Engineering Mathematics II 9789387432345, 9387432343

Table of contents :
Title
Contents
Unit 1 Matrices
Unit 2 Vector Calculus
Unit 3 Analytic Functions
Unit 4 Complex Integration
Unit 5 Laplace Transforms

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Engineering Mathematics II

About the Author

T Veerarajan retired as Dean, Department of Mathematics, Velammal College of Engineering and Technology, Viraganoor, Madurai, Tamil Nadu. A gold medalist from Madras University, he has had a brilliant academic career all through. He has 50 years of teaching experience at undergraduate and postgraduate levels in various established engineering colleges in Tamil Nadu including Anna University, Chennai.

Engineering Mathematics II

T Veerarajan Former Dean, Department of Mathematics Velammal College of Engineering and Technology Viraganoor, Madurai Tamil Nadu

McGraw Hill Education (India) Private Limited Chennai McGraw Hill Education Offices Chennai new York St Louis San Francisco auckland Bogotá Caracas Kuala Lumpur Lisbon London Madrid Mexico City Milan Montreal San Juan Santiago Singapore Sydney Tokyo Toronto

McGraw Hill Education (India) Private Limited Published by McGraw Hill Education (India) Private Limited 444/1, Sri Ekambara Naicker Industrial Estate, Alapakkam, Porur, Chennai 600 116 Engineering Mathematics II Copyright © 2018 by McGraw Hill Education (India) Private Limited. No part of this publication may be reproduced or distributed in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise or stored in a database or retrieval system without the prior written permission of the publishers. The program listings (if any) may be entered, stored and executed in a computer system, but they may not be reproduced for publication. This edition can be exported from India only by the publishers, McGraw Hill Education (India) Private Limited. 1 2 3 4 5 6 7 8 9

D102739

22 21 20 19 18

Printed and bound in India. ISBN (13): 978-93-87432-34-5 ISBN (10): 93-87432-34-3 Managing Director: Kaushik Bellani Director—Science & Engineering Portfolio: Vibha Mahajan Senior Portfolio Manager—Science & Engineering: Hemant K Jha Associate Portfolio Manager—Science & Engineering: Mohammad Salman Khurshid Production Head: Satinder S Baveja Assistant Manager—Production: Anuj K Shriwastava General Manager—Production: Rajender P Ghansela Manager—Production: Reji Kumar Information contained in this work has been obtained by McGraw Hill Education (India), from sources believed to be reliable. However, neither McGraw Hill Education (India) nor its authors guarantee the accuracy or completeness of any information published herein, and neither McGraw Hill Education (India) nor its authors shall be responsible for any errors, omissions, or damages arising out of use of this information. This work is published with the understanding that McGraw Hill Education (India) and its authors are supplying information but are not attempting to render engineering or other professional services. If such services are required, the assistance of an appropriate professional should be sought. Typeset at Text-o-Graphics, B-1/56, Aravali Apartment, Sector-34, Noida 201 301, and printed at Cover Printer:

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Contents Preface Unit 1. Matrices

ix 1.1–1.80

1.1 1.2 1.3 1.4

Introduction 1.3 Vectors 1.4 Linear Dependence and Linear Independence of Vectors 1.5 Methods of Testing Linear Dependence or Independence of a Set of Vectors 1.5 1.5 Consistency of a System of Linear Algebraic Equations 1.5 Worked Example 1(a) 1.7 Exercise 1(a) 1.24 1.6 Eigenvalues and Eigenvectors 1.27 Worked Example 1(b) 1.32 Exercise 1(b) 1.44 1.7 Cayley – Hamilton Theorem 1.46 1.8 Property 1.47 1.9 Calculation of Powers of a Matrix A 1.49 1.10 Diagonalisation by Orthogonal Transformation or Orthogonal Reduction 1.49 Worked Example 1(c) 1.49 Exercise 1(c) 1.62 1.11 Quadratic Forms 1.64 Worked Example 1(d) 1.67 Exercise 1(d) 1.74 Answers 1.75 Unit 2. Vector calcUlUs 2.1 Introduction 2.3 2.2 Vector Differential Operator ∇ 2.3 Worked Example 2(a) 2.6 Exercise 2(a) 2.11 2.3 The Divergence of a Vector 2.12 Worked Example 2(b) 2.18 Exercise 2(b) 2.26 2.4 Line Integral of Vector Point Functions 2.27

2.1–2.59

vi

Contents

Worked Example 2(c) 2.30 Exercise 2(c) 2.37 2.5 Integral Theorems 2.39 Worked Example 2(d) 2.42 Exercise 2(d) 2.56 Answers 2.58 Unit 3. analytic FUnctions

3.1–3.97

3.1 Introduction 3.3 3.2 The Complex Variable 3.4 Worked Example 3(a) 3.12 Exercise 3(a) 3.24 3.3 Properties of Analytic Functions 3.26 Worked Example 3(b) 3.32 Exercise 3(b) 3.43 3.4 Conformal Mapping 3.45 3.5 Some Simple Transformation 3.48 3.6 Some Standard Transformations 3.51 Worked Example 3(c) 3.59 Exercise 3(c) 3.76 3.7 Bilinear and Schwarz-Christoffel Transformations 3.79 3.8 Schwarz-Christoffel Transformations 3.82 Worked Example 3(d) 3.84 Exercise 3(d) 3.93 Answers 3.95 Unit 4. coMplex integration

4.1–4.73

4.1 Introduction 4.3 4.2 Cauchy’s Integral Theorem or Cauchy’s Fundamental Theorem 4.4 Worked Example 4(a) 4.9 Exercise 4(a) 4.19 4.3 Series Expansions of Functions of Complex Variable-Power Series 4.22 4.4  Classification of Singularities  4.27 Worked Example 4(b) 4.33 Exercise 4(b) 4.45 4.5 Contour Integration—Evaluation of Real Integrals 4.49 Worked Example 4(c) 4.52 Exercise 4(c) 4.69 Answers 4.70

Contents Unit 5. laplace transForMs

vii 5.1–5.120

5.1 Introduction 5.3 5.2 Linearity Property of Laplace and Inverse Laplace Transforms 5.4 5.3 Laplace Transforms of Some Elementary Functions 5.5 5.4 Laplace Transforms of Some Special Functions 5.9 5.5 Properties of Laplace Transforms 5.12 Worked Example 5(a) 5.14 Exercise 5(a) 5.35 5.6 Laplace Transform of Periodic Functions 5.38 5.7 Derivatives and Integrals of Transforms 5.39 Worked Example 5(b) 5.42 Exercise 5(b) 5.62 5.8 Laplace Transforms of Derivatives and Integrals 5.65 5.9 Initial and Final Value Theorems 5.68 5.10 The Convolution 5.70 Worked Example 5(c) 5.72 Exercise 5(c) 5.94 5.11 Solutions of Differential and Integral Equations 5.98 Worked Example 5(d) 5.98 Exercise 5(d) 5.110 Answers 5.112

Preface

I am deeply gratified with the enthusiastic response shown by students and faculty  members alike to all my books on Engineering Mathematics and Mathematics in general throughout the country. The motivation behind writing this book is to meet the requirements of students of first-year undergraduate course on Engineering Mathematics offered to the students  of engineering. The contents have been covered in adequate depth for semester II of various universities/ deemed universities across the country. It offers a balanced coverage of both theory and problems. Lucid writing style supported by step-bystep solutions to all problems enhances understanding of the concepts. The book has ample number of solved and unsolved problems of different variety to help students and teachers learning and teaching this subject. I hope that the book will be received by both the faculty and the students as enthusiastically as my other books on Engineering Mathematics. Critical evaluation and suggestions for the improvement of the book will be highly appreciated and acknowledged. T VEERARAJAN

Publisher’s Note McGraw Hill Education (India) invites suggestions and comments from you, all of which can be sent to [email protected] (kindly mention the title and author name in the subject line). Piracy-related issues may also be reported.

Unit-1 Matrices

Unit

1

Matrices 1.1

intRODUCtiOn

In the lower classes, the students have studied a few topics in Elementary Matrix theory. They are assumed to be familiar with the basic definitions and concepts of matrix theory as well as the elementary operations on and properties of matrices. Though the concept of rank of a matrix has been introduced in the lower classes, we briefly recall the definition of rank and working procedure to find the rank of a matrix, as it will be of frequent use in testing the consistency of a system of linear algebraic equations, that will be discussed in the next section.

1.1.1

Rank of a Matrix

Determinant of any square submatrix of a given matrix A is called a minor of A. If the square submatrix is of order r, then the minor also is said to be of order r. Let A be an m × n matrix. The rank of A is said to be ‘r’, if (i) there is at least one minor of A of order r which does not vanish and (ii) every minor of A of order (r + 1) and higher order vanishes. In other words, the rank of a matrix is the largest of the orders of all the nonvanishing minors of that matrix. Rank of a matrix A is denoted by R(A) or ρ(A). To find the rank of a matrix A, we may use the following procedure: We first consider the highest order minor (or minors) of A. Let their order be r. If any one of them does not vanish, then ρ(A) = r. If all of them vanish, we next consider minors of A of next lower order (r – 1) and so on, until we get a non-zero minor. The order of that non-zero minor is ρ(A). This method involves a lot of computational work and hence requires more time, as we have to evaluate many determinants. An alternative method to find the rank of a matrix A is given below: Reduce A to any one of the following forms, (called normal forms) by a series of elementary operations on A and then find the order of the unit matrix contained in the normal form of A:  I   I |O

[ I r ]; [ I r | O ];  r  ;  r   O   O | O 

Mathematics II

1.4

Here Ir denotes the unit matrix of order r and O is zero matrix. By an elementary operation on a matrix (denoted as E-operation) we mean any one of the following operations or transformations: (i) Interchange of any two rows (or columns). (ii) Multiplication of every element of a row (or column) by any non-zero scalar. (iii) Addition to the elements of any row (or column), the same scalar multiples of corresponding elements of any other row (or column).

Note

The alternative method for finding the rank of a matrix is based on the property that the rank of a matrix is unaltered by elementary operations. Finally we observe that we need not necessarily reduce a matrix A to the normal form to find its rank. It is enough we reduce A to an equivalent matrix, whose rank can be easily found, by a sequence of elementary operations on A. The methods are illustrated in the worked examples that follow.

1.2 VECtORS A set of n numbers x1, x2, . . ., xn written in a particular order (or an ordered set of n numbers) is called an n-dimensional vector or a vector of order n. The n numbers are called the components or elements of the vector. A vector is denoted by a single letter X or Y etc. The components of a vector may be written in a row as X = (x1, x2, ..., xn)  x1    x  or in a column as X =  2  . These are called respectively row vector and  x   n  column vector. We note that a row vector of order n is a 1 × n matrix and a column vector of order n is an n × 1 matrix.

1.2.1 Addition of Vectors The sum of two vectors of the same dimension is obtained by adding the corresponding components. i.e., if X = (x1, x2, . . ., xn) and Y = (y1, y2, . . ., yn), then X + Y = (x1 + y1, x2 + y 2. . ., xn + yn).

1.2.2

Scalar Multiplication of a Vector

If k is a scalar and X = (x1 , x2, . . ., xn) is a vector, then the scalar multiple kX is defined as kX = (kx1, kx2, . . ., kxn).

1.2.3

Linear Combination of Vectors

If a vector X can be expressed as X = k1X1 + k2X2 + . . . + krXr then X is said to be a linear combination of the vectors X1, X2, . . ., Xr.

Matrices

1.3

1.5

LinEAR DEPEnDEnCE AnD LinEAR inDEPEnDEnCE OF VECtORS

The vectors X1, X2, . . ., Xr are said to be linearly dependent if we can find scalars k1, k2, . . . kr, which are not all zero, such that k1Xl + k2X2 + . . . + krXr = 0. A set of vectors is said to be linearly independent if it is not linearly dependent, i.e. the vectors X1, X2, . . ., Xr are linearly independent, if the relation k1X1 + k2X2+ . . . krXr = 0 is satisfied only when kl = k2 = . . . = kr = 0.

Note

When the vectors X1, X2, . . ., Xr are linearly dependent, then k1X1 + k2X2 + . . . + krXr = 0, where at least one of the k’s is not zero. Let km ≠ 0. Xm = −

Thus

k1 k k ⋅ X 1 − 2 X 2 −⋅⋅⋅− r X r . km km km

Thus at least one of the given vectors can be expressed as a linear combination of the others.

1.4

MEtHODS OF tEStinG LinEAR DEPEnDEnCE OR inDEPEnDEnCE OF A SEt OF VECtORS

Method 1

Using the definition directly.

Method 2

We write the given vectors as row vectors and form a matrix. Using elementary row operations on this matrix, we reduce it to echelon form, i.e. the one in which all the elements in the rth column below the rth element are zero each. If the number of non-zero row vectors in the echelon form equals the number of given vectors, then the vectors are linearly independent. Otherwise they are linearly dependent.

Method 3

If there are n vectors, each of dimension n, then the matrix formed as in method (2) will be a square matrix of order n. If the rank of the matrix equals n, then the vectors are linearly independent. Otherwise they are linearly dependent.

1.5

COnSiStEnCY OF A SYStEM OF LinEAR ALGEBRAiC EQUAtiOnS

Consider the following system of m linear algebraic equations in n unknowns: a11x1 + a12x2 + . . . + a1nxn = b1 a21x1 + a22x2 + . . . + a2nxn = b2 am1x1 + am2x2 + . . . + amnxn = bm This system can be represented in the matrix form as AX = B, where a  11 A =  a21   am1

 b1  x   1   … a1n  a12  x2  b     … a2 n  , X =   , B =  2  a22    am 2   amn  x  b    n   m 

Mathematics II

1.6

The matrix A is called the coefficient matrix of the system, X is the matrix of unknowns and B is the matrix of the constants. If B ≡ O, a zero matrix, the system is called a system of homogeneous linear equations; otherwise, the system is called a system of linear non-homogeneous equations. The m × (n + 1) matrix, obtained by appending the column vector B to the coefficient matrix A as the additional last column, is called the augmented matrix of the system and is denoted by [A, B] or [A | B]. a  11 A , B = [ ]  a21   am1

i.e.

a12 … a1n b1  a22 … a2 n b2   am 2 … amn bm  

1.5.1  Definitions A set of values of x1, x2 . . ., xn. which satisfy all the given m equations simultaneously is called a solution of the system. When the system of equations has a solution, it is said to be consistent. Otherwise the system is said to be inconsistent. A consistent system may have either only one or infinitely many solutions. When the system has only one solution, it is called the unique solution. The necessary and sufficient condition for the consistency of a system of linear non-homogeneous equations is provided by a theorem, called Rouches’s theorem, which we state below without proof.

1.5.2

Rouche’s theorem

The system of equations AX = B is consistent, if and only if the coefficient matrix A and the augmented matrix [A, B] are of the same rank. Thus to discuss the consistency of the equations AX = B (m equations in n unknowns), the following procedure is adopted: We first find R(A) and R(A, B). (i) If R(A) ≠ R(A, B), the equations are inconsistent (ii) If R(A) = R(A, B) = the number of unknowns n, the equations are consistent and have a unique solution. In particular, if A is a non-singular (square) matrix, the system AX = B has a unique solution. (iii) If R(A) = R(A, B) < the number of unknowns n, the equations are consistent and have an infinite number of solutions.

1.5.3

System of Homogeneous Linear Equations

Consider the system of homogeneous linear equations AX = O (m equations in n unknowns) i.e a11x1 + a12x2 + . . . a1nxn = 0 a21x1 + a22x2 + . . . + a2nxn = 0 –––––––––––––––– am1x1 + am2x2 + . . . amnxn = 0

Matrices

1.7

This system is always consistent, as R(A) = R(A, O). If the coefficient matrix A is non-singular, the system has a unique solution, namely, x1 = x2 = . . . = xn = 0. This unique solution is called the trivial solution, which is not of any importance. If the coefficient matrix A is singular, i.e. if | A| = 0, the system has an infinite number of non-zero or non-trivial solutions. The method of finding the non-zero solution of a system of homogeneous linear equations is illustrated in the worked examples that follow. WORKED EXAMPLE 1(a) Example 1.1 Show that the vectors X1 = (1, 1, 2), X2 = (1, 2, 5) and X3 = (5, 3, 4) are linearly dependent. Also express each vector as a linear combination of the other two.

Method 1 Let i.e. ∴

k1X1 + k2X2 + k3X3 = 0 k1(1, 1, 2) + k2(1, 2, 5) + k3(5, 3, 4) = (0, 0, 0) k1 + k2 + 5k3 = 0 k1 + 2k2 + 3k3 = 0 2k1 + 5k2 + 4k3 = 0

(2) – (1) gives k2 – 2k3 = 0

or

Using (4) in (3),

k2 =2k3

(4)

k1 = – 7k3

(5)

Taking k3 = 1, we get k1 = – 7 and k2 = 2. Thus –7X1 + 2X2 + X3 = 0 ∴ The vectors X1, X2, X3 are linearly dependent. From (6), we get

Method 2

(1) (2) (3)

X1 =

2 1 X2 + X3 , 7 7

X2 =

7 1 X1 − X 3 2 2

(6)

and X3 = 7X1 – 2X2

Writing X1, X2, X3 as row vectors, we get  1 1 2  1 1 2     A = 1 2 5  ∼  0 1 3 ( R2′ = R2 − R1 , R3′ = R3 − 5 R1 )      5 3 4   0 −2 −6       1 1 2   ∼  0 1 3 ( R3′′ = R3′ + 2 R2′ )    0 0 0   In the echelon form of the matrix, the number of non-zero vectors = 2 (< the number of given vectors). ∴ X1, X2, X3 are linearly dependent.

Mathematics II

1.8 Now

0 = R3′′ = R3′ + 2 R2′ = ( R3 − 5 R1 ) + 2 ( R2 − R1 ) = − 7 R1 + 2 R2 + R3

i.e.

–7X1 + 2X2 + X3 = 0

As before,

X1 =

2 1 7 1 X 2 + X 3 , X 2 = X1 − X 3 7 7 2 2

and X3 = 7X1 – 2X2.

Method 3 |A| = 0 ∴ R (A) ≠ 3; R (A) = 2 ∴ The vectors X1, X2, X3 are linearly dependent. Example 1.2 Show that the vectors X1 = (1, –1, –2, –4), X2 = (2, 3, –1, –1), X3 = (3, 1, 3, –2) and X4 = (6, 3, 0, –7) are linearly dependent. Find also the relationship among them.  X 1   1 −1 −2 −4   1 −1 −2 −4         X 2  2 5 3 7 ( R2′ = R2 − 2 R1 , R3′ = 3 −1 −1  0       A=   =   R − 3R , R ′ = R − 6 R )  ∼ 0 X 4 9 10 − 3 1 3 2 1 4 4 1 3     3    X  6    0 9 12 17 − 3 0 7     4    1 −1 −2 −4     3 7 0   1 1   R2′′ = R2′ ; R3′′ = R3′; R4′′ = R4′  ∼ 5 5     5 0 4 9 10   0 9 12 17   1 −1 −2 −4    3 7 0  1  5 5   ∼ 33 22  = ( R3′′′= R3′′ − 4 R2′′ ; R4′′′= R4′′ − 9 R2′′) 0  0  5 5    33 22  0 0    5 5   1 −1 −2 −4     3 7 0  1  5 5  ′′′ ′ ∼ ( R4 = R4′′′ − R3′′′)  33 22  0  0  5 5    0 0 0 0 Number of non-zero vectors in echelon form of the matrix A = 3. ∴ The vectors X1, X2, X3, X4 are linearly dependent.

Matrices Now

1.9

0 = R4′′′′= R4′′′− R3′′′ = ( R4′′ − 9R2′′) − ( R3′′ − 4R2′′) = R4′ − R3′ − R2′ = ( R4 − 6R1) − ( R3 − 3R1) − ( R2 − 2R1) = − R1 − R2 − R3 + R4

∴ The relation among Xl, X2, X3, X4 is – X1 – X2 – X3 + X4 = 0

or

X1 + X2 + X3 – X4 = 0.

Example 1.3 Show that the vectors X1 = (2, –2, 1), X2 = (1, 4, –1) and X3 = (4, 6, –3) are linearly independent.

Method 1 Let

k1 X1 + k2 X2 + k3 X3 = 0

i.e. k1 (2, –2, 1) + k2 (1, 4, –1) + k3 (4, 6, –3) = (0, 0, 0) ∴

From (1) and (2), From (2) and (3), ∴

2k1 + k2 + 4k3 = 0 –2k1 + 4k2 + 6k3 = 0 k1 – k2 –3k3 = 0 k2 + 2k3 = 0 k2 = 0 k1 = 0 = k2 = k3.

(1) (2) (3) (4) (5)

∴ The vectors X1, X2, X3 are linearly independent.

Method 2  X 1   2 −2 1  1 4 −1            A = X2 = 1 4 −1 ∼  2 −2 1 ( R1′ = R2 ; R2′ = R1 )  X  4 6 −3  4 6 −3  3  1 4 −1    ∼  0 −10 3 R2′′ = R2′ − 2 R1′ ; R3′′ = R3′ − 4 R1′  0 −10 1  1 4 −1    ∼  0 −10 3 R3′′′ = R3′′ − R2′′ 0 0 −2  Number of non-zero vectors in the echelon form of A = number of given vectors, ∴ X1, X2, X3 are linearly independent.

(

(

)

)

Example 1.4 Show that the vectors X1 = (1, −1, −1, 3), X2 = (2, 1, −2, −1) and X3 = (7, 2, −7, 4) are linearly independent.

(

 X 1   1 −1 −1 3  1 −1 −1 3 R2′ = R2 − 2 R1 ;            A =  X 2  = 2 1 −2 −1 ∼  0 3 0 −7  X  7 2 −7 4  0 9 0 −17 R3′ = R3 − 7 R1  3 

)

Mathematics II

1.10

 1 −1 −1 3    ∼ 0 3 0 −7 R3′′ = R3′ − 3R2′ 0 0 0 4  Number of non-zero vectors in the echelon form of A = number of given vectors. ∴ X1, X2, X3 are linearly independent.

(

)

Example 1.5 Test for the consistency of the following system of equations: x1 + 2x2 + 3x3 + 4x4 = 5 6x1 + 7x2 + 8x3 + 9x4 = 10 11x1 + 12x2 + 13x3 + 14x4 =15 16x1 + 17x2+ 18x3 + 19x4 = 20 21x1 + 22x2 + 23x3 + 24x4 = 25 The given equations can be put as 5 1 2 3 4    x1     6 7 8 9    10       11 12 13 14   x2  = 15       16 17 18 19   x3   20   x       4    21 22 23 24  25   i.e. AX = B (say) Let us find the rank of the augmented matrix [A, B] by reducing it to the normal form 1 2 3 4 5 1 2 3 4 5    ( R → R2 − R1   6 7 8 9 10   5 5 5 5 5  2    R →R −R  3 1 [ A, B ] = 11 12 13 14 15  ∼ 10 10 10 10 10  3 R → R − R 4 1 16 17 18 19 20 15 15 15 15 15  4    R →R −R )     5  5 1  21 22 23 24 25  20 20 20 20 20

Note

If two matrices A and B are equivalent, i.e. are of the same rank, it is denoted as A ~ B. 1  1  ∼ 1 1   1

2 3 4 5  1 1 1 1 1 1 1  R2 → R2 , R3 → R3 , R4 → R4 ,  10 15 5 1 1 1 1  1 R5  1 1 1 1 R5 → 20   1 1 1 1

1  0  ∼  0 0   0

2 −1 −1 −1 −1

3 −2 −2 −2 −2

4 −3 −3 −3 −3

5  −4  ( R → R2 − R1 , R3 → R3 − R1 , −4  2 R → R4 − R1 , R5 → R5 − R1 ) −4  4  −4 

Matrices 1  0  ∼  0 0    0 1  0  ∼  0 0    0

0 0 0 0  −1 −2 −3 −4  (C2 → C2 − 2C1 , C3 → C3 − 3C1 , −1 −2 −3 −4  C4 → C4 − 4C1 , C5 → C5 − 5C1 ) −1 −2 −3 −4   −1 −2 −3 −4  0 0 0 0  1 1 1 1 [C2 → − C2 , C3 → C3 ÷ (−2), 1 1 1 1 C → C4 ÷ (−3) , C5 → C5 ÷ (−4) 1 1 1 1 4  1 1 1 1

1  0  ∼ 0 0   0

0 1 0 0

1  0  ∼  0 0    0

0 0 0 0  1 0 0 0 (C → C3 − C2 , C4 → C4 − C2 , 0 0 0 0 3 C → C5 − C2 ) 0 0 0 0 5  0 0 0 0

0  1 ( R → R3 − R2 , R4 → R4 − R2 , 0 3 R → R5 − R2 ) 0 5  0 0 0 0 0 1 0 0

0 1 0 0

 I2 0    Now [A, B] has been reduced to the normal form  0 0    The order of the unit matrix present in the normal form = 2. Hence the rank of [A, B] = 2. The rank of the coefficient matrix A can be found as 2, in a similar manner. Thus R (A) = R [A, B] = 2 ∴ The given system of equations is consistent and possesses many solutions. Example 1.6 Test for the consistency of the following system of equations: x1 − 2x2 − 3x3 = 2; 3x1 − 2x2 = −1; −2x2 − 3x3 = 2; x2 + 2x3 = 1. The system can be put as  2  1 −2 −3    x1     3 −2 0   −1     0 −2 −3  x2  =  2   x    3 0 1 2    1 

1.11

Mathematics II

1.12 i.e. AX = B (say) 1  3 [ A, B ] =  0 0  1  0 ∼  0 0 

2   1 −2 −3 2 −2 −3      0 −1  0 4 9 −7 −2 ∼ ( R2 → R2 − 3R1 ) 2  0 −2 −3 2 −2 −3 1 2 1  0 1 2 1 0 0 0  4 9 −7 (C2 → C2 + 2C1 , C3 → C3 + 3C1 , C4 → C4 − 2C1 ) −2 −3 2 1 2 1

1 0 0 0   0  1 2 1  (R → R , R → R ) ∼  2 4 4 2 2  0 −2 −3 0 4 9 −7  1  0 ∼  0 0 

0 0 1 2

1  0 ∼  0 0 

0 0 1 0

1  0 ∼  0 0 

0 0 1 0

1  0 ∼  0 0 

0  1 0 0 (C4 → C4 − 4C3 ) 0 1 0 0 0 −15

1  0 ∼  0 0 

0 1 0 0

0 0

0 0

0  1 ( R3 → R3 + 2 R2 , R4 → R4 − 4 R2 ) 1 4 1 −11

0  0 (C3 → C3 − 2C2 , C4 → C4 − C1 ) 1 4 1 −11

0  0 ( R4 → R4 − R3 ) 0 1 4 0 0 −15 0 0

0 0 1 0

0   0   R4 → − 1 R4    0  15   1

Matrices

1.13

∴ R[A, B] = 4 But R (A) ≠ 4, as A is a (4 × 3) matrix. In fact R (A) = 3, as the value of the minor 3 −2 0 0 −2 −3 ≠ 0 0 1 2 Thus R (A) ≠ R [A, B] ∴The given system is inconsistent. Example 1.7 Test for the consistency of the following system of equations and solve them, if consistent, by matrix inversion. x − y + z + 1 = 0; x − 3y + 4z + 6 = 0; 4x + 3y − 2z + 3 = 0; 7x − 4y + 7z+ 16 = 0.  1 −1 1    1 −3  4  A =  3 −2 4  7 −4 7   1 −1 1 −1 1 −1  1 −1  ( R 2 → R2 − R1 ,       1 −3  0 − 2 3 − 5 4 − 6  R → R − 4R , ∼ [ A, B ] =  3 3 1 7 −6 1 3 −2 −3  0 4 → − R R R 7 4 4 1)  7 −4 7 −16  0 3 0 −9  1 0 0 0    0 −2 3 −5  ∼ (C2 → C2 + C1 , C3 → C3 − C1 , C4 → C4 + C1 ) 7 −6 1 0 0 3 0 −9  1 0 0 0   0 1 −6 7  ∼ ( R2 ↔ R3 and then C2 ↔ C4 ) 3 −2  0 −5  0 −9 0 3  1  0 ∼  0 0 

0 0 0  ( R3 → R3 + 5 R2 , R4 → R4 + 9 R2 ) 1 0 0 and then 0 −27 33 (C3 → C3 + 6C2 , C4 → C4 − 7C2 ) 0 −54 66

Mathematics II

1.14 1  0 ∼  0 0  1  0 ∼  0 0 

0 0 0   1 0 0  C3 → − 1 C3 , C4 → 1 C4    0 1 1  27 33   0 2 2 0 0 0  1 0 0 ( R 4 → R4 − 2 R3 and theen 0 1 0 C4 → C4 − C3 ) 0 0 0 

∴ R [A, B] = 3. Also R (A) = 3 ∴ The given system is consistent and has a unique solution. To solve the system, we take any three, say the first three, of the given equations.

i.e.

 1 −1 1  x  −1       1 −3   y  = −6 4      4 3 −2  z  −3 

i.e. ∴

Let

AX = B, say X = A−1 B  1 −1 1  a11    4 ≡  a21 A =  1 −3 4 3 −2  a31 

(1) a12 a22 a32

a13   a23  a33 

Now A11 = co-factor of a11 in |A| = −6 A12 = 18; A13 = 15; A21 = 1; A22 = −6; A23 = −7; A31 = −1; A32 = −3; A33 = −2. ∴

−6 1 −1    Adj ( A) =  18 −6 −3  15 −7 −2   |A| = a11 A11 + a12 A12 + a13 A13 = −9

∴

−1

A

−6 1 −1  1 1  adj A = −  18 −6 −3 = A 9   15 −7 −2

Using (2) in (1),  x −6 1 −1 −1      1  y  = −  18 −6 −3 −6     9   z      15 −7 −2 −3

(2)

Matrices

1.15

 1    3  −3      1 = −  27 =  −3    9   33 − 11  3  1 11 ∴ Solution of the system is x = − , y = − 3, z = − 3 3 Example 1.8 Test for the consistency of the following system of equations and solve them, if consistent: 3x + y + z = 8; − x + y − 2z = − 5; x + y + z = 6; – 2x + 2y − 3z = − 7.

Note

As the solution can be found out by any method, when the system is consistent, we may prefer the triangularisation method (also known as Gaussian elimination method) to reduce the augmented matrix [A, B] to an equivalent matrix. Using the equivalent matrix, we can test the consistency of the system and also find the solution easily when it exists. In this method, we use only elementary row operations and convert the elements below the principal diagonal of A as zeros.  3   −1 [ A, B ] =   1 −2 

1 1 8  1   1 −2 −5  −1 ∼ 1 1 6  3 2 −3 −7 −2

1 1 6  1 −2 −5 ( R1 ↔ R3 ) 1 1 8 2 −3 −7

1 1 1 6   0 2 −1 1  ∼  ( R2 → R2 + R1 , R3 → R3 − 3R1 , R4 → R4 + 2 R1 )  0 −2 −2 −10 0 4 −1 5  1  0 ∼  0 0 

1 1 6  2 −1 1 ( R3 → R3 + R2 , R4 → R4 − 2 R2 ) 0 −3 −9 0 1 3

1  0 ∼  0 0 

1 1 6   2 −1 1   R4 → R4 + 1 R3   0 −3 −9  3  0 0 0

(1)

Now, Determinant of [A, B] = − Determinant of the equivalent matrix = 0. (∴ Two rows interchanged in the first operation) ∴ R [ A, B ] ≤ 3

Mathematics II

1.16 1 1 1 Now 0 2 −1 = − 6 ≠ 0 0 0 −3

∴ R [A, B] = R (A) = 3 = the number of unknowns. ∴ The system is consistent and has a unique solution. A system of equations equivalent to the given system is also obtained from the equivalent matrix in (1). The equivalent equations are x + y + z = 6,

2y − z= 1

and

− 3z = − 9

Solving them backwards, we get x = 1, y = 2, z = 3. Example 1.9 Examine if the following system of equations is consistent and find the solution if it exists. x + y + z = 1, 2 x − 2 y + 3 z = 1; x − y + 2 z = 5; 3 x + y + z = 2. 1 1 1 1  1 1 1 1  ( R2 → R2 − 2 R1 ,     2 −2 3 1  0 −4 1 −1  R3 → R3 − R1 , ∼ [ A, B ] =    1 4  1 −1 2 5  0 −2 R4 → R4 − 3R1 ) 3 1 1 2  0 −2 −2 −1  1 1 1 1    0 −4 1 −1     1  1 1 9   R3 → R3 − R2 , R4 → R4 − R2  ∼ 0 0   2  2 2 2      5 1 0 −  0 −  2 2  1  1 1 1   0 −4 1 −1   ∼ 1 9  ( R4 → R4 + 5 R3 ) 0 0  2 2   0 0 0 22  It is obvious that det [A, B] = 4 and det [A] = 3 ∴ R [A, B] ≠ R [A]. ∴ The system is inconsistent.

Note

The last row of the equivalent matrix corresponds to the equation 0 ⋅ x + 0 ⋅ y + 0 ⋅ z = 22, which is absurd. From this also, we can conclude that the system is inconsistent.

Matrices

1.17

Example 1.10 Solve the following system of equations, if consistent: x + y + z = 3, x + y − z = 1; 3x + 3y − 5z = 1. 1  [ A, B ] =  1 3  1  ∼ 0  0 

1 1 3  1   1 −1 1 ∼  0   3 −5 1  0 1 1 3  0 −2 −2 ( R3  0 0 0

1 1 3  0 −2 −2 ( R2 → R2 − R1 , R3 → R3 − 3R1 )  0 −8 −8 → R3 − 4 R2 )

∴ All the third order determinants vanish ∴ R [A, B] ≠ 3 1 1 , which is a minor of both A and [A, B]. 0 −2 The value of this minor = − 2 ≠ 0 ∴ R (A) = R [A, B] < the number of unknowns. ∴ The system is consistent with many solutions. From the first two rows of the equivalent matrix, we have x + y + z = 3 and − 2z = − 2 i.e. z=1 and x + y = 2. Consider

∴ The system has a one parameter family of solutions, namely x = k, y = 2 − k, z = 1, where k is the parameter. Giving various values for k, we get infinitely many solutions. Example 1.11 Solve the following system of equations, if consistent: x1 + 2 x2 − x3 − 5 x4 = 4; x1 + 3 x2 − 2 x3 − 7 x4 = 5; 2 x1 − x2 + 3 x3 = 3. 1 2 −1 −5 4     [ A, B ] =  1 3 −2 −7 5  2 −1 3 0 3  1 2 −1 −5 4   ∼ 0 1 −1 −2 1 ( R2 → R2 − R1 , R3 → R3 − 2 R1 )    0 −5 5 10 −5   1 2 −1 −5 4    ∼  0 1 −1 −2 1 ( R3 → R3 + 5 R2 )   0 0  0 0 0   ∴ ∴ R [A, B] ≠ 3 ( the last row contains only zeros) Similarly R (A) ≠ 3. 1 2 ≠ 0 , R (A) = R [A, B] = 2 < the number of unknowns. 0 1 ∴ The given system is consistent with many solutions. Since

Mathematics II

1.18

From the first two rows of the equivalent matrix, we have x1 + 2 x2 − x3 − 5 x4 = 4 x2 − x3 − 2 x4 = 1

and

(1) (2)

As there are only 2 equations, we can solve for only 2 unknowns. Hence the other 2 unknowns are to be treated as parameters. Taking x3 = k and x4 = k′, we get x2 = 1 + k + 2k' [from (2)] and x1 = 4 − 2 (1+ k + 2k') + k + 5k' [from (1)] i.e. x1 = 2 − k + k' ∴ The given system possesses a two parameter family of solutions. Note From the Examples (10) and (11), we note that the number of parameters in the solution equals the difference between the number of unknowns and the common rank of A and [A, B]. Example 1.12 Find the values of k, for which the equations x + y + z = 1, x + 2y + 3z = k and x + 5y + 9z = k2 have a solution. For these values of k, find the solutions also. 1  [ A, B ] = 1 1  1  0  0 

1 1 1 1 1 1 1     ( R → R2 − R1 , 2 3 k  ∼ 0 1 2 k − 1  2  R →R −R)   3 1 5 9 k 2   0 4 8 k 2 − 1 3  1 1 1  1 2 k − 1  ( R3 → R3 − 4 R2 )  0 0 k 2 − 4k + 3

1 1 1   A ∼  0 1 2    0 0 0  

(1)

∴ R ( A) = 2

If the system possesses a solution, R [A, B] must also be 2. ∴ The last row of the matrix in (1) must contain only zeros. ∴ k2 − 4k + 3 = 0 i.e. k = 1 or 3. For these values of k, R (A) = R [A, B] = 2 < the number of unknowns. ∴ The given system has many solutions. Case (i) k=1 The first two rows of (1) give the equivalent equations as x+y+z=1 and y + 2z = 0

(2) (3)

Puting z = λ, the one-parameter family of solutions of the given system is x = λ + 1, y = − 2λ and z=λ Case (ii) k=3 The equivalent equations are x+y+z=1

(2)

Matrices and y + 2z = 2 Putting z = μ, the one-parameter family of solutions of the given system is

1.19 (4)

x = μ − 1, y = 2 − 2 μ, z = μ. Example 1.13 Find the condition satisfied by a, b, c, so that the following system of equations may have a solution: x + 2y − 3z = a; 3x − y + 2z = b; x − 5y + 8z = c. 1  [ A, B ] = 3 1  1  ∼ 0  0 

2 −3 a   −1 2 b  −5 8 c  a  2 −3  −7 11 b − 3a  ( R2 → R2 − 3R1 , R3 → R3 − R1 )  −7 11 c − a 

1  2 −3 a   ∼  0 −7 11 b − 3a  ( R3 → R3 − R2 )   0  − + 0 0 2 a b c   1 2 −3   A ∼  0 −7 11 ∴ R ( A) = 2   0 0 0 

(1)

If the given system possesses a solution, R [A, B] = 2. ∴ The last row of (1) should contain only zeros. ∴ 2a − b + c = 0. Only when this condition is satisfied by a, b, c, the system will have a solution. Example 1.14 Find the value of k such that the following system of equations has (i) a unique solution, (ii) many solutions and (iii) no solution. kx + y + z = 1; x + ky + z = 1; x + y + kz = 1.

∴

k 1  A = 1 k  1 1  A = k (k 2

1  1  k  − 1) + (1 − k ) + (1 − k )

= (k − 1) (k 2 + k − 2) 2

= (k − 1) (k + 2) |A| = 0, when k = 1 or k = –2 ∴ R(A) = 3 When k ≠ 1 and k ≠ −2, |A| ≠ 0 Then the system will have a unique solution.

Mathematics II

1.20

When k = 1, the system reduces to the single equation x + y + z = 1. In this case, R(A) = R[A, B] = 1. ∴ The system will have many solutions. (i.e. a two parameter family of solutions) When k = − 2, −2 1 1 1  1 −2 1 1     [ A, B ] =  1 −2 1 1 ∼ −2 1 1 1 ( R1 ↔ R2 )  1 1 −2 1  1 1 −2 1   1 −2 1 1    ∼ 0 −3 3 3 ( R2 → R2 + 2 R1 , R3 → R3 − R1 )   0 3 −3 0 

Now

 1 −2 1  ∼  0 −3 3  0 0 0  1 −2 0 −3 0

1  3 ( R3 → R3 + R2 )  3 1 3 = 0 ∴ R ( A) < 3 0 0

1 −2 ≠ 0 ∴ R ( A) = 2 0 −3 −2

1 1 −3 3 3 = a minor of [ A, B ] ≠ 0 0 0 3

∴ R[A, B] = 3. Thus R(A) ≠ R[A, B]. ∴ The system has no solution. Example 1.15 Investigate for what values of λ, μ, the equations x + y + z = 6, x + 2y + 3z = 10 and x + 2y + λz = μ have (i) no solution, (ii) a unique solution, (iii) an infinite number of solutions. 1 1 1 6 1 1 1 6      ( R → R2 − R1 , 4  2 [ A, B ] = 1 2 3 10 ∼ 0 1 2  1 2 λ µ  0 1 λ − 1 µ − 6 R3 → R3 − R1 )     1 1 1 6    ∼ 0 1 2 4  ( R3 → R3 − R2 )    0 0 λ − 3 µ − 10   1 1 1     ∴ A∼ 0 1 2  and A = λ − 3    0 0 λ − 3  

Matrices

1.21

∴ R(A) = 3 If λ ≠ 3, |A| ≠ 0 ∴ When λ ≠ 3 and μ takes any value, the system has a unique solution. If λ = 3, |A| = 0 and a second order minor of A, i.e.

1 1 ≠0 0 1

∴ R (A) = 2. 1 1 1 6    4  [ A, B ] ∼ 0 1 2   0 0 0 µ − 10  

When λ = 3,

(1)

When λ = 3 and μ = 10, the last row of (1) contains only zeros. ∴ R[A, B] ≠ 3 and clearly R[A, B] = 2. Thus, when λ = 3 and μ = 10, R(A) = R[A, B] = 2. ∴ The system has an infinite number of solutions. When λ = 3 and μ ≠ 10, a third order minor of [A, B], i.e. 1 1 1 2

6 4

= µ − 10 ≠ 0

0 0 µ − 10 ∴ R [A, B] = 3 Thus, when λ = 3 and μ ≠ 10, R(A) ≠ R[A, B]. ∴ The given system has no solution. Example 1.16 Test whether the following system of equations possess a non-trivial solution. x1 + x2 + 2x3 + 3x4 = 0; 3x1 + 4x2 + 7x3 + 10x4 = 0; 5x1 + 7x2 + 11x3 + 17x4 = 0;

6x1 + 8x2 + 13x3 + 16x4 = 0.

The given system is a homogeneous linear system of the form AX = 0. 1 2 3  1 1 2 3  ( R2 → R2 − 3R1 ,   4 7 10  0 1 1 1  R3 → R3 − 5 R1 , ∼ 7 11 17  0 2 1 2 R4 → R4 − 6 R1 ) 8 13 16  0 2 1 −2 1 1 2 3   0 1 1 1  ( R3 → R3 − 2 R2 , R4 → R4 − 2 R2 ) ∼  0  0 0 −1  0 0 −1 −4    1 1 2 3   0 1 1 1   ( R4 → R4 − R3 ) ∼  − 0 0 1 0   0 0 0 −4  ∴ |A| = 4 i.e. A is non-singular 1  3 A =  5 6 

Mathematics II

1.22

R(A) = R[A, 0] = 4 ∴ The system has a unique solution, namely, the trivial solution. Example 1.17 Find the non-trivial solution of the equations x + 2y + 3z = 0, 3x + 4y + 4z = 0, 7x + 10y + 11z = 0, if it exists. 1 2  A = 3 4   7 10  1 2  ∼  0 −2  0 0  ∴ |A| = 0

and

3  1 2 3    ( R → R2 − 3R1 , 4  ∼  0 −2 −5 2    R → R3 − 7 R1 ) 11  0 −4 −10 3 3  −5 ( R3 → R3 − 2 R2 )  0

1 2 ≠0 0 −2

(1)

∴ R(A) = 2

∴ The system has non-trivial solution. From the first two rows of (1), we see that the given equations are equivalent to

and

x + 2y + 3z = 0

(2)

− 2y − 5z = 0

(3)

5 Putting z = k, we get y = − k from (3) and x = 2k. 2 Thus the non-trivial solution is x = 4k, y = –5k and z = 2k. Example 1.18 Find the non-trivial solution of the equations x − y + 2z − 3w = 0, 3x + 2y – 4z + w = 0, 5x – 3y + 2z + 6w = 0, x – 9y + 14z − 2w = 0, if it exists.  1 −1 2 −3  1 −1 2 −3  ( R2 → R2 − 3R1 ,    3 2 −4 1  0 5 −10 10   R3 → R3 − 5 R1 ,   A=    ∼ 0 5 3 2 6 2 8 21 − −  R →R −R)    1 4 4  1 −9 14 −2 0 −8 12 1     1 −1 2 −3   0 1 −2 2     R2 → 1 R2  ∼   2 −8 21  5  0  0 −8 12 1   1 −1 2 −3   0 1 −2 2   ( R3 → R3 − 2 R2 , R4 → R4 + 8 R2 ) ∼  0 0 − 4 17   0 0 −4 17 

Matrices    ~    

1.23

1 −1 2 −3  0 1 −2 2 ( R4 → R4 − R3 ) 0 0 −4 17 0 0 0 0

∴ |A| = 0 i.e. R(A) < 4 ∴ The system has a non-trivial solution. The system is equivalent to x − y + 2z − 3w = 0

(1)

y − 2z +2w = 0

(2)

−4z + 17w = 0

(3)

Putting w = 4k, we get z = 17k from (3), y = 26k from (2) and x = 4k. Thus the non-trivial solution is x = 4k, y = 26k, z = 17k and w = 4k. Example 1.19 Find the values of λ for which the equations x +(λ + 4) y + (4λ + 2)z = 0, x + 2(λ + 1) y + (3λ + 4) z = 0, 2x + 3λy + (3λ + 4) z = 0 have a non-trivial solution. Also find the solution in each case. 1 λ + 4  A = 1 2λ + 2  2 3λ  1 λ + 4  ~ 0 λ − 2  0 λ − 8 

4λ + 2  3λ + 4   3λ + 4  4λ + 2   ( R → R2 − R1 , −λ + 2 2  R → R − 2R ) 3 1 −5λ  3

(1)

For non-trivial solution, |A| = 0 i.e.

−5λ (λ − 2) − (λ − 8) (2 − λ) = 0

i.e.

−4λ2 + 16 = 0

∴ λ=±2 When λ = 2, the system is equivalent to x + 6y + 10z = 0 −6y − 10z = 0,

from (1)

Putting z = 3k, we get y = −5k and x = 0 i.e. the solution is x = 0, y = −5k and z = 3k. When λ = −2, the system is equivalent to x + 2y − 6z = 0 −4y + 4z = 0, Putting z = k, we get y = k and x = 4k. i.e. the solution is x = 4k, y = k and z = k.

from (1)

Mathematics II

1.24

EXERCiSE 1(a) Part A (Short Answer Questions) 1. Define the linear dependence of a set of vectors. 2. Define the linear independence of a set of vectors. 3. If a set of vectors is linearly dependent, show that at least one member of the set can be expressed as a linear combination of the other members. 4. Show that the vectors X1 = (1, 2), X2 = (2, 3) and X3 = (4, 5) are linearly dependent. 5. Show that the vectors X1 = (0, 1, 2), X2 = (0, 3, 5) and X3 = (0, 2, 5) are linearly dependent. 6. Express X1 = (1, 2) as a linear combination of X2 = (2, 3) and X3 = (4, 5). 7. Show that the vectors (1, 1, 1), (1, 2, 3) and (2, 3, 8) are linearly independent. 8. Find the value of a if the vectors (2, −1, 0), (4, 1, 1) and (a, −1, 1) are linearly dependent. 9. What do you mean by consistent and inconsistent systems of equations. Give examples. 10. State Rouche’s theorem. 11. State the condition for a system of equations in n unknowns to have (i) one solution, (ii) many solutions and (iii) no solution. 12. Give an example of 2 equations in 2 unknowns that are (i) consistent with only one solution and (ii) inconsistent. 13. Give an example of 2 equations in 2 unknowns that are consistent with many solutions. 14. Find the values of a and b, if the equations 2x − 3y = 5 and ax + by = −10 have many solutions. 15. Test if the equations x + y + z = a, 2x + y + 3z = b, 5x + 2y + z = c have a unique solution, where a, b, c are not all zero. 16. Find the value of λ, if the equations x + y − z = 10, x − y + 2z = 20 and λx − y + 4z = 30 have a unique solution. 17. If the augmented matrix of a system of equations is equivalent to 1 2 1 2    0 −5 −3 −2 , find the value of λ, for which the system has a unique   0  0 0 λ   solution. 18. If the augmented matrix of a system of equations is equivalent to  1 −2 1 3    0 2 2 −2  , find the values of λ and μ for which the system has   0 0 λ + 1 µ − 3  only one solution.

Matrices

1.25

19. If the augmented matrix of a system of equations is equivalent to  1 −2 3 4    0 5 4 2  , find the values of λ and μ for which the system   0  0 − 2 − 3 λ µ   has many solutions. 20. If the augmented matrix of a system of equations is equivalent to 1 1 2 3     0 −3 −1 −2  , find the values of λ and μ for which the system   0  0 − 8 − 11 λ µ   has no solution. 21. Do the equations x − 3y − 8z = 0, 3x + y = 0 and 2x + 5y + 6z = 0 have a nontrivial solution? Why? 22. If the equations x + 2y + z = 0, 5x + y –z = 0 and x + 5y + λz = 0 have a nontrivial solution, find the value of λ. 23. Given that the equations x + 2y − z = 0, 3x + y − z = 0 and 2x – y = 0 have non-trivial solution, find it. Part B Show that the following sets of vectors are linearly dependent. Find their relationship in each case: 24. 25. 26. 27. 28. 29.

30. 31. 32. 33. 34.

X1 = (1, 2, 1), X2 = (4, 1, 2), X3 = (6, 5, 4), X4 = (−3, 8, 1). X1 = (3, 1, −4), X2 = (2, 2, −3), X3 = (0, −4, 1), X4 = (−4, −4, 6) X1 = (1, 2, −1, 3), X2 = (0, −2, 1, −1), X3 = (2, 2, −1, 5) X1 = (1, 0, 4, 3), X2 = (2, 1, −1, 1), X3 = (3, 2, −6, −1) X1 = (1, −2, 4, 1), X2 = (1, 0, 6, −5), X3 = (2, −3, 9, −1) and X4 = (2, −5, 7, 5). Determine whether the vector x5 = (4, 2, 1, 0) is a linear combination of the set of vectors X1 = (6, − 1, 2, 1), X2 = (1, 7, − 3, −2), X3 = (3, 1, 0, 0) and X4 = (3, 3,−2,−1). Show that each of the following sets of vectors is linearly independent. X1 = (1, 1, 1); X2 = (1, 2, 3); X3 = (2, −1, 1). X1 = (1, −1, 2, 3); X2 = (1, 0, −1, 2); X3 = (1, 1, −4, 0) X1 = (1, 2, −1, 0) X2 = (1, 3, 1, 2); X3 = (4, 2, 1, 0); X4 = (6, 1, 0, 1). X1 = (1, −2, −3, −2, 1); X2 = (3, −2, 0, −1, −7); X3 = (0, 1, 2, 1, −6); X4 = (0, 2, 2, 1, −5). Test for the consistency of the following system of equations: 7 3 4 5 6   x     4 5 6 7   1  8        5 6 7 8   x2  =  9       10 11 12 13  x3  14   x       4    15 16 17 18 19  

1.26

Mathematics II

Test for the consistency of the following systems of equations and solve, if consistent: 35. 2x − 5y + 2z = −3; −x − 3y + 3z = −1; x + y − z = 0; −x + y = 1. 36. 3x + 5y − 2z = 1; x − y + 4z = 7; −6x − 2y + 5z = 9; 7x − 3y + z = 4. 37. 2x + 2y + 4z = 6; 3x + 3y + 7z = 10; 5x + 7y + 11z = 17; 6x + 8y + 13z = 16. Test for the consistency of the following systems of equations and solve, if consistent: 38. x + 2y + z = 3; 2x + 3y + 2z = 5; 3x − 5y + 5z = 2; 3x + 9y − z = 4. 39. 2x + 6y − 3z = 18; 3x − 4y + 7z = 31; 5x + 3y + 3z = 48; 8x − 3y + 2z = 21. 40. x + 2y + 3z = 6; 5x − 3y + 2z = 4; 2x + 4y − z = 5; 3x + 2y + 4z = 9. 41. x + 2y = 4; 10y + 3z = −2; 2x − 3y − z = 5; 3x + 3y + 2z = 1. 42. 2x1 + x2 + 2x3 + x4 = 6; x1 − x2 + x3 + 2x4 = 6; 4x1 + 3x2 + 3x3 − 3x4 = −1; 2x1 + 2x2 − x3 + x4= 10 43. 2x + y + 5z + w = 5; x + y + 3z − 4w = − 1; 3x + 6y − 2z + w = 8; 2x + 2y + 2z − 3w = 2. Test for the consistency of the following systems of equations and solve, if consistent: 44. x − 3y − 8z = − 10; 3x + y = 4z; 2x + 5y + 6z = 13. 45. 5x + 3y + 7z = 4; 3x + 26y + 2z = 9; 7x + 2y + 10z = 5. 46. x − 4y − 3z + 16 = 0; 2x + 7y + 12z = 48; 4x − y + 6z = 16; 5x − 5y + 3z = 0. 47. x − 2y + 3w = 1; 2x − 3y + 2z + 5w = 3; 3x − 7y − 2z + 10w = 2. 48. x1 + 2x2 + 2x3 − x4 = 3; x1 + 2x2 + 3x3 + x4 = 1; 3x1 + 6x2 + 8x3 + x4 = 5. 49. Find the values of k, for which the equations x + y + z = 1, x + 2y + 4z = k and x + 4y + 10z = k2 have a solution. For these values of k, find the solutions also. 50. Find the values of λ, for which the equtions x + 2y + z = 4, 2x − y − z = 3λ and 4x − 7y − 5z = λ2 have a solution. For these values of λ, find the solutions also. 51. Find the condition on a, b, c, so that the equations x + y + z = a, x + 2y + 3z = b, 3x + 5y + 7z = c may have a one-parameter family of solutions. 52. Find the value of k for which the equations kx − 2y + z = 1, x − 2ky + z = −2 and x − 2y + kz = 1 have (i) no solution, (ii) one solution and (iii) many solutions. 53. Investigate for what values of λ, μ the equations x + y + 2z = 2, 2x − y + 3z = 2 and 5x − y + λz = μ have (i) no solution, (ii) a unique solution, (iii) an infinite number of solutions. 54. Find the values of a and b for which the equations x + y + 2z = 3, 2x − y + 3z = 4 and 5x − y + az = b have (i) no solution, (ii) a unique solution, (iii) many solutions. 55. Find the non-trivial solution of the equations x + 2y + z = 0; 5x + y − z = 0 and x + 5y + 3z = 0, if it exists. 56. Find the non-trivial solution of the equations x + 2y + z + 2w = 0; x + 3y + 2z + 2w = 0; 2x + 4y + 3z + 6w = 0 and 3x + 7y + 4z + 6w = 0, if it exists. 57. Find the values of λ for which the equations 3x + y − λz = 0, 4x − 2y − 3z = 0 and 2λx + 4y + λz = 0 possess a non-trivial solution. For these values of λ, find the solution also. 58. Find the values of λ for which the equations (11 − λ) x − 4y − 7z = 0, 7x − (λ + 2) y − 5z = 0, 10x − 4y − (6 + λ) z = 0 possess a non-trivial solution. For these values of λ, find the solution also.

Matrices

1.6

1.27

EiGEnVALUES AnD EiGEnVECtORS

1.6.1  Definition Let A = [aij] be a square matrix of order n. If there exists a non-zero column vector X and a scalar λ, such that AX = λX then λ is called an eigenvalue of the matrix A and X is called the eigenvector corresponding to the eigenvalue λ. To find the eigenvalues and the corresponding eigenvectors of a square matrix A, we proceed as follows: Let λ be an eigenvalue of A and X be the corresponding eigenvector. Then, by definition, AX = λX = λIX, where I is the unit matrix of order n. i.e.

i.e.

i.e.

(A − λI) X = 0  a    11   a21     a    n1

a12 a22 an 2

(1)

    1 0 … … 0   x1   0 … a1n      x   0 … a2 n  − λ  0 1 … … 0   2  =             … ann   0 0 0 … 1        xn   0

(a11 − λ ) x1 + a12 x2 + ⋅⋅⋅ + a1n xn = 0 a21 x1 + (a22 − λ ) x2 + ⋅⋅⋅ + a2 n xn = 0

(2)

..................................................... an1 x1 + an 2 x2 + ⋅⋅⋅ (ann − λ ) xn = 0 Equations (2) are a system of homogeneous linear equations in the unknowns x1, x2, . . . , xn.  x1    x  Since X =  2  is to be a non-zero vector,  x   n  x1, x2, . . . , xn should not be all zeros. In other words, the solution of the system (2) should be a non-trivial solution. The condition for the system (2) to have a non-trivial solution is

i.e.

a11 − λ a12 a1n ⋅⋅⋅ a21 a22 − λ ⋅⋅⋅ a2 n = 0 an1 an 2 ⋅⋅⋅ ann − λ

(3)

|A − λ I| = 0

(4)

The determinant |A − λI| is a polynomial of degree n in λ and is called the characteristic polynomial of A. The equation |A − λ I| = 0 or the equation (3) is called the characteristic equation of A.

Mathematics II

1.28

When we solve the characteristic equation, we get n values for λ. These n roots of the characteristic equation are called the characteristic roots or latent roots or eigenvalues of A. Corresponding to each value of λ, the equations (2) possess a non-zero (nontrivial) solution X. X is called the invariant vector or latent vector or eigenvector of A corresponding to the eigenvalue λ.

Notes 1. Corresponding to an eigenvalue, the non-trivial solution of the system (2) will be a one-parameter family of solutions. Hence the eigenvector corresponding to an eigenvalue is not unique. 2. If all the eigenvalues λ1, λ2, . . . , λn of a matrix A are distinct, then the corresponding eigenvectors are linearly independent. 3. If two or more eigenvalues are equal, then the eigenvectors may be linearly independent or linearly dependent.

1.6.2

Properties of Eigenvalues

1. A square matrix A and its transpose AT have the same eigenvalues. Let A = (aij); i, j = 1, 2, . . . , n. The characteristic polynomial of A is a11 − λ a12 ⋅⋅⋅ a1n A − λI = a21 a22 − λ ⋅⋅⋅ a2 n an1 an 2 ⋅⋅⋅ ann − λ

(1)

The characteristic polynomial of AT is … a11 − λ a21 an1 A − λI = a12 a22 − λ … an 2 a1n a2 n … ann − λ T

(2)

Determinant (2) can be obtained by changing rows into columns of determinant (1). ∴ |A − λ I| = |AT − λ I| ∴ The characteristic equations of A and AT are identical. ∴ The eigenvalues of A and AT are the same. 2. The sum of the eigenvalues of a matrix A is equal to the sum of the principal diagonal elements of A. (The sum of the principal diagonal elements is called the Trace of the matrix.) The characteristic equation of an nth order matrix A may be written as n

λ n − D1λ n−1 + D2 λ n−2 − ⋅⋅⋅ + (−1) Dn = 0,

(1)

where Dr is the sum of all the rth order minors of A whose principal diagonals lie along the principal diagonal of A.

Matrices

1.29

(Note

Dn = |A|). We shall verify the above result for a third order matrix.  a11 a12 a13    Let A =  a21 a22 a23  a   31 a32 a33  The characteristic equation of A is given by  a11 − λ a12 a13     a21 a22 − λ a23  = 0 (2)    a  a32 a33 − λ   31 Expanding (2), the characteristic equation is   a a  (a11 − λ)λ 2 − (a22 + a33 ) λ + 22 23   a32 a33         a21 a22  a21 a23   − a12  −a21λ +  + a13  a31λ + = 0   a31 a33  a31 a32        i.e.

−λ 3 + (a11 + a22 + a33 ) λ 2

  a11 a12 a a13 a a23  − + 11 + 22  λ+ A =0  a31 a33 a32 a33    a21 a22 i.e. λ3 − D1 λ2 + D2 λ − D3 = 0, using the notation given above. This result holds good for a matrix of order n.

Note

This form of the characteristic equation provides an alternative method for getting the characteristic equation of a matrix. Let λ1, λ2, . . . , λn be the eigenvalues of A. ∴ They are the roots of equation (1). ∴

λ1 + λ 2 + ⋅⋅⋅ + λ n =

− (− D1 )

= D1 1 = a11 + a22 + ⋅⋅⋅ + ann = Trace of the matrix A.

3. The product of the eigenvalues of a matrix A is equal to |A|. If λ1, λ2, . . . , λn are the eigenvalues of A, they are the roots of n

λ n − D1λ n−1 + D2 λ n−2 − ⋅⋅⋅ + (−1) Dn = 0 n

∴ Product of the roots

=

n

(−1) ⋅ (−1) Dn

i.e. λ1, λ2 . . . λn = Dn = |A|.

1.6.3 Aliter λ1, λ2, . . . , λn are the roots of |A − λI| = 0

1

.

Mathematics II

1.30

∴ |A − λI| ≡ (−1)n (λ − λ1) (λ − λ2) . . . (λ − λn), since L.S. is a nth degree polynomial in λ whose leading term is (−1)n λn. Putting λ = 0 in the above identity, we get |A| = (−1)n (−λ1) (−λ2) . . . (−λn) i.e. λ1 λ2 . . . λn = |A|.

1.6.4

Corollary

If |A| = 0, i.e. A is a singular matrix, at least one of the eigenvalues of A is zero and conversely. 4. If λ1, λ2, . . ., λn are the eigenvalues of a matrix A, then (i) kλ1, kλ2, . . . kλn are the eigenvalues of the matrix kA, where k is a nonzero scalar. (ii) λ1p , λ 2p , … , λ np are the eigenvalues of the matrix Ap, where p is a positive integer. (iii) 1 , 1 ,⋅⋅⋅ 1 are the eigenvalues of the inverse matrix A−1, provided λr λ1 λ 2 λn ≠ 0 i.e. A is non-singular. (i) Let λr be an eigenvalue of A and Xr the corresponding eigenvector. Then, by definition, AXr = λr xr (1) Multiplying both sides of (1) by k, (kA)Xr = (kλr) Xr (2) From (2), we see that kλr is an eigenvalue of kA and the corresponding eigenvector is the same as that of λr, namely Xr. (ii) Premultiplying both sides of (1) by A, A2 X r = A( AX r ) = A (λ r X r ) = λ r ( AX r ) = λ 2r X r 3 3 Similarly A X r = λ r X r and so on.

In general, A p X r = λ rp X r

(3)

p From (3), we see that λ r is an eigenvalue of Ap with the corresponding eigenvector equal to Xr, which is the same for λr. (iii) Premultiplying both sides of (1) by A−1,

A−1 (AXr) = A−1 (λr Xr) i.e. ∴

Xr = λr (A−1 Xr) A−1 X r =

1 Xr λr

(4)

Matrices

1.31

1 is an eigenvalue of A–1 with the corresponding λr eigenvector equal to Xr, which is the same for λr. 5. The eigenvalues of a real symmetric matrix (i.e. a symmetric matrix with real elements) are real. Let λ be an eigenvalue of the real symmetric matrix and X be the corresponding eigenvector. Then AX = λ (1) From (4), we see that

Premultiplying both sides of (1) by X T (the transpose of the conjugate of X), we get X T AX = λ X T X (2) Taking the complex conjugate on both sides of (2), X T A X = λ X T X (assuming that λ may be complex ) X T A X = λ X T X (∵ A = A, as A is real) i.e. Taking transpose on both sides of (3), X T AT X = λ X T X i.e.

XT AX =λ XT X

∵( AB)T = BT AT    T ∵( A) = A, as A issymmetric  

(3)

(4)

From (2) and (4), we get λ XT X =λ XT X i.e.

(λ − λ ) X T X = 0

X T X is an 1 × 1 matrix, i.e. a single element which is positive λ −λ = 0 i.e. λ is real. Hence all the eigenvalues are real. 6. The eigenvectors corresponding to distinct eigenvalues of a real symmetric matrix are orthogonal.  y1   x1       y2   x2   Note Two column vectors X =   and Y =   are said to be             yn   xn  … orthogonal, if their inner product (x1y1 + x2y2 + xnyn) = 0 i.e. if XTY = 0. Let λ1, λ2 be any two distinct eigenvalues of the real symmetric matrix A and X1, X2 be the corresponding eigenvectors respectively. Then AX1 = λ1X1 (1) ∴

and

AX2 = λ2X2

Premultiplying both sides of (1) by, X 2T we get X 2T AX 1 = λ1 X 2T X 1

(2)

Mathematics II

1.32

Taking the transpose on both sides, X 1T AX 2 = λ1 X 1T X 2

(∵ AT = A)

(3)

Premultiplying both sides of (2) by X1T , we get X 1T AX 2 = λ 2 X 1T X 2

(4)

From (3) and (4), we have

i.e.

λ1 X 1T X 2 = λ 2 X 1T X 2 (λ1 − λ 2 ) X 1T X 2 = 0 λ1 ≠ λ 2 , X 1T X 2 = 0

Since

i.e. the eigenvectors X1 and X2 are orthogonal.

WORKED EXAMPLE 1(b) 5 4 Example 1.1 Given that A =   , verify that the eigenvalues of A2 are the  1 2  squares of those of A. Verify also that the respective eigenvectors are the same. The characteristic equation of A is

5− λ 4 =0 1 2−λ

i.e.

(5 − λ) (2 − λ) − 4 = 0

i.e.

λ2 − 7λ + 6 = 0

∴ The eigenvalues of A are λ = 1, 6. The eigenvector corresponding to any λ is given by (A − λ I) X = 0 i.e.

5− λ 4 1 2−λ

 x1   =0  x2   

When λ = 1, the eigenvector is given by the equations 4x1 + 4x2 = 0 and x1 + x2 = 0, which are one and the same. Solving, x1 = − x2. Taking x1 = 1, x2 = −1. ∴ The eigenvector is  1  −1

Matrices

1.33

When λ = 6, the eigenvector is given by −x1 + 4x2 = 0 and x1 − 4x2 = 0 Solving, x1 = 4x2 Taking x2 = 1, x1 = 4 1 ∴ The eigenvector is    4   5 4   5 4   29 28   =  A2 =  Now  1 2   1 2   7 8  29 − λ 28 The characteristic equation of A2 is =0 7 8− λ i.e. (29 − λ) (8 − λ) − 196 = 0 i.e. λ2 − 37λ + 36 = 0 i.e. (λ − 1)(λ − 36) = 0 ∴ The eigenvalues of A2 are 1 and 36, that are the squares of the eigenvalues of A, namely 1 and 6. When λ = 1, the eigenvector of A2 is given by  28 28   x1      = 0. i.e. 28x1 + 28 x2 = 0 7 x1 + 7 x2 = 0 and  7 7   x2  Solving, x1 = –x2. Taking x1 = 1, x2 = –1. When λ = 36, the eigenvector of A2 is given by  − 7 28   x1      = 0. i.e. − 7 x1 + 28 x2 = 0  7 − 28   x2 

and

7 x1 − 28 x2 = 0.

Solving, x1 = 4x2. Taking x2 = 1, x1 = 4. Thus the eigenvectors of A2 are  1   4   and   , which are the same as the respective eigenvectors of A.  −1  1  Example 1.2 Find the eigenvalues and eigenvectors of the matrix  1 1 3   A = 1 5 1    3 1 1  The characteristic equation of A is 1− λ 1 1 5− λ

3 1 =0

3 1 1− λ i.e. (l – λ) {λ – 6λ + 4} – (1 – λ – 3) + 3(1 – 15 + 3λ) = 0 i.e. –λ3 + 7λ2 – 36 = 0 or λ3 – 7λ2 + 36 = 0 i.e. (λ + 2) (λ2 – 9λ + 18) = 0 i.e. (λ + 2) (λ – 3) (λ – 6) = 0 ∴ The eigenvalues of A are λ = –2, 3, 6. Case (i) λ = –2. The eigenvector is given by 2

(1) [∴ λ= –2 satisfies (1) ]

Mathematics II

1.34

 3 1 3   x1      1 7 1   x2  = 0     3 1 3   x3  i.e. x1 + 7x2 + x3 = 0 3x1 + x2 + 3x3 = 0 Solving these equations by the rule of cross-multiplication, we have

(2)

x1 x x = 2 = 3 21−1 3 − 3 1− 21 x1 x2 x = = 3 20 0 − 20

i.e.

(3)

Note To solve for x1, x2, x3, we have taken the equations corresponding to the second and third rows of the matrix in step (2). The proportional values of x1, x2, x3 obtained in step (3) are the co-factors of the elements of the first row of the determinant of the matrix in step (2). This provides an alternative method for finding the eigenvector. From step (3), x1 = k, x2 = 0 and x3 = –k. Usually the eigenvector is expressed in terms of the simplest possible numbers, corresponding to k = 1 or – 1. x1 = 1,

∴

x2 = 0,

x3 = – 1

Thus the eigenvector corresponding to λ = – 2 is  1   X1 =  0     −1 Case (ii)

λ = 3.

 −2 1 3   x1      The eigenvector is given by 1 2 1  x2  = 0.     3 1 −2   x3  Values of x1, x2, x3 are proportional to the co-factors of –2, 1, 3 (elements of the first row i.e. –5, 5, –5. i.e.

x1 x x = 2 = 3 or −5 5 −5

∴

 1   X 2 =  −1     1

Case (iii)

x1 x2 x3 = = 1 −1 1

λ = 6.

 −5 1 3   x1     The eigenvector is given by  1 −1 1  x2  = 0     3 1 −5   x3 

Matrices

∴ or ∴

1.35

x1 x2 x3 = = 4 8 4 x1 x2 x3 = = 1 2 1 1   X3 =  2    1 

Note

Since the eigenvalues of A are distinct, the eigenvectors X1, X2, X3 are linearly independent, as can be seen from the fact that the equation k1X1 + k2X2 + k3X3 = 0 is satisfied only when k1 = k2 = k3 = 0. Example 1.3 Find the eigenvalues and eigenvectors of the matrix 0 1 1   A = 1 0 1    1 1 0  The characteristic equation is given by λ3 – D1λ2 + D2λ – D3 = 0, where D1 = the sum of the first order minors of A that lie along the main diagonal of A =0+0+0 =0 D2 = the sum of the second order minors of A whose principal diagonals lie along the principal diagonal of A. 0 1 0 1 0 1 = + + 1 0 1 0 1 0 =–3 D3= |A| = 2 Thus the characteristic equation of A is λ3 – 3λ – 2 = 0 i.e.

(λ + l)2 (λ – 2) = 0

∴ The eigenvalues of A are λ = –1, –1, 2. Case (i) λ = –1. The eigenvector is given by − λ 1 1   x1      1 −λ 1   x2  = 0     1 1 − λ   x3   All the three equations reduce to one and the same equation x1 + x2 + x3 = 0. There is one equation in three unknowns. ∴ Two of the unknowns, say, x1 and x2 are to be treated as free variables (parameters). Taking x1 = 1 and x2 = 0, we get x3 = –1 and taking x1 = 0 and x2= 1, we get x3= – l

Mathematics II

1.36

 1   X 1 =  0 and −1  

∴

 0   X 2 =  1 −1  

Case (ii) λ = 2. The eigenvector is given by −2 1 1    1 −2 1    1 1 −2 

 x1     x2  = 0   x   3

Values of x1, x2, x3 are proportional to the co-factors of elements in the first row. x x1 x2 = = 3 3 3 3 x3 x1 x2 or = = 1 1 1 1  ∴ x3 = 1  1  Note Though two of the eigenvalues are equal, the eigenvectors X1, X2, X3 are found to be linearly independent. i.e.

Example 1.4 Find the eigenvalues and eigenvectors of the matrix  2 −2 2   A = 1 1 1   1  − 3 1   The characteristic equation of A is 2 − λ −2 1 1− λ i.e. i.e.

2 1

=0

1 3 −1− λ (2 – λ)(λ – 4) + 2(– l – λ – l) + 2(3 – l + λ) = 0 (2 – λ)(λ – 2)(λ + 2) = 0 2

∴ The eigenvalues of A are λ = –2, 2, 2. Case (i) λ=–2 The eigenvector is given by  4 −2 2  x1     1 3 1  x2  = 0    1 3 1  x3   ∴

x x1 x = 2 = 3 (by taking the co-factors of elements of the third row) −8 −2 14

Matrices

1.37

x x1 x = 2 = 3 −4 −1 7

i.e.

−4   X 1 =  −1    7  

∴ Case (ii) λ = 2. The eigenvector is given by

0 −2 2  x1    1 −1 1  x2  = 0   1 3 −3  x3  ∴

x1 x2 x3 = = 0 4 4

∴

 0   X 2 = X 3 =  1    1  

Note

or

x1 x2 x3 = = 0 1 1

Two eigenvalues are equal and the eigenvectors are linearly dependent.

Example 1.5 Find the eigenvalues and eigenvectors of the matrix 11 −4 −7   A =  7 −2 −5   10 −4 −6   Can you guess the nature of A from the eigenvalues? Verify your answer. The characteristic equation of A is 11 − λ −4 7 −2−λ 10 i.e.

−4

−7 −5 −6−λ

(11 – λ)( λ2 + 8λ – 8) + 4(8 – 7λ)–7(10λ – 8) = 0

i.e. λ3 – 3λ2 + 2λ = 0 ∴ The eigenvalues of A are λ = 0, 1, 2. Case (i) λ = 0. 11 −4 −7   The eigenvector is given by  7 −2 −5   10 −4 −6   ∴

=0

x x1 x = 2 = 3 −8 −8 −8

 x1     x2  = 0   x   3

Mathematics II

1.38 or

x x1 x2 = = 3 1 1 1

∴

1  X 1 = 1  1  λ = 1.

Case (ii)

10 −4 −7   The eigenvector is given by  7 −3 −5   10 −4 −7  

 x1     x2  = 0   x   3

x x1 x = 2= 3 −1 1 −2

∴

 1   X 2 = −1    2  

∴ Case (iii)

λ = 2.  9 −4 −7     7 −4 −5   10 −4 −8   x x1 x = 2= 3 12 6 12

The eigenvector is given by

∴

 x1     x2  = 0   x   3

x x1 x2 = = 3 2 1 2  2   ∴ X 3 =  1    2   Since one of the eigenvalues of A is zero, product of the eigenvalues = |A| = 0, i.e. A is non-singular. It is verified below: or

11 −4 −7 7 −2 −5 = 11(12 − 20) + 4 (− 42 + 50) − 7 (− 28 + 20) = 0. 10 −4 −6 Example 1.6 Verify that the sum of the eigenvalues of A equals the trace of A and that their product equals |A|, for the matrix 1 0 0   A = 0 3 −1    0 −1 3 

Matrices

1.39

The characteristic equation of A is 1− λ 0 0 0 3 − λ −1 = 0 0 −1 3 − λ (l – λ)(λ2 – 6λ + 8) = 0

i.e.

∴ The eigenvalues of A are λ = 1, 2, 4. Sum of the eigenvalues = 7. Trace of the matrix = 1 + 3 + 3 = 7 Product of the eigenvalues = 8. |A| = 1 × (9 – l) = 8. Hence the properties verified. Example 1.7 Verify that the eigenvalues of A2 and A–1 are respectively the squares and reciprocals of the eigenvalues of A, given that  3 1 4   A =  0 2 6    0 0 5   The characteristic equation of A is 3−λ 1 0 2−λ 0 i.e.

0

4 6

=0

5−λ

(3 – λ) (2 – λ) (5 – λ) = 0

∴ The eigenvalues of A are λ = 3, 2, 5.  3 1 4  3 1 4 9 5 38      A =  0 2 6  0 2 6 =  0 4 42 Now       0 0 5  0 0 5  0 0 25      The characteristic equation of A2 is 2

9−λ 5 0 4−λ 0 i.e.

0

38 42

=0

25 − λ

(9 – λ)(4 – λ)(25 – λ) = 0

∴ The eigenvalues of A2 are 9, 4, 25, which are the squares of the eigenvalues of A.  3 1 4  a11    A =  0 2 6 ≡  a21     0 0 5  a    31 A11 = Co-factor of a11 = 10; A12 = 0; A13 = 0; Let

a12 a22 a32

a13   a23   a33 

Mathematics II

1.40

A21 = – 5; A22 = 15; A23 = 0; A31 = – 2; A32 = – 18; A33 = 6 |A| = 30. 10 − 5 − 2    0 15 − 18 ∴ A    0  0 6   1 1 1  − −  3 6 15     1 3 = 0 −   2 5   1   0 0  5   The characteristic equation of A–1 is 1 1 1 −λ − − 3 6 15 1 3 0 −λ − =0 2 5 1 0 0 −λ 5   1   − λ  1 − λ  1 − λ = 0 i.e.    5   2  3 −1

1 = 30

1 1 1 , , , which are the reciprocals of the eigenvalues of A. 3 2 5 Hence the properties verified. ∴ The eigenvalues of A–1 are

Example 1.8 Find the eigenvalues matrix  2  A=  0  − 1  The characteristic equation of A is

and eigenvectors of (adj A), given that the 0 − 1  2 0  0 2

2−λ 0 0 2−λ −1 i.e.

0

−1 0 =0 2−λ

(2 – λ)3 – (2 – λ) = 0

i.e. (2 – λ) (λ2 – 4λ + 3) = 0 ∴ The eigenvalues of A are λ = 1, 2, 3. Case (i) λ = 1.  1 0 − 1  x1     0  x2  = 0 The eigenvector is given by  0 1    − 1 0  x  1    3

Matrices x x1 x = 2 = 3 1 0 1  1   X1 =  0  1  

∴

∴ Case (ii)

1.41

λ = 2.

 0 0 −1  x1     The eigenvector is given by  0 0 0  x2  = 0 −1 0 0  x3   i.e. –x3 = 0 and –x1 = 0 ∴ x1 = 0, x3 = 0 and x2 is arbitrary. Let x2 = 1  0   X 2 = 1  0  

∴ Case (iii)

λ = 3.

−1 0 −1  x1      0  x2  = 0. The eigenvector is given by  0 −1 −1 0 −1  x3   x x1 x = 2 = 3 −1 1 0  1   ∴ X 3 =  0 −1   1 1 The eigenvalues of A–1 are 1, , with the eigenvectors X1, X2, X3. 2 3 ∴

Now

adj A = A−1 A

i.e. adj A = |A| · A–1 = 6A–1 (∵ A = 6 for the given matrix A) ∴ The eigenvalues of (adj A) are equal to 6 times those of A–1, namely, 6, 3, 2. The corresponding eigenvectors are X1, X2, X3 respectively. Example 1.9 Verify that the eigenvectors of the real symmetric matrix  3 −1 1   A = −1 5 −1  1 −1 3 

are orthogonal in pairs. The characteristic equation of A is

Mathematics II

1.42

3 − λ −1 1 −1 5 − λ −1 = 0 1 −1 3 − λ i.e.

(3 – λ)(λ2 – 8λ + 14) + (λ – 3 + 1) + (1 + λ – 5) = 0

i.e.

λ3 – 11λ2 + 36λ – 36 = 0

i.e.

(λ – 2) (λ – 3) (λ – 6) = 0

∴ The eigenvalues of A are λ = 2, 3, 6. Case (i) λ = 2.  1 −1 1  x1     The eigenvector is given by −1 3 −1  x2  = 0   1 −1 1  x3   x x1 x = 2 = 3 −2 2 0  1   X1 =  0 −1  

∴

∴

Case (ii)

or

x x1 x = 2 = 3 −1 1 0

λ = 3.

 0 −1 1  x1     The eigenvector is given by −1 2 −1  x2  = 0   1 −1 0  x3   x x1 x = 2 = 3 − 1 − 1 −1 1  X 2 = 1 1 

∴

∴ Case (iii)

λ = 6.

−3 −1 1    The eigenvector is given by  −1 −1 −1  1 −1 −3  

 x1     x2  = 0   x   3

x x1 x = 2 = 3 2 −4 2  1   X 3 = −2  1  

Matrices Now X1T

X 2T

X 3T

1.43

1  X 2 = [1 0 −1] 1 = 0 1   1   X 3 = [1 1 1] −2 = 0  1    1   X 1 = [1 −2 1]  0 = 0 −1  

Hence the eigenvectors are orthogonal in pairs. Example 1.10 Verify that the matrix  2 2 1   −2  1 2    1 −2 2   1 is an orthogonal matrix. Also verify that is an eigenvalue of A, if λ is an eigenvalue λ and that the eigenvalues of A are of unit modulus. 1 A= 3

Note

A square matrix A is said to be orthogonal if AAT = ATA = I.

Now

 2  2 −2 1 2 1   1  −2 1 −2 1 2 ×  2   1 −2 2  3  1 2 2     9 0 0  1 0 0    1  =  0 9 0 =  0 1 0 = I 9     0 0 9  0 0 1

1 AA = 3 T

Similarly we can prove that ATA = I. Hence A is an orthogonal matrix. The characteristic equation of 3A is 2−λ 2 1 −2 1 − λ 2 =0 1 −2 2 − λ i.e.

(2 – λ) (λ2 – 3λ + 6) – 2(2λ – 4 – 2) + (4 – 1 + λ) = 0

i.e.

λ3 – 5λ2 + 15λ – 27 = 0

i.e.

(λ – 3) (λ2 – 2λ + 9) = 0

∴ The eigenvalues of 3A are given by

Mathematics II

1.44

λ=3

λ=

and

2 ± 4 − 36 =1±i 2 2 2

∴ The eigenvalues of A are λ1 = 1, λ 2 =

Now

1+ i 2 2 1− i 2 2 , λ3 = 3 3

1 = 1 = λ1 λ1 1 3 3 (1− i 2 2 ) 1− i 2 2 = = = = λ3 3 λ 2 1 + i 2 2 (1+ i 2 2 ) (1 − i 2 2 )

and similarly

1 = λ2 . λ3

Thus, if λ is an eigenvalue of an orthogonal matrix,

1 is also an eigenvalue. λ

Also |λ1| = |1| =1. λ2 =

1 i2 2 1 8 + = + =1 3 3 9 9

Similarly, |λ3| = 1. Thus the eigenvalues of an orthogonal matrix are of unit modulus.

EXERCiSE 1(b) Part A (Short Answer Questions) 1. Define eigenvalues and eigenvectors of a matrix. 2. Prove that A and AT have the same eigenvalues.  4 1 . 3. Find the eigenvalues of 2A2, if A =    3 2

 1 2 . 4. Prove that the eigenvalues of (–3A–1) are the same as those of A =   2 1    1 2 −2    0 3 . 5. Find the sum and product of the eigenvalues of the matrix A =  1 −2 −1 −3    3 1 4   6. Find the sum of the squares of the eigenvalues of A =  0 2 6 .  0 0 5  

Matrices

1.45

 8 −6 2    7 −4 . 7. Find the sum of the eigenvalues of 2A, if A = −6  2 −4 3  8. 9. 10.

11.

 2 2 1   Two eigenvalues of the matrix A =  1 3 1 are equal to 1 each. Find the  1 2 2 third eigenvalue.   If the sum of two eigenvalues and trace of a 3 × 3 matrix A are equal, find the value of |A|.  1 2 . Find the eigenvectors of A =   0 3    3 0 0   Find the sum of the eigenvalues of the inverse of A = 8 4 0 .  6 2 5  

 6 −2 2   12. The product of two eigenvalues of the matrix A = −3 3 −1 is 16. Find  the third eigenvalue.  2 −1 3  Part B 13. Verify that the eigenvalues of A–1 are the reciprocals of those of A and that the respective eigenvectors are the same with respect to the matrix  1 −2  . A=  −5 4   a b 1 1  are   and   . 14. Show that the eigenvectors of the matrix A =  −b a  i  −i       Find the eigenvalues and eigenvectors of the following matrices:  2 2 0    1 15.  2 1 −7 2 −3  

−1 2 −2     2 1 16.  1 −1 −1 0 

 2 2 −7    2 17.  2 1  0 1 −3  

−2 2 −3   1 −6 18.  2  −1 −2 0 

 2 2 1   19.  1 3 1  1 2 2  

 6 −2 2   3 −1 20. −2  2 −1 3 

 3 10 5    21. −2 −3 −4    3 5 7 

 2 −2 2     22. 1 1 1   1 3 −1 

 2 1 0   23.  0 2 1    0 0 2  

Mathematics II

1.46 5  2 24.  0 0 

2 0 0  2 0 0 0 5 −2 0 −2 2

 8 −6 2   25. Find the eigenvalues and eigenvectors of A = −6 7 −4 .  2 −4 3  What can you infer about the matrix A from the eigenvalues? Verify your answer. −15 4 3   26. Given that A =  10 −12 6 , verify that the sum and product of the eigen 20 −4 2   values of A are equal to the trace of A and |A| respectively. 27. Verify that the eigenvalues of A2 and A–1 are respectively the squares and  3 0 0   reciprocals of the eigenvalues of A, given that A = 8 4 0 .  6 2 5    2 −1 1   28. Find the eigenvalues and eigenvectors of (adj A), when A = −1 2 −1 .  1 −1 2   2 1 −1    1 −2 . 29. Verify that the eigenvectors of the real symmetric matrix A =  1 −1 −2 1  are orthogonal in pairs.  −1 2 −2   1   is orthogonal and that its 30. Verify that the matrix A = −2 1 2  3  1  2 2 eigenvalues are of unit modulus.

1.7

CAYLEY-HAMiLtOn tHEOREM

This theorem is an interesting one that provides an alternative method for finding the inverse of a matrix A. Also any positive integral power of A can be expressed, using this theorem, as a linear combination of those of lower degree. We give below the statement of the theorem without proof:

1.7.1

Statement of the theorem

Every square matrix satisfies its own characteristic equation.

Matrices

1.47

This means that, if c0 λn + c1 λn –1 + … + cn –1 λ + cn = 0 is the characteristic equation of a square matrix A of order n, then c0 An + c1 Αn –1 + … + cn –1 A + cn I = 0 (1)

Note:

When λ is replaced by A in the characteristic equation, the constant term cn should be replaced by cn I to get the result of Cayley-Hamilton theorem, where I is the unit matrix of order n. Also 0 in the R.S. of (1) is a null matrix of order n.

1.7.2

Corollary

(1) If A is non-singular, we can get A–1, using the theorem, as follows: Multiplying both sides of (1) by A–1 we have c An –1 +c An –2 + … + c I + c A–1 = 0 0

1

n –1

n

1 c0 An−1 + c1 An−2 +⋅⋅⋅+ cn−1I . cn (2) If we multiply both sides of (1) by A, c0 An + 1 + c1 An + … + cn –1 A2 + cn A = 0 ∴

A−1 = −

(

)

1 c1 An + c2 An−1 +⋅⋅⋅+ cn + 1 A2 + cn A c0 Thus higher positive integral powers of A can be computed, if we know powers of A of lower degree. ∴

1.7.3

(

An+1 = −

)

Similar Matrices

Two matrices A and B are said to be similar, if there exists a non-singular matrix P such that B= P–1 AP. When A and B are connected by the relation B = P–1 AP, B is said to be obtained from A by a similarity transformation. When B is obtained from A by a similarity transformation, A is also obtained from B by a similarity transformation as explained below: B= P–1 AP Premultiplying both sides by P and postmultiplying by P-1, we get PBP–1 = PP–1 APP–1 =A Thus A = PBP–1 Now taking P–1 = Q, we get A = Q–1 BQ.

1.8

PROPERtY

Two similar matrices have the same eigenvalues. Let A and B be two similar matrices. Then, by definition, B = P–1 AP ∴

B – λΙ = P–1AP – λ I = P–1 AP – P–1 λ IP

Mathematics II

1.48

= P–1 (A – λI) P ∴ |B – λI| = |P–1| |A – λI| |P| = |A – λI| |P–1P| = |A – λI| |I| = |A – λI| Thus A and B have the same characteristic polynomials and hence the same characteristic equations. ∴ A and B have the same eigenvalues.

1.8.1

Diagonalisation of a Matrix

The process of finding a matrix M such that M–1 AM = D, where D is a diagonal matrix, is called diagonalisation of the matrix A. As M–1 AM = D is a similarity transformation, the matrices A and D are similar and hence A and D have the same eigenvalues. The eigenvalues of D are its diagonal elements. Thus, if we can find a matrix M such that M–1 AM = D, D is not any arbitrary diagonal matrix, but it is a diagonal matrix whose diagonal elements are the eigenvalues of A. The following theorem provides the method of finding M for a given square matrix whose eigenvectors are distinct and hence whose eigenvectors are linearly independent.

1.8.2 theorem If A is a square matrix with distinct eigenvalues and M is the matrix whose columns are the eigenvectors of A, then A can be diagonalised by the similarity transformation M–1 AM = D, where D is the diagonal matrix whose diagonal elements are the eigenvalues of A. Let λ1, λ2, . . ., λn be the distinct eigenvalues of A and X1, X2, . . . , Xn be the corresponding eigenvectors. Let M = [X1, X2, …, Xn], which is an n × n matrix, called the Modal matrix. ∴ AM = [AX1, AX2,…, AXn] [Note Each AXr is a (n × 1) column vector] Since Xr is the eigenvector of A corresponding to the eigenvalue λr, ∴

AXr=λXr (r=l,2,…n) AM = [λ1X1, λ2X2,…, λnXn]  λ1 0 0 −− 0     0 λ 2 0 −− 0    = [ X1 , X 2 , …, X n ]   − − − − −    0 0 0 −− λ   n 

= MD (1) As X1, X2, . . . , Xn are linearly independent column vectors, M is a non-singular matrix Premultiplying both sides of (1) by M–1, we get M–1 AM = M–1 MD = D.

Note

For this diagonalisation process, A need not necessarily have distinct eigenvalues. Even if two or more eigenvalues of A are equal, the process holds good, provided the eigenvectors of A are linearly independent.

Matrices

1.9

1.49

CALCULAtiOn OF POWERS OF A MAtRiX A

Assuming A satisfies the conditions of the previous theorem, D = M−1 AM ∴ A = M D M−1 A2 = (M D M−1) (M D M−1)

Similarly, Extending,

1.10

= MD(M−1 M)DM−1 = MD2M−1 3 A = MD3 M−1 Ak = M Dk M−1  λ1k 0 0 −− 0     0 λ k 0 −− 0  2   M −1 =M  − − − −− −    k  0 0 0 −− λ n 

DiAGOnALiSAtiOn BY ORtHOGOnAL tRAnSFORMAtiOn OR ORtHOGOnAL REDUCtiOn

If A is a real symmetric matrix, then the eigenvectors of A will be not only linearly independent but also pairwise orthogonal. If we normalise each eigenvector Xr, i.e. divide each element of Xr by the square-root of the sum of the squares of all the elements of Xr and use the normalised eigenvectors of A to form the normalised modal matrix N, then it can be proved that N is an orthogonal matrix. By a property of orthogonal matrix, N−1 = NT. ∴ The similarity transformation M−1 A M = D takes the form NT AN = D. Transforming A into D by means of the transformation NT AN = D is known as orthogonal transformation or orthogonal reduction. Note: Diagonalisation by orthogonal transformation is possible only for a real symmetric matrix.

WORKED EXAMPLE 1(c)  1 3 7   Example 1.1 Verify Cayley-Hamilton theorem for the matrix A =  4 2 3  1 2 1   and also use it to find A−1. The characteristic equation of A is 1− λ 3 7 4 2−λ 3 =0 1 2 1− λ

Mathematics II

1.50

i.e. (1 − λ)( λ2 − 3 λ − 4) − 3(4 − 4 λ − 3) + 7(8 − 2 + λ) = 0 i.e. λ3 − 4λ2 − 20λ – 35 = 0 Cayley-Hamilton theorem states that A3 − 4A2 − 20A − 35 I=0 which is to be verified. Now,

1  A =  4 1  1  A3 = A ⋅ A2 =  4 1  2

3 7  1 3 7  20 23    2 3  4 2 3 = 15 22 2 1  1 2 1 10 9 3 7  20 23 23 135    2 3 15 22 37 = 140 2 1 10 9 14   60

(1)

23  37 14  152 232  163 208 76 111

Substituting these values in (1), we get, 135 152 232 80 92 92   20 60 140 35 0 0          L.S. = 140 163 208 −  60 88 148 − 80 40 60  −  0 35 0   60 76 111  40 36 56   20 40 20   0 0 35          0 0 0   =  0 0 0  0 0 0   = R.S. Thus Cayley-Hamilton theorem is verified. Premultiplying (1) by A−1, A2 – 4A – 20I – 35A−1 = 0 ∴

1 2 A − 4 A − 20 I 35   20 23 23  4 12 28  20 0 0          1   6 8 12  −  0 20 0   =  15 22 37 − 16  35         10 9 14   4 8 4   0 0 20  −4 11 −5  1  25 = −1 −6  35  1 −10  6

A−1 =

(

)

 2 −1 2   2 −1 satisfies its characteristic Example 1.2 Verify that the matrix A = −1  1 −1 2  4 equation and hence find A . The characteristic equation of A is 2 − λ −1 2 −1 2 − λ −1 = 0 1 −1 2 − λ

Matrices

1.51

i.e. (2 − λ) (λ2 − 4λ + 3) + (λ − 2 + 1) + 2(l − 2 + λ) = 0 i.e. λ3 − 6λ2 + 8λ − 3 = 0 According to Cayley-Hamilton theorem, A satisfies (1), i.e.

(1)

A3 − 6A2 + 8A − 3I = 0

(2)

which is to be verified.  2 −1 2  2    A = − 1 2 − 1  − 1 Now    1 −1 2   1  2 −1 2  7   3 2 A = A ⋅ A =  − 1 2 − 1  − 5    1 −1 2   5 Substituting these values in (2), 2

−1 2   7 −6 9      2 −1 = − 5 6 − 6    −1 2   5 − 5 7  −6 9   29 − 28 38     6 − 6  =  − 22 23 − 28     −5 7   22 − 22 29 

 29 − 28 38   42 − 36 54   16 − 8 16   3 0        L.S. =  − 22 23 − 28  −  − 30 36 − 36  +  − 8 16 − 8  −  0 3         22 − 22 29   30 − 30 42   8 − 8 16   0 0 0 0 0   =  0 0 0  = R.S.    0 0 0  Thus A satisfies its characteristic equation. Multiplying both sides of (2) by A, we have, A4 − 6A3 + 8A2 − 3A = 0 ∴ A4 = 6A3 − 8A2 + 3A = 6(6A2 − 8A + 3 I) − 8A2 + 3A, using (2) = 28A2 − 45A+18I 4 A can be computed by using either (3) or (4). From (4),

0  0  3 

(3) (4)

 196 −168 252   90 −45 90   18 0 0           A = −140 168 −168 − −45 90 −45  +  0 18 0         140 −140 196   45 −45 90   0 0 18   124 −123 162    =  −95 96 −123     95 −95 124  4

Example 1.3 Use Cayley-Hamilton theorem to find the value of the matrix given by (A8 − 5A7 + 7A6 − 3A5 + 8A4 − 5A3 + 8A2 − 2A + I), if the matrix 2 1 1   A = 0 1 0.    1 1 2 

Mathematics II

1.52 The characteristic equation of A is

2−λ 1 1 0 1− λ 0 =0 1 1 2−λ i.e. (2 − λ) (λ2 − 3λ + 2) + λ − 1 = 0 i.e. λ3 − 5λ2 + 7λ − 3 = 0 3 ∴ A − 5A2 + 7A − 3I = 0, by Cayley-Hamilton theorem Now the given polynomial in A = A5(A3 − 5A2 + 7A − 3I) + A(A3 − 5A2 + 8A − 2I) + I = 0 + A(A3 − 5A2 + 7A − 3I) + A2 + A +I, by (1) = A2 + A +I, again using (1)

(1)

(2)

 2 1 1  2 1 1  5 4 4      Now A = 0 1 0 0 1 0 = 0 1 0       1 1 2   1 1 2   4 4 5  Substituting in (2), the given polynomial 2

 5 4 4  2 1 1 1 0 0       =  0 1 0 +  0 1 0 + 0 1 0        4 4 5   1 1 2   0 0 1  8 5 5   = 0 3 0    5 5 8  Example 1.4 Find the eigenvalues of A and hence find An (n is a positive integer), 1 2 given that A =  .  4 3  The characteristic equation of A is 1− λ 2 =0 4 3− λ i.e. λ2 − 4λ − 5 = 0 ∴ The eigenvalues of A are λ = − 1, 5 When λn is divided by (λ2 − 4λ − 5), let the quotient be Q(λ) and the remainder be (aλ + b). Then λn ≡ (λ2 − 4λ − 5) Q(λ) + (aλ + b) (1) Put λ = −1 in (1).

− a + b = (−l)n

(2)

Put λ = 5 in (l). Solving (2) and (3), we get

5a + b = 5n

(3)

a=

5n − ( −1) n 6

and

b=

5n + 5( −1)n 6

Matrices

1.53

Replacing λ by the matrix A in (1), we have An = ( A2 − 4 A − 5 I ) Q( A) + aA + bI = 0 × Q( A) + aA + bI (by Cayley-Hamilton theorrem)   5n + 5(−1) n   1 0   5n − (−1) n  1 2     +  =        0 1   6 6     4 3     For example, when n = 3,  125 +1   1 2   125 − 5   1 0   +  A3 =     6   4 3   6   0 1   21 42   20 0  +  =  84 63   0 20   41 42   =  84 83   2 2 −7    Example 1.5 Diagonalise the matrix A =  2 1 2  by similarity transforma 0 1 − 3  tion and hence find A4. The characteristic equation of A is 2−λ 2 2 1− λ 0

1

−7 2

=0

− 3− λ

2

i.e. (2 − λ) (λ + 2λ − 5) −2 (−6 − 2λ + 7) = 0 i.e. λ3 − 13λ + 12 = 0 i.e. (λ − l) (λ − 3) (λ + 4) = 0 ∴ Eigenvalues of A are λ = 1, 3, −4. Case (i) λ = 1.  1 2 − 7   x1     The eigenvector is given by  2 0 2   x2  = 0     0 1 − 4   x3  ∴

x1 x x = 2= 3 −2 8 2

∴

 1    X1 =  − 4     −1 

Case (ii)

λ = 3.

Mathematics II

1.54

 −1 2 − 7   x1      2   x2  = 0 The eigenvector is given by 2 − 2     0 1 − 6   x3  x1 x2 x3 ∴ = = 10 12 2 ∴ Case (iii)

5   X2 = 6    1  λ = −4.

 6 2 − 7   x1     2   x2  = 0 The eigenvector is given by  2 5     0 1 1  x3  x1 x x ∴ = 2 = 3 3 −2 2

∴

 3   X3 = −2    2 

 1 5 3   Hence the modal matrix is M =  − 4 6 − 2     −1 1 2   a11 a12 a13    M ≡  a21 a22 a23  Let    a31 a32 a33  Then the co-factors are given by A11 = 14,

A12 = 10, A13 = 2, A21 = −7, A22 = 5, A23 = −6,

A31= −28, and

A32 = −10, A33 = 26. |M| = a11A11 + a12A12 + a13A13 = 70.

 14 − 7 − 28   1   10 5 −10   70   2 − 6 26  The required similarity transformation is M–1 A M = D(1, 3, −4) which is verified as follows:  2 2 −7   1 5 3    AM =  2 1 2  −4 6 −2      0 1 − 3   −1 1 2  ∴

M −1 =

(1)

Matrices

1.55

 1 15 −12    =  − 4 18 8    −1 3 − 8  M

−1

1 AM = 70

 14 − 7 − 28   1 15 −12      10 5 −10   − 4 18 8     2 − 6 26   −1 3 − 8 

 70 0 0  1   0 210 0 =   70  0 0 − 280  1 0 0   0 =0 3    0 0 − 4  A4 is given by

A4 = M D4 M–1 1 0 0   1 D 4 M −1 =  0 81 0 ×   70  0 0 256 

(2)

 14 − 7 − 28     10 5 −10     2 − 6 26 

 14 − 7 − 28   1   810 405 −810  =  70   512 −1536 6656   1 5  14 3 − 7 − 28     1  4 −1    810 405 −810  M D M = −4 6 −2 ×   70    −1 1  512 −1536 6656  2   5600 − 2590 15890   1   = 3780 5530 −18060    70  1820 − 2660 12530  i.e.

 80 − 37 227    A =  54 79 − 258     26 − 38 179  4

2 2 1   Example 1.6 Find the matrix M that diagonalises the matrix A =  1 3 1  by    1 2 2  means of a similarity transformation. Verify your answer. The characteristic equation of A is

Mathematics II

1.56

2−λ 2 1 1 3−λ 1 =0 1 2 2−λ i.e.

(2 − λ ) (λ 2 − 5λ + 4) − 2 (1 − λ) + (λ − 1) = 0

i.e.

λ 3 − 7λ 2 + 11λ − 5 = 0

i.e.

(λ − 1) (λ − 5) = 0

2

∴ The eigenvalues of A are λ = 5, 1, 1. Case (i) λ = 5. −3 2 1  x1     1  x2  = 0 The eigenvector is given by  1 −2    1 2 −3  x3   ∴

x x1 x2 = = 3 4 4 4

∴

1  X 1 = 1  1 

Case (ii)

λ = 1.

1 2 1  x1     The eigenvector is given by 1 2 1  x2  = 0 1 2 1  x     3 All the three equations are one and the same, namely, x1 + 2x2 + x3 = 0 Two independent solutions are obtained as follows: Putting x2 = –1 and x3 = 0, we get x1 = 2 Putting x2 = 0 and x3 = -1, we get x1 = 1  2   X 2 = −1 and    0  

∴

 1   X 3 =  0   −1  

Hence the modal matrix is 1 2 1  a11     M = 1 −1 0 ≡  a21    1 0 −1  a31  Then the co-factors are given by A11 = 1, A12 = 1, A13 = 1, A21 = 2, A31 = 1, A32 = 1, A33 = –3 and

A22 = –2,

a12 a22 a32

a13   a23   a33 

A23 = 2

|M| = a11A11 + a12A12 + a13A13 = 4

Matrices

M

∴

−1

1 = 4

1.57

1 2 1   1 −2 1   1 2 −3 

The required similarity transformation is M −1 A M = D (5, 1, 1)

(1)

We shall now verify (1). 2  AM =  1  1  5  = 5  5 

M −1

2 1 1 2 1     3 1 1 −2 1   2 2 1 2 −3 2 1  −2 1  2 −3 1 2 1 5 2 1   1  0   5 −2 1 A M = 1 −1   4  0 −1 5 2 −3 1  20 0 0  1  = 0 4 0   4  0 0 4    5 0 0   =  0 1 0    0 0 1   = D (5, 1, 1) .

 2 1 −1   Example 1.7 Diagonalise the matrix A =  1 1 −2    −1 −2 1  orthogonal transformation. The characteristic equation of A is 2−λ 1 −1 1 1 − λ −2 = 0 −1 −2 1 − λ i.e.

(2 − λ) (λ 2 − 2λ − 3) − (−λ − 1) − (−λ − 1) = 0

i.e..

λ 3 − 4λ 2 − λ + 4 = 0

i.e.

(λ + 1)(λ − 1) (λ − 4) = 0

∴ The eigenvalues of A are 1 = –1, 1, 4.

by means of an

Mathematics II

1.58 Case (i)

λ = –1.

 3 1 −1  x1      2 −2  x2  = 0 The eigenvector is given by  1   −1 −2 2  x3   ∴

x x1 x2 = = 3 0 5 5

∴

 0   X 1 =  1    1  

Case (ii)

λ = 1.

 1 1 −1  x1     The eigenvector is given by  1 0 −2  x2  = 0    −1 −2 0  x3   ∴

x x1 x = 2= 3 −4 2 −2

∴

 2   X 2 = −1    1  

Case (iii)

λ = 4.

−2 1 −1  x1     The eigenvector is given by  1 −3 −2  x2  = 0  −1 −2 −3  x     3 ∴

x x1 x2 = = 3 5 5 −5

∴

 1   X 3 =  1   −1  

0 2 1    1 Hence the modal matrix M =  1 −1  1  − 1 1   Normalising each column vector of M, i.e. dividing each element of the first column by 2 , that of the second column by 6 and that of the third column by 3 , we get the normalised modal matrix N.

Matrices      N=      

Thus

1   3  1  3  1  −  3

2

0 1

−

2 1 2

1.59

6 1 6 1 6

The required orthogonal transformation that diagonalises A is NT A N = D (–1, 1, 4)

(1)

which is verified below:            

 2 1 −1   1  AN= 1 1 −2   2 −1 −2 1  1      = −   −        T N A N =      

2

0 1 2 1

−

6 1 6 1

2

6

2

0 −

2

1

6 1 6

4  3  4  3  4  −  3 

1   2 2   2 1 1   − − 6 6 6   1 1 1  −  − 3 3 3   0

6 1

1  3  1  3  1  −  3 

2

0 1 2 1 2

−

6 1 6 1 6

4  3  4  3  4  −  3 

−1 0 0   =  0 1 0    0 0 4   = D (−1, 1, 4) .  2 0 4   Example 1.8 Diagonalise the matrix A =  0 6 0 by means of an orthogonal  4 0 2   transformation.

Mathematics II

1.60 The characteristic equation of A is

2−λ 0 4 0 6−λ 0 =0 4 0 2−λ i.e.

(2 − λ) (6 − λ) (2 − λ) − 16 (6 − λ) = 0

i.e.

(6 − λ) (λ 2 − 4λ − 12) = 0

i.e.

(6 − λ ) (λ − 6)(λ + 2) = 0

∴ The eigenvalues of A are λ = −2, 6, 6. Case (i) λ = −2.  4 0 4  x1     The eigenvector is given by  0 8 0  x2  = 0     4 0 4  x     3 x x1 x ∴ = 2= 3 32 0 −32  1   ∴ X 1 =  0   −1   Case (ii) λ = 6. −4 0 4  x1     0  x2  = 0 The eigenvector is given by  0 0    4 0 −4   x     3 We get only one equation, i.e. x1 − x3 = 0 (1) From this we get, x1 = x3 and x2 is arbitrary. x2 must be so chosen that X2 and X3 are orthogonal among themselves and also each is orthogonal with X1.  1   Let us choose X2 arbitrarily as  0    1   Note This assumption of X2 satisfies (1) and x2 is taken as 0. a   Let X 3 = b   c   X3 is orthogonal to X1 ∴ a−c=0 (2) X3 is orthogonal to X2 ∴ a+c=0 (3)

Matrices

1.61

Solving (2) and (3), we get a = c = 0 and b is arbitrary.  0   Taking b = 1, X 3 = 1    0  

Note

Had we assumed X2 in a different form, we should have got a different

X3.  1  1     For example, if X 2 =  2 , then X 3 = −1 ⋅      1  1      1 1 0   The modal matrix is M =  0 0 1 −1 1 0   The normalised model matrix is  1   2  N =  0  1 −  2 

1 2 0 1 2

 0   1  0 

The required orthogonal transformation that diagonalises A is NT AN = D(−2, 6, 6)

(1)

which is verified below:

2 0  AN =  0 6 4 0   2 −  2  =  0  2   2 

 1  4  2  0  0 2  1 − 2   6 0  2  0 6  6 0 2 

1 2 0 1 2

 0   1  0 

Mathematics II

1.62      T N AN =       

1 2 1 2 0

0 − 0 1

1   2 2  − 2 1   0 2   2 0    2 

6 2 0 6 2

 0   6  0 

−2 0 0   =  0 6 0  0 0 6   = D (−2 , 6 , 6)

Note

From the above problem, it is clear that diagonalisation of a real symmetric matrix is possible by orthogonal transformation, even if two or more eigenvalues are equal.

EXERCiSE 1(c) Part A (Short Answer Questions) 1. State Cayley-Hamilton theorem. 2. Give two uses of Cayley-Hamilton theorem. 3. When are two matrices said to be similar? Give a property of similar matrices. 4. What do you mean by diagonalising a matrix? 5. Explain how you will find Ak, using the similarity transformation M−1 AM = D. 6. What is the difference between diagonalisation of a matrix by similarity and orthogonal transformations? 7. What type of matrices can be diagonalised using (i) similarity transformation and (ii) orthogonal transformation? 5 3 . 8. Verify Cayley-Hamilton theorem for the matrix A =  1 3    7 3 . 9. Use Cayley-Hamilton theorem to find the inverse of A =    2 6 −1 3 . 10. Use Cayley-Hamilton theorem to find A3, given that A =    2 4

Matrices

1.63

11. Use Cayley-Hamilton theorem to find (A4 – 4A3 – 5A2 + A + 2I), when  1 2 . A=   4 3   5 3 ⋅ 12. Find the modal matrix that will diagonalise the matrix A =  1 3   Part B a b   satisfies its own characteristic equation and 13. Show that the matrix A =  c d   hence find A−1.  7 2 −2    14. Verify Cayley-Hamilton theorem for the matrix A = −6 −1 2 and  6 2 −1  −1 hence find A 1 1 1    15. Verify Cayley-Hamilton theorem for the matrix A =  1 2 −3 and hence  2 −1 3  find A−1. 1 2 3    16. Verify that the matrix A =  2 −1 4 satisfies its own characteristic 3 1 −1  4 equation and hence find A . 1 0 3    17. Verify that the matrix A = 2 1 −1 satisfies its a own characteristic   1 −1 1  4 equation and hence find A . 5 3  . Hence find A4. 18. Find An, using Cayley-Hamilton theorem, when A =  1 3    7 3  . Hence find A3. 19. Find An, using Cayley-Hamilton theorem, when A =   2 6   1 0 3   20. Given that A =  2 1 −1 , compute the value of (A6 − 5A5 + 8A4 − 2A3 −  1 −1 1  9A2 + 31A − 36I), using Caylay-Hamilton theorem. Diagonalise the following matrices by similarity transformation:  2 2 0    21.  2 1 1 −7 2 −3    1 1 1   22.  0 2 1 ; find also the fourth power of this matrix.   −4 4 3  

Mathematics II

1.64  3 −1 1    5 −1 23. −1  1 −1 3 

 1 −3 3   24.  3 −5 3  6 −6 4  

 6 −2 2    3 −1 25. −2  2 −1 3 

 0 1 1   26.  1 0 1  1 1 0   Diagonalise the following matrices by orthogonal transformation:  10 −2 −5   2 3 27. −2 − 5 3 5 

 3 −1 0   2 −1 28. −1  0 −1 3 

 2 −1 1   2 −1 29. −1  1 −1 2 

−2 2 −3    30.  2 1 −6  −1 −2 0 

1.11

QUADRAtiC FORMS

A homogeneous polynomial of the second degree in any number of variables is called a quadratic form. For example, x12 + 2 x22 − 3 x32 + 5 x1 x2 − 6 x1 x3 + 4 x2 x3 is a quadratic form in three variables. The general form of a quadratic form, denoted by Q in n variables is Q = c11 x12 + c12 x1 x2 + ⋅⋅⋅ + c1n x1 xn + c21 x2 x1 + c22 x22 + ⋅⋅⋅ + c2 n x2 xn + c311 x3 x1 + c32 x3 x2 + ⋅⋅⋅ + c3n x3 xn +(

)

−−−−−−−−−−−−−−−−−−−−−−−

+ cn1 xn x1 + cn 2 xn x2 + ⋅⋅⋅ + cnn xn2 n

n

Q = ∑ ∑ cij xi x j

i.e.

j =1 i =1

In general, cij ≠ cji. The coefficient of xi xj = cij + cji. Now if we define aij =

1 (cij + c ji ) , for all i and j, then aii = cii, aij = aji and 2

aij + aji = 2aij = cij + cji. n

n

∴ Q = ∑ ∑ aij xi x j , where aij = aji and hence the matrix A = [aij] is a symmetric j =1 i =1

matrix. In matrix notation, the quadratic form Q can be represented as Q = XT AX, where

Matrices

1.65

 x1      A =  aij  , X =  x2  and X T = [ x1 , x2 , … , xn ]⋅  x   n  .  a11 a12  a1n     a21 a22  a2 n   is called the matrix of The symmetric matrix A =  aij  =            a  a a  n1 n2 nn  the quadratic form Q.

Note

To find the symmetric matrix A of a quadratic form, the coefficient of 1  xi2 is placed in the aii position and  × coefficient xi x j  is placed in each of the aij   2 and aji positions. For example, (i) if Q = 2 x12 − 3 x1 x2 + 4 x22 , then  3  2 −   2 A=    3  4 −   2 2 2 2 (ii) if Q = x1 + 3 x2 + 6 x3 − 2 x1 x2 + 6 x1 x3 + 5 x2 x3 ,

   1 −1 3     5  then A = −1 3  2     3 5 6   2   Conversely, the quadratic form whose matrix is   1 3 0   2   1  0 6 is Q = 3 x12 − 7 x32 + x1 x2 + 12 x2 x3  2     0 6 −7     

1.11.1  Definitions If A is the matrix of a quadratic form Q, |A| is called the determinant or modulus of Q. The rank r of the matrix A is called the rank of the quadratic form. If r < n (the order of A) or |A| = 0 or A is singular, the quadratic form is called singular. Otherwise it is non-singular.

Mathematics II

1.66

1.11.2

Linear transformation of a Quadratic Form

Let Q = XT AX be a quadratic form in the n variables x1, x2, . . . , xn. Consider the transformation X = PY, that transforms the variable set X = [x1, x2, . . . , xn]T to a new variable set Y = [y1, y2, . . . , yn]T, where P is a non-singular matrix. We can easily verify that the transformation X = PY expresses each of the variables x1, x2, . . . , xn as homogeneous linear expressions in y1, y2, . . . , yn. Hence X = PY is called a non-singular linear trans formation. By this transformation , Q = X T AX is transformed to T Q = ( PY ) A( PY ) = Y T ( PT AP )Y = Y T BY , where B = PT AP T

BT = ( PT AP ) = PT AT P = PT AP (∵ A is sysmmetric) =B ∴ B is also a symmetric matrix. Hence B is the matrix of the quadratic form YT BY in the variables y1, y2, . . . , yn. Thus YT B Y is the linear transform of the quadratic form XT AX under the linear transformation X = PY, where B = PT AP. Now

1.11.3

Canonical Form of a Quadratic Form

In the linear transformation X = PY, if P is chosen such that B = PT A P is a diagonal  λ1 0   0 λ2 matrix of the form    0 0 

0  0  , then the quadratic form Q gets reduced as     λ n   

Q = Y T BY  λ1 0  0   y1      0 λ 2  0   y2    = [ y1 , y2 ,… , yn ]             0 0  λ  y   n   n  2 2 2 = λ1 y1 + λ 2 y2 +  + λ n yn This form of Q is called the sum of the squares form of Q or the canonical form of Q.

1.11.4

Orthogonal Reduction of a Quadratic Form to the Canonical Form

If, in the transformation X = PY, P is an orthogonal matrix and if X = PY transforms the quadratic form Q to the canonical form then Q is said to be reduced to the canonical form by an orthogonal transformation.

Matrices

1.67

We recall that if A is a real symmetric matrix and N is the normalised modal matrix of A, then N is an orthogonal matrix such that NT AN = D, where D is a diagonal matrix with the eigenvalues of A as diagonal elements. Hence, to reduce a quadratic form Q = XT AX to the canonical form by an orthogonal transformation, we may use the linear transformation X = NY, where N is the normalised modal matrix of A. By this orthogonal transformation, Q gets transformed into YT DY, where D is the diagonal matrix with the eigenvalues of A as diagonal elements.

1.11.5

nature of Quadratic Forms

When the quadratic form XT AX is reduced to the canonical form, it will contain only r terms, if the rank of A is r. The terms in the canonical form may be positive, zero or negative. The number of positive terms in the canonical form is called the index (p) of the quadratic form. The excess of the number of positive terms over the number of negative terms in the canonical form i.e. p − (r -p) = 2p − r is called the signature(s) of the quadratic form i.e. s = 2p − r. The quadratic form Q = XT A X in n variables is said to be (i) positive definite, if r = n and p = n or if all the eigenvalues of A are positive. (ii) negative definite, if r = n and p = 0 or if all the eigenvalues of A are negative. (iii) positive semidefinite, if r < n and p = r or if all the eigenvalues of A ≥ 0 and at least one eigenvalue is zero. (iv) negative semidefinite, if r < n and p = 0 or if all the eigenvalues of A ≤ 0 and at least one eigenvalue is zero. (v) indefinite in all other cases or if A has positive as well as negative eigenvalues.

WORKED EXAMPLE 1(d)

Example 1.1 Reduce the quadratic form 2 x12 + x22 + x32 + 2 x1 x2 − 2 x1 x3 − 4 x2 x3 to canonical form by an orthogonal transformation. Also find the rank, index, signature and nature of the quadratic form.  2 1 −1   Matrix of the Q.F. is A =  1 −1 −2   −1 −2 −1   Refer to the worked example (7) in section 1(c). The eigenvalues of A are −1, 1, 4. The corresponding eigenvectors are [0, 1, 1]T [2, −1, l]T and [1, 1, −l]T respectively.

1.68

Mathematics II

0 2 1    1 The modal matrix M =  1 −1  1 1 −1   2 1   0    6 3    1 1 1   The normalised modal matrix N = −  2 6 3    1 1 1   −   2 6 3  T Hence N AN = D (−1, 1, 4), where D is a diagonal matrix with −1, 1, 4 as the principal diagonal elements. ∴ The orthogonal transformation X = NY will reduce the Q.F. to the canonical form − y12 + y22 + 4 y32 ⋅ Rank of the Q.F. = 3. Index = 2 Signature = 1 Q.F. is indefinite in nature, as the canonical form contains both positive and negative terms. Example 1.2 Reduce the quadratic form 2 x12 + 6 x22 + 2 x32 + 8 x1 x3 to canonical form by orthogonal reduction. Find also the nature of the quadratic form.  2 0 4   Matrix of the Q.F. is A =  0 6 0    4 0 2   Refer to worked example (8) in section 1(c). The eigenvalues of A are −2, 6, 6. The corresponding eigenvectors are [1, 0,−1]T, [1, 0, l]T and [0, 1, 0]T respectively.

Note

Though two of the eigenvalues are equal, the eigenvectors have been so chosen that all the three eigenvectors are pairwise orthogonal.  1 1 0   The modal matrix M =  0 0 1   −1 1 0   The normalised modal matrix is given by     N=   −  

1 2 0 1 2

 0  2  0 1   1 0  2  1

Matrices

1.69

Hence NT AN = Diag (−2, 6, 6) ∴ The orthogonal transformation X = NY 1 1 i.e. x1 = y1 + y2 2 2 x2 = y2 x3 = −

1 2

y1 +

1 2

y2

will reduce the given Q.F. to the canonical form −2 y12 + 6 y22 + 6 y32 . The Q.F. is indefinite in nature, as the canonical form contains both positive and negative terms. Example 1.3 Reduce the quadratic form x12 + 2 x22 + x32 − 2 x1 x2 + 2 x2 x3 to the canonical form through an orthogonal transformation and hence show that it is positive semidefinite. Give also a non-zero set of values (x1, x2, x3) which makes this quadratic form zero.  1 −1 0    Matrix of the Q.F. is A = −1 2 1    0 1 1  1 − λ −1 0 The characteristic equation of A is −1 2 − λ 1 =0 0 1 1− λ i.e.

(1− λ) {(2 − λ) (1− λ ) −1} − (1− λ) = 0 (1− λ) (λ 2 − 3λ ) = 0

i.e. ∴ The eigenvalues of A are λ = 0, 1, 3. When λ = 0, the elements of the eigenvector are given by x1 − x2 = 0, − x1 +2x2 + x3 = 0 and x2 + x3 = 0. Solving these equations, x1 = 1, x2 = 1, x3 = −1 ∴ The eigenvector corresponding to λ = 0 is [1, 1, − 1]T When λ = 1, the elements of the eigenvector are given by −x2 = 0, − x1 + x2 + x3 = 0 and x2 = 0. Solving these equations, x1 = 1, x2 = 0, x3 = 1. ∴ When λ = 1, the eigenvector is [1, 0, 1]T When λ = 3, the elements of the eigenvector are given by −2x1 − x2 = 0, − x1 − x2 + x3 = 0 and x2 − 2x3 = 0 Solving these equation, x1 = −1, x2 = 2, x3 = 1. ∴ When λ = 3, the eigenvector is [−1, 2, 1]T.  1 1 −1   Now the modal matrix is M =  1 0 2   −1 1 1 

Mathematics II

1.70 The normalised modal matrix is

     N=    −  

1

1

3 1

2

1   6  2  6  1  6 

−

0

3 1

1

3

2

Hence NT AN = Diag (0, 1, 3) ∴ The orthogonal transformation X = NY. x1 =

i.e.

x2 =

1

y1 +

3 1 3

x3 = −

y1 +

1 3

1 2 2

y1 +

6

y2 −

1 6

y3

y3 1 2

y2 +

1 6

y3

will reduce the given Q.F. to the canonical form 0 ⋅ y12 + y22 + 3 y32 = y22 + 3 y32 . As the canonical form contains only two terms, both of which are positive, the Q.F. is positive semi-definite. The canonical form of the Q.F. is zero, when y2 = 0, y3 = 0 and y1 is arbitrary. Taking y1 = 3 , y2 = 0 and y3 = 0, we get x1 = 1, x2 = 1 and x3 = −1. These values of x1, x2, x3 make the Q.F. zero. Example 1.4 Determine the nature of the following quadratic forms without reducing them to canonical forms:

(i) x12 + 3x22 + 6 x32 + 2 x1 x2 + 2 x2 x3 + 4 x3 x1 (ii) 5 x12 + 5 x22 + 14 x32 + 2 x1 x2 − 16 x2 x3 − 8 x3 x1 (iii) 2 x12 + x22 − 3x32 + 12 x1 x2 − 8 x2 x3 − 4 x3 x1 .

Note

We can find the nature of a Q.F. without reducing it to canonical form. The alternative method uses the principal sub-determinants of the matrix of the Q.F., as explained below: Let A = (aij)n×n be the matrix of the Q.F. Let

D2 =

D1 = |a11| = a11, a11 D3 = a21 a31

a12 a22 a32

a13 a23 etc. a33

a11 a21

a12 , a22

and Dn = |A|

Matrices

1.71

D1, D2, D3, … Dn are called the principal sub-determinants or principal minors of A. (i) The Q.F. is positive definite, if D1, D2, …, Dn are all positive i.e. Dn > 0 for all n. (ii) The Q.F. is negative definite, if D1, D3, D5, … are all negative and D2, D4, D6, … are all positive i.e. (−1)n Dn > 0 for all n. (iii) The Q.F. is positive semidefinite, if Dn ≥ 0 and least one Di = 0. (iv) The Q.F. is negative semidefinite, if (−1)n Dn ≥ 0 and at least one Di = 0. (v) The Q.F. is indefinite in all other cases. (i) Q = x12 + 3 x22 + 6 x32 + 2 x1 x2 + 2 x2 x3 + 4 x3 x1  1 1 2   Matrix of the Q.F. is A =  1 3 1    2 1 6   Now

D1 = |1| = 1; D2 =

1 1 =2 ; 1 3

D3 = 1 · (18 − 1) − 1 · (6 − 2) + 2(1 − 6) = 3. D1, D2, D3 are all positive. ∴ The Q.F. is positive definite. (ii) Q = 5 x12 + 5 x22 + 14 x32 + 2 x1 x2 − 16 x2 x3 − 8 x3 x1 .  5 1 −4    A=  1 5 −8   −4 −8 14   Now D1 = 5; D2 =

5 1 = 24; 1 5

D3 = A = 5 ⋅ (70 − 64) − 1 ⋅ (14 − 32) − 4 ⋅ (−8 + 20) = 30 + 18 − 48 = 0 D1 and D2 are > 0, but D3 = 0 ∴ The Q.F. is positive semidefinite. (iii) Q = 2 x12 + x22 − 3 x32 + 12 x1 x2 − 8 x2 x3 − 4 x3 x1  2 6 −2     A= 6 1 −4    −2 −4 −3   Now D1 = |2| = 2; D2 =

2 6 = − 34; 6 1

Mathematics II

1.72

D3 = A = 2 ⋅ (−3 − 16) − 6 ⋅ (−18 − 8) − 2 (−24 + 2) = − 38 + 156 + 44 = 162 ∴ The Q.F. is indefinite. Example 1.5 Reduce the quadratic forms

6 x12 + 3 x22 + 14 x32 + 4 x1 x2 + 4 x2 x3

+ 18 x3 x1 and 2 x12 + 5 x22 + 4 x1 x2 + 2 x3 x1 simultaneously to canonical forms by a real non-singular transformation.

Note

We can reduce two quadratic forms XT AX and XT BX to canonical forms simultaneously by the same linear transformation using the following theorem, (stated without proof): If A and B are two symmetric matrices such that the roots of |A − λB| = 0 are all distinct, then there exists a matrix P such that PT AP and PT BP are both diagonal matrices. The procedure to reduce two quadratic forms simultaneously to canonical forms is given below: (1) Let A and B be the matrices of the two given quadratic forms. (2) Form the characteristic equation |A – λB| = 0 and solve it. Let the eigenvalues (roots of this equation) be λ1, λ2, …, λn. (3) Find the eigenvectors Xi (i = 1, 2, …, n) corresponding to the eigenvalues λi, using the equation (A − λi B) Xi = 0. (4) Construct the matrix P whose column vectors are X1, X2, …, Xn. Then X = PY is the required linear transformation. (5) Find PT AP and PT BP, which will be diagonal matrices. (6) The quadratic forms corresponding to these diagonal matrices are the required canonical forms. The matrix of the first quadratic form is 6 2 9    A =  2 3 2   9 2 14   The matrix of the second quadratic form is  2 2 1   B =  2 5 0  1 0 0   The characteristic equation is |A − λB| = 0 i.e. Simplifying, i.e. 1 ∴ λ = −1, , 1 5

6 − 2λ 2 − 2λ 9 − λ 2 − 2λ 3 − 5λ 2 =0 9−λ 2 14 5λ3 − λ2 − 5λ+ 1 = 0 (λ − 1)(5λ − 1)(λ + 1) = 0

Matrices

1.73

When λ = −1, (A − λB) X = 0 gives the equations. 8x1 + 4x2 + 10x3 = 0; 4x1 + 8x2 + 2x3 = 0; 10x1 + 2x2 + 14x3 = 0. Solving these equations, ∴

x x1 x = 2 = 3 −72 24 48 X1 = [−3, 1, 2]T

1 When λ = , ( A − λB ) X = 0 gives the equations. 5 28x1 + 8x2 + 44x3 = 0; Solving these equations, ∴

8x1 + 10x2 + 10x3 = 0;

44x1 + 10x2 + 70x3 = 0.

x x1 x = 2 = 3 −360 72 216 X2 = [−5, 1, 3]T

When λ = 1,(A − λB) X = 0 gives the equations 4x1 + 8x3 = 0; − 2x2 + 2x3 = 0; 8x1 + 2x2+ 14x3 = 0 ∴

X3 = [2, −1, −1]T.

−3 −5 2   1 −1 Now P = [ X1 , X 2 , X 3 ] =  1  2 3 −1  − 3 1 2  6 2 9 −3 −5 2 PT AP = −5 1 3  2 3 2  1 1 −1  2 −1 −1  9 2 14  2 3 −1 Now  2 1 3 −3 −5 2  1 0 0  = −1 −1 −1  1 1 −1 =  0 1 0  1 −1 2  2 3 −1  0 0 1 Hence the Q.F. XT AX is reduced to the canonical form y12 + y 22 + y32 . −3 1 2  2 2 1 −3 −5 2 T     Now P B P = −5 1 3  2 5 0  1 1 −1  2 −1 −1  1 0 0  2 3 −1 −2 −1 −3 −3 −5 2 −1 0 0 = −5 −5 −5  1 1 −1 =  0 5 0  1 −1 2  2 3 −1  0 0 1 Hence the Q.F. XT B X is reduced to the canonical form − y12 + 5 y22 + y32 . Thus the transformation X = PY reduces both the Q.F.’s to canonical forms.

Note

X = PY is not an orthogonal transformation, but only a linear non-singular transformation.

1.74

Mathematics II EXERCiSE 1(d)

Part A (Short answer questions) 1. Define a quadratic form and give an example for the same in three variables: 2 2 2 2. Write down the matrix of the quadratic form 3 x1 + 5 x2 + 5 x3 − 2 x1 x2 + 2 x2 x3 + 6 x3 x1 .  2 1 −2     3. Write down the quadratic form corresponding to the matrix 1 2 −2  .   −2 −2  3   4. When is a Q.F. said to be singular? What is its rank then? 5. If the Q.F. XT AX gets transformed to YT BY under the transformation X = PY, prove that B is a symmetric matrix. 6. What do you mean by canonical form of a quadratic form? State the condition for X = PY to reduce the Q.F. XT AX into the canonical form. 7. How will you find an orthogonal transformation to reduce a Q.F. XT AX to the canonical form? 8. Define index and signature of a quadratic form. 9. Find the index and signature of the Q.F. x12 + 2 x22 − 3 x32 . 10. State the conditions for a Q.F. to be positive definite and positive semidefinite. Part B 11. Reduce the quadratic form 2 x12 + 5 x22 + 3 x32 + 4 x1 x2 to canonical form by an orthogonal transformation. Also find the rank, index and signature of the Q.F. 12. Reduce the Q.F. 3 x12 − 3 x22 − 5 x32 − 2 x1 x2 − 6 x2 x3 − 6 x3 x1 to canonical form by an orthogonal transformation. Also find the rank, index and signature of the Q.F. 13. Reduce the Q.F. 6 x12 + 3 x22 + 3 x32 − 4 x1 x2 − 2 x2 x3 + 4 x3 x1 to canonical form by an orthogonal transformation. Also state its nature. 14. Obtain an orthogonal transformation which will transform the quadratic form 2 x12 + 2 x22 + 2 x32 − 2 x1 x2 − 2 x2 x3 + 2 x3 x1 into sum of squares form and find also the reduced form. 15. Find an orthogonal transformation which will reduce the quadratic form 2 x1 x2 + 2 x2 x3 + 2 x3 x1 into the canonical form and hence find its nature. 16. Reduce the quadratic form 8 x12 + 7 x22 + 3 x32 − 12 x1 x2 − 8 x2 x3 + 4 x3 x1 to the canonical form through an orthogonal transformation and hence show that it is positive definite. Find also a non-zero set of values for x1, x2, x3 that will make the Q.F. zero. 17. Reduce the quadratic form 10 x12 + 2 x22 + 5 x32 + 6 x2 x3 − 10 x3 x1 − 4 x1 x2 to a canonical form by orthogonal reduction. Find also a set of non-zero values of x1, x2, x3, which will make the Q.F. zero. 18. Reduce the quadratic form 5 x12 + 26 x22 + 10 x32 + 6 x1 x2 + 4 x2 x3 + 14 x3 x1 to a canonical form by orthogonal reduction. Find also a set of non-zero values of x1, x2, x3, which will make the Q.F. zero.

Matrices

1.75

19. Determine the nature of the following quadratic forms without reducing them to canonical forms: (i) 6 x12 + 3 x22 + 14 x32 + 4 x2 x3 + 18 x1 x3 + 4 x1 x2 (ii) x12 − 2 x1 x2 + x22 + x32 (iii) x12 + 2 x22 + 3 x32 + 2 x1 x2 + 2 x2 x3 − 2 x3 x1 20. Find the value of λ so that the quadratic form λ(x12 + x22 + x32 ) + 2x1 x2 − 2 x2 x3 + 2 x3 x1 may be positive definite. 21. Find real non-singular transformations that reduce the following pairs of quadratic forms simultaneously to the canonical forms. (i) 6 x12 + 2 x22 + 3 x32 − 4 x1 x2 + 8 x3 x1 and 5 x12 + x22 + 5 x32 − 2 x1 x2 + 8 x3 x1 . (ii) 3 x12 + 3 x22 − 3 x32 + 2 x1 x2 − 2 x2 x3 + 2 x3 x1 and 4 x1 x2 + 2 x2 x3 − 2 x3 x1 . (iii) 2 x12 + 2 x22 + 3 x32 + 2 x1 x2 − 4 x2 x3 − 4 x3 x1 and 2 x2 x3 − 2 x1 x2 − x22 . (iv) 3 x12 + 6 x22 + 2 x32 + 8 x1 x2 − 4 x2 x3 and 5 x12 + 5 x22 + x32 − 8 x1 x2 − 2 x2 x3 .

AnSWERS

Exercise 1(a) Part A 1 3 (6) X 1 = − X 2 + X 3 2 2 (8) a = 8 (12) x + 2y = 3 and 2x – y = 1; x + 2y = 3 and 2x + 4y = 5 (13) x + 2y = 3 and 2x + 4y = 6 (14) a = −4, b = 6 (15) Have a unique solution. (16) λ ≠ 5 (17) No unique solution for any value of λ. (18) λ ≠ −1 and μ = any value (19) λ = 2 and μ = 3 (20) λ = 8 and μ ≠ 11 (21) No, as |A| ≠ 0 (22) λ = 3 (23) x = k, y = 2k, z = 5k Part B (24) −7X1 + X2 + X3 + X4 = 0 (25) 2X1 − X2 − X3 + X4 = 0 (26) 2X1 + X2 − X3 = 0 (27) X1 − 2X2 + X3 = 0 (28) X1 − X2 + X3 − X4 = 0 (29) Yes. X5 = 2X1 + X2 − 3X3 + 0.X4 (34) R(A) = R[A, B] = 2; Consistent with many solutions.

Mathematics II

1.76 (35) (36) (37) (38) (40) (42) (43) (44) (45) (46) (47) (48) (49) (50)

R(A) = 3 and R[A, B] = 4; Inconsistent R(A) = 3 and R[A, B] = 4; Inconsistent R(A) = 3 and R[A, B] = 4; Inconsistent Consistent; x = –1, y = 1, z = 2 (39) Consistent; x = 3, y = 5, z = 6 Consistent; x = 1, y = 1, z = 1 (41) Consistent; x = 2, y = 1, z = −4 Consistent; x1 = 2, x2 = 1, x3 = −1, x4 = 3 1 4 Consistent; x = 2 , y = , z = 0 , w = 5 5 Consistent; x = 2k − 1, y = 3 − 2k, z = k 7 −16k k +3 Consistent; x = ,y= ,z=k 11 11 Consistent; x = 16 − 9 k , y = 16 − 6 k , z = k 3 5 3 5 Consistent; x = 3 − 4k − k′, y = 1 − 2k + k′, z = k, w = k′ Consistent; x1 = −2k + 5k′ + 7, x2 = k, x3 = −2k′ − 2, x4 = k′ k = 1, 2: When k = 1, x = 2λ + 1, y = −3λ, z = λ When k = 2, x = 2μ, y = 1 − 3μ, z = μ λ = 1, 8: When λ = 1, x = k + 2, y = 1 − 3k, z = 5k 1 1 When λ = 8, x = (k + 52) , y = − (3k + 16) , z = k 5 5

(51) a + 2b − c = 0 (52) No solution, when k = 1; one solution, when k ≠ 1 and −2; many solutions, when k = −2. (53) No solution when λ = 8; and μ ≠ 6; unique solution, when λ ≠ 8 and μ = any value; many solutions when λ = 8 and μ = 6. (54) If a = 8, b ≠ 11 no solution; If a ≠ 8 and b = any value, unique solution; If a = 8 and b = 11, many solutions (55) x = k, y = −2k, z = 3k (56) x = −4k, y = 2k, z = −2k, w = k (57) λ = 1, −9; When λ = 1, x = k, y = −k, z = 2k and when λ = −9, x = 3k, y = 9k, z = −2k (58) λ = 0, 1, 2; When λ = 0, solution is (k, k, k); When λ = 1, solution is (k, −k, 2k); When λ = 2, solution is (2k, k, 2k).

Exercise 1(b) (3) 2, 50 (6) 38 (8) 5  1 (10)   and  0 (12) 2

(5) −2, −1 (7) 36 (9) 0 1  1 

(11)

47 60

(15) 1, 3 −4; (−2, 1, 4)T, (2, 1, −2)T, (1, −3, 13)T T

(16) 1, 5 , − 5 ; (1, 0, −1) ,

(

T

) (

5 −1, 1, −1 ,

(17) 1, 3, −4; (−1, 4, 1)T, (5, 6, 1)T, (3, −2, 2)T

T

)

5 + 1, −1, 1

Matrices (18) (19) (20) (21) (22) (23) (24) (25) (26) (28) (29)

1.77

5, −3, −3; (1, 2, −1)T, (2, −1, 0)T , (3, 0, 1)T 5, 1, 1, ; (1, 1, 1)T, (2, −1, 0)T, (1, 0, −1)T 8, 2, 2; (2, −1, 1)T, (1, 2, 0)T, (1, 0, −2)T 3,.2, 2; (1, 1, −2)T, (5, 2, −5)T −2, 2, 2 ; (4, 1, −7)T, (0, 1, 1)T 2, 2, 2; (1, 0, 0)T 1, 1, 6, 6; (0, 0, 1, 2)T, ( 1, −2, 0, 0)T, (0, 0, 2, −1)T and (2, 1, 0, 0)T 0, 3, 15; (1, 2, 2)T, (2, 1, −2)T, (2, −2, 1)T; A is singular Eigenvalues are 5, −10, −20; Trace = −25; |A| = 1000 1, 4, 4; (1, −1, 1)T, (2, −1, 0)T, (1, 0, −1)T −1, 1, 4; (0, 1, 1)T, (2, −1, 1)T, (1, 1, −1)T

Exercise 1(c) 1  6 −3 7 36 −2  3 2  (11)   4 5   (9)

 d −b 1   (13) a  ad − bc −c  3 −4 −5  1  1 4 (15) − −9 11  3 1 −5  7 −30 42   (17)  18 −13 46 −6 −14 17  

−19 57  (10)    38 76  1 (12) M =  −1   −3 1  (14)  6 3 −6  248  (16)  272   104 

3  1

−2 2  5 −2 −2 5 101 218  109 50 98 204

 6n − 2n  5 3  3.2n − 6n  1 0 976 960 n  ⋅     +  ;   (18) A =    0 1 320 336  4  1 3  2      9n − 4n   7 3  9.4n − 4.9n  1 0  463 399 n       +  ;   (19) A =    0 1  266 330  5   2 6  5      0 0 0  2 2 1     1 −3 (20)  0 0 0 (21) D (1, 3, −4); M =  −1 −4 −2 13  0 0 0      1 1 1  −99 115 65     (22) D (1, 2, 3); M =  2 1 1 ; A4 = −100 116 65     −2 0 1 −160 160 81    

Mathematics II

1.78

(23)

(24)

(25)

(26)

(27)

 1 1 1    D (2, 3, 6); M =  0 1 −2 −1 1 1  1 1 1    D (4, − 2, − 2); M =  1 1 0  2 0 −1    2  1 1   D (8, 2, 2); M = −1 0 2  1 −2 0    1 1 1   D (2, −1, −1); M = 1 −1 0 1 0 −1    1 1  −  42 3   −5 1 D (0, 3, 14); N =  42 3   4 1   3  42

     (28) D (1, 3, 4); N =      

1  2 3  1  0 − 3  1 1  −  2 3 

1

1

6 2 6 1

     (29) D (4, 1, 1); N = −     

3  14  1  14  2   14 

6 1

1

3 1

2 1

3 1

2 0

3

     (30) D (5, − 3, − 3); N =    −  

−

1  6  1  6  2   6 

1

2

6 2

5 1

6 1 6

−

5 0

1  30  2  30  5   30 

Matrices

1.79

Exercise 1(d) (2) (3) (4) (6) (9)

(11)

(12)

(13)

(14)

(15)

 3 −1 3   −1 5 1   3 1 5  2 x12 + 2 x22 + 3 x32 + 2 x1 x2 − 4 x2 x3 − 4 x3 x1 Singular, when |A| = 0; Rank r < n PT AP must be a diagonal matrix index = 2 and signature = 1  1 2    0 −  5  5    2 1   N= ; Q = y12 + 3 y22 + 6 y32 ; r = 3; p = 3; s = 3 0  5  5    0 1 0       3 1 1  −  10 35 14    5 2  2 2 2 0 − N =   ; Q = 4 y1 − y2 − 8 y3 ; r = 3; p = 1; s = −1 35 14     1 3 3     10 35 14    2 1   0  6 3    1 1 1  2 2 2 N = −  ; Q = 8 y1 + 2 y2 + 2 y3 ; Q is positive definite 6 2 3    1 1 1   −   6 2 3    1 1 1   −  3 2 6    1 1 1  2 2 2 N = −  ; Q = 4 y1 + y2 + y3 3 2 6    1 2   0   3 6    1 1 1   − −  3 2 5    1 1 1  N =  − ; Q = 2 y12 − y22 − y32 ; Q is indefinite  2 5  3   1 2   0   5   3

Mathematics II

1.80

1 2 2   3 3 3   2 1 2 (16) N =  −  ; Q = 3 y22 + 15 y32 ; x1 = 1, x2 = 2, x3 = 2 3 3 3   2 1 2 −   3 3   3      (17) N = −           (18) N = −   −  

1

1

42 5

3 1

42 4

3 1

42

3

16 378 1 378 11 378

−

−

−

2 14 1 14 3 14

3  14  1  ; Q = 3 y22 + 14 y32 ; x1 = 1, x2 = − 5, x3 = 4 14  2   14  1  27  5  ; Q = 14 y22 + 27 y32 ; x1 = 16, x2 = −1, x3 = −11 27  1   27 

(19) (i) positive definite; (ii) positive semidefinite; (iii) indefinite (20) λ > 2  1 1 0   (21) (i) P =  1 1 1 ; Q1 = − y12 + 4 y22 + 2 y32 ; Q2 = y12 + 4 y22 + y32   −1 0 0    1 −1 1   (ii) P =  −1 1 1 ; Q1 = −16 y12 + 4 y22 + 8 y32 ; Q2 = 4 y12 − 4 y22 + 4 y32  −2 0 0  1 0 1   (iii) P =  0 1 1 ; Q1 = y12 + y22 + y32 ; Q2 = y22 − y32   1 1 1   0 0 1   (iv) P =  0 1 1 ; Q1 = 2 y12 + 4 y22 − y32 ; Q2 = y12 + 4 y22 + y32  1 1 −1  

Unit-2 Vector calculus

UNIT

2

Vector Calculus

2.1

INTRODUCTION

In Vector Algebra we mostly deal with constant vectors, viz. vectors which are constant in magnitude and fixed in direction. In Vector Calculus we deal with variable vectors i.e. vectors which are varying in magnitude or direction or both. Corresponding to each value of scalar variable t, if there exists a value of the vector F, then F is called a vector function of the scalar variable t and is denoted as F = F (t ) . For example, the position of a particle that moves continuously on a curve in space varies with respect to time t. Hence the position vector r of the particle with respect to a fixed point is a function of time t, i.e. r = r (t ) . Also the position vector of the particle varies from point to point. Hence it can also be considered as a function of the point. If the points are specified by their rectangular cartesian co-ordinates (x, y, z), then r is a function of the scalar variables x, y, z, i.e. r = r (x, y, z ). A physical quantity, that is a function of the position of a point in space, is called a scalar point function or a vector point function, according as the quantity is a scalar or vector. Temperature at any point in space and electric potential are examples of scalar point functions. Velocity of a moving particle and gravitational force are examples of vector point functions. When a point function is defined at every point of a certain region of space, then that region is called a field. The field is called a scalar field or a vector field, according as the point function is a scalar point function or vector point function. In Vector Calculus, though differentiation and integration of vector time functions and vector point functions are dealt with, we will be concerned with the latter only.

2.2 VECTOR DIFFERENTIAL OPERATOR ∇ We now consider an operator ∇ (to be read as ‘del’) which is useful in defining three quantities known as the gradient, the divergence and the curl, that are useful in engineering applications.

Mathematics II

2.4 The operator ∇ is defined as

∇= i

∂ ∂ ∂ + j +k ∂x ∂y ∂z

∇ is called the vector differential operator, as it behaves like a vector (though not a vector) with ∇=∑i

∂ ∂ ∂ as coefficients of i , j , k , , ∂x ∂y ∂z

respectively. When writing

∂ ∂ , it should be noted that i and are written as the first and second ∂x ∂x

factors respectively.

2.2.1

Gradient of a Scalar Point Function

Let f (x, y, z) be a scalar point function defined in a certain region of space. Then the vector point function given by  ∂ ∂ ∂  ∂φ ∂φ ∂φ φ = i ∇φ =  i + j +k + j +k ∂y ∂z  ∂x ∂y ∂z  ∂x is defined as the gradient of f and shortly denoted as grad f.

Note

1. ∇ f should not be written as f ∇ . 2. When ∇ combines with f, neither . nor × should be put between ∇ and f. 3. If f is a constant, ∇ f = 0. 4. ∇ (c1f1 ± c2f2) = c1 ∇f1 ± c2 ∇ f2 where c1 and c2 are constants and f1, f2 are scalar point functions. 5. ∇ (f1 f2) = f1 ∇ f2 + f2 ∇ f1.  6. ∇  φ  φ

1

2

 φ ∇φ − φ ∇φ  = , if φ ≠ 0 .  φ 2

1

1

2

2

2

2

7. If v = f (u), then ∇ v = f ′ (u) ∇ u.

2.2.2

Directional Derivative of a Scalar Point Function f (x, y, z)

Let P and Q be two neighbouring points whose position vectors with respect to the origin O be r (= OP) and r + ∆r ( = OQ) respectively, so that PQ = ∆r and PQ =∆r. Let f and f + ∆f be the values of a scalar point function f at the points P and Q respectively.  ∆φ  dφ Then = lim   is called the directional derivative of f in the direction OP. dr ∆r → 0  ∆r 

i.e. dφ gives the rate of change of f with respect to the distance measured in the dr direction of r .

Vector Calculus

2.5

∂φ ∂φ ∂φ are the directional derivatives of f at P(x, y, z) in the , , ∂x ∂y ∂z directions of the co-ordinate axes. In particular,

2.2.3

Gradient as a Directional Derivative

Let f (x, y, z) be a scalar point function. Then f (x, y, z) = c represents, for various values of c, a family of surfaces, called the level surfaces of the function f. Consider two level surfaces of f, namely f = c1 and f = c2 passing through two neighbouring points P ( r ) and Q ( r + ∆ r ) . [Refer to Fig. 2.1] R Let the normal at P to the level surface f = c1 Q meet f = c2 at R. = c2 ∆n θ ∆r Clearly the least distance between the two surfaces = PR = ∆n.  = θ , then, from the figure PQR, which If QPR P = c1 is almost a right angled triangle, Fig. 2.1

∆n = ∆r cos θ (1) If n is the unit vector along PR, i.e. in the direction of outward drawn normal at P to the surface f = c1, then (1) can be written as where ∆r = PQ . ∆n = n ⋅ ∆r , or

dn = n ⋅ dr

∴

dφ = =

(2)

dφ dn dn dφ  n ⋅ dr dn

[by using (2)]

(3)

∂φ ∂φ ∂φ dx + dy + dz [Refer to Chapter 4 of Part I] ∂x ∂y ∂z  ∂φ ∂φ ∂φ  =  i+ j+ k ⋅ (dx i + dy j + dz k ) ∂y ∂z   ∂x

dφ =

Also

= ∇φ ⋅ dr

(4)

From (3) and (4), ∇φ ⋅ dr = Since

PQ = ∆r (or d r ) is arbitrary, ∇φ =

From (1),

dφ  n.dr dn

dφ  n dn

dφ dφ = dn dr (cos θ )

(5)

Mathematics II

2.6

dφ dφ = cos θ dr dn

i.e.

dφ dφ ≤ [∵ cos θ ≤ 1] dr dn dφ i.e. the maximum directional derivative of f is , that is the directional derivative dn  f in the direction of n. Thus, from (5), we get the following interpretation of ∇f : ∇f is a vector whose magnitude is the greatest directional derivative of f and whose direction is that of the outward drawn normal to the level surface f = c. i.e.

WORKED EXAMPLE 2(a) Example 2.1 Find the directional derivative of f = x 2yz + 4xz2 at the point P (1, −2, −1), (i) that is maximum, (ii) in the direction of PQ, where Q is (3, −3, −2). φ = x 2 yz + 4 xz 2 ∇φ =

∂φ ∂φ ∂φ i+ j+ k ∂x ∂y ∂z

= (2 xyz + 4 z 2 ) i + x 2 z j + (x 2 y + 8 xz ) k \ (∇φ) (1, −2 , −1) = 8 i − j − 10k The magnitude of (∇f)P is the greatest directional derivative of f at P. Thus the maximum directional derivative of f at (1, − 2 , −1) = 64 + 1 + 100 = 165 units. PQ = OQ − OP = 2 i − j − k Directional derivative of f in the direction of PQ = Component (or projection) of ∇f along PQ . =

∇φ ⋅ PQ PQ

= =

(8 i − j − 10k ) ⋅ (2 i − j − k ) 4 +1+1 27

units. 6 Example 2.2 Find the unit normal to the surface x3 − xyz + z3 = 1 at the point (1, 1, 1).

Note

Unit normal to a surface f = c at a point means the unit vector n in the direction of the outward drawn normal to the surface at the given point.

Vector Calculus

2.7

The given equation of the surface x3 − xyz + z3 = 1 is taken as f (x, y, z) = c. ∴

f = x3 − xyz + z3

∇f is a vector acting in the direction of the outward drawn normal to the surface f = c. Now

∴

∇φ = (3 x 2 − yz ) i − xzj + (3 z 2 − xy ) k (∇φ)(1,1,1) = 2 i − j + 2k = n (= a vector in the direction of the normal) n n = |n | 1 = ( 2 i − j + 2k ) 3

Example 2.3 Find the directional derivative of the function f = xy2 + yz2 at the point (2, −1, 1) in the direction of the normal to the surface x log z − y2 + 4 = 0 at the point (− 1, 2, 1). The equation of the surface x log z − y2 + 4 = 0 is identified with ψ (x, y, z) = c. \

ψ(x, y, z) = x log z − y2 and c = − 4.

The direction of the normal to this surface is the same as that of ∇ψ. x Now ∇ψ = (log z ) i − 2 y j + k z ∴

(∇ψ ) (−1,2 ,1) = −4 j − k = b (say) φ = xy 2 + yz 3 ∇φ = y 2 i + (2 xy + z 3 ) j + 3 yz 2 k (∇φ)( 2 ,−1,1) = i − 3 j − 3k

Directional derivative of f in the direction of b = = =

∇φ ⋅ b |b | ( i − 3 j − 3k ) ⋅ ( − 4 j − k ) 16 + 1 15 17

units.

Example 2.4 Find the angle between the normals to the surface xy = z2 at the points (−2, −2, 2) and (1, 9, −3). Angle between the two normal lines can be found out as the angle between the vectors acting along the normal lines. Identifying the equation xy = z2 with f (x, y, z) = c, we get f = xy − z2 and c = 0. ∇φ = yi + x j − 2 zk

Mathematics II

2.8

(∇φ)(−2 ,−2 ,2) =− 2 i − 2 j − 4k = n1 (say) (∇φ)(1,9 ,−3) = 9 i + j + 6k = n2 (say) n1 and n2 are vectors acting along the normals to the surface at the given points.

∴ If q is the required angle, cos θ =

∴

n1 ⋅ n2 = n1 n2

−44

=−

24 ⋅ 118

11 177

  11    θ = cos−1  −      177  

Example 2.5 Find the angle between the surfaces x2 − y2 − z2 = 11 and xy + yz − zx = 18 at the point (6, 4, 3). Angle between two surfaces at a point of intersection is defined as the angle between the respective normals at the point of intersection. Identifying x2 − y2 − z2 = 11 with f = c, we have f = x2 − y2 − z2 and c = 11. ∴ ∇φ = 2 xi − 2 y j − 2 zk (∇φ)(6 ,4 ,3) =12 i − 8 j − 6k = n1 Identifying xy + yz − zx = 18 with ψ = c´, we have ψ = xy + yz − zx and c´ = 18 ∇ψ = (y − z ) i + (z + x) j + (y − x)k

∴

∇ψ (6 ,4 ,3) = i + 9 j − 2k

If q is the angle between the surfaces at (6, 4, 3), then n ⋅n cos θ = 1 2 n1 n2 =

∴

−48 244 86

=−

24 61×86

  −24    θ = cos−1      5246    

Example 2.6 Find the equation of the tangent plane to the surface 2xz2 − 3xy − 4x = 7 at the point (1, −1, 2). Identifying 2xz2 − 3 xy − 4x = 7 with f = c, we have f = 2xz2 − 3xy − 4x and c = 7.

∴

∇φ = (2 z 2 − 3 y − 4) i − 3 x j + 4 xzk (∇φ)(1,−1,2) = 7 i − 3 j + 8k

(∇f)(1,−1,2) is a vector in the direction of the normal to the surface f = c.

Vector Calculus

2.9

∴ D.R.’s of the normal to the surface (f = c) at the point (1,−1, 2) are (7, −3, 8). Now the tangent plane is the plane passing through the point (1, −1, 2) and having the line whose D.R.’s are (7, − 3, 8) as a normal. ∴ Equation of the tangent plane is 7(x − 1) − 3(y + 1) + 8(z − 2) = 0 i.e. 7x − 3y + 8z − 26 = 0. Example 2.7 Find the constants a and b, so that the surfaces 5x2 − 2yz − 9x = 0 and ax2y + bz3 = 4 may cut orthogonally at the point (1,−1, 2). Two surfaces are said to cut orthogonally at a point of intersection, if the respective normals at that point are perpendicular. Identifying 5x2 − 2yz − 9x = 0 with f1 = c, we have ∴

∇φ1 = (10 x − 9) i − 2 z j − 2 yk (∇φ1 )(1,−1,2) = i − 4 j + 2 k = n1 (say)

Identifying ax2y + bz3 = 4 with f2 = c´. we have ∴

(∇φ2 ) = 2axyi + ax 2 j + 3bz 2 k (∇φ2 )(1,−1,2) = − 2ai + a j + 12bk = n2 (say)

Since the surfaces cut orthogonally, n1 ⊥ n2 . n1 ⋅ n2 = 0 i.e. i.e. −6a + 24b = 0 i.e. −a + 4b = 0 (1) Since (1,−1, 2) is a point of intersection of the two surfaces, it lies on ax2y + bz3 = 4 ∴ −a + 8b = 4 Solving (1) and (2), we get a = 4 and b = 1.

(2)

Example 2.8 If r is the position vector of the point (x, y, z), a is a constant vector and φ = x2 + y2 + z2, prove that (i) grad (r ⋅ a ) = a and (ii) r ⋅ grad φ = 2φ . r = xi + yj + zk Let ∴

a = a1 i + a2 j + a3 k r ⋅ a = a1 x + a2 y + a3 z

∴

grad (r ⋅ a )= a1 i + a2 j + a3 k = a

∴

f = x2 + y2 + z2 grad φ = 2 xi + 2 yj + 2 zk

∴

r ⋅ grad φ = 2(x 2 + y 2 + z 2 ) = 2φ.

Example 2.9 If r is the position vector of the point (x, y, z) with respect to the origin, prove that ∇(r n ) = nr n−2 r .

Mathematics II

2.10

r = xi + yj + zk r 2 =| r 2 |= x 2 + y 2 + z 2

∴

∂ n ∂ ∂ (r ) i + (r n ) j + (r n )k ∂x ∂y ∂z  ∂r ∂r ∂r  = nr n−1  i + j + k  ∂y ∂z   ∂x

(1)

∇(r n ) =

∴

From (l),

2r

i.e. Similarly,

(2)

∂r = 2x ∂x ∂r x = ∂x r ∂r y = ∂y r

and

∂r z = ∂z r

(3)

Using (3) in (2), we have x y z  ∇(r n ) = nr n−1  i + j + k   r r r  = nr n−2 ( xi + yj + zk ) = nr n−2 r . Example 2.10 Find the function f, if grad f = ( y 2 − 2 xyz 3 ) i + (3 + 2 xy − x 2 z 3 ) j + (6 z 3 − 3 x 2 yz 2 )k . ∇φ = ( y 2 − 2 xyz 3 ) i + (3 + 2 xy − x 2 z 3 ) j + (6 z 3 − 3 x 2 yz 2 )k

By definition,

∇φ =

∂φ ∂φ ∂φ i+ j+ k ∂x ∂y ∂z

(1)

(2)

Comparing (1) and (2), we get ∂φ = y 2 − 2 xyz 3 ∂x

(3)

∂φ = 3 + 2 xy − x 2 z 3 ∂y

(4)

∂φ = 6 z 3 − 3 x 2 yz 2 ∂z

(5)

Integrating both sides of (3) partially with respect to x (i.e. treating y and z as constants), f = xy2 − x2yz3 + a function not containing x

(6)

Vector Calculus

2.11

Note

When we integrate both sides of an equation ordinarily with respect to x, we usually add an arbitrary constant in one side. When we integrate partially, we add an arbitrary function of the other variables y and z, i.e. an arbitrary function independent of x. Integrating (4) partially with respect to y. f = 3y + xy2 − x2yz3 + a function not containing y (7) Integrating (5) partially with respect to z, 3 (8) φ = z 4 − x 2 yz 3 + a function not containing z 2 (6), (7) and (8) give only particular forms of f. The general form of f is obtained as follows: The terms which do not repeat in the R.S’s of (6), (7) and (8) should necessarily be included in the value of f. The terms which repeat should be included only once in the value of f. The last terms indicate that there is a term of f which is independent of x, y, z, i.e. a constant. 3 ∴ φ = 3 y + z 4 + xy 2 − x 2 yz 3 + c . 2 EXERCISE 2(a) Part A (Short Answer Questions) 1. Define grad f and give its geometrical meaning. 1 2. If r is the position vector of the point (x, y, z), prove that ∇(r ) = r . r 3. If r is the position vector of the point (x, y, z), prove that ∇( | r |2 ) = 2r . 1 4. If r is the position vector of the point (x, y, z), prove that ∇f (r ) = f ′(r )r . r 5. Find grad f at the point (1, –2, –1) when f = 3x2y − y3z2. 6. Find the maximum directional derivative of f = x3y2z at the point (1, 1, 1). 7. Find the directional derivative of f = xy + yz + zx at the point (1, 2, 3) along the x-axis. 8. In what direction from (3, 1, −2) is the directional derivative of f = x2y2z4 maximum? 9. If the temperature at any point in space is given by T = xy + yz + zx, find the direction in which the temperature changes most rapidly with distance from the point (1, 1, 1). 10. The temperature at a point (x, y, z) in space is given by T(x, y, z) = x2 + y2 − z. A mosquito located at (1, 1, 2) desires to fly in such a direction that it will get warm as soon as possible. In what direction should it fly? Part B 11. If f = xy + yz + zx and F = x 2 yi + y 2 zj + z 2 xk , find F ∙ grad f and F × grad f at the point (3, −1, 2).

Mathematics II

2.12

12. Find the directional derivative of f = 2xy + z2 at the point (1, −1, 3) in the direction of i + 2 j + 2k . 13. Find the directional derivative of f = xy2 + yz3 at the point P ( 2, −1, 1) in the direction of PQ where Q is the point (3, 1, 3). 14. Find a unit normal to the surface x2y + 2xz = 4 at the point (2, −2, 3). 15. Find the directional derivative of the scalar function f = xyz in the direction of the outer normal to the surface z = xy at the point (3, 1, 3). 16. Find the angle between the normals to the surface xy3z2 = 4 at the points (−1, −1, 2) and (4, 1, −1). 17. Find the angle between the normals to the surface x2 = yz at the points (1, 1, 1) and (2, 4, 1). 18. Find the angle between the surfaces z = x2 + y2 − 3 and x2 + y2 + z2 = 9 at the point (2, −1, 2). 19. Find the angle between the surfaces xy2z = 3x + z2 and 3x2 − y2 + 2z = 1 at the point (1, −2, 1). 20. Find the angle between the tangent planes to the surfaces x log z − y2 = −1 and x2y + z = 2 at the point (1, 1, 1). 21. Find the equation of the tangent plane to the surface xz2 + x2y = z − 1 at the point (1, −3, 2). 22. Find the values of l and μ, if the surfaces lx2 − μyz = (λ + 2)x and 4x2y + z3 = 4 cut orthogonally at the point (1, −1, 2). 23. Find the values of a and b, so that the surfaces ax3 − by2z = (a + 3)x2 and 4x2y − z3 = 11 may cut orthogonally at the point (2, −1, −3). 24. Find the scalar point function whose gradient is (2 xy − z 2 ) i + (x 2 + 2 yz ) j +(y 2 − 2 zx) k . 25. If ∇φ = 2 xyz 3 i + x 2 z 3 j + 3 x 2 yz 2 k , find f (x, y, z), given that f (1, −2, 2) = 4.

2.2 THE DIVERGENCE OF A VECTOR If F (x , y , z ) is a differentiable vector point function defined at each point (x, y, z) in some region of space, then the divergence of F , denoted as div F , is defined as div

F = ∇⋅ F  ∂ ∂ ∂ =  i +j + k ⋅ F ∂y ∂z   ∂x =i⋅

∂F ∂F ∂F + j⋅ +k ⋅ ∂x ∂y ∂zz

Formula for Ñ× F , when F = F1 i + F2 j + F3 k  ∂ ∂ ∂ ∇⋅ F =  i +j + k ⋅ (F1 i + F2 j + F3 k ) ∂y ∂z   ∂x

Vector Calculus =

2.13

∂F1 ∂F2 ∂F3 + + ∂x ∂y ∂z

Note

Since F is a vector point function, F1, F2 and F3 are scalar point functions and hence ∇⋅ F is also a scalar point function.

2.3.1

Physical meaning of ∇⋅ F

(i) If V = Vx i + Vy j + Vz k is a vector point function representing the instantaneous velocity of a moving fluid at the point (x, y, z), then ∇⋅V represents the rate of loss of the fluid per unit volume at that point. (ii) If the vector point function V represents an electric flux, then ∇⋅V represents the amount of flux that diverges per unit volume in unit time. (iii) If the vector point function V represents heat flux, then ∇⋅V represents the rate at which heat is issuing from the concerned point per unit volume. In general, if F represents any physical quantity, then ∇⋅ F gives at each point the rate per unit volume at which the physical quantity is issuing from that point. It is due to this physical interpretation of ∇⋅ F , it is called the divergence of F .

2.3.2

Solenoidal Vector

If F is a vector such that ∇⋅ F = 0 at all points in a given region, then it is said to be a solenoidal vector in that region.

2.3.3

Curl of a Vector

If F (x, y, z) is a differentiable vector point function defined at each point (x, y, z) in some region of space, then the curl of F or the rotation of F , denoted as curl F or rot F is defined as Curl F = ∇ × F  ∂ ∂ ∂  =i +j + k × F ∂y ∂z   ∂x ∂F ∂F ∂F =i× + j× +k × ∂x ∂y ∂z

Note

Curl F is also a vector point function.

Formula for Ñ´F, when F = F1 i + F2 j + F3 k (where F1, F2 and F3 are scalar point functions): ∂ (F1 i + F2 j + F3 k ) ∂x ∂F ∂F ∂F = ∑ 1 (i × i ) + 2 (i × j ) + 3 (i ×k ) ∂x ∂x ∂x

∇× F = ∑ i ×

Mathematics II

2.14  ∂F ∂F = ∑  2 k − 3  ∂x ∂x

 j  

 ∂F ∂F   ∂F ∂F   ∂F ∂F  =  3 − 2  i +  1 − 3  j +  2 − 1  k  ∂y  ∂z ∂x  ∂z  ∂y   ∂x i ∂ = ∂x F1

2.3.4

j ∂ ∂y F2

k ∂ ∂z F3

Physical Meaning of Curl F

If F represents the linear velocity of the point (x, y, z) of a rigid body that rotates about a fixed axis with constant angular velocity ω , then curl F at that point represents 2ω .

2.3.5

Irrotational Vector

If F is a vector such that ∇× F = 0 at all points in a given region, then it is said to be an irrotational vector in that region.

2.3.6

Scalar Potential of an Irrotational Vector

If F is irrotational, then it can be expressed as the gradient of a scalar point function. F = F1 i + F2 j + F3 k

Let

Since F is irrotational, Curl F = 0 i ∂ ∂x F1

i.e.

i.e.

∴

j ∂ ∂y F2

k ∂ =0 ∂z F3

 ∂F ∂F   ∂F3 ∂F2   ∂F ∂F    i +  1 − 3  j +  2 − 1  k = 0 −  ∂x  ∂y  ∂z ∂y  ∂z  ∂x  ∂F3 ∂F2 ∂F1 ∂F3 ∂F2 ∂F1 ; ; = = = ∂y ∂z ∂z ∂x ∂x ∂y

Equations (1) are satisfied when F1 =

∂φ ∂φ , F2 = ∂x ∂y

and

F3 =

∂φ ∂z

(1)

Vector Calculus

2.15

∂φ ∂φ ∂φ +j +k ∂x ∂y ∂z = ∇φ

∴

F=i

If F is irrotational and F = ∇φ , then f is called the scalar potential of F .

2.3.7

Expansion Formulae Involving Operations by ∇

Expansion formulas involving operations by one or two ∇’s are given below: The proofs of some of them are given and those of the rest are left as exercise to the students. 1. If u and v are vector point functions, then ∇⋅ (u ± v ) = ∇⋅ u ± ∇⋅ v . 2. If u and v are vector point functions, then ∇×(u ± v ) =∇×u ±∇× v . 3. If f is a scalar point function and F is a vector point function, then ∇⋅ (φ F ) = φ(∇⋅ F ) + (∇φ )⋅ F Proof:

∂ (φ F ) ∂x  ∂F ∂φ  = ∑ i ⋅φ + F   ∂x ∂x 

∇⋅ (φ F ) = ∑ i ⋅

= φ∑ i ⋅

∂ F  ∂φ  + ∑ i ⋅ F  ∂x  ∂x 

= φ (∇⋅ F ) + (∇φ )⋅ F 4. If f is a scalar point function and F is a vector point function, then ∇×(φ F ) = φ(∇× F ) + (∇ φ )× F 5. If u and v are vector point functions, then ∇⋅(u ×v )= v ⋅ curl u − u ⋅ curl v. Proof:

∂ (u ×v ) ∂x  ∂u ∂v  = ∑ i ⋅ ×v + u ×   ∂x ∂x    ∂v   ∂u = ∑ i ⋅ ×v − ∑ i ⋅ ×u   ∂x  ∂x    ∂v   ∂u  = ∑  i × ⋅ v − ∑  i × ⋅ u   ∂x  ∂x 

∇⋅ (u ×v ) = ∑ i ⋅

[ the value of a scalar triple product is unaltered, when dot and cross are interchanged]

Mathematics II

2.16

  ∂v  ∂u  = ∑ i × ⋅ v −∑ i × ⋅ u     ∂x  ∂x = v ⋅ Curl u − u ⋅ Curl v 6. If u and v are vector point functions, then ∇×(u ×v ) = (∇⋅ v )u + (v ⋅∇)u − (∇⋅ u )v − (u ⋅∇)v In this formula, ∇⋅ v and v ⋅∇ are not the same, ∇⋅v means div v , but ∂ ∂ ∂ + vy + vz , if v = vx i + v y j + vz k ⋅ v ⋅∇ represents the operator vx ∂x ∂y ∂z ∂u ∂u ∂u Thus (v ⋅∇) u represents vx + vy + vz ⋅ ∂x ∂y ∂z

Note

7. If u and v are vector point functions, then ∇ (u ⋅ v )= v ×curl u + u × curl v + (v ⋅∇)u + (u ⋅∇)v . 8. If f is a scalar point function, then div (grad f) = ∇ . (∇f) = ∇2f, ∂2 ∂2 ∂2 + + is called the Laplacian operator and ∂x 2 ∂y 2 ∂z 2 ∂ 2φ ∂ 2φ ∂ 2φ ∇2φ = 2 + 2 + 2 is called the Laplacian of f. ∇2f = 0 is called the ∂x ∂y ∂z

where ∇2 =

Laplace equation.

Note

∇2 can also operate on a vector point function F resulting in

∂2 F ∂2 F ∂2 F . + + ∂x 2 ∂y 2 ∂z 2 9. If f is a scalar point function, then curl (grad f) = ∇ × (∇f) = 0.  ∂φ   ∂φ   ∂φ  Proof grad φ =   i +   j +   k  ∂z   ∂x   ∂y  ∇2 F =

i

j

k

∂ ∂ ∂ ∴ curl (grad f) = ∂x ∂y ∂z ∂φ ∂φ ∂φ ∂x ∂y ∂z  ∂ 2φ  ∂ 2φ  ∂ 2φ ∂ 2φ  ∂ 2φ  ∂ 2φ   j +    i +  =  − − −  ∂z ∂x ∂x ∂z   ∂x ∂y ∂y ∂x  k  ∂y ∂z ∂z ∂y  = 0.

Note

This result means that (grad f) is always an irrotational vector.

Vector Calculus

2.17

10. If F is a vector point function, then div (curl F ) =∇⋅ (∇× F ) =0 . Let F = F1 i + F2 j + F3 k

Proof:

curl F =

∴

i

j

k

∂ ∂x F1

∂ ∂y F2

∂ ∂z F3

 ∂F ∂F   ∂F ∂F   ∂F ∂F  =  3 − 2  i +  1 − 3  j +  2 − 1  k  ∂z ∂x  ∂z   ∂x ∂y   ∂y

∴ div (curl F )=

∂  ∂F3 ∂F2  ∂  ∂F1 ∂F3  ∂  ∂F2 ∂F1   +  − − − +   ∂x  ∂y ∂z  ∂y  ∂z ∂x  ∂z  ∂x ∂y 

∂ 2 F3 ∂ 2 F3 ∂ 2 F2 ∂ 2 F2 ∂ 2 F1 ∂ 2 F1 − + − − + ∂ x ∂ y ∂x ∂z ∂y ∂z ∂y ∂x ∂z ∂x ∂z ∂y = 0.

=

Note

This result means that (curl F ) is always a solenoidal vector.

11. If F is a vector point function, then 2 curl (curl F ) =∇×(∇× F ) =∇ (∇⋅ F ) −∇ F .

Proof: Let Then

F = F1 i + F2 j + F3 k

 ∂F ∂F   ∂F3 ∂F2   ∂F ∂F   i +  1 − 3  j +  2 − 1  k − curl F =   ∂x ∂y    ∂z ∂z  ∂x   ∂y

∴ curl (curl F )

=

i

j

k

∂ ∂x

∂ ∂y

∂ ∂z

 ∂F3 ∂F2     ∂y − ∂z 

 ∂F1 ∂F3   −   ∂z ∂x 

 ∂F2 ∂F1     ∂x − ∂y 

 ∂  ∂F ∂F  ∂  ∂F ∂F  = ∑   2 − 1 −  1 − 3  i   ∂x   ∂y  ∂x ∂y  ∂z  ∂z   ∂2 F ∂ 2 F3   ∂ 2 F1 ∂ 2 F1  2 − 2 + 2  i = ∑   +  ∂z    ∂y ∂x ∂z ∂x   ∂y  ∂2 F ∂ 2 F3   ∂ 2 F1 ∂ 2 F1 ∂ 2 F1  ∂ 2 F2 − + 2 + 2  i = ∑   21 + + ∂x ∂y ∂x ∂z   ∂x 2 ∂y ∂z    ∂x

Mathematics II

2.18

 ∂  ∂F ∂F ∂F   ∂ 2 ∂2 ∂2   = ∑   1 + 2 + 3 − 2 + 2 + 2  F1  i ∂y ∂z   ∂x ∂y ∂z    ∂x  ∂x ∂  = ∑  (∇⋅ F )−∇2 F1  i   ∂x  ∂  ∂ ∂ = i ∇⋅ F ) + k ∇⋅ F ) −∇2  F1 i + F2 j + F3 k  ∇⋅ F ) + j ( ( (  ∂x  ∂y ∂z   2 =∇ (∇⋅ F )−∇ F . 12. If F is a vector point function, then grad (div F ) =∇(∇⋅ F ) =∇×(∇× F ) +∇2 F .

Note

Rewriting the formula (11), this result is obtained. WORKED EXAMPLE 2(b)

Example 2.1 When f = x3 + y3 + z3 − 3xyz, find ∇f, ∇ . ∇f and ∇ × ∇f at the point (1, 2, 3). f = x3 + y3 + z3 − 3x yz  ∂φ  ∇φ = ∑   i  ∂x  = 3(x 2 − yz ) i + 3(y 2 − zx)j + 3(z 2 − xy )k ∂ ∂ ∂ 3(x 2 − yz )] + [3(y 2 zx)] + [3(z 2 − xy )] [3 ∂x ∂y ∂z = 6(x + y + z )

∇⋅∇φ =

i j k ∂ ∂ ∂ ∇×∇φ = ∂y ∂z ∂x 2 2 2 3(x − yz ) 3(y − zx) 3(z − xy ) = ( − 3 x + 3 x) i − ( − 3 y + 3 y )j + ( − 3 z + 3 z )k

Note

∇ × ∇f = 0, for any f, as per the expansion formula (9).

(∇φ)(1, 2 ,3) =−15 i + 3 j + 21k (∇ . ∇f)(1, 2, 3) = 36

and

(∇ × ∇f)(1, 2, 3) = 0.

2 2 2 2 Example 2.2 If F = (x − y + 2 xz ) i + (xz − xy + yz )j + (z + x )k , find ∇⋅ F ,∇

(∇⋅ F ) ,∇× F ,∇⋅ (∇× F ) and ∇×(∇× F ) at the point (1, 1, 1).

Vector Calculus

2.19

F = (x 2 − y 2 + 2 xz ) i + (xz − xy + yz ) j + (z 2 + x 2 )k ∂ 2 ∂ ∂ (x − y 2 + 2 xz ) + (xz − xy + yz ) + (z 2 + x 2 ) ∂x ∂y ∂z = (2 x + 2 z ) + ( − x + z ) + 2 z = x + 5z ∂ ∂ ∂ ∇(∇⋅ F ) = (x + 5 z ) i + (x + 5 z ) j + (x + 5 z )k ∂y ∂z ∂x ∇⋅ F =

= i + 5k i ∂ ∇× F = ∂x 2 x − y 2 + 2 xz

j

k

∂ ∂y xz − xy + yz

∂ ∂z 2 z + x2

=− (x + y ) i − (2 x − 2 x) j + (y + z )k =− (x + y ) i + (y + z )k ∂ ∂ ∂ [ − (x + y )] + (0) + (y + z ) ∂x ∂y ∂z =−1+ 0 +1= 0

∇⋅ (∇× F ) =

Note

∇⋅ (∇× F ) = 0 , for any F , as per the expansion formula i

j

∂ ∂ ∇×(∇× F ) = ∂x ∂y −( x + y ) 0

k ∂ ∂z y+z

=i +k

∴

(∇⋅ F )(1,1,1) = 6; [∇(∇⋅ F )](1,1,1) = i + 5k ; (∇× F )(1,1,1) =− 2 i + 2k ; [∇⋅ (∇× F )](1,1,1) = 0; [∇×(∇× F )](1,1,1) = i + k

Example 2.3 If ā is a constant vector and r is the position vector of the point (x, y, z,) with respect to the origin, prove that (i) grad (a ⋅ r ) = a , (ii) div (a ×r ) = 0 and (iii) curl (a ×r ) = 2a . r = xi + yj + zk Let a = a1 i + a2 j + a3 k , where a1, a2, a3 are constants.

∴

a ⋅ r = a1 x + a2 y + a3 z

Mathematics II

2.20

∴

∂ (a1 x + a2 y + a3 z ) i ∂x = a1 i + a2 j + a3 k

grad (a ⋅ r ) = ∑

=a i a ×r = a1 x

j a2 y

k a3 z

= (a2 z − a3 y ) i + (a3 x − a1 z ) j + (a1 y − a2 x)k

∴

∂ ∂ ∂ (a2 z − a3 y ) + (a3 x − a1 z ) + (a1 y − a2 x) ∂x ∂y ∂z =0

div (a ×r ) =

i j k ∂ ∂ ∂ curl (a ×r ) = ∂x ∂y ∂z a2 z − a3 y a3 x − a1 z a1 y − a2 x = (a1 + a1 ) i − ( − a2 − a2 ) j + (a3 + a3 )k = 2(a1 i + a2 j + a3 k ) = 2a

2 2 3 2 2 3 Example 2.4 Show that u = (2x + 8 xy z ) i + (3 x y − 3 xy ) j − (4 y z + 2 x z )k is

not solenoidal, but v = xyz 2 u is solenoidal. ∇⋅ u =

∂ ∂ ∂ (2 x 2 + 8 xy 2 z ) + (3 x 3 y − 3 xy ) + { − (4 y 2 z 2 + 2 x 3 z )} ∂x ∂y ∂z

= (4x + 8y2z) + (3x3 − 3x) − (8y2z + 2x3) = x3 + x ≠ 0, for all points (x, y, z) ∴ u is not solenoidal. v = xyz 2 u = (2 x 3 yz 2 + 8 x 2 y 3 z 3 ) i + (3 x 4 y 2 z 2 − 3 x 2 y 2 z 2 ) j − (4 xy 3 z 4 + 2 x 4 yz 3 )k ∴

∇⋅ v = (6 x 2 yz 2 +16 xy 3 z 3 ) + (6 x 4 yz 2 − 6 x 2 yz 2 ) − (16 xy 3 z 3 + 6 x 4 yz 2 )

= 0, for all points (x, y, z) v ∴ is solenoidal. Example 2.5 Show that F = (y 2 − z 2 + 3 yz − 2 x) i + (3 xz + 2 xy ) j + (3 xy − 2 xz + 2z )k is both solenoidal and irrotational.

Vector Calculus ∇⋅ F =

2.21

∂ 2 ∂ ∂ (y − z 2 + 3 yz − 2 x) + (3 xz + 2 xy ) + (3 xy − 2 xz + 2 z ) ∂x ∂y ∂z

= −2 + 2x − 2x + 2

= 0, for all points (x, y, z) ∴ F is a solenoidal vector.

∇× F =

i

j

k

∂ ∂x

∂ ∂y

∂ ∂z

(y 2 − z 2 + 3 yz − 2 x) (3 xz + 2 xy ) (3 xy − 2 xz + 2 z ) = (3 x − 3 x) i − (3 y − 2 z + 2 z − 3 y ) j + (3 z + 2 y − 2 y − 3 z ) k

= 0, for all points (x, y, z) :. F is an irrotational vector. Example 2.6 Show that F = (y 2 + 2 xz 2 ) i + (2 xy − z ) j + (2 x 2 z − y + 2 z ) k is irrotational and hence find its scalar potential.

∇× F =

i

j

k

∂ ∂x

∂ ∂y

∂ ∂z

(y 2 + 2 xz 2 ) (2 xy − z ) (2 x 2 z − y + 2 z ) = ( −1 + 1) i − (4 xz − 4xxz ) j + (2 y − 2 y ) k

= 0, for all points (x, y, z) ∴ F is irrotational. Let the scalar potential of F be f. ∴

F = ∇φ =

∂φ ∂φ ∂φ i + j+ k ∂x ∂y ∂z

∂φ = y 2 + 2 xz 2 ∂x Integrating partially w.r.t. x; f = xy2 + x2z2 + a function independent of x ∴

(1)

∂φ = 2 xy − z ∂y Integrating partially w.r.t. y; f = xy2 − yz + a function independent of y ∂φ = 2 x2 z − y + 2 z ∂z

(2)

Mathematics II

2.22

Integrating partially w.r.t. z; f = x2z2 − yz + z2 + a function independent of z From (1), (2), (3), we get f = xy2 + x2z2 − yz + z2 + c. Example 2.7

(3)

3 Find the values of the constants a, b, c, so that F = (axy + b z ) i

+ (3 x 2 − cz ) j + (3 xz 2 − y ) k may be irrotational. For these values of a, b, c, find also the scalar potential of F . F is irrotational. ∴

i.e.

∇× F = 0 i ∂ ∂x

j ∂ ∂y

k ∂ ∂z

=0

(axy + bz 3 ) (3 x 2 − cz ) (3 xz 2 − y ) i.e.

∴

( −1+ c) i − (3 z 2 − 3bz 2 ) j + (6 x − ax) k = 0 c − 1 = 0,

∴

3z2 (1 − b) = 0, x (6 − a) = 0 a = 6, b= 1, c = 1.

Using these values of a, b, c, F = (6 xy + z 3 ) i + (3 x 2 − z ) j + (3 xz 2 − y ) k Let f be the scalar potential of F . ∂φ ∂φ ∂φ i+ j+ k ∂x ∂y ∂z ∂φ ∂φ ∂φ = 6 xy + z 3 , = 3x 2 − z, = 3 xz 2 − y ∂x ∂y ∂z Integrating partially w.r.t. the concerned variables, ∴

F =∇φ =

f = 3x2y + xz3 + a function independent of x

(1)

f = 3x2y − yz + a function independent of y

(2)

f = xz3 − yz + a function independent of z

(3)

From(l), (2) and (3), we get f = 3x2y + xz3 − yz + c Example 2.8 If f and ψ are scalar point functions, prove that (i) f ∇f is irrotational and (ii) ∇f × ∇ψ is solenoidal. ∇ × f∇f = f (∇ × ∇f) + ∇f × ∇f , by expansion formula = f (0) + 0 , by expansion formula =0 ∴ f∇f is irrotational

Vector Calculus

2.23

∇⋅(u ×v )= v ⋅ curl u − u ⋅ curl v , by expansion formula ∴ ∇ ⋅ (∇f × ∇ψ) = ∇ψ ⋅ curl (∇f) − ∇f ⋅ curl (∇ψ ) = ∇ψ ⋅ 0 − ∇f ⋅ 0 = 0. ∴ (∇f × ∇ψ) is solenoidal. If r = | r | , where r is the position vector of the point (x, y, z), prove 1 that ∇2 (rn ) = n(n + 1) rn−2 and hence deduce that satisfies Laplace equation. r We have already proved, in worked example (9) of the previous section, that ∇ (r n ) = nr n − 2 r . Example 2.9

Now

∇2 (r n ) =∇⋅ (∇r n ) =∇⋅ (nr n − 2 r ) = n ∇ (r n − 2 ) ⋅ r + r n − 2 (∇⋅ r ) , by expansion formula   n−4 n−2   = n  (n − 2) r r ⋅ r + 3r  ,  [since ∇⋅ r =∇⋅ (xi + yj + zk ) ∂ ∂ ∂ (x) + (y ) + (z ) ∂x ∂y ∂z = 3] =

∴

∇2(rn)

= n [(n − 2) rn−4 r2 + 3rn−2] = n (n+1) rn−2

Taking n = −1 in the above result,

i.e.

1 ∇2   = ( −1) (0) r −3 = 0  r  1 satisfies Laplace equation. r

Example 2.10 If u and v are scalar point functions, prove that ∇ ⋅ (u∇v − v∇u) = u∇2 v − v∇2 u. ∇ ⋅ (u∇v − v∇u) = ∇ ⋅ (u∇v) − ∇ ⋅ (v∇u) = (∇u ⋅ ∇v + u∇2 v) − (∇v ⋅ ∇u + v∇2 u) [by the expansion formula] = u∇2v − v∇2 u. Example 2.11 If u and v are scalar point functions and F is a vector point function such that uF =∇v , prove that F ⋅ curl F =0 . Given uF =∇v

∴

1 F = ∇v u

Mathematics II

2.24

1  Curl F =∇× ∇v  u 

∴

1 1 = (∇×∇v) +∇  ×∇v , by expansion formula  u  u 1 =∇  ×∇v , since ∇ × ∇v = 0.  u   1    1 F ⋅ curl F =  ∇v ⋅∇ ×∇v     u u    

Now

1 = ( 0) , u

[ Two factors are equal in the scalar triple product]

= 0. Example 2.12 If r = r , where r is the position vector of the point (x, y, z) with f ′ (r ) respect to the origin, prove that (i) ∇f (r) = r and r 2 (ii) ∇2 f (r ) = f ′′ (r ) + f ′ (r ). r r2 = x2 + y2 + z2 ∴ ∴

2r

∂r = 2x ∂x

∂r z ∂r x ∂r y = . = . Similarly, = and ∂z r ∂x r ∂y r

Now

∇f (r ) =

∂

∑ i ∂x f (r ) ∂r

= f ′ (r )

∑ ∂x i

= f ′ (r )

∑ri x

f ′(r ) r r ∇2 f (r ) =∇⋅∇ f (r )   f ′ (r )   =∇⋅ r     r     =

  f ′ (r )   f ′ (r )  =∇  ∇⋅ r  ⋅ r +   r r       r f ′′ (r ) − f ′ (r )   f ′ (r )  =   ∇ (r ) ⋅ r + 3 2   r r    

[∵∇⋅ r = 3]

Vector Calculus

2.25

 r f ′′ (r ) − f ′ (r )    1 3 f ′ (r )  =   r ⋅r + 2   r r    r  ∵∇ (r ) = 

∂r

1 

∑ ∂x i = ∑ r i = r r  x

  r f ′′ (r ) − f ′ (r )   1 2 3 f ′ (r )  =   (r ) + 2   r r r     2 = f ′′ (r ) + f ′ (r ) r Example 2.13 Find f (r) if the vector f (r) r is both solenoidal and irrotational. f (r) r is solenoidal ∴

∇⋅ { f (r ) r } = 0

i.e.

∇ f (r ) ⋅ r + f (r ) ∇⋅ r = 0 f ′ (r ) r ⋅ r + 3 f (r ) = 0 r rf ′ (r ) + 3f (r) = 0

i.e. i.e. i.e.

[by expansion formula] [Refer to the previous problem]

f ′ (r ) 3 + =0 f (r ) r

Integrating both sides w.r.t. r, log f (r) + 3 log r = log c i.e. log r3 f (r) = log c c ∴ f (r )= 3 r f (r) r is also irrotational

∴ i.e.

(1)

∇×{ f (r ) r } = 0 ∇ f (r )×r + f (r ) ∇×r = 0 [by expansion formula]

 i   ′ f (r ) ∂ i.e. (r ×r )+ 0 = 0  ∵∇×r = r ∂ x   x  f ′(r ) i.e. ( 0) + 0 = 0 r This is true for all values of f (r)

j ∂ ∂y y

   ∂ = 0 ∂z   z  k

From (1) and (2), we get that f (r) r is both solenoidal and irrotational if f (r) =

(2) c

r3 If f is a scalar point function, prove that ∇f is both solenoidal and Example 2.14 irrotational, provided f is a solution of Laplace equation. ∇ ⋅ ∇f = 0, only when ∇2f = 0. i.e, ∇f is solenoidal, only when ∇2f = 0 (1)

Mathematics II

2.26

∇ × ∇f = 0, always [by expansion formula] i.e. ∇f is irrotational always (2) From (1) and (2), ∇f is both solenoidal and irrotational, when ∇2f = 0, i.e. when f is a solution of Laplace equation. Example 2.15 If F is solenoidal, prove that curl curl curl curl F =∇4 F . Since F is solenoidal, ∇⋅ F = 0 (1) By the expansion formula, Curl curl F =∇ (∇⋅ F ) −∇2 F (2) =−∇2 F [by (1)] ∴ Curl curl curl curl F

= curl curl ( −∇2 F ) = − curl curl (∇2 F ) =− [∇{∇⋅∇2 F }−∇2 (∇2 F )] , by using (2) =− [∇{∇2 (∇⋅ F )}−∇4 F ] ,

by interchanging the operations ∇ ⋅ and ∇2

= ∇4 F {by using (1)}

EXERCISE 2(b)

Part A (Short Answer Questions) 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15.

Define divergence and curl of a vector point function. Give the physical meaning of ∇⋅ F Give the physical meaning of ∇×F When is a vector said to be (i) solenoidal, (ii) irrotational ? Prove that the curl of any vector point function is solenoidal. Prove that the gradient of any scalar point function is irrotational. If r is the position vector of the point (x, y, z) w.r.t. the origin, find div r and curl r . If F = 3 xyz 2 i + 2 xy 3 j − x 2 yz k , find ∇⋅ F at the point (1,−1, 1). If F = (x 2 + yz ) i + (y 2 + 2 zx) j + (z 2 + 3 xy ) k , find ∇×F at the point (2,−1, 2). If F = (x + y +1) i + j − (x + y ) k , show that F is perpendicular to curl F . If F = zi + x j + yk , prove that curl curl F =0 . Show that F = (x + 2 y ) i + (y + 3 z ) j + (x − 2 z )k is solenoidal. Show that F = (sin y + z ) i + (x cos y − z ) j + (x − y )k is irrotational. Find the value of l, so that F = λy 4 z 2 i + 4 x3 z 2 j + 5 x 2 y 2 k may be solenoidal. Find the value of l, if F = ( 2 x − 5 y ) i + (x + λy ) j + (3 x − z )k is solenoidal.

Vector Calculus

2.27

3 2 2 16. Find the value of a, if F = (a x y − z ) i + (a − 2)x j + (1 − a )xz k is irrotational. 17. Find the values of a, b, c, so that the vector F = (x + y + a z ) i + (bx + 2 y − z ) j + ( − x + cy + 2 z )k may be irrotational. 18. If ū and v are irrotational, prove that (u ×v ) is solenoidal. 19. If f1 and f2 are scalar point functions, prove that ∇ × (f1 ∇f2) = − ∇ × (f2∇f1). 20. If ∇f is a solenoidal vector, prove that f is a solution of Laplace equation.

Part B 21. If u = x2yz and v = xy − 3z2, find (i) ∇ ⋅ (∇u × ∇v) and ∇ × (∇u × ∇v) at the point (1, 1, 0). 22. Find the directional drivative of ∇ ⋅ (∇f) at the point (1, −2, 1) in the direction of the normal to the surface xy2 z = 3x + z2 , where f = x2 y2 z2 23. If F = 3 x 2 i + 5 xy 2 j + x y z 3 k , find ∇⋅ F , ∇ (∇⋅ F ), ∇× F , ∇⋅ (∇× F ) and ∇×(∇× F ) at the point (1, 2, 3) 24. If ā is a constant vector and r is the position vector of (x, y, z) w.r.t. the origin, prove that ∇×[(a ⋅ r ) r ] = a ×r . 25. Prove that F = 3 yzi + 2 zx j + 4 xy k is not irrotational, but (x2 y z3) F is irrotational. Find also its scalar potential. 26. Show that F = (z 2 + 2 x + 3 y ) i + (3 x + 2 y + z ) j + (y + 2 z x) k is irrotational, but not solenoidal. Find also its scalar potential. 27. Find the constants a, b, c, so that F = (x + 2 y + az ) i + (bx − 3 y − z ) j + (4 x + cy + 2 z ) k may be irrotational. For these values of a, b, c, find its scalar potential also. 28. Find the smallest positive integral values of a, b, c, if F = axyz 3 i + bx 2 z 3 j +cx 2 yz 2 k is irrotational. For these values of a, b, c, find its scalar potential also. 29. If r is the position vector of the point (x, y, z) w.r.t. the origin, prove that   1  2 1  2 (i) ∇⋅ r  = and (ii) ∇ ∇⋅ r  = − 3  r  r   r r   30. Find the value of n, if rn r is both solenoidal and irrotational, when r = xi + y j + zk .

2.4

LINE INTEGRAL OF VECTOR POINT FUNCTIONS

Let F (x, y, z) be a vector point function defined at all points in some region of space and let C be a curve in that region (Fig. 2.2). Let the position vectors of two neighbouring points P and Q on C be r and r +∆ r .Then PQ =∆ r If F acts at P in a direction that makes an angle θ with PQ ,

Q

F θ P

C ∆r r + ∆r

r Fig. 2.2

Mathematics II

2.28

then F ⋅ ∆r = F (∆r ) cos θ In the limit, F ⋅ dr = F dr cos θ . Note Physically F ⋅ dr means the elemental work done by the force F through the displacement dr . Now the integral

∫ F ⋅ dr

is defined as the line integral of F along the curve C.

C

Since

∫ F ⋅ dr = ∫ F cos θ dr , it is also called the line integral of the tangential C

C

component of F along C. B

Note

∫ F ⋅ dr

(1)

depends not only on the curve C but also on the terminal

A (C )

points A and B. B

∫ F ⋅ dr

(2) Physically

denotes the total work done by the force F in

A (C )

displacing a particle form A to B along the curve C. B

∫ F ⋅ dr

(3) If the value of

does not depend on the curve C, but only

A

on the terminal points A and B, F is called a Conservative vector. Similarly, if the work done by a force F in displacing a particle from A to B does not depend on the curve along which the particle gets displaced but only on A and B, the force F is called a Conservative force. (4) If the path of integration C is a closed curve, the line integral is denoted as ∫  F ⋅ dr . C

(5) When F = F i + F j + F k , 1 2 3

∫ F ⋅ dr = ∫ (F

1

C

i + F2 j + F3 k ) ⋅ (dx i + dy j + dz k )

C

( ∵ r = xi + yj + zk ) = ∫ (F1dx + F2 dy + F3 dz ) , which is evaluated as in the probC lems in Chapter 5 of Part I. (6)

∫ φdr , where f is a scalar point function and ∫ F × dr C

line integrals.

2.4.1

Condition for F to be Conservative

If F is an irrotational vector, it is conservative.

C

are also

Vector Calculus Proof:

2.29

Since F is irrotational, it can be expressed as ∇f. i.e., F = ∇φ B

∫

B

F ⋅ dr = ∫ ∇φ ⋅ dr

A

A

 ∂φ ∂φ ∂φ = ∫  i + j+ ∂y ∂z A  ∂x B

 k  ⋅ (dx i + dy j + dz k ) 

B  ∂φ ∂φ ∂φ  = ∫  dx + dy + dz  ∂y ∂z  A  ∂x

B

= ∫ dφ A

= [φ]BA , whatever be the path of integration = φ(B) − φ(A) ∴ F is conservative. Note If F is irrotational (and hence conservative) and C is a closed curve, then ∫ F ⋅ dr = 0 . C

[∵ φ (A) = φ (B ) , as A and B coincide]

2.4.2

Surface Integral of Vector Point Function

Let S be a two sided surface, one side of which is considered arbitrarily as the positive side. Let F be a vector point function defined at all points of S. Let dS be the typical elemental surface area in S surrounding the point (x, y, z). Let n be the unit vector normal to the surface S at (x, y, z) drawn in the positive side (or outward direction) Let θ be the angle between F and n . ∴ The normal component of F = F ⋅ n = F cos θ

nˆ

F θ

ds

(x, y, z)

S

Fig. 2.3

The integral of this normal component through the elemental surface area dS over the surface S is called the surface integral of F over S and denoted as  F cos θ dS or ∫ F ⋅ n dS .

∫

S

S

If dS is a vector whose magnitude is dS and whose direction is that of n , then dS = n dS . ∴

∫ F ⋅ n dS S

can also be written as

∫ F ⋅ dS . S

Mathematics II

2.30

Note

(1) If S is a closed surface, the outer surface is usually chosen as the positive side. (2) φ dS and ∫ F × dS , where f is a scalar point function are also

∫

S

S

surface integrals. (3) When evaluating

∫ F ⋅ n dS , the surface integral is first expressed in S

the scalar form and then evaluated as in problems in Chapter 5 of part I. To evaluate a surface integral in the scalar form, we convert it into a double integral and then evaluate. Hence the surface integral ∫ F ⋅ dS is also denoted as

∫∫ F ⋅ dS .

S

S

WORKED EXAMPLE 2(c)

Example 2.1 Evaluate

∫ φ dr , where C is the curve x = t, y = t , z = (1 − t) and 2

C

f = x2 y (1 + z) from t = 0 to t = 1. r = xi + yj + zk dr = dx i + dy j + dz k

∴

Hence the given line integral I = ∫ x 2 y (1 + z ) (dx i + dy j + dz k ) C

= i ∫ x y (1 + z ) dx + j ∫ x 2 y (1 + z ) dy + k ∫ x 2 y (1 + z ) dz 2

C

C

1

C

1

1

= i ∫ t 4 (2 − t ) dt + j ∫ t 4 (2 − t ) 2t dt + k ∫ t 4 (2 − t ) ( − dt ) 0

0

0

 t6  t5 t6   t5 t  t  = i  2 −  + j  4 − 2  + k −2 +   6   5 6 7  0 5 6  0    0 7 8 7 = i + j − k. 30 21 30 6 1

Example 2.2

1

7 1

If F = xyi − zj + x 2 k , evaluate

∫ F × dr , where C is the curve C

x = t2, y = 2t, z = t3 from (0, 0, 0) to (1, 2, 1). i j k F × dr = xy −z x 2 dx dy d z = − (z dz + x 2 dy ) i − (xy dz − x 2 dx) j + (xy dy + zd x) k

Vector Calculus

2.31

∴ The given line integral = ∫ [ − (zdz + x 2 dy ) i − (xy dz − x 2 dx) j + (xy dy + zdx) k ] C 1

= ∫ [ − (t 3 ⋅ 3t 2 ⋅+ t 4 ⋅ 2)) dt i − (2t 3 ⋅ 3t 2 − t 4 ⋅ 2t ) dt j + (2t 3 ⋅ 2 + t 3 2 t ) dt k ] 0

[ (0, 0, 0, ) corresponds to t = 0 and (1, 2, 1) corresponds to t = 1] 1

1

1

= −i ∫ (3t 5 + 2t 4 ) dt − j ∫ (6t 5 − 2t 5 ) dt + k ∫ (4t 3 + 2t 4 ) dt 0

0 5 1

0 6 1

5 1

  t   t t  t  = −i 3 + 2  − j 4  + k t 4 + 2      6   6 5 0 5 0 0 9 2 7 =− i − j + k 10 3 5 6

Example 2.3 Find the work done when a force F = (x 2 − y 2 + x) i − (2xy + y ) j displaces a particle in the xy-plane from (0, 0) to (1, 1) along the curve (i) y = x, (ii) y2 = x. Comment on the answer. W = Work done by F = ∫ F ⋅ dr C

= ∫ [(x − y + x) i − (2 xy + y ) j ] ⋅ (dx i + dy j + dz k ) 2

2

C

= ∫ [(x 2 − y 2 + x) dx − (2 xy + y ) dy ] C

Case (i) C is the line y = x.

∴

∫

W1 =

[(x 2 − y 2 + x) dx − (2 xy + y ) dy ]

y=x (dy = dx ) 1

= ∫ ( − 2 x 2 ) dx 0

2 3 Case (ii) C is the curve y2 = x. =−

∴

∫

W2 =

[(x 2 − y 2 + x) dx − (2 xy + y ) dy ] 2

x= y (dx = 2 y dy ) 1

= ∫ (2 y 5 − 2 y 3 − y ) dy 0

2 3 Comment As the works done by the force, when it moves the particle along two different paths from (0, 0) to (1, 1), are equal, the force may be a conservative force. In fact, F is a conservative force, as F is irrotational. =−

Mathematics II

2.32

It can be verified that the work done by F when it moves the particle from (0, 0) 2 to (1, 1) along any other path (such as x2 = y) is also equal to − . 3 Example 2.4

Find the work done by the force F = zi + xj + yk , when it moves

a particle along the arc of the curve r = cos ti + sin t j + tk from t = 0 to t = 2p. From the vector equation of the curve C, we get the parametric equations of the curve as x = cos t, y = sin t, z = t. Work done by

F = ∫ F ⋅ dr C

= ∫ (zi + xj + yk ) ⋅ (dx i + dy j + dz k ) C

= ∫ (z dx + x dy + y dz ) C 2π

= ∫ t ( − sin t ) + cos 2 t + sin t  dt 0   1 sin 2t  = t cos t − sin t + t +  − cos t       2 2

2π

0

= (2π + π − 1) − ( −1) = 3π

Example 2.5

Evaluate

∫ F ⋅ dr , where

F = (sin y ) i + x (1 + cos y ) j + zk and

C

C is the circle x2 + y2 = a2 in the xy-plane. Given integral = ∫  (sin y ) i + x(1 + cos y ) j + zk  ⋅ (dx i + dy j + dz k ) C

=

[sin y dx + x(1 + cos y ) dy + z dz ]

∫

 x 2 + y 2 = a 2      z=0 

=

[sin y dx + x(1 + cos y ) dy ]

∫

x2 + y2 = a2

Since C is a closed curve, we use the parametric equations of C, namely x = a cos θ, y = a sin θ and the parameter θ as the variable of integration. To move around the circle C once completely, θ varies from 0 to 2 π.

∫ [(sin y dx + x cos y dy) + x dy ] = ∫ [d(x sin y ) x dy ]

Now, given integral =

2π

= ∫ {d [ a cos θ ⋅ sin (a sin θ )] + a 2 cos θ dθ } 0

Vector Calculus

2.33

2π  a 2  sin 2θ   =  a cos θ ⋅ sin (a sin θ ) + θ +  2  2  0 

= πa 2 Example 2.6 Find the work done by the force F = (2 xy + z 3 ) i + x 2 j + 3 xz 2 k , when it moves a particle from (1, −2, 1) to (3, 1, 4) along any path. To evaluate the work done by a force, the equation of the path and the terminal points must be given. As the equation of the path is not given in this problem, we guess that the given force F is conservative. Let us verify whether F is conservative, i.e. irrotational. i ∂ ∂x

j ∂ ∂y

k ∂ ∂z

2 xy + z 3

x2

3 xz 2

∇× F =

= (0 − 0) i − (3 z 2 − 3 z 2 ) j + (2 x − 2 x) k =0 ∴ F is irrotational and hence conservative. ∴ Work done by F depends only on the terminal points. Since F is irrotational, let F =∇φ . It is easily found that φ = x2 y + z3 x + c. (3 ,1,4 )

Work done by

F=

∫

F ⋅ dr

(1,−2 ,1) (3,1, 4)

=

∫

∇φ ⋅ dr

(1, −2, 1) ( 3, 1, 4 )

=

∫

dφ

(1, −2, 1)

= [φ (x, y, z ) ](1, −2, 1) (3, 1, 4 )

( 3, 1, 4) =  x 2 y + z 2 x + c    (1, −2, 1)

= (201 + c) − ( −1 + c) = 202 . Example 2.7

Find the work done by the force F = y (3 x 2 y − z 2 ) i + x(2 x 2 y − z 2 )

j −2 x y zk , when it moves a particle around a closed curve C. To evaluate the work done by a force, the equation of the path C and the terminal points must be given.

Mathematics II

2.34

Since C is a closed curve and the particle moves around this curve once completely, any point (x0, y0, z0 ) can be taken as the initial as well as the final point. But the equation of C is not given. Hence we guess that the given force F is conservative, i.e. irrotational. Actually it is so, as verified below. i j k ∇× F =

∂ ∂x 2 2 3 y x − yz 2

∂ ∂ ∂y ∂z 3 2 2 x y − z x −2 xyz

= ( − 2 xz + 2 xz ) i − ( − 2 yz + 2 yz ) j + (6 x 2 y − z 2 − 6 x 2 y + z 2 ) k =0 Since F is irrotational, let F = ∇φ . ∴ Work done by F = ∫  F ⋅ dr C

=∫  ∇φ ⋅ dr C

(x0 , y0 , z0 )

= ∫ dφ (x0 , y0 , z0 )

= φ(x0 , y0 , z0 ) − φ (x0 , y0 , z0 ) =0 Example 2.8

Evaluate

∫∫ A ⋅ dS , where

A = 12 x 2 yi − 3 yz j + 2 zk and S is

S

the portion of the plane x + y + z = 1 included in the first octant (Fig. 2.4). Given integral I = A⋅ n dS, where n is the unit normal to the surface S given by

∫∫

z

S

f = c, i.e. x + y + z = 1 ∴ φ = x+ y+z

C

∇φ = i + j + k 1  n= (i + j + k ) 3

B

O A Fig. 2.4

∴

I = ∫∫ (12 x 2 yi − 3 yzj + 2 zk ) ⋅

1

S

=

1

∫∫ (12 x 3 S

2

y − 3 yz + 2 z ) dS

3

( i + j + k ) dS

y x

Vector Calculus

2.35

To convert the surface integral as a double integral, we project the surface S on the xoy - plane. Then dS cos γ= dA, where γ is the angle between the surface S and the xoy - plane, i.e. the angle between n and k . ∴cos γ =  n⋅k ∴

dS =

∴

I=

dA dx dy = n ⋅ k 1 3 1

dx dy 1 3 [ the projection of S on the xoy plane is ΔOAB]

∫∫ (12 x 3 ∆OAB

2

y − 3 yz + 2 z )

= ∫∫ {12 x 2 y − 3 y (1 − x − y ) + 2(1 − x − y )} dx dy ∆OAB

Note

To express the integrand as a function of x and y only, z is expressed as a function of x and y from the equation of S. 1 1−y

I=∫

∫

0

0

(12 x 2 y + 3 xy + 3 y 2 − 5 y − 2 x + 2) dx dy

  3y = ∫  4 y (1 − y )3 + (1 − y ) 2 + 3 y 2 (1 −y ) − 5 y (1 − y ) − (1 − y ) 2 + 2(1 − y ) dy   2 0  49 = 120 1

Example 2.9

Evaluate

∫∫ F ⋅ dS , where F = yzi

+ zxj + xyk and S is the part

S

of the sphere x2 + y2 + z2 = 1 that lies in the first octant. F ⋅ n dS , where n is the unit normal to the surface S given by Given integral I =

∫∫ S

f = c i.e. x2 + y2 + z2 = 1. f = x2 + y2 + z2 ∴

∇φ = 2 xi + 2 yj + 2 zk

∴

2 xi + 2 yj + 2 zk n = 4(x 2 + y 2 + z 2 ) =

2(xi + yj + zk ) 4 ×1

[ the point (x, y, z) lies on S]

= xi + yj + zk . ∴

I = ∫∫ (yzi + zxj + xyk ) ⋅ (xi + yj + zk ) dS S

Mathematics II

2.36 = ∫∫ 3 xyz dS

z

S

= ∫∫ 3 xyz R

dx dy , n ⋅ k

where R is the region in the xy-plane bounded by the circle x2 + y2 = 1 and lying in the first quadrant.

B

O

y

A

dx dy I = ∫∫ 3 xyz z R 1

=∫ 0

x

1 − y2

Fig. 2.5

∫ 3 xy dx dy 0

 x2  = ∫ 3 y    2  0

1 − y2

1

dy 0

=

3 1 2 ∫ y (1 − y ) dy 2 0

=

y4  3  y 2  −  2 2 4 0

=

3 8

1

Example 2.10

Evaluate

∫∫ F ⋅ dS if ,

F = yzi + 2 y 2 j + xz 2 k and S is the sur-

S

face of the cylinder x2 + y2 = 9 contained in the first octant between the planes z = 0 and z = 2. Given integral I = ∫∫ F ⋅ n dS , where n is the unit normal to the surface S given by S

f = c, i.e. x2 + y2 = 9

∴

f = x2 + y2 ∇φ = 2 xi + 2 yj

∴

2 xi + 2 yj n = 4(x 2 + y 2 ) =

=

2(xi + yj ) 4×9 1 (xi +yj ) 3

[∴ the point (x, y, z) lies on S]

Vector Calculus

2.37

1 ∴ I = ∫∫ (yzi + 2 y 2 j + xz 2 k ) ⋅ (xi + yj ) dS 3 s =

1 (xyz + 2 y 3 ) dS 3 ∫∫ s

=

1 dx dz (xyz + 2 y 3 ) ∫∫ n ⋅ j 3 R

z C

B O

where R is the rectangular region OABC in the xoz-plane, got by projecting the cylindrical surface S on the xoz-plane (Fig. 2.6).

y

A x Fig. 2.6

1 dx dz I = ∫∫ (xyz + 2 y 3 ) y 3 R 3 = ∫∫ (xz + 2 y 2 ) dx dz R 2

=∫ 0

3

∫ [xz + 2(9 − x

2

)] dx dz

0

2 9  = ∫  z + 18×3 − 2×9 dz  2  0

9 z2  =  ⋅ + 36 ⋅ z  = 81   2 2 2

0

EXERCISE 2(c) Part A (Short Answer Questions) 1. Define line integral of a vector point function. 2. When is a force said to be conservative? State also the condition to be satisfied by a conservative force. 3. If F is irrotational, prove that it is conservative. 4. Define surface integral of a vector point function. 5. Explain how ∫∫ F ⋅ dS is evaluated. S

6. Evaluate (2, 2).

∫ r ⋅ dr , where C is the line y = x in the xy-plane from (1, 1) to C

7. Find the work done by the force F = xi + 2 yj when it moves a particle on the curve 2y = x2 from (0, 0) to (2, 2). 8. Prove that the force F = (2 x + yz ) i + (xz − 3)j + xyk is conservative.

Mathematics II

2.38 9. Evaluate

∫∫ (yzi + zxj + xyk ) ⋅ dS , where S is the region bounded by x = 0, S

x = a, y = 0, y = b and lying in the xoy-plane. 10. Find the work done by a conservative force when it moves a particle around a closed curve. Part B 11. Evaluate

∫ φ dr , where f = 2xyz

2

and C is the curve given by x = t2, y = 2t,

C

z = t3 from t = 0 to t = 1. 12. Evaluate ∫ F ×dr along the curve x = cos t, y = 2 sin t, z = cos t from t = 0 C

π to t = , given that F = 2 xi + yj + zk . 2 13. Evaluate

∫ F ×dr

along the curve x = cos θ, y = sin θ, z = 2 cos θ from θ = 0

C

to θ =

π , given that F = 2 yi − zj + xk 2

14. Evaluate

∫ F ⋅ dr

along the curve x = t2, y = 2t, z = t3 from t = 0 to t = 1,

C

given that F = xyi − zj + x 2 k . 15. Find the work done by the force F = 3 xyi − y 2 j , when it moves a particle along the curve y = 2x2 in the xy-plane from (0, 0) to (1, 2). 16. Find the work done by the force F = (y + 3 z ) i + (2 z + x)j + (3 x + 2 y )k , when it moves a particle along the curve x = a cos t, y = a sin t, z =

2at π

between the points (a, 0, 0) and (0, a, a). 17. Find the work done by the force F = (2 y + 3) i + xzj + (yz − x)k when it moves a particle along the line segment joining the origin and the point (2, 1, 1). 18. Evaluate

∫ F ⋅ dr , where C is the circle x

2

+ y2 = 4 in the xy-plane, if

C

F = (2 x − y − z ) i + (x + y − z 2 )j + (3 x − 2 y + 4 z )k . 19. Find the work done by the force F = (x 2 + y 2 ) i + (x 2 + z 2 )j + yk , when it moves a particle along the upper half of the circle x2 + y2 = 1 from the point (−1, 0) to the point (1, 0). 20. Find the work done by the force F = (e x z − 2 xy ) i + (1− x 2 )j + (e x + z )k , when it moves a particle from (0, 1, −1) to (2, 3, 0) along any path. 2 3 21. Find the work done by the force F = (y cos x + z ) i + (2 y sin x − 4)j + π  (3 xz 2 + 2)k ,when it moves a particle from (0, 1, −1) to  , −1, 2 along any  2  path.

Vector Calculus

2.39

22. Find the work done by the force F = y 2 i + 2(xy + z ) j + 2 yk , when it moves a particle around a closed curve C. 23. Evaluate

∫∫ F ⋅ dS , where

F = xyi − x 2 j + (x + z )k and S is the part of

S

the plane 2x + 2y + z = 6 included in the first octant. 24. Evaluate

∫∫ F ⋅ dS , where F = yi − xj + 4k

and S is the part of the sphere

S 2

x2 + y2 + z = a2 that lies in the first octant. 25. Evaluate

∫∫ F ⋅ dS , where

F = zi + xj −3 y 2 zk and S is the surface of the

S

cylinder x2 + y2 = 16 included in the first octant between the planes z = 0 and z = 5.

2.5

INTEGRAL THEOREMS

The following three theorems in Vector Calculus are of importance from theoretical and practical considerations: 1. Green’s theorem in a plane 2. Stoke’s theorem 3. Gauss Divergence theorem Green’s theorem in a plane provides a relationship between a line integral and a double integral. Stoke’s theorem, which is a generalisation of Green’s theorem, provides a relationship between a line integral and a surface integral. In fact, Green’s theorem can be deduced as a particular case of Stoke’s theorem. Gauss Divergence theorem provides a relationship between a surface integral and a volume integral. We shall give the statements of the above theorems (without proof) below and apply them to solve problems:

2.5.1

Green’s Theorem in a Plane

If C is a simple closed curve enclosing a region R in the xy-plane and P(x, y), Q (x, y) and its first order partial derivatives are continuous in R, then  ∂Q ∂P  ∫ (P dx + Q dy) = ∫∫  ∂ x − ∂ y  d x dy C R where C is described in the anticlockwise direction.

2.5.2

Stoke’s Theorem

If S is an open two sided surface bounded by a simple closed curve C and if F is a vector point function with continuous first order partial derivatives on S, then

∫ F ⋅ dr = ∫∫ curl F ⋅ dS , C

S

where C is described in the anticlockwise direction as seen from the positive tip of the outward drawn normal at any point of the surface S.

Mathematics II

2.40

2.5.3

Gauss Divergence Theorem

If S is a closed surface enclosing a region of space with volume V and if F is a vector point function with continuous first order partial derivatives in V, then

∫∫ F ⋅ dS = ∫∫∫ (div F ) dv . S

2.5.4

V

Deduction of Green’s Theorem from Stoke’s Theorem

∫ F ⋅ dr = ∫∫ curl F ⋅ dS

Stoke’s theorem is

C

(1)

S

Take S to be a plane surface (region) R in the xy-plane bounded by a simple closed curve C. take F = P(x, y ) i + Q(x, y ) j

Also

Then curl

F=

i

j

k

∂ ∂x P

∂ ∂y Q

∂ ∂z 0

 ∂Q ∂P  k =  − ∂y   ∂x

[ P and Q are functions of x and y]

Inserting all these in (1), we get

∫ (Pi C

i.e.

 ∂Q ∂P   k ⋅ dS + Q j ) ⋅ (dxi + dyj + dzk ) = ∫∫  −  ∂ x ∂ y  R (P dx + Q dy ) = (Q − P ) k ⋅  n dS

∫

∫∫

C

R

x

y

(2)

Now n is the unit vector in the outward drawn normal direction to the surface R. Normal at any point of R, that lies in the xy-plane, is parallel to the z-axis. Taking the positive direction of the z-axis as the positive (outward) direction of n , we get n = k . Using this in (2), we get

∫ (P dx + Q dy) = ∫∫ (Q

x

C

− Py ) dx dy

R

[ dS = elemental plane surface area in the xy-plane = dx dy]

Note

If the surface S is a plane surface, problems in Stoke’s theorem will reduce to problems in Green’s theorem in a plane.

2.5.5

Scalar form of Stoke’s Theorem

Take F = Pi + Q j + Rk in Stoke’s theorem, where P, Q, R are functions of x, y, z.

Vector Calculus

2.41

F ⋅ dr = P d x + Q dy + R dz

Then

(1)

Curl F = (Ry − Qz ) i + (Pz − Rx ) j + (Qx − Py ) k ∴

Curl F ⋅ dS = curl F ⋅  n dS = (Ry − Qz )( n ⋅ i )dS + (Pz − Rx )( n ⋅ j )dS + (Qx − Py )( n ⋅ k )dS = (Ry − Qz ) dy dz + (Pz − Rx ) dz dx + (Qx − Py ) dx dy

(2)

[ dS cos γ = dx dy i.e. dS (n ⋅ k ) = dx dy] Inserting (1) and (2) in Stoke’s theorem, it reduces to the scalar form

∫ (P dx + Q dy + R dz ) = ∫∫ [(R C

y

− Qz ) dy dz + (Pz − Rx ) dz dx + (Qx − Py ) dx dy ]

S

(3)

Note

If we take P and Q as functions of x and y only and R = 0 in (3), we get Green’s theorem in a plane.

2.5.6

Scalar form of Gauss Divergence Theorem

Take F = Pi + Qj + Rk in Divergence theorem, where P, Q, R are functions of x, y, z. ∂P ∂Q ∂R + + ∂x ∂ y ∂z F ⋅ dS = F ⋅ n dS = P (n ⋅ i ) dS + Q (n ⋅ j ) dS + R (n ⋅ k ) dS = P dy dz + Q dz dx + R dx dy F=

Then div

(1)

(2)

Inserting (1) and (2) in Divergence theorem, we get

∫∫ (Pdy dz + Qdz dx + Rdx dy) = ∫∫∫ (P

x

S

+ Qy + Rz ) dx dy dz

(3)

V

which is the scalar form of Divergence theorem.

Note 2.5.7

(3) is also called Green’s theorem in space.

Green’s Identities

In Divergence theorem, take F = zd}, where f and y are scalar point functions. Then div (f ∇ ψ) = ∇⋅ (f ∇ ψ) = f ∇2 ψ + ∇f ⋅ ∇y ∴ Divergence theorem becomes

∫∫ φ∇ψ ⋅ dS = ∫∫∫ (φ∇ ψ + ∇φ ⋅ ∇ψ) dV 2

S

V

(1)

Mathematics II

2.42 Interchanging f and y,

## }dz $ dS = ### (}d z + dz $ d})d V 2

S

(2)

V

(l) - (2) gives,

∫∫ (φ∇ψ − ψ∇φ) ⋅ dS = ∫∫∫ (φ∇ ψ − ψ∇ φ) dV 2

S

Note

2

(3)

V

(1) or (2) is called Green’s first identity and (3) is called Green’s second

identity. WORKED EXAMPLE 2(d)

∫ (x

Example 2.1 Verify Green’s theorem in a plane with respect to

2

− y 2 ) dx +

C

2xy dy, where C is the boundary of the rectangle in the xoy-plane bounded by the lines x = 0, x = a, y = 0 and y = b. (OR) Verify Stoke’s theorem for a vector field defined by F = (x − y ) i + 2 xy j in the rectangular region in the xoy-plane bounded by the lines x = 0, x = a, y = 0 and y = b. 2

2

∫ F ⋅ dr = ∫∫ curl F ⋅ dS

Stoke’s theorem is

C

S

i ∂ curl F = ∂x x −y

Now

2

2

j ∂ ∂y 2 xy

k ∂ ∂z 0

= (2 y + 2 y ) k ∴ We have to verify that

∫ [(x

2

− y 2 ) i + 2 xy j ] ⋅ (dxi + dy j + dzk ) = ∫∫ 4 yk ⋅  n dS

C

i.e.

S

∫ [(x

2

− y 2 ) dx + 2 xy dy ] = ∫∫ 4 y dx dy [Fig. 2. 7]

C

R

(1) is also the result for the given function as per Green’s theorem. y

y=b

D

B

x=0

x=a

O

y=0 Fig. 2.7

A

x

(1) ∵ n = k   

Vector Calculus

∫

L.S. of (1) =

(

∫

+

OA y =0 dy =0

)

(

∫

+

AB x= a d x =0

(

)

2.43

∫

+

BD y =b dy =0

)

(

[(x 2 − y 2 ) dx + 2 x y dy ]

DO x=0 d x=0

)

[∵ the boundary C consists of 4 lines] a

=

b

0

∫ x d x + ∫ 2ay d y +∫ (x 2

0

0

2

− b2 ) d x + 0

a

Note

To simplify the line integral along each line, we make use of the equation of the line and the corresponding value of dx or dy  x3  a  x3  a =   + a (y 2 ) b0 −  − b 2 x   3   3 0 0 = 2 ab 2 b

R.S. of (1) =

∫ ∫ 4 y dx d y 0 b

=

a

0

∫ 4 y (x)

a 0

dy = 2a (y 2 )

b 0

0

= 2 a b2 . Since L.S. of (1) = R.S. of (1), Stoke’s theorem (Green’s theorem) is verified. Example 2.2 Verify Green’s theorem in a plane for

∫ [(3x

2

− 8y 2 ) d x + (4y − 6 xy )

C

dy ] , where C is the boundary of the region defined by the lines x = 0, y = 0 and x + y = 1. Green’s theorem is

2

x

C

∴ For the given integral,

∫ [3x

∫ (P dx + Q dy) = ∫∫ (Q

− Py ) dx dy

R

− 8y 2 ] d x + (4y − 6 x y ) dy ] =

∫∫ 10 y d x dy [Fig. 2.8]

C

R

y B x+y=1

x=0

y=0

O

A

x

Fig. 2.8

∫

L.S. of (1) =

(

+

OA y =0 dy = 0

)

∫

AB  x + y =1  x 1− y  =  d x =−dy

+

∫

BO

(dxx==00)

[(3x 2 − 8y 2 )d x + (4y − 6 x y ) dy ]

(1)

Mathematics II

2.44 1

=

1

∫ 3x d x + ∫ [{3(1− y) 0 1

=

0

2

2

− 8y }( − dy ) + {4y − 6 y (1− y )}dy ]+ 2

0 1

∫ 3x

2

d x+

1 1

∫ (11y

0

∫ 4 y dy

2

+ 4y − 3) dy −

0

11



∫ 4 y dy = 1+  3 + 2 − 3 − 2 0

5 = 3 1 1− y

R.S. of (1) =

∫ ∫ 10 y dx dy 0 1

=

0

∫ 10 y (1− y) dy 0

1  5 y 3  = 5 y 2 −10  =  3  3 0

Since L.S. of (1) = R.S. of (1), Green’s theorem is verified. Example 2.3 Use Stoke’s theorem to evaluate

∫ F ⋅ dr , where

F = (sin x − y )

C

π  i − cos xj and C is the boundary of the triangle whose vertices are (0, 0),  , 0  2   π  and  ,1 .  2 

Note

Evaluating

∫ F ⋅ dr

by using Stoke’s theorem means expressing the line

C

integral in terms of its equivalent surface integral and then evaluating the surface integral. By Stoke’s theorem, F ⋅ dr = curl F ⋅ dS , where S is any open two-sided

∫

∫∫

C

S

surface bounded by C. [Fig. 2.9] To simplify the work, we shall choose S as the plane surface R in the xoy-plane bounded by C. ∴

∫ F ⋅ dr = ∫∫ curl F ⋅ k dx dy [ ∵ For the xoy-plane, n = k and dS = d x dy] C

R

For this problem, i j ∂ ∂ curl F = ∂x ∂y (sin x − y ) −cos x = (sin x + 1) k

k ∂ ∂z 0

Vector Calculus ∴ Τhe given line integral =

2.45 y

∫∫ (1 + sin x) d x d y 1

=

π 2

y=

∫ ∫ (1 + sin x) d x dy 0 πy 2 1

=

∫ ( x − cos x) 0 1

=

∫ 0

π 2 πy 2

π ,1 2

B

R

2 x π

x= O

y=0

π 2

π ,0 2

A

dy

x

Fig. 2.9

π πy π y  + cos  dy  − 2 2 2 

1 π π y2 2 π y  =  y − + sin   2 4 π 2  0 π 2 = + . 4 π Example 2.4 Use Green’s theorem in a plane to evaluate

∫ [(2x − y) dx + C

(x + y ) dy ] , where C is the boundary of the circle x2 + y2 = a2 in the xoy-plane. By Green’s theorem in a plane,

∫ (P d x + Q dy) = ∫∫ (Q

x

C

∴

y

− Py ) d x dy

R

∫ [(2x − y) d x + (x + y) dy] = ∫∫ [1− ( −1)] d x dy C

x O

R

=2

∫∫ d x dy[Fig. 2.10] R

Fig. 2.10

= 2 × area of the region R = 2π a 2

Example 2.5 Verify Stoke’s theorem when F = (2 x y − x 2 ) i − (x 2 − y 2 ) j and C is the boundary of the region enclosed by the parabolas y2 = x and x2 = y. Stoke’s theorem is

∫ F ⋅ dr = ∫∫ curl F ⋅ n d S C

i ∂ Now curl F = ∂x 2 xy − x 2

S

j ∂ ∂y −x 2 + y 2

= ( − 2 x − 2 x)k = −4 xk

k ∂ ∂z 0

Mathematics II

2.46 ∴ Stoke’s theorem becomes

∫ [(2 x y − x C

2

) d x − (x 2 − y 2 ) dy ] = ∫∫ −4 x k ⋅ k dx dy R

= ∫∫ −4 x d x dy [Fig. 2.11]

(1)

R

B(1, I)

D

y = x2

A O x = y2

Fig. 2.11

∫

L.S. of (1) =

∫

+

OAB  y = x 2  dy = 2 x d x  

BDO  x = y 2  d x = 2 y dy   0

1

=

∫

[(2 x y − x 2 ) dx − (x 2 − y 2 ) dy ]

(2 x5 − x 2 ) d x +

0

∫ (3 y

4

− 2 y 5 + y 2 ) dy

1

[ the coordinates of B are found as (1, 1) by solving the equations y = x2 and x = y2] −3 = 5 1

R.S. of (1) =

y

∫ ∫ −4 x d x d y 0 y2 1

= −2

∫ (x ) 2

y y2

dy

0 1

= −2

∫ (y − y ) dy 4

0

3 =− 5 Since L.S. of (1) = R.S. of (1), Stoke’s theorem is verified. Example 2.6 Prove that the area bounded by a simple closed curve C is given by 1 (x dy − y d x) Hence find the area bounded (i) by the parabola y2 = 4a x and its 2 C x2 y 2 latus rectum and (ii) by the ellipse 2 + 2 = 1 . a b

∫

Vector Calculus By Green’s theorem,

∫ (P dx + Q dy) = ∫∫ (Q

x

C

Taking P = −

2.47 − Py )dx dy

R

y x and Q = , we get 2 2 1 (x dy − y dx) = ∫∫ 2 ∫C R

 1  1   − −  dx dy  2  2 

= ∫∫ dx dy R

=Area of the region R enclosed by C. (i) Area bounded by the parabola and its latus 1 rectum = ∫ (x dy − y dx) [Fig. 2.12] 2 C y

(a, 2a)

L

=

 1   + ( x d y − y d x ) ∫ ∫  2  LOL  ′ ′ L SL

x=a

         1 =  ∫ (x dy − y dx) + ∫ (x dy − y dx)   2  y2  x=a  x=   dx = 0  4 a     dx= y dy   2 a   =

S

O

L'

x

(a, –2a)

Fig. 2.12

2a −2 a  1  y2 d d − y + a y ∫  2  ∫ 4 a 2a −2 a  2a

=∫ 0

2a

y2 dyy + ∫ a dy 4a 0

8 = a2 . 3 y

(ii) Area bounded by the ellipse =

1 (x dy − y dx) 2 ∫C

=

1 (x dy − y dx) [Fig. 2.13] 2  x=∫ a cos θ     y =b sin θ 

θ=

θ=0 x θ = 2π

θ=π O

θ=

2π

=

1 ab (cos 2 θ + sin 2 θ ) dθ 2 ∫0

= πab.

π 2

Fig. 2.13

3π 2

Mathematics II

2.48

Example 2.7 Use Stoke’s theorem to prove that (i) curl (grad f) = 0 and (ii) div (curl F ) = 0. (i) Stoke’s theorem is

∫∫ curl F ⋅ dS = ∫ F ⋅ dr S

S2

S1

C

S3

Taking F = grad φ , we have

∫∫ curl (grad φ) ⋅ dS = ∫ grad φ ⋅ dr S

C

C

=∫  dφ

Fig. 2.14

C

=0 The above result is true for any open two-sided surface S, provided it is bounded by the same simple closed curve C. [Fig. 2.14] ∴ curl (grad φ) ⋅ dS = 0, [for any S and hence for any dS ] ∴ curl (grad φ) =0 (ii) Gauss divergence theorem is

∫∫∫ (div F ) dv = ∫∫ F ⋅ dS , where S is a closed surface enclosing a volume V. V

S

Replacing F by curl F , we have

∫∫∫ div (curl F ) dv = ∫∫ curl F ⋅ dS V

(1)

S

The surface integral in (1) appears to be the same as that in Stoke’s theorem. However Stoke’s theorem cannot be used to simplify the R.S. of (1), as S is a closed surface. In order to enable us to use Stoke’s theorem, we divide S into two parts S1 and S2 by a plane. S1 and S2 are open two-sided surfaces, each of which is bounded by the same S1 closed curve C as given in Fig. 2.15. Now (1) becomes C ∫∫∫ div (curl F ) dv = ∫∫ curl F ⋅ dS + ∫∫ curl F ⋅ dS V

S1

S2

=∫  F ⋅ dr + ∫ F ⋅ dr C

S2

C

[by Stoke’s theorem]

Note

Fig. 2.15

If C is described in the anticlockwise sense as seen from the positive tip of the outer normal to S1, it will be described in the clockwise sense as seen from the positive tip of the outer normal to S2. ∴

∫∫∫ div (curl F ) dv = ∫ F ⋅ dr − ∫ F ⋅ dr V

C

= 0. Since V is arbitrary, div (curl F ) = 0 .

C

Vector Calculus Example 2.8 Evaluate

2.49

∫ (sin z dx − cos x dy + sin y dz ) , by using Stoke’s theorem, C

where C is the boundary of the rectangle defined by 0 ≤ x ≤ π, 0 ≤ y ≤ 1, z = 3. The scalar form of Stoke’s theorem is

∫ (P dx + Q dy + R dz ) = ∫∫ [(R C

y

− Qz )

S

dy dz + (Pz − Rx ) dz dx + (Qx − Py ) dx dy ] Taking P = sin z, Q = −cos x, R = sin y, we get

∫ (sin z dx − cos x dy + sin y dz ) C

= ∫∫ (cos y dy dz + cos z dz dx + sin x dx dy ) S

= ∫∫ sin x dx dy

[  S is the rectangle in the z = 3 plane and hence dz = 0]

S 1

=∫ 0

π

∫ sin x dx dy 0

= −(cos x)π0

=2 Example 2.9 Evaluate

∫∫ (x dy dz + 2 y dz dx + 3z dx dy)

where S is the closed

S

surface of the sphere x2 + y2 + z2 = a2. Scalar from of divergence theorem is

∫∫ (P dy dz + Q dz dx + R dx dy) = ∫∫∫ (P + Q x

S

y

+ Rz ) dV

V

Taking P = x, Q = 2y, R = 3z, the given surface integral = ∫∫∫ 6 dv = 6V V

4 = 6× πa 3 3

= 8πa3. Example 2.10 Verify Stoke’s theorem for F = xyi − 2 yz j − zxk where S is the open surface of the rectangular parallelopiped formed by the planes x = 0, x = 1, y = 0, y = 2 and z = 3 above the xoy-plane. Stoke’s theorem is

∫ F ⋅ dr = ∫∫ curl F ⋅ dS C

Here

i j k ∂ ∂ ∂ curl F = ∂x ∂y ∂z xy −2 yz −xz

S

Mathematics II

2.50 = 2 yi + zj + xk ∴ Stoke’s theorem takes form

∫ (xy dx − 2 yz dy − zx dz ) = ∫∫ (2 yi + zj − xk ) ⋅ dS C

(1)

S

z C

A'

B'

O' ˆn = j

nˆ = – j O

y

B

A C'

x Fig. 2.16

The open cuboid S is made up of the five faces x = 0, x = 1, y = 0, y = 2 and z = 3 and is bounded by the rectangle OAC′ B lying on the xoy-plane. [Fig. 2.16] L.S. of (1)

= ∫

(xy dx − 2 yz dy − zx dz )

OAC ′B

=

∫

xy dx [Fig. 2.17]

∫

+

( the boundary C lies on the plane z = 0) y

OAC ′B

=

∫

AC ′  x =1  dx = 0

OA  y = 0  dy = 0

+

∫

C ′B  y = 2  dy = 0

+

∫

(xy dx)

C'

B

BO  x = 0  dx = 0

0

= ∫ 2 x dx

O

A

1

x

Fig. 2.17

= −1. R.S. of (1) =

## + ## + ## + ## + ## (2yir + zjr - xkr ) $ nt ds

`ntx==-0ij

1 ` xnt = = ij

y=0 ct m n = - rj

y=2 ct rm n=j

z=3 ct rm n=k

Note

nt is the unit normal at any point of the concerned surface. For example, at any point of the plane surface y = 0, the outward drawn normal is parallel to the y-axis, but opposite in direction. ∴ nt at any point of y = 0 is equal to − j . Similarly nt for y = 2 is equal to j and so on. Using the relevant value of nt and simplifying the integrand, we have R. S. of (1) = −∫∫ 2 y dS + ∫∫ 2 y dS −∫∫ z dS + ∫∫ z dS − ∫∫ x dS x=0

x =1

y =0

y=2

z =3

Vector Calculus

2.51

2

3

2

1

3

1

3

2

= −∫

∫

2 y dy d z + ∫

∫

2 y dy dz − ∫

∫

z dz dx + ∫

∫

z dz dx − ∫

0

0

0

0

0

0

0

0

0

3

1

∫ x dx dy 0

[ Elemental plane (surface) area dS on x = 0 and x = 1 are equal, each equal to dy dz etc.] 2

= −∫

1

∫ x dx d y

0

( the other integrals cancel themselves)

0 1

2  x2  = −∫   dy  2 0 0

= −1. Thus Stoke’s theorem is verified. Example 2.11 Verify Stoke´s theorem for F = y 2 zi + z 2 xj + x 2 yk where S is the open surface of the cube formed by the planes x = ±a, y = ±a and z = ±a, in which the plane z = −a is cut. Stoke’s theorem is

∫ F ⋅ dr = ∫∫ curl F ⋅ dS C

i ∂ curl F = ∂x y2 z

Here

j ∂ ∂y z2 x

S

k ∂ ∂z x2 y

= (x 2 − 2 zx) i + (y 2 − 2 xy )j + (z 2 − 2 yz )k ∴ Stoke’s theorem takes the form

∫ (y

2

z dx + z 2 x dy + x 2 y dz ) = ∫∫ [(x 2 − 2 zx) i + (y 2 − 2 xy )j + (z 2 − 2 yz )k ]⋅ dS

C

(1)

S

F

E

O

H

z 0 D

x

y C

A

B Fig. 2.18

The open cube S is bounded by the square ABCD that lies in the plane z = −a (Fig. 2.18) 2 2 2 L.S. of (1) = ∫ (y z dx + z x dy + x y dz ) (z =− a )

=

∫ ABCD

( − ay 2 dx + a 2 x dy )

[ dz = 0, as z = −a]

Mathematics II

2.52

y=a dy = 0

C

B

x=a dx = 0

x=–a dx = 0

D

A

y=–a dy = 0 Fig. 2.19

2 2 L.S. of (1) = ∫ + ∫ + ∫ + ∫ ( − ay dx + a x dy ) (Fig. 2.19) AB

BC

CD

DA −a

−a

a

a

= ∫ a dy − ∫ a dx − ∫ a dy − ∫ a 3 dx 3

3

−a

3

a

−a

a

=4a . 4

R.S. of (1) =

∫∫

∫∫ + ∫∫

+

 x = −a    n = − i 

 x = a     n = i 

∫∫ [(x

+

2

+

 y = −a    n = − j 

∫∫

 y = a     n = j 

− 2 zx) i + (y 2 − 2 xy ) j + (z 2 − 2 yz ) k ] ⋅ ndS

 z = a    n = k 

=

∫∫

x = −a

(2 zx − x 2 ) dS + ∫∫ (x 2 − 2 zx) dS + x=a

∫∫

(2 xy − y 2 ) dS

y = −a

+ ∫∫ (y 2 − 2 xy ) dS + ∫∫ (z 2 − 2 yz ) dS y=a

z=a

= ∫∫ ( − 2az − a 2 ) dy dz + ∫∫ (a 2 − 2az ) dy dz + ∫∫ ( − 2ax − a 2 ) dz dx + ∫∫ (a 2 − 2ax) dz dx +∫∫ (a 2 − 2ay ) dx dy a

=∫

a

a

[ using the equations of the planes to simplify the integrands] a

∫ − 4a z dy dz + ∫ ∫ − 4ax dz dx

−a −a

−a −a

a

+∫

a

∫ (a

2

− 2ay ) dx dy

−a −a

= 0 + 0 + 4a4 = 4a4. Thus Stokes’ theorem is verified.

Vector Calculus

2.53

2 Example 2.12 Verify Stokes’ theorem for F = − yi + 2 yz j + y k , where S is the upper half of the sphere x2 + y2 + z2 = a2 and C is the circular boundary on the xoy-plane.

j ∂ ∂y 2 yz

i ∂ Curl F = ∂x −y

k ∂ ∂z y2

=k. Stoke’s theorem is

∫ F ⋅ dr = ∫∫ F ⋅ dS C

Here

S

∫ ( − y d x + 2 yz dy + y

2

dz ) = ∫∫ k ⋅ dS

C

(1)

S

C is the circle in the xoy-plane whose equation is x2 + y2 = a2 and whose parametric equations are x = a cos θ and y = a sin θ. ∴

∫

L.S. of (1) =

− y dx

( C lies on z = 0)

x2 + y2 = a2

2π

=

∫a

2

sin 2 θ dθ

0

=

a2 2

  θ − sin 2θ   2 0

2π

= πa2. R.S. of (1) =

∫∫ k ⋅ n dS , where S

∇φ n = , where φ = x 2 + y 2 + z 2 ∇φ =

2 (xi + yj + zk ) 4 (x 2 + y 2 + z 2 )

xi + yj + zk a z R.S. of (1) = ∫∫ dS a S =

∴

= ∫∫ R

[ the point (x, y, z) lies on f = a2]

z dx dy , where R is the projection of S on the xoya  n ⋅ k plane.

= ∫∫ dx dy , where R is the region enclosed by x2 + y2 = a2. R

= πa2 Thus Stoke’s theorem is verified.

Mathematics II

2.54

Example 2.13 If S is the surface of the sphere x2 + y2 + z2 = 1, evaluate

∫∫ (xi

+ 2 yj + 3 zk ) ⋅ dS .

S

By Divergence theorem,

∫∫ F ⋅ dS = ∫∫∫ (div F ) dv S

∴

V

∂

∂

∂



∫∫ F ⋅ dS = ∫∫∫  ∂x (x) + ∂y (2 y) + ∂z (3z ) dV S

V

= 6 ∫∫∫ dV V

= 6V, where V is the volume enclosed by S. 4 = 6× π 3 = 8π. 2 2 2 Example 2.14 Verify Gauss divergence theorem for F = x i + y j + z k , where S is the surface of the cuboid formed by the planes x = 0, x=a, y=0, y=b, z = 0 and z = c. Divergence theorem is (1) ∫∫ F ⋅ dS = ∫∫∫ (div F ) dV S

V

S is made up of six plane surfaces. ∴

L.S. of (1) =

∫∫

+

x=0 n = − i

∫∫ + ∫∫ x=a n = i

y=0 n = − j

+ ∫∫ + y=b n = j

∫∫

z=0 n = −k

+ ∫∫ (x 2 i + y 2 j + z 2 k ) ⋅ n dS z=c n = k

= ∫∫ − x 2 dS + ∫∫ x 2 dS + ∫∫ − y 2 dS + x=0

x=a

y=0

∫∫

y 2 dS

y=b

− ∫∫ z 2 dS + ∫∫ z 2 dS (on using the relevant values z=0 z=c of n ) c

b

∫∫

= a2

0

a

dy dz + b 2

0

c

∫∫ 0

b

dz d x + c 2

0

= abc (a + b + c) R.S. of (1) = ∫∫∫ (2 x + 2 y + 2 z ) dx dy dz V c

=∫ 0

b

a

∫ ∫ (2 x + 2 y + 2 z ) dx dy dz 0

0

a

∫ ∫ dx dy 0

0

Vector Calculus c

2.55

b

=∫

∫ (x

0

2

+ 2 y ⋅ x + 2 z ⋅ x)0a dy d z

0

c

= ∫ (a 2 y + ay 2 + 2 a z ⋅ y )b0 d z 0 c

= ∫ (a 2 b + ab 2 + 2ab z ) dz 0

= a 2 bc + + ab 2 c + abc 2 = abc (a + b + c) Thus divergence theorem is verified. 2 Example 2.15 Verify divergence theorem for F = x i + z j + yz k over the cube

formed by x = ±1,

y = ±1,

z = ±1.

∫∫ F ⋅ dS = ∫∫∫ (div F ) dV

Divergence theorem is

S

L.S. of (1) =

(1)

V

∫∫

∫∫

+

x =−1 n =− i

+

x =1 n = i

∫∫

+

y =−1 n =− j

∫∫

+

y =1 n = j

∫∫

z =−1 n =− k

+ ∫∫ (x 2 i + z j + yz k ) ⋅ n dS z =1 n = k

=

∫∫ −x dS + ∫∫ x dS + ∫∫ −z dS + ∫∫ z dS 2

x = −1

2

x =1

∫∫ − yz dS + ∫∫

z = −1

y = −1

yz dS

z =1

=0 R.S. of (1) = ∫∫∫ (2 x + y ) d x dy dz v 1

=∫

1

1

∫ ∫ (2 x + y) dx dy dz

− 1 − 1 −1 1

=∫

1

∫ ( 2 y dy d z )

− 1 −1

=0 Thus divergence theorem is verified.

y =1

(using the relevant values of n )

Mathematics II

2.56

EXERCISE 2(d) Part A (Short Answer Questions) 1. State Green’s theorem in a plane or the connection between a line integral and a double integral. 2. State Stoke’s theorem or the connection between a line integral and a surface integral. 3. State Gauss divergence theorem or the connection between a surface integral and a volume integral. 4. Deduce Green’s theorem in a plane from Stoke’s theorem. 5. State the scalar form of Stoke’s theorem. 6. Give the scalar form of divergence theorem. 7. Derive Green’s identities from divergence theorem. 8. Use Stoke’s theorem to prove that ∇ × ∇ f = 0. 9. Use the integral theorems to prove ∇⋅ (∇× F ) = 0 . 10. Evaluate ∫  (yz dx + zx dy + xy dz ) where C is the circle given by x2 + y2 + C

z2 = 1 and z = 0. 11. Evaluate

∫∫ (x dy d z + y d z d x + z d x dy)

over the surface of the sphere

S

x2+ y2 + z2 = a2. 12. If C is a simple closed curve and r = xi + yj + zk , prove that

∫ r ⋅ dr = 0 . C

13. If C is a simple closed curve and f is a scalar point function, prove that ∫ φ∇φ ⋅ dr = 0 . C 1 14. If S is a closed surface and r = xi + yj + zk , find the value ∫∫ ∇   ⋅ dS . r S 15. If S is a closed surface enclosing a volume V, evaluate

∫∫ ∇(r

2

) ⋅ dS .

S

16. Evaluate square.

∫ [(x − 2 y)d x + (3x − y)dy] , where C is the boundary of a unit

17. Evaluate

∫ (x dy − y d x) , where C is the circle x

C

2

+ y2 = a2.

C

18. If A = curl F , prove that

∫∫ A ⋅ dS = 0 , where S is any closed surface. S

19. If S is any closed surface enclosing a volume V and if A = a xi + by j + c zk , prove that

∫∫ A ⋅ dS = (a + b + c)V . S

20. If r = x i + yj + zk and S the surface of a sphere of unit radius, find

∫∫ r ⋅ dS . S

Vector Calculus

2.57

Part B 21. Verify Green’s theorem in a plane with respect to

∫ (x

2

dx − xy dy ) , where C

C

is the boundary of the square formed by x = 0, y = 0, x = a, y = a. 22 Verify Stoke’s theorem for F = x 2 i + xyj in the square region in the xyplane bounded by the lines x = 0, y = 0, x = 2, y = 2. 23. Use Green’s theorem in a plane to evaluate

∫  x

2

(1 + y ) dx + (x 3 + y 3 ) dy 

C

where C is the square formed by x = ± 1 and y = ± 1.

∫ F ⋅ dr , when

24. Use Stoke’s theorem to find the value of

F = (xy − x 2 ) i +

C

x 2 y j and C is the boundary of the triangle in the xoy-plane formed by x = 1, y = 0, and y = x. 25. Verify Green’s theorem in a plane with respect to

∫ (2 x

2

− y 2 ) dx +

(x 2 + y 2 ) dy  , where C is the boundary of the region in the xoy-plane enclosed by the x-axis and the upper half of the circle x2 + y2 = 1. Verify Stoke’s theorem for F = (xy + y 2 ) i + x 2 j in the region in the xoyplane bound by y = x and y = x2. Use Green’s theorem in a plane to find the finite area enclosed by the parabolas y2 = 4ax and x2 = 4ay. Use Green’s theorem in a plane to find the area of the region in the xoy-plane bounded by y3 = x2 and y = x. Verify Stoke’s theorem for F = (y − z + 2) i + (yz + 4) j − xzk , where S is the open surface of the cube formed by x = 0, x = 2, y = 0, y = 2 and z = 2. C

26. 27. 28. 29.

30. Verify Stoke’s theorem for F = (x 2 − y 2 ) i + 2 xyj + xyzk over the surface of the box bounded by the planes x = 0, x = a, y = 0, y = b and z = c. 31. Verify Stoke’s theorem for F = (2 x − y ) i − yz 2 j − y 2 zk where S is the upper half of the sphere x2 + y2 + z2 = 1 and C is the circular boundary in the xoy-plane. 2 2 32. Verify Gauss divergence theorem for F = (x − yz ) i + (y − zx) j + (z 2 − xy ) k and the closed surface of the rectangular parallelopiped formed by x = 0, x = 1, y = 0, y = 2, z = 0 and z = 3. 33. Verify

divergence

theorem

for

F = 4 xzi − y 2 j + yzk

(i)

and

(ii) F = (2 x − z ) i + x 2 yj − xz 2 k , when S is the closed surface of the cube formed by x = 0, x = 1, y = 0, y = 1, z = 0, z = 1. 34. Use divergence theorem to evaluate

∫∫ (yz S

2

i + zx 2 j + 2 z 2 k ) ⋅ dS , where

S is the closed surface bounded by the xoy-plane and the upper half of the sphere x2 + y2 + z2 = a2 above this plane.

Mathematics II

2.58

∫∫ (4 xi − 2 y

35. Use divergence theorem to evaluate

2

j + z 2 k ) ⋅ dS , where S is

s

the closed surface bounded by the cylinder x2 + y2 = 4 and the planes z = 0 and z = 3. ANSWERS

Exercise 2(a) (5) −12 i − 9 j −16k (8)

1 19

( i + 3 j − 3k )

(6)

(7) 5

14 1

(9)

3

1 (2 i + 2 j − k ) 3 11 (13) − 3

(i + j + k )

(10)

(11) 25; −56 i + 30 j − 47 k

(12)

14 3

1 (14) ± ( − i + 2 j + 2k ) 3

(15)

−

 13  -1  (17) cos   3 22 

 8  (18) cos -1    3 21 

 1  -1   (20) cos −  30 

(21) 2x − y − 3z + 1 = 0

7 64 (23) a = − ; b = 3 9

  -1  −45  (16) cos    2299   3  -1  (19) cos −  7 6 

27 11

(24) x2y − xz2 + y2z + c

(22) λ =

5 ; µ =1 2

2 3 (25) φ = x yz + 20

Exercise 2(b) (7) 3; 0

(8) 4

(9) 2( i + j + k )

(14) λ can take any value

(15) λ = −1

(16) a = 4

(17) a = −1, b= l, c = − l

(21) 0; 3k

(22)

124 21

(23) 80; 80 i + 37 j + 36k ; 27 i − 54 j + 20k ; 0; 74 i + 27 j (25) x3y2z4 (26) x2 + y2 + 3xy + yz + z2x (27) a = 4, b = 2, c = − 1; (28) x2yz3

x2 3 y 2 − + z 2 + 2 xy − yz + 4 zx 2 2 (30) −3

Exercise 2(c) (6) 3

(7) 6

(9)

a 2b2 2

Vector Calculus

(10) 0

(11)

8 4 i + j +k 11 5

  π 1 (13) 2 −  i + π −  j   4 2 (15) −

7 6

2.59 (12) −π i + (14)

51 70

(16) 2a2

(17) 8

(18) 8π

(19) 2

(20) −

(21) 4π + 5

(22) 0

(24) πa2

(23) 27 4

(25) 90

19 2

Exercise 2(d) (10) 0 (16) 5

(11) 4πa3 (17) 2πa2

(23)

8 3

(24) −

(28)

1 10

(34) πa4

1 12

(15) 6V (20) 4π (27) 16 a 2 3 (35) 84π

1 3π j+ k 2 2

Unit-3 AnAlytic Functions

UNIT

3

Analytic Functions 3.1

INTRODUCTION

Before we introduce the concept of complex variable and functions of a complex variable, the definition and properties of complex numbers, which the reader has studied in the lower classes are briefly recalled. If x and y are real numbers and i denotes −1, then z = x + iy is called a Complex number. x is called the real part of z and is denoted by Re(z) or simply R(z); y is called the imaginary part of z and is denoted by Im(z) or simply I(z). The Complex number x − iy is called the conjugate of the Complex number z = x + iy and is denoted by z . Clearly z z = x 2 + y 2 = r 2 , where r is z , viz., the modulus of the Complex number z. Also z = z ; R(z ) =

z+z z−z , I (z ) = . 2 2i

Y P (x, y) (r, θ)

r y

θ O

x

M

X

Fig. 3.1

The complex number z = x + iy is geometrically represented by the point P (x, y) with reference to a pair of rectangular coordinate axes OX and OY, which are called the real axis and the imaginary axis respectively (Fig. 3.1). Corresponding to each complex number, there is a unique point in the XOY−plane and conversely, corresponding to each point in the XOY-plane, there is a unique complex number. The XOY-plane, the points in which represent complex numbers, is called the Complex plane or Argand plane or Argand diagram.

Mathematics II

3.4

If the polar coordinates of the point P are (r, q ), then r = OP = x 2 + y 2 =  y modulus of z or z and θ = ∠MOP = tan −1   = amplitude of z or amp (z).  x Since x = r cos q and y = r sin q, z = x + iy = r (cos q + i sin q ) or re iq. r (cos q + i sin q) is called the modulus-amplitude form of z and re iq is called the polar form of z.

3.2

THE COMPLEX VARIABLE

The quantity z = x + iy is called a complex variable, when x and y are two independent real variables. The Argand plane in which the variables z are represented by points is called the z-plane. The point that represents the complex variable z is referred to as the point z.

3.2.1

Function of a Complex Variable

If z = x + iy and w = u + iv are two complex variables such that there exists one or more values of w, corresponding to each value of z in a certain region R of the z-plane, then w is called a function of z and is written as w = f (z) or w = f (x + iy). When w = u + iv = f (z) = f (x + iy), clearly u and v are functions of the variables x and y. For example, if w = z2, then u + iv = (x + iy)2 Thus ∴

= (x2 − y2) + i(2xy) u = x2 − y2 and v = 2xy. w = f (z) = u(x, y) + iv(x, y)

If z is expressed in the polar form, u and v are functions of r and q. If, for every value of z, there corresponds a unique value of w, then w is called a single-valued function of z. 1 are single-valued functions of z. z If, for every value of z, there correspond more than one value of w, then w is called a multiple valued function of z. For example, w = z1/4 and w = amp(z) are multiple valued functions of z. w = z1/4 is four valued and w = amp(z) is infinitely many valued for z ≠ 0. For example, w = z2 and w =

3.2.2

Limit of a Function of a Complex Variable

The single valued function f (z) is said to have the limit l (= a + ib) as z tends to z0, if f (z) is defined in a neighbourhood of z0 (except perhaps at z0) such that the values of f (z) are as close to l as desired for all values of z that are sufficiently close to z0, but different from z0. We express this by writing lim { f (z )} = l . z → z0

Analytic Functions

3.5

Note

Neighbourhood of z0 is the region of the z-plane consisting of the set of points z for which |z − z0| < ρ, where r is a positive real number, viz., the set of points z lying inside the circle with the point z0 as centre and r as radius. Mathematically we say lim { f (z )} = l , if, for every positive number ∈ (however z → z0

small it may be), we can find a positive number d, such that | f (z) − l| < ∈, whenever 0 < | z − z0 | < d.

Note

In real variables, x → x0 implies that x approaches x0 along the x-axis or a line parallel to the x-axis (in which x is varying) either from left or from right. In complex variables, z → z0 implies that z approaches z0 along any path (straight or curved) joining the points z and z0 that lie in the z-plane, as shown in the Fig. 3.2. Y

z

z0 X O

Fig. 3.2

Thus, in order that lim { f (z )} may exist, f (z) should approach the same value l, z → z0

when z approaches z0 along all paths joining z and z0.

3.2.3

Continuity of f (z)

The single valued function f (z) is said to be continuous at a point z0, if lim { f (z )} = f ( z0 ) .

z → z0

This means that if a function f (z) is to be continuous at the point z0, the value of f (z) at z0 and the limit of f (z) as z → z0 must exist (as per the definition given above) and these two values must be equal. A function f (z) is said to be continuous in a region R of the z-plane, if it is continuous at every point of the region. If f (z) = u (x, y) + iv (x, y) is continuous at z0 = x0 + iy0, then u (x, y) and v (x, y) will be continuous at (x0, y0) and conversely. The function f (x, y) is said to be continuous at the point (x0, y0 ), if lim [ φ (x, y )] = φ (x0 , y0) , in whatever manner x → x0 and y → y0. x → x0 y → y0

3.2.4

Derivative of f (z)

The single valued function f (z) is said to be differentiable at a point z0, if

Mathematics II

3.6

 f ( z 0 + ∆z ) − f ( z 0 )  lim   exists ∆ z→ 0 ∆z  

(1)

This limit is called the derivative of f (z) at z0 and is denoted as f ′ (z0). On putting z0 + Δz = z or Δz = z − z0, we may write  f ( z) − f ( z0 )  f ′ (z0 ) = lim  . z → z0 z − z0   On putting z0 = z in (1), we get  f ( z + ∆z ) − f ( z )  f ′ (z ) = lim   ∆ z→ 0 ∆z  

3.2.5

Analytic Function

A single valued function f (z) is said to be analytic at the point z0, if it possesses a derivative at z0 and at every point in some neighbourhood of z0. A function f (z) is said to be analytic in a region R of the z-plane, if it is analytic at every point of R. An analytic function is also referred to as a regular function or a holomorphic function. A point, at which a function f (z) is not analytic, viz., does not possess a derivative, is called a singular point or singularity of f (z).

3.2.6

Cauchy-Riemann Equations

We shall now derive two conditions (usually referred to as necessary conditions) that are necessarily satisfied when a function f (z) is analytic in a region R of the z-plane. Theorem If the function f (z) = u (x, y) + iv (x, y) is analytic in a region R of the z-plane, then (i) ∂u , ∂u , ∂v and ∂v exist and (ii) ∂u = ∂v and ∂u = − ∂v at every point in ∂x ∂y ∂y ∂x ∂x ∂y ∂x ∂y that region. Proof f (z) = u (x, y) + iv (x, y) is analytic in R. ∴ f ' (z) exists at every point z in R(by definition) i.e.,

 f ( z + ∆z ) − f ( z )  L = lim   exists ∆ z→ 0 ∆z  

(1)

i.e., L takes the same value, when Δz → 0 along all paths. In particular, L takes the same value when Δz → 0 along two specific paths QRP and QSP shown in Fig. 3.3. (2)

Analytic Functions

3.7

Y Q(z + Δz)

S

Δy P(z)

Δx

R X

O Fig. 3.3

It is evident that when z takes the increment Δz, x and y take the increments Δx and Δy respectively; hence we may write L=

 {u ( x + ∆x, y + ∆y ) + iv ( x + ∆x, y + ∆y )} − {u ( x, y ) + iv ( x, y )}    ( ∆ x + i ∆ y )→ 0 ∆ x + i∆ y   lim

Let us now find the value of L, say L1, corresponding to the path QRP, viz., by letting Δy → 0 first and then by letting Δx → 0. Thus

 {u ( x + ∆x, y ) + iv ( x + ∆x, y )} − {u ( x y ) + iv ( x y )}  L1 = lim   ∆ x→ 0 ∆x  

  u ( x + ∆x, y ) − u ( x, y )   v ( x + ∆x, y ) − v ( x, y )  = lim    + i  ∆ x→ 0 ∆x ∆x     ∂u ∂v (3) = +i ∂x ∂x (by the definition of partial derivatives.) Now we shall find the value of L, say L2, corresponding to the path QSP, viz., by letting Δx → 0 first and then by letting Δy → 0. Thus

∴ i.e.

 {u ( x, y + ∆y ) + iv ( x, y + ∆y )} − {u ( x, y ) + iv ( x, y )}  L2 = lim   ∆ y→ 0 i∆y   1  u ( x, y + ∆y ) − u ( x, y )   v ( x, y + ∆y ) − v ( x, y )  = lim   +  ∆ y→ 0 i ∆y ∆y      ∂u ∂v (4) = −i + ∂y ∂y (by the definition of partial derivatives.) L exists [by (1)] L1 and L2 exist at every point in R. ∂u ∂u ∂v and ∂v exist at every point in R , , ∂x ∂y ∂x ∂y

L exists uniquely [by (2)]

(i)

Mathematics II

3.8 ∴

L1 = L2

i.e.,

∂u ∂v ∂v ∂u , by (3) and (4). +i = −i ∂x ∂x ∂y ∂ y

∴

∂u ∂v = ∂x ∂y

and

∂u ∂v at every point in R =− ∂y ∂x

(ii)

Note 1. The two equations given in (ii) above are called Cauchy-Riemann equations which will be hereafter referred to as C.R. equations. 2. When f (z) is analytic, f ′ (z) exists and is given by L, viz., by L1 or L2. Thus f ′(z) = ux + ivx or vy − iuy , where ux , uy , vx, vy denote the partial derivatives. 3. If w = f (z), then f ′ (z) is also denoted as d w . dz Thus

Also

d w ∂u ∂v = +i d z ∂x ∂x ∂ = (u + iv) ∂x ∂w = ∂x dw ∂u ∂v =−i + dz ∂y ∂y ∂ (u + iv) ∂y ∂w =−i ∂y =−i

4. In the above theorem, we have assumed that w = f (z) is analytic and then derived the two conditions that necessarily followed. However the two conditions do not ensure the analyticity of the function w = f (z). In other words, the two conditions are not sufficient for the analyticity of the function w = f (z). The sufficient conditions for the function w = f (z) to be analytic in a region R are given in the following theorem. Theorem The single valued continuous function w = f (z) = u (x, y) + iv (x, y) is analytic in a region R of the z-plane, if the four partial derivatives ux, vx, uy and vy have the following features: (i) They exist, (ii) They are continuous and (iii) They satisfy the C.R. equations ux = vy and uy = −vx at every point of R.

Analytic Functions Proof Consider

3.9

Δu = u (x + Δx, y + Δy) − u (x, y) = [u (x + Δx, y + Δy) − u (x, y + Δy) + u (x, y + Δy) − u (x, y)] = Δ x . u (x + q Δx, y + Δy) + Δy . u (x, y + q Δy), x

1

y

2

where 0 < q1, q2 < 1, (by using the mean-value theorem.) ux is a continuous function of x and y, by (ii). ux(x + q1∆x, y + ∆y) = ux(x, y) + ∈1, where

∴

∈1 → 0 as Δx and Δy → 0 or as Δz → 0. uy is a continuous function of x and y, by (ii). uy (x, y + q2∆y) = uy (x, y) + ∈2, where

∴

∈2 → 0 as Δz → 0 ∴

∆u = ∆x [ux (x, y) + ∈1] + ∆y[uy (x, y) + ∈2],

where ∈1 and ∈2 → 0 as Δz → 0. Similarly, using the continuity of the partial derivatives vx and vy, we have

(1)

∆v = ∆x [vx(x, y) + ∈3] + ∆y [vy(x, y) + ∈4], where Now

∈3 and ∈4 → 0 as Δz → 0

(2)

Δw = Δu + iΔv = (ux + ivx) Δx + (uy + ivy) Δy + (∈1 + i∈3) ∆x + (∈2 + i∈4 ) ∆y,

using (1) and (2). = (ux + ivx) Δx + (uy + ivy) Δy + h1∆x + h2∆y, where h1 = (∈1 + i∈3) and h2 = (∈2 + i∈4) → 0 as ∆z → 0

(3)

Using (iii) in (3), i.e., putting vy = ux and uy = − vx in (3), we have Δw = (ux + ivx) Δx + (−vx + iux) Δy + h1Δx + h2Δy = ux(Δx + iΔy) + ivx(Δx + iΔy) + h1Δx + h2Δy = (ux + ivx) Δz + h1Δx + h2Δy ∴

∆w ∆x ∆y = u x + ivx + η1 + η2 ∆z ∆z ∆z

(4)

Now ∆x ≤ ∆z and ∆y ≤ ∆z ∆x ≤1 ∆z

∴

∴

η1

and

∆x ∆y + η2 ≤ η1 + η2 ∆z ∆z

∆y ≤1 ∆z

Mathematics II

∴

∆y   ∆x + η2  → 0 as ∆z → 0 [  η1 ∆z ∆z 

∴

3.10

h1 and h2 → 0 as ∆z →0]

(5)

Now taking limits on both sides of (4) and using (5), we have dw = u x + ivx . dz i.e., the derivative of w = f (z) exists at every point in R. i.e., w = f (z) is analytic in the region R.

3.2.7

C.R. Equations in Polar Coordinates

When z is expressed in the polar form reiq, we have already observed that u and v, where w = u + iv, are functions of r and q. In this case, we shall derive the C.R. equations satisfied by u(r, q) and v (r, q), assuming that w = u(r, q ) + iv (r, q ) is analytic. Theorem If the function w = f (z) = u(r, q) + iv (r, q) is analytic in a region R of the z-plane, then (i) ∂u , ∂u , ∂v , ∂v exit and (ii) they satisfy the C.R. equations, viz., ∂r ∂θ ∂r ∂θ ∂u 1 ∂v = ∂r r ∂θ

and

∂v 1 ∂u at every point in that region. =− ∂r r ∂θ

Proof f (z) = u (r, q) + iv (r, q) is analytic in R. ∴ f ′ (z) exists at every point z in R (by definition)  f ( z + ∆z ) − f ( z )  i.e., L = lim   exists ∆ z→ 0 ∆z  

(1)

 {u ( r + ∆r , θ + ∆θ ) + iv ( r + ∆r , θ + ∆θ )} − {u ( r ,θ ) + iv ( r ,θ )}  i.e., L = lim   iθ iθ ∆ (r e ) → 0  ∆ re  

( )

exists. i.e., L take the same value, in whatever manner ∆z = ∆(reiq ) → 0. In particular, L takes the same value, corresponding to the two ways given below in which ∆z → 0. (2) ∆z = ∆(reiq) = eiq · ∆r, if q  is kept fixed ∴ When ∆z → 0, ∆r → 0, if q is kept fixed. Let us find the value of L, say L1, corresponding to this way of ∆z tending to zero.

Analytic Functions   u ( r + ∆r ,θ ) − u ( r , θ )   v ( r + ∆r , θ ) − v ( r , θ )  Thus L1 = lim   +i   ∆ r→0 eiθ ∆r eiθ ∆r      ∂u ∂v  = e − iθ  + i   ∂r ∂r 

3.11

(3)

(by the definition of partial derivatives.) Now

∆z = ∆ (reiq ) = reiq i∆q, if r is kept fixed.   

∴ When ∆z → 0, ∆q → 0, if r is kept fixed. We shall now find the value of L = L2, corresponding to this way of ∆z tending to zero. Thus

  u ( r , θ + ∆θ ) − u ( r ,θ )   v ( r ,θ + ∆θ ) − v ( r ,θ )  L2 = lim   +i   ∆θ → 0 r eiθ ⋅ i∆θ r eiθ ⋅ i∆θ     1  ∂u ∂v  = e − iθ  −i + (4)  ∂θ ∂θ  r (by the definition of partial derivatives.)

Since L exists [by (1)], L1 and L2 exist at every point in R. ∂u ∂u ∂v ∂v exists at every point in R. , , , ∂ r ∂θ ∂ r ∂θ

i.e.

(i)

Since L exists uniquely [by (2)], L 1 = L2  ∂u ∂v  1  ∂u ∂v  e − i θ  + i  = e − i θ  −i , from (3) and (4). +  ∂r  ∂θ ∂θ  ∂r  r

i.e. ∴

∂u 1 ∂v = ∂r r ∂θ

and

∂v 1 ∂u at every point in R =− ∂r r ∂θ

Note 1. When f (z) is analytic, f ′ (z) exists and is given by L, viz., by L1 or L2. Thus, when f (z) = u(r, q) + iv (r, q) is analytic, 1 −iθ e (vθ − iuθ ), r where ur , uq , vr , vq denote the partial derivatives. 2. If w = f (z), then f ′ (z)= e−iq(ur + ivr )

or

(ii)

Mathematics II

3.12

dw ∂ ∂w . = e − iθ ⋅ (u + iv) = e − iθ dz ∂r ∂r dw i = − e − iθ (uθ + ivθ ) dz r i ∂ = − e− iθ (u + iv) r ∂θ i ∂w . = − e− iθ r ∂θ

Also

3. The two conditions derived in the theorem do not ensure the analyticity of w = f (z). The sufficient conditions for the analyticity of w = f (z) =u (r, q ) + iv (r, q ) in R are given below without proof. (i) ur , uq , vr, vq must exist (ii) they must be continuous; (ii) they must satisfy the C.R. equations in polar co-ordinates at every point in the region R. WORKED EXAMPLE 3(a) Example 3.1 Find lim f (z ) , when f (z ) = z→0

x2 y . x2 + y 2

Let us find the limit of f (z) corresponding to any one manner, say, by letting y → 0 first and then by letting x → 0. Thus

 0 lim [ f ( z ) ] = lim  2  = lim ( 0) = 0 . y→ 0 x→ 0  x  x→ 0 x→0

This does not mean that lim [ f ( z ) ] exists and is equal to 0. z→0

However, we proceed to verify whether lim [ f ( z ) ] can be 0, as per the mathez→0

matical definition, which states that lim [ f ( z ) ] = 0 , if we can find a d such that z→ 0

f ( z ) − 0 0 cosh 2 c sinh 2 c [When − p/2 ≤ x ≤ p/2 and c > 0, cos x > 0 and sinh c > 0]. i.e. the image of BC is the upper half B′ C′ of the ellipse u2 v2 + =1 cosh 2 c sinh 2 c

(3)

Similarly the images of the line segments CS2, S2D and DA of the z-plane are the line segments C′S′2, S′2D′ (which is the same as S′2C′) and the lower half D′A′ of the ellipse (3). Thus the image of the boundary of the rectangle consists of the ellipse and the two segments of the u-axis. Hence the image of the rectangular region is the interior of the ellipse in the w-plane. Example 3.11 Show that the transformation w = cos z maps the segment of the x-axis given by 0 ≤ x ≤ p/2 into the segment 0 ≤ u ≤ 1 of the u-axis. Show also that it maps the strip y ≥ 0, 0 ≤ x ≤ p/2 into the fourth quadrant of the w-plane. The transformation equation of w = cos z are given by u + iv = cos (x + iy) = cos x cosh y − i sin x sinh y i.e.

u = cos x cosh y (1) v = − sin x sinh y (2) The image of the x-axis, i.e. y = 0 is given by u = cos x and v = 0 Since we consider the segment of the x−axis given by 0 ≤ x ≤ p/2, 0 ≤ u ≤ 1 Thus the image of the segment of y = 0, 0 ≤ x ≤ p/2, is the segment of v = 0, 0 ≤ u ≤ 1. The boundaries of the given strip are x = 0, x = p/2 and y = 0. The image of x = 0 is given by u = cosh y and v = 0, i.e. u ≥ 1 and v = 0. The image of x = p/2 is given by u = 0 and v = sinh y. Since y ≥ 0 for the given strip, v ≤ 0 Thus the image of x = p/2, y ≥ 0 is u = 0, v ≤ 0 The image of y = 0, 0 ≤ x ≤ p/2 is v = 0, 0 ≤ u ≤ 1. Thus the image of the given region is the region bounded by v = 0, 0 ≤ u ≤ l; v = 0, u ≥ 1 and u = 0, v ≤ 0 i.e. v = 0, u ≥ 0 and u = 0, v ≤ 0 i.e. the fourth quadrant in the w-plane. The corresponding regions are shown in the Figs. 3.24 (a) and (b)

Analytic Functions

3.73 v

y

x = π/2

x=0

u

O

x

O y=0 (a)

v' (b)

Fig. 3.24

Example 3.12 Find the image of the region defined by 0 ≤ x ≤ 3, − p/4 < y < p/4 under the transformation w = sinh z. The transformation equations of w = sinh z are given by u + iv = sinh (x + iy) = − i sin i (x + iy) = − i sin i (ix − y) = sinh x cos y + i cosh x sin y i.e. u = sinh x cos y (1) and v = cosh x sin y (2) The image of x = 0 is given by u = 0 and v = sin y i.e. u = 0 and −1 2 ≤ v ≤ 1 2 ( −p/4 ≤ y ≤ p/4)

∴

The image of x = 3 is given by u = sinh 3 cos y and v = cosh 3 . sin y 2 u v2 i.e. the ellipse + = 1, u > 0, since cos y ≥ 0. 2 sinh 3 cosh 2 3 The image of y = − p/4 is given by u=

1

sinh x

v=−

and

1

cosh x 2 2 1 i.e. the lower part of rectangular hyperbola v 2 − u 2 = ( v < 0) 2 Similarly the image of y = p/4 is the upper part of the rectangular hyperbola v2 − u2 = 1/2. Thus the image of the given region is the region in the right part of the v-axis, bounded by the segment of the v-axis, the ellipse and the rectangular hyperbola as shown in the Figs 3.25 (a) and (b).

∴

y D

y = π/4

v C B' D'

x=3

x=0

x

O A

O A'

B y = – π/4 (a)

(b)

Fig. 3.25

C' u B'

Mathematics II

3.74

Show that the transformation w = z +

Example 3.13

a 2 − b2 4z

transforms the

1 (a + b) in the z-plane into an ellipse of semi-axes a, b in the w-plane. 2 k2 In the discussion of the transformation w = z + , we have proved that the z image of the circle |z| = c, i.e., r = c is the ellipse whose equation is circle z =

u2  k2   c + c 

2

+

v2  k2   c − c 

k2 =

Putting c+

2

=1

(1)

a 2 − b2 1 and c = (a + b) in (1), we get 4 2

1 k2 1 = (a + b) + (a − b) = a and 2 c 2

1 k2 1 = (a + b) − (a − b) = b 2 c 2 1 1 ∴ The image of z = (a + b) , i.e., r = (a + b) under the transformation 2 2 c−

 a 2 − b2  u 2 v2 / z is the ellipse 2 + 2 = 1 , whose semi-axes are a and b. w= z+  a b  4  Example 3.14 Prove that the region outside the circle |z| = 1 maps onto the 1 whole of the w-plane under the transformation w = z + . z k2 Proceeding as in the discussion of the transformation w = z + , we can prove z that the image of the circle |z| = 1, i.e. r = 1 is the segment of the real axis, given by − 2 ≤ u ≤ 2 and that the image of the circle |z| = a, i.e. r = a is the ellipse u2 1   a +  a

2

+

v2 1   a −  a

2

=1

The semi axes of this ellipse are f (a) = a + 1/a and g (a ) = a − Now f ′ (a ) = 1 − 1 a 2 =

(1)

1 . a

a2 − 1 > 0 , when a > 1 and g′(a) = 1 + 1/a2 > 0, for all a2

a and hence when a > 1. ∴ f (a) and g(a) are increasing functions of a, when a > 1. Hence, when a increases, the semi axes of the ellipse (1) increase. The region outside the circle r = 1 may be regarded as the area swept by the circle r = a, where 1 < a < ∞.

Analytic Functions Similarly the area swept by the ellipse

3.75

u2 1   a +  a

2

+

v2 1   a −  a

2

= 1, 1 < a < ∞ , is

the entire w-plane. Hence the region outside the circle r = 1 maps onto the entire w-plane. Example 3.15 Show that the transformation w =

1  −α eα  ze +  , where a is real, 2  z

maps the upper half of the interior of the circle |z| = ea onto the lower half of the w-plane. 1 1  The transformation equations of the mapping function w =  ze −α + −α  are 2 ze  given by u + iv =

1  iθ −α 1 − iθ   re e + −α e  2 re

i.e.

u=

1  −α 1  r e + −α  cos θ  2 re 

(1)

and

v=

1  −α 1  r e + −α  sin θ  2 re 

(2)

The boundary of interior of the circle |z| = ea or r = ea lying in the upper half of the z-plane consists of θ = 0, θ = π and r = eα. The image of θ = 0 is given by u=

1 u= 2

i.e.

i.e.

1  −α 1  r e + −α  and v = 0, from (1) and (2).  2 re 

u≥

  r e −α / 2 −  

2    and v = 0 2 +   r e −α / 2  

1

1 (0 + 2) and v = 0 or u ≥ 1 and v = 0. 2

Similarly the image of θ = p is given by u ≤ − 1 and v = 0. The image of r = eα is given by i.e.,

u = cos θ and v = 0, from (1) and (2). − 1 ≤ u ≤ 1 and v = 0.

Thus the boundary of the semi-circle r = eα lying in the upper half of the z-plane is the entire u-axis. The interior of the semi-circle is given by r < eα and 0 < θ < p. When r < ea, re−a < 1 ∴

−

1 < −1 re −α

Mathematics II

3.76

 −α 1   r e − r e −α  < 0

∴

∴

1  −α 1  r e − −α  sin θ < 0 , since sin θ is positive when 0 < θ < p.  2 re 

i.e. v < 0. Hence the interior of the semi-circle |z| = eα lying in the upper half of the z-plane maps onto the lower half of the w-plane. The corresponding images are shown in the Figs 3.26 (a) and (b). v

y |z| = eα

O (a)

O

u

x (b) Fig. 3.26

EXERCISE 3(c) Part A (Short Answer Questions) 1. What do you mean by conformal mapping? 2. When is a transformation said to be isogonal? Prove that the mapping w = z is isogonal. 3. State the conditions for the transformation w = f (z) to be conformal at a point. 4. Define critical point of a transformation. 5. Find the critical points of the transformation 1 eα  w =  ze −α +  . 2 z 6. Find the critical points of the transformation w2 = (z − a) (z − b). 7. Find the magnification factor of small lengths near z = p/4 under the transformation w = sin z. Find the image of the circle |z| = a under the following transformations 8. w = z + 2 + 3i 9. w = 2z 10. w = (3 + 4i) z i 11. w = (1 + i) z + 2 − i 12. w = 2z 13. Find the image of the infinite strip 0 ≤ x ≤ 2 under the transformation w = iz. 14. Find the image of the region that lies on the right of the y-axis under the transformation w = iz + i.

Analytic Functions

3.77

15. Find the image of the region y > 1 under the transformation w = (1 − i)z. 16. Find the image of the hyperbola xy = 1 under the transformation w = 3z. 17. Find the image of the hyperbola x2 − y2 = a2 under the transformation w = (1 + i)z. 18. Find the image of the half-plane y > c, when c > 0 under the transformation 1 w= . z 19. Find the image of the half-plane x > c, when c < 0 under the transformation w = 1/z. i 20. Find the image of the line y = mx under the transformation w = . z 21. Find the image of the region 0 < θ < p/n in the z-plane under the transformation w = zn, where n is a positive integer. 22. Find the images of the lines x = 1 and y = 1 under the transformation w = iz2. 23. Find the image of the region 2 < | z | < 3 under the transformation w = z2. π π 24. Find the image of the region < arg z < under the transformation w = z2. 4 2 25. Find the level curves of u and v under the transformation w = − iz2. 26. Find the image of the infinite strip 0 ≤ x ≤ 1 under the transformation w = ez. 27. Find the image of the x-axis under the transformation w = cos z. 28. Find the image of the y-axis under the transformation w = cos z. 29. Find the image of the x-axis under the transformation w = sinh z. 30. Find the image of the y-axis under the transformation w = sinh z. 31. Find the equations of transformation for the mapping given by w =

a 2

1   z +  . z

32. Find the images of the points whose polar co-ordinates are (1, 0) and (1, p) 1 under the transformation w = z + . z Part B 33. Find the image of the triangle with vertices at z = i, z = 1 − i, z = 1 + i under the transformation (i) w = 3z + 4 − 2i and (ii) w = iz + 2 − i. 34. Find the map of the square whose vertices are z = 1 + i, − 1 + i, − 1 − i and 1 − i by the transformation w = az + b, where a = 2 (1 + i ) and b = 3 + 3i. 35. Prove that the interior of the circle |z| = a maps onto the interior of an ellipse ib y in the w-plane under the transformation w = x + , 0 < b < a. Is the a transformation conformal? 1 36. Show that the transformation w = maps the circle |z − 3| = 5 onto the circle z 3 5 w+ = . What is the image of the interior of the given circle in the 16 16 z- plane?

Mathematics II

3.78

37. Find the image of (i) the infinite strip 0 < y < y > 1, under the transformation w =

1 and (ii) the quadrant x > 0, 2c

1 . z

38. Find the image of the triangle formed by the lines y = x, y = − x and y = 1 under the transformation w = z2. 39. Find the image of the region of the square whose vertices are z = 0, 1, 1 + i and i under the transformation w = z2. 40. When a > 0, show that w = exp (p z/a) transforms the infinite strip 0 ≤ y ≤ a onto the upper half of the w-plane. 41. Discuss the transformation w = log z. Also prove that the whole of the z-plane maps onto the horizontal strip − p ≤ v ≤ p. (The principal value of log z is to be considered.)

Note The properties of the transformation w = log z are identical with those of w = ez, if we interchange z and w. 42. Discuss the transformation w = cos z. 43. Discuss the transformation w = sinh z. πz maps the region of the z-plane a given by y ≥ 0 and − a/2 ≤ x ≤ a/2 onto the upper half of the w-plane. 45. Find the image of the semi-infinite strip 0 ≤ x ≤ π, y ≥ 0 under the transformation w = cos z. 46. Find the image of the region defined by − p/4 ≤ x ≤ p/4, 0 ≤ y ≤ 3 under the transformation w = sin z. 47. Show that the transformation w = cosh z maps (i) the segment of the y-axis 44. Show that the transformation w = sin

from 0 to iπ onto the segment 0 ≤ u ≤ 1 of the u-axis, (ii) the semi-infinite 2 strip x > 0, 0 ≤ y ≤ p/2 onto the first quadrant of the w-plane. 48. Find the image of the region in the z-plane given by 1 ≤ x ≤ 2 and − p/2 ≤ y ≤ p/2 under the transformation w = sinh z. 49. Show that the transformation w =

a 2

  z +

1  where a is a positive constant z

maps (i) the semi-circle |z| = 1 in the upper z-plane onto the segment − a ≤ u ≤ a of the u-axis, (ii) the exterior of the unit circle in the upper z-plane onto the upper w-plane. 50. Show that the transformation w=

1  −α eα  ze +  , where a is real, maps the interior of the circle |z| = 1 2  z

onto the exterior of an ellipse whose major and minor axes are of lengths 2 cosh a and 2 sinh a respectively.

Analytic Functions

3.79

u2 v2  Hint: The image of the circle |z| = 1 is the ellipse cosh 2 α + sinh 2 α = 1 . The  image of any point inside |z| = 1, for example, the point (a, q ), where a < 1 is a  point outside the ellipse, since |u (a, q )| > |u (1, q )| and |v(a, q)| >|v(1, q)|  . 

3.7 3.7.1

BILINEAR AND SCHWARZ−CHRISTOFFEL TRANSFORMATIONS Bilinear Transformation

az + b , where a, b, c, d are complex constants such that cz + d ad − bc ≠ 0 is called a bilinear transformation. It is also called Mobius or linear fractional transformation. The transformation w =

Now d w = ad − bc . If ad − bc = 0, every point of the z-plane becomes a critical d z (cz + d ) 2 point of the bilinear transformation. The transformation can be rewritten as a (ad − bc) . Hence, if ad − bc = 0, w= − c c(cz + d ) a , which has no meaning as a mapping c function. Due to these reasons, we assume that ad − bc ≠ 0. The expression (ad − bc) is called the determinant of the bilinear transformation. the transformation takes the form w =

The inverse of the transformation w =

az + b − d w+b , which is also a is z = cz + d cw − a

bilinear transformation. The images of all points in the z-plane are uniquely found, except for the point d z = − . Similarly the image of every point in the w-plane is a unique point, except c a d a for the point w = . If we assume that the images of the points z = − and w = c c c are the points at infinity in the w- and z-planes respectively, the bilinear transformation becomes one-to-one between all the points in the two planes. To discuss the transformation w =

az + b , cz + d

(1)

we express it as the combination of simple transformations discussed in the previous section. When c ≠ 0, (1) can be expressed as

Mathematics II

3.80

a  bc − ad  1 w= + ⋅ c  c  cz + d If we make the substitutions w1 = cz + d 1 w2 = w1 then

w = Aw2 + B

where

A=

bc − ad c

(2) (3) (4) and

a B= . c

The substitutions (2), (3) and (4) can be regarded as transformations from the z-plane onto the w1-plane, from the w1-plane onto the w2-plane and from the w2-plane onto the w-plane respectively. We know that each of the transformations (2), (3) and (4) maps circles and straight lines into circles and straight lines (since straight lines may be regarded as circles of infinite radii). Hence the bilinear transformation (1) maps circles and straight lines onto circles and  a  b straight lines, in general. When c = 0, (1) becomes w =   z +   (d ≠ 0) i.e. d d (1) reduces to the form w = Az + B. This transformation also maps circles into circles. Thus the bilinear transformation always maps circles into circles with lines as limiting cases.

3.7.2  Definition If the image of a point z under a transformation w = f(z) is itself, then the point is called a fixed point or an invariant point of the transformation. Thus a fixed point of the transformation w = f(z) is given by z = f(z). az + b az + b =z . are given by cz + d cz + d As this is a quadratic equation in z, we will get two fixed points for the bilinear transformation. The fixed points of the bilinear transformation w =

Note 1. A bilinear transformation can be uniquely found, if the images w1, w2, w3 of any three points z1, z2, z3 of the z-plane are given. Let the bilinear transformation required be w=

az + b cz + d

 a  b   z +   d d (1) can be re-written as w =  c   z +1 d

(1)

or

Az + B C z +1

Analytic Functions

3.81

Since the images of z1, z2 and z3 are w1, w2 and w3 respectively, we have

and

w1 =

Az1 + B C z1 + 1

(2)

w2 =

Az2 + B C z2 + 1

(3)

w3 =

Az3 + B C z3 + 1

(4)

Equations (2), (3) and (4) are three equations in three unknowns A, B, C. Solving them we get the values of A, B, C uniquely and hence the bilinear transformation (1) uniquely. 2. If a set of three points and their images by a bilinear transformation are given, it can be found out by using the cross-ratio property of the bilinear transformation, which is given below.

3.7.3  Definition If z1, z2, z3, z4 are four points in the z-plane, then

( z1 − z2 ) ( z3 − z4 ) ( z1 − z4 ) ( z3 − z2 )

is called the

cross-ratio of these points. Cross-ratio property of a bilinear transformation The cross-ratio of four points is invariant under a bilinear transformation. i.e. if w1, w2, w3, w4 are the images of z1, z2, z3, z4 respectively under a bilinear transformation, then

( w1 − w2 ) ( w3 − w4 ) = ( z1 − z2 ) ( z3 − z4 ) . ( w1 − w4 ) ( w3 − w2 ) ( z1 − z4 ) ( z3 − z2 ) Proof az +b Let the bilinear transformation be w = cz + d

( (

wi − w j =

a zi + b a z j + b ( ad − bc) zi − z j − = c z i + d c z j + d ( c zi + d ) c z j + d

∴

(w1 − w2 ) (w3 − w4 ) =

( ad − bc)2 ( z1 − z2 ) ( z3 − z4 ) (c z1 + d ) (c z2 + d ) (c z3 + d ) (c z4 + d )

and

( ad − bc)2 ( z1 − z4 ) ( z3 − z2 ) (w1 − w4 ) (w3 − w2 ) = (c z1 + d ) (c z2 + d ) (c z3 + d ) (c z4 + d )

∴

( w1 − w2 ) ( w3 − w4 ) = ( z1 − z2 ) ( z3 − z4 ) ( w1 − w4 ) ( w3 − w2 ) ( z1 − z4 ) ( z3 − z2 )

Then

) )

Mathematics II

3.82

Note To get the bilinear transformation that maps z2, z3, z4 of the z-plane onto w2, w3, w4, we assume that the image of z under this transformation is w and make use of the invariance of the cross-ratio of the four points z, z2, z3, z4. Thus

( w − w2 ) ( w3 − w4 ) = ( z − z2 ) ( z3 − z4 ) ( w − w4 ) ( w3 − w2 ) ( z − z4 ) ( z3 − z2 )

(1)

(1) ensures that the images of z = z2, z3, z4 are respectively w = w2, w3, w4. Now, simplifying (1) and solving for w, we get the required bilinear transformation in the form w =

3.8

az + b . cz + d

SCHWARZ-CHRISTOFFEL TRANSFORMATION

3.8.1  Definition The transformation that maps the boundary of a given polygon in the w-plane onto the x-axis (and hence maps the vertices of the polygon onto points on the x-axis) and the interior of the polygon onto the upper half of the z-plane is called SchwarzChristoffel transformation. Specifically, if x1, x2, . . . ., xn that are points on the x-axis such that x1 < x2 < x3 < . . . . . < xn, are the images of the vertices w1, w2, . . . ., wn of a polygon in the w-plane and a1, a2, . . . . an are the corresponding interior angles of the polygon, then the required Schwarz-Christoffel transformation is given by αn α1 α2 dw −1 −1 −1 = A ( z − x1 ) π ⋅ ( z − x2 ) π  ( z − xn ) π , dz

where A is an arbitrary complex constant. Proof [The proof is in the nature of verification of the mapping of the transformation.] v

α4

A4 (w4)

y

α3 A3 (w3)

P (w) α2

A2 (w2)

α1 A1 (w1)

π – α1 u

O

P'

A1'

A2'

z

x1

x2

0

Fig. 3.27

The given transformation is α1

d w = A ( z − x1 ) π

−1

α2

⋅ ( z − x2 ) π

−1

αn

 ( z − xn ) π

−1

⋅dz

A3'

A4'

x3

x4

x

Analytic Functions

3.83

α  α  …+ ∴ amp (d w) = amp (A) +  1 − 1 amp (z − x1 ) +  2 − 1 amp (z − x2 ) + … π  π   αn   π − 1 amp (z − xn )

(1)

amp (z1z2 ...) = amp (z1) + amp (z2 ) + .... and amp (zk) = k amp (z)] Let the image of P(w) be P′(z) as shown in Fig. 3.27. When P moves towards A1, P′ moves towards A′1. As long as P′(z) is on the left of A′1 (x1), z − x1, z − x2, . . . ., z − xn are all negative real numbers. ∴ amp (z − x1) = amp (z − x2) = . . . . = amp (z − xn) = p But once P′(z) has crossed A′1(x1), i.e., when x1 < z < x2, z − x1 is a positive real number and hence amp (z − x1) = 0, while amp (z − x2) = amp (z − x3) = . . . . = amp (z − xn) = p Also amp(A) and amp(dz) do not change. [ A is a constant and dz is positive hence amp (dz) = 0] Thus when z croses A′1, amp (z − x1) suddenly changes from p to 0 or undergoes an increment of −p. ∴ From (1), we get ∴

[

∴

α  Increase in amp (dw) =  1 − 1 ( − π ) = π − α 1 . π  This increase is in the anticlockwise direction. This means that when P(w), moving along PA1, reaches A1, it changes its direction through an angle p−a1 in the anticlockwise sense and then starts moving along A1A2. Similarly when P(w) reaches A2, it turns through an angle p−a 2 and then starts moving along A2 A3. Proceeding further, we find that as P(w) moves along the boundary of the polygon, P′(z) moves along the x-axis and conversely. Now when a person walks along the boundary of the polygon in the w-plane in the anticlockwise sense, the interior of the polygon lies to the left of the person. Hence the corresponding area in the z-plane should lie to the left of the person, when he or she walks along the corresponding path in the z-plane, i.e. along the x-axis from left to right. Clearly the corresponding area is the upper half of the z-plane. Thus the interior of the polygon in the w-plane is mapped onto the upper half of the z-plane.

Note 1. Integrating (1), the Schwarz-Christoffel transformation can also be expressed α1

as w = A∫ ( z − x1 ) π

−1

α2

⋅ ( z − x1 ) π

−1

αn

⋅⋅⋅ ( z − xn ) π

−1

⋅ dz + B

where B is a complex constant of integration. 2. The transformation which maps a polygon in the z-plane onto the real axis of the w-plane is got by interchanging z and w in the above transformation. 3. It is known that not more than three of the points x1, x2, . . ., xn can be chosen arbitrarily.

Mathematics II

3.84

4. It is advantageous to choose one point, say xn, at infinity, as explained below. If we take A =

λ

(− xn )

αn −1 π

, where l is a constant, the transformation can be

written as

(

α1 α2 dw −1 −1 = λ ( z − x1 ) π ⋅ ( z − x2 ) π ⋅⋅⋅ z − xn − 1 dz

)

αn −1 π

αn

−1

 x − z π ⋅ n  x 

−1

n

As xn → ∞ , the transformation reduces to

(

α1 α2 dw −1 −1 = λ ( z − x1 ) π ⋅ ( z − x2 ) π ⋅⋅⋅⋅ z − xn −1 dz

)

αn −1 π

−1

αn

This means that, if xn is at infinity, the factor ( z − xn ) π is absent in the transformation. Thus the R.H.S. of the transformation contains one factor less than the original form. 5. Infinite open polygons can be considered as limiting cases of closed polygons. −1

WORKED EXAMPLE 3(d) Example 3.1 Find the invariant points of the transformation w = −

2 z + 4i . Prove iz + 1

also that these two points together with any point z and its image w, form a set of four points having a constant cross ratio. The invariant points of the transformation are given by z=− i.e. i.e.

iz2 + 3z + 4i = 0 (z − 4i) (z + i) = 0

or

2 z + 4i i z +1

z2 − 3iz + 4 = 0

∴ The invariant points are 4i and − i. Taking z1 = z , z2 = w = −

2 z + 4i , z3 = 4i and i z +1

z4 = − i, the cross-ratio of the four points z1, z2, z3 and z4 is given by

( z1 , z2 , z3 , z4 ) = =

 2 z + 4i   z + i z + 1  ( 4i + i )  

( z + i )  4i +

(

2 z + 4i  i z + 1 

5i i z 2 + 3 z + 4i

)

( z + i ) ( − 2 z + 8i)

Analytic Functions

=

( 2i (i z

3.85

) + 3 z + 4i )

5i i z 2 + 3 z + 4i 2

= 5/2 = a constant, independent of z. Example 3.2 Find the bilinear transformation that maps the points 1 + i, − i, 2 − i of the z-plane into the points 0, 1, i of the w-plane. Taking z1 = 1 + i, z2 = − i, z3 = 2 − i and w1 = 0, w2 = 1, w3 = i and using the invariance of the cross-ratio (z, z1, z2, z3), we have

( w − 0) (1 − i) = ( z − 1 − i ) ( −2) ( w − i ) (1 − 0) ( z − 2 + i ) ( −1 − 2i) i.e.

w − i ( z − 2 + i ) ( −1 − 2i ) (1 − i ) = w ( z − 1 − i) ( −2)

i.e.

1−

i (3 + i ) ( z − 2 + i ) = w 2 ( z − 1 − i)

(3 + i ) z − 7 + i i = 1− w 2 z − 2 − 2i − (1 + i ) z + 5 − 3i = 2 z − 2 − 2i

∴

w=

∴

( 2 z − 2 − 2i ) −i {− (1 + i ) z + 5 − 3i}

i.e. the required bilinear transformation is w=

2 z − 2 − 2i (i − 1) z − 3 − 5i

Example 3.3 Find the bilinear transformation which maps the points (i) i, − 1, 1 of the z-plane into the points 0, 1, ∞ of the w-plane respectively (ii) z = 0, z = 1 and z = ∞ into the points w = i, w = 1 and w = − i. (i) i.e.,

(w, w1, w2, w3) = (z, z1, z2, z3)

( w − w1 ) ( w2 − w3 ) = ( z − z1 ) ( z2 − z3 ) ( w − w3 ) ( w2 − w1 ) ( z − z3 ) ( z2 − z1 )

To avoid the substitution of w3 = ∞ in (1) directly, we put w3 = then put w′3 = 0. Thus (1) becomes

(1) 1 simplify and w3′

Mathematics II

3.86

( w − w1 ) ( w2 w3′ − 1) ( z − z1 ) ( z2 − z3 ) = ( w w3′ − 1) ( w2 − w1 ) ( z − z3 ) ( z2 − z1 )

(2)

Using the given values z1 = i, z2 = − 1, z3 = 1, w1 = 0, w2 = 1 and w′3 = 0, we get w ( −1) ( z − i ) ( −2) , i.e., w = (1 − i ) ( z − i ) = − 1 1 z z −1 ( ) ( − 1) ( −1 − i )

( w − w1 ) ( w2 − w3 ) = ( z − z1 ) ( z2 − z3 ) ( w − w3 ) ( w2 − w1 ) ( z − z3 ) ( z2 − z1 ) ( z − z1 ) ( z2 z3′ − 1) , where z = 1 = 3′ z3 ( z z3′ − 1) ( z2 − z1 )

(ii)

Using the values z1 = 0, z2 = 1, z’3 = 0 and w1 = i, w2 = 1 and w3 = − i, we get

( w − i ) (1 + i ) = z ( −1) ( w + i) (1 − i ) ( −1)⋅1 w − i (1 − i ) z = w+i 1+ i

i.e.

2 w (1 − i ) z + (1 + i )  Nr + Dr  = =  2i (1 + i ) − (1 − i ) z  Dr − Nr 

i.e.

w=

∴

(1 + i ) z + (i − 1) (1 + i ) − (1 − i) z

or

w=

z+i . iz +1

Example 3.4 If a, b are the two fixed points of a bilinear transformation, show that w−a  z − a  , where k is a constant, if a ≠ b; it can be written in the form (i) = k  z − b  w−b (ii)

1 1 = + c , where c is a constant, if a = b. w−a z −a

(i) Since the images of a and b are a and b respectively, (w, a, w3, b) = (z, a, z3, b) i.e.

( w − a) ( w3 − b) ( z − a) ( z3 − b) = ( w − b) ( w3 − a ) ( z − b) ( z3 − a )

i.e.

w − a  ( z3 − b ) ( w3 − a )   z − a  =   w − b  ( z3 − a ) ( w3 − b )   z − b 

Analytic Functions i.e.

3.87

w−a  z − a = k , where k is a constant.  z − b  w−b

(ii) Since the image of z = a is w = a, the bilinear transformation can assumed as w−a =

z−a cz + d

(1)

The fixed points of (1) are given by (z − a) [(cz + d) − 1] = 0 i.e.

d − 1  =0 c (z − a )  z + c  

(2)

Since both the roots of (2) are equal to a, d −1 = − a ∴ d = −ca + 1 c Using (3) in (1), the required bilinear transformation is w − a = i.e.

1 c (z − a ) + 1 = w−a z−a

i.e.

1 1 = + c, where c is a constant. w−a z −a

(3) z−a cz − ca +1

Example 3.5 Show that the transformation w = z − 1 maps the unit circle in the z +1 w-plane onto the imaginary axis in the z-plane. Find also the images of the interior and exterior of the unit circle. The image of the unit circle |w| = 1 is given by z −1 = 1, i.e. |z − 1| = |z + 1| z +1 i.e. i.e.

|(x − 1) + iy| = |(x + 1) + iy| (x − 1)2 + y2 = (x + 1)2 + y2

i.e. −2x = 2x or x = 0, which is the imaginary axis. The image of the interior of the unit circle i.e. |w| < 1 is given by |z − 1| < | z + 1| i.e. −2x < 2x or x > 0, which is the right half of the z-plane. Similarly the image of the exterior of the circle |w| = 1 is the left half of the z-plane. z −i maps (i) the interior of the 1 − iz circle | z | = 1 onto the lower half of the w-plane and (ii) the upper half of the z-plane onto the interior of the circle | w | = 1. Example 3.6 Show that the transformation w =

Mathematics II

3.88 w=

z−i 1 − iz

∴

w − iwz = z − i

i.e.

z (1 + iw) = w + i z=

∴

(1)

w+i 1 + iw

(2)

(2) is the inverse transformation of (1). The interior of the circle | z | = 1 is given by | z | < 1. From (2), the image of | z | < 1 is given by w+i < 1 , i.e. | w + i | < | 1 + iw | 1+ iw i.e.

| u + i (v + 1)| < | (1 − v) + iu |

i.e.

u2 + (v + 1)2 < (1 − v)2 + u2

i.e.

2v < − 2v or

4v < 0

or v < 0,

i.e., the lower half of the w-plane. (2) can be written as x + iy = = =

u + i (v + 1) (1 − v) + iu

[u + i (v + 1)] [(1 − v) − iu ] (1 − v) 2 + u 2

u [1 − v + v + 1] + i (1 − v 2 − u 2 ) u 2 + (1 − v) 2

∴

x=

2u u + (1 − v) 2

(3)

and

y=

1 − u 2 − v2 u 2 + (1 − v) 2

(4)

2

The upper half of the z-plane is given by y > 0. Its image is given by 1 − u 2 − v2 > 0 [from (4)] u 2 + (1 − v) 2 i.e.

u2 + v2 < 1 or | w |2 < 1

i.e.

|w| 0 has to be | w | < 1, the boundaries of the two regions must correspond i.e. the image of y = 0 must be | w | = 1. Since three points determine a circle uniquely, we shall make three convenient points lying on y = 0 map into three points on | w | = 1. Let us assume that the three points z = 0, z = ∞ and z = 1 map onto points on the circle | w | = 1. Thus, when z = 0, | w | = 1. ∴ From (1), we get

1=

b , i.e. b = d d

(2)

Rewriting (1), we have

w=

a+b/ z c+d /z

(1)′

When

z = ∞, | w | = 1.

a = 1 , i.e. | a | = | c | (3) c If a = 0, then, from (3), we see that c = 0 b In this case, the transformation (1) becomes w = , which will map the whole d b of z-plane onto a single point w = , which is not true. Hence a ≠ 0 and so c ≠ 0, d from (3). ∴ From (1)′, we get

a a = 1 , from (3) ≠ 0 , such that c c

∴

∴

a may be taken as eiθ, where θ is real. c a (z + b/a ) Again, re-writing (1), w = c (z + d /c)

i.e.

Putting

 z + b/a  w = eiθ   z + d /c  b d = − α and = − β , where a and b are complex, the required a c

transformation becomes  z −α w = eiθ   z − β 

(4)

Mathematics II

3.90

From (2) and (3), we have b = d a c |a|=|b|

i.e.

(5)

We have assumed that the point z = 1 also maps onto a point on the circle | w | = 1. 1−α 1− β i.e. |1 − a | = |1 − b | From (5) and (6), we find that either a = b or α = β . ∴ From (4), 1 = eiθ

(6)

If we assume that a = b, the transformation reduces to w = e iq, which will map the whole of the z-plane into a single point, which is not true. Hence β = α .  z −α ∴ The required transformation is w = eiθ  . Since α is arbitrary, it can be  z − α  taken as any point in the upper half of the z-plane. The image of z = α is w = 0, which lies inside | w | = 1. Thus the upper half of the z-plane maps onto | w | < 1 by the transformation  z −α w = eiθ  , where α is any point in y > 0.  z − α  Example 3.8 Find the transformation that will map the strip x ≥ 1 and 0 ≤ y ≤ 1 of the z-plane into the half-plane (i) v ≥ 0 and (ii) u ≥ 0. (i) Consider the isosceles triangle BAC, which is a three sided polygon. [Fig. 3.28] v y=1

B

C

x=1

O

A

(a)

y=0

x

u

O (b)

Fig. 3.28

When C → ∞, the interior of this triangle becomes the region of the given strip. π π In the limit, the interior angles of the polygon are , and 0 . 2 2 Let us assume that the images of B, A, C are w = − 1, 1 and ∞ respectively. Then the required Schwarz-Christoffel transformation is π π dz −1 −1 = A (w + 1) 2π ⋅ (w − 1) 2π , dw

Analytic Functions

3.91

omitting the factor corresponding to w = ∞ , as per Note (4) under the discussion of the transformation.

Note

z and w are interchanged in the Schwarz-Christoffel transformation formula, as we are mapping a polygon in the z-plane onto the upper half of the w-plane. dz = A ( w + 1) −1 / 2 . ( w − 1) −1 / 2 dw A = w2 − 1

i.e. ∴

Integrating with respect to w, we get z = A cosh−1 w + B When When i.e.

z = 1, w = 1, ∴ B = 1 z = 1 + i, w = −1 ∴ 1 + i = A cosh−1 (−1) + 1 i = A . ip [ cosh ip = cos p = − 1] ∴

A=

1 π

1 cosh −1 w + 1 π cosh−1 w = p (z − 1) or w = cosh p (z − 1) Put w′ = u′ + iv′ = i (u + iv) = iw

∴ Required transformation is z = i.e. (ii)

This means that u − v system is rotated about the origin through π in the 2 positive direction giving u′ − v′ system. Hence v′ ≥ 0 corresponds to u ≥ 0. ∴ The required transformation that maps the given region in the z-plane onto u ≥ 0 is i w = cosh p (z − 1) i.e. w = − i cosh p (z − 1) Example 3.9 Find the transformation that maps the semi-infinite strip y ≥ 0, − a ≤ x ≤ a onto the upper half of the w-plane. Make the points z = −a and z = a correspond to the points w = –1 and w = 1. v y

C

A

(a)

B

x

A'

Fig. 3.29

O

(b)

B'

u

Mathematics II

3.92

Proceeding as in the previous example, the required transformation is dz A iA k [Fig. 3.29] = = = 2 2 dw 1 − w2 w −1 i w −1

Note

This change is made to simplify the evaluation of constants after integration. dw z=k ∫ +B 1 − w2

∴

When ∴

= k sin−1 w + B z = − a, w = − 1 and when z = a, w = 1 k (− π/2) + B = −a and k (π/2) + B = a

(1)

2a and B = 0. π 2a Using these values in (1), the required transformation is z = sin −1 w or π  π z w = sin    2a  k=

Solving, we get

Example 3.10 Find the transformation that maps the infinite strip 0 ≤ v ≤ π of the w-plane onto the upper half of the z-plane. u A B

y

v=π C

O (a)

–1 u

A'

1 O(B') (b)

x

Fig. 3.30

Consider the rhombus ABOC, which is a four sided polygon. [Fig. 3.30] When B and C → ∞, the interior of the this rhombus becomes the region of the given strip. In the limit, the interior angles at A, B, O and C are p, O, p and O respectively. Let us assume that the images of A, B, O, C are the points −1, 0, 1 and ∞ of the z-plane. Then the required transformation is r 1 r 1 0 1 dw = A (z + 1) r $ (z - 0) r $ (z - 1) r dz i.e.

dw A ∴ w = A log z + B = dz z

When w = 0, z = 1 ∴B = 0 When w = p i, z = −1 ∴ A log (−1) = p i i.e. A log (e ip ) = pi i.e. A (ip) = ip or A = 1 Using these values in (1), the required transformation is w = log z.

(1)

Analytic Functions

3.93

EXERCISE 3(d)

Part A (Short Answer Questions) 1. Define a bilinear transformation and its determinant. 2. Find the value of the determinant of the transformation w =

1− i z . z −i

3. What do you mean by the fixed point of a transformation? Does the fixed point exist only for a bilinear transformation? 4. Define the cross ratio of four points in a complex plane. 5. State the cross ratio property of a bilinear transformation. 6. What is Schwarz-Christoffel transformation? State the formula for the same. 7. What is the advantage of choosing the image of one vertex of the polygon at infinity in Schwarz-Christoffel transformation? 8. Find the invariant points of the transformations (i) w = iz2 and (ii) w = z3 Find the invariant points of the transformations 9.

w=

2z + 6 z+7

10. w =

3z − 5 i i z −1

11. w =

z −1− i z+2

12. Find the condition for the invariant points of the transformation w =

az +b to cz + d

be equal. 13. Find all linear fractional transformations whose fixed points are − 1 and 1. 14. Find all linear fractional transformations whose fixed points are − i and i. 15. Find all linear fractional transformations without fixed points in the finite plane. 1 16. Find the image of the real axis of the z-plane by the transformation w = . z+i Part B 17. Find the bilinear transformation which maps the points (i) z = 0, − i, − 1 into w = i, 1, 0 respectively, (ii) z = − i, 0, i into w = − 1, i, 1 respectively. 18. Find the bilinear transformation that maps the points (i) z = 0, − 1, ∞, into the points w = − 1, −2 − i, i respectively, (ii) z = 0, − i, 2 i into the points w = 5i, ∞ − i/3 respectively. 19. Prove that w =

z maps the upper half of the z-plane onto the upper half 1− z

of the w-plane. What is the image of the circle | z | = 1 under this transformation? 20. When the point z moves along the real axis of the z-plane from z = − 1 to z = + 1, find the corresponding movement of the point w in the w-plane, if w =

1− i z . z −i

3.94

Mathematics II

21. Prove that, under the transformation w =

z −i , the region Im(z) ≥ 0 is mapped i z −1

onto the region | w | ≤ 1. Into what region is Im (z) ≤ 0 mapped by this transformation? 22. Find the images of (i) the segment of the real axis between z = + 1 and z = − 1 (ii) the interior of the circle | z | = 1 and (iii) the exterior of the circle | z | = 1 under the transformation w =

1+ i z . z+i

23. Show that the transformation w =

i −z maps the circle | z | = 1 onto the i+ z

imaginary axis of the w-plane. Find also the images of the interior and exterior of this circle. 24. Find the bilinear transformation that maps the upper half of the z-plane onto the interior of the unit circle of the w-plane in such a way that the points z = i, ∞ are mapped onto w = 0, −1. [Hint: Use Worked Example (3.7).] 25. Find the transformation which maps the area in the z-plane within an infinite sector of angle π onto the upper half of the w-plane. m  Hint: The sectoral region bounded by OX and OP may be regarded as an  open polygon with vertex at O and interior angle π . Make the origin of the m  z-plane correspond to the origin of the w-plane.  

26. Find the transformation which maps the semi-infinite strip (i) x ≥ 0, 0 ≤ y ≤ c onto v ≥ 0 (ii) u ≥ 0, 0 ≤ v ≤ p onto y ≥ 0 27. Find the transformation which maps the semi-infinite strip y ≥ 0, 0 ≤ x ≤ a onto the upper half of the w-plane. Make the points z = 0 and z = a correspond to w = +1 and w = –1. 28. Find the transformation that maps the infinite strip 0 ≤ y ≤ k onto the upper half of the w-plane. Make the points z = 0, z = ik correspond to w = 1, –1. 29. Find the transformation that maps the region in the z-plane above the line y = b, when x < 0 and that above the x-axis, when x > 0 into the upper half of the w-plane. Make the points z = ib and z = 0 correspond to w = –1 and w = 1. 30. Find the transformation that maps the region in the w-plane above the u-axis when u < 0 and that above v = b when u > 0 into the upper half of the z-plane. Make the points w = 0 and w = ib correspond to z = 0 and z = 1.

Analytic Functions

3.95

ANSWERS

Exercise 3(a) 16. 18. 20. 22. 39.

at all points on the line y = x 17. at all points nowhere 19. only at the origin at all points except z = −1 21. a = 2, b = −1, c = −1, d = 2 p=−1 36. 2z 37. ez 38. −sin z cosh z

Exercise 3(b) 5. y + c

6. Yes

7. Yes

9. Yes

10. No

11. 2 xy

12. ex sin y

13. cos x sinh y

14. x2 − y2 + 2y;

 y 15. 2 tan −1    x

16. log z

17. −z2

18. eiz

19. cosh z

20. 1 z

21. v = 3x2y − y3 + 2xy + y2 − x2 + c;

8. No

f (z) = z3 + z2 − iz2 + ic

22. u = x3 − 3xy2 − 2xy + c; f (z) = z3 + iz2 + c y 1 23. ψ = 2 xy − 2 + c; f (z ) = z 2 + + ic z x + y2 24. f = x2 − y2 − 2x + 3y − 2xy + c

f(z) = z2 − 2z + i (z2 −3z)

z4 z2 x4 − 6x2y2 + y4 + 2 (x2 − y2) + 4 (x − y) = c′ + + (1 + i ) z 4 2 26. w = −iz . ez + ic ; v = ex (y sin y − x cos y) + c 25.

27. w = ize−z + c ; u = e−x (x sin y − y cos y) +c 2

28. w = −ieiz +ic ; v = e−2xy cos (x2 − y2) + c 29. w = zei2z + c ; u = e−2y (x cos 2x − y sin 2x) + c 1+ i −1 30. f (z ) = 31. f (z) = iz2 − z z cot z 1 +c 32. f (z ) = 33. f (z ) = (1 + sec z ) 1+ i 2 37. 2y + y3 − 3x2y = b

Exercise 3(c) 5. z = ± ea 2

6. z = a, β   and 2

8. (u − 2) + (v − 3) = a

2

1 (α + β ) 2

9. | w | = 2a

7.

1 2

10. | w | = 5a

Mathematics II

3.96

1 2a 16. u v = 9

12. w =

11. (u − 2)2 + (v + 1)2 = 2a2 14. v > 1

15. u + v > 2

2

1   1 18. Interior of the circle u 2 +  v +  =     2c  2c  2

13. 0 < v < 2 17. u v = a2

2

1 1 19. Exterior of the circle  u −  + v 2 =     2c  2c 

2

20. u = mv 21. Upper half of the w-plane 22. u2 = − 4 (v − 1); u2 = 4 (v + 1) 23. 4 < |w| < 9 π < arg (w) < π 24. 25. x y = c1 and y2 − x2 = c2 2 26. 27. 28. 29.

The annular region 1 ≤ R ≤ e Segment of v = 0, given by − 1 ≤ u ≤ 1 Part of v = 0, given by u ≥ 1 u-axis 30. Segment of u = 0, given by −1 ≤ v ≤ 1.

31. u =

a 2

a  1  1  r +  cos θ , v =  r −  sin θ r 2 r

32. (± 2, 0) 33. (i) Triangle with vertices w = 4 + i, 7 + i, 7 − 5i (ii) Triangle with vertices w = 1 − i, 1, 3 − 2i

(

) (

)

34. Square with vertices w = 3 + 3 ± 2 2 i, 3 ∓ 2 2 + 3i 35. No 36. The exterior of the image circle 37. (i) u2 + (v + c)2 > c2, v < 0 (ii) u2 + v2 + v < 0, u > 0 38. The region enclosed by v2 = 4 (u + 1) and u = 0 39. The region above the u-axis, bounded by the u-axis, v2 = 4 (1 − u) and v2 = 4 (1 + u). 45. The lower half of the w-plane 46. The region in the upper half of the w-plane bounded by the part of the u-axis given by −

1 2

≤u≤

1 2

, the ellipse

1 . 2 48. The elliptic annular region bounded by

u2 v2 + = 1 and the 2 cosh 3 sinh 2 3

hyperbola u 2 − v 2 =

u2 v2 + = 1 and 2 sinh 1 cosh 2 1

u2 v2 + = 1 , lying in the right half of the w-plane. 2 sinh 2 cosh 2 2

Analytic Functions

3.97

Exercise 3(d) 2. −2

3. No

8.

z = −i

(i) z = 0 and (ii) z = 0, ± 1.

9. z = 1, − 6

10. z = i, − 5i

12. (a − d)2 + 4bc = 0 13. w =

az +b bz + a

11. z = − i, 14. w =

z=−1+i

az +b a −bz

15. w = z + a 16. u2 + v2 + v = 0  z + 1 17. (i) w = − i   z − 1 (ii) w =

i −iz z +1

18. (i) w =

iz −2 z+2

(ii) w =

3 z − 5i i z −1

19. The line u = – 1/2 20. w moves along the upper half of the circle | w | = 1 from w = −1 to w = +1 in the clockwise sense. 21. | w | ≥ 1 22. The lower half of the circle | w | = 1; v < 0; v > 0 24. w =

23. u > 0; u < 0 26. (i) w = cosh

z c

i−z i+z

25. w = kzm

(ii) z = cosh w 27. w = cos π z a 29. z =

b ( w2 − 1 + cosh −1 w ) π

28. w = ep z/k 30. w =

(

2i b sin −1 z + π

z (1 − z )

)

Unit-4 Complex IntegratIon

UNIT

4

Complex Integration 4.1

INTRODUCTION

The concept of a real line integral, with which the reader is familiar is extended to that of a complex line integral as given below in Fig. 4.1. y C

αk

zn –1

B (zn)

zk zk –1

A (z0)

z1

O

x Fig. 4.1

Let f (z) be a continuous function of the complex variable z (= x + iy) and bet C be any continuous curve connecting two points A(z0) and B(zn) on the z-plane. Let C be divided into n parts by means of the points z1, z2, ...zn–1. Let zk –zk–1 = Δzk. Let αk be an arbitrary point in the arc zk–1 zk (k = 1, 2, ..., n). n

Let

Sn = ∑ f ( k ) ( zk − zk −1 ) or k =1

n

∑ f ( k ) ∆zk . k =1

If the limit of Sn exists as n → ∞ in such a way that each Δzk → 0 and if the limit is independent of the mode of subdivision of C and the choice of the points αk, then it is called the complex line integral of f (z) along C from A to B and denoted as

∫ f ( z ) dz . Practically a complex line integral is expressed in terms of two real line

C

integrals and evaluated.

Mathematics II

4.4

f ( z ) = u ( x, y ) + iv( x, y ), then

i.e. If

∫ f ( z )dz = ∫ (u + iv) (dx + idy)

C

C

= ∫ (udx − vdy ) + i ∫ (vdx + udy ) C

C

Note

When C is an open arc (of finite length) of a continuous curve, the sense of description of C is obvious, as it is traversed from A(z0) to B(zn). When C is a simple closed curve, i.e. a continuous closed curve which does not intersect itself and which encloses a finite region in the Argand plane, then C is traversed in the direction indicated by the arrows drawn on C. In this case, the complex line integral is called contour integral and denoted by the special symbol

∫ f ( z )dz. C

If the sense of description of C is not indicated by arrows, C is traversed in the positive sense or direction, i.e. the direction in which a person, walking along C, has the interior region of C to his or her left. Practically we shall take the anticlockwise direction of traversal of C as the positive direction.

4.1.1

Simply and Multiply Connected Regions

A region R is called simply connected, if any simple closed curve which lies in R can be shrunk to a point without leaving R. A region R which is not simply connected is called multiply connected. Obviously, a simply connected region is one which does not have any “holes” in it, whereas a multiply connected region is one which has. A multiply connected region can be converted into a simply connected region by introducing cross-cuts as shown in Figs 4.2 (a), (b) and (c)

(a) Simply connected region

(b) Multiply connected region

(c) Multiply connected region converted into simply connected region by cross-cuts

Fig. 4.2

4.2

CAUCHY’S INTEGRAL THEOREM OR CAUCHY’S FUNDAMENTAL THEOREM

If f (z) is analytic and its derivative f ' (z) is continuous at all points on and inside a simple closed curve C, then ∫ f ( z )dz = 0. C

Complex Integration Proof Let

4.5

f (z) = u (x, y) + iv (x, y) = u + iv

∫ f ( z ) dz = ∫ (u + iv ) (dx + idy )

Then

C

C

= ∫ (udx − vdy ) + i ∫ (vdx + udy )

(1)

Since f ' (z) is continuous, the four partial derivatives

∂u ∂u ∂v ∂v are also , , , ∂x ∂y ∂x ∂y

C

C

continuous on C and in the region R enclosed by C. Hence Green’s theorem in a plane, namely,  ∂Q

∂P 

∫ ( Pdx + Qdy) = ∫∫  ∂x − ∂y  dx dy

C

R

can be applied to each of the lines integral in the R.H.S. of (1).  ∂v

∂u 

 ∂u

∂v 

∫ f ( z )dz = ∫∫  − ∂x − ∂y  dx dy + i ∫∫  ∂x − ∂y  dx dy

∴

C

R

R

Since the function f (z) = u + iv is analytic, u and v satisfy the Cauchy-Riemann equations in R. ∂u ∂v ∂u ∂v = and =− ∂x ∂y ∂y ∂x

i.e

∂u ∂v ∂v ∂u − = 0 and − − =0 ∂x ∂y ∂x ∂y Using (3) in (2), we get i.e

∫

(3)

f ( z )dz = 0 + i 0 = 0

C

Note 1. The above theorem can be proved without assuming that f ' (z) is continuous, as was done by a French mathematician E. Goursat. We state below the modified form of the above theorem, called Cauchy-Goursat theorem without proof. “If f (z) is analytic at all points on and inside a simple closed curve C, then

∫ f ( z )dz = 0”

C

2. We have proved Cauchy's integral theorem for a simply connected region. It can be extended to a multiply connected region as follows.

4.2.1

Extension of Cauchy’s Integral Theorem

If f (z) is analytic on and inside a multiply connected region whose outer boundary is C and inner boundaries are C1, C2,..., Cn, then

Mathematics II

4.6

∫ f ( z ) dz = ∫

C

∫

f ( z ) dz +

C1

f ( z ) dz +  +

C2

∫

f ( z )dz where all the integrals are taken in

Cn

the same sense. We shall prove the extension for a doubly connected region for simplicity.

B B

A A

C1 C

Fig. 4.3

We convert the doubly connected region into a simply connected region by introducing the cross-cut AB (Fig. 4.3).

∫ f ( z )dz = 0

By Cauchy's integral theorem,

where C' includes C described in

C′

anticlockwise sense, C1 described in clockwise sense, AB and BA ∴

f (z)dz = f (z)dz + C'

C

f (z)dz + C1

f (z)dz + AB

f (z)dz = 0 BA

The last two integrals in the R.H.S. are equal in value but opposite in sign and hence cancel each other. ∴

f (z)dz +

i.e.

f (z)dz = –

f (z)dz C1

C

f (z)dz

=

Note

f (z)dz = 0 C1

C

C1

By introducing as many cross-cuts as the number of inner boundaries, we can give the proof in a similar manner for the extension of Cauchy's integral theorem stated above.

4.2.2

Cauchy’s Integral Formula

If f (z) is analytic inside and on a simple closed curve C that encloses a simply 1 f ( z) dz , connected region R and if ‘a’ is any point in R, then f (a ) = ∫  2i C z − a where C is described in the anticlockwise sense.

Complex Integration

4.7

y C' a P C

O

x Fig. 4.4

Proof f ( z) Since f (z) is analytic on and inside C, is also analytic on and inside C, z−a except at the point z = a. Now we draw a small circle C' with centre at z = a and f ( z) radius ρ, lying completely inside C. The function is analytic in the doubly z−a connected region bounded by C and C'. ∴

By Cauchy's Extended theorem, we have

∴

∫ z − a dz = ∫ z − a dz

f ( z)

f ( z)

C

(1)

C'

If z is any point on C', then |z – a| = ρ and hence z – a = ρeiθ or z = a + ρeiθ ∴

∴

dz = iρeiθ dθ

∫ C

f ( z) dz = z−a

2

∫

f (a + ei ) ei

0

i ei d

( When z moves around C' once completely, θ varies from 0 to 2π) 2

= i ∫ f (a + ei )d

(2)

0

(2) is true, however small the circle C' is and hence true when ρ → 0. Taking limits of (2) as ρ → 0, we get 2

f ( z) ∫ z − a dz = i ∫ f (a)d = if (a) ⋅ 2  C' 0

(3)

Mathematics II

4.8 Using (3) in (1), we have

f ( z)

∫ z − a dz = 2i f (a) C

f (a) =

∴

4.2.3

1 f ( z) ∫ z − a dz. 2i  C

Extension of Cauchy’s Integral Formula to a Doubly Connected Region

If f (z) is analytic on C1 and C2 (C2 lies completely within C1) and in the annular region R between C1 and C2 and if ‘a’ is any point in R, then f (a) =

1 1 f ( z) f ( z) dz − dz ∫ ∫   2 i C z − a 2 i C z − a 1

B B

A A

2

C2

‘a’ C1

Fig. 4.5

Proof We convert the doubly connected region to a simply connected region by introducing the cross-cut AB. ‘a’ lies in this region (Fig. 4.5). By Cauchy's integral formula, f (a) =

1 f ( z) ∫ z − a dz, 2i  C

where C includes C1 described in the anticlockwise sense, C2 in the clockwise sense, AB and BA. ∴

f (a) =

=

1 2 πi

C1

f (z) f (z) f (z) f (z) dz + dz + dz + dz z–a z – a z – a z –a C AB BA 1

f (z) 1 dz + 1 2 πi C z – a 2 πi 1

C2

f (z) dz ( the last two integrals cancel each z–a other)

Complex Integration =

f ( z) f ( z) 1 1 ∫ z − a dz − 2 i ∫ z − a dz 2 i C C 1

4.2.4

4.9

2

Cauchy’s Integral Formulas for the Derivatives of an Analytic Function

By Cauchy's integral formula, we have f (a) =

1 f ( z) ∫ z − a dz 2i  C

(1)

Differentiating partially both sides of (1) with respect to ‘a’ and performing the differentiation within the integration symbol in the R.H.S., we get f ( z) 1! (2) f '(a ) = dz  ∫ 2 i C ( z − a )2 Proceeding further, we get f ′′ (a ) = f

(n)

2! f ( z) dz etc. ∫  2 i C ( z − a )3

(a) =

n! f ( z) ∫ ( z − a )n+1 dz 2 i  C

WORKED EXAMPLE 4(a)

Example 4.1 If f (z) is analytic in a simply connected region R, show that z1

∫

f ( z )dz

is independent of the path joining the points z0 and z1 in R and lying within

z0

R. Verify this by evaluating ( z 2 + 3 z )dz along (i) the circle |z| = 2 from (2, 0) to (0, 2) ∫ C

in the anticlockwise direction (ii) the straight line from (2, 0) to (0, 2) and (iii) the straight lines (2, 0) to (2, 2) and then from (2, 2) to (0, 2). Let C1 (ADB) and C2 (AEB) be any two curves joining A(z0) and B(z1) in the region R. (Fig. 4.6) Now ADBEA may be regarded as a simple closed curve in R. ∴ By Cauchy's integral theorem,

∫

ADBEA

f ( z )dz = 0 , i.e.

∫

ADB

f ( z ) dz +

∫

BEA

f ( z ) dz = 0

Mathematics II

4.10

B

C2

Z1

E C1 A Z0

D

x Fig. 4.6

∫

i.e.

f ( z ) dz = −

ADB

∫

i.e.

C1

∫

f ( z ) dz =

BEA

∫

f ( z ) dz

AEB

f ( z ) dz = ∫ f ( z ) d z C2

2

∫ (z

2

∫

+ 3 z ) dz =

2

( z + 3 z )d z =

| z |= 2

C1

(2ei ) 2 + 3.2ei  2ei id

∫ 0

iθ

( on the circle |z| = 2, z = 2e and the end points are given by θ = 0 and θ = π/2 )  2

 e3i e 2i   = 8i + 12i ⋅  2i  0  3i 8 = (ei 3 2 − 1) + 6(ei − 1) 3

8 44 8 = ( −i − 1) + 6( −2 + i.0) = − − i 3 3 3 The equation of C2, the line joining (2, 0) and (0, 2) is x + y = 2. [Fig. 4.7] y D (0, 2)

C3

C1

B (2, 2)

C3

C2 A (2, 0)

O

x

Fig. 4.7

∫ (z

C2

2

∫

+ 3 z ) dz =

[ x 2 − y 2 + i 2 xy + 3( x + iy )](dx + idy )

C2 ( x + y = 2 ) 0

= ∫ [ x 2 − (2 − x) 2 + i 2 x(2 − x) + 3{x + i (2 − x)}](dx − idx) 2

Complex Integration

4.11

0

= ∫ [( −2 x 2 + 8 x + 2) + i ( −2 x 2 − 6 x + 10)]dx 2

0

 x3  2  =  −2 + 4 x 2 + 2 x + i  − x3 − 3 x 2 + 10 x   3  2 3 

=−

44 −8 i 3 3

∫ (z

2

+ 3 z ) dz =

C3

∫

+

AB ( x = 2)

∫

( z 2 + 3 z ) dz

BD ( y = 2)

2

0

0

2

= ∫ [(2 + iy ) 2 + 3(2 + iy )]idy + ∫ [( x + i 2)2 + 3( x + i 2)]dx [ on AB, dx = 0 and on BD, dy = 0] 2

0

 (2 + iy )3 3(2 + iy ) 2   ( x + i 2)3 3( x + i 2) 2  + = +   + 3 2 3 2 2 0  

44 8  8  8 =  − − 6 − i − 6 = − − i.  3  3 3 3 The values of the given integral are the same, irrespective of the curve joining the two points, since f (z) = z2 + 3z is analytic everywhere. If the curve is not specified, the integral can be evaluated easily as follows, provided the integrand is an analytic function. ( 0 , 2)

2i

0 + 2i

 z 3 3z 2  44 8 f ( z ) d z = ( z + 3 z ) d z =  +  = − − i. ∫ ∫ 3 2 2 3 3 2+ i 0 ( 2 , 0) 2

1+ i

Example 4.2 Evaluate

∫ ( x − y + ix

2

)dz along (i) the line joining z = 0 and z = 1 + i,

0

(ii) the parabola y = x2 and (iii) the curve x = t, y = 2t – t2.

(i) The line joining the points z = 0 and z = 1 + i, i.e. the points (0, 0) and (1, 1) is y = x 1+ i

∫ 0

1

( x − y + ix 2 )(dz + idy ) = ∫ ( x − x + ix 2 )(1 + i )dx 0

1

 x3  1 = ( −1 + i )   = ( −1 + i ).  3 0 3

(ii) When y = x2, dy = 2x dx

Mathematics II

4.12 1

2 2 The given integral = ∫ ( x − x + ix )(dx + i 2 xdx)

∴

0

1

= ∫ [( x − x 2 − 2 x3 ) + i (3 x 2 − 2 x3 )]dx 0

1 i 1 1 2  2 =  − −  + i 1 −  = − + 3 2 2 3 4  4

(iii) (0, 0) corresponds to t = 0 and (1, 1) corresponds to t = 1. ∴ The given integral 1

= ∫ (t − 2t + t 2 + i ⋅ t 2 )[dt + i (2 − 2t )dt ] 0 1

= ∫ [(2t 3 − t 2 − t ) + i (−2t 3 + 5t 2 − 2t )] dt 0

1 1 2 1 1  2 5  =  − −  + i  − + − 1 = − + i 4 3 2 4 3 3 6    

Note The values of the integral along three different curves are different, as the integrand is not an analytic function of z. Example 4.3 Evaluate (i)

dz

∫ z−2

and

(ii)

∫ ( z − 2)

n

dz

C

C

(n ≠ –1), where C is the circle whose centre is 2 and radius 4. (i) The equation of the circle whose centre is 2 and radius 4 is |z –2| = 4 ∴

z – 2 = 4eiθ

and dz = 4eiθ idθ

To describe C once completely, θ has to vary from 0 to 2π. dz ∫ z−2 = C

∴ (ii)

∫ ( z − 2)

C

2

∫ 0

4eiid . = i[]02  = 2 i 4ei

2 n

dz =

∫4

n in

e .4ei id

0

= 4n +1

2

∫e

i ( n +1)

d

0

2

 ei ( n +1)   = 4n +1   = 0, since n ≠ −1  i (n + 1)  0

Complex Integration

4.13

Aliter for (ii) (z – 2)n is analytic on and inside C. ∴ By Cauchy's integral theorem,

∫ ( z − 2)

n

dz = 0

C

z+2 dz , where (i) C is the semicircle |z| = 2 in the upper z C half of the z-plane, (ii) C is the semicircle |z| = 2 in the lower half of the z-plane and (iii) C is the entire circle |z| = 2. Example 4.4 Evaluate

∫

(i) On the semicircle |z| = 2, z = 2eiθ and θ varies from 0 to π in the upper half. ∴

∫

C

( 2ei + 2) i z+2 dz = ∫ ⋅ 2e id z 2ei 0 



 ei  = 2i  +  = 2 ( ei − 1) + 2 i  i 0 = – 4 + 2πi (ii) On the semicircle in the lower half, θ varies from π to 2π 2

∴

 ei  z+2 ∫ z dz = 2i  i +   = 4 + 2 i C

(iii) On the entire circle, θ varies from 0 to 2π 2

∴

∫

C

 ei   z+2 +   = 4 i dz = 2i  z  i 0 = The sum of values of the integral along the two semi-circles.

Aliter for (iii) z+2 is of the form z z =2

∫

f ( z)

∫ z − a dz , where f (z) is analytic on and inside C

C

that contains the point a. Here f (z) = z + 2 is analytic on and inside |z| = 2, contains the point z = 0. ∴ By Cauchy's integral formula, we have f ( 0) = ∴

f ( z) 1 dz ∫ 2 i C z

z+2 dz = 2 i ( 0 + 2) = 4 i z C

∫

Mathematics II

4.14

Example 4.5 Evaluate

ze z

∫ ( z − a )3 dz , where z = a lies inside the closed curve C,

C

using Cauchy’s integral formula. By Cauchy’s integral formula, f (z)

∫ ( z − a )3 dz =

C

2 i f " (a) 2!

(1)

where f(z) is analytic on and inside C and the point z = a lies inside C. Comparing the given integral with the L.H.S. of (1), we find that f (z) = zez. ze is analytic everywhere and hence analytic on and inside C. Also ‘a’ lies inside C. ze z 2 i  d 2 z  z = d  2 ze  ∴ ∫ ( z − a )3 2 !  dz  z =a C z

( )

= πi {(z + 2)ez}z = a = (a + 2)eaπi. z tan 2 dz = ( −2 < a < 2) ,  where C is the boundary of the Example 4.6 Evaluate ∫ 2 C ( z − a) square whose sides lie along x = ± 2 and y = ± 2 described in the positive sense (Fig. 4.8). y

y= 2

x x=–2

x= 2 y=–2 Fig. 4.8

z f (z) 2 ∫ ( z − a )2 dz ≡ ∫ ( z − a )2 dz C C tan

Let

z is analytic on and inside C, since 1 z f ' ( z ) = sec 2 does not exist at z 2 2 2 = ± π, ± 3π etc. which lie outside C. f ( z ) = tan

Also, since – 2 < a < 2, the point z = a lies inside C.

Complex Integration

4.15

Hence Cauchy’s integral formula holds good. z 2 i  d  z  2 ∫ ( z − a )2 dz = 1!  dz  tan 2   ( z =a ) tan

∴

z a 1 = 2 i ⋅  sec 2  = i sec 2 . 2  z =a 2 2 Example 4.7 Use Cauchy's integral formula to evaluate sin z 2 + cos z 2 ∫ z − 2 ) ( z − 3) dz , where C is the circle |z| = 4. C ( 1

( z − 2)( z − 3) ∴

=

1 1 − ( z − 3) ( z − 2) sin z 2 + cos z 2 sin z 2 + cos z 2 dz − ∫ dz z −3 z−2 C C

Given integral = ∫ =∫

C

f ( z) f ( z) dz − ∫ dz , say z −3 z −2 C

(1)

f (z) = sin πz2 + cos πz2 is analytic on and inside C. The points z = 2 and z = 3 lie inside C (Fig.4.9). ∴ By Cauchy's integral formula, form (1), we get sin z 2 + cos z 2 2 2 ∫ ( z − 2) ( z − 3) dz = 2 i sin z + cos z C

(

)

z =3

– 2πi(sin πz2 + cos πz2)z=2 = 2πi (sin 9π + cos 9π) – 2πi (sin 4π + cos 4π) = – 2πi – 2πi = –4πi. y |z|=4

x O

2 3

Fig. 4.9

Mathematics II

4.16

Example 4.8

7z −1

∫ z 2 − 3z − 4 dz,

Evaluate

where C is the ellipse x2 + 4y2 = 4,

C

using Cauchy’s integral formula. The ellipse x 2 + 4 y 2 = 4 or

x2 y 2 + = 1 is the 22 12

standard ellipse as shown in the Fig. 4.10. 7z −1

7z −1

∫ z 2 − 3z − 4 dz = ∫ ( z − 4 ) ( z + 1) dz

C

C

Of the two points z = 4 and z = –1 the point z = –1 only lies inside C. Re-writing the given integral, we have

∫

C

f ( z) = ∴

 7z −1    7z −1  z − 4  dz ≡ f ( z ) dz d z = ∫ z + 1) ∫ z +1 z 2 − 3z − 4 C ( C

(1)

7z −1 is analytic inside C and the point z = –1 lies inside C. z−4

By Cauchy's integral formula, form (1),  −8 

7z −1

∫ z 2 − 3z − 4 dz = 2 i f ( −1) = 2 i  −5  =

C

16 i. 5

y C

–1

x

O

Fig. 4.10

Example 4.9 Evaluate

z dz

∫ ( z − 1) ( z − 2)2

C

, where C is the circle z − 2 =

1 , using 2

Cauchy’s integral formula. 1 is the circle with centre at z = 2 and radius equal to 1/2. The point z = 2 2 lies inside this circle (Fig. 4. 11). z−2 =

Complex Integration

4.17

The given integral can be re-written as  z    f (z)  z −1  ≡ ∫ ( z − 2 )2 ∫ ( z − 2 )2 dz, say, C C f ( z) = ∴

z is analytic on and inside C and the point z = 2 lies inside C. z −1

By Cauchy's integral formula, z

∫ ( z − 1)( z − 2)2 dz =

C

2 i f ′ ( 2) 1!

 d  z  = 2 i     dz  z − 1  z = 2  −1  = −2 i = 2 i  2  ( z − 1)  z = 2 y

2

1

|z–2|= 1 2 O

1

x

2

Fig. 4.11

Example 4.10 Use Cauchy’s integral formula evaluate

z +1

∫ z 3 − 2 z 2 dz ,

where C

C

is the circle |z − 2 − i| = 2. The circle |z − (2 + i)| = 2 is the circle whose centre is the point z = 2 + i and radius is 2, as shown in Fig. 4.12. The point z = 2 lies inside this circle. The given integral can be re-written as  z +1  2  f (z)  z  ∫ z − 2 dz ≡ ∫ z − 2 dz, say. C C z +1 is analytic on and inside C and the point z = 2 lies inside C. z2 ∴ By Cauchy’s integral formula, f ( z) =

z +1

3  z + 1 = i. 2  z  z=2 2

∫ z 3 − 2 z 2 dz = 2 i f ( 2) = 2 i 

C

Mathematics II

4.18 y 2

O

(2 + i) x

2 Fig. 4.12

Example 4.11 Evaluate

z+4

∫ z 2 + 2 z + 5 dz,

where C is the circle |z + 1 + i| = 2,

C

using Cauchy’s integral formula. |z + 1 + i| = 2 is the circle whose centre is z = −1 −i and radius is 2. z+4 z+4 = z + 2 z + 5 ( z + 1 + 2i ) ( z + 1 − 2i ) 2

∴ The integrand is not analytic at z = −1 −2i and z = −1 + 2i. Of these, the point z = −1 −2i lies inside C. (Fig. 4.13) Noting this, we rewrite the given integral as  z+4    f (z)  z + 1 − 2i  ∫ z + 1 + 2i dz ≡ ∫ z − ( −1 − 2i ) dz, say. C C f(z) is analytic on and inside C and the point (−1 − 2i) lies inside C. ∴ By Cauchy’s integral formula. z+4

∫ z 2 + 2 z + 5 dz = 2 i f ( −1 − 2i )

C

 −1 − 2i + 4  = 2 i    −1 − 2i + 1 − 2i  =−

 (3 − 2i ) 2 (–1 + 2i)

y

O (–1 – i) 2

(–1 – 2i)

Fig. 4.13

x

Complex Integration

4.19

3z 2 + 7 z + 1 dz , where C is the circle |z| = 2, find the z−a C values of f (3), f ′(1 − i) and f″ (1 − i) If f ( a ) = ∫

Example 4.12

Note

In Cauchy’s integral formula, f (z) was used to denote the numerator of the

integrand and

f ( z)

∫ z − a dz = 2i f ( a ) , but in this problem f (a) is used to denote the

C

value of the integeral. ( z ) dz , where φ (z) = 3z2 + 7z + 1 z a − C

f (a) = ∫ ∴

( z ) ( z ) dz and is analytic on and inside C. z −3 z −3 C

f (3) = ∫

∴ By Cauchy’s integral theorem, f (3) = 0 f ′ (a) = ∫

Now

C

( z )

( z − a )2

dz

2 i ′ ( a ) , if the point ‘a’ lies inside C 1! (by Cauchy’s integral formula) The point (l − i) lies within the circle C (Fig. 4.14). =

∴

d  f ′ (1 − i ) = 2 i  3z 2 + 7 z + 1   dz  z =1−i

(

)

y |z|=2

= 2πi [(6(1 − i) + 7] = 2π (6 + 13i)

z=3 O

Also

f ′′ ( a ) = 2 ∫ C

( z)

( z − a)

3

dz = 2.

2 i ′′ ( a ) 2!

  d2 = 12 i = 2 i  2 3 z 2 + 7 z + 1   z =1−i  dz

(

)

EXERCISE 4(a)

Part A (Short Answer Questions) 1. Define simply and multiply connected regions. 2. State Cauchy’s integral theorem.

z=1–i

Fig. 4.14

x

Mathematics II

4.20

3. State Cauchy-Goursat theorem. 4. State Cauchy’s extended integral theorem as applied to a multiply connected region. 5. State Cauchy’s integral formula. 6. State Cauchy’s extended integral formula, as applied to a doubly connected region. 7. State Cauchy’s integral formula for the nth derivative of an analytic function. 3+ i

8. Evaluate

∫

2

z dz , along the lines 3y = x.

0

9. Evaluate ∫ (x2 − iy2) dz along the straight line from (0, 0) to (0, 1) and then from (0, 1) to (2, 1). 10. Evaluate

∫z

2

dz , where C is circle |z – 1| = 1.

C

2 −i

11. Evaluate

∫ ( 3xy + iy i

12. Evaluate

1

∫ (x

2

)

+ iy dz along the parabola (i) y = x2

0

14. Evaluate

) dz along the line joining the points z = i and z = 2 – i.

∫ z dz, where C is the semi-circular are |z| = 1 above the real axis.

C 1+ i

13. Evaluate

2

and

(ii) x = y2.

∫ log z dz, where C is the circle |z| = 2.

C

15. Evaluate

∫ z dz along the curve z = t2 + it from 0 to 4 + 2i.

C

Part B 16. Evaluate

∫ (5z

4

)

− z 3 + 2 dz around (a) the circle |z| = 1, (b) the square with

C

vertices at (0, 0), (1, 0) (1, 1) and (0,1) (c) the parabola y = x2 from (0, 0) to (1, 1) and then the parabola y2 = x from (1, 1) to (0, 0). 17. Evaluate

∫(x

2

− iy 2 ) dz (i) the parabola y = 2x2 from (1, 1) to (2, 8), (ii) the

C

straight lines from (1, 1) to (1, 8) and then from (1, 8) to (2, 8), (iii) the straight line from (1, 1) to (2, 8). 18. Evaluate

∫(z

C

2

)

2

+ 1 dz , along the arc of the cycloid x = a (q − sin q), y =

a (1 − cos q) from the point q = 0 to the point q = 2p.

Complex Integration

4.21

(Hint: The end points are z = 0 and z = 2πa. The integrand is an analytic function and hence the value of the integral does not depend on the curve joining the end points.) 19. Evaluate

dz

∫ z − 2 − i , where C is the boundary of (i) the square bounded by the

C

real and imaginary axes and the lines x = 1 and y = 1, (ii) the rectangle bounded by the real and imaginary axes and the lines x = 3 and y = 2, described in the counter clockwise sense. Evaluate the following integrals using Cauchy’s integral formula. 20.

∫

sinh 2 z

dz where C is the boundary of the square whose sides lie along x = z4 ± 2 and y = ± 2, described in the positive sense. C

sin z 2 + cos z 2

21.

∫ ( z + 1)( z + 2) C

22.

∫ z 2 − 1 dz, where C is the circle of unit radius with centre at (i) z = 1 and

dz , where C is |z| = 3.

z2 + 1

C

(ii) z = i. 23.

z3 + 1 ∫ z 2 − 3iz dz, where C is |z| = 1. C

24.

1 e zt dz , where C is |z| = 3. 2 i C∫ z 2 + 1

25.

∫ ( z + 1)4 dz,

e2 z

where C is |z| = 2.

C

26.

∫

C

27.

dz

(

2

z +4

)

2

, where C is |z – i| = 2.

z+4

∫ z 2 + 2 z + 5 dz, where C is |z + 1–i| = 2.

C

28.

z3 + z + 1 2 2 ∫ z 2 − 7 z + 6 dz, where C is the ellipse 4x + 9y = 1. C 4z2 + z + 5 dz , where C is |z| = 2, find the values of f (1), f (i), z−a C f ′ (−1) and f″ (−i).

29. If f ( a ) = ∫

Mathematics II

4.22

30. If f (z) is analytic on and inside a simple closed curve C and z0 is a point not f ′ ( z) f ( z ) dz lying on C, show that  ∫ z − z0 dz = ∫ z − z 2 . ) C C ( 0

4.3 4.3.1

SERIES EXPANSIONS OF FUNCTIONS OF COMPLEX VARIABLE-POWER SERIES Power Series ∞

A series of the form

∑ cn ( z − a)

n=0

n

= c0 + c1 ( z − a ) + c2 ( z − a ) +  is called a power 2

series in powers of (z – a), where z is a complex variable, the constants c0, c1, c2, … are called the coefficients and the constant a is called the centre of the series. Most of the definitions and theorems relating to convergence of infinite series of real terms, with which the reader is familiar, hold good for series of complex terms also. It can be proved that there exists a positive number R such that the power series given above converges for |z – a| < R and diverges for |z – a| > R, while it may or may not converge for |z – a| = R. This means that the power series converges at all points inside the circle |z – a| = R, diverges at all points outside the circle and may or may not converge on the circle. Due to this interpretation. R is called the radius of convergence of the above series and the circle |z – a| = R is called the circle of convergence. Power series play an important role in complex analysis, since they represent analytic functions and conversely every analytic function has a power series representation, called Taylor series that are similar to Taylor series in real calculus. Analytic functions can also be represented by another type of series, called Laurent’s series, which consist of positive and negative integral powers of the independent variable. They are useful for evaluating complex and real integrals, as will be seen later.

4.3.2 Taylor’s Series (Taylor’s Theorem) If f (z) is analytic inside a circle C0 with centre at ‘a’ and radius r0, then at each point z inside C0, f ( z ) = f (a) +

f ′ (a) f ′′ ( a ) f ′′′ ( a ) ( z − a) + ( z − a )2 + ( z − a)3 +  ∞ 1! 2! 3!

Proof Let C1 be any circle with centre at a and radius r1 < r0, containing the point z (Fig. 4.15). Let w be any point on C1. Then, by Cauchy’s integral formula,

Complex Integration f ( z) =

f ( w) 1 dw ∫ 2 i C w − z

(1)

f ( w) n! dw ∫ 2 i C ( w − a )n+1 1

(2)

4.23 C0 C1

1

f ( n) ( a ) =

and

Now

z a

1 1 = w − z (w − a) − ( z − a) =

1   z − a  1− ( w − a )   w − a  

−1

Fig. 4.15

 2 n −1 1    z − a   z − a   z − a    1 +  =  + + +       w − a    w − a   w − a  w − a    2

1

n

 z−a + ⋅  w − a 

  1   z−a 1−   w − a  

 1 − n 2 n −1 ∵ 1 +  +  +   = and so 1  1−   −

and so 1 +  + 2 +  n −1 +

n 1  = 1 −  1 −  

n −1

( z − a) +  + ( z − a) +  z − a  n 1 z−a 1 1 = + +   w − z w − a ( w − a )2 ( w − a )3 ( w − a )n  w − a  w − z 2

i.e.

−

f ( w) f ( w) f ( w) 1 1 1 dw = dw + ( z − a ) dw 2 i C∫ w − z 2 i C∫ ( w − a ) 2 i C∫ ( w − a )2 1 1 1

∴

+  + ( z − a)

n −1

f ( w) 1 dw + Rn , ∫ 2 i C ( w − a )n 1

f ( w) 1  z−a dz   . 2 i C∫  w − a  ( w − z ) n

where

Rn =

1

f ( z) = f (a) +

i.e.

! Let

!

f ′(a) 1!

f ′′ ( a )

( z − a) +

+ ... +

f(

n −1)

2!

(a)

( n − 1)!

|z – a| = r and |w – a| = r1.

( z − a )2 + ... +

( z − a )n −1 + Rn ' by (1) and ( 2 )

(3)

Mathematics II

4.24 ∴

|w – z| = | (w – a) – (z – a)| ≥ r1 – r Let M be the maximum value of |f (w)| on C1 Rn ≤

Then

rn 2

2

0

r1M  r  r1 − r  r1 

(

M r1d ∵ w = r1ei r1n

∫ r1 − r

)

n

i.e.

=

Since

r r < 1, lim   = 0 ∴ Rn → 0 and so Rn → 0 as n → ∞. n →∞  r1  r1

n

Taking limits on both sides of (3) as n → ∞, we get f ( z ) = f (a) +

f ′ (a) f ′′ ( a ) ( z − a) + ( z − a )2 +  ∞ 1! 2!

(4)

Note

For any point z inside C0, we can always find C1. So the Taylor series representation of f (z) is valid for any z inside C0. The largest circle with centre at ‘a’ such that f (z) is analytic at every point inside it is the circle of convergence of the Taylor’s series and its radius is the radius of convergence of the Taylor’s series. Clearly the radius of convergence is the distance between ‘a’ and the nearest singularity of f (z). 2. Putting a = 0 in the Taylor’s series, we get f ( z ) = f ( 0) +

f ′ ( 0) f ′′ ( 0) 2 z+ z + ∞ 1! 2!

(5)

This series is called the Maclaurin’s series of f (z). 3. The Maclaurin’s series of some elementary functions, which can be derived by using (5), are given below: (i) e z = 1 +

z z 2 z3 + + + ..., when z < ∞. 1! 2 ! 3!

(ii) sin z = z −

z3 z5 + ..., when z < ∞. 3! 5!

(iii) cos z = 1 −

z2 z4 + ..., when z < ∞. 2! 4!

(iv) sinh z = z +

z3 z5 + ...+, when z < ∞. 3! 5!

(v) cosh z = 1 +

z2 z4 + ..., when z < ∞. 2! 4!

Complex Integration

4.25

(vi) (1 – z)–1 = 1+ z + z2 + z3 +…, when |z| < 1 (vii) (1 – z)–2 = 1+ 2z + 3z2 + 4z3 +…, when |z| < 1.

4.3.3

Laurent’s Series (Laurent’s Theorem)

If C1 and C2 are two concentric circles with centre at ‘a’ and radii r1 and r2 (r1>r2) and if f(z) is analytic on C1 and C2 and throughout the annular region R between them, then at each point z in R, f ( z) = an =

where

∞

∞

∑ an ( z − a ) + ∑ bn ( z − a ) n

n=0

−n

,

n =1

f ( w) 1 ∫ ( w − a)n+1 dw, n = 0,1, 2,... 2 i C 1

and

bn =

f ( w) 1 ∫ ( w − a)− n+1 dw, n = 1, 2,... 2 i C 2

Proof By Extension of Cauchy’s integral formula to a doubly connected region, f ( z) =

f ( w) f ( w) 1 1 ∫ ( w − z ) dw − 2 i ∫ w − z dw [Fig. 4.16] 2 i C C 1

2

C1

= I1+I2, say In I1 ,

C2

1 1 = w − z (w − a) − ( z − a) 1   z −a = 1 −   w − a   w − a 

(1)

z −1

a

( z − a) 1 z−a + + 2 w − a ( w − a) ( w − a)3 2

=

Fig. 4.16

+ +

since |z – a| < |w –a| when w is on C1 In I2,

−

1 1 = w − z ( z − a) − ( w − a) =

1   w − a   1 − ( z − a )   z − a  

−1

( z − a)n −1 + ( z − a)n ( w − a )n ( w − a )n

1 w− z

( 2)

Mathematics II

4.26

n −1

( w − a) + ( w − a) , w−a 1 = + ++ z − a ( z − a )2 ( z − a)n ( z − a)n ( z − w) since |w – a| < |z – a|, when w is on C2. Using the expansion (2) in I1, we have n

f ( w) f ( w) 1 z−a dw + dw +  + ∫ ∫   2 i C w − a 2 i C ( w − a )2

I1 =

1

1

( z − a)n −1 2 i = a0 + a1 ( z − a ) +  + an −1 ( z − a ) Rn

where

(3)

n z − a) ( =

2 i

n −1

f ( w)

∫ ( w − a)n dw + Rn

C1

+ Rn ,

f ( w) dw

∫ ( w − a)n ( w − z )

(4)

C1

Using the expansion (3) in I2, we have f ( w) ( w − a) 1 1 ∫ z − a dw + 2 i ∫ ( z − a)2 f ( w) dw +  2 i C C

I2 =

2

2

n −1

( w − a) f ( w) dw + S . 1 + n ∫  2 i C ( z − a )n 2

−1

= b1 ( z − a ) + b2 ( z − a ) Sn =

where

1 2 i ( z − a )

n

∫

−2

+  + bn ( z − a )

−n

+ Sn

( w − a)n f ( w) dw

C2

z−w

(5)

If |z – a| = r, then r2 < r < r1. Rn → 0 as n → ∞, as in Taylor’s theorem. Let M be the maximum value of |f(w)| on C2. Then

| Sn | ≤

1 2 r n

2

∫ 0

r2n n r2 d ∵on C2 , w = r2 ei r − r2

(

)

n

i.e

= M r2  r2    r − r2  r 

∴

| Sn | → 0 and so Sn → 0 as n → ∞, since r2 < r.

Using (4) and (5) in (1) and taking limits as n → ∞, we have f ( z ) = ∞

∑ b ( z − a) n

n =1

−n

⋅

∞

∑ an ( z − a ) + ∑ b

n=0

n

1

−

Complex Integration

4.27

Note 1. If f (z) is analytic inside C2, then the Laurent’s series reduces to the Taylor series of f (z) with centre a, since in this case all the coefficients of negative powers in Laurent’s are zero. 2. As the Taylor’s and Laurent’s expansions in the given regions are unique, they are not usually found by the theorems given above, but by other simpler methods such as use of binomial series. ∞

∑ an ( z − a ) , consisting of positive integral powers of (z – a), is ∞ n=0 −n the analytic part of the Laurent’s series, while ∑ bn ( z − a ) ,

3. The part called

n

n =1

consisting of negative integral powers of (z – a) is called the principal part of the Laurent’s series.

4.4

CLASSIFICATION OF SINGULARITIES

We have stated, in Chapter 3, that a point at which f (z) is not analytic is called a singular point or singularity of f (z). We now consider various types of singularities.

1. Isolated Singularity The point z = a is called an isolated singularity of f (z), if there is no other singularity in its neighbourhood. In other words, z = a is called an isolated singularity of f (z), if we can find a δ > 0 such that the circle |z – a| = δ encloses no singularity other than a. If we cannot find any such δ, then ‘a’ is called a non-isolated singularity. [If ‘a’ is not a singularity and we can find δ > 0 such that |z – a| = δ encloses no singularity, then ‘a’ is called a regular point or ordinary point of f (z).]

Note 1. If a function has only a finite number of singularities in a region, those singlarities are necessarily isolated. 1 For example, z = 1 is an isolated singularity of f ( z ) = and z = 0 and ( z − 1)2 ± i are isolated singularities of f ( z ) =

z +1

z

3

( z + 1) 2

.

2. If ‘a’ is an isolated singularity of f (z), then f (z) can be expanded in Laurent’s series valid throughout some neighbouhood of z = a (except at z = a itself), i.e., valid in 0 < |z – a| < r1. Here r2 is chosen arbitrarily small.

2. Pole If z = a is an isolated singularity of f (z) such that the principal part of the Laurent’s expansion of f (z) at z = a valid in 0 < |z – a| < r1 has only a finite number of terms, then z = a is called a pole.

Mathematics II

4.28

∞

∞

f ( z ) = ∑ an ( z − a ) + ∑ bn ( z − a )

i.e. if in

n

n =0

−n

,

n =1

bm ≠ 0, bm+1=0 = bm+2...., then z = a is called a pole of order m. A pole of order 1 is called a simple pole. 1 For example, z = 0 is a simple pole of f ( z ) = , as the Laurent’s expansion. 2 z ( z − 1) 1 (1 − z )−2 z

of f (z) valid in 0 < |z| < 1 is given by f ( z ) = i.e.

f ( z) =

(

1 1 + 2 z + 3z 2 +  ∞ z

(

)

)

= 2 + 3z + 4 z 2 +  ∞ +

1 z

It has only one term i.e. 1/z in the principal part. Similarly, z =1 is a pole of order 2 of f (z), as the Laurent’s expansion of f (z), valid in 0 < |z – 1| < 1, is given by f (z) = =

1

( z − 1)

−1

{1 + ( z − 1)}

2

1

( z − 1)

2

{1 − ( z − 1) + ( z − 1)

2

}

− ( z − 1) +  3

 1 1  2 = 1 − ( z − 1) + ( z − 1) −  ∞ + − +  z − 1 ( z − 1)2   

{

}

Here b2 ≠ 0 and b3 = b4= ...= 0.

3. Essential Singularity If z = a is an isolated singularity of f (z) such that principal part of the Laurent’s expansion of f (z) at z = a, valid in 0 < |z – a| < r1, has an infinite number of terms, then z= a is called an essential singularity. For example, z = 1 is an essential singularity of f(z) = e1/(z – 1), as the Laurent’s expansion is given by f ( z ) = 1 +

1 1 1 1 1 + + +  ∞. 1! z − 1 2 ! ( z − 1)2 3!( z − 1)3

4. Removable Singularity If a single-valued function f (z) is not defined at z = a, but  lim f ( z )  exists, then z  z →a  = a is called a removable singularity. For example, z=0 is a removable singularity of

Complex Integration f ( z) =

4.29

sin z , as f (0) is not defined, but lim  sin z  = 1. The Laurent’s expansion of z z →0  z 

f (z) is given by  sin z 1  z3 z5 =  z − + −  ∞ z z 3! 5!  = 1−

z2 z4 + − ∞ 3! 5 !

Note 1. If f (z) has a pole of order m at z = a, then its Laurent’s expansion is ∞ bm b b2 n f ( z ) . = ∑ an ( z − a ) + 1 + ++ , 2 z − a ( z − a) ( z − a )m n=0 where bm ≠ 0. =

1

( z − a )m

∞ n+ m + bm + bm −1 ( z − a ) +  + b1  ∑ an ( z − a ) n=0 +b1 ( z − a )

=

1

( z − a )m

m −1

  

 ( z ) ,say.

Clearly, φ (z) is analytic everywhere that includes z = a and φ (a) = bm ≠ 0. ( z) Thus for a function of the form , z = a is a pole of order m, provided that ( z − a )m φ (z) is analytic everywhere and φ (a) ≠ 0. 2. A function f (z) which is analytic everywhere in the finite z-plane is called an entire function or integral function. An entire function can be represented by a Taylor’s series whose radius of convergence is ∞ and conversely a power series whose radius of convergence is ∞ represents an entire function. The functions ez, sin z, cosh z are examples of entire functions. 3. A function f (z) which is analytic everywhere in the finite plane except at a finite number of poles is called a meromorphic function. For example, f ( z ) =

1

is a mermorphic function, as it has only two poles 2 z ( z − 1) —a simple pole at z = 0 and a double pole at z =1.

4.4.1

Residues and Evaluation of Residues

If ‘a’ is an isolated singularity of any type for the function f (z), then the coefficient 1 of (viz. b1) in the Laurent’s expansion of f (z) at z = a valid in 0 < |z – a| < r1 is z−a called the residue of f (z) at z = a.

Mathematics II

4.30

C1 C z r2

a

C2

Fig. 4.17

1 ∫ f (w)dw, where C2 is any 2 i C

We know, from Laurent’s theorem, that b1 =

2

circle |z – a| = r2< r1, described in the anticlockwise sense. [Fig. 4.17] Now if C is any closed curve around ‘a’ such that f (z) is analytic on and inside C except at ‘a’ itself, then by Extension of Cauchy’s integral theorem,

∫ f ( w) dw = ∫ f ( w) dw.

C2

C

1 ∫ f ( z ) dz, 2i  C where C is any closed curve around ‘a’ such that f (z) is analytic on and inside it except at z = a itself. Hence the residue of f (z) at z = a is also given by [Res. f (z)]z=a =

4.4.2

Formulas for the Evaluation of Residues

1. If z = a is a simple pole of f (z), then  Res. f ( z ) z = a = lim {( z − a ) f ( z )} . z→a Since z = a is a simple pole of f (z), then the Laurent’s expansion of f (z) is of the following form: f ( z) =

∴

∞

∑ an ( z − a )

n=0

n

+

b1 z−a

∞

( z − a ) f ( z ) = ∑ an ( z − a )n +1 + b1 n=0

∴

lim {( z − a ) f ( z )} = b1 =  Res. f ( z )  z = a .

z→a

Complex Integration 2. If z = a is a simple pole of f ( z ) =

4.31

P ( z) then [Ref.f (z)]z=a Q ( z)

 P ( z)  . = lim  . z→a  Q ′ ( z ) By the previous formula, ( z − a) P ( z)  0 [Res. f ( z )]z = a = lim   → form z →a  Q z ( )  0   ( z − a) P′ ( z ) + P ( z )  = lim   , by L’Hospital’s rule z→a Q′ ( z)   =

P (a) Q ′ (a)

or

 P ( z)  lim  ( z ) 

z→a  Q ′

3. If z = a is a pole of order m of f (z), then  d m −1 1  Res. f ( z )  z = a = lim  m −1 ( m − 1)! z → a  dz

{( z − a)

m

 f ( z)  

}

Since z = a is a pole of order m of f (z), then the Laurent’s expansion of f (z) is of the following form: f ( z) =

∞

∑ an ( z − a)

n

+

n=0

∴ ( z − a) f ( z ) = m

∞

bm −1 bm b1 b2 + ++ + m −1 2 z − a ( z − a) ( z − a) ( z − a )m

∑ an ( z − a)

n+ m

n=0

+ b1 ( z − a )

m −1

+ b2 ( z − a )

m−2

++

bm −1 ( z − a ) + bm .  d m −1 ∴ lim  m −1 z → a dz 

∴ b1 =  Res. f ( z )  z = a where

Note

D≡

 f ( z )  = ( m − 1)!b1  1 m , lim  D m −1 ( z − a ) f ( z ) = ( m − 1)! z → a 

{( z − a)

m

}

{

}

d . dz

The residue at an essential singularity of f (z) is found out using the Laurent’s expansion of f (z) directly.

Mathematics II

4.32

4.4.3

Cauchy’s Residue Theorem

If f (z) is analytic on and inside a simple closed curve C, except for a finite number of singularities a1, a2,…, an lying inside C and if R1, R2,…, Rn are the residues of f (z) at these singularities respectively, then

∫ f ( z ) dz = 2 i ( R1 + R2 +  + Rn ). C

Proof We enclose the singularities ai by simple closed curves Ci, which are not overlapping. f (z) is analytic in the multiply connected region bounded by the outer curve C and the inner curves C1, C2,… Cn. (Fig. 4.18) ∴ By Extension of Cauchy’s Integral theorem, we have

∫ f ( z ) dz = ∫ f ( z ) dz + ∫ f ( z ) dz +  ∫ f ( z ) dz C

C1

C2

Cn

= 2 i ( R1 + R2 +  + Rn ) , by the property of residue at a point.

a2

C’2

a1

Cn

C’1

an C Fig. 4.18

Note

Cauchy’s integral formulas for f (a) and f(n) (a) can be deduced from Cauchy’s residue theorem. Consider,

f ( z)

∫ z − a dz , where f (z) is analytic on and inside C enclosing the point ‘a’.

C

Now

 f ( z )  f ( z) has a simple pole at z = a with residue = lim ( z − a )  = f (a). z→a  z−a ( z − a )  

∴ By Cauchy’s residue theorem,

f ( z)

∫ ( z − a) dz = 2if ( a) .

C

f ( z)

( z − a )n +1

has a ploe of order (n + 1) at z = a

Complex Integration

4.33

 f ( z)  1 n +1  lim D n ( z − a ) n! z→a  ( z − a )n +1  1 ( n) f (a) = n!

with residue =

f ( n) ( a ) dz = 2 i ∫ n +1 n! C ( z − a) f ( z)

∴

f n (a) =

or

f ( z) n! dz. 2 i C∫ ( z − a )n +1 WORKED EXAMPLE 4(b)

Example 4.1 Expand each of the following functions in Taylor’s series about the indicated point and also determine the region of convergence in each case: (i) e2z about z = 2i; (ii) cos z about z = –π/2; (iii) (i) Let ∴ Also

z+3 ( z − 1) ( z − 4) at z = 2;

1+ z  (iv) log  at z = 0.  1 − z 

e2z = e2(z –2i) + 4i = e4i. e2(z – 2i) f (z) = e2(z – 2i). Then f (n) (z) = 2n e2(z – 2i) (n) f (2i) = 2n; n = 1, 2,…∞. f (2i) = 1.

Taylor's series of f (z) at z = 2i is given by f ( z ) = f ( 2i ) +

f ′ ( 2i ) f ′′ ( 2i ) ( z − 2i ) + ( z − 2i )2 +  1! 2!

 2 ( z − 2i ) 22 ( z − 2i )2 23 ( z − 2i )3  2z 4i 1 + e = e + + +  ∴ 1! 2! 3!   The series converges in a circle whose centre is 2i and radius is the distance between 2i and the nearest singularity of f (z) which is ∞. ∴The region of convergence of the Taylor’s series in this case is |z – 2i| < ∞. (ii) Let f ( z ) = cos z; f ′ ( z ) = − sin z; f ′′ ( z ) = − cos z; ∴

f ′′′ ( z ) = sin z; f iv ( z ) = cos z , etc. f ( −  / 2 ) = 0; f ′ ( −  / 2 ) = 1; f ′′ ( −  / 2 ) = 0;

f ′′′ ( −  / 2) = −1; f iv ( −  / 2) = 0, etc. Taylor’s series of f (z) at z = − π/2 is given by f ( z ) = f ( −  / 2) +

f ' ( −  / 2) f ′′ ( −  / 2) ( z +  / 2) + ( z +  / 2) 2 +  1! 2!

Mathematics II

4.34

3 5 7 z +  / 2) ( z +  / 2) ( z +  / 2) ( z +  / 2) ( cos z = − + − +

∴

1!

2!

5!

7!

The region of convergence is |z + π/2| < ∞. z+3 ( z − 1) ( z − 4) The Taylor’s series of f (z) at z = 2 is a series of powers of (z – 2). Putting z – 2 = u or z = u + 2, we have (iii)

f ( z) =

f ( z) =

u+5 −4 / 3 7 / 3 = + , (u + 1) (u − 2) u + 1 u − 2

on resolving into partial fractions 4 7 −1 −1 = − (1 + u ) − (1 − u / 2) 3 6  4 7  u u 2 1 − u + u 2 − u 3 +  − 1 + + 2 +  3 6  2 2  n ∞ ∞ 4 7 u n = − ∑ ( −1) u n − ∑ n 3 n=0 6 n=0 2

{

=−

∞

7 1 n − . n  ( z − 2) 6 2  n=0 The Taylor’s series expansion was obtained using binomial series, which are convergent in |u| < 1 and |u/2| < 1, i.e., in |u| < 1. ∴ The region of convergence of the Taylor’s series of f (z) is |z – 2| < 1. =

4

}

∑  3 ( −1)

n +1

Note

The singularities of f (z) are at z = 1 and z = 4 which lie outside the region of convergence. (iv)

1+ z  log  = log (1 + z) – log (1 – z)  1 − z 

Log (1 ± z) are many valued functions. We consider only those branches (values) which take the value zero when z = 0. Now consider f (z) = log (1 + z) −1

−2

−3

f ′ ( z ) = (1 + z ) ; f ′′ ( z ) = − (1 + z ) ; f ′′′ ( z ) = ( −1) ( −2) (1 + z ) etc. ∴

f ( 0 ) = 0; f ′ ( 0 ) = 1; f ′′ ( 0 ) = −1; f ′′′ ( 0 ) = 2 !, etc.

Taylor’s series of f (z) at z = 0 [or Maclaurin’s series of f (z)] is f ( z ) = f ( 0) + ∴

log (1 + z ) = z −

f ′ ( 0) f ′′ ( 0) 2 z+ z + 1! 2!

z 2 z3 z 4 + − + 2 3 4

(1)

Complex Integration

4.35

The only singularity of log (1 + z) is – 1 and its distance from the origin, i.e. the radius of convergence of series (1) is 1. ∴ The region of convergence of series (1) is given by |z| < 1. Changing z into – z in (1), the Taylor series of log (1 – z) at z = 0 is given by log (1 – z) = – z – z2/2 – z3/3 – z4/4 – … The region of convergence being |z| < 1 Using (1) and (2), we get 3 5   log  1 + z  = 2  z + z + z +  , which converges for |z| < 1.  1− z  3 5  

Example 4.2 Expand each of the following functions in Laurent’s series about z = 0. Identify the type of singularity also. (i) ze − z

2

(iii) z–1 e–2z

(ii) (1 – cosz)/z (v) (z – 1) sin 1 z

(iv)

1 z

3

ez

2

2 2 z4 z6  z  ze − z = z 1 − + − +   1! 2 ! 3! 

(i)

z3 z5 z7 (1) + − + 1! 2 ! 3! The Laurent’s series (1) does not contain negative powers of z and the circle of convergence |z| = ∞ does not include any singularity ∴ z = 0 is an ordinary point = z−

of ze − z (ii)

2

 1 − cos z 1   z 2 z 4 z 6 = 1 − 1 − + − + . z z   2 ! 4 ! 6 !  =

z z3 z5 − + − 2! 4! 6!

(2)

1 − cos z , the Laurent’s series of 1 − cos z z z at z = 0 does not contain negative powers of z. Though z = 0 appears to be a singularity of

∴ z = 0 is a removable singularity of

Note

(iii)

1 − cos z . z

 1 − cos z   1 − cos z  The value of   at z = 0 is not defined; but lim   = 0.  z →0 z z  z −1e −2 z = =

 1  2 z 22 z 2 23 z 3 + − +  1 − 2! 3! z  1!  1 22 z 23 z 2 −2+ − + z 2! 3!

(3)

Mathematics II

4.36

The principal part in the Laurent’s series (3) contains the only term

1 . z

∴ z = 0 is a simple pole of z–1 e–2z. 1

(iv)

z

2

ez = 3 =

1  z 2 z 4 z 6  1+ + + +  3  z  1! 2 ! 3!  1 z3

+

1 z z3 + + + z 2 ! 3!

(4)

The last non-vanishing term with negative powers of z in the principal part of (4) is 1 . z3 ∴ z = 0 is pole of order 3 for the given function. 1 1 1 1 1 1 (v) ( z − 1) sin   = ( z − 1)  − 3 + 5 −  5! z z   z 3! z 1 1 1 1 1 1 1 = 1− − + + − z 3! z 2 3! z 3 5! z 4

(5)

The principal part consists of an infinite number of terms in the Laurent’s series (5). ∴ z = 0 is an essential singularity. Example 4.3 Find the Laurent’s series of f ( z ) =

1 valid in the region (i) z (1 − z )

|z + 1| < 1, (ii) 1 < |z + 1| < 2 and (iii) |z + 1| > 2. 1 ; Since Laurent’s series in powers of (z + 1) are required, f ( z) = z (1 − z ) put z + 1 = u or z = u – 1. ∴

1 1 + (1) (u − 1) ( 2 − u ) u − 1 2 − u (i) Since the region of convergence of required Laurent’s series is |u| < 1, the two terms in the R.H.S. of (1) should be rewritten as standard binomials whose first term is 1 and the numerator of the second term is u. Accordingly. f ( z) =

f ( z) = −

1

1 1 1  u −1 + = − (1 − u ) + 1 −  1− u 2  2  u 2 1 −   2

= −∑ un +

∞

1 ∞ un ∑ 2 n = 0 2n



1 

=

=

n=0 ∞

∑  −1 + 2n+1  ( z + 1)

n

−

(2)

n=0

The Laurent’s expansion (2) is valid, if |u| < 1 and |u| < 2, i.e. |z + 1| < 1.

Complex Integration

4.37

(ii) Since the region of convergence of the required Laurent’s series is given by |u| >1 and |u| < 2, the first term in the R.H.S. (1) should be re-written as a standard binomial in such a way that u occurs in the denominator of the second member and the second term as a standard binomial in such a way that u occurs in the numerator of the second member. Accordingly, 1 1 f ( z) = +  1  u u 1 −  2 1 −   u  2 =

1 1 u (1 − 1 / u )−1 + 1 −  u 2 2

=

1 ∞ 1 1 ∞ un ∑ + ∑ u n = 0 u n 2 n = 0 2n ∞

=∑

n=0

∞

1

( z + 1)

n +1

+∑

n=0

1 2

n +1

−1

( z + 1)n

(3)

The Laurent’s expansion (3) is valid, if

i.e.

1 u < 1 and < 1 i.e., u > 1 and u < 2, u 2 1< |z + 1| < 2.

(iii) Since the region of convergence of the required Laurent’s series is |u| > 2, both the terms in the R.H.S. of (1) should be rewritten as standard binomials in which u occurs in the denominator of the second member. Accordingly, f ( z) =

1  1 u 1 −   u

−

−1

1  2 u 1 −   u

=

1  1 1 −  u  u

=

1 ∞ 1 1 ∞ 2n ∑ − ∑ u n=0 u n u n=0 u n ∞

= ∑ (1 − 2n ) n=0

1  2 − 1 −  u  u

−1

1

( z + 1)n +1

The Laurent’s expansion (4) is valid, if 1 2 < 1and < 1, i.e., u > 1 and u > 2 i.e., z + 1 > 2. u u

(4)

Mathematics II

4.38

Example 4.4 Find all possible Laurent’s expansions of the function

f (z) =

4 − 3z about z =0. Indicate the region of convergence in each case. Find z (1 − z ) ( 2 − z ) also the residue of f (z) at z = 0, using the appropriate Laurent’s series. f (z) =

4 − 3z 2 1 1 = + + , z (1 − z ) ( 2 − z ) z 1 − z 2 − z

on resolving into partial fractions. Various Laurent’s series about z = 0 will be series 2 z unaltered and rewriting the other two terms as standard binomials in three different ways and expanding in powers of z. of ascending and descending powers of z. They can be obtained by keeping

Case (i) 2 1 z −1 f ( z ) = + (1 − z ) + 1 −  z 2 2

−1

=

2 ∞ n 1 ∞ zn +∑z + ∑ n 2 n=0 2 z n=0

=

2 ∞  1  + ∑ 1 + n +1  z n if z < 1 and z < 2, i.e. z < 1.  z n=0 2 

(1)

But |z| < 1 includes z = 0, which is a singularity of f (z). Laurent’s series expansion is valid in an annular region which does not contain any singularity of the function expanded. Hence the region of convergence of the Laurent’s series, in this case, is 0 < |z| 1 and z < 2 z 2

i.e.1< |z| < 2, which represents the region of convergence of series (2).

(2)

Complex Integration

4.39

Case (iii) f (z) =

2 1  1 − 1 −  z z  z

−1

1  2 − 1 −  z z

=

∞ 2 ∞ 1 2n − ∑ n +1 − ∑ n +1 z n=0 z n=0 z

=

2 ∞ ( 1 − ∑ 1 − 2n ) n +1 z n=0 z

−1

(3)

1 2 < 1 and < 1, i.e. z >1 and z > 2 z z i.e. |z| > 2 or 2 < |z| < ∞, which represents the region of convergence of series (3). Since z = 0 is an isolated singularity of f (z), the residue at z = 0 is the coefficient 1 of in the Laurent’s expansion of f (z),valid in the region 0 < |z| < 1, i.e. in the series z (1) above. if

Thus

[Res. f (z)]z = 0 = 2.

Example 4.5 Find the residues of f ( z ) =

z2

( z − 1) ( z + 2)2

at its isolated singularities

using Laurent’s series expansions. f (z) =

1/ 9 8 / 9 4 / 3 , on resolving into partial fractions. z2 = + − 2 z − 1 z + 2 ( z + 2 )2 ( z − 1) ( z + 2 )

Both z = 1 and z = – 2 are isolated singularities of f (z). To find the residue of f (z) at z = 1, we have to expand f (z) in series of powers of (z – 1), valid in 0 < |z – 1| < r and find the coefficient of Thus

f (z) =

1/ 9 8/9 4/3 + − z − 1 3 + ( z − 1) {3 + ( z − 1)}2 −1

=

1 in it. z −1

1 / 9 8  z − 1 4  z − 1 + 1 +  − 1 +  z − 1 27  3  27  3 

−2

∞ ( )n ( )n 1/ 9 8 ∞ ( −1)n z −n1 − 4 ∑ ( −1)n (n + 1) z −n1 + ∑ z − 1 27 n = 0 27 n = 0 3 3 z − 1 The expansion (1) is valid in the region < 1, i.e. 0 < z − 1 < 3. 3

=

[Residue of f (z) at z =1] = coefficient of 1 = . 9

1

( z − 1)

in (1)

(1)

Mathematics II

4.40

To find the residue of f (z) at z = –2, we have to expand f (z) in series of powers of (z + 2), valid in 0 < |z + 2|< r and find the coefficient of f ( z) =

Thus

1/ 9

( z + 2) − 3

+

8/9 4/3 − z + 2 ( z + 2 )2

1 in it. z+2

−1

=−

1  z + 2 8/9 4/3 − 1 −  + 27  3  z + 2 ( z + 2 )2

=−

1 ∞ ( z + 2) 8/9 4/3 + − ∑ 27 n = 0 3n z + 2 ( z + 2 )2 n

(2)

The expansion (2) is valid in the region z+2 < 1, i.e.0 < z + 2 < 3. 3 [Residue of f (z) at z = –2] = Coefficient of 8 = . 9

1 in (2) z+2

Example 4.6 Find the singularities of f ( z ) =

z2 + 4 and the corresponding z + 2z2 + 2z 3

residues. The singularities of f (z) are given by z 3 + 2 z 2 + 2 z = 0, z ( z 2 + 2 z + 2) = 0.i.e, z = 0, −1 ± i

i.e

For these values of z, the numerator (z2 + 4) does not vanish. Also it is analytic everywhere. ∴ z = 0, −1 ± i are simple poles of f ( z ) . 

z2 + 4   = 2. z2 + 2z + 2  z2 + 4  = [( z + 1 − i ) f ( z )]z = −1+ i =    z ( z + 1 + i )  z = −1+i

[Res. f ( z )]z =0 =  z ⋅ f ( z ) z =0 =  [Res. f ( z )]z = −1+i

2−i 1 = ( −1 + 3i ) −1 − i 2 Since z = –1– i is the conjugate of z = –1+ i, =

−1  Res. f ( z )  z =−1−i = (1 + 3i ) . 2 Example 4.7 Identify the singularities of f ( z ) =

z2

( z − 2 )2 ( z 2 + 9 )

residue at each singularity. The singularities of f (z) are given by

and also find the

Complex Integration

4.41

( z − 2)2 ( z 2 + 9) = 0, i.e. z = 2, ± i3. Of these z = 2 is a double pole and z = ± i3 are simple poles. 1d 2  Res. f ( z ) z =2 =  ( z − 2 ) f ( z )  1!  dz 

 Res. f ( z )  z =i 3

 d  z2   36  18 z  = =  2   =  2 2   dz z + 9  z = 2  ( z + 9)  z = 2 169   z2  = ( z − i3) f ( z )  z =i 3 =   ( z − 2 )2 ( z + i3)  z =i 3 =

−9 i 6 (i 3 − 2 )

=−

2

=−

3i 2 (5 + 12i )

3 (12 + 5i ) 338

3  Res. f ( z )  z =i 3 = conjugate of − (12 + 5i ) 338 =

−3 (12 − 5i ) 338

z2 and also find the residue at z 4 + a4 each singularity. The singularities of f (z) are given by z4 + a4 = 0, i.e., z4 = (−1) a4 = e i(2r + 1)π. a4 Example 4.8 Find the singularities of f ( z ) =

i.e.

z = ei(2r + 1)π/4 a, where r = 0, 1, 2, 3.

∴ The singularities are z = a ⋅ ei  / 4 , a ⋅ e3 i / 4 ,a ⋅ e5i/4 and a ⋅ e7 i / 4 , all of which are simple poles.  P (z)  P (z) . lim  , as f ( z ) is of the form  Q (z) z → ei  / 4  Q ′ ( z )  1 −i / 4 1  1 (1 − i ) e =  = =  4 z  z = aei  / 4 4a 4 2a

[Res. f ( z )]z = a⋅ei / 4 =

[ Re s. f ( z )]z = aei3 / 4 = 41a e−i3 / 4 = −

1 4 2a

[ Re s. f ( z )]z = aei5  / 4 = 41a e−i5  / 4 =

1 4 2a

[ Re s. f ( z )]z = aei 7  / 4 = 41a e−i 7  / 4 =

4 2a

1

(1 + i )

( −1 + i ) (1 + i )

Mathematics II

4.42

Example 4.9 Evaluate the following integrals, using Cauchy’s residue theorem. cos z 2 + sin z 2 (i) ∫ ( z + 1) ( z + 2) dz , where C is | z |= 3. C (ii) (iii)

∫( ∫

C

(i)

e z dz

, where C is z = 4. 2 z 2 + 2 ) dz , where C is z = 1 z sin z

cos z 2 + sin z 2 ∫ ( z + 1) ( z + 2) dz = ∫ f ( z ) dz, say. C C

The singularities of f (z) are z = –1 and z = −2, which are simple poles lying within the circle |z| = 3.  cos z 2 + sin z 2  Re s. f ( z ) z =−1 =  = −1   z =−1  z+2  cos z 2 + sin z 2  Re s. f ( z ) z =−2 =  = −1   z =−2  z +1 By Cauchy’s residue theorem,

∫ f ( z ) dz = 2i ×sum of the residues

C

= − 4πi. (ii)

Let

∫

C

f ( z ) dz = ∫ C

e z dz

( z 2 +  2 )2

=∫ C

e z dz

( z − i  )2 ( z + i  )2

The singularities of f (z) are z = iπ and − iπ each of which is a double pole of f (z) and lies within the circle |z| = 4. 1d 2  Res. f ( z ) z =i  = R1 =  ( z − i  ) f ( z )  1!  dz 

z =i 

 d  e z  =  2   dz  ( z + i  )  z =i   ( z + i  ) 2 e z − 2e z ( z + i  )  =  ( z + i  )4  z = i   =

ei ( 2i  − 2 ) 3 3

8i 

=

( −1) ( i  − 1) −4i 

3

=

1 4 3

( + i)

Complex Integration

[ Res. f ( z )]z =−i = R2 =

4.43

1 ( − i) 4 3

By Cauchy’s residue theorem,

∫ f ( z ) dz = 2 i ( R1 + R2 )

C

= (iii)

Let

i 2 i × 2 = 3  4

dz

∫ z sin z = ∫ f ( z ) dz C

C

The singularity of f (z) is given by   z3 z5 z sin z = 0, i.e.z  z − + − = 0 3! 5 !    z2 z4  z 2 1 − + − = 0  3! 5! 

i.e.

i.e. z = 0, which is a double pole of f (z) lying within the circle |z| = 1. 



2



z [ Res. f ( z )]z = 0 = 11!  ddz  z sin  z 

 z=0

 sin z − z cos z  0  = →  form  2    0 sin z z=0  = 

z   , by L’Hospital’s rule 2 cos z  z = 0

=0 By Cauchy’s residue theorem.

∫ f ( z ) dz = 2 i × 0 = 0.

C

Example 4.10 Evaluate the following integrals, using Cauchy’s residue theorem. (i)

( z + 1) dz

∫ z 2 + 2 z + 4 , where C is |z + 1 + i| = 2

C

(ii)

(12 z − 7 ) dz

∫ ( z − 1)2 (2 z + 3), where C is z + i =

C

(iii)

∫(

C

dz z + 9) 2

3

, where C is z − i = 3

3

Mathematics II

4.44

(i) Let

( z + 1) dz

∫ f ( z ) dz = ∫ z 2 + 2 z + 4

C

C

(– 1 + i 3 )

y

x

O (– 1 – i) (– 1 – i 3 )

Fig. 4.19

The singularities of f (z) are given by z2 + 2z + 4 = 0, i.e. (z + 1)2= −3, i.e. z = −1± i 3 , each of which is a simple pole. Of the two poles z = −1−i 3 alone lies inside the circle |z−(−1−i)| = 2, as shown in the Fig.4.19. 1 = {Res. f ( z )}z =−1−i 3 =  z + 1   z + 1 − i 3  z =−1−i 3 2 By Cauchy’s residue theorem, 1 ∫ f ( z ) dz = 2 i × 2 = i C (ii) Let

(12 z − 7 )

∫ ( z − 1)2 (2 z + 3) dz = ∫ f ( z ) dz.

C

C

The singularities of f (z) are z = 1 and z = −3/2. z = 1 is a double pole and z = −3/2 is a simple pole. Only the pole z = 1 lies inside the circle |z−(−i)|= 3 , as shown in the Fig. 4.20.

[ Res. f ( z )]z =1 = 11!  ddz  122 zz+−37   

y

2 O

z = – 3/2

2 z=1

3

3 z=–i

 z =1

 50  = 2. = 2  ( 2 z + 3)  z =1 By Cauchy’s residue theorem,

Fig. 4.20

x

Complex Integration

∫f

4.45

( z )dz = 2 i × 2 = 4 i.

C

(iii) Let

dz

∫ f ( z )dz = ∫ ( z 2 + 9)3

C

C

The singularities of f (z) are z = ±3i, of which z = 3i lies inside the circle |z − i| = 3, as shown in Fig. 4.21. y

z = 3i is a triple pole of f (z). ∴

[Res. f ( z )]z =3i = =

=

 1  d2 1  2 3 2 !  dz ( z + 3i )  z =3i 1  12    2 !  ( z + 3i )5  z = 3i 6 5 5

6 i

=

z = 3i

z=i x

O

1 1296i z = – 3i

By Cauchy's residue theorem, Fig. 4.21

1



∫ f ( z )dz = 2 i × 1296i = 648

C

EXERCISE 4(b) Part A (Short Answer Questions) 1. Define radius and circle of convergence of a power series. 2. State Taylor's theorem. 3. State Laurent's theorem. 4. What do you mean by analytic part and principal part of Laurent's series of a function of z? 5. Define regular point and isolated singularity of f (z). Give one example for each. 6. Define simple pole and multiple pole of a function f (z). Give one example for each. 7. Define essential singularity with an example. 8. Define removable singularity with an example. 9. Define entire function with an example. 10. Define meromorphic function with an example. 11. Define the residue of a function at an isolated singularity. 12. Express the residue of a function at an isolated singularity as a contour integral.

Mathematics II

4.46

13. State two different formulas for finding the residue of a function at a simple pole. 14. State the formula for finding the residue of a function at a multiple pole. 15. State Cauchy's residue theorem. 16. Derive Cauchy's integral formula as a particular case of Cauchy's residue theorem. Find the Taylor series for each of the following functions about the indicated point: 17. sin z about z = π/4 18. cos z about z = π/3 19. ez about z = –i 20. e–z about z = 1 21.

z −1

about z = 1 z2 If each of the following functions is expanded as a Taylor's series about the indicated point, find the region of convergence in each case, without actually expanding. z −1 22. about z = 0 23. sin z about z = 0 z +1 z2 + 4 24. sec πz about z = 1

25.

z+3 about z = 2 ( z − 1)( z − 4)

e z about z = 4i z ( z −1) (Hint: The radius of convergence is the distance between the centre of the Taylor's series and the nearest singularity of the concerned function.) Find the Laurent's series of each of the following functions valid in the indicated regions: 26.

27.

1 2

z ( z − 2)

,valid in 0 < | z | < 2

28.

1 , valid in 0 < | z – 1 | < 1 z ( z − 1)

29.

1 , valid in | z | > 1 z (1 − z ) 3

30. z −1 , valid in | z – 1 | > 1 z2 Find the residue at the essential singularity of each of the following functions, using Laurent's expansions: z sinh z cos z 31. e1/z 32 1 − e 33. 34. 35. 1 − cosh z 4 2 z z z3 z

Complex Integration

4.47

Find the residues at the isolated singularities of each of the following functions: z z.2 36. 37. ( z + 1)( z − 2) z 2 + a2 ze z 38. cot z (at z = 0) 39. ( z −1) 2

40. z sin z . ( z −  )3 Evaluate the following integrals using Cauchy's residue theorem. 41.

z +1

∫ z ( z − 1)

dz, where C is |z| = 2

C

42.

z2

∫ z 2 + 1 dz, where C is |z| = 2

C

43.

dz

∫ sin z

dz, where C is |z| = 1

C

44.

∫ tan zdz , where C is |z| = 2

C

45.

e− z dz , where C is |z| = 1 2 z C

∫

Part B 46. Find the Taylor's series expansion of f ( z ) =

1 about z = –1. State also z (1 − z )

the region of convergence of the series. z about z = i. z ( z + 1)( z + 2) State also the region of convergence of the series.

47. Find the Taylor's series expansion of f ( z ) =

48. Find the Laurent's series of f ( z ) =

z2 −1 2

z + 5z + 6

valid in the region (i) |z| < 2,

(ii) 2 < |z| < 3, and (iii) |z| > 3. 49. Find the Laurent's series of f ( z ) =

z , valid in the region (i) |z + 2| ( z − 1)( z − 2)

< 3, (ii) 3 < |z + 2| < 4, and (iii) |z + 2| > 4. 7z − 2 z ( z − 2)( z + 1) about z = –1. Indicate the region of convergence in each case. Find also the residue of f (z) at z = –1.

50. Find all possible Laurent's expansions of the function f ( z ) =

Mathematics II

4.48

all possible Laurent's expansion of the function f ( z ) = z − 6z −1 about z = 3. Indicate the region of convergence ( z − 1)( z − 3)( z + 2)

51. Find

3

in each case. Find also the residue of f (z) at z = 3. 52. Find the residues of f ( z ) =

z2 ( z − 1) 2 ( z − 2)

at its isolated singularities, using

Laurent's series expansions. 53. By finding appropriate Laurent's expansions for f ( z ) =

1 2

z ( z 2 + 1)

, find the

residue at the poles of f (z). 54. Indentify the singularities of f (z) and find the corresponding residues, when (i) f ( z ) =

z

(ii) f ( z ) =

2

z + 2z + 5

1 2

2

z ( z + 2 z + 2) f ( z) =

55. Find the singularities of the following: (i)

z ( z + 1)( z 2 − 1)

2

z − 2z

(ii) f ( z ) =

; also find the corresponding residues. ( z + 1) 2 ( z 2 + 4) 56. Indentify the singularities of f (z) and the corresponding residues, when f ( z) =

(i)

57. Evaluate (i)

1 z + 16

(ii) f ( z ) =

4

∫ ze

1/ z

dz ; and (ii)

C

z z + a4 4

dz

∫ z 2 sin z ,

C

using Cauchy's residue theorem, where C is the circle |z| = 1 in both cases. 58. Use Cauchy's residue theorem to evaluate the following (i)

dz

∫ sinh z , where C is the circle | z | = 4

C

(ii)

z

∫ cos z dz, where C is the circle

C

z−

  = 2 2

59. Use Cauchy's residue theorem to evaluate the following: (i)

z dz

∫ ( z − 1)( z − 2)2 , where C is circle |z – 2| = 1/2

C

(ii)

(3 z 2 + z )dz

∫ ( z − 1)( z 2 + 9) , where C is the circle |z – 2| = 2

C

(iii)

(3 z 2 + z + 1)dz

∫ ( z 2 − 1)( z + 3) , where C is the circle | z – i | = 2

C

Complex Integration

4.49

60. Use Cauchy's residue theorem to evaluate the following: ( z − 3)dz , where C is the circle | z + 1 – i | = 2 (i) ∫ 2 C z + 2z + 5 (ii)

dz

∫ ( z 2 + 4)3 , where C is the circle |z – i| = 2

C

(iii)

( z − 1)dz

∫ ( z + 1)2 ( z − 2) , where C is the circle |z + i| = 2

C

4.5

CONTOUR INTEGRATION—EVALUATION OF REAL INTEGRALS

The evaluation of certain types of real definite integrals can be done by expressing them in terms of integrals of complex functions over suitable closed paths or contours and applying Cauchy's residue theorem. This process of evaluation of definite integrals is known as contour integration. We shall consider only three types of definite integrals which are commonly used in applications.

4.5.1 Type 1 ∞

Integrals of the form

P( x)

∫ Q( x) dx, where P(x) and Q(x) are polynomials in x. This

−∞

integral converges (exists), if (i) The degree of Q(x) is at least 2 greater than the degree of P(x). (ii) Q(x) does not vanish for any real value of x. P( z ) dz , To evaluate this integral, we evaluate ∫ Q( z ) y C |z|=R where C is the closed contour consisting of the S real axis from – R to + R and the semicircle S above the real axis having this line as diameter, R by using Cauchy's residue theorem and then letting R → ∞ . The contour C is shown in Fig. –R O R 4.22. Fig. 4.22 Now we consider a result, known as Cauchy's Lemma, which will be used in evaluation of ∫ f ( z )dz , where S is the semicircle |z| = R above the real axis.

y

S

Cauchy's Lemma If f (z) is a continuous function such that |zf (z)| → 0 uniformly as |z| → ∞ on S, then ∫ f ( z )dz → 0, as R → ∞, where S is the S

semicircle |z| = R above the real axis.

Mathematics II

4.50 Proof 1

∫ f ( z )dz = ∫ zf ( z ) ⋅ z dz S

S

≤ ∫ zf ( z ) ⋅ S



1 ⋅ dz |z|

1

∫ | zf ( z ) | ⋅ R ⋅ Rd

i.e.

0

[∵ When |z| = R, z = Reiθ and dz = Rieiθdθ] ∴

lim

R →∞

| zf ( z ) | ∫ f ( z )dz ≤  × Rlim →∞ S

i.e. ∴

≤ 0, by the given condition.

∫ f ( z ) dz

= 0 and hence

S

Note

∫ f ( z )dz = 0, as R → ∞. S

Similarly, we can shown that

∫ f ( z )dz → 0 as r → 0, where S1 is the

S1

semicircle |z – a| = r above the real axis, provided that f (z) is a continuous function such that |(z – a) f (z) | → 0 as r → 0 uniformly.

4.5.2 Type 2 ∞

Integrals of the form polynomials in x.

∞

P ( x) sin m x P( x) cos mx ∫ Q( x) dx or ∫ Q( x) dx, where P(x) and Q (x) are −∞ −∞

This integral converges (exists), if (i) m > 0 (ii) the degree of Q(x) is greater than the degree of P(x) (iii) Q(x) does not vanish for any real value of x. To evaluate this integral, we P( z ) imz e dz , where C is the same contour as in Type 1, by using evaluate ∫ Q ( z) C ∞

Cauchy's residue theorem and letting R→ ∞. After getting

P( x) imx e dx, Q ( x) −∞

∫

we find the value of the real part or the imaginary part of this integral as per the requirement.

Complex Integration

4.51

A result, known as Jordan's Lemma, which will be used in the evaluation of

∫ f ( z )e

imz

dz ,

S

where S is the semicircle |z| = R above the real axis, is given as follows.

Jordan’s Lemma If f (z) is a continuous function such that |f (z)| → 0 uniformly as |z| → ∞ on S, then

∫e

imz

f ( z )dz → 0 as R → ∞, where S is he semicircle |z| = R above the real axis and

S

m > 0. Proof |z| R, z = Reiθ.

On the circle ∴

∫e

imz

∫e

f ( z ) dz =

S

imR (cos  + i sin  )

. f ( z ).iRei d

S



≤ ∫ eimR (cos  + i sin  ) | f ( z ) | Rd 0 

≤ ∫ e − mRsin  | f ( z )| Rd

i.e.

0

/ 2

≤2

i.e.

∫

e − mRsin  | f ( z ) | Rd

(1)

0

Now

sin  ≥

2 , for 0 ≤  ≤  / 2  2mR , for 0 ≤  ≤  / 2 

∴

− mR sin  ≤ −

∴

e − mR sin  ≤ e −2 mR /  , for 0 ≤  ≤  / 2

(2)

Using (2) in (1), we have imz ∫ e f ( z ) dz ≤ 2 S

∴

lim

R →∞

/ 2

∫ 0

imz ∫ e f ( z )dz ≤ S

e −2 mR /  | f ( z )| Rd / 2

∫ 0

lim (2 Re −2 mR /  ) lim [ f ( z )]d

R →∞

R →∞

/ 2

i.e.

i.e.

≤ lim [ f ( z )] × lim R →∞

R →∞

∫

2 Re −2 mR /  d

0

π  ≤ lim [ f ( z )] × lim  1 − e − mR  R →∞ R →∞  m 

(

)

Mathematics II

4.52 ≤

i.e.

∫e

∴

imz

S

 × 0, by the given condition. m

f ( z )dz = 0 and hence ∫ eimz f ( z )dz = 0 S

as R → ∞.

4.5.3 Type 3 2

Integrals of the form

P (sin , cos )

∫ Q (sin , cos ) d, where P and Q are polynomials in sin θ 0

and cos θ. To evaluate this integral, we take the unit circle |z| = 1 as the contour. When |z| = 1, z = eiθ and so sin  =

z + z −1 z − z −1 and cos  = . Also dz = eiθ 2 2i

idθ or d = dz . When θ varies from 0 to 2π, the point z moves once around the unit iz circle |z| = 1. 2

Thus

P(sin , cos )

∫ Q(sin , cos ) d = ∫ f ( z ) dz, 0

C

where f (z) is a rational function of z and C is |z| = 1. Now applying Cauchy’s residue theorem, we can evaluate the integral on the R.H.S.

WORKED EXAMPLE 4(c) ∞

Example 4.1

Evaluate

x 2 dx

∫ ( x2 + a 2 ) ( x2 + b2 ) ,

using contour integration, where

−∞

a > b > 0. Consider

z 2 dz

∫ ( z 2 + a 2 ) ( z 2 + b2 ) , where C is the contour consisting of the segment

C

of the real axis from − R to + R and the semicircle S above the real axis having this segment as diameter. The singularities of the integrand are given by (z2 + a2) (z2 + b2) = 0 i.e. z = ± ia and z = ± ib, which are simple poles. Of these poles, only z = ia and z = ib lie inside C. (Fig. 4.23)

Complex Integration

4.53

y S

x R

O

R

Fig. 4.23

  z2  R1 = (Res. of the integrand)z = ia =  2 2  ( z + ia ) ( z + b )  z =ia

Now

=

−a2

=

2ia (b 2 − a 2 )

a

2i ( a 2 − b 2 )

R2 = Res. at (z = ib) = −

Similarly

b

2i ( a − b 2 ) 2

By Cauchy’s residue theorem, z2

∫ ( z 2 + a 2 )( z 2 + b2 ) dz = 2 i( R1 + R2 )

C

= R

i.e.,

Now Similarly ∴

2

 (a − b) 2

a −b

2

=

 a+b

z 2 dz

π (1) ( ) (z + b ) a + b −R S z +a ( on the real axis, z = x and so dz = dx) |z2 + a2| ≥ |z|2 − a2 = R2 − a2 on |z| = R. |z2 + b2| ≥ R2 − b2 on |z| = R. x dx

∫ ( x2 + a 2 ) ( x2 + b2 )

z.

(z

z2 2

+a

2

) (z

2

+b

2

+∫

2

2

≤

2

2

) (R

2

2

=

R3 −a

) ( R 2 − b2 )

Since the limit of the R.H.S. is zero as R → ∞, lim z.

R→∞

∴ By Cauchy’s lemma,

z2

( z 2 + a 2 )( z 2 + b2 )

= 0, on | z |= R.

z 2 dz

∫ ( z 2 + a 2 ) ( z 2 + b2 ) = 0 S

Using (2) in (1) and letting R → ∞, we get ∞

x 2 dx



∫ ( x2 + a 2 ) ( x2 + b2 ) = a + b

−∞

(2)

Mathematics II

4.54 ∞

Example 4.2 Evaluate

Consider ∫

C

x 2 dx , using contour integration. ∫ 2 2 2 −∞ ( x + 1 )( x + 2 x + 2)

z 2 dz

( z 2 + 12 ) ( z 2 + 2 z + 2)

, where C is the same contour as in the previous

example. The singularities of f ( z ) =

z2

( z 2 + 1)2 ( z 2 + 2 z + 2)

are given by (z2 + 1)2

(z2 + 2z + 2) = 0 i.e. z = ± i and z = − 1 ± i, of which z = i and z = − 1 + i lie inside C. z = i is double pole and z = − 1 + i is a simple pole. R1 = [Re s. f ( z ) z = i =

 1  d  ( z − i)2 ⋅ z 2    2 2 2 1!  dz  ( z + i ) ( z − i ) ( z + 2 z + 2) 

z =i

 ( z + i ) 2 ( z 2 + 2 z + 2).2 z − z 2 {2( z + i )( z 2 + 2 z + 2) + ( z + i )2 (2 z + 2)}   =  2   ( z + i )4 ( z 2 + 2 z + 2 ) z =i =

9i − 12 100

  z2 3 − 4i = R2 = [Re s. f ( z )]z =−1+i =   2 25  ( z 2 + 1) ( z + 1 + i )  z =−1+i By Cauchy’s residue theorem,

∫(

C

z 2 dz z 2 + 1)

2

( z 2 + 2z + 2)

= 2 i ( R1 + R2 )  9i − 12 3 − 4i  = 2 i  +  25   100 7π 50

= R

∫(

i.e.

−R

lim

R →∞

x 2 dx x 2 + 1)

2

( x2 + 2 x + 2)

z. z 2

( z 2 + 1)2 ( z 2 + 2 z + 2 )

+∫ S

z 2 dz

( z 2 + 1) ( z 2 + 2 z + 2 ) 2

= 0 on | z |= R.

=

7 50

(1)

Complex Integration

4.55

∴ By Cauchy’s Lemma, when R → ∞

∫( S

z 2 dz

=0

z 2 + 1) ( z + 2 z + 2 ) 2

(2)

Letting R → ∞ in (1) and using (2), we get ∞

∫(

−∞

x2d x x + 1) 2

( x + 2x + 2)

2

2

∞

Example 4.3 Evaluate

∫( 0

Consider

∫(

C

dz z 2 + a2 )

3

=

dx x + a2 )

3

2

7 50 , using contour integration, where a > 0.

, where C is the same contour as in example 4.1.

The singularities of f ( z ) =

1

(z

+ a2 ) the two poles, z = ia alone lies inside C.

3

2

are z = ± ia,which are poles of order 3. Of

1  d2  2 !  dz 2

R = [Re s. f ( z )]z =ia =

  1  3   ( z + ia )  z =ia

  6 3 = = 5 5  ( z + ia )  z =ia 16a i ∴ By Cauchy’s residue theorem,

∫(

C R

∫(

i.e.

−R

dx 2

x +a

)

2 3

dz 2

z +a

+∫ S

z.

Now

lim z.

R →∞

= 2 iR = 2 i .

)

2 3

dz

(z

2

+a

)

2 3

1

(z

2

(z

2

+a

)

2 3

1 + a2 )

3

=

≤

3 16a 5i

=

3 8a 5

3 (1)

8a 5 R

(R

2

− a2 )

3

and so

= 0 on |z| = R

∴ By Cauchy’s Lemma, when R → ∞,

∫( S

dz z 2 + a2 )

3

=0

(2)

Mathematics II

4.56

Letting R → ∞ in (1) and using (2), we get ∞

∫(

−∞

dx x2 + a

=

)

2 3

3 8a 5

.

1 Since the integrand ∞ dx 3 ∫ ( 2 2 )3 = 16a5 0 x +a

(x

2

is an even function of x, we have

+ a2 )

3

Example 4.4 Use contour integration to prove that ∞

x 2 dx

∫ x4 + a4 = 2



0

Consider

2a

, where a > 0.

z 2 dz

∫ z 4 + a 4 , where C is the same contour as in Example 4.1.

C

z2

The singularities of f ( z ) =

4

4

are given by z4 = − a4 = ei(2r + 1)π. a4

i.e.

z +a z = ei(2r + 1)π/4. a, where r = 0, 1, 2, 3.

i.e.

z = aeiπ/4, aei3π|4, aei5π/4, aei7π/4, all of which are simple poles.

Of these poles, only z = aeiπ/4 and z = aei3π/4 lie inside C.  z2   P( z )  R1 = [Res. f ( z )]z = aei  / 4 = lim  3   = lim 1  i / 4 z → ae  4 z   z →a Q ( z )  1 − i/ 4 1 e = (1 − i ) 4a 4 2a 1 −i3 / 4 1 ( −1 − i ) R2 = [Res. f ( z ) z = aei 3  / 4 = e = 4a 4 2a ∴ By Cauchy’s residue theorem, =

z 2 dz

∫ z 4 + a 4 = 2 i ⋅ 4

C R

i.e.

Now

∫

x 2 dx 4

−R x + a

4

+∫ S

lim z.

r →∞

z 2 dz 4

z + a4 z2 4

z + a4

=

1 2a

(1 − i − 1 − i ) =

 2a

 2a (1)

= 0 on | z |= R.

∴ By Cauchy’s Lemma, when R → ∞, z 2 dz

∫ z 4 + a4 = 0 s

(2)

Complex Integration

4.57

Letting R → ∞ in (1) and using (2), we get ∞

∫

−∞

x 2 dx 4

x +a

4

=

 2a

Since the integrand is even, ∞

x 2 dx

∫ x4 + a4 = 2 0

 2a ∞

Example 4.5 Evaluate

∫

−∞

x4 6

x − a6

dx, using contour integration where a > 0.

z4 ∫ z 6 − a6 dz, where C is the same contour as in example 4.1, with some C modifications explained below. Consider

The singularities of f ( z ) =

z4 are given by z − a6 6

r

i

z6 = a6 = ei2rπ a6, i.e. z = ae 3 , r = 0, 1, 2, 3, 4, 5. i.e. z = a, aeπi/3, ae2πi/3, − a, ae4πi/3, ae5πi/3. Of these singularities (simple poles), z = aeπi/3 and z = ae2πi/3 lie inside C, but z = a and z = − a lie on the real axis. But for the evaluation of the integrals of type 1, no singularity of f (z) should lie on the real axis. To avoid them, we modify C by introducing small indents, i.e., semi-circle of small radius at z = ± a, as shown in Fig. 4.24. y S S2

S1 –R

z=–a

O

z= a

R

x

Fig. 4.24

Now the modified contour C contains only the simple poles z = aeπi/3 and ae2πi/3 R1 = [Re s. f ( z )]z = ae  / 3i  z 4  1 − i / 3 e = 5 =  6 z  6a =

1 1 3 −i  6a  2 2 

Mathematics II

4.58

1 −2 i 3 1  3 = e  − 1 2 − i  6a 6a 2  By Cauchy’s residue theorem, R2 =

Similarly,

z4

∫ z 6 − a6 dz =

C

= –a–r

i.e.

–R

2 i  3 3 −1 2 − i  1 2 − i 6a 2 2   3a a – r'

x4 z4 x4 z4 dx + dz+ dx+ dz 6 6 6 6 6 6 6 x –a z –a x –a z – a6 S –a+r S 1

2

R

∫

+

a+r′

x4 6

x −a

6

dx +  ∫ S

z4 6

z −a

6

dz =

 3a

(1)

where r and r' are the radii of the semicircles S1 and S2 whose equations are |z + a| = r and |z – a| = r'. These two integrals taken along S1 and S2 vanish as r → 0 and r' → 0, by the note under Cauchy’s Lemma. z4 ∫ z 6 − a6 = 0 , as R→∞, by Cauchy’s lemma. S Now, letting r → 0, r' → 0 and R → ∞ in (1), −a

∫

We get

−∞

∞

a

+ ∫ +∫ −a

a

x 4 dx  = x − a6 3a 6

∞

 x4 ∫ x6 − a6 dx = 3a −∞

i.e. ∞

Example 4.6 Evaluate ∫ 0

x sin x x2 + a2

dx , by contour integration.

zeiz dz , where C is the same contour as in Example 4.1. z 2 + a2 C

Consider ∫

zeiz . are z = ± ia, which are simple poles. Of z 2 + a2 these poles, only z = ia lies inside C. The singularities of f ( z ) =

 zeiz  1 = e− a . [Res. f (z)]z = ia =   z ia +   z = ia 2

Complex Integration

4.59

By Cauchy’s residue theorem, zeiz

1

∫ z 2 + a 2 dz = 2 i . 2 e

−a

= ie − a

C

R

zeiz xeix −a x + d ∫ x 2 + a 2 ∫ z + a 2 dz = ie −R S

i.e.

(1)

z R ≤ 2 2 z +a R − a2 Since the limit of the R.H.S. is zero as R → ∞, Now

2

lim

R →∞

z 2

z + a2

= 0 on |z| = R.

∴ By Jordan’s Lemma,

zeiz

∫ z 2 + a 2 dz

→ 0 as R → ∞,

(2)

S

Letting R→∞ in (1) and using (2), we get ∞

xeix dx = i e − a 2 2 x +a −∞

∫

∞

x (cos x + i sin x)dx = i e − a . 2 x a + −∞

∫

i.e.

2

Equating the imaginary parts on both sides, ∞

x sin x dx = e − a . 2 2 x +a −∞

∫

we get

Since the integrand is an even function of x, ∞

x sin x



∫ x 2 + a 2 dx = 2 e

−a

.

0

∞

Example 4.7 Evaluate

−∞

a > b > 0. Consider

cos xdx

∫ ( x 2 + a 2 ) ( x 2 + b2 ) ., using contour integration, where

eiz

∫ ( z 2 + a 2 ) ( z 2 + b2 ) dz , where C is the same contour as in Example 4.1.

C

The singularities of the integrand f (z) are given by z = ± ia and z = ± ib, which are simple poles. Of these poles, z = ia and z = ib only lie inside C.

Mathematics II

4.60

 eiz R1 = [Res. f ( z )]z =ia =   ( z + ia ) z 2 + b 2 

(

e− a

=

(

)

    z =ia

)

−2ia a 2 − b 2

R2 = [Res. f ( z )]z =ib =

e −b 2ib(a 2 − b 2 )

By Cauchy’s residue theorem,

∫

C

(z

eiz 2

+ a2

)( z

2

+ b2

dz = 2 i ( R1 + R2 )

)

= R

∫

i.e.

−R

(x

eix 2

+ a2

)( x

2

+ b2

)

dx + ∫ S

=

Now

1

(z

2

+a

)( z

2

(a

 2

(z

−b

2

 e−b e− a  −   b a 

)

eiz 2

+ a2

)( z

)

dz

  e−b e− a  −   a  a − b2  b 1

)

≤

2

+ b2

)

= 0 on |z| = R.

2

+ b2

)

+b

+ b2

(1)

2

2

2

2

2

2

(

( R − a ) R 2 − b2

)

R.H.S. → 0 as R → ∞. Hence lim

R→∞

∴ By Jordan’s Lemma,

∫ S

(z (z

1

2

+a

2

)( z

eiz 2

+ a2

)( z

dz → 0 , as R → ∞

Letting R→∞ in (1) and using (2), we get ∞

∫

−∞

(x

eix 2

+ a2

)( x

2

+ b2

)

dx =

(a

 2

− b2

)

 e−b e− a  −   b a 

Equating the real parts on both sides, we get ∞

∫

−∞

(x

cos x 2

+ a2

)( x

2

+ b2

)

dx =

(a

 2

− b2

)

 e−b e− a  −   b a .

(2)

Complex Integration ∞

Example 4.8 Evaluate

4.61



cos ax

∫ ( x 2 + b2 ) dx = 4b3 (1 + ab)e

− ab

, where a > 0 and b > 0.

0

Consider

∫(

C

eiaz z 2 + b2 )

2

dz , where C is the same contour as in Example 4.1.

The singularities of f ( z ) =

eiaz

are z = ± ib, which are double poles. Of these

( z 2 + b 2 )2

poles, only z = ib lies inside C. 1 d 

[Res. f ( z )]z =ib = 1! dz 

eiaz



2

 ( z + ib )   z = ib

 ( z + ib)ia eiaz − 2eiaz  =  ( z + ib )3  z = ib  1 (ab + 1) e − ab . 4ib3

= By Cauchy's residue theorem, R

∫(

−R

eiax x +b 2

dx + ∫

)

2 2

S

eiaz

(z

2

+b

)

2 2

dz = 2 i ×

= Now

1

≤

(z + b ) 2

2 2

1 (ab + 1)e − ab 4ib3

 (ab + 1)e − ab 2b3

(1)

1

( R − b 2 )2 2

Since the R.H.S.→ 0 as R → ∞, L.H.S. also → 0 as R → ∞ on |z| = R. ∴ By Jordan's Lemma,

∫( S

eiaz z 2 + b2 )

2

dz → 0 as R → ∞, since a > 0

Letting R → ∞ in (1) and using (2), we get ∞

∫(

−∞

eiax 2

x +b

)

2 2

dx =

 ( ab + 1)e − ab . 2b3

(2)

Mathematics II

4.62

cos ax Equating the real parts on both sides and noting that is an even function of x, we get ( x 2 + b 2 )2 ∞

∫( 0

cos ax 2

x +b

)

2 2

dx =

 4b

3

( ab + 1) e− ab .

Example 4.9 Use contour integration to prove that ∞

 sin mx dx = , when m > 0. x 2 0

∫

eimz dz , where C is the usual semicircular contour, but with an indent i.e. z C

Consider ∫

a small semicircle at the origin, which is introduced to avoid the singularity z = 0, which lies on the real axis. The modified contour is shown in Fig. 4.25. y S S' –R

–r O r

R

x

Fig. 4.25

eimz . The modified contour C does not include any singularity of f ( z ) = z ∴ By Cauchy's residue theorem, –r –R

e imx dx e imz dz + + x z S'

R r

e imx dx e imz dz = 0 + x z S

(1)

The Eqn. of S′ is |z| = r. ∴ z = reiθ and dz = reiө idθ. As S′ is described in the clockwise sense, θ varies from πto 0. 0

Thus ∴

i

eimz eimre i ∫ z dz = ∫ rei re id  S′ 0

i eimz dz = ∫  lim ( eimr e )  id = − i    r →0  r →0 z S′ 

lim ∫

(2)

1 1 = → 0 as R → ∞ on |z| = R. z R imz ∴ By Jordan’s Lemma, e dz → 0 as R → ∞, since m > 0 ∫ z S

(3)

Complex Integration

4.63

Letting r → 0 and R → ∞ in (1) and using (2) and (3), ∞

0

eimx eimx d −  + x i ∫ x ∫ x dx = 0 0 −∞

we get

∞

eimx dx = i  x −∞

∫

i.e.

Equating the imaginary parts on both sides and noting that

sin mx is an even x

∞

function of x, we get

 sin mx dx = . x 2 0

∫

∞

Example 4.10 Evaluate ∫ 0

Consider ∫

C

eiz

z ( z 2 + a2 )

sin x dx

x ( x2 + a2 )

, using contour integrations where a > 0.

dz , where C is the same modified contour as in Example

4.9, which includes the only pole z = ia of f ( z ) =

eiz

z ( z 2 + a2 )

 eiz  e− a [Res. f ( z )]z =ia =  =   z ( z + ia )  z =ia −2a 2 By Cauchy's residue theorem, –r

–R

eix eiz dx + dz + 2 2 x x2 + a2 s' z z + a

R

r

=− eiz

∫ z ( z 2 + a 2 ) S′

∴

eix eiz dx + dz 2 2 x x 2 + a2 s z z +a i − a e a2 i

0

dz = ∫



(1)

eir e .rei id

(

rei r 2 ei 2  + a 2

)

 eir ei   2 i 2  id z lim ∫ d = lim ∫ r →0 z ( z 2 + a 2 ) r →0  r e + a 2   S′  0

eiz

=− 1

z(z + a 2

2

)

≤

i a2 1

R ( R − a2 ) 2

(2)

Mathematics II

4.64

Since R.H.S. → 0 as R → ∞ on |z| = R, L.H.S. also tends to 0 as R → ∞. ∴ By Jordan's Lemma, eiz

∫ z ( z 2 + a 2 ) dz → 0, as R → ∞

(3)

S

Letting r → 0 and R → ∞ in (1) and using (2) and (3), we get ∞

eix

i

i

∫ ( 2 2 ) dx = a 2 − a 2 e −∞ x x + a

−a

.

Equating imaginary parts on both sides and noting that function of x, we get,

∞



sin x

∫ x ( x 2 + a 2 ) dx = 2a 2 (1 − e

−a

sin x

x ( x2 + a2 )

is an even

).

0

2

d

∫ 1 − 2 x sin  + x 2 (0 < x < 1),

Example 4.11 Evaluate

using contour integration.

0

1 z− 2 dz z = z − 1. On the circle |z| = 1, z = e , dz = ie dθ or d = and sin  = iz 2i 2iz iθ

iθ

dz iz , where C is |z| = 1  z 2 − 1 2 C +x 1 − 2x   2iz 

∴ The given integral I = ∫

i.e.

dz iz − xz + x + ix 2 z C

I=∫

=−

2

1 x C∫

dz 1 dz =− ∫ 1 x    C ( z − ix )  z − z2 − i  x +  z −1    x

The singularities of

1 ( z − ix )  z − 

i  x

are z = ix and z =

i  x

i , which are simple poles x

Now |ix| = |x| < 1, as 0 < x < 1 ∴ The pole z = ix lies inside C, but z =   Res. 

(1)

i lies outside C. x

1   1 ix = = (  ) z − ix z − i x 1 ( )   x2 1 −    z =ix i  x −   x

Complex Integration

4.65

Using Cauchy's residue theorem, from (1) we get 1 ix 2 = I = − × 2 i × . x 1 − x2 1 − x2 Note

We have integrated w.r.t. θ and x is a parameter. 2

Example 4.12 Evaluate

d

∫ a + b cos  (a > b > 0),

using contour integration.

0

2

Deduce the value of

d

∫ (a + b cos )2 .

On the circle |z| = 1, z = eiθ, d =

0

dz and iz

2

cos  = z + 1 . 2z 2

∴I =

d

∫ a + b cos  = ∫ 0

C

dz iz 2

z +1 a+b 2z

=−

2i dz ∫ b C  2 2a  z + 1 z + b  

(1)

where C is |z| = 1. The singularities of the integrand are given by z 2 +

2a z + 1 = 0, b

a a2 i.e. z = − ± 2 − 1, which are simple poles. b b a a2 a a a2 Since a > b, > 1 and hence − − 2 − 1 > 1 and so z = − − 2 − 1 lies b b b b b outside |z| = 1. a a2 z = − + 2 − 1 lies inside |z| = 1. b b 1     Res. of   2a  z + 1   z2 + b   z = − a +  b

=

1     a   2  z +   a b z =− + b

=

b 2

2 a − b2

a2 b2

−1

a2 b2

−1

(2)

Mathematics II

4.66

By Cauchy's residue theorem and using (2) in (1), we get I=−

2

d

∫ a + b cos  =

i.e.

2

2i b × 2 i = 2 b 2 a − b2

0

2

a − b2

2

(3)

2

a − b2

Differentiating both sides of (3) partially with respect to ‘a’, we get 2

0

2

0

2

Example 4.13 Evaluate

a 2 − b2 )

32

2 a

d

∫ (a + b cos )2 = (

∴

2 a

d

∫ − (a + b cos )2 = − (

a − b2 )

32

2

sin 2 

∫ 5 − 3 cos  d, using contour integration. 0

2 2 On the circle |z| = 1, z = eiθ, d = dz , sin  = z − 1 and cos  = z + 1 . iz 2z 2iz

2

I=

∴

∫ 0

( z 2 − 1)2

dz sin  iz , where C is |z| = 1 d = ∫ −4 z 2 5 − 3 cos  ( 3 z + 1) C 5− 2z 2

2

.

( z 2 − 1) dz i ∫ 2( 6 C z z − 3) ( z − 1 3) 2

=−

The singularities of the integrand which lie within C are z = 0 and z = 1/3. 1 z = 0 is a double pole and z = is a simple pole. 3 R1 = [Res. of the integrand f (z)]z = 0 =

2   1  d  ( z 2 − 1)  10 =  1!  dz  2 10 3   z − 3 z + 1  z =0

R2 = [Res. f ( z )]z =1 3

 ( z 2 − 1)2  8  = 2 =− 3  z ( z − 3)  z =1 3

(1)

Complex Integration

4.67

By Cauchy's residue theorem, from (1), we get i   10 8  2  I = − × 2 i ( R1 + R2 ) =  −  = . 6 3  3 3 9 2

Example 4.14

Evaluate

cos 2 d

∫ 1 − 2a cos  + a 2 ,

using contour integration, where

0

a2 < 1. ei 2 

dz z2 + 1 ei 2  + e − i 2  On the circle |z| = 1, z = eiθ, d = , cos  = and coss2 = iz 2z 2 e i 2 z 4 + 1 = . 2z2 z 4 + 1 dz 2 z 2 iz ∴ The given integral I = ∫ , where C is |z| = 1  z 2 + 1 2 C +a 1− a  z 

( z 4 + 1) dz

=

i 2a C∫

=

( z 4 + 1) dz i ∫ 2( 2a C z z − a ) ( z − 1 / a )

 1   z 2  z 2 −  a +  z + 1  a   (1)

The singularities of the integrand which lie inside C are z = 0 and z = a (∵ a2 < 1) z = 0 is a double pole and z = a is a simple pole. R1 = [Res. of the integrand]z = 0 =

 1d  z4 + 1    2 1!  dz  z − ( a + 1 / a ) z + 1 

= a+ R2 = (Res. of the integrand) z = a =

∴

R1 + R2 = a +

1 a a4 + 1

a 2 (a − 1 / a )

1 a4 + 1 + 2 a a (a − 1 / a )

2a 3 a2 − 1 By Cauchy's residue theorem and from (1), we get =

I=

i 2a 3 2 a 2 × 2 i × 2 = 2a a −1 1 − a2

z =0

.

Mathematics II

4.68 2

Example 4.15 Evaluate

cos 3

∫ 5 + 4 cos  d,

using contour integration

0

On the circle |z| = 1, z = eiθ, d =

ei 3θ + e − i 3θ z 6 + 1 z2 + 1 dz and cos3θ = . = , cos  = 2 2z3 2z iz

z 6 + 1 dz . 3 ∴ The given integral I = ∫ 2 z iz , where C is |z| = 1.  2  C 5 + 4 z +1    2z 

i.e.

I =−

( z 6 + 1) dz i ∫ 3( 4 C z z + 2) ( z + 1 / 2)

The singularities of the integrand f (z) which lie inside C are the simple pole z = –1/2 and the triple pole z = 0. 6 65 R1 = (Res. of the integrand)z = –1/2 =  z + 1  =− 3 12  z ( z + 2)  z =−1/ 2

R2, the residue at z = 0 is found out as the coefficient of 1 in the Laurent’s z expansion of the integrand. 1  z  f ( z ) =  z 3 + 3  1 +  (1 + 2 z )    2 z  1   z z2  =  z 3 + 3  1 − + −  {1 − 2 z + 4 z 2 − }    z  2 4

{

}

1 5 21  =  z3 + 3  1 − z + z 2 −    2 4 z ∴ R2 = Coefficient of

21 1 in this expansion = . 4 z

By Cauchy’s residue theorem.   65 21 I = − i / 4 × 2 i  − +  = −  12 4  12

Complex Integration

4.69

EXERCISE 4(c)

Part A (Short Answer Questions) 1. Give the forms of the definite integrals which can be evaluated using the infinite semi-circular contour above the real axis. 2

2. Explain how to convert

∫

f (sin ,cos )d into a contour integral, where f is

0

a rational function.

∞

sin mx dx. x 0

∫

3. Sketch the contour to be used for the evaluation of

Part B Evaluate the following integrals by contour integration technique. ∞

4.

∞

x 2 dx

∫ ( x 2 + 1)( x 2 + 4)

5.

0

0

x2 − x + 2 ∫ x 4 + 10 x 2 + 9 dx −∞ ∞

8.

∫( 0

∞

10.

7.

x 2 + 1)

∞

x 4 dx

∫ x6 + 1 ∞

11.

∫ ( x 2 + 4) ( x 2 + 9)

∞

x sin x

∫ ( x 2 + 1)( x 2 + 4)

13.

16.

∫

−∞ ∞

18.

sin x dx 2

x + 4x + 5 sin x

∫ (2 ) 0 x x +1

x sin x

∫ x 4 + 1 dx 0

∞

dx

15.

∫(

−∞

−∞ ∞

cos ax dx (a > 0; b > 0) x2 + b2 −∞

∫

∞

cos x dx

−∞

14.

x2

∫ x 6 + 1 dx 0

0

12.

dx

0

9.

3

2

∫ x 4 + a 4 dx

∞

x 2 dx

x 2 + 4) ( x 2 + 9)

∞

∞

6.

∫(

x 2 dx

x 2 + 1)

2

dx

∞

17.

x cos x dx x + 2x + 5 −∞

∫

2

dx 2

x sin x

19.

2

d

∫ 1 − 2 p cos  + p 2 ,0 < p < 1 0

Mathematics II

4.70 2

20.

∫

0 2

22.

∫ 0

2

24.

2

d ( a > b > 0) a + b sin  a d 2

2

a + sin 

21.

0 2

( a > 0)

23.

sin 2 

∫ a + b cos  d(a > b > 0) cos 2 d 

∫ 5 + 4 cos  0

2

cos 3 d

∫ 5 − 4 cos 

25.

cos 2 3

∫ 5 − 4 cos 2d  0

0

ANSWERS

Exercise 4(a) 8.

10 (3 + i ) 3

9.

3 – i2

11. − 4 + 8 i 3 3

10. 4πi 12. πi 13. (i) –1/6 + i5/6; (ii) − 1 6 + i 13 15 14. 4πi

15.

10 − i

8 3

16. 0 in all cases 17. (i) 511 − 49 i; (ii) 518 − 57i; (iii) 518 − 8i 3 3 5 3 1 5 5 3 3 18. 19. 96  a + 80  a + 30 a 15

(

20.

)

8 i 3

(i) 0; (ii) 2πi

21.

–4πi

22. (i) 2πi; (ii) 0

23.

−

24. sin t

25.

 16 28. 0

27.

26.

29.

2 3 8 i 3e 2

 (3 + 2i ) 2 20πi; 2π(i – 1); –14πi; 16πi

Exercise 4(b) 17.

sin z =

( z −  4) 2 − ( z −  4)3 +  1  1 + ( z −  4) − 2! 3!  2

Complex Integration

4.71

1 1 3 2 1 − 3 ( z −  3) − ( z −  3) + ( z −  3)3 2  2! 3! 1  4 + ( z −  3) −  4! 

18.

cos z =

19.

 ( z + i ) ( z + i )2  e − i 1 + + +  1! 2!  

20.

 ( z − 1) ( z − 1)2 ( z − 1)3  e −1 1 − + − +  1! 2! 3!  

21. (z – 1) –2(z–1)2 + 3(z–1)3 – 4(z–1)4 + ···· 22. |z| < 1

23.

|z| < 2

1 2

25.

|z – 2| < 1

26. |z – 4i| < 4

27.

−

2 2  1  −  + 1−  2 z − 1 z −1  ( z − 1)  31. 1

32.

–1

33. 1

34.

1 6

1 2

36.

1 2 R( z = −1) = ; R( z = 2) = 3 3

ia 2 39. 2e

38.

1

40.

–1

41. 2πi

42.

0

43. 2πi

44.

–4πi

24.

z −1
3. Res. at ( z = −1) = −3

45 1   1 n + ∑ ( −1)n  n+1 + n+ 2  ( z − 3) , in 0 < |z – 3|< 2 z −3 5  2  2n  45 1 1+ + ∑ ( −1)n  + n+ 2 .( z − 3)n  , in 2 < |z – 3|< 5 + 1 n z −3 5  ( z − 3) 

1+

(iii)

1+

52. (i)

−

45 1 4 + ∑ ( −1)n {2n + 5n −1} , in |z – 3| > 5. Res. at ( z = 3) = n z −3 ( z − 3) 5

3 1 − − 4∑ ( z − 1) n ; Res. at (z = 1) = –3; 0 < |z – 1| < 1 z − 1 ( z − 2) 2

4 n − ∑ ( −1) (n + 4) ( z − 2) n ; Res. at (z = 2) = 4; 0 < |z – 2| < 1 z−2 53. (i) 1 − 1 ∑ {( −1)n + 1} i n z n ,;; Res. at (z = 0) = 0 z2 2 i 2 1  i  (ii) − ∑ (n + 1) + n+ 2  i n (z–i)n; Res. at ( z = i ) = z −i 2 2   (ii)

(iii) 54. (i)

i −i 2 1   n n − ∑ ( n + 1) + n+ 2  ( −1) ( z + i ) ; Res. at ( z = −i ) = − 2 z+i 2   Res. at the simple pole ( z = −1 + 2i ) = and that at ( z = −1 − 2i ) =

1 (2 − i ) 4

1 (2 + i ) 4

Complex Integration

4.73

(ii) z = 0 is a double pole with Res. = –1/2; z = –1 + i is a simple pole with Res. = 1/4; z = –1 – i is a simple pole with Res. = 1/4. 55. (i) z = 1 is a simple pole; Res = 1/4; z = –1 is a double pole; Res. = –1/4 14 (ii) z = –1 is a double pole; Res = − ; z = 2i is a simple pole with 25 1 Res. = (7 + i ) ; z = –2i is a simple pole with Res. = 1 (7 − i) 25 25 56. (i) z = 2eiπ/4, 2ei3π/4, 2ei5π/4, 2ei7π/4 are simple poles. Cor. residues are, 1 1 1 1 ( −1 − i ), (1 − i ), (1 + i ), ( −1 + i ) 32 2 32 2 32 2 32 2 (ii) z = aeiπ/4, aei3π/4, aei5π/4, aei7π/4 are simple poles. Cor. residues are −i i −i i , , , . 4a 2 4a 2 4a 2 4a 2 57. (i) πi; (ii) i ; 3 59. (i) –2πi; (ii) πi; (iii) − i 4 (iii) −2 i 9

2

58. –2πi; (ii) −  i 2 60. (i) π(i – 2); (ii) 3 256

Exercise 4(c) 4. 8. 12.

 6  16   3 2  −  30  e 2 e3 

5. 9. 13.

14.

 (e − 1) 3e 2

17.

 −2  18. e 2 2 a − a 2 − b2 b2

21. 24.

{

 12

15.

 200  6  −1 e 2

10. 2

 2e  3 − 2 4e

}

25.

6.

7. 11.

 2 2a 3  − ab e b

 1  sin   2  16. 19. 22.

3 8

5 12  3

 sin 2 e 2 1 − p2 2 −

1 + a2

2

20. 23.

a − b2 2

 6

Unit-5 LapLace Transforms

UNIT

5

Laplace Transforms 5.1

INTRODUCTION

The Laplace transform is a powerful mathematical technique useful to the engineers and scientists, as it enables them to solve linear differential equations with given initial conditions by using algebraic methods. The Laplace transform technique can also be used to solve systems of differential equations, partial differential equations and integral equations. Starting with the definition of Laplace transform, we shall discuss below the properties of Laplace transforms and derive the transforms of some functions which usually occur in the solution of linear differential equations.

5.1.1  Definition

∞

If f (t) is a function of t defined for all t ≥ 0,

∫e

− st

f (t ) dt is defined as the Laplace

0

transform of f(t), provided the integral exists. Clearly the integral is a function of the parameters s. This function of s is denoted as f (s) or F(s) or f(s). Unless we have to deal with the Laplace transforms of more than one function, we shall denote the Laplace transform of f (t) as f(s). Sometimes the letter ‘p’ is used in the place of s. The Laplace transform of f (t) is also denoted as L{f (t)}, where L is called the Laplace transform operator. ∞

Thus

L { f (t )} = φ ( s ) = ∫ e−st f (t ) dt. 0

The operation of multiplying f (t) by e−st and integrating the product with respect to t between 0 and • is called Laplace transformation. The function f (t) is called the Laplace inverse transform of f(s) and is denoted by L−1{φ (s)}. Thus

f (t) = L−1{f (s)}, when

L {f (t)} = f (s)

Mathematics II

5.4

Note 1. The parameter s used in the definition of Laplace transform is a real or complex number, but we shall assume it to be a real positive number sufficiently large to ensure the existence of the integral that defines the Laplace transform. 2. Laplace transforms of all functions do not exist. For example, L (tan t) and 2

L (et ) do not exist. We give below the sufficient conditions (without proof) for the existence of Laplace transform of a function f (t): Conditions for the existence of Laplace transform If the function f (t) defined for t ≥ 0 is (i) piecewise continuous in every finite interval in the range t ≥ 0, and (ii) of the exponential order, then L{ f (t)} exists.

Note 1. A function f (t) is said to be piecewise continuous in the finite interval a ≤ t ≤ b, if the interval can be divided into a finite number of sub-intervals such that (i) f (t) is continuous at every point inside each of the sub-intervals and (ii) f (t) has finite limits as t approaches the end points of each sub-interval from the interior of the sub-interval. 2. A function f (t) is said to be of the exponential order, if | f (t)| ≤ M eαt, for all t ≥ 0 and some constants M and α or equivalently, if lim{e−αt f (t )} = a finite quantity. t →∞

Most of the functions that represent physical quantities and that we encounter in differential equations satisfy the conditions stated above and hence may be assumed to have Laplace transforms.

5.2

LINEARITY PROPERTY OF LAPLACE AND INVERSE LAPLACE TRANSFORMS

L{k1 f1(t) ± k2 f2(t)} = k1L{f1(t)} ± k2 L{ f2(t)}, where k1 and k2 are constants. ∞

Proof: L{k1 f1 (t ) ± k2 f 2 (t )}= ∫ {k1 f1 (t ) ± k2 f 2 (t )}e−st dt 0 ∞

∞

= k1 ∫ f1 (t ) ⋅ e−st dt ± k2 ∫ f 2 (t ) ⋅ e−st dt 0

0

= k1 ⋅ L{f1 (t )} ± k2 ⋅ L{ f 2 (t )}. Thus L is a linear operator. As a particular case of the property, we get L{k f (t)} = kL{f(t)}, where k is a constant.

Laplace Transforms

5.5

If we take L{f1(t)} = φ1(s) and L{ f2(t)} = φ2(s), the above property can be written in the following form. L{k1 f1(t) ± k2 f2(t)} = k1φ1(s) ± k2φ2(s).

∴

L−1 {k1φ1(s) ± k2 φ2(s)} = k1∙ f1(t) ± k2∙ f2(t) = k1∙L−1{φ1(s) ± k2∙L−1{φ2(s)} −1 Thus L is also a linear operator As a particular case of this property, we get L−1{kf(s)} = k L−1{f (s)}, where k is a constant.

Note 1. L{f1(t) . f2(t)} ≠ L{f1(t)} . L{f2(t)} and L−1{φ (s) . φ (s)} ≠ L−1(φ (s)} × L−1{φ (s)} 1

2

1

2

2. Generalising the linearity properties,  n  n   L ∑ kr f r (t )  = ∑ kr ⋅ L{ f r (t )}     r = 1  r =1 n   n   −1 (ii) L−1  ∑ kr φr ( s )  = ∑ kr ⋅ L {φr ( s )}      r =1  r =1 Using (i), we can find Laplace transform of a function which can be expressed as a linear combination of elementary functions whose transforms are known. Using (ii), we can find inverse Laplace transform of a function which can be expressed as a linear combination of elementary functions whose inverse transforms are known. we get

5.3

(i)

LAPLACE TRANSFORMS OF SOME ELEMENTARY FUNCTIONS

k 1. L{k}= , s > 0, where k is a constant, s ∞

L{k}= ∫ k e−st dt ,by definition. 0 ∞

 e−st   =k   −s   0 =

k (0 −1)[ e−st → 0 as t → ∞, if s > 0] −s

=

k. s

5.6

Mathematics II

In particular, L(0) = 0 and L(1) =

1 s

∴

1   L−1    =1.     s − at 2. L{e } =

1 , where a is a constant, s+a ∞

L{e−at } = ∫ e−st ⋅ e−at dt 0

∞

 e−(s + a )t   dt =   −(s + a )   0

∞

=∫ e

−(s + a )t

0

=

1 (0 −1) , if (s + a) > 0 −(s + a )

=

1 , if s > −a. s+a

 1   = e−at . Inverting, we get L−1   s + a  3. L{e at } =

1 , where a is a constant, if s −a > 0 or s > a. s−a

Changing a to − a in (2), this result follows. The corresponding inverse result is  1  = e at L−1   s − a  4. L(t n ) =

(n + 1) , if s > 0 and n > − 1. sn + 1 ∞

L(t ) = ∫ e−st ⋅ t n dt n

0 ∞  x d x , on putting st = x = ∫ e− x   ⋅  s  s 0 n

=

1 sn + 1

∞

∫e

−x

0

x n dx

5.7

Laplace Transforms =

(n + 1) , if s > 0 and n + 1 > 0 sn + 1 [by definition of Gamma function]

In particular, if n is a positive integer, (n + 1) = n! ∴

n! , if s > 0 and n is a positive integer. sn + 1  n!    Inverting, we get L−1  n + 1  = t n or    s   L(t n ) =

 1   1  L−1  n + 1  = t n    s   n! 1 −1  Changing n to n − 1, we get L  n   s 1  1 n− 1  If n > 0, then L−1  n  = t    s   (n) In particular, L(t ) = 5. L(sin at ) =

1 s2

or

 1  t n− 1 , if n is a positive integer. =  − ( 1 )! n  

1 L−1  2  = t .  s 

a s2 + a2 ∞

L(sin at ) = ∫ e−st sin at dt 0

∞

 e−st  = 2 ( − s sin at − a cos at )  s + a2   0

=−

=

∞ ∞ s a ( e−st sin at)0 − s 2 + a 2 (e−st cos at )0 s2 + a2

a s2 + a2

[ e−st sin at and e−st cos at tend to zero at t → ∞, if s > 0]  a   = sin at . Inverting this result we get L−1  2 2  s + a 

Mathematics II

5.8 6. L(cos at ) =

s s2 + a2 ∞

L(cos at ) = ∫ e−st cos at dt 0 ∞

 e−st  = 2 ( − s cos at + a sin at )  s + a2   0 ∞ ∞ s a e−st cos at ) + 2 e−st sin at ) =− 2 2 ( 2 ( 0 0 s +a s +a s , as per the results stated above. s + a2  s   = cos at . Inverting the above result we get L−1  2 2  s + a  =

2

Aliter L(cos at + i sin at) = L(eiat)

i.e., L(cos at ) + iL (sin at ) =

=

1 , by result (3). s − ia

=

s + ia s2 + a2

s a +i 2 , by linearity property. 2 s +a s + a2 2

Equating the real parts, we get L(cos at ) =

s . s2 + a2

Equating the imaginary parts, we get L(sin at ) = 7. L(sinh at ) =

a s + a2 2

a s − a2 2

1 L(sinh at ) = L  (e at − e−at  2

 ) 

=

1 L (e at ) − L (e−at ) , by linearity property. 2 

=

1  1 1   , if s > a and s > − a. −  2  s − a s + a 

=

a , if s > |a| s − a2 2

Laplace Transforms

Aliter ∴

L(sinh at) = − iL(sin i at) [ = −i ⋅ =

ia , by result (5) s +i a 2

a s −a 2

sin iθ = i sinh θ]

2

2

2

 a    = sinh at. Inverting the above result we get L−1  2 2   − s a     s 8. L(cosh at ) = s −a 2

2

1  L(cosh at ) = L  (e a t + e−a t )  2  =

1 L (e a t ) + L (e−a t ), by linearity property,    2

1 1 1  =  +  , if s > |a|.  2  s − a s + a  =

s , if s > |a|. s − a2 2

Aliter L(cosh at) = L(cos iat) [ cos iq = cosh q] =

s , by result (6) s + i2a2

=

s s − a2

2

2

 s    = cosh at. Inverting the above result we get L−1  2 2    s −a  

5.4

LAPLACE TRANSFORMS OF SOME SPECIAL FUNCTIONS

5.4.1  Definition 0, when t < a  The function f (t ) =  , is called Heavyside’s unit step    1, when t > a, where a ≥ 0 function and is denoted by ua (t) or u (t − a)

5.9

Mathematics II

5.10

0, when t < 0  In particular, u0 (t ) =     1, when t > 0 ∞

Now

L{ua (t )} = ∫ e−s t ua (t ) dt 0 ∞

a

= ∫ e−s t ua (t ) dt + ∫ e−s t ua (t ) dt 0

a

∞

= ∫ e−s t dt , by the definition of ua (t) a ∞

 e−s t  e−a s =  −S  = S , assuming that s > 0.   a In particular, L{u0 (t )} =

1 S

, which is the same as L(1).

 e−as     Inverting the above result, we get L−1    = ua (t ) .   s    

5.4.2  Definition lim { f (t )} , where f (t) is defined by h→0

   1 , when a − h ≤ t ≤ a + h f (t ) =  h 2 2 is called Unit Impulse   , otherwise 0   Function or Dirace Delta Function and is denoted by δa(t) or δ (t − a). Now L{δa (t )} = L  lim { f (t )} ,where f(t) is taken as given in the definition  h→0  = lim L{ f (t )} h→ 0

∞

= lim ∫ e−s t f (t ) dt h→ 0

0 a+

= lim h→ 0

∫ a−

h 2

e−s t ⋅ h 2

1 dt h

h  a+   1  e−s t  2    = lim  ⋅  h → 0  h  −s   h  a−   2

Laplace Transforms  h  −s a − h  − s a +    e  2  − e  2    = lim   h→ 0 sh     s h / 2 −  s h/2  −e e  = e−a s lim   h→ 0 sh     sh    2 sinh     2     = e−a s lim  h→ 0   sh    

  sh    s cosh     2     , by L’ Hospital’s rule. = e−a s lim  h→ 0   s      sh  = e−a s lim  cosh  = e−a s h→ 0  2  

Aliter f (t ) =

u

a−

h 2

 1  u h (t ) − u h (t ) , since  a+ h  a− 2 2 

(t ) = 1, when t > a −

h and u h (t ) = 0 , a+ 2 2

h and hence u h (t ) − u h (t ) = 1 a− a+ 2 2 2

when

t

π ω

π ω

for 0 < t < 4 for t > 4

∞

L{f (t )}= ∫ e−st f (t ) dt 0 ∞

1

(i)

L{f (t )}= ∫ e−st ⋅ odt + ∫ (t −1) e−st dt 2

0

1 ∞

− st    e−st   e−st  2 e  − 2 (t −1) 2  + 2  3  = (t −1) ⋅  s   − s   −s   1

=0+

2 −s 2 −s e = 3e s3 s

Aliter 

2



(t −1)

2

for t >1

f (t) = (t − 1)2 u1 (t)

Thus

L{f (t)} = e−s · L(t2), by the second shifting property.

∴

=

2 −s e s3 ∞

a

(ii)

for 0 < t a

f (t) = e k t − e k t ua (t) = e k t −e k (t − a) + ka · ua (t)

∴

L{f (t)} = L(e k t) −e ak · L{e k (t − a) · ua(t)} =

1 − e ak ⋅ e−as ⋅L{e kt } , by the second shifting property s−k

=

1 1 −a s − k −e ( ) ⋅ s−k s−k

=

1 −a s − k {1− e ( ) }. s−k π ω

(iii)

L{f (t )}= ∫ e−st sin ωt dt 0

 e−st  = 2 − s sin ωt − ω cos ωt )  s + ω2 (   0 ω = 2 (1+ e−π s ω ) s +ω2

π ω

4

(iv)

∞

L{f (t )}= ∫ te−st dt + ∫ 5e−st dt. 0

4

∞   e−st   e−st   e−st            = t ⋅ − 2  + 5 − s   4   − s   s  0 1 5 4 1 =− e−4 s − 2 e−4 s + 2 + e−4 s s s s s 1 −4 s 1 −4 s 1 = e − 2e + 2 s s s 4

Aliter t, t{1− u4 (t )} + 5u4 (t ) =  5, Thus

f (t) = t − (t − 5) u4(t) = t − (t − 4) u4(t) + u4(t)

for 0 < t < 4 for t > 4

5.17

Laplace Transforms L (f (t )}= L(t ) − e−4 s ⋅ L(t ) + L{u4 (t )}

∴

=

1 1 −4 s 1 −4 s − e + e . s s2 s2

Example 5.2 Find the Laplace transforms of the following functions: (i)

1+ 2 t cos t , (ii) sin t , (iii) . t t

(i)

 1+ 2 t   −1 2 12  L   = L(t ) + 2 L(t )   t     =

(1 2) 12

s

+2

(3 2) s3 2

1 π = + 2 s s s π  1  = 1+ . s  s  π

2⋅

∵ (1 2)= π 

( t) +( t) t− 3

sin t =

(ii)

3!

and

(n +1)= n (n) 

5

5!

−....∞

1 1 = t1 2 − t 3 2 + t 5 2 −....∞ 3! 5! ∴

(

)

1 1 L sin t = L(t1 2 ) − L (t 3 2 ) + L(t 5 2 ) −....∞ 3! 5! =

(3 2) 1 (5 2) 1 (7 2) s3 2

−

3! s 5 2

+

5! s 7 2

 1 1 1 3 1 1 15 3 1 1  (1 2)− ⋅ ⋅ (1 2)⋅ + ⋅ ⋅ (1 2)⋅ 2 −....∞ 32   s 5! 2 2 2 3! 2 2 s 2 s 2  π  11 1 1 = 3 2 1− ⋅  + ⋅  −......∞ 2 s  1!  4 s  2!  4 s    π = 3 2 e−1 4 s 2s 2 4   t t  cos t 1  1− + −.....∞ =   2! 4! t t    1 1 = t −1 2 − t1 2 + t 3 2 −.....∞ 2! 4! =

(iii)

−....∞

( ) ( )

Mathematics II

5.18

∴

 cos t L   t

  = L(t −1/ 2 ) − 1 L (t1 2 ) + 1 L (t 3 2 )−.....∞  2! 4! =

(1 2) 1 (3 2) 1 (5 2) s1 2

−

2! s 3 2

+

4! s 5 2

−.....∞

  1− 1 + 1 3 ⋅1 2 −.....∞  2! 2s 4! 2 ⋅ 2 s    2   π  1  1  1  1   − 1 + ⋅ = − ⋅ . . ...∞      s  1!  4 s  2!  4s   π −1 4 s e = s π s

=

Example 5.3 Find the Laplace transforms of the following functions: (i) (t3 + 3e2t − 5 sin 3t)e−t (ii) (1 + te−t)3 −2t 3 (iii) e cosh 2t (iv) cosh at cos at 3 t (v) sinh sin t 2 2 (i)

L{(t3 + 3e2t − 5 sin 3t)e−t} = L(t3 + 3e2t − 5 sin 3t)s → s + 1, by first shifting property.  3! 3 3   =  + −5⋅ 2  s 4 s − 2 s + 9 s → s +1 =

(ii)

6

( s +1)

4

+

3 15 − 2 ( s −1) s + 2 s +10

L(1 + te−t)3 = L(1 + 3t e−t + 3t2 e−2t + t3 e−3t) = L(1) + 3L(t)s → s + 1 + 3L(t2)s → s + 2 + L(t3)s → s + 3 1 3 6 6 = + + + s (s +1) 2 (s + 2)3 (s + 3) 4

(iii)

 2t −2 t 3    −2t  e + e   L(e cosh 2t ) = L e ⋅      2       1 = L{e−2t (e6t + 3e 2t + 3e−2t + e−6t )} 8 1 = L{e 4t + 3 + 3e−4t + e−8t } 8 3 3 1  1 1  =  + + + 8  s − 4 s s + 4 s + 8  −2t

3

Laplace Transforms

(iv)

  e a t + e−a t     cos at  L (cosh at cos at )= L         2    1 =  L (cos at )s → s − a + L (cos at )s → s + a   2   1 s −a s +a  + =  2  ( s − a )2 + a 2 ( s + a )2 + a 2     1  s −a s +a  = + 2 2 2 2  2  s + 2a − 2as s + 2a + 2as    2 2 2 2 1  (s − a ) ( s + 2a + 2as ) + (s + a ) ( s + 2a − 2as ) =   2 2 2 2 2  2 s 2 a 4 a s + − ( )   =

(v)

s3 . s 4 + 4a 4

3 t L(sinh sin t) 2 2   et 2 − e−t 2   3   sin = L t      2 2           1  3  3   − L sin =  L sin t t  2   2 s→s − 1 2 s→s + 1  2 2    3 2 1 3 2  =  −  2 2  ( s −1 2) + 3 4 ( s +1 2)2 + 3 4      3 1 1  = − 4  s 2 − s +1 s 2 + s +1 =

s 3 ⋅ 4 2 s + s 2 +1

Example 5.4 Find the Laplace transforms of the following functions: (i) eat cos (bt + c) (ii) e−2t cos2 3t (iii) et sin3 2t (iv) e−t sin 2t cos 3t (v) e3t sin 2t sin t (i) L{eatcos (bt + c)}= L{cos (bt + c)}s → s − a = L{cos c cos bt − sin c sin bt}s→ s − a (s − a ) cos c b sin c = − (s − a ) 2 + b 2 (s − a ) 2 + b 2

5.19

Mathematics II

5.20 (ii)

L{e−2t cos2 3t}= L(cos2 3t)s → s + 2 1    = L  (1+ cos 6t )     2 s → s + 2  s+2 1 1  =  +  2 2  s + 2 ( s + 2) + 36   

(iii)

L{et sin3 2t} = L(sin3 2t)s → s − 1   1 3  = L  sin 2t − sin 6t    4 4    s → s −1 3  2 1 6  =  ⋅ 2 − ⋅ 2  4 s + 4 4 s + 36 s → s −1  3 1 1  =  2 − 2 2  s − 2 s + 5 s − 2 s + 37 

(iv)

L{e−t sin 2t cos 3t} = L(sin 2t cos 3t)s → s + 1    1 = L  (sin 5t − sin t )    s → s +1 2   1 5 1  =  2 − 2  2  s + 25 s +1 s → s +1   1 5 1 =  − 2  2  2   s + 2 s + 26 s + 2 s + 2 

(v)

L{e3t sin 2t sin t} = L(sin 2t sin t)s → s − 3   1  = L  (cos t − cos 3t ) s→s−3    2   1 s s  =   2 − 2   + + 2 1 9 s s    s→s −3  s −3  1 s −3   =  − 2  2   2 s 6 s 10 s 6 s 18 − + − +    

Example 5.5 Find the Laplace transforms of the following functions: (i) (t − 1)2 · u1(t) (ii) sin t · uπ(t) (iii) e−3t · u2(t) (i) L{(t − 1)2 u1(t)} = e−s · L{t2}, by the second shifting property = (ii)

2 −s e s3

L{sint · uπ(t)} = L{sin (t − π + π) · uπ(t)} = −L{sin (t − π) · uπ(t)}

5.21

Laplace Transforms = − e−sp L(sin t) =− (iii)

e −sπ s 2 +1

L{e−3t u2(t)} = L{e−3(t−2)−6 ⋅ u2(t)} = e−6 ⋅ e−2 s ⋅ L {e−3t }=

e

−(2 s + 6)

s +3

Example 5.6 Find the Laplace transforms of the following functions: (i) t sin at (ii) t cos at (iii) te−4t sin 3t (i)

L (t sin at) = L{t × I.P. of e iat} = I.P. of L{t e iat} = I.P. of L(t)s → s − ia = I.P. of

( s + ia )

2

1

= I.P. of

( s − ia )

2

(s 2 + a 2 )

2

 2   s − a2 2 as  = I.P. of  +i  2  ( s 2 + a 2 )2 (s 2 + a 2 )   2 as = 2 2 (s + a 2 ) (ii)

L(t cos at) = R.P. of L{te iat} =

s2 − a2

(s 2 + a 2 )

2

(iii) L(t e−4t sin 3t) = I.P. of L {t e −4t e i3t} = I.P. of L {t · e−(4 − i3)t} = I.P. of L(t)s → (s + 4 − i3) = I.P. of

( s + 4 + i3)

2

1

= I.P. of 2

( s + 4 − i3)

{(s + 4) + 9}

{(s + 4) −9} + i 6(s + 4) = I.P. of {(s + 4) + 9} 2

2

=

6 ( s + 4)

{(s + 4) + 9} 2

2

2

2

2

Mathematics II

5.22 Example 5.7

 t (i) Assuming L(sin t) and L(cos t), find the Laplace transforms of Lsin  and  2  L(cos 3t). 2s s 2 −1 (ii) Given that L(t sin t )= and L(t cost )= , find L(t sin at) and 2 2 (s 2 +1) (s 2 +1)  t Lt cos  .  a (i) L (sin t )=

∴

1 s +1 2

 t 1 L sin = 2 ⋅ 2  2  (2s ) +1     t   ∵ L   f   = a ⋅ L { f (t )}s→as , by the charge of sccale property       a    =

L (cos t )= ∴

2 4 s 2 +1 s s +1 2

s/3 1 L (cos3t )= ⋅ 3 ( s / 3)2 +1   1 ∵ L { f (a t )} = L { f (t )} s , by the charge of scale property s→   a a =

s . s +9 2

(ii) Given that L (t sin t )=

∴

2s

(s

2

+1)

2

1 L (t sin at ) = L(at sin at ) a 1 1 2s/ a = ⋅ ⋅ , by change of scale property⋅ a a ( s 2 / a 2 +1)2 =

2 as

(s 2 + a 2 )

2

5.23

Laplace Transforms Given that L (t cos t )=

s 2 −1

(s 2 +1)

2

t  t t L t cos  = a ⋅ L  cos      a a a

∴

(as ) −1 2

=a⋅a⋅

=

(as )2 +1  

a4 s2 − a2

(a 2 s 2 +1)

2

2

, by change of scale property

1 2 a or . 2  2 1   s +   a 2  s2 −

Example 5.8 Find the inverse Laplace transforms of the following functions: (i)

(v)

1

( s + 2)

5/ 2

( s + 4) s2 − 4

e−s (ii)

e−2 s

(2s − 3)

3

(iii)

s e−s

( s − 3)

5

(iv)

(3a − 4s ) s2 + a2

e−bs

e−4 s

(i) From the second shifting property, we have L−1 {e−as φ (s)} = L−1 {φ (s)t → t − a ⋅ ua(t)       1 1   −1   −s ∴ L−1  ⋅ u1 (t ) =L  e ⋅ 5/ 2  5/ 2      2 2 s + s + ( ) ( )         t → t −1        1  1    −2 t −1  Now L−1  =e ⋅ L  5/ 2   5/ 2      s         ( s + 2)  

(1)

[ L−1 {φ (s + 2) = e−2t L−1 {φ (s)}, by the first shifting property.] = e−2t ⋅ = e−2t ⋅

=

4 3 π

1

(5 / 2)

   1 n−1   −1  1 t  ∵L  n =  (n)   s 

⋅t3/ 2

1 3 1 ⋅ ⋅ (1/ 2) 2 2

⋅t3/ 2

3

t 2 e−2t

Inserting (2) in (1), we have     1 4 3/ 2   L−1 e−s = (t −1) e−2(t −1) ⋅ u1 (t ) 5/ 2        ( s + 2)   3 π

∵ (n)= (n −1) (n −1)   ∵ (1/ 2)= π   

(2)

Mathematics II

5.24

or

0, when t < 1    = 4  (t −1)3/2 ⋅ e−2 (t −1) , when t > 1    3 π    e−2 s     1  = L−1    (ii) L−1  ⋅u2 (t )   3 3     ( 2 − 3 ) ( 2 − 3 ) s s      t→ t −2  

(1)

       1 1  = 1 L−1    Now L−1    3 3     8 ( ) ( ) 2 3 3 2 − s s −         3  1 2 t −1  1   = e L  3   8 s    1 3t 1 1 3t = e2 ⋅ t2 = e2 t2 8 2! 16

(2)

Using (2) in (1), we get 3   e−2 s    = 1 e 2 (t −2 ) (t − 2) 2 ⋅ u (t ) L−1   2 3     (2 s − 3)   16

 s e−s      s   = L−1    (iii) L−1  ⋅ u1 (t )   5 5     ( − 3 ) ( − 3 ) s s      t → t −1  

(1)

  s     (s − 3) + 3  = L−1    Now L−1    5 5      ( − 3 ) ( 3 ) s s −           + 3 s = e3t L−1   5       s   1  3  = e3t L−1  4 + 5    s s       1 1   = e 3t  t 3 + 3 ⋅ t 4   3!!  4!     1 3t 3 = e (4t + 3t 4 ) 24 Using (2) in (1), we have

(2)

 1 3(t −1)  s  L−1  e−s  {4(t −1)3 + 3(t −1) 4 }⋅ u1 (t ) = e  (s − 3)5  24    (3a − 4s ) −bs     3a − 4 s   (iv) L−1  e  = L−1  2 ⋅ ub (t )  2 2 2     + + s a s a        t → t −b

(1)

5.25

Laplace Transforms  3a − 4 s    a    s     = 3L−1  2 − 4 L−1  2 Now L−1  2 2 2 2      + + s a s a           s + a  = 3 sin at − 4 cos at Using (2) in (1), we have

(2)

 (3a − 4s ) −bs    L−1  e   2  = [3 sin a (t − b) − 4 cos a (t − b)]ub (t ) 2      s +a   ( s + 4) −4 s    s+4     (v) L−1  2 e  = L−1  2 ⋅ u4 (t )      − 4 − 4 s s        t → t −4 Now

(1)

 s+4    s    2   −1  −1    L−1   2 = L   2 + 2 L   2           s − 4   s − 4   s − 4 

= cosh 2t + 2 sinh 2t Using (2) in (1), we have

(2)

  ( s + 4 ) −4 s   L−1  e   2  ={cosh 2(t − 4) + 2 sinh 2(t − 4)}⋅ u4 (t )   − 4 s     Example 5.9 Find the inverse Laplace transforms of the following functions: (i)

e−s ( s − 2) ( s + 3)

(v)

( s +1)e−π s s 2 + 2s + 5

(ii)

e−2 s s ( s 2 +1) 2

(iii)

e−s s ( s 2 + 4)

(iv)

      1 e−s −1  −1     ⋅ u1 (t ) = L    (i) L        ( s − 2) ( s + 3)  t→ t −1  ( s − 2) ( s + 3)  

e−3 s s + 4 s + 13 2

(1)

    1 1  Now to find L−1  into partial fractions   , we resolve   (s − 2) (s + 3)   (s − 2) (s + 3)   and then use the linearity property of L−1 operator. 1 A B Let = + ( s − 2) ( s + 3) s − 2 s + 3 Then

A (s + 3) + B (s − 2) = 1

1 1 By the usual procedure, we get A= , B =− 5 5       1 1 5 −1  1 5   ∴ L−1  −  = L        s s s s − 2 + 3 2 3 − + )( )      (

Mathematics II

5.26

 1  1 −1  1  1 − L   = L−1   s + 3   s − 2  5 5 1 = (e 2 t − e−3 t ) 5

(2)

Putting (2) in (1), we have     1 2 (t −1) e−s  − e−3(t −1) }⋅ u1 (t ) L−1    = {e   ( − 2 ) ( + 3 ) 5 s s         e−2 s    1 −1    (ii) L−1  ⋅ u2 (t )  2 2 = L   2 2      ( + 1 ) ( + 1 ) s s s s      t → t −2   Now

(1)

1 1 1 1 1 1 = = − = 2− 2 2 s ( s +1) u (u +1) u u +1 s s +1 2

 1      1 −1  1  −1  L−1   2 2  = L  2 − L  2    s   s +1   s ( s +1)  

∴

= t − sin t Inserting (2) in (1), we get

(2)

  e−2 s   L−1  2 2  ={(t − 2) − sin (t − 2} u2 (t )  s ( s +1)     e−s      1 −1  −1    ⋅ u1 (t ) (iii) L   2 = L    2     + s s s s ( + 4 ) ( 4 )      t →t −1  

Let Then ∴

∴

(1)

1 A B s +C = + 2 s ( s 2 + 4) s s +4 A(s2 + 4) + s (Bs + c) = 1 1 1 A = , B =− and C = 0 4 4 1 1   s      1 −1  4 4   L−1   2 = L  − 2    s s + 4    s ( s + 4)       1 = (1− cos 2t ) 4

(2)

Laplace Transforms

5.27

Using (2) in (1), we have  e−s    1  L−1   2  = {1− cos 2 (t −1)}⋅ u1 (t )     s ( s + 4)   4       e−3 s 1 −1   L−1  ⋅ u3 (t )  2 = L    2       s + 4 s +13t →t −3   s + 4 s +13

(iv)

(1)

      1 1 −1   L−1   2 = L   2 2       s + 4 s +13   ( s + 2) + 3 

Now

  1    , by the shifting property = e−2t ⋅ L−1   2 2   3 s +      3    1  = e−2t L−1   2 2   3 3 s +     1 = e−2t sin 3t 3

(2)

Using (2) in (1), we get     1 −2 (t − 3) e−3 s  ⋅ sin 3(t − 3) ⋅ u3 (t ) L−1   2 = e     s + 4 s +13  3 −π s    s +1    −1  ( s +1) e −1    ⋅ uπ (t ) (v) L   2 = L    2        s + 2s + 5 t →t −π   s + 2s + 5  s +1    ( s +1)    −1    Now L−1   2 = L   2 2     + 2 + 5 ( + 1 ) + 2 s s s           s   , by the shifting property = e− t ⋅ L−1   2 2    s +2   = e−t cos 2t Using (2) in (1), we have   ( s +1) e−π s   −( t − π )  ⋅ cos 2(t − π ) ⋅ uπ (t ) L−1   2 = e     s + 2s + 5  Example 5.10 Find the inverse Laplace transforms of the following functions: (i)

(iv)

s2 + s − 2 s ( s + 3) ( s − 2)

(ii)

1 s ( s + 1) ( s 2 + 9) 2

2

2 s 2 + 5s + 2 ( s − 2) 4 (v)

(iii)

s ( s + 1) 2 ( s 2 + 1)

s . ( s + 1) ( s + 4) ( s 2 + 9) 2

2

(1)

(2)

Mathematics II

5.28

 s2 + s − 2     (i) To find L−1    , we resolve the given function of s into partial   ( + 3 ) ( − 2 ) s s s     fractions and then use the linearity property of L−1 operator. Let

s2 + s − 2 A B C = + + s ( s + 3) ( s − 2) s s + 3 s − 2

We find, by the usual procedure, that 1 4 A= , B = 3 15

2 and C = . 5

 s2 + s − 2     4 / 15 2 / 5  −1 1 / 3  + + L−1    = L   s + 3 s − 2     s  s ( s + 3) ( s − 2)  

∴

1 4 2 = + e−3 t + e 2 t 3 15 5 (ii) To resolve

2 s2 + 5 s + 2 into partial fractions, we put s − 2 = x, so that ( s − 2) 4 s = x + 2.

2 s 2 + 5 s + 2 2 ( x + 2) 2 + 5 ( x + 2) + 2 = ( s − 2) 4 x4

Then

2 x 2 +13 x + 20 x4 2 13 20 = 2+ 3+ 4 x x x 2 13 20 = + + ( s − 2) 2 ( s − 2) 3 ( s − 2 ) 4 =

∴

 2 s 2 + 5 s + 2   1      1   1  −1  −1   +13L−1   + 20 L  = 2L      4 2 3        ( s − 2) 4      ( s − 2)    ( s − 2)   ( s − 2) 

L−1  

 1  2t 2t −1  1  −1  1       s 2  +13 e ⋅ L  s 3  + 20 e ⋅ L  s 4  13 20  2t  = e  2⋅ t + t 2 + t 3   2! 3!  13 2 10 3  2t  = e 2 t + t + t   2 3  = 2 e ⋅ L−1  2t

5.29

Laplace Transforms =

s

(iii) Let ∴

1 2t e (12t + 39t 2 + 20t 3 ) 6

(s + 1) 2 ( s 2 + 1)

=

A B C s+D + + 2 s + 1 (s + 1) 2 s +1

A(s + 1) (s2 + 1) + B(s2 + 1) + (Cs + D) (s + 1)2 = s By the usual procedure, we find that 1 A = 0; B = − ; C = 0 2

∴

and D =

1 2

     1 / 2  s   −1  − 1 / 2  L−1  + 2    = L  2 2 2   s + 1 (s + 1) (s + 1)     (s + 1)      1  1 1 1 = − e−t ⋅ L−1  2  + L−1  2  s + 1    2 2 s 1 t = − e−t + sin t 2 2 1 (iv) Since 2 2 is a function of s2, we put s2 = u, we resolve s ( s + 1)( s 2 + 9) 1 into partial fractions and then replace u by s2. u (u + 1) (u + 9) 1 A B C = + + u (u + 1) (u + 9) u u + 1 u + 9

Now let

1 1 1 By the usual procedure, we find that A = , B = − and C = 9 8 72     1 1/ 8 1 / 72    −1 1 / 9  ∴ L−1  + 2  2 2  2 − 2  = L  2   s s + 1 s + 9  s ( s + 1)( s + 9)         1 1 1 = t − sin t + sin 3t 9 8 216 s (v) To resolve 2 into partial fractions, we first resolve 2 (s +1)(s + 4)(s 2 + 9) 1

(s 2 +1)(s 2 + 4)(s 2 + 9) Thus

into partial fractions as shown in (iv).

1

(s 2 +1)(s 2 + 4)(s 2 + 9) =

=

1 (u + 1) (u + 4) (u + 9)

A B C + + , say. u +1 u + 4 u + 9

Mathematics II

5.30 We find that A =

1 1 1 , B =− and C = . 24 15 40

1

∴

(s 2 +1)(s 2 + 4)(s 2 + 9)

∴

(s

s 2

+ 1)( s + 4)( s + 9) 2

2

=

1/ 24 1/15 1/ 40 − + s 2 +1 s 2 + 4 s 2 + 9

=

s s s 1 1 1 ⋅ 2 − ⋅ 2 + ⋅ 2 24 s + 1 15 s + 4 40 s + 9

    1 −1  s  1 −1  s  s    ∴ L−1  + − L  2 L  2  2 = 2 2  s + 4   s + 1 15   s + 1)( s + 4)( s + 9) 24    (  1 −1  s  L  2  s + 9  40 =

1 1 1 cos t − cos 2t + cos 3t 24 15 40

Example 5.11 Find the inverse Laplace transforms of the following functions: (i)

14 s + 10 49 s 2 + 28s + 13

(iii)

1 s3 − a3

(v)

s s 4 + s 2 +1

(i)

(ii) (iv)

2s3 + 4s 2 − s +1 s 2 ( s 2 − s + 2) 1 s4 + 4

  5    s+    14 + 10 14 s    −1  7 = L−1  L     49 s 2 + 28s + 13    4 13 2   49   + + s s    7 49        5     s +   2 −1  7  = L  2 2  7    3   2   s +  +        7   7               2 3     s +  +  2 −1  7 7     = L  2 2  7        s + 2  +  3       7   7            3      s+ 2   − t 2 7 −1  7  = e ⋅L  2   7   3    s 2 +      7       

5.31

Laplace Transforms =

2 − 72 t  3 3  e cos t + sin t    7 7 7 

2s3 + 4s 2 − s +1 A B Cs + D = + 2+ 2 s s s 2 (s 2 − s + 2) s −s+2

(ii) Let

As(s2 − s + 2) + B(s2 − s + 2) + (Cs + D)s2 = 2s3 + 4s2 − s + 1.

∴

By the usual procedure, we find that 13 1 1 9 A= , B= ,C = and D = 4 4 2 4 13   9 s+  3   2 s + 4 s 2 − s + 1 1 −1  1  1 −1  1  −1  4   4    L  2 2   = − L   + L  2  + L  2  s − s + 2   s  s 2 4 s ( s − s + 2)            −1

∴

    9  1  35      s −  +  1 t 2 8   4  = − + + L−1  2 2  4 2   1   7       s −  +      2   2          5 7 7    ⋅  1 t s 9 4 2  + = − + + et / 2 L−1  ⋅ 2 2 4 4 2  7    7   2 2  s +   s +       2    2     9 1 t 7 5 7 7  t+ t sin = − + + et / 2  cos  4 4 2 2 4 2  (iii)

∴ Let ∴

1 1 = s 3 − a 3 (s − a ) ( s 2 + as + a 2 ) 1 A Bs + C = + 2 3 s − a s + as + a 2 s −a 3

A(s2 + as + a2) + (s − a)(Bs + C) = 1 By the usual procedure, we find that A=

1 1 ,B =− 2 3a 2 3a

and C = −

2 . 3a

Mathematics II

5.32

 1   − 2 s − 2   1   1 −1  1  −1     3a a a 3 = 2 L  L  3  + L  2 2  3       s as a 3 − + + s a a s−a            −1

∴

=

1 3a 2

=

1 3a 2

=

1 3a 2

         1 2 s a +  e at − 2 L−1  2 2  3a  3a      a    s +  +      2   2          3    s + a  1 − at2 −1    at 2 e − 2 e ⋅L  2  3a  3a       s 2 +     2       at   3 3    e at − e− 2  cos + sin 3 at at     2 2    

1 1 1 = 4 = 2 2 2 s + 4 ( s + 4 s + 4) − 4 s ( s 2 + 2) − ( 2 s ) 2

(iv)

4

=

1

(s

2

+ 2 s + 2)( s 2 − 2 s + 2)

∴Let

1 As + B Cs + D = 2 + 2 s + 4 s + 2s + 2 s − 2s + 2

∴

(As + B) (s2 − 2s + 2) + (Cs + D)(s2 + 2s + 2) = 1

4

By the usual procedure, we get A = 1/8, B = 1/4, C = −1/8 and D = 1/4

∴

 1  1  1  1       8s+ 4   8 s − 4    1     −1  −1  L  4 = L  2 − L  2     s + 2s + 2  s − 2 s + 2     s + 4                   −1

 (s + 1) + 1    1 1 −1  (s −1) −1   = L−1     − L  2 2    8    (s −1) + 1  (s + 1) + 1  8  s +1   t −1  s −1  1  =  e−t ⋅ L−1  2  − e ⋅ L  2   s + 1  8    s + 1 

Laplace Transforms

5.33

1 =  e−t (cos t + sin t ) − et (cos t − sin t ) 8 =

  et − e−t  1  et + e−t     − sin t cos t    2   4  2 

1 (sin t cosh t − cos t sinh t ) 4 s s s = = s 4 + s 2 + 1 ( s 4 + 2 s 2 + 1) − s 2 ( s 2 + 1)2 − s 2 =

(v)

=

∴Let ∴

s

(s 2 + s +1)(s 2 − s +1)

s As + B Cs + D = 2 + 2 2 s + s +1 s + s +1 s − s +1 4

(As + B) (s2 − s + 1) + (Cs + D) (s2 + s + 1) = s By the usual procedure, we get −1 1 , C = 0 and D = . 2 2   1 −1    −1 −1  1 1 s     L−1  4 L  2 = L  2   2         s + s + 1  s + s + 1  2  s − s + 1  2 A = 0, B =

∴

                1 −1  1 1 1    = L  − L−1  2 2 2 2   2         2    s + 1  +  3    s − 1  +  3          2   2   2   2                           1  t / 2 −1  1 1    −t / 2 −1  e L − ⋅ = e L    2 2   2  3    3     2 2     s +     s +        2      2               t/2 1 2  3  −t / 2  = ⋅ t (e − e ) sin 2 3 2    2 3 t nh . = sin t sin 2 2 3 Example 5.12            s t s −1    = sin t , find L  (i) If L  2 2 2 2 2    2 s + 1)  s + a )       (  (  −1

.

Mathematics II

5.34

 2   2    s +4   −1  s + 1   L cosh 2 (ii) Given that L−1  t t , find =  2   2 2  s −1 2   s − 4)    )  (   (          1 1 1 −1   . (iii) Given that L−1  t t t find L (sin 2 2 cos 2 ), = −   2 2 2 2    16 s + 4)   ( s + 9)    (    (i) By change of scale property, 1 s L{ f (at )}= φ   a  a  ∴

L−1 {f (s/a)} = a L−1{f (s)}t → at

      s   = t sin t Given L  2 2  2 s + 1)      (              / s a −1   = a ⋅ at sin at , by ∴ L  2 2  2 s     + 1      2         a        s a2 3 −1   i.e., a L  t sin at =  2  2 s2 + a2 )    (          s t ∴ L−1  sin at .  2 = 2 2  2 a + s a   ( )    (ii) By change of scale property,

(1)

−1

∴ Given

∴

i.e.,

∴

L{ f (t/a)} = a f (as) 1 L−1{φ (as )}= L−1 (φ (s )}t→t / a a      2  −1  s + 4  L   = t cosh 2t 2 2  s − 4)    (     2     ( 2 ) + 4 s  = 1 ⋅ t cosh t , by L−1   2 2    [(2 s ) − 4]   2 2    2  1 −1  s +1  t   L   = cosh t 2 2   4 4 s −1)     (   2    s +1    = t cosh t . L−1   2 2   − 1 s   ( )    

(1)

(2)

(2)

Laplace Transforms

(iii)

5.35

      1  = 1 (sin 2t − 2t cos 2t ) Given L−1   2 2  16 s + 4)      ( 

∴

                1 3 −1  1  −1     , by (2) L  = L  2 2 2 2     2  2 s    4 s +   ( )   t → 3 t   + 4    2  3            

i.e.

     81 −1  1  = 3 (sin 3t − 3t cos 3t ) L   2 2  16 32 s + 9)      (        1   = 1 (sin 3t − 3t cos 3t ) L  2 2  54 s + 9)      (  −1

∴

EXERCISE 5(a) Part A (Short Answer Questions) 1. Define Laplace transform. 2. State the conditions for the existence of Laplace transform of a function. 3. Give two examples for a function for which Laplace transform does not exist. 4. State the change of scale property in Laplace transformation. 5. State the first shifting property in Laplace transformation. 6. State the second shifting property in Laplace transformation. 7. Find the Laplace transform of unit step function. 8. Find the Laplace transforms of unit impulse function.  e 2t , for 0< t < 1 9. Find L{ f (t)}, if f (t ) =     0, for t > 1 0, for0 < t < 1    10. Find L{ f (t)}, if f (t ) =  t , for 1 < t < 2   0, for t > 2   11.

sin 2t , for 0< t < π  Find L{ f (t)}, if f (t ) =    for t > π  0,

cos t , for 0< t < 2π  12. Find L{ f (t)}, if f (t ) =    for t > 2π  0,

Mathematics II

5.36

 2π   0, for 0< t <   3 13. Find L{ f (t)}, if f (t )=      2 π 2 π  cos t −  , for t >    3 3     sin t , for 0< t < π 14. Find L{ f (t)}, if f (t )=    for t > π  t ,  1    15. Find L t and L   π t 

( )

16. Find L{(t − 3) u3(t)} and L{u1(t) sin p (t − 1)}. 17. Find L{t2u2 (t)} Find the Laplace transforms of the following functions: 18. (at + b)3 19. sin (wt + q) 20. sin2 3t 21. cos3 2t 22. sin 2t cos t 23. cos 3t cos 2t 24. sinh3 t 25. cosh2 2t  2 −2 t  26. (t + 1)2 e−t 27. e cos 3t − sin 3t    3  1 t 28. e cosh 2t + sinh 2t  29. sin t sinh t   2 e3(t + 2) 30. t2 cosh t 31. Find the Laplace inverse transforms of the following functions: 32. e−as/s2 (a > 0). 33. (e−2s − e−3s)/s 34.

e−2 s s −3

35.

se−s s2 + 9

1 + e−π s s 2 +1 Find f (t) if L{ f (t)} is given by the following functions: 36.

37. 39.

1 (s + 1)3/2 1

(2s − 3)

4

38. 40.

s 2 + 2s + 3 s3 s

( s − 2)

5

41.

2s + 3 s2 + 4

42.

s+6 s2 − 9

43.

1 s (s + a )

44.

1 s 2 + 2 s +5

45.

s −3 s 2 − 6 s + 10

5.37

Laplace Transforms

Part B Find the Laplace transforms of the following functions: at 46. e3t(2t + 3)3 47. e (1+ 2at ) / t −t 3 48. e sinh t 49. sin at cosh at − cos at sinh at  cos 2t  51.  t   e  53. e−t sin 3t cos t 55. e−2t sin 2t sin 3t sin 4t 57. te−t cos t 3

50. (et sin t)2 52. 54. 56. 58.

e−kt sin (wt + q) et cost cos 2t cos 3t t cos 2t t e2t sin 3t

59. Given that L(t sin 2t ) =

60. Given that L(t cos 3t ) =

4s

(s

2

+ 4)

2

s2 −9

( s 2 + 9)

2

, find L(t sin t).

, find L(t cos 2t).

Find the Laplace inverse transforms of the following function: s 2 +1 4 s 2 − 3s + 5 61. s 3 + 3s 2 + 2 s 62. ( s +1)( s 2 − 3s + 2) 63.

65.

67.

69. 71.

s 2 − 3s + 5

64.

( s +1)

3

1

(s 2 + s)

2

2s

(s

2

+1)( s + 2)( s + 3) 2

s +1 s + s +1 2

3s 2 −16 s + 26 s ( s + 4 s +13) 2

2

7 s −11

( s +1)( s − 2)

2

s 4 − 8s 2 + 31

66.

(s 2 +1)(s 2 + 4)(s 2 + 9)

68.

2s − 9 s + 6 s + 34

70.

ls + m as + bs + c

72.

1 s 3 +1 s s +4

2

2

73.

1 s + 4a 4

74.

75.

s2 s 4 + 64

4s3 76. 4 s 4 +1

77.

s 2 +1 s + s 2 +1

4

4

4

Mathematics II

5.38

       s    1 s −1 −1    t t sinh = 78. If L  , find L  2  2 2   2 2  2  s − a )   ( s −1)      (          1  1 1 −1   L 79. If L−1  t t t find (sin 3 3 cos 3 ), = −  2  .  2 2  s +1 2    54 9 + s   ( ) ( )                 s t s −1    = sin 3t , find L  80. Given that L  2 2 2 2   s + 4  6 s + 9)    )   (  (  −1

5.6

LAPLACE TRANSFORM OF PERIODIC FUNCTIONS

5.6.1  Definition A function f (t) is said to be a periodic function, if there exists a constant P (> 0) such that f (t + P) = f (t), for all values of t. Now f (t + 2P) = f (t + P + P) = f (t + P) = f (t), for all t. In general, f (t + nP) = f (t), for all t, when n is an integer (positive or negative). P is called the period of the function. Unlike other functions whose Laplace transforms are expressed in terms of an integral over the semi-infinite interval 0 ≤ t < ∞, the Laplace transform of a periodic function f (t) with period P can be expressed in terms of the integral of e−st f (t) over the finite interval (0, P), as established in the following theorem.

Theorem If f (t) is a piecewise continuous periodic function with period P, then P 1 L { f (t )} = ⋅ e−st f (t ) dt. 1− e− Ps ∫0 Proof: ∞

− st By definition, L { f (t )} = ∫ e f (t ) dt 0 ∞

P

= ∫ e−st f (t ) dt + ∫ e−st f (t ) dt 0

In the second integral in (1), put t = x + P, ∞

∴

∞

∫e

− st

P

(1)

P

− s( x + P )

f (t ) dt = ∫ e

∴ dt = dx and the limits for x become 0 and ∞.

f (x + P) dx

0 ∞

= e−sP ⋅ ∫ e−sx f ( x) dx 0

[ f(x + P) = f(x)]

Laplace Transforms

5.39

∞

= e−sP ∫ e−st f (t ) dt , on changing the dummy variable x to t. 0

= e−sP ⋅ L { f (t )}

(2)

By putting (2) in (1), we have P

L { f (t )} = ∫ e−st f (t ) dt + e−sP ⋅ L { f (t )} 0

P

∴

(1− e− Ps ) L { f (t )} = ∫ e−st f (t ) dt 0

∴

5.7

L { f (t )} =

1 1− e− Ps

P

∫e

− st

f (t ) dt

0

DERIVATIVES AND INTEGRALS OF TRANSFORMS

The following two theorems, in which we differentiate and integrate the transform 1  function f (s) = L{f (t)} with respect to s, will help us to find L{t f (t)} and L  f (t )  t  respectively. Repeated differentiation and integration of f (s) will enable us to find    1 L{tn f (t)} and L  n f (t ) , where n is a positive integer.  t   

Theorem If

L{f (t)} = f (s),

then L{t f (t)} = − f' (s).

Proof : Given: i.e.

L{f (t)} = f (s) ∞

∫e

− st

(1)

f (t ) dt = φ (s )

0

Differentiating both sides of (1) with respect to s, ∞

d d e−st f (t ) dt = φ (s ) ∫ ds 0 ds Assuming that the conditions for interchanging the two operations of integration with respect to t and differentiation with respect to s in (2) are satisfied, we have ∞

d ∫ ds {e } f (t ) dt = φ ′(s) − st

0

(2)

Mathematics II

5.40 ∞

∫ −t e

− st

i.e.

f (t ) dt = φ ′(s )

0 ∞

∫

i.e.

e−st [t f (t )]dt = −φ ′(s )

0

L{t f (t )} = −φ ′(s )

i.e.

Corollary Differentiating both sides of (1) n times with respect to s, we get L{t n f (t )} = ( −1) n

dn φ (s ) ds n

or

( −1) nφ (n ) (s ) .

Note 1. The above theorem can be rewritten as a working rule in the following manner d φ (s ) ds d = − L{ f (t )} ds

L{t f (t )} = −

Thus, to find the Laplace transform of the product of two factors, one of which is ‘t’, we ignore ‘t’ and find the Laplace transform of the other factor as a function of s; then we differentiate this function of s with respect to s and multiply by (− 1). Extending the above rule, L{t 2 f (t )} = ( −1) 2

d2 L{ f (t )} and in general ds 2

dn L{ f (t )} . ds n 2. The above theorem can be stated in terms of the inverse Laplace operator as follows: L{t n f (t )} = ( −1) n

If

L−1{φ (s)} = f (t),

then L−1{φ ′(s)} = −t f (t). From this form of the theorem, we get the following working rule: 1 L−1{ φ (s )} = − L−1{ φ ′(s )} t This rule is applied when the inverse transform of the derivative of the given function can be found out easily. In particular, the inverse transforms of functions of s that contain logarithmic functions and inverse tangent and cotangent functions can be found by the application of this rule.

5.41

Laplace Transforms

Theorem    t

1  t

∞

   

 

  If L{ f (t )} = φ (s) , then L  1 f (t ) = ∫ φ (s ) ds , provided lim  f (t ) exists. t →0 s

Proof: L{ f (t)} = φ (s)

Given: ∞

i.e.

∫e

− st

f (t ) dt = φ (s )

(1)

0

Integrating both sides of (1) with respect to s between the limits s and ∞ , we have ∞  (2) f (t ) dt  ds = ∫ φ (s ) ds  s  0 s Assuming that the conditions for the change of order of integration in the double integral on the left side of (2) are satisfied, we have ∞ ∞ ∞   e−st ds  f (t ) dt = φ (s ) ds ∫  ∫ ∫   0  s s ∞

∞

∫  ∫ e

∞

∫

i.e.

0

− st

s =∞

∞  e−st    f (t ) dt = ∫ φ (s ) ds  −t    s=s s

∞

i.e.

∞

1 − st ∫ − t (0 − e ) f (t ) dt = ∫ φ (s) ds , 0 s

assuming that s > 0 ∞

i.e.

∫ 0

i.e.

∞  f (t )   dt = ∫ φ (s ) ds e−st   t  s

 f (t )   ∞  L  = ∫ φ (s ) ds     t   s

Corollary 1     1  1  L  2 f (t ) = L   f (t )  t      t  t     ∞ ∞  = ∫  ∫ φ (s ) d s  ds  s  s ∞ ∞

=∫ s

Generalising this result, we get 1  ∞   L  n f (t ) = ∫     t  s

∫ φ (s)ds ds s

∞

∞

∫ …∫ φ (s) (ds) s

s

n

Mathematics II

5.42

Note 1. The above theorem can be rewritten as a working rule as given below: ∞   1  L  f (t ) = ∫ L{ f (t )}d s     t  s

Thus, to find the Laplace transform of the product of two factors, one of which is

1 , t

1 , find the Laplace transform of the other factor as a function of s and t integrate this function of s with respect to s between the limits s and ∞ . Extending the above rule. We get; we ignore

∞   1  L  2 f (t ) = ∫   t  s  

1  ∞   L  n f (t ) = ∫     t  s

∞

∫ L{ f (t )}ds ds

and in general

s ∞

∞

n ∫ …∫ L{ f (t )}(ds) . s

s

2. The above theorem can be stated in terms of the inverse Laplace operator as follows: If then

L−1{ φ (s )} = f (t ) ,  1 ∞ L−1  ∫ φ (s ) ds  = f (t ) .  t  s

From this form of the theorem, we get the following working rule:  ∞ L { φ (s )} = t ⋅ L  ∫ φ (s ) d s    s −1

−1

This rule is applied when the inverse transform of the integral of the given function with respect to s between the limits s and ∞ can be found out easily. In particular, the inverse transforms of proper rational functions whose numerators are first degree expressions in s and denominators are squares of second degree expressions in s can be found by applying this rule WORKED EXAMPLE 5(b)

Example 5.1 Find the Laplace transform of the “saw-tooth wave” function f (t) which is periodic with period 1 and defined as f (t) = kt, in 0 < t < 1. The graph of f (t) is shown in Fig. 5.1 below. If the period of the function f (t) is P, the function will be defined as f (t ) =

k t in 0 < t < P. P

5.43

Laplace Transforms f (t) k

t 0

1

2

3

Fig. 5.1

By the formula for the Laplace transform of a periodic function f (t) with period P, L{f (t )} =

P

1 1− e− Ps

∫e

− st

f (t ) dt

0

∴ For the given function, L{f (t )} =

1 1− e−s

1

∫ kt e

− st

dt

0

− st    e−st   −1⋅  e  t     −s   s 2    0 − s − s 1 k  e e = − − 2 + 2 −s  1− e  s s s  −s   k  (1− e ) e−s  = −   s  1− e−s  s 2   −s k ke = 2− s s (1− e−s ) 1

k = 1− e−s

Example 5.2 Find the Laplace transform of the “square wave” function f (t) defined by f (t) = k in 0 ≤ t ≤ a = − k in a ≤ t ≤ 2a and

f (t +2a) = f (t) for all t.

f (t + 2a) = f (t) means that f (t) is periodic with period 2a. The graph of the function is shown in Fig. 5.2. For a periodic function f (t) with period P,

L{f (t )} =

1 1− e− Ps

P

∫e

− st

0

f (t ) dt

Mathematics II

5.44 f (t) k

t O

a

2a

3a

4a

–k

Fig. 5.2

∴ For the given function; 2a  a  k e−st dt + ( − k ) e−st dt  ∫ ∫     0 a a a 2  e−st   e−st       −       −s 0  −s a 

L{f (t )} =

1 1− e−2 as

=

k 1− e−2aas

=

k [1 − e−as − e−as + e−2 as ] s (1− e−2 as )

=

k (1 − e−as ) 2 s (1 − e−as ) (1 + e−as )

k (1 − e−as ) k (e as / 2 − e−as / 2 ) = s (e as / 2 + e−as / 2 ) s (1 + e−as )  as  k = tanh    2  s =

Example 5.3 Find the Laplace transform of “triangular wave function f (t) whose graph is given below in Fig. 5.3. y = [= f (t)] A

a

B t O

a

2a

3a

4a

5a

Fig. 5.3

From the graph it is obvious that f (t) is periodic with period 2a. Let us find the value of f (t) in 0 ≤ t ≤ 2a, by finding the equations of the lines OA and AB.

5.45

Laplace Transforms OA passes through the origin and has a slope 1. ∴ Equation of OA is y = t, in 0 ≤ t ≤ a AB passes through the point B(2a, 0) and has a slope − 1. ∴

Equation of AB is y − 0 = (− 1) (t − 2a)

or

y = 2a − t in a ≤ t ≤ 2a.

Thus the definition of f(t) [= y] can be taken as f(t) = t, in 0 ≤ t ≤ a = 2a − t, in a ≤ t ≤ 2a and

f(t + 2a) = f(a).

Now L{f (t )} =

1 1− e−2 a s

=

1 1− e−2 a s

=

1 1− e−2 a s

=

1 1− e−2 a s

2a

∫e

− st

f (t ) dt

0 2a  a  te−st dt + (2a − t )e−st dt  ∫  ∫   0 a a 2a     e−st    e−s t   e−st   e−st             − 1⋅  2   + (2a − t )     t   −s  + 1⋅  s 2    s     −s      a     0  a s a s a s − − − 2  a −a s e  − e − 2 + 12 + e 2 + a e−a s − e 2   s s s s s s  

1 − 2e−a s + e−2 a s (1 − e−a s ) 2 = s 2 (1 − e−a s ) (1 + e−a s ) s 2 (1 − e−2 a s ) 1 (1 − e−a s ) 1  e a s / 2 − e− a s / 2   = 2 = 2  a s / 2 −a s s (1 + e ) s  e + e− a s / 2 

=

=

 as  1 tanh   2  2  s

Example 5.4 Find the Laplace transform of the “half-sine wave rectifier” function f (t) whose graph is given in Fig. 5.4. f (t)

a O

π 2ω

π/ω

2π/ω

3π/ω Fig. 5.4

4π/ω

t

Mathematics II

5.46

From the graph, it is obvious that f(t) is a periodic function with period 2π/ω. The π  graph of f (t) in 0 ≤ t ≤ π/ω is a sine curve that passes through (0, 0),  , a and  2ω   π   , 0  ω  ∴ The definition of f (t) is given by f (t) = a sin ω t, in 0 ≤ t ≤ π/ ω = 0, in π/ ω ≤ t ≤ 2π/ ω  2π  f t +  = f (t ) . and  ω L { f (t )}=

Now

=

2π ω

1 1− e−2 π s ω a 1− e−2 π s ω

∫

e−st f (t ) dt

0 π ω

∫e

− st

sin ωt dt

0

=

 e−st  a ⋅ ( − s sin ωt − ω cos ωt  −2 π s ω  2 2  1− e  s +ω 0

=

a  ωe−π s ω + ω   (s + ω ) (1− e−2 π s ω ) 

π ω

=

2

2

ω a (1+ e−π s ω ) (s + ω ) (1− e 2

−2 π s ω

2

)

=

ωa (s + ω ) (1− e−π s ω ) 2

2

Example 5.5 Find the Laplace transform of the “full-sine wave rectifier” function f (t), defined as f (t) = |sin ω t|, t ≥ 0 We note that

f (t + π/ ω) = |sin ω (t + π/ ω)| = |sin ω t| = f (t)

∴ f (t) is periodic with period π/ ω. Also f (t) is always positive. The graph of f (t) is the sine curve as shown in Fig. 5.5.

1 O

π/ω

2π/ω

3π/ω Fig. 5.5

t

5.47

Laplace Transforms

Now

L { f (t )}= =

1 1− e−π s ω 1 1− e−π s ω

π ω

∫e

− st

| sin ω t | dt

0 π ω

∫e

− st

sin ω t dt [ ∵sin ωt > 0 in 0 ≤ t ≤ π ω ]

0

 e−st   2 ( − s sin ωt − ω cos ωt ) 2   s +ω  0

π ω

=

1 1− e−π s ω

=

1 ω 1+ e−π s ω   ωe−π s ω + ω )= 2  −π s ω ( (s + ω ) (1− e ) s + ω 2 1− e−π s ω  2

2

=

ω  eπ s 2 ω + e−π s 2 ω   , on integration and simplification  s 2 + ω 2  eπ s 2 ω − e−π s 2 ω 

=

 πs  ω coth   2  2ω  s +ω 2

Example 5.6 Find the Laplace transforms of the following functions: (i) t cosh3 t (ii) t cos 2t cos t (iii) t sin3 t (iv) (t sin at)2   et + e−t 3     3   = L { t cosh t } L t   (i)   2          1 = L {t (e3t + 3et + 3e−t + e−3t )} 8

(1)

1 d L (e3t + 3et + 3e−t + e−3t ) 8 ds  1 d 1 3 3 1   =− + + +   8 ds   s − 3 s −1 s +1 s + 3 =−

 1 1 3 3 1  =  + + +   2 2 2 8 (s −1) (s +1) (s + 3) 2    (s − 3)

Note

After getting step (1), we could have applied the first shifting property and got the same result.   t  (ii) L(t cos 2t cos t ) = L  (cos 3t + cos t )   2      1 d = − L (cos 3t +ccos t )   2  ds =−

1 d  s s  + 2   2 2 ds  s + 9 s +1

Mathematics II

5.48

1  s 2 + 9 − 2 s 2 s 2 +1− 2 s 2  =−  2 + 2  (s + 9) 2 (s 2 +1) 2  1  s2 −9 s 2 −1  =  2 + 2  (s + 9) 2 (s 2 +1) 2  3  1   L(t sin 3 t ) = L  sin t − sin 3t    4   4  1 d =− {L (3 sin t − sin 3t )} 4 ds 1 d  3 3   =− − 2  2 4 ds  s +1 s + 9 

(iii)

 3 2s 2 s  =−  + 2 − 2  2 4 (s + 9) 2    (s +1)   3  1 1 = s − 2 − 2 2 2 2  (s + 9)    (s +1)  1− cos 2at   L {(t sin at ) 2 }= L t 2     2  2 1 d = ( −1) 2 2 {L(1− cos 2at )} 2 ds    1 d2  s 1 −  = 2  2 2  2 ds    s s + 4a  

(iv)

=

 1 d 1 s 2 − 4a 2   − 2 + 2  2 ds  (s + 4a 2 ) 2    s

1  2 (s 2 + 4a 2 ) 2 ⋅ 2 s − (s 2 − 4a 2 ) ⋅ 2 (s 2 + 4a 2 ) ⋅ 2 s  =  3+  2  s (s 2 + 4a 2 ) 4  2 2 1  2 2 s (12a − s )  =  3+ 2 2  s (s + 4a 2 )3  =

1 s (12a 2 − s 2 ) + s 3 (s 2 + 4a 2 )3

Example 5.7 Use Laplace transforms to evaluate the following; ∞

(i)

∫

∞

te−2t sin 3t dt

0

(ii)

∫ te

−3 t

cos 2t dt

0 ∞

(i)

∫e

− st

0

(t sin 3t ) dt = L(t sin 3t )

(by definition)

(1)

Laplace Transforms

5.49

d L (sin 3t ) ds d  3   =−  2 ds  s + 9 

L (t sin 3t ) =−

Now

6s ( s 2 + 9) 2

=

(2)

Putting (2) in (1), we have ∞

∫ te

− st

6s ,s > 0 ( s + 9) 2

sin 3t dt =

2

0

(3)

Putting s = 2 in (3), we get ∞

∫ te

−2 t

sin 3t dt =

0

12 169

.

∞

∫e

− st

(ii)

(t cos 2 t ) dt = L(t cos 2t ) , by definition

(1)

0

d L(cos 2t ) ds d  s   =−  2 d s  s + 4 

Now

L(t cos 2t ) =−

=

s2 − 4 ( s 2 + 4) 2

(2)

Inserting (2) in (1), we have ∞

∫ te

− st

cos 2t dt =

0

s2 − 4 ,s > 0 ( s 2 + 4) 2

(3)

Putting s = 3 in (3), we get ∞

∫ te

−3 t

0

cos 2t dt =

5 . 169

Example 5.8 Find the Laplace transforms of the following functions: (i) te−4t sin 3t; (ii) t cosh t cos t; (iii) te−2t sinh 3t (iv) t 2e−t cos t. (i) L {te−4t sin 3t}= [ L(t sin 3t )] s → s + 4 (by the first shifting property)

(1)

Mathematics II

5.50 Now

d L(sin 3t ) ds d  3   =−  2 d s  s + 9 

L(t sin 3t ) =−

=

6s ( s + 9) 2 2

(2)

Using (2) in (1), we have  6s   L {te−4t sin 3t}=  2  (s + 9) 2   s→s+4 6 (s + 4) = 2 (s + 8s + 25) 2

Note

The same problem has been solved by using an alternative method in Worked Example (6) in Section 5(a). t    (ii) L {t cosh t cos t}= L  (et + e−t ) cos t      2  1 (1) = [ L(t cos t ) s → s −1 + L (t cos t ) s → s +1 ] 2 Now

L (t cos t ) =− =

d  s    ds  s 2 +1

s 2 −1 ( s 2 +1) 2

(2)

Using (2) in (1), we have 1  ( s −1) 2 −1 ( s +1) 2 −1  L (t cosh t cos t ) =  2 + 2 2 2  ( s − 2 s + 2) ( s + 2 s + 2) 2  s 2 + 2 s  1  s2 − 2 s =  2 + 2 2 2  ( s − 2 s + 2) ( s + 2 s + 2) 2  (iii) L{t e−2t sinh 3t}= L{t sinh 3 t}s → s + 2 Now

(1)

d L (sinh 3t ) ds d  3   =−  2 ds  s − 9 

L (t sinh 3t ) =−

= Using (2) in (1), we have

6s ( s − 9) 2 2

(2)

5.51

Laplace Transforms L {te−2t sinh 3t}=

6 ( s + 2) 6 ( s + 2) = {( s + 2) 2 − 9}2 ( s −1) 2 ( s + 5) 2

Aliter   1   L {te−2t sinh 3t}= L te−2t ⋅ (e3t − e−3t )   2     1 = L { tet − te−5t } 2  1  1 1   =  −  2 2  2 s s ( 2 ) ( 5 ) − +     =

  1 12 s + 24  6 ( s + 2)   =  2 2  2 s s ( 1 ) ( 5 ) ( s − 1 ) 2 ( s + 5) 2 − +    

(iv) L{t2 e−t cos t} = [L (t2 cos t)]s → s + 1

(1)

d2 L (cos t ) ds 2 d2  s  = 2  2  ds  s +1

L (t 2 cos t ) = (−1) 2

Now

=

  d  1− s 2     2 2   ds  ( s + 1 )   

=

2 ( s 3 − 3s ) ( s 2 +1)3

(2)

Using (2) in (1), we have L {t 2 e−t cos t}=

2 {( s +1)3 − 3 ( s +1)} ( s 2 + 2 s + 2) 3

Example 5.9 Find the inverse Laplace transforms of the following functions:  a (i) log 1−   s  s +1 s ( s +1) 2

(iii) log

s +a (ii) log   s + b 2

2

2

2

  

 s −1  + k (k is a constant) (iv) s log   s +1

1 (i) L−1 { φ ( s )}=− L−1 { φ ′ ( s )} t  s − a   a  s − a  1 −1  d  log   =− L ∴  L−1 log 1−  = L−1 log    s   s   s   ds t 

(1)

5.52

Mathematics II 1 d =− L−1 {log (s − a ) − log s} t ds 1  1 1 =− L−1  −  t  s − a s  1 1 =− (e at −1) = (1− e at ) t t

(ii) By rule (1),  s 2 + a 2  1 d  =− L−1 L−1 log 2 [log(s 2 + a 2 ) − log(s 2 + b 2 )] 2 ds t  s + b   2s  1  2s   =− L−1  − 2  2 2 2   t + + s a s b     2 = (cos bt − cos at ) t (iii) By rule (1),

 s 2 +1   =−1 L−1 d [log (s 2 +1) − log s − log(s +1)]] L−1 log   s (s +1)  ds t    2s  1  1 1   =− L−1  − −  2    + t  1 s s 1 + s    1 =− (2 cos t −1− e−t ) t 1 = (1+ e−t − 2 cos t ) t (iv) By rule (1),   s −1  1 d  + k  =− L−1 [s log(s −1) − s log (s +1)] + L−1 (k ) L−1  s log    ds t  s +1    1  s s =− L−1  + log(s −1) − − log(s +1) + k δ(t )   t s +1  s −1  s  1 −1 1  s =− L−1  − − L [log(s −1) − log(s +1)] + k δ(t )   t  s −1 s +1 t 1  2 s  1  1  1 1  =− L−1  2  − − L−1  − + k δ(t )      t  s −1 t  t   s −1 s +1

5.53

Laplace Transforms

Note  −1  s   and L−1  s  do not exist, as s and s are improper rational [ L    s −1   s +1  s +1 s−1   s 2s , s   as 2 − functions. Hence we have simplified  which is a proper  s −1  s −1 s +1  rational function. ] 2 1 cosh t + 2 (et − e−t ) + k δ (t ) t t 2 2 = 2 sinh t − cosh t + k δ (t ) t t =−

Example 5.10 Find the inverse Laplace transforms of the following functions: −1  s + a   (i) cot–1 (as) (ii) tan   b   2   (iii) cot −1   s +1  (i)

 2  (iv) tan −1  2  s 

1 L−1 { φ (s )} = − L−1{ φ ′ (s )} t

(1)

d  1 L−1{cot −1 (as )} = − L−1  cot −1 (as )   t d s 

∴

 −a 1 = − L−1  2 2 t  1+ a s  1  a  = L−1  2 2 t 1+ a s 

  

 1/ a    1  = 1 sin t . = L−1   2 2  a t    s + (1/ a )   t (ii)

   −1  s + a   1 −1  d −1  s + a    L−1   tan   = − L  tan    b   b   t     d s  1 1/ b = − L−1  2  t    1 +  s + a   b     1 b = − L−1  2 2  t  (s + a ) + b

         

Mathematics II

5.54

1 = − e−at sin bt t . (iii)

(iv)

d   2   −1  2   1  .  = − L−1  cot −1  L−1  cot      + s t s s + 1 1 d                         1 1 2  =− L−1  − −     4  (s +1) 2  t   1+     (s +1) 2      1 2  =− L−1  2   t  (s +1) + 4  1 =− e−t sin 2t. t    −1  2   1 −1  d −1  2  L−1  tan  2   =− L  tan  2    s   s  t     d s     1 −1  1  −4  =− L  . 3  t 1+ 4  s    s4     4 s  = L−1  4  s + 4 t   s s Consider 4 = s + 4 (s 2 + 2) 2 − (2 s ) 2 s = 2 (s − 2 s + 2) (s 2 + 2 s + 2)

(2)

 1 1 1 , =  2 − 2  4  s − 2 s + 2 s + 2 s + 2   1 1 1  =  − 2 2  4  (s −1) +1 (s +1) +1

by resolving into partial fractions

 s  1 t ∴ L−1  4  =  e sin t − e−t sin t    s + 4  4  1 = sin t sinh t 2 Using (3) in (2), we have  −1  2   2  L−1  tan  2   = sin t sinh t   s      t

(3)

5.55

Laplace Transforms Example 5.11 Find the Laplace transforms of the following functions: (i)

sinh t e at −cosbt e−at − e−bt ; (ii) ; (iii) ; (iv) t t t

2 sin 2t sin t ; t

 sin t  (v)  .  t  2

(i)

 f (t )   ∞  L  = ∫ L {f (t )}ds     t   s

∴

 sinh t   ∞  L  = ∫ L (sinh t ) ds     t   s ∞

=∫ s

(1)

1 ds s −1 2

∞

1  s −1  =  log   s +1 s  2 1  s −1   s −1  1   =  log  − log    s +1  s → ∞ 2  s +1   2   1  1−    s +1  1 1   + log  =  log  s    1  − 2 s 1 2  1+   s  s→∞   s +1 1  s +1 1 1  = log   = log 1 + log   s −1  2 2  s −1 2

(ii)

∞  e−at − ebt    − at −bt  L   = ∫ L (e − e ) d s , by rule   t     s

∞  1 1  d s = ∫  −  s + a s + b  s

  s + a   s + a    =  log  − log     s + b    s + b  s → ∞   a   1+   s + b    s   + log  =  log     1+ b   s + a     s  s → ∞ 

(1)

Mathematics II

5.56

 s + b   = log1+ log   s + a   s + b   = log   s + a 

(iii)

∞   e at −cosbt   at  = L   ∫ L (e − cos bt ) ds , by rule (1)   t     s ∞  1 s   ds = ∫  −  s − a s 2 + b 2  s ∞

  1 =  log ( s − a ) − log ( s 2 + b 2 )   s 2

      s − a    s − a  − log  =  log     s 2 + b 2    s 2 + b 2      s →∞     s 2 + b 2     1− a /s   =  log  + log   2 2    s − a    1+ b /s      s →∞ = log 1 + log

s2 + b2 ( s − a )2

  s2 + b2   1  = log   2  2   ( s−a )   ∞   2 sin 2t sin t    (iv) L    = ∫ L(2 sin 2t sin t ) ds, by rule (1)   t     s ∞

= ∫ L( cos t − cos3t ) ds s

 s s  ds = ∫  2 −  s +1 s 2 + 9  s ∞

∞

1  s 2 +1   =  log  2  s + 9  s  2 =

 s 2 +1  1   s 2 +1  1    − log log    s 2 + 9  2   s 2 + 9  s → ∞ 2

5.57

Laplace Transforms   1    1+ 2   s 2 + 9  1 1   s    =  log  + log   s 2 +1  2 2  1 + 9/ s 2         s→∞  s 2 + 9  1 1  = log1 + log  2 2 2  s +1   s 2 + 9  1  = log  2 2  s +1  (v)

∴

 f (t )   ∞  L 2  = ∫     t   s ∞   sin 2 t   L  2  = ∫     t   s

∞

∫ L{ f (t )} ds ds ∞

∫ L (sin

s

2

t ) ds d s

s

∞ ∞

=∫

(2)

s

1− cos 2t     ds d s  2  

∫ L  s

∞ ∞

1

s   ds ds + 4 

=

1 2

=

 s2 + 4 1 1 log  2  ds, ∫ 2 s 2  s 

∫ ∫  s − s s

s

2

∞

by putting a = 0 and b = 2 in (iii) above. ∞   ∞  s 2 + 4    2 s 1  2      = s log  2  − ∫ s  2 −  ds  s + 4 s    s  4   s  s   by integrating by parts.

s  s2 + 4 1 ∞ 8  s 2 + 4  s − log  2 + ∫ 2 =  log  2  ds  s  4 s s + 4  4  s  s →∞ 4  s 2   −1 s  − cot = L + log  2  s + 4   4

∞

s   , say 2 s

 s 2  s s  + cot −1   = L + log  2  2   s + 4  4  4 L = log 1+ 2   s 

s/4

Now

(3)

Mathematics II

5.58

1

s 2 /4  s  4     = log 1+ 2    s  

1  s 2 /4   s     4   lim ( L) = log  lim 1+ 2     s    s →∞  s →∞        0 = log (e ) = log 1 = 0

(4)

Using (4) in (3), we have  sin 2 t  s  s 2  s  + cot −1   . L  2  = log  2  2     t  4  s + 4  Example 5.12 Use Laplace transforms to evaluate the following: ∞

∫

(i)

0 ∞

0

∞

∫ 0

(ii)

∫ 0 ∞

 cos at −cos bt  d t  t

∫ 

(iii)

(i)

∞

e−t sin 3 t dt t

(iv)

sin 2 t dt t et

 e−2t − e−4t   d t .  t

∫  0

∞  sin 3 t   e−t sin 3 t  d t  d t =  ∫ e−st   t   t  0  s =1  sin 3 t   = L    t s =1

(1)

 sin 3 t  ∞  = ∫ L(sin 3 t ) ds L    t s

Now

∞

=∫ s

3 s + ( 3 )2 2

ds ∞

  1   s  =  3 −  cot −1      3  3   s   s = 0 + cot −1    3  Using (2) in (1), we have ∞

∫ s

 1  π e−1 sin 3t d t = cot −1   = . t  3  3

(2)

Laplace Transforms ∞

∫

(ii)

0

5.59

∞  2  sin 2 t −t  sin t  d t e = ∫  t  dt t et 0

∞  sin 2 t    dt  =  ∫ e−st   t   s =1  0  sin 2 t   = L   t 

(1)

s =1

 sin 2 t  1− cos 2t   = L   L   2 t   t 

Now

= =

1 2 1 2

∞

∫ L(1− cos 2t ) ds s ∞

1

∫  s − s s

2

s  d s + 4  ∞

   s 1   =  log   2   s 2 + 4   s   2   1 s  1 s   − log  = log 2   s 2 + 4  2  2 s + 4     s→ ∞  s 2 + 4   1 1 1    + = log log   2   2  s 1 + 4 /s s → ∞ 2   =

 s 2 + 4  1 1   log 1+ log    2 s 2  

 s 2 + 4  1   = log   s 2  

(2)

Using (2) in (1), we have ∞

∫ 0

sin 2 t 1 1 dt = log 5 = log 5 2 4 t et

∞   cos at − cos bt    e−st  cos at − cos bt dt   d t =  ∫  ∫    t t  0 0 s = 0    cos at − cos bt  =L   s = 0 t   ∞

(iii)

(1)

Mathematics II

5.60

∞   cos at − cosbt   L  = ∫ L(cos at − cos bt ) ds   t     s ∞  s s   ds = ∫  2 − 2 2  s + a s + b 2 

Now

s

∞

1  s 2 + a 2   =  log  2 2  s + b  s  2 1  s 2 + a 2   s 2 + a 2  1     − =  log  2 log  s 2 + b 2   s + b 2  2  2  s →∞ = log

s 2 + b2 s2 + a2

(2)

Using (2) in (1), we have ∞

b  cos at − cos bt   dt = log   a  t

∫  0

−4 t   ∞  −2 t  e−2t − e−4t    e−st  e − e  dtt   = d t   ∫  t   ∫  t   0  s=0 0  e−2t − e−4t   = L    t ∞

(iv)

(1)

s=0

 e−2t − e−4t  ∞  = ∫ L (e−2t − e−4t ) ds L   t  s

Now

∞  1 1   ds = ∫  −  s + 2 s + 4  s ∞

  s + 2   =  log   s + 4   s    s + 2   s + 2    − log  =  log      s + 4    s + 4  s →∞  s + 4   = log   s + 2  Using (2) in (1), we have ∞

 e−2t − e−4t   dt = log 2  t

∫  0

(2)

Laplace Transforms

5.61

Example 5.13 Find the inverse Laplace transforms of the following functions: s (s + a 2 ) 2

(ii)

(iii)

4(s −1) (s 2 − 2 s + 5) 2

(iv)

(v)

s +1 (s + 2 s − 8) 2

(i)

2

s (s − 4) 2 2

s2 −3 (s 2 + 4 s + 5) 2

2

∞

L−1{φ (s )} = t ⋅ L−1 ∫ φ (s ) ds

(i)

(1)

s ∞     s s  = t ⋅ L−1 ds L−1   2 2 2 2 ∫   (s + a 2 ) 2   (s + a )   s

∴

∞

= t L−1

1 dx , on putting s2 + a2 = x 2 x2 s2 + a2

∫

∞

(ii)

=

t −1  1  L −   x s2 + a2 2

=

t −1  1   L  2 2 2  s + a 

=

t sin at. 2a

∞     s s  = t ⋅ L−1 ds L−1   2 2 2 ∫   (s − 4) 2   (s − 4)   s  1  t  , as in (i) above. = L−1  2 2  s − 4 

=

t sinh 2t 4

Note

The inverse transform in this case can also found out by resolving the given function into partial fractions. (iii)

       4(s −1)  s −1 −1   = 4L  L  2 2 2 2 2  (s − 2 s + 5)    (s −1) + 2           −1

    s  , by the first shifting property = 4et L−1   2 2 2   ( + ) s 2    

Mathematics II

5.62 = 4et

t sin 2t , as in problem (i) 4

= tet sin 2t. 2     s2 −3  = L−1  (s + 4 s + 5) − (4 s + 8)  (iv) L−1   2 2    (s 2 + 4 s + 5) 2    (s + 4 s + 5)    

      1 s+2 −1   = L−1   2 − 4 L   2 2       s + 4s + 5   (s + 4 s + 5)        1 s+2 −1   = L−1    − 4 L  2 2 2     (s + 2) +1   {(s + 2) +1}  t = e−2t sin t − 4e−2t sin t , as in problem (i) 2 = e−2t (1−2t) sin t. (v)

      s +1 s +1  = L−1    L−1   2  2 2 2 2        (s + 2 s − 8)   [(s +1) − 3 ]       s  , = e−t ⋅ L−1   2 2 2   ( 3 ) s −     t = e−t ⋅ sinh 3t , proceeding as in problem (ii) 6 t = e−t sinh 3t 6

EXERCISE 5(b) Part A (Short Answer Questions) 1. State the formula for the Laplace transform of a periodic function. 2. Find the Laplace transform of f(t) = t, in 0 < t < 1 if f (t + 1) = f(t) 3. State the relation between the Laplace transforms of f(t) and t · f(t). 4. State the relation between the inverse Laplace transforms of φ (s) and φ′ (s). 1 5. State the relation between the Laplace transforms of f(t) and f (t ) . t 6. State the relation between the inverse Laplace transform of φ (s) and its integral. Find the Laplace transforms of the following functions: 1 t sin at 7. 8. t cos at 9. sin kt − kt cos kt 2a

5.63

Laplace Transforms 10. sin kt + kt cos kt

11.

t (1− cos at ) a2

1 kt sin kt . 2 Find the inverse Laplace transforms of the following functions: 12. cos kt −

 s + 1 13. log    s −1

 s + 1 14. log   s 

 s + a  16. log   s + b 

 s  17. log   s −1

 a 15. log 1+   s  s 2 + 1   18. log  2  s + 4 

a s Find the Laplace transforms of the following functions: 1− et 1− e−t sin at 21. 22. 23. t t t 20. tan −1

19. cot−1 s

24.

1− cos at t

25.

sin 2 t t

Part B Find the Laplace transforms of the following periodic functions: 1 f (t ) = E , in 0 ≤ t < 26. E 1 2π = 0, in ≤ t < E n  2π  f t +  = f (t ) given that  n T 2 = − E, in T/2 ≤ t < T given that f (t + T) = f (t) 28. f (t) = et, in 0 < t < 2π and f (t + 2π) = f (t) t  29. f (t ) = sin   , in 0 < t < 2 π and f (t + 2π) = f (t)  2 30. f (t) = |cos ωt|, t ≥ 0 27.

f (t ) = E , in 0 ≤ t
0 L (s + a ) 2 + w 2

i (t ) =

If

E - at ◊ e sin w t Lw

ω = 0, i (s )=

∴

i (t )= ω2 < 0

If

E 1 ⋅ L ( s + a)2 E −at te L

a n d ω2 = − k 2 ,

∴

i (s ) =

E 1 ⋅ L (s + a ) 2 − k 2

i (t ) =

E −at e sinh kt . Lk

Example 5.17 Solve the simultaneous equations 3x′ + 2y′ + 6x =0 and t

y ′ + y + 3∫ x dt = cos t + 3 sin t ,

x (0) = 2 and y (0) = − 3.

0

Taking Laplace transforms of both the equations, we get 3[s x (s ) − 2] + 2[s y (s ) + 3] + 6 x (s ) = 0 and 3 3 s + s y (s ) + 3 + y (s ) + x (s ) = 2 s s + 1 s 2 +1 i.e.

(3s + 6) x (s ) + 2 s y (s ) = 0

3 s +3 x (s ) + (s +1) y (s ) = −3 s s +1 Solving (1) and (2) for x (s ), we have

and

2

[3(s + 2) (s +1) − 6] x (s ) =−

i.e.

2 s (s + 3) + 6s s +1 2

2 1 2 x (s )=− ⋅ + 3 s +1 s + 3 2

(1) (2)

Laplace Transforms

5.109

2 x =− sin t + 2e−3t . 3

∴

Solving (1) and (2) for y (s ) , we have y (s ) =

 1 2 s −1 =−  + 2   (s + 3) ( s +1)  s + 3 s +1  3s 2 + 5s − 2 2

y = − e−3t − 2 cos t + sin t.

∴

Example 5.18 Solve the integral equation t

y (t ) =

t2 − u y (t − u ) du 2 ∫0

Noting that the integral in the given equation is a convolution type integral and taking Laplace transforms, we get 1 − L(t ) ⋅ L{ y (t )} s3 1 1 = 3 − 2 y (s ) s s

y (s ) =

1+ s 2  1 1   y (s ) = 3 or y (s ) = 2   s  s (1+ s 2 ) s

∴

t

y (t ) = ∫ sin t dt

∴

0

= 1 − cos t. Example 5.19 Solve the integral equation t

y (t ) = a sin t − 2 ∫ y (u ) cos (t − u ) du. 0

Taking Laplace transforms, y (s ) =

a − 2 L { y (t )}⋅ L (cos t ) s 2 +1

i.e.

y (s ) =

2s a − ⋅ y (s ) s 2 + 1 s 2 +1

i.e.

(s +1) 2 a y (s ) = 2 2 s +1 s +1

Mathematics II

5.110 a

i.e.

y (s ) =

∴

y (t ) = a t e−t .

( s +1)

2

Example 5.20 Solve the integro-differential equation t

y ′ (t ) = t + ∫ y (t − u ) cos u d u , y (0) = 4 0

Taking Laplace transforms, s y (s ) − 4 =

1 s + 2 y (s ) 2 s s +1

 1  1 s 1− 2  y (s ) = 2 + 4  s +1 s

i.e.

y (s ) =

i.e.

(s 2 +1)(1+ 4 s 2 ) s5

4 5 1 = + 3+ 5 s s s 5 1 y (t ) = 4 + t 2 + t 4 . 2 24

∴

EXERCISE 5(d) Part A (Short Answer Questions) Using Laplace transforms, solve the following equations: 1. x ′ + x = 2 sin t, x (0) = 0 2. x ′ − x = et , x (0) = 0 3. y ′ − y = t, y (0) = 0 4. y ′ + y = 1, y (0) = 0 t

5.

y + ∫ y (t ) d t = e−t 0 t

2 6. x + ∫ x (u ) d u = t + 2t 0

Laplace Transforms

5.111

t

7. x + ∫ x (t ) d t = cos t + sin t 0 t

8. x − 2 ∫ x (t ) d t =1 0

t

9. y =1+ 2 ∫ e−2u y (t − u ) d u 0 t

10. y =1+ ∫ y (u ) sin (t − u ) d u 0 t −u 11. f (t ) = cos t + ∫ e f (t − u ) d u 0 t

12. y (t ) = t + ∫ sin u y (t − u ) d u 0

Part B Solve the following differential equations, using Laplace transforms: 13. x″ + 3x′ +2x = 2 (t2 +t + 1), x (0) = 2,x′ (0) = 0 14. y″ − 3y′ − 4y = 2e−t, y (0) = y′ (0)=1 15. x″ + 4x′ + 3x = 10 sin t, x (0) = x′ (0) = 0 16. (D2 + D − 2) y = 20 cos 2t, y = − 1, Dy = 2 at t = 0 17. x″+ 4x′ + 5x = e−2t (cos t − sin t), x (0) = 1, x′ (0) = −3 18. y″ + 2y′ + 2y = 8 et sin t, y (0)= y′ (0) = 0 19. x″ − 2x′ + x = t 2 et, x (0) = 2, x′ (0) = 3 20. y″ + y = t cos 2t, y (0) = y′ (0) = 0 21. x″ + 9x=18t, x (0) = 0, x (π/2) = 0 22. y″ + 4y′ = cos 2t, y (π) = 0, y′ (π) = 0 23. (D2 + a2) x = f (t) 24. (D3 − D) y = 2 cos t, x = 3, Dx = 2, D2x = 1 at t = 0 25. x′″ − 3x″ + 3x′ − x =16 e3t, x (0) = 0, x′ (0) = 4, x″ (0) = 6 26. (D4 − a4) y = 0, y (0) = 1, y′ (0) = y″(0) = y′″ (0) = 0 Solve the following simultaneous equations, using Laplace transforms: 27. x′ − y = et; y′ + x = sin t, given that x (0)= 1 and y (0) = 0 28. x′ − y = sin t; y′ − x = − cos t, given that x = 2 and y = 0 for t = 0 29. x′ + 2 x − y = − 6 t; y′ − 2x + y = − 30 t, given that x = 2 and y = 3 at t = 0 30. Dx + Dy + x − y = 2; D2x + Dx − Dy = cos t, given that x = 0, Dx = 2 and y = 1 at t = 0 31. D2x + y = − 5 cos 2t; D2y + x = 5 cos2t, given that x = Dx = Dy = 1 and y = − 1 at t = 0

Mathematics II

5.112

t

32. x ′ + y ′ − x = 2et + e−t ; 2x ′ + y + 3 ∫ y dt = 2et (t + 3) , given that x (0) = 0

− 1 and y (0) =2 Solve the following integral equations, using Laplace transforms: t

33. x ′ + 3 x + 2 ∫ x dt = t , x (0) = 0 0 t −t 34. y ′ + 4 y + 5 ∫ y dt = e , y (0) = 0 0 t

35. x ′ + 2 x + ∫ x dt = cos t , x (0) = 1 0 t −2 t 36. y ′ + 4 y + 13 ∫ y dt = 3e sin 3t , y (0) = 3 0 t

37. x (t ) = 4t − 3 ∫ x (u ) sin (t − u ) du 0 t −t 38. y (t ) = e − 2 ∫ y (u ) cos (t − u ) du 0 t

39.

∫ 0

x(u ) t −u

du = 1 + t + t 2

t

40.

∫

y (u ) y (t − u ) du = 2 y (t ) + t − 2

0

ANSWERS

Exercise 5(a) t 3. tan t ; e

2

9.

1 {1 − e−(s − 2 ) } s −2

2 1 1 1 10.  + 2  e−s −  + 2  e−2 s s s  s s  11. (1 − e−π s ) ⋅

2 s +4 2

12. (1 − e− 2 π s ) ⋅

2 s +1 2

Laplace Transforms

13.

s ⋅ e−2 π s / 3 s2 + 1

15.

1 2s

π 1 ; s s

2 4 4  −2 s 17.  3 + 2 +  e s s s

14.

π 1 1 (1 + e− πs ) +  + 2  e−πs  s s  s2 + 1

16.

1 −3 s π e ; 2 e−s s2 s + π2

6a 3 6a 2b 3ab 2 b3 + 3 + 2 + s4 s s s   1 1 s  1  (ω cos θ + s sin θ ) 20.  − 2 19. 2 36  s s +  s +ω 2

18.

2

21.

1  3s s   − 2  2 4  s + 4 s + 36 

22.

1  3 1   − 2  2 2  s + 9 s + 1

23.

1  s s   + 2  2 2  s + 25 s + 1

24.

1  1 1 3 1   − + −  8  s − 3 s − 1 s + 1 s + 3 

25.

1  1 3 + +  4 s − 2 s +2

26.

2 2 1 + + (s + 1)3 (s + 1) 2 s + 1

27.

s (s + 2) 2 + 9

28.

 1 3 1   +   4   s − 3 s + 1

29.

2s s +4

30.

1 1 + 3 (s − 1) (s + 1)3

31.

2e3 2s − 3

32. (t − a) ua (t)

4

33. u2 (t) − u3 (t) 35. cos 3 (t −1) u1 (t) 37. 2 39.

t −t e π

1 3 3t / 2 t e 96

41. 2 cos 2t + 43.

3 sin 2t 2

1 (1 − e−at ) a

45. e3t cos t

47.

π s ⋅ s−a s−a

2   s 

34. e3(t −2). u2 (t) 36. sin t + sin (t − π) uπ(t) 38.

1 (2 + 4t + 3t 2 ) 2

40.

1 3 t (2 + t ) e 2t 12

42. cosh 3t + 2 sinh 3t 44.

1 −t e sin 2t 2

46.

8 12 9 + + 3 2 (s − 3) (s − 3) s −3

48.

 1 1 3 3 1   − + −   8   s − 2 s s + 2 s + 4 

5.113

Mathematics II

5.114

2 (s − 2) ( s 2 − 4 s + 8)

49.

4a 3 s 4 + 4a 4

51.

   (s + 3)  3 1   +   2 2  ( s − ) + ( s + ) + 4  3 4 3 36    

52.

50.

w cos q + (s + k ) sin q (s + k ) 2 w 2

53.

3 ( s 2 + 2 s + 9)

(s 2 + 2s + 5)(s 2 + 2s + 17)

54.

  1 1 1 1 (s − 1)   + + +   2 2 2 2 4  (s − 1) + 4 (s −1) + 16 (s + 1) + 36    (s − 1)

55.

  1 1 3 5 9  + + −   2 2 2 2 4 (s + 2) + 9 (s + 2) + 25 (s + 2) + 81   (s + 2) s2 − 4

56.

(s

2

58.

(s

2

60.

+ 4)

2

6 (s − 2) − 4 s + 13)

2

s2 − 4

(s

2

(s

2

59.

(s

2

61.

+ 4)

2

s (s + 2)

57.

+ 2 s + 2)

2

2s + 1)

2

1 5 − 2et + e− 2t 2 2

62. 2e − t − 3et + 5e2t

 9  63. 1 − 5t + t 2  e−t  2 

64. − 2e − t + 2e2t + te2t

65. − 2 + t + 2e−t + t e−t

66. sin t −

1 1 sin 2t + sin 3t 2 3

68. e − 3t (2 cos 5t − 3 sin 5t)

70.

67. cos t − 2 cos 2 t + cos 3 t 69.

 3 3 e− t / 2  3 cos t + sin  2 2 3

1

 λt    λt   m 1 − bt / 2 a  b  2a ⋅ sin   e cos   +  − ⋅   2a  a 2a  λ    2a   l if b2 − 4ac < 0 and = − λ2  kt   kt   m  1 − bt / 2 a  b  2a ⋅ sinh   , e cosh   +  − ⋅  2a    2a   l 2a  k a    if b2 − 4ac > 0 and = k2 1 − bt / 2 a e a

  m  b 2  1 +  −  t   , if b − 4ac = 0    2a  l    

 t  

5.115

Laplace Transforms 71. 2 + e − 2t (cos 3t + 2 sin 3t) 72.

 1  − t 3 3 t/2  cos e − e t − 3 sin   3  2 2

73.

1 (sin at cosh at − cos at sinh at) 4a 3

74.

1 sin t sinh t 2

75.

1 (sinh 2t cos 2t + cosh 2t sin 2t) 4

76. cos

t t cosh 2 2

78.

1 t sinh at 2a

80.

1 t sin 2t 4

77. 79.

2 3

 t   

sin

3 t t cosh 2 2

1 (sin t − t cos t ) 2

Exercise 5(b) 2.

1 1 − 2 s s s (e − 1)

s 7.

s2 − a2 8. 10.

12.

2

2ks 2

(s 2 + k 2 )

2

s3

(s

+k

2

2k 3

(s 2 + a 2 )

2

(s 2 − a 2 )

)

2 2

9. 11.

(s 2 + k 2 )

2

3s 2 + a 2 s 2 (s 2 + a 2 )

2

13.

2sinht t

14.

1− e− t t

15.

1− e− at t

16.

e− bt − e− at . t

17.

et −1 t

18.

2 (cos 2t − cos t ) t

19.

sin t t

20.

sin at t

21. cot − 1 (s/a)

Mathematics II

5.116

 s −1  23. log   s 

 s + 1 22. log    s  24.

26.

1 log 2

 s 2 + a 2     s 2 

E (1− e−s / E ) s (1− e

−2 π s / n

)

 s 2 + 4     s 2 

25.

1 log 4

27.

 sT  E tanh    4  s

28. [1 − e−2π(s − 1)]/(s − 1) (1 − e−2πs) 29.

2 coth (πs) 4 s +1

31.

  1 π 1  1− e−π s ) − e−π s  −2 π s  2 (   s 1− e s    

30.

2

32. 1/(s2 + 1) (1 − e−πs) 34.

Ï Ê ps ˆ¸ ˝ Ì s + w cosech Á Ë 2w ˜¯ ˛ s +w Ó 1

2

2

33. ω/(s2 + ω2) (eπ s/ω − 1)

 πs  1 tanh   2 2 s

1  π −π s 2 35.  2 (e−π s −1) + e (e−π s −1) / (1− e−2 π s )  s  s 36.

3 3 1  1 1   − + − 2 2 2  (s −1) (s + 1) (s + 3) 2  8  (s − 3)

37.

1  3( s 2 − 4) ( s 2 − 36)  + 2 2 4  ( s 2 + 4) ( s 2 + 36) 

38.

( 2 ) 1  ( s 2 − 4) − s − 64 2  2 2  ( s 2 + 4) ( s 2 + 64) 

  s s  + 39. s  2 2  ( s 2 + 36) ( s 2 + 4)   41.

43.

9 ( s 2 − 3)

( s 2 + 9)

3

+

1− 3s 2

( s 2 + 1)

s 2 − 6s − 7 (s 2 − 6 s + 25) 2

3

40.

1 s (48 − s 2 ) − s 3 ( s 2 + 16)3

42.

18(s 2 + 4 s + 1) (s 2 + 4 s + 13)3

44.

1 1 + 3 (s + 1) (s + 5)3

  s−2 s+2  − 45. 3  2  (s − 4 s + 13) 2 (s 2 + 4 s + 13) 2 

5.117

Laplace Transforms

46.

2 (1− cos a t ) t

47.

2 −bt (e − cos a t ) t

48.

2 (cos t − e 2t ) t

49.

2 2a sinh a t − cosh at + a δ (t ) t t2

50.

t  1 sin    2 t

1 −bt 52. − e sin at t

1 −2 t 51. − ⋅ e sin 3t t 53. −

2 t t sin sinh t 2 2

54. 0

55.

1 2

57.

1 log 2 2

59.

π 8

61.

1 log 2

 s 2 + a 2    s 2 

62.

63.

1 log 4

 s 2 + 16     s 2 + 4 

64. s log

65.

t sin t 2

66.

t sinh at 2a

67.

t 2t e sin t 2

68.

1 sin at − t sin a t a

69. e−3t(1 + t) sin t

70.

t −4t e sinh t. 2

56. log 3 b 58. log   a  s + 1  60. log   s  1 log 4

 s 2 + 16    s 2  s s +1 2

+ cot −1 s

Exercise 5(c) 8. No; 1* t = 12.

15.

t2 ≠t 2

2 s ( s 2 + 4) 1 at e (a sin b t + b cos b t ) b

10.

1 s−a

13. t (1 − t) e−2t

11.

s s + a2 2

2  t t 14. e  + 2 t + 1 2 

16. 2 e−2t − e−t

Mathematics II

5.118

17.

1 cot −1 s s

18.

 s −1 1 log    s  s

19.

20.

1 s (s + 1) 2

21.

1 . s (s 2 + 2 s + 2)

22.

23.

1 (1− e−at ). a

24. e−at + t − 1

26. 1 − cos t 1 t (e − e−3t ). 4 34. e −t + t − 1

37.

2a s

2

(s

2

+ 1)

2

1 (cosh a t −1) a2

1 ; 0 30. sin t 2 1 33. 2 (1− cos a t ) a

27. 1;0

28.

32. t et

31.

36.

25.

 s 2 + 1 1 log    s  s

35. sin t − cos t + e −t 2s3

; 2

; 2

2a s 2

(s 2 + a 2 ) (s 2 + a 2 ) (s 2 + a 2 ) s2 + a2

; 2

2a s 2

2

2a 3

; 2

(s 2 − a 2 ) (s 2 − a 2 ) (s 2 − a 2 )

2

38.

1 1 (e−bt − e−at ); (a e−at − b e−bt ) a −b a −b

39.

1 t 1 1 9 e − e 2 t + e 3 t ; e t − 4e 2 t + e 3 t 2 2 2 2

40.

 1  1 1 1 (cos b t − cos a t );  sin b t − sin a t  ; 2 a 2 − b 2  b a − b2 a 1 (a sin at − b sin bt ) a − b2 2

41.

1 t (sin 2 t + 2 t sin 2 t ); (cos 2 t − t sin 2 t ); e−3t sin 2 t 4 4

42.

1 1 t −at (sinh a t + a t cosh a t ); (a t sinh a t + 2 cosh a t ); e sinh at 2a 2 2a

43.

1 1 sin t sinh t ; (sin t cosh t + cos t sinh t ); cos t cosh t 2 2

44.

2 (s −1) s ( s 2 − 2 s + 2)

2

45.

2

( s 2 − 2s + 2)

2

47.

1 −1  s + 2  cot    3  s

49.

 s 2 + 4 s + 13  6 3 −1  s + 2   log   − cot  2   3    13 26 s

48.

 s + 2  1 cot −1    3  s+2

46.

3s 2 − 4 s + 2 s 2 ( s 2 − 2 s + 2)

2

5.119

Laplace Transforms 50. 2 − t − 2 e−t 4 −3t /2 3 −5t /3 7 73 e − e + t− 9 25 15 225 1 [4 − e−3t (3 sin 4 t + 4cos 4 t )] 52. 100 51.

53. e − t (1 − cos t)

1 (1− 2t − cos 2 t + sin 2 t ) 4 1 (2 − 2 cos 3 t − 3 t sin 3 t ) 56. 162 54.

1 (sin 3t − 3 t cos 3 t ) 54 1 1 −3 t (a t cosh a t − sinh a t ) e (sin t − t cos t ) 57. 58. 2a 3 2 1 59. (2 − 2 cosh a t + a t sinh a t ) 2a 4 1 −2 t e (2 t cosh 2 t − sinh 2 t ) 60. 16 55.

63.

1 (cos t − sin t − e−t ) 2

64.

1 (sin a t + sinh a t ) 2a

65.

1 (cos 2t − cos 5 t ) 5

66.

1 (sin a t + a t cos a t ) 2a

67.

1 (sin 2 t − 2 t cos 2 t ) 16

68. 2 e − t + sin 2 t − 2 cos 2 t

69.

e−t 2 (t + 4 t + 6) + t − 3 2

70.

1 (sin t cosh t − cos t sinh t ) 4

Exercise 5(d) 1. x = e − t − cos t + sin t 4. y = 1 − e −t 7. x = cos t 10. y = 1 +

t2 2

13. x = t 2 − 2t + 3 − e −2t

2. x = t e t 5. y = (1 − t)e −t 8. x = e 2 t

3. y = e t − t − 1 6. x = 2t 9. y = 1 + 2t

11. f (t) = cos t + sin t 12. y = t + 14. y =

t3 6

1 (13e−t −10t e−t + 12e 4t ) 25

15. x =

5 −t 1 − 3 t e − e − 2 cos t + sin t 2 2

16. y =

2 −2 t 4 t e + e − 3 cos 2t + sin 2t 3 3

 3 t −2 t  17. x = e  cos t − sin t + (sin t + cos t )   2 2

Mathematics II

5.120

18. y = 2 (sin t cosh t − cos t sinh t)  t4  19. x =  + t + 2 e t 12 

5 4 t 20. y =− sin t + sin 2 t − cos 2 t 9 9 3

21. x = 2 t + π sin 3 t

22. y =

1 (t − π ) sin 2 t 4

t

1 23. x = A cos a t + B sin a t + ∫ sin a u ⋅ f (t − u ) d u a 0 24. y = 3 sinh t + cosh t − sint + 2 25. x = 2e 3 t − 5t 2 e t − 2e t

26. y =

1 (cos a t + cosh a t ). 2

1 t 27. x = (e +2sin t + cos t − t cos t ); 2 1 y = ( − e−t − sin t + cos t + t sin t ) 2 28. 29. 30. 31. 32.

x = 2 cosh t; y = 2 sinh t − sin t x = 1 + 2t − 6t 2 + e−3t; y = 4 − 2t − 12t 2 − e −3t x = t + sin t; y = t + cos t x = sin t + cos 2t; y = sin t − cos 2 t x = t e t − e −t; y = e t + e −t

1 −2 t −t 33. x = (1+ e ) − e 2

1 − t 1 −2 t 34. y =− e + e (cos t + 3 sin t ) 2 2

1 35. x = {(1− t ) e−t + cos t} 2  7 3  −2 t  3 cos 3t − sin 3t + t sin 3t + t cos 3t  36. y = e    3 2   37. x = t +

3 sin 2 t 2

39. x (t )=

1  −1 2 8 3 2 12 t + 2t + t  π 3 

40. y (t) = 1

38. y (t) = e −t (1− t)2