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SOLUTIONS MANUAL FOR
R . L . H E R M A N  V E R S I O N D AT E : N O V E M B E R 1 4 , 2 0 1 3
SOLUTIONS MANUAL FOR A COURSE IN MATHEMATICAL METHODS FOR PHYSICISTS by
Russell L. Herman
Boca Raton London New York
CRC Press is an imprint of the Taylor & Francis Group, an informa business
CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 334872742 © 2014 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Printed on acidfree paper Version Date: 20140114 International Standard Book Number13: 9781466584709 (Ancillary) This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 9787508400. CCC is a notforprofit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com
ii
Contents Preface 1 Introduction and Review
iii 1
2 Free Fall and Harmonic Oscillators
25
3 Linear Algebra
69
4 Nonlinear Dynamics
111
5
141
The Harmonics of Vibrating Strings
6 Nonsinusoidal Harmonics
173
7 Complex Representations of Functions
219
8 Transform Techniques in Physics
245
9 Vector Analysis and EM Waves
291
10 Extrema and Variational Calculus
327
11 Problems in Higher Dimensions
357
A Review of Sequences and Infinite Series
379
B Quick Answers Ch.1 Introduction . Ch.2 Free Fall and Harmonic Oscillators Ch.3 Linear Algebra Ch.4 Nonlinear Dynamics Ch.5 The Harmonics of Vibrating Strings Ch.6 Nonsinusoidal Harmonics Ch.7 Complex Representations of Functions Ch.8 Transform Techniques in Physics Ch.9 Vector Analysis and EM Waves Ch.10 Extrema and Variational Calculus Ch.11 Problems in Higher Dimensions Ch. A Review of Sequences and Infinite Series
395 395 398 403 408 412 415 419 422 427 431 434 436
Preface This is the solution manual for the textbook, A Course in Mathematical Methods for Physicists, by CRC Press. It has been provided for use by instructors only. It only contains solutions to problems in the text. The solutions are worked out in detail in most cases and a list of answers to many of the problems are provided at the end of this manual. Hopefully, it is free from error. However, any noted errors will be posted at the website http://www.russherman.com/cmmp/ along with errata from the textbook.1 An important use of this manual is in scanning problem solutions before assigning them to students. While many problems are relatively straight forward, there are a few challenging problems, or problems for which instructors might decide to add hints for students depending on the level of the course. Of course, instructors are free to add their own favorite problems to accompany the text. The book was originally written to use in an intermediate course on mathematical methods in physics, but due to the addition of a few chapters and topics, this book can certainly be used in other courses such as applied mathematics, engineering mathematics, senior and first year graduate courses in mathematical methods. Amongst some standard topics the book also offers: • A quick review of mathematical prerequisites, proceeding to applications of differential equations and linear algebra • Classroomtested explanations of complex and Fourier analysis for trigonometric and special functions • Coverage of vector analysis and curvilinear coordinates for solving higher dimensional problems • Sections on nonlinear dynamics, variational calculus, numerical solutions of differential equations, and Green’s functions
There are a couple of errors in textbook problems that were missed before printing: 1
• Problem 19 in Chapters 4 and 5: During editing, Problem 19 on finding the circumference of an ellipse in Chapter 4 was overwritten by an intended correction for Problem 19 Chapter 5. The correct problems are listed in this manual.
1 Introduction and Review 1. Prove the following identities using only the definitions of the trigonometric functions, the Pythagorean identity, or the identities for sines and cosines of sums of angles. a. cos 2x = 2 cos2 x − 1. cos 2x
= cos2 x − sin2 x = cos2 x − (1 − cos2 x ) = 2 cos2 x − 1.
b. sin 3x = A sin3 x + B sin x, for what values of A and B? sin 3x
= sin x cos 2x + sin 2x cos x = sin x (cos2 x − sin2 x ) + 2 sin x cos2 x = 3 sin x (1 − sin2 x ) − sin3 x = 3 sin x − 4 sin3 x.
So, A = −4, B = 3. θ π c. sec θ + tan θ = tan + . 2 4 tan
θ π + 2 4
sin
cos
=
θ 2
+
π 4
θ 2
+
π 4
=
sin 2θ cos π4 + sin π4 cos 2θ
=
sin 2θ + cos 2θ
=
sin 2θ + cos 2θ
= =
cos 2θ cos π4 − sin π4 sin 2θ cos 2θ − sin 2θ cos 2θ + sin 2θ
!
cos 2θ − sin 2θ cos 2θ + sin 2θ 2 cos 2θ + sin 2θ cos2 cos2
θ 2
θ 2
− sin2
θ 2
+ sin2 2θ + 2 sin 2θ cos 2θ cos θ
2
mathematical methods for physicists
= =
1 + sin θ cos θ sec θ + tan θ.
2. Determine the exact values of π a. sin . 8 π 8
sin2
= = =
Therefore, sin b. tan 15o .
π 1 = 8 2
q 2−
tan 15o
√
1 π 1 − cos 2 4 √ ! 1 2 1− 2 2 √ 1 2− 2 . 4
2.
= tan(60o − 45o ) tan 60o − tan 45o = 1 + tan 60o tan 45o √ 3−1 √ = 1+ 3 √ √ ! 3−1 1− 3 √ √ = 1+ 3 1− 3 √ = 2 − 3.
One can get the same answer using
√ √ 2 1 − cos 30o sin2 15o 2− 3 √ = = ( 2 − tan 15 = = 3) . 1 + cos 30o cos2 15o 2+ 3 2
o
c. cos 105o . 1 (1 + cos 210o ) 2 1 = (1 − cos 30o ) 2 √ 1 = (2 − 3). 4 p √ Therefore, cos 105o = − 21 2 − 3. One could also use cos2 105o
=
cos2 105o = sin2 15o =
1 (1 − cos 30o ) . 2
Note that this answer can be denested: q q √ √ 1 1 − 2− 3 = − 8−4 3 2 4
introduction and review
q √ √ 1 6−2 2 6+2 4q √ √ 1 − ( 6 − 2)2 4 √ 1 √ ( 2 − 6). 4
= − = =
3. Denest the following if possible. p √ 3 − 2 2. a. p √ √ Assume that 3 − 2 2 = a + b 2. Then, √ √ 3 − 2 2 = ( a + b 2)2
√ = a2 + 2b2 + 2ab 2.
This would hold if a2 + 2b2 = 3 and 2ab = −2. These simultaneous equations have a = −b = ±1. Therefore, the positive p the√solutions √ solution is 3 − 2 2 = 2 − 1. p √ b. 1 + 2. Following the reasoning from the last problem, one tries to solve the system a2 + 2b2 = 1 and 2ab = 1. Solving the second equation for b = 1/2a and inserting this into the first equation, we have 2a4 + a2 − 2 = 0. Solving, we obtain
√ 2
a =
17 − 1 . 4
p √ Therefore, 1 + 2 cannot be denested. p √ c. 5 + 2 6. √ √ √ √ √ Note that 5 + 2 6 = 2 + 3 + 2 2 3 = ( 2 + 3)2 . Therefore, p √ √ √ 5 + 2 6 = 2 + 3. p p √ √ 3 3 d. 5+2− 5 − 2. p √ √ 3 We first guess the form 5 ± 2 = a + b 5. Cubing both sides of this equation, we find √ √ √ 5 ± 2 = a3 + 3a2 b 5 + 15ab2 + 5b3 5. This leads to the system of equations a3 + 15ab2 2
3a b + 5b
3
= ±2 = 1.
Solving the second equation for a, we have p 3b(1 − 5b3 ) . a=± 3b Substituting this into the first equation of the system, we obtain p 3b(1 − 5b3 )(40b3 + 1) = ±2. 9b2
3
4
mathematical methods for physicists
Now, square both sides of the equation. Rearranging and factoring, we have (8b3 − 1)(1000b6 − 25b3 + 1) = 0. The only real solutions come from b3 = 1/8. Thus, b = 1/2. Inserting this result into the expression for a, p 3/2(1 − 5/8) 3/4 1 a=± =± =± . 3/2 3/2 2 √ √ Therefore, 5 ± 2 = 12 ( 5 ± 1). Using this result, we can proceed to solve the problem: q√ q√ 1 √ 1 √ 3 3 5+2− 5 − 2 = ( 5 + 1) − ( 5 − 1) = 1. 2 2 √ e. Find the roots of x2 + 6x − 4 5 = 0 in simplified form. p p √ √ From the quadratic formula, x = −3 ± 9 + 4 5. Setting 9 + 4 5 = √ a + b 5, we need to solve the equations a2 + 5b2
= 9
2ab
= 4.
Solving for b = 2a , and substituting into the first equation, we have a4 − 9a2 + 20 = 0. Solving for a2 , 9±1 = 4, 5. 2 p √ √ √ So, a = 2, 5 and, therefore, b = 2/ 5, 1. Both cases give 9 + 4 5 = √ 2 + 5. a2 =
Using this result, we find the solutions of the problem, x = −1 + √ √ 5, x = −5 − 5. 5 4 θ 3 Figure 1.1: Triangle for Problem 4a.
4. Determine the exact values of 3 a. sin cos−1 . 5 3 3 This problem takes the form sin θ for θ = cos−1 , or cos θ = . 5 5 One can draw a triangle with base of length 3 and hypotenuse 4 of length 5 as shown in Figure 1.1. Then, sin θ = . This is also 5 obtainable from the Pythagorean identity, 2 3 + sin2 θ = 1. 5
7 x θ √ 49 − x2 Figure 1.2: Triangle for Problem 4b.
x b. tan sin−1 . 7 Similar to the last problem one can use the triangle in Figure 1.2. x Let sin θ = . Then, tan θ = 49−x x2 . 7
introduction and review
From identities, 1 + cot2 θ = csc2 θ = Then cot2 θ =
49 . x2
49 − x2 49 −1 = . 2 x x2
Therefore, tan θ = 49−x x2 . 3π . c. sin−1 sin 2 This is a direct computation. 3π π −1 sin sin = sin−1 (−1) = − . 2 2 5. Do the following. a. Write (cosh x − sinh x )6 in terms of exponentials. First note that cosh x − sinh x =
e x + e− x e x − e− x − = e− x . 2 2
Therefore, (cosh x − sinh x )6 = e−6x . b. Prove cosh( x − y) = cosh x cosh y − sinh x sinh y using the exponential forms of the hyperbolic functions.
cosh x cosh y − sinh x sinh y
= =
=
e x + e− x ey + e−y e x − e− x ey − e−y − 2 2 2 2 1 x +y x −y − x +y − x −y e +e +e +e 4 1 − e x +y − e x −y − e− x +y + e− x −y 4 1 x −y e + e−(x−y) = cosh( x − y). 2
c. Prove cosh 2x = cosh2 x + sinh2 x. From the last problem with y = − x we have cosh 2x = cosh2 x − sinh2 x. However, this can be obtained using the definition in terms of exponentials, cosh2 x + sinh2 x
2 x 2 e − e− x e x + e− x + 2 2 i 1h i 1 h 2x e + 2 + e−2x + e2x − 2 + e−2x 4 4 1 2x −2x e +e = cosh 2x. 2
= = =
5
6
mathematical methods for physicists
13 and x < 0, find sinh x and tanh x. 12 Since cosh2 x − sinh2 x = 1, 2 13 25 . sinh2 x = cosh2 x − 1 = −1 = 12 144
d. If cosh x =
5 and tanh x = Then, sinh x = − 12
sinh x cosh x
5 = − 13 .
e. Find the exact value of sinh(arccosh 3). Let cosh x = 3 for x ≥ 0. Then, we seek sinh x. From sinh2 x = √ cosh2 x − 1 = 8. Therefore, sinh x = 2 2. 6. Prove that the inverse hyperbolic functions are the following logarithms: √ a. cosh−1 x = ln x + x2 − 1 . Let y = cosh−1 x. Then, x = cosh y =
ey + e−y . 2
Rewriting,
(ey )2 − 2xey + 1 = 0. Solving for ey , ey =
p 1 (2x + 4x2 − 4). 2
√ Then, y = ln( x + x2 − 1). 1 1+x b. tanh−1 x = ln . 2 1−x −1 Let y = tanh x, or
x = tanh y =
ey − e−y . ey + e−y
Solving for ey , we have x (ey + e−y )
= ey − e−y ,
(1 + x ) e − y
= (1 − x ) e y , 1+x = . 1−x
e2y So, y =
1 2
x ln 11+ −x .
7. Write the following in terms of logarithms: a. cosh−1 43 .
cosh−1
4 3
=
= =
2 4 4 ln + − 1 3 3 √ ! 4 3 ln + 3 3 √ ln(4 + 7) − ln 3. s
introduction and review
b. tanh−1 12 . tanh−1
1 1 3/2 1 = ln = ln 3. 2 2 1/2 2
c. sinh−1 2. sinh−1 2 = ln(2 +
p
22 + 1) = ln(2 +
√
5).
8. Solve the following equations for x. a. cosh( x + ln 3) = 3. Therefore, x + ln 3 = cosh−1 3 = ln(3 + √ Solving, x = ln(3 + 2 2) − ln 3. b. 2 tanh−1
x −2 x −1
√
8).
= ln 2.
Rearranging x−2 x−1
= tanh(ln = = =
eln
√ √
2
√
2)
− e− ln
eln 2 + e− ln √ √ 2 − 22 √ √ 2 + 22 1 3
√ √
2 2
Solving for x, one find x = 25 . c. sinh2 x − 7 cosh x + 13 = 0. Writing sinh2 x = cosh2 x − 1, we have
0
= sinh2 x − 7 cosh x + 13 = (cosh2 x − 1) − 7 cosh x + 13 = cosh2 x − 7 cosh x + 12 = (cosh x − 3)(cosh x − 4).
This gives two sets of solutions,
√ x = ± cosh−1 3 = ± ln(3 + 2 2), √ x = ± cosh−1 4 = ± ln(4 + 15). The negative solutions could be rewritten,
√ x = − ln(3 + 2 2) = ln(
√ √ ) = ln(3 − 2 2), 3+2 2 √ √ 1 √ ) = ln(4 − 15). x = − ln(4 + 15) = ln( 4 + 15 1
7
8
mathematical methods for physicists
9. Compute the following integrals. R 2x2 a. xe dx R 2x2 2 xe dx = 41 e2x + C. R3 5x b. 0 √ dx. 2 x + 16 Z 3
√
0
c.
5x x2
+ 16
dx = 5
p
3 x2 + 16 0 = 5.
x3 sin 3x dx. (Do this using integration by parts, the Tabular Method, and differentiation under the integral sign.)
R
Z
1 = − x3 cos 3x + x2 cos 3x dx 3 Z 1 2 1 = − x3 cos 3x + x2 sin 3x − x sin 3x dx 3 3 3 1 1 = − x3 cos 3x + x2 sin 3x 3 3 Z 1 1 2 cos 3x dx . − − x cos 3x + 3 3 3 Z
x3 sin 3x dx
Thus, Z
x3 sin 3x dx =
2 1 1 2 2 x − x3 cos 3x + x − sin 3x + C. 9 3 3 27
This result is also found from the Tabular Method in Table 2.3. Table 1.1: Tabular Method for Problem 9c.
D
I
x3
+
sin 3x
3x2
−
− 13 cos 3x
6x
+
− 19 sin 3x
6
−
0
1 27
cos 3x
1 81
sin 3x
Finally, we apply differentiation under the integral. First, let I = R cos βx dx = β1 sin βx. Then, Z
x3 sin 3x dx
d3 I dβ3 β=3 d2 1 1 − 2 x cos βx − 2 sin βx dβ β β β =3
= − =
introduction and review
d 2 1 2 − x2 sin βx − 2 x cos βx + 3 sin βx dβ β β β β =3 1 3 3 2 6 6 − − x cos βx + 2 x sin βx + 3 x cos βx − 4 sin βx β β β β β =3
= − =
1 1 2 2 = − x3 cos 3x + x2 sin 3x + x cos 3x − sin 3x + C 3 3 9 27 1 1 2 2 2 x − x3 cos 3x + x − = sin 3x + C. 9 3 3 27 d.
cos4 3x dx.
R
Z
cos4 3x dx
1 (1 + cos 6x )2 dx 4 Z 1 (1 + 2 cos 6x + cos2 6x ) dx 4 Z 1 1 1 + 2 cos 6x + (1 + cos 12x ) dx 4 2 1 1 3 x+ sin 6x + sin 12x + C. 8 12 96 Z
= = = =
e.
R π/4 0
sec3 x dx.
Z π/4 0
sec3 x dx
= =
f.
R
1 [sec x tan x + ln  sec x + tan x ]0π/4 2 √ 1 √ ( 2 + ln(1 + 2)). 2
e x sinh x dx. Z
e x sinh x dx
= = =
g.
R√
Z 1 2x e − 1 dx 2 Z 1 1 2x e −x +C 2 2 1 x x e (cosh x + sinh x ) − + C. 4 2
9 − x2 dx
Using the trigonometric substitution x = 3 sin θ, along with the associated expressions, p dx = 3 cos θ dθ, 9 − x2 = 3 cos θ, we have Z p
9 − x2 dx
= = =
Z
9 cos2 θ dθ
9 (1 + cos 2θ ) dθ 2 9 1 (θ + sin 2θ ) + C 2 2
Z
9
10
mathematical methods for physicists
= = = h.
9 (θ + sin θ cos θ ) + C 2 !! √ 9 x 9 − x2 −1 x sin + + C. 2 3 3 3 √ x 9 − x2 9 −1 x sin + + C. 2 3 2
dx , using the substitution x = 2 tanh u. (4 − x 2 )2 Using the hyperbolic substitution x = 2 tanh u, along with the associated expressions,
R
dx = 2 sech2 u du,
4 − x2 = 4 − 4 tanh2 u = 4 sech2 u,
we have dx (4 − x 2 )2
Z
= = = = =
2 sech2 u du
Z
(4 sech2 u)2 1 cosh2 u du 8 Z 1 (1 + cosh 2u) du 16 1 1 (u + sinh 2u) + C 16 2 1 (u + sinh u cosh u) + C. 16 Z
We need to write sinh u and cosh u in terms of the original variable, x. First, recall that 4 sech2 u = 4 − x2 . Then, cosh2 u = 1/ sech2 u = 2 4 . Also, sinh2 u = cosh2 u − 1 = 4−x x2 . So, 4− x 2 cosh u = √
2 4−
x2
and sinh u = √
x 4 − x2
.
We can now finish the computation. Z
dx (4 − x 2 )2
= = =
i.
R4
√
1 (u + sinh u cosh u) + C 16 x 2 1 −1 x √ tanh +√ +C 16 2 4 − x2 4 − x2 1 2+x x ln + +C 32 2 − x 8(4 − x2 )
dx
, using a hyperbolic function substitution. 9 + x2 Using the hyperbolic substitution x = 3 sinh u, along with the associated expressions, p dx = 3 cosh u du, 9 + x2 = 3 cosh u, 0
we have Z 4 0
√
dx 9 + x2
=
Z sinh−1 0
4 3
du
introduction and review
= sinh−1 = ln
j.
4 3
4 + 3
r
16 1+ 9
!
= ln 3.
dx , using the substitution x = tanh u. 1 − x2 Using the hyperbolic substitution x = tanh u, along with the associated expressions,
R
dx = sech2 u du,
1 − x2 = 1 − tanh2 u = sech2 u,
we have dx 1 − x2
Z
Z
=
sech2 u du sech2 u
Z
=
du
= u + C = tanh−1 x + C. 1 1+x ln + C. = 2 1−x This can be verified using a partial fraction decomposition, Z Z dx 1 1 1 + dx. = 2 1−x 1+x 1 − x2 k.
dx , using the substitutions x = 2 tan θ and x = 2 sinh u. + 4)3/2 First we do the trigonometric substitution,
R
( x2
x = 2 tan θ,
Z
dx = 2 sec2 θ dθ,
dx ( x2 + 4)3/2
= = = =
Here we used sin θ =
√ x x 2 +4
x2 + 4 = 4 sec2 θ,
2 sec2 θ dθ (4 sec2 θ )3/2
Z
1 cos θ dθ 4 1 sin θ + C 4 x √ + C. 4 x2 + 4 Z
which can be seen in Figure 1.3.
√
x2 + 4
x
θ 2
For the hyperbolic function substitution,
Figure 1.3: Triangle for Problem 9k.
x = 2 sinh u,
dx = 2 cosh u du,
x2 + 4 = 4 cosh2 u,
we have Z
dx ( x2 + 4)3/2
=
Z
2 cosh u du (2 cosh u)3
11
12
mathematical methods for physicists
= = =
l.
R
√
1 sech2 u du 4 1 tanh u + C 4 x √ + C. 4 x2 + 4 Z
dx
. − 6x + 4 For this problem, we complete the square and make a hyperbolic function substitution. First, we note that 3x2
3x2 − 6x + 4 = 3( x2 − 2x + 1 − 1) + 4 = 3( x − 1)2 + 1. Next, we let 3( x − 1)2 = sinh2 u. Then, 3x2 − 6x + 4 = sinh2 u + 1 = cosh2 u.
Z
√
dx 3x2 − 6x + 4
1 √ du 3 1 √ u+C 3 √ 1 √ sinh−1 3( x − 1) + C 3 √ p 1 √ ln 3( x − 1) + 3x2 − 6x + 4 + C. 3 Z
= = = =
10. Find the sum for each of the series: a. 5 +
25 7
+
125 49
+
625 343
5+ b. ∑∞ n =0
+···.
25 125 625 5 + + +··· = 7 49 343 1−
=
35 . 2
(−1)n 3 . 4n ∞
(−1)n 3 3 = n 4 1+ n =0
∑
c. ∑∞ n =2
5 7
1 4
=
12 . 5
2 . 5n ∞
2 2/25 1 = = . n 1 5 10 1− 5 n =2
∑
n +1 d. ∑∞ n=−1 (−1)
e n . π
∞
∑
n=−1
(−1)n+1
π e n = e π 1+
e π
=
π2 . (π + e)e
introduction and review
e. ∑∞ n =0
1 5 + n n 2 3
∞
∑
.
n =0
5 1 + n 2n 3
=
5 1−
1 2
+
1 1−
1 3
=
23 . 2
3 . n ( n + 3) The Nth partial sum is
f. ∑∞ n =1
N
N
3 ∑ n ( n + 3) n =1
= =
=
1 1 − ∑ n n+3 n =1 1 1 1 1 1 1 1 1 1 C C 1− + − + − + − + − 2 5 3 6 4CC 7 5CC 8 4 Z1 Z1 1 1 1 1 A +···+ + + − Z − Z − N −Z 2 N+1 N −Z 1 N+2 NAA N + 3 1 1 1 1 1 + + . 1+ + − 2 3 N+1 N+2 N+3
Here forward crossed terms like 14 cancel with terms later in sum and backward crossed terms like A41A cancel with earlier terms. 3 11 Letting N → ∞, we have ∑∞ = . n =1 n ( n + 3) 6 ¯ g. What is 0.569?
0.569¯
= 0.56 + 0.009 + 0.0009 + . . . n 1 ∞ = 0.56 + ∑ = 0 (0.009) 10 n .009 = 0.56 + 1 1 − 10 = 0.57.
11. A superball is dropped from a 2.00 m height. After it rebounds, it reaches a new height of 1.65 m. Assuming a constant coefficient of restitution, find the (ideal) total distance the ball will travel as it keeps bouncing. The ball drops 2.00 m and rebounds 1.65 m or 82.5% of the height it fell. Then, it drops 1.65 m and rebounds 82.5% of that height. This continues, giving the total distance d = 2.00 + 2(2.00)(.825) + 2(2.00)(.825)2 + 2(2.00)(.825)3 + . . . = 2.00 + 12. Here are some telescoping series problems. a. Verify that ∞
∞ 1 ∑ (n + 2)(n + 1) = ∑ n =1 n =1
n+1 n − n+2 n+1
.
2(1.65) = 20.9m. 1 − .825
13
14
mathematical methods for physicists
One just adds the terms to verify the sum. n+1 n − n+2 n+1
= =
( n + 1)2 − n ( n + 2) (n + 2)(n + 1) 1 . (n + 2)(n + 1)
Note that partial fractions does not give this representation, but instead gives ∞ 1 ∑ (n + 2)(n + 1) = ∑ n =1 n =1 ∞
1 1 − n+1 n+2
.
This is also summed as a telescoping series. n n+1 b. Find the Nth partial sum of the series − and n+2 n+1 use it to determine the sum of the resulting telescoping series. ∑∞ n =1
The Nth partial sum of the series is N
sN
= = =
n n+1 − ∑ n+2 n+1 n =1 2 1 4 C3 N + 1 ZN 3 C2 − + − + − +···+ − Z N+2 N +Z 1 4 3CC 3 2 5 4CC N+1 1 . − + 2 N+2
Here forward crossed terms like 23 cancel with terms later in sum and backward crossed terms like A23A cancel with earlier terms. n n+1 ∞ − = − 12 + 1 = 21 . Letting N → ∞, we have ∑n=1 n+2 n+1 −1 c. Sum the series ∑∞ n − tan−1 (n + 1) by first writing the n=1 tan Nth partial sum and then computing lim N →∞ s N . The Nth partial sum is N
=
sN
∑
h
tan−1 n − tan−1 (n + 1)
i
n =1
=
XX−X 1 1X − tan−1 1 − tan 2 + tan 2 − tan−1 3 X 1 X + · · · + tanX−X N − tan−1 ( N + 1)
= tan−1 2 − tan−1 ( N + 1) Letting N → ∞, we have ∞
∑
n =1
h
i π π π tan−1 n − tan−1 (n + 1) = − = − . 4 2 4
13. Determine the radius and interval of convergence of the following infinite series:
introduction and review
n a. ∑∞ n=1 (−1)
( x − 1) n . n q  x − 1 lim n  an  = lim √ =  x − 1 < 1. n n→∞ n→∞ n
Thus, R = 1, and one has absolute convergence for x ∈ (0, 2). Looking at the endpoints, there is conditional convergence for x = 2. xn b. ∑∞ n =1 n . 2 n!
 a n +1 2n n! x  x  n +1 = lim = lim n+1 = 0. n → ∞ 2( n + 1) n→∞  an  n→∞ 2 ( n + 1) !  x  n lim
Thus, R = ∞, and one has absolute convergence for x ∈ (−∞, ∞). 1 x n . c. ∑∞ n =1 n 5 lim
q
n→∞
n
 an  = lim
n→∞
x x √ = < 5. 5 5nn
Thus, R = 5, and one has absolute convergence for x ∈ (−5, 5). Looking at the endpoints, there is conditional convergence for x = −5. n n x d. ∑∞ n=1 (−1) √ . n √  a n +1  x  n +1 n lim = lim √ =  x  < 1. n→∞  an  n→∞ n + 1  x n Thus, R = 1 and one has absolute convergence for x ∈ (−1, 1). Looking at the endpoints, there is conditional convergence for x = 1. 14. Use the partition function Z for the quantum harmonic oscillator to find the average energy, h Ei. We have that ∂ ln Z < E >= − , ∂β where β = 1/kT and the partition function is given by Z = (2 sinh Thus,
< E >= −
β¯hω −1 2 ) .
∂ h¯ ω β¯hω β¯hω (2 sinh β¯hω/2)−1 = coth csch . ∂β 4 2 2
15. Find the Taylor series centered at x = a and its corresponding radius of convergence for the given function. In most cases, you need not employ the direct method of computation of the Taylor coefficients. a. f ( x ) = sinh x, a = 0.
15
16
mathematical methods for physicists
Using the Maclaurin series expansion for f ( x ) = e x , we have sinh x
= = =
e x − e− x 2 1 x2 x3 x2 x3 1+x+ + +... − 1−x+ − +... 2 2! 3! 2! 3! 1 x3 x5 2x + 2 + 2 + . . . 2 3! 5!
= x+ b. f ( x ) =
√
∞ x3 x5 x2k−1 + +... = ∑ . 3! 5! (2k − 1)! k =1
1 + x, a = 0.
Using the binomial expansion, √ 1 + x = (1 + x )1/2 ∞ 1 1 −1 ··· 1 −r+1 2 2 2 xr , = ∑ r! r =0 1 1 1 = 1 + x − x2 + x3 + . . . , 2 8 16
 x  < 1,
 x  < 1.
1+x , a = 0. 1−x 1+x Noting that ln = ln(1 + x ) − ln(1 − x ) and using Maclaurin 1−x series expansions for ln(1 + x ),  x  < 1,
c. f ( x ) = ln
ln(1 + x )
=
ln(1 − x )
=
ln
1+x 1−x
x2 x3 + −..., 2 3 x2 x3 −x − − −..., 2 3 x−
= ln(1 + x ) − ln(1 − x ) 2 2 = 2x + x3 + x5 + . . . , 3 5
 x  < 1.
d. f ( x ) = xe x , a = 1. First replace x = x − 1 + 1 and then expand the exponential in powers of x − 1. xe x
= ( x − 1 + 1 ) e x −1+1 = [( x − 1)e + e]e x−1 1 1 = [( x − 1)e + e] 1 + ( x − 1) + ( x − 1)2 + ( x − 1)3 + . . . 2 3! 1 1 1 = e + 2e( x − 1) + 1 + e ( x − 1)2 + e ( x − 1)3 + . . . + 2! 2! 3! ∞ 1 1 = e+ ∑ e ( x − 1 ) n +1 + n! ( n + 1 ) ! n =0 ∞
= e+
( n + 2) e ( x − 1 ) n +1 , ( n + 1 ) ! n =0
∑
 x − 1 < ∞.
introduction and review
Since  x − 1 < ∞ is the same as  X  < ∞, we have xe x = e +
∞
( n + 2) e ( x − 1 ) n +1 , ( n + 1 ) ! n =0
∑
 x  < ∞.
1 e. f ( x ) = √ , a = 1. x 1 1 Since √ = √ = [1 + ( x − 1)]−1/2 , we can use the binox x−1+1 mial expansion for  x − 1 < 1 to obtain ∞ −1 −1 − 1 · · · −1 − r + 1 2 2 2 1 √ = ∑ ( x − 1)r ,  x − 1 < 1. r! x r =0 1 3 5 = 1 − ( x − 1) + ( x − 1)2 − ( x − 1)3 + . . . , 2 8 16
 x − 1 < 1.
f. f ( x ) = x4 + x − 2, a = 2. We replace x = x − 2 + 2 to obtain f (x)
=
x4 + x − 2
= ( x − 2 + 2)4 + ( x − 2) = ( x − 2)4 + 8( x − 2)3 + 24( x − 2)2 + 33( x − 2) + 16. x−1 , a = 1. 2+x Write 2 + x = 3 + ( x − 1) = 3(1 + geometric series.
g. f ( x ) =
x−1 2+x
x −1 3 )
and write the result as a
1 3 + ( x − 1) x−1 1 1 3 1 + x− 3 " # x−1 ∞ 1−x n ∑ 3 3 n =0 ∞ 1 − x n +1  x − 1 −∑ , < 1. 3 3 n =0
= ( x − 1) = = =
16. Consider Gregory’s expansion tan−1 x = x −
∞ x3 x5 (−1)k 2k+1 + −··· = ∑ x . 3 5 2k + 1 k =0
a. Derive Gregory’s expansion by using the definition tan−1 x =
Z x 0
dt , 1 + t2
expanding the integrand in a Maclaurin series, and integrating the resulting series term by term.
17
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mathematical methods for physicists
Since
1 1+ t2
2 n ∑∞ n=0 (− t ) , we have
tan−1 x
=
Z x 0
=
dt 1 + t2
Z x ∞
∑ (−t2 )n
0 n =0 ∞
=
(−1)n x2n+1 . 2n + 1 n =0
∑
b. From this result, derive Gregory’s series for π by inserting an appropriate value for x in the series expansion for tan−1 x. Setting x = 1 in the last result, tan−1 (1) π 4 Therefore,
∞
=
(−1)n 2n + 1 n =0
∑
= 1−
1 1 + −... 3 5
1 1 π = 4 1− + −... . 3 5
17. In the event that a series converges uniformly, one can consider the derivative of the series to arrive at the summation of other infinite series. a. Differentiate the series representation for f ( x ) = 1−1 x to sum the n series ∑∞ n=1 nx ,  x  < 1. ∞ 1 For this problem, we need f ( x ) = = ∑ x n ,  x  < 1. Then, 1−x n =0 ∞ 1 n −1 0 = ∑ nx ,  x  < 1. Through observation, we f (x) = (1 − x )2 n =1 ∞ x n ,  x  < 1. see that ∑ nx = (1 − x )2 n =1 n b. Use the result from part a to sum the series ∑∞ n =1 n . 5 1 n 5 ∞ 5 . ∑ n =1 n = 2 = 5 16 1 1− 5 n c. Sum the series ∑∞ n=2 n ( n − 1) x ,  x  < 1.
∞ 2 = ∑ n ( n − 1 ) x n −2 , 3 (1 − x ) n =2 ∞ 2x2 n  x  < 1. Therefore, we see that ∑ n(n − 1) x = ,  x  < 1. (1 − x )3 n =2
Further differentiation gives f 00 ( x ) =
∞
n2 − n ∑ 5n . n =2 We rewrite the given expression and use the previous results to find 2 ∞ ∞ 2 15 n2 − n n ( n − 1) 5 ∑ 5n = ∑ 5n = 4 3 = 32 . n =2 n =2
d. Use the result from part c to sum the series
5
introduction and review
n2 . 5n We rewrite the given expression and use the previous results to find ∞ ∞ n2 n2 4 9 = − + ∑ 5n ∑ 5n 25 125 n =2 n =4 ! ∞ 5 n 4 9 = + − + n 32 n∑ 25 125 =2 5 ! ∞ 5 1 n 4 9 = +∑ n− − + 32 n=1 5 5 25 125 5 5 1 4 9 = + − − + 32 16 5 25 125 147 = . 4000
e. Use the results from this problem to sum the series ∑∞ n =4
18. Evaluate the integral
R π/6 0
sin2 x dx by doing the following:
a. Compute the integral exactly.
Z π/6 0
1 sin x dx = 2 2
Z π/6 0
√ π 1 π π 3 (1 − cos 2x ) dx = − sin = − . 12 2 3 12 8
b. Integrate the first three terms of the Maclaurin series expansion of the integrand and compare with the exact result. Using sin x = x − Z π/6 0
x3 3!
sin2 x dx
+ =
x5 5!
+ O( x7 ), we find Z π/6 0
= ≈ =
x−
x3 x5 + + O( x7 ) 3! 5!
2
x4 2x6 x − + + O( x8 ) 3 45 0 3 π/6 x x5 2x7 − + 3 15 315 0 Z π/6
2
dx dx
π3 π5 π7 − + ≈ 0.0453. 648 116640 44089920
19. Determine the next term in the time dilation example,1.39. That is, find 4 the vc2 term and determine a better approximation to the time difference of 1 ns. The time difference is ! r v2 ∆t − ∆τ = 1 − 1 − 2 ∆t c v2 v4 = 1− 1− 2 − 4 +... ∆t 2c 8c
≈
v2 ∆t 2 2c{z }
1.0125×10−9 s
+
v4 ∆t 4 8c{z }
1.4×10−22 s
19
20
mathematical methods for physicists
Therefore, the next term in the approximation is 1.4 × 10−13 ns which is far too small to be measured as compared to the first term. 20. Evaluate the following expressions at the given point. Use your calculator or your computer (such as Maple). Then use series expansions to find an approximation of the value of the expression to as many places as you trust. Note: In this problem we are considering that all values of the independent variable are exact. Otherwise, we would need to propagate the errors and find that the accuracy reported below is far too precise. Using ideas from Chapter 9, we can compute the absolute errors of computing f ( x ) in terms of ∆x as ∂f ∆f = ∆x. ∂x We report the results below. a. √
Table 1.2: Values of √ x = 0.015.
1 1 + x3
1 1 + x3
− cos x2 at x = 0.015.
Tool TI83 Plus Maple (10 Digits) Maple (20 Digits) MATLAB (short) MATLAB (long)
− cos x2 at
Value −1.66218325E − 6 −1.6617 × 10−6 −1.66218322863445 × 10−6 −1.6622e − 06 −1.662183228723356e − 06
Computation:
√
1 1 + x3
− cos x
2
=
1 3 3 6 1 4 1 8 1− x + x −... − 1− x + x −... 2 8 2 4! x3 −  {z2}
=
+
x4 3x6 x8 + − +... 2 8 24
−1.6875×10−6
 
{z
−1.6621875×10−6
{z
} }
−1.662183228515625×10−6

{z
−1.6621832286344256592×10−6
}
For x = 0.015 ± 0.0005, f ( x ) = (1.66 ± 0.17) × 10−6 . r 1+x b. ln − tan x at x = 0.0015. 1−x r Table 1.3: Values of ln x = 0.0015.
1+x − tan x at 1−x
Tool TI83 Plus Maple (10 Digits) Maple (20 Digits) MATLAB (short) MATLAB (long)
Value 0 1.19 × 10−10 5.062439 × 10−16 6.0607e − 16 6.060690144193970e − 16
introduction and review
Computation: r 1+x ln − tan x 1−x
1 2
=
2 2 1 2 2x + x3 + x5 + . . . − x + x3 + x5 + . . . 3 5 3 15
x5 15 {z}
=
21
+
253x9 4x7 ++ +... 45 2835
5.0625×10−16
{z

}
5.0625151875×10−16
{z

5.0625151875343074777×10−16
}
For x = 0.0015 ± 0.00005, f ( x ) = (5.06 ± 0.84) × 10−6 . 1 − 1 + x2 at x = 5.00 × 10−3 . c. f ( x ) = √ 1 + 2x2 Tool TI83 Plus Maple (10 Digits) Maple (20 Digits) MATLAB (short) MATLAB (long)
Value 9.3747E − 10 0.6 × 10−9 9.3746093916 × 10−10 9.3746e − 10 9.374608339397285e − 10
Table 1.4: Values of f ( x ) = √ 1 + x2 at x = 5.00 × 10−3 .
1 1 + 2x2
−
Computation:
√
1 1 + 2x2
− 1 + x2
= =
3 5 35 1 − x2 + x4 − x6 + x8 + . . . − 1 + x2 2 2 8 5 6 35 8 3 4 − x + x +... x 2 2 8 {z}
9.375×10−10

{z
}
9.374609375×10−10

{z
}
9.37460939208984375×10−10
For x = 0.005 ± 0.0005, f ( x ) = (9.37 ± 3.75) × 10−10 . √ d. f ( R, h) = R − R2 + h2 for R = 1.374 × 103 km and h = 1.00 m. Tool TI83 Plus Maple (10 Digits) Maple (20 Digits) MATLAB (short) MATLAB (long)
Computation: First write f ( R, h) = R − Since tain
h R
√
r R2 + h2 = R − R
1+
2 h R
≈ 7.278 × 10−7 , we can use the binomial expansion to ob
f ( R, h)
Table 1.5: Values of f ( R, h) = R − √ R2 + h2 for R = 1.374 × 103 km and h = 1.00 m.
Value 0 0 −.36390010 × 10−6 −3.6391e − 07 −3.639142960309982e − 07
=
R 1 −
s
2 h 1+ R
.
22
mathematical methods for physicists
= =
!! 6 1 h 2 1 h 4 1 h R 1− 1+ − + +... 2 R 8 R 16 R ! 6 1 h 2 1 h 4 1 h R − + − +... 2 R 8 R 16 R 1 h2 − 2 R}  {z
=
+
1 h4 1 h6 − +... 3 8R 16 R5
−0.3639010892285298399×10−6
 
{z
}
−0.3639010892280479478×10−6
{z
−0.3639010892280479478×10−6
}
= −0.00036390097 . . . . Therefore, for h = 1.00 ± 0.005 m and R = (1.374 ± 0.0005) × 103 km we have f ( R, h) = (−3.639 ± 0.038) × 10−7 . 1 for x = 2.5 × 10−13 . e. f ( x ) = 1 − √ 1−x Table 1.6: Values of f ( x ) = 1 − √ for x =
2.5 × 10−13 .
1 1−x
Tool TI83 Plus Maple (10 Digits) Maple (20 Digits) MATLAB (short) MATLAB (long)
Value 0 0 −0.1250000 × 10−12 −1.2501e − 13 −1.250111125727926e − 13
Computation: 1− √
1 1−x
1 3 5 35 = 1 − 1 + x + x2 + x3 + +... 2 8 16 128 3 5 1 − x3 + . . . = − x − x2 2 8 16  {z } −1.25×10−13

{z
}
−1.250000000000234375×10−13

{z
−1.250000000000234375×10−13
}
For x = (2.5 ± 0.05) × 10−13 , f ( x ) = (−1.250 ± 0.025) × 10−13 . 21. Use dimensional analysis to derive a possible expression for the drag force FD on a soccer ball of diameter D moving at speed v through air [ M] of density ρ and viscosity µ. [Hint: Assuming viscosity has units [ L][T ] , there are two possible dimensionless combinations: π1 = µD α ρ β vγ and π2 = FD D α ρ β vγ . Determine α, β, and γ for each case and interpret your results.] Since these are dimensionless combinations, we can simply check that the dimensions cancel for particular values of the exponents. First, we consider the dimensions of π1 :
[π1 ] = [µ][ D ]α [ρ] β [v]γ β γ = [ M][ L]−1 [ T ]−1 [ L]α [ M][ L]−3 [ L][ T ]−1 = [ M]1+ β [ L]−1+α−3β+γ [ T ]−1−γ .
introduction and review
All exponents need to cancel. So, we immediately have β = −1 and γ = −1. This leaves −1 + α − 3β + γ = 0, or α = 1 + 3β − γ = −1. Therefore, π1 = µD −1 ρ−1 v−1 . Next we consider the dimensions of π2 :
[π2 ] = [ FD ][ D ]α [ρ] β [v]γ β γ = [ M][ L][ T ]−2 [ L]α [ M][ L]−3 [ L][ T ]−1 = [ M]1+ β [ L]1+α−3β+γ [ T ]−2−γ . In this case we have β = −1 and γ = −2. This leaves 1 + α − 3β + γ = 0, or α = −1 + 3β − γ = −2. Therefore, π2 = FD D −2 ρ−1 v−2 . Therefore, we have found π1 = µD −1 ρ−1 v−1 and π2 = FD D −2 ρ−1 v−2 . From π2 we have FD = π2 D2 ρv2 . In Chapter 2 we discuss the drag force in the form FD = 12 CD ρAv2 where A is the cross sectional area, thus proportional to D2 , and CD is the drag coefficient. Thus, π2 is effectively the drag coefficient. π1 is possibly more elusive, but also in Chapter 2 we will discuss the ρLU Reynolds number, Re = η , where L is a characteristic length, U a characteristic speed and η is the viscosity. Thus, π1 = 1/Re. 22. Challenge: Read the section on dimensional analysis. In particular, look at the results of Example 1.44. Using measurements in/on Figure 1.9, obtain an estimate of the energy of the blast in tons of TNT. Explain your work. Does your answer make sense? Why? The energy formula in the text takes the form E≈
r5 ρ . t2
The image shows the cloud 0.016 s after the blast. From the picture scale, one can estimate r = 100 m. The density of air is roughly 1.0 kg/m3 . So, the energy is on the order of E ≈ 4 × 1013 J. Since a kiloton of TNT is 4.184 × 1012 J, E ≈ 10 ktons TNT. Since the yield is given as 20 ktons TNT, this answer is of the same order as the reported yield.
23
2 Free Fall and Harmonic Oscillators 1. Find all the solutions of the first order differential equations. When an initial condition is given, find the particular solution satisfying that condition. a.
dy ex = . dx 2y Using separation of variables, we have 2
Z
y dy
y2 ( x )
Z
=
e x dx
= e x + C.
This is an implicit solution. b.
dy = y2 (1 + t2 ), y(0) = 1. dt Using separation of variables, we have Z
dy y2 1 − y
Z
=
(1 + t2 ) dt
1 = t + t3 + C. 3
The initial condition, y(0) = 1, implies C = −1, giving the solution y(t) =
1 3 = . 1 3 3 − 3t − t3 1 − t − 3t
p 1 − y2 dy c. = . dx x Using separation of variables, we have dy
Z
p
1 − y2
sin−1 y y( x ) d. xy0 = y(1 − 2y),
y(1) = 2.
=
Z
dx x
= ln  x  + C = sin(ln  x  + C )
26
mathematical methods for physicists
Using separation of variables and partial fraction decomposition, we have dx x
=
ln  x  + C
=
Z
dy y(1 − 2y) Z 1 2 + dy y 1 − 2y y ln 1 − 2y Z
= Exponentiating,
y ln  x +C ≡ A  x . 1 − 2y = e Using the initial condition, y(1) = 2, we find that A = solutions near y = 2 and x = 1, this gives y 2y − 1 3y
= =
2 3.
For
2 x 3 2x (2y − 1)
(3 − 4x )y = −2x 2x y( x ) = . 4x − 3 e. y0 − (sin x )y = sin x. This is a linear first order differential equation. The integrating factor is Z µ( x ) = exp − sin x dx = ecos x . This gives
(ecos x y( x ))0 ecos x y( x )
= sin xecos x Z
=
sin xecos x dx + C
= −ecos x + C y( x )
= Cecos x − 1.
f. xy0 − 2y = x2 , y(1) = 1. This is a linear first order differential equation. The integrating factor is Z dx 1 µ( x ) = exp −2 = e−2 ln x = 2 . x x This gives
1 y( x ) x2
0
y( x )
=
1 x
=
x2 (ln  x  + C ).
The initial condition, y(1) = 1, gives C = 1, or y( x ) = x2 (ln  x  + 1).
free fall and harmonic oscillators
g.
27
ds + 2s = st2 , s(0) = 1. dt This is a linear first order differential equation, however it is also separable. Rewriting the problem in separable form, we have ds dt Z ds s
= s ( t2 − 2)
ln s
=
=
Z
(t2 − 2) dt
1 3 t − 2t + C. 3
Using the initial condition, s(0) = 1, C = 0. The solution satisfying this initial condition is then 1 3 −2t
s(t) = e 3 t
.
h. x 0 − 2x = te2t . This is a linear first order differential equation. The integrating factor is µ(t) = e−2t . This gives
( xe−2t )0
= t 1 2 t +C xe−2t = 2 1 2 t + C e2t . x (t) = 2
i.
dy + y = sin x, y(0) = 0. dx This is a linear first order differential equation. The integrating factor is µ( x ) = e x . This gives
(y( x )e x )0 y( x )e x
= e x sin x = =
y( x )
=
Z
e x sin x dx + C
1 x e (sin x − cos x ) + C. 2 1 (sin x − cos x ) + Ce− x . 2
The initial condition gives C = 12 . Thus, y( x ) = j.
1 (sin x − cos x + e− x ). 2
dy 3 − y = x3 , y(1) = 4. dx x The integrating factor is Z dx µ( x ) = exp −3 = e−3 ln x = x −3 . x This gives
( x −3 y( x ))0 x −3 y ( x ) y( x )
= 1 = x+C = x 3 ( x + C ).
D
I
ex
+
sin x
ex
−
− cos x
ex
− sin x
Figure 2.1: Tabular R Method for Problem 1i for computing e x sin x dx.
28
mathematical methods for physicists
For y(1) = 4, C = 3. Thus, y( x ) = x4 + 3x3 . 2. Find all the solutions of the second order differential equations. When an initial condition is given, find the particular solution satisfying that condition. a. y00 − 9y0 + 20y = 0. This is a second order, constant coefficient differential equation. The characteristic equation is 0 = r2 − 9r + 20 = (r − 4)(r − 5). Thus, the roots are r = 4, 5 and the general solution is y( x ) = c1 e4x + c2 e5x . b. y00 − 3y0 + 4y = 0,
y(0) = 0,
y0 (0) = 1.
The characteristic equation is 0 = r2 − 3r + 4. The roots are found √ as r =
3± 7i 2
and the general solution is " √ ! 7 3x/2 x + c2 sin y( x ) = e c1 cos 2
√
7 x 2
!# .
The initial conditions are y(0) = 0, y0 (√0) = 1. The first condition 7 3x/2 gives c1 = 0. Thus, y( x ) = c2 e sin 2 x . Noting that " y0 ( x ) = c2 e3x/2
3 sin 2
√
7 x 2
!
√
7 + cos 2
√
7 x 2
!# ,
√
we have y0 (0) = 27 c2 = 1. This gives c2 = √2 and the particular 7 solution is given as √ √ ! 2 7 3x/2 7 y( x ) = e sin x . 7 2 c. x2 y00 + 5xy0 + 4y = 0,
x > 0.
This is a second order, constant coefficient differential equation. This is a CauchyEuler type of differential equation. The characteristic equation is 0 = r (r − 1) + 5r + 4 = r2 + 4r + 4. It has one real root r = −2. The general solution is y( x ) = x −2 (c1 + c2 ln  x ). d. x2 y00 − 2xy0 + 3y = 0,
x > 0.
This is a CauchyEuler type of differential equation. The characteristic equation is 0 = r (r − 1) − 2r + 3 = r2 − 3r + 3. It has complex √ conjugate roots r = "
3 2
±
y( x ) = x3/2 c1 cos
3 2 i.
The general solution is ! !# √ √ 3 3 ln  x  + c2 sin ln  x  . 2 2
3. Consider the differential equation dy x x = − . dx y 1+y
free fall and harmonic oscillators
a. Find the 1parameter family of solutions (general solution) of this equation. First, we note that this equation is separable, dy x x x = − = . dx y 1+y y (1 + y ) Separating variables, we find Z
(y + y2 ) dy = 1 2 1 3 y + y 2 3
=
Z
x dx
1 2 x + C, 2
or 3y2 + 2y3 = 3x2 + k. b. Find the solution of this equation satisfying the initial condition y(0) = 1. Is this a member of the 1parameter family? Inserting x = 0 into the implicit solution, we find k = 5. Yes, 3y2 + 2y3 = 3x2 + 5 is a member of the 1parameter family. 4. The initial value problem dy y2 + xy , = dx x2
y (1) = 1
does not fall into the class of problems considered in this chapter. However, if one substitutes y( x ) = xz( x ) into the differential equation, one obtains an equation for z( x ) that can be solved. Use this substitution to solve the initial value problem for y( x ). Let y( x ) = xz( x ). Then, y0 = z + xz0 . From the original equation, we have y0 = z2 + z. Equating these expressions, we obtain xz0 = z2 . This is separable and can be solved, Z
dz z2 1 − z
z( x )
=
Z
dx x
= ln  x  + C, =
−1 . ln  x  + C
This gives,
−x . ln  x  + C Using the initial condition, C = −1, the particular solution becomes y( x ) = xz( x ) =
y( x ) =
x . 1 − ln  x 
5. Consider the nonhomogeneous differential equation x 00 − 3x 0 + 2x = 6e3t . a. Find the general solution of the homogenous equation. The characteristic equation is 0 = r2 − 3r + 2 = (r − 1)(r − 2). The roots are r = 1, 2. This gives the solution of the homogeneous equation as xh (t) = c1 et + c2 e2t .
29
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mathematical methods for physicists
b. Find a particular solution using the Method of Undetermined Coefficients by guessing x p (t) = Ae3t . We seek the particular solution satisfying x 00p − 3x 0p + 2x p = 6e3t . We guess x p (t) = Ae3t . Inserting this into the differential equation, (9A − 9A + 2A)e3t = 6e3t , or A = 3. Therefore, x p (t) = 3e3t . c. Use your answers in the previous parts to write the general solution for this problem. The general solution is x (t) = xh (t) + x p (t) = c1 et + c2 e2t + 3e3t . 6. Find the general solution of the given equation by the method given. a. y00 − 3y0 + 2y = 10. Method of Undetermined Coefficients.
D
I
−3x2
ex
+
−6x
−
−6
+
0
ex ex ex
The solution to the homogeneous problem is of the same form as in the last problem, yh ( x ) = c1 e x + c2 e2x . We obtain the particular solution of the nonhomogeneous problem from the guess y p ( x ) = A. Inserting this guess, we obtain A = 5. This gives the general solution as y( x ) = c1 e x + c2 e2x + 5. b. y00 + y0 = 3x2 . Variation of Parameters. We first solve the homogeneous problem, y00h + y0h = 0. The characteristic equation is 0 = r2 + r = r (r + 1). The roots are r = 0, −1, giving yh ( x ) = c1 + c2 e− x . In order to apply the Method of Variation of Parameters, we seek a solution of the form
Figure 2.2: Tabular Method for Problem R 6b for computing −3 x2 e x dx.
y p ( x ) = c1 ( x ) + c2 ( x ) e − x . The unknown coefficients satisfy c10 + c20 e− x
−c20 e− x
= 0 = 3x2 .
Solving the second equation for c2 , c2 = −
Z
3x2 e x dx = (−3x2 + 6x − 6)e x + k2 .
Using the second equation for c20 in the first equation, c1 can be found as Z c1 = 3x2 dx = x3 + k1 . Inserting these results into the form for y p ( x ), y p (x)
= c1 ( x ) + c2 ( x ) e − x = ( x3 + k1 ) + (−3x2 + 6x − 6)e x + k2 )e− x = k1 + k2 e− x + x3 − 3x2 + 6x − 6.
free fall and harmonic oscillators
k1 and k2 can be anything. Setting them to zero gives one solution y p ( x ) = x3 − 3x2 + 6x − 6. Note that leaving the k’s arbitrary takes care of the homogeneous part of the solution. In fact, the constant term can be absorbed into k1 , giving the general solution to the original problem as y( x ) = k1 + k2 e− x + x3 − 3x2 + 6x. 7. Find the general solution of each differential equation. When an initial condition is given, find the particular solution satisfying that condition. a. y00 − 3y0 + 2y = 20e−2x ,
y(0) = 0,
y0 (0) = 6.
The solution to the homogeneous problem is yh ( x ) = c1 e x + c2 e2x . The particular solution is found using the guess y p ( x ) = Ae−2x . Inserting this guess into the equation gives A = 53 . So, the general solution is 5 y( x ) = c1 e x + c2 e2x + e−2x . 3 Inserting the initial conditions, we have 5 = c1 + c2 + , 3 10 6 = c1 + 2c2 − . 3 0
Solving these equations, we obtain c1 = − 38 3 , c2 = 11. Thus, the solution to the initial value problem is y( x ) = −
38 x 5 e + 11e2x + e−2x . 3 3
b. y00 + y = 2 sin 3x. The solution to the homogeneous problem is yh ( x ) = c1 cos x + c2 sin x. We guess a particular solution of the form y p ( x ) = A sin 3x since there is no first derivative term. Then, A = 5. The general solution is then y( x ) = c1 cos x + c2 sin x + 5 sin 3x. c. y00 + y = 1 + 2 cos x. The solution to the homogeneous problem is yh ( x ) = c1 cos x + c2 sin x. In this problem the forcing term, 1 + 2 cos x, involves a solution to the homogeneous problem. Therefore, we need to use a modification of the Method of Undetermined Coefficients by making the guess y p ( x ) = A + Bx sin x. Computing the derivatives, y0p ( x )
=
y00p ( x )
= 2B cos x − Bx sin x.
B sin x + Bx cos x,
Then, y00 + y = A + 2B cos x = 1 + 2 cos x. This gives A = B = 1. As a result, the general solution is y( x ) = c1 cos x + c2 sin x + 1 + x sin x.
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d. x2 y00 − 2xy0 + 2y = 3x2 − x,
x > 0.
This is a CauchyEuler type of differential equation. The characteristic equation is 0 = r (r − 1) − 2r + 2 = r2 − 3r + 2 = (r − 1)(r − 2). The roots are r = 1, 2. Thus, the solution to the homogeneous problem is y h ( x ) = c1 x + c2 x 2 . There are two ways this could be solved. We will first use the Method of Variation of Parameters and then use a modification of the Method of Undetermined Coefficients since the forcing terms are solutions of the homogeneous problem. Variation of Parameters We assume y p ( x ) = c1 ( x ) x + c2 ( x ) x 2 . The unknown coefficients satisfy c10 x + c20 x2 c10 + 2c20 x
= 0 =
3x2 − x . x2
Multiplying the second equation by x, the system becomes c10 x + c20 x2 c10 x + 2c20 x2
= 0 = 3x − 1.
Eliminating c10 from the new system, we have c20 = grating, we find c2 = 3 ln  x  + 1x . This gives
3 x
−
1 . x2
Inte
1 c10 = −c20 x = −3 + . x Therefore, c1 = −3x + ln  x . Using these results, we find y p (x)
= c1 ( x ) x + c2 ( x ) x 2 1 = (−3x + ln  x ) x + (3 ln  x  + ) x2 x 2 2 = −3x + x ln  x  + 3x ln  x  + x.
The first and last terms are solution of the homogeneous problem, so we can take y p ( x ) = x ln  x  + 3x2 ln  x  and write the general solution as y( x ) = c1 x + c2 x2 + x ln  x  + 3x2 ln  x .
free fall and harmonic oscillators
Undetermined Coefficients The solution using the Method of Variation of Parameters suggests an approach to the modified Method of Undetermined Coefficients for CauchyEuler equations. We can prove this in general. Consider the problem ax2 y00 + bxy0 + cy = Cx p , where x p is a solution to the homogeneous problem. Then, ap( p − 1) + bp + c = 0. We assume a particular solution of the form y p ( x ) = Ax p ln  x . Inserting this guess into the differential equation, we obtain
( ap( p − 1) + bp + c) Ax p ln  x  + (2ap − a + b) Ax p = Cx p . Since, ap( p − 1) + bp + c = 0, we can solve for A if 2ap − a + b 6= 0, A=
C . 2ap − a + b
Now, consider the problem at hand. There are two forcing terms and due to linearity, we can make the guess y p ( x ) = ( Ax + Bx2 ) ln  x . According to the theory just developed, we have (for a = 1, b = −2, and C = −1, 3 with p = 1, 2, respectively) A
=
B
=
−1 = 1, 2(1)(1) − 1 + (−2) 3 = 3. 2(1)(2) − 1 + (−2)
So, y p ( x ) = ( x + 3x2 ) ln  x . This is the same solution as we had obtained using Variation of Parameters. 8. Verify that the given function is a solution and use Reduction of Order to find a second linearly independent solution. a. x2 y00 − 2xy0 − 4y = 0,
y1 ( x ) = x 4 .
Verification is simple, x2 y00 − 2xy0 − 4y = x2 (12x2 ) − 2x (4x3 ) − 4x4 = 0. For Reduction of Order, we let y2 ( x ) = v( x )y1 ( x ) and determine v( x ). First, compute the derivatives, y2
= x4 v,
y20
= 4x3 v + x4 v0 ,
y200
= 12x2 v + 8x3 v0 + x4 v00 .
Inserting these into the differential equation, we find 6x5 v0 + x6 v00 = 0.
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This is a separable equation for v0 ( x ). Defining z( x ) = v0 ( x ), we obtain a first order, separable equation, 6x5 z + x6 z0 = 0. Solving for z, we have Z dx z = exp −6 = e−6 ln  x = x −6 , x
x > 0.
A further integration gives v( x ) = − 15 x −5 . Finally, we see that y2 ( x ) = x4 v( x ) = x −1 (up to a multiplicative constant). b. xy00 − y0 + 4x3 y = 0,
y1 ( x ) = sin( x2 ).
Let y2 = v sin( x2 ). The derivatives are given by y20
= v0 sin( x2 ) + 2xv cos( x2 ),
y200
= v00 sin( x2 ) + 4xv0 cos( x2 ) − 4x2 v sin( x2 ) + 2v cos( x2 ).
Inserting these into the differential equation, we find
( x sin( x2 ))v00 + (4x2 cos( x2 ) − sin( x2 ))v0 = 0. Let z = v0 . Then,
( x sin( x2 ))z0 + (4x2 cos( x2 ) − sin( x2 ))z = 0. This is a separable first order differential equation. z0 z
=
ln z
=
4x2 cos( x2 ) − sin( x2 ) x sin( x2 ) Z 1 − 4x cot( x2 ) dx x
= ln  x  − 2 ln  sin( x2 ) z = x (sin( x2 ))−2 Since z = v0 , one further integration gives v( x ), v( x ) =
Z
x (sin( x2 ))−2 dx =
1 2
Z
1 csc2 u du = − cot2 ( x2 ). 2
So, up to a multiplicative constant, we have y2 ( x ) = sin( x2 )v( x ) = sin( x2 )
cos( x2 ) = cos( x2 ). sin( x2 )
9. Use the Method of Variation of Parameters to determine the general solution for the following problems. a. y00 + y = tan x.
free fall and harmonic oscillators
The linearly independent solutions of the homogeneous problem are y( x ) = cos x, sin x. So, we consider y p ( x ) = c1 ( x ) cos x + c2 ( x ) sin x. The coefficients satisfy the equations c10 cos x + c20 sin x
= 0,
−c10 sin x + c20 cos x = tan x. Multiplying the first equation by sin x, the second by cos x, and adding the equations, we find c20 = sin x. Thus, c2 ( x ) = − cos x. Substituting this result into the first equation, we have c10 = − tan x sin x. So, c1 ( x )
= −
tan x sin x dx
sin2 x dx cos x Z 1 − cos2 x − dx cos x Z
= − =
Z
= −
Z
(sec x − cos x ) dx
= − ln  sec x + tan x  + sin x. Thus, y p (x)
= c1 ( x ) cos x + c2 ( x ) sin x = (− ln  sec x + tan x  + sin x ) cos x − cos x sin x = − ln  sec x + tan x .
Therefore, the general solution is y( x ) = c1 cos x + c2 sin x − ln  sec x + tan x . b. y00 − 4y0 + 4y = 6xe2x . The homogeneous equation is a constant coefficient equation, y00 − 4y0 + 4y = 0. The characteristic equation is r2 − 4r + 4 = (r − 2)2 = 0. There is one root, r = 2, therefore the general solution of the homogeneous problem is yh ( x ) = (c1 + c2 x )e2x . We seek a particular solution of the form y p ( x ) = c1 ( x )e2x + c2 ( x ) xe2x . The coefficients satisfy the system of equations c10 e2x + c20 xe2x 2c10 e2x
+ c20 (1 + 2x )e2x
= 0, = 6xe2x .
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This system of equations can be solved for c10 and c20 using Cramer’s Rule (from next chapter): 0 xe2x 2x 2x 6xe ( 1 + 2x ) e −6x2 e4x 0 = = −6x2 , c1 = (1 + 2x )e4x − 2xe4x xe2x e2x 2x 2e (1 + 2x )e2x e2x 0 2x 2e 6xe2x 6xe4x 0 = c2 = = 6x. (1 + 2x )e4x − 2xe4x xe2x e2x 2x 2e (1 + 2x )e2x Integrating these results, gives c1 = −2x3 and c2 = 3x2 . Therefore, y p ( x ) = −2x3 e2x + (3x2 ) xe2x = x3 e2x and the general solution is y( x ) = (c1 + c2 x )e2x + x3 e2x . 10. Instead of assuming that c10 y1 + c20 y2 = 0 in the derivation of the solution using Variation of Parameters, assume that c10 y1 + c20 y2 = h( x ) for an arbitrary function h( x ) and show that one gets the same particular solution. We begin with the nonhomogeneous equation a( x )y00 ( x ) + b( x )y0 ( x ) + c( x )y( x ) = f ( x ). The solution of the homogeneous equation can be written in terms of two linearly independent solutions, y h ( x ) = c1 y1 ( x ) + c2 y2 ( x ). Let the particular solution be y p ( x ) = c1 ( x ) y1 ( x ) + c2 ( x ) y2 ( x ). The first derivative is given by y0p ( x ) = c1 ( x )y10 ( x ) + c2 ( x )y20 ( x ) + c10 ( x )y1 ( x ) + c20 ( x )y2 ( x ). At this point we assume that c10 ( x )y1 ( x ) + c20 ( x )y2 ( x ) = h( x ), where h( x ) is an arbitrary function. This gives y0p ( x )
= c1 ( x )y10 ( x ) + c2 ( x )y20 ( x ) + h( x )
y00p ( x )
= c1 ( x )y100 ( x ) + c2 ( x )y200 ( x ) + c10 ( x )y10 ( x ) + c20 ( x )y20 ( x ) + h0 ( x ).
Now, we insert these expressions into the differential equation and use the fact that y1 ( x ) and y2 ( x ) are solutions to the homogeneous problem.
free fall and harmonic oscillators
Then, we have f (x)
= a( x )y00p ( x ) + b( x )y0p ( x ) + c( x )y p ( x ) = a c1 y100 + c2 y200 + c10 y10 + c20 y20 + h0 +b c1 y10 + c2 y20 + h + c [c1 y1 + c2 y2 ] . = c1 ay100 + by10 + cy1 + c2 ay200 + by20 + cy2 + a c10 y10 + c20 y20 + h0 + b [h] = a c10 y10 + c20 y20 + h0 + bh.
Thus, we have the equations c10 y1 + c20 y2
= h, f − bh c10 y10 + c20 y20 = − h0 . a This system of equations can be solved for c10 and c20 using Cramer’s Rule (from the next chapter): h y2 f −bh f −bh a − h0 y20 hy20 − ( a − h0 )y2 0 = c1 = y y y1 y20 − y10 y2 1 2 0 y1 y20
c20
bhy2 fy = − 2+ + aW aW y h 1 0 f −bh 0 y1 a −h = y y 1 2 0 y1 y20
=
y p (x)
(hy2 )0 aW f −bh y1 ( a − h0 ) − y10 h = y1 y20 − y10 y2
f y1 bhy1 (hy1 )0 − − aW aW aW
fy bhy2 (hy2 )0 − 2+ + dξ aW aW aW Z x f y1 bhy1 (hy1 )0 + y2 ( x ) − − dξ aW aW aW Z x Z x f ( ξ ) y1 ( ξ ) f ( ξ ) y2 ( ξ ) dξ − y1 ( x ) dξ + H ( x ), = y2 ( x ) a ( ξ )W ( ξ ) a ( ξ )W ( ξ )
= y1 ( x )
Z x
where H ( x ) = y1 ( x )
Z x bhy
2
aW
+
(hy2 )0 aW
dξ − y2 ( x )
Z x bhy
1
aW
+
(hy1 )0 aW
dξ.
H ( x ) depends on the arbitrary function h( x ) and, more importantly, is independent of f ( x ). Therefore, when f ( x ) = 0, y p ( x ) = H ( x ) is a linear combination of y1 ( x ) and y2 ( x ), H ( x ) = k1 y1 ( x ) + k2 y2 ( x ). So, we can absorb H ( x ) into the homogeneous equation and we obtain the same solution as we had for the derivation of the Method of Variation of Parameters. Z x Z x f ( ξ ) y1 ( ξ ) f ( ξ ) y2 ( ξ ) dξ − y1 ( x ) dξ. y ( x ) = c1 y1 ( x ) + c2 y2 ( x ) + y2 ( x ) a ( ξ )W ( ξ ) a ( ξ )W ( ξ )
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11. Find the solution of each initial value problem using the appropriate initial value Green’s function. For these problems we determine the initial value Green’s function G ( x, ξ ) =
y1 ( ξ ) y2 ( x ) − y1 ( x ) y2 ( ξ ) a ( ξ )W ( ξ )
and then find the solution of the initial value problem using the integral form Z x y( x ) = yh ( x ) + G ( x, ξ ) f (ξ ) dξ, 0
where y1 ( x ), y2 ( x ), and yh ( x ) are solutions of the homogeneous equation satisfying y1 (0) = 0, y2 (0) 6= 0, y10 (0) 6= 0, y20 (0) = 0, yh (0) = y0 , y0h (0) = v0 . a. y00 − 3y0 + 2y = 20e−2x ,
y(0) = 0,
y0 (0) = 6.
First one finds y1 ( x ) and y2 ( x ). The general solution to the homogeneous problem, y00 − 3y0 + 2y = 0, is yh ( x ) = c1 e x + c2 e2x . Requiring y1 (0) = 0 and y20 (0) = 0, we obtain y1 ( x ) = e x − e2x
and
y2 ( x ) = e2x − 2e x .
The Wronskian is given by W ( y1 , y2 )
= y1 y20 − y10 y2 = (e x − e2x )(2e2x − 2e x ) − (e x − 2e2x )(e2x − 2e x ) = −e3x .
and a( x ) = 1. We also need the solution of the homogeneous problem, yh ( x ), which satisfies the given initial conditions, y(0) = 0, y0 (0) = 6. The first condition gives yh ( x ) = c1 (e x − e2x ). For the second condition, we have y0h (0) = c1 (e0 − 2e2(0) ) = 6. Therefore, c1 = −6 and yh ( x ) = 6(e2x − e x ). We construct the Green’s function, G ( x, ξ )
=
y1 ( ξ ) y2 ( x ) − y1 ( x ) y2 ( ξ ) a ( ξ )W ( ξ )
(eξ − e2ξ )(e2x − 2e x ) − (e x − e2x )(e2ξ − 2eξ ) −e3x 2( x − ξ ) x −ξ = e −e . = .
Finally, we obtain the particular solution through integration: y p (x)
=
Z x 0
=
Z x 0
= 20 =
G ( x, ξ ) f (ξ ) dξ
[e2(x−ξ ) − e x−ξ ](20e−2ξ ) dξ
Z x 0
(e2x e−4ξ − e x e−3ξ ) dξ
5 −2x 20 e + 5e2x − e x . 3 3
free fall and harmonic oscillators
The solution to the original problem is
= 6(e2x − e x ) + 5e2x −
y( x )
= 11e2x − b. y00 + y = 2 sin 3x,
y(0) = 5,
20 x 5 −2x e + e 3 3
38 x 5 −2x e + e . 3 3 y0 (0) = 0.
First one finds y1 ( x ) and y2 ( x ). The general solution to the homogeneous problem, y00 + y = 0, is yh ( x ) = c1 cos x + c2 sin x. Requiring y1 (0) = 0 and y20 (0) = 0, we obtain y1 ( x ) = cos x and y2 ( x ) = sin x. The Wronskian is given by W (y1 , y2 ) = y1 y20 − y10 y2 = −1 and a( x ) = 1. We also need yh ( x ) to satisfy the initial conditions, y(0) = 5, y0 (0) = 0. The solution is easily found as yh ( x ) = 5 cos x. We construct the Green’s function, G ( x, ξ )
y1 ( ξ ) y2 ( x ) − y1 ( x ) y2 ( ξ ) a ( ξ )W ( ξ ) cos ξ sin x − cos x sin ξ = sin( x − ξ ).
= =
Finally, we obtain the particular solution through integration: y p (x)
=
Z x
G ( x, ξ ) f (ξ ) dξ
0
=
Z x 0
= 2
[cos ξ sin x − cos x sin ξ ](2 sin 3ξ ) dξ
Z x 0
[cos ξ sin x − cos x sin ξ ][3 sin ξ − 4 sin3 ξ ] dξ
= (3 sin2 x − 2 sin4 x ) sin x − 2 sin3 x cos2 x = sin3 x. The solution to the original problem is y( x ) = 5 cos x + sin3 x. c. y00 + y = 1 + 2 cos x,
y(0) = 2,
y0 (0) = 0.
This problem is similar to the last problem. The Green’s function is G ( x, ξ ) = cos ξ sin x − cos x sin ξ = sin( x − ξ ). The particular solution is then y p (x)
=
Z x 0
=
Z x 0
=
Z x 0
+2
G ( x, ξ ) f (ξ ) dξ
[cos ξ sin x − cos x sin ξ ](1 + 2 cos ξ ) dξ [cos ξ sin x − cos x sin ξ ] dξ Z x 0
[cos2 ξ sin x − cos x sin ξ cos ξ ] dξ
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=
sin ξ sin x + cos ξ cos x + (ξ +
sin 2ξ ) sin x + cos2 ξ cos x 2
x 0
= (1 + x sin x + sin2 x cos x + cos3 x ) − 2 cos x = 1 + x sin x − cos x. The solution to the homogeneous problem [y00 + y = 0, y(0) = 2, y0 (0) = 0] is given by yh ( x ) = 2 cos x. This gives the solution of the nonhomogeneous initial value problem as y( x ) = 1 + cos x + x sin x. d. x2 y00 − 2xy0 + 2y = 3x2 − x,
y(1) = π,
y0 (1) = 0.
We first determine the solution of the CauchyEuler equation, x2 y00h − 2xy0h + 2yh = 0. The characteristic equation is 0 = r (r − 1) − 2r + 2 = r2 − 3r + 2 = (r − 2)(r − 1). Thus, yh ( x ) = c1 x + c2 x2 . We need yh ( x ) to satisfy the given initial conditions, y(1) = π, y0 (1) = 0. Therefore, π
= c1 + c2
0
= c1 + 2c2 .
Subtracting equations, we find c2 = −π = −c1 /2. So, yh ( x ) = π (2x − x2 ). The solution y1 ( x ) = x2 − x satisfies y1 (1) = 0. For y20 (1) = 0, we consider y2 ( x ) = c1 x + c2 x2 . Then, y20 ( x ) = c1 + 2c2 x and this gives y20 (1) = c1 + 2c2 = 0. Choosing c2 = 1 and c1 = −2, we obtain y2 ( x ) = x2 − 2x. The Wronskian is given by
= y1 y20 − y10 y2
W ( y1 , y2 )
= ( x2 − x )(2x − 2) − (2x − 1)( x2 − 2x ) = x2 . and a( x ) = x2 . We construct the Green’s function, G ( x, ξ )
= = =
y1 ( ξ ) y2 ( x ) − y1 ( x ) y2 ( ξ ) a ( ξ )W ( ξ )
(ξ 2 − ξ )( x2 − 2x ) − ( x2 − x )(ξ 2 − 2ξ ) . ξ4 ξx2 − xξ 2 x(x − ξ ) = 4 ξ3 ξ
Finally, we obtain the particular solution through integration: y p (x)
=
Z x
G ( x, ξ ) f (ξ ) dξ Z x x(x − ξ ) (3ξ 2 − ξ ) dξ ξ3 1 1
=
free fall and harmonic oscillators
Z x Z x 3 1 1 = x2 − 2 dξ − x 3− dξ ξ ξ ξ 1 1 x 1 x2 − (3ξ − ln ξ ) x = 3 ln ξ  + ξ 1
= 4x − 4x2 + (3x2 + x ) ln  x . The solution to the original problem is y( x ) = π (2x − x2 ) + 4( x − x2 ) + (3x2 + x ) ln  x . 12. Use the initial value Green’s function for x 00 + x = f (t), x (0) = 4, x 0 (0) = 0, to solve the following problems. As noted in Problem 11b, the initial value Green’s function is given by G (t, τ ) = cos τ sin t − cos t sin τ. The corresponding solution of the homogeneous problem is yh (t) = 4 cos t. For the problems below we only need to find the particular solutions. a. x 00 + x = 5t2 .
y p (t)
Z t
=
0
Z t
=
0
G (t, τ ) f (τ ) dτ
(cos τ sin t − cos t sin τ )(5τ 2 ) dτ
= 5[(τ 2 sin τ − 2 sin τ + 2τ cos τ ) sin t]0t −[(−τ 2 cos τ + 2 cos τ + 2τ sin τ ) cos t]0t = 10 cos t − 10 + 5t2 . The solution is y(t) = 14 cos t − 10 + 5t2 . b. x 00 + x = 2 tan t.
y p (t)
=
Z t 0
G (t, τ ) f (τ ) dτ
Z t
(cos τ sin t − cos t sin τ )(2 tan τ ) dτ ! Z t sin2 τ = 2 sin τ sin t − cos t dτ cos τ 0
=
0
= 2
Z t 0
(sin τ sin t − (sec τ − cos τ ) cos t) dτ
= 2 [− cos τ sin t − (ln  sec τ + tan τ  − sin τ ) cos t]0t = −2 ln  sec t + tan t cos t + 2 sin t The solution to the nonhomogeneous initial value problem is y(t) = 4 cos t + 2 sin t − 2 cos t ln  sec t − tan t.
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13. For the problem y00 − k2 y = f ( x ), y(0) = 0, y0 (0) = 1, The general solution to the homogeneous problem, y00 − k2 y = 0, is yh ( x ) = c1 ekx + c2 e−kx . The solution satisfying the given initial conditions has to satisfy c1 + c2 = 0, k(c1 − c2 ) = 1. Therefore c1 = −c2 = 1/2k. So, yh ( x ) =
1 kx 2k ( e
− e−kx ) =
1 k
sinh kx.
a. Find the initial value Green’s function. The linearly independent solutions needed for construction of the Green’s function are y1 ( x ) = sinh kx and y2 ( x ) = cosh kx. The Wronskian of these solutions is given by W ( y1 , y2 )
= y1 y20 − y10 y2 = (sinh kx )(k sinh kx ) − (k cosh kx )(cosh kx ) = k(sinh2 kx − cosh2 kx ) = −k.
This gives the Green’s function as G ( x, ξ )
= =
y1 ( ξ ) y2 ( x ) − y1 ( x ) y2 ( ξ ) a ( ξ )W ( ξ ) (sinh kξ )(cosh kx ) − (sinh kx )(cosh kξ ) sinh k( x − ξ ) = . −k k
b. Use the Green’s function to solve y00 − y = e− x . For this problem, k = 1 and yh ( x ) = sinh x. The particular solution is y p (x)
=
Z x
G ( x, ξ ) f (ξ ) dξ
0
=
Z x
sinh( x − ξ )e−ξ dξ
0
= = = = =
1 x ( x −ξ ) (e − e(ξ − x) )e−ξ dξ 2 0 Z 1 x ( x−2ξ ) (e − e− x ) dξ 2 0 !x 1 e( x−2ξ ) −x − ξe ) 2 −2 0 x 1 e− x e − xe− x + 2 −2 2 1 1 sinh x − xe− x . 2 2 Z
Then, the general solution is y( x ) =
3 2
− 12 xe− x .
c. Use the Green’s function to solve y00 − 4y = e2x . For this problem, k = 2 and yh ( x ) = tion is y p (x)
=
Z x 0
1 2
sinh 2x. The particular solu
G ( x, ξ ) f (ξ ) dξ
free fall and harmonic oscillators
= = = = = =
1 x sinh 2( x − ξ )e2ξ dξ 2 0 Z 1 x 2( x − ξ ) (e − e2(ξ − x) )e2ξ dξ 4 0 Z 1 x 2x (e − e4ξ −2x ) dξ 4 0 x e4ξ −2x 1 2x ξe − 4 4 0 2x − 1 e e 2x xe2x − + 4 4 4 1 2x 1 xe − sinh 2x. 4 8 Z
Then, the general solution is y( x ) = 14 xe2x + 38 sinh 2x. 14. Find and use the initial value Green’s function to solve x2 y00 + 3xy0 − 15y = x4 e x , y(1) = 1, y0 (1) = 0. We first need the solution of x2 y00h + 3xy0h − 15yh = 0. The characteristic equation is 0 = r (r − 1) + 3r − 15 = r2 + 2r − 15 = (r + 5)(r − 3). Thus, yh ( x ) = c1 x3 + c2 x −5 . We need yh ( x ) to satisfy the given initial conditions, y(1) = 1 and y0 (1) = 0. Therefore, 1
= c1 + c2
0
= 3c1 − 5c2 .
Since c2 = 35 c1 , 1 = c1 + c2 = or c1 =
5 8
8 c , 5 1
and c2 = 83 . So, yh ( x ) =
5 3 3 −5 x + x . 8 8
Next, we construct the Green’s function. The solution y1 ( x ) = x3 − x −5 satisfies y1 (1) = 0. For y20 (1) = 0, we consider y2 ( x ) = c1 x3 + c2 x −5 . Then, y20 ( x ) = 3c1 x2 − 5c2 x −6 and this gives y20 ( x ) = 3c1 − 5c2 = 0 Choosing c2 = 3 and c1 = 5, we obtain y2 ( x ) = 5x3 + 3x −5 . The Wronskian is given by W ( y1 , y2 )
= y1 y20 − y10 y2 = ( x3 − x −5 )(15x2 − 15x −6 ) − (3x2 + 5x −6 )(5x3 + 3x −5 ) = −64x −3 .
and a( x ) = x2 .
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mathematical methods for physicists
D ξ8 8ξ 7 56ξ 6 336ξ 5 1680ξ 4 6720ξ 3 20160ξ 2 40320ξ 40320 0
I
+ − + − + − + − +
eξ
The Green’s function can now be found. We have G ( x, ξ )
=
eξ
= eξ
=
eξ eξ
y1 ( ξ ) y2 ( x ) − y1 ( x ) y2 ( ξ ) a ( ξ )W ( ξ )
(ξ 3 − ξ −5 )(5x3 + 3x −5 ) − ( x3 − x −5 )(5ξ 3 + 3ξ −5 ) . ξ 2 (−64ξ −3 ) 1 x3 ξ4 − . 8 ξ4 x5
Finally, we obtain the particular solution through integration:
eξ
y p (x)
= = =
eξ eξ
Figure R 2.3: Tabular Method for the integral ξ 8 eξ dξ in Problem 14.
G ( x, ξ ) f (ξ ) dξ Z 1 x x3 ξ4 − ξ 4 eξ dξ 8 1 x5 ξ4 Z 1 x ξ8 x3 − 5 eξ dξ 8 1 x Z x 1 3 ξ x 1 x e − 5 ξ 8 eξ dξ. 8 8x 1 1 1
eξ eξ
Z x
= From Table 2.3 we have Z x 1
ξ 8 eξ dξ
= (40320 − 40320ξ + 20160ξ 2 − 6720ξ 3 x +1680ξ 4 − 336ξ 5 + 56ξ 6 − 8ξ 7 + ξ 8 )eξ 1
= (40320 − 40320x + 20160x2 − 6720x3 +1680x4 − 336x5 + 56x6 − 8x7 + x8 )e x − 14833e. Then, y p ( x )
= =
=
Z x 1 3 ξ x 1 ξ 8 eξ dξ x e − 5 8 8x 1 1 1 3 x 1 x (e − e) − 5 [(40320 − 40320x + 20160x2 − 6720x3 8 8x +1680x4 − 336x5 + 56x6 − 8x7 + x8 )e x − 14833e] 14833e e − x3 + (−5040x −5 + 5040x −4 − 2520x −3 + 840x −2 5 8 8x −210x −1 + 42 − 7x + x2 )e x .
The solution to the original problem is y( x )
= =
5 3 3 −5 x + x + y p (x) 8 8 3 + 14833e 5 − e 3 x + (−5040x −5 + 5040x −4 − 2520x −3 + 8 8x5 +840x −2 − 210x −1 + 42 − 7x + x2 )e x .
15. A ball is thrown upward with an initial velocity of 49 m/s from 539 m high. How high does the ball get, and how long does in take before it hits the ground? [Use results from the simple free fall problem, y00 = − g.]
free fall and harmonic oscillators
Starting with y(t) = −4.9t2 + 49t + 539 m, one can compute the time to reach maximum height (v = 0) 0=
dy = −9.8t + 49, dt
or t = 5.0 s. The height at this time is y(5.0) = 760 m. The time in flight is twice the time it takes to reach maximum height. Thus, the ball returns in 10 s. 16. Consider the case of free fall with a damping force proportional to the velocity, f D = ±kv with k = 0.1 kg/s. a. Using the correct sign, consider a 50 kg mass falling from rest at a height of 100 m. Find the velocity as a function of time. Does the mass reach terminal velocity? We start with v˙ = − g − αv, where α = separable, leading to Z
dv g + αv
1 ln  g + αv α g + αv
k m
> 0. This equation is
= −t + C, = −t + C, =
Ae−αt ,
A = eC ,
Using the initial condition v(0) = 0, we find A = g. This gives the solution g v(t) = − (1 − e−αt ). α g
As t → ∞, v → − α . Using the numbers in the problem, we have v(t) = −(98m/s)(1 − e−0.002t ) and vterm = −98 m/s. b. Let the mass be thrown upward from the ground with an initial speed of 50 m/s. Find the velocity as a function of time as it travels upward and then falls to the ground. How high does the mass get? What is its speed when it returns to the ground? As with the first part of the problem, we start with v˙ = − g − αv, where α = mk > 0. After integrating, we found g + αv = Ae−αt . In this case we have v(0) = v0 . Thus, A = g + αv0 and we can solve for the velocity as a function of time, g g v(t) = − + + v0 e−αt . α α Further integration gives the position as a function of time, y ( t ) = y0 −
g 1 g − ( + v0 )e−αt . α α α
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mathematical methods for physicists
For y(0) = 0, y0 =
g+αv0 α2
y(t) =
and gt g + αv0 (1 − e−αt ) − . 2 α α
The maximum height occurs for v(t) g g − + + v0 e−αT = α α g + v0 e−αT = α eαT
=
T
=
= 0. Solving for T, we find 0, g , α αv0 1+ , g 1 αv0 ln 1 + . α g
Inserting this into y(t), we have y( T ) =
g αv v0 − 2 ln 1 + 0 . α g α
Using the values in the problem, we can determine the time to the maximum as T = 5.076 s leading to a height y(5.076) = 127 m. We also need the time it takes to return to the ground. Thus, we seek y(τ ) = 0. This equation is transcendental and one needs technology to solve for this. The value obtained is τ = 10.17 s. The speed at this time is then v(10.17) = −49.7 m/s. 17. A piece of a satellite falls to the ground from a height of 10,000 m. Ignoring air resistance, find the height as a function of time. [Hint: For free fall from large distances, GM h¨ = − . ( R + h )2 ˙ show that Multiplying both sides by h, d 1 ˙2 d GM h = . dt 2 dt R + h ˙ Further integrating gives h(t).] Integrate and solve for h. We begin with GM h¨ = − ( R + h )2 ˙ and multiply both sides by h,
d dt
h¨ h˙
1 ˙2 h 2
GM ˙ h ( R + h )2 d GM . dt R + h
= − =
˙ We can integrate and solve for h. 1 ˙2 h (t) − 2
1 ˙ 2=0 GM h (0) = . 2 R+h
free fall and harmonic oscillators
h˙ = −
r
2GM . R+h
Here we note that the velocity is negative for falling bodies. This equation is separable and can be integrated. Z √
R + h dh
2 ( R + h)3/2 3
Z √ = − 2GM dt
√ = − 2GMt + C.
At t = 0, h(0) = h0 = 10, 000 m. So, C = 23 ( R + h0 )3/2 . Thus,
√ 2 2 ( R + h)3/2 = ( R + h0 )3/2 − 2GMt, 3 3 or
2/3 3√ − R, h(t) = ( R + h0 )3/2 − 2GMt 2
18. The problem of growth and decay is stated as follows: The rate of change of a quantity is proportional to the quantity. The differential equation for such a problem is dy = ±ky. dt The solution of this growth and decay problem is y(t) = y0 e±kt . Use this solution to answer the following questions if 40 percent of a radioactive substance disappears in 100 years. Before answering the questions, one can find the decay constant. Since 0.6 60% is left after 100 years, 0.6y0 = y0 e−100k . Therefore, k = − ln100 = .0051. a. What is the halflife of the substance? The halflife is the time, τ, for which 50% of the initial substance decays. Thus, 0.5y0 = y0 e−kτ . Therefore, −kτ = ln 0.5, or τ=−
ln 0.5 ln2 = = 136 yr. k k
b. After how many years will 90% be gone? For 90% gone, there is 10% left. Therefore, we need to solve 0.1y0 = y0 e−kt for t. The result is t = − ln k0.1 = 451 yr. 19. A spring fixed at its upper end is stretched 6 inches by a 10pound weight attached at its lower end. The springmass system is suspended in a viscous medium so that the system is subjected to a damping force of 5 dx dt lbs. Describe the motion of the system if the weight is drawn down an additional 4 inches and released. What would happen if you changed the coefficient “5” to “4”? [You may need to consult your introductory physics text.] The key equation governing the oscillation of the mass on the spring is m x¨ + b x˙ + kx = 0.
47
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mathematical methods for physicists
We need to determine the constants in the equation. First, we note that adding the block to the spring allows one to determine the spring constant from the equilibrium equation mg = kx. In this problem one needs to be careful with units. The mass is not 10pounds. That is mg. So, k=
mg 10lb = = 20lb/ft. x 0.5 f t
The mass is given by m =
mg g .
In these units the mass is in slugs and
m/s2 .
g = 32.2 [Note that if one takes g = 32 m/s2 , then the qualitative answer will be off for the first part of the problem.] So, m = 0.311 slugs. When the damping force is 5 dx dt lbs, we have b2 − 4km = 52 − 4(20)(.311) > 0. Therefore, the system is overdamped and the mass’s motion will decay monotonically to zero. When the damping force is 4 dx dt lbs, we have b2 − 4km = 42 − 4(20)(.311) < 0. Therefore, the system is underdamped and the mass will oscillate with a decreasing amplitude. 20. Consider an LRC circuit with L = 1.00 H, R = 1.00 × 102 Ω, C = 1.00 × 10−4 F, and V = 1.00 × 103 V. Suppose that no charge is present and no current is flowing at time t = 0 when a battery of voltage V is inserted. Find the current and the charge on the capacitor as functions of time. Describe how the system behaves over time. In this problem we need to solve the equation Lq¨ + Rq˙ +
1 q = V ( t ). C
The given values lead to the initial value problem q¨ + 100q˙ + 10000q = 1000,
q(0) = q˙ (0) = 0.
The solution of the homogeneous problem: The roots of the characteristic equation are r=
−100 ±
√
√ 104 − 4 × 104 = −50 ± 50 3i. 2
Therefore, the solution is given by
√ √ qh (t) = (c1 cos 50 3t + c2 sin 50 3t)e−50t .
free fall and harmonic oscillators
49
The particular solution is relatively simple using the Method of Undetermined Coefficients. Let q p (t) = A. Then, we find y p (t) = 0.10. This gives the general solution to the original problem as
√ √ 1 q(t) = (c1 cos 50 3t + c2 sin 50 3t)e−50t + . 10 Requiring q(0) = 0, gives C1 = −0.1 C. Noting that h√ √ √ √ √ i q˙ (t) = 50 3(c2 cos 50 3t − c1 sin 50 3t) − (c1 cos 50 3t + c2 sin 50 3t) e−50t . and
√ q˙ (0) = 50( 3(c2 ) − (c1 )) = 0, √ we find c2 = c1 /sqrt3 = − 3/30. So, √ √ √ 1 3 1 − ( cos 50 3t + sin 50 3t)e−50t . q(t) = 10√ 10 10 √ 20 3 −50t I (t) = e sin 50 3t. 3 In Figure 2.4 we show the behavior of these solutions. We see that as t → ∞, q(t) → 0.1 C and i (t) → 0 A. 21. Consider the problem of forced oscillations as described in Section 2.7.2. a. Derive the general solution in Equation (2.77). The problem is given by x¨ + ω02 x = Fm0 cos ωt. In the text we solved the ω 6= ω0 case. So, we consider the problem x¨ + ω02 x =
F0 cos ω0 t. m
The solution to the homogeneous problem was found as xh (t) = c1 cos ω0 t + c2 sin ω0 t. Because the driving term is a solution of the homogeneous problem, we need to use the Modified Method of Undetermined Coefficients. The guess x p (t) and its derivatives are given by x p (t)
= t( A cos ω0 t + B sin ω0 t),
x˙ p (t)
= ( A cos ω0 t + B sin ω0 t) + ω0 t(− A sin ω0 t + B cos ω0 t),
x¨ p (t)
= 2ω0 (− A sin ω0 t + B cos ω0 t) − ω02 t( A cos ω0 t + B sin ω0 t).
Then, x¨p + ω02 x p = 2ω0 (− A sin ω0 t + B cos ω0 t) = From this result we see that A = 0 and B = solution for this case is x (t) = c1 cos ω0 t + c2 sin ω0 t +
F0 cos ωt. m
F0 2mω0 .
Therefore, the
F0 t sin ω0 t. 2mω0
Figure 2.4: Plots of the charge and the current as functions of time for Problem 20.
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mathematical methods for physicists
b. Plot the solutions in Equation (2.77) for the following cases: Let c1 = 0.5, c2 = 0, F0 = 1.0 N, and m = 1.0 kg for t ∈ [0, 100]. i. ω0 = 2.0 rad/s, ω = 0.1 rad/s. ii. ω0 = 2.0 rad/s, ω = 0.5 rad/s. iii. ω0 = 2.0 rad/s, ω = 1.5 rad/s. iv. ω0 = 2.0 rad/s, ω = 2.2 rad/s. v. ω0 = 1.0 rad/s, ω = 1.2 rad/s. vi. ω0 = 1.5 rad/s, ω = 1.5 rad/s. In this problem we plot the solution 1 cos ωt, 2 ( ω0 − ω 2 ) x (t) = 0.5 cos ω0 t + 1 t sin ω0 t, 2ω0
ω 6 = ω0 , ω = ω0 .
For the different cases we plot the solutions in Figures 2.52.10 Figure 2.5: Plot from Problem 21b [i.] ω0 = 2 rad/s, ω = 0.1 rad/s.
Figure 2.6: Plot from Problem 21b [ii.] ω0 = 2 rad/s, ω = 0.5 rad/s.
Figure 2.7: Plot from Problem 21b [iii.] ω0 = 2 rad/s, ω = 1.5 rad/s.
c. Derive the form in Equation (2.78). We consider the case that ω 6= ω0 , and choose initial conditions such that c1 = − F0 /(m(ω02 − ω 2 )), c2 = 0. x (t)
= =
F0 F0 cos ωt − cos ω0 t m(ω02 − ω 2 ) m(ω02 − ω 2 ) F0 (cos ωt − cos ω0 t) m(ω02 − ω 2 )
free fall and harmonic oscillators
51
Figure 2.8: Plot from Problem 21b [iv.] ω0 = 2 rad/s, ω = 2.2 rad/s.
Figure 2.9: Plot from Problem 21b [v.] ω0 = 1 rad/s, ω = 1.2 rad/s.
=
F0 ( ω0 + ω ) t ( ω0 − ω ) t ( ω0 − ω ) t ( ω0 + ω ) t − cos cos − + 2 2 2 2 m(ω02 − ω 2 )
=
( ω0 − ω ) t ( ω0 + ω ) t 2F0 sin . sin 2 2 2 2 m ( ω0 − ω )
d. Confirm that the solution in Equation (2.78) is the same as the solution in Equation (2.77) for F0 = 2.0 N, m = 10.0 kg, ω0 = 1.5 rad/s, and ω = 1.25 rad/s, by plotting both solutions for t ∈ [0, 100]. Using these values, the functions obtained are x (t) = 0.581 sin(0.125t) sin(1.375t)
and
x (t) = 0.290(cos(1.25t) − cos(1.5t)).
These solutions are the same and plots of these solutions give the plot as in Figure ??. 22. A certain model of the motion of a light plastic ball tossed into the air is given by mx 00 + cx 0 + mg = 0,
x (0) = 0,
x 0 (0) = v0 .
Here m is the mass of the ball, g=9.8 m/s2 is the acceleration due to gravity and c is a measure of the damping. Since there is no x term, we can write this as a first order equation for the velocity v(t) = x 0 (t) : mv0 + cv + mg = 0.
Figure 2.10: Plot from Problem 21b [vi.] ω0 = 1.5 rad/s, ω = 1.5 rad/s.
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mathematical methods for physicists
Figure 2.11: Plot from Problem 21d.
?? This problem was essentially solved in Problem 16 with α = mc > 0. However, care needs to be exercised in this problem since the ball moves upward and then downward. The resistive force is f d = −cv, where c > 0. When the ball is traveling upward, v > 0 and f d < 0. When the ball is traveling downward, v < 0 and f d > 0. Therefore, the resistive force acts in the correct direction. a. Find the general solution for the velocity v(t) of the linear first order differential equation above. g g + + v0 e−αt . α α b. Use the solution of part a to find the general solution for the position x (t). v(t) = −
gt g + αv0 (1 − e−αt ) − . α α2 c. Find an expression to determine how long it takes for the ball to reach it’s maximum height? x (t) =
1 αv0 T = ln 1 + . α g d. Assume that α = c/m = 5 s−1 . For v0 = 5, 10, 15, 20 m/s, plot the solution, x (t), versus the time. The solutions are shown in Figure 2.12. e. From your plots and the expression in part c, determine the rise time. Do these answers agree? The rise times are v0 (m/s) 5 10 15 20
T (s) 0.2534 0.3617 . 0.4316 0.4833
The values trise = T are confirmed in Figure 2.12.
free fall and harmonic oscillators
53
Figure 2.12: Plot of Position vs. Time for the plastic ball in Problem 22 for α = 5 s−1 and v0 = 5, 10, 15, 20 m/s.
f. What can you say about the time it takes for the ball to fall as compared to the rise time? From the plots we see that the fall time is longer than the rise time. These can be determined using the total flight time. The total flight time is found by numerically solving x (t) = 0 for t f inal . Then, we compute t f all = t f inal − trise using the previous values for the rise time. The values of these times are indicated below. v0 (m/s) 5 10 15 20
t f inal (s) 0.6874 1.2176 1.7303 2.2408
trise (s) 0.2534 0.3617 0.4316 0.4833
t f all (s) 0.4340 0.8559 . 1.2987 1.7575 Figure 2.13: Plot from Problem 23a.
23. Use i) Euler’s Method and ii) the Midpoint Method to determine the given value of y for the following problems. For each problem the Euler and Midpoint Methods were executed and the results plotted in the Figures 2.132.15 with circles designating Euler’s Method and diamonds the Midpoint Method. Plots of the exact solutions are given by the solid curves. The sought values for each problem are provided below. a.
b.
dy = 2y, y(0) = 2. Find y(1) with h = 0.1. dx y( x ) = 2e2x , yexact = 14.78, y Euler = 12.38, y Midpoint = 14.61. dy = x − y, y(0) = 1. Find y(2) with h = 0.2. dx y( x ) = x − 1 + 2e− x , yexact = 1.271, y Euler = 1.215, y Midpoint = 1.275.
Figure 2.14: Plot from Problem 23b.
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mathematical methods for physicists
c.
Figure 2.15: Plot from Problem 23c.
q dy = x 1 − y2 , y(1) = 0. Find y(2) with h = 0.2. dx 2 y( x ) = sin x 2−1 , yexact = 0.998, y Euler = 1.061, y Midpoint = 0.978.
24. Numerically solve the nonlinear pendulum problem using the EulerCromer method for a pendulum with length L = 0.5 m using initial angles of θ0 = 10o , and θ0 = 70o . In each case run the routines long enough and with an appropriate h such that you can determine the period in each case. Compare your results with the linear pendulum period. As shown in the Figure 2.16 there is little difference for 10o . The period of the linear pendulum for this problem gives T = 1.4192 s. From the plots in Figure 2.17 we find five cycles at t = 7.096 s for the linear pendulum and t = 7.815 s for the linear pendulum. This gives T = 1.4192 s for the linear pendulum and t = 1.563 s for the linear pendulum.
Figure 2.16: Plot comparing the nonlinear and linear pendulum for θ0 = 10o in Problem 24.
10.5
10
10 θ (degrees)
θ (degrees)
5
0
9 8.5
5 EulerCromer Method Linear Pendulum
10 0
Figure 2.17: Plot comparing the nonlinear and linear pendulum for θ0 = 70o in Problem 24.
9.5
2
4 t (s)
EulerCromer Method Linear Pendulum
8 6
8
6.8
60
6.9
7
7.1 t (s)
7.2
7.3
7.4
70
20
θ (degrees)
θ (degrees)
40
0 20
65
60
55
40 EulerCromer Method Linear Pendulum
60 0
2
4 t (s)
EulerCromer Method Linear Pendulum
50 6
8
6.8
7
7.2
7.4 t (s)
7.6
7.8
25. For the Baumgartner sky dive we had obtained the results for his position as a function of time. There are other questions that could be asked. a. Find the velocity as a function of time for the model developed in the text. See Figure 2.18. b. Find the velocity as a function of altitude for the model developed in the text. See Figure 2.18.
free fall and harmonic oscillators
55
c. What maximum speed is obtained in the model? At what time and position? The maximum speed is about 390 m/s at an altitude of 27 km and time of 52 s. d. Does the model indicate that terminal velocity was reached? No, it does not. e. What speed is predicted for the point at which the parachute opened? It is roughly 64.01 m/s at 238.8 s, or 1585 m. f. How do these numbers compare with reported data? The data was released in February, 2013 at http://www.redbullsratos. The maximum vertical speed was reported as 1,357.6 kmh. The jump altitude was adjusted as 38,969.4 m and he experienced free fall 36,402.6 m. He reached up to 560 kmh before the parachute opened. At 34s, he was traveling 1115 kmh at 33,446 m. Maximum speed was reached at 50 s, with speed 1,357,6 kmh at 27,833 m. com/science/scientificdatareview/.
0
50
50
100
100
150
150 v (m/s)
v (m/s)
Figure 2.18: The velocity as a function of time and altitude in Problem 25.
Velocity vs Altitude
Velocity vs Time
0
200
200
250
250
300
300
350
350 400
400 0
50
100
150 t (s)
200
250
300
0
10
20 h (km)
30
40
26. Consider the flight of a golf ball with mass 46 g and a diameter of 42.7 mm. Assume it is projected at 30o with a speed of 36 m/s and no spin. We begin with the equations dv x dt dvz dt
= −α(CD v x + CL vz )(v2x + v2z )1/2 , = − g − α(CD vz − CL v x )(v2x + v2z )1/2 .
a. Ignoring air resistance, analytically find the path of the ball and determine the range, maximum height, and time of flight for it to land at the height that the ball had started. In this case, dv x dt dvz dt
= 0, = − g.
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mathematical methods for physicists
Integrating this system yields the projectile motion equations encountered in introductory physics. v x = v x0 , vz = vz0 − gt, 1 z(t) = z0 + vz0 t − gt2 . 2
x (t) = x0 + v x0 t,
The maximum height occurs for vz (t) = 0 which occurs when t = vgz0 . This gives a height, H = z(t) − z0 , of H
= vz0 =
vz0 g
1 − g 2
vz0 g
2
v2z0 . 2g
The time of flight is twice this time, t f = 2vgz0 and the range is given by v2 sin(2θ ) 2vz0 2v v R = v x0 ( ) = x0 z0 = 0 . g g g For the numbers in the problem, we have R = 115 m, H = 16.5 m, and t f = 3.673 s. b. Now consider a drag force f D = 21 CD ρπr2 v2 , with CD = 0.42 and ρ = 1.21 kg/m3 . Determine the range, maximum height, and time of flight for the ball to land at the height that it had started.
z (m)
Figure 2.19: The flight of the golf ball in Problem 26.
10
CD = 0 CL =0 CD = 0.42 CL =0
0 0
20
40
60
80
100
x (m)
6
Flight with drag needs to be solved numerically. In Figure 2.19 are plots for this problem and the previous part of the problem. Reading from the plot we have R = 115 m and H = 16.5 m.
Reynolds Number vs Time
x 10 4 3.5
t f = 3.673 s.
Re
3 2.5
c. Plot the Reynolds number as a function of time. [Take the kinematic viscosity of air, ν = 1.47 × 10−5 .]
2 1.5 1 0
1
2 t (s)
3
4
Figure 2.20: The plot of the Reynolds number as a function of time for the flight of the golf ball in Problem 26. Change Part 26d.
d. Based on the plot in part c, create a model to incorporate the change in Reynolds number and repeat part b. Compare the results from parts a, b, and d. The Reynolds number is too high to make any difference. 27. Consider the flight of a tennis ball with mass 57 g and a diameter of 66.0 mm. Assume the ball is served 6.40 m from the net at a speed of 50.0 m/s down the center line from a height of 2.8 m. It needs to just clear the net (0.914 m).
free fall and harmonic oscillators
57
a. Ignoring air resistance and spin, analytically find the path of the ball assuming it just clears the net. Determine the angle to clear the net and the time of flight. The path of the ball just clearing the net is shown in Figure 2.21. In order to find the angle and time for the ball to clear the net, we use the projectile motion equations from the last problem with v x0 = v0 cos θ and vz0 = v0 sin θ : x (t) = x0 + (v0 cos θ )t,
1 z(t) = z0 + (v0 sin θ )t − gt2 . 2
For the ball to barely clear the net, we set x (t) = 6.40 m and z(t) = 0.914 m. Then, 6.40 0.914
= (50 cos θ )t 1 = 2.8 + (50 sin θ )t − gt2 . 2
A solution gives t = 0.1330 s and θ = −0.27408 rad. b. Find the angle to clear the net assuming the tennis ball is given a topspin with ω = 50 rad/s. We turn to the system of equations for two dimensional motion incorporating both drag and lift. These are given by dv x dt dvz dt
= −α(CD v x + CL vz )(v2x + v2z )1/2 , = − g − α(CD vz − CL v x )(v2x + v2z )1/2 .
For this part we set CD = 0. One can obtain CL from ω using CL =
1 2+
v vspin
,
q where vspin = rω. However, v = v2x0 + (vy0 − gt)2 varies from 50 to 50.5 m/s over the short time the ball is in the air. Then, CL =
1 v 2 + rω
varies from 0.0310 to 0.0307 for this range of speeds. So, taking CL = −0.031 for top spin, we numerically solve the system. In Figures 2.222.23 we indicate the paths for topspin and bottom spin for an angle θ = −0.27408. Numerically changing the initial angle one can have the ball with spin just clear the net. For the topspin in this part of the problem, we find θ = −0.2702 radians. c. Repeat part b assuming the tennis ball is given a bottom spin with ω = 50 rad/s. The only change from part b is letting CL = 0.031. For the bottom spin in this part of the problem, we find θ = −0.2778 radians. This is shown in the plots in Figure 2.23.
2.18m 0.914m 6.40m Figure 2.21: In Problem 2.27 a ball is served 6.40 meters from the net from a height of 2.8 m.
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mathematical methods for physicists
Figure 2.22: The flight of the tennis ball in Problem 27 as it clears the net for CD = 0.
CD = 0 CL =0 CD = 0 CL =0.031 CD = 0 CL =0.031
z (m)
2 1 0
1 z (m)
Figure 2.23: The (zoomed in) flight of the tennis ball in Problem 27 as it clears the net for CD = 0.
2
4 x (m)
CD = 0 CL =0 CD = 0 CL =0.031 CD = 0 CL =0.031
0.8 0.6
6
8
0.95 z (m)
0
CD = 0 CL =0 CD = 0 CL =0.031 CD = 0 CL =0.031
0.9
0.4 5.5
6
6.5
7
7.5
8
0.85 6.3
6.35
6.4
x (m)
6.45 6.5 x (m)
6.55
6.6
6.65
d. Repeat parts a, b, and c with a drag force, taking CD = 0.55. In Figures 2.242.25 we indicate the paths for topspin and bottom spin for an angle θ = −0.27408 and CD = 0.55. This angle no longer works without spin. Numerically changing the initial angle one can have the ball with drag just clear the net. For the topspin in the problem, we find θ = −0.2692 radians. For no spin in the part of the problem, we find θ = −0.2728 radians. For the bottom spin in the problem, we find θ = −0.2766 radians. Figure 2.24: The flight of the tennis ball in Problem 27 as it clears the net for CD = 0.55.
CD = 0.55 CL =0 CD = 0.55 CL =0.031 CD = 0.55 CL =0.031
z (m)
2 1 0 0
1
2
3
4 x (m)
5
6
7
28. In Example 2.32 a(t) was determined for a curved universe with nonrelativistic matter for Ω0 > 1. Derive the parametric equations for Ω0 < 1, a
=
t
=
Ω0 (cosh η − 1), 2(1 − Ω0 ) Ω0 (sinh η − η ) , 2H0 (1 − Ω0 )3/2
for η ≥ 0. We begin with the Friedman equation in the form r Ω0 a˙ = ± H0 + (1 − Ω0 ). a Let α =
1− Ω0 Ω0 .
We will see that only the positive sign applies in this case, so
free fall and harmonic oscillators
Figure 2.25: The (zoomed in) flight of the tennis ball in Problem 27 as it clears the net for CD = 0.55
1.2 CD = 0.55 CL =0 CD = 0.55 CL =0.031 CD = 0.55 CL =0.031
0.8
0.95 z (m)
z (m)
1
0.6
CD = 0.55 CL =0 CD = 0.55 CL =0.031 CD = 0.55 CL =0.031
0.9
59
0.4 6
6.5
7 x (m)
7.5
0.85 6.3
8
6.35
6.4
6.45 6.5 x (m)
6.55
6.6
6.65
the equation becomes Ω0 √ 1 + αa. a This differential equation is separable and can be integrated with a hyperbolic function substitution. We let r
a˙ = H0
αa = sinh2 u,
α du = 2 sinh u cosh u du.
Then, we have
√
H0
p
Ω0 dt
a 1 + αa sinh u(2 sinh u cosh u du) √ α α cosh u
√
= =
= 2α−3/2 sinh2 u du = α−3/2 [cosh 2u − 1]. Integrating, H0
p
Ω0 t + C =
1 sinh u − u. 2
For t = 0, a(0) = 0, therefore u(0) = 0. Defining η = 2u, we have the parametric form for t : t
= =
1 (sinh η − η ) 2H0 Ω0 α3/2 ω0 (sinh η − η ). 2H0 (1 − Ω0 )3/2
√
the parametric form for a can be obtained from the original hyperbolic substitution: 1 η a = sinh2 α 2 ω0 = (cosh η − 1). 2(1 − Ω0 )
29. Find numerical solutions for other models of the universe.
25 20 a
In Figure 2.30 we give both the exact and the numerical solution for Ω0 = 0.8, 1.1.
30
15 10 5
a. A flat universe with nonrelativistic matter only with Ωm,0 = 1. In Figure 2.26 we show the numerical solution for this problem. The exact solution to this problem was found as !2/3 t a(t) = 2 . 3 H0
0 0
20
40
60
80
100
t
Figure 2.26: A plot of a(t) vs t for a flat universe with nonrelativistic matter only in Problem 2.29a.
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mathematical methods for physicists
b. A curved universe with radiation only with curvature of different types. In Figure 2.27 we show the numerical solution for this problem. The exact solution can also be determined. We insert ΩΛ,0 = 0 = Ωm,0 and Ωr,0 = Ω0 , into the Friedmann equation, 2 Ωr,0 Ωm,0 a˙ 1 − Ω0 + Ω + . = H02 + Λ,0 a a3 a2 a4
600 500 Ω0 = 0.8
a
400
and obtain
r
300
a˙ = ± H0
Ω0 = 0
200
Ω0 + 1 − Ω0 . a2
This equation is separable and can be integrated,
100 Ω0 = 1.1
0 0
20
40
60
80
100
H0 t = ±
t
Figure 2.27: A plot of a(t) vs t for a curved universe with radiation only in Problem 2.29b.
a da
Z
p
Ω0 + (1 − Ω0 ) a2
.
For the case Ω0 < 1, we have (a˙ > 0) H0 t
= =
where α =
1− Ω0 Ω0
1 a da √ √ , Ω0 1 + αa2 1 p √ 1 + αa2 + C, α Ω0 Z
> 0. For a(0) = 0, H0 t =
i hp 1 1 + αa2 − 1 . 1 − Ω0
For the case Ω0 > 1, we have the possibility that a˙ changes sign. Thus, we have s Ω0 amax = Ω0 − 1 and for a < amax , 1.4
H0 t
=
1.2
1 √ Ω0
Z a 0
a da q
1
a
0.8
=
" a2max √ 1− Ω0
=
√
0.6 0.4 0.2
1 Ω0 − 1
1− s
a2 a2max
1−
,
a2
#
a2max q [ am ax − a2max − a2 ].
0 0
0.2
0.4
0.6
0.8
1
t
Figure 2.28: A plot of a(t) vs t for a flat universe with nonrelativistic matter and radiation in Problem 2.29c.
c. A flat universe with nonrelativistic matter and radiation with several values of Ωm,0 and Ωr,0 + Ωm,0 = 1. In Figure 2.28 we show the numerical solution for this problem with the exact solution superimposed. The exact solution can be found in Ryden’s book, Introduction to Cosmology (2003).
free fall and harmonic oscillators
61
We begin with the Friedmann equation in the form 2 Ωm,0 a˙ 1 − Ω0 2 Ωr,0 + 3 + ΩΛ,0 + = H0 . a a a2 a4 For this problem we have ΩΛ,0 = 0 and Ω0 = 1. Therefore, 2 Ωm,0 a˙ 2 Ωr,0 = H0 + 3 . a a a4 Defining α =
Ωr,0 Ωm,0 ,
we have 2 Ω h ai a˙ 1 + . = H02 r,0 a α a4
Since the differential equation is separable, we write H0 dt
= p
=
a da q , Ωr,0 1 + αa
a u = 1+ , α
α2 u − 1 p √ du. u Ωr,0
Integrating with the condition a(0) = 0, we have
= = =
α2 p Ωr,0
Z 1+ a/α u−1
√
du. u α2 a 3/2 a 1/2 4 2 p 1+ −2 1+ + α α 3 Ωr,0 3 2 a 1/2 a 4α p 1+ 1+ −1 . α 2α 3 Ωr,0 1
2500
2000
1500 a
H0 t
1000
d. Look up the current values of Ωr,0 , Ωm,0 , ΩΛ,0 , and κ. Use these values to predict future values of a(t).
e. Investigate other types of universes of your choice, but different from the previous problems and examples. In Figure 2.30 we show the numerical solution for a simple case of nonrelativistic matter plus curvature. The solutions were presented in the text with the Ω0 < 1 case derived in the Problem 28. In Figure 2.30 we also show the exact solution superimposed. 30. Consider the system
0 0
2
4
6
8
Figure 2.29: A plot of a(t) vs t for the numerical solution for Ω0 = 1, Ωr,0 = 8.4 × 10−5 , Ωm,0 = 0.30 in Problem 2.29d. 50 Ω0 = 0.8
40
30
20 Ω0 = 1.1
10
x 0 = −4x − y, y0 = x − 2y. a. Determine the second order differential equation satisfied by x (t).
10
t
a
In Figure 2.29 we show the numerical solution for Ω0 = 1, Ωr,0 = 8.4 × 10−5 , Ωm,0 = 0.30, and ΩΛ,0 = Ω0 − (Ωr,0 + Ωm,0 ).
500
0 0
20
40
60
80
100
t
Figure 2.30: A plot of a(t) vs t for a flat universe with nonrelativistic matter and curvature for Problem 2.29d.
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mathematical methods for physicists
Differentiate the first equation with respect to t and then replace y0 = x − 2y and y = − x 0 − 4x to find x 00
= −4x 0 − y0 = −4x 0 − ( x − 2y) = −4x 0 − ( x − 2(− x 0 − 4x )) = −6x 0 − 9x.
Therefore, x 00 + 6x 0 + 9 = 0. b. Solve the differential equation for x (t). The characteristic equation is r2 + 6r + 9 = 0, whose solution is r = −3. This has the general solution x (t) = (c1 + c2 t)e−3t . c. Using this solution, find y(t). Using the first equation of the system, we find y(t) as y(t)
= − x 0 − 4x = −(c2 − 3(c1 + c2 t))e−3t − 4(c1 + c2 t)e−3t = (−c1 − c2 − c2 t)e−3t .
So, y(t) = (−c1 − c2 − c2 t)e−3t . d. Verify your solutions for x (t) and y(t). Insert x (t) = (c1 + c2 t)e−3t and y(t) = (−c1 − c2 − c2 t)e−3t into the system of differential equations. Then we have x0
= (c2 − 3(c1 + c2 t))e−3t = (3c1 − 3c2 t + c2 )e−3t .
−4x − y = −4(c1 + c2 t)e−3t − (−c1 − c2 − c2 t)e−3t = (3c1 − 3c2 t + c2 )e−3t . y0
= (2c2 − 3(−c1 − c2 − c2 t))e−3t = (3c1 + 2c2 + 3c2 t)e−3t .
− x − 2y = −(c1 + c2 t)e−3t − 2(−c1 − c2 − c2 t)e−3t = (3c1 + 2c2 + 3c2 t)e−3t . e. Find a particular solution to the system given the initial conditions x (0) = 1 and y(0) = 0. From x (t) = (c1 + c2 t)e−3t , x (0) = c1 = 1. From y(t) = (−c1 − c2 − c2 t)e−3t , y(0) = −c1 − c2 = −1 − c2 = 0. Therefore, c2 = −1. Thus, the particular solution to the system is x (t) = (1 − t)e−3t and y(t) = te−3t .
free fall and harmonic oscillators
63
31. Consider the following systems. Determine the families of orbits for each system and sketch several orbits in the phase plane and classify them by their type (stable node, etc.). a. x0
= 3x,
y0
= −2y.
This is an uncoupled systems and is easily solved: x (t) = c1 e3t , y(t) = c2 e−2t . For initial conditions in which c1 = 0, solutions tend to the origin, otherwise they tend to infinity. Thus, this system has an unstable saddle as seen in Figure 2.31. One could also obtain a family solution curves dy y0 2y = 0 =− . dx x 3x Integrating this separable first order equation, Z
dy y
ln y we find y =
A . x2/3
2 dx , 3 x 2 − ln  x  + C, 3
= − =
Z
Sketching these curves also leads to a saddle.
b. x0 y
0
= −y, = −5x.
Differentiating the first equation, x 00 = −y0 = 5x. The roots of the √ characteristic equation are r = ± 5. Thus, the solution for x is √
x ( t ) = c1 e
5t
+ c2 e −
√
5t
Figure 2.31: The direction field for the system in Problem 31a.
.
Inserting this solution for x (t) back into the system, gives √ √ √ y(t) = − 5c1 e 5t − c2 e− 5t . √ Based upon the solutions, or that the roots are r = ± 5, we see that this system has an unstable saddle. This can also be seen in Figure 2.32 One could also obtain a family solution curves dy y0 x = 0 =5 . dx x y Integrating this separable first order equation, Z
y dy
y2 + C
= 5
Z
Figure 2.32: The direction field for the system in Problem 31b.
x dx,
= 5x2 , (2.1)
64
mathematical methods for physicists
we find 5x2 − y2 = C. This is a family of hyperbolae for real C, showing that the equilibrium point is a saddle point. c. x0
= 2y,
y0
= −3x.
Differentiating the first equation, x 00 = 2y0 = −6x. The roots of the √ characteristic equation r2 + 6 = 0, are r = ±i 6. Thus, the solution for x is √ √ x (t) = c1 cos 6t + c2 sin 6t. Inserting this solution into the system of differential equations, gives √ √ √ 5 y(t) = c2 cos 6t − c1 sin 6t . 2 Thus, this system has solutions that follow elliptical paths. One could also obtain a family solution curves y0 3x dy = 0 =− . dx x 2y Integrating this separable first order equation, we find 3x2 + 2y2 = C. This is a family of ellipses for C > 0 showing that the equilibrium point is a center as seen in Figure 2.33. d. x0 y
Figure 2.33: The direction field for the system in Problem 31c.
0
= x − y, = y.
The second equation can be solved directly. y(t) = c1 et . Inserting this into the first equation, x 0 − x = − c1 e t . This can be solved using the integrating factor µ = e−t . Then, (e−t x )0 = −c1 . Thus, x ( t ) = ( c2 − c1 t ) e t . In Figure 2.34 the direction field for this system is shown. This system is similar to Example 2.43, which has a line of unstable equilbria. e. x0
Figure 2.34: The direction field for the system in Problem 31d.
y
0
= 2x + 3y, = −3x + 2y.
free fall and harmonic oscillators
65
Differentiating the first equation, x 00
= 2x 0 + 3y0 = 2x 0 + 3(−3x + 2y) = 2x 0 − 9x + 2( x 0 − 2x ) = 4x 0 − 13x
So, we need to solve x 00 − 4x 0 + 13x = 0. The roots of the characteristic equation are r = 2 ± 3i. So, x (t) = (c1 cos 3t + c2 sin 3t)e2t . Inserting this into the system, we find y(t) = (c2 cos 3t − c1 sin 3t)e2t . The direction field plot in Figure 2.35 as well as the solution indicate that the orbits are spirals. This problem could also be approached in polar coordinates. Recall that 0
=
θ0
=
r
xx 0 + yy0 , r 0 0 xy − yx . r2
The radial equation is obtained by multiplying the first equation by x, the second equation by y, xx 0
= 2x2 + 3xy,
yy0
= −3xy + 2y2 ,
and then adding these expressions to obtain rr 0 = xx 0 + yy0 = 2( x2 + y2 ) = 2r2 . Similarly, the equation for θ is obtained by multiplying the first equation by y, the second equation by x, xy0
= −3x2 + 2xy,
yx 0
= 2xy + 3y2 ,
and subtracting, to find r2 θ 0 = xy0 yx 0 = −3( x2 + y2 ) = −3r2 . As a result we have the system in polar as r 0 = 2r and θ 0 = −3. The second equation indicates that the spirals move in a clockwise direction as time increases. The radial equation indicates that the spiral is exponentially growing since the solution is r = r0 e2t . This is consistent with the previous analysis of the problem.
Figure 2.35: The direction field for the system in Problem 31e.
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mathematical methods for physicists
32. Use the transformations relating polar and Cartesian coordinates to prove that dθ 1 dy dx = 2 x −y . dt dt dt r We begin with tan θ = time,
y x.
Differentiating this expression with respect to
sec2 θ θ 0 =
y0 x − yx 0 . x2
Note that sec2 θ = 1 + tan2 θ =
x 2 + y2 . x2
This gives y0 x − yx 0 θ = x2 0
x2 x 2 + y2
=
y0 x − yx 0 . r2
33. Consider the system of equations in Example 2.46. a. Derive the polar form of the system. The system of equation under consideration is x0
= − y + x (1 − x 2 − y2 )
y0
= x + y (1 − x 2 − y2 ).
We compute the forms xx 0 + yy0
= − xy + x2 (1 − x2 − y2 ) + ( xy + y2 (1 − x2 − y2 )) = ( x2 + y2 )(1 − x2 − y2 ).
xy0 − yx 0
= x2 + xy(1 − x2 − y2 ) − (−y2 + xy(1 − x2 − y2 )) = ( x 2 + y2 ).
These reduce to rr 0 r2 θ 0
= r 2 (1 − r 2 ), = r2 ,
or r0
= r (1 − r 2 ),
θ0
= 1,
b. Solve the radial equation, r 0 = r (1 − r2 ), for the initial values r (0) = 0, 0.5, 1.0, 2.0. Next, we solve the radial problem. Note that r (t) = 0, 1 are equilibrium solutions (r 0 = 0.) So, we consider solutions with r (0) = r0 6= 0, 1. The equation is separable, leading to t+C =
Z
dr . r (1 − r 2 )
free fall and harmonic oscillators
67
The integration is done using a partial fraction decomposition. Let A 1 B C A(1 − r2 ) + Br (1 + r ) + Cr (1 − r ) = + . + = 2 r 1−r 1+r r (1 − r ) r (1 − r 2 ) Solving for the constants, we find A = 1, B = −C = 12 . Finishing the integration, t+C
= ln r + = ln p
Aet
=
A2 e2t
=
1 1 ln(1 + r ) − ln 1 − r  2 2 r
1 − r 2  r
p
1 − r 2  r2 . 1 − r 2 
Solving for r2 (t), we find that for 0 < r < 1, r2 (t) =
1 , 1 + Ce−2t
r2 (t) =
1 . 1 − Ce−2t
and for r > 1,
Using the initial conditions, one can obtain C in terms of r0 . We find that for 0 < r < 1, r (t) = q
1 1 + ( r12 0
− 1)e−2t
= q
r0 r02 + (1 − r02 )e−2t
,
and for r > 1, r (t) = q
1 1 − (1 −
1 −2t )e r02
= q
r0 r02
− (r02 − 1)e−2t
.
Solutions for the initial values r (0) = 0.5, 1.0, 2.0 are shown in the figure. c. Based upon these solutions, plot and describe the behavior of all solutions to the original system in Cartesian coordinates. Typical solutions are shown in Figure 2.36. These indicate that the origin, r = 0, is unstable and the unit circle, r = 1, is stable. The unit circle is a stable limit cycle with nearby orbits spiraling in towards the orbit.
Figure 2.36: Solutions for the initial conditions in Problem 33.
3 Linear Algebra 1. Express the vector v = (1, 2, 3) as a linear combination of the vectors a1 = (1, 1, 1), a2 = (1, 0, −2), and a3 = (2, 1, 0). The linear combination would take the form
(1, 2, 3) = a(1, 1, 1) + b(1, 0, −2) + c(2, 1, 0) = ( a + b + 2c, a + c, a − 2b). Thus, one needs to solve the system of equations a + b + 2c
= 1
a+c
= 2
a − 2b
= 3.
Subtracting the second and third equations from the first, gives b+c
= −1
3b + 2c
= −2.
Solving these equations gives b = 0, c = −1, and therefore a = 2 − c = 3. The linear combination is then (1, 2, 3) = 3(1, 1, 1) − (2, 1, 0). 2. A symmetric matrix is one for which the transpose of the matrix is the same as the original matrix, A T = A. An antisymmetric matrix is one that satisfies A T = − A. a. Show that the diagonal elements of an n × n antisymmetric matrix are all zero. The diagonal elements of A are aii , i = 1, 2, . . . , n. The diagonal elements of A T are aii , i = 1, 2, . . . , n. Since A T = − A, we have aii = − aii , i = 1, 2, . . . , n. This is only true if aii = 0, i = 1, 2, . . . , n. b. Show that a general 3 × 3 antisymmetric matrix has three independent offdiagonal elements. From part a we know that a general has zero diagonal elements, 0 a12 A = a21 0 a31 a32
3 × 3 antisymmetric matrix a13 a23 . 0
70
mathematical methods for physicists
However, by definition A T = − A, so 0 a21 a31 0 a12 0 a32 . = − a21 a13 a23 0 − a31
− a12 0 − a32
− a13 − a23 . 0
We see that a21 = − a12 , a31 = − a13 , and a32 = − a23 . Thus, we can see that a general 3 × 3 antisymmetric matrix has three independent elements and takes the form 0 a12 a13 A = − a12 0 a23 . − a13 − a23 0 c. How many independent elements does a general 3 × 3 symmetric matrix have? By a similar argument we have matrix takes the form. a11 A = a12 a13
that a general 3 × 3 symmetric a12 a22 a23
a13 a23 . a33
It has three more independent elements (along the diagonal) than an antisymmetric matrix The sum, S = ∑nk=1 k, is a wellknown sum. Writing out this sum, S = 1 + 2 + . . . + (n − 1) + n. writing the sum in reverse, S = n + (n − 1) + . . . + 2 + 1, and adding these sums term by term, we find 2S
= =
( n + 1) + ( n + 1) + . . . + ( n + 1)  {z } n terms n ( n + 1).
This gives S =
n ( n +1) . 2
d. How many independent elements does a general n × n symmetric matrix have? A general n × n symmetric matrix takes the form a11 a12 . . . a1n a12 a22 . . . a2n A= .. .. .. . .. . . . . a1n
a2n
...
ann
The number of independent elements are the n diagonal elements plus the number of entries in the upper triangular portion of the matrix. There are n independent elements in the first row, n − 1 in the second row, etc. According to the side note, the total is n
∑k=
k =1
n ( n + 1) . 2
e. How many independent elements does a general n × n antisymmetric matrix have? A general n × n antisymmetric matrix takes the form 0 a12 . . . a1n 0 . . . a2n − a12 A= .. .. .. .. . . . . .
− a1n
− a2n
...
0
linear algebra
It does not have independent elements along the diagonal, so there are n fewer independent elements than for a symmetric matrix. This gives n ( n + 1) n2 + n − 2n n ( n − 1) −n = = 2 2 2 independent elements for an antisymmetric matrix. 3. Consider the matrix representations for twodimensional rotations of vectors by angles α and β, denoted by Rα and R β , respectively. In this problem we are considering active rotations in two dimensions which are represented by the rotation matrices ! cos α − sin α Rα = . sin α cos α 1 T a. Find R− α and Rα . How do they relate?
The transpose is given by RαT =
cos α − sin α
sin α cos α
! .
Rα is a 2 × 2 matrix and the inverse is easily obtained as ! cos α sin α −1 Rα = − sin α cos α since det Rα = 1. 1 T T Note that R− α = Rα , or Rα Rα = I. This is one of the properties of a rotation matrix, namely that Rα is an orthogonal matrix. [The other condition is det Rα = 1.]
b. Prove that Rα+ β = Rα R β = R β Rα . We see that Rα R β
= =
R β Rα
= =
cos α sin α
− sin α cos α
!
cos β sin β
− sin β cos β
!
cos α cos β − sin α sin β − cos α sin β − sin α cos β sin α cos β + cos α sin β − sin α sin β + cos α cos β ! ! cos β − sin β cos α − sin α sin β cos β sin α cos α
!
cos β cos α − sin β sin α cos β sin α + sin β cos α
!
− sin β cos α − cos β sin α − sin β sin α + cos β cos α
Therefore, we have Rα R β
= =
R β Rα cos(α + β) sin(α + β)
− sin(α + β) cos(α + β)
!
= Rα+ β .
71
72
mathematical methods for physicists
4. Consider the matrix A=
1 2 − √1 2 1 2
√1 2
0
− √1
2
1 2 √1 2 1 2
.
a. Verify that this is a rotation matrix. One needs to show that this is an orthogonal A T A = I and that detA = 1. First of all, 1 1 1 1 √1 − √1 2 2 2 2 2 2 √1 √1 √1 √1 − 0 0 − A= 2 2 2 2 1 1 1 1 √1 √1 − 2 2 2 2 2
detA
=
= = =
2
1 1 √1 2 2 2 − √1 √1 0 2 2 1 1 − √1 2 2 2 √1 1 0 1 − √12 √12 2 − √ 1 1 2 − √1 2 12 2 2 2 1 1 1 1 1 1 −√ −√ + 2 2 2 2 2 2 1.
matrix, AA T =
1 = 0 0
1 − √1 + 1 2 2 2
0 0 . 1
0 1 0
0
− √1
2
b. Find the angle and axis of rotation. The idea that this is doable is known as Euler’s Theorem for orthogonal matrices. Namely, in three dimensions the composition of rotations about different axes can be written as a rotation about a single direction. In this case it means that one can seek a direction v which is left invariant under the rotation, Av = v. Thus, there exists a real eigenvalue λ = 1 of A. We verify this as
=
=
det( A − λI ) 1 −1 √1 2 2 2 1 − √1 √ −1 2 2 1 − √1 − 12 2 2 √1 1 −1 1 − √12 2 − − √ 1 2 − √1 − 12 2 2 2
= =
1 1 1 − (1) − √ (0) + (1) 2 2 2 0.
√1 2 − 12
1 − √1 + 1 2 2 2
−1 − √1 2
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Next, we seek the fixed direction of rotation, the eigenvector corresponding to λ = 1. The eigenvector v = (v1 , v2 , v3 ) satisfies the equation Av = v, which gives the system 1 1 1 − v1 + √ v2 + v3 2 2 2 1 1 − √ v1 − v2 + √ v3 2 2 1 1 1 v − √ v2 − v3 2 1 2 2
= 0 = 0 = 0.
√ Dividing the middle equation by 2 and adding to the third equation, we find v2 = 0. The remaining equations reduce to one equation, v1 = v3 . Therefore, any eigenvector is proportional to v = (1, 0, 1). For any of the simple three dimensional rotations about the coordinate axes in the text, we note that the trace is trRα = 1 + 2 cos α. The trace is invariant under rotations. Another approach is to first note that the eigenvalues of a rotation matrix (orthogonal, det R = 1) are 1, eiα , and e−iα . The diagonalized form is 1 0 0 Rˆ v (α) = 0 eiα 0 . 0 0 e−iα Again, the trace is given by tr Rˆ v (α) = 1 + eiα + e−iα = 1 + 2 cos α. In this problem tr Rˆ v (α) = 1. Therefore, cos α = 0, or α = ±π/2. Thus, this matrix represents a rotation of 90o about the axis v = (1, 0, 1) either clockwise or counterclockwise. The direction still needs to be determined. It should be noted that this transformation can be performed through the Euler transformation A
= =
π Rˆ (0.6154797087, , 0.6154797087) 3 π ˆ Rz (0.6154797087) Rˆ x ( ) Rˆ z (0.6154797087). 3
c. Determine the corresponding similarity transformation using the results from part b. We can determine a similarity transformation that converts A into a rotation about the z−axis: Namely, SAS−1 = Rˆ z . A represents a rotation of α about the axis v = (1, 0, 1). We next need to pick two vectors orthogonal to v1 . Let these vectors be v2 = (0, 1, 0) and v3 = (1, 0, −1). We construct a matrix using these vectors as
In Example 1.37 and in Chapter 7 we introduce eiα = cos α + i sin α.
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the columns. The resulting matrix is an orthogonal matrix. If these vectors are first normalized, then the matrix will be a rotation matrix, S. Then, we find that SAS−1 = 1 1 1 √1 √ √1 √1 0 0 √1 2 2 2 2 2 2 2 − √1 √1 0 0 0 1 0 0 1 2 2 1 1 √1 √1 0 − √1 0 √1 − √1 2 2 2 2 2 2 2 1 0 0 = 0 0 −1 = R z . 0 1 0 Noting that this matrix takes the form 1 0 Rˆ x (θ ) Rˆ z (φ) = 0 cos θ 0 − sin θ
0
sin θ cos θ
with θ = − π2 . This indicates a rotation about the vector v = (1, 0, 1) of α = −π/2. 5. Consider the matrix
−0.8124 −0.5536 −0.1830 A = −0.3000 0.6660 −0.6830 . 0.5000 −0.5000 −0.7071
We have chosen the definition presented in the text, Rˆ (φ, θ, ψ) = Rˆ z (ψ) Rˆ x (θ ) Rˆ z (φ), which agrees with classics like Goldstein’s Classical Mechanics (1965).
This matrix represents the active rotation through three Euler angles. Determine the possible angles of rotation leading to this matrix. The three dimensional Euler rotation matrix Rˆ (φ, θ, ψ) = Rˆ z (ψ) Rˆ x (θ ) Rˆ z (φ) needs to be computed. We first compute cos φ sin φ 0 1 0 0 Rˆ x (θ ) Rˆ z (φ) = 0 cos θ sin θ − sin φ cos φ 0 0 0 1 0 − sin θ cos θ cos φ sin φ 0 = − cos θ sin φ cos θ cos φ sin θ sin θ sin φ − sin θ cos φ cos θ Rˆ (φ, θ, ψ)
=
=
Rˆ z (ψ) Rˆ x (θ ) Rˆ z (φ) cos ψ sin ψ − sin ψ cos ψ 0 0
0
cos φ
0 − cos θ sin φ 1 sin θ sin φ
sin φ cos θ cos φ
− sin θ cos φ
0
sin θ cos θ sin ψ sin θ
= − sin ψ cos φ − cos ψ cos θ sin φ − sin ψ sin φ + cos ψ cos θ cos φ cos ψ sin θ sin θ sin φ − sin θ cos φ cos θ
cos ψ cos φ − sin ψ cos θ sin φ
cos ψ sin φ + sin ψ cos θ cos φ
linear algebra
Looking at various elements of this matrix, one can determine that Rˆ = − 31 , Rˆ q 32 Rˆ 231 + Rˆ 232 tan θ = , Rˆ 33 Rˆ 13 tan ψ = . Rˆ 23 tan φ
Applying these to the given matrix, we find tan φ = −
0.5000 = 1, −0.5000
p
(0.5000)2 + (−0.5000)2 = −1, −0.7071 −0.1630 tan ψ = = 0.2679355783. −0.1830 However, one needs to be careful of the ambiguity in the computation of the inverse tangent. Part of this is remedied by only letting 0 < θ < Π. Then, we need only pay attention to the quadrants determined by the signs of the entries in A. In particular, we note that tan θ =
cos θ
=
cos φ
=
sin φ
=
cos ψ
=
sin ψ
=
R33 = −0.7071. R32 −0.5000 = = 0.7071. − sin θ −.7071 R31 0.5000 = = 0.7071. sin θ .7071 R23 −0.6830 = = −0.9659. sin θ .7071 R13 −0.1830 = = −0.2588. sin θ .7071
So, φ is in quadrant one, θ is in quadrant two, and ψ is in quadrant three. Paying attention to these quadrants, we have that φ=
π , 4
θ=
3π , 4
ψ=
13π . 12
6. Consider the three dimensional Euler rotation matrix Rˆ (φ, θ, ψ) = Rˆ z (ψ) Rˆ x (θ ) Rˆ z (φ). From the last problem we obtained Rˆ (φ, θ, ψ) =
cos ψ cos φ − sin ψ cos θ sin φ
− sin ψ cos φ − cos ψ cos θ sin φ sin θ sin φ
cos ψ sin φ + sin ψ cos θ cos φ
− sin ψ sin φ + cos ψ cos θ cos φ cos ψ sin θ . − sin θ cos φ cos θ
a. Find the elements of Rˆ (φ, θ, ψ). See above.
sin ψ sin θ
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b. Compute Tr ( Rˆ (φ, θ, ψ).
Tr ( Rˆ )
= cos ψ cos φ − sin ψ cos θ sin φ − sin ψ sin φ + cos ψ cos θ cos φ + cos θ = cos(ψ + φ) + cos(ψ + φ) cos θ + cos θ.
c. Show that Rˆ −1 (φ, θ, ψ) = Rˆ T (φ, θ, ψ). T Recall that Rˆ −1 = C ˆ , where C is the matrix of cofactors. detR The determinant of each rotation matrix is one and therefore det Rˆ T (φ, θ, ψ) =  Rˆ z (ψ) Rˆ x (θ ) Rˆ z (φ) = 1. The cofactors are given by
C11
C12
C13
C21
C22
C23
C31
C32
C33
=
− sin ψ sin φ + cos ψ cos θ cos φ − sin θ cos φ
cos ψ sin θ cos θ
= cos ψ cos φ − sin ψ cos θ sin φ. − sin ψ cos φ − cos ψ cos θ sin φ cos ψ sin θ = − sin θ sin φ cos θ
= cos ψ sin φ + sin ψ cos θ cos φ. − sin ψ cos φ − cos ψ cos θ sin φ − sin ψ sin φ + cos ψ cos θ cos φ = sin θ sin φ − sin θ cos φ = sin ψ sin θ. cos ψ sin φ + sin ψ cos θ cos φ sin ψ sin θ = − − sin θ cos φ cos θ = − cos ψ cos θ sin φ − sin ψ cos φ. cos ψ cos φ − sin ψ cos θ sin φ sin ψ sin θ = sin θ sin φ cos θ
= cos ψ cos θ cos φ − sin ψ sin φ. cos ψ cos φ − sin ψ cos θ sin φ cos ψ sin φ + sin ψ cos θ cos φ = − sin θ sin φ − sin θ cos φ = cos ψ sin θ. cos ψ sin φ + sin ψ cos θ cos φ sin ψ sin θ = − sin ψ sin φ + cos ψ cos θ cos φ cos ψ sin θ
= sin θ sin φ. cos ψ cos φ − sin ψ cos θ sin φ sin ψ sin θ = − − sin ψ cos φ − cos ψ cos θ sin φ cos ψ sin θ
= − sin θ cos φ. cos ψ cos φ − sin ψ cos θ sin φ cos ψ sin φ + sin ψ cos θ cos φ = − sin ψ cos φ − cos ψ cos θ sin φ − sin ψ sin φ + cos ψ cos θ cos φ = cos θ.
linear algebra
Therefore, the matrix of cofactors is given by C = cos ψ cos φ − sin ψ cos θ sin φ cos ψ sin φ + sin ψ cos θ cos φ − cos ψ cos θ sin φ − sin ψ cos φ cos ψ cos θ cos φ − sin ψ sin φ sin θ sin φ − sin θ cos φ
sin ψ sin θ cos ψ sin θ . cos θ
Seeing that C = Rˆ (φ, θ, ψ), we have CT Rˆ −1 (φ, θ, ψ) = = Rˆ T (φ, θ, ψ). det Rˆ d. Show that Rˆ −1 (φ, θ, ψ) = Rˆ (−ψ, −θ, −φ). One only needs to compute Rˆ (−ψ, −θ, −φ) and compare this to the last result, Rˆ −1 (φ, θ, ψ) =
cos ψ cos φ − sin ψ cos θ sin φ cos ψ sin φ + sin ψ cos θ cos φ sin ψ sin θ
− cos ψ cos θ sin φ − sin ψ cos φ − sin θ cos φ cos ψ cos θ cos φ − sin ψ sin φ − sin θ cos φ . cos ψ sin θ cos θ
Rˆ (−ψ, −θ, −φ) =
cos φ cos ψ − sin φ cos θ sin ψ
sin φ cos ψ + cos φ cos θ sin ψ sin θ sin ψ
− cos φ sin ψ − sin φ cos θ cos ψ
sin φ sin θ
− sin φ sin ψ + cos φ cos θ cos ψ − cos φ sin θ . sin θ cos ψ cos θ
7. The Pauli spin matrices are given by the following ! ! ! in quantum mechanics 1 0 0 −i 0 1 . Show , and σ3 = , σ2 = matrices: σ1 = 0 −1 i 0 1 0 that a. σ12 = σ22 = σ32 = I. 1 0
!
−i 0
!
=
0 i
0 −1
!
=
1 0
0 1
σ12 =
σ22
!
0 1
1 0
0 i
−i 0
!
1 0
0 −1
!
1 0
=
0 1
! .
0 1
!
=
1 0
0 1
!
=
1 0
.
and σ32
.
b. {σi , σj } ≡ σi σj + σj σi = 2δij I, for i, j = 1, 2, 3 and I the 2 × 2 identity matrix. {, } is the anticommutation operation. First, we note that {σi , σi } = 2σi2 = 2I, i = 1, 2, 3.
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Next, we compute specific anticommutations:
{σ1 , σ2 } = σ1 σ2 + σ2 σ1 = {σ2 , σ1 } ! ! ! ! 0 1 0 −i 0 −i 0 1 = + 1 0 i 0 i 0 1 0 ! ! ! i 0 −i 0 0 0 = + = 0 −i 0 i 0 0 {σ1 , σ3 } = σ1 σ3 + σ3 σ1 = {σ3 , σ1 } ! ! ! ! 0 1 1 0 1 0 0 1 = + 1 0 0 −1 0 −1 1 0 ! ! ! 0 −1 0 1 0 0 = + = 1 0 −1 0 0 0 {σ2 , σ3 } = σ2 σ3 + σ3 σ2 = {σ3 , σ2 } ! ! ! 0 −i 1 0 1 0 = + i 0 0 −1 0 −1 ! ! ! 0 0 0 −i 0 i = + = 0 0 −i 0 i 0
0 i
−i 0
!
From these, we have shown that {σi , σj } = 2δij I, for i, j = 1, 2, 3. c. [σ1 , σ2 ] ≡ σ1 σ2 − σ2 σ1 = 2iσ3 , and similarly for the other pairs. [, ] is the commutation operation. First, we note that [σi , σi ] = 0, i = 1, 2, 3. Next, we compute
[σ1 , σ2 ] = σ1 σ2 − σ2 σ1 = −[σ2 , σ1 ] ! ! ! ! 0 1 0 −i 0 −i 0 1 − = 1 0 i 0 i 0 1 0 ! ! ! 1 0 −i 0 i 0 = 2iσ3 = 2i − = 0 −1 0 i 0 −i
[σ1 , σ3 ] = σ1 σ3 − σ3 σ1 = −[σ3 , σ1 ] ! ! ! ! 0 1 1 0 1 0 0 1 = − 1 0 0 −1 0 −1 1 0 ! ! ! 0 −1 0 1 0 −i = − = −2i = −2iσ2 1 0 −1 0 i 0
[σ2 , σ3 ] = σ2 σ3 − σ3 σ2 = −[σ3 , σ2 ] ! ! ! ! 0 −i 1 0 1 0 0 −i = − i 0 0 −1 0 −1 i 0 ! ! ! 0 i 0 −i 0 1 = 2i = 2iσ1 = − −i 0 1 0 i 0
linear algebra
d. Show that an arbitrary 2 × 2 matrix M can be written as a linear combination of Pauli matrices, M = a0 I + ∑3j=1 a j σj , where the a j ’s are complex numbers. First, we write out the elements of M. ! m11 m12 M = m21 m22 3
= a0 I + ∑ a j σj j =1
= a0 =
1 0
!
0 1
+ a1
a0 + a3 a1 + ia2
a1 − ia2 a0 − a3
0 1 1 0 !
!
+ a2
0 i
−i 0
!
+ a3
1 0
0 −1
!
.
So, given matrix M one can determine the ai ’s by solving the system of equations m11 = a0 + a3 ,
m12 = a1 − ia2 ,
m22 = a0 − a3 ,
m21 = a1 + ia2 ,
to obtain m12 + m21 , 2 m − m21 a0 = 12 . 2i
m11 + m22 , 2 m − m22 a0 = 11 , 2
a1 =
a0 =
8. Use Cramer’s Rule to solve the system: 2x − 5z
= 7
x − 2y
= 1
3x − 5y − z
= 4.
First one computes the determinant of the coefficient matrix, 2 0 −5 −2 0 1 −2 1 −2 0 = 2 −5 −1 − 5 3 −5 = 2(2) − 5(1) = −1. 3 −5 −1 Next, one computes the three matrices obtained by replacing the columns with (7, 1, 4) T . This gives 7 0 −5 −2 0 1 −2 1 −2 0 = 2 −5 −1 − 5 3 −5 = 2(2) − 5(1) = −1, 4 −5 −1 2 1 3
−5 = 2 1 0 1 0 4 −1 4 −1 7
1 7 4
0 −1
1 −5 3
1 4
= 2(−1) − 7(−1) − 5(1) = 0,
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and 2 1 3
0
−2 −5
7 = 2 −2 1 −5 4
1 4
1 +7 3
−2 −5
= 2(−3) + 7(1) = 1.
Dividing these determinants by the determinant of the coefficient matrix, we obtain the solution x = 1, y = 0, z = −1. 9. Find the eigenvalue(s) and eigenvector(s) for the following: ! 4 2 a. 3 3 The eigenvalues in these problems are determined from the eigenvalue equation, 0
= det( A − λI ) 4−λ 2 = 3 3−λ
= (4 − λ)(3 − λ) − 6 = λ2 − 7λ + 6 = (λ − 6)(λ − 1). Therefore, the eigenvalues are λ = 1, 6. For each eigenvalue, one determines nontrivial eigenvectors v satisfying ( A − λI )v = 0. For λ = 1, 0
2 3−λ
!
v1 v2 !
!
=
4−λ 3
=
3 3
=
3v1 + 2v2 3v1 + 2v2
2 2
!
v1 v2
!
Thus, we can pick any vector whose components satisfy 3v1 + 2v2 = 0. For example, v = (2, −3) T . For λ = 6, 0
=
4−λ 3
=
−2 2 3 −3
=
−2v1 + 2v2 3v1 − 3v2
2 3−λ !
!
!
v1 v2
v1 v2 !
!
linear algebra
Thus, we can pick any vector whose components satisfy v1 = v2 . For example, v = (1, 1) T . The eigenvalues are λ = 1, 6 with corresponding eigenvectors (2, −3) T and (1, 1) T . ! 3 −5 b. 1 −1 The eigenvalue equation is, 0
= det( A − λI ) 3−λ −5 = 1 −1 − λ
= (3 − λ)(−1 − λ) + 5 = λ2 − 2λ + 2 = ( λ − 1)2 + 1). Therefore, the eigenvalues are λ = 1 ± i. The corresponding eigenvectors are determined next. For λ = 1 ± i, ! ! v1 3 − (1 ± i ) −5 0 = v2 1 −1 − (1 ± i ) ! ! v1 2∓i −5 = v2 1 −2 ∓ i ! (2 ∓ i )v1 − 5v2 = v1 + (−2 ∓ i )v2 Thus, we can pick any vector whose components satisfy (2 ∓ i )v1 − 5v2 = 0. Setting v2 = 1, v1 =
5 = 2 ± i. 2∓i
Therefore, the eigenvectors corresponding to λ = 1 ± i are found as v = (2 ± i, 1) T . ! 4 1 c. 0 4 The eigenvalue equation is (4 − λ)2 = 0. Therefore, there is one eigenvalue, λ = 4. The corresponding eigenvector satisfies the equation ! ! 0 1 v1 0 = 0 0 v2 Thus, we can pick any vector whose components satisfy v2 = 0. Setting v1 = 1, the eigenvector corresponding to λ = 4 is v = (1, 0)T .
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1 −1 4 d. 3 2 −1 2 1 −1 We first find the eigenvalue equation: 1−λ −1 4 0 = 3 2−λ −1 2 1 −1 − λ 2−λ −1 3 = (1 − λ ) + 1 −1 − λ 2
−1 −1 − λ
3 +4 2
2−λ 1
= (1 − λ)((2 − λ)(−1 − λ) + 1) +3(−1 − λ) + 2 + 4(3 − 2(2 − λ)) = (1 − λ)(λ − 3)(λ + 2). For each eigenvalue, we solve the following systems 0 −1 4 v1 0 λ = 1, 3 1 −1 v2 = 0 , 2 1 −2 v3 0 0 v1 3 −1 4 λ = −2, 3 4 −1 v2 = 0 , 0 v3 2 1 1 0 −2 −1 4 v1 λ = 3, 3 −1 −1 v2 = 0 0 2 1 −4 v3 to obtain for the eigenvalues λ = 1, −2, 3. The corresponding eigenvectors are found as (−1, 4, 1) T , (−1, 1, 1) T , and (1, 2, 1) T . 10. For the matrices in the previous problem, compute the determinants and find the inverses, if they exist. 4 2 a. = 6. 3 3 ! −1 ! 1 4 2 − 13 2 = . 2 3 3 − 12 3 3 −5 b. = 2. 1 −1 ! −1 ! 3 −5 − 21 52 . = − 12 32 1 −1 4 1 c. = 16. 0 4 ! −1 ! 1 1 4 1 − 16 4 = . 1 0 4 0 4
linear algebra
d. 1 3 2
−1 4 2 −1 2 −1 = 1 −1 1 −1
3 + 2
−1 −1
3 +4 2
2 1
= −1 − 1 + 4(−1) = −6.
1 3 2
1 −6
−1
−1 4 2 −1 1 −1
=
1 = − 6
=
1
61 −6 1 6
3 −1 2 −1 − 2 −1 1 −1 1 4 −1 4 − 2 −1 1 −1 1 4 −1 4 − 3 −1 2 −1 T −1 1 −1 3 9 −3 −7 13 5 7 − 12 6 3 − 13 2 6 . 1 − 56 2
3 2 1 − 2 1 3
11. Consider the conic 5x2 − 4xy + 2y2 = 30. a. Write the left side in matrix form. 2
2
5x − 4xy + 2y =
x
5 −2
y
−2 2
!
x y
! .
b. Diagonalize the coefficient matrix, finding the eigenvalues and eigenvectors. The eigenvalue equation is 0 = (5 − λ)(2 − λ) − 4 = λ2 − 7λ + 6 = (λ − 6)(λ − 1). Thus, the eigenvalues are λ = 1, 6. For each eigenvalue, one determines nontrivial eigenvectors v satisfying ( A − λI )v = 0. For λ = 1, 0
−2 1
!
=
4 −2
=
4v1 − 2v2 −2v1 + v2
v1 v2 ! .
Thus, v2 = 2v1 . A unit eigenvector is given by ! 1 1 v1 = √ . 2 5
!
2 1
−1 1 −1 2
T
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For λ = 6, 0
!
=
−1 −2 −2 −4
=
−v1 − 2v2 −2v1 − 4v2
v1 v2
!
! .
Thus, v1 = −2v2 . A unit eigenvector is given by ! 1 −2 . v2 = √ 1 5 c. Construct the rotation matrix from the information in part b. What is the angle of rotation needed to bring the conic into standard form? The equation of the conic was shown in part a to be of the form xT Qx = 30, where xT = ( x, y). We seek a rotation R to new coordinates yT = ( x 0 , y0 ) such that the conic equation takes the form yT Λy = 30, where Λ is a diagonal matrix. Letting y = Rx, or x = R−1 y = R T y, we have 30
= xT Qx = ( R T y) T Q ( R T y) = yT ( RQR T )y.
Figure 3.1: Conics in Problem 12.
Therefore, we seek rotation matrix R such that RQR T = Λ is diagonal. This is accomplished by letting R T = S be the matrix whose columns are the eigenvectors of Q and S−1 QS = Λ provides the similarity transformation which diagonalizes Q. Therefore, ! 1 1 − 2 RT = S = √ 2 1 5 and 1 R= √ 5
1 −2
2 1
! .
Since this is an active rotation matrix, we have from the first col√ √ umn cos θ = 1/ 5 and sin θ = −2/ 5. Thus, θ = −63o . d. What is the conic? The equation of the rotated conic is yT Λy = 30, or ! ! 1 0 x0 0 0 30 = x y = x 02 + 6y02 . 0 6 y0 Thus, this is an ellipse.
linear algebra
12. In Equation (3.109), the exponential of a matrix was defined. a. Let A=
2 0
0 0
! .
Compute e A . We begin with the definition eA = I + A +
∞ 1 2 1 1 1 n A + A3 + A4 + . . . = ∑ A . 2! 3! 4! n! n =0
Since the matrix is diagonal, it is easy to compute ! 2n 0 n A = . 0 0 Then, ∞
1 n A = e = ∑ n! n =0 A
1 n ∑∞ n=0 n! 2 0
0 0
!
b. Give a definition of cos A and compute cos
e2 0
= 1 0
0 2
!
0 0 !
.
in simplest
form. Since cos θ = 21 (eiθ + e−iθ ), we have cos A = I −
1 0
For A =
0 2
∞ 1 2 1 (−1)n 2n A + A4 + . . . = ∑ A . 2! 4! (2n)! n =0
!
diagonal, it is easy to compute ! cos 1 0 . Therefore, cos A = 0 cos 2
c. Using the definition of e A , prove e PAP
e PAP
−1
= = = = =
−1
An
=
1 0
0 2n
! .
= Pe A P−1 for general A.
1 1 ( PAP−1 )2 + ( PAP−1 )3 + . . . 2! 3! 1 1 −1 −1 −1 I + PAP + PAP PAP + PAP−1 PAP−1 PAP−1 + . . . 2! 3! 1 1 −1 2 −1 3 −1 I + PAP + PA P + PA P + . . . 2! 3! 1 2 1 3 P ( I + A + A + A + . . . ) P −1 2! 3! A −1 Pe P . I + PAP−1 +
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13. Prove the following for matrices A, B, and C. a. ( AB)C = A( BC ). Let A, B, and C, be m × n, n × p, and p × q matrices, respectively. We will show that the i `th elements of each side of the equation are the same for i = 1, 2..., m and ` = 1, 2..., q. For clarity, we will not use Einstein’s summation convention. p
[( AB)C ]i`
∑ ( AB)ij Cj`
=
j =1 p
n
j =1
k =1
∑ ∑ Aik Bkj
=
n
p
k =1
j =1
! Cj`
∑ Aik ∑ Bkj Cj`
=
!
n
∑ Aik ( BC)k`
=
k =1
= [ A( BC )]i` . b. ( AB) T = B T A T Let A be an m × n matrix and B an n × p matrix. Then, one can form the product AB, which is an m × p matrix. Furthermore, A T is an n × m matrix and B T is a p × n matrix. Then, one can form B T A T , a p × m matrix. The ijth element of ( AB) T can be manipulated to prove that it is equal to the ijth element of B T A T .
[( AB)T ]ij
= ( AB) ji =
∑
A jk Bki
k =1n
=
∑n ( BT )ik ( AT )kj
k =1
= [ B T A T ]ij . c. tr( A) is invariant under similarity transformations. We first note that tr( AB) = tr( BA) for two n × n matrices. This can be proven n
tr( AB)
=
∑ ( AB)ii
i =1 n n
=
∑ ∑ Aik Bki
i =1 k =1 n n
=
∑ ∑ Bki Aik
k =1 i =1 n
=
∑ ( AB)kk
k =1
= tr( BA).
linear algebra
It then follows that tr(S−1 AS) = tr( ASS−1 ) = tr( AI ) = tr( A). d. If A and B are orthogonal, then AB is orthogonal. Since A and B are orthogonal, we have AA T = I and BB T = I. Then, we have AB( AB) T = ABB T A T = AI A T = AA T = I. 14. Consider the following systems. For each system determine the coefficient matrix. When possible, solve the eigenvalue problem for each matrix and use the eigenvalues and eigenfunctions to provide solutions to the given systems. Finally, in the common cases that you investigated in Problem 31, make comparisons with your previous answers, such as what type of eigenvalues correspond to stable nodes. a. x0 y
0
= 3x − y, = 2x − 2y.
The matrix form is given by !0 x = y
3 2
−1 −2
!
x y
! .
The eigenvalue equation is 0 = (3 − λ)(−2 − λ) + 2 = λ2 − λ − 4 √ Thus, the eigenvalues are λ = 12 (1 ± 17). For each eigenvalue, one determines nontrivial eigenvectors v satisfying ( A − λI )v = 0. √ For λ = 21 (1 ± 17), ! ! √ 1 ( 5 ∓ 17 ) − 1 v 1 2 √ 0 = . 2 − 21 (5 ± 17) v2 Thus, v2 = 12 (5 ∓
√
17)v1 . The eigenvectors are given by ! 1 √ v1 = . 1 2 (5 ∓ 17)
We can write the general solution ! ! √ 1 x 1 √ = c1 e 2 (5+ 17)t + c2 1 y 2 (5 − 17)
1 √ 1 ( 5 + 17) 2
!
c1 e 2 (5+√17)t + c2 e 2 (5− 17)t √ √ √ 1 (5+ 17) t 1 c1 c2 2 2 (5− 17) t ( 5 − 17 ) e + ( 5 + 17 ) e 2 2
!
1
=
√
1
√
1
e 2 (5− .
√
17)t
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b. x0 y
= −y,
0
= −5x.
The matrix form is given by x y
!0
=
!
−1 0
0 −5
!
x y
.
The eigenvalue equation is 0 = λ2 − 5. Thus, the eigenvalues are √ λ = ± 5. √ For λ = ± 5, ! ! √ ∓ 5 −1 v1 √ 0 = . v2 −5 ∓ 5
√ Thus, v2 = ∓ 5v1 . The eigenvectors are given by ! 1 √ v= . ∓ 5 We can write the general solution ! ! √ 1 x √ = c1 e 5t + c2 − 5 y √
√
c1 e 5t + c2 e− 5t √ √ √5t 5 c1 e − c2 e− 5t
=
1 √ 5 !
! e−
√
5t
.
c. x0 y
0
= x − y, = y.
The matrix form is given by x y
!0
=
1 0
−1 1
!
x y
! .
The eigenvalue equation is 0 = (1 − λ)2 . Thus, there is one eigenvalue, λ = 1. The eigenvector satisfies ! ! 0 −1 v1 0 = . 0 0 v2 Thus, v2 = 0. The eigenvector is v=
1 0
! .
linear algebra
This is insufficient to produce the general solution to the problem. However, we have seen that a second linearly independent solution can be written as x = teλt v1 + eλt v2 , where v1 and v2 satisfy the equations
( A − λI )v1 = 0, ( A − λI )v2 = v1 . We have already found v1 = (1, 0) T . Letting v2 = (u1 , u2 ) T , we have ! ! ! 0 −1 u1 1 = . 0 0 u2 0 This gives u2 = −1. We are free to pick u1 = 0. Then, the general solution takes the form !# ! ! " ! 0 1 1 x t et t+ e + c2 = c1 −1 0 0 y ! ( c1 + c2 t ) e t = . − c2 e t d. x0 y
0
The matrix form is given by !0 x = y
= 2x + 3y, = −3x + 2y.
2 3 −3 2
!
x y
! .
The eigenvalue equation is 0 = (2 − λ)2 + 9. Thus, λ = 2 ± 3i. The corresponding eigenvectors are determined next. ! ! 2 − (2 ± 3i ) 3 v1 0 = −3 2 − (2 ± 3i ) v2 ! ! ∓3i 3 v1 = −3 ∓3i v2 ! (∓3i )v1 + 3v2 = . −3v1 + (∓3i )v2 Thus, we can pick any vector whose components satisfy v2 = ±iv1 . Setting v1 = 1, v2 = ±i. Therefore, the eigenvectors corresponding to λ = 2 ± 3i are v = (1, ±i )T .
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The general solution is formed from the real and imaginary parts of ! 1 λt ve = e(2+3i)t , i " Re " Im
1 i
!
1 i
!
# e
(2+3i )t
e
(2+3i )t
! e2t cos 3t −e2t sin 3t ! e2t sin 3t . e2t cos 3t
= #
=
Thus, x y
!
= =
e2t cos 3t −e2t sin 3t
!
e2t sin 3t c1 + c2 e2t cos 3t ! e2t (c1 cos 3t + c2 sin 3t) . e2t (−c1 sin 3t + c2 cos 3t)
!
e. x0
= −4x − y,
y0
= x − 2y.
The matrix form is given by !0 x = y
!
−4 −1 1 −2
x y
! .
The eigenvalue equation is 0 = (−4 − λ)(−2 − λ) + 1 = λ2 + 6λ + 9 = (λ + 3)2 . Thus, λ = −3. The eigenvector satisfies 0
−1 −1 1 1
=
!
v1 v2
! .
Thus, v2 = −v1 and one eigenvector of the system is v = (1, −1) T . Since there is only one root of the eigenvalue equation, then we also need to solve the problem ! ! ! −1 −1 u1 1 = . 1 1 u2 −1 This gives u1 + u2 = −1. We are free to pick u1 = 0, yielding u2 = −1. x y
!
= c1 =
1 −1
!
" e
−3t
+ c2
(c1 + c2 t)e−3t −(c1 + c2 + c2 t)e−3t
1 −1 ! .
! t+
0 −1
!# e−3t
linear algebra
f. x0
= x − y,
0
= x + y.
y
The matrix form is given by x y
!0
=
1 1
−1 1
!
x y
! .
The eigenvalue equation is 0 = (1 − λ)2 + 1. Thus, λ = 1 ± i. The corresponding eigenvectors are determined next. 1 − (1 ± i ) −1 1 1 − (1 ± i ) ! ! v1 ∓ i −1 v2 1 ∓i ! (∓i )v1 − v2 . v1 + (∓i )v2
=
0
= =
!
v1 v2
!
Thus, we can pick any vector whose components satisfy v2 = ∓iv1 . Setting v1 = 1,, v2 = ∓i. Therefore, the eigenvectors corresponding to λ = 1 ± i are v = (1, ∓i )T . The general solution is formed from the real and imaginary parts of ! 1 λt e (1+ i ) t , ve = −i " Re " Im
1 −i
!
1 −i
!
# e
(1+ i ) t
= =
et sin t −et cos t
# e (1+ i ) t
!
et cos t et sin t
! .
Thus, x y
!
= =
et cos t et sin t
!
et sin t c1 + c2 −et cos t ! et (c1 cos t + c2 sin t) . et (c1 sin t − c2 cos t)
!
15. In Example 3.22 we investigated a couple massspring system as a pair of secondorder differential equations.
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q
a. In that problem, we used
√ 3± 5 2
√
=
5±1 2 .
Prove this result.
The simplest way to prove this is to square both sides
√
5±1 2
!2
=
√ √ 5±2 5+1 3± 5 = . 4 2
b. Rewrite the system as a system of four first order equations. The firstorder system for this example (k1 = k2 = k, m1 = m2 = m) is given by x˙ 1
= x3
x˙ 2
= x4
x˙ 3
= −
x˙ 4
=
k k x + ( x2 − x1 ) m 1 m k − ( x2 − x1 ). m
c. Find the eigenvalues and eigenfunctions for the system of equations in part b to arrive at the solution found in Example 3.22. The coefficient matrix for this system becomes A= where ω02 =
0
=
=
=
k m.
0 0 −2ω02 ω02
0 0 ω02 −ω02
1 0 0 1 0 0 0 0
,
The eigenvalue equation is given by
−λ 0 1 0 0 −λ 0 1 −2ω02 ω02 −λ 0 ω2 −ω02 0 −λ 0 −λ 0 −λ 0 1 −λ ω02 −λ 0 + −2ω02 ω02 −ω 2 −ω02 0 −λ ω02 0 ! −λ 0 ω 2 −λ 0 −λ −λ + 0 −λ −ω02 0 −2ω 2 −2ω 2 ω 2 0 0 0 0 +λ + ω02 −λ ω02 −ω02
1 0 −λ
= λ4 + 3ω02 λ2 + ω04 3 5 = (λ2 + ω02 )2 − ω04 . 2 4 √
The solutions of this quadratic are λ2 = − 3±2 5 ω02 . Using part a, we have √ 5±1 λ=± iω0 . 2
linear algebra
The eigenvectors satisfy the system 0 0 1 0 0 0 0 1 −2ω02 ω02 0 0 ω02 −ω02 0 0
v1 v2 v3 v4
=
λv1 λv2 λv3 λv4
.
The first two equations give v3 = λv1 and v4 = λv2 . Inserting into the remaining equations, we obtain ω02 v2
= (λ2 + 2ω02 )v1
ω02 v1
= (λ2 + ω02 )v2 .
√
For λ2 = − 3±2 5 ω02
√ √ 3± 5 1∓ 5 v2 = (− + 2) v1 = v1 2 2 √
√
So, for v1 = 1, v3 = λ, v2 = 1∓2 5 , and v4 = λ( 1∓2 5 ). Therefore, the eigenvectors are given as 1√ 1√ 1− 5 1+ 5 , v2 = √ 2 , √2 v1 = 1+ 5 5−1 2 iω0 2 iω0 −iω0 iω0 1√ 1√ 1+ 5 1− 5 2 . √2 √ v3 = 1+ 5 , v4 = 1− 5 − 2 iω0 2 iω0 iω0 −iω0 We construct solutions from the real and imaginary parts of the fundamental solutions, veλt . Writing out the first and second eigenvectors, we have √ √ 5+1 5+1 cos ( ω t ) + i sin ( ω t ) 0 0 2 2 √ √ 1− √5 5+1 5+1 √ 2 cos ( ω t ) + i sin ( ω t ) 0 0 5+1 2 2 √ √ v1 e 2 iω0 t = 1+√5 2 i cos( 52+1 ω0 t) − sin( 52+1 ω0 t) √ √ ω0 −i cos( 52+1 ω0 t) + sin( 52+1 ω0 t) √
v2 e
5−1 2 iω0 t
=
√
√
cos( 52−√1 ω0 t) + i sin( 52−√1 ω0 t) √ 1+ 5 cos( 52−1 ω0 t) + i sin( 52−1 ω0 t) √2 √ √ 5−1 5−1 5−1 ) 2 i cos( 2 ω0 t ) − sin( 2 ω0 t √ √ ω0 i cos( 52−1 ω0 t) − sin( 52−1 ω0 t)
The general solution is obtained using linear combinations of the real and imaginary parts. Therefore, the general solution is found as
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√
x1
=
x2
=
x3
=
x4
=
√ 5+1 5+1 c1 cos( ω0 t) + c2 sin( ω0 t ) √2 √2 5−1 5−1 +c3 cos( ω0 t) + c4 sin( ω0 t ) 2 2 ! √ √ √ 5+1 5+1 1− 5 c1 cos( ω0 t) + c2 sin( ω0 t ) 2 2 2 ! √ √ √ 1+ 5 5−1 5−1 c3 cos( ω0 t) + c4 sin( ω0 t ) + 2 2 2 ! √ √ √ 1+ 5 5+1 5+1 ω0 −c1 sin( ω0 t) + c2 cos( ω0 t ) 2 2 2 ! √ √ √ 1− 5 5−1 5−1 + ω0 −c3 sin( ω0 t) + c4 cos( ω0 t ) 2 2 2 ! √ √ 5+1 5+1 ω0 t) − c2 cos( ω0 t ) ω0 c1 sin( 2 2 ! √ √ 5−1 5−1 +ω0 −c3 sin( ω0 t) + c4 cos( ω0 t ) . 2 2
d. Let k = 5.00 N/m and m = 0.250 kg. Assume that the masses are initially at rest and plot the positions as a function of time if initially i) x1 (0) = x2 (0) = 10.0 cm and ii) x1 (0) = − x2 (0) = 10.0 cm. Describe the resulting motion. √ √ For k = 5.00 N/m and m = 0.250 kg we have ω0 = k/m = 2 5 Hz. Evaluating the solution at t = 0, we obtain x1 (0)
= c1 + c3 √ √ 1− 5 1+ 5 x2 (0) = c + c3 2√ 1 2 √ 1+ 5 1− 5 ω0 c 2 + ω0 c 4 0 = 2 2 0 = ω0 (−c2 + c4 ). We see that c2 = c4 = 0. So, the solution for the positions of the masses is given by x1 x2
√ √ = c1 cos(5 + 5)t + c3 cos(5 − 5)t √ √ √ √ 1− 5 1+ 5 = c1 cos(5 + 5)t + c3 cos(5 − 5)t. 2 2
We still need c1 and c3 . Solving the first equation for c3 , c3 = x1 (0) − c1 .
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95
Then, x2 (0)
√ √ 1− 5 1+ 5 c1 + ( x1 (0) − c1 ) 2 √2 √ 1+ 5 − c1 5 + x1 (0). 2
= =
Then, c1
=
√ 1+ 5 0) − x2 (0) 2 x1 (√
√
= and
5
√ 5+ 5 5 x1 (0) − x2 (0) 10 5
√ √ 5 5− 5 x1 (0) + x2 (0). c3 = 10 5
√ For i) x1 (0) = x2 (0) = 10.0 cm we have c1 = 5 − 5 and c3 = √ 5 + 5. The solution satisfying these initial conditions is given by √ √ √ √ x1 = (5 − 5) cos(5 + 5)t + (5 + 5) cos(5 − 5)t √ √ √ √ x2 = (5 − 3 5) cos(5 + 5)t + (5 + 3 5) cos(5 − 5)t. The solutions are shown in Figure 3.2. Figure 3.2: Plot of the position of each block in Problem 3.15e for x1 (0) = x2 (0) = 10.0 cm.
Figure 3.3: Plot of the position of each block in Problem 3.15e for x1 (0) = − x2 (0) = 10.0 cm.
√ For ii) x1 (0) = − x2 (0) = 10.0 cm we have c1 = 5 + 3 5 and √ c3 = 5 − 3 5. The solution satisfying these initial conditions is given by √ √ √ √ x1 = (5 + 3 5) cos(5 + 5)t + (5 − 3 5) cos(5 − 5)t √ √ √ √ x2 = (−5 − 5) cos(5 + 5)t + (−5 + 5) cos(5 − 5)t. The solutions are shown in Figure 3.3.
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16. Add a third spring connected to mass 2 in the coupled system shown in Figure 3.15 to a wall on the far right. Assume that the masses are the same and the springs are the same. Figure 3.4: System of two masses and three springs.
m1
m2
k1
k3
k2
x1
x2
a. Model this system with a set of firstorder differential equations. The modified system is shown in Figure 3.4. The third spring only affects the second block and adds the force F = −k3 x2 to the second equation assuming that all three springs are in relaxed mode beforehand. The governing equations are given as m1 x¨1
= − k 1 x1 + k 2 ( x2 − x1 )
m2 x¨2
= − k 2 ( x2 − x1 ) − k 3 x2 .
These lead to the first order system x˙ 1
= x3
x˙ 2
= x4
x˙ 3
= −
x˙ 4
=
k1 k2 x + ( x2 − x1 ) m1 1 m1 k k − 2 ( x2 − x1 ) − 3 x2 . m2 m2
The coefficient matrix for this system is 0 0 0 0 A = k1 +k2 k2 − m m k2 m2
1
1
− k2m+2k3
1 0 0 1 0 0 0
.
0
b. If the masses are all 2.0 kg and the spring constants are all 10.0 N/m, then find the general solution for the system. The coefficient matrix for this system becomes 0 0 1 0 0 0 0 1 A= . −10 5 0 0 5 −10 0 0
linear algebra
The eigenvalue equation is given by −λ 0 1 0 0 −λ 0 1 0 = −10 5 −λ 0 5 −10 0 −λ −λ −λ 1 0 1 0 = −λ 5 5 0 −λ 0 + −10 −10 0 −λ 5 −10 −λ ! −λ 0 5 −λ = −λ −λ + 0 −λ −10 0 −10 0 −10 5 +λ + 5 −10 −λ 5
= λ4 + 20λ2 + 75 = (λ2 + 5)(λ2 + 15)
√ √ The solutions of the eigenvalue equation are therefore λ = ± 5i, ± 15i. The eigenvectors satisfy the system 0 0 1 0 0 0 0 1 −10 5 0 0 5 −10 0 0
v1 v2 v3 v4
=
λv1 λv2 λv3 λv4
.
The first two equations give v3 = λv1 and v4 = λv2 . Inserting these vslues into the remaining equations, we obtain 5v2
= (λ2 + 10)v1
5v1
= (λ2 + 10)v2 .
√ √ For λ = ± 5i, λ2 = −5 and v1 = v2 . For λ = ± 15i, λ2 = −15 and v1 = −v2 . Therefore, the eigenvectors are given as 1 1 1 1 v1 = √ , v2 = √ , 5i − 5i √ √ 5i − 5i 1 1 −1 −1 v3 = √ , v4 = √ . − 15i 15i √ √ − 15i 15i We construct solutions from the real and imaginary parts of the fundamental solutions, veλt . Writing out the first and third eigenvectors, we have √ √ cos 5t + i sin 5t √ √ √ cos 5t + i sin 5t √ √ v1 e 5it = √ 5(i cos 5t − sin 5t) √ √ √ 5(i cos 5t − sin 5t)
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√
v3 e
15it
=
√ √ cos 15t + i sin 15t √ √ − cos 15t − i sin 15t √ √ √ 15(i cos 15t − sin 15t) √ √ √ 15(−i cos 15t + sin 15t)
The general solution is obtained using linear combinations of the real and imaginary parts. Therefore, the general solution is found as
√
√
√
√
x1
= c1 cos
x2
= c1 cos 5t + c2 sin 5t − c3 cos 15t − c4 sin 15t √ √ √ √ √ √ 5(c2 cos 5t − c1 sin 5t) + 15(c4 cos 15t − c3 sin 15t) = √ √ √ √ √ √ = 5(c2 cos 5t − c1 sin 5t) + 15(−c4 cos 15t + c3 sin 15t)
x3 x4
√
5t + c2 sin
√
5t + c3 cos
√
15t + c4 sin
√
15t
c. Move mass 1 to the left (of equilibrium) 10.0 cm and mass 2 to the right 5.0 cm. Let them go. find the solution and plot it as a function of time. Where is each mass at 5.0 seconds? At t = 0, we have from the solution the following system for the unknown constants:
−10 = c1 + c3 = c1 − c3 √ √ 0 = 5c2 + 15c4 √ √ 0 = 5c2 − 15c4 . 5
Solving this system, we find c1 = −2.50 cm, c2 = 0, c3 = −7.50 cm, c4 = 0. This gives the positions and velocities of each block as
√
√
x1
= −2.50 cos
x2
= −2.50 cos 5t + 7.50 cos 15t √ √ √ √ = 2.50 5 sin 5t + 7.50 15 sin 15t √ √ √ √ = 2.50 5 sin 5t − 7.50 15 sin 15t
x3 x4
√
5t − 7.50 cos
√
15t
At 5.0s the masses are at −6.99 cm and 6.07 cm, respectively. This is seen in the figure. d. Model this initial value problem with a set of two second order differential equations. Set up the system in the form Mx¨ = −Kx and solve using the values in part b. The equations of motion for the two masses are given by m1 x¨1
= − k 1 x1 + k 2 ( x2 − x1 ),
m2 x¨2
= − k 2 ( x2 − x1 ) − k 3 x2 .
We can write the governing equations in matrix form as Mx¨ = −Kx,
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99
Figure 3.5: Plot of the position of each block in Problem 3.16c.
where M=
m1 0
0 m2
! ,
k1 + k2 −k2
K=
−k2 k2 + k3 .
! .
Since the masses are the same, we can write x¨ = −Ωx, where Ω=
10 −5
−5 10.
! .
We could guess x = aeiωt and obtain
(Ω − ω 2 I )a = 0. This is just an eigenvalue problem. We have the eigenvalues ω 2 = 5, 15 with eigenvectors (1, 1) T and (1, −1) T , respectively. The general solution is constructed as √ √ √ √ x(t) = c1 a1 cos 5t + c2 a1 sin 5t + c3 a2 cos 15t + c4 a2 sin 15t, from the eigenvectors. This is the same as the solution previously obtained. 17. Consider the series circuit in Figure 2.7 with L = 1.00 H, R = 1.00 × 102 Ω, C = 1.00 × 10−4 F, and V0 = 1.00 × 103 V. a. Set up the problem as a system of two firstorder differential equations for the charge and the current.
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This series LRC circuit is governed be the equation Lq¨ + Rq˙ +
1 q = V0 . C
The given values lead to the i problem q¨ + 100q˙ + 10000q = 1000. ˙ this second order equation Letting the current be written as I = q, can be written as a system of two first order differential equations. ! ! ! d q I 0 = + dt I −100I − 10000q 1000 ! ! ! 0 1 q 0 = + . −10000 −100 I 1000 Note that this is a nonhomogeneous system. b. Suppose that no charge is present and no current is flowing at time t = 0 when V0 is applied. Find the current and the charge on the capacitor as functions of time. The initial condition is (q(0), I (0)) T = (0, 0) T . Writing x = (q, I ) T and f = (0, 1000) T , we seek the solution as a sum of the solution of the homogeneous system plus a particular solution. x = xh + x p . The particular solution can be found using the Method of Undetermined Coefficients. Let x p = ( A, B) T . Then, ! ! ! ! 0 1 A 0 0 + = . −10000 −100 B 1000 0 So, A = 0.10 and B = 0. We construct the solution of the homogeneous problem from the eigenvalues and eigenvectors of the coefficient matrix. The eigenvalue equation is given by
(−λ)(−100 − λ) + 10000 = λ2 + 100λ + 10000 = 0. Solving for λ, we obtain λ=
−100 ±
√ 2
−30000
√ = −50 ± 50 3i.
The eigenfunctions satisfy the equation ! ! ! √ 50 ∓ 50 3i 1 v1 0 √ = . −10000 −50 ∓ 50 3i v2 0 √ Thus, v2 = (−50 ± 50 3i )v1 and the solution is a linear combination of the real and imaginary parts of the vector ! √ 1 (−50±50 3i )t √ xh = e −50 ± 50 3i
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The solution to the homogeneous problem is ! ! √ √ q e−50t (c1 cos(50 3t) + c2 sin(50 3t)) √ √ √ √ = . I 50e−50t (( 3c2 − c1 ) cos(50 3t) − (c2 + 3c1 ) sin(50 3t)) The full general solution is √ √ q(t) = (c1 cos 50 3t + c2 sin 50 3t)e−50t + 0.10, √ √ √ √ I (t) = 50(( 3c2 − c1 ) cos(50 3t) − (c2 + 3c1 ) sin(50 3t))e−50t . Applying the initial condition, (q(0), I (0)) T = (0, 0) T , we have √ 0 = c1 + 0.10 and 0 = 50( 3c2 − c1 ). Therefore, c1 = −0.10, √ c2 = −0.10/ 3. The solution of the initial value problem is
√ √ 1 q(t) = −0.10(cos 50 3t + √ sin 50 3t)e−50t + 0.10, 3 √ 20 I (t) = √ sin(50 3t)e−50t . 3 These are plotted in Figure 3.6. Figure 3.6: Plot of the charge and current in Problem 3.17.
c. Plot your solutions and describe how the system behaves over time. See Figure 3.6. 18. Consider the series circuit in Figure 3.16 with L = 1.00 H, R1 = R2 = 1.00 × 102 Ω, C = 1.00 × 10−4 F, and V0 = 1.00 × 103 V. a. Set up the problem as a system of first order differential equations for the charges and the currents in each loop. In section 3.5.2 the first order system for this circuit was found as q˙ 1
=
q˙ 2
=
q˙ 3
=
q˙ 4
=
q V − 2 R1 R1 C q V − 2 − q4 R1 R1 C q4 q2 R − 2 q4 . LC L
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Inserting the circuit element values, we have q˙ 1
= 10 − 1000q2
q˙ 2
= 10 − 1000q2 − q4
q˙ 3
= q4
q˙ 4
= 10000q2 − 100q4 .
b. We note that we can solve for q2 and q4 and then determine q1 and q3 . Thus, we focus on the nonhomogeneous system ! ! ! ! d 10 −1000 −1 q2 q2 + . = dt 0 10000 −100 q4 q4 We seek the solution as a sum of the solution of the homogeneous system plus a particular solution, x = xh + x p . The particular solution can be found using the Method of Undetermined Coefficients. Let x p = ( A, B) T . Then, ! ! ! ! 0 10 A −1000 −1 . = + 0 0 B 10000 −100 So, A = 1/110 and B = 10/11. We construct the solution of the homogeneous problem from the eigenvalues and eigenvectors of the coefficient matrix. The eigenvalue equation is given by
(−1000 − λ)(−100 − λ) + 10000 = λ2 + 1100λ + 110000 = 0. Solving for λ, we obtain
√ λ = −550 ± 50 77. The eigenfunctions satisfy the equation ! ! √ v1 0 −450 ∓ 50 77 −1 √ = 10000 450 ∓ 50 77 v2 0 √ Thus, v2 = (−450 ∓ 50 77)v1 and the solution is ! ! √ q2 1 √ = c1 e(−550+50 77)t q4 −450 − 50 77 ! √ 1 √ + c2 e(−550−50 77)t + −450 + 50 77 √ √ )t + 1 c1 e(−550+50 77)t + c2 e(−550−50 77 110 √ √ 77)t = (−450 − 50 77)c1 e(−550+50 √ √ +(−450 + 50 77)c2 e(−550−50 77)t + 10 11
! .
1 − 110 10 − 11
!
(3.1)
We can use these solutions to find the other charges, q3 = (4 +
√ √ 5√ 5√ 10 77)c1 e(−550+50 77)t + (4 − 77)c2 e(−550−50 77)t + t, 11 11 11
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103
√ √ 5√ 5√ 10 77)c1 e(−550+50 77)t + (5 − 77)c2 e(−550−50 77)t + t. 11 11 11 Suppose that no charge is present at time t = 0 when V0 is applied. Find the current and the charge on the capacitor as functions of time.
q1 = (5 +
Setting t = 0, one finds that 1 √ 1 1 √ 1 1 + , c3 = 0, c4 = − . 77, c2 = − 77 − c1 = − 220 1540 1540 220 110 The charge and the current across the capacitor are given by √ √ 1 1 √ 1 √ 10 q2 = (− + ) 77)e(−550+50 77)t + (− ) 77 − 1/220)e(−550−50 77)t + 220 1540 1540 11 √ √ √ 45 √ 45 I2 = (5 − 77)e(−550+50 77)t + ( 77 + 5)e(−550−50 77)t . 77 77 c. Plot your solutions and describe how the system behaves over time. These solutions are shown in Figure 3.7. Figure 3.7: Plot of the charge and current in Problem 3.18.
19. Initially, a 200gallon tank is filled with pure water. At time t = 0, a salt concentration with 3 pounds of salt per gallon is added to the container at the rate of 4 gallons per minute, and the wellstirred mixture is drained from the container at the same rate. a. Find the number of pounds of salt in the container as a function of time. Let x (t) be the number of pounds of salt in the container. dx gal lb gal x lb = 4 3 − 4 dt min gal min 200 gal x = 12 − . 50 This is a separable equation and the solution is found as Z
dt 50
t +C 50 x (t)
=
Z
dx 600 − x
= − ln(600 − x ) = 600 − Ae−t/50 .
(3.2)
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Using the initial condition, x (0) = 0, the number of pounds of salt as a function of time is x (t) = 600(1 − e−t/50 ). b. How many minutes does it take for the concentration to reach 2 pounds per gallon? For the 200gallon tank to reach a concentration of 2 pounds per gallon, one would need 400 lbs of salt in the mixture. Therefore, we seek the time for which 400 = 600(1 − e−t/50 ). 2 3
= 1 − e−t/50
e−t/50
=
t
=
1 3 50 ln 3 = 54.9 min.
c. What does the concentration in the container approach for large values of time? Does this agree with your intuition? For large t x approaches 600 lbs. This makes sense because asymptotically the tank concentration will match that of the incoming mixture. d. Assuming that the tank holds much more than 200 gallons, and everything is the same except that the mixture is drained at 3 gallons per minute, what would the answers to parts a and b become? In this case the volume in the tank changes. There is a net increase of one gallon per minute. Thus, at time t the volume is V (t) = 200 + t.
dx dt
=
4
gal min
= 12 −
3
lb gal
gal x lb − 3 min 200 + t gal
3x . 200 + t
(3.3)
The resulting equation is a linear first order differential equation. The integrating factor is Z 3dt µ(t) = exp = exp [3 ln(200 + t)] = (200 + t)3 . 200 + t
0 x (200 + t)3 x (200 + t)3 x
= 12(200 + t)3 = 3(200 + t)4 + C = 3(200 + t) +
C . (200 + t)3
Applying the initial condition, we have 0 = 600 + C = −3(2004 ) and x = 3(200 + t) −
3(2004 ) . (200+t)3
C . 2003
Therefore,
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The time at which the concentration is 2 lb/gal, or x = 2(200 + t) lbs is determined from 2(200 + t) 200 + t
= 3(200 + t) − =
3(2004 ) (200 + t)3
3(2004 ) (200 + t)3
(200 + t)4
= 3(2004 ) √ 4 t = 200( 3 − 1).
The time is t = 63.2 min. This can be verified in Figure 3.8 20. You make 2 quarts of salsa for a party. The recipe calls for 5 teaspoons of lime juice per quart, but you had accidentally put in 5 tablespoons per quart. You decide to feed your guests the salsa anyway. Assume that the guests take a quarter cup of salsa per minute and that you replace what was taken with chopped tomatoes and onions without any lime juice. [1 quart = 4 cups and 1 Tb = 3 tsp.] a. Write the differential equation and initial condition for the amount of lime juice as a function of time in this mixturetype problem. Let x (t) be the number of teaspoons of lime juice in the salsa. The rate of change of lime juice in the salsa is given by the rate of lime juice in less the rate of lime juice out. There is only lime juice leaving the salsa mixture. So, the The rate of change of lime juice in the salsa is given by dx x tsp 1/16 qt = − dt 2 qt 1 min x (3.4) = − . 32 Initially, there was 5 Tb per quart. For two quarts, this give the initial amount of lime as x (0) = 15 tsp. b. Solve this initial value problem. This is a simple exponential model whose solution is given by x (t) = 15e−t/32 . c. How long will it take to get the salsa back to the recipe’s suggested concentration? The recipe called for 5 tsp of lime juice. So, we seek the time, t, when x (t) = 5. Therefore, e−t/32 = 1/3, or t = 32 ln 3 = 35 min. 21. Consider the chemical reaction leading to the system in 3.149. Let the rate constants be k1 = 0.20 ms−1 , k2 = 0.05 ms−1 , and k3 = 0.10 ms−1 . What do the eigenvalues of the coefficient matrix say about the behavior of the system? Find the solution of the system assuming [ A](0) = A0 = 1.0 µmol, [ B](0) = 0, and [C ](0) = 0. Plot the solutions for t = 0.0 to 50.0 ms and describe what is happening over this time.
Figure 3.8: Plot of the amount of salt in the tank in Problem 3.19.
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The system of equations are given by d[ A] dt d[ B] dt d[C ] dt
= −0.20[ A] + 0.10[ B], = 0.20[ A] − 0.05[ B] − 0.10[ B], = 0.05[ B].
(3.5)
−0.20 0.10 0 The coefficient matrix is C = 0.20 −0.15 0 0 0.05 0 We first find the eigenvalue equation: −0.20 − λ 0.10 0 0 = 0.20 −0.15 − λ 0 0 0.05 −λ 0.20 0 −0.15 − λ 0 = (−0.20 − λ) − 0.10 0 −λ 0.05 −λ
= (−0.20 − λ)(λ(0.15 + λ)) − 0.10(−0.2λ) = −(λ + 0.3186)(λ + 0.03139)λ. Thus, λ = 0, −0.3186, −0.03139. d Note that dt ([ A] + [ B] + [C ]) = 0, so [ A] + [ B] + [C ] is constant. So, we need only solve the first two equations for [ A] and [ B]. The eigenvalue equation then reduces to
(−0.20 − λ)(−0.15 − λ) − 0.02 = 0. The solutions are
√ −7 ± 33 λ= = −0.3186, −0.03139. 40
For each eigenvalue, we solve the following system ! ! −0.20 − λ 0.10 v1 = 0.20 −0.15 − λ v2
0 0
! ,
in order to get the eigenvectors. √ We see that v2 = √ 10(−0.20 − λ ) v1 . We can write [ A] = c1 e(−7+ 33)t/40 + c2 e(−7− 33)t/40 . Then, from the first equation in the system, we have d[ A] + 2.0[ A] dt√ √ √ √ 1 + 33 1 − 33 (−7+ 33)t/40 ( ) c1 e +( )c2 e(−7− 33)t/40 . 4 4 1.686c1 e−0.03139t − 1.186c2 e−0.3186t .
[ B] = 10 = ≈
From the initial conditions, [ A](0) = A0 = 1.0 µmol, [ B](0) = 0, we have 1.0 = c1 + c2 , 0 = 1.686c1 − 1.186c2 . Solving, we obtain c1 = 0.4129, c2 = p p 0.5870. [Exact values are c1 = 1/2 − (33)/66, c2 = 1/2 + (33)/66.]
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This gives the solutions for the chemical concentrations as
[ A] = 0.4129e−0.03139t + 0.5870e−0.3186t [ B] = 0.6963e−0.03139t − 0.6963e−0.3186t [C ] = 1.0 − [ A] − [ B] = 1.0 − 1.109271796e−0.03139t + .1092717960e−0.3186t Plots of these concentrations are shown in Figure 3.9. Figure 3.9: Plot of the chemical concentrations in Problem 3.21.
22. Consider the epidemic model leading to the system in 3.153. Choose the constants as a = 2.0 days−1 , d = 3.0 days−1 , and r = 1.0 days−1 . What are the eigenvalues of the coefficient matrix? Find the solution of the system assuming an initial population of 1, 000 and one infected individual. Plot the solutions for t = 0.0 to 5.0 days and describe what is happening over this time. Is this model realistic? The linear SID model is given by S −a r 0 S d I = a −d − r 0 I dt D 0 d 0 D −2 1 0 S = 2 −4 0 I . 0 3 0 D Similar to the last example, we can concentrate on solving the first two equations and note that d (S + I + D ) = 0, dt or S + I + D = constant. ! ! ! d S −2 1 S = . (3.6) dt I 2 −4 I The eigenvalue equation is found as 0 = (−2 − λ)(−4 − λ) − 2 = λ2 + 6λ + √ 6. The solutions are λ = −3 ± 3.
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We let S = c1 e(−3+ I
= =
√
3) t
+ c2 e(−3−
√
3) t .
Then,
dS + 2S dt √ √ √ √ (−1 + 3)c1 e(−3+ 3)t + (−1 − 3)c2 e(−3− 3)t .
The initial conditions are S(0) = 999 and I (0) = 1. Then 999
= c1 + c2
√ √ = (−1 + 3)c1 + (−1 − 3)c2 . √ √ 500 500 The solution is c1 = 999 3, c2 = 999 3. 2 + 3 2 − 3 The full solution of the SID model is found as 999 500 √ (−3+√3)t 999 500 √ (−3−√3)t S = ( + +( − 3) e 3) e 2 3 2 3 1 1997 √ (−3−√3)t 1 1997 √ (−3+√3)t 3) e +( − 3) e I = ( + 2 6 2 6 999 √ (−3+√3)t 999 √ (−3−√3)t D = 1000 − (500 + 3) e 3) e − (500 − . 2 2 1
Figure 3.10: Plot of the populations in the epidemic model in Problem 3.22.
In Figure 3.10 these populations are shown for 5 days. 23. Find and classify any equilibrium points in the Romeo and Juliet problem for the following cases. Solve the systems and describe their affections as a function of time. The linear Romeo and Juliet problem is given by dR = aR + bJ dt dJ = cR + dJ. dt
(3.7)
This is a relatively simple system. In each case we provide the coefficient matrix, the eigenvalues and the solution of the initial value problem. Also, the direction field for the system is shown. a. a = 0, b = 2, c = −1, d = 0, R(0) = 1, J (0) = 1. The coefficient matrix is 0 2 −1 0 Figure 3.11: Plot of the phase plane for the Romeo and Juliet affection model in Problem 3.23a.
!
The eigenvalue equation is given by λ2 + 2 = 0. The solutions are √ λ = ± 2i The solution of the initial value problem is found to be √ √ √ R(t) = cos( 2t) + 2 sin( 2t), √ √ √ 2 J (t) = cos( 2t) − sin( 2t). 2 In Figure 3.11 we see that the origin is an equilibrium point and is a center. The relationship is cyclical with their affections for each other going back an forth between like and dislike.
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b. a = 0, b = 2, c = 1, d = 0, R(0) = 1, J (0) = 1. The coefficient matrix is 0 1
2 0
!
The eigenvalue equation is given by λ2 − 2 = 0. The eigenvalues √ are found as λ = ± 2 The solution of the initial value problem is found to be R(t)
=
J (t)
=
√ √ 1 1 + 2 e 2t + 2 √ √ 1 2 + 2 e 2t + 4
√ √ 1 1 − 2 e− 2t , 2 √ √ 1 2 − 2 e− 2t . 2
In Figure 3.12 we see that the origin is a saddle point. The relationship is unstable. For the given initial conditions, they find their affections for each other getting ever stronger. c. a = −1, b = 2, c = −1, d = 0, R(0) = 1, J (0) = 1. The coefficient matrix is
−1 2 −1 0
!
Figure 3.12: Plot of the phase plane for the Romeo and Juliet affection model in Problem 3.23b.
The eigenvalue equation is given by λ2 + λ + 2 = 0. The eigenval√ ues are found as λ = 21 (−1 ± 7i ) The solution of the initial value problem is found to be √ √ ! 3√ 7 7 −t/2 R(t) = e cos( t) + t) , 7 sin( 2 7 2 √ √ √ ! 7 7 7 −t/2 J (t) = e cos( t) − sin( t) . 2 7 2 In Figure 3.11 we see that the origin is a stable spiral. The relationship is cyclical with the affections for each other going back an forth between like and dislike until finally there is no feeling one way or the other.
Figure 3.13: Plot of the phase plane for the Romeo and Juliet affection model in Problem 3.23c.
4 Nonlinear Dynamics 1. Solve the general logistic problem, dy = ky − cy2 , dt
y (0) = y0
using separation of variables. This can be accomplished using partial fraction decomposition. Z
dt
= =
kt + C
=
Aekt
=
dy y(k − cy) Z c 1 1 + dy. k k − cy y y . ln k − cy y , A = ±eC . k − cy Z
Solving for y(t), we have y(t) =
kAekt . 1 + cAekt
From the initial condition, we have A=
y0 . k − cy0
Inserting this into the solution, we find y(t)
= = =
kAekt 1 + cAekt ky0 ekt k − cy0 + cy0 ekt ky0 . cy0 + (k − cy0 )e−kt
2. Find the equilibrium solutions and determine their stability for the following systems. For each case, draw representative solutions and phase lines.
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a. y0 = y2 − 6y − 16. For this problem, the equilibrium solutions satisfy the equation y
y0 = (y − 8)(y + 2) = 0.
u
y=8
s
t y = −2
Figure 4.1: Solutions for Problem 4.2a. y
u
y=
3π 2
s
y=
π 2
u
y = − π2
s
y = − 3π 2
Thus, y = 8, −2 are the equilibrium solutions. In Figure 4.1 the sign of the derivative is given in the phase line and typical solutions are indicated in the regions between the solutions. y = 8 is unstable and y = −2 is stable. b. y0 = cos y. For this problem, the equilibrium solutions satisfy cos y = 0, Thus, y = 2k2−1 π, for k an integer, are the equilibrium solutions. In Figure 4.2 the sign of the derivative is given in the phase line and typical solutions are indicated in the regions between the solutions. Solutions y = 2k2−1 π, for k an even integer are unstable and the others are stable. c. y0 = y(y − 2)(y + 3). For this problem, the equilibrium solutions satisfy are y = 0, 2, −3. In Figure 4.3 the signs of the derivatives are given on the phase line and typical solutions are indicated in the regions between the solutions. y = 2, −3 are unstable and y = 0 is stable. d. y0 = y2 (y + 1)(y − 4). For this problem, the equilibrium solutions are y = 0, −1, 4. In Figure 4.4 the signs of the derivatives are given on the phase line and typical solutions are indicated in the regions between the solutions. Solution y = 4 is unstable, y = −1 is stable, and y = 0 is neither.
Figure 4.2: Solutions for Problem 4.2b. y
y=2
u
3. For y0 = y − y2 , find the general solution corresponding to y(0) = y0 . y = 0 Provide specific solutions for the following initial conditions and sketch them: a. y(0) = 0.25, b. y(0) = 1.5, and c. y(0) = −0.5. This is a special case of problem 4.1. This equation is separable and can y = −3 be integrated using partial fraction decomposition.
s
u
Z Figure 4.3: Solutions for Problem 4.2c.
dt
y
u
s
= =
y=4
y=0 y = −1
t+C
=
Aet
=
y(t)
=
dy y (1 − y ) Z 1 1 + dy. 1−y y y . ln 1 − y y , A = ±eC . 1−y Z
Aet . 1 + Aet
From the initial condition, we have A= Figure 4.4: Solutions for Problem 4.2d.
y0 . 1 − y0
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113
Inserting this into the solution, we find y(t)
= =
y
Aet 1 + Aet y0 y0 + (1 − y0 ) e − t
y=1
The particular solutions for the given initial conditions are a. y(0) = 0.25, y(t) = b. y(0) = 1.5, y(t) =
t
1 . 1+3e−t
Figure 4.5: Solution behavior for Problem 4.3
3 . 3− e − t
c. y(0) = −0.5. y(t) =
y=0
1 . 1−3e−t
These solutions are shown in Figure 4.5. 4. For each problem determine equilibrium points, bifurcation points, and construct a bifurcation diagram. Discuss the different behaviors in each system. a. y0 = y − µy2 The equilibrium points are given by y = 0 and y = 1/µ. Letting f (y) = y − µy2 , then f 0 (y) = 1 − 2µy. So, f 0 (0) = 1 > 0, and f 0 (1/µ) = −1 < 0. Therefore, y = 0 is unstable and and y = 1/µ is stable. This is seen in Figure 4.6. b. y0 = y(µ − y)(µ − 2y)
Figure 4.6: Bifurcation diagram for Problem 4.4a y=µ
The equilibrium points are y = 0, µ/2, µ. Let f (y) = y(µ − y)(µ − 2y). Then, f 0 (y) = µ2 − 6µy + 6y2 and f 0 (0) = µ2 , f 0 (µ/2) = −µ2 /2, f 0 (µ) = µ2 . Therefore, y = 0, µ are unstable and y = µ/2 is stable. This is seen in Figure 4.7.
y=
µ 2
y=0
c. x 0 = µ − x3
√ The equilibrium point is at x = 3 µ. Let f ( x ) = µ − x3 . Then √ f 0 ( x ) = −3x2 and x = 3 µ is stable as seen in Figure 4.8.
d.
x0
= x−
µx 1+ x 2
p The equilibrium points are x = 0, x = ± µ − 1. There is one µx equilibrium point for µ < 1. For f ( x ) = x − 1+ x2 , f 0 (x) = 1 −
µ 2µ ∗ x2 + 1 + x2 (1 + x 2 )2
x=
√ 3
µ
Figure 4.8: Bifurcation diagram for Problem 4.4c
and f 0 (0) = 1 − µ. So, x = 0 is stable for µ > 1 and unstable for p 2( µ −1) µ < 1. Since f 0 (± µ − 1) = µ > 0 for µ > 1, the equilibrium p points x = ± µ − 1 are unstable. This is seen in Figure 4.9. 5. Consider the family of differential equations x 0 = x3 + δx2 − µx. a. Sketch a bifurcation diagram in the xµplane for δ = 0.
Figure 4.7: Bifurcation diagram for Problem 4.4b
x=
p
x=−
√
For δ = 0, x 0 = ( x2 − µ) x. The equilibrium points are x = 0, ± µ.
µ−1
p
µ−1
Figure 4.9: Bifurcation diagram for Problem 4.4d
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b. Sketch a bifurcation diagram in the xµplane for δ > 0. From f ( x ) = x ( x2 + δx − µ) we have x = 0 is an equilibrium p point and x = 12 (−δ ± δ2 + 4µ) is a pair of equilibria provided δ2 + 4µ > 0. Noting that f 0 ( x ) = 3x2 + 2δx − µ, we have f 0 (0) = −µ and x = 0 is stable for µ > 0 and unstable for µ < 0. q q 1 1 f ( (−δ ± δ2 + 4µ)) = (δ2 + 4µ ∓ δ (δ2 + 4µ)) 2 2 p p 1 So, x = 2 (−δ − δ2 + 4µ) is always unstable. x = 21 (−δ + δ2 + 4µ) is stable when q 2 δ + 4µ < δ (δ2 + 4µ) 2
or − δ4 < µ < 0. x=
Figure 4.10: Bifurcation diagrams for Problem 4.5b for (a) δ = 0 and (b) δ > 0.
√
µ
x=0
√ x=− µ
(a)
(b)
Hint: Pick a few values of δ and µ in order to get a feel for how this system behaves. 6. System (4.63) can be solved exactly. Integrate the requation using separation of variables. For initial conditions (a) r (0) = 0.25, θ (0) = 0, and (b) r (0) = 1.5, θ (0) = 0, and µ = 1.0, find and plot the solutions in the xyplane showing the approach to a limit cycle. System 4.63 is given by r 0 = µr − r3 ,
θ 0 = 1.
The radial equation is separable. Integrating, we obtain Z
dt
=
t+C
= =
Aeµt
=
dr r (µ − r2 Z 1 1 r + dr µ r µ − r2 1 1 ln r − ln µ − r2  µ 2 r p , A = ±eC . µ − r2 Z
nonlinear dynamics
Solving for r, we find
√ r (t) = √
Letting r (0) = r0 , we find A = √ r0
µAeµt
1 + A2 e2µt . So, 2
115
.
µ −r0
√ r (t)
=
=
=
q
µr0 eµt
µ − r02 + r02 e2µt √ µr0
q
r02 + (µ − r02 )e−2µt √ µ r . µ −r 2 1 + r2 0 e−2µt 0
Setting µ = 1, the following initial conditions yield the given solutions. √ a. r (0) = 0.25, r (t) = 1/ 1 + 15e−2t . √ b. r (0) = 1.5, r (t) = 3/ 9 − 5e−2t . Letting t → ∞ we see these solutions tend to the limit cycle r = 1, which is the unit circle. This is shown in Figure 4.11 where the parametric curves (r (t) cos t, r (t) sin t). 7. Consider the system x0 y0
h i = − y + x µ − x 2 − y2 , h i = x + y µ − x 2 − y2 .
Rewrite this system in polar form. Look at the behavior of the r equation and construct a bifurcation diagram in µr space. What might this diagram look like in the threedimensional µxy space? (Think about the symmetry in this problem.) This leads to what is called a Hopf bifurcation. We first recall the expressions for r 0 and θ 0 : r0
=
θ0
=
xx 0 + yy0 , r xy0 − yx 0 . r2
The numerators can be evaluated using the system of differential equations, giving xx 0 + yy0
= − xy + x2 (µ − x2 − y2 ) + ( xy + y2 (µ − x2 − y2 )) = ( x2 + y2 )(µ − x2 − y2 ).
xy0 − yx 0
= x2 + xy(µ − x2 − y2 ) − (−y2 + xy(µ − x2 − y2 )) = x 2 + y2 .
Since x2 + y2 = r2 , rr 0 r2 θ 0
= r 2 ( µ − r 2 ), = r2 .
Figure 4.11: The solutions in Problem 4.6 tend towards the unit circle.
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Thus, the system in polar form is given by the equations x
µ y
Figure 4.12: A three dimensional bifurcation diagram for Problem 4.7.
r0
= r ( µ − r 2 ),
θ0
= 1,
Focusing on the radial equation, we find equilibrium solutions at r = 0 √ and r = µ, for µ ≥ 0. The stability is determined from differentiating the right hand side of the radial equation, f 0 (r ) = µ − 3r2 . For the origin, f 0 (0) = µ. So, the origin is stable for µ < 0 and unstable for µ > 0. This is shown in Figure 4.12. Thus, the stable origin goes unstable as µ changes from positive to negative. So, there is a bifurcation when µ = 0. When the √ origin goes unstable, a stable equilibrium is born (appears) at r = µ. In µrspace this is a pitchfork bifurcation. However, since r = sqrtx2 + y2 , we find in µxyspace a circular paraboloid exhibiting a Hopf bifurcation. 8. Find the fixed points of the following systems. Linearize the system about each fixed point and determine the nature and stability in the neighborhood of each fixed point, when possible. Verify your findings by plotting phase portraits using a computer. a. x0
= x (100 − x − 2y),
0
= y(150 − x − 6y).
y
The equilibrium solutions are found from solving the system 0
= x (100 − x − 2y),
0
= y(150 − x − 6y).
From the first equation, either x = 0 or x + 2y = 100. If x = 0, then either y = 0 or y = 150/6 = 25. If x + 2y = 100, then y = 0 or x + 6y = 150. In the first case, we have y = 0 and x = 100. In the second case, We need to solve x + 2y
= 0,
x + 6y
= 150.
The solution of this system is x = 75, y = equilibrium points
25 2 .
(0, 0), (0, 25), (100, 0), (75,
Thus, there are four
25 ). 2
The Jacobian matrix is given by J ( x, y) =
100 − 2x − 2y −y
−2x 150 − x − 12y
! .
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117
! 100 0 a. For (0, 0) the Jacobian matrix is J (0, 0) = and 0 150 the corresponding eigenvalues are λ = 100, 150. This indicates that (0, 0) is an unstable node. ! 50 0 b. For (0, 25) the Jacobian matrix is J (0, 25) = −25 −150 and the corresponding eigenvalues are λ = 50, −150. This indicates that (0, 25) is a saddle. ! −100 −200 c. For (100, 0) the Jacobian matrix is J (100, 0) = 0 50 and the corresponding eigenvalues are λ = −100, 50. This indicates that (100, 0) is a saddle. Figure 4.13: The phase portrait for the system in Problem 4.8a.
! −75 150 d. For (75, the Jacobian matrix is J (75, = − 25 2 √−75 and the corresponding eigenvalues are λ = −75 ± 25 3. This indicates that (75, 25 2 ) is a stable node. 25 2 )
25 2 )
The phase portrait for the system is shown in Figure 4.13. b. x0
= x + x3 ,
y0
= y + y3 .
There is only one equilibrium point for this system, (0, 0). The Jacobian matrix is given by J ( x, y) =
1 + 3x2 0
0 1 + 3y2
! .
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! 1 0 For (0, 0) the Jacobian matrix is J (0, 0) = and the eigen0 1 value is λ = 1. The phase portrait for the system is shown in Figure 4.14. This indicates that (0, 0) is an unstable node. Figure 4.14: The phase portrait for the system in Problem 4.8b.
c. x0
= x − x2 + xy,
y0
= 2y − xy − 6y2 .
The equilibrium solutions are found from solving the system 0
= x (1 − x + y ),
0
= y(2 − x − 6y).
From the first equation, either x = 0 or x − y = 1. If x = 0, then either y = 0 or y = 31 . If x − y = 1, then y = 0 or x + 6y = 2. In the first case, we have y = 0 and x = 1. In the second case, We need to solve x−y
= 1,
x + 6y
= 2.
The solution of this system is x = equilibrium points
8 7,
y =
1 7.
Thus, there are four
1 8 1 (0, 0), (1, 0), (0, ), ( , ). 3 7 7 The Jacobian matrix is given by J ( x, y) =
1 − 2x + y −y
x 2 − x − 12y
! .
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119
! 1 0 a. For (0, 0) the Jacobian matrix is J (0, 0) = and the 0 2 corresponding eigenvalues are λ = 1, 2. This indicates that (0, 0) is an unstable node. ! −1 1 b. For (1, 0) the Jacobian matrix is J (1, 0) = and the 0 1 corresponding eigenvalues are λ = ±1. This indicates that (1, 0) is a saddle. ! 4 0 3 c. For (0, 31 ) the Jacobian matrix is J (0, 31 ) = and − 13 −2 the corresponding eigenvalues are λ = −2, 34 . This indicates that (0, 31 ) is a saddle. ! 8 8 − 7 7 d. For ( 78 , 71 ) the Jacobian matrix is J ( 87 , 17 ) = and − 17 − 67 √
the corresponding eigenvalues are λ = −1 ± that ( 87 , 17 ) is a stable focus.
3 7 i.
This indicates
The phase portrait for the system is shown in Figure 4.15. Figure 4.15: The phase portrait for the system in Problem 4.8c.
d. x0 y
0
= −2xy, = − x + y + xy − y3 .
The equilibrium solutions are solutions of the system 0
= −2xy,
0
= − x + y + xy − y3 .
Therefore, either x = 0 or y = 0. If x = 0, the second equation becomes y − y3 = 0. This gives y = 0, ±1. If y = 0, then the second
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equation becomes − x = 0. So, we have found three equilibrium solutions, (0, 0), (0, 1), (0, −1). The Jacobian matrix is given by J ( x, y) =
−2y −2x y − 1 1 + x − 3y2
! .
! 0 0 a. For (0, 0) the Jacobian matrix is J (0, 0) = and the −1 1 corresponding eigenvalues are λ = 0, 1. This indicates that (0, 0) is an unstable node. ! −2 0 b. For (0, 1) the Jacobian matrix is J (1, 0) = and the 0 −2 corresponding eigenvalue is λ = −2. This indicates that (0, 1) is a stable node. ! 2 0 1 c. For (0, −1) the Jacobian matrix is J (0, 3 ) = and −2 −2 the corresponding eigenvalues are λ = ±2. This indicates that (0, −1) is a saddle. The phase portrait for the system is shown in Figure 4.16. Figure 4.16: The phase portrait for the system in Problem 4.8d.
9. Plot phase portraits for the Lienard system x0
= y − µ ( x 3 − x ),
y0
= − x,
for a small and a not so small value of µ. Describe what happens as one varies µ. Phase portraits for different values of µ are shown in Figures 4.174.19. As the magnitude of µ gets larger the limit cycle grows and becomes less
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Figure 4.17: The phase portrait for the system in Problem 4.9 for µ = 0.05 (left) and µ = 0.2 (right).
Figure 4.18: The phase portrait for the system in Problem 4.9 for µ = 1.0 (left) and µ = 2.0 (right).
Figure 4.19: The phase portrait for the system in Problem 4.9 for µ = −1.0 (left) and µ = −2.0 (right).
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circular. The limit cycle is stable for µ > 0 and the origin is unstable. The opposite occurs for µ < 0. ! µ 1 This is confirmed by looking at the Jacobian matrix J (0, 0) = . −1 0 p The corresponding eigenvalues are λ = 21 (µ ± µ2 + 4). This indicates that (0, 0) is unstable for µ > 0 and stable for µ < 0. 10. Consider the period of a nonlinear pendulum. Let the length be L = 1.0 m and g = 9.8 m/s2 . Sketch T versus the initial angle θ0 , and compare the linear and nonlinear values for the period. For what angles can you use the linear approximation confidently? The exact period for the nonlinear pendulum is given by s L θ0 T=4 K sin g 2 and the period for the linear pendulum is given by s L . T = 2π g Using a CAS package, one can plot the Elliptic function and the results are shown in Figure 4.20. For an error of about 1% the angle needs to be below 18o . The desired error will depend on the typical error desired in the experiment. Figure 4.20: The period (in seconds) for the nonlinear pendulum in Problem 4.10 as a function of the angle (in degrees).
11. Another population model is one in which species compete for resources, such as a limited food supply. Such a model is given by x 0 = ax − bx2 − cxy, y0 = dy − ey2 − f xy. In this case, assume that all constants are positive. a. Describe the effects/purpose of each terms. • ax is the growth rate of species x if there is no competition with either species.
nonlinear dynamics
• bx2 is logistic growth term which contributes when species x gets large. • c is the term that accounts for competition with species y. • dy is the growth rate of species y if there is no competition with either species. • ey2 is logistic growth term which contributes when species y gets large. • f s the term that accounts for competition with species x. b. Find the fixed points of the model. Equilibrium solutions satisfy the system 0
= x ( a − bx − cy),
0
= y(d − ey − f x ).
From the first equation, either x = 0 or bx + cy = a. If x = 0, then either y = 0 or y = de . If bx + cy = a, then y = 0 or f x + ey = d. In the first case, we have y = 0 and x = ba . In the second case, We need to solve bx + cy
= a,
f x + ey
= d.
The solution of this system is x = are four equilibrium points
ae−cd be−c f
and y =
bd− a f be−c f
. Thus, there
d a ae − cd bd − a f (0, 0), (0, ), ( , 0), ( , ). e b be − c f be − c f c. Linearize the system about each fixed point and determine the stability. The Jacobian matrix is given by J ( x, y) =
a − 2bx − cy −fy
!
−cx d − 2ey − f x
a. For (0, 0) the Jacobian matrix is J (0, 0) =
.
a 0
0 d
! and the
corresponding eigenvalues are λ = a, d. b. For (0,
d e)
the Jacobian matrix is J (0,
d e)
=
0 −d
!
− a − acb af 0 d− b
!
a − cd e − def
and
the corresponding eigenvalues are λ = −d, cd−e ae . c. For ( ba , 0) the Jacobian matrix is J ( ba , 0) = the corresponding eigenvalues are λ = − a, d. For
bd− a f b
.
ae−cd bd− a f ( be −c f , be−c f
) the Jacobian matrix is (cd− ae)b ae − cd a f − bd cf J( , ) = (abef −−bd )f be − c f be − c f be−c f
(cd− ae)b be−c f ( a f −bd)e be−c f
and
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and the corresponding eigenvalues are fairly complicated. The eigenvalue equation is given by
(be − c f )λ2 − ( ae f + bcd − abe − bde)λ + ( abde + acd f − a2 e f − bcd2 ) = 0. d From the above, describe the types of solution behavior you might expect, in terms of the model. The model is to general to pinpoint exact behaviors for each equilibrium point. However, a few things can be said. • (0, 0) has two positive eigenvalues. This is an unstable node. • (0, de ) has one negative eigenvalue, λ = −d. The other eigenvalue, λ = cd−e ae , is positive for cd − ae > 0, leading to a saddle and for cd − ae < 0 we have a stable node. • Similarly, ( ba , 0) has one negative eigenvalue, λ = − a. The other bd− a f
eigenvalue, λ = b , is positive for bd − a f > 0, leading to a saddle and for bd − a f < 0 we have a stable node. Other types of behavior are possible at the fourth equilibrium point. For example, we can determine the trace and determinant as tr( J )
=
ae f + bcd − abe − bde , be − c f
det( J )
=
abde + acd f − a2 e f − bcd2 . be − c f
Then, we have different behaviors for the different parameters: • det( J ) < 0, saddle points. • det( J ) > 0, tr( J ) > 0, unstable spirals. • det( J ) > 0, tr( J ) > 0, stable spirals. • det( J ) > 0, tr( J ) = 0, centers. 12. Consider a model of a food chain of three species. Assume that each population on its own can be modeled by logistic growth. Let the species be labeled by x (t), y(t), and z(t). Assume that population x is at the bottom of the chain. That population will be depleted by population y. Population y is sustained by x’s, but eaten by z’s. A simple, but scaled, model for this system can be given by the system x0
= x (1 − x ) − xy,
y
0
= y(1 − y) + xy − yz,
z
0
= z(1 − z) + yz.
a. Find the equilibrium points of the system. We determine the equilibria from the solution of three equations and three unknowns, 0
= x (1 − x − y )
0
= y (1 − y + x − z )
0
= z (1 − z + y ).
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125
In Figure 4.21 we indicate the progression through the system in Problem 12 to find the equilibrium solutions. At the top level we list the implications of the first equation. At the third and fifth levels are the effects on the second and third equations followed by the results at each step. x (1 − x − y ) = 0 x=0
x = 1−y
y (1 − y − z )
y (2(1 − y ) − z )
y = 1−z 2z(1 − z) = 0 z=0
z=1
(0, 1, 0)
(0, 0, 1)
y = 1−
y=0
z=0
(0, 1, 0)
(0, 0, 1)
z=
b. Find the Jacobian matrix for the system and evaluate it at the equilibrium points.
0 1 − 2z + y . −y
Evaluation at each equilibrium yields 1 0 0 1 J (0, 0, 0) = 0 0 1 . J (0, 0, 1) = 0 0 1 0 0 −1 −1 0 J (1, 0, 0) = 0 0 1 . J (1, 0, 1) = 0 2 0 0 0 0 21 4 J (0, 1, 0) = 0 0 2 . J( , ) = 33 3 1 −1 −1
z=1
(1, 0, 0)
(1, 0, 1)
Figure 4.21: Progression through system in Problem 12 to find the equilibrium solutions.
2 1 4 (0, 0, 0), (0, 0, 1), (1, 0, 0), (1, 0, 1), (0, 1, 0), ( , , ). 3 3 3
z=0 4 3
( 32 , 13 , 43 )
For this problem there are six equilibria:
The Jacobian matrix for the system is 1 − 2x − y −x J ( x, y, z) = 0 z y 1 − 2y + x − z
y=0
z(2 − 32 z) = 0
z=1
z=0
(0, 0, 0)
z 2
0 1 0
0 −1 . 0
−1 −1 0 0 1 −1 . 0 1 0 − 23 − 23 0 4 0 − 43 . 3 1 − 13 − 31 3
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c. Find the eigenvalues and eigenvectors. • (0, 0, 0), λ = ±1,
0 (v1 , v2 , v3 ) = −1 1
0 1 1 0 . 1 0
• (0, 0, 1), λ = 1,
0 (v1 , v2 , v3 ) = 1 1
1 0 0
0 1 . 0
√ • (1, 0, 0), λ = −1, ± 2, (v1 , v2 , v3 ) =
1 2 (−2√+ 2 2
√
2)
√
1 2 (−2 − √ − 22
1 • (0, 0, 1), λ = −1, 12 (1 ±
√
0 (v1 , v2 , v3 ) = 0 0 • (0, 1, 0), λ = 0, 12 (−1 ±
√
2)
1
0 . 0
1
3i ),
√
√
1 6 (−3 −√ 3i ) 1 2 (1 + 3i )
1 6 (−3 +√ 3i ) 1 2 (1 − 3i )
1
1
.
7),
0 √ 1 (v1 , v2 , v3 ) = 2 (−1 − 7i ) 1
0 √ 1 2 (−1 + 7i ) 1
1 0 . 0
• ( 32 31 , 43 ), λ = −0.6464 ± 0.3589i, 1.6261
−.7156 −.7156 0.2732 (v1 , v2 , v3 ) = 0.02176 + 0.3853i 0.02176 − 0.3853i −0.9396 . 0.1360 + 0.5662i 0.1360 − 0.5662i 0.2063 d. Describe the solution behavior near each equilibrium point. For each equilibrium point one looks at the eigenvalues to determine the behavior in the neighborhood of the point. This gives the following results. • (0, 0, 0), λ = ±1, This is a saddle. • (0, 0, 1), λ = 1,
√ • (1, 0, 0), λ = −1, ± 2, This is a saddle √ • (0, 0, 1), λ = −1, 12 (1 ± 3i ), √ • (0, 1, 0), λ = 0, 12 (−1 ± 7), • ( 23 31 , 43 ), λ = −0.6464 ± 0.3589i, 1.6261
nonlinear dynamics
e. Which of these equilibria are important in the study of the population model and describe the interactions of the species in the neighborhood of these point(s). The only equilibrium point with all three populations as nonzero is ( 32 13 , 43 ). This would be the most important to study. 13. Derive the first integral of the LotkaVolterra system, a ln y + d ln x − cx − by = C. The LotkaVolterra system is given by x˙
= ax − bxy,
y˙
= −dy + cxy,
where we can eliminate t by writing dy −dy + cxy y = = dx ax − bxy x
−d + cx a − by
.
This equation is separable and can be integrated. a − by dy y Z a − b dy y a ln y − by Z
−d + cx dx x Z d = dx c− x = cx − d ln  x  + C. =
Z
For populations with x, y > 0, the result follows. 14. Show that the system x 0 = x − y − x3 , y0 = x + y − y3 , has at least one limit cycle by picking an appropriate ψ( x, y) in Dulac’s Criteria. In general, appropriate ψ( x, y)’s are hard to find. In this case, we can try ψ( x, y) ≡ 1 defined in the annular region a < x2 + y2 < b. Also, we have f ( x, y) = x − y − x3 , g( x, y) = x + y − y3 . Then, ∂ ∂ (ψ f ) + (ψg) = 2 − 3( x2 + y2 ) ∂x ∂y for a < x2 + y2 < b. Therefore 2 − 3b
0. Since tr( A) < 0, we have stable nodes or stable spirals. To determine which, we need to consider tr2 ( A) − 4det( A)
= (σ + 1)2 − 4σ(1 − ρ) = (σ − 1)2 + 4σρ > 0.
Therefore, the origin is a stable node confirming the above. We can consider the stability of the other equilibria. The eigenvalue equation in both cases is given by λ3 + (1 + σ + β)λ2 + (σβ + β + β(ρ − 1))λ + 2β(ρ − 1)σ = 0. However, this is unwieldy. So, we fix σ = 10, β = 8/3, and vary ρ. The eigenvalue equation for the origin is given by (8 + 3λ)(−λ2 − 11λ − 10 + 10ρ) = 0. The roots are given by p 8 1 λ = − , (11 ± 81 + 40ρ). 3 2
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For ρ < 1 we confirm that the origin is a stable node. For ρ > 1, the origin is unstable and now we can look at the other equilibria. The solution of the eigenvalue equation is the root of the cubic f ( λ ) = λ3 +
41 2 8 160 λ + (10 + ρ)λ + ( ρ − 1). 3 3 3
We can plot this function for different values of ρ. As ρ varies, we expect the number of real roots to change, leading to the change in the behavior of the solutions. Figure 4.22: A plot of f (λ) = λ3 + 41 2 8 160 3 λ + 3 (10 + ρ ) λ + 3 ( ρ − 1) in Problem 4.15 as a function of λ for ρ = 1.0, 1.1, 1.2, 1.3, 1.4.
In Figure 4.22 we show this for ρ = 1.0, 1.1, 1.2, 1.3, and 1.4. We see that for 1 < ρ ≤ 1.3, that there are three real and negative roots. Thus, the equilibrium points are stable nodes. For ρ ≥ 1.4 there are two complex roots, leading to stable spirals. We can find the bifurcation point by determining when the maximum lies on the axis. df 82 8 = 3λ2 + λ + (10 + ρ) = 0. dλ 3 3 The locations for the extrema of f (λ) are λ=
p 1 (−41 ± 961 − 72ρ). 9
Inserting these into f (λ) and finding ρ for which the maximum is zero, we have ρ = 1.3456.... It is at this point that the origin is unstable and the first stable oscillations appear at the other equilibria. Further analysis can be done numerically. In Figures 4.234.26 we plot representative solutions for different ρvalues. In these cases we have two initial conditions near the origin such as x (0) = 0.01, y(0) = 0.01, z(0) = 0.01 and x (0) = −0.01, y(0) = −0.01, z(0) = 0.01. In Figure 4.23 the solutions tend to the origin. In Figure 4.24 the solutions move away from the origin. In the next figures we see the nodes turn into spirals. As
nonlinear dynamics
Figure 4.23: A plot of the solution of Problem 4.15 for ρ = 0.5 (left) and ρ = 1.1 (right).
Figure 4.24: A plot of a solution of Problem 4.15 for ρ = 5.0 (left) and ρ = 13.0 (right).
Figure 4.25: A plot of a solution of Problem 4.15 for ρ = 15.0 (left) and ρ = 20.0 (right).
Figure 4.26: A plot of a solution of Problem 4.15 for ρ = 24.0 (left) and ρ = 26.0 (right).
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the parameter ρ increases, we see the orbits no longer tend to the closes equilibrium point and limit cycles appear to form. In Figures 4.274.29 are shown plots of x (t) vs t for different ρvalues. In Figure 4.27 solutions for ρ = 0.5, 1.2, 1.5, 10, 13, 14 are shown. The lower values of ρ tend towards constant x values. However, larger values show the osciallation about an equilibrium point with x > 0. Figure 4.27: A plot of solutions x (t) vs t in Problem 4.15 for ρ = 0.5, 1.2, 1.5, 10, 13, 14.
In Figure 4.28 a plot of x (t) vs t for ρ = 20.0 is shown. Note that in this case the orbit spends time circling both equilibria before finally settling on one of them. In Figure 4.29 similar behaviors are shown for ρ = 22.0 and ρ = 24.3. In the latter case we see that the orbit has not settled on a particular equilibrium point. Figure 4.28: A plot of solutions x (t) vs t in Problem 4.15 for ρ = 20.0.
Figure 4.29: A plot of solutions x (t) vs t in Problem 4.15 for ρ = 22.0, 24.3
d. This system is known to exhibit chaotic behavior. Lorenz found a socalled strange attractor for parameter values σ = 10, β = 8/3, and ρ = 28. Using a computer, locate this strange attractor. In Figure 4.30 we show a typical plot of the strange attractor for this system. 16. The MichaelisMenten kinetics reaction is given by E+S o
k3 k1
/ ES
k2
/ E + P.
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133
Figure 4.30: A plot solutions in Problem 4.15d for ρ = 28.0 showing the strange attractor of the Lorenz system.
The resulting system of equations for the chemical concentrations is d[S] dt d[ E] dt d[ ES] dt d[ P] dt
= −k1 [ E][S] + k3 [ ES], = −k1 [ E][S] + (k2 + k3 )[ ES], = k1 [ E][S] − (k2 + k3 )[ ES], = k3 [ ES].
In chemical kinetics, one seeks to determine the rate of product formation (v = d[ P]/dt = k3 [ ES]). Assuming that [ ES] is a constant, find v as a function of [S] and the total enzyme concentration [ ET ] = [ E] + [ ES]. As a nonlinear dynamical system, what are the equilibrium points? If [ ES] is a constant, then k1 [ E][S] − (k2 + k3 )[ ES] = 0, or [ ES] = k k+1k [ E][S] Furthermore, we have
[ ES] =
d dt [ ET ]
k1 k1 [ E][S] = ([ E]0 − [ ES])[S] k2 + k3 k2 + k3
or
(1 +
k1 k1 [S])[ ES] = [ E ]0 [ S ] k2 + k3 k2 + k3
[ ES] = where K =
k2 +k3 k1
2
3
= 0. Let [ ET ] = [ E]0 be the constant. Then,
k1 k2 +k3 [ E ]0 [ S ] (1 + k2k+1k3 [S])
=
[ E ]0 [ S ] , K + [S]
is the MichaelisMenten constant. This gives v = k3 [ ES] =
k 3 [ E ]0 [ S ] . K + [S]
The equilibria are found as solutions of the system 0
= −k1 [ E][S] + k3 [ ES],
0
= −k1 [ E][S] + (k2 + k3 )[ ES],
0
= k1 [ E][S] − (k2 + k3 )[ ES],
0
= k3 [ ES].
Thus, [ ES] = 0 and either[ E] = 0 or [S] = 0.
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17. In Equation (3.153), we saw a linear version of an epidemic model. The commonly used nonlinear SIR model is given by dS = − βSI, dt dI = βSI − γI, dt dR = γI, dt where S is the number of susceptible individuals, I is the number of infected individuals, and R is the number who have been removed from the the other groups, either by recovering or dying. a. Let N = S + I + R be the total population. Prove that N = constant. Thus, one need only solve the first two equations and find R = N − S − I afterward. A simple computation yields d (S + I + R) = − βSI + βSI − γI + γI = 0, dt we have that S + I + R = constant. b. Find and classify the equilibria. Describe the equilibria in terms of the population behavior. The equilibria satisfy 0
= − βSI
0
=
0
= γI,
βSI − γI
The only equilibria are those with I = 0 and arbitrary S and R. Starting in the state (S, I, R) = ( N, i0 , 0), the system ends in an equilibrium state (S, I, R) = (0, 0, N ). This can be seen by looking at the linearized system about (S, 0, R). dS = − βS dt dI = βS − γI dt dR = γI, dt Thus, S and I will tend to zero. Since S + I + R = N, then R tends to N. c. Let β = 0.05 and γ = 0.2. Assume that in a population of 100, there is one infected person. Numerically solve the system of equations for S(t) and I (t) and describe the solution being careful to determine the units of population and the constants. In Figure 4.31 we show the plot of the solution with these parameters. The susceptible population decreases quickly as the infected population first increases and then decreases leaving most of the population in the recovered state.
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135
Figure 4.31: A plot of solutions S(t) (blue), I (t) (red), and R(t) (black) for the SIR model in Problem 4.17 for the initial values of S(0) = 99, I (0) = 1, and R(0) = 0. The parameters are β = 0.05, and γ = 0.2.
d. The equations can be modified by adding constant birth and death rates. Assuming these are the same, one would have a new system. dS dt dI dt dR dt
= − βSI + µ( N − S), =
βSI − γI − µI,
= γI − µR.
How does this affect any equilibrium solutions? Now the equilibria are more interesting. Setting the left hand side to zero, one finds two equilibria, (S, I, R) = ( N, 0, 0) and γ + µ µ( Nβ − γ − µ) γ( Nβ − γ − µ) (S, I, R) = , , . β β(γ + µ) β(γ + µ) Thus, we typically would start near the equilibrium point with some nonzero small Ivalue and S near N. Then, we expect that this equilibrium would be unstable and the system would be driven towards the second equilibrium. This is born out in the stability analysis of the system. The Jacobian matrix is given by − βI − µ J ( x, y, z) = βI 0
− βS 0 βS − γ − µ 0 . γ −µ
The Jacobian matrix at the equilibrium point ( N, 0, 0) is −µ − βN 0 J= 0 βN − γ − µ 0 . 0 γ −µ The eigenvalue equation is
(µ + λ)2 (−γ − µ + βN − λ) = 0. Note that the eigenvalues are λ = −µ, βN − γ − µ. Since µ > 0 and βN is typically larger than γ + µ, we have that this equilibrium point is a saddle point.
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γ+µ µ( Nβ−γ−µ) γ( Nβ−γ−µ) The Jacobian matrix at the equilibrium point , β(γ+µ) β , β(γ+µ) is γ−µ) − µ( Nβγ− − µ − γ − µ 0 +µ µ( Nβ−γ−µ) J= 0 0 . γ+µ 0 γ −µ The eigenvalue equation is given by
−1 (µ + λ)((γ + µ)λ2 + µβNλ + µ(γ + µ)(−γ − µ + βN )) = 0. γ+µ The eigenvalues are given as λ = −µ, −
1
2(γ + µ) µβN ±
p
µ2 β2 N 2 + 4µ(γ + µ)3 − 4µβN (γ + µ)2
Note that the third term in the radical is larger than the second term when βN > γ + µ. This was the same assumption used to say that ( N, 0, 0) is a saddle. Furthermore, the discriminant is positive in this case and therefore all three eigenvalues are real and negative and the equilibrium point is a stable node. e. Again, let β = 0.05 and γ = 0.2. Let µ = 0.1 For a population of 100 with one infected person, numerically solve the system of equations for S(t) and I (t) and describe the solution being careful to determine the units of population and the constants. In Figure 4.32 we show the plot of the solution with these parameters. We note that the equilibrium point gives the asymptotic values of S
=
I
=
R
=
γ+µ =6 β µ( Nβ − γ − µ) = 31, β(γ + µ) γ( Nβ − γ − µ) = 62. β(γ + µ)
Also, the eigenvalues corresponding to this point are λ = −0.1000, −0.3606, −1.2894. This verifies that the solution is being driven towards the stable node for this set of parameters. In this modified problem the number of infected and susceptible people do not tend to zero in the end and the number of recovered do not tend to 100% of the population. 18. An undamped, unforced Duffing Equation, x¨ + ω 2 x + ex3 = 0, can be solved exactly in terms of elliptic functions. Using the results of Example 4.18, determine the solution of this equation and determine if there are any restrictions on the parameters.
.
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137
Figure 4.32: A plot of solutions S(t) (blue), I (t) (red), and R(t) (black) for the modified SIR model in Problem 4.17 for the initial values of S(0) = 100, I (0) = 1, and R(0) = 0. The parameters are β = 0.05, γ = 0.2, and µ = 0.1
In Exercise 4.18 we had shown that y(u) = sn(u) satisfies the differential equation y00 + (1 + k2 )y = 2k2 y3 . Comparing this equation to the unforced Duffing equation, we assume a solution of the form x (t) = ay(b(t − t0 )) for real a and b. Then, x˙ = aby0 and x¨ = ab2 y00 . The differential equation for x (t) becomes ab2 y00 + ω 2 ay + ea3 y = 0. Dividing by ab2 , we have y00 +
ω2 ea2 y = − y. b2 b2
Comparing this to the equation for y(u), ω2 b2 ea2 − 2 b
= 1 + k2 , = 2k2 .
We can solve this system for b and k in terms of ω, e, and a. The result is r r 1 2 −e 2 a. b = ω + ea , k = 2 2 2ω + ea2 Note that for real b we need 2ω 2 + ea2 > 0. For real k, we further need e < 0. Then, a solution of the undamped, unforced Duffing equation is given by ! r r 1 − e x (t) = a sn ω 2 + ea2 (t − t0 ); a . 2 2ω 2 + ea2 19. Determine the circumference of an ellipse in terms of an elliptic integral. One needs to know about arclength and parametrization of curves to carry out this computation. The parametric form of a standard ellipse centered at the origin is given by x = a cos θ, y = b sin θ, for 0 ≤ θ ≤ 2π. The infinitesimal arclength is s 2 2 dx dy ds = + dθ dθ dθ p = a2 sin2 θ + b2 cos2 θ dθ.
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Using symmetry, the circumference of an ellipse is given by the integral C=4
Z π/2 p 0
a2 sin2 θ + b2 cos2 θ dθ.
We now rewrite the integral as an elliptic integral. We will consider a > b, and define k2 = 1 − b2 /a2 . A little manipulation of the integral will yield the result. Z π/2 q C = 4 a2 (1 − cos2 θ ) + b2 cos2 θ dθ 0 Z π/2 q = 4 a2 − ( a2 − b2 ) cos2 θ dθ 0 Z π/2 r π = 4 a2 − ( a2 − b2 ) sin2 ( − θ ) dθ 2 0 Z π/2 p = 4a 1 − k2 sin2 t dt 0
= 4aE(k), where E(k) is the complete elliptic integral of the second kind. 20. Evaluate the following in terms of elliptic integrals, and compute the values to four decimal places. Recall the definitions of elliptic integrals of the first kind:
≡
F (φ, k)
0
=
dθ
Z φ
p
Z sin φ 0
≡
K (k)
dθ
R π/4 0
q
p
Z 1 0
a.
dθ
Z π/2 0
=
1 − k2 sin2 θ dz p 2 (1 − z )(1 − k2 z2 )
p
1 − k2 sin2 θ dz
(1 − z2 )(1 − k2 z2 )
.
.
1− 12 sin2 θ
By direct comparison with F (φ, k) we have dθ
Z π/4
q
0
b.
dθ
R π/2 0
q
1− 14 sin2 θ
π 1 = F ( , √ ) ≈ 0.8260. 4 2 1 − 12 sin2 θ
.
By direct comparison with K (k) we have dθ
Z π/2 0
c.
R2 0
√
dx . (9− x2 )(4− x2 )
q
1 = K ( ) ≈ 1.6858. 2 1 − sin θ 1 4
2
nonlinear dynamics
This integral needs to be rewritten in the form K (k) =
Z 1
dz p
0
(1 − z2 )(1 − k2 z2 )
.
This can be done using the substitution x = 2y. Z 2 0
dx p
(9 −
x2 )(4 − x2 )
=
0
= = d.
R π/2 0
Z 1
2dy p
Z 1 1
3
0
(9 − 4y2 )(4 − 4y2 ) dy q (1 − y2 )(1 − 49 y2 )
1 2 K ( ) ≈ 0.6023. 3 3
√ dθ . cos θ
In this problem we need to write the cosine in terms of a sine. We first note that sin2 2θ = 12 (1 − cos θ ), or cos θ = 1 − 2 sin2 2θ . √ Next, we consider the simple substitution 2 sin 2θ = sin t. Then, 1 − 2 sin2
θ = 1 − sin2 t = cos2 t 2
and the denominator is simplified. For θ = 0, t = 0. For θ = π/2, √ sin t = 2 sin π4 = 1. So, t = π/2. The differentials can be related after some manipulation: √ 2 θ cos dθ = cos t dt 2 2 √ 2 cos t dt dθ = cos 2θ √ 2 cos t dt = q 1 − sin2 2θ √ 2 cos t dt = q . 1 − 12 sin2 t Under this transformation, the integral now becomes √ Z π/2 Z π/2 2 cos t dt dθ q √ = 0 0 cos θ cos t 1 − 12 sin2 t
=
=
√ Z 2 √
dt
π/2
q
0
√ 2K (
1 − 21 sin2 t
2 ) ≈ 2.6221. 2
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mathematical methods for physicists
e.
R∞ 1
√ dx . x 4 −1
We can make a few transformations to turn the integral into the one in part d. First, we let x = 1/z, and dx = −dz/z2 . Z ∞
√
1
dx x4
−1
= −
Z 0 1 z2
Z 1
=
√
0
√
dz z −4 − 1
dz 1 − z4
Then, we introduce z2 = sin θ. So, 2z dz = cos θ dθ and this leads to the result. Z ∞ 1
√
dx x4
−1
=
Z 1
√
0
= = =
= =
dz 1 − z4
Z 1 π/2
2
0
Z 1 π/2
2
0
Z 1 π/2
2
0
√ √
cos θ dθ p sin θ 1 − sin2 θ dθ
sin θ dθ q cos( π2 − θ )
dt √ cos t √ 2 2 K( ) ≈ 1.3110. 2 2
1 2 √
Z π/2 0
5 The Harmonics of Vibrating Strings 1. Solve the following boundary value problems directly, when possible. a. x 00 + x = 2,
x (0) = 0,
x 0 (1) = 0.
The general solution of the differential equation is x (t) = c1 cos t + c2 sin t + 2. The first boundary condition, x (0) = 0, implies c1 = −2. This gives x (t) = −2 cos t + c2 sin t + 2 and x 0 (t) = 2 sin t + c2 cos t. The second boundary condition is x 0 (1) = 2 sin 1 + c2 cos 1 = 0. This gives c2 = −2 sin 1/ cos 1. Inserting c1 and c2 into the general solution we have x (t)
= b. y00 + 2y0 − 8y = 0,
sin 1 sin t + 2 cos 1 cos(1 − t) 2− . cos 1
= −2 cos t − 2
y(0) = 1,
y(1) = 0.
The characteristic equation is 0 = r2 + 2r − 8 = (r + 4)(r − 2). So, the general solution is y( x ) = c1 e−4x + c2 e2x . The boundary conditions imply c1 + c2
−4e
−4
= 1
2
c1 + 2e c2
= 0.
The second equation gives c2 = 2e−6 c1 . Substituting this result into the first equation, we obtain
(1 + 2e−6 )c1 c1
= 1 =
1 1 + 2e−6
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mathematical methods for physicists
and c2
= 1 − c1 2e−6 = 1 + 2e−6
Therefore, the solution is y( x ) = c. y00 + y = 0,
y(0) = 1,
e−4x + 2e−6 e2x . 1 + 2e−6
y(π ) = 0.
The general solution is y( x ) = c1 cos x + c2 sin x. The boundary condition y(0) = 1 gives c1 = 1 and the condition y(π ) = 0 implies that c2 = 0. So, the solution is y( x ) = cos x. 2. Find product solutions, u( x, t) = b(t)φ( x ), to the heat equation satisfying the boundary conditions u x (0, t) = 0 and u( L, t) = 0. Use these solutions to find a general solution of the heat equation satisfying these boundary conditions. We consider the heat equation of the form ut = ku xx for 0 < x < L and t > 0. Inserting u( x, t) = b(t)φ( x ), we use the method of separation of variables to obtain b0 φ00 = = − λ2 , kb φ where λ2 is the separation constant. This gives two ODEs b0
= −kλ2 b,
0
= φ00 + λ2 φ. 2
The temporal equation is readily solved, b(t) = b0 e−kλ t . The spatial function, φ( x ), satisfies the boundary conditions φ0 (0) = 0 and φ( L) = 0. The general solution of the spatial ODE is given by φ( x ) = c1 cos λx + c2 sin λx. The first boundary condition implies that c2 = 0, or φ( x ) = c1 cos λx. The second boundary condition gives φ( L) = c1 cos λL = 0. This is true for λn L =
2n − 1 π, 2
n = 1, 2, . . . ,
or
2n − 1 π, n = 1, 2, . . . . 2L Therefore, the eigenfunctions are 2n − 1 φn ( x ) = cos πx , n = 1, 2, . . . . 2L λn =
The product solutions are given by the products of the eigenfunctions and b(t), 2 2n − 1 un ( x, t) = cos πx e−kλn t , n = 1, 2, . . . . 2L
the harmonics of vibrating strings
A linear combination gives the general solution of the heat equation satisfying the given boundary conditions, ∞ 2n−1 2 2n − 1 u( x, t) = ∑ an cos πx e−k( 2L π ) t . 2L n =1 3. Find product solutions, u( x, t) = b(t)φ( x ), to the wave equation satisfying the boundary conditions u(0, t) = 0 and u x (1, t) = 0. Use these solutions to find a general solution of the heat equation satisfying these boundary conditions. We consider the wave equation of the form utt = c2 u xx for 0 < x < 1 and t > 0. Inserting u( x, t) = b(t)φ( x ), we use the method of separation of variables to obtain b00 φ00 = = − λ2 , 2 φ c b where λ2 is the separation constant. This gives two ODEs b00 + c2 λ2 b 00
2
φ +λ φ
= 0, = 0.
We write the temporal equation in the form b00 + ω 2 b = 0, where ω = cλ. Then, the general solution is given by b(t) = a cos ωt + b sin ωt. The spatial function φ( x ) satisfies the boundary conditions φ(0) = 0 and = 0. The general solution of the spatial ODE is given by
φ 0 (1)
φ( x ) = c1 cos λx + c2 sin λx. The first boundary condition implies that c1 = 0, or φ( x ) = c2 sin λx. The second boundary condition gives φ0 (1) = c2 λ cos λ = 0. This is true for λn =
2n − 1 π, 2
n = 1, 2, . . . .
Therefore, the eigenfunctions are 2n − 1 φn ( x ) = sin πx , 2
n = 1, 2, . . . .
The product solutions are given by the products of the eigenfunctions and b(t), 2n − 1 un ( x, t) = sin πx ( a cos ωn t + b sin ωn t), n = 1, 2, . . . , 2 where
2n − 1 πc, n = 1, 2, . . . . 2 A linear combination gives the general solution of the wave equation satisfying the given boundary conditions, ∞ 2n − 1 u( x, t) = ∑ ( an cos ωn t + bn sin ωn t) sin πx . 2 n =1 ωn =
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mathematical methods for physicists
4. Consider the following boundary value problems. Determine the eigenvalues, λ, and eigenfunctions, y( x ) for each problem. a. y00 + λy = 0,
y(0) = 0,
y0 (1) = 0.
The general solution is y( x ) = c1 cos
√
λx + c2 sin
√ λx.
The first boundary condition implies c1 = 0, giving √ y( x ) = c2 sin λx. The second boundary condition gives √ √ λ cos λ = 0. Therefore,
2n − 1 π, n = 1, 2, . . . . 2 So, the corresponding eigenfunctions are 2n − 1 πx . yn ( x ) = sin 2 p
b. y00 − λy = 0,
λn =
y(−π ) = 0,
y0 (π ) = 0.
The boundary conditions cannot be satisfied for λ > 0. So, we assume λ < 0. The general solution in this case is then √ √ y( x ) = c1 cos −λx + c2 sin −λx. The first boundary condition, y(−π ) = 0, implies √ √ c1 cos −λπ − c2 sin −λπ = 0. Solving for c2 , we have
√ cos −λπ √ c2 = c1 . sin −λπ Inserting this into the solution, we find √ √ y( x ) = c1 cos −λx + c2 sin −λx ! √ √ √ cos −λπ √ = c1 cos −λx + sin −λx sin −λπ √ √ √ √ cos −λx sin −λπ + sin −λx cos −λπ √ = c1 sin −λπ √ sin −λ( x + π ) √ = c1 . sin −λπ Note that we could have started with the guess √ √ y( x ) = c1 cos −λ( x + π ) + c2 sin −λ( x + π )
the harmonics of vibrating strings
145
do to the nature of the first boundary condition. The second boundary condition, y0 (π ) = 0, gives √ √ cos 2π −λ √ y 0 ( π ) = c1 − λ = 0. sin −λπ This is true for 2π
p
−λn =
2n − 1 π, 2
or λn = −
2n − 1 4
n = 1, 2, . . . ,
2 n = 1, 2, . . . .
,
The eigenfunctions are 2n − 1 yn ( x ) = sin (x + π) , 4 c. x2 y00 + xy0 + λy = 0,
y(1) = 0,
n = 1, 2, . . . .
y(2) = 0.
This is a CauchyEuler type differential equation. The characteristic equation is 0 = r (r − 1) + r + λ = r2 + λ. The roots of this √ equation are r = ± λi when λ > 0. So, the general solution for x > 0 is √ √ y( x ) = c1 cos( λ ln x ) + c2 sin( λ ln x ). The first boundary condition, y(1) = 0, implies c1 = 0, giving √ y( x ) = c2 sin( λ ln x ). The second boundary condition, y(2) = 0, implies √ y(2) = c2 sin( λ ln 2) = 0. This is true for p
λn ln 2 = nπ,
n = 1, 2, . . . .
The eigenfunctions are therefore given by yn ( x ) = sin(
nπ ln x ), ln 2
n = 1, 2, . . . .
d. ( x2 y0 )0 + λy = 0, y(1) = 0, y0 (e) = 0. Expanding the derivatives, we obtain a CauchyEuler type differential equation, x2 y00 + 2xy0 + λy = 0. The characteristic equation is given by 0
= r (r − 1) + 2r + λ = r2 + r + λ.
In problem 4d we will not get exact eigenvalues. We obtain a transcendental equation for the eigenvalues in the form tan z = βz. We will determine the first three eigenvalues numerically.
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mathematical methods for physicists
√ The roots of the equation are r = 12 (−1 ± 1 − 4λ). The boundary conditions can only be satisfied if the roots are complex. This restricts λ to the interval 1 − 4λ < 0. For these values the roots can be written as √ 1 r = (−1 ± i 4λ − 1). 2 The corresponding general solution for x > 0 is given by " # √ √ 4λ − 1 4λ − 1 y( x ) = c1 cos( ln x ) + c2 sin( ln x ) e− x/2 . 2 2 The first boundary condition, y(1) = 0, implies c1 = 0. This leaves √ 4λ − 1 ln x ). y( x ) = c2 e− x/2 sin( 2 Now, we apply the second boundary condition, y0 (e) = 0. Noting that, # " √ √ √ 4λ − 1 4λ − 1 4λ − 1 1 0 − x/2 ln x ) + cos( ln x ) , y ( x ) = c2 e − sin( 2 2 2x 2 we have " 0
y ( e ) = c2 e
−e/2
or
1 − sin( 2
√
√
4λ − 1 ) 2 √ 4λ − 1 tan( ) 2
1 sin( 2
4λ − 1 )+ 2
√
4λ − 1 cos( 2e
√
# 4λ − 1 ) = 0, 2
√
= =
√ 4λ − 1 4λ − 1 cos( ) 2e 2 ! √ 2 4λ − 1 . e 2
We see that λ values satisfy a transcendental equation. This √
equation is of the form tan z = 2e z where z = 4λ2 −1 . In Figure 5.1 the functions f (z) = tan z and f (z) = 2e z are shown. The intersections are the solutions of tan z = 2e z. We see there are an infinite number of solutions. Since −z is a solution when z is a solution, we need only look at positive values. The first three positive values can be obtained numerically as z = 4.414, 7.679, 10.87. These give the values of the corresponding eigenvalues as λ = 19.73, 59.21, 118.4. The first three eigenfunctions are
Figure 5.1: A plot of f (z) = tan z and f (z) = 2e z for problem 5.4d.
y1 ( x )
= e− x/2 sin(4.414 ln x )
y2 ( x )
= e− x/2 sin(7.679 ln x )
y3 ( x )
= e− x/2 sin(10.87 ln x ).
5. Consider the boundary value problem for the deflection of a horizontal beam fixed at one end, d4 y = C, dx4
y(0) = 0,
y0 (0) = 0,
y00 ( L) = 0,
y000 ( L) = 0.
the harmonics of vibrating strings
Solve this problem assuming that C is a constant. The solution is obtained through simple integration and employing the boundary conditions at each step. d4 y dx4 y000
= C. y000 ( L) = 0 ⇒ A = −CL,
= Cx + A,
= C ( x − L ). 1 C ( x − L)2 + B, y00 ( L) = 0 ⇒ B = 0, y00 = 2 1 = C ( x − L )2 . 2 1 1 y0 = C ( x − L)3 + D, y0 (0) = 0 ⇒ D = CL3 , 6 6 1 1 3 3 = C ( x − L) + CL . 6 6 1 1 4 y = C ( x − L) + CL3 x + E. 24 6 4
Since y(0) = 0, we have E = − CL 24 and the final solution is y( x ) =
1 1 CL4 C ( x − L)4 + CL3 x − . 24 6 24
6. Write y(t) = 3 cos 2t − 4 sin 2t in the form y(t) = A cos(2π f t + φ). We can determine the constants by expanding the cosine function, y(t) = A cos(2π f t + φ) = A cos φ cos 2π f t − A sin φ sin 2π f t. Comparing this to y(t) = 3 cos 2t − 4 sin 2t, we see 2π f = 2 and A cos φ
= 3,
A sin φ
= 4.
Adding the squares of these equations, 25 = A2 cos2 φ + A2 sin2 φ = A2 , we obtain A = 5. Dividing the first equation into the second, tan φ = 4/3. So, we find y(t)
= 3 cos 2t − 4 sin 2t −1 4 = 5 cos 2t + tan ( ) 3 ≈ 5 cos (2t + 0.927) .
7. Derive the coefficients bn in Equation (5.24). This derivation parallels that for the an ’s. We begin with the Fourier series ∞ a0 f (x) ∼ + ∑ [ an cos nx + bn sin nx ] . 2 n =1
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mathematical methods for physicists
We multiply this Fourier series by sin mx for some positive integer m and then integrate: Z 2π 0
f ( x ) sin mx dx
= +
Z 2π a0
sin mx dx
0
2
0
n =1
Z 2π ∞
∑ [an cos nx + bn sin nx] sin mx dx.
Integrating term by term, the right side becomes Z 2π 0
=
f ( x ) sin mx dx
Z a0 2π
2
+
0 ∞
∑
sin mx dx
an
Z 2π 0
n =1
cos nx sin mx dx + bn
Z 2π 0
sin nx cos mx dx .
R 2π
We have shown that 0 sin mx dx = 0, which implies that the first term vanishes. Also, we have that Z 2π 0
cos nx sin mx dx = 0
for integers n and m. R 2π We still need to evaluate 0 sin nx sin mx dx which was not done in the book. We compute this integral by using the product identity for sines. We have for m 6= n that Z 2π 0
sin nx sin mx dx
1 2
Z 2π
[cos(m − n) x − cos(m + n) x ] dx 1 sin(m − n) x sin(m + n) x 2π = − 2 m−n m+n 0 = 0.
=
0
For n = m, we have Z 2π
sin2 mx dx
1 2
Z 2π
(1 − cos 2mx ) dx 2π 1 1 = x− sin 2mx 2 2m 0 1 = (2π ) = π. 2 Now, we can finish the derivation. We have shown the orthogonality of the sines, Z 0
=
0
2π
0
sin nx sin mx dx = πδnm .
So, Z 2π 0
∞
f ( x ) sin mx dx
=
∑
bn
n =1 ∞
=
Z 2π 0
∑ bn πδnm
n =1
= bm π.
sin nx cos mx dx
the harmonics of vibrating strings
Solving for bm , we have bm =
1 π
Z 2π 0
f ( x ) sin mx dx.
8. Let f ( x ) be defined for x ∈ [− L, L]. Parseval’s identity is given by 1 L
Z L −L
f 2 ( x ) dx =
∞ a20 + ∑ a2n + bn2 . 2 n =1
Assuming the the Fourier series of f ( x ) converges uniformly in (− L, L), prove Parseval’s identity by multiplying the Fourier series representation by f ( x ) and integrating from x = − L to x = L. [In Section 8.6.3, we will encounter Parseval’s equality for Fourier transforms, which is a continuous version of this identity.] We begin with the Fourier series ∞ a0 2πnx 2πnx f (x) ∼ + ∑ an cos + bn sin . 2 L L n =1 Multiplying this Fourier series by f ( x ) and integrating over x ∈ [− L, L], we obtain Z L Z L Z L ∞ a0 2πnx 2πnx f 2 ( x ) dx = f ( x ) dx + + b sin f ( x ) dx a cos n ∑ n L L −L −L 2 − L n =1
=
a0 2
+
Z L
f ( x ) dx
−L ∞
∑
an
n =1
Z L −L
2πnx f ( x ) cos dx + bn L
Z L −L
2πnx f ( x ) sin dx L
Recall the Fourier coefficients are given by a0
=
an
=
bn
=
1 L f ( x ) dx L −L Z L 1 2πnx dx f ( x ) cos L −L L Z 1 L 2πnx f ( x ) sin dx. L −L L Z
Replacing the integrals in the integrated Fourier series with Fourier coefficients, we have Z L −L
f 2 ( x ) dx =
∞ a0 ( a0 L) + ∑ [ an ( an L) + bn (bn L)] , 2 n =1
leading to the sought result, 1 L
Z L −L
f 2 ( x ) dx =
∞ a20 + ∑ a2n + bn2 . 2 n =1
9. Consider the square wave function ( 1, 0 < x < π, f (x) = −1, π < x < 2π.
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mathematical methods for physicists
a. Find the Fourier series representation of this function and plot the first 50 terms. Since f ( x ) is an odd function on a symmetric interval, an = 0, n = 0, 1, . . . . We need only compute the bn ’s. 1 π f ( x ) sin nx dx π −π Z 2 π sin nx dx π 0 2 − (cos nπ − 1). nπ Z
=
bn
= = The Fourier series is then ∞
f (x)
∑
∼
−
n =1
4 π
=
2 (cos nπ − 1) sin nx nπ
∞
sin(2k − 1) x . 2k − 1 k =1
∑
The first 50 terms (n = 50) are shown in Figure 5.2. b. Apply Parseval’s identity in Problem 8 to the result in part a. Parseval’s identity states
Figure 5.2: A plot of first terms of the Fourier series of f ( x ) in Problem 5.9.
1 π
Z π −π
f 2 ( x ) dx =
∞
∑ bn2 .
n =1
From part a we have bn = −
2 (cos nπ − 1). nπ
Noting that f 2 ( x ) = 1, we have 1 π
Z π −π
f 2 ( x ) dx
1 π
Z π −π
∞
∑ bn2
=
n =1 ∞
dx
=
2
=
2 ∑ nπ (cos nπ − 1) n =1 2 ∞ 4 ∑ (2k − 1)π k =1 16 π2
∞
1 (2k − 1)2 k =1 16 1 1 1 1+ 2 + 2 + 2 +··· . π2 3 5 7
= =
∑
c. Use the result of part b to show
π2 8
∞
=
1 . ( 2n − 1)2 n =1
∑
Multiplying the series in part b, 2=
2
−
16 π2
by π 2 /16 yields this result.
∞
1 , ( 2k − 1)2 k =1
∑
the harmonics of vibrating strings
10. For the following sets of functions: (i) show that each is orthogonal on the given interval, and (ii) determine the corresponding orthonormal set. [See page 288.] Orthogonality and normalization can be done using simple substitutions and relating the integrals to the basic orthogonality relations between sines and cosines, Z 2π 0
cos nx cos mx dx =
a. {sin 2nx },
Z 2π 0
n = 1, 2, 3, . . . ,
sin nx sin mx dx = πδnm .
0 ≤ x ≤ π.
Let y = 2x, dy = 2 dx. Then, Z π 0
Z 2π
1 2
sin 2nx sin 2mx dx =
0
sin ny sin my dy =
π δnm . 2
These can be normalized by letting φn ( x ) = A sin 2nx. Then 1=
Z π 0
Z π
φn2 ( x ) dx =
0
A2 sin2 2nx dx =
π 2 A . 2
q
So, we have A = π2 and the orthonormal set is given by { n = 1, 2, 3, . . . , 0 ≤ x ≤ π. b. {cos nπx },
q
2 π
sin 2nx },
0 ≤ x ≤ 2.
n = 0, 1, 2, . . . ,
Let y = πx, dy = π dx. Then, Z 2 0
cos nπx cos mπx dx =
1 π
Z 2π 0
cos ny cos my dy = δnm .
These functions are already orthonormal. c. {sin nπx L },
n = 1, 2, 3, . . . ,
x ∈ [− L, L].
Let y = πx/L, dy = π/L dx. Then, Z L −L
sin
nπx mπx L sin dx = L L π
Z π −π
sin ny sin my dy = Lδnm .
These can be normalized by letting φn ( x ) = A sin nπx L . Then 1=
Z −L
So, we have A =
Lφn2 ( x ) dx = √1 L
Z L −L
A2 sin2
nπx dx = LA2 . L
and the orthonormal set is given by { √1 sin nπx L }, L
n = 1, 2, 3, . . . , x ∈ [− L, L]. 11. Consider f ( x ) = 4 sin3 2x. a. Derive the trigonometric identity giving sin3 θ in terms of sin θ and sin 3θ using DeMoivre’s Formula. We note that e3iθ = (eiθ )3 . Writing both sides of this equation in terms of trigonometric functions, we have e3iθ cos 3θ + i sin 3θ
= (eiθ )3 = (cos θ + i sin θ )3 = cos3 θ + 3i cos2 θ sin θ − 3 cos θ sin2 θ − i sin3 θ
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mathematical methods for physicists
Equating the real and imaginary parts we have cos 3θ
= cos3 θ − 3 cos θ sin2 θ,
sin 3θ
= 3 cos2 θ sin θ − sin3 θ.
The second equation can be rearranged to get the result.
= 3 cos2 θ sin θ − sin3 θ
sin 3θ
= 3(1 − sin2 θ ) sin θ − sin3 θ = 3 sin θ − 4 sin3 θ. Therefore, sin3 θ =
3 1 sin θ − sin 3θ. 4 4
b. Find the Fourier series of f ( x ) = 4 sin3 2x on [0, 2π ] without computing any integrals. We need only let θ = 2x in part a. Then, f ( x ) = 4 sin3 2x = 3 sin 2x − sin 6x. 12. Find the Fourier series of the following: The a0 ’s are computed separately from an ’s when determining the Fourier coefficients.
a. f ( x ) = x, x ∈ [0, 2π ]. We first compute the Fourier coefficients. Note that we compute a0 separately from an . a0
=
an
= =
bn
=
x2 2π 1 2π x dx = = 2π. π 0 2π 0 Z 2π 1 x cos nx dx π 0 2π 1 1 1 = 0. x sin nx + 2 cos nx π n n 0 Z
1 π
Z 2π 0
x sin nx dx
=
1 1 1 − x cos nx + 2 sin nx π n n
2π 0
2 =− . n
The Fourier series is given by ∞
f (x) ∼ π − 2
Figure 5.3: A plot of first terms of the Fourier series of f ( x ) in Problem 5.12a.
sin nx . n n =1
∑
A plot of the first terms of the series is given in Figure 5.3. 2
b. f ( x ) = x4 ,  x  < π. f ( x ) is an even function on  x  < π. So, bn = 0 for all n. We only need the an ’s. a0
=
1 π
Z π 2 x −π
4
dx =
2 4π
Z π 0
x2 dx =
π2 . 6
the harmonics of vibrating strings
an
153
1 π x2 cos nx dx π −π 4 Z π 1 x2 cos nx dx 2π 0 π 1 1 2 2 2 x sin nx + 2 x cos nx − 3 sin nx 2π n n n 0 1 2 cos nπ (−1)n π cos nπ = = . 2π n2 n2 n2 Z
= = = =
Then,the Fourier series is given as f (x) ∼
∞ π2 (−1)n cos nx. +∑ 12 n=1 n2
A plot of the first terms of the series is given in Figure 5.4. ( π 0 < x < π, 2, c. f ( x ) = − π2 , π < x < 2π. We first compute the Fourier coefficients. a0
= = =
an
= = =
bn
= = = =
Figure 5.4: A plot of first terms of the Fourier series of f ( x ) in Problem 5.12b.
1 2π f ( x ) dx π 0 Z Z 1 π π 1 0 π dx + dx − π 0 2 π −π 2 1 π 1 0 x − x = 0. 2 0 2 −π Z 2π 1 f ( x ) cos nx dx π 0 Z Z 1 π π 1 0 π cos nx dx + cos nx dx − π 0 2 π −π 2 π 0 1 1 sin nx − sin nx = 0. 2n 2n 0 −π Z 2π 1 f ( x ) sin nx dx π 0 Z Z 1 π π 1 0 π sin nx dx + − sin nx dx π 0 2 π −π 2 π 0 1 1 cos nx − cos nx + 2n 2n 0 −π 1 − cos nπ . n Z
The resulting Fourier series is ∞
f (x) ∼
∞ 1 − cos nπ sin(2k − 1) x sin nx = 2 ∑ . n 2k − 1 n =1 k =1
∑
A plot of the first terms of the series is given in Figure 5.5. 13. Find the Fourier series of each function f ( x ) of period 2π. For each series, plot the Nth partial sum, SN =
N a0 + ∑ [ an cos nx + bn sin nx ] , 2 n =1
Figure 5.5: A plot of first terms of the Fourier series of f ( x ) in Problem 5.12c.
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for N = 5, 10, 50 and describe the convergence (Is it fast? What is it converging to?, etc.) [Some simple Maple code for computing partial sums is shown in the notes.] a. f ( x ) = x,  x  < π. Since f ( x ) = x is an odd function on  x  < π, the an ’s vanish for all n. So, we just compute the bn ’s. bn
= = =
1 π x sin nx dx π −π Z 2 π x sin nx dx π 0 π 2 1 1 − x cos nx + 2 sin nx π n n 0 Z
= −
2 cos nπ (−1)n+1 =2 . n n
The resulting Fourier series is ∞
f (x) ∼ 2
∑ (−1)n+1
n =1
sin nx . n
As seen in Figure 5.6 the convergence is slow as the terms decay like n1 . The discontinuities in the periodic extension also are an indication. Figure 5.6: A plot of first terms of the Fourier series of f ( x ) in Problem 5.13a.
b. f ( x ) =  x ,  x  < π. Since f ( x ) =  x  is an even function on  x  < π, the bn ’s vanish for all n. We compute a0 and an . a0
= =
an
= =
1 π  x  dx π −π Z π 2 x dx = π π 0 Z 1 π  x  cos nx dx π −π Z π 2 x cos nx dx π 0 Z
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155
π 2 1 1 x sin nx + 2 cos nx π n n 0 2 (cos nπ − 1). πn2
= =
The resulting Fourier series is f (x)
∼
∞ π 2(cos nπ − 1) cos nx +∑ 2 n =1 πn2
=
π 4 − 2 π
∞
cos(2k − 1) x . (2k − 1)2 k =1
∑
The convergence is fast as the terms decay like n12 . There are not discontinuities in the periodic extension. See Figure 5.7. Figure 5.7: A plot of first terms of the Fourier series of f ( x ) in Problem 5.13b.
(
0, −π < x < 0, 1, 0 < x < π. We need to compute all of the Fourier coefficients.
c. f ( x ) =
a0
= =
an
= = =
bn
= = = =
1 π f ( x ) dx π −π Z π 1 dx = 1. π 0 Z π 1 f ( x ) cos nx dx π −π Z π 1 cos nx dx π 0 π 1 sin nx = 0. nπ 0 Z 1 π f ( x ) sin nx dx π −π Z 1 π sin nx dx π 0 π 1 − cos nx nπ 0 1 − cos nπ . nπ Z
The resulting Fourier series is f (x)
∼
∞ π 1 − cos nπ +∑ sin nx 2 n =1 nπ
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mathematical methods for physicists
π 2 + 2 π
=
∞
sin(2k − 1) x . 2k − 1 k =1
∑
The convergence is slow as the terms decay like n1 . There are discontinuities in the periodic extension as seen in Figure 5.8. Figure 5.8: A plot of first terms of the Fourier series of f ( x ) in Problem 5.13c.
14. Find the Fourier series of f ( x ) = x on the given interval. Plot the Nth partial sums and describe what you see. a. 0 < x < 2. We compute the Fourier coefficients: Z x2 2 2 2 x dx = a0 = = 2. 2 0 2 0 Z 2 2 2nπx an = x cos dx 2 0 2
=
Z 2 0
1 1 x sin nπx + 2 2 cos nπx nπ n π
= bn
= =
x cos nπx dx
Z 2
2 2
0
Z 2 0
x sin
2
= 0. 0
2nπx dx 2
x sin nπx dx
= =
1 1 x cos nπx + 2 2 sin nπx nπ n π 2 − . nπ
2
−
0
The Fourier series representation is given by f (x) ∼ 1 −
2 π
∞
sin nπx . n n =1
∑
A plot of the Fourier series is shown in Figure 5.9. Figure 5.9: A plot of first terms of the Fourier series of f ( x ) in Problem 5.14a.
b. −2 < x < 2. Since f ( x ) = x is an odd function on −2 < x < 2, we only need the bn ’s. bn
=
2 4
Z 2 −2
x sin
2nπx dx 4
the harmonics of vibrating strings
=
Z 2 0
x sin
nπx dx 2
= =
157
nπx 2 nπx 4 x cos + 2 2 sin nπ 2 2 n π 4 cos nπ. − nπ
2
−
0
The Fourier series is f (x) ∼
∞
(−1)n+1 nπx sin . n 2 n =1
4 π
∑
A plot of this Fourier series is shown in Figure 5.10. c. 1 < x < 2. The Fourier coefficients are found as Z 2 2 a0 = 2 x dx = x2 = 3. 1
1
Z 2
an
= 2 x cos 2nπx dx 1 2 1 1 x sin 2nπx + 2 2 cos 2nπx = 0. = 2nπ 4n π 1
bn
= 2 x sin 2nπx dx 1 2 1 1 x cos 2nπx + 2 2 sin 2nπx = − 2nπ 4n π 1 1 = − . nπ
Figure 5.10: A plot of first terms of the Fourier series of f ( x ) in Problem 5.14b.
Z 2
This fives the Fourier series f (x) ∼
3 1 − 2 π
∞
sin 2nπx . 2 n =1
∑
A plot of this Fourier series is shown in Figure 5.11. 15. The result in Problem 12b above gives a Fourier series representation 2 of x4 . By picking the right value for x and a little arrangement of the series, show that [See Example 5.6.] a.
π2 1 1 1 = 1+ 2 + 2 + 2 +···. 6 2 3 4 Using the results in Problem 5.12b, one has that ∞ π2 (−1)n x2 = +∑ cos nx. 4 12 n=1 n2
Letting x = π, we have π2 4
=
∞ π2 (−1)n +∑ cos nπ 12 n=1 n2
=
∞ π2 1 + ∑ 2. 12 n=1 n
Figure 5.11: A plot of first terms of the Fourier series of f ( x ) in Problem 5.14c.
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mathematical methods for physicists
Then,
∞
1 π2 π2 π2 = − = . 2 4 12 6 n =1 n
∑
b.
π2 1 1 1 = 1+ 2 + 2 + 2 +···. 8 3 5 7 Hint: Consider how the series in part a. can be used to do this. Let S = 1+
1 1 1 + 2 + 2 +···. 32 5 7
Note that ∞
π2 6
1 2 n n =1
∑
=
1 1 1 + 2 + 2 +··· 2 2 3 4 1 1 1 S+ 2 + 2 + 2 +··· 2 4 6 1 1 1 S+ 1+ 2 + 2 +··· 4 2 3 2 1 π S+ 4 6
= 1+ = = =
= S+
π2 . 24
Therefore, we have S
1 1 1 + 2 + 2 +··· 2 3 5 7 π2 π2 π2 − = . 6 24 8
= 1+ =
16. Sketch (by hand) the graphs of each of the following functions over four periods. Then sketch the extensions each of the functions as both an even and odd periodic function. Determine the corresponding Fourier sine and cosine series and verify the convergence to the desired function using Maple. For these problems we make use of the Fourier cosine series, f (x) = where an =
2 L
∞ a0 nπx + ∑ an cos , 2 L n =1
Z L 0
f ( x ) cos
nπx dx, L
and the Fourier sine series, ∞
f (x) =
∑ bn sin
n =1
where bn =
2 L
Z L 0
f ( x ) sin
nπx , L nπx dx. L
the harmonics of vibrating strings
159
a. f ( x ) = x2 , 0 < x < 1. The given function is shown in Figure 5.12. In Figure 5.13 this function is reflected about the yaxis and the new function is then periodically extended to give the even periodic extension. In Figure 5.14 the function is reflected about the origin and the new function is then periodically extended to give the odd periodic extension.
f (x)
The Fourier cosine series coefficients are given by a0 an
= 2 = 2
Z 1 0
Z 1 0
x 0
2 x2 dx = . 3
Figure 5.12: Function given in Problem 16a.
x2 cos nπx dx
= =
1
2 1 2 2 sin nπx 2 x sin nπx + 2 2 x cos nπx + nπ n π (nπ )3 4(−1)n . n2 π 2
1 0
The resulting Fourier cosine series is given by f (x) =
1 4 + 2 3 π
∞
(−1)n cos nπx. 2 n =1 n
∑
A plot of this series representation is shown in Figure 5.15. Figure 5.13: Sketch of the even periodic extension of the function given in Problem 16a.
f (x) 1
−4
−3
−2
−1
0
2
1
3
4
x
Figure 5.14: Sketch of the odd periodic extension of the function given in Problem 16a.
f (x) 1
−4
−3
−2
−1
0
2
1
3
4
x
−1 The Fourier sine series coefficients are given by bn
= 2
Z 1 0
x2 sin nπx dx
1 1 2 2 2 cos nπx = 2 − x cos nπx + 2 2 x sin nπx + nπ n π (nπ )3 0 2 4 = − cos nπ + (cos nπ − 1). nπ (nπ )3 The resulting Fourier sine series is given by ∞
f (x) =
2
4
∑ (− nπ cos nπ + (nπ )3 (cos nπ − 1)) sin nπx.
n =1
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mathematical methods for physicists
Figure 5.15: A plot of first terms of the Fourier cosine series of f ( x ) in Problem 5.16a.
A plot of this series representation is shown in Figure 5.16. Figure 5.16: A plot of first terms of the Fourier sine series of f ( x ) in Problem 5.16a.
f (x)
b. f ( x ) = x (2 − x ), 0 < x < 2. x 0
1
2
Figure 5.17: Function given in Problem 16b.
The given function is shown in Figure 5.17. In Figure 5.18 this function is reflected about the yaxis and the new function is then periodically extended to give the even periodic extension. In Figure 5.19 the function is reflected about the origin and the new function is then periodically extended to give the odd periodic extension.
Figure 5.18: Sketch of the even periodic extension of the function given in Problem 16b.
f (x) 1
−4 −3 −2 −1 0 1 2 3 4
x
The Fourier cosine series coefficients are given by 2 Z 2 x3 4 2 a0 = x (2 − x ) dx = x − = . 3 0 3 0 an
=
Z 2 0
x (2 − x ) cos
nπx dx 2
the harmonics of vibrating strings
Figure 5.19: Sketch of the odd periodic extension of the function given in Problem 16b.
f (x) 1
−8 −7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 7 8 −1 "
= =
2 nπx x (2 − x ) sin − 2(1 − x ) nπ 2
2 nπ
2
161
x
nπx cos +2 2
2 nπ
3
# nπx sin _02 2
8 (cos nπ − 1). (nπ )2
The resulting Fourier cosine series is given by ∞ 2 8 nπx 2 16 f (x) = + ∑ (cos nπ − 1) cos = − 2 3 n=1 (nπ )2 2 3 π
(2k−1)πx
∞
cos ∑ (2k − 21)2 . k =1
A plot of this series representation is shown in Figure 5.20. Figure 5.20: A plot of first terms of the Fourier cosine series of f ( x ) in Problem 5.16b.
The Fourier sine series coefficients are given by bn
=
Z 2
"0
=
x (2 − x ) sin
nπx dx 2
2 nπx − x (2 − x ) cos + 2(1 − x ) nπ 2
= −
2 nπ
2
nπx sin −2 2
∞
(2k−1)πx
16 (cos nπ − 1). (nπ )3
The resulting Fourier sine series is given by ∞
f (x) = −
16 nπx 32 (cos nπ − 1) sin = 3 3 2 ( nπ ) π n =1
∑
sin 2 . (2k − 1)3 k =1
∑
A plot of this series representation is shown in Figure 5.21. ( 0, 0 < x < 1, c. f ( x ) = 1, 1 < x < 2.
2 nπ
3
# nπx cos _02 2
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mathematical methods for physicists
Figure 5.21: A plot of first terms of the Fourier sine series of f ( x ) in Problem 5.16b.
f (x) 1
x 0
1
2
Figure 5.22: Function given in Problem 16c.
The given function is shown in Figure 5.22. In Figure 5.23 this function is reflected about the yaxis and the new function is then periodically extended to give the even periodic extension. In Figure 5.24 the function is reflected about the origin and the new function is then periodically extended to give the odd periodic extension.
Figure 5.23: Sketch of the even periodic extension of the function given in Problem 16c.
f (x) 1
−8 −7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 7 8
Figure 5.24: Sketch of the odd periodic extension of the function given in Problem 16c.
x
f (x) 1
−8 −7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 7 8 −1
x
The Fourier cosine series coefficients are given by a0
=
an
=
Z 2 0
Z 2 0
=
Z 2
f ( x ) cos
Z 2 1
dx = 1.
nπx dx 2
nπx dx 2 2 nπx 2 2 nπ sin sin . =− nπ 2 1 nπ 2 1
=
f ( x ) dx =
cos
The resulting Fourier cosine series is given by f (x) =
∞ 1 2 nπ nπx −∑ sin cos . 2 n=1 nπ 2 2
A plot of this series representation is shown in Figure 5.25.
the harmonics of vibrating strings
163
Figure 5.25: A plot of first terms of the Fourier cosine series of f ( x ) in Problem 5.16c.
The Fourier sine series coefficients are given by bn
=
Z 2 0
= =
Z 2
f ( x ) sin
nπx dx 2
nπx dx 2 1 2 nπx 2 2 nπ − cos cos − cos nπ . = nπ 2 1 nπ 2 sin
The resulting Fourier sine series is given by ∞
f (x) =
∑
n =1
cos
nπ nπx − cos nπ sin . 2 2
A plot of this series representation is shown in Figure 5.26. Figure 5.26: A plot of first terms of the Fourier sine series of f ( x ) in Problem 5.16c.
f (x) π
( d. f ( x ) =
π, 2π − x,
0 < x < π, π < x < 2π.
The given function is shown in Figure 5.27. In Figure 5.28 this function is reflected about the yaxis and the new function is then periodically extended to give the even periodic extension. In Figure 5.29 the function is reflected about the origin and the new function is then periodically extended to give the odd periodic extension.
x 0
π
2π
Figure 5.27: Function given in Problem 16d.
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mathematical methods for physicists
Figure 5.28: Sketch of the even periodic extension of the function given in Problem 16d.
f (x) π
−8π −7π −6π −5π −3π −2π−π 0 π 2π 3π 4π 5π 6π 7π 8π −4π Figure 5.29: Sketch of the odd periodic extension of the function given in Problem 16d.
x
f (x) π
−8π −7π −6π −5π −3π −2π−π 0 π 2π 3π 4π 5π 6π 7π 8π −4π −π
x
The Fourier cosine series coefficients are given by
=
a0
= = =
an
= = =
2π 2 f ( x ) dx 2π 0 Z Z 1 π 1 2π π dx + (2π − x ) dx π 0 π π 3π . 2 Z 1 2π nx f ( x ) cos dx π 0 2 Z π Z 1 nx 1 2π nx π cos dx + (2π − x ) cos dx π 0 2 π π 2 2 nx π 1 2 nx 4 nx 2π sin (2π − x ) sin − 2 cos + n 2 0 π n 2 2 π n 4 nπ − 2 (cos nπ − cos ). 2 n
Z
The resulting Fourier cosine series is given by f (x) =
∞ cos nπ − cos nπ 3π nx 2 −4 ∑ . cos 2 4 2 n n =1
A plot of this series representation is shown in Figure 5.30. Figure 5.30: A plot of first terms of the Fourier cosine series of f ( x ) in Problem 5.16d.
The Fourier sine series coefficients are given by bn
= =
1 2π nx f ( x ) sin dx π 0 2 Z Z 1 π nx 1 2π nx π sin dx + (2π − x ) sin dx π 0 2 π π 2 Z
the harmonics of vibrating strings
= =
165
1 2 2 nx π nx 4 nx 2π − cos + − (2π − x ) cos + 2 sin n 2 0 π n 2 2 π n 2 4 nπ + sin . n πn2 2
The resulting Fourier sine series is given by ∞
f (x) =
2
4
∑ ( n + πn2 sin
n =1
nπ nx ) sin . 2 2
A plot of this series representation is shown in Figure 5.31. Figure 5.31: A plot of first terms of the Fourier sine series of f ( x ) in Problem 5.16d.
17. Consider the function f ( x ) = x, −π < x < π. n+1 sin nx . a. Show that x = 2 ∑∞ n=1 (−1) n
Since f ( x ) = x is an odd function on  x  < π, the an ’s vanish for all n. The Fourier sine series coefficients are bn
= = =
1 π x sin nx dx π −π Z π 2 x sin nx dx π 0 π 2 1 1 − x cos nx + 2 sin nx π n n 0
= −
Z
2 cos nπ (−1)n+1 =2 . n n
This gives the Fourier sine series representation ∞
f (x) ∼ 2
∑ (−1)n+1
n =1
sin nx . n
b. Integrate the series in part a and show that x2 = For f ( x ) = x, we consider the integral Z x 0
f (ξ ) dξ =
x2 . 2
π2 3
n+1 cos nx . − 4 ∑∞ n=1 (−1) n2
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mathematical methods for physicists
Integrating the series as well, we have Z x 0
∞
∼ 2
f (ξ ) dξ
∑ (−1)n+1
n =1 ∞
= 2
n
0
dξ.
∑ (−1)n
cos nξ x n2 0
∑ (−1)n
∞ cos nx 1 − 2 (−1)n 2 ∑ n2 n n =1
n =1 ∞
= 2
Z x sin nξ
n =1
Therefore, x2 = C + 4
∞
∑ (−1)n
n =1
where
∞
C = −2
cos nx , n2 1
∑ (−1)n n2 .
n =1
We still need to evaluate C. We can use the results in Problem 15 to do this. ∞
C
= −4
1
∑ (−1)n n2
n =1
1 1 1 1 1 1 4 1− 2 + 2 − 2 + 2 − 2 + 2 −··· 2 3 4 5 6 7 1 1 1 1 1 1 4 1+ 2 + 2 + 2 +··· −4 + 2 + 2 +··· 3 5 7 22 4 6 2 2 2 π π π 4 −4 = . 8 24 3
= = =
Therefore, we have that x2 =
∞ cos nx π2 + 4 ∑ (−1)n . 3 n2 n =1
c. Find the Fourier cosine series of f ( x ) = x2 on [0, π ] and compare to the result in part b. We first determine the Fourier cosine series coefficients. a0
=
an
= = =
2 π 2 2π 2 . x dx = π 0 3 Z π 2 x2 cos nx dx π 0 π 2 1 2 2 2 x sin nx + 2 x cos nx − 3 sin nx π n n n 0 2 2 4 4 π cos nπ = 2 cos nπ = 2 (−1)n . π n2 n n Z
This gives the Fourier cosine series representation x2 =
∞ π2 cos nx . + 4 ∑ (−1)n 3 n2 n =1
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167
18. Consider the function f ( x ) = x, 0 < x < 2. a. Find the Fourier sine series representation of this function and plot the first 50 terms. The Fourier sine series coefficients are bn
Z 2
=
0
"
=
x sin
nπx dx 2
2 nπx − x cos + πn 2
2 nπ
2
nπx sin 2
#2 0
4 4 cos nπ = (−1)n+1 . − nπ nπ
=
This gives the Fourier sine series representation f (x) ∼
4 π
∞
(−1)n+1 nπx sin . n 2 n =1
∑
The first few terms of this series are shown in Figure 5.32. Figure 5.32: A plot of first terms of the Fourier sine series of f ( x ) in Problem 5.18a.
b. Find the Fourier cosine series representation of this function and plot the first 50 terms. The Fourier cosine series coefficients are a0
=
an
=
Z 2 0
Z 2 0
"
x dx = 2. x cos
nπx dx 2
=
2 nπx x sin + πn 2
=
4 (cos nπ − 1). n2 π 2
2 nπ
2
nπx cos 2
#2 0
This gives the Fourier cosine series representation f (x)
∞
∼ 1+
4 π2
cos nπ − 1 nπx cos 2 2 n n =1
= 1−
8 π2
1 (2k − 1)πx cos . 2 2 ( 2k − 1 ) k =1
∑ ∞
∑
The first few terms of this series are shown in Figure 5.33.
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mathematical methods for physicists
Figure 5.33: A plot of first terms of the Fourier cosine series of f ( x ) in Problem 5.18b.
c. Extend and apply Parseval’s identity in Problem 8 to the result in part b. Parseval’s identity can be extended to Fourier cosine series by slightly modifying the derivation in Problem 5.8. We begin with the Fourier cosine series f (x) = where an =
1 L
Z L −L
∞ a0 nπx + ∑ an cos , 2 L n =1
f ( x ) cos
nπx dx, L
n = 0, 1, . . . .
Then, we multiply this Fourier cosine series by f ( x ), integrate over x ∈ [0, L], and replace the integrals with Fourier coefficients. This gives Z L 0
f 2 ( x ) dx
Z L a0
=
2
0
Z a0 L
=
2
0
f ( x ) dx + f ( x ) dx
Z L ∞
∑ an cos
0 n =1 Z L ∞ + an 0 n =1
∑
nπx f ( x ) dx L
f ( x ) cos
nπx dx L
∞ a0 a0 L an L ( ) + ∑ an ( ). 2 2 2 n =1
=
This gives the Parseval identity 2 L
Z L 0
f 2 ( x ) dx =
∞ a20 + ∑ a2n . 2 n =1
We now apply this result to f ( x ) = x, x ∈ [0, 2]. From part b, we have a0 = 2 and an =
4 n2 π 2
(cos nπ − 1),
n = 1, 2, . . . .
Then, 2 L
Z L 0
f 2 ( x ) dx Z 2 0
x2 dx 8 3
= =
∞ a20 + ∑ a2n 2 n =1 2 ∞ 22 4 +∑ ( cos nπ − 1 ) 2 2 2 n =1 n π
= 2+
64 4 ∞ 1 . ∑ π k=1 (2k − 1)4
the harmonics of vibrating strings
This gives the sum ∞
π4 96
1 ( 2k − 1)4 k =1
∑
=
= 1+
1 1 1 + 4 + 4 +···. 4 3 5 7
d. Use the result of part c. to find the sum ∑∞ n =1
1 . n4
The result in part c is not quite the sum we seek as the terms involve only the odd terms. We can rearrange the series to make 1 use of the result in part c and solve for S = ∑∞ n =1 n4 : ∞
S
=
1 4 n =1 n
∑
1 1 1 1 1 = 1+ 4 + 4 + 4 + 5 + 6 +··· 3 4 2 3 4 1 1 1 1 1 = 1+ 4 + 4 +··· + + 4 + 4 +··· 3 5 24 4 6 1 π4 1 1 = + 4 1+ 4 + 4 +··· 96 2 2 3 4 π 1 = + S 96 16 15 π4 S = 16 96 π4 S = . 90 Therefore, we have shown that ∑∞ n =1
1 n4
=
π4 90 .
19. Differentiate the Fourier sine series term by term in the last problem. Show that the result is not the derivative of f ( x ) = x.. The Fourier sine series is given by f (x) ∼
4 π
∞
(−1)n+1 nπx sin . n 2 n =1
∑
A simple term by term differentiation gives 4 π
∞
∞ nπx nπx (−1)n+1 d sin = 2 (−1)n+1 cos . ∑ n dx ∑ 2 2 n =1 n =1
However, this is a divergent series and cannot sum to f 0 ( x ) = 1. 20. Rewrite the solution to Problem 2 and identify the initial value Green’s function. The general solution of the heat equation satisfying the boundary conditions in Problem 2 is given by ∞
u( x, t) =
∑ an cos
n =1
(2n − 1)πx −k( 2n−1 π )2 t 2L e . 2L
169
170
mathematical methods for physicists
Figure 5.34: Plot of first 50 terms of (left) the Fourier sine series of f ( x ) = x and (b) the derivative of these terms from Problem 5.19.
The Fourier coefficients, an , n = 1, 2, . . . , are determined using the initial condition, u( x, 0) = f ( x ), x ∈ [0, L]. Setting t = 0 in the solution, we have ∞
f (x) =
∑ an cos
n =1
(2n − 1)πx . 2L
We need to establish the orthogonality of the functions {cos x ∈ [0, L] in order to solve for the an ’s. π Let y = πx L and dy = L dx. Then, for n 6 = m Z L
= =
on
(2n − 1)πx (2m − 1)πx cos dx 2L 2L Z L π (2m − 1)y (2n − 1)y cos dy cos π 0 2 2 Z π L [cos(m + n)y + cos(m − n)y] dy 2π 0 L sin(m + n)y sin(m − n)y π + = 0. 2π m+n m−n 0 0
=
(2n−1)πx } 2L
cos
Thus, these functions are orthogonal on this interval. Furthermore, they are not orthonormal. For n = m, we have Z L 0
cos2
(2n − 1)πx dx = 2L = =
(2n − 1)y L π cos2 dy π 0 2 Z π L (1 + cos(2n − 1)y) dy 2π 0 L sin(2n − 1)y π L y+ = . 2π 2n − 1 2 0 Z
So, we can extract the Fourier coefficients of f ( x ) in this basis. Multi(2m−1)πx plying the series for f ( x ) by cos and integrating over the interval 2L x ∈ [0, L], we have an =
2 L
Z L 0
f ( x ) cos
(2n − 1)πx dx. 2L
Now we can determine the Green’s function. Inserting the Fourier coefficients into the general solution, we find ∞
u( x, t)
=
∑ an cos
n =1
(2n − 1)πx −k( 2n−1 π )2 t 2L e 2L
the harmonics of vibrating strings
∞
∑
=
n =1
Z L
=
0
Z L
=
0
Z L
2 L
0
f (ξ )
(2n − 1)πξ (2n − 1)πx −k( 2n−1 π )2 t 2L dξ cos e 2L 2L ! 2 ∞ (2n − 1)πx (2n − 1)πξ −k( 2n−1 π )2 t 2L cos cos e dξ. L n∑ 2L 2L =1 f (ξ ) cos
G ( x, ξ; t, 0) f (ξ ) dξ.
Here we have identified the initial value Green’s function as 2 L
G ( x, ξ; t, 0) =
∞
∑ cos
n =1
(2n − 1)πx (2n − 1)πξ −k( 2n−1 π )2 t 2L cos e . 2L 2L
21. Rewrite the solution to Problem 3 and identify the initial value Green’s functions. The general solution of the wave equation satisfying the given boundary conditions in Problem 3 is given by ∞
∑ (an cos ωn t + bn sin ωn t) sin
u( x, t) =
n =1
(2n − 1)πx . 2
where ωn = 2n2−1 πc, n = 1, 2, . . . . The Fourier coefficients, an , bn , n = 1, 2, . . . , are determined using the initial conditions, u( x, 0) = f ( x ), ut ( x, 0) = g( x ), x ∈ [0, 1]. Setting t = 0 in the solution and its derivative, we have ∞
f (x)
=
∑ an sin
n =1 ∞
g( x )
=
(2n − 1)πx , 2
∑ ωn bn sin
n =1
(2n − 1)πx . 2
We need to establish the orthogonality of the functions {sin x ∈ [0, 1] in order to solve for the Fourier coefficients. Then, for n 6= m, Z 1 0
= =
1 2
sin
(2n−1)πx } 2
on
(2n − 1)πx (2m − 1)πx sin dx 2 2
Z 1
[cos(m − n)πx − cos(m + n)πx ] dx 1 sin(m − n)πx sin(m + n)πx 1 − = 0. 2π m−n m+n 0 0
Thus, these functions are orthogonal on this interval. For n = m, we have Z 1 0
sin2
(2n − 1)πx dx = 2 =
1 2
Z 1
(1 − cos(2n − 1)πx ) dx sin(2n − 1) x 1 1 1 x− = . 2 (2n − 1)π 0 2 0
As usual, we can determine the Fourier coefficients of f ( x ) in this basis (2n−1)πx by multiplying by sin and integrating over the interval x ∈ [0, 1]. 2
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For the solution of the wave equation, we have an ω n bn
Z 1
(2n − 1)πx dx, 2 0 Z 1 (2n − 1)πx = 2 g( x ) sin dx. 2 0
= 2
f ( x ) sin
Now we can determine the Green’s function. Inserting the Fourier coefficients into the general solution and rearranging integration and summation, we find ∞
u( x, t)
(2n − 1)πx 2 n =1 ∞ Z 1 (2n − 1)πx (2n − 1)πξ dξ cos ωn t sin = ∑ 2 f (ξ ) sin 2 2 0 n =1 Z ∞ 1 (2n − 1)πξ 2 (2n − 1)πx g(ξ ) sin +∑ dξ sin ωn t sin ω 2 2 0 n n =1 ! Z 1 ∞ (2n − 1)πx (2n − 1)πξ 2 ∑ sin = sin cos ωn t f (ξ ) dξ 2 2 0 n =1 ! Z 1 ∞ (2n − 1)πx (2n − 1)πξ sin ωn t + 2 ∑ sin sin g(ξ ) dξ 2 2 ωn 0 n =1
=
=
∑ (an cos ωn t + bn sin ωn t) sin
Z 1 0
Gc ( x, ξ; t, 0) f (ξ ) dξ +
Z 1 0
Gs ( x, ξ; t, 0) g(ξ ) dξ.
Here we have identified the two initial value Green’s functions as related to the initial profile and velocity as ∞
Gc ( x, ξ; t, 0)
= 2
∑ sin
(2n − 1)πx (2n − 1)πξ sin cos ωn t, 2 2
∑ sin
(2n − 1)πx (2n − 1)πξ sin ωn t sin . 2 2 ωn
n =1 ∞
Gs ( x, ξ; t, 0)
= 2
n =1
6 Nonsinusoidal Harmonics and Special Functions 1. Consider the set of vectors (−1, 1, 1), (1, −1, 1), (1, 1, −1). a. Use the GramSchmidt process to find an orthonormal basis for R3 using this set in the given order. The GramSchmidt orthogonalization process is given by e1 = a1 and n −1 a · e n j e j , n = 2, 3, . . . , N. en = an − ∑ 2 ej j =1 For this problem we have a1 = (−1, 1, 1), a2 = (1, −1, 1), and a3 = (1, 1, −1). Next, we apply the process to obtain e1 , e2 , and e3 . e1 e2
e3
= a1 = (−1, 1, 1) a ·e = a2 − 2 2 1 e1 e1 −1 = (1, −1, 1) − (−1, 1, 1) 3 2 2 4 = ( ,− , ) 3 3 3 a ·e a ·e = a3 − 3 2 1 e1 − 3 2 2 e2 e1 e2 −1 −4/3 2 2 4 = (1, 1, −1) − (−1, 1, 1) − ( ,− , ) 3 8/3 3 3 3 = (1, 1, 0).
For a normalized set of basis vectors we need to divide by the lengths of each vector. This gives eˆ 1 eˆ 2 eˆ 3
1 1 1 1 √ (−1, 1, 1) = (− √ , √ , √ ). 3 3 3 3 r 3 2 2 4 1 1 2 = ( , − , ) = ( √ , − √ , √ ). 8 3 3 3 6 6 6 1 1 1 = √ (1, 1, 0) = ( √ , √ , 0). 2 2 2
=
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mathematical methods for physicists
b. What do you get if you do reverse the order of these vectors? For this part, a1 = (1, 1, −1), a2 = (1, −1, 1), a3 = (−1, 1, 1). Applying the GramSchmidt process, we obtain e1 , e2 , and e3 as e1 e2
e3
= a1 = (1, 1, −1) a ·e = a2 − 2 2 1 e1 e1 −1 = (1, −1, 1) − (1, 1, −1) 3 4 2 2 = ( ,− , ) 3 3 3 a ·e a ·e = a3 − 3 2 1 e1 − 3 2 2 e2 e1 e2 −1 −4/3 4 2 2 = (−1, 1, 1) − (1, 1, −1) − ( ,− , ) 3 8/3 3 3 3 = (0, 1, 1).
For a normalized set of basis vectors we need to divide by the lengths of each vector. This gives eˆ 1 eˆ 2 eˆ 3
1 1 1 1 √ (1, 1, −1) = ( √ , √ , − √ ). 3 3 3 3 r 3 4 2 2 2 1 1 = ( , − , ) = ( √ , − √ , √ ). 8 3 3 3 6 6 6 1 1 1 = √ (0, 1, 1) = (0, √ , √ ). 2 2 2
=
2. Use the GramSchmidt process to find the first four orthogonal polynomials satisfying the following: The orthogonal polynomials are constructed from the set of monomials Rb f n ( x ) = x n , n ∈ N0 , for x ∈ ( a, b) and inner product h f , gi = a f ( x ) g( x )σ ( x ) dx using weight function σ ( x ). The GramSchmidt process consists in finding the functions φ0 ( x ) = 1, and n −1 h f , φ i n j φn ( x ) = f n ( x ) − ∑ φj ( x ), n = 1, 2, . . . . k φ j k2 j =0 2
a. Interval: (−∞, ∞) Weight Function: e− x . In order to carry out the process, we need to evaluate the integrals I = hxn , xm i =
Z ∞ −∞
2
x n x m e− x dx.
If n + m is odd, then I = 0 because the integrand is an odd function. Therefore, we have the inner products ( Z ∞ 0, n + m, odd, n m n m − x2 hx , x i = x x e dx = Ik , n + m = 2k, even, −∞ where Ik =
Z ∞ −∞
2
x2k e− x dx,
k = 0, 1, . . . .
nonsinusoidal harmonics
175
We can compute these integrals. First note that for k = 0, we have the Gaussian integral I0 =
Z ∞ −∞
2
e− x dx =
√ π.
For k > 0, we can carry out an integration by parts. This is done by first borrowing an x to go with the exponential in order to find an antiderivative. Ik
= =
Z ∞ −∞
Z ∞
−∞
2
x2k e− x dx 2
x2k−1 xe− x dx,
Z 2 ∞ 2k − 1 ∞ 2(k−1) − x2 1 dx = − x2k−1 e− x x e + 2 2 −∞ −∞ 2k − 1 = Ik−1 . 2
We next use this difference equation to find needed values of the integrals: I0
=
I1
=
I2
=
√ π. 1 1√ I0 = π. 2 2 3 3√ I1 = π. 2 4
The GramSchmidt orthogonalization process can now be carried out. Inner products like h x3 , 1i vanish since the exponents sum to an odd integer. Inner products like h x2 , 1i = I1 and can be computed. φ0 ( x )
= 1.
φ1 ( x )
= =
φ2 ( x )
= = = =
φ3 ( x )
h x, φ0 i φ0 ( x ) kφ0 k2 h x, 1i x− 1 = x. h1, 1i h x2 , φ0 i h x2 , φ1 i x2 − φ ( x ) − φ (x) 0 kφ0 k2 kφ1 k2 1 h x2 , x i h x 2 , 1i x2 − 1− x h1, 1i h x, x i I2 x2 − I0 1√ π 1 x − 2√ = x2 − 2 π x−
h x3 , φ0 i h x3 , φ1 i h x3 , φ2 i φ ( x ) − φ ( x ) − φ2 ( x ) 0 1 kφ0 k2 kφ1 k2 kφ2 k2 h x3 , x2 − 12 i h x 3 , 1i h x3 , x i 1 2 1− x− x − = x3 − h1, 1i h x, x i 2 h x2 − 12 , x2 − 12 i
=
x3 −
There are two hints needed for students. First, they need the k = 0 value. The second is the trick of factoring out an x in order to carry out the integration by parts.
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= x3 −
I2 x I1 √ 3
π 3 = x3 − 41 √ x = x3 − x. 2 π 2 Thus, the first four orthogonal polynomials in this system are 3 1 1, x, x2 − , x3 − x. 2 2 These are the Hermite polynomials up to a constant factor. Note: One can obtain a general expression for Ik , k > 0. One just needs to iterate ` times. Then, 2k − 1 Ik−1 2 2k − 3 2k − 1 Ik−2 = 2 2 .. . 2k − 3 2k − 2` + 1 2k − 1 ··· Ik−` = 2 2 2  {z } ` terms (2k−1)!! √ Letting ` = k, we obtain Ik = 2k π. Ik
=
b. Interval: (0, ∞) Weight Function: e− x . In order to carry out the process, we need to evaluate the integrals Z I = hxn , xm i =
∞
0
x n x m e− x dx.
Consider the integrals Ik =
Z ∞ 0
x k e− x dx,
k = 0, 1, . . . .
For k = 0, we have I0 =
Z ∞
∞ e− x dx = −e− x = 1. 0
0
For k > 0, we can carry out an integration by parts. Ik
Z ∞
x k e− x dx Z ∞ k −x −x e + k
=
0
=
0
∞ 0
x k−1 e− x dx
= kIk−1 . We can use this difference equation to find needed values of the integrals: I0
= 1.
I1
=
I2
= 2I1 = 2.
I3
= 3I2 = 3(2) = 6.
I4
= 4I3 = 4(6) = 24.
I0 = 1.
nonsinusoidal harmonics
We see a pattern and can deduce that Ik = k!. We can prove this using mathematical induction, or just iterate the first order difference equation, obtaining a general expression for Ik , k > 0. One just needs to iterate ` times. Then, Ik
= kIk−1 = k(k − 1) Ik−2 .. . = k(k − 1) · · · (k − ` + 1) Ik−`  {z } ` terms
For ` = k, we have Ik = k!I0 = k!. The GramSchmidt orthogonalization process can now be carried out. The inner products are easily computed using
h x n , x m i = In+m = (n + m)!. The first four classical orthogonal polynomials on x ∈ (0, ∞) with weight σ = e− x are found as φ0 ( x )
= 1.
φ1 ( x )
= x− = =
φ2 ( x )
= = = =
h x, φ0 i φ0 ( x ) kφ0 k2 h x, 1i x− 1 h1, 1i I x − 1 = x − 1. I0 h x2 , φ0 i h x2 , φ1 i x2 − φ ( x ) − φ (x) 0 kφ0 k2 kφ1 k2 1 h x 2 , 1i h x 2 , x − 1i x2 − 1− x h1, 1i h x − 1, x − 1i I2 I3 − I2 x2 − − ( x − 1) I0 I2 − 2I1 + I0 x 2 − 2 − 4( x − 1)
= x2 − 4x + 2. h x3 , φ0 i h x3 , φ1 i h x3 , φ2 i φ3 ( x ) = x3 − φ ( x ) − φ ( x ) − φ2 ( x ) 0 1 kφ0 k2 kφ1 k2 kφ2 k2 h x 3 , x − 1i h x3 , x2 − 4x + 2i h x 3 , 1i 1− ( x − 1) − 2 ( x2 − 4x + 2) = x3 − h1, 1i h x − 1, x − 1i h x − 4x + 2, x2 − 4x + 2i I I4 − I3 I5 − 4I4 + 2I3 = x3 − 3 + ( x − 1) + ( x2 − 4x + 2) I1 I2 − 2I1 + I0 I4 − 8I3 + 20I2 − 16I1 + 4I0 = x3 − 6 − 18( x − 1) − 9( x2 − 4x + 2) = x3 − 9x2 + 18x − 6. Thus, the first four orthogonal polynomials in this system are 1, x − 1, x2 − 4x + 2, x3 − 9x2 + 18x − 6. These are the Laguerre polynomials up to a constant factor.
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mathematical methods for physicists
3. Find P4 ( x ) using a. The Rodrigues Formula in Equation (6.20). Rodrigues’ Formula is given by Pn ( x ) =
1 dn 2 ( x − 1) n . 2n n! dx n
One simply grinds through the dervatives to obtain the result. P4 ( x )
= = = = = = = =
d4 2 ( x − 1)4 dx4 i 1 d3 h 2 3 8x ( x − 1 ) 24 4! dx3 i 1 d2 h 2 2 2 2 3 48x ( x − 1 ) + 8 ( x − 1 ) 24 4! dx2 i 1 d h 3 2 2 2 192x ( x − 1 ) + 144x ( x − 1 ) 24 4! dx i 1 d h 5 3 336x − 480x + 144x 24 4! dx 1 (1680x4 − 1440x2 + 144) 24 4! 1 (105x4 − 90x2 + 9) 24 1 (35x4 − 30x2 + 3). 8 1
24 4!
b. The threeterm recursion formula in Equation (6.22). The threeterm recursion formula is given by
(n + 1) Pn+1 ( x ) = (2n + 1) xPn ( x ) − nPn−1 ( x ),
n = 1, 2, . . . .
We can obtain P4 ( x ) if we know P2 ( x ) = 12 (3x2 − 1) and P3 ( x ) = 12 (5x3 − 3x ). Setting n = 3, we have 4P4 ( x )
= 7xP3 ( x ) − 3P2 ( x ) 1 1 = 7x (5x3 − 3x ) − 3 (3x2 − 1) 2 2 1 = (35x4 − 30x2 + 3) 2
Therefore,
1 (35x4 − 30x2 + 3). 8 This is the same result as in part a but with fewer computations. P4 ( x ) =
4. In Equations (6.35) through (6.42) we provide several identities for Legendre polynomials. Derive the results in Equations (6.36) through (6.42) as described in the text. Namely, a. Differentiating Equation (6.35) with respect to x, derive Equation (6.36).
nonsinusoidal harmonics
In this problem we use the identity
(n + 1) Pn+1 ( x ) = (2n + 1) xPn ( x ) − nPn−1 ( x ) to prove that
(n + 1) Pn0 +1 ( x ) = (2n + 1)[ Pn ( x ) + xPn0 ( x )] − nPn0 −1 ( x ). This follows easily from a simple differentiation. b. Derive Equation (6.37) by differentiating g( x, t) with respect to x and rearranging the resulting infinite series. The derivative identity Pn ( x ) = Pn0 +1 ( x ) − 2xPn0 ( x ) + Pn0 −1 ( x ) is found using the generating function g( x, t) = √
1 1 − 2xt + t2
∞
=
∑ Pn (x)tn ,
 x  ≤ 1, t < 1.
n =0
Differentiating g( x, t) with respect to x, we have ∂g( x, t) t = . ∂x (1 − 2xt + t2 )3/2 Multiplying by 1 − 2xt + t2 , we can write this as tg( x, t) = (1 − 2xt + t2 )
∂g( x, t) . ∂x
Inserting the infinite series for g( x, t, we obtain ∞
t
∑ Pn (x)tn
= (1 − 2xt + t2 )
n =0 ∞
∑ Pn (x)tn+1
∞
∑ Pn0 (x)tn
n =0
=
n =0
∞
∞
∞
n =0
n =0
n =0
∑ Pn0 (x)tn − 2x ∑ Pn0 (x)tn+1 + ∑ Pn0 (x)tn+2 .
We can now reindex each series as a sum over terms proportional to tk . This gives ∞
∑
k =1
Pk−1 ( x )tk =
∞
∑
Pk0 ( x )tk − 2x
k =0
∞
∑
Pk0 −1 ( x )tk +
k =1
∞
∑ Pk0 −2 (x)tk .
k =2
Next, we equate the coefficients of tk on both sides of the equation. Not all sums have constant or linear terms. So, we consider k = 0, k = 1, and k ≥ 2 terms separately. For k = 0, we have P00 ( x ) = 0, which we already know to be true. For k = 1, we have P10 ( x ) = P0 ( x ), which is we also know to be true. This leaves the coefficients of tk for k ≥ 2. These are found to be Pk−1 ( x ) = Pk0 ( x ) − 2xPk0 −1 ( x ) + Pk0 −2 ( x ),
k ≥ 2.
Replacing k = n + 1, we have Pn ( x ) = Pn0 +1 ( x ) − 2xPn0 ( x ) + Pn0 −1 ( x ),
n ≥ 1.
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c. Combining the previous result with Equation (6.35), derive Equations (6.38) and (6.39). We seek to prove Pn0 −1 ( x )
=
xPn0 ( x ) − nPn ( x ),
Pn0 +1 ( x )
=
xPn0 ( x ) + (n + 1) Pn ( x ).
We begin with the identity from the last part: Pn ( x ) = Pn0 +1 ( x ) − 2xPn0 ( x ) + Pn0 −1 ( x ) Multiply this equation by n + 1 to obtain
(n + 1) Pn ( x ) = (n + 1) Pn0 +1 ( x ) − 2x (n + 1) Pn0 ( x ) + (n + 1) Pn0 −1 ( x ). One can differentiate Equation (6.35) to get Equation (6.36), which is the result from part a:
(n + 1) Pn0 +1 ( x ) = (2n + 1)[ Pn ( x ) + xPn0 ( x )] − nPn0 −1 ( x ). Then, insert this result into the previous equation and rearrange to obtain the first identity.
(n + 1) Pn ( x ) = (n + 1) Pn0 +1 ( x ) − 2x (n + 1) Pn0 ( x ) + (n + 1) Pn0 −1 ( x ) = (n + 1) Pn0 +1 ( x ) −2x (2n + 1)( Pn ( x ) + xPn0 ( x )) − nPn0 −1 ( x ) +(n + 1) Pn0 −1 ( x ) nPn ( x )
=
xPn0 ( x ) − Pn0 −1 ( x ).
Pn0 −1 ( x )
=
xPn0 ( x ) − nPn ( x ).
Inserting this result into the identity from part b, we have Pn ( x )
=
Pn0 +1 ( x ) − 2xPn0 ( x ) + Pn0 −1 ( x ) Pn0 +1 ( x ) − 2xPn0 ( x ) + xPn0 ( x ) − nPn ( x )
=
Pn0 +1 ( x ) − xPn0 ( x ) − nPn ( x ).
=
Rearranging, we obtain the second identity Pn0 +1 ( x ) = xPn0 ( x ) + (n + 1) Pn ( x ). d. Adding and subtracting Equations (6.38) and (6.39), obtain Equations (6.40) and (6.41). We want to prove the identities Pn0 +1 ( x ) + Pn0 −1 ( x )
= 2xPn0 ( x ) + Pn ( x ),
Pn0 +1 ( x ) − Pn0 −1 ( x )
= (2n + 1) Pn ( x ).
Adding and subtracting the identities from part c, Pn0 −1 ( x )
= xPn0 ( x ) − nPn ( x ),
Pn0 +1 ( x )
= xPn0 ( x ) + (n + 1) Pn ( x ),
nonsinusoidal harmonics
we obtain Pn0 +1 ( x ) + Pn0 −1 ( x )
= ( xPn0 ( x ) − nPn ( x )) + ( xPn0 ( x ) + (n + 1) Pn ( x )) = 2xPn0 ( x ) + Pn ( x ).
Pn0 +1 ( x ) − Pn0 −1 ( x )
= ( xPn0 ( x ) − nPn ( x )) − ( xPn0 ( x ) + (n + 1) Pn ( x )) = (2n + 1) Pn ( x ).
e. Derive Equation (6.42) using some of the other identities. Equation (6.42),
( x2 − 1) Pn0 ( x ) = nxPn ( x ) − nPn−1 ( x ), can be obtained using the identities Pn0 −1 ( x )
= xPn0 ( x ) − nPn ( x ),
Pn0 +1 ( x )
= xPn0 ( x ) + (n + 1) Pn ( x ).
Multiply the first equation by x, xPn0 −1 ( x ) = x2 Pn0 ( x ) − nxPn ( x ). Replace n with n − 1 in the second equation, Pn0 ( x ) = xPn0 −1 ( x ) + nPn−1 ( x ). Now, eliminate the xPn0 −1 ( x ) terms between the two equations to obtain x2 Pn0 ( x ) − nxPn ( x ) = Pn0 ( x ) − nPn−1 ( x ). Rearranging this equation gives the sought identity. R1 5. Use the recursion relation (6.22) to evaluate −1 xPn ( x ) Pm ( x ) dx, n ≤ m. This integral would be simpler without the x, as we would have the orthogonality property, Z 1 −1
Pn ( x ) Pm ( x ) dx =
2 δnm . 2n + 1
The recursion relation
(n + 1) Pn+1 ( x ) = (2n + 1) xPn ( x ) − nPn−1 ( x ),
n = 1, 2, . . . ,
can help get around the appearance of the x factor. Solving for xPn ( x ), we have 1 xPn ( x ) = [(n + 1) Pn+1 ( x ) + nPn−1 ( x )] . 2n + 1 Inserting this into the integral and making using of the orthogonality relation, we have Z 1 −1
xPn ( x ) Pm ( x ) dx
= = =
1 1 [(n + 1) Pn+1 ( x ) + nPn−1 ( x )] Pm ( x ) dx 2n + 1 −1 1 2 2 (n + 1)δn+1,m + nδ 2n + 1 2(n + 1) + 1 2(n − 1) + 1 n−1,m 2( n + 1) 2n δ + δ . (2n + 1)(2n + 3) n+1,m (2n + 1)(2n − 1) n−1,m
Z
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Since the problem states that n ≤ m, then δn−1,m = 0 and we have Z 1 −1
xPn ( x ) Pm ( x ) dx =
2( n + 1) δ , (2n + 1)(2n + 3) n+1,m
n ≤ m.
6. Expand the following in a FourierLegendre series for x ∈ (−1, 1). a. f ( x ) = x2 . Since f ( x ) is a polynomial of degree two, then the FourierLegendre series has a finite number of terms. Namely, ∞
f (x) =
∑ cn Pn (x) = c0 P0 (x) + c1 P1 (x) + c2 P2 (x).
n =0
Furthermore, f ( x ) is an even function, so c1 = 0. This gives x2
= c0 P0 ( x ) + c2 P2 ( x ) 1 = c0 + c2 (3x2 − 1) 2 1 3 = c0 − c2 + c2 x 2 . 2 2
Comparing both sides of the equation, we have c2 = c0 =
2 3
and
1 1 c2 = . 2 3
Therefore, the FourierLegendre series representation of f ( x ) is the finite sum 1 2 x2 = P0 ( x ) + P2 ( x ). 3 3 This can be verified as 2 1 2 1 1 P0 ( x ) + P2 ( x ) = + ( (3x2 − 1)) = x2 . 3 3 3 3 2 b. f ( x ) = 5x4 + 2x3 − x + 3. Since f ( x ) is a polynomial of degree four, then the FourierLegendre series has a finite number of terms. Namely, ∞
f (x) =
∑ cn Pn (x) = c0 P0 (x) + c1 P1 (x) + c2 P2 (x) + c3 P3 (x) + c4 P4 (x).
n =0
Inserting the polynomials, we have 5x4 + 2x3 − x + 3
= c0 P0 ( x ) + c1 P1 ( x ) + c2 P2 ( x ) + c3 P3 ( x ) + c4 P4 ( x ) 3x2 − 1 5x3 − 3x = c0 + c1 x + c2 + c3 2 2 35x4 − 30x2 + 3 + c4 8 1 3 3 = c0 − c2 + c4 + c1 − c3 x 2 8 2 3 15 5 35 + c2 − c4 x 2 + c3 x 3 + c4 . 2 4 2 8
nonsinusoidal harmonics
Equating coefficients of powers of x, we obtain 5
=
2
=
0
=
−1 = 3
=
35 c , 8 4 5 c3 , 2 3 15 c2 − c4 , 2 4 3 c1 − c3 , 2 1 3 c0 − c2 + c4 . 2 8
This set of equations can be solved, giving c4 =
8 4 20 1 , c3 = , c2 = , c1 = , c0 = 4. 7 5 7 5
Therefore, the FourierLegendre series representation of f ( x ) is 1 20 4 8 f ( x ) = 4P0 ( x ) + P1 ( x ) + P2 ( x ) + P3 ( x ) + P4 ( x ). 5 7 5 7 This can be verified: 20 3x2 − 1 4 5x3 − 3x 8 35x4 − 30x2 + 3 1 +( ) +( ) 4+ x+( ) 5 7 2 5 2 7 8 = 5x4 + 2x3 − x + 3. ( −1, −1 < x < 0, c. f ( x ) = 1, 0 < x < 1. This is similar to the Heaviside example in the book. A slight modification of that example yields the result. We seek the coefficients in the expansion of this function, ∞
f (x) =
∑ cn Pn (x).
n =0
We have to compute
=
cn
=
2n + 1 1 f ( x ) Pn ( x ) dx 2 −1 Z Z 2n + 1 1 2n + 1 0 Pn ( x ) dx − Pn ( x ) dx. 2 2 0 −1 Z
We can make use of identity Pn0 +1 ( x ) − Pn0 −1 ( x ) = (2n + 1) Pn ( x ),
n > 0.
We have for n > 0 cn
= =
1 1 0 1 0 0 [ Pn+1 ( x ) − Pn0 −1 ( x )] dx − [ P ( x ) − Pn0 −1 ( x )] dx 2 0 2 −1 n +1 1 1 [ P ( x ) − Pn−1 ( x )]10 − [ Pn+1 ( x ) − Pn−1 ( x )]0−1 . 2 n +1 2 Z
Z
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These terms can be evaluated using the special values Pn (1) = 1 and Pn (−1) = −1. Since n + 1 and n − 1 are either both even or both odd, then all of these terms cancel leaving cn = Pn+1 (0) − Pn−1 (0). Since Pn (0) = 0 for n odd, the cn ’s vanish for n even as noted in the text. For n = 0, we have c0 =
1 2
Z 1 0
dx −
1 2
Z 0 −1
dx = 0.
This leads to the expansion ∞
f (x)
∑ [ Pn−1 (0) − Pn+1 (0)] Pn (x)
∼
n =1 ∞
(2k − 3)!! 4k − 1 P2k−1 ( x ), 2k
∑ (−1)k+1 (2k − 2)!!
=
k =1
where we used the result from the book that for n = 2k − 1, P2k−2 (0) − P2k (0) = (−1)k+1
(2k − 3)!! 4k − 1 . (2k − 2)!! 2k
In Figure 6.1 we show a plot of the sum of the first 20 terms. ( x, −1 < x < 0, d. f ( x ) = 0, 0 < x < 1. We seek the coefficients in the expansion of this function, ∞
Figure 6.1: A plot of first 20 terms of the FourierLegendre series of f ( x ) in Problem 6.6c.
f (x) =
∑ cn Pn (x).
n =0
We have to compute cn
= =
2n + 1 1 f ( x ) Pn ( x ) dx 2 −1 Z 2n + 1 0 xPn ( x ) dx 2 −1 Z
We first get rid of the x factor by using the identity
(n + 1) Pn+1 ( x ) = (2n + 1) xPn ( x ) − nPn−1 ( x ),
cn
= =
2n + 1 0 xPn ( x ) dx 2 −1 Z 1 0 [(n + 1) Pn+1 ( x ) + nPn−1 ( x )] dx. 2 −1 Z
We can make use of identity Pn ( x ) = Pn0 +1 ( x ) − 2xPn0 ( x ) + Pn0 −1 ( x )
nonsinusoidal harmonics
in order to integrate the Pn ( x )’s. Integrating, we find Z
Pn ( x ) dx
Z
Pn ( x ) dx
=
Z P0
=
Pn+1 ( x ) + Pn−1 ( x ) − 2
=
Z Pn+1 ( x ) + Pn−1 ( x ) − 2 xPn ( x ) − Pn ( x ) dx
0 n+1 ( x ) − 2xPn ( x ) +
Z
Pn0 −1 ( x ) dx xPn0 ( x ) dx
= 2xPn ( x ) − Pn+1 ( x ) − Pn−1 ( x ) + C.
Integrating from x = −1 to x = 0, we have Z 0 −1
Pn ( x ) dx
= [2xPn ( x ) − Pn+1 ( x ) − Pn−1 ( x )]0−1 = (2Pn (−1) + Pn+1 (−1) + Pn−1 (−1)) − Pn+1 (0) − Pn−1 (0) = (2(−1)n + (−1)n+1 + (−1)n−1 ) − Pn+1 (0) − Pn−1 (0) = − Pn+1 (0) − Pn−1 (0).
This expression can be evaluated using the fact that Pn (0) = 0 for n odd and (2k − 1)!! . P2k (0) = (−1)k (2k)!! Thus, for n = 2k − 1, we find
− Pn+1 (0) − Pn−1 (0) = − ( P2k (0) + P2k−2 (0)) (2k − 1)!! (2k − 3)!! = − (−1)k − (−1)k (2k)!! (2k − 2)!! ( 2k − 3 ) !! = (−1)k (2k)!! Returning to the computation of the FourierLegendre coefficients, for n > 1 cn
= = =
2n + 1 0 xPn ( x ) dx 2 −1 Z 1 0 [(n + 1) Pn+1 ( x ) + nPn−1 ( x )] dx. 2 −1 1 − [(n + 1)( Pn+2 (0) + Pn (0)) + n( Pn (0) + Pn−2 (0))] 2 Z
This vanishes for n odd and for n = 2k, c2k
1 = − [(2k + 1)( P2k+2 (0) + P2k (0)) + 2k( P2k (0) + P2k−2 (0))] 2 1 k (2k − 1) !! k (2k − 3) !! = − (2k + 1) (−1) − 2k (−1) 2 (2k + 2)!! (2k)!! 1 ( 2k + 1 ) !! ( 2k − 3 ) !! = − (−1)k − 2 (2k + 2)!! (2k − 2)!! 1 (2k − 3)!! k (2k + 1)(2k − 1) = − (−1) −1 2 2k(2k + 2) (2k − 2)!! 4k + 1 (2k − 3)!! = (−1)k . 8k(k + 1) (2k − 2)!!
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For n = 0, 1 1 1 f ( x ) P0 ( x ) dx 2 −1 Z 1 1 0 x dx = − 2 −1 4 Z 3 1 f ( x ) P1 ( x ) dx 2 −1 Z 3 0 2 1 x dx = . 2 −1 2 Z
=
c0
= =
c1
=
This leads to the FourierLegendre series f (x)
∞ 1 x ∼ − + + ∑ c2k P2k ( x ) 4 2 k =0
=
∞ 1 x 4k + 1 − + + ∑ (−1)k 4 2 k =0 8k(k + 1)
(2k − 3)!! (2k − 2)!!
P2k ( x ).
In Figure 6.2 we show a plot of the sum of the first 20 terms. Note that there is no Gibbs phenomena seen in the sum and that the extra power of k in the denominator of the general term in the series leads to faster convergence of the series. 7. Use integration by parts to show Γ( x + 1) = xΓ( x ).
Γ ( x + 1)
Figure 6.2: A plot of first 20 terms of the FourierLegendre series of f ( x ) in Problem 6.6d.
=
Z ∞
t x e− x dx Z ∞ −t x e− x + x 0
=
= x
Z ∞ 0
0
∞ 0
t x−1 e− x dx
t x−1 e− x dx
= xΓ( x ). 8. Prove the double factorial identities:
(2n)!! = 2n n! and
(2n − 1)!! =
(2n)! . 2n n!
One way to prove these is using mathematical induction. Another is to write out the factors. The first identity follows from
(2n)!! = (2n)(2n − 2)(2n − 4) · · · 4(2) = 2( n )2( n − 1)2( n − 2) · · · 2(2)2(1) = 2n (n)(n − 1)(n − 2) · · · (2)(1) = 2n n!
nonsinusoidal harmonics
The second can be proven using the first.
(2n)! = (2n)(2n − 1)(2n − 2) · · · 2(1) = [(2n)(2n − 2)(2n − 4) · · · (4)(2)][(2n − 1)(2n − 3) · · · (3)(1)] = (2n)!!(2n − 1)!! = 2n n!(2n − 1)!! 9. Express the following as Gamma functions. Namely, noting the form R∞ Γ( x + 1) = 0 t x e−t dt and using an appropriate substitution, each expression can be written in terms of a Gamma function. In each case we need to relate the given integral to the Gamma function integral. Sometimes this is accomplished through a substitution in the integral. R∞ a. 0 x2/3 e− x dx. Z ∞ 0
b.
R∞ 0
2 5 x2/3 e− x dx = Γ( + 1) = Γ( ). 3 3
2
x5 e− x dx.
Let t = x2 and dt = 2x dx. Then, Z ∞ 0
2
x5 e− x dx
= = =
c.
R1h
1 ∞ 4 − x2 x e 2x dx 2 0 Z ∞ 1 t2 e−t dt 2 0 1 Γ (3). 2 Z
in 1 x
dx. Let t = ln 1x , or x = e−t . Then, dx = −e−t dt. Then, 0
ln
Z 1 0
n 1 dx ln x
=
Z ∞ 0
tn e−t dt
= Γ ( n + 1). p
10. The coefficients Ck in the binomial expansion for (1 + x ) p are given by p
Ck =
p ( p − 1) · · · ( p − k + 1) . k!
p
a. Write Ck in terms of Gamma functions. p
Ck
= = = =
p ( p − 1) · · · ( p − k + 1) k! p( p − 1) · · · ( p − k + 1) ( p − k )( p − k − 1) · · · (2)1 k! ( p − k)( p − k − 1) · · · (2)1 p ( p − 1) · · · (2)1 k!( p − k)( p − k − 1) · · · (2)1 Γ ( p + 1) . Γ ( p − k + 1) Γ ( k + 1)
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b. For p = 1/2, use the properties of Gamma functions to write Ck1/2 in terms of factorials. Using the result from part a, we have Ck1/2
= =
Γ( 32 ) Γ( 23 − k)Γ(k + 1)
.
1 1 2 Γ( 2 ) . Γ( 23 − k)k!
This can be rewritten in terms of factorials or double factorials. We need the identities from the text: √ 1 Γ = π, 2 1 (2n − 1)!! √ π, Γ(n + ) = 2 2n π Γ ( x ) Γ (1 − x ) = . sin πx From the last identity we have 3 3 Γ( − k)Γ(1 − ( − k)) 2 2
3 1 = Γ( − k)Γ(k − ) 2 2 π = sin π ( 23 − k)
= (−1)k+1 π. So far, we have Ck1/2
= =
Γ( 32 )
. Γ( 23 − k )Γ(k + 1) √ π 1 (−1)k+1 Γ ( k − ). 2πk! 2
We can rewrite the Gamma function using an identity, 1 Γ(n + ) 2
= =
1 Γ(k − ) 2
= =
(2n − 1)!! √ π, 2n (2n)! √ π 4n n! (2k − 3)!! √ π, 2k −1 (2k − 2)! √ π. k 4 −1 ( k − 1 ) !
Here we used the result from Problem 8,
(2n − 1)!! =
(2n)! . 2n n!
Putting this all together, we can write Ck1/2 in terms of double factorials and factorials, Ck1/2
=
Γ( 23 ) Γ( 23 − k)Γ(k + 1)
nonsinusoidal harmonics
√
= (−1)k+1
π 1 Γ(k − ) 2πk! 2
(−1)k+1 (2k − 3)!!, 2k k!
=
or without double factorials Ck1/2
Γ( 32 )
=
Γ( 32 − k )Γ(k + 1) √ 1 π k +1 (−1) Γ(k − ) 2πk! 2 k + 1 (−1) (2k − 2)! 2k! 4k−1 (k − 1)!
= = =
(−1)k+1 (2k − 2)! 2k(2k − 1) 2k! 4k−1 (k − 1)! 2k (2k − 1)
=
(−1)k+1 (2k)! . 4k (k!)2 (2k − 1)
c. Confirm your answer in part b. by deriving the Maclaurin series expansion of (1 + x )1/2 . We can set up a table in order to find the Maclaurin series coefficients. n
f (n) ( x )
f ( n ) (0)
0
(1 + x )1/2
1
1 0!
1 2
1 (2)1!
1 2 3 4
1 2 (1 +
x )−1/2
− 14 (1 + x )−3/2 3 8 (1 +
x )−5/2
−7/2 − 15 16 (1 + x )
− 14 3 8
− 15 16
cn
=1 =
1 2
− (41)2! = − 18 3 (8)3!
=
1 16
5 − (1615)4! = − 128
Collecting terms, we have shown that 1 1 1 5 4 (1 + x )1/2 = 1 + x − x2 + x3 − x +···. 2 8 16 128 This can also be determined using the general binomial expansion,
(1 + x ) p =
∞
∑
k =0
p ( p − 1) · · · ( p − k + 1) k x , k!
 x  < 1.
Inserting p = 21 , we have
(1 + x )1/2
=
∞ 1(1 2 2
∑
r =0
− 1) · · · ( 12 − k + 1) k x , k!
 x  < 1.
1 1 1 5 4 7 5 = 1 + x − x2 + x3 − x + x +···. 2 8 16 128 256
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Checking the first terms with the values in the last part, we find they agree. 11. The Hermite polynomials, Hn ( x ), satisfy the following: R∞ √ 2 i. < Hn , Hm >= −∞ e− x Hn ( x ) Hm ( x ) dx = π2n n!δn,m . ii. Hn0 ( x ) = 2nHn−1 ( x ). iii. Hn+1 ( x ) = 2xHn ( x ) − 2nHn−1 ( x ). 2 2 dn e− x . iv. Hn ( x ) = (−1)n e x dx n Using these, show that a. Hn00 − 2xHn0 + 2nHn = 0. [Use properties ii. and iii.] First, we differentiate iii, Hn0 +1 ( x ) = 2Hn ( x ) + 2xHn0 ( x ) − 2nHn0 −1 ( x ), and differentiate ii, Hn00 ( x ) = 2nHn0 −1 ( x ). Eliminating the 2nHn0 −1 ( x ) terms between these equations, we have Hn0 +1 ( x ) = 2Hn ( x ) + 2xHn0 ( x ) − Hn00 ( x ). We need to rewrite Hn0 +1 ( x ) in terms of Hn ’s. Replacing n with n + 1 in expression ii, gives Hn0 +1 ( x ) = 2(n + 1) Hn ( x ). Then, Hn0 +1 ( x )
= 2Hn ( x ) + 2xHn0 ( x ) − Hn00 ( x )
2(n + 1) Hn ( x )
= 2Hn ( x ) + 2xHn0 ( x ) − Hn00 ( x )
0
= −2nHn ( x ) + 2xHn0 ( x ) − Hn00 ( x ).
A slight rearrangement gives the differential equation satisfied by the Hermite polynomials. R∞ √ 2 b. −∞ xe− x Hn ( x ) Hm ( x ) dx = π2n−1 n! [δm,n−1 + 2(n + 1)δm,n+1 ] . [Use properties i. and iii.] We need to replace the xHn ( x ) in terms of Hn ’s and use the orthogonality relation. From iii we have 1 xHn ( x ) = nHn−1 ( x ) + Hn+1 ( x ). 2 Therefore, Z ∞
= = = =
2
xe− x Hn ( x ) Hm ( x ) dx −∞ Z ∞ 1 − x2 e nHn−1 ( x ) + Hn+1 ( x ) Hm ( x ) dx 2 −∞ Z ∞ Z 1 ∞ − x2 − x2 n e Hn−1 ( x ) Hm ( x ) dx + e Hn+1 ( x ) Hm ( x ) dx 2 −∞ −∞ h√ i 1 h√ i n π2n−1 (n − 1)!δn−1,m + π2n+1 (n + 1)!δn+1,m 2 √ n −1 π2 n! [δm,n−1 + 2(n + 1)δm,n+1 ] .
nonsinusoidal harmonics
( c. Hn (0) =
n odd,
0,
[Let x = 0 in iii. and iterate. n = 2m. Note from iv. that H0 ( x ) = 1 and H1 ( x ) = 2x. ] )! (−1)m (2m m! ,
From iii, Hn+1 ( x ) = 2xHn ( x ) − 2nHn−1 ( x ), setting x = 0, we obtain Hn+1 (0) = −2nHn−1 (0). Before carrying out the iteration in general, one might try a few cases knowing that H0 (0) = 1 and H1 (0) = 0. For example, Hn (0)
= −2(n − 1) Hn−2 (0)
H2 (0)
= −2(1) H0 (0) = −2
H3 (0)
= −2(2) H1 (0) = 0
H4 (0)
= −2(3) H2 (0) = −2(3)(−2) = 12
H5 (0)
= −2(4) H3 (0) = 0
H6 (0)
= −2(5) H3 (0) = −2(5)(12) = −120.
Thus, we have Hn (0) = 0 for n odd. For n even, we need to look closer. Consider H6 (0), H6 (0) = [−2(5)][] − 2(3)][−2(1)] = (−2)3 (5!!) = −8(15) = −120. This suggests that H2m = (−2)m (2m − 1)!!. We can prove this using mathematical induction, or just carry out the iteration carefully. Thus, H2m (0)
= −2(2m − 1) H2m−2 (0) = −2(2m − 1)[−2(2m − 3) H2m−4 (0)] = (−2)2 (2m − 1)(2m − 3) H2m−4 (0) = (−2)3 (2m − 1)(2m − 3)(2m − 5) H2m−6 (0)
.. .
= (−2)k (2m − 1)(2m − 3)(2m − 5) · · · (2m − 2k + 1) H2m−2k (0). We continue until k = m, to obtain H2m (0) = (−2)m (2m − 1)(2m − 3)(2m − 5) · · · (3)(1) H0 (0) = (−2)m (2m − 1)!!. While we have a general form for Hn (0) for n = 2m even, it is not the sought form. However, from Problem 8, we have
(2m − 1)!! =
(2m)! . 2m m!
So, H2m = (−2)m (2m − 1)!! = (−1)m
(2m)! . m!
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12. In Maple, one can type simplify(LegendreP(2*n2,0)LegendreP(2*n,0)); to find a value for P2n−2 (0) − P2n (0). It gives the result in terms of Gamma functions. However, in Example 6.10 for FourierLegendre series, the value is given in terms of double factorials! So, we have √ π (4n − 1) (2n − 3)!! = (−1)n−1 (4n − 1) P2n−2 (0) − P2n (0) = . (2n)!! 2Γ(n + 1)Γ 32 − n You will verify that both results are the same by doing the following: a. Prove that P2n (0) = (−1)n and a binomial expansion.
(2n−1)!! (2n)!!
using the generating function
The generating function is g( x, t) = √
∞
1 1 − 2xt + t2
=
∑ Pn (x)tn ,
 x  ≤ 1, t < 1.
n =0
Inserting x = 0, we obtain g(0, t) = √
∞
1 1 + t2
=
∑ Pn (0)tn ,
 x  ≤ 1, t < 1.
n =0
The binomial expansion is given by ∞
(1 + x ) p =
∑
n =0
p ( p − 1) · · · ( p − n + 1) n x , n!
 x  < 1.
Letting p = − 21 ,
√
∞
1 1 + t2
=
(− 12 )(− 12 − 1) · · · ( 21 − n) 2n t n! n =0
=
(− 12 )(− 32 ) · · · (−( 2n2−1 )) 2n t n! n =0
=
∑ (−1)n
1(3)(5) · · · (2n − 1) 2n t 2n n!
∑ (−1)n
(2n − 1)!! 2n t , (2n)!!
∑ ∞
∑ ∞
n =0 ∞
=
n =0
where we have used (2n)!! = 2n n! from Problem 8. Therefore, P2n (0) = (−1)n b. Prove that Γ n + 21 = and iteration.
(2n−1)!! √ π 2n
(2n − 1)!! . (2n)!! using Γ( x ) = ( x − 1)Γ( x − 1)
We repeatedly use Γ( x ) = ( x − 1)Γ( x − 1) in Γ n + 12 to obtain
1 Γ n+ 2
= =
1 1 n− Γ n− 2 2 1 3 3 n− n− Γ n− 2 2 2
nonsinusoidal harmonics
1 2
1 n− 2
n−
= .. .
=
3 2
n−
3 n− 2
n−
5 2
5 Γ n− 2
1 1 ··· n− −k Γ n− −k . 2 2
Letting k = n − 1, we have 1 Γ n+ 2
= = =
1 n− 2 1 n− n− 2 (2n − 1)!! √ π. 2n
3 1 1 ··· n− −k Γ n− −k 2 2 2 3 1 1 ··· Γ 2 2 2
n−
√
c. Verify the result from Maple that P2n−2 (0) − P2n (0) =
π (4n−1) . 2Γ(n+1)Γ( 32 −n)
We use the result from part a, P2n (0) = (−1)n
(2n − 1)!! , (2n)!!
to compute P2n−2 (0) − P2n (0)
(2n − 3)!! (2n − 1)!! − (−1)n (2n − 2)!! (2n)!! (2n − 3)!! 2n − 1 = (−1)n−1 1+ (2n − 2)!! 2n (2n − 3)!! = (−1)n−1 (4n − 1). (2n)!!
= (−1)n−1
Note that this is the result presented in the problem statement. Using the result from part b, we have P2n−2 (0) − P2n (0)
(2n − 3)!! (4n − 1) (2n)!! 2n−1 Γ n − 21 4n − 1 √ = (−1)n−1 n 2 n! π 1 Γ n− 2 4n − 1 √ = (−1)n−1 . 2n! π
= (−1)n−1
Noting the identity Γ( x )Γ(1 − x ) = 3 1 Γ( − n)Γ(n − ) 2 2
π sin πx ,
=
π sin π ( 32 − n)
= (−1)n+1 π. Then, we obtain the sought expression, P2n−2 (0) − P2n (0)
=
1 Γ n − 2 4n − 1 √ (−1)n−1 2n! π
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= (−1)n−1 √
=
4n − 1 (−1)n+1 π √ 2n! π Γ( 32 − n)
π (4n − 1) . 2Γ(n + 1)Γ 23 − n
d. Can either expression for P2n−2 (0) − P2n (0) be simplified further? There is not much more that can be done. Removing the double factorials, we have P2n−2 (0) − P2n (0)
(2n − 3)!! 4n − 1 (2n − 2)!! 2n (2n − 3)!! = ((−1)n−1 (4n − 1) (2n)!! = (−1)n−1
= (−1)
n −1
(4n − 1)
= (−1)n−1 (4n − 1)
(2n−2)! 2n −1 ( n −1 ) ! 2n n!
(2n − 2)! . 22n−1 n!(n − 1)!
13. A solution of Bessel’s equation, x2 y00 + xy0 + ( x2 − n2 )y = 0, can be j+n . One obtains the recurrence relafound using the guess y( x ) = ∑∞ j =0 a j x − 1 tion a j = j(2n+ j) a j−2 . Show that for a0 = (n!2n )−1 , we get the Bessel function of the first kind of order n from the even values j = 2k: ∞
Jn ( x ) =
(−1)k x n+2k . k!(n + k)! 2 k =0
∑
One just needs to iterate the recurrence relation.
=
aj
= =
=
−1 a j(2n + j) j−2 1 1 (−1)2 a j(2n + j) ( j − 2)(2n + j − 2) j−4 1 1 1 (−1)3 a j(2n + j) ( j − 2)(2n + j − 2) ( j − 4)(2n + j − 4) j−6 .. . 1 1 1 ··· a . (−1)3 j(2n + j) ( j − 2)(2n + j − 2) ( j − 2`)(2n + j − 2`) j−2−2`
Picking j = 2k and ` = k − 1, this becomes a2k
1 1 1 ··· a , j(2n + j) ( j − 2)(2n + j − 2) ( j − 2`)(2n + j − 2`) j−2−2` 1 1 1 (−1)k ··· a0 j(2n + 2k) (2k − 2)(2n + 2k − 2) 2(2n + 2)
= (−1)`+1 = = =
(−1)k 1 2k 2 k! (n + k )(n + k − 1) · · · (n + 1)2n n! (−1)k , k ≥ 1. 22k k!
nonsinusoidal harmonics
Note that setting k = 0 into this result gives a0 =
1 2n n! .
Therefore,
∞
Jn ( x ) =
(−1)k x n+2k . k!(n + k)! 2 k =0
∑
14. Use the infinite series in Problem 13 to derive the derivative identities (6.59) and (6.60): a.
d dx
[ x n Jn ( x )] = x n Jn−1 ( x ).
Starting with ∞
(−1)k x Jn ( x ) = ∑ k!(n + k)! k =0 n
n+2k 1 x2n+2k , 2
we differentiate to obtain the result, d n [ x Jn ( x )] dx
∞
(−1)k 2(n + k) 2n+2k−1 x k!(n + k )! 2n+2k k =0 x (n−1)+2k ∞ (−1)k = xn ∑ k!(n + k − 1)! 2 k =0
=
∑
= x n Jn−1 ( x ). b.
d dx
[ x −n Jn ( x )] = − x −n Jn+1 ( x ),
Starting with x
−n
∞
(−1)k Jn ( x ) = ∑ k!(n + k )! k =0
n+2k 1 x2k , 2
we differentiate to obtain the result, d −n [ x Jn ( x )] dx
∞
(−1)k 1 x2k−1 n + ( k − 1 ) ! ( n + k ) ! 2 2k−1 k =1 ∞ (−1)`+1 x n+2`+1 = xn ∑ ` ! ( n + ` + 1) ! 2 `=0
=
∑
= x −n Jn+1 ( x ). 15. Prove the following identities based on those in Problem 14. We first note that d n [ x Jn ( x )] dx d −n x Jn ( x ) dx
= nx n−1 Jn ( x ) + x n Jn0 ( x ) = x n Jn−1 ( x ), = −nx −n−1 Jn ( x ) + x −n Jn0 ( x ) = − x −n Jn+1 ( x ).
This gives, n Jn ( x ) + Jn0 ( x ) x n − Jn ( x ) + Jn0 ( x ) x
=
Jn−1 ( x ),
= − Jn+1 ( x ).
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a. J p−1 ( x ) + J p+1 ( x ) =
2p x J p ( x ).
Subtracting the last system of equations with n = p, we obtain the result. b. J p−1 ( x ) − J p+1 ( x ) = 2J p0 ( x ). Adding the last system of equations with n = p, we obtain the result. 16. Use the derivative identities of Bessel functions,(6.59) and (6.60), and integration by parts to show that Z
x3 J0 ( x ) dx = x3 J1 ( x ) − 2x2 J2 ( x ) + C.
We first note that d [ xJ ( x )] = xJ0 ( x ), dx 1
d 2 [ x J2 ( x )] = x2 J1 ( x ). dx
Integration by parts gives Z
x3 J0 ( x ) dx
=
Z
x2
d [ xJ ( x )] dx dx 1 Z
= x3 J1 ( x ) − 2
d 2 [ x J2 ( x )] dx dx x3 J1 ( x ) − 2x2 J2 ( x ) + C.
= x3 J1 ( x ) − 2 =
x2 J1 ( x ) dx
Z
17. Use the generating function to find Jn (0) and Jn0 (0). The generating function is given by 1
g( x, t) = e x(t− t )/2 =
∞
∑
n=−∞
Jn ( x )tn ,
x > 0, t 6= 0.
i. Evaluating g( x, t) at x = 0, g(0, t) = e0 =
∞
∑
n=−∞
Jn (0)tn .
Therefore, we have Jn (0) = 0 for n > 0 and J0 (0) = 1. ii. Evaluating gx ( x, t) at x = 0, ∞ ∂g(0, t) 1 = (t − t−1 )e0 = ∑ Jn0 (0)tn . ∂x 2 n=−∞
0 (0) = − 1 , and J 0 (0) = 0 for n 6 = ±1. This gives J10 (0) = 12 , J− n 1 2
18. Bessel functions J p (λx ) are solutions of x2 y00 + xy0 + (λ2 x2 − p2 )y = 0. Assume that x ∈ (0, 1) and that J p (λ) = 0 and J p (0) is finite. a. Show that this equation can be written in the form d dy p2 x + (λ2 x − )y = 0. dx dx x
nonsinusoidal harmonics
This is the standard SturmLiouville form for Bessel’s equation. A simple verification comes from expanding the derivative and multiplying by x, d dy p2 0 = x + ( λ2 x − ) y dx dx x dy d2 y p2 2 + + ( λ x − )y dx x dx2 d2 y dy x2 2 + x + (λ2 x2 − p2 )y. dx dx
= x = b. Prove that
Z 1 0
xJ p (λx ) J p (µx ) dx = 0,
λ 6= µ
by considering Z 1 d d d d J p (µx ) x J p (λx ) − J p (λx ) x J p (µx ) dx. dx dx dx dx 0 Thus, the solutions corresponding to different eigenvalues (λ, µ) are orthogonal. Let I=
Z 1 0
d J p (µx ) dx
d d d x J p (λx ) − J p (λx ) x J p (µx ) dx. dx dx dx
From the differential equation we have d p2 d x J p (λx ) = −(λ2 x − ) J p (λx ). dx dx x Therefore, I
=
Z 1 0
J p (µx )
= ( µ2 − λ2 )
Z 1 0
p2 − λ2 x x
J p (λx ) − J p (λx )
p2 − µ2 x x
J p (µx ) dx
xJ p (λx ) J p (µx ) dx.
We also note that the integrand, d d d d x J p (λx ) − J p (λx ) x J p (µx ) , J p (µx ) dx dx dx dx in the definition of I is a perfect derivative. This is the essence of Lagrange’s identity, uLv − vLu = [ p(uv0 − vu0 )]0 d for L a SturmLiouville operator. Adding and subtracting x dx J p (µx ) we have d d d d J p (µx ) x J p (λx ) − J p (λx ) x J p (µx ) dx dx dx dx d d d d = J p (µx ) x J p (λx ) + x J p (µx ) J p (λx ) dx dx dx dx d d d d −x J p (µx ) J p (λx ) − J p (λx ) x J p (µx ) dx dx dx dx d d d d = xJ p (µx ) J p (λx ) − xJ p (λx ) J p (µx ) dx dx dx dx
d dx J p ( λx ),
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Noting from the Chain Rule that d J p (λx ) = λJ p0 (λx ), dx this becomes d d d J p (λx ) − J p (λx ) x J p (µx ) dx dx dx h i d λxJ p (µx ) J p0 (λx ) − µxJ p (λx ) J p0 (µx ) . dx
J p (µx )
=
d dx
x
Inserting this last result into the integral I and integrating, we obtain h i1 I = λxJ p (µx ) J p0 (λx ) − µxJ p (λx ) J p0 (µx ) 0
= λJ p (µ) J p0 (λ) − µJ p (λ) J p0 (µ) = 0, since we assumed J p (λ) = 0. So, we have found from the two expressions for I that
( µ2 − λ2 )
Z 1 0
xJ p (λx ) J p (µx ) dx = 0.
For λ 6= µ, this gives the sought result, Z 1 0
xJ p (λx ) J p (µx ) dx = 0.
c. Prove that Z 1 0
2 1 1 x J p (λx ) dx = J p2+1 (λ) = J p02 (λ). 2 2
In the previous part we assumed that J p (λ) = 0. However, this is not true in general. So, a more general result from the part b for λ 6= µ is Z 1 0
xJ p (λx ) J p (µx ) dx =
λJ p (µ) J p0 (λ) − µJ p (λ) J p0 (µ) µ2 − λ2
.
For the case λ = µ, we need to take the limit µ → λ of this result. However, this in an indeterminate form for which we need to apply L’Hopital’s Rule. Therefore, we differentiate the numerator and denominator with respect to µ to find Z 1 0
xJ p2 (λx ) dx
= =
lim
µ→λ
lim
µ→λ
λJ p (µ) J p0 (λ) − µJ p (λ) J p0 (µ) µ2 − λ2 λJ p0 (µ) J p0 (λ) − J p (λ) J p0 (µ) − µJ p (λ) J p00 (µ) 2µ
.
In order to evaluate the numerator, we need an expression for J p00 (µ). Bessel’s equation and the Chain Rule give d2 d 0 = x2 2 + x + (µ2 x2 − p2 ) J p (µx ) dx dx
= µ2 x2 J p00 (µx ) + µxJ p0 (µx ) + (µ2 x2 − p2 ) J p (µx ).
nonsinusoidal harmonics
Setting x = 1, µ2 J p00 (µ) + µJ p0 (µ) + (µ2 − p2 ) J p (µ) = 0, or J p00 (µ)
1 0 µ2 − p2 J (µ) + J p (µ) , =− µ p µ2
Therefore, Z 1 0
=
= = =
xJ p2 (λx ) dx
lim
λJ p0 (µ) J p0 (λ) − J p (λ) J p0 (µ) − µJ p (λ) J p00 (µ) 2µ
µ→λ
lim
λJ p0 (µ) J p0 (λ) − J p (λ) J p0 (µ) + µJ p (λ)
h
.
µ2 − p2 1 0 µ J p ( µ ) + µ2 J p ( µ )
2µ
µ→λ
λJ p0 (λ) J p0 (λ) +
λ2 − p2 λ J p (λ)
2λ λ2 − p2 2 1 0 2 [ J p (λ)] + J ( λ ). 2 2λ2 p
.
For the case that J p (λ) = 0, we have Z 1 0
xJ p2 (λx ) dx =
1 0 [ J (λ)]2 . 2 p
p 0 x J p ( x ) − J p ( x ).
From Problem 15b, J p+1 ( x ) = also have 12 J p02 (λ) = 12 J p2+1 (λ).
So, for this case, we
19. We can rewrite Bessel functions, Jν ( x ), in a form that will allow the order to be noninteger by using the gamma function. You will need the results from Problem 12b for Γ k +
1 2
.
a. Extend the series definition of the Bessel function of the first kind of order ν, Jν ( x ), for ν ≥ 0 by writing the series solution for y( x ) in Problem 13 using the gamma function. The series solution is given by ∞
Jn ( x ) =
(−1)k x n+2k . ∑ k!(n + k)! 2 k =0
Replacing factorials by Gamma functions, this is rewritten as ∞
Jν ( x ) =
x ν+2k (−1)k , Γ ( k + 1) Γ ( ν + k + 1) 2 k =0
∑
ν ≥ 0.
b. Extend the series to J−ν ( x ), for ν ≥ 0. Discuss the resulting series and what happens when ν is a positive integer. We can replace ν with −ν in the series representation, obtaining ∞
J−ν ( x ) =
x 2k−ν (−1)k , ∑ Γ ( k + 1) Γ ( k − ν + 1) 2 k =0
ν ≥ 0.
i .
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However, the Gamma function is not defined for negative integer values of a zero value. Therefore, we have to restrict ν. If ν is an integer, then ν ≥ k + 1 for all integers k ≥ 0. So, ν cannot be an integer in this case. c. Use these results to obtain the closed form expressions r 2 J1/2 ( x ) = sin x, πx r 2 J−1/2 ( x ) = cos x. πx Let ν = 1/2 in the series for Jν ( x ) and rearrange. ∞
J1/2 ( x )
= = = = = = =
(−1)k x 2k+ 12 , ∑ 3 2 k=0 k!Γ ( k + 2 ) r x 2k+1 2 ∞ (−1)k , ∑ 1 1 x k=0 k!(k + 2 )Γ(k + 2 ) 2 r 2 ∞ (−1)k 2k x2k+1 , ∑ 1 πx k=0 k!(k + 2 )22k+1 (2k − 1)!! r 2 ∞ (−1)k x2k+1 , ∑ πx k=0 k!(2k + 1)2k (2k − 1)!! r (−1)k 2 ∞ x2k+1 , ∑ πx k=0 (2k + 1)(2k )! r 2 ∞ (−1)k 2k+1 x , πx k∑ (2k + 1)! =0 r 2 sin x. πx
Next we consider J−1/2 ( x ). Since the order is not a negative integer, this is computable. Let ν = 1/2 in the series for J−ν ( x ) and rearrange. ∞
J−1/2 ( x )
(−1)k x 2k− 12 , 1 2 k =0 k!Γ ( k + 2 ) r 2 ∞ (−1)k 2k x2k = , πx k∑ k!(2k − 1)!! 22k =0 r 2 ∞ (−1)k 2k = x , πx k∑ (2k)! =0 r 2 = cos x. πx =
∑
d. Use the results in part c with the recursion formula for Bessel functions to obtain a closed form for J3/2 ( x ). The threeterm recursion formula for Bessel functions is given by Jν+1 ( x ) + Jν−1 ( x ) =
2ν Jν ( x ). x
nonsinusoidal harmonics
201
Setting ν = 1/2, we have J3/2 ( x )
= = =
1 J ( x ) − J−1/2 ( x ), x 1/2 ! r r 1 2 2 sin x − cos x, x πx πx r 2 sin x − cos x . πx x
(6.1)
20. In this problem you will derive the expansion x2 =
∞ J (α x ) c2 0 j +4∑ 2 , 2 α j=2 j J0 ( α j c )
0 < x < c,
where the α0j s are the positive roots of J1 (αc) = 0, by following the below steps. a. List the first five values of α for J1 (αc) = 0 using Table 6.3 and Figure 6.8. [Note: Be careful in determining α1 .] According to Table 6.3, the first five zeros are αc = 3.832, 7.016, 10.173, 13.324, 16.4710. However, αc = 0 is also a zero as seen in Figure 6.8 in the text is also and seen in Figure 6.3. This gives α1 = 0 for this problem. b. Show that k J0 (α1 x )k2 =
c2 2.
Recall that
k J0 (α j x )k2 =
Z c 0
xJ02 (α j x ) dx.
For j = 1 and α1 = 0,
Figure 6.3: A plot of J1 ( x ) showing the zeros for Problem 6.20a.
k J0 (α1 x )k2 =
Z c 0
xJ02 (0) dx =
Z c 0
x dx =
c2 . 2
2 2 c. Show that k J0 (α j x )k2 = c2 J0 (α j c) , j = 2, 3, . . . . (This is the most involved step.) First note from Problem 18 that y( x ) = J0 (α j x ) is a solution of x2 y00 + xy0 + α2j x2 y = 0. Note: One could also use the results from the Problem 18 directly. Namely, Z 1 0
xJ p2 (λx ) dx =
1 0 λ2 − p2 2 [ J p (λ)]2 + J ( λ ). 2 2λ2 p
i. Verify the SturmLiouville form of this differential equation: ( xy0 )0 = −α2j xy. Expanding the derivative, we have
( xy0 )0 = xy00 + y0 =
1 2 00 ( x y + xy0 ) = −α2j xy. x
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ii. Multiply the equation in part i. by y( x ) and integrate from x = 0 to x = c to obtain Z c 0
( xy0 )0 y dx = −α2j = −α2j
Z c 0
Z c 0
xy2 dx xJ02 (α j x ) dx.
Verification of this step is just a repeat of the step. iii. Noting that y( x ) = J0 (α j x ), integrate the lefthand side by parts and use the following to simplify the resulting equation. 1. J00 ( x ) = − J1 ( x ) from Equation (6.60). 2. Equation (6.63). 3. J2 (α j c) + J0 (α j c) = 0 from Equation (6.61). Z c 0
c Z c ( xy0 )0 y dx = xy0 y − xy02 dx 0 0 2 c Z c d d = x J0 (α j x ) J0 (α j x ) − J0 (α j x ) dx x dx dx 0 0 Z αc 2 j = − y J00 (y) dy 0
= − = −
Z αc j 0 c2
y [− J1 (y)]2 dy
J2 (α j c) 2 Here we used items 1. and 2., Z a 0
2
.
2 a2 x x xJ p ( j pn ) J p ( j pm ) dx = J p+1 ( j pn ) δn,m , a a 2
where j pn is the nth root of J p ( x ), J p ( j pn ) = 0, n = 1, 2, . . . . Furthermore, Equation (6.61) [see Problem 15], 2p J p ( x ), x
J p −1 ( x ) + J p +1 ( x ) = for p = 1 and x = α j c, gives J0 (α j c) + J2 (α j c) =
2 J (α c) = 0. αj c 1 j
Therefore, Z c 0
( xy0 )0 y dx = −
2 2 c2 c2 J2 (α j c) = − J0 (α j c) . 2 2
iv. Now you should have enough information to complete part c. 2 2 We are trying to prove k J0 (α j x )k2 = c2 J0 (α j c) , j = 2, 3, . . . . Therefore,
k J0 (α j x )k2
=
Z αc j 0
= −
Z c 0
=
yJ02 (y) dy
( xy0 )0 y dx
2 c2 J0 (α j c) . 2
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203
d. Use the results from parts b. and c. to derive the expansion coefficients for x2 =
∞
∑ c j J0 (α j x)
j =1
in order to obtain the desired expansion. The expansion coefficients can be obtained in the usual way: Z c
x2 ( xJ0 (αk x )) dx
0
=
∞
Z c
j =1
0
∑ cj
= ck
Z c 0
xJ0 (α j x ) J0 (αk x ) dx
xJ02 (αk x ) dx c2 2,
(
= ck
c2 2
J0 (α j c)
2
,
k=1 k > 1.
We can solve for the ck ’s if we can do the integral on the left. For k = 1, we have Z c 0
So, c1 =
c2 2
x3 J0 (0) dx =
c4 . 4
.
Problem 16 gives the result of the integration for k > 1. We have Z c 0
x3 J0 (αk x ) dx
= =
1 αk 3 cy J0 (y) dy α4k 0 1 3 α c [y J1 (y) − 2y2 J2 (y)]0 k α4k
= −
Z
2c2 J2 (αk c), α2k
k > 1.
This gives the FourierBessel coefficients 2
− 2c J (α c) α2 2 k 4J2 (αk c) ck = 2 k 2 = − 2 2 . c J ( α c ) α J ( α c ) 0 0 j j 2 k This gives x2 =
∞ c2 4J (α c) − 4 ∑ 2 22 k J0 (α j x ), 2 α j=2 k J0 ( α j c )
0 < x < c.
Using iii above gives the series expansion in the problem. A plot of the sum of the first ten terms of the FourierBessel series is shown in Figure 6.4. 21. Prove that if u( x ) and v( x ) satisfy the general homogeneous boundary conditions α1 u ( a ) + β 1 u 0 ( a )
= 0,
α2 u ( b ) + β 2 u 0 ( b )
= 0
Figure 6.4: A plot of the sum of the first ten terms of the FourierBessel series of f ( x ) = x2 for 0 < x < 1 for Problem 6.20d.
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at x = a and x = b, then b p( x )[u( x )v0 ( x ) − v( x )u0 ( x )] xx= = a = 0.
Evaluating the left hand side at the limits, we have b p( x )[u( x )v0 ( x ) − v( x )u0 ( x )] xx= =a =
p(b)[u(b)v0 (b) − v(b)u0 (b)] − p( a)[u( a)v0 ( a) − v( a)u0 ( a)]. It is sufficient to look at one of the bracketed expression: α α u( a)v0 ( a) − v( a)u0 ( a) = − 1 u( a)v( a) − − 1 v( a)u( a) β1 β1 = 0. A similar computation for the second bracket gives the result. b 0 − vu0 )] for the ( u L v − v L u ) dx = [ p ( uv a a general SturmLiouville operator L. Lagrange’s identity was proven in the text, 22. Prove Green’s identity
Rb
uLv − vLu = [ p(uv0 − vu0 )]0 . So, one needs only integrate Lagrange’s identity for x = a to x = b. Thus, Z b a
Z b
[ p(uv0 − vu0 )]0 dx b = [ p(uv0 − vu0 )] .
(uLv − vLu) dx =
a
a
23. Find the adjoint operator and its domain for Lu = u00 + 4u0 − 3u, u0 (0) + 4u(0) = 0, u0 (1) + 4u(1) = 0. The adjoint operator operator and its domain are found by forcing the relation hv, Lui = h L† v, ui for all u in the domain of L and v in the domain of L† . This is accomplished using integration by parts. Z 1 0
vLu dx
=
Z 1 0
v(u00 + 4u0 − 3u) dx 0
= [vu − v
0
u + 4vu]10
+
Z 1 0
(v00 − 4v0 − 3v)u dx
= v(1)u0 (1) − v0 (1)u(1) + 4v(1)u(1) −[v(0)u0 (0) − v0 (0)u(0) + 4v(0)u(0)] +
Z 1 0
(v00 − 4v0 − 3v)u dx
= v 0 (0) u (0) − v 0 (1) u (1) +
Z 1 0
(v00 − 4v0 − 3v)u dx.
Thus, we identify L† v = v00 − 4v0 − 3v and require that v0 (0) = 0, and = 0.
v 0 (1)
nonsinusoidal harmonics
24. Show that a SturmLiouville operator with periodic boundary conditions on [ a, b] is selfadjoint if and only if p( a) = p(b). [Recall that periodic boundary conditions are given as u( a) = u(b) and u0 ( a) = u0 (b).] We need to show that the periodic boundary conditions give b A = [ p(uv0 − vu0 )] = 0. a
We insert the conditions u( a) = u(b), u0 ( a) = u0 (b), v( a) = v(b), and 0 v ( a) = v0 (b) into A. Then, b A = [ p(uv0 − vu0 )] a
p(b)[u(b)v0 (b) − v(b)u0 (b)] − p( a)[u( a)v0 ( a) − v( a)u0 ( a)]
=
= ( p(b) − p( a))[u(b)v0 (b) − v(b)u0 (b)]. This expression vanishes if p( a) = p(b). If there are no other conditions on the solution, then A = 0 implies p( a) = p(b). Note that d v uv0 − vu0 = . dx u u2 Therefore, if A = 0 and p( a) 6= p(b), then v(b) ∝ v( a). would have to be an additional condition. 25. The Hermite differential equation is given by y00 − 2xy0 + λy = 0. Rewrite this equation in selfadjoint form. From the SturmLiouville form obtained, verify that the differential operator is self adjoint on (−∞, ∞). Give the integral form for the orthogonality of the eigenfunctions. To put this equation in selfadjoint, or SturmLiouville, form, we need to find p( x ). This is found through integration, p( x ) = e
R
(−2x ) dx
2
= e− x .
Then, the selfadjoint form becomes 2
2
2
(e− x y0 )0 = e− x [y00 − 2xy0 ] = −λe− x y. We then have the form of a SturmLiouville problem, d d p( x ) y( x ) + q( x )y( x ) = −λσ( x )y( x ), dx dx 2
2
where p( x ) = e− x , q( x ) = 0, and σ ( x ) = e− x . This is a selfadjoint operator. We need to show
hv, Lui = hLv, ui for L =
d dx
e− x
Z ∞ −∞
2
d dx
2
. Two integrations by parts will to the trick:
v(e− x u0 )0 dx
Z ∞ ∞ 2 2 = ve− x u0 − e− x u0 v0 dx −∞ −∞ Z ∞ ∞ 0 2 − x2 0 = −e uv + u e− x v0 dx −∞ −∞ Z ∞ 0 2 = u e− x v0 dx. −∞
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mathematical methods for physicists
26. Find the eigenvalues and eigenfunctions of the given SturmLiouville problems. a. y00 + λy = 0, y0 (0) = 0 = y0 (π ). The general solution of the differential equation is √ √ y( x ) = c1 cos λx + c2 sin λx It’s derivative is y0 ( x ) =
√
λ(−c1 sin
√
λx + c2 cos
√
λx ).
The condition y0 (0) = 0 gives c2 = 0 and the condition y0 (π ) = 0 √ √ gives sin λπ = 0. Therefore, λ = n, and n = 1, 2, . . . . Therefore, solutions of the eigenvalue problem are yn ( x ) = cos nx,
λ = n2 , n = 1, 2, . . . .
In this problem we need to check for λ = 0 solutions. In this case, y00 = 0. The general solution is y( x ) = Ax + B. The boundary conditions force A = 0. So, y( x ) = B is the λ = 0 eigenfunction. b. ( xy0 )0 + λx y = 0, y(1) = y(e2 ) = 0. Expanding the derivatives and multiplying by x, we have x2 y00 + xy0 + λy = 0. The characteristic equation is r (r − 1) + r + λ = r2 + λ = 0. For λ > 0, the general solution is √ √ y( x ) = c1 cos( λ ln x ) + c2 sin( λ ln x ). Applying the boundary condition y(1) = 0, we obtain c1 = 0. The boundary condition y(e2 ) = 0 implies that
√ √ n2 π 2 sin(2 λ) = 0 ⇒ 2 λ = 0 ⇒ λ = , n = 1, 2, . . . . 4 Thus, the solution of the eigenvalue problem is yn ( x ) = sin(
nπ ln x ), 2
λn =
n2 π 2 , n = 1, 2, . . . . 4
27. The eigenvalue problem x2 y00 − λxy0 + λy = 0 with y(1) = y(2) = 0 is not a SturmLiouville eigenvalue problem. Show that none of the eigenvalues are real by solving this eigenvalue problem. The characteristic equation is given by 0 = r (r − 1) − λr + λ = r2 − (1 + λ)r + λ.
nonsinusoidal harmonics
The roots of the equation are p
(1 + λ )2 − 4 . 2 To obtain solutions satisfying the boundary conditions, we will need (1 + 2 λ) − 4 < 0. For real λ, we have −3 < λ < 1. The general solution to the differential equation is " ! !# p p 1+ λ 4 − (1 + λ )2 4 − (1 + λ )2 y( x ) = x 2 c1 cos ln x + c2 sin ln x . 2 2 r± =
1+λ±
Applying the boundary conditions, c1 = 0 and ! p 1+ λ 4 − (1 + λ )2 ln 2 = 0, 2 2 sin 2 or
p
4 − (1 + λ )2 ln 2 = nπ, n = 1, 2, . . . . 2 However, the largest value of the left side of this equation for the eigenvalues is when λ = −1. This maximum is ln 2 which is smaller than any positive integer multiple of π. Therefore, there are no real eigenvalues. 28. In Example 6.20 we found a bound on the lowest eigenvalue for the given eigenvalue problem. In Example 6.20, the eigenvalue problem is given by φ00 + λφ = 0,
φ(0) = 0, φ(1) = 0.
In this case, the lowest eigenvalue is λ1 = π 2 and a bound can be found on this eigenvalue using R b dy 2 dy b 2 − py dx + a p dx − qy dx a λ1 ≤ Rb 2 a y σ dx and a nice function, y( x ), satisfying the boundary conditions. For this problem we have p( x ) = 1, q( x ) = 0, and σ ( x ) = 1. a. Verify the computation in the example. In Example 6.20 the trial function is y( x ) = x − x2 and therefore y0 ( x ) = 1 − 2x. The bound is found as R b dy 2 dy b 2 − py dx + a p dx − qy dx a λ1 ≤ Rb 2 a y σ dx R1 (1 − 2x )2 dx = R01 2 2 0 ( x − x ) dx R1 (1 − 4x + 4x2 ) dx = R 10 2 3 4 0 ( x − 2x + x ) dx
=
4 3 1 1 2 + 5
1−2+ 1 3
−
= 10.
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mathematical methods for physicists
Of course, 10 > π 2 . b. Apply the method using ( y( x ) =
0 < x < 12 , 1 2 < x < 1.
x, 1 − x,
Is this an upper bound on λ1 ? Again, the computation leads to R1 λ1 ≤ R01 0
2
y0 dx y2 dx
.
For the trial function, we have Z 1
2
y0 dx =
0
Z 1 0
2
y0 dx
= = =
Z 1/2
dx +
0
Z 1/2
Z 1
x2 dx +
0 x3 1/2
3 0 1 . 12
+
1/2
Z 1
dx = 1,
(1 − x )2 dx
1/2 (1 − x )3 1
3
1/2
This gives λ1 ≤ 12. 12 > π 2 as an upper bound on the first eigenvalue. c. Use the Rayleigh quotient to obtain a good upper bound for the lowest eigenvalue of the eigenvalue problem: φ00 + (λ − x2 )φ = 0, φ(0) = 0, φ0 (1) = 0. For this problem we have p( x ) = 1, q( x ) = − x2 , and σ ( x ) = 1. There are different trial functions on could pick, such as (a) φ( x ) = x (2 − x ), (b) φ( x ) = sin πx 2 , (c) φ( x ) = sin2
πx 2 .
The corresponding eigenvalue estimates for these are 3.0179, 3.0034, 3.9103, respectively. We will provide the computation for the trial function φ( x ) = sin πx 2 .
λ1
1 R 2 1 dφ 2 φ2 dx −φ dφ + + x dx 0 dx 0 R1 2 0 φ dx R 1 02 2 2 0 ( φ + x φ ) dx . R1 2 0 φ dx
≤
=
The integrals can be computed Z 1 0
φ2 dx =
Z 1 0
sin2
πx 1 dx = , 2 2
nonsinusoidal harmonics
Z 1 0
2
(φ0 + x2 φ2 ) dx = =
Z 1 2 π 0
= and we have λ1 ≤
πx πx + x2 sin2 2 2
Z 1 2 π 0
=
4
cos2
dx
1 2 (1 + cos πx ) + x (1 − sin πx ) dx 8 2 2
1 π2 + 8 6 1 1 1 2 2 2 − x sin πx + 2 x cos πx − 3 sin πx 2 π π π 0 π2 1 1 + + 2, 8 6 π π2 1 2 + + 2 ≈ 3.0034. 4 3 π
29. Use the method of eigenfunction expansions to solve the problems: a. y00 = x2 ,
y(0) = y(1) = 0.
We seek solutions that can be written as an expansion in the eigenfunctions, ∞
y( x ) =
∑ cn φn (x).
n =1
The eigenvalue problem is given by φ00 = −λφ,
φ(0) = φ(1) = 0.
For this problem, the eigenfunctions are φn ( x ) = sin eigenvalues λn = n2 π 2 , n = 1, 2, . . . .
√
λn x, with the
Inserting this solution into the differential equation, we have ∞
∞
n =1
n =1
∑ cn φn00 (x) = − ∑ n2 π2 cn sin nπx.
x2 =
Thus, we have a Fourier sine series and can determine the coefficients using
− n2 π 2 c n
= 2
Z 1 0
x2 sin nπx dx
x2 x 2 2 − cos nπx + 2 2 sin nπx + 3 3 cos nπx nπ n π n π 1 2 2 2 − cos nπ + 3 3 cos nπ − 3 3 nπ n π n π i h 2 2 − (2 − n2 π 2 )(−1)n . n5 π 5
= = cn
=
The series solution of this problem is ∞
y( x ) =
i 2 h 2 2 n 2 − ( 2 − n π )(− 1 ) sin nπx. ∑ 5 5 n =1 n π
1 0
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One can also solve this problem using direct integration. Namely, successive integrations give y0 ( x )
=
y( x )
=
1 3 x + C1 3 1 4 x + C1 x + C2 . 12
Setting x = 0 gives C2 = 0. The other boundary condition at x = 1 gives C1 = −1. So, the closed from solution is 1 4 ( x − x ). 12
y( x ) =
In Figure 6.5 both solutions are shown to agree. b. y00 + 4y = x2 ,
y0 (0) = y0 (1) = 0.
We seek solutions that can be written as an expansion in the eigenfunctions, ∞
Figure 6.5: A plot of the sum of the first ten terms of the series solution in Problem 6.29a.
∑ cn φn (x).
y( x ) =
n =1
The eigenvalue problem is given by φ00 + 4φ = −λφ,
φ0 (0) = φ0 (1) = 0.
For this problem, the eigenfunctions are p φn ( x ) = cos λn + 4x = cos nπx,
n = 0, 1, . . . ,
where λn = n2 π 2 − 4. Note that for n = 0 the eigenfunction is the constant solution corresponding to λ0 = −4. Inserting this solution into the differential equation, we have x2 =
∞
∞
n =1
n =1
∑ cn [φn00 (x) + 4φ(x)] = ∑ (4 − n2 π2 )cn cos nπx.
Thus, we have a Fourier cosine series and can determine the coefficients. For n = 0, and noting c0 = a0 /2 in the typical Fourier cosine series, 4c0 = or c0 =
Z 1 0
x2 dx =
1 , 3
1 12 .
For n > 0,
(4 − n2 π 2 ) c n
= 2
Z 1 0
x2 x 2 2 sin nπx + 2 2 cos nπx − 3 3 sin nπx nπ n π n π 2 cos nπ 2 2 2 n π 4 cos nπ . n2 π 2 (4 − n2 π 2 )
Figure 6.6: A plot of the sum of the first ten terms of the series solution in Problem 6.29b.
= = cn
=
x2 cos nπx dx 1 0
nonsinusoidal harmonics
The series solution of this problem is y( x ) =
i ∞ 1 2 h + ∑ 5 5 2 − (2 − n2 π 2 )(−1)n cos nπx. 12 n=1 n π
In Figure 6.6 we show the sum of the first 10 terms. This solution can be compared to the closed form solution. This is a nonhomogeneous equation. The general solution to the homogeneous equation is yh ( x ) = c1 cos 2x + c2 sin 2x. A particular solution is of the form y p ( x ) = Ax2 + Bx + C. Inserting this solution into the nonhomogeneous differential equation, 2A + 4( Ax2 + Bx + C ) = x2 . Therefore, A = 41 , B = 0, and C = − 18 , and the general solution becomes 1 1 y( x ) = c1 cos 2x + c2 sin 2x + x2 − . 4 8 Applying the boundary conditions to 1 y0 ( x ) = −2c1 sin 2x + 2c2 cos 2x + x, 2 1 we find c2 = 0 and 0 = −2c1 sin 2 + 12 , or c1 = 4 sin 2 . Therefore, the closed form of the solution of the boundary value problem is
y( x ) =
cos 2x 1 2 1 + x − . 4 sin 2 4 8
The plot of this solution agrees with that in Figure 6.6. 30. Determine the solvability conditions for the nonhomogeneous boundary value problem: u00 + 4u = f ( x ), u(0) = α, u0 (π/4) = β. The solution of the boundary value problem Lu = f with boundary conditions Bu = g exists if and only if
< f , v > −S(u, v) = 0 for all v satisfying L† v = 0 and B† v = 0. In this case the adjoint problem is v00 + 4v = 0,
v(0) = 0, v0 (π/4) = 0.
The solution of this problem is v( x ) = A sin 2x. Thus, the condition on f ( x ) is given by
< f , v > = S(u, v) Z π/4 0
f ( x ) sin 2x dx
= [vu0 − uv0 ]0π/4 = u0 (π/4)v(π/4) + u(0)v0 (0) =
β + α.
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31. Consider the problem y00 = sin x, y0 (0) = 0, y(π ) = 0. a. Solve by direct integration. Integrating twice and using the boundary conditions, we obtain y00 ( x )
= sin x.
0
y (x)
= − cos x + C1 ,
y( x )
= x − sin x + C2 ,
y0 (0) = 0 ⇒ C1 = 1. y(π ) = 0 ⇒ C2 = −π.
= x − π − sin x. b. Determine the Green’s function. The Green’s function satisfies the problem y00 = δ( x − ξ ),
y0 (0) = 0, y(π ) = 0.
The solution of the homogeneous problem is of the form y( x ) = Ax + B. Solutions satisfying the boundary conditions separately, are y1 ( x ) = C and y2 ( x ) = C ( x − π ). The Green’s function can be constructed from these solutions as ( C ( x − π ), 0 ≤ ξ ≤ x, G ( x, ξ ) = C (ξ − π ), x ≤ ξ ≤ π. In order to find C, we make use of the jump condition, ∂G (ξ + , ξ ) ∂G (ξ − , ξ ) − ∂x ∂x C−0
= =
1 p(ξ ) 1.
Therefore, we have ( G ( x, ξ ) =
x − π, ξ − π,
0 ≤ ξ ≤ x, x ≤ ξ ≤ π.
c. Solve the boundary value problem using the Green’s function. The solution can be found through integration, y( x )
=
Z π 0
G ( x, ξ ) f (ξ ) dξ
Z x
Z π
( x − π ) sin ξ dξ + (ξ − π ) sin ξ dξ 0 x x π = −( x − π ) cos ξ + (−(ξ − π ) cos ξ + sin ξ ) =
x
0
= x − π − sin x. d. Change the boundary conditions to y0 (0) = 5, y(π ) = −3. i. Solve by direct integration. Integrating twice and using the boundary conditions, we obtain y00 ( x )
= sin x.
0
y (x)
= − cos x + C1 ,
y( x )
= 6x − sin x + C2 ,
y0 (0) = 5 ⇒ C1 = 6. y(π ) = −3 ⇒ C2 = −6π − 3.
= 6x − 6π − 3 − sin x.
nonsinusoidal harmonics
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ii. Solve using the Green’s function. The solution can be determined from the sum of a solution for the nonhomogeneous problem with homogeneous boundary conditions and a solution to the homogeneous problem with nonhomogeneous boundary conditions. Therefore, y( x ) = x − π − sin x + y2 ( x ), where y200 ( x ) = 0,
y20 (0) = 5, y2 (π ) = −3.
The general solution is y2 ( x ) = Ax + B. Inserting the boundary conditions gives y2 ( x ) = 5x − 5π − 3. So, y( x ) = 6x − 6π − 3 − sin x. 32. Consider the problem: ∂2 G = δ ( x − x0 ), ∂x2
∂G (0, x0 ) = 0, ∂x
G (π, x0 ) = 0.
a. Solve by direct integration. Formally, this can be done by integrating. However, we need to integrate the delta function. Let y00 ( x ) = δ( x − x0 ). Then, 0
y (x)
=
Z x
(
= =
δ( x − x0 ) dx 0, 1,
0 ≤ x ≤ x0 , + C1 x0 ≤ x ≤ π.
H ( x − x0 ) + C1 .
This is a piecewise defined function since the delta function support is outside the region of integration when x0 > x. A second integration leads to the solution y( x ). The integral of the Heaviside function is the area under the function as shown in Figure 6.7 if x0 < x. Otherwise the area is zero. This gives the solution as y( x )
Z x
( H ( x − x0 ) + C1 ) dx ( 0, 0 ≤ x ≤ x0 , = C1 x + C2 + x − x0 , x0 ≤ x ≤ π, =
= C1 x + C2 + ( x − x0 ) H ( x − x0 ). The boundary conditions can be applied. The first boundary condition, y0 (0) = 0, implies that y0 (0) = H (− x0 ) + C1 = 0, or since 0 < x0 < π, C1 = 0.
H ( x − x0 ) 1
x0
x
Figure 6.7: The area under the Heaviside step function, H ( x − x0 ), for x > x0 .
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mathematical methods for physicists
For the second boundary condition, y(π ) = 0, we have y(π ) = (π − x0 ) H (π − x0 ) + C2 . Since 0 < x0 < π, we have 0 < π − x0 < π. This gives y(π ) = (π − x0 ) + C2 = 0, or C2 = x0 − π. The solution is then found as ( y( x )
= x0 − π + (
=
x0 − π, x − π,
0, x − x0 ,
0 ≤ x ≤ x0 , x0 ≤ x ≤ π
0 ≤ x ≤ x0 , x0 ≤ x ≤ π.
b. Compare this result to the Green’s function in part b. of Problem 31. The Green’s function in the previous problem was found as ( x − π, 0 ≤ ξ ≤ x, G ( x, ξ ) = ξ − π, x ≤ ξ ≤ π. Replacing ξ = x0 , we have (
x − π, x0 − π,
0 ≤ x0 ≤ x, x ≤ x0 ≤ π.
(
x − π, x0 − π,
x0 ≤ x ≤ π, 0 ≤ x ≤ x0 .
G ( x, x0 ) = or G ( x, x0 ) =
Clearly, this is the same as the Green’s function just found. c. Verify that G is symmetric in its arguments. For
(
x − π, ξ − π,
0 ≤ ξ ≤ x, x ≤ ξ ≤ π.
(
ξ − π, x − π,
0 ≤ x ≤ ξ, ξ ≤ x ≤ π.
G ( x, ξ ) = we have G (ξ, x ) =
Comparing these, we see G (ξ, x ) = G ( x, ξ ). 33. Consider the boundary value problem: y00 − y = x, x ∈ (0, 1), with boundary conditions y(0) = y(1) = 0. a. Find a closed form solution without using Green’s functions. The general solution of the homogeneous problem is yh ( x ) = c1 e x + c2 e− x . A particular solution is y p ( x ) = − x using the Method of Undetermined Coefficients. Therefore, the general solution of the nonhomogeneous problem is y( x ) = c1 e x + c2 e− x − x.
nonsinusoidal harmonics
The boundary conditions give 0
= c1 + c2
0
= c1 e + c2 e−1 − 1.
The first equation gives c2 = −c1 and the second equation gives c1 =
1 1 = . 2 sinh 1 e − e −1
The solution of the boundary value problem is now y( x )
= c1 e x + c2 e − x − x e x − e− x = −x 2 sinh 1 sinh x = − x. sinh 1
b. Determine the closed form Green’s function using the properties of Green’s functions. Use this Green’s function to obtain a solution of the boundary value problem. Solutions of the homogeneous equation, y00 − y = 0, can be written in terms of hyperbolic functions. Those satisfying the boundary conditions independently are y1 ( x ) = sinh x and y1 ( x ) = sinh(1 − x ). Then, the Green’s function would take the form ( C sinh(1 − x ) sinh ξ, 0 ≤ ξ ≤ x ≤ 1, G ( x, ξ ) = C sinh(1 − ξ ) sinh x, 0 ≤ x ≤ ξ ≤ 1. In order to find C, we make use of the jump condition, ∂G (ξ + , ξ ) ∂G (ξ − , ξ ) − ∂x ∂x −C cosh(1 − ξ ) sinh ξ − C cosh ξ sinh(1 − ξ )
= =
1 p(ξ ) 1.
−C sinh 1 = 1 C
= −
1 . sinh 1
Therefore, we have ( G ( x, ξ ) =
− x ) sinh ξ − sinh(1sinh , 0 ≤ ξ ≤ x ≤ 1, 1 sinh(1−ξ ) sinh x − , 0 ≤ x ≤ ξ ≤ 1. sinh 1
The solution is then obtained through integration. y( x )
=
Z 1 0
=
G ( x, ξ ) f (ξ ) dξ
Z x sinh(1 − x ) sinh ξ
ξ dξ +
Z 1 sinh(1 − ξ ) sinh x
ξ dξ − sinh 1 − sinh 1 x Z Z sinh(1 − x ) x sinh x 1 = − ξ sinh ξ dξ − ξ sinh(1 − ξ ) dξ sinh 1 sinh 1 x 0 sinh(1 − x ) = − [ξ cosh ξ − sinh ξ ]0x sinh 1 0
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sinh x [−ξ cosh(1 − ξ ) − sinh(1 − ξ )]1x sinh 1 1 − [− sinh(1 − x ) sinh x + sinh x sinh(1 − x )] sinh 1 sinh x −x sinh 1 sinh x − x. sinh 1
− =
=
c. Determine a series representation of the Green’s function. Use this Green’s function to obtain a solution of the boundary value problem. To obtain a series representation one need the eigenfunctions satisfying φ00 − φ = −λφ, φ(0) = φ(1) = 0. The general solution is φ( x ) = c1 cos
√
λ − 1x + c2 sin
√
λ − 1x.
The first boundary condition gives c1 = 0 and the second leads to √ sin λ − 1 = 0. So, the solutions to the eigenvalue problems are λn = n2 π 2 + 1, n = 1, 2, . . . .
φn ( x ) = sin nπx,
The series representation of the Green’s function is given as ∞
G ( x, ξ )
=
φn ( x )φn (ξ ) −λn Nn n =1
∑
∞
= −
2 sin nπx sin nπξ . ( n2 π 2 + 1) n =1
∑
The solution is then obtained through integration. y( x )
=
Z 1 0
= −
G ( x, ξ )ξ dξ ∞
2 sin nπx ∑ ( n2 π 2 + 1) n =1
Z 1 0
ξ sin nπξ dξ
∞
= −
2 sin nπx 1 1 [− cos nπξ + 2 2 sin nπξ ]10 2 2 nπ n π n =1 ( n π + 1 )
∑
∞
=
2(−1)n sin nπx. 2 2 n=1 nπ ( n π + 1)
∑
d. Confirm that all of the solutions obtained give the same results. The solutions can be graphed as in Figure 6.8 in order to verify that the series solution agrees with the closed form solution. Also, we can find the Fourier sine series representation of Figure 6.8: A plot of the sum of the first ten terms of the series solution in Problem 6.33.
y( x ) =
sinh x − x. sinh 1
Carrying out the integration, we confirm that the solutions are the same.
nonsinusoidal harmonics
bn
= 2 = =
= =
Z 1
y( x ) sin nπx dx Z 1 sinh x 2 − x sin nπx dx sinh 1 0 cosh x sin nπx − nπ sinh x cos nπx 2 sinh 1(n2 π 2 + 1) sin nπx − nπx cos nπx 1 − n2 π 2 0 −nπ cos nπ −nπ cos nπ 2 − n2 π 2 + 1 n2 π 2 n 2(−1) . nπ (n2 π 2 + 1) 0
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7 Complex Representations of Functions 1. Write the following in standard form. a. (4 + 5i )(2 − 3i ) = 8 − 12i + 10i + 15 = 23 − 2i. b. (1 + i )3 = 2i (1 + i ) = −2 + 2i. c.
5+3i 1− i .
5 + 3i 5 + 3i = 1−i 1−i
1+i 1+i
=
2 + 8i = 1 + 4i. 2
2. Write the following in polar form, z = reiθ . √ a. i − 1 = 2e3πi/4 . p √ In this problem r = 12 + (−1)2 = 2, and tan θ = −1. Since the point is in the second quadrant, θ = 3π 4 . b. −2i = 2e−πi/2 . √ √ c. 3 + 3i = 2 3eπi/3 . Here we note that
√
√
3 + 3i = 2 3
√ ! 1 3 + i . 2 2
3. Write the following in rectangular form, z = a + ib. a. 4eiπ/6 . Using Euler’s formula,
√ 4eiπ/6
b.
√
π π = 4 cos + i sin =4 6 6
3 1 + i 2 2
!
√ = 2 3 + 2i.
2e5iπ/4 = −1 − i.
Using Euler’s formula,
√
2e
5iπ/4
√ √ ! √ √ 5π 5π 2 2 = 2 cos + i sin = 2 − − i = −1 − i. 4 4 2 2
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c. (1 − i )100 . Note that 1 − i =
√
2eiπ/4 . Then,
(1 − i )100 =
√
2eiπ/4
100
= −250 .
4. Find all z such that z4 = 16i. Write the solutions in rectangular form, z = a + ib, with no decimal approximation or trig functions. First, we write z4 = 16i = 16eiπ/2 . Then, we compute the fourth root to get four solutions: z
= =
1/4 16eiπ/2 2 eiπ/2 e2kπi ,1/4
k = 0, 1, 2, 3,
= 2eiπ/8+kπ/2 π π π π = 2[cos( + k ) + i sin( + k )]. 8 2 8 2 For k = 0, z = 2(cos π8 + i sin π8 ). The computation of these trigonometric functions exactly is done using identities:
cos2
π 8
= = =
π 1 Therefore, cos = 8 2 Similarly, we have
q 2+
√
sin2
1 π 1 + cos 2 4 √ ! 1 2 1+ 2 2 √ 1 2+ 2 . 4
2.
π 8
= = =
π 1 1 − cos 2 4 √ ! 1 2 1− 2 2 √ 1 2− 2 . 4
q p √ √ 1 π Therefore, sin = 2 − 2. Combining these results, we have 2 + 2 + 8 2 p √ i 2 − 2. The other roots can be found by relating to these. In fact, locating these points in the complex plane as seen p simplifies pthe computation p p in Fig√ √ √ √ ure 7.1. These roots are 2 + 2 + i 2 − 2, − 2 − 2 + i 2 + 2, p p p p √ √ √ √ − 2 + 2 − i 2 − 2, and 2 − 2 − i 2 + 2. 5. Show that sin( x + iy) = sin x cosh y + i cos x sinh y using trigonometric identities and the exponential forms of these functions.
complex representations of functions
y
Figure 7.1: The location of the roots in Problem 7.4.
p √ p √ − 2 − 2, 2 + 2 p
2+
√ p √ 2, 2 − 2
x p p √ √ − 2 + 2, − 2 − 2 p
2−
√
2, −
p
2+
√ 2
From the addition formlua, sin( x + iy)
= = sin x cos(iy) + sin(iy) cos x 2y
2
2
2
+ e −i y ei y − e −i y + cos x 2 2i e−y + ey e−y − ey = sin x + cos x 2 2i = sin x cosh y + i cos x sinh y. = sin x
ei
6. Find all z such that cos z = 2, or explain why there are none. You will need to consider cos( x + iy) and equate real and imaginary parts of the resulting expression similar to Problem 5. First, we note that cos z
= cos( x + iy) = cos x cos(iy) − sin x sin(iy) = cos x cosh y − i sin x sinh y.
Therefore, we seek x and y values such that cos x cosh y − i sin x sinh y = 2. This is true only if cos x cosh y
= 2,
sin x sinh y
= 0.
The second equation implies that either y = 0 or x = nπ, where n an integer. If y = 0, then the first equation gives cos x = 2 for x real. Since this cannot happen, y 6= 0. If x = nπ, then cos nπ cosh y cosh y
221
= 2, = 2(−1)n .
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mathematical methods for physicists
Since cosh y ≥ 1, then n cannot be odd. So, for n even, y = ± cosh−1 2 = ± ln(2 +
√
3)
since cosh−1 = ln( x + √ The solutions y = − ln(2 + 3) can be rewritten: √ y = − ln(2 + 3) 1 √ ) = ln( 2+ 3 √ 2− 3 = ln( . )
p
x 2 − 1).
Therefore, the solutions are z = nπi ± i ln(2 +
√
3),
n is an even integer.
7. Find the principal value of ii . Rewrite the base, i, as an exponential first. First, we have i = eiπ/2 . Then, we see that ii = (eiπ/2 )i = e−π/2 .
y
8. Consider the circle z − 1 = 1. ( x, y) x
a. Rewrite the equation in rectangular coordinates by setting z = x + iy. We insert z = x + iy into z − 12 = 1 to obtain 1
Figure 7.2: The circle in Problem 7.8.
=  z − 12 =  x − 1 + iy2 = ( x − 1)2 + y2 .
Therefore, this is a unit circle centered at (1, 0). b. Sketch the resulting circle using part a. The circle is shown in Figure 7.2. c. Consider the image of the circle under the mapping f (z) = z2 , given by z2 − 1 = 1. i. By inserting z = reiθ = r (cos θ + i sin θ ), find the equation of the image curve in polar coordinates. We insert z = reiθ into z2 − 12 = 1 to obtain 1
=  z2 − 12 = r2 e2iθ − 12 = (r2 e2iθ − 1)(r2 e−2iθ − 1). = r4 − r2 (e2iθ + e−2iθ ) + 1
0
= r2 (r2 − 2 cos 2θ ).
This gives the equation r2 = 2 cos 2θ. ii. Sketch the image curve. You may need to refer to your Calculus II text for polar plots. [Maple might help.]
complex representations of functions
Th plot of r2 = 2 cos 2θ is shown in Figure 7.3. It is obtained using the parametrization
( x (θ ), y(θ )) = (r (θ ) cos θ, r (θ ) sin θ ),
θ ∈ [0, 2π ].
One could also parametrize z2 − 1 = 1 using z = x + iy. The resulting implicit equation is given by x4 + 2x2 y2 − 2x2 + y4 + 2y2 = 0, or
( x 2 + y2 )2 = 2( x 2 − y2 ). This is the lemniscate of Bernoulli. 9. Find the real and imaginary parts of the functions: a. f (z) = z3 . Inserting z = x + iy, z3
= ( x + iy)3 = x3 + 3ix2 y − 3xy2 − iy3 = ( x3 − 3xy2 ) + i (3x2 y − y3 )
Therefore, u( x, y) = x3 − 3xy2 and v( x, y) = 3x2 y − y3 . b. f (z) = sinh(z). Inserting z = x + iy, sinh z
= sinh( x + iy) = sinh x cosh(iy) + sinh(iy) cosh x = sinh x cos y + i sin y cosh x.
Therefore, u( x, y) = sinh x cos y and v( x, y) = sin y cosh x. c. f (z) = cos z. Inserting z¯ = x − iy, cos z¯
= cos( x − iy) = cos x cos(iy) + sin x sin(iy) = cos x cosh y + i sin x sinh y.
Therefore, u( x, y) = cos x cosh y and v( x, y) = sin x sinh y. 10. Find the derivative of each function in Problem 9 when the derivative exists. Otherwise, show that the derivative does not exist. a. f (z) = z3 .
Figure 7.3: A plot of polar equation r2 = 2 cos 2θ in Problem 7.8
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mathematical methods for physicists
We have u( x, y) = x3 − 3xy2 and v( x, y) = 3x2 y − y3 . The partial derivatives are given by ∂u ∂x ∂v ∂y ∂u ∂y ∂v ∂x
= 3x2 − 3y2 . = 3x2 − 3y2 . = −6xy. = 6xy.
From these we have ∂v ∂u = ∂x ∂y
∂v ∂u =− . ∂y ∂x
Therefore, f (z) = z3 is differentiable. The derivative can be found as df dz
= =
∂u ∂v +i ∂x ∂x 3x2 − 3y2 + 6ixy
= 3( x + iy)2 = 3z2 . b. f (z) = sinh(z). We have u( x, y) = sinh x cos y and v( x, y) = sin y cosh x. The partial derivatives are given by ∂u ∂x ∂v ∂y ∂u ∂y ∂v ∂x
= cosh x cos y. = cosh x cos y. = − sinh x sin y. = sinh x sin y
From these we have ∂u ∂v = ∂x ∂y
∂u ∂v =− . ∂y ∂x
Therefore, f (z) = sinh(z) is differentiable. The derivative can be found as df dz
= =
∂u ∂v +i ∂x ∂x cosh x cos y + i sinh x sin y
= cosh x cos(iy) + sinh x sin(iy) = cosh( x + iy)2 = cosh z.
complex representations of functions
c. f (z) = cos z. We have u( x, y) = cos x cosh y and v( x, y) = sin x sinh y. The partial derivatives are given by ∂u ∂x ∂v ∂y ∂u ∂y ∂v ∂x
= − sin x cosh y. = sin x cosh y. = cos x sinh y. = cos x sinh y.
From these we have ∂u ∂v 6= ∂x ∂y
∂u ∂v 6= − . ∂y ∂x
Therefore, f (z) = cos z is not differentiable. 11. Let f (z) = u + iv be differentiable. Consider the vector field given by F = vi + uj. Show that the equations ∇ · F = 0 and ∇ × F = 0 are equivalent to the CauchyRiemann Equations. [You will need to recall from multivari∂ ∂ ∂ + j ∂y + k ∂z .] able calculus the del operator, ∇ = i ∂x For F = vi + uj, we have
∇·F =
∇×F =
= =
∂v ∂u + = 0. ∂x ∂y i j k ∂ ∂ ∂ ∂x ∂y ∂z v u 0 ∂ ∂ ∂x ∂y k v u ∂u ∂v − k = 0. ∂x ∂y
These equations are the CauchyRiemann equations, ∂v ∂u = ∂x ∂y
∂u ∂v =− . ∂y ∂x
12. What parametric curve is described by the function γ(t) = (t − 3) + i (2t + 1), 0 ≤ t ≤ 2? [Hint: What would you do if you were instead considering the parametric equations x = t − 3 and y = 2t + 1?] This curve is in the parametric form γ(t) = x (t) + iy(t) for x (t) = t − 3 and y(t) = 2t + 1. Eliminating t, we can relate x and y. We have t = x + 3. Then, y = 2t + 1 = 2( x + 3) + 1 = 2x + 7.
225
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mathematical methods for physicists
This is a linear relationship between x and y. It gives the line y = 2x + 7 with slope 2 and yintercept 7. 13. Write the equation that describes the circle of radius 3 that is centered at z = 2 − i in (a) Cartesian form (in terms of x and y); The equation of the circle takes the form z − (2 − i ) = 3. Inserting z = x + iy, we find 9
= z − (2 − i )2 =  x + iy − (2 − i )2 = ( x − 2)2 + ( y + 1)2 .
The Cartesian form of the equation of the circle is
( x − 2)2 + (y + 1)2 = 9. (b) polar form (in terms of θ and r); The polar form is determined from the parametric equations x (θ ) − 2
= 3 cos θ,
y(θ ) + 1
= 3 sin θ,
or x (θ )
= 3 cos θ + 2,
y(θ )
= 3 sin θ − 1.
(c) complex form (in terms of z, r, and eiθ ). From the last part, we can write z = x + iy = 3 cos θ + 2 + i (3 sin θ − 1) = 3eiθ + 2 − i. Another form would be z − (2 − i ) = 3eiθ . 14. Consider the function u( x, y) = x3 − 3xy2 . a. Show that u( x, y) is harmonic; that is, ∇2 u = 0. Computing ∇2 u = u xx + uyy , we have u xx + uyy = 6x − 6x = 0.
complex representations of functions
227
b. Find its harmonic conjugate, v( x, y). We use the CauchyEquations. First, ∂v ∂y ∂v ∂x
= =
∂u = 3x2 − 3y2 , ∂x ∂u − = 6xy. ∂y
Integrating the second equation, which is the simplest one, with respect to x, we have Z
v( x, y) =
6xy dx = 3x2 y + c(y),
where c(y) is an arbitrary function of y. From the first equation, we have 3x2 + c0 (y) = 3x2 − 3y2 . Therefore, c0 (y) − 3y2 and c(y) = −y3 + k for k a constant. So, we have found v( x, y) = 3x3 y − y3 + k. c. Find a differentiable function, f (z), for which u( x, y) is the real part. We insert u( x, y) and v( x, y) into f (z) = u + iv, f (z)
= u( x, y) + iv( x, y) = x3 − 3xy2 + i (3x3 y − y3 + k) = ( x + iy)3 = z3 .
d. Determine f 0 (z) for the function in part c. [Use f 0 (z) = and rewrite your answer as a function of z.]
∂u ∂x
∂v + i ∂x
The derivative is given by df dz
= =
∂u ∂v +i ∂x ∂x 3x2 − 3y2 − 6ixy
= 3( x + iy)2 = 3z2 . 15. Evaluate the following integrals: R a. C z dz, where C is the parabola y = x2 from z = 0 to z = 1 + i. The contour is shown in Figure 7.4. The parabolic path can be parametrized by z = x + ix2 , z ∈ [0, 1]. So, dz = (1 + 2ix ) dx.
y 1+i
i
x 0
1
The integration is then Z C
z dz
=
Z 1 0
=
Z 1 0
=
( x − ix2 )(1 + 2ix ) dx (ix2 + x + 2x3 ) dx
1 1 1 i i+ + = 1+ . 3 2 2 3
Figure 7.4: The contour needed for Problem 7.15a
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mathematical methods for physicists
b.
R
C f ( z ) dz, where f ( z ) = 2z − z and C is the path from z = 0 to z = 2 + i consisting of two line segments from z = 0 to z = 2 and then z = 2 to z = 2 + i.
The contour is shown in Figure 7.5. This integration is carried out using two separate parametrizations over the horizontal and vertical parts of the path. For the horizontal path,
y 2+i
i γ1
dz = dx,
f (z) = 2z − z = x.
x 2
1
z = x, x ∈ [0, 2],
γ1 :
γ2
For the vertical path,
Figure 7.5: The contour needed for Problem 7.15b
γ2 :
z = 2 + iy, y ∈ [0, 1],
dz = idy,
f (z) = 2z − z = 2 + 3iy. Using these parametrizations, we evaluate the integral as Z C
f (z) dz
Z
=
f (z) dz +
γ1 Z 2
=
0
x dx +
c. C
x
−1
0
1
1 C z2 +4
0
(2 + 3iy)i dy
3 2
dz for C the positively oriented circle, z = 2. [Hint: Parametrize
the circle as z = 2eiθ , multiply numerator and denominator by e−iθ , and put in trigonometric form.]
i
−2
R
Z 1
1 + 2i. 2
=
2i
f (z) dz γ2
= 2 + 2i − y
Z
2
−i
The contour is shown in Figure 7.4. We let z = 2eiθ , θ ∈ [0, 2π ]. This gives dz = 2ieiθ dθ. The integration using this parametrization gives Z
−2i
C
1 dz z2 + 4
= =
Figure 7.6: The contour needed for Problem 7.15c
= = = =
2ieiθ dθ 4e2iθ + 1 0 Z i 2π dθ iθ 2 0 e + e−iθ Z i 2π dθ 4 0 cos θ Z i 2π sec θ dθ 4 0 i ln  sec θ + tan θ 2π 0 = 0. 4
Z 2π
16. Let C be the positively oriented ellipse 3x2 + y2 = 9. Define F ( z0 ) =
Z C
z2 + 2z dz. z − z0
complex representations of functions
Find F (2i ) and F (2). [Hint: Sketch the ellipse in the complex plane. Use the Cauchy Integral Theorem with an appropriate f (z), or Cauchy’s Theorem if z0 is outside the contour.] Cauchy’s Integral Theorem gives the value of a function at a point using 1 f ( z0 ) = 2πi
I C
y 4i
3i
f (z) dz z − z0
2i
provided f (z) is analytic inside and on C and z0 is inside the region bounded f (z) by the contour. However, if z0 is outside the region and not on C, then z−z0 is analytic inside and on C. In that case, the integral vanishes due to Cauchy’s Theorem.
C i x
−2
−1
For this problem the contour is the ellipse in Figure 7.8. The ellipse equation is easily rewritten as
z2
where f (z) = have
Z C
z2 + 2z dz = 2πi f (z0 ), z − z0
0
1
2
−i
x2 y2 + = 1. 3 9 The semimajor and semiminor axes can then be identified. The given function can be written as F ( z0 ) =
229
−2i −3i Figure 7.7: The contour needed for Problem 7.16
+ 2z. For z0 = 2i, which is inside the elliptical region, we
F (2i ) = 2πi f (2i ) = 2πi (−4 + 4i ) = −8π (1 + i ). For z0 = 2, which is outside the elliptical region, F (2) = 0. 17. Show that Z C
dz = ( z − 1 − i ) n +1
(
0, 2πi,
n 6= 0, n = 0,
for C the boundary of the square 0 ≤ x ≤ 2, 0 ≤ y ≤ 2 taken counterclockwise. [Hint: Use the fact that contours can be deformed into simpler shapes (like a circle) as long as the integrand is analytic in the region between them. After picking a simpler contour, integrate using parametrization.] The rectangular contour in Figure 7.8 can be deformed in into a circular contour C 0 with z0 = 1 + i in the center as shown in the figure. Then, Z C
dz = ( z − 1 − i ) n +1
Z C0
dz . ( z − 1 − i ) n +1
We parametrize the new contour using z − (1 + i ) = eiθ , 0 ≤ θ ≤ 2π. This gives dz = ieiθ dθ and the integral is computed for n 6= 1 as Z C0
dz ( z − 1 − i ) n +1
=
Z 2π iθ ie
einθ
0
= i = =
Z 2π 0
dθ
ei(n−1)θ dθ
ei(n−1)θ 2π n−1 0 cos(n − 1)θ − sin(n − 1)θ 2π = 0. n−1 0
y 2i C i C0 x
−2
−1
0
1
2
−i −2i Figure 7.8: The contour needed for Problem 7.17
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mathematical methods for physicists
For n = 1, the integral becomes Z C0
dz ( z − 1 − i ) n +1
=
Z 2π iθ ie 0
=
Z 2π 0
eiθ
dθ
dθ = 2πi.
Combining these results confirms the answer. 18. Show that for g and h analytic functions at z0 , with g(z0 ) 6= 0, h(z0 ) = 0, and h0 (z0 ) 6= 0, g(z) g ( z0 ) Res ; z0 = 0 . h(z) h ( z0 ) This is shown by carrying out the computation of the residue noting that there is a simple pole at z = z0 . Also, one needs to apply L’Hopital’s Rule g(z) since evaluation of (z − z0 ) h(z) at z = z0 is indeterminate. Carrying out the computation, we have g(z) g(z) Res ; z0 = lim (z − z0 ) z → z0 h(z) h(z)
= = =
lim
z → z0
d dx (( z − z0 ) g ( z )) h0 (z)
g ( z ) + ( z − z0 ) g 0 ( z ) z → z0 h0 (z) g ( z0 ) . h 0 ( z0 ) lim
19. For the following, determine if the given point is a removable singularity, an essential singularity, or a pole (indicate its order). In some cases one could make this determination by observation. However, we will express each function as a Laurent series and draw conclusions based on the expansion. a.
1−cos z , z2
z = 0.
Using the Maclaurin series expansion for the cosine, we have 1 − cos z 1 z2 z4 z6 = 1− 1− + − +··· 2! 4! 6! z2 z2 2 4 6 1 z z z = − + −··· 4! 6! z2 2!
=
z2 z4 1 − + −···. 2! 4! 6!
Since there are no negative powers of z, the singularity at z = 0 is a removable singularity. b.
sin z , z2
z = 0.
Using the Maclaurin series expansion for the sine, we have sin z 1 z3 z5 z7 = z − + − + · · · 3! 5! 7! z2 z2
=
1 z z3 z5 − + − +···. z 3! 5! 7!
complex representations of functions
The most negative power of z is −1 from the 1z term. Therefore, the singularity at z = 0 is a pole or order one, or simple pole. c.
z2 −1 , ( z −1)2
z = 1.
In this example we need to write the function as an expansion in powers of z − 1. The denominator is of the appropriate form. The numerator could a expansion using Taylor series methods. However, a simple algebraic manipulation will suffice to write the numeration as a function of z − 1. z2 − 1 ( z − 1)2
( z − 1 + 1)2 − 1 ( z − 1)2 ( z − 1)2 + 2( z − 1) = ( z − 1)2 2 = 1+ . z−1 =
The most negative power of z − 1 is −1 from the second term. Therefore, the singularity at z = 1 is a pole or order one, or a simple pole. d. ze1/z ,
z = 0.
Using the Maclaurin series expansion for the exponential, we have 1 1 1 1 1/z + + +··· ze = z 1+ + z 2!z2 3!z3 4!z4 1 1 1 = z+1+ + +···. + 2!z 3!z2 4!z3 Since there are an infinite number of negative powers of z, the singularity at z = 0 is an essential singularity. e. cos z−ππ ,
z = π.
Using the Maclaurin series expansion for cos x, we have cos x cos
π z−π
x2 x4 x6 + − +··· 2! 4! 6! 2 4 1 π 1 π + −··· 1− 2! z − π 4! z − π
= 1− =
= 1−
π2 π4 ( z − π ) −2 + ( z − π ) −4 − · · · . 2! 4!
Since there are an infinite number of negative powers of z, the singularity at z = π is an essential singularity. z 20. Find the Laurent series expansion for f (z) = sinh about z = 0. [Hint: z3 You need to first do a MacLaurin series expansion for the hyperbolic sine.] The MacLaurin series expansion for the hyperbolic sine is similar to that of the sine can be found from either the sine function or the exponential function. Using the MacLaurin series expansion for the sine, we have
sinh z
= −i sin(iz)
231
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mathematical methods for physicists
(iz)3 (iz)5 (iz)7 = −i iz − + − +··· 3! 5! 7! 3 5 7 z z z = −i iz + i + i + i + · · · 3! 5! 7! 5 7 3 z z z = z+ + + +···. 3! 5! 7! Using the MacLaurin series expansion for the exponential, we have sinh z
= =
1 z (e − e−z ) 2 1 1 1 1 1 1+ + + + + · · · 2 z 2!z2 3!z3 4!z4 1 1 1 1 + + − 1+ + +··· z 2!z2 3!z3 4!z4
z3 z5 z7 + + +···. 3! 5! 7! Now, we can find the Laurent series expansion 1 sinh z z3 z5 z7 = z+ + + +··· 3! 5! 7! z3 z3
= z+
1 z2 z4 1 + + +···. + 3! 5! 7! z2 21. Find series representations for all indicated regions. In this problem we employ the geometric series
=
∞
1
∑ x n = 1 + x + x2 + · · · = 1 − x ,
x < 1
n =0
for small x expansions and ∞ n 1 = ∑ x n =0
1+
= a. f (z) =
z z −1 ,
1 1 + 2 +··· x x
1 , 1 − 1x
1   < 1, x
or  x  > 1.
z < 1, z > 1.
The small z expansion is obtained by rewriting the rational function so that one can employ the above geometric series. z z−1
1 1−z − z 1 + z + z2 + · · ·
= −z =
∞
= −
∑ zn ,
z < 1.
n =1
For the large z expansion, one needs to rewrite the rational function as a function of 1/z. 1 z = z−1 1 − 1z ∞ n 1 = ∑ , z > 1. z n =0
complex representations of functions
233
b. f (z) = (z−i)(1 z+2) , z < 1, 1 < z < 2, z > 2. [Hint: Use partial fractions to write this as a sum of two functions first.] Either through observation, or using partial fraction decomposition, we have 1 1 1 1 = − . (z − i )(z + 2) 2+i z−i z+2 In order to determine the series representations of f (z) in all indicated regions, we need the small z and large z expansions of 1 1 z−i and z+2 . These are found by rewriting the fractions so that a geometric series can be obtained. 1 z−i
=
1 z−i
1 z+2
∑ (−iz)
n
,
=
1 1 21+
=
1 ∞ z (− )n , ∑ 2 n =0 2
=
1 1 z1−
=
1 ∞ i n ( ) , z n∑ =0 z
=
1 1 z1+ ∞
=
A(z + 2) + B(z − i ) = 1.
z < 1.
Equating powers of z, A + B = 0 and 2A − iB = 1. Solving this system, we obtain 1 . A = −B = 2+i
z 2
z < 2.
i z
y
z > 1.
2i 3
2 z
1
z > 2. −2
These series are combined to give series expansions in the regions indicated in Figure 7.9. Each region is labeled by the order of appearance of the four series. For z < 1, we combined the first and second series (in order of the above series): ! ∞ 1 1 1 ∞ z n n f (z) = = i ∑ (−iz) − ∑ (− ) . (z − i )(z + 2) 2+i 2 n =0 2 n =0 For 1 < z < 2, we combine the second and third series: 1 ∞ i n 1 ∞ z ( ) − ∑ (− )n ∑ z n =0 z 2 n =0 2
! .
For 2 < z, we combine the last two series: 1 1 f (z) = = (z − i )(z + 2) 2+i
i
3
2 1 (− )n , ∑ z n =0 z
1 1 = f (z) = (z − i )(z + 2) 2+i
Adding the terms on the right side gives
or
∞
n =0
1 z+2
1 A B = + . (z − i )(z + 2) z−i z+2
1 A ( z + 2) + B ( z − i ) = , (z − i )(z + 2) (z − i )(z + 2)
i 1 + iz
= i
The partial fraction decomposition in Problem 7.21b would first assume
2 1 ∞ i n 1 ∞ ( ) − ∑ (− )n z n∑ z z z =0 n =0
! .
−1 2
2
2
1
x
−i 4
−2i Figure 7.9: Regions of convergence for Laurent expansions of f (z) = (z−i)(1 z+2) for Problem 7.21b. Each number indicates the series number.
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mathematical methods for physicists
22. Find the residues at the given points: a.
2z2 +3z z −1
at z = 1.
The point z = 1 is a simple pole. The residue is 2 2z + 3z 2z2 + 3z Res ;z = 1 = lim (z − 1) z−1 z−1 z →1
=
lim (2z2 + 3z) = 5.
z →1
The residue can also be determined as the coefficient of (z − 1)−1 in the Laurent series expansion. The expansion is found as 2z2 + 3z z−1
= = =
2( z − 1 + 1)2 + 3( z − 1 + 1) z−1 2 2[(z − 1) + 2(z − 1) + 1] + 3(z − 1) + 3 z−1 5 . 2( z − 1) + 7 + z−1
The coefficient of (z − 1)−1 is 5. b.
ln(1+2z) z
at z = 0.
ln(1 + 2z) Res ;z = 0 z
= =
ln(1 + 2z) z z →0 lim ln(1 + 2z) = 0. lim (z − 0)
z →0
The Laurent series expansion is found using the Maclaurin series expansion x3 x2 + −.... ln(1 + x ) = x − 2 3 Then, ln(1 + 2z) 1 4z2 8z3 = 2z − + −... z z 2 3 8 2 = 2 − 2z + z − . . . . 3 There is no z−1 term, so the residue is zero. c.
cos z (2z−π )3
at z =
π 2.
cos z π Res ;z = 2 (2z − π )3
= = = =
d π 2 cos z lim (z − ) 2 (2z − π )3 z→ π2 dz cos z d limπ π z→ 2 dz 8( z − 2 ) 1 −(z − π2 ) sin z − cos z lim (z − π2 )2 z→ π2 8 1 −(z − π2 ) cos z limπ = 0. 2(z − π2 ) z→ 2 8
complex representations of functions
The Laurent series is found by noting that sin(z − So, cos z (2z − π )3
π 2)
235
= − cos z.
sin(z − π2 ) 8(z − π2 )3 1 π 1 π 3 1 π 5 − (z − ) − (z − ) + (z − ) − · · · 2 3! 2 5! 2 8(z − π2 )3 1 1 π 1 + − ( z − )2 − · · · . − 48 960 2 8(2z − π )2
= − = =
There is no (2z − π )−1 term, so the residue is again found to be zero. R 2π 23. Consider the integral 0 5−4dθcos θ . a. Evaluate this integral by making the substitution 2 cos θ = z + 1z , z = eiθ , and using complex integration methods. For the substitution 2 cos θ = z + 1z , z = eiθ , we have dz = ieiθ dθ = iz dθ and
1 5 − 4 cos θ = 5 − 2 z + z
=
−2z2 + 5z − 2 . z
This gives the integral in the form dθ =i 5 − 4 cos θ
I C 2z2
dz , − 5z + 2
where C is the unit circle centered about the origin. The roots of 2z2 − 5z + 2 = 0 are z = 2, 12 . There is only one (simple) pole inside the region of integration as seen in Figure 7.10. The integral can be computed by computing a residue at this pole. Z 2π 0
dθ 5 − 4 cos θ
dz 2z2 − 5z + 2 I i dz 2 C (z − 2)(z − 12 ) " # i 1 1 (2πi ) Res ;z = 2 2 (z − 2)(z − 12 )
= i = =
y i
−1
1
2
x
−i
I
C
1 1 = −π lim (z − ) 1 2 (z − 2)(z − 12 ) z→ 2 2π = . 3 b. In the 1800s, Weierstrass introduced a method for computing integrals involving rational functions of sine and cosine. One makes the substitution t = tan 2θ and converts the integrand into a rational function of t. Note that the integration around the unit circle corresponds to t ∈ (−∞, ∞).
Figure 7.10: Contour for Problem 7.23
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mathematical methods for physicists
i. Show that sin θ =
2t , 1 + t2
cos θ =
1 − t2 . 1 + t2
Starting with t = tan 2θ , we write t2 =
sin2
θ 2 cos2 2θ
1 − cos θ . 1 + cos θ
=
This can be solved for cos θ. t2
=
(1 + cos θ )t2
=
2
1 − cos θ 1 + cos θ 1 − cos θ
= 1 − t2 1 − t2 = . 1 + t2
(1 + t ) cos θ cos θ
This is now used to find the sin θ, sin2 θ
= 1 − cos2 θ (1 − t2 )2 . = 1− (1 + t2 )2 4t2 = . (1 + t2 )2
2t Taking the square root, we have sin θ = 1+ . Note, since t can t2 be negative, we need not specify positive or negative signs.
ii. Show that dθ =
2dt . 1 + t2
Differentiating t = tan 2θ , we have
=
dt
= = = =
1 θ sec2 dθ 2 2 1 2 dθ cos2
θ 2 1 2 dθ
1 2 (1 + cos θ )
dθ 1− t2 1+ t2 1 + t2
1+ 2
dθ.
Solving for dθ gives the result. iii. Use the Weierstrass substitution to compute the above integral. Substituting the transformation into the integral gives dθ 5 − 4 cos θ
=
Z ∞ −∞
2 1+ t2
5−4
dt 1− t2 1+ t2
complex representations of functions
Z ∞
=
237
2 dt
− ∞ 5(1 + t2 ) − 4(1 − t2 ) Z ∞ 2 dt
=
9t2 + 1 ∞ 2 tan−1 3t 3 −∞ 2 π π 2π − (− ) = . 3 2 2 3 −∞
= = 24. Do the following integrals: a. I
z−i =3 z2
ez dz. + π2
The contour for this integral is shown in Figure 7.11. There are two simple poles at z = ±iπ and only one is inside the region of integration. The value of the integral is found using the residue at this simple pole: I z−i =3 z2
ez dz + π2
= 2πi lim (z − iπ ) z→iπ
z2
y 4i
ez + π2
C 2i
ez z→iπ z + πi = −1.
= 2πi lim = eiπ
iπ
3i
i x
−3
−2
−1
0
1
2
3
−i
b. I z−i =3
z2 − 3z + 4 dz. z2 − 4z + 3
The contour for this integral is shown in Figure 7.12. There are two simple poles at z = 1, 3 and only one is inside the region of integration. The value of the integral is found using the residue at this simple pole: I z−i =3
z2 − 3z + 4 dz z2 − 4z + 3
= = =
c.
Z ∞ −∞
[Hint: This is Im
R∞
−∞
eix x 2 +4
−∞
−3i
z2 − 3z + 4 dz z−i =3 ( z − 3)(z − 1) z2 − 3z + 4 2πi lim ( z − 3) z →1 2 2πi = −2πi. −2
y 4i 3i C 2i i x
sin x dx. x2 + 4
−3
−2
−1
0
1
2
3
−i −2i
dx.]
sin x dx = Im x2 + 4
−iπ
Figure 7.11: The contour needed for Problem 7.24a
I
The real integral can be written in terms of a contour integral in the complex plane. First, we note that Z ∞
−2i
Z ∞ −∞
eix dx. x2 + 4
Figure 7.12: The contour needed for Problem 7.24b
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mathematical methods for physicists
Then, we consider the integral eiz dz, +4
Z CR
z2
where the contour for this integral is shown in Figure 7.13. Letting R → ∞, we can obtain the answer since Z C
eiz dz 2 z +4
=
lim
R → ∞ CR
=
lim
Z ∞
=
ΓR 2i
−R
x
R
−2i
Z
R→∞
y
−∞
eiz dz +4
Z
z2 R
−R
eiz dz + z2 + 4
Z ΓR
eiz dz z2 + 4
eix dx. x2 + 4
By Jordan’s Lemma the last integral over Γ R vanishes as R gets large. The remaining integral is the desired integral which is found by applying the Residue Theorem to the contour integral over C. There are two simple poles at z = ±2i and only one is inside the region of integration. The value of the integral is found using the residue at this simple pole in the upper half plane.
Figure 7.13: The contour needed for Problem 7.24c
Z C
eiz dz z2 + 4
eiz z2 + 4 z→2i iz π e = 2. 2πi lim e z→2i z + 2i
= 2πi lim (z − 2i ) =
We can now obtain the value of the integral. − R + πi
−R Figure 7.14: Problem 25.
y
Z ∞ R + πi
R
−∞
x
Rectangular contour for
π sin x = 0. dx = Im x2 + 4 e2
R ∞ (ln x)2 25. Evaluate the integral 0 1+ x2 dx. [Hint: Replace x with z = et and use the rectangular contour in Figure 7.14 with R → ∞.] Letting x = et , dx = et dt, the integral becomes Z ∞ (ln x )2 0
1+
x2
dx =
Z ∞ −∞
t2 e t dt = 1 + e2t
Z ∞ −∞
t2 dt. 2 cosh t
So, we consider the integral I C
z2 dz 2 cosh z
using the suggested contour. We note from cosh z = cosh x cos y + i sinh x sin y that cosh(iπ/2) = 0 and cosh( x + iπ ) = − cosh x. The first indicates that there is one pole inside the contour at z = iπ/2. Then, from the Residue Theorem we have " 2 # I z − i π2 1 i π2 π3 z2 dz = 2πi =− . 2 cosh z 4 C 2 cosh z π z =i 2
complex representations of functions
239
H z2 The integral C 2 cosh z dz can also be computed by integrating over each part of the path. Let the finite R contributions be given by z2 dz 2 cosh z
I CR
=
Z R −R
z2 dz + 2 cosh z z2
Z − R+πi
+
Z R+πi
2 cosh z
R+πi
R
dz +
z2 dz 2 cosh z
Z −R − R+πi
z2 dz. 2 cosh z
The two vertical paths will not contribute as R → ∞ since for the vertical path z = ± R + iy, z2 (± R + iy)2 R2 dz = dy → R dy → 0. 2 cosh z 2 cosh(± R + iy) 2e The first integral is the desired integral in the limit R → ∞. So, that leaves the third integral, which can be transformed using z = w + πi. Under this transformation, we have Z − R+πi R+πi
z2 dz 2 cosh z
= −
Z −R (w + πi )2
dw 2 cosh w Z R Z R w2 2πiw π2 dw + dw − dw. 2 cosh w − R 2 cosh w − R 2 cosh w
R
=
Z R −R
The first integral is a copy of the sought integral. The second integral vanishes as the integrand is an odd function. This leaves the lim
Z R
R→∞ − R
π2 dw 2 cosh w
= =
π2 ∞ sech w dw 2 −∞ ∞ π3 π2 tan−1 sinh w = . 2 2 −∞ Z
Combining the results so far, we have I C
Z ∞ −∞
Therefore,
z2 dz 2 cosh z π3 − 4 2 z dz 2 cosh z
R∞ 0
(ln x )2 1+ x 2
dx =
= 2 = 2 =
Z ∞ −∞
Z ∞ −∞
z2 π2 dz − 2 cosh z 2 2 π3 z dz − 2 cosh z 2
Z ∞ −∞
sech w dw
π3 . 8
π3 8 .
26. Do the following integrals for fun! a. For C the boundary of the square  x  ≤ 2, y ≤ 2, I C
dz . z(z − 1)(z − 3)2
In this problem there are three poles. Two simple poles are inside the region of bounded by the contour as seen in Figure 7.15. I C
dz z(z − 1)(z − 3)2
2
=
1 2πi ∑ Res[ ; z = zi ] z(z − 1)(z − 3)2 i =1
y 2i C
i
−3 −2 −1 0 −i
x 1
2
3
−2i Figure 7.15: Contour for Problem 7.26a
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mathematical methods for physicists
1 1 + (z − 1)(z − 3)2 z=0 z(z − 3)2 z=1 1 1 2πi − + 9 4 5π i. 18
= 2πi = = b. Z π 0
This integral can be done using by transforming it to a contour integration over the unit circle in the complex plane. We let z = eiθ , which gives
y i
−1
sin2 θ dθ. 13 − 12 cos θ
cos θ =
z = 1
1
x
1 1 ( z + ), 2 z
sin θ =
1 1 ( z − ), 2i z
and dz = ieiθ dθ = iz dθ,
−i
Z π Figure 7.16: Contour for Problem 7.26b
0
sin2 θ dθ 13 − 12 cos θ
=
1 2
Z π
=
1 2
I
=
1 8i
I
=
1 8i
I
sin2 θ dθ 13 − 12 cos θ 2 1 1 2i ( z − z ) dz
−π
z=1
z=1
13 − 6(z + 1z ) iz
( z2 − 1)2 dz. z2 (6z2 − 13z + 6) ( z2 − 1)2
z=1 z2 (3z − 2)(2z − 3)
dz.
There are three poles, z = 0, 23 , 32 . Only the first two contribute to the value of the integral. The residues are computed,
( z2 − 1)2 2 Res 2 ;z = 3 z (3z − 2)(2z − 3)
= = =
Res
( z2 − 1)2 ;z = 0 z2 (3z − 2)(2z − 3)
lim
(z − 23 )(z2 − 1)2 z2 (3z − 2)(2z − 3)
lim
( z2 − 1)2 3z2 (2z − 3)
= =
=
z→ 23
z→ 23
( 49 − 1)2 5 =− , 4 4 36 3 9 ( 3 − 3) d ( z2 − 1)2 z→0 dz (6z2 − 13z + 6) 2 (z − 1)(4z) lim z→0 6z2 − 13z + 6 (z2 − 1)2 (12z − 13) − (6z2 − 13z + 6)2 13 , 36 lim
complex representations of functions
241
and the Residue Theorem gives Z π 0
sin2 θ dθ 13 − 12 cos θ
c.
( z2 − 1)2 dz. z=1 z2 (3z − 2)(2z − 3) 1 5 13 = 2πi − + 8i 36 36 π . = 18 =
1 8i
Z ∞ −∞ x2
I
dx . + 5x + 6
We consider the integral I=
I C
dz = z2 + 5z + 6
I C
dz (z + 2)(z + 3)
y ΓR
using the contour shown in Figure 7.17. The integrand has simple poles at z = −2, −3 which lie on the real axis. There are no singularities enclosed by the contour, so I = 0. We then need to look at the contributions to the contour from its pieces. For finite R we integrate over CR to obtain 0
Cε
− R− 4 − 3 − 2 − 1 0 1 2 3 4 R x Figure 7.17: Contour for Problem 7.26c
dz ( z + 2 )(z + 3) CR Z Z −3− ε Z dz dz dz = + + ( z + 2 )( z + 3 ) ( z + 2 )( z + 3 ) ( z + 2 )(z + 3) ΓR −R Cε Z −2− ε Z Z R dz dz dz + + + . 0 ( z + 2 )( z + 3 ) ( z + 2 )( z + 3 ) ( z + 2 )(z + 3) −3+ ε Cε −2+ ε I
=
We note that the integral vanishes on Γ R as R → ∞. This can be seen by using z = Re−iθ and letting R get large. Z ΓR
dz (z + 2)(z + 3)
= →
iRe−iθ dθ Γ R ( Re−iθ + 2)( Re−iθ + 3) Z 1 i dθ = 0, as R → ∞. R ΓR e−iθ
Z
In the limit as R → ∞ and e → 0 the second, fourth, and sixth terms give the desired integral. So, we have Z −3− ε Z ∞ dz dx = lim 2 (z + 2)(z + 3) R→∞,e→0 −R −∞ x + 5x + 6 Z −2− ε Z R dz dz + + −3+ε (z + 2)(z + 3) −2+ε ( z + 2)(z + 3) Z Z dz dz = − lim + . e →0 Cε ( z + 2)( z + 3) Cε0 ( z + 2)( z + 3) The integration around the two smaller semicircles, Cε and Cε0 , contribute a half residue each. Thus, we have Z ∞ −∞ x2
dx = −πi (1 − 1) = 0. + 5x + 6
Cε0
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mathematical methods for physicists
d.
Z ∞ cos πx
1 − 9x2
0
dx.
First, because the integrand is an even function, we can write Z ∞ cos πx 0
1 − 9x2
1 ∞ cos πx dx 2 −∞ 1 − 9x2 Z ∞ 1 eiπx Re dx. 2 −∞ 1 − 9x2 Z
=
dx
=
Therefore, we will consider the integral I= y
−R
Cε0 R
Figure 7.18: Contour for Problem 7.26d
C
eiπz dz 1 − 9z2
using the contour shown in Figure 7.18. The integrand has simple poles at z = ± 13 . These poles lie on the real axis just like in the last example. Jordan’s Lemma can be used to dispel of the large semicircular arc.
ΓR Cε
I
x
Since the contour does not contain any singularities, the integral over the closed contour vanishes. The contour can be written as a sum of six contours, 0
= =
CR
eiπz dz 1 − 9z2
ΓR
eiπz dz + 1 − 9z2
I
Z
+
Z
1 −ε 3
eiπz
− 13 +ε
1 − 9z2
Z − 1 −ε 3 −R
dz +
eiπz dz + 1 − 9z2 eiπz
Z Cε0
1 − 9z2
Z Cε
dz +
eiπz dz 1 − 9z2
Z R 1 +ε 3
eiπz dz. 1 − 9z2
Taking the limits R → ∞ and e → 0, the second, fourth, and sixth terms give the principal value integral. So, we have Z ∞ −∞
eiπx dx 1 − 9x2
=
Z − 1 −ε 3
lim
R→∞,e→0
+
1 −ε 3
Z
− 13 +ε
= − lim
−R
eiπz dz + 1 − 9z2
Z
e →0
eiπz dz 1 − 9z2
Cε
Z R 1 +ε 3
eiπz dz + 1 − 9z2
Z Cε0
eiπz dz 1 − 9z2
!
eiπz dz . 1 − 9z2
Each integral around the semicircles gives a value −πiRes [ f (z); z]. The residues are found as iπz e 1 Res ;z = ± 3 1 − 9z2
=
1 eiπz lim (z − (± )) 3 1 − 9z2 z→± 13
= ∓
√ e±iπ/3 1 = (∓1 − i 3). 6 12
Inserting the residues gives √ Z ∞ √ √ eiπx 1 1 3 dx = πi (−1 − i 3) + (1 − i 3) = π 2 12 12 12 −∞ 1 − 9x
complex representations of functions
243
and the final result is
√
Z ∞ cos πx 0
e.
1 − 9x2
dx = Re
Z ∞ −∞ ( x2
3 π 12
√
!
=
3 π. 24
dx . + 9)(1 − x )2
The integrand in this problem has poles at z = ±3i and z = 1. The first two are simple poles and the third on is a double pole. Enclosing the contour in the upper half complex plane, we will R consider C (z2 +9dz for the contour shown in Figure 7.19. )(1−z)2 We only need the residues for z = 3i and z = 1. These are 1 1 Res ; z = 3i = lim (z − 3i ) 2 (z2 + 9)(1 − z)2 (z + 9)(1 − z)2 z→3i 1 = lim (z − 3i ) (z + 3i )(1 − z)2 z→3i 1 1 = + i, 100 75 and Res
1 ;z = 1 2 (z + 9)(1 − z)2
= =
y
1 z2 + 9 1 2z =− . lim − 50 z →1 ( z 2 + 9 )2 d lim z→1 dz
3i ΓR Cε
−R
dz 2 ( z + 9 )(1 − z)2 CR Z Z 1− ε dz dz = + 2 2 2 ( z + 9 )( 1 − z ) ( z + 9 )(1 − z)2 ΓR −R Z Z R dz dz + + . 2 2 2 Cε ( z + 9)(1 − z ) 1+ε ( z + 9)(1 − z )2
1
Rx
Z
The integral over CR in the limits R → ∞ and ε → 0 is 2πi times the residue of the included pole. The integral over Cε will give −πi times the residue of the bypassed pole. The remaining integrals combine to give the principal value integral sought. Thus, Z ∞ dx 1 πi 2π 1 + i − =− . = 2πi 100 75 50 75 −∞ ( x2 + 9)(1 − x )2 f.
Z ∞ 0
√
x dx. (1 + x )2
In this problem there is a branch point at z = 0. We will choose the branch cut along the positive real axis as shown in Figure 7.20. The contour will consist of a large circle of radius R two horizontal
−3i Figure 7.19: Contour for Problem 7.26e
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mathematical methods for physicists
paths above and below the branch cut, connected by a small circle of radius ε. The value of the integral around the two circles as R → ∞ and ε → 0 vanish. This is easily seen using a parametrization of the circles, z = reiθ . Then, √ Z Z 2π √ iθ/2 z re ireiθ dθ. dz = 2 ( 1 + z ) ( 1 + reiθ )2 C 0
y CR Ce
−1
x
Figure 7.20: Contour for Problem 7.26f
For r = R, as R → ∞, this integral is of order R−1/2 . So, the integral vanishes in the limit. Similarly, For r = ε, as ε → 0, this integral is of order ε3/2 . The integral also vanishes on this path. We next consider the horizontal paths. For the top path, z = reiθ for θ = 0. So, the integral is √ √ Z ∞ Z ∞ z r dz = dr. 2 ( 1 + z ) ( 1 + r )2 0 0 For the bottom path, z = reiθ for θ = 2π. So, the integral is Z 0 ∞
√
z dz = (1 + z )2
Z 0 √ iπ re ∞ (1 + r )2
dr =
Z ∞ 0
√
r dr. (1 + r )2
Finally, there is one pole enclosed by the contour at z = −1. The residue is √ d √ 1 r Res ; z = − 1 = [ z]z=eiπ = e−iπ/2 . 2 dz 2 (1 + z ) Putting these results together, we have √ √ √ I Z Z ε z z z dz = dz + dz 2 2 CR (1 + z ) C (1 + z ) − R (1 + z )2 √ √ Z Z R z z + dz + dz. Cε (1 + z )2 ε (1 + z )2 √ Z ∞ 1 −iπ/2 z 2πi e = 2 dz 2 0 (1 + z )2 Therefore,
Z ∞ 0
√
x π dx = . 2 2 (1 + x )
8 Transform Techniques in Physics 1. In this problem you will show that the sequence of functions 1 n f n (x) = π 1 + n2 x 2
n π n π
approaches δ( x ) as n → ∞. Use the following to support your argument: x
a. Show that limn→∞ f n ( x ) = 0 for x 6= 0. n lim n→∞ π
1 1 + n2 x 2
=
n 1 = 0, π n2 x 2
Figure 8.1:Plot of the functions f n ( x ) = 1 n for Problem 8.1 π 1+ n2 x 2
x 6= 0.
b. Show that the area under each function is one. n π
Z ∞ −∞
dx 1 + n2 x 2
= =
1 −1 ∞ tan nx π −∞ 1 π π + = 1. π 2 2
2. Verify that the sequence of functions { f n ( x )}∞ n=1 , defined by f n ( x ) = n −n x  , approaches a delta function. 2e First, we have n lim f n ( x ) = lim e−n x = 0. n→∞ n→∞ 2 Next, we need to compute the area under each function in the sequence. Z ∞ −∞
f n ( x ) dx
= = = =
n 2 n 2 n 2 n 2
Z ∞ −∞ Z 0
e−n x dx Z ∞
e−nx dx −∞ 0 nx e 0 e−nx ∞ + n −∞ −n 0 1 1 − − = 1. n n enx dx +
3. Evaluate the following integrals: Rπ a. 0 sin xδ x − π2 dx. Z π 0
π dx = 1. sin xδ x − 2
n π n 2
x Figure 8.2: Plot of the functions f n ( x ) = n −n x  for Problem 8.2 2e
246
mathematical methods for physicists
b.
R∞
−∞
x −5 2x 3 e
δ
3x2 − 7x + 2 dx.
The argument of the Dirac delta function is not simple. It is of the 5 2x form δ( f ( x )) where f ( x ) = x− 3 e . The Dirac delta function can be written as
δ ( x −5)  f 0 (5)
δ ( x −5) .  13 e10 
=
Z ∞
x − 5 2x 2 e 3x − 7x + 2 dx 3 δ ( x − 5) 2 3x − 7x + 2 dx 1 10 3e
δ
−∞
Z ∞
=
−∞
126 . e10
= c.
Rπ 0
x2 δ x +
Since x =
π 2 π −2
dx.
is not in the integration interval, [0, π ], Z π 0
d.
R∞ 0
Therefore,
π dx = 0. x2 δ x + 2
e−2x δ( x2 − 5x + 6) dx. [See Problem 4.]
We note that δ( x2 − 5x + 6)
= δ(( x − 2)( x − 3)) δ ( x − 2) δ ( x − 3) = + 1 1 = δ ( x − 2) + δ ( x − 3).
The denominators come from using f ( x ) = x2 − 5x + 6 and  f 0 ( x ) = 2x − 5 = 1 at x = 2, 3. Now we evaluate the integral. Z ∞ 0
e−2x δ( x2 − 5x + 6) dx
=
Z ∞
= e e.
R∞
−∞ ( x
2
0 −4
e−2x [δ( x − 2) + δ( x − 3)] dx
+ e −6 .
− 2x + 3)δ( x2 − 9) dx. [See Problem 4.] We note that δ ( x 2 − 9)
= δ(( x − 3)( x + 3)) δ ( x − 3) δ ( x + 3) + . = 6 6
The denominators come from using f ( x ) = x2 − 9 and  f 0 ( x ) = 2x  = 6 at x = ±3. Now we evaluate the integral. Z ∞ −∞
=
1 6
( x2 − 2x + 3)δ( x2 − 9) dx
Z ∞ −∞
( x2 − 2x + 3)[δ( x − 3) + δ( x + 3)] dx = 4.
transform techniques in physics
4. For the case that a function has multiple roots, f ( xi ) = 0, i = 1, 2, . . . , it can be shown that n δ ( x − xi ) δ( f ( x )) = ∑ .  f 0 ( xi ) i =1 R∞ Use this result to evaluate −∞ δ( x2 − 5x − 6)(3x2 − 7x + 2) dx. We note that δ( x2 − 5x + 6)
= δ(( x + 1)( x − 6)) δ ( x + 1) δ ( x − 6) = + 7 7 1 = [δ( x − 2) + δ( x − 3)]. 7 The denominators come from using f ( x ) = x2 − 5x − 6 and  f 0 ( x ) = 2x − 5 = 1 at x = −1, 6. Now we evaluate the integral. Z ∞ −∞
δ( x2 − 5x − 6)(3x2 − 7x + 2) dx
1 ∞ (δ( x + 1) + δ( x − 6))(3x2 − 7x + 2) dx 7 −∞ 1 80 = [12 + 68] = . 7 7 5. Find a Fourier series representation of the Dirac delta function, δ( x ), on [− L, L]. We seek the representation
=
Z
δ( x ) ∼
∞ a0 nπx + ∑ [ an cos + bn sin nπxL]. 2 L n =1
The Fourier coefficients are Z 1 L 1 a0 = δ( x ) dx = . L −L L Z nπx 1 1 L δ( x ) cos an = dx = . L −L L L Z 1 L nπx bn = δ( x ) sin dx = 0. L −L L Inserting the Fourier coefficients into the series gives δ( x ) ∼
1 1 + 2L L
∞
∑ cos
n =1
nπx . L
6. For a > 0, find the Fourier transform, fˆ(k), of f ( x ) = e− a x . A direct computation of the Fourier transform gives fˆ(k)
= = = =
Z ∞ −∞ Z 0
e−a x eikx dx e(ik+ a) x dx +
−∞ e(ik+a) x 0
Z ∞
e(ik− a) x dx
0 ( ik − a) x ∞ e
+ ik + a −∞ ik − a 0 1 1 2a − = 2 . ik + a ik − a a + k2
247
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mathematical methods for physicists
7. Use the result from Problem 6 plus properties of the Fourier transform to find the Fourier transform, of f ( x ) = x2 e− a x for a > 0. We make use of the property
F [ x2 f ( x )] = −
d2 dk2
Z ∞ −∞
f ( x )eikx dx.
Then, 2 − a x 
F [x e
] = = =
d2 2a − 2 2 dk a + k2 4ak d dk ( a2 + k2 )2 4a( a2 − 3k2 ) . ( a2 + k 2 )3
2 8. Find the Fourier transform, fˆ(k), of f ( x ) = e−2x + x . The Fourier transform integral is
fˆ(k)
Z ∞
=
−∞ Z ∞
=
−∞
e−2x
2 +x
e−2x
2 +(1+ik ) x
eikx dx dx.
This integral can be computed by rewriting the integral as a Gaussian integral and using r Z ∞ π −2y2 . e dy = 2 −∞ We begin by rewriting the exponent, 1 + ik x] 2 # 1 + ik 1 + ik 2 1 + ik 2 2 −2 x − x+ − 2 4 4 " 2 2 # 1 + ik 1 + ik − −2 x− 4 4 2 1 + ik (1 + ik)2 −2 x − + . 4 8
−2x2 + (1 + ik) x = −2[ x2 − " = = =
The Fourier transform integral becomes fˆ(k )
= =
Z ∞ −∞
(1+ik)2 e 8
= e =
(1+ik)2 8
2 +(1+ik ) x
dx
"
Z ∞
1 + ik exp −2 x − 4 −∞
Z ∞−(1+ik)/4 −∞−(1+ik)/4
(1+ik)2 e 8
r
=
e−2x
Z ∞ −∞
2
e−2y dy
π (1+ik)2 e 8 . 2
2
e−2y dy
2 # dx
transform techniques in physics
249
In Problem 12 we show Z ∞ −∞
e
−αt2 + βt
r dt =
π β2 e 4α . α
Inserting α = 2 and β = 1 + ik, we obtain the result in this problem. 9. Prove the Second Shift Property in the form h i F eiβx f ( x ) = fˆ(k + β). This follows quickly from a direct computation: h i Z F eiβx f ( x ) =
∞
−∞
f ( x )ei( β+k) x dx = fˆ(k + β).
10. A damped harmonic oscillator is given by ( Ae−αt eiω0 t , t ≥ 0, f (t) = 0, t < 0. . a. Find fˆ(ω ) A durect computation of the Fourier transform in frequency space gives fˆ(ω )
= = = =
Z ∞ −∞
A A
f (t)eiωt dt
Z ∞
e−αt eiω0 t eiωt dt
0 e(−α+i(ω +ω0 ))t
− α + i ( ω + ω0 ) A α − i ( ω + ω0 )
∞ 0
b. and the frequency distribution  fˆ(ω )2 . One computes the square of the modulus by multiplying fˆ(ω ) by its complex conjugate,  fˆ(ω )2 = fˆ(ω ) fˆ(ω ) :
 fˆ(ω )2
= =
A α − i ( ω + ω0 )
α2
A α + i ( ω + ω0 )
A2 . + ( ω + ω0 ) 2
c. Sketch the frequency distribution. The sketch of  fˆ(ω )2 is shown in Figure 8.3. Note that for large ω the distribution tends to zero. There is a peak in the middle when the denominator is the least. This is seen to occur for ω = −ω0 .
 fˆ(ω )2 A2 /α2
− ω0
11. Show that the convolution operation is associative: ( f ∗ ( g ∗ h))(t) = (( f ∗ g) ∗ h)(t).
ω
Figure 8.3: Plot of the functions  fˆ(ω )2 for Problem 8.10c
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mathematical methods for physicists
It is sometimes best to write out both sides of this identity using the convolution definition. This then allows for one to see what variable transformation is needed to complete the proof. Then, write the proof in the form
( f ∗ ( g ∗ h))(t) = = = = = =
Z ∞ −∞
Z ∞
−∞
f ( x )( g ∗ h)(t − x ) dx Z ∞ f (x) g(y)h(y − (t − x )) dy dx −∞
Z ∞ Z ∞ −∞ −∞
Z ∞ Z ∞
f ( x ) g(y)h(y − (t − x )) dydx f (z − y) g(y)h(z − t) dydz f (z − y) g(y) dy h(z − t) dz
−∞ −∞ Z ∞ Z ∞ −∞
Z ∞
−∞
−∞
( f ∗ g)(z)h(z − t) dz
= (( f ∗ g) ∗ h)(t). Note that going from line three to four the variable transformation z = x + y was made. 12. In this problem, you will directly compute the convolution of two Gaussian functions in two steps. a. Use completing the square to evaluate Z ∞ −∞
e−αt
2 + βt
dt.
First, we complete the square by factoring out −α. β −αt2 + βt = −α t2 − t α " 2 2 # β β β 2 = −α t − t + − α 2α 2α β 2 β2 = −α t − + . 2α 4α Inserting this in the integral and making a substitution x = t − we have Z ∞ −∞
e−αt
2 + βt
dt
=
Z ∞ β 2 β2 −α t− 2α + 4α
−∞ β2
e
Z ∞
β 2α ,
dt
2
= e 4α e−αx dx −∞ r π β2 = e 4α . α b. Use the result from part a. to directly compute the convolution in Example 8.16:
( f ∗ g)( x ) = e−bx
2
Z ∞ −∞
e−(a+b)t
2 +2bxt
dt.
transform techniques in physics
251
Comparing this integral with the previous part, we have α = a + b and β = 2bx. Therefore,
( f ∗ g)( x ) = e−bx
2
Z ∞ −∞
e−(a+b)t
2 +2bxt
dt
r 2 π 44b(a2+xb2) = e−bx e a+b r π − ab x2 = e a+b . a+b 13. You will compute the (Fourier) convolution of two box functions of the same width. Recall that the box function is given by ( 1,  x  ≤ a f a (x) = 0,  x  > a. Consider ( f a ∗ f a )( x ) for different intervals of x. A few preliminary sketches will help. In Figure 8.4, the factors in the convolution integrand are show for one value of x. The integrand is the product of the first two functions. The convolution at x is the area of the overlap in the third figure. Think about how these pictures change as you vary x. Plot the resulting areas as a function of x. This is the graph of the desired convolution. Figure 8.4: Sketch used to compute the convolution of the box function with itself. In the top figure is the box function. The middle figure shows the box shifted by x. The bottom figure indicates the overlap of the functions.
f a (t) 1 a
−a f a ( x − t)
x 1
−a + x
a+x
f a (t) f a ( x − t)
a
−a The convolution integral takes the form
( f a ∗ f a )( x ) =
Z ∞ −∞
f a (t) f a ( x − t) dt.
Thus,the integrand is the product of the functions ( ( 1, t ≤ a 1, f a (t) = and f a ( x − t) = 0, t > a. 0,
 x − t ≤ a  x − t > a.
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mathematical methods for physicists
Figure 8.5: Sketch used to compute the areas of overlap. (a) f a (t) is shown for reference. (b) The nonzero part of f a (t) f a ( x − t) is shaded for x > 0. It has area 2a − x. (c) The nonzero part of f a (t) f a ( x − t) is shaded for x < 0. It has area 2a −  x  = 2a + x.
f a (t) 1
( a)
a
−a f a (t) f a ( x − t)
x
(b)
a
−a + x f a (t) f a ( x − t) x
−a
(c)
a+x
Looking at the Figure 8.4, we note that f a (t) f a ( x − t) = 1 when t ≤ a and  x  ≤ 2a. Using Figure 8.5 we can compute the needed areas. In both cases the area of f a (t) is 2a. The area of f a (t) f a ( x − t) is  x  less or A = 2a −  x . We can actually compute these areas through integration for x > 0 and x < 0 : Z a x−a Z x+a
−a
dt
= a − ( x − a) = 2a − x,
dt
= x + a − (− a) = 2a + x. 0 < x < 2a,
0 < x < 2a,
Otherwise the area is zero. In summary, we have Z ∞ fa ∗ fa
−∞
2a
−2a
2a
x
Figure 8.6: The convolution in Problem 8.13.
0, f a (t) f a ( x − t) dt = 2a + x, 2a − x,
 x  ≥ 2a −2a ≤ x ≤ 0 0 ≤ x ≤ 2a.
Thus, the area changes linearly as x goes from x = 0 to x = 2a. A similar behavior can be seen for decreasing xvalues. This leads to a triangular shape for the convolution. This is shown in Figure 8.6. R∞ √ 2 14. Define the integrals In = −∞ x2n e− x dx. Noting that I0 = π, a. Find a recursive relation between In and In−1 . In =
Z ∞ −∞
2
x2n e− x dx,
n = 0, 1, . . . .
For n = 0, we have the Gaussian integral I0 =
Z ∞ −∞
2
e− x dx =
√ π.
transform techniques in physics
For n > 0, we can carry out an integration by parts. This is done by first borrowing an x to go with the exponential in order to find an antiderivative. In
= = = =
Z ∞ −∞
2
x2n e− x dx
Z ∞
2
x2n−1 xe− x dx, −∞ Z 2 ∞ 1 2n − 1 ∞ 2(n−1) − x2 − x2n−1 e− x + x e dx 2 2 −∞ −∞ 2n − 1 In−1 . 2
b. Use this relation to determine I1 , I2 , and I3 . Just insert n = 1, 2, 3. to find I0
=
I1
=
I2
=
I3
=
√ π. 1 I0 = 2 3 I = 2 1 5 I = 2 1
1√ π. 2 3√ π. 4 15 √ π. 8
c. Find an expression in terms of n for In . We obtain a general expression for In , n > 0 by iterating ` times. This gives In
= =
=
2n − 1 In−1 2 2n − 1 2n − 3 In−2 2 2 .. . 2n − 3 2n − 2` + 1 2n − 1 In−` ··· 2 2 2  {z } ` terms
Letting ` = n, we obtain In =
(2n−1)!! √ π. 2n
15. Find the Laplace transform of the following functions: a. f (t) = 9t2 − 7.
L[9t2 − 7] =
18 7 − . s s3
b. f (t) = e5t−3 .
L[e5t−3 ] =
e −3 . s−5
c. f (t) = cos 7t.
L[cos 7t] =
s2
s . + 49
253
There are two hints needed for students. First, they need the k = 0 value. The second is the trick of factoring out an x in order to carry out the integration by parts.
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mathematical methods for physicists
d. f (t) = e4t sin 2t. 2 s2 +4
One can use the First Shift Theorem plus L[sin 2t] =
L[e4t sin 2t] =
to obtain
2 2 = 2 . ( s − 4)2 + 4 s − 8s + 20
e. f (t) = e2t (t + cosh t). One can use the First Shift Theorem plus L[t] = s , formation L[cosh t] = s2 − 1
L[e2t (t + cosh t)] = =
1 s2
and the trans
1 s−2 + ( s − 2)2 ( s − 2)2 − 1 1 s−2 + . (s − 2)2 s2 − 4s + 3
f. f (t) = t2 H (t − 1). The Second Shift Theorem can be used,
L[ f (t − 1) H (t − 1)] = e−s F (s). However, we need to identify f (t). We rewrite t2 as t2 = (t − 1 + 1)2 = (t − 1)2 + 2(t − 1) + 1. Then, f (t) = t2 + 2t + 1, giving 2 2 1 + 2+ . s s3 s The Laplace transform becomes F (s) =
L[t2 H (t − 1)] = e−s ( g. f (t) =
sin t, sin t + cos t,
2 + 2s + s2 . s3
t < 4π, . t > 4π
This problem can be rewritten in terms of a Heaviside step function and then prepared for the Second Shift Theorem. Both branches have a sin t term and what remains is cos t times a step function. ( sin t, t < 4π, f (t) = sin t + cos t, t > 4π ( 0, t < 4π, = sin t + cos t, t > 4π ( 0, t < 4π, = sin t + cos t 1, t > 4π
= sin t + cos tH (t − 4π ) = sin t + cos(t − 4π ) H (t − 4π ).
transform techniques in physics
The last step made use of the periodicity of the cosine function. The transforms of the sine and cosine can now be used to obtain
L[sin t + cos tH (t − 4π )] = h. f (t) =
Rt
0 (t − u)
1 + se−4πs . s2 + 1
2 sin u du.
This is a convolution integral of the functions t2 and sin t. We apply the convolution theorem,
L[
Z t 0
f (u) g(t − u) du] = F (s) G (s),
to f (t) = sin t and g(t) = t2 to obtain Z t 2 1 2 2 L[ (t − u) sin u du] = 3 . = 3 2 2 s s +1 s ( s + 1) 0 i. f (t) = (t + 5)2 + te2t cos 3t and write the answer in the simplest form. This can be done in pieces. First,
L[(t + 5)2 ] = L[t2 + 10t + 25] 2 10 25 = 3+ 2 + s s s 2 + 10s + 25s2 = . s3 The second term, L[te2t cos 3t], take a little more effort. Looking up only properties and the transform of the cosine, we would need the First Shift Theorem because of the exponential factor and
L[t f (t)] = −
d F (s) ds
because of the t factor. Here, we have F (s) = L[e2t cos 3t] =
s−2 s−2 = 2 . ( s − 2)2 + 9 s − 4s + 13
Therefore,
L[t f (t)] = −
s−2 d ds s2 − 4s + 13
=
s2 − 4s + 13 − (s − 2)(2s − 4) (s2 − 4s + 13)2
=
s2 − 4s − 5 . (s2 − 4s + 13)2
So, the full answer is given by
L[(t + 5)2 + te2t cos 3t] = = =
10 25 (s − 2)2 − 9 2 + 2 + + 3 s s s ( s − 2)2 + 9
2 + 10s + 25s2 s2 − 4s − 5 + s3 (s2 − 4s + 13)2
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mathematical methods for physicists
16. Find the inverse Laplace transform of the following functions using the properties of Laplace transforms and the table of Laplace transform pairs. a. F (s) =
18 7 + . s s3
L −1 [ b. F (s) =
2 1 . − 2 s−5 s +4
L −1 [ c. F (s) =
18 7 + ] = 9t2 + 7. s s3
1 2 − 2 ] = e5t − sin 2t. s−5 s +4
s+1 . s2 + 1
L −1 [
s+1 ] = cos t + sin t. s2 + 1
3 . s2 + 2s + 2 The inverse transform is found after first completing the square,
d. F (s) =
3 3 = . s2 + 2s + 2 ( s + 1)2 + 1 The s + 1 term suggests applying a shift theorem to obtain
L −1 [ 1 . ( s − 1)2 Note that
3 ] = 3e−t sin t. ( s + 1)2 + 1
e. F (s) =
d 1 1 =− . ds s − 1 ( s − 1)2
d Using L[t f (t)] = − ds F (s), we have
L −1 [
1 ] = tet . ( s − 1)2
e−3s . s2 − 1 The exponential factor suggests using the Second Shift Theorem 1 combined with L−1 [ 2 ] = sinh t. Therefore, s −1
f. F (s) =
L −1 [
e−3s ] = sinh(t − 3) H (t − 3). s2 − 1
Note, the partial fraction decomposition 1 1 1 1 = − 2 s−1 s+1 s2 − 1 also leads to the sinh t function.
transform techniques in physics
1 . + 4s − 5 Completing the square, we have
g. F (s) =
s2
s2
1 1 = . + 4s − 5 ( s + 2)2 − 9
This is a shifted version of F (s) = 1 3
1 3 3 s2 −9
whose inverse Laplace
transform is f (t) = sinh 3t. Using the First Shift Theorem, we obtain 1 1 L −1 [ 2 ] = e−2t sinh 3t. 3 s + 4s − 5 s+3 . s2 + 8s + 17 Completing the square and writing the result as F (s + 4) leads to the answer. First, we see
h. F (s) =
s+3 s+3 ( s + 4) − 1 = = = F ( s + 4) s2 + 8s + 17 ( s + 4)2 + 1 ( s + 4)2 + 1 for F (s) =
s −1 s +1 .
Then, the inverse transform becomes
L −1 [
s+3 ] = e−4t (cos t − sin t). s2 + 8s + 17
17. Compute the convolution ( f ∗ g)(t) (in the Laplace transform sense) and its corresponding Laplace transform L[ f ∗ g] for the following functions: a. f (t) = t2 , g(t) = t3 . The convolution is found directly as
( f ∗ g)(t) = =
Z t 0
Z t 0
=
Z t 0
=
Z t 0
= =
f (u) g(t − u) du u2 (t − u)3 du u2 (t3 − 3t2 u + 3tu2 − u3 ) du
(t3 u2 − 3t2 u3 + 3tu4 − u5 ) du
1 6 3 6 3 6 1 6 t − t + t − t 3 4 5 6 t6 . 60
One can also do this computation using f (t) = t3 and g(t) = t2 . The computation is given by
( f ∗ g)(t) = =
Z t 0
Z t 0
=
Z t 0
f (u) g(t − u) du u3 (t − u)2 du
(t2 u3 − 2tu4 + u5 ) du
257
258
mathematical methods for physicists
1 6 2 6 1 6 t − t + t 4 5 6 t6 . 60
= =
The Laplace transform of the convolution is the product of the transforms F (s) and G (s).
L[ f ∗ g] = =
F (s) G (s) 2 3! 12 = 7. s3 s4 s
This agrees with a direct computation. 6 t 12 L[ f ∗ g] = L = 7. 60 s b. f (t) = t2 , g(t) = cos 2t. The convolution is given by
( f ∗ g)(t) = =
Z t
f (u) g(t − u) du
0
Z t
u2 cos 2(t − u) du
0
= =
1 1 1 − u2 sin 2(t − u) + u cos 2(t − u) + sin 2(t − u) 2 2 4 1 1 t − sin 2t. 2 4
One can also do this computation using f (t) = cos t and g(t) = t2 . The computation is given by
( f ∗ g)(t) = =
Z t 0
Z t 0
f (u) g(t − u) du
(t − u)2 cos 2u du
t 1 1 1 2 (t − u) sin 2u − (t − u) cos 2u − sin 2u = 2 2 4 0 1 1 = t − sin 2t. 2 4 The Laplace transform of the convolution is the product of the transforms F (s) and G (s).
L( f ∗ g) = =
F (s) G (s) 2 s 2 = 2 . 3 2 s s +4 ( s + 4) s2
This agrees with a direct computation. 1 1 L[ f ∗ g] = L − sin 2t + t 4 2 1 1 = − 2 + 2(s + 4) 2s2 2 = . ( s2 + 4) s2
t 0
transform techniques in physics
c. f (t) = 3t2 − 2t + 1, g(t) = e−3t . The convolution is given by
( f ∗ g)(t) = =
Z t 0
Z t 0
f (u) g(t − u) du
(3u2 − 2u + 1)e−3(t−u) du
= e−3t
0
(3u2 − 2u + 1)e3u du
1 (3u2 − 2u + 1)e3u 3 t 6 1 − (6u − 2)e3u + e3u 9 27 0 7 −3t 7 4 2 − e + − t+t . 9 9 3
= e
=
Z t
−3t
The Laplace transform of the convolution is the product of the transforms F (s) and G (s).
L( f ∗ g) = =
F (s) G (s) 2 1 1 s2 − 2s + 6 6 − + = . 3 2 s s+3 s s ( s + 3) s3
This agrees with a direct computation. 7 −3t 7 4 2 L[ f ∗ g] = L − e + − t + t 9 9 3 7 2 7 4 = − + 3 + − 9(s + 3) 9s 3s2 s
= = d. f (t) = δ t −
π 4
−7s3 + 7(s + 3)s2 − 12s(s + 3) + 18(s + 3) 9s3 (s + 3) 2 s − 2s + 6 . ( s + 3) s3 , g(t) = sin 5t.
( f ∗ g)(t) = =
Z t 0
Z t
(0
= = =
f (u) g(t − u) du δ u−
π sin 5(t − u) du 4
0, sin 5(t −
π 4 ),
π 4 π 4
6∈ [0, t] ∈ [0, t],
( π 0, t < π4 sin 5(t − ) × 4 1, t ≥ π4 , π π sin 5(t − ) H (t − ). 4 4
The Laplace transform of the convolution is the product of the transforms F (s) and G (s).
L( f ∗ g) =
F (s) G (s)
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260
mathematical methods for physicists
=
5 s2 + 25
e−πs/4 =
5e−πs/4 . s2 + 25
This agrees with a direct computation. h π π i L[ f ∗ g] = L sin 5(t − ) H (t − ) 4 4 − πs/4 5e = . s2 + 25 18. For the following problems, draw the given function and find the Laplace transform in closed form. n a. f (t) = 1 + ∑∞ n=1 (−1) H ( t − n ).
A plot of this function in shown in Figure 8.7. We compute the Laplace transform and sum the resulting geometric series. Figure 8.7: A plot of f (t) in Problem 8.18a.
∞
F (s)
= L[1 +
∑ (−1)n H (t − n)]
n =1
∞ 1 e−ns + ∑ (−1)n s n =1 s
=
∞
(−e−s )n s n =0 1 1 1 = = . s 1 + e−s s (1 + e − s )
∑
=
b. f (t) = ∑∞ n=0 [ H ( t − 2n + 1) − H ( t − 2n )]. A plot of this function in shown in Figure 8.8. We compute the Laplace transform and sum the resulting geometric series. ∞
F (s)
= L[ ∑ [ H (t − 2n + 1) − H (t − 2n)]] n =0
Figure 8.8: A plot of f (t) in Problem 8.18b.
∞
= = = = =
∑
"
e−(2n−1)s e−2ns − s s
#
n =0 es − 1 ∞
s
∑ (e−2s )n
n =0
es − 1 1 s 1 − e−2s s e −1 1 s e−2s (e2s − 1) e2s . s (1 + e s )
It is interesting to note that the solutions in the first two parts of this problem are related. The answer to part b can be written as e2s 1 = es . s (1 + e s ) s (1 + e − s )
transform techniques in physics
261
Thus, we have for t > 0, and ∞
g(t) = 1 +
∑ (−1)n H (t − n),
n =1
f (t)
=
H ( t + 1) g ( t + 1)
=
H ( t + 1) +
=
H ( t + 1) − H ( t ) + H ( t − 1) − . . .
=
∑ [ H (t − 2n + 1) − H (t − 2n)].
∞
∑ (−1)n H (t − n + 1)
n =1 ∞
n =0
c. f (t) = ∑∞ n=0 ( t − 2n )[ H ( t − 2n ) − H ( t − 2n − 1)] + (2n + 2 − t )[ H ( t − 2n − 1) − H (t − 2n − 2)]. A plot of this function in shown in Figure 8.9. For this problem we need e− as L[(t − a) H (t − a)] = 2 . s Then, we have
Figure 8.9: A plot of f (t) in Problem 8.18c.
∞
F (s)
= L[ ∑ (t − 2n)[ H (t − 2n) − H (t − 2n − 1)] n =0
+(2n + 2 − t)[ H (t − 2n − 1) − H (t − 2n − 2)]] ∞
= L[ ∑ [(t − 2n) H (t − 2n) − 2H (t − (2n + 1)) n =0
+(t − (2n + 2)) H (t − (2n + 2))]] ∞
= = =
e−2ns e−(2n+1)s e−(2n+2)s − 2 + 2 s s2 n =0 s 1 2e−s e−2s ∞ − + 2 ∑ (e−2s )n 2 s s s n =0
∑
1 − 2se−s + e−2s . (1 − e−2s )s2
19. Use the Convolution Theorem to compute the inverse transform of the following: In each case we identify the product F (s) = G (s) H (s). Then, we find g(t) and h(t) and compute the convolution integral f (t) = ( g ∗ h)(t) = 2 . s2 ( s2 + 1) We choose 2 G (s) = 2 s
Z t 0
g(u)h(t − u) du.
a. F (s) =
and
H (s) =
1 . ( s2 + 1)
Thus, g(t) = 2t,
and
h(t) = sin t.
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mathematical methods for physicists
This gives f (t)
= ( g ∗ h)(t) Z t
=
0
Z t
=
0
h(u) g(t − u) du 2(t − u) sin u du
= [−2(t − u) cos u − 2 sin u]0t = 2t − 2 sin t. e−3s . s2 We choose
b. F (s) =
e−3s s
G (s) =
H (s) =
and
1 . s
Thus, g ( t ) = H ( t − 3),
and
h(t) = 1.
This gives f (t)
= ( g ∗ h)(t) Z t
=
0
Z t
=
0
g(u)h(t − u) du H (u − 3) du.
Care is needed to compute this integral. If t < 3, then u < 3 and the integrand vanishes. If t > 3, then we compute Z t 0
H (u − 3) du =
Z t 3
H (u − 3) du =
Z t 3
du = t − 3.
So, f ( t ) = ( t − 3) H ( t − 3). 1 . s(s2 + 2s + 5) We choose 1 2 H (s) = 2 ((s + 1)2 + 4)
c. F (s) =
Thus, g(t) =
1 −t e sin 2t, 2
and
and
G (s) =
1 . s
h(t) = 1.
This gives f (t)
= ( g ∗ h)(t) =
Z t 0
= = =
g(u)h(t − u) du
Z t 1 −u e sin 2u du
0 2 1 2 1 [−( cos 2u + sin 2u)e−u ]0t 2 5 5 1 1 − (2 cos 2t + sin 2t)e−t . 5 10
transform techniques in physics
263
y
20. Find the inverse Laplace transform in two different ways: (i) Use Tables. (ii) Use the Bromwich Integral. 1 . s3 ( s + 4)2 i. Using Tables and Properties The Convolution Theorem can be used to do this problem. We choose 1 1 H (s) = 3 and G (s) = . s ( s + 4)2
c + iR CR
a. F (s) =
Thus, h(t) =
1 2 t , 2
and
g(t) = te−4t .
Z t 0
= = =
h(u) g(t − u) du
1 t 2 u (t − u)e−4(t−u) du 2 0 Z 1 −4t t 2 (tu − u3 )e4u du e 2 0 1 −4t e [(4t + 3 − 16tu − 12u 256 it +32t ∗ u2 + 24u2 − 32u3 )e4u Z
0
=
1 2 3 1 4t + 3 −4t t + − t− e . 32 256 32 256
ii. Using the Bromwich Integral. We set up the Bromwich integral and use the Residue Theorem. f (t)
= =
c+i∞ 1 F (s)est ds 2πi c−i∞ Z c+i∞ 1 est ds. 3 2πi c−i∞ s (s + 4)2
Z
The poles are at s = 0 and s = −4. The contour is shown in Figure 8.10. The point s = −4 is a double pole and the residue of st G (s) = s3 (se+4)2 is Res [ G (s); s = −4]
= =
d est s3 s→−4 ds st − 3 st 4t + 3 −4t lim e =− e 4 256 s→−4 s lim
The point s = 0 is a pole of order three and the residue of st G (s) = s3 (se+4)2 is Res [ G (s); s = 0]
=
1 d2 lim s→0 2 ds2
est ( s + 4)2
c
x
Figure 8.10: The contour used for applying the Bromwich integral to the Laplace 1 transform F (s) = 3 in Problem s ( s + 4)2 8.20a.
= ( g ∗ h)(t) =
0
c − iR
This gives f (t)
4
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mathematical methods for physicists
t2 s2 + 8t2 s + 16t2 − 4st − 16t + 6 st e s →0 ( s + 4)4 1 2 3 1 t + − t. 32 256 32
= lim = y
c + iR
CR
From the Residue Theorem, we have f (t)
= =
1
5 c
x
c − iR Figure 8.11: The contour used for applying the Bromwich integral to the Laplace 1 transform F (s) = 2 in Problem s − 4s − 5 8.20b.
c+i∞ 1 est ds 3 2πi c−i∞ s (s + 4)2 4t + 3 −4t 1 3 1 − e + t2 + − t. 256 32 256 32
Z
This agrees with the result obtained using the Laplace transform pairs. 1 . b. F (s) = 2 s − 4s − 5 i. Using Tables and Properties This one can be computed by using partial fraction decomposition. 1 1 1 1 1 = = − . (s + 1)(s − 5) 6 s−5 s+1 s2 − 4s − 5 This gives the inverse Laplace transform as f (t) =
1 5t 1 (e − e−t ) = e2t sinh 3t. 6 3
ii. Using the Bromwich Integral. We set up the Bromwich integral and use the Residue Theorem. f (t)
= =
c+i∞ 1 F (s)est ds 2πi c−i∞ Z c+i∞ 1 est ds. 2 2πi c−i∞ s − 4s − 5
Z
The roots of s2 − 4s − 5 = 0 are s = 5, −1. The poles are simple poles. The residues of G (s) = are Res [ G (s); s = −1]
= =
Res [ G (s); s = 5]
=
lim (s + 1)
s→−1
est (s + 1)(s − 5)
est e−t =− . 6 s→−1 s − 5 est lim (s − 5) (s + 1)(s − 5) s →5 lim
est e5t = . 6 s →5 s + 1
= lim
From the Residue Theorem, we have f (t)
= =
c+i∞ 1 est ds 2 2πi c−i∞ s − 4s − 5 e5t e−t 1 − = e2t sinh 3t. 6 6 3
Z
est s2 −4s−5
transform techniques in physics
265
y
This agrees with the result obtained using the Laplace transform pairs. s+3 c. F (s) = 2 . s + 8s + 17 i. Using Tables and Properties The inverse Laplace transform can be found after completing the square. We note that s2 + 8s + 17 = (s + 4)2 + 1. Then, s2
s+3 s+3 ( s + 4) − 1 = = . 2 + 8s + 17 ( s + 4) + 1 ( s + 4)2 + 1
This takes the form F (s + 4), where s−1 . s2 + 1 So, g(t) = cos t − sin t. The First Shift Theorem then gives G (s) =
f (t) = (cos t − sin t)e−4t . ii. Using the Bromwich Integral. We set up the Bromwich integral and use the Residue Theorem. Z c+i∞ 1 f (t) = F (s)est ds 2πi c−i∞ Z c+i∞ 1 (s + 3)est = ds. 2πi c−i∞ s2 + 8s + 17 The roots of s2 + 8s + 17 = 0 are s = −4 ± i. (s+3)est The poles are simple poles. The residues of G (s) = s2 +8s+17 are (s + 3)est Res [ G (s); s = −4 + i ] = lim (s + 4 − i ) 2 s + 8s + 17 s→−4+i (s + 3)est = lim s→−4+i s + 4 + i (−1 + i )e−4t e−it = . 2i (s + 3)est Res [ G (s); s = −4 − i ] = lim (s + 4 + i ) 2 s + 8s + 17 s→−4−i st ( s + 3) e = lim s→−4−i s + 4 − i (−1 − i )e−4t eit = . 2i From the Residue Theorem, we have f (t)
= = =
c+i∞ ( s + 3) est 1 ds 2πi c−i∞ s2 + 8s + 17 (−1 + i )e−4t e−it (−1 − i )e−4t eit + 2i 2i it − it it − it e +e e −e − e−4t 2 2
Z
= (cos t − sin t)e−4t .
c + iR CR
c
x
c − iR Figure 8.12: The contour used for applying the Bromwich integral to the Laplace s+3 transform F (s) = 2 in Probs + 8s + 17 lem 8.20c.
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y
This agrees with the result obtained using the Laplace transform pairs.
c + iR
CR
d. F (s) =
4
2
c x
s+1 . ( s − 2)2 ( s + 4)
i. Using Tables and Properties One could use partial fraction decomposition. Assume a decomposition of the form
= =
c − iR Figure 8.13: The contour used for applying the Bromwich integral to the Laplace s+1 transform F (s) = in ( s − 2)2 ( s + 4) Problem 8.20d.
s+1 ( s − 2)2 ( s + 4) B A C + + ( s − 2)2 s − 2 s + 4 A(s + 4) + B(s − 2)(s + 4) + C (s − 2)2 . ( s − 2)2 ( s + 4)
Then, we determine coefficients such that A(s + 4) + B(s − 2)(s + 4) + C (s − 2)2 = s + 1 is true for all s. For s = 2, 6A = 3, or A = 1/2. For s = −4, 36C = −3, or C = −1/12. Finally, picking s = 0, we have 1 = 4A − 8B + 4C = 2 − 8B −
1 5 = −8B + . 3 3
Solving for B, we have B = 1/12. This gives the partial fraction decomposition s+1 1 1 1 = + − . 12(s − 2) 12(s + 4) ( s − 2)2 ( s + 4) 2( s − 2)2 Then, the inverse Laplace transform can be found as f (t) =
1 2t 1 1 te + e2t − e−4t 2 12 12
ii. Using the Bromwich Integral. We set up the Bromwich integral and use the Residue Theorem. f (t)
= =
c+i∞ 1 F (s)est ds 2πi c−i∞ Z c+i∞ 1 (s + 1)est ds. 2πi c−i∞ (s − 2)2 (s + 4)
Z
The poles are at s = 2 and s = −4. The point s = −4 is a simple pole. The residue of G (s) =
(s+1)est ( s −2)2 ( s +4)
is
Res [ G (s); s = −4]
(s + 1)est s→−4 ( s − 2)2 −3 −4t 1 = e = − e−4t . 36 12
=
lim
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267
The point s = 2 is a double pole and the residue of G (s) =
(s+1)est ( s −2)2 ( s +4)
is
(s + 1)est Res [ G (s); s = 2] = s+4 3 + ts2 + 5st + 4t st = lim e s →2 ( t + 4)2 3 + 18t 2t 1 2t 1 2t = e = e + te . 36 12 2 d lim s→2 ds
From the Residue Theorem, we have f (t)
= =
y
c+i∞ 1 (s + 1)est ds 2πi c−i∞ (s − 2)2 (s + 4) 1 1 2t 1 2t e + te − e−4t . 12 2 12
Z
c + iR CR
i
This agrees with the result obtained using the Laplace transform pairs.
c
1
x
s2
+ 8s − 3 . (s2 + 2s + 1)(s2 + 1) i. Using Tables and Properties One could use partial fraction decomposition. Assume a decomposition of the form
e. F (s) =
( s2 = = =
i
c − iR Figure 8.14: The contour used for applying the Bromwich integral to the Laplace s2 + 8s − 3 transform F (s) = (s2 + 2s + 1)(s2 + 1) in Problem 8.20e.
s2 + 8s − 3 + 2s + 1)(s2 + 1)
s2 + 8s − 3 ( s + 1)2 ( s2 + 1) A B Cs + D + + 2 ( s + 1)2 s + 1 s +1 A(s2 + 1) + B(s + 1)(s2 + 1) + Cs(s + 1)2 + D (s + 1)2 . ( s + 1)2 ( s2 + 1)
Then, we determine the coefficients such that A(s2 + 1) + B(s + 1)(s2 + 1) + Cs(s + 1)2 + D (s + 1)2 = s2 + 8s − 3 is true for all s. For s = −1, 2A = −10, or A = −5. Then, B(s + 1)(s2 + 1) + Cs(s + 1)2 + D (s + 1)2 = s2 + 8s − 3 + 5(s2 + 1) or
( B + C )s3 + B + 2C + D )s2 + ( B + C + 2D )s + B + D = 6s2 + 8s + 2. We see that C = − B. So,
( D − B)s2 + 2Ds + B + D = 6s2 + 8s + 2.
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Therefore, D = 4 and B = −2. This gives the partial fraction decomposition s2 + 8s − 3 −5 2 2s + 4 = − . + 2 2 2 2 s+1 (s + 2s + 1)(s + 1) ( s + 1) s +1 Then, the inverse Laplace transform can be found as f (t) = 2 cos t + 4 sin t − (2 + 5t)e−t . ii. Using the Bromwich Integral. We set up the Bromwich integral and use the Residue Theorem. f (t)
= =
c+i∞ 1 F (s)est ds 2πi c−i∞ Z c+i∞ 2 (s + 8s − 3)est 1 ds. 2πi c−i∞ (s + 1)2 (s2 + 1)
Z
The poles are s = −1 and s = ±i. The points s = ±i are simple poles. The residues of G (s) =
(s2 +8s−3)est ( s +1)2 ( s2 +1)
at s = ±i are
Res [ G (s); s = i ]
= lim s →i
=
(s2 + 8s − 3)est ( s + 1)2 ( s + i )
(8i − 4)eit (i + 1)2 (2i )
= (−2i + 1)eit . (s2 + 8s − 3)est Res [ G (s); s = −i ] = lim s→−i ( s + 1)2 ( s − i ) =
(−8i − 4)e−it (1 − i )2 (−2i )
= (2i + 1)e−it . The point s = −1 is a double pole and the residue of G (s) =
= =
(s2 +8s−3)est ( s +1)2 ( s2 +1)
is
Res [ G (s); s = −1] d (s2 + 8s − 3)est lim ( s2 + 1) s→−1 ds 8s − 8s2 + 8 + ts4 − 2ts2 + 8ts3 + 8st − 3t st e ( s + 1)2 s→−1 lim
= −(2 + 5t)e−t . From the Residue Theorem, we have f (t)
=
1 2πi
Z c+i∞ 2 (s + 8s − 3)est c−i∞
( s + 1)2 ( s2 + 1)
ds
= 2 cos t + 4 sin t − (2 + 5t)e−t . This agrees with the result obtained using the Laplace transform pairs.
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269
21. Use Laplace transforms to solve the following initial value problems. Where possible, describe the solution behavior in terms of oscillation and decay. In these problems we transform the differential equation using
df L dt 2 d f L dt2
= sF (s) − f (0), = s2 F ( s ) − s f (0) − f 0 (0).
a. y00 − 5y0 + 6y = 0, y(0) = 2, y0 (0) = 0. We transform the equation and solve for Y (s) to obtain s2 Y (s) − 2s − 5(sY (s) − 2) + 6Y (s)
= 0.
2
(s − 5s + 6)Y (s) = 2s − 10. 2s − 10 Y (s) = s2 − 5s + 6 6 4 = − . s−2 s−3 The inverse Laplace transform gives the solution, y(t) = 6e2t − 4e3t . This solution is shown in Figure 8.15. The solution grows exponentially in time.
Figure 8.15: A plot of the solution of the initial value problem in Problem 8.21a.
b. y00 − y = te2t , y(0) = 0, y0 (0) = 1. We transform the equation and solve for Y (s) to obtain s2 Y ( s ) − 1 − Y ( s )
=
( s 2 − 1 )Y ( s ) = Y (s)
= =
1 . ( s − 2)2 1 1+ . ( s − 2)2 1 1 + 2 2 s − 1 (s − 1)(s − 2)2 2 2s + 7 1 4 1 + − . 9 s2 − 1 3( s − 2)2 9s−2
The inverse Laplace transform gives the solution, y(t) = or
14 4 1 sinh(t) + cosh(t) + (3t − 4)e2t , 9 9 9
5 1 y(t) = − e−t + et + (3t − 4)e2t . 9 9
This solution is shown in Figure 8.16. The solution has an exponential behavior. c. y00 + 4y = δ(t − 1), y(0) = 3, y0 (0) = 0.
Figure 8.16: A plot of the solution of the initial value problem in Problem 8.21b.
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mathematical methods for physicists
We transform the equation and solve for Y (s) to obtain
= e−s .
s2 Y (s) − 3s + 4Y (s)
(s2 + 4)Y (s) = 3s + e−s . 3s + e−s Y (s) = . s2 + 4 The first term gives
L −1
3s = 3 cos 2t. s2 + 4
We use the Second Shift Theorem on the second term,
e−s , s2 +4
L [ f (t − a) H (t − a)] = e−as F (s), with
L to obtain
L
−1
−1
1 1 = sin 2t 2 2 s +4
e−s 1 = sin 2(t − 1) H (t − 1). 2 s2 + 4
Combining these inverse Laplace transforms gives the solution, y(t) = 3 cos 2t + Figure 8.17: A plot of the solution of the initial value problem in Problem 8.21a.
1 sin 2(t − 1) H (t − 1). 2
This solution is shown in Figure 8.17. Without the delta function impulse, this is a simple harmonic motion problem possibly modeling a mass on a spring. The impulse is applied at t = 1 in such a way as to appose the motion as seen by the lower amplitude of oscillation after t = 1. d. y00 + 6y0 + 18y = 2H (π − t), y(0) = 0, y0 (0) = 0. We first need the transform of the step function whose argument appears backwards compared to the usual argument. A direct computation gives
L
−1
[ H (π − t)] = =
Z ∞ 0
Z π 0
=
H (π − t)e−st dt H (π − t)e−st dt
1 − e−πs . s
We can now transform the equation and solve for Y (s) to obtain
(s2 + 6s + 18)Y (s) = Y (s)
=
1 − e−s . s 1 − e−s . s(s2 + 6s + 18)
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271
For the first term, we need to compute the inverse Laplace transform of 1 1 = . 2 s(s + 6s + 18) s((s + 3)2 + 9) This can be done using the Convolution Theorem, f (t) = ( g ∗ h)(t) =
Z t
g(u)h(t − u) du.
0
We choose G (s) =
1 ( s + 3)2 + 9
and
H (s) =
g(t) =
1 −3t e sin 3t, 3
and
h(t) = 1.
Thus,
1 . s
This gives f (t)
= ( g ∗ h)(t) =
Z t 0
=
1 3
g(u)h(t − u) du
Z t 0
e−3u sin 3u du
t 1 −3u e (cos 3u + sin 3u) 18 0 1 1 −3t − e (cos 3t + sin 3t). 18 18
= − =
The second term is the same as the first but with an extra exponential factor. Therefore, we can apply the Second Shift Theorem. The solution to the initial value problem is then y(t)
=
=
1 1 −3t − e (cos 3t + sin 3t) 9 9 1 1 −3( t − π ) − e (cos 3(t − π ) + sin 3(t − π )) H (t − π ) − 9 9 1 1 −3t − e (cos 3t + sin 3t) 9 9 1 1 −3( t − π ) − + e (cos 3t + sin 3t) H (t − π ). 9 9
This solution is shown in Figure 8.18. The original underdamped solution is disturbed by the removal of a background force at t = π. 22. Use Laplace transforms to convert the following system of differential equations into an algebraic system, and find the solution of the differential equations. x 00
= 3x − 6y,
x (0) = 1,
x 0 (0) = 0,
y00
= x + y,
y(0) = 0,
y0 (0) = 0.
Figure 8.18: A plot of the solution of the initial value problem in Problem 8.21a.
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mathematical methods for physicists
We transform each of the equations obtaining s2 X ( s ) − s
= 3X (s) − 6Y (s).
2
s Y (s)
= X ( s ) + Y ( s ).
This gives the system of equations
(s2 − 3) X (s) + 6Y (s) = s. − X (s) + (s2 − 1)Y (s) = 0. We solve this system using Cramer’s Rule, s 6 0 s2 − 1 s ( s2 − 1) = X (s) = 2 (s − 3)(s2 − 1) + 6 6 s2 − 3 −1 s2 − 1 s2 − 3 s −1 0 s = Y (s) = 2 2 (s − 3)(s2 − 1) + 6 6 s −3 −1 s2 − 1 Therefore, we have X (s) =
s ( s2 − 1) , s4 − 4s2 + 9
Y (s) =
s . s4 − 4s2 + 9
The denominator can be factored √ √ s4 − 4s2 + 9 = s2 − (2 + 5i ) s2 − (2 − 5i ) . p √ √ The roots can be found if one assumes 2 + 5i = a + b 5. Also, the approximation is given as q √ 2 + 5i = 1.58113883... + (0.707106781...)i. Thus, a better guess might be a=
√ √5 . 2
p
2+
√
√
5i = a +
2 2 i.
Solving for a, we find
Therefore, the four roots of g(s) = s4 − 4s2 + 9 are √ ! √ ! √ √ 5 2 5 2 ± √ + i and ± √ − i . 2 2 2 2 These can be used to find the inverse Laplace transform using the Bromwich integral. √ √ Letting a = s4 − 4s2 + 9
√5 2
and b =
2 2 ,
we can write
= (s − ( a + ib))(s − ( a − ib))(s − (− a + ib))(s − (− a − ib)) = ((s − a)2 + b2 )((s + a)2 + b2 ).
transform techniques in physics
Then, we note that 1 4as 1 − = . ( s − a )2 + b2 ( s + a )2 + b2 ((s − a)2 + b2 )((s + a)2 + b2 ) This suggests the partial fraction decomposition for s 1 1 1 Y (s) = − . = 4a (s − a)2 + b2 ((s − a)2 + b2 )((s + a)2 + b2 ) ( s + a )2 + b2 Then, y(t) =
1 1 at 1 e sin bt − e− at sin bt . 4a b b
The partial fraction decomposition for X (s) is a little messier, but doable by the persistent student. Of course, at this point one can use the second equation in the system of differential equations to solve for x (t) using the solution for y(t). The more general partial fraction decomposition is of the form
((s −
a )2
Cs + D s ( s2 − 1) As + B + . = 2 2 2 2 2 + b )((s + a) + b ) (s − a) + b ( s + a )2 + b2
Putting the terms over a common denominator and equating numerators, we find that the constants satisfy the equations (from coefficients of highest to lowest powers of s) 1
=
A+C
0
= ( A + C )b2 + 2a( B − D ) + ( A + C ) a2
−1 = 2a( A − C ) + B + D 0
= ( B + D ) a2 + ( B + D ) b2 .
The last equation gives B + D = 0 and this leaves 1 1 − 2a
=
A+C
=
A−C
1 1 Adding and subtracting, A = 21 (1 − 2a ), C = 12 (1 + 2a ). The second equation and the result from the fourth become
−
b2 + a2 2a 0 2
2
=
B−D
=
B+D 2
2
+a , D = b +a . Adding and subtracting, B = − b 4a 4a Thus, we have found that 1 1 + a2 + b2 + 2as 1 + a2 + b2 − 2as X (s) = − 4a ( s + a )2 + b2 ( s − a )2 + b2
and x (t) =
1 + a2 + b2 − at 1 (e sin bt − e at sin bt) + (e−at cos bt + e at cos bt). 4ab 2
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mathematical methods for physicists
Inserting a = x (t)
=
= y(t)
= =
√ √5 2
√
and b =
2 2
into these solutions, we have
1 √ 1 √10t 1√ 1√ 1 √ − 1 √10t sin( sin( 5e 2 2t) − 5e 2 2t) 10 2 10 2 √ √ 1 1 1 1 1√ 1√ 2t) + e− 2 10t cos( 2t) + e 2 10t cos( 2 2 2 √2 1√ 1√ 1√ 1√ 5 10t) sin( 2t) + cosh( 10t) cos( 2t), sinh( 5 2 2 2 2 √ √ 1 √ 1 10t 1√ 1 √ − 1 10t 1√ 5e 2 sin( 2t) − 5e 2 sin( 2t) 10 2 10 2 √ 5 1√ 1√ sinh( 10t) sin( 2t). 5 2 2
23. Use Laplace transforms to convert the following nonhomogeneous systems of differential equations into an algebraic system, and find the solutions of the differential equations. These problems are relatively easy to set up, but the finding the inverse Laplace transform can be messy and a CAS such as Maple might be the easiest route to solution. The inversion can be done using partial fraction decomposition or using the Browich integral. Answers can be confirmed using methods from Chapters 2 and 3. a. x0
= 2x + 3y + 2 sin 2t,
y0
= −3x + 2y,
x (0) = 1,
y(0) = 0.
The Laplace transform of the system yields sX (s) − 1
= 2X (s) + 3Y (s) +
sY (s)
= −3X (s) + 2Y (s).
4 , s2 + 4
This can be rewritten as
(s − 2) X (s) − 3Y (s) = 3X (s) + (s − 2)Y (s)
=
s2 + 8 , s2 + 4 0.
One can use Cramer’s Rule to obtain the solutions of the algebraic equations, 2 s +8 −3 s2 +4 0 s−2 (s − 2)(s2 + 8) = X (s) = . ((s − 2)2 + 9)(s2 + 4) s − 2 −3 3 s−2 s − 2 s2 +8 2 +4 s 3 0 3( s2 + 8) = Y (s) = . ((s − 2)2 + 9)(s2 + 4) s − 2 −3 3 s−2
transform techniques in physics
One can invert these expressions after using a partial fraction decomposition. One can seek a decomposition of the form f (s) A(s − 2) + B Cs + D = + 2 , ((s − 2)2 + 9)(s2 + 4) ( s − 2)2 + 9 s +4 where f (s) = (s − 2)(8 + s2 ) or f (s) = 3(8 + s2 ). Thus, we need f (s) = ( A(s − 2) + B)(s2 + 4) + (Cs + D )((s − 2)2 + 9). The partial fraction decomposition is X (s)
=
Y (s)
=
1 141(s − 2) + 144 1 4s − 136 + , 145 s2 + 4 145 (s − 2)2 + 9 1 48s + 108 1 48(s − 2) − 423 − + . 145 s2 + 4 145 (s − 2)2 + 9
An inverse Laplace transform of these expressions gives the solution x (t)
=
y(t)
=
4 68 3 2t cos 2t − sin 2t + e (47 cos 3t + 16 sin 3t), 145 145 145 54 3 2t 48 cos 2t − sin 2t + e (16 cos 3t − 47 sin 3t). − 145 145 145
As a note, one can also employ the Bromwich integral. The integrals would take the form f (t) =
1 2πi
Z c+i∞ c−i∞
f (s)(8 + s2 ) ds. ((s − 2)2 + 9)(s2 + 4)
for f (s) = s − 2 or f (s) = 3. The integrand has simple poles at s = ±2i, 2 ± 3i. The residues are founds as Res[z = 2i ]
=
Res[z = −2i ]
=
Res[z = 2 + 3i ]
=
Res[z = 2 − 3i ]
=
1 (−8 + 9i )e2it f (2i ), 145 1 (−8 − 9i )e−2it f (−2i ), 145 1 47 (8 + i )e2t+3it f (2 + 3i ), 145 2 1 47 2t−3it (8 − i ) e f (2 − 3i ). 145 2
The solutions are found by adding the residues. b. x0 y
0
= −4x − y + e−t , = x − 2y + 2e
−3t
,
x (0) = 2, y(0) = −1.
The Laplace transform of the system yields 1 , s+1 2 X (s) − 2Y (s) + . s+3
sX (s) − 2
= −4X (s) − Y (s) +
sY (s) + 1
=
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mathematical methods for physicists
This can be rewritten as
( s + 4) X ( s ) + Y ( s ) = − X ( s ) + ( s + 2 )Y ( s ) =
2s + 3 , s+1 s−1 . s+3
One can use Cramer’s Rule to obtain the solutions of the algebraic equations, 2s+3 1 s +1 s −1 s +3 s + 2 2s3 + 14s2 + 29s + 19 = X (s) = . (s + 1)(s + 3)3 1 s+4 −1 s + 2 s + 4 2s+3 s +1 s−1 −1 (s − 1)(s2 + 5 ∗ s + 5) s +3 =− . Y (s) = (s + 1)(s + 3)3 1 s+4 −1 s + 2 The partial fraction decomposition is 2 1 1 7 − + + , 3 2 4( s + 1) 4( s + 3) ( s + 3) 2( s + 3) 2 5 1 5 + + − . ( s + 3)3 2 ∗ ( s + 3)2 4( s + 1) 4( s + 3)
X (s)
= −
Y (s)
=
An inverse Laplace transform of these expressions gives the solution x (t)
=
y(t)
=
1 −t e − 4 1 −t e + 4
1 −3t 2 e (4t + 2t − 7), 4 1 −3t 2 e (4t + 10t − 5). 4
c. x0
= x − y + 2 cos t,
x (0) = 3,
0
= x + y − 3 sin t,
y(0) = 2.
y
The Laplace transform of the system yields 2s , s2 + 1 3 X (s) + Y (s) − 2 . s +1
sX (s) − 3
= X (s) − Y (s) +
sY (s) − 2
=
This can be rewritten as
( s − 1) X ( s ) + Y ( s ) = − X ( s ) + ( s − 1 )Y ( s ) =
3s2 + 2s + 3 , s2 + 1 2s2 − 1 . s2 + 1
transform techniques in physics
One can use Cramer’s Rule to obtain the solutions of the algebraic equations, 2 3s +2s+3 1 s2 +1 2s2 −1 s−1 3s3 − 3s2 + s − 2 2 +1 s = X (s) = . 2 s−1 (s + 1)((s − 1)2 + 1) 1 −1 s − 1 s − 1 3s2 +2s+3 2 +1 s 2s2 −1 −1 2s3 + s2 + s + 4 2 +1 s =− Y (s) = . 2 s−1 (s + 1)((s − 1)2 + 1) 1 −1 s − 1 The partial fraction decomposition is X (s)
=
Y (s)
=
1 3( s − 1) − 1 + , + 1 ( s − 1)2 + 1 ( s − 1) + 3 s+1 + . 2 s + 1 ( s − 1)2 + 1 s2
An inverse Laplace transform of these expressions gives the solution x (t)
= 3et cos t − sin t(et − 1),
y(t)
= cos t(1 + et ) + sin t(1 + 3et ).
24. Consider the series circuit in Problem 2.20 (and 3.17) and in Figure 2.7 with L = 1.00 H, R = 1.00 × 102 Ω, C = 1.00 × 10−4 F, and V0 = 1.00 × 103 V. a. Write the secondorder differential equation for this circuit. In this problem we need to solve the equation Lq¨ + Rq˙ +
1 q = V ( t ). C
The given values lead to the equation q¨ + 100q˙ + 10000q = 1000. b. Suppose that no charge is present and no current is flowing at time t = 0 when V0 is applied. Use Laplace transforms to find the current and the charge on the capacitor as functions of time. The given values lead to the initial value problem q¨ + 100q˙ + 10000q = 1000,
q(0) = q˙ (0) = 0.
Let Q(s) = L[q(t)]. Then, the Laplace transform of the initial value problem yields
(s2 + 100s + 10000) Q(s) =
1000 . s
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mathematical methods for physicists
Solving this equation, we have Q(s)
= = =
1000 s(s2 + 100s + 10000) 1 1 s + 100 − 10s 10 s2 + 100 ∗ s + 10000 1 1 s + 100 − . 10s 10 (s + 50)2 + 7500
The inverse Laplace transform gives the solution for the charge. q(t) =
√ √ √ 1 1 − e−50t (3 cos(50 3t) + 3 sin(50 3t)). 10 30
Differentiating, we obtain the current √ √ 20 3 −50t e sin 50 3t. I (t) = 3
Figure 8.19: Plots of the charge and the current as functions of time for Problem 8.24b.
c. Replace the battery with the alternating source V (t) = V0 sin 2π f t with V0 = 1.00 × 103 V and f = 150 Hz. Again, suppose that no charge is present and no current is flowing at time t = 0 when the AC source is applied. Use Laplace transforms to find the current and the charge on the capacitor as functions of time. The given values lead to the initial value problem q¨ + 100q˙ + 10000q = 1000 sin 300πt,
q(0) = q˙ (0) = 0.
Then, the Laplace transform of the initial value problem yields
(s2 + 100s + 10000) Q(s) =
300000π . s2 + 00000π 2
Solving this equation, we have Q(s)
300000π (s2 + 00000π 2 )(s2 + 100s + 10000) s + 900π 2 s − 100 + 900π 2 = 3πA 2 − 2 s + 100s + 10000 s + 90000π 2 (s + 50) + (900π 2 − 50) s − 100 + 900π 2 = 3πA , − (s + 50)2 + 7500 s2 + 90000π 2
=
where A=
810π 4
1 . − 90π 2 + 10
The inverse Laplace transform gives the solution for the charge and current. h q(t) = A −3π cos(300πt) − sin(300πt)(9π 2 − 1) i √ √ √ +π 3 cos(50 3t) + 3 sin(50 3t)(18π 2 − 1) e−50t , h I (t) = 10π A (3 − 27π 2 ) cos(300πt) + 9π sin(300πt)− i √ √ √ (3 − 27π 2 ) cos(50 3t) + 3(9π 2 + 1) sin(50 3t) e−50t .
transform techniques in physics
279
d. Plot your solutions, and describe how the system behaves over time. The solutions for parts b and c are shown in Figures 8.19 and 8.20, respectively. In the DC case we see that as t → ∞, q(t) → 0.1 C and i (t) → 0 A. For the AC case the charge and current eventually settle into an oscillation with frequency. 25. Use Laplace transforms to sum the following series as a closed expression or as a simple integral to be evaluated. For these problems we can use Laplace transforms to sum ∑∞ n=0 F ( n ) by using Z ∞
F (s) =
0
f (t)e−st dt.
In many cases, we can use the Feynman trick. In a series of the form 1 ∑∞ n=0 a(n)b(n) we can insert 1 = ab
Z 1 0
du
Z ∞ 0
te−t[ a(1−u)+bu] dt
and attempt to sum the series before integrating. ∞
a.
(−1)n . 1 + 2n n =0 Since, L[1] = 1/s, we set s = 1 + 2n and obtain
∑
∞ Z ∞
∞
(−1)n ∑ 1 + 2n n =0
∑
=
n =0 0
=
Z ∞
"
0
=
(−1)n e−(1+2n)t dt
∞
∑ (−e
# −2t n
e−t dt
)
n =0 e−t
Z ∞
dt 1 + e−2t du , u = e−t , 0 1 + u2 h i1 π tan−1 u = . 4 0 0
= =
Z 1
∞
b.
1 . n ( n + 3) n =1 Here we let a = n and b = n + 3. This gives
∑
a(1 − u) + bu = n(1 − u) + (n + 3)u = n + 3u. Then, we evaluate the sum as ∞
1 ∑ n ( n + 3) n =1
=
∞ Z 1
∑
du
Z 1
Z ∞
n =1 0
=
0
=
du
0
Z ∞ Z 1 0
Z ∞
0
0
te−t[ a(1−u)+bu] dt "
te
−3ut
∞
∑e
# −nt
dt
n =1
e
−3ut
du
te−t dt 1 − e−t
Figure 8.20: Plots of the charge and the current as functions of time for Problem 8.24c.
280
mathematical methods for physicists
1 ∞ e−t (1 − e−3t ) dt 3 0 1 − e−t Z 1 1 (1 − z3 ) dz, z = e−t , 3 0 1−z Z 1 1 (1 + z + z2 ) dz 3 0 1 1 11 1 1+ + = . 3 2 3 18 Z
= = = = ∞
c.
(−1)n . n ( n + 3) n =1 Here we let a = n and b = n + 3. This gives
∑
a(1 − u) + bu = n(1 − u) + (n + 3)u = n + 3u. Then, we evaluate the sum as ∞
∞
(−1)n n ( n + 3) n =1
∑
∑ (−1)n
=
Z 1 0
n =1
Z 1
=
0
du
Z ∞ 0
0
0
" te
−3ut
te−t[ a(1−u)+bu] dt
∞
∑ (−1)
# n −nt
e
dt
n =1
Z ∞ Z 1
= −
Z ∞
du
0
e−3ut du
te−t dt 1 + e−t
Z 1 ∞ e−t (1 − e−3t )
= −
dt 3 0 1 + e−t Z 1 1 (1 − z3 ) − dz, z = e−t , 3 0 1+z Z 2 1 1 2 − z + z − 1 dz − 3 0 1+z 1 1 3 1 2 − 2 ln(1 + z) − z + z − z dz 3 3 2 5 2 − ln 2. 18 3
= = = = ∞
d.
(−1)n . 2 2 n =0 n − a Here we replace a by n − a and b by n + a in the expression
∑
1 = ab
Z 1 0
du
Z ∞ 0
te−t[ a(1−u)+bu] dt.
The exponent becomes
−t[ a(1 − u) + bu] = −t[(n − a)(1 − u) + (n + a)u] = −nt + ( a − 2au)t. Then, the sum has the integral representation ∞
(−1)n ∑ n2 − a2 n =0
∞
=
∑ (−1)n
n =0
Z 1 0
du
Z ∞ 0
te−t[ a(1−u)+bu] dt
transform techniques in physics
=
Z 1 0
=
du
Z ∞ 0
Z ∞ Z 1 0
1 2a
=
0
e
te
−2aut
du
Z ∞ − at e − e at 0
∞
" ( a−2au)t
∑ (−1)
n =0 te at
1 + e−t
# n −nt
e
dt
dt
dt.
1 + e−t
This is typically far as a student can go using what they know at this level. Letting z = e−t , some might even be write the result as Z ∞ (−1)n 1 1 ( z a −1 − z − a −1 ) dz. ∑ 2 2 = 2a 0 1+z n =0 n − a Turning to a computer algebra system, like Maple, a student would find that the sum of this series given in terms of the digamma function, Z 1 x −1 t −1
d ψ( x ) = ln(Γ( x )) = −γ − dx
1−t
0
dt,
where γ is the EulerMascheroni constant. Maple gives ∞ (−1)n 1 1+a a 1−a a = ( ψ ( ) − ψ ( ) − ψ ( ) − ψ (− ) ∑ 2 2 4a 2 2 2 2 n =0 n − a πa 2
1 πa + 2 cos sin πa cos πa 4a2 2 sin 2 π 1 − csc πa + . 2a πa
= − =
πa 2
(−1)n ∑ n2 − a2 n =0
= = = = = =
1 2a
Z 1 a −1 (z − z − a −1 ) 0
1+z
.
dz
1 ( β( a) − β(− a)) 2a 1 1+a a 1−a a (ψ( ) − ψ( ) − ψ( ) − ψ(− ) 4a 2 2 2 2 1 1+a 1−a a a ψ( ) − ψ( ) + ψ(− ) − ψ( ) 4a 2 2 2 2 1 πa πa 2 −π cot − π cot − 4a 2 2 a π 1 − csc πa + . 2a πa
∞
e.
Integrals of the forms Z 1 a u − ub 0
1−u
du = ψ(b + 1) − ψ( a + 1)
and Z 1 a u − ub 0
1+u
du = β(b + 1) − β( a + 1)
are evaluated in terms of the digamma, ψ( x ), and incomplete beta, β( x ), functions. Useful relations, found in Gradshteyn and Rhyzik’s tables, are ψ ( z + 1)
≡
Z 1 1 − tz
1−t 1 ψ(z) + , z 0
= ψ(−z)
=
β(z)
≡
Z 1 z −1 x
1 . ( 2n + 1 )2 − a2 n =0 This is another example in which the use of digamma and beta functions are useful. [See the previous problem.] Noting that
∑
(2n + 1)2 − a2 = (2n + 1 − a)(2n + 1 + a),
1 , z
dx 1+x z 1 z+1 ) − ψ( ) . ψ( 2 2 2 0
=
dt − γ
ψ(z) + π cot πz +
Using known identities for the digamma and incomplete beta function, this form can be derived. ∞
281
282
mathematical methods for physicists
we can apply the Feynman trick to obtain an integral representation for the series. ∞
1 ∑ (2n + 1)2 − a2 n =0
= =
∞ Z 1
∑
du
∑
du
Z 1
Z ∞
n =1 0 ∞ Z 1 n =0 0
=
0
=
Z ∞ 0
=
= =
te
0
0
Z ∞ 0
te−t[ a(1−u)+bu] dt te−t[(2n+1− a)(1−u)+(2n+1+ a)u] dt "
te
( a−2au−1)t
∞
∑e
# −2nt
dt
n =0
( a −1) t
1
Z 0
e
(−2au)t
du
1 dt 1 − e−2t
Z ∞ 1 − e−2at te(a−1)t 0
=
du
Z ∞
1 2a 1 2a 1 2a
dt 2at 1 − e−2t Z ∞ (1 − e−2at )e(a−1)t dt 1 − e−2t 0 Z ∞ at e − e− at −t e dt 1 − e−2t 0 Z 1 a u − u− a du, u = e−t . 1 − u2 0
Once again, a student might stop at this point. However, a partial fraction expansion yields ∞
1 ∑ (2n + 1)2 − a2 n =0
= = =
1 2a
Z 1 a u − u− a
1 4a 1 4a 1 4a 1 4a π 4a
Z 1 a u − u− a
0
0
1 − u2 1−u
du du −
1 4a
Z 1 a u − u− a 0
1+u
du
[ψ(1 − a) − ψ(1 + a) − β(1 − a) + β(1 + a)] .
[ψ( a) − ψ(1 + a) + π cot πa − β(1 − a) + β(1 + a)] 1 = − + π cot πa − β(1 − a) + β(1 + a) a 1 = cot πa − sin πa πa π = − tan . 4a 2 =
∞
The sum of this series is ∞
f.
1 − an e . n n =1 Note that
1 π = − tan(πa/2). 2 2 4a n=0 (2n + 1) − a
∑
∑
e− an = n
Z ∞ 0
H (t − a)e−st dt.
Using this Laplace transform of the Heaviside function, we have ∞
1 ∑ n e−an n =1
=
∞ Z ∞
∑
n =1 0
H (t − a)e−nt dt
transform techniques in physics
Z ∞
=
0
∞
H (t − a)
∑ (e−t )n dt
n =1
Z ∞
e−t dt −t a 1−e ∞ ln(1 − e−t ) = − ln(1 − e− a ) = ln(
= =
a
ea ). ea − 1
26. Use Laplace transforms to prove ∞
1 1 = ( n + a )( n + b ) b − a n =1
∑
Z 1 a u − ub
1−u
0
du.
Use this result to evaluate the following sums: The proof is follows from using partial fractions and Laplace transforms: 1 1 1 1 = − (n + a)(n + b) b−a n+a n+b Z ∞h i 1 e−(n+ a)t − e−(n+b)t dt. = b−a 0 Then, the sum becomes ∞
1 (n + a)(n + b) n =1
∑
∞
= = = =
1 b−a n =1
∑
1 b−a
Z ∞ 0
Z ∞h 0
i e−(n+ a)t − e−(n+b)t dt "
(e
− at
−e
−bt
)
∞
∑ (e
# −t n
)
dt
n =1 e−t
∞ 1 (e−at − e−bt ) dt b−a 0 1 − e−t Z 1 a 1 u − ub du, u = e−t . b−a 0 1−u
Z
∞
a.
1 . n ( n + 1) n =1
∑
For this problem a = 0 and b = 1. This gives ∞
1 = n ( n + 1) n =1
∑
Z 1 1−u 0
1−u
du = 1.
∞
b.
1 . ( n + 2 )( n + 3) n =1 For this problem a = 2 and b = 3. This gives
∑
∞
1 = (n + 2)(n + 3) n =1
∑
Z 1 2 u − u3 0
1−u
du =
Z 1 0
u2 du =
1 . 3
27. Do the following: a. Find the first four nonvanishing terms of the Maclaurin series exx pansion of f ( x ) = x . e −1 The computation of the first few coefficients of the Taylor series expansion of this function is shown the the table below.
283
284
mathematical methods for physicists
n
f (n) ( x )
f ( n ) (0)
cn
0
x e x −1
1
1
1
(1− x ) e x −1 ( e x −1)2
− 12
− 12
2
( x −2)e2x +( x +2)e x ( e x −1)3
1 6
1 12
3
(3− x )e3x −4xe2x −( x +3)e x ( e x −1)4
0
0
4
( x −4)e4x +(11x −12)e3x +(11x +12)e2x +( x +4)e x ( e x −1)5
1 − 30
1 − 720
This give the series expansion x 1 1 1 4 = 1 − x + x2 − x +.... ex − 1 2 12 720 b. Use the result in part a. to determine the first four nonvanishing Bernoulli numbers, Bn . Since
∞ x Bn n = x , ∑ x e −1 n! n =0
we see that 1 B1 = − , 2
B0 = 1,
B2 =
1 , 6
B4 = −
1 . 30
c. Use these results to compute ζ (2n) for n = 1, 2, 3, 4. We need to compute ζ (2n) =
22n−1 π 2n Bn , (2n)!
n = 1, 2, 3, 4.
From the previous results on the Bernoulli numbers, the values of the Riemann zeta function are ζ (2)
=
ζ (4)
=
ζ (6)
=
ζ (8)
=
21 π 2 B 2! 1 23 π 4 B2 4! 25 π 6 B3 6! 27 π 8 B 8! 4
π2 . 6 π4 = . 90 π6 = . 945 π8 = . 9450
=
28. Given the following Laplace transforms, F (s), find the function f (t). Note that in each case there are an infinite number of poles, resulting in an infinite series representation.
transform techniques in physics
a. F (s) =
1 s2 (1 + e − s )
285
.
We set up the Bromwich integral and use the Residue Theorem. f (t)
Z c+i∞ 1
=
5πi
F (s)est ds
c + iR
CR
2πi c−i∞ Z c+i∞ 1 est ds. 2πi c−i∞ s2 (1 + e−s )
=
y
3πi
The poles occur for s = 0 and 1 + e−s = 0. Since 1 + e(2k+1)πi = 0 for k an integer, there are an infinite number of poles at s = (2k + 1)πi. The contour and poles are shown in Figure 8.21.
πi c
x
πi
st
The point s = 0 is a double pole and the residue of G (s) = s2 (1e+e−s ) is d est Res [ G (s); s = 0] = lim 1 + e−s s→0 ds 1 1 t + te−s + e−s st e = t+ . = lim − s 2 2 4 s →0 (1 + e ) The other poles are simple poles. The residues of G (s) = are Res [ G (s); s = (2k + 1)πi ]
= =
lim
(s − (2k + 1)πi )
lim
−
s→(2k +1)πi
s→(2k +1)πi
= −
est s2 (1+ e − s )
est s2 (1 + e − s )
est s2 e − s
e(2k+1)πit . (2k + 1)2 π 2
From the Residue Theorem, we have f (t)
Z c+i∞
est
=
1 2πi
=
1 e 1 t+ − ∑ 2 4 k=−∞ (2k + 1)2 π 2
=
∞ 1 1 cos(2k − 1)πt t+ −2 ∑ . 2 4 (2k − 1)2 π 2 k =1
ds
c−i∞ s2 (1 + e−s ) ∞ (2k+1)πit
1 . s sinh s We set up the Bromwich integral and use the Residue Theorem.
b. F (s) =
f (t)
= =
c+i∞ 1 F (s)est ds 2πi c−i∞ Z c+i∞ 1 est ds. 2πi c−i∞ s sinh s
Z
The poles occur for s = 0 and sinh s = 0. Since sinh( x + iy) = sinh x cos y + i cosh x sin y,
3πi
5πi
c − iR
Figure 8.21: The contour used for applying the Bromwich integral to the Laplace 1 transform F (s) = 2 in Probs (1 + e − s ) lem 8.28a.
286
mathematical methods for physicists
then sinh nπi = 0 for n an integer and therefore there are an infinite number of poles at s = nπi. The contour and poles are shown in Figure 8.22. st
y 5πi
c + iR
4πi CR
3πi 2πi πi c
x
πi 2πi
e The point s = 0 is a double pole and the residue of G (s) = s sinh s is st se d Res [ G (s); s = 0] = lim sinh s s→0 ds sinh s + st sinh s − s cosh s st = lim e s →0 sinh2 s t sinh s + st cosh s − s sinh s st = lim e 2 sinh s cosh s s →0 2t cosh s + st sinh s − sinh s − s cosh s st = lim e s →0 2 cosh2 s + 2 sinh2 s = t.
The other poles are simple poles. The residues of G (s) =
3πi 4πi 5πi
Res [ G (s); s = nπi ]
c − iR
=
Figure 8.22: The contour used for applying the Bromwich integral to the Laplace 1 in Problem transform F (s) = s sinh s 8.28b.
= = =
lim (s − nπi )
s→nπi
lim
s→nπi
est s sinh s
are
est s sinh s
est s cosh s
enπit nπi cosh nπi enπit (−1)n . nπi
From the Residue Theorem, we have y 5π 2 i
CR
f (t)
c + iR
= =
3π 2 i
est ds c−i∞ s sinh s ∞ enπit t+ ∑ (−1)n nπi n=−∞,n6=0 1 2πi
Z c+i∞
∞
= t+2
π 2i
c
x
 π2 i
c. F (s) =
sinh s s2 cosh s
.
We set up the Bromwich integral and use the Residue Theorem.
 3π 2 i  5π 2 i
(−1)n sin nπt. nπ n =1
∑
c − iR
Figure 8.23: The contour used for applying the Bromwich integral to the Laplace sinh s transform F (s) = 2 in Problem s cosh s 8.28c.
f (t)
= =
c+i∞ 1 F (s)est ds 2πi c−i∞ Z c+i∞ 1 sinh s st e ds. 2πi c−i∞ s2 cosh s
Z
The poles occur for s = 0 and cosh s = 0. We note that cosh( x + iy) = cosh x cos y + i sinh x sin y = 0, when x = 0 and cos y = 0. So, there are poles at s = 2k2−1 π for k an integer. The contour and poles are shown in Figure 8.23.
transform techniques in physics
All poles are simple poles. The residues of G (s) = Res [ G (s); s = 0]
Res G (s); s =
2k − 1 π 2
are
sinh sest s→0 s cosh s cosh sest = 1. lim s→0 cosh s 2k − 1 sinh sest lim (s − π) 2 2 s cosh s s→ 2k2−1 π
= lim =
sinh sest s2 cosh s
= =
lim
s→ 2k2−1 π
sinh sest 4e(2k−1)πit/2 . = s2 sinh s (2k − 1)2 π 2 y
From the Residue Theorem, we have f (t)
=
1 2πi
Z c+i∞ c−i∞ ∞
c + iR
sinh s st e ds 2 s cosh s
= 1+
4e(2k−1)πit/2 (2k − 1)2 π 2 k =−∞
= 1−
8 π2
CR
∑
∞
cos((2k − 1)πt/2) . (2k − 1)2 k =1
∑
√ sinh( β sx ) √ d. F (s) = . s sinh( β sL) We set up the Bromwich integral and use the Residue Theorem. c+i∞ 1 F (s)est ds 2πi c−i∞ √ Z c+i∞ sinh( β sx ) st 1 √ = e ds. 2πi c−i∞ s sinh( β sL) √ The poles occur for s = 0 and sinh( β sL) = 0. The latter gives √ β sL = nπi and there are poles at s = −n2 π2/β2 L2 for n an integer. The contour and poles are shown in Figure 8.24.
f (t)
=
287
2
 β4π 2 L2
2
 βπ2 L2
c
x
Z
√ sinh( β sx )
All poles are simple poles. The residues of G (s) = s sinh( β√sL) est are √ sinh( β sx ) st √ Res [ G (s); s = 0] = lim e s→0 sinh( β sL ) √ √ 2βx s cosh( β sx ) st √ √ = lim e s→0 2βL s cosh( β sL ) x = . L √ n2 π 2 n2 π 2 sinh( β sx ) st √ Res G (s); s = − 2 2 = lim (s + 2 2 ) e 2 2 β L β L s sinh( β sL) s→− n π β2 L2
=
lim
2 2
s→− n 2π2
√ √ 2 z sinh( β sx ) st √ e sβL cosh( β sL)
β L
2nπi sinh( nπix L ) −n2 π 2 t/βL e 2 2 n π cosh(nπi ) 2(−1)n sin nπx 2 2 L e−n π t/βL . nπ
= − =
c − iR Figure 8.24: The contour used for applying the Bromwich integral√to the Laplace sinh( β sx ) √ in Probtransform F (s) = s sinh( β sL) lem 8.28d.
288
mathematical methods for physicists
From the Residue Theorem, we have f (t)
√ Z c+i∞ sinh( β sx ) st √ e ds
=
1 2πi
=
x 2 + L π
c−i∞ ∞
s sinh( β sL)
(−1)n nπx −n2 π2 t/βL sin e . n L n =1
∑
Note that f (t) looks like a solution of the heat equation on x ∈ [0, L]. Setting t = 0, the Fourier sine series that results is familiar from Chapter 4. x 2 ∼− L π
∞
(−1)n nπx sin n L n =1
∑
29. Consider the initial boundary value problem for the heat equation: ut = 2u xx , 0 < t, 0 ≤ x ≤ 1, u( x, 0) = x (1 − x ), 0 < x < 1, u(0, t) = 0, t > 0, u(1, t) = 0, t > 0. Use the finite transform method to solve this problem. Namely, assume that the solution takes the form u( x, t) = ∑∞ n=1 bn ( t ) sin nπx and obtain an ordinary differential equation for bn and solve for the bn ’s for each n. Inserting u( x, t) = ∑∞ n=1 bn ( t ) sin nπx into the heat equation, we have ∞
∞
n =1
n =1
∑ b˙ n (t) sin nπx = 2 ∑ (−n2 π2 )bn (t) sin nπx.
Due to linear independence of the set of functions sin nπx, the coefficients can be equated, b˙ n (t) = −2n2 π 2 bn (t),
n = 1, 2, . . . .
This is readily solved to find bn (t) = bn (0)e−2n
2 π2 t
n = 1, 2, . . . .
,
So far, the solution of the heat equation is given by ∞
u( x, t) =
∑ bn (0)e−2n π t sin nπx. 2
2
n =1
We note that the eigenfunction expansion was chosen to satisfy the boundary conditions. It remains to find the bn (0)’s such that the initial condition is satisfied, ∞
u( x, 0) =
∑ bn (0) sin nπx = x(1 − x).
n =1
transform techniques in physics
This is a Fourier sine series and the coefficients are given by bn ( 0 )
Z 1
= 2 x (1 − x ) sin nπx dx 0 1 (1 − 2x ) 2 x (1 − x ) cos nπx + sin nπx − cos nπx = − nπ n2 π 2 n3 π 3 0 2(1 − cos nπ ) = , n = 1, 2, . . . . n3 π 3
Thus, the solution is given by ∞
u( x, t)
=
2(1 − cos nπ ) −2n2 π2 t e sin nπx n3 π 3 n =1
=
2 2 4 e−2(2k−1) π t sin((2k − 1)πx ) 3 3 (2k − 1) π k =1
∑ ∞
∑
289
9 Vector Analysis and EM Waves 1. Compute u × v using the permutation symbol. Verify your answer by computing these products using traditional methods. a. u = 2i − 3k, v = 3i − 2j. Using the permutation symbol to compute this cross product, we have u×v
= e123 u1 v2 k + e231 u2 v3 i + e312 u3 v1 j +e213 u2 v1 k + e132 u1 v3 j + e321 u3 v2 i = 2(−2)k + 0(0)i − 3(3)j −(0(3)k + 2(0)j + −3(−2)i) = −6i − 9j − 4k.
We can also compute the cross product using determinants. i j k u × v = 2 0 −3 3 −2 0 2 0 2 −3 0 −3 = k j+ i− 3 −2 3 0 −2 0
= −6i − 9j − 4k. b. u = i + j + k, v = i − k. Using the permutation symbol to compute this cross product, we have u×v
= e123 u1 v2 k + e231 u2 v3 i + e312 u3 v1 j +e213 u2 v1 k + e132 u1 v3 j + e321 u3 v2 i = 1(0)k + 1(−1)i + 1(1)j −(1(1)k + 1(−1)j + 1(0))i = −i + 2j − k.
We can also compute the cross product using determinants. i j k u×v = 1 1 1 1 0 −1
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=
1 0
1 −1
1 i− 1
1 −1
1 j+ 1
1 0
k
= −i + 2j − k. c. u = 5i + 2j − 3k, v = i − 4j + 2k. Using the permutation symbol to compute this cross product, we have u×v
= e123 u1 v2 k + e231 u2 v3 i + e312 u3 v1 j +e213 u2 v1 k + e132 u1 v3 j + e321 u3 v2 i = 5(−4)k + 2(2)i − 3(1)j −(2(1)k + 5(2)j − 3(−4))i = 16i − 13j − 22k.
We can also compute the cross product using determinants. i j k u × v = 5 2 −3 1 −4 2 5 2 5 −3 2 −3 = k j+ i− 1 −4 1 2 −4 2
= 16i − 13j − 22k. 2. Compute the following determinants using the permutation symbol. Verify your answer. 3 2 0 a. 1 4 −2 −1 4 3 Using the permutation symbol, we have 3 2 0 1 4 −2 = e123 (3)(4)(3) + e231 (2)(−2)(−1) + e312 (0)(1)(4) −1 4 3
+e213 (2)(1)(3) + e132 (3)(−2)(4) + e321 (0)(4)(−1) = 36 + 4 + 0 − 6 − (−24) − 0 = 58. We can compute the determinant using the usual expansion in 2 × 2 determinants. 3 2 0 4 −2 1 −2 1 4 −2 = 3 −2 4 3 −1 3 −1 4 3
= 3(20) − 2(1) = 58. 1 b. 4 2
2 −6 3
2 3 1
vector analysis and em waves
Using the permutation symbol, we have 1 2 2 4 −6 3 = e123 (1)(−6)(1) + e231 (2)(3)(2) + e312 (2)(4)(3) 2 3 1
+e213 (2)(4)(1) + e132 (1)(3)(3) + e321 (2)(−6)(2) = −6 + 12 + 24 − 8 − 9 − (−24) = 37. Using an expansion in 2 × 2 determinants, we find 1 2 2 4 4 3 −6 3 +2 −2 4 −6 3 = 2 2 1 3 1 2 3 1
−6 3
= −15 − 2(−2) + 2(24) = 37. 3. For the given expressions, write out all values for i, j = 1, 2, 3. 1, i = 1, j = 3. a. ei2j = −1, i = 3, j = 1, 0 otherwise. ( −1, i = 2, b. ei13 0 otherwise. c. eij1 ei32 . Summing over i, we have eij1 ei32
= e1j1 e132 + e2j1 e232 + e3j1 e332 = e1j1 e132 = 0.
4. Show that a. δii = 3. δii = δ11 + δ22 + δ33 = 3. b. δij eijk = 0. δij eijk = 0 since δij = 1 when i = j and then eiik = 0. This can also be confirmed by writing out the double sum over i and j as δij eijk
= δ11 e11k + δ12 e12k + δ13 e13k +δ21 e21k + δ22 e22k + δ23 e23k +δ31 e31k + δ32 e32k + δ33 e33k = 0.
c. eimn e jmn = 2δij . First, we note that m 6= n. Otherwise, we would have e jmn = 0. Furthermore, i cannot be m or n for the same reason. Similarly, j cannot be m or n. The only way this can happen is if i = j. Then,
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there are only two nonvanishing terms with the same factors. For example, if i = j = 1, then e1mn e1mn = e123 e123 + e132 e132 = 2. Since this is true for i = j = 2 and i = j = 3, we have eimn e jmn = 2δij . d. eijk eijk = 6. It might be helpful to insert the suppressed summations and expanding each one in turn. Then, we obtain 3
eijk eijk
=
3
3
∑ ∑ ∑ eijk eijk
i =1 j =1 k =1 3
=
3
∑ ∑ (eij1 eij1 + eij2 eij2 + eij3 eij3 )
i =1 j =1 3
=
2 2 2 2 2 2 + ei31 + ei12 + ei32 + ei13 + ei23 ) ∑ (ei21
i =1
2 2 2 2 2 2 = 6. + e123 + e213 + e132 + e312 + e231 = e321
5. Show that the vector (a × b) × (c × d) lies on the line of intersection of the two planes: (1) the plane containing a and b and (2) the plane containing c and d. We assume that a and b are linearly independent and that c and d are linearly independent. Furthermore, we assume that the planes defined by these pairs of vectors intersect. The vector a × b is perpendicular to both a and b. Therefore, a × b is normal to the plane defined by a and b. Similarly, the vector c × d is normal to the plane containing c and d. Using the same argument, the vector (a × b) × (c × d) is perpendicular to the vectors a × b and c × d. Therefore, it is perpendicular to the normals of the planes defined by a and b and defined by c and d. So, the vector (a × b) × (c × d) lies parallel to the line of intersection of the two planes. 6. Prove the following vector identities: a. (a × b) · (c × d) = (a · c)(b · d) − (a · d)(b · c). We begin by noting that the cross product can be written 3
a×b =
∑
eijk ai b j ek = eijk ai b j ek ,
i,j,k =1
using the Einstein summation convention. The components on the left hand side give
(a × b) · (c × d) = (eijk ai b j ek ) · (emnp cm dn e p ) = eijk emnp δkp ai b j cm dn = eijk emnk ai b j cm dn .
vector analysis and em waves
Now, we use the identity eijk emnk = δim δjn − δin δjm . This gives
(a × b) · (c × d) = eijk emnk ai b j cm dn = (δim δjn − δin δjm ) ai b j cm dn = δim δjn ai b j cm dn − δin δjm ai b j cm dn = ai ci b j d j − ai di b j c j = (a · c)(b · d) − (a · d)(b · c). b. (a × b) × (c × d) = (a · b × d)c − (a · b × c)d. We begin by noting that the cross product can be written 3
a×b =
∑
eijk ai b j ek = eijk ai b j ek ,
i,j,k =1
using the Einstein summation convention. The components on the left hand side give
(a × b) × (c × d) = (eijk ai b j ek ) × (emnp cm dn e p ) = eijk emnp ai b j cm dn ek × (e p = eijk emnp ai b j cm dn ekpq eq . We next use the identity emnp ekqp = δmk δnq − δmq δnk on the second and third occurrences of the permutation symbol, this gives
(a × b) × (c × d) = −eijk emnp ekqp ai b j cm dn eq = −(δmk δnq − δmq δnk )eijk ai b j cm dn eq = δmq δnk eijk ai b j cm dn eq − δmk δnq eijk ai b j cm dn eq = eijn ai b j cm dn em − eijk ai b j ck dn en = ai (eijn b j dn )(cm em ) − ai (eijk b j ck )(dn en ) = (a · b × d)c − (a · b × c)d. If we had used the identity eijk e pqk = δip δjq − δiq δjp on the first and third occurrences of the permutation symbol, this gives another form of the identity.
(a × b) × (c × d) = eijk emnp e pqk ai b j cm dn eq = (δip δjq − δiq δjp )emnp ai b j cm dn eq = δip δjq emnp ai b j cm dn eq − δiq δjp emnp ai b j cm dn eq = emni ai b j cm dn e j − emnj ai b j cm dn ei = (emni cm dn ai )(b j e j ) − (emnj cm dn b j )( ai ei ) = (c · d × a)b − (c · d × b)a.
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7. Use Problem 6a to prove that a × b = ab sin θ. In Problem 6a set c = a and d = b. Denoting the angle between a and b as θ, we have
a × b2
= (a × b) · (a × b) = (a · a)(b · b) − (a · b)(b · a) = a2 b2 − a2 b2 cos2 θ = a2 b2 sin2 θ.
Taking the square root yields the result. 8. A particle moves on a straight line, r = tu, from the center of a disk. If the disk is rotating with angular velocity ω, then u rotates. Let u = (cos ωt)i + (sin ωt)j. a. Determine the velocity, v. We have r = tu = (t cos ωt)i + (t sin ωt)j. Then, v=
dr = (cos ωt − tω sin ωt)i + (sin ωt + tω cos ωt)j. dt
b. Determine the acceleration, a. From the previous part, we have a=
dv = (−2ω sin ωt − tω 2 cos ωt)i + (2ω cos ωt − tω 2 sin ωt)j. dt
c. Describe the resulting acceleration terms identifying the centripetal acceleration and Coriolis acceleration. We can rearrange the terms in the acceleration result as a
= (−2ω sin ωt − tω 2 cos ωt)i + (2ω cos ωt − tω 2 sin ωt)j = ω [(−2 sin ωt)i + (2 cos ωt)j] − ω 2 [(t cos ωt)i + (t sin ωt)j] = 2ω θˆ − ω 2 rˆr.
The second term is the centripetal acceleration, ac = −ω 2 rˆr and ˆ the other term is the Coriolis term, acorr = 2ω θ. 9. Compute the gradient of the following: a. f ( x, y) = x2 − y2 . The gradient is found as
∇f
= =
∂f ∂f i+ j ∂x ∂y 2xi − 2yj.
b. f ( x, y, z) = yz + xy + xz. The gradient is found as
∇f
= =
∂f ∂f ∂f i+ j+ k ∂x ∂y ∂z (y + z)i + (z + x )j + (y + x )k.
vector analysis and em waves
c. f ( x, y) = tan−1
y x .
The gradient is found as
∇f
∂f ∂f i+ j ∂x ∂y 1 y 1 1 − i+ j y 2 x 2 y 2 x 1+ x 1+ x x y i+ 2 j. − 2 2 x +y x + y2
= = =
d. f ( x, y, z) = 2y x cos z − 5 sin z cos y. The gradient is found as
∇f
=
∂f ∂f ∂f i+ j+ k ∂x ∂y ∂z
= (2y x ln x cos z)i + (2xy x−1 cos z + 5 sin z sin y)j + (−2y x sin z − 5 cos z cos y)k. 10. Find the directional derivative of the given function at the indicated point in the given direction. The directional derivative is found as Dn f = ∇ f · n, where n is a unit vector in the direction given. a. f ( x, y) = x2 − y2 , (3, 2), u = i + j. The directional derivative is Dn f ( x, y)
= ∇ f ( x, y) · n
1 = (2xi − 2yj) · √ (i + j) 2 1 Dn f (3, 2) = (6i − 4j) · √ (i + j) 2 √ 2 = √ = 2. 2 y
b. f ( x, y) = x , (2, 1), u = 3i + 4j. The directional derivative is
= ∇ f ( x, y) · n y 1 1 = (− 2 i + j) · (3i + 4j) x 5 x 1 1 3 4 Dn f (2, 1) = (− i + j) · ( i + j) 4 2 5 5 1 3 1 4 1 = (− )( ) + ( )( ) = . 4 5 2 5 4
Dn f ( x, y)
c. f ( x, y, z) = x2 + y2 + z2 , (1, 0, 2), u = 2i − j. The directional derivative is Dn f ( x, y)
= ∇ f ( x, y) · n 1 = (2xi + 2yj + 2zk) · √ (2i − j) 5
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Dn f (1, 0, 2)
2 1 = (2i + 4k) · ( √ i − √ j) 5 5 4 4 = √ − √ = 0. 5 5
11. Zaphod Beeblebrox was in trouble after the infinite improbability drive caused the Heart of Gold, the spaceship Zaphod had stolen when he was President of the Galaxy, to appear between a small insignificant planet and its hot sun. The temperature of the ship’s hull is given by T ( x, y, z) = 2 2 2 e−k( x +y +z ) Nivleks. He is currently at (1, 1, 1), in units of globs, and k = 2 globs−2 . (Check the Hitchhikers Guide for the current conversion of globs to kilometers and Nivleks to Kelvin.) a. In what direction should he proceed so as to decrease the temperature the quickest? The direction to go is parallel to the negative of the gradient, u = −∇ T. In general, we have at each point ( x, y, z) this direction is given by u
= −∇ T ∂T ∂T ∂T i+ j+ k = − ∂x ∂y ∂z = 2k( xi + yj + zk)e−k(x
2 + y2 + z2 )
(9.1) It is clear that a simpler expression of the direction is u = xi + yj + zk. At the point at (1, 1, 1), this gives the direction of fastest decrease is u = i + j + k = (1, 1, 1). b. If the Heart of Gold travels at e6 globs per second, then how fast will the temperature decrease in the direction of fastest decline? This requires looking at the units. We have Nivleks Nivleks globs = . s globs s This implies that we are looking for dT dr dr = Dn T = − ∇ T  . dt dt dt From the previous part, we have
√ Nivleks dT = −2ki + j + ke−6k × e6 = −2k 3 . dt s 12. A particle moves under the force field F = −∇V, where the potential function is given by V ( x, y, z) = x3 + y3 − 3xy + 5. Find the equilibrium points of F and determine if the equilibria are stable or unstable.
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The equilibrium points are where the force vanishes, or ∇V = 0. Thus, 0 = ∇V = (3x2 − 3y)i + (3y2 − 3x )j. This gives the system of equations 0
= 3x2 − 3y
0
= 3y2 − 3x.
From the first equation, y = x2 . Inserting this into the second equation, we have 0 = x4 − x = x ( x3 − 1). Therefore, y = x = 0 or x = 1, y = 1. So, the equilibrium points are (0, 0) and (1, 1). From Figure 9.1 we can see that the origin is unstable as it is a saddle. From the next chapter we learn that we can classify the stationary points by 2 = 36xy − 9 at each point. This also indicates that computing Vxx Vyy − Vxy the origin is a saddle point and (1, 1) is a local minimum and is stable. 13. For the given vector field, find the divergence and curl of the field. a. F = xi + yj. The divergence is ∂F1 ∂F ∂F + 2+ 3 ∂x ∂y ∂z ∂x ∂y + ∂x ∂y 2.
∇·F = = = The curl is
i j k ∂ ∂ ∂ ∂x ∂y ∂z x y 0 ∂y ∂x − k = 0. ∂x ∂y
∇×F =
= y
b. F = r i − xr j, for r =
p
x 2 + y2 .
The divergence is
∇·F = = =
∂F1 ∂F ∂F + 2+ 3 ∂x ∂y ∂z ∂ y ∂ x − ∂x r ∂y r x 3 x −3 r + 3 r3 = 0. y y
The curl is
∇×F =
i ∂ ∂x y r
j ∂ ∂y − xr
k ∂ ∂z 0
Figure 9.1: A plot of the potential function showing that the origin is a saddle and (1, 1) is a stable point.
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∂ x ∂ y = − − k ∂x r ∂y r 1 x2 1 y2 = − +3 3 − +3 3 k r r r r 1 2 1 = 3 − k = k. r r r c. F = x2 yi + zj + xyzk. The divergence is
∇·F = = =
∂F1 ∂F ∂F + 2+ 3 ∂x ∂y ∂z ∂z ∂( xyz) ∂( x2 y) + + ∂x ∂y ∂z 2xy + xy = 3xy.
The curl is
∇×F =
=
i j k ∂ ∂ ∂ ∂x ∂y ∂z 2 x y z xyz ∂( xyz) ∂z ∂( xyz) ∂( x2 y) − i− − j ∂y ∂z ∂x ∂z ∂z ∂( x2 y) + − k ∂x ∂y
= xzi − yzj − x2 k. 14. Write the following using eijk notation and simplify if possible. a. C × (A × (A × C)). We first consider A × (A × C)
= A × (eijk Ai Cj eˆ k ) =
Am (eijk Ai Cj (eˆ m × eˆ k )
= eijk Am Ai Cj (emkn eˆ n ) = −eijk emnk Am Ai Cj eˆ n = −[δim δjn − δin δjm ] Am Ai Cj eˆ n = −[ Ai Ai Cj eˆ j − A j Ai Cj eˆ i ] =
A j Cj ( Ai eˆ i ) − Ai Ai (Cj eˆ j )
= (A · C)A − (A · A)C. Now we have C × (A × (A × C))
= C × ((A · C)A − (A · A)C) = (A · C)C × A.
vector analysis and em waves
b. ∇ × (∇ × A). Using ∇ × A = eijk ∂x∂ A j eˆ k , we have i ∂ ∇ · (∇ × A) = ∇ · eijk A j eˆ k ∂xi ∂ ∂ eˆ ` · eijk = A j eˆ k ∂x` ∂xi ∂ ∂ A. = eij` ∂x` ∂xi j From the definition of the LeviCivita symbol, we know that the only nonvanishing terms are those for i, j, and ` all different. ∂ ∂ A ∂x` ∂xi j ∂ ∂ ∂ ∂ ∂ ∂ A + ei2` A2 + ei3` A3 ei1` ∂x` ∂xi 1 ∂x` ∂xi ∂x` ∂xi ∂ ∂ ∂ ∂ + e312 e213 A1 ∂x3 ∂x2 ∂x2 ∂x3 ∂ ∂ ∂ ∂ + e321 A2 + e123 ∂x3 ∂x1 ∂x1 ∂x3 ∂ ∂ ∂ ∂ + e231 A3 + e132 ∂x2 ∂x1 ∂x1 ∂x2 0.
∇ · (∇ × A) = eij` = =
= c. ∇ × ∇φ.
From the definitions of the curl and the gradient, we have ! ∂φ ∂ eˆ k ∇ × ∇φ = eijk ∂xi ∂x j ! ! ! ∂2 φ ∂2 φ ∂2 φ = eij1 eˆ 1 + eij2 eˆ 2 + eij3 eˆ 3 ∂xi ∂x j ∂xi ∂x j ∂xi ∂x j ∂2 φ ∂2 φ eˆ 1 = e231 + e321 ∂x2 ∂x3 ∂x3 ∂x2 ∂2 φ ∂2 φ + e132 + e312 eˆ 2 ∂x1 ∂x3 ∂x3 ∂x1 ∂2 φ ∂2 φ + e123 + e213 eˆ 3 ∂x1 ∂x2 ∂x2 ∂x1 = 0. 15. Prove the identities: a. ∇ · (∇ × A) = 0. Using ∇ × A = eijk ∂x∂ A j eˆ k , we have i ∂ ∇ · (∇ × A) = ∇ · eijk A j eˆ k ∂xi ∂ ∂ eˆ ` · eijk A j eˆ k = ∂x` ∂xi ∂ ∂ = eij` A. ∂x` ∂xi j
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From the definition of the LeviCivita symbol, we know that the only nonvanishing terms are those for i, j, and ` all different. ∂ ∂ A ∂x` ∂xi j ∂ ∂ ∂ ∂ ∂ ∂ A + ei2` A2 + ei3` A3 ei1` ∂x` ∂xi 1 ∂x` ∂xi ∂x` ∂xi ∂ ∂ ∂ ∂ e213 + e312 A1 ∂x3 ∂x2 ∂x2 ∂x3 ∂ ∂ ∂ ∂ + e123 + e321 A2 ∂x3 ∂x1 ∂x1 ∂x3 ∂ ∂ ∂ ∂ + e231 A3 + e132 ∂x2 ∂x1 ∂x1 ∂x2 0.
∇ · (∇ × A) = eij` = =
=
b. ∇ · ( f ∇ g − g∇ f ) = f ∇2 g − g∇2 f . We can use the identity
∇ · ( f A) = f ∇ · A + A · ∇ f . Then, we have
∇ · ( f ∇ g − g∇ f ) = = c. ∇r n = nr n−2 r,
f ∇ · ∇ g + ∇ g · ∇ f − ( g∇ · ∇ f + ∇ f · ∇ g) f ∇2 g − g ∇2 f .
n ≥ 2.
This is a direct computation.
∇r n
= nr n−2 r ∂r n ∂r n ∂r n = i+ j+ k ∂x ∂y ∂z ∂r ∂r ∂r n −1 = nr i+ j+ k ∂x ∂y ∂z = nr n−2 [ xi + yj + zk] = nr n−2 r,
n ≥ 2.
If one knows about the gradient operator in spherical coordinates, then an easier computation would be
∇r n =
∂r n rˆ = nr n−1 rˆ = nr n−2 r. ∂r
16. For r = xi + yj + zk and r = r, simplify the following. a. ∇ × (k × r). We begin with k × r = k × ( xi + yj + zk) = −yi + xj. Then,
i ∂ ∇ × (k × r) = ∂x −y
j ∂ ∂x
x
k ∂ = 2k. ∂x 0
vector analysis and em waves
r r
b. ∇ ·
.
∇·
r
=
r
= = c. ∇ ×
∇×
r r
.
r
=
r
= = = d. ∇ ·
r r3
∇·
i j k ∂ ∂ ∂ ∂x ∂x ∂x x y z r r r ∂ z ∂ y ∂ z ∂ x ∂ y ∂ x − − − i− j+ k ∂y r ∂z r ∂x r ∂z r ∂x r ∂y r h yz zy i h xz zx i h xy yx i − 3 + 3 i− − 3 + 3 j+ − 3 + 3 k r r r r r r 0.
.
r r3
= = =
e. ∇ ×
∂ x ∂ y ∂ z + + ∂x r ∂y r ∂z r 2 2 1 x 1 y 1 z2 − 3 + − 3 + − 3 r r r r r r 2 . r
r r3
r ∇× 3 r
∂ x ∂ y ∂ z + + 3 3 ∂x r ∂y r ∂z r3 1 3x2 1 3y2 1 3z2 − 5 + 3− 5 + 3− 5 r3 r r r r r 0.
.
=
= = =
i j k ∂ ∂ ∂ ∂x ∂x ∂x x y z 3 3 3 r r r ∂ z ∂ y ∂ z ∂ x ∂ y ∂ x − i− − j+ − k ∂y r3 ∂z r3 ∂x r3 ∂z r3 ∂x r3 ∂y r3 3xz 3zx 3xy 3yx 3yz 3zy − 5 + 5 i− − 5 + 5 j+ − 5 + 5 k r r r r r r 0.
17. Newton’s Law of Gravitation gives the gravitational force between two masses as GmM F = − 3 r. r
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a. Prove that F is irrotational. We first need to compute (or do the previous problem) i j k r ∂ ∂ ∂ ∇× 3 = ∂x ∂x ∂x r y z x3 3 3 r r r ∂ z ∂ y ∂ z ∂ x = − i− − j ∂y r3 ∂z r3 ∂x r3 ∂z r3 ∂ y ∂ x + − k ∂x r3 ∂y r3 3yz 3zy 3xz 3zx 3xy 3yx = − 5 + 5 i− − 5 + 5 j+ − 5 + 5 k r r r r r r = 0. Then, we have
∇ × F = − GmM∇ ×
r = 0. r3
b. Find a scalar potential for F. The scalar potential satisfies F = −∇φ. Therefore, we have ∂φ ∂x ∂φ ∂y ∂φ ∂z
x , r3 y − GmM 3 , r z − GmM 3 . r
= − GmM = =
We can integrate the first equation to obtain φ( x, y, z) = −
GmM + c(y, z). r
Inserting this result into the second equation gives
− GmM
y y ∂c = − GmM 3 . + ∂y r3 r
Thus, c(y, z) = c(z). Inserting this result into the third equation gives
− GmM
z ∂c z + = − GmM 3 . ∂z r3 r
Thus, c(z) = const. Picking the constant to be zero, we have φ( x, y, z) = − GmM r . 18. Consider a constant electric dipole moment p at the origin. It produces p·r an electric potential of φ = 4πe r3 outside the dipole. Noting that E = −∇φ, 0 find the electric field at r. This problem is best done using spherical coordinates for which
∇φ =
∂φ 1 ∂φ 1 ∂φ eˆ ρ + eˆ + eˆ φ . ∂ρ ρ ∂θ θ ρ sin θ ∂φ
vector analysis and em waves
Orienting the coordinate system such that the angle between p and r is θ, the electric potential can be written as φ
= = =
p·r 4πe0 r3 p · rˆ 4πe0 r2 p cos θ . 4πe0 r2
The electric potential is independent of φ. Letting ρ = r, we find
∇φ = = =
∂φ 1 ∂φ ˆ rˆ + θ ∂r r ∂θ 2p cos θ p sin θ ˆ − rˆ − θ 4πe0 r3 4πe0 r2 p − (2 cos θˆr + sin θ θˆ ). 4πe0 r3
This gives the electric field as E = −∇φ =
p (2 cos θˆr + sin θ θˆ ). 4πe0 r3
To put it in the prescribed coordinate free form, one needs ˆ p = p cos θˆr − p sin θ θ, or p sin θ θˆ = p cos θˆr − p. Then, E
= = = = =
p (2 cos θˆr + sin θ θˆ ) 4πe0 r3 1 (2p cos θˆr + p cos θˆr − p) 4πe0 r3 1 (3p cos θˆr − p) 4πe0 r3 1 (3(p · rˆ )rˆ − p) 4πe0 r3 1 3(p · r)r p ( − 3 ). 4πe0 r5 r
19. In fluid dynamics, the Euler equations govern inviscid fluid flow and provide quantitative statements on the conservation of mass, momentum, and energy. The continuity equation is given by ∂ρ + ∇ · (ρv) = 0, ∂t where ρ( x, y, z, t) is the mass density and v( x, y, z, t) is the fluid velocity. The momentum equations are given by ∂ρv + v · ∇(ρv) = f − ∇ p. ∂t Here, p( x, y, z, t) is the pressure and f is the external force per volume.
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a. Show that the continuity equation can be rewritten as ∂ρ + ρ∇ · (v) + v · ∇ρ = 0. ∂t This immediately follows from the identity ∇ · ( f A) = f ∇ · A + A · ∇ f . Thus, 0
∂ρ + ∇ · (ρv) ∂t ∂ρ + ρ∇ · v + v · ∇ρ. ∂t
= =
b. Prove the identity 12 ∇v2 = v · ∇v for irrotational v. We begin with the identity
∇(A · B) = A × (∇ × B) + B × (∇ × A) + (A · ∇)B + (B · ∇)A with A = B = v. This gives
∇(v · v) = v × (∇ × v) + v × (∇ × v) + (v · ∇)v + (v · ∇)v Since v is irrotational, ∇ × v = 0, and we have the result 1 1 ∇v2 = ∇(v · v) = v · ∇v. 2 2 c. Assume that • the external forces are conservative (f = −ρ∇φ), • the velocity field is irrotational (∇ × v = 0), • the fluid is incompressible (ρ =const), and • the flow is steady,
∂v ∂t
= 0.
Under these assumptions, prove Bernoulli’s Principle: p 1 2 v + φ + = const. 2 ρ We insert the assumptions and obtain ∂ρv + v · ∇(ρv) ∂t ρv · ∇v 1 ρ ∇ v2 2
= f − ∇p = −ρ∇φ − ∇ p = −∇(ρφ + p)
1 ∇( ρv2 + ρφ + p) = 0. 2 Next, we consider the time dependence of 21 ρv2 + ρφ + p. First we note the Chain Rule, d dx ∂ dy ∂ dz ∂ = + + = v · ∇. dt dt ∂x dt ∂y dt ∂z
vector analysis and em waves
Then, 1 d 1 2 ( ρv + ρφ + p) = v · ∇( ρv2 + ρφ + p) = 0. dt 2 2 Therefore, we have 1 2 p v + φ + = const. 2 ρ Note that for the usual gravitational potential, f = −ρgk. Therefore, f = −ρ∇( gz), or φ = gz and Bernoulli’s Principle becomes 1 2 ρv + ρgz + p = const. 2 20. Find the lengths of the following curves: a. y( x ) = x for x ∈ [0, 2]. The length is computed as Z bq
=
L
a
Z 2√
=
0
1 + [y0 ( x )]2 dx
√ 2 dx = 2 2.
√
b. ( x, y, z) = (t, ln t, 2 2t) for 1 ≤ t ≤ 2. The length is computed as L
Z t 1
=
t0 Z 2
=
r0 (t) dt q
1
Z 2
x˙ 2 (t) + y˙ 2 (t) + z˙ 2 (t) dt
r
1 + 8 dt t2 Z 2r 1 9 + 2 dt t 1 Z 2√ 2 9t + 1 dt. t 1
=
1
= =
1+
The integral can be performed using the trigonometric substitution 3t = tan u,
3 dt = sec2 u du,
9t2 + 1 = sec2 u.
Then, Z √ 2 9t + 1
t
dt
= = =
sec3 u du tan u Z tan2 u + 1 sec u du tan u Z Z
(tan u sec u + csc u) du
= sec u − ln(csc u + cot u) ! √ p 1 9t2 + 1 2 = 9t + 1 − ln + + C. 3t 3t
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This gives L
Z 2√ 2 9t + 1
=
t
1
dt.
√
" p
=
9t2 + 1 − ln
9t2 + 1 1 + 3t 3t
√
√
!#2
√
37 + 1 37 − ln ) − 10 + ln( 6 ! √ √ √ 2 10 + 2 37 − 10 + ln √ . 37 + 1
= =
1
√
10 + 1 3
!
c. y( x ) = cosh x, x ∈ [−2, 2]. (Recall the hanging chain example from classical dynamics.) The length is computed as L
=
Z bq a
= =
Z 2 −2
Z 2 −2
1 + [y0 ( x )]2 dx q 1 + sinh2 x dx
cosh x dx
2 = sinh x
−2
= 2 sinh 2. 21. Consider the integral following curves:
R C
y2 dx − 2x2 dy. Evaluate this integral for the
a. C is a straight line from (0, 2) to (1, 1). We first write the equation of the line. The slope is m = Since the yintercept is (0, 2), we have y = − x + 2. Z C
y2 dx − 2x2 dy
=
Z 1 0
=
Z 1 0
[(2 − x )2 + 2x2 ] dx [3x2 − 4x + 4] dx
= 1 − 2 + 4 = 3. b. C is the parabolic curve y = x2 from (0, 0) to (2, 4). Since y = x2 , we have dy = 2xdx. So, Z
2
C
2
y dx − 2x dy
=
Z 2 0
= =
Z 2 0 25
5
( x4 − 2x2 (2x )) dx ( x4 − 4x3 ) dx −4
24 48 =− . 4 5
1−2 1−0
= −1.
vector analysis and em waves
c. C is the circular path from (1, 0) to (0, 1) in a clockwise direction. In this case, we need x = cos θ and y = sin θ for − 3π 2 ≤ θ ≤ 0. Then, y2 dx − 2x2 dy = − sin3 θ dθ − 2 cos3 θ dθ. Therefore, Z
2
C
2
y dx − 2x dy
= −
Z −3π/2
sin3 θ + 2 cos3 θ dθ
0
= −
Z −3π/2 0
+2
(1 − cos2 θ ) sin θ dθ
Z −3π/2 0
= [(cos θ −
(1 − sin θ ) cos θ dθ
1 1 cos3 θ ) − 2(sin θ − sin3 θ )]0−3π/2 3 3
2 4 = − − = −2. 3 3
R 22. Evaluate C ( x2 − 2xy + y2 ) ds for the curve x (t) = 2 cos t, y(t) = 2 sin t, 0 ≤ t ≤ π. We note that r(t) = (2 cos t, 2 sin t) and thus ds = r0 (t) dt = 2 dt.
Z C
( x2 − 2xy + y2 ) ds =
Z π 0
2(4 − 8 cos t sin t) dt
= 8[t − sin2 t]0π = 8π. 23. Prove that the magnetic flux density, B, satisfies the wave equation. This follows a similar derivation as in the text. We will use Maxwell’s equations for a vacuum in the form
∇ · E = 0, ∇ · B = 0, ∂B , ∂t ∂E µ 0 e0 . ∂t
∇×E = − ∇×B = We start with an identity
∇ × (∇ × B) = ∇(∇ · B) − ∇2 B From Gauss’ Law for magnetism, we have
∇ × (∇ × B) = −∇2 B Using the AmpèreMaxwell Law on the left hand side of this equation, we obtain ∂E ∇ × (∇ × B) = ∇ × µ0 e0 . ∂t
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Interchanging the time and space derivatives, and using the Faraday’s Law, we find ∂ (∇ × E) ∂t ∂ ∂B − µ 0 e0 ∂t ∂t
∇ × (∇ × B) = µ0 e0 =
= − e0 µ 0
∂2 B . ∂t2
(9.2)
Combining the two expressions for ∇ × (∇ × B), we have the wave equation for the magnetic field: ∂2 B = c2 ∇2 B, ∂t2 where c =
(9.3)
√1 . e0 µ 0
24. Let C be a closed curve and D the enclosed region. Prove the identity Z C
φ∇φ · n ds =
Z D
(φ∇2 φ + ∇φ · ∇φ) dA.
First, we consider the identity
∇ · (φ∇φ) = ∇φ · ∇φ + φ∇2 φ. Then, we apply the Divergence Theorem. Z V
(φ∇2 φ + ∇φ · ∇φ) dV
= =
Z ZV S
∇ · (φ∇φ) dV φ∇φ · n dS.
25. Let S be a closed surface and V the enclosed volume. Prove Green’s first and second identities, respectively. R R a. S φ∇ψ · n dS = V (φ∇2 ψ + ∇φ · ∇ψ) dV. First, we consider the identity
∇ · (φ∇ψ) = ∇φ · ∇ψ + φ∇2 ψ. Then, we apply the Divergence Theorem. Z V
(φ∇2 ψ + ∇φ · ∇ψ) dV
= =
Z ZV S
b.
∇ · (φ∇ψ) dV φ∇ψ · n dS.
R = V (φ∇2 ψ − ψ∇2 φ) dV. We can either use Green’s first identity in part a, or derive this identity from scratch. The simplest approach would be to consider the two forms of Green’s first identity: R
S [ φ ∇ ψ − ψ ∇ φ ] · n dS
Z S
Z S
φ∇ψ · n dS
=
ψ∇φ · n dS
=
Z V
Z V
(φ∇2 ψ + ∇φ · ∇ψ) dV (ψ∇2 φ + ∇ψ · ∇φ) dV.
vector analysis and em waves
Subtracting these integrals give Green’s second identity. Z
=
ZS
[φ∇ψ − ψ∇φ] · n dS
V
=
Z V
[(φ∇2 ψ + ∇φ · ∇ψ) − (ψ∇2 φ + ∇ψ · ∇φ)] dV (φ∇2 ψ − ψ∇2 φ) dV.
A second approach might be a derivation independent of part a but similar to that derivation. We consider the identity
∇ · ( φ ∇ ψ − ψ ∇ φ ) = ∇ φ · ∇ ψ + φ ∇2 ψ − ∇ ψ · ∇ φ − ψ ∇2 φ = φ∇2 ψ − ψ∇2 φ. Then, we apply the Divergence Theorem. Z V
(φ∇2 ψ − ψ∇2 φ) dV
= =
Z ZV S
∇ · (φ∇ψ − ψ∇φ) dV
[φ∇ψ − ψ∇φ] · n dS.
26. Let C be a closed curve and D the enclosed region. Prove Green’s identities in two dimensions. a. First prove Z D
(v∇ · F + F · ∇v) dA =
Z C
(vF) · ds.
For v = 1 this is the two dimensional version of the Divergence Theorem, Z Z ∇ · F dA = F · ds. D
C
This can be shown using Green’s Theorem in the plane. We note that ds = nˆ ds points in the direction of the outward normal to C. Since dr = (dx, dy) is tangent to C, then ds = (dy, −dx ) is perpendicular and to the right of dr. Letting F = ( F1 , F2 ), then Z C
F · ds
= =
Z ZC C
( F1 , F2 ) · (dy, −dx ) − F2 dx + F1 dy.
Now, we apply Green’s Theorem in the Plane and find Z C
F · ds
Z
− F2 dx + F1 dy. Z ∂F1 − F2 = − dA ∂x ∂y D Z ∂F1 F = + 2 dA ∂x ∂y D =
C
=
Z D
∇ · F dA.
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Now, we only need to replace F with vF and use the identity
∇ · ( f A) = f ∇ · A + A · ∇ f . This gives the final result, Z C
(vF) · ds = =
Z ZD D
∇ · (vF) dA [v∇ · F + F · ∇v] dA.
b. Let F = ∇u and obtain Green’s first identity, Z
(v∇2 u + ∇u · ∇v) dA =
D
Z C
(v∇u) · ds.
We simply insert F = ∇u into the result in part a. Z D
Z D
Z
(v∇ · F + F · ∇v) dA =
ZC
(v∇ · ∇u + ∇u · ∇v) dA = Z D
ZC
(v∇2 u + ∇u · ∇v) dA =
C
(vF) · ds (v∇u) · ds (v∇u) · ds.
c. Use Green’s first identity to prove Green’s second identity, Z D
(u∇2 v − v∇2 u) dA =
Z C
(u∇v − v∇u) · ds.
We consider two form of Green’s first identity by interchanging u and v in the first. Z ZD D
(v∇2 u + ∇u · ∇v) dA = (u∇2 v + ∇v · ∇u) dA =
Z ZC C
(v∇u) · ds (u∇v) · ds.
Now, subtracting these expressions, we obtain the second identity. Z
=
ZC
(v∇u − u∇v) · ds
D
=
Z D
[(v∇2 u + ∇u · ∇v) − (u∇2 v + ∇v · ∇u)] dA [v∇2 u − u∇2 v] dA.
27. Compute the work done by the force F = ( x2 − y2 )i + 2xyj in moving a particle counterclockwise around the boundary of the rectangle R = [0, 3] × [0, 5]. R The work done is given by W = C F · dr. One can compute this directly by carrying out four integrations along each side of the rectangle, or one can apply Stoke’s Theorem (which is Green’s Theorem in the Plane for this problem).
vector analysis and em waves
First, the curl is given by i ∂ 2 ∂x 2 x −y
∇×F =
k ∂ ∂z 0
j ∂ ∂y
2xy
∂(2xy) ∂( x2 − y2 ) − k ∂x ∂y 4yk.
= =
Now, we apply Stoke’s Theorem to find Z C
F · dr
Z
=
ZD
=
D
= 4
∇ × F · nˆ dS 4yk · k dS
Z 5Z 3
= 12
0
0
Z 5 0
y dxdy
y dy = 150.
Directly applying Green’s Theorem gives the same integral: W
Z
=
ZC
=
C
Z
=
S
= 4
F · dr
( x2 − y2 ) dx + 2xy dy (2y + 2y) dxdy
Z 5Z 3 0
0
y dxdy.
A direct computation of the line integral gives sum of the four integrals as Z C
F · dr
=
Z 3 0
x2 dx +
Z 5 0
6y dy +
Z 0 3
x2 − 25 dx +
Z 0 5
0 dy
= 9 + 75 + (−9 + 75) = 150. 28. Compute the following integrals: R a. C ( x2 + y) dx + (3x + y3 ) dy for C the ellipse x2 + 4y2 = 4. We just apply Green’s Theorem in the plane to obtain the result. Z
( x2 + y) dx + (3x + y3 ) dy Z ∂(3x + y3 ) ∂( x2 + y) = − dxdy ∂x ∂y D C
= 2
Z D
dxdy
= 2(Area of ellipse) = 4π. b.
R
S ( x − y ) dydz + ( y
2
oriented unit sphere.
+ z2 ) dzdx + (y − x2 ) dxdy for S the positively
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R This can be written as S F · dS for F = ( x − y, y2 + z2 , y − x2 ). We can use the Divergence Theorem, noting that ∇ · F = 1 + 2y. Then, we have Z S
= = = =
( x − y) dydz + (y2 + z2 ) dzdx + (y − x2 ) dxdy
Z ZS ZV
F · dS
∇ · F dV (1 + 2y) dV
V Z 2π Z π Z 1 0
= c.
0
0
(1 + 2ρ sin θ sin φ)ρ2 sin θ dρdθdφ
4π . 3
R
C ( y − z ) dx + (3x + z ) dy + ( x + 2y ) dz, where C is the curve of intersection between z = 4 − x2 − y2 and the plane x + y + z = 0. R This can be written as S F · dr for F = (y − z, 3x + z, x + 2y). We can employ Stoke’s Theorem in which case we need the curl. i j k ∂ ∂ ∂ ∇ × F = ∂x ∂y ∂z y − z 3x + z x + 2y
= i − 2j + 2k The intersecting curve lives in the plane x + y + z = 0 as seen in (1,1,1) Figure 9.2. The unit normal to this surface is nˆ = √ . Stoke’s 3 Theorem gives Z Figure 9.2: A plot of the surfaces in Problem 9.28c.
= =
ZC
(y − z) dx + (3x + z) dy + ( x + 2y) dz
ZS D
= =
F · dr ˆ (∇ × F) · ndS
1 √ (i − 2j + 2k) · (i + j + k) dS 3 D Z 1 √ dS. 3 D Z
The intersection of the plane with the paraboloid gives an ellipse. The projection of this ellipse to the xyplane is a circle. Instead of integrating over the ellipse, we can integrate over the circle by first relating area elements. The area element in the ellipse is dS =
dA , cos θ
where dA is the area element of unit circle in the xyplane and θ is the angle between the normal to the intersecting plane and k.
vector analysis and em waves
Thus,
1 cos θ = nˆ · k = √ . 3
In order to complete the evaluation of the integral, we need the area of the circle. This is found from the equation of the circle. Eliminating z from the equations of the plane and paraboloid and completing the square, we have
− x − y = 4 − x 2 − y2 x 2 − x + y2 − y = 4 2 1 1 2 9 x− + y− = . 2 2 2 √ Thus, we have a circle of radius 3/ 2 and centered at (1/2, 1/2).
We can now complete the computation. Z C
(y − z) dx + (3x + z) dy + ( x + 2y) dz
1 √ 3
=
Z
=
A
d.
Z D
dS.
dA = area of circle =
9π . 2
x2 y dx − xy2 dy for C a circle of radius 2 centered about the origin.
R
C
Applying Green’s Theorem in the plane, we have Z Z ∂(− xy2 ) ∂( x2 y) x2 y dx − xy2 dy = − dxdy ∂x ∂y C D
= − = −
Z
( x2 + y2 )dxdy
D Z 2π Z 2 0
e.
0
r2 drdθ = −8π.
x2 y dydz + 3y2 dzdx − 2xz2 dxdy, where S is the surface of the cube [−1, 1] × [−1, 1] × [−1, 1]. R This can be written as S F · dS for F = ( x2 y, 3y2 , −2xz2 ). We can use the Divergence Theorem, noting that ∇ · F = 2xy + 6y − 4xz. Then, we have
R
S
Z
x2 y dydz + 3y2 dzdx − 2xz2 dxdy
S Z 1 Z 1 Z 1
=
−1 −1 −1 Z 1 Z 1
= 12
−1 −1
(2xy + 6y − 4xz) dxdydz
y dydz = 0.
29. Use Stokes’ Theorem to evaluate the integral Z C
−y3 dx + x3 dy − z3 dz
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for C the (positively oriented) curve of intersection between the cylinder x2 + y2 = 1 and the plane x + y + z = 1. R This can be written as S F · dr for F = (−y3 , x3 , −z3 ). We can employ Stoke’s Theorem in which case we need the curl. i j k ∂ ∂ ∂ ∇ × F = ∂x ∂y ∂z − y3 x 3 − z3
= 3( x 2 + y2 )k The intersecting curve lives in the plane x + y + z = 1. The unit normal (1,1,1) to this surface is nˆ = √ . Stoke’s Theorem gives 3
Z C
Z
−y3 dx + x3 dy − z3 dz =
Figure 9.3: A plot of the surfaces in Problem 9.29.
ZS
=
D
F · dr ˆ (∇ × F) · ndS
3 √ 3
=
Z D
( x2 + y2 ) dS.
The intersection of the plane with the cylinder gives an ellipse. The projection of this ellipse to the xyplane is a unit circle. Instead of integrating over the ellipse, we can integrate over the unit circle by first relating area elements. The area element in the ellipse is dS = dA/ cos θ where dA is the area element of unit circle in the xyplane and θ is the angle between the normal to the intersecting plane and k. Thus, 1 cos θ = nˆ · k = √ . 3 This gives Z C
3 √ 3
−y3 dx + x3 dy − z3 dz =
= 3 Figure 9.4: A plot of the surfaces in Problem 9.29.
Z
( x2 + y2 ) dS
D Z 2π Z 1 0
0
r3 drdθ =
3π . 2
30. Use Stokes’ Theorem to derive the integral form of Faraday’s Law, Z C
E · ds = −
∂ ∂t
Z D
B · dS
from the differential form of Maxwell’s equations. Noting that Faraday’s Law is given by
∇×E = −
∂B , ∂t
we have Z C
E · ds
=
Z S
=
Z
(∇ × E) · nˆ dS (−
S
=
∂ ∂t
∂B ) · nˆ dS ∂t
Z D
B · dS.
vector analysis and em waves
317
31. For cylindrical coordinates, x
= r cos θ,
y
= r sin θ,
z
= z,
find the scale factors and derive the following expressions:
∇f =
∂f 1 ∂f ∂f eˆ r + eˆ θ + eˆ z . ∂r r ∂θ ∂z
1 ∂(rFr ) 1 ∂Fθ ∂Fz ∇·F = + + . r ∂r r ∂θ ∂z 1 ∂Fz ∂F ∂Fr ∂Fz 1 ∂(rFθ ) ∂Fr ∇×F = − θ eˆ r + − eˆ θ + − eˆ z . r ∂θ ∂z ∂z ∂r r ∂r ∂θ 1 ∂ ∂f 1 ∂2 f ∂2 f ∇2 f = r + 2 2 + 2. r ∂r ∂r r ∂θ ∂z ∂r We first need the scale factors, hi = ∂u for u1 = r, u2 = θ, and u3 = z. i Let r = r cos θi + r sin θj + zz.
z
P
z
θ
r
y
x
Computing the tangent vectors, ∂r ∂r ∂r ∂θ ∂r ∂θ
Figure 9.5: Cylindrical coordinate system.
= cos θi + sin θj, = −r sin θi + r cos θj, = k.
we find the scale factors as hr = 1,
hθ = r,
hz = 1.
We can use the scale factors to find the gradient, divergence, curl, and Laplacian in cylindrical coordinates.
∇f
= = =
∇·F = = =
1 ∂f 1 ∂f 1 ∂f eˆ 1 + eˆ 2 + eˆ 3 . h1 ∂u1 h2 ∂u2 h3 ∂u3 1 ∂f 1 ∂f 1 ∂f eˆ r + eˆ + eˆ z . hr ∂r hθ ∂θ θ hz ∂z ∂f 1 ∂f ∂f eˆ r + eˆ + eˆ z . ∂r r ∂θ θ ∂z
∂ ∂ ∂ 1 (h2 h3 F1 ) + (h1 h3 F2 ) + (h1 h2 F3 ) . h1 h2 h3 ∂u1 ∂u2 ∂u3 1 ∂ ∂ ∂ (hθ hz Fr ) + (hr hz Fθ ) + (hr hθ Fz ) . hr hθ hz ∂r ∂θ ∂z 1 ∂(rFr ) 1 ∂Fθ ∂Fz + + . r ∂r r ∂θ ∂z
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h eˆ ˆ 2 h3 eˆ 3 1 1 h2 e 1 ∂ ∂ ∂ . ∂u1 ∂u ∂u 2 3 h1 h2 h3 F1 h1 F2 h2 F3 h3 h eˆ h eˆ ˆ z r r θ θ hz e 1 ∂ ∂ ∂ . ∂θ ∂z hr hθ hz ∂r Fr hr Fθ hθ Fz hz 1 ∂Fz ∂Fθ ∂Fr ∂Fz 1 ∂(rFθ ) ∂Fr − eˆ r + − eˆ θ + − eˆ z . r ∂θ ∂z ∂z ∂r r ∂r ∂θ
∇×F =
=
=
∇2 f
=
=
=
h2 h3 ∂ f ∂ 1 ∂ h1 h3 ∂ f + h1 h2 h3 ∂u1 h1 ∂u1 ∂u2 h2 ∂u2 ∂ h1 h2 ∂ f + ∂u3 h3 ∂u3 1 ∂ hθ hz ∂ f ∂ hr h z ∂ f + hr hθ hz ∂r hr ∂r ∂θ hθ ∂θ ∂ hr h θ ∂ f + ∂z hz ∂z 1 ∂ ∂f 1 ∂2 f ∂2 f r + 2 2 + 2. r ∂r ∂r r ∂θ ∂z
32. For spherical coordinates, z
x
= ρ sin θ cos φ,
y
= ρ sin θ sin φ,
z
= ρ cos θ,
θ
find the scale factors and derive the following expressions: ρ
∇f = φ
∂f 1 ∂f 1 ∂f eˆ ρ + eˆ + eˆ φ . ∂ρ ρ ∂θ θ ρ sin θ ∂φ
y
x
∇·F =
Figure 9.6: Definition of spherical coordinates for Problem 32.
∇×F =
1 ∂(ρ2 Fρ ) 1 ∂(sin θFθ ) 1 ∂Fφ + + . 2 ∂ρ ρ sin θ ∂θ ρ sin θ ∂φ ρ
∂(sin θFφ ) ∂Fθ ∂(ρFφ ) 1 1 1 ∂Fρ − eˆ ρ + − eˆ θ ρ sin θ ∂θ ∂φ ρ sin θ ∂φ ∂ρ 1 ∂(ρFθ ) ∂Fρ − + eˆ φ . ρ ∂ρ ∂θ
1 ∂ ∂f 1 ∂2 f sin θ + 2 2 . 2 ∂θ ρ sin θ ∂θ ρ sin θ ∂φ2 ∂r We first need the scale factors, hi = ∂u for u1 = ρ, u2 = θ, and u3 = φ. i Let r = ρ sin θ cos φi + ρ sin θ sin φj + ρ cos θz.
∇2 f =
1 ∂ ρ2 ∂ρ
ρ2
∂f ∂ρ
+
vector analysis and em waves
We first find the tangent vectors, ∂r ∂ρ ∂r ∂θ ∂r ∂φ
= sin θ cos φi + sin θ sin φj + cos θz, = ρ cos θ cos φi + ρ cos θ sin φj − ρ sin θz, = −ρ sin θ sin φi + ρ sin θ cos φj.
Then, we find the scale factors as 2 ∂r 2 hρ = ∂ρ
=  sin θ cos φi + sin θ sin φj + cos θz2 = sin2 θ cos2 φ + sin2 θ sin2 φ + cos2 θ
h2θ
= sin2 θ + cos2 θ = 1. 2 ∂r = ∂θ = ρ cos θ cos φi + ρ cos θ sin φj − ρ sin θz2 = ρ2 (cos2 θ cos2 φ + cos2 θ sin2 φ + sin2 θ ) = ρ2 (cos2 θ + sin2 θ ) = ρ2 . h2φ
=
2 ∂r ∂φ
=  − ρ sin θ sin φi + ρ sin θ cos φj2 = ρ2 (sin2 θ sin2 φ + sin2 θ cos2 φ) = ρ2 sin2 θ. Therefore, the scale factors are hρ = 1,
hθ = ρ,
hφ = ρ sin θ.
We can use the scale factors to find the gradient, divergence, curl, and Laplacian in spherical coordinates.
∇f
= = =
∇·F = = = =
1 ∂f 1 ∂f 1 ∂f eˆ + eˆ 2 + eˆ 3 . h1 ∂u1 1 h2 ∂u2 h3 ∂u3 1 ∂f 1 ∂f 1 ∂f eˆ ρ + eˆ θ + eˆ φ . hρ ∂ρ hθ ∂θ hφ ∂φ ∂f 1 ∂f 1 ∂f eˆ ρ + eˆ θ + eˆ φ . ∂ρ ρ ∂θ ρ sin θ ∂φ
1 ∂ ∂ ∂ (h2 h3 F1 ) + (h1 h3 F2 ) + (h1 h2 F3 ) . h1 h2 h3 ∂u1 ∂u2 ∂u3 1 ∂ ∂ ∂ hθ hφ Fρ + hρ hφ Fθ + hρ hθ Fφ . hρ hθ hφ ∂ρ ∂θ ∂φ ∂ ∂ 1 ∂ 2 ρ sin θF + ρ sin θF + ρF . ( ) ρ φ θ ∂θ ∂φ ρ2 sin θ ∂ρ 1 ∂(ρ2 Fρ ) 1 ∂(sin θFθ ) 1 ∂Fφ + + . ρ sin θ ∂θ ρ sin θ ∂φ ρ2 ∂ρ
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h eˆ ˆ 2 h3 eˆ 3 1 1 h2 e 1 ∂ ∂ ∂ ∂u2 ∂u3 . h1 h2 h3 ∂u1 F1 h1 F2 h2 F3 h3 hρ eˆ ρ h eˆ ˆ φ θ θ hφ e 1 ∂ ∂ ∂ ∂θ ∂φ . hρ hθ hφ ∂ρ Fρ hρ Fθ hθ Fφ hφ eˆ ρ ρˆeθ ρ sin θ eˆ φ 1 ∂ ∂ ∂ . ∂φ ρ2 sin θ ∂ρ ∂θ Fρ ρFθ ρ sin θFφ ∂(sin θFφ ) ∂Fθ ∂(ρFφ ) 1 1 1 ∂Fρ − − eˆ ρ + eˆ θ ρ sin θ ∂θ ∂φ ρ sin θ ∂φ ∂ρ 1 ∂(ρFθ ) ∂Fρ − eˆ φ + ρ ∂ρ ∂θ
∇×F =
=
=
=
∇2 f
=
=
=
=
1 ∂ h2 h3 ∂ f ∂ h1 h3 ∂ f + h1 h2 h3 ∂u1 h1 ∂u1 ∂u2 h2 ∂u2 ∂ h1 h2 ∂ f + ∂u3 h3 ∂u3 ∂ hρ hφ ∂ f 1 ∂ hθ hφ ∂ f + hρ hθ hφ ∂ρ hρ ∂ρ ∂θ hθ ∂θ hρ hθ ∂ f ∂ + ∂φ hφ ∂φ ∂ 1 ∂f ∂ ∂f 2 hρ sin θ + sin θ ∂ρ ∂θ ∂θ ρ2 sin θ ∂ρ 1 ∂f ∂ + ∂φ sin θ ∂φ 1 ∂ 1 ∂f 1 ∂ ∂2 f 2∂f ρ + sin θ + . ∂ρ ∂θ ρ2 ∂ρ ρ2 sin θ ∂θ ρ2 sin2 θ ∂φ2
33. The moments of inertia for a system of point masses are given by sums instead of integrals. For example, Ixx = ∑i mi (y2i + z2i ) and Ixy = − ∑i mi xi yi . Find the inertia tensor about the origin for m1 = 2.0 kg at (1, 0, 1), m2 = 5.0 kg at (1, −1, 0), and m3 = 1.0 kg at (1, 1, 1). One first computes each element of the inertia tensor. Ixx
∑ mi (y2i + z2i )
=
i
= 2.0(02 + 12 ) + 5.0((−1)2 + 02 ) + 1.0(12 + 12 ) = 9.0 kg m2 . Iyy
=
∑ mi (xi2 + z2i ) i
= 2.0(12 + 12 ) + 5.0(12 + 02 ) + 1.0(12 + 12 ) = 11.0 kg m2 .
vector analysis and em waves
Izz
∑ mi (xi2 + y2i )
=
i
= 2.0(12 + 02 ) + 5.0(12 + (−1)2 ) + 1.0(12 + 12 ) = 14.0 kg m2 . One first computes each element of the inertia tensor. Ixy
= − ∑ mi xi yi = Iyx i
= −[2.0(1)(0) + 5.0(1)(−1) + 1.0(1)(1)] = 4.0 kg m2 . Iyz
= − ∑ mi xi yi = Izy i
= −[2.0(0)(1) + 5.0(−1)(0) + 1.0(1)(1)] = −1.0 kg m2 . Izx
= − ∑ mi xi yi = Ixz i
= −[2.0(1)(1) + 5.0(0)(1) + 1.0(1)(1)] = −3.0 kg m2 . Therefore, the inertia tensor is Ixx Ixy Ixz 9.0 I = Iyx Iyy Iyz = 4.0 Izx Izy Izz −3.0
4.0 11.0 −1.0
−3.0 −1.0 . 14.0
34. Consider the octant of a uniform sphere of density 5 lying in the first octant. a. Find the inertia tensor about the origin. The moments of inertia about the origin for a continuous distribution of mass are Ixx
=
Iyy
=
Izz
=
Ixy
= =
Iyz
=
Izx
Z
y2 + z2 dm,
Z
x2 + z2 dm,
Z
x2 + y2 dm,
Iyx = − Izy = − Ixz = −
Z Z Z
xy dm, yz dm, xz dm.
For a uniform distribution in spherical coordinates, we have dm = 5dV = 5ρ2 sin θ dρdθdφ. Recall that spherical coordinates are defined by x
= ρ sin θ cos φ,
y
= ρ sin θ sin φ,
z
= ρ cos θ.
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The components of the inertia tensor can be computed. Ixx
=
Z
y2 + z2 dm, Z π/2 Z π/2 Z a
= 5
0
0
Z π/2 Z π/2 Z a
= 5
0
0
0
= a5
Z π/2 Z π/2
5
Z π/2 Z π/2
= a
0
0
0
0
= a5
Z π/2 Z π/2
= a5
Z π/2
= = = =
=
0
0
0
Z
= 5 = 5
Z π/2
(sin θ − (1 − cos2 θ ) sin θ cos2 φ) dθdφ, (sin θ − (sin θ − cos2 θ sin θ ) cos2 φ) dθdφ, 1 cos3 θ ) cos2 φ]0π/2 dφ, 3
x2 + z2 dm, Z π/2 Z π/2 Z a 0
0
0
Z π/2 Z π/2 Z a 0
0
0
= a5
Z π/2 Z π/2
= a5
Z π/2 Z π/2
= a5
Z π/2
0
0
0
0
0
0
0
(ρ2 − y2 )ρ2 sin θ dρdθdφ, (1 − sin2 θ sin2 φ)ρ4 sin θ dρdθdφ,
(1 − sin2 θ sin2 φ) sin θ dθdφ, (sin θ − (1 − cos2 θ ) sin θ sin2 φ) dθdφ, (sin θ − (sin θ − cos2 θ sin θ ) sin2 φ) dθdφ,
[− cos θ − (− cos θ +
1 cos3 θ ) sin2 φ]0π/2 dφ, 3
2 sin2 φ) dθdφ, 3 Z π/2 1 a5 (1 − (1 − cos 2φ)) dφ, 3 0 Z π/2 2 1 a5 ( + cos 2φ) dφ, 3 3 0 1 5 2 a [ θ + sin 2φ]0π/2 , 3 6
= a
=
(1 − sin2 θ cos2 φ) sin θ dθdφ,
(1 −
Z π/2 Z π/2
=
(1 − sin2 θ cos2 φ)ρ4 sin θ dρdθdφ,
[− cos θ − (− cos θ +
= a5
=
(ρ2 − x2 )ρ2 sin θ dρdθdφ,
2 cos2 φ) dθdφ, 3 0 Z π/2 1 5 a (1 − (1 + cos 2φ)) dφ, 3 0 Z π/2 2 1 5 ( − cos 2φ) dφ, a 3 3 0 2 1 a5 [ θ − sin 2φ]0π/2 , 3 6 1 5 πa . 3
= a5
Iyy
0
5
Z π/2 0
(1 −
vector analysis and em waves
1 5 πa . 3
=
Z
=
Izz
x2 + y2 dm, Z π/2 Z π/2 Z a
= 5
0
0
= a5
Z π/2
= −5 = −5
0
0
0
Z
0
0
0
0
0
0
Z π/2 Z π/2
=
(ρ sin θ cos φ)(ρ sin θ sin φ)ρ2 sin θ dρdθdφ,
Z π/2 Z π/2 Z a
= − a5
=
1 cos3 θ ]0π/2 dφ, 3
xy dm,
Z π/2 Z π/2
=
(sin θ − cos2 θ sin θ ) dθdφ,
Z
= − a5
=
(1 − cos2 θ )ρ4 sin θ dρdθdφ,
[− cos θ +
Z π/2 Z π/2 Z a
5
(ρ2 − z2 )ρ2 sin θ dρdθdφ,
2 5 π/2 dφ, a 3 0 1 5 πa . 3
Iyx = −
= −a
0
Z π/2 Z π/2
=
Iyz
0
= a5
=
=
0
Z π/2 Z π/2 Z a
= 5
Ixy
0
0
0
0
0
Z π/2 0
ρ4 cos φ sin φ sin3 θ dρdθdφ,
cos φ sin φ sin3 θ dθdφ, cos φ sin φ(1 − cos2 θ ) sin θ dθdφ,
cos φ sin φ[− cos θ +
1 cos3 θ ]0π/2 dφ, 3
π/2 2 cos φ sin φ dφ, − a5 3 0 1 − a5 [sin2 φ]0π/2 , 3 1 5 − a . 3
Z
Izy = −
= −5 = −5
Z
yz dm,
Z π/2 Z π/2 Z a 0
0
0
Z π/2 Z π/2 Z a
= − a5
0
0
0
Z π/2 Z π/2 0
0
(ρ sin θ sin φ)(ρ cos θ )ρ2 sin θ dρdθdφ, ρ4 sin φ cos θ sin2 θ dρdθdφ,
sin φ cos θ sin2 θ dθdφ,
π/2 1 = − a5 sin φ[sin2 θ ]0π/2 dφ, 3 0 1 = − a5 [− cos φ]0π/2 , 3 1 = − a5 . 3
Z
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Izx
=
Ixz = −
= −5 = −5
Z
xz dm,
Z π/2 Z π/2 Z a 0
0
0
Z π/2 Z π/2 Z a 0
0
0
= − a5
Z π/2 Z π/2
= − a5
Z π/2 Z π/2
0
0
0
0
(ρ sin θ cos φ)(ρ cos θ )ρ2 sin θ dρdθdφ, ρ4 cos φ cos θ sin2 θ dρdθdφ,
cos φ cos θ sin2 θ dθdφ, cos φ cos θ sin2 θ dθdφ,
π/2 1 cos φ[sin2 θ ]0π/2 dφ, = − a5 3 0 1 = − a5 [sin φ]0π/2 , 3 1 = − a5 . 3
Z
Therefore, the inertia tensor is Ixx Ixy Ixz I = Iyx Iyy Iyz Izx Izy Izz − 13 a5 = − 13 a5 . 1 5 3 πa π −1 −1 a15 a15 = M. −1 π −1 ≡ 9 9 −1 −1 π
1 5 3 πa − 13 a5 − 13 a5
− 13 a5 1 5 3 πa 1 5 −3a
b. What are the principal moments of inertia? These are found as the eigenvalues of the inertia matrix. We seek the eigenvalues of matrix M obtained in the previous part and the 5 eigenvalues, λ, of I will be multiples of these, a3 λ. π−λ −1 −1  M − λI3  = −1 π−λ −1 −1 −1 π−λ π−λ −1 −1 −1 = (π − λ) + −1 π − λ −1 π − λ −1 π − λ − −1 −1
= (π − λ)[(π − λ)2 − 1] − 2[π − λ + 1] = −λ3 + 3πλ2 + 3(1 − π 2 )λ − 2 − 3π = (π − λ − 2)(1 + π − λ)2 . Therefore, there are two eigenvalues, λ = π − 2, π + 1, with the second one a repeated eigenvalue.
vector analysis and em waves
c Find the principal axes of inertia, that is, the eigenvectors of the inertia tensor. The eigenvectors of I are the same as those of M. For λ = π − 2, we have to solve 2 −1 −1 v1 0 −1 2 −1 v2 = 0 −1 −1 2 v3 0 Setting v1 = 1, we have v2 + v3
= 2,
2v2 − v3
= 1,
−v2 + 2v3
= 1,
The solution is v2 = v3 = 1. Thus, the eigenvector associated with λ = π − 2 is vT = (1, 1, 1). For λ = π + 1, we have to solve 0 −1 −1 −1 v1 −1 −1 −1 v2 = 0 0 −1 −1 −1 v3 All three equations give v1 + v2 + v3 = 0. Thus, we are free to choose two linearly independent eigenvectors in the plane v1 + v2 + v3 = 0. For example, vT = (−1, 1, 0) and vT = (−1, 0, 1). 35. Let T α be a contravariant vector and Sα be a covariant vector. a. Show that R β = gαβ T α is a covariant vector. We transform R β by using 0
R α = Λασ T σ and 0 gαβ = gγδ (Λ−1 )α (Λ−1 )δβ . γ
Then, we have R0β
0
=
0 gαβ Tα
=
gγδ (Λ−1 )α (Λ−1 )δβ Λασ T σ
=
gγδ (Λ−1 )δβ δσ T σ
=
gγδ T γ (Λ−1 )δβ
γ
γ
Therefore, R β = gαβ T α is a covariant vector.
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b. Show that R β = gαβ Sα is a contravariant vector. We transform R β by using Sα0 = and
0
g αβ = . Then, we have 0
Rβ
=
0
g αβ R0α
= Λαγ Λδ gγδ T σ (Λ−1 )σα β
= δγσ Λδ gγδ T σ β
= Λδ gγδ T γ β
Therefore, R β = gαβ Sα is a contravariant vector. 36. Show that T αβγδρ S βρ is a tensor. What is its rank? Transforming this expression, we have 0
T αβγδρ S0βρ
= Λαµ Λν Λγe Λδη Λζ T µνeηζ Sνζ (Λ−1 )νβ (Λ−1 )ζρ β
ρ
= Λαµ Λγe Λδη T µνeηζ Sνζ Therefore, T αβγδρ S βρ is a rank three contravariant tensor. 37. The line element in terms of the metric tensor, gαβ is given by ds2 = gαβ dx α dx β . 0
0
Show that the transformed metric for the transformation x α = x α ( x β ) is given by ∂x α ∂x β 0 gγδ = gαβ 0 γ 0 δ ∂x ∂x 0
We first note that for x α = x α ( x µ ) we have dx α =
∂x α 0 µ 0 dx . ∂x µ
Then, ds2
= = =
gαβ dx α dx β ∂x α 0 µ ∂x β 0 ν 0 dx )( 0 dx ) ∂x ν ∂x µ 0 0 ∂x α ∂x β [ gαβ 0 µ 0 ν ]dx µ dx ν . ∂x ∂x gαβ (
Thus, we can identify 0 gµν = gαβ
to have ds2 = gµν dx µ dx ν .
∂x α ∂x β 0 0 ∂x µ ∂x ν
10 Extrema and Variational Calculus 1. For each problem, locate the critical points and classify each one using the second derivative test. a. f ( x, y) = ( x + y)2 . The gradient is given by
∇ f = 2( x + y)(i + j). 2 = 0, The critical points lie along the line y = − x. Since f xx f yy − f xy the second derivative test gives no information.
b. f ( x, y) = x2 y + xy2 . The gradient is given by
∇ f = (2xy + y2 )i + ( x2 + 2xy)j. The critical points are found from solving the system 0
= (2x + y)y
0
= ( x + 2y) x.
From the first equation, we have y = 0 or y = 2x. Both cases lead to x = 0. Therefore the origin is the critical point. The second derivative test gives 2 f xx f yy − f xy = (2y)(2x ) − (2y)2 = 0.
The second derivative test gives no information. c. f ( x, y) = x4 y + xy4 − xy. The gradient is given by
∇ f = (4x3 y + y4 − y)i + ( x4 + 4xy3 − x )j. The critical points are found from solving the system 0
= 4x3 y + y4 − y = (4x3 + y3 − 1)y
0
= x4 + 4xy3 − x = ( x3 + 4y3 − 1) x.
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From the first equation, we have y = 0 or y3 = 1 − 4x3 . Inserting y = 0 into the second equation, we have ( x3 − 1) x = 0. Solutions are x = 0, 1. This gives two critical points, (0, 0), (1, 0). Inserting y3 = 1 − 4x3 into the second equation, we have 0 = ( x3 + 4(1 − 4x3 ) − 1) x = 3(1 − 5x3 ) x. Solutions are x = 0, 5−1/3 . The corresponding values for y are found as y = 1, 5−1/3 . This gives two more critical points, (0, 1), (5−1/3 , 5−1/3 ). From the second derivative test, we have detH ( x, y)
=
2 f xx f yy − f xy
= (12x2 y)(12xy2 ) − (4x3 + 4y3 − 1)2 = 112x3 y3 − 16x6 + 8x3 − 16y6 + 8y3 − 1. Evaluating det H ( x, y) at the critical points, we have det H (0, 0)
= −1.
det H (0, 1)
= −9.
= −9. 27 det H (5−1/3 , 5−1/3 ) = , 5 det H (1, 0)
f xx (5−1/3 , 5−1/3 ) =
12 > 0. 5
Therefore, there are three saddle points and one local minimum. d. f ( x, y) = x2 − 3xy + 2x + 10y + 6y2 + 12. The gradient is given by
∇ f = (2x − 3y + 2)i + (−3x + 10 + 12y)j. The critical points are found from solving the system 2x − 3y
= −2
3x − 12y
= 10.
Solving this system of equations, we find one critical point, ( x, y) = (− 85 , − 26 15 ). The second derivative test gives 2 f xx f yy − f xy = 24 − (−3)2 > 0
and f xx > 0. Therefore, (− 85 , − 26 15 ) is a local minimum. e. f ( x, y) = ( x2 − y2 )e−y . The gradient is given by
∇ f = 2xe−y i + (y2 − x2 − 2y)e−y j. The critical points are found from solving the system 0
= 2xe−y
0
= ((y2 − x2 − 2y)e−y .
extrema and variational calculus
We immediately have that x = 0 from the first equation. Then, the second equation gives 0 = y2 − 2y = y(2 − y). So, there are two critical points, (0, 0) and (0, 2). The Hessian matrix is H( f )
=
∂2 f ∂x2 ∂2 f ∂x∂y
=
2e−y −2xe−y
∂2 f ∂x∂y ∂2 f ∂y2
,
−2xe−y . ( x2 − y2 + 4y − 2)e−y
!
Then, det H ( x, y)
= 2( x2 − y2 + 4y − 2)e−2y − 4x2 e−2y = −2( x2 + y2 − 4y + 2)e−2y .
Evaluating det H ( x, y) at the critical points, we have det H (0, 0)
= −4 < 0.
det H (0, 2)
= 4e−4 > 0,
f xx (0, 2) = 2e−2 > 0.
Therefore, (0, 0) is a saddle point and (0, 2) is a local minimum. 2. For each problem, locate the critical points and evaluate the Hessian matrix at each critical point. The Hessian matrix is 2 2 H( f ) =
∂ f ∂x2 ∂2 f ∂x∂y
∂ f ∂x∂y ∂2 f ∂y2
.
a. f ( x, y, z) = ( x + y)2 . The isolated critical point is the origin as seen in Problem 1a. The Hessian matrix is ! 2 2 H ( f )(0, 0) = . 2 2 b. f ( x, y) = x2 y + xy2 . The critical point is the origin as seen in Problem 1b. The Hessian matrix is ! 2y 2( x + y ) H ( f )(0, 0) = . 2( x + y ) 2x ! 0 0 = . 0 0
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c. f ( x, y) = x4 y + xy4 − xy. In Problem 1c we found the four critical points (0, 0), (1, 0), (0, 1), and (5−1/3 , 5−1/3 ). The Hessian matrix is given by ! 12x2 y 4x3 + 4y3 − 1 H( f ) = . 4x3 + 4y3 − 1 12xy2 Evaluating the Hessian matrix at the critical points, gives ! 0 −1 . H ( f )(0, 0) = −1 0 ! 0 3 H ( f )(0, 1) = . 3 0 ! 0 3 H ( f )(1, 0) = . 3 0 ! H ( f )(5−1/3 , 5−1/3 )
=
12 5 3 5
3 5 12 5
.
d. f ( x, y, z) = xy + xz + yz. The gradient is given by
∇ f = (y + z)i + ( x + z)j + ( x + y)k. The critical points are found from solving the system 0
= y+z
0
= x+z
0
= x + y.
From the first two equations, x = y = −z. The third equation then gives z = 0. So, (0, 0, 0) is the critical point. The Hessian matrix is H ( f )(0, 0, 0)
=
0 1 1
1 0 1
1 1 . 0
e. f ( x, y, z) = x2 + y2 + xz + 2z2 . The gradient is given by
∇ f = (2x + z)i + 2yj + ( x + 4z)k. The critical points are found from solving the system 0
= 2x + z
0
= 2y
0
= x + 4z.
extrema and variational calculus
It is easy to see that (0, 0, 0) is the critical point. The Hessian matrix is H ( f )(0, 0, 0)
=
2 0 1
0 2 0
1 0 . 4
3. Find the absolute maxima and minima of the function f ( x, y) = x2 + xy + y2 on the unit circle. The critical points satisfy the equation
∇ f = (2x + y)i + ( x + 2y)j = 0. Thus, the origin is the critical point. The second derivative test gives 2 f xx f yy − f xy = 2(2) − (1)2 > 0.
Thus, there is only one local minimum. A plot of the function shows that it is a global minimum. 4. A thin plate has a temperature distribution of T ( x, y) = x2 − y3 − x2 y + y + 20 for 0 ≤ x, y, ≤ 2. Find the coldest and hottest points on the plate. We need to locate the global maxima and minima and determine the temperature at those points. First we determine the critical points. The critical points satisfy the equation
∇ T = (2x − 2xy)i + (−3y2 − x2 + 1)j = 0, or 0
= 2x (1 − y)
1
= x2 + 3y2 .
The first equation gives x = 0 or y = 1. Inserting these into the second equation we have no real solutions for y = 1 and 3y2 = 1 for x = 0. Therefore, there are two critical points, (0, √1 ) and (0, − √1 ). Evaluating the 3 3 temperature at these points, we have 1 1 1 2√ T (0, ± √ ) = 20 − (± √ )3 + ± √ , = 20 ± 3 ≈ 19.62, 20.38. 9 3 3 3 The temperature along the four sides of the domain, 0 ≤ x, y, ≤ 2, take the form T ( x, 0)
= x2 + 20,
T (2, y)
= 16 − y3 − 3y,
T ( x, 2)
= − x2 + 14,
T (0, y)
= 20 − y3 + y.
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We need to find the maximum and minimum temperatures of these boundary functions. This is done by finding the locations where the derivatives of each function vanish. T 0 ( x, 0) 0
= 2x = 0,
T (2, y)
= −3y2 − 3 = 0,
T 0 ( x, 2)
= −2x,
0
T (0, y)
= −3y2 + 1 = 0.
Solving these, the only new points of interest are at the corners (0, 0) and (0, 2). For these points T (0, 0) = 20 and T (0, 2) = 14. Therefore, the global maximum is T = 30.38 and the global minimum is T = 14. 5. Find the extrema of the given function subject to the given constraint. a. f ( x, y) = ( x + y)2 , x2 + y = 1. We employ the method of Lagrange multipliers where the constraint is g( x, y) = x2 + y = 1. Then, ∇ f = λ∇ g plus the constraint equation give the system of equations Figure 10.1: A plot of surfaces in Problem 10.5a.
2( x + y )
= 2λx,
2( x + y )
= λ,
2
= 1
x +y
Comparing the first two equations, we see that x = 1/2 or λ = 0. If x = 1/2, the y = 3/4. If λ = 0, then y = − x. Inserting this into the third equation, we find √ x2 − x = 1. The solution of this quadratic gives x = 21 (1 ± 5) = −y. The values of f ( x, y) at these points are
Figure 10.2: A plot of the function h ( x ) = ( x + 1 − x 2 )2 in Problem 10.5a.
1 3 f( , ) 2 4 √ √ 1 1 f ( (1 ± 5), − (1 ± 5)) 2 2
1 3 5 = ( + )2 = . 2 4 4
= 0.
Thus, we have a maximum at the point ( 12 , 34 ) and a minimum at √ √ the point ( 12 (1 ± 5), − 12 (1 ± 5)). One could also solve the constraint for y = 1 − x2 and substitute it into f ( x, y) to obtain h( x ) = ( x + 1 − x2 )2 . Now, this becomes an optimization problem for a function of one variable. The plot of this function is shown in Figure 10.2. Noting that the critical points satisfy g0 ( x ) = 2( x + 1 − x2 )(1 − 2x ) = 0, we see that the extrema are readily found. b. f ( x, y) = x2 y + xy2 , x2 + y2 = 2.
extrema and variational calculus
333
We employ the method of Lagrange multipliers where the constraint is g( x, y) = x2 + y2 = 1. Then, ∇ f = λ∇ g plus the constraint equation give the system of equations 2xy + y2 2
x + 2xy 2
x +y
2
= 2λx, = 2λy, = 1
In this problem we can see that interchanging x and y in the equations does not change the system. So, we let y = x and have x2 + √ √ √ √ y2 = 2x2 = 1. This gives the points (1/ 2, 1/ 2) and (−1/ 2, −1/ 2). The values of f ( x, y) = x2 y + xy2 , at these points are easily found since y = x. Then, f ( x, x ) = x ( x2 + y2 ) = x, √ √ √ f (1/ 2, 1/ 2) = 1/ 2. √ √ √ f (−1/ 2, −1/ 2) = −1/ 2. √ √ Thus, we have a maximum at the point (1/ 2, 1/ 2) and a min√ √ imum at the point (−1/ 2, −1/ 2). This can be seen in Figure 10.4.
Figure 10.3: A plot of the surfaces in Problem 10.5b.
c. f ( x, y) = 2x + 3y, 3x2 + 2y2 = 3. We employ the method of Lagrange multipliers where the constraint is g( x, y) = 3x2 + 2y2 = 3. The function and constraint are shown in Figure 10.4. We need ∇ f = λ∇ g plus the constraint equation to give the system of equations
2
2
= 6λx,
3
= 4λy,
2
= 3.
3x + 2y
This is an example of a problem where one can solve directly for x and y in terms of λ. Thus, we have x=
1 , 3λ
y=
3 4λ
Inserting these into equation three, we have 3
= 3x2 + 2y2 2 2 3 1 = 3 +2 3λ 4λ 1 9 35 = + 2 = . 3λ2 8λ 24λ2
This gives,
√
λ
=
λ
=
70 2√ 9√ , x= 70, y = 70, 12 35 70 √ 70 2√ 9√ − , x=− 70, y = − 70. 12 35 70
Figure 10.4: A plot of the surfaces in Problem 10.5c.
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. Evaluating f ( x, y) = 2x + 3y, we easily see the first point makes the function positive and is the maximum point. The other point is a minimum. d. f ( x, y, z) = x2 + y2 + z2 , xyz = 1. We employ the method of Lagrange multipliers where the constraint is g( x, y) = xyz = 1. Then, ∇ f = λ∇ g plus the constraint equation give the system of equations 2x
= λyz,
2y
= λxz,
2z
= λxy, = 1.
xyz
Since xyz = 1, we can multiply the first three equations by a variable which will give the product xyz on the right hand side. This simplifies the equations as 2x2
= λ,
2y
2
= λ,
2z
2
= λ,
Therefore, x2 = y2 = z2 . Since xyz = 1, 1 = x2 y2 z2 = x6 . So, x = ±1, y = ±1, and z = ±1, in such a way that xyz = +1. Therefore, we have the points (1, 1, 1), (1, −1, −1), (−1, −1, 1), (−1, 1, −1) are the locations of the extrema and f ( x, y, z) = x2 + y2 + z2 = 1 at each point. One can argue that these are all minima be thinking about a few simple cases. However, we can actually show that they are by tackling the problem without Lagrange multipliers. Substituting z = 1/xy into f ( x, y, z), we have the new function g( x, y) = x2 + y2 +
1 . x 2 y2
The critical points satisfy the equation
∇ g = (2x −
2 2 )i + (2y − 2 3 )j = 0, x 3 y2 x y
or x
=
y
=
1 x 3 y2 1 . 2 x y3
Multiplying the first equation by x and the second equation by y, we find that 1 x 2 = y2 = 2 2 . x y
extrema and variational calculus
Once again, we have x = y = ±1. The Hessian is given by H( f )
6 x 4 y2 4 x 3 y3
2+
=
4 x 3 y3 2 + x26y4
! .
Evaluating the Hessian matrix at the critical points, gives for (1, 1) and (−1, −1) ! 8 4 H = . 4 8 For the points (1, −1) and (−1, 1) the Hessian is ! 8 −4 H = . −4 8 In all cases the determinant of the Hessian is positive and gxx > 0. Therefore, all of the points are local minima. e. f ( x, y, z) = xy + xz, x2 + y2 = 1, xz = 1. We employ the method of Lagrange multipliers where there are two contraint equations, g1 ( x, y, z) = x2 + y2 = 1
and
g2 ( x, y, z) = xz = 1.
Then, we introduce two Lagrange multipliers
∇ f = λ 1 ∇ g1 + λ 2 ∇ g2 . The components of this equation and the constraint equations give the system of equations y+z
2
= 2λ1 x + λ2 z,
x
= 2λ1 y,
x
= λ2 x,
2
= 1
xz
= 1
x +y
From the third equation, we have that x = 0 or λ2 = 1. However, x 6= 0 since the last equation would not be satisfied. So, λ2 = 1. This leaves the first two equations in the form y
= 2λ1 x,
x
= 2λ1 y,
Eliminating x, we have y = (2λ1 )2 y. So, y = 0 or λ1 = ±1/2. Since x 6= 0, y 6= 0. So, λ1 = ±1/2. The first equations now give y = ± x.
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mathematical methods for physicists
√ Inserting in the first constraint, 2x2 = 1, or x = ±1/ 2. Therefore, the extrema are given by 1 √ 1 1 √ 1 ( √ , √ , 2), ( √ , − √ , 2), 2 2 2 2
√ √ 1 1 1 1 (− √ , √ , − 2), (− √ , − √ , − 2). 2 2 2 2 6. A particle moves under the force field F = −∇V, where the potential function is given by V ( x, y) = x3 + y3 − 3xy + 5. Find the equilibrium points of F and determine if the equilibria are stable or unstable. The force is given by F = −∇V and equilibrium points are where the force is zero, or where the gradient vanishes. We have 0 = F = −∇V = 3(y − x2 )i + 3( x − y2 )j. Thus, we have to solve the system, y = x2 ,
x = y2 .
Eliminating y, between these equations, we have x ( x3 − 1) = 0. So, x = 0 or x = 1. This gives the critical points as (0, 0) and (1, 1). The determinant of the Hessian of V ( x, y) is given by 6x −3 det H = = 36xy − 9. −3 6y Evaluating this determinant at the critical points, we have det H (0, 0) < 0 and det H (1, 1) > 0. Therefore, the origin is a saddle point and (1, 1) is a local minimum since Vxx = 6 > 0. This can be seen in Figure 10.5. Figure 10.5: A plot of the potential function showing that the origin is a saddle and (1, 1) is a local minimum.
Table 10.1: The data from Hubble’s 1929 paper.
7. In 1929 E. Hubble confirmed the linear dependence of the velocity on the distance by using the observed values of these quantities. He proposed that the observed radial velocity v, and the estimated distance d, satisfied the equation v = H0 d. This equation is known as Hubble’s Law and the constant H0 is known as Hubble’s constant. Object S. Mag. L. Mag. NGC6822 NGC598 NGC221 NGC224 NGC5457 NGC4736 NGC5194 NGC4449 NGC4214 NGC3031
Dist (Mpc) 0.032 0.034 0.214 0.263 0.275 0.275 0.45 0.5 0.5 0.63 0.8 0.9
Vel (km/s) +170 +290 130 70 185 220 +200 +290 +270 +200 +300 30
Object NGC3627 NGC4826 NGC5236 NGC1068 NGC5055 NGC7331 NGC4258 NGC4151 NGC4382 NGC4472 NGC4486 NGC4649
Dist (Mpc) 0.9 0.9 0.9 1.0 1.1 1.1 1.4 1.7 2.0 2.0 2.0 2.0
Vel (km/s) +650 +150 +500 +920 +450 +500 +500 +960 +500 +850 +800 +1090
extrema and variational calculus
Consider the set of data in Table 10.1, which was extracted from Hubble’s 1929 paper to find the proposed linear relationship between the radial velocities and distances of several nebulae.1 a. Plot the velocity vs. distance data and find the best fit line through the data using your favorite graphing application.
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“A Relation Between Distance and Radial Velocity Among ExtraGalactic Nebulae”, E. Hubble, Proc. Nat. Acad. 15, 168 (1929) 1
The data is shown in Figure 10.6 with the best fit line, v = 454.1584 − 40.7836d.
Figure 10.6: A plot of the velocity vs the distance for the Hubble data.
1200 1000 800 v (km/s)
600 400 200 0 200 400 0
0.5
1 d (Mpc)
1.5
2
b. Carry out the computation of the means of the distance and velocities, variance, and covariance for the data set. The brute force computation would give something similar to Table 10.2. c. Determine the slope, the intercept, and the correlation coefficient for fit. Plot this line on the graph of the data. The best fit line is shown in Figure 10.6. The slope and intercept can be computed from the data in Table 10.2. a=
Cov( x, y) = 454.158, Var( x )
b = y − ax = −40.784. d. From the previous results, compute the sums of squares of the errors, the residuals and the total sums of squares. The sums of the squares of the errors, SSE =
∑(vi − vˆi )2 = 1193442 i
and the sums of the squares of the residuals, SSR =
∑(vˆi − µv )2 = 1976648 i
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Table 10.2: Computation of variance and covariance for the Hubble data.
d (Mpc) 0.032 0.034 0.214 0.263 0.275 0.275 0.45 0.5 0.5 0.63 0.8 0.9 0.9 0.9 0.9 1 1.1 1.1 1.4 1.7 2 2 2 2 d (Mpc) 0.911
v (km/s) (km/s) 170 290 130 70 185 220 200 290 270 200 300 30 650 150 500 920 450 500 500 960 500 850 800 1090 v (km/s) 373
( d − d )2 (Mpc2 ) 0.773 0.770 0.486 0.420 0.405 0.405 0.213 0.169 0.169 0.079 0.012 0.000 0.000 0.000 0.000 0.008 0.036 0.036 0.239 0.622 1.185 1.185 1.185 1.185 Var(d) (Mpc2 ) 1.917
( v − v )2 (km2 /s2 ) 4.13E+04 6.91E+03 2.53E+05 1.96E+05 3.12E+05 3.52E+05 3.00E+04 6.91E+03 1.06E+04 3.00E+04 5.35E+03 1.63E+05 7.67E+04 4.98E+04 1.61E+04 2.99E+05 5.91E+03 1.61E+04 1.61E+04 3.44E+05 1.61E+04 2.27E+05 1.82E+05 5.14E+05 Var(v) (km2 /s2 ) 6.34E+05
(d − d)(v − v) (Mpc km/s) 178.6 72.9 350.9 287.3 355.2 377.4 79.9 34.2 42.4 48.7 8.1 4.6 3.1 2.5 1.4 48.5 14.5 23.9 62.0 462.8 138.1 519.1 464.7 780.4 Cov(d, v) (Mpc km/s) 870
extrema and variational calculus
339
are readily computed from the tabulated sums. The total sums of squares is given by SST = SSE + SSR = NVar(v) = 3170090.625. From Excel the LINEST output is given in Table 10.3.
value seslope R2 F SSR
slope 454.158 75.237 0.624 36.438 1976648
intercept 40.784 83.439 232.911 22 1193442
value seint sey df SSE
e. Compute the coefficient of determination and the standard errors in the slope and the intercept. A computation based on the above is given by SSR = 0.623530521, SST r SSE sev = = 0.085, N−2 sey = 0.267, seslope = p NVar(y) sey seint = q = 83.44. N (1 − x2 /x2 ) R2 =
This agrees with Excel’s LINEST output as given in Table 10.3. f. What linear relationship did you find between the velocities and the distances? The best fit line is the best fit line, v = 454.1584 − 40.7836d. g. What is your estimate of Hubble’s constant? [Show that kilometers per second per megaparsec has units of 1/time]. How does this compare to the currently accepted value of Hubble’s constant? Based on the slope result, we have H0
km/s Mpc km/s 1Mpc 1pc 454 Mpc 106 pc 3.086 × 1013 km
= 454 =
= 1.47 × 10−17 s−1 . h. Estimate the age of the universe (in years) using your data. How does this compare to the current estimates? A simple estimate at Hubble’s time would give t = H0−1 = 6.8 × 1016 s == 2.0 × 109 yr.
Table 10.3: Output from Excel’s LINEST function for the Hubble data.
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mathematical methods for physicists
8. For each of the following, find a path that extremizes the given integral. R2 a. 1 (y02 + 2yy0 + y2 ) dy, y(1) = 0, y(2) = 1. Identifying F ( x, y, y0 ) = y02 + 2yy0 + y2 , we apply the Euler equation ∂F d ∂F 0 = − ∂y dx ∂y0 d = (2y0 + 2y) − (2y0 + 2y) dx = y00 − y. Next, we seek solutions of this differential equation satisfying the fixed conditions y(1) = 0, y(2) = 1. The solutions are simple exponentials, e± x , and the boundary values indicate that is is best to use hyperbolic functions. So, to satisfy the first condition, we choose y( x ) = A sinh( x − 1). The second condition gives y(2) = A sinh 1 = 1. Therefore, A = 1/ sinh 1 and the path that extremizes the integral is sinh( x − 1) y( x ) = . sinh 1 R2 b. 0 y2 (1 − y02 ) dy, y(0) = 1, y(2) = 2. For this problem we will use the second from of Euler’s equation, ∂F d ∂F 0 − F − 0 y = 0. ∂x dx ∂y Since the first term vanishes, we have d ∂F − F − 0 y0 = 0 dx ∂y ∂F F − 0 y 0 = c1 , ∂y where c1 is an integration constant. Using F (y, y0 ) = y2 (1 − y02 ), we have c1
=
F−
∂F 0 y ∂y0 0
= y2 (1 − y02 ) − y2 (−2y 2 ) 0
= y2 + y2 y 2 . We can solve for y0 to obtain a separable differential equation which can be integrated. c1 0
y2 dy dx
0
= y2 + y2 y 2 c1 − y2 = y2 p c1 − y2 = ± y
extrema and variational calculus
Z
y
Z
dx
= ±
x + c2
= ∓
y2 + ( x + c2 )2
= c1 .
p q
c1 − y2
dy
c1 − y2
Thus, the path that extremizes this integral is a circle. The specific circle sought satisfies the conditions y(0) = 1, y(2) = 2. Imposing these values on the solution, we have 1 + c22 4 + (2 + c2 )
2
= c1 = c1 .
Eliminating c1 , 1 + c22 = 4 + (2 + c2 )2 = 8 + 4c2 + c22 . Therefore, c2 = −7/4 and this in turn gives c1 = 65/16. So, the path that extremizes this integral is the circle 7 2 65 y2 + x − = . 4 16 c.
R1
−1 5y
02
+ 2yy0 dy, y(−1) = 1, y(1) = 0.
Identifying F ( x, y, y0 ) = 5y02 + 2yy0 , we apply the Euler equation ∂F d ∂F 0 = − ∂y dx ∂y0 d = 2y0 − (10y0 + 2y) dx = 10y00 . This is easily integrated to give the path as a straight line, y( x ) = Ax + B. The end point conditions y(−1) = 1, y(1) = 0, then give the path that extremizes this integral, y( x ) = 12 (1 − x ). 9. A bead slides frictionlessly down a wire from the point (0, 0) to (1, 1). a. Determine the equation of the path (shape of the wire) such that the bead takes the least time between the given points. The full theory was developed in the text. The path is a cycloid, given by x
= a( ϕ − sin ϕ) + k,
y
= a(1 − cos ϕ).
Recall that the positive yaxis points downward. We just need to specify the appropriate constants and the range of the variable ϕ.
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mathematical methods for physicists
It is easiest to set ϕ = 0 at the origin. Then, k = 0 and we still need to determine a. Next we consider the point (1, 1). We seek the angle ϕ0 and constanta satisfying the equations 1
= a( ϕ0 − sin ϕ0 ),
1
= a(1 − cos ϕ0 ).
Equating these expressions, we obtain a transcendental equation for the angle, ϕ0 − sin ϕ0 = 1 − cos ϕ0 . This can be solved numerically to obtain ϕ0 ≈ 2.412. Then, we have a=
1 ≈ .5729. 1 − cos ϕ0
A plot of this path is shown in Figure 10.7 on top of the underlying cycloid. Figure 10.7: Plot of the brachistochrone path on top of a cycloid for Problem 9b.
2
1
3
4
x
−1 y
Therefore, the shape of the wire such that the bead takes the least time is given by the cycloid x
= .5729( ϕ − sin ϕ),
y
= .5729(1 − cos ϕ)
for 0 ≤ ϕ ≤ 2.412. b. How much time did it take the bead to follow this path? The time is given by the integral T
=
Z C
=
ds v
Z 1 0
ds p
2gy( x )
We can write ds
= = =
q
( x 0 ( ϕ))2 + (y0 ( ϕ))2 dt
q
( a(1 − cos ϕ))2 + ( a sin ϕ)2 dt
q
2a2 (1 − cos ϕ) dt
= 2a sin ϕ dt.
extrema and variational calculus
Also, v=
p
2gy =
q
√ 2ga(1 − cos ϕ) = 2 ag sin ϕ.
So, ds 2a sin ϕ dt = √ = v 2 ag sin ϕ
r
a dt g
Therefore, we have T=
Z C
ds = v
Z ϕ0 r a
r dt =
g
0
a ϕ0 ≈ 0.583 s. g
c. If the path between these points is given by the line y = x, then how much time does the bead take? Since y = y( x ), we compute the time using T [x]
= =
ds C v Z 1p Z
0
=
1 + [y0 ( x )]2 p dx 2gy( x )
Z 1√ 1 + 22
p
0
s
=
5 2g
2gx
Z 1 0
dx
x −1/2 dx
s
5 √ 1 2 x 0 2g
s
10 ≈ 1.01 s. g
= =
d. If the path between these points is given by the parabola y = x2 , then how much time does the bead take? Since y = y( x ), we compute the time using T [x]
= =
Z C Z 1 0
=
1 + [y0 ( x )]2 p dx 2gy( x )
Z 1p 1 + [2x ]2
dx 2gx2 Z 1√ 1 1 + 4x2 p dx. x 2g 0 0
=
ds v p
p
This integral can be evaluated using either a trigonometric substitution (2x = tan θ) or a hyperbolic function substitution (2x = sinh u). The trigonometric substitution gives Z √ Z 1 + 4x2 sec θ 1 dx = sec2 θ dθ 1 x 2 tan θ 2
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mathematical methods for physicists
=
Z
csc θ sec2 θ dθ
= csc θ tan θ + = sec θ +
Z
Z
csc θ cot θ tan θ dθ
csc θ dθ
= sec θ + ln  csc θ − cot θ  + C √ 1 + 4x2 − 1 p 1 + 4x2 + ln = + C. 2x For comparison, the hyperbolic substitution is presented. Z √
1 + 4x2 dx x
= = = = =
Z
cosh u 1 cosh u du sinh u 2
1 2
cosh2 u du sinh u Z 1 + sinh2 u du sinh u cosh u − 1 +C cosh u + ln sinh u √ 1 + 4x2 − 1 p 2 1 + 4x + ln +C 2x Z
These results give T [x]
= = = =
=
1
x
−1 y Figure 10.8: Plot of parabolic path for Problem 9d.
ds C v Z 1√ 1 1 + 4x2 p dx x 2g 0 √ # " 1 + 4x2 − 1 1 p 1 p 1 + 4x2 + ln 2x 2g 0 " √ √ 1 5−1 p 5 + ln 2 2g √ !# 1 + 4x2 − 1 p − lim 1 + 4x2 + ln 2x x →0 " √ # √ 2x2 − O( x4 ) 5−1 1 p 5 + ln − lim 1 + ln . 2 2x x →0 2g
Z
The limit as x → 0 is infinite! Thus, it takes forever to follow this path. However, a plot of y( x ) = x2 should reveal the problem. The bead never moves as seen in Figure 10.8. The bead sits on an unstable equilibrium point. 10. A light ray travels from point A in a medium with index of refraction n1 toward point B in a medium with index of refraction n2 . Assume that the rays travel in straight lines and bend at a point at the interface as shown in
extrema and variational calculus
Figure 10.9. The time functional is
A
T [y] =
Z
ds . v
θ1
a. Write the time functional in terms of the travel path y( x ).
n1
The time is given by the integral Z
=
T
C Z 1
=
0
ds v p
n2
θ2
1 + [y0 ( x )]2 dx. v
b. Apply Fermat’s Principle of least time to write the Euler Equation for this functional. Fermat’s principle of least time suggests that the path of the light in each medium, separately, minimizes the time functional. Assuming v does not depend explicitly on time except at y = 0, we have from Euler’s Equation √ 1+[y0 ( x )]2 d ∂ v = 0. dx ∂y0 c. Solve the equation in part b. and show that n sin θ is a constant. One integration gives
√ ∂
1+[y0 ( x )]2 v ∂y0
= const.
A further integration gives y0 v
p
1 + [y0 ( x )]2
= const.
Since the slope is related to the angle, we have y0 = tan θ. Furthermore, 1 + [y0 ( x )]2 = 1 + tan2 θ = sec2 θ. This gives y0 v
p
1 + [y0 ( x )]2
345
=
tan θ n = sin θ = const. v sec θ c
Therefore, n sin θ = const. d. Let the point at which the light is incident to the interface be at ( x, 0). Write an expression for the total time to travel from point A at ( x1 , y1 ) to point B at ( x2 , y2 ) in terms of the indices of refraction and the coordinates x, x1 , y1 , x2 , and y2 The time along each straight line can be computed using the distances between the points and the speed in each medium. From
B
Figure 10.9: A light ray travels between two media from point A towards point B.
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mathematical methods for physicists
( x1 , y1 ) A q
( x − x1 )2 + y21
 x − x1 
n1 n2
( x, 0)
Figure 10.10 we can determine these distance and compute the time of flight in each medium. The total time as the light ray goes from point A to point B is given by q q ( x − x1 )2 + y21 ( x − x2 )2 + y22 + T (x) = v1 v q q 2 2 2 n1 ( x − x1 ) + y1 n2 ( x − x2 )2 + y22 = + c c e. Treating the time as a function of x, minimize this function as a function of one variable and derive Snell’s law of refraction.
B ( x2 , y2 )
Figure 10.10: The original diagram is labeled with coordinates for Problem 10d.
To find the least time, we seek the extremum of the function in part d. Setting the derivative to zero, we have 0
= =
=
dT ( x ) dx q q 2 2 n2 ( x − x2 )2 + y22 d n1 ( x − x1 ) + y1 + dx c c 1 n1 ( x − x1 ) n2 ( x − x2 ) q +q c ( x − x )2 + y2 ( x − x )2 + y2 1
=
2
1
2
1 [n sin θ1 − n2 sin θ2 ] . c 1
Therefore, n1 sin θ1 = n2 sin θ2 . In the derivation we used the relations sin θ1 = q
x − x1
( x − x1 )2 + y21
,
sin θ2 = q
x2 − x
( x − x2 )2 + y22
.
This can be seen in Figure 10.10. 11. Given the cylinder defined by x2 + y2 = 4, find the path of shortest length connecting the given points. To find the geodesics on a cylinder, we need to write the length of a curve in cylindrical coordinates with r = 2. Then, we have the parametrization x
= 2 cos θ,
y
= 2 sin θ,
z
= z.
The corresponding differentials become dx
= −2 sin θ dθ,
dy
= 2 cos θ dθ,
dz
= dz.
extrema and variational calculus
347
So, the line element becomes ds2
= dx2 + dy2 + dz2 = 4 sin2 θ dθ 2 + 4 cos2 θ dθ 2 + dz2 = 4dθ 2 + dz2 .
The length of a curve on the surface of the cylinder is now written as L
=
Z b
ds
a
=
Z bp a
=
4dθ 2 + dz2
s
Z b
4
a
dθ dz
2
+ 1 dz.
Now, we find the geodesics by applying the Euler equation to the function p 0 F (z, θ, θ 0 ) = 4θ 2 + 1. The Euler equation yields 0
c1
=
∂F d − ∂θ dz
=
d − dz
=
√
∂F ∂θ 0
! √ 0 ∂ 4θ 2 + 1 ∂θ 0
4θ 0 0
4θ 2 + 1
.
Now we solve for θ 0 to find 0
r
θ =
1 c1 ≡ . 16 − 4c1 A
Integrating, we have θ=
1 ( z − c2 ), A
or z = Aθ + c2 . Thus, we have the parametrization of a helical path on the cylinder, r = (2 cos θ, 2 sin θ, Aθ + c2 ). This path is shown in Figure 10.11. For the given points below, we can determine the two integration constants. In both cases x = 2 for θ = 0. So, r = (2 cos θ, 2 sin θ, Aθ ). since c2 = 0.
Figure 10.11: A plot of paths on the cylinder in Problem 10.9.
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mathematical methods for physicists
a. (2, 0, 0) and (0, 2, 5). For the final point, we want r = (2 cos θ, 2 sin θ, Aθ ) = (0, 2, 5). This occurs for θ = π/2. So, r = (0, 2, π2 A). Therefore, A = resulting path is 10 r = (2 cos θ, 2 sin θ, θ ). π b. (2, 0, 0) and (2, 0, 5).
10 π.
The
5 2π .
The
For the final point, we want r = (2 cos θ, 2 sin θ, Aθ ) = (2, 0, 5). This occurs for θ = 2π. So, r = (2, 0, 2π A). Therefore, A = resulting path is r = (2 cos θ, 2 sin θ,
5 θ ). 2π
12. The shape of a hanging chain between the points (− a, b) and ( a, b) is such that the gravitational potential energy Z a q V [y] = ρg y 1 + y02 dx −a
is minimized subject to the length of the chain remaining constant, Z a q L[y] = 1 + y02 dx. −a
Find the shape, y( x ) of the hanging chain. This problem is solved by introducing a Lagrange multiplier to handle the constraint. The path can be determined from the solution of the Euler Equation. Let q q q F ( x, y, y0 ) = ρgy 1 + y02 − λ 1 + y02 = (ρgy − λ) 1 + y02 . Then, we can use the Euler equation ∂( F d − ∂y dx
∂F ∂y0
= 0,
or the second form of the Euler equation, ∂F d ∂F 0 − F − 0 y = 0. ∂x dx ∂y Since F = V − λL does not explicitly depend on x, we have d ∂F 0 0 = F− 0y dx ∂y q q d 0 ∂ 0 2 0 2 = (ρgy − λ) 1 + y − y 0 (ρgy − λ) 1 + y dx ∂y ! 02 q d ( ρgy − λ ) y = (ρgy − λ) 1 + y02 − p dx 1 + y 02 ! d (ρgy − λ) p = . dx 1 + y 02
extrema and variational calculus
Therefore,
(ρgy − λ) p = const. 1 + y 02 Letting the integration constant be c1−1 , we can solve for y0 to obtain dy =± dx
q
[c1 (ρgy − λ)]2 − 1.
This equation is separable and can be integrated with the hyperbolic function substitution c1 (ρgy − λ) = cosh u, ρc1 g dy = sinh u du. x − x0
= ±
Z
p
1 ρc1 g u ± . ρc1 g u
Z
= ± = ±ρc1 g( x − x0 ) =
dy [c1 (ρgy − λ)]2 − 1 du
cosh(ρc1 g( x − x0 ))
= cosh u = c1 (ρgy − λ) 1 λ y = cosh(ρc1 g( x − x0 )) + . ρc1 g ρc1 g
So, if we let ρc1 g = A−1 , then the path is given in the form y( x ) = A(λ + cosh
x − x0 ). A
We now apply the conditions that the curve passes through the points (± a, b), obtaining a − x0 = − a − x0 . Therefore, x0 = 0. The value of the path at x = ± a is then given by b = A(λ + cosh
a ). A
Thus, we have λ=
b a − cosh . A A
There is still one unknown, A. However, we first need to specify the length of the chain. The length is L
= = = =
Z a q −a Z a −a Z a
1 + y02 dx
r 1 + (sinh
x 2 ) dx A
x dx A −a a 2A sinh . A cosh
Therefore, we can in principle solve this transcendental equation for A once we have a value for a.
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So, the path is given as y( x )
= = =
where 2A sinh
a A
x − x0 ) A a x − x0 b ) A( − cosh + cosh A A A x − x0 a b + A(cosh − cosh ), A A A(λ + cosh
= L.
13. In Example 10.34, the geodesic equations for geodesics in the plane in polar coordinates were found as 2 dφ d2 r r − 2 = 0 ds ds d dφ r2 = 0. ds ds a. Solve this set of equations for r = r (s) and φ = φ(s). The second equation can be integrated to obtain dφ = `. ds
r2 This gives the first equation as d2 r ds2
dφ 2 r ds 2 ` r 2 r
= =
`2 . r3
= Multiplying by dr/ds, we have dr d2 r ds ds2 2 1 d dr 2 ds ds 2 dr ds
= = =
dr `2 . ds r3 1 d ` 2 − 2 ds r 2 ` − + C. r
Since ds2 = dr2 + r2 dφ2 , 2 2 dφ dr + r2 = 1, ds ds or
dr ds
2
2 ` + = 1. r
Therefore, C = 1 and dr =± ds
r 1−
`2 . r2
extrema and variational calculus
In order to find a relation between r and φ, we can eliminate s using the Chain Rule. Note that dφ dφ/ds ` = =± q dr dr/ds r2 1 −
`2 r2
.
Therefore, we can integrate this result to find φ − φ0
= ±
`
Z
r2
= ±`
Z
q
1−
`2 r2
dr
dr √ . r r 2 − `2
This integral is evaluated using the substitution r = ` sec θ, dr = ` sec θ tan θ dθ. Then, Z dr √ φ − φ0 = ±` r r 2 − `2 Z ` sec θ tan θ dθ p = ±` ` sec θ `2 (sec2 θ − 1)
= ±
Z
dθ
r = ±θ = ± sec−1 . ` Therefore, we have found that cos(φ − φ0 ) = `r . b. Prove that the solutions obtained in part a. are the familiar straight lines in the plane. We can cast the solution found in the previous part in Cartesian coordinates using the polar coordinate transformation. Thus,
` = cos(φ − φ0 ) r ` = r cos φ cos φ0 + r sin φ sin φ0 = x cos φ0 + y sin φ0 . We have obtained the equation of a line in Cartesian coordinates, thus proving that the geodesics in the plane are straight lines. 14. Use a Lagrange multiplier to find the curve y( x ) of length L = π on R1 the interval [0, 1] which maximizes the integral I = 0 y( x ) dx and passes through the points (0, 0) and (1, 0). We introduce a Lagrange multiplier and seek to extremize the integral p R1 J = 0 (y( x ) + λ 1 + [y0 ( x )]2 ) dx. p We apply the Euler equation to the function F ( x, y, y0 ) = y( x ) + λ 1 + [y0 ( x )]2 to find d ∂F ∂F 0 = − ∂y dx ∂y0 ! p ∂ 1 + [y0 ( x )]2 d = 1−λ dx ∂y0 ! d y0 p = 1−λ . dx 1 + [y0 ( x )]2
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Integrating this equation and solving for y0 ( x ), we obtain ! d y0 p 1 = λ dx 1 + [y0 ( x )]2 x + c1
=
λy0
=
λ2 y
λy0 p
1 + [y0 ( x )]2 q ( x + c) 1 + [y0 ( x )]2
02
= ( x + c )2 (1 + y 2 )
0
0
= ( x + c )2 ( x + c )2 = 2 λ − ( x + c )2 x+c = ±p . λ2 − ( x + c )2
( λ2 − ( x + c )2 ) y 2 0
y2 dy dx
Further integrating this result, we have dy dx y( x )
= ±p = ± = ∓
Z
q
x+c λ2
− ( x + c )2 x+c p dx 2 λ − ( x + c )2
λ2 − ( x + c)2 + k.
Rearranging, we have that the curve is part of the circle
( x + c )2 + ( y − k )2 = λ2 . The circular arc passes through the points (0, 0) and (1, 0), which implies that c2 + k 2
= λ2 ,
(1 + c )2 + k 2
= λ2 .
Therefore, c2 = (1 + c)2 , or ±c = 1 + c. This is true if c = − 12 and λ2 = k2 + 41 . This gives the form of the path in terms of k as 1 1 ( x − )2 + ( y − k )2 = k 2 + . 2 4 The curve has length L = π. So, we need to compute L = L
=
Z 1q 0
1 + [y0 ( x )]2 dx
( x + c )2 ]1/2 dx − ( x + c )2 0 Z 1 λ p = dx 0 λ2 − ( x + c )2 −1 x + c 1 = λ sin 0 λ =
Z 1
[1 +
λ2
R1p 0
1 + [y0 ( x )]2 dx.
extrema and variational calculus
353
c 1+c − λ sin−1 λ λ 1 1 = λ sin−1 + λ sin−1 2λ 2λ 1 π = 2λ sin−1 2λ π 1 . = sin 2λ 2λ This is a transcendental equation of the form sin πx = x whose solutions are x = 0, ±0.7365.... Therefore, λ = 0.6789 and k = 0.4592. This gives the path as the circle
= λ sin−1
( x − 0.500)2 + (y − 0.4592)2 = 0.4609.
m1 r
15. A mass m lies on a table and is connected to a string of length ` as shown in Figure 10.12. The string passes through a hole in the table and is connected to another mass M that is hanging in the air. We assume that the string remains taught and that mass M can only move vertically. a. The Lagrangian for this setup is 1 1 Mr˙ 2 + m(r˙ 2 + r2 θ˙ 2 ) + Mg(` − r ), 2 2 where r and θ are polar coordinates describing where mass m is on the table with respect to the hole. Explain why the terms in the Lagrangian are appropriate.
L=
The mass on the table, m, moves in the plane of the table and one can use polar coordinates to describe the motion. It only contributes to the kinetic energy, 1 m(r˙ 2 + r2 θ˙ 2 ), 2 where the first term comes from the radial component of the ve˙ ˙ and the second term from the angular component, r θ. locity, r, KE1 =
Mass M moves vertically. Its position from the table is y = ` − r. Therefore, the vertical velocity is given by y˙ = −r˙ and the kinetic energy is KE2 = 12 Mr˙ 2 . As mass M moves, its gravitational potential energy changes. Thus, there is a contribution U = Mgy = Mg(` − r ). b. Derive the equations of motion for r (t) and θ (t). There are only two degrees of freedom, so there are two EulerLagrange equations: d ∂L ∂L 0 = − , dt ∂r˙ ∂r d = ( Mr˙ + mr˙ ) − mr θ˙ 2 + Mg. dt d ∂L ∂L − , 0 = dt ∂θ˙ ∂θ d 2 ˙ = mr θ . dt
m2 Figure 10.12: Two masses connected by a string.
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Therefore, the equations of motion are given by 0
=
0
=
M¨r + m¨r − mr θ˙ 2 + Mg, d 2 ˙ mr θ . dt
c. What angular velocity is needed for mass m to maintain uniform circular motion? For uniform circular motion, we have r˙ = 0 and ω = θ˙ =const. From the equations of motion we find mr θ˙ 2 = Mg. Therefore, we p find that ω = θ˙ = Mg/mr. x1 ⇐
m ⇒ x2 M
θ
Figure 10.13: The weight of a small block on an inclined plane.
16. An inclined plane of mass M and inclination θ lies on a frictionless horizontal surface. A small block of mass m is placed carefully on the plane. Find the acceleration of the inclined plane using the EulerLagrange equations for the indicated horizontal coordinates in Figure 10.13 Let the small block and the inclined plane start at horizontal position (with respect to the left ends) x = 0 corresponding to x1 = x2 = 0. Then, as the small block moves down the incline, x1 < 0, and x2 > 0, the horizontal distance between the blocks (left ends) will be x1 − x2 . In order for the small block to be on the incline at this position, the distance it has fallen vertically will be y = ( x1 − x2 ) tan θ. Therefore, the kinetic energy of the small block is 1 1 T = m( x˙ 2 + y˙ 2 ) = m( x˙ 22 + ( x˙ 1 − x˙ 2 )2 tan2 θ ). 2 2 The gravitational potential energy is U = mgy = −mg( x1 − x2 ) tan θ. The kinetic energy of the incline is simple, so the Lagrangian for the system is given by
L=
1 1 M x˙ 12 + m( x˙ 22 + ( x˙ 1 − x˙ 2 )2 tan2 θ ) + mg( x1 − x2 ) tan θ. 2 2
There are two EulerLagrange equations: ∂L d ∂L 0 = − , dt ∂ x˙ 1 ∂x1 d = M x˙ 1 + m( x˙ 1 − x˙ 2 ) tan2 θ − mg tan θ. dt d ∂L ∂L 0 = − , dt ∂ x˙ 2 ∂x2 d = m x˙ 2 − m( x˙ 1 − x˙ 2 ) tan2 θ + mg tan θ. dt Therefore, we have the equations of motion are M x¨1 + m( x¨1 − x¨2 ) tan2 θ 2
m x¨2 − m( x¨1 − x¨2 ) tan θ
= mg tan θ. = −mg tan θ.
extrema and variational calculus
355
This is an algebraic system of equations for x¨1 and x¨2 . Namely, we have ! ! ! M + m tan2 θ −m tan2 θ x¨1 mg tan θ = . −m tan2 θ m + m tan2 θ x¨2 −mg tan θ The solution of this system is mg tan θ −m tan2 θ −mg tan θ m + m tan2 θ x¨1 = −m tan2 θ M + m tan2 θ −m tan2 θ m + m tan2 θ
=
x¨2
=
mg tan θ M sec2 θ + m tan2 θ M + m tan2 θ mg tan θ −m tan2 θ −mg tan θ M + m tan2 θ −m tan2 θ −m tan2 θ m + m tan2 θ
= −
Mg tan θ M sec2 θ + m tan2 θ
Therefore, the acceleration of the inclined plane is given by x¨1 =
mg tan θ mg sin θ cos θ = . 2 2 M sec θ + m tan θ M + m sin2 θ
Note, we can solve this system of second order differential equations. Adding the two equations, we have M x¨1 + m x¨2 = 0. Integrating and assuming the initial velocities are zero, M x˙ 1 + m x˙ 2 = 0. This is just the conservation of linear momentum. Further integrating, we have Mx1 + mx2 = 0 since the masses started at x1 = x2 = 0. Solving for x2 = −
M x , m 1
we have
M x¨ . m 1 This is consistent with the above result. x¨2 = −
17. Two similar masses are connected by a string of fixed length and hung over two pulleys that are the same height as depicted in Figure 10.14. One mass is set into pendular motion and the other only moves vertically. Find the equations of motion of the length and angle from vertical of the swinging pendulum.
m
m
Figure 10.14: Two masses are connected to two pulleys and one is set into motion.
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mathematical methods for physicists
Let the string length be `, the distance of the left mass be y, the separation of the pulleys d, and the string length for the right mass be r. Then, ` = y + d + r. There are two free coordinates, r and θ as seen in Figure 10.15. First, we find the kinetic and potential energies. The kinetic energy of the left mass is 1 1 KE1 = my˙ 2 = mr˙ 2 . 2 2 That of the second mass is KE2 =
1 m(r˙ 2 + r2 θ˙ 2 ). 2
Both masses have varying heights leading to the total potential energy U = mgy + mgr cos θ = mg(` − d − r ) + mgr cos θ.
d y
m
θ
r
The Lagrangian then takes the form m
Figure 10.15: Coordinate labels are indicated for Problem 10.17.
L = =
1 2 1 mr˙ + m(r˙ 2 + r2 θ˙ 2 ) − mg(` − d − r ) − mgr cos θ, 2 2 1 mr˙ 2 + mr2 θ˙ 2 + mgr (1 − cos θ ) − mg(` − d). 2
The EulerLagrange equations for this system yield d ∂L ∂L 0 = − , dt ∂r˙ ∂r d = (2mr˙ ) + mr θ˙ 2 − mg(1 − cos θ ). dt d ∂L ∂L 0 = − , dt ∂θ˙ ∂θ d 2 ˙ = mr θ − mgr sin θ. dt Therefore, we have the equations of motion are
= 2m¨r + mr θ˙ 2 − mg(1 − cos θ ), 0 = 2mrr˙ θ˙ + mr2 θ¨ − mgr sin θ. 0
11 Problems in Higher Dimensions 1. A rectangular plate 0 ≤ x ≤ L 0 ≤ y ≤ H with heat diffusivity constant k is insulated on the edges y = 0, H and is kept at constant zero temperature on the other two edges. Assuming an initial temperature of u( x, y, 0) = f ( x, y), use separation of variables to find the general solution. For this problem the heat equation takes the form ut = k(u xx + uyy ). Let u( x, y, t) = X ( x )Y (y) T (t). Then, XYT 0 T0 kT
k( X 00 YT + XY 00 T ). X 00 Y 00 + X Y
= = ≡ −λ − µ.
This gives the system of equations plus boundary conditions as X 00 + λX 00
Y + µY T0
= 0,
X (0) = X ( L) = 0,
= 0,
Y 0 (0) = Y 0 ( H ) = 0,
= −k(λ + µ) T.
The last equation is easily integrated, giving T (t) = Ae−k(λ+µ)t . The solutions to the two boundary values problems follow the examples in the text. The general solutions are sines and cosines. The conditions on te left end of the interval determine which function and the condition on the right side of the interval give the eigenvalues. Thus, we have nπ 2 nπx Xn ( x ) = sin , λn = , n = 1, 2, . . . , L L mπ 2 mπy Ym ( x ) = cos , µm = , m = 0, 1, 2, . . . . H H From these results we have the product solutions unm ( x, y, t) = sin
nπx mπy −k(λn +µm )t cos e , L H
n = 1, 2, . . . ,
m = 0, 1, 2, . . . ,
and the general solution takes the form ∞
u( x, y, t) =
∞
∑ ∑
n =1 m =0
Anm sin
nπx mπy −k(λn +µm )t cos e . L H
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mathematical methods for physicists
Next, we use the initial condition u( x, y, 0) = f ( x, y), ∞
f ( x, y) =
∞
∑ ∑
nπx mπy cos . L H
Anm sin
n =1 m =0
This is a double Fourier sine series. Given f ( x, y) we can obtain the Fourier coefficients as Anm =
4 LH
Z HZ L 0
f ( x, y) sin
0
nπx mπy cos dxdy, L H
where n = 1, 2, . . . , m = 0, 1, 2, . . . . 2. Solve the following problem: u xx + uyy + uzz = 0,
0 < x < 2π,
u( x, y, 0) = sin x sin y,
0 < y < π,
0 < z < 1,
u( x, y, z) = 0 on the other faces.
Let u( x, y, z) = X ( x )Y (y) Z (z). Then, separation of variables gives X 00 YZ + XY 00 Z + XYZ 00 Y 00 X 00 + X Y
= 0 Z 00 Z Z 00 − . Z
= −
−λ − µ ≡
This gives the system of equations plus boundary conditions as X 00 + λX
= 0,
X (0) = X (2π ) = 0,
Y + µY
= 0,
Y (0) = Y (π ) = 0,
Z − (λ + µ) Z
= 0,
Z (1) = 0.
00
00
The first two give eigenvalue problems whose solutions are nx , 2 sin my,
Xn ( x )
= sin
Ym ( x )
=
λn =
n 2
2 µ m = m2 ,
,
n = 1, 2, . . . ,
m = 1, 2, . . . .
The third equation can be solved in terms of exponentials or hyperbolic functions. The boundary condition, Z (1) = 0, gives the solution as p Z (z) = sinh( λn + µm (1 − z)). From these results we have the product solutions unm ( x, y, z) = sin
p nx sin my sinh( λn + µm (1 − z)), 2
n, m = 1, 2, . . . ,
and the general solution takes the form ∞
u( x, y, t) =
∞
∑ ∑
n =1 m =1
Anm sin
p nx sin my sinh( λn + µm (1 − z)). 2
The other boundary condition is u( x, y, 0) = sin x sin y. We could evaluate the series at z = 0 and get the Fourier coefficients for a double Fourier sine
problems in higher dimensions
359
series. However, this boundary condition is an eigenfunction corresponding to n = 2 and m = 1 with eigenvalues λ2 = 1, and µ1 = 1. Thus, u( x, y, 0)
= sin x sin y =
p A21 sin x sin y sinh( λ2 + µ1 ) √ A21 sin x sin y sinh 2.
= √ Therefore, A21 = 1/ sinh 2 and all of the other coefficients vanish. The solution is then given as √ nx sinh( 2(1 − z)) √ u( x, y, z) = sin sin my . 2 sinh 2 3. Consider Laplace’s Equation on the unit square, u xx + uyy = 0, 0 ≤ x, y ≤ 1. Let u(0, y) = 0, u(1, y) = 0 for 0 < y < 1 and uy ( x, 0) = 0 for 0 < y < 1. Carry out the needed separation of variables and write the product solutions satisfying these boundary conditions. We insert the solution u( x, y) = X ( x )Y (y) into the Cartesian form of Laplace’s Equation, u xx + uyy = 0, and perform a separation of variables. We obtain X 00 Y + XY 00 Y 00 − Y
= 0 = +
X 00 = − λ2 . X
This gives the system X 00 + λ2 X 00
2
Y −λ Y
= 0,
X (0) = X (1) = 0,
= 0. Y 0 (0) = 0.
The first boundary value problem gives the usual eigenfunctions and eigenvalues Xn ( x ) = sin nπx,
λn = nπ,
n = 1, 2, . . . .
Solutions of the second problem can be written as Yn (y) = a cosh λn y + b sinh λn y. The boundary condition at y = 0 is satisfied by Yn (y) = cosh nπy. Therefore, the product solutions are un ( x, y) = sin nπx cosh nπy,
n = 1, 2, . . . .
4. Consider a cylinder of height H and radius a. a. Write Laplace’s Equation for this cylinder in cylindrical coordinates. Laplace’s equation in Cylindrical coordinates is given by 0
= ∇2 u ∂u 1 ∂2 u ∂2 u 1 ∂ = r + 2 2 + 2. r ∂r ∂r r ∂θ ∂z
y 1
∇2 u = 0
u=0 0
u=0
1
0
x
uy = 0 Figure 11.1: The square plate for problem 10.3.
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mathematical methods for physicists
b. Carry out the separation of variables and obtain the three ordinary differential equations that result from this problem. We insert u(r, θ, z) = R(r )Θ(θ ) Z (z) into Laplace’s equation and rearrange the equation. 0
−
Z 00 Z
= ∇2 ( RΘZ ) 1 ∂ ∂( RΘZ ) 1 ∂2 ( RΘZ ) ∂2 ( RΘZ ) = + r + 2 r ∂r ∂r r ∂θ 2 ∂z2 ΘZ RZ = (rR0 )0 + 2 Θ00 + RΘZ 00 r r 1 1 Θ00 0 0 = (rR ) + 2 . rR r Θ
The left hand side of the equation is a function of z and the right hand side is a function of r and θ. Therefore, these expressions equal a constant. We let the constant be −λ. Then, we have Z 00 − λZ
= 0,
r (rR0 )0 + λr2 R
= −
Θ00 . Θ
The last equation is in separated form. We have a function of r is equal to a function of θ. Therefore, the left and right side of the equation are constant functions, which we set to µ2 . The resulting system of equations is Z 00 − λZ
= 0,
00
= 0,
2
Θ +µ Θ 2
00
0
2
2
r R + rR + (λr − µ ) R
= 0.
c. What kind of boundary conditions could be satisfied in this problem in the independent variables? The boundary conditions we need to impose on these equations are periodic boundary conditions in θ, and finite solutions at the origin (r = 0). The periodic boundary conditions will give µ = m as integers. Anything else depends on the sign of λ. If homogeneous boundary conditions are given on the top and bottom of the cylinder, then λ < 0 and Z (z) will be a combination of sines and cosines. The radial equation, which we recognize as a Bessel equation with the wrong sign for λ, will have solutions known as modified Bessel functions. Modified Bessel functions, which are not in the current version of the book, satisfy the differential equation x2 y00 + xy0 − ( x2 + ν2 )y = 0. Solutions of the first and second kind are given by Iν ( x ) = i−ν Jν (ix ) and Kν ( x ), respectively. If a homogeneous boundary condition is given at r = a, then λ > 0 and the radial solutions will be Bessel functions of the first kind. One
problems in higher dimensions
361
then can impose a homogeneous solution on the top r bottom surface and Z (z) would be of hyperbolic function form. 5. Consider a square drum of side s and a circular drum of radius a. a. Rank the modes corresponding to the first six frequencies for each. For the rectangular membrane we obtained the angular frequencies r nπ 2 mπ 2 ωnm = c + , n, m = 1, 2, . . . . L H For a square of side s, the frequencies become r c nπ 2 mπ 2 νnm = + , 2π s s cp 2 n + m2 n, m = 1, 2, . . . . = 2s The modes are given in Table 11.1. n 1 1 2 2 3 1 3 2 4 1
√
m 1 2 1 2 1 3 2 3 1 4
n2 + m2 √ 2 √ 5 √ 5 √ 8 √ 10 √ 10 √ 13 √ 13 √ 17 √ 17
νnm ν0 1.5811ν0 1.5811ν0 2.0000ν0 2.2361ν0 2.2361ν0 2.5495ν0 2.5495ν0 2.9155ν0 2.9155ν0
Table 11.1: The first modes of a square membrane in Problem 10.5
For a circular membrane we have the angular frequencies are proportional to the zeros of Bessel functions, ωmn =
jmn c, a
or
jmn c. 2πa The first six zeros and the corresponding modes are shown in Table 11.2. νmn =
m 0 1 2 0 1 2
n 1 1 1 2 2 2
jmn 2.405 3.832 5.136 5.520 7.016 8.417
νmn ν0 1.5933ν0 2.1356ν0 2.2952ν0 2.9173ν0 3.4998ν0
Table 11.2: The first modes of a circular membrane in Problem 10.5
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mathematical methods for physicists
b. Write each frequency (in Hz) in terms of the fundamental (i.e., the lowest frequency.) These are shown in Tables 11.1 and 11.2. For the rectangular membrane we define √ c ν0 = 2 . 2s For the circular membrane we define c ν0 = 2.405 . 2πa c. What must the lengths of the sides of the square drum be to have the same fundamental frequency? (Assume that c = 1.0 for each one.) We just need to equate the lowest modes and solve for s. For the square membrane it is the (1, 1) mode, for the circular membrane it is the (0, 1) mode. j01 c 2πa
=
s
=
c√ 2 2s √ 2π a. j01
6. We presented the full solution of the vibrating rectangular membrane in Equation (11.37). Finish the solution to the vibrating circular membrane by writing out a similar full solution. The product solutions for the vibrating circular membrane were given by ) )( ( jmn cos mθ cos ωmn t Jm ( u(r, θ, t) = r ), a sin mθ sin ωmn t leading to four different types of product solutions. The angular frequency depends on the zeros of the Bessel functions, jmn c, m = 0, 1, . . . , n = 1, 2, . . . . a Therefore, the general solution can be written in the form ωmn =
u(r, θ, t) ∞ ∞
=
jmn r jmn r ∑ ∑ Amn Jm a cos mθ + Bmn Jm a sin mθ cos ωmn t m =0 n =1 ∞ ∞ jmn r jmn r cos mθ + Dmn Jm sin mθ sin ωmn t + ∑ ∑ Cmn Jm a a m =0 n =1
In order to determine the expansion coefficients, we need to specify initial conditions. Let u(r, θ, 0) = f (r, θ ),
ut (r, θ, 0) = g(r, θ ).
These conditions give ∞ ∞ jmn r jmn r f (r, θ ) = ∑ ∑ Amn Jm cos mθ + Bmn Jm sin mθ . a a m =0 n =1 ∞ ∞ jmn r jmn r g(r, θ ) = ∑ ∑ ωmn Cmn Jm cos mθ + Dmn Jm sin mθ . a a m =0 n =1
problems in higher dimensions
In order to extract the coefficients from the first double sum, we write ∞ a0 (r ) + ∑ [ am (r ) cos mθ + bm (r ) sin mθ ], 2 m =1
f (r, θ ) = where
∞
a0 (r )
= 2
∑
A0n J0
n =1 ∞
a m (r )
=
bm ( r )
=
j0n r a
jmn r , ∑ a n =1 ∞ jmn r ∑ Bmn Jm a , n =1
m = 1, 2, . . . ,
Amn Jm
m = 1, 2, . . . .
Since f (r, θ ) is given by a Fourier series expansion, we have a m (r )
=
bm ( r )
=
1 2π f (r, θ ) cos mθ dθ, π 0 Z 1 2π f (r, θ ) sin mθ dθ, π 0 Z
m = 0, 1, . . . m = 1, 2, . . . .
Knowing these coefficients, we just need to recall how to obtain FourierBessel coefficients. For f (r ) defined on r ∈ [0, a], the FourierBessel expansion is given by ∞ jmn r f (r ) = ∑ cn Jm ( ), a n =1 where the FourierBessel coefficients are given as cn =
Z a
2 a2 [ Jm+1 ( jmn )]
2
0
r f (r ) Jm (
jmn r )r dr. a
Therefore,
=
A0n
= =
Amn
= =
Bmn
=
Z a
1 a2
[ J1 ( j0n )] 1
πa2 a2
[ J1 ( j0n )] 2
0
ra0 (r ) J0 (
Z 2π Z a 2
[ Jm+1 ( jmn )] 2
πa2 a2
2
0
0
Z a 2
0
r f (r, θ ) J0 (
ram (r ) Jm (
Z 2π Z a 2
[ Jm+1 ( jmn )] Z 2
[ Jm+1 ( jmn )] 2
j0n r ) dr a
2
0
0
rbm (r ) Jm (
Z 2π Z a
πa2 [ Jm+1 ( jmn )]2
0
0
jmn r ) dr a
r f (r, θ ) Jm (
a 0
j0n r ) drdθ. a
jmn r ) cos mθ drdθ. a
jmn r ) dr a
r f (r, θ ) Jm (
jmn r ) sin mθ drdθ. a
Similarly, one can use the second initial condition to obtain the remaining coefficients. C0n
=
1 πa2 ω0n [ J1 ( j0n )]2
Z 2π Z a 0
0
rg(r, θ ) J0 (
j0n r ) drdθ. a
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z ux = 0 u = −10 uy = 0
uy = 0 ux = 0
y
x u=0 Figure 11.2: The cube for problem 10.7.
Cmn
=
Dmn
=
Z 2π Z a 2
rg(r, θ ) Jm (
jmn r ) cos mθ drdθ. a
2
rg(r, θ ) Jm (
jmn r ) sin mθ drdθ. a
2
0 πa2 ωmn [ Jm+1 ( jmn )] 0 Z 2π Z a 2
πa2 ω
mn
[ Jm+1 ( jmn )]
0
0
7. A copper cube 10.0 cm on a side is heated to 100◦ C. The block is placed on a surface that is kept at 0◦ C. The sides of the block are insulated, so the normal derivatives on the sides are zero. Heat flows from the top of the block to the air governed by the gradient uz = −10◦ C/m. Determine the temperature of the block at its center after 1.0 minute. Note that the thermal diffusivity is given by k = ρcKp , where K is the thermal conductivity, ρ is the density, and c p is the specific heat capacity. This is a heat conduction problem and can be formulated using the heat equation ut = k u xx + uyy + uzz . The boundary conditions are given by uz ( x, y, s, t) = −0.1◦ C/cm,
u( x, y, 0, t) = 0,
0 ≤ x, y ≤ s,
u x (0, y, z, t) = 0,
u x (0, y, z, t) = 0,
0 ≤ y, z ≤ s,
uy ( x, 0, z, t) = 0,
uy ( x, 0, z, t) = 0,
0 ≤ x, z ≤ s,
and the initial condition is given by u( x, y, z, 0) = 100◦ C. Note that a hint might be needed for this problem.
The problem does not have homogeneous boundary conditions. However, this can be remedied by noting that the solution takes the form u( x, y, z, t) = uh ( x, y, z, t) + f (z). Inserting this form into the initialboundary value problem, we have that uh ( x, y, z, t) satisfies the differential equation uht = k uh xx + uhyy + uhzz + f 00 (z) with boundary conditions uh ( x, y, 0, t) = − f (0),
uhz ( x, y, s, t) = − f 0 (0) − 0.1,
0 ≤ x, y ≤ s,
uh x (0, y, z, t) = 0,
uh x (0, y, z, t) = 0,
0 ≤ y, z ≤ s,
uhy ( x, 0, z, t) = 0,
uhy ( x, 0, z, t) = 0,
0 ≤ x, z ≤ s,
and the initial condition is given by uh ( x, y, z, 0) = 100 − f (z) We can make the boundary conditions homogeneous if f (0) = 0 and = −0.1. Also, if f 00 (z) = 0, the heat equation is homogeneous. So, we assume that f (z) = Az + B. The boundary conditions give B = 0 and A = −0.1. This yields f (z) = −0.1z. f 0 (0)
problems in higher dimensions
So far, the general solution is given by u( x, y, z, t) = uh ( x, y, z, t) − 0.1z, where uh ( x, y, z, t) satisfies the initialboundary value problem uht = k uh xx + uhyy + uhzz , 0 < x, y, z < s, t > 0, uh ( x, y, 0, t)
= 0,
uhz ( x, y, s, t) = 0,
0 ≤ x, y ≤ s,
uh x (0, y, z, t)
= 0,
uh x (0, y, z, t) = 0,
0 ≤ y, z ≤ s,
uhy ( x, 0, z, t)
= 0,
uhy ( x, 0, z, t) = 0,
0 ≤ x, z ≤ s,
uh ( x, y, z, 0)
= 100 + 0.1z.
Now, we can proceed to solve this heat equation using separation of variables. Let uh ( x, y, z, t) = X ( x )Y (y) Z (z) T (t). Then, we find Y 00 Z 00 X 00 1 T0 + + . = k T X Y Z {z} {z} {z} − λ2
µ2
− ν2
Setting the individual spatial terms equal to the given constants, we have the set of ordinary differential equations T0
= −k(λ2 + µ2 + ν2 ) T,
X 00 + λ2 X
= 0,
X 0 (0) = X 0 (s) = 0,
Y 00 + µ2 Y
= 0,
Y 0 (0) = Y 0 (s) = 0,
Z 00 + ν2 Z
= 0,
Z (0) = Z 0 (s) = 0.
The boundary conditions give the eigenfunctions Xn ( x )
nπx mπy , Ym (y) = cos , n, m, = 0, 1, . . . , s s (2` + 1))πz , ` = 0, 1, . . . , sin 2s
= cos
Z (z)
=
and eigenvalues λn = given by Tnm` (t)
nπ s ,
µm = 2
2
mπ s ,
ν` =
(2`+1)π . 2s
The time dependence is
2
= e−k(λn +µm +ν` )t = e−(n
2 + m2 +(2`+1)2 /4) π 2 kt/s2
,
n, m, ` = 0, 1, . . . .
We can now form the general solution of the homogeneous problem. ∞
uh ( x, y, z, t) =
∑
Anm` cos
n,m,`=0
mπy (2` + 1)πz −(n2 +m2 +(2`+1)2 /4)π2 kt/s2 nπx cos sin e . s s 2s
The initial condition then can be used to find the expansion coefficients. Namely, we have for t = 0, ∞
100 + 0.1z =
∑
n,m,`=0
Anm` cos
nπx mπy (2` + 1)πz cos sin . s s 2s
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Since this contains a double cosine series in x and y and the left hand side is independent of x and y, only the Aoo` terms survive and the triple sum reduces to a single sum, ∞
100 + 0.1z =
∑ A00` sin
`=0
Since
Z s
(2` + 1)πz . 2s
(2` + 1)πz s dz = , 2s 2 0 we can obtain the Fourier coefficients as Z (2` + 1)πz 2 s (100 + 0.1z) sin dz, ` = 0, 1, . . . A00` = s 0 2s 2 2s (2` + 1)πz = − (100 + 0.1z) cos s (2` + 1) π 2s s 0.4s2 (2` + 1)πz + sin 2 2 2s (2` + 1) π 0 (2` + 1) π 200s 2 0.4s2 sin = + s (2` + 1) π 2 (2` + 1)2 π 2 sin2
So, the solution of the original problem can be written as u( x, y, z, t) = −0.1z +
2 ∞ (2` + 1))πz −(2`+1)2 )π2 kt/4s2 e , ∑ b` sin s `= 2s 0
where
200s 0.4s2 . + (−1)` (2` + 1) π (2` + 1)2 π 2 In order to evaluate the temperature of the block at its center after 1.0 minute, we need the thermal diffusivity. We can look it up, to find it is around 1.15 cm2 /s, or we can use k = ρcKp , where K = 400 W/m◦ C is the b` =
Figure 11.3: A plot of the temperature at the center of the cube in Problem 11.7 for t = 0, 1, . . . , 9 s.
thermal conductivity, ρ = 8.96 g/cm3 is the density, and c p = 0.385 J/g◦ C is the specific heat capacity. In tis case, we find essentially the same value for temperatures around 25◦ C to 100◦ C. Thus, we can use k = 1.15cm2 /s = 1.15 × 10−4 m2 /s. As most of the units are in centimeters and seconds, we will stick with those. Thus, we want to find u(5, 5, 5, 60). ∞
u(5, 5, 5, 60) = −0.5 + 0.2
∑ b` sin
`=0
where
2000 40 + (−1)` . (2` + 1) π (2` + 1)2 π 2 In Figure 11.3 is shown the temperature at the center of the cube in Problem 11.7 for t = 0, 1, . . . , 9 s for 201 terms. The initial temperature is seen to be 100◦ C, though the convergence is slow. However, convergence for t > 0 seems to be quick. In Figure 11.4 is shown the temperature at the center of the cube in Problem 11.7 for t = 10, 20 . . . , 100 s for 201 terms. One can find the value of the temperature at the center after one minute using only one or two terms. It is u(5, 5, 5, 60) = 17.0◦ C. b` =
Figure 11.4: A plot of the temperature at the center of the cube in Problem 11.7 for t = 10, . . . , 100 s.
(2` + 1))π −1.70(2`+1)2 )π2 e , 4
problems in higher dimensions
8. Consider a spherical balloon of radius a. Small deformations on the surface can produce waves on the balloon’s surface. a. Write the wave equation in spherical polar coordinates. (Note: ρ is constant!) The wave equation in spherical coordinates for ρ = a is constant and u = u(θ, φ) is given by utt = c
2
1 ∂ 2 a sin θ ∂θ
∂u sin θ ∂θ
1 ∂2 u . + 2 2 a sin θ ∂φ2
b. Carry out a separation of variables and find the product solutions for this problem. Let u(θ, φ) = Θ(θ )Φ(φ) T (t). Then, we separate out the time dependence to find 1 1 00 2 0 0 00 ΘΦT = c sin θΘ Φ + ΘΦ T a2 sin θ a2 sin2 θ 1 1 Φ00 a2 T 00 0 0 = sin θΘ + = − λ2 . sin θΘ c2 T sin2 θ Φ Therefore, T 00 + ω 2 T = 0, If we set
ω=
cλ . a
Φ00 = − µ2 , Φ
then
(sin θΘ0 )0 − µ2 Θ = −λ2 sin2 θΘ and Φ00 + µ2 Φ = 0. Imposing periodic boundary conditions, Φ(0) = Φ(2π ) and Φ0 (0) = Φ0 (2π ), we have that µ = m = 0, 1, 2, . . . . The solutions are cos mφ, sin mφ. The equation for Θ is then recognized as the equation for associated Legendre polynomials. The eigenvalue problem has solutions Θ(θ ) = Pnm (cos θ ),
n = 0, 1, . . . ,
, m = −n, . . . , n,
λ n = n ( n + 1).
Thus, the product solutions are ( u(θ, φt) = where ωnm =
c a
p
cos ωnm t sin ωnm t
n ( n + 1).
)(
cos mθ sin mθ
) Pnm (cos θ ),
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Table 11.3: In these figures we show the first six nodal curves of Ynm (θ, φ)2 Along the first column (m = 0) are the zonal harmonics seen as n horizontal circles. Along the top diagonal (m = n) are the sectional harmonics. These look like orange sections formed from m vertical circles. The remaining harmonics are tesseral harmonics. They look like a checkerboard pattern formed from intersections of n − m horizontal circles and m vertical circles.
m=0
m=1
m=2
n=0
n=1
n=2 c. Describe the nodal curves for the first six modes. The nodal curves are typically ordered by the frequency. The frequencies only depend on n, so there is a degeneracy in the counting. One needs only look at the nodal curves for spherical harmonics. The nodal curves of the spherical harmonics are shown in Table 11.3. d. For each mode determine the frequency of oscillation (in Hz) assuming c = 1.0 m/s. For c = 1.0 m/s and a = 0.2 m the frequencies are q √ √ ωnm = 5 n(n + 1) = 0, 5 2, 5 6 for n = 0, 1, 2. 9. Consider a circular cylinder of radius R = 4.00 cm and height H = 20.0 cm that obeys the steadystate heat equation 1 urr + ur + uzz = 0. r
u = 20◦ C
z R = 4.00 cm
Find the temperature distribution, u(r, z), given that u(r, 0) = 0◦ C, u(r, 20) = 20◦ C, and heat is lost through the sides due to Newton’s Law of Cooling,
[ur + hu]r=4 = 0, ur + hu = 0
H = 20.0 cm
y x
u = 0◦ C
Figure 11.5: Geometry for the cylinder in Problem 9.
for h = 1.0 cm−1 . Let u(r, z) = φ(r ) Z (z). Then, we have 1 φ00 Z + φ0 Z + φZ 00 = 0. r Separating variables and introducing a separation constant, we obtain 1 00 1 0 Z 00 (φ + φ ) = − = − λ2 . φ r Z This gives the differential equations Z 00 − λ2 Z 1 φ00 + φ0 + λ2 φ r
= 0,
Z (0) = 0,
= 0,
[φ0 + hφ]r= R = 0.
The second equation is a Bessel equation, x2 y00 + xy0 + (λ2 x2 − µ)y = 0
problems in higher dimensions
369
for µ = 0. Thus, the solutions finite at the origin are φ(r ) = J0 (λr ) satisfying λJ00 (λR) + J0 (λR) = 0. The solution of this transcendental equation will give the eigenvalues, λn . In Figure 11.6 we show the plot of f ( x ) = xJ00 ( x ) + J0 ( x ) = xJ1 ( x ) + J0 ( x ), where x = λR. If we call the roots jn , satisfying jn J00 ( jn ) + J0 ( jn ) = 0, then the eigenvalues are given in terms of these roots, λn = jn /R. The first four roots are λ = 1.2558, 4.0795, 7.1558, 10.2710. Figure 11.6: A plot of f ( x ) = xJ00 ( x ) + J0 ( x ) = xJ1 ( x ) + J0 ( x ) in Problem 11.9 showing the location of the zeros.
The first differential equation and boundary condition can be solved to obtain Z (z) = sinh λz. Therefore, the general solution is given by ∞
u(r, z) =
∑
An J0 (λn r ) sinh λn z.
n =1
The only condition left to be satisfied is the condition on the top of the cylinder, ∞
∑ ( An sinh λn H ) J0 (λn r).
u(r, H ) =
n =1
This is a FourierBessel series and we need to find the Fourier coefficients. The Fourier coefficients can be found by using Green’s Identity to verify orthogonality. The normalization constant is a little more challenging. This is similar to the derivation in Problem 6.18c. We begin using the SturmLiouville form d d L J0 (λx ) = x J0 (λx ) = −λ2 xJ0 (λx ). dx dx Then
( µ2 − λ2 ) =
Z R 0
=
Z R 0
xJ0 (µx ) J0 (λx ) dx
[ J0 (µx )L( J0 (λx )) − J0 (λx )L( J0 (µx ))] dx
Z R d
[λxJ0 (µx ) J00 (λx ) − µxJ0 (λx ) J00 (µx )] dx dx [λxJ0 (µx ) J00 (λx ) − µxJ0 (λx ) J00 (µx )]0R 0
=
= λRJ0 (µR) J00 (λR) − µRJ0 (λR) J00 (µR).
The determination of the normalization constant might be challenging for students. One first proves the general relation Z R i 1h 2 0 xJ02 (λx ) dx = R [ J0 (λR)]2 + R2 J02 (λR) . 2 0 Using the boundary condition for this problem, we will obtain Z R R2 1 xJ02 (λx ) dx = 1 + 2 J02 (λR). 2 λ 0
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Using the boundary condition, λJ00 (λR) = − J0 (λR), gives for µ 6= λ
( µ2 − λ2 )
Z R 0
xJ0 (µx ) J0 (λx ) dx
= λRJ0 (µR) J00 (λR) − µRJ0 (λR) J00 (µR) = − RJ0 (µR) J0 (λR) + RJ0 (λR) J0 (µR) = 0.
Therefore,
Z R 0
xJ0 (µx ) J0 (λx ) dx = 0,
We still need to evaluate cients. We just found that Z R 0
RR 0
xJ0 (λx ) J0 (µx ) dx =
µ 6= λ.
xJ02 (λx ) dx before finding the Fourier coeffiλRJ0 (µR) J00 (λR) − µRJ0 (λR) J00 (µR) . µ2 − λ2
We next take the limit µ → λ of this result. However, this in an indeterminate form for which we need to apply L’Hopital’s Rule. Therefore, we differentiate the numerator and denominator with respect to µ to find Z R 0
xJ02 (λx ) dx
=
lim
λRJ0 (µR) J p0 (λR) − µRJ0 (λR) J00 (µR) µ2 − λ2
µ→λ
=
lim
µ→λ
λR2 J00 (µR) J00 (λR) − RJ0 (λR) J00 (µR) − µR2 J0 (λR) J000 (µR) . 2µ
In order to evaluate the numerator, we need an expression for J000 (µR). Bessel’s equation and the Chain Rule give 0
=
d d2 2 2 x + x + µ x J0 (µx ) dx dx2 2
= µ2 x2 J000 (µx ) + µxJ00 (µx ) + µ2 x2 J0 (µx ). Setting x = R, µ2 R2 J000 (µR) + µRJ00 (µR) + µ2 R2 J0 (µR), or J000 (µ)
1 0 =− J (µR) + J0 (µR) . µR 0
Inserting this result into the computation of the normalization constant, we have Z R 0
=
= =
xJ02 (λx ) dx
lim
µ→λ
lim
λR2 J00 (µR) J00 (λR) − RJ0 (λR) J00 (µR) − µR2 J0 (λR) J000 (µR) 2µ h i 1 0 λR2 J00 (µR) J00 (λR) − RJ0 (λR) J00 (µR) + µR2 J0 (λR) µR J0 (µR) + J0 (µR) 2µ
µ→λ
1h 2
i
R2 [ J00 (λR)]2 + R2 J02 (λR) .
problems in higher dimensions
So far, this is general and applies for a variety of boundary conditions. Since λJ00 (λR) = − J0 (λR), we have the result Z R 0
xJ02 (λx ) dx
In general, if
R2 = 2
1 1+ 2 λ
J02 (λR).
∞
f (r ) =
∑ an J0 (λn r)r dr
n =1
jn J00 ( jn )
where + J0 ( jn ) = 0 and λn = jn /R, then the FourierBessel coefficients are given by an =
2 jn2 R2 J02 ( jn ) jn2 + R
Z R 0
f (r ) J0 (
jn r )r dr. R
Returning to the problem, we have the FourierBessel series ∞
u(r, H ) =
∑ ( An sinh λn H ) J0 (λn r) = 20.
n =1
Therefore, we can solve for the FourierBessel coefficients An sinh λn H =
jn2 40 R2 J02 ( jn ) jn2 + R
Z R 0
J0 (
jn r )r dr. R
The integral can be done. We make a substitution y = λn r in the integral, Z R 0
J0 (λn r )r dr
= = =
λn R 1 J0 (y)y dy 2 λn 0 Z jn 1 d [yJ1 (y)] dy 2 λn 0 dy
Z
R2 J ( jn ). jn 1
The coefficients can be found as An sinh λn H
= = = =
2 R 40 jn2 J1 ( jn ) R2 J02 ( jn ) jn2 + R jn 40jn J1 ( jn ) 2 ( jn + R) J02 ( jn ) 40jn J ( jn ) ] [ 0 ( jn2 + R) J02 ( jn ) jn 40 . ( jn2 + R) J0 ( jn )
This give the solution to the problem as ∞
u(r, z) =
∑
2 n=1 ( jn
jn jn J0 ( r ) sinh( z). jn R R R) J0 ( jn ) sinh( R H ) 40
+
10. The spherical surface of a homogeneous ball of radius 1.0 is maintained at zero temperature. It has an initial temperature distribution u(ρ, 0) = 100◦
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mathematical methods for physicists
C. Assuming a heat diffusivity constant k, find the temperature throughout the sphere, u(ρ, t). As indicated, the temperature is independent of the angles and the problem in reduced spherical coordinates is given by k 2 2 ∂u ut = k ∇ u = 2 ρ , ∂ρ ρ u(1, t) = 0, t > 0, u(ρ, 0) = 100◦ C,
0 < ρ < 1.
We perform a separation of variables using u(ρ, t) = R(ρ) T (t). Then, we have RT 0
=
T0 kT
=
k 2 0 0 (ρ R ) T ρ2 1 ( ρ2 R 0 ) 0 = − λ2 . 2 ρ R
This gives the differential equations T0 0
= −λ2 kT, = (ρ2 R0 )0 + λ2 ρ2 R,
R(1) = 0. 2
This problem involves spherical Bessel functions.
The first equation is readily solved, T (t) = e−λ kt . The radial equation is a special case of the spherical Bessel function equation, r2 R00 + 2rR0 + (λ2 r2 − n(n + 1)) R = 0. Solutions are found in terms of Bessel functions of half integral order. Those well behaved at the origin are given by r π jn (λr ) = J 1 (λr ). 2λr n+ 2 For our problem, n = 0. So, r R(ρ) = j0 (λr ) =
π J 1 (λr ). 2λr 2
The eigenvalues satisfy J 1 (λn ) = 0. 2 The general solution is given as ∞
u(ρ, t) =
∑ an j0 (λn ρ)e−λn kt . 2
n =1
The initial condition gives the series expansion ∞
100 =
∑ an j0 (λn ρ).
n =1
We need to obtain the expansion coefficients. Since the radial equation is in SturmLiouville form, similar to that of the Bessel function, then we expect the expansion coefficients to be found as R1 100j0 (λn ρ)ρ2 dρ an = 0R 1 . 2 2 0 j0 ( λn ρ ) ρ dρ
problems in higher dimensions
From Problem 6.19, we found that r J1/2 ( x ) =
2 sin x. πx
Therefore,
sin x . x Furthermore, we obtain the eigenvalues from j0 ( x ) = 0, or sin x = 0. So, we find that λn = nπ, n = 1, 2, . . . . Letting x = nπρ, we have j0 ( x ) =
R1
an
100j0 (nπρ)ρ2 dρ R1 2 2 0 j0 ( nπρ ) ρ dρ R 1 sin(nπρ) ρ dρ nπ = 100 R0 1 sin2 (nπρ) dρ 0 n2 π 2 R1 sin(nπρ)ρ dρ = 100nπ R01 2 0 sin ( nπρ ) dρ
=
0
= 200nπ
Z 1 0
sin(nπρ)ρ dρ
=
1 1 200nπ − ρ cos(nπρ) + 2 2 sin(nπρ) nπ n π
1 0
= −200 cos(nπ ) = 200(−1)n+1 . The solution is therefore, ∞
u(ρ, t) = 200
∑ (−1)n+1 j0 (nπρ)e−n π kt , 2
2
n =1
or
∞
u(ρ, t) = 200
∑ (−1)n+1
n =1
sin(nπρ) −n2 π2 kt e . nπρ
11. Determine the steadystate temperature of a spherical ball maintained at the temperature u( x, y, z) = x2 + 2y2 + 3z2 ,
ρ = 1.
[Hint: Rewrite the problem in spherical coordinates and use the properties of spherical harmonics.] From the text we have that solutions can be written as ∞
u(ρ, θ, φ) =
`
∑ ∑
a`m ρ` P`m (cos θ )eimφ .
`=0 m=−`
[One can take real part to get a real valued solution.] At ρ = 1, we have x2 + 2y2 + 3z2 =
∞
`
∑ ∑
`=0 m=−`
a`m P`m (cos θ )eimφ .
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mathematical methods for physicists
So, we need only write the Cartesian form of the initial condition in terms of associated Legendre functions. Using the the spherical coordinate transformation and the table of associated Legendre functions, we have x2 + 2y2 + 3z2
= sin2 θ cos2 φ + 2 sin2 θ sin2 φ + 3 cos2 θ 1 2 2 = P (cos θ ) cos2 φ + P22 (cos θ ) sin2 φ + 2P20 (cos θ ) + P00 (cos θ ) 3 2 3 1 2 2 = P (cos θ )(1 + cos 2φ) + P22 (cos θ )(1 − cos 2φ) + 2P20 (cos θ ) + P00 (cos θ ) 6 2 6 1 2 1 2 = P (cos θ ) − P2 (cos θ ) cos 2φ + 2P20 (cos θ ) + P00 (cos θ ). 2 2 6 Comparing this with the expansion for the boundary condition, we can identify a finite number of nonzero coefficients. The general solution is formed by inserting factors of ρ` in each term. Therefore, we have u(ρ, θ, φ)
= =
1 2 2 1 ρ P (cos θ ) − ρ2 P22 (cos θ ) cos 2φ + 2ρ2 P20 (cos θ ) + P00 (cos θ ) 2 2 6 3 1 ρ2 − cos2 φ sin2 θ + ρ2 (3 cos2 θ − 1) + 1. 2 2
12. Find the Green’s function for homogeneous fixed values on the boundary of the quarter plane x > 0, y > 0, for Poisson’s Equation using the infinite plane Green’s function for Poisson’s equation. Use the method of images. This is similar to finding the Green’s function for the half plane. We imagine that the problem is one for determining the electric potential due to a finite charge distribution. We locate image charges so that the potential is constant along the axes, or field lines are perpendicular to the axes. Figure 11.7: The source “charge” and image “charges” for the Green’s function for the quarter plane. Imagine each pair of two opposite charges forming dipoles. The resulting electric field lines are depicted indicating that the electric potential, or Green’s function, is constant along y = 0 and x = 0.
y G (0, y; ξ, η ) = 0
(− x, y)
( x, y) −
+ G ( x, 0; ξ, η ) = 0 x
+ (− x, −y)
− ( x, −y)
In Figure 11.7 we show the needed source charge and its image charges for the Green’s function for the quarter plane. The resulting electric field lines are depicted indicating that the electric potential, or Green’s function,
problems in higher dimensions
is constant along y = 0 and x = 0. Three image charges are needed. We place the first charge at ( x, y). The second charge at ( x, −y) gives G ( x, 0; ξ, η ) = 0. Adding a third charge at (− x, y) is not enough to force G (0, y; ξ, η ) = 0 due to the presence of the second charge. Therefore, we need a fourth charge at (− x, −y) to complete the symmetry as shown. Now we can construct the Green’s function. Adding terms of the form
±
1 ln((ξ − x0 )2 + (η − y0 )2 ) 4π
with the sign corresponding to the sign of the charge and ( x0 , y0 ) giving the charge location, we write the Green’s function for this problem as G ( x, y, ξ, η )
=
1 1 ln((ξ − x )2 + (η − y)2 ) − ln((ξ − x )2 + (η + y)2 ) 4π 4π 1 1 ln((ξ + x )2 + (η − y)2 ) + ln((ξ + x )2 + (η + y)2 ). − 4π 4π
We can verify that this Green’s function satisfies the boundary conditions. Namely, we have G (0, y, ξ, η )
=
G ( x, 0, ξ, η )
=
1 1 ln(ξ 2 + (η − y)2 ) − ln(ξ 2 + (η + y)2 ) 4π 4π 1 1 ln(ξ 2 + (η − y)2 ) + ln(ξ 2 + (η + y)2 ) = 0. − 4π 4π 1 1 ln((ξ − x )2 + η 2 ) − ln((ξ − x )2 + η 2 ) 4π 4π 1 1 − ln((ξ + x )2 + η 2 ) + ln((ξ + x )2 + η 2 ) = 0. 4π 4π
13. Find the Green’s function for the onedimensional heat equation with boundary conditions u(0, t) = 0 u x ( L, t) = 0, t > 0. We construct the Green’s functions from the eigenfunctions of the spatial part of the problem. Separation of variables would give X 00 + λ2 X = 0,
X (0) = 0, X 0 ( L) = 0.
The first boundary conditions give X ( x ) = sin λx. The second boundary condition gives cos λL = 0. So, λL = 2n2−1 π, n = 1, 2, . . . . Therefore, the normalized eigenfunctions are r Xn ( x ) =
2 (2n − 1)πx sin . L 2L
We can now construct the initial value Green’s function for this problem in the spirit of Problem 5.20. G ( x, ξ; t, 0) =
2 L
∞
∑ sin
n =1
(2n − 1)πx (2n − 1)πξ −k( 2n−1 π )2 t 2L sin e . 2L 2L
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14. Consider Laplace’s Equation on the rectangular plate in Figure 11.8. Construct the Green’s function for this problem. We found the solution to this problem as ∞
u( x, y)
∑ an sin
=
n =1
Z L
2 L sinh nπH L
=
an
nπx nπ ( H − y) sinh , L L 0
f ( x ) sin
nπx dx. L
Inserting an into the expression for u( x, t) and interchanging the sum and integration, we have ! Z L ∞ nπξ 2 nπx nπ ( H − y) u( x, y) = ∑ f (ξ ) sin dξ sin sinh , nπH 0 L L L L sinh L n =1 ! nπ ( H −y) Z L nπx nπξ sinh 2 ∞ L sin f (ξ ) dξ. = sin L n∑ L L 0 sinh nπH =1 L Therefore, we can identify the Green’s function as G ( x, ξ; y) =
2 L
nπ ( H −y)
∞
∑
nπx nπξ sinh L sin L L sinh nπH L
sin
n =1
and the solution can be written as u( x, y) =
Z L 0
G ( x, ξ; y) f (ξ ) dξ.
15. Construct the Green’s function for Laplace’s Equation in the spherical domain in Figure 11.18. We consider a sphere of radius r with the boundary condition u(r, θ, φ) = g(θ, φ) where u(r, θ, φ) satisfies Laplace’s equation. The general solution was found as ∞
`
∑ ∑
u(ρ, θ, φ) =
a`m ρ` P`m (cos θ )eimφ .
`=0 m=−`
A brute force method for finding the Green’s function is to solve for the a`m ’s using the boundary condition and inserting the results into the general solution. The boundary condition gives ∞
g(θ, φ) =
`
∑ ∑
a`m r ` P`m (cos θ )eimφ .
`=0 m=−` 0
0
Multiplying both sides by P`m0 (cos θ )e−im φ sin θ and integrating, we have Z π Z 2π 0 ∞
=
0 `
∑ ∑
`=0 m=−`
0
0
g(θ, φ) P`m0 (cos θ )e−im φ sin θ dφ dθ a`m r `
Z π Z 2π 0
0
0
0
P`m (cos θ ) P`m0 (cos θ )ei(m−m )φ sin θ dφ dθ.
problems in higher dimensions
We can use the orthogonality of the spherical harmonics to compute the integrals on the right side of the equation. Recall that the spherical harmonics were defined as s 2` + 1 (` − m)! m m Y`m (θ, φ) = (−1) P (cos θ )eimφ . 4π (` + m)! ` These satisfy the orthogonality relation Z π Z 2π 0
0
Y`m (θ, φ)Y`∗0 m0 (θ, φ) sin θ dφ dθ = δ``0 δmm0 .
Therefore, we have Z 2π 0
0
0
P`m (cos θ ) P`m0 (cos θ )ei(m−m )φ sin θ dφ dθ =
4π (` + m)! δ 0δ 0. 2` + 1 (` − m)! `` mm
Inserting this into the above integration, we have Z π Z 2π 0
0
g(θ, φ) P`m (cos θ )e−imφ sin θ dφ dθ = a`m r `
4π (` + m)! . 2` + 1 (` − m)!
Therefore, a`m =
2` + 1 (` − m)! −` r 4π (` + m)!
Z π Z 2π 0
0
g(θ, φ) P`m (cos θ )e−imφ sin θ dφ dθ.
Inserting these coefficients into the general solution and rearranging, we obtain ∞
u(ρ, θ, φ)
=
`
∑ ∑
a`m ρ` P`m (cos θ )eimφ
∞
∑ ∑
Z π Z 2π
"
`=0 m=−`
=
`
`=0 m=−`
=
0
0
2` + 1 (` − m)! 4π (` + m)! ∞
Z π Z 2π 0
0
g(θ 0 , φ0 ) P`m (cos θ 0 )e−imφ sin θ 0 dφ0 dθ 0
∞
`
` 2` + 1 (` − m)! m m 0 im(φ−φ0 ) ρ P ( cos θ ) P ( cos θ ) e ∑ ∑ 4π (` + m)! ` ` r` `=0 m=−`
and the solution is in the form u(ρ, θ, φ) =
Z π Z 2π 0
ρ` m P (cos θ )eimφ r` `
# ` 0 2` + 1 (` − m)! m ρ ∑ ∑ 4π (` + m)! P` (cos θ ) P`m (cos θ 0 )eim(φ−φ ) r` g(θ 0 , φ0 ) sin θ 0 dφ0 dθ 0 . `=0 m=−` `
Therefore, we can identify the Green’s function G (ρ; θ, θ 0 ; φ, φ0 ) =
0
G (ρ; θ, θ 0 ; φ, φ0 ) g(θ 0 , φ0 ) sin θ 0 dφ0 dθ 0 .
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A Review of Sequences and Infinite Series 1. For those sequences that converge, find the limit limn→∞ an . a. an =
n2 +1 . n3 +1
n2 + 1 n → ∞ n3 + 1 n2 lim 3 n→∞ n 1 lim = 0. n→∞ n
=
lim an
n→∞
lim
= = b. an =
3n+1 n +2 .
=
lim an
n→∞
= c. an =
lim
n→∞
3n + 1 n+2 3 + n1 1+
2 n
= 3.
3 1/n . n lim an
n→∞
= =
d. an =
lim
n→∞
1/n 3 n→∞ n ! 31/n lim = 1. n→∞ n1/n lim
2n2 √ +4n3 . n3 +5 2+ n6
lim an
n→∞
= = =
2n2 + 4n3 √ n → ∞ n3 + 5 2 + n6 2 +4 nq lim n→∞ 1 + 5 n26 + 1 lim
lim
n→∞
2 4 = . 1+5 3
e. an = n ln 1 + n1 .
lim an
n→∞
=
1 lim n ln 1 + n→∞ n
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1 n lim ln 1 + n→∞ n ln e = 1.
= = f. an = n sin
1 n
. lim an
n→∞
=
= = g. an =
1 lim n sin n→∞ n 1 sin n lim 1 n→∞
n
sin x lim = 1. x →0 x
(2n+3)! . ( n +1) !
(2n + 3)! ( n + 1) ! (2n + 3)(2n + 2) · · · (n + 2)(n + 1)! = lim n→∞ ( n + 1) ! = ∞ =
lim an
n→∞
lim
n→∞
2. Find the sum for each of the series: a. ∑∞ n =0
(−1)n 3 4n .
This is a geometric series with a = 3 and r = − 14 . ∞
(−1)n 3 3 ∑ 4n = 1 + n =0 b. ∑∞ n =2
1 4
=
12 . 5
2 5n .
This is a geometric series with a =
2 52
and r = 15 .
∞
2 2/25 1 = = . n 1 5 10 1− 5 n =2
∑
c. ∑∞ n =0
5 2n
+
1 3n
.
This is the sum of two geometric series with a = 5, r = a = 1, r = 31 . ∞
∑
n =0
d. ∑∞ n =1
5 1 + n 2n 3
=
5 1−
1 2
+
1 1−
1 3
=
1 2
and
23 . 2
3 . n ( n +3)
Using a partial fraction decomposition, this can be written as a telescoping series. The Nth partial sum is N
3 ∑ n ( n + 3) n =1
N
=
∑
n =1
1 1 − n n+3
review of sequences and infinite series
1 1 1 1 1 1 1 1 − + − + − + C − 2 5 3 6 4CC 7 4 Z1 1 1 1 + C − + · · · + Z − 5CC 8 N −Z 2 N+1 Z1 1 1 1 + A − + Z − N −Z 1 N+2 NAA N + 3 1 1 1 1 1 1+ + − + + . 2 3 N+1 N+2 N+3
=
=
Here forward crossed terms like 14 cancel with terms later in sum and backward crossed terms like A41A cancel with earlier terms. 11 3 Letting N → ∞, we have ∑∞ = . n =1 n ( n + 3) 6 3. Determine if the following converge, or diverge, using one of the convergence tests. If the series converges, is it absolute or conditional? n +4 a. ∑∞ n=1 2n3 +1 . Since the tail of the series determines the convergence, we note that for large n 1 n+4 n ∼ 3 = 2. 3 2n + 1 2n 2n
So, we can compare the original series to ∑∞ n =1 We compute the following limit n +4 2n3 +1 1 n→∞ n2
lim
= =
lim
n→∞
1 . n2
2n3 + 1 n2 ( n + 4)
(2n3 = 2. n → ∞ n3 lim
Therefore, these series either both converge or both diverge. Since 1 ∑∞ n=1 n2 converges, then the original series converges by the Limit Comparison Test. Also, since the terms are all positive, it converges absolutely. sin n b. ∑∞ n =1 n2 . We note that
∞
∑
n =1
sin n n2 ≤
∞
1 < ∞. 2 n n =1
∑
Therefore, this series converges absolutely. 2 n n c. ∑∞ . n =1 n +1 We apply the nth Root Test. lim
n→∞
√ n
n n n→∞ n + 1 1 1 lim n = < 1. n→∞ e 1 + n1
an
= =
lim
Therefore, the series converges by the nth Root Test. Also, since the terms are all positive, it converges absolutely.
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mathematical methods for physicists
n n −1 d. ∑∞ n=1 (−1) 2n2 −3 .
We first note that this is an alternating series whose terms have magnitude decreasing to zero, ∞
n−1
1
2
3
∑ (−1)n 2n2 − 3 = 0 + 5 − 15 + 38 − . . . .
n =1
Therefore, the series converges by the Liebniz Test. n −1 Dropping the signs, we have that ∑∞ n=1 2n2 −3 has terms that behave like 1 n−1 ∼ . 2n 2n2 − 3
Therefore, this series behaves like the harmonic series, which diverges. So, the original series converges conditionally. e. ∑∞ n =1
ln n n .
We note that
ln n n
> n1 , n ≥ 3. Therefore, ∞
∞ ln n 1 > ∑ n ∑ n > ∞. n =3 n =1
Therefore, this series diverges by the Comparison Test. f. ∑∞ n =1
100n . n200
We apply the nth Root Test. lim
n→∞
√ n
an = lim
n→∞
100 = 100 > 1 1
Therefore, the series diverges by the nth Root Test. n n g. ∑∞ n=1 (−1) n+3 .
For this series the terms do not go to zero for large n. Namely, lim
n→∞
n = 1. n+3
So, by the nth term divergence test, this series diverges. √
n 5n h. ∑∞ n=1 (−1) n+1 .
The magnitudes of the terms goes to zero for large n. √ √ 5n 5 lim = lim √ = 0. n→∞ n + 1 n→∞ n Therefore, the series converges by the Liebniz Test. √
5n Dropping the signs, we have that ∑∞ n=1 n+1 has terms that behave like √ √ 5n 5 ∼ √ . n+1 n
Therefore, this series diverges according to the p test. So, the original series converges conditionally.
review of sequences and infinite series
4. Do the following: a. Compute: limn→∞ n ln 1 − n3 . 3 lim n ln 1 − = n→∞ n
3 lim ln 1 − n→∞ n
n
= ln e−3 = −3. x b. Use L’Hopital’s Rule to evaluate L = limx→∞ 1 − 4x . [Hint: Consider ln L.] We use ln L and L’Hopital’s Rule to find 4 x ln L = lim ln 1 − x →∞ x 4 = lim x ln 1 − x →∞ x
= =
lim
x →∞
lim
ln(1 − 4x ) 1 x 4 x2
− x12 (1 − 4x ) 4 = lim = −4. x →∞ −(1 − 4 ) x x →∞
Since ln L = −4, L = e−4 . n2 n c. Determine the convergence of ∑∞ . n=1 3n+2 We apply the nth Root Test. n √ n n lim an = lim n→∞ n→∞ 3n + 2 1 = lim n→∞ 3 + 2 n n 1 = lim n = 0 < 1. n→∞ n 3 1 + 2/3 n Therefore, by the nth Root Test this series converges. −1 d. Sum the series ∑∞ n − tan−1 (n + 1) by first writing the n=1 tan Nth partial sum and then computing lim N →∞ s N . The Nth partial sum is N
sN
=
∑
h
tan−1 n − tan−1 (n + 1)
i
n =1
=
XX−X 1 1X − tan−1 1 − tan 2 + tan 2 − tan−1 3 X 1 X + · · · + tanX−X N − tan−1 ( N + 1)
= tan−1 2 − tan−1 ( N + 1) Letting N → ∞, we have i ∞ h π π π ∑ tan−1 n − tan−1 (n + 1) = 4 − 2 = − 4 . n =1
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mathematical methods for physicists
5. Consider the sum ∑∞ n =1
1 . (n+2)(n+1)
a. Use an appropriate convergence test to show that this series converges. Since the tail of the series determines the convergence, we note that for large n 1 1 ∼ 2. (n + 2)(n + 1) n 1 So, we can compare the original series to ∑∞ n=1 n2 . Since the latter series converges, so does this one by the Limit Comparison Test.
One can also use the following ∞
∞ 1 1 < ∑ (n + 2)(n + 1) ∑ n2 < ∞. n =1 n =1
Therefore, the series converse by the Comparison Test as well. b. Verify that ∞ 1 ∑ (n + 2)(n + 1) = ∑ n =1 n =1 ∞
n n+1 − n+2 n+1
.
One just adds the terms to verify the sum. n+1 n − n+2 n+1
= =
( n + 1)2 − n ( n + 2) (n + 2)(n + 1) 1 . (n + 2)(n + 1)
Note that partial fractions does not give this representation, but instead gives ∞ ∞ 1 1 1 ∑ (n + 2)(n + 1) = ∑ n + 1 − n + 2 . n =1 n =1 This may also be summed as a telescoping series but the instructions for part c apply to the other representation. n +1 n c. Find the nth partial sum of the series ∑∞ − n =1 n +2 n+1 and use it to determine the sum of the resulting telescoping series. ∞
sN
= = =
n n+1 ∑ n+2 − n+1 n =1 2 1 3 2 4 3 N + 1 ZN − + − C + − C + · · · + − Z N+2 N +Z 1 4 3CC 3 2 5 4CC 1 N+1 − + . 2 N+2
Here forward crossed terms like 23 cancel with terms later in sum and backward crossed terms like A23A cancel with earlier terms. n+1 n ∞ Letting N → ∞, we have ∑n=1 − = − 12 + 1 = 21 . n+2 n+1
review of sequences and infinite series
385
6. Recall that the alternating harmonic series converges conditionally. a. From the Taylor series expansion for f ( x ) = ln(1 + x ), inserting x = 1 gives the alternating harmonic series. What is the sum of the alternating harmonic series? The Taylor series expansion for f ( x ) = ln(1 + x ) is given by ln(1 + x ) = x −
∞ x2 x3 xn + − . . . = ∑ (−1)n+1 . 2 3 n n =1
Inserting x = 1 gives the sum of the alternating harmonic series ln 2 = 1 −
∞ 1 1 1 + − . . . = ∑ (−1)n+1 . 2 3 n n =1
b Because the alternating harmonic series does not converge absolutely, a rearrangement of the terms in the series will result in series whose sums vary. One such rearrangement in alternating p 1 positive terms and n negative terms leads to the following sum1 : This is discussed by Lawrence H. Riddle in the Kenyon Math. Quarterly, 1(2), 621. 1 1 1 4p 1 1 1 − ln = 1+ +···+ + +···+ 2 n 3 2p − 1 2 4 2n {z }  {z }  n terms p terms 1 1 1 1 + +···+ − +···+ +···. 2p + 1 4p − 1 2n + 2 4n {z }  {z }  n terms p terms Find rearrangements of the alternating harmonic series to give the following sums; that is, determine p and n for the given expression and write down the above series explicitly; that is, determine p and n leading to the following sums. i.
5 2
ln 2. In each case we need to rewrite the given expression in the form 4p 1 2 ln n and select values for n and p. In the first problem we have 1 4p ln 2 n
= =
Note: This problem does not have unique solutions.
5 ln 2 2 1 1 ln 25 = ln 32. 2 2
For this problem we can pick n = 1 and p = 8. Then, we would have 1 1 1 1 1 1 1 1 5 ln 2 = 1+ + + + + + + − 2 3 5 7 9 11 13 15 2 1 1 1 1 1 1 1 1 1 + + + + + + + + − +.... 17 19 21 23 25 27 29 31 4
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mathematical methods for physicists
ii. ln 8. In this problem we have 1 4p ln 2 n
= ln 8 =
1 ln 64. 2
For this problem we can pick n = 1 and p = 16. The rearranged series is then 1 1 1 1 1 1 1 ln 8 = 1+ + + + + + + 3 5 7 9 11 13 15 1 1 1 1 1 1 1 1 1 + + + + + + + + − +.... 17 19 21 23 25 27 29 31 2 iii. 0. 4p In this problem we have 12 ln n = 0. Therefore, n = 4p. We can pick p = 1 and n = 4. The rearranged series is then 1 1 1 1 1 1 1 1 1 0 = 1− + + + + − + + + . 2 4 6 8 3 10 12 14 16 1 1 1 1 1 1 + − + + + + −.... 5 18 20 22 24 7 iv. A sum that is close to π. 4p 4p For this problem we have π ≈ 21 ln n , or n ≈ e2π ≈ 535.4916560. So, one choice is n = 1 and p = 535.49/4 ≈ 134. Thus, there is one positive term followed by 134 negative terms, etc. 7. Determine the radius and interval of convergence of the following infinite series: ( x −1) n
n a. ∑∞ . n=1 (−1) n Using the nth Root Test, we have q  x − 1 lim n  an  = lim √ =  x − 1 < 1. n n→∞ n→∞ n
Thus, R = 1, and one has absolute convergence for x ∈ (0, 2). Looking at the endpoints, there is conditional convergence for the endpoint x = 2. x b. ∑∞ n=1 2n n! . Using the Ratio Test, we have n
 a n +1  x  n +1 2n n! x = lim n+1 = lim = 0. n→∞  an  n→∞ 2 n → ∞ 2( n + 1) ( n + 1) !  x  n lim
Thus, R = ∞, and one has absolute convergence for x ∈ (−∞, ∞). 1 x n c. ∑∞ n =1 n 5 Using the nth Root Test, we have q x x lim n  an  = lim √ = < 5. n→∞ n→∞ 5 n n 5
review of sequences and infinite series
Thus, R = 5, and one has absolute convergence for x ∈ (−5, 5). Looking at the endpoints, there is conditional convergence for the endpoint x = −5. x n√ d. ∑∞ . n=1 (−1) n n
Using the Ratio Test, we have
√  a n +1  x  n +1 n lim =  x  < 1. = lim √ n→∞  an  n→∞ n + 1  x n Thus, R = 1 and one has absolute convergence for x ∈ (−1, 1). Looking at the endpoints, there is conditional convergence for the endpoint x = 1. 8. Find the Taylor series centered at x = a and its corresponding radius of convergence for the given function. In most cases, you need not employ the direct method of computation of the Taylor coefficients. a. f ( x ) = sinh x, a = 0. Using the Maclaurin series for e x , we have sinh x
e x − e− x 2 1 x2 x3 x2 x3 1+x+ !+ !+... − 1−x+ !− !+... 2 2 3 2 3 3 5 1 x x 2x + 2 ! + 2 ! + . . . 2 3 5
= = =
= x+ b. f ( x ) =
√
∞ x5 x2k−1 x3 !+ !+... = ∑ . 3 5 (2k − 1)! k =1
1 + x, a = 0.
Using the binomial expansion, we have
√
1+x
= (1 + x )1/2 ∞ 1 1 −1 ··· 1 −r+1 2 2 2 = ∑ xr , r! r =0 1 1 1 = 1 + x − x2 + x3 + . . . , 2 8 16
 x  < 1,
 x  < 1.
c. f ( x ) = xe x , a = 1. We first rewrite f ( x ) as a function of x − 1 and then use the Maclaurin series expansion for e x . This gives xe x
= ( x − 1 + 1 ) e x −1+1 = [( x − 1)e + e]e x−1 1 1 2 3 = [( x − 1)e + e] 1 + ( x − 1) + ( x − 1) + ( x − 1) + . . . 2 3! 1 1 1 2 = e + 2e( x − 1) + 1 + e ( x − 1) + + e ( x − 1)3 + . . . 2! 2! 3!
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mathematical methods for physicists
∞
= e+
∑
1 1 + n! (n + 1)!
n =0 ∞
= e+
e ( x − 1 ) n +1
( n + 2) e ( x − 1 ) n +1 , ( n + 1 ) ! n =0
∑
 x − 1 < ∞.
Thus, the series converges for  x  < ∞. d. f ( x ) =
x −1 2+ x ,
a = 1.
We rewrite f ( x ) as a function of x − 1 and then use a geometric series. This gives x−1 2+x
1 3 + ( x − 1) x−1 1 1 3 1 + x− 3 " # x−1 ∞ 1−x n ∑ 3 3 n =0 ∞ 1 − x n +1  x − 1 −∑ , < 1. 3 3 n =0
= ( x − 1) = = =
Thus, the series converges for  x  < 3. 9. Test for pointwise and uniform convergence on the given set. [The Weierstraß MTest might be helpful.] ln nx a. f ( x ) = ∑∞ n=1 n2 , x ∈ [1, 2]. For x ∈ [1, 2] we have ln nx = ln n + ln x ≤ ln 2n. Therefore, ln nx ln 2n n2 ≤ n2 .
So, we pick Mn =
ln 2n n2
in the Weierstraß MTest.
Next we determine if ∑∞ n=1 Mn converges. By the Integral Test, we have Z ∞ ln 2x ln 2x 1 ∞ dx = − − = ln 2 + 1. x x 1 x2 1 Therefore, ∑ Mn is a convergent series and the original series converges uniformly on [1, 2] by the Weierstraß MTest. b. f ( x ) = ∑∞ n =1 We note that
1 3n
cos 2xn on R. 1 cos x ≤ 1 . 3n 2n 3n
So, we pick Mn = 31n in the Weierstraß MTest. Furthermore, since ∑ Mn is a convergent geometric series, the original series converges uniformly on R by the Weierstraß MTest. 10. Consider Gregory’s expansion tan−1 x = x −
∞ x5 (−1)k 2k+1 x3 + −··· = ∑ x . 3 5 2k + 1 k =0
review of sequences and infinite series
a. Derive Gregory’s expansion using the definition tan
−1
x=
Z x 0
dt , 1 + t2
expanding the integrand in a Maclaurin series, and integrating the resulting series term by term.
tan−1 x
=
Z x 0
=
dt 1 + t2
Z x ∞
∑ (−t2 )n
0 n =0 ∞
=
(−1)n x2n+1 . 2n + 1 n =0
∑
b. From this result, derive Gregory’s series for π by inserting an appropriate value for x in the series expansion for tan−1 x.
tan−1 (1) π 4 Therefore,
∞
=
(−1)n 2n + 1 n =0
∑
= 1−
1 1 + −... 3 5
1 1 π = 4 1− + −... . 3 5
11. Use deMoivre’s Theorem to write sin3 θ in terms of sin θ and sin 3θ. [Hint: Focus on the imaginary part of e3iθ .] We note that e3iθ = (eiθ )3 . Writing both sides of this equation in terms of trigonometric functions, we have e3iθ cos 3θ + i sin 3θ
= (eiθ )3 = (cos θ + i sin θ )3 = cos3 θ + 3i cos2 θ sin θ − 3 cos θ sin2 θ − i sin3 θ
Equating the real and imaginary parts we have cos 3θ
= cos3 θ − 3 cos θ sin2 θ,
sin 3θ
= 3 cos2 θ sin θ − sin3 θ.
The second equation can be rearranged to get the result. sin 3θ
= 3 cos2 θ sin θ − sin3 θ = 3(1 − sin2 θ ) sin θ − sin3 θ = 3 sin θ − 4 sin3 θ.
Therefore, sin3 θ =
3 1 sin θ − sin 3θ. 4 4
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mathematical methods for physicists
12. Evaluate the following expressions at the given point. Use your calculator and your computer (such as Maple). Then use series expansions to find an approximation of the value of the expression to as many places as you trust. Note: In this problem we are considering that all values of the independent variable are exact. Otherwise, we would need to propagate the errors and find that the accuracy reported below is far too precise. Using ideas from Chapter 9, we can compute the absolute errors of computing f ( x ) in terms of ∆x as ∂f ∆f = ∆x. ∂x We report the results below. a. √
Table A.1: Values of √ x = 0.015.
1 1 + x3
1 1 + x3
− cos x2 at x = 0.015.
Tool TI83 Plus Maple (10 Digits) Maple (20 Digits) MATLAB (short) MATLAB (long)
− cos x2 at
Value −1.66218325E − 6 −1.6617 × 10−6 −1.66218322863445 × 10−6 −1.6622e − 06 −1.662183228723356e − 06
Computation:
√
1 1 + x3
− cos x2
=
1 3 1 1 1 − x3 + x6 − . . . − 1 − x4 + x8 − . . . 2 8 2 4! x3 −  {z2}
=
+
x4 3x6 x8 − +... + 2 8 24
−1.6875×10−6

{z
−1.6621875×10−6
{z

} }
−1.662183228515625×10−6

{z
−1.6621832286344256592×10−6
}
For x = 0.015 ± 0.0005, f ( x ) = (1.66 ± 0.17) × 10−6 . r 1+x b. ln − tan x at x = 0.0015. 1−x r Table A.2: Values of ln x = 0.0015.
1+x − tan x at 1−x
Tool TI83 Plus Maple (10 Digits) Maple (20 Digits) MATLAB (short) MATLAB (long) Computation: r 1+x ln − tan x 1−x
=
Value 0 1.19 × 10−10 5.062439 × 10−16 6.0607e − 16 6.060690144193970e − 16
1 2
2 3 2 5 1 3 2 5 2x + x + x + . . . − x + x + x + . . . 3 5 3 15
review of sequences and infinite series
x5 15 {z}
=
+
391
4x7 253x9 +... ++ 45 2835
5.0625×10−16

{z
}
5.0625151875×10−16

{z
5.0625151875343074777×10−16
}
For x = 0.0015 ± 0.00005, f ( x ) = (5.06 ± 0.84) × 10−6 . 1 c. f ( x ) = √ − 1 + x2 at x = 5.00 × 10−3 . 1 + 2x2 Tool TI83 Plus Maple (10 Digits) Maple (20 Digits) MATLAB (short) MATLAB (long)
Value 9.3747E − 10 0.6 × 10−9 9.3746093916 × 10−10 9.3746e − 10 9.374608339397285e − 10
Table A.3: Values of f ( x ) = √ 2
1 + x at x =
5.00 × 10−3 .
1 1 + 2x2
−
Computation:
√
1 1 + 2x2
− 1 + x2
= =
3 5 35 1 − x2 + x4 − x6 + x8 + . . . − 1 + x2 2 2 8 3 4 5 6 35 8 x − x + x +... 2 2 8 {z}
9.375×10−10

{z
}
9.374609375×10−10

{z
}
9.37460939208984375×10−10
For x = 0.005 ± 0.0005, f ( x ) = (9.37 ± 3.75) × 10−10 . √ d. f ( R, h) = R − R2 + h2 for R = 1.374 × 103 km and h = 1.00 m. Tool TI83 Plus Maple (10 Digits) Maple (20 Digits) MATLAB (short) MATLAB (long)
Computation: First write f ( R, h) = R − Since tain
h R
√
r R2
+ h2
= R−R 1+
2 h R
≈ 7.278 × 10−7 , we can use the binomial expansion to ob
f ( R, h)
Table A.4: Values of f ( R, h) = R − √ R2 + h2 for R = 1.374 × 103 km and h = 1.00 m.
Value 0 0 −.36390010 × 10−6 −3.6391e − 07 −3.639142960309982e − 07
=
= =
R 1 −
s
2 h 1+ R
!! 6 1 h 2 1 h 4 1 h R 1− 1+ − + +... 2 R 8 R 16 R ! 6 1 h 2 1 h 4 1 h R − + − +... 2 R 8 R 16 R
.
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mathematical methods for physicists
1 h2 − 2 R}  {z
=
+
1 h4 1 h6 +... − 3 8R 16 R5
−0.3639010892285298399×10−6

{z
}
−0.3639010892280479478×10−6

{z
−0.3639010892280479478×10−6
}
= −0.00036390097 . . . . Therefore, for h = 1.00 ± 0.005 m and R = (1.374 ± 0.0005) × 103 km we have f ( R, h) = (−3.639 ± 0.038) × 10−7 . 1 e. f ( x ) = 1 − √ for x = 2.5 × 10−13 . 1−x Table A.5: Values of f ( x ) = 1 − √ for x = 2.5 × 10−13 .
1 1−x
Tool TI83 Plus Maple (10 Digits) Maple (20 Digits) MATLAB (short) MATLAB (long)
Value 0 0 −0.1250000 × 10−12 −1.2501e − 13 −1.250111125727926e − 13
Computation: 1 1− √ 1−x
= =
1 3 5 35 1 − 1 + x + x2 + x3 + +... 2 8 16 128 1 3 5 − x − x2 − x3 + . . . 2 8 16  {z }
−1.25×10−13

{z
}
−1.250000000000234375×10−13

{z
−1.250000000000234375×10−13
}
For x = (2.5 ± 0.05) × 10−13 , f ( x ) = (−1.250 ± 0.025) × 10−13 . 13. Determine the order, O( x p ), of the following functions. You may need to use series expansions in powers of x when x → 0, or series expansions in powers of 1/x when x → ∞. p a. x (1 − x ) as x → 0. √ We factor out a x and use the binomial expansion. Then, for small x we have q √ √ 1 x (1 − x ) = x (1 − x + O( x2 )) = O( x ). 2 b.
x5/4 1−cos x
as x → 0.
In this problem, one pays attention to the denominator. A Maclaurin series expansion gives x5/4 1 − cos x
=
=
x5/4
1 − 1 − 21 x2 + O( x4 ) 1 2 2x
x5/4 + O( x4 )
review of sequences and infinite series
= O c.
x x 2 −1
x5/4 x2
!
= O( x −3/4 ).
as x → ∞.
For large x we use expansions in 1/x. We rewrite the expression accordingly: 1 1 x x = O ( ). = 1 x x2 − 1 1 − x2 √ x2 + x − x as x → ∞. d. For large x we use expansions in 1/x. We rewrite the expression using a binomial expansion. p
x2
1 1/2 x 1+ −x x 1 11 + O( 2 ) − x x 1− 2x x 1 1 + O ( ) = O ( x 0 ) = O (1). 2 x
+x−x = = =
393
B Quick Answers Ch.1
Introduction
1. a. Proof. b. Proof, leading to A = −4, B = 3. c. Proof. 2.
√ 2 − 2. √ b. 2 − 3. p √ √ √ c. − 21 2 − 3 = 14 ( 2 − 6). p
a.
1 2
a.
p
3.
√ √ 3 − 2 2 = 2 − 1.
b. Cannot denest. p √ √ √ 5 + 2 6 = 2 + 3. c. p p √ √ 3 3 d. 5+2− 5 − 2 = 1. √ √ e. x = −1 + 5, x = −5 − 5. 4. a. sin(cos−1 35 ) =
4 . 5
b. tan(sin−1 7x ) = 49−x x2 . 3π − 1 c. sin sin = − π2 . 2 5. a. (cosh x − sinh x )6 = e−6x . b. Proof. c. Proof. 5 5 d. sinh x = − 12 and tanh x = − 13 .
396
mathematical methods for physicists
√ e. sinh(arccosh 3) = 2 2. 6. a. Proof. b. Proof. 7. a. cosh−1 b.
4 3 tanh−1 12 −1
= ln(4 + =
1 2
ln 3.
2 = ln(2 +
c. sinh
√
√
7) − ln 3.
5).
8.
√ a. x = ln(3 + 2 2) − ln 3. b. x = 52 .
√ √ c. x = ± cosh−1 3 = ± ln(3 + 2 2) = ln(3 ± 2 2), √ √ x = ± cosh−1 4 = ± ln(4 + 15) = ln(4 ± 15). 9. a. b. c. d. e. f. g. h. i. j. k. l.
2
2
xe2x dx = 14 e2x + C. R3 5x √ dx = 5. 0 2 R 3 x + 16 2 x sin 3x dx = ( 92 x − 13 x3 ) cos 3x + ( 31 x2 − 27 ) sin 3x + C. R 3 1 1 cos4 3x dx = 8 x + 12 sin 6x + 96 sin 12x + C. √ √ R π/4 sec3 x dx = 12 ( 2 + ln(1 + 2)). 0 R x e sinh x dx = 14 e x (cosh x + sinh x ) − 2x + C. √ R√ 2 9 − x2 dx = 92 sin−1 3x + x 92− x . R dx 1 2+x x = ln + + C. 32 2 − x 8(4 − x2 ) (4 − x 2 )2 R4 dx √ = ln 3. 0 9 + x2 R dx 1 1+x + C. = ln 2 1−x 1 − x2 R dx x + C. = √ 2 3/2 ( x + 4) 4 x2 + 4 √ p R dx 1 √ = √ ln 3( x − 1) + 3x2 − 6x + 4 + C. 3 3x2 − 6x + 4
R
10. 625 + 125 49 + 343 + · · · = ∞ (−1)n 3 12 b. ∑ = . n 4 5 n =0
a. 5 +
∞
c.
25 7
2 1 = . n 5 10 n =2
∑
35 2 .
quick answers
∞
e n π2 = . π (π + e)e n=−1 ∞ 5 1 23 e. ∑ + = . n n 2 3 2 n =0
∑
d.
(−1)n+1
∞
3 11 = . n ( n + 3 ) 6 n =1 g. 0.569¯ = 0.57. f.
∑
11. The total distance traveled is 20.9 m. 12. a. Proof. ∞ n 1 n+1 − = . b. ∑ n + 2 n + 1 2 n =1 i ∞ h π c. ∑ tan−1 n − tan−1 (n + 1) = − . 4 n =1 13. a. R = 1, (0, 2]. b. R = ∞, (−∞, ∞). c. R = 5, [−5, 5). d. R = 1, (−1, 1]. 14. < E >=
h¯ ω 4
coth
β¯hω 2
csch
β¯hω 2 .
15. a. sinh x = x +
√
∞ x3 x5 x2k−1 !+ !+... = ∑ ,  x  < ∞. 3 5 (2k − 1)! k =1
1 3 x + . . . ,  x  < 1. 1 + x = 1 + 12 x − 18 x2 + 16 1+x 2 2 c. ln = 2x + x3 + x5 + . . . ,  x  < 1. 1−x 3 5 ∞ ( n + 2 ) e ( x − 1)n+1 ,  x  < ∞. d. xe x = e + ∑ ( n + 1 ) ! n =0
b.
e. f. g.
√1 = 1 − 1 ( x − 1) + 3 ( x − 1)2 − 5 ( x − 1)3 + . . . ,  x − 1 < 1. 2 8 16 x 4 4 3 x + x − 2 = ( x − 2) + 8( x − 2) + 24( x − 2)2 + 33( x − 2) + 16. ∞ x−1 = −∑ 2+x n =0
1−x 3
n +1
16. ∞
(−1)n x2n+1 . 2n + 1 n =0 b. π = 4 1 − 31 + 15 − . . . .
a. tan−1 x =
∑
,
 x − 1 < 3.
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mathematical methods for physicists
17. ∞
a.
∑ nxn =
n =1 ∞
b.
x ,  x  < 1. (1 − x )2
n = 5n n =1
∑
1 5
1−
1 5
2 =
∞
c.
2x2 ,  x  < 1. (1 − x )3
∑ n ( n − 1) x n =
n =2 ∞
5 . 16
d.
n2 − n 5 = . n 5 32 n =2
e.
147 n2 = . n 5 4000 n =2
a.
R π/6
b.
R π/6
∑ ∞
∑
18. 0 0
sin2 x dx = sin2 x dx =
√ 3 π − 12 8 . 3 π5 π 648 − 116640
+
π7 44089920
≈ 0.0453.
19. The next term in the approximation is 1.4 × 10−13 ns which is far too small to be measured as compared to the first term. 20. a. For x = 0.015 ± 0.0005, f ( x ) = (1.66 ± 0.17) × 10−6 . b. For x = 0.0015 ± 0.00005, f ( x ) = (5.06 ± 0.84) × 10−6 . c. For x = 0.005 ± 0.0005, f ( x ) = (9.37 ± 3.75) × 10−10 . d. For h = 1.00 ± 0.005 m and R = (1.374 ± 0.0005) × 103 km, f ( R, h) = (−3.639 ± 0.038) × 10−7 . e. For x = (2.5 ± 0.05) × 10−13 , f ( x ) = (−1.250 ± 0.025) × 10−13 . 21. π1 = µD −1 ρ−1 v−1 and π2 = FD D −2 ρ−1 v−2 . 22. E is on the order of 4 × 1013 J which is 10 ktons TNT.
Ch.2
Free Fall and Harmonic Oscillators
1. a. y2 ( x ) = e x + C. b. y(t) =
1 . 1−t− 13 t3
c. y( x ) = sin(ln  x  + C ). e. y( x ) =
2x 4x −3 . Cecos x
f. y( x ) =
x2 (ln  x  + 1).
d. y( x ) =
1 3
− 1.
g. s(t) = e 3 t −2t . h. x (t) = 12 t2 + C e2t .
quick answers
i. y( x ) = 21 (sin x − cos x + e− x ). j. y( x ) = x4 + 3x3 . 2. a. y( x ) = c1 e4x + c2 e5x . √ √ b. y( x ) = 2 7 7 e3x/2 sin 27 x . c. y( x ) = x −2 (c1 + c2 ln  x ). √ √ d. y( x ) = x3/2 c1 cos 23 ln  x  + c2 sin 23 ln  x  . 3. a. 3y2 + 2y3 = 3x2 + k. b. 3y2 + 2y3 = 3x2 + 5. 4. y( x ) =
x . 1−ln  x 
5. a. xh (t) = c1 et + c2 e2t . b. x p (t) = 3e3t . c. x (t) = c1 et + c2 e2t + 3e3t . 6. a. y( x ) = c1 e x + c2 e2x + 5. b. y( x ) = k1 + k2 e− x + x3 − 3x2 + 6x. 7. 5 −2x x 2x a. y( x ) = − 38 . 3 e + 11e + 3 e
b. y( x ) = c1 cos x + c2 sin x + 5 sin 3x. c. y( x ) = c1 cos x + c2 sin x + 1 + x sin x. d. y( x ) = c1 x + c2 x2 + ( x + 3x2 ) ln  x . 8. a. y2 ( x ) = x −1 . b. y2 ( x ) = cos( x2 ). 9. a. y( x ) = c1 cos x + c2 sin x − ln  sec x + tan x . b. y( x ) = (c1 + c2 x )e2x + x3 e2x . 10. Proof. 11. a. y( x ) = 11e2x −
38 x 3 e
+ 53 e−2x .
399
400
mathematical methods for physicists
b. y( x ) = 5 cos x + sin3 x. c. y( x ) = 1 + cos x + x sin x. d. y( x ) = π (2x − x2 ) + 4( x − x2 ) + (3x2 + x ) ln  x . 12. a. y(t) = 14 cos t − 10 + 5t2 . b. y(t) = 4 cos t + 2 sin t − 2 cos t ln  sec t − tan t. 13. sinh k ( x −ξ ) . k 3 1 −x 2 − 2 xe . 1 2x 3 4 xe + 8 sinh 2x.
a. G ( x, ξ ) = b. y( x ) = c. y( x ) = 14.
y( x )
=
3 + 14833e 5 − e 3 x + (−5040x −5 + 5040x −4 − 2520x −3 + 8 8x5 +840x −2 − 210x −1 + 42 − 7x + x2 )e x .
15. 760 m and 10.0 s. 16. a. v(t) = −(98m/s)(1 − e−0.002t ) and vterm = −98 m/s. g g b. v(t) = − α + α + v0 e−αt , the maximum height is y(5.076) = 127 m, and the speed when it hits the ground is v(10.17) = −49.7 m/s. h i2/3 √ − R. 17. h(t) = ( R + h0 )3/2 − 32 2GMt 18. a. 136 yr. b. 451 yr. 19. When the damping force is 5 dx dt lbs, there is overdamped motion. When dx the damping force is 4 dt lbs, there is underdamped motion. 20. q(t)
=
I (t)
=
21. a. Derivation.
√ √ √ 1 1 3 − ( cos 50 3t + sin 50 3t)e−50t . 10√ 10 10 √ 20 3 −50t e sin 50 3t. 3
quick answers
b. Plots. c. Derivation. d. Plots. 22. g
a. v(t) = − α +
g α
+ v0 e−αt , α =
c m.
g+αv0 (1 − e−αt ) − gt 2 α. α αv0 1 . α ln 1 + g
b. x (t) = c. T =
d. Plots. e. The rise times are found from trise = T. v0 (m/s) 5 10 15 20
T (s) 0.2534 0.3617 . 0.4316 0.4833
f. The total and fall times are given as v0 (m/s) 5 10 15 20
t f inal (s) 0.6874 1.2176 1.7303 2.2408
trise (s) 0.2534 0.3617 0.4316 0.4833
t f all (s) 0.4340 0.8559 . 1.2987 1.7575
23. a. y( x ) = 2e2x , yexact = 14.78, y Euler = 12.38, and y Midpoint = 14.61. b. y( x ) = x − 1 + 2e− x , yexact = 1.271, y Euler = 1.215, and y Midpoint = 1.275. 2 c. y( x ) = sin x 2−1 , yexact = 0.998, y Euler = 1.061, and y Midpoint = 0.978. 24. T = 1.4192 s for the linear pendulum and t = 1.563 s for the linear pendulum. 25. a. Figure needed. b. Figure needed. c. The maximum speed is about 390 m/s at an altitude of 27 km and time of 52 s. d. No. e. It is roughly 64.01 m/s at 238.8 s, or 1585 m. f. The data was released in February, 2013 at http://www.redbullsratos. com/science/scientificdatareview/.
401
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mathematical methods for physicists
26. a. R = 115 m, H = 16.5 m, and t f = 3.673 s. b. R = 115 m and H = 16.5 m, and t f = 3.673 s. c. Plot needed. d. The Reynolds number is too high to make any difference. 27. a. A solution gives t = 0.1330 s and θ = −0.27408 rad. b. θ = −0.2702 radians. c. θ = −0.2778 radians. d. For the topspin in the problem, we find θ = −0.2692 radians. For no spin in the part of the problem, we find θ = −0.2728 radians. For the bottom spin in the problem, we find θ = −0.2766 radians. 28. Derivation. 29. a. Plots. b. Plots. c. Plots. d. Plots. e. Plots. 30. a. x 00 + 6x 0 + 9 = 0. b. x (t) = (c1 + c2 t)e−3t . c. y(t) = (−c1 − c2 − c2 t)e−3t . d. Verify solution. e. x (t) = (1 − t)e−3t and y(t) = te−3t . 31. a. x (t) = c1 e3t , y(t) = c2 e−2t , saddle point, hyperbolae. √ √ √ √ √ b. x (t) = c1 e 5t + c2 e− 5t , y(t) = − 5c1 e 5t − c2 e− 5t , saddle point, hyperbolae. √ √ √ √ √ c. x (t) = c1 cos 6t + c2 sin 6t, y(t) = 25 c2 cos 6t − c1 sin 6t , center, ellipses. d. x (t) = (c2 − c1 t)et , y(t) = c1 et . e. x (t) = (c1 cos 3t + c2 sin 3t)e2t , y(t) = (c2 cos 3t − c1 sin 3t)e2t . 32. Derivation.
quick answers
33. For 0 < r < 1, r (t) = q
r0 r02
+ (1 − r02 )e−2t
and for r > 1, r (t) = q
Ch.3
r0 r02 − (r02 − 1)e−2t
,
.
Linear Algebra
1. (1, 2, 3) = 3(1, 1, 1) − (2, 1, 0). 2. a. Derivation. b. Derivation. c. 6. e.
n ( n +1) . 2 n ( n −1) . 2
a.
RαT
d.
3.
=
1 R− α
=
cos α − sin α
sin α cos α
!
b. Proof. 4. a. Proof. b. A rotation of −90o about the axis v = (1, 0, 1). 1 √ 0 − √1 2 2 c. SAS−1 for S = 0 1 0 . √1 2
5. φ =
π 4,
θ=
3π 4 ,
ψ=
√1 2
1 2
13π 12 .
6. a. Rˆ (φ, θ, ψ) =
cos ψ cos φ − sin ψ cos θ sin φ
− sin ψ cos φ − cos ψ cos θ sin φ sin θ sin φ
cos ψ sin φ + sin ψ cos θ cos φ
− sin ψ sin φ + cos ψ cos θ cos φ − sin θ cos φ
b. Tr ( Rˆ ) = cos(ψ + φ) + cos(ψ + φ) cos θ + cos θ. c. Derivation. d. Derivation.
sin ψ sin θ
cos ψ sin θ . cos θ
403
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mathematical methods for physicists
7. a. Derivation. b. Derivation. c. Derivation. d. Derivation. 8. x = 1, y = 0, z = −1. 9. a. λ = 1, 6 with corresponding eigenvectors (2, −3) T and (1, 1) T . b. Eigenvectors corresponding to λ = 1 ± i are v = (2 ± i, 1) T . c. Eigenvector corresponding to λ = 4 is v = (1, 0) T . d. λ = 1, −2, 3 the corresponding eigenvectors (−1, 4, 1) T , (−1, 1, 1) T , and (1, 2, 1) T . 10. a. det = 6,
− 12
b. det = 2, c. det = 16,
.
2 3
!
− 12 − 21
5 2 3 2
1 4
1 − 16
0
!
− 31
1 2
. ! .
1 4
1 6
d. det = −6, − 61
− 12 3 2 1 2
1 6
7 6 − 13 6 − 56
.
11. a.
5x2
− 4xy + 2y2
b. λ = 1, 6, v1 = c. θ = −63o .
=
12. !
a. b.
cos 1 0 0 cos 2
c. Proof. 13. a. Proof.
0 0
x
√1 (1, 2) T , 5
d. x 02 + 6y02 = 30.
e2 0
. !
y
v2 =
5 −2
−2 2
!
√1 (−2, 1) T . 5
x y
! .
quick answers
b. Proof. c. Proof. d. Proof. 14. a. √
1
1
√
= c1 e 2 (5+ 17)t + c2 e 2 (5− 17)t , √ √ √ √ 1 1 c1 c (5 − 17)e 2 (5+ 17)t + 2 (5 + 17)e 2 (5− 17)t . y(t) = 2 2
x (t)
b. √
√
= c1 e 5t + c2 e− 5t , √ √ √5t y(t) = − c2 e− 5t . 5 c1 e
x (t)
c. x (t)
= ( c1 + c2 t ) e t ,
y(t)
= − c2 e t .
d. x (t)
= e2t (c1 cos 3t + c2 sin 3t) ,
y(t)
= e2t (−c1 sin 3t + c2 cos 3t) .
e. x (t)
= (c1 + c2 t)e−3t ,
y(t)
= −(c1 + c2 + c2 t)e−3t .
x (t)
= et (c1 cos t + c2 sin t) ,
y(t)
= et (c1 sin t − c2 cos t) .
f.
15. a. Proof. b. The system of equations is x˙ 1
= x3
x˙ 2
= x4
x˙ 3
= −
x˙ 4
=
k k x1 + ( x2 − x1 ) m m k − ( x2 − x1 ). m
405
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mathematical methods for physicists
√
c. The eigenvalues are λ = ± 52±1 iω0 . The corresponding eigenvectors are 1√ 1√ 1− 5 1+ 5 2 √ 2 , √ v1 = 1+ 5 , v2 = 5−1 2 iω0 2 iω0 −iω0 iω0
1√
1− 5 2 √ v3 = 1+ 5 − 2 iω0 iω0
,
v4 =
1√
1+ 5 √2 1− 5 2 iω0
−iω0
.
d. i. The solutions for the first initial conditions are
√
√ √ √ 5) cos(5 + 5)t + (5 + 5) cos(5 − 5)t √ √ √ √ (5 − 3 5) cos(5 + 5)t + (5 + 3 5) cos(5 − 5)t.
x1
= (5 −
x2
=
ii. The solutions for the second set of initial conditions are
√ √ √ √ = (5 + 3 5) cos(5 + 5)t + (5 − 3 5) cos(5 − 5)t √ √ √ √ = (−5 − 5) cos(5 + 5)t + (−5 + 5) cos(5 − 5)t.
x1 x2 16.
0 0
0 0
k2 m2
k2 m1 − k2m+2k3
a. A = k1 +k2 − m1
1 0 0 1 0 0 0
.
0
b. The general solution is
√
√
√
√
x1
= c1 cos
x2
= c1 cos 5t + c2 sin 5t − c3 cos 15t − c4 sin 15t √ √ √ √ √ √ = 5(c2 cos 5t − c1 sin 5t) + 15(c4 cos 15t − c3 sin 15t) √ √ √ √ √ √ = 5(c2 cos 5t − c1 sin 5t) + 15(−c4 cos 15t + c3 sin 15t)
x3 x4
√
5t + c2 sin
√
5t + c3 cos
√
15t + c4 sin
√
15t
c. The particular solution is
√
= −2.50 cos
x2
= −2.50 cos 5t + 7.50 cos 15t √ √ √ √ = 2.50 5 sin 5t + 7.50 15 sin 15t √ √ √ √ = 2.50 5 sin 5t − 7.50 15 sin 15t
x3 x4
√
5t − 7.50 cos
√
x1
√
15t
d. The general solution is x(t) = c1 a1 cos
√
5t + c2 a1 sin
√
5t + c3 a2 cos
where a1 = (1, 1) T and a2 = (1, −1) T .
√
15t + c4 a2 sin
√
15t,
quick answers
17. a. The system of equations is d dt
q I
!
=
0 −10000
1 −100
!
q I
!
+
0 1000
! .
b. The particular solution satisfying the given initial conditions is
√ √ 1 = −0.10(cos 50 3t + √ sin 50 3t)e−50t + 0.10, 3 √ 20 I (t) = √ sin(50 3t)e−50t . 3
q(t)
c. Plot the above solutions. 18. a. The system is given by q˙ 1
= 10 − 1000q2
q˙ 2
= 10 − 1000q2 − q4
q˙ 3
= q4
q˙ 4
= 10000q2 − 100q4 .
b. The charge and the current across the capacitor are given by √ 1 1 √ + ) 77)e(−550+50 77)t 220 1540 √ 1 √ 10 +(− ) 77 − 1/220)e(−550−50 77)t + 1540 11 √ √ √ 45 45 √ 77)e(−550+50 77)t + ( 77 + 5)e(−550−50 77)t . (5 − 77 77
q2
= (−
I2
=
c. Plot the solution in part b. 19. a. x (t) = 600(1 − e−t/50 ). b. 54.9 min. c. 600 lbs. d. 63.2 min, unbounded. 20. a.
dx dt
= − x/32, x (0) = 15 tsp.
b. x (t) = 15e−t/32 . c. 35 min.
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21. The concentration as found as
[ A] = 0.4129e−0.03139t + 0.5870e−0.3186t [ B] = 0.6963e−0.03139t − 0.6963e−0.3186t [C ] = 1.0 − [ A] − [ B] = 1.0 − 1.109271796e−0.03139t + .1092717960e−0.3186t 22. The eigenvalues for this model are λ = −3 ± the initial conditions are given by
√
3. The populations given
999 500 √ (−3+√3)t 999 500 √ (−3−√3)t + +( − 3) e 3) e 2 3 2 3 √ √ √ √ 1 1997 1 1997 3)e(−3+ 3)t + ( − 3)e(−3− 3)t ( + 2 6 2 6 999 √ (−3+√3)t 999 √ (−3−√3)t 1000 − (500 + 3) e 3) e − (500 − . 2 2
S
= (
I
=
D
=
23. In all cases the origin is the equilibrium point.
√ a. The eigenvalues are λ = ± 2i. This is a center with solutions to the initial value problem given by √ √ √ = cos( 2t) + 2 sin( 2t), √ √ √ 2 sin( 2t). J (t) = cos( 2t) − 2 √ b. The eigenvalues are λ = ± 2. This is a saddle with solutions to the initial value problem given by R(t)
√ √ √ √ 1 1 1 + 2 e 2t + 1 − 2 e− 2t , 2 2 √ √2t 1 √ √ 1 J (t) = 2+ 2 e + 2 − 2 e− 2t . 4 2 √ c. The eigenvalues are λ = 12 (−1 ± 7i ). This is a stale focus with solutions to the initial value problem given by √ √ ! 7 3√ 7 −t/2 R(t) = e cos( t) + 7 sin( t) , 2 7 2 √ √ √ ! 7 7 7 −t/2 t) − sin( t) . J (t) = e cos( 2 7 2 R(t)
Ch.4
=
Nonlinear Dynamics
1. y(t) =
ky0 . cy0 +(k −cy0 )e−kt
2. a. y = −2 is stable and y = 8 is unstable.
quick answers
b. y = 2k2−1 π, k = 1, 2, . . . . Solutions are stable for k even and unstable for k odd. c. y = 0 stable and y = 2, −3 unstable. d. y = −1 stable, y = 4 unstable, and y = 0 is neither. 3. The general solution is y(t) =
y0 . y0 + (1 − y0 ) e − t
1 . 1+3e−t 3 y(0) = 1.5, y(t) = 3−e−t . 1 y(0) = −0.5. y(t) = 1−3e −t .
a. y(0) = 0.25, y(t) = b. c. 4.
a. y = 0 is stable and y =
1 µ
is unstable.
b. y = 0, µ is stable and y = √ c. x = 3 µ is stable.
µ 2
is unstable.
d. x = 0 is stable for µ > 1 and unstable for µ < 1. x = ± is unstable for µ > 1.
p
µ−1
5. a. Plots b. Plots 6. The general solution is
√ r (t)
=
=
µr0
q
r02 + (µ − r02 )e−2µt √ µ r . µ −r 2 1 + r2 0 e−2µt 0
√ a. r (0) = 0.25, r (t) = 1/ 1 + 15e−2t . √ b. r (0) = 1.5, r (t) = 3/ 9 − 5e−2t . 7. r 0 = r (µ − r2 ), θ 0 = 1. 8. a. (0, 0) is an unstable node, (0, 25) is a saddle, (100, 0) is a saddle, and (75, 25/2) is a stable node. b. (0, 0) is an unstable node. c. (0, 0) is an unstable node, (1, 0) is a saddle, (0, 1/3) is a saddle, and (8/7, 1/7) is a stable focus. d. (0, 0) is an unstable node, (0, 1) is a stable node, and (0, −1) is a saddle.
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9. (0, 0) is stable for µ < 0 and unstable for µ > 0. The limit cycle is unstable for µ < 0 and stable for µ > 0. 10. Plot 11. a. Description of terms. b. There are four equilibrium points d a ae − cd bd − a f (0, 0), (0, ), ( , 0), ( , ). e b be − c f be − c f c. For parts c and d, • (0, 0) is an unstable node. • (0, de ) is a saddle for cd − ae > 0 or a stable node for cd − ae < 0. • ( ba , 0) is a saddle for bd − a f > 0 or a stable node for bd − a f < 0. bd− a f
ae−cd • ( be −c f , be−c f ) can have a variety of behaviors depending on the constants.
12. a. The equilibria are 2 1 4 (0, 0, 0), (0, 0, 1), (1, 0, 0), (1, 0, 1), (0, 1, 0), ( , , ). 3 3 3 b. The Jacobian matrices evaluated at each point are
1 J (0, 0, 0) = 0 0
0 0 1
0 1 . 0
−1 −1 0 J (1, 0, 0) = 0 0 1 0 2 0 0 0 0 J (0, 1, 0) = 0 0 2 1 −1 −1
. .
1 J (0, 0, 1) = 0 0
0 1 0
0 −1 . 0
−1 −1 0 J (1, 0, 1) = 0 1 −1 . 0 1 0 − 23 − 32 0 21 4 4 J( , ) = 0 − 43 . 3 33 3 1 1 1 −3 −3 3
c. The eigenvalues and eigenvectors are • (0, 0, 0), λ = ±1, 0 0 1 (v1 , v2 , v3 ) = −1 1 0 . 1 1 0
quick answers
• (0, 0, 1), λ = 1,
0 (v1 , v2 , v3 ) = 1 1
√ • (1, 0, 0), λ = −1, ± 2, 1 (v1 , v2 , v3 ) =
2 (−2√+ 2 2
√
2)
1 0 0
√
1 2 (−2 − √ − 22
1 • (0, 0, 1), λ = −1, 21 (1 ± 0 (v1 , v2 , v3 ) = 0 0 • (0, 1, 0), λ = 0, 21 (−1 ±
√
0 1 . 0
2)
1
0 . 0
1
3i ),
√
√
1 6 (−3 −√ 3i ) 1 2 (1 + 3i )
1 6 (−3 +√ 3i ) 1 2 (1 − 3i )
1
1
√
.
7),
0 √ (v1 , v2 , v3 ) = 12 (−1 − 7i ) 1
0 √ 1 2 (−1 + 7i ) 1
1 0 . 0
• ( 32 13 , 34 ), λ = −0.6464 ± 0.3589i, 1.6261 −.7156 −.7156 (v1 , v2 , v3 ) = 0.02176 + 0.3853i 0.02176 − 0.3853i 0.1360 + 0.5662i 0.1360 − 0.5662i
0.2732 −0.9396 . 0.2063
d. • (0, 0, 0), λ = ±1, This is a saddle. • (0, 0, 1), λ = 1,
√ • (1, 0, 0), λ = −1, ± 2, This is a saddle √ • (0, 0, 1), λ = −1, 21 (1 ± 3i ), √ • (0, 1, 0), λ = 0, 21 (−1 ± 7), • ( 23 13 , 34 ), λ = −0.6464 ± 0.3589i, 1.6261
e. The only equilibrium point with all three populations as nonzero is ( 32 13 , 34 ). 13. Derivation. 14. Proof. 15. a. The equilibrium points are q q (0, 0, 0), ( β(ρ − 1), β(ρ − 1), ρ − 1),
(−
q
β ( ρ − 1), −
q
β ( ρ − 1), ρ − 1).
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mathematical methods for physicists
b. The Jacobian matrices at the equilibrium points are (for p ξ = β ( ρ − 1)) σ −σ 0 J (0, 0, 0) = ρ −1 0 . 0 0 −β
−σ σ 0 J (ξ, ξ, ρ − 1) = 1 −1 −ξ . ξ ξ −β −σ σ 0 J (−ξ, −ξ, ρ − 1) = 1 −1 ξ . −ξ −ξ − β c. This is done for special values of the parameters. d. Plots. 16. [ ES] = 0 and either [ E] = 0 or [S] = 0. v = k3 [ ES] = 17. a. Proof. b. (0, 0, N ) c. Plot. d. Equilibrium is a stable node. e. S = 6, I = 31, R = 62. q q − e 1 2 2 ω + 2 ea (t − t0 ); 2ω2 +ea2 a . 18. x (t) = a sn 19. C = 4aE(k). 20. a. F ( π4 , √1 ) ≈ 0.8260. 2
b. K ( 12 ) ≈ 1.6858. c. d. e.
Ch.5
1 2 3 K( 3 )
√
≈ 0.6023.
√
2 ) ≈ 2.6221. √2 2 2 2 K ( 2 ) ≈ 1.3110.
√
2K (
The Harmonics of Vibrating Strings
1. cos(1−t) cos 1 . e−4x +2e−6 e2x . 1+2e−6
a. x (t) = 2 − b. y( x ) =
c. y( x ) = cos x.
k 3 [ E ]0 [ S ] . K +[S]
quick answers
2. The product solutions are un ( x, t) = cos
2 2n − 1 πx e−kλn t , 2L
n = 1, 2, . . . .
The general solution is ∞
∑ an cos
u( x, t) =
n =1
2n−1 2 2n − 1 πx e−k( 2L π ) t . 2L
3. The product solutions are 2n − 1 un ( x, t) = sin πx ( a cos ωn t + b sin ωn t), 2 where
2n − 1 πc, 2
ωn =
n = 1, 2, . . . ,
n = 1, 2, . . . .
The general solution is ∞
u( x, t) =
∑ (an cos ωn t + bn sin ωn t) sin
n =1
2n − 1 πx . 2
4.
√
a.
λn =
2n−1 2 π,
n = 1, 2, . . . , yn ( x ) = sin
2
2n−1 2 πx
.
b. λn = − 2n4−1 , n = 1, 2, . . . , yn ( x ) = sin 2n4−1 ( x + π ) . √ c. λn ln 2 = nπ, n = 1, 2, . . . , yn ( x ) = sin( nπlnln2 x ).
√
tan(
5. y( x ) =
1 24 C ( x
4λ−1 ) 2
=
2 e
√
√
4λ−1 2 4λ−1 . 2
d. y( x ) = c2 e− x/2 sin(
− L)4 + 16 CL3 x −
ln x ), where the λ’s are the roots of
CL4 24 .
6. y(t) = 3 cos 2t − 4 sin 2t = 5 cos(2t + tan−1 34 ). 7. Derivation. 8. Proof. 9. a. f (x) ∼ b. 2=
16 π2
4 π
1+
∞
sin(2k − 1) x . 2k − 1 k =1
∑
1 1 1 + + + · · · . 32 52 72
c. Proof. 10. a. {
q
2 π
sin 2nx },
n = 1, 2, 3, . . . ,
0 ≤ x ≤ π.
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mathematical methods for physicists
b. {cos nπx }, c.
{ √1 L
sin
0 ≤ x ≤ 2.
n = 0, 1, 2, . . . ,
nπx L },
n = 1, 2, 3, . . . ,
x ∈ [− L, L].
11. 3 4
a. sin3 θ =
sin θ − 41 sin 3θ.
b. 4 sin3 2x = 3 sin 2x − sin 6x. 12. ∞
a. x ∼ π − 2 b.
sin nx . n n =1
∑
∞ x2 π2 (−1)n cos nx. ∼ +∑ 2 12 n=1 n2 ∞
c. f ( x ) ∼
∞ 1 − cos nπ sin(2k − 1) x sin nx = 2 ∑ . n 2k − 1 n =1 k =1
∑
13. n+1 sin nx ,  x  < π. a. f ( x ) ∼ 2 ∑∞ n=1 (−1) n
b. f ( x ) ∼
π 2
c. f ( x ) ∼
1 2
− +
∑∞ n =1
cos(2k −1) x ,  x  < π. (2k−1)2 ∞ sin(2k −1) x 2 ,  x  < π. π ∑ n =1 2k −1 4 π
14. ∞ sin nπx 2 , 0 < x < 2. π ∑ n =1 n n +1 (− 1 ) ∞ 4 sin nπx π ∑ n =1 n 2 , −2 < x ∞ sin 2nπx 3 1 , 1 < x < 2. 2 − π ∑ n =1 n
a. f ( x ) ∼ 1 − b. f ( x ) ∼ c. f ( x ) ∼
< 2.
15. a. Proof. b. Proof. 16. (−1)n
a. f ( x ) ∼ 13 + π42 ∑∞ n=1 n2 cos nπx, 0 < x < 1. ∞ 2 f ( x ) = ∑n=1 (− nπ cos nπ + (nπ4 )3 (cos nπ − 1)) sin nπx, 0 < x < 1. b. f ( x ) ∼
d.
−
16 π2
∑∞ k =1
(2k−1)πx
cos 2 (2k−1)2
(2k−1)πx sin 2 (2k−1)3
. 0 < x < 2.
∞ 32 , 0 < x < 2. π 3 ∑ k =1 2 nπ nπx f ( x ) ∼ 12 − ∑∞ n=1 nπ sin 2 cos 2 , 0 < x < 2. ∞ nπ f ( x ) = ∑n=1 cos 2 − cos nπ sin nπx 2 ,0< x 0 and J0 (0) = 1. 0 (0) = − 1 , and J 0 (0) = 0 for n 6 = ±1. ii. J10 (0) = 12 , J− n 1 2
18. a. Derivation. b. Proof. c. Proof. 19. a.
∞
Jν ( x ) =
x ν+2k (−1)k , Γ ( k + 1) Γ ( ν + k + 1) 2 k =0
b.
∑
ν ≥ 0.
∞
J−ν ( x ) =
x 2k−ν (−1)k , ∑ Γ ( k + 1) Γ ( k − ν + 1) 2 k =0
ν cannot be an integer. c. Derivations. q 2 d. J3/2 ( x ) = πx
sin x x
− cos x .
ν ≥ 0.
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mathematical methods for physicists
20. a. αc = 0, 3.832, 7.016, 10.173, 13.324. b. Derivation. c. Derivation. d. x2 =
∞ 4J (α c) c2 − 4 ∑ 2 22 k J0 (α j x ), 2 α j=2 k J0 ( α j c )
0 < x < c.
21. Proof. 22. Proof. 23. L† v = v00 − 4v0 − 3v and v0 (0) = 0, and v0 (1) = 0. 24. Proof. 25. Derivation. 26. a. yn ( x ) = cos nx, eigenfunction.
λ = n2 , n = 1, 2, . . . and y( x ) = 1 is the λ = 0
b. yn ( x ) = sin( nπ 2 ln x ),
λn =
n2 π 2 4 ,n
= 1, 2, . . . .
27. Derivation. 28. a. Derivation. b. λ1 ≤ 12. c. For φ( x ) = sin πx 2 , λ1 ≤ 29.
a. y( x ) = ∑∞ n =1
2 n5 π 5
π2 4
+ 31 +
2 π2
≈ 3.0034.
2 − (2 − n2 π 2 )(−1)n sin nπx.
b. For the case in which y0 (0) = y0 (1) = 0, we have y( x ) =
i ∞ 1 2 h + ∑ 5 5 2 − (2 − n2 π 2 )(−1)n cos nπx. 12 n=1 n π
For the case in which y(0) = y(1) = 0, we have y( x ) =? 30. For u0 (π/4) = β, the condition is Z π/4 0
f ( x ) sin 2x dx = β + α.
31. a. y( x ) = x − π − sin x.
quick answers
b.
(
x − π, ξ − π,
G ( x, ξ ) =
0 ≤ ξ ≤ x, x ≤ ξ ≤ π.
c. y( x ) = x − π − sin x. d. y( x ) = 6x − 6π − 3 − sin x. 32. a.
( y( x ) =
x0 − π, x − π,
0 ≤ x ≤ x0 , x0 ≤ x ≤ π.
b. Verification. c. Verification. 33. a. y( x ) = b.
sinh x sinh 1
− x.
− sinh(1 − x ) sinh ξ , sinh 1 G ( x, ξ ) = − sinh(1 − ξ ) sinh x , sinh 1
c.
2 sin nπx sin nπξ . ( n2 π 2 + 1) n =1
∑
∞
y( x ) =
2(−1)n sin nπx. 2 2 n=1 nπ ( n π + 1)
∑
d. Verification.
Complex Representations of Functions
1. a. (4 + 5i )(2 − 3i ) = 23 − 2i. b. (1 + i )3 = −2 + 2i. c.
0 ≤ x ≤ ξ ≤ 1.
∞
G ( x, ξ ) = −
Ch.7
0 ≤ ξ ≤ x ≤ 1,
5+3i 1− i
= 1 + 4i.
2. a. i − 1 =
√
2e3πi/4 .
b. −2i = 2e−πi/2 . √ √ c. 3 + 3i = 2 3eπi/3 . 3.
√ a. 4eiπ/6 = 2 3 + 2i. √ b. 2e5iπ/4 = −1 − i. c. (1 − i )100 = −250 .
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p p p p p p √ √ √ √ √ √ 4. p 2 + 2 + ip 2 − 2, − 2 − 2 + i 2 + 2, − 2 + 2 − i 2 − 2, √ √ 2 − 2 − i 2 + 2. 5. Proof.
√ 6. z = 2πk + i ln(2 ± 3), k is any integer. Note: This includes solutions √ √ with imaginary part − ln(2 + 3) = ln(2 − 3). 7. ii = e−π/2 . 8. a. ( x − 1)2 + y2 = 1 b. Sketch. c. r2 = 2 cos 2θ, ( x2 + y2 )2 = 2( x2 − y2 ), and sketch. 9. a. u( x, y) = x3 − 3xy2 , v( x, y) = 3x2 y − y3 . b. u( x, y) = sinh x cos y, v( x, y) = cosh x sin y. c. u( x, y) = cos x cosh y, v( x, y) = sin x sinh y. 10. a. f 0 (z) = 3z2 . b. f 0 (z) = cosh z. c. This function is not holomorphic. 11. Proof. 12. Line, y = 2x + 7. 13. a. ( x − 2)2 + (y + 1)2 = 9. b. x (θ ) = 3 cos θ + 2, y(θ ) = 3 sin θ − 1. c. z = 3eiθ + 2 − i. 14. a. Proof. b. v( x, y) = 3x2 y − y3 + k, k is an arbitrary constant. c. f (z) = z3 + k. d. f 0 (z) = 3z2 . 15. a. b. c.
R RC RC
z dz = 1 + 3i . f (z) dz =
1 C z2 +4
1 2
dz = 0.
+ 2i.
quick answers
16. F (2i ) = −8π (1 + i ), F (2) = 0. 17. Proof. 18. Proof. 19. a. Removable. b. Pole, order 1 (simple pole). c. Pole, order 1 (simple pole). d. Essential. e. Essential. 20.
sinh z z3
=
1 z2
+ 16 +
1 2 120 z
+ O ( z3 ).
21. ∞
a. f (z) = − ∞
f (z) =
∑ zn+1 , z < 1,
n= 0
1 n , z > 1. z "
∑
n =0
# ∞ 1 1 ∞ z n n b. f (z) = i (−iz) − ∑ (− ) , z < 1, 2 + i n∑ 2 n =0 2 = 0 # " ∞ i n 1 ∞ z n 1 1 ( ) − ∑ (− ) , 1 < z < 2, f (z) = 2 + i z n∑ z 2 n =0 2 " =0 # 1 1 ∞ i n 1 ∞ 2 n f (z) = ( ) − ∑ ( ) , z > 2, 2 + i z n∑ z n =0 z =0 z 22. 2z2 + 3z ; z = 1] = 5. z−1 ln(1 + 2z) b. Res[ ; z = 0] = 0. z cos z π c. Res[ ; z = ] = 0. 2 (2z − π )3 a. Res[
23. Z 2π
dθ 2π = . 5 − 4 cos θ 3 0 b. Proofs and verifying answer in part a.
a.
24. a. b. c.
ez
I
dz = −1.
z−i =3 z2 + π 2 I z2 − 3z + 4 z−i =3 z2 Z ∞ sin x −∞
− 4z + 3
x2 + 4
dx = 0.
dz = −2πi.
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mathematical methods for physicists
25.
Z ∞ (ln x )2 0
1+
x2
dx =
π3 . 8
26. a. b. c. d. e. f. g.
Ch.8
dz
5π i. 18 π dθ = . 18 0 13 − 12 cos θ Z ∞ dx = 0. 2 x + 5x + 6 −∞ √ Z ∞ cos πx 3 dx = π. 2 24 1 − 9x 0 Z ∞ dx 2π =− . 2 2 75 −∞ ( x + 9)(1 − x ) √ Z ∞ π x dx = . 2 0 (1 + x )2 √ Z ∞ π x dx = . Repeated problem, 2 2 0 (1 + x ) I
C z ( z − 1)( z − 3)2 Z π sin2 θ
=
Transform Techniques in Physics
1. a. Proof. b. Proof. 2. Proof. 3. a. b. c. d. e. 4.
R∞
−∞
Rπ
sin xδ x − π2 dx = 1. Z ∞ x − 5 2x 2 126 δ e 3x − 7x + 2 dx = 10 . 3 e −∞ Rπ 2 π x δ x + dx = 0. R0∞ −2x 2 2 e δ( x − 5x + 6) dx = e−4 + e−6 . R 0∞ 2 2 −∞ ( x − 2x + 3) δ ( x − 9) dx = 4. 0
δ( x2 − 5x + 6)(3x2 − 7x + 2) dx = 8.
5. δ( x ) = . 6. fˆ(k) =
2a . a2 + k 2
4a( a2 − 3k2 ) 7. fˆ(k) = . ( a2 + k 2 )3 r π ( k + i )2 exp − 8. fˆ(k) = . 2 8 9. Proof.
quick answers
10. A . α − i ( ω + ω0 ) A2 b.  fˆ(ω )2 = 2 . α + ( ω + ω0 ) 2 c. Add plot. a. fˆ(ω ) =
11. Proof. 12.
Z ∞
r
π β2 /4α dt = e . α −∞ r π − ab x2 b. ( f ∗ g)( x ) = e a+b . a+b a.
e
−αt2 + βt
13. The sketch of the convolution is the triangular function 0,  x  ≥ 2a Z ∞ f a (t) f a ( x − t) dt = 2a + x, −2a ≤ x ≤ 0 −∞ 2a − x, 0 ≤ x ≤ 2a. 14. 2n − 1 In−1 , √ √2 √ π 3π 15 π , I2 = , I3 = . b. I1 = 2 4 8 (2n − 1)!! √ c. In = π. 2n a. In =
15. a. L[9t2 − 7] =
18 7 − . s s3
e −3 . s−5 s c. L[cos 7t] = 2 . s + 49
b. L[e5t−3 ] =
2 . ( s − 4)2 + 4 1 s−2 L[e2t (t + cosh t)] = + . 2 ( s − 2) ( s − 2)2 + 1 2 + 2s + s2 L[t2 H (t − 1)] = e−s . s3 1 + se−4πs L[sin t + cos tH (t − 4π )] = . 2 s + 1 Rt 2 1 L[ 0 (t − u)2 sin u du] = 3 . s s2 + 1 2 10 25 (s − 2)2 − 9 L[(t + 5)2 + te2t cos 3t] = 3 + 2 + + . s s s ( s − 2)2 + 9
d. L[e4t sin 2t] = e. f. g. h. i. 16. a. L−1 [
18 7 + ] = 9t2 + 7. s s3
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mathematical methods for physicists
1 2 − 2 ] = e5t − sin 2t. s−5 s +4 s+1 ] = cos t + sin t. L −1 [ 2 s +1 3 L −1 [ 2 ] = 3e−t sin t. s + 2s + 2 1 ] = tet . L −1 [ ( s − 1)2 e−3s L −1 [ 2 ] = sinh(t − 3) H (t − 3). s −1 1 1 L −1 [ 2 ] = e−2t sinh 3t. 3 s + 4s − 5 s + 3 L −1 [ 2 ] = e−4t (cos t − sin t). s + 8s + 17
b. L−1 [ c. d. e. f. g. h. 17.
t6 12 , L[ f ∗ g] = 7 . 60 s b. ( f ∗ g)(t) = 12 t − 14 sin 2t, L[ f ∗ g] =
a. ( f ∗ g)(t) =
2 . ( s2 +4) s2
s2 − 2s + 6 . s3 ( s + 3) s2 − 2s + 6 − 43 t + t2 , L[ f ∗ g] = . ( s + 3) s3
c. ( f ∗ g)(t) = 97 (1 − e−3t ) + t2 − 43 t, L[ f ∗ g] = d. ( f ∗ g)(t) = − 79 e−3t +
7 9
18. 1 . s (1 + e − s ) e2s b. F (s) = . s (1 + e s ) 1 − 2se−s + e−2s . c. F (s) = (1 − e−2s )s2
a. F (s) =
19. 2 ] = 2t − 2 sin t. s2 ( s2 + 1) e−3s b. L−1 [ 2 ] = (t − 3) H (t − 3). s 1 1 1 ] = − (2 cos 2t + sin 2t)e−t . c. L−1 [ 4 2 5 10 s ( s + 1) a. L−1 [
20. 1 1 3 1 4t + 3 −4t ] = t2 + − t− e . 32 256 32 256 s3 ( s + 4)2 1 e5t e−t 1 b. L−1 [ 2 ]= − = e2t sinh 3t. 6 6 3 s − 4s − 5 s + 3 c. L−1 [ 2 ] = (cos t − sin t)e−4t . s + 8s + 17 s+1 1 1 d. L−1 [ ] = − e−4t + (1 + 6t)e2t . 12 12 ( s − 2)2 ( s + 4)
a. L−1 [
quick answers
e. L−1 [
( s2
s2 + 8s − 3 ] = 2 cos t + 4 sin t − (2 + 5t)e−t . + 2s + 1)(s2 + 1)
21. a. y(t) = −4e3t + 6e2t . b. y(t) = − 59 e−t + et + 19 (3t − 4)e2t = 4)e2t .
4 9
cosh t +
14 9
sinh t + 91 (3t −
c. y(t) = 3 cos 2t + 21 sin 2(t − 1) H (t − 1). h i d. y(t) = − 19 + 19 e−3(t−π ) (cos 3t + sin 3t) H (t − π ). 22. x (t)
=
y(t)
=
1 + a2 + b2 − at 1 (e sin bt − e at sin bt) + (e−at cos bt + e at cos bt), 2 √ 4ab 5 1√ 1√ sinh( 10t) sin( 2t). 5 2 2
23. a. x (t)
=
y(t)
=
68 3 2t 4 cos 2t − sin 2t + e (47 cos 3t + 16 sin 3t), 145 145 145 54 3 2t 48 cos 2t − sin 2t + e (16 cos 3t − 47 sin 3t). − 145 145 145
b. 1 −t e − 4 1 −t e + 4
1 −3t 2 e (4t + 2t − 7), 4 1 −3t 2 e (4t + 10t − 5). 4
x (t)
=
y(t)
=
x (t)
= 3et cos t − sin t(et − 1),
y(t)
= cos t(1 + et ) + sin t(1 + 3et ).
c.
24. a. q¨ + 100q˙ + 10000q = 1000. b. q(t)
=
I (t)
=
√ √ √ 1 1 − e−50t (3 cos(50 3t) + 3 sin(50 3t)), 10√ 30 √ 20 3 −50t e sin 50 3t. 3
c. q(t)
=
I (t)
=
h A −3π cos(300πt) − sin(300πt)(9π 2 − 1) i √ √ √ +π 3 cos(50 3t) + 3 sin(50 3t)(18π 2 − 1) e−50t , h 10π A (3 − 27π 2 ) cos(300πt) + 9π sin(300πt)− i √ √ √ (3 − 27π 2 ) cos(50 3t) + 3(9π 2 + 1) sin(50 3t) e−50t .
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d. Plots. 25. ∞
a.
(−1)n π = . 1 + 2n 4 n =0
b.
1 11 = . n ( n + 3 ) 18 n =1
c.
(−1)n 5 2 = − ln 2. n ( n + 3) 18 3 n =1
∑ ∞
∑ ∞
∑
d. ∞
(−1)n 2 2 n =0 n − a
∑
=
1 2a
= −
Z 1 a −1 (z − z − a −1 )
1+z
0
dz
π 1 csc πa + . 2a πa
e. ∞
1 ∑ (2n + 1)2 − a2 n =0
= =
1 2a
Z 1 a u − u− a
1 − u2 π πa − tan . 4a 2 0
∞
f.
ea 1 − an e = ln ( ). ∑n ea − 1 n =1
a.
1 = 1. n ( n + 1) n =1
b.
1 1 = . ( n + 2 )( n + 3 ) 3 n =1
a.
x e x −1
26. ∞
∑ ∞
∑
27.
= 1 − 12 x +
b. B0 = 1,
B1 =
1 2 12 x − 21 ,
1 4 720 x + . . . . B2 = 16 , B4 =
−
1 − 30 .
c. ζ (2)
=
ζ (4)
=
ζ (6)
=
ζ (8)
=
π2 . 6 π4 . 90 π6 . 945 π8 . 9450
28. a. L−1 [
∞ 1 1 1 cos(2k − 1)πt ] = .. t + − 2 ∑ 2 − s 2 4 s (1 + e ) (2k − 1)2 π 2 k =1
du
quick answers
b. L−1 [
∞ 1 (−1)n ] = t+2 ∑ sin nπt. s sinh s nπ n =1
sinh s 8 ∞ cos((2k − 1)πt/2) . ] = 1 − s2 cosh s π 2 k∑ (2k − 1)2 =1 √ sinh( β sx ) x 2 ∞ (−1)n nπx −n2 π2 t/βL √ d. L−1 [ ]= + ∑ sin e . L π n =1 n L s sinh( β sL) c. L−1 [
29. ∞
u( x, t)
=
2(1 − cos nπ ) −2n2 π2 t e sin nπx n3 π 3 n =1
=
2 2 4 e−2(2k−1) π t sin((2k − 1)πx ) 3 π3 ( 2k − 1 ) k =1
∑ ∞
Ch.9
∑
Vector Analysis and EM Waves
1. a. u × v = −6i − 9j − 4k. b. u × v = −i + 2j − k. c. u × v = 16i − 13j − 22k. 2. a.
b.
3 1 −1 1 4 2
3. 1, i = 1, j = 3. a. ei2j = −1, i = 3, j = 1, 0 otherwise. ( −1, i = 2, b. ei13 0 otherwise. c. eij1 ei32 = 0. 4. a. Proof. b. Proof. c. Proof. d. Proof.
2 4 4
0 −2 3
= 58.
2 2 −6 3 = 37. 3 1
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mathematical methods for physicists
5. Proof. 6. a. Proof. b. Proof. 7. Proof. 8. a. v = (cos ωt − tω sin ωt)i + (sin ωt + tω cos ωt)j. b. a = (−2ω sin ωt − tω 2 cos ωt)i + (2ω cos ωt − tω 2 sin ωt)j. c. a = 2ω θˆ − ω 2 rˆr. 9. a. ∇ f = 2xi − 2yj. b. ∇ f = (y + z)i + (z + x )j + (y + x )k. y
c. ∇ f = − x2 +y2 i +
x j. x 2 + y2
d. ∇ f = (2y x ln x cos z)i + (2xy x−1 cos z + 5 sin z sin y)j + (−2y x sin z − 5 cos z cos y)k. 10. a. Dn f ( x, y) =
√
b. Dn f ( x, y) =
2.
1 4.
c. Dn f ( x, y) = 0. 11. a. u = i + j + k = (1, 1, 1). √ Nivleks b. dT . s dt = −2k 3 12. (0, 0) is a saddle point. 13. a. ∇ · F = 2, ∇ × F = 0. b. ∇ · F = 0, ∇ × F = 1r k. c. ∇ · F = 2xy + xy = 3xy, ∇ × F = xzi − yzj − x2 k. 14. a. C × (A × (A × C)) = (A · C)C × A. b. ∇ · (∇ × A) = 0. c. ∇ × ∇φ = 0.
quick answers
15. a. Proof. b. Proof. c. Proof. 16. a. ∇ × (k × r) = 2k. r 2 r = r. r c. ∇ × r = 0. d. ∇ · rr3 = 0. e. ∇ × rr3 = 0.
b. ∇ ·
17. a. ∇ × F = 0. b. φ( x, y, z) = − GmM r . 18. E =
3(p·r)r 1 4πe0 ( r5
−
p ). r3
19. a. Proof. b. Proof. c. Proof. 20.
√ dt = 2 2. √ √ √ R2 √ 2 b. 1 9tt +1 dt. = 37 − 10 + ln 2√ 10+2 . 37+1 R2 p 2 c. −2 1 + sinh x dx = 2 sinh 2. a.
R2q
a.
R
1
9+
1 t2
21.
b. c. 22.
R
C (x
2
RC RC C
y2 dx − 2x2 dy = 3. y2 dx − 2x2 dy = − 48 5 . y2 dx − 2x2 dy = −2.
− 2xy + y2 ) ds = 8π.
23. Proof. 24. Proof. 25. a. Proof. b. Proof.
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mathematical methods for physicists
26. a. Proof. b. Proof. c. Proof. 27.
R C
F · dr = 150.
28. a. b. c. d. e. 29.
R C
R RC RS
( x2 + y) dx + (3x + y3 ) dy = 4π. ( x − y) dydz + (y2 + z2 ) dzdx + (y − x2 ) dxdy =
(y − z) dx + (3x + z) dy + ( x + 2y) dz RC 2 x y dx − xy2 dy = −8π. RC 2 2 2 S x y dydz + 3y dzdx − 2xz dxdy = 0.
−y3 dx + x3 dy − z3 dz =
=
4π 3 .
9π 2 .
3π 2 .
30. Derivation. 31. hr = 1,
hθ = r,
hz = 1 plus derivations.
32. hρ = 1,
hθ = ρ,
hφ = ρ sin θ plus derivations.
33.
Ixx I = Iyx Izx
Ixy Iyy Izy
Ixz 9.0 Iyz = 4.0 Izz −3.0
4.0 11.0 −1.0
−3.0 −1.0 . 14.0
34. a.
π I= −1 9 −1 a15
−1 −1 π −1 −1 π
b. λ = π − 2, π + 1, π + 1. c. vT = (1, 1, 1), vT = (−1, 1, 0), and vT = (−1, 0, 1). 35. a. Proof. b. Proof. 36. Proof. 37. Proof.
quick answers
Ch.10
Extrema and Variational Calculus
1. a. The critical points lie along the line y = − x. Test indecisive. b. (0, 0). Test indecisive. c. There are three saddle points (0, 0), (1, 0), (0, 1), and one local minimum, (5−1/3 , 5−1/3 ).. d. (− 85 , − 26 15 ) is a local minimum. e. (0, 0) is a saddle point and (0, 2) is a local minimum. 2. a. 2 2
!
H ( f )(0, 0) =
2 2
0 0
!
H ( f )(0, 0) =
0 0
b.
.
.
c. H ( f )(0, 0)
=
0 −1
H ( f )(0, 1)
=
0 3
3 0
=
0 3
3 0
=
12 5 3 5
H ( f )(1, 0) H ( f )(5−1/3 , 5−1/3 ) d.
.
! . 3 5 12 5
1 0 1
1 1 . 0
0 2 0
1 0 . 4
2 H ( f )(0, 0, 0) = 0 1
!
.
0 H ( f )(0, 0, 0) = 1 1 e.
−1 0 !
! .
3. (0, 0) is a global minimum. 4. The global maximum is T = 30.38 and the global minimum is T = 14. 5. a. Maximum at the point ( 21 , 34 ) and minimum at the point √ √ ( 12 (1 ± 5), − 21 (1 ± 5)).
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mathematical methods for physicists
√ √ b. Maximum at the point (1/ 2, 1/ 2) and a minimum at the √ √ point (−1/ 2, −1/ 2). √ √ 2 9 c. Maximum at the point ( 35 70, 70 70) and a minimum at the √ √ 2 9 point (− 35 70, − 70 70). d. The points (1, 1, 1), (1, −1, −1), (−1, −1, 1), (−1, 1, −1) are local minima. e. The extrema are given by 1 1 1 √ 1 √ ( √ , √ , 2), ( √ , − √ , 2), 2 2 2 2
√ √ 1 1 1 1 (− √ , √ , − 2), (− √ , − √ , − 2). 2 2 2 2 6. The origin is a saddle point and (1, 1) is a local maximum 7. a. Plot with v = 454.1584 − 40.7836d. b. d = 0.911, v = 373, Var(d) = 1.917, Var(v) = 6.34E + 05, c. The slope and intercept can be computed from the data in Table 10.2. Cov( x, y) a= = 454.158, Var( x ) b = y − ax = −40.784. d. The results are in Table B.1. Table B.1: Output from Excel’s LINEST function for the Hubble data.
value seslope R2 F SSR e.
slope 454.158 75.237 0.624 36.438 1976648
intercept 40.784 83.439 232.911 22 1193442
value seint sey df SSE
SSR = 0.623530521, SST r SSE sev = = 0.085, N−2 sey seslope = p = 0.267, NVar(y) R2 =
seint = q
sey N (1 − x2 /x2 )
f. v = 454.1584 − 40.7836d. g. H0 = 454 km/s = 1.47 × 10−17 s−1 . Mpc h. 2.0 × 109 yr.
= 83.44.
quick answers
8. sinh( x −1) sinh 1 . 2 y2 + x − 74 = 65 16 . 1 y ( x ) = 2 (1 − x ).
a. y( x ) = b. c. 9.
a. The shape of the path is x
= .5729( ϕ − sin ϕ),
y
= .5729(1 − cos ϕ)
for 0 ≤ ϕ ≤ 2.412. q R a b. T = C ds v = g ϕ0 ≈ 0.583 s. q c. T = 10 g ≈ 1.01 s. d. It takes forever. 10. Z 1p 1 + [y0 ( x )]2
a. T = dx. v 0 √ 1+[y0 ( x )]2 d ∂ v = 0. b. dx ∂y0 c. Derivation.q q n1 ( x − x1 )2 + y21 n2 ( x − x2 )2 + y22 d. T ( x ) = + . c c e. Derivation. 11. a. r = (2 cos θ, 2 sin θ, 10 π θ ). 5 b. r = (2 cos θ, 2 sin θ, 2π θ ).
12. The path is given as y( x ) = b + A(cosh L. 13. a. cos(φ − φ0 ) = `r . b. Proof. 14. ( x − 0.500)2 + (y − 0.4592)2 = 0.4609. 15. a. Description.
x − x0 A
− cosh Aa ), where 2A sinh
a A
=
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mathematical methods for physicists
b.
c. ω = θ˙ = 16. a =
p
0
=
0
=
M¨r + m¨r − mr θ˙ 2 + Mg, d 2 ˙ mr θ . dt
Mg/mr.
mg sin θ cos θ . M +m sin2 θ
17.
= 2m¨r + mr θ˙ 2 − mg(1 − cos θ ), 0 = 2mrr˙ θ˙ + mr2 θ¨ − mgr sin θ. 0
Ch.11 Problems in Higher Dimensions ∞
∞
nπx mπy −k(λn +µm )t cos e , L H n =1 m =0 Z HZ L mπy 4 nπx cos dxdy, where Anm = f ( x, y) sin LH 0 0 L H for n = 1, 2, . . . , m = 0, 1, 2, . . . . √ nx sinh( 2(1 − z)) √ 2. u( x, y, z) = sin . sin my 2 sinh 2 1. u( x, y, t) =
∑ ∑
Anm sin
3. un ( x, y) = sin nπx cosh nπy,
n = 1, 2, . . . .
4. a. 1 ∂ r ∂r
r
∂u ∂r
+
1 ∂2 u ∂2 u + 2 = 0. r2 ∂θ 2 ∂z
b. Z 00 − λZ
= 0,
00
= 0,
2
Θ +µ Θ 2
00
0
2
2
r R + rR + (λr − µ ) R
= 0.
c. Description. 5. a. The modes for the rectangular membrane are in Table B.2 and the modes for the circular membrane are in Table B.3 b. These results are i Tables B.2 and B.3. √
c. s =
2π j01 a.
quick answers
n 1 1 2 2 3 1 3 2 4 1
√
m 1 2 1 2 1 3 2 3 1 4 m 0 1 2 0 1 2
n 1 1 1 2 2 2
n2 + m2 √ 2 √ 5 √ 5 √ 8 √ 10 √ 10 √ 13 √ 13 √ 17 √ 17 jmn 2.405 3.832 5.136 5.520 7.016 8.417
Table B.2: The first modes of a square membrane in Problem 10.5
νnm ν0 1.5811ν0 1.5811ν0 2.0000ν0 2.2361ν0 2.2361ν0 2.5495ν0 2.5495ν0 2.9155ν0 2.9155ν0
Table B.3: The first modes of a circular membrane in Problem 10.5
νmn ν0 1.5933ν0 2.1356ν0 2.2952ν0 2.9173ν0 3.4998ν0
6. u(r, θ, t) ∞ ∞
jmn r jmn r ∑ ∑ Amn Jm a cos mθ + Bmn Jm a sin mθ cos ωmn t m =0 n =1 ∞ ∞ jmn r jmn r + ∑ ∑ Cmn Jm cos mθ + Dmn Jm sin mθ sin ωmn t a a m =0 n =1
=
where
=
A0n
= =
Amn
= =
Bmn
=
C0n Cmn
= =
Z a
1 a2
[ J1 ( j0n )] 1
πa2 a2
[ J1 ( j0n )] 2
Z 2π Z a 2
0 2
0
2
[ Jm+1 ( jmn )] Z 2 2
[ J1 ( j0n )] 2
r f (r, θ ) J0 (
ram (r ) Jm (
0
0
rbm (r ) Jm (
Z 2π Z a 2
0
0
Z 2π Z a 2
0
πa2 ωmn [ Jm+1 ( jmn )]2
0
jmn r ) dr a
rg(r, θ ) J0 (
0
jmn r ) cos mθ drdθ. a
jmn r ) dr a
r f (r, θ ) Jm (
Z 2π Z a 0
j0n r ) drdθ. a
r f (r, θ ) Jm (
a 0
[ Jm+1 ( jmn )]
0n
j0n r ) dr a
Z 2π Z a
1 πa2 ω
0
Z a
[ Jm+1 ( jmn )] 2
πa2
ra0 (r ) J0 (
0
[ Jm+1 ( jmn )] 2
πa2 a2
2
435
jmn r ) sin mθ drdθ. a
j0n r ) drdθ. a
rg(r, θ ) Jm (
jmn r ) cos mθ drdθ. a
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mathematical methods for physicists
Dmn
=
Z 2π Z a
2 πa2 ωmn [ Jm+1 ( jmn )]
2
0
7. u(5, 5, 5, 60) = −0.5 + 0.2 ∑∞ `=0 b` sin 40 ` b` = (22000 + (− 1 ) . `+1)π (2`+1)2 π 2
0
rg(r, θ ) Jm (
jmn r ) sin mθ drdθ. a
(2`+1))π −1.70(2`+1)2 )π 2 e , 4
where
8. i 1 ∂2 u sin θ ∂u ∂θ + a2 sin2 θ ∂φ2 . ( )( ) cos ωnm t cos mθ b. u(θ, φt) = Pnm (cos θ ), where ωnm = sin ωnm t sin mθ p c n ( n + 1). a
a. utt = c2
h
1
a2 sin θ
∂ ∂θ
c. Description. p √ √ d. ωnm = 5 n(n + 1) = 0, 5 2, 5 6 for n = 0, 1, 2. 9. u(r, z) = ∑∞ n =1
j j 40 J0 ( Rn r ) sinh( Rn z). j ( jn2 + R) J0 ( jn ) sinh( Rn H )
−n π kt . n +1 10. u(ρ, t) = 200 ∑∞ n=1 (−1) nπρ e 11. u(ρ, θ, φ) = ρ2 23 − 21 cos2 φ sin2 θ + ρ2 (3 cos2 θ − 1) + 1. sin(nπρ)
2
2
12. G ( x, y, ξ, η )
13. G ( x, ξ; t, 0) = 14. G ( x, ξ; y) =
1 1 ln((ξ − x )2 + (η − y)2 ) − ln((ξ − x )2 + (η + y)2 ) 4π 4π 1 1 − ln((ξ + x )2 + (η − y)2 ) + ln((ξ + x )2 + (η + y)2 ). 4π 4π
=
2 L
2 L
∑∞ n=1 sin
(2n−1)πx 2L
nπξ nπx ∑∞ n=1 sin L sin L
` 15. G (ρ; θ, θ 0 ; φ, φ0 ) = ∑∞ `=0 ∑m=−`
Ch. A
sin
(2n−1)πξ −k( 2n−1 π )2 t 2L . e 2L nπ ( H −y) L nπH L
sinh sinh
` 2`+1 (`−m)! m m 0 im(φ−φ0 ) ρ 4π (`+m)! P` (cos θ ) P` (cos θ ) e r`
Review of Sequences and Infinite Series
1. a. limn→∞ an = 0. b. limn→∞ an = 3. c. limn→∞ an = 1. d. limn→∞ an = 23 . e. limn→∞ an = 1. f. limn→∞ an = 1. g. limn→∞ an = ∞.
quick answers
2. a. b. c. d.
12 5 . 1 10 . 23 2 . 11 6 .
3. a. Converges absolutely. b. Converges absolutely. c. Converges absolutely. d. Converges conditionally. e. Diverges. f. Diverges. g. Diverges. h. Converges conditionally. 4. a. −3. b. −4. c. Converges. d. − π4 . 5. a. Derivation. b. Derivation. c.
1 2.
6. a. ln 2. b. i.
5 ln 2 2
1 1 1 1 1 1 1 1 1+ + + + + + + − 3 5 7 9 11 13 15 2 1 1 1 1 1 1 1 1 1 + + + + + + + + − +.... 17 19 21 23 25 27 29 31 4
=
ii. ln 8
=
1 1 1 1 1 1 1 + + + + + + 3 5 7 9 11 13 15 1 1 1 1 1 1 1 1 1 + − +.... + + + + + + + 17 19 21 23 25 27 29 31 2 1+
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iii. 1 1 1 1 1 1 1 1 1 + + + + − + + + . 1− 2 4 6 8 3 10 12 14 16 1 1 1 1 1 1 + − + −.... + + + 5 18 20 22 24 7
0
=
iv. One positive term followed by 134 negative terms 7. a. (0, 2], R = 1. b. (−∞, ∞), R = ∞. c. [−5, 5), R = 5. d. (−1, 1], R = 1. 8. 2k−1
x a. sinh x = x + x3 ! + x5 ! + . . . = ∑∞ k =1 (2k −1)! ,  x  < ∞. √ 1 3 x + . . . ,  x  < 1. b. 1 + x = 1 + 12 x − 18 x2 + 16 3
5
n+1 ,  x  < ∞. c. xe x = e + ∑∞ n =0 ( n +1) ! ( x − 1 ) n +1 1− x −1 = − ∞ ,  x − 1 < 3. d. 2x+ ∑ n = 0 x 3
( n +2) e
9. a. The series converges uniformly on [1, 2]. b. The series converges uniformly on R. 10. a. Derivation. b. π = 4 1 − 11. sin3 θ =
3 4
1 3
+ 51 − . . . .
sin θ − 14 sin 3θ.
12. a. For x = 0.015 ± 0.0005, f ( x ) = (1.66 ± 0.17) × 10−6 . b. For x = 0.0015 ± 0.00005, f ( x ) = (5.06 ± 0.84) × 10−6 . c. For x = 0.005 ± 0.0005, f ( x ) = (9.37 ± 3.75) × 10−10 . d. For h = 1.00 ± 0.005 m and R = (1.374 ± 0.0005) × 103 km, f ( R, h) = (−3.639 ± 0.038) × 10−7 . e. For x = (2.5 ± 0.05) × 10−13 , f ( x ) = (−1.250 ± 0.025) × 10−13 . 13.
√ a. O( x ). b. O( x −3/4 ). c. O( 1x ). d. O(1).
K19053 ISBN: 9781466584709
90000
w w w. c r c p r e s s . c o m
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