C + P (t), so after substitution into the system of differential equations we ﬁnd that Φ (t)C + P (t) = AΦ(t)C + AP(t) + b(t). However, Φ (t)C = A(t)Φ(t)C, and by deﬁnition P(t) is any solution of x = A(t)x + b(t), so P (t) = AP(t) + b(t), showing that substitution of the general solution into the equation leads to an identity, so the theorem is proved. It is important to recognize that solutions of nonhom*ogeneous linear systems do not have the linear superposition property of solutions of hom*ogeneous systems, and so they do not form a vector space.

EXAMPLE 6.30

Find the solution of the initial value problem for the nonhom*ogeneous system of equations x1 + 2x1 + 4x2 = 1 + 2t,

x2 + x1 − x2 = 3t

subject to the initial conditions x1 (0) = 56/9 and x2 (0) = −13/9, and verify the results of Theorem 6.8. Solution Using the elimination method, the solution of the system can be shown to be x1 (t) = 2/9 + (7/3)t + 2e2t + 4e−3t

x2 (t) = −4/9 − (2/3)t − 2e2t + e−3t ,

and

and in matrix form this becomes 2t e 2/9 + (7/3)t 4e−3t x1 (t) 2 = + . −4/9 − (2/3)t x2 (t) 1 −e2t e−3t $ %& ' $ %& ' %& ' $%&' $ x(t)

Φ(t)

C

P(t)

Inspection of this form of solution identiﬁes the fundamental matrix Φ(t) containing exponentials, a column vector C with elements C1 = 2 and C2 = 1, and a particular solution P(t) of the nonhom*ogeneous system represented by the last matrix vector. It is easily checked that the vector P(t), which contains no constants, is a particular solution of the system.

Matrix Methods of Solution We now describe a number of matrix methods for the solution of both hom*ogeneous and nonhom*ogeneous constant coefﬁcient systems of linear ﬁrst order differential equations.

340

Chapter 6

Second and Higher Order Linear Differential Equations and Systems

(a) Solution by diagonalization when A has real eigenvalues solution by diagonalization when eigenvalues are real

Having already illustrated the elementary elimination method for solving a small system of equations, we now describe the powerful and systematic matrix diagonalization method that can be used with systems involving any number of differential equations. Consider a general nonhom*ogeneous constant coefﬁcient system x = Ax + b(t)

(108)

where A is a constant coefﬁcient n × n matrix with real eigenvalues and n linearly independent eigenvectors. The approach we will use is to try to ﬁnd a transformation of the dependent variables x1 , x2 , . . . , xn forming the elements of vector x, which creates a new set of variables u1 , u2 , . . . , un that form the elements of a vector u with the property that system (108) can be written as u = Du + h,

(109)

where D is a diagonal matrix and h is an n-element column vector with elements that depend on the elements in the nonhom*ogeneous term b(t). If such a transformation can be found, the equations in the system will have been uncoupled, because each equation for u1 , u2 , . . . , un can then be solved individually. When u1 , u2 , . . . , un are known, reversing the transformation will give the solution x1 (t), x2 (t), . . . , xn (t) of system (108). Such a transformation has already been provided by Theorem 4.6. It was shown there that if a matrix P is constructed with the n eigenvectors of A as its columns, then P−1 AP = D, where D is a diagonal matrix with the eigenvalues of A arranged along its leading diagonal in the same order as the corresponding eigenvectors appear in P. Adopting this approach, setting x = Pu,

(110)

Pu = APu + b(t),

(111)

and substituting in (108) gives

where when differentiating x(t) use has been made of the fact that P is a constant matrix. The linear independence of the n eigenvectors forming the columns of P ensures the existence of the inverse matrix P−1 , so premultiplying (111) by P−1 gives u = P−1 APu + P−1 b(t), but P−1 AP = D, so system (108) has been transformed into the uncoupled system u = Du + P−1 b(t).

(112)

The required solution vector x(t) follows from the result x(t) = Pu. Before giving an example, it is necessary to consider whether systems exist for which this method will fail. The answer to this question is not difﬁcult to ﬁnd,

Section 6.11

Nonhom*ogeneous Systems

341

because the method depends for its success on the diagonalization of A, and this in turn requires that A have n linearly independent eigenvectors. Consequently, we see that the method will fail if the n × n matrix A has fewer than n linearly independent eigenvectors, because then the diagonalizing matrix P cannot be constructed. This situation occurs when A has multiple eigenvalues but an eigenvalue with multiplicity r has associated with it fewer than r linearly independent eigenvectors. A typical matrix with this property is ⎡ ⎤ 1 5 7 1 1⎦. A = ⎣0 0 −1 −1 In this case the eigenvalue λ = 1 occurs with multiplicity 1 and the eigenvalue λ = 0 with multiplicity 2, but the matrix has only the two linearly independent eigenvectors ⎤ ⎡ ⎤ ⎡ 1 −2 (λ = 1), x1 = ⎣ 0 ⎦ and (λ = 0, twice), the single eigenvector x2 = ⎣ −1 ⎦ . 1 0 EXAMPLE 6.31

Use diagonalization to solve the nonhom*ogeneous system x1 (t) + 2x1 + 4x2 = 2t − 1,

x2 (t) + x1 − x2 = sin t.

Solution The system can be written in the form x = Ax + b(t) with −2 −4 2t − 1 x1 x= , A= , and b(t) = . x2 −1 1 sin t Matrix A has the two eigenvalues and eigenvectors −1 4 λ1 = 2, x1 = , λ2 = −3, x2 = . 1 1 The diagonalizing matrix is thus −1 4 P= , so 1 1

P−1 =

−1/5 4/5 , 1/5 1/5

and from the order in which the eigenvectors have been entered as the columns of P, it follows without further computation that 2 0 D = P−1 AP = . 0 −3 We have

−1

P b(t) =

1/5 − (2/5)t + (4/5) sin t −1/5 + (2/5)t + (1/5) sin t

,

so, corresponding to (112) the transformed system becomes u1 1/5 − (2/5)t + (4/5) sin t 2 0 u1 = + . u2 0 −3 −1/5 + (2/5)t + (1/5) sin t u2 In component form these are seen to be the uncoupled equations u1 = 2u1 + 1/5 − (2/5)t + (4/5) sin t

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Chapter 6

Second and Higher Order Linear Differential Equations and Systems

and u2 = −3u2 − 1/5 + (2/5)t + (1/5) sin t. The solution of the uncoupled equations is easily shown to be u1 (t) = C1 e2t − (4/25) cos t − (8/25) sin t + (1/5)t u2 (t) = C2 e−3t − (1/50) cos t + (3/50) sin t + (2/15)t − 1/9, where C1 and C2 are arbitrary constants. If we use these as the elements of the column vector U, the required solution is given by x(t) = Pu, and so C1 e2t − (4/25) cos t − (8/25) sin t + (1/5)t −1 4 x1 (t) . = 1 1 C2 e−3t − (1/50) cos t + (3/50) sin t + (2/15)t − 1/9 x2 (t) In component form the solution becomes x1 (t) = −4/9 + (1/3)t + (2/25) cos t + (14/25) sin t − C1 e2t + 4C2 e−3t x2 (t) = −1/9 + (1/3)t − (9/50) cos t − (13/50) sin t + C1 e2t + C2 e−3t .

(b) Solution by diagonalization when A has complex eigenvalues solution by diagonalization when eigenvalues are complex

EXAMPLE 6.32

When the diagonalization method is used to solve a system in which A has pairs of complex conjugate eigenvalues, the approach only differs from the case involving real eigenvalues in that the arbitrary constants introduced at the integration stage are complex. When A has real coefﬁcients and complex eigenvalues exist, they must do so in complex conjugate pairs, so after integrating an equation corresponding to the complex eigenvalue λ = α + iβ, we must introduce a complex integration constant C1 + iC2 . Then, to make the solution real, when integrating the equation corresponding to the complex conjugate eigenvalue λ¯ = α − iβ the complex conjugate integration constant C1 − iC2 must be introduced. Use diagonalization to solve the system of nonhom*ogeneous equations x1 (t) = x1 + 2x2 + x3 + 1 x2 (t) = x2 + x3 + t x3 (t) = 2x1 + x3 + 2t. Solution The matrix A is

⎡

1 A = ⎣0 2

2 1 0

⎤ 1 1⎦, 1

and its eigenvalues and eigenvectors are ⎤ ⎡ ⎤ ⎡ 2 −i 1 ⎦, λ1 = 3, x1 = ⎣ 1 ⎦ , λ2 = i, x2 = ⎣ 2 −1 + i

⎤ i x3 = ⎣ 1 ⎦ . −1 − i ⎡

λ3 = −i,

Section 6.11

Nonhom*ogeneous Systems

343

The diagonalizing matrix ⎡ ⎤ 2 −i i 1 1 ⎦ and P = ⎣1 2 −1 + i −1 − i ⎡ ⎤ 1/5 1/5 1/5 ⎢ ⎥ P−1 = ⎣ −1/10 + 3i/10 2/5 − i/5 −1/10 − i/5 ⎦ . −1/10 − 3i/10 2/5 + i/5 −1/10 + i/5 The order in which the columns of P are arranged shows without further computation that when diagonalized, A will become the matrix ⎡ ⎤ 3 0 0 0⎦. D = ⎣0 i 0 0 −i This is because D can be written down immediately without the need to calculate D = P−1 AP, because the order in which the eigenvalues are arranged along the leading diagonal of D is the order in which their corresponding eigenvectors form the columns of A. If we write the system as x = Ax + b(t), with

⎤ x1 (t) x = ⎣ x2 (t) ⎦ x3 (t)

⎡

⎡

and

⎤ 1 b(t) = ⎣ t ⎦ , 2t

and set x(t) = Pu, the system becomes Pu = APu + b(t), so u = P−1 APu + P−1 b(t)

or

u = Du + P−1 b(t).

A simple calculation then gives ⎡

⎤ 1/5 + 3t/5 P−1 b(t) = ⎣ −1/10 + 3i/10 + t/5 − 3it/5 ⎦ , −1/10 − 3i/10 + t/5 + 3it/5

so writing u = Du + P−1 b(t) in component form shows that the uncoupled equations become u1 (t) = 3u1 + 1/5 + 3t/5 u2 (t) = iu2 − 1/10 + 3i/10 + t/5 − 3it/5 u3 (t) = −iu3 − 1/10 − 3i/10 + t/5 + 3it/5. Solving the ﬁrst equation involves no complex numbers and so gives rise to the solution u1 (t) = −2/15 − t/5 + C1 e3t .

344

Chapter 6

Second and Higher Order Linear Differential Equations and Systems

However, the other two equations are complex, so remembering that the complex integration constant in the third equation must be the complex conjugate of the one in the second equation leads to the results u2 (t) = 3t/5 − 1/10 − 7i/10 + it/5 + (C2 + iC3 )(cos t + isin t) u3 (t) = 3t/5 − 1/10 + 7i/10 − it/5 + (C2 − iC3 )(cos t − isin t). Combining these results gives ⎡

⎤ −2/15 − t/5 + C1 e3t u = ⎣ 3t/5 − 1/10 − 7i/10 + it/5 + (C2 + iC3 )(cos t + isin t) ⎦ , 3t/5 − 1/10 + 7i/10 − it/5 + (C2 − iC3 )(cos t − isin t)

so ﬁnally, using x(t) = Pu, we arrive at the required solution x1 (t) = −5/3 + 2C1 e3t + 2C2 sin t + 2C3 cos t x2 (t) = −1/3 + t + C1 e3t + 2C2 cos t − 2C3 sin t x3 (t) = 4/3 − 2t + 2C1 e3t + (2C3 − 2C2 ) sin t − (2C2 + 2C3 ) cos t.

(c) Solution of a hom*ogeneous system by the matrix exponential solution using the matrix exponential

For the sake of completeness, we now show how, when A is diagonalizable, the solution of the hom*ogeneous constant coefﬁcient system x = Ax can be solved by means of the matrix exponential, and we indicate how the method can be extended to enable the solution to be found when A is not diagonalizable. As the Laplace transform method to be described later deals with the solution of initial value problems for linear equations automatically and is simpler to use, the ideas involved will only be outlined. Nevertheless, the matrix exponential is both useful and important when working with systems of equations, so it is necessary to make some mention of it here. We consider the initial value problem x = Ax

subject to the initial condition

x(t0 ) = v,

(113)

where A is an n × n constant matrix and v is an arbitrary n-element constant column vector. Then the existence and uniqueness theorem guarantees that a solution certainly exists in some open interval containing t0 . If we deﬁne a vector x(t) = etA v, and set etA = In + tA +

t2 2 t3 3 A + A + ..., 2! 3!

then dx/dt = d(etA )/dt v = AetA v = Ax,

Section 6.11

Nonhom*ogeneous Systems

345

so the solution of the initial value problem in (113) can be represented in the form x(t) = etA v.

(114)

We saw in Section 4.5 that etA is easily computed when A is diagonalizable, but before using this result we ﬁrst review the ideas that are involved. If A is diagonalizable to a matrix D, a matrix P exists such that A = PDP−1 , where the columns of P are the eigenvectors of A, and the elements of D are the corresponding eigenvalues of A. Thus, A2 = PDP−1 PDP−1 = PD2 P−1 , and by extending this argument we have the general result Am = PDmP−1 , for m = 1, 2 . . . . Using this property in the deﬁnition of the matrix exponential etA given above allows it to be written etA = P[In + tD Consequently, if

t2 2 t3 3 D + D + . . .]P−1 . 2! 3!

⎡

λ1 0 ⎢ 0 λ2 D=⎢ ⎣ . . . . 0 0 then

⎡

j

λ1

0 0 . . 0

⎤ ... 0 ... 0⎥ ⎥, . . . . ⎦ · · · λn ...

⎤

⎢ ⎥ j ⎢ 0 λ2 0 . . . 0 ⎥ Dj = ⎢ ⎥, ⎣ . . . . . . . . . . ⎦ j 0 0 0 . . . λn and so

⎡ e

tA

⎢ ⎢ = P⎢ ⎢ ⎣

j

∞ λ1 t j j=0 j!

0 0

∞

...

j

λ2 t j j!

0 ... . . . . . . . . . . . . .

∞ 0 0 j=0 j=0

j

⎤

⎥ 0⎥ ⎥ P−1 , ⎥ ⎦

λn t j j!

and this shows that ⎡

exp(λ1 t)

⎢ ⎢0 etA = P ⎢ ⎣ 0

...

exp(λ2 t)

...

. . . . . . . . . . . . . 0 0 . . . exp(λn t)

⎤ ⎥ ⎥ −1 ⎥P . ⎦

(115)

We have shown that the matrix exponential etA is simply another way of representing a fundamental matrix for system (113). So, provided A can be diagonalized and has real eigenvalues, etA can be written down immediately by using result (115).

346

Chapter 6

Second and Higher Order Linear Differential Equations and Systems

EXAMPLE 6.33

Use the matrix exponential to solve the system x1 (t) = −2x1 + 6x2

x2 (t) = −2x1 + 5x2 .

Solution The matrix A is

6 , 5

−2 A= −2

and its eigenvalues and eigenvectors are 2 3/2 (λ1 = 1) x1 = , (λ2 = 2) x2 = . 1 1 The diagonalizing matrix 2 3/2 P= , 1 1 and

−1

P

1 D= 0

2 −3 = , −2 4

0 . 2

So from (115) we have

e

tA

2 3/2 = 1 1

et 0

0 e2t

2 −3 , −2 4

and after evaluating the matrix products we obtain t 4e − 3e2t , −6et + 6e2t etA = . 2et − 2e2t , −3et + 4e2t Deﬁning a two-element column matrix C with the arbitrary constants C1 and C2 as elements allows the general solution to be written as t C1 4e − 3e2t , −6et + 6e2t tA x(t) = e C = , 2et − 2e2t , −3et + 4e2t C2 so

(4C1 − 6C2 )et + (6C2 − 3C1 )e2t . x(t) = (2C1 − 3C2 )et + (4C2 − 2C1 )e2t

In component form the solution is x1 (t) = (4C1 − 6C2 )et + (6C2 − 3C1 )e2t x2 (t) = (2C1 − 3C2 )et + (4C2 − 2C1 )e2t . The method applies equally well to the situation in which matrix A is real but the eigenvalues occur in complex conjugate pairs, as shown by the next example.

Section 6.11

EXAMPLE 6.34

Nonhom*ogeneous Systems

347

Use the matrix exponential to solve the system x1 (t) = −3x1 − 4x2 Solution The matrix A is

A=

x2 (t) = 2x1 + x2 .

and

−3 −4 2 1

and its eigenvalues λ1 , λ2 and eigenvectors x1 and x2 are −1 + i , λ2 = −1 − 2i, λ1 = −1 + 2i, x1 = 1 So

x2 =

−1 − i . 1

−i/2 1/2 − i/2 −1 − i , , P−1 = i/2 1/2 + i/2 1 −1 + 2i 0 D= , 0 −1 − 2i

P=

and consequently e

tA

−1 + i 1

e−t (cos 2t + isin 2t) 0 =P P−1 0 e−t (cos 2t − isin 2t) −2e−t sin 2t e−t (cos 2t − sin 2t) = . e−t sin 2t e−t (cos 2t + sin 2t)

If we use this expression for etA in x(t) = etA C, the general solution becomes e−t (cos 2t − sin 2t) −2e−t sin 2t C1 x(t) = . −t −t e sin 2t e (cos 2t + sin 2t) C2 In component form this reduces to x1 (t) = C1 e−t cos 2t − (C1 + 2C2 )e−t sin 2t x2 (t) = (C1 + C2 )e−t sin 2t + C2 e−t cos 2t. When A is not diagonalizable, it is still possible to compute eA by writing eA = eK eL , where A is the sum of a diagonal matrix K and a nilpotent matrix L (a square matrix that when raised to a ﬁnite power becomes the null matrix), because under these circ*mstances the matrices eK and eL commute and eK+L = eK eL . The next example illustrates this approach. EXAMPLE 6.35

Find etA given that

A=

4 0

1 4

and use it to solve the hom*ogeneous system x1 (t) = 4x1 + x2

and

x2 (t) = 4x2 .

Solution Matrix A is not diagonalizable, because the repeated eigenvalue λ = 4 only gives rise to a single eigenvector. However, tA can be written as the sum of

348

Chapter 6

Second and Higher Order Linear Differential Equations and Systems

the following diagonal matrix tK and nilpotent matrix tL: 4t 0 0 tA = tK + tL, where tK = and tL = 0 4t 0

t . 0

It is easily checked that (tL)2 = 0 and the matrices tK and tL commute, so etA = etK etL . It follows from this that 4t e 1 0 0 t 1 t 0 tK tL , and e = + = , e = 0 0 0 1 0 1 0 e4t so we arrive at the result e

tA

e4t = 0

0 e4t

1 0

4t t e = 1 0

te4t . e4t

The exponential matrix etA is a fundamental matrix for the system, so as the general solution is given by x(t) = etA C, 4t C1 e4t + C2 te4t e te4t C1 x(t) = = . C2 0 e4t C2 e4t In component form the solution becomes x1 (t) = C1 e4t + C2 te4t

and

x2 (t) = C2 e4t .

The nilpotent matrix L in the last example was seen to give rise to the second linearly independent solution te4t corresponding to the eigenvalue λ = 4 that occurred with multiplicity 2. If in a larger system with a repeated eigenvalue λ and a nondiagonalizable matrix A it had been necessary to raise a nilpotent matrix to the power r before it became the null matrix, then in addition to a term of the form eλt appearing in etA , the repeated eigenvalue would also give rise to the linearly independent terms teλt , t 2 eλt , . . . , t (r −1) eλt .

(d) Variation of parameters

the matrix exponential and variation of parameters

A particular integral can be found from the general solution of the hom*ogeneous form of a constant coefﬁcient system by a direct generalization of the method of variation of parameters described in Section 6.6. If the system is x = Ax + b(t),

(116)

to ﬁnd a particular integral x p (t) we set x p (t) = etA u(t),

(117)

where the vector u(t) is to be determined. Then, as xp (t) = AetA u(t) + etA u (t), substituting for x p (t) in system (106) gives AetA u(t) + etA u (t) = AetA u(t) + b(t), so after cancelling the terms AetA u(t) and premultiplying the result by e−tA , the inverse of etA because etA and e−tA commute, we ﬁnd that u (t) = e−tA b(t), from which u(t) now follows.

(118)

Section 6.11

Nonhom*ogeneous Systems

349

In the equation for u(t) the matrix exponential e−tA is determined from etA by changing the sign of t. The expression on the right of (118) is simply a column vector with elements that are known functions of t, so the components of (118) can be integrated separately to ﬁnd the elements u1 (t), u2 (t), . . . , un (t) of U(t). Then, when U(t) is known, the particular integral follows from (117). The general solution of (116) is the sum of the solution of the hom*ogeneous form of the system and the particular integral x p (t). EXAMPLE 6.36

Use the method of variation of parameters to solve the nonhom*ogeneous system. x1 (t) = −2x1 + 6x2 + t x2 (t) = −2x1 + 5x2 − 1. Solution The hom*ogeneous form of this system was obtained in Example 6.33, where it was shown that t 4e − 3e2t −6et + 6e2t tA , e = 2et − 2e2t −3et + 4e2t so e As

b(t) =

−tA

4e−t − 3e−2t = 2e−t − 2e−2t

−6e−t + 6e−2t . −3e−t + 4e−2t

t 2(3 + 2t)e−t − 3(2 + t)e−2t , we have e−tA b(t) = , −1 (3 + 2t)e−t − 2(2 + t)e−2t

but u (t) = e−tA b(t), so u1 (t) = 2(3 + 2t)e−t − 3(2 + t)e−2t u2 (t) = (3 + 2t)e−t − 2(2 + t)e−2t . When these equations are integrated, the arbitrary constants of integration can be set equal to zero, because if they are nonzero the terms they introduce are of the same type as the solution of the hom*ogeneous system, so they can be absorbed into it. As a result, integration gives 3 5 + t e−2t u1 (t) = −2(5 + 2t) + 2 2 and u2 (t) = −(5 + 2t)e−t +

5 + t e−2t . 2

Finally, if we set x p (t) = etA u(t) with u1 (t) and u2 (t) taken as the elements of u(t), the particular integral becomes ⎡ ⎤ 25 5 − t − ⎢ 4 2 ⎥ ⎥. x p (t) = ⎢ ⎣ ⎦ 5 − −t 2

350

Chapter 6

Second and Higher Order Linear Differential Equations and Systems

The solution xc (t) of the hom*ogeneous system (the complementary function) found in Example 6.33 was (4C1 − 6C2 )et + (6C2 − 3C1 )e2t xc (t) = , (2C1 − 3C2 )et + (4C2 − 2C1 )e2t so the solution of the nonhom*ogeneous system x(t) = xc (t) + x p (t) is given by 25 5 − t 4 2 5 x2 (t) = (2C1 − 3C2 )et + (4C2 − 2C1 )e2t − − t. 2 x1 (t) = (4C1 − 6C2 )et + (6C2 − 3C1 )e2t −

General Remark The way in which combinations of arbitrary constants appear when we multiply functions in the general solution of a hom*ogeneous system of differential equations is determined by the method of solution. So, for example, when we solve a system by elimination, the choice of variable to be eliminated ﬁrst will inﬂuence the form of the result, as will the ordering of the eigenvectors when diagonalizing the matrix A. A combination of arbitrary constants is simply an arbitrary constant, though the ratio of all similar combinations of constants multiplying corresponding functions in different forms of the solution must be the same. This can be illustrated by considering the solution of the hom*ogeneous form of the equation in Example 6.36 that was found to be x1 (t) = (4C1 − 6C2 )et + (6C2 − 3C1 )e2t x2 (t) = (2C1 − 3C2 )et + (4C2 − 2C1 )e2t . This solution can be written in an equivalent but different-looking form by setting K1 = 2C1 − 3C2 and K2 = 6C2 − 3C1 , where K1 and K2 are themselves arbitrary constants. After changing the constants in this manner the solution becomes x1 (t) = 2K1 et + K2 e2t and x2 (t) = K1 et +

2 K2 e2t , 3

and other equivalent forms are also possible. The above remarks should be remembered when comparing solutions to problem sets with the solutions given at the end of the book. As a particular integral contains no arbitrary constants, its form remains the same irrespective of the manner in which it has been determined. An account of the material in this section is to be found in references [3.5] and [3.15].

Summary

The structure of the solution of a linear nonhom*ogeneous system of equations was explained, and a matrix method of solution was developed for constant coefﬁcient systems that depended on the diagonalization of the coefﬁcient matrix. The cases of real and complex eigenvalues of the coefﬁcient matrix were examined separately, and it was shown how systems of equations with real coefﬁcient matrices can lead to solutions involving trigonometric functions. A different method of solution was then developed using the concept of the matrix exponential.

Section 6.12

Autonomous Systems of Equations

351

EXERCISES 6.11 In Exercises 1 through 6 ﬁnd a fundamental matrix and the general solution of the system. 1. 2. 3. 4. 5. 6.

x1 x1 x1 x1 x1 x1

= −x2 , x2 = 2x1 . = −x1 − 5x2 , x2 = x1 − 5x2 . = −3x1 − 4x2 , x2 = 2x1 + x2 . = −x1 − 4x2 , x2 = x1 + 4x2 . = 2x2 , x2 = −2x3 , x3 = 2x2 . = −3x2 , x2 = −3x3 , x3 = 3x2 .

In Exercises 7 through 18 ﬁnd the general solution of the system by diagonalization. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18.

x1 = −10x1 − 18x2 + t, x2 = 6x1 + 11x2 + 3. x1 = −2x2 + sin t, x2 = −2x1 − t. x1 = x1 − x2 + cos t, x2 = −x1 + x2 + e3t . x1 = x2 + e−t , x2 = −x1 + 2x2 − 4. x1 = 2x1 + 3x2 − sin t, x2 = x1 − 2x2 . x1 = −x1 − 2x2 + cos t, x2 = x1 + x2 + 4. x1 = −2x1 + 2x2 + 2x3 + sin t, x2 = −x2 + 3, x3 = −2x1 + 4x2 + 3x3 . x1 = x1 + 2x2 + 3 + 2t, x2 = x2 + t, x3 = 2x1 + x3 + 1. x1 = x1 + 2x2 + x3 + t, x2 = x2 − x3 + 2, x3 = 2x1 + x3 + 2t. x1 = x2 + t, x2 = x3 , x3 = x2 . x1 = x1 + 2x2 + x3 + 2e−t , x2 = x2 + x3 + t, x3 = 2x1 + x3 + 2t. x1 = x2 + 5, x2 = x3 + t, x3 = x2 + 2t.

Solve Exercises 19 through 26 by means of the matrix exponential. 19. 2x1 = x1 − x2 , 2x2 = 3x1 + 5x2 .

6.12

20. 21. 22. 23. 24. 25. 26.

x1 x1 x1 x1 x1 x1 x1

= −10x1 − 18x2 , x2 = 6x1 + 11x2 . = −x2 , x2 = 2x1 . = 2x1 − 12x2 , 2x2 = 3x1 − 8x2 . = 7x1 − 34x2 , x2 = 2x1 − 9x2 . = −x1 − 5x2 , x2 = x1 − 5x2 . = −3x1 − 4x2 , x2 = 2x1 + x2 . = −x1 + 2x2 , x2 = x1 + x2 .

Solve Exercises 27 through 30 by the method of variation of parameters. 27. 28. 29. 30.

x1 x1 x1 x1

= 10x1 + 18x2 + sin t, x2 = −6x1 − 11x2 + t. = −x2 + 3e4t , x2 = −2x1 + x2 − 2. = 3x1 + 4x2 , x2 = −2x1 − x2 − t 2 . = −x2 + 5, x2 = x1 + 2x2 − 1.

Solve the initial value problems 31 through 36 by any of the methods in this chapter. 31. x1 = x2 + 1, x2 = 2x1 − x2 + t, with x1 (0) = 1, x2 (0) = 0. 32. x1 = 3x2 + t, x2 = 2x1 + x2 − 3, with x1 (0) = 1, x2 (0) = 1. 33. x1 = 2x1 + x2 − et , x2 = − 2x1 − x2 − 3, with x1 (0) = 0, x2 (0) = 1. 34. x1 = −3x1 − x2 + 3t, x2 = x1 − x2 − 3, with x1 (0) = 1, x2 (0) = 3. 35. x1 = −3x1 − 5x2 − 12x3 + sin t, x2 = −2x1 + 1, x3 = x1 + x2 + 2x3 − t, with x1 (0) = 1, x2 (0) = 0, x3 (0) = −1. 36. x1 = −2x1 + 2x2 + 2x3 + 3et , x2 = −x1 − x2 − 2x3 + 1, x3 = x1 + 2x2 + 3x3 − 3, with x1 (0) = 1, x2 (0) = 1, x3 (0) = 0.

Autonomous Systems of Equations Autonomous Systems, the Phase Plane, Stability, and Linear Systems The general form of a nonlinear system of two simultaneous ﬁrst order differential equations for the functions x(t), y(t) that depend on the time t is dx = f1 (x, y, t) dt dy = g1 (x, y, t). dt

(119)

352

Chapter 6

Second and Higher Order Linear Differential Equations and Systems

This system is linear and nonhom*ogeneous if f1 (x, y, t) = a(t)x(t) + b(t)y(t) + h(t) and g1 (x, y, t) = c(t)x(t) + d(t)y(t) + k(t), and hom*ogeneous if, in addition, h(t) = k(t) ≡ 0. If the dependence of the functions f1 and g1 on the time t is only through the functions x(t) and y(t), the time dependence is implicit and f1 = f (x, y) and g1 = g(x, y), causing the system of equations in (119) to become dx = f (x, y) dt dy = g(x, y). dt autonomous and nonautonomous systems

equilibrium or critical point

trajectories or paths

phase portrait

(120)

Systems of this type are called autonomous, and they describe physical phenomena such as chemical reactions that, provided all conditions remain the same, will yield identical results whenever the reactions are repeated. It is because of this that autonomous systems are sometimes said to be time invariant systems. This situation should be contrasted with the nonautonomous behavior of an electrical circuit containing temperature-dependent elements that will cause its behavior to vary as the ambient temperature changes with time. A point (x0 , y0 ) where both of the derivatives dx/dt and dy/dt in (120) vanish, so that

dx dt

2

+

dy dt

2 = 0,

is called an equilibrium point or a critical point of the system. If the differential equations in (120) are solved subject to the initial conditions x0 = x(t0 ), y0 = y(t0 ) imposed at time t = t0 , it is convenient to regard (x(t), y(t)) as a point in the (x, y)-plane that traces out a curve as t increases. Such curves, along which the time t can be regarded as a parameter, are called trajectories or paths, and sometimes orbits in the (x, y)-plane. The (x, y)-plane itself is then called the phase plane. Associated with each trajectory is the direction in which the point (x(t), y(t)) moves as t increases, and in the phase plane these directions are usually indicated by adding arrows to trajectories. The pattern of trajectories associated with a given autonomous system of equations is called the phase portrait of the system. The reason why in autonomous systems the time t can be regarded as a parameter can be seen by dividing the second equation in (120) by the ﬁrst to obtain the differential equation g(x, y) dy = , dx f (x, y)

(121)

in which t is absent. Had the nonautonomous system of equations in (119) been treated in similar fashion, dy/dx would have exhibited an explicit dependence on the time.

Section 6.12

Autonomous Systems of Equations

353

FIGURE 6.13 A depression in a surface surrounded by an elevated rim.

stability, instability, and asymptotic stability

At an equilibrium point (x0 , y0 ) of the system in (120), the vanishing of both f and g causes dy/dx in (121) to become indeterminate at that point, so initial conditions imposed at an equilibrium point cannot determine a unique solution. This has the effect that on passing through an equilibrium point, a point moving along one trajectory can move onto a different trajectory. At an equilibrium point of an autonomous system, a physical system represented by the equations is in an equilibrium state. This state is said to be stable if, when the system is subjected to arbitrarily small disturbances, it always remains in the neighborhood of the same equilibrium state. If, however, the result of arbitrarily small disturbances is to make the system change to a different equilibrium state, to make the displacement grow unrestrictedly, or, depending on the displacement, to make the system sometimes return to the original equilibrium state and sometimes to cause the displacement increase unrestrictedly, the state is said to be unstable. A dynamical analogy illustrating stable and unstable situations is provided by considering Fig. 6.13, which represents a depression in a surface surrounded by an elevated rim, beyond which the level of the surface falls away steadily. A ball placed at the bottom of the depression is in a stable equilibrium state, because after any small displacement gravity will cause it to try to return to the equilibrium state. If, however, the displacement is large the motion will be unstable, because the ball will leave the depression and roll away indeﬁnitely as time increases. Every point on the top of the rim represents an unstable equilibrium state because, depending on the direction of the displacement, the ball may move to another point on the rim, return to the depression, or roll away indeﬁnitely. So this system has one stable equilibrium state at the bottom of the depression, and an inﬁnite number of unstable states around the top of the rim.

Stability and asymptotic stability The notion of stability can be made more precise by introducing the function (t) that measures the distance in the phase plane of a point (x(t), y(t)) on a trajectory at time t from an equilibrium point at (x0 , y0 ), where (t) =

(x(t) − x0 )2 + (y(t) − y0 )2 .

354

Chapter 6

Second and Higher Order Linear Differential Equations and Systems

(i) The equilibrium point (x0 , y0 ) is said to be stable if for every arbitrarily small number ε > 0, a number δ > 0 can be found such that if x02 + y02 < δ, then (t) < ε for all time t. (ii) The equilibrium point (x0 , y0 ) is said to be asymptoticallystable if it is stable in the sense of (i) and a number α can be found such that if x02 + y02 < α, then (t) → 0 as t → ∞.

predator–prey problem

The implication of these deﬁnitions is that when an equilibrium point (x0 , y0 ) is stable, a trajectory starting close to (x0 , y0 ) will remain close to it, but if the point is asymptotically stable any trajectory starting close to (x0 , y0 ) will eventually converge to the equilibrium point as t → ∞. Asymptotically stable equilibrium points can be said to attract trajectories, so such points are called attractors, whereas equilibrium points from which the distance function (t) increases without bound as t increases are said to repel trajectories. In the dynamical example just given, in the absence of friction, the point at the bottom of the depression will be a stable state, because after a small displacement the ball will forever move around the lowest point. If, however, friction is present, the lowest point of the depression will be an asymptotically stable state, because after any small displacement the ball will eventually come to rest at the lowest point. Interest in autonomous systems centers around the fact that trajectories in phase space provide qualitative information about the entire class of solutions of the system and, in particular, about properties of solutions when f and g are nonlinear and no analytical solution can be found. A classical example of a nonlinear autonomous system is the predator–prey system of equations introduced and studied by Volterra and Lotka around 1930. They considered the ecological situation in which an isolated colony of foxes and rabbits coexist, with the foxes eating the rabbits and the rabbits feeding on a plentiful supply of vegetation. When the rabbits are numerous, the foxes are well fed and their numbers will grow, but when the number of foxes increases to the point where the rabbit population declines, the number of foxes will begin to fall, giving the rabbit population an opportunity to regenerate. This process, it was postulated, could explain the nonlinear cyclic variation in fox and rabbit populations that is observed in nature. This predator–prey model involving foxes and rabbits will become nonautonomous if some external factors are introduced that reduce the fox and rabbit populations by some other means. To derive the predator–prey equations, let x(t) be the number of rabbits present at time t. Then, as vegetation is plentiful, without foxes the rabbit population will grow at a rate proportional to the number of rabbits, so we can write dx = ax, dt where a > 0 is a constant. Assuming that the rate at which foxes eat rabbits is proportional to the product of the number of rabbits x(t) and the number of foxes y(t) present at time t, the rabbit population described by the preceding equation must be modiﬁed to allow for this reduction, and so it becomes dx = ax − bxy, dt where b > 0 is a constant.

Section 6.12

Autonomous Systems of Equations

355

The differential equation governing the fox population y(t) is derived in a similar manner, but now the number of foxes decreases as the rabbit population decreases, leading to a differential equation of the form dy = −cy + dxy, dt where c > 0 and d > 0 are constants. The classical predator–prey equations are the two nonlinear autonomous equations dx = x(a − by) dt dy = y(xd − c). dt

linearization

(122)

This nonlinear autonomous system has no analytical solution, so either individual solutions must be found by numerical computation (see Section 19.7), or phase-plane methods must be used to determine the qualitative behavior of solutions of this system. An obvious feature of the predator–prey system of equations is that an equilibrium state exists when dx/dt = dy/dt = 0, and this occurs at the origin (0, 0) and when x = c/d and y = a/b. The ﬁrst equilibrium state is of no interest because then neither rabbits nor foxes are present, but in the other equilibrium state the rabbit and fox populations will remain static, though deviations from this situation can be expected to initiate nonlinear oscillations in the population numbers. The predator–prey model, although simple and developed initially for ecological reasons, can be modiﬁed and applied to other situations such as the spread of an infectious disease, the competition between industries for a raw material that is in limited supply, or when industries compete for the same market. When the functions f (x, y) and g(x, y) in (120) are nonlinear, or are complicated in other ways, to help understand the behavior of the system the functions f and g are often linearized about an equilibrium point at (x0 , y0 ) that is of interest. This involves expanding f and g about (x0 , y0 ) as two-variable Taylor series expansions, and then replacing f and g in (120) by the linear terms in these expansions. If, for example, (x0 , y0 ) is an equilibrium point of the system of equations in (120), then f (x0 , g0 ) = 0 and g(x0 , y0 ) = 0, and expanding f and g about the point (x0 , y0 ) gives f (x, y) = fx (x0 , y0 )(x − x0 ) + fy (x0 , y0 )(y − y0 ) + higher order terms and g(x, y) = gx (x0 , y0 )(x − x0 ) + g y (x0 , y0 )(y − y0 ) + higher order terms. Substituting only the ﬁrst order terms from these expansions into system (120) simpliﬁes it to the constant coefﬁcient linear autonomous system d(x − x0 ) = fx (x0 , y0 )(x − x0 ) + fy (x0 , y0 )(y − y0 ) dt and d(y − y0 ) = gx (x0 , y0 )(x − x0 ) + g y (x0 , y0 )(y − y0 ). dt

(123)

356

Chapter 6

Second and Higher Order Linear Differential Equations and Systems

Setting X = x − x0 and Y = y − x0 , we can write these equations in the matrix form dz = J(x0 , y0 )z, dt

(124)

where z= Jacobi matrix of the system

X Y

and

J(x0 , y0 ) =

fx (x0 , y0 ) gx (x0 , y0 )

fy (x0 , y0 ) . g y (x0 , y0 )

The matrix J(x0 , y0 ) is called the Jacobi matrix of the system at the point (x0 , y0 ), and we will see later how the eigenvalues of J(x0 , y0 ) determine the nature of the equilibrium point at (x0 , y0 ). It is reasonable to suppose that when the neglected remainder terms in the Taylor series expansions of f and g are suitably small, the behavior of this linearized system of equations in some neighborhood of the equilibrium point at (x0 , y0 ) will be qualitatively similar to that of the original nonlinear system. As an illustration of the linearization process, let us now linearize the predator– prey equations in (122) about the equilibrium point at x = c/d and y = a/b. Identifying f (x, y) with x(a − by) and g(x, y) with y(dx − c), substituting into the Jacobian J(x0 , y0 ) with x0 = c/d and y0 = a/b, and setting X = x − c/d and Y = y − a/b leads to the linearized predator–prey equations bc dX =− Y dt d dY ad = X. dt b

(125)

These equations are easily integrated to give the following equation for the trajectories in the (X, Y) phase plane: X 2 + (cb2 /ad2 )Y 2 = k2 ,

where k is an integration constant.

Reverting to the original variables shows that after linearization, each trajectory in the (x, y) phase plane that is close to the equilibrium point is a member of the family of ellipses (x − c/d)2 + (cb2 /ad2 )(y − a/b)2 = k 2 ,

(126)

which have their common center at the point (c/d, a/b) in the (x, y) phase plane. This shows that in a neighborhood of the equilibrium point the phase portrait of the predator–prey system can be expected to be approximated by this family of ellipses. This result indicates that close to the equilibrium condition, the rabbit and fox populations can be expected to exhibit a cyclic variation with respect to time. This conclusion follows from the fact that as the time t increases, starting at an initial point on a trajectory where x0 = x(t0 ), y0 = y(t0 ) at a time t = t0 , the point (x(t), y(t)) will move around the ellipse that passes through this point until after a suitable interval of time it returns to its starting point. In this case linearization has produced elliptical trajectories centered on the equilibrium point, so in the nonlinear case the trajectories can be expected to be distorted ellipses. Before considering nonlinear autonomous systems we will determine the nature of the equilibrium points associated with the general linear two variable

Section 6.12

Autonomous Systems of Equations

357

autonomous system, which in standard notation can be written dx = ax + by dt dy = cx + yd, dt

(127)

where a, b, c, and d are constants, and the second term dy on the right of the second equation is not to be confused with the differential dy. Setting dx/dt = dy/dt = 0 in (127) and solving for x and y shows the origin to be the only equilibrium point if a b (128) c d = ad − bc = 0. When the (127) is integrated once, it yields what is called a ﬁrst integral of the system. A ﬁrst integral is not a solution of the system, because although it is an equation that connects x(t) and y(t), it does not express either function explicitly in terms of t. First integrals are useful because they are easier to obtain than solutions of general autonomous systems, and they provide qualitative information about the general behavior of the set of all solutions. This can be seen from the ﬁrst integral of the linearized predator–prey system in (125) because, although this did not yield a solution in terms of t, it did conﬁrm that the linearized system exhibits a periodic behavior of the two populations in a neighborhood of the equilibrium point. A simple example of a linear autonomous system can be derived from any physical system, be it electrical, mechanical, or otherwise, that can be represented by the hom*ogeneous constant-coefﬁcient second order equation d2 y dy + by = 0. (129) +a dt 2 dt Setting dy/dt = x, we can write the second order equation as the linear autonomous system dx = −ax − by dt dy = x, dt with t as a parameter, or to the equivalent variables separable equation dy x =− , dx ax + by

(130)

(131)

where now only x and y are present. As a special case, when a = 0 and b = n2 , result (131) becomes x dy =− 2 , dx n y for which a ﬁrst integral is seen to be x 2 + n2 y2 = k2 . This represents a family of elliptical trajectories all centered on the equilibrium point of the system that is located at the origin. The argument used earlier in connection with the linearized predator–prey equations shows that solutions of the system (131) when a = 0 and b = n2 must be periodic. This is to be expected, because

358

Chapter 6

Second and Higher Order Linear Differential Equations and Systems

with these values of a and b equation (129) describes undamped simple harmonic oscillations. In this simple case, as x = dy/dt,

dy k2

− n2 y2

= dt,

and after integration this gives y(t) = (k/n) sin[n(t + t0 )], which is the general solution of (129) for a = 0 and b = x 2 . When we considered the linearized predator–prey equations, the family of ellipses around the equilibrium point that were found represented an approximation to the phase portrait of the system in a neighborhood of the equilibrium point. In this case, however, system (130) is linear, so no linearization is involved and the family of elliptical trajectories forms the true phase portrait of system (130). The linear autonomous system (127) can be written in the matrix form dx/dt = Jx,

(132)

where x x= , y

a J= c

b . d

(133)

This system was studied in detail in Section 6.10, where it was seen that its solution depends on the eigenvalues of J determined by the characteristic equation a − λ b = λ2 − (a + d)λ + (ad − bc) = 0. c d − λ

(134)

Setting α = a + d and β = ad − bc, the characteristic equation in (134) becomes λ2 − αλ + β = 0,

(135)

with the discriminant = (a − d)2 + 4bc. The pattern of the trajectories of the autonomous system in (132), equivalently (127), is determined completely by the eigenvalues λ1 and λ2 of J and their associated eigenvectors: that is to say, by the fundamental solutions of the system. If the eigenvalues are real and λ1 = λ2 , a matrix P can always be found that simpliﬁes the system by reducing J to a diagonal matrix D through the result P−1 JP = D, with λ1 and λ2 the elements on the leading diagonal of D (see Section 4.2). The transformation x = Pu with u = [u, v]T then reduces (132) to the simpler form du/dt = Du, showing that du/dt = λ1 u and dv/dt = λ2 v. These equations have the general solution u = Aeλ1 t

and

ν = Beλ2 t ,

(136)

so the form of the trajectories about the equilibrium point at the origin in the (u, v) phase plane is seen to depend on both the signs of the eigenvalues λ1 and λ2 and their magnitudes.

Section 6.12

Autonomous Systems of Equations

359

When the discriminant > 0, the eigenvalues λ1 and λ2 will be real, and then there are three cases to consider.

(i) Unstable nodes: λ1 and λ2 are positive Examination of the solution in (136) shows that the trajectories must take one of the two forms illustrated in Figs. 6.14a and 6.14b. In this case the equilibrium point at v

v

u

(a)

u

(b) v

v

u

(c)

u

(d)

v

v

u

(e)

u

(f)

FIGURE 6.14 (a,b) Unstable nodes. (c,d) Stable nodes. (e,f) Saddle points.

360

Chapter 6

Second and Higher Order Linear Differential Equations and Systems

types of critical point

the origin is called a node. As the eigenvalues are both positive, a point (u(t), v(t)) on a trajectory moves away from the origin as t increases, so this type of equilibrium point is called an unstable node.

(ii) Stable nodes: λ1 and λ2 are negative Examination of the solution in (136) shows that the trajectories must take one of the two forms illustrated in Figs. 6.14c and 6.14d, where the equilibrium point at the origin is again called node. This time, as the eigenvalues are both negative, a point (u(t), v(t)) on a trajectory will move toward the origin as t increases, so in this case the equilibrium point is called a stable node.

(iii) Saddle points: λ1 and λ2 have opposite signs Examination of the solution in (136) shows that the trajectories take one of the two forms illustrated in Figs. 6.14e and 6.14f, where the equilibrium point is called a saddle point. The eigenvalues are real and have opposite signs, so as t increases a point (u(t), v(t)) on a branch of a hyperbola will move toward the origin and then away again, showing that a saddle point represents an instability. The two diagonal straight lines that form degenerate hyperbolas are each called a separatrix in the phase portrait, because they separate the phase plane into four distinct regions, and a solution in any one of these regions cannot be related to a solution in a different region.

(iv) Degenerate node: Equal eigenvalues λ = λ1 = λ2 When the discriminant = 0 the eigenvalues coincide, so λ = λ1 = λ2 . In this case the Jacobi matrix J cannot be diagonalized, but system (132) can always be reduced to the form du/dt = Su, where S=

λ 1

0 λ

and

u=

u , v

and this has the general solution u = Aeλt

and

ν = (At + B)eλt .

(137)

An examination of solution (137) shows that when λ > 0, the trajectories are qualitatively similar to the general pattern seen in case (i), corresponding to an equilibrium point that is an unstable node. When λ < 0 the trajectories are qualitatively similar to the general pattern seen in case (ii), corresponding to a stable node. Equilibrium points with nodes of this type that arise from coincident eigenvalues are called degenerate nodes, so the ones where λ > 0 are called unstable degenerate nodes, and the ones where λ < 0 are called stable degenerate nodes. Typical patterns of trajectories at unstable degenerate nodes are shown in Figs. 6.15a and 6.15b and at stable degenerate nodes in Figs. 6.15c and 6.15d.

Section 6.12

Autonomous Systems of Equations

v

361

v

u

u

(a)

(b)

v

v

u

u

(c)

(d)

FIGURE 6.15 (a,b) Unstable degenerate nodes. (c,d) Stable degenerate nodes.

(v) Focus or spiral point: Complex conjugate eigenvalues If the discriminant < 0, the eigenvalues will be the complex conjugates with λ1 = ξ + iη and λ2 = ξ − iη. Diagonalization of J then produces a system of equations of the form du/dt = Cu, where

ξ C= −η

η ξ

and

u u= . v

This system is easily shown to have the general solution u = eξ t (A sin ηt − B cos ηt)

and

v = eξ t (B sin ηt + A cos ηt), (138)

which deﬁnes spiral trajectories about the equilibrium point. In this case the equilibrium point is called a focus or a spiral point. The direction in which a point (u(t), v(t)) along a spiral as t increases is determined by the sign of ξ . When ξ > 0 the point moves away from the origin as t increases, so the equilibrium point is then called either an unstable focus or an unstable spiral point. Conversely, when ξ < 0, the point moves toward the origin as t increases, so in this case the equilibrium point is called a stable focus or a stable spiral point. Figure. 6.16a shows an unstable focus and Figure. 6.16b a stable focus. Spirals may evolve in either a clockwise or a counterclockwise direction, and this can be determined by the direction of the vector with components (dx/dt, dy/dt) at any point on the spiral (see Example 6.39).

362

Chapter 6

Second and Higher Order Linear Differential Equations and Systems v

v

u

u

(a)

(b)

FIGURE 6.16 (a) An unstable focus. (b) A stable focus.

v

u

FIGURE 6.17 A center located at the origin.

(vi) Center: Purely imaginary complex conjugate eigenvalues If in the characteristic equation (135) α = a − d = 0 and the discriminant < 0, the eigenvalues will be purely imaginary complex conjugates. Setting ξ = 0 in (138) shows that the trajectories become a family of ellipses centered on the origin, as shown in Fig. 6.17. In this case the equilibrium point at the origin is called a center, and the corresponding solutions are considered to be stable because they remain bounded for all time. It follows from this that the equilibrium point in the linearized predator–prey system is a center. EXAMPLE 6.37

Locate and identify the nature of the equilibrium point of the system dx = −x, dt and draw some typical trajectories.

dy = −x − 2y, dt

Solution The equilibrium point is located at the origin, and its nature can be identiﬁed by examining the eigenvalues of the Jacobi matrix J that follows by setting f (x, y) = −x and g(x, y) = −x − 2y. We have −1 0 J= , −1 −2 and this has the eigenvalues λ1 = −1 and λ2 = −2. As the eigenvalues are real, and both are negative, it follows from Case (ii) that the equilibrium point at the origin is a stable node. To draw trajectories it is necessary to solve this system, and a routine

Section 6.12

Autonomous Systems of Equations

363

y

x

FIGURE 6.18 Trajectories in the neighborhood of the stable node at the origin.

calculation shows that x = −C1 e−t and y = C1 e−t + C2 e−2t . Eliminating t, we ﬁnd that the equation of the trajectories is y = −x + (C2 /C12 )x 2 . This equation describes a family of parabolas that at the origin are all tangent to the degenerate parabola y = −x that forms a separatrix marking a boundary between phase curves with different properties. Some typical trajectories are shown in Fig. 6.18, where the arrows indicate that the node is stable. It is important to recognize that as the node is a singularity of the system where dy/dx is indeterminate, a point moving along a trajectory that passes through the node cannot leave it on a different trajectory. EXAMPLE 6.38

Locate and identify the nature of the equilibrium point of the system dx = −x − y − 2, dt and draw some typical trajectories.

dy = −x + y − 4, dt

Solution The equilibrium point occurs when −x − y − 2 = 0 and −x + y − 4 = 0, corresponding to x = −3, y = 1. For convenience we shift the equilibrium point to the origin in the (X, Y) phase plane by making the change of variables X = x + 3 and Y = y − 1, when the system becomes dX = −X − Y, dt

dY = −X + Y. dt

The nature of the equilibrium point that is now located at the origin in the (X, Y) phase plane can be identiﬁed by examining the eigenvalues of the Jacobi matrix −1 −1 J= , −1 1 √ √ which are easily seen to be λ1 = − 2 and λ2 = 2. As the eigenvalues are real, and opposite in sign, it follows from Case (iii) that the equilibrium point at the

364

Chapter 6

Second and Higher Order Linear Differential Equations and Systems

origin is a saddle point. To draw trajectories it is necessary to solve this system of equations. After some calculations, the equation of the family of trajectories determined by dY/dX = (X − Y)/(X + Y) is found to be given by Y 2 + 2XY − X 2 = c, where the constant c is determined by the point in the phase plane through which a trajectory is required to pass. The general equation of a conic is AX 2 + 2BXY + CY 2 + DX + EY + F = 0, and this represents an ellipse if B2 − AC < 0, a parabola if B2 − AC = 0, and a hyperbola if B2 − AC > 0. So comparing the equation of the trajectories with the general form of a conic, we see that B2 − AC > 0, so it describes a family of hyperbolas. This family of hyperbolas with parameter c is centered on the origin, and solving for Y gives Y = −X + 2X 2 + c and Y = −X − 2X 2 + c, where for any given value of c, each equation represents one pair of hyperbolas. Some typical hyperbolas are shown in Fig. 6.19, where the upper and lower branches correspond to different values of c in the ﬁrst equation, and the left and right branches correspond to other values of c in the second equation. The asymptotes, which represent degenerate hyperbolas, are √ √ seen by inspection of these equations to be given by Y = ( 2 − 1)X and Y = −( 2 + 1)X. Each of these is a separatrix in the phase portrait of the system, and a solution in any one of the four regions into which these lines divides the phase plane cannot connect with a solution in any other region. The simplest way to determine the direction along the upper and lower hyperbolic trajectories as t increases is to ﬁnd the direction of the vector (dX/dt, dY/dt) on a trajectory. For example, when X = 0, we see from the differential equations that the direction of the vector along a trajectory that crosses the Y-axis has the components (−Y, Y). This shows that when Y > 0 the vector is directed upward and toward the left, whereas when Y < 0 it is directed downward and toward the

Y

X

FIGURE 6.19 Trajectories around the saddle point at the origin in the (X, Y) phase plane.

Section 6.12

Autonomous Systems of Equations

365

right. The direction of the arrows on the left and right hyperbolic trajectories are determined in similar fashion by ﬁnding the direction of the vector (dX/dt, dY/dt) that crosses the X-axis where Y = 0. The pattern of the trajectories around the saddle point in the original coordinate system is obtained by translating the picture in Fig. 6.19 to the point (−3, 1). EXAMPLE 6.39

Locate and identify the equilibrium point of the system dx = −x + 2y + 1, dt

dy = −2x − y + 2, dt

and sketch some trajectories. Solution The equilibrium point occurs when −x + 2y + 1 = 0 and −2x − y + 2 = 0, corresponding to x = 1 and y = 0. For convenience we shift the equilibrium point to the origin in the (X, Y) phase-plane by making the change of variables X = x − 1 and Y = y, when the system becomes dX = −X + 2Y, dt

dY = −2X − Y. dt

The nature of the equilibrium point that is now located at the origin in the (X, Y) phase plane can be identiﬁed by examining the eigenvalues of the Jacobi matrix −1 2 J= , −2 −1 which follows from setting f (X, Y) = −X + 2Y and g(X, Y) = −2X − Y. The eigenvalues are λ1 = −1 + 2i and λ2 = −1 − 2i, so as these are complex conjugates with negative real parts, it follows from Case (v) that the equilibrium point at the origin in the (X, Y) phase plane is a stable focus. This means that the trajectories spiral into the origin as t increases, so the only question that remains is whether the spiral is clockwise or counterclockwise. Figure 6.20 shows two possible spirals, where in Fig. 6.20a the direction around the spiral is conterclockwise, while in Fig. 6.20b it is clockwise. Arguing as in Example 6.38, and considering the vector with components (dX/dt, dY/dt) where the spiral crosses the X-axis, by setting Y = 0 we ﬁnd that the vector has components (−X, −2X). As this vector is directed downward and for x > 0 to the left, it

Y

Y

X

(a)

X

(b)

FIGURE 6.20 Two stable foci in the (X, Y) phase plane. (a) Counterclockwise spiral. (b) Clockwise spiral.

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Chapter 6

Second and Higher Order Linear Differential Equations and Systems

follows that the trajectories must spiral clockwise into the origin, so Fig. 6.20b is the only possible phase portrait for this system. This information is sufﬁcient to enable trajectories to be sketched, but as the general solution of the system is easily found to be X(t) = e−t (c1 sin 2t − c2 cos 2t),

Y(t) = e−t (c1 cos 2t + c2 sin 2t),

it is not difﬁcult to construct accurate spiral trajectories. The pattern of trajectories for the original autonomous system is obtained by translating the pattern in Fig. 6.20b to the point (1, 0) in the (x, y) phase plane. If it is only necessary to identify the nature of the equilibrium point at the origin belonging to the linear autonomous system, dx = ax + by dt dy = cx + dy, dt identiﬁcation of critical points

results (i) to (vi) can be summarized as follows: (a) A node if (a + d)2 ≥ 4(ad − bc) > 0; stable if a + d < 0 and unstable if a + d > 0. (b) A saddle point if ad − bc < 0. (c) A focus if (a + d)2 < 4(ad − bc); stable if a + d < 0 and unstable if a + d > 0. (d) A center if a + d = 0 and ad − bc > 0.

(vii) Nonlinear autonomous systems nonlinear autonomous systems

If the nonlinear autonomous system ⎧ dx ⎪ ⎪ ⎨ dt = f (x, y) ⎪ ⎪ ⎩ dy = g(x, y) dt

(139)

has an equilibrium point at (x0 , y0 ), the transformation X = x − x0 , Y = y − y0 will shift it to the origin in the (X, Y) phase plane. Accordingly, when considering an equilibrium point of system (139), we will always assume that such a translation has been made. It is plausible to expect that when the nonlinear system in (139) has an equilibrium point at the origin, and in some sense the system is close to a linear system, then the nature of the equilibrium point at the origin will be the same in both systems. To make more precise the meaning of the term close, we restrict consideration to functions f and g that can be written f (x, y) = ax + by + F(x, y) g(x, y) = cx + dy + G(x, y),

(140)

Section 6.12

Autonomous Systems of Equations

367

where ad − bc = 0 and the nonlinear terms F and G are such that

lim

x→0,y→0

F(x, y) =0 x 2 + y2

and

lim

x→0,y→0

G(x, y) = 0. x 2 + y2

(141)

This conjecture concerning the relationship between the equilibrium points of a nonlinear and a related linear autonomous system can be shown to be correct, subject only to a single qualiﬁcation. Speciﬁcally, if the linearized system dx = ax + by dt dy = cx + dy dt

(142)

has a node, a saddle point, or a focus at the origin, then so also has the nonlinear system in (140). The qualiﬁcation that must be added is that if the equilibrium point at the origin of the linearized system in (142) is a center, then the corresponding nonlinear system in (140) has an equilibrium point at the origin that is either a center or a focus. The reason why a center of the linear system (142) may be either a center of a focus of the nonlinear system (140) is not difﬁcult to understand. Conditions (c) and (d) at the end of section (vi) show that the criteria identifying a focus and a center in the linear case are closely related, and it is due to the insensitivity of the linearization process that it fails to distinguish between them when a nonlinear autonomous system is considered. No proof of these statements will be offered here, as this involves methods that do not belong to this ﬁrst account of autonomous systems. However, a detailed proof of the nature of the relationship between the types of equilibrium points in nonlinear and linearized systems, together with other important results due to Liapunov, Poincare, ´ and others, can be found in the references at the end of the book. Nonlinear autonomous systems possess an important property that is not shared by linear systems. This is that in the phase plane a curve may exist, not enclosing an equilibrium point, with the property that a trajectory starting from a point either inside or outside is attracted to and spirals into it as t increases. A curve of this type, to which trajectories are attracted, is called a limit cycle for the system. Clearly, although a limit cycle represents a stable oscillatory solution, it is not one that is ´ asymptotically stable. This statement is essentially the substance of the Poincare– Bendixson theorem, the details of which can be found in the references at the end of the book. HENRI POINCARE´ (1854–1912) An outstanding French mathematician who studied in the Ecole Polytechnique in France before proceeding to study in the Ecole Nationale Superieure des Mines in Paris and receiving his doctorate from the University of Paris in 1879. He was appointed to the chair of physical and experimental mechanics at the Sorbonne and later to the chairs of mathematical physics and then the chair of mathematical astronomy. He made fundamental contributions to almost all of mathematics and was probably the last of the mathematical geniuses about whom it could truly be said that he knew all that was then known about mathematics.

368

Chapter 6

Second and Higher Order Linear Differential Equations and Systems

It was proved separately by Bendixson that if in system (139) the functions f and g have continuous partial derivatives for all x and y, and fx + fy is either positive or negative in some region of the phase plane, then the system has no limit cycle in . Although the proof of this result is not difﬁcult, it will not be given here. The result is useful for establishing the nonexistence of limit cycles in given regions of the phase plane. A theorem that gives sufﬁcient, though not necessary, conditions for the existence of a limit cycle for a special type of autonomous system is Lienard’s ´ theorem. The theorem is now stated without proof. THEOREM 6.9 conditions identifying a limit cycle: Lienard’s ´ theorem

´ Lienard’s theorem Write the linear equation d2 x dx + g(x) = 0 + f (x) 2 dt dt as the ﬁrst order Lienard ´ system dx =y dt dy = −g(x) − f (x)y. dt Let f (x) and g(x) satisfy the following conditions: (i) f (x) and g(x) are continuous functions with continuous ﬁrst derivatives for all x. (ii) g(x) is an odd function that is positive for x > 0 and f (x) is an even function. x (iii) the function F(x) = 0 f (ξ )dξ , which is an odd function, has precisely one positive root at x = α, with F(x) < 0 for 0 < x < α, F(x) > 0 and nondecreasing for x > α, and F(x) → ∞ as x → ∞. Then the Lienard ´ system possesses a unique closed curve enclosing the origin in the phase plane, with the property that every trajectory spirals toward as t → ∞.

van der Pol equation and phase portraits

An application of this theorem will be made later to the van der Pol equation d2 x dx + x = 0, + ε(x 2 − 1) dt 2 dt

(143)

which provides a classical example of a limit cycle. The equation itself was derived in the 1920s by Balthazar van der Pol when studying self-sustained oscillations in vacuum tubes, and it was his work that prompted Lienard ´ to study corresponding problems in nonlinear mechanics. The task of ﬁnding the complete phase portrait of a nonlinear autonomous system, usually called the global phase portrait, can be difﬁcult. This is because nonlinear systems may have more than one equilibrium point, and while linearization techniques provide information in a neighborhood of each of these points (with the exception of centers), they provide very little information about the general phase portrait or any separatrix that may occur, and no information at all about the existence of a limit cycle, though Lienard’s ´ theorem helps in the linear case.

Section 6.12

Autonomous Systems of Equations

369

The Predator–Prey Problem

more on the predator–prey problem

The predator–prey equations have been shown to have a single physically meaningful equilibrium point at (c/d, a/b) in the phase plane, where the linearized form of the equations has a center with elliptical trajectories surrounding it. In view of the fact that when the linearized form of a nonlinear system identiﬁes an equilibrium point as a center, the associated nonlinear system may have either a center or a focus, a more careful examination is necessary in the predator–prey case before it is possible to state with certainty that (c/d, a/b) is a center and that cyclic variations in the populations take place. In more advanced accounts of nonlinear autonomous systems, theorems exist that can resolve this ambiguity, but here we will make use of a simple device that in this and other straightforward cases will sufﬁce to distinguish between the two possibilities. The idea is simple, and it involves asking how many times a trajectory will intersect a straight line drawn through the equilibrium point at (c/d, a/b). If the equilibrium point is a center, a trajectory can only intersect this line twice, but if it is a focus (a spiral point) it will intersect it inﬁnitely many times. Dividing the second of the predator–prey equations in (122) by the ﬁrst equation, rearranging terms, and integrating gives (a − by) (xd − c) dy = dx, y x and so a ln y + c ln x − by − xd = k, where k is an integration constant. To proceed further, we consider a typical case where a = 1, b = 1, c = 2, and d = 1, when the predator–prey system will have an equilibrium point at (2, 1) in the phase plane, and the equation determining the trajectories becomes ln y + 2 ln x − y − x = k. Let us now select a convenient trajectory through any point in the ﬁrst quadrant that does not coincide with the equilibrium point. It is convenient to choose the point (1, 1), when it follows from the above equation that k = −2, so the equation of the trajectory through this point becomes ln y + 2 ln x − x = −3. We may choose any test line through the equilibrium point, but it is simplest to choose the line y = 1 that passes through the equilibrium point of the system at (2, 1) in the phase plane. Setting y = 1 in the preceding equation reduces it to 2 ln x − x = −3, so if this equation has only two real roots the equilibrium point will be a center, but if it has inﬁnitely many it will be a focus. Graphing y = 2 ln x − x and y = −3 to determine where they intersect, we ﬁnd that only two intersections occur, with one at x ≈ 0.25 and the other at x ≈ 6.85. This shows that in this model of the predator–prey system the equilibrium point at (2, 1) must be a center. A similar argument applies to any other choice of nonnegative coefﬁcients a, b, c, and d. This demonstrates that the equilibrium point of the predator–prey system located at (2, 1) in the ﬁrst quadrant of the phase plane is, indeed, a center.

370

Chapter 6

Second and Higher Order Linear Differential Equations and Systems

y

x

2.5

5

2

4

1.5

3 P

1

2 1

0.5 0

1

2

3

5 x

4

2

4

6

(a)

8

10 t

(b) y 7 6 5 4 3 2 P

1 0

2

4

6

8

x

(c) FIGURE 6.21 (a) The phase plane for the system through the point (1, 1) with an equilibrium point at (2, 1). (b) The variation of x(t) showing the cycle time to be approximately 4.7 time units. (c) A general family of trajectories, each with the same equilibrium point.

Negative rabbit and fox populations have no physical signiﬁcance, so no attention need be paid to the saddle point located at the origin of the phase plane, but notice that each axis is a separatrix belonging to the saddle point. Accordingly, the computer-generated phase portrait in the ﬁrst quadrant is shown in Fig. 6.21a, with a = 1, b = 1, c = 2, and d = 1, the rabbit population along the horizontal axis, and the fox population along the vertical axis. The equilibrium point is shown as P. To ﬁnd the period of this cycle of events, it is sufﬁcient to ﬁnd the period of either x(t) or y(t). The variation of x(t) is shown in Fig. 6.21b with t along the horizontal axis and x along the vertical axis, from which the period is seen to be approximately T ≈ 4.7 time units. Figure 6.21c shows a general family of trajectories for this system, each with a different period.

The Undamped and Damped Simple Pendulum study of the undamped and damped pendulum

The geometry of the simple pendulum is illustrated in Fig. 6.22, where a mass m is attached to the end of a light rigid rod of length l that is pivoted at the end opposite to the mass and allowed to oscillate under gravity. The equation of motion, when damping proportional to dθ/dt is present, can be written ml 2

d2 θ dθ + 2mlk + mgl sin θ = 0, 2 dt dt

Section 6.12

Autonomous Systems of Equations

371

m mg

θ l

m

m

mg

mg

(a)

(b)

(c)

FIGURE 6.22 (a) Small oscillations. (b) Stable equilibrium. (c) Inverted pendulum—unstable equilibrium.

where k > 0 is a constant. Here, to simplify the associated characteristic equation, the constant of proportionality for wind resistance has been set equal to 2ml k. This is equivalent to setting μ = 2mk/l in the equation of motion for a damped pendulum derived at the start of Section 6.1.

The undamped pendulum Let us start by considering the undamped case k = 0. Introducing the new variable x = dθ/dt, we see the nonlinear autonomous system determining the motion to be )g* dx =− sin θ dt l

and

dθ = x, dt

with equilibrium points on the θ-axis where sin θ = 0. This shows there are inﬁnitely many equilibrium points along the θ -axis at θ = ±nπ, for n = 0, 1, . . . . Accordingly, because of the periodicity of sin θ , only the interval −π ≤ θ ≤ π need be considered. If we write sin θ = θ + (sin θ − θ ), the system becomes )g* )g* dx =− θ− (sin θ − θ ) dt l l dθ = x. dt

372

Chapter 6

Second and Higher Order Linear Differential Equations and Systems

The nonlinear term (g/l) (sin θ − θ) satisﬁes the condition in (141), so when the equilibrium point at the origin is considered, the Jacobi matrix becomes 0 −g/l J= . 1 0 This has the purely imaginary eigenvalues λ1 = −i (g/l) and λ2 = i (g/l), so the equilibrium point of the linearized system located at the origin is a center. An argument similar to the one used with the predator–prey equations can be used to show that any trajectory starting at a point on the line θ = 0 in the interval −π < θ < π will intersect the x-axis twice, so the equilibrium point of the nonlinear system is also a center. This conﬁrms the expected result that the pendulum will perform periodic oscillations. Next we must consider the equilibrium point at (π, 0), and to do this we shift the origin of the system to this point by setting u = θ − π . This causes the equation dx/dt = −(g/l) sin θ to become dx/dt = (g/l) sin u, so the system can now be written du =x dt )g* dx ) g * = u+ (sin u − u). dt l l The nonlinear term again satisﬁes the conditions in (141), so the nature of this equilibrium point is determined by the eigenvalues of the Jacobi matrix J, which now becomes 0 g/l . J= 1 0 This has the real eigenvalues λ1 = − (g/l) and λ2 = (g/l), so as these are of opposite sign the equilibrium point at (π, 0) is seen to be a saddle point. An analogous argument shows that the equilibrium point at (−π, 0) is also a saddle point, so the nonlinear system also has saddle points at (±π, 0). A repetition of these arguments shows the equilibrium points at (±2nπ, 0) all to be centers, and the equilibrium points at ((2n + 1)π, 0) all to be saddle points. A computer plot of some typical trajectories is shown in Fig. 6.23a. An examination of Fig. 6.23a explains the signiﬁcance of these centers and saddle points. As the angular displacement of the pendulum is indeterminate up to a multiple of 2π , each center represents the stable nonlinear oscillations that occur in Fig. 6.22a when the pendulum never becomes inverted. Similarly, each saddle point represents the unstable position of the inverted pendulum shown in Fig. 6.22c. As the oscillations are nonlinear, each different closed curve about a center represents a nonlinear oscillation with a different period. Each dashed curve is a separatrix forming a boundary between phase curves with different properties. An important and useful result is obtained by writing d2 θ dx dθ dx dx = = =x . 2 dt dt dt dθ dθ

Section 6.12

Autonomous Systems of Equations

373

•

θ

−π

π

2π

3π θ

(a) •

θ

θ

(b) FIGURE 6.23 (a) The phase portrait for the undamped pendulum. (b) The phase portrait for the damped pendulum.

Using this result the equation of motion becomes ml 2 x

dx + mglsin θ = 0, dθ

so after integration we have 1 2 ml 2

dθ dt

2 − mglcos θ = C,

where C is an integration constant. This ﬁrst integral of the equation of motion expresses the conservation of energy in the system, which is possible because when k = 0 there is no dissipation of energy due to friction.

The damped pendulum When damping occurs (k > 0), the nonlinear autonomous system governing the oscillations of the pendulum becomes dθ =x dt

and

dx −2kx ) g * = − sin θ. dt l l

374

Chapter 6

Second and Higher Order Linear Differential Equations and Systems

Considering the equilibrium point that again occurs at the origin, we write the system as dθ =x dt )g* dx −2kx ) g * = − θ− (sin θ − θ). dt l l l Then, proceeding as before, we see that the nature of the equilibrium point at the origin is determined by the eigenvalues of the Jacobi matrix J=

−2k/l 1

−g/l . 0

The characteristic equation of J is λ2 + (2k/l)λ + g/l = 0, so as λ = −k/l ± k2 − lg/l, and as k > 0, the eigenvalues are real and negative when k > g/l, corresponding to overdamped oscillations. When (k/l)2 < g/l the eigenvalues are complex conjugates with negative real parts, corresponding to the asymptotically stable oscillatory case. So, when friction is present, the equilibrium point at the origin is seen to be an asymptotically stable focus. In time, friction will cause the oscillations to decay to zero, causing the pendulum to come to rest in the positions shown in Fig. 6.23b. EXAMPLE 6.40

Locate and classify the equilibrium points of the nonlinear autonomous system dx = 4 − x 2 − 4y2 dt

and

dy = xy. dt

Solution The equilibrium points occur when 4 − x 2 − 4y2 = 0 and xy = 0, so the points are located at (0, −1), (0, 1), (2, 0), and (−2, 0). Let us consider the equilibrium point at (0, 1) and shift the origin to this point by setting Y = y − 1 and X = x. The system now becomes dX = −8Y − X 2 − 4Y 2 dt

and

dY = X + XY. dt

Setting X = r cos θ , Y = r sin θ , we easily see that conditions (141) are satisﬁed, so the nature of the equilibrium point at (0, 1) will be determined by the eigenvalues of the Jacobi matrix 0 −8 J= . 1 0 These satisfy the characteristic equation λ2 + 8 = 0, so as they are purely imaginary, the equilibrium point of the linearized system that is located at (0, 1) must be a center, and arguments similar to those used with the pendulum problem conﬁrm that the nonlinear system also has a center at (0, 1).

Section 6.12

Autonomous Systems of Equations

y

375

y

1.5 0.2

1

0.1

0.5

−1.5

−1

−0.5

0.5

1

1.5 x

−0.2

−0.1

−0.5

0.1

0.2

x

−0.1

−1

−0.2

−1.5 (a)

(b)

FIGURE 6.24 (a) The origin is a center. (b) The origin is an unstable focus.

It is left as an exercise to use similar arguments to show that the equilibrium point at (0, −1) is also a center and the equilibrium points at (−2, 0) and (2, 0) are saddle points. The inability of a linearized system to reﬂect the difference between a center and a focus in the nonlinear system from which it is derived is best illustrated by means of computer-generated phase portraits. The following two systems only differ in the power of x associated with dx/dt, and each has the same linearized form that indicates the existence of a center at the origin of the phase plane: (i)

dx = −4y + x 2 dt

and

dy = 4x + y2 dt

(ii)

dx = −4y + x 3 dt

and

dy = 4x + y2 . dt

and

However, the nonlinear phase portrait of system (i) in Fig. 6.24a shows that the system does, indeed, have a center located at the origin in the phase plane, but the nonlinear phase portrait of system (ii) in Fig. 6.24b shows that the system has an unstable focus at the origin. A typical example of a limit cycle is provided by the van der Pol equation dx d2 x + x = 0. + ε(x 2 − 1) dt 2 dt If we set f (x) = ε(x 2 − 1) and g(x) = x in Lienard’s ´ theorem, it is easily seen that x the conditions of the theorem are satisﬁed provided F(x) = 0 ε(ξ 2 − 1)dξ has precisely one positive root x = α with F(x) < 0 for 0 < x < α, and F(x) is such that it is positive and nondecreasing for x > α with F(x) → ∞ as x → ∞. This

376

Chapter 6

Second and Higher Order Linear Differential Equations and Systems

−3

−2

−1

y

y

3

3

2

2

1

1

1

2

3

x

−3

−1

−2

−1

−1

−2

−2

−3

−3

(a)

1

2

3

x

(b)

y 3

x 3

2 2 1

−3

−2

−1

0 −1 −2

1 1

2

3

x

5

10

15

20

25

30 t

−1 −2 −3

−3 (c)

(d)

FIGURE 6.25 Phase portraits for the van der Pol equation with ε = 0.9 and the variation of x(t) with t. (a) A trajectory starting outside the limit circle. (b) A trajectory starting inside the limit cycle. (c) The limit cycle. (d) The periodicity of x(t) as a function of t.

is seen to be the case, because F(x) = 13 ε(x 3 − 3x), so the theorem ensures the existence of a limit cycle for the van der Pol equation provided ε > 0. Figure 6.25a shows a computer-generated phase portrait for the van der Pol equation with = 0.9, where the trajectory starting from an initial point at t = 0 outside the limit cycle (the parallelogram-shaped closed curve) is attracted inward toward the limit cycle. Figure 6.25b shows the corresponding situation when the initial point lies inside the limit cycle, where here the trajectory is attracted outward toward the limit cycle. Figure 6.25c shows the limit cycle itself. A plot of x(t) against t is shown in Fig. 6.25d, from which the solution is seen to become periodic, with a period of approximately 6.5 time units, after the time t = 5. More examples of the phase plane are to be found in references [3.3] to [3.5], whereas a more extensive and advanced account is to be found in references [3.1], [3.2], and [3.13].

Section 6.12

Summary

Autonomous Systems of Equations

377

An autonomous system involving the variables x(t) and y(t), where the parameter t is usually the time, are systems of the form dx/dt = f (x, y) and dy/dt = g(x, y), where the dependence of the f and g on t is implicit. Critical points of such systems were deﬁned and the concept of a trajectory, or path, was introduced leading to the notion of a phase portrait. Stability, instability, and asymptotic stability were deﬁned, and the classical predator– prey problem was used to illustrate ideas. Linearization of the functions f and g led to the identiﬁcation of different types of critical points for linear autonomous systems. These ideas were extended to nonlinear autonomous systems where it was possible for trajectories to spiral in or out until they entered a closed loop called a limit cycle, where the solution became periodic, though nonlinear. These ideas were illustrated by application to the full nonlinear predator–prey problem, the pendulum problem, and the van der Pol equation.

EXERCISES 6.12 In Exercises 1 through 6, locate and identify the nature of the equilibrium point and sketch the pattern of the trajectories. 1. 2. 3. 4. 5. 6.

dx/dt dx/dt dx/dt dx/dt dx/dt dx/dt

= y, dy/dt = x. = x + 2, dy/dt = −x + 2y − 8. = x − 2y, dy/dt = 4x − 3y. = x − y, dy/dt = 2x − y. = x + 3y − 4, dy/dt = −6x − 5y + 22. = 2y − x, dy/dt = 3x + 6.

In Exercises 7 through 9 locate the equilibrium points of the given nonlinear autonomous system and, where possible, use linearization to identify their nature. 7. dx/dt = x 2 − y2 − 4, dy/dt = y. 8. dx/dt = 2 + y − x 2 , dy/dt = x 2 − xy.

9. dx/dt = x + y + y2 , dy/dt = 2x + y. 10. Locate and identify the equilibrium points of dx/dt = −x + xy,

dy/dt = 3y − 2xy + x.

11. Show that the only equilibrium point of the van der Pol equation dx d2 x +x=0 + ε(x 2 − 1) dt 2 dt is located at the origin. By linearizing the equation about the origin, ﬁnd conditions that must be imposed on ε in order that (a) the equilibrium point be an unstable spiral, (b) that it be an unstable node, and (c) that it be a center. Relate your results to the phase portraits in Fig. 6.25.

378

Chapter 6

Second and Higher Order Linear Differential Equations and Systems

CHAPTER 6

TECHNOLOGY PROJECTS The purpose of the ﬁrst two projects is to use a computer algebra phase portrait package to construct the phase portraits for linear and nonlinear systems, and to examine the nature of the limit cycles in the van der Pol equation for different choices of the parameter ε and the initial conditions. Project 1 Phase Portraits Use a computer phase portrait package to construct the phase portraits for the following systems about the origin: (a) (b) (c) (d) (e)

dx dt dx dt dx dt dx dt dx dt

dy = x + 2y. dt dy = x + 2y, = x 3y. dt dy = 2x 3y, = x + 2y. dt dy = 2x 4y, = 4x 2y. dt dy = x + 3y2 , = x + 2y. dt = 2x 2

3y,

Project 2 The Limit Cycle of the van der Pol Equation Use a computer algebra phase portrait package to construct integral curves for the van der Pol equation x + ε(x 2

1)x + x = 0

for ε = 0.5, 1.0, and 1.5, starting trajectories from points inside and outside the limit cycle shown in Fig. 6.25. Project 3 Period of Oscillation of a Nonlinear Pendulum The nonlinear equation of motion of a simple pendulum when the mass of the pendulum rod is neglected is mφ + (mg/l) sin φ = 0,

378

where a prime denotes differentiation with respect to the time t, m is the mass of the pendulum bob, g is the acceleration due to gravity, l is the length of the pendulum, and φ is the angle of deﬂection of the pendulum from the vertical. When the maximum angle of deﬂection of the pendulum from the vertical is θ , the period of oscillation T is given by the complete elliptic integral 0 l π/2 du T=4 . (I) 2 g 0 (1 sin (u) sin2 ( 12 θ ))1/2 1. Use the numerical integration facility of MAPLE to ﬁnd (T/4) (g/l) for some speciﬁc θ . 2. Expand the integrand of (I) as a Maclaurin series in u and integrate term byterm to ﬁnd a series representation for (T/4) (g/l) in terms of powers of sin θ . 3. Set θ = 2π/5 and approximate the result in Part 2 by taking the ﬁrst N terms, with N = 2m and m = 1, 2, . . . . By repeatedly doubling N and comparing the estimate of (T/4) (g/l) with the result obtained in Part 1, ﬁnd how many terms must be used in the approximation if the result is to agree to four decimal places.

C H A P T E R

7

The Laplace Transform

M

any problems in engineering and physics can be described in terms of the evolution of solutions of linear differential equations subject to initial conditions. An important group of these problems involves constant coefﬁcient differential equations, and equations like these can be solved very easily by using the Laplace transform. The Laplace transform is an integral transform that changes a real variable function f (t) into a function F (s) of a variable s through ∞ F (s) = e−st f (t) dt, 0

where in general s is a complex variable. The importance of the Laplace transform in the study of initial value problems for linear constant coefﬁcient differential equations is that it replaces the operation of integrating a differential equation in f (t) by much simpler algebraic operations involving F (s). Unlike previous methods, where ﬁrst a general solution is found, and then the constants in the complementary function are chosen to match the initial conditions, when the Laplace transform method is used the initial conditions are incorporated from the start. The task of ﬁnding the function f (t) from its Laplace transform F (s) is called inverting the transform, and when working with constant coefﬁcient equations we can accomplish this by appeal to tables of Laplace transform pairs—that is, to a table listing a function f (t) and its corresponding Laplace transform F (s). The fundamental ideas underlying the Laplace transform are derived, along with its operational properties, which are illustrated by examples. Initial value problems for ordinary differential equations are solved by the Laplace transform, which is then applied to systems of equations and to certain variable coefﬁcient equations. The chapter concludes with applications of the Laplace transform to a variety of problems, the last of which is the heat equation.

7.1

Laplace Transform: Fundamental Ideas et the real function f (t) be deﬁned for a ≤ t ≤ b, and let the function K(t, s) of the variables t and s be deﬁned for a ≤ t ≤ b and some s. When it exists, the

L

379

380

Chapter 7

The Laplace Transform

b integral a f (t)K(t, s) dt is a function of the single variable s, so denoting the integral by F(s) we can write F(s) =

b

K(t, s) f (t)dt.

(1)

a

The function F(s) in (1) is called an integral transform of f (t), the function K(t, s) is the kernel of the transform, and s is the transform variable. The limits a and b may be ﬁnite or inﬁnite, and when at least one limit is inﬁnite the integral in (1) becomes an improper integral. When it exists, the Laplace transform F(s) of a real function f (t) with domain of deﬁnition 0 ≤ t < ∞ is deﬁned as the integral transform (1) with the kernel K(t, s) = e−st , the interval of integration 0 ≤ t < ∞, and s a complex variable such that Re s < c for some nonnegative constant c, so that ∞ e−st f (t)dt. (2) F(s) = 0

Throughout the present chapter the transform variable s will be considered to be a real variable, and c will be chosen such that the integral in (2) converges. However, when the general problem of recovering a function f (t) from its Laplace transform F(s) is considered in Chapter 16, it will be seen that s must be allowed to be a complex variable. The advantage of restricting s to the real variable case in this chapter is that the recovery of many useful and frequently occurring functions f (t) from their Laplace transforms F(s) can be accomplished in a very simple manner without the use of complex variable methods. The reason for interest in integral transforms in general, and the Laplace transform in particular, will become clear when the solution of initial value problems for differential equations is considered. It will then be seen that the Laplace transform replaces integrations with respect to t by simple algebraic operations involving F(s), so provided f (t) can be recovered from F(s) in a simple manner, the solution of an initial value problem can be found by means of straightforward algebraic operations. Clearly the kernel e−st will only decrease as t increases if s > 0, and the Laplace transform of f (t) will only be deﬁned for functions f (t) that decrease sufﬁciently rapidly as t → ∞ for the integral in (2) to exist. In general, if the function to be transformed is denoted by a lowercase letter such as f , then its Laplace transform will be denoted by the corresponding uppercase letter F, as in (2). It is convenient to denote the Laplace transform operation by the symbol L, so that symbolically F(s) = L{ f (t)}. The Laplace transform formal deﬁnition of the Laplace transform

Let f (t) be deﬁned for 0 ≤ t < ∞. Then, when the improper integral exists, the Laplace transform F(s) of f (t), written symbolically F(s) = L{ f (t)}, is deﬁned as F(s) = 0

∞

e−st f (t)dt.

Section 7.1

EXAMPLE 7.1

Laplace Transform: Fundamental Ideas

381

Find L{eat } where a is real. Solution From (2) we have

∞

L{e } = at

e−st eat dt

t→∞ −e−(s−a)t = s−a 0 −(s−a)t −e 1 = lim + t→∞ s−a s−a =

1 , s−a

provided s > a, for only then will the limit in the ﬁrst term vanish. This has shown that L{eat } = F(s) = 1/(s − a) for s > a, where it is necessary to include the inequality s > a to ensure the convergence of the integral. PIERRE SIMON LAPLACE (1749–1827) A French mathematician of remarkable ability who made contributions to analysis, differential equations, probability, and celestial mechanics. He used mathematics as a tool with which to investigate physical phenomena, and made fundamental contributions to hydrodynamics, the propagation of sound, surface tension in liquids, and many other topics. His many contributions had a wide-ranging effect on the development of mathematics. Laplace transform pair and inverse transform

The two functions f (t) and F(s) are called a Laplace transform pair, and for all ordinary functions, given F(s) the corresponding function f (t) is determined uniquely, just as f (t) determines F(s) uniquely. This relationship is expressed symbolically by using the symbol L−1 to denote the operation of ﬁnding a function f (t) with a given Laplace transform F(s). This process is called ﬁnding the inverse Laplace transform of F(s). In terms of the foregoing example, we have L{eat } = 1/(s − a) and L−1 {1/(s − a)} = eat . This is a particular case of the general result that, by deﬁnition, the inverse Laplace transform acting on the Laplace transform of the function returns the original function, so we can write L−1 {L{ f (t)}} = f (t).

how to be sure a Laplace transform exists

A sufﬁcient condition for the existence of the Laplace transform of a function f (t) is that the absolute value of f (t) can be bounded for all t ≥ 0 by | f (t)| ≤ Mekt ,

(3)

for some constants M and k. This means that if numbers M and k can be found such that |e−st f (t)| ≤ Me(k−s)t , then

L{ f (t)} = 0

∞

e−st f (t)dt ≤ M

∞

e(k−s)t dt = M/(s − k).

382

Chapter 7

The Laplace Transform

TABLE 7.1 Laplace Transform Pairs f (t) 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14.

F(s) = L{ f (t)}

Condition on s s>0 s>0 s>0 s>a s>a s>a s≥a s > 0, a > 0 s>0 s>0 s>0 s>0 s>a s>a

1 t t n (n = 1, 2, . . .) t a (a > −1) eat t n eat (n = 1, 2, . . .) H(t − a) δ(t − a) sin at cos at t sin at t cos at eat sin bt eat cos bt 1 1 15. sin at − 2 t cos at 2a 3 2a

1/s 1/s 2 n!/s n+1 (a + 1)/s a+1 1/(s − a) n!/(s − a)n+1 e−as /s e−as a/(s 2 + a 2 ) s/(s 2 + a 2 ) 2as/(s 2 + a 2 )2 (s 2 − a 2 )/(s 2 + a 2 )2 b/[(s − a)2 + b2 ] (s − a)/[(s − a)2 + b2 ] 1/(s 2 + a 2 )2

s>0

1 1 sin at + t cos at 2a 2 17. 1 − cos at 18. at − sin at

s 2 /(s 2 + a 2 )2

s>0

a 2 /[s(s 2

a 3 /[s 2 (s 2 + a 2 )]

s>0 s>0

a/(s 2 − a 2 ) s/(s 2 − a 2 )

s > |a| s > |a|

1/(s 2 − a 2 )2

s > |a|

s/(s 2 − a 2 )2

s > |a|

1 1 sinh at + t cosh at 2a 2 24. sinh at − sin at

s 2 /(s 2 − a 2 )2

s > |a|

2a 3 /(s 4

s > |a|

25. cosh at − cos at

2a 2 s/(s 4 − a 4 )

16.

19. sinh at 20. cosh at 1 1 21. sinh at + 2 t cosh at 2a 3 2a 22.

1 t sinh at 2a

23.

+ a 2 )]

− a4)

s > |a|

The integral on the right will be convergent provided s > k > 0, so when this is true the Laplace transform F(s) = L{ f (t)} will exist. It should be clearly understood that (3) is only a sufﬁcient condition for the existence of a Laplace transform, and not a necessary one, because Laplace transforms can be found for functions that do not satisfy condition (3). For example, the function f (t) = t −1/4 does not satisfy condition (3), but its Laplace transform exists and is a special case of entry 4 in Table 7.1. The preceding inequality implies that when L{ f (t)} exists, F(s) must be such that lims→∞ F(s) = 0. In addition, the condition L{ f (t)} ≤ M/(s − k) implies that F(s) cannot be the Laplace transform of on ordinary function f (t) unless F(s) → 0 as s → ∞. For example, F(s) = (s 2 − 1)/(s 2 + 1) is not a Laplace transform of an ordinary function. Exceptions to this condition are functions like the delta function, which is deﬁned in Section 7.2, though there the delta function will be seen to involve integration, and so it is not a function in the usual sense. The Laplace transform is a linear operation, and the consequence of this important and useful property is expressed in the following theorem.

Section 7.1

THEOREM 7.1 fundamental linearity property

Laplace Transform: Fundamental Ideas

383

Linearity of the Laplace transformation Let the functions f1 (t), f2 (t), . . . , fn (t) have Laplace transforms, and let c1 , c2 , . . . , cn be any set of arbitrary constants. Then L{c1 f1 (t) + c2 f2 (t) + · · · + cn fn (t)} = c1 L{ f1 (t)} + c2 L{ f2 (t)} + · · · + cn L{ fn (t)}. Proof The proof is simple and follows directly from the fact that integration is a linear operation, so the integral of a sum of functions is the sum of their integrals. Thus, ∞ e−st {c1 f1 (t) + c2 f2 (t) + · · · + cn fn (t)}dt 0

= c1 0

∞

f1 (t)e−st dt + c2

∞

f2 (t)e−st dt + · · · + cn

∞

fn (t)e−st dt

= c1 L{ f1 (t)} + c2 L{ f2 (t)} + · · · + cn L{ fn (t)}. This theorem has many applications and its use is essential when working with the Laplace transform. EXAMPLE 7.2

some examples

Find the Laplace transform of f (t) = c1 eat + c2 e−at , and use the result to ﬁnd L{sinh at} and L{cosh at}. Solution Applying Theorem 7.1 and the result L{eat } = 1/(s − a) from Example 7.1, we ﬁnd that L{c1 eat + c2 e−at } = c1 L{eat } + c2 L{e−at } = c1 /(s − a) + c2 /(s + a). As sinh at = (eat − e−at )/2 and cosh at = (eat + e−at )/2, L{sinh at} is obtained from the preceding result by setting c1 = 1/2 and c2 = −1/2, and L{cosh at} is obtained by setting c1 = c2 = 1/2, when we obtain L{sinh at} = a/(s 2 − a 2 )

and

L{cosh at} = s/(s 2 − a 2 ),

for s > |a| ≥ 0. Notice that because s must be be positive, but in sinh at and cosh at the number a may be either positive or negative, the relationship between s and a necessary to ensure that the convergence of the integrals must be s > |a| ≥ 0, and not s > a > 0. The process of ﬁnding an inverse Laplace transformation involves reversing the foregoing argument and seeking a function f (t) that has the required Laplace transform F(s). Where possible, this is accomplished by simplifying the algebraic structure of F(s) to the point at which it can be recognized as the sum of the Laplace transforms of known functions of t. EXAMPLE 7.3

Find the inverse Laplace transform of F(s) =

4s + 10 . s 2 + 6s + 8

384

Chapter 7

The Laplace Transform

Solution Expanding the Laplace transform in terms of partial fractions gives 1 3 4s + 10 = + , + 6s + 8 s+2 s+4

s2 so

L−1 {F(s)} = L−1

4s + 10 s 2 + 6s + 8

(

= L−1

( ( 1 1 + 3L−1 . s+2 s+4

Using the result of Example 7.1 we ﬁnd that ( 4s + 10 = e−2t + 3e−4t . f (t) = L−1 2 s + 6s + 8

EXAMPLE 7.4

Find (a) L{1} and (b) L{t}. Solution (a) By deﬁnition,

L{1} =

∞

1 , s

e−st dt =

(b) By deﬁnition, L{t} = 0

EXAMPLE 7.5

∞

for s > 0.

∞ t e−st 1 e−st tdt = − e−st − 2 = 2, s s s t=0

for s > 0.

Find L{sin at}. Solution By deﬁnition,

L{sin at} =

∞

= lim

k→∞

=

e−st sin atdt = lim

k

k→∞ 0

e−st sin atdt

−e−sk(a cos ak + s sin ak) s2 + a2

a s2 + a2

+

s2

a + a2

for s > 0,

where the condition s > 0 is required to ensure that the limit is ﬁnite as k → 0. This has shown that a for s > 0. L{sin at} = 2 s + a2 In the next example we ﬁnd L{t n }, and in the process introduce an integral that will be useful later in Chapter 8 when ﬁnding series solutions of linear second order variable coefﬁcient differential equations. EXAMPLE 7.6

Find L{t n } for n = 1, 2, . . . . Solution By deﬁnition

L{t n } = 0

∞

e−st t n dt.

Section 7.1

Laplace Transform: Fundamental Ideas

385

To evaluate this integral we will make use of integration by parts to establish a recursion (recurrence) relation from which the result for arbitrary positive integral n can be found. Accordingly, we deﬁne I(n, s) as ∞ k n −t d −st I(n, s) = (e )dt e−st t n dt = lim k→∞ s dt 0 0 and use integration by parts to express this as n −st k −t e n ∞ n−1 −st = lim + t e dt k→∞ s s 0 t=0 n = I(n − 1, s), for s > 0. s This has established the recursion relation I(n, s) = (n/s)I(n − 1, s), satisﬁed by the integral I(n, s). ∞ As I(0, s) = 0 e−st dt = 1/s, by setting n = 1 in the recursion relation we ﬁnd that I(1, s) = (1/s)I(0, s) = 1/s 2 ,

for s > 0.

Similarly, setting n = 2 in the recursion relation shows that I(2, s) = (2/s)I(1, s) = 2 · 1/s 3 = 2!/s 3 ,

for s > 0,

and an inductive argument shows that I(n, s) = n!/s n+1 . In terms of the Laplace transform notation, we have shown that L{t n } = n!/s n+1

for n = 0, 1, 2, . . . ,

for s > 0.

Notice that setting s = 1 in the general result of Example 7.3 enables n! to be expressed as the integral ∞ n! = e−t t n dt, for n = 0, 1, 2, . . . . 0

ﬁrst encounter with the Gamma function

This provides a way of representing factorial n in terms of an integral, and it is our ﬁrst encounter with a special case of the Gamma function that will be required later. The gamma function, denoted by (x) for x > 0, is deﬁned by the integral ∞ (x) = e−t t x−1 dt. (4) 0

In terms of the earlier notation, when the restriction that n is an integer is removed, and n is replaced by a positive real variable x, we can write ∞ (x + 1) = e−t t x dt = I(x, 1), 0

but I(x, 1) = x I(x − 1, 1) = x(x)

for x > 0,

386

Chapter 7

The Laplace Transform

so combining results shows that the gamma function satisﬁes the fundamental relation (x + 1) = x(x)

for x > 0.

(5)

It is easily seen from this that (n + 1) = n!

for n = 0, 1, 2, . . . ,

so as (x) is deﬁned for all positive x the gamma function provides a generalization of the factorial function n! for positive non-integer values of n. It will be seen later that the gamma function, which belongs to the general class of functions called higher transcendental functions, occurs frequently throughout mathematics.

Discontinuous Functions Because the Laplace transform is deﬁned in terms of an integral, it is possible to ﬁnd Laplace transforms of discontinuous functions. Suppose, for example, that a function g(t) is discontinuous at t = a, as in Fig. 7.1. Then, provided it converges, the integral deﬁning the Laplace transform of g(t) is given by L{g(t)} = lim

a−ε

ε→0 0

Heaviside step function

e−st g(t)dt + lim

∞

δ→0 a+δ

e−st g(t)dt,

(6)

where ε and δ are both positive. For simplicity, the upper limit in the ﬁrst integral is usually denoted by a− and the lower limit in the second integral by a+ . These are, respectively, the limits of integration to the left and right of t = a. An important discontinuous function that ﬁnds numerous applications in connection with the Laplace transform, and elsewhere, is the unit step function f (t) = H(t − a) with a ≥ 0, known also as the Heaviside step function. The unit step function is deﬁned as

H(t − a) =

⎧ ⎨0 ⎩

1

if t < a if t > a

(a ≥ 0).

(7)

A related function that is also of considerable importance is the unit pulse function,

y y(a − 0)

y = g(t)

y(a + 0)

a

FIGURE 7.1 A discontinuous function g(t).

t

Section 7.1 y

Laplace Transform: Fundamental Ideas

387

y y = H(t − a) − H(t − b)

y = H(t − a)

1

1

a

t

a

b

t

(b)

(a)

FIGURE 7.2 (a) The unit step function y = H(t − a). (b) The unit pulse function y = p(t) = H(t − a) − H(t − b).

y

y

y

y = f (t)

a

y = H(t − a) f (t)

t

a

(a)

y = [H(t − a) − H(t − b)] f (t)

t

(b)

a

b

t

(c)

FIGURE 7.3 The effect on f (t) of multiplication by H(t − a) and H(t − a) − H(t − b).

deﬁned as p(t) = H(t − a) − H(t − b),

switching functions on and off with the Heaviside step function

EXAMPLE 7.7

with b > a ≥ 0.

(8)

The function p(t) operates like a “switch,” because it switches on at t = a and off at t = b. Graphs of these two functions are shown in Fig. 7.2. If a function f (t) is multiplied by a unit step function, the function f (t) can be considered to be “switched on” at time t = a, in the sense that the product H(t − a) f (t) is zero for t < a and f (t) for t > a. Similarly, multiplication of f (t) by a unit pulse function “switches on” the function f (t) at time t = a and “switches it off” at time t = b. This property is illustrated in Fig. 7.3, where Fig. 7.3(a) shows the original function f (t), Fig. 7.3(b) shows the product H(t − a) f (t), and Fig. 7.3(c) the product {H(t − a) − H(t − b)} f (t). In the next example we make use of result (6) to ﬁnd the Laplace transforms of the unit step function and the unit pulse function. Find (a) L{H(t − a)} and (b) L{H(t − a) − H(t − b)}. Solution (a) By deﬁnition

L{H(t − a)} =

∞

e−st dt

a

−st ∞ e e−as = − = s t=a s

for s > a ≥ 0.

388

Chapter 7

The Laplace Transform

(b) Using result (a) we have

b

L{H(t − a) − H(t − b)} =

e−st dt

a

∞

=

e−st dt −

a

= EXAMPLE 7.8

∞

e−st dt

b

e−as − e−bs s

for s > b > a ≥ 0.

Find (a) L{t 3 − 4t + 5 + 3 sin 2t} and (b) L−1 {(s 4 + 5s 2 + 2)/[s 3 (s 2 + 1)]. Solution (a) Using Theorem 7.1 together with the Laplace transform pairs found in the previous examples, we have L{t 3 − 4t + 5 + 2 sin 3t} = L{t 3 } − 4L{t} + L{5} + 3L{sin 2t} = 6/s 4 − 4/s 2 + 5/s + 6/(s 2 + 4) = (5s 5 + 2s 4 + 20s 3 − 10s 2 + 24)/[s 4 (s 2 + 4)]. (b) Simplifying the transform by means of partial fractions gives s 4 + 5s 2 + 2 2 s 3 = 3 + −2 2 . 3 2 s (s + 1) s s s +1 Taking the inverse Laplace transform of each term on the right and using the linearity property of the Laplace transform, we ﬁnd that L−1

s 4 + 5s 2 + 2 s 3 (s 2 + 1)

= L−1

2 s3

(

+ L−1

( ( 3 s − 2L−1 2 . s s +1

Finally, using the transform pairs established in the previous examples, we have −1

L

s 4 + 5s 2 + 2 s 3 (s 2 + 1)

( = t 2 + 3 − 2 cos t.

To make further progress with the Laplace transform it is necessary to have available a table of Laplace transform pairs for the most commonly occurring functions. Theorems to be developed later will enable such a table to be extended in a straightforward manner, so that transforms and inverse Laplace transforms of more complicated functions can be found. Table 7.1 provides a list of the most useful Laplace transform pairs involving elementary functions. All of these entries can be established either by means of routine integration, or by the combination of simpler results, with the sole exception of the delta function δ(t − a) in entry 8. The derivation of this result is to be found in Section 7.2 after the delta function has been deﬁned. The example that now follows illustrates how entry 15 can be found from entries 9 through 12.

Section 7.1

EXAMPLE 7.9

Laplace Transform: Fundamental Ideas

389

Find L−1 {1/(s 2 + a 2 )2 } by combining related entries in Table 7.1. Solution Our objective will be to use the linearity property of the Laplace transform to express 1/(s 2 + a 2 )2 as a linear combination of terms that we hope will be found listed in the column F(s) of Table 7.1. If this is possible, the inverse Laplace transform can then be found by adding the inverse transform of each expression in partial fraction representation of F(s). A routine calculation shows that F(s) can be written as 2 a 1 s − a2 1 1 − , = (s 2 + a 2 )2 2a 3 s 2 + a 2 2a 2 (s 2 + a 2 )2 so from using entries 9 and 12 in Table 7.1 we have ( 1 1 1 −1 = 3 sin at − 2 t cos at, L 2 2 2 (s + a ) 2a 2a and this is entry 15 in the table.

Summary

The Laplace transform of a function f (t) has been deﬁned. A condition has been given that ensures the existence of the transform, and the concept of a Laplace transform pair has been introduced. The transform has been shown to have the fundamental property of linearity, and some simple transform pairs have been found directly from the deﬁnition. The Heaviside unit step function H (t − a), which jumps from zero for 0 ≤ t < a to unity for t > a, has been introduced and used. The section closed with a table of useful Laplace transform pairs.

EXERCISES 7.1 In Exercises 1 through 4 use the deﬁnition of the Laplace transform to obtain the stated result. 1. Show that L{t 2 } = 2/s 3 for s > 0. 2. Show that L{teat } = 1/(s − a)2 for s > a. 3. Find L{eiat }, and by equating the real and imaginary parts show that L{sin at} = a/(s 2 + a 2 ) and L{cos at} = s/(s 2 + a 2 ) for s > 0. 4. Show that L{sinh at} = a/(s 2 − a 2 ) for s > |a|. In Exercises 5 through 20 use Table 7.1 of Laplace transform pairs to ﬁnd L{ f (t)}. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14.

f (t) = te2t . f (t) = 2 sin 3t − cos 3t. f (t) = t − t 2 + t 3 . f (t) = e3t (sin t − cos t). f (t) = e−2t (cos 2t − sin 2t). f (t) = t(sin 2t − cos 2t). f (t) = tcosh 3t − sinh 3t. f (t) = sinh t − t cos t. f (t) = e−t cos 2t − t. f (t) = 2t 2 − 3t + 4 cos 3t.

15. 16. 17. 18. 19. 20.

f (t) = H(t − π/2)et sin t. f (t) = H(t − 3π/2)(sin t − 3 cos t). f (t) = [H(t − π/2) − H(t − π )]t. f (t) = [1 − H(t − π/2)]t. f (t) = H(t − π/2)e−t cos t. f (t) = [1 − H(t − π/2)]e3t .

In Exercises 21 through 30 use Table 7.1 of Laplace transform pairs to ﬁnd L−1 {F(s)}. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30.

F(s) = (s 2 − 1)/[s(s 2 + 4)]. F(s) = (s 2 + 3s + 1)/[s(s 2 − 4)]. F(s) = (3s + 5)/[s(s 2 + 9)]. F(s) = (s 2 − 4)/[(s 2 + 1)(s 2 − 1)]. F(s) = (s 3 − 1)/[(s + 2)2 (s 2 − 9)]. F(s) = (s 2 + s + 1)/[(s 2 + 4)(s 2 − 9)]. F(s) = s 2 /[(s − 1)2 (s + 1)]. F(s) = s/(s − 1)3 . F(s) = (s 2 + 4)/[(s 2 − 9)(s − 1)]. F(s) = (s 2 + 1)/[(s + 1)(s + 2)(s + 3).

In Exercises 31 through 36 ﬁnd the Laplace transform of the function f (t) shown in graphical form.

390

Chapter 7

The Laplace Transform 34.

31. f(t )

f(t) f (t ) = 0, t > 2a

1

f (t) = 0, t > π 1 0

a

2a

−1

π/2

π

t

FIGURE 7.7

FIGURE 7.4

35.

32.

f (t)

f(t ) f (t ) = 0, t > 3π/2

1

f (t) = 0, t > 2a

f (t ) = sin t

f (t) = sin t

t

π/2

π

k

3π/2

t 0

a

t

2a

FIGURE 7.8

−1

36. f (t)

FIGURE 7.5

ka

f (t) = 0, t > 2a

33. f (t) = k, t > 1

f(t) k

a

2a

t

f(t) = kt 0 FIGURE 7.6

7.2

1

t

−ka FIGURE 7.9

Operational Properties of the Laplace Transform In the previous section the Laplace transform of a basic list of commonly occurring functions f (t) was recorded as the list of Laplace transform pairs in Table 7.1. To use the Laplace transform to solve initial value problems for linear differential equations and systems it is necessary to establish a number of fundamental properties of the transform known as its operational properties. This name is given to properties of the transform itself that relate to the way it operates on any function f (t) that is transformed, rather than to the effect these properties of the transform have on speciﬁc functions f (t). This means that operational properties are general properties of the Laplace transform that are not speciﬁc to any particular function f (t) or to its transform

Section 7.2

Operational Properties of the Laplace Transform

391

F(s). An important example of an operational property has already been encountered in Theorem 7.1, where the linearity property of the transformation was established. Some operational properties, such as the scaling and shift theorems that will be proved later, save effort when ﬁnding the Laplace transform of a function or inverting a transform, whereas others such as the transform of a derivative are essential when applying the Laplace transform to solve initial value problems for differential equations. The way derivatives transform is used to ﬁnd how the hom*ogeneous part of a linear differential equation is transformed, and we will see later that it also shows how the initial conditions for the differential equation enter into the transformed equation. Table 7.1 of Laplace transform pairs is needed when transforming the nonhom*ogeneous term in the differential equation. THEOREM 7.2 transforming derivatives

Transform of a derivative Let f (t) be continuous on 0 ≤ t < ∞, and let f (t) be piecewise continuous on every ﬁnite interval contained in t ≥ 0. Then if L{ f (t)} = F(s), L{ f (t)} = s F(s) − f (0). Proof Using integration by parts, and assuming that f satisﬁes the sufﬁciency condition for the existence of a Laplace transform, we have ∞ k L{ f (t)} = e−st f (t)dt = lim e−st f (t)dt k→∞ 0

= lim [e−st f (t)]k0 − lim k→∞

k

k→∞ 0

−se−st f (t)dt

= lim [e−sk f (k) − f (0)] + s F(s) k→∞

= s F(s) − f (0), where limk→∞ e−sk f (k) = 0 because of condition (3). THEOREM 7.3

Transform of a higher derivative Let f (t) be continuous on 0 ≤ t < ∞, and let f (t), f (t), . . . , f (n−1) (t) be piecewise continuous on every ﬁnite interval contained in t ≥ 0. Then if L{ f (t)} = F(s), L{ f (n) (t)} = s n F(s) − s n−1 f (0) − s n−2 f (0) − · · · − s f (n−2) (0) − f (n−1) (0). Proof The proof uses repeated integration by parts, but otherwise is analogous to the one used in Theorem 7.2, so the details are left as an exercise. The two most frequently used results are those of Theorem 7.2 and the result from Theorem 7.3 corresponding to n = 2, so for convenience we record these here. The Laplace transform of ﬁrst and second derivatives L{ f (t)} = s F(s) − f (0).

(9a)

L{ f (t)} = s F(s) − s f (0) − f (0). 2

(9b)

392

Chapter 7

The Laplace Transform

THEOREM 7.4

Transform of f when f is discontinuous at t = a Let f (t) be continuous on 0 ≤ t < a and on a < t < ∞, and let it have a simple jump discontinuity at t = a with the value f− (a) to the immediate left of a at t = a− and the value f+ (a) to the immediate right of t = a at a+. Then if L{ f (t)} = F(s), L{ f (t)} = s F(s) − f (0) + [ f− (a) − f+ (a)]e−as . Proof

Using integration by parts, as in Theorem 7.2, we have a− ∞ L{ f (t)} = e−st f (t)dt + lim e−st f (t)dt k→∞ a+

= [e

−st

f (t)]a− 0

+ lim [e−sk f (k) − e−as f+ (a)] + s F(s) k→∞

= s F(s) − f (0) + [ f− (a) − f+ (a)]e−as . The next example illustrates the application of results (8) and (9) to a simple initial value problem. EXAMPLE 7.10

Solve the initial value problem y + 3y + 2y = sin 2t,

where y(0) = 2

y (0) = −1.

and

Solution Because of the linearity of the equation and of the Laplace transform operation, taking the Laplace transform of the differential equation we have L{y } + 3L{y } + 2L{y} = L{sin 2t}. Setting L{y(t)} = Y(s), and using the initial conditions y(0) = 2 and y (0) = −1, we ﬁnd from (9a,b) that L{y } = s 2 Y(s) − 2s + 1, and L{y } = sY(s) − 2. Entry 9 in Table 7.1 shows that L{sin 2t} = 2/(s 2 + 4), so combining these results enables the transformed differential equation to be written s 2 Y(s) − 2s + 1 + 3[sY(s) − 2] + 2Y(s) =

2 , s2 + 4

or as (s 2 + 3s + 2)Y(s) =

2s 3 + 5s 2 + 8s + 22 . s2 + 4

Solving for the Laplace transform of the solution gives Y(s) =

2s 3 + 5s 2 + 8s + 22 . (s 2 + 4)(s 2 + 3s + 2)

When expressed in partial fraction form, Y(s) becomes Y(s) =

−5 1 17 1 1 2 3 s + − − . 4 s+2 5 s + 1 20 s 2 + 4 20 s 2 + 4

Section 7.2

Operational Properties of the Laplace Transform

393

Using the linearity property when taking the inverse Laplace transform, we have ( ( 1 1 5 17 L−1 {Y(s)} = − L−1 + L−1 4 s+2 5 s+1 ( ( 2 s 1 3 − L−1 2 − L−1 2 , 20 s +4 20 s +4 so using Table 7.1 to identify the four transforms involved shows that the solution of the initial value problem is 3 17 1 5 sin 2t − cos 2t, y(t) = − e−2t + e−t − 4 5 20 20

for t > 0.

This example illustrates a fundamental difference between the solution of an initial value problem obtained by using the Laplace transform and that obtained by the previous methods that have been developed. In the other methods, when solving an initial value problem, ﬁrst a general solution was found, and then the arbitrary constants were matched to the initial conditions. However, in the Laplace transform approach the initial conditions are incorporated when the equation is transformed, so the inversion of Y(s) gives the required solution of the initial value problem immediately. As the structure of the solution in Example 7.10 is typical of the structure obtained when solving all initial value problems for ordinary differential equations by means of the Laplace transform, a closer examination of it will help understand how the solution is generated. Returning to the point where the equation was transformed, the result can be rewritten as 2 = 2s + 5' (s 2 + 3s + 2) Y(s) + 2 $ %& %& ' $ + 2' $s %& Transformed hom*ogeneous equation with y , y , and y replaced, respectively, by s2 , s, and 1

Transformed initial conditions

Transformed nonhom*ogeneous term

Setting G(s) = 1/(s 2 + 3s + 2), and denoting the transformed initial conditions by I(s) and the transformed nonhom*ogeneous term by R(s), the above result can be solved for Y(s) and written in the form Y(s) = G(s)I(s) + G(s)R(s). transfer function

(10)

This shows how the transform G(s), called in engineering applications the transfer function associated with the differential equation, modiﬁes the transform of the initial conditions and the transform of the nonhom*ogeneous term to arrive at the transform Y(s) of the solution. The name transfer function comes from the fact that when all the initial conditions are zero, so I(s) = 0, the only term generating a solution is the forcing function (the nonhom*ogeneous term), so (10) describes how the effect of the input is transferred to the output (the solution). In terms of Example 7.10 we can write G(s) = =

Y(s) L{y(t)} = R(s) L{sin 2t} L{output} . L{input}

(11)

394

Chapter 7

The Laplace Transform

In control theory the transfer function of a system characterizes the behavior of the entire system. We now develop the most important operational properties of the Laplace transform, starting with the ﬁrst shift theorem, also called the s-shift theorem. THEOREM 7.5 the s-shift theorem

The ﬁrst shift theorem or the s-shift theorem Let L{ f (t)} = F(s) for s > γ . Then the Laplace transform of eat f (t) is obtained from F(s) by replacing s by s − a, where s − a > γ . Thus, L{eat f (t)} = F(s − a)

for s − a > γ .

Conversely, the inverse transform L−1 {F(s − a)} = eat f (t). Proof L{e−at

∞ From the conditions of the theorem, L{ f (t)} = 0 e−st f (t)dt for s > γ , so ∞ ∞ f (t)} = e−st eat f (t)dt = e−(s−a)t f (t)dt = F(s − a) for s − a > γ . 0

The converse result follows by reversing this argument to arrive at the result L−1 {F(s − a)} = eat f (t). EXAMPLE 7.11

Use Theorem 7.5 to ﬁnd L{eat t n }, L{eat cos bt}, and L{eat t sin bt}. Solution Using the Laplace transforms of t n , cos bt, and t sin bt listed as entries 3, 10, and 11 in Table 7.1, with a replaced by b in entries 10 and 11, and then replacing s by s − a we ﬁnd that L{eat t n } =

n! (s − a)n+1

for s > 0,

L{eat cos bt} =

(s − a) [(s − a)2 + b2 ]

for s > a,

and L{eat t sin bt} = EXAMPLE 7.12

2b(s − a) [(s − a)2 + b2 ]2

for s > a.

Use Theorem 7.5 to ﬁnd L−1 {1/(s 2 + 4s + 13)}. Solution Completing the square in the denominator we have ( ( 1 1 −1 −1 L . =L s 2 + 4s + 13 (s + 2)2 + 32 A comparison with entry 13 in Table 7.1 shows that L−1 {1/(s 2 + 4s + 13)} =

1 −2t e sin 3t. 3

We now derive the second shift theorem, also called the t-shift theorem, in which use will be made of the unit step function H(t − a).

Section 7.2

Operational Properties of the Laplace Transform

y

f(t )

k

395

k

y = f(t)

t

y = H(t − a) f(t − a)

a

(a)

t (b)

FIGURE 7.10 The relationship between f (t) and H(t − a) f (t − a).

THEOREM 7.6

Let L{ f (t)} = F(s). Then

The second shift theorem or the t-shift theorem

the t-shift theorem

L{H(t − a) f (t − a)} = e−as F(s) and, conversely, L−1 {e−as F(s)} = H(t − a) f (t − a). Proof Before proving the theorem it is necessary to understand the precise meaning of H(t − a) f (t − a). This can be seen by examining Fig. 7.10. The unit step function H(t − a) is zero until t = a, when it jumps to the value 1 and thereafter remains constant for t > a. The function f (t − a) is simply the function f (t) with its origin shifted to t = a, so it can be considered to be the function f (t) translated to the right by an amount a. Thus, H(t − a) f (t − a) is a function that is zero until t = a, after which it reproduces the function f (t) translated to the right by an amount a. The result of the theorem is obtained as follows: ∞ ∞ −st L{H(t − a) f (t − a)} = e H(t − a) f (t − a)dt = e−st f (t − a)dt. 0

a

If we make the change of variable τ = t − a, this becomes ∞ e−sτ f (τ )dτ L{H(t − a) f (t − a)} = e−as 0

and so L{H(t − a) f (t − a)} = e−as F(s). The converse result follows by reversing this argument. EXAMPLE 7.13

Use Theorem 7.6 to ﬁnd (a) L{H(t − 4) sin(t − 4)}, (b) to show that L{H(t − a)} = e−as /s in agreement with entry 7 in Table 7.1, and (c) to ﬁnd L−1 {se−as /(s 2 + b2 )}. Solution (a) From entry 9 in Table 7.1 we have L{sin t} = 1/(s 2 + 1), so applying Theorem 7.6 with a = 4 gives L{H(t − 4) sin(t − 4)} = e−4s /(s 2 + 1). (b) Setting f (t) = 1 in Theorem 7.6 and using the fact that L{1} = 1/s gives L{H(t − a)} = e−as /s.

396

Chapter 7

The Laplace Transform

(c) Entry 10 in Table 7.1 shows that L{cos bt} = s/(s 2 + b2 ), so using this in Theorem 7.6 gives L−1 {se−as /(s 2 + b2 )} = H(t − a) cos[b(t − a)]. The next example makes use of Theorem 7.6 when solving an initial value problem. EXAMPLE 7.14

Solve the initial value problem y + 3y + 2y = H(t − π ) sin 2t

with

y(0) = 1

and

y (0) = 0.

Solution Setting L{y(t)} = Y(s), transforming the differential equation, and incorporating the initial conditions as in Example 7.10 gives s 2 Y(s) − s + 3(sY(s) − 1) + 2Y(s) =

2e−π s , s2 + 4

or (s 2 + 3s + 2)Y(s) = s + 3 +

2e−π s . s2 + 4

As s 2 + 3s + 2 = (s + 1)(s + 2), this last result can be written in the form Y(s) =

2e−π s s+3 + 2 . (s + 1)(s + 2) (s + 4)(s + 1)(s + 2)

It is now necessary to invert Y(s), and to accomplish this some algebraic manipulation will be necessary if we are to identify terms on the right with entries in Table 7.1. When expressed in terms of partial fractions, after a little manipulation Y(s) becomes 2 1 1 1 1 1 3 2 2 s − + e−π s − − − Y(s) = . s+1 s+2 5 s + 1 4 s + 2 20 s 2 + 4 20 s 2 + 4 Each term can now be identiﬁed as the transform of an entry in Table 7.1, though as the last four terms are multiplied by e−πs their inverse Laplace transforms will need to be obtained by using Theorem 7.6. As a result, y(t) = L−1 {Y(s)} becomes y(t) = 2e−t − e−2t + H(t − π ) 2 3 1 1 sin 2(t − π ) − cos 2(t − π ) , × e−(t−π ) − e−2(t−π ) − 5 4 20 20 for t > 0. A graph of this solution is shown in Fig. 7.11, from which it can be seen that in the interval 0 < t < π the solution y(t) only involves the ﬁrst two terms, and so decays exponentially. At t = π the forcing function sin 2t is switched on, after which all the exponential terms decay to zero as t → ∞, leaving only the periodic steady state solution. THEOREM 7.7 differentiating a transform

Differentiation of a transform: Multiplication of f (t) by t n Let L{ f (t)} = F(s). Then L{t n f (t)} = (−1)n

dn F(s) . ds n

Section 7.2

Operational Properties of the Laplace Transform

397

y 1 0.8 0.6 0.4 0.2 0 −0.2

2

4

6

8

10

12

14

t

−0.4 FIGURE 7.11 The solution y(t) showing the inﬂuence of the forcing function after t = π .

Proof

By deﬁnition

∞

e−st f (t)dt = F(s),

so differentiating under the integral sign with respect to s gives ∞ ∂(e−st ) dF(s) = f (t)dt, ds ∂s 0 and so dF(s) = ds

∞

(−t)e−st f (t)dt = −

∞

e−st t f (t)dt,

which is the result of the theorem when n = 1. Each subsequent differentiation will introduce a further factor (−t) into the integrand, leading the general result of the theorem. EXAMPLE 7.15

Use Theorem 7.7 to ﬁnd (a) L{t sin at} and (b) L{t eat cos bt}. Solution (a) Entry 9 in Table 7.1 shows that L{sin at} = a/(s 2 + a 2 ) for s > 0, so from Theorem 7.7 a 2as d = 2 for s > 0, L{t sin at} = (−1) ds (s 2 + a 2 ) (s + a 2 )2 in agreement with entry 11 in Table 7.1. (b) Entry 14 in Table 7.1 shows that L{eat cos bt} = (s − a)/[(s − a)2 + b2 ] for s > a, so from Theorem 7.7 L{t eat cos bt} = (−1) =

(s − a) d ds [(s − a)2 + b2 ]

(s − a)2 − b2 [(s − a)2 + b2 ]2

for s > a.

These examples show that, in many cases, less effort is involved ﬁnding transforms by means of Theorem 7.7 than by direct use of the deﬁnition of the Laplace transform.

398

Chapter 7

The Laplace Transform

THEOREM 7.8

Scaling theorem Let L{ f (t)} = F(s). Then if k > 0,

scaling a transform

L{ f (kt)} =

s 1 F . k k

Proof The result follows by setting u = kt in the deﬁnition of the Laplace transform, because ∞ { f (kt)} = e−st f (kt)dt 0

= =

1 k 1 k

∞

e−s(u/k) f (u)du

∞

e−(s/k)u du

s 1 . = F k k EXAMPLE 7.16

If L{ f (t)} = e−3s (1 − 2s)/(2s 2 − s + 1), ﬁnd { f (3t)}. Solution In this case k = 3 > 0, so from Theorem 7.8, replacing s by s/3 in L{ f (t)} and multiplying the result by 1/3 gives 1 e−s (1 − 2s/3) 3 (2(s/3)2 − s/3 + 1) e−s (3 − 2s) . = 2 2s − 3s + 9

L{ f (3t)} =

Many functions whose Laplace transform is required are periodic functions with period T, though they are not necessarily continuous functions for all t > 0. In the Laplace transform, where only the behavior of a function f (t) for t > 0 is involved, a periodic function with period T is deﬁned as a function f (t) with the property that T is the smallest value for which f (t + T) = f (t)

for all t > 0.

(12)

An example of a piecewise continuous function f (t) with period T that is deﬁned for t > 0 is shown in Fig. 7.12.

y

y = f (t)

T

2T

3T

FIGURE 7.12 A function f (t) with period T.

t

Section 7.2

THEOREM 7.9 transforming a periodic function

Operational Properties of the Laplace Transform

399

Transform of a periodicfunction with period T Let f (t) be a periodic function T with period T such that 0 e−st f (t)dt is ﬁnite. Then 1 L{ f (t)} = 1 − e−Ts

T

e−st f (t)dt

for s > 0.

Proof In the deﬁnition of the Laplace transform we divide the interval of integration into subintervals of length T and write 2T T e−st f (t)dt + e−st f (t)dt + · · · · L{ f (t)} = 0

T

Then, because of the periodicity of f (t), the function f (t) will be the same in each integral. Consequently, changing the variable in the (r + 1)th integral to t = τ + r T with r = 0, 1, 2, . . . gives T T e−s(τ +r T) f (τ )dτ = e−r sT e−sτ f (τ )dτ for r = 0, 1, 2, . . . 0

= e−r sT

T

e−st f (t)dt,

where the dummy variable τ has been replaced by t. Substituting this result into the original integral gives T −Ts −2Ts +e + · · ·] e−st f (t)dt, L{ f (t)} = [1 + e 0

T

which is ﬁnite because we have assumed that 0 e−st f (t)dt is ﬁnite. The bracketed terms form a geometrical series with the common ratio e−Ts < 1, so its sum is 1/(1 − e−Ts ), and thus T 1 L{ f (t)} = e−st f (t)dt, for s > 0, 1 − e−Ts 0 and the proof is complete. T The necessity of the condition in Theorem 7.9 that 0 e−st f (t)dt is ﬁnite arises because periodic functions exist for which this integral is divergent. EXAMPLE 7.17

Find the Laplace transform of the square wave shown in Fig. 7.13. Solution As the function is discontinuous with period 2a we compute the integral in Theorem 7.9 in two parts as a 2a 2a e−st f (t)dt = ke−st dt + (−k)e−st dt 0

a

k k = (1 − e−as ) + (e−2as − e−as ) s 5 k = (1 + e−2as − 2e−as ). s

400

Chapter 7

The Laplace Transform

f (t) k

a

2a

3a

4a t

−k FIGURE 7.13 A square wave with period 2a.

Then from Theorem 7.9 we have k(1 + e−2as − 2e−as ) s(1 − e−2as ) k(1 − e−as ) = s(1 + e−as ) k(eas/2 − e−as/2 ) = s(eas/2 + e−as/2 ) k k sinh(as/2) = tanh(as/2) = s cosh(as/2) s

L{ f (t)} =

EXAMPLE 7.18

for s > 0.

Use Theorem 7.9 to show that L{sin t} = 1/(s 2 + 1) and Theorem 7.8 to show that L{sin at} = a/(s 2 + a 2 ). Solution The function f (t) = sin t is periodic with period 2π and is ﬁnite, so from Theorem 7.9 we have 2π 1 L{sin t} = e−st sin tdt (1 − e−2π s ) 0 e−2π s 1 1 − = (1 − e−2π s ) s 2 + 1 s 2 + 1 1 for s > 0. = 2 s +1

2π 0

e−st sin tdt

Setting k = a in Theorem 7.8 and using the preceding result gives 1 1 a [(s/a)2 + 1] a for s > 0. = 2 s + a2 Find the Laplace transform of the solution of the initial value problem L{sin at} =

EXAMPLE 7.19

y + 3y + 2y = f (t),

where y(0) = y (0) = 0

and f (t) is the square wave in Example 7.17. Solution Transforming the equation as in Examples 7.10 and 7.14 and using the result of Example 7.17 gives s 2 Y(s) + 3sY(s) + 2Y(s) =

k tanh(as/2), s

Section 7.2

Operational Properties of the Laplace Transform

401

so Y(s) =

k tanh(as/2) . s(s 2 + 3s + 2)

The convolution operation Let the functions f (t) and g(t) be deﬁned for t ≥ 0. Then the convolution of the functions f and g denoted by ( f ∗ g)(t), and in abbreviated form by ( f ∗ g), is deﬁned as the integral ( f ∗ g)(t) =

t

f (τ )g(t − τ )dτ.

convolution and the convolution theorem

The change of variable v = t − τ followed by the replacement of the dummy variable v by t shows that the convolution operation is commutative, so ( f ∗ g)(t) = (g ∗ f )(t).

EXAMPLE 7.20

(13)

Find (t 2 ∗ cos t) and (cos t ∗ t 2 ) and hence conﬁrm the equality of these two convolution operations. Compare the effort required in each case. Solution We have

t

(t ∗ cos t) = 2

τ 2 cos(t − τ )dτ

t

=

τ 2 [cos t cos τ + sin t sin τ ]dτ

t

= cos t

t

τ cos τ dτ + sin t 2

τ 2 sin τ dτ

= 2(t − sin t). Similarly, (cos t ∗ t 2 ) = 0

=t

t

cos τ (t − τ )2 dτ

2

t

cos τ dτ − 2t

t

t

τ cos τ dτ +

τ 2 cos τ dτ

= 2(t − sin t). While conﬁrming that the convolution operation is commutative, this example also shows that sometimes calculating ( f ∗ g)(t) is simpler than calculating (g ∗ f )(t).

The convolution operation has various uses, one of the most important of which occurs in the following important theorem that expresses the relationship between the product of two Laplace transforms F(s) and G(s) and the convolution of their transform pairs f (t) and g(t).

402

Chapter 7

The Laplace Transform

THEOREM 7.10

The convolution theorem Let L{ f (t)} = F(s) and L{g(t)} = G(s). Then L{( f ∗ g)(t)} = F(s)G(s) or, equivalently, L

t

( f (τ )g(t − τ )dτ

= F(s)G(s).

Conversely, L−1 {F(s)G(s)} =

t

f (τ )g(t − τ )dτ.

Proof From the deﬁnition of the Laplace transform and the convolution operation, we have t ∞ −st L{( f ∗ g)(t)} = e f (τ )g(t − τ )dτ dt. 0

Inspection of Fig. 7.14 shows that interchanging the order of integration allows the integral to be written as ∞ ∞ −st L{( f ∗ g)(t)} = f (τ ) e g(t − τ )dt dτ. τ

Using the second shift theorem reduces the inner integral to e−st G(s), so that ∞ L{( f ∗ g)(t)} = G(s)e−sτ f (τ )dτ 0

= G(s)

∞

e−sτ f (τ )dτ

= G(s)F(s). The converse result follows if we reverse the argument to ﬁnd the inverse Laplace transform of F(s)G(s).

τ

τ=

t

0 FIGURE 7.14 Region of integration for Theorem 7.10.

t

Section 7.2

EXAMPLE 7.21

Operational Properties of the Laplace Transform

403

Use Theorem 7.10 to ﬁnd (a) L{t 2 ∗ cos t} and (b) L−1 {s/(s 2 + a 2 )2 }. Solution (a) L{t 2 } = 2/s 3 and L{cos t} = s/(s 2 + a 2 ), so from Theorem 7.10 L{t 2 ∗ cos t} = L{t 2 } L {cos t} =

2s . (s 2 + a 2 )

(b) Writing s s 1 = 2 (s 2 + a 2 )2 (s + a 2 ) (s 2 + a 2 ) shows that in Theorem 7.10 we may take F(s) =

1 (s 2 + a 2 )

and

G(s) =

s . (s 2 + a 2 )

So as L−1 {F(s)} = (1/a) sin at and L−1 {G(s)}= cos at, it follows from Theorem 7.10 that L−1 {s/(s 2 + a 2 )2 } = (1/a)(sin at ∗ cos at) 1 t sin aτ cos a(t − τ )dτ = a 0 1 t sin at, 2a in agreement with entry 11 in Table 7.1. =

When evaluating convolution integrals of this type, instead of expanding a term such as cos a(t − τ ) and sin a(t − τ ) using integration by parts, it is often quicker to replace sin at and cos at by sin at = (eiat − e−iat )/(2i) and cos a(t − τ ) = ei(t−τ ) + e−i(t−τ ) /2 before performing the integrations, and again using these identities to interpret the result in terms of trigonometric functions. EXAMPLE 7.22

Solve the initial value problem y + 4y + 13y = 2e−2t sin 3t

with y(0) = 1

and

y (0) = 0.

Solution Before we solve this initial value problem, it should be noted that the complementary function is yc (t) = e−2t (C1 cos 3t + C2 sin 3t), so the nonhom*ogeneous term 2e−2t sin 3t is contained in yc (t). It will be seen that, unlike the special cases that arise when determining a particular integral by the method of undetermined coefﬁcients, this situation does not give rise to a special case when the solution is obtained by means of the Laplace transform. Transforming the equation in the usual way gives s 2 Y(s) − s + 4(sY(s) − 1) + 13Y(s) =

6 , s 2 + 4s + 13

404

Chapter 7

The Laplace Transform

and so Y(s) =

6 s+4 + . s 2 + 4s + 13 (s 2 + 4s + 13)2

Writing s + 4 = s + 2 + (2/3)3 allows Y(s) to be rewritten as Y(s) =

s+2 2 6 3 + + . (s + 2)2 + 32 3 (s + 2)2 + 32 [(s + 2)2 + 32 ]2

Taking the inverse Laplace transform of Y(s) and using entries 13 and 14 of Table 7.1 leads to the result 2 y(t) = e−2t cos 3t + sin 3t + L−1 {6/[(s + 2)2 + 32 ]2 }. 3 To ﬁnd L−1 {6/[(s + 2)2 + 32 ]2 }, we ﬁrst write this as 3 3 6 2 , = [(s + 2)2 + 32 ]2 3 (s + 2)2 + 32 (s + 2)2 + 32 and then, from entry 13 in Table 7.1, we ﬁnd that L−1 {3/[(s + 2)2 + 32 ]} = e−2t sin 3t. An application of Theorem 7.10 shows that 2 −2t (e sin 3t ∗ e−2t sin 3t) 3 2 t −2τ = e sin 3τ e−2(t−τ ) sin 3(t − τ )dτ 3 0 2 −2t t = e sin 3τ sin 3(t − τ )dτ 3 0 1 2 −2t 1 sin 3t − t cos 3t . = e 3 6 2

L−1 {6/[(s + 2)2 + 32 ]2 } =

Substituting this result in the expression for y(t) shows that the solution of the initial value problem is 7 1 y(t) = e−2t cos 3t + sin 3t − t cos 3t , for t > 0. 9 3

integral equation

Although the previous example could have been solved by the method of undetermined coefﬁcients, the next two examples cannot be solved in this manner. The ﬁrst involves a special type of equation called an integral equation, and the second an integro-differential equation. An equation of the form y(t) = f (t) + λ

t

K(t, τ )y(τ )dτ

(14)

is called a Volterra integral equation, where λ is a parameter and K(t, τ ) is called the kernel of the integral equation. Equations of this type are often associated with the solution of initial value problems. The Laplace transform is well suited to the solution of such integral equations when the kernel K(t, τ ) has a special form that depends on t and τ only through the difference t − τ , because then K(t, τ ) = K(t − τ ) and the integral in (14) becomes a convolution integral.

Section 7.2

Operational Properties of the Laplace Transform

405

An examination of the Volterra integral equation in (14) shows it to be essentially the integral form of an initial value problem, and it relates the solution y(t) at the current time t to an integral of the past history of the solution over the interval [0, t]. The following is a simple example of a problem that leads to a Volterra integral equation. Determine the amount of a manufactured material contained in a store from time t = 0 until time t, if the only supply of material comes immediately from the manufacturer and it begins degrading exponentially with time from the moment it enters the store. Let the amount of material present at time t = 0 be Q and the amount present in the store at time t be y(t), and suppose it degrades exponentially as e−kt with k > 0. Then, by time t, the amount of material that entered the store at time τ but has not degraded is e−k(t−τ ) y(τ ). Thus the amount of material present at time t is determined by the solution of the Volterra integral equation t y(t) = Qe−kt + e−k(t−τ ) y(τ )dτ. 0

By using the method of solution explained in the next example, the solution of this problem is easily shown to be y(t) = Qe−(k−1)t . EXAMPLE 7.23

Solve the Volterra integral equation y(t) = 2e−t +

t

sin(t − τ )y(τ )dτ.

Solution The Laplace transform of the integral equation is t 2 +L sin(t − τ )y(τ )dτ, Y(s) = s+1 0 and after applying Theorem 7.10 to the last term the equation for Y(s) becomes Y(s) =

Y(s) 2 + . s + 1 s2 + 1

Solving for Y(s) and expanding the result in partial fractions shows that Y(s) =

2(s 2 + 1) 2 4 2 = 2− + . 2 s (s + 1) s s s+1

Taking the inverse Laplace transform shows the solution to be y(t) = 2t − 2 + 4e−t ,

integro-differential equation

for t > 0.

The next example is a differential equation of an unusual type, because the function y(t) occurs not only as the dependent variable in the differential equation, but also inside a convolution integral that forms the nonhom*ogeneous term. Equations of this type that involve both the integral of an unknown function and its derivative are called integro-differential equations. These equations occur in many applications of mathematics, one of which arises in the continuum mechanics of polymers, where the dynamical response y(t) of certain types of material at time t depends on a derivative of y(t) and the time-weighted cumulative effect of what has happened to the material prior to time t. For obvious reasons materials of this type are called materials with memory.

406

Chapter 7

The Laplace Transform

An example of an integro-differential equation was obtained in Section 5.3(d) when considering the R–L–C circuit in Fig. 5.4, though at the time this was not recognized. When the circuit was closed, and the charge q on the capacitor was allowed to ﬂow causing a current i(t) in the circuit, the equation determining i(t) was shown to be q di L + Ri + = 0. dt C To recognize that this t is an integro-differential equation, we use the result that at time t we have q = 0 i(τ )dτ , so the equation determining i(t) becomes the integro-differential equation 1 t di L + Ri + i(τ )dτ. dt C 0 In this case it was possible to reduce this to a second order constant coefﬁcient differential equation for i(t), but in other more complicated cases a reduction of this type may not be possible. EXAMPLE 7.24

Solve the equation y + y =

t

sin τ y(t − τ )dτ,

subject to the initial conditions y(0) = 1 and y (0) = 0. Solution Taking the Laplace transform in the usual way gives t 2 s Y(s) − s + Y(s) = L sin τ y(t − τ )dτ. 0

The last term is the Laplace transform of a convolution integral, so from Theorem 7.10 it follows that ( t sin τ y(t − τ )dτ = L{sin t}L{y(t)} L 0

=

Y(s) . s2 + 1

Using this result in the transformed equation, solving for Y(s), and expanding the result using partial fractions gives s2 + 1 11 1 s = + . 2 2 s(s + 2) 2s 2 (s + 2)

Y(s) =

After the inverse Laplace transform is taken, the solution becomes y(t) = THEOREM 7.11 transforming an integral

√ 1 (1 + cos 2t), 2

for t > 0.

The transform of an integral Let f (t) be a piecewise continuous function such that | f (t)| ≤ Mekt for k > 0 and all t ≥ 0. Then, if L{ f (t)} = F(s), L 0

(

t

f (τ )dτ

=

F(s) s

for

s > k,

Section 7.2

Operational Properties of the Laplace Transform

407

and, conversely,

−1

t

L {F(s)/s} =

f (τ )dτ. 0

Proof The condition | f (t)| ≤ Mekt is sufﬁcient to ensure the existence of the t Laplace transform F(s), so writing h(t) = 0 f (τ )dτ we have |h(t)| ≤

t

| f (τ )|dτ ≤ M

t

ekτ dτ ≤ M

ekt k

for

t ≥ 0.

This result shows that |h(t)| grows no faster than | f (t)| as t → ∞, so the existence of the Laplace transform Y(s) ensures the existence of the Laplace transform of h(t). Using the fundamental result from the calculus that h (t) = f (t) together with Theorem 7.2 means that, apart from points where f (t) is discontinuous, F(s) = L{ f (t)} = L{h (t)} = sL{h(t)} = sL

t

( f (τ )dτ ,

and so L

(

t

f (τ )dτ 0

=

F(s) . s

The converse result follows by taking the inverse Laplace transform and the proof is complete. EXAMPLE 7.25

t

Find (a) L{

τ cos aτ dτ } and (b) L−1 {1/[s(s 2 + a 2 )]}.

Solution (a) As L{t cos at} = (s 2 − a 2 )/(s 2 + a 2 )2 for s > 0, an application of Theorem 7.11 shows that ( t s2 − a2 τ cos aτ dτ = for s > 0. L s(s 2 + a 2 )2 0 (b) We can write s(s 2

1 1 1 = 2 . 2 +a ) s + a2 s

So if we set F(s) = 1/(s 2 + a 2 ), for which f (t) = L−1 F(s) = (1/a) sin at, it follows from Theorem 7.11 that ( ( t F(s) 1 1 sin aτ dτ L−1 = L−1 = 2 2 s s(s + a ) 0 a 1 = 2 (1 − cos at), a in agreement with entry 17 of Table 7.1.

408

Chapter 7

The Laplace Transform

THEOREM 7.12 integrating a transform

The integral of a transform Let f (t)/t be piecewise continuous, deﬁned for t ≥ 0 and such that | f (t)/t| ≤ Me−kt for t ≥ 0. Then if L{ f (t)/t} = G(s) for s > k, and L{ f (t)} = F(s),

f (t) t

L

(

∞

=

F(u)du s

and, conversely, L−1 {G(s)} = Proof

We have

G(s) =

∞

e−st

−1 −1 L {G (s)}. t

f (t) t

s > k.

for

However, from Theorem 7.7, ∞ ∞ f (t) −st dt = − G (s) = e (−t) e−st f (t)dt = −F(s), t 0 0 so after integration we have ∞ F(u)du = − s

∞

G (u)du = G(s) − G(∞)

s

To proceed further we now make use of the fact that the condition | f (t)/t| ≤ Me−kt implies that G(s)lim s→∞ = 0, showing that ∞ G(s) = L{ f (t)/t} = F(u)du for s > k. s

The converse result follows by taking the inverse Laplace transform and using the fact that L−1 {G(s)} = f (t)/t together with the result L{ f (t)} = F(s) = −G (s).

EXAMPLE 7.26

Find

sin at (a) L t

( and

−1

(b) L

s+a ln s+b

( .

Solution (a) The function (sin at)/t is deﬁned and ﬁnite for all t > 0, so Theorem 7.12 can be applied. If we use the fact that L{sin at} = a/(s 2 + a 2 ), it follows from the ﬁrst part of Theorem 7.12 that ( ∞ sin at a L du = 2 + a2 t u s = π/2 − Arctan (s/a) = Arctan (a/s). (b) If we set

G(s) = ln

s+a , s+b

Section 7.2

Operational Properties of the Laplace Transform

409

differentiation gives G (s) =

1 1 b−a = + , (s + a)(s + b) s+a s+b

from which we see that L−1 {G (s)} = e−at − e−bt . From the second part of Theorem 7.11 we have ( s+a −1 −1 L−1 {G(s)} = L−1 ln L {G (s)} = s+b t = (e−bt − e−at )/t. The conditions of Theorem 7.11 assert that method used to derive this result is permissible if L−1 {G(s)} is deﬁned and ﬁnite for t ≥ 0. We see from the preceding result that L−1 {G(s)} is deﬁned and ﬁnite for t > 0 and limt→0 [(e−bt − e−at )/t] = a − b, so the conditions of the theorem are satisﬁed and we have shown that ( s+a L−1 ln = (e−bt − e−at )/t. s+b The theorem that follows shows how the initial values f (0), f (0), . . . , of a suitably differentiable function f (t) can be found directly from its Laplace transform F(s). An example of the use of the theorem is to be found in Section 7.3(d) when determining the Laplace transform of a function known only as the solution of a differential equation. THEOREM 7.13 relating initial values and the transform

The initial value theorem Let L{ f (t)} = F(s) be the Laplace transform of an n times differentiable function f (t). Then 1 2 f (r ) (0) = lim s r +1 F(s) − s r f (0) − s r −1 f (0) − · · · − s f (r −1) (0) , s→∞

r = 0, 1, . . . , n. In particular, f (0) = lim {s F(s)},

f (0) = lim {s 2 F(s) − s f (0)}

2

s→∞

s→∞

f (0) = lim {s F(s) − s f (0) − s f (0)}. 3

s→∞

Proof The theorem follows directly from Theorem 7.3 by ﬁrst replacing n by r + 1 and rewriting the result as 1 2 f (r ) (0) = s r +1 F(s) − s r f (0) − · · · − s f (r −1) (0) − L f (r +1) (t) . Then, provided f (r +1) (t) satisﬁes the sufﬁciency condition for the existence of a Laplace transform given in (3), it follows that for some M > 0 and k > 0 1 2 L f (r +1) (t) < M/(s − k) for s > k and r = 0, 1, . . . , n. As a result, lim

s→∞

and the theorem is proved.

1

2 f (r +1) (t) = 0,

410

Chapter 7

The Laplace Transform y

y(t) = (1/h)[H(t − a) − H(t − a − h)]

1/h Area = h(1/h) = 1

a

a+h

t

FIGURE 7.15 δ(t − a) = limh→0 y(t).

EXAMPLE 7.27

Given that F(s) = 2as/(s 2 + a 2 )2 , use Theorem 7.13 to ﬁnd f (0), f (0), and f (0). Use f (t) = L−1 {F(s)} = t sin at to conﬁrm the results by direct differentiation. Solution From Theorem 7.13 2as 2 = 0, s→∞ + a 2 )2 2as 3 f (0) = lim {s 2 F(s) − s f (0)} = lim 2 = 0, s→∞ s→∞ (s + a 2 )2 2as 4 f (0) = lim {s 3 F(s) − s 2 f (0) − s f (0)} = lim 2 = 2a. s→∞ s→∞ (s + a 2 )2 f (0) = lim {s F(s)} = lim

s→∞ (s 2

These results are easily conﬁrmed by differentiation of f (t) = t sin at. The last operational property to be considered concerns the Dirac delta function, usually abbreviated to the delta function and sometimes called the unit impulse function. The Dirac delta function, named after the Oxford University Nobel laureate mathematical physicist P. A. M. Dirac and denoted by δ(t − a), is actually a limiting mathematical operation, and not a function as its name implies. For our purposes the delta function can be considered to be the limit of a rectangular “pulse” of height h and width 1/ h in the limit as h → ∞. Thus the area of the graph representing the pulse remains constant at 1 as h → ∞, while its height increases to inﬁnity and its width decreases to zero. The graphical representation of such a pulse f (t) = δ(t − a) located at t = a, before proceeding to the limit, is shown in Fig. 7.15. We adopt the following deﬁnition of the delta function in terms of the unit step function. The delta function the delta or impulse function

The delta function located at t = a and denoted by δ(t − a) is deﬁned as the limit δ(t − a) = lim

h→0

1 [H(t − a) − H(t − a − h)]. h

Section 7.2

Operational Properties of the Laplace Transform

411

The operational property of the delta function, usually called its ﬁltering property and sometimes its sifting property, is represented by the following theorem. THEOREM 7.14 a useful property of the delta function

Filtering property of the delta function Let f (t) be deﬁned and integrable over all intervals contained within 0 ≤ t < ∞, and let it be continuous in a neighborhood of a. Then for a ≥ 0

∞

f (t)δ(t − a)dt = f (a).

Proof

From the deﬁnition of the delta function, a+h ∞ f (t) dt, f (t)δ(t − a)dt = lim h→0 a h 0

so applying the mean value theorem for integrals we have ∞ 1 f (t)δ(t − a)dt = lim h f (th ) , h→0 h 0 where a < th < a + h. In the limit as h → 0 the variable th → a, showing that ∞ f (t)δ(t − a)dt = f (a), 0

and the theorem is proved. Consideration of the deﬁnition of the delta function suggests that, in a sense, δ(t − a) is the derivative of the unit step function H(t − a), though the justiﬁcation of this conjecture requires arguments involving generalized functions that are beyond the scope of this account. In mechanical problems the delta function is used to represent an impulse, deﬁned as the integral of a large force applied locally for a very short time. The delta function has many other applications, such as the distribution of point masses along a supporting beam, whereas in electrical systems it can be used to represent the brief application of a very large voltage, or the sudden discharge of energy contained in a capacitor. A purely formal derivation of the Laplace transform of the delta function proceeds as follows. By deﬁnition, ∞ e−st δ(t − a)dt. L{δ(t − a)} = 0

An application of the ﬁltering property of Theorem 7.14 reduces this to L{δ(t − a)} = e−as .

(15)

L{δ(t)} = 1.

(16)

As a special case we have

412

Chapter 7

The Laplace Transform y 0.25 0.2 0.15 0.1 0.05 0 −0.05

2

4

6

8

t

−0.1 FIGURE 7.16 The solution y(t) as a function of the time t. EXAMPLE 7.28

Solve the initial value problem y + 3y + 2y = δ(t − 1) − δ(t − 2)

with y(0) = y (0) = 0.

Solution Taking the Laplace transform in the usual way and using result (15) gives (s 2 + 3s + 2)Y(s) = e−s − e−2s , and so Y(s) =

e−s − e−2s e−s − e−2s e−s − e−2s = − . 2 s + 3s + 2 s+1 s+2

Inverting the transform using Theorem 7.6 (the t-shift theorem) shows that y(t) = H(t − 1)[e1−t − e2−2t ] − H(t − 2)[e2−t − e4−2t ]. A graph of this solution is given in Fig. 7.16. The graph shows that a physical system represented by the given differential equation subject to the equilibrium initial conditions y(0) = y (0) = 0 is at rest until it is excited by the delta function at time t = 1 and then, after peaking just before t = 2, it is excited in the opposite sense by the delta function at time t = 2, after which the solution decays to zero as t increases, corresponding to the system returning to rest. The Laplace transform is also discussed in references [3.4], [3.8], [3.9], [3.17], and [3.20]; tables of Laplace transform pairs are to be found in references [G.1], [G.3], [3.11], and [3.14]. An advanced account of the Laplace transform is to be found in reference [3.19]. PAUL ADRIEN DIRAC (1902–1984) An English mathematical physicist who introduced the delta function in a fundamental paper on quantum mechanics presented to the Royal Society of London in 1927. Together with the German physicist Erwin Schrodinger he shared the Nobel Prize for physics because of contributions made to quantum mechanics.

Summary

This section has been concerned with what are known as the operational properties of the Laplace transform. These are general properties of the transform itself that can be applied to any function f (t) that possesses a Laplace transform, or to any function F (s) that is the Laplace transform of a function f (t). It will be seen later that these properties can be used to extend the table of Laplace transforms given at the end of Section 7.1, and when using the Laplace transform to solve differential equations.

Section 7.2

Operational Properties of the Laplace Transform

413

EXERCISES 7.2 Exercises involving the transformation of derivatives 1. Prove that L{ f (t)} = s 2 F(s) − s f (0) − f (0). 2. Prove that L{ f (t)} = s 3 F(s) − s 2 f (0) − s f (0) − f (0). 3. Given that f (0) = 1, f (0) = 0, f (0) = 1, ﬁnd L{ f (t)}. 4. Given that f (0) = 0, f (0) = 2, f (0) = 2, f (0) = −4, ﬁnd L{ f (4) (t)}. ⎧ ⎨sin t, 0 ≤ t < π/2 , ﬁnd L{ f (t)}. 5. Given that f (t) = ⎩ t = 0, t ≥ π/2 ⎧ ⎨sin t, 0 ≤ t < π/2 , ﬁnd L{ f (t)}. 6. Given that f (t) = ⎩ 1, t ≥ π/2 7. Solve y − 3y + 2y = cos t, with y(0) = 1, y (0) = −1. 8. Solve y + 5y + 4y = exp(−t), with y(0) = 1, y (0) = 0. 9. Solve y + 8y − 9y = t, with y(0) = 2, y (0) = 1. 10. Solve y + 5y + 6y = 1 + t 2 , with y(0) = 0, y (0) = 0.

Exercises involving the ﬁrst shift theorem (s-shift) 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24.

Exercises involving graphing functions with a t-shift 25. 26. 27. 28. 29. 30.

Sketch Sketch Sketch Sketch Sketch Sketch

Exercises involving the second shift theorem (t-shift) 33. 34. 35. 36. 37. 38. 39. 40. 41. 42. 43. 44. 45. 46.

−2t

Find L{(2 + t )e }. Find L{e−3t cos 2t}. Find L{e−t t sin 2t}. Find L{(1 + t 2 )e−4t }. Find L{e2t sin 3t}. Find L{e−4t sinh 3t}. Find L−1 {1/(s 2 − 4s + 13)}. Find L−1 {s/(s 2 + 4s + 13)}. Find L−1 {(1 − 3s)/(s 2 + 2s + 5)}. Find L−1 {1/[s(s 2 − 2s + 5)]}. Find L−1 {s/[(s + 1)(s 2 − 4s + 13)]}. Find L−1 {3/(s 2 + 6s + 25)}. Find L−1 {3(s 2 + 4)/[s(s 2 + 4s + 8)]}. Find L−1 {2/[(s + 3)2 (s 2 + 8s + 20)]}. 3

31. Sketch f (t) = [H(t − 1) − H(t − 2)](t − 1)2 . 32. Sketch f (t) = H(t − π/2) cos(t − π/2).

f (t) = H(t − 2)(1 + t). f (t) = H(t − π) sin t + H(t − 2π). f (t) = [H(t − π) − H(t − 2π )] cos t.

f (t) = r4=0 H(t − r ). f (t) = H(t − π) cos(t − π). f (t) = H(t − 1)(t − 1)2 .

47. 48. 49. 50.

Find L{H(t − 3)(t − 3)3 }. Find L{H(t − 1) sin(t − 1)}. Find L{H(t − 3π/2) sin 2(t − 3π/2)}. Find L{H(t − π/2)(t − π/2)3 − H(t − 3π/2) × (t − 3π/2)3 }. Find L{H(t − 4) sinh 3(t − 4)}. Find L{H(t − 1)(t − 1) sin(t − 1)}. Find L−1 {s e−2s /(s 2 + 4)}. Find L−1 {e−πs/3 /(s 2 + 9)}. Find L−1 {e−πs/2 (s + 1)/(s 2 + 4s + 5)}. Find L−1 {e−2s (s 2 + s + 1)/[s(s + 2)2 ]}. Find L−1 {e−4s (s + 3)/(s 2 + 4s + 13)}. Find L−1 {e−3s s 2 /[s(s 2 + 4s + 8)]}. Solve y + 5y + 6y = H(t − π) cos(t − π ), with y(0) = 1, y (0) = 0. Solve y − 5y + 6y = t H(t − 1), with y(0) = 0, y (0) = 0. Solve y − 5y + 6y = 1 + t H(t − 2), with y(0) = 0, y (0) = 1. Solve y − 6y + 10y = t H(t − 3), with y(0) = 1, y (0) = 1. Solve y + 2y + 10y = e−t H(t − 1), with y(0) = −1, y (0) = 0. Solve y − y − 2y = e−t H(t − 1), with y(0) = 1, y (0) = 0.

Exercises involving differentiation of transforms 51. Find L{t 2 e3t sin t}. 52. Find L{te−t sin 4t}.

53. Find L{t 3 e2t sin 2t}. 54. Find L{t 2 e3t cos 2t}.

Exercises involving scaling 55. 56. 57. 58.

If L{ f (t)} = e−3s (s 2 − 1)/(s 4 − a 4 ), ﬁnd L{ f (2t)}. If L{ f (t)} = (s + 1)(s 2 + 2)/(s 2 + 4)2 , ﬁnd L{ f (3t)}. If L{ f (t)} = 1/[s 2 (s 2 + 4)], ﬁnd L{ f (t/3)}. If L{ f (t)} = (s 2 − 4)/[(s 2 + 4)2 ], ﬁnd L{ f (t/2)}.

Exercises involving the Laplace transform of periodic functions In Exercises 59 through 66 ﬁnd the Laplace transform of the periodic function f (t).

414

Chapter 7

The Laplace Transform 64.

59. f(t )

f (t)

periodic with period 2k

3k

1

2k 0

2k

k

2k

2k

2k

t

k

FIGURE 7.17

60.

a

2a

3a

t

FIGURE 7.22

f(t )

periodic with period 2π/a

f(t ) = sin(αt ) 1

65. f(x)

π/a

2π/a

3π/a

t

periodic with period 2a

k

FIGURE 7.18

61.

a

2a

3a

4a

5a

t

f(t ) −k 2k periodic with period 4k

FIGURE 7.23

66. 0

2k

4k

6k

8k

t

f(t )

FIGURE 7.19

62.

k3 k2

f(t ) f (t ) = ⎢sin kt⎥

k 1

1 periodic with period π/k

0 π/k

2π/k

a

2a

3a

4a

5a

t

FIGURE 7.24

t

FIGURE 7.20

Exercises involving the convolution operation

63. f(t )

67. Find (e−t ∗ e−2t ). 68. Find (t ∗ sin t). 69. Find (t 2 ∗ sin t).

periodic with period a

k

70. Find (t ∗ e−t ). 71. Find (cos t ∗ cos t). 72. Find (sin 2t ∗ sin 2t).

Exercises involving the convolution theorem

a

FIGURE 7.21

2a

3a

4a

t

73. 74. 75. 76.

Find L{t ∗ e−2t }. Find L{2t ∗ cos 2t}. Find L{e−t sin t ∗ t}. Find L{e−2t cos t ∗ et }.

77. 78. 79. 80.

Find L−1 {1/[s 2 (s 2 + 4)]}. Find L−1 {1/(s 2 − 9)2 }. Find L−1 {s 2 /(s 2 − 1)2 }. Find L−1 {s/(s 2 − 4)2 }.

Section 7.3

Systems of Equations and Applications of the Laplace Transform

Exercises involving integral equations

t

81. Solve y(t) = sin t + 82. Solve y(t) = cos t + 83. Solve y(t) = t +

sin[2(t − τ )]y(τ )dτ .

0 t

cos(t − τ )y(τ )dτ .

2

sin(t − τ )y(τ )dτ .

0 t

84. Solve y(t) = e−2t +

t

cos(t − τ )y(τ )dτ .

Exercises involving integro-differential equations

t 85. Solve y + 4y = 4 sin τ y(t − τ )dτ, with y(0) = 1. t 0 e−2τ y(t − τ )dτ, with y(0) = 3. 86. Solve y + y = 0 t sinh τ y(t − τ )dτ, with y(0) = 1, 87. Solve y − y = 0 y (0) = 0. t sinh 2τ y(t − τ )dτ, with y(0) = 1, 88. Solve y − 4y = 2 0 y (0) = 0.

2 ( s + a2 96. Find L−1 ln . s2

Exercises involving the initial value theorem In Exercises 97 through 100 use the initial value theorem to ﬁnd f (0), f (0), and f (0) from F(s), and verify the result by differentiation of f (t) = L−1 {F(s)}. 97. 98. 99. 100.

F(s) = (s 2 + 6)/{s(s 2 + 9)}. F(s) = s/(s 2 + 6s + 9). F(s) = (s − 1)/(s 2 − 4s + 4). F(s) = (2s 2 + s − 12)/{s(s + 2)(s + 3)}.

Exercises involving the delta function

∞

101. Evaluate 0

102. Evaluate

4

∞

103. Evaluate 0

t

89. Find L

(

104. Evaluate

τ 2 sin 2τ dτ . 0 ( t 2τ e cos τ dτ . 90. Find L

Exercises involving an integral of a transform

7.3

δ(t − π/2)dt.

∞

3 ( 3 π4 sin nt δ t − (2n + 1) dt. t 2 n=1 {[H(t − 1) − H(t − 2)]t +

cos(t − 3π)δ(t − 3π)}dt.

91. Find L−1 {1/(s 2 + a 2 )2 }. 92. Find L−1 {s/(s 2 + a 2 )}.

( sinh 2t . 93. Find L t ( 1 − cos 3t 94. Find L . t 2 ( s − a2 . 95. Find L−1 ln s2

1 − 3 sin2 t t

sin2 tδ(t − 2π )dt.

Exercises involving the transform of an integral

415

105. Solve y + 9y = 1 + δ(t − 1), with y(0) = 0, y (0) = 0. 106. Solve y + 4y + 4y = δ(t − 1), with y(0) = 1, y (0) = 1. 107. Solve y + 2y + y = sin t + δ(t − π ), with y(0) = y (0) = 0. 108. Solve y − 4y + 3y = e−t + 3δ(t − 2), with y(0) = y (0) = 0. 109. Solve y + 4y = 1 − H(t − 1) + δ(t − 2), with y(0) = 1, y (0) = 0. 110. Solve y + 3y + 2y = δ(t − 1), with y(0) = 0, y (0) = 1.

Systems of Equations and Applications of the Laplace Transform (a) Solution of Systems of Linear First Order Equations by the Laplace Transform The Laplace transform can be used to solve initial value problems for systems of linear ﬁrst order differential equations by introducing the Laplace transform of

416

Chapter 7

The Laplace Transform

solving systems of equations

EXAMPLE 7.29

each dependent variable that is involved, solving the resulting algebraic equations for each transformed dependent variable, and then inverting the results. As a system of linear higher order differential equations can always be reduced to a system of ﬁrst order equations by introducing higher order derivatives as new dependent variables, the solution of a system of linear ﬁrst order equations can be considered to be the most general case. The example that follows, involving two simultaneous ﬁrst order equations, illustrates the approach to be used in all cases, but by restricting the number of equations and using simple nonhom*ogeneous terms (forcing functions) the algebra is kept to a minimum. Solve the initial value problem x − 2x + y = sin t y + 2x − y = 1, with x(0) = 1, y(0) = −1. Solution We deﬁne the transforms of the dependent variables x(t) and y(t) to be L{x(t)} = X(s),

L{y(t)} = Y(s).

Transforming the system of equations in the usual way leads to the following system of linear algebraic equations for X(s) and Y(s): s X(s) − 1 − 2X(s) + Y(s) = 1/(s 2 + 1) sY(s) + 1 + 2X(s) − Y(s) = 1/s. Solving these for X(s) and Y(s) gives X(s) =

(s − 1)(s 3 + s 2 + 2s + 1) s 2 (s − 3)(s 2 + 1)

and

Y(s) =

−(s 4 − s 3 + 3s 2 + s + 2) . s 2 (s − 3)(s 2 + 1)

Expressing these results in terms of partial fractions, we ﬁnd that X(s) =

2 s 43 1 1 1 41 1 1 + − + − 2 2 2 9s 3s 5 s + 1 5 s + 1 45 s − 3

Y(s) =

3 s 43 1 1 1 51 2 1 + − − . + 9s 3 s2 5 s 2 + 1 5 s 2 + 1 45 s − 3

and

Finally, taking the inverse transform gives the solution x(t) =

1 2 43 4 1 + t − sin t − cos t + e3t 9 3 5 5 45

and y(t) =

1 3 43 5 2 + t + sin t − cos t − e3t 9 3 5 5 45

for t > 0.

This method can be used for any number of simultaneous linear differential equations, though the complexity of both the algebraic manipulation and the associated inversion problem increases rapidly when more than two equations are involved.

Section 7.3

Systems of Equations and Applications of the Laplace Transform

417

A typical example of the way systems of ﬁrst order equations arise in practice is provided by considering a chemical reaction that converts a raw chemical into an end product, via several intermediate reactions. The simplest situation involves chemical reactions that are irreversible, so that once a product has been produced the chemical process cannot be reversed, causing the new product to revert to a previous one. Let us derive the system of equations governing such a process when three intermediate reactions are involved, each of which is irreversible, with each reaction proceeding at a rate that is proportional to the amount of material to be converted from one stage to the next. Denote the raw chemical by A and the end product by E, with the intermediate products denoted by B, C, and D, and let the reaction rates (the constants of proportionality) from A → B, B → C, C → D, and D → E be k1 , k2 , k3 , and k4 , respectively. Then if the amounts of chemicals A, B, C, D, and E present at time t are x, y, u, v, and w, the production and removal of the chemical products involved is described as follows. Reaction

Reaction Rate of Removal dx = −k1 x dt A→B dy = −k2 y dt B→C du = −k3 u dt C→D dv = −k4 v dt D→E

A→ B B→C C→D D→ E

Reaction Rate of Production dy = k1 x dt A→B du = k2 y dt B→C dv = k3 u dt C→D dw = k4 v dt D→E

Combining these results gives dx dt dy dt du dt dv dt

dx = −k1 x dt A→B dy dy = + = k1 x − k2 y dt A→B dt B→C du du = + = k2 y − k3 u dt B→C dt C→D dv dv = + = k3 u − k4 v. dt C→D dt D→E =

If the amount of raw material A present at the start is Q, the initial conditions for the system are seen to be x(0) = Q,

y(0) = 0,

u(0) = 0,

v(0) = 0,

and

w(0) = 0.

Provided no additional by-products are produced during the reactions, it follows from the conservation of mass that x + y + u + v + w = Q, and so w = Q − x − y − u − v.

418

Chapter 7

The Laplace Transform

Taking the Laplace transform of this system of ﬁrst order linear equations and using the stated initial conditions leads to the transformed system s X(s) + k1 X(s) = Q sY(s) − k1 X(s) + k2 Y(s) = 0 sU(s) − k2 Y(s) + k3 U(s) = 0 sV(s) − k3 U(s) + k4 V(s) = 0, where L{x(t)} = X(s), L{y(t)} = Y(s), L{u(t)} = U(s), and L{v(t)} = V(s). Solving for the Laplace transforms, we have X(s) =

Q , s + k1

Y(s) =

k1 Q , (s + k1 )(s + k2 )

U(s) =

k1 k2 Q , (s + k1 )(s + k2 )(s + k3 )

and V(s) =

k1 k2 k3 Q . (s + k1 )(s + k2 )(s + k3 )(s + k4 )

After expressing these Laplace transforms in terms of partial fractions the required solutions are seen to be x(t) = Qe−k1 t , and

y(t) =

k1 Q (e−k1 t − e−k2 t ) k1 − k2

u(t) = k1 k2 Q

1 1 e−k1 t + e−k2 t (k2 − k1 )(k3 − k1 ) (k1 − k2 )(k3 − k2 ) 1 −k3 t e + (k1 − k3 )(k2 − k3 )

with v(t) similarly deﬁned. The amount of the end product w(t) produced at time t follows from w(t) = Q − x(t) − y(t) − u(t) − v(t).

solving systems of equations in matrix form

We now outline a matrix method of solution of initial value problems for systems of linear ﬁrst order differential equations, of which Example 7.29 is a typical case. Let us consider the system d x(t) = Ax(t) + b(t), dt where

⎡

⎤ x1 (t) ⎢ x2 (t)⎥ ⎢ ⎥ ⎥ x(t) = ⎢ ⎢ · ⎥, ⎣ · ⎦ xn (t)

⎡

a11 a12 · · ⎢a21 a22 · · ⎢ ........... A=⎢ ⎢ ⎣ ........... an1 an2 · ·

· · ·

⎤ a1n a2n ⎥ ⎥ ⎥, ⎥ ⎦ ann

(17) ⎡

⎤ b1 (t) ⎢b2 (t)⎥ ⎢ ⎥ ⎥ b(t) = ⎢ ⎢ · ⎥, ⎣ · ⎦ bn (t)

subject to the initial conditions x1 (0) = x1 , x2 (0) = x2 , . . . , xn (0) = xn .

Section 7.3

Systems of Equations and Applications of the Laplace Transform

419

Deﬁne L{x1 (t)} = X1 (s), L{x2 (t)} = X2 (s) . . . , L{xn (t)} = Xn (s), L{b1 (t)} = B1 (s), L{b2 (t)} = B2 (s), . . . , L{bn (t)} = Bn (s), and set ⎡ ⎡ ⎡ ⎤ ⎤ ⎤ X1 (s) B1 (s) x1 ⎢ B2 (s)⎥ ⎢ x2 ⎥ ⎢ X2 (s)⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎥ ⎢ ⎥ ⎢ ⎥ Z(s) = ⎢ ⎢ · ⎥ , c(s) = ⎢ · ⎥ and v = ⎢ · ⎥ . ⎣ · ⎦ ⎣ · ⎦ ⎣·⎦ xn Xn (s) Bn (s) Then taking the Laplace transform of (17) and using the result L{xr (t)} = s X(s) − xr , for r = 1, 2, . . . , n, we arrive at the system sZ(s) − v = AZ(s) + c(s) or, equivalently, (sI − A)Z(s) = v + c(s), where I is the n × n unit matrix. Premultiplying this last result by (sI − A)−1 gives Z(s) = [sI − A]−1 [v + c(s)].

(18)

Finally, taking the inverse Laplace transform of (18) we obtain the solution x(t) of the initial value problem in the form x(t) = L−1 {[sI − A]−1 [v + c(s)]}. EXAMPLE 7.30

Solve the initial value problem of Example 7.29 by using result (19). Solution Making the necessary identiﬁcations we have 1 0 2 −1 1 1/(s 2 + 1) I= , A= , v= , c(s) = , 0 1 −2 1 −1 1/s so (18) becomes 1 Z(s) = s 0 or

0 2 − 1 −2

s−2 Z(s) = 2

−1 1

1 s−1

−1

1 1/(s 2 + 1) , + −1 1/s

(s 2 + 2)/(s 2 + 1) . (1 − s)/s

−1

The inverse of the ﬁrst matrix in this product is ⎡ s−1 −1 ⎢ s−2 1 ⎢ s(s − 3) =⎢ 2 s−1 ⎣ −2 s(s − 3) so

⎡

s−1 ⎢ s(s − 3) ⎢ Z(s) = ⎢ ⎣ −2 s(s − 3)

⎤ −1 s(s − 3) ⎥ ⎥ ⎥, s−2 ⎦ s(s − 3)

⎤⎡ ⎤ −1 s2 + 2 s(s − 3) ⎥ s2 + 1 ⎥ ⎥⎢ ⎥. ⎥⎢ s − 2 ⎦⎣ 1 − s ⎦ s(s − 3) s

(19)

420

Chapter 7

The Laplace Transform

After forming the matrix product this becomes ⎡ ⎤ (s − 1)(s 3 + s 2 + 2s + 1) ⎢ ⎥ ⎢ ⎥ s 2 (s − 3)(s 2 + 1) ⎢ ⎥. Z(s) = ⎢ ⎥ 4 3 2 ⎣ −(s − s + 3s + s + 2) ⎦ s 2 (s − 3)(s 2 + 1) The inverse transforms involved are, of course, the same as the ones in Example 7.29, so, as would be expected, the solution is the same as before, apart from a change of notation involving the replacement of x(t) and y(t) by x1 (t) and x2 (t) giving x1 (t) =

1 2 43 4 1 + t − sin t − cos t + e3t 9 3 5 5 45

and x2 (t) =

1 3 43 5 2 + t + sin t − cos t − e3t 9 3 5 5 45

for t > 0.

(b) Determination of etA by Means of the Laplace Transform The matrix solution of system (17) given in (19) has an interesting and useful consequence, because it provides a different and efﬁcient way of ﬁnding the matrix exponential etA . To see how this comes about, notice that from equation (114) in Section 6.10(c) the solution of the hom*ogeneous system of equations x = Ax,

(20)

subject to the initial condition x(0) = v, can be written x(t) = etA v.

(21)

Setting c(s) = 0 (corresponding to b(t) = 0) reduces solution (19) to x(t) = L−1 {[sI − A]−1 }v,

(22)

so comparison of (21) and (22) shows that e tA = L−1 {[sI − A]−1 }.

(23)

We have established the following theorem. THEOREM 7.15 ﬁnding the matrix exponential by the Laplace transform

Determination of etA by means of the Laplace transform Let A be a real n × n matrix with constant elements. Then the exponential matrix etA = L−1 {[sI − A]−1 }. The following examples show how Theorem 7.15 determines etA in the cases when A is diagonalizable with real eigenvalues, when it is diagonalizable with complex conjugate eigenvalues, and also when it is not diagonalizable.

EXAMPLE 7.31

Use Theorem 7.15 to ﬁnd etA when A=

−2 −2

6 . 5

Section 7.3

Systems of Equations and Applications of the Laplace Transform

421

Solution Matrix A has the distinct eigenvalues 1 and 2, and so is diagonalizable. s+2 −6 [sI − A] = 2 s−5 so ⎡

[sI − A]−1

s−5 ⎢ s 2 − 3s + 2 ⎢ =⎢ ⎣ −2 s 2 − 3s + 2

⎤ 6 s 2 − 3s + 2 ⎥ ⎥ ⎥. s+2 ⎦ s 2 − 3s + 2

Expressing each element of this matrix in terms of partial fractions and taking the inverse Laplace transform gives t 4e − 3e2t −6et + 6e2t tA , e = 2et − 2e2t −3et + 4e2t in agreement with the result in Example 6.33. EXAMPLE 7.32

Use Theorem 7.14 to ﬁnd etA when

−3 A= 2

−4 . 1

Solution Matrix A has the complex conjugate eigenvalues −1 ± 2i. s+3 4 [sI − A] = , −2 s−1 so ⎡

[sI − A]−1

s−1 ⎢ s 2 + 2s + 5 ⎢ =⎢ ⎣ 2 s 2 + 2s + 5

⎤ −4 s 2 + 2s + 5 ⎥ ⎥ ⎥. s+3 ⎦ s 2 + 2s + 5

Expressing each element of this matrix in terms of partial fractions and taking the inverse Laplace transform gives −t e (cos 2t − sin 2t) −2e−t sin 2t etA = , e−t (cos 2t + sin 2t) e−t sin 2t in agreement with the result of Example 6.34. EXAMPLE 7.33

Use Theorem 7.14 to ﬁnd etA when

4 A= 0

1 . 4

Solution Matrix A has the repeated eigenvalue 4 and is not diagonalizable. s−4 −1 [sI − A] = , 0 s−4

422

Chapter 7

The Laplace Transform

so

⎡

[sI − A]−1

1 ⎢s − 4 =⎢ ⎣ 0

⎤ 1 (s − 4)2 ⎥ ⎥. ⎦ 1 s−4

Taking the inverse of the elements of this matrix, we ﬁnd that 4t te4t e tA e = , 0 e4t in agreement with the result of Example 6.35.

(c) The Weighting Function To introduce the concept of a weighting function, which has important engineering applications, we consider the differential equation a0

weighting function and its uses

dn y dn−1 y + a + · · · + an y = f (t), 1 dt n dt n−1

(24)

subject to the initial conditions y(0) = y (0) = · · · = y(n−1) (0) = 0. We shall denote by w(t) the solution of equation (24) when f (t) = δ(t), and call it the weighting function associated with the equation. Thus the solution w(t) can be regarded as the output from a system described by equation (24) that is produced by the impulsive input (nonhom*ogeneous term) δ(t) applied at time t = 0 when the system is at rest. The weighting function w(t) is the solution of the equation a0

dn w dn−1 w + a1 n−1 + · · · + an w = δ(t), n dt dt

(25)

with w(t) = 0 for t < 0. Let us now consider the output y(t) from a system described by (24) produced by an arbitrary input f (t), subject to the hom*ogeneous initial conditions y(0) = y (0) = · · · = y(n−1) (0) = 0. Taking the Laplace transform of (24) we ﬁnd that G(s)Y(s) = F(s),

(26)

where G(s) = a0 s n + a1 s n−1 + · · · + an−1 s + an , Y(s) = L{y(t)} and

F(s) = L{ f (t)}.

Setting W(s) = L{w(t)}, taking the Laplace transform of (25), and using the fact that w(t) and all its derivatives vanish for t < 0 leads to the result G(s)W(s) = 1.

(27)

Eliminating G(s) between (26) and (27) relates the Laplace transform of the output Y(s) to the Laplace transform F(s) of the input by the equation Y(s) = W(s)F(s).

(28)

Section 7.3

Systems of Equations and Applications of the Laplace Transform

423

Taking the inverse Laplace transform of (28) and using the convolution theorem gives y(t) =

t

w(τ ) f (t − τ )dτ.

(29)

This form of the solution of (24) explains why w(t) is called the weighting function, because (29) shows how the input y(t − τ ) at time t − τ is weighted by the function w(τ ) over the interval 0 ≤ τ ≤ t in the integral determining y(t). The determination of the weighting function has the advantage that once it has been found, the solution of (24), subject to the conditions that y(0) = y (0) = · · · = y(n−1) (0) = 0, is always expressible as result (29) for every nonhom*ogeneous term f (t). It is instructive to compare this result, which applies to a linear differential equation of any order, to the one in (76) of Section 6.6, which was obtained by applying the method of variation of parameters to a second order equation with hom*ogeneous initial conditions when t = a. The weighting function is also sometimes called the Green’s function for an initial value problem for a hom*ogeneous differential equation. The modiﬁcation that must be made to result (29) to take account of initial conditions for y(t) that are not all zero at t = 0 is to be found in Exercise 25 at the end of this section. EXAMPLE 7.34

Find the weighting function for the equation y + 2y + 5y = sin t and use it to solve the equation subject to the initial condition y(0) = y (0) = 0. Solution The weighting function w(t) is the solution of w + 2w + 5w = δ(t) with w(0) = w (0) = 0. Taking the Laplace transform and setting L{w(t)} = W(s) gives s 2 W(s) + 2sW(s) + 5W(s) = 1, so W(s) =

1 . s 2 + 2s + 5

Taking the inverse Laplace transform, we ﬁnd that 1 −t e sin 2t for t ≥ 0. 2 The solution of the differential equation with y(0) = y (0) = 0 now follows from (29) as t w(τ ) sin(t − τ )dτ y(t) = w(t) = L−1 {W(s)} =

=

1 2

t

e−τ sin 2τ sin(t − τ )dτ

1 e−t 1 cos t + (2 cos 2t − sin 2t). = sin t − 5 10 20

424

Chapter 7

The Laplace Transform

The concept of a weighting function can be generalized to include systems of equations, though then more than one weighting function must be introduced, and the solution of each dependent variable becomes the sum of convolution integrals of the type given in (29). The ideas involved are illustrated by considering the following system of equations involving x(t) and y(t): x + ax + by = f1 (t) y + cx + dy = f2 (t),

(30)

subject to the initial conditions x(0) = y(0) = 0. It is necessary to introduce a weighting function for each of the variables x(t) and y(t) corresponding ﬁrst to f1 (t) = δ(t) and f2 (t) = 0, and then to f1 (t) = 0 and f2 (t) = δ(t). Let w x1 (t) and w y1 (t) be the weighting functions corresponding to + aw x1 + bw y1 = δ(t) w x1 w y1 + cw x1 + dw y1 = 0,

(31)

and w x2 (t) and w y2 (t) be the Green’s functions corresponding to w x2 + aw x2 + bw y2 = 0 w y2 + cw x2 + dw y2 = δ(t),

(32)

where w x1 (0) = w x2 (0) = w y1 (0) = w y2 (0) = 0. The notation used here indicates that w x1 (t) is the x response and w y1 (t) the y response to the input f1 (t) = δ(t) and f2 (t) = 0, and w x2 (t) is the x response and w y2 (t) the y response to the input f1 (t) = 0 and f2 (t) = δ(t). Then, because the equations are linear, to obtain the solution x(t) subject to the initial conditions x(0) = y(0) = 0, it is necessary to add the contribution due to w x1 (t) to the one due to w x2 (t), and similarly for the solution y(t). This leads to the solution in the form t t x(t) = w x1 (τ ) f1 (t − τ )dτ + w x2 (τ ) f2 (t − τ )dτ (33a) 0

and

y(t) =

t

t

w y1 (τ ) f1 (t − τ )dτ +

w y2 (τ ) f2 (t − τ )dτ.

(33b)

Once the weighting functions have been found, equations (33) give the solution of system (30) for any choice of functions f1 (t) and f2 (t), subject to the initial conditions x(0) = y(0) = 0. EXAMPLE 7.35

Find weighting functions for the equations x + 2x − y = f1 (t) y − 2x + y = f2 (t) and use them to solve the system subject to the initial conditions x(0) = y(0) = 0 when (a) f1 (t) = sin t and f2 (t) = 2 and (b) f1 (t) = cos t and f2 (t) = 0. Solution (a) From (31) the functions w x1 (t) and w y1 (t) satisfy + 2w x1 − w y1 = δ(t) w x1 w y1 − 2w x1 + w y1 = 0,

Section 7.3

Systems of Equations and Applications of the Laplace Transform

425

so taking the Laplace transform of these equations we have (s + 2)L{w x1 (t)} − L{w y1 (t)} = 1 (s + 1)L{w y1 (t)} − 2L{w x1 (t)} = 0. Solving for L{w x1 (t)} and L{w y1 (t)} gives L{w x1 (t)} =

s+1 s(s + 3)

and

L{w y1 (t)} =

2 . s(s + 3)

Taking the inverse Laplace transforms, we ﬁnd that w x1 (t) =

1 2 −3t + e 3 3

and

w y1 (t) =

2 2 −3t − e 3 3

for t ≥ 0.

Similarly, solving the equations for w x2 (t) and w y2 (t) corresponding to (32), we obtain w x2 (t) =

1 1 −3t − e 3 3

and

w y2 (t) =

2 1 −3t + e 3 3

for t ≥ 0.

The solution of the system subject to the initial conditions x(0) = y(0) = 0, f1 (t) = sin t, and f2 (t) = 2 now follows from (33) as t t x(t) = w x1 (τ ) sin(t − τ )dτ + 2 w x2 (τ )dτ 0

and

t

y(t) =

t

w y1 (τ ) sin(t − τ )dτ + 2

w y2 (τ )dτ. 0

After the integrations are performed, the solution is found to be x(t) =

13 2 1 2 1 + t + e−3t + sin t − cos t 9 3 45 5 5

and y(t) =

8 4 13 3 1 + t − e−3t − sin t − cos t 9 3 45 5 5

for t > 0.

(b) Similarly, the solution when f1 (t) = cos t and f2 (t) = 0 is given by t x(t) = w x1 (τ ) cos(t − τ )dτ 0

and

y(t) =

t

w y1 (τ ) cos(t − τ )dτ,

so after performing the integrations, 1 1 2 x(t) = − e−3t + sin t + cos t 5 5 5 and y(t) =

1 −3t 3 1 e + sin t − cos t 5 5 5

for t > 0.

426

Chapter 7

The Laplace Transform

(d) Differential Equations with Polynomial Coefﬁcients special variable coefﬁcient differential equations

The Laplace transform can be applied to linear differential equations with polynomial coefﬁcients to ﬁnd the solution of an initial value problem in the usual way, and also to deduce the Laplace transform of a function from its deﬁning differential equation. This last situation is useful when the integral deﬁning the Laplace transform of a function f (t) cannot be evaluated directly. First, however, we use Theorems 7.3 and 7.7 to ﬁnd the transform of a product of a power of t and a derivative of f (t).

THEOREM 7.16

L{t m f (n) (t)} Let f (t) be n times differentiable with L{ f (t)} = F(s). Then 1 2 dm L t m f (n) (t) = (−1)m m [s n F(s) − s n−1 f (0) − s n−2 f (0) ds − s n−3 f (0) − · · · − f (n−1) (0)]. Useful special cases are: (i)

L{t f (t)} = −F (s)

(ii)

L{t f (t)} = −s F (s) − F(s)

(iii)

L{t f (t)} = −s 2 F (s) − 2s F(s) + f (0)

(iv)

L{t 2 f (t)} = s F (s) + 2F(s)

(v)

L{t 2 f (t)} = s 2 F (s) + 4s F (s) + 2F(s)

Proof The results of the theorem are direct consequences of Theorems 7.3 and 7.7. We prove the general result, from which the special cases all follow. From Theorem 7.3 we have 2 1 L f (n) (t) = s n F(s) − s n−1 f (0) − s n−2 f (0) − s n−3 f (0) − · · · − f (n−1) (0), m

d whereas from Theorem 7.7 L{t m g(t)} = (−1)m ds m G(s), where L{g(t)} = G(s). The main result of the theorem now follows by setting g(t) = f (n) (t) in this last result.

(i) L{ exp(−t 2 )} and its connection with the error function Laplace transform of the error function

We will use the differential equation satisﬁed by y(t) = exp(−t 2 ) to show that L{exp(−t 2 )} =

1√ π exp(s 2 /4)[1 − erf(s/2)], 2

where 2 erf s = √ π

s

exp(−u2 )du 0

Section 7.3

Systems of Equations and Applications of the Laplace Transform

427

is a special function called the error function. The error function arises in the theory of heat conduction (see Section 7.3(f) and Chapter 18), in chemical diffusion processes, statistics, and elsewhere. An attempt to ﬁnd L{exp(−t 2 )} directly from the deﬁnition fails because the integral cannot be evaluated in terms of elementary functions, so some other method must be used. If we set y(t) = exp(−t 2 ), it is easily shown that y(t) satisﬁes the ﬁrst order variable coefﬁcient equation dy + 2t y = 0, dt subject to the initial condition y(0) = exp(0) = 1. Setting L{y(t)} = Y(s) and taking the Laplace transform of the differential equation gives sY(s) − y(0) + 2L{t y(t)} = 0. However, y(0) = 1, and from result (i) of Theorem 7.15 (or directly from Theorem 7.7) L{t y(t)} = −Y (s), so using these results in the preceding equation shows that the Laplace transform satisﬁes the differential equation dY 1 1 − sY = − . ds 2 2 The integrating factor for this linear ﬁrst order equation is μ(s) = exp(−s 2 /4), so after multiplication of the equation by μ(s) the result becomes 1 d [exp(−s 2 /4)Y(s)] = − exp(−s 2 /4). ds 2 Integrating over the interval 0 ≤ u ≤ s gives (after the introduction of the dummy variable u) s d 1 s [exp(−u2 /4)Y(u)]du = − exp(−u2 /4)du, 2 0 0 du or

1 s exp(−u2 /4)du. exp(−s /4)Y(s) − Y(0) = − 2 0 ∞ ∞ From the deﬁnition Y(s) = 0 e−st exp(−t 2 )dt, weﬁnd that Y(0) = 0 exp(−t 2 )dt. √ ∞ The integral determining Y(0) is a standard result, 0 exp(−t 2 )dt = π /2, so making use of this we ﬁnd that √ s π 1 2 2 exp(s /4) 1 − √ exp(−u /4)du . Y(s) = 2 π 0 2

The change of variable u = 2v brings this last result into the form √ s/2 π 2 exp(s 2 /4) 1 − √ Y(s) = exp(−v2 )dv . 2 π 0 If we now deﬁne the error function as 2 erf(x) = √ π

x

exp(−v2 )dv, 0

428

Chapter 7

The Laplace Transform

the Laplace transform Y(s) becomes Y(s) = L{exp(−t 2 )} =

√ π exp(s 2 /4)[1 − erf(s/2)]. 2

The function erfc (x), deﬁned as erfc(x) = 1 − erf(x), is called the complementary error function, so in terms of this function the transform Y(s) becomes √ π exp(s 2 /4)erfc(s/2). Y(s) = 2 This method of determining the Laplace transform was successful because the differential equation satisﬁed by Y(s) happened to be simpler than the differential equation satisﬁed by y(t).

(ii) Laplace transform of the Bessel function J 0 (t) and the series expansion of J 0 (t) Laplace transform of a Bessel function

The following linear second order differential equation, called Bessel’s equation, d2 y dy + (t 2 − v2 )y = 0, +t dt 2 dt contains a parameter v that is a constant. It has many applications, one of which is to be found in Chapter 18, where it enters into the solution of a vibrating circular membrane. The properties of its solutions are developed in some detail in Sections 8.6 and 8.7 of Chapter 8. For each constant value v, Bessel’s equation has two linearly independent solutions denoted by Jv (t) and Yv (t), called, respectively, Bessel functions of order v of the ﬁrst and second kind. We now use the Laplace transform to ﬁnd L{J0 (t)}, and then to ﬁnd a power series expansion for J0 (t) that will be obtained in a completely different way in Section 8.6. When v = 0, Bessel’s equation reduces to t2

t

d2 J0 d J0 + t J0 = 0, + 2 dt dt

and we will now ﬁnd L{J0 (t)} subject to the initial condition J0 (0) = 1. A second initial condition follows by setting t = 0 in the differential equation that gives J0 (0) = 0, though this result will not be needed in what is to follow as the condition is implied later when the initial value Theorem 7.13 is used. Taking the Laplace transform of Bessel’s equation of order zero, setting L{J0 (t)} = Y(s), and using the results of Theorem 7.16, we obtain −s 2 Y (s) − 2sY(s) + 1 + sY(s) − 1 − Y (s) = 0, and after simpliﬁcation this shows that Y(s) satisﬁes the ﬁrst order differential equation dY s + 2 Y(s) = 0. ds s +1

Section 7.3

Systems of Equations and Applications of the Laplace Transform

429

Separating the variables and integrating gives dY s =− ds, 2 Y s +1 and so Y(s) =

C . (s 2 + 1)1/2

We now know the form of Y(s), apart from the magnitude of the constant C. To ﬁnd the constant we use the initial value theorem (Theorem 7.13), which shows that we must have J0 (0) = lim [sY(s)], s→∞

but from the initial condition J0 (0) = 1, so 1 = lim

s→∞ (s 2

sC = C, + 1)1/2

and thus L{J0 (t)} =

(s 2

1 + 1)1/2

for s > 0.

This result can be used to obtain a series expansion for J0 (t) by ﬁrst writing it as

1 1 −1/2 L{J0 (t)} = , 1+ 2 s s

and then expanding the result by the binomial theorem to obtain L{J0 (t)} =

1 1 1 3 1 5 1 − + − + ···· 3 5 s 2s 8s 16 s 7

Finally, taking the inverse Laplace transform of each term and adding the results, we arrive at the series expansion of J0 (t): J0 (t) = 1 −

t4 t6 t2 + − + ···· 4 64 2304

If the general term in the expansion of 1s (1 + s12 )−1/2 is found, and the result is combined with entry 3 of Table 7.1, it is not difﬁcult to show that J0 (t) can be written as J0 (t) =

∞ (−1)n t 2n n=0

(iii) L{ sin

√

t}

22n (n!)2

.

√ We now show how L{sin t} √= Y(s) can be found from the differential equation satisﬁed by the function sin t, and how in this case a different form of argument from the one used in (ii) must be employed to determine the constant of integration

430

Chapter 7

The Laplace Transform

in the expression for Y(s). It is easily seen that y(t) = sin 4t

√ t is a solution of

d2 y dy + y = 0, +2 2 dt dt

and clearly y(0) = 0. Writing L{y(t)} = Y(s), transforming the equation using result (iii) of Theorem 7.16, and incorporating the initial condition y(0) = 0 leads to the following ﬁrst order differential equation for Y(s): 1 − 6s dY = Y. ds 4s 2 Integration of this variables separable equation gives Y(s) = Cs −3/2 exp[−1/(4s)], so it only remains to determine the value of the constant C. In this case the initial value theorem is of no help in determining C, so to accomplish this we return to the deﬁnition of the Laplace transform: ∞ √ √ e−st sin tdt. L{sin t} = Y(s) = 0

The intuitive argument we now use can be made rigorous, but as the details of its justiﬁcation are not appropriate here, they will be omitted. Inspection of the √ integrand shows that as | sin t| ≤ 1 for all t, when s is large and positive the expo√ nential function will only√ be signiﬁcant close to the origin where the function sin t can be approximated by t. So for large s the integral can be approximated by ∞ √ e−st t 1/2 dt, L{sin t} ≈ 0 √ (3/2) π = = , 3/2 3/2 s 2s where entry√4 of Table 7.1 has been used together with the result (3/2) = 1 (1/2) = 12 π that will be proved later in Section 8.5 of Chapter 8. 2 Comparing √ the original expression for Y(s) when s is large with this last result gives C = 12 π, so √ L{sin t} =

√ π exp[−1/(4s)], 2s 3/2

for s > 0.

This form of argument used to determine the behavior of the integral as s → ∞, where the approximation approaches arbitrarily close to the exact value as s increases, is called an asymptotic argument (see, for example, reference [3.3]).

(e) Two-Point Boundary Value Problems: Bending of Beams boundary value problems and the bending of beams

The Laplace transform is ideally suited to the solution of initial value problems because of the way the initial values of a function enter into the Laplace transform of its derivatives. It can, however, also be used to solve certain types of two-point boundary value problems, as we now show. It will be helpful to use a simple physical example to illustrate the method of approach, so we will consider the case of a

Section 7.3

Systems of Equations and Applications of the Laplace Transform

431

Q

a

x

2a/3 y FIGURE 7.25 Clamped beam supporting a point load.

uniform horizontal beam of mass M and length a that is clamped at each end and supports a point load Q at a distance 2a/3 from one end, as illustrated in Fig. 7.25. The beam equation was introduced in Section 5.2(f) and is EI

d4 y = w(x). dx 4

Here x is measured along the axis of the undeﬂected beam, y(x) is the vertical deﬂection, E is the Young’s modulus of the material of the beam, I is the second moment of the area of the beam about an axis normal to the x- and y-axes, and w(x) is the transverse load per unit length of the beam, which in this case is an isolated point mass Q located at x = 2a/3. The boundary conditions for a clamped beam are y(0) = y (0) = 0

and

y(a) = y (a) = 0,

because neither deﬂection nor bending can occur at the ends, so both y(x) and y (x) vanish at x = 0 and x = a. The function w(x) can be expressed as M + Qδ(x − 2a/3), for 0 ≤ x ≤ a, a where the point load Q is represented by the delta function that only makes a contribution at x = 2a/3. Transforming the equation, setting L{y(x)} = Y(s), and this time writing x in place of t, because it is conventional to denote a length by x, we ﬁnd w(x) =

EI [s 4 Y(s) − s 3 y(0) − s 2 y (0) − sy (0) − y (0)] = L{w(x)}. However, M + Qe−2as/3 , as so using this in the preceding equation, incorporating the two known initial conditions y(0) = y (0) = 0, and rearranging terms, we ﬁnd that L{w(x)} =

M 1 Q e−2as/3 1 1 + + 3 y (0) + 4 y (0). aEI s 5 EI s 4 s s Taking the inverse Laplace transform of this expression gives Y(s) =

y(x) =

M Q 1 1 x4 + (x − 2a/3)3 H(x − 2a/3) + x 2 y (0) + x 3 y (0). 24aEI 6EI 2 6

432

Chapter 7

The Laplace Transform

We must now solve for the unknown initial conditions y (0) and y (0) by requiring this expression to satisfy the two remaining boundary conditions at x = a, namely, y(a) = y (a) = 0. The condition y(a) = 0 gives Qa Ma + + 3y (0) + ay (0), 4EI 27EI and the condition y (a) = 0 gives 0=

Q Ma 2 1 + + y (0) + ay (0), 6EI 18EI 2 so solving for y (0) and y (0), we obtain 0=

a (9M + 8Q) and 108EI The required solution is then given by y (0) =

y(x) =

y (0) = −

1 (27M + 14Q). 54EI

M Q a x4 + (x − 2a/3)3 H(x − 2a/3) + (9M + 8Q)x 2 24a EI 6EI 216EI 1 (27M + 14Q), − 324EI

for 0 ≤ x ≤ a. This same form of approach can be used for other two-point boundary value problems, but its success depends on the ability to solve for the unknown initial values in terms of the given boundary conditions.

(f) An Application of the Laplace Transform to the Heat Equation

a ﬁrst encounter with a partial differential equation: the heat equation

The Laplace transform can also be used to solve certain types of partial differential equation, involving two or more independent variables. Although the solution of partial differential equations (PDEs) forms the topic of Chapter 18, it will be instructive at this early stage to introduce a simple example that illustrates how the transform can be used for this purpose, and the way the result of Section 7.3d(i) enters into the solution. The one-dimensional heat equation is the partial differential equation ∂2T 1 ∂T = , κ ∂t ∂ x2 where T(x, t) is the temperature in a one-dimensional heat-conducting solid at position x at time t, and κ is a constant that describes the thermal conductivity property of the solid. This is a partial differential equation because it is a differential equation that involves the partial derivatives of the dependent variable T(x, t). The physical situation modeled by this equation can be considered to be a semi-inﬁnite slab of metal with a plane face on which the origin of the x-axis is located, with the positive half of the axis directed into the slab. This situation is illustrated in Fig. 7.26. We will consider the situation where for t < 0 all of the metal in the slab is at the temperature T = 0 and then, at time t = 0, the plane face of the slab is suddenly brought up to and maintained at the constant temperature T = T0 . The problem is to ﬁnd the temperature inside the slab on any plane x = constant at any time t > 0, knowing that physically the temperature must remain ﬁnite for all x > 0 and t > 0.

Section 7.3

Systems of Equations and Applications of the Laplace Transform

T = T0

433

1 ∂T = ∂2T k ∂t ∂x2 x

FIGURE 7.26 A semi-inﬁnite metal slab.

The approach will be to take the Laplace transform of the dependent variable T(x, t) in the heat equation with respect to the time t, as a result of which an ordinary differential equation with x as its independent variable will be obtained for the transformed variable that will then depend on both the Laplace transform variable s and x. After this ordinary differential equation has been solved for the transformed variable, the inverse Laplace transform will be used to recover the time variation, and so to arrive at the required solution as a function of x and t. Before proceeding with this approach we notice ﬁrst that if the Laplace transform is applied to the independent variable t in the function of two variables T(x, t), the variable x will behave like a constant. Consequently, the rules for transforming derivatives of functions of a single independent variable also apply to a function of two independent variables. So, using the notation T(x, s) = tL{T(x, t)} to denote the Laplace transform of T(x, t) with respect to the time t, it follows directly from the formula for the transform of a derivative in (9a) that t L{∂ T(x, t)}

= sT(x, s) − T(x, 0).

To proceed further we must now use the condition that at time t = 0 the material of the slab is at zero temperature, so T(x, 0) = 0, as a result of which t L{∂ T(x, t)/∂t}

= sT(x, s).

Next, as x is regarded as a constant, we have ∂ 2 T(x, s) . ∂ x2 Using these results when taking the Laplace transform of the heat equation with respect to t, and making use of the linearity property of the transform, gives t L{∂

2

T(x, t)/∂ x 2 } =

d2 T(x, s) , dx 2 where we now use an ordinary derivative with respect to x because in this differential equation s appears as a parameter so x can be considered to be the only independent variable. When the differential equation is written s T − T = 0, κ using a prime to denote a derivative with respect to x, it is seen to have the general solution / / s s T(x, s) = Aexp x + B exp − x . κ κ sT(x, s) = κ

434

Chapter 7

The Laplace Transform

As a Laplace transform must vanish in the limit s → +∞, we must set A = 0, so the Laplace transform of the temperature is seen to be given by / s x . T(x, s) = B exp − κ In this case, the rejection of the term with the positive exponent in the general solution for T(x, s) corresponds to the physical requirement that the temperature remain ﬁnite for x > 0 and t > 0. To determine B we now make use of the boundary condition on the plane face of the slab that requires T(0, t) = T0 , from which it follows that t L{T(0, t)} = T0 /s. Thus, the Laplace transform of the solution with respect to the time t is seen to be / T0 s exp − x . T(x, s) = s κ To recover the time variation from this Laplace transform it is necessary to ﬁnd −1 L {T(x, s)}. As T(x, s) is not the Laplace transform of an elementary function t listed in our table of transform pairs, the solution T(x, t) must be found by means of the Laplace inversion integral. In Chapter 16 on the Laplace inversion integral, it is shown in Example 16.6 that 2( √ k k −1 −k s L {e } = exp − . √ t 3 4t 2 πt So, setting k = x/κ 2 in this result and using it with Theorem 7.11 to invert the Laplace transform T(x, s) shows that the solution is ( x , for x > 0, t > 0. T(x, t) = T0 erfc √ 2 κt The use of integral transforms is discussed in reference [4.4].

Summary

The Laplace transform has been applied to systems of differential equations, and the results extended to systems in matrix form. Various applications have been made to some useful variable coefﬁcient ordinary differential equations, and to the important partial differential equation that describes one-dimensional unsteady heat ﬂow.

EXERCISES 7.3 (a) Exercises involving systems of equations

3. Solve x + x + y = 2

1. Solve x + 5x − 2y = 1

and

y − 5x + 2y = 3

given x(0) = 0, y(0) = 2.

and

y + x − y = 1

given x(0) = −1, y(0) = 1. 4. Solve x + x + 2y = e−t

and

y + 2x + y = 1

given

x(0) = 0, y(0) = 0.

2. Solve 5. Solve

x − x − y = cos t

and

y + x + y = cos t

given x(0) = 1, y(0) = 1.

x − x + 3y = 1 + t

and

y + x − y = 2

given

x(0) = 2, y(0) = −2.

Section 7.3

Systems of Equations and Applications of the Laplace Transform can be written in the form t y(t) = w(τ )[y0 (t − τ ) − h(t − τ )]dτ.

6. Solve x + x + y = sin 2t

and

y + x − y = 1

given

x(0) = 0, y(0) = 0.

Here y0 (t) is the solution of the equation with the initial conditions y(0) = y (0) = · · · = y(n−1) (0) = 0, and h(t) = {H(s)/G(s)}, with H(s) the polynomial produced by the nonvanishing initial values of the derivatives, so that the transformed equation corresponding to (26) becomes

7. Solve x + x − z = 1,

y − x + y = 1,

z + y − x = 0,

given that x(0) = 1, y(0) = 0, z(0) = 1. 8. Solve x + x − y = 1,

y − y + 2z = 0,

z + x − y = sin t,

given x(0) = 1, y(0) = 0, z(0) = 2. 9. Solve x − z = et ,

y − z = 2,

z − x = 1,

given x(0) = 0,

y(0) = 1, z(0) = 0.

G(s)Y(s) + H(s) = F(s). 26. 27. 28. 29. 30.

y − 4y + 3y = cos t, given y(0) = 0 and y (0) = 0. y + 2y + 2y = e2t , given y(0) = 0 and y (0) = 0. y + 4y + 13y = cos 2t, given y(0) = 0 and y (0) = 0. y + 6y + 5y = e−t , given y(0) = 0 and y (0) = 0. Use the result of Exercise 25 to solve

y − 2y − 3y = 1 + sin t,

10. Solve x + z = 3,

and y + x = 1,

z − x = sin t,

given

x(0) = 1, y(0) = 0, z(0) = 1.

(b) Exercises involving et A In Exercises 11 through 24 ﬁnd etA for the given matrix A. 6 −1 1 3 . 19. A = . 11. A = 0 6 1 −1 2 3 −2 4 . 20. A = . 12. A = 0 4 3 2 −2 4 3 6 . 21. A = . 13. A = 0 −2 2 −1 1 4 3 7 . 22. A = . 14. A = 3 0 3 −1 ⎤ ⎡ 5 10 7 4 −5 . 15. A = 4 0 23. A = ⎣0 −1 −1⎦. 0 2 2 3 4 ⎤ ⎡ . 16. A = 1 0 0 3 −1 24. A = ⎣1 3 2⎦. 2 −4 . 17. A = 1 2 3 1 2 −2 3 . 18. A = 5 0

(c) Exercises involving the weighting function In Exercises 26 through 32 ﬁnd the weighting function when a single equation is involved, and the four weighting functions when a pair of equations is involved. Use the weighting function(s) to solve the given differential equation(s). 25. Show that if the initial conditions for equation (24) are y(0) = y0 , y (0) = y1 , . . . , y(n−1) = yn−1 , the solution

435

given y(0) = 1

y (0) = −1.

31. x − 3x + 2y = e−t , y + 3x − 4y = 3, with x(0) = y(0) = 0. 32. x + 2x − y = sin t, y − 2x + y = 2, with x(0) = y(0) = 0.

(d) Differential equations with polynomial coefﬁcients 33. Use the fact that y(x) = sin ax satisﬁes the differential equation y + a 2 y = 0

with y(0) = 0, y (0) = a

to derive L{sin ax} from the differential equation. 34. Use the fact that y(x) = 1 − cos ax satisﬁes the differential equation y + a 2 y = a 2

with y(0) = 0, y (0) = 0

to derive L{1 − cos ax} from the differential equation. 35.* The Laguerre equation xy + (1 − x)y + ny = 0, with n = 0, 1, 2, . . . a parameter, has polynomial solutions y(x) = Ln (x) called Laguerre polynomials. These polynomials are used in many branches of mathematics and physics, and also in connection with numerical integration. By taking the Laplace transform of the differential equation ﬁnd L{Ln (x)} and hence show that L4 (x) = 24 − 96x + 72x 2 − 16x 3 + x 4 . 36.* The Hermite equation y − 2xy + 2ny = 0,

436

Chapter 7

The Laplace Transform

with n = 0, 1, 2, . . . a parameter, has polynomial solutions y(x) = Hn (x) called Hermite polynomials. Like the Laguerre polynomials, these polynomials are also used in mathematics and physics, and in connection with numerical integration. By transforming the equation and using the initial conditions y(0) = H4 (0) = 12 and y (0) = 0, ﬁnd L{H4 (x)}, and hence show that H4 (x) = 16x 4 − 48x 2 + 12.

41. Using the notation of Section 7.3(e), solve the beam equation d4 y EI 4 = w(x) dx for the uniform beam of mass M and length a with clamped ends shown in Fig. 7.28, where a point mass Q is located at a distance 3a/4 from the left-hand end. The boundary conditions to be used are y(0) = y (0) = 0

37.* The Bessel function y(x) = J0 (ax) satisﬁes the differential equation

and

y(a) = y (a) = 0.

xy + y + axy = 0

Q

subject to the initial conditions y(0) = J0 (0) = 0. Derive L{J0 (ax)} from the differential equation and conﬁrm the result by using L{J0 (x)} = 1/(s 2 + 1)1/2 in conjunction with the scaling theorem. 38.* The Bessel function y(x) = J1 (x) satisﬁes the differential equation x 2 y + xy + (x 2 − 1)y = 0

with J1 (0) = 0 and J1 (0) = 1/2.

By taking the Laplace transform of the differential equation show that L{J1 (x)} = C{1 − s/(s 2 + 1)1/2 }, and deduce that C = 1.

(e) Exercises involving two-point boundary value problems 39. Solve x + x = sin 2t with x(0) = 0 and x(π/2) = 1. 40. Using the notation of Section 7.3(e), solve the beam equation d4 y = w(x) dx 4 for the uniform cantilevered beam of mass M and length a shown in Fig. 7.27, where a point mass Q is located at a distance a/3 from the clamped end. The boundary conditions to be used are EI

y(0) = y (0) = 0

and

a

0 3a/4 y FIGURE 7.28 Supported beam with clamped ends and a point load.

42. Using the notation of Section 7.3(e), solve the beam equation d4 y = w(x) dx 4 for the uniform beam of mass M and length a shown in Fig. 7.29 that is clamped at the end x = 0 and supported at the end x = a, where a point mass Q is located at a distance a/4 from the right-hand end. The boundary conditions to be used are EI

y(0) = y (0) = 0

and

y(a) = y (a) = 0.

Q a

y (a) = y (a) = 0.

x

0 3a/4

Q y

a x

00

x

FIGURE 7.29 Beam clamped at one end and supported at the other with a point load.

a/3

( f ) Physical problems to be solved by computer algebra y FIGURE 7.27 Cantilevered beam with a point load.

43. In an R–L–C circuit the current i(t) and charge q(t) resulting from a constant voltage E0 applied at time

Section 7.4

The Transfer Function, Control Systems, and Time Lags

t = 0, when i(0) = 0 and q(0) = 0, are determined by the equations q di = E0 L + Ri + dt C

dq . and i = dt

Find i(t), and comment on its form depending on the sign of R2 C − 4L. Choose representative values of R, L, C corresponding to each of the foregoing cases and plot i(t) in a suitable interval 0 ≤ t ≤ T. 44. Figure 6.10 in Section 6.3 illustrates three particles of equal mass joined by identical springs that oscillate in a straight line, with each end of the system clamped. In a representative case, the nondimensional equations determining the magnitudes of the displacements y1 (t), y2 (t), and y3 (t) are d2 y1 = y2 − 2y1 + y3 , dt 2 d2 y3 3 2 = y1 − 2y3 + y2 . dt

3

3

d2 y2 = y3 − 2y2 + y1 , dt 2

tive reaction rates k1 , k2 , and k3 are dx = −k1 x, dt

dy = k1 x − k2 y, dt

and

dz = k2 y − k3 z, dt

where x, y, and z are the number of molecules of A, B, and C present at time t. If Q molecules of Aare present at time t = 0, the number of molecules of D present at time t is w(t) = Q − x(t) − y(t) − z(t). Find w(t)/Q as a function of t given that k1 = 2, k2 = 3, and k3 = 3, and plot the result for 0 ≤ t ≤ 5. Find the percentage of chemical A that has been transformed into chemical D at the instants of time t = 1, 2, and 3. 46. In the following nondimensional equations, x(t) and y(t) represent the magnitudes of the currents ﬂowing in the primary and secondary windings of a transformer, when initially x(0) = 0, y(0) = 0 and at time t = 0 the primary winding is subjected to an exponentially decaying voltage of magnitude e−t :

Find y1 (t), y2 (t), and y3 (t) given that y1 (0) = 1, y1 (0) = 0, y2 (0) = 2, y2 (0) = 1, y3 (0) = 3, y3 (0) = 0. 45. If, similar to the example in Section 7.3(a), an irreversible reaction converts a molecule of chemical A into a molecule of chemical D, via molecules of chemicals B and C, the governing equations in terms of the respec-

7.4

437

1 dy dx + + 3x = e−t , dt 3 dt

dy dx +3 + 9y = 0. dt dt

Find x(t) and y(t), and by plotting the magnitudes of the currents show that x(t) is always positive and after peaking decays to zero, while y(t) is initially negative, but after becoming positive it decays to zero faster than x(t).

The Transfer Function, Control Systems, and Time Lags The study of engineering systems of all types whose behavior is determined by linear ordinary differential equations is often carried out by examining what is called the system transfer function. Typically, a system is governed by a linear nth order constant coefﬁcient ordinary differential equation whose solution or output, also called the response of the system, we will denote by u0 (t) and whose forcing function, or input, is a known function we will denote by ui (t), where t is the time. A typical example of a simple system has already been encountered in Fig. 6.2, where the spring-mounted and damped vibrating machine has an input F(t) and an output y(t) that are related by d2 y dy + by = F(t). +a dt 2 dt An nth order system may be governed by the equation dn u0 dn−1 u0 + a + · · · + a0 u0 = ui , n−1 dt n dt n−1 which can be represented graphically as in Fig. 7.30, where F[.] is the differential operator an

F[.] ≡ an

dn dn−1 + a + · · · + a0 . n−1 dt n dt n−1

(34)

438

Chapter 7

The Laplace Transform

ui(t)

F[ui(t)]

u0(t)

FIGURE 7.30 Block-diagram representation of equation (34).

More generally, in linear systems the input itself may be the solution of another linear differential equation, in which case the system relating the response u0 (t) to the input ui (t) becomes an

dn u0 dn−1 u0 dmui dm−1 ui + a + · · · + a u = b + b + · · · b0 ui , n−1 0 0 m m−1 dt n dt n−1 dt m dt m−1

(35)

where n ≥ m and the coefﬁcients ar and bs are constants. The transfer function of a system is deﬁned as the quotient of the Laplace transforms of the system output and the system input, when all of the initial conditions are taken to be zero. This last condition means that when the Laplace transform is used to transform a differential equation we may set L{dr u/dt r } = s r U(s). So, after transforming (35), we obtain (an s n + an−1 s n−1 + · · · + a0 )U0 (s) = (bms m + bm−1 s m−1 + · · · b0 )Ui (s),

(36)

where U0 (s) = L{u0 (t)} and Ui (s) = L{ui (t)}. The transfer function G(s) = U0 (s)/ Ui (s) becomes the rational function of the transform variable s G(s) =

bms m + bm−1 s m−1 + · · · b0 . an s n + an−1 s n−1 + · · · a0

(37)

Let us now set G(s) = N(s)/D(s), where N(s) is the polynomial in s of degree m in the numerator of G(s), and D(s) is the polynomial in s of degree n in the denominator. The polynomial D(s) is called the characteristic polynomial of the system, and D(s) = 0 is called the characteristic equation of the system. The order of the system in (37) is the degree n of the polynomial D(s). As the coefﬁcients of D(s) are real, it follows that the roots of the characteristic equation, called the poles of the transfer function G(s), either are all real or, if complex, they must occur in complex conjugate pairs. When G(s) is expressed in partial fraction form, this last observation implies that the system will be stable provided all the roots of the characteristic equation have negative real parts. Here, by stability, we mean that any bounded input to a system that is stable will result in an output that is also bounded for all time, and this will be the case when every root of D(s) = 0 has a negative real part. The requirement that n ≥ m imposed on (35) is necessary in order to prevent unbounded behavior of the output caused by the occurrence of delta functions. It is important to recognize that systems describing quite different physical phenomena can have the same transfer function, so transfer functions provide a means of examining a class of similar systems independently of their physical origin. It follows that for any given input with Laplace transform Ui (s), the Laplace transform of the output U0 (s) is given by U0 (s) = G(s)Ui (s).

(38)

The time variation of the output of the system then follows by taking the inverse Laplace transform of (38).

Section 7.4

EXAMPLE 7.36

The Transfer Function, Control Systems, and Time Lags

439

Find the transfer function of the system with input ui (t) and output u0 (t) described by d2 u0 (t) du0 (t) dui (t) + 25u0 (t) = 3 + 2ui (t), + 16 dt 2 dt dt and show it is stable. 4

Solution Taking the Laplace transform of the governing equation and assuming all initial conditions to be zero gives (4s 2 + 16s + 25)U0 (s) = (3s + 2)Ui (s), so the system transfer function is G(s) =

3s + 2 U0 (s) = 2 . Ui (s) 4s + 16s + 25

The system is of order 2, and its characteristic equation is 4s 2 + 16s + 25 = 0. The characteristic equation has the roots s1 = −2 − 32 i and s2 = −2 + 32 i, so as their real parts are negative, the system is stable. Systems that compare the difference between an input and an output, and attempt to reduce the difference to zero to make the output follow the input, are called control systems. A typical example is a temperature control system for a chemical reactor in which the temperature is required to remain constant, but where as the reaction progresses heat is released at variable rates, causing cooling to become necessary. A simple control system is illustrated in Fig. 7.31, where F is the system differential equation. The idea here is that an input ui is compared with the output u0 , called the feedback, and the difference ε = ui − u0 , called the error signal, is then used as an input to system F. The result is that u0 = ui when ε = 0. It is often necessary to modify the feedback by passing u0 through another system G with output v = G[u0 ], and then to use the the difference v − ui to drive F. The reason for this is to improve the overall performance of a system, whose physical characteristics may be difﬁcult to alter, by using an easily modiﬁed feedback to make the system more responsive and to reduce any tendency it may have for excessive oscillation. EXAMPLE 7.37

A steering mechanism for a small boat comprises an input heading θi from the helm, an ampliﬁer for the error signal, and a servomotor to drive the rudder with moment of inertia I that produces a resisiting torque proportional to the rate of change of the output angle θ0 . Derive the differential equation governing the system and ﬁnd its transfer function given that the feedback is the unmodiﬁed output θ0 . u0(t)

ui

ε = ui − u0

ε(t)

F[ε(t)]

FIGURE 7.31 A typical feedback control system.

u0(t)

440

Chapter 7

The Laplace Transform

Solution If the resisting torque is kdθ0 /dt and the ampliﬁer increases the magnitude of the error signal by a factor K, the system can be represented as in Fig. 7.31 with the governing differential equation dθ0 d2 θ0 = K(θi − θ0 ). +k dt 2 dt Taking the Laplace transform of this equation gives I

(Is 2 + ks + K)L{θ0 } = L{θi }, and so L{θ0 } =

1 L{θi }. Is 2 + ks + K

This result shows that the transfer function G(s) = 1/(Is 2 + ks + K), so the system will be stable provided the roots of the characteristic equation Is 2 + ks + K = 0 have negative real parts. This will be the case since I > 0 and K > 0, but the steering will oscillate about the required heading if 4I K > k2 . As the design of the boat determines I and k, any improvement of the steering response can only be obtained by using a modiﬁed feedback signal instead of the direct feedback θ0 . We close this section by mentioning an important consequence of the introduction of a delay into an equation governing the response of a system. Consider a vibrating system characterized by y(t) in which instantaneous damping proportional to the velocity dy/dt occurs with coefﬁcient of proportionality a1 , and where there is also present an additional time retarded damping of a similar type but with a time lag τ and a coefﬁcient of proportionality a2 . Then, when a springlike restoring effect is present with constant of proportionality a3 , the governing equation takes the form dy(t) dy(t − τ ) d2 y(t) + a2 + a3 y(t) = 0. + a1 2 dt dt dt

(39)

Because of the presence of the time-delayed derivative dy(t − τ )/dt, an equation of this type is called a differential-difference equation. If we now seek a solution of this equation by using the Laplace transform (or by seeking solutions of the form y(t) = Aexp(λt), where Aand λ are constants) we arrive at a characteristic equation of the form s 2 + a1 s + a2 s exp(−τ s) + a3 = 0.

(40)

This is called an exponential polynomial in s, and its root will determine both the stability and response of the system. Without going into detail, by using Rouche’s theorem from complex analysis it is not difﬁcult to prove that exponential polynomials have an inﬁnite number of zeros. Consequently, the response of a system with a characteristic polynomial in the form of an exponential polynomial will only be stable if all of its zeros have negative real parts, and this can only be shown analytically. Methods exist that can be used to determine when all the zeros of such exponential polynomials have negative real parts. An interested reader will ﬁnd a valuable discussion of this subject in Section 13 of Differential-Difference Equations by R. Bellman and K. Cooke, published by Academic Press in 1963. It is necessary to ask in what way the inﬁnite number of zeros of an exponential polynomial of degree n approximate the n zeros of the ordinary polynomial of

Section 7.4

The Transfer Function, Control Systems, and Time Lags

441

degree n when time lags are absent. This is a simpler question, and it can be answered by appeal to Hurwitz’s theorem from complex analysis, though again the arguments used go beyond this ﬁrst account of the subject.

A result on exponential polynomials Let Pτ (s) be an exponential polynomial of degree n in s with a time lag τ , and let P0 (s) be the corresponding constant coefﬁcient polynomial when τ = 0. Then, as τ → 0, so each of the n zeros si of P0 (s) is approached arbitrarily closely by a number of zeros of Pτ (s) equal in number to its multiplicity, and the remaining inﬁnite number of zeros of Pτ (s) can be made to lie outside a circle of arbitrarily large radius centered on the origin. As this result says nothing about how the zeros move as τ → 0, it is possible for the system to be stable when τ lies in certain intervals and unstable otherwise.

EXERCISES 7.4 1. Find the transfer function for each of the following systems. Determine the order of each system and ﬁnd which is stable. (a)

(b)

2.* For safety reasons, a control system is often duplicated, with the sensors for each system located in different positions, and in such cases the possibility of interaction between the control systems must be considered. A typical case is illustrated in Fig. 7.32, where two identical control systems are shown between which there is assumed to be linear cross-coupling of the error signals. This means that the respective actuating error signals are ε1 = a11 ε1 + a12 ε2 and ε2 = a21 ε1 + a22 ε2 , with the coefﬁcients ai j constants. Derive and discuss the equations governing the response of the system when

d3 u0 d2 u0 du0 − 20u0 + 3 2 + 16 3 dt dt dt 2 d ui dui − 6ui . =2 2 + dt dt d2 u0 du0 d3 u0 + 20u0 + 4 2 + 14 3 dt dt dt d2 ui dui + 6ui . = 6 2 − 13 dt dt

F(u0 ) =

d2 u0 du0 d2 ui dui + 10u0 = 6 2 + 5 − 6ui . (c) 9 2 + 6 dt dt dt dt

d2 u0 du0 + 2 u0 , + 2ζ 2 dt dt

with ζ > 0 and > 0.

u01(t)

ui2(t)

−

−

ε1 = ui1 − u01

ε2 = ui2 − u02

ε'1(t) Error signal cross coupling

ui1(t)

F[ε'1(t)]

u01(t)

u02(t)

ε'2(t)

FIGURE 7.32 Two interacting control systems.

F[ε'2(t)]

u02(t)

442

Chapter 7

The Laplace Transform

CHAPTER 7

TECHNOLOGY PROJECTS The purpose of these projects is to use a computer algebra differential equation solver to ﬁnd the analytical solutions of initial value problems involving linear constant coefﬁcient differential equations, some of which contain either the Dirac delta function or the Heaviside step function. As all the initial conditions are given at t = 0, the Laplace transform can also be used to solve these problems. Project 1

take the Laplace transform of the equation, (b) to ﬁnd the Laplace transform X(s) of the solution, and (c) to invert the transform to ﬁnd x(t).

Solving a Third Order Initial Value Problem Use a computer algebra Laplace solver to solve the initial value problem x + 2x

x

2x = e

sin t, with x(0) = 1, x (0) = 1, and x (0) = 0. t

Verify the result by using computer algebra (a) to take the Laplace transform of the equation, (b) to ﬁnd the Laplace transform X(s) of the solution, and (c) to invert the transform to ﬁnd x(t). Project 2 Solving an Equation with the Heaviside Step Function in the Nonhom*ogeneous Term Use a computer algebra Laplace solver to solve the initial value problem x + 3x + 2x = {H(t 1) H(t 2)}t, with x(0) = 1, x (0) =

1.

Verify the result by using computer algebra (a) to take the Laplace transform of the equation, (b) to ﬁnd the Laplace transform X(s) of the solution, and (c) to invert the transform to ﬁnd x(t). Plot the solution for 0 ≤ t ≤ 6. Project 3 Solving an Equation with the Dirac Delta Function in the Nonhom*ogeneous Term Use a computer algebra Laplace solver to solve the initial value problem x + 3x + 2x = 3e

t

+ δ(t

2),

with x(0) = 1, x (0) = 2.

Verify the result by using computer algebra (a) to 442

Project 4 Solving a System Solve the initial value problem for the system dx = x(t) + 2y(t) + 3, dt

dy = 1 x(t) + y(t), dt with x(0) = 1, y(0) = 0.

Verify the result by using computer algebra (a) to take the Laplace transform of the system, (b) to solve for the Laplace transforms X(s) and Y(s) of x(t) and y(t), and then (c) to invert the transforms to ﬁnd x(t) and y(t). Project 5 Examining the Properties of a Spring Damper System In an experiment, a wheel of mass M is mounted vertically below a rigid plate to which it is attached by a spring with spring constant k and a damper whose resisting force is μ times the speed of its displacement. If at time t the vertical displacement of the wheel from its equilibrium position is x(t), and a force F(t) is applied to the wheel, its equation of motion is M

d2 x dx +μ + kx = F(t). dt 2 dt

Set = (k/M)1/2 , = 2√μkM , and assume the wheel is initially at rest, so that x(0) = 0 and (dx/dt)t=0 = 0. If a constant load F(t) = F0 is suddenly applied to the wheel at the time t = 0, ﬁnd an expression for x(t)k/F0 . Plot this expression for several values of in the interval 0 < < 2 and comment on the results.

8

C H A P T E R

Series Solutions of Differential Equations, Special Functions, and Sturm–Liouville Equations

L

inear second order variable coefﬁcient equations arise in many applications, but only in a few special cases is it possible to express their general solution of a ﬁnite linear combination of elementary functions. As analytical, rather than purely numerical, information about solutions is often essential, some other way must be found to represent the solutions of such equations. The approach developed in this chapter involves seeking solutions of certain types of equation in the form of power series, and in other cases using an approach due to Frobenius that involves seeking solutions in the form of power series multiplied by a factor x c , where c is not an integer. Applications are made to a number of typical linear variable coefﬁcient equations, and then to the important Legendre, Chebyshev, and Bessel equations that lead in turn to Legendre and Chebyshev polynomials and to Bessel functions. Two-point boundary value problems, called Sturm–Liouville systems, that are deﬁned over an interval a ≤ x ≤ b and contain a parameter λ are introduced. It is shown that their solutions only exist for an inﬁnite number of special values of the parameter λ1 , λ2 , . . . , called the eigenvalues of the problem. Each solution ϕn (x) corresponding to an eigenvalue λn is called an eigenfunction, and the eigenfunctions are shown to have the special property of orthogonality with respect to a function w(x) called the weight function. This means b that if the set of eigenfunctions is {ϕn (x)}∞ n=1 , the integral a ϕm (x)ϕn (x)w(x)dx is positive when n = m and zero when n = m. This property will be used extensively in Chapter 18 when solving partial differential equations. Fundamental properties of eigenfunctions and eigenvalues are established for general Sturm–Liouville systems, after which a number of frequently occurring and important special cases are examined.

8.1

A First Approach to Power Series Solutions of Differential Equations

T

he solutions of many differential equations can be expressed in terms of elementary functions such as sine, cosine, exponential, and logarithm, all of whose mathematical properties are well known. When required, the analytical behavior of solutions that involve elementary functions can be explored by making use of 443

444

Chapter 8

Series Solutions of Differential Equations, Special Functions, and Sturm–Liouville Equations

their familiar properties. Numerical solutions are obtained easily, either by using a pocket calculator to ﬁnd the values of the elementary functions involved, or through the use of standard subroutines that form a part of all basic mathematical software packages. With either a pocket calculator or a software package, the method of calculating functional values is usually based on a series expansion of the function concerned. Most differential equations cannot be solved in terms of elementary functions, yet some form of analytical solution is often needed rather than a purely numerical one, so the fundamental question that then arises is how to obtain a solution in the form of a series, when only the differential equation is known. It is the purpose of this chapter to answer this question, and in the process to show how the form of series solution obtained depends on what are called the singular points of the differential equation. We begin our approach to this problem by showing how series solutions can be found for ﬁrst and second order linear differential equations with initial conditions speciﬁed at x = x0 . The series we obtain will be in powers of x − x0 , and they will be said to be expanded about the point x0 . The ﬁrst order linear differential equation will be assumed to be of the form y + p(x)y = r (x)

with y(x0 ) = y0 ,

(1)

and the second order linear differential equation will be assumed to be of the form y + P(x)y + Q(x)y = R(x)

analytic in a neighborhood

how to ﬁnd a power series solution

with y(x0 ) = y0 , y (x0 ) = y1 ,

(2)

where the functions p(x), r (x), P(x), Q(x), and R(x) can all be expanded as Taylor series about the point x0 . Functions with this property are said to be analytic in a neighborhood of the point x0 or, more simply, to be analytic at x0 . The method to be developed will be seen to be capable of extension to a higher order linear differential equation in an obvious manner, provided only that the coefﬁcients of y and its derivatives that are involved and the nonhom*ogeneous term are analytic at x0 . The approach is best illustrated by considering equation (1), and seeking a solution about x0 of the form y(x) = y(x0 ) + (x − x0 )y (x0 ) + =

∞ (x − x0 )n n=0

n!

y(n) (x0 ),

(x − x0 )2 (x − x0 )3 y (x0 ) + y (x0 ) + · · · 2! 3! with y(n) (x) = dn y/dx n . (3)

Setting x = x0 in (1) gives y(1) (x0 ) + p(x0 )y(x0 ) = r (x0 ), but y(x0 ) = y0 , so y(1) (x0 ) = r (x0 ) − p(x0 )y(x0 ) = r (x0 ) − p(x0 )y0 .

Section 8.1

A First Approach to Power Series Solutions of Differential Equations

445

To determine y(2) (x) we differentiate equation (1) once with respect to x to obtain y(2) (x) + p(1) (x)y(x) + p(x)y(1) (x) = r (1) (x), where p(1) (x) = p (x) and r (1) (x) = r (x). Then, after setting x = x0 and using the fact that y(1) (x0 ) = r (x0 ) − p(x0 )y0 , we ﬁnd that y(2) (x0 ) = r (1) (x0 ) − p(1) (x0 )y0 − p(x0 )[r (x0 ) − p(x0 )y0 ]. Higher order derivatives y(n) (x0 ) can be computed in similar fashion by repeated differentiation of the original differential equation coupled with the use of lower order derivatives that have already been determined. Once the values of y(k) (x0 ) have been found for k = 1, 2, . . . , N, for some given integer N, substitution into series (3) provides the required approximation to the power series solution of the initial value problem for the differential equation up to terms of order (x − x0 ) N . The existence and uniqueness of the solution are guaranteed by Theorem 5.2. This method generates the Taylor series expansion of y(x) about the point x0 when x0 = 0, and its Maclaurin series expansion when x0 = 0, though these series are often simply called power series about x0 = 0 and x0 = 0, respectively. EXAMPLE 8.1

Find the ﬁrst ﬁve terms in the series solution of y + (1 + x 2 )y = sin x,

with y(0) = a.

Solution As the initial condition is speciﬁed at x = 0, the power series solution is an expansion about the origin and so is, in fact, a Maclaurin series. The functions 1 + x 2 and cos x are analytic for all x, so the series expansion can certainly be found about the origin. Setting x = 0 in the equation and substituting for the initial conditions shows that y (0) = y(1) (0) = −a. Differentiation of the differential equation gives y(2) + 2xy + (1 + x 2 )y(1) = cos x, where y(2) = y , so setting x = 0 this becomes y(2) (0) + y(1) (0) = 1, but y(1) (0) = −a and so y(2) (0) = 1 + a. Repeating this process to ﬁnd higher order derivatives leads to the results y(3) (0) = −(1 + 3a), y(4) (0) = 9a, . . . . Substituting these results into series (3) shows that, to terms of order x 4 , the required solution takes the form y(x) = a − ax + (1 + a) EXAMPLE 8.2

x2 x3 x4 − (1 + 3a) + 9a + · · · . 2! 3! 4!

Find the ﬁrst ﬁve terms in the series solution of y + 4xy = 3e x−1 ,

with y(1) = 1.

Solution In this case the functions x and e x−1 are analytic for all x, but as the expansion is about x = 1, the power series solution that is obtained will be a Taylor series expansion about the point x = 1. Setting x = 1 in the differential equation and using the initial condition y(1) = 1 shows that y(1) (1) = −1.

446

Chapter 8

Series Solutions of Differential Equations, Special Functions, and Sturm–Liouville Equations

Differentiation of the differential equation gives y(2) + 4y + 4xy(1) = 3e x−1 , so setting x = 1 and using the result y(1) (1) = −1 shows that y(2) (1) = 3. Repeating this process leads to the results that y(3) (1) = −1 and y(4) (1) = −29, so substituting into (3) shows that the Taylor series expansion of the solution up to terms of order (x − 1)4 is 1 29 3 y(x) = 1 − (x − 1) + (x − 1)2 − (x − 1)3 − (x − 1)4 + · · · . 2 6 24 This same method can be applied to a second order equation of the type shown in (2), though a more general approach will be developed later to deal with the case in which the ﬁrst term is of the form a(x)y (x), and the expansion is about a point x0 where a(x0 ) = 0. EXAMPLE 8.3

Find the terms up to x 5 in the series solution of y + xy + (1 − x 2 )y = x

with y(0) = a, y (0) = b.

Solution The coefﬁcients x and (1 − x 2 ) and the nonhom*ogeneous term x are analytic for all x, so as the initial data is given at x = 0, a Maclaurin series solution can be found. Setting x = 0 in the equation and using the initial conditions y(0) = a and y (0) = b gives y(2) (0) = −a. Differentiating the differential equation we have y(3) + y(1) + xy(2) − 2xy + (1 − x 2 )y(1) = 1, so setting x = 0 and using the results y(2) (0) = −a and y(1) (0) = b shows that y(3) (0) = 1 − 2b. A repetition of this process leads to the results y(4) (0) = 5a, y(5) (0) = 14b − 4, . . . , so substituting into (3) shows that to terms of order x 5 the Maclaurin series expansion of the solution is 1 1 − 2b 3 5a 4 7b − 2 y(x) = a + bx − ax 2 + x + x + x5 + · · · . 2 6 24 60

Summary

Often a variable coefﬁcient equation cannot be solved in terms of known functions, though some form of analytical solution is still required. This section has shown how to overcome this difﬁculty in some cases by ﬁnding a solution in terms of a power series expanded about a point of interest x = a. The method was seen to work provided the functions in the equation have Taylor series expansions about x = a. It will be shown later how to ﬁnd series solutions in a systematic manner, and also how to generalize this approach to other types of equation.

EXERCISES 8.1 Find the ﬁrst ﬁve terms in the power series solution of the following initial value problems. 1. 2. 3. 4.

y + (1 + x 2 )y = x 2 , with y(0) = 1. 2y + xy = 1 − x, with y(0) = 2. y + (1 − 2x)y = x, with y(0) = −1. 4y + (1 + x + x 2 )y = x, with y(0) = 3.

5. 6. 7. 8. 9. 10.

y + (x − 2x 2 )y = 1, with y(0) = 1. y − 2xy = 1 − x, with y(0) = 2. 3y + (1 − x 2 )y = 1, with y(0) = 2. y + (1 + x)y = 1 + x 2 , with y(0) = 1. y − 2xy + x 2 y = 0, with y(0) = a, y (0) = b. 2y + 2(1 + x)y − y = 0, with y(0) = a, y (0) = b.

Section 8.2

A General Approach to Power Series Solutions of hom*ogeneous Equations

11. (1 + x 2 )y + 3xy + (1 − x 2 )y = 1 + x, with y(0) = a, y (0) = b. 12. (1 + 3x 2 )y + 2xy + 2xy = 1, with y(0) = a, y (0) = b. 13. y + 7y + x 2 y = 0, with y(0) = a, y (0) = b.

8.2

447

14. xy + (1 + x)y + xy = b, with y(0) = a, y (0) = 0. 15. 2y + 3x 2 y + (1 − x 2 )y = 2x, with y(0) = a, y (0) = b. 16. 3y + 2xy + (1 − 2x 2 )y = 1 + 2x, with y(0) = a, y (0) = b.

A General Approach to Power Series Solutions of hom*ogeneous Equations The method developed in Section 8.1 works satisfactorily if only the ﬁrst few terms in a power series solution are required, but it has the disadvantage that a separate calculation is required each time a coefﬁcient is determined. The present section shows how in many cases this difﬁculty can be overcome by introducing a systematic and simple way of generating arbitrarily many terms in a power series solution of the hom*ogeneous linear differential equation a(x)y + b(x)y + c(x)y = 0

(4)

about a point x0 , when a(x), b(x), and c(x) are polynomials with a(x0 ) = 0. The approach enables the coefﬁcients of the power series solution to be determined by means of a recurrence relation that relates a few consecutive coefﬁcients in the series. This has the advantage that once the ﬁrst few coefﬁcients in the series expansion have been found, the rest can be generated by means of the recurrence relation. There will be no loss of generality if the approach is based on an expansion about the origin, because if one is required about an arbitrary point x = x0 , the change of variable X = x − x0 will shift the point x = x0 to X = 0. For example, suppose a solution of y + (2 + 3x)y + x 2 y = 0 is required about the point x = 1, corresponding to the speciﬁcation of the initial conditions for y(1) and y (1) at x = 1. Setting X = x − 1 and y(x) = Y(x − 1) = Y(X), it follows that y(1) = Y(0), dy/dx = dY/dX, d2 y/dx 2 = d2 Y/dX 2 , and x = X + 1, so in terms of the new variables X and Y the equation and initial conditions become Y + (5 + 3X)Y + (1 + X)2 Y = 0,

with Y(0) = y(1), Y (0) = y (1).

Setting X = x − 1 in the power series solution of this equation expanded about X = 0 reduces it to the solution of the original equation expanded about x = 1. The approach we now describe involves seeking a solution in the form of a general power series y(x) =

∞

an x n

(5)

n=0

and ﬁnding a relationship between the coefﬁcients an by substituting (5) into the

448

Chapter 8

Series Solutions of Differential Equations, Special Functions, and Sturm–Liouville Equations

hom*ogeneous differential equation a(x)y + b(x)y + c(x)y = 0.

(6)

We will assume that the coefﬁcients a(x), b(x), and c(x) in the differential equation are polynomials in x, and so are analytic at x = 0, and also that a(0) = 0. If (5) is to be a solution of (6), it must satisfy the differential equation for all x, but this will only be possible if, after combining terms, the coefﬁcient of each power of x in the new power series is zero. It will be seen later that it is this last requirement that leads to the determination of the coefﬁcients an in terms of a recurrence relation. Before illustrating the approach by means of an example, we ﬁrst ﬁnd expressions for the derivatives y (x) and y (x) that will be needed in the calculation. Writing out the ﬁrst few terms of y(x) in (5) gives

y(x) = a0 + a1 x + a2 x 2 + a3 x 3 + · · · =

∞

an x n .

(7)

n=0

Differentiating this expression term by term with respect to x, which is permitted for x inside the interval of convergence of the series, we arrive at the result y (x) = a1 + 2a2 x + 3a3 x 2 + · · · =

∞

nan x n−1 ,

(8)

n(n − 1)an x n−2 .

(9)

n=1

and after a further differentiation we have y (x) = 2a2 + 2 · 3a3 x + 3 · 4a4 x 2 + · · · =

∞ n=2

In what is to follow it will be important to remember that the summation in (8) starts at n = 1, whereas the summation in (9) starts at n = 2. EXAMPLE 8.4

Find the recurrence relation that must be satisﬁed by coefﬁcients in the series solution of the differential equation y + 2xy + (1 + x 2 )y = 0 when the expansion is about the origin. Solve the initial value problem for this differential equation given that y(0) = 3 and y (0) = −1. Solution Substituting y(x) = (8) and (9) gives ∞ n=2

∞ n=0

n(n − 1)an x n−2 + 2x

an x n into the differential equation and using

∞ n=1

nan x n−1 + (1 + x 2 )

∞ n=0

an x n = 0.

Section 8.2

A General Approach to Power Series Solutions of hom*ogeneous Equations

449

Taking the factor 2x in the second term and the factor x 2 in the third term under their respective summation signs allows the equation to be written in the form ∞

n(n − 1)an x n−2 +

n=2

∞

2nan x n +

∞

an x n +

n=0

n=1

∞

an x n+2 = 0.

n=0

The powers of x in the ﬁrst and last summations are different from those in the middle two summations, so before combining the summations in order to ﬁnd the coefﬁcient of each power of x, it will ﬁrst be necessary to change the power of x in the ﬁrst and last terms from n − 2 and n + 2 to n. In the ﬁrst summation we set m = n − 2, causing the summation to become ∞

(m + 2)(m + 1)am+2 x m.

m=0

However, m is simply a summation index that can be replaced by any other symbol, so we will replace it by n to obtain the equivalent expression ∞ (n + 2)(n + 1)an+2 x n . n=0

Similarly, by setting m = n + 2 in the last summation, and then replacing m by n, we ﬁnd that ∞

an x n+2

becomes

n=0

∞

an−2 x n .

n=2

We now substitute these last two results into the series solution of the differential equation to obtain ∞ ∞ ∞ ∞ (n + 2)(n + 1)an+2 x n + 2nan x n + an x n + an−2 x n = 0, n=0

n=1

n=0

n=2

where now each summation involves x , though not all summations start from n = 0. Separating out the terms corresponding to n = 0 and n = 1, and collecting all the remaining terms under a single summation sign in which the summation starts from n = 2, this becomes n

2a2 + a0 + (6a3 + 3a1 )x +

∞ [(n + 2)(n + 1)an+2 + 3an + an−2 ]x n = 0. n=2

deriving and using a recurrence relation

As already remarked, if this power series is to be a solution of the differential equation it must satisfy the equation identically for all x, but this will only be possible if in the foregoing expression the coefﬁcient of each power of x vanishes. Applying this condition to the preceding series we ﬁnd that for it to vanish identically for all x, (coefﬁcient of x 0 ) 2a2 + a0 = 0 (coefﬁcient of x) 6a3 + 3a1 = 0 and (coefﬁcient of x n )

(n + 2)(n + 1)an+2 + 3an + an−2 = 0,

for n ≥ 2.

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Chapter 8

Series Solutions of Differential Equations, Special Functions, and Sturm–Liouville Equations

The ﬁrst condition shows that 1 a2 = − a0 , 2 while the second condition shows that 1 a3 = − a1 , 2 where a0 and a1 are arbitrary constants. The third condition is a recurrence relation (also called a recursion relation or an algorithm) that in this case relates three coefﬁcients whose indices differ by 2, so given an−2 and an we can ﬁnd an+2 for n = 2, 3, 4, . . . . We now show how to determine the ﬁrst few coefﬁcients an by writing the recursion relation in the form [(2n + 1)an + an−2 ] an+2 = − (n + 1)(n + 2) and setting n = 2, 3, 4, . . . . For n = 2, after using a2 = − 12 a0 , we ﬁnd that a4 = −

a0 (5a2 + a0 ) = , 12 8

whereas for n = 3, after using a3 = − 12 a1 , we ﬁnd that a1 (7a3 + a1 ) = . 20 8 Continuing this process generates the coefﬁcients a0 a1 a0 a1 , a9 = ,.... a6 = − , a7 = − , a8 = 48 48 384 384 Thus, all the coefﬁcients with even sufﬁxes are determined in terms of the arbitrary constant a0 , whereas all the coefﬁcients with odd sufﬁxes are determined in terms of the arbitrary constant a1 .

n Substituting these coefﬁcients into the power series y(x) = ∞ n=0 an x and grouping terms gives 1 6 1 8 1 2 1 4 x − ··· y(x) = a0 1 − x + x − x + 2 8 48 384 1 1 1 9 1 x − ··· . + a1 x − x 3 + x 5 − x 7 + 2 8 48 384 a5 = −

As the coefﬁcients a0 and a1 are arbitrary, the functions represented by the series 1 1 1 1 8 x − ··· y1 (x) = 1 − x 2 + x 4 − x 6 + 2 8 48 384 and 1 1 1 1 9 x − ··· y2 (x) = x − x 3 + x 5 − x 7 + 2 8 48 384 are seen to be the two linearly independent solutions known to be associated with a hom*ogeneous linear second order equation. So all possible solutions of the differential equation can be written in the form y(x) = C1 y1 (x) + C2 y2 (x),

Section 8.2

A General Approach to Power Series Solutions of hom*ogeneous Equations

451

with C1 and C2 arbitrary constants, where to reconcile this result with our previous notation we notice that C1 and C2 have been written in place of a0 and a1 . To solve the initial value problem the constants C1 and C2 must be chosen such that y(0) = 3 and y (0) = −1, so 3 = C1 y1 (0) + C2 y2 (0)

−1 = C1 y1 (0) + C2 y2 (0),

and

but y1 (0) = 1, y2 (0) = 0, and differentiation of the expressions for y1 (x) and y2 (x) shows that y1 (0) = 0 and y2 (0) = 1, so solving for C1 and C2 gives C1 = 3 and C2 = −1, showing that the required solution to the initial value problem is y(x) = 3y1 (x) − y2 (x). The coefﬁcients of the power series expansions for y1 (x) and y2 (x) in the last example were sufﬁciently complicated that no attempt was made to deduce their general forms and they were merely generated from the recurrence relation. The next example is simpler, and we use it to illustrate the type of argument that is necessary when attempting to arrive at the form of the general term in a power series solution of a hom*ogeneous linear differential equation. There are no speciﬁc rules to follow when seeking the form of a general term in a series, and success depends on experience and the ability to recognize the pattern of signs and numbers forming the coefﬁcients. EXAMPLE 8.5

Find two linearly independent solutions of y + xy + y = 0, when the series expansion is about the origin, and hence solve the initial value problem for which y(0) = 1 and y (0) = 0. Solution Substituting results (7) to (9) into the differential equation gives ∞

n(n − 1)an x n−2 + x

n=2

∞

nan x n−1 +

∞

an x n = 0.

n=0

n=1

Shifting the summation index in the ﬁrst term, taking the factor x under the second summation and separating out the constant term, as in Example 8.4, gives 2a2 + a0 +

∞ [(n + 2)(n + 1)an+2 + (n + 1)an ]x n = 0. n=1

Equating the coefﬁcient of each power of x to zero, as in Example 8.4, shows that 2a2 + a0 = 0,

so a2 = −

a0 , 2

and (n + 2)(n + 1)an+2 + (n + 1)an = 0

for n ≥ 1,

but as n + 1 = 0 this last condition reduces to the simpler recurrence relation an+2 = −

an , n+2

for n = 1, 2, . . . .

452

Chapter 8

Series Solutions of Differential Equations, Special Functions, and Sturm–Liouville Equations

It follows directly from the recurrence relation that all even coefﬁcients are multiples of a0 and all odd coefﬁcients are multiples of a1 with a0 a1 a0 a1 a2 a3 a4 , a5 = − = , a6 = − = − , a3 = − , a4 = − = 3 4 2·4 5 3·5 6 2·4·6 a1 a0 a1 a5 a6 a7 a7 = − = − , a8 = − = , a9 = − = ,..., 7 3·5·7 8 2·4·6·8 9 3·5·7·9 where a0 and a1 are arbitrary constants. It is apparent that the pattern of coefﬁcients with even sufﬁxes differs from the one for coefﬁcients with odd sufﬁxes, so each must be considered separately. Starting with the coefﬁcients with even sufﬁxes, we use the fact that if m = 1, 2, . . . , then 2m is an even number. A little experimentation shows that the signs of the terms with even sufﬁxes are given by the factor (−1)m. Noticing that a2 , a4 , a6 , and a8 can be written in the form a2 =

(−1)a0 , 2

a4 =

1 (−1)2 a0 , 2 · 4 22 2!

a6 =

(−1)3 a0 −a0 = , 2·4·6 23 3!

(−1)4 a0 (−1)4 a0 = 2·4·6·8 24 4! suggests that if we set n = 2m, for m = 0, 1, 2, . . . , the even numbered terms can be written a8 =

a2m =

(−1)m a0 . 2mm!

A formal proof that this is the general coefﬁcient in the series involving even powers of x can be obtained by mathematical induction, but we leave this as an exercise. It is now necessary to consider the coefﬁcients with odd sufﬁxes, and to do this we use the fact that if m = 1, 2, 3, . . . , then 2m + 1 is an odd number. Noticing that the coefﬁcients a3 , a5 , a7 , and a9 can be written a3 =

(−1)2a1 −a1 = , 3 3!

a7 =

(−1)3 2 · 4 · 6a1 (−1)3 23 3!a1 −a1 = = , 3·5·7 1·2·3·4·5·6·7 7!

a5 =

(−1)2 2 · 4a1 (−1)2 22 2! a1 = = , 3·5 1·2·3·4·5 5!

(−1)4 2 · 4 · 6 · 8a1 (−1)4 24 4!a1 a1 = = 3·5·7·9 9! 9! suggests that the coefﬁcients in the series of odd powers of x can be written a9 =

a2m+1 =

(−1)m 2mm! a1 . (2m + 1)!

Here again we leave as an exercise the task of giving an inductive proof that this is, indeed, the coefﬁcient of the general term in the series involving odd powers of x. The solution of the differential equation has now separated into two series, one multiplied by a0 containing only even powers of x and the other multiplied by a1 containing only odd powers of x, so the solution becomes y(x) = a0

∞ ∞ (−1)m x 2m (−1)m 2mm!x 2m+1 + a . 1 2mm! (2m + 1)! m=0 m=0

Section 8.2

A General Approach to Power Series Solutions of hom*ogeneous Equations

453

As a0 and a1 are arbitrary constants, and the two series are not proportional, it follows that two linearly independent solutions of the differential equation are y1 (x) =

∞ (−1)m x 2m 2mm! m=0

and

y2 (x) =

∞ (−1)m 2mm!x 2m+1 , (2m + 1)! m=0

so the general solution is y(x) = C1 y1 (x) + C2 y2 (x), where C1 and C2 are arbitrary constants. Using the series for y1 (x) and y2 (x), simple calculation gives y1 (0) = 1, y1 (0) = 0, y2 (0) = 0, and y2 (0) = 1, so the initial conditions y(0) = 1, y (0) = 0 will be satisﬁed if the constants C1 and C2 are such that 1 = C1 y1 (0) + C2 y2 (0)

and

0 = C1 y1 (0) + C2 y2 (0).

This pair of equations has the solution C1 = 1 and C2 = 0, so the solution of the initial value problem becomes y(x) =

∞ (−1)m x 2m . 2mm! m=0

y(x) =

∞ (−x 2 /2)m , m! m=0

Rewriting this as

we recognize that the solution is simply y(x) = exp(−x 2 /2), so this series is known to converge for all x. Finally, to complete our examination of the two linearly independent solutions, let us ﬁnd the radius of convergence of the second solution y2 (x). The formula for the radius of convergence R based on the ratio test requires all powers of x to be present, whereas the series y2 (x) only contains odd powers of x, so we must modify the series before using the test. All that is necessary is to set z = x 2 and to write the series in the form ∞ (−1)m 2mm! m y2 (x) = x z , (2m + 1)! m=1 for now the radius of convergence of the series in z can be found. The coefﬁ