Modern Engineering Mathematics 4th Edition Glyn James Pdf

Advanced Modern Engineering Mathematics Glyn James Solutions Manual 4th Edition

User Manual:

Open the PDF directly: View PDF .
Page Count: 688

Solutions Manual  Advanced Modern Engineering Mathematics fourth edition  Glyn James  Solutions Manual Advanced Modern Engineering Mathematics th  4 edition  Glyn James  ISBN 978-0-273-71925-0 c Pearson Education Limited 2011  Lecturers adopting the main text are permitted to download the manual as required.  c Pearson Education Limited 2011   i  Pearson Education Limited Edinburgh Gate Harlow Essex CM20 2JE England and Associated Companies throughout the world Visit us on the World Wide Web at: www.pearsoned.co.uk This edition published 2011 c Pearson Education Limited 2011  The rights of Glyn James, David Burley, Dick Clements, Phil Dyke, John Searl, Nigel Steele and Jerry Wright to be identiﬁed as authors of this work have been asserted by them in accordance with the Copyright, Designs and Patents Act 1988. ISBN: 978-0-273-71925-0 All rights reserved. Permission is hereby given for the material in this publication to be reproduced for OHP transparencies and student handouts, without express permission of the Publishers, for educational purposes only. In all other cases, no part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise without either the prior written permission of the Publishers or a licence permitting restricted copying in the United Kingdom issued by the Copyright Licensing Agency Ltd, 90 Tottenham Court Road, London W1T 4LP. This book may not be lent, resold, hired out or otherwise disposed of by way of trade in any form of binding or cover other than that in which it is published, without the prior consent of the Publishers.  ii  TABLE OF CONTENTS  Page Chapter 1. Matrix Analysis  1  Chapter 2. Numerical Solution of Ordinary Diﬀerential Equations  86  Chapter 3. Vector Calculus  126  Chapter 4. Functions of a Complex Variable  194  Chapter 5. Laplace Transforms  270  Chapter 6. The z Transform  369  Chapter 7. Fourier Series  413  Chapter 8. The Fourier Transform  489  Chapter 9. Partial Diﬀerential Equations  512  Chapter 10. Optimization  573  Chapter 11. Applied Probability and Statistics  639  iii  1 Matrix Analysis Exercises 1.3.3 1(a) Yes, as the three vectors are linearly independent and span threedimensional space.  1(b) No, since they are linearly dependent ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 1 3 ⎣2⎦ − 2⎣0⎦ = ⎣2⎦ 3 1 5 1(c)  No, do not span three-dimensional space. Note, they are also linearly  dependent.  2  Transformation matrix is ⎡ 1 1 1 ⎣ 1 −1 A= √ 2 0 0  ⎤ ⎡ 1 ⎤⎡ √ 0 1 0 0 2 1 ⎦ ⎣ ⎦ ⎣ 0 0 1 0 = √ √ 2 0 0 1 2 0  √1 2 − √12  0  ⎤ 0 0⎦ 1  Rotates the (e1 , e2 ) plane through π/4 radians about the e3 axis.  3  By checking axioms (a)–(h) on p. 10 it is readily shown that all cubics  ax3 + bx2 + cx + d form a vector space. Note that the space is four dimensional. 3(a)  All cubics can be written in the form ax3 + bx2 + cx + d  and {1, x, x2 , x3 } are a linearly independent set spanning four-dimensional space. Thus, it is an appropriate basis.  c Pearson Education Limited 2011   2  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  3(b)  No, does not span the required four-dimensional space. Thus a general  cubic cannot be written as a linear combination of (1 − x), (1 + x), (1 − x3 ), (1 + x3 ) as no term in x2 is present.  3(c) Yes as linearly independent set spanning the four-dimensional space a(1 − x) + b(1 + x) + c(x2 − x3 ) + d(x2 + x3 ) = (a + b) + (b − a)x + (c + a)x2 + (d − c)x3 ≡ α + βx + γx2 + δx3  3(d) Yes as a linear independent set spanning the four-dimensional space a(x − x2 ) + b(x + x2 ) + c(1 − x3 ) + d(1 + x3 ) = (a + b) + (b − a)x + (c + d)x2 + (d − c)x3 ≡ α + βx + γx2 + δx3  3(e)  No not linearly independent set as (4x3 + 1) = (3x2 + 4x3 ) − (3x2 + 2x) + (1 + 2x)  4 x + 2x3 , 2x − 3x5 , x + x3 form a linearly independent set and form a basis for all polynomials of the form α + βx3 + γx5 . Thus, S is the space of all odd quadratic polynomials. It has dimension 3.  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Exercises 1.4.3 5(a) Characteristic polynomial is λ3 − p1 λ2 − p2 λ − p3 with p1 = trace A = 12 ⎤ ⎡ −9 2 1 B1 = A − 12I = ⎣ 4 −7 −1 ⎦ 2 3 −8 ⎤ ⎡ −17 −5 −7 7 ⎦ A2 = A B1 = ⎣ −18 −30 2 −5 −33 p2 =  1 trace A2 = −40 2 ⎤ 23 −5 −7 7⎦ B2 = A2 + 40I = ⎣ −18 10 2 −5 7 ⎤ ⎡ 35 0 0 A3 = A B2 = ⎣ 0 35 0 ⎦ 0 0 35 ⎡  p3 =  1 trace A3 = 35 3  Thus, characteristic polynomial is λ3 − 12λ2 + 40λ − 35 Note that B3 = A3 − 35I = 0 conﬁrming check. 5(b)  Characteristic polynomial is λ4 − p1 λ3 − p2 λ2 − p3 λ − p4 with  p1 = trace A = 4 ⎡  −2 −1 ⎢ 0 −3 B1 = A − 4I = ⎣ −1 1 1 1 ⎡ −3 4 ⎢ −1 −2 A2 = A B1 = ⎣ 2 0 −3 −3  ⎤ 1 2 1 0⎥ ⎦ −3 1 1 −4 ⎤ 0 −3 1 −2 1⎥ ⎦ ⇒ p2 = trace A2 = −2 −2 −5 2 −1 3  c Pearson Education Limited 2011   3  4  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition ⎤ −1 4 0 −3 0 −2 1⎥ ⎢ −1 = A2 + 2I = ⎣ ⎦ 2 0 0 −5 −3 −3 −1 5 ⎤ ⎡ −5 2 0 −2 1 0 −2 −4 ⎥ ⎢ 1 = A B2 = ⎣ ⎦ ⇒ p3 = trace A3 = −5 −1 −7 −3 4 3 0 4 −2 −7 ⎤ ⎡ 0 0 0 −2 5 −2 −4 ⎥ ⎢ 1 = A3 + 5I = ⎣ ⎦ −1 −8 2 4 0 4 −2 −2 ⎤ ⎡ −2 0 0 0 1 0 0⎥ ⎢ 0 −2 = A B3 = ⎣ ⎦ ⇒ p4 = trace A4 = −2 0 0 −2 0 4 0 0 0 −2 ⎡  B2  A3  B3  A4  Thus, characteristic polynomial is λ4 − 4λ3 + 2λ2 + 5λ + 2 Note that B4 = A4 + 2I = 0 as required by check.  6(a)    1  2 Eigenvalues given by  1−λ 1 1−λ = λ − 2λ = λ(λ − 2) = 0  so eigenvectors are λ1 = 2, λ2 = 0 Eigenvectors given by corresponding solutions of (1 − λi )ei1 + ei2 = 0 ei1 + (1 − λi )ei2 = 0 Taking i = 1, 2 gives the eigenvectors as e1 = [1 1]T , e2 = [1 − 1]T  6(b)    2  2 Eigenvalues given by  1−λ 3 2−λ = λ − 3λ − 4 = (λ + 1)(λ − 4) = 0  so eigenvectors are λ1 = 4, λ2 = −1 Eigenvectors given by corresponding solutions of (l − λi )ei1 + 2ei2 = 0 3ei1 + (2 − λi )ei2 = 0  c Pearson Education Limited 2011   (1)  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition Taking i = 1, 2 gives the eigenvectors as e1 = [2 3]T , e2 = [1 − 1]T 6(c) Eigenvalues given by   1 − λ 0 −4    0 5−λ 4  = λ3 + 9λ2 + 9λ − 81 = (λ − 9)(λ − 3)(λ + 3) = 0   −4 4 3 − λ So the eigenvalues are λ1 = 9, λ2 = 3, λ3 = −3. The eigenvectors are given by the corresponding solutions of (1 − λi )ei1 + 0ei2 − 4ei3 = 0 0ei1 + (5 − λi )ei2 + 4ei3 = 0 −4ei1 + 4ei2 + (3 − λi )ei3 = 0 Taking i = 1, λi = 9 solution is e12 e13 e11 =− = = β1 8 16 −16  ⇒ e1 = [−1 2 2]T  Taking i = 2, λi = 3 solution is e21 e22 e23 =− = = β2 −16 16 8  ⇒ e2 = [2 2 − 1]T  Taking i = 3, λi = −3 solution is e32 e33 e31 =− = = β3 32 16 32 6(d)  Eigenvalues given by  1 − λ   0   −1  ⇒ e3 = [2 − 1 2]T   1 2  2−λ 2  = 0 1 3 − λ  Adding column 1 to column 2 gives   1 − λ 2 − λ 2    0 2−λ 2  = (2 − λ)   −1 0 3 − λ   1 − λ 0 0   R1 −R2 (2 − λ)  0 1 2   −1 0 3 − λ    1− λ 1 2    0 1 2    −1 0 3 − λ = (2 − λ)(1 − λ)(3 − λ)  c Pearson Education Limited 2011   5  6  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  so the eigenvalues are λ1 = 3, λ2 = 2, λ3 = 1. Eigenvectors are the corresponding solutions of (A − λi I)ei = 0 When λ = λ1 = 3 we have ⎡  −2 1 ⎣ 0 −1 −1 1  ⎤ 2 2⎦ 0  ⎡  ⎤ e11 ⎣ e12 ⎦ = 0 e13  leading to the solution e12 e13 e11 =− = = β1 −2 2 −1 so the eigenvector corresponding to λ2 = 3 is e1 = β1 [2 2 1]T , β1 constant. When λ = λ2 = 2 we have ⎡  −1 ⎣ 0 −1  1 0 1  ⎤ 2 2⎦ 1  ⎡  ⎤ e21 ⎣ e22 ⎦ = 0 e23  leading to the solution e21 e22 e23 =− = = β3 −2 2 0 so the eigenvector corresponding to λ2 = 2 is e2 = β2 [1 1 0]T , β2 constant. When λ = λ3 = 1 we have ⎡  0 ⎣ 0 −1  1 1 1  ⎤ 2 2⎦ 2  ⎡  ⎤ e31 ⎣ e32 ⎦ = 0 e33  leading to the solution e32 e33 e31 =− = = β1 0 2 1 so the eigenvector corresponding to λ3 = 1 is e3 = β3 [0 − 2 1]T , β3 constant.  6(e)  Eigenvalues given by     5 − λ 0 6    = λ3 − 14λ2 − 23λ − 686 = (λ − 14)(λ − 7)(λ + 7) = 0  0 11 − λ 6    6 6 −2 − λ  so eigenvalues are λ1 = 14, λ2 = 7, λ3 = −7 c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition Eigenvectors are given by the corresponding solutions of (5 − λi )ei1 + 0ei2 + 6ei3 = 0 0ei1 + (11 − λi )ei2 + 6ei3 = 0 6ei1 + 6ei2 + (−2 − λi )ei3 = 0 When i = 1, λ1 = 14 solution is −e12 e13 e11 = = = β1 ⇒ e1 = [2 6 3]T 12 −36 18 When i = 2, λ2 = 7 solution is −e22 e23 e21 = = = β2 ⇒ e2 = [6 − 3 2]T −72 −36 −24 When i = 3, λ3 = −7 solution is e31 −e32 e33 = = = β3 ⇒ e3 = [3 2 − 6]T 54 −36 −108 6(f)  Eigenvalues given by     −1 − λ 0 −1 − λ  −1 0   1 2−λ 1  2−λ 1  R1 +R2   −2 1 −1 − λ  1 −1 − λ     −1 0 0   0  = 0, i.e. (1 + λ)(2 − λ)(1 − λ) = 0 = (1 + λ)  1 2 − λ  −2 1 1 − λ   1 − λ   1   −2  so eigenvalues are λ1 = 2, λ2 = 1, λ3 = −1 Eigenvectors are given by the corresponding solutions of (1 − λi )ei1 − ei2 + 0ei3 = 0 ei1 + (2 − λi )ei2 + ei3 = 0 −2ei1 + ei2 − (1 + λi )ei3 = 0 Taking i = 1, 2, 3 gives the eigenvectors as e1 = [−1 1 1]T , e2 = [1 0 − 1]T , e3 = [1 2 − 7]T  c Pearson Education Limited 2011   7  8  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  6(g)  Eigenvalues given by    5− λ 5 − λ 5 − λ 1 1    5−λ 4  5 − λ 4  R1 + (R2 + R3 )  2  −1 −1 −λ  −1 −λ     1 0 0   2  = (5 − λ)(3 − λ)(1 − λ) = 0 = (5 − λ)  2 3 − λ  −1 0 1− λ   4− λ   2   −1  so eigenvalues are λ1 = 5, λ2 = 3, λ3 = 1 Eigenvectors are given by the corresponding solutions of (4 − λi )ei1 + ei2 + ei3 = 0 2ei1 + (5 − λi )ei2 + 4ei3 = 0 −ei1 − ei2 − λi ei3 = 0 Taking i = 1, 2, 3 and solving gives the eigenvectors as e1 = [2 3 − 1]T , e2 = [1 − 1 0]T , e3 = [0 − 1 1]T  6(h)  Eigenvalues given by      1 − λ 2 − 2λ 0 −4 −2    3−λ 1  3−λ 1  R1 +2R2  0  1 2 4 − λ 2 4 − λ   1 0 0   1  = (1 − λ)(3 − λ)(4 − λ) = 0 = (1 − λ)  0 3 − λ 1 0 4 − λ   1− λ   0   1  so eigenvalues are λ1 = 4, λ2 = 3, λ3 = 1 Eigenvectors are given by the corresponding solutions of (1 − λi )ei1 − 4ei2 − 2ei3 = 0 2ei1 + (3 − λi )ei2 + ei3 = 0 ei1 + 2ei2 + (4 − λi )ei3 = 0  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  9  Taking i = 1, 2, 3 and solving gives the eigenvectors as e1 = [2 − 1 − 1]T , e2 = [2 − 1 0]T , e3 = [4 − 1 − 2]T  Exercises 1.4.5 7(a)  Eigenvalues given by      1 − λ −1 + λ 2 1  0   3−λ 1  R1 −R2  0 3−λ 1   1 2 2 − λ 2 2− λ   1 0 0   1  = (1 − λ)[λ2 − 6λ + 5] = (1 − λ)(λ − 1)(λ − 5) = 0 = (1 − λ)  1 4 − λ 1 3 2 − λ   2 − λ   1   1  so eigenvalues are λ1 = 5, λ2 = λ3 = 1 The eigenvectors are the corresponding solutions of (2 − λi )ei1 + 2ei2 + ei3 = 0 ei1 + (3 − λi )ei2 + ei3 = 0 ei1 + 2ei2 + (2 − λi )ei3 = 0 When i = 1, λ1 = 5 and solution is −e12 e13 e11 = = = β1 ⇒ e1 = [1 1 1]T 4 −4 4 When λ2 = λ3 = 1 solution is given by the single equation e21 + 2e22 + e23 = 0 Following the procedure of Example 1.6 we can obtain two linearly independent solutions. A possible pair are e2 = [0 1 2]T , e3 = [1 0 − 1]T  c Pearson Education Limited 2011   10  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  7(b)  Eigenvalues given by   −λ   −1   −1  −2 1−λ −1   −2  2  = −λ3 + 3λ2 − 4 = −(λ + 1)(λ − 2)2 = 0 2 − λ  so eigenvalues are λ1 = λ2 = 2, λ3 = −1 The eigenvectors are the corresponding solutions of −λi ei1 − 2ei2 − 2ei3 = 0 −ei1 + (1 − λi )ei2 + 2ei3 = 0 −ei1 − ei2 + (2 − λi )ei3 = 0 When i = 3, λ3 = −1 corresponding solution is e31 −e32 e33 = = = β3 ⇒ e3 = [8 1 3]T 8 −1 3 When λ1 = λ2 = 2 solution is given by −2e21 − 2e22 − 2e23 = 0  (1)  −e21 − e22 + 2e23 = 0  (2)  −e21 − e22 = 0  (3)  From (1) and (2) e23 = 0 and it follows from (3) that e21 = −e22 . We deduce that there is only one linearly independent eigenvector corresponding to the repeated eigenvalues λ = 2. A possible eigenvector is e2 = [1 − 1 0]T 7(c)  Eigenvalues given by   1 − λ 6 6    3−λ 2  R1 −3R3  1  −1 −5 −2 − λ     1 −3 0    2  = (1 − λ) = (1 − λ)  1 3 − λ  −1 −5 −2 − λ    4− λ   1   −1   −3 + 3λ 0  3−λ 2  −5 −2 − λ     1 0 0   1 6 − λ 2   1 −8 −2 − λ   = (1 − λ)(λ2 + λ + 4) = (1 − λ)(λ − 2)2 = 0 c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  11  so eigenvalues are λ1 = λ2 = 2, λ3 = 1. The eigenvectors are the corresponding solutions of (4 − λi )ei1 + 6ei2 + 6ei3 = 0 ei1 + (3 − λi )ei2 + 2ei3 = 0 −ei1 − 5ei2 − (2 + λi )ei3 = 0 When i = 3, λ3 = 1 corresponding solution is −e32 e33 e31 = = = β3 ⇒ e3 = [4 1 − 3]T 4 −1 −3 When λ1 = λ2 = 2 solution is given by 2e21 + 6e22 + 6e23 = 0 e21 + e22 + 2e23 = 0 −e21 − 5e22 − 4e23 = 0 so that  e21 −e22 e23 = = = β2 6 −2 −4 leading to only one linearly eigenvector corresponding to the eigenvector λ = 2. A possible eigenvector is e2 = [3 1 − 2]T 7(d)  Eigenvalues given by     1− λ  7 − λ −2 −4     R1 −2R2  3  3 −λ −2     6  6 −2 −3 − λ     1 −2 0   −2  = (1 − λ) = (1 − λ)  3 −λ  6 −2 −3 − λ    −2 + 2λ 0  −λ −2  −2 −3 − λ     1 0 0   3 6− λ −2   6 10 −3 − λ   = (1 − λ)(λ − 2)(λ − 1) = 0 so eigenvalues are λ1 = 2, λ2 = λ3 = 1. The eigenvectors are the corresponding solutions of (7 − λi )ei1 − 2ei2 − 4ei3 = 0 3ei1 − λi ei2 − 2ei3 = 0 6ei1 − 2ei2 − (3 + λi )ei3 = 0 c Pearson Education Limited 2011   12  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  When i = 1, λ2 = 2 and solution is e11 −e12 e13 = = = β1 ⇒ e1 = [2 1 2]T 6 −3 6 When λ2 = λ3 = 1 the solution is given by the single equation 3e21 − e22 − 2e23 = 0 Following the procedures of Example 1.6 we can obtain two linearly independent solutions. A possible pair are e2 = [0 2 − 1]T , e3 = [2 0 3]T  8  ⎡  −4 (A − I) = ⎣ 2 1  ⎤ −7 −5 3 3⎦ 2 1  Performing ⎤a series of row and column operators this may be reduced to the form ⎡ 0 0 0 ⎣ 0 0 1 ⎦ indicating that (A − I) is of rank 2. Thus, the nullity q = 3 − 2 = 1 1 0 0 conﬁrming that there is only one linearly independent eigenvector associated with the eigenvalue λ = 1. The eigenvector is given by the solution of −4e11 − 7e12 − 5e13 = 0 2e11 + 3e12 + 3e13 = 0 e11 + 2e12 + e13 = 0 giving  9  e11 −e12 e13 = = = β1 ⇒ e1 = [−3 1 1]T −3 −1 1 ⎡  1 (A − I) = ⎣ −1 −1  ⎤ 1 −1 −1 1⎦ −1 1  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  13  Performing ⎤a series of row and column operators this may be reduced to the form ⎡ 1 0 0 ⎣ 0 0 0 ⎦ indicating that (A − I) is of rank 1. Then, the nullity of q = 3 − 1 = 2 0 0 0 conﬁrming that there are two linearly independent eigenvectors associated with the eigenvalue λ = 1. The eigenvectors are given by the single equation e11 + e12 − e13 = 0 and two possible linearly independent eigenvectors are e1 = [1 0 1]T and e2 = [0 1 1]T  Exercises 1.4.8 10  These are standard results.  11(a) (i) (ii)  Trace A = 4 + 5 + 0 = 9 = sum eigenvalues; det A = 15 = 5 × 3 × 1 = product eigenvalues; ⎡  (iii)  A−1  4 1 ⎣ −4 = 15 3  ⎤ −1 −1 1 −14 ⎦ . Eigenvalues given by 3 18   −1 −1  1 − 15λ −14  C3 −C2 3 18 − 15λ      4 − 15λ −1 0   1 − 15λ −1  = (15 − 15λ)  −4  3 3 1    4 − 15λ   −4   3    4 − 15λ   −4   3  −1 1 − 15λ 3    0  −15 + 15λ  15 − 15λ   = (15 − 15λ)(15λ − 5)(15λ − 3) = 0  conﬁrming eigenvalues as 1, 13 , 15 . c Pearson Education Limited 2011   14  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition ⎤ 2 −1 5 −1 ⎦ having eigenvalues given by 4 0  ⎡  (iv)  4 AT = ⎣ 1 1  4 − λ   1   1   2 −1  5 − λ −1  = (λ − 5)(λ − 3)(λ − 1) = 0 4 −λ   that is, eigenvalue as for A .  ⎡  11(b) (i)  8 2A = ⎣ 4 −2  2 10 −2  ⎤ 2 8 ⎦ having eigenvalues given by 0    8− λ 2 2    4 10 − λ 8  C1 −C2   −2 −2 −λ     1 2 2   = (6 − λ)  −1 10 − λ 8   0 −2 −λ    2 2  10 − λ 8  −2 −λ    1 2 2   = (6 − λ)  0 12 − λ 10  0 −2 −λ     6−λ   −6 + λ   0  = (6 − λ)(λ − 10)(λ − 2) = 0 Thus eigenvalues are 2 times those of A; namely 6, 10 and 2. ⎡  (ii)  ⎤ 6 1 1 A + 2I = ⎣ 2 7 4 ⎦ having eigenvalues given by −1 −1 2   6 − λ 1   2 7−λ   −1 −1   1  4  = −λ3 + 15λ2 − 71λ + 105 = −(λ − 7)(λ − 5)(λ − 3) = 0 2 − λ  conﬁrming the eigenvalues as 5 + 2, 3 + 2, 1 + 2. Likewise for A − 2I c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition ⎡  (iii)  17 A2 = ⎣ 14 −6  8 23 −6  ⎤ 8 22 ⎦ having eigenvalues given by −5      25 − λ 25 − λ 25 − λ  8 8    23 − λ 22 − λ  R1 + (R2 ) + R3 )  14 23 − λ 22   −6 −6 −5 − λ  −6 −5 − λ     1 0 0    8  = (25 − λ)(9 − λ)(1 − λ) = 0 = (25 − λ)  14 9 − λ  −6 0 1 − λ    17 − λ   14   −6  that is, eigenvalues A2 are 25, 9, 1 which are those of A squared.  12  Eigenvalues of A given by  −3 −3  1−λ −1  R3 +R2 −1 1 − λ    −3 − λ −3 −3   1 − λ −1  = (λ − 2)  −3  0 1 −1     −3 − λ   −3   −3   −3 −3  1−λ −1  −2 + λ 2 − λ    −3 − λ −3   C3 +C2 (λ − 2)  −3 (1 − λ)  0 1    −3 − λ   −3   0  = −(λ − 2)(λ + 6)(λ − 3) = 0 so eigenvalues are λ1 = −6, λ2 = 3, λ3 = 2 Eigenvectors are given by corresponding solutions of (−3 − λi )ei1 − 3ei2 − 3ei3 = 0 −3ei1 + (1 − λi )ei2 − ei3 = 0 −3ei1 − ei2 + (1 − λi )ei3 = 0 Taking i = 1, 2, 3 gives the eigenvectors as e1 = [2 1 1]T , e2 = [−1 1 1]T , e3 = [0 1 − 1]T It is readily shown that eT1 e2 = eT1 e3 = eT2 e3 = 0 so that the eigenvectors are mutually orthogonal.  c Pearson Education Limited 2011    −6  −λ  0   15  16 13  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition Let the eigenvector be e = [a b c]T then since the three vectors are mutually  orthogonal  a + b − 2c = 0 a+b−c=0  giving c = 0 and a = −b so an eigenvector corresponding to λ = 2 is e = [1 −1 0]T .  Exercises 1.5.3 14  Taking x(0) = [1 1 1]T iterations may then be tabulated as follows:  Iteration k x(k)  A x(k) λ   0 1 1 1 9 10 5 10  1 0.9 1 0.5 7.6 8.7 4.3 8.7  2 0.874 1 0.494 7.484 8.61 4.242 8.61  3 0.869 1 0.493 7.461 8.592 4.231 8.592  4 0.868 1 0.492 7.457 8.589 4.228 8.589  Thus, estimate of dominant eigenvalue is λ  8.59 and corresponding eigenvector x  [0.869 1 0.493]T or x  [0.61 0.71 0.35]T in normalised form. 15(a)  Taking x(0) = [1 1 1]T iterations may then be tabulated as follows:  Iteration k x(k)  A x(k) λ   0 1 1 1 3 4 4 4  1 0.75 1 1 2.5 3.75 3.75 3.75  2 0.667 1 1 2.334 3.667 3.667 3.667  3 0.636 1 1 2.272 3.636 3.636 3.636  4 0.625 1 1 2.250 3.625 3.625 3.625  5 0.620 1 1 2.240 3.620 3.620 3.620  6 0.619 1 1  Thus, correct to two decimal places dominant eigenvalue is 3.62 having corresponding eigenvectors [0.62 1 1]T .  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  17  Taking x(0) = [1 1 1]T iterations may be tabulated as follows:  15(b)  Iteration k x(k)  A x(k) λ   0 1 1 1 4 6 11 11  1 0.364 0.545 1 2.092 3.818 7.546 7.546  2 0.277 0.506 1 1.831 3.566 7.12 7.12  3 0.257 0.501 1 1.771 3.561 7.03 7.03  4 0.252 0.493 1 1.756 3.49 6.994 6.994  5 0.251 0.499 1  Thus, correct to two decimal places dominant eigenvalue is 7 having corresponding eigenvector [0.25 0.5 1]T .  15(c)  Taking x(0) = [1 1 1 1]T iterations may then be tabulated as follows:  Iteration k x(k)  A x(k) λ   0 1 1 1 1 1 0 0 1 1  1 1 0 1 1 2 −1 −1 2 2  2 1 −0.5 − 0.5 1 2.5 −1.5 −1.5 2.5 2.5  3 1 −0.6 −0.6 1 2.6 −1.6 −1.6 2.6 2.6  4 1 −0.615 −0.615 1 2.615 −1.615 −1.615 2.615 2.615  5 1 −0.618 − 0.618 1 2.618 −1.618 −1.618 2.618 2.618  6 1 − 0.618 − 0.618 1  Thus, correct to two decimal places dominant eigenvalue is 2.62 having corresponding eigenvector [1 − 0.62 − 0.62  16  1]T .  The eigenvalue λ1 corresponding to the dominant eigenvector e1 = [1 1 2]T  is such that A e1 = λ1 e1 so ⎤⎡ ⎤ ⎡ ⎤ 1 3 1 1 1 ⎣ 1 3 1 ⎦ ⎣ 1 ⎦ = λ1 ⎣ 1 ⎦ 2 1 1 5 2 ⎡  so λ1 = 6. c Pearson Education Limited 2011   18  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Then 1 1 2 A1 = A − 6ê1 êT1 where ê1 = √ √ √ 6 6 6 so  ⎤ ⎡ 1 3 1 1 ⎦ ⎣ ⎣ A1 = 1 3 1 − 1 2 1 1 5 ⎡  1 1 2  ⎤ ⎡ 2 2 ⎦ ⎣ 0 2 = −1 4  T  ⎤ 0 −1 2 −1 ⎦ −1 1  Applying the power method with x(0) = [1 1 1]T ⎤ 1 = ⎣ 1 ⎦ = x(1) −1 ⎤ ⎤ ⎡ ⎡ 1 3 = ⎣ 3⎦ =3 ⎣ 1 ⎦ −1 −3 ⎡  y(1) = A1 x(0)  y(2) = A1 x(1)  1 Clearly, λ2 = 3 and ê2 = √ [1 1 − 1]T . 3 Repeating the process ⎡  2 T ⎣ 0 A2 = A1 − λ2 ê2 ê2 = −1  0 2 −1  ⎤ ⎡ 1 −1 ⎦ ⎣ 1 −1 − −1 1  1 1 −1  ⎤ ⎡ 1 −1 ⎦ ⎣ −1 = −1 0 1  Taking x(0) = [1 − 1 0]T the power method applied to A2 gives ⎤ ⎤ ⎡ 1 2 = ⎣ −2 ⎦ = 2 ⎣ −1 ⎦ 0 0 ⎡  y(1) = A2 x(0)  1 and clearly, λ3 = 2 with ê3 = √ [1 − 1 0]T . 2  17  The three Gerschgorin circles are | λ − 5 |= 2, | λ |= 2, | λ + 5 |= 2  c Pearson Education Limited 2011   ⎤ −1 0 1 0⎦ 0 0  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  19  which are three non-intersecting circles. Since the given matrix A is symmetric its three eigenvalues are real and it follows from Theorem 1.2 that 3 < λ1 < 7 , −2 < λ2 < 2 , −7 < λ3 < 7 (i.e., an eigenvalue lies within each of the three circles).  18  The characteristic equation of the matrix A is   10 − λ   −1   0  that is  −1 2−λ 2   0  2  = 0 3 − λ  (10 − λ)[(2 − λ)(3 − λ) − 4] − (3 − λ) = 0 or  f(λ) = λ3 − 15λ2 + 51λ − 17 = 0  Taking λ0 = 10 as the starting value the Newton–Raphson iterative process produces the following table: f(λi ) f (λi )  i  λi  f(λi )  f (λi )  −  0  10  7  −51.00  0.13725  1  10.13725  − 0.28490  −55.1740  − 0.00516  2  10.13209  − 0.00041  −55.0149  −0.000007  Thus to three decimal places the largest eigenvalue is λ = 10.132 Using Properties 1.1 and 1.2 of section 1.4.6 we have 3  3  λi =| A |= 17  λi = trace A = 15 and i=1  i=1  c Pearson Education Limited 2011   20  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Thus, λ2 + λ3 = 15 − 10.132 = 4.868 λ2 λ3 = 1.67785 λ2 (4.868 − λ2 ) = 1.67785  so  λ22 − 4.868λ2 + 1.67785 = 0 λ2 = 2.434 ± 2.0607 that is 19(a)  λ2 = 4.491 and  λ3 = 0.373  If e1 , e2 , . . . , en are the corresponding eigenvectors to λ1 , λ2 , . . . , λn then  (KI − A)ei = (K − λi )ei so that A and (KI − A) have the same eigenvectors and eigenvalues diﬀer by K . Taking x(o) =  n  αr ei then i=1 n  x  (p)  = (KI − A)x  (p−1)  = (KI − A) x  2 (p−2)  αr (K − λr )p er  = ... = r=1  Now K − λn > K − λn−1 > . . . > K − λ1 and n  x  (p)  = αn (K − λn ) en +  αr (K − λr )p er  p  r=1 n−1  = (K − λn )p [αn en +  αr r=1  K − λr K − λn  p  er ]  → (K − λn ) αn en = Ken as p → ∞ p  Also (p+1)  xi  (p)  xi  →  (K − λn )p+1 αn en = K − λn (K − λn )p αn en  Hence, we can ﬁnd λn  19(b)  Since A is a symmetric matrix its eigenvalues are real. By Gerschgorin's  theorem the eigenvalues lie in the union of the intervals | λ − 2 |≤ 1, | λ − 2 |≤ 2, | λ − 2 |≤ 1 i.e.  | λ − 2 |≤ 2 or 0 ≤ λ ≤ 4. c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  21  Taking K = 4 in (a) ⎡  2 KI − A = 4I − A = ⎣ 1 0  1 2 1  ⎤ 0 1⎦ 2  Taking x(0) = [1 1 1]T iterations using the power method are tabulated as follows: Iteration k x(k)  A x(k) λ   0 1 1 1 3 4 3 4  1 0.75 1 0.75 2.5 3.5 2.5 3.5  2 0.714 1 0.714 2.428 3.428 2.428 3.428  3 0.708 1 0.708 2.416 3.416 2.416 3.416  4 0.707 1 0.707 2.414 3.414 2.414 3.414  5 0.707 1 0.707  Thus λ3 = 4 − 3.41 = 0.59 correct to two decimal places.  Exercises 1.6.3 20  Eigenvalues given by   −1 − λ  Δ =  0  0   −12  30  = 0 20 − λ   6 −13 − λ −9      −13 − λ 30  = (−1 − λ)(λ2 − 7λ + 10) Now Δ = (−1 − λ)  −9 20 − λ  = (−1 − λ)(λ − 5)(λ − 2) so Δ = 0 gives λ1 = 5, λ2 = 2, λ3 = −1 Corresponding eigenvectors are given by the solutions of (A − λi I)ei = 0 When λ = λ1 = 5 we have ⎡  −6 6 ⎣ 0 −18 0 −9  ⎤ −12 30 ⎦ 15  ⎡  ⎤ e11 ⎣ e12 ⎦ = 0 e13  c Pearson Education Limited 2011   22  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  leading to the solution  e11 −e12 e13 = = = β1 −36 −180 108  so the eigenvector corresponding to λ1 = 5 is e1 = β1 [1 − 5 − 3]T When λ = λ2 = 2, we have ⎡ −3 6 ⎣ 0 −15 0 −9 leading to the solution  ⎤ −12 30 ⎦ 18  ⎡  ⎤ e21 ⎣ e22 ⎦ = 0 e23  −e22 e23 e21 = = = β2 0 −90 45  so the eigenvector corresponding to λ2 = 2 is e2 = β2 [0 2 1]T When λ = λ3 = −1, we have ⎡ 0 6 ⎣ 0 −12 0 −9 leading to the solution  ⎤ −12 30 ⎦ 21  ⎡  ⎤ e31 ⎣ e32 ⎦ = 0 e33  −e32 e33 e31 = = = β3 18 0 0  so the eigenvector corresponding to λ3 = −1 is e3 = β3 [1 0 0]T A modal matrix M ⎡ 1 0 ⎣ M = −5 2 −3 1 ⎡ 0 1 −1 ⎣ M = 0 3 1 −1 21  and spectral matrix Λ are ⎤ ⎤ ⎡ 5 0 0 1 0⎦ 0⎦ Λ = ⎣0 2 0 0 −1 0 ⎤ −2 −5 ⎦ and matrix multiplication conﬁrms M−1 A M = Λ 2  From Example 1.9 the eigenvalues and corresponding normalised eigenvectors  of A are λ1 = 6, λ2 = 3, λ3 = 1 1 1 ê1 = √ [1 2 0]T , ê2 = [0 0 1]T , ê3 = √ [−2 1 0]T , 5 5 ⎡ ⎤ 1 0 −2 1 ⎣ 2 0 1⎦ M̂ = √ 5 0 √5 0 c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition ⎡ M̂T A M =  1 5  1 = 5 =  22  1 5  1 2 ⎣ 0 0 −2 1 ⎡ 6 12 ⎣ 0 0 −2 1 ⎡ 30 0 ⎣ 0 15 0 0  ⎤⎡ 2 2 √0 ⎦ ⎣ 2 5 5 0 0 0 ⎤⎡ 1 0 √ ⎦ ⎣ 2 3 5 0 0 ⎤ ⎡ 6 0 ⎦ ⎣ 0 = 0 0 5  ⎤⎡ 1 0 ⎦ ⎣ 2 0 0 3  0 √0 5 ⎤  23  ⎤ −2 1⎦ 0  0 −2 1 ⎦ √0 5 0 ⎤ 0 0 3 0⎦ = Λ 0 1  The eigenvalues of A are given by   5 − λ 10   10 2−λ   8 −2   8  −2  = −(λ3 −18λ2 −81λ+1458) = −(λ−9)(λ+9)(λ−18) = 0 11 − λ   so eigenvalues are λ1 = 18, λ2 = 9, λ3 = −9 The eigenvectors are given by the corresponding solutions of (5 − λi )ei1 + 10ei2 + 8ei3 = 0 10ei1 + (2 − λi )ei2 − 2ei3 = 0 8ei1 − 2ei2 + (11 − λi )ei3 = 0 Taking i = 1, 2, 3 and solving gives the eigenvectors as e1 = [2 1 2]T , e2 = [1 2 − 2]T , e3 = [−2 2 1]T Corresponding normalised eigenvectors are ê1 =  1 1 1 [2 1 2]T , ê2 = [1 2 − 2]T , ê3 = [−2 2 1]T 3 3 3 ⎡  2 1 ⎣ 1 M̂ = 3 2  ⎤ 1 −2 2 2 ⎦, −2 1  ⎡  2 1 1 ⎣ T M̂ = 1 2 3 −2 2  c Pearson Education Limited 2011   ⎤ 2 −2 ⎦ 1  24  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition ⎡ M̂T A M =  =  1 9 1 9 ⎡  2 ⎣ 1 −2 ⎡ 36 ⎣ 9 18  4 2 ⎣ 2 = 1 2 −2 ⎡ 18 0 ⎣ = 0 9 0 0  ⎤⎡ ⎤⎡ 2 5 10 8 2 ⎦ ⎣ ⎦ ⎣ 1 10 2 −2 −2 2 8 −2 11 1 ⎤ ⎤⎡ 2 1 −2 18 36 ⎦ ⎣ 1 2 2⎦ 18 −18 2 −2 1 −18 −9 ⎤ ⎤⎡ 2 1 −2 4 ⎦ ⎣ 1 2 2⎦ −2 2 −2 1 −1 ⎤ 0 0⎦ =Λ −9 1 2 2  ⎡  23  1 1 ⎣ A = −1 2 0 1  1 2 −2  ⎤ −2 2⎦ 1  ⎤ −2 1⎦ −1  Eigenvalues given by  1 − λ  0 =  −1  0  1 2−λ 1   −2  1  = −(λ3 − 2λ2 − λ + 2) = −(λ − 1)(λ − 2)(λ + 1) = 0 −1 − λ   so eigenvalues are λ1 = 2, λ2 = 1, λ3 = −1. The eigenvectors are given by the corresponding solutions of (1 − λi )ei1 + ei2 − 2ei3 = 0 −ei1 + (2 − λi )ei2 + ei3 = 0 0ei1 + ei2 − (1 + λi )ei3 = 0 Taking i = 1, 2, 3 and solving gives the eigenvectors as e1 = [1 3 1]T , e2 = [3 2 1]T , e3 = [1 0 1]T ⎤ ⎡ 2 1 ⎦ ⎣ 0 , Λ= 0 0 1 ⎤ 2 −2 −2 1 ⎣ −3 0 −3 ⎦ =− 6 1 2 −7 ⎡  1 ⎣ M= 3 1 M−1  3 2 1 ⎡  0 1 0  c Pearson Education Limited 2011   ⎤ 0 0 ⎦ −1  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  25  Matrix multiplication then conﬁrms M−1 A M = Λ  and A = M Λ M−1  24 Eigenvalues given by    3 − λ −2 4    −2 −2 − λ 6  = −λ3 + 63λ − 162 = −(λ + 9)(λ − 6)(λ − 3) = 0   4 6 −1 − λ  so the eigenvalues are λ1 = −9, λ2 = 6, λ3 = 3.  The eigenvectors are the  corresponding solutions of (3 − λi )ei1 − 2ei2 + 44ei3 = 0 −2ei1 − (2 + λi )ei2 + 6ei3 = 0 4ei1 + 6ei2 − (1 + λi )ei3 = 0 Taking i = 1, 2, 3 and solving gives the eigenvectors as e1 = [1 2 − 2]T , e2 = [2 1 2]T , e3 = [−2 2 1]T Since eT1 e2 = eT1 e3 = eT2 e3 = 0 the eigenvectors are orthogonal ⎡  1 1 ⎣ 2 L = [ê1 ê2 ê3 ] = 3 −2 ⎡  25  L̂ A L =  1 9  =  1 9  =  1 9  1 ⎣ 2 −2 ⎡ −9 ⎣ 12 −6 ⎡ −81 ⎣ 0 0  2 1 2  ⎤ −2 2⎦ 1  ⎤⎡ ⎤⎡ 1 2 3 −2 4 −2 6 ⎦⎣ 2 1 2 ⎦ ⎣ −2 −2 −2 2 4 6 −1 1 ⎤ ⎤⎡ 1 2 −2 −18 18 2⎦ 6 12 ⎦ ⎣ 2 1 −2 2 1 6 3 ⎤ ⎤ ⎡ −9 0 0 0 0 54 0 ⎦ = ⎣ 0 6 0 ⎦ = Λ 0 0 3 0 27  2 1 2  Since the matrix A is symmetric the eigenvectors e1 = [1 2 0]T , e2 = [−2 1 0]T , e3 = [e31 e32 e33 ]T c Pearson Education Limited 2011   ⎤ −2 2⎦ 1  26  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  are orthogonal. Hence, eT1 e3 = e31 + 2e32 = 0 and eT2 e3 = −2e31 + e32 = 0 Thus, e31 = e32 = 0 and e33 arbitrary ⎡ 6 Using A = M̂ Λ M̂T where Λ = ⎣ 0 0 ⎡ A= ⎣ ⎡  − √25  √1 5 √2 5  2 ⎣ = 2 0  √1 5  0  2 5 0  0 ⎤ 0 0⎦ 3  so a possible eigenvector is e3 = [0 0 1]T . ⎤ 0 0 1 0 ⎦ gives 0 3  ⎤⎡ 0 6 ⎦ ⎣ 0 0 0 1  0 1 0  ⎤⎡ 1 √ 0 5 0 ⎦ ⎣ − √2 5 3 0  √2 5 √1 5  0  ⎤ 0 0⎦ 1  ⎤ ⎤ ⎡ 0 0 0 −4 −7 −5 3 3 ⎦ ∼ ⎣ 0 −1 0 ⎦ is of rank 2 26 A − I = ⎣ 2 1 0 0 1 2 1 Nullity (A − I) = 3 − 2 = 1 so there is only one linearly independent vector ⎡  corresponding to the eigenvalue 1. The corresponding eigenvector e1 is given by the solution of (A − I)e1 = 0 or −4e11 − 7e12 − 5e13 = 0 2e11 + 3e12 + 3e13 = 0 e11 + 2e12 + 212 = 0 that is, e1 = [−3 1 1]T . To obtain the generalised eigenvector e∗1 we solve (A − I)e∗1 = e1 or ⎤ ⎤⎡ ∗ ⎤ ⎡ ⎡ e11 −3 −4 −7 −5 ⎣ 2 3 3 ⎦ ⎣ e∗12 ⎦ = ⎣ 1 ⎦ 1 1 2 1 e∗13 giving e∗1 = [−1 1 0]T . To obtain the second generalised eigenvector e∗∗ 1 we solve ∗ (A − I)e∗∗ 1 = e1 or ⎤ ⎤ ⎡ ∗∗ ⎤ ⎡ ⎡ e11 −1 −4 −7 −5 ⎦ = ⎣ 1⎦ ⎣ 2 3 3 ⎦ ⎣ e∗∗ 12 ∗∗ 0 1 2 1 e13  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  27  T giving e∗∗ 1 = [2 − 1 0] .  ⎤ −3 −1 2 ⎣ 1 1 −1 ⎦ M = [e1 e∗1 e∗∗ 1 ]= 1 0 0 ⎤ ⎡ ⎡ 0 0 0 −1 −1 ⎦ ⎣ ⎣ det M = −1 and M = − −1 −2 −1 = 1 1 −1 −1 −2 ⎡  0 2 1  ⎤ 1 1⎦ 2  Matrix multiplication then conﬁrms ⎤ 1 1 0 A M = ⎣0 1 1⎦ 0 0 1 ⎡  M−1  27  Eigenvalues are given by | A − λI |= 0  that is, λ4 − 4λ3 − 12λ2 + 32λ + 64 = (λ + 2)2 (λ − 4)2 = 0 so the eigenvalues are − 2, − 2, 4 and 4 as required. Corresponding to the repeated eigenvalue λ1 , λ2 = −2 ⎡  3 ⎢ 0 (A + 2I) = ⎣ −0.5 −3  0 3 −3 0  0 −3 3 0  ⎤ ⎡ 1 0 −3 0⎥ ⎢0 1 ⎦ ∼ ⎣ 0 0 0.5 0 0 3  0 0 0 0  ⎤ 0 0⎥ ⎦ is of rank 2 0 0  Thus, nullity (A+2I) is 4−2 = 2 so there are two linearly independent eigenvectors corresponding to λ = −2. Corresponding to the repeated eigenvalues λ3 , λ4 = 4 ⎡  −3 0 ⎢ (A − 4I) = ⎣ −0.5 −3  0 −3 −3 0  0 −3 −3 0  ⎤ ⎤ ⎡ 1 0 0 0 −3 0⎥ ⎢0 1 0 0⎥ ⎦ is of rank 3 ⎦ ∼ ⎣ 0 0 0 0 0.5 0 0 0 1 −3  Thus, nullity (A − 4I) is 4 − 3 = 1 so there is only one linearly independent eigenvector corresponding to λ = 4. c Pearson Education Limited 2011   28  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  When λ = λ1 = λ2 = −2 the eigenvalues are given by the solution of (A+2I)e = 0 giving e1 = [0 1 1 0]T , e2 = [1 0 0 1]T as two linearly independent solutions. When λ = λ3 = λ4 = 8 the eigenvectors are given by the solution of (A − 4I)e = 0 giving the unique solution e3 = [0 1 − 1 0]T . The generalised eigenvector e∗3 is obtained by solving (A − 4I)e∗3 = e3 giving e∗3 = (6 − 1  0 − 6]T . The Jordan canonical form is ⎡  ⎤  ⎢ −2 0 ⎢ ⎢ 0 −2 ⎢ J= ⎢ ⎢ ⎢ 0 0 ⎣ 0  0  0 0 4 0  0⎥ ⎥ 0⎥ ⎥ ⎥ ⎥ 1⎥ ⎦ 4  Exercises 1.6.5 28 [ x1  The quadratic form may be written in the form V = xT Ax where x = T x2 x3 ] and ⎤ ⎡ 2 2 1 A = ⎣2 5 2⎦ 1 2 2  The eigenvalues of A are given by    2− λ 2 1   =0  2 5 − λ 2    1 2 2 − λ ⇒ (2 − λ)(λ2 − 7λ + 6) + 4(λ − 1) + (λ − 1) = 0 ⇒ (λ − 1)(λ2 − 8λ + 7) = 0 ⇒ (λ − 1)2 (λ − 7) = 0 giving the eigenvalues as λ1 = 7, λ2 = λ3 = 1 Normalized eigenvector corresponding to λ1 = 7 is ê1 =  √1 6  √2 6  √1 6  T  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  29  and two orthogonal linearly independent eigenvectors corresponding to λ − 1 are √1 2  ê2 =  − √12  0  ê3 = − √13  √1 3  T  − √13  T  Note that ê2 and ê3 are automatically orthogonal to ê1 . The normalized orthogonal modal matrix M̂ and spectral matrix Λ are ⎡ ⎢ M̂ = ⎣  √1 6 √2 6 √1 6  √1 2  − √13  0  √1 3 − √13  − √12  ⎤  ⎡  7 ⎥ ⎣ ⎦,Λ = 0 0  0 1 0  ⎤ 0 0⎦ 1  such that M̂T AM̂ = Λ. Under the orthogonal transformation x = M̂y the quadratic form V reduces to V = yT M̂T AM̂y = yT Λy ⎤⎡ ⎤ ⎡ y1 7 0 0 = [ y1 y 2 y 3 ] ⎣ 0 1 0 ⎦ ⎣ y 2 ⎦ 0 0 1 y3 = 7y21 + y22 + y23  ⎤ 1 −1 2 2 −1 ⎦ and its leading 29(a) The matrix of the quadratic form is A = ⎣ −1 2 −1 7 principal minors are    1 −1   = 1, det A = 2  1,  −1 2 ⎡  Thus, by Sylvester's condition (a) the quadratic form is positive deﬁnite. ⎡  29(b)  1 −1 ⎣ 2 Matrix A = −1 2 −1   1 1,  −1  ⎤ 2 −1 ⎦ and its leading principal minors are 5  −1  = 1, det A = 0 2  Thus, by Sylvester's condition (c) the quadratic form is positive semideﬁnite.  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition ⎤ ⎡ 1 −1 2 2 −1 ⎦ and its leading principal minors are 29(c) Matrix A = ⎣ −1 2 −1 4 30    1 1,  −1   −1  = 1, det A = −1. 2  Thus, none of Sylvester's conditions are satisﬁed and the quadratic form is indeﬁnite.  a −b and its leading 30(a) The matrix of the quadratic form is A = −b c principal minors are a and ac − b2 . By Sylvester's condition (a) in the text the quadratic form is positive deﬁnite if and only if   a > 0 and ac − b2 > 0 that is, a > 0 and ac > b2 ⎤ 2 −1 0 30(b) The matrix of the quadratic form is A = ⎣ −1 a b ⎦ having principal 0 b 3 2 minors 2, 2a − 1 and det A = 6a − 2b − 3. Thus, by Sylvester's condition (a) in ⎡  the text the quadratic form is positive deﬁnite if and only if 2a − 1 > 0 and 6a − 2b2 − 3 > 0 or 2a > 1 and 2b2 < 6a − 3 31  The eigenvalues of the matrix A are given by       2 − λ 3 − λ 3 − λ 1 −1 0     2−λ 1  R1 +R3  1 2−λ 1  0 =  1  −1  −1 1 2− λ 1 2 − λ    1 1 0   1  = (3 − λ)  1 2 − λ  −1 1 2− λ     1 0 0   1  = (3 − λ)(λ2 − 3λ) = (3 − λ)  1 1 − λ  −1 2 2− λ  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  31  so the eigenvalues are 3, 3, 0 indicating that the matrix is positive semideﬁnite. The principal minors of A are  2 2,  1   1  = 3, det A = 0 2  conﬁrming, by Sylvester's condition (a), that the matrix is positive semideﬁnite. ⎤ K 1 1 32 The matrix of the quadratic form is A = ⎣ 1 K −1 ⎦ having principal 1 −1 1 minors   K 1   = K2 − 1 and det A = K2 − K − 3 K,  1 K ⎡  Thus, by Sylvester's condition (a) the quadratic form is positive deﬁnite if and only if K2 − 1 = (K − 1)(K + 1) > 0 and K2 − K − 3 = (K − 2)(K + 1) > 0 i.e. K > 2 or K < −1. If K = 2 then det A = 0 and the quadratic form is positive semideﬁnite.  33  Principal minors of the matrix are  3+ a 3 + a,  1   1  = a2 + 3a − 1, det A = a3 + 3a2 − 6a − 8 a  Thus, by Sylvester's condition (a) the quadratic form is positive deﬁnite if and only if  3 + a > 0, a2 + 3a − 1 > 0 and a3 + 3a2 − 6a − 8 > 0 or (a + 1)(a + 4)(a − 2) > 0 3 + a > 0 ⇒ a > −3 a2 + 3a − 1 > 0 ⇒ a < −3.3 or a > 0.3 (a + 1)(a + 4)(a − 2) > 0 ⇒ a > 2 or − 4 < a < −1  Thus, minimum value of a for which the quadratic form is positive deﬁnite is a = 2.  c Pearson Education Limited 2011   32  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  ⎤ 1 2 −2 λ −3 ⎦ 34 A = ⎣ 2 −2 −3 λ Principal minors are ⎡   2  = λ − 4, det A = λ2 − 8λ + 15 = 0 λ   1 1,  2  Thus, by Sylvester's condition (a) the quadratic form is positive deﬁnite if and only if λ−4>0 ⇒ λ>4 and (λ − 5)(λ − 3) > 0 ⇒ λ < 3 or λ > 5 Thus, it is positive deﬁnite if and only if λ > 5.  Exercises 1.7.1 35  The characteristic equation of A is    5− λ 6  = λ2 − 8λ + 3 = 0   2 3 − λ  2  Now A =  A − 8A + 3I = 2  37 16  5 2 48 21  6 3      5 2   −  6 3 40 16     =  48 24  27 16     +  48 21 3 0  so that A satisﬁes its own characteristic equation.  36  The characteristic equation of A is   1 − λ 2   = λ2 − 2λ − 1 = 0  1 1 − λ  c Pearson Education Limited 2011    so 0 3     =  0 0  0 0    Glyn James, Advanced Modern Engineering Mathematics, 4th Edition By Cayley–Hamilton theorem A2 − 2A − I = 0   36(a)  2 Follows that A = 2A + I = 2  36(b)  6 A = 2A + A = 4  36(c)  14 A = 2A + A = 10  37(a)  The characteristic equation of A is    3  2    4  3  2  8 6     +  20 14  that is,  1 1   +   +  2 1    2 − λ   1    4 2  2    1 0 0 1   =  3 2  4 3  7 5       =  10 7   =  3 2  4 3      17 12  24 17     1  =0 2 − λ  λ2 − 4λ + 3 = 0  Thus, by the Cayley–Hamilton theorem  so that  37(b)  A2 − 4A + 3I = 0 1 I = [4A − A2 ] 3 1 A−1 = [4I − A] 3    1 2 4 0 − = 1 0 4 3  1 2    1 = 3  The characteristic equation of A is  1 − λ 1   3 1 − λ   2 3 that is,   2  1  = 0 1 − λ  λ3 − 3λ2 − 7λ − 11 = 0  c Pearson Education Limited 2011     2 −1 −1 2    33  34  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition ⎡  1 2 ⎣ A = 3 2  1 1 3  ⎤ ⎡ 1 2 ⎦ ⎣ 3 1 2 1  1 1 3  ⎤ ⎡ 8 2 ⎦ ⎣ 8 1 = 13 1  8 7 8  ⎤ 5 8⎦ 8  Using (1.44) 1 (A2 − 3A − 7I) 11 ⎡ ⎤ −2 5 −1 1 ⎣ −1 −3 5 ⎦ = 11 7 −1 −2  A−1 =  ⎤ ⎡ 2 3 2 3 1 38 A2 = ⎣ 3 1 2 ⎦ ⎣ 3 1 1 2 1 2 3 The characteristic equation of A ⎡  ⎤ ⎡ 14 1 2 ⎦ = ⎣ 11 11 3 is  11 14 11  ⎤ 11 11 ⎦ 14  λ2 − 6λ2 − 3λ + 18 = 0 so by the Cayley–Hamilton theorem A3 = 6A2 + 3A − 18I giving A4 = 6(6A2 + 3A − 18I) + 3A2 − 18A = 39A2 − 108I A5 = 39(6A2 + 3A − 18I) + 108A = 234A2 + 9A − 702I A6 = 234(6A2 + 3A − 18I) + 9A2 − 702A = 1413A2 − 4212I A7 = 1413(6A2 + 3A − 18I) + 4212A = 8478A2 + 27A − 25434I Thus, A7 − 3A6 + A4 + 3A3 − 2A2 + 3I = 4294A2 + 36A − 12957I ⎤ ⎡ 47231 47342 47270 = ⎣ 47342 47195 47306 ⎦ 47270 47306 47267  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 39(a)  Eigenvalues A are λ = 1 (repeated). Thus, eAt = α0 I + α1 A with  = α0 + α1 ⇒ α1 = tet , α0 = (1 − t)et = α1  t e At t t so e = (1 − t)e I + te A = tet  et tet  39(b)    Eigenvalues A are λ = 1 and λ = 2. Thus,  et e2t  40  0 et  eAt = α0 I + α1 A with  = α0 + α1 ⇒ α0 = 2et − e2t , α1 = e2t − et = α0 + 2α1  At t 2t 2t t so e = (2e − e )I + (e − e )A =  Eigenvalues of A are λ1 = π, λ2 =  Thus,  et 2t e − et  π π , λ3 = . 2 2  sin A = α0 A + α1 A + α2 A2 with sin π = 0 = α0 + α1 π + α2 π2 π π2 π = 1 = α0 + α1 + α2 2 2 4 π cos = 0 = α1 + πα2 2 4 4 Solving gives α0 = 0, α1 = , α2 = − 2 so that π π ⎤ ⎡ 0 0 0 4 4 sin A = A − 2 A2 = ⎣ 0 1 0 ⎦ π π 0 0 1 sin  41(a) dA = dt    d 2 dt (t + 1) d dt (5 − t)  d dt (2t − 3) d 2 dt (t − t + 3)     =  2t −1  c Pearson Education Limited 2011   2 2t − 1    0 e2t    35  36  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  41(b)   1  2  2 2 (t + 1)dt Adt = 12 (5 − t)dt 1   42 2  A =   t2 + 1 5  2  (2t − 3)dt 21 2 (t − t + 3)dt 1  t−1 0     t2 + 1 5  ⎡ 10    = ⎣  t−1 0  0  7 2  23 6  ⎤ ⎦    t3 − t2 + t − 1 = 5t − 5   3 d 4t + 4t + 5 3t2 − 2t + 1 2 (A ) = 10t 5 dt   3 2 dA 4t + 4t 2t + 1 2A = 20t 0 dt Thus,  t4 + 2t2 + 5t − 4 5t2 + 5  3    d dA (A2 ) = 2A . dt dt  Exercises 1.8.4 43(a)  row rank ⎡  ⎤ ⎡ ⎤ 3 4 row2 − 3row1 1 2 1 2 3 4 ⎣ 0 −2 −2 −2 ⎦ → A = ⎣ 3 4 7 10 ⎦ row3 − 2row1 0 −3 −1 −1 2 1 5 7 ⎤ ⎤ ⎡ ⎡ 1 2 3 4 1 2 4 4 1 − 2 row2 ⎣ row3 + 3row2 ⎣ 0 1 1 1⎦ 0 1 1 1 ⎦ → → 0 0 2 2 0 −3 −1 −1 echelon form, row rank 3  column rank ⎡ ⎤ 0 0 col3 − col2 1 ⎣0 → −2 2 ⎦ 2 0 col4 − col2 2 ⎤ ⎡ 1 0 0 0 col4 − col3 ⎣ 3 −2 0 0 ⎦ → 2 −3 2 0  col2 − 2col1 ⎡ 1 → ⎣ 3 A col3 − 3col1 2 col4 − 4col1  0 −2 −3  echelon form, column rank3 Thus row rank(A) = column rank(A) = 3 c Pearson Education Limited 2011   ⎤ 0 0 0 −2 0 0 ⎦ −3 2 2  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  37  (b) A is of full rank since rank(A)=min( m, n) =min(3,4) = 3 ⎤     4 8 37 9 333 81 4 11 14 ⎣ T ⎦ =9 11 7 = 44(a) AA = 9 13 81 117 8 7 −2 14 −2 The eigenvalues λi of AAT are given by the solutions of the equations      8    333 − λ T  = 0 ⇒ λ2 − 450λ + 32400 = 0 AA − λI =  81 117 − λ      ⎡  ⇒ (λ − 360)(λ − 90) = 0 giving the eigenvalues as λ1 = 360, λ2 = 90. Solving the equations. (AAT − λi I)ui = 0 gives the corresponding eigenvectors as u1 = [ 3  T  T  1 ] , u2 = [ 1 −2 ]  with the corresponding normalized eigenvectors being √1 10  √3 10  û1 =  T  √1 10  , û2 =  − √310  T  leading to the orthogonal matrix  Û = ⎡  4 T ⎣ A A = 11 14  ⎤ 8  4 7 ⎦ 8 −2  √3 10 √1 10  11 7  √1 10 − √310      ⎡  80 14 ⎣ = 100 −2 40     T   80 − μ   Solving A A − μI =  100  40  100 170 − μ 140  ⎤ 40 140 ⎦ 200  40  140  = 0 200 − μ  100 170 140  gives the eigenvalues μ1 = 360, μ2 = 90, μ3 = 0 with corresponding normalized eigenvectors v̂1 = [ 13  2 3  2 T 3 ]  , v̂2 = [ − 32  − 13  2 T 3 ]  , v̂3 = [ 23  c Pearson Education Limited 2011   − 23  1 T 3 ]  38  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  leading to the orthogonal matrix ⎡1 V̂ = ⎣  The singular values of A are σ1 =  √  − 23 − 13  3 2 3 2 3  2 3 − 23 1 3  2 3  ⎤ ⎦  √ √ √ 360 = 6 10 and σ2 = 90 = 3 10 giving   √ 6 10 Σ= 0  √0 3 10    0 0  Thus, the SVD form of A is  A = ÛΣV̂T =  √3 10 √1 10  √1 10 − √310   √ 6 10 0   4 (Direct multiplication conﬁrms A = 8  √0 3 10  0 0    ⎡  2 3 − 13 − 23  1 3 ⎣ −2 3 2 3  2 3 2 3 1 3  ⎤ ⎦   14 ) −2  11 7  (b) Using (1.55) the pseudo inverse of A is ⎡  √1 6 10  A† = V̂Σ∗ Û, Σ∗ = ⎣ 0 0  3  †  AA =   1 180  4 8  0  ⎤  ⎡1  ⎦⇒⎣  − 23 − 13  3 2 3 2 3  2 3 − 23 1 3  ⎤⎡  √1 6 10  ⎦⎣ 0 2 0 0 3 ⎤ ⎡  −1 13 √ √1 † 1 ⎣ 10 10 4 8 ⎦ ⇒ A = 180 √1 √3 − 10 10 10 −10 ⎤ ⎡  −1 13   14 ⎣ 180 0 1 4 8 ⎦ = 180 =I −2 0 180 10 −10  11 7  √2 3 10  0 √1 3 10  ⎤ ⎦  0  (c) Rank( A) = 2 so A is of full rank. Since number of rows is less than the number of columns A† may be determined using (1.58b) as ⎡  A† = AT (AAT )−1  4 ⎣ = 11 14  ⎤ 8  333 7 ⎦ 81 −2  81 117  −1  which conﬁrms with the value determined in (b).  c Pearson Education Limited 2011   ⎤ −1 13 1 ⎣ 4 8 ⎦ = 180 10 −10 ⎡  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  39  ⎤ ⎡ ⎤ ⎤ ⎡ 1 1 1 1 1 row3 + row2 row2 − 3row1 0⎥ ⎢ 0 −3 ⎥ ⎢ 0 −3 ⎥ ⎥ row4 + 23 row2 ⎢ ⎥ ⎥ row3 + 2row1 ⎢ 1⎥ ⎢0 3 ⎥ ⎢0 0 ⎥ → → ⎦ ⎣ ⎦ ⎦ ⎣ 0 2 0 0 2 row5 + row1 row5 + row2 0 3 0 0 2 echelon form so row rank = 2 = column rank  ⎡  1 ⎢ 3 ⎢ 45 A = ⎢ −2 ⎣ 0 −1  Thus, rank A = 2 =min(5,2) and so A is of full rank. Since A is of full rank and number of rows is greater than number of columns we can determine the pseudo inverse using result (1.58a)  †  T  A = (A A)  −1  −1  1 3 −2 0 15 −3 A = 1 0 1 2 −3 10   1 3 −2 0 10 3 1 = 141 1 0 1 2 3 15   13 30 −17 6 −4 1 = 141 18 9 9 30 27   T    †  A A=  1 141  13 18  30 9  −17 9  6 30  1 T ⎣ (b) AA = −2 2 The eigenvalues λi  ⎤ −1  1 2 2 ⎦ −1 2 −2 are given by   2 − λ −4   −4 2−λ   4 −8    −1 2     ⎡  1 ⎢ 3 −4 ⎢ ⎢ −2 27 ⎣ 0 −1  ⎤ 1 0⎥ ⎥ 1⎥ = ⎦ 2 2   1 141  141 0   0 =I 141  ⎤ −1 0 ⎦ 0  ⎡ ⎤ 1 −1 row2 + 2row1 1 ⎣0 → 46(a) A = ⎣ −2 2 ⎦ 2 −2 row3 − 2row1 0 Thus, rank A = 1and is not of full rank ⎡  ⎡  −1 2  ⎡  2 2 ⎣ = −4 −2 4  −4 8 −8  ⎤ 4 −8 ⎦ 8   4  −8  = 0 ⇒ λ2 (−λ + 18) = 0 8− λ  giving the eigenvalues as λ1 = 18, λ2 = 0, λ3 = 0. The corresponding eigenvectors and normalized eigenvectors are c Pearson Education Limited 2011   40  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition T  u1 = [ 1 −2  2 ] ⇒ û1 = [ 13  u2 = [ 0  1  T 1 ] ⇒ û2 = 0  u3 = [ 2  1  0 ] ⇒ û3 =  T  − 32 √1 2  √2 5  √1 5  2 T 3 ] T √1 2 T  0  leading to the orthogonal matrix ⎡ Û =   AT A =  1 −1  1 3 ⎢ −2 ⎣ 3 2 3    √2 5 √1 5  0 √1 2 √1 2  ⎡  1 −2 2 ⎣ −2 2 −2 2  ⎤ ⎥ ⎦  0 ⎤   −1 1 −1 2 ⎦=9 −1 1 −2  having eigenvalues μ1 = 18 and μ2 = 0 and corresponding eigenvectors T  √1 2  v1 = [ 1 −1 ] ⇒ v̂1 = T  √1 2  √1 2  v2 = [ 1 1 ] ⇒ v̂2 =  − √12  T  T  leading to the orthogonal matrix  V̂ =  √1 2 √1 2  √1 2 − √12    √ √ single ⎤ (equal to its rank) singular value σ1 = 18 = 3 2 so that 0 0 ⎦ and the SVD form of A is 0 ⎡ 1 ⎤⎡ √ ⎤ √2  0 3 3 2 0 5 √1 √1 − ⎢ ⎥ 1 1 2 2 A = ÛΣV̂T = ⎣ − 3 √2 √5 ⎦ ⎣ 0 0 ⎦ 12 √ √1 2 2 2 0 0 √1 0 3 2 ⎤ ⎡ 1 −1 Direct multiplication conﬁrms that A = ⎣ −2 2 ⎦ 2 −2 A has √ ⎡ the 3 2 Σ=⎣ 0 0  (c) Pseudo inverse is given by  †  ∗  T  A = V̂Σ Û =  √1 2 − √12  √1 2 √1 2    1 √ 3 2  0  0 0    ⎡  1 3  0 ⎣ 0 0 2 √  5  − 23 √1 2 √1 5  c Pearson Education Limited 2011   2 3 √1 2  0  ⎤ ⎦=   1 18  1 −2 −1 2  2 −2    Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  41  Direct multiplication conﬁrms AA† A = AandA† AA† = A† (d) Equations may be written as ⎡  1 ⎣ −2 2  ⎡ ⎤ ⎤ 1 −1   x ⎣ ⎦ = 2 ⎦ ≡ Ax = b 2 y 3 −2    x † The least squares solution is x = A b ⇒ = y  1  6 giving x = 16 and y = − 16 − 61   1 18  1 −1  −2 2  ⎡ ⎤ 1 2 ⎣ ⎦ 2 = −2 3   (e) Minimize L = (x − y − 1)2 + (−2x + 2y − 2)2 + (2x − 2y − 3)2 ∂L = 0 ⇒ 2(x − y − 1) − 4(−2x + 2y − 2) + 4(2x − 2y − 3) = 18x − 18y − 6 = 0 ∂x ⇒ 3x − 3y − 1 = 0 ∂L = 0 ⇒ −2(x − y − 1) + 4(−2x + 2y − 2) − 4(2x − 2y − 3) = −18x + 18y + 6 = 0 ∂y ⇒ −3x + 3y + 1 = 0 Solving the two simultaneous equations gives the least squares solution x =  1 6,  y = − 16 conﬁrming the answer in (d)  47(a) Equations may be written as ⎡  3 ⎣1 1  ⎡ ⎤ ⎤ 1 −1   x ⎣ ⎦ = 2 ⎦ ≡ Ax = b 3 y 3 1  Using the pseudo inverse obtained in Example 1.39, the least squares solution is   x † x=A b⇒ = y giving x = y =   1 60  17 −7  4 16  ⎡ ⎤  1 2 5 ⎣ ⎦ 2 = 32 5 3 3  2 3  c Pearson Education Limited 2011   42  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  (b) Minimize L = (3x − y − 1)2 + (x + 3y − 2)2 + (x + y − 3)2 ∂L = 0 ⇒ 6(3x − y − 1) + 2(x + 3y − 2) + 2(x + y − 3) = 0 ∂x ⇒ 11x + y − 8 = 0 ∂L = 0 ⇒ −2(3x − y − 1) + 6(x + 3y − 2) + 2(x + y − 3) = 0 ∂y ⇒ x + 11y − 8 = 0 Solving the two simultaneous equations gives the least squares solution x = y =  2 3  conﬁrming the answer in (a) 48(a) ⎡  1 ⎢ 0 A=⎣ −1 2  0 1 1 −1  ⎡ ⎤ 1 −2 row3 + row1 −1 ⎥ ⎢0 → ⎣ ⎦ 0 1 row4 − 2row1 0 2  0 1 1 −1  ⎡ ⎤ 1 −2 row3 − row2 −1 ⎥ ⎢0 → ⎣ ⎦ 0 −1 row4 + row2 0 6  0 1 0 0  ⎤ −2 −1 ⎥ ⎦ 0 5  Thus, A is of rank 3 and is of full rank as 3=min(4,3) (b) Since A is of full rank ⎤−1 ⎡ 6 −3 1 1 † T −1 T ⎦ ⎣ ⎣ A = (A A) A = −3 3 −2 0 1 −2 10 −2 ⎤ ⎤⎡ ⎡ 1 0 −1 2 26 28 3 1 ⎣ 1 1 −1 ⎦ = ⇒ A† = 75 28 59 9 ⎦ ⎣ 0 −2 −1 1 2 3 9 9 ⎡  ⎤ 0 −1 2 1 1 −1 ⎦ −1 1 2 ⎤ ⎡ 4 5 1 6 1 ⎣ 2 10 8 3 ⎦ 15 −3 0 3 3  (c) Direct multiplication conﬁrms that A† satisﬁes the conditions AAT and AT A are symmetric, AA† A = A and A† AA† = A†  ⎡  ⎤ 1 2 ⎦ is of full rank 2 so pseudo inverse is 1   0.6364 −0.3636 0.0909 † T −1 T A = (A A) A = −0.3636 0.6364 0.0909  2 49(a) A = ⎣ 1 1  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  43  Equations (i) are consistent with unique solution ⎡ ⎤   3 x †⎣ ⎦ =A 3 ⇒x=y=1 y 2 Equations (ii) are inconsistent with least squares solution ⎡ ⎤   3 x †⎣ ⎦ = A 3 ⇒ x = 1.0909, y = 1.0909 y 3 ⎤  2 1 0.5072 † ⎦ ⎣ with pseudo inverse A = (b) A = 1 2 −0.4928 10 10 Equations (i) are consistent with unique solution ⎡  −0.4928 0.5072  0.0478 0.0478    ⎡ ⎤   3 x = A† ⎣ 3 ⎦ ⇒ x = y = 1 y 20 Equations (ii) are inconsistent and have least squares solution ⎡ ⎤   3 x = A† ⎣ 3 ⎦ ⇒ x = y = 1.4785 y 30 ⎤  2 1 0.5001 † 2 ⎦ with pseudo inverse A = (c) A = ⎣ 1 −o.4999 100 100 Equations (i) are consistent with unique solution ⎡  ⎤ ⎡   3 x = A† ⎣ 3 ⎦ ⇒ x = y = 1 y 200 Equations (ii) are inconsistent with least squares solution ⎤ ⎡   3 x = A ⎣ 3 ⎦ ⇒ x = y = 1.4998 y 300 c Pearson Education Limited 2011   −0.4999 0.5001  0.0050 0.0050    44  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Since the sets of equations (i) are consistent weighting the last equation has no eﬀect on the least squares solution which is unique. However, since the sets of equations (ii) are inconsistent the solution given is not unique but is the best in the least squares sense. Clearly as the weighting of the third equation increases from (a) to (b) to (c) the better is the matching to the third equation, and the last case (c) does not bother too much with the ﬁrst two equations.  50 Data may be represented in the matrix form ⎡ ⎤ ⎤ 1 1   1⎥ ⎢1⎥ ⎢ ⎥ ⎥ m = ⎢2⎥ 1⎥ c ⎣ ⎦ ⎦ 2 1 3 1  ⎡  0 ⎢1 ⎢ ⎢2 ⎣ 3 4 Az = Y MATLAB gives the pseudo inverse   −0.2 A = 0.8 †  −0.1 0.4  0 0.2  0.1 0  0.2 −0.2  and, the least squares solution      m 0.5 † =A y= c 0.8  leads to the linear model y = 0.5x + 0.8  Exercises 1.9.3 51(a)  Taking x1 = y dy dt d2 y ẋ2 = x3 = 2 dt d3 y ẋ3 = 3 = u(t) − 4x1 − 5x2 − 4x3 dt  ẋ1 = x2 =  c Pearson Education Limited 2011     Glyn James, Advanced Modern Engineering Mathematics, 4th Edition Thus, state space form is ⎡ ⎤ ⎡ ẋ1 0 1 0 ẋ = ⎣ ẋ2 ⎦ = ⎣ 0 −4 −5 ẋ3  45  ⎤ ⎡ ⎤ ⎡ ⎤ x1 0 0 1 ⎦ ⎣ x2 ⎦ + ⎣ 0 ⎦ u(t) −4 1 x3  y = x1 = [1 0 0] [x1 x2 x3 ]T 51(b) x1 = y dy dt d2 y x3 = ẋ2 = 2 dt d3 y x4 = ẋ3 = 3 dt 4 d y ẋ4 = 4 = −4x2 − 2x3 + 5u(t) dt  x2 = ẋ1 =  Thus, state space form is ⎡ ⎤ ⎡ ẋ1 0 ⎢ ẋ ⎥ ⎢0 ẋ = ⎣ 2 ⎦ = ⎣ 0 ẋ3 0 ẋ4  ⎤ ⎡ ⎤ ⎡ ⎤ x1 1 0 0 0 0 1 0 ⎥ ⎢ x2 ⎥ ⎢0⎥ ⎦ ⎣ ⎦ + ⎣ ⎦ u(t) 0 0 1 0 x3 −4 −2 0 5 x4  y = x1 = [1 0 0 0] [x1 x2 x3 x4 ]T  52(a)  Taking A to be the companion matrix of the LHS ⎤ ⎡ 0 1 0 0 1 ⎦ A=⎣ 0 −7 −5 −6  and taking b = [ 0  0  T  1]  and then using (1.67) in the text c = [ 5 3  Then from (1.84) the state-space form of the dynamic model is ẋ = Ax + bu, y =cx (b) Taking A to be the companion matrix of ⎡ 0 1 ⎣ A= 0 0 0 −3  the LHS ⎤ 0 1 ⎦ −4  c Pearson Education Limited 2011   1 ].  46  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  and taking b = [ 0  0  T  1 ] then using (1.67) in the text c = [ 2  3  1 ]. Then  from (1.84) the state-space form of the dynamic model is ẋ = Ax + bu, y =cx  53  Applying Kirchhoﬀ's second law to the individual loops gives di1 1 , v̇c = (i1 + i2 ) dt C di2 + R2 i2 e = R1 (i1 + i2 ) + vc + L2 dt  e = R1 (i1 + i2 ) + vc + L1  so that, di1 R1 vc e R1 i2 − + = − i1 − dt L1 L1 L1 L1 di2 R1 (R1 + R2 ) vc e = − i1 − i2 − + dt L2 L2 L2 L2 dvc 1 = (i1 + i2 ) dt C Taking x1 = i1 , x2 = i2 , x3 = vc , u = e(t) gives the state equation as ⎡  ⎤ ⎡ R1 − L1 ẋ1 ⎣ ẋ2 ⎦ = ⎣ − R1 L2 1 ẋ3 C  2 −R L1 2) − (R1L+R 2  1 C  ⎤ ⎡ 1 ⎤ x1 L1 1 ⎦ ⎣ x ⎦ + ⎣ 1 ⎦ u(t) − L2 2 L2 x 0 3 0 − L11  ⎤ ⎡  (1)  The output y = voltage drop across R2 = R2 i2 = R2 x2 so that y = [0 R2 0] [x1 x2 x3 ]T  (2)  Equations (1) and (2) are then in the required form ẋ = A x + bu , y = cT x  54  The equations of motion, using Newton's second law, may be written down  for the body mass and axle/wheel mass from which a state-space model can be deduced. Alternatively a block diagram for the system, which is more informative for modelling purposes, may be drawn up as follows c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  47  where s denotes the Laplace 's' and upper case variables X, Y, Y1 denote the corresponding Laplace transforms of the corresponding lower case time domain variables x(t), y(t), y1 (t); y1 (t) is the vertical displacement of the axle/wheel mass. Using basic block diagram rules this block diagram may be reduced to the input/output transfer function model  X −→  (M1  s2  K1 (K + Bs) + K1 )(Ms2 + Bs + K) + Ms2 (K + Bs)  Y −→  or the time domain diﬀerential equation model d4 y d3 y d2 y M1 M 4 + B(M1 + M) 3 + (K1 M + KM1 + KM) 2 dt dt dt dy dx + K1 B + K1 Ky = K1 K2 x + K1 B dt dt A possible state space model is ⎡  ż1  ⎤  ⎡  −B(M1 + M)  ⎢ ⎢ ⎥ ⎢ −(K1 M +KM1 +KM ) ⎢ ⎥ ż ⎢ ⎢ 2⎥ M M1 ⎢ ⎢ ⎥ ⎢ ⎥ = ⎢ ⎢ ⎢ ⎥ −K1 B ⎢ ⎢ ż3 ⎥ M1 M ⎣ ⎣ ⎦ ż4  −K1 K M1 M  1  0 0  ⎤ ⎡  ⎥ ⎥ 0 1 0⎥ ⎥ ⎥ ⎥ 0 0 1⎥ ⎦ 0 0 0  z1  ⎤  ⎡  ⎢ ⎥ ⎢ ⎢ ⎥ ⎢ z ⎢ 2⎥ ⎢ ⎢ ⎥ ⎢ ⎢ ⎥ + ⎢ ⎢ ⎥ ⎢ ⎢ z3 ⎥ ⎢ ⎣ ⎦ ⎣ z4  0 0 K1 B M1 M  ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ x(t) ⎥ ⎥ ⎦  K1 K 2 M M1  y = [1 0 0 0]z(t), z = [z1 z2 z3 z4 ]T . Clearly alternative forms may be written down, such as, for example, the companion form of equation (1.66) in the text. Disadvantage is that its output y is not one of the state variables.  c Pearson Education Limited 2011   48  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  55  Applying Kirchhoﬀ's second law to the ﬁrst loop gives x1 + R3 (i − i1 ) + R1 i = u that is, (R1 + R3 )i − R3 i1 + x1 = u  Applying it to the outer loop gives x2 + (R4 + R2 )i1 + R1 i = u Taking α = R1 R3 + (R1 + R3 )(R4 + R2 ) then gives αi = (R2 + R3 + R4 )u − (R2 + R4 )x1 − R3 x2 and αi1 = R3 u + R1 x1 − (R1 + R3 )x2 Thus, α(i − i1 ) = (R4 + R2 )u − (R1 + R2 + R4 )x1 + R1 x2 1 (i − i1 ) C1 1 [−(R1 + R2 + R4 )x1 + R1 x2 + (R4 + R2 )u](1) = αC1  Voltage drop across C1 : ẋ1 =  1 i1 C2 1 = [R1 x1 − (R1 + R3 )x2 + R3 u] αC2  Voltage drop across C2 : ẋ2 =  (R1 + R3 ) R1 R3 x− x2 + u α α α (R4 + R2 ) R3 R3 R1 y2 = R2 (i − i1 ) = − (R1 + R2 + R4 )x1 + x2 + R3 u α α α  y1 = i1 =  (2)  (3) (4)  Equations (1)–(4) give the required state space model. Substituting the given values for R1 , R2 , R3 , R4 , C1 and C2 gives the state matrix A as  ⎡ A= ⎣  −9 35.10−3  1 35.10−3  1 35.10−3  −4 35.10−3  ⎤ 3 ⎦ = 10 35    −9 1  c Pearson Education Limited 2011   1 −4    Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 103 then eigenvalues are solutions of Let β = 35     −9β − λ β  = λ2 + 13βλ + 35β2 = 0   β −4β − λ  giving −13 ± λ= 2  √ 29  β  −2.6 × 102 or − 1.1 × 102  Exercises 1.10.4  1 0 56 Φ (t) = e where A = 1 1 Eigenvalues of A are λ = 1, λ = 1 so   At  eAt = α0 (t)I + α1 (t)A where α0 , α1 satisfy eλt = α0 + α1 λ,  λ=1  teλt = α1 giving α1 = tet , α0 = et − tet Thus,   Φ (t) = e  At   56(a)  Φ (0) =  et − tet = 0  1 0  0 1  0 t e − tet     +  tet tet  0 tet     =  et tet  0 et     =I  56(b)   t  e1 0 0 et2 e−t1 Φ(t1 ) = Φ(t2 − t1 )Φ (t2 − t1 )et2 e−t1 et2 e−t1 t1 et1 et1   t   e2 0 0 et2 = = Φ(t2 ) = (t2 − t1 )et2 + t1 et2 et2 t2 et2 et2   c Pearson Education Limited 2011   49  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  t   −t  1 e e 0 0 −1 = = Φ (−t) 56(c) Φ = 2t −tet et −te−t e−t e 50  dy d2 y , ẋ2 = 2 = −x1 − 2x2 so in vector–matrix 57 Take x1 = y, x2 = ẋ1 = dt dt form the diﬀerential equation is   0 1 x, y = [1 0]A ẋ = −1 −2   0 Taking A = −1  1 −2   its eigenvalues are λ = −1, λ = −1  eAt = α0 I + α1 A where α0 , α1 satisfy eλt = α0 + α1 λ, λ = −1 teλt = α1 giving α0 = e−t + te−t , α1 = te−t . Thus,  −t e + te−t At e = −te−t  te−t −t e − te−t    Thus, solution of diﬀerential equation is x(t) = eAt x(0), x(0) = [1 1]T   −t e + 2te−t = e−t − 2te−t giving y(t) = x1 (t) = e−t + 2te−t The diﬀerential equation may be solved directly using the techniques of Chapter 10 of the companion text Modern Engineering Mathematics or using Laplace transforms. Both approaches conﬁrm the solution y = (1 + 2t)e−2t  58  Taking A =  1 0 1 1   then from Exercise 56   e  At  et = tet  0 et    c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition and the required solution is   x(t) = e  At  et x(0) = tet  0 et       et 1 = 1 (1 + t)et   0 1 its eigenvalues are λ1 = −3, λ2 = −2. 59 Taking A = −6 −5 Thus, eAt = α0 I + α1 A where α0 , α1 satisfy   e−3t = α0 − 3α1 , e−2t = α0 − 2α1 α0 = 3e−2t − 2e−3t , α1 = e−2t − e−3t   so e  At  3e−2t − 2e−3t = 6e−3t − te−2t  e−2t − e−3t 3e−3t − 2e−2t    Thus, the ﬁrst term in (6.73) becomes  e  At  x(0) = e  At  [1 − 1] = T  2e−2t − e−3t 3e−3t − 4e−2t    and the second term is   e 0   6e−2(t−τ ) − 6e−3(t−τ ) dτ bu(τ)dτ = 2 18e−3(t−τ ) − 12e−2(t−τ ) 0  −2(t−τ ) t 3e − 2e−3(t−τ ) =2 6e−3(t−τ ) − 6e−2(t−τ ) 0   1 − 3e−2t + 2e−3t =2 6e−2t − 6e−3t   t A(t−τ )  t    Thus, required solution is   2e−2t − e−3t + 2 − 6e−3t + 4e−3t x(t) = 3e−3t − 4e−2t + 12e−2t − 12e−3t   2 − 4e−2t + 3e−3t = 8e−2t − 9e−3t    that is, x1 = 2 − 4e−2t + 3e−3t , x2 = 8e−2t − 9e−3t  c Pearson Education Limited 2011   51  52  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  60  In state space form,     2 0 1 u(t), u(t) = e−t , x(0) = [0 1]T x + ẋ = 0 −2 −3   0 1 its eigenvalues are λ1 = −2, λ2 = −1 so Taking A = −2 −3 eAt = α0 I + α1 A where α0 , α1 satisfy e−2t = α0 − 2α1 , e−t = α0 − α1 ⇒ α0 = 2e−t − e−2t , α1 = e−t − e−2t Thus,    2e−t − e−2t e = −2e−t + 2e−2t   −t e − e−2t At and e x(0) = −e−t + 2e−2t At      t (t−τ )  A 0  e−t − e−2t −e−t + 2e−2t      4e−(t−τ ) − 2e−2(t−τ ) bu(τ)dτ = −4e−(t−τ ) + 4e−2(t−τ ) 0   t  4e−t − 2e−2t eτ dτ = −4e−t + 4e−2t eτ 0  t 4τe−t − 2e−2t eτ = −4τe−t + 4e−2t eτ 0   4te−t − 2e−t + 2e−2t = −4te−t + 4e−t − 4e−2t t    e−τ dτ  We therefore have the solution   t  At  eA(t−τ ) bu(τ)dτ x(t) = e x(0) + 0   −t 4te + e−2t − e−t = −4te−t + 3e−t − 2e−2t that is, x1 = 4te−t + e−2t − e−t , x2 = −4te−t + 3e−t − 2e−2t   61  3 Taking A = 2  4 1   its eigenvalues are λ1 = 5, λ2 = −1.  eAt = α0 I + α1 A where α0 , α1 satisfy e5t = α0 + 5α1 , e−t = α0 − α1 ⇒ α0 =  1 5t 5 −t 1 1 e + e , α1 = e5t + e−t 6 6 6 6  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition Thus, transition matrix is 1 e  At  =  −t + 23 e5t 3e 1 5t 1 −t 3e − 3e  2 5t 3e 1 5t 3e  − 23 e−t + 23 e−t     5t −t − e 2e and eAt x(0) = eAt [1 2]T = e5t + e−t      t  t 4 0 1 A(t−τ ) A(t−τ ) dτ e Bu(τ)dτ = e 3 1 1 0 0    t 3 t−τ = dτ A 7 0  t  20 5(t−τ ) 11 −(t−τ )  − 3e 3 e = 10 5(t−τ ) −(t−τ ) dτ + 11 0 3 e 3 e  4 5(t−τ ) 11 −(t−τ ) t − 3e −3e = 2 5(t−τ ) − e + 11 e−(t−τ ) 0  3 11 −t 3 4 5t  −5 + 3 e + 3 e = −t 3 − 11 + 23 e5t 3 e   Thus, solution is    t  At  eA(t−τ ) Bu(t)dτ x(t) = e x(0) + 0   8 −t 5t −5 + 3 e + 10 e 3 = 3 − 83 e−t + 53 e5t  Exercises 1.10.7  62  Eigenvalues of matrix A =  − 23 1  3 4 − 52   are given by  | A − λI |= λ2 + 4λ + 3 = (λ + 3)(λ + 1) = 0 that is, λ1 = −1, λ2 = −3 having corresponding eigenvectors e1 = [3 2]T, e2 = [1 − 2]T. Denoting the reciprocal basis vectors by r1 = [r11 r12 ]T , r2 = [r21 r22 ]T c Pearson Education Limited 2011   53  54  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  and using the relationships rTi ej = δij (i, j = 1, 2) we have 3r11 + 2r12 = 1 r11 − 2r12 = 0 3r21 + 2r22 = 0 r21 − 2r22 = 1     r1 = [ 14  1 T 8]  r2 = [ 14 − 38 ]T  Thus, rT1 x(0) =  1 1 1 3 + = 1, rT2 x(0) = − = −1 2 2 2 2  so the spectral form of solution is x(t) = e−t e1 − e−3t e2 The trajectory is readily drawn showing that it approaches the origin along the eigenvector e1 since e−3t → 0 faster than e−t . See Figure 1.9 in the text.  −2 2 eigenvalues are λ1 = −6, λ2 = −1 having 63 Taking A = 2 −5 corresponding eigenvectors e1 = [1 − 2]T , e2 = [2 1]T .   Denoting the reciprocal basis vectors by r1 = [r11 r12 ]T, r2 = [r21 r22 ]T and using the relationships rTi ej = δij (i, j = 1, 2) we have r11 − 2r12 = 1 2r11 + r12 = 0 r21 − 2r22 = 0 2r21 + r22 = 1 Thus,     ⇒ r11 = 15 , r12 = − 25 ⇒ r1 = 15 [1 − 2]T ⇒ r21 = 25 , r22 = − 15 ⇒ r2 = 15 [2 1]T   4 1 2 =− = [1 − 2] 3 5 5   7 1 2 = rT2 x(0) = [2 1] 3 5 5  rT1 x(0)  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  55  then response is  2  rTi x(0)eλit ei  x(t) = i=1  4 = − e−6t 5         7 −t 2 1 −4e−6t + 14e−t 1 + e = −2 1 8e−6t + 7e−t 5 5  Again, following Figure 1.9 in the text, the trajectory is readily drawn and showing that it approaches the origin along the eigenvector e2 since e−6t → 0 faster than e−t .   0 −4 eigenvalues are λ1 = −2 + j2, λ2 = −2 − j2 having 64 Taking A = 2 −4 corresponding eigenvectors e1 = [2 1 − j]T , e2 = [2 1 + j]T .   Let r1 = r1 + jr1 be reciprocal base vector to e1 then rT1 e1 = 1 = [r + jr1 ]T [e1 + je1 ]T where e1 = e1 + je1 rT1 e2 = 0 = [r1 + jr1 ]T [e1 − je1 ]T since e2 = conjugate e1  Thus, [(r1 )T e1 − (r1 )T e1 ] + j[(r1 )T e1 + (r1 )T e1 ] = 1 and [(r1 )T e1 − (r1 )T e1 ] + j[(r1 )T e11 − (r1 )T e1 ] = 0  c Pearson Education Limited 2011   56  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  giving (r1 )T e1 =  1 1 , (r1 )T e1 = , (r1 )T e1 = (r1 )T e1 = 0 2 2  Now e1 = [2 1]T , e1 = [0 − 1]T Let r1 = [a b]T and r1 = [c d]T then from above 1 1 2a + b = , −b = 0 and −d = − , 2c + d = 0 2 2 1 1 1 giving a = , b = 0, c = − , d = so that 4 4 2 r1 = r1 + jr1 =  1 [1 − j 2j]T 4  Since r2 is the complex conjugate of r1 r2 =  1 [1 + j − 2j]T 4  so the solution is given by x(t) = rT1 x(0)eλ1 t e1 + rT2 x(0)eλ2 t e2 and since rT1 x(0) =  x(t) = e  −2t  −2t      =e  =e  −2t  1 1 (1 + j), rT2 x(0) = (1 − j) 2 2  1 (1 + j)e2jt 2    2 1−j    1 + (1 − j)e−2jt 2    2 1+j        0 2 − (cos 2t + sin 2t) (cos 2t − sin 2t) −1 1   (cos 2t −  sin 2t)e1  − (cos 2t +  sin 2t)e1    where e1 = e1 + je1  To plot the trajectory, ﬁrst plot e1 , e1 in the plane and then using these as a frame of reference plot the trajectory. A sketch is as follows c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  65  57  Following section 1.10.6 if the equations are representative of ẋ = A x + bu , y = cT x  then making the substitution x = M ξ , where M is the modal matrix of A, reduces the system to the canonical form ξ̇ξ = Λ ξ + (M−1 b)u , y = (cT M)ξξ where Λ is the spectral matrix of A. Eigenvalues of A are given by  1 − λ   −1   0   1 −2  2−λ 1  = λ3 − 2λ2 − λ + 2 = (λ − 1)(λ + 2)(λ + 1) = 0 1 −1 − λ   so the eigenvalues are λ1 = 2, λ2 = 1, λ3 = −1. The corresponding eigenvectors are readily determined as e1 = [1 3 1]T , e2 = [3 2 1]T , e3 = [1 0 1]T ⎡  1 ⎣ Thus, M = 3 1  3 2 1  ⎤ 1 0⎦ 1  ⎡  2 ⎣ and Λ = 0 0  ⎤ 0 0 1 0 ⎦ 0 −1  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition ⎤ ⎡ 2 −2 −2 1 1 adj M = − ⎣ −3 0 3 ⎦ so required canonical form is M−1 = det M 6 1 2 −7  58  ⎡  ⎤ ⎡ ξ̇1 2 ⎣ ξ̇2 ⎦ = ⎣ 0 0 ξ̇3  0 1 0  ⎤ ⎡ ⎤ ⎡ 1 ⎤ ξ1 0 3 0 ⎦ ⎣ ξ2 ⎦ + ⎣ 0 ⎦ u −1 − 43 ξ3  y = [1 − 4 − 2] [ξ1 ξ2 ξ3 ]T  66  Let r1 = [r11 r12 r13 ]T , r2 = [r21 r22 r23 ]T , r3 = [r31 r32 r33 ]T be the  reciprocal base vectors to e1 = [1 1 0]T , e2 = [0 1 1]T , e3 = [1 2 3]T . ⎫ rT1 e1 = r11 + r12 = 1 ⎬ rT1 e2 = r11 + r13 = 0 ⎭ rT1 e3 = r11 + 2r12 + 3r13 = 0 ⎫ rT2 e1 = r21 + r22 = 0 ⎬ rT2 e2 = r22 + r23 = 1 ⎭ rT2 e3 = r21 + 2r22 + 3r23 = 0 ⎫ rT3 e1 = r31 + r32 = 0 ⎬ rT3 e2 = r32 + r33 = 0 ⎭ rT3 e3 = r31 + 2r32 + 3r33 = 1  ⇒ r1 =  1 [1 1 − 1]T 2  ⇒ r2 =  1 [−3 3 1]T 2  ⇒ r3 =  1 [1 − 1 1]T 2  Then using the fact that x(0) = [1 1 1]T α0 = rT1 x(0) = − 12 , α1 = rT2 x(0) =  67  1 2  , α3 = rT3 x(0) =  1 2  The eigenvectors of A are given by    5− λ 4  = (λ − 6)(λ − 1) = 0   1 2 − λ  so the eigenvalues are λ1 = 6, λ2 = 1. The corresponding eigenvectors are readily determined as e1 = [4 1]T , e2 = [1 − 1]T .   4 1 then substituting x = Mξξ Taking M to be the modal matrix M = 1 −1 into ẋ = Ax(t) reduces it to the canonical form ξ̇ξ = Λ ξ c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition   6 0 . Thus, the decoupled canonical form is where Λ = 0 1   ξ̇1 ξ̇2     =  6 0  0 1     ξ1 ξ2  59   or ξ̇1 = 6ξ1 and ξ̇2 = ξ2  which may be individually solved to give ξ1 = αe6t and ξ1 = βet       1 −1 −1 1 1 −1 = Now ξ (0) = M x(0) = − −3 4 4 5 −1 so ξ1 (0) = 1 = α and ξ2 (0) = −3 = β giving the solution of the uncoupled system as  6t  e ξ= −3et The solution for x(t) as  x=M ξ =  4 1 1 −1     e6t −3et     =  4e6t − 3et e6t + 3et     3 4 its eigenvalues are λ1 = 5, λ2 = −1 having 68 Taking A = 2 1 corresponding e1 = [2 1]T , e2 = [1 − 1]T .   eigenvectors 2 1 be the modal matrix of A, then ẋ = M ξ reduces the Let M = 1 −1 equation to     5 0 0 1 −1 ξ +M ξ̇ξ (t) = u(t) 0 −1 1 1   1 1 1 1 −1 we have, Since M = adj M = det M 3 1 −2    ξ̇ξ (t) =  5 0  0 −1    1 ξ+ 3    1 −2  2 −1    With u(t) = [4 3]T the decoupled equations are 10 3 11 ξ̇2 = −ξ2 − 3 ξ̇1 = 5ξ1 +  c Pearson Education Limited 2011   u(t)  60  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  which can be solved independently to give 2 11 , ξ2 = βe−t − 3 3       1 1 1 1 1 −1 so = We have that ξ (0) = M x(0) = −1 2 3 1 −2 ξ1 = αe5t −  2 5 ⇒ α= 3 3 8 11 ⇒ β= −1 = β − 3 3 1=α−  giving   ξ=  5 5t 3e − 8 −t − 3e  2 3 11 3      5 5t 2    −5 + 83 e−t + 10 e −3 e5t 2 1 3 3 = and x = M ξ = 8 −t 1 −1 − 11 3 − 83 e−t + 53 e5t 3e 3 which conﬁrms Exercises 57 and 58.   Exercises 1.11.1 (Lyapunov) 69  Take tentative Lyapunov function V(x) = xT Px giving V̇(x) = xT (AT P + PA)x = −xT Qx where AT P + PA = −Q  (i)  Take Q = I so that V̇(x) = −(x21 + x22 ) which is negative deﬁnite. Substituting in (i) gives   −4 3 2 −2    p11 p12    p p12 + 11 p22 p12  p12 p22      −1 −4 2 = 0 3 −2  0 −1    Equating elements gives −8p11 + 6p12 = −1, 4p12 − 4p22 = −1, 2p11 − 6p12 + 3p22 = 0 5 2  5 2 11 3 Principal minors Solving gives p11 = 8 , p12 = 3 , p22 = 12 so that, P = 82 11 of P are:  5 8  3  12  55 > 0 and det P = ( 96 − 49 ) > 0 so P is positive deﬁnite and the system  is asymptotically stable c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 5 2 4 11 2 8 x1 + 3 x1 x2 + 12 x2 4 11 2 2 3 x1 ẋ2 + 6 x2 ẋ2 = −x1 − x2  Note that, in this case, we have V(x) = deﬁnite and V̇(x) =  5 4 4 x1 ẋ1 + 3 ẋ1 x2 +  61  which is positive which is negative  deﬁnite. 70  Take tentative Lyapunov function V(x) = xT Px giving V̇(x) = xT (AT P + PA)x = −xT Qx where  (i)  AT P + PA = −Q  Take Q = I so that V̇(x) = −(x21 + x22 ) which is negative deﬁnite. Substituting in (i) gives   −3 −1 2 −1      p p12 + 11 p22 p12  p11 p12  p12 p22      −1 −3 2 = 0 −1 −1  0 −1    Equating elements gives −6p11 − 2p12 = −1, 4p12 − 2p22 = −1, 2p11 − 4p12 − p22 = 0  Solving gives p11 =  7 40 , p12  Principal minors of P are:  =  1 − 40 , p22  7 40  =  18 40  so that P =  > 0 and det P =  5 64  7 40 1 − 40  1 − 40    18 40  > 0 so P is positive deﬁnite  and the system is asymptotically stable.  71  Take tentative Lyapunov function V(x) = xT Px giving V̇(x) = xT (AT P + PA)x = −xT Qx where AT P + PA = −Q  (i)  Take Q = I so that V̇(x) = −(x21 + x22 ) which is negative deﬁnite. Substituting in (i) gives   0 −a 1 −b    p11 p12    p p12 + 11 p22 p12  p12 p22    0 −a    −1 1 = 0 −b  0 −1  Equating elements gives −8p12 = −1, 2p12 − 2bp22 = −1, p11 − bp12 − ap22 = 0 c Pearson Education Limited 2011     62  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Solving gives p12 =  1 2a , p22  =  a+1 2ab , p11  =  b2 +a2 +a 2ab   b2 +a2 +a so that, P =  2ab 1 2a  1 2a a+1 2ab    For asymptotic stability the principal minors of P must be positive. Thus, b2 + a2 + a >0 2ab  (ii)  and (b2 + a2 + a)(a + 1) > b2  (iii)  Case 1 ab > 0 (ii) ⇒ a2 + b2 + a > 0 so (iii) ⇒ a + 1 >  b2 b2 + a2 + a  ⇒ a[a2 + (a + 1)2 ] > 0 ⇒ a > 0. Since ab > 0 ⇒ b > 0 it follows that (ii) and (iii) are satisﬁed if a, b > 0 Case 2 ab < 0 No solution to (ii) and (iii) in this case. Thus, system is asymptotically stable when both a > 0 and b > 0. Note: This example illustrates the diﬃculty in interpretating results when using the Lyapunov approach. It is a simple task to conﬁrm this result using the Routh– Hurwitz criterion developed in Section 5.6.2.  72(a) ẋ1 = x2  (i)  ẋ2 = −2x2 + x3  (ii)  ẋ3 = −kx1 − x3  (iii)  If V̇(x) is identically zero then x3 is identically zero ⇒ x1 is identically zero from (iii) ⇒ x2 is identically zero from (i) Hence V̇(x) is identically zero only at the origin. (b) AT P + PA = −Q ⇒ ⎡  0 ⎣1 0  0 −2 1  ⎤⎡ p11 −k ⎦ ⎣ 0 p12 −1 p13  p12 p22 p23  ⎤ ⎡ p11 p13 ⎦ ⎣ p23 + p12 p33 p13  p12 p22 p23  ⎤⎡ 0 1 p13 ⎦ ⎣ 0 −2 p23 −k 0 p33  c Pearson Education Limited 2011   ⎤ ⎡ 0 0 ⎦ ⎣ 1 = 0 0 −1  0 0 0  ⎤ 0 0 ⎦ −1  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  63  Equating elements and solving for the elements of P gives the matrix ⎡ k2 +12k P=⎣  6k 12−2k 3k 12−2k k 12−2k  12−2k 6k 12−2k  0  0 k 12−2k 6 12−2k  ⎤ ⎦  (c) Principal minors of Pare: k2 + 12k > 0 if k > 0and(12 − 2k) > 0 ⇒ 0 < k < 6  122− 2k   36k2 3k k + 12k 3k3 − Δ2 = > 0 if k > 0 = 12 − 2k 12 − 2k 12 − 2k (12 − 2k)2 (k2 + 12k)(8k − k2 ) 216k2 − > 0if (6k3 − k4 ) > 0 ⇒ 0 < k < 6 Δ3 = (12 − 2k)3 (12 − 2k)3  Δ1 =  Thus system asymptotically stable for 0 < k < 6.  73  State-space form is   ẋ1 ẋ = ẋ2      0 = −k  1 −a    x1 x2   (i)  Take V(x) = kx21 + (x2 + ax1 )2 then V̇(x) = 2kx1 ẋ1 + 2(x2 + ax1 )(ẋ2 + aẋ1 ) = 2kx1 (x2 ) + 2(x2 + ax1 )(−kx1 − ax2 + ax1 )using (i) = −2kax21 Since k>0 and a>0 then V̇(x) is negative semideﬁnite but is not identically zero along any trajectory of (i). Consequently, this choice of Lyapunov function assures asymptotic stability.  Review Exercises 1.13 1(a)  Eigenvalues given by   −1 − λ 6   0 −13 − λ   0 −9   12  30  = (1 + λ)[(−13 − λ)(20 − λ) + 270] = 0 20 − λ  c Pearson Education Limited 2011   64  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  that is, (1 + λ)(λ − 5)(λ − 2) = 0 so eigenvalues are λ1 = 5, λ2 = 2, λ3 = −1 Eigenvectors are given by corresponding solutions of ⎤⎡ ⎤ ⎡ ei1 6 12 −1 − λi ⎣ 0 −13 − λi 30 ⎦ ⎣ ei2 ⎦ = 0 0 −9 20 − λi ei3 When i = 1, λi = 5 and solution given by e11 −e12 e13 = = = β1 198 −90 54 so e1 = [11 5 3]T When i = 2, λi = 2 and solution given by e21 −e22 e23 = = = β2 216 −54 27 so e2 = [8 2 1]T When i = 3, λi = −1 and solution given by e31 −e32 e33 = = = β3 1 0 0 so e3 = [1 0 0]T 1(b)  Eigenvalues given by       2 − λ 0 1   4 − λ −1     +  −1  −1 4−λ −1  =    −1  2 −λ  −1 2 0 − λ   4 − λ  =0 2   that is, 0 = (2 − λ)[(4 − λ)(−λ) + 2] + [−2 + (4 − λ)] = (2 − λ)(λ2 − 4λ + 3) = (2 − λ)(λ − 3)(λ − 1) = 0 so eigenvalues are λ1 = 3, λ2 = 2, λ3 = 1 Eigenvectors are given by the corresponding solutions of (2 − λi )ei1 + 0ei2 + ei3 = 0 −ei1 + (4 − λi )ei2 − ei3 = 0 −ei1 + 2ei2 − λi ei3 = 0 c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  65  Taking i = 1, 2, 3 gives the eigenvectors as e1 = [1 2 1]T , e2 = [2 1 0]T , e3 = [1 0 − 1]T  1(c)  Eigenvalues given by  1 − λ   −1   0  −1 2−λ −1    −1 −1   that is, λ  −1 2 − λ  0 −1   0  −1  R1 + (R2 + R3 ) 1 − λ  −1  −1  = λ 1− λ    −λ   −1   0    −1 0   −1 3 − λ   0 −1  −λ 2−λ −1   −λ  −1  = 0 1 − λ   0  0  = λ(3 − λ)(1 − λ) = 0 1 − λ  so eigenvalues are λ1 = 3, λ2 = 1, λ3 = 0 Eigenvalues are given by the corresponding solutions of (1 − λi )ei1 − ei2 − 0ei3 = 0 −ei1 + (2 − λi )ei2 − ei3 = 0 0ei1 − ei2 + (1 − λi )ei3 = 0 Taking i = 1, 2, 3 gives the eigenvectors as e1 = [1 − 2 1]T , e = [1 0 − 1]T , e3 = [1 1 1]T  2  Principal stress values (eigenvalues) given by    6 − λ 3− λ 2 1      2  2 R 3 − λ 1 + (R + R ) 1 2 3     1  1 1 4− λ   1 1 1    1  = 0 = (6 − λ)  2 3 − λ 1 1 4 − λ  1  that is, (6 − λ)  2 1   6 − λ 6 − λ  3−λ 1  1 4 − λ   0 0  1−λ −1  = (6 − λ)(1 − λ)(3 − λ) = 0 0 3 − λ  so the principal stress values are λ1 = 6, λ2 = 3, λ3 = 1. c Pearson Education Limited 2011   66  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Corresponding principal stress direction e1 , e2 and e3 are given by the solutions of  (3 − λi )ei1 + 2ei2 + ei3 = 0 2ei1 + (3 − λi )ei2 + ei3 = 0 ei1 + ei2 + (4 − λi )ei3 = 0  Taking i = 1, 2, 3 gives the principal stress direction as e1 = [1 1 1]T, e2 = [1 1 − 2]T, e3 = [1 − 1 0]T It is readily shown that eT1 e2 = eT1 e3 = eT2 e3 = 0 so that the principal stress directions are mutually orthogonal. 3  Since [1 0 1]T is an eigenvector of A ⎡  2 ⎣ −1 0  −1 3 b  ⎡ ⎤ ⎤ ⎡ ⎤ 1 1 0 b⎦ ⎣0⎦ = λ ⎣0⎦ 1 1 c  so 2 = λ, −1 + b = 0, c = λ giving b = 1 and c = 2. Taking these values A has eigenvalues given   2 − λ −1 0    −1 3−λ 1  = (2 − λ)   0 1 2− λ  by  3− λ   1   1  − (2 − λ) 2 − λ  = (2 − λ)(λ − 1)(λ − 4) = 0 that is, eigenvalues are λ1 = 4, λ2 = 2, λ3 = 1 Corresponding eigenvalues are given by the solutions of (2 − λi )ei1 − ei2 + 0ei3 = 0 −ei1 + (3 − λi )ei2 + ei3 = 0 0ei1 + ei2 + (2 − λi )ei3 = 0 Taking i = 1, 2, 3 gives the eigenvectors as e1 = [1 − 2 − 1]T , e2 = [1 0 1]T , e3 = [1 1 − 1]T  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 4  67  The three Gerschgorin circles are | λ − 4 | =| −1 | + | 0 |= 1 | λ − 4 | =| −1 | + | −1 |= 2 |λ−4|=1  Thus, | λ − 4 |≤ 1 and | λ − 4 |≤ 2 so | λ − 4 |≤ 2 or 2 ≤ λ ≤ 6. Taking x(o) = [−1 1 − 1]T iterations using the power method may be tabulated as follows Iteration k x(k)  A x(k) λ  0 −1 1 −1 −5 6 −5 6  1 −0.833 1 −0.833 −4.332 5.666 −4.332 5.666  2 −0.765 1 −0.765 −4.060 5.530 −4.060 5.530  3 −0.734 1 −0.734 −3.936 5.468 −3.936 5.468  4 −0.720 1 −0.720 −3.88 5.44 3.88 5.44  5 −0.713 1 −0.713 −3.852 5.426 −3.852 5.426  6 −0.710 1 −0.710  Thus, correct to one decimal place the dominant eigenvalue is λ = 5.4 5(a)  Taking x (o) = [1 1 1]7 iterations may be tabulated as follows  Iteration k x(k)  A x(k) λ  0 1 1 1 4 4.5 5 5  1 0.800 0.900 1 3.500 4.050 4.700 4.700  2 0.745 0.862 1 3.352 3.900 4.607 4.607  3 0.728 0.847 1 3.303 3.846 4.575 4.575  4 0.722 0.841 1 3.285 3.825 4.563 4.563  5 0.720 0.838 1 3.278 3.815 4.558 4.558  6 0.719 0.837 1 3.275 3.812 4.556 4.556  7 0.719 0.837 1  Thus, estimate of dominant eigenvalues is λ  4.56 with associated eigenvector x = [0.72 0.84 1]T  5(b)  3 i=1  λi = trace A ⇒ 7.5 = 4.56 + 1.19 + λ3 ⇒ λ3 = 1.75  c Pearson Education Limited 2011   68  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  5(c) (i)  det A =  3   λi = 9.50 so A−1 exists and has eigenvalues  i=1  1 1 1 , , 1.19 1.75 4.56 so power method will generate the eigenvalue 1.19 corresponding to A. (ii)  A − 3I has eigenvalues 1.19 − 3, 1.75 − 3, 4.56 − 3 that is, −1.91, −1.25, 1.56  so applying the power method on A − 3I generates the eigenvalues corresponding to 1.75 of A. 6  ẋ = αλeλt , ẏ = βλeλt , ż = γλeλt so the diﬀerential equations become αλeλt = 4αeλt + βeλt + γeλt βλeλt = 2αeλt + 5βeλt + 4γeλt γλeλt = −αeλt − βeλt  Provided eλt = 0 (i.e. non-trivial solution) we have the eigenvalue problem ⎡  4 ⎣ 2 −1 Eigenvalues given  4 − λ 1   2 5 − λ   −1 −1  ⎡ ⎤ ⎤ ⎡ ⎤ α α 1 1 ⎦ ⎣ ⎦ ⎣ β = λ β⎦ 5 4 γ γ −1 0  by   4− λ 1   4  C2 −C3  2  −1 0  0 1−λ λ−1    4 − λ 1   4  = (λ − 1)  2  −1 −λ   0 −1 1   1  4  −λ   = −(λ − 1)(λ − 5)(λ − 3) so its eigenvalues are 5, 3 and 1. When λ = 1 the corresponding eigenvector is given by 3e11 + e12 + e13 = 0 2e11 + 4e12 + 4e13 = 0 −e11 − e12 − e13 = 0 c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition −e12 e13 e11 = = = β1 0 2 2 Thus, corresponding eigenvector is β[0 − 1 1]T having solution  7  Eigenvalues are given by   8− λ −8 −2   −3 − λ −2  = 0 | A − λI | =  4  3 −4 1 − λ  Row 1 − (Row 2 + Row 3) gives  1 − λ  | A − λI | =  4  3     1 −1 −1  −1 + λ   −2  −2  = (1 − λ)  4 −3 − λ 3 −4 1 − λ 1−λ   0  0 1−λ 2  = (1 − λ)[(1 − λ)(4 − λ) + 2] −1 4 − λ  −1 + λ −3 − λ −4   1  = (1 − λ)  4 3  = (1 − λ)(λ − 2)(λ − 3) Thus, eigenvalues are λ1 = 3, λ2 = 2, λ3 = 1. Corresponding eigenvectors are given by (8 − λ)ei1 − 8ei2 − 2ei3 = 0 4ei1 − (3 + λ)ei2 − 2ei3 = 0 3ei1 − 4ei2 + (1 − λ)ei3 = 0 When i = 1, λi = λ1 = 3 and solution given by e11 −e12 e13 = = = β1 4 −2 2 so a corresponding eigenvector is e1 = [2 1 1]T . When i = 2, λi = λ2 = 2 and solution given by e21 −e22 e23 = = = β2 −3 2 −1 so a corresponding eigenvector is e2 = [3 2 1]T . When i = 3, λi = λ3 = 1 and solution given by e31 −e32 e33 = = = β3 −8 6 −4 c Pearson Education Limited 2011   69  70  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  so a corresponding eigenvector is e3 = [4 3 2]T . Corresponding modal and spectral matrices are ⎤ ⎤ ⎡ ⎡ 3 0 0 2 3 4 M = ⎣ 1 2 3 ⎦ and Λ = ⎣ 0 2 0 ⎦ 1 0 1 1 1 2 ⎤ ⎡ 1 −2 1 M−1 = ⎣ 1 0 −2 ⎦ and matrix multiplication conﬁrms M−1 A M = Λ −1 1 1 8  Eigenvectors of A are given by    1 − λ 0 −4    =0  0 5 − λ 4    −4 4 3 − λ  that is, λ3 − 9λ2 − 9λ + 81 = (λ − 9)(λ − 3)(λ + 3) = 0 so the eigenvalues are λ1 = 9, λ2 = 3 and λ3 = −3. The eigenvectors are given by the corresponding solutions of (1 − λi )ei1 + 0ei2 − 4ei3 = 0 0ei1 + (5 − λi )ei2 + 4ei3 = 0 −4ei1 + 4ei2 + (3 − λi )ei3 = 0 Taking i = 1, 2, 3 the normalized eigenvectors are given by ê1 = [ 13  −2 −2 T 3 3 ] , ê2  The normalised modal matrix  = [ 23 ⎡  1 1 ⎣ −2 M̂ = 3 −2 so  2 −1 T 3 3 ] , ê3  2 2 −1  = [ 23  −1 2 T 3 3]  ⎤ 2 −1 ⎦ 2  ⎤ ⎡ ⎤ ⎡ 1 1 0 −4 1 −2 −2 1 ⎣ M̂T A M̂ = 2 2 −1 ⎦ ⎣ 0 5 4 ⎦ ⎣ −2 9 −2 −4 4 3 2 −1 2 ⎤ ⎡ 9 0 0 = ⎣0 3 0 ⎦ = Λ 0 0 −3 ⎡  c Pearson Education Limited 2011   2 2 −1  ⎤ 2 −1 ⎦ 2  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  71  ⎤ −6 0 0 0 0 0⎥ ⎢ 6 −4 9 Ṅ = ⎣ ⎦ N, N = [N1 N2 N3 N4 ]T 0 4 −2 0 0 0 2 0 Since the matrix A is a triangular matrix its eigenvalues are the diagonal elements. ⎡  Thus, the eigenvalues are λ1 = −6, λ2 = −4, λ3 = −2, λ4 = 0 The eigenvectors are the corresponding solutions of (−6 − λi )ei1 + 0ei2 + 0ei3 + 0ei4 = 0 6ei1 + (−4 − λi )ei2 + 0ei3 + 0ei4 = 0 0ei1 + 4ei2 + (−2 − λi )ei3 + 0ei4 = 0 0ei1 + 0ei2 + 2ei3 − λi ei4 = 0 Taking i = 1, 2, 3, 4 and solving gives the eigenvectors as e1 = [1 − 3 3 − 1]T , e2 = [0 1 − 2 1]T e3 = [0 0 1 − 1]T , e4 = [0 0 0 1]T Thus, spectral form of solution to the equation is N = αe−6t e1 + βe−4t e2 + γe−2t e3 + δe4 Using the given initial conditions at t = 0 we have ⎤ ⎡ ⎤ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ 0 0 0 1 C ⎢0⎥ ⎢ 0⎥ ⎢ 1⎥ ⎢ −3 ⎥ ⎢0⎥ ⎦ +δ ⎣ ⎦ ⎦ +γ ⎣ ⎦ +β ⎣ ⎣ ⎦ =α ⎣ 0 1 −2 3 0 1 −1 1 −1 0 ⎡  so C = α, 0 = −3α + β, 0 = 3α − 2β + γ, 0 = −α + β − γ + δ which may be solved for α, β, γ and δ to give α = C, β = 3C, γ = 3C, δ = C Hence,  N4 = −αe−6t + βe−4t − γe−2t + δ = −Ce−6t + 3Ce−4t − 3Ce−2t + C  c Pearson Education Limited 2011   72  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  10(a) (i)  Characteristic equation of A is λ2 − 3λ + 2 = 0 so by the Cayley–Hamilton   theorem  4 A = 3A − 2I = 3  0 1  2     8 0 A = 3(3A − 2I) − 2A = 7A − 6I = 7 1   16 0 4 A = 7(3A − 2I) − 6A = 15A − 14I = 15 1   32 0 5 A = 15(3A − 2I|) − 14A = 31A − 30I = 31 1   64 0 6 A = 31(3A − 2I) − 30A = 63A − 62I = 63 1   128 0 7 A = 63(3A − 2I) − 62A = 127A − 126I = 127 1   −29 0 7 6 4 3 2 Thus, A − 3A + A + 3A − 2A + 3I = −32 3   3  (ii)  Eigenvalues of A are λ1 = 2, λ2 = 1. Thus, Ak = α0 I + α1 A where α0 and α1 satisfy 2k = α0 + 2α1 , 1 = α0 + α1 α1 = 2k − 1, α0 = 2 − 2k   Thus, Ak =  10(b)  α0 + 2α1 α1  0 α0 + α1     =  2k k 2 −1  0 1    Eigenvalues of A are λ1 = −2, λ2 = 0. Thus,  Thus, eAt  eAt = α0 I + α1 A where α0 and α1 satisfy 1 e−2t = α0 − 2α1 , 1 = α0 ⇒ α0 = 1, α1 = (1 − e−2t ) 2     1 12 (1 − e−2t ) α1 α0 = = 0 e−2t 0 α0 − 2α1  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  73  ⎤ 1 2 3 11 The matrix A = ⎣ 0 1 4 ⎦ has the single eigenvalue λ = 1 (multiplicity 3) 0 0 1 ⎤ ⎤ ⎡ ⎡ 0 1 0 0 2 3 (A − I) = ⎣ 0 0 4 ⎦ ∼ ⎣ 0 0 1 ⎦ is of rank 2 so has nullity 3 − 2 = 1 0 0 0 0 0 0 indicating that there is only one eigenvector corresponding to λ = 1. ⎡  This is readily determined as e1 = [1 0 0]T The corresponding Jordan canonical form comprises a single block so ⎤ 1 1 0 J = ⎣0 1 1⎦ 0 0 1 ⎡  Taking T = A − I the triad of vectors (including generalized eigenvectors) has ⎤ ⎡ 0 0 8 the form {T2 ω, T ω, ω} with T2 ω = e1 . Since T2 = ⎣ 0 0 0 ⎦ , we may take 0 0 0 1 T 2 1 T ω = [0 0 8 ] . Then, T ω = [ 8 8 0] . Thus, the triad of vectors is e1 = [1 0 0]T , e∗1 = [ 38  1 2  1 T 0]T , e∗∗ 1 = [0 0 8 ]  The corresponding modal matrix is ⎡  1 M = ⎣0 0 ⎡  1 16  M−1 = 16 ⎣ 0 0 ⎡ 1  16  M−1 A M = 16 ⎣ 0 0 ⎡ 1 1 = ⎣0 1 0 0  3 − 64 1 8  0 3 − 64 1 8  0 ⎤  3 8 1 2  0  ⎤ 0 0⎦ 1 8  ⎤ 0 0 ⎦ and by matrix multiplication 1 2  ⎤ ⎡ 1 0 ⎦ ⎣ 0 0 1 0 2  2 1 0  ⎤ ⎡ 1 3 ⎦ ⎣ 4 0 1 0  0 1⎦ = J 1  c Pearson Education Limited 2011   3 8 1 2  0  ⎤ 0 0⎦ 1 8  74 12  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition Substituting x = X cos ωt, y = Y cos ωt, z = Z cos ωt gives −ω2 X = −2X + Y −ω2 Y = X − 2Y + Z −ω2 Z = Y − 2Z  or taking λ = ω2  (λ − 2)X + Y = 0 X + (λ − 2)Y + Z = 0 Y + (λ − 2)Z = 0  For non-trivial solution  λ − 2   1   0  1 λ−2 1   0  1  = 0 λ − 2  that is, (λ − 2)[(λ − 2)2 − 1] − (λ − 2) = 0 (λ − 2)(λ2 − 4λ + 2) = 0 √ so λ = 2 or λ = 2 ± 2 When λ = 2 , Y = 0 and X = −Z so X : Y : Z = 1 : 0 : −1 √ √ √ When λ = 2 + 2 , X = Z and Y = − 2X so X : Y : Z = 1 : − 2 : 1 √ √ √ When λ = 2 − 2 , X = Z and Y = 2X so X : Y : Z = 1 : 2 : 1 13  In each section A denotes the matrix of the quadratic form.  ⎤   2 −1 0  2 −1   = 1 and 1 −1 ⎦ has principal minors of 2,  13(a) A = ⎣ −1 −1 1  0 −1 2 det A = 0 ⎡  so by Sylvester's condition (c) the quadratic form is positive-semideﬁnite. ⎡  3 13(b) A = ⎣ −2 −2 det A = 6  −2 7 0  ⎤  −2  3 0 ⎦ has principal minors of 3,  −2 2   −2  = 17 and 7   so by Sylvester's condition (a) the quadratic form is positive-deﬁnite.  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition ⎡  16 13(c) A = ⎣ 16 16 det A = −704  16 36 8  ⎤  16  16 ⎦ 8 has principal minors of 16,  16 17  75   16  = 320 and 36   so none of Sylvester's conditions are satisﬁed and the quadratic form is indeﬁnite. ⎡  −21 13(d) A = ⎣ 15 −6 and det A = 0  15 −11 4  ⎤  −6  −21 ⎦ 4 has principal minors of −21,  15 −2   15  =6 −11   so by Sylvester's condition (d) the quadratic form is negative-semideﬁnite. ⎤   −1 1 1  −1 1   = 2 and 1 ⎦ has principal minors of −1,  13(e) A = ⎣ 1 −3 1 −3  1 1 −5 det A = −4 so by Sylvester's condition (b) the quadratic form is negative-deﬁnite. ⎡  ⎡ 14  7 2  A e1 = ⎣ 4 − 32  − 21 −1 3 2  ⎡ ⎤ ⎤ ⎡ ⎤ 1 1 − 12 ⎦ ⎣ ⎦ ⎣ 2 = 2⎦ 0 1 3 3 2  Hence, e1 = [1 2 3]T is an eigenvector with λ1 = 1 the corresponding eigenvalue. Eigenvalues are given by  7 − − λ − 12  2 4 −1 − λ 0 =  3  −3 2 2   − 12  0  = −λ3 + 3λ2 + λ − 3 1  2 −λ = (λ − 1)(λ2 + 2λ + 3) = −(λ − 1)(λ − 3)(λ + 1)  so the other two eigenvalues are λ2 = 3, λ3 = −1. Corresponding eigenvectors are the solutions of (− 27 − λi )ei1 − 12 ei2 − 12 ei3 = 0 4ei1 − (1 + λi )ei2 + 0ei3 = 0 − 23 ei1 + 32 ei2 + ( 21 − λi )ei3 = 0 c Pearson Education Limited 2011   76  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Taking i = 2, 3 gives the eigenvectors as e2 = [1 1 0]T , e3 = [0 − 1 1]T The diﬀerential equations can be written in the vector–matrix form ẋ = A x , x = [x y z]T so, in special form, the general solution is x = αeλ1 t e1 + βeλ2 t e2 + γeλ3 t e3 ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ 1 1 0 = αet ⎣ 2 ⎦ + βe3t ⎣ 1 ⎦ + γe−t ⎣ −1 ⎦ 3 0 1 With x(0) = 2, y(0) = 4, z(0) = 6 we have α = 2, β = 0, γ = 0 ⎡ ⎤ 1 x = 2et ⎣ 2 ⎦ 3  so  that is, x = 2et , y = 4et , z = 6et .  15(a)    1.2 AA = 1.6 T  ⎤  1.2 1.6 18.25 −4 ⎣ 0.9 1.2 ⎦ = −9 3 −4 3   0.9 1.2  ⎡  −9 13    Eigenvalues λi given by (18.25 − λ)(13 − λ) − 81 = 0 ⇒ (λ − 25)(λ − 6.25) = 0 ⇒ λ1 = 25, λ2 = 6.25 having corresponding eigenvectors u1 = [ −4 u2 = [ 3  T  3 T 5 ]  3 ] ⇒ û1 = [ − 45 T  4 ] ⇒ û2 = [ 35  4 5  ]  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  77  leading to the orthogonal matrix  Û =  3 5  ⎤ 1.6  1.2 1.2 ⎦ 1.6 3  ⎡  1.2 AT A = ⎣ 0.9 −4  − 54  0.9 1.2  3 5 4 5    ⎡    4 −4 = ⎣3 3 0  3 2.25 0  ⎤ 0 0 ⎦ 25  Eigenvalues μi given by (25 − μ) [(4 − μ)(2.25 − μ) − 9] = 0 ⇒ (25 − μ)μ(μ − 6.25) = 0 ⇒ μ1 = 25, μ2 = 6.25, μ3 = 0 with corresponding eigenvalues v1 = v̂1 = [ 0 v2 = [ 4  1]  T  T  0 ] ⇒ v̂2 = [ 45  3  v3 = [ −3  0  4  3 5  0]  T  0 ] ⇒ v̂3 = [ − 35  4 5  T T  0]  leading to the orthogonal matrix ⎡  0 ⎣ V̂ = 0 1  4 5 3 5  − 35  0  0  4 5  ⎤ ⎦  √ √ = 25 = 5 and σ = 6.25 = 2.5 so that The singular values of A are σ 1 2  5 0 0 giving the SVD form of A as Σ= 0 2.5 0  T    A = ÛΣV̂ =  −0.8 0.6  0.6 0.8     5 0  0 2.5  1.2 0.9 (Direct multiplication conﬁrms A = 1.6 1.2    ⎡  0 0 ⎣ 0.8 0 −0.6  −4 ) 3  c Pearson Education Limited 2011   0 0.6 0.8  ⎤ 1 0⎦ 0  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  78  ⎡  ⎤⎡ 1 4 3 0 − 5 5 5 ∗ 4 ⎦⎣ 0 (b) A† = V̂Σ ÛT = ⎣ 0 35 5 1 0 0 0 ⎤ ⎡ 0.192 0.256 = ⎣ 0.144 0.192 ⎦ −0.16 0.12  0 2 5  ⎤ ⎦    0  − 54 3 5  3 5 4 5    ⎡  24 1 ⎣ 18 = 125 −20  ⎤ 32 24 ⎦ 15  AA† = I CHECK  LHS =  1 125  1.2 0.9 1.6 1.2  ⎤ 24 32 −4 ⎣ 18 24 ⎦ = 1 −24 15   ⎡   1 125  125 0   0 = I = RHS 125  (c) Since A is of full rank 2 and there are more columns than rows ⎤ ⎤ ⎡ −1 1.2 1.6  1.2 1.6  18.25 −9 1 ⎣ 0.9 1.2 ⎦ 13 = ⎣ 0.9 1.2 ⎦ = 156.25 −9 13 9 −4 ⎡3 −4 ⎤ 3 ⎤ ⎡ 0.192 0.256 30 40 1 ⎣ 22.5 = 156.25 30 ⎦ = ⎣ 0.144 0.192 ⎦ −0.16 0.12 −25 18.25 ⎡  †  T  T −1  A = A (AA )  9 18.25  which checks with the answer in (b).  16 (a)  Using partitioned matrix multiplication the SVD form of A may be  expressed in the   form T  A = ÛΣV̂ = [ Ûr  S Ûm−r ] 0  0 0    V̂Tr V̂Tn−r    T  = Ûr SV̂r  (b) Since the diagonal elements in S are non-zero the pseudo inverse may be expressed in the form  ∗  A† = V̂Σ ÛT = V̂r S−1 ÛTr  ⎤ 1 −1 (c) From the solution to Q46, exercises 1.8.4, the matrix A = ⎣ −2 2 ⎦ has a single 2 −2 √ √ singularity σ1 = 18 so r = 1 and S is a scalar 18; Ûr = Û1 = û1 = [ 13 − 23 23 ]T ⎡  and V̂r = V̂1 = v̂1 =  √1 2  − √12  T  c Pearson Education Limited 2011     Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  79  The SVD form of A is ⎡ A = û1 Sv̂T1 =  1 3 ⎣ −2 3 2 3  ⎤  √ ⎦ 18 ⎡  1 ⎣ with direct multiplication conﬁrming A = −2 2 Thus, the pseudo inverse is  A† = v̂1 S−1 ûT1 =  √1 2 √ − 12  =  √1 2  ⎤ −1 2 ⎦ −2     √1 18    1 18  − √12  [  1 −1  1 3  − 23 −2 2  2 3  2 −2  ]=   1 6 − 16   [ 13  − 32  2 3  ]  which agrees with the answer obtained in Q46, Exercises 1.8.4  17  ẋ = A x + bu , y = cT x  Let λi , ei , i = 1, 2, . . . , n, be the eigenvalues and corresponding eigenvectors of A. Let M = [e1 , e2 , . . . , en ] then since λi 's are distinct the ei 's are linearly independent and M−1 exists. Substituting x = M ξ gives M ξ̇ξ = A M ξ + bu Premultiplying by M−1 gives ξ̇ξ = M−1 A M ξ + M−1 bu = Λ ξ + b1 u where Λ = M−1 A M = (λi δij ), i, j = 1, 2, . . . , n, and b1 = M−1 b Also, y = cT x ⇒ y = cT Mξξ = cT1 ξ , cT1 = cT M. Thus, we have the desired canonical form. If the vector b1 contains a zero element then the corresponding mode is uncontrollable and consequently (A1 b1 c) is uncontrollable. If the matrix cT has a zero element then the system is unobservable. The eigenvalues of A are λ1 = 2, λ2 = 1, λ3 = −1 having corresponding eigenvectors e1 = [1 3 1]T , e2 = [3 2 1]T and e3 = [1 0 1]T . c Pearson Education Limited 2011   80  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  The modal matrix ⎡  1 ⎣ M = [e1 e2 e3 ] = 3 1  3 2 1  so canonical form is ⎡  ⎤ ⎡ ξ̇1 2 ⎣ ξ̇2 ⎦ = ⎣ 0 0 ξ̇3  ⎤ ⎡ 1 2 1 ⎣ −1 ⎦ 0 with M = − −3 6 1 1  0 1 0  ⎤ −2 −2 0 3 ⎦ 2 −7  ⎤ ⎡ ⎤ ⎡ 1 ⎤ ξ1 0 3 0 ⎦ ⎣ ξ2 ⎦ + ⎣ 0 ⎦ u −1 − 43 ξ3  y = [1 − 4 − 2][ξ1 ξ2 ξ3 ]T We observe that the system is uncontrollable but observable. Since the system matrix A has positive eigenvalues the system is unstable. Using Kelman matrices ⎤ ⎡ ⎤ ⎡ ⎤ 2 0 1 1 0 (i) A2 = ⎣ −3 4 3 ⎦ , A b = ⎣ 2 ⎦ , A2 b = ⎣ 4 ⎦ 2 −1 1 2 0 ⎤ ⎤ ⎡ ⎡ 1 0 0 −1 2 0 Thus, [b A b A2 b] = ⎣ 1 2 4 ⎦ ∼ ⎣ 0 1 0 ⎦ and is of rank 2 0 0 0 −1 2 0 so the system is uncontrollable. ⎡  ⎡  −2 (ii) [c AT c (AT )2 c] = ⎣ 1 0 so the system is observable.  18  −3 0 5  ⎤ ⎡ 0 −3 2 ⎦ ∼ ⎣1 0 1  0 0 1  ⎤ 1 0 ⎦ and is of full rank 3 0  Model is of form ẋ = Ax + Bu and making the transformation x = Mz gives Mż = AMz + Bu ⇒ ż = M−1 AMz + M−1 Bu ⇒ ż = Λz + M−1 Bu  where M and Λ are respectively the modal and spectral matrices ofA. The eigenvalues of A are given by   −2 − λ   0   0   −2 0  −λ 1  = 0 ⇒ −(2 − λ)(4λ + λ2 + 3) = 0 −3 −4 − λ  ⇒ (λ + 2)(λ + 1)(λ + 3) = 0 ⇒ λ1 − 1, λ2 = −2, λ3 = −3 c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  81  with corresponding eigenvectors e1 = [ −2  T  1 −2 ] , e2 = [ 1  0  T  0 ] and e3 = [ −2  −1  3]  T  Thus, the modal and spectral matrices are ⎤ ⎡ −1 1 −2 ⎦ ⎣ 0 −4 andΛ = 0 0 0 −1  ⎡  −2 ⎣ M= 1 −1 ⎡  and det M = −2 ⇒ M−1 ⎡1 2  =⎣3 1 2  ⎤  0 ⎣ = 1 0  1 2  3 2  0 −2 0 ⎡  ⎤  0 −1 ⎣ ⎦ 2 ⇒M B= 1 1 0 2  4 1 2  2 6 ⎦ leading to the canonical form 1 ⎤⎡ ⎤ ⎡ 1 ⎡ ⎤ ⎡ z1 −1 0 0 ż1 2 ⎦ ⎣ ⎦ ⎣ ⎣ z2 ⎦ + ⎣ 3 ż = ż2 = 0 −2 0 1 0 0 −3 ż3 z3 2  ⎤ 0 0 ⎦ −3 1 2  3 2  ⎤⎡  1 ⎦ ⎣ 0 2 1 1 2  4 1 2  ⎤ 0 1⎦ 1  ⎤ 2   u 6⎦ 1 u2 1  From (1.99a) the solution is given by ⎡  ⎤ ⎡ −t z1 e ⎣ z2 ⎦ = ⎣ 0 0 z3  ⎤ ⎤⎡ ⎡  t e−(t−τ ) z1 (0) 0 ⎣ 0 0 ⎦ ⎣ z2 (0) ⎦ + 0 e−3t z3 (0) 0  0  e−2t 0  ⎤   2 2 ⎣ 3 6 ⎦ τ dτ 1 1 1 2 ⎤⎡ ⎤ 1 3 10 2 2 ⎦ ⎣ 5 ⎦ = [ 17 4 2 2 1 1 2 2 2  0  e−2(t−τ ) 0  e−3(t−τ )  ⎡1  ⎡  −1  with z(0) = M  ⎡  0 ⎣ x(0) = 1 0 17 t 2 e −2t  z = ⎣ 34e 7 −3t 2e ⎡ +⎣  ⎤  34  7 T 2 ]  . Thus,  ⎡  ⎤ ⎡ 17 t ⎤ 1 −(t−τ ) τ)e (2 + 2 2 e  ⎦ + t ⎣ (6 + 3τ)e−2(t−τ ) ⎦ dτ ⇒ z = ⎣ 34e−2t ⎦ 0 7 −3t (1 + 12 τ)e−3(t−τ ) 2e  1 3 3 −t 2t + 2 − 2e 9 9 −2t 3 2t + 4 − 4e 5 5 −3t 1 6 t + 18 − 18 e  ⎤  ⎡  ⎤ + 32 + 7e−t ⎦ ⇒ z = ⎣ 3 t + 9 − 127 e−2t ⎦ 2 4 4 5 29 −3t 1 6 t + 18 + 9 e 1 2t  c Pearson Education Limited 2011   0 0  ⎤ ⎦  82  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  ⎤⎡ 1 ⎤ 3 −t −2 1 −2 t + + 7e 2 2 −2t ⎦ giving x = Mz = ⎣ 1 0 −1 ⎦ ⎣ 32 t + 94 − 127 4 e 5 29 −3t 1 −1 0 3 6 t + 18 + 9 e ⎡ ⎤ 58 −3t 1 47 −2t −14e−t + 127 e − e + t − 4 9 6 36 −3t ⎦ 7e−t − 29 + 13 t + 11 ⇒ x(t) = ⎣ 9 e 9 29 −3t 2 −t −7e + 3 e −3 ⎡  19(a)  Eigenvalues of the matrix given by  −1  −9  C1 −C2 1 − λ      3−λ 2 −1    −3 + λ 6 − λ −9    0 1 1− λ    1 2 −1    = (3 − λ)  0 8 − λ −10  0 1 1 − λ   5 − λ 2  6−λ 0 =  3  1 1  = (3 − λ)(λ2 − 9λ + 18) = (3 − λ)(λ − 3)(λ − 6) λ3 = 3⎡ so the eigenvalues are λ⎡1 = 6, λ2 = ⎤ ⎤ 0 0 1 2 2 −1 When λ = 3, A − 3I = ⎣ 3 3 −9 ⎦ ∼ ⎣ 1 0 0 ⎦ is of rank 2 0 0 0 1 1 −2 so there is only 3 − 2 = 1 corresponding eigenvectors. The eigenvector corresponding to λ1 = 6 is readily determined as e1 = [3 2 1]T . Likewise the single eigenvector corresponding to λ2 = 6 is determined as e2 = [1 −1 0]T The generalized eigenvector e∗2 determined by (A − 2I)e∗2 = e2 or  3e∗21 + 2e∗22 − e∗23 = 1 3e∗21 + 3e∗22 − 9e∗23 = −1 e∗21 + e∗22 − 2e∗23 = 0  giving e∗2 = [  1 1 1 T 3 3 3]  .  For convenience, we can take the two eigenvectors corresponding to λ = 3 as e2 = [3 − 3 0]T, e∗2 = [1 1 1]T c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition ⎡  6 The corresponding Jordan canonical form being J = ⎣ 0 0  19(b)  83  ⎤ 0 1⎦ 3  0 3 0  The generalised modal matrix is then ⎡  3 −3 ⎣ M = 2 −3 1 0 ⎡  5 ⎣ AM= 3 1 ⎡ 3 M J = ⎣2 1  ⎤ −1 −9 ⎦ 1 ⎤ 3 1 −3 1 ⎦ 0 1  2 6 1  ⎡  3 ⎣2 1 ⎡ 6 ⎣0 0  −3 −3 0 0 3 0  ⎤ 1 1⎦ 1 ⎤ ⎤ ⎡ 18 9 6 1 1 ⎦ = ⎣ 12 −9 0 ⎦ 6 0 3 1 ⎤ ⎤ ⎡ 13 9 6 0 1 ⎦ = ⎣ 12 −9 0 ⎦ 6 0 3 3  so A M = M J ⎡  19(c)  M−1  −3 −3 1 ⎣ −1 2 =− 9 3 3  so  ⎤ ⎡ 6t e 6 Jt ⎦ ⎣ 0 −1 , e = −15 0  ⎤ ⎡ 6t e 3 3 1 1 x(t) = − ⎣ 2 −3 1 ⎦ ⎣ 0 9 1 0 1 0 ⎤ ⎡ 6t 3t 9e − 9(1 + t)e 1 ⎣ 6t = 6e + (3 + 9t)e3t ⎦ 9 3e6t − 3e3t ⎡  20  0 e3t 0  0 e3t 0  ⎤ 0 te3t ⎦ e3t  ⎤ ⎡ −3 −3 0 2 te3t ⎦ ⎣ −1 3t 3 3 e  ⎤ ⎡ ⎤ 0 6 −1 ⎦ ⎣ 1 ⎦ 0 −15  Substituting x = eλt u, where u is a constant vector, in x = A x gives λ2 u = A u or (A − λ2 I)u = 0  (1)  so that there is a non-trivial solution provided | A − λ2 I |= 0 c Pearson Education Limited 2011   (2)  84  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  If λ21 , λ22 , . . . , λ2n are the solutions of (2) and u1 , u2 , . . . , un the corresponding solutions of (1) deﬁne M = [u1 u2 . . . un ] and S = diag (λ21 λ22 . . . λ2n ) Applying the transformation x = M q , q = [q1 q2 . . . qn ] gives M q̈ = A M q giving  q̈ = M−1 A M q provided u1 , u2 , . . . , un are linearly independent  so that q̈ = S q since M−1 A M = S This represents n diﬀerential equations of the form q̈i = λ2i qi , i = 1, 2, . . . , n When λ2i < 0 this has the solution of the form qi = Ci sin(ωi t + αi ) where Ci and αi are arbitrary constants and λi = jωi The given diﬀerential equations may be written in the vector–matrix form       x1 ẍ1 −3 2 ẋ = = 1 −2 ẍ2 x2 which is of the above form ẍ = A x 0 =| A − λ2 I | gives (λ2 )2 + 5(λ2 ) + 4 = 0 or λ21 = −1, λ22 = −4. Solving the corresponding equation (A − λ2i I) ui = 0 we have that u1 = [1 1]T and u2 = [2 − 1]T . Thus, we take     −1 0 1 2 and S = M= 0 −4 1 −1 The normal modes of the system are given by       q̈1 q1 −1 0 = 0 −4 q̈2 q2 c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition giving  85  q1 (t) = C1 sin(t + α1 ) ≡ γ1 sin t + β1 cos t  q2 (t) = C2 sin(2t + α2 ) ≡ γ2 sin 2t + β2 cos 2t  5      1 −1 −2 1 −1 3 = Since x = M q we have that q(0) = M x(0) = − 2 − 13 3 −1 1 also q̇(0) = M−1 ẋ(0) so that q̇1 (0) = 2 and q̇2 (0) = 0 Using these initial conditions we can determine γ1 , β1 , γ2 and β2 to give 5 cos t + 2 sin t 3 1 q2 (t) = − cos 2t 3 q1 (t) =  The general displacements x1 (t) and x2 (t) are then given by x = M q so 2 5 cos t + 2 sin t − cos 2t 3 3 1 5 x2 = q1 − q2 = cos t + 2 sin t − cos 2t 3 3  x1 = q1 + 2q2 =  c Pearson Education Limited 2011   2 Numerical Solution of Ordinary Differential Equations Exercises 2.3.4 1  Euler's method for the solution of the diﬀerential equation  dx = f(t, x) is dt  Xn+1 = Xn + hFn = Xn + hf(tn , Xn ) dx Applying this to the equation = − 21 xt with x(0) = 1 and a step size of h = 0.1 dt yields x0 = x(0) = 1 X1 = x0 + hf(t0 , x0 ) = x0 + h (− 12 x0 t0 ) = 1 − 0.1 ×  1 2  × 1 × 0 = 1.0000  X2 = X1 + hf(t1 , X1 ) = X1 + h (− 12 X1 t1 ) = 1.0000 − 0.1 ×  1 2  × 1.0000 × 0.1 = 0.9950  X3 = X2 + hf(t2 , X2 ) = X2 + h (− 12 X2 t2 ) = 0.9950 − 0.1 ×  1 2  × 0.9950 × 0.2 × 0.98505  Hence Euler's method with step size h = 0.1 gives the estimate X(0.3) = 0.98505.  2  Euler's method for the solution of the diﬀerential equation  Xn+1 = Xn + hFn = Xn + hf(tn , Xn )  c Pearson Education Limited 2011   dx = f(t, x) is dt  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition Applying this to the equation h = 0.025 yields  87  dx = − 12 xt with x(1) = 0.1 and a step size of dt  x0 = x(1) = 0.1 X1 = x0 + hf(t0 , x0 ) = x0 + h (− 12 x0 t0 ) = 0.1 − 0.025 ×  1 2  × 0.1 × 1 = 0.09875  X2 = X1 + hf(t1 , X1 ) = X1 + h (− 12 X1 t1 ) = 0.09875 − 0.025 ×  1 2  × 0.09875 × 1.025 = 0.09748  X3 = X2 + hf(t2 , X2 ) = X2 + h (− 12 X2 t2 ) 1 = 0.09748 − 0.025 × × 0.09748 × 1.050 = 0.09621 2 X4 = X3 + hf(t3 , X3 ) = X3 + h (− 12 X3 t3 ) = 0.09621 − 0.025 ×  1 2  × 0.09621 × 1.075 = 0.09491  Hence Euler's method with step size h = 0.1 gives the estimate X(1.1) = 0.09491.  3  Euler's method for the solution of the diﬀerential equation  dx = f(t, x) is dt  Xn+1 = Xn + hFn = Xn + hf(tn , Xn ) Applying this to the equation  x dx = with x(0.5) = 1 and a step of h = 0.1 dt 2(t + 1)  yields  x0 = x(0.5) = 1 1 x0 = 1 + 0.1 = 1.0333 2(t0 + 1) 2(0.5 + 1) 1.0333 X1 X2 = X1 + hf(t1 , X1 ) = X1 + h = 1.0333 + 0.1 = 1.0656 2(t1 + 1) 2(0.6 + 1) X1 = x0 + hf(t0 , x0 ) = x0 + h  (Note that tn = t0 + nh = 0.5 + 0.1n.) X3 , X4 and X5 may be computed in similar fashion. It is usually easier to set out numerical solutions in a systematic tabular form such as the following: n 0 1  tn 0.5 0.6  Xn 1.0000 1.0333  f(tn , Xn ) 0.3333 0.3229  Xn + hf(tn , Xn ) 1.0333 1.0656  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  88 2 3 4 5  0.7 0.8 0.9 1.0  1.0656 1.0969 1.1274 1.1571  0.3134 0.3047 0.2967  1.0969 1.1274 1.1571  Hence Euler's method with step size h = 0.1 gives the estimate X(1) = 1.1571. 4  Euler's method for the solution of the diﬀerential equation  dx = f(t, x) is dt  Xn+1 = Xn + hFn = Xn + hf(tn , Xn ) Applying this to the equation yields  4−t dx = with x(0) = 1 and a step size of h = 0.05 dt t+x  x0 = x(0.0) = 1 4 − t0 4−0 = 1.2000 = 1 + 0.05 t0 + x0 0+1 4 − t1 4 − 0.05 = 1.2000 + 0.05 X2 = X1 + hf(t1 , X1 ) = X1 + h = 1.3580 t1 + X1 0.05 + 1.2000 X1 = x0 + hf(t0 , x0 ) = x0 + h  (Note that tn = t0 + nh = 0.0 + 0.05n.) X3 , X4 , . . . , X10 may be computed in similar fashion. It is usually easier to set out numerical solutions in a systematic tabular form such as the following: tn 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50  Xn 1.0000 1.2000 1.3580 1.4917 1.6090 1.7140 1.8095 1.8972 1.9784 2.0541 2.1250  f(tn , Xn ) 4.0000 3.1600 2.6749 2.3451 2.1006 1.9093 1.7540 1.6242 1.5136 1.4177  Xn + hf(tn , Xn ) 1.2000 1.3580 1.4917 1.6090 1.7140 1.8095 1.8972 1.9784 2.0541 2.1250  Hence Euler's method with step size h = 0.05 gives the estimate X(0.5) = 2.1250. 5  Figure 2.1 shows a suitable pseudocode program for computing the estimates  Xa (2) and Xb (2) . Figure 2.2 shows a Pascal implementation of the pseudocode program. c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  89  procedure deriv (t, x → f) f ← x∗ t/(t∗ t + 2) endprocedure t start ← 1 x start ← 2 t end ← 2 write (vdu, "Enter step size") read (keyboard, h) write (printer, t start, x start ) t ← t start x ← x start repeat deriv (t, x → f) t←t+h x ← x + h∗ f write (printer, t,x) until t >= t end Figure 2.1: Pseudocode algorithm for Exercise 5 Using this program the results Xa (2) = 2.811489 and Xb (2) = 2.819944 were obtained. Using the method described in Section 2.3.6, the error in Xb (2) will be approximately equal to Xb (2) − Xa (2) = −0.008455 and so the best estimate of X(2) is 2.819944 + 0.008455 = 2.828399. The desired error bound is 0.1% of this value, 0.0028 approximately. Since Euler's method is a ﬁrst-order method, the error in the estimate of X(2) varies like h ; so, to achieve an error of 0.0028, a step size of no more than (0.0028/0.008455) × 0.05 = 0.0166 is required. We will choose a sensible step size which is less than this, say h = 0.0125. This yields an estimate X(2) = 2.826304. The exact solution of the diﬀerential equation may be obtained by separation:    xt dx t dt dx 2 1 = 2 ⇒ = ⇒ ln x = ln(t + 2) + C ⇒ x = ±D t2 + 2 2 dt t +2 x t2 + 2  √ t2 + 2 x(1) = 2 ⇒ 2 = ± 3D ⇒ x = 2 3 √ Hence x(2) = 2 2 = 2.828427 and the true errors in Xa (2), Xb (2) and the ﬁnal estimate of X(2) are 0.016938, 0.008483 and 0.002123 respectively. The estimate, X(2), derived using the step size h = 0.0125 is comfortably within the 0.1% error requirement. c Pearson Education Limited 2011   90  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  var t start, x start, t end, h,x,t,f:real; procedure deriv (t,x:real;var f:real); begin f := x∗ t/(t∗ t + 2) end; begin t start := 1; x start := 2; t end := 2; write (  Enter step size ==>  ); readln (h); writeln (t start : 5 : 2, x start : 10 : 6) ; t := t start ; x := x start ; repeat deriv(t,x,f); t : = t + h; x : = x + h∗ f ; writeln (t:5:2, x:10:6); until t > = t end ; end. Figure 2.2: Pascal program for Exercise 5 6  The programs shown in Figures 2.1 and 2.2 may readily be modiﬁed to solve  this problem. Estimates Xa (2) = 1.573065 and Xb (2) = 1.558541 should be obtained. Using the method described in Section 2.3.6, the error in Xb (2) will be approximately equal to Xb (2) − Xa (2) = −0.014524 and so the best estimate of X(2) is 1.558541 − 0.014524 = 1.544017. The desired error bound is 0.2% of this value, 0.0031 approximately. Since Euler's method is a ﬁrst-order method, the error in the estimate of X(2) varies like h so, to achieve an error of 0.0031, a step size of no more than (0.0031/0.014524) × 0.05 = 0.0107 is required. We will choose a sensible step size which is less than this, say h = 0.01. This yields an estimate X(2) = 1.547462. The exact solution of the diﬀerential equation may be obtained by separation:    dt 1 dx = ⇒ xdx = ⇒ 12 x2 = ln t + C ⇒ x = ± 2(ln t + C) dt xt t √ x(1) = 1 ⇒ 1 = 2C ⇒ x(t) = 2 ln t + 1 c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Hence x(2) =  √  91  2 ln 2 + 1 = 1.544764 and the true errors in Xa (2), Xb (2) and  the ﬁnal estimate of X(2) are −0.028301, −0.013777 and − 0.002698 respectively. The estimate, X(2), derived using the step size h = 0.01 is comfortably within the 0.2% error requirement.  7  The programs shown in Figures 2.1 and 2.2 may readily be modiﬁed to solve  this problem. Estimates Xa (1.5) = 2.241257 and Xb (1.5) = 2.206232 should be obtained. Using the method described in Section 2.3.6, the error in Xb (1.5) will be approximately equal to Xb (1.5) − Xa (1.5) = −0.035025 and so the best estimate of X(1.5) is 2.206232 − 0.035025 = 2.171207. The desired error bound is 0.25% of this value, 0.0054 approximately. Since Euler's method is a ﬁrst-order method, the error in the estimate of X(1.5) varies like h; so, to achieve an error of 0.0054, a step size of no more than (0.0054/0.035025) × 0.025 = 0.0039 is required. If we choose h = 0.04, this yields an estimate X(1.5) = 2.183610. The exact solution of the diﬀerential equation may be obtained by separation:   1 dx = ⇒ ln xdx = dt ⇒ x ln x − x = t + C ct ln x x(1) = 1.2 ⇒ 1.2 ln 1.2 − 1.2 = 1 + C ⇒ C = −1.981214 ⇒ x ln x − x = t − 1.981214  Hence, by any non-linear equation solving method (e.g. Newton–Raphson), we may obtain x(1.5) = 2.179817 and the true errors in Xa (1.5), Xb (1.5) and the ﬁnal estimate of X(1.5) are 0.061440, 0.026415 and 0.003793 respectively. The estimate, X(1.5), derived using the step size h = 0.04 is comfortably within the 0.25% error requirement.  c Pearson Education Limited 2011   92  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Exercises 2.3.9 8  The starting process, using the second-order predictor–corrector method, is X̂1 = x0 + hf(t0 , x0 )   X1 = x0 + 12 h f(t0 , x0 ) + f(t1 , X̂1 )  and the second-order Adams–Bashforth method is Xn+1 = Xn + 12 h (3f(tn , Xn ) − f(tn−1 , Xn−1 ))  8(a)  Applying this method to the problem  h = 0.1, we have  dx = x2 sin t − x, x(0) = 0.2 with dt  X̂1 = x0 + hf(t0 , x0 ) = 0.2 + 0.1 × (0.22 sin 0 − 0.2) = 0.1800   X1 = x0 + 12 h f(t0 , x0 ) + f(t1 , X̂1 ) = 0.2 + 12 0.1 × (0.22 sin 0 − 0.2 + 0.182 sin 0.1 − 0.18) = 0.1812 X2 = X1 + 12 h (3f(t1 , X1 ) − f(t0 , x0 ))   = 0.1812 + 12 0.1 × 3(0.18122 sin 0.1 − 0.1812) − (0.22 sin 0 − 0.2) = 0.1645 X3 , X4 and X5 are obtained as X2 . The computation is most eﬃciently set out as a table. n 0 1 2 3 4 5  tn 0.0 0.1 0.2 0.3 0.4 0.5  Xn 0.2000 0.1812 0.1645 0.1495 0.1360 0.1238  f(tn , Xn ) −0.2000 −0.1779 −0.1591 −0.1429 −0.1288  − f(tn−1 , Xn−1 )) (use predictor–corrector) −0.016685 −0.014970 −0.013480 −0.012175  1 2 h(3f(tn , Xn )  Hence X(0.5) = 0.1238.  c Pearson Education Limited 2011   Xn+1 0.1812 0.1645 0.1495 0.1360 0.1238  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  8(b)  Applying this method to the problem  0.1, we have  93  dx = x2 etx , x(0.5) = 0.5 with h = dt  X̂1 = x0 + hf(t0 , x0 ) = 0.5 + 0.1 × 0.52 e0.5×0.5 = 0.5321   X1 = x0 + 12 h f(t0 , x0 ) + f(t1 , X̂1 ) = 0.5 + 12 0.1 × (0.52 e0.5×0.5 + 0.53212 e0.6×0.5321 ) = 0.5355 X2 = X1 + 12 h (3f(t1 , X1 ) − f(t0 , x0 ))   = 0.5355 + 12 0.1 × 3 × 0.53552 e0.6×0.5355 − 0.52 e0.5×0.5 = 0.5788 X3 , X4 , X5 , X6 and X7 are obtained as X2 . The computation is most eﬃciently set out as a table. n 0 1 2 3 4 5 6 7  tn 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2  Xn 0.5000 0.5355 0.5788 0.6344 0.7095 0.8191 0.9998 1.3740  f(tn , Xn ) 0.3210 0.3955 0.5024 0.6685 0.9534 1.5221 3.0021  − f(tn−1 , Xn−1 )) (use predictor–corrector) 0.043275 0.055585 0.075155 0.109585 0.180645 0.374210  1 2 h(3f(tn , Xn )  Xn+1 0.5355 0.5788 0.6344 0.7095 0.8191 0.9998 1.3740  Hence X(1.2) = 1.3740.  9  The starting process, using the second-order predictor–corrector method, is X̂1 = x0 + hf(t0 , x0 )   X1 = x0 + 12 h f(t0 , x0 ) + f(t1 , X̂1 ) X̂2 = X1 + hf(t1 , X1 )   1 X2 = X1 + 2 h f(t1 , X1 ) + f(t2 , X̂2 )  and the third-order Adams–Bashforth method is Xn+1 = Xn +  1 12  h (23f(tn , Xn ) − 16f(tn−1 , Xn−1 ) + 5f(tn−2 , Xn−2 ))  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  94  Applying this method to the problem have  dx  2 = x + 2t, x(0) = 1 with h = 0.1, we dt   X̂1 = x0 + hf(t0 , x0 ) = 1.0 + 0.1 × 12 + 2 × 0 = 1.100   X1 = x0 + 12 h f(t0 , x0 ) + f(t1 , X̂1 )    = 1.0 + 12 0.1 × 12 + 2 × 0 + 1.12 + 2 × 0.1 = 1.1094  X̂2 = X1 + hf(t1 , X1 ) = 1.1094 + 0.1 × 1.10942 + 2 × 0.1 = 1.2290   X2 = X1 + 12 h f(t1 , X1 ) + f(t2 , X̂2 )    = 1.1094 + 12 0.1 × 1.10942 + 2 × 0.1 + 1.22902 + 2 × 0.2 = 1.2383 h (23f(t2 , X2 ) − 16f(t1 , X1 ) + 5f(t0 , x0 ))    1 = 1.2383 + 12 0.1 × 23 1.23832 + 2 × 0.2 − 16 1.10942 + 2 × 0.1   + 5 1.02 + 2 × 0 = 1.3870  X3 = X2 +  1 12  X4 and X5 are obtained as X3 . The computation is most eﬃciently set out as a table. n  tn  Xn  f(tn , Xn )  h(23f(tn , Xn ) − 16f(tn−1 , Xn−1 )  Xn+1 1.1094  0  0.0  1.0000  1.0000  + 5f(tn−2 , Xn−2 ))/12 (use predictor–corrector)  0  0.1  1.1094  1.1961  (use predictor–corrector)  1.2383  2 3  0.2 0.3  1.2383 1.3870  1.3905 1.5886  0.1487 0.1689  1.3870 1.5559  4  0.4  1.5559  1.7947  0.1901  1.7460  5  0.5  1.7460  Hence X(0.5) = 1.7460.  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 10  95  The second-order predictor–corrector method is X̂n+1 = Xn + hf(tn , Xn ) Xn+1 = Xn + 12 h(f(tn , Xn ) + f(tn+1 , X̂n+1 ))  10(a)  Applying this method to the problem  h = 0.05, we have  dx = (2t + x) sin 2t, x(0) = 0.5 with dt  X̂1 = x0 + hf(t0 , x0 ) = 0.5 + 0.05 × (2 × 0 + 0.5) sin 0 = 0.5 X1 = x0 + 12 h(f(t0 , x0 ) + f(t1 , X1 )) = 0.5 + 12 0.05 × ((2 × 0 + 0.5) sin 0 + (2 × 0.05 + 0.5) sin(2 × 0.05)) = 0.5015 X2 to X10 are obtained as X1 . The computation is most eﬃciently set out as a table. n 0 1 2 3 4 5 6 7 8 9 10  tn 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50  Xn 0.5000 0.5015 0.5065 0.5160 0.5311 0.5527 0.5820 0.6198 0.6673 0.7254 0.7948  f(tn , Xn ) 0.0000 0.0150 0.0497 0.1034 0.1752 0.2637 0.3670 0.4832 0.6098 0.7442  X̂n+1 0.5000 0.5045 0.5135 0.5281 0.5492 0.5780 0.6153 0.6623 0.7199 0.7890  f(tn+1 , X̂n+1 ) 0.0599 0.1400 0.2404 0.3614 0.5030 0.6651 0.8474 1.0490 1.2689 1.5054  Xn+1 0.5015 0.5065 0.5160 0.5311 0.5527 0.5820 0.6198 0.6673 0.7254 0.7948  Hence X(0.5) = 0.7948.  10(b)  Applying this method to the problem  1+x dx =− , x(0) = −2 with dt sin(t + 1)  h = 0.1, we have 1−2 = −1.8812 X̂1 = x0 + hf(t0 , x0 ) = −2 + 0.1 × − sin(0 + 1)   X1 = x0 + 12 h f(t0 , x0 ) + f(t1 , X̂1 ) = −2 + 12 0.1 × −  1 − 1.8812 1−2 − sin(0 + 1) sin(0.1 + 1)  = −1.8911  X2 to X10 are obtained as X1 . The computation is most eﬃciently set out as a table. c Pearson Education Limited 2011   96 n 0 1 2 3 4 5 6 7 8 9 10  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition tn 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0  Xn −2.0000 −1.8911 −1.7987 −1.7189 −1.6489 −1.5867 −1.5309 −1.4803 −1.4339 −1.3910 −1.3511  f(tn , Xn ) −1.3072 −1.2343 −1.1802 −1.1416 −1.1162 −1.1028 −1.1005 −1.1092 −1.1295 −1.1624  X̂n+1 −1.8812 −1.7912 −1.7130 −1.6443 −1.5830 −1.5279 −1.4778 −1.4318 −1.3893 1.3497  f(tn+1 , X̂n+1 ) 0.9887 0.8488 0.7400 0.6538 0.5845 0.5281 0.4818 0.4434 0.4114 0.3846  Hence X(1.0) = −1.3511. 11  Taylor's theorem states that  df h3 d 3 f h4 d 4 f h2 d 2 f (t) + (t) + (t) + K (t) + dt 2! dt2 3! dt3 4! dt4 dx dx (t − h) and (t − 2h) yields Applying this to dt dt f(t + h) = f(t) + h  dx dx d2 x h2 d 3 x (t − h) = (t) − h 2 (t) + (t) + O(h3 ) dt dt dt 2! dt3 dx dx d2 x 4h2 d3 x (t − 2h) = (t) − 2h 2 (t) + (t) + O(h3 ) dt dt dt 2! dt3 Multiplying the ﬁrst equation by 2 and subtracting the second yields dx dx (t − h) − (t − 2h) = dt dt d3 x dx that is, h2 3 (t) = −2 (t − h) + dt dt 2  d3 x dx (t) − h2 3 (t) + O(h3 ) dt dt dx dx (t − 2h) + (t) + O(h3 ) dt dt  Multiplying the ﬁrst equation by 4 and subtracting the second yields dx dx d2 x dx (t − h) − (t − 2h) = 3 (t) − 2h 2 (t) + O(h3 ) dt dt dt dt dx dx dx d2 x (t − 2h) + 3 (t) + O(h3 ) that is, 2h 2 (t) = −4 (t − h) + dt dt dt dt 4  Now Taylor's theorem yields h2 d 2 x h3 d 3 x dx + (t) + O(h3 ) x(t + h) = x(t) + h (t) + 2 3 dt 2! dt 3! dt c Pearson Education Limited 2011   Xn+1 −1.8911 −1.7987 −1.7189 −1.6489 −1.5867 −1.5309 −1.4803 −1.4339 −1.3910 1.3511  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition Hence, substituting for h  3 d2 x 2d x (t) and h (t) yields dt2 dt3  h dx dx dx dx (t) + −4 (t − h) + (t − 2h) + 3 (t) + O(h3 ) dt 4 dt dt dt dx dx dx h −2 (t − h) + (t − 2h) + (t) + O(h3 ) + O(h3 ) + 6 dt dt dt dx dx dx h 23 (t) − 16 (t − h) + 5 (t − 2h) + O(h3 ) = x(t) + 12 dt dt dt  x(t + h) = x(t) + h  12  Taylor's theorem states that f(t + h) = f(t) + h  Applying this to  h3 d 3 f h4 d 4 f df h2 d 2 f (t) + (t) + (t) + K (t) + dt 2! dt2 3! dt3 4! dt4  dx dx (t + h) and (t − h) yields dt dt  dx (t + h) = dt dx (t − h) = dt  dx d2 x (t) + h 2 (t) + dt dt dx d2 x (t) − h 2 (t) + dt dt  h2 d 3 x (t) + O(h3 ) 2! dt3 h2 d 3 x (t) + O(h3 ) 2! dt3  Summing the two equations yields d3 x dx dx dx (t + h) + (t − h) = 2 (t) + h2 3 (t) + O(h3 ) dt dt dt dt 3 d x dx dx dx (t + h) + (t − h) − 2 (t) + O(h3 ) that is, h2 3 (t) = dt dt dt dt Subtracting the second from the ﬁrst yields dx d2 x dx (t + h) − (t − h) = 2h 2 (t) + O(h3 ) dt dt dt 2 dx d x dx that is, 2h 2 (t) = (t + h) − (t − h) + O(h3 ) dt dt dt Now Taylor's theorem gives h2 d 2 x h3 d 3 x dx + (t) + O(h4 ) x(t + h) = x(t) + h (t) + 2 3 dt 2! dt 3! dt c Pearson Education Limited 2011   97  98  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Hence, substituting for h  3 d2 x 2d x (t) and h (t), we have dt2 dt3  dx h dx dx (t) + (t + h) + (t − h) + O(h3 ) dt 4 dt dt dx dx h dx (t + h) + (t − h) − 2 (t) + O(h3 ) + O(h4 ) + 6 dt dt dt dx dx dx h 5 (t + h) + 8 (t) − (t − h) + O(h4 ) = x(t) + 12 dt dt dt dx h dx dx 5 + O(h4 ) = xn + +8 − 12 dtn+1 dtn dtn−1  x(t + h) = x(t) + h  that is, xn+1  13  Taylor's theorem states that f(t + h) = f(t) + h  h2 d 2 f df h3 d 3 f h4 d 4 f (t) + (t) + (t) + (t) + K dt 2! dt2 3! dt3 4! dt4  Applying this to x(t − h) and  dx (t − h) yields dt  h2 d 2 x dx h3 d 3 x (t) + (t) − (t) + O(h4 ) dt 2! dt2 3! dt3 dx dx d2 x h2 d 3 x (t − h) = (t) − h 2 (t) + (t) + O(h3 ) dt dt dt 2! dt3  x(t − h) = x(t) − h  Multiplying the ﬁrst equation by 2, the second equation by h and adding yields dx dx h3 d3 x (t) + O(h4 ) (t − h) = 2x(t) − h (t) + dt dt 6 dt3 dx dx h3 d 3 x (t) = 2x(t − h) + h (t − h) − 2x(t) + h (t) + O(h4 ) that is, 3 6 dt dt dt 2x(t − h) + h  Multiplying the ﬁrst equation by 3, the second equation by h and adding yields dx h2 d2 x dx (t − h) = 3x(t) − 2h (t) + (t) + O(h4 ) dt dt 2 dt2 dx h2 d 2 x dx (t) = 3x(t − h) + h (t − h) − 3x(t) + 2h (t) + O(h4 ) that is, 2 2 dt dt dt 3x(t − h) + h  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  99  Now Taylor's theorem gives  x(t + h) = x(t) + h  Hence, substituting for h2  dx h2 d 2 x h3 d 3 x + (t) + O(h4 ) (t) + dt 2! dt2 3! dt3  3 d2 x 3d x (t) and h (t), we have dt2 dt3  dx dx dx (t) + 3x(t − h) + h (t − h) − 3x(t) + 2h (t) + O(h4 ) dt dt dt dx dx + 2x(t − h) + h (t − h) − 2x(t) + h (t) + O(h4 ) + O(h4 ) dt dt dx dx = −4x(t) + 5x(t − h) + 4h (t) + 2h (t − h) + O(h4 ) dt dt  x(t + h) = x(t) + h  that is, xn+1 = −4xn + 5xn−1 + 2h 2  dx dx + dtn dtn−1  + O(h4 )  This gives rise to the approximate scheme Xn+1 = −4Xn +5Xn−1 +2h(2Fn +Fn−1 ) which may equally be written as Xn+1 = 5[Xn−1 +2hFn ]−4[Xn + 12 h(3Fn −Fn−1 )]; in other words, this scheme gives an approximation for Xn+1 which is 5 times the central diﬀerence approximation minus 4 times the second-order Adams–Bashforth approximation. Because of the inclusion of the central diﬀerence approximation, this scheme will be unstable whenever the central diﬀerence scheme is.  14  The predictor–corrector scheme speciﬁed is X̂n+1 = Xn + 12 h(3Fn − Fn−1 ) = Xn + 12 h(3f(tn , Xn ) − f(tn−1 , Xn−1 )) Xn+1 = Xn +  1 12  h(5Fn+1 + 8Fn − Fn−1 ) = Xn +  1 12  h(5f(tn+1 , X̂n+1 )  + 8f(tn , Xn ) − f(tn−1 , Xn−1 )) This is obviously not self-starting and requires that one initial step is taken using a dx self-starting method. For the problem = x2 + t2 , x(0.3) = 0.1 with a step size dt c Pearson Education Limited 2011   100  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  of h = 0.05, using the fourth-order Runge–Kutta method for the initial step we obtain c1 = hf(t0 , x0 ) = h(x20 + t20 ) = 0.05 × (0.12 + 0.32 ) = 0.0050   c2 = hf(t0 + 12 h, x0 + 12 c1 ) = 0.05 × (0.1 + 12 0.0050)2 + (0.3 + 12 0.05)2 = 0.0058    c3 = hf(t0 + 12 h, x0 + 12 c2 ) = 0.05 × (0.1 + 12 0.0058)2 + (0.3 + 12 0.05)2 = 0.0058    c4 = hf(t0 + h, x0 + c3 ) = 0.05 × (0.1 + 0.0058)2 + (0.3 + 0.05)2 = 0.0067 X1 = x0 + 16 (c1 + 2c2 + 2c3 + c4 ) = 0.1 + 16 (0.0050 + 2 × 0.0058 + 2 × 0.0058 + 0.0067) = 0.1058 Now we can say X̂2 = X1 + 12 h (3f(t1 , X1 ) − f(t0 , x0 ))   = 0.1058 + 12 0.05 × 3(0.10582 + 0.352 ) − (0.12 + 0.32 ) = 0.1133   1 X2 = X1 + 12 h 5f(t2 , X̂2 ) + 8f(t1 , X1 ) − f(t0 , x0 )   1 = 0.1058 + 12 0.05 × 5(0.11332 + 0.42 ) + 8(0.10582 + 0.352 ) − (0.12 + 0.32 ) = 0.1134 Computing X3 and X4 in a similar manner and setting the computation out in tabular fashion we obtain: n  tn  Xn  f(tn , Xn )  X̂n+1  f(tn+1 , X̂n+1 )  0  0.30  0.1000  0.1000  1  0.35  0.1058  0.1337  0.1133  0.1728  0.1134  2  0.40  0.1134  0.1729  0.1230  0.2176  0.1231  3  0.45  0.1231  0.2177  0.1351  0.2683  0.1352  4  0.50  0.1352  (done by Runge–Kutta method)  Hence X(0.5) = 0.1352.  c Pearson Education Limited 2011   Xn+1 0.1058  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  101  15  The fourth-order Runge–Kutta method for the solution of the diﬀerential dx = f(t, x) using a stepsize of h is given by equation dt c1 = hf(tn , Xn ) c2 = hf(tn + 12 h, Xn + 12 c1 ) c3 = hf(tn + 12 h, Xn + 12 c2 ) c4 = hf(tn + h, Xn + c3 ) Xn+1 = Xn + 16 (c1 + 2c2 + 2c3 + c4 )  15(a)  To solve the equation  h = 0.15, we write  dx = x + t + xt, x(0) = 1 , using a stepsize of dt  c1 = hf(t0 , x0 ) = 0.15 × (1 + 0 + 1 × 0) = 0.1500 c2 = hf(t0 + 12 h, x0 + 12 c1 ) = 0.15 × ((1 + 12 0.1500) + (0 + 12 0.15) + (1 + 12 0.1500) × (0 + 12 0.15)) = 0.1846 c3 = hf(t0 + 12 h, x0 + 12 c2 ) = 0.15 × ((1 + 12 0.1846) + (0 + 12 0.15) + (1 + 12 0.1846) × (0 + 12 0.15)) = 0.1874 c4 = hf(t0 + h, x0 + c3 ) = 0.15 × ((1 + 0.1874) + (0 + 0.15) + (1 + 0.1874) × (0 + 0.15)) = 0.2273 X1 = x0 + 16 (c1 + 2c2 + 2c3 + c4 ) = 1 + 16 (0.1500 + 2 × 0.1846 + 2 × 0.1874 + 0.2273) = 1.1869  X2 , X3 , X4 and X5 may be computed in a similar manner. computation out in tabular fashion we obtain: c Pearson Education Limited 2011   Setting the  102  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  n 0 1 2 3 4 5  tn 0.00 0.15 0.30 0.45 0.60 0.75  Xn 1.0000 1.1869 1.4630 1.8603 2.4266 3.2345  c1 0.1500 0.2272 0.3303 0.4721 0.6724  c2 0.1846 0.2727 0.3921 0.5583 0.7954  c3 0.1874 0.2769 0.3984 0.5681 0.8109  c4 0.2273 0.3304 0.4724 0.6728 0.9623  Xn+1 1.1869 1.4630 1.8603 2.4266 3.2345  Hence X(0.75) = 3.2345.  15(b)  To solve the equation  h = 0.1 we write c1 = hf(t0 , x0 ) = 0.1 ×  dx 1 = , x(1) = 2 , using a stepsize of dt x+t  1 = 0.0333 2+1  c2 = hf(t0 + 12 h, x0 + 12 c1 ) = 0.1 × c3 = hf(t0 + 12 h, x0 + 12 c2 ) = 0.1 × c4 = hf(t0 + h, x0 + c3 ) = 0.1 ×  1 (2 +  1 2 0.0333)  (2 +  1 2 0.0326)  + (1 + 12 0.1)  1 + (1 + 12 0.1)  = 0.0326 = 0.0326  1 = 0.0319 (2 + 0.0326) + (1 + 0.1)  X1 = x0 + 16 (c1 + 2c2 + 2c3 + c4 ) = 2 + 16 (0.0333 + 2 × 0.0326 + 2 × 0.0326 + 0.0319) = 2.0326 X2 , X3 , . . . , X10 may be computed in a similar manner. Setting the computation out in tabular fashion we obtain: n 0 1 2 3 4 5 6 7 8 9 10  tn 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0  Xn 2.0000 2.0326 2.0639 2.0939 2.1228 2.1507 2.1777 2.2037 2.2289 2.2534 2.2771  c1 0.0333 0.0319 0.0306 0.0295 0.0284 0.0274 0.0265 0.0256 0.0248 0.0241  c2 0.0326 0.0313 0.0300 0.0289 0.0279 0.0269 0.0260 0.0252 0.0244 0.0237  c3 0.0326 0.0313 0.0300 0.0289 0.0279 0.0269 0.0260 0.0252 0.0244 0.0237  Hence X(2) = 2.2771.  c Pearson Education Limited 2011   c4 0.0319 0.0306 0.0295 0.0284 0.0274 0.0265 0.0256 0.0248 0.0241 0.0234  Xn+1 2.0326 2.0639 2.0939 2.1228 2.1507 2.1777 2.2037 2.2289 2.2534 2.2771  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  16  In this exercise the diﬀerential equation problem  is solved using a variety of methods.  16(a)  103  3 dx = x2 + t 2 , x(0) = −1 , dt  Figure 2.3 shows a pseudocode algorithm for solving the equation  using the second-order Adams–Bashforth method with a second-order predictor– corrector starting step and Figure 2.4 shows a Pascal program derived from it.  procedure deriv (t, x → f) f ← x∗ x+sqrt (t)∗ t endprocedure t start ← 0 x start ← −1 t end ← 2 write (vdu, "Enter step size") read (keyboard, h) write (printer, t start , x start) deriv (t start, x start → f) x hat ← x start + h∗ f deriv(t start + h, x hat → f hat) t ← t start + h x ← x start + h∗ (f + f hat)/2 write (printer, t,x) f n minus one ← f repeat deriv (t, x → f) t←t+h x ← x + h∗ (3∗ f − f n minus one)/2 f n minus one ← f write (printer, t,x) until t >= t end Figure 2.3: Pseudocode algorithm for Exercise 16(a) c Pearson Education Limited 2011   104  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  var x start, x hat, x,t start, t end; t:real; h,f,f hat, f n minus one:real; procedure deriv (t,x:real;var f:real); begin f := x*x+sqrt(t)*t end; begin t start := 0; x start := -1; t end := 2; write(  Enter step size ==> ); readln(h); writeln(t start:10:3, x start:10:6); deriv(t start, x start, f); x hat := x start + h*f; deriv(t start + h, x hat, f hat); t :=t start + h; x :=x start + h *(f+f hat)/2; writeln(t:10:3, x:10:6); f n minus one := f; repeat deriv(t,x,f); t := t + h; x := x + h*(3*f-f n minus one)/2; f n minus one := f; writeln(t:10:3, x:10:6); until t >= t end ; end. Figure 2.4: Pascal program for Exercise 16(a)  Using this program with h = 0.2 gives X(2) = 2.242408 and, with h = 0.1, X(2) = 2.613104. The method of Richardson extrapolation given in Section 2.3.6 gives the estimated error in the second of these as (2.613104 − 2.242408)/3 = 0.123565. For 3 decimal place accuracy in the ﬁnal estimate we need error ≤ 0.0005; in other words, the error must be reduced by a factor of 0.123565/0.0005 = 247.13. Since Adams–Bashforth is a second-order method, the required step length will √ be 0.1/ 247.13 = 0.0064. Rounding this down to a suitable size suggests that a step size of h = 0.005 will give a solution accurate to 3 decimal places. In fact the program of Figure 2.4 yields, with h = 0.005, X(2) = 2.897195. With c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  105  h = 0.0025 it gives X(2) = 2.898175. Richardson extrapolation predicts the error in the h = 0.0025 solution as 0.000327 and therefore that in the h = 0.005 as 0.001308. The required accuracy was therefore not achieved using h = 0.005 but was achieved with h = 0.0025. 16(b) Figure 2.5 shows a pseudocode algorithm for solving the equation using the second-order predictor–corrector method and Figure 2.6 shows a Pascal program derived from it. procedure deriv (t, x → f) f ← x∗ x+sqrt (t)∗ t endprocedure t start ← 0 x start ← −1 t end ← 2 write (vdu, "Enter step size") read (keyboard, h) write (printer, t start , x start) t ← t start x ← x start repeat deriv (t, x → f) x hat ← x + h∗ f derive (t + h, x hat → f hat) t←t+h x ← x + h∗ (f + f hat)/2 write (printer, t,x) until t >= t end Figure 2.5: Pseudcode algorithm for Exercise 2.16(b) Using this program with h = 0.2 gives X(2) = 2.788158 and, with h = 0.1, X(2) = 2.863456. The method of Richardson extrapolation given in Section 2.3.6 gives the estimated error in the second of these as (2.863456 − 2.788158)/3 = 0.025099. For 3 decimal place accuracy in the ﬁnal estimate we need error ≤ 0.0005; in other words, the error must be reduced by a factor of 0.025099/0.0005 = 50.20. Since the second-order predictor–corrector method is being used, the required step size √ will be 0.1/ 50.20 = 0.014. Rounding this down to a suitable size suggests that a step size of h = 0.0125 will give a solution accurate to 3 decimal places. In fact the program of Figure 2.6 yields, with h = 0.0125, X(2) = 2.897876. With c Pearson Education Limited 2011   106  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  h = 0.00625 it gives X(2) = 2.898349. Richardson extrapolation predicts the error in the h = 0.00625 solution as 0.000158 and therefore that in the h = 0.0125 as 0.000632. The required accuracy was therefore not quite achieved using h = 0.0125 but was achieved with h = 0.00625.  var x start,x hat,x,t start,t end, t:real; h,f,f hat:real; procedure deriv(t,x:real;var f:real); begin f := x*x+sqrt(t)*t end; begin t start := 0; x start := -1; t end := 2; write('Enter step size ==> '); readln(h); writeln(t start:10:3,x start:10:6); t := t start; x := x start; repeat deriv(t,x,f); x hat := x + h*f; deriv(t + h,x hat,f hat); t := t + h; x := x + h*(f + f hat)/2; writeln (t:10:3,x:10:6); until t >= t end; end. Figure 2.6: Pascal program for Exercise 16(b)  16(c)  Figure 2.7 shows a pseudocode algorithm for solving the equation using  the fourth-order Runge–Kutta method and Figure 2.8 shows a Pascal program derived from it. Using this program with h = 0.4 gives X(2) = 2.884046 and, with h = 0.2, X(2) = 2.897402. The method of Richardson extrapolation given  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  107  in Section 2.3.6 gives the estimated error in the second of these as (2.897402 − 2.884046)/15 = 0.000890. For 5 decimal place accuracy in the ﬁnal estimate we need error ≤ 0.000005; in other words, the error must be reduced by a factor of 0.000890/0.000005 = 178. Since Runge–Kutta is a fourth-order method the required step size will be 0.2/(178)1/4 = 0.0547. Rounding this down to a suitable size suggests that a step size of h = 0.05 will give a solution accurate to 5 decimal places. In fact the program of Figure 2.8 yields, with h = 0.05, X(2) = 2.89850975. With h = 0.025 it gives X(2) = 2.89850824. Richardson extrapolation predicts the error  procedure deriv (t,x → f) f → x∗ x + sqrt(t)∗ t endprocedure t start ← 0 x start ← −1 t end ← 2 write(vdu, "Enter step size") read(keyboard, h) write (printer,t start,x start) t ← t start x ← x start repeat deriv(t, x, → f) c1 ← h∗ f deriv(t + h/2, x + c1/2 → f) c2 ← h∗ f deriv(t + h/2, x + c2/2 → f) c3 ← h∗ f deriv(t + h, x + c3 → f) c4 ← h∗ f t←t+h x ← x + (c1 + 2∗ c2 + 2∗ c3 + c4)/6 write(printer,t,x) until t >= t end Figure 2.7: Pseudocode algorithm for Exercise 16(c) c Pearson Education Limited 2011   108  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  {Program for exercise 2.16c} var x start,x,t start,t end,t:real; h,f,c1,c2,c3,c4:real; procedure deriv (t,x:real;var f:real); begin f := x ∗ x+sqrt(t) ∗ t end; begin t start := 0; x start := -1; t end := 2; write('Enter step size ==> '); readln(h); writeln(t start:10:3,x start:10:6); t := t start; x := x start; repeat deriv(t,x,f); c1 := h*f; deriv(t + h/2,x + c1/2, f); c2 := h*f; deriv(t + h/2,x + c2/2, f); c3 := h*f; deriv(t + h,x + c3, f); c4 := h*f; t := t + h; x := x + (c1 + 2*c2 + 2*c3 + c4)/6; writeln(t:10:3,x:10:6); until t >= t end; end. Figure 2.8: Pascal program for Exercise 16(c) in the h = 0.025 solution as 0.000000101 and therefore that in the h = 0.05 as 0.00000161. The required accuracy was therefore achieved using h = 0.05.  17  The pseudocode algorithm shown in Figure 2.7 and the Pascal program in  Figure 2.8 may easily be modiﬁed to solve this problem. With a step size of h = 0.5 the estimate X(3) = 1.466489 is obtained, whilst with a step size of h = 0.25, the estimate is X(3) = 1.466476. Richardson extrapolation suggests that the step c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  109  size of h = 0.25 gives an error of 0.00000087 which is comfortably within the 0.000005 range, which 5 decimal place accuracy requires. Hence, X(3) = 1.46648 to 5 decimal places. In fact, of course, the analytic solution of the problem is e and so x(3) = 1.466474. 1−t e +e−1  Exercises 2.4.3 18  In each part of this question, the technique is to introduce new variables to  represent each derivative of the dependent variable up to one less than the order of the equation. This can be done by inspection. 18(a) dx = v, dt dv = 4xt − 6(x2 − t)v, dt  x(0) = 1 v(0) = 2  18(b) dx = v, dt dv = −4(x2 − t2 ), dt  x(1) = 2 v(1) = 0.5  18(c) dx = v, dt dv = − sin v − 4x, dt  x(0) = 0 v(0) = 0  18(d) dx = v, dt dv = w, dt dw = e2t + x2 t − 6et v − tw, dt  x(0) = 1 v(0) = 2 w(0) = 0  18(e) dx = v, dt dv = w, dt dw = sin t − tw − x2 , dt c Pearson Education Limited 2011   x(1) = 1 v(1) = 0 w(1) = −2  110  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  18(f) dx = v, dt dv = w, dt dw = (tw + t2 x2 )2 , dt  x(2) = 0 v(2) = 0 w(2) = 2  18(g) dx dt dv dt dw dt du dt  = v,  x(0) = 0  = w,  v(0) = 0  = u,  w(0) = 4  = ln t − x2 − xw,  u(0) = −3  18(h) dx dt dv dt dw dt du dt 19  = v,  x(0) = a  = w,  v(0) = 0  = u,  w(0) = b  = t2 + 4t − 5 +  √  xt − v − (v − 1)tu,  u(0) = 0  First, we recast the equation as a pair of coupled ﬁrst-order equations in the  form dx = f1 (t, x, v), dt dv = f2 (t, x, v), dt  x(0) = x0 v(0) = v0  This yields dx = v, dt dv = sin t − x − x2 v, dt c Pearson Education Limited 2011   x(0) = 0 v(0) = 1  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  111  Now, applying Euler's method to the two equations we have  X1 = x0 + h1 f(t0 , x0 , v0 ) that is, X1 = 0 + 0.1 × 1 = 0.10000 X2 = X1 + hf1 (t1 , X1 , V1 ) that is, X2 = 0.10000 + 0.1 × 1.00000  V1 = v0 + hf2 (t0 , x0 , v0 ) V1 = 1.00000 = 1 + 0.1 × (sin 0 − 0 − 02 × 1) V2 = V1 + hf2 (t1 , X1 , V1 ) V2 = 1 + 0.1 × (sin 0.10000 − 0.10000 − 0.100002 × 1.00000)  = 0.20000  = 0.99898 X3 = X2 + hf1 (t2 , X2 , V2 ) that is, X2 = 0.20000 + 0.1 × 0.99898 = 0.29990  20 The second-order Adams–Bashforth method applied to a pair of coupled equations is Xn+1 = Xn + 12 h (3f1 (tn , Xn , Yn ) − f1 (tn−1 , Xn−1 , Yn−1 )) Yn+1 = Yn + 12 h (3f2 (tn , Xn , Yn ) − f2 (tn−1 , Xn−1 , Yn−1 )) First, we recast the diﬀerential equation as a pair of coupled ﬁrst-order equations. dx = v, x(0) = 0 dt dv v(0) = 1 = sin t − x − x2 v, dt Now, since the Adams–Bashforth method is a two-step process, we need to start the computation with another method. We use the second-order predictor–corrector. This has the form X̂n+1 = Xn + hf1 (tn , Xn , Yn )Ŷn+1 = Yn + hf2 (tn , Xn , Yn )  1  Xn+1 = Xn + h f1 (tn+1 , X̂n+1 , Ŷn+1 ) + f1 (tn , Xn , Yn ) 2  1  Yn+1 = Yn + h f2 (tn+1 , X̂n+1 , Ŷn+1 ) + f2 (tn , Xn , Yn ) 2 Hence we have X̂1 = 0 + 0.1 × 1 = 0.10000V̂1 = 1 + 0.1 × (0 − 0 − 02 × 1)1 = 1.00000 X1 = 0 + 0.05 × (1 + 1) = 0.10000V1 = 1 + 0.05 × (−0.01017 + 0) = 0.99949 c Pearson Education Limited 2011   112  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Now, continuing using the Adams–Bashforth method we have X2 = X1 + 12 h (3f1 (t1 , X1 , Y1 ) − f1 (t0 , x0 , v0 )) V2 = V1 + 12 h (3f2 (t1 , X1 , V1 ) − f2 (t0 , x0 , v0 )) that is,  X2 = 0.10000 + 0.05 × (3 × 0.99949 − 1) = 0.19992 V2 = 0.99949 + 0.05 × (3 × −0.01016 − 0) = 0.99797 X3 = X2 + 12 h (3f1 (t2 , X2 , V2 ) − f1 (t1 , X1 , V1 ))  that is,  21  X3 = 0.19992 + 0.05 × (3 × 0.99797 − 0.99949) = 0.29964  First, we formulate the problem as a set of 3 coupled ﬁrst-order diﬀerential  equations dx = u, x(0.5) = −1 dt du = v, u(0.5) = 1 dt dv = x2 − v(x − t) − u2 , v(0.5) = 2 dt We can then solve these by the predictor–corrector method. Notice that we need to compute the predicted values for all three variables before computing the corrected values for any of them.  X̂1 = x0 + hf1 (t0 , x0 , u0 , v0 ) = −1 + 0.05(1) = −0.95000 Û1 = u0 + hf2 (t0 , x0 , u0 , v0 ) = 1 + 0.05(2) = 1.10000 V̂1 = v0 + hf3 (t0 , x0 , u0 , v0 ) = 2 + 0.05((−1)2 − 2(−1 − 0.5) − 1) = 2.15000         X1 = x0 + h f1 t1 , X̂1 , Û1 , V̂1 + f1 (t0 , x0 , u0 , v0 ) 1 2  = −1 + 0.025 × (1.10000 + 1.00000) = −0.94750     U1 = u0 + 12 h f2 t1 , X̂1 , Û1 , V̂1 + f2 (t0 , x0 , u0 , v0 ) = 1 + 0.025 × (2.15000 + 2.00000) = 1.10375     1 V1 = v0 + 2 h f3 t1 , X̂1 , Û1 , V̂1 + f3 (t0 , x0 , u0 , v0 ) = 2 + 0.025 × (2.91750 + 3.00000) = 2.14794 Continuing the process we obtain c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  113  X̂2 = −0.94750 + 0.05 × 1.10375 = −0.89231 Û2 = 1.10375 + 0.05 × 2.14794 = 1.21115 V̂2 = 2.14794 + 0.05 × 2.89603 = 2.29274 X2 = −0.94750 + 0.025 × (1.21115 + 1.10375) = −0.88963 U2 = 1.10375 + 0.025 × (2.29274 + 2.14794) = 1.21477 V2 = 2.14794 + 0.025 × (2.75083 + 2.89603) = 2.28911 X̂3 = −0.88963 + 0.05 × 1.21477 = −0.82889 Û3 = 1.21477 + 0.05 × 2.28911 = 1.32923 V̂3 = 2.28911 + 0.05 × 2.72570 = 2.42539 X3 = −0.88963 + 0.025 × (1.32923 + 1.21477) = −0.82603 22  The ﬁrst step in solving this problem is to convert the problem to a pair of  coupled ﬁrst-order diﬀerential equations dx = v, dt dv = sin t − x2 v − x, dt  x(0) = 0 v(0) = 1  A pseudocode algorithm to compute the value of X(1.6) is shown in Figure 2.9. Using a program derived from this algorithm with a step size h = 0.4 gives X(1.6) = 1.220254 and, with a step size h = 0.2, gives X(1.6) = 1.220055. The method of Richardson extrapolation, given in Section 2.3.6, is equally applicable to problems such as this one involving coupled equations. Since the Runge– Kutta method has a local error of O(h)5 the global error will be O(h4 ) . The method therefore gives the estimated error in the second value of X(1.6) as (1.220254 − 1.220055)/15 = 0.000013. For 6 decimal place accuracy in the ﬁnal estimate we need error ≤ 0.0000005; in other words, the error must be reduced by a factor of 0.000013/0.0000005 = 26. Since Runge–Kutta is a fourth-order √ method the required step length will be 0.2/4 26 = 0.088. Rounding this down to a suitable size suggests that a step size of h = 0.08 will give a solution accurate to 6 decimal places. In fact the program yields, with h = 0.08, X(1.6) = 1.2200394. With h = 0.04 it gives X(1.6) = 1.2200390. Richardson extrapolation predicts the c Pearson Education Limited 2011   114  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  error in the h = 0.04 solution as 0.00000003 and therefore that in the h = 0.08 as 0.0000005. The required accuracy was therefore just achieved using h = 0.08. {program solves a pair of ordinary differential equations by the 4th order Runge-Kutta Method} procedure f1(t, x, v → f1) f1 ← v endprocedure procedure f2(t, x, v → f2) f2 ← sin(t) − x∗ x∗ v − x endprocedure {procedure computes values of x and v at the next time step} procedure rk4(t, x, v, h → xn, vn) c11 ← h∗ f1(t, x, v) c21 ← h∗ f2(t, x, v) c12 ← h∗ f1(t + h/2, x + c11/2, v + c21/2) c22 ← h∗ f2(t + h/2, x + c11/2, v + c21/2) c13 ← h∗ f1(t + h/2, x + c12/2, v + c22/2) c23 ← h∗ f2(t + h/2, x + c12/2, v + c22/2) c14 ← h∗ f1(t + h, x + c13, v + c23) c24 ← h∗ f2(t + h, x + c13, v + c23) xn ← x + (c11 + 2∗ (c12 + c13) + c14)/6 vn ← v + (c21 + 2∗ (c22 + c23) + c24)/6 endprocedure t start ← 0 t end ← 1.6 x0 ← 0 v0 ← 1 write(vdu, "Enter step size") read(keyboard, h) write(printer,t start,x0) t ← t start x ← x0 v ← v0 repeat rk4(t, x, v, h → xn, vn) x ← xn v ← vn t←t+h write(printer,t,x) until t >= t end Figure 2.9: Pseudocode algorithm for Exercise 22 c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 23  115  The ﬁrst step in solving this problem is to convert the problem to a set of  coupled ﬁrst-order diﬀerential equations dx = v, dt dv = w, dt dw = x2 − v2 − (x − t)w, dt  x(0.5) = −1 v(0.5) = 1 w(0.5) = 2  A pseudocode algorithm to compute the value of X(2.2) is shown in Figure 2.11. The procedures used in the algorithm are deﬁned in Figure 2.10. {procedures for pseudocode algorithm in figure 2.11} procedure f1(t, x, v, w → f1) f1 ← v endprocedure procedure f2(t, x, v, w → f2) f2 ← w endprocedure procedure f3(t, x, v, w → f3) f3 ← x∗ x − v∗ v − (x − t)∗ w endprocedure {procedure computes values of x, v and w at the next time step using the 4th order Runge-Kutta procedure} procedure rk4(t, x, v, w, h → xn, vn, wn) c11 ← h∗ f1(t, x, v, w) c21 ← h∗ f2(t, x, v, w) c31 ← h∗ f3(t, x, v, w) c12 ← h∗ f1(t + h/2, x + c11/2, v + c21/2, w + c31/2) c22 ← h∗ f2(t + h/2, x + c11/2, v + c21/2, w + c31/2) c32 ← h∗ f3(t + h/2, x + c11/2, v + c21/2, w + c31/2) c13 ← h∗ f1(t + h/2, x + c12/2, v + c22/2, w + c32/2) c23 ← h∗ f2(t + h/2, x + c12/2, v + c22/2, w + c32/2) c33 ← h∗ f3(t + h/2, x + c12/2, v + c22/2, w + c32/2) c14 ← h∗ f1(t + h, x + c13, v + c23, w + c33) c24 ← h∗ f2(t + h, x + c13, v + c23, w + c33) c34 ← h∗ f3(t + h, x + c13, v + c23, w + c33) (Continued) c Pearson Education Limited 2011   116  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  xn ← x + (c11 + 2∗ (c12 + c13) + c14)/6 vn ← v + (c21 + 2∗ (c22 + c23) + c24)/6 wn ← w + (c31 + 2∗ (c32 + c33) + c34)/6 endprocedure {procedure computes values of x, v and w at the next time step using the 3rd order predictor-corrector procedure} procedure pc3(t, xo, vo, wo, x, v, w, h → xn, vn, wn) xp ← x + h∗ (3∗ f1(t, x, v, w) − f1(t − h, xo, vo, wo))/2 vp ← v + h∗ (3∗ f2(t, x, v, w) − f2(t − h, xo, vo, wo))/2 wp ← w + h∗ (3∗ f3(t, x, v, w) − f3(t − h, xo, vo, wo))/2 xn ← x + h∗ (5∗ f1(t + h, xp, vp, wp) + 8∗ f1(t, x, v, w) -f1(t-h,xo,vo,wo))/12 ∗ ∗ vn ← v + h (5 f2(t + h, xp, vp, wp) + 8∗ f2(t, x, v, w) -f2(t-h,xo,vo,wo))/12 ∗ ∗ wn ← w + h (5 f3(t + h, xp, vp, wp) + 8∗ f3(t, x, v, w) -f3(t-h,xo,vo,wo))/12 endprocedure Figure 2.10: Pseudocode procedures for algorithm for Exercise 23 Using a program derived from this algorithm with a step size h = 0.1 gives X(2.2) = 2.923350 and, with a step size h = 0.05, gives X(2.2) = 2.925418. The method of Richardson extrapolation given in Section 2.3.6, is equally applicable to problems such as this one involving coupled equations. Since the third-order predictor–corrector method used has a local error of O(h4 ) , the global error will be O(h3 ) . The method therefore gives the estimated error in the second value of X(2.2) as 2.923350−2.925418)/7 = −0.000295. For 6 decimal place accuracy in the ﬁnal estimate we need error ≤ 0.0000005; in other words, the error must be reduced by a factor of 0.000295/0.0000005 = 590. Since we are using a third-order method, √ the required step length will be 0.05/3 590 = 0.00596. Rounding this down to a suitable size suggests that a step size of h = 0.005 will give a solution accurate to 6 decimal places. In fact the program yields, with h = 0.005, X(2.2) = 2.92575057. With h = 0.0025 it gives X(2.2) = 2.92575089. Richardson extrapolation predicts the error in the h = 0.0025 solution as 0.000000046 and therefore that in the  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition h = 0.005 as 0.00000037.  117  The required accuracy was therefore comfortably  achieved using h = 0.005. {program solves three ordinary differential equations by the 3rd order predictor-corrector method} t start ← 0.5 t end ← 2.2 x start ← −1 v start ← 1 w start ← 2 write(vdu, "Enter step size") read(keyboard, h) write(printer, t start, x start) t ← t start xo ← x start vo ← v start wo ← w start rk4(t, xo, vo, wo, h → x, v, w) t←t+h write(printer,t,x) repeat pc3(t, xo, vo, wo, x, v, w, h → xn, vn, wn) xo ← x vo ← v wo ← w x ← xn v ← vn w ← wn t←t+h write (printer, t, x) until t >= t end Figure 2.11: Pseudocode algorithm for Exercise 23  Review exercises 2.7 1  Euler's method for the solution of the diﬀerential equation Xn+1 = Xn + hFn = Xn + hf(tn , Xn ) c Pearson Education Limited 2011   dx = f(t, x) is dt  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  118  Applying this to the equation yields  dx √ = x with x(0) = 1 and a step size of h = 0.1 dt  x0 = x(0) = 1  √ √ X1 = x0 + hf(t0 , x0 ) = x0 + h x0 = 1 + 0.1 × 1 = 1.1000  √ X2 = X1 + hf(t1 , X1 ) = X1 + h X1 = 1.1000 + 0.1 1.1000 = 1.2049  √ X3 = X2 + hf(t2 , X2 ) = X2 + h X2 = 1.2049 + 0.1 1.2049 = 1.3146  √ X4 = X3 + hf(t3 , X3 ) = X3 + h X3 = 1.3146 + 0.1 1.3146 = 1.4293  √ X5 = X4 + hf(t4 , X4 ) = X4 + h X4 = 1.4293 + 0.1 1.4293 = 1.5489 Hence Euler's method with step size h = 0.1 gives the estimate X(0.5) = 1.5489.  2  Euler's method for the solution of the diﬀerential equation  dx = f(t, x) is dt  Xn+1 = Xn + hFn = Xn + hf(tn , Xn ) Applying this to the equation yields  dx = −ext with x(1) = 1 and a step size of h = 0.05 dt  x0 = x(1) = 1 X1 = x0 + hf(t0 , x0 ) = x0 + h(−ex0 t0 ) = 1 − 0.05 exp(1 × 1) = 0.86409 X2 = X1 + hf(t1 , X1 ) = X1 + h(−eX1 t1 ) = 0.86409 − 0.05 exp(0.86409 × 1.05) = 0.74021 X3 = X2 + hf(t2 , X2 ) = X2 + h(−eX2 t2 ) = 0.74021 − 0.05 exp(0.74021 × 1.10) = 0.62733 X4 = X3 + hf(t3 , X3 ) = X3 + h(−eX3 t3 ) = 0.62733 − 0.05 exp(0.62733 × 1.15) = 0.52447 Hence Euler's method with step size h = 0.05 gives the estimate X(1.1) = 0.52447.  3  This question could be solved by hand computation or using a computer  program based on a simple modiﬁcation of the pseudocode algorithm given in Figure 2.1. With a step size of h = 0.1 it is found that X(0.4) = 1.125584 and, with a step size of h = 0.05, X(0.4) = 1.142763. Using the Richardson c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  119  extrapolation method, because Euler's method is a ﬁrst-order method and the global error is therefore of O(h), the error in the estimate of X(0.4) is approximately 1.142763 − 1.125584 = 0.017179. To obtain X(0.4) accurate to 2 decimal places we need error ≤ 0.005. To achieve this we would need to reduced step size by a factor of 0.017179/0.005 = 3.44. This suggests a step size of 0.05/3.44 = 0.0145. Rounding this down to a sensible ﬁgure suggests trying a step size of 0.0125 or 0.01.  4  This question could be solved by hand computation or using a computer  program based on a simple modiﬁcation of the pseudocode algorithm given in Figure 2.1. With a step size of h = 0.05 it is found that X(0.25) = 2.003749 and, with a step size of h = 0.025, X(0.25) = 2.004452. Using the Richardson extrapolation method, because Euler's method is a ﬁrst-order method and the global error is therefore of O(h) , the error in the estimate of X(0.25) is approximately 2.004452−2.003749 = 0.000703. To obtain X(0.25) accurate to 3 decimal places we need error ≤ 0.0005. To achieve this we would need to reduce step size by a factor of 0.0007/0.0005 = 1.4. This suggests a step size of 0.025/1.4 = 0.0179. Rounding this down to a sensible ﬁgure suggests trying a step size of 0.0166667 or 0.0125.  5  This question could be solved by hand computation or using a computer  program based on a simple modiﬁcation of the pseudocode algorithm given in Figure 2.5. By either method it is found that X1 (1.2) = 2.374037, X2 (1.2) = 2.374148 and X3 (1.2) = 2.374176. The local error of the second-order predictor– corrector is O(h3 ) so the global error is O(h2 ) . Hence it is expected |X − x| ∝ h2  that is,  therefore, |X1 − X2 | ≈ x − αh − 2  and |X2 − X3 | ≈ x − α  Hence  h 2  x ≈ X + αh2  x−α  2  −  |X1 − X2 | ≈ |X2 − X3 |  2  h 2  x−α  h 4  3 2 4 ah 3 2 16 ah  ≈4  c Pearson Education Limited 2011   = 2  =  3 2 αh 4  3 αh2 16  120  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  In fact we ﬁnd that  |2.374037 − 2.374148| | − 0.000111| |X1 − X2 | = = = 3.97 ≈ 4 |X2 − X3 | |2.374148 − 2.374176| | − 0.000028|  6 This question is best solved using a computer program based on a simple modiﬁcation of the pseudocode given Figure 2.7. Let X1 denote the solution using a step size of h = 0.2, X2 that using h = 0.1 and X3 that using h = 0.05. By either method it is found that X1 (2) = 5.19436687, X2 (2) = 5.19432575 and X3 (2) = 5.19432313. The local error of the fourth-order Runge–Kutta method is O(h5 ) so the global error is O(h4 ) . Hence, by Richardson extrapolation, we may expect 4  x(2) = X2 (2) + αh = X3 (2) + α  h 2  4  therefore αh4 ≈ 16(x(2) − X3 (2)) therefore x(2) = X2 (2) + 16(x(2) − X3 (2)) 16 × 5.19432313 − 5.19432575 16X3 (2) − X2 (2) = = 5.19432296 15 15 and the most accurate estimate of x(2) is 5.19432296.  Hence x(2) =  7 The boundary conditions for this problem are p(r0 ) = p0 and p(r1 ) = 0. Hence we have p+r  dp = 2a − p dr  ⇒  dp 2p 2a + = dr r r  This is a linear diﬀerential equation, so we ﬁrst ﬁnd the integrating factor.  2 2lnr = r2 r dr = 2 ln r so the integrating factor is e therefore, r2  dp + 2rp = 2ar dr  that is, r2 p = ar2 + C c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Now p(r0 ) = p0 p(r1 ) = 0  ⇒ ⇒  r20 p0 = ar20 + C  ⇒  121  C = r20 (p0 − a)  0 = ar21 + C = ar21 + r20 (p0 − a)  ⇒  a=  r20 p0 r20 − r21  r20 r21 r20 p0 r20 p0 r20 p0 Hence p(r) = 2 + p0 − 2 = 2 −1 r0 − r21 r0 − r21 r2 r1 − r20 r2 22 12 1 7 − 1 = 13 ( 16 and p(1.5) = 2 9 − 1) = 27 2 − 12 1.52 2(3p + 1) dp =− , p(1) = 1 . This may easily The problem to solve numerically is dr 3r be solved using a modiﬁcation of the pseudocode algorithm of Figure 2.7 or of the program of Figure 2.8.  We ﬁnd that, using a step size of h = 0.05,  p(1.5) = 0.25925946. 8 The ﬁrst step is to recast the problem as a set of three coupled (linked) ﬁrst-order ordinary diﬀerential equations dx = v, x(1) = 0.2 dt dv = w, v(1) = 1 dt dw w(1) = 0 = sin(t) + xt − 4v2 − w2 , dt Figure 2.12 shows a pseudocode algorithm for the solution of these three equations by Euler's Method. {program solves three ordinary differential equations by the Euler method} procedure f1(t, x, v, w → f1) f1 ← v endprocedure procedure f2(t, x, v, w → f2) f2 ← w endprocedure procedure f3(t, x, v, w → f3) f3 ← sin(t) + x∗ t − 4∗ v∗ v − w∗ w endprocedure (Continued) c Pearson Education Limited 2011   122  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  {procedure computes values of x, v and w at the next time step using the Euler procedure} procedure euler(t, x, v, w, h → xn, vn, wn) xn ← x + h∗ f1(t, x, v, w) vn ← v + h∗ f2(t, x, v, w) wn ← w + h∗ f3(t, x, v, w) endprocedure t start ← 1.0 t end ← 2.0 x start ← 0.2 v start ← 1 w start ← 0 write(vdu, "Enter step size") read(keyboard, h) write(printer, t start, x start) t ← t start x ← x start v ← v start w ← w start repeat euler(t, x, v, w, h → xn, vn, wn) x ← xn v ← vn w ← wn t←t+h write(printer, t,x) until t >= t end Figure 2.12: Pseudocode algorithm for Review Exercise 8 Using a program derived from this algorithm with a step size h = 0.025 gives X(2) = 0.847035 and, with a step size h = 0.0125, gives X(2) = 0.844067. The method of Richardson extrapolation, given in Section 2.3.6, is equally applicable to problems such as this one involving coupled equations. Since Euler's method has a local error of O(h2 ) the global error will be O(h) . The method therefore gives the estimated error in the second value of X(2) as (0.847035−0.844067)/1 = 0.002968. This is less than 5 in the third decimal place, so we have two signiﬁcant ﬁgures of accuracy. The best estimate we can make is that x(2) = 0.84 to 2dp.  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  123  9 The ﬁrst step is to recast the problem as a set of two coupled (linked) ﬁrst-order ordinary diﬀerential equations dx = v, dt dv = (1 − x2 )v − 40x, dt  x(0) = 0.02 v(0) = 0  Figure 2.13 shows a pseudocode algorithm for the solution of these equations by the second-order predictor–corrector method. Using a program derived from this algorithm with a stepsize h = 0.02 gives X(4) = 0.147123 and, with a step size h = 0.01, gives X(4) = 0.146075. The method of Richardson extrapolation, given in Section 2.3.6, is applicable to problems such as this one involving coupled equations. Since the second-order predictor–corrector method has a local error of O(h3 ) , the global error will be O(h2 ) . The method therefore gives the estimated error in the second value of X(4) as (0.147123 − 146075)/3 = 0.001048. For 4 decimal place accuracy in the ﬁnal estimate we need error ≤ 0.00005; in other words, the error must be reduced by a factor of 0.001048/0.00005 = 20.96. Since this predictor–corrector is a second-order method the required step length will be √ 0.01/ 20.96 = 0.0022. Rounding this down to a suitable size suggests that a step size of h = 0.002 will give a solution accurate to 4 decimal places. In fact the program yields, with h = 0.002, X(4) = 0.145813. With h = 0.001 it gives X(4) = 0.145807. Richardson extrapolation predicts the error in the h = 0.001 solution as 0.000002 and therefore that in the h = 0.002 as 0.000008. The required accuracy was therefore comfortably achieved using h = 0.002. We can be conﬁdent that x(4) = 0.1458 to 4dp. {program solves two ordinary differential equations by the second order predictor-corrector method} procedure f1(t, x, v → f1) f1 ← v endprocedure procedure f2(t, x, v → f2) f2 ← (1 − x∗ x)∗ v − 40∗ x endprocedure (Continued) c Pearson Education Limited 2011   124  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  {procedure computes values of x, v and w at the next time step using the second order predictor-corrector procedure} procedure pc2(t, x, v, h → xn, vn) xp ← x + h∗ f1(t, x, v) vp ← v + h∗ f2(t, x, v) xn ← x + h∗ (f1(t + h, xp, vp) + f1(t, x, v))/2 vn ← v + h∗ (f2(t + h, xp, vp) + f2(t, x, v))/2 endprocedure t start ← 0.0 t end ← 4.0 x start ← 0.02 v start ← 0 write(vdu, "Enter step size") read(keyboard, h) writeprinter, t start, x start t ← t start x ← x start v ← v start repeat pc2(t, x, v, h → xn, vn) x ← xn v ← vn t←t+h write(printer,t,x) until t >= t end Figure 2.13: Pseudcode algorithm for Review Exercise 9  10  The ﬁrst step is to recast the problem as a set of three coupled (linked)  ﬁrst-order ordinary diﬀerential equations. dx = v, x(1) = −1 dt dv = w, v(1) = 1 dt  dw = sin(t) + xt − 4v3 − |w|, w(1) = 2 dt A minor modiﬁcation of the pseudocode algorithm shown in Figure 2.9 provides an algorithm for the solution of this problem. Using a program derived from this algorithm with a step size h = 0.1 gives X(2.5) = −0.651076 and, c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition with a step size h = 0.05, gives X(2.5) = −0.653798. of Richardson extrapolation, given in Section 2.3.6.  125  We use the method  Since the Runge–Kutta  5  method has a local error of O(h ) the global error will be O(h4 ) . The method therefore gives the estimated error in the second value of X(2.5) as (0.651076 − 0.653798)/15 = −0.000181. For 4 decimal place accuracy in the ﬁnal estimate we need error ≤ 0.00005; in other words, the error must be reduced by a factor of 0.000181/0.00005 = 3.63. Since the Runge–Kutta is a fourth-order method the √ required step length will be 0.054 3.63 = 0.036. Rounding this down to a suitable size suggests that a stepsize of h = 0.025 will give a solution accurate to 4 decimal places. In fact the program yields, with h = 0.025, X(2.5) = −0.653232. With h = 0.0125 it gives X(2.5) = −0.653217. Richardson extrapolation predicts the error in the h = 0.0125 solution as 0.0000009 and therefore that in the h = 0.025 as 0.000015. The required accuracy is therefore easily achieved using h = 0.025. We can be conﬁdent that x(2.5) = −0.6532 to 4dp.  c Pearson Education Limited 2011   3 Vector Calculus Exercises 3.1.2 1(a)  f(x, y) = c ⇒ x2 + y2 = 1 + ec  Contours are a family of concentric circles, centre (0,0) and radius > 1.  1(b)  f(x, y) = c ⇒ y = (1 + x) tan c  Contours are a family of straight lines whose y intercept equals their slope and pass through (-1,0).  2(a)  Flow lines are given by  dx dt  = y and  dy dt  = 6x2 − 4x .  Thus, 6x2 − 4x dy = dx y   y dy = (6x2 − 4x) dx + c 1 2 y = 2x3 − 2x2 + c 2 y2 = 4x2 (x − 1) + C  2(b) Flow lines are given by  dx dt  = y and  dy dt  = 16 x3 − x .  Thus, −x y   1 y dy = ( x3 − x) dx + c 6 1 2 1 4 1 2 y = x − x +c 2 24 2 1 2 2 2 x (x − 12) + C y = 12 dy = dx  1 3 6x  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 3(a)  127  Level surfaces are given by f(r) = c ⇒ z = c + xy .  3(b)  Level surfaces are given by f(r) = c ⇒ z = c − e−xy .  4(a)  Field lines are given by  dx dt  = xy ,  dy dt  = y2 + 1,  dz dt  = z.  dz = z ⇒ z = Aet dt dy = 1 + y2 ⇒ y = tan(t + α) dt dx = xy ⇒ ln x = C − ln(cos(t + α)) dt B cos(t + α)  x 2 is a hyperbola, the curve is on a Since 1 + tan2 θ = sec2 θ ⇒ 1 + y2 = B x=  hyperbolic cylinder.  4(b) Field lines are given by  dx dt  = yz ,  dy dt  dz dt  = zx ,  = xy .  Hence, dy x = ⇒ y 2 = x2 − c dx y dz x = ⇒ z 2 = x2 + k dx z The curve is the intersection of these hyperbolic cylinders.  5(a) ∂f ∂x ∂2 f ∂x2 ∂2 f ∂x∂y ∂2 f ∂z∂x  = yz − 2x = −2 =z =y  ∂f ∂y ∂2 f ∂y2 ∂2 f ∂y∂x ∂2 f ∂y∂z  = xz + 1 =0 =z =x  ∂f = xy − 1 ∂z ∂2 f =0 ∂z2 ∂2 f =y ∂x∂z ∂2 f =x ∂z∂y  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  128 5(b)  ∂f ∂x ∂2 f ∂x2 ∂2 f ∂x∂y ∂2 f ∂z∂x  = 2xyz3 = 2yz3 = 2xz3 = 6xyz2  ∂f ∂y ∂2 f ∂y2 ∂2 f ∂y∂x ∂2 f ∂y∂z  ∂f = 3x2 yz2 ∂z ∂2 f = 6x2 yz 2 ∂z ∂2 f = 6xyz2 ∂x∂z ∂2 f = 3x2 z2 ∂z∂y  = x2 z 3 =0 = 2xz3 = 3x2 z2  5(c)  −y  z ∂f −yz = = y 2 ∂x 1 + (x) x2 x2 + y 2  ∂f z = ∂y 1 + ( xy )2    xz 1 = 2 x x + y2  ∂2 f −2xyz = 2 2 ∂y (x + y2 )2  ∂2 f 2xyz = 2 2 ∂x (x + y2 )2   ∂f −1 y = tan ∂z x 2 ∂ f =0 ∂z2  ∂2 f z 2x2 z z(x2 + y2 ) − 2x2 z (y2 − x2 )z = 2 − = = ∂x∂y x + y2 (x2 + y2 )2 (x2 + y2 )2 (x2 + y2 )2 ∂2 f 2y2 z 2y2 z − z(x2 + y2 ) (y2 − x2 )z −z + = = = 2 ∂y∂x x + y2 (x2 + y2 )2 (x2 + y2 )2 (x2 + y2 )2  −y  ∂2 f ∂2 f −y 1 −y = = = y 2 ∂x∂z 1 + (x) x2 x2 + y 2 ∂z∂x x2 + y 2   ∂2 f ∂2 f x 1 1 x = = = 2 y 2 2 2 ∂y∂z 1 + (x) x x +y ∂z∂y x + y2 6(a)  ∂f ∂x  = 2x ,  ∂f ∂y  = 2y ,  ∂f ∂z  = −1,  dx dt  = 3t2 ,  df = 2(t3 − 1)(3t2 ) + 2(2t)(2) + (−1) dt  dy dt    = 2,  dz dt  −1 (t − 1)2  =  −1 (t−1)2    = {2t(3t4 − 3t + 4)(t − 1)2 + 1}/(t − 1)2 = {2t(3t6 − 6t5 + 3t4 − 3t3 + 10t2 − 11t + 4) + 1}/(t − 1)2 c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  129  6(b) ∂f ∂f ∂f dx = yz, = xz, = xy, = e−t (cos t − sin t), ∂x ∂y ∂z dt dy dz = −e−t (cos t + sin t), =1 dt dt df = te−t cos t.e−t (cos t − sin t) − te−t sin t.e−t (cos t + sin t) + e−2t sin t cos t.(1) dt = te−2t (cos2 t − sin2 t − 2 sin t cos t) + e−2t sin t cos t 1 = te−2t (cos 2t − sin 2t) + e−2t sin 2t 2  7  r2 = x2 + y2 + z2 , tan φ = xy , tan θ =  (x2 +y 2 )1/2 z  ∂r 1 y ∂φ 1 x cos φ = = sin θ sin φ, = = 2 = y 2 2 ∂y r ∂y 1 + (x) x x +y r sin θ ∂θ ∂ yz (x2 + y2 )1/2 sin φ cos θ = {tan−1 }= 2 = 2 2 2 2 1/2 ∂y ∂y z r (x + y + z )(x + y ) ∂f ∂f cos φ ∂f sin φ cos θ ∂f = sin θ sin φ + + ∂y ∂r r sin θ ∂φ r ∂θ z ∂φ ∂r = = cos θ, =0 ∂z r ∂z ∂θ 1 (x2 + y2 )1/2 −(x2 + y2 )1/2 − sin θ = − = = 2 2 2 2 2 2 x +y ∂z z x +y +z r 1 + z2 ∂f ∂f sin θ ∂f = cos θ − ∂z ∂r r ∂θ 8  ∂u ∂x  =  df ∂r ∂r dr ∂x , ∂x  =  x r  ⇒  ∂u ∂x  =  x df r dr    ∂2 u ∂  x  df x ∂ df + = ∂x2 ∂x r dr r ∂x dr x r − x( r ) df x d2 f x + = r2 dr r dr2 r 2 2 2 2 y + z df x d f + 2 2 = r.r2 dr r dr c Pearson Education Limited 2011   130  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Similarly (by symmetry),  ∂2u ∂y 2  =  x2 +y 2 df r.r 2 dr  +  y2 ∂ 2 f d2 u r2 ∂r2 , dz 2  =  x2 +z 2 df r.r 2 dr  +  2 z2 d f r2 dr2  ∂2 u ∂2 u ∂2 u 2(x2 + y2 + z2 ) df x2 + y2 + z2 d2 f + ⇒ 2+ 2 + 2 = ∂x ∂y ∂z r.r2 dr r2 dr2 2 df d2 f + 2 = r dr dr Hence, the result.  9  V(x, y, z) = ∂V ∂x ∂2 V ∂x2 ∂V ∂y ∂2 V ∂y2 ∂V ∂z  1 z  2  +y exp − x 4z  2   1 x2 + y2  −x  = exp − z 4z 2z   2 2 x +y −x 2 1 1 x2 + y 2 = exp − − 2 exp − z 4z 2z 2z 4z  1 x2 + y2  −y  = exp − z 4z 2z   2 2 −y 2 1 1 x2 + y 2 x +y = exp − − 2 exp − z 4z 2z 2z 4z   1 x2 + y 2 x2 + y 2 1 x2 + y 2 + exp − = − 2 exp − z 4z z 4z 4z2 ⇒  10  ∂V ∂2 V ∂ 2 V + 2 = 2 ∂x ∂y ∂z  V = sin 3x cos 4y cosh 5z ∂2 V = −9 sin 3x cos 4y cosh 5z ∂x2 ∂2 V = −16 sin 3x cos 4y cosh 5z ∂y2 ∂2 V = 25 sin 3x cos 4y cosh 5x ∂z2 ⇒  ∂2 V ∂2 V ∂2 V + 2 + 2 =0 ∂x2 ∂y ∂z  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Exercises 3.1.4 11  x + y = u,  y = uv ∂x ∂y + = 1, ∂u ∂u  ∂y ∂x =v⇒ =1−v ∂u ∂u  ∂y ∂x ∂x ∂y + = 0, =u⇒ = −u ∂v ∂v ∂v ∂v   ∂(x, y)  1 − v v  = u − uv − (−uv) = u = ∂(u, v)  −u u  12  x + y + z = u, y + z = uv, z = uvw ∂x ∂y ∂z ∂y ∂z ∂z + + = 1, + = v, = vw ∂u ∂u ∂u ∂u ∂u ∂u ⇒  ∂y ∂x ∂z = vw, = v(1 − w), =1−v ∂u ∂u ∂u  ∂x ∂y ∂z ∂y ∂z ∂z + + = 0, + = u, = uw ∂v ∂v ∂v ∂v ∂v ∂v ∂z ∂y ∂x = uw, = u − uw, = −u ∂v ∂v ∂v ∂x ∂y ∂z ∂y ∂z ∂z + + = 0, + = 0, = uv ∂w ∂w ∂w ∂w ∂w ∂w ∂y ∂x ∂z = uv, = −uv, =0 ∂w ∂w ∂w    1 − v v − vw vw    ∂(x, y, z) ⇒ =  −u u − uw uw  ∂(u, v, w)  0 −uv uv       1 − v v vw  1 − v v       =  −u u uw  = uv   −u u  0 0 uv      2 1− v v = u v = u2 v −1 1  ⇒  c Pearson Education Limited 2011   131  132  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  13  x = eu cos v, y = eu sin v ∂x = eu cos v, ∂u ∂x = −eu sin v, ∂v  ∂(x, y)  eu cos v = ⇒ ∂(u, v)  −eu sin v  ∂y = eu sin v ∂u ∂y = eu cos v ∂v   eu sin v  = e2u (cos2 v + sin2 v) = e2u eu cos v   1 ln(x2 + y2 ) 2 y y = tan v ⇒ v = tan−1 x x  x2 + y2 = e2u ⇒ u =  ∂u ∂v x y y x ∂u ∂v = 2 = 2 =− 2 = 2 , , 2 2 2 ∂x x +y ∂y x + y ∂x x + y ∂y x + y2  x  y 2 2  − x2 +y ∂(u, v)  x2 +y 1 1 2 2 = x +y = = y = 2u x  2 2 2 2 2 ∂(x, y) (x + y ) x +y e x2 +y 2 x2 +y 2 Hence, the result.  14   ∂(u, v)  (− sin x cos y − λ cos x sin y) (cos x cos y − λ sin x sin y)  = ∂(x, y)  (− cos x sin y − λ sin x cos y) (− sin x sin y + λ cos x cos y)  = − (sin x cos y + λ cos x sin y)(− sin x sin y + λ cos x cos y) + (cos x sin y + λ sin x cos y)(cos x cos y − λ sin x sin y) = − [− sin2 x sin y cos y + λ sin x cos x cos2 y − λ sin x cos x sin2 y + λ2 cos2 x sin y cos y] + [cos2 x sin y cos y − λ sin x cos x sin2 y + λ sin x cos x cos2 y − λ2 sin2 x sin y cos y] = sin y cos y − λ2 sin y cos y ∂(u, v) = 0 ⇒ λ2 = 1 ⇒ λ = −1 or ∂(x, y) c Pearson Education Limited 2011   1  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 15  133     2Kx 3 (3z + 6y)   ∂(u, v, w)  =  8y 2 (2z + 6x)  ∂(x, y, z)  2z 1 (2y + 3x)     Kx − 3z 0 (3z − 9x)    = 2  4y − 2z 0 (2z − 4y)   z 1 (2y + 3x)     Kx − 3z 3z − 9x   = 4(z − 2y)(−Kx + 9x)  = −2  4y − 2z 2z − 4y  ∂(u, v, w) =0⇒K=9 ∂(x, y, z) u = 9x2 + 4y2 + z2 v2 = 9x2 + 4y2 + z2 + 12xy + 6xz + 4yz 2w = 12xy + 6xz + 4yz u = v2 − 2w  16  ∂u ∂x ∂u ∂y + (diﬀerentiating u = g(x, y) with respect to u) ∂x ∂u ∂y ∂u ∂v ∂x ∂v ∂y 0= + (diﬀerentiating v = h(x, y) with respect to u) ∂x ∂u ∂y ∂u    1 ∂u  ∂y    0 ∂v  ∂v ∂x  = ⇒ =  ∂u ∂y /J ∂u  ∂x ∂y  ∂u ∂y    ∂v ∂v  1=  ∂x  ∂y  ∂y ∂v = − /J ∂u ∂x Similarly, diﬀerentiating u = g(x, y) and v = h(x, y) with respect to v obtains the other two expressions. 17 u = ex cos y, v = e−x sin y ∂u ∂v = ex cos y = u, = −e−x sin y = −v ∂x ∂x ∂u ∂v = −ex sin y, = e−x cos y ∂y ∂y c Pearson Education Limited 2011   134  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition ∂x ∂u ∂x ∂v ∂y ∂u ∂y ∂v  e−x cos y cos2 y − sin2 y ex sin y = cos2 y − sin2 y e−x sin y = cos2 y − sin2 y ex cos y = cos2 y − sin2 y =  Since 2uv = 2 sin y cos y = sin 2y , it is possible to express these results in terms of u and v.  √ 1 (1 + 1 − 4u2 v2 ) 2 √ 1 cos y = (1 − 1 − 4u2 v2 ) 2 √ 1 (1 + 1 − 4u2 v2 ) ex = 2u sin y =  Exercises 3.1.6 18(a) ∂ 2 (y + 2xy + 1) = 2y + 2x ∂y ∂ (2xy + x2 ) = 2y + 2x ∂x Therefore, it is an exact diﬀerential. Let  ∂f ∂x  = y2 + 2xy + 1, then f(x, y) = xy2 + x2 y + x + c(y) and  But,  ∂f ∂y  ∂f dc = 2xy + x2 + ∂y dy  = 2xy + x2 from the question. Hence,  dc dy  = 0; so c is independent of x  and y ⇒ f(x, y) = x y + y x + x + c. 2  2  18(b) ∂ (2xy2 + 3y cos 3x) = 4xy + 3 cos 3x ∂y ∂ (2x2 y + sin 3x) = 4xy + 3 cos 3x ∂x Therefore, it is an exact diﬀerential. c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition Let  ∂f ∂x  135  = 2xy2 + 3y cos 3x , then f(x, y) = x2 y2 + y sin 3x + c(y) and  Hence,  dc dy  ∂f dc = 2x2 y + y sin 3x + ∂y dy  = 0 and c is a constant with respect to both x and y ⇒ f(x, y) = x2 y2 + y sin 3x + c  18(c) ∂ (6xy − y2 ) = 6x − 2y ∂y ∂ (2xey − x2 ) = 2ey − 2x ∂x Not equal, so not an exact diﬀerential. 18(d) ∂ 3 (z − 3y) = −3 ∂y ∂ (12y2 − 3x) = −3 ∂x Hence, exact. ∂f ∂y  = −3x +  Let  ∂c ∂y  ∂ (12y2 − 3x) = 0 ∂z ∂ (3xz2 ) = 0 ∂y  ∂ 3 (z − 3y) = 3z2 ∂z ∂ (3xz2 ) = 3z2 ∂x  = z3 − 3y , then f(x, y, z) = z3 x − 3xy + c(y, z) and  ∂f ∂x  . This inturn implies that  ∂c ∂y  = 12y2 and c(y, z) = 4y3 + k(z) .  ∂f ∂c dk = 3z2 x + = 3z2 x + . ∂z ∂z dz This inturn implies that  19  dk dz  = 0 and so f(x, y, z) = z3 x − 3xy + 4y3 + K.  ∂ (y cos x + λ cos y) = cos x − λ sin y ∂y ∂ (x sin y + sin x + y) = sin y + cos x ∂x  Equal, if λ = −1. Let ∂f ∂y  ∂f ∂x  =  y cos x − cos y , then f(x, y)   =  y sin x − x cos y + c(y) and    = sin x + x sin y + c (y) so that c (y) = y and c(y) = 12 y2 + k . c Pearson Education Limited 2011   136  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Hence, f(x, y) = y sin x − x cos y + 12 y2 + k . f(0, 1) = 0 ⇒ 0 = 0 + 0 +  1 1 +k⇒k=− 2 2  and f(x, y) = y sin x − x cos y + 12 (y2 − 1) .  20 ∂ (10x2 + 6xy + 6y2 ) = 6x + 12y ∂y ∂ (9x2 + 4xy + 15y2 ) = 18x + 4y ∂x Hence, not exact. ∂ [(2x + 3y)m (10x2 + 6xy + 6y2 )] = 3m(2x + 3y)m−1 (10x2 + 6xy + 6y2 ) ∂y + (2x + 3y)m (6x + 12y) ∂ [(2x + 3y)m (9x2 + 4xy + 15y2 )] = 2m(2x + 3y)m−1 (9x2 + 4xy + 15y2 ) ∂x + (2x + 3y)m (18x + 4y) Hence, exact if 3m(10x2 +6xy+6y2 )+(2x+3y)(6x+12y) = 2m(9x2 +4xy+15y2 )+(2x+3y)(18x+4y) Comparing coeﬃcients of x2 gives m = 2. Let ∂f = (2x + 3y)2 (10x2 + 6xy + 6y2 ) = 40x4 + 144x3 y + 186x2 y2 + 126xy3 + 54y4 ∂x ⇒ f(x, y) = 8x5 + 36x4 y + 62x3 y2 + 63x2 y3 + 54xy4 + c(y) ∂f = 36x4 + 124x3 y + 99x2 y2 + 216xy3 + c (y) ∂y ⇒ c (y) = 9y2 × 15y2 ⇒ c(y) = 27y5 + k Hence, f(x, y) = 8x5 + 36x4 y + 62x3 y2 + 63x2 y3 + 5xy4 + 27y5 + k .  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  137  Exercises 3.2.2 21  grad f = (2xyz2 , x2 z2 , 2x2 yz) .  At (1,2,3), grad f = (36, 9, 12) = 3(12, 3, 4) .  21(a)  Unit vector in direction of (−2, 3, −6) is √(−2,3,−6) = 17 (−2, 3, −6) . (4+9+36)  Directional derivative of f in direction of (−2, 3, −6) at (1,2,3) is 3(12, 3, 4) · (−2, 3, −6)/7 = −117/7  21(b)   Maximum rate of change is |grad f| = 3 (144 + 9 + 16) = 39 and is in  the direction of grad f, that is, (12,3,4)/13.  22(a)  ∇(x2 + y2 − z) = (2x, 2y, −1)  22(b)    ∇ z tan  −1   y  x   =      −zy zx −1 y , , tan x2 + y 2 x2 + y 2 x  22(c)  ∇  e−x−y+z  x3 + y 2      −e−x−y+z 1 3x2 e−x−y+z −e−x−y+z  − ,  2 (x3 + y2 )3/2 x3 + y 2 x3 + y 2  1 2ye−x−y+z e−x−y+z − , 2 (x3 + y2 )3/2 x3 + y 2   e−x−y+z 3 2 3 2 3 2 3 2 −x − y − x , −x − y − y, x + y = 3 2 (x + y2 )3/2  =  22(d) ∇(xyz sin π(x + y + z)) =(yz sin π(x + y + z) + πxyz cos π(x + y + z), xz sin π(x + y + z) + πxyz cos π(x + y + z), xy sin π(x + y + z) + πxyz cos π(x + y + z))  c Pearson Education Limited 2011   138  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  23  grad (x2 + y2 − z) = (2x, 2y, −1) .  At (1,1,2), grad f = (2, 2, −1) . 1 3 (2, 2, −1) . 1 1 3 (2, 2, −1) = 3 (4 + 4  Unit vector in the direction of (4, 4, −2) is Directional derivative is (2, 2, −1) ·  24  + 1) = 3.  ∇(xy2 − 3xz + 5) = (y2 − 3z, 2xy, −3x) .  At (1, −2, 3) , grad f = (−5, −4, −3) .  √ Unit vector in the direction of grad f is (−5, −4, −3)/ 50 .  √ Unit normal to surface xy2 − 3xz + 5 = 0 at (1, −2, 3) is (5, 4, 3)/ 50 .  25(a)  r=   x2 + y 2 + z 2    ∇r =  x  y  z   , , x2 + y 2 + z 2 x2 + y 2 + z 2 x2 + y 2 + z 2  1 (x, y, z) r r = r =  25(b)     −x 1 −y −z = ∇ , , r (x2 + y2 + z2 )3/2 (x2 + y2 + z2 )3/2 (x2 + y2 + z2 )3/2 r =− 3 r 26 ∂φ = 2xy + z2 ⇒ φ(x, y, z) = x2 y + xz2 + f(y, z) ∂x ∂φ ∂f = x2 + z ⇒ x2 + z = x2 + ⇒ f(y, z) = zy + g(z) ∂y ∂y ∂φ dg = y + 2xz ⇒ y + 2xz = 2xz + y + ∂z dz Hence,  dg dz  = 0 ⇒ g(z) = c, a constant.  Hence, φ(x, y, z) = x2 y + xz2 + zy + c.  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 27  φ(x, y, z) = x2 y − 3xyz + z3  grad φ = (2xy − 3yz, x2 − 3xz, −3xy + 3z2 ) At (3,1,2), grad φ = (0, −9, 3) .  √ Unit vector in direction of (3, −2, 6) = (3, −2, 6)/ 49 .  Directional derivative at (3,1,2) in the direction of (3, −2, 6) is (0, −9, 3) · (3, −2, 6)/7 = 36/7 28  ∇(x2 + y2 + z2 − 9) = (2x, 2y, 2z) .  At (2, −1, 2) , grad (x2 + y2 + z2 − 9) = (4, −2, 4) . Unit normal to surface at (2, −1, 2) is (2, −1, 2)/3. ∇(x2 + y2 − z − 3) = (2x, 2y, −1). At (2, −1, 2) , grad (x2 + y2 − z − 3) = (4, −2, −1) . √ Unit normal to surface at (2, −1, 2) is (4, −2, −1)/ 21 . Let angle between normals be θ, then cos θ =  (2, −1, 2) (4, −2, −1) √ · 3 21  8 ⇒ cos θ = √ , hence, θ = 54.41◦ 3 21 29(a)  ∇(x2 + 2y2 + 3z2 − 6) = (2x, 4y, 6z) .  At (1,1,1), grad f = (2, 4, 6) , so tangent plane at (1,1,1) is (1, 2, 3) · (x − 1, y − 1, z − 1) = 0 i.e. x + 2y + 3z = 6 and normal line is  29(b)  y−1 z−1 x−1 = = 1 2 3  ∇(2x2 + y2 − z2 + 3) = (4x, 2y, −2z)  At (1,2,3), grad f = (4, 4, −6) , so the tangent plane at (1,2,3) is (2, 2, −3) · (x − 1, y − 2, z − 3) = 0 i.e. 2x + 2y − 3z = −3 and the normal line is  y−2 3−z x−1 = = 2 2 3 c Pearson Education Limited 2011   139  140  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  29(c)  ∇(x2 + y2 − z − 1) = (2x, 2y, −1)  At (1,2,4), grad f = (2, 4, −1) , so that the tangent plane is (2, 4, −1) · (x − 1, y − 2, z − 4) = 0 i.e. 2x + 4y − z = 6 and the normal line is x−1 y−2 z−4 = = 2 4 −1  30  The change Δr in the vector r can be resolved into the three directions ur ,  uθ , uφ . Thus, Δr = Δrur + rΔθuθ + r sin θΔφuφ Hence, f(r + Δr) − f(r) Δr→0 |Δr| 1 ∂f 1 ∂f ∂f ur + uθ + uφ = ∂r r ∂θ r sin θ ∂φ  grad f = lim  Exercises 3.3.2 31(a)  div (3x2 y, z, x2 ) = 6xy + 0 + 0 = 6xy  31(b)  div (3x + y, 2z + x, z − 2y) = 3 + 0 + 1 = 4  32  div F = 2y2 − 2yz3 + 2yz − 3xz2 .  At (−1, 2, 3) , div F = −61.  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 33  141  ∇(a · r) = ∇(a1 x + a2 y + a3 z) = (a1 , a2 , a3 ) = a   ∂ ∂ ∂ (x, y, z) (a · ∇)r = a1 , a2 , a3 ∂x ∂y ∂z = (a1 , a2 , a3 ) = a a(∇ · r) = a(1 + 1 + 1) = 3a  34   ∇·v=  ∂ since ∂x  1 x2 − 3 r r   x  x2 + y 2 + z 2     +  =  1 y2 − 3 r r     +  1 x2 + y 2 + z 2  1 z2 − r r3  −    x.(2x) 1 2 2 (x + y2 + z2 )3/2  3 x2 + y 2 + z 2 2 Hence, ∇.v = − = r r3 r     − 12 .(2x) − 12 (2y) − 12 (2z) 2 =2 , , ∇ r (x2 + y2 + z2 )3/2 (x2 + y2 + z2 )3/2 (x2 + y2 + z2 )3/2 2 2r = − 3 (x, y, z) = − 3 r r 35 div F = 4xy2 + 9xy2 + λxy2 = (4 + 9 + λ)xy2 div F = 0 ⇒ λ = −13 36  In spherical polar coordinates, an element of volume has side Δr in the ur  direction, rΔθ in the uθ direction and r sin θΔφ in the uφ direction. The total ﬂow out of the elementary volume is ∂ ∂ ∂ (v · ur r2 sin θΔθΔφ)Δr + (v · uθ r sin θΔφΔr)Δθ + (v · uφ rΔθΔr)Δφ ∂r ∂θ ∂φ + terms of order |Δr|2 c Pearson Education Limited 2011   142  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Dividing by the volume of the element, r2 sin θΔθΔφΔr , and proceeding to the limit, we obtain div v =  ∂ 1 ∂ 1 1 ∂ 2 (r (r sin θv (vφ ) v ) + ) + r θ r2 ∂r r sin θ ∂θ r sin θ ∂φ  37 div   x y z  r = div , , r3 r3 r3 r3 3x2 1 3y2 1 3z2 1 = 3− 5 + 3− 5 + 3− 5 r r r r r r 3 (x2 + y2 + z2 ) = 3 −3 =0 r r5  Exercises 3.3.4 38  39  40    i  ∂ curl v =  ∂x  3xz2   i  ∂ curl v =  ∂x  yz  j ∂ ∂y  xz    i  ∂  curl v =  ∂x  2x + yz  j  k  ∂ ∂y  ∂ ∂z  −yz  x + 2z      = (y, 6xz − 1, 0)     k   ∂  ∂z  = (x − x, y − y, z − z) = 0 xy      ∂ ∂  = (0, 0, 0) = 0 ∂y ∂z  2y + zx 2z + xy    ∂f ∂f ∂f ∂f grad f = , , ⇒ = 2x + yz ∂x ∂u ∂z ∂x ∂f = 2x + yz ⇒ f(x, y, z) = x2 + xyz + g(y, z) ∂x ∂f ∂f ∂g = 2y + zx and = xz + ⇒ g(y, z) = y2 + h(z) ∂y ∂y ∂y ∂f ∂f dh = 2z + xy and = xy + ⇒ h(z) = z2 + c ∂z ∂z dz j  k  Hence, f(x, y, z) = x2 + y2 + z2 + xyz + C.  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  41      i j k     ∂ ∂ ∂ ∇ × (fv) =   = (x, 5x3 − 2y, z) ∂x ∂y ∂z    zx3 − zy 0 −x4 + xy     i j k    ∂ ∂ ∂  f(∇ × v) = (x3 − y)  ∂x  ∂y ∂z    z 0 −x  = (x3 − y)(0, 2, 0) = (0, 2x3 − 2y, 0) (∇f) × v = (3x2 , −1, 0) × v    i j k   2 =  3x −1 0  = (x, 3x3 , z)  z 0 −x   42   i   ∂ ∇×F= ∂x   4xy + az3  j ∂ ∂y      ∂  = (c − 3, 3az2 − 6z2 , 2bx − 4x) ∂z  6xz2 + cy  k  bx2 + 3z ∇ × F = 0 ⇒ c = 3, a = 2, b = 2   ∂φ ∂φ ∂φ , , grad φ = ∂x ∂y ∂z ∂φ = 4xy + 2z3 ⇒ φ(x, y, z) = 2x2 y + 2xz3 + f(y, z) ∂x ∂φ ∂φ ∂f ∂f = 2x2 + 3z and = 2x2 + ⇒ = 3z ∂y ∂y ∂y ∂y Hence, f(y, z) = 3yz + g(z) .  dg ∂φ dg ∂φ = 6xz2 + 3y and = 6xz2 + 3y + ⇒ =0 ∂z ∂z dz dz Hence, φ(x, y, z) = 2x2 y + 2xz3 + 3yz + C.  c Pearson Education Limited 2011   143  144  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition   i 1 1  ∂ ω = curl u =  ∂x 2 2  −y  43  =  j ∂ ∂y  x   k  ∂  ∂z  xyz   1 (xz, −yz, 2) 2  At (1,3,2), ω = 12 (2, −6, 2) = (1, −3, 1) ⇒  ω| = |ω  √ 11  44 div v = a + d   i j   ∂ ∂ curl v =  ∂x ∂y   ax + by cx + dy div v = 0 ⇒ a = −d   k  ∂  ∂z  = (0, 0, c − b) 0   curl v = 0 ⇒ c = b v = (ax + by)i + (bx − ay)j = grad φ ∂φ ∂φ = ax + by and = bx − ay ∂x ∂y 1 ⇒ φ(x, y) = ax2 + bxy + f(y) 2 ⇒  ∂φ 1 = bx + f (y) ⇒ f (y) = −ay ⇒ f(y) = − ay2 + K ∂y 2 Hence, φ(x, y) =  45  1 2 1 ax + bxy − ay2 + K. 2 2  In spherical polar coordinates, an element of volume has side Δr in the ur  direction, rΔθ in the uθ direction and r sin θΔφ in the uφ direction. Setting v · ur = vr , v · uθ = vθ and v · uφ = vφ , we see that the circulation around the ur direction is ∂ ∂ (vφ r sin θΔφ)Δθ − (vrΔθ)Δφ + terms of order Δθ2 etc. ∂θ ∂φ c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  145  The area around which this circulation takes place is r2 sin θΔθΔφ , so, proceeding to the limit we have (curl v) · ur = Similarly  (curl v) · uθ =  and  (curl v) · uφ =   ∂ ∂ (vφ r sin θ) − (vθ r) /(r2 sin θ) ∂θ ∂φ  ∂ ∂ (vr ) − (r sin θvφ ) /(r2 sin θ) ∂φ ∂r  ∂ ∂ (rvθ ) − (vr ) /r ∂r ∂θ  Hence, the result.  Exercises 3.3.6 46   ∂g ∂g ∂g , , grad g = ∂x ∂y ∂z   dg ∂r dg ∂r dg ∂r , , = dr ∂x dr ∂y dr ∂z dg  x y z  = , , since r2 = x2 + y2 + z2 dr r r r 1 dg = r r dr   div [(u × r)g] = (u × r) · grad g + g∇ · (u × r) ∇ · (u × r) = r · (∇ × u) − u · (∇ × r) ∇×r=0  ⇒  But (u × r) is perpendicular to grad g =  (3.19d) (3.19f)  ∇ · (u × r) = r · curl u 1 dg r, so r dr  (u × r) · grad g = 0 Hence, div ((u × r)g) = r · curl u.  47  φ(x, y, z) = x2 y2 z3 ,  F (x, y, z) = (x2 y, xy2 z, −yz2 )  c Pearson Education Limited 2011   146  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  47(a)  ∇2 φ = 2y2 z3 + 2x2 z3 + 6x2 y2 z  47(b) grad div F = grad (2xy + 2xyz − 2yz) = (2y + 2yz, 2x + 2xz − 2z, 2xy − 2y)  47(c)    i   ∂ curl F =  ∂x   x2 y  j ∂ ∂y  xy2 z   k    ∂ ∂z   −yz2   = i(−z2 − xy2 ) + j(0) + k(y2 z − x2 )     i j k     ∂ ∂ ∂ curl (curl F) =   ∂x ∂y ∂z    −z2 − xy2 0 y2 z − x2  = i(2yz) + j(2x − 2z) + k(2xy)  48 grad [(r · r)(a · r)] = [grad (r · r)](a · r) + (r · r)grad (a · r)  (3.19b)  = 2r(a · r) + (r · r)a  div {grad [(r · r)(a · r)]} = 2div [r(a · r)] + div [(r · r)a] = 2{[div r](a · r) + r · grad (a · r)} + [grad (r · r)] · a + (r · r)div a = 2{3(a · r) + r · a} + 2r · a + 0 = 10(r · a)  c Pearson Education Limited 2011   (3.19d)  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  49 v = x3 yi + x2 y2 j + x2 yzk ∇2 v = 6xyi + 2(x2 + y2 )j + 2yzk grad div v = grad (3x2 y + 2x2 y + x2 y) = grad (6x2 y) = 12xyi + 6x2 j    i j k     ∂ ∂ ∂  2 2 3 curl v =  ∂x ∂y ∂z  = x zi − 2xyzj + (2xy − x )k    x3 y x2 y2 x2 yz     i  j k    ∂  ∂ ∂ curl(curl v) =  ∂x  ∂y ∂z    x2 z −2xyz 2xy2 − x3  = (4xy + 2xy)i + (x2 − 2y2 + 3x2 )j + (−2yz)k grad div v − curl curl v = 6xyi + 2(x2 + y2 )j + 2yzk as required.  50    i  u × v =  0  xy  j xy 0   k  xz  = xy2 zi + x2 yzj − x2 y2 k yz   div (u × v) = y2 z + x2 z = (x2 + y2 )z     i j k    ∂ ∂ ∂  curl u =  ∂x ∂y ∂z  = 0i − zj + yk   0 xy xz  v · curl u = y2 z   i   ∂ curl v =  ∂x   xy  j ∂ ∂y  0   k  ∂  ∂z  = zi + 0j − xk yz   u · curl v = −x2 z ⇒ v · curl u − u · curl v = (x2 + y2 )z c Pearson Education Limited 2011   147  148  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition   i   ∂ curl (u × v) =  ∂x   xy2 z  j ∂ ∂y  x2 yz      ∂  = −3x2 yi + 3xy2 j + 0k ∂z  −x2 y2  k  udiv v = (xyj + xzk)(y + y) = 2xy2 j + 2xyzk vdiv u = (xyi + yzk)(x + x) = 2x2 yi + 2xyzk   ∂ ∂ + yz (xyj + xzk) = xy2 j + 2xyzk (v · ∇)u = xy ∂x ∂z   ∂ ∂ (xyi + yzk) = x2 yi + 2xyzk (u · ∇)v = xy + xz ∂y ∂z   2 2 ⇒ udiv v − vdiv u + (v · ∇)u − (u · ∇)v = −3x yi + 3xy j + 0k  51(a)    r 1 =− 3 grad r r  div      r 1 1 1 = −div 3 = − 3 div r − r · grad grad r r r r3 3 =− 3 −r· r    −3r r5   =−  3 3r2 + =0 r3 r5  51(b)  curl         −k × r 1 1 = curl = curl (yi − xj) 3 k × grad 3 r r r 1 = [curl (yi − xj)] 3 + grad r   i   ∂ =  ∂x   y  j ∂ ∂y  −x    1 r3   × (yi − xj)   k  3r ∂  1 ∂z  r3 − r5 × (yi − xj) 0   c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  149    i j k   3 1 z  = (0i + 0j − 2k) 3 − 5  x y r r  y −x 0  2k 3 = − 3 + 5 (−xzi − yzj + (x2 + y2 )k) r  r      −z  k·r 1 = grad − 3 = grad grad k · grad r r r3   1 1 − (grad z) 3 = −zgrad 3 r r   1 3r = −z − 5 − 3 k r r  curl       1 3k 1 3 + grad k · grad = − 3 + 5 (−xzi − yzj k × grad r r r r + (x2 + y2 )k) 3 + 5 (xzi + yzj + z2 k) r =0  52(a)  grad  A·r r3     = grad  A r3   ·r        A A A A + (r · ∇) 3 + · ∇ r (3.19c) = 3 × curl r + r × curl r r3 r r3     1 A 1 × A + A(r · ∇) 3 + 3 = 0 + r × grad 3 r r r     A 3r 3r2 =r× − 5 ×A +A − 5 + 3 r r r  Now, a × (b × c) = (a · c)b − (a · b)c   so  3r r× A× 5 r     Hence  grad   =  A·r r3  3r r· 5 r   =   A − (A · r)  A (A · r) − 3r 3 r r5  c Pearson Education Limited 2011   3r r5  150  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  52(b)  curl  A×r r3           A A A A = (r · ∇) 3 − r ∇ · 3 − · ∇ r + 3 (∇ · r) 3 r r r r   A A·r A 3A − 3 +3 3 = − 3 − r −3 5 r r r r =  3 A (A · r)r − 3 5 r r  (A × r) × r = (A · r)r − (r · r)A (A · r)r = (A × r) × r + Ar2   3 A×r 3Ar2 A = 5 [(A × r) × r] + 5 − 3 curl 3 r r r r =2  53(a)  A 3 + 5 (A × r) × r 3 r r    i   ∂ Δ × r = curl r =  ∂x   x  53(b)   (a · ∇)r =   k  ∂  ∂z  = 0 z   j ∂ ∂y  y  ∂ ∂ ∂ + a2 + a3 a1 ∂x ∂y ∂z   (xi + yj + zk)  = a1 i + a2 j + a3 k = a  53(c) ∇ × [(a · r)b − (b · r)a] = ∇ × [(a × b) × r] = (a × b)(∇ · r) − [(a × b) · ∇]r = 3(a × b) − a × b = 2a × b c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  151  53(d) ∇ · [(a · r)b − (b · r)a] = ∇ · [(a × b) × r] = (a × b) · (∇ × r) = (a × b) · (0) = 0 54 1 ∂f 1 ∂f ∂f ur + uθ + uφ (Exercise 30) ∂r r ∂θ r sin θ ∂φ       1 ∂ sin θ ∂f 1 ∂ 1 ∂f 1 ∂ 2 ∂f r + + ∇ · (∇f) = 2 r ∂r ∂r r sin θ ∂θ r ∂θ r sin θ ∂φ r sin θ ∂φ ∇f =  (using Exercise 36)     1 ∂ ∂f ∂2 f 1 ∂ 1 2 ∂f r + 2 2 sin θ = 2 + 2 r ∂r ∂r r sin θ ∂θ ∂θ r sin θ ∂φ2 55 1 div H = c     div  ∂Z curl ∂t   =0  (3.22)  div E = div (curl curl Z) = 0 1 ∂E curl H = becomes c ∂t ∂Z 1 curl H = curl curl c ∂t 1 ∂E 1 ∂ 1 ∂Z = (curl curl Z) = curl curl c ∂t c ∂t c ∂t curl E = curl curl curl Z 1 ∂2 Z 1 ∂H = curl c ∂t c ∂t2 1 ∂2 Z 1 ∂H ⇒ curl curl curl Z = − curl curl E = − c ∂t c ∂t2 1 ∂2 Z ⇒ curl curl Z = − c ∂t2 1 ∂2 Z ⇒ grad (div Z) − ∇2 Z = − c ∂t2 Hence, ∇2 Z =  1 ∂2 Z when div Z = 0 c ∂t2  c Pearson Education Limited 2011   (3.22)  152  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Exercises 3.4.2 56      B  24  √  (2 x) 1 + 1/x dx  24  √ 4 (x + 1)3/2 2 x + 1 dx = 3  y ds = A  3   = 3  =  57  24 3  4 [125 − 8] = 156 3  B  [2xy dx + (x2 − y2 ) dy] = I  ∂S A  x2 + y2 = 1 ⇒ x dx = −y dy   y=1  [−2y2 + (1 − 2y2 )] dy  I= y=0    1  4 (1 − 4y ) dy = y − y3 3  1  =−  2  = 0  0  1 3  58 r = (t3 , t2 , t) dr = (3t2 , 2t, 1) dt     1  [(2yz + 3x2 )(3t2 ) + (y2 + 4xz)2t + (2z2 + 6xy)1] dt  V· dr = 0  C    1  [6t5 + 9t8 + 2t5 + 8t5 + 2t2 + 6t5 ] dt  = 0    1  (22t5 + 9t8 + 2t2 ) dt =  = 0  2 16 11 +1+ = 3 3 3  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 59  153  A = (2y + 3, xz, yz − x) .  A· dr where r = (2t2 , t, t3 ) and dr = (4t, 1, 3t2 ) dt  59(a) C      1  [(2t + 3)4t + (2t5 )1 + (t4 − 2t2 )3t2 ] dt  A· dr = C    0 1  8 6 1 3 − + + 3 5 3 7  (12t + 8t2 − 6t4 + 2t5 + 3t6 ) dt = 6 +  = 0  288 = 35     59(b)    Q  A · dr +  A· dr = P  C    R  S  A · dr + Q  A · dr R  where P = (0, 0, 0) , Q = (0, 0, 1) , R = (0, 1, 1) , S = (2, 1, 1) (using straight lines) On PQ  A = 3i  (x = y = 0)  On QR  A = (2y + 3)i + yk  (x = 0, z = 1) r = yj + k  On RS  A = 5i + xj + (1 − x)k (y = 1, z = 1) r = xi + j + k      3i · k dz +  A· dr = C    1 0    1  2  [(2y + 3)i + yk] · j dy + 0  r = zk  [5i + xj + (1 − x)k] · i dx 0  = 10 since i · k = 0 etc.     59(c)  S  A · dr  A· dr = C  P  where C is a straight line, P = (0, 0, 0) and S = (2, 1, 1) . Parametrically, straight line is r = (2, 1, 1)t , so     1  [(2t + 3)i + 2t2 j + (t2 − 2t)k] · (2i + j + k) dt  A· dr = 0  C      1  1  [4t + 6 + 2t + t − 2t] dt = 2  = 0  2  (2t + 6 + 3t2 ) dt 0  = [1 + 6 + 1] = 8 c Pearson Education Limited 2011   154  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  60  F is conservative if there exists a φ such that   ∂φ ∂φ ∂φ , , F = (y cos x + z , 2y sin x − 4, 3xz + z) = − ∂x ∂y ∂z 2  3  2  Such a φ is readily determined giving   1 2 2 3 F = −grad 4y − y sin x − xz − 2 z Hence, work done in moving an object is   1 F· dr = − 4y − y sin x − xz − z2 2 C 2   1 2 3 3 4t , 8t    2  F· dr = 0  C   , so that dr =  61(a) Curve is r = t,   1 2 3 3 2 3 4 F = 3t , 4 t − 4 t , 8 t   (0,1,−1)  1 − [−5 − 4π − 2] = 4π + 10.5 2  = 4−   (π/2,−1,2)  3   1,  1 9 2 2 t, 8 t  dt and  3 1 27 3t2 + t5 − t3 + t5 dt 8 8 64  3 1 9 6 t = t + t6 − t4 + 48 32 128  2  3  =8+4−  61(b)  0  1 9 + = 16 2 2  Curve is r = (2t, t, 3t), 0 ≤ t ≤ 1, so that  dr = (2, 1, 3) dt and F = (12t2 , 12t2 − t, 3t) .     1  (24t2 + 12t2 − t + 9t) dt  F· dr = 0  C    1  (36t2 + 8t) dt = 12 + 4 = 16  = 0  c Pearson Education Limited 2011     Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 61(c) No.  If F is conservative, there is a function U(x, y, z) such that  F = −grad U Test for existence of U: F · dr has to be an exact diﬀerential ∂ ∂ (3x2 ) = (2xz − y) ∂y ∂x Hence, not exact and F is not conservative.  62  F = (3x2 − y, 2yz2 − x, 2y2 z) ∂ ∂ (3x2 − y) = −1 = (2yz2 − 1) ∂y ∂x ∂ ∂ (3x2 − y) = 0 = (2y2 z) ∂z ∂x ∂ ∂ (2y2 z) = 4yz = (2yz2 − x) ∂y ∂z  Hence, conservative and F = −grad U where U = −x3 + xy − y2 z2 . div F = 6x + 2z2 + 2y2 = 0, hence not solenoidal.  C  63  155  (1,2,3)  F· dr = x3 − xy + y2 z2 (0,0,0) = 1 − 2 + 36 = 35  F = (2t3 , −t3 , t4 ) , r = (t2 , 2t, t3 ) , dr = (2t, 2, 3t2 ) dt   i   F × dr =  2t3   2t  j −t3 2   k    4  t  dt  2 3t  = [(−3t5 − 2t4 )i + (−4t5 )j + (4t3 + 2t4 )k] dt     1  [(−3t5 − 2t4 )i − 4t5 j + (4t3 + 2t4 )k] dt  F× dr = C  0  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  156   = =−     4 2 1 2 i− j+ 1+ k − − 2 5 6 5  2 7 9 i− j+ k 10 3 5  64   i  A × B =  3x + y  2   j k  −x y − z  = (3y − 3z − x)i + (y − 2z − 3x)j + (−3y − 7x)k −3 1  r = (2 cos θ, 2 sin θ, 0) dr = (−2 sin θ, 2 cos θ, 0) dθ  On circle, z = 0 and A × B = (6 sin θ − 2 cos θ)i + (2 sin θ − 6 cos θ)j − (6 sin θ + 14 cos θ)k   i   (A × B) × dr =  6 sin θ − 2 cos θ  −2 sin θ       k  −6 sin θ − 14 cos θ  dθ  0  2π  (A × B) × dr = C  j 2 sin θ − 6 cos θ 2 cos θ  {(6 sin θ + 14 cos θ)2 cos θi + (6 sin θ + 14 cos θ)2 sin θj 0  + [(6 sin θ − 2 cos θ)(2 cos θ) + (2 sin θ − 6 cos θ)(2 sin θ)]} dθ     2π  2π  sin θ cos θ dθ = 0  0      2π  2π  =0 0  2π  2  cos2 θ dθ = π  sin θ dθ = 0  1 1 sin 2θ dθ = cos 2θ 2 4  0   (A × B) × dr = 28πi + 12πj + 0k C  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Exercises 3.4.4 65(a)    3      2  3  1 2 2 1 3 x y + xy 2 3  xy(x + y) dy dx = 0  1    0 3  2  dx 1  1 2 1 x (4 − 1) + x(8 − 1) dx 2 3  = 0  3  1 3 7 2 x + x = 2 6 0 27 21 + = 24 = 2 2  65(b)    3      5  5  2  x y dy dx = 2    3  2  x dx  1  2  y dy 1  3  5  1 2 1 3 = x y 3 2 2 1 1 1 = (27 − 8) (25 − 1) 3 2 = 76  65(c)   1      2  −1  66  2  1  1 2x y + y3 (2x + y ) dy dx = dx 3 −2 −1 −2   1 16 32 16 2 dx = + = 16 = 8x + 3 3 3 −1 2  2   1  2  2   0  2   2 1 dy x2 dx y 1  0 8 8 = ln 2 = (ln 2) 3 3  x2 dx dy = y    2  c Pearson Education Limited 2011   157  158  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  67    1      1−x 2  1  2  (x + y ) dy dx = 0  0  1 x y + y3 3    0 1  = 0  68(a)    2  2x  dx   1 2    1  x2  x  1 4 1 x − (1 − x)4 4 12 1 1 − − = 12 6  dy + y2 2x  y 1 = dx tan−1 x x x 1  2 1 = (tan−1 2 − tan−1 1) dx 1 x = (tan−1 2 − tan−1 1)[ln x]21   1 −1 ln 2 = tan 3  68(b)   dx  0      1−x  (x + y) dy 0  1  = 0 1  = 0  1 xy + y2 2  1−x  dx 0  1 x(1 − x) + (1 − x)2 2  x3 1 x2 − − (1 − x)3 2 3 6 1 1 1 1 − + = = 2 3 6 3  dx 0  1 x2 (1 − x) + (1 − x)3 dx 3  1 3 = x − 3 1 1 = − 3 4   1−x  2  dx  1  = −  0  c Pearson Education Limited 2011   1 0  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition     1  68(c)  dx 0    √ √  1−x2  x−x2  159  y    1 1 − x2 − y2 √  1  dy  1 Part circle x 2 + y2 = 1  1−x2  y sin−1 √ dx Part circle 1 − x2 √x−x2  0  x2 − x + y 2 = 0   1 2 x−x sin−1 1 − sin−1 dx = 1 − x2 0   1 0 1/2 1 x x −1 −1 = dx sin 1 − sin 1+x 0  1 x π −1 = − x sin 2 1+x 0  1 √ x 1 dx + 2 0 (1 + x) √ √ 1 1 π = − sin−1 √ + x − tan−1 x 0 , using substitution x = tan2 θ 2 2   π π =1 = + 1− 4 4 =  69 y 2  (1,2)  (2,1)  1  0     1  2  x  1 sin π(x + y) dx dy 2  2  1  x  3−x 1 1 = dx sin π(x + y) dy + dx sin π(x + y) dy 2 2 0 x/2 1 x/2  1  2 x 3−x 1 1 2 2 − cos π(x + y) − cos π(x + y) = dx + dx π 2 π 2 0 1 x/2 x/2  2  1 3 2 3 2 3π 2 2 cos πx − cos πx dx + cos πx − cos dx = π 4 π π 4 π 2 0 1 c Pearson Education Limited 2011   160  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 1  2 3 8 3 8 sin πx − 2 sin πx + sin πx = 2 2 3π 4 π 3π 4 0 8 3 3π 3π 8 8 = 2 sin π + 2 sin − 2 sin 3π 4 3π 2 3π 4 8 =− 2 3π  70(a)    1  1  dx      x    π/2  70(b) 0 π/2 =  cos 2y  1 0  1 − k2 sin2 x dx  0 π/2 dx  0    y  dy    1  xy  dy 1 + y4  y  1 xy  dy dx = 1 + y4 0 0  1  1 2 y x y 2 = dy 1 + y4 0 0  1 1 3 y 1 2 2 . dy = 1 + y4 = 8 1 1 + y4 0 1 √ = ( 2 − 1) 4 0  2   cos 2y  1 − k2 sin2 x dy  x   1 = dx sin 2y 1 − k2 sin2 x 2 0 x  π/2  = − sin x cos x 1 − k2 sin2 x dx 0 −1  1 1 + k2 t dt = 2 0 (Let t = − sin2 x , then π/2  π/2  dt = −2 sin x cos x dx ) −1 1 2 3/2 (1 + k t) = 3k2 0 1 2 3/2 = 2 ((1 − k ) − 1) 3k  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition     1  71  dy  √  0    1 y    1  x2  dx  = 1  =   √  0  0    dx  y(1 + x2 )  √  2 1 √ [2 y ]x0 dx 1 + x2  √  2x dx 1 + x2  0 1  =  1 1 √ dy 2 y 1+x  0   √ = [2 1 + x2 ]10 = 2( 2 − 1)  72  1  √    x−x2  dx 0  0  x  dy x2 + y 2  Equation of circle in polar coordinates is r = cos θ   √    1  dx 0  0  x−x2      x x2  +  y2  π/2    cos θ  dy = 0    0 π/2    r cos θ r2 cos2 θ + r2 sin2 θ  r dr dθ  cos θ  =  (cos θ)r dr dθ   0  0 π/2  =   0 π/2  = 0  =    1 cos θ r2 2  cos θ  dθ 0  1 1 2 cos3 θ dθ = . 2 2 3  1 3  c Pearson Education Limited 2011     π/2  cos θ dθ 0  161  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  162  73  √    1  1−x2  dx 0  0  x+y  dy x2 + y 2  Change to polar coordinates     1  √  1−x2  dx 0  0  x+y  dy = x2 + y 2    π/2    1  r cos θ + r sin θ r dr dθ r 0 0  π/2  1 = (cos θ + sin θ) dθ r dr 0 π/2  = [sin θ − cos θ]0  1 2 r 2  0 1 0  =1    x+y dx dy + y2 + a2 over ﬁrst quadrant of circle 74  x2  using polar coordinates      π/2    a  r cos θ + r sin θ r dr dθ r2 + a2 0 0  π/2  a r2 = (cos θ + sin θ) dθ dr 2 2 0 0 r +a   a a2 1− 2 dr =2 r + a2 0   r a π = 2a 1 − = 2 r − a tan−1 a 0 4  x+y dx dy = 2 x + y2 + a2  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 75  Using polar coordinates, parabola becomes r2 sin2 θ = 4 − 4r cos θ r2 = 4 − 4r cos θ + r2 cos2 θ = (2 − r cos θ)2 r = (2 − r cos θ), 2 = 1 + cos θ    positive root because r = 2 at θ =  x2 − y 2 dx dy = x2 + y 2     π/2    2 1+cos θ  (cos2 θ − sin2 θ)r dr dθ  0  0 π/2  2(cos2 θ − sin2 θ)  =   0 π/2  = 0  1 dθ (1 + cos θ)2  2(2 cos2 θ − 1) dθ (1 + cos θ)2  = (6π − 20)/3 = −0.3835 (use the substitution t = tan 12 θ).  76  Circles are r = a cos θ, r = b sin θ and intersect at θ = tan−1 ab .    (x2 + y2 )2 dx dy = (xy)2    tan−1  a b     π/2  1 r dr dθ sin θ cos2 θ 2  0  0    b sin θ  a cos θ  + tan−1    tan−1  a b  tan−1  a b  =   0  = 0  0   π/2 1 b2 sin2 θ 1 a2 cos2 θ dθ + dθ 2 sin2 θ cos2 θ 2 sin2 θ cos2 θ tan−1 a b  π/2 1 2 1 2 2 b sec θ dθ + a cosec2 θ dθ a 2 2 −1 tan b  1 2 = b tan θ 2 =  a b  1 r dr dθ sin θ cos2 θ 2  tan−1 0  a b  1 + − a2 cot θ 2  π/2 tan−1  a b  1 1 ab + ab = ab 2 2 c Pearson Education Limited 2011   π 2  163  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  164  Exercises 3.4.6  77  [sin y dx + (x − cos y) dy]   C1 =  1 1 1 sin πx + (x − cos πx) π dx 2 2 2 0  0 1 + sin π dx 2 1 0 + [− cos y] dy π/2  1  1 1 2 1 = − cos πx − πx2 − sin πx π 2 4 2 0 0 1 + x sin π + [− sin y]0π/2 2 1 2 1 = −1 + + π − 1 + 1 π 4 2 π = −1 + + π 4   [1 − cos y] dx dy [sin y dx + (x − cos y) dy] = C  A      1  π/2  dx  = 0  1 2 πx  (1 − cos y) dy   1 π π − x − 1 + sin πx dx = 2 2 2 0 2 π 2 π π = − −1+ = −1+ 2 4 π 4 π   1     78  [(x2 y − y) dx + (x + y2 ) dy]  (1 − x2 + 1) dx dy = A      2  (2 − x2 ) dy  dx  = 0 2  x  0  1 = (2x − x ) dx = x − x4 4 0 =4−4=0 3  2  2  0  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  165   79  (xy dx + x dy) C   1  (x3 + 2x2 ) dx 0   0 1 1/2 3/2 x + x dx + 2 1 11 2 1 1 2 + = = + − 4 3 5 3 60 =     (1 − x) dx dy  (xy dx + x dy) = C  A      1  √  x  (1 − x) dy  dx  =  x2  0    1  √ (1 − x)( x − x2 ) dx  1  √  = 0   =  x − x3/2 − x2 + x3 dx  0  =  11 2 2 1 1 − − + = 3 5 3 4 60  80         (e − 3y ) dx + (e + 4x ) dy = x  2  y  2  c  (8x + 6y) dx dy A      2π  =  2  dθ   0  (8r cos θ + 6r sin θ)r dr dθ 0    2π  2  (8 cos θ + 6 sin θ) dθ  = 0  r dr 0  = 0(2) = 0  81    a  2a−x  dx 0  x  y−x dy = I 4a2 + (y + x)2   u=x+y v=x−y  ⇒  x = (u + v)/2 y = (u − v)/2  and  ∂(x, y) ∂(u, v)  =  c Pearson Education Limited 2011   1 2  1 2  1 2 − 12       =  −  1 2  166  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  y=x  ⇒  u−v=u+v  ⇒  v=0  y = 2a − x  ⇒  u − v = 4a − u − v  ⇒  u = 2a  x=0  ⇒  u = −v  ⇒  (0, 0)  (0, 0), (a, a)      2a  0  du  I=  −u  0    ⇒  −v 2 4a + u2  (2a, 0), (0, 2a)  ⇒  (2a, −2a)     2a  1 u2 −  dv = 1 du  2 4 0 4a2 + u2  2a  1 4a2 1− 2 du 4 0 4a + u2 2a π a 1 −1 u u − 2a tan 1− = = 4 2a 0 2 4  =   82    1  dy 0  y  2−y  x + y x+y e dx x2   u=x+y v = y/x  ⇒  ∂(u, v) ∂(x, y)  =   1  1   − xy2   1  x  =  1 (x + y) x2  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition    y=x  ⇒  v=1  y=0  ⇒  v=0  y=2−x  ⇒  u=2    1  2−y  dy 0  y  x + y x+y e dx = x2      2  1  eu dv  du 0 2  0  =e −1  Exercises 3.4.8       83  Surface area =  1+  ∂z ∂x  2    ∂z ∂y  +  2 dx dy  A  where A is the domain x2 + y2 ≤ 2, z = 0. z = 2 − x2 − y2 ⇒  ∂z ∂z = −2x, = −2y ∂x ∂y    1 + 4x2 + 4y2 dx dy Surface area = A  Set x = r cos θ, y = r sin θ, then  Surface area =  √    2π  dθ 0  2   1 + 4r2 r dr  0  1 (1 + 4r2 )3/2 = 2π 12 1 (27 − 1) = 2π 12  √  2  0  = 52π/12 = 13π/3  84(a)  Direction cosines of normal to S are ( 32 , 13 , 23 ) , so that    2  2  (x2 + y2 )  (x + y ) dS = S  dx dy 2/3  A  c Pearson Education Limited 2011   167  168  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  where A is the area between y = 6−2x and y = 2−2x lying between y = 0 and y = 3.    2  Thus    3  2  (x + y ) dS =  dy 2−y 2  0  S  6−y 2    3 2 (x + y2 ) dx 2 6−y  3  2 3 x3 + y2 x dy = 2−y 3 0 2 2   3 15 2 183 13 − 6y + y dy = = 4 4 0     z dS =  84(b) S    1  dx    √ + x−x2  z  √ − x−x2  0  x ∂z =− , ∂x z  ∂z y =− ∂y z        1+   dx  0  S    1  =2  2   +  ∂z ∂y  2 dy  and x2 + y2 + z2 = 1  1  z dS =  ∂z ∂x  √  √ + x−x2 √ − x−x2  1 dy  x − x2 dx  0  π = 4 (Use the substitution x = radius  1 2  1 2  +  1 2  sin t . Alternatively, recognize area of circle of  .)  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 85(a)  ⇒  169  v = (xy, −x2 , x + z)   2 2 1 dS dS = n dS = , , 3 3 3     2 2 2 1 v · dS = xy − x + (x + z) dS 3 3 3 S  S      3  3−x  [xy − 2x2 + (x + 6 − 2x − 2y)] dy  dx  =   0    0  0 3  = 3  = 0  3−x  1 2 xy − 2x2 y − xy + 6y − y2 2  dx 0  x (3 − x)2 − 2x2 (3 − x) − x(3 − x) + 6(3 − x) 2  dx  = 27/4  85(b)  Use cylindrical polar coordinates, then 2  3  dS = (i cos φ + j sin φ) dφ dz on  2  cylinder and v = (3y, 2x , z ) = (3 sin φ, 2 cos φ, z3 )  v · dS S      2π  = 0  1  (3 sin φ cos φ + 2 cos2 φ sin φ) dz  dφ 0  3 2 sin2 φ − cos3 φ = 2 3    2  z dS =  86 S  2π  =0 0   z2 / (1 − x2 − y2 ) dx dy  A  where x2 + y2 + z2 = 1 and A is the interior of the circle x2 + y2 = 1,  2  z dS = S     1 − x2 − y2 dx dy  A      2π  =  1  dθ 0  √  1 − r2 r dr  0  1 = 2π − (1 − r2 )3/2 3  1  = 0  c Pearson Education Limited 2011   2π 3  z=0  170  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition    1 dS =  87(a) S   1 1 + 4x2 + 4y2 dx dy  A  where A is the interior of the circle x2 + y2 = 2    dS =  √    2π  2  dθ 0  S   1 + 4r2 r dr  0 √  1 (1 + 4r2 )3/2 = 2π 12 = 13π/3  87(b)     2  (x + y ) dS =  √    2  r2  dθ 0  S  = 0  2π (27 − 1) 12  Surface Area  2π  2  2    1 + 4r2 r dr  0 √  2π = 16 π = 8  2  1 1 (1 + 4r2 )5/2 − (1 + 4r)3/2 5 3 0 1 1 (243 − 1) − (27 − 1) = 149π/30 5 3  2nd moment of surface area about z -axis.  87(c)    z dS =  2  (2 − r2 )  dθ 0  S  √    2π   1 + 4r2 r dr  0   =2  √    2π  2  dθ 0     1 + 4r2 r dr −  0  √ 2    2π  r2  dθ 0  0  111π 37π 26π 149π − = = = 3 30 30 10  88  Direction cosines of normal to S are ( 23 , 13 , 23 ) so that    dS =  S  dx dy 2/3  A  c Pearson Education Limited 2011    1 + 4r2 dr  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition where A is the interior of x2 + y2 = 64 lying in the ﬁrst quadrant.       π/2  dS = 0  S   89  8  dθ 0  3 π 3 1 2 r dr = × r 2 2 2 2  8  = 24π 0     x 2  y 2 dS = 1+ + dx dy 2 2  S  A  where A is the annulus between x2 + y2 = 4 and x2 + y2 = 12    dS =  12    dθ 0  S  √    2π  2  1 1 + r2 r dr 4  √   3/2  12 4 1 8π 3/2 = 2π 1 + r2 [4 − 23/2 ] = 3 4 3 2  16π (4 − 21/2 ) = 3 90  Using cylindrical polar coordinates, and Thus  and  dS = (4i cos φ + 4j sin φ) dφ dz  V = zi + 2 cos φj − 12 sin2 φzk  V · dS = (4z cos φ + 8 sin φ cos φ) dφ dz  π/2  5 V · dS = dφ (4z cos φ + 8 sin φ cos φ) dz 0  S    0 π/2  (50 cos φ + 40 sin φ cos φ) dφ  = 0  π/2  = [50 sin φ + 40 sin2 φ]0 91  = 90  F = yi + (x − 2xz)j − xyk   i   ∂ curl F =  ∂x   y  j  k  ∂ ∂y  ∂ ∂z  x − 2xz  −xy       = (x, y, −2z)    On sphere, x = a sin θ cos φ , y = a sin θ sin φ , z = a cos θ c Pearson Education Limited 2011   171  172  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  and  dS = a2 (sin θ cos φ, sin θ sin φ, cos θ) sin θ dθ dφ curl F · dS = a3 (sin2 θ cos2 φ + sin2 θ sin2 φ − 2 cos2 θ) sin θ dθ dφ = a3 (sin2 θ − 2 cos2 θ) sin θ dθ dφ  2π  π  curl F · dS = dφ a3 [sin3 θ − 2 cos2 θ sin θ] dθ 0  S    0    2π 3  =  π  1 3 sin θ − sin 3θ − 2 cos2 θ sin θ 4 4 0 π 1 2 3 3 cos 3θ + cos θ = 0 − cos θ + 4 12 3 0  a dφ 0  2π  = a3 φ 0  dθ  Exercises 3.4.10 92(a)      1  dx 0    2    3  dy    2    3    xyz dz dy dx = 1  2    2  2  0  z dz  0  1  1 8 1 1 . (4) (32 − 1) = 3 2 2 3    4  3  y dy  0  =  92(b)  x dx  1    2  2  x yz dz =  0    1  2    3  dx 0    1  2    2  3  x dx  = 0  4  xyz2 dz  dy    z2 dz  y dy 1  4  2  1 1 1 (4) (32 − 1) (43 − 23 ) 2 2 3 448 = 3  =  93     1  dz −1    z  dx 0    x+z  (x + y + z) dy =  z  {2z(x + z) + 2xz} dx  dz −1  1  x−z    1  0  3z3 dz = 0  = −1  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  173  94       π  sin(x + y + z) dx dy dz =  dx   0    π  = 0  π−x  [− cos(x + y + z)]π−x−y dy 0 0    π  π−x  dx  =   sin(x + y + z) dz 0  dx   π−x−y  dy  0  V    π−x  0  [1 + cos(x + y)] dy 0  π  [y + sin(x + y)]π−x dx 0  = 0 π  (π − x + sin π − sin x) dx  = 0  x2 + cos x = πx − 2  π  = 0  1 2 π −2 2  95       1  xyz dx dy dz =  dx 0  V    dy 0  x dx 0  0 1  = 0    1−x−y  xyz dz 0    1  =     1−x  1−x  1 y(1 − x − y)2 dy 2  1 1 2 1 x (1 − x)2 y2 − (1 − x)y3 + y4 2 2 3 4  1  1 x(1 − x)4 dx 24 0  1 1 (x − 4x2 + 6x3 − 4x4 + x5 ) dx = 24 0 1 = 720  =  c Pearson Education Limited 2011   1−x  dx 0  174  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  96      V=  dV =   V    1  √  dx dy dz V  dx  =   1  =  dz 0  x  (2 − x − y) dy  dx   2−x−y  dy x2  √  0    x  x2  0  √ 1 1 2 x − x3/2 − x − 2x2 + x3 + x4 = 2 2 0 1 11 4 2 1 2 1 − − − + + = = 3 5 4 3 4 10 30 1   dx  97    2  2    π/2  2  (x + y + z )x dx dy dz =  dφ   1  r2 r sin θ cos φ sin θr2 dr  dθ  0  V    π/2  0  0    π/2  0  r5 dr  sin θ dθ 0  =1×  1  2  cos φ dφ  =    π/2  0  1 π 1 π × × = 2 2 6 24   x2 y2 z2 (x + y + z) dx dy dz  98 V     3 2 2  x y z dx dy dz +  =   V 1   3  =  2  x dx 0    0  z3 dz  + 0    1    x3 dx  =3 0    1−z  0    y2 dy 0  V 1−y  3  y dy 0   2  0  y2 dy  0 1−x−y  z2 dz from symmetry 0  c Pearson Education Limited 2011   1−z−y  x2 dx  z dz  1−z−x  x2 dx  1−x    1  z dz + 0     2  y dy  1  x2 y3 z2 dx dy dz  x y z dx dy dz +  V 1−x−y    1−x   2 2 3  0  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition   1−x  x dx 0 1      0 1−x  {y2 [(1 − x)3 − 3(1 − x)2 y + 3(1 − x)y2 − y3 ]} dy  x3 dx  =   1 2 y (1 − x − y)3 dy 3  3  =3  0  0 1  1 3 3 1 (1 − x)6 − (1 − x)6 + (1 − x)6 − (1 − x)6 3 4 5 6  x3  = 0  1 = 60 =    1  1 60    1  x3 (1 − x)6 dx 0    1  x6 (1 − x)3 dx 0    1 1 x6 (1 − 3x + 3x2 − x3 ) dx = 60 0   1 1 1 3 3 − + − = 60 7 8 9 10 1 = 50400  99  ⎫ u = x + y + z⎬ uv = y + z ⎭ uvw = z  x = u − uv y = uv − uvw z = uvw      1 − v v − vw vw   1 − v    ∂(x, y, z) =  −u u − uw uw  =  −u ∂(u, v, w)  0 −uv uv   0    1 v vw    =  0 u uw  = u2 v  0 0 uv    v vw  u uw  0 uv    exp(−(x + y + z)3 ) dx dy dz  I=   V 1  =   dx  0    1−x  1−x−y  exp(−(x + y + z)3 ) dz  dy 0  0  x+y+z=1 x=0  ⇒  ⇒  u=y+z uv = y + z  u=1  ⇒  c Pearson Education Limited 2011   v=1  dx  175  176  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  ⎫ u = x + z⎬ uv = z ⎭ uvw = z  ⇒  y=0  ⎫ u = x + y⎬ uv = y ⎭ uvw = 0  ⇒  z=0      1  I=  1  du   0  2 −u3  u e  =  1    0 1  dv 0  1    du  0  ⇒  w=1  ⇒  w=0  e−u u2 v dv 3    0 1  1  v dv  3 1 2 1 v = − e−u 3 0 2 1 = [1 − e−1 ] 6  dw 0  1  [w]10 0  100        1  yz dx dy dz =  dx  dy  0  V    0    1  = 0    1    0    0    0  0  yz dz  1−x  1 y(2 − x − y)2 dy 2  1−x  1 [y(2 − x)2 − 2y2 (2 − x) + y3 ] dy 2  0  dx  =  2−x−y  0  dx     1−x  1  1 1 2 1 (1 − x)2 (2 − x)2 − (1 − x)3 (2 − x) + (1 − x)4 dx 2 2 3 4  1  1 1 2 2 1 t (1 + t)2 − t3 (1 + t) + t4 2 2 3 4  = =  dt  1  1 1 2 1 2 2 1 t + t3 + t 4 − t 3 − t 4 + t 4 2 3 3 4 0 2 2 1 1 2 1 1 + + = = 2 6 12 60 15  =  c Pearson Education Limited 2011   dt  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  177   Volume of prism =  dx dy dz   V 1  =   dx    0    1  2−x−y  dy  dz  0  0 1−x  (2 − x − y) dy  dx  =     1−x  0  0 1  1 (2 − x)(1 − x) − (1 − x)2 dx 2 0   1 1 2 1 2 2 − 3x + x − + x − x dx = 2 2 0 1 2 3 = −1+ = 2 6 3 =  Let the coordinates of the centroid be ( x, y , z), the taking moments about appropriate axes, we have 2 x= 3      2 x dx dy dz, y = 3  y dx dy dz,  V  2 z= 3  V   z dx dy dz V  From symmetry y = x . 3 x= 2 3 = 2      1  dx   0  0  x dz 0  1−x  x(2 − x − y)dy  dx 0  2−x−y  dy   1    1−x  0   1 3 13 x − 2x2 + x3 dx = 2 0 2 2 5 3 3 2 1 = = − + 2 4 3 8 16  c Pearson Education Limited 2011   178  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  3 z= 2 3 = 2 =  3 2  3 = 2      1  dx 0      1−x  0  1 (2 − x − y)2 dy 2 y=1−x  −1 (2 − x − y)3 6  1  0    z dz 0  dx 0  2−x−y  dy 0  1      1−x  1  − 0  dx y=0   1 1 − (2 − x)3 dx 6  1 1 3 − x − (2 − x)4 2 6 24 11 3 11 = = × 2 24 16  1  =  0  101       2π  z dx dy dz =    π/2  dφ 0  V    r3 cos θ sin θ dr  π/4    2π  0  π/2  1 sin 2θ 2  dφ  =  1  dθ  0  π/4  1 = [2π] − cos 2θ 4  π/2    1  r3 dr 0  π 1 = 4 8  π/4  102    x dx dy dz =    π/2  dφ 0  V    π/2   0  r sin2 θ.r cos φ dr  dθ 0  0    π/2    π/2 2  cos φ dφ  =  a  r3 dr  sin θ dθ 0  π 1 a4 = πa4 /16 = [1] 4 4  c Pearson Education Limited 2011   a  0  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  179  Exercises 3.4.13  F · dS  103  F = (4xz, −y2 , yz)  S  S has six faces and the integral can be evaluated as the sum of six integrals.         F · dS =  F · dS + on x=0  S  F · dS + on x=1        F · dS + on y=1 1  F · dS on y=0     +      F · dS + on z=0  F · dS on z=1  1  (0, −y2 , yz) · (−i dy dz)  = 0  0 1 1    (4z, −y2 , yz) · (i dy dz)  +   0    0    0  1    0    0    0  1  (4xz, 0, 0) · (−j dx dz)  + 1  1  (4xz, −1, yz) · (j dx dz)  + 1  1  (0, −y2 , 0) · (−k dx dy)  +   0 1    0 1  (4x, −y2 , y) · (k dx dy)  + 0    0  1      1    0  =2 − 1 +  0    1  4z dy dz + 0 +  =0 +  1  0    S  div F dV V   (z + z + 2z) dx dy dz  =   V 2π    0    π/2  dφ  =  1  y dx dy 0  1 3 = 2 2  F · dS =    (−1) dx dz + 0 + 0   104  1  2  4r cos θ.r2 sin θ dr  dθ 0  0 π/2  = [2π] [− cos 2θ]0  1 4 r 4  2  = 16π 0  c Pearson Education Limited 2011   0  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  180  dS = −k dx dy,  105 On z = 0, On z = 3,  F · dS = 0  F · dS = 9 dx dy  F = (4x, −2y , 9), 2  dS = k dx dy,  On x2 + y2 = 4,  F = (4x, −2y2 , 0),  F = (8 cos φ, −8 sin φ, z2 )  dS = (i cos φ + j sin φ)2 dφ dz,  and F · dS = 16(cos2 φ − sin3 φ) dφ dz     F · dS =      2π  9 dx dy + 0  S  (z=3)    3  16(cos2 φ − sin3 φ) dz  dφ 0  2π  cos2 φ − sin3 φ dφ  = 36π + 48 0  = 84π  div F dV = (4 − 4y + 2z) dx dy dz   V    V 2π    =    2  dφ 0    0    2π  (21 − 12r sin φ)r dr  0    (4 − 4r sin φ + 2z)r dz 0  2  dφ  =  3  dr  0 2π  (42 − 32 sin φ) dφ = 84π  = 0  106  div (F × grad φ) = grad φ · curl F − F · curl (grad φ) and curl (grad φ) ≡  0 for all φ. grad φ · curl F dV =  Hence V  107     F · dS =  S  (F × grad φ) · dS S    F · dS +  on x=0  F · dS + on y=0           F · dS +  on z=0  F · dS on z=1  F · dS  + on x2 +y 2 =4  On x = 0, F = (y2 , 0, 0), dS = −i dy dz,  F · dS = on x=0  2 0  dy  1 0  −y2 dz = − 83  On y = 0, F = (0, 0, 0) , so contribution is zero On z = 0, F = (xy + y2 , x2 y, 0) , 2  2  On z = 1, F = (xy + y , x y, 0) ,  dS = −k dx dy , so contribution is zero dS = k dx dy , so contribution is zero  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition On x2 + y2 = 4, F = (4(sin φ cos φ + sin2 φ), 8 cos2 φ sin φ, 0) ,  181  dS = 2(cos φi +  sin φj) dφ dz         π/2  F · dS = 0  on x2 +y 2 =4  1  8 sin φ cos2 φ + 8 sin2 φ cos φ + 16(cos2 φ sin2 φ) dz  dφ 0  8 1 8 = − cos3 φ + sin3 φ + 2φ − sin 4φ 3 3 2 16 = +π 3       V π/2    0    0  π/2    = 0  0 2  (r2 sin φ + r3 cos2 φ) dr  dφ 0  1  (r sin φ + r2 cos2 φ)r dz  dr   π/2  =     2  dφ  =  =  0  (y + x2 ) dx dy dz  div F dV = V  π/2  0  8 sin φ + 4 cos2 φ 3   dφ  8 +π 3  Hence, result  108    i   ∂ curl F =  ∂x   36xz + 6y cos x  j ∂ ∂y      ∂  ∂z  2 18x − cos y  k  3 + 6 sin x + z sin y = i(sin y − sin y) + j(36x − 36x) + k(6 cos x − 6 cos x) =0  Hence, there is a function φ(x, y, z) , such that F = grad φ  c Pearson Education Limited 2011   182  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition     109  curl A · dS  A · dr = C  S    i j k  ∂ ∂ ∂ curl A =  ∂x ∂y ∂z  −y x 0   2k · dS A · dr = C  Let S be the ellipse Then,  2  x a2  +      = 2k    S 2  y b2  = 1, z = 0  dS = k · dx dy , and    A · dr = 2  dx dy = 2πab S  110   i   ∂ curl F =  ∂x   2x − y =k      ∂ ∂y  −yz2   2π  curl F · dS = 16  π/2  k · (sin θ cos φi + sin θ sin φj + cos θk) sin θ dθ  dφ 0  S   k   ∂ ∂z  = i(−2yz + 2yz) + j(0) + k(1)  −y2 z   j    0    2π  π/2  dφ  = 16 0  sin θ cos θ dθ = 16[2π] 0  1 sin2 θ 2  π/2 0  = 16π  F · dr C  On circle x2 + y2 = 16, z = 0, x = 4 cos φ , y = 4 sin φ , r = 4(cos φ, sin φ, 0) F = (8 cos φ − 4 sin φ, 0, 0),     2π  F · dr = C  dr = 4(− sin φ, cos φ, 0) dφ  (−32 cos φ sin φ + 16 sin2 φ + 0) dφ 0  = 16π c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 111  curl (af(r)) = −a × grad f(r)  −(a × grad f(r)) · n dS = af(r)· dr    C  S     −(grad f(r) × n) · a dS =  ⇒   a·   (n × grad f(r)) dS = a ·  S     n × grad f(r) dS =  f(r) dr C  S  ⇒  f(r) = 3xy2  1  f(r) dr C  ⇒    af(r)· dr C  S  ⇒  183  grad f(r) = (3y2 , 6xy, 0)  n × grad f(r) = k × grad f(r) = (−6xy, 3y2 , 0)  2  1 2 dx (−6xy, 3y , 0) dy = (−12x, 8, 0) dx  0  0  0  = (−6, 8, 0)  1  f(r) dr = 0.i dx +    0  C    2  3y j dy + 0    0  2  0  12xi dx + 1  0.j dy 2  = (−6, 8, 0)    curl F · dS =  112  F · dr C  S    i   ∂ curl F =  ∂x   2y + z   j ∂ ∂y  x−z      ∂ ∂z  = (2, 2, −1) y − x k   curl F · dS =  S  (2, 2, −1) · (sin θ cos φ, sin θ sin φ, cos θ) sin θ dφ dθ S      π/2  (2 sin θ cos φ + 2 sin θ sin φ − cos θ) sin θ dθ  dφ  = 0   =  0  π/2    π/2  0  1 π sin φ − 2   dφ =  3π 4  c Pearson Education Limited 2011   184  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Here, C has three portions: ⇒  On z = 0, r = (cos φ, sin φ, 0)  dr = (− sin φ, cos φ, 0) dφ  and F = (2 sin φ, cos φ, sin φ − cos φ)     π/2  (−2 sin2 φ + cos2 φ) dφ = −  F · dr = 0  On y = 0, r = (sin θ, 0, cos θ)  ⇒  π 4  dr = (cos θ, 0, − sin θ) dθ  and F = (cos θ, sin θ − cos θ, − sin θ)     π/2  F · dr =  (cos2 θ + sin2 θ) dθ = 0  ⇒  On x = 0, r = (0, sin θ, cos θ)  π 2  dr = (0, cos θ, − sin θ) dθ  and F = (2 sin θ + cos θ, − cos θ, sin θ)     0  (− cos2 θ − sin2 θ) dθ =  F · dr = π/2  π 2  Review Exercises 3.7 1   −y  ∂u = nxn−1 f(t) + xn f (t) ∂x x2   ∂u 1 = xn f (t) ∂y x x  ∂u ∂u +y = nxn f(t) = nu ∂x ∂y  (1)  Diﬀerentiate (1) w.r.t. x ∂u ∂2 u ∂u ∂2 u +x 2 +y =n ∂x ∂x ∂x∂y ∂x  (2)  Diﬀerentiate (1) w.r.t. y x  ∂u ∂2 u ∂2 u ∂u + +y 2 =n ∂x∂y ∂y ∂y ∂y  c Pearson Education Limited 2011   (3)  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition x × (2) + y × (3)  185  ⇒    2 2 ∂u ∂u ∂u ∂2 u ∂u 2∂ u 2∂ u +y +x +y +y x + 2xy =n x ∂x ∂y ∂x2 ∂x∂y ∂y2 ∂x ∂y ⇒  x2  2 ∂2 u ∂2 u 2∂ u + 2xy = n(n − 1)u + y ∂x2 ∂x∂y ∂y2  u(x, y) = x4 + y4 + 16x2 y2 ∂u = x(4x3 + 32xy2 ) x ∂x ∂u = y(4y3 + 32x2 y2 ) y ∂y x x2  ∂u ∂u +y = 4(x4 + 16x2 y2 + y4 ) ∂x ∂y  2 ∂2 u ∂2 u 2∂ u + y + 2xy = x2 (12x2 + 32y2 ) + 2xy(64xy) + y2 (12y2 + 32x2 ) ∂x2 ∂x∂y ∂y2 = 12(x4 + y4 + 16x2 y2 )  2  ∂f ∂f ∂f ∂2 f ∂2 f ∂2 f ∂2 f = + , 2 = + + 2 ∂x ∂u ∂v ∂x ∂u2 ∂u∂v ∂v2 ∂2 f ∂2 f ∂2 f ∂2 f ∂2 f =a 2 +b + a b+ ∂x∂y ∂u ∂u∂v ∂v2 ∂v∂u ∂f ∂2 f 2 ∂2 f ∂f ∂f ∂2 f ∂2 f 2 a + 2 b = a + b, 2 = ab + ∂y ∂u ∂v ∂y ∂u2 ∂u∂v ∂v2 9  ⇒  2 2 ∂2 f ∂2 f ∂2 f 2 ∂ f 2 ∂ f + 2 − 9 =(9 − 9a + 2a ) + (9 − 9b + 2b ) ∂x2 ∂x∂y ∂y2 ∂u2 ∂v2  2  ∂ f 9 + 2 9 − (a + b) + 2ab 2 ∂u∂v  ⎫ 9 − 9a + 2a2 = 0 ⎪ ⎪ ⎬ 2 9 − 9b + 2b = 0 ⎪ ⎪ ⎭ 9 9 − 2 (a + b) + 2ab = 0  ⇒  a = b  ⇒  ∂2 f =0 ⇒ f = F(u) + G(v) ∂u∂v i.e. f(x, y) = F(x + 3y) + G(x + 3y/2) c Pearson Education Limited 2011   a = 3,  b=  3 2  186  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition f(x, 0) = F(x) + G(x) = sin x ∂f 3 (x, 0) = 3F (x) + G (x) = 3 cos x ∂y 2 1 F(x) + G(x) = sin x + k 2  ⇒  1 G(x) = −k and F(x) = sin x + 2k 2 ⇒ f(x, y) = sin(x + 3y)  ⇒  3    i j k    ∂ ∂   ∂ ∇ × (P, Q, R) =  ∂x ∂y ∂z   ∂f ∂f ∂f    ∂x ∂y ∂z  2   2   2  ∂ f ∂ f ∂ f ∂2 f ∂2 f ∂2 f =i +j +k − − − ∂y∂z ∂z∂y ∂x∂z ∂z∂x ∂x∂y ∂y∂x =0 ⇒ 4(a)  ∇ × (∇f) ≡ 0  xy = c, hyperbolas grad f = (y, x) =  y= That is, tangent in direction of t · grad f =  4(b)  x2  x =c + y2  c x   c x  ⇒   , x on hyperbola dy c =− 2 dx x   1, − xc2 = t  c c − =0 x x  that is, orthogonal  circles, centres on x -axis, through (0,0)  grad f =  −2y y2 − x2 , 2 2 2 2 (x + y ) (x + y2 )2  c Pearson Education Limited 2011     Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 5(a) ω · r) = ω × (∇ × r) + r × (∇ × ω) + (ω ω · ∇)r + (r · ∇)ω ω grad(ω =ω×0+0+ω+0 =ω 5(b) ω × r) = −(ω ω · ∇)r + ω(∇ · r)(+r · ∇)ω ω − r(∇ · ω) curl (ω ω + 3ω ω+0+0 = −ω ω = 2ω 6(a)  See problem 3 above.  6(b) div v = div {grad [zf(r)] + αf(r)k} = div {kf(r) + z grad f(r)} + αk · ∇f(r) = k · ∇f(r) + k · grad f(r) + z∇2 f(r) + αk · ∇f(r) ∂f = (2 + α) ∂z 2 ∇ v = ∇(∇ · v) − ∇ × (∇ × v)   ∂f − ∇ × (∇ × (∇(zf) + αfk)) = (2 + α)∇ ∂z ∇ × ∇(zf) ≡ 0 ∇ × (αfk) = α∇f × k ∇ × (∇ × αfk) = α(k · ∇)∇f − αk(∇2 f)   ∂f ∂ = α (∇f) = α∇ ∂z ∂z   ∂f ⇒ ∇2 v = 2∇ ∂z 7  F = (x2 − y2 + x)i − (2xy + y)j   i j   ∂ ∂ ∇×F= ∂x ∂y   x2 − y2 + x −2xy − y   k  ∂  ∂z  = (0, 0, −2y + 2y) = 0  0   c Pearson Education Limited 2011   187  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  188   ∇f =  ∂f ∂f ∂f , , ∂x ∂y ∂z  ⇒    = (x2 − y2 + x, −2xy − y, 0)  f(x, y, z) =   (2,1)  (2,1)  F · dr = (1,2)  x3 x2 − y 2 − y2 x + +c 3 2 (2,1)  grad f · dr = [f](1,2) =  (1,2)  22 3  dr = i dx + j dy = (i − j) dx as on y = 3 − x ,   dy = − dx   2  2  (x − y + x + 2xy + y) dx = 2  (x2 − (3 − x)2 + x + 2(3 − x) + 3 − x) dx  2  1  1  22 = 3  8   W=  F· dr C  r = (1 − cos θ)i + sin θj dr = (sin θi + cos θj) dθ  8(a)  F = 2 sin 12 θi      θ θ cos dθ 2 2 0 π θ 8 8 sin3 = = 3 2 0 3  F· dr = C  8(b)  π  4 sin2  F = 2 sin θ2 n̂ = 2 sin θ2 (sin θi + cos θj)     π  F· dr = C  0  θ θ 2 sin dθ = 4 − cos 2 2  c Pearson Education Limited 2011   π  =4 0  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 9  0≤t≤1  r = (i + j + k)t dr = (i + j + k) dt  1 F · dr W= 0  F = (xy, −y, 1) = (t2 , −t, 1)  1 1 1 5 (t2 − t + 1) dt = − + 1 = W= 3 2 6 0  ⇒   10  dr × B  F=I C  θ r = sin θi + cos θj + sin k 2   θ 1 dr = cos θi − sin θj + cos k dθ 2 2 B = sin θi − cos θj + k     2π  θ 1 θ 1 cos cos θ − sin θ + j cos sin θ − cos θ i F=I 2 2 2 2 0  4 + k(sin2 θ − cos2 θ) dθ = Ij 3 11   v · dr  Circulation =  C  −1  = 1  −y dx + x2 + y 2    −1 1  x dy + 2 x + y2    1 −1  y dx + 2 x + y2    1 −1  x dy + y2  x2  on y = 1 on x = −1 on y = −1 on x = 1  −1  1  1  1 1 dy dx dy dx + − + = 2 2 2 2 1+y −1 1 + x 1 −1 1 + x −1 1 + y =0   12  ρ(x2 + y2 ) dA,  Iz = A      c  c  (x2 + y2 )kxy dy  dx  = 0  where density ρ = kxy  x2 /c  c Pearson Education Limited 2011   189  190  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition   c  c  1 kx(x2 + y2 )2 = dx 4 0 x2 /c     c 4 2 1 x 1 = − kx x2 + 2 + kx(x2 + c2 )2 dx 4 c 4 0   c  9 1 x 2x7 3 2 4 dx = k − + 2 − 2x c − xc 4 0 c4 c 13 6 = kc 80  13  Equation of cone is x2 + y2 =  V=2  dx   c  z dy 0    a  √ a2 −x2  h 2 h+ x + y2 a  dx  =2   0  c  − h)2  √ a2 −x2    a  a2 h2 (z   dy √  a −x hx2 hy  2 −1 y 2 =2 hy + x +y dx sinh − 2a x 2a c 0  √  a √ 2 2 − x2 a h 2 hx sinh−1 dx a − x2 + =2 2 2a x c  √   hc √ 3 2 − c2 a 2ha2  π hc −1 −1 c − sin − tanh = a2 − c2 − 3 2 a 3 3a a  14  a  Volume is 8  2  2  dV x y z≥0  x2 + y2 = a2 is a cylinder with z -axis as axis of symmetry and radius a. z2 + y2 = a2 is a cylinder with x -axis as axis of symmetry and radius a.  ⇒  a   √a2 −y2   √a2 −y2  dy dx dz 0 0  a √ √ 2 2 a2 −y 2 a −y =8 [x]0 [z]0 dy 0  a a 1 16a3 (a2 − y2 ) dy = 8 a2 y − y3 = =8 3 3 0 0  V=8  0  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 15  Elastic energy of ΔV is q2 ΔV/(2EI) where q = q0 ρ/r and ρ is the distance  from the centre and r is radius of cylinder.     2π  Total energy =  dφ 0    =  On x = 0,  dS = −i dy dz  On y = 0,  v · dS = 0  On z = 1,  v · dS = 0  ⇒  On  0 2 2 πq0 r l  0  l  q20 ρ3 dz 2EIr2  q20 ρ3 dρ 2EIr2  4EI and  v · dS = −3x2 y dy dz ≡ 0  x + y = 1, dS =   1  dx  0    dρ  z = 0, v · dS = 0  On    ⊂⊃ v · dS =    r  0 r  = 2πl  16  0  1  √1 (i 2  + j) dS   √ 3 2 1 2 √ x (1 − x) + √ x(1 − x) 2 dz 2 2  1  (2x2 + x)(1 − x) dx  = 0  1 = 3  17  191   ⊂⊃ v · dS S  dS = (i sin θ cos φ + j sin θ sin φ + k cos θ)a2 sin θ dθ dφ  On S,  and v = i2a sin2 θ cos φ sin φ − ja2 sin2 θ sin2 φ + k(a sin θ cos φ + a sin θ sin φ)   ⊂⊃ v · dS =  2π  {2a3 sin4 θ cos2 φ sin φ − a4 sin4 θ sin3 φ  dθ 0  S    π  0  + a3 sin2 θ cos θ cos φ + a3 cos θ sin2 θ sin φ} dφ =0  F · dr  18 C  C is the circle x2 + y2 = 16, z = 0, so that, on the circle   F = x2 + y − 4, 3xy, 0 ,  r = 4 (cos θ, sin θ, 0)  c Pearson Education Limited 2011   192  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  and d r = (−4 sin θ, 4 cos θ, 0) dθ     2π    θ=0 2π       −16 4 cos2 θ + sin θ − 1 sin θ + 192 cos θ sin2 θ dθ  F · dr = C  −16 sin2 θ dθ  =  (from symmetries)  0  = −16π   i   ∂ curl F =  ∂x   x2 + y − 4      ∂  = (0, −2z, 3y − 1) ∂z  2 2xz + z  j  k  ∂ ∂y  3xy  On the hemisphere r = 4 (sin θ cos φ, sin θ sin φ, cos θ) dS = 16 (sin θ cos φ, sin θ sin φ, cos θ) sin θ dφ dθ   curl F . dS = S      π/2  2π  16(−8 sin2 θ cos θ sin φ + 12 sin2 θ cos θ cos φ − cos θ sin θ)dφ  dθ 0  0    π/2  −16 cos θ sin θ [2π] dθ = −16π  = 0  19            a · dS =  div a dV −  S  V  a · dS S1  where V is the hemisphere x2 + y2 + z2 = a2 (diﬀerent a from the vector a), S1 is the circle x2 + y2 = a2 , z = 0. div a = 0 and d S = −k dx dy on S1      a · dS =  S1  Hence  (xi + yj) · (−k dx dy) = 0 S    a · dS = 0 S  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 20        1    1−x    xyz dV = V  2−x  xyz dz dy dx   0    0  1    0  0 1−x  = 0  1 2 xy (2 − x) dy dx 2  1  1 2 2 x (1 − x) (2 − x) dx 0 4  1  1 5 x − 6x4 + 13x3 − 12x2 + 4x dx = 0 4 1 3 13 1 13 = − + −1+ = 24 10 16 2 240  =  21  −u  −v  y + ∆y  v  x  u  y x + ∆x  Net circulation (anti clockwise) is −u(x, y)Δx − v(x + Δx, y)Δy + u(x, y + Δy)Δx + v(x, y)Δy If net circulation is zero then, dividing by ΔxΔy , u(x, y + Δy) − u(x, y) v(x + Δx, y) − v(x, y) − =0 Δy Δx Δx, Δy → 0 gives ∂u ∂v − = 0. ∂y ∂x Since u = − ∂ψ ∂y and v =  ∂ψ ∂x  we obtain ∂2 ψ ∂2 ψ + 2 =0 ∂x2 ∂y  Laplace equation.  c Pearson Education Limited 2011   193  4 Functions of a Complex Variable Exercises 4.2.2 1(a)  If | z − 2 + j |=| z − j + 3 |, so that | x + jy − 2 + j |=| x + jy − j + 3 |  or  (x − 2)2 + (y + 1)2 = (x + 3)2 + (y − 1)2 x2 − 4x + 4 + y2 + 2y + 1 = x2 + 6x + 9 + y2 − 2y + 1  Then cancelling the squared terms and tidying up, y=  1(b)  5 5 x+ 2 4  z + z∗ + 4j(z − z∗ ) = 6  Using, z + z∗ = 2x, z − z∗ = 2jy gives 2x + 4j2jy = 6 3 1 y= x− 4 4  2  The straight lines are | z − 1 − j | =| z − 3 + j | | z − 1 + j | =| z − 3 − j |  which, in Cartesian form, are (x − 1)2 + (y − 1)2 = (x − 3)2 + (y + 1)2 that is,  x2 − 2x + 1 + y2 − 2y + 1 = x2 − 6x + 9 + y2 + 2y + 1 y=x−2 c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  195  and (x − 1)2 + (y + 1)2 = (x − 3)2 + (y − 1)2 i.e.  x2 − 2x + 1 + y2 + 2y + 1 = x2 − 6x + 9 + y2 − 2y + 1 y = −x + 2  These two lines intersect at π/2 (the products of their gradients is −1) and y = 0, x = 2 at their intersection, that is, z = 2 + j0.  3  w = jz + 4 − 3j can be written as w = ejπ/2 z + 4 − 3j (since j = cos  π π + j sin = ejπ/2 ) 2 2  which is broken down as follows:  z  −→  −→  ejπ/2 z  rotate  translation  anticlockwise by 12 π  (0, 0) → (4, −3)  ejπ/2 z + 4 − 3j = w  Let w = u + jv and z = x + jy so that,  u + jv = j(x + jy) + 4 − 3j = jx − y + 4 − 3j  that is,  u = −y + 4  (1)  v=x−3  (2)  Taking 6 times equation (2) minus equation (1) gives, 6v − u = 6x + y − 22 so that, if 6x + y = 22, we must have 6v − u = 0 so that, u = 6v is the image of the line 6x + y = 22  c Pearson Education Limited 2011   196 4  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition Splitting the mapping w = (1 − j)z into real and imaginary parts gives  u + jv = (1 − j)(x + jy) = x + y + j(y − x) that is,  u=x+y v=y−x  so that,  u + v = 2y  Therefore y > 1 corresponds to u + v > 2.  5  Since w = jz + j x = v − 1, y = −u  so that x > 0 corresponds to v > 1.  6  Since w = jz + 1 v=x u = −y + 1  so that x > 0 ⇒ v > 0 and 0 < y < 2 ⇒ −1 < u < 1 or | u |< 1. This is illustrated below  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  7 Given w = (j + parts,  √  197  √ 3)z + j 3 − 1, we obtain, on equating real and imaginary  √ √ √ u = x 3 − y − 1, v = x + y 3 + 3 √ √ or v 3 − u = 4y + 4, and v + u 3 = 4x  on rearranging. Thus 7(a)  √ √ y = 0 corresponds to v 3 − u = 4 or u = v 3 − 4  7(b)  √ √ x = 0 corresponds to v + u 3 = 0 or v = −u 3  7(c)  √ √ √ Since u + 1 = x 3 − y and v − 3 = x + y 3 squaring and adding gives (u + 1)2 + (v −  √ 2 √ √ 3) = (x 3 − y)2 + (x + y 3)2 = 4x2 + 4y2  Thus, x2 + y2 = 1 ⇒ (u + 1)2 + (v − 7(d)  √ 2 3) = 4  √ √ Since v 3 − u = 4y + 4 and v + u 3 = 4x , squaring and adding gives 4v2 + 4u2 = 16(y + 1)2 + 16x2 or  u2 + v2 = 4(x2 + y2 + 2y + 1)  Thus, x2 + y2 + 2y = 1 corresponds to u2 + v2 = 8  c Pearson Education Limited 2011   198 8(a)  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition w = αz + β  Inserting z = 1 + j, w = j and z = −1, w = 1 + j gives the following two equations for α and β j = α(1 + j) + β  1 + j = −α + β  or  from which, by subtraction, −1 = (2 + j)α or so that, β = 1 + j + α =  8(b) gives  α=  1 (−2 + j) 5  1 (3 + 6j) gives 5w = (−2 + j)z + 3 + 6j 5  Writing w = u + jv, z = x + jy and equating real and imaginary parts 5u = −2x − y + 3 5v = x − 2y + 6  Eliminating y yields 5v − 10u = 5x  or  v − 2u = x  5u + 10v = −5y + 15  or  u + 2v = −y + 3  Eliminating x yields  so that y > 0 corresponds to u + 2v < 3  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 8(c)  From part (b), x = v − 2u y = 3 − u − 2v  Squaring and adding gives x2 + y2 = (v − 2u)2 + (3 − u − 2v)2 = 5(u2 + v2 ) − 6u − 12v + 9 | z |< 2 ⇒ x2 + y2 < 4 so that, 5(u2 + v2 ) − 6u − 12v + 5 < 0 or (5u − 3)2 + (5v − 6)2 < 20  4  2 √ 2 6 2 20 3 2  = = + v− < 5 that is, v − 5 5 25 5 5  8(d)  The ﬁxed point(s) are given by 5z = (−2 + j)z + 3 + 6j so that,  3 + 6j (3 + 6j)(7 + j) = 7−j 50 3 (1 + 3j) = 10  z=  Exercises 4.2.5 9  Writing w =  1 1 , z= z w u − jv 1 = 2 u + jv u + v2 v y=− 2 u + v2  ⇒ x + jy = so that  c Pearson Education Limited 2011   199  200  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  v >c + v2 v or rearranging u2 + v2 + < 0 c  If y > c then, −  u2    1 u + v+ 2c  2  2   <  1 2c  2  v >c⇒v<0 + v2 v If c < 0, put c = −d and − 2 > −d u + v2  1 2  1 2 > or, on rearranging, u2 + v − 2d 2d If c = 0, −  u2  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition   1  3 7 3 1 in z + + j = gives  + + j = 10 Putting z = w 4 4 w 4    7 3 1 + + j w = | w | which, writing w = u + jv and expanding, gives 4 4  2  2 3 3 49 2 1+ u−v + v+u = (u + v2 ) 4 4 16  201 7 or 4  or, on rearranging gives 2 4 u2 + v2 − u + v − = 0 3 3  2  2  2 1 2 7 u− + v+ = 2 3 6 a circle centre  1 2 7 ,− and radius . 2 3 6  1 1 in | z − a |= a gives | 1 − aw |= a | w | from which u = w 2a can be obtained (on writing w = u2 + v2 ). 1 1 1 under w = · · · = ; that is, the half Hence | z − a |= a maps to Re{w} = 2a z 2a 1 . plane Re{w} > 2a 1 . The Moreover, the interior of | z − a |= a maps to right of the line Re{w} = 2a 2 1 point z = a mapping to w = conﬁrms this. 2 a 11  Putting z =  12  The general bilinear mapping is w=  az + b cz + d  with z = 0, w = j ⇒ b = jd , c Pearson Education Limited 2011   202  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  z = −j, w = 1 ⇒ d − jc = b − ja and z = −1, w = 0 ⇒ a = b Hence b = a, d = −ja and c = ja and the mapping is w=  z+1 j(z − 1)  Making z the subject of this formula, we obtain z=  jw + 1 jw − 1  Writing z = x + jy, w = u + jv and equating real and imaginary parts  x=  −2u u2 + v2 − 1 , y= 2 2 2 u + (v + 1) u + (v + 1)2  Lines x = constant = k , say, transform to  or  k[u2 + (v + 1)2 ] = u2 + v2 − 1 1 2k v=− u2 + v2 + k−1 k−1  This can be rewritten as   k u + v+ k−1 2  2 =  1 k 1 = − 2 (k − 1) k−1 (k − 1)2  which are circles (except k = 1 which is v = −1). Lines y = constant = l, say, transform to  or  2u u2 + (v + 1)2 + =0 l 2  1 1 + (v + 1)2 = 2 u+ l l  which are circles (except l = 0 which is u = 0). c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition The ﬁxed points are given by z+1 jz − j 2 jz − (j + 1)z − 1 = 0 z=  or  (j + 1) ±    (j + 1)2 + 4j 2j √ (j + 1) ± 6j = 2j √ 1 = (j − 1)(−1 ± 3) 2  z=  √ √ 6j = ±(1 + j) 3 ).  (since  13  w=  1+j z  13(a) z=1⇒w=1+j 1+j z=1−j⇒w= =j 1−j z=0⇒w=∞ c Pearson Education Limited 2011   203  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition √ 2 |1+j| 13(b) | w |= = |z| √ |z| √ 2 < 1 ⇒| w |> 2 so that | z |= |w| that is, interior of the unit circle maps to the exterior of the circle, centre as the √ origin and radius 2. 204  13(c)  z=  1+j w (1 + j) (u − jv) so that u2 + v2 u+v u−v x= 2 , y= 2 2 u +v u + v2  ⇒ x + jy =  Therefore x = y corresponds to v = 0 (the real axis) and x + y = 1 corresponds 2u to 2 = 1 that is (u − 1)2 + v2 = 1 a circle, centre (1.0) and radius 1. u + v2 The ﬁxed point of the mapping is given by z2 = 1 + j . Using the polar √ form 1 + j = 2eπj/4 , so z = ±21/4 eπj/8  13(d)  14  The bilinear transformation w=  z+1 z−1  Writing z = x + jy, w = u + jv and equating real and imaginary parts gives x2 + y2 − 1 2y u= , v= 2 2 (1 + x) + y (1 + x)2 + y2 Hence, all points on the circle x2 + y2 = 1 correspond to u = 0. From the point (0, −1) to the point (0, 1) on the circle x2 + y2 = 1 we use the Parameterization x = cos θ, y = sin θ, π/2 ≤ θ ≤ 3π/2. Using v = 2y 2y y = we note that v = on x2 + y2 = 1, so that 2 2 2 2 (1 + x) + y 1 + x + y + 2x 1+x 2 sin 12 θ cos 12 θ 1 sin θ π 3π 1 = tan v= = θ and between θ = and θ = , tan θ 1 1 + cos θ 2 2 2 2 2 cos2 2 θ ranges from 1 to ∞ and from −∞ to −1 hence | v |≥ 1.  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  205  15(a)  With w = u + jv and z = x + jy 3w + j z+j implies z = from which we deduce that The transformation w = z−3 w−1 3(u2 + v2 ) − 3u + v , y= (u − 1)2 + v2 x2 + y2 − 3x + y u= , v= (x − 3)2 + y2  x= and  u − 3v − 1 (u − 1)2 + v2 x − 3y − 3 (x − 3)2 + y2  The line y = 0 corresponds to the line u − 3v − 1 = 0 in the w plane. The line x = y corresponds to the curve  that is,  3(u2 + v2 ) − 3u + v = u − 3v − 1 2  2  2 5 2 + v+ = u− 3 3 9  (1)  2 2 1√ ,− and radius 5 in the w plane. 3 3 3 The origin in the z plane (the intersection of the line y = 0 and x = y ) corresponds 1 to the point w = −j in the w plane. The point at inﬁnity in the z plane (the 3 other 'intersection') corresponds to the point w = 1 in the w plane. a circle centre  The origin (in the w plane) lies outside the circle (1), and is also outside the wedge shaped region in the z plane (z = −j3 is its image). So, the following ﬁgure can be drawn:  2 lies inside the shaded region in the w plane, and corresponds to 3 3· 2 + j = −3(2 + j) = −6 − 3j inside the shaded region of the z the point z = 2 3 3 −1 plane. (This is a useful check.) The point w =  c Pearson Education Limited 2011   206  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  15(b)  The fact that w = 1 does not correspond to any ﬁnite value of z has  already been established. z+j . z−3 Taking the modulus of both sides gives Consider the equation w =  z + j  | w |=  z−3 If  | w |= 1 ⇒| z + j | =| z − 3 | or  x2 + (y + 1)2 = (x − 3)2 + y2  x2 + y2 + 2y + 1 = x2 − 6x + 9 + y2 so that, or  2y = −6x + 8  y + 3x = 4  Hence the unit circle in the w plane, | w |= 1, corresponds to the line y + 3x = 4. z−j z+j −wj − j then z = w−1 w + 1 . so that | z |=  w−1 So if | z |= 2, | w + 1 |= 2 | w − 1 |  16  If w =  or (u + 1)2 + v2 = 4(u − 1)2 + 4v2 which simpliﬁes to   5 u− 3  2  16 , a circle centre +v = 9 2     5 4 , 0 and radius 3 3  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 17  If w = ejθ0  taking modulus  207  z − z0 then, z − z∗0  z − z0   since | ejθ0 |= 1 | w |=  z − z∗0  If z is real (i.e. z is on the real axis) then  | z − z0 |=| z − z∗0 |= (x − x0 )2 + y20  1/2  and z0 = x0 + jy0  Hence | w |= 1. Thus the real axis in the z plane corresponds to the unit circle | w |= 1 in the w plane. Making z the subject of the transformation gives z=  wz∗0 − ejθ0 z0 w − ejθ0  Hence the origin in the w plane maps to z = z0 . Thus the inside of the unit circle in the w plane corresponds to the upper half of the z plane provided Im{z0 } > 0 Since w = 0 maps to z = z0 and z0 = j and z = ∞ maps to w = ejθ0 = −1 it gives θ0 = π.  18  For the transformation w=  2jz z+j  the ﬁxed points are given by 2jz z+j 2 z + jz = 2jz z=  or  z(z − j) = 0, z = 0 or j  Hence circular arcs or straight lines through z = 0, j are transformed to circular arcs or straight lines through w = 0, j by the properties of bilinear transformation (section 4.2.4). The inverse transformation is z=  jw 2j − w  c Pearson Education Limited 2011   208  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition   jw 1 1 1 1 |< becomes  −  < which simpliﬁes to | w − 1 |< 1 (use 2 2 2j − w 2 2 w = u + jv and split into real and imaginary parts). 1 1 4 2 Similarly, | z − |< becomes | w − |> 2 2 3 3 | z−  19  The general bilinear mapping is w=  az + b cz + d  if w = 0 corresponds to z = z0 then w=  (z − z0 )ejθ0 cz + d   z − z0   = 1 and the inverse If, additionally, | w |= 1 is mapped to | z |= 1 then  cz + d of z0 is also mapped to the inverse of w = 0 that is, w = ∞. Hence cz + d can be replaced by z∗0 z − 1 giving the mapping as   z − z0 jθ0 w=e z∗0 z − 1 where θ0 is any real number.  Exercises 4.2.7 20  Under the mapping w = z2 , u = x2 − y2 , v = 2xy  It is not possible to achieve formulae of the type x = φ(u, v), y = ψ(u, v) , however we can use u = x2 − y2 , v = 2xy to determine images. Points (0 + j0), (2 + j0) and (0 + j2) transform to (0 + j0), (4 + j0) and (−4 + j0) respectively. The positive real axis y = 0, x ≥ 0 transforms to the (positive) real axis v = 0, u = x2 . The positive imaginary axis x = 0, y ≥ 0 transforms to the (negative) real axis v = 0, u = −y2 . The line joining the point 2 + j0 to the point 0 + j2 has equation x + y = 2. By using the equations u = x2 − y2 and v = 2xy we obtain u = 4(1 − y), v = 2y(2 − y) from which, eliminating y we get 8v = 16 − u2 c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  209  Hence we deduce the following picture:  21  Under the transformation w = z2 , u = x2 − y2 and v = 2xy .  Hence the line y = x transforms to u = 0, v > 0 and the line y = −x transforms to u = 0, v < 0. 2m u. The line y = mx transforms to v = 1 − m2 2m = tan 2θ0 . Putting m = tan θ0 , 1 − m2 Hence y = x tan θ0 transforms to v = u tan 2θ. Thus lines through the origin of slope θ0 in the z plane transform to lines through the origin of slope 2θ0 in the w plane. Hence the angle between the lines through the origin in the z plane is doubled by the transformation w = z2 .  22  w = zn  Writing z = rejθ , w = rn enjθ 22(a)  Circles | z |= r are transformed to circles | w |= rn  22(b)  Straight lines passing through the origin intersecting with angle θ0 are  θ = k and θ = k + θ0 . These are transformed to w = rn enjk and w = rn enj(k+θ0 ) that is, lines φ = nk and φ = nk + nθ0 as required. 1 + z2 1 z∗ =z+ =z+ z z | z |2 x y then u = x + 2 and v = y − 2 2 x +y x + y2 23  If w =  c Pearson Education Limited 2011   210  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition    1 1 If | z |= r , then u = x 1 + 2 and v = y 1 − 2 r r x=  r2 u r2 v , y = r2 + 1 r2 − 1   r2 u 2  r2 v 2 + 2 = r2 (r = 1) (I) . Squaring and adding gives 2 r +1 r −1 If r = 1 and v = 0, | x |≤ 1 (because x2 = 1 − y2 ) and u = 2x . Hence the image of the unit circle | z |= 1, that is, −2 ≤ u ≤ 2, v = 0 and the portion of the real axis in the w plane is between −2 and +2. | r2 − 1 | 1 + r2 and minor axis if r is The curves ( I) are ellipses, major axis r2 r2 very large, and both of these quantities tend to 1. Hence the image curve I tends to a circle u2 + v2 = r2 .  Exercises 4.3.3 24(a) zez = (x + jy)ex+jy = ex (x + jy)(cos y + j sin y) = ex (x cos y − y sin y) + jex (y cos y + x sin y) so u = (x cos y − y sin y)ex , v = (y cos y + x sin y)ex We need to check the Cauchy–Riemann equations ∂u ∂x ∂u ∂y ∂v ∂x ∂v ∂y  = (x cos y − y sin y + cos y)ex = (−x sin y − y cos y − sin y)ex = (y cos y + x sin y + sin y)ex = (−y sin y + cos y + x cos y)ex  ∂v ∂u ∂v ∂u = and =− ∂x ∂y ∂y ∂x Thus the Cauchy–Riemann equations are valid and Hence  d (zez ) = (z + 1)ez dz  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 24(b)  211  Following the same procedure as in (a), we deduce that sin 4z is analytic  with derivative 4 cos 4z . This time, zz∗ = x2 + y2 which is real. ∂u ∂v Obviously, therefore, (= 2x) = (= 0) . ∂x ∂y Thus zz∗ is not analytic. 24(c)  24(d)  25  Similarly to part (a), cos 2z is analytic with derivative −2 sin 2z .  w = x2 + ay2 − 2xy + j(bx2 − y2 + 2xy) = u + jy ∂u ∂u = 2x − 2y , = 2ay − 2x ∂x ∂y ∂v ∂v = 2bx + 2y , = −2y + 2x ∂x ∂y  ∂u ∂v ∂u ∂v =− , = . ∂x ∂y ∂y ∂x The second is satisﬁed and the ﬁrst only holds if a = −1, b = 1. Since w(z) = w(x+jy) we simply put y = 0 which gives w(x) = x2 +jx2 and hence w(z) = z2 +jz2 dw = 2(1 + j)z and dz  The Cauchy–Riemann equations are  26  With u = 2x(1 − y) = 2x − 2xy ∂u ∂u = 2 − 2y , = −2x ∂x ∂y  The Cauchy–Riemann equations demand ∂v ∂v = 2 − 2y , = 2x ∂y ∂x Integrating and comparing, these give v = x2 − y2 + 2y + C (take C = 0) Form u + jv = 2x − 2xy + j(x2 − y2 + 2y) = w(z) . c Pearson Education Limited 2011   212  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Since z = x + jy , if we put y = 0, we can ﬁnd w(x) which will give the functional form of w. Thus w(x) = 2x + jx2 Hence w(z) = 2z + jz2  27  hence  φ(x, y) = ex (x cos y − y sin y) ∂φ = ex (x cos y − y sin y + cos y) ∂x ∂φ = ex (−x sin y − y cos y − sin y) ∂y ∂2 φ = ex (x cos y − y sin y + 2 cos y) ∂x2 ∂2 φ = ex (−x cos y + y sin y − 2 cos y) ∂y2 ∂2φ ∂x2  +  ∂2φ ∂y 2  = 0 and φ is harmonic.  Writing z = φ(x, y) + jψ(x, y) , the Cauchy–Riemann equations demand ∂φ ∂ψ =− = ex (x sin y + y cos y + sin y) ∂x ∂y ∂ψ ∂φ = = ex (x cos y − y sin y + cos y) ∂y ∂x Integrating  ∂ψ ∂x x  with respect to x (using integration by parts for the ﬁrst term)  gives ψ = e (x sin y + y cos y) + f(y) . Examining φ(x, y) demands that f(y) = 0 because all terms will be multiplied by ex . Hence w(z) = φ(x, y) + jψ(x, y) = ex (x cos y − y sin y) + jex (x sin y + y cos y) and w(x + j0) = w(x) = xex . Hence w(z) = zez .  28  Here we have u(x, y) = sin x cosh y so that hence  so that ∇2 u =  ∂2u ∂x2  +  ∂u ∂u = cos x cosh y and = sin x sinh y ∂x ∂y ∂2 u ∂2 u = − sin x cosh y and = sin x cosh y ∂x2 ∂y2 ∂2u ∂y 2  = 0 and u is harmonic. c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  213  Using the Cauchy–Riemann equations gives v = cos x sinh y so that u + jv = w(z) = sin x cosh y + j cos x sinh y . Putting y = 0 gives w(x + j0) = sin x so that w(z) = sin z .  29  The orthogonal trajectories of a family of curves φ(x, y) = α are ψ(x, y) = β  where φ and ψ are conjugate functions: that is, φ(x, y) + jψ(x, y) = w(z) which is an analytic function. Proceeding as in the previous examples. 1 4 3 (x + y4 ) − x2 y2 4 2  29(a)  If φ(x, y) = x3 y − xy3 then ψ(x, y) =  29(b)  1 If φ(x, y) = e−x cos y + xy then ψ(x, y) = e−x sin y + (x2 − y2 ) . 2  Hence the orthogonal trajectories are, 29(a)  x4 − 6x2 y2 + y4 = β , a constant  29(b)  2e−x sin y + x2 − y2 = β , a constant.  30(a)  z2 e2z  = (x + jy)2 e2(x+jy) = (x2 − y2 + 2jxy)(e2x (cos 2y + j sin 2y)) = e2x ((x2 − y2 ) cos 2y − 2xy sin 2y) + je2x ((x2 − y2 ) sin 2y + 2xy cos 2y) 30(b)  sin 2z  = sin(2x + j2y) = sin 2x cosh 2y + j cos 2x sinh 2y Straightforward calculus reveals that both functions obey the Cauchy–Riemann equations and are thus analytic. Their derivatives are (a) (2z2 + 2z)e2z and (b) 2 cos 2z respectively.  31  Writing w = sin−1 z we can say that z = sin w = sin(u + jv) = sin u cosh v + j cos u sinh v c Pearson Education Limited 2011   214  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  so that, equating real and imaginary parts, x = sin u cosh v and y = cos u sinh v Squaring and adding gives x2 + y2 = sin2 u cosh2 v + cos2 u sinh2 v = sin2 u cosh2 v + (1 − sin2 u)(cosh2 v − 1) = sin2 u + cosh2 v − 1 from which x2 + y2 + 1 = sin2 u +  x2 sin2 u  (I)  Solving for sin2 u gives sin2 u =  1 1 (1 + x2 + y2 )2 − 4x2 (1 + x2 + y2 ) − 2 2  where the minus sign is taken, since with u = π/2 (i.e. inconsistencies result otherwise. From cosh2 v = cosh2 v =  2  x sin2 u  we obtain  1 1 (1 + x2 + y2 ) + (1 + x2 + y2 )2 − 4x2 2 2  (This is most easily found by solving equation (I) for x2 sin2 u  x = cosh v, y = 0)  1 sin2 u  then using cosh2 v =  .)  Square rooting and inverting give u and v in terms of x and y . It can be shown that the expression under the square root sign is positive, for 1 + x2 + y2 − 2x = (x − 1)2 + y2 ≥ 0 for all real x and y thus (1 + x2 + y2 )2 ≥ 4x2 . Hence w = sin−1 z is an analytic function with derivative  32  √ 1 1−z 2  .  | sin z |2 =| sin x cosh y + j cos x sinh y |2 = sin2 x cosh2 y + cos2 x sinh2 y = cosh2 y − cos2 x = sinh2 y + sin2 x  The result follows immediately from the last two expressions.  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  215  Exercises 4.3.5 33  Mappings are not conformal at the points where  dw =0 dz  dw = 2z = 0 when z = 0. z = 0 is the only point where the mapping dz fails to be conformal.  33(a)  dw = 6z2 − 42z + 72 = 0 when z2 = 7z + 12 = 0 that is, non-conformal dz points are z = 4, z = 3 (both real).  33(b)  33(c) points.  34  √ 1 1 1±j 3 dw 1 3 = 8 − 3 = 0 when z = giving , as non-conformal dz z 8 2 4  Proceeding as in Example 4.13, the mapping w=z−  1 z  has a ﬁxed point at z = ∞, is analytic everywhere except at z = 0 and conformal dw =0 except where dz 1 that is, 1 + 2 = 0, z = ±j z Since both of these occur on the imaginary axis, consideration of this axis is adequate to completely analyse this mapping. The image of z = j is w = 2j , and the image of z = −j is w = −2j . Writing z = j + jε, ε real, we ﬁnd that 1 j + jε = j[1 + ε − (1 + ε)−1 ]  w = jε −  = j[1 + ε + 1 − ε + ε2 + . . .] j[2 + ε2 ] So, no matter whether ε > 0 or ε < 0, the image point of z = j + jε is above w = j2 on the imaginary axis. c Pearson Education Limited 2011   216  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  that is, points Q and P in the z plane both map to R in the w plane in a manner to Example 4.13, the non-conformality of z = ±j is conﬁrmed and as the imaginary axis (in the z plane) is traversed from −jz to 0, the imaginary axis (in similar the w plane) is traversed from −jz to −j2 and back to −j∞ (when z = −j, w reaches −j2). Similarly, as the imaginary axis (in the z plane) is traversed from +j∞ to 0, the imaginary axis (in the w plane) is traversed from +j∞ to +j2 and back to +j∞ again. Finally, points on the imaginary axis in the w plane such that w = aj, −2 < a < 2, do not arise from any points on the imaginary axis in the z plane. This point is obvious once one solves aj = z − to obtain z=  35  1 z  1√ 1 aj ± 4 − a2 2 2  If w = ez  then u = ex cos y and v = ex sin y Hence the expressions u2 + v2 = e2x and v = u tan y can be derived. 35(a)  0 ≤ x < ∞ is mapped to the exterior of the unit circle u2 + v2 = 1  35(b)  0 ≤ x ≤ 1 is mapped to the annulus 1 ≤ u2 + v2 ≤ e2  0 ≤ y ≤ 1 is mapped to the region between the radiating lines v = 0 and v = u tan 1. c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  217  1 35(c) 2 π ≤ y ≤ π is mapped to the region between u = 0(v > 0) and v = 0(u < 0)  that is,  Thus if 0 ≤ x < ∞ then the image region in the w plane is in the shaded quadrant, but outside the unit circle.  36  If w = sin z then  dw dz  = cos z .  Since cos z = 0 when z = (2n + 1)π/2 these are the points where the mapping is not conformal w = sin z ⇒ u + jv = sin x cosh y + j cos x sinh y c Pearson Education Limited 2011   218  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Hence v = sin x cosh y, v = cos x sinh y 2  2  thus lines x = k transform to sinu k − cosv k = 1 (hyperbolae)  u 2  v 2 + sinh l = 1 (ellipses) and lines y = k transform to cosh l  a2 jθ ζ and ζ = Re 2 then z = Rejθ + aR e−jθ  2 so that x = R + aR cos θ and y  37  If z = ζ +   = R−  a2 R    sin θ.  If R = a, x = 2a cos θ and y = 0 then the real line between ±2a is traversed. Length of line segment = 4a. For a circle of radius b,   a2  a2  cos θ, y = b − sin θ x= b+ b b Hence the image in the z plane is an ellipse of the form b2 x2 b2 y2 + 2 =1 (a2 + b2 )2 (b − a2 )2  Exercises 4.4.2 38(a)  38(b)     1 z −1 z  z 2 = (z − j)−1 = j 1 − =j 1+ + + ... z−j j j j 2 = j + z − jz − z3 + jz4 . . .   j −1 j  j 2 1 1 1 = 1+ + + ... = 1− z−j z z z z z j 1 j 1 = + 2 − 3 − 4 + ... z z z z c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 38(c)  In order that | z − 1 − j |<  219  √ 2 we write  1 1 = = (1 + z − 1 + j)−1 z−j z−1−j+1 = 1 − (z − 1 − j) + (z − 1 − j)2 − (z − 1 − j)3 + . . . Which is valid inside | z − 1 − j |<| 1 − j |=  √ 2.  39 1 = (z2 + 1)−1 = 1 − z2 + z4 − z6 + . . . z2 + 1 where | z |< 1. Using the fact that we can diﬀerentiate power series term-by-term and the radius of convergence remains unaltered −  (z2  2z = −2z + 4z3 − 6z5 + . . . 2 + 1)  so 39(a) 1 = 1 − 2z2 + 3z4 − 4z6 + 5z8 + . . . 2 2 (z + 1) | z |< 1  Operating on  (z2  1 in a similar fashion gives + 1)2 −  4z = −4z + 12z3 − 24z5 + 40z7 . . . (z2 + 1)3  so 39(b) (z2  1 = 1 − 3z2 + 6z4 − 10z6 + 15z8 + . . . + 1)3 | z |< 1  c Pearson Education Limited 2011   220  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Exercises 4.4.4 40  Taylor's theorem is f(z) = f(a) + (z − a)f (a) +  (z − a)2  f (a) + . . . 2!  We thus compute f(z) and its ﬁrst few derivatives then evaluate them at z = a.  40(a) f(z) =  1 1 2 6   , f (z) = − , f (z) = and f (z) = − 1+z (1 + z)2 (1 + z)3 (1 + z)4  Hence f(1) =  1 1  1 2 3 , f (1) = − , f (1) = = and f (1) = − 2 4 8 4 8  thus 1 1 1 1 1 = − (z − 1) + (z − 1)2 − (z − 1)3 + . . . 1+z 2 4 8 16 The radius of convergence is the distance between the nearest singularity of f(z) to the point about which the expansion is made. The point z = −1 is the only singularity and the distance between this and z = 1 is 2 (along the real axis).  40(b) f(z) = f (z) = f (z) = f (z) = fiv (z) = fv (z) =  j 1 1  1 = − using partial fractions z(z − 4j) 4 z z − 4j  1  j 1 1 − 2+ ; f (1) = 0 f(1) = 4 z (z − 4j)2 4  j 2 2 1   − 3+ ; f (1) = 0 f (1) = − 4 z (z − 4j)3 8  j 6 6 3 v iv − 4+ f (1) = + ; f (1) = 0 4 z (z − 4j)4 8  j  24 24 45 vi − 4 + f (1) = − 4 z (z − 4j)5 16 j  120 120  − 6 + etc. 4 z (z − 4j)6 c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  221  Thus 1 1 1 1 1 = − (z − 2j)2 + (z − 2j)4 − (z − 2j)6 + . . . z(z − 4j) 4 16 64 256 The radius of convergence is 2 since z = 2j is 2 from the singularities at z = 0 and z = 4j.  40(c)  f(z) =  1 z2  gives f (z) = − z23 , f (z) =  6 z4  and f (z) = − 24 z5 .  Putting z = 1 + j gives j  2 1+j f(1 + j) = − , f (1 + j) = − = 3 2 (1 + j) 2   6 3 24 f (1 + j) = = and f (1 + j) = − = −3(1 − j) 4 (1 + j) 2 (1 + j)5 Hence 1 j 1 1 3 = − + (1 + j)(z − 1 − j) + (z − 1 − j)2 − (1 − j)(z − 1 − j)3 + . . . 2 z 2 2 4 2 The radius of convergence is the distance between the origin (a double pole) and √ 1 + j that is, 2.  1 1 + z + z2 we could use the binomial expansion 41  With f(z) =  f(z) = (1 + z + z2 )−1 gathering terms to O(z3 ) This is certainly more eﬃcient than using the derivatives of f(z) . However, the best way is to use the fact that (z3 − 1) = (z − 1)(z2 + z + 1) . That is 1 1−z z−1 = = 3 2 1+z+z z −1 1 − z3 = (1 − z)(1 − z3 )−1 c Pearson Education Limited 2011   222  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition = 1 − z + z3 . . . to order z3  valid in the region | z |< 1.  42  If f(z) =  1 z 4 −1  the singularities are at the points where z4 = 1 that is,  z = 1, −1, −j and j. The radii of convergence are the minimum distances of the points z = 0, 1 + j and 2 + j2 from these singularities. z = 0 is equidistant (1) from each radius of convergence = 1. z = 1 + j is distance (1) from z = 1 and z = j with radius of convergence = 1. z = z + j2 is a distance | 2 + j2 − 1 | from 1 and a distance | 2 + j2 − j | from j. √ Both of these distances = [22 + (2 − 1)2 ]1/2 = 5.  43  If f(z) = tan z then f (z) = sec2 (z) and f = 2 sec2 z tan z but subsequent  derivatives get cumbersome to compute (except by using a Computer Algebra sin z , we can use the series for sin z and cos z as follows: package). Since tan z = cos z tan z =  z3 z5 6 + 120 2 4 − z2 + z24  z−  1  −1   z2 z2 z4 z4  1− =z 1− + − 6 120 2 24   2 z4 z2 z4  z2 z4  z2 + 1+ − + − + ... =z 1− 6 120 2 24 2 24   1 1 1 5 1 1 3 − − + z + ... =z+ z + 3 120 12 24 4 2 1 tan z = z + z3 + z5 + · · · 3 15 Since z = π/2 is the closest singularity, the radius of convergence is  π 2  .  Exercises 4.4.6 1 has a simple pole at z = 0 and a double pole at z(z − 1)2 z = 1. In order to ﬁnd the Laurent expansions, we simply ﬁnd the following 44  The function  binomial expansions c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  223  1 1 (1 − z)−2 = (1 + 2z + 3z2 + 4z3 + . . .) z z 1 = + 2 + 3z + 4z2 + . . . about z = 0 z valid for 0 <| z |< 1 and  1 (1−(1 − z))−1 (z − 1)2 1 = [1 + (1 − z) + (1 − z)2 + (1 − z)3 + . . .] (z − 1)2 1 1 + 1 + (1 − z) + (1 − z)2 + . . . + = 2 (1 − z) 1−z valid for 0 <| 1 − z |< 1    With f(z) = z2 sin z1 , there is a singularity at z = 0 and another at   z = ∞. Expanding sin z1 as a power series in z1 we ﬁnd   1 1 1 2 2 1 =z + − ··· z sin − z z 3!z3 5!z5 1 1 + − ··· =z− 3!z 5!z3 1 11 = ··· + − + z. 3 5!z 3! z 45(a)  Since the principal part is inﬁnite, there must be an essential singularity at z = 0. (b)  in order to investigate z = ∞ we obtain   1 1 w5 w3 1 2 = 2 sin w = 2 w − + − ··· z sin z w w 3! 5! 1 w w3 = − + − ··· w 3! 5! 1 1 + 3 − ··· =z− z3! z 5!  Writing z =  1 w  which implies a simple pole at z = ∞. (The expansion is the same as that about z = 0, but re-interpreted.) At any other point z2 sin z1 is regular and has a Taylor series of the   form f(z) = a0 + a1 z + a2 z2 + . . . . speciﬁcally, about z = a, z2 sin z1 =  (c)  a2 sin a1 + a1 z + a2 z2 + . . . where a1 = f (a), a2 = f (a) , etc. c Pearson Education Limited 2011   224  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  46  With f(z) =  z (z−1)(2−z)  there are simple poles at z = 1 and z = 2. 46(a)  Inside the unit circle | z |= 1, therefore there is a Taylor series  z(1 − z)−1 (2 − z)−1  z −1 z = (1 − z)−1 1 − 2 2   z z  z 2  z 3 2 3 = (1 + z + z + z + . . .) 1 + + + + ··· 2 2 2 2     z 3 1 3 1 3 1 3 z 3 2 z 2 1 2 1 2 z + z + z + ... + z + z + z + z + ... = + z + 2 4 2 2 4 2 2 4 8 3 7 15 1 = z + z2 + z3 + z4 + · · · | z |< 1 2 4 8 16 46(b)  In the annulus 1 <| z |< 2 we rearrange f(z) to obtain a Laurent series  as follows −1  −1  1 1 z 2 1 z 1− − = − =− 1− (z − 1)(z − 2) z−2 z−1 2 z z     2 1 1 1 z z + ··· − 1 + + 2 + ··· =− 1+ + 2 4 z z z 1 1 1 z z2 = ... − 3 − 2 − − 1 − − − ··· z z z 2 4 46(c)  For | z |> 2 we rearrange as follows  z 2 1 = − (z − 1)(z − 2) z−2 z−1 2 −1 1  1 −1 2 = 1− − 1− z z z z    2 4 1 1 2 8 1 1 1 + + 2 + 3 + ··· − 1 + + 2 + 3 + ··· = z z z z z z z z 3 7 15 1 = + 2 + 3 + 4 + ··· z z z z 46(d) If w =  For | z − 1 |> 1 we write 1 z−1  then wz − w = 1 or z  1 z−1 = = 1+w w  w and ﬁnd a Taylor's series in w.  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition so that  225    z 1+w =w (z − 1)(z − 2) 1−w = w(1 + w)(1 + w + w2 + w3 + . . .) = w + 2w2 + 2w3 + . . . 2 2 1 + + + ··· = 2 z − 1 (z − 1) (z − 1)3  46(e)  For 0 <| z − 2 |< 1 we write w = z − 2 hence  z w+2 2 = = 1+ (1 + w)−1 (z − 1)(z − 2) w(w + 1) w  2 (1 − w + w2 − w3 + . . .) = 1+ w 2 = − 1 + w − w2 + w3 − . . . w 2 − 1 + (z − 2) − (z − 2)2 + (z − 2)3 . . . = (z − 2)  Exercises 4.5.2 47  The point at inﬁnity is ignored in this question. Most if not all can be found  immediately by inspection. 47(a)  47(b)  cos z : z2  double pole at z = 0, zeros whenever cos z = 0 that is, z = 12 (2n + 1)π, n = integer.  1 : has a double pole at z = −j , a simple pole at z = j (z + j)2 (z − j) and no zeros in the ﬁnite z plane.  47(c)  z : simple poles at z4 = 1 that is, z = 1, −1, j, −j and a zero at 4 z −1 z = 0.  47(d)  cosh z : since coth z =  cosh z this has simple poles at those points where sinh z z = jnπ and zero at those points where z = 12 j(2n + 1)π, n = integer.  c Pearson Education Limited 2011   226  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  47(e)  sin z : simple poles at z = ±jπ and zeros at z = nπ, n = integer. z2 + π2  47(f)  ez/(1−z) : this has an essential singularity at z = 1 and no zeros.  47(g)  47(h)  47(i)  z−1 : this has simple poles at z = ±j and a zero at z = 1. z2 + 1 z+j : this has a triple pole at z = −2, a simple pole at z = 3 (z + 2)3 (z − 3) and a zero at z = −j.  z2 (z2  1 : this has simple poles at z2 − 4z + 5 = 0, that is, − 4z + 5) z = 5, −1 and a double pole at z = 0.  1 − cos z . In order to investigate this, we expand cos z . Only z = 0 is a z2 possible (ﬁnite) singularity  48(a)   1− 1−  z2 2!  +  z2  z4 4!  − ···   =  z2 1 − + ··· 2! 4!  The RHS is a power series, thus the singularity at z = 0 is removable. 2  48(b)  ez 2 . Using the power series for ez gives the expansion 3 z   2 ez 1 z4 2 + ··· = 3 1+z + z3 z 2! z 1 1 = 3 + + + ··· z z 2!  z = 0 is thus a pole of order 3. 48(c)  1 z 1+     1 1 z cosh z . 1 1 z 2 2! + z 4 4! +  Obviously the point z = 0 is a problem.  · · · is the Laurent series which indicates that z = 0 is an  essential singularity. c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 48(d)  227  tan−1 (z2 + 2z + 2) . For this problem the easiest way to proceed is to ﬁnd  the Maclaurin series from ﬁrst principles. At z = 0, tan−1 (z2 + 2z + 2) = tan−1 2 which is ﬁnite. This means that z = 0 is a regular point, hence it is not actually necessary to ﬁnd the Laurent series (in this case Maclaurin series) for the function. In fact 6 2 tan−1 (z2 + 2z + 2) = tan−1 2 + z − z2 + · · · 5 25  49  If f(z) =  p(z) q(z)  where p(z) and q(z) are polynomials, then the only singularities  of f(z) are the algebraic zeros of q(z) . These zeros are either distinct or multiple. The distinct zeros give rise to simple poles of f(z) whereas the multiple zeros give rise to poles of higher order. f(z) can only have these kinds of singularity, although it may have none if q divides p, so that f(z) is polynomial. f(z) therefore cannot have an essential singularity.  Exercises 4.5.4 2z + 1 2z + 1 = hence the singularities are simple poles at − z − 2) (z − 2)(z + 1) z = 2, z = −1.  50(a)  (z2  Using the formula residue = lim [(z − z0 )f(z)] the residues are z→z0  at z = −1.  5 3  at z = 2 and  1 has a simple pole at z = 1 and a double pole at z = 0. − z)  1 The residue at z = 1 is lim − 2 = −1 z→1 z  50(b)  z2 (1  1 1 1 1 = 2 (1 + z + z2 + . . .) = 2 + + 1 + · · · − z) z z z  z2 (1  Hence the residue at z = 0 is 1.  c Pearson Education Limited 2011   1 3  228  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  50(c)  3z2 + 2 3z2 + 2 = (z − 1)(z2 + 9) (z − 1)(z + j3)(z − j3)  Hence there are simple poles at z = 1, j3, −j3 5 1 3+2 = = (1 + j3)(1 − j3) 10 2 −3 × 9 + 2 −25 5 at z = j3, residue = = = (3 − j) (j3 − 1)j6 6(−3 − j) 12 5 (3 + j) by symmetry. at z = −j3, residue = 12 At z = 1, residue =  z3 − z2 + z − 1 z3 − z2 + z − 1 (z − 1)(z2 + 4) = = z3 + 4z z(z + j2)(z − j2) z(z + j2)(z − j2) which has simple poles at z = 0, j2, −j2.  50(d)  At z = 0, residue = −  1 4  (z − 1)(z2 + 4) 3 = (−1 + 2j) z→j2 (z + j2) 8 3 at z = −j2, residue = (−1 − 2j) similarly. 8 at z = j2, residue = lim  z6 + 4z4 + z3 + 1 has a pole of order 5 at z = 1. (z − 1)5 The formula for calculating residues is convenient for this problem.  50(e)  d4 1 lim 4 (z6 + 4z4 + z3 + 1) 4! z→1 dz 1 (6.5.4.3 + 4.4.3.2) = 24 456 = 19 = 24  Residue =  50(f)   z + 1 2 has a double pole at z = 1. z−1 d (z + 1)2 = 4 z→1 dz  Residue = lim  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 50(g)  229  z+1 has a simple pole at z = −3 and a double pole at z = 1. (z − 1)2 (z + 3) 1 8 1 d  z + 1  = Residue at z = 1 is z=1 dz z + 3 8 Residue at z = −3 is −  3 + 4z 3 + 4z = has simple poles at z = −2, −1 and 0. 2 + 3z + 2z z(z + 1)(z + 2) 5 3 Residues are, respectively, − , 1 and following the same procedure as part (c). 2 2  50(h)  z3  cos z at z = 0 is simple, thus the residue is cos(0) = 1. z  51(a)  The pole of  51(b)  The poles of  sin z are all simple, and the residue at z = eπj/3 is + z2 + 1   z − eπj/3 sin z πj/3 πj/3 = sin e lim (z − e ) 4 lim z + z2 + 1 z4 + z2 + 1 z→eπj/3 z→eπj/3 z4  Using L'Hôpital's rule, the limit is 1 + 2eπj/3 √ 1 1 1 √  = √ = (−3 − j 3) = 1 12 −3 + j 3 −4 + 2 2 + j 23  4eπj  √ √   1 (3 + j 3) sin 12 (1 + j 3) giving the residue − 12  51(c)  The pole of  part. The residue is  z4 − 1 at z = eπj/4 is simple and we proceed as in the last z4 + 1   z − eπj/4 − 2 lim z4 + 1 z→eπj/4   1 j 1 1 1 1 √ +√ . = = −2 3πj/4 = − 2 − √1 + j √1 2 4e 2 2 2 2   √  Hence the residue is  2 4 (1  + j) .  c Pearson Education Limited 2011   230  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  51(d)  z has a simple pole at z = π sin z Residue = π lim  z→π  51(e)  (z2  z − π  = −π.  sin z  1 has a double pole at z = j + 1)2 d 1 (z − j)2 2 z→j dz (z − j) (z + j)2 2 = lim − z→j (z + j)3 2 j =− 3 =− . 8j 4  Residue = lim  52(a)  cos z has a triple pole at z = 0. z3 1 cos z 1 1 + z − ··· = 3− 3 z z 2z 24  52(b)  residue = −  1 2  z2 − 2z has a double pole at z = −1. (z + 1)2 (z2 + 4) d (z + 1)2 (z2 − 2z) at z = −1 dz (z + 1)2 (z2 + 4) (2z − 2)(z2 + 4) − 2z(z2 − 2z)  14 = =− 2 2 z=−1 (z + 4) 25  Residue =  52(c)  The function  ez has a double pole wherever sin2 z  sin z = 0 that is at z = nπ, n = an integer In order to ﬁnd the residue, we need to compute lim  z→nπ  d  (z − nπ)2 ez dz sin2 z  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition Now  d (z − nπ)2 ez dz sin2 z  = ez  231  d (z − nπ)2 (z − nπ)2 z + e dz sin2 z sin2 z  (I)  and      d (z − nπ)2 z − nπ d z − nπ =2 · dz sin z dz sin z sin2 z 2(z − nπ) sin z − (z − nπ) cos z = · sin z sin2 z z − nπ → 1 and As z → nπ, sin z cos z − cos z + (z − nπ) sin z sin z − (z − nπ) cos z (using L'Hôpital's rule) → 2 2 sin z cos z sin z → 0 as z → nπ Hence the RHS of equation (I) → enπ as z → nπ. Thus the residue is enπ .  Exercises 4.6.3 53    (z2 + 3z)dz with z = x + jy and dz = dx + jdy  C  hence (z2 + 3z)dz = (x2 − y2 + j2xy + 3x + j3y)(dx + jdy) = (x2 − y2 + 3x)dx − (2xy + 3y)dy + j[(x2 − y2 + 3x)dy + (2xy + 3y)dx]  53(a)  The straight line joining 2 + j0 to 0 + j2 has equation x + y = 2 in  Cartesian coordinates. This has parametric equation x = t and y = 2 − t from which dx = dt and dy = −dt , and using the above expression for (z2 + 3z)dz (z2 + 3z)dz = (t2 − (2 − t)2 + 3t)dt + (2t(2 − t) + 3(2 − t))dt + j[−(t2 − (2 − t)2 + 3t)dt + (2t(2 − t) + 3(2 − t))dt] and the range of integration is from t = 2 to t = 0. c Pearson Education Limited 2011   232  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Hence     0  (8t − 2t2 + 2)dt  2  (z + 3z)dz = C  2    0  (−6t − 2t2 + 10)dt  +j 2   2 = 4t2 − t3 + 2t 3  8 44 so (z2 + 3z)dz = − − j 3 3 C  53(b)   2 + j −3t2 − t3 + 10t 3 2 0  0 2  On the straight line from 2 + j0 to 2 + j2, x = 2 and y goes from 0 to  2, so that dx = 0. Therefore      2  −(4t + 3t)dt  2  (z + 3z)dz = C1  0    2  (4 − t2 + 6)dt  +j 0 2   1 + j 10t − t3 3 0 52 = −14 + j 3   7 = − t2 2  2 0  On the straight line from 2 + j2 to 0 + j2, y = 2 and x goes from 2 to 0, so that dy = 0. Therefore      0  (t2 − 4 + 3t)dt  2  (z + 3z)dz = C2  2    0  +j  (4t + 6)dt 2  1 3 3 t − 4t + t2 3 2 2 = − − j14 3  0  =  Thus   C  (z2 + 3z)dz =   C1  (z2 + 3z)dz +     + j 2t2 + 3t  2  (z2 + 3z)dz = −  C2  c Pearson Education Limited 2011   0 2  8 44 −j . 3 3  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 53(c)  233  For this part, we use z = 2ejθ on | z |= 2 and θ varies between 0 and π/2  on the quarter circle joining 2+j0 to 0+j2. Thus (z2 +3z)dz = (4e2jθ +6ejθ )2jejθ dθ so that      π/2  2    8je3jθ + 12je2jθ dθ  (z + 3z)dz = 0  C  π/2  8 3jθ e + 6e2jθ = 3 0 8 8 +6 = − −6 − 3 3 8j 44 =− − 3 3 Hence the integrals are all the same. 54(a)  On | z |= 1, z = ejθ , 0 ≤ θ ≤ 2π  so that (5z4 − z3 + 2)dz  2π (5e4jθ − e3jθ + 2)jejθ dθ = 0  =  5 5jθ 1 4jθ e − e + 2ejθ 5 4  2π  =0 0  hence e2πj = e0 = 1 54(b)  Integrating around the square in the order 0 + j0, 1 + j0, 1 + j1 and  0 + j1 gives the answers  j 11 11 4 , 3 + 4, − 4  and − 3 − 4j . Adding these together gives  0. 54(c)  On the parabola y = x2 , x = t and y = t2 so that z = t + jt2 and  dz = (1 + 2jt)dt . On the parabola y2 = x, x = t2 and y = t so that z = t2 + jt and dz = (2t + j)dt .  The computation of (5z4 −z3 +2)dz is extremely long winded but straightforward C  and gives the answer 0. 55  In order to evaluate   C  dz (z − z0 )n  c Pearson Education Limited 2011   234  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  we surround the point z = z0 with a circle of radius ε on which z = z0 + εejθ , 0 ≤ θ < 2π. Using equation (4.45) the integral around C is the same as the integral around the circle on which z = z0 + εejθ dθ. Thus  C  dz = (z − z0 )n    2π  0  jεejθ dθ εn enjθ  If n = 1, then the integral integrates to jε  (1−n)  e(1−n)jθ (1 − n)j  2π  =0 0  as in Example 4.30. If n = 1,   C  56(a)  dz = (z − z0 )n  C  jdθ = 2πj 0  dz =0 z−4  If z = 4 is inside C, by problem 55  C  57  2π  If z = 4 is outside C, by Cauchy's theorem,   56(b)    dz = 2πj z−4  In order to use Cauchy's integral theorem, we split into partial fractions 2/5 4/5 2z = + (2z − 1)(z + 2) 2z − 1 z + 2  57(a)  If C is the circle | z |= 1  C    2zdz dz dz 2 4 = + (2z − 1)(z + 2) 5 C 2z − 1 5 C z + 2 2 4 = · 2πj + · 0 5 5 c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition since z = =  1 2  is inside | z |= 1 whereas z = −2 is outside. Hence  4 πj 5  57(b)  C  2zdz (2z − 1)(z + 2)  If C is the circle | z |= 3, both singularities (poles) are inside C, and   hence  2zdz 2 4 = 2πj + · 2πj (2z − 1)(z + 2) 5 5 12 πj = 5  C  58    This follows a pattern similar to Exercise 57.  Using partial fractions gives 5z = (z + 1)(z − 2)(z + 4j) 58(a)  1 − 4j) (1 − 2j) + 3 + z+1 z−2  5 51 (−1  2 17 (−2  + 9j) z + 4j  Only the ﬁrst two poles (z = −1, z = 2) are inside | z |= 3 hence  C  58(b)    5zdz 5 1 = 2πj (−1 − 4j) + (1 − 2j) (z + 1)(z − 2)(z + 4j) 51 3 4π (9 + 2j) = 17  All three poles are inside | z |= 5 hence  C   5zdz 5 1 = 2πj (−1 − 4j) + (1 − 2j) (z + 1)(z − 2)(z + 4j) 51 3  2 + (−2 + 9j) 17 =0  59  235  Equation (4.48) gives the general form of Cauchy's integral theorem  C  f(z) 2πj (n) f (z0 ) dz = n+1 (z − z0 ) n!  where C is a contour enclosing the point z = z0 . c Pearson Education Limited 2011   236  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  59(a)  C  z3 + z 1 dz = (2z + 1)3 8    z3 + z 1 3 dz C (z + 2 )  1 2πj d2  = (z3 + z)(z = − 12 is inside | z |= 1) 8 2 dz2  1 −2  1  3πj π  j 6(− ) = − 8 2 8  =  59(b)  First of all we need to separate the integrand using partial fractions 4 4 8 4z 9 9 3 − + = (z − 1)(z + 2)2 z − 1 z + 2 (z + 2)2    Hence  C  4zdz 4 4 = 2πj − 2πj + 0 = 0 2 (z − 1)(z + 2) 9 9  using Cauchy's integral theorem (the derivative of  8 3  is of course zero). All poles  of the integrand are inside the circle | z |= 3.  Exercises 4.6.6 60  z2  z has poles at z = ±j +1 z 1 is regular inside | z |= +1 2  zdz 1 = 0 if C is the circle | z |= 2 2 C z +1  60(a)  Since  60(b)  The residues of  z2  z at z = ±j (both inside | z |= 2) are z2 + 1  lim  (z − j)z z 1 = lim = (z + j)(z − j) z→+j z + j 2  lim  (z + j)z z 1 = lim = (z + j)(z − j) z→−j z − j 2  z→+j  and z→−j  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  237  Hence, using the residue theorem  C    zdz 1 1 = 2πj = 2πj + z2 + 1 2 2  z2 + 3jz − 2 are at z3 + 9z = 0, that is, z = 0, 3j, −3j z3 + 9z (all simple poles). Only z = 0 is inside | z |= 1 but all three are inside | z |= 4.  61  The singularities of  Hence we shall ﬁnd all the residues.  z2 + 3jz − 2  2 =− At z = 0, residue is lim 2 z→0 z +9 9 At z = 3j, the residue is (z − 3j)(z2 + 3jz − 2) z→3j z(z − 3j)(z + 3j) −9 − 9 − 2 10 (3j)2 + 3j3j − 2 = = = 3j(3j + 3j) −18 9 lim  At z = −3j , the residue is (z + 3j)(z2 + 3jz − 2) z→−3j z(z − 3j)(z + 3j) −9 + 9 − 2 1 (−3j)2 + (3j)(−3j) − 2 =− = = (−3j)(−3j − 3j) −18 9 lim  61(a)  For this part, since only the residue at z = 0 is inside C(| z |= 1)  ∴ C  61(b)   2 z2 + 3jz − 2 4πj dz = 2πj − = − 3 z + 9z 9 9  For this part, all residues need to be taken into account since all the poles  of f(z) are inside C(| z |= 4)  ∴ C   2 10 1  z2 + 3jz − 2 = 2πj dz = 2πj − + + z3 + 9z 9 9 9  Note that in this case, all the zeros of the denominator were obviously poles. In general, we would need to check if they were not removable by factorizing the numerator. c Pearson Education Limited 2011   238  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  62  f(z) =  √ (z2 + 2)(z2 + 4) has poles at z = ±j, z = ±j 6. (z2 + 1)(z2 + 6)  Residue at z = j is lim  z→j  (z − j)(z2 + 2)(z2 + 4) (−1 + 2)(−1 + 4) 3j = =− 2 (z − j)(z + j)(z + 6) 2j(−1 + 6) 10  Residue at z = −j is lim  z→−j  (z + j)(z2 + 2)(z2 + 4) (−1 + 2)(−1 + 4) 3j = = 2 (z + j)(z − j)(z + 6) (−2j)(−1 + 6) 10  √ Residue at z = j 6 is √ (z − j 6)(z2 + 2)(z2 + 4) (−6 + 2)(−6 + 4) 8 √ √ √ lim√ = = √ 2j 6(−6 + 1) 2j 6(−5) z→j 6 (z − j 6)(z + j 6)(z2 + 1) √ 2 = j 6 15 √ 2 √ Residue at z = −j 6 is thus = − j 6 15 The circle | z |= 2 contains the poles at z = ±j but not those at √ 3j 3j = 0. z = ±j 6 . The sum of the residues inside C = − + 10 10 Hence the integral = 0. 62(a)  62(b) Hence  The circle | z − j |= 1 contains the residue only at z = j.  C  62(c)   3j  3π (z2 + 2)(z2 + 4) dz = 2πj − = (z2 + 1)(z2 + 6) 10 5  The circle | z |= 4 contains all the poles. Since the sum of the residues is  zero, so is the integral.  63  The function  1 has double poles at z = 0 and z = ±j . z2 (1 + z2 )2  Residue at z = 0 is     d 1 4z   =− =0 dz (1 + z2 )2 z=0 (1 + z2 )3 z=0 c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  239  Residue at z = j is   d 1  = − 2(2z + j) = − 3 j 2 2 dz z (z + j) z=j (z2 + jz)3 4 Residue at z = −j is  63(a)  3 j 4  If C is the circle | z |=  C  63(b)  1 , only the residue at z = 0 is in C. Thus 2  dz = 2πj(0) = 0 z2 (1 + z2 )2  All the singularities are inside | z |= 2, but since they sum to 0,  C  z2 (1  dz =0 + z2 ) 2  3z2 + 2 are at z = 1, ±2j . 64(a) The singularities of (z − 1)(z2 + 4) They are all simple poles. Using the formula (4.37) the residues are :at z = 1 : 1  1 (2 − j) 8 1 at z = −2j : (2 + j) 8 at z = 2j :  (i) If C is | z − 2 |= 2 only the residue at z = 1 is included, hence  C  3z2 + 2 dz = 2πj (z − 1)(z2 + 4)  (ii) If C is | z |= 4, all the residues are included, hence  C  3z2 + 2 5 dz = πj 2 (z − 1)(z + 4) 2  z2 − 2z are at z = −1, ±2j . A double pole (z + 1)2 (z2 + 4) is at z = −1 and simple poles are at z = ±2j.  64(b)  The singularities of  c Pearson Education Limited 2011   240  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Residues are:-  14 25 at z = 2j : −1 + j  at z = −1 : −  at z = −2j : −1 − j (i) If C is | z |= 3, all singularities are inside C.   Hence  C    z2 − 2z 14 128 dz = 2πj − − 2 = − πj 2 (z + 1) (z + 4) 25 25  (ii) If C is | z + j |= 2 then z = −1 and z = −2j are inside C, but z = 2j is not. Hence  C  64(c)    z2 − 2z 14 2π dz = 2πj − − 1 − j = (25 − j39) 2 2 (z + 1) (z + 4) 25 25  The function  (z +  1)3 (z  1 − 1)(z − 2)  Simple poles at z = 1, 2, triple poles at z = −1. Residues :-  1 z=1 : − 8 1 z=2 : 27 1 19  1 − =− z = −1 : 27 8 216 1 (i) The circle | z |= 2 contains none of the singularities and therefore  C  dz =0 (z + 1)3 (z − 1)(z − 2)  (ii) The circle | z + 1 |= 1 contains the singularity z = −1 and therefore   19  dz 19πj = 2πj − =− 3 216 108 C (z + 1) (z − 1)(z − 2) (iii) The rectangle and vertices ±j, 3 ± j, contains the singularities at z = 1, z = 2 and therefore   C  dz 19πj = − (z + 1)3 (z − 1)(z − 2) 108  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  64(d)  The function  poles at z = ±2. Residue at z = 2 is  (z2  1 324  Residue at z = −2 is −  241  z−1 has a pole of order 4 at z = −1 and simple − 4)(z + 1)4  3 4  1 d3  z − 1  3! dz3 z2 − 4 This residue is calculated by using partial fractions  Residue at z = −1 is given by  1 3 z−1 4 4 = + z2 − 4 z−2 z+2  whence  1 d3  z − 1  1 d 3  1  1 d3  1  = + 3! dz3 z2 − 4 24 dz3 z − 2 8 dz3 z + 2 1 3 =− − 4 4(z − 2) 4(z + 2)4  putting z = −1 gives the residue − (i) The circle | z |=  1 2  contains none of the singularities hence the integral  (z2 C  (ii) The circle | z +  3 2  61 1 3 − =− 4 4.3 4 81  (z − 1) dz = 0 − 4)(z + 1)4  |= 2 contains the singularities at z = −1 and z = −2 but  not that at z = 2 Hence   C   3 61  (z − 1) 487πj =− dz = 2πj − − 2 4 (z − 4)(z + 1) 4 81 162  3 3 (iii) The triangle with vertices − + j, − − j, 3 + j0 contains all the singularities, 2 2 hence  C    z−1 1 3 3 1 dz = 2πj − − − (z2 − 4)(z + 1)4 4.34 4 4.34 4 = −3πj  c Pearson Education Limited 2011   242  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition ∞  dx +x+1 −∞ Since the integrand satisﬁes the condition for Type 1 inﬁnite real integrals, given  65(a)  x2  in section 4.6.5, we consider  C  z2  dz where C is a semicircle with radius +z+1  R and centre the origin in the upper half z -plane  By the residue theorem  C  z2  dz = 2πj {sum of residues of poles of integrand inside C} +z+1  z +z+1=0⇒z= 2  −1 ±  √ √ 1−4 = − 12 ± j 21 3 2  Only one of these simple poles lies inside C (the one with positive imaginary part) Residue there = lim (z − z0 ) z→z0  That is, residue = limz→z0 Thus   C    √ 1 1 1 = − + j 3 , z 0 2 2 z2 + z + 1  1 1  using L'Hôpital's rule (for simplicity) = √ 2z + 1 j 3  dz 1 2π √ √ = 2πj · = z2 + z + 1 j 3 3  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition Now,        = and, as R → ∞, On  R  C    243  R  + −R  Γ  →0  Γ  , z = x = real.  −R  Thus, letting R → ∞ we ﬁnd that   ∞ −∞  65(b)  x2  dx 2π =√ +x+1 3  This integral is done in precisely the same way as that of part (a). This  time, the poles are at ±j but they are both double.  −  (z2  C  dz 1 π = 2πj × residue at j = 2πj· = 2 + 1) 4j 2  Thus    ∞ −∞  65(c)  To evaluate  ∞ 0  (x2  dx π = (x2 + 1)2 2  dx we use the same semicircular contour, + 1)(x2 + 4)2  except that we note   ∞ 0  dx 1 = (x2 + 1)(x2 + 4)2 2    ∞ −∞  dx (x2 + 1)(x2 + 4)2  1 has two poles inside C this time, the simple + 1)(z2 + 4)2 1 and residue pole at z = j and the double pole at z = 2j . Residue at z = j is 18j 11 at z = 2j is − . 288j Thus  ∞  1 dx 11  5π 1 − = = 2πj 2 2 2 (x + 1)(x + 4) 2 18j 288j 288 0 plus the fact that  (z2  c Pearson Education Limited 2011   244  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  2π   cos 3θ dθ 0 5 − 4 cos θ we follow Example 4.39 and put z = ejθ so that  65(d)  In order to evaluate  cos 3θ =  1 1 3 (z + z−3 ) and cos θ = (z + z−1 ) 2 2  With dz = jejθ dθ. Hence we consider   + z−3 ) dz jz(5 − 2(z + z−1 )) 1 3 2 (z  C  The function under the integral can be written 1 1 + z6 1 + z6 = 2j 5z4 − 2z5 − 2z3 2j(z − 2)(1 − 2z)z3 The poles inside | z |= 1 are a triple pole at z = 0 and a simple pole at z =  1 . 2  65 1 gives . Using the Laurent expansion 2 48j 1 21 . The sum is . Hence about z = 0 yields the residue − 16j 24j Using the formula for the residue at z =   0  2π  cos 3θ 1 dθ = 5 − 4 cos θ 2j   C  1 + z6 1 2πj dz = (z − 2)(1 − 2z)z3 24j π = 12  65(e) 2π  4dθ 5 + 4 sin θ  0  This follows in the same way as part (d). Putting z = ejθ yields dz = jejθ dθ, that is, dθ =  1 dz and sin θ = (z − z−1 ) jz 2j  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition Thus    2π 0  4dθ = 5 + 4 sin θ    245  4dz  C  jz(5 +  4 2j (z  − z−1 ))  Thus   2π 0  4dθ = 5 + 4 sin θ   C  4dz 4 8π = 2πj· = (2z + j)(z + 2j) 3j 3 (C is the unit circle | z |= 1)  65(f)    ∞ −∞  x2 dx (x2 + 1)(x2 + 2x + 2)  This follows along similar lines to parts (a) and (b). Consider the semicircular contour and centre the origin radius R on the upper half plane and labelled C  C  z2 dz = 2πj{sum of residues inside C} (z2 + 1)(z2 + 2z + 2)  Double pole is at z = j, simple pole is at z = −1 + j 1 3 (−4 + j3) ; residue at −1 + j is (3 − j4) Residue at z = j is 20 5 7 7 π Sum = − j giving the integral as 20 10  65(g) 2π 0  dθ 3 − 2 cos θ + sin θ  Once again let z = ejθ and consider the integral around the unit circle   2π 0  dθ = 3 − 2 cos θ + sin θ   2 1 C z (2  dz − j) + 3jz − j −  1 2  j 5 and − . Only the ﬁrst is inside | z |= 1. 1 − 2j 1 − 2j 1 1 and so the value of the integral is 2πj = π. The residue is 2j 2j  The poles are at −  c Pearson Education Limited 2011   246  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  65(h)  ∞  dx x4 + 1  0  We have a choice here, let us choose a quarter circle contour as shown below.  1+j Only the root of z4 + 1 = 0 in the positive quadrant, that is z = √ , needs to 2 be taken into account. 1 −1 − j Residue at this point is | 1+j = √ 2 √ 4z z= 2 4 2 Hence  dz π jπ (−1 − j) √ = √ − √ = 2πj 4 4 2 2 2 2 2 C 1+z  0  R   dz π π = + + = √ −j √ 4 2 2 2 2 C z +1 Γ jR 0 on the imaginary axis, z = jy , and on the real axis z = x . Therefore   0  R  dz jdy dx dz = + + 4 4 4 4 C z +1 Γ z +1 R y +1 0 x +1 where we have used (jy)4 = y4 . Letting R → ∞, the last integral → 0. Thus  ∞  0 jdy dx π π √ √ + = − j 4 x4 + 1 2 2 2 2 ∞ y +1 0 Equating real (or indeed imaginary) parts gives  ∞ dx π = √ 4 x +1 2 2 0 c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  247  65(i) ∞ (x2  −∞  dx + 4x + 5)2  The semicircular contour is used and the poles of  (z2  1 are at + 4z + 5)2  z = −2 + j, −2 − j both double. Only the residue at z = −2 + j , which is  1 , 4j  needs to be taken into account. Hence    ∞ −∞  (x2  dx π 1 = 2πj = 2 + 4x + 5) 4j 2  65(j) 2π  cos θdθ 3 + 2 cos θ  0  z2 + 1 Again we use the unit circle on which z = e . The integrand is 2j(z3 + 3z2 + z) √ √ 3 3 1 1 with simple poles at z = 0, − + 5, − − 5. Only the ﬁrst two are inside 2 2 2 2  3  3 C and residues are 1 and − √ . Hence the integral has the value π 1 − √ . 5 5 jθ  Exercises 4.8.3 1 1 x − jy , u + jv = = 2 z x + jy x + y2 x 1 , x2 + y2 = 2ax . Thus u = 2 if u = 2 x +y 2a For the two wires shown in Figure 4.41 potentials are centred at V0 or −V0 66  Since w =  and are tangent to the imaginary (y) axis. They are thus circles of the form x2 + y2 = 2aV0 x . The equipotential curves are shown c Pearson Education Limited 2011   248  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  67(a) j+1 =j 1−j 1 49 + j7 (7 + j)(1 + j7) z= (24 + j7) ⇒ w = =7 = j7 25 1 − j7 1 + 49 3 7 +1 3 z = − ⇒ w = 4 3 = 41 = 7 4 1− 4 4 z = −1 ⇒ w − 0, z = j ⇒ w =  giving images as (0,0),(0,1),(0,7),(7,0). 67(b)  If w =  w−1 z+1 then z = 1−z w+1 (u − 1 + jv)(u + 1 − jv) u − 1 + jv = u + 1 + jv (u + 1)2 + v2 (u2 − 1) + v2 + j(vu + v − uv + v) or x + yj = (u + 1)2 + v2 u2 + v2 − 1 + 2vj x + yj = (u + 1)2 + v2  so that x + yj =  Hence y = 0 corresponds to v = 0.  If x2 + y2 = 1 ⇒| z |= 1 w − 1 w−1 =1 Since z = this means that  w+1 w+1 or (u − 1)2 + v2 = (u + 1)2 + v2 from which u = 0  67(c)  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  249  In order to progress, note that the image of the semicircular conductor in the w plane is the positive quadrant u ≥ 0, v ≥ 0. Instead of temperature T o C we πT π consider since this function has the value on u = 0 (where T = 100o C ). The 200 2 mapping w = ez (Example 4.14) provides a means of eliminating the singularity at w = 0. The complex variable z is already deﬁned, therefore write w = eζ (complex variable ζ). The imaginary part of ζ is identiﬁed with the (scaled) temperature πT . 200 v   πT 2y Thus = tan−1 = tan−1 as required. 2 2 200 u 1−x −y  68(a)  G(x, y) = 2x − 2xy ; thus  ∂G ∂H = = 2 − 2y ∂y ∂x  ∂H ∂G =− = 2x ∂x ∂y Integrating this gives H = x2 − y2 + 2y and  Hence W = G + jH = 2z − jz2  68(b)  If w = lnz , then z = ew  Given H(z) = 2z + jz2 = 2ew + je2w equating real parts gives G(x, y) = 2eu cos v − e2u sin 2v as required  68(c)  If w = f(z) then the real and imaginary parts of f(z) are harmonic  functions. Hence if ζ = g(w) , then the real and imaginary parts of g are harmonic. So ζ = g(w) = g(f(z)) implies that harmonic functions (the real and imaginary c Pearson Education Limited 2011   250  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  parts of f(z) ) transform to harmonic functions (the real and imaginary parts of g(w) ).  69  If w =  z+3 then | w |= k transforms to z−3 z + 3  =k z−3  that is, (x + 3)2 + y2 = k2 (x − 3)2 + k2 y2 1 + k2 or x2 + y2 + 6 x + 9 = 0 as required. 1 − k2 If the centre of the circle is to be (−5, 0) , then  1 + k2  = −10 or k = 2 6 1 − k2 We thus (following section 4.8) require the potential V to be a harmonic function which has a constant value on a circle u2 + v2 = 4. Hence V has the general form V = A ln(u2 + v2 ) V0 on u2 + v2 = 4, V = V0 hence A = ln 4 V0 so that V = ln(u2 + v2 ) ln 4  z + 3 2 (x + 3)2 + y2 2 2 2  = Now u + v =| w | =  z−3 (x − 3)2 + y2 Thus V0 {ln[(x + 3)2 + y2 ] − ln[(x − 3)2 + y2 ]} V= ln 4 70  This problem follows a similar pattern to Exercise 67.  The points A, x = 1, y = 0; B, x = 0, y = 1; C, x = 0, y = −1 under the j(1 − z) transform to mapping of w = 1+z 70(a)  w = 0 that is (0, 0), w =  and w =  j(1 − j) = 1 that is (1, 0) 1+j  j(1 + j) = −1 that is (−1, 0) respectively 1−j c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 70(b)  w=  251  j(1 − z) 1+z  1−x j(1 − x) is purely imaginary. Since If z = x = real (i.e. y = 0) then w = 1+x 1+x can take all real values, points on y = 0 correspond to points on u = 0, the imaginary axis.  j−w j(1 − z) then z = 1+z j+w So that | z |= 1 ⇒| j − w |=| j + w | or v = 0 70(c)  If w =  For the last part we note the following property of the mapping that is, w = j(1 − z) (from (a), (b), and (c)) 1+z  πT which is (in the z plane), 100 π on the negative real axis, and 0 on the positive real axis. The mapping of w = eζ πT as imaginary part). ( ζ - complex variable which has the values of 100 Thus v  1 − x2 − y2  πT = tan−1 = tan−1 100 u 2y In a way similar to Exercise 67, identify the function  (using w =  71  j(1 − z) to ﬁnd u and v). This gives the result. 1+z  This problem is similar to the last part of Exercise 69. The successive  mappings are z1 =  z + j4 w = ln z z − j4  c Pearson Education Limited 2011   252  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  In order for the circle centre 5j to be mapped to a circle centred at the origin in the z plane we require | z1 |2 =|  z + j4 2 | = k2 for some constant k z − j4  2  1+k that is, x2 + y2 + 8 1−k 2 y + 16 = 0 needs a centre at 5j  Therefore, 8(1 + k2 ) = 10(1 − k)2 ) or h =  1 3  In the w plane, | w |=| ln z1 |=| ln(x21 + y21 ) |= 2 ln 3 on the boundary of the circle on which T = 100. Thus writing T = A | ln(x21 + y21 ) | gives T = 100 when x21 + y21 = Thus  1 9  if  100 = A· 2 ln 3 or A =  50 ln 3  50 ln(x21 + y21 ) ln 3   x2 + (4 + y)2 50 ln as required = ln 3 x2 + (4 − y)2  T=  Note that T = 0 corresponds to x21 + y21 = 1 or y = 0 as is also required (| z + j4 |2 =| z − j4 |2 is y = 0). 1 was studied in Example 4.13. Writing, as usual, z w = u + jv and z = x + jv leads to 72  The mapping w = z +  u=x+  x y and v = y − x2 + y 2 x2 + y 2  Hence the unit circle x2 + y2 = 1 in the z plane corresponds to v = 0 (the real axis) in the w plane. Points ejπ/3 and e2jπ/3 ( P and Q of this problem) correspond to u = 2 cos π3 and 2 cos 2π 3 respectively. The arc PQ thus corresponds to −1 ≤ u ≤ 1 in the w plane. w+1 takes this portion of the real axis (−1 ≤ u ≤ 1) The further mapping ζ = w−1 to −∞ ≤ Re{ζ} ≤ 0 . This negative real axis corresponds to T = 100. Hence in a similar fashion to Exercise 70, we identify the variable c Pearson Education Limited 2011   πT 100  (which = π on this  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  253  line) with the imaginary part of ln ζ which is argζ. We cannot use tan−1 here because the argument of the logarithm function is quadratic and so     πT w+1 z + 1/z + 1 = argζ = arg = arg 100 w−1 z + 1/z − 1  100 arg(z2 + z + 1) − arg(z2 − z + 1) that is, T = π as required.  Review Exercises 4.9 1(a)  z = 1 + j, w = (1 + j)z + j = (1 + j)2 + j = 1 + 2j − 1 + j = 3j  1(b)  z = 1 − j2, w = j3z + j + 1 = j(1 − j2)3 + j + 1 = 3j + 6 + j + 1 = 4j + 7  1(c)  z = 1, w = 12 (1 − j)z + 12 (1 + j) = 1  1(d)  z = j2, w = 12 (1 − j)z + 12 (1 + j) = (1 − j)j + 12 (1 + j) = 32 (1 + j)  2(a)  y = 2x  (b)  x+y=1  For the mapping w = (1 + j)z + j u = x − y, v = x + y + 1 so y = 2x ⇒ v + 3u = 1 and x + y = 1 ⇒ v = 2 For the mapping w = j3z + j + 1 u = −3y + 1, v = 3x + 1 so y = 2x ⇒ u + 2v = 3 and x + y = 1 ⇒ v − u = 3 For the mapping w = 12 (1 − j)z + 12 (1 + j) u=  1 1 (x + y + 1), v = (x − y + 1) 2 2  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  254  so y = 2x ⇒ 3v − u = 1 and x + y = 1 ⇒ u = 1 3  w = αz + β  when z = 2 − j, w = 1, and when z = 0, w = 3 + j 3(a)  Solving the simultaneous equations gives α = − 15 (3 + j4), β = 3 + j .  3(b)  Since 1 w = − (3 + 4j)z + 3 + j 5 1 z = (3 + j − w)(3 − j4) 5 so x = 13 − 3u − 4v  and Re{z} ≤ 0 corresponds to 3u + 4v ≥ 13 3(c) | z |=  1 | 3 + j − w |5 ≤ 1 5 ⇒| w − 3 − j |≤  3(d)  Fixed point is given by z = αz + β or z =  4  1 5  w=  j z  ⇒x=  v u2 +v 2 ,  4(a)  x=y+1⇒  4(b)  y = 3x ⇒ u = 3v  y=  v u2 +v 2  =  1 β = (7 − j) 1−α 4  u u2 +v 2 u u2 +v 2  + 1 or u2 + v2 + u − v = 0  Line joining A(1 + j) to B(z + j3) or (1, 1) to (2, 3) is y = 2x − 1 which u 2v transforms to 2 = − 1 or u2 + v2 + u − 2v = 0 u + v2 u2 + v2  4(c)  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 4(d)  255  y = 4 ⇒ 4(u2 + v2 ) = u  The following Argand diagram shows all these curves.  5  w=  w+1 z+1 ⇒z= z−1 w−1  from which  u2 + v2 − 1 −2v x= , y= 2 2 (u − 1) + v (u − 1)2 + v2  lines x = k and y = l map to circles u2 + v2 −  k+1 2k 2v u+ = 0 and u2 + v2 − 2u + +1=0 k−1 k−1 l  Fixed points are z =  z+1 z−1  in z2 − 2z − 1 = 0, that is, z = 1 +  ﬁxed points.  c Pearson Education Limited 2011   √ √ 2, 1 − 2 are the  256  6  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  w=  1 − z2 z  Fixed points occur at z =  1 − z2 2 or  z 2 = 1 − z2  z2 = 1/2 √ Hence z = ± 2/2 z∗ 1 −z= − z ( z∗ = complex conjugate). z | z |2 1  x −y Whence u = 2 − x , v = − − y and v = −y + 1 x + y2 x2 + y 2 r2 2 2 r u −r v so 2 = x, 2 =y r −1 r +1 Squaring and adding gives Writing w =    ur2 r2 − 1  2   +  vr2 r2 + 1  2 = x2 + y 2 = 1  the required ellipses. Since u =  7  x x2 +y 2  − x , if x2 + y2 = 1 then u = 0 (imaginary axis in the w plane).  w = z3  = (x + jy)3 = x3 + 3jx2 y + 3j2 xy + j3 y3 so u = x3 − 3xy2 v = 3x2 y − y3 are the real and imaginary parts. ∂u = 3x2 − 3y2 , ∂x ∂v = 3x2 − 3y2 , ∂y  ∂u = −6xy ∂y ∂v = 6xy ∂x  hence verifying the Cauchy–Riemann equations: ∂u ∂v ∂u ∂v = , =− ∂x ∂y ∂y ∂x  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 8  257  u(x, y) = x sin x cosh y − y cos x sinh y hence and  ∂u = sin x cosh y + x cos x cosh y + sin x sinh y ∂x ∂u = x sin x sinh y − y cos x cosh y − cos x sinh y ∂y  By the Cauchy–Riemann equations, ∂u ∂v ∂u ∂v = and =− ∂x ∂y ∂y ∂x hence, integrating  ∂u with respect to y gives: ∂x  v = sin x sinh y + x cos x sinh y + y sin x cosh y − sin x sinh y + f1 (y) Integrating −  ∂u with respect to x gives: ∂y  v = (x cos x − sin x) sinh y + y sin x cosh y + sin x sinh y + f2 (x) where f1 and f2 are arbitrary functions. Comparing these gives v = y sin x cosh y + x cos x sinh y (ignoring the additive constant). Thus  w = u + jv =x sin x cosh y − y cos x sinh y + j(y sin x cosh y + x cos x sinh y)  Since this is f(z) , we put y = 0 to ﬁnd f(x) which will give the functional form of f, namely f(x) = x sin x. Thus f(z) = z sin z. az + b (the general bilinear mapping) since z = 0 ⇒ w = ∞ we cz + d αz + β must have d = 0, hence (relabelling the constants) w = . z Writing this as wz = αz + β and inserting the pairs of values z = j, w = −j and 9  Writing w =  z = 12 (1 + j), w = 1 − j gives 1 = αj + β, 1 = 12 (1 + j)α + β from which α = 0, β = 1. Hence w =  1 . z  c Pearson Education Limited 2011   258  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  x y and v = − 2 2 +y x + y2 the real axis (in the z plane) maps to the real axis (in the w plane).  9(a)  Since u =  x2  y If y > 0 then v = − x2 +y 2 < 0 and vice versa.  Thus the lower half of the z plane maps to the upper half of the w plane. 9(b) The circle | z − 12 j |=   y or v = −1 v = − x2 +y 2 If | z − 12 j |<  1 2  is x2 + y2 − y = 0  1 2  then x2 + y2 − y < 0 or v < −1, that is, the interior of | z − 12 j |=  1 2  maps to Im(w) < −1 as required. a2 10 The mapping z = ζ + 4ζ maps R = constant (where ζ = Rejθ ) to curves z = Rejθ +  a2 −jθ e 4R  which describe ellipses in the z plane as can be seen by writing   a2  a2  cos θ, y = R − sin θ x= R+ 4R 4R whence  x2 (R +  when R =  1 2 a,  2  a 2 4R )  +  y2 (R −  a2 2 4R )  =1  y = 0 (real axis). This mapping is used together with bilinear  mappings to map an aerofoil shape onto the unit circle. aeronautical engineering. 1 = (1 + z3 )−1 1 + z3 using the binomial expansion gives 11  1 = 1 − z3 + z 6 − z 9 + · · · 3 1+z Similarly 1 = 1 − 2z3 + 3z6 − 4z9 + · · · (1 + z3 )2 and both are valid in the disc | z |< 1. c Pearson Education Limited 2011   This is useful in  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  259  12(a)  2−1−z 2 1−z = = −1 1+z 1+z 1+z Using the binomial series again gives 1−z = 1 − 2z + 2z2 − 2z3 + . . . 1+z  since the nearest singularity of the function to z = 0 is at z = −1, the radius of convergence is 1.  12(b)  f(z) =  z2  use Taylor's series  1 . This time we need to expand about the point z = 1. We +1  2z 2 8z2 1   , f (z) = − , f (z) = − + z2 + 1 (z2 + 1)2 (z2 + 1)2 (z2 + 1)3 24z 48z3 24 288z2 384z4 iv f (z) = 2 − , f (z) = − + (z + 1)3 (z2 + 1)4 (z2 + 1)3 (z2 + 1)4 (z2 + 1)5 f(z) =  At z = 1 these have values  1 1 1 2 , − 2 , 2 , 0, −3  giving the expansion z2  1 1 1 1 1 = − (z − 1) + (z − 1)2 − (z − 1)4 + . . . +1 2 2 4 6  √ 1 are at z = ±j which are a distance 2 from z = 1, 2 1+z √ hence the radius of convergence is 2 . The singularities of  12(c) 1 z =1− = f(z) 1+z 1+z 1 2 6 f (z) = , f = − , f (z) = + 2 3 (1 + z) (1 + z) (1 + z)4 1 1 1 z 1 = (1 + j) + j(z − j) − (1 + j)(z − j)2 − (z − j)3 + . . . 1+z 2 2 8 √4 The radius of convergence is again 2.  Thus  1 has singularities at z = 0, j, −j . The radius of + 1) convergence is the distance of the centre of the point of expansion from the nearest  13  The function  z(z2  c Pearson Education Limited 2011   260  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  of these singularities. These are found straightforwardly either by inspection or by using Pythagoras' theorem z = 1; 1; z = −1; 1, z = 1 + j; 1, z = 1 + j 12 ;  14  f(z) =  14(a)  (z2  1 2  √  √ 5, z = 2 + j3; 2 2 .  1 + 1)z  The Laurent expansion is 1 (1 + z2 )−1 z 1 = (1 − z2 + z4 − z6 + . . .) z 1 = − z + z3 − z 5 + . . . z valid for 0 <| z |< 1  14(b)  Since f(z) is regular at z = 1, f(z) has a Taylor expansion : 1 (z − 1)2   = f(1) + (z − 1)f f (1) + . . . (1) + z(1 + z2 ) 2 1 + 3z2 1 , f (1) = −1 f(1) = , f (z) = − 2 (z + z3 )2 6z 2(1 + 3z2 )2 + f (z) = − (z + z3 )2 (z + z3 )3 5 4 2 × 16 = so f (1) = − + 4 8 2  so that  1 1 5 = −(z − 1) + (z − 1)2 + . . . 2 z(1 + z ) 2 4 valid for | z − 1 |< 1  15  f(z) = ez sin  15(a)   1  1−z  At z = 0, f(z) is regular. Thus the principal part is zero and f(0) =  sin 1, f(z) = sin 1 + q1 z + q2 z2 + . . . | z |< 1, Taylor series.  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  261  15(b)  At z = 1 , f(z) has an essential singularity.  15(c)  At z = ∞, ez has an essential singularity. Hence for parts (b) and (c) the  principal part has inﬁnitely many terms. 16(a) ez sinh z = = = Real part = Imaginary part =  1 z z 1 e (e + e−z ) = (e2z + 1) 2 2 1 (1 + e2x (e2jy )) 2 1 (1 + e2x cos 2y + je2x sin 2y) 2 1 (1 + e2x cos 2y) 2 1 2x e sin 2y 2  16(b) cos 2z = cos(2x + j2y) = cos 2x cosh 2y − j sin 2x sinh 2y 16(c) sin z x − jy = 2 sin(x + jy) z x + y2 (x − jy)(sin x cosh y + j cos x sinh y) = x2 + y 2 x sin x cosh y + y cos x sinh y + j(x cos x sinh y − y sin x cosh y) = x2 + y 2 16(d) tan x + tan jy 1 − tan x tan jy tan x + j tanh y 1 + j tan x tanh y · = 1 − j tan x tanh y 1 + j tan x tanh y  tan z = tan(x + jy) =  from which tan z =  tan x(1 − tanh2 y) + j tanh y(1 + tan2 x) 1 + tan2 x tanh2 y  c Pearson Education Limited 2011   262  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  17(a)  Since  2 dw = − 3 = 0, this mapping is conformal. dz z  17(b) dw = 6z2 + 6z + 6(1 − j) dz = 0 when z2 + z + 1 − j = 0 or (z − j)(z + j + 1) = 0 so z = j, −1 − j are the points where the mapping fails to be conformal. 17(c)  w = 64z +  1 z3 dw 3 = 64 − 4 = 0 dz z 3 where z4 = 64 so z = 0.465, −0.465, j0.465, −j0.465  are the points where the mapping fails to be conformal.  18  w = cos z ,  dw = sin z = 0 when z = nπ, n = an integer. dz u + jv = cos(x + jy) = cos x cosh y − j sin x sinh y u = cos x cosh y v = − sin x sinh y  Lines x = k will thus transform to u2 v2 − = 1 − hyperpolae cos2 k sin2 k Lines y = l will thus transform to u2 v2 + = 1 − ellipses cosh2 l sinh2 l c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  19(a) at z = 0. 19(b)  263  sin z sin z sin z 1 · . Since → 1 is z → 0 this function has a simple role = 2 z z z z  (z3  1 has double poles when z3 = 8, that is, at 2, 2e2πj/3 , 2e4πj/3 . − 8)2  19(c) z+1 1 z+1 = 2 = 4 2 z −1 (z − 1)(z + 1) (z − 1)(z2 + 1) The singularity at z = −1 was removable, those at z = 1, ±j are simple poles. 1 which has simple poles wherever cosh z = 0 that is, cosh z z = 12 jπ(2n + 1), n = 0, ±1, ±2, . . . 19(d)  sech z =  19(e)  sinh z is entire (no singularities in the ﬁnite plane).  19(f)  Essential singularity at z = 0.  19(g)  zz = ez ln z  which has an essential singularity at z = 0.  20(a)  e2z double pole at z = −1 residue given by (1 + z)2 d 2z  (e ) z=−1 dz = 2e−2  c Pearson Education Limited 2011   264  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  20(b)  cos z simple pole at z = π/2 residue is 2z − π lim (z − π/2)  z→π/2  cos z 1 π = cos = 0 (2z − π) 2 2  sin z 1 tan z = which has a double pole at z = π 2z − π cos z(2z − π) 2 Writing w = z − π/2  20(c)  1 − 12 w2 + . . . tan z cos w cos w =− =− =− 2z − π 2w 2w sin w 2w(w − 16 w3 + . . .) 1 1 = − 2 + + ... 2w 4 Hence residue is 0.  z has a triple pole at z = −8 (z + 8)3 z 1 1 8 Writing w = z + 8 and z = w − 8 gives = 3 (w − 8) = 2 − 3 3 (z + 8) w w w Hence the residue is 0. 20(d)  21 f(z) =  (z2 − 1)(z2 + 3z + 5) z(z4 + 1)  Zeros are at z = ±1 and at the roots of z2 + 3z + 5 = 0 z=−  3±  √ 9 − 20 3 1√ =− ±j 11 2 2 2  Simple poles are at z = 0 and where z4 = −1 √ that is, z = (±1 ± j)/ 2 The residue at z = 0 is given by (z2 − 1)(z2 + 3z + 5) = −5 z→0 z4 + 1 lim  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  265  The residue at z = z0 (where z40 + 1 = 0) is  (z − z0 )(z2 − 1)(z2 + 3z + 5) lim z→z0 z(z4 + 1)   z − z0 (z20 − 1)(z20 + 3z0 + 5) = lim z→z0 z0 z4 + 1 (z2 − 1)(z20 + 3z0 + 5) 1 1 = 0 = − (z20 − 1)(z20 + 3z0 + 5) 3 z0 4z0 4   (using z40 = −1) √ , − 1+j √ , − 1−j √ , 1−j √ in turn gives the residues Putting z0 = 1+j 2 2 2 2 √ √ √ 3 3 3 3 3 3 2 − 4 2 + j, 2 − 4 2 − j and 2 + 4 2 + j respectively.  22  f(z) =  3 2  +  3 4  √  2 − j,  z7 + 6z5 − 30z4 has a triple pole at z = 1 + j (z − 1 − j)3 1 d2 7 (z + 6z5 − 30z4 ) is evaluated at z = 1 + j 2! dz2  = 21z5 + 60z3 − 180z2 z=1+j  Residue =  = 21(−4 − 4j) + 60(−2 + 2j) − 360j = −204 − j324  z has poles at z = −6 and z = −1. Only the z2 + 7z + 6 z 1 second is inside C. Residue = |z=−1 = − z+6 5 2πj Integral = − 5  23(a)  The integrand  23(b)  The integrand  (z2 + 1)(z2 + 3) has four simple poles ±2j, ±3j all (z2 + 9)(z2 + 4)  inside C.  3 3 8 8 j, j, j and − j the sum of which is 0. 20 20 5 5 The integral is thus 0. Residues are −  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  266  23(c) The integrand  1 has double poles at z = 0, 1 and −1. z2 (1 − z2 )2  Residues are, at z = 0,   1 d 2z  =+  =0 2 2 2 3 z=0 dz (1 − z ) (1 − z ) z=0  1 2(2z + 1)  3 d − 2 = = At z = 1,  2 2 3 dz z (1 + z) (z + z ) z=1 4  1 −2(1 − 2z)  d 3 = At z = −1, =  2 2 2 3 dz z (1 − z) (z − z ) z=−1 4 (i) If C is | z |=  1 2  then only z = 0 is inside C, so integral = 0  (ii) If C is | z |= 2 then all poles are inside C, so integral = 3πj. 1 3j has simple poles at and −j . (2z − 3j)(z + j) 2 1 1 3j and −j are, respectively, − j and j. Residues at 2 5 5 (i) Both poles are inside | z |= 2, but since their sum is zero so is the integral 1 (ii) Inside | z − 1 |= 1 the function is regular. By Cauchy's (2z − 3j)(z + j) theorem the integral = 0. 23(d)  The integrand  z3 has simple poles at 23(e) The integrand 2 (z + 1)(z2 + z + 1) √ z = ±j, − 12 ± j 12 3 . C is the circle | z − j |= 12 which contains the pole at z = j but not the other three poles. Residue at z = j is  1 2j  hence the integral = −π.  z−1 has simple poles at z = 0, 3 and a double pole at z = 2. z(z − 2)2 (z − 3) 1 and − 34 . Residues at z = 0, 2 are respectively 12  23(f)  (i) If C is | z |= 1 only the residue at z = 0 is considered : integral = πj  1 63  (ii) If C is | z |= 52 , residues at 0 at 2 are summed; integral = πj 12 −4 = −4πj/3.  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  267  ∞  x2 dx 2 2 2 −∞ (x + 1) (x + 2x + 2) we use the semicircular contour C on the upper half plane (see Exercise 65). The  24(a)  To evaluate  integral along the curved portion → 0 as the radius of the semicircle → ∞. The residues in the upper half plane are at the (double) pole at j and the (simple) pole at −1 + j.  d z2 z2 at z = j is which is Residue of (z2 + 1)2 (z2 + 2z + 2) dz (z + j)2 (z2 + 2z + 2) 9j − 12 . evaluated at z = j. This is (after some algebra) 100 2 z evaluated at z = −1 + j which is Residue at z = −1 + j is 2 (z + 1)(z + 1 + j) 3 − 4j 7j . Sum of residues is − 25 100   ∞ x2 dx z2 dz −7j 7π = = 2πj = 2 2 2 2 2 2 100 50 C (z + 1) (z + 2z + 2) −∞ (x + 1) (x + 2x + 2) ∞ x2 dx one can either use a quarter circle contour (as in 4 0 x + 16 ∞ x2 dx ∞ x2 dx 1 = and use Exercise 65(h)) or note that, by symmetry, 2 4 4 −∞ x + 16 0 x + 16 the same semicircular contour as above. 24(b)  To evaluate  The disadvantage of doing this is that there are two poles inside the semicircular contour, but only one in the quarter circle. However, this is compensated by the easier manipulation of the integral. We shall thus use the semicircle. The poles inside C, both simple, are at z=  √  2(−1 + j) and z =  √  2(1 + j)  The way to avoid unnecessary arithmetic/algebra is to determine the residue at z = z0 where z0 is one of the above poles. This is given by lim  z→z0  (z − z0 )z2 z4 + 16  Since z = z0 is a root of z4 + 16, we can use L'Hôpital's rule to obtain  1 3z2 − 2zz0  = Residue =  3 4z 4z0 z=z0 c Pearson Education Limited 2011   268  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Sum of residues is thus 1 1 j √ + √ =− √ 4 2(−1 + j) 4 2(1 + j) 4 2   Thus  ∞ −∞  x2 dx = x4 + 16   0   −j  z2 dz π π√ √ √ = 2πj = = 2 z4 + 16 4 4 2 2 2    Hence  ∞ 0  24(c)  x2 dx π√ = 2 x4 + 16 8  To evaluate 2π  sin2 θdθ 5 + 4 cos θ  0  we follow Exercise 65(d) and put z = ejθ so that z2 − z4 − 1 (2z3 + 5z2 + 2z)4 Hence  0  2π  sin2 θdθ 1 = 5 + 4 cos θ 4j   C  sin2 θ dz = dθ and = jz 5 + 4 cos θ  z2 − z4 − 1 dz z2 (2z + 1)(z + 2)  where C is the unit circle. Residues at z = 0 and z = − 12 (not that at z = −2) are summed.  5 d  z 2 − z4 − 1  [evaluated at z = 0] is . Residue at dz 2z2 + 5z + 2 4 z2 − z4 − 1  1 13 ) z = − is ( 2 =− 2 z (z + 2) z=− 1 6 2 Hence    2π sin2 θdθ 1 5 13 11π = 2πj − =− 5 + 4 cos θ 4j 4 6 24 0  Residue at z = 0 is  24(d)  The integral 2π 0  cos 2θ dθ 5 − 4 cos θ  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  269  is evaluated similarly   2π 0  cos 2θ 1 dθ = 5 − 4 cos θ 2j   C    z4 + 1 17 5 19π 1 − = dz = 2πj 3 4 2 5z − 2z − 2z 2j 6 4 12  (In part (c) the negative sign arises from the choice of direction of the line integral. sin2 θ is always positive it can be ignored.) Since the integrand 5 + 4 cos θ  c Pearson Education Limited 2011   5 Laplace Transforms Exercises 5.2.6 1(a)  1  s 1 1 1 + = 2 , Re(s) > 2 L{cosh 2t} = L{ (e2t + e−2t ) = 2 2 s−2 s+2 s −4  1(b)  L{t2 } =  1(c)  L{3 + t} =  1 3 3s + 1 + 2 = , Re(s) > 0 s s s2  1(d)  L{te−t } =  1 , Re(s) > −1 (s + 1)2  2(a)  5  (g) 0  2 , Re(s) > 0 s3  (b) -3  (c) 0  (h) 0  (i) 2  (d) 3  (e) 2  (f) 0  (j) 3  5 5s − 3 3 , Re(s) > 0 − 2 = s s s2  3(a)  L{5 − 3t} =  3(b)  L{7t3 − 2 sin 3t} = 7.  3(c)  L{3 − 2t + 4 cos 2t} =  3(d)  L{cosh 3t} =  3(e)  L{sinh 2t} =  s2  s2  42 6 3 6 = 4 − 2 , Re(s) > 0 − 2. 2 4 s s +9 s s +9 3 s 4s 2 3s − 2 + 2 − 2 + 4. 2 = , Re(s) > 0 2 s s s +4 s s +4  s , Re(s) > 3 −9  2 , Re(s) > 2 −4 c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 3(f)  L{5e−2t + 3 − 2 cos 2t} =  3(g)  L{4te−2t } =  3(h)  L{2e−3t sin 2t} =  3(i)  L{t2 e−4t } =  5 3 s + − 2. 2 , Re(s) > 0 s+2 s s +4  4 , Re(s) > −2 (s + 2)2 4 4 = , Re(s) > −3 (s + 3)2 + 4 s2 + 6s + 13  2 , Re(s) > −4 (s + 4)3  3(j) 36 6 4 2 − 3+ 2− 4 s s s s 36 − 6s + 4s2 − 2s3 = , Re(s) > 0 s4  L{6t3 − 3t2 + 4t − 2} =  3(k)  3(l)  L{2 cos 3t + 5 sin 3t} = 2.  3 2s + 15 s + 5. = , Re(s) > 0 s2 + 9 s2 + 9 s2 + 9  s +4 s2 − 4 d s  = L{t cos 2t} = − , Re(s) > 0 ds s2 + 4 (s2 + 4)2 L{cos 2t} =  s2  3(m) 6s d 3  = 2 2 ds s + 9 (s + 9)2   (s2 + 9)2 6 − 6s(s2 + 9)2 4s  6s d = − L{t2 sin 3t} = − ds (s2 + 9)2 (s2 + 9)4 18s2 − 54 , Re(s) > 0 = 2 (s + 9)3 L{t sin 3t} = −  3(n)  L{t2 − 3 cos 4t} =  2 3s , Re(s) > 0 − 2 3 s s + 16  c Pearson Education Limited 2011   271  272  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  3(o) 3 2 (s + 1) + + 3 2 (s + 2) (s + 1) + 4 s 3 2 s+1 + , Re(s) > 0 = + 2 3 (s + 2) s + 2s + 5 s  L{t2 e−2t − e−t cos 2t + 3} =  Exercises 5.2.10   4(a)  L−1  4(b)  L−1  4(c)  L−1  4(d)  L−1    1    14 1 1 = L−1 − 4 = [e−3t − e−7t ] (s + 3)(s + 7) s+3 s+7 4    −1 s+5 2  = L−1 = −e−t + 2e3t + (s + 1)(s − 3) s+1 s−3  1 4   s−1  4 4 1 4 3 9 −1 9 = L − = − t − e−3t − 2 2 s (s + 3) s s s+3 9 3 9   2s + 6   s 2  = L−1 2. 2 = 2 cos 2t + 3 sin 2t + 3. 2 2 2 s +4 s +2 s + 22  4(e) −1  L  4(f)  L−1      1 1    1 −1 0 16 16 = L + − s2 (s2 + 16) s s2 s2 + 16 1 1 1 t− sin 4t = [4t − sin 4t] = 16 64 64    s+8  −1 (s + 2) + 6 = L = e−2t [cos t + 6 sin t] s2 + 4s + 5 (s + 2)2 + 1  4(g) L−1     1 − 18 s + 12  s+1 −1 8 = L + s2 (s2 + 4s + 8) s (s + 2)2 + 22  1 1 1 (s + 2) − 3(2)  = L−1 . − 8 s 8 (s + 2)2 + 22 1 = [1 − e−2t cos 2t + 3e−2t sin 2t] 8 c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 4(h) L−1      1  1 2 4s −1 − + = L (s − 1)(s + 1)2 s − 1 (s + 1) (s + 1)2 = et − e−t + 2tet  4(i)  L−1      s+7  −1 (s + 1) + 3(2) = L = e−t [cos 2t + 3 sin 2t] s2 + 2s + 5 (s + 1)2 + 22  4(j) −1  L    4(k) L−1  1    12 3 s2 − 7s + 5 −1 =L − + 2 (s − 1)(s − 2)(s − 3) s−1 s−2 s−3 1 11 = et − 3e2t + e3t 2 2      −2 2s − 1  5s − 7 −1 = L + (s + 3)(s2 + 2) s + 3 s2 + 2 √ √ 1 = −2e−3t + 2 cos 2t − √ sin 2t 2  4(l) −1  L  4(m)  L−1        15 1 s−2  s −1 = L − (s − 1)(s2 + 2s + 2) s − 1 5 s2 + 2s + 2  15 1 (s + 1) − 3  − = L−1 s − 1 5 (s + 1)2 + 1 1 1 = et − e−t (cos t − 3 sin t) 5 5   (s + 1) − 2  s−1  −1 = L = e−t (cos 2t − sin 2t) s2 + 2s + 5 (s + 1)2 + 22  4(n) L−1    3    12 2 s−1 = L−1 − + 2 (s − 2)(s − 3)(s − 4) s−2 s−3 s−4 1 3 = e2t − 2e3t + e−4t 2 2  c Pearson Education Limited 2011   273  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  274  4(o)      3s 3s −1 = L (s − 1)(s2 − 4) (s − 1)(s − 2)(s + 2) 3 1   −1 + 2 − 2 = L−1 s−1 s−2 s+2 3 1 = −et + e2t − e−2t 2 2    9 1    36 −1 4 2s 2s = L − + 2 2 2 2 s(s + 1)(s + 9) s s +1 s +9 1 9 = 4 − cos t + cos 3t 2 2  L−1  4(p) L−1  4(q) L−1       9 7s + 9 2s2 + 4s + 9 = L−1 − 3 2 2 (s + 2)(s + 3s + 3) s + 2 (s + 2 ) + 3/4 √ √  9 7(s + 32 ) − 3. 3/2  −1 √ − =L s+2 (s + 32 )2 + ( 3/2)2 √ √ √ 3  3 3  t − 3 sin t = 9e−2t − e− 2 t 7 cos 2 2  4(r) L−1    1 1 1    19 1 −1 10 90 s + 9 = L − − 2 2 (s + 1)(s + 2)(s + 2s + 10) s + 1 s + 2 s + 2s + 10  1  1  1 s + 10 −1 9 10 − − =L s + 1 s + 2 90 (s + 1)2 + 32  1  1 1  (s + 1) + 3(3)  −1 9 10 − − =L s + 1 s + 2 90 (s + 1)2 + 32 1 1 1 = e−t − e−2t − e−t (cos 3t + 3 sin 3t) 9 10 90  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Exercises 5.3.5 5(a) 2s + 5 1 = s+2 s+2 2s + 5 1 1 X(s) = = + (s + 2)(s + 3) s+2 s+3  (s + 3)X(s) = 2 +  x(t) = L−1 {X(s)} = e−2t + e−3t 5(b) 2 s2 + 6 = s2 + 4 s2 + 4 35 3 4 s2 + 6 26 26 s + 26 = − X(s) = (3s − 4)(s2 + 4) 3s − 4 s2 + 4 35 4 t 3 2 x(t) = L−1 {X(s)} = e 3 − (cos 2t + sin 2t) 78 26 3 (3s − 4)X(s) = 1 +  5(c) (s2 + 2s + 5)X(s) =  1 s  1 1 s+2 1 5 = − · X(s) = s(s2 + 2s + 5) s 5 s2 + 2s + 5 1 1 (s + 1) + 12 (2) = 5 − s 5 (s + 1)2 + 22 1 1 x(t) = L−1 {X(s)} = (1 − e−t cos 2t − e−t sin 2t) 5 2  5(d) 4s 2s2 + 4s + 8 = s2 + 4 s2 + 4 2s2 + 4s + 8 X(s) = (s + 1)2 (s2 + 4) 12 6 1  12s − 32  25 5 = − + (s + 1) (s + 1)2 25 s2 + 4 16 12 −t 6 −t 12 e + te − cos 2t + sin 2t x(t) = L−1 {X(s)} = 25 5 25 25 (s2 + 2s + 1)X(s) = 2 +  c Pearson Education Limited 2011   275  276  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  5(e) (s2 − 3s + 2)X(s) = 1 +  X(s) =  x(t) = L−1 {X(s)} =  2 s+6 = s+4 s+4  1 7 4 s+6 = 15 − 5 + 3 (s + 4)(s − 1)(s − 2) s+4 s−1 s−2  1 −4t 7 t 4 2t e − e + e 15 5 3  5(f) (s2 + 4s + 5)X(s) = (4s − 7) + 16 +  X(s) =  3 s+2  3 (s + 2) + 1 4s2 + 17s + 21 = + 2 (s + 2)(s + 4s + 5) s + 2 (s + 2)2 + 1  x(t) = L−1 {X(s)} = 3e−2t + e−2t cos t + e−2t sin t  5(g) (s2 + s − 2)X(s) = s + 1 +  5(2) (s + 1)2 + 4  s3 + 3s2 + 7s + 15 X(s) = (s + 2)(s − 1)(s2 + 2s + 5) 13 1 5 − 31 12 4s − 4 = + + s + 2 s − 1 s2 + 2s + 5 13 − 31 1  (s + 1) − 3(2)  = + 12 + s + 2 s − 1 4 (s + 1)2 + 22  1 13 1 3 x(t) = L−1 {X(s)} = − e−2t + et + e−t cos 2t − e−t sin 2t 3 12 4 4  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 5(h) (s2 + 2s + 3)Y(s) = 1 +  3 s2  2 4 − 32 1 s2 + 3 3s + 3 = + 2+ 2 Y(s) = 2 2 s (s + 2s + 3) s s s + 2s + 3 √ 1 − 23 2  (s + 1) − √2 ( 2)  1 √ = + 2+ s s 3 (s + 1)2 + ( 2)2  √ √ 2 1 2 y(t) = L−1 {Y(s)} = − + t + e−t (cos 2t + √ sin 2t) 3 3 2  5(i) (s2 + 4s + 4)X(s) = X(s) =  = x(t) = L−1 {X(s)} =  1 1 2 s+2+ 3 + 2 s s+2 s5 + 6s4 + 10s3 + 4s + 8 2s3 (s + 2)3 3 8  s  −  1 2 s2  +  1 2 s3  +  1 8  s+2  +  3 4  (s +  2)2  +  1 (s + 2)3  3 1 1 3 1 1 − t + t2 + e−2t + te−2t + t2 e−2t 8 2 4 8 4 2  5(j) (9s2 + 12s + 5)X(s) = X(s) =  = x(t) = L−1 {X(s)} =  1 s 1 1 4 1 5 5 s + 15 = − s 9s(s2 + 43 s + 59 ) (s + 23 )2 + 1 5  s  −  1 [(s + 23 ) + 23 ] 5 (s + 23 )2 + ( 31 )2  1 1 1 −2t 1 − e 3 (cos t + 2 sin t) 5 5 3 3  c Pearson Education Limited 2011   1 9  277  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  278 5(k)  1 4 −s3 − 6s2 − 16s + 32 (s2 + 8s + 16)X(s) = − s + 1 − 4 + 16· 2 = 2 s + 16 2(s2 + 16) −s3 − 6s2 − 16s + 32 X(s) = 2(s + 4)2 (s2 + 16) 1 1 0 2s + = − 2 2 s + 4 (s + 4) s + 16 1 x(t) = L−1 {X(s)} = te−4t − cos 4t 2 5(l) (9s2 + 12s + 4)Y(s) = 9(s + 1) + 12 +  1 s+1  9s2 + 30s + 22 Y(s) = (3s + 2)2 (s + 1) 0 18 1 + + = s + 1 3s + 2 (3s + 2)2 2  y(t) = L−1 {Y(s)} = e−t + 2te− 3 t 5(m) 1 2 + 2 s s 3 s − 2s2 + 2s + 1 X(s) = 2 s (s − 1)(s − 2)(s + 1)  (s3 − 2s2 − s + 2)X(s) = s − 2 +  5 4  1 2 s2  5 2 1 12 + − 3 = + − s s−1 s−2 s+1 5 1 5 2 x(t) = L−1 {X(s)} = + t − et + e2t − e−t 4 2 12 3  5(n) s +9 9 3 2 s + 2s + 10s + 18 1 7s − 25 1 s+9 20 X(s) = 2 = − − (s + 9)(s + 1)(s2 + 1) s + 1 16 s2 + 1 80 s2 + 9 25 1 3 9 −t 7 e − cos t + sin t − cos 3t − sin 3t x(t) = L−1 {X(s)} = 20 16 16 80 80 (s3 + s2 + s + 1) = (s + 1) + 1 +  s2  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  279  6(a) 1 1 2sX(s) − (2s + 9)Y(s) = − + 2 s+2 (2s + 4)X(s) + (4s − 37)Y(s) = 1 Eliminating X(s) 1 1 )(2s + 4) − 2s = −3s [−(2s + 9)(2s + 4) − 2s(4s − 37)]Y(s) = (− + 2 s+2 3s 1 s Y(s) = = · 2 12s − 48s + 36 4 (s − 3)(s − 1) 3 1  1 2 − 2 = 4 s−3 s−1 1  3 3t 1 t  3 3t 1 t y(t) = L−1 {Y(s)} = e − e = e − e 4 2 2 8 8 Eliminating  dx from the two equations dt  dy + 4x − 28y = −e−2t dt 1 dy  1  −2t 21 3t 7 t 27 3t 3 t  = −e x(t) = −e−2t + 28y − 6 + e − e − e + e 4 dt 4 4 2 3 4 1  15 3t 11 t 1 e − e − e−2t , y(t) = (3e3t − et ) i.e. x(t) = 4 4 4 8 6  6(b) 5 s2 + 1 1 (2s + 1)X(s) + (3s − 1)Y(s) = s−1 (s + 1)X(s) + (2s − 1)Y(s) =  Eliminating X(s) s+1 5 (2s + 1) − +1 s−1 s+1 10s + 5 − Y(s) = 2 s(s + 1)(s − 2) s(s − 1)(s − 2) 5 3   − 25 5   12 2 + 2 − 2 − − + 2 = s s−2 s +1 s s−1 s−2 5 5 1 3 y(t) = L−1 {Y(s)} = − + e2t − 5 sin t − + 2et − e2t 2 2 2 2 2t t = −3 + e + 2e − 5 sin t [(2s − 1)(2s + 1) − (3s − 1)(s + 1)]Y(s) =  s2  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  280  Eliminating  dx dt  from the original equations  dy + x − y = 10 sin t − et dt x(t) = 10 sin t − et − 3 + e2t + 2et − 5 sin t − 2e2t − 2et + 5 cos t = 5 sin t + 5 cos t − 3 − et − e2t  6(c) (s + 2)X(s) + (s + 1)Y(s) = 3 +  1 3s + 10 = s+3 s+3  5X(s) + (s + 3)Y(s) = 4 +  4s + 13 5 = s+2 s+2  Eliminating X(s) [5(s + 1) − (s + 2)(s + 3)]Y(s) =  15s + 50 −4s2 − 10s + 11 − (4s + 13) = s+3 s+3  9 7 − 12 4s2 + 10s − 11 2s − 2 = + 2 Y(s) = (s + 3)(s2 + 1) s+3 s +1  7 1 9 y(t) = L−1 {Y(s)} = − e−3t + cos t − sin t 2 2 2 From the second diﬀerential equation 21 3 3 27 cos t + sin t − e−3t 5x = 5e−2t + e−3t − 2 2 2 2 +  7 9 sin t + cos t 2 2  x(t) = 3 sin t − 2 cos t + e−2t  6(d) (3s − 2)X(s) + 3sY(s) = 6 + sX(s) + (2s − 1)Y(s) = 3 +  1 6s − 5 = s−1 s−1 1 3s + 1 = s s  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition Eliminating X(s) [3s2 − (3s − 2)(2s − 1)]Y(s) = Y(s) =  s(6s − 5) (3s − 2)(3s + 1) − s−1 s 6s2 − 5s 9s2 − 3s − 2 − s(3s − 1)(s − 2) (s − 1)(3s − 1)(s − 2)  18 14   1 5 = − + + 5 s 3s − 1 s − 2 9 14   − 12 10 − − + 5 s − 1 3s − 1 s − 2 1 1 =− + 2 s s−1 1 y(t) = L−1 {Y(s)} = −1 + et + 2  Eliminating  +  9 2  3s − 1  3 t e3 2  dx from the original equations dt 1 3 9 t 3 3 t 3 − et − 3 + e t + e 3 − e t − e 3 2 2 2 2 2 t 3 1 = e 3 − et 2 2  x(t) =  6(e) (3s − 2)X(s) + sY(s) = −1 +  3 5s −s2 + 5s + 2 + = s2 + 1 s2 + 1 s2 + 1  2sX(s) + (s + 1)Y(s) = −1 +  s −s2 + s 1 + = s2 + 1 s2 + 1 s2 + 1  Eliminating Y(s) [(3s − 2)(s + 1) − 2s2 ]X(s) = X(s) = =  s2  1 [(−s2 + 5s + 2)(s + 1) − (−s2 + s)s] +1  3s2 + 7s + 2 3s + 1 = 2 (s + 2)(s − 1)(s + 1) (s − 1)(s2 + 1) 2 2s − 1 − 2 s−1 s +1  x(t) = L−1 {X(s)} = 2et − 2 cos t + sin t c Pearson Education Limited 2011   281  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  282  Eliminating  dy from the original equation dt dx dt = −2 sin t − 4 cos t − 4et + 4 cos t − 2 sin t + 2et + 2 sin t + cos t  y(t) = −2 sin t − 4 cos t − 2x +  that is, y(t) = −2et − 2 sin t + cos t, x(t) = 2et − 2 cos t + sin t  6(f) 1 s2 + 1 = s2 s2 s+1 1 (s + 1)X(s) + 4sY(s) = 1 + = s s sX(s) + (s + 1)Y(s) = 1 +  Eliminating Y(s)  s2 + 1 3s2 − 2s + 3 (s + 1)2 = − [4s2 − (s + 1)2 ]X(s) = 4s s2 s s 2 −3 1 9 3s − 2s + 3 = − + X(s) = s(s − 1)(3s + 1) s s − 1 3s + 1 t  x(t) = L−1 {X(s)} = −3 + et + 3e− 3 Eliminating  dy from the original equation dt  dx  1 4t − 1 + x + 3 4 dt t t 1 = 4t − 1 − 3 + et + 3e− 3 − 3et + 3e− 3 4 t 1 3 t that is, y(t) = t − 1 − et + e− 3 , x(t) = −3 + et + 3e− 3 2 2 y=  6(g) 14 + 7s 12 7 + = 2 s s s2 14 14 14 − 14s (5s + 4)X(s) − (3s − 6)Y(s) = 2 − = s s s2 (2s + 7)X(s) + 3sY(s) =  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  283  Eliminating Y(s) 1 [(3s − 6)(14 + 7s) + 3s(14 − 14s)] s2 21 21(s2 + s − 2)X(s) = 21(s + 2)(s − 1)X(s) = 2 (−s2 + 2s − 4) s 2 −s + 2s − 4 X(s) = 2 s (s + 2)(s − 1) −1 1 0 2 = + + + 2 s−1 s+2 s s −1 x(t) = L {X(s)} = −et + e−2t + 2t  [(2s + 7)(3s − 6) + (5s + 4)(3s)]X(s) =  Eliminating  dy from the original equations dt  dx dt t = 28t − 7 + 7e + 14e−2t − 14 + 11et − 11e−2t − 22t 1 7 giving y(t) = t − + 3et + e−2t , x(t) = −et + e−2t + 2t. 2 2 6y = 28t − 7 − 11x − 7  6(h) (s2 + 2)X(s) − Y(s) = 4s −X(s) + (s2 + 2)Y(s) = 2s Eliminating Y(s) [(s2 + 2)2 − 1]X(s) = 4s(s2 + 2) + 2s (s4 + 4s2 + 3)X(s) = 4s3 + 10s 3s s 4s3 + 10s = 2 + 2 2 2 (s + 1)(s + 3) s +1 s +3 √ −1 x(t) = L {X(s)} = 3 cos t + cos 3t X(s) =  From the ﬁrst of the given equations √ √ d2 x = 6 cos t + 2 cos 3t − 3 cos t − 3 cos 3t 2 dt √ √ that is, y(t) = 3 cos t − cos 3t, x(t) = 3 cos t − cos 3t y(t) = 2x +  c Pearson Education Limited 2011   284  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  6(i)    35 + 12 = (5s2 + 6)X(s) + 12s2 Y = s 4  35  2 2 5s X(s) + (16s + 6)Y(s) = s + 16 = 4  83 s 4 99 s 4  Eliminating X(s) s [83(5s2 ) − 99(5s2 + 6)] 4 s 4 2 [−20s − 126s − 36]Y(s) = [−80s2 − 594] 4  [60s4 − (5s2 + 6)(16s2 + 6)]Y(s) =  s(40s2 + 297) 4(s2 + 6)(10s2 + 3) 25 − 14 s 2 s + = 2 s + 6 10s2 + 3 √ 1 5 y(t) = L−1 {Y(s)} = − cos 6t + cos 4 4  Y(s) =  Eliminating  3 t 10  d2 x from the original equations dt2 √  3 3 d2 y  15 3 − cos t + − + 3 cos = 6t dt2 4 4 10 4 √ √ 3 3 5 3 1 t + cos 6t, x(t) = cos t − cos 6t. 10 4 4 10 4  3x = 3y + 3 i.e. x(t) = cos  6(j) (2s2 − s + 9)X(s) − (s2 + s + 3)Y(s) = 2(s + 1) − 1 = 2s + 1 (2s2 + s + 7)X(s) − (s2 − s + 5)Y(s) = 2(s + 1) + 1 = 2s + 3 Subtract (−2s + 2)X(s) − (2s − 2)Y(s) = −2 ⇒ X(s) + Y(s) =  1 s−1  ⇒ x(t) + y(t) = et  (i)  Add (4s2 + 16)X(s) − (2s + 8)Y(s) = 4(s + 1) 2(s + 1) ⇒ 2x(t) − y(t) 2X(s) − Y(s) = 2 s +4 = 2 cos 2t + sin 2t c Pearson Education Limited 2011   (ii)  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  285  Then from (i) and (ii) x(t) =  1 t 2 1 2 1 2 e + cos 2t + sin 2t, y(t) = et − cos 2t − sin 2t 3 3 3 3 3 3  Exercises 5.4.3 7  1μF = 10−6 F so 50μ = 5.105 F  Applying Kirchhoﬀ's second law to the left hand loop  di1 di2 = E. sin 100t i1 dt + 2 − dt dt  1 5.105 Taking Laplace transforms  2.104 100 I1 (s) + 2s[I1 (s) − I2 (s)] = E. 2 s s + 104 50s (104 + s2 )I1 (s) − s2 I2 (s) = E. 2 s + 104  (i)  Applying Kirchhoﬀ's law to the right hand loop  di1 di2 − =0 100i2 (t) − 2 dt dt which on taking Laplace transforms gives sI1 (s) = (50 + s)I2 (s) Substituting in (i) 50s2 s2 + 104 Es2 2 4 (s + 200s + 10 )I2 (s) = 2 s + 104   s2 I2 (s) = E 2 (s + 104 )(s + 100)2   s(50 + s) then from (ii) I1 (s) = E 2 (s + 104 )(s + 100)2  (104 + s2 )(50 + s)I2 (s) − s2 I2 (s) = E.  c Pearson Education Limited 2011   (ii)  286  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Expanding in partial functions 1 1 1   − 200 2 200 s + I2 (s) = E + s + 100 (s + 100)2 s2 + 104    1 −100t 1 −100t 1 e cos 100t i2 (t) = L−1 {I2 (s)} = E − + te + 200 2 200  8  Applying Kirchhoﬀ's second law to the primary and secondary circuits  respectively gives 2i1 +  di2 di1 +1 = 10 sin t dt dt  2i2 + 2  di1 di2 + =0 dt dt  Taking Laplace transforms (s + 2)I1 (s) + sI2 (s) =  10 +1  s2  sI1 (s) + 2(s + 1)I2 (s) = 0 Eliminating I1 (s) [s2 − 2(s + 1)(s + 2)]I2 (s) =  10s s2 + 1  I2 (s) = −  (s2  10s 10s =− 2 2 + 1)(s + 7s + 6) (s + 1)(s + 6)(s + 1)  12 25  −1 s + 35  + 37 + 37 2 37 s+1 s+6 s +1  =−  i2 (t) = L−1 {I2 (s)} = e−t −  9  35 12 −6t 25 e cos t − sin t − 37 37 37  Applying Kirchhoﬀ's law to the left and right hand loops gives d (i1 + i2 ) + 1 i1 dt = E0 = 10 dt di2 − 1 i1 dt = 0 i2 + dt  (i1 + i2 ) +  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  287  Applying Laplace transforms 1 10 (s + 1)I1 (s) + (s + 1)I2 (s) + I1 (s) = s s 1 (s + 1)I2 (s) − I1 (s) = 0 ⇒ I1 (s) = s(s + 1)I2 (s) s Substituting back in the ﬁrst equation s(s + 1)2 I2 (s) + (s + 1)I2 (s) + (s + 1)I2 (s) = (s2 + s + 2)I2 (s) = I2 (s) =  10 s 10 s(s + 1) 10 s(s + 1)(s2 + s + 2)  Then from (i) 10 10 = 1 +s+2 (s + 2 )2 + √ 1 7 20 t i1 (t) = L−1 {I1 (s)} = √ e− 2 t sin 2 7 I1 (s) =  10  s2  Applying Newton's law to the motion of each mass ẍ1 = 3(x2 − x1 ) − x1 = 3x2 − 4x1 ẍ2 = −9x2 − 3(x2 − x1 ) = −12x2 + 3x1  giving ẍ1 + 4x1 − 3x2 = 0, x1 (0) = −1, x2 (0) = 2 ẍ2 + 12x2 − 3x1 = 0 Taking Laplace transforms (s2 + 4)X1 (s) − 3X2 (s) = −s −3X1 (s) + (s2 + 12)X2 (s) = 2s c Pearson Education Limited 2011   7 4  (i)  288  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Eliminating X2 (s) [(s2 + 4)(s2 + 12) − 9]X1 (s) = −s(s2 + 12) + 6s (s2 + 13)(s2 + 3)X1 (s) = −s3 − 6s 3 7 − 10 s −s3 − 6s 10 s = − (s2 + 13)(s2 + 3) s2 + 3 s2 + 13 √ √ 3 7 cos 13t x1 (t) = L−1 {X1 (s)} = − cos 3t − 10 10  X1 (s) =  From the ﬁrst diﬀerential equation 3x2 = 4x1 + ẍ1 √ √ √ √ 6 14 9 91 = − cos 3t − cos 13t + cos 3t + cos 13t 5 5 10 10 √ √ 1 [21 cos 13t − cos 3t] x2 (t) = 10 √ √ √ √ 1 1 (3 cos 3t + 7 cos 13t), x2 (t) = [21 cos 13t − cos 3t] 10 10 √ √ Natural frequencies are 13 and 3 . Thus, x1 (t) = −  11  The equation of motion is Mẍ + bẋ + Kx = Mg ; x(0) = 0 , ẋ(0) =  2gh  The problem is then an investigative one where students are required to investigate for diﬀerent h values either analytically or by simulation.  12 By Newton's second law of motion M2 ẍ2 = −K2 x2 − B1 (ẋ2 − ẋ1 ) + u2 M1 ẍ1 = B1 (ẋ2 − ẋ1 ) − K1 x1 + u1 Taking Laplace transforms and assuming quiescent initial state (M2 s2 + B1 s + K2 )X2 (s) − B1 sX1 (s) = U2 (s) −B1 sX2 (s) + (M1 s2 + B1 s + K1 )X1 (s) = U1 (s) c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  289  Eliminating X1 (s) [(M1 s2 + B1 s + K1 )(M2 s2 + B1 s + K2 ) − B21 s2 ]X2 (s) = (M1 s2 + B1 s + K1 )U2 (s) + B1 sU1 (s) B1 s (M1 s2 + B1 s + K1 ) U1 (s) + U2 (s) Δ Δ   (M1 s2 + B1 s + K1 ) −1 −1 B1 s and x2 (t) = L {X2 (s)} = L U1 (s) + U2 (s) Δ Δ i.e. X2 (s) =  Likewise eliminating X2 (t) from the original equation gives x1 (t) = L−1 {X1 (s)} = L−1   (M1 s + B1 s + K2 )  B1 s U1 (s) + U2 (s) Δ Δ  Exercises 5.5.7 13 f(t) = tH(t) − tH(t − 1) = tH(t) − (t − 1)H(t − 1) − 1H(t − 1) Thus, using theorem 2.4 L{f(t)} =  1 1 1 1 − e−s 2 − e−s = 2 (1 − e−s ) − e−s 2 s s s s  14(a) f(t) = 3t2 H(t) − (3t2 − 2t + 3)H(t − 4) − (2t − 8)H(t − 6) = 3t2 H(t) − [3(t − 4)2 + 22(t − 4) + 43]H(t − 4) − [2(t − 6) + 4]H(t − 6) Thus, 6 − e−4s L[3t2 + 22t + 43] − e−6s L[2t + 4] s3 6 6 22 43  −4s  2 4 e = 3− 3+ 2 + − 2 + e−6s s s s s s s  L{f(t)} =  c Pearson Education Limited 2011   290  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  14(b) f(t) = tH(t) + (2 − 2t)H(t − 1) − (2 − t)H(t − 2) = tH(t) − 2(t − 1)H(t − 1) − (t − 2)H(t − 2) Thus, 1 − 2e−s L{t} + e−2s L{t} s2 1 = 2 [1 − 2e−s + e−2s ] s  L{f(t)} =   e−5s  1 −1 −5s = L {e F(s)} where F(s) = and by the ﬁrst (s − 2)4 (s − 2)4 1 shift theorem f(t) = L−1 {F(s)} = t3 e2t . 6 Thus, by the second shift theorem 15(a)  L−1  L−1  15(b)  L−1     e−5s  = f(t − 5)H(t − 5) (s − 2)4 1 = (t − 5)3 e2(t−5) H(t − 5) 6   3e−2s = L−1 {e−2s F(s)} where (s + 3)(s + 1) 3 − 32 3 = + 2 (s + 3)(s + 1) s+3 s+1 3 3 f(t) = L−1 {F(s)} = e−t − e−3t 2 2 −2s   3e = f(t − 2)H(t − 2) L−1 (s + 3)(s + 1) 3 = [e−(t−2) − e−3(t−2) ]H(t − 2) 2  F(s) =  so  15(c)  L−1     s+1 e−s = L−1 {e−s F(s)} where + 1)  s2 (s2  F(s) =  1 1 s+1 s+1 = + 2− 2 + 1) s s s +1  s2 (s2  f(t) = L−1 {F(s)} = 1 + t − cos t − sin t c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  291  so L−1     s+1 e−s = f(t − 1)H(t − 1) + 1)  s2 (s2  = [1 + (t − 1) − cos(t − 1) − sin(t − 1)]H(t − 1) = [t − cos(t − 1) − sin(t − 1)]H(t − 1)  15(d)  L−1     s+1 −πs = L−1 {e−πs F(s)} where e s2 + s + 1 √ √1 ( 3 ) 3 2 √ 1 2 3 2 ) + ( 2 2 )  (s + 12 ) +  s+1 = + s + 1) (s + √ √ 1  3 3  1 −2t t + √ sin t cos f(t) = e 2 2 3  F(s) =  (s2  so  L−1  √ √ √ 1    3 3 1 s+1 (t−π) − −πs 2 √ e (t − π) + sin (t − π) .H(t − π) e = 3 cos s2 + s + 1 2 2 3  15(e)  L−1     s −4πs/5 = L−1 {e−4πs/5 F(s)} where e s2 + 25 F(s) =  so L−1  15(f)  L−1   s2  s ⇒ f(t) = L−1 {F(s)} = cos 5t s2 + 25   4π s 4π  H t− e−4πs/5 = f t − + 25 5 5  4π = cos(5t − 4π)H t − 5  4π = cos 5t H t − 5   e−s (1 − e−s )  = L−1 {(e−s − e−2s )F(s)} where s2 (s2 + 1) F(s) = −1  f(t) = L  1 s2 (s2  + 1)  =  1 1 − 2 2 s s +1  {F(s)} = t − sin t  c Pearson Education Limited 2011   292  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  so L−1 {(e−s − e−2s )F(s)} =f(t − 1)H(t − 1) − f(t − 2)H(t − 2) =[(t − 1) − sin(t − 1)]H(t − 1) − [(t − 2) − sin(t − 2)]H(t − 2)  16  dx 1 + x = f(t), L{f(t)} = 2 (1 − e−s − se−s ) dt s  Taking Laplace transforms with x(0) = 0 (1 + s) 1 − e−s 2 s s2 1 1 X(s) = 2 − e−s 2 s (s + 1) s 1 1 1 − e−s L{t} =− + 2 + s s s+1  (s + 1)X(s) =  Taking inverse transforms x(t) = −1 + e−t + t − (t − 1)H(t − 1) = e−t + (t − 1)[1 − H(t − 1)] or x(t) = e−t + (t − 1) for t ≤ 1 x(t) = e−t for t ≥ 1 Sketch of response is  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  17  293  d2 x dx + x = g(t), x(0) = 1, ẋ(0) = 0 + dt2 dt  with L{g(t)} =  1 (1 − 2e−s + e−2s ) s2  Taking Laplace transforms (s2 + s + 1)X(s) = s + 1 +  1 (1 − 2e−s + e−2s ) 2 s  1 s+1 + (1 − 2e−s + e−2s ) (s2 + s + 1) s2 (s2 + s + 1)   1 (s + 1) s 1 [1 − 2e−s + e−2s ] = 2 + − + 2+ 2 (s + s + 1) s s s +s+1 √ √ (s + 12 ) + √13 ( 23 ) (s + 12 ) − √13 ( 23 )  1 1 √ √ = [1 − 2e−s + e−2s ] + − + 2+ s s 3 2 3 2 ) ) (s + 12 )2 + ( (s + 12 )2 + ( 2 2 √ √ 1  3 3 1 −2t −1 cos x(t) = L {X(s)} = e t + √ sin t 2 2 3 √ √ 1  3 3 1 −2t t − √ sin t cos +t−1+e 2 2 3 √ 1  3 − 2 (t−1) (t − 1) − 2H(t − 1) t − 2 + e cos 2 √   3 1 − √ sin (t − 1) 2 3 √ 1  3 − 2 (t−2) (t − 2) cos + H(t − 2) t − 3 + e 2 √   3 1 − √ sin (t − 2) 2 3  X(s) =  that is, 1  x(t) = 2e− 2 t cos  √  3 2 t  +t−1  √   3 3 1 (t − 1) − √ sin (t − 1) cos − 2H(t − 1) t − 2 + e 2 2 3 √ √  1   3 3 1 − 2 (t−2) + H(t − 2) t − 3 + e cos (t − 2) − √ sin (t − 2) 2 2 3 1  − 2 (t−1)  √  c Pearson Education Limited 2011   294 18  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition   π  π π = cos t − H t − f(t) = sin tH t − 2 2 2  π since cos t − = sin t. 2  Taking Laplace transforms with x(0) = 1, ẋ(0) = −1   π  π  (s2 + 3s + 2)X(s) = s + 2 + L cos t − H t − 2 2 π  = s + 2 + e− 2 s L{cos t} π s = s + 2 + e− 2 s · 2 s +1 π   s 1 X(s) = + e− 2 s s+1 (s + 1)(s + 2)(s2 + 1) 2 π  −1 1 1 s+3  2 = + e− 2 s + 5 + · 2 s+1 s + 1 s + 2 10 s + 1 π   1 1 2 1 = + e− 2 s L − e−t + e−2t + (cos t + 3 sin t) s+1 2 5 10 π π  π 1 2 1 so x(t) = L−1 {X(s)} = e−t + − e−(t− 2 ) + e−2(t− 2 ) + (cos t − 2 5 10 2    π π + 3 sin t − ) H t − 2 2 π   π 1 sin t − 3 cos t + 4eπ e−2t − 5e 2 e−t H t − = e−t + 10 2  19  f(t) = 3H(t) − (8 − 2t)H(t − 4) = 3H(t) + 2(t − 4)H(t − 4) 2 3 3 L{f(t)} = + 2e−4s L{t} = + 2 e−4s s s s  Taking Laplace transforms with x(0) = 1, ẋ(0) = 0 2 3 + 2 e−4s s s s 3 2 X(s) = 2 + + 2 2 e−4s 2 s + 1 s(s + 1) s (s + 1) 1 3 3 s 1  −4s + − 2 +2 2 − 2 e = 2 s +1 5 s +1 s s +1 2 3 + 2e−4s L{t − sin t} = − 2 5 s +1  (s2 + 1)X(s) = s +  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  295  Thus, taking inverse transforms x(t) = 3 − 2 cos t + 2(t − 4 − sin(t − 4))H(t − 4)  20 θ̈0 + 6θ̇0 + 10θ0 = θi θi (t) = 3H(t) − 3H(t − a) 3 3 3 so L{θi } = − e−as = (1 − e−as ) s s s Taking Laplace transforms in (1) with θ0 = θ̇0 = 0 at t = 0 3 (1 − e−as ) s  Φ0 (s) = 3(1 − e−as )  (s2 + 6s + 10)Φ0 (s) =   1 s(s2 + 6s + 10) 1 3 (s + 3) + 3  = (1 − e−as ) − 10 s (s + 3)2 + 1   3 (1 − e−as )L 1 − e−3t cos t − 3e−3t sin t = 10  Thus, taking inverse transforms θ0 (t) =  3 [1 − e−3t cos t − 3e−3t sin t]H(t) 10 3 − [1 − e−3(t−a) cos(t − a) − 3e−3(t−a) sin(t − a)]H(t − a) 10  If T > a then H(T) = 1, H(T − a) = 1 giving 3 −3T cos T − e−3(T −a) cos(T − a)] [e 10 3 − [3e−3T sin T − 3e−3(T −a) sin(T − a)] 10 3 −3T {cos T + 3 sin T − e3a [cos(T − a) + 3 sin(T − a)]} =− e 10  θ0 (T) = −  21  θi (t) = f(t) = (1 − t)H(t) − (1 − t)H(t − 1) = (1 − t)H(t) + (t − 1)H(t − 1) c Pearson Education Limited 2011   (1)  296  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  so  1 1 − 2 + e−s L{t} s s 1 1 1 s−1 1 = − 2 + 2 e−s = + 2 e−s 2 s s s s s  L{θi (t)} =  Then taking Laplace transforms, using θ0 (0) = θ̇0 (0) = 0 (s2 + 8s + 16)Φ0 (s) =  s−1 1 −s + e s2 s2    1 s−1 −s + e s2 (s + 4)2 s2 (s + 4)2  2 10  e−s  3 2 2 3 1 1 3 − 2− − + + + + = 2 s s s s + 4 (s + 4)2 32 s s2 s + 4 (s + 4)2  Φ0 (s) =  which on taking inverse transforms gives 1 [3 − 2t − 3e−4t − 10te−4t ] 32 1 + [−1 + 2(t − 1) + e−4(t−1) + 2(t − 1)e−4(t−1) ]H(t − 1) 32 1 [3 − 2t − 3e−4t − 10te−4t ] = 32 1 + [2t − 3 + (2t − 1)e−4(t−1) ]H(t − 1) 32  θ0 (t) = L−1 {Φ0 (s)} =  22  e(t) = e0 H(t − t1 ) − e0 H(t − t2 ) e0 L{e(t)} = (e−st1 − e−st2 ) s c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  297  By Kirchhoﬀ's second law current in the circuit is given by Ri +  1 C  idt = e  which on taking Laplace transforms RI(s) +  I(s) =  1 e0 I(s) = (e−st1 − e−st2 ) Cs s  e0 C (e−st1 − e−st2 ) RCs + 1  =  e0 /R −st1 − e−st2 ) 1 (e s + RC  =  e0 /R −st2 e0 /R −st1 − 1 e 1 e s + RC s + RC  then i(t) = L−1 {I(s)}  e0  −(t−t1 )/RC e H(t − t1 ) − e−(t−t2 )/RC H(t − t2 ) = R  23  Sketch over one period as shown and readily extended to 0 ≤ t < 12.  c Pearson Education Limited 2011   298  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  f1 (t) = 3tH(t) − (3t − 6)H(t − 2) − 6H(t − 4) = 3tH(t) − 3(t − 2)H(t − 2) − 6H(t − 4) 3 3 6 L{f1 (t)} = F1 (s) = 2 − 2 e−2s − e−4s s s s Then by theorem 5.5 1 F1 (s) 1 − e−4s 1 = 2 (3 − 3e−2s − 6se−4s ) s (1 − e−4s )  L{f(t)} = F(s) =  24  Take K t, T = 0,  f1 (t) =  then f1 (t) =  0T  Kt K K K tH(t) − H(t − T) = tH(t) − (t − T)H(t − T) − KH(t − T) T T T T  L{f1 (t)} = F1 (s) =  K K K K K = 2 (1 − e−sT ) − e−sT − e−sT 2 − e−sT 2 Ts Ts s Ts s  Then by theorem 5.5 L{f(t)} = F(s) =  K K e−sT 1 F (s) = − 1 1 − e−sT Ts2 s 1 − e−sT  Exercises 5.5.12 25(a) 2s2 + 1 10s + 11 9 19 =2− =2+ − (s + 2)(s + 3) (s + 2)(s + 3) s+2 s+3 2   2s + 1 = 2δ(t) + 9e−2t − 19e−3t L−1 (s + 2)(s + 3)  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  25(b) 5 s2 − 1 =1− 2 2 s +4 s +4 2 s − 1 5 = δ(t) − sin 2t L−1 2 s +4 2  25(c) 2s + 3 s2 + 2 =1− 2 2 s + 2s + 5 s + 2s + 5  2(s + 1) + 12 (2)  =1− (s + 1)2 + s2  s2 + 2   1 = δ(t) − e−t 2 cos 2t + sin 2t L−1 2 s + 2s + 5 2  26(a) (s2 + 7s + 12)X(s) =  2 + e−2s s    −2s 1 2 + e s(s + 4)(s + 3) (s + 4)(s + 3) 1 2 1  1 1  −2s 6 3 e = − + 2 + − s s+3 s+4 s+3 s+4   1 2 −3t 1 −4t − e + e−3(t−2) − e−4(t−2) H(t − 2) x(t) = L−1 {X(s)} = + e 6 3 2 X(s) =  26(b) (s2 + 6s + 13)X(s) = e−2πs 1 e−2πs (s + 3)2 + 22 1  = e−2πs L e−3t sin 2t 2 1 so x(t) = L−1 {X(s)} = e−3(t−2π) sin 2(t − 2π).H(t − 2π) 2 1 6π −3t sin 2t.H(t − 2π) = e e 2 X(s) =  c Pearson Education Limited 2011   299  300  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  26(c) (s2 + 7s + 12)X(s) = s + 8 + e−3s  X(s) =  =    −3s 1 s+8 + e (s + 4)(s + 3) (s + 4)(s + 3)  5 4   1 1  −3s − + − e s+3 s+4 s+3 s+4  x(t) = L−1 {X(s)} = 5e−3t − 4e−4t + [e−3(t−3) − e−4(t−3) ]H(t − 3)  27(a)  Generalized derivative is f (t) = g (t) − 43δ(t − 4) − 4δ(t − 6) where  ⎧ 6t, ⎪ ⎨ g (t) = 2, ⎪ ⎩ 0,  0≤t<4 4≤t<6 t≥6  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 27(b)  ⎧ ⎨ 1, f (t) = g (t) = −1, ⎩ 0,  0≤t<1 1≤t<2 t≥2  27(c)  f (t) = g (t) + 5δ(t) − 6δ(t − 2) + 15δ(t − 4) where  ⎧ ⎨  2, −3, g (t) = ⎩ 2t − 1,   0≤t<2 2≤t<4 t≥4  c Pearson Education Limited 2011   301  302  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  28  (s2 + 7s + 10)X(s) = 2 + (3s + 2)U(s) = 2 + (3s + 2) X(s) =  = x(t) = L−1 {X(s)} =  29 f(t) =  ∞   5s + 6 1 = s+2 s+2  5s + 6 (s + 2)2 (s + 5) 19 9  s+2  −  4 3  (s + 2)2  −  19 9  (s + 5)  19 −2t 4 −2t 19 −5t e − te − e 9 3 9  δ(t − nT)  n=0  Thus, F(s) = L{f(t)} =  ∞   L{δ(t − nT)} =  n=0  ∞   e−snT  n=0  This is an inﬁnite GP with ﬁrst term 1 and common ratio e−sT and therefore having sum (1 − e−sT )−1 . Hence, F(s) =  1 1 − e−sT  Assuming zero initial conditions and taking Laplace transforms the response of the harmonic oscillator is given by 1 1 − e−sT ∞   −snT  1 X(s) = e 2 s + w2 n=0  (s2 + w2 )X(s) = F(s) =   1 sin wt = [1 + e−sT + e−2sT + . . .]L w  giving x(t) = L−1 {X(s)} =  1 [sin wt + H(t − T). sin w(t − T) + H(t − 2T). w  sin w(t − 2T) + . . .] ∞ 1  or x(t) = H(t − nT) sin w(t − nT) . w n=0  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  303  29(a) π ; T= w  ∞ 1   nπ x(t) = sin(wt − nπ) H t− w n=0 w    π 2π 1 sin wt − sin wt. H t − + sin wt. H t − + ... = w w w    and a sketch of the response is as follows  29(b) T=  2π ; w  x(t) =  ∞ 2πn 1   sin(wt − 2πn) H t− w n=0 w    2π 4π 1 sin wt + sin wt.H t − + sin wt.H t − + ... = w w w and the sketch of the response is as follows  c Pearson Education Limited 2011     304  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  30  The charge q on the LCR circuit is determined by L  d2 q dq 1 + R + q = e(t) dt2 dt C  where e(t) = Eδ(t), q(0) = q̇(0) = 0. Taking Laplace transforms  2 1 Q(s) = L{Eδ(t)} = E Ls + Rs + C Q(s) = =  s2  E/L +R Ls +  1 LC  E/L  = (s +  R 2 2L )  1 + ( LC −  E/L R ,η = ,μ = 2 2 (s + μ) + η 2L  R2 4L2 )  1 R2 − LC 4L2  E −μt e sin ηt Lη E −μt e (η cos ηt − μ sin ηt) and current i(t) = q̇(t) = Lη Thus, q(t) =  Exercises 5.5.14 31    M 1 H(x) + Wδ x − − R1 δ(x) , where R1 = (M + W)  2 2 so the force function is  Load W(x) =   M + W  M H(x) + Wδ x − − δ(x)  2 2  W(x) = having Laplace transform  W(s) =  M (M + W) + We−s/2 − s 2  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  305  Since the beam is freely supported at both ends y(0) = y2 (0) = y() = y2 () = 0 and the transformed equation (2.64) of the text becomes  y1 (0) y3 (0) 1 M W −s/2  M + W 1 + Y(s) = + e − + 4 EI s5 s4 2 s4 s2 s Taking inverse transforms gives   1  1 1 M 4 1 x + W(x − )3 · H x − − (M + W)x3 y(x) = EI 24  6 2 2 12    1 + y1 (0)x + y3 (0)x3 6 for x >   2   1 1 M 4 1  x + W x− y(x) = EI 24  6 2  3   1 1 3 − (M + W)x + y1 (0)x + y3 (0)x3 12 6    1 1 1M 2  x +W x− − (M + W)x + y3 (0)x y2 (x) = EI 2  2 2 y2 () = 0 then gives y3 (0) = 0 and y() = 0 gives  W3 1 1 M3 1 3 3 + − M − W + y1 (0) 0= EI 24 24 12 2 1 1 1 M2 + W2 y1 (0) = EI 24 16       2 1   Mx4 + 8W(x − )3 H x − − 4(M + W)x3 + (2M + 3W)2 x so y(x) = 48EI  2 2  c Pearson Education Limited 2011   306  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  32  Load W(x) = w(H(x − x1 ) − H(x − x2 )) − R1 δ(x), R1 = w(x2 − x1 ) so the force function is W(x) = w(H(x − x1 ) − H(x − x2 )) − w(x2 − x1 )δ(x) having Laplace transform 1 1 W(s) = w e−x1 s − e−x2 s − w(x2 − x1 ) s s with corresponding boundary conditions y(0) = y1 (0) = 0, y2 () = y3 () = 0 The transformed equation (2.64) of the text becomes  y2 (0) y3 (0) 1 −x2 s (x2 − x1 ) w 1 −x1 s + 3 + 4 e − 5e − Y(s) = 5 4 EI s s s s s which on taking inverse transforms gives y(x) =  w1 1 (x − x1 )4 H(x − x1 ) − (x − x2 )4 H(x − x2 ) EI 24 24  x2 x3 1 3 − (x2 − x1 )x + y2 (0) + y3 (0) 6 2 6  For x > x2  w1 1 1 x2 x3 (x − x1 )4 − (x − x2 )4 − (x2 − x1 )x3 + y2 (0) + y3 (0) EI 24 24 6 2 6  w1 1 y2 (x) = (x − x1 )2 − (x − x2 )2 − (x2 − x1 )x + y2 (0) + y3 (0)x EI 24 2  w (x − x1 ) − (x − x2 ) − (x2 − x1 ) + y3 (0) ⇒ y3 (0) = 0 y3 (x) = EI y(x) =  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition The boundary condition y2 () = 0 then gives  w 1 2 1 ( − 2x1 + x21 ) − (2 − 2x2 + x22 ) − x2  + x1  + y2 (0) EI 2 2 w 2 2 (x − x1 ) ⇒ y2 (0) = 2EI 2 w  (x − x1 )4 H(x − x1 ) − (x − x2 )4 H(x − x2 ) − 4(x2 − x1 )x3 y(x) = 24EI  + 6(x22 − x21 )x2 0=  When x1 = 0, x2 = , max deﬂection at x =  ymax =  w w4 {4 − 44 + 64 } = 24EI 8EI  33  Load W(x) = Wδ(x − b) − R1 δ(x), R1 = W so the force function is W(x) = Wδ(x − b) − Wδ(x) having Laplace transform W(s) = We−bs − W with corresponding boundary conditions y(0) = y1 (0) = 0, y2 () = y3 () = 0 The transformed equation (2.64) of the text becomes Y(s) = −  1  W −bs W  y2 (0) y3 (0) e − 4 + 3 + 4 EI s4 s s s  which on taking inverse transforms gives 1 3 x2 x3 W 1 3 y(x) = − (x − b) H(x − b) − x + y2 (0) + y3 (0) EI 6 6 2 6 c Pearson Education Limited 2011   307  308  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  For x > b y(x) = −  1  x2 x3 W 1 (x − b)3 − x3 + y2 (0) + y3 (0) EI 6 6 2 6  y2 (x) = −   W (x − b) − x + y2 (0) + y3 (0)x EI  y3 (x) = −   W 1 − 1 + y3 (0) ⇒ y3 (0) = 0 EI  Using the boundary condition y2 () = 0 0=−  Wb W (−h) + y2 (0) ⇒ y2 (0) = − EI EI  giving (x − b)3 bx2  W  x3 − H(x − b) − EI 6 6 2 ⎧ 2 Wx ⎪ ⎨− (3b − x), 0 < x ≤ b 6EI = 2 ⎪ ⎩ − Wb (3x − b), b < x ≤  6EI  y(x) =  Exercises 5.6.5 34(a) gives  Assuming all the initial conditions are zero taking Laplace transforms (s2 + 2s + 5)X(s) = (3s + 2)U(s)  so that the system transfer function is given by  G(s) =  3s + 2 X(s) = 2 U(s) s + 2s + 5  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  34(b)  309  The characteristic equation of the system is s2 + 2s + 5 = 0  and the system is of order 2.  34(c)  The transfer function poles are the roots of the characteristic equation s2 + 2s + 5 = 0  which are s = −1 ± j. That is, the transfer function has single poles at s = −1 + j and s = −1 − j. The transfer function zeros are determined by equating the numerator polynomial 2 to zero; that is, a single zero at s = − . 3  35  Following the same procedure as for Exercise 34  35(a)  The transfer function characterizing the system is  G(s) =  s3 + 5s + 6 s3 + 5s2 + 17s + 13  c Pearson Education Limited 2011   310  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  35(b)  The characteristic equation of the system is s3 + 5s2 + 17s + 13 = 0  and the system is of order 3.  35(c)  The transfer function poles are given by s3 + 5s2 + 17s + 13 = 0 that is, (s + 1)(s2 + 4s + 13) = 0  That is, the transfer function has simple poles at s = −1, s = −2 + j3, s = −2 − j3 The transfer function zeros are given by s2 + 5s + 6 = 0 (s + 3)(s + 2) = 0 that is, zeros at s = −3 and s = −2.  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 36(a)  311  Poles at (s + 2)(s2 + 4) = 0; that is, s = −2, s = +2j, s = −j.  Since we have poles on the imaginary axis in the s-plane, system is marginally stable. Poles at (s + 1)(s − 1)(s + 4) = 0; that is, s = −1, s = 1, s = −4.  36(b)  Since we have the pole s = 1 in the right hand half of the s-plane, the system is unstable.  36(c)  Poles at (s + 2)(s + 4) = 0; that is, s = −2, s = −4.  Both the poles are in the left hand half of the plane so the system is stable.  2 2 Poles √ at (s + s + 1)(s + 1) = 0; that is, s = −1 (repeated), 3 1 . s=− ±j 2 2 Since all the poles are in the left hand half of the s-plane the system is stable.  36(d)  √ 39 1 36(e) Poles at (s + 5)(s − s + 10) = 0; that is, s = −5, s = ± j . 2 2 Since both the complex poles are in the right hand half of the s-plane the system 2  is unstable.  37(a)  s2 − 4s + 13 = 0 ⇒ s = 2 ± j3.  Thus, the poles are in the right hand half s-plane and the system is unstable.  37(b) 5s3 + 13s2 + 31s + 15 = 0 a1 a0 a3 a2 Routh–Hurwitz (R-H) determinants are:   13 Δ1 = 13 > 0, Δ2 =  15   5  > 0, Δ3 = 15Δ2 > 0 31   so the system is stable.  c Pearson Education Limited 2011   312  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  37(c)  s3 + s2 + s + 1 = 0  R–H determinants are  1 Δ1 = 1 > 0, Δ2 =  1   1  = 0, Δ3 = 1Δ2 = 0 1  Thus, system is marginally stable. This is readily conﬁrmed since the poles are at s = −1, s = ±j  37(d)  24s4 + 11s3 + 26s2 + 45s + 36 = 0  R–H determinants are   11 Δ1 = 11 > 0, Δ2 =  45   24  <0 26   so the system is unstable.  37(e)  s3 + 2s2 + 2s + 1 = 0  R–H determinants are  2 Δ1 = 2 > 0, Δ2 =  1   3  = 1 > 0, Δ3 = 1Δ2 > 0 2  √ 3 1 conﬁrming the and the system is stable. The poles are at s = −1, s = − ± j 2 2 result. d3 x d2 x dx + Krx = 0; m, K, r, c > 0 38 m 3 + c 2 + K dt dt dt R–H determinants are Δ1 = c > 0    c m  = cK − mKr > 0 provided r < c Δ2 =  Kr K  m Δ3 = KrΔ2 > 0 provided Δ2 > 0 Thus, system stable provided r <  c m  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 39  s4 + 2s2 + (K + 2)s2 + 7s + K = 0 a2 a1 a0 a3  R–H determinants are Δ1 =| a3 |= 9 > 0    2  a3 a4     = Δ2 =  7 a1 a2     a3 a4 0    Δ3 =  a1 a2 a3  =  0 a0 a1    3 1  = 2K − 3 > 0 provided K >  K+2 2   2 1 0    7 K + 2 2  = 10K − 21 > 0 provided K > 2   0 K 7  Δ4 = KΔ3 > 0 provided Δ3 > 0 Thus, the system is stable provided K > 2.1.  40  s2 + 15Ks2 + (2K − 1)s + 5K = 0, K > 0  R–H determinants are Δ1 = 15K > 0     15K 1  = 30K2 − 20K  Δ2 =  5K (2K − 1)  Δ3 = 5KΔ2 > 0 provided Δ2 > 0 Thus, system stable provided K(3K − 2) > 0 that is K >  41(a)  2 , since K > 0. 3  Impulse response h(t) is given by the solution of d2 h dh + 15 + 56h = 3δ(t) 2 dt dt  with zero initial conditions. Taking Laplace transforms (s2 + 15s + 56)H(s) = 3 H(s) =  3 3 3 = − (s + 7)(s + 8) s+7 s+8  so h(t) = L−1 {H(s)} = 3e−7t − 3e−8t Since h(t) → 0 as t → ∞ the system is stable.  c Pearson Education Limited 2011   313  314  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  41(b)  Following (a) impulse response is given by (s2 + 8s + 25)H(s) = 1 1 (s + 4)2 + 32 1 so h(t) = L−1 {H(s)} = e−4t sin 3t 3 H(s) =  Since h(t) → 0 as t → ∞ the system is stable. 41(c)  Following (a) impulse response is given by (s2 − 2s − 8)H(s) = 4 4 2 1 2 1 = − (s − 4)(s + 2) 3s−4 3s+2 2 so h(t) = L−1 {H(s)} = (e4t − e−2t ) 3 H(s) =  Since h(t) → ∞ as t → ∞ system is unstable. 41(d)  Following (a) impulse response is given by (s2 − 4s + 13)H(s) = 1 H(s) = so h(t) = L−1 {H(s)} =  s2  1 1 = − 4s + 13 (s − 2)2 + 32  1 2t e sin 3t 3  Since h(t) → ∞ as t → ∞ system is unstable. 7 dx 2 = e−t − 3e−2t + e−4t 42 Impulse response h(t) = dt 3 3 System transfer function G(s) = L{h(t)}; that is, 3 2 7 − + 3(s + 1) s + 2 3(s + 4) s+8 = (s + 1)(s + 2)(s + 4)  G(s) =  The original unit step response can be reconstructed by evaluating Note:  1 L−1 G(s) . s c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  f(t) = 2 − 3 cos t , F(s) =  43(a)  315  3s 2 − 2 s s +1  sF(s) = 2 −  3 3s2 = 2 − s2 + 1 1 + s12  Thus, lim (2 − 3 cos t) = 2 − 3 = −1 t→0+  and lim sF(s) = 2 − s→∞  3 = −1 so conﬁrming the i.v. theorem. 1  43(b) f(t) = (3t − 1)2 = 9t2 − 6t + 1, lim f(t) = 1 t→0+    18 6 18 6 1 + 1 =1 F(s) = 3 − 2 + so lim sF(s) = lim − s→∞ s→∞ s2 s s s s thus, conﬁrming the i.v. theorem.  43(c) f(t) = t + 3 sin 2t , lim = 0 t→0+  1 6  1 6 so lim sF(s) = lim + F(s) = 2 − 2 =0 s→∞ s→∞ s s s +4 s + 4s thus, conﬁrming the i.v. theorem.  44(a) f(t) = 1 + 3e−t sin 2t , lim f(t) = 1 t→∞  F(s) =    6 1 6s + and lim =1 1 + sF(s) = lim s→0 s→0 s (s + 1)2 + 4 (s + 1)2 + 4  thus conﬁrming the f.v. theorem. Note that, sF(s) has its poles in the left half of the s-plane so the theorem is applicable.  c Pearson Education Limited 2011   316  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  44(b)  f(t) = t2 3e−2t , lim f(t) = 0 t→∞   2s  2 F(s) = =0 and lim sF(s) = lim s→0 s→0 (s + 2)3 (s + 2)3 thus conﬁrming the f.v. theorem. Again note that sF(s) has its poles in the left half of the s-plane.  44(c)  f(t) = 3 − 2e−3t + e−t cos 2t , lim f(t) = 3 t→∞  2 (s + 1) 3 + F(s) = − s s + 3 (s + 1)2 + 4  2s s(s + 1)  =3 lim sF(s) = lim 3 − + s→0 s→0 s + 3 (s + 1)2 + 4 conﬁrming the f.v. theorem. Again sF(s) has its poles in the left half of the s-plane. 45  For the circuit of Example 5.28 I2 (s) =  1.22 4.86 3.64 + − s s + 59.1 s + 14.9  Then by the f.v. theorem  4.86s  1.22s − lim i2 (t) = lim sI2 (s) = lim 3.64 + t→∞ s→0 s→0 s + 59.1 s + 14.9 = 3.64 which conﬁrms the answer obtained in Example 5.28. Note that, sI2 (s) has all its poles in the left half of the s-plane. 46  For the circuit of Example 5.29 sI2 (s) =  28s2 (3s + 10)(s + 1)(s2 + 4)  and since it has poles at s = ±j2 not in the left hand half of the s-plane the f.v. theorem is not applicable.  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 47  Assuming quiescent initial state taking Laplace transforms gives 1 4 + +2 s s+3 4 1 2 Y(s) = + + s(7s + 5) (s + 3)(7s + 5) 7s + 5 s 2s 4 + + sY(s) = 7s + 5 (s + 3)(7s + 5) 7s + 5  (7s + 5)Y(s) =  By the f.v. theorem, lim y(t) = lim sF(s) = lim  t→∞  s→0    s→0  =  s 2s  4 + + 7s + 5 (s + 3)(7s + 5) 7s + 5  4 5  By the i.v. theorem, lim y(t) = y(0+) = lim sF(s) = lim s→∞  t→0+  s→∞  =    s 2  4 + + 7s + 5 (1 + 3s )(7s + 5) 7 + 5s  2 7  2 Thus, jump at t = 0 = y(0+) − y(0−) = 1 . 7  Exercises 5.6.8 48(a) t  f ∗ g(t) =  τ cos(3t − 3τ)dτ 0  t  1 1 = − τ sin(3t − 3τ) + cos(3t − 3τ) 3 9 0 1 = (1 − cos 3t) 9 t  (t − τ) cos 3τdτ  g ∗ f(t) = 0  =  t t 1 τ 1 sin 3τ − sin 3τ − cos 3τ = (1 − cos 3t) 3 3 9 9 0 c Pearson Education Limited 2011   317  318  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  48(b) t  f ∗ g(t) =  (τ + 1)e−2(t−τ ) dτ  0  t 1 1 = (τ + 1)e−2(t−τ ) − e−2(t−τ ) 2 4 0 1 1 1 = t + − e−2t 2 4 4 t  g ∗ f(t) =  (t − τ + 1)e−2τ dτ  0  t  1 1 = − (t − τ + 1)e−2τ + e−2τ 2 4 0 1 1 −2t 1 = t+ − e 2 4 4 48(c)  Integration by parts gives t  t  τ2 sin 2(t − τ)dτ =  (t − τ)2 sin 2τdτ  0  0  1 1 1 = cos 2t + t2 − 4 2 4 48(d)  Integration by parts gives t  t  e−τ sin(t − τ)dτ =  0  e−(t−τ ) sin τdτ  0  1 = (sin t − cos t + e−t ) 2  49(a)  Since L−1 L−1  1   1 2 −3t 1 = 1 = f(t) and L−1 = t e s (s + 3)3 2  1  1 · = 3 s (s + 3)  t  f(t − τ)g(τ)dτ 0 t  1 1. τ2 e−3τ dτ 2 0 t 2 2 1 = −τ2 e−3τ − τe−3τ − e−3τ 4 3 9 0 1 = [2 − e−3t (9t2 + 6t + 2)] 54  =  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  319  Directly L−1  L−1  49(b)  L−1    2 1  1 18 6 2  −1 1 · − = L − − s (s + 3)3 54 s (s + 3)3 (s + 3)2 (s + 3) 1 [2 − e−3t (9t2 + 6t + 2)] = 54      1 1 = te2t = f(t), L−1 = te−3t = g(t) 2 2 (s − 2) (s + 3)   1 1 = · 2 2 (s − 2) (s + 3) =e  t  (t − τ)e2(t−τ ) .τe−3τ dτ  0 t  −2t  (tτ − τ2 )e−5τ dτ  0   1 1 2 −5τ t e = e2t − (tτ − τ2 )e−5τ − (t − 2τ)e−5τ + 5 25 125 0   t 2 t 2 = e2t e−5t + e−5t + − 25 125 25 125  1  2t e (5t − 2) + e−3t (5t + 2) = 125 Directly −2 1 2 1 1 125 25 125 25 + + = + (s − 2)2 (s + 3)2 s − 2 (s − 2)2 (s + 3) (s + 3)2   1 −2 2t 1 2 −3t 1 e + te2t + e = ∴ L−1 + te−3t 2 2 (s − 2) (s + 3) 125 25 125 25 1 2t [e (5t − 2) + e−3t (5t + 2)] = 125  49(c)  L−1  1  1  = e−4t = g(t) = t = f(t), L−1 2 s (s + 4) −1  L  1 1  = · s2 s + 4  t  (t − τ)e−4t dτ  0  t  1 1 = − (t − τ)e−4τ + e−4τ 4 16 0 1 −4t 1 1 = + t− e 16 4 16 c Pearson Education Limited 2011   320  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Directly L−1  50      1 1 1 1 1 1 −1 1 = L · − · + · s2 (s + 4) 16 s + 4 16 s 4 s2 1 1 1 −4t e + t − = 16 16 4  Let f(λ) = λ and g(λ) = e−λ so F(s) =  1 1 and G(s) = s2 s+1  Considering the integral equation t  y(t) =  λe−(t−λ) dλ  0  By (5.80) in the text −1  L  t  {F(s)G(s)} =  f(λ)g(t − λ)dλ 0 t  =  λe−(t−λ) dλ = y(t)  0  so   1 + 1)  1 1 1  = L−1 − + 2 + s s s+1 −t = (t − 1) + e  y(t) = L−1 {F(s)G(s)} = L−1  51    s2 (s  Impulse response h(t) is given by the solution of d2 h 7dh + 12h = δ(t) + dt2 dt  subject to zero initial conditions. Taking Laplace transforms (s2 + 7s + 12)H(s) = 1 H(s) =  1 1 1 = − (s + 3)(s + 4) s+3 s+4  giving h(t) = L−1 {H(s)} = e−3t − e−4t c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  321  Response to pulse input is  x(t) = A  t   [e−3(t−τ ) − e−4(t−τ ) ]dτ H(t)  0    t −3(t−τ ) [e − e−4(t−τ ) ]dτ H(t − T) −A  T  1 1 1 −3t 1 −4t  − − e =A H(t) + e 3 4 3 4   1 1 1 −3(t−T ) 1 −4(t−T )  − − − e H(t − T) − e 3 4 3 4  1  = A 1 − 4e−3t + 3e−4t − (1 − 4e−3(t−T ) + 3e−4(t−T ) )H(t − T) 12 or directly u(t) = A[H(t) − H(t − T)] so U(s) = L{u(t)} =  A [1 − e−sT ] s  Thus, taking Laplace transforms with initial quiescent state A [1 − e−sT ] s 1 1 1 1 1 1  (1 − e−sT ) X(s) = A · − · + · 12 s 3 s + 3 4 s + 4 A [1 − 4e−3t + 3e−4t − (1 − 4e−3(t−T ) + 3e−4(t−T ) )H(t − T)] x(t) = L−1 {X(s)} = 12  (s2 + 7s + 12)X(s) =  52  Impulse response h(t) is the solution of d2 h dh + 5h = δ(t), h(0) = ḣ(0) = 0 +4 2 dt dt  Taking Laplace transforms (s2 + 4s + 5)H(s) = 1 H(s) =  s2  1 1 = + 4s + 5 (s + 2)2 + 1  so h(t) = L−1 {H(s)} = e−2t sin t. c Pearson Education Limited 2011   322  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  By the convolution integral response to unit step is t  θ0 (t) =  e−2(t−τ ) sin(t − τ).1dτ  0 t  = e−2t  e2τ sin(t − τ)dτ 0  which using integration by parts gives t e−2t  2τ e [2 sin(t − τ) + cos(t − τ)] 0 5 1 1 −2t = − e (2 sin t + cos t) 5 5  θ0 (t) =  Check Solving d2 θ0 dθ0 + 5θ0 = 1 , θ̇0 (0) = θ0 (0) = 0 +4 2 dt dt gives 1 s 1 1 s+4 1 = − · Φ0 (s) = 2 s(s + 4s + 5) 5s 5 (s + 2)2 + 1 1 1 so θ0 (t) = L−1 {Φ0 (s)} = − [cos t + 2 sin t]e−2t . 5 5 (s2 + 4s + 5)Φ0 (s) =  Exercises 5.7.2 53 State–space form of model is   −5 = 3  x y = cT x ⇒ y = [ 1 2 ] 1 x2  ẋ1 ẋ = Ax + bu ⇒ ẋ2  −1 −1      x1 2 u + 5 x2  System transfer function is G(s) = c (sI − A) T  −1  b = [1  s+5 1 2] −3 s + 1  c Pearson Education Limited 2011   −1  2 5    Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  s+5 1 = Δ = (s + 5)(s + 1) + 3 = (s + 2)(s + 4), det −3 s + 1 −1  1 s + 1 −1 s+5 1 = −3 s + 1 3 s+5 Δ   2 s + 1 −1 1 1 =Δ Thus, G(s) = Δ [ 1 2 ] (12s + 59) 5 3 s+5 so the system transfer function is  323  12s + 59 (s + 2)(s + 4)  G(s) =  54 In this case, the denominator can be factorized G(s) =  s+1 Y(s) = U(s) (s + 1)(s + 6)  and care must be taken not to cancel the common factor, to avoid the system being mistaken for a ﬁrst order system. To proceed it is best to model the system by the diﬀerential equation ÿ + 7ẏ + 6y = u̇ + u from which  ẋ1 = −7x1 + x2 + u ẋ2 = −6x1 + u y = x1  so that a state–space model is   −7 = −6  x1 T y = c x ⇒ y = [1 0] x2  ẋ1 ẋ = Ax + bu ⇒ ẋ2  1 0      x1 1 u + 1 x2  Check G(s) = cT (sI − A)−1 b  s + 7 det(sI − A) = Δ =  6 G(s) =  1 [1 Δ   −1  = s2 + 7s + 6 s    s+1 s+1 1 s 1 = 0] = 2 1 −6 s + 7 Δ s + 7s + 6  c Pearson Education Limited 2011   324  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  55(a)  Taking A to be the companion matrix ⎡ 0 1 0 A= ⎣ 0 −7 −5  ⎤ 0 1⎦ −6  then b = [0 0 1]T c = [5 3 1]T Then from equation (5.84) in the text the state space form of the dynamic model is ẋ = A x + bu y = cT x 55(b)  Taking A to be the companion matrix ⎤ ⎡ 0 1 0 0 1⎦ A = ⎣0 0 −3 −4 then b = [0 0 1]T c = [2 3 1]T  And state space model is ẋ = A x + bu, y = cT x  56  We are required to express the transfer function in the state space form ẋ = A x + bu y = cT x ⎡  ⎤ 0 1 0 0 1 ⎦ and y = [1 0 0]x . To where A is the companion matrix A = ⎣ 0 −6 −11 −6 determine b, we divide the denominator into the numerator as follows 5s−1 − 29s−2 + 120s−3 s3 + 6s2 + 11s + 6 5s2 + s + 1 5s2 + 30s + 55 | neglect these terms −29s − 54 | −29s − 174 | 120 | c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  325  giving b = [5 − 29 − 120]T . Thus, state space form is ⎡  ⎤ ⎡ ẋ1 0 ⎣ ⎦ ⎣ 0 ẋ(t) = ẋ2 = −6 ẋ3  1 0 −11  ⎤ ⎤ ⎡ 5 0 1 ⎦ x(t) + ⎣ −29 ⎦ u(t) 120 −6  y = [1 0 0]x(t) It is readily checked that this is a true representation of the given transfer function.  57  s−3 −2  [sI − A] =   −4 , det[sI − A] = (s − 5)(s + 1) s−1  Thus,     2/3 2/3 1/3 + s+1 − 4 s−5 s−5 = [sI − A] 1/3 1/3 1/3 s−3 − s−5 − s+1 s−5 +  2 5t e + 1 e−t 2 e5t − 2 e−t ∴ L−1 [sI − A]−1 = 31 5t 31 −t 31 5t 32 −t = Φ1 3e − 3e 3e + 3e  4  1 0 1 s−1 4 −1 s [sI − A] B U(s) = 7 1 1 2 5−3 (s − 5)(s + 1) s    11/3 4/3 5 − + + 1 3s + 25 s s+1 s−5 = = 11/3 2/3 3 7s − 15 s(s − 5)(s + 1) s − s+1 + s−5 −1  −1  ∴L  1 = (s − 5)(s + 1)  −1  {[sI − A]  s−1 2  −t + 43 e5t −5 + 11 3 e B U(s)} = −t 3 − 11 + 23 e5t 3 e   = Φ2  Thus, solution is  x(t) = Φ1 x(0) + Φ2 =  5t −5 + 83 e−t + 10 3 e 3 − 83 e−t + 53 e5t    which conﬁrms the answer obtained in Exercise 61 in Chapter 1.  c Pearson Education Limited 2011   2/3 s+1 2/3 s+1    326 58  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition   s−1 3 1 −3 , sI − A = A= −2 s + 4 2 −4 | sI − A | = Δ = (s − 1)(s + 4) + 6 = (s + 2)(s + 1)  1 s + 4 −3 −1 [sI − A] = 2 s−1 Δ 3 2 3 3  − s+2 − s+1 + s+2 s+1 = 2 2 2 3 − s+1 + s+2 s+1 − s+2  3e−t − 2e−2t −3e−t + 3e−2t −1 −1 giving L [sI − A] = 2e−t − 2e−2t −2e−t + 3e−2t  e2t −1 −1 so L [sI − A] x(0) = e−2t   1 1 s + 4 −3 1 −1 [sI − A] bU(s) = 1 s+3 2 s−1 Δ  1 1 1  − s+3 (s+2)(s+3) s+2 = = 1 1 1 (s+2)(s+3) s+2 − s+3  e−2t − e−3t so L−1 {[sI − A]−1 bU(s)} = e−2t − e−3t  Thus, x(t) =  e−2t e−2t   +  e−2t − e−3t e−2t − e−3t    giving x1 (t) = x2 (t) = 2e−2t − e−3t   2 0 1 , u(t) = e−t H(t), x0 = [1 0]T , b= 59 A = 0 −2 −3 with X(s) and the solution x(t) given by equations (5.88) and (5.89) in the text   1 s+3 1 s −1 −1 giving (sI − A) = (sI − A) = −2 5 2 s+3 (s + 1)(s + 2) so that, (sI − A)  −1  1 x0 = (s + 1)(s + 2)  s+3 −2   =  2 1 s+1 − s+2 2 2 − s+1 + s+2    and (sI − A)  −1  1 bU(s) = (s + 1)(s + 2)  2(s + 3) −4    1 = s+1  c Pearson Education Limited 2011   2 s+2 4 s+1  + −  4 (s+1)2 4 (s+1)2  − −  2 s+1 4 s+2    Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Thus, X(s) = giving x(t) =  60  4 (s+1)2  +  327    1 (s+2)  2 4 − (s+2) − (s+1) 2 −t −2t 4te + e 2e−t − 2e−2t − 4te−t 2 (s+1)  Taking A as the companion matrix following the procedure of Example 5.61  we have  ⎡  0 ⎣ 0 A= −6  1 0 −11  ⎤ 0 1 ⎦, b = [0 0 1]T , c = [1 2 3]T −6  Eigenvalue of A given by   −λ 1   0 −λ   −6 −11   0  1  = −(λ3 + 6λ2 + 11λ + 6) = −(λ + 1)(λ + 2)(λ + 3) = 0 −6 − λ   so the eigenvalues are λ1 = −3, λ2 = −2, λ3 = −1. The eigenvectors are given by the corresponding solutions of −λi ei1 + ei2 + 0ei3 = 0 0ei1 − λi ei2 + ei3 = 0 −6ei1 − 11ei2 − (6 + λi )ei3 = 0 Taking i = 1, 2, 3 and solving gives the eigenvectors as e1 = [1 − 3 9]T , e2 = [1 − 2 4]T , e3 = [1 − 1 1]T ⎤ 1 1 1 Taking M to be the modal matrix M = ⎣ −3 −2 −1 ⎦ then the transformation 9 4 1 X = M ξ will reduce the system to the canonical form ⎡  ξ̇ξ = Λ ξ + M−1 bu , y = cT M ξ ⎡  M−1  2 1 ⎣ −6 = 2 6  ⎤ ⎤ ⎡ 3 1 1 1 ⎣ −8 −2 ⎦ , M−1 b = −2 ⎦ , cT M = [22 9 2] 2 5 1 1 c Pearson Education Limited 2011   328  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Thus, canonical form is ⎤ ⎡ −3 ξ̇1 ⎣ ξ̇2 ⎦ = ⎣ 0 0 ξ̇3 ⎡  0 −2 0  ⎤ ⎡ ⎤ ⎡ 1 ⎤ ξ1 0 2 0 ⎦ ⎣ ξ2 ⎦ + ⎣ −1 ⎦ u 1 −1 ξ3 2  y = [22 9 2] [ξ1 ξ2 ξ3 ]T Since the eigenvalues of A are negative the system is stable. Since vector M−1 b has no zero elements the system is controllable and since cT M has no zero elements the system is also observable. b = [0 0 1]T, A b = [0 1 − 6]T, A2 b = [1 6 25]T ⎡  0 [b A b A2 b] = ⎣ 0 1  0 1 −6  ⎤ ⎤ ⎡ 0 0 1 1 6 ⎦ ∼ ⎣ 0 1 0 ⎦ which is of full rank 3 1 0 0 25  so the system is controllable. c = [1 2 3]T, AT c = [−18 − 32 − 16], ⎤ ⎡ ⎡ 1 1 −18 96 T T 2 ⎦ ⎣ ⎣ [c A c (A ) c] = 2 −32 158 ∼ 0 0 3 −16 63  (AT )2 c = [96 158 63]T ⎤ 0 0 1 0 ⎦ which is of full rank 3 0 1  so the system is observable. ⎤ 0 1 0 0 1 ⎦, b = [0 0 1]T, c = [5 3 1]T 61 A = ⎣ 0 0 −5 −6 Eigenvalues of A given by ⎡    −λ 1  0 =  0 −λ  0 −5   0  1  = −λ(λ + 5)(λ + 1) = 0 −6 − λ   so eigenvalues are λ1 = −5, λ2 = −1, λ3 = 0. The corresponding eigenvectors are determined as e1 = [1 − 5 25]T , e2 = [1 − 1 1] , e3 = [1 0 0]T c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 329 ⎤ ⎡ 1 1 1 Take M to be the modal matrix M = ⎣ −5 −1 0 ⎦ then the transformation 25 1 0 x = Mξξ will reduce the system to the canonical form  M−1  ξ̇ξ = Λ ξ + M−1 bu , y = cT M ξ ⎤ ⎤ ⎡ ⎡ 0 1 1 1 1 ⎣ 1 ⎣ 0 −25 −5 ⎦, M−1 b = −5 ⎦, cT M = [15 3 5] = 20 20 20 24 4 4  Thus, canonical form is ⎡ ⎤ ⎡ ξ̇1 −5 ⎣ ξ̇2 ⎦ = ⎣ 0 0 ξ̇3  ⎤ ⎡ ⎤ ⎡ 1 ⎤ ξ1 0 0 20 −1 0 ⎦ ⎣ ξ2 ⎦ + ⎣ − 41 ⎦ u 1 0 0 ξ3 5  y = [15 3 5] [ξ1 ξ2 ξ3 ]T Since A has zero eigenvalues the system is marginally stable, since M−1 b has no zero elements the system is controllable and since cT M has no zero elements the system is observable. Again as in Exercise 60 these results can be conﬁrmed by using the Kalman matrices.  Exercises 5.7.4      1 1 −1 0 −2 0 , x0 = , u= , B= 62 A = t 1 1 1 −3 1 The solution x(t) is given by equation (5.97) in the text.  (sI − A) =  s 2 −1 s + 3   giving (sI − A)  −1  1 = (s + 1)(s + 2)  s+3 1  −2 s  so that, (sI − A)  −1  1 x0 = (s + 1)(s + 2)  −2 s   =  1 s + 3 −2 1 s (s + 1)(s + 2)   (s+5) 1 − 2 s (s+1)(s+2) = s(s+2) s−1 1 + s(s+2) s2 (s+1)(s+2)  (sI − A)−1 B U(s) =  c Pearson Education Limited 2011   2 s+2 2 s+2    − − 1 1  2 s+1 1 s+1  −1 1     ⎡ ⎣  1 s 1 s2  ⎤ ⎦    Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  330 Thus,  2 2 1 (s + 5) − + − 2 s + 2 s + 1 s(s + 2) s (s + 1)(s + 2) 6 15/4 5/2 9/4 − + − 2 = s+2 s+1 s s  X1 (s) = L{x1 (t)} =  giving x1 (t) =  63  9 −2t 15 5 e − t − 6e−t + 4 4 2  ÿ1 + ẏ1 − ẏ2 + y1 = u1  (i)  ÿ2 + ẏ2 − ẏ1 + y2 = u2  (ii)  Let x = [ x1  x2  T  x3  x4 ] = [ y 1  ẏ1  y2  T  ẏ2 ] then  ẋ1 = ẏ1 = x2 (i) ⇒ ÿ1 = ẋ2 = −ẏ1 + ẏ2 − y1 + u1 ⇒ ẋ2 = −x2 + x4 − x1 + u1 ẋ3 = ẏ2 = x4 (ii) ⇒ ÿ2 = ẋ4 = xx − ẏ2 + ẏ1 − y2 + u2 ⇒ ẋ4 = −x4 + x2 − x3 + u2 giving the state–space representation ⎤⎡ ⎤ ⎡ x1 1 0 0 0 −1 0 1 ⎥ ⎢ x2 ⎥ ⎢ 1 ⎦⎣ ⎦ + ⎣ 0 0 1 0 x3 1 −1 −1 0 x4 ⎡ ⎤  x1  1 0 0 0 ⎢ x2 ⎥ y y = Cx ⇒ 1 = ⎣ ⎦ 0 0 1 0 y2 x3 x4 ⎤ ⎡ 0 ẋ1 ⎢ ẋ ⎥ ⎢ −1 ẋ = Ax + Bu ⇒ ⎣ 2 ⎦ = ⎣ 0 ẋ3 0 ẋ4 ⎡  Transfer function = G(s) =  ⎤ 0  0 ⎥ u1 ⎦ 0 u2 1  Cadj(sI−A)B det(sI−A)    0   s −1 0    1 s + 1 0 −1  det(sI − A) = Δ =   = s4 + 2s3 + 2s2 + 2s + 1 0 0 s −1     0 −1 1 s + 1 ⇒ Δ = (s + 1)2 (s2 + 1) so that 1 s2 + s + 1 G(s) = s Δ   1 s2 + s + 1 s = 2 s +s+1 s (s + 1)2 (s2 + 1) c Pearson Education Limited 2011   s 2 s +s+1    Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  331  Poles given by Δ = (s + 1)2 (s2 + 1) = 0 Eigenvalues associated matrix A given by det(λI − A) = (λ + 1)2 (λ2 + 1) = 0 Thus, poles and eigenvalues are identical.  64 (a)  u2 = R1 (u1 + x1 + x2 + x3 ) + L1 ẋ1 ; R1 = 1, L1 = 1 ⇒ ẋ1 = −x1 − x2 − x3 − u1 + u2 u2 = R1 (u1 + x1 + x2 + x3 ) + R2 (x2 + x3 ) + L2 ẋ2 ; R2 = 2, L2 = 1 ⇒ ẋ2 = −x1 − 3x2 − 3x3 − u1 + u2 u2 = R1 (u1 + x1 + x2 + x3 ) + R2 (x2 + x3 ) + R3 x3 + L3 ẋ3 ; R3 = 3, L3 = 1 ⇒ ẋ3 = −x1 − 3x2 − 6x3 − u1 + u2  giving the state–space model ⎤⎡ ⎤ ⎡ ⎤ ⎡ x1 ẋ1 −1 −1 −1 −1 ⎣ ẋ2 ⎦ = ⎣ −1 −3 −3 ⎦ ⎣ x2 ⎦ + ⎣ −1 −1 −3 −6 −1 ẋ3 x3 ⎡  y1 y2  (b) Transfer matrix = G(s) =    0 = 0  Y(s) U(s)  2 0  2 1    x1 x2  ⎤  1 u 1 1⎦ u2 1    = C(sI − A)−1 B = C adj(sI−A) det(sI−A) B    s + 1 1 1   s+3 3  = s3 + 10s2 + 16s + 6 det(sI − A) = Δ =  1  1 3 s + 6 ⎡ 2 ⎤ s + 9s + 9 −(s + 3) −s s2 + 7s + 5 −(3s + 2) ⎦ adj(sI − A) = ⎣ −(s + 3) −s −(3s + 2) s2 + 4s + 2 ⎡ ⎤⎡  s2 + 9s + 9 −1 −(s + 3) −s 1 0 2 2 ⎣ 2 ⎦ ⎣ −1 −(s + 3) s + 7s + 5 −(3s + 2) G(s) = Δ 0 0 1 −1 −s −(3s + 2) s2 + 4s + 2  1 −2s(2s + 3) 2s(2s + 3) = with Δ = s3 + 10s2 + 16s + 6 −s2 s2 Δ  c Pearson Education Limited 2011   ⎤ 1 1⎦ 1  332  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  (c) Y(s) = G(s)U(s) where U(s) = 1 −2s(2s + 3) = −s2 Δ  2 1 −4s −2s+6 s = Δ −s + 1  1 s 1 s2  2s(2s + 3) s2     1 s 1 s2    To obtain response express in partial fractions and take inverse Laplace transforms Factorizing Δ gives Δ = (s + 8.12)(s + 0.56)(s + 1.32) so  Y1 (s) =  1 0.578 1.824 0.246 −4s2 − 2s + 6 = + − + s(s + 8.12)(s + 0.56)(s + 1.32) s s + 8.12 s + 0.56 s + 1.32  ⇒ y1 (t) = 1 + 0.578e−8.12t − 1.824e−0.56t + 0.246e−1.32t 0.177 0.272 0.449 −s + 1 = + − Y2 (s) = (s + 8.12)(s + 0.56)(s + 1.32) s + 8.12 s + 0.56 s + 1.32 ⇒ y2 (t) = 0.177e−8.12t + 0.272e−0.56t − 0.449e−1.32t  Exercises 5.9.3 65  dy then dt   1 0 u(t), y(t) = [1 0]x(t) x(t) + 1 1 2  Choose x1 (t) = y(t), x2 (t) = ẋ1 (t) =  ẋ(t) =  0 1 2  Taking u(t) = K1 x1 (t) + K2 x2 (t) + uext (t) 0 ẋ(t) = K1 +  1 2  1 K2 +   1 2  x(t) +   0 u 1 ext  The eigenvalues of the matrix are given by   0−λ   K1 + 1 2    1  =0 1 K2 + 2 − λ   or λ2 − (K2 + 12 )λ − (K1 + 12 ) = 0 If the poles are to be at λ = −4 then we require the characteristic equation to be λ2 + 8λ + 16 = 0 c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition By comparison, we have −K2 − 12 = 8 and −K1 −  33 17  x + uext so u(t) = − − K2 = − 17 2 2 2 66 ẋ(t) =  0  1  − 14  − 54  1 2  333  = 18 giving K1 = − 33 2 ,   0 u(t) , y(t) = [0 2]x(t) 1   x(t) +  Setting u = KT x + uext , K = [K1 K2 ]T gives the system matrix A=  0 K1 −  1 4  1 K2 −   5 4  whose eigenvalues are given by λ2 − (K2 − 54 )λ − (K1 − 54 ) = 0. Comparing with the desired characteristic equation (λ + 5)2 = λ2 + 10λ + 25 = 0 35 gives K1 = − 99 4 , K2 = − 4 . Thus,   99 35  x(t) + uext u(t) = − − 4 4  67  With u1 = [K1 K2 ]x(t) the system matrix A becomes  1 + K2 K1 A= 6 + K1 1 + K2  having characteristic equation λ2 − λ(1 + K1 + K2 ) − 6(1 − K2 ) = 0 which on comparing with λ2 + 10λ + 25 = 0  35 31  35 31 so that u1 (t) = − − x(t) gives K1 = − , K2 = − 6 6 6 6 0 1 Using u2 (t) the matrix A becomes where KT = [−31 − 11] 6 + K1 1 + K2 68  See p. 472 in the text.  c Pearson Education Limited 2011   334  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  69  For the matrix of Exercise 68 the Kalman matrix is   1 0 2 −2 ∼ M= 0 0 1 −1  which is of rank 1. Thus, the system is uncontrollable. For the matrix of Exercise 65 the Kalman matrix is   0 1 0 1 M= ∼ 1 12 1 0 which is of full rank 2. Thus, the system is controllable.  70 M = (b A b) = 8 35  v A = [4 − 1] T    1 0 0 −1 , vT = [4 − 1] , M = 4 −1 −1   4 −1 −2 1 , T−1 = = [−3 1] so T = −3 1 −9 3  1 4  1 4    Taking z(t) = Tx or x = T−1 z(t) then equation reduces to    1 T z(t) + u T ż(t) = 4   8 −2 1 −1 T z(t) + T u or ż(t) = I 35 −9 4     4 −1 1 1 8 −2 4 −1 z(t) + = −3 1 3 4 35 −9 −3 1   0 0 1 u z(t) + = 1 2 −1 −1  8 35  −2 −9  −1  1 4   u  Clearly both system matrices have the same eigenvalues λ = −2, λ = 1. This will always be so since we have carried out a singularity transformation.  Review Exercises 5.10 1(a)  d2 x dx + 5x = 8 cos t, x(0) = ẋ(0) = 0 Taking Laplace transforms +4 2 dt dt (s2 + 4s + 5)X(s) =  8s +1  s2  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  335  8s (s2 + 1)(s2 + 4s + 5) s+1 s+5 = 2 − 2 s + 1 s + 4s + 5 1 (s + 2) + 3 s + 2 − = 2 s + 1 s + 1 (s + 2)2 + 1  X(s) =  giving x(t) = L−1 {X(s)} = cos t + sin t − e−2t [cos t + 3 sin t] d2 x dx − 2x = 6, x(0) = ẋ(0) = 1 −3 2 dt dt Taking Laplace transforms  1(b) 5  5s2 + 2s + 6 6 (5s − 3s − 2)X(s) = 5(s + 1) − 3(1) + = s s 5s2 + 2s + 6 X(s) = 5s(s + 25 )(s + 1) 2  13 15 3 =− + 7 + 7 2 5 s−1 s+ 5  giving x(t) = L−1 {X(s)} = −3 +  13 t 15 − 2 t e + e 5 7 7  2(a)  Thus , L−1 2(b)  1 1 1 1 1 s+2 = − · − · 2 2 (s + 1)(s + 2)(s + 2s + 2) s + 1 2 s + 2 2 s + 2s + 2 1 1 1 (s + 1) + 1 1 − · − · = s + 1 2 s + 2 2 (s + 1)2 + 1   1 1 1 = e−t − e−2t − e−t (cos t + sin t) 2 (s + 1)(s + 2)(s + 2s + 2) 2 2  From equation (5.26) in the text the equation is readily deduced.  Taking Laplace transforms (s2 + 3s + 2)I(s) = s + 2 + 3 + V.  1 (s + 1)2 + 1    1 s+5 +V (s + 2)(s + 1) (s + 2)(s + 1)(s2 + 2s + 2)   3 4 − + V extended as in (a) = s+1 s+2  I(s) =  c Pearson Education Limited 2011   336  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Thus, using the result of (a) above   1 1 i(t) = L−1 {I(s)} = 4e−t − 3e−2t + V e−t − e−2t − e−t (cos t + sin t) 2 2 3  Taking Laplace transforms 1 s2 2 −2sX(s) + (s2 − 4)Y(s) = − s (s2 − 1)X(s) + 5sY(s) =  Eliminating Y(s) [(s2 − 1)(s2 − 4) + 2s(5s)]X(s) =  s2 − 4 11s2 − 4 + 10 = s2 s2  11s2 − 4 s2 (s2 + 1)(s2 + 4) 1 5 4 =− 2 + 2 − 2 s s +1 s +4  X(s) =  giving x(t) = L−1 {X(s)} = −t + 5 sin t − 2 sin 2t From the ﬁrst diﬀerential equation 1 d2 x  dy = t+x− 2 dt 5 dt 1 = [t − t + 5 sin t − 2 sin 2t + 5 sin t − 8 sin 2t] 5 = (2 sin t − 2 sin 2t) then y = −2 cos t + cos 2t + const. and since y(0) = 0, constant = 1 giving y(t) = 1 − 2 cos t + cos 2t x(t) = −t + 5 sin t − 2 sin 2t  4  Taking Laplace transforms (s2 + 2s + 2)X(s) = sx0 + x1 + 2x0 + c Pearson Education Limited 2011   s2  s +1  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  337  s sx0 + x1 + 2x0 + 2 2 2 s + 2s + 2 (s + 1)(s + 2s + 2) 1 s+4 x0 (s + 1) + (x1 + x0 ) 1 s + 2 + · 2 − · = 2 (s + 1) + 1 5 s + 1 5 (s + 1)2 + 1  X(s) =  giving x(t) = L−1 {X(s)} 1 = e−t (x0 cos t + (x1 + x0 ) sin t) + (cos t + 2 sin t) 5 1 − e−t (cos t + 3 sin t) 5 that is, x(t) =    1 1 3 (cos t + 2 sin t) + e−t (x0 − ) cos t + (x1 + x0 − ) sin t 5 5 5 ↑ ↑ steady state  transient  2 1 Steady state solution is xs (t) = cos t + sin t ≡ A cos(t − α) 5 5  1 1 2 2 2 having amplitude A = ( 5 ) + ( 5 ) = √ 5 −1 ◦ and phase lag α = tan 2 = 63.4 .  5  Denoting the currents in the primary and secondary circuits by i1 (t) and i2 (t)  respectively Kirchoﬀ's second law gives di1 di2 + = 100 dt dt di1 di2 + =0 20i2 + 3 dt dt 5i1 + 2  Taking Laplace transforms 100 s sI1 (s) + (3s + 20)I2 (s) = 0 (5 + 2s)I1 (s) + sI2 (s) =  Eliminating I1 (s) [s2 − (3s + 20)(2s + 5)]I2 (s) = 100 c Pearson Education Limited 2011   338  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 20 −100 = − 5s2 + 55s + 100 s2 + 11s + 20 1 20  20 √ = − =− 11 2 41 11 (s + 2 ) − 4 41 (s + 2 −  I2 (s) =  √  41 2 )  −  1 (s +  11 2  +  √    41 2 )  giving the current i2 (t) in the secondary loop as √  20  −(11+√41)t/2 e − e−(11− 41)t/2 i2 (t) = L−1 {I2 (s)} = √ 41  6(a) (i) L{cos(wt + φ)} = L{cos φ cos wt − sin φ sin wt} s w = cos φ 2 − sin φ 2 2 s +w s + w2 = (s cos φ − w sin φ)/(s2 + w2 ) (ii) L{e−wt sin(wt + φ)} = L{e−wt sin wt cos φ + e−wt cos wt sin φ} s+w w + sin φ = cos φ 2 2 (s + w) + w (s + w)2 + w2 = [sin φ + w(cos φ + sin φ)]/(s2 + 2sw + 2w2 ) 6(b)  Taking Laplace transforms s +4 3 2 2s + 9s + 9s + 36 = 2 (s + 4)(s2 + 4s + 8) 1 39s + 172 1 s+4 · 2 + · 2 = 20 s + 4 20 s + 4s + 8 1 s+4 1 39(s + 2) + 47(2) = · 2 + · 20 s + 4 20 (s + 2)2 + (2)2  (s2 + 4s + 8)X(s) = (2s + 1) + 8 +  giving x(t) = L−1 {X(s)} = 7(a) L−1    s2  1 1 (cos 2t + 2 sin 2t) + e−2t (39 cos 2t + 47 sin 2t) . 20 20     s−4 −1 (s + 2) − 2(3) = L s2 + 4s + 13 (s + 2)2 + 32 = e−2t [cos 3t − 2 sin 3t] c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 7(b)  Taking Laplace transforms (s + 2)Y(s) = −3 +  4 2s 4 + 2 + 2 s s +1 s +1  Y(s) =  −3s3 + 6s2 + s + 4 s(s + 2)(s2 + 1)  =  5 2 2 − + 2 s s+2 s +1  Therefore, y(t) = L−1 {Y(s)} = 2 − 5e−2t + 2 sin t  8  Taking Laplace transforms (s + 5)X(s) + 3Y(s) = 1 +  5X(s) + (s + 3)Y(s) =  2s s2 − 2s + 6 5 − = s2 + 1 s2 + 1 s2 + 1  6 3s 6 − 3s − = s2 + 1 s2 + 1 s2 + 1  Eliminating Y(s) [(s + 5)(s + 3) − 15]X(s) =  (s2 + 8s)X(s) =  X(s) =  (s + 3)(s2 − 2s + 6) 3(6 − 3s) − 2 s2 + 1 s +1 s3 + s2 + 9s s2 + 1 1 1 s2 + s + 9 = + 2 2 (s + 8)(s + 1) s+8 s +1  so x(t) = L−1 {X(s)} = e−8t + sin t From the ﬁrst diﬀerential equation 3y = 5 sin t − 2 cos t − 5x −  dx = 3e−8t − 3 cos t dt  Thus, x(t) = e−8t + sin t, y(t) = e−8t − cos t .  c Pearson Education Limited 2011   339  340 9  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition Taking Laplace transforms 100 + 104 2 (s + 100)(s + 200)Q(s) = 104 · 2 s + 104 2.104 Q(s) = (s + 100)(s + 200)(s2 + 104 ) 1 2 1 1 3s − 100 1 · − · − · = 100 s + 100 500 s + 200 500 s2 + 104  (s2 + 300s + 2 × 104 )Q(s) = 200·  giving q(t) = L−1 {Q(s)} = that is, q(t) =  s2  1 −100t 2 −200t 1 e e (3 cos 100t − sin 100t) − − 100 500 500  1 1 [5e−100t − 2e−200t ] − [3 cos 100t − sin 100t] 500 500 ↑ ↑ transient  steady state  1 3 sin 100t + cos 100t ≡ A sin(100t + α) 5o 5 18 12 .  Steady state current = where α = tan−1  1 5  o  Hence, the current leads the applied emf by about 18 12 . 10 dx + 6x + y = 2 sin 2t dt d2 x dy = 3e−2t +x− 2 dt dt 4  dx = −2 when t = 0 so from (i) y = −4 when t = 0. dt Taking Laplace transforms Given x = 2 and  8s2 + 36 4 = s2 + 4 s2 + 4 2s2 + 6s + 7 3 = (s2 + 1)X(s) − sY(s) = 2s − 2 + 4 + s+2 s+2 (4s + 6)X(s) + Y(s) = 8 +  Eliminating Y(s) [s(4s + 6) + (s2 + 1)]X(s) =  8s2 + 36 2s2 + 6s + 7 + s2 + 4 s+2  c Pearson Education Limited 2011   (i) (ii)  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  X(s) = = =  341  2s2 + 6s + 7 8s2 + 36 + s(s2 + 4)(s + 1)(s + 15 ) 5(s + 2)(s + 1)(s + 15 ) 11 5  s+1 29 20  s+1  − +  227 505  s+  1 5  1 3  s+2  1 3 49 1 76s − 96 3 4 · 2 + − + 60 1 − 505 s + 4 s+2 s+1 s+ 5  +  445 1212 s + 15  −  1  76s − 96  505 s2 + 4  giving x(t) = L−1 {X(s)} =  11(a)  29 −t 1 −2t 445 − 1 t 1 e + e e 5 − (76 cos 2t − 48 sin 2t) + 20 3 1212 505  Taking Laplace transforms (s2 + 8s + 16)Φ(s) =  s2  2 +4  Φ(s) =  2 4)2 (s2  (s + + 4) 1 1 2s − 3 1 1 1 = − · 2 · + · 2 25 s + 4 10 (s + 4) 50 s + 4  1 −4t 1 1 e (4 cos 2t − 3 sin 2t) so θ(t) = L−1 {Φ(s)} = + · te−4t − 25 10 100 1 that is, θ(t) = (4e−4t + 10te−4t − 4 cos 2t + 3 sin 2t) 100  11(b)  Taking Laplace transforms (s + 2)I1 (s) + 6I2 (s) = 1 I1 (s) + (s − 3)I2 (s) = 0  Eliminating I2 (s) [(s + 2)(s − 3) − 6]I1 (s) = s − 3 I1 (s) = giving i1 (t) = L−1 {I1 (s)} =  1 6 s−3 = 7 + 7 (s − 4)(s + 3) s−4 s+3  1 4t (e + 6e−3t ) 7  c Pearson Education Limited 2011   342  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Then from the ﬁrst diﬀerential equation 6 di1 6 = − e4t + e−3t dt 7 7 1 −3t 1 giving i2 (t) = (e − e4t ), i1 (t) = (e4t + 6e−3t ). 7 7 6i2 = −2i1 −  12  The diﬀerential equation di d2 i LCR 2 + L + Ri = V dt dt  follows using Kirchhoﬀ's second law. Substituting V = E and L = 2R2 C gives 2R3 C2  which on substituting CR =  d2 i 2 di + 2R C + Ri = E dt2 dt  1 leads to 2n 1 d2 i 1 di E + +i= 2 2 2n dt n dt R  and it follows that  d2 i E di + 2n + 2n2 i = 2n2 2 dt dt R  Taking Laplace transforms (s2 + 2ns + 2n2 )I(s) =  2n2 E 1 · R s   2n2 E 2 2 R s(s + 2ns + 2n )  E1 s + 2n = − 2 2 R s (s + n) + n  I(s) =  so that i(t) =  E [1 − e−nt (cos nt + sin nt)] R  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  13  343  The equations are readily deduced by applying Kirchhoﬀ's second law to the  left- and right-hand circuits. Note that from the given initial conditions we deduce that i2 (0) = 0. Taking Laplace transforms then gives E s −RI1 (s) + (sL + 2R)I2 (s) = 0 (sL + 2R)I1 (s) − RI2 (s) =  Eliminating I2 (s) E (sL + 2R) s E (sL + 3R)(sL + R)I1 (s) = (sL + 2R) s [(sL + 2R)2 − R2 ]I1 (s) =  I1 (s) =  s + 2R E L L s(s + R )(s + L   3R L ) 1 6  1  E  23 − 2R − R s s+ L s + 3R L R 3R   E 1 4 − 3e− L t − e− L t giving i1 (t) = L−1 {I1 (s)} = 6R  =  For large t , the exponential terms are approximately zero and i1 (t)  2E 3R  From the ﬁrst diﬀerential equation Ri2 = 2Ri1 + L  di1 −E dt  Ignoring the exponential terms we have that for large t i2  4E E 1E − = 3R R 3R  c Pearson Education Limited 2011   344  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  14  Taking Laplace transforms (s2 + 2)X1 (s) − X2 (s) =  s2  2 +4  −X1 (s) + (s2 + 2)X2 (s) = 0 Eliminating X1 (s) [(s2 + 2)2 − 1]X2 (s) =  X2 (s) = = so x2 (t) = L−1 {X2 (s)} =  s2  2 +4  2 (s2 + 4)(s2 + 1)(s2 + 3) 2 3  s2  +4  +  1 3  s2  +1  −  s2  1 +3  √ 1 1 1 sin 2t + sin t − √ sin 3t 3 3 3  Then from the second diﬀerential equation √ √ √ 2 2 1 d 2 x2 2 4 √ sin 2t + sin t − sin 2t − sin t + sin = 3t − 3 sin 3t dt2 3 3 3 3 3 √ 1 1 2 or x1 (t) = − sin 2t + sin t + √ sin 3t 3 3 3  x1 (t) = 2x2 +  15(a) (i) L−1   s2    (s + 1) + 3  s+4 = L−1 + 2s + 10 (s + 1)2 + 32 = e−t (cos 3t + sin 3t)  (ii) L−1      1 2 s−3 1  = L−1 + − 2 2 (s − 1) (s − 2) (s − 1) (s − 1) s−2 = et + 2tet − e2t  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  15(b)  Taking Laplace transforms (s2 + 2s + 1)Y(s) = 4s + 2 + 8 + L{3te−t } 3 (s + 1)2 Y(s) = 4s + 10 + (s + 1)2 4s + 10 3 Y(s) = + 2 (s + 1) (s + 1)4 4 3 6 = + + 2 s + 1 (s + 1) (s + 1)4 1 giving y(t) = L−1 {Y(s)} = 4e−t + 6te−t + t3 e−t 2 1 −t that is, y(t) = e (8 + 12t + t3 ) 2  16(a) 5 2 5 = · − 14s + 53 2 (s − 7)2 + 22 5 Therefore, f(t) = L−1 {F(s)} = e7t sin 2t 2 F(s) =  16(b)  s2  d2 θ dθ n2 i 2 + n , θ(0) = θ̇(0) = 0, i const. + 2K θ = dt2 dt K  Taking Laplace transforms  (s2 + 2Ks + n2 )Φ(s) = Therefore, Φ(s) =  n2 i Ks n2 i Ks(s2 + 2Ks + n2 )  For the case of critical damping n = K giving 1 1   K12 Ki K2 K = Ki − − Φ(s) = s(s + K)2 s s + K (s + K)2  c Pearson Education Limited 2011   345  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  346 Thus,  θ(t) = L−1 {Φ(s)} =  i [1 − e−Kt − Kte−Kt ] K  17(a) (i) L{sin tH(t − α)} = L{sin[(t − α) + α]H(t − α)} = L{[sin(t − α) cos α + cos(t − α) sin α]H(t − α)} cos α + s sin α −αs .e = s2 + 1 (ii) L−1  17(b)   se−αs (s + 1) − 1  = L−1 e−αs 2 s + 2s + 5 (s + 1)2 + 4   1 = L−1 eαs L[e−t (cos 2t − sin 2t)] 2 1 = e−(t−α) [cos 2(t − α) − sin 2(t − α)]H(t − α) 2  Taking Laplace transforms   −e−sπ  1 − by (i) above in part (a) s2 + 1 s2 + 1 1 + e−πs = 2 s +1 −πs   1 s−2 1 s 1+e = − + (1 + e−πs ) Y(s) = 2 (s + 1)(s2 + 2s + 5) 10 s2 + 1 10 s2 + 2s + 5 (s2 + 2s + 5)Y(s) =  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  347  giving 1 [2 sin t − cos t + e−(t−π) [2 sin(t − π) − cos(t − π)]H(t − π)] 10 1 + e−t (cos 2t − sin 2t) + e−(t−π) [cos 2(t − z) 2 1 − sin 2(t − π)]H(t − π)] 2 1  −t 1 e (cos 2t − sin 2t) + 2 sin t − cos t = 10 2  1 + [e−(t−π) (cos 2t − sin 2t) + cos t − 2 sin t]H(t − π) 2  y(t) = L−1 {Y(s)} =  18  By theorem 5.5 L{v(t)} = V(s) =  T  1 1 − e−sT  1 = 1 − e−sT  0  T /2  e 0  e−st v(t)dt  −st  T  dt −  e  −st   dt  T /2     1 −st T /2  1 −st T 1 − e − − e = 1 − e−sT s s 0 T /2 1 1 = · (e−sT − e−sT /2 − e−sT /2 + 1) s 1 − e−sT (1 − e−sT /2 )2 1  1 − e−sT /2  1 = = s (1 − e−sT /2 )(1 + e−sT /2 ) s 1 + e−sT /2  Equation for current ﬂowing is 250i +  1 (q0 + C  t  i(τ)dτ) = v(t),  q0 = 0  0  Taking Laplace transforms 250I(s) +  1  1 − e−sT /2  1 1 · I(s) = V(s) = · 10−4 s s 1 + e−sT /2 1  1 − e−sT /2  (s + 40)I(s) = 250 1 + e−sT /2 1 − e−sT /2 1 · or I(s) = 250(s + 40) 1 + e−sT /2 c Pearson Education Limited 2011   348  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  3 1 (1 − e−sT /2 )(1 − e−sT /2 + e−sT − e− 2 sT + e−2sT . . .) 250(s + 40) 3 1 = [1 − 2e−sT /2 + 2e−sT − 2e− 2 sT + 2e−2sT . . .] 250(s + 40)   1 −40t 1 = e Since L−1 using the second shift theorem gives 250(s + 40) 250  I(s) =  i(t) =   T −40(t−T /2) 1 e−40t − 2H t − e + 2H(t − T)e−40(t−T ) 250 2   3T −40(t−3T /2) e −2H t − + ... 2  If T = 10−3 s then the ﬁrst few terms give a good representation of the steady state 1 of the circuit is large compared to the period T. since the time constant 4 19  The impulse response h(t) is the solution of d2 h 2dh + 2h = δ(t) + dt2 dt  subject to the initial conditions h(0) = ḣ(0) = 0. Taking Laplace transforms (s2 + 2s + s)H(s) = L{δ(t)} = 1 1 H(s) = (s + 1)2 + 1 that is, h(t) = L−1 {H(s)} = e−t sin t. Using the convolution integral the step response xs (t) is given by t  h(τ)u(t − τ)dτ  xs (t) = 0  with u(t) = 1H(t) ; that is, t  xs (t) =  1.e−τ sin τdτ  0  1 = − [e−τ cos τ + e−τ sin τ]t0 2 1 that is, xs (t) = [1 − e−t (cos t + sin t)]. 2 c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition Solving  d 2 xs 2dxs + 2xs = 1 directly we have taking Laplace transforms + 2 dt dt (s2 + 2s + 2)Xs (s) =  1 s  1 + 2s + 2)  s+2 1 1 1 = · − 2 2 s 2 (s + 1) + 1  Xs (s) =  s(s2  giving as before xs (t) =  1 1 −t − e (cos t + sin t) 2 2  20  d4 y = 12 + 12H(x − 4) − Rδ(x − 4) dx4 y(0) = y (0) = 0, y(4) = 0, y (5) = y (5) = 0 EI  With y (0) = A, y (0) = B taking Laplace transforms 12 12 −4s + e − Re−4s s s A B 12 1 12 1 −4s R 1 −4s · 5+ · 5e · e Y(s) = 3 + 4 + − s s EI s EI s EI s4  EIs4 Y(s) = EI(sA + B) +  giving A B 1 4 1 x + (x − 4)4 H(x − 4) y(x) = L−1 {Y(s)} = x2 + x3 + 2 6 2EI 2EI R − (x − 4)3 H(x − 4) 6EI  c Pearson Education Limited 2011   349  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  350 or  EIy(x) =  1 1 1 1 R A1 x2 + B1 x3 + x4 + (x − 4)4 H(x − 4) − (x − 4)3 H(x − 4) 2 6 2 2 6  32 B1 + 128 ⇒ 3A1 + 4B1 = −48 3 y (5) = 0 ⇒ 0 = A1 + 5B1 + 6(25) + 6 − R ⇒ A1 + 5B1 − R = −156 y(4) = 0 ⇒ 0 = 8A1 +  y (5) = 0 ⇒ 0 = B1 + 12(5) + 12 − R ⇒ B1 − R = −72 which solve to give A1 = 18, B1 = −25.5, R = 46.5 Thus, ⎧ ⎨  1 4 x − 4.25x3 + 9x2 , 0 ≤ x ≤ 4 2 y(x) = 1 ⎩ x4 − 4.25x3 + 9x2 + 1 (x − 4)4 − 7.75(x − 4)3 , 4 ≤ x ≤ 5 2 2 R0 = −EIy (0) = 25.5kN, M0 = EIy (0) = 18kN.m Check :  R0 + R = 72kN , Total load = 12 × 4 + 24 = 72kN  Moment about x = 0 is 12 × 4 × 2 + 24 × 4.5 − 4R = 18 = M0  21(a)  f(t) = H(t − 1) − H(t − 2) e−s e−2s and L{f(t)} = F(s) = − s s c Pearson Education Limited 2011   √  √  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  351  Taking Laplace transforms throughout the diﬀerential equation 1 −s (e − e−2s ) s 1 (e−s − e−2s ) X(s) = s(s + 1) 1 1  −s  1 1  −2s − e − − e = s s+1 s s+1  (s + 1)X(s) =  giving x(t) = L−1 {X(s)} = [1 − e−(t−1) ]H(t − 1) − [1 − e−(t−2) ]H(t − 2)  21(b) I(s) =  E s[Ls + R/(1 + Cs)]  (i) By the initial value theorem E =0 s→∞ Ls + R/(1 + Cs)  lim i(t) = lim sI(s) = lim  t→0  s→∞  (ii) Since sI(s) has all its poles in the left half of the s-plane the conditions of the ﬁnal value theorem hold so lim i(t) = lim sI(s) =  t→∞  22  s→0  E R  We have that for a periodic function f(t) of period T L{f(t)} =  1 1 − e−sT  T  e−sT f(t)dt  0  Thus, the Laplace transform of the half-rectiﬁed sine wave is π 1 e−sT sin tdt L{v(t)} = −2πs 1−e 0 π   1 (j−s)t = Im e dt 1 − e−2πs 0 π   e(j−s)t   1 = Im 1 − e−2πs j − s 0   (−e−πs − 1)(−j − s)  1 + e−πs 1 = = Im 1 − e−2πs (j − s)(−j − s) (1 − e−2πs )(1 + s2 ) 1 that is, L{v(t)} = 2 (1 + s )(1 − e−πs )  c Pearson Education Limited 2011   352  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Applying Kirchoﬀ's law to the circuit the current is determined by di + i = v(t) dt which on taking Laplace transforms gives (s + 1)I(s) =  1  (1 + − e−πs )  1 s+1  1 1 − · I(s) = 1 − e−πs s + 1 s2 + 1 2  s + 1  1 1 − 2 1 + e−πs + e−2πs + . . . = 2 s+1 s +1  Since L−1  s2 )(1  1 1 s + 1  1 − 2 = (sin t − cos t + e−t )H(t) = f(t) 2 s+1 s +1 2  we have by the second shift theorem that i(t) = f(t) + f(t − π) + f(t − 2π) + . . . =  ∞   f(t − nπ)  n=0  The graph may be plotted by computer and should take the form  1 1 , L{te−t } = 2 s (s + 1)2 −t taking f(t) = t and g(t) = te in the convolution theorem 23(a)  Since L{t} =  L−1 [F(s)G(s)] = f ∗ g(t) c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  353  gives L−1  1  1 = · 2 2 s (s + 1)  t  f(t − τ)g(τ)dτ 0 t  =   (t − τ)τe−τ  0  = −(t − τ)τe−τ − (t − 2τ)e−τ + 2e−τ 1  1 = t − 2 + 2e−t + te−t . i.e. L−1 2 · 2 s (s + 2)  23(b)  y(t) = t + 2  t 0  y(u) cos(t − u)du  t 0  s Taking f(t) = y(t), g(t) = cos t ⇒ F(s) = Y(s), G(s) = 2 giving on taking s +1 transforms s 1 + 2Y(s) 2 2 s s +1 2 s +1 (s2 + 1 − 2s)Y(s) = s2 2 1 2 s +1 2 2 + + or Y(s) = 2 = − s (s − 1)2 s s2 s − 1 (s − 1)2 Y(s) =  and y(t) = L−1 {Y(s)} = 2 + t − 2et + 2tet . Taking transforms (s2 Y(s) − sy(0) − y (0))(sY(s) − y(0)) = Y(s) or (s2 Y(s) − y1 )(sY(s)) = Y(s) 1 y1 giving Y(s) = 0 or Y(s) = 2 + 3 s s which on inversion gives y(t) = 0 or y(t) =  1 2 t + ty1 2  In the second of these solutions the condition on y (0) is arbitrary.  c Pearson Education Limited 2011   354  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  24  Equation for displacement is EI  d4 y = −Wδ(x − ) dx4  with y(0) = 0, y(3) = 0, y (0) = y (3) = 0 with y (0) = A, y (0) = B then taking Laplace transforms gives EIs4 Y(s) = EI(sA + B) − We−s −W −s A B Y(s) = e + 3+ 4 4 EIs s s −W A B giving y(x) = (x − )3 .H(x − ) + x2 + x3 6EI 2 6 −3W B (x − )2 + Ax + x2 6EI 2  so y (3) = 0 and y(3) = 0 gives For x > , y (x) =  2 2W2 + 3A + 9B EI 2 3 9 4W 9 + A2 + B3 0=− 3EI 2 2 20 W 4W and B = giving A = − 9EI 27 EI 0=−  Thus, deﬂection y(x) is y(x) = −  2 W 2 10 W 3 W (x − )3 H(x − ) − x + x 6EI 9 EI 81 EI  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  355  With the added uniform load the diﬀerential equation governing the deﬂection is EI  25(a)  d4 y = −Wδ(x − ) − w[H(x) − H(x − )] dx4  Taking Laplace transforms (s2 − 3s + 3)X(s) =  1 −as e s  1 1  −as  16 1 −as 6s − 2 · e − ·e = s(s2 − 3s + 3) s s2 − 3s + 3 √ √ 1  1 (s − 32 ) − 3( 23 )  −as √ − e = 6 s (s − 3 )2 + ( 3 )2  X(s) =  2  2  √ √   √ 3  3 3 e−as L 1 − e− 2 t cos t − 3 sin t = 6 2 2 giving −1  x(t) = L  √ √  √ 3  3 3 1 − 2 (t−a) 1−e (t − a) − 3 sin (t − a) H(t − a) cos {X(s)} = 6 2 2  25(b) X(s) = G(s)L{sin wt} = G(s) =  s2  w + w2  w G(s) (s + jw)(s − jw)  Since the system is stable all the poles of G(s) have negative real part. Expanding in partial fractions and inverting gives x(t) = 2Re   F(jw)w jwt  + terms from G(s) with negative exponentials ·e 2jw  Thus, as t → ∞ the added terms tend to zero and x(t) → xs (t) with xs (t) = Re   ejwt F(jw)  j  c Pearson Education Limited 2011   356  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  26(a)  In the absence of feedback the system has poles at s = −3 and s = 1  and is therefore unstable.  26(b) G1 (s) =  26(c)  1 1 G(s) = = 2 1 + KG(s) (s − 1)(s + 3) + K s + 2s + (K − 3)  Poles G1 (s) given by s = −1 ±  √ 4 − K.  These may be plotted in the s-plane for diﬀerent values of K. Plot should be as in the ﬁgure  26(d)  Clearly from the plot in (c) all the poles are in the left half plane when  K > 3. Thus system stable for K > 3.  a1 a0 a2 1s2 + 2s + (K − 3) = 0 Routh–Hurwitz determinants are 26(e)  Δ1 = 2 > 0    2  a1 a2    =  Δ2 =   0 0 a0   1  = 2(K − 3) > 0 if K > 3 K − 3  thus, conﬁrming the result in (d).  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 27(a)  Closed loop transfer function is G1 (s) =  2 G(s) = 2 1 + G(s) s + αs + 5    2 = h(t) = 2e−2t sin t Thus L−1 2 s + αs + 5  α   2 −2t −1 i.e. L sin (5 − = 2e 2 (s + α2 )2 + (5 − α4 ) giving α = 4  27(b)  357  α2 4 )t  = 2e−2t sin t  Closed loop transfer function is G(s) = 1  10 s(s−1) − (1+Ks)10 s(s−1)  =  s2  10 + (10K − 1)s + 10  Poles of the system are given by s2 + (10K − 1)s + 10 = 0 which are both in the negative half plane of the s-plane provided (10K − 1) > 0; that is, K > is K =  1 10  1 10  . Thus the critical value of K for stability of the closed loop system  .  28(a)  c Pearson Education Limited 2011   358  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  28(b) L{e  At  1 } = [sI − A] = (s + 2)(s + 3) 6 3 2 6  − s+3 s+2 − s+3 s+2 = 1 −2 1 1 s+3 − s+2 s+2 − s+3 −1  s+5 −1  6 8    Taking inverse transforms gives e  28(c)  At  3e−2t − 2e−3t e−3t − e−2t  =  6e−2t − 6e−3t 3e−3t − 2e−2t    Taking Laplace transforms [sI − A]X(s) = x(0) + bU(s) ; Y(s) = cT X(s)  With x(0) = 0 and U(s) = 1 the transform Xδ (s) of the impulse response is Xδ (s) = [sI − A]−1 b , Yδ (s) = cT [sI − A]−1 b Inverting then gives the impulse response as yδ (t) = [1 1]  6e−2t − 6e−3t 3e−3t − 2e−2t  With x(0) = [1 0]T and X(s) =  Y(s) = [1 1] = [1 1] so y(t) = [1 1]    = 4e−2t − 3e−3t , t ≥ 0  1 s    1 0 1 −1 + [sI − A] [sI − A] 0 1 s  1/6 1/2 1/3  3 2 s+2 − s+3 + 6 s − s+2 + s+3 1 1 1 1 s+3 − s+2 + s+2 − s+3  3e−2t − 2e−3t + 1 − 3e−2t + 2e−3t e−3t − e−2t + e−2t − e−3t −1  that is, y(t) = 1, t ≥ 0  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  29  L{e  At  −1  } = [sI − A]  s+2 −2  =  1 s  −1  −1 (s+1)2 +1 s+2 (s+1)2 +1  s (s+1)2 +1 2 (s+1)2 +1  =  −e−t sin t e−t (cos t − sin t) Thus e = −t −t 2e sin t e (cos t + sin t) At and e x(0) = 0 since x(0) = 0 1 With U(s) = L{u(t)} = we have s  359      At  −1  [sI − A]  bU(s) =  =  −1  so L  {(sI − A)  −1  Thus  bU(s)} =  s s2 +2s+2 2 s2 +2s+2  −1 s2 +2s+2 s+2 s2 +2s+2  1 (s+1)2 +1 1 s+2 s − (s+1)2 +1    1 0    1 = s  1 s2 +2s+2 2 s(s2 +2s+2)    e−t sin t −t 1 − e (cos t + sin t)    x(t) = eAt x(0) + L−1 {(sI − A)−1 bU(s)}  e−t sin t = 1 − e−t (cos t + sin t)  For the transfer function, we have, Y(s) = c X(s) and when x(0) = 0 Y(s) = c[s I − A]−1 bU(s) = HU(s) where H = c[s I − A]−1 b For this system, we have H(s) = [1 1]  s s2 +2s+1 2 s2 +2s+1   =  s+2 (s + 1)2 + 1  When u(t) = δ(t), U(s) = 1 and so the impulse response yδ (t) is given by L{yδ (t)} = Yδ (s) =  s+1 1 s+2 = + 2 2 (s + 1) + 1 (s + 1) + 1 (s + 1)2 + 1  ⇒ yδ (t) = e−t (cos t + sin t)  c Pearson Education Limited 2011     360  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  30  The controllability question can be answered by either reducing to canonical  form as in section 6.7.8 of the text or by using the Kalman matrix criterion given in Exercise 61 of the text. Adopting the Kalman matrix approach ⎤ ⎡ 1 2 0 A = ⎣ 0 −1 0 ⎦ and −3 −3 −2 ⎤ ⎡ ⎡ ⎤ ⎡ ⎤ 2 0 0 2 ⎦ ⎦ ⎣ ⎣ ⎣ b = 1 , A b = −1 , A b = 1 ⎦ −3 9 0 so the controllability Kalman matrix is ⎡  0 [b A b A2 b] = ⎣ 1 0  ⎤ 2 0 −1 1 ⎦ = C −3 9  Since det C = 0, rank C = 3 so the system is controllable. The eigenvalues of A are given by    1 − λ 2 0    0 −1 − λ 0  = (1 − λ)   −3 −3 −2 − λ       −(1 + λ) 0    −3 −(2 + λ)   = (1 − λ)(1 − λ)(2 + λ) = 0 so that the eigenvalues are λ1 = −2, λ2 = −1, λ3 = 1. The system is therefore unstable with λ3 = 1 corresponding to the unstable mode. The corresponding eigenvectors of A are given by (A − λi I)ei = 0 and are readily determined as e1 = [0 0 1]T e2 = [1 − 1 0]T e3 = [1 0 − 1]T To determine the control law to relocate λ3 = 1 at −5 we need to determine the eigenvector v3 of AT corresponding to λ3 = 1. This is readily obtained as v3 = [1 1 0]T c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition Thus, the required control law is u(t) = KvT3 x(t) = K[1 1 0]T x(t) 6 (−5) − 1 p3 − λ3 ⎡ ⎤ = = −6 = where K = T 1 v3 b 0 [1 1 0] ⎣ 1 ⎦ 0 So u(t) = −6(x1 (t) + x2 (t))  31(a)  Let x [ x1  T  x2 ] then state − space model is   −2 = 0  x y = cT x ⇒ y = [ 1 0 ] 1 x2  ẋ1 ẋ = Ax + bu ⇒ ẋ2  (b) G(s) =  Y (s) U (s)  4 1      x1 1 u + 1 x2  = cT (sI − A)−1 b     s + 2 −4   = (s + 2)(s − 1) det(sI − A) = Δ =  0 s − 1  s−1 4 adj(sI − A) = 0 s+2   s+3 1 1 s−1 4 = G(s) = [1 0] 1 0 s+2 Δ (s + 2)(s − 1) System has positive pole s = 1 and is therefore is unstable. (c) u(t) = r(t) − ky(t) ⇒ ẋ = Ax + b(r − kcT x) = (A − kbcT )x + br   −2 − k 1 4 [1 0] = −k −k 1 1   λ + 2 + k 4  =0⇒ ⇒ Eigenvalues given by  −k λ − 1  −2 (A − kbc ) = 0  4 1  T  λ2 + (k + 1)λ + 3k − 2 = 0 so system is stable if and only if 3k − 2 > 0 or k >  c Pearson Education Limited 2011   2 3    361  362  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  (d) r(t) = H(t) ⇒ R(s) = Since k >  2 3  1 s  ⇒ Y(s) = cT (sI − A + kbcT )−1 b 1s  , system stable, the ﬁnal value theorem gives  lim y(t) = lim sY(s) = lim [cT (sI − A + kbcT )−1 b] = −cT (A − kbcT )−1 b s→0 s→0  −1  1 1 −4 −2 − k 4 1 = −[1 0] = −[1 0] −k 1 1 3k − 2 k −2 − k 3 = 3k − 2  t→∞  Thus, lim y(t) = 1 if and only if t→∞  32(a)  =1⇒k=  5 3  Overall closed loop transfer function is G(s) =  32(b)  3 3k−2  1+  K s(s+1) K s(s+1) (1 +  K1 s)  =  K s2 + s(1 + KK1 ) + K  Assuming zero initial conditions step response x(t) is given by X(s) = G(s)L{1.H(t)} =  K s[s2 + s(1 + KK1 ) + K]  wn + 2ξwn s + w2n ] 1 s + 2ξwn = − 2 s s + 2ξwn s + w2n  (s + ξwn ) + ξwn 1 = − s (s + ξwn )2 + [w2n (1 − ξ2 )]  (s + ξwn ) + ξwn 1 = − s (s + ξwn )2 + w2d   ξ sin wd t , t ≥ 0. giving x(t) = L−1 {X(s)} = 1 − e−ξwn t cos wd t + 1 − ξ2 =  32(c)  s[s2  The peak time tp is given by the solution of  dx = e−ξwn t ξwn − dt = e−ξwn t  wn 1 − ξ2  ξwd 1 − ξ2  dx  =0 dt t=tp  cos wd t   ξ2 wn 1 − ξ2  sin wd t  c Pearson Education Limited 2011   + wd sin wd t    Glyn James, Advanced Modern Engineering Mathematics, 4th Edition Thus, tp given by the solution of e−ξwn tp  wn 1 − ξ2  sin wd tp = 0  i.e. sin wd tp = 0 Since the peak time corresponds to the ﬁrst peak overshoot wd tp = π or tp =  π wd  The maximum overshoot Mp occurs at the peak time tp . Thus Mp = x(tp ) − 1 = e  =  ξw π − wn  d cos π  ξwn π − e wd  =e  ξ  +  −ξπ/  1 − ξ2   sin π  √  1−ξ 2  π  We wish Mp to be 0.2 and tp to be 1s, thus e−ξπ/  √  1−ξ 2  = 0.2 giving ξ = 0.456  and π = 1 giving wd = 3.14 wd wd = 3.53 from which we deduce that 1 − ξ2  tp = Then it follows that wn =  K = w2n = 12.5 and K1 =  32(d)  2wn ξ − 1 = 0.178. K  The rise time tr is given by the solution of  x(tr ) = 1 = 1 − e−ξwn tr cos wd tr +  ξ 1 − ξ2  Since e−ξwn tr = 0 cos wd tr +  ξ 1 − ξ2  sin wd tr = 0  c Pearson Education Limited 2011   sin wd tr    363  364  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  giving tan wd tr = −  or  1 − ξ2 ξ tr =   1 tan−1 − wd  1 − ξ2 π − 1.10 = = 0.65s. ξ wd  The response x(t) in (b) may be written as x(t) = 1 −  so the curves 1 ±  e−ξwn t 1 − ξ2  e−ξwn t 1 − ξ2   sin wα t + tan−1  1 − ξ2  ξ  are the envelope curves of the transient response to  1 . The settling time ts may ξwn be measured in terms of T. Using the 2% criterion ts is approximately 4 times a unit step input and have a time constant T =  the time constant and for the 5% criterion it is approximately 3 times the time constant. Thus, 4 = 2.48s ξwn 3 = 1.86s 5% criterion : ts = 3T = ξwn 2% criterion : ts = 4T =  Footnote :  This is intended to be an extended exercise with students being  encouraged to carry out simulation studies in order to develop a better understanding of how the transient response characteristics can be used in system design.  33  As for Exercise 32 this is intended to be an extended problem supported by  simulation studies. The following is simply an outline of a possible solution. Figure 5.67(a) is simply a mass-spring damper system represented by the diﬀerential equation d2 x dx + K1 x = sin wt M1 2 + B dt dt Assuming that it is initially in a quiescent state taking Laplace transforms X(s) =  M1  s2  w 1 · 2 + Bs + K1 s + w2  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  365  The steady state response will be due to the forcing term and determined by the αs + β term in the partial fractions expansion of X(s) . Thus, the steady state s2 + w2 response will be of the form A sin(wt + δ) ; that is, a sinusoid having the same frequency as the forcing term but with a phase shift δ and amplitude scaling A. In the situation of Figure 5.67(b) the equations of motion are d2 x dx + K2 (y − x) + sin wt = −K1 x − B 2 dt dt d2 y M2 2 = −K2 (y − x) dt  M1  Assuming an initial quiescent state taking Laplace transforms gives [M1 s2 + Bs + (K1 + K2 )]X(s) − K2 Y(s) = w/(s2 + w2 ) −K2 X(s) + (s2 M2 + K2 )Y(s) = 0 Eliminating Y(s) gives X(s) =  w(s2 M2 + K2 ) (s2 + w2 )p(s)  where p(s) = (M1 s2 + Bs + K1 + K2 )(s2 M2 + K2 ) . Because of the term (s2 + w2 ) in the denominator x(t) will contain terms in sin wt and cos wt . However, if (s2 M2 + K2 ) exactly cancels (s2 + w2 ) this will be avoided. Thus choose K2 = M2 w2 . This does make practical sense for if the natural frequency of the secondary system is equal to the frequency of the applied force then it may resonate and therefore damp out the steady state vibration of M1 . It is also required to show that the polynomial p(s) does not give rise to any undamped oscillations. That is, it is necessary to show that p(s) does not possess purely imaginary roots of the form jθ, θ real, and that it has no roots with a positive real part. This can be checked using the Routh–Hurwitz criterion. To examine the motion of the secondary mass M2 solve for Y(s) giving Y(s) =  (s2  K2 w + w2 )p(s)  Clearly due to the term (s2 + w2 ) in the denominator the mass M2 possesses an undamped oscillation. Thus, in some sense the secondary system has absorbed the energy produced by the applied sinusoidal force sin wt . c Pearson Education Limited 2011   366  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  34  Again this is intended to be an extended problem requiring wider exploration  by the students. The following is an outline of the solution.  34(a)  Students should be encouraged to plot the Bode plots using the steps  used in Example 5.65 of the text and using a software package. Sketches of the magnitude and phase Bode plots are given in the ﬁgures below.  34(b)  With unity feedback the ampliﬁer is unstable. Since the −180◦ crossover  gain is greater than 0dB (from the plot it is +92dB) .  34(c)  Due to the assumption that the ampliﬁer is ideal it follows that for 1 must be 92dB (that is, the plot is eﬀectively marginal stability the value of β lowered by 92dB) . Thus 20 log  34(d)  1 = 92 β  92 1 ⇒β = antilog β 20  2.5 × 10−5  From the amplitude plot the eﬀective 0dB axis is now drawn through  the 100dB point. Comparing this to the line drawn through the 92dB point, corresponding to marginal stability, it follows that Gain margin = −8dB and Phase margin = 24◦ .  34(e) G(s) =  K (1 + sτ1 )(1 − sτ2 )(1 + sτ3 )  Given low frequency gain K = 120dB so 20 log K = 120 ⇒ K = 106 Ti =  1 where fi is the oscillating frequency in cycles per second of the pole. fi c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  367  Since 1MHz = 10 cycles per second 1 1 = 6 since f1 = 1MHz f1 10 1 1 = since f2 = 10MHz τ2 = f2 10.106 1 1 = since f3 = 25MHz τ3 = f3 25.106 τ1 =  Thus, G(s) = =  106 (1 +  s 106 )(1  (s +  106 )(s  +  s 10.106 )(1 24  +  s 25.106 )  250.10 + 107 )(s + 52 .107 )  The closed loop transfer function G1 (s) is  G(s) =  34(f)  G(s) 1 + βG(s)  The characteristic equation for the closed loop system is (s + 106 )(s + 107 )(s + 52 .107 ) + β25.1025 = 0  or s3 + 36(106 )s2 + (285)1012 s + 1019 (25 + 25β106 ) = 0 ↓ ↓ ↓ A1 A2 A3 By Routh–Hurwitz criterion system stable provided A1 > 0 and A1 A2 > A3 . If β = 1 then A1 A2 < A3 and the system is unstable as determined in (b). For marginal stability A1 A2 = A3 giving β = 1.40−5 (compared with β = 2.5.10−5 using the Bode plot).  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  368  Magnitude vs Frequency Plot 120  Data margin – 8 dB  100 80 60 Gain dB  1/β = 92 dB 40 20 0 Corresponds to 180° phase lag  To phase plot  1  10  25  Log freq. MHz  Phase vs Frequency Plot 0°  – 90°  – 180°  Phase margin 24°  16 MHz – 270°  1  10  c Pearson Education Limited 2011   25  ln freq. MHz  6 The  Transform  Z  Exercises 6.2.3 1(a) F(z) =  ∞  (1/4)k k=0  1(b) F(z) =  zk  ∞  3k k=0  1(c) F(z) =  ∞  (−2)k k=0  1(d) F(z) =  zk  zk  ∞  −(2)k k=0  zk  =  4z 1 = 1 − 1/4z 4z − 1  =  z 1 = 1 − 3/z z−3  =  z 1 = 1 − (−2)/z z+2  if | z |> 2  z 1 =− 1 − 2/z z−2  if | z |> 2  =−  if | z |> 1/4  if | z |> 3  1(e) Z{k} =  z (z − 1)2  if | z |> 1  from (6.6) whence Z{3k} = 3  2  z (z − 1)2  if | z |> 1  k  uk = e−2ωkT = e−2ωT  whence U(Z) =  z z − e−2ωT  c Pearson Education Limited 2011   370  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Exercises 6.3.6 3 Z{sin kωT} = = 4  z z 1 1 − jωT 2j z − e 2j z − e−jωT  z2  z sin ωT − 2z cos ωT + 1   k 2z 1 }= Z{ 2 2z − 1  so  2 1 2z = 2 × 3 z 2z − 1 z (2z − 1)  Z{yk } = Proceeding directly Z{yk } =  ∞  xk−3  zk  k=3  5(a)    =  1 Z − 5   =  ∞  xr 2 1 } {x = = × Z k zr+3 z3 z2 (2z − 1) r=0  r ∞   −1 r=0  5z  =  5z 5z + 1  | z |>  5(b) {cos kπ} = (−1)k so Z {cos kπ} = 6   k 1 Z 2  z z+1  =  By (3.5) Z (ak ) = so Z (kak−1 ) =  | z |> 1  2z 2z − 1  z z−a z (z − a)2  c Pearson Education Limited 2011   1 5  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition thus Z (kak ) = whence   k 1 Z k 2  az (z − a)2  =  2z (2z − 1)2  7(a) sinh kα = so  1 Z {sinh kα} = 2  1 α k 1 −α k (e ) − (e ) 2 2   z z sinh α z = 2 − α −α z−e z−e z − 2z cosh α + 1  7(b) cosh kα =  1 α k 1 −α k (e ) + (e ) 2 2  then proceed as above.  8(a)    k  uk = e−4kT = e−4T ;  8(b)  Z {uk } =  z z − e−4T   1  j kT e − e−j kT 2j  z z sin T z = 2 − j T −j T z−e z−e z − 2z cos T + 1 uk =  Z {uk } =  1 2j  8(c) uk =   1  j 2kT e + e−j 2kT 2  then proceed as in 8(b) to give Z{uk } =  9  z2  z(z − cos 2T) − 2z cos 2T + 1  Initial value theorem: obvious from deﬁnition.  c Pearson Education Limited 2011   371  372 9  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition Final value theorem (1 − z−1 )X(z) =  ∞  xr − xr−1  zr  r=0  = x0 + As z → 1 and if  lim r→∞  x2 − x1 x1 − x0 xr − xr−1 + + . . . + + ... z z2 zr xr exists, then lim (1 − z−1 )X(z) = lim xr r→∞  z→1  10  Multiplication property (6.19): Let Z {xk } = ∞  ak xk  Z a xk = k  k=0  10  zk  ∞ xk k=0 k = X(z) then z  = X(z/a)  Multiplication property (6.20) −z  ∞  ∞  k=0  k=0   kxk d d  xk = = Z {kxk } X(z) = −z dz dz zk zk  The general result follows by induction.  Exercises 6.4.2 11(a)  11(b)  11(c)  z ; z−1  from tables uk = 1  z z = ; z+1 z − (−1)  z ; z − 1/2  from tables uk = (−1)k  from tables uk = (1/2)k  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 11(d) 1 z 1 z = ←→ (−1/3)k 3z + 1 3 z + 1/3 3 11(e) z ; z−j 11(f)  from tables uk = (j )k  √ z z √ = √ ←→ (−j 2)k z+j 2 z − (−j 2)  11(g) 1 1 z = ←→ z−1 zz−1    0; 1;  k=0 k>0  using ﬁrst shift property.  11(h)   z+2 1 z 1; =1+ ←→ k−1 (−1) ; z+1 zz+1  1; k=0 = k+1 (−1) ; k>0  k=0 k>0  12(a) Y(z)/z = so Y(z) =  12(b) 1 Y(z) = 7    1 1 1 1 − 3z−1 3z+2   1 z 1 1 z − ←→ 1 − (−2)k 3z−1 3z+2 3  z z − z − 3 z + 1/2   ←→   1 k (3) − (−1/2)k 7  12(c) Y(z) =  1 z 1 z 1 1 + ←→ + (−1/2)k 3 z − 1 6 z + 1/2 3 6 c Pearson Education Limited 2011   373  374  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  12(d) 2 z 2 z 2 2 − ←→ (1/2)k − (−1)k 3 z − 1/2 3 z + 1 3 3  Y(z) =  = 12(e)   z z − z−j z − (−j )  z z − z − ej π/2 z − e−j π/2  1 (ej π/2 )k − (e−j π/2 )k = sin kπ/2 ←→ 2j  1 Y(z) = 2j 1 = 2j  12(f)  2 2 (1/2)k + (−1)k+1 3 3    z √ √   Y(z) =  z − ( 3 + j) z − ( 3 − j)   z z 1 √ √ − = 2j z − ( 3 + j ) z − ( 3 − j )  z z 1 − = 2j z − 2ej π/6 z − 2e−j π/6  1 2k ej kπ/6 − 2k e−j kπ/6 = 2k sin kπ/6 ←→ 2j  12(g) Y(z) =  z 1 z 5 1 z − + 2 2 (z − 1) 4z−1 4z−3 ←→   5 1 1 − 3k k+ 2 4  12(h) Y(z)/z = so Y(z) =  z (z −  1)2 (z2  − z + 1)  =  1 1 − 2 2 (z − 1) z −z+1  ⎛  ⎞  z 1 ⎜ ⎜ √ − (z − 1)2 3j ⎝  ⎟ z z √ √ ⎟ − 1 + 3j 1 − 3j ⎠ z− z− 2 2  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  =  1 z −√ 2 (z − 1) 3j  375   z z − z − ej π/3 z − e−j π/3  2 2 ←→ k − √ sin kπ/3 = k + √ cos(kπ/3 − 3π/2) 3 3 13(a) Y(z) =  ∞  xk k=0  zk  =  2 1 + 7 z z  whence x0 = 0, x1 = 1, x2 = x3 = . . . = x6 = 0, x7 = 2 and xk = 0, k > 7; giving Y(z) ↔ {0, 1, 0, 0, 0, 0, 0, 2, · · ·} 13(b)  Proceed as in Exercise 13(a) to give Y(z) ↔ {1, 0, 3, 0, 0, 0, 0, 0, 0, −2, · · ·}  13(c) Observe that 3z + z2 + 5z5 1 3 =5+ 3 + 4 5 z z z and proceed as in Exercise 13(a) to give Y(z) ↔ {5, 0, 0, 1, 3, · · ·} 13(d) Y(z) =  1 1 z + + z2 z3 z + 1/3  ←→ {0, 0, 1, 1} + {(−1/3)k } 13(e) 1 1/2 3 + 2− z z z + 1/2  1 0, k = 0 ←→ {1, 3, 1} − 2 (−1/2)k , k ≥ 1 ⎧ ⎧ 1, k = 0 1, k = 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ 5/2, k = 1 ⎨ 5/2, k = 1 = 5/4, k = 2 = 5/4, k = 2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ − 1 (−1/2)k−3 , k ≥ 3 ⎩ − 1 (−1/2)k−1 , k ≥ 3 2 8 Y(z) = 1 +  c Pearson Education Limited 2011   376  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  13(f) Y(z) =   2 1 1 − + z − 1 (z − 1)2 z−2  0, k = 0 1 − 2(k − 1) + 2k−1 , k ≥ 1  0, k = 0 = 3 − 2k + 2k−1 , k ≥ 1  ←→  13(g) Y(z) =  ←→  1 2 − z−1 z−2  0, k = 0 2 − 2k−1 , k ≥ 1  Exercises 6.5.3 14(a)  If the signal going into the left D-block is wk and that going into the right  D-block is vk , we have yk+1 = vk ,  1 vk+1 = wk = xk − vk 2  so 1 yk+2 = vk+1 = xk − vk 2 1 1 = xk − vk = xk − yk+1 2 2 that is, 1 yk+2 + yk+1 = xk 2 14(b)  Using the same notation yk+1 = vk ,  1 1 vk+1 = wk = xk − vk − yk 4 5  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition Then 1 1 yk+2 = xk − yk+1 − yk 4 5 or 1 1 yk+2 + yk+1 + yk = xk 4 5 15(a) z2 Y(z) − z2 y0 − zy1 − 2(zY(z) − zy0 ) + Y(z) = 0 with y0 = 0, y1 = 1 Y(z) =  z (z − 1)2  so yk = k, k ≥ 0.  15(b)  Transforming and substituting for y0 and y1 , Y(z)/z =  2z − 15 (z − 9)(z + 1)  so Y(z) =  3 z 17 z − 10 z − 9 10 z + 1  thus yk =  3 k 17 9 − (−1)k , k ≥ 0 10 10  15(c) Transforming and substituting for y0 and y1 , Y(z) = 1 = 4j thus yk =  z (z − 2j )(z + 2j )   z z − z − 2ej π/2 z − 2e−j π/2   1 k j kπ/2 2 e − e−j kπ/2 = 2k−1 sin kπ/2, k ≥ 0 4j c Pearson Education Limited 2011   377  378  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  15(d)  Transforming, substituting for y0 and y1 , and rearranging Y(z)/z =  so Y(z) = 2  6z − 11 (2z + 1)(z − 3)  z z + z + 1/2 z − 3  thus yk = 2(−1/2)k + 3k , k ≥ 0 16(a) 6yk+2 + yk+1 − yk = 3,  y0 = y 1 = 0  Transforming with y0 = y1 = 0, (6z2 + z − 1)Y(z) = so Y(z)/z = and Y(z) =  3z z−1  3 (z − 1)(3z − 1)(2z + 1)  9 z 2 z 1 z − + 2 z − 1 10 z − 1/3 5 z + 1/2  Inverting yk = 16(b)  9 1 2 − (1/3)k + (−1/2)k 2 10 5  Transforming with y0 = 0, y1 = 1, (z2 − 5z + 6)Y(z) = z + 5  whence Y(z) =  z z−1  7 z z 5 z + −6 2z−1 2z−3 z−2  so yk =  5 7 k + (3) − 6 (2)k 2 2  16(c) Transforming with y0 = y1 = 0, (z2 − 5z + 6)Y(z) =  z z − 1/2  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition so Y(z) =  z 2 z 2 z 4 − + 15 z − 1/2 3 z − 2 5 z − 3  whence yn = 16(d)  4 2 2 (1/2)k − (2)k + (3)k 15 3 5  Transforming with y0 = 1, y1 = 0, (z2 − 3z + 3)Y(z) = z2 − 3z +  so  so  z z−1  z z − 2 Y(z) = z − 1 z − 3z + 3 ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ ⎬ z 1 ⎨ z z √ − √ −√ = z−1 3j ⎪ 3 − 3j ⎪ ⎪ ⎪ ⎩ z − 3 + 3j ⎭ z− 2 2   z z z 1 √ √ = − −√ z−1 3j z − 3ejπ/6 z − 3e−jπ/6 √ 2 √ ejnπ/6 − e−jnπ/6 = 1 − 2( 3)n−1 sin nπ/6 yn = 1 − √ ( 3)k 2j 3  16(e) Transforming with y0 = 1, y1 = 2, (2z2 − 3z − 2)Y(z) = 2z2 + z + 6 so  z +z Y(z) = z−2 =    z z + 2 (z − 1) z−1  z+5 (z − 1)2 (2z + 1)(z − 2)    12 z 2 z z z − − −2 5 z − 2 5 z + 1/2 z − 1 (z − 1)2  so yn =  12 n 2 (2) − (−1/2)n − 1 − 2n 5 5  16(f) Transforming with y0 = y1 = 0, (z2 − 4)Y(z) = 3  z z −5 2 (z − 1) z−1  c Pearson Education Limited 2011   379  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  380 so  z 1 z 1 z z − − − 2 z − 1 (z − 1) 2z−2 2z+2  Y(z) = and  1 1 yn = 1 − n − (2)n − (−2)n 2 2  17(a)  Write the transformed equations in the form      z − 3/2 1 c(z) zC0 = zE0 −0.21 z − 1/2 e(z)  Then    c(z) e(z)    1 = 2 z − 2z + 0.96  Solve for c(z) as c(z) = 1200    z − 1/2 −1 0.21 z − 3/2    zC0 zE0    z z + 4800 z − 1.2 z − 0.8  and Ck = 1200(1.2)k + 4800(0.8)k This shows the 20% growth in Ck in the long term as required. (b) Then Ek = 1.5Ck − Ck+1 = 1800(1.2)k + 7200(0.8)k − 1200(1.2)k+1 − 4800(0.8)k+1 Diﬀerentiate wrt k and set to zero giving 0.6 log(1.2) + 5.6x log(0.8) = 0 where x = (0.8/1.2)  k  Solving, x = 0.0875 and so k=  log 0.0875 = 6.007 log(0.8/1.2)  The nearest integer is k = 6, corresponding to the seventh year in view of the labelling, and C6 = 4841 approximately. 18  Transforming and rearranging Y(z)/z =  1 z−4 + (z − 2)(z − 3) (z − 1)(z − 2)(z − 3)  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition so Y(z) =  z 1 z 1 z + − 2z−1 z−2 2z−3  thus yk = 19  1 1 + 2k − 3k 2 2  Ik = Ck + Pk + Gk = aIk−1 + b(Ck − Ck−1 ) + Gk = aIk−1 + ba(Ik−1 − Ik−2 ) + Gk  so Ik+2 − a(1 + b)Ik+1 + abIk = Gk+2 Thus substituting 1 Ik+2 − Ik+1 + Ik = G 2 Using lower case for the z transform, we obtain 1 z (z2 − z + )i(z) = (2z2 + z)G + G 2 z−1 ⎡  whence  ⎢ i(z)/z = G ⎣ ⎡  ⎤ 1 1 z2 − z + 2  +  2 ⎥ ⎦ z−1 ⎤  ⎢ 2 + = G⎣ z−1  so  Thus  1 ⎥ 1+j 1−j ⎦ (z − )(z − ) 2 2 ⎫⎤ ⎧ ⎡ ⎪ ⎪ ⎪ ⎪ ⎬⎥ ⎨ ⎢ z 2 z z ⎥ ⎢ + − i(z) = G ⎣2 ⎦ 1 1 ⎪ z − 1 2j ⎪ ⎪ z − √ e−j π/4 ⎪ ⎭ ⎩ z − √ ej π/4 2 2 #  &  % 2 1 k $ j kπ/4 (√ ) e − e−j kπ/4 Ik = G 2 + 2j 2 ( ' k  1 sin kπ/4 = 2G 1 + √ 2  c Pearson Education Limited 2011   381  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  382 20  Elementary rearrangement leads to in+2 − 2 cosh α in+1 + in = 0  with cosh α = 1 + R1 /2R2 . Transforming and solving for I(z)/z gives I(z)/z =  zi0 + (i1 − 2i0 cosh α) (z − eα )(z − e−α )  # α & i0 e + (i1 − 2i0 cosh α) i0 e−α + (i1 − 2i0 cosh α) 1 = − 2 sinh α z − eα z − e−α Thus ik =  (i0 eα + (i1 − 2i0 cosh α))enα − (i0 e−α + (i1 − 2i0 cosh α))e−nα 2 sinh α =  1 {i1 sinh nα − i0 sinh(n − 1)α} sinh α  Exercises 6.6.5 21  Transforming in the quiescent state and writing as Y(z) = H(z)U(z) , then  21(a) H(z) =  z2  1 − 3z + 2  z2  z−1 − 3z + 2  21(b) H(z) = 21(c) H(z) = 22  z3  1 + 1/z − z2 + 2z + 1  For the ﬁrst system, transforming from a quiescent state, we have (z2 + 0.5z + 0.25)Y(z) = U(z)  The diagram for this is the standard one for a second-order system and is shown in Figure 6.1 and where Y(z) = P(z) , that is yk = pk . c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  383  Figure 6.1: The block diagram for the basic system of Exercise 22. Transforming the second system in the quiescent state, we obtain (z2 + 0.5z + 0.25)Y(z) = (1 − 0.6)U(z) Clearly (z2 + 0.5z + 0.25)(1 − 0.6z)P(z) = (1 − 0.6z)U(z) indicating that we should now set Y(z) = P(z) − 0.6zP(z) and this is shown in Figure 6.2.  Figure 6.2: The block diagram for the second system of Exercise 22.  c Pearson Education Limited 2011   384  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  23(a)  Yδ (z)/z =  so  z 1 z 1 − 2 z + 1/4 2 z + 1/2  Yδ (z) = yk = 23(b)  1 (4z + 1)(2z + 1)  1 1 (−1/4)k − (1/2)k 2 2  Yδ (z)/z =  z2  z − 3z + 3  whence √ 3 + 3j √ Yδ (z) = 2 3j so  √ z z 3 − 3j √ √ − √ 2 3j (3 + 3j ) (3 − 3j ) z− z− 2 2  √ √ 3 + 3j √ k j kπ/6 3 − 3j √ k −j kπ/6 √ ( 3) e ( 3) e − √ yk = 2 3j 2 3j ( '√ √ k 3 1 sin kπ/6 + cos kπ/6 = 2( 3) 2 2 √ = 2( 3)k sin(k + 1)π/6  23(c) Yδ (z)/z = so Yδ (z) = then yk =  z (z − 0.4)(z + 0.2)  1 z 2 z + 3 z − 0.4 3 z + 0.2 2 1 (0.4)k + (−0.2)k 3 3  23(d) Yδ (z)/z = so Yδ (z) =  5z − 12 (z − 2)(z − 4)  z z +4 z−2 z−4  and yk = (2)k + (4)k+1 c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  385  24(a) Yδ (z) = =  yk =  z2  1 − 3z + 2  1 1 − z−2 z−1 0, k = 0 2k−1 − 1, k ≥ 1  24(b) Yδ (z) = so   yk =  25  0, k−1  2  1 z−2  k=0 , k≥1  Examining the poles of the systems, we ﬁnd  25(a)  Poles at z = −1/3 and z = −2/3, both inside | z |= 1 so the system is  stable.  25(b)  Poles at z = −1/3 and z = 2/3, both inside | z |= 1 so the system is  stable. √ 25(c) Poles at z = 1/2 ± 1/2j , | z |= 1/ 2 , so both inside | z |= 1 and the system is stable.  25(d) Poles at z = −3/4 ± system is unstable.  √  17/4, one of which is outside | z |= 1 and so the  25(e) Poles at z = −1/4 and z = 1 thus one pole is on | z |= 1 and the other is inside and the system is marginally stable.  c Pearson Education Limited 2011   386 26  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition To use the convolution result, calculate the impulse response as yδ,k − (1/2)k .  Then the step response is  yk =  k   1 × (1/2)  k−j  = (1/2)  k  j=0  k   1 × (2)j = (1/2)k  j=0  1 − (2)k+1 1−2  = (1/2)k (2k+1 − 1) = 2 − (1/2)k Directly, Y(z)/z =  2 1 z = − (z − 1/2)(z − 1) z − 1 z − 1/2  so yk = 2 − (1/2)k 27 t k k k = − ⇒ f(t) = k − ke− τ 1 s(sτ + 1) s s+ τ & # T z kz(1 − e− τ ) k − = ⇒ G(z) = k T z − 1 z − e− Tτ (z − 1)(z − e− τ )  G(s) =  Characteristic equation is 1 + G(z) = 0 ⇒ (z − 1)(z − e− τ ) + kz(1 − e− τ ) = 0 T  T  T τ = 0, where K = k(1 − e−a ) − (1 + e−a )  ⇒ z2 + [k(1 − e−a ) − (1 + e−a )]z + e−a = 0, where a = ⇒ z2 + Kz + e−a Using Jury's procedure: F(z) = z2 + Kz + e−a  F(1) = 1 + K + e−a = k(1 − e−a ) > 0 since k > 0, (1 − e−a ) > 0 (−1)2 F(−1) = 1 − Kz + e−a > 0 ⇒ 2(1 + e−a ) − k(1 − e−a ) > 0   T 2(1 + e−α ) a = 2coth ⇒k< = 2coth −a (1 − e ) 2 2τ ) ) ) 1 e−a )) = 1 − e−2a > 0 Δ1 = )) −a e 1 ) T  Thus system is stable if and only if 0 < k < 2coth 2τ  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 28  Substituting yn+1 − yn + Kyn−1 = K/2n  or yn+2 − yn+1 + Kyn = K/2n+1 Taking z transforms from the quiescent state, the characteristic equation is z2 − z + K = 0 with roots  1 1√ 1 1√ + 1 − 4K and z2 = − 1 − 4K 2 2 2 2 For stability, both roots must be inside | z |= 1 so if K < 1/4 then z1 =  z1 < 1 ⇒ and z2 > −1 ⇒  1 1√ + 1 − 4K < 1 ⇒ K > 0 2 2 1 1√ 1 − 4K > −1 ⇒ k > −2 − 2 2  If K > 1/4 then  1 1√ +j 4K − 1 |2 < 1 ⇒ K < 1 2 2 The system is then stable for 0 < K < 1. |  When k = 2/9, we have 2 1 yn+2 − yn+1 + yn = 9 9 Transforming with a quiescent initial state, 1 z 2 (z2 − z + )Y(z) = 9 9 z − 1/2 so  # & 1 1 Y(z) = z 9 (z − 1/2)(z − 1/3)(z − 2/3) =2  z z z +2 −4 z − 1/3 z − 2/3 z − 1/2  which inverts to yn = 2(1/3)n + 2(2/3)n − 4(1/2)n c Pearson Education Limited 2011   387  388  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  29 z2 + 2z + 2 = (z − (−1 + j ))(z − (−1 + j )) establishing the pole locations. Then Yδ (z) = So since (−1 ± j ) =  z 1 z 1 − 2j z − (−1 + j ) 2j z − (−1 − j )  √ ±3j π/4 2e etc., √ yk = ( 2)k sin 3kπ/4  Exercises 6.8.3 30(a) # zI − A = (zI − A)−1 k  −1  A =Z  {z(zI − A)  −1  z −1 −4 z  &  1 = (z − 2)(z + 2) −1  }=Z  #1  1/2 2 z−2 z z−2   # 1 k 2 2 = 4 4 30(b)  # zI − A =  #  z 4  1 z  '  & =  1/2 z−2 1 z−2  + −  1/2 z+2 1 z+2  & 1/2 1 z 1 z + 12 z+2 4 z−2 − 4 z+2 z z 1 z − z+2 2 z−2 + z+2 & & # 1 2 −1 k + (−2) 2 −4 2  z + 1 −3 −3 z + 1  1/4 z−2 1/2 z−2  − +  &  & # 1 z+1 3 (zI − A) = 3 z+1 (z + 4)(z − 2) ' ( 1/2 1/2 1/2 −1/2 + + z+4 z−2 z+4 z−2 = 1/2 1/2 1/2 1/2 − z+4 + z−2 + z−2 z+2 # 1 z z z z & + 12 z−2 − 21 z+4 + 12 z−2 k −1 −1 −1 2 z+4 A = Z {z(zI − A) } = Z z z 1 z 1 z − 12 z+4 + 12 z−2 2 z−2 + 2 z−2 & &  # # 1 1 −1 1 1 k k + (2) (−4) = −1 1 1 1 2 −1  c Pearson Education Limited 2011   1/4 z+2 1/2 z+2  (  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition & z + 1 −1 zI − A = 0 z+1 & # 1 # & 1 1 z+1 1 −1 z+1 (z+1)2 = (zI − A) = 1 0 z+1 (z + 1)2 0 z+1 & # z & # z 1 −k 2 = (−1)k Ak = Z−1 {z(zI − A)−1 } = Z−1 z+1 (z+1) z 0 1 0 z+1 #  30(c)  31  Taking x1 = x and x2 = y we can express equations in the form # x(k + 1) =  & 4 x(k) with x(0) = [1 2] 1  −7 −8  The solution is given by # k  x(k + 1) = A x(0), A =  −7 −8  4 1  &  where Ak = α1 I + α1 A . That is, the solution is # x(k) =  α0 − 7α1 −8α1  4α1 α0 + α1  # & # & & α0 + α1 1 = 2 2α0 − 6α1  The eigenvalues of A are given by λ2 − 6λ + 25 = 0 so λ = 3 ± j4 or, in polar form, λ1 = 5ejθ, λ2 = 5e−jθ where θ = cos−1 (− 53 ) . Thus, α0 and α1 are given by 5k ejkθ = α0 + α1 5ejθ, 5k e−jkθ = α0 + α1 5e−jθ which are readily solved to give α0 = −  sin kθ (5)k sin(k − 1)θ 1 , α1 = (5)k sin θ 5 sin θ  c Pearson Education Limited 2011   389  390  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  # & (5)k 1 sin kθ − sin(k − 1)θ α0 + α1 = sin θ 5 & # 3 4 5 k 1 sin kθ + sin kθ + cos kθ = (5) 4 5 5 5  Then,  = (5)k [sin kθ + cos kθ] # & 6 (5)k −2(sin kθ cos θ − cos kθ sin θ) − sin kθ 2α0 − 6α1 = sin θ 5 = (5)k [2 cos(kθ)] Thus, solution is x(k) = (5)k [sin kθ + cos kθ] y(k) = (5)k [2 cos(kθ)] We have x(1) = 1 , y(1) = −6 , x(2) = −31 , y(2) = −14  # 32  A=  0 −0.16  1 −1  &  # zI − A = '  (zI − A)−1 = #  −1  A =Z  {z(zI − A)  −1  z+1 (z+0.2)(z+0.8) −0.16 (z+0.2)(z+0.8)  # }=  A x(0) = A [1 − 1] = k  1 (z+0.2)(z+0.8) z (z+0.2)(z+0.8)  T  (zI − A)  (  5 1 5 1 3 · z+0.2 − 3 · z+0.8 1 1 − 13 z+0.2 + 43 · z+0.8  1 bU(z) = (z + 0.2)(z + 0.8)  #  z+1 −0.16  c Pearson Education Limited 2011   &  5 5 k k 3 (−0.2) − 3 (−0.8) − 31 (−0.2)k + 43 (−0.8)k  − 31 (−0.2)k + 43 (−0.8)k 0.2 3.2 k k 3 (−0.2) − 3 (−0.8)  U(z) = Z{u(k)} = z/(z − 1) −1  &  4 1 k k 3 (−0.2) − 3 (−0.8) −0.8 0.8 k k 3 (−0.2) + 3 (−0.8)  # k  −1 z+1  4 1 1 1 3 · z+0.2 − 3 · z+0.8 −0.8 1 1 0.8 3 z+0.2 + 3 z+0.8  =  k  z 0.16  1 z  &  & # & z 1 1 z−1  &  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  391  & # z z+2 = (z + 0.2)(z + 0.8)(z − 1) z − 0.16 ' 5 z z 25 z ( − 2 z+0.2 + 10 9 z+0.8 + 18 z−1 = 8 7 z z z 1 2 z+0.2 − 9 z+0.8 + 18 z−1 # 5 25 & k − 2 (−0.2)k + 10 −1 −1 9 (−0.8) + 18 Z {(zI − A) bU(z)} = 1 8 7 k k 2 (−0.2) − 9 (−0.8) + 18 Thus, solution is x(k) = Ak x(0) + Z−1 {(zI − A)−1 bU(t)} ( ' 17 25 k − 6 (−0.2)k + 22 9 (−0.8) + 18 = 17.6 7 3.4 k k 6 (−0.2) − 9 (−0.8) + 18 33  Let x1 (k) = y(k), x2 (k) = x1 (k + 1) = y(k + 1) then the diﬀerence equation  may be written # x(k + 1) =  x1 (k + 1) x2 (k + 1)  &  # =  0 1  1 1  & #  & x1 (k) , x(0) = [0 1]T x2 (k)  √ √ & 1+ 5 1 5 0 1 its eigenvalues are λ1 = , λ2 = Taking A = 1 1 2 2 Ak = α1 I + α1 A where α0 and α1 satisfy #  √ √ k √ √ k 1 + 5 1 − 5 1 − 5 1 + 5 α1 , α1 = α0 + = α0 + 2 2 2 2 giving  √ k √ k& # 1 − 5 1 1 + 5 − α1 = √ 2 2 5 √ k √ k √ √ & # 1 1 + 5  5 − 1 1 − 5 1 + 5 + α0 = √ 2 2 2 2 5  Solution to the diﬀerence equation is & # # α0 y(k) = x(k) = α1 y(k + 1)  α1 α0 + α1  # & # & & α1 0 = 1 α0 + α1  √ k √ k& # 1 − 5 1 1 + 5 so y(k) = α1 = √ − 2 2 5 c Pearson Education Limited 2011   392  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  √ k √ k+1 & # 1 − 5 1 1 + 5 using above [Note that, y(k + 1) = α0 + α1 = √ − 2 2 5 values.] √ √ k 1 − 5 ( 1+2 5 )k+1 1 √ √ As k → ∞, → 0 and y(k + 1)/y(k) → = ( 5 + 1) 2 2 ( 1+ 5 )k 2  Exercises 6.9.3 & # & 0 0 1 B= A= 1 0 −2 #  34  &  #  #  &  ⎡  1 1 s −1 s + 2 1 ⎢ ⇒ (sI − A)−1 = sI − A = =⎣s 0 s+2 0 s s(s + 2) 0 & # 1 12 (1 − e−2T ) ⇒ G = e−AT = L−1 {(sI − A)−1 } = 0 e−2T &T # & # 1 # * T 1 −2T 0 − t 12 t + 14 e−2t At 2T + 4e = e Bdt = H= 1 −2t 1 −2T 1 1 0 −2e 0 2 − 2e 0  1 2  s  1 4  −  1 2  s+2 ⎥ ⎦ 1 s+2  &  Discretized form is: #  & # x1 [(k + 1)T] 1 = x2 [(k + 1)T] 0  ! 2 (1  − e−2T ) e−2T  &#  & #1 & x1 (kT) T + 14 (e−2T − 1) 2 + u(kT) 1 −2T x2 (kT) ) 2 (1 − e  In the particular case, when sampling period is T = 1, this reduces to #  & # x1 (k + 1) 1 = 0 x2 (k + 1)  0.432 0.135  &#  & & # x1 (k) 0.284 u(k) + 0.432 x2 (k)  In MATLAB the commands: A = [0, 1; 0, −2]; B = [0; 1]; [G, H] = c2d(A, B, 1) return 1.0000 0 0.2838 H= 0.4324 G=  0.4323 0.1353  which check with the answer. c Pearson Education Limited 2011   ⎤  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition #  35  (a)  0 A= −1  & # & 0 1 B= 1 −1  Using (6.89),  #  1 T G1 = (TA + I) = −T 1 − T # & 0 H1 = TB = T  &  giving the Euler discretized form of state-space model as & # 1 x1 [(k + 1)T] = x[(k + 1)T] = −T x2 [(k + 1)T] #  y(kT) = [ 1 (b)  T 1−T  &#  & # & x1 (kT) 0 u(kT) + T x2 (kT)  0 ] x(kT)  & & # 1 −1 s+1 1 −1 ⇒ (sI − A) = 2 s+1 s + s + 1 −1 s ⎡ ⎤ 1 1 (s + 2 ) + 2 1 √ √ ⎢ 1 2 3 2 1 2 3 2 ⎥ (s + ) + ( ) (s + ) + ( ⎢ 2 2 2 2 ) ⎥ ⇒ (sI − A)−1 = ⎢ ⎥ (s + 12 ) − 12 ⎣ ⎦ −1 #  s sI − A = 1  (s + 12 )2 + (  √ 3 2 2 )  ⇒ G = L−1 {(sI − A)−1 } √ √ ' 3 √1 sin( 3 T) cos( T) + T 2 2 3√ = e− 2 2 3 √ − 3 sin( 2 T)  (s + 12 )2 + (  √  3 2 2 )  √ ( √2 sin( 3 T) 2 √ 3 √ 3 1 √ cos( 2 T) − 3 sin( 23 T)  Since det(A) = 0 the matrix H is best determined using (6.95), −1  #  −1 B = (G − I) 0  &  H = (G − I)A √ √ ( ' T T 1 − e− 2 cos( 23 T) − √13 e− 2 sin( 23 T) √ ⇒H= T √2 e− 2 sin( 3 T) 2 3 giving the step-invariant discretized form of the state-space model as x[(k + 1)] = Gx(kT) + Hu(kT) y(kT) = [ 1  0 ]x(kT)  c Pearson Education Limited 2011   393  394  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  36(a)  The eigenvalues of the matrix A are given by ) ) ) )λ + 1 −1 )=0 det(λI − A) = 0 ⇒ )) 1 λ + 2) √ √ 3 3 ⇒ λ2 + 3λ + 1 = 0 ⇒ λ1 = − + j 3, λ2 = − − j 3 2 2  Both eigenvalues have negative real parts so the matrix A represents a stable system. (b) From (6.89), the corresponding Euler discretized state matrix Ad is given by #  1−T T Ad = G1 = I + TA = −T 1 − 2T (c)  &  The eigenvalues of the matrix Ad are given by ) )λ − 1 + T ) ) T  ) ) −T )=0 λ − 1 + 2T )  ⇒ λ2 + (3T − 2)λ + (3T2 − 3T + 1) = 0 Let F(λ) = λ2 + (3T − 2)λ + (3T2 − 3T + 1) then F(1) = 1 + (3T − 2) + (1 − 3T + 3T2 ) = 3T2 > 0 if T > 0 (since T is non-negative by deﬁnition). (−1)2 F(−1) = 1 − ((3T − 2) + 1 − 3T + 3T2 = 3T2 − 6T + 4 > 0 all T. Taking a1 = (3T − 2) and a0 = (1 − 3T + 3T2 ) gives F(λ) = λ2 + a1 λ + a0 leading to Jury table: 1 a0 Δ1 = 1 − a20 a1 (1 − a0 )  a1 a1  a0 1  a1 (1 − a0 ) 1 − a20  Δ2 = (1 − a20 )2 − a21 (1 − a0 )2 = (1 − a0 )2 [(1 + a0 )2 − a21 ] with Δ1 > 0 if 1 − a20 > 0 ⇒ |a0 | < 1 and Δ2 > 0 if [(1 + a0 )2 − a21 ] > 0 ⇒ 1 + a0 > |a1 | or a0 > −1 + a1 and a0 > −1 − a1 . In terms of T these conditions become: c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  395  Δ1 > 0 ⇒ 3T2 − 3T + 1 < 1 ⇒ T(T − 1) < 0 ⇒ T < 1 Δ2 > 0 ⇒ 3T2 − 3T + 1 > 3T − 3 ⇒ 3T2 − 6T + 4 > 0 all T (as above); and 3T2 − 3T + 1 > −3T − 1 ⇒ T2 > 0 Thus discrete system stable provided 0 < T < 1.  37(a) #  −1 A= 1  & # k 0 B= 1 0 0  0 −1  & ⎡  ⎤ 1 0 1 s+1 0 s 0 ⎢ ⎥ ⇒ (sI − A)−1 = = ⎣ 1 s + 11 sI − A = ⎦ 1 −1 s s(s + 1) 1 s + 1 − s s+1 s & # −T 0 e ⇒ G = eAT = L−1 {(sI − A)−1 } = −T (1 − e ) 1 & # & & # * T T # k1 (1 − e−T ) k1 0 0 −e−t c At = H= e B dt = t + e−t t 0 0 −1 k1 (e−T + T − 1) −T 0 #  &  #  &  Thus discrete form of model is & # k1 (1 − e−T ) 0 x(kT) + k1 (e−T + T − 1) 1  #  e−T x[(k + 1)T] = (1 − e−T )  & 0 u(kT) −T  In the particular case T = 1, the model becomes #  0.368 x(k + 1) = 0.632  & # 0.632 k1 0 x(k) + 1 0.368k1  & 0 u(k) −1  (b) Taking sampling period T = 1 and feedback control policy, u1 (k) = kc − x2 (k) #  0.368 x(k + 1) = 0.632 # 0.368 ⇒ x(k + 1) = 0.632  &#  & # &# & x1 (k) 0.632k1 0 kc − x2 (k) + x2 (k) 0.368k1 −1 u2 (k) &# & # &# & x1 (k) 0.632k1 0 kc −0.632k1 + x2 (k) 0.368k1 −1 1 u2 (k)  0 1  Given u2 = 1.1x1 (0) and k1 =  3 16  , the discrete-time state-equation becomes  c Pearson Education Limited 2011   396  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition #  0.368 x(k + 1) = 0.632  −0.1185 1  &#  & # x1 (k) 0.1185 + 0.069 x2 (k)  0 −1  &#  kc 1.1x1 (0)  &  (c) Adopting the feedback control policy u1 (t) = kc − x2 (t) the given continuous-time state model becomes & # k −1 k1 x+ 1 ẋ = 0 1 0 #  Taking k1 =  3 16  0 −1  &#  kc u2  &  and u2 = 1.1x1 (0) this reduces to  &# & # 3 & 3 k 0 −1 − 16 c x + 16 ẋ = 0 −1 1 0 1.1x1 (0) & & # # 3 3 1 s − 16 s + 1 16 −1 ⇒ (sI − Ac ) = 2 (sI − Ac ) = 3 −1 s 1 s+1 s + s + 16 ⎡ −1 ⎤ 3 3 3 −8 2 2 8 + + ⎢ s+ 1 s + 34 s + 14 s + 34 ⎥ −1 4 ⎢ ⎥ ⇒ (sI − Ac ) = ⎣ 3 1 ⎦ 2 2 2 2 − − s + 14 s + 34 s + 14 s + 34 #  giving e  Ac t  −1  =L  {(sI − Ac )  −1  #  − 12 e− 4 t + 32 e− 4 t }= 1 3 2e− 4 t − 2e− 4 t 1  − 38 e− 4 t + 38 e− 4 t 3 3 − 14 t − 12 e− 4 t 2e  3  1  The response of the continuous feedback system is # x(t) = e  Ac t  & & * t # x1 (0) kc A(t−τ ) + e dτB kc 1.1x1 (0) 0  Carrying out the integration and simplifying gives the response x1 (t) = x1 (0)[1.1 − 2.15e− 4 t + 2.05e− 4 t ] 1  3  x2 (t) = kc + x1 (0)[−5.867 + 8.6e− 4 t − 2.714e− 4 t ] 1  c Pearson Education Limited 2011   3  3  &  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  397  Exercises 6.11.6 38 H(s) = Replace s with  s2  1 + 3s + 2  2 z−1 to give Δ z+1 H̃(z) =  Δ2 (z + 1)2 4(z − 1)2 + 6Δ(z2 − 1) + 2Δ2 (z + 1)2  Δ2 (z + 1)2 = (4 + 6Δ + 2Δ2 )z2 + (4Δ2 − 8)z + (4 − 6Δ + 2Δ2 ) This corresponds to the diﬀerence equation (Aq2 + Bq + C)yk = Δ2 (q2 + 2q + 1)uk where A = 4 + 6Δ + 2Δ2  B = 4Δ2 − 8  C = 4 − 6Δ + 2Δ2  Now put q = 1 + Δδ to get (AΔ2 δ2 + (2A + B)Δδ + A + B + C)yk = Δ2 (Δ2 δ2 + 4Δδ + 4)uk With t = 0.01 in the q form the system poles are at z = 0.9048 and z = 0.8182, inside | z |= 1. When t = 0.01 these move to z = 0.9900 and z = 0.9802, closer to the stability boundary. Using the δ form with t = 0.1, the poles are at ν = −1.8182 and ν = −0.9522, inside the circle centre (−10, 0) in the ν-plane with radius 10. When t = 0.01 these move to ν = −1.9802 and ν = −0.9950, within the circle centre (−100, 0) with radius 100, and the closest pole to the boundary has moved slightly further from it.  39  The transfer function is H(s) =  s3  +  1 + 2s + 1  2s2  c Pearson Education Limited 2011   398  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  To discretize using the bi-linear form use s →  2z−1 to give Tz+1  T3 (z + 1)3 H̃(z) = Az3 + Bz2 + Cz + D and thus the discrete-time form (Aq3 + Bq2 + Cq + D)yk = T3 (q3 + 3q2 + 3q + 1)uk where A = T3 + 4T2 + 8T + 8, C = 3T3 − 4T2 − 8T + 3, To obtain the δ form use s →  B = 3T3 + 4T2 − 8T − 3, D = T3 − 4T2 + 8T − 1  2δ giving the δ transfer function as 2 + Δδ (2 + Δδ)3 Aδ3 + Bδ2 + Cδ + D  This corresponds to the discrete-time system (Aδ3 + Bδ2 + Cδ + D)yk = (Δ3 δ3 + 2Δ2 δ2 + 4Δδ + 8)uk where A = Δ3 + 4Δ2 + 8Δ + 8, C = 12Δ + 16,  B = 6Δ2 + 16Δ + 16, D=8  40 Making the given substitution and writing the result in vector–matrix form, we obtain # & # & 0 1 0 ẋ(t) = x(t) + u(t) −2 −3 1 and y(t) = [1, 0]x(t) This is in the general form ẋ(t) = Ax(t) + bu(t) y = cT x(t) + d u(t) c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  399  The Euler discretization scheme gives at once x((k + 1)Δ) = x(k Δ) + Δ [Ax(k Δ) + bu(k Δ)] Using the notation of Exercise 38 write the simpliﬁed δ form equation as & # , 8 12 + 8Δ 1 + 2 2 2 δ+ yk = Δ δ + 4Δδ + 4 uk δ + A A A Now, as usual, consider the related system & # 12 + 8Δ 8 2 pk = uk δ + δ+ A A and introduce the state variables x1 (k) = pk , x2 (k) = δpk together with the redundant variable x3 (k) = δ2 pk . This leads to the representation ⎡  0 δx(k) = ⎣ 8 − A # yk =  8Δ2 4 − 2 A A  ⎤ # & 1 0 u(k) 12 + 8Δ ⎦ x(k) + 1 − A    & 4Δ (12 + 8Δ)Δ2 Δ2 − u(k) , x(k) + A A2 A  or x(k + 1) = x(k) + Δ [A(Δ)x(k) + bu(k)] yk = cT (Δ)x(k) + d(Δ)uk Since A(0) = 4 it follows that using A(0) , c(0) and d(0) generates the Euler Scheme when x(k) = x(kΔ) etc.  41(a)  In the z form substitution leads directly to H(z) =  12(z2 − z) (12 + 5Δ)z2 + (8Δ − 12)z − Δ  When Δ = 0.1, this gives H(z) =  12(z2 − z) 12.5z2 + −11.2z − 0.1  c Pearson Education Limited 2011   400  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  (b)  The γ form is given by replacing z by 1 + Δγ.  rearrangement gives H̃(γ) =  γ2 Δ(12  12γ(1 + Δγ) + 5Δ) + γ(8Δ − 12) + 12  when Δ = 0.1, this gives H̃(γ) =  12γ(1 + 0.1γ) 1.25γ2 − 11.2γ + 12  Review exercises 6.12 1 Z {f(kT)} = Z {kT} = TZ {k} = T  2   Z a sin kω = Z k  z (z − 1)2  ak (ej kω − e−j kω ) 2j    1 Z (aej ω )k − (ae−j ω )k 2j  z z 1 − = 2j z − aej ω z − ae−j ω az sin ω = 2 z − 2az cos ω + a2 =  3 Recall that Z ak =  z (z − a)2  Diﬀerentiate twice wrt a, then put a = 1 to get the pairs k ←→  z (z − 1)2  k(k − 1) ←→  2z (z − 1)3  then Z k2 =  2z z z(z + 1) + = (z − 1)3 (z − 1)2 (z − 1)3  c Pearson Education Limited 2011   Substitution and  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 4 H(z) =  2z 3z + z − 1 (z − 1)2  so inverting, the impulse response is {3 + 2k} 5  z (z + 1)(z + 2)(z − 1) 1 z 1 z 1 z + + =− 2z+1 3z+2 6z−1  YSTEP (z) =  Thus  1 1 1 ySTEP,k = − (−1)k + (−2)k + 2 3 6  6 F(s) =  1 1 1 = − s+1 s s+1  which inverts to f(t) = (1 − e−t )ζ(t) where ζ(t) is the Heaviside step function, and so F̃(z) = Z {f(kT)} =  z z − z − 1 z − e−T  Then e−sT F(s) ←→ f((t − T)) which when sampled becomes f((k − 1)T) and Z {f((k − 1)T)} =  ∞  f((k − 1)T) k=0  That is e−sT F(s) →  zk  =  1 F̃(z) z  1 F̃(z) z  So the overall transfer function is z−1 z   1 − e−T z z − = z − 1 z − e−T z − e−T  c Pearson Education Limited 2011   401  402  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  7  H(s) =  2 1 s+1 = − (s + 2)(s + 3) s+3 s+2  yδ (t) = 2e−3t − e2t −→ {2e−3kT − e2kT } so H̃(z) = 2 8(a)  z z − −3T z−e z − e−2T  Simple poles at z = a and z = b. The residue at z = a is lim (z − a)zn−1 X(z) = lim (z − a)  z→a  z→a  The residue at z = b is similarly of these, that is  zn an = (z − a)(z − b) a−b  bn and the inverse transform is the sum b−a   n a − bn a−b  8(b) (i) There is a only double pole at z = 3 and the residue is d zn (z − 3)2 = n3n−1 z→3 dz (z − 3)2 √ 3 1 j . The individual residues are (ii) There are now simple poles at z = ± 2 2 thus given by n √ 1 3 ± j 2 2 √ ± lim√ 3j z→(1/2± 3/2j ) lim  Adding these and simplifying in the usual way gives the inverse transform   as   2 √ sin nπ/3 3  9 H(z) =  z z − z+1 z−2  so  YSTEP (z) =  z z − z+1 z−2    z z−1  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  403  3z .z (z − 1)(z + 1)(z − 2) 3 z 1 z z = + −2 2z−1 2z+1 z−2  =−  so ySTEP,k =  3 1 + (−1)k − 2k+1 2 2    1 z z2 × 1− = Y(z) = (z + 1)(z − 1) z z+1  10  so yk = (−1)k   α + β αβ z2 × 1− + 2 =1 Y(z) = (z − α)(z − β) z z  11  so yk = {δk } = {1, 0, 0, . . .} z is clearly given by The response of the system with H(z) = (z − α)(z − β) Y(z) = 1/z , which transforms to yk = {δk−1 } = {0, 1, 0, 0, . . .}  12  From H(s) =  s the impulse response is calculated as (s + 1)(s + 2) yδ (t) = (2e−2t − e−t ) t ≥ 0  Sampling gives {yδ (nT)} = 2e−2nT − enT t with z transform Z {yδ (nT)} = 2  z z − = D(z) −2T z−e z − e−T  Setting Y(z) = TD(z)X(z) gives . z z X(z) − Y(z) = T 2 z − e−2T z − e−T -  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  404  Substituting for T and simplifying gives # & z − 0.8452 1 X(z) Y(z) = z 2 2 z − 0.9744z + 0.2231 so (z2 − 0.9744z + 0.2231)Y(z) = (0.5z2 − 0.4226z)X(x) leading to the diﬀerence equation yn+2 − 0.9744yn+1 + 0.2231yn = 0.5xn+2 − 0.4226xn+1 As usual (see Exercise 22), draw the block diagram for pn+2 − 0.9744pn+1 + 0.2231pn = xn then taking yn = 0.5pn+2 − 0.4226pn+1 yn+2 − 0.9744yn+1 + 0.2231yn = 0.5pn+4 − 0.4226pn+3 −0.9774(0.5pn+3 − 0.4226pn+2 ) + 0.2231(0.5pn+2 − 0.4226pn+1 ) = 0.5xn+2 − 0.4226xn+1 13 yn+1 = yn + avn vn+1 = vn + bun = vn + b(k1 (xn − yn ) − k2 vn ) = bk1 (xn − yn ) + (1 − bk2 )vn so yn+2 = yn+1 + a[bk1 (xn − yn ) + (1 − bk2 )vn ] (a) Substituting the values for k1 and k2 , we get 1 yn+2 = yn+1 + (xn − yn ) 4 or 1 1 yn+2 − yn+1 + yn = xn 4 4 c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  405  Transforming with relaxed initial conditions gives Y(z) =  (b) When X(z) =  1 X(z) (2z − 1)2  A , z−1  # & z z z A 4 −4 −2 Y(z) = 4 z−1 z − 1/2 (z − 1/2)2 then yn = 14  , A+ 4 − 4(1/2)n − 2n(1/2)n−1 4  Substitution leads directly to yk − 2yk−1 + yk−2 yk − yk−1 + 2yk = 1 + 3 T2 T Take the z transform under the assumption of a relaxed system to get [(1 + 3Tz + 2T2 )z2 − (2 + 3T)z + 1]Y(z) = T2  z3 z−1  The characteristic equation is thus (1 + 3Tz + 2T2 )z2 − (2 + 3T)z + 1 = 0 with roots (the poles) 1 1 , z= 1+T 1 + 2T The general solution of the diﬀerence equation is a linear combination of these z=  together with a particular solution. That is  yk = α  1 1+T  k   +β  1 1 + 2T  k +γ  This can be checked by substitution which also shows that γ = 1/2. The yk − yk−1 , y (0) = 0 condition y(0) = 0 gives y0 = 0 and since y (t) → T implies yk−1 = 0. Using these we have 1 =0 2 1 α(1 + T) + β(1 + 2T) + = 0 2 α+β+  c Pearson Education Limited 2011   406  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  with solution α = −1, β = 1/2 so  yk = −  1 1+T  k  1 + 2    1 1 + 2T  k +  1 2  The diﬀerential equation is simply solved by inverting the Laplace transform to give y(t) =  1 −2t (e − 2e−t + 1), t ≥ 0 2  T = 0.1  Figure 6.3: Response of continuous and discrete systems in Review Exercise 14 over 10 seconds when T = 0.1.  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  407  T = 0.05  Figure 6.4: Response of continuous and discrete systems in Review Exercise 14 over 10 seconds when T = 0.05.  15 Substitution for s and simplifying gives [(4 + 6T + 2T2 )z2 + (4T2 − 8)z + (4 − 6T + 2T2 )]Y(z) = T2 (z + 1)2 X(x) The characteristic equation is (4 + 6T + 2T2 )z2 + (4T2 − 8)z + (4 − 6T + 2T2 ) = 0 with roots z= That is, z=  8 − 4T2 ± 4T 2(4 + 6T + 2T2 )  2−T 1−T and z = 1+T 2+T  c Pearson Education Limited 2011   408  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition The general solution of the diﬀerence equation is then  yk = α  1−T 1+T  k   +β  2−T 2+T  k +γ  This can be checked by substitution which also shows that γ = 1/2. The yk − yk−1 , y (0) = 0 condition y(0) = 0 gives y0 = 0 and since y (t) → T implies yk−1 = 0. Using these we have α+β+ α  2+T 1 1+T +β + =0 1−T 2−T 2  with solution α= Thus, 1−T yk = 2  1 =0 2    1−T 2  1−T 1+T  k  β=−  2−T 2  2−T +− 2    2−T 2+T  k +  1 2  T = 0.05  Figure 6.5: Response of continuous and discrete systems in Review Exercise 15 over 10 seconds when T = 0.1. c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  409  T = 0.05  Figure 6.6: Response of continuous and discrete systems in Review Exercise 14 over 10 seconds when T = 0.05. 16  f(t) = t2 ,  {f(kΔ)} = k2 Δ2 , k ≥ 0  Now Z{k2 } = −z  z d z(z + 1) = 2 dz (z − 1) (z − 1)3  So Z{k2 Δ2 } =  z(z + 1)Δ2 (z − 1)3  To get D -transform, put z = 1 + Δγ to give   FΔ (γ) =  (1 + Δγ)(2 + Δγ)Δ2 Δ3 γ3  Then the D -transform is   FΔ (γ) = ΔFΔ (γ) =  (1 + Δγ)(2 + Δγ) γ3  c Pearson Education Limited 2011   410  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  17  Eigenvalues are given by ) )1− λ 1 ) ) 2−λ 0 = ) −1 ) 0 1  ) −2 )) 1 )) = (1 − λ)(λ2 − λ − 3) + (1 − λ) −1 − λ ) = (1 − λ)(λ − 2)(λ + 1)  so eigenvalues are λ1 = 2, λ2 = 1, λ3 = −1 Corresponding eigenvectors are the solutions of (1 − λi )ei1 + ei2 − 2ei3 = 0 −ei1 + (2 − λi )ei2 + ei3 = 0 0ei1 + ei2 − (1 + λi )ei3 = 0 Taking i = 1, 2, 3 the eigenvectors are e1 = [1 3 1]T , e2 = [3 2 1]T , e3 = [1 0 1]T The modal matrix M and spectral matrix Λ are ⎡  1 M = ⎣3 1 ⎡ 2 M Λ = ⎣6 2 ⎡ 2 = ⎣6 2  ⎤ ⎡ 2 1 0 ⎦, Λ = ⎣ 0 0 1 ⎤ 3 −1 2 0 ⎦, A M = 1 −1 ⎤ 3 −1 2 0⎦ =M Λ 1 −1 3 2 1  ⎤ 0 0 1 0 ⎦ 0 −1 ⎤ ⎤ ⎡ ⎡ 1 3 1 1 1 −2 ⎣ −1 2 1 ⎦ ⎣3 2 0⎦ 1 1 1 0 1 −1  Substituting x = M y in x(k + 1) = A x(k) gives M y(k + 1) = A M y(k) or y(k + 1) = M−1 A M y(k) = Λ y(k) so y(1) = Λ y(0), y(2) = Λ y(1) = Λ2 y(0) ⇒ y(k) = Λk y(0) c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition Thus,  411  ⎤ ⎡ ⎤ ⎡ ⎤ 2k 0 0 1 3 1 α ⎦ ⎣ ⎣ ⎦ ⎣ 0 1 0 β ⎦ say x(k) = 3 2 0 k 0 0 (−1) 1 1 1 γ ⎤ ⎡ ⎡ ⎤ k α2 1 3 1 ⎦ ⎣ ⎣ = 3 2 0 β ⎦ 1 1 1 γ(−1)k ⎤ ⎡ k α2 + 3β + γ(−1)k ⎦ 3α2k + 2β = ⎣ k k α2 + β + γ(−1) ⎡  ⎤ ⎡ ⎡ ⎤ α + 3β + γ 1 ⎣ 0 ⎦ = ⎣ 3α + 2β ⎦ α+β+γ 0  When k = 0  1 1 1 which gives α = − , β = , γ = 3 2 6 so that ⎡ 1 k 3 1 ⎤ − 3 (2) + 2 + 6 (−1)k ⎦ x(k) = ⎣ −(2)k + 1 − 31 (2)k + 12 + 16 (−1)k  18  x1 (k + 1) = u(k) − 3x1 (k) − 4x2 (k) x2 (k + 1) = −2x1 (k) − x2 (k) y(k) = x1 (k) + x2 (k)  or in vector-matrix form # −3 x(k + 1) = −2 −1  D(x) = c[zI − A]  #  1 0  −3 −2  −4 −1  &  # & 1 u(k), y(k) = [1 − 1]x(k) x(k) + 0  1 [1 − 1] b= 2 z + 4z − 5 z+3 = 2 z + 4z − 5  #  z + 1 −4 −2 z + 3  &  (i)  Mc =  (ii)  det Mc = −2 = 0 so Mc is of rank 2 c Pearson Education Limited 2011   & # & 1 0  (1)  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition & # # & 1 − 32 1 −2 3 −1 = (iii) Mc = − 2 0 1 0 − 12 1 T so v = [0 − 2 ] 412  # (iv)  T=  0 1  − 21  &  1 2  # (v)  −1  det T = 0 and T  =2  1 2  −1  1 2  &  0  # =  1 1 −2 0  &  Substituting z(k) = T x(k) in (1) gives T−1 z(k + 1) = A T−1 z(k) + bu(k) or z(k + 1) = T A # 0 = 1 # 0 thatis, z(k + 1) = 5  T−1 z(k) + T bu(k) & # & # 1 −3 −4 − 12 1 −2 −2 −1 2 & # & 0 1 u(k) z(k) + 1 −4  1 0  &  # z(k) +  0 1  − 12 1 2  & # & 1 u(t) 0  Thus C and bc are of the required form with α = −5, β = 4 which are coeﬃcients in the characteristic polynomial of D(z) .  c Pearson Education Limited 2011   7 Fourier series Exercises 7.2.6 1(a) 1 a0 = π      0 −π  −πdt +    π  tdt 0  2 π     t π 1 π2 1 0 2 (−πt)−π + −π + =− = = π 2 0 π 2 2  0   π 1 an = −π cos ntdt + t cos ntdt π −π 0  π 0 t 1 1  π − sin nt sin nt + 2 cos nt + = π n n n 0 −π 2 1 − 2 , n odd (cos nπ − 1) = = πn πn2 0, n even  0   π 1 bn = −π sin ntdt + t sin ntdt π −π 0  π 0  t 1 1 π cos nt + − cos nt + 2 sin nt = π n n n 0 −π ⎧ 3 ⎪ ⎨ , n odd 1 = (1 − 2 cos nπ) = n 1 ⎪ n ⎩ − , n even n Thus, the Fourier expansion of f(t) is   2   1  3 π + sin nt − sin nt − 2 cos nt + 4 πn n n n even n odd n odd ∞ ∞ ∞  sin(2n − 1)t  π 2  cos(2n − 1)t sin 2nt f(t) = − − − +3 2 4 π n=1 (2π − 1) (2n − 1) 2n n=1 n=1 f(t) = −  i.e.  c Pearson Education Limited 2011   414  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  1(b)  0 1 t2 π + πt (t + π)dt = = π 2 2 −π −π  0  0 sin nt cos nt 1 1 (t + π) + an = (t + π) cos ntdt = π −π π n n2 −π 0, n even 1 2 = (1 − cos nπ) = , n odd πn2 πn2  0  cos nt sin nt 1 0 1 1 −(t + π) + bn = (t + π) sin ntdt = = − π −π π n n2 −π n 1 a0 = π    0  Thus, the Fourier expansion of f(t) is  i.e.  1(c)  ∞  2  1 π sin nt cos nt − f(t) = + 2 4 πn n n=1 n odd ∞ ∞ π 2  cos(2n − 1)t  sin nt f(t) = + − 4 π n=1 (2n − 1)2 n n=1  From its graph we see that f(t) is an odd function; so it has Fourier  expansion f(t) =  ∞   bn sin nt  n=1  with    t 2 π 2 π sin ntdt 1− f(t) sin nt = bn = π 0 π 0 π  π 1 t 1 2 2 − 1− cos nt − sin nt = = 2 π n π πn nπ 0  Thus, the Fourier expansion of f(t) is ∞ 2  sin nt f(t) = π n=1 n  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 1(d)  From its graph f(t) is seen to be an even function; so its Fourier  expansion is  ∞  a0  + an cos nt f(t) = 2 n=1  with 2 a0 = π 2 an = π 2 = π    π    0    0  π  2 f(t)dt = π    π/2  2 cos tdt = 0  2 f(t) cos ntdt = π    2 4 π/2 [2 sin t]0 = π π  π/2  2 cos t cos ntdt 0  π/2  [cos(n + 1)t + cos(n − 1)t]dt 0   π/2 2 sin(n + 1)t sin(n − 1)t + = π (n + 1) (n − 1) 0   1 2 π 1 π = sin(n + 1) + sin(n − 1) π (n + 1) 2 (n − 1) 2 ⎧ 0, n odd ⎪ ⎪ ⎪ 1 4 ⎨− , n = 4, 8, 12, . . . = π (n2 − 1) ⎪ ⎪ 4 1 ⎪ ⎩ , n = 2, 6, 10, . . . 2 π (n − 1) Thus, the Fourier expansion of f(t) is f(t) =  ∞ 4  (−1)n+1 cos 2nt 2 + π π n=1 4n2 − 1  1(e)  π 1 t t 4 2 sin cos dt = = 2 π 2 −π π −π   π  π  1 1 1 1 t an = cos(n + )t + cos(n − )t dt cos cos ntdt = π −π 2 2π −π 2 2   2 1 2 1 2 sin(n + )π + sin(n − )π = 2π (2n + 1) 2 (2n − 1) 2 ⎧ 4 ⎪ , n = 1, 3, 5, . . . ⎨ π(4n2 − 1) = 4 ⎪ ⎩− , n = 2, 4, 6, . . . 2 π(4n − 1) bn = 0 1 a0 = π    π  c Pearson Education Limited 2011   415  416  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Thus, the Fourier expansion of f(t) is ∞ 4  (−1)n+1 cos nt 2 f(t) = + π π n=1 (4n2 − 1)  1(f)  Since f(t) is an even function, it has Fourier expansion ∞  a0  an cos nt f(t) = + 2 n=1   2 π 2 π | t | dt = tdt = π a0 = π 0 π 0  π  1 2 π 2 t sin nt + 2 cos nt t cos ntdt = an = π 0 π n n 0 0, n even 2 4 = (cos nπ − 1) = − 2 , n odd πn2 πn Thus, the Fourier expansion of f(t) is  with  π 4  1 − cos nt 2 π n2 n odd ∞ π 4  cos(2n − 1)t that is, f(t) = − 2 π n=1 (2n − 1)2 f(t) =  1(g)  1 π 1 2 π t − πt 0 = 0 a0 = (2t − π)dt = π 0 π  π  π 2 1 1 (2t − π) sin nt + 2 cos nt (2t − π) cos ntdt = an = π 0 π n n 0 4 2 = (cos nπ − 1) = − πn2 , n odd πn2 0, n even  π  π (2t − π) 2 1 1 − cos nt + 2 sin nt bn = (2t − π) sin ntdt = π 0 π n n 0 0, n odd 1 2 = − (cos nπ + 1) = − , n even n n c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  417  Thus, the Fourier expansion of f(t) is   f(t) =  −   4 2 cos nt + − sin nt 2 πn n n even  odd ∞ ∞ 4  cos(2n − 1)t  sin 2nt f(t) = − − π n=1 (2n − 1)2 n n=1 n  that is,  1(h) 1 a0 = π = = an = =  = =  1 bn = π      0  t  (−t + e )dt + −π    π  t  (t + e )dt 0   0 π  t2 1  t2 t t − +e + +e π 2 2 −π 0 1 2 2 π + (eπ − e−π ) = π + sinh π π π  0   π 1 t t (−t + e ) cos ntdt + (t + e ) cos ntdt π −π 0  0 t  t 1 1 1 − sin nt + 2 cos nt ne sin nt + et cos nt + 2 π n n (n + 1) −π  π  t t 1 1 π ne sin nt + et cos nt 0 + sin nt + 2 cos nt + 2 n n (n + 1) 0 2 cos nπ  eπ − e−π 2 (−1 + cos nπ) + πn2 π(n2 + 1) 2   2 (cos π − 1) cos nπ sinh π , cos nπ = (−1)n + 2 2 π n (n + 1)      0  π  t     (t + e ) sin ntdt t  (−t + e ) sin ntdt + −π  0 −π  0  0 π  t 1 1 1 t cos nt − 2 sin nt + − cos nt + 2 sin nt = π n n n n −π 0  t π 2 t e cos nt e sin nt n − + + 2 π +1 n n2 −π 2n n cos nπ(eπ − e−π ) = − cos nπ sinh π, cos nπ = (−1)n =− 2 π(n + 1) π(n2 + 1)  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  418  Thus, the Fourier expansion of f(t) is  ∞   2 π 1 (−1)n − 1 (−1)n sinh π + sinh π + cos nt + f(t) = 2 π π n=1 n2 n2 + 1 −  2  ∞ 2  n(−1)n sinh π sin nt π n=1 n2 + 1  Since the periodic function f(t) is an even function, its Fourier expansion is ∞  a0  + an cos nt f(t) = 2 n=1 with  π 1 2 2 3 − (π − t) (π − t) dt = = π2 π 3 3 0 0  π  π 2(π − t) 2 2 (π − t)2 2 2 sin nt − an = (π − t) cos ntdt = cos nt − 3 sin nt π 0 π n n2 n 0 4 = 2 n 2 a0 = π    π  2  Thus, the Fourier expansion of f(t) is ∞  1 π2 +4 cos nt f(t) = 3 n2 n=1  Taking t = π gives 0=  ∞  1 π2 +4 (−1)n 2 3 n n=1  so that ∞ 1 2  (−1)n+1 π = 12 n2 n=1  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 3  Since q(t) is an even function, its Fourier expansion is ∞  q(t) = with  a0  + an cos nt 2 n=1   2 π a0 = π 0  2 π an = π 0  Qt dt = Q π  π Qt 2Q t 1 cos ntdt = 2 sin nt + 2 cos nt π π n n 0 0, n even 2Q 4Q = 2 2 (cos nπ − 1) = − 2 2 , n odd π n π n Thus, the Fourier expansion of q(t) is   ∞ 4  cos(2n − 1)t 1 q(t) = Q − 2 2 π n=1 (2n − 1)2 4   1 π 1 10 5 sin tdt = [−5 cos t]π0 = a0 = π 0 π π  π  π 5 5 sin t cos ntdt = [sin(n + 1)t − sin(n − 1)t]dt an = π 0 2π 0  π cos(n + 1)t cos(n − 1)t 5 − + , n = 1 = 2π (n + 1) (n − 1) 0   1 5  cos nπ cos nπ 1 − − = − + 2π n+1 (n − 1) n+1 n−1 0, n odd, n = 1 5 10 =− (cos nπ + 1) = , n even − π(n2 − 1) π(n2 − 1)  Note that in this case we need to evaluate a1 separately as   π 1 π 5 5 sin t cos tdt = sin 2tdt = 0 a1 = π 0 2π 0   π 5 π 5 bn = sin t sin ntdt = − [cos(n + 1)t − cos(n − 1)t]dt π 0 2π 0  π 5 sin(n + 1)t sin(n − 1)t − , n = 1 =− 2π (n + 1) (n − 1) 0 = 0 , n = 1 c Pearson Education Limited 2011   419  420  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Evaluating b1 separately,   π 5 π 5 sin t sin tdt = (1 − cos 2t)dt b1 = π 0 2π 0 π 1 5  5 t − sin 2t = = 2π 2 2 0 Thus, the Fourier expansion of f(t) is ∞ 10  cos 2nt 5 5 + sin t − π 2 π n=1 4n2 − 1  f(t) =  5  a0 = = an = = =  1 bn = π   0   π 1 2 2 π dt + (t − π) dt π −π 0  π 1 4 1  2 0 3 π t −π + (t − π) = π2 π 3 3 0  0   π 1 2 2 π cos ntdt + (t − π) cos ntdt π −π 0  0  (t − π)2 2(t − π) 2 1  π2 sin nt sin nt + + cos nt − 3 sin nt 2 π n n n n −π 2 n2     0 2  0   (t − π) sin ntdt 2  π sin ntdt + −π  π  π  0   0  (t − π)2 (t − π) 2 1  π2 − cos nt cos nt + 2 + − sin nt + cos nt = π n n n2 n3 −π  π2 1  π2 − + (−1)n = π n n 2 π [1 − (−1)n ] = (−1)n − n πn3 c Pearson Education Limited 2011   π 0  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition Thus, the Fourier expansion of f(t) is  ∞  ∞ 2 2  2 (−1)n 4  sin(2n − 1)t f(t) = π + cos nt + π sin nt − 3 n2 n π n=1 (2n − 1)3 n=1  5(a)  Taking t = 0 gives ∞  2 π2 + π2 2 = π2 + 2 3 n2 n=1  and hence the required result  5(b)  ∞  1 1 = π2 2 n 6 n=1  Taking t = π gives ∞  2 π2 + 0 2 = π2 + (−1)n 2 2 3 n n=1  and hence the required result ∞  (−1)n+1 1 2 π = 2 n 12 n=1  6(a)  c Pearson Education Limited 2011   421  422  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  6(b)  The Fourier expansion of the even function (a) is given by ∞  f(t) =  a0  + an cos nt 2 n=1  with 2 a0 = π      π/2  (π − t)dt  tdt + 0    π π/2   π π 2  1 2 π/2  1 t + − (π − t)2 = = π 2 0 2 2 π/2  π/2   π 2 t cos ntdt + (π − t) cos ntdt an = π 0 π/2  π/2 π − t 1 1 2 t sin nt + 2 cos nt sin nt − 2 cos nt + = π n n n n 0   1 nπ 2 2 − 2 (1 + (−1)n ) cos = 2 π n 2 n ⎧ n odd ⎪ ⎨ 0, 8 = − 2 , n = 2, 6, 10, . . . ⎪ ⎩ πn 0, n = 4, 8, 12, . . . Thus, the Fourier expansion of f(t) is ∞ π 2  cos(4n − 2)t f(t) = − 4 π n=1 (2n − 1)2  Taking t = 0 where f(t) = 0 gives the required result.  c Pearson Education Limited 2011   π π/2  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  423  7  1 a0 = π    π 0  t (2 − )dt + π      2π  t/πdt π   π  t2 2π t2 1  2t − =3 + = π 2π 0 2π π  π   2π t 1 t cos ntdt an = (2 − ) cos ntdt + π 0 π π π  π 2π  t t 1 1 1 2 sin nt − sin nt − sin nt + = cos nt + cos nt π n πn πn2 πn πn2 0 π 2 = 2 2 [1 − (−1)n ] π n 0, n even 4 = , n odd π2 n2  π   2π t 1 t bn = (2 − ) sin ntdt + sin ntdt π 0 π π π  π  t t 1 1 1  2 − cos nt + sin nt + − sin nt cos nt − cos nt + = 2 π n πn πn πn πn2 0 =0 Thus, the Fourier expansion of f(t) is ∞ 3 4  cos(2n − 1)t f(t) = + 2 2 π n=1 (2n − 1)2  Replacing t by t − 12 π gives ∞ 1 3 4  cos(2n − 1)(t − π) f(t − π) = + 2 2 2 π n=1 (2n − 1)2  c Pearson Education Limited 2011   2π π  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  424 Since  1 π π cos(2n − 1)(t − π) = cos(2n − 1)t cos(2n − 1) + sin(2n − 1)t sin(2n − 1) 2 2 2 n+1 sin(2n − 1)t = (−1) ∞ 3 4  (−1)n+1 sin(2n − 1)t 1 f(t − π) − = 2 2 2 π n=1 (2n − 1)2  The corresponding odd function is readily recognized from the graph of f(t) .  Exercises 7.2.8 8  Since f(t) is an odd function, the Fourier expansion is f(t) =  ∞   bn sin  n=1  with 2 bn =      0  nπt     t nπt nπt 2 nπt   2 − t sin sin dt = cos +   nπ  nπ  0  2 cos nπ =− nπ Thus, the Fourier expansion of f(t) is ∞ 2  (−1)n+1 nπt f(t) = sin π n=1 n   9  Since f(t) is an odd function (readily seen from a sketch of its graph) its  Fourier expansion is f(t) =  ∞   bn sin  n=1  with 2 bn =    0    nπt   K nπt ( − t) sin tdt      K nπt Kt nπt K nπt 2 − cos + cos − sin =  nπ  nπ  (nπ)2  0 2K = nπ c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition Thus, the Fourier expansion of f(t) is ∞ nπt 2K  1 sin f(t) = π n=1 n   10    1 a0 = 5  5  3dt = 3 0   5 nπt 1 15 nπt 3 cos =0 dt = sin 5 5 nπ 5 0 0  5  1 15 nπt 1 5 nπt dt = − cos bn = 3 sin 5 0 5 5 nπ 5 0 6 3 = [1 − (−1)n ] = nπ , n odd nπ 0, n even   1 an = 5  5  Thus, the Fourier expansion of f(t) is ∞ 3 6 1 (2n − 1) f(t) = + sin πt 2 π n=1 (2n − 1) 5  11  π/ω A ω 2A − cos ωt A sin ωtdt = = π ω π 0 0  π/ω  π/ω Aω Aω sin ωt cos nωtdt = [sin(n + 1)ωt − sin(n − 1)ωt]dt an = π 0 2π 0  π/ω cos(n + 1)ωt cos(n − 1)ωt Aω − + = , n = 1 2π (n + 1)ω (n − 1)ω 0   A A 2(−1)n+1 2 = = − 2 [(−1)n+1 − 1] 2 2 2π n − 1 n −1 π(n − 1) ⎧ n odd , n = 1 ⎨ 0, 2A = − , n even ⎩ π(n2 − 1) 2ω a0 = 2π    π/ω  Evaluating a1 separately, Aω a1 = π    π/ω 0  A sin ωt cos ωtdt = 2π    π/ω  sin 2ωtdt = 0 0  c Pearson Education Limited 2011   425  426  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Aω bn = π    π/ω 0  Aω sin ωt sin nωtdt = − 2π    π/ω  [cos(n + 1)ωt − cos(n − 1)ωt]dt 0   π/ω Aω sin(n + 1)ωt sin(n − 1)ωt − =− , n = 1 2π (n + 1)ω (n − 1)ω 0  = 0, n = 1   Aω π/ω 2 Aω π/ω A b1 = sin ωtdt = (1 − cos 2ωt)dt = π 0 2π 0 2 Thus, the Fourier expansion of f(t) is   ∞  cos 2nωt π A 1 + sin ωt − 2 f(t) = π 2 4n2 − 1 n=1  12  Since f(t) is an even function, its Fourier expansion is ∞  f(t) =  nπt a0  + an cos 2 T n=1  with 2 a0 = T an = =  2 T    T   T 2 1 3 2 t t dt = = T2 T 3 0 3  2 T 2 Tt nπt 2tT2 2T3 nπt nπt nπt 2 dt = sin + − t cos cos sin T T nπ T (nπ)2 T (nπ)3 T 0 2  0    T  0 2  4T (−1)n (nπ)2  Thus, the Fourier series expansion of f(t) is  f(t) =  ∞ 4T2  (−1)n nπt T2 + 2 cos 3 π n=1 n2 T  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 13 2 a0 = T 2 an = T    T  0    T  0   T E 2E 1 2 tdt = 2 t =E T T 2 0 E 2πnt t cos dt T T  T 2E tT 2πnt 2πnt  T 2 = 2 cos =0 sin + T 2πn T 2πn T 0  2E T 2πnt bn = 2 t sin dt T 0 T T  2πnt  T 2 tT 2πnt E 2E cos + sin =− = 2 − T 2πn T 2πn T 0 πn   Thus, the Fourier expansion of e(t) is ∞ 2πnt E E1 sin e(t) = − 2 π n=1 n T  Exercises 7.3.3 14  Half range Fourier sine series expansion is given by f(t) =  ∞   bn sin nt  n=1  with 2 bn = π    π 0   π 1 2 − cos nt 1 sin ntdt = π n 0  2 [(−1)n − 1] nπ 0, n even 4 = , n odd nπ Thus, the half range Fourier sine series expansion of f(t) is =−  ∞ 4  sin(2n − 1)t f(t) = π n=1 (2n − 1)  Plotting the graphs should cause no problems.  c Pearson Education Limited 2011   427  428  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  15  Half range Fourier cosine series expansion is given by ∞  a0  + an cos nπt f(t) = 2 n=1 with 2 a0 = 1    1  (2t − 1)dt = 0   0 1  (2t − 1) cos nπtdt  an = 2 0   1 2 (2t − 1) sin nπt + =2 cos nπt nπ (nπ)2 0 4 = [(−1)n − 1] (nπ)2 0, n even 8 = − , n odd (nπ)2 Thus, the half range Fourier cosine series expansion of f(t) is ∞ 1 8  cos(2n − 1)πt f(t) = − 2 π n=1 (2n − 1)2  Again plotting the graph should cause no problems.  16(a)   1  a0 = 2   0   1 (1 − t2 )dt = 2 t − t3 3  1  = 0  4 3  1  (1 − t2 ) cos 2nπtdt  an = 2 0   1 (1 − t2 ) 2 2t =2 cos 2nπt + sin 2nπt sin 2nπt − 2nπ (2nπ)2 (2nπ)3 0 1 =− (nπ)2 c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition   1  (1 − t2 ) sin 2nπtdt  bn = 2 0   1 2t (1 − t2 ) 2 cos 2nπt − =2 − sin 2nπt − cos 2nπt 2nπ (2nπ)2 (2nπ)3 0 1 = nπ Thus, the full-range Fourier series expansion for f(t) is f(t) = f1 (t) =  16(b)  ∞ ∞ 1  1 2 11 − 2 sin 2nπt cos 2nπt + 3 π n=1 n2 π n=1 n  Half-range sine series expansion is f2 (t) =  ∞   bn sin nπt  n=1  with    1  (1 − t2 ) sin nπtdt  bn = 2 0   (1 − t2 ) =2 − cos nπt − nπ  2 1 =2 − (−1)n + 3 (nπ) nπ ⎧ 2 ⎪ ⎨ , n = nπ  1 4 ⎪ ⎩2 + , n nπ (nπ)3  1 2 2t sin nπt − cos nπt (nπ)2 (nπ)3 0  2 + (nπ)3 even odd  Thus, half-range sine series expansion is  ∞ ∞  2 4 1 11 f2 (t) = sin 2nπt + + sin(2n − 1)πt π n=1 n π n=1 (2n − 1) π2 (2n − 1)3  16(c)  Half-range cosine series expansion is ∞  a0  f3 (t) = an cos nπt + 2 n=1 c Pearson Education Limited 2011   429  430  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  with    1  (1 − t2 )dt =  a0 = 2   0  4 3  1  (1 − t2 ) cos nπtdt  an = 2 0   1 2t (1 − t2 ) 2 sin nπt − =2 cos nπt + sin nπt nπ (nπ)2 (nπ)3 0 −4(−1)n = (nπ)2 Thus, half-range cosine series expansion is ∞ 2 4  (−1)n+1 cos nπt f3 (t) = + 2 3 π n=1 n2  Graphs of the functions f1 (t), f2 (t), f3 (t) for −4 < t < 4 are as follows:  17  Fourier cosine series expansion is ∞  a0  an cos nt f1 (t) = + 2 n=1 c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  with   2 π 1 (πt − t2 )dt = π2 a0 = π 0 3  π 2 an = (πt − t2 ) cos ntdt π 0  π (π − 2t) 2 2 (πt − t2 ) sin nt + cos nt + 3 sin nt = π n n2 n 0 2 = − 2 [1 + (−1)n ] n 0, n odd 4 = − 2 , n even n  Thus, the Fourier cosine series expansion is ∞ 1 2  1 f1 (t) = π − cos 2nt 2 6 n n=1  Fourier sine series expansion is  f2 (t) =  ∞   bn sin nt  n=1  with bn = = = =   2 π (πt − t2 ) sin ntdt π 0  π (πt − t2 ) (π − 2t) 2 2 − cos nt + sin nt − 3 cos nt π n n2 n 0 4 [1 − (−1)n ] πn3 0, n even 8 , n odd πn3  Thus, the Fourier sine series expansion is ∞ 1 8 sin(2n − 1)t f2 (t) = π n=1 (2n − 1)3  c Pearson Education Limited 2011   431  432  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Graphs of the functions f1 (t) and f2 (t) for −2π < t < 2π are  18   2a x, 0              n, then m + n > 2n which implies that Dn+m (t2 − 1)n = 0 so that Im,n = 0 If m = n then  Im,n = In,n = (−1)  1  n −1  = (2n)!(−1)  (t2 − 1)n D2n (t2 − 1)n dt   1  n −1    (t2 − 1)n dt  1  (1 − t2 )n dt  = 2(2n)! 0  Making the substitution t = sin θ then gives   π/2  cos2n+1 θdθ = 2(2n)!  In,n = 2(2n)! 0  2 2 ... 2n + 1 3  22n+1 (n!)2 = 2n + 1 and the result follows. 41(c)  f(t) = c0 P0 (t) + c1 P1 (t) + c2 P2 (t) + . . .  Multiplying by P0 (t)     1 −1  giving    f(t)P0 (t)dt = c0   1  (−1)1dt + −1  1 −1  P20 (t)dt = 2c0  1  (1)1dt = 0 = 2c0 so that c0 = 0 0  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  461  Multiplying by P1 (t) ,     1 −1  giving    f(t)P1 (t)dt = c1   0  −1  P21 (t)dt =  1  (−1)tdt + −1  1  2 3 c1 , so that c1 = 3 2  (1)tdt = 1 = 0  Likewise,      1 −1  f(t)P2 (t)dt = c2  1 −1  2 a1 3  P22 (t)dt =  2 c2 5  giving 1 2      0  1 (−1)(3t − 1)dt + 2 −1  0  (1)(3t2 − 1)dt = 0 =  2  and    1    1 −1  f(t)P3 (t)dt = c3  1 −1  P23 (t)dt =  2 c2 , so that c2 = 0 5  2 c3 7  giving 1 2  42    0  1 (−1)(5t − 3t)dt + 2 −1    1  (1)(5t3 − 3t)dt = −  3  0  1 7 2 = c3 , so that c3 = − 4 7 8  Taking f(x) = c0 P0 (x) + c1 P1 (x) + c2 P2 (x) + c3 P3 (x) + . . .  and adopting the same approach as in 41(c) gives   −1  giving    1  f(x)P0 (x)dx = c0   1 −1  P20 (x)dx = 2c0  1  1 1 = 2c0 , so that c0 = 2 4 0  1  1 2 f(x)P1 (x)dx = c1 P21 (x)dx = c1 3 −1 −1 xdx =  c Pearson Education Limited 2011   462  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  giving    1  1 1 2 = c1 , so that c1 = 3 3 2 0  1  1 2 f(x)P2 (x)dx = c2 P22 (x)dx = c2 5 −1 −1 x2 dx =  giving    1 2  1  2 1 5 = c2 , so that c2 = 8 5 16 0  1  1 2 f(x)P3 (x)dx = c3 P23 (x)dx = c3 7 −1 −1 x(3x2 − 1)dx =  giving 1 2    1  x(5x3 − 3x)dx = 0 = 0  43(a)  2 c3 , so that c3 = 0 7  L0 (t) = et (t0 e−t ) = 1 L1 (t) = et (−te−t + e−t ) = 1 − t  Using the recurrence relation, L2 (t) = (3 − t)L1 (t) − L0 (t) = t2 − 4t + 2 L3 (t) = (5 − t)L2 (t) − 4L1 (t) = (5 − t)(t2 − 4t + 2) − 4(1 − t) = 6 − 18t + 9t2 − t3  43(b) This involves evaluating the integral combinations of m and n.  43(c)  ∞   If f(t) =  ∞ 0  e−t Lm (t)Ln (t)dt for the 10  cr Lr (t) to determine cn , multiply throughout by e−t Ln (t)  r=0  and integrate over (0, ∞)   ∞  −t    ∞ ∞  e Ln (t)f(t)dt = 0  0  cr e−t Lr (t)Ln (t)dt  r=0  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  463  Using the orthogonality property then gives   ∞    −t  ∞  e Ln (t)f(t)dt = cn 0  0  giving cn =  44(a)  e−t Ln (t)Ln (t)dt  1 (n!)2  = cn (n!)2  ∞ e−t Ln (t)f(t)dt, n = 0, 1, 2, . . . 0  By direct use of formula, 2  H0 (t) = (−1)0 et  /2 −t2 /2  e =1 2 2 d H1 (t) = (−1)et /2 e−t /2 = t dt Using recurrence relation, Hn (t) = tHn−1 (t) − (n − 1)Hn−2 (t) H2 (t) = t.t − 1.1 = t2 − 1 H3 (t) = t(t2 − 1) − 2(t) = t3 − 3t H4 (t) = t(t3 − 3t) − 3(t2 − 1) = t4 − 6t2 + 3  44(b) This involves evaluating the integral combinations of n and m .  44(c)  If f(t) =  ∞   ∞  e−t  2  −∞  /2  Hn (t)Hm (t)dt for the 10  cr Hr (t) to determine cn , multiply throughout by  r=0  e−t  2  /2  Hn (t) and integrate over (−∞, ∞) giving   ∞  e −∞  −t2 /2   Hn (t)f(t)dt =  ∞  ∞   −∞ r=0  ∞  = cn = cn  cr e−t  2  e−t  −∞  2  /2  /2  Hn (t)Hr (t)dt  Hn (t)Hn (t)dt   (2π)n!  c Pearson Education Limited 2011   464  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  so that 1 cn =  n! (2π)  45(a)    ∞  e−t  2  /2  −∞  f(t)Hn (t)dt  Directly from the formula, T0 (t) = cos 0 = 1 T1 (t) = cos(cos−1 t) = t  then from the recurrence relationship T2 (t) = 2t(t) − 1 = 2t2 − 1 T3 (t) = 2t(2t2 − 1) − t = 4t3 − 3t T4 (t) = 2t(4t3 − 3t) − (2t2 − 1) = 8t4 − 8t2 + 1 T5 (t) = 2t(8t4 − 8t2 + 1) − (4t3 − 3t) = 16t5 − 20t3 + 5t  45(b)  Evaluate the integral   1 Tn (t)Tm (t)  dt for the 10 combinations of n −1 (1 − t2 )  and m . ∞  If f(t) = cr Tr (t) to obtain cn , r=0  cn Tn (t)/ (1 − t2 ) and integrate over (−1, 1) giving  45(c)    1 −1    multiply throughout by  ∞  cr Tn (t)Tr (t)  dt (1 − t)2 −1 r=0  1 Tn (t)Tn (t)  = cn dt Tn = 0, 1, 2, 3, . . . (1 − t2 ) −1  T (t)f(t) n dt = (1 − t2 )  1   =  c0 π, cn π2 ,  n=0 n = 0  Hence the required results.  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  465  46(a)  To show that they are orthonormal on (0, T) , evaluate the integral  T 0  Wn (t)  Wm (t)dt for the 10 combinations of n and m . For example,     T  T  W0 (t)W0 (t)dt = 0  0  and it is readily seen that this extends to     T  W1 (t)W2 (t)dt = 0  0  T /4  1 dt + T    T /2  T /4  T 0  1 at = 1 T  W2n (t)dt = 1  (−1) dt + T    3T /4  T /2  c Pearson Education Limited 2011   1 dt + T    T  3T /4  (−1) dt = 0 T  466  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  46(b) f(t) = c0 W0 (t) + c1 W1 (t) + c2 W2 (t) + . . . where f(t) is the square wave of Exercise 40. In this case T = 2π. Multiplying throughout by the appropriate Walsh function and integrating over (0, 2π) gives     2π  2π  W0 (t)f(t)dt = c0 0  0  1 W20 (t)dt = c0 , W0 (t) = √ 2π  giving   2π 1  π 1f(t)dt = √ dt − dt = 0 2π 0 0 π  2π  2π √1 , 2 2π W1 (t)f(t)dt = c1 W1 (t)dt = c1 , W1 (t) = √1 , − 0 0 2π 1 c0 = √ 2π    2π  0                T1 T2 4MK · π T1 + T2  c Pearson Education Limited 2011   8 The Fourier Transform Exercises 8.2.4   1  0  at −jωt  e e  F(jω) =    ∞  dt +  −∞  e−at e−jωt dt  0  2a = 2 a + ω2  2    0  Ae  F(jω) =   −jωt    −T T  =  T  dt +  −Ae−jωt dt  0  2jA sin ωt dt 0  2jA (1 − cos ωT) ω 4jA ωT = sin2 ω 2   ωT = jωAT2 sinc2 2  =  3     T At At −jωt +A e + A e−jωt dt F(jω) = − dt + T T −T 0   T At + A cos ωt dt − =2 T 0   ωT 2 = AT sinc 2   0    Exercise 2 is T × derivative of Exercise 3, so result 2 follows as (jω × T) × result 3. Sketch is readily drawn.  c Pearson Education Limited 2011   490  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  4    2  2Ke−jωt dt = 8K sinc(2ω)  F(jω) = −2 1    Ke−jωt dt = 2K sinc(ω)  G(jω) = −1  H(jω) = F(jω) − G(jω) = 2K(4 sinc(2ω) − sinc(ω)) 5    −1  e  F(jω) =  −jωt    1  dt +  −2  e  −jωt    2  dt +  −1  −e−jωt dt  1   1  jω 2(e − e−jω ) − (e2jω − e−2jω ) = jω = 4 sinc(ω) − 2 sinc(2ω) 6 1 F(jω) = 2j 1 f̄(a) = 2j =    π a  π −a  π a  π −a sin ω πa  (ejat − e−jat )e−jωt dt  e  jat −jωt  e  1 dt = 2j    π a π −a  ej(a−ω)t dt  j(a − ω)  F(jω) = f̄(a) + f̄(−a) =  7   F(jω) =  ∞  π 2jω sin ω ω2 − a2 a  e−at . sin ω0 t.e−jωt dt  0  = f̄(ω0 ) − f̄(−ω0 )  1 ∞ (−a+j(ω0 −ω)t) where f̄(ω0 ) = e dt 2j 0     1 1 1 1 = = 2j a − j(ω0 − ω) 2j (a + jω) − jω0 ω0 ∴ F(jω) = (a + jω)2 + ω20  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 8  1 Fc (x) = 4  deﬁne g(x, b) =    a  (ejt + e−jt )(ejxt + e−jxt ) dt  0 a  ej(b+x)t dt  0  1 [ej(b+x)a − 1] j(b + x) 1 Fc (x) = [g(x, 1) + g(x, −1) + g(−x, 1) + g(−x, −1)] 4  1 sin(1 + x)a sin(1 − x)a = + 2 1+x 1−x =  9  Consider F(x) =  a 0  1.ejxt dt  −j (cos ax + j sin ax − 1) x sin ax Fc (x) = Re F(x) = x 1 − cos ax Fs (x) = Im F(x) = x =  10  Consider F(x) =  ∞ −at jxt e e 0  =  dt  a + jx a2 + x2  a + x2 x Fs (x) = Im F(x) = 2 a + x2 Fc (x) = Re F(x) =  a2  Exercises 8.3.6 11  Obvious  12  (jω)2 Y(jω) + 3jωY(jω) + Y(jω) = U(jω) 1 U(jω) (1 − ω2 ) + 3jω 1 H(jω) = (1 − ω2 ) + 3jω Y(jω) =  c Pearson Education Limited 2011   491  492  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  13 → sinc  ω 2  → e−iω3/2 + eiω3/2 sinc  ω 2  2 (sin(2ω) − sin(ω)) ω = 4 sinc(2ω) − 2 sinc(ω)  =    14 F(jω) =  T 2 T −2  cos(ω0 t)e−iωt dt  1 1 T T sin(ω0 − ω) + sin(ω0 + ω) ω0 − ω 2 ω0 + ω 2  T T  sin(ω0 + ω) 2 T sin(ω0 − ω) 2 = + T 2 (ω0 − ω) 2 (ω0 + ω) T2  =  ω = ±ω0  Evaluating at ω = ±ω0 ⇒   T T T sinc(ω0 − ω) + sinc(ω0 + ω) F(jω) = 2 2 2 15   F(jω) =  T  cos ω0 t.e−jωt dt  0  1 = [f̄(ω0 ) + f̄(−ω0 )] 2  where  T  f̄(ω0 ) =  ej(ω0 −ω)t dt  0  =  1 [ej(ω0 −ω)T − 1] j(ω0 − ω)  c Pearson Education Limited 2011   ω = ω0  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  1 1 (ej(ω0 −ω)T − 1) F(jω) = 2 j(ω0 − ω)  1 −j(ω0 +ω)T (e − − 1) j(ω0 − ω)  jω0 T /2 T −jωT /2 e sin(ω0 − ω) =e ω0 − ω 2  −jω0 T /2 e T sin(ω0 + ω) + ω0 + ω 2  ω = ω0  ω = ±ω0  Checking at ω = ±ω0 gives   T −jωT /2 jω0 T /2 T T −jω0 T /2 F(jω) = e e sinc(ω0 − ω) + e sinc(ω0 + ω) 2 2 2  16    1  F(jω) =  sin 2t.e−jωt dt  −1   1 1 −j(ω−2)t = e − e−j(ω+2)t dt 2j −1  1 f̄(a) = e−j(ω−a)t dt = 2 sinc(ω − a) −1  1 1 F(jω) = f̄(a) − f̄(−a), a = 2 2j 2j = j[sinc(ω + 2) − sinc(ω − 2)]  Exercises 8.4.3 17 I H(s) =  s2    1 + 3s + 2  ∞  H(jω) =  h(t) = (e−t − e−2t )ξ(t)  (e−t − e−2t )e−jωt dt =  0  =  2−  1 as required. + 3jω  ω2  s+2 II H(s) = 2 s +s+1  h(t) = e−1/2t    1 1 − 1 + jω 2 + jω √  √  √ 3 3 t + 3 sin t ξ(t) cos 2 2  c Pearson Education Limited 2011   493  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  ∞ e−(1/2tjω−jω0 )t dt Consider G(ω0 ) =  494  0  =  1 2  1 + j(ω − ω0 )  √ √ 3 3 1 1 (G(ω0 ) − G(−ω0 )), ω0 = H(jω) = G(ω0 ) + G(−ω0 ) + 2 2 2j 2 2 + 4jω 6 So H(jω) = + 2 4 + 4jω − 4ω 4 + 4jω − 4ω2 2 + jω = 1 − ω2 + jω 18  P(jω) = 2ATsinc ωT  So  F(jω) = (e−jωτ + eiωτ )P(jω) = 4AT cos ωτ sinc ωT  19  G(s ) =  (s )2 √ (s )2 + 2s + 1  G(jω) =  =  −ω2 √ 1 − ω2 + 2jω 1 ω2  1 √ − 1 + 2 ωj  Thus, | G(jω) |→ 0 as ω → 0 and | G(jω) |→ 1 as ω → ∞ High-pass ﬁlter.  20  g(t) = e−a|t| −→ G(jω) = f(jt) =  a2  2a + ω2  1 G(jt) −→ πg(−ω) = πe−a|ω| 2 c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 21  F{f(t) cos ω0 t} =  1 1 F(j(ω − ω0 )) + F(j(ω + ω0 )) 2 2 F(jω) = 2T sinc ωT  ∴ F{PT (t) cos ω0 t}   = T sinc(ω − ω0 )T + sinc(ω + ω0 )T  Exercises 8.5.3 22  1 2π    ∞ −∞  πδ(ω − ω0 )e  jωt  1 dω + 2π    ∞  −∞  πδ(ω + ω0 )ejωt dω  1 jω0 t (e + e−jω0 t ) 2 = cos ω0 t =  23  F{e±jω0 t } = 2πδ(ω ∓ ω0 ) 1 ∴ F{sin ω0 t} = {2πδ(ω − ω0 ) − 2πδ(ω + ω0 )} 2j = jπ[δ(ω + ω0 ) − δ(ω − ω0 )]  ∞ 1 jπ[δ(ω + ω0 ) − δ(ω − ω0 )]ejωt dω 2π −∞ j = [e−jω0 t − e+jω0 t ] = sin ω0 t 2 ∞  24  G(jω) =  g(t)e−jωt dt; G(jt) =  −∞   So  ∞ −∞  g(ω)e−jωt dω  ∞  f(t)G(jt) dt  ∞   ∞ −jωt f(t) g(ω)e dω dt = −∞ −∞  ∞   ∞ −jωt g(ω) f(t)e dt dω = −∞ −∞  ∞  ∞ g(ω)F(jω)dω = g(t)F(jt) dt = −∞  −∞  −∞  c Pearson Education Limited 2011   495  496  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  25  Write result 24 as  ∞    ∞  f(ω)F{g(t)}dω = −∞ ∞    −∞  ∞  f(ω)F{G(jt)}dω =  so −∞  −∞  F{f(t)}g(ω)dω F{f(t)}G(jω)dω  ⎫ g(t) → G(jω) ⎬ Now G(jt) → 2πg(−ω) symmetry ⎭ G(−jt) → 2πg(ω)  ∞  ∞ f(ω).2πg(ω)dω = F(jω)G(−jω)dω Thus, −∞ −∞  ∞  ∞ 1 f(t)g(t) dt = F(jω)G(−jω)dω or 2π −∞ −∞ 26  F{H(t) sin ω0 t}    ∞   1 1 du = πj δ(ω − u + ω0 ) − δ(ω − u − ω0 ) πδ(u) + 2π −∞ ju    1 1 j 1 = πδ(ω + ω0 ) − πδ(ω − ω0 ) + − 2 2 ω + ω0 ω − ω0  πj  ω0 δ(ω + ω0 ) − δ(ω − ω0 ) − 2 = 2 ω − ω20  27 A an = T    d/2  e−jnω0 t dt =  −d/2  f(t) =  nω0 d Ad sinc , T 2  ω0 = 2π/T  ∞ Ad  nω0 d jnω0 t e sinc , T n=−∞ 2  ∞ nω0 d 2πAd  δ(ω − nω0 ) sinc F(jω) = T n=−∞ 2  Exercises 8.6.6 28 T = 1,  N = 4,  Δω = 2π/(4 × 1) =  c Pearson Education Limited 2011   π 2  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  G0 =  G1 =  G2 =  3  n=0 3  n=0 3   gn e−j ×n×0×π/2 = 2 gn e−j ×n×1×π/2 = 0 gn e−j ×n×2×π/2 = 2  n=0  G3 =  3   gn e−j ×n×3×π/2 = 0  n=0  G = {2, 0, 2, 0} 29 N = 4, Wn = e−j nπ/2 ⎡  1 0 ⎢ gn = ⎣ 1 0  0 1 0 1  ⎡  ⎤ ⎡ G00 1 ⎢G ⎥ ⎢1 G = ⎣ 10 ⎦ = ⎣ 1 G01 0 G11 Bit reversal gives  1 0 −1 0 1 −1 0 0  ⎤⎡ ⎤ ⎡ ⎤ 2 1 0 1⎥⎢0⎥ ⎢0⎥ ⎦⎣ ⎦ = ⎣ ⎦ 0 1 0 0 0 −1 0 0 1 1  ⎤⎡ ⎤ ⎡ ⎤ 2 2 0 0⎥⎢0⎥ ⎢2⎥ ⎦⎣ ⎦ = ⎣ ⎦ 0 0 −j 0 0 j  ⎡ ⎤ 2 ⎢0⎥ G=⎣ ⎦ 2 0  30 Computer experiment.  31 Follows by direct substitution.  c Pearson Education Limited 2011   497  498  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Exercises 8.9.3 32 We have θc =  π 2  so   jθ  D(e ) =  1,  |θ| ≤  0,  |θ| >  π 2 π 2  The ﬁlter coeﬃcients are given by 1 hd (n) = 2π =  1 2π    π  D(ejθ )ejnθ dθ   −π π 2 π −2  ejnθ dθ   nπ  1 , n = 0 = sin nπ 2   1 nπ = sinc 2 2 Hence,  h±5 = 0.06366, h±4 = 0, h±3 = −0.10610 h±2 = 0, h±1 = 0.31831, h0 = 0.5  Thus, the non-causal transfer function is D̃(z) = 0.06366z−5 − 0.1066z−3 + 0.31832z−1 + 0.5 + 0.31831z − 0.10610z3 + 0.06366z5 and the causal version is D(z) = 0.06366 − 0.10660z−2 + 0.31831z−4 + 0.5z−5 + 0.31831z−6 − 0.010660z−8 + 0.06366z−10 33 The Hamming window coeﬃcients are given by   πk , |k| ≤ 5 wH (k) = 0.54 + 0.46 cos 5 Note that wH (±4)andwH (±2)are not needed. Now wH (±5) = 0.08000, wH (±3) = 0.39785, wH (±1) = 0.91215, wH (0) = 1 and the causal transfer function is found by multiplying the ﬁlter coeﬃcients by the appropriate wH giving D(z) = 0.00509(1 + z−10 ) − 0.04221(z−2 + z−8 ) + 0.29035(z−4 + z−6 ) + 0.5z−5 c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  499  Plots of the frequency responses for both Exercises 32 and 33 are given in the following ﬁgure. 1.4 Rectangular Window Hamming Window  1.2  Frequency response  1  0.8  0.6  0.4  0.2  0  −3pi  −2pi  −pi  0  pi  2pi  q in radians  Review exercises 8.10 1   FS (x) =  2  t sin xt dt + 0  2    1  sin xt dt = 1  sin x cos 2x − x2 x  πt π π + H(t − 2) f(t) = − H(−t − 2) + (H(t + 2) − H(t − 2)) 2 4 2 1 + πδ(ω) jω   1 −2jω F{H(t − 2)} = e + πδ(ω) jω F{H(t)} =  c Pearson Education Limited 2011   3pi  500  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition    −1 2jω −1 + πδ(−ω) = e + πδ(ω) F{H(−t − 2)} = e jω jω   π π 2 −jωt π H(t − 2) te dt + F F{f(t)} = F{− H(−t − 2)} + 2 4 −2 2   2jω  =  −πj sinc 2ω ω  3 F {H(t + T/2) − H(t − T/2)} = T sinc  ωT 2  F {cos ω0 t} = π [δ(ω + ω0 ) + δ(ω − ω0 )] Using convolution, π F {f(t)} = 2π    ∞  T T sinc (ω − u) (δ(u + ω0 ) + δ(u − ω0 ))du 2 −∞    T T T sinc(ω − ω0 ) + sinc(ω + ω0 ) = 2 2 2 4    1 1 [πδ(ω − ω0 ) + πδ(ω + ω0 )] ∗ πδ(ω) + F {cos ω0 tH(t)} = 2π jω 1 = 2π      1 {πδ(ω − u − ω0 ) + πδ(ω − u + ω0 )} πδ(u) + du ju −∞ ∞  =  π jω [δ(ω − ω0 ) + δ(ω + ω0 )] + 2 2 ω 0 − ω2  5 F{f(t) cos ωc t cos ωc t} =  F(jω + jωc ) + F(jω − jωc ) ∗ π [δ(ω − ωc ) + δ(ω + ωc )] 2  1 ∞ [F(j(u + ωc )) + F(j(u − ωc ))] × = 4 −∞ [δ(ω − u − ωc ) + δ(ω − u + ωc )] du 1 1 = F(jω) + [F(jω + 2jωc ) + F(jω − 2jωc )] 2 4 c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  501  Or write as 1 f(t) (1 + cos 2ωc t) 2  etc.  6 H(t + 1) − H(t − 1) ↔ 2 sinc ω By symmetry, 2 sinc t ↔ 2π(H(−ω + 1) − H(−ω − 1)) = 2π(H(ω + 1) − H(ω − 1)) 7(a) Simple poles at s = a and s = b. Residue at s = a is eat /(a − b) , at s = b it is ebt /(b − a) , thus 1 eat − ebt H(t) a−b  f(t) =  7(b) Double pole at s = 2, residue is d lim s→2 ds    es t (s − 2) (s − 2)2 2   = te2t  So f(t) = te2t H(t)  7(c) Simple pole at s = 1, residue e−t , double pole at s = 0, residue d lim s→0 ds    es t s+1   = (t − 1)H(t)  Thus, f(t) = (t − 1 + e−t )H(t)  8(a)    ∞  y(t) = −∞  Thus,   − sin ω0 t =  ∞ −∞  h(t − τ)u(τ) dτ  h(t − τ) cos ω0 τdτ = f(t), say  c Pearson Education Limited 2011   502  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  If u(τ) = cos ω0 (τ + π/4)   ∞  y(t) = −∞    ∞  = −∞  h(t − τ) cos ω0 (τ + π/4) dτ  h(t − (τ − π/4)) cos ω0 τ dτ = f(t + π/4) = − sin ω0 (t + π/4)  8(b) Since sin ω0 t = cos ω0 (t − π/2ω0 )   ∞  y(t) = −∞    ∞  =   −∞ ∞  = −∞  h(t − τ) sin ω0 t dτ  h(t − τ) cos ω0 (τ − π/2ω0 ) dτ h(t − (τ + π/2ω0 )) cos ω0 τ dτ  = f(t − π/2ω0 ) = − sin(ω0 t − π/2) = cos ω0 t 8(c) ejω0 t = cos ω0 t + j sin ω0 t This is transformed from above to − sin ω0 t + j cos ω0 t = j ejω0 t 8(d) Proceed as above using e−jω0 t = cos ω0 t − j sin ω0 t 9 F(sgn(t)) = F(f(t)) = F(jω) =  2 , obvious jω  Symmetry, F(jt) =  2 ↔ 2/πf(−ω) = 2πsgn(−ω) jt  That is, 1 ↔ −πsgn(ω) jt c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition   or − 10  1 1 g(t) = − ∗ f(t) = − πt π  so    1 πt  ∞    ↔ jsgn(ω)  f(τ) 1 dτ = t−τ π  −∞  1 g(x) = π    ∞ −∞    ∞  −∞  f(τ) dτ = FHi (t) τ−t  f(t) dt = FHi (x) t−x  So from Review Exercise 9 FHi (jω) = jsgn(ω) × F(jω) so |FHi (jω)| = |jsgn(ω)| |F(jω)| = |F(jω)| and arg(FHi (jω)) = arg(F(jω)) + π/2, ω ≥ 0 Similarly arg(FHi (jω)) = arg(F(jω)) − π/2, ω < 0 11 First part, elementary algebra. 1 FHi (x) = π 1 1 = π x2 + a2    ∞      −∞  12(a)  ∞ −∞  t (t2  +  a2 )(t  − x)  dt   xt a2 t − dt + t2 + a2 t − x t2 + a2 a = 2 x + a2   ∞  f(t) dt = FHi (x) −∞ t − x  1 ∞ f(a + t) dt H{f(a + t)} = π −∞ t − x  f(t) 1 ∞ dt = FHi (a + x) = π −∞ t − (a + x) 1 H{f(t)} = π  c Pearson Education Limited 2011   503  504  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  12(b)   1 ∞ f(at) H{f(at)} = dt π −∞ t − x  1 ∞ f(t) dt = FHi (ax), a > 0 = π −∞ t − ax  12(c)   1 ∞ f(−at) dt H{f(−at)} = π −∞ t − x  1 ∞ f(t) =− dt = −FHi (−ax), a > 0 π −∞ t + ax  12(d)   H  1 = π  df dt    1 = π    ∞ −∞  ∞  f(t)  + t − x −∞  ∞ −∞  f (t) dt t−x   f(t) dt (t − x)2  Provided lim f(t)/t = 0, then |t|→∞    df dt  H    1 = π    ∞  −∞  f(t) 1 d dt = 2 (t − x) π dx  x π    ∞  −∞  f(t) 1 dt + t−x π  ∞ −∞    ∞  1 f(t) dt = π −∞    ∞ −∞  = H{tf(t)} 13  f(t) dt t−x  d FHi (x) dx  =  12(e)    From Review Exercise 10 FHi (t) = −  1 ∗ f(t) πt  So from Review Exercise 9, F{FHi (t)} = jsgn (ω) × F(j ω) c Pearson Education Limited 2011   tf(t) dt t−x  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition so F(jω) = −jsgn (ω) × F{FHi (t)} Thus,    ∞  f(t) = −∞  1 1 FHi (τ)dτ = − π(t − τ) π    ∞ −∞  1 FHi (x)dx (x − τ)  14 fa (t) = f(t) − jFHi (t) F{fa (t)} = F(jω) − j(jsgn (ω))F(jω) = F(jω) + sgn (ω)F(jω)  =  2F(jω), ω > 0 0, ω<0  15 F{H(t)} =  1 + πδ(ω) = F(jw) jω  Symmetry, F(j t) =  1 + πδ(t) ↔ 2πH(−ω) = 2π[1 − H(ω)] jt = 2π[Fδ(t) − H(ω)]  or H(ω) ↔  1 j + δ(t) 2πt 2  Thus, F−1 {H(ω)} = Then  j 1 + δ(t) 2πt 2     j 1 1 ∗ f(t) = f(t) − j − ∗ f(t) f̂(t) = 2 δ(t) + 2 2πt πt   = f(t) − jFHi (t) When f(t) = cos ω0 t, ω0 > 0, then F(jω) = π[δ(ω − ω0 ) + δ(ω + ω0 )] so F{f̂(t)} = 2πδ(ω − ω0 ) c Pearson Education Limited 2011   505  506  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  whence f̂(t) = f(t) − jFHi (t) = ejω0 t = cos ω0 t + j sin ω0 t and so FHi (t) = − sin ω0 t When g(t) = sin ω0 t, ω0 > 0 G(jω) = jπ[δ(ω + ω0 ) − δ(ω − ω0 )] and thus ĝ(t) = −jejω0 t = −j(cos ω0 t + j sin ω0 t) so H{sin ω0 t} = cos ω0 t 16 If h̄(t) = 0, t < 0, then when t < 0 h̄e (t) =  1 1 h̄(−t), and h̄o (t) = − h̄(−t) 2 2 that is, h̄o (t) = −h̄e (t)  When t > 0, then h̄e (t) =  1 1 h̄(t), and h̄o (t) = h̄(t) 2 2  that is, h̄o (t) = h̄e (t) That is, h̄o (t) = sgn (t)h̄e (t) ∀t Thus, h̄(t) = h̄e (t) + sgn (t)h̄e (t) When h(t) = sin t H(t) ,  ⎧ ⎪ ⎨  1 sin t, t > 0 2 h̄e (t) = ⎪ ⎩ − 1 sin t, t < 0 2 and since sgn (t) h̄e (t) =  1 sin t ∀t 2  the result is conﬁrmed. c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition Then taking the FT of the result, # $ H̄(jω) = H̄e (jω) + F sgn (t)h̄e (t)   2 ∗ H̄e (jω) jω # $ = H̄e (jω) + jH H̄e (jω)  1 = H̄e (jω) + 2π    When  ∞  a ω − j a2 + ω2 a2 + ω2  e−at e−jwt dt =  H̄(jω) = −∞    then H  a 2 a + ω2    or H  a 2 a + t2     =−  a2  ω + ω2  a2  x + x2   =−  Finally,  H  at a2 + t2    x 1 = −x 2 + a + x2 π   So H  t 2 a + t2    ∞  −∞   =  a2  a a2 dt = a2 + t2 a2 + x2  a + x2  17(a)   ∞  FH (s) =  e−at (cos 2πst + sin 2πst) dt =  0  17(b)   FH (s) =    18  T  (cos 2πst + sin 2πst) dt = −T    ∞ −∞    E(s) − j O(s) =  1 sin 2πst πs  ∞  f(t) cos 2πst dt O(s) =  E(s) =  a + 2πs + 4π2 s2  a2  f(t) sin 2πst dt −∞  ∞  f(t)e−j2πst dt = F(js)  −∞  c Pearson Education Limited 2011   507  508  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  From Review Exercise 17(a) FH (s) =  1 + πs 2 + 2π2 s2  1 , 2 + 2π2 s2  O(s) =  F(j s) =  1 − jπs 2 + 2π2 s2  whence E(s) = so  πs 2 + 2π2 s2  agreeing with the direct calculation,   ∞  F(js) =  e−2t e−j2πst dt =  0    19 H{f(t − T)} =   ∞  −∞  1 − jπs 2 + 2π2 s2  f(t − T) cas 2πst dt  ∞  =  f(τ) [cos 2πsτ(cos 2πsT + sin 2πsT)+ −∞  sin 2πsτ(cos 2πsT − sin 2πsT)] dt = cos 2πsTFH (s) + sin 2πsTFH (−s) 20 The Hartley transform follows at once since FH (s) = {F(js)} − {F(js)} =  1 1 δ(s) + 2 sπ  From time shifting,    1 1 1 1 + cos πs δ(s) + FH (s) = sin πs δ(−s) − 2 sπ 2 sπ   =  cos πs − sin πs 1 δ(s) + 2 πs   21  ∞  δ(t) cas 2πst dt = 1  H{δ(t)} = −∞  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  509  From Review Exercise 18, it follows that the inversion integral for the Hartley   transform is  ∞  f(t) = −∞  FH (s) cas 2πstds  and so the symmetry property is simply f(t) ↔ FH (s) =⇒ FH (t) ↔ f(s) Thus, H{1} = δ(s) At once,   H{δ(t − t0 )}  ∞ −∞  δ(t − t0 ) cas 2πst dt = cas 2πst0  By symmetry, H{cas 2πs0 t} = δ(s − s0 )  22  1 = 2    1 1 FH (s − s0 ) + FH (s + s0 ) 2 2 ∞  −∞  f(t) {cos 2π(s − s0 )t + sin 2π(s − s0 )t  + cos 2π(s + s0 )t + sin 2π(s + s0 )t} dt  ∞ f(t) cos 2πs0 t [cos 2πst + sin 2πst] dt = −∞  = H{f(t) cos 2πs0 t} From Review Exercise 21, setting f(t) = 1 H{cos 2πs0 t} =  1 (δ(s − s0 ) + δ(s + s0 )) 2  also H{sin 2πs0 t} = H{cas 2πs0 t} − H{cos 2πs0 t} 1 1 = δ(s − s0 ) − (δ(s − s0 ) + δ(s + s0 )) = (δ(s − s0 ) − δ(s + s0 )) 2 2 c Pearson Education Limited 2011   510  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  23    t  (1 + τ2 )−1 dτ = tan−1 t +  −∞  Thus, F{tan  −1   t} = F  t  2 −1  (1 + τ )  π 2   dτ − F  π  2  ∞ π 2 −1 =F (1 + τ ) H(t − τ)dτ − F 2 −∞   π 1 =F ∗ H(t) − F 1 + t2 2     1 1 π =F × + πδ(ω) − × 2πδ(ω) 2 1+t jω 2 −∞    But from Review Exercise 1   F e−|t| = and so by symmetry,   F  whence  #  F tan and so  −1  $  t = πe  1 1 + t2  −|ω|   ×    2 1 + ω2  = πe−|ω|   π 1 + πδ(ω) − × 2πδ(ω) jω 2  # $ πe−|ω| F tan−1 t = jω  24 1 1 [1 + cos ω0 t] ↔ [2πδ(ω) + πδ(ω − ω0 ) + πδ(ω + ω0 )] 2 2 and H(t + T/2) − H(t − T/2) ↔ 2T sinc ω   so F {x(t)} =  ∞ −∞  2T sinc (ω − u)    1 × πδ(u) + (δ(ω − ω0 ) + δ(u + ω0 )) du 2 c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition   1 1 = T sinc ω + sinc (ω − ω0 ) + sinc (ω + ω0 ) 2 2 25  511  1 2πνr H(ν) = f(r) cas 4 r=0 4 3  1 [f(0) + f(1) + f(2) + f(3)] 4 1 H(1) = [f(0) + f(1) − f(2) − f(3)] 4 1 H(0) = [f(0) − f(1) + f(2) − f(3)] 4 1 H(0) = [f(0) − f(1) − f(2) + f(3)] 4  H(0) =  so  ⎡  1 1 ⎢1 T= ⎣ 4 1 1  1 1 −1 −1  1 −1 1 −1  ⎤ 1 −1 ⎥ ⎦ −1 1  By elementary calculation, T2 = 1/4T and if T−1 exists, T−1 = 4T. Since T−1 T = I , it does. Then ⎡  1 ⎢1 T−1 H = ⎣ 1 1  1 1 −1 −1  ⎤ ⎤ ⎡ ⎤⎡ f(0) H(0) 1 1 −1 −1 ⎥ ⎢ H(1) ⎥ ⎢ f(1) ⎥ ⎦ ⎦=⎣ ⎦⎣ f(2) H(2) 1 −1 f(3) H(3) −1 1  c Pearson Education Limited 2011   9 Partial Differential Equations Exercises 9.2.6 1  Diﬀerentiating ∂2 u ∂2 u 2 = −a cos at sin bx and 2 = −b2 cos at sin bx 2 ∂t ∂x  and hence a2 = c2 b2 2 Since the function is a function of a single variable only, on diﬀerentiating ∂2 u ∂2 u 2  = α f and = f and hence α2 = c2 . 2 2 ∂t ∂x 3  Veriﬁed by diﬀerentiation.  4  Diﬀerentiating 1 1 cos(r − ct) − sin(r − ct) 2 r r 2 2 1 Zrr = 3 cos(r − ct) + 2 sin(r − ct) − cos(r − ct) r r r 2 c Ztt = − cos(r − ct) r  Zr = −  and it can be checked that the equation is satisﬁed. 5  Applying the given expression into the equation gives  α αt α e V = eαt V  or V  = V κ κ and the solution clearly depends on the sign of α. α = 0 ⇒ V  = 0  and hence V = A + Bx  α > 0 ⇒ V  = a2 V and hence V = A sinh ax + B cosh ax where a2 =  α κ  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  513  α < 0 ⇒ V = −b2 V and hence V = A cos bx + B sin bx α where b2 = − κ  6  Substituting the expression into the LHS of the equation, ∂V ∂ = nrn−1 (3 cos2 θ − 1) and ∂r ∂r    2 ∂V r = n(n + 1)rn (3 cos2 θ − 1) ∂r  and in the RHS, ∂V ∂ = −rn 6 cos θ sin θ and ∂θ ∂θ    ∂V sin θ ∂θ   = −rn 6(− sin3 θ + 2 cos2 θ sin θ)  Applying these expressions into the equation, n(n + 1)rn (3 cos2 θ − 1) − rn 6(− sin2 θ + 2 cos2 θ) = 0 or n(n + 1)rn (3 cos2 θ − 1) − rn 6(−1 + 3 cos2 θ) = 0 and hence n(n + 1) − 6 = 0 with roots −3 and 2. 7  Now, c2  ∂2 u = −c2 m2 e−kt cos mx cos nt ∂x2  ∂u = −ke−kt cos mx cos nt − ne−kt cos mx sin nt ∂t and ∂2 u = k2 e−kt cos mx cos nt + 2kne−kt cos mx sin nt − n2 e−kt cos mx cos nt ∂t2 Thus, ∂2 u ∂u + 2k = [k2 −n2 +2k(−k)]e−kt cos mx cos nt + [2kn + 2k(−n)]e−kt cos mx sin nt 2 ∂t ∂t and comparing with the LHS gives k2 + n2 = c2 m2  8  Diﬀerentiating Vx = 3x2 + ay2 and Vy = 2axy c Pearson Education Limited 2011   514  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  and evaluating x  ∂V ∂V +y = 3x3 + axy2 + 2axy2 = 3(x3 + axy2 ) = 3V ∂x ∂y  gives the required result. Now Vxx + Vyy = 6x + 2ax ⇒ rhs = 0 if a = −3 Putting r2 = x2 + y2 , ﬁrst note that 2r  ∂r ∂r = 2x and 2r = 2y ∂x ∂y  so x u = r3 V ⇒ ux = r3 Vx + 3r2 V = r3 Vx + 3rxV r and diﬀerentiating again x x2 uxx = r3 Vxx + 3r2 Vx + 3rxVx + 3rV + 3 V r r Similarly for uyy and adding the two expressions and using the two previous results uxx + uyy = r3 (Vxx + Vyy ) + 6r(xVx + yVy ) + 6rV + 3  (x2 + y2 ) V r  the quoted answer is proved.  9  Diﬀerentiating φxx = Φxx e−kt/2 and φ t = Φt e  −kt/2  k − Φe−kt/2 2   φtt =   k2 Φtt − kΦt + Φ e−kt/2 4  and substituting gives   1 1 −kt/2 2 k2 k2 c Φxx − Φtt + kΦt − Φ − kΦt + Φ 0 = φxx − 2 (φtt + kφt ) = 2 e c c 4 2 Neglecting terms in k2 , the RHS is just the wave equation for Φ .  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 10(a)  515  With r = g = 0, the equations become − Ix = cvt − vx = LIt  ⇒ −Ixx = cvxt = c(−LIt )t = −cLItt  and hence satisfy the wave equation. 10(b)  When L = 0, − Ix = gv + cvt − vx = rI  ⇒ vxx = r(gv + cvt ) = rgv + rcvt  and the result is a heat conduction equation with an additional forcing term rgv. Applying W = vegt/c it may be noted that  g  Wxx = vxx egt/c and Wt = vt + v egt/c c and hence comparing with the previous equation, Wxx = (rc)Wt which satisﬁes the usual heat conduction equation. The exponential damps the solution to zero over a long time. 10(c) First eliminate I −vxx = rIx + LIxt = r(−gv − cvt ) + L(−gv − cvt )t vxx = Lcvtt + (rc + Lg)vt + rgv Apply in the expression for a 1 rg vxx = vtt + 2avt + v Lc Lc and substitute v = we−at 1 rg −at wxx e−at = (wtt − 2awt + a2 w)e−at + 2a(wt − aw)e−at + we Lc Lc  1 rg  2 w wxx = wtt + −a + Lc Lc c Pearson Education Limited 2011   516  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  But   rg 1 rg  = − −a + Lc Lc 4 2    g2 r2 rg + + 2 L2 Lc c2    1 =− 4    g2 r2 rg + − 2 L2 Lc c2   =0  from the condition rc = gL and hence the variable w satisﬁes the wave equation. Such a transmission line is called a balanced line and transmits the signal exactly in shape, though damped by the exponential.  11  Applying the expression into the equation −a2 f sin(ay + b) = (f − 2af ) sin(ay + b)  so f must satisfy f − 2af + a2 f = 0 which is a second-order constant coeﬃcient equation with equal roots a. Thus, f = (A + Bx)eax and agrees with the given result.  12  The given formula can be checked by diﬀerentiation.  The method of Section 9.2.5 solves the equations dy df dx = = 2 2 x y 4x y dx dy yields = −→ x = Ay x y df dy = 2 2 y 4x y  yields  yields  −→ df = 4A2 y3 dy −→ f = A2 y4 + B  The arbitrary constants A and B can be isolated as A=  x and B = f − x2 y2 y  x When x = x(t), y = y(t) are given on a curve with f = f(t) then A(t) = y   x for some function F. Putting this into B(t) gives and hence t = F y c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 517      x x =g for some function g and thus f is of the required form B F y y   x f(x, y) = x y + g y 2 2  The MAPLE code produces this solution also. Given that x = 1 − t, y = t, f = t2   x and t = (1 − t) t + g y  x 1−t = y t  2  2 2    y3 (y + 2x) x = Eliminating t gives g . y (x + y)4 Use MAPLE to solve, as follows: with (PDEtools): Q12:=x ∗ diff(u(x,y),x)+y ∗ diff(u(x,y),y)-4 ∗ xˆ2 ∗ yˆ2; sol:=pdsolve(Q12,u(x,y)); # this instruction gives the solution sol:= u(x,y) = x 2 y 2 + F1(y/x) eval (sol,{x=1-t,y=t,u(x,y)=t^2}); simplify (eval(%,t=1/(1+z))); solve(%,_F1(1/z)); # gives the solution (1 + 2z)/(1 + z)4   ∂u ∂ ∂u +u =0⇒ + u = f(y) 13 Write as ∂x ∂y ∂y where f is an arbitrary function. Using an integrating factor ey , this partial diﬀerential equation can now be written as ∂ (uey ) = ey f(y) ∂y which can be integrated to give u = e−y [H(x) + G(y)] where H(x) and G(y) are arbitrary functions.  c Pearson Education Limited 2011   518 14  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition The method of Section 9.2.5 solves the equations dx dy du = 2 = 2 x y (x + y)u  These equations yield dx dy = 2 2 x y du = u    2 A + y 1 − Ay    yields  −→  1 1 = −A x y  yields  dy −→ u =  By2 = Bxy 1 − Ay  Hence from any starting curve, with parameter s, u x−y and B(s) = xy xy   x−y , where F is an arbitrary function Eliminating s gives u = xyF xy determined by the conditions on the starting curve. A(s) =  MAPLE gives this general solution. Putting in the data x = s, y = 1, f = s2 , the arbitrary function becomes 1 and the given result for u follows. F(z) = 1−z  Exercises 9.3.4 15  From the separated solutions (9.25) choose u = sin(λx) cos(λct)  Clearly, both initial conditions (a) and (b) are satisﬁed for λ = 1. The d'Alembert solution is obtained from equation (9.19) as 1 u = [sin(x + ct) + sin(x − ct)] 2 which gives the same result when the sines are expanded. 16 First note that sin x(1 + cos x) = sin x + 12 sin 2x The two initial conditions imply that the solution is of the form u = A sin x sin ct + B sin 2x sin 2ct and matching the conditions gives A = 1/c and B = 1/4c.  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 17  519  The MAPLE implementation is as follows:  f:=(x-c ∗ t)/(1+(x-c ∗ t)ˆ2)+(x+c ∗ t)/(1-(x+c ∗ t)ˆ2); simplify (f);  # gives the simplification - nearly  simplify(diff(f,x,x)-diff(f,t,t)/c^2);  18  Let    # gives zero as required  denote diﬀerentiation with respect to (ct − r) and 'dot' with respect to  (ct + r) ; then the terms of the spherically symmetric wave equation are 1 1 utt = [f (ct − r) + g̈(ct + r)] 2 c r and 1 1 [f(ct − r) + g(ct + r)] + [−f (ct − r) + ġ(ct + r)] 2 r r 2 2 = 3 [f(ct − r) + g(ct + r)] − 2 [−f (ct − r) + ġ(ct + r)] r r 1  + [f (ct − r) + g̈(ct + r)] r  ur = − urr  Collecting terms together 1 2 1  utt − urr − ur = 3 r2 (f + g̈) − 2(f + g) − 2r(f − ġ ) 2 c r r  −r2 (f + g̈) + 2(f + g) + 2r(f − ġ) = 0 so the equation is satisﬁed for any functions f and g. The two terms represent an outward spherical wave emanating from the origin and an inward wave converging into the origin. Note the singular behaviour at r = 0.  19  The equation (9.28) is split by the trigonometric formula into two parts 2π2 u π π 1 3π 1 3π = sin (x − ct) + sin (x + ct) − sin (x − ct) − sin (x + ct) . . . (4l) l l 9 l 9 l 1 3π 1 5π π sin (x − ct) + . . .] = [sin (x − ct) − sin (x − ct) + l 9 l 25 l π 1 3π 1 5π + [sin (x + ct) − sin (x + ct) + sin (x + ct) + . . .] l 9 l 25 l c Pearson Education Limited 2011   520  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  The two terms depend on (x−ct) and (x+ct) respectively and represent travelling waves in the +x and −x directions. 20  The d'Alembert solution is obtained from equation (9.19) as 1 u= 2c  x+ct   x exp(−x2 )dx x−ct  which on integration gives the quoted result. 21  Again the d'Alembert solution is obtained from equation (9.19) as u = [F(x − ct) + F(x + ct)]/2  where F is the function given in the exercise. 22  Try a solution of the form u = f(x + ky) , so the equation gives  3f + 6kf + k2 f = 0 ⇒ 3 + 6k + k2 = 0 √ which has solutions k = −3 ± 6 and hence the characteristics are x + (−3 + x + (−3 − 23  √ √  6)y = const 6)y = const  Substituting 6f − λf − λ2 f = 0 ⇒ λ = 2, −3  and hence a solution of the form u = f(x + 2t) + g(x − 3t) The initial conditions give x2 − 1 = f(x) + g(x) and 2x = 2f (x) − 3g (x) Integrating and solving for f and g produces the solution u=  1 [4(x + 2t)2 + (x − 3t)2 − 5] 5  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 24  521  Diﬀerentiating g ∂u g = − 2 cos ωt + cos ωt ∂r r r  and ∂2 u = ∂r2    2g 2g g − + r3 r2 r   cos ωt  Applying these expressions into the equation gives   2g 2g 2 g g g + − − + + r3 r2 r r r2 r   cos ωt = −  ω2 g cos ωt c2 r  and cancelling produces the equation for g as ω2 g=0 c2 This simple harmonic equation has sine and cosine solutions which are written in g +  the form ω ω (b − r) + B sin (b − r) c c The second boundary condition is now satisﬁed by applying A = 0 and the ﬁrst condition gives ω B u(a, t) = β cos ωt = sin (b − a) cos ωt a c and hence B is known and the required solution is g = A cos  u(r, t) =  25  aβ cos ωt sin ωc (b − r) r sin ωc (b − a)  This question is similar to Example 9.11 but initially the velocity is given and  the displacement is zero. On the initial line t = 0 the solution u = f(x + t) + g(x − t) satisﬁes condition (a) only if f = −g and therefore the condition (b) gives ut (x, 0) = 2f (x) = exp(−|x|) Integrate to obtain f c Pearson Education Limited 2011   522  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 1 x 2e  for x < 0 1− for x > 0 where it has been arranged that the function goes to zero at inﬁnity and matches f=  1 −x 2e  at x = 0. The numerical solution can now be computed from the values on the initial line given by f. The values at subsequent times t = 0.5, 1.0, 1.5, 2.0, . . . can be computed easily from u(x, 0.5) = f(x + 0.5) − f(x − 0.5) u(x, 1) = f(x + 1) − f(x − 1) u(x, 1.5) = f(x + 1.5) − f(x − 1.5) etc. to give the quoted solution. On a spreadsheet, applying f (x) into column B corresponding to values of x = −3, −2.5, . . . , 2.5, 3, then a typical entry, which can be copied onto the other entries in the column, in column D, D7 reads  +B8 − B6  in column E, E7 reads  +B9 − B5  in column F, F7 reads  +B10 − B4  etc. It is instructive to derive the exact solution and then compare with the numerical solution.  u(x, t) =  26  for x < −t ex sinh t 1 − e−t cosh x for − t < x < t e−x sinh t for x > t  From the possible separated solutions, the conditions (a) and (b) imply that u = cos λct sin λx  is the only one that satisﬁes these conditions. The condition (c) gives sin λπ = 0 ⇒ λ = N which is an integer. Thus, a superposition of these solutions gives ∞  aN cos Nct sin Nx  u(x, t) = N =1  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  523  and the condition (d) gives the standard Fourier problem of evaluating the coeﬃcients in ∞  πx − x = 2  aN sin Nx N =1  The coeﬃcients are obtained from the usual integral and the result follows by two integrations by parts aN = 27  4 (1 − cos Nπ) πN3  Taking the Laplace transform with respect to t, in equation (9.33) both u(x,0)  and ut (x, 0) are zero from conditions (a) and (b); so the equation is 2d  2  U = s2 U with solution U = Aesxlc + Be−sxlc 2 dx From condition (d), the constant A = 0 since the solution must be bounded for all c  x > 0. The condition (c) is transformed to U(0, s) =  s2  aω + ω2  and hence the solution for U takes the form aω e−sxlc 2 +ω and the exponential just shifts the solution as U(x, s) =  s2    x  x    H ω t− u(x, t) = a sin ω t − c c It is easily checked that all the conditions are satisﬁed by the function. For x > ct the wave has not reached this value of x, so u = 0 beyond this point.  Exercises 9.3.6 28  The problem is best solved by using MATLAB.  Explicit n=5;L=0.25;x=[0:1/(n-1):1];z=zeros(1,5); zz=.25 ∗ [0 .25 .5 .25 0] c Pearson Education Limited 2011   524  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition zzz=[0,2 ∗ zz([2:n-1])-z([2:n-1])+L ∗ (zz([1:n-2]) -2 ∗ zz([2:n-1])+zz([3:n])),0] % gives  0  0.1250 0.2188  0.1250  0  z=zz;zz=zzz; zzz=[0,2 ∗ zz([2:n-1])-z([2:n-1])+L ∗ (zz([1:n-2]) -2 ∗ zz([2:n-1])+zz([3:n])),0] % gives  0  0.1797  0.2656  0.1797  0  Implicit n=5;L=0.25; a=[-L 2 ∗ (1+L)-L];A=eye(n);for i=2:n-1,A(i,i-1:i+1)=a;end b=[L/2 1-L L/2];C=eye(n);for i=2:n-1,C(i,i-1:i+1)=b;end u=[0 0 0 0 0]';v=C ∗ u+.25 ∗ [0 .25 .5 .25 0]'; B=inv(A); w=4 ∗ B ∗ v-u;w' % gives 0 0.1224 0.2245 0.1224 0 u=v;v=w;w=4 ∗ B ∗ v-u;w' % gives 0 0.1741 0.2815 0.1741 0 29  Again MATLAB is a convenient method for the explicit calculation. n=6;L=0.01;delt=0.02; format long z=eye(1,6);zz=[sin(delt/2 ∗ pi),0 0 0 0 0] % gives 0.0314107 0 0 0 0 0 zzz=[sin(2 ∗ delt/2 ∗ pi),2 ∗ zz([2:n-1])-z([2:n-1]) +L ∗ (zz([1:n-2])-2 ∗ zz([2:n-1])+zz([3:n])),0] % gives 0.062790 0.000314 0 0 0 0 z=zz;zz=zzz; zzz=[sin(3 ∗ delt/2 ∗ pi),2 ∗ zz([2:n-1])-z([2:n-1]) +L ∗ (zz([1:n-2])-2 ∗ zz([2:n-1])+zz([3:n])),0] % gives 0.094108 0.001249 0.000003 0 0 0  30  Care must be taken to include the '+2' term but the MATLAB  implementation is quite straightforward. c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  525  Explicit n=6;L=0.25;delt=0.2;x=[0:0.2:1];z=x. ∗ (1-x) zz=[0,(1-L) ∗ z([2:n-1])+L ∗ (z([1:n-2]) +z([3:n]))/2,0]+[0,deltˆ2 ∗ ones(1,4),0] % gives 0 0.1900  0.2700  0.2700  0.1900  0  zzz=[0,2 ∗ zz([2:n-1])-z([2:n-1])+L ∗ (zz([1:n-2])-2 ∗ zz([2:n-1]) +zz([3:n])),0] +[0,2 ∗ deltˆ2 ∗ ones(1,4),0] % gives 0 0.2725 z=zz;zz=zzz;  0.3600  0.3600  0.2725  0  zzz=[0,2 ∗ zz([2:n-1])-z([2:n-1])+L ∗ (zz([l:n-2])-2 ∗ zz([2:n-1]) +zz([3:n])),0]+[0,2 ∗ deltˆ2 ∗ ones(1,4),0] % gives 0 0.3888 0.5081 0.5081 0.3888 0 Implicit n=6;L=0.25;delt=0.2;x=[0:0.2:1]; a=[-L2 ∗ (1+L)-L];A=eye(n);for i=2:n-1,A(i,i-1:i+1)=a;end b=[L/2 1-L L/2];C=eye(n);for i=2:n-1,C(i,i-1:i+1)=b;end u=(x. ∗ (1-x))';v=C ∗ u+[0;deltˆ2 ∗ ones(4,1);0] % gives 0 0.1900 0.2700 0.2700 0.1900 0 B=inv(A); w=B ∗ (4 ∗ v+[0;2 ∗ deltˆ2 ∗ ones(4,1);0])-u %gives 0 0.2319 0.3191 0.3191 0.2319 0 u=v;v=w;w=B ∗ (4 ∗ v+[0;2 ∗ deltˆ2 ∗ ones(4,1);0])-u %gives 31  0  0.2785  0.3849  0.3849  0.2785  0  The problem is now more diﬃcult since there is an inﬁnite region. The  simplest way to cope with this diﬃculty for small times is to impose boundaries some distance from the region of interest. Hopefully, the eﬀect of any sensible boundary condition would only aﬀect the solution marginally. For longer times, an alternative strategy must be sought. In the current problem, the region x = −1 to 2 is chosen with the solution quoted in the region x = 0 to 1. Explicit n=16;L=0.25;delt=0.2; x=[-1:0.2:2];z=x. ∗ (1-x); c Pearson Education Limited 2011   526  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition zz=[-2,(1-L) ∗ z([2:n-1])+L ∗ (z([l:n-2]) +z([3:n]))/2,-2]+deltˆ2 ∗ ones(1,16) % gives 0.0300 0.1900 0.2700 0.2700  0.1900  0.0300  zzz=[-2,2 ∗ zz([2:n-1])-z([2:n-1])+L ∗ (zz([1:n-2])-2 ∗ zz([2:n-1]) +zz([3:n])),-2]+2 ∗ deltˆ2 ∗ ones(1,16) % gives 0.1200 0.2800 0.3600 0.3600  0.2800  0.1200  -0.1200  z=zz;zz=zzz; zzz=[-2,2 ∗ zz([2:n-1])-z([2:n-1])+L ∗ (zz([1:n-2])-2 ∗ zz([2:n-1]) +zz([3:n])),-2]+2 ∗ deltˆ2 ∗ ones(1,16) % gives 0.2700 0.4300 0.5100 0.5100 0.4300  0.2700  Implicit n=16;L=0.25;delt=0.2; x=[-1:0.2:2]; a=[-L 2 ∗ (1+L)-L];A=eye(n);for i=2:n-1,A(i,i-1:i+1)=a;end b=[L/2 1-L L/2];C=eye(n);for i=2:n-1,C(i,i-1:i+1)=b;end u=(x. ∗ (1-x))';v=C ∗ u+deltˆ2 ∗ ones(16,1) % gives 0.0300 0.1900 0.2700 0.2700 0.1900 0.0300 B=inv(A); w=B ∗ (4 ∗ v+2 ∗ deltˆ2 ∗ ones(16,1))-u % gives 0.0800 0.2400 0.3200 0.3200 u=v;v=w;w=B ∗ (4 ∗ v+2 ∗ deltˆ2 ∗ ones(16,1))-u % gives 0.1495 0.3099 0.3900 0.3900  0.2400  0.0800  0.3099  0.1495  Exercises 9.4.3 32  From the set of separated solutions in equation (9.40), the only ones that  satisfy condition (a) are u = e−αt cos λx and the second condition (b) implies   cos λ = 0 ⇒ λ = n + 12 π where n is an integer. The third condition (c) can be rewritten as a cos u= 2    3πx 2   + cos  πx 2  for 0 ≤ x ≤ 1 when t = 0  Thus, the complete solution is u=  a [exp(−κπ2 t/4) cos(πx/2) + exp(−9κπ2 t/4) cos(3πx/2)] 2  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 33  527  If v = ru , then diﬀerentiating produces vr = u + rur vrr = 2ur + rurr  and hence urr + 2 urr = 1r vrr . Applying these expressions into the spherically symmetric heat conduction equation gives 1 1 vrr = vt r κr Cancelling out the r, it is seen that v satisﬁes the standard heat conduction equation. If v remains bounded it may be noted that u → 0 as r → ∞. Since u = v/r , the boundary conditions for v are yields  u(a, t) = T0 −→ v(a, t) = aT0 for t > 0 yields  u(b, t) = 0 −→ v(b, t) = 0 for t > 0 yields  u(r, 0) = 0 −→ v(r, 0) = 0 for a < r < b The ﬁrst two of these conditions are satisﬁed by the given expression  v(r, t) = aT0  b−r − b−a  ∞  AN e−κλ t sin 2  r − a  N =1  b−a   Nπ  and the third gives the Fourier problem b−r = b−a  ∞  AN sin  r − a  N =1  b−a  Nπ  The coeﬃcients can be obtained from integration or from standard tables of Fourier series as AN = 2/πN .  34  Substituting into the partial diﬀerential equation gives the ODE 4ηF + (2 + η)F − αF = 0  and applying F = exp(κη) gives the equation c Pearson Education Limited 2011   528  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  4ηκ2 + (2 + η)κ − α = 0 which is clearly satisﬁed by κ = − 14 and α = − 21 and produces the classic similarity solution.  35  Diﬀerentiating ∂u ∂u = −βf(x) cos(x − βt) and = f (x) sin(x − βt) + f(x) cos(x − βt) ∂t ∂x  and  ∂2 u = f (x) sin(x − βt) + 2f (x) cos(x − βt) − f(x) sin(x − βt) ∂x2  Putting these expressions into the heat conduction equation and equating the sine and the cosine terms gives −βf = 2f and f − f = 0 Both equations can be satisﬁed only if β = 2 and f = Ae−x ; the solution is then u = Ae−x sin(x − 2t) Physically, the slab of material is given an initial temperature of Ae−x sin x , the temperature is zero at inﬁnity and at the end x = 0 the temperature is periodic taking the form u(0, t) = −A sin 2t and hence A = −u0 .  36  The suggested substitution gives, on diﬀerentiation, θ − θ0 = ue−ht θt = (ut − hu)e−ht θxx = uxx e−ht  Putting the expressions into the given equation and cancelling the exponential gives ut − hu = κuxx − hu ⇒ ut = κuxx c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  529  and produces the standard equation for u. The term h(θ − θ0 ) is a heat loss term proportional to the excess temperature over an ambient temperature θ0 ; this is the usual Newton cooling through a surface.  37  First it is clear that the ﬁnal steady solution is U = 0. The general separated  solution in equation (9.40) is u = e−αt (A sin λx + B cos λx) where  λ2 = α/κ  Condition (a) can only be satisﬁed at x = 0 if A = 0. Condition (b) then implies   that cos λl = 0 so that λl =  1 n+ 2   π  and hence the solution takes the form  2 2     ∞ π t 1 πx 1 cos n + u(x, t) = an exp −κ n + 2 l2 2 l n=0 The initial condition given in (c) leads to the Fourier problem of evaluating the coeﬃcients in the expression  u0  1 x − 2 l      ∞  =  an cos n=0  1 n+ 2    πx l  These can be evaluated by standard integration or using the standard series ∞  (−1)n cos 2n + 1 n=0 and  ∞  1 cos 2 (2n + 1) n=0    1 n+ 2    πx π = for − l < x < l l 4    π2  x 1 πx = 1− for 0 < x < 2l n+ 2 l 8 l  Thus, the coeﬃcients can be calculated as a combination of the last two expressions as an = u0  8 2 (−1)n − π2 (2n + 1)2 π 2n + 1  c Pearson Education Limited 2011   530 38  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition At any time t , the sine term ensures that the sum is zero at x = 0 and L so  only the ﬁrst term survives at the end points, and therefore v = v0 at x = 0 and v = 0 at x = L. From the basic solutions obtained in the text, or by inspection, it is clear that the heat equation is also satisﬁed. The additional condition at t = 0 leads to the Fourier series problem  nπx  x + 1− cn sin L L n=1   0 = v0  ∞  with the coeﬃcients evaluated from   L    1−  0 = v0 0   nπx  L x sin dx + cn L L 2  0 An integration by parts gives cn = − 2v nπ as required.  39  At the ends of the bar the conditions are  (a) u = 0 at x = 0 for t > 0, (b) u = 0 at x = l for t > 0 and the initial condition is (c) u = 10 for 0 < x < l at t = 0. From the set of separated solutions in equation (9.40), the only ones that satisfy condition (a) are u = e−αt sin λx . The condition (b) then gives sin λl = 0 ⇒ λ =  nx l  where n is an integer. The  solution is therefore of the form   ∞  an exp  u(x, t) = n=1  −κn2 π2 t l2   sin   nπx  l  The third condition (c) reduces the problem to a Fourier series, namely ∞  an sin  10 =   nπx   n=1  l  Integrating in the usual way over the interval l 10 sin   nπx  l  dx =  l an 2  0  gives an =  20 nπ (1  − cos nπ) and the required result.  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 40  531  Taking Laplace transforms of the equation with respect to t leads to sφ̄ − φ(x, 0) = aφ̄ +  b s  so using condition (b) the required equation is φ̄ =  s b φ̄ − a as  This equation has the obvious particular integral φ̄ = b/s2 and it is convenient to write the complementary function in terms of sinh and cosh functions. The solution is thus b φ̄ = 2 + A sinh s       s s x + B cosh x a a  The boundary conditions in (a) transform to φ̄(±h, s) = 0 and hence      s s h + B cosh h a a     s s b h + B cosh h 0 = 2 − A sinh s a a  b 0 = 2 + A sinh s  Clearly, A = 0 and B is easily calculated to give   s   cosh x b  as  φ̄ = 2 1 − s cosh ah To transform back to the real plane needs either some tricky integrations or the use of advanced tables of Laplace transform pairs. Tables give the solution, φ 1 2 16h2 2 = (h − x ) + b 2a aπ3  ∞  (−1)n (2n − 1)πx (2n − 1)2 π2 at cos exp − 3 2 (2n − 1) 4h 2h n=1  Exercises 9.4.5 41 In the explicit formulation equation (9.48), the MATLAB implementation can be written using the 'colon' notation to great eﬀect. n=6;L=0.5;x=0:0.2:1,u=x.^2 c Pearson Education Limited 2011   532  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition v=[0,L ∗ (u([1:n-2])+u([3:n]))+(1-2 ∗ L) ∗ u([2:n-1]),1] % gives  0  0.0800  0.2000  0.4000  0.6800  1.0000  u=v;v=[0,L ∗ (u([l:n-2])+u([3:n]))+(1-2 ∗ L) ∗ u([2:n-1]),1] % gives 0  0.1000  0.2400  0.4400  0.7000  1.0000  Repeating the last line gives successive time steps.  42  Again a MATLAB formulation solves the problem very quickly; lamda (L in  the program) is chosen to be 0.4 and time step 0.05. Explicit n=6;L=0.4;u=[0 0 0 0 0 1]; v=[0,L ∗ (u([1:n-2])+u([3:n]))+(1-2 ∗ L) ∗ u([2:n-1]),exp(-0.05)] % gives 0 0 0 0 0.4000 0.9512 for p=2:20,u=v;v=[0,L ∗ (u([1:n-2])+u([3:n]))+(1-2 ∗ L) ∗ u([2:n-1]),exp(-p ∗ 0.05)];end v % gives the values at t = 1 as 0 0.1094 0.2104 0.2939 0.3497 0.3679  Repeating the last two lines produces the solution at successive times. Implicit There are some slight diﬀerences in the solution depending on how the right hand boundary is treated. Equation (9.49) is constructed in MATLAB again using the 'colon' notation L=0.4;M=2 ∗ (1+L);N=2 ∗ (1-L); n=6;u=zeros(n,1);u(n)=1; p=[-L M -L];A=eye(n);for i=2:n-1,A(i,i-1:i+1)=p;end q=[L N L];B=eye(n);for i=2:n-1,B(i,i-1:i+1)=q;end DD=inv(A) ∗ B; v=DD ∗ u;v(n)=exp(-0.05); % for first step for p=2:20,u=v;v=DD ∗ u;v(n)=exp(-p ∗ 0.05);end % gives for t =1  0 0.1082 0.2096 0.2955 0.3551 0.3679  Repeat the last line of code for further time steps.  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 43  533  The equations are easily produced in MATLAB. Because of the derivative  boundary condition, the region is extended to x = −0.2 and u(−0.2, t) is obtained from u(0.2, t) − u(−0.2, t) = 0.4 L=0.5;M=2 ∗ (1+L);N=2 ∗ (1-L);n=7; p=[-L M -L];A=eye(n);for i=2:n-1,A(i,i-1:i+1)=p;end A(1,3)=1;A(1,1)=-1 % gives LHS matrix q=[L N L];B=eye(n);for i=2:n-1,B(i,i-1:i+1)=q;end B(1,1)=0  % gives RHS matrix  rhs=[0.4 0 0 0 0 0 0]'; % gives vector from derivative condition at x=0 AA=inv(A); x=0:0.2:1,u=[-0.24,x. ∗ (1-x)]' % starting data v=AA ∗ (B ∗ u+rhs) % produces next time step -0.2800 -0.0400 0.1200 0.2002 0.2012 0.1269 0 u=v;v=AA ∗ (B ∗ u+rhs) % produces next time step -0.3197 -0.0799 0.0803 0.1613 0.1657 0.1034 0 Repeating the last line produces further time steps.  Exercises 9.5.2 44  From equation (9.52), the only separated solution to satisfy u → 0 as y → ∞  is u = (A sin μx + B cos μx)(Ceμy + De−μy ) with C = 0 Thus u = (a sin μx + b cos μx)e−μy To satisfy the boundary conditions, u = 0 at x = 0 ⇒ b = 0 u = 0 at x = 1 ⇒ sin μ = 0 ⇒ μ = nπ where n is an integer. The condition at y = 0 can be satisﬁed by a sum of terms over n. c Pearson Education Limited 2011   534  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  On y = 0, u =  ∞   an sin nπx =  n=1  1 16 (10 sin πx  − 5 sin 3πx + sin 5πx)  and the an can be obtained by inspection to give the required solution.  45  The four boundary conditions are satisﬁed by inspection and the Laplace  equation is satisﬁed by straightforward diﬀerentiation.  46 It can easily be checked that the function x2 y satisﬁes the given Poisson equation. The boundary conditions on u(x, y) become u(x, 0) = 0 for 0 ≤ x ≤ 1 u(x, 1) = sin πx for 0 ≤ x ≤ 1 u(0, y) = 0 for 0 ≤ y ≤ 1 u(1, y) = 0 for 0 ≤ y ≤ 1 The only solution in equation (9.52d) that satisﬁes these conditions is u = sin πx  sinh πy sinh π  and hence the ﬁnal result φ = x2 y + sin πx  47  sinh πy sinh π  Diﬀerentiating ur = Bnrn−1 sin nθ urr = Bn(n − 1)rn−2 sin nθ uee = −Bn2 rn sin nθ  and substitution gives LHS = B sin nθrn−2 [n(n − 1) + n − n2 ] = 0 = RHS and hence the Laplace equation in plane polars is satisﬁed. To be periodic in θ the constant n must be an integer. A solution of the equation is a sum of the expressions given, so that c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  535  ∞  Bn rn sin nθ  u(r, θ) = n=1  Putting the condition on the rim, r = a, gives the Fourier problem to calculate Bn as B1 = 3/4a and B3 = −1/4a2 and otherwise zero. Thus, 1  r 3 3 r sin θ − sin 3θ u(r, θ) = 4 a 4 a  48  Let D = x2 + y2 + 2x + 1; then the derivatives can be computed as ux =  2y(2x + 2) 4y 4y(2x + 2)2 and u = − xx D2 D2 D3  −2 2y2y 4y 8y 4y2 4y and u = + − + yy D D2 D2 D2 D3 Adding the two second derivatives gives uy =   1  2 2 2 3 16y(x + y + 2x + 1) − 16y(x + 1) − 16y D3 The RHS can easily be checked to be zero and hence the Laplace equation is ∇2 u =  satisﬁed. A similar process shows that v also satisﬁes the Laplace equation. The u and v come from the complex variable expression u + jv =  j(x − 1 + jy)(x + 1 − jy) j(x − 1 + jy) = x + 1 + jy (x + 1 + jy)(x + 1 − jy)  Multiplying out u + jv =  −2y + j(x2 + y2 − 1) x2 + y2 + 2x + 1  gives the expressions quoted in the question. A check can be made by using MAPLE. u:=2 ∗ y/(xˆ2+yˆ2+2 ∗ x+1); simplify(diff(u,x,x)+diff(u,y,y)); # gives zero as required -v follows similarly h:=I ∗ (x+I ∗ y-1)/(x+I ∗ y+1); simplify(evalc(h)); # gives the u and v of the question c Pearson Education Limited 2011   536  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  For ﬁxed u and v the two expressions can be rearranged as   2  v 2 1 1 1 2 2 x+ +y = and (x + 1) + y + = 2 2 v−1 (v − 1) u u  which are circles with radii  1 v−1  and  1 u  −v , centres ( v−1 , 0) and (−1, −1 u ) respectively.  Note that all the circles pass through the point ( −1,0). 49  This is an important example that illustrates that sensible solutions can only  be obtained if correct boundary conditions are set. First, it is a matter of simple diﬀerentiation to verify that the given function satisﬁes the Laplace equation. Again, since the sinh function is zero at x = 0 the ﬁrst condition is satisﬁed. Diﬀerentiating with respect to x, 1 ∂u 1 ∂u = cosh nx sin ny, so at x = 0, = sin ny ∂x n ∂x n The solution therefore satisﬁes all the conditions of the problem. It is known that the solution is unique. For any given n, however large, sinh nx can be made as large as required and even when divided by n2 it is still large; for instance, n = 10, x = 5 and y = π/200 gives u = 4.1 × 1018 . The 'neighbouring' problem has a boundary condition ux = 0 and solution u identically zero. For the values chosen for illustration, the maximum change at the boundary is 0.1; yet the solution changes by 1018 . Such behaviour is very unstable; these boundary conditions give a unique solution; yet small changes in the boundary produce huge changes in the solution. Figure 9.59 should be referred for a summary of the 'correctness' of boundary conditions.  50  It is useful in solution by separation to try to modify the problem so that the  function is zero on two opposite boundaries. Apply u = x + f(x, y) ; then f satisﬁes the Laplace equation and the four boundary conditions become f(0, y) = 0  f(1, y) = 0  for 0 < y < 1  f(x, 0) = −x f(x, 1) = 1 − x for 0 < x < 1 The solution given in equation (9.52d) is the appropriate one and the cosine can be omitted since it cannot satisfy the ﬁrst of the four conditions. Thus, c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  537  f = sin μx(a cosh μy + sinh μy) The second condition now gives sin μ = 0 ⇒ μ = Nπ where N is an integer. The solution therefore takes the form ∞  sin Nπx(aN cosh Nπy + bN sinh Nπy)  f= N =1  and the coeﬃcients are derived as Fourier series from the other two sides of the boundary as ∞  −x =  aN sin Nπx N =1 ∞  1−x=  sin Nπx(aN cosh Nπ + bN sinh Nπ) N =1  Straightforward integration gives aN =  2 cos Nπ 2 and (aN cosh Nπ + bN sinh Nπ) = Nπ Nπ  The ﬁnal solution is obtained by substituting back ∞ 1 2 sinh Nπy u=x+ sin Nπx cos Nπ cosh Nπy + (1 − cos Nπ cosh Nπ) π N sinh Nπ N =1 which can be tidied up to the given solution. 51 and  The boundary conditions on the four sides are u(0, y) = 0  u(a, y) = 0  u(x, 0) = 0  u(x, a) = u0  for  0                  0 where the equation  is elliptic. (b) y = 0 where the equation is parabolic and (c) y < 0 where the equation is hyperbolic. From equation (9.83), the characteristics are obtained from the solution of the equation  √ −y dy =± dx y  The equation only makes sense in the hyperbolic region where y < 0. Substitute z = −y and the diﬀerential equation becomes √ z 1 dz =± = ±√ dx z z which is easily integrated to 3 3 3 3 z 2 = ± x + K ⇒ (−y) 2 ± x = K 2 2  as the characteristics. 72  Diﬀerentiating    2B B 2 3 fx = 3Ax − 3 y(1 − y ) and fy = Ax + 2 (1 − 3y2 ) x x     B B 2 3 fxx = 6 Ax + 4 y(1 − y ) and fyy = Ax + 2 (−6y) x x   2  The given equation can be checked by substitution. Now 'AC − B2 ' = x2 (1 − y2 ) so ⎧ ⎨ elliptic if |y| < 1 parabolic if x = 0 or y = ±1 ⎩ hyperbolic if |y| > 1 c Pearson Education Limited 2011   558  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  73  Calculating 'AC − B2 ' = −4p2 +4q2 , it can be deduced that the equations are (a) p > q or p < −q then the equations are hyperbolic (b) p = q then the equations are parabolic  (c) −q < p < q then the equations are elliptic Using the substitution −p2 +q2 = − 41 (x4 −2x2 y2 +y4 −x4 −2x2 y2 −y4 ) = x2 y2 > 0, hence leads to the elliptic region when it is expected that the equation will look like the Laplace equation. vxx = vp + vq + x2 (vpp + 2vpq + vqq )  vx = xvp + xvq vy = −yvp + yvq  and  vyy = −vp + vq + y2 (vpp − 2vpq + vqq ) vxy = xy(−vpp + vqq )  It may be noted that vxx + vyy = 2vq + (x2 + y2 )vpp + 2(x2 − y2 )vpq + (x2 + y2 )vqq from which the required transformation to the Laplace follows immediately.  74  In this case, 'AC − B2 ' = −(xy)2 so the equation is hyperbolic away from  the axes. The characteristics are computed from equation (9.83) as dy =± dx   x2 y 2 y =± 2 x x  and the solution is obtained by integration as ln y = ± ln x + K ⇒ y = ax and yx = b so one set of characteristics are straight lines through the origin and the other are rectangular hyperbolas. The domain of dependence and the range of inﬂuence can now be sketched.  Review exercises 9.11 1  The boundary and initial conditions on y(x, t) are y=0  String initially at rest Initial displacement  ⇒  ∂y ∂t  =0  y = f(x) =  at  x=0  at  t=0 for 0 ≤ x ≤ b at t = 0 for b ≤ x ≤ a  εx b ε(a−x) a−b  and a  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  559  Out of the possible separated solutions (9.25) that satisfy the wave equation, the only one that satisﬁes the conditions at x = 0 and y = 0 is y = sin λx cos λct . The condition at x = a gives λa = nπ where n is an integer. Thus, the solution is a sum of such terms ∞  An sin  y=   nπx   n=1  a   cos  nπct a    and the ﬁnal condition produces the Fourier series ∞  f(x) =  An sin   nπx  a  n=1  The coeﬃcients are evaluated from the integral 1 aAn = 2  b   nπx  εx sin dx + b a  0  a   nπx  ε(a − x) sin dx a−b a  b  which, after some careful integration by parts, gives the required coeﬃcient 2εa2 An = 2 2 sin n π b(a − b)    nπb a    It is interesting to look at the solution for various values of b since the solution gives strengths of the harmonics for diﬀerent musical instruments. It is these values that give the characteristic sound of the instrument. For example, a violin has b = a/7; it is seen that A7 = 0 and sevenths do not occur for this instrument. 2 Taking the Laplace transform of the equation and the boundary conditions gives φ̄ = s2 φ̄ − sx2 2l s The most convenient form for the complementary function is and  φ̄(0, s) = 0, φ̄ (l, s) =  φ̄ = A cosh s(x − l) + B sinh s(x − l) The particular integral is a quadratic in x which when substituted into the equation gives c Pearson Education Limited 2011   560  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  particular integral =  x2 2 + 3 s s  The complete solution is therefore φ̄ = A cosh s(x − l) + B sinh s(x − l) +  2 x2 + 3 s s  Putting the two conditions on φ̄ into the solution φ̄ (0, s) = 0 ⇒ 0 = A cosh(−sl) + B sinh(−sl) +  2 s3  2l 2l 2l ⇒ = sB + ⇒B=0 s s s The Laplace transform of the solution is φ̄ (l, s) =  2 x2 2 cosh s(x − l) + 3− 3 s s s cosh sl The solution in real space can be obtained from advanced tables of transform pairs. φ̄ =  3  Take the separated solutions that are quoted and ﬁrst note that the conditions  y(0) = y(l) = 0 are satisﬁed. Secondly, substitute into equation (9.92) 1 c2    Tn  1 + Tn τ   = −Tn   nπ 2 l  This equation can then be solved in the standard way by looking for solutions of the form Tn = eαt which produces the quadratic equation  cnπ 2 1 α2 + α + =0 τ l The equation has roots ⎞ ⎛ #  # 2  2  cnπ 2 cnπ 1 1⎝ 1 l 1 ⎠=− ±j α= −4 1− − ± 2 τ τ l 2τ l 2τcnπ and hence the solution is   t Tn (t) = exp − 2τ   (an cos ωn t + bn sin ωn t)  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition and general solution  ∞  y=  Tn (t) sin n=1  The condition  ∂y ∂t (x, 0)  561   nπx  l  = 0 implies, on diﬀerentiation and substituting t = 0,  1 an + bn ωn 2τ From the other condition it is seen that the only term to survive is when n = 3. 0=−  Thus,  4 sin  3πx l      0 = an exp − 2τ       1 3πx cos ω3 0 + sin ω3 0 sin 2τω3 l  and hence the solution satisfying all the conditions is   t y(x, t) = 4exp − 2τ       1 3πx cos ω3 t + sin ω3 t sin 2τω3 l  and ω3 given in the question. 4  This problem is the extension of the wave equation to beams.  Simply  substituting the given form into the beam equation gives the equation (9.93) for V and the end conditions follow immediately. Again simply substituting sin, sinh, cos and cosh into the equation shows they are solutions of (9.93). A linear combination of the functions is also a solution and since it contains four arbitrary constants, it is the general solution. To satisfy the end conditions V(0) = 0 ⇒ A + B = 0 V (0) = 0 ⇒ C + D = 0 V(l) = 0  ⇒  A cosh αl + B cos αl + C sinh αl + D sin αl = 0  V (l) = 0  ⇒  A sinh αl − B sin αl + C cosh αl + D cos αl = 0  Thus, A(cosh αl − cos αl) = D(sinh αl − sin αl) A(sinh αl + sin αl) = D(− cos αl + cosh αl) c Pearson Education Limited 2011   562  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Dividing these two equations, and using the trigonometric and hyperbolic identities, produces the equation cos αl cosh αl = 1 from which the natural frequencies of vibration of the beam can be calculated.  5 The separated solutions in equation (9.40) that satisfy the condition at x = 0 are θ = e−αt cos λx . To satisfy the condition at x = l requires that  λl =  1 n+ 2   π where n is an integer  Thus, the solution takes the form   2  (2n + 1)π (2n + 1)πx θ(x, t) = exp − A2n+1 cos t 2l 2al n=0 ∞  and the initial condition now produces the usual Fourier series problem with 1 A2n+1 = 2  l f(x) cos  (2n + 1)πx dx 2l  0  Integration by parts is required to obtain the coeﬃcient for the given function f(x) = θ0 (l − x) as A2n+1 =  8θ0 l + 1)2  π2 (2n  and the temperature can be obtained from the series solution.  6 Evaluating the partial derivatives ∂φ ∂φ ∂2 φ 1 1 1 x = f = √ f , = − 3/2 f and 2 ∂x ∂t 2t ∂x t t Substituting the derivatives back into the equation 1 x 1 1 κ f = − 3/2 f ⇒ f = − zf t 2t 2κ c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  563  the problem has been reduced to an ordinary diﬀerential equation. Rearranging the equation df 1 z2  = − = − zdz which integrates to ln f +C f 2κ 4κ  2 z which on further integration gives or f = A exp − 4κ z f=A   2 v dv + B exp − 4κ  0  Substitute u = z √ 2 κ  f=a  &  v √ 2 κ  then the equation reduces to  e−u2 du + b where a and b are new arbitrary constants.  0  The solution is as required  f = a erf  z √ 2 κ   +b  For the particular problem, at t = 0  φ(x, 0) = 0  for all x > 0  at x = 0  φ(0, t) = φ0  for t > 0  In the expression containing the error function, these conditions give respectively  0 = a erf(∞) + b and φ0 = a erf(0) + b Since erf(0) = 0 and erf (∞) = 1, the solution can be constructed as  T(x, t) = T0 + φ0 1 − erf  7  x √ 2 κt    The problem is standard except for the treatment of the derivative boundary  conditions. Note the way that they are handled in the MATLAB implementation. c Pearson Education Limited 2011   564  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Explicit u=[1 1 1 1 1 1]; u=0.1 ∗ [u([2:6]),u(5)-0.4 ∗ u(6)]+0.8 ∗ u([1:6]) +.1 ∗ [u(2)-0.4 ∗ u(1),u([1:5])] % gives 0.9600 1.0000 1.0000 1.0000 1.0000 0.9600 u=0.1 ∗ [u([2:6]),u(5)-0.4 ∗ u(6)] +0.8 ∗ u([1:6])+.1 ∗ [u(2)-0.4 ∗ u(1),u([1:5])] % gives 0.9296 0.9960 1.0000 1.0000 0.9960 0.9296 u=0.1 ∗ [u([2:6]),u(5)-0.4 ∗ u(6)]+ 0.8 ∗ u([1:6])+.1 ∗ [u(2)-0.4 ∗ u(1),u([1:5])] % gives 0.9057 0.9898 0.9996 0.9996 0.9898 0.9057 Repeating the last line of code produces subsequent time steps. Implicit  A=[-2.4 2 0 0 0 0;1 -2 1 0 0 0;0 1 -2 1 0 0;0 0 1 -2 1 0; 0 0 0 1 -2 1;0 0 0 0 2 -2.4] u=[1;1;1;1;1;1]; B=2 ∗ eye(6)-0.1 ∗ A C=2 ∗ eye(6)+0.1 ∗ A E=inv(B) ∗ C u=E ∗ u % gives 0.9641 0.9984 0.9999 0.9999 0.9984 0.9641 u=E ∗ u % gives 0.9354 0.9941 0.9996 0.9996 0.9941 0.9354 u=E ∗ u % gives 0.9120 0.9881 0.9988 0.9988 0.9881 0.9120 Repeating the last line of code produces subsequent time steps.  8 From the possible solutions in equation (9.52), one that can be chosen to satisfy the conditions on x = 0 and y = a is u = sin μx sinh μ(a − y) c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  565  On x = a, u = 0 ⇒ sin μa = 0 ⇒ μa = nπ where n is an integer. The ﬁnal condition on y = 0 can be satisﬁed by taking a sum of such solutions ∞  x(a − x) =  bn sin   nπx   n=1  a  sinh nπ  The problem is the usual evaluation of the Fourier coeﬃcients a bn sinh nπ = 2  a (a − x) sin   nπx  a  dx  0  and after two integrations by parts gives  a 3 a bn = 2 (1 − cos nπ) 2 nπ All the even terms go to zero and putting n = 2r + 1, the expression quoted is recovered.  9 This exercise is a harder one since the regular mesh points do not lie on the boundary. y  y2 = x c 7 b a  4  5  6  1  2  3  x  At a point such as node 4 the lengths of the mesh are not uniform, so the second derivative needs to be approximated as c Pearson Education Limited 2011   566  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition      f =  f3 − f0 Δx     −  f0 − f1 Δx    2 Δx + Δx  for a typical conﬁguration x  x′ 1  0  3  The curve does not pass through the mesh points and the lengths are calculated √ √ as a = 0.25, b = 0.5 − 0.5 and c = 1.5 − 1. Working through the equations one at a time, using this formula where appropriate 1 + u2 + 2u4 − 4u1 = 0 u1 + u3 + 2u5 − 4u2 = 0 2 0.5 + a    u2 + 1 + 2u6 − 4u3 = 0    u5 − u4 1 − u4 2 1 − u4 u1 − u4 + + + = 0.5(0.5)2 0.5 a 0.5 + b b 0.5 u4 + u2 + 1 + u6 − 4u5 = 0.0625 u5 + u3 + u7 + 1 − 4u6 = 0.09375   1 − u7 u6 − u7 1 + 1 − 2u7 2 + = 1.5 + 0.52 c + 0.5 c 0.5  The equations can be transformed to matrix form as ⎡  −4 ⎢ 1 ⎢ ⎢ 0 ⎢ a = ⎢ 5.6569 ⎢ ⎢ 0 ⎣ 0 0 bT = [ −1  1 −4 1 0 1 0 0 0  0 1 −4 0 0 1 0  −1  2 0 0 −35.3137 1 0 0  −24.1985  0 2 0 5.3333 −4 1 0  −0.9375  0 0 2 0 1 −4 5.5192 −0.9063  ⎤ 0 0 ⎥ ⎥ 0 ⎥ ⎥ 0 ⎥ ⎥ 0 ⎥ ⎦ 1 −25.7980 −18.7788 ]  and the solution of Ax = b can be obtained from any package, for example MATLAB, as xT = [ 0.9850  0.9648  0.9602  0.9876  0.9570  c Pearson Education Limited 2011   0.9380  0.9286 ]  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  567  10 Clearly, at z = 0 the velocity u = U cos ωt so the given solution has the correct velocity on the wall. It also satisﬁes the equation of motion since substituting the derivatives ut = −Uωe−αz sin(ωt − αz) uz = −αUe−αz cos(ωt − αz) − αUe−αz sin(ωt − αz) uzz = α2 Ue−αz cos(ωt − αz) − 2α2 Ue−αz sin(ωt − αz) −α2 Ue−αz cos(ωt − αz) into the equation gives −Uω = −ν2α2 U which agrees precisely with the deﬁnition of α.  11  Diﬀerentiating    2 −1 −r2 r +t exp − Ut = kt 2 4 t 4t   2  r −2r k exp Ur = t 4t 4t    2  ∂ 2 r −2r −tk−1 2 3 (r Ur ) = 3r + r exp ∂r 2 4t 4t   k−1  r2 exp − 4t      k  and applying into the spherically symmetric heat equation tk−1 − 2    r2 3− 2t     2   2  r r r2 k−1 exp − =t exp − k+ 4t 4t 4t  gives the relation k = −3/2.  12 The equation is the same as Example 9.7 and can be dealt with in the same way, see also equation (9.14). However, the MAPLE solution is very straightforward.  with (PDEtools): rev 12:=diff(z(x,y),x) + diff(z(x,y),y); sol:=pdesolve (rev12,z(x,y)); # gives the solution  sol:=z(x,y)=_F1(y-x)  c Pearson Education Limited 2011   568  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  The boundary conditions x = s, y = −s, z = 2s for s > 0 give the solution z(x, y) = x − y  for  x>y  and not deﬁned otherwise. 13  The separated solutions in equation (9.52) that tend to zero for large y must  be of the form φ = (A sin μx + B cos μx)e−μy The condition φ(0, y) = 0 ⇒ B = 0 and φ(π, y) = 0 ⇒ sin μπ = 0; thus, the required solution is  ∞  φ=  cn e−ny sin nx  n=1  Yet again, the ﬁnal condition requires the evaluation of the coeﬃcients by Fourier analysis. Integration by parts gives π cn = 2  π x(π − x) sin nxdx =  2 (1 − cos nπ) n3  0  When n is even, the coeﬃcient is zero and the odd values give the result in the exercise. 14  First observe that the function χ = a2 − x2 satisﬁes ∇2 χ = −2 and second  that a separated solution of the Laplace equation derived in equation (9.52) is cosh μy cos μx and the overall solution is a combination of terms of these types.   The conditions that χ = 0 on x = ±a ⇒ cos μa = 0 ⇒ μa = n + 12 π where n is an integer. Thus, an appropriate solution is   ∞  χ(x, y) = a − x + 2  2  A2n+1 cosh n=0  1 n+ 2    y π cos a    1 n+ 2   π  x a  The function is even, so only the condition at y = b needs to be considered since the other boundary is satisﬁed by symmetry. Therefore,   ∞  0=a −x + 2  2  A2n+1 cosh n=0  1 n+ 2    b π cos a  c Pearson Education Limited 2011     1 n+ 2   π  x a  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  569  and the coeﬃcients are derived by Fourier analysis. From tables of Fourier series it may be noted that  (−1)n π3 πx  = cos (2n + 1) (a2 − x2 ) for − a < x < a 3 2 (2n + 1) 2a 32a n=0 ∞  and thus the coeﬃcient can be identiﬁed as A2n+1 =  (−1)n+1 32a2   π3 (2n + 1)3 cosh (2n+1)πb 2a  15  The possible separated solutions of the wave equation are given in equation  (9.25) and to satisfy conditions (a) and (b), the solution sin λx cos λct must be chosen. The condition at x = 1 implies that sin λ = 0 ⇒ λ = nπ where n is an integer. Thus, the solution takes the form ∞  an sin nπx cos nπt  u(x, t) = n=1  and condition (c) is substituted to give the Fourier series ∞  1−x=  an sin nπx n=1  The coeﬃcients are obtained from 1 an = 2  1 (1 − x) sin nπxdx 0  which can be integrated by parts to give an =  2 nx  and agrees with the quoted  answer. 16  Complete drainage at top and bottom implies u = 0 at z = 0 and z = h .  Since there are no sources the pressure tends to zero as time becomes inﬁnite. Thus the solution that is relevant to this problem is u = e−cα t sin αz . It is readily 2  checked that this function is a solution of the consolidation equation. The only c Pearson Education Limited 2011   570  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  boundary condition not satisﬁed is at z = h ; to do this choose αh = mπ where m is an integer. Therefore, ∞  an sin  u(z, t) =   mπz  h  m=1    cm2 π2 t exp − h2  and the initial uniform pressure leads to the Fourier series problem of evaluating the coeﬃcients in  ∞  A=  am sin   mπz  h  m=1  so h am = 2  h A sin   mπz  h  dz =  Ah (1 − cos mπ) mπ  0  For m, even the coeﬃcient is zero and substituting m = 2n+1 provides the solution quoted in the exercise. 17  Substituting φ = X(x)T(t) into the equation 1 T̈ + KṪ X = 2 = −λ2 X c T  where the separation constant has been chosen to be negative, to ensure periodic solutions for X . The variable X satisﬁes X + λ2 X = 0 with solution  X = P cos λx + Q sin λx  From condition (a) the sine term must be zero so Q = 0. For the T equation, T̈ + KṪ + c2 λ2 T = 0 so trying a solution of the form T = exp(at) gives the quadratic   1 2 2 −K ± K − 4(cλ) a + Ka + (cλ) = 0 with solution a = 2 2  2  The constant λ can be identiﬁed as p, so the condition that (cp)2 > 14 K2 gives an overall solution of the type   Kt φ = cos px exp − 2   (M sin bt + N cos bt)  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  571  where b2 = (cp)2 − 14 K2 . Condition (a) requires that N = A Condition (b) requires the derivative   K Kt ∂φ = − cos px exp − (M sin bt + N cos bt) ∂t 2 2   Kt (M cos bt − N sin bt) + b cos px exp − 2 At x = t = 0  1 1 − AK = − NK + bM 2 2 and since N = A then M = 0. The required solution is therefore   Kt φ(x, t) = A cos px exp − 2    1 cos bt where b2 = (cp)2 − K2 4  Results can be checked easily in MAPLE. f:=cos(p ∗ x) ∗ exp(-k ∗ t/2) ∗ cos(b ∗ t); simplify(diff(f,x,x)-(diff(f,t,t)+k ∗ diff(f,t))/cˆ2); gives    −4p2 c2 + k2 + 4b2 1 cos(px) exp − kt cos(bt) 4c2 2  18 Substituting the expressions for vr and vθ into the continuity equation, it is satisﬁed for any stream function ∂ LHS = ∂r    ∂ψ ∂θ    ∂ + ∂θ    ∂ψ − = 0 = RHS ∂r  since the cross partial derivatives are equal for all diﬀerentiable functions. For the given stream function,     a2 a2 1 = U cos θ 1 − 2 vr = U cos θ r − r r r   a2 vθ = −U sin θ 1 + 2 r  c Pearson Education Limited 2011   572  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  On the circle r = a, the radial velocity vr = 0 and there is no ﬂow into the circle. As r gets very large vr → U cos θ and vθ → −U sin θ The velocities parallel and perpendicular to the axes are V = vr cos θ − vθ sin θ  y  →U and  vθ  W  vr  W = vr sin θ + vθ cos θ →0  r  V  θ x  and hence the ﬂow far downstream is a steady ﬂow in the x direction with velocity U. Overall the ﬂow represents inviscid, irrotational ﬂow past a circular obstacle in uniform ﬂow. 19, 20, and 21 are intended to be open exercises extending the work of the chapter to more investigative work. Consequently, no advice is oﬀered for these problems. For Exercises 19 and 20 the text quotes sources for the work and for Exercise 21 many books on heat transfer will have a version of this problem.  c Pearson Education Limited 2011   10 Optimization Exercises 10.2.4 1  The required region is shaded on the graph and the lines of constant cost are all  parallel to the lines labelled f = 9 and f = 12. The point where f is a maximum in the region is at the point x = 1 and y = 1 giving a maximum cost of f = 9. 3.0 y  2.0 f = 12 f=9 1.0  3x + 7y = 10  2x + y = 3 0.5  2  1.0  1.5  x  A graphical or a tabular solution is possible but the MAPLE solution is given  here. with(simplex): con2:={2 ∗ x-y<=6,x+2 ∗ y<=8,3 ∗ x+2 ∗ y<=18,y<=3}; obj2:=x+y; maximize(obj2,con2,NONNEGATIVE); # gives the solution {y=2, x=4} 3  Let x be number of type 1 and y the number of type 2. The proﬁt from these  numbers is f = 24x + 12y c Pearson Education Limited 2011   574  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  T2 5x + 2y = 200 f = 1320  80  60 4x + 5y = 400 40 5x + 3y = 250  f = 840 20  f = 1080 10  20  30  T1  40  T2 f = 1320  80  60 4x + 5y = 400 min at (5,75) 40 5x + 3y = 250  5x + 2y = 175  20  f = 1020 10  20  30  T1  c Pearson Education Limited 2011   40  Glyn James: Advanced Modern Engineering Mathematics, 4th edition  575  and the constraints are, in the appropriate units, chipboard  4x + 5y ≤ 400  veneer  5x + 2y ≤ 200  labour  5x + 3y ≤ 250  and the obvious constraints x ≥ 0 and y ≥ 0. The ﬁrst ﬁgure shows the feasible region bounded by the axes and the three lines 4x + 5y = 400, 5x + 2y = 200 and 5x + 3y = 250. These lines intersect at x = 20, y = 50 and give the optimum proﬁt of £1080. Reducing the available amount of oak veneer to 175 m , the diagram changes as in the second ﬁgure. It can be seen that the same two constraints are active. They give the solution x = 5, y = 75 and a reduced optimum proﬁt of £1020. 4  Let n and s be the number of kg of nails and screws respectively. The proﬁt  made is therefore z = 2n + 3s The constraints are labour  3n + 6s ≤ 24  material  2n + s ≤ 10  The ﬁgure shows the feasible region. The point of intersection of the two constraints gives the maximum proﬁt of 14p making 4 kg of nails and 2 kg of screws. screws(Kg) 5 profit = 14  2n + s = 10  3  profit = 6 3n + 6s = 24  1  2  4  6  nails(Kg)  c Pearson Education Limited 2011   576 5  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition Let C1 and C2 be the number of cylinders CYL1 and CYL2 produced. The  proﬁt is 4C1+3C2 and the constraints given by the availability of the materials are: M1  C1 + 5C2 ≤ 45  M2  C1 + 2C2 ≤ 21  M3  2C1 + C2 ≤ 24  It is clear from the ﬁgure that the optimum is z = 54 with C1 = 9 and C2 = 6. The constraints M2 and M3 are active so all these materials are used up. The constraint M1 has some slack, it may be checked that 6 units remain unused. C2 2C1 + C2 = 24  12 profit = 36  10 8  C1 + 5C2 = 45 6 C1 + 2C2 = 21 4 profit = 54  2 2  6  4  6  8  10  12  C1  Let y be the number of Yorks and w the number of Wetherbys, then the proﬁt  made is z = 25y + 30w The constraints are cloth  3y + 4w ≤ 400  labour  3y + 2w ≤ 300  and the problem is a straightforward LP problem that can be solved graphically. It can be seen from the ﬁgure that the optimum is at y = 66.67 and w = 50 with a proﬁt of £3166.67. Note that the solution must be integral to make sense, c Pearson Education Limited 2011   Glyn James: Advanced Modern Engineering Mathematics, 4th edition  577  two-thirds of a jacket is not much good to anyone. However, the solution is an approximate one and, to proceed correctly, it is necessary to undertake the problem as an Integer Programming problem. This is much harder and can be found in any advanced book on Mathematical Programming. If the amount of cloth is increased then the line 3y + 4w = C is moved from its present C = 400 parallel and upwards. The intersection point moves upwards and the proﬁt is increased. This situation continues until the solution is y = 0 and w = 150 and all the labour is used on the Wetherbys. The amount of cloth required would be 600 m and the proﬁt in this case is £4500. This is the maximum possible proﬁt even if unlimited cloth is available because of the labour constraint. 160 140 120 100 3y + 2w = 300  f = 4500  f = 3000  3y + 4w = 400  w  80 60 40 20 f = 1500  0 −20 −40  7  0  10  20  30  40  50 y  60  70  80  90  100  The initial tableau is z x3 x4  x1 −k 1 3  x2 −20 2 1  x3 0 1 0  x4 0 0 1  Soln 0 20 25  x3 10 0.5 −0.5  x4 0 0 1  Soln 200 10 15  Eliminating the x2 column ﬁrst z x2 x4  x1 10−k 0.5 2.5  x2 0 1 0  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  578  If k < 10 then the tableau is optimal and the solution is x1 = 0, x2 = 10 and the value of z is 200. If, however, k > 10 then the method continues with the x1 column cleared. x1 0 0 1  z x2 x1  x2 0 1 0  x3 (60 − k)/5 0.6 −0.2  x4 2(k − 10)/5 0.2 0.4  Soln 140 + 6k 7 6  If 10 < k < 60 then the solution is optimal and the solution is x1 = 6, x2 = 7 and z takes the value 140 + 6k . If, however, k > 60 then the solution is not optimal and a further tableau needs to be formed using the x3 column. x1 0 0 1  z x3 x1  x2 −(60 − k)/5 5/3 1/3  x3 0 1 0  x4 (7k − 120)/15 1/3 7/15  Soln 25k/3 35/3 25/3  For k > 60, the z row is positive and therefore optimal with x1 = 25/3, x2 = 0 and z = 25k/3. x2  3x1 + x2 = 25  10 A  optimum for k = 90  8 optimum for k = 5 B 6 x1 + 2x2 = 20  4  optimum for k = 30 2 C 2  4  6  8  10  12  x1  The situation for this problem is illustrated in the ﬁgure. The feasible region is shown together with three diﬀerent cost functions corresponding to the three c Pearson Education Limited 2011   Glyn James: Advanced Modern Engineering Mathematics, 4th edition  579  diﬀerent cases. Note that for small k the optimum is at the corner A; as the cost steepens, the corner B is optimum; and ﬁnally as the cost steepens further with the highest values of k, the point C is the optimum.  8  This problem has four variables and so it has to be solved using the simplex  algorithm.  There are no diﬃculties with this problem and the tableaux are  presented, to two decimal places, without comment. z x5 x6 x7  x1 −2 2 1 0  x2 −1 0 0 4  z x5 x3 x7  x1 −0.67 1.67 0.33 −0.33  x2 −1 0 0 4  z x5 x3 x2  x1 −0.75 1.67 0.33 −0.08  x2 0 0 0 1  z x1 x3 x2  x1 0 1 0 0  x2 0 0 0 1  x3 −4 1 3 1 x3 0 0 1 0 x3 0 0 1 0  x3 0 0 1 0  x4 −1 0 1 1 x4 0.33 −0.33 0.33 0.67 x4 0.5 −0.33 0.33 0.17  x4 0.35 −0.2 0.4 0.15  x5 0 1 0 0  x6 0 0 1 0  x5 0 1 0 0 x5 0 1 0 0  x5 0.45 0.6 − 0.2 0.05  x6 1.33 −0.33 0.33 −0.33 x6 1.25 −0.33 0.33 −0.08  x6 1.1 − 0.2 0.4 − 0.1  x7 0 0 0 1  Soln 0 3 4 3  x7 0 0 0 1  Soln 5.33 1.67 1.33 1.67  x7 0.25 0 0 0.25  Soln 5.75 1.67 1.33 0.42  x7 0.25 0 0 0.25  Soln 6.5 1 1 0.5  The solution is read oﬀ as x1 = 1, x2 = 0.5, x3 = 1 and x4 = 0; the maximum is at z = 6.5. The MAPLE instructions con8:= {2 ∗ x1+x3<=3,x1+3 ∗ x3+x4<=4,4 ∗ x2+x3+x4<=3}; obj8:=2 ∗ x1+x2+4 ∗ x3+x4; maximize(obj8,con8,NONNEGATIVE); and the MATLAB instructions c Pearson Education Limited 2011   580  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  f=[-2;-1;-4;-1];A=[2,0,1,0;1,0,3,1;0,4,1,1];b=[3;4;3]; options=optimset('LargeScale','off','Simplex','on'); [x,fval]=linprog(f,A,b,[],[],zeros(4,1),[],[],options) also produce the solution {x4=0, x1=1, x3=1, x2=1/2}. 9  If b1,b2,b3 are the respective number (×1000) of books printed then the proﬁt  will be z = 900b1 + 800b2 + 300b3 and the constraints are sales restriction  b1 + b2 ≤ 15  paper  3b1 + 2b2 + b3 ≤ 60  The ﬁrst tableau is easily set up and the other tableaux follow by the usual rules. z b4 b5  b1 −900 1 3  z b1 b3  b1 0 1 0  b2 100 1 −1  z b1 b3  b1 0 1 0  b2 −200 1 −1  z b2 b3  b2 −800 1 2  b1 200 1 1  b3 −300 0 1 b3 −300 0 1 b3 0 0 1  b2 0 1 0  b3 0 0 1  b4 0 1 0 b4 900 1 −3 b4 0 1 −3  b4 200 1 −2  b5 0 0 1  Soln 0 15 60  b5 0 0 1  Soln 13,500 15 15  b5 300 0 1  Soln 18,000 15 15  b5 300 0 1  Soln 21,000 15 30  Thus the optimal proﬁt is made if none of book 1, 15,000 of book 2 and 30,000 of book 3 are printed giving a proﬁt of £21,000.  c Pearson Education Limited 2011   Glyn James: Advanced Modern Engineering Mathematics, 4th edition  581  10 Let L, M, S be the respective number of long, medium and short range aircraft purchased. The proﬁt is z = 0.4L + 0.3M + 0.15S and the constraints are money available  4L + 2M + S ≤ 60  pilots  L + M + S ≤ 25  maintenance  2L + 1.5M + S ≤ 30  The initial tableau is z P Q R  L −0.4 4 1 2  M −0.3 2 1 1.5  S −0.15 1 1 1  P 0 1 0 0  Q 0 0 1 0  R 0 0 0 1  Soln 0 60 25 30  The ﬁrst column is chosen, since it has the most negative entry, and when the ratios are calculated, the P and R rows produce the same value of 15. From the two, the P row is chosen arbitrarily. z L Q R  L 0 1 0 0  M −0.1 0.5 0.5 0.5  S −0.05 0.25 0.75 0.5  P 0.1 0.25 −0.25 −0.5  Q 0 0 1 0  R 0 0 0 1  Soln 6 15 10 0  In this tableau the M, R element is the pivot and performing the elimination gives  z L Q M  L 0 1 0 0  M 0 0 0 1  S 0.05 −0.25 0.25 1  P 0 0.75 0.25 −1  Q 0 0 1 0  R 0.2 −1 −1 2  Soln 6 15 10 0  Note that the ﬁnal column has not changed between the last two tableaux because of the zero in the very last entry. The optimal solution is to purchase 15 long range aircraft and no medium or short range aircraft; the estimated proﬁt is £6 m. c Pearson Education Limited 2011   582  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  In the problem, it would seem that some operational constraints have been omitted. In many optimisation problems it takes several steps to achieve a sensible cost and set of constraints.  11  The ﬁrst tableau is set up from the data. Choosing the pivot in the usual way,  x1 and x5 are interchanged and the elimination follows the basic rules. Again, the pivot is found and x3 and x6 are interchanged; the elimination produces a top row that is all positive so the solution is optimal. The solution is read from the tableau as x1 = 1.5, x2 = 0, x3 = 2.5, x4 = 0 and f = 14.  z x5 x6 x7  z x1 x6 x7  z x1 x3 x7  x1 −6 2 1 1  x1 0 1 0 0  x1 0 1 0 0  x2 −1 1 0 1  x3 −2 0 1 3  x4 −4 1 1 2  x2 2 0.5 −0.5 0.5  x3 −2 0 1 3  x4 −1 0.5 0.5 1.5  x2 1 0.5 −0.5 2  x3 0 0 1 0  x4 0 0.5 0.5 0  x5 0 1 0 0  x5 3 0.5 −0.5 −0.5  x5 2 0.5 − 0.5 1  x6 0 0 1 0  x7 0 0 0 1  x6 0 0 1 0  x6 2 0 1 −3  Soln 0 3 4 10  Ratio 1.5 4 10  x7 0 0 0 1  Soln 9 1.5 2.5 8.5  Ratio  x7 0 0 0 1  Soln 14 1.5 2.5 1  Ratio  2.5 2.83  The MAPLE implementation just gives the 'answer'. Some of the detail can be extracted from MAPLE and the various tableau can be identiﬁed. con11:={2 ∗ x1+x2+x4<=3,x1+x3+x4<=4,x1+x2+3 ∗ x3+2 ∗ x4<=10}; obj11:=6 ∗ x1+x2+2 ∗ x3+4 ∗ x4; c Pearson Education Limited 2011   Glyn James: Advanced Modern Engineering Mathematics, 4th edition  583  maximize(obj11,con11,NONNEGATIVE); # gives the same solution {x4=0, x2=0, x1=3/2, x3=5/2} # for the detail z:=setup(con11); z:={_SL1=3-2 x1-x2-x4,_SL2=4-x1-x3-x4, _SL3=10-x1-x2-3 x3-2 x4} piv:=pivoteqn(z,x1); piv:=[_SL1=3-2 x1-x2-x4] z:=pivot(z,x1,piv); z:={x1=3/2-1/2_SL1-1/2 x2-1/2 x4, _SL2=5/2+1/2_SL1+1/2 x2-1/2 x4-x3, _SL3=17/2+1/2 _SL1-1/2 x2-3/2 x4-3 x3}  obj11:=eval(obj11,z); obj11:= 9-3_SL1 - 2x2 + x4 + 2 x3 # compare with second tableau piv:= pivoteqn(z,x3); piv:=[_SL2=5/2 + 1/2_SL1 + 1/2x2 - 1/2x4-x3] z:=pivot(z,x3,piv); z:= {_SL3=-_SL1 + 1 + 3_SL2 - 2 x2, x1 = 3/2 - 1/2_SL1 - 1/2 x2 - 1/2 x4, x3=1/2_SL1 + 5/2 -_SL2 + 1/2 x2 - 1/2 x4} obj11:=eval(obj11,z); obj11:=-2_SL2 + 14 - 2_SL1-x2 # compare with the final tableau Note that the entry in position (z, x4 ) is zero so we expect many solutions – this is easily identiﬁed from the tableau but not from the MAPLE results. Continuing with MAPLE code piv:=pivoteqn(z,x4); piv:=[x1=3/2 - 1/2_SL1 - 1/2 x2 - 1/2 x4] z:=pivot(z,x4,piv); c Pearson Education Limited 2011   584  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  z:={_SL3 = -_SL1 + 1 + 3_SL2 - 2 x2, x4 = -2 x1+3 -_SL1 - x2, x3 =_SL1 + 1 -_SL2 + x2 + x1} obj11:=eval(obj11,z); obj11:=-2_SL2+14-2_SL1-x2 The new solutions are x1 = x2 = 0, x3 = 1, x4 = 3 and f = 14, giving, as expected, the same function value. The MATLAB instructions give the same result but no detail f=[-6;-1;-2;-4];A=[2,1,0,1;1,0,1,1;1,1,3,2];b=[3;4;10]; options=optimset('LargeScale','off','Simplex','on'); [x,fval]=linprog(f,A,b,[],[],zeros(4,1),[],[],options)  Exercises 10.2.6 12 The problem can be easily plotted and it is clear that x = 1, y = 4 gives the solution f = x + 2y = 9. 6 5 4 y 3 2 1 0 1  2  3 x  4  5  c Pearson Education Limited 2011   6  Glyn James: Advanced Modern Engineering Mathematics, 4th edition  585  The MAPLE check is as follows: with (simplex): con12:={y>=1,y<=4,x+y<=5}; obj12:=x+2 ∗ y; maximize(obj12,con12,NONNEGATIVE); # gives {y=4,x=1} The graph can be plotted using the instructions with(plots): F:=inequal(con12,x=0..6,y=0..6): G:=plot([(-x+7)/2,(-x+9)/2,(-x+11)/2],x=0..6, thickness=3,labels=['x','y'],color=yellow): display(F,G);  13 The problem has 'greater than' constraints, so the two-phase method is required. Note that a surplus variable x4 and an artiﬁcial variable x7 need to be introduced. At the end of phase 1 the artiﬁcial variable x7 will be eliminated from the tableau. The cost function in phase 1 is −x7 , but recall that this cost has to be modiﬁed to ensure that the tableau is in standard form. Phase 1  z x3 x7 x5 x6  z x3 x7 x1 x6  x1 −2 4 2 2 −2 x1 0 0 0 1 0  x2 −1 1 1 −1 1  x2 −2 3 2 −0.5 0  x3 0 1 0 0 0  x3 0 1 0 0 0  x4 1 0 −1 0 0  x4 1 0 −1 0 0  x5 0 0 0 1 0  x6 0 0 0 0 1  x5 1 −2 −1 0.5 1  x6 0 0 0 0 1  c Pearson Education Limited 2011   x7 0 0 1 0 0  x7 0 0 1 0 0  Soln −12 32 12 4 8  Soln −8 24 8 2 12  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  586  x1 0 0 0 1 0  z x3 x2 x1 x6  x2 0 0 1 0 0  x3 0 1 0 0 0  x4 0 1.5 − 0.5 − 0.25 0  x5 0 − 0.5 −0.5 0.25 1  x6 0 0 0 0 1  x7 1 − 1.5 0.5 0.25 0  Soln 0 12 4 4 12  The artiﬁcial cost has been driven to zero so the artiﬁcial variable x7 can be eiiminated and the original cost function reinstated. Phase 2 Note that the new cost is negative because the problem is a minimisation problem. x1 0 0 0 1 0  z x3 x2 x1 x6  x2 0 0 1 0 0  x3 0 1 0 0 0  x4 3 1.5 −0.5 −0.25 0  x5 −2 −0.5 −0.5 0.25 1  x6 0 0 0 0 1  x2 f = 40 30  f = 30 – 2x1 – x2 = 8  20 f = 20  2x1 – x2 = 4 10  4x1 + x2 = 32 2x1 + x2 = 12 2  4  6  8  c Pearson Education Limited 2011   x1  Soln −44 12 4 4 12  Glyn James: Advanced Modern Engineering Mathematics, 4th edition x1 0 0 0 1 0  z x3 x2 x1 x5  x2 0 0 1 0 0  x3 0 1 0 0 0  x4 3 1.5 − 0.5 − 0.25 0  x5 0 0 0 0 1  x6 2 0.5 0.5 − 0.25 1  587  Soln − 20 18 10 1 12  The solution is read oﬀ as x1 = 1, x2 = 10 with a minimal cost of 20. 14  Let S be the number of shoes produced and B the number of boots, then the  problem is to maximize the proﬁt z = 8B + 5S production  2B + S ≤ 250  sales  B + S ≤ 200  customer  B ≥ 25  The tableaux are constructed in the usual way. Phase 1 z P Q T  B −1 2 1 1  S 0 1 1 0  B 0 0 0 1  z P Q B  P 0 1 0 0  S 0 1 1 0  Q 0 0 1 0  P 0 1 0 0  Q 0 0 1 0  R 1 0 0 −1  R 0 2 1 −1  T 0 0 0 1  T 1 −2 −1 1  Soln −25 250 200 25  Soln 0 200 175 25  Phase 2 z P Q B  B 0 0 0 1  S −5 1 1 0  P 0 1 0 0  Q 0 0 1 0  c Pearson Education Limited 2011   R −8 2 1 −1  Soln 200 200 175 25  588  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition B 0 0 0 1  z R Q B  S −1 0.5 0.5 0.5  B 0 0 0 1  z R S B  P 4 0.5 −0.5 0.5  S 0 0 1 0  P 3 1 −1 1  Q 0 0 1 0  Q 2 −1 2 −1  R 0 1 0 0  R 0 1 0 0  Soln 1000 100 75 125  Soln 1150 25 150 50  Thus, the manufacturer should make 50 pairs of boots and 150 pairs of shoes and the maximum proﬁt is £1150. A graphical solution is possible since the problem has only two variables. It is of interest to follow the progress of the simplex solution on the graph. boots 150  profit = 1150 2B + S = 250  100  B + S = 200  profit = 600 50  B = 25  30  15  60  90  120  150  180  shoes  There are techniques for adding a constraint to an existing solution and thence  determining the solution. However, here the whole problem will be recomputed. From Exercise 9 the initial tableau can be constructed, but the additional constraint is a 'greater than' constraint so the procedure for entering phase 1 must be followed. c Pearson Education Limited 2011   Glyn James: Advanced Modern Engineering Mathematics, 4th edition  z b4 b5 b7  b1 −1 1 3 1  b2 −1 1 2 1  b3 −1 0 1 1  b4 0 1 0 0  b5 0 0 1 0  b6 1 0 0 −1  b7 0 0 0 1  589  Soln −50 15 60 50  Note that the last row gives the new constraint and that the z row involves the artiﬁcial variable b7, which has been eliminated to bring the tableau to standard form. Phase I  z b1 b5 b7  b1 0 1 0 0  b2 0 1 −1 0  z b1 b3 b7  b1 0 1 0 0  b2 −1 1 −1 1  b1 0 1 0 0  z b1 b3 b4  b3 −1 0 1 1  b3 0 0 1 0  b2 0 0.5 0.5 0.5  b3 0 0 1 0  b4 1 1 −3 −1 b4 −2 1 −3 2  b5 0 0 1 0  b6 1 0 0 −1  b7 0 0 0 1  Soln −35 15 15 35  b5 1 0 1 −1  b6 1 0 0 −1  b7 0 0 0 1  Soln −20 15 15 20  b4 0 0 0 1  b5 0 0.5 − 0.5 − 0.5  b3 0 0 1 0  b4 0 0 0 1  b6 0 0.5 −1.5 −0.5  b7 1 −0.5 1.5 0.5  b5 300 0.5 −0.5 −0.5  b6 0 0.5 −1.5 −0.5  Soln 0 5 45 10  Phase 2 z b1 b3 b4  b1 0 1 0 0  b2 −200 0.5 0.5 0.5  c Pearson Education Limited 2011   Soln 18,000 5 45 10  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  590  b1 400 2 −1 1  z b1 b3 b4  b2 0 1 0 0  b3 0 0 1 0  b4 0 0 0 1  b5 500 1 −1 −1  b6 200 1 −2 −1  Soln 20,000 10 40 5  The solution is optimal with the production schedule as none for book 1, 10,000 of book 2 and 40,000 of book 3. The proﬁt is down to £20,000 because of the additional constraint. 16  The MAPLE solution is presented. with (simplex): con16:={x>=1,x+2*y<=3,y+3*z<=4}; obj16:=x+y+z; maximize(obj16,con16,NONNEGATIVE); # gives the solution {y=0,z=4/3,x=3}, f=x+y+z=13/3  Note that there is no detail of the two-phase method and just the 'answer' is given. It is not easy to rewrite phase 1 into MAPLE so that the detail can be extracted. The same is true of MATLAB, the following instructions produce the correct result f=[-1;-1;-1];A=[1,2,0;0,1,3;-1,0,0];b=[3;4;-1]; options=optimset('LargeScale','off,'Simplex','on'); [x,fval]=linprog(f,A,b,[],[],zeros(3,1),[],[],options)  17  This is a standard two-phase problem with surplus variables x5 , x6 and  artiﬁcial variables x7 , x8 . With the cost constructed from the artiﬁcial variables x7 , x8 , and the tableau reduced to standard form, the sequence of tableaux is presented. Phase 1  z x7 x9  x1 −1 1 0  x2 −1 0 1  x3 0 −1 1  x4 1 −1 0  x5 1 −1 0  x6 1 0 −1  c Pearson Education Limited 2011   x7 0 1 0  x8 0 0 1  Soln −2 0 2  Glyn James: Advanced Modern Engineering Mathematics, 4th edition  591  There is an arbitrary choice of the columns since two columns have the value −1; the second one is chosen. x1 −1 1 0  z x7 x2  x1 0 1 0  z x1 x2  x2 0 0 1  x3 1 −1 1  x4 1 −1 0  x5 1 −1 0  x6 0 0 −1  x7 0 1 0  x8 1 0 1  x2 0 0 1  x3 0 −1 1  x4 0 −1 0  x5 0 −1 0  x6 0 0 −1  x7 1 1 0  x8 1 0 1  Soln 0 0 2  Soln 0 0 2  The solution is optimal so phase 1 ends and the tableau is reconstituted for phase 2. Phase 2 z x1 x2  x1 0 1 0  x2 0 0 1  x1 0 1 0  z x1 x3  x3 −1 −1 1  x2 1 1 1  x3 0 0 1  x4 7 −1 0  x5 2 −1 0  x4 7 −1 0  x5 2 −1 0  x6 7 0 −1  x6 6 −1 −1  Soln −14 0 2  Soln −12 2 2  The solution is now optimal with x1 = 2, x2 = 0, x3 = 2, x4 = 0 and the cost function being 12. 18  Let the company buy a,b,c litres of the products A,B,C, then the cost of the  materials will be Cost  1.8a + 0.9b + 1.5c  For a total of 100 litres  a + b + c = 100  Glycol  0.65a + 0.25b + 0.8c ≥ 50  Additive  0.1a + 0.03b ≥ 5  There are two 'greater than' constraints and an equality constraint. Recall that an equality constraint is dealt with via an artiﬁcial variable. Thus, phase 1 is entered c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  592  with two surplus variables and three artiﬁcial variables; the cost row is manipulated until the tableau has the standard form. Phase 1 a −1.75 1 0.65 0.1  z r s t  b −1.28 1 0.25 0.03  a z −0.29 r 0.19 0.81 c 0.1 t  z p c t  a −0.1 0.15 1 0.1  z p c a  a 0 0 0 1  c −1.8 1 0.8 0  b −0.72 0.69 0.31 0.03  c 0 0 1 0  b −0.03 0.55 1 0.03  b 0 0.51 0.7 0.3  c 0 0 1 0  p −1.25 1.25 −1.25 0 c 0 0 1 0  p 0 1 0 0  p 1 0 −1 0  p 0 1 0 0  q 0 1.5 10 − 10  q 1 0 0 −1  q 1 0 0 −1  r 0 1 0 0  r 0 1 0 0  q 1 0 0 −1  s 2.25 −1.25 1.25 0 r 1 0.8 1 0  r 1 0.8 1 0  s 0 0 1 0  t 0 0 0 1  Soln −155 100 50 5  t 0 0 0 1  Soln −42.5 37.5 62.5 5  s 1 −1 0 0  s 1 −1 0 0  t 1 − 1.5 − 10 10  t 0 0 0 1  Soln −5 30 100 5  Soln 0 22.5 50 50  The artiﬁcial cost has been driven to zero so phase 1 is complete and the three artiﬁcial columns are deleted and the actual cost introduced. Recall that this is a minimisation problem so the phase 2 tableau can now be extracted from the tableau above. Phase 2 z p c a  a 0 0 0 1  b −0.69 0.51 0.7 0.3  c 0 0 1 0  p 0 1 0 0  q 3 1.5 10 −10  c Pearson Education Limited 2011   Soln −165 22.5 50 50  Glyn James: Advanced Modern Engineering Mathematics, 4th edition a 0 0 0 1  z b c a  b 0 1 0 0  c 0 0 1 0  p 1.37 1.98 − 1.39 −0.59  q 5.05 2.97 7.92 − 10.89  593  Soln −134.26 44.55 18.81 36.63  The tableau is optimal so the solution can be read oﬀ as a = 36.63%, b = 44.55%, c = 18.81% and a minimum cost of £134.26. MATLAB solves the problem as follows f=[1.8;0.9;1.5];A=[-0.65,-0.25,-0.8;-0.1,-0.03,0];b=[-50;-5]; options=optimset('LargeScale','off,'Simplex','on'); [x,fval]=linprog(f,A,b,[1,1,1],[100],zeros(3,1),[],[],options) % Note how MATLAB deals with the equality constraint  19  Let s1 , s2 , s3 be the number of houses of the three styles the builder decides  to construct. His proﬁt (× £100) is 10s1 + 15s2 + 25s3 and the constraints are plots  s1 + 2s2 + 2s3 ≤ 40  facing stone  s1 + 2s2 + 5s3 ≤ 58  weather boarding  3s1 + 2s2 + s3 ≤ 72  local authority  − s1 + s2 ≥ 5  The solution requires the two-phase method. The tableaux are listed, to two decimal places, without comment. Phase 1 z s4 s5 s6 s8  s1 1 1 1 3 −1  s2 −1 2 2 2 1  s3 0 2 5 1 0  s4 0 1 0 0 0  s5 0 0 1 0 0  s6 0 0 0 1 0  c Pearson Education Limited 2011   s7 1 0 0 0 −1  s8 0 0 0 0 1  Soln −5 40 58 72 5  594  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition s1 0 3 3 5 −1  z s4 s5 s6 s2  s2 0 0 0 0 1  s3 0 2 5 1 0  s4 0 1 0 0 0  s5 0 0 1 0 0  s6 0 0 0 1 0  s7 0 2 2 2 −1  s8 1 −2 −2 −2 1  Soln 0 30 48 62 5  s5 0 0 1 0 0  s6 0 0 0 1 0  s7 −15 2 2 2 −1  Soln 75 30 48 62 5  s5 0 0 1 0 0  s6 0 0 0 1 0  s7 1.67 0.67 0 −1.33 −0.33  Soln 325 10 18 12 15  Phase 2  z s4 s5 s6 s2  z s1 s5 s6 s2  z s1 s3 s6 s2  s1 −25 3 3 5 −1  s1 0 1 0 0 0  s2 0 0 0 0 1  s2 0 0 0 0 1  s1 0 1 0 0 0  s3 −25 2 5 1 0  s3 −8.33 0.67 3 −2.33 0.67  s2 0 0 0 0 1  s3 0 0 1 0 0  s4 0 1 0 0 0  s4 8.33 0.33 −1 −1.67 0.33  s4 5.56 0.56 − 0.33 − 2.44 0.56  s5 2.78 − 0.22 0.33 0.78 − 0.72  s6 0 0 0 1 0  s7 1.67 0.67 0 − 1.33 − 0.33  Soln 375 6 6 26 11  The tableau is optimal, so the solution should build 6,11 and 6 houses of styles 1,2 and 3 respectively and make a proﬁt of £37,500. 20 Let x1 x2 , x3 be the amounts (×1000 m2 ) of carpet of type C1,C2,C3 produced, then the maximum of 2x1 + 3x2 is required subject to the constraints c Pearson Education Limited 2011   Glyn James: Advanced Modern Engineering Mathematics, 4th edition  M1  x1 + x2 + x3 ≤ 5  M2  x1 + x2 ≤ 4  policy1  x1 ≥ 1  policy2  x1 − x2 + x3 ≥ 2  595  Phase 1 z x4 x5 x6 x9  x1 2 1 1 1 1  x2 1 1 1 0 −1  x3 −1 1 0 0 1  z x4 x5 x1 x9  x1 0 0 0 1 0  x2 1 1 1 0 −1  x3 1 1 0 0 1  z x4 x5 x1 x3  x1 0 0 0 1 0  x2 0 2 1 0 −1  x4 0 1 0 0 0 x4 0 1 0 0 0  x3 0 0 0 0 1  x4 0 1 0 0 0  x5 0 0 1 0 0 x5 0 0 1 0 0  x5 0 0 1 0 0  x6 1 0 0 −1 0 x6 −1 1 1 −1 1  x6 0 0 1 −1 1  x7 1 0 0 0 −1 x7 1 0 0 0 −1  x7 0 1 0 0 −1  x8 0 0 0 1 0  x9 0 0 0 0 1  Soln −3 5 4 1 2  x8 2 −1 −1 1 −1  x9 0 0 0 0 1  Soln −1 4 3 1 1  x9 1 −1 0 0 1  Soln 0 3 3 1 1  x8 1 0 −1 1 −1  Phase 2 z x4 x5 x1 x3  x1 0 0 0 1 0  x2 3 8 1 0 1  x3 0 0 0 0 1  x4 0 1 0 0 0  x5 0 0 1 0 0  z x2 x4 x1 x3  x1 0 0 0 1 0  x2 0 1 0 0 0  x3 0 0 0 0 1  x4 1.5 0.5 0.5 0 0.5  x5 0 0 1 0 0  x6 −2 0 1 −1 1 x6 −2 0 1 0 0  c Pearson Education Limited 2011   x7 0 1 0 0 −1  Soln 2 9 3 1 1  x7 1.5 0.5 0.5 0 −0.5  Soln 6.5 1.5 1.5 1.5 2.5  596  z x2 x6 x1 x3  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition x1 0 0 0 1 0  x2 0 1 0 0 0  x3 0 0 0 0 1  x4 0.5 0.5 −0.5 −0.5 1  x5 2 0 1 1 −1  x6 0 0 1 0 0  x7 0.5 0.5 − 0.5 − 0.5 0  Soln 9.5 1.5 1.5 2.5 1  Hence, the optimum is 2500 m2 of C1, 1500 m2 of C2 and 1000 m2 of C3 giving a proﬁt of £9500.  Exercises 10.3.3 21  Introducing a Lagrange multiplier, the minimum is obtained from 2x + y + λ = 0 x + 2y + λ = 0 x+y = 1  with solution x = y = 1/2. To prove that it is a minimum, put x=  1 2  + ε and y =  1 2  −ε  and substitute into f to give f=  3 4  + ε2  and hence a minimum. 22  The answer to this exercise is geometrically obvious since it is just the length  of the minor axis which in this case, is a. It is required to ﬁnd the minimum of D2 = x2 + y2 subject to  x2 y2 + =1 a2 b2  The Lagrange equations are 0 = 2x +  2λy 2λx and 0 = 2y + a2 b2  Since λ cannot equal −a2 and −b2 at the same time then either x = 0 or y = 0. If a < b the it is clear that y = 0 and x = ± a gives the minimum c Pearson Education Limited 2011   Glyn James: Advanced Modern Engineering Mathematics, 4th edition 23  The area of the rectangle is  597  y  A = 4xy  (x,y)  and since the point must lie on the ellipse  x  x2 y2 + =1 a2 b2 and hence a Lagrange multiplier problem is produced. The two equations are 2λx 2λy and 0 = 4x + 2 2 a b √ √ It is soon checked that the required solution is x = a/ 2, y = b/ 2, A = 2ab. 0 = 4y +  24  The necessary conditions for an optimum are ∂f ∂g +λ = 0 = y2 z + λ ∂x ∂x ∂g ∂f +λ = 0 = 2xyz + 2λ ∂y ∂y ∂f ∂g +λ = 0 = xy2 + 3λ ∂z ∂z  (1) (2) (3)  together with the given constraint x + 2y + 3z = 6  (4)  Dividing equation (2) by two and then subtracting the ﬁrst two equations gives yz(x − y) = 0 Thus, there are three cases y = 0 Note that all the equations except the constraint are satisﬁed since λ = 0. Thus, a possible solution is (6 − 3α, 0, α) for any α. z = 0 Again, this implies that λ = 0 so from equation (3) x = 0 and hence from (4) y = 3. The case when y = 0 in equation (3) has been covered in the ﬁrst case. x = y Equations (2) and (3) give  1 3 3x  = x2 z so there are two cases  either x = 0 and hence y = 0 and z = 2 c Pearson Education Limited 2011   598  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition or x = 3z and hence from (4) z = 1/2 and x = 3/2, y = 3/2  Equations (1) and (2) have been used to eliminate λ; so a similar exercise must be undertaken with (2) and (3) to check whether any new solutions arise. They give the equation xy(y − 3z) = 0 and again three cases must be studied. x = 0 In this case λ = 0 so it reduces to one of the above. y = 0 Again, it reduces to the ﬁrst case above. y = 3z From equations (1) and (2) it can be seen that the same cases arise and correspond to the solutions already obtained. Note that in the solutions quoted, (6,0,0) is included in the solution (6 − 3α, 0, α) with α = 0. 25  Using λ and μ as the Lagrange multipliers, the equations that give the  optimum are as quoted  0 = 2x + λ + (2z − 2y)μ 0 = 2y + λ + (z − 2x)μ 0 = 2z − λ + (y + 2x)μ  together with the two given constraints. Adding the last two equations 2(y + z) + (y + z)μ = 0 ⇒ either y = −z or μ = −2 Adding the ﬁrst and last equations 2(x + z) + (2z + 2x − y)μ = 0 So, λ has been eliminated. In the ﬁrst constraint x = −y + z = 2z . Thus, (2t, −t, t) gives, in √ the second constraint, 7t2 = 1 so t = ±1/ 7 and two of the quoted solutions are obtained. y+z=0  μ = −2  From above 2(x + z) + (2z + 2x − y)(−2) = 0 and so this expression  and the second constraint give a pair of equations 0 = −x + y − z 0=x+y−z c Pearson Education Limited 2011   Glyn James: Advanced Modern Engineering Mathematics, 4th edition  599  and hence, x = 0 and y = z . Putting these values into the second constraint gives y = ±1 and possible optimum points at (0, 1, 1) and (0,− 1,− 1). 26  The volume of the ﬁgure is c  V = abc and the surface area is A = 2ab + 2ac + bc  a  The problem is a Lagrange multiplier problem and the three equations are  b  0 = bc + λ(2b + 2c) 0 = ac + λ(2a + c) 0 = ab + λ(2a + b) The last two equations give a(c − b) + λ(c − b) = 0 ⇒ either c = b or λ = −a c = b Putting back into the ﬁrst equation gives b2 + 4bλ = 0 so either b = 0 or λ = −b/4 The case b = 0 can be dismissed as geometrically uninteresting. In the third equation this value of λ gives 0 = ab − 14 b(2a + b) ⇒ b = 2a since again the case b = 0 is uninteresting. Calculating a gives the possible maximum as  c = b = 2a with a =  A and V = 4a3 12  λ = −a Putting this value back into the third of the Lagrange equations gives a = 0 so gives zero volume and is geometrically uninteresting again. 27  This is a very tough problem but more typical of realistic problems in  engineering rather than most of the illustrative exercises in the book. c Pearson Education Limited 2011   600  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  The exercise will be solved using MAPLE. Note the eﬃcient way it performs the algebra and integrations for the simplest approximation z:=A ∗ cos(Pi ∗ x/2)ˆ2 ∗ cos(Pi ∗ y/2)ˆ2; I1:int(int(z^2,x=-1..1),y=-1..1); # gives I1:=9/16 A 2 f:=diff(z,x,x)+diff(z,y,y); I2:=int(int(f^2,x=-1..1),y=-1..1); # gives I2:=1/2 A 2 Pi 4 freq:=I2/I1; # gives freq:=8/9 Pi 4 = 86.58 as required. The ﬁrst approximation could be done by hand but the next approximation needs a package like MAPLE. zz:=cos(Pi ∗ x/2)ˆ2 ∗ cos(Pi ∗ y/2)ˆ2 ∗  (A+B ∗ cos(Pi ∗ x/2) ∗ cos(Pi ∗ y/2)); I3:=evalf(int(int(zz^2,x=-1..1),y=-1..1)); # gives I3:=.5624999999 A 2 + .9222479291 A B + .3906249999 B 2 ff:=diff(zz,x,x)+diff(zz,y,y); I4:=evalf(int(int(ff^2,x=-1..1),y=-1..1)); # gives I4:=58.21715209 B 2 + 48.70454554 A 2 + 92.64268668 A B dLbydA:=diff(I4+lam ∗ I3,A); dLbydB:=diff(I4+lam ∗ I3,B); fsolve({dLbydA,dLbydB,I3=1}, {A,B,lam}); # gives the solution {lam=-81.38454660, A=-1.829151961, B=.6086242001} The frequency is given by −lam which should be compared with the ﬁrst approximation.  c Pearson Education Limited 2011   Glyn James: Advanced Modern Engineering Mathematics, 4th edition 28  601  The problem asks for a minimum, which is equivalent to the maximum of −2x21 − x22 − 2x1 x2  The Kuhn–Tucker conditions for this problem are −4x1 − 2x2 − μ = 0 −2x1 − 2x2 + μ = 0 x1 − x2 − α ≤ 0 μ(x1 − x2 − α) = 0 μ≥0 There are two cases μ=0  The ﬁrst two equations give x1 = x2 = 0 and these can only give the  optimum if the inequality is satisﬁed, namely α ≥ 0. x1 − x2 = α  Adding the ﬁrst two equations gives the solution in terms of the  simultaneous linear equations 3x1 + 2x2 = 0 x1 − x2 = α which clearly have the solution x1 = 2α/5 and x2 = −3α/5. Calculation from the ﬁrst or second equation gives μ = −2α/5 which is only optimal if α ≤ 0.  Exercises 10.4.2 29(a)  The MATLAB version of this problem (note it deals with the maximum)  is as follows:  a=0.1;h=0.2;nmax=10;n=0; zold=q29(a);a=a+h;z=q29(a);c=[z]; while(z(2)>zold(2))&(nb(2) if xstar0 an=c;bn=b; else bn=c;an=a;end % an and bn contain the new bracket values  c Pearson Education Limited 2011   603  604  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  30 If the derivatives are not used, then the bracket is 0 < x < 3, whilst using the derivatives gives the much narrower range of 0 < x < 1 but at the expense of twice the number of function evaluations. 30(a) The calculation is identical to Exercise 29 with q29 replaced by q30 in qapp  p=[q30(0),q30(1),q30(3)]; u=p(1,:),v=p(2,:) % gives  u= 0  1  3  v= 0  0.4207 0.0141  [u,v]=qapp(u,v) % gives  u= 0  1.0000 1.5113  v= 0 0.4207 0.3040 [u,v]=qapp(u,v) % gives  u= 0  0.9898 1.0000  v= 0  0.4222 0.4207  where the function q30 is in the M-ﬁle function a=q30(x) a=[x;sin(x)/(1+xˆ2);cos(x)/(1+xˆ2)-2ˆ∗ xˆ∗ sin(x)/(1+xˆ2)ˆ2];  30(b) Again, as in Exercise 29, with q29 replaced by q30 in cufit p=q30(0);q=q30(1);p  ,q  % gives  x=0  f=0  f = 1  % and  x=1.0000  f=0.4207  f  =-0.1506  [p,q]=cufit(p,q);p  ,q  % gives  x=0  f=0  f = 1  % and  x=0.8667  f=0.4352  f  =-0.0612  [p,q]=cufit(p,q);p  ,q  % gives  x=0  f=0  f = 1  % and  x=0.8242  f=0.4371  f  =-0.0247  The built in MATLAB procedure fminbnd gives x = 0.7980, f = 0.4374 in 9 iterations c Pearson Education Limited 2011   Glyn James: Advanced Modern Engineering Mathematics, 4th edition  605  f=@(x)-sin(x)/(1+x^2); options=optimset('display','iter'); x=fminbnd(f,0,1,options) 31  The problem is a standard exercise to illustrate the use of the two basic  algorithms for a single variable search. 31(a) An adaptation of Exercise 29 with the appropriate function used gives  p=[q31(1),q31(5/3),q31(3)];u=p(1,:),v=p(2,:) % gives  u=1.0000  1.6667  3.0000  v=0.2325  0.2553  0.1419  u=1.000  1.6200  1.6667  v=0.2325  0.2571  0.2553  1.4784 0.2602  1.6200 0.2571  [u,v]=qapp(u,v) % gives  [u,v]=qapp(u,v) % gives  u=1.0000 v=0.2325  where q31 is the M-ﬁle function a=q31(x) a=[x;x ∗ (exp(-x)-exp(-2 ∗ x));(exp(-x)-exp(-2 ∗ x)) -x ∗ (exp(-x)-2 ∗ exp(-2 ∗ x))];  31(b)  Use the same bracket and adapt the algorithm in Exercise 29. p=q31(1);q=q31(3);p  ,q  % gives  x=1.0000  f=0.2325  f  = 0.1353  % and  x=3.0000  f=0.1419  f  = -0.0872  [p,q]=cufit(p,q);p  ,q  % gives  x=1.0000  f=0.2325  f  = 0.1353  % and  x=1.5077  f=0.2599  f  = -0.0136  [p,q]=cufit(p,q);p  ,q  c Pearson Education Limited 2011   606  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition % gives  x=1.0000  f=0.2325  f  = 0.1353  % and  x=1.4462  f=0.2603  f  = -0.0001  The next iteration gives the three-ﬁgure accuracy required in (c). x = 1.4456 f = 0.2603 f  = 0.0000 32 The diﬃculty in this problem is that the eigenvalues must be computed at each function evaluation. With a package such as MATLAB this is less of a problem since the instruction eig(A) will give them almost instantly. The M-ﬁle q32 performs this computation. Note the -max(eig(A)) since qapp looks for a maximum function a=q32(x) A=[x -1 0;-1 0 -1;0 -1 x^2]; a=[x;-max(eig(A))]; The calculations are performed as in Exercise 29 p=[q32(-1),q32(0),q32(1)];u=p(1,:),v=p(2,:) % gives  u=-1  0  v=-1.7321  1  -1.4142  -2.0000  [u,v]=qapp(u,v) % gives  u=-1.0000  -0.1483  0  v=-1.7321  -1.3854  -1.4142  [u,v]=qapp(u,v) % gives u=-1.0000  -0.2356  -0.1483  v=-1.7321  -1.3769  -1.3854  See the text for the MATLAB fminbnd code for this problem.  33  To establish the formula is a matter of simple substitution into (10.10).  To ﬁnd when f (x) = 0, from the Newton method, requires the iteration of the formula xnew = x −  f (x) f (x)  If the function is known at x − h, x, x + h with values f1 , f2 , f3 respectively, then the derivatives can be replaced by their approximations c Pearson Education Limited 2011   Glyn James: Advanced Modern Engineering Mathematics, 4th edition f3 − f1 f1 − 2f2 + f3 and f (x) ≈ 2h h2 and the formula follows immediately. f (x) ≈  34(a)  To obtain the Golden Section ratio, the ﬁgure gives A  l  C  B  D  l  L  AC AD = AD AB  L−l l = L−l L  yields  −→  Solve for a = l/L to obtain a = 12 (3 −  √  5) .  The Golden Section algorithm can be written in MATLAB as alpha = (3-sqrt(5))/2;a=[0;1;1;2.5]; f=@ (x)x ∗ sin(x); a=[0;(1-alpha) ∗ a(1)+alpha ∗ a(4);alpha ∗ a(1)+(1-alpha) ∗ a(4);2.5]; F=[f(a(1));f(a(2));f(a(3));f(a(4))];aa=[a];FF=[F]; for i=1:5 if F(2)>F(3)a(4)=a(3);a(3)=a(2);F(4)=F(3);F(3)=F(2); a(2)=(1-alpha) ∗ a(1)+alpha ∗ a(4);F(2)=f(a(2)); else a(1)=a(2);a(2)=a(3);F(1)=F(2);F(2)=F(3); a(3)=alpha ∗ a(1)+(1-alpha) ∗ a(4);F(3)=f(a(3)); end, aa=[aa,a];FF=[FF,F]; end The instructions aa, FF prints out x and f at ﬁve successive iterations  aa =  0  0.9549  1.5451  1.9098  1.9098  1.9098  0.9549  1.5451  1.9098  2.1353  2.0492  1.9959  1.5451  1.9098  2.1353  2.2746  2.1353  2.0492  2.5000  2.5000  2.5000  2.5000  2.2746  2.1353  c Pearson Education Limited 2011   607  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  608 FF =  0  0.7795  1.5446  1.8011  1.8011  1.8011  0.7795  1.5446  1.8011  1.8040  1.8191  1.8183  1.5446  1.8011  1.8040  1.7341  1.8040  1.8191  1.4962  1.4962  1.4962  1.4962  1.7341  1.8040  with the best value at x = 2.0492 and f = 1.8191.  34(b)  The algorithm is the same as part (a) except the function is now  f=@ (x)(log(x)+2 ∗(1-x)/(1+x))/(1-x)∧ 2; The ﬁve iterations give aa = 1.5000  1.8820  1.8820  2.0279  2.1180  2.1180  1.8820  2.1180  2.0279  2.1180  2.1738  2.1525  2.1180  2.2639  2.1180  2.1738  2.2082  2.1738  2.5000  2.5000  2.2639  2.2639  2.2639  2.2082  FF = 0.0218604  0.0260436  0.0260436  0.0265463  0.0266783  0.0266783  0.0260436  0.0266783  0.0265463  0.0266783  0.0267059  0.0266998  0.0266783 0.0262879  0.0266780 0.0262879  0.0266783 0.0266780  0.0267059 0.0266780  0.0267051 0.0266780  0.0267059 0.0267051  Note that the function is so ﬂat that it has been quoted to six ﬁgures; the best result is x = 2.1738, f = 0.0267059.  Exercises 10.4.4 35  The gradient is given by    1 1 + G= −1 0  −1 2    x1 x2    so, at the ﬁrst step     1 1 , f = 1.5, G = a= 1 1 c Pearson Education Limited 2011   Glyn James: Advanced Modern Engineering Mathematics, 4th edition  609  The ﬁrst search is for min [f(1 − μ, 1 − μ)] and a simple calculation gives μ = 2. μ The second iteration can be started as    1 −1 , f = −0.5, G = a= −1 −1   Note that the two gradients are perpendicular. The search min [f(−1 − μ, −1 + μ)] μ is easily performed (exactly in this problem) to give μ = 0.4 and the problem is ready for the next iteration.    1/5 −7/5 , f = −0.9, G = a= 1/5 −3/5   36  For this function the gradient is   6x + 2y + 3 G= 2x + 6y + 2    and it easily checked that x = −7/16, y = −3/16 gives zero gradient and hence the minimum at f = −0.84375. The steepest descent method is easily written in MATLAB. For more complicated functions two M-ﬁles are needed, one to set up the function and its gradient and the second to set up the line search routine. Anonymous functions can be used for more .straightforward functions. The following lines of code solve the problem. f=@ (x)2 ∗ (x(1)+x(2))ˆ2+(x(1)-x(2))ˆ2+3 ∗ x(1)+2 ∗ x(2); g=@(x)[4∗(x(1)+x(2))+2∗(x(1)-x(2))+3;4∗(x(1)+x(2))-2∗(x(1)-x(2))+2]; fm=@ (t,xx,G)f(xx-t ∗ G); % The function, its derivative and the line search are now set xx=[0,0];ff=f(xx);G=g(xx);XX=[xx];FF=[ff];GG=[G];% Start values [t,fval]=fminbnd@(t)fm(t,xx,G),0,2);% Note how the minimisation is done xx=xx-t ∗ G;G=g(xx);ff=f(xx);XX=[XX,xx];GG=[GG,G];FF=[FF,ff]; %Next point %Repeat the last two lines to iterate c Pearson Education Limited 2011   610  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  The ﬁrst ﬁve values are produced in XX,FF and GG as XX=  0  -0.3824  -0.4296  -0.4365  -0.4374  0  -0.2549  -0.1841  -0.1887  -0.1874  FF=  0  -0.8284  -0.8435  -0.8437  -0.8437  GG=  3.0000 2.0000  0.1961 -0.2941  0.0545 0.0363  0.0036 -0.0053  0.0010 0.0007  Because the function is a quadratic, the Newton method, which is based on the assumption that the function is approximated by a quadratic, must converge in one iteration. 37  The MATLAB implementation is similar to Exercise 36.  The program is identical except for the instructions that generate the functions, which are now f=@ (x)(x(1)-x(2)+x(3)) ∧ 2+(2 ∗ x(1)+x(3)-2) ∧ 2+(x(3) ∧ 2-1) ∧ 2; g=@ (x)[2 ∗ (x(1)-x(2)+x(3))+4 ∗ (2 ∗ x(1)+x(3)-2);2 ∗ (x(1)-x(2)+x(3));... 2 ∗ (x(1)-x(2)+x(3))+2 ∗ (2 ∗ x(1)+x(3)-2)+4 ∗ x(3) ∗ (x(3) ∧ 2-1)]; The ﬁrst few iterations are XX =  2.0000  1.1518  0.4985  0.6175  0.4941  0.5178  2.0000  2.1696  1.8171  1.7505  1.6616  1.6358  2.0000  0.4733  0.7984  0.9653  1.0178  1.0300  FF =  29.0000  1.5023  0.4440  0.0729  0.0237  0.0158  GG =  20.0000  2.0184  -1.8592  0.4657  -0.2758  0.0859  -4.0000  1.0891  1.0405  0.3354  0.2995  0.1762  36.0000  -1.0044  -2.6078  -0.1980  -0.1414  0.2054  To use the built in routine fminunc, it is easier to use an M ﬁle. function [f,g] = q37(x) %Question 37 f=(x(1)-x(2)+x(3)) ∧ 2+(2 ∗ x(1)+x(3)-2) ∧ 2+(x(3) ∧ 2-1) ∧ 2; g=[2 ∗ (x(1)-x(2)+x(3))+4 ∗ (2 ∗ x(1)+x(3)-2);2 ∗ (x(1)-x(2)+x(3));... 2 ∗ (x(1)-x(2)+x(3))+2 ∗ (2 ∗ x(1)+x(3)-2)+4 ∗ x(3) ∗ (x(3) ∧ 2-1)]; end c Pearson Education Limited 2011   Glyn James: Advanced Modern Engineering Mathematics, 4th edition  611  The main progam is x=[2;2;2]; options=optimset('GradObj','on','display','iter','HessUpdate', 'steepdesc','LargeScale','off'); [x,fval]=fminunc(@q37,x,options) The ﬁrst three and last three iterations are Iteration 0  Func-count 1  f(x) 29  1  2  3.67901  2  4  2.35628  with starting point x=(2,2,2)  . . . . . . . . . . . . .  38  59  119  8.6485e-010  60  121  6.17926e-010  61  123  4.41502e-010  with final point x=(0.5,1.5,1.0)  The function, gradient and Hessian matrix are computed in the M-ﬁle  newton38 function [a,agrad,ajac]=newton38(z) t1=z(1)-z(2)+z(3);t2=2 ∗ z(1)+z(3)-2; a=t1ˆ2+t2ˆ2+(z(3)ˆ2-1)ˆ2; agrad(1)=2 ∗ t1+4 ∗ t2; agrad(2)=-2 ∗ t1; agrad(3)=2 ∗ t1+2 ∗ t2+4 ∗ z(3)ˆ3-4 ∗ z(3); ajac(1,1)=10;ajac(1,2)=-2;ajac(1,3)=6; ajac(2,1)=-2;ajac(2,2)=2;ajac(2,3)=-2; ajac(3,1)=6;ajac(3,2)=-2;ajac(3,3)=12 ∗ z(3)ˆ2; and the Newton iteration proceeds as a=[2;2;2];E=[0;0;0;0]; for i=1:5 [f,G,J]=newton38(a);E=[E,[f;a]]; c Pearson Education Limited 2011   612  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition a=a-J\G  ; end % gives  39  f  29.0000  1.2448  0.1056  0.0026  0.0000  x  2.0000  0.2727  0.4245  0.4873  0.4995  y  2.0000  1.7273  1.5755  1.5127  1.5005  z  2.0000  1.4545  1.1510  1.0253  1.0009  The ﬁgure illustrates the cost of the road from (0,0) to (1,a) to ( b,11-2b) is 1/2  C = 2(1 + a2 )1/2 + (b − 1)2 + (11 − 2b − a)2  and any of the minimization methods give a = 0.2294, b = 4.5083 and c = 5.9743, giving equations of lines as y = 0.2294x, y = 0.5x − 0.2706 and cost = 5.974.  y (b,11– 2b)  (1,a) y = 11 – 2x  x  (0,0)  Exercises 10.4.7 40(a)  The gradient can be calculated as G =    10x − 2y − 8 −2x + 2y  choice of H is the unit matrix. c Pearson Education Limited 2011    and the initial  Glyn James: Advanced Modern Engineering Mathematics, 4th edition  613  Iteration 1 The computation commences       2 8 1 0 f = 4 G1 = H1 = a1 = 2 1 0 0 1   2 − 8λ . The minimum and the minimization takes place in the direction a = 2 in this direction may be obtained as λ = 0.1. The new values are    1.2 0 a2 = f2 = 0.8 G2 = 2 1.6   and the values of h1 and y1 are calculated as   −0.8 h1 = a2 − a1 = 0      −8 and y1 = G2 − G1 = 1.6    Finally, the H is updated as      1 −0.8 1 −8 1 0 [−8 1.6] + [−0.8 0] − H2 = 0 0 1 66.56 1.6 6.4       0.1 0 0.9615 −0.1923 1 0 + − = 0 0 −0.1923 0.0385 0 1   0.1385 0.1923 = 0.1923 0.9615   Iteration 2 The iteration starts with the variables computed from iteration 1     1.2 0 0.1385 f2 = 0.8 G2 = H2 = a2 = 2 1.6 0.1923   0.1923 0.9615    The method follows the same pattern as iteration 1 and so the computations are not written down in the same detail.      1 0 0.1250 f = 0 G3 = H3 = a3 = 1 3 0 0.1250  0.1250 0.6250    The minimum has been achieved; this is expected since the function is quadratic and it is known that the method converges in n steps for an n -dimensional quadratic, provided the minimizations are performed exactly.  c Pearson Education Limited 2011   614  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  40(b)  The problem has three variables and is not quadratic. The one-dimensional  minimizations need a numerical procedure; so it is better to use a package such as MATLAB. The M-ﬁle required for the function is function [f,G]=q40b(z) t1=z(1)-z(2)+z(3);t2=2 ∗ z(1)+z(3)-2; f=t1ˆ2+t2ˆ2+z(3)ˆ4; G(1)=2 ∗ t1+4 ∗ t2; G(2)=-2 ∗ t1; G(3)=2 ∗ t1+2 ∗ t2+4 ∗ z(3)ˆ3; The main DFP segment is 1. 2.  a=[0;0;0];H=eye(3);[f,G]=q40b(a); fm=@(t,xx,GG)q40b(xx-t ∗ GG);  3.  aa=[a];ff=[f];gg=[G];HH=[H];  4.  for n=1:4  5.  D=H ∗ G;[t,fval]=fminbnd(@ (t)fm(t,a,D),0,2);  6.  aold=a;Gold=G;a=aold-t ∗ D;[f,G]=q40b(a);  7.  h=a-aold;y=G-Gold;  8.  H=H-H ∗ y ∗ y' ∗ H/(y' ∗ H ∗ y)+h ∗ h'/(h' ∗ y);  9.  aa=[aa,a];ff=[ff,f];gg=[gg,G];HH=[HH,H];  10. end The instructions give the results aa=  0  0.5853  1.0190  1.0186  1.0184  0  0  0.9813  0.9814  0.9815  0  0.2926  -0.0372  -0.0372  -0.0369  ff=  4.0000  1.0662  0.0000  0.0000  0.0000  gg=  -8.0000  -0.3916  0.0047  -0.0000  -0.0000  0 -4.0000  -1.7558 0.7823  -0.0012 0.0027  -0.0001 -0.0002  -0.0000 -0.0002  The convergence looks good but slows because H tends to a singular matrix.  c Pearson Education Limited 2011   Glyn James: Advanced Modern Engineering Mathematics, 4th edition 41  615  Evaluating H i+1 y i it is readily shown that equation (10.18) is satisﬁed.  The update is one that is quite eﬀective but suﬀers from the problem that the denominator can become zero when line searches are not exact. A great deal of remedial action must be taken to ensure that the problem is overcome. 41(a)  The function is quadratic so the solution is obtained in two steps. The  program in Exercise 39, suitably adapted, was used to compute the solution. Iteration 1      1 2 1 H= f=9G= a= 0 8 2  0 1    Iteration 2     0.9948 0.9697 0.4848 H= f = 0.2424 G = a= −0.0619 −0.2424 −0.0606   41(b)  −0.0619 0.2577    The exercise is similar to Exercise 40(b) but with a diﬀerent function and  updating methods. function [f,G]=q41b(z) t1=z(1);t2=z(1)-z(2)+1; f=t1ˆ2+t2ˆ2+z(2)ˆ2 ∗ z(3)ˆ2; G(1)=2 ∗ t1+2 ∗ t2; G(2)=-2 ∗ t2+2 ∗ z(2) ∗ z(3)ˆ2; G(3)=2 ∗ z(3) ∗ z(2)ˆ2; In the lines 1,2 and 6 of Exercise 40(b) the M ﬁle q41b replaces q40b. In line 8, the Rank 1 and the BFGS updates replace the one in the program. Also, a diﬀerent start point is used. The results for Rank 1 update (i) are aa=  0.5000  -0.0732  -0.1628  -0.00592  -0.0593  0.5000  0.8344  0.7747  0.9235  0.9234  0.5000  0.4522  0.0525  -0.0640  -0.0640  ff=  1.3125  0.1563  0.0321  0.0073  0.0073  gg=  3.0000  0.0386  -0.2006  -0.0839  -0.0839  -1.7500  0.1564  -0.1207  -0.0270  -0.0270  0.2500  0.6296  0.0630  -0.1091  -0.1092  c Pearson Education Limited 2011   616  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  The results for BFGS update (ii) are  aa=  0.5000  -0.0732  0.0969  -0.0726  -0.0348  0.5000  0.8344  0.7384  0.9408  0.9344  0.5000  0.4522  0.0659  0.0141  0.0071  ff=  1.3125  0.1563  0.0389  0.0056  0.0022  gg=  3.0000  0.0386  0.1355  -0.1718  -0.0080  -1.7500 0.2500  0.1564 0.6296  -0.3229 0.0718  0.0271 0.0250  -0.0614 0.0124  Note that BFGS does better than Rank 1 and is the choice of method for the built-in function fminunc. 42  It is easy to check that equation (10.18) is satisﬁed but note that the notation  has changed and y,h have been replaced by u,v respectively. H u = Hu + vpT u − HuqT u = Hu + v − Hu = v since pT u = qT u = 1 To match with the Davidon formula (10.19) ﬁrst choose β = α = 0 and hence pT u = αvT u = 1 and qT u = β uT Hu = 1 Substituting gives H = H −  HuuT H vvT + T uT Hu v u  as required. The formula in Exercise 41(i) is obtained by putting β = β = −α = −α . Thus p = q = α(V − Hu) and hence pT u = α(v − Hu)T u = 1 Substituting gives the formula H = H +  (v − Hu)(v − Hu)T (v − Hu)T u  This whole class of solutions was devised by Huang. Many general results can be proved for this class and many of the commonly used formulae are included in it.  c Pearson Education Limited 2011   Glyn James: Advanced Modern Engineering Mathematics, 4th edition 43 (a)  The Fletcher-Reeves method is easily written as a MATLAB segment global a b p a=1;b=1;g=[3 ∗ a;b];p=-g; lam=fminbnd(@ffr,0,2) % gives  lam = 0.3571 ∗  a=a+lam p(1),b=b+lam ∗ p(2),gold=g;g=[3 ∗ a;b] % gives ∗  p=-g+p (g % gives  a= -0.0714 ∗  g)/(gold  ∗  b= 0.6429  g= -0.2143  0.6429  gold)  p= 0.0765  -0.6888  lam=fminbnd(@ffr,0,2) % gives  lam = 0.9333  a=a+lam ∗ p(1),b=b+lam ∗ p(2),gold=g;g=[3 ∗ a;b] % gives  a=-4.0246e-016  b=-1.1102e-016  which is the minimum point. The M-ﬁle used is function v=ffr(x) global a b p v=3 ∗(a+x ∗ p(1))ˆ2+(b+x ∗ p(2))ˆ2;  43(b)  617  The three-variable problem is handled in a similar manner. global a b c p a=0.5;b=0.5;c=0.5; f=(a-b+1)ˆ2+aˆ2 ∗ bˆ2+(c-1)ˆ2; g=[2 ∗ (a-b+1)+4 ∗ a ∗ bˆ2;-2 ∗ (a-b+1)+2 ∗ aˆ2 ∗ b;2 ∗ (c-1)]; p=-g;W=[f;a;b;c]; for i=1:5 gold =g;lam=fminbnd(@ffr2,-2,2); a=a+lam ∗ p(1);b=b+lam ∗ p(2);c=c+lam ∗ p(3); f=(a-b+1)ˆ2+aˆ2 ∗ bˆ2+(c-1)ˆ2;W=[W[f;a;b;c]]; g=[2 ∗ (a-b+1)+4 ∗ a ∗ bˆ2;-2 ∗ (a-b+1)+2 ∗ aˆ2 ∗ b;2 ∗ (c-1)]; p=-g+p ∗ (g  ∗ g)/(gold ∗ gold); end c Pearson Education Limited 2011   618  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  W % gives the sequence of iterations f 1.3125 0.0764 0.0072 0.0007  0.0004  0.0000  a  0.5000  -0.0950  0.0057  -0.0251  -0.0079  -0.0032  b  0.5000  0.9165  0.9276  0.9633  0.9742  0.9978  c  0.5000  0.7380  0.9674  1.0009  1.0044  1.0014  Review Exercises 10.7 1  The successive tableaux are as follows: z x3 x4 x5  x1 −12 1 2 1  z x3 x1 x5  x1 0 0 1 0  z x2 x1 x5  x2 −8 1 1 3 x2 −2 0.5 0.5 2.5  x1 0 0 1 0  x3 0 1 0 0 x3 0 1 0 0  x2 0 1 0 0  x4 0 0 1 0 x4 6 −0.5 0.5 −0.5  x3 4 2 −1 −5  x4 4 −1 1 2  x5 0 0 0 1  Soln 0 350 600 900  x5 0 0 0 1  Soln 3600 50 300 600  x5 0 0 0 1  Soln 3800 100 250 350  Hence the solution is read from the table as x1 = 250, x2 = 100 and F = 3800 . 2  Let x1 , x2 , x3 be the numbers of sailboard constructed of types 1,2,3. The  proﬁt is 10x1 + 15x2 + 25x3 and the constraints are 5x1 + 10x2 + 25x3 ≤ 290 3x1 + 2x2 + x3 ≤ 72 10x1 + 20x2 + 30x3 ≤ 400 c Pearson Education Limited 2011   Glyn James: Advanced Modern Engineering Mathematics, 4th edition  619  The system only has 'less than' inequalities so the initial and subsequent tableaux can be written down immediately.  z x4 x5 x6  x1 −10 5 3 10  z x3 x5 x6  x1 −5 0.2 2.8 4  z x3 x5 x1  x1 0 0 0 1  z x3 x4 x1  x1 0 0 0 1  x2 −15 10 2 20  x3 −25 25 1 30  x2 −5 0.4 1.6 8  x3 0 1 0 0  x2 5 0 −4 2  x3 0 1 0 0  x2 2.5 0.5 −5 0.5  x3 0 1 0 0  x4 0 1 0 0  x5 0 0 1 0  x4 1 0.04 −0.04 −1.2 x4 −0.5 0.1 0.8 −0.3  x4 0 0 1 0  x5 0 0 1 0 x5 0 0 1 0  x5 0.62 − 0.12 1.25 0.37  x6 0 0 0 1 x6 0 0 0 1 x6 1.25 −0.05 −0.7 0.25  x6 0.81 0.04 − 0.88 −0.01  Soln 0 290 72 400 Soln 290 11.6 60.4 52 Soln 355 9 24 13  Soln 370 6 30 22  The solution is x1 = 22, x2 = 0, x3 = 6 and maximum proﬁt is £ 370. 3  Let x1 , x2 , x3 be the number of standard, super and deluxe cars respectively,  then the proﬁt function is 100x1 + 300x2 + 400x3 and the constraints are 10x1 + 20x2 + 30x3 ≤ 1600 10x1 + 15x2 + 20x3 ≤ 1500 x2 + x3 ≤ 50 x1 + x2 + x3 ≥ 70 c Pearson Education Limited 2011   620  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  The problem includes a 'greater than' inequality, so the two-phase approach is needed. An artiﬁcial variable is introduced and the artiﬁcial cost is introduced in phase 1. The z row of the tableau is manipulated to bring it to standard form. z x4 x5 x6 x8  x1 −1 10 10 0 1  x2 −1 20 15 1 1  x1 0 0 0 0 1  z x4 x5 x6 x1  x2 0 10 5 1 1  x3 −1 30 20 1 1  x3 0 20 10 1 1  x4 0 1 0 0 0  x4 0 1 0 0 0  x5 0 0 1 0 0  x5 0 0 1 0 0  x6 0 0 0 1 0  x7 1 0 0 0 −1  x6 0 0 0 1 0  x7 0 10 10 0 −1  x8 0 0 0 0 1  x8 1 − 10 − 10 0 1  Soln −70 1600 1500 50 70  Soln 0 900 800 50 70  A feasible solution has been obtained so the method moves to phase 2. The z row is ﬁrst re-calculated z x4 x5 x6 x1  x1 0 0 0 0 1  x2 −200 10 5 1 1  z x3 x5 x6 x1  x1 0 0 0 0 1  x2 −50 0.5 0 0.5 0.5  z x3 x5 x2 x1  x1 0 0 0 0 1  x2 0 0 0 1 0  x3 −300 20 10 1 1 x3 0 1 0 0 0  x3 0 1 0 0 0  x4 0 1 0 0 0  x5 0 0 1 0 0  x6 0 0 0 1 0  x7 −100 10 10 0 −1  Soln 7000 900 800 50 70  x4 15 0.05 −0.5 −0.05 −0.05  x5 0 0 1 0 0  x6 0 0 0 1 0  x7 50 0.5 5 −0.5 −1.5  Soln 20500 45 350 5 25  x4 10 0.1 −0.5 −0.01 0  x5 0 0 1 0 0  x6 100 −1 0 2 −1  c Pearson Education Limited 2011   x7 0 1 5 −1 −1  Soln 21,000 40 350 10 20  Glyn James: Advanced Modern Engineering Mathematics, 4th edition  621  The solution is read oﬀ the table as x1 = 20, x2 = 10, x3 = 40 and the maximum proﬁt is £21,000. Note that the (z, x7 ) entry is zero; so a non-unique solution is expected. Interchanging the x3 and x7 entries and completing one further tableau gives the alternative solution x1 = 60, x2 = 50, x3 = 0 and proﬁt is still £21,000. The MAPLE solution with(simplex): cor3:={10 ∗ x+20 ∗ y+30 ∗ z<=1600,10 ∗ x+15 ∗ y+20 ∗ z<=1500, y+z<=50,x+y+z>=70}; obr3:=100 ∗ x+300 ∗ y+400 ∗ z; maximize(obr3,cor3,NONNEGATIVE); # gives the solution {x=20, z=40,y=10} MAPLE provides the same solution but does not identify the alternative solution. Similarly, for the corresponding MATLAB code, which is f=[-100,-300,-400];A=[10,20,30;10,15,20;0,1,1;-1,-1,-1];b=[1600;1500;50;-70]; options=optimset('LargeScale','off','Simplex','on'); [x,fval]=linprog(f,A,b,[],[],zeros(3,1),[],[],options)  4 Let the student buy x1 kg of bread and x2 kg of cheese, then the cost to be minimized is 60x1 + 180x2 and the two constraints are 1000x1 + 2000x2 ≥ 3000 25x1 + 100x2 ≥ 100 There are two surplus and two artiﬁcial variables in the tableau and the artiﬁcial cost function which has been processed to standard form. Phase 1 z x5 x6  x1 −1025 1000 25  x2 −2100 2000 100  x3 1 −1 0  x4 1 0 −1  x5 0 1 0  c Pearson Education Limited 2011   x6 0 0 1  Soln −3100 3000 100  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  622  z x5 x2  x1 −500 500 0.25  z x1 x2  x1 0 1 0  x2 0 0 1  x2 0 0 1  x3 1 −1 0  x4 −20 20 −0.01  x3 0 −0.002 0.0002  x4 0 0.04 − 0.02  x5 0 1 0  x5 1 0.002 − 0.0002  x6 21 −20 0.01  x6 1 − 0.04 0.02  Soln −1000 1000 1  Soln 0 2 0.5  Phase 1 is completed so the tableau is reconstituted as z x1 x2  x1 0 1 0  x2 0 0 1  x3 0.03 − 0.002 0.0002  x4 1.2 − 0.04 0.02  Soln − 210 2 0.5  It may be noted that the z row entries are positive; so the tableau is optimal and there is no need to enter phase 2. Thus the minimum cost of the diet is 210p and is made up of 2 kg of bread and 0.5 kg of cheese. 5  The square of the distance from the origin to the point ( x, y ) is f = x2 + y 2  so the problem is to optimize this function subject to the condition that it lies on the curve g = x2 − xy + y2 − 1 = 0 so ∂g ∂f +λ = 2x + λ(2x − y) ∂x ∂x ∂f ∂g 0= +λ = 2y + λ(−x + 2y) ∂y ∂y 0=  Subtracting these two equations gives 0 = 2(x − y) + λ(3x − 3y)  ⇒  x=y  or  c Pearson Education Limited 2011   λ=−  2 3  Glyn James: Advanced Modern Engineering Mathematics, 4th edition  623  Adding the two equations gives ⇒  0 = 2(x + y) + λ(x + y)  x = −y  or  λ = −2  The ﬁrst possibility, x = y, gives the points, (1,1) and ( −1,−1) with f = 2 in each case and the second possibility, x = −y , gives   1 −1 √ ,√ 3 3     and  −1 1 √ ,√ 3 3   with f =  2 3  2 and λ = −2 reproduce the identical solutions. 3 Although it is not proved, the ﬁrst solution is the maximum and the second the In this problem the cases λ = −  minimum.  6  The volume of the solid is V = x3 + y 3  where the sides of the two cubes are x and y . The surface area has essentially the area of two faces of the smaller cube removed so S = 7 = 6x2 + 4y2 The Lagrange multiplier equations are 3x2 + 12xλ = 0 3y2 + 8yλ = 0 The solutions when x = y = 0 can be dismissed since the volume is zero. There are three other cases Case 1  x = 0,  −8 y= λ= 3  Case 2  y = 0,  x = −4λ =  Case 3  x = −4λ,  y=  −8 λ, 3    32   7 7 ,V = 4 4   32   7 7 ,V = 6 6 ⇒  7 = 96λ2 +  c Pearson Education Limited 2011   256 2 λ 9  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  624  3 2 35 ±3 √ so the sides have lengths √ and √ and volume 3 (4 10) 10 10 (10) 2 The cases when either x = 0 or y = 0 imply that the problem has collapsed to a  and hence λ =  single cube so these solutions are omitted as geometrically uninteresting.  7  The problem is to maximize the distance (x − 1)2 + y2 + z2  subject to 2x + y2 + z = 8 The Lagrange equations are 2(x − 1) + 2λ = 0 ⇒ x = 1 − λ 2y + 2yλ = 0 ⇒ y = 0 or λ = −1 −1 λ 2z + λ = 0 ⇒ z = 2 There are two cases y=0 gives λ =  −12 5  so  x=  17 6 , y = 0, z = 5 5  λ = −1 1 gives x = 2, z = 2   and y = ±  7 2  The ﬁrst of these possibilities gives the maximum distance.  8  The Lagrange equations for this problem are 1 + 2λx = 0 2 + 2λy = 0 3 + 2λz = 0  and putting back into the constraint gives 2λ = ±1. The local extrema are therefore at (1,2,3) and (−1, −2, −3) with corresponding F = 14 and F = −14. c Pearson Education Limited 2011   Glyn James: Advanced Modern Engineering Mathematics, 4th edition  625  To obtain the global extremum all the points on each of the boundaries must be examined. However, in this case, the geometry is suﬃciently simple to establish √ the result. The region is the inside of the sphere of radius 14 in the region where all the variables are positive. The cost function comprises a series of parallel planes, so the global maximum in the region is the local maximum at (1,2,3) and the global minimum in the region is when all the variables are zero, namely (0,0,0) and F = 0. 9 (i) We are given that 2s = a + b + c, so maximizing A2 with respect to b and c, together with this constraint, gives ∂(A2 ) ∂(A2 ) = s(s − a)(s − c) + λ = 0, = s(s − a)(s − b) + λ = 0 ∂b ∂c Clearly b = c and the triangle is isosceles. (ii) Now, as a so to the above equations we add ∂(A2 ) = s(s − b)(s − c) + λ = 0 ∂a and we see that we must have a = b = c and the triangle is equilateral.  10  We need to consider the problem of maximizing  a 2 π 2 + =b r h Using the Lagrange multiplier approach we have the two equations V = πr2 h subject to the constraint  ∂V 2a2 = 2πrh − 3 λ = 0 ∂r r ∂V 2π2 = πr2 − 3 λ = 0 ∂h h together with the constraint itself. Eliminating λ and solving gives 3π2 3a2 and r2 = b 2b Note that it has not been proved that these values give the maximum but this can h2 =  be inferred from physical or geometric reasoning.  c Pearson Education Limited 2011   626  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  11  First notice that as k → ∞ then F → 0 and a careful Taylor expansion  shows that lim F = 0. Take any k > 1, then F takes a positive value so we know k→1  there must be a maximum for k > 1. We start the bracket procedure with k = 1.1 and initial increment of 0.1. k F/A  1.1 0.00721  1.2 0.01258  1.4 0.01962  1.8 0.02556  2.6 0.02602  4.2 0.01995  The maximum has been bracketed by 1.8 ≤ k ≤ 4.2 and the quadratic algorithm is given in terms of anonymous functions as qr11=@(x)[x;(log(x)-2∗(x-1)/(x+1))/(x-1)∧2]; %x and the function value zz=[qr11(1.8),qr11(2.6),qr11(4.2)] % start values zz =  1.8000  2.6000  4.2000  0.0256  0.0260  0.0200  % repeat from here p=polyfit(zz(1,:),zz(2,:),2); xstar=-0.5 ∗ p(2)/p(1);zstar=qr11(xstar); if zstar(2)>zz(2,2) if zstar(1)zold(2))&(n                          μA is rejected.  9  From the data, respective sample averages X̄A = 7281 and X̄B = 6885,  standard deviations SA,n−1 = 419.1 and SB,n−1 = 402.6, and sizes nA = nB = 8, the pooled estimate of standard deviation is  Sp =  7 × (419.12 + 402.62 ) = 410.9 14  Using t.05,14 = 1.761 and t.025,14 = 2.145, conﬁdence intervals for μA − μB are   2 = (34, 758) 90% : 7281 − 6885 ± 1.761 × 410.9 × 8  2 95% : 7281 − 6885 ± 2.145 × 410.9 × = (−45, 837) 8 The hypothesis that μA = μB is rejected at the 10% level but accepted at the 5% level.  10  The sample proportion p̂ =  38 540  = 0.0704 and n = 540.  Using z.05 = 1.645 and z.025 = 1.96, conﬁdence intervals for the true proportion p are   0.0704 × (1 − 0.0704) = (0.052, 0.089) 540  0.0704 × (1 − 0.0704) = (0.049, 0.092) 95% : 0.0704 ± 1.96 × 540  90% : 0.0704 ± 1.645 ×  The hypothesis that p < 0.05 is rejected at the 10% level but accepted at the 5% level. Alternatively, the test statistic Z=   0.0704 − 0.05 = 21.8 0.05 × (1 − 0.05)/540  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  642  leads to rejection at both 5% and 10% levels. The test statistic is more accurate (the conﬁdence interval is approximate).  11  Using z.05 = 1.645, the 90% conﬁdence interval for proportion is  p̂ ± 1.645  p̂(1 − p̂) n  Thus,     p̂(1 − p̂) = 0.9 P | p − p̂ |≤ 1.645 n  so with probability 0.9 the maximum error is 1.645 p̂(1 − p̂)/n. Although p̂ is  unknown before the experiment, a ﬁgure in the region of 0.25 is expected, hence we require  0.25 × 0.75 ≤ 0.05 1.645 n from which n≥  0.25 × 0.75 = 203 (0.05/1.645)2  If in fact n = 200, the sample proportion is p̂ = conﬁdence interval for p is  0.275 ± 1.645  12  55 200  = 0.275, and the 90%  0.275 × 0.725 = (0.223, 0.327) 200  Using sample proportions p̂1 = 0.31 and p̂2 =  74 150  = 0.493, and z.05 = 1.645  and z.025 = 1.96, conﬁdence intervals for p1 − p2 are 1/2  p̂1 (1 − p̂1 ) p̂2 (1 − p̂2 ) + 90% : p̂1 − p̂2 ± 1.645 = (−0.28, −0.08) 100 150  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  643  1/2  p̂1 (1 − p̂1 ) p̂2 (1 − p̂2 ) = (−0.30, −0.06) 95% : p̂2 − p̂1 ± 1.96 + 100 150 The hypothesis that p1 ≤ p2 − 0.08 is therefore accepted at the 10% level but rejected at the 5% level. 13  Using sample proportions p̂1 =  30 180  = 0.1667 and p̂2 =  32 500−180  = 0.1, and  z.025 = 1.96, the 95% conﬁdence interval for p1 − p2 is 1/2  p̂1 (1 − p̂1 ) p̂2 (1 − p̂2 ) + = (0.003, 0.130) p̂1 − p̂2 ± 1.96 180 320 The hypothesis that p1 > p2 is therefore accepted at the 5% level. Alternatively, the test statistic p̂1 − p̂2  1 p̂(1 − p̂) 180 +  Z=  (where p̂ =  30+32 500  1 320    1/2  = 2.17 > z.025  = 0.124) again leads to the hypothesis that p1 > p2 is accepted.  Exercises 11.4.7 14 Y 1 2 3 total  14(a)  1 0 0.20 0.14 0.34  X 2 0.17 0.11 0.25 0.53  3 0.08 0 0.05 0.13  total 0.25 0.31 0.44 1  Marginal distributions of X and Y (summing rows and columns) are P(X = 1) = 0.34, P(X = 2) = 0.53, P(X = 3) = 0.13 P(Y = 1) = 0.25, P(Y = 2) = 0.31, P(Y = 3) = 0.44  c Pearson Education Limited 2011   644  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  14(b) P(Y = 3 | X = 2) =  14(c)  P(X = 2 ∩ Y = 3) 0.25 = = 0.472 P(X = 2) 0.53  The mean and variance of X are given by E(X) = 1 × 0.34 + 2 × 0.53 + 3 × 0.13 = 1.79 E(X2 ) = 1 × 0.34 + 4 × 0.53 + 9 × 0.13 = 3.63 so σ2X = 3.63 − 1.792 = 0.426  Similarly the mean and variance of Y are given by E(Y) = 1 × 0.25 + 2 × 0.31 + 3 × 0.44 = 2.19 E(Y2 ) = 1 × 0.25 + 4 × 0.31 + 9 × 0.44 = 5.45 so σ2X = 5.45 − 2.192 = 0.654 The expected value of the product XY is given by E(XY) =1 × 0 + 2 × 0.17 + 3 × 0.08 + 2 × 0.2 + 4 × 0.11 + 9 × 0 + 3 × 0.14 + 6 × 0.25 + 9 × 0.05 = 3.79 Hence the correlation coeﬃcient is 3.79 − 1.79 × 2.19 = −0.246 ρX,Y = √ 0.426 × 0.654  15 1/2  E(X) =  xdx = 0 −1/2 1/2  3  x3 dx = 0  E(X ) = −1/2 2  Cov (X, X ) = E(X3 ) − E(X)E(X2 ) = 0  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 16  From cx ≤ y ≤ cx + 1 we infer  y−1 c  ≤x≤  y c  , so  fY (y) =  fX,Y (x, y)dx ⎧  y/c y 1dx = ⎪ 0 ⎪ ⎨ c 1 1dx = 1 = 0 ⎪ ⎪ 1 ⎩1 1dx = 1 − (y − 1) (y−1)/c c  if 0 ≤ y ≤ c if c ≤ y ≤ 1 if 1 ≤ y ≤ 1 + c  17(a) 1 ∞ ∞ −(x+y)/2 xe dydx 8 1 1 1 ∞ −x/2 = xe [−2e−y/2 ]∞ 1 dx 8 1 ∞ 1 = e−1/2 xe−x/2 dx 4 1 1 −1/2 −x/2 ∞ 1 −1/2 =− e [xe ]1 + e 2 2 1 1 = e−1 + e−1/2 [−2e−x/2 ]∞ 1 2 2 3 = 0.552 = 2e  P(X > 100, Y > 100) =  ∞  e−x/2 dx  1  17(b) fY (y) =  1 8  ∞  xe−(x+y)/2 dx =  0  e−y/2  −xe−x/2 |∞ = 0 + 4 Hence  ∞  P(Y > 2) = 2  1 −y/2 e 8 ∞  ∞  xe−x/2 dx  0  e−x/2 dx =  0  1 −y/2 e 2  1 −y/2 1 e = 0.368 dy = e−y/2 |∞ 2 = 2 e  c Pearson Education Limited 2011   645  646  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  18  From the data, if X denotes height and Y denotes weight, X̄ = 174.26,  SX = 7.184, Ȳ = 75.7, SY = 11.703, XY = 13270; so the sample correlation coeﬃcient is r=  19  13270 − 174.26 × 75.7 = 0.934 7.184 × 11.703  From the data, X̄ = 14.23, SX = 2.457, Ȳ = 16.68, SY = 3.4, XY = 243.47  so the sample correlation coeﬃcient is r=  20  243.47 − 14.23 × 16.68 = 0.732 2.457 × 3.4  Given a sample correlation coeﬃcient r = 0.7 with n = 30, and using  z.025 = 1.96,   c = exp  2 × 1.96 √ 27   = 2.126  and the 95% conﬁdence interval for correlation is   21  1 + r − c(1 − r) 1 + r − (1 − r)/c , 1 + r + c(1 − r) 1 + r + (1 − r)/c   = (0.45, 0.85)  From Exercise 18, r = 0.934 with n = 8. Using z.025 = 1.96,  c = exp  2 × 1.96 √ 5   = 5.772  and the 95% conﬁdence interval for correlation is   1 + r − c(1 − r) 1 + r − (1 − r)/c , 1 + r + c(1 − r) 1 + r + (1 − r)/c   = (0.67, 0.99)  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  647  22 From the data, if X and Y denote mathematics and computer studies marks respectively, X̄ = 56.80, SX = 8.880, Ȳ = 54.40, SY = 12.05, XY = 3137.4 so the sample correlation coeﬃcient is r=  3137.4 − 56.8 × 54.4 = 0.444 8.88 × 12.05  Using z.05 = 1.645,   c = exp  2 × 1.645 √ 17   = 2.221  and the 90% conﬁdence interval is   1 + r − c(1 − r) 1 + r − (1 − r)/c , 1 + r + c(1 − r) 1 + r + (1 − r)/c   = (0.08, 0.70)  Similarly using z.025 = 1.96,  c = exp  2 × 1.96 √ 17   = 2.588  and the 95% conﬁdence interval is (0.00, 0.74) . This suggests that the correlation coeﬃcient is signiﬁcant at the 10% level but is marginal at the 5% level. The test statistic  √ Z=  17 1 + 0.444 ln = 1.968 2 1 − 0.444  leads to a similar conclusion. The ranks of the data are as follows:  Math. Comp.  3.5  20  2  16.5  13  16.5  13  7  19  3.5  10.5  5  18  10.5  7  7  15  13  9  1  16.5  8  20  4  15  4  18.5  6.5  16.5  9  18.5  14 6.5  10  11.5  11.5  c Pearson Education Limited 2011   2  13  4  1  648  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  The rank correlation is rs = 0.401, and the test statistic √ Z = rs n − 1 = 1.748 This is signiﬁcant at the 10% level. (The approximate formula for rank correlation gives a value 0.405, from which Z = 1.767 with the same result.) 23(a) 1  1  1  1  c(1 − y)dydx  fX,Y (x, y)dydx = 0  x  0  x 1  2  c[y − y2 ]1x dx 0  1  1 1 2 c = − x + x dx 2 2 0  1 1 1 2 1 3 =c x− x + x 2 2 6 0 c = 6 Since the joint density must integrate to unity, we must have c = 6. =  23(b) P(X <  1 3 ,Y > ) = 4 2  1  3/4  fX,Y (x, y)dxdy 1/2  0  1  min{3/4,y}  c(1 − y)dxdy  = 1/2  0  3/4  y  1  3/4  c(1 − y)dxdy +  = 1/2  0  3/4  c(1 − y)dxdy 3/4  1  0  3 c(1 − y)dy 1/2 3/4 4  2 3/4 1  y 3 y3 y2 =c + c y− − 2 3 1/2 4 2 3/4     27 1 1 9 1 3 9 9 − − + + 1− − + =6 32 192 8 24 2 2 4 32  =  c(1 − y)ydy +  = 0.484  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  23(c) 1  fX (x) = x y  1    x2 y2 1 −x+ for 0 ≤ x ≤ 1 c(1 − y)dy = 6 y − =6 2 x 2 2 c(1 − y)dx = 6(1 − y)y for 0 ≤ y ≤ 1  fY (y) = 0  24  Individual density functions for X and Y are   2 for 29.8 < x < 30.3  0 otherwise 2 for 30.1 < y < 30.6 fY (y) = 0 otherwise  fX (x) =  By independence, the joint density function is therefore  fX,Y (x, y) =  4 for 29.8 < x < 30.3 and 30.1 < y < 30.6 0 otherwise  The required probability is therefore 30.3  min{30.6,x+0.6}  P(0 ≤ Y − X ≤ 0.6) =  4dydx 29.8 30.0  max{30.1,x} x+0.6  29.8 30.1  30.1 30.6  =  4dydx 30.3  30.6  4dydx +  + 30.0   =4  30.1 30.0  4dydx 30.1  x 30.1  (x − 29.5)dx + 29.8 30.3    0.5dx 30.0  (30.6 − x)dx  + 30.1   2 x − 29.5x =4 2  30.0    x2 + 0.5 × 0.1 + 30.6x − 2 29.8  = 0.84  c Pearson Education Limited 2011   30.3  30.1  649  650  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Alternatively, the shaded area must be excluded from the square. The two  parts of  shaded area together form a square of side 0.2,  so  P(0 ≤ Y − X ≤ 0.6) = 0.5 × 0.5 − 0.2 × 0.2 = 0.84. 0.5 × 0.5  Exercises 11.5.5 25  From the data, X̄ = 16.370, Sx = 6.789, Ȳ = 36.110, Sy = 14.576, XY = 689.343  Hence  689.343 − 16.37 × 36.11 = 2.13 6.7892 â = 36.11 − 2.13 × 16.37 = 1.22 b̂ =  26  From the data, X̄ = 6.5, SX = 3.452, Ȳ = 101.5, SY = 50.74, XY = 834.25  Hence, the regression coeﬃcients are 834.25 − 6.5 × 101.5 = 14.64 3.4522 â = 101.5 − 14.64 × 6.5 = 6.315 b̂ =  When load (X) is 15 kg , a deﬂection of 226 mm is predicted.  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 27  651  From the results in Example 11.17 we ﬁnd â = −2.294 and b̂ = 0.811, and  when 15 V is measured a tension of 9.88 kN is predicted. Using t.025,12 = 2.179 and  SE =  S2Y − b̂2 S2X =  √  16.25 − 0.8112 × 34.51 = 0.360  (remember that X and Y are essentially reversed compared with Example 11.17), the 95% conﬁdence interval for tension when 15 V are measured is  1 + (15 − 12.07)2 /24.51 = (9.62, 10.14) 9.88 ± 2.179 × 0.36 × 12  28(a)  From the data, X̄ = 34.17, SX = 11.70, Ȳ = 453.8, SY = 59.34, XY = 15944  and the regression coeﬃcients are 15944 − 34.17 × 453.8 = 3.221 (using unrounded ﬁgures) 11.72 â = 453.8 − 3.221 × 34.17 − 343.7 b̂ =  For advertising x = 6 (£ 6000), sales of £ 537,000 are predicted.  28(b)  SE =  √ 59.342 − 3.2212 × 11.72 = 45.8 and using t.025,10 = 2.228 the  95% conﬁdence interval for regression slope is b̂ ± t.025,10  SE √  SX 10  = (0.46, 5.98)  The hypothesis that b = 0 is rejected at the 5% level. 28(c)  The 95% conﬁdence interval for mean sales when x = 60 is  537 ± 2.228 × 45.8 ×  1 + (60 − 34.17)2 /136.8 = (459, 615) 10  c Pearson Education Limited 2011   652  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  29  From the data, X̄ = 11.5, SX = 2.291, Ȳ = 13.25, SY = 2.99, XY = 158.38  so the regression coeﬃcients are 158.38 − 11.5 × 13.25 = 1.143 2.2912 â = 13.25 − 1.143 × 11.5 = 0.107 b̂ =  and Y = 16.1 is predicted when x = 14. Also using  SE = S2Y − b̂2 S2X = 1.442 and t.05,6 = 1.943, the 90% conﬁdence interval for mean number of defectives per hour when x = 14 is  16.1 ± 1.943 × 1.442 ×  30  1 + (14 − 11.5)2 /5.25 = (14.4, 17.8) 6  Given a model (with no constant) of form Yi = bXi + Ei  and minimizing Q=  n  [Yi − bXi ]2 i=1  we have  hence  n  dQ = −2 Xi [Yi − b̂Xi ] = 0 db i=1  n Xi Y i b̂ = i=1 n 2 i=1 Xi  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 31  If X denotes voltage and Y = X/R denotes current then from the data, Σi Xi Yi = 5397, Σi X2i = 650  so that (using the result of the previous exercise) b̂ = 8.30 The estimated resistance is R =  32  1000 = 120Ω . 8.3  Taking logs we have ln Pi + λ ln Vi = ln C  This is of the form Yi = a + bXi with Yi = ln Pi , a = ln C, b = −λ, Xi = ln Vi From the data, X̄ = 4.272, SX = 0.254, Ȳ = 3.423, XY = 14.452 so that b̂ = −2.664 and â = 14.80 Hence Ĉ = eâ = 2.69 × 106 and λ̂ = −b̂ = 2.66 When V = 80cm 3 , a pressure P = 22.9kg/cm 3 is predicted. 33  Taking logs we have ln Yi = ln a + b ln Xi or Yi = a + bXi  From the data, X̄ = 6.183, SX  = 1.515, Y¯ = 2.377, X Y  = 12.266  c Pearson Education Limited 2011   653  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  654 so  b̂ = −1.059, â = 8.927 and â = 7533 For a lot size X = 300, a unit cost of 17.9 is predicted.  Exercises 11.6.3 Under the hypothesis, P(A) = 47 , P(B) = 27 , P(C) =  34  observed 63 22 15  A B C Total  probability 4/7 2/7 1/7  expected 57.14 28.57 14.29  1 7  .  χ2 contribution 0.601 1.511 0.035 2.147  No parameters were estimated ( t = 0), so we compare χ2 = 2.147 with χ20.05,2 = 5.991. The hypothesis is accepted. 35  The total number of books is 640, so a uniform number would be 128 per  day. Hence Obs. (fk ) : Exp. (ek ) :  153  108  120  114  145  128  128  128  128  128  χ2  4.9  3.1  0.5  1.5  2.3  χ  contribution:  The total chi-square value is χ2 = 12.3 and this is greater than χ2.05,4 = 9.49 (signiﬁcant at 5% , but not signiﬁcant at 1%). 36  The observed mean number of ﬂaws per sample is (12 + 6 × 2)/50 or 0.48.  Setting λ = 0.48, the Poisson probabilities are given by λk e−λ /k! and hence expected values are as follows: number of flaws 0 1 ≥ 2  Obs. (fk ) 32 12 6  probability 0.619 0.297 0.084  exp. (ek ) 30.9 14.9 4.2  c Pearson Education Limited 2011   χ2 contribution 0.036 0.547 0.761  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  655  The total chi-square value is χ2 = 1.35 which is very small, so the Poisson hypothesis is accepted.  37  The observed average number of α-particles per time interval (taking class  > 10 as 11 for this calculation) is (203 + 2 × 383 + . . . + 11 × 6)/(57 + 203 + . . . + 6) = 3.87 Using this value for λ in the Poisson probabilities and proceeding as in the previous exercise a total chi-square value χ2 = 12.97 is obtained. One parameter has been estimated and used for predicting the expected values, so the comparison is with χ2.05,10 = 18.3, and the Poisson hypothesis is accepted.  38  Using the measured average and standard deviation, probabilities can be  obtained from the normal table as follows:  X − 10 6.5 − 10  < = Φ(−1.75) = 1 − Φ(1.75) = 0.0401 2 2 P(X < 7.5) = 1 − Φ(1.25) = 0.1056  P(X < 6.5) = P  P(X < 8.5) = 1 − Φ(0.75) = 0.2266 P(X < 9.5) = 1 − Φ(0.25) = 0.4013 P(X < 10.5) = Φ(0.25) = 0.5987 P(X < 11.5) = Φ(0.75) = 0.7734 P(X < 12.5) = Φ(1.25) = 0.8944 P(X < 13.5) = Φ(1.75) = 0.9599 P(X < 13.5) = 0.0401  c Pearson Education Limited 2011   656  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  The class probabilities can now be inferred by P(6.5 < X < 7.5) = P(X < 7.5) − P(X < 6.5) = 0.0655 and so on; hence, the following table:  Class < 6.5 6.5–7.5 7.5–8.5 8.5–9.5 9.5–10.5 10.5–11.5 11.5–12.5 12.5–13.5 > 13.5  Probability 0.0401 0.0655 0.1210 0.1747 0.1974 0.1747 0.1210 0.0655 0.0401  Expected 4.01 6.55 12.10 17.47 19.74 17.47 12.10 6.55 4.01  Sample 1 4 6 16 16 17 20 12 6 3  Sample 2 3 6 16 13 26 7 19 5 5  For sample 1, χ2 = 2.48 which is not signiﬁcant. For sample 2, χ2 = 15.51 which exceeds χ2.025,6 = 14.45 (signiﬁcant at 2.5% level) but does not exceed χ2.01,6 = 16.81 (not signiﬁcant at 1% level). The second subscript is m − t − 1 with m = 9 (classes) and t = 2 (parameters estimated).  39  A B C Total  The contingency table is as follows:  Perfect 89 (89.04) 62 (58.88) 119 (122.07) 270  Intermediate 23 (21.44) 12 (14.18) 30 (29.39) 65  Unacceptable 12 (13.52) 8 (8.94) 21 (18.54) 41  The expected values are shown in brackets, 270 × 124 = 89.04 376  c Pearson Education Limited 2011   Total 124 82 170 376  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  657  and so on. Hence (89 − 89.04)2 (21 − 18.54)2 + ··· + 89.04 18.54 = 1.30  χ2 =  This is less than χ2.05,(3−1)(3−1) = 9.49, so there is no signiﬁcant diﬀerence in quality.  40  The contingency table (with the expected values in brackets) is as follows:  o.k. 442 536 544 397 593 442 434 195 438 594 585 541 5741  (441.6) (539.8) (539.8) (392.5) (588.8) (441.6) (441.6) (196.3) (441.6) (588.8) (588.8) (539.8)  defective 8 14 6 3 7 8 16 5 12 6 15 9 109  (8.38) (10.25) (10.25) (7.45) (11.18) (8.38) (8.38) (3.73) (8.38) (11.18) (11.18) (10.25)  total 450 550 550 400 600 450 450 200 450 600 600 550 5850  (442 − 441.6)2 (9 − 10.25)2 + ··· + = 20.56 which exceeds χ2.05,11 = 441.6 2 10.25 19.68 but is less than χ.025,11 = 21.92. The variation is signiﬁcant at the 5% Hence, χ2 =  level. For a 2 × c table of this form, eﬀectively, a comparison of c proportions, it is quicker (and equivalent) to compute c  (fj − nj p̂)2 χ = nj p̂(1 − p̂) j=1 2  where fj is the number of defectives in column j (total nj ) and p̂ =  c  j=1  is the overall sample proportion of defectives.  c Pearson Education Limited 2011   fj /  c  j=1  nj  658  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  41  The contigency table (with expected values in brackets and adjusted residuals  underneath) is as follows:  Spending level Low Medium High Total  Jacket  Shirt  Trousers  Shoes  Total  21(36) −3.0 66(66) 0.0 58(43) 2.9 145  94(92) 0.4 157(169) −1.5 120(110) 1.3 371  57(47) 1.8 94(87) 1.0 41(57) −2.8 192  113(110) 0.4 209(204) 0.7 125(133) − 1.1 447  285 526 344 1155  Chi-square = 20.7, d.f. = (4 − 1)(3 − 1) = 6, so compare with χ20.005,6 = 18.5 : signiﬁcant at 0.5% level. High-spending customers are tending to buy more of the jacket and less of the trousers. For low-spending customers, it is the other way round. 42  If p is the proportion requiring adjustments, the number of such sets in  a sample of size n is binomial with parameters n, p. With n = 4, to test the hypothesis that p = 0.1 we have P(0) = 0.94 = 0.6561   4 P(1) = 0.93 0.1 = 0.2916 1   4 P(2) = 0.92 0.12 = 0.0486 2   4 P(3) = 0.9 × 0.13 = 0.0036 3 Hence the following table:  k 0 1 2 3  fk 110 73 16 1  pk 0.6561 0.2916 0.0486 0.0036  ek = 200pk 131.22 58.32 9.72 0.72  χ2 contribution 3.43 3.70 4.06 0.11  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  659  The total chi-square value is χ2 = 11.30, which is signiﬁcant at 5% (χ2.05,3 = 7.82) but not quite signiﬁcant at 1% (χ2.01,3 = 11.35) . Using proportions, the total number of sets requiring adjustments is 73 + 2 × 16 + 3 × 1 = 108, so 108 = 0.135 p̂ = 800 Using z.025 = 1.96 and z.005 = 2.576, conﬁdence intervals for the proportion p of sets requiring adjustments are   p̂(1 − p̂) = (0.111, 0.159) 800  p̂(1 − p̂) = (0.104, 0.166) p̂ ± 2.576 800 p̂ ± 1.96  95% : 99% :  This indicates that p > 0.1, signiﬁcant (just) at the 1% level.  Exercises 11.7.4 43  We must have ∞  1=  ∞  fX (x)dx = c 0  xe−2x dx  0  1 1 = − c[xe−2x ]∞ 0 + c 2 2 ∞  1 1 = c − e−2x 2 2 0 c = 4  ∞  e−2x dx  0  so c = 4. The m.g.f. is then ∞  MX (t) =  ∞  tx  e fX (x)dx = 4 0  xe(t−2)x dx  0  4 4 [xe(t−2)x ]∞ 0 − t−2 t−2 4 provided t < 2 = (t − 2)2  ∞  =  0  c Pearson Education Limited 2011   e(t−2)x dx  660  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Hence    8  =− =1 E(X) = 3 (t − 2) t=0   24 3  = E(X2 ) = MX (0) =  4 (t − 2) t=0 2 1 Var(X) = E(X2 ) − [E(X)]2 = 2 MX (0)  44  If X1 , . . . , Xn are independent Poisson random variables with parameters  λ1 , . . . , λn and if Y = X1 + . . . + Xn then MY (t) = MX1 (t) . . . MXn (t) = exp[λ1 (et − 1)] . . . exp[λn (et − 1)] = exp[(λ1 + . . . + λn )(et − 1)] Hence, Y is another Poisson random variable with parameter λ = λ1 + . . . + λn . 45  Numbers of breakdowns in one hour are separately binomial with parameters  n1 = 30, p1 = 0.01 and n2 = 40, p2 = 0.005 respectively, and hence approximately Poisson with parameters λ1 = 0.3 and λ2 = 0.2 respectively. The total number of breakdowns per hour is therefore also approximately Poisson with λ = λ1 + λ2 = 0.5, and  46   λ2  = 0.014 P (three or more)  1 − e−λ 1 + λ + 2!  Let the proportion defective be p. By the Poisson approximation, P (k defective in 100)   λk e−λ k!  where λ = 100p. The requirement is P (k ≤ 1)  e−λ (1 + λ) > 0.9 from which λ < 0.531 (solving this as a nonlinear equation) and hence p <  0.531 100  0.0053. Therefore, at least 99.47% of servomechanisms must be satisfactory.  c Pearson Education Limited 2011   =  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 47  661  With 2 1 fZ (z) = √ e−z /2 , −∞ < z < ∞ 2π  the m.g.f. is 1 MZ (t) = √ 2π 1 =√ 2π 2  = et  /2  ∞ −∞ ∞ −∞ ∞ −∞  etz e−z 1  e− 2 (z  2  2  /2  dz  −2tz+t2 ) t2 /2  e  dz  1 2 1 √ e− 2 (z−t) dz 2π  The integrand is the p.d.f. of a normal random variable with mean t and standard deviation equalling one, hence 2  MZ (t) = et  /2  Exercises 11.9.7 48  From the table (Figure 11.29), with n = 50, p = 0.1 so that np = 5, we read  oﬀ the Shewhart warning limit as 9.5 and the action limit as 13.5. For the given data, the warning limit is exceeded at samples 3, 9 and 11, and the action limit is exceeded at sample 12. The upper control limit is given by  UCL = np + 3 np(1 − p) = 11.4 and this is ﬁrst exceeded at sample 9.  49  From Figure 11.29, with n = 100, p = 0.02 so that np = 2, we have Shewhart  warning and action limits 5.5 and 7.5 respectively. The warning limit is exceeded three times (samples 20, 22 and 25) before the action limit is exceeded at sample 28. The upper control limit is given by  UCL = np + 3 np(1 − p) = 6.2 and this is ﬁrst exceeded at sample 25.  c Pearson Education Limited 2011   662  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  50  Using σ = 3 and n = 10, the Shewhart warning and action limits are √ cW = 1.96σ/ n = 1.86 √ cA = 3.09σ/ n = 2.93  (above and below the mean). Relative to the mean, the warning limit is exceeded at samples 3 and 9, and the action limit at sample 12.  51  Using σ = 3 and n = 10, the Shewhart warning and action limits above the  design mean μ = 12 are  √ μ + 1.96σ/ n = 13.86 √ μ + 3.09σ/ n = 14.93  The warning limit is exceeded several times (at samples 6, 12, 15, 17 and 18) before the action limit is crossed at sample 19.  52(a)  Using σ = 3 and n = 10, the cusum parameters are σ r = √ = 0.474 (relative to the mean) 2 n σ h = 5 √ = 4.74 n  The chart is built up in the following table:  value cusum  -0.2 1.3 2.1 0 0.83 2.45  0.3 -0.8 2.28 1.00  1.7 1.3 0.6 2.5 1.4 1.6 3.0 2.23 3.05 3.18 5.21 6.13 7.26 9.78  The decision interval (h) is exceeded at sample 9.  52(b)  Using r = 0.3 and σ, n as above, the GMA action limits are  r σ √ = ±1.23 ±3.09 2−r n  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  663  (relative to the mean). The chart is built up in the following table:  value GMA  -0.2 1.3 2.1 0.3 -0.8 1.7 1.3 0.6 2.5 1.4 1.6 3.0 -0.06 0.35 0.87 0.70 0.25 0.69 0.87 0.79 1.30 1.33 1.41 1.89  The action limit is exceeded at sample 9. 53  For the cusum chart we have μ = 12, σ = 3 and n = 10, so σ r = μ + √ = 12.47 2 n σ h = 5 √ = 4.74 n  For the GMA chart with r = 0.3 we have action limits at  r σ √ = 10.77 and 13.23 μ ± 3.09 2−r n The control charts are built up in the following table: X̄m cusum GMA X̄m cusum GMA  12.8 11.2 13.4 12.1 13.6 13.9 12.3 12.9 13.8 13.1 0.33 0 0.93 0.55 1.68 3.10 2.93 3.35 4.68 5.31 12.24 11.93 12.37 12.29 12.68 13.05 12.82 12.85 13.13 13.12 12.9 14.0 13.7 13.4 14.2 13.1 14.0 14.0 15.1 14.3 5.73 7.26 8.48 9.41 11.13 11.76 13.28 14.81 17.44 19.26 13.06 13.34 13.45 13.43 13.66 13.49 13.65 13.75 14.16 14.20  The cusum chart indicates action at sample 10, the GMA chart at sample 12. 54  Using n = 50, p = 0.1 so that np = 5, the cusum parameters from Figure  11.31 are r = 7, h = 8.5 (nearest values).  count cusum  5 8 11 5 6 4 9 7 12 9 10 0 1 5 3 2 0 2 2 7 9 12  14 19  The chart indicates action at sample 10.  c Pearson Education Limited 2011   664  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  55  Using n = 100, p = 0.02 so that np = 2, the cusum parameters from Figure  11.31 are r = 3, h = 5.5 (nearest values). count cusum count cusum  3 0 3 6  3 0 4 7  5 3 5 0 3 1 3 5 4 2 2 4 1 1 0 0 2 3 5 6 5 6 4 4 7 5 4 9 12 14 17 18 19 23 25 26  2 4 3 5 4 2 3 3 5 6 8 5 6 6 7 31 33 36 39 43  The chart indicates action at sample 16. 56 For the Shewhart chart we have n = 12, σ = 1, and hence warning and action limits given by √ cW = 1.96σ/ n = 0.57 √ cA = 3.09σ/ n = 0.89 For the cusum chart we have σ r = √ = 0.144 (relative to the mean) 2 n σ h = 5 √ = 1.443 n For the GMA chart with r = 0.2, the action limit is  r σ √ = 0.297 3.09 2−r n X̄m cusum GMA X̄m cusum GMA X̄m cusum GMA  0.1 0 .020 0.1 .623 .197 0.5 3.535 .379  0.3 .156 .076 0.4 .878 .238 0.7 4.091 .443  -0.2 0 .021 0.6 1.334 .310 0.3 4.246 .415  0.4 .256 .097 0.3 1.490 .308 0.1 4.202 .352  0.1 .211 .097 0.4 1.745 .326 0.6 4.658 .401  0 .067 .078 0.3 1.901 .321 0.5 5.013 .421  0.2 .123 .102 0.6 2.357 .377 0.6 5.469 .457  -0.1 0 .062 0.5 2.712 .402 0.7 6.025 .505  0.2 .056 .089 0.4 2.968 .401 0.4 6.280 .484  0.4 0.5 .311 .667 .152 .221 0.2 0.3 3.024 3.179 .361 .349 0.5 6.636 .488  For the Shewhart chart there are several warnings but no action indicated. For the cusum and GMA charts, action is indicated at samples 15 and 14 respectively.  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 57  665  The GMA Sm is deﬁned recursively by S0 = μX Sm = rX̄m + (1 − r)Sm−1 for m ≥ 1  Substituting for Sm−1 , then Sm−2 and so on gives Sm = rX̄m + (1 − r)[rX̄m−1 + (1 − r)Sm−2 ] = r[X̄m + (1 − r)X̄m−1 ] + (1 − r)2 [rXm−2 + (1 − r)Sm−3 ] and eventually Sm = r  m−1   (1 − r)i X̄m−i + (1 − r)m μX  i=0  But E(X̄m−i ) = μX and Var(X̄m−i ) = Using the result m−1   xi =  i=0  we have E(Sm ) = rμX  σ2X for all i, m . n  1 − xm for | x |< 1 1−x  m−1   (1 − r)i + (1 − r)m μX  i=0  1 − (1 − r)m + (1 − r)m μX = rμX r = μX m−1 σ2X 2  σ2 r2 1 − (1 − r)2m Var(Sm ) = [(1 − r)2 ]i = X r n n 1 − (1 − r)2 i=0  σ2X r [1 − (1 − r)2m ] n 2−r 2  r  σX as m → ∞ −→ 2−r n  =  58  Let Xm = count of defectives for sample m , n = sample size, p = probability  of defective. Deﬁne S0 = np Sm = rXm + (1 − r)Sm−1 for m ≥ 1 c Pearson Education Limited 2011   666  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Substituting for Sm−1 , Sm−2 and so on (as in the previous exercise) leads to Sm = r  m−1   (1 − r)i Xm−i + (1 − r)m np  i=0  From the mean and variance of the binomial distribution, E(Xm−i ) = np Var(Xm−i ) = np(1 − p) hence E(Sm ) = rnp  m−1   (1 − r)i + (1 − r)m np  i=0  = np[1 − (1 − r)m + (1 − r)m ] = np Var(Sm ) = np(1 − p)r  2  m−1   [(1 − r)2 ]i  i=0   r  [1 − (1 − r)2m ]np(1 − p) = 2−r  r  np(1 − p) as m → ∞ −→ 2−r The upper control limit for n = 50, p = 0.05, r = 0.2 is   r  np(1 − p) = 4.04 UCL = np + 3 2−r  Xm Sm Xm Sm  3 2.6 7 4.36  5 3.08 4 4.29  2 2.86 5 4.43  2 2.69 5 4.55  1 2.35 8 5.24  6 3.08 6 5.39  4 3.27 5 5.31  4 3.41 9 6.05  2 3.13 7 6.24  6 3.70 8 6.59  The chart indicates action after 11 samples. 59 For the Shewhart chart we have n = 10, μ = 6, σ = 0.2 and hence warning and action limits given by √ cW = μ ± 1.96σ/ n = 5.88 and 6.12 √ cA = μ ± 3.09σ/ n = 5.80 and 6.20 c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  667  For the cusum chart we have σ r = μ + √ = 6.032 2 n σ h = 5 √ = 0.316 n For the GMA chart with r = 0.2, the action limits are  r σ √ = 5.935 and 6.065 μ ± 3.09 2−r n  X̄m cusum GMA X̄m cusum GMA X̄m cusum GMA  6.04 0.01 6.01 6.17 0.28 6.07 6.10 0.65 6.06  6.12 0.10 6.03 6.03 0.28 6.06 6.01 0.62 6.05  5.99 0.06 6.02 6.13 0.38 6.07 5.96 0.55 6.03  6.02 0.04 6.02 6.05 0.40 6.07 6.12 0.64 6.05  6.04 0.05 6.03 6.17 0.54 6.09 6.02 0.63 6.05  6.11 0.13 6.04 5.97 0.47 6.07 6.20 0.80 6.08  5.97 0.07 6.03 6.07 0.51 6.07 6.11 0.88 6.08  6.06 0.10 6.03 6.14 0.62 6.08 5.98 0.82 6.06  6.05 0.12 6.04 6.03 0.62 6.07 6.02 0.81 6.05  6.06 0.14 6.04 5.99 0.58 6.05 6.12 0.90 6.07  The Shewhart chart indicates action after 26 samples, the cusum chart after 13 samples and the GMA chart after 11 samples.  Exercises 11.10.6 60  If gales occur at rate  15 12  = 1.25 per month, and occur independently, then  the number of gales in any one month has a Poisson distribution, so P (more than two in one month) = 1 − P(0) − P(1) − P(2)   λ2 T2 −λT 1 + λT + =1−e 2! = 0.132 (with λ = 1.25, T = 1).  c Pearson Education Limited 2011   668  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  61  If λ = 30 calls per hour (on average) arrive at the switchboard, and do so  independently, then the number of calls has a Poisson distribution. Hence P (no calls in T = 3 min) = e−λT /60 = 0.223 P (more than ﬁve calls in T = 5 min) −λT /60  =1−e    (λT)5 λT + ... + 5 1+ 60 60 5!    = 0.042  62  The steady-state distribution for the number (N) in the system is P(n in system) = pn = (1 − ρ)ρn , n ≥ 0  Now  ∞ ∞ d  n  n−1 d (1 − ρ)−1 = (1 − ρ)−2 ρ = nρ = dρ n=0 dρ n=0  Hence, the mean number in the system is NS =  ∞   n(1 − ρ)ρ = ρ(1 − ρ) n  n=0  ∞   nρn−1  n=0  ρ(1 − ρ) ρ = = (1 − ρ)2 1−ρ If there are n > 0 customers in the queue then there are n + 1 in the system, so P(n > 0 in queue) = (1 − ρ)ρn+1 , n ≥ 1 (we do not need the probability for n = 0). Mean number in queue is NQ = (1 − ρ)  ∞   nρ  n+1  = ρ (1 − ρ)  n=1  ∞  n=0  ρ (1 − ρ) ρ = 2 (1 − ρ) 1−ρ 2  =  2  2  c Pearson Education Limited 2011   nρn−1  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 63  For a single-channel queue with λ = 3 arrivals per hour and μ = 4 patients  treated per hour, 63(a)  P(0 in system) = 1 −  63(b)  NQ =  1 λ = μ 4  (λ/μ)2 9 = 1 − λ/μ 4  63(c) P(> 3 in queue) = P(> 4 in system) = 1 − P(0) − P(1) − P(2) − P(3) − P(4)  λ  λ 2  λ 3  λ 4  λ  1+ + + + =1− 1− μ μ μ μ μ = 0.237  3 λ/μ = hour. μ−λ 4  63(d)  WQ =  63(e)  P (wait more than one hour) =  64  669  λ −(μ−λ) = 0.276 e μ  Mean number of aircraft on ground is NS =  λ μ−λ  so, total mean cost per hour (waiting time plus servicing) is E [total cost per hour] =  c1 λ + c2 μ μ−λ  c Pearson Education Limited 2011   670  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  Minimizing this with respect to μ: d c1 λ E [total cost per hour] = − + c2 = 0 dμ (μ − λ)2 from which (μ − λ)2 = so μ = λ +  65    c1 λ c2  c1 λ/c2  Breakdown rate is λ = 3 per hour, and machine idle time is costed at £ 60  per hour per machine. For option A, service rate is μ = 4 per hour at £ 20 per hour, so mean hourly cost is 60NS + 20 =  60λ + 20 = 200 μ−λ  For option B, service rate is μ = 5 at £ 40 per hour, so mean hourly cost is 60NS + 40 =  60λ + 40 = 130 μ−λ  Option B is preferred.  66  Ship arrival rate is λ =  1 3  per hour, and service rate per berth is μ =  1 12  , so  ρ = λ/μ = 4. Mean waiting time in the queue is  −1   c−1 n ρ ρc+1 ρc 1 + WQ = λ (c − 1)!(c − ρ)2 n=0 n! (c − 1)!(c − ρ) where c is the number of berths. For c = 5 berths we ﬁnd WQ = 6.65 hours, which exceeds the required minimum, so c = 6 berths are needed (WQ = 1.71) .  67  Arrival rate is λ = 2 per minute, and basic service rate per cashier is μ =  5 4  per minute. If this service rate is doubled (by providing a packer) then mean queueing time is 4/5 ρ = = 1.6 min WQ = μ−λ 5/2 − 2 c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  671  Alternatively, if an additional cash desk is provided then (using WQ as in the previous exercise, and ρ = 8/5)  −1  ρ3 ρ2 1 1+ρ+ = 0.076 min WQ = λ (2 − ρ)2 2−ρ Clearly, a second cash desk is preferable.  Exercises 11.11.3 68  We have P(agent) = P(agent|option 1)P(option 1) + P(agent|option 2)P(option 2) + P(agent|option 3)P(option 3) = 0.28 × 0.45 + 0.41 × 0.32 + 0.16 × 0.23 = 0.294  69  Total probability of explosion is P(E) = P(E | (a))P((a)) + P(E | (b))P((b)) + P(E | (c))P((c)) + P(E | (d))P((d)) = 0.25 × 0.2 + 0.2 × 0.4 + 0.4 × 0.25 + 0.75 × 0.15 = 0.3425  Hence, by Bayes' Theorem, P((a) | E) = P(E | (a))P((a))/P(E) = 0.146 P((b) | E) = 0.234 P((c) | E) = 0.292 P((d) | E) = 0.328 and sabotage is therefore the most likely cause of the explosion.  c Pearson Education Limited 2011   672  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  70  If two bullets hit the target then they could be ﬁred by each possible pair of  marksmen, so P (two hits) = P(A ∩ B ∩ C̄) + P(A ∩ B̄ ∩ C) + P(Ā ∩ B ∩ C) (where A denotes 'bullet from A hits target', etc) = 0.6 × 0.5 × 0.6 + 0.6 × 0.5 × 0.4 + 0.4 × 0.5 × 0.4 = 0.38 Hence P(C | two hits) =  P(C ∩ two hits) P (two hits)  P(A ∩ B̄ ∩ C) + P(Ā ∩ B ∩ C) P ( two hits) 0.6 × 0.5 × 0.4 + 0.4 × 0.5 × 0.4 = 0.38 = 0.526 =  Thus, it is more probable than not that C hit the target.  71  Prior probabilities are P(A) = 13 , P(B) =  2 3  . Also P( Smith | A) = 0.1 and  P( Smith | B) = 0.05. Hence P (Smith | A)P(A) P (Smith | A)P(A) + P (Smith | B)P(B) 0.1 × 13 = 0.1 × 13 + 0.05 × 23 1 = 2  P(A | Smith) =  72  Let D denote 'has disease' and + denote 'positive diagnosis', so that P(D) =  0.08, P(+ | D) = 0.95 and P(+ | D̄) = 0.02 72(a)  P(+) = P(+ | D)P(D) + P(+ | D̄)P(D̄) = 0.95 × 0.08 + 0.02 × 0.92 = 0.0944  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  72(b)  73  P(D | +) =  673  0.95 × 0.08 P(+ | D)P(D) = = 0.81 P(+) 0.0944  Let G denote 'good stock', B = Ḡ denote 'bad stock', g denote 'stockbroker says good', b denote 'stockbroker says bad',  so that P(G) = 0.5, P(g | G) = 0.6, P(b | B) = 0.8. 73(a) P(g | G)P(G) P(g | G)P(G) + P(g | B)P(B) 0.6 × 0.5 3 = = 0.6 × 0.5 + 0.2 × 0.5 4  P(G | g) =  73(b)  Let E denote 'k out of n stockbrokers say good'. Since the stockbrokers  are independent, by the binomial distribution   n P(E | G) = [P(g | G)]k [P(b | G)]n−k k   n P(E | B) = [P(g | B)]k [P(b | B)]n−k k Hence P(E | G)P(G) P(E | G)P(G) + P(E | B)P(B) n  k G)]n−k P(G) k [P(g | G)] [P(b|  = n k n−k P(G) + n [P(g | B)]k [P(b | B)]n−k P(B) k [P(g | G)] [P(b | G)] k  P(G | E) =  0.6k × 0.4n−k × 0.5 0.6k × 0.4n−k × 0.5 + 0.2k × 0.8n−k × 0.5 −1   1 k n−k = 1+ 2 3  =  c Pearson Education Limited 2011   674  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  74  Given that the probability of correct reception of a letter is 0.6, and the error  probabilities are 0.2 for the two alternatives, we have P(ABCA received | AAAA transmitted) = 0.6 × 0.2 × 0.2 × 0.6 = 0.0144 P(ABCA received | BBBB transmitted) = 0.2 × 0.6 × 0.2 × 0.2 = 0.0048 P(ABCA received | CCCC transmitted) = 0.2 × 0.2 × 0.6 × 0.2 = 0.0048 Also P(AAAA transmitted) = 0.3 etc. Hence P(ABCA received) = 0.0144 × 0.3 + 0.0048 × (0.4 + 0.3) = 0.00768 and 0.0144 × 0.3 = 0.5625 0.00768 P(BBBB transmitted | ABCA received) = 0.25 P(AAAA transmitted | ABCA received) =  P(CCCC transmitted | ABCA received) = 0.1875  75  Average number of accidents per day =  1×12+2×4 100  =  1 5  First hypothesis (H1 ) is for a Poisson distribution, so set λ = pi = P(i accidents in one day) =  1 5  and probabilities  λi e−λ i!  Hence, p0 = 0.8187, p1 = 0.1637, p2 = 0.0164. Second hypothesis (H2 ) is for a binomial distribution with n = 3, so set np =  1 1 (hence p = ) and probabilities 5 15   3 i p (1 − p)3−i qi = P(i accidents in one day) = i  Hence, q0 = 0.8130, q1 = 0.1742, q2 = 0.0124. If E denotes the evidence then the odds are updated by ln  P(E | H1 ) P(H1 ) P(H1 | E) = ln + ln P(H2 | E) P(E | H2 ) P(H2 ) c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  675  where 12 4 P(E | H1 ) = p84 0 p1 p2 12 4 P(E | H2 ) = q84 0 q1 q2 1 P(H1 )/P(H2 ) = (initial odds) 2  Hence ln  p0 p1 p2 1 P(H1 | E) = 84 ln + 12 ln + 4 ln + ln = 0.247 P(H2 | E) q0 q1 q2 2  and the updated odds are therefore 1.28 to 1 in favour of the Poisson distribution.  76  If the probabilities of the evidence (E) under H1 and H2 are n! pn1 . . . pnk k n1 ! . . . nk ! 1 n! P(E | H2 ) = qn1 1 . . . qnk k n1 ! . . . nk !  P(E | H1 ) =  then the log-likelihood ratio becomes    k  p1 n1  pk nk P(E | H1 ) pi = ln ln = ··· ni ln P(E | H2 ) q1 qk qi i=1  77  Under hypothesis H1 we have p1 = 0.92, p2 = 0.05, p3 = 0.02, p4 = 0.01  and under H2 q1 = 1 − 0.05 − q3 − q4 , q2 = 0.05, q3 , q4 unknown (where q3 = pB and q4 = PAB ). The likelihood of the evidence E under H2 is P(E | H2 ) =  n! (0.95 − q3 − q4 )n1 0.05n2 qn3 3 qn4 4 n1 ! . . . n4 !  c Pearson Education Limited 2011   676  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  where n1 = 912, n2 = 45, n3 = 27, n4 = 16. Thus ln P(E | H2 ) = n1 ln(0.95 − q3 − q4 ) + n3 ln q3 + n4 ln q4 + constant ∂ ln P(E | H2 ) n1 n3 n3 (0.95 − q3 − q4 ) − n1 q3 =− + = ∂q3 0.95 − q3 − q4 q3 (0.95 − q3 − q4 )q3 = 0 if (n1 + n3 )q3 + n3 q4 = 0.95n3 ∂ ln P(E | H2 ) n1 n4 n4 (0.95 − q3 − q4 ) − n1 q4 =− + = ∂q4 0.95 − q3 − q4 q4 (0.95 − q3 − q4 )q4 = 0 if n4 q3 + (n1 + n4 )q4 = 0.95n4 From the simultaneous equations 939q3 + 27q4 = 25.65 16q3 + 928q4 = 15.2 we ﬁnd q3 = 0.0269, q4 = 0.0159 and therefore q1 = 0.9072. It follows that (using the result of the previous exercise) pi P(E | H1 )  = ni ln ln P(E | H2 ) qi i=1 4  0.92 0.05 0.02 0.01 + 45 ln + 27 ln + 16 ln 0.9072 0.05 0.0269 0.0159 = −2.645  = 912 ln  If initial odds are P(H1 )/P(H2 ) = 5 then updated odds are P(H1 | E) = e−2.645 × 5 = 0.355 P(H2 | E) that is, 2.8 to 1 in favour of H2 . 78  Under hypothesis H1 we have separate estimates as follows: 24 = 4.0 6 36 λB = mean number of defects for B = = 7.2 5 λA = mean number of defects for A =  Under hypothesis H2 we have a single estimate as follows: λ = overall mean number of defects =  60 = 5.455 11  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  677  The evidence E can be expressed as follows: E = {A : 2 with 3 defects, 2 with 4 defects, 2 with 5 defects; B : 1 with 5 defects, 1 with 6 defects, 2 with 8 defects, 1 with 9 defects} The log-likelihood ratio is ln  λ3 e−λA λ4 e−λA λ5 e−λA P(E | H1 ) = 2 ln A3 −λ + 2 ln A4 −λ + 2 ln A5 −λ P(E | H2 ) λ e λ e λ e + ln  λ5B e−λB λ6B e−λB λ8B e−λB λ9B e−λB + ln + 2 ln + ln λ5 e−λ λ6 e−λ λ8 e−λ λ9 e−λ λB λA + (5 + 6 + 16 + 9)ln λ λ + (2 + 2 + 2)(λ − λA ) + (1 + 1 + 2 + 1)(λ − λB )  = (6 + 8 + 10)ln  = 2.551 and the updated odds (with no initial preference) are P(H1 | E) = e2.551 P(H2 | E) or 12.8 to 1 in favour of H1 .  Review Exercises 11.12 1  For the standard corks, sample proportion oxidized is p̂1 =  for the plastic bungs, sample proportion oxidized is p̂2 = proportion oxidized is p̂ =  9 96  = 0.0938. Test statistic  z=   p̂1 − p̂2 1 1 p̂(1 − p̂)( 60 ) + ( 36 )  = 0.271  The hypothesis is accepted. 2  The model is d = d0 e−λt  or equivalently ln d = ln d0 − λt c Pearson Education Limited 2011   3 36  6 60  = 0.1, whereas  = 0.0833. Overall  678  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  which is of the form Y = a + bX From the data, X̄ = 5.746, SX = 3.036, Ȳ = −0.2811, SY = 0.721, XY = −3.775 so b̂ =  XY − X̄Ȳ = −0.234 S2X  â = Ȳ − b̂X̄ = 1.065 From these we infer λ̂ = −b̂ = 0.234 d̂0 = eâ = 2.90 Also the error variance is (using unrounded results) S2E = S2Y − b̂2 S2X = 0.01418 and the 95% conﬁdence interval for λ is λ̂ ± t.025,8  3  SE √ = (0.202, 0.266) SX 8  If position P and load X are related by P = a + bX  then by linear regression on the data we ﬁnd â = 6.129 b̂ = 0.0624 But extension Y and load X are related by b̂ =  L Y = X ÊA  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  679  where L = 101.4 and A = 1.62 × 10−5 . The estimate of Young's modulus is therefore Ê =  L  = 1.003 × 108  b̂A Also from the linear regression the error standard deviation is SE = 0.00624, so the 95% conﬁdence interval for b is b̂ ± t.025,6  SE √ = (0.0597, 0.0651) SX 6  We infer the 95% conﬁdence interval for E = L/bA as (96.1 × 106 , 104.9 × 106 )  4 From the data, the mean time between arrivals is estimated as 9.422 hours, and this is 1/λ for the exponential distribution. If we form a histogram of the data, the expected probability of a class (a, b) under the exponential distribution with parameter λ is P(a < X < b) = FX (b) − FX (a) = 1 − e−λb − (1 − e−λa ) = e−λa − e−λb Using class intervals of ﬁve hours we obtain the table as follows: Class (k) 0-5 5-10 10-15 15-20 20-25 25-30 > 30  Observations (fk ) 48 22 13 12 3 3 4  Expected (ek ) 43.24 25.43 14.96 8.80 5.18 3.04 4.35  Probability 0.4118 0.2422 0.1425 0.0838 0.0493 0.0290 0.0414  The value of χ2 = 3.35 is less than χ2.05,5 = 11.07 (seven classes with one parameter estimated) so the ﬁt to the exponential distribution is good. 5  The maximum value is Xmax = 72 and the total is y=  Xmax = 0.0728 Σ i Xi  c Pearson Education Limited 2011    i  Xi = 989.3, so  680  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  and [1/y] = 13. Hence P(Y ≤ 0.0728) =  13    (−1)  k=0  k   105 (1 − 0.0728k)104 k  = 0.9599 The probability of there occurring such a large value is therefore around 4% , so the value 72 can be regarded as an outlier at the 5% signiﬁcance level. With the outlier included we have X̄ = 9.422, SX = 10.77 so the 95% conﬁdence interval for mean inter-arrival time is SX = (7.36, 11.48) X̄ ± 1.96 √ 105 With the outlier excluded we have X̄ = 8.820, SX = 8.90 so the conﬁdence interval is (7.11, 10.53).  6  The contingency table (with expected values in brackets and adjusted residuals  underneath) is as follows: Grade Very satisﬁed Fairly satisﬁed Neutral Fairly dissatisﬁed Very dissatisﬁed Total  French 16 (15) 0.5 63 (50) 2.8 40 (42) −0.5 10 (18) −2.5 3 (7) −1.8 132  German 6 (7) −0.5 13 (24) −3.2 27 (20) 1.9 13 (9) 1.6 5 (3) 64  Spanish 22 (22) −0.1 76 (77) −0.2 60 (64) −1.0 32 (28) 1.2 12 (10) 0.8 202  Total 44 152 127 55 20 398  Chi-square = 20.0, d.f. = (5 − 1)(3 − 1) = 8, and so compare with χ20.025,8 = 17.54: signiﬁcant at 2.5% level. The French course scores highest, followed by Spanish and then German.  c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 7  681  Let D denote 'has disease' and O denote 'operation performed', then 1 2 1 P(survive | D ∩ Ō) = 20 4 P(survive | D̄ ∩ O) = 5  P(survive | D ∩ O) =  and we can assume that P(survive | D̄ ∩ Ō) = 1. If the operation is performed, then using the hint we have P(survive | O) = P(survive | D ∩ O)P(D) + P(survive | D̄ ∩ O)P(D̄) 4 4 3 1 = p + (1 − p) = − p 2 5 5 10 (where p = P(D) ). If the operation is not performed, P(survive | Ō) = P(survive | D ∩ Ō)P(D) + P(survive | D̄ ∩ Ō)P(D̄) 19 1 p + (1 − p) = 1 − p = 20 20 These probabilities are equal when 3 19 4 − p=1− p 5 10 20 from which p =  4 13  . The surgeon will operate if the assessment of P(D) exceeds  this value.  8  With 200 machines each becoming misaligned every 200 hours on average, the  rate at which machines become misaligned is λ0 =  200 200  = one per hour on average.  The total cost per hour for each option is the sum of three components: the ﬁxed cost per hour, the cost of correcting the output and the cost of lost production. For option A, the ﬁxed cost is £1 per hour per machine, hence £200 per hour. The average run length ARL0 for a misaligned machine is 20 hours, and this amount of output must be corrected, so the cost per hour of correcting the output is λ0 × ARL0 × 10 = £200. Lost production occurs while a machine is in the queue and being serviced, and this occurs whether the machine is actually misaligned or c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  682  not (false alarm). Actual misalignments occur at the rate λ0 and are detected by the control chart. False alarms occur at the rate λ1 =  200 = 0.2 per hour ARL1  where ARL1 = 1000 is the mean time between false alarms for a well-adjusted machine. These two kinds of action are independent, so the total rate of actions is λ = λ0 + λ1 = 1.2. Also, the service rate μ = 2 per hour so ρ = λ/μ = 0.6. With σs =  1 4  , the mean number of machines out of production is  NS = ρ +  (λσs )2 + ρ2 = 1.163 2(1 − ρ)  and the cost per hour of lost production is 200NS = £ 232.5. The total cost for option A is therefore £ 200 + £ 200 + £ 232.5 = £ 632.5 per hour. For option B, the ﬁxed cost is £ 1.50 per hour per machine, hence £ 300 per hour. With ARL0 = 4, the cost per hour of correcting the output is λ0 × ARL0 × 10 = £ 40. With ARL1 = 750, false alarms occur at the rate λ1 = 200/750 = 0.267 per hour, so machines are taken out of production at the total rate λ = λ0 +λ1 = 1.267, hence ρ = λ/μ = 0.633. The mean number of machines out of production is therefore NS = 1.317 at a cost per hour 200NS = £ 263.4. The total cost for option B is therefore £ 300 + £ 40 + £ 263.4 = £ 603.4 per hour. This is less than for option A.  9  For the source, P(in = 0) = α and P(in = 1) = 1 − α. For the channel,  P(out = 0 | in = 1) = P(out = 1 | in = 0) = p.  9(a) P(out = 0) = P(out = 0 | in = 0)P(in = 0) + P(out = 0 | in = 1)P(in = 1) = (1 − p)α + p(1 − α) = p̄α + pᾱ c Pearson Education Limited 2011   Glyn James, Advanced Modern Engineering Mathematics, 4th Edition  683  (where p̄ = 1 − p, ᾱ = 1 − α). Hence p̄α P(out = 0 | in = 0)P(in = 0) = p̄α + pᾱ P(out = 0) pᾱ P(in = 1 | out = 0) = p̄α + pᾱ pα P(in = 0 | out = 1) = pα + p̄ᾱ p̄ᾱ P(in = 1 | out = 1) = pα + p̄ᾱ P(in = 0 | out = 0) =  9(b)  P( in = 0 | out = 0) > P( in = 1 | out = 0) if p̄α > pᾱ  from which p̄α > p(1 − α) hence (p̄ + p)α = α > p Similarly, P( in = 1 | out = 1) > P( in = 0 | out = 1) if p̄ᾱ > pα from which p̄(1 − α) > pα hence (p + p̄)α = α < p̄ The source symbol is assumed to be the same as the received symbol if p < α < p̄. 10  For the binary symmetric channel, X = {0, 1} with P(X = 0) = α, and  Y = {0, 1} with P(Y = 0) = p̄α + pᾱ, P(Y = 1) = pα + p̄ᾱ(p̄ = 1 − p, ᾱ = 1 − α, using the results of the previous exercise). Also P(X = 0 ∩ Y = 0) = P(Y = 0 | X = 0)P(X = 0) = p̄α P(X = 0 ∩ Y = 1) = P(Y = 1 | X = 0)P(X = 0) = pα c Pearson Education Limited 2011   684  Glyn James, Advanced Modern Engineering Mathematics, 4th Edition P(X = 1 ∩ Y = 0) = P(Y = 0 | X = 1)P(X = 1) = pᾱ P(X = 1 ∩ Y = 1) = P(Y = 1 | X = 1)P(X = 1) = p̄ᾱ  The mutual information between X and Y is as follows: I(X; Y) =  1 1    P(x, y) log2  x=0 y=0  P(x, y) P(X = x)P(Y = y)  p̄α pα + pα log2 α(p̄α + pᾱ) α(pα + p̄ᾱ) p̄ᾱ pᾱ + p̄ᾱ log2 + pᾱ log2 ᾱ(p̄α + pᾱ) ᾱ(pα + p̄ᾱ) = p̄α log2  = p̄(α + ᾱ) log2 p̄ + p(α + ᾱ) log2 p − (p̄α + pᾱ) log(p̄α + pᾱ) − (pα + p̄ᾱ) log(pα + p̄ᾱ) = H(p) − H(p̄α + pᾱ) where H(t) = t log2 t + (1 − t) log2 (1 − t) is called the 'entropy function'. In particular, when α =  1 2  we have p̄α + pᾱ =  and H( 21 ) =  1 2  log2  1 2  +  1 2  log2  1 2  1 1 (p̄ + p) = 2 2  = −1  so that I(X; Y) = 1 + H(p) = 1 + p log2 p + (1 − p) log2 (1 − p) When p = 12 , I(X; Y) = 1 +  1 2  log2  1 2  +  1 2  log2  1 2  = 0.  When p → 0, p log2 p → 0 and p̄ log2 p̄ → 0 so that I(X; Y) → 1 and similarly when p → 1. Full information is transmitted through the channel when either every bit is correct (p = 0) or every bit is inverted (p = 1) . No information is transmitted when the bits are uniformly randomized (p = 12 ) .  c Pearson Education Limited 2011

Source Exif Data:

File Type                       : PDF File Type Extension             : pdf MIME Type                       : application/pdf PDF Version                     : 1.7 Linearized                      : Yes Author                          : dynstab2/ThePirateBay Create Date                     : 2011:08:26 12:38:18+02:00 Keywords                        : - Modify Date                     : 2011:09:19 06:21:14-04:00 Has XFA                         : No XMP Toolkit                     : Adobe XMP Core 5.2-c001 63.139439, 2010/09/27-13:37:26 Metadata Date                   : 2011:09:19 06:21:14-04:00 Creator Tool                    : - Format                          : application/pdf Title                           : - Creator                         : dynstab2/ThePirateBay Description                     : - Subject                         : - Document ID                     : uuid:806f6716-8e1d-4388-91ec-a3027283f743 Instance ID                     : uuid:2185403b-0f1e-427a-9c39-6987b86285d4 Producer                        : - Page Layout                     : SinglePage Page Mode                       : UseOutlines Page Count                      : 688

EXIF Metadata provided by EXIF.tools

Modern Engineering Mathematics 4th Edition Glyn James Pdf

Source: https://usermanual.wiki/Document/Advanced20Modern20Engineering20Mathematics2020Glyn20James20Solutions20Manual20204th20edition.1008768827/help

Posted by: woodbeetch.blogspot.com

Modern Engineering Mathematics 4th Edition Glyn James Pdf

Advanced Modern Engineering Mathematics Glyn James Solutions Manual 4th Edition

Modern Engineering Mathematics 4th Edition Glyn James Pdf

0 Response to "Modern Engineering Mathematics 4th Edition Glyn James Pdf"

Post a Comment

Iklan Atas Artikel

Iklan Tengah Artikel 1

Iklan Tengah Artikel 2

Iklan Bawah Artikel