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*65*
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*English*
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*Year 2005*

Instructor’s Solutions Manual

PARTIAL DIFFERENTIAL EQUATIONS with FOURIER SERIES and BOUNDARY VALUE PROBLEMS

Second Edition

´ H. ASMAR NAKHLE University of Missouri

Contents Preface Errata

1

v vi

A Preview of Applications and Techniques

1

1.1 What Is a Partial Differential Equation? 1 1.2 Solving and Interpreting a Partial Differential Equation 4

2

Fourier Series 2.1 2.2 2.3 2.4 2.5 2.6 2.7

13

Periodic Functions 13 Fourier Series 21 Fourier Series of Functions with Arbitrary Periods 35 Half-Range Expansions: The Cosine and Sine Series 51 Mean Square Approximation and Parseval’s Identity 58 Complex Form of Fourier Series 63 Forced Oscillations 73

Supplement on Convergence 2.9 Uniform Convergence and Fourier Series 79 2.10 Dirichlet Test and Convergence of Fourier Series 81

3

Partial Differential Equations in Rectangular Coordinates

82

3.1 Partial Differential Equations in Physics and Engineering 82 3.3 Solution of the One Dimensional Wave Equation: The Method of Separation of Variables 87 3.4 D’Alembert’s Method 104 3.5 The One Dimensional Heat Equation 118 3.6 Heat Conduction in Bars: Varying the Boundary Conditions 128 3.7 The Two Dimensional Wave and Heat Equations 144 3.8 Laplace’s Equation in Rectangular Coordinates 146 3.9 Poisson’s Equation: The Method of Eigenfunction Expansions 148 3.10 Neumann and Robin Conditions 151

4

Partial Differential Equations in Polar and Cylindrical Coordinates 4.1 The Laplacian in Various Coordinate Systems 155

155

Contents

4.2 Vibrations of a Circular Membrane: Symmetric Case

iii

228

4.3 Vibrations of a Circular Membrane: General Case 166 4.4 Laplace’s Equation in Circular Regions 175 4.5 Laplace’s Equation in a Cylinder 191 4.6 The Helmholtz and Poisson Equations 197

Supplement on Bessel Functions

4.7 Bessel’s Equation and Bessel Functions

204

4.8 Bessel Series Expansions 213 4.9 Integral Formulas and Asymptotics for Bessel Functions 228

5

Partial Differential Equations in Spherical Coordinates

231

5.1 Preview of Problems and Methods 231 5.2 Dirichlet Problems with Symmetry 233 5.3 Spherical Harmonics and the General Dirichlet Problem

236

5.4 The Helmholtz Equation with Applications to the Poisson, Heat, and Wave Equations 242

Supplement on Legendre Functions

5.5 Legendre’s Differential Equation 245 5.6 Legendre Polynomials and Legendre Series Expansions

6

Sturm–Liouville Theory with Engineering Applications

6.1 Orthogonal Functions

257

6.2 Sturm–Liouville Theory 6.3 The Hanging Chain

259

263

6.4 Fourth Order Sturm–Liouville Theory 6.6 The Biharmonic Operator 6.7 Vibrations of Circular Plates

267 269

265

251

257

iv Contents

7

The Fourier Transform and Its Applications 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.9 7.10

The Fourier Integral Representation 271 The Fourier Transform 276 The Fourier Transform Method 286 The Heat Equation and Gauss’s Kernel 294 A Dirichlet Problem and the Poisson Integral Formula The Fourier Cosine and Sine Transforms 306 Problems Involving Semi-Infinite Intervals 310 Generalized Functions 315 The Nonhomogeneous Heat Equation 325 Duhamel’s Principle 327

271

303

Preface This manual contains solutions with notes and comments to problems from the textbook Partial Differential Equations with Fourier Series and Boundary Value Problems Second Edition Most solutions are supplied with complete details and can be used to supplement examples from the text. There are also many figures and numerical computions on Mathematica that can be very useful for a classroom presentation. Certain problems are followed by discussions that aim to generalize the problem under consideration. I hope that these notes will sevre their intended purpose: • To check the level of difficulty of assigned homework problems; • To verify an answer or a nontrivial computation; and • to provide worked solutions to students or graders. As of now, only problems from Chapters 1–7 are included. Solutions to problems from the remaining chapters will be posted on my website www.math.missouri.edu/ nakhle as I complete them and will be included in future versions of this manual. I would like to thank users of the first edition of my book for their valuable comments. Any comments, corrections, or suggestions from Instructors would be greatly appreciated. My e-mail address is [email protected] Nakhl´e H. Asmar Department of Mathematics University of Missouri Columbia, Missouri 65211

Errata The following mistakes appeared in the first printing of the second edition. Corrections in the text and figures p. 224, Exercise #13 is better done after Section 4.4. p. 268, Exercise #8(b), n should be even. p.387, Exercise#12, use y2 = I0 (x) not y2 = J1 (x). p. 425 Figures 5 and 6: Relabel the ticks on the x-axis as −π, −π/2, π/2, π, instead of −2π, −π, π, 2π. p. 467, l. (−3): Change reference (22) to (20). p. 477 l. 10: (xt) ↔ (x, t). p. 477 l. 19: Change ”interval” to ”triangle” p. 487, l.1: Change ”is the equal” to ”is equal” p. 655, l.13: Change ln | ln(x2 + y2 )| to ln(x2 + y2 ). Corrections to Answers of Odd Exercises Section 7.2, # 7: Change i to −i. Section 7.8, # 13: f(x) = 3 for 1 < x < 3 not 1 < x < 2. q −iw −1) P3 2 (e Section 7.8, # 35: j=1 j sin(jw) π w q Section 7.8, # 37: i π2 w13 Section 7.8, # 51: √32π [δ1 − δ0 ]. Section 7.8, # 57: The given answer is the derivative of the real answer, which should be 1 √ (x + 2) U−2 − U0 + (−x + 2) U0 − U1 + U1 − U3 + (−x + 4) U3 − U4 2π Section 7.8, # 59: The given answer is the derivative of the real answer, which should be 1 √1 2 2π (x + 3) U−3 − U−2 + (2x + 5) U−2 − U−1 + (x + 4) U−1 − U0 +(−x + 4) U0 − U1 + (−2x + 5) U1 − U2 + (−x + 3) U2 − U3 Section 7.10, # 9: 12 t sin(x + t) + 12 cos(x + t) − 12 cos(x − t) . Any suggestion or correction would be greatly appreciated. Please send them to my e-mail address [email protected] Nakhl´e H. Asmar Department of Mathematics University of Missouri Columbia, Missouri 65211

Section 1.1

What Is a Partial Differential Equation?

1

Solutions to Exercises 1.1 1. If u1 and u2 are solutions of (1), then ∂u1 ∂u1 + = 0 and ∂t ∂x

∂u2 ∂u2 + = 0. ∂t ∂x

Since taking derivatives is a linear operation, we have ∂ ∂ (c1 u1 + c2 u2) + (c1 u1 + c2u2 ) ∂t ∂x

= c1

= c1

∂u1 ∂u2 ∂u1 ∂u2 + c2 + c1 + c2 ∂t ∂t ∂x ∂x z

=0

}|

∂u1 ∂u1 + ∂t ∂x

{

+c2

z

=0

}| { ∂u2 ∂u2 ++ = 0, ∂t ∂x

showing that c1 u1 + c2u2 is a solution of (1). 2. (a) General solution of (1): u(x, t) = f(x − t). On the x-axis (t = 0): u(x, 0) = 2 2 xe−x = f(x − 0) = f(x). So u(x, t) = f(x − t) = (x − t)e−(x−t) . 3. (a) General solution of (1): u(x, t) = f(x − t). On the t-axis (x = 0): u(0, t) = t = f(0 − t) = f(−t). Hence f(t) = −t and so u(x, t) = f(x − t) = −(x − t) = t − x. 4. Let α = ax + bt, β = cx + dt, then ∂u ∂x ∂u ∂t

∂u ∂α ∂u ∂β ∂u ∂u + =a +c ∂α ∂x ∂β ∂x ∂α ∂β ∂u ∂α ∂u ∂β ∂u ∂u + =b +d . ∂α ∂t ∂β ∂t ∂α ∂β

= =

So (1) becomes (a + b)

∂u ∂u + (c + d) = 0. ∂α ∂β

Let a = 1, b = 1, c = 1, d = −1. Then 2

∂u =0 ∂α

∂u = 0, ∂α

⇒

which implies that u is a function of β only. Hence u = f(β) = f(x − t). 5. Let α = ax + bt, β = cx + dt, then ∂u ∂x ∂u ∂t

= =

∂u ∂α ∂u ∂β ∂u ∂u + =a +c ∂α ∂x ∂β ∂x ∂α ∂β ∂u ∂α ∂u ∂β ∂u ∂u + =b +d . ∂α ∂t ∂β ∂t ∂α ∂β

Recalling the equation, we obtain ∂u ∂u − =0 ∂t ∂x

⇒

(b − a)

∂u ∂u + (d − c) = 0. ∂α ∂β

Let a = 1, b = 2, c = 1, d = 1. Then ∂u =0 ∂α

⇒

u = f(β)

⇒

u(x, t) = f(x + t),

where f is an arbitrary differentiable function (of one variable). 6. The solution is very similar to Exercise 5. Let α = ax + bt, β = cx + dt, then ∂u ∂x ∂u ∂t

= =

∂u ∂α ∂u ∂β ∂u ∂u + =a +c ∂α ∂x ∂β ∂x ∂α ∂β ∂u ∂α ∂u ∂β ∂u ∂u + =b +d . ∂α ∂t ∂β ∂t ∂α ∂β

2 Chapter 1 A Preview of Applications and Techniques Recalling the equation, we obtain 2

∂u ∂u +3 =0 ∂t ∂x

⇒

(2b + 3a)

∂u ∂u + (2d + 3c) = 0. ∂α ∂β

Let a = 0, b = 1, c = 2, d = −3. Then the equation becomes 2

∂u =0 ∂α

⇒

u = f(β)

⇒

u(x, t) = f(cx + dt) = f(2x − 3t),

where f is an arbitrary differentiable function (of one variable). 7. Let α = ax + bt, β = cx + dt, then ∂u ∂x ∂u ∂t

∂u ∂α ∂u ∂β ∂u ∂u + =a +c ∂α ∂x ∂β ∂x ∂α ∂β ∂u ∂α ∂u ∂β ∂u ∂u + =b +d . ∂α ∂t ∂β ∂t ∂α ∂β

= =

The equation becomes (b − 2a)

∂u ∂u + (d − 2c) = 2. ∂α ∂β

Let a = 1, b = 2, c = −1, d = 0. Then 2

∂u =2 ∂β

∂u = 1. ∂β

⇒

Solving this ordinary differential equation in β, we get u = β + f(α) or u(x, t) = −x + f(x + 2t). 8. Let α = c1 x + d1 t, β = c2 x + d2t, then ∂u ∂x ∂u ∂t

= =

∂u ∂u + c2 ∂α ∂β ∂u ∂u d1 + d2 . ∂α ∂β

c1

The equation becomes (ad1 + bc1 )

∂u ∂u + (ad2 + bc2) = u. ∂α ∂β

Let c1 = a, d1 = −b, c2 = 0, d2 =

1 a

∂u =u ∂β

(a 6= 0). Then ⇒

u = f(α)eβ .

t

Hence u(x, t) = f(ax − bt)e a . 9. (a) The general solution in Exercise 5 is u(x, t) = f(x + t). When t = 0, we get u(x, 0) = f(x) = 1/(x2 + 1). Thus u(x, t) = f(x + t) = (c) As t increases, the wave f(x) = 1 3 2

0 -2 -1

1 0 1 20

Figure for Exercise 9(b).

1 1+x2

1 . (x + t)2 + 1

moves to the left.

10. (a) The directional derivative is zero along the vector (a, b). b ∂u (b) Put the equation in the form ∂u ∂x + a ∂y = 0 (a 6= 0). The characteristic curves are obtained by solving dy b = dx a

⇒

y=

b x+C a

⇒

b y − x = C. a

Section 1.1

What Is a Partial Differential Equation?

3

Let φ(x, y) = y − ab x. The characteristic curves are the level curves of φ. ∂u b (c) The solution of a ∂u ∂x + b ∂y = 0 is of the form u(x, y) = f(φ(x, y)) = f(y − a x), where f is a differentiable function of one variable. 11. The characteristic curves are obtained by solving dy = x2 dx

⇒

y=

1 3 x +C 3

1 y − x3 = C. 3

⇒

Let φ(x, y) = y − 13 x3. The characteristic curves are the level curves of φ. The solution of is of the form u(x, y) = f(φ(x, y)) = f(y − 13 x3), where f is a differentiable function of one variable. 12. We follow the method of characteristic curves. Let’s find the characteristic curves. For x 6= 0, ∂u ∂x dy dx

=

+ y x

y ∂u x ∂y

= 0;

⇒ y = Cx,

where we have used Theorem 1, Appendix A.1, to solve the last differential equation. Hence the characteristic curves are yx = C and the solution of the partial differential y equation is u(x, y) = f x . To verify the solution, we use the chain rule and get ux = − xy2 f 0 yx and uy = x1 f 0 xy . Thus xux + yuy = 0, as desired. dy 13. To find the characteristic curves, solve dx = sin x. Hence y = − cos x + C or y + cos x = C. Thus the solution of the partial differential equation is u(x, y) = f (y + cos x). To verify the solution, we use the chain rule and get ux = − sin xf 0 (y + cos x) and uy = f 0 (y + cos x). Thus ux + sin xuy = 0, as desired. −x2 ∂u 14. Put the equation in the form ∂u ∂x + xe ∂y = 0. The characteristic curves are obtained by solving 2 dy = xe−x dx

⇒ 2

2 1 y = − e−x + C 2

⇒

2 1 y + e−x = C. 2

Let φ(x, y) = y + 12 e−x . The characteristic curves are the level curves of φ. 2 The solution is of the form u(x, y) = f(φ(x, y)) = f(y + 12 e−x ), where f is a differentiable function of one variable.

4

Chapter 1

A Preview of Applications and Techniques

Exercises 1.2 1. We have ∂ ∂t

∂u ∂t

So

∂ =− ∂t

∂v ∂x

∂2u ∂ 2v = − ∂t2 ∂t∂x 2

2

and

and

∂2v ∂2u = − 2. ∂x∂t ∂x 2

∂v ∂t

∂ ∂x

∂ =− ∂x

∂u ∂x

.

2

∂ v ∂ v Assuming that ∂t∂x = ∂x∂t , it follows that ∂∂tu2 = ∂∂xu2 , which is the one dimensional wave equation with c = 1. A similar argument shows that v is a solution of the one dimensional wave equation. 2. (a) For the wave equation in u, the appropriate initial conditions are u(x, 0) = f(x), as given, and ut (x, 0) = −vx (x, 0) = h0 (x). (b) For the wave equation in v, the appropriate initial conditions are v(x, 0) = h(x), as given, and vt (x, 0) = −ux (x, 0) = f 0 (x). 3. uxx = F 00(x + ct) + G00(x + ct), utt = c2 F 00(x + ct) + c2 G(x − ct). So utt = cuxx , which is the wave equation. 4. (a) Using the chain rule in two dimensions:

∂u ∂x ∂2u ∂x2

= = = =

∂u ∂α ∂u ∂β ∂u ∂u + = + ∂α ∂x ∂β ∂x ∂α ∂β ∂ ∂u ∂u + ∂x ∂α ∂β ∂2u ∂2u ∂2u ∂2u + + + ∂α2 ∂β∂α ∂α∂β ∂β 2 2 2 2 ∂ u ∂ u ∂ u +2 . + 2 ∂α ∂β∂α ∂β 2

Similarly ∂u ∂t ∂2u ∂t2

= = = =

∂u ∂α ∂u ∂β ∂u ∂u + =c −c ∂α ∂t ∂β ∂t ∂α ∂β ∂ ∂u ∂u c −c ∂t ∂α ∂β 2 ∂ u ∂2u ∂2u ∂2u c2 2 − c2 − c2 + c2 2 ∂α ∂β∂α ∂α∂β ∂β 2 2 2 ∂ u ∂ u ∂ u c2 2 − 2c2 + c2 2 . ∂α ∂β∂α ∂β

Substituting into the wave equation, it follows that c2

∂2u ∂2u ∂2u ∂2u ∂2u ∂2u +2 + c2 2 = c2 2 − 2 + c2 2 2 ∂α ∂β∂α ∂β ∂α ∂β∂α ∂β

(b) The last equation says that

∂u ∂β

⇒

∂2u = 0. ∂α∂β

is constant in α. So ∂u = g(β) ∂β

where g is an arbitrary differentiable function. (c) Integrating the equation in (b) with respect to β, we find that u = G(β) + F (α), where G is an antiderivative of g and F is a function of α only. (d) Thus u(x, t) = F (x + ct) + G(x − ct), which is the solution in Exercise 3. 5. (a) We have u(x, t) = F (x + ct) + G(x − ct). To determine F and G, we use the initial data: u(x, 0) =

1 1 + x2

⇒

F (x) + G(x) =

1 ; 1 + x2

(1)

Section 1.2

Solving and Interpreting a Partial Differential Equation

∂u (x, 0) = 0 ⇒ cF 0 (x) − cG0 (x) = 0 ∂t ⇒ F 0 (x) = G0(x) ⇒ F (x) = G(x) + C,

5

(2)

where C is an arbitrary constant. Plugging this into (1), we find 1 1 1 2G(x) + C = ⇒ G(x) = −C ; 1 + x2 2 1 + x2 and from (2) F (x) =

1 1 + C . 2 1 + x2

Hence u(x, t) = F (x + ct) + G(x − ct) =

1 1 1 . + 2 1 + (x + ct)2 1 + (x − ct)2

6. We have u(x, t) = F (x + ct) + G(x − ct). To determine F and G, we use the initial data: 2 2 u(x, 0) = e−x ⇒ F (x) + G(x) = e−x ; (1) ∂u (x, 0) = 0 ⇒ cF 0 (x) − cG0 (x) = 0 ∂t ⇒ F 0 (x) = G0(x) ⇒ F (x) = G(x) + C,

(2)

where C is an arbitrary constant. Plugging this into (1), we find 2

2G(x) + C = e−x and from (2)

F (x) =

⇒

G(x) =

i 1 h −x2 −C ; e 2

i 1 h −x2 +C . e 2

Hence u(x, t) = F (x + ct) + G(x − ct) =

i 2 1 h −(x+ct)2 + e−(x−ct) . e 2

7. We have u(x, t) = F (x + ct) + G(x − ct). To determine F and G, we use the initial data: u(x, 0) = 0 ⇒ F (x) + G(x) = 0; (1) 2 ∂u (x, 0) = −2xe−x ∂t

2

⇒ cF 0(x) − cG0(x) = −2xe−x Z 2 2 ⇒ cF (x) − cG(x) = −2xe−x dx = e−x + C 2

⇒ F (x) − G(x) =

e−x + C, c

(2)

where we rewrote C/c as C to denote the arbitrary constant. Adding (2) and (1), we find 2 i e−x 1 h −x2 2F (x) = +C ; + C ⇒ F (x) = e c 2c and from (1) i 1 h −x2 G(x) = − +C . e 2c Hence i 2 1 h −(x+ct)2 u(x, t) = F (x + ct) + G(x − ct) = − e−(x−ct) . e 2c

6

Chapter 1

A Preview of Applications and Techniques

8. We have u(x, t) = F (x + ct) + G(x − ct). To determine F and G, we use the initial data: u(x, 0) = 0

⇒

F (x)+G(x) = 0

∂u x (x, 0) = ∂t (1 + x2 )2

⇒

F (x) = −G(x)

⇒ cF 0(x) − cG0(x) =

⇒

F 0(x) = −G0(x);

(1)

x (1 + x2)2

x x ⇒ 2cF 0(x) = ⇒ F 0(x) = , (from (1)) (1 + x2)2 2c(1 + x2 )2 Z x −1 ⇒ F (x) = dx = + C, 2 2 2c(1 + x ) 4c(1 + x2 ) where C is an arbitrary constant. From (1), G(x) = −F (x) =

1 − C; 4c(1 + x2)

where the C here is the same as the C in the definition of F (x). So 1 1 −1 u(x, t) = F (x + ct) + G(x − ct) = . + 4c (1 + (x + ct))2 (1 + (x − ct))2 9. As the hint suggests, we consider two separate problems: The problem in Exercise 5 and the one in Exercise 7. Let u1(x, t) denote the solution in Exercise 5 and u2(x, t) the solution in Exercise 7. It is straightforward to verify that u = u1 + u2 is the desired solution. Indeed, because of the linearity of derivatives, we have utt = (u1)tt + (u2 )tt = c2 (u1 )xx + c2 (u2 )xx, because u1 and u2 are solutions of the wave equation. But c2(u1 )xx + c2 (u2)xx = c2 (u1 + u2)xx = uxx and so utt = c2 uxx, showing that u is a solution of the wave equation. Now u(x, 0) = u1(x, 0)+u2(x, 0) = 1/(1+x2)+0, because u1 (x, 0) = 1/(1+x2) and u2 (x, 0) = 0. 2 Similarly, ut(x, 0) = −2xe−x ; thus u is the desired solution. The explicit formula for u is i 2 1 1 h −(x+ct)2 1 1 u(x, t) = + + − e−(x−ct) . e 2 2 2 1 + (x + ct) 1 + (x − ct) 2c 10. Reasoning as in the previous exercise, we find the solution to be u = u1 + u2 , wehre u1 is the solution in Exercise 6 and u2 is the solution in Exercise 8. Thus, i 2 1 h −(x+ct)2 1 1 −1 u(x, t) = . + e−(x−ct) + + e 2 4c (1 + (x + ct))2 (1 + (x − ct))2 11. We have

∂I ∂2I ∂I −L − RI = −L −R ; ∂t ∂x∂t ∂x ∂I ∂V ∂V −1 ∂I = −C − GV ⇒ = + GV ; ∂x ∂t ∂t C ∂x so ∂2V −1 ∂ 2 I ∂V = + G . ∂t2 C ∂t∂x ∂t To check that V verifies (1), we start with the right side ∂2V ∂ = ∂x2 ∂x

∂2V ∂V + (RC + LG) + RGV ∂t2 ∂t 2 −1 ∂ I ∂V −1 ∂I = LC +G + (RC + LG) + GV + RGV C ∂t∂x ∂t C ∂x

LC

=)

2

= =

∂ I ∂I −R ∂t∂x ∂x ∂2I ∂I −L −R ∂t∂x ∂x −L

z }| { ∂V LG ∂I C − + + GV C ∂t ∂x ∂2V , = ∂x2

Section 1.2

Solving and Interpreting a Partial Differential Equation

7

which shows that V satisfies (1). To show that I satisfies (1), you can proceed as we did for V or you can note that the equations that relate I and V are interchanged if we interchange L and C, and R and G. However, (1) remains unchanged if we interchange L and C, and R and G. So I satisfies (1) if and only if V satisfies (1). 12. The function being graphed is the solution (2) with c = L = 1: u(x, t) = sin πx cos πt. √

In the second frame, t = 1/4, and so u(x, t) = sin πx cos π/4 = 22 sin πx. The maximum of this function (for 0 < x < π is attained at x = 1/2 and is equal to √ 2 , which is a value greater than 1/2. 2 13. The function being graphed is 1 1 sin 2πx cos 2πt + sin 3πx cos 3πt. 2 3

u(x, t) = sin πx cos πt − In frames 2, 4, 6, and 8, t = u(x, t), we find u(x, t) = sin πx cos

m 4,

where m = 1, 3, 5, and 7. Plugging this into

mπ 1 mπ 1 3mπ − sin 2πx cos + sin 3πx cos . 4 2 2 3 4

For m = 1, 3, 5, and 7, the second term is 0, because cos times, we have, for, m = 1, 3, 5, and 7, u(x,

mπ 2

= 0. Hence at these

m 1 ) = sin πx cos πt + sin 3πx cos 3πt. 4 3

To say that the graph of this function is symmetric about x = 1/2 is equivalent m to the assertion that, for 0 < x < 1/2, u(1/2 + x, m 4 ) = u(1/2 − x, 4 ). Does this equality hold? Let’s check: u(1/2 + x,

m ) = 4 =

mπ 1 3mπ + sin 3π(x + 1/2) cos 4 3 4 mπ 1 3mπ cos πx cos − cos 3πx cos , 4 3 4

sin π(x + 1/2) cos

where we have used the identities sin π(x + 1/2) = cos πx and sin 2π(x + 1/2) = − cos 3πx. Similalry, u(1/2 − x,

m ) = 4 =

So u(1/2 + x,

m ) 4

mπ 1 3mπ + sin 3π(1/2 − x) cos 4 3 4 mπ 1 3mπ cos πx cos − cos 3πx cos . 4 3 4

sin π(1/2 − x) cos

= u(1/2 − x,

m ), 4

as expected.

14. Note that the condition u(0, t) = u(1, t) holds in all frames. It states that the ends of the string are held fixed at all time. It is the other condition on the first derivative that is specific to the frames in question. The function being graphed is u(x, t) = sin πx cos πt −

1 1 sin 2πx cos 2πt + sin 3πx cos 3πt. 2 3

As we just stated, the equality u(0, t) = u(1, t) holds for all t. Now ∂ u(x, t) = π cos πx cos πt − π cos 2πx cos 2πt + π cos 3πx cos 3πt. ∂x ∂ ∂ To say that ∂x u(0, t) = 0 and ∂x u(1, t) means that the slope of the graph (as a function of x) is zero at x = 0 and x = 1. This is a little difficult to see in the

8

Chapter 1

A Preview of Applications and Techniques

frames 2, 4, 6, and 8 in Figure 8, but follows by plugging t = 1/4, 3/4, 5/4 and 7/4 and x = 0 or 1 in the derivative. For example, when x = 0 and t = 1/4, we obtain: ∂ u(0, 1/4) = π cos π/4 − π cos π/2 + π cos 3π/4 = π cos π/4 + π cos 3π/4 = 0. ∂x 15. Since the initial velocity is 0, from (10), we have

u(x, t) =

∞ X

bn sin

n=1

The initial condition u(x, 0) = f(x) = ∞ X

1 2

nπx cnπt cos . L L

sin πx L +

1 4

sin 3πx L implies that

nπx 2πx = sin . L L

bn sin

n=1

The equation is satisfied with the choice b1 = 0, b2 = 1, and all other bn’s are zero. This yields the solution u(x, t) = sin

2πx 2cπt cos . L L

Note that the condition ut(x, 0) = 0 is also satisfied. 16. Since the initial velocity is 0, from (10), we have

u(x, t) =

∞ X

bn sin

n=1

The initial condition u(x, 0) = f(x) = ∞ X

bn sin

n=1

1 2

nπx cnπt cos . L L

sin πx L +

1 4

sin 3πx L implies that

nπx 1 πx 1 3πx = sin + sin . L 2 L 4 L

Clearly, this equation is satisfied with the choice b1 = 12 , b3 = 14 , and all other bn’s are zero. This yields the solution u(x, t) =

1 πx cπt 1 3πx 3cπt sin cos + sin cos . 2 L L 4 L L

Note that the condition ut(x, 0) = 0 is also satisfied. 17. Same reasoning as in the previous exercise, we find the solution u(x, t) =

1 πx cπt 1 3πx 3cπt 2 7πx 7cπt sin cos + sin cos + sin cos . 2 L L 4 L L 5 L L

18. Since the initial displacement is 0, we use the functions following (1):

u(x, t) =

∞ X n=1

b∗n sin

nπx cnπt sin . L L

The initial condition u(x, 0) = 0 is clearly satisfied. To satisfy the second intial

Section 1.2

Solving and Interpreting a Partial Differential Equation

9

condition, we proceed as follows: g(x)

∂ πx = u(x, 0) L ∂t ∞ nπx ∂ X ∗ cnπt bn sin sin ∂t n=1 L L ∞ X nπx ∂ cnπt b∗n sin sin L ∂t L t=0 n=1 ! ∞ X nπx cnπ cnπt b∗n sin cos L L L t=0 n=1

=

sin

= =

=

∞ X

=

b∗n sin

n=1

Thus sin

nπx cnπ . L L

∞ πx X ∗ cnπ nπx cnπ bn = sin . L L L L n=1

This equality holds if we take b∗1 cπ = 1 or b∗1 = L u(x, t) =

L , cπ

and all other b∗n = 0. Thus

L πx cπt sin sin . cπ L L

This solution satisfies both initial conditions. 19. Reasoning as in the previous exercise, we satrt with the solution u(x, t) =

∞ X

b∗n sin

n=1

nπx cnπt sin . L L

The initial condition u(x, 0) = 0 is clearly satisfied. To satisfy the second intial condition, we must have " !# ∞ 1 nπx 3πx 1 6πx ∂ X ∗ cnπt b sin sin − sin = sin 4 L 10 L ∂t n=1 n L L t=0

∞ X cnπ ∗ nπx bn sin . L L n=1

= Thus 1 3cπ ∗ = b 4 L 3 1 6cπ ∗ − = b 10 L 6

L ; 12cπ L ⇒ b∗6 = − ; 60cπ ⇒ b∗3 =

and all other b8n are 0. Thus u(x, t) =

L 3πx 3cπt L 6πx 6cπt sin sin − sin sin . 12cπ L L 60cπ L L

20. Write the initial condition as u(x, 0) = 15 or 16 and you will get u(x, t) =

1 4

sin 4πx , then proceed as in Exercises L

1 4πx 4cπt sin cos . 4 L L

21. (a) We have to show that u( 12 , t) is a constant for all t > 0. With c = L = 1, we have u(x, t) = sin 2πx cos 2πt ⇒ u(1/2, t) = sin π cos 2πt = 0

for all t > 0.

10

Chapter 1

A Preview of Applications and Techniques

(b) One way for x = 1/3 not to move is to have u(x, t) = sin 3πx cos 3πt. This is the solution that corresponds to the initial condition u(x, 0) = sin 3πx and ∂u (x, 0) = 0. For this solution, we also have that x = 2/3 does not move for ∂t all t. 22. (a) Reasoning as in Exercise 17, we find the solution to be u(x, t) =

1 1 sin 2πx cos 2πt + sin 4πx cos 4πt. 2 4

(b) We used Mathematica to plot the shape of the string at times t = 0 to t = 1 by increments of .1. The string returns to some of its previous shapes. For example, when t = .1 and when t = .9, the string has the same shape. u x_, t_ 1 2 Sin 2 Pi x Cos 2 Pi t tt Table u x, t , t, 0, 1, .1 ; Plot Evaluate tt , x, 0, 1 1 Cos 2 2

t Sin 2

x

1 Cos 4 4

1 4 Sin 4 Pi x Cos 4 Pi t

t Sin 4

x

0.6 0.4 0.2 0.2

0.4

0.6

0.8

1

-0.2 -0.4 -0.6 The point x = 1/2 does not move. This is clear: If we put x = 1/2 in the solution, we obtain u(1/2, t) = 0 for all t. 23. The solution is u(x, t) = sin 2πx cos 2πt. The motions of the points x = 1/4, 1/3, and 3/4 are illustrated by the following graphs. Note that the point x = 1/2 does not move, so the graph that describes its motion is identically 0. u x_, t_ Sin 2 Pi x Cos 2 Pi t Plot u 1 4, t , u 1 3, t , u 3 4, t

, t, 0, Pi

= 1/4 1 0.5

= 1/3 0.5 -0.5 -1

= 3/4

1

1.5

2

2.5

3

Section 1.2

Solving and Interpreting a Partial Differential Equation

11

In each case, we have a cosine wave, namely u(x0, t) = sin 2πx0 cos 2πt, scaled by a factor sin 2πx0 . 24. The solution in Exercise 22 is u(x, t) =

1 1 sin 2πx cos 2πt + sin 4πx cos 4πt. 2 4

The motions of the points x = 1/4, 1/3, 1/2, and 3/4 are illustrated by the following graphs. As in the previous exercise, the point x = 1/2 does not move, so the graph that describes its motion is identically 0. u x_, t_ 1 2 Sin 2 Pi x Cos 2 Pi t 1 4 Sin 4 Pi x Cos 4 Pi t Plot u 1 4, t , u 1 3, t , u 3 4, t , t, 0, Pi 1 Cos 2 2

t Sin 2

x

1 Cos 4 4

t Sin 4

x

0.4 0.2 0.5

1

1.5

2

2.5

3

-0.2 -0.4 -0.6

Unlike the previous exercises, here the motion of a point is not always a cosine wave. In fact, it is the sum of two scaled cosine waves: u(x0, t) = 12 sin 2πx0 cos 2πt + 1 4 sin 4πx0 cos 4πt. 25. The solution (2) is u(x, t) = sin Its initial conditions at time t0 =

πx πct cos . L L

3L 2c

are 3L πx πc 3L πx 3π u(x, ) = sin cos · = sin cos = 0; 2c L L 2c L 2

and ∂u 3L πc πx (x, ) = − sin sin ∂t 2c L L

πc 3L · L 2c

=−

πc πx 3π πc πx sin sin = sin . L L 2 L L

26. We have u(x, t + (3L)/(2c))

= =

πx πc 3L cos · (t + ) L 2c L πx πct 3π πx πct sin cos + = sin sin , L L 2 L L

sin

where we used the indentity cos(a + 3π ) = sin a. Thus u(x, t + (3L)/(2c)) is equal 2 to the solution given by (7). Call the latter solution v(x, t). We know that v(x, t) represents the motion of a string that starts with initial shape f(x) = 0 and intial πx velocity g(x) = πc L sin L . Now, appealing to Exercise 25, we have that at time t0 = (3L)/(2c), the shape of the solution u is u(x, (3L)/(2c)) = 0 and its velocity is

12

Chapter 1

A Preview of Applications and Techniques ∂u 3L ∂t (x, 2c )

πx = πc L sin L . Thus the subsequent motion of the string, u, at time t + t0 is identical to the motion of a string, v, starting at time t = 0, whenever v has initial shape u(x, t0) and the initial velocity ∂u ∂t (x, t0 ).

27. (a) The equation is equivalent to −

1 ∂u κ ∂u − = u. r ∂t r ∂x

The solution of this equation follows from Exercise 8, Section 1.1, by taking a = − 1r and b = − κr . Thus 1 κ u(x, t) = f(− x + t)e−rt . r r Note that this equivalent to u(x, t) = f(x − κt)e−rt , by replacing the function x 7→ f(x) in the first formula by x 7→ f(−rx). This is acceptable because f is arbitrary. R∞ (b) The number of particles at time t ≥ 0 is given by −∞ u(x, t) dx. We have R∞ R∞ M = −∞ u(x, 0) dx. But u(x, 0) = f(x), so M = −∞ f(x) dx. For t > 0, the number of particles is Z ∞ Z ∞ u(x, t) dx = f(x − κt)e−rt dx −∞ −∞ Z ∞ Z ∞ = e−rt f(x − κt) dx = e−rt f(x) dx = M e−rt , −∞

−∞

R∞ where, in evaluating the integral −∞ f(x − κt) dx, we used the change of variables R∞ x ↔ x − κt, and then used M = −∞ f(x) dx.

Section 2.1

Periodic Functions

13

Solutions to Exercises 2.1 1. (a) cos x has period 2π. (b) cos πx has period T = 2π (c) cos 23 x has π = 2. 2π period T = 2/3 = 3π. (d) cos x has period 2π, cos 2x has period π, 2π, 3π,.. ˙ A common period of cos x and cos 2x is 2π. So cos x + cos 2x has period 2π. 2π (b) sin nπx has period T = nπ = n2 . 2. (a) sin 7πx has period T = 2π 7π = 2/7. Since any integer multiple of T is also a period, we see that 2 is also a period of sin nπx. (c) cos mx has period T = 2π m . Since any integer multiple of T is also a period, we see that 2π is also a period of cos mx. (d) sin x has period 2π, cos x has period 2π; cos x + sin x so has period 2π. (e) Write sin2 2x = 12 − cos24x . The 2 π π function cos 4x has period T = 2π 4 = 2 . So sin 2x has period 2 . 3. (a) The period is T = 1, so it suffices to describe f on an interval of length 1. From the graph, we have 0 if − 12 ≤ x < 0, f(x) = 1 if 0 ≤ x < 12 . For all other x, we have f(x + 1) = f(x). (b) f is continuous for all x 6= k2 , where k is an integer. At the half-integers, + − x = 2k+1 2 , using the graph, we see that limh→x f(h) = 0 and limh→x f(h) = 1. At the integers, x = k, from the graph, we see that limh→x+ f(h) = 1 and limh→x− f(h) = 0. The function is piecewise continuous. (c) Since the function is piecewise constant, we have that f 0 (x) = 0 at all x 6= k2 , where k is an integer. It follows that f 0 (x+) = 0 and f 0 (x−) = 0 (Despite the fact that the derivative does not exist at these points; the left and right limits exist and are equal.) 4. The period is T = 4, so it suffices to describe f on an interval of length 4. From the graph, we have x+1 if − 2 ≤ x ≤ 0, f(x) = −x + 1 if 0 < x < 2. For all other x, we have f(x + 4) = f(x). (b) The function is continuous at all x. (c) (c) The function is differentiable for all x 6= 2k, where k is an integer. Note that f 0 is also 4-periodic. We have 1 if − 2 < x ≤ 0, 0 f (x) = −1 if 0 < x < 2. For all other x 6= 2k, we have f(x + 4) = f(x). If x = 0, ±4, ±8, . . . , we have f 0 (x+) = 1 and f 0 (x−) = −1. If x = ±2, ±6, ±10, . . . , we have f 0 (x+) = −1 and f 0 (x−) = 1. 5. This is the special case p = π of Exercise 6(b). 6. (a) A common period is 2p. (b) The orthogonality relations are Z p mπx nπx cos cos dx = 0 if m 6= n, m, n = 0, 1, 2, . . . ; p p −p Z p mπx nπx sin sin dx = 0 if m 6= n, m, n = 1, 2, . . . ; p p −p Z p mπx nπx cos sin dx = 0 for all m = 0, 1, 2, . . . , n = 1, 2, . . . . p p −p These formulas are established by using various addition formulas for the cosine and sine. For example, to prove the first one, if m 6= n, then Z p mπx nπx cos cos dx p p −p Z 1 p (m + n)πx (m − n)πx = cos + cos dx 2 −p p p 1 p (m + n)πx p (m − n)πx p = sin + sin = 0. 2 m + n)π p m − n)π p −p

14 Chapter 2 Fourier Series 7. Suppose that Show that f1 , f2 , . . ., fn, . . . are T -periodic functions. This means that fj (x + T ) = f(x) for all x and j = 1, 2, . . . , n. Let sn (x) = a1 f1 (x) + a2 f2 (x) + · · · + an fn (x). Then sn (x + T )

= =

= a1 f1 (x + T ) + a2 f2 (x + T ) + · · · + anfn (x + T ) a1f1 (x) + a2f2 (x) + · · · + an fn (x) = sn (x); P∞ which means that sn is T -periodic. In general, if s(x) = j=1 aj fj (x) is a series that converges for all x, where each fj is T -periodic, then s(x + T ) =

∞ X

aj fj (x + T ) =

j=1

∞ X

aj fj (x) = s(x);

j=1

and so s(x) is T -periodic. 8. (a) Since | cos x| ≤ 1 and | cos πx| ≤ 1, the equation cos x + cos πx = 2 holds if and only if cos x = 1 and cos πx = 1. Now cos x = 1 implies that x = 2kπ, k and integer, and cos πx = 1 implies that x = 2m, m and integer. So cos x + cos πx = 2 implies that 2m = 2kπ, which implies that k = m = 0 (because π is irrational). So the only solution is x = 0. (b) Since f(x) = cos x + cos πx = 2 takes on the value 2 only at x = 0, it is not periodic. 9. (a) Suppose that f and g are T -periodic. Then f(x + T ) · g(x + T ) = f(x) · g(x), and so f · g is T periodic. Similarly, f(x + T ) f(x) = , g(x + T ) g(x) and so f/g is T periodic. (b) Suppose that f is T -periodic and let h(x) = f(x/a). Then x x + aT h(x + aT ) = f =f +T a a x = f (because f is T -periodic) a = h(x). Thus h has period aT . Replacing a by 1/a, we find that the function f(ax) has period T /a. (c) Suppose that f is T -periodic. Then g(f(x + T )) = g(f(x)), and so g(f(x)) is also T -periodic. 10. (a) sin x has period 2π, so sin 2x has period 2π/2 = π (by Exercise 9(b)). (b) cos 12 x has period 4π and sin 2x has period π (or any integer multiple of it). So a common period is 4π. Thus cos 12 x + 3 sin 2x had period 4π (by Exercise 7) 1 (c) We can write 2+sin x = g(f(x)), where g(x) = 1/x and f(x) = 2 + sin x. Since 1 f is 2π-periodic, it follows that 2+sin x is 2π-periodic, by Exercise 9(c). (d) Same as part (c). Here we let f(x) = cos x and g(x) = ex . Then ecos x is 2π-periodic. 11. Using Theorem 1, Z

π/2

f(x) dx =

Z

−π/2

π

f(x) dx =

Z

0

π

sin x dx = 2. 0

12. Using Theorem 1, Z

π/2

f(x) dx = −π/2

Z

π

f(x) dx = 0

Z

π

cos x dx = 0. 0

Section 2.1

13.

Z

π/2

f(x) dx =

Z

−π/2

14. Using Theorem 1, Z π/2

F (x + 2π) =

Z

Rx a

15

π/2

1 dx = π/2. 0

Z

f(x) dx =

−π/2

15. Let F (x) =

Periodic Functions

π

f(x) dx =

Z

0

π

x2 dx = 0

π3 . 3

f(t) dt. If F is 2π-periodic, then F (x) = F (x + 2π). But

x+2π

f(t) dt =

Z

a

x

f(t) dt +

Z

a

x+2π

f(t) dt = F (x) +

Z

x

x+2π

f(t) dt. x

Since F (x) = F (x + 2π), we conclude that Z x+2π f(t) dt = 0. x

Applying Theorem 1, we find that Z x+2π Z f(t) dt = x

2π

f(t) dt = 0. 0

The above steps are reversible. That is, Z 2π Z x+2π f(t) dt = 0 ⇒ f(t) dt = 0 0

⇒

Z

x x

f(t) dt = a

Z

x

f(t) dt +

Z

a

x+2π

f(t) dt = x

Z

x+2π

f(t) dt a

⇒ F (x) = F (x + 2π); and so F is 2π-periodic. 16. We have F (x + T ) =

Z

x+T

f(t) dt = a

Z

x

f(t) dt + a

Z

x+T

f(t) dt. x

So F is T periodic if and only if F (x + T ) = F (x) ⇔

Z

x

f(t) dt +

⇔

Z

a

x a+T

⇔

Z

Z

x+T

f(t) dt = x

Z

x

f(t) dt a

x+T

f(t) dt = 0 f(t) dt = 0, a

where the last assertion follows from Theorem 1. R2 17. By Exercise 16, F is 2 periodic, because 0 f(t) dt = 0 (this is clear from the graph of f). So it is enough to describe F on any interval of length 2. For 0 < x < 2, we have Z x t2 x x2 F (x) = (1 − t) dt = t − = x − . 2 0 2 0 For all other x, F (x+2) = F (x). (b) The graph of F over the interval [0, 2] consists of the arch of a parabola looking down, with zeros at 0 and 2. Since F is 2-periodic, the graph is repeated over and over.

16 Chapter 2 Fourier Series 18. (a) We have Z

(n+1)T

f(x) dx

=

nT

= =

Z

T

f(s + nT ) ds

Z

0

Z

0

(let x = s + nT, dx = ds)

T

f(s) ds

(because f is T -periodic)

T

f(x) dx. 0

(b) Z

a+T

f(x) dx

=

(n+1)T

= =

Z

a

f(s + T ) ds

nT Z a

ZnT a

f(s) ds

(let x = s + T, dx = ds)

(because f is T -periodic)

f(x) dx.

nT

(c) Z

a+T

f(x) dx

=

a

= =

Z Z Z

(n+1)T

f(x) dx + a (n+1)T

f(x) dx + a (n+1)T

f(x) dx = nT

Z Z

a+T

f(x) dx (n+1)T a

f(x) dx

nT Z T

(by (b))

f(x) dx (by (a)).

0

19. (a) The plots are shown in the following figures.

(b) Let us show that f(x) = x − p f(x + p)

h i x p

is p-periodic.

x+p x x = x+p−p = x+p−p +1 = x+p−p +1 p p p x = x−p = f(x). p

From the graphs it is clear that f(x) = x for all 0 < x < p. To see this from the formula, use the fact that [t] = 0 if 0 ≤ t < 1. So, if 0 ≤ x < p, we have 0 ≤ xp < 1, h i so xp = 0, and hence f(x) = x. h i 20. (a) Plot of the function f(x) = x − 2p x+p for p = 1, 2, and π. 2p

Section 2.1

p 1 Plot x 2 pFloor

x

p

2p

, x, 2,2

Periodic Functions

17

p p

2

(b)

(x + 2p) + p x+p = (x + 2p) − 2p = (x + 2p) − 2p +1 2p 2p x+p x+p = (x + 2p) − 2p + 1 = x − 2p = f(x). 2p 2p

f(x + 2p)

So f is 2p-periodic. For −p < x < p, we have 0 < h i so f(x) = x − 2p x+p = x. 2p

x+p 2p

< 1, hence

h

x+p 2p

i

= 0, and

21. (a) With p = 1, the function f becomes f(x) = x−2 x+1 , and its graph is the 2 first one in the group shown in Exercise 20. The function is 2-periodic and is equal to x on the interval −1 < x < 1. By Exercise 9(c), the function g(x) = h(f(x) is 2periodic for any function h; in particular, taking h(x) = x2, we see that g(x) = f(x)2 is 2-periodic. (b) g(x) = x2 on the interval −1 < x < 1, because f(x) = x on that 2 interval. (c) Here is a graph of g(x) = f(x)2 = x − 2 x+1 , for all x. 2 Plot

x

2 Floor

x

1

2

^ 2, x,

3, 3

1

-3

-2

-1

1

2

3

is 2-periodic and is equal 22. (a) As in Exercise 21, the function f(x) = x − 2 x+1 2 to x on the interval −1 < x < 1. So, by Exercise 9(c), the function x + 1 g(x) = |f(x)| = x − 2 2 is 2-periodic and is clearly equal to |x| for all −1 < x < 1. Its graph is a triangular wave as shown in (b).

18 Chapter 2 Fourier Series Plot Abs x

2 Floor

x

1

2

, x,

3, 3

1

-3

-2

-1

1

2

3

(c) To obtain a triangular wave of arbitrary period 2p, we use the 2p-periodic function x+p f(x) = x − 2p , 2p which is equal to x on the interval −p < x < p. Thus, x + p g(x) = x − 2p 2p is a 2p-periodic triangular wave, which equal to |x| in the interval −p < x < p. The following graph illustrates this function with p = π. p Pi Plot Abs x

2 p Floor

x

p

2p

, x,

2 Pi, 2 Pi

23. (a) Since f(x + 2p) = f(x), it follows that g(f(x + 2p)) = g(f(x)) and so g(f(x)) is 2p-periodic. For −p < x < p, f(x) = x and so g(f(x)) = g(x). (b) The function eg(x) , with p = 1, is the 2-periodic extension of the function which equals ex on the interval −1 < x < 1. Its graph is shown in Figure 1, Section 2.6 (with a = 1). 24. Let f1 , f2 , . . . , fn be the continuous components of f on the interval [0, T ], as described prior to the exercise. Since each fj is continuous on a closed and bounded interval, it is bounded: That is, there exists M > 0 such that |fj (x)| ≤ M for all x in the domain of fj . Let M denote the maximum value of Mj for j = 1, 2, . . . , n. Then |f(x)| ≤ M for all x in [0, T ] and so f is bounded.

Section 2.1

Periodic Functions

19

25. We have |F (a + h) − F (a)| =

=

Z Z a+h a f(x) dx − f(x) dx 0 0 Z a+h f(x) dx ≤ M · h, a

where M is a bound for |f(x)|, which exists by the previous exercise. (In deriving the last inequality, we used the following property of integrals: Z b f(x) dx ≤ (b − a) · M, a which is clear if you interpret the integral as an area.) As h → 0, M · h → 0 and so |F (a+h)−F (a)| → 0, showing that F (a+h) → F (a), showing that F is continuous at a. Ra (b) If f is continuous and F (a) = 0 f(x) dx, the fundamental theorem of calculus implies that F 0(a) = f(a). If f is only piecewise continuous and a0 is a point of continuity of f, let (xj−1, xj ) denote the subinterval on which f is continuous and a0 is in (xj−1, xj ). Recall that f = fj on that subinterval, where fj Ris a continuous a component of f. For a in (xj−1, xj ), consider the functions F (a) = 0 f(x) dx and Ra R xj−1 G(a) = xj−1 fj (x) dx. Note that F (a) = G(a) + 0 f(x) dx = G(a) + c. Since fj is continuous on (xj−1, xj ), the fundamental theorem of calculus implies that G0(a) = fj (a) = f(a). Hence F 0(a) = f(a), since F differs from G by a constant. 26. We have F (a) =

Z

a+T

f(x) dx − 0

Z

a

f(x) dx. 0

By the previous exercise, F is a sum of Rtwo continuous and piecewise smooth a functions. (The first term is a translate of 0 f(x) dx by T , and so it is continuous and piecewise smooth.) Thus F is continuous and piecewise smooth. Since each term is differentiable the points of continuity of f, we conclude that F is also differentiable at the points of continuity of f. (b) By Exercise 25, we have, at the points where f is continuous, F 0(a) = f(a + T ) − f(a) = 0, because f is periodic with period T . Thus F is piecewise constant. (c) A piecewise constant function that is continuous is constant (just take left and right limits at points of discontinuity.) So F is constant. 27. (a) The function sin x1 does not have a right or left limit as x → 0, and so it is not piecewise continuous. (To be piecewise continuous, the left and right limits must exist.) The reason is that 1/x tends to +∞ as x → 0+ and so sin 1/x oscillates between +1 and −1. Similarly, as x → 0− , sin 1/x oscillates between +1 and −1. See the graph. (b) The function f(x) = x sin x1 and f(0) = 0 is continuous at 0. The reason for this is that sin 1/x is bounded by 1, so, as x → 0, x sin 1/x → 0, by the squeeze theorem. The function, however, is not piecewise smooth. To see this, let us compute its derivative. For x 6= 0, f 0 (x) = sin

1 1 1 − cos . x x x

As x → 0+ , 1/x → +∞, and so sin 1/x oscillates between +1 and −1, while 1 cos x1 oscillates between +∞ and −∞. Consequently, f 0 (x) has no right limit at x 0. Similarly, it fails to have a left limit at 0. Hence f is not piecewise smooth. (Recall that to be piecewise smooth the left and right limits of the derivative have to exist.) (c) The function f(x) = x2 sin x1 and f(0) = 0 is continuous at 0, as in case (b).

20 Chapter 2 Fourier Series Also, as in (b), the function is not piecewise smooth. To see this, let us compute its derivative. For x 6= 0, f 0 (x) = 2x sin

1 1 − cos . x x

As x → 0+ , 1/x → +∞, and so 2x sin 1/x → 0, while cos x1 oscillates between +1 and −1. Hence, as x → 0+ , 2x sin x1 − cos x1 oscillates between +1 and −1, and so f 0 (x) has no right limit at 0. Similarly, it fails to have a left limit at 0. Hence f is not piecewise smooth. (d) The function f(x) = x3 sin x1 and f(0) = 0 is continuous at 0, as in case (b). It is also smooth. We only need to check the derivative at x = 0. We have h3 sin h1 f(h) − f(0) 1 = lim lim h2 sin = 0. h→0 h→0 h→0 h h h

f 0 (0) = lim For x 6= 0, we have

1 1 − x cos . x x Since f 0 (x) → 0 = f 0 (0) as x → 0, we conclude that f 0 exists and is continuous for all x. f 0 (x) = 3x2 sin

Plot Sin 1 x , x, .3, .3 , PlotPoints

5

Plot x Sin 1 x , x, .3, .3 , PlotPoints

1

0.1

-0.3 -0.3

10

0.3

0.3

-0.2

-1

x ^ 2 Sin 1 x , x, .1, .1

Plot x ^ 3 Sin 1 x , x, .1, .1

0.0002

0.004

-0.1

0.1

-0.004

-0.1

0.1

-0.0002

Section 2.2

Fourier Series

21

Solutions to Exercises 2.2 1. The graph of the Fourier series is identical to the graph of the function, except at the points of discontinuity where the Fourier series is equal to the average of the function at these points, which is 12 .

2. The graph of the Fourier series is identical to the graph of the function, except at the points of discontinuity where the Fourier series is equal to the average of the function at these points, which is 0 in this case.

3. The graph of the Fourier series is identical to the graph of the function, except at the points of discontinuity where the Fourier series is equal to the average of the function at these points, which is 3/4 in this case.

4. Since the function is continuous and piecewise smooth, it is equal to its Fourier series.

5. We compute the Fourier coefficients using he Euler formulas. Let us first note that since f(x) = |x| is an even function on the interval −π < x < π, the product f(x) sin nx is an odd function. So

1 bn = π

Z

π

odd function

z }| { |x| sin nx dx = 0,

−π

because the integral of an odd function over a symmetric interval is 0. For the other

22 Chapter 2 Fourier Series coefficients, we have a0

= = =

Z π 1 f(x) dx = |x| dx 2π −π −π Z 0 Z π 1 1 = (−x) dx + x dx 2π −π 2π 0 Z 1 π 1 2 π π x dx = x = . π 0 2π 0 2 1 2π

Z

π

In computing an (n ≥ 1), we will need the formula Z cos(a x) x sin(a x) x cos ax dx = + +C a2 a

(a 6= 0),

which can be derived using integration by parts. We have, for n ≥ 1, Z Z 1 π 1 π an = f(x) cos nx dx = |x| cos nx dx π −π π −π Z π 2 = x cos nx dx π 0 2 x 1 π = sin nx + 2 cos nx π n n 0 n 2 2 (−1) 1 (−1)n − 1 = − 2 = π n2 n πn2 0 if n is even = 4 − πn if n is odd. 2 Thus, the Fourier series is ∞ π 1 4X cos(2k + 1)x . − 2 π (2k + 1)2 k=0

s n_, x_ : Pi 2

4 Pi Sum 1

partialsums Table s x 2 Pi Floor f x_ g x_ Abs f x Plot g x , x, 3 Pi, Plot Evaluate g x ,

2k

1 ^ 2 Cos

n, x , n, 1, 7 x Pi 2 Pi 3 Pi partialsums

2k

1 x , k, 0, n

;

, x,

2 Pi, 2 Pi

The function g(x) = | x | and its periodic extension

Partial sums of the Fourier series. Since we are summing over the odd integers, when n = 7, we are actually summing the 15th partial sum.

Section 2.2

Fourier Series

23

6. We compute the Fourier coefficients using he Euler formulas. Let us first note that f(x) is an odd function on the interval −π < x < π, so Z π 1 a0 = f(x) dx = 0, 2π −π because the integral of an odd function over a symmetric interval is 0. A similar argument shows that an = 0 for all n. This leaves bn . We have, for n ≥ 1, Z 1 π bn = f(x) sin nx dx π −π Z Z 1 0 1 π/2 = (−1) sin nx dx + sin nx dx π −π/2 π 0 Z 2 π/2 = sin nx dx π 0 2 1 π/2 = − cos nx π n 0 2 2 (−1)n 1 nπ = − 2 = 1 − cos 2 π n n πn 2 Thus the Fourier series is ∞ 2 X 1 − cos π n=1 n

nπ 2

sin nx.

(b) The new part here is defining the periodic function. We willuse the same construction as in the previous exercise. We will use the Which command to define a function by several formulas over the interval - 2 Pi to 2 Pi. We then use the construction of Exrcises 20 and 22 of Section 2.1 to define the periodic extension. s n_, x_ : 2 Pi Sum partialsums Table s Which x Pi f x_ g x_ x 2 Pi Floor h x_ f g x Plot h x , x, 3 Pi, Plot Evaluate h x ,

1

Cos k Pi 2

k Sin k x , k, 1, n

n, x , n, 1, 10, 2 ; 2, 0, Pi 2 x 0, 1, 0 x Pi 2 Pi 3 Pi partialsums

, x,

x

Pi 2, 1, x

Pi 2, 0

2 Pi, 2 Pi

1 1

-1 -1 The partial sums of the Fourier series converge to the function at all points of continuity. At the points of discontinuity, we observe a Gibbs phenomenon. At the points of discontinuity, the Fourier series is converging to 0 or ±1/2. depending on the location of the point of discontinuity: For all x = 2kπ, the Fourier series converges to 0; and for all x = (2k + 1)π/2, the Fourier series converges to (−1)k /2.

24 Chapter 2 Fourier Series 7. f is even, so all the bn’s are zero. We have Z π Z π 1 1 π 1 2 a0 = f(x) dx = sin x dx = − cos x = . 0 2π −π π 0 π π We will need the trigonometric identity sin a cos b =

1 sin(a − b) + sin(a + b) . 2

For n ≥ 1, an

= = = = =

Z Z 1 π 2 π f(x) cos nx dx = sin x cos nx dx π −π π 0 Z 2 π1 sin(1 − n)x + sin(1 + n)x dx π 0 2 1 −1 1 π (if n 6= 1) cos(1 − n)x − cos(1 + n)x π 1−n (1 + n) 0 1 −1 1 1 1 (−1)1−n − (−1)1+n + + π 1−n (1 + n) 1−n 1+n 0 if n is odd 4 if n is even. π(1−n2 )

If n = 1, we have a1 =

2 π

Z

π

sin x cos x dx = 0

Thus, the Fourier series is : | sin x| =

1 π

Z

π

sin(2x) dx = 0. 0

∞ 2 1 4X − cos 2kx . π π (2k)2 − 1 k=1

8. You can compute the Fourier series directly as we did in Exercise 7, or you can use Exercise 7 and note that | cos x| = | sin(x + π2 )|. This identity can be verified by comparing the graphs of the functions or as follows: =0

=1

z }| { z }| { π π π sin(x + ) = sin x cos + cos x sin = cos x. 2 2 2 So | cos x| = = =

π )| 2 ∞ 2 1 4X π − cos 2k(x + ) 2 π π (2k) − 1 2

| sin(x +

2 4 − π π

k=1 ∞ X k=1

(−1)k cos 2kx, (2k)2 − 1

where we have used π cos 2k(x + ) = cos 2kx cos kπ − sin 2kx sin kπ = (−1)k cos 2kx. 2 9. Just some hints: (1) f is even, so all the bn ’s are zero. (2) Z 1 π 2 π2 a0 = x dx = . π 0 3

Section 2.2

Fourier Series

25

(3) Establish the identity Z

x2 cos(ax) dx =

−2 + a2 x2 sin(a x) 2 x cos(a x) + +C a2 a3

(a 6= 0),

using integration by parts. 10. The function f(x) = 1 − sin x + 3 cos 2x is already given by its own Fourier series. If you try to compute the Fourier coefficients by using the Euler formulas, you will find that all an s and bn’s are 0 except a0 = 1, a2 = 3, and b1 = −1. This is because of the orthogonality of the trigonometric system. Let us illustrate this by computing the bn ’s (n ≥ 1): Z π 1 bn = (1 − sin x + 3 cos 2x) sin nx dx 2π −π 1 2π

zZ

=0 for all n π

}|

{

Z

π

1 3 = sin nx dx − sin x sin nx dx + 2π −π 2π −π Z π 1 = − sin x sin nx dx 2π −π ( −1 if n = 1 = 0 otherwise,

zZ

=0 for all n π

}|

{

cos 2x sin nx dx −π

which shows that b1 = −1, while all other bns are 0. 11. We have f(x) = 12 − 12 cos 2x and g(x) = given by their Fourier series.

1 2

+

1 2

cos 2x. Both functions are

12. The function is clearly odd and so, as in Exercise 6, all the ans are 0. Also, Z 2 π 2 bn = (π x − x3 ) sin nx dx. π 0 To compute this integral, we use integration by parts, as follows: u v Z 2 π z 2 }| 3 { z }| { (π x − x ) sin nx dx π 0 Z 2 2 − cos nx π 2 π 2 − cos nx (π − 3x2) (π x − x3) dx − π n π 0 n 0 0

bn

= =

= =

v0

u z }| { zcos}|nx{ 2 2 (π − 3x2) dx π 0 n Z 2 2 sin nx π 12 π sin nx x 2 dx (π − 3x2) 2 − π n π 0 n 0

Z

π

=0

= =

zZ }| { 12 (− cos nx) π 12 π cos nx dx x + π n3 π 0 n3 0 (− cos nπ) (−1)n+1 12 = 12 3 n n3

Thus the Fourier series is 12

∞ X (−1)n+1 sin nx. n3 n=1

13. You can compute directly as we did in Example 1, or you can use the result of Example 1 as follows. Rename the function in Example 1 g(x). By comparing

26 Chapter 2 Fourier Series graphs, note that f(x) = −2g(x + π). Now using the Fourier series of g(x) from Example, we get ∞ ∞ X X sin n(π + x) (−1)n+1 =2 sin nx. f(x) = −2 n n n=1 n=1

14. The function is even. Its graph is as follows:

Rπ 1 Since a0 = 2π −π f(x) dx, by interpreting the integral as an area, we see that a0 is 1/(2π) times the area under the graph of f(x), above the x-axis, from x = −d to 1 x = d. Thus a0 = 2π cd. To compute an, we use integration by parts: an

= = =

2c dπ

Z

d

u

0

v z }| { z }| { (x − d) cos nx dx

0

Z 2c sin nx d 2 d sin nx − (x − d) dx dπ n 0 π 0 n 2c − cos nx d 2c (1 − cos(nd)) . = dπ n2 dπn2 0

Using the identity 1 − cos α = 2 sin2 α2 , we find that bn =

nd 4c sin2 . 2 dπn 2

Thus the Fourier series is ∞ cd 4c X sin2 nd 2 cos nx. + 2π dπ n=1 n2

15. f is even, so all the bn’s are zero. We have Z π Z 1 1 π −x e−x π 1 − e−π a0 = f(x) dx = e dx = − . = 2π −π π 0 π 0 π We will need the integral identity Z a ea x cos(b x) b ea x sin(b x) eax cos(bx) dx = + +C a2 + b2 a2 + b2

(a2 + b2 6= 0),

which can be established by using integration by parts; alternatively, see Exercise 17, Section 2.6. We have, for n ≥ 1, Z 2 π −x an = e cos nx dx π 0 2 1 n π −x −x = sin nx − cos nx e e 2 2 0 π n +1 n +1 n −π 2 2(1 − (−1) e ) = −e−π (−1)n + 1 = . π(n2 + 1) π(n2 + 1) Thus the Fourier series is

∞ eπ − 1 2 X 1 + (eπ − (−1)n ) cos nx . πeπ πeπ n=1 n2 + 1

Section 2.2

Fourier Series

27

16. In general, the function is neither even nor odd. Its graph is the following:

1 As in Exercise 14, we interpret a0 as a scaled area and obtain a0 = 2π (2c)/(2c) = 1/(2π). For the other coefficients, we will compute the integrals over the interval (d − π, d + π), whose length is 2π. This is possible by Theorem 1, Section 2.1, because the integrands are 2π-periodic. We have Z d+c 1 an = cos nx dx 2cπ d−c d+c 1 1 = = sin nx (sin n(d + c) − sin n(d − c)) 2cnπ 2cnπ d−c 1 = cos(nd) sin(nc). cnπ

Similarly, bn

= = =

1 2cπ

Z

d+c

sin nx dx d−c

d+c 1 1 = cos nx (cos n(d − c) − cos n(d + c)) 2cnπ 2cnπ d−c 1 sin(nd) sin(nc). cnπ

−

Thus the Fourier series is ∞ 1 1 X cos(nd) sin(nc) sin(nd) sin(nc) + cos nx + sin nx . 2π cπ n=1 n n Another way of writing the Fourier series is as follows: ∞ 1 1 X sin(nc) + (cos(nd) cos nx + sin(nd) sin nx) 2π cπ n=1 n ∞ 1 1 X sin(nc) = + cos n(x − d) . 2π cπ n=1 n 17. Setting x = π in the Fourier series expansion in Exercise 9 and using the fact that the Fourier series converges for all x to f(x), we obtain π2 = f(π) =

∞ ∞ X X (−1)n 1 π2 π2 cos nπ = , +4 + 4 2 2 3 n 3 n n=1 n=1

where we have used cos nπ = (−1)n. Simplifying, we find ∞ X π2 1 . = 2 6 n n=1

18. Take x = π2 in the Fourier series of Exercise 13, note that sin(n π2 ) = 0 if n = 2k and sin(n π2 ) = (−1)k if n = 2k + 1, and get ∞

∞

k=0

k=0

X (−1)2k+2 (−1)k X (−1)k π =2 =2 , 2 2k + 1 2k + 1

28 Chapter 2 Fourier Series and the desired series follows upon dividing by 2. 19. (a) Let f(x) denote the function in Exercise 1 and w(x) the function in Example 5. Comparing these functions, we find that f(x) = π1 w(x). Now using the Fourier series of w, we find 1 f(x) = π

"

# ∞ ∞ X sin(2k + 1)x π 1 2 X sin(2k + 1)x +2 = + . 2 2k + 1 2 π 2k + 1 k=0

k=0

(b) Let g(x) denote the function in Exercise 2 and f(x) the function in (a). Comparing these functions, we find that g(x) = 2f(x) − 1. Now using the Fourier series of f, we find g(x) =

∞ 4 X sin(2k + 1)x . π 2k + 1 k=0

(c) Let k(x) denote the function in Figure 13,and let f(x) be as in (a). Comparing these functions, we find that k(x) = f x + π2 . Now using the Fourier series of f, we get

k(x) =

∞ 1 2 X sin(2k + 1)(x + π2 ) + 2 π 2k + 1 k=0

=0

=

=(−1)k

z }| { z }| { ∞ 1 π π 2X 1 + sin[(2k + 1)x] cos[(2k + 1) ] + cos[(2k + 1)x] sin[(2k + 1) ] 2 π 2k + 1 2 2 k=0

=

∞ 1 2 X (−1)k + cos[(2k + 1)x]. 2 π 2k + 1 k=0

(d) Let v(x) denote the function in Exercise 3, and let k(x) be as in (c). Comparing these functions, we find that v(x) = 12 k(x) + 1 . Now using the Fourier series of k, we get v(x) =

∞ 3 1 X (−1)k + cos[(2k + 1)x]. 4 π 2k + 1 k=0

20. (a) Let f(x) denote the function in Figure 14 and let | sin x|. By comparing graphs, we see that | sin x| + sin x = 2f(x). So

f(x) =

1 (| sin x| + sin x) . 2

Now the Fourier series of sin x is sin x and the Fourier series of | sin x| is computed in Exercise 7. Combining these two series, we obtain

f(x) =

∞ 1 1 1 2X sin x + − cos 2kx. 2 π π (2k)2 − 1 k=1

In particular, f(x) has only one nonzero bn; namely, b1 = 12 . All other bn s are 0. (b) Let g(x) denote the function in Figure 15. Then g(x) = f(x + π2 ). Hence, using

Section 2.2

Fourier Series

29

sin(a + π2 ) = cos a, we obtain g(x)

π 2

=

f x+

=

∞ h 1 1 π 1 2X π i sin x + + − cos 2k x + 2 2 2 π π (2k) − 1 2

=

1 1 + cos x π 2

k=1

(−1)k

=0

z }| { z }| { i ∞ 2 Xh π π 1 − cos(2kx) cos(2k ) − sin(2kx) sin(2k ) π (2k)2 − 1 2 2 k=1

=

∞ i 1 1 2 X h (−1)k + cos x − cos(2kx) . π 2 π (2k)2 − 1 k=1

21. (a) Interpreting the integral as an area (see Exercise 16), we have a0 =

1 1 π 1 · · = . 2π 2 2 8

To compute an, we first determine the equation of the function for π2 < x < π. From Figure 16, we see that f(x) = π2 (π − x) if π2 < x < π. Hence, for n ≥ 1, an

= = = =

Z

u

0

v 2 z }| { z }| { (π − x) cos nx dx π/2 π Z π 2 sin nx sin nx π 2 (π − x) + 2 dx 2 π n π/2 π π/2 n π 2 −π nπ 2 cos nx sin − π/2 π2 2n 2 π 2 n2 2 π 1 nπ nπ (−1)n − 2 − cos sin + . π 2n 2 n2 n2 2

1 π

π

Also,

bn

= = =

Z

u

v0

2 z }| { z }| { (π − x) sin nx dx π/2 π Z π 2 cos nx cos nx π 2 − 2 (π − x) − dx π n π/2 π2 π/2 n 2 π nπ nπ 1 sin cos + . π2 2n 2 n2 2 1 π

π

Thus the Fourier series representation of f is ∞ 1 π 1 nπ 2 X nπ (−1)n f(x) = − − 2 cos + sin + cos nx 8 π2 n=1 2n 2 n2 n 2 π nπ nπ 1 + cos + 2 sin sin nx . 2n 2 n 2

30 Chapter 2 Fourier Series (b) Let g(x) = f(−x). By performing a change of variables x ↔ −x in the Fourier series of f, we obtain (see also Exercise 24 for related details) Thus the Fourier series representation of f is g(x) =

∞ 1 nπ π 2 X nπ (−1)n 1 − cos − + 2 sin + cos nx 8 π n=1 2n 2 n2 n2 2 π nπ nπ 1 − cos + 2 sin sin nx . 2n 2 n 2

22. By comparing graphs, we see that the function in Exercise 4 (call it k(x)) is the sum of the three functions in Exercises 19(c) and 21(a) and (b) (call them h, f, and g, respectively). Thus, adding and simplifying, we obtain k(x) =

=

∞ 1 2 X sin nπ 2 + cos nx 2 π n=1 n ∞ 1 π 1 nπ 2 X nπ (−1)n + + 2 − − 2 cos sin + cos nx 8 π n=1 2n 2 n2 n 2 π nπ nπ 1 + cos + 2 sin sin nx 2n 2 n 2 ∞ 1 π 1 nπ 2 X nπ (−1)n + + 2 − − cos sin + cos nx 8 π n=1 2n 2 n2 n2 2 π nπ nπ 1 − cos + 2 sin sin nx 2n 2 n 2 ∞ 3 (−1)n 1 nπ 4 X − cos − 2 cos nx . 4 π n=1 n2 n2 2

We illustrate the convergence of the Fourier series in the following figure. s n_, x_ : 3 4 4 Pi ^ 2 Sum 1 ^ k Cos k Pi 2 partialsums Table s n, x , n, 1, 7 ; Plot Evaluate partialsums , x, Pi, Pi Plot Evaluate partialsums , x, 3 Pi, 3 Pi

Cos k x

k ^ 2, k, 1, n

23. This exercise is straightforward and follows from the fact that the integral is linear. 24. (a) This part follows by making appropriate changes of variables. We have a(g, 0)

= =

1 2π 1 2π

Z

π

g(x) dx =

−π Z −π π

1 2π

Z

π

f(−x) dx −π

f(x) (−1)dx =

1 2π

Z

π

f(x) dx = a(f, 0). −π

Section 2.2

Fourier Series

31

Similarly,

a(g, n) = = b(g, n) = =

1 π

Z

π

g(x) cos nx dx =

−π Z −π

1 π

Z

π

f(−x) cos nx dx −π

Z 1 1 π f(x) cos(−nx) (−1)dx = f(x) cos nx dx = a(f, n); π π π −π Z π Z π 1 1 g(x) sin nx dx = f(−x) sin nx dx π −π π −π Z Z 1 −π 1 π f(x) sin(−nx) (−1)dx = − f(x) sin nx dx = −b(f, n). π π π −π

(b) As in part (a),

a(h, 0) = = = a(h, n) = = = =

= b(h, n) = = = =

1 2π

Z

π

h(x) dx =

−π Z π−α

1 2π

Z

π

f(x − α) dx −π

1 f(x) dx 2π −π−α Z π 1 f(x) dx = a(f, 0); (by Theorem 1, Section 2.1) 2π −π Z π Z 1 1 π h(x) cos nx dx = f(x − α) cos nx dx π −π π −π Z 1 π−α f(x) cos(n(x + α)) dx π −π−α Z Z cos nα π−α sin nα π−α f(x) cos nx dx − f(x) sin nx dx π π −π−α −π−α Z Z cos nα π sin nα π f(x) cos nx dx − f(x) sin nx dx π π −π −π (by Theorem 1, Section 2.1) a(f, n) cos nα − b(f, n) sin nα; Z Z 1 π 1 π h(x) sin nx dx = f(x − α) sin nx dx π −π π −π Z 1 π−α f(x) sin(n(x + α)) dx π −π−α Z Z cos nα π−α sin nα π−α f(x) sin nx dx + f(x) cos nx dx π π −π−α −π−α b(f, n) cos nα + a(f, n) sin nα.

25. For (a) and (b), see plots. Pn (c) We have sn (x) = k=1 sinkkx . So sn (0) = 0 and sn (2π) = 0 for all n. Also, π limx→0+ f(x) = 2 , so the difference between sn (x) and f(x) is equal to π/24 at x = 0. But even we look near x = 0, where the Fourier series converges to f(x), the difference |sn(x)−f(x)| remains larger than a positive number, that is about .28 and does not get smaller no matter how large n. In the figure, we plot |f(x) − s150(x)|. As you can see, this difference is 0 everywhere on the interval (0, 2π), except near the points 0 and 2π, where this difference is approximately .28. The precise analysis of this phenomenon is done in the following exercise.

32 Chapter 2 Fourier Series

26. (a) We have sN

π N

=

N X sin n=1

nπ N

n

N X π sin nπ N . = · nπ N N n=1

(b) Consider the Riemann integral I=

Z

π 0

sin x dx, x

sin x x

where the integrand f(x) = may be considered as a continuous function on the interval [0, π] if we define f(0) = 0. This is because limx→0 sinx x = 1. Since the definite integral of a continuous function on a closed interval is the limit of Riemann sums, it follows that the integral I is the limit of Riemann sums corresponding to the function f(x) on the interval [0, π]. We now describe these Riemann sums. π Let ∆x = N and partition the interval [0, π] into N subintervals of equal length ∆x. Form the N th Riemann sum by evaluating the function f(x) at the right endpoint of each subinterval. We have N of these endpoints and they are nπ N , for n = 1, 2, . . . , N . Thus, I

=

= =

N nπ X f ∆x N →∞ N n=1

lim

N N X X sin nπ π sin nπ N N lim = lim · nπ N →∞ N →∞ N n N n=1 n=1 π lim sN . N →∞ N

Hence lim sN

N →∞

π N

=

Z

π 0

sin x dx, x

Section 2.2

33

Fourier Series

as desired. (c) From the Taylor series sin x =

∞ X n=0

(−1)n

x2n+1 , (2n + 1)!

(−∞ < x < ∞),

we obtain, for all x 6= 0, ∞ sin x X x2n (−1)n = , x (2n + 1)! n=0

(−∞ < x < ∞, x 6= 0).

The right side is a power series that converges for all x 6= 0. It is also convergent for x = 0 and its value for x = 0 is 1. Since the left side is also equal to 1 (in the limit), we conclude that ∞ sin x X x2n (−1)n = , x (2n + 1)! n=0

(−∞ < x < ∞).

(d) A power series can be integrated term-by-term within its radius of convergence. Thus ! Z π Z π X ∞ sin x x2n n (−1) dx = dx x (2n + 1)! 0 0 n=0 ∞ Z π X x2n = (−1)n dx (2n + 1)! n=0 0 ∞ ∞ X (−1)n x2n+1 π X (−1)n π2n+1 = . = (2n + 1)! 2n + 1 0 n=0 (2n + 1)! 2n + 1 n=0 We have thus expressed the integral I as an alternating series. Using a property of an alternating series with decreasing coefficients, we know that I is greater than any partial sum that is obtained by leaving out terms starting with a negative term; and I is smaller than any partial that is obtained by leaving out terms starting Psum ∞ with a positive term. So if I = n=0 (−1)n an, where an are positive and decreasing PN PN to 0, then I > n=0 (−1)n an if N is even and I < n=0 (−1)n an if N is odd. So 5 4 X X (−1)n π2n+1 (−1)n π2n+1 1, bn

= = = =

Z 2 π cos x sin nx dx π 0 2 cos(1 − n)x cos(1 + n)x π − 0 π 2(1 − n) 2(1 + n) n−1 n+1 1 (−1) (−1) 1 1 − − + π (1 − n) (1 + n) (1 − n) (1 + n) 2n 1 + (−1)n . π n2 − 1

For n = 1, you can easily show that b1 = 0. Thus the sine Fourier series is ∞ ∞ 2 X 1 + (−1)n k 4X n 2 sin nx = sin(2kx). π n=2 n −1 π (2k)2 − 1 k=1

b k_ 2 k Pi 1 1 ^k k^2 1 ss n_, x_ : Sum b k Sin k x , k, partialsineseries Table ss n, x , Cos x f x_ Plot Evaluate partialsineseries, f

; 2, n ; n, 5, 30, 5 x

;

, x, 0, Pi

1 0.5 0.5

1

1.5

2

2.5

3

-0.5 -1

7. The even extension is the function | cos x|. This is easily seen by plotting the graph. The cosine series is (Exercise 8, Section 2.2):

| cos x| =

∞ 2 4 X (−1)n − cos(2nx). π π n=1 (2n)2 − 1

Section 2.4

Half-Range Expansions: The Cosine and Sine Series

55

Sine series: bn

= =

4 π 2 π

Z Z

π 2

cos x sin 2nx dx 0 π 2

[sin(1 + 2n)x − sin(1 − 2 n)x)] dx 0 =0

z

= =

=0

}| { z }| { (−1 + 2 n) π (1 + 2 n) π cos( ) cos( )i 2h 1 1 2 2 + − − π −1 + 2 n 1 + 2 n −1 + 2 n 1+2n 8 n ; π 4n2 − 1

thus the odd extension has the sine series ∞ n 8X fo (x) = sin 2nx. π n=1 4n2 − 1 8. The even 2π-periodic extension of the function f(x) = x sin x for 0 < x < π is given on the interval (−π, π) by f1 (x) = x sin x. Its Fourier series is computed in Exercise 11, Section 2.3. We have f1 (x) = 1 −

∞ X 2(−1)n+1 cos x +2 cos nx. 2 n2 − 1 n=2

The odd extension is given by f2 (x) = |x| sin x for −π < x < π. For the sine series, we have, for n 6= 1, Sine series: Z 2 π bn = x sin x sin nx dx π 0 Z π 1 = x − cos(n + 1)x + cos(n − 1)x dx π 0 Z 1 π = x cos(n + 1)x + cos(n − 1)x dx π 0 1 x −1 = cos(n + 1)x − sin(n + 1)x 2 π (n + 1) n+1 1 x π + cos(n − 1)x + cos(n − 1)x + sin(n − 1)x + cos(n − 1)x 2 0 (n − 1) n−1 n−1 1 (−1) 1 1 −1 = (−1)n+1 + + − π (n + 1)2 (n − 1)2 (n + 1)2 (n − 1)2 1 −4n = , [1 + (−1)n ] 2 π (n − 1)2 and

2 b1 = π

Z

Thus f2 (x) =

π

1 x sin x dx = π 2

0

Z

π

x(1 − cos 2x) dx = 0

π . 2

∞ π n 4X [1 + (−1)n ] 2 sin nx. sin x − 2 π n=2 (n − 1)2

9. We have bn = 2

Z

1

x(1 − x) sin(nπx) dx. 0

To evaluate this integral, we will use integration by parts to derive the following two formulas: for a 6= 0, Z x cos(a x) sin(a x) x sin(ax) dx = − + C, + a a2

56 Chapter 2 Fourier Series and

Z

x2 sin(ax) dx =

2 cos(a x) x2 cos(a x) 2 x sin(a x) − + C. + a3 a a2

So Z

x(1 − x) sin(ax) dx =

−2 cos(a x) x cos(a x) x2 cos(a x) sin(a x) 2 x sin(a x) − − + C. + + a3 a a a2 a2

Applying the formula with a = nπ, we get Z

1

x(1 − x) sin(nπx) dx 0

−2 cos(nπ x) x cos(nπ x) x2 cos(nπ x) sin(nπ x) 2 x sin(nπ x) 1 − − + + 0 (nπ)3 nπ nπ (nπ)2 (nπ)2 n n n n −2 ((−1) − 1) (−1) (−1) −2 ((−1) − 1) = − + = (nπ)3 nπ nπ (nπ)3 4 if n is odd, (nπ)3 = 0 if n is even. =

Thus bn = Hence the sine series in

8 (nπ)3

0

if n is odd, if n is even,

∞ 8 X sin(2k + 1)πx . π3 (2k + 1)3 k=0

b k_ 8 Pi ^ 3 1 2 k 1 ^ 3; ss n_, x_ : Sum b k Sin 2 k 1 Pi x , k, 0, n ; partialsineseries Table ss n, x , n, 1, 5 ; x 1 x f x_ Plot Evaluate partialsineseries, f x , x, 0, 1

0.25 0.2 0.15 0.1 0.05 0.2

0.4

0.6

0.8

1

Perfect! 10. We use the fact that the function is the sum of three familiar functions to derive its sine series without excessive computations. Write 1 − x2 = 1 − x + x(1 − x) = g1(x) + g2(x) + g3 (x), where g1 (x) = 1, g2 (x) = −x, and g3(x) = x(1 − x). For g1 , use Exercise 1; for g2 , use Example 1; and and for g3, use Exercise 9. Putting all this together, we find

Section 2.4

Half-Range Expansions: The Cosine and Sine Series

57

the sine coefficient bn

2 2(−1)n+1 −4 ((−1)n − 1) (1 − (−1)n ) − + nπ nπ (nπ)3 n 2 ((−1) − 1) . −4 nπ (nπ)3

= =

Thus, we have the sine series ∞ X 1 ((−1)n − 1) 2 sin nπx. −2 nπ (nπ)3 n=1 11. The function is its own sine series. 12. Sine series expansion: Z 1 π bn = 2 sin( x) sin(nπx) dx 2 0 sin(nπ − π/2)x sin(nπ + π/2)x 1 = − nπ − π/2 nπ + π/2 0 sin(nπ − π/2) sin(nπ + π/2) = − nπ − π/2 nπ + π/2 cos(nπ) 1 cos(nπ) 1 = − − = −(−1)n + nπ − π/2 nπ + π/2 nπ + π/2 nπ − π/2 n+1 2nπ (−1) 8n = (−1)n+1 = . (nπ)2 − (π/2)2 π 4n2 − 1 Thus the sine series is

∞ 8 X (−1)n+1 n sin nπx. π n=1 4n2 − 1

13. We have

1 sin 2πx. 2 This yields the desired 2-periodic sine series expansion. sin πx cos πx =

14. We have

1 sin 2πx. 2 This yields the desired 2-periodic sine series expansion. (1 + cos πx) sin πx = sin πx +

15. We have bn = 2

Z

1

1 2ex (sin nπx − nπ cos nπx) 1 + (nπ)2 0 2nπ 2e n+1 nπ(−1) + 1 + (nπ)2 1 + (nπ)2 2nπ 1 + e(−1)n+1 . 2 1 + (nπ)

ex sin nπx dx = = 0

= = Hence the sine series is ∞ X

2nπ 1 + e(−1)n+1 sin nπx. 2 1 + (nπ) n=1 16. Just add the sine series of 1 and the sine series of −ex for 0 < x < 1 and you will get the sine series ∞ X 2 2nπ n n+1 1 + e(−1) sin nπx. (1 − (−1) ) − πn 1 + (nπ)2 n=1

58 Chapter 2 Fourier Series Here is an illustration of the since series convergence. b k_ 2 k Pi 1 1 ^k 2 k Pi 1 k Pi ^ 2 1 E 1 ^ k 1 ss n_, x_ : Sum b k Sin k Pi x , k, 1, n ; partialsineseries Table ss n, x , n, 3, 10, 2 ; 1 E^x f x_ , x, 0, 1 , PlotRange All Plot Evaluate partialsineseries, f x

0.2

0.4

0.6

0.8

1

-0.25 -0.5 -0.75 -1 -1.25 -1.5 -1.75 17. (b) Sine series expansion: Z Z 2 ah nπx 2 p h nπx bn = x sin dx + (x − p) sin dx p 0 a p p a a−p p Z a 2h h p nπx i p nπx a = cos −x cos dx + ap nπ p 0 nπ 0 p Z p 2h h p (−p) n π x p nπx i + (x − p) cos( ) + cos dx (a − p)p nπ p p a a nπ 2h h −ap nπa i nπa p2 = sin cos + pa nπ p (nπ)2 p h 2h nπa i p nπa p2 + sin (a − p) cos − (a − p)p nπ p (nπ)2 p 2hp nπa 1 1 = sin − (nπ)2 p a a−p 2hp2 nπa = sin . 2 (nπ) (p − a)a p Hence, we obtain the given Fourier series.

Solutions to Exercises 2.5 1. We have f(x) =

1 −1

if 0 < x < 1, if − 1 < x < 0;

The Fourier series representation is f(x) =

∞ 4X 1 sin(2k + 1)πx. π 2k + 1 k=0

The mean square error (from (5)) is Z N 1 1 2 1X 2 2 EN = f (x) dx − a0 − (a + b2n). 2 −1 2 n=1 n 4 In this case, an = 0 for all n, b2k = 0, b2k+1 = π(2k+1) , and Z Z 1 1 2 1 1 f (x) dx = dx = 1. 2 −1 2 −1

;

Section 2.5

Mean Square Approximation and Parseval’s Identity

59

So

1 8 E1 = 1 − (b21 ) = 1 − 2 ≈ 0.189. 2 π Since b2 = 0, it follows that E2 = E1. Finally, 8 1 8 E1 = 1 − (b21 + b23) = 1 − 2 − ≈ 0.099. 2 π 9 π2 2. We have x if − π < x < π, f(x) = f(x + 2π) otherwise. Its Fourier series representation is f(x) = 2

∞ X (−1)n+1 sin nx. n n=1

The mean square error (from (5)) is Z π N 1 1X 2 EN = f 2 (x) dx − a20 − (a + b2n) 2π −π 2 n=1 n Z π N 1 1X 2 = x2 dx − a20 − (a + b2n) 2π −π 2 n=1 n =

N 1 3 π 1X 2 2 (a + b2n) x − a0 − 6π −π 2 n=1 n

=

N X π2 1 , −2 2 3 n n=1 n+1

where, in the last equality, we used an = 0 and bn = 2 (−1)n

for all n. So

2

E1 E2 E3

π − 2 ≈ 1.29; 3 = E1 − 2/(22 ) ≈ .79; = E2 − 2/(32 ) ≈ .57. =

3. We have f(x) = 1 − (x/π)2 for −π < x < π. Its Fourier series representation is f(x) =

∞ X 2 (−1)n+1 cos nx. +4 3 (nπ)2 n=1

n+1

Thus a0 = 23 , an = 4 (−1) (nπ)2 , and bn = 0 for all n ≥ 1. The mean square error (from (5)) is Z π N 1 1X 2 EN = f 2 (x) dx − a20 − (a + b2n) 2π −π 2 n=1 n Z π N 2 1 4 8 X 1 = 1 − (x/π)2 dx − − 4 2π −π 9 π n=1 n4 Z π N x2 4 1 x4 8 X 1 1 − 2 2 + 4 dx − − 4 = 2π −π π π 9 π n=1 n4 =

N 1 x5 π 4 2 x3 8 X 1 + − x− − π 3π2 5π4 0 9 π4 n=1 n4

=

1−

=

N 4 8 X 1 . − 4 45 π n=1 n4

N 2 1 4 8 X 1 + − − 4 3 5 9 π n=1 n4

60 Chapter 2 Fourier Series So 4 8 − 45 π4 8 = E1 − 4 π 8 = E2 − 4 π

E1

=

E2 E3

≈ .0068; 1 ≈ .0016; 24 1 ≈ .0006. 34

4. We have f(x) = x2 for −1 < x < 1. Its Fourier series representation is f(x) =

∞ 4 X (−1)n 1 + 2 cos nπx. 3 π n=1 n2

n

Thus a0 = 13 , an = 4 (−1) , and bn = 0 for all n ≥ 1. The mean square error (from (nπ)2 (5)) is EN

=

=

1 2 Z

Z

1 2

f (x) dx −

a20

−1 1

x4 dx −

0

N 1X 2 − (a + b2n ) 2 n=1 n

N 1 8 X 1 − 4 9 π n=1 n4

=

N 1 1 8 X 1 − − 4 5 9 π n=1 n4

=

N 4 8 X 1 . − 4 45 π n=1 n4

So 4 8 − 45 π4 8 = E1 − 4 π 8 = E2 − 4 π

E1

=

E2 E3

≈ .0068; 1 ≈ .0016; 24 1 ≈ .0006. 34

5. We have EN

=

=

1 2

Z

1

f 2 (x) dx − a20 − −1

1−

N 1X 2 8 b = 1− 2 2 n=1 n π

N 1X 2 (a + b2n ) 2 n=1 n

X 1≤n odd≤N

1 . n2

With the help of a calculator, we find that E39 = .01013 and E41 = .0096. So take N = 41. 6. This exercise is a well-known application of the definite integral and can be found in any calculus textbook. 7. (a) Parseval’s identity with p = π: Z π ∞ 1 1X 2 2 2 f(x) dx = a0 + (a + b2n). 2π −π 2 n=1 n Applying this to the given Fourier series expansion, we obtain Z π 2 ∞ ∞ ∞ X 1 x 1 π2 π2 1X 1 1X 1 ⇒ ⇒ . dx = = = 2 2π −π 4 2 n=1 n2 12 2 n=1 n2 6 n n=1

Section 2.5

Mean Square Approximation and Parseval’s Identity

61

For (b) and (c) see the solution to Exercise 17, Section 2.3. 8. Applying Parseval’s identity, we obtain 1 2π

Z

π

x4 dx = −π

∞ π4 1 X 16 + 9 2 n=1 n4

∞ X π4 1 π4 = +8 4 5 9 n n=1

⇒

⇒

9. We have f(x) = π2 x − x3 for −π < x < π and, for n ≥ 1, bn = Parseval’s identity 2 ∞ 1 X 12 2 n=1 n3

= = = =

∞ X 1 π4 . = 4 n 90 n=1 12 n+1 . n3 (−1)

By

Z π 2 1 π 2 x − x3 2π −π Z 1 π 4 2 π x − 2π2x4 + x6 dx π 0 1 π4 3 2π2 5 x7 π x − x + π 3 5 7 0 1 2 1 8 π6 − + = π6 . 3 5 7 105

Simplifying, we find that ζ(6) =

∞ X 1 (8)(2) π6 = π6 = . 6 n (105)(144) 945 n=1

13. For the given function, we have bn = 0 and an = n12 . By Parseval’s identity, we have Z π Z π ∞ ∞ X 1 1 1X 1 π5 2 f 2 (x) dx = ⇒ f (x) dx = π = πζ(4) = , 2π −π 2 n=1 n4 n4 90 −π n=1 where we have used the table preceding Exercise 7 to compute ζ(4). 15. Let us write the terms of the function explicitly. We have f(x) =

∞ X cos nx cos x cos 2x = 1+ +··· . + n 2 2 22 n=0

Thus for the given function, we have bn = 0 for all n, a0 = 1, an =

1 for n ≥ 1. 2n

By Parseval’s identity, we have 1 2π

Z

π

f 2 (x) dx =

a20 +

−π

∞ 1X 2 a 2 n=1 n

∞ ∞ 1X 1 1 1 1X 1 = + + 2 n=1 (2n )2 2 2 2 n=1 4n

=

1+

=

∞ 1 1X 1 . + 2 2 n=0 4n

To sum the last series, we use a geometric series: if |r| < 1, ∞ X n=0

rn =

1 . 1−r

62 Chapter 2 Fourier Series Hence

∞ X 1 1 = n 4 1 − n=0

and so

Z

1 4

=

4 , 3

π

1 14 7π f 2 (x) dx = 2π( + )= . 2 23 3 −π

17. For the given function, we have a0 = 1, an =

1 1 , bn = for n ≥ 1. n 3 n

By Parseval’s identity, we have 1 2π

Z

π 2

f (x) dx = −π

a20

∞ 1X 2 + (a + b2n) 2 n=1 n ∞ 1X 1 1 ( n 2 + 2) 2 n=1 (3 ) n

=

1+

=

∞ ∞ 1 1 1X 1 1X 1 + + + 2 2 2 n=1 9n 2 n=1 n2

=

∞ ∞ 1 1X 1 1X 1 + . + 2 2 n=0 9n 2 n=1 n2

Using a geometric series, we find ∞ X 1 1 = n 9 1 − n=0

1 9

=

9 . 8

By Exercise 7(a), ∞ X 1 π2 = . 2 n 6 n=1

So

Z

π

1 1 9 1 π2 17π π3 f 2 (x) dx = 2π( + + )= + . 2 28 2 6 8 6 −π

Section 2.6

Complex Form of Fourier Series

63

Solutions to Exercises 2.6 1. From Example 1, for a 6= 0, ±i, ±2i, ±3i, . . ., ∞ sinh πa X (−1)n inx e π a − in n=−∞

eax =

(−π < x < π);

consequently, −ax

e

∞ sinh πa X (−1)n inx = e π a + in n=−∞

(−π < x < π),

and so, for −π < x < π, eax + e−ax 2 ∞ sinh πa X 1 1 (−1)n + einx 2π n=−∞ a + in a − in

cosh ax = =

∞ a sinh πa X (−1)n inx e . π n2 + a2 n=−∞

=

2. From Example 1, for a 6= 0, ±i, ±2i, ±3i, . . ., ∞ sinh πa X (−1)n inx e π a − in n=−∞

eax =

(−π < x < π);

consequently, ∞ sinh πa X (−1)n inx e π a + in n=−∞

e−ax =

(−π < x < π),

and so, for −π < x < π, sinh ax = =

=

eax − e−ax 2 ∞ sinh πa X 2π

(−1)

n=−∞ ∞ X

n

1 1 − a − in a + in

einx

n i sinh πa (−1)n 2 einx . 2 π n + a n=−∞

3. In this exercise, we will use the formulas cosh(iax) = cos ax and sinh(iax) = i sin ax, for all real a and x. (An alternative method is used in Exercise 4.) To prove these formulas, write cosh(iax) =

eiax + e−iax = cos ax, 2

by Euler’s identity. Similarly, eiax − e−iax = i sin ax. 2 If a is not an integer, then ia 6= 0, ±i, ±2i, ±3i, . . ., and we may apply the result of Exercise 1 to expand eiax in a Fourier series: sinh(iax) =

cos(ax)

∞ (ia) sinh(iπa) X (−1)n inx e π n2 + (ia)2 n=−∞

=

cosh(iax) =

=

∞ −a sin(πa) X (−1)n inx e . π n2 − a2 n=−∞

64 Chapter 2 Fourier Series 4. We have

eiax − e−iax . 2 Since a 6= 0, ±1, ±2, . . ., we can use Example 1 to expand eiax and e−iax . We have, for −π < x < π, sin ax =

eiax

=

∞ sinh(iπa) X (−1)n inx e π (ia) − in n=−∞

=

∞ eiπa − e−iπa X (−1)n inx e 2iπ a−n n=−∞

=

∞ sin(πa) X (−1)n inx e . π a−n n=−∞

Similarly, e−iax =

∞ sin(πa) X (−1)n inx e . π a+n n=−∞

Hence eiax − e−iax 2i ∞ sin πa X 1 1 (−1)n − einx 2iπ n=−∞ a−n a+n

sin ax = =

=

−

∞ n i sin πa X (−1)n 2 einx. 2 π a − n n=−∞

5. Use identities (1); then e2ix + e−2ix e3ix − e−3ix +2 2 2i −2ix 2ix e e ie−3ix + + − ie3ix . 2 2

cos 2x + 2 sin 3x = = 6. Use identities (1); then sin 3x = =

e3ix − e−3ix 2i e−3ix e3ix i −i . 2 2

7. You can use formulas (5)–(8) to do this problem, or you can start with the Fourier series in Exercise 3 and rewrite it as follows: ∞ −a sin(πa) X (−1)n einx cos ax = π n2 − a2 n=−∞ =

−1 ∞ −a sin(πa) X (−1)n einx a sin(πa) 1 a sin(πa) X (−1)n einx − − π n2 − a2 π −a2 π n2 − a2 n=−∞ n=1

=

∞ ∞ sin(πa) a sin(πa) X (−1)−n e−inx a sin(πa) X (−1)n einx − − πa π (−n)2 − a2 π n2 − a2 n=1 n=1 =2 cos nx

= =

z }| { inx + e−inx sin(πa) a sin(πa) ne (−1) − πa π n2 − a2 n=1 ∞ X

∞ cos nx sin(πa) a sin(πa) X (−1)n 2 . −2 πa π n − a2 n=1

Section 2.6

Complex Form of Fourier Series

65

8. You can use formulas (5)–(8) to do this problem, or you can start with the Fourier series in Exercise 4 and rewrite it as follows: sin ax =

∞ n i sin πa X (−1)n 2 einx − 2 π a − n n=−∞

=

−1 ∞ i sin(πa) X (−1)n neinx i sin(πa) X (−1)n neinx + π n2 − a2 π n2 − a2 n=−∞ n=1

=

∞ ∞ i sin(πa) X (−1)n (−n)e−inx i sin(πa) X (−1)n neinx + π (−n)2 − a2 π n2 − a2 n=1 n=1

=

∞ z }| { i sin(πa) X (−1)n n inx −inx e ( − e ) π n2 − a2 n=1

=

−

=2i sin nx

∞ 2 sin(πa) X (−1)n n sin nx. π n2 − a2 n=1

9. If m = n then Z p Z p Z p mπ nπ mπ mπ 1 1 1 ei p xe−i p x dx = ei p x e−i p x dx = dx = 1. 2p −p 2p −p 2p −p If m 6= n, then Z p mπ nπ 1 ei p x e−i p x dx = 2p −p

1 2p

Z

p

ei

(m−n)π p

x

dx

−p

(m−n)π p −i ei p x −p 2(m − n)π −i ei(m−n)π − e−i(m−n)π 2(m − n)π −i (cos[(m − n)π] − cos[−(m − n)π]) = 0. 2(m − n)π

= = = 10. From (6), we have

cn + c−n = an, and cn − c−n = −ibn ⇒ bn = i(cn − c−n ). 11. The function in Example 1 is piecewise smooth on the entire real line and continuous at x = 0. By the Fourier series representation theorem, its Fourier series converges to the value of the function at x = 0. Putting 4x=04 in the Fourier series we thus get ∞ sinh πa X (−1)n f(0) = 1 = (a + in). π a2 + n2 n=−∞ The doubly infinite sum is to be computed by taking symmetric partial sums, as follows: ∞ X

(−1)n (a + in) a2 + n2 n=−∞

=

=

N X

lim

N →∞

n=−N

a lim

N →∞

(−1)n (a + in) a2 + n2

N X n=−N

N X (−1)n (−1)n n + i lim . 2 2 N →∞ a +n a2 + n2

But N X (−1)n n = 0, a2 + n2

n=−N

n=−N

66 Chapter 2 Fourier Series because the summand is an odd function of n. So 1=

∞ ∞ sinh πa X (−1)n a sinh πa X (−1)n (a + in) = , π a2 + n2 π a2 + n2 n=−∞ n=−∞

which is equivalent to the desired identity. 12. Starting with the Fourier series of Example 1, we have, for −π < xπ and a 6= 0, eax

=

= =

=

∞ sinh πa X (−1)n (a + in)einx 2 + n2 π a n=−∞ ∞ sinh πa X (−1)n (a + in) cos nx + i sin nx 2 + n2 π a n=−∞ ∞ sinh πa X (−1)n (a cos nx − n sin nx) + i(n cos nx + a sin nx) π a2 + n2 n=−∞ ∞ sinh πa X (−1)n (a cos nx − n sin nx) , π a2 + n2 n=−∞

where in the last step we have used the fact that the symmetric partial sums of an odd integrand in n adds-up to 0 (see the previous exercise). 13. (a) At points of discontinuity, the Fourier series in Example 1 converges to the average of the function. Consequently, at x = π the Fourier series converges to eaπ +e−aπ = cosh(aπ). Thus, plugging x = π into the Fourier series, we get 2 =(−1)n

∞ ∞ z}|{ sinh(πa) X (−1)n sinh(πa) X (a + in) inπ cosh(aπ) = e (a + in) = . π a2 + n2 π a2 + n2 n=−∞ n=−∞

The sum

P∞

in n=−∞ a2 +n2

is the limit of the symmetric partial sums N X

i

n=−N

Hence

P∞

in n=−∞ a2 +n2

cosh(aπ) =

n = 0. a2 + n2

= 0 and so

∞ sinh(πa) X a 2 π a + n2 n=−∞

⇒

coth(aπ) =

∞ 1 a X , 2 π n=−∞ a + n2

upon dividing both sides by sinh(aπ). Setting t = aπ, we get coth t =

∞ ∞ X t X 1 t = , π2 n=−∞ ( πt )2 + n2 n=−∞ t2 + (πn)2

which is (b). Note that since a is not an integer, it follows that t is not of the form kπi, where kis an integer. 14. Start with the expansion from Exercise 13(b): For t 6= 0, ±iπ ± 2iπ, . . ., coth t =

∞ X

t . 2 + (πn)2 t n=−∞

Take t = ix with x 6= kπ; then coth(ix) =

∞ X

∞ X ix x = i . 2 + (πn)2 2 − x2 (ix) (πn) n=−∞ n=−∞

Section 2.6

But coth(ix) = So −i cot x =

Complex Form of Fourier Series

67

2 cos x eix + e−ix = = −i cot x. ix −ix e −e 2i sin x

∞ X

∞ X ix x = i ; 2 + (πn)2 2 − x2 (ix) (πn) n=−∞ n=−∞

equivalently, for x 6= kπ, cot x =

∞ X

x . 2 − (πn)2 x n=−∞

Additional remarks: For a fixed x in the domain of convergence of the series, the series converges absolutely. By writing the n = 0 term separately and taking symmetric partial sums, we get, for x 6= kπ, cot x =

∞ X 1 x . +2 2 x x − (πn)2 n=1

This is a pretty good approximation of the cotangent function by a sum of simple rational functions. See the figure for the approximation obtained by using only one term from the series on the interval (0, π). The approximation of cot x that we derived tells you how badly the cotangent behaves around kπ. For example, around x = 0, cot x is as bad (or as good) as 1/x. The approximation of cot x by rational functions is also useful in solving equations like cot x = x, which arise frequently in applications (see Section 3.6). Plot

Cot x , 1 x

2x

x^2

1.5

2

Pi ^ 2

, x, 0, Pi

60 40 20 0.5

1

2.5

3

-20 -40 -60

15. (a) This is straightforward. Start with the Fourier series in Exercise 1: For a 6= 0, ±i, ±2i, ±3i, . . ., and −π < x < π, we have cosh ax = On the left side, we have Z π 1 cosh2(ax) dx = 2π −π =

∞ a sinh πa X (−1)n inx e . π n2 + a2 n=−∞

Z 1 π cosh(2ax) + 1 dx π 0 2 π 1 1 1 1 x+ sinh(2ax) = π+ sinh(2aπ) . 0 2π 2a 2π 2a

On the right side of Parseval’s identity, we have ∞ (a sinh πa)2 X 1 . 2 + a2 )2 π2 (n n=−∞

68 Chapter 2 Fourier Series Hence

∞ (a sinh πa)2 X 1 1 1 . π+ sinh(2aπ) = 2 + a2 )2 2π 2a π2 (n n=−∞

Simplifying, we get ∞ X n=−∞

(n2

1 1 π π+ = sinh(2aπ) . 2 2 2 +a ) 2(a sinh πa) 2a

(b) This part is similar to part (a). Start with the Fourier series of Exercise 2: For a 6= 0, ±i, ±2i, ±3i, . . ., and −π < x < π, we have sinh ax =

∞ i sinh πa X n (−1)n 2 einx . 2 π n + a n=−∞

On the left side, we have Z π Z 1 1 π cosh(2ax) − 1 2 sinh (ax) dx = dx 2π −π π 0 2 π 1 1 1 1 = −x+ sinh(2ax) = −π+ sinh(2aπ) . 2π 2a 2π 2a 0 On the right side of Parseval’s identity, we have ∞ sinh2 πa X n2 . π2 (n2 + a2)2 n=−∞

Hence

∞ sinh2 (πa) X 1 n2 1 . −π+ sinh(2aπ) = 2 2 2π 2a π (n + a2 )2 n=−∞

Simplifying, we get ∞ X n=−∞

(n2

n2 π 1 = sinh(2aπ) . −π+ 2 2 2 +a ) 2a 2 sinh (πa)

16. (a) Following the hint, for a given w, we have d ez+w ez ez+w − ez ez+w = = 0. z dz e (ez )2 z+w

d u d e (This calculation uses the fact that dz e = eu du .) Since dz = 0, we conclude dz ez ez+w z+w z that ez = C or e = Ce for all z, where C is a constant (that depends on w). Take z = 0. It follows that C = ew and hence ez+w = ez ew . (b) We have

iθ

e

= = =

∞ X (iθ)n θ2 θ3 θ4 = 1 + iθ + i2 + +i3 + i4 + · · · n! 2! 3! 4! n=0 θ2 θ4 θ3 θ5 1− + −···+i θ − + −··· 2! 4! 3! 5! cos θ + i sin θ.

(c) Replace θ by −θ in (b) and you get e−iθ = cos(−θ) + i sin(−θ) = cos θ − i sin θ, which is the complex conjugate of cos θ + i sin θ. Thus e−iθ = eiθ .

Section 2.6

Complex Form of Fourier Series

69

(d) For any complex number, |z|2 = z · z. Thus iθ 2 e = eiθ · eiθ = eiθ · e−iθ = 1. So eiθ = 1. (e) eiθ +e−iθ = (cos θ+i sin θ)+(cos θ−i sin θ) = 2 cos θ. Thus cos θ = (eiθ +e−iθ )/2. Similarly, sin θ = (eiθ − e−iθ )/(2i) (f) eiπ/2

=

cos(π/2) + i sin(π/2) = i;

e3iπ/4

=

cos(3π/4) + i sin(3π/4) = −

eiπ

=

cos π + i sin π = −1;

2iπ

= =

cos(2π) + i sin(2π) = 1; cos(16π) + i sin(16π) = 1.

e e16iπ

√

√ 2 2 +i ; 2 2

(g) We have ez+2πi = ez e2πi = ez · 1 = ez . 17. (a) In this exercise, we let a and b denote real numbers such that a2 + b2 6= 0. Using the linearity of the integral of complex-valued functions, we have Z Z I1 + iI2 = eax cos bx dx + i eax sin bx dx Z = (eax cos bx + ieax sin bx) dx

= =

Z

eibx

z }| { eax (cos bx + i sin bx) dx Z Z ax ibx e e dx = ex(a+ib) dx

1 ex(a+ib) + C, a + ib R where in the last step we used the formula eαx dx = α1 eαx + C (with α = a + ib), which is valid for all complex numbers α 6= 0 (see Exercise 19 for a proof). (b) Using properties of the complex exponential function (Euler’s identity and the fact that ez+w = ez ew ), we obtain =

I1 + iI2

= = = =

1 ex(a+ib) + C a + ib (a + ib) eax eibx + C (a + ib) · (a + ib) a − ib ax cos bx + i sin bx + C e 2 2 a +b eax a cos bx + b sin bx + i − b cos bx + a sin bx + C. 2 2 a +b

(c) Equating real and imaginary parts in (b), we obtain I1 = and I2 =

eax a cos bx + b sin bx 2 +b

a2

eax − b cos bx + a sin bx . a2 + b2

18. (a) Use Euler’s identity and properties of the exponential as follows: n cos nθ + i sin nθ = einθ = eiθ = (cos θ + i sin θ)n .

70 Chapter 2 Fourier Series (b) With n = 2, the identity in (a) becomes cos 2θ + i sin 2θ

2

= =

(cos θ + i sin θ) (cos θ)2 + (i sin θ)2 + 2i cos θ sin θ

=

cos2 θ − sin2 θ + 2i cos θ sin θ.

Equating real and imaginary parts, we obtain cos 2θ = cos2 θ − sin2 θ

and

sin 2θ = 2 sin θ cos θ.

(c) Proceeding as in (b) with n = 3, we obtain cos 3θ + i sin 3θ

= = =

3

(cos θ + i sin θ) (cos θ)3 + 3 cos θ(i sin θ)2 + 3(cos θ)2 (i sin θ) + (i sin θ)3 (cos θ)3 − 3 cos θ(sin θ)2 + i 3(cos θ)2 sin θ − (sin θ)3 .

Equating real and imaginary parts, we obtain cos 3θ = cos3 θ − 3 cos θ sin2 θ

and

sin 3θ = 3 cos2 θ sin θ − sin3 θ.

But sin2 θ = 1 − cos2 θ, so cos 3θ = cos3 θ − 3 cos θ(1 − cos2 θ) = 4 cos3 θ − 3 cos θ. Similarly, using cos2 θ = 1 − sin2 θ, we obtain sin 3θ = −4 sin3 θ + 3 sin θ. 19. The purpose of this exercise is to show you that the familiar formula from calculus for the integral of the exponential function, Z 1 eax dx = eax + C, a holds for all nonzero complex numbers a. Note that this formula is equivalent to d ax e = aeax , dx where the derivative of

d ax e dx

means

d ax d d d e = Re (eax ) + i Im (eax ) = Re (eax ) + i Im (eax ). dx dx dx dx Write a = α + iβ, where α and β are real numbers. Then eax

= ex(α+iβ) = eαx eiβx = eαx cos βx + ieαx sin βx.

So d ax e dx

= = = = =

as claimed.

d αx d e cos βx + i eαx sin βx dx dx αeαx cos βx − βeαx sin βx + i αeαx sin βx + βeαx cos βx eαx α cos βx − β sin βx + i α sin βx + β cos βx eαx (cos βx + i sin βx)(α + iβ) aeαx eiβx = aeαx+iβx aeiax ,

Section 2.6

Complex Form of Fourier Series

71

20. By definition of the integral of a complex-valued function, we have Z b Z b Z b (h(t) + g(t)) dt = Re (h(t) + g(t)) dt + i Im (h(t) + g(t)) dt. a

a

a

But Re (h(t) + g(t)) = Re (h(t)) + Re (g(t)) and Im (h(t) + g(t)) = Im (h(t)) + Im (g(t)). So Z b Z b Z b (h(t) + g(t)) dt = Re (h(t)) dt + i Im (h(t)) dt a

a

+ =

a

Z

b

Re (g(t)) dt + i a

Z

b

h(t) dt + a

Z

b

Im (g(t)) dt a

Z

b

g(t) dt. a

In a similar way, we can show that Z b Z αh(t) dt = α a

b

h(t) dt a

for any complex number α. Combining the two properties, we obtain that Z b Z b Z b (αh(t) + βg(t)) dt = α h(t) dt + β g(t) dt. a

a

21. By Exercise 19, Z 2π eit + 2e−2it dt

=

0

a

1 it 2 −2it 2π e + e i −2i 0 =1

=1

z}|{ z }| { = −i e2πi +i e−2πi −(−i + i) = 0. Of course, this result follows from the orthogonality relations of the complex exponential system (formula (11), with p = π). 22. Taking a hint from the real-valued case, and using Exercise 19, we integrate by parts and obtain =dv

Z

π

=u z}|{ z }| { t e2it dt =

0

= = 23. First note that

Hence

Z

1 cos t − i sin t

Z 1 2it π 1 π 2it t e − e dt 2i 2i 0 0 π 1 π −i e2iπ + e2it 0 2 4 π 1 π −i + (1 − 1) = −i 2 4 2

1 1 = −it = eit . cos t − i sin t e Z dt = eit dt = −ieit + C = sin t − i cos t + C.

24. Using linearity and integrating term by term, we have Z 2π (3t − 2 cos t + 2i sin t) dt = 6π. 0

Hence

Z

2π

(3t − 2 cos t + 2i sin t) dt = 6π = 6π. 0

72 Chapter 2 Fourier Series 25. First note that 1 − t2 2t 1 + it (1 + it)2 1 − t2 + 2it = +i . = = 1 − it (1 − it)(1 + it) 1 + t2 1 + t2 1 + t2 Hence Z

Z 2t 1 − t2 dt + i dt 2 1+t 1 + t2 Z Z 2 2t (−1 + dt + i dt 1 + t2 1 + t2 −t + 2 tan−1 t + i ln(1 + t2) + C.

Z

1 + it dt = 1 − it = =

22. For n 6= 0, using the orthogonality relations, we have =0

Z

=dv zZ }| { π z }| { z }| { 1 −int π 1 −int −int (1 − t) e dt = (1 − t) e dt e + −π −in −in −π −π i (1 − π)e−inπ − (1 + π)einπ = n i 2iπ = (−1)n − 2π = (−1)n+1 . n n π

=u

If n = 0, the integral becomes Z

π

(1 − t) dt = 2π. −π

Section 2.7

Forced Oscillations

73

Solutions to Exercises 2.7 1. (a) General solution of y00 + 2y0 + y = 0. The characteristic equation is λ2 + 2λ + 1 = 0 or (λ + 1)2 = 0. It has one double characteristic root λ = −1. Thus the general solution of the homogeneous equation y00 + 2y0 + y = 0 is y = c1e−t + c2te−t . To find a particular solution of y00 + 2y0 + y = 25 cos 2t, we apply Theorem 1 with µ = 1, c = 2, and k = 1. The driving force is already given by its Fourier series: We have bn = an = 0 for all n, except a2 = 25. So αn = βn = 0 for all n, B2 a2 2 2 a2 except α2 = AA2 +B = −3 and B2 = 4. 2 and β2 = A2 +B 2 , where A2 = 1 − 2 2

2

2

2

100 Thus α2 = −75 25 = −3 and β2 = 25 = 4, and hence a particular solution is yp = −3 cos 2t + 4 sin 2t. Adding the general solution of the homogeneous equation to the particular solution, we obtain the general solution of the differential equation y00 + 2y0 + y = 25 cos 2t

y = c1 e−t + c2te−t − 3 cos 2t + 4 sin 2t. (b) Since limt→∞ c1 e−t + c2 te−t = 0, it follows that the steady-state solution is ys = −3 cos 2t + 4 sin 2t. 3. (a) General solution of 4y00 + 4y0 + 17y = 0. The characteristic equation is 4λ2 + 4λ + 17 = 0. Its characteristic roots are p √ −2 ± 4 − (4)(17) −2 ± −64 1 λ= = = − ± 2i. 4 4 2 Thus the general solution of the homogeneous equation is y = c1e−t/2 cos 2t + c2e−t/2 sin 2t. It is easy to see that y = 1/17 is a particular solution of 4y00 + 4y0 + 17y = 1. (This also follows from Theorem 1.) Hence the general solution is y = c1e−t/2 cos 2t + c2e−t/2 sin 2t +

1 . 17

(b) Since limt→∞ c1 e−t/2 cos 2t + c2e−t/2 sin 2t = 0, it follows that the steady-state solution is 1 ys = . 17 5. (a) To find a particular solution (which is also the steady-state solution) of y00 + 4y0 + 5y = sin t − 12 sin 2t, we apply Theorem 1 with µ = 1, c = 4, and k = 5. The driving force is already given by its Fourier series: We have bn = an = 0 for all n, except b1 = 1 and b2 = −1/2. So αn = βn = 0 for all n, except, possibly, α1, α2, β1 , and beta2. We have A1 = 4, A2 = 1, B1 = 4, and B2 = 8. So α1

=

α2

=

β1

=

β2

=

−B1 b1 A21 + B12 −B2 b2 A22 + B22 A1b1 A21 + B12 A2b2 A22 + B22

−4 1 =− , 32 8 4 4 = = , 65 65 4 1 = = , 32 8 −1/2 1 = =− . 65 130 =

74 Chapter 2 Fourier Series Hence the steady-state solution is 1 1 4 1 yp = − cos t + sin t + cos 2t − sin 2t. 8 8 65 130 (b) We have yp

=

(yp )0

=

(yp )00

=

(yp )00 + 4(yp )0 + 5yp

=

=

1 1 4 1 − cos t + sin t + cos 2t − sin 2t, 8 8 65 130 1 1 8 1 sin t + cos t − sin 2t − cos 2t, 8 8 65 65 1 1 16 2 cos t − sin t − cos 2t + sin 2t, 8 8 65 65 1 4 5 1 4 5 + − cos t + − + + sin t 8 8 8 8 8 8 2 32 5 16 4 20 + − − sin 2t + − − + cos 2t 65 65 130 65 65 65 2 32 5 1 sin t + − − sin 2t − sin 2t, 75 65 130 2

which shows that yp is a solution of the nonhomogeneous differential equation. 9. (a) Natural frequency of the spring is s k √ ω0 = = 10.1 ≈ 3.164. µ (b) The normal modes have the same frequency as the corresponding components of driving force, P in the following sense. Write the driving force as a Fourier series F (t) = a0 + ∞ n=1 fn (t) (see (5). The normal mode, yn (t), is the steady-state response of the system to fn (t). The normal mode yn has the same frequency as fn . In our case, F is 2π-periodic, and the frequencies of the normal modes are computed in Example 2. We have ω2m+1 = 2m + 1 (the n even, the normal mode is 0). Hence the frequencies of the first six nonzero normal modes are 1, 3, 5, 7, 9, and 11. The closest one to the natural frequency of the spring is ω3 = 3. Hence, it is expected that y3 will dominate the steady-state motion of the spring. 13. According to the result of Exercise 11, we have to compute y3 (t) and for this 4 purpose, we apply Theorem 1. Recall that y3 is the response to f3 = 3π sin 3t, the component of the Fourier series of F (t) that corresponds to n = 3. We have a3 = 0, 4 b3 = 3π , µ = 1, c = .05, k = 10.01, A3 = 10.01 − 9 = 1.01, B3 = 3(.05) = .15, α3 =

−B3 b3 −(.15)(4)/(3π) A3 b3 = ≈ −.0611 and β3 = ≈ .4111. 2 2 2 2 A3 + B3 (1.01) + (.15) A3 + B32

So y3 = −.0611 cos 3t + .4111 sin 3t. √ The amplitude of y3 is .06112 + .41112 ≈ .4156. 17. (a) In order to eliminate the 3rd normal mode, y3 , from the steady-state solution, we should cancel out the component of F that is causing it. That is, we 3t 4 sin 3t must remove f3 (t) = 4 sin from the input function. The 3π . Thus subtract 3π modified input function is 4 sin 3t F (t) − . 3π Its Fourier series is he same as the one of F , without the 3rd component, f3 (t). So the Fourier series of the modified input function is ∞ 4 4 X sin(2m + 1)t sin t + . π π m=2 2m + 1

Section 2.7

Forced Oscillations

75

(b) The modified steady-state solution does not have the y3 -component that we found in Exercise 13. We compute its normal modes by appealing to Theorem 1 and using as an input function F (t) − f3 (t). The first nonzero mode is y1 ; the second nonzero normal mode is y5 . We compute them with the help of Mathematica. Let us first enter the parameters of the problem and compute αn and βn , using the definitions from Theorem 1. The input/output from Mathematica is the following Clear a, mu, p, k, alph, bet, capa, capb, b, y mu 1; c 5 100; k 1001 100; p Pi; a0 0; 0; a n_ 2 Pi n 1 Cos n Pi ; b n_ alph0 a0 k; k mu n Pi p ^ 2 capa n_ c n Pi p capb n_ alph n_ capa n a n capb n b n capa n ^ 2 capb n ^ 2 bet n_ capa n b n capb n a n capa n ^ 2 capb n ^ 2 1001 100

n2

n 20 1 10 2

Cos n

n2 400

1001 100 n2 n 400

1001 100

n2

1 1001 100

n2

2

Cos n n2

2

It appears that

αn = 10

− (1 − cos(n π)) n2 1001 2 2 π + − n 400 100

and

βn =

2 n

− n2 (1 − cos(n π)) n2 1001 2 2 π + − n 400 100

1001

100

Note how these formulas yield 0 when n is even. The first two nonzero modes of the modified solution are

y1 (t) = α1 cos t + β1 sin t = −.0007842 cos t + .14131 sin t

and y5 (t) = α5 cos 5t + β5 sin 5t − .00028 cos 5t − .01698 sin 5t.

(c) In what follows, we use 10 nonzero terms of the original steady-state solution and compare it with 10 nonzero terms of the modified steady-state solution. The graph of the original steady-state solution looks like this:

76 Chapter 2 Fourier Series

steadystate t_ Sum alph n Cos n t bet n Sin n t , n, 1, 20 Plot Evaluate steadystate t , t, 0, 4 Pi

;

0.4 0.2 2

4

6

8

10

12

-0.2 -0.4

The modified steady-state is obtained by subtracting y3 from the steady-state. Here is its graph. modifiedsteadystate t_ steadystate t alph 3 Cos 3 t Plot Evaluate modifiedsteadystate t , t, 0, 4 Pi

bet 3 Sin 3 t

;

0.1 0.05 2

4

6

8

10

12

-0.05 -0.1

In order to compare, we plot both functions on the same graph. Plot Evaluate

steadystate t , modifiedsteadystate t

, t, 0, 4 Pi

0.4 0.2 2

4

6

8

10

12

-0.2 -0.4

It seems like we were able to reduce the amplitude of the steady-state solution by a factor of 2 or 3 by removing the third normal mode. Can we p do better? Let us analyze the amplitudes of the normal modes. These are equal to α2n + βn2 . We

Section 2.7

Forced Oscillations

77

have the following numerical values: amplitudes

N Table Sqrt alph n ^ 2

bet n ^ 2 , n, 1, 20

0.141312, 0., 0.415652, 0., 0.0169855, 0., 0.00466489, 0., 0.00199279, 0., 0.00104287, 0., 0.000616018, 0., 0.000394819, 0., 0.000268454, 0., 0.000190924, 0.

It is clear from these values that y3 has the largest amplitude (which is what we expect) but y1 also has a relatively large amplitude. So, by removing the first component of F , we remove y1 , and this may reduce the oscillations even further. Let’s see the results. We will plot the steady-state solution ys , ys − y3 , and ys − y1 − y3 .

modifiedfurther t_ modifiedsteadystate t alph 1 Cos t bet 1 Sin t ; Plot Evaluate modifiedfurther t , steadystate t , modifiedsteadystate t , t, 0, 4 Pi

0.4 0.2 2

4

6

8

10

12

-0.2 -0.4

21. (a) The input function F (t) is already given by its Fourier series: F (t) = 2 cos 2t + sin 3t. Since the frequency of the component sin 3t of the input function is 3 and is equal to the natural frequency of the spring, resonance will occur (because there is no damping in the system). The general solution of y00 +9y = 2 cos 2t+sin 3t is y = yh + yp , where yh is the general solution of y00 + 9y = 0 and yp is a particular solution of the nonhomogeneous equation. We have yh = c1 sin 3t + c2 cos 3t and, to find yp , we apply Exercise 20 and get a2 b2 yp = cos 2t + sin 2t + R(t), A2 A2 where a2 = 2, b2 = 0, A2 = 9 − 22 = 5, an0 = 0, bn0 = 1, and t R(t) = − cos 3t. 6 Hence yp =

2 t cos 2t − cos 3t 5 6

and so the general solution is y = c1 sin 3t + c2 cos 3t +

2 t cos 2t − cos 3t. 5 6

78 Chapter 2 Fourier Series (b) To eliminate the resonance from the system we must remove the component of F that is causing resonance. Thus add to F (t) the function − sin 3t. The modified input function becomes Fmodified (t) = 2 cos 2t. 25. The general solution is y = c1 sin 3t + c2 cos 3t + 25 cos 2t − 6t cos 3t. Applying the initial condition y(0) = 0 we get c2 + 25 = 0 or c2 = − 25 . Thus y = c1 sin 3t −

2 2 t cos 3t + cos 2t − cos 3t. 5 5 6

Applying the initial condition y0 (0) = 0, we obtain y0

=

3c1 cos 3t +

6 6 1 t sin 3t − sin 2t − cos 3t + sin 3t, 5 5 6 2

1 3c1 − , 6 1 0 y (0) = 0 ⇒ c1 = . 18 y0 (0)

=

Thus y=

1 2 2 t sin 3t − cos 3t + cos 2t − cos 3t. 18 5 5 6

Section 2.9

Uniform Convergence and Fourier Series

79

Solutions to Exercises 2.9 1.

sin nx 1 |fn(x)| = √ ≤ √ → 0 as n → ∞. n n

The sequence converges uniformly to 0 for all real x, because independently of x.

√1 n

controls its size

5. If x = 0 then fn (0) = 0 for all n. If x 6= 0, then applying l’Hospital’s rule, we find n 1 lim |fn(x)| = |x| lim −nx = |x| lim = 0. n→∞ n→∞ e n→∞ |x|e−n The sequence does not converge uniformly on any interval that contains 0 because fn ( n1 ) = e−1 , which does not tend to 0. 6. As in Exercise 5, the sequence converges to pointwise to 0 for all x, but not uniformly on any interval that contains 0 because fn ( n1 ) = e−1 − e−2 , which does not tend to 0. 7. The sequence converges to 0 for all x because the degree of n in the denominator is greater than its degree in the numerator. For each n, the extreme points of fn (x) occur when fn0 (x) = 0 or n − n 3 x2 (1 +

n2

2 x2 )

=0

⇒

n − n3 x = 0

⇒

x=

±1 . n

Since fn( n1 ) = 12 the sequence does not converge to 0 uniformly on the interval [0, ∞) (or any interval that contains 0). P 9. cosk2kx ≤ k12 = Mk for all x. Since Mk < ∞ (p-series with p > 1), the series converges uniformly for all x. x P ≤ ( 9 )k = Mk for all |x| ≤ 9. Since ( 9 )k < ∞ (geometric series with 14. 10 10 10 ratio 9/10 < 1), the series converges uniformly for all |x| ≤ 9. P (−1)k 1 17. |x|+k Mk < ∞ (p-series with p > 1), the 2 ≤ k 2 = Mk for all x. Since series converges uniformly for all x. P sin kx −kx 24. (a) If x = 0, the series ∞ e is obviously convergent. If x > 0, say k=1 k x ≥ δ > 0, then sin kx −kx e−kδ ≤ = Mk . k e k P P e−kδ The series Mk = is convergent by comparison to the geometric series k P −δ k P∞ e < ∞. By the Weierstrass M -test, the series k=1 sinkkx e−kx converges uniformly for all x ≥ δ > 0. Since x > 0 is arbitrary, it follows that the series converges for all x > 0. P∞ (b) The differentiated series is k=1 (cos kx − sin kx)e−kx. For x ≥ δ > 0, we have (cos kx − sin kx)e−kx ≤ 2e−kδ = Mk .

k P P P The series k Mk = 2 k e−kδ is convergent (geometric series k e−δ < ∞. By P∞ −kx the Weierstrass M -test, the series k=1 (cos kx − sin kx)e converges uniformly P∞ sin kx −kx for all x ≥ δ > 0. We proved in (a) that the series converges k=1 k e uniformly for all x ≥ δ > 0. Hence by Theorem 4, since the series and the differentiated series are uniformly convergent for x ≥ δ > 0, it follows that we can differentiate the series term by term and get !0 0 X ∞ ∞ ∞ X X sin kx −kx sin kx −kx = = (cos kx − sin kx)e−kx. e e k k k=1

k=1

k=1

80 Chapter 2 Fourier Series Since x > 0 as arbitrary, we see that term-by-term differentiation is justified for all x > 0. Note that this process can be repeated as often as we wish, because, as we differentiate a term in the series, we obtain polynomials in k times e−kx . Since e−kx dominates the polynomial, just like it did with the first derivative, the differentiated series will converge uniformly.

Section 2.10

Dirichlet Test and Convergence of Fourier Series

81

Solutions to Exercises 2.10 3. By TheoremP 2(c), the series converges for all x 6= 2nπ. For x = 2nπ, the series P ∞ cos 2knπ ∞ √ √1 = k=1 k=1 k is divergent. k 4.

This series converges uniformly for all x by the Weierstrass M -test, because sin 3kx 1 k2 ≤ k2 (for all x) P and k k12 is convergent.

5. The cosine part converges uniformly for all x, by the Weierstrass M -test. The sine part converges for all x by Theorem 2(b). Hence the given series converges for all x. 9.

(a) If limk→∞ sin kx = 0, then lim sin2 kx = 0

⇒

k→∞

lim (1 − cos2 kx) = 0

⇒

k→∞

lim cos2 kx = 1 (∗).

k→∞

Also, if limk→∞ sin kx = 0, then limk→∞ sin(k + 1)x) = 0. But sin(k + 1)x = sin kx cos x + cos kx sin x, so →0

z }| { 0 = lim (sin kx cos x + cos kx sin x)

⇒

k→∞

lim cos kx sin x = 0

k→∞

⇒

lim cos kx = 0 or sin x = 0.

k→∞

By (*), cos kx does not tend to 0, so sin x = 0, implying that xP = mπ. Consequently, ∞ if x 6= mπ, then lim k → ∞ sin kx is not 0 and the series k=1 sin kx does not converge by the nth term test, which proves (b). P∞ 10. (a) The series k=1 coskkx converges uniformly on [0.2, π2 ] by 2(a). If PTheorem ∞ we differentiate the series term by term, we obtain the series − k=1 sin kx, which does not converge for any x in the interval [0.2, π2 ] by Exercise 9. So the given series is uniformly convergent on [0.2, π2 ], but cannot be differentiated term by term. P∞ sin nx (b) A similar situation occurs with the series on any interval [a, b] n=1 n where 0 < a < b < 2π. We omit the proof.

82 Chapter 3 Partial Differential Equations in Rectangular Coordinates

Solutions to Exercises 3.1 1. uxx + uxy = 2u is a second order, linear, and homogeneous partial differential equation. ux(0, y) = 0 is linear and homogeneous. 2. uxx + xuxy = 2, is a second order, linear, and nonhomogeneous (because of the right side) partial differential equation. u(x, 0) = 0, u(x, 1) = 0 are linear and homogeneous. 3. uxx − ut = f(x, t) is a second order, linear, and nonhomogeneous (because of the right side) partial differential equation. ut(x, 0) = 2 is linear and nonhomogeneous. 4. uxx = ut is a second order, linear, and homogeneous partial differential equation. u(x, 0) = 1 and u(x, 1) = 0 are linear and nonhomogeneous (because of the first condition. 5. ut ux +uxt = 2u is second order and nonlinear because of the term utux . u(0, t)+ ux (0, t) = 0 is linear and homogeneous. 6. uxx +et utt = u cos x is a second order, linear, and homogeneous partial differential equation. u(x, 0) + u(x, 1) = 0 is linear and homogeneous. 7. (a) uxx = uyy = 0, so uxx + uyy = 0. (b) uxx = 2, uyy = −2, so uxx + uyy = 0. (c) ux =

−x2 + y2 (x2

+

2 y2 )

,

We have u =

uy =

−2 x y (x2 + y2 )

2

x x2 +y 2 .

So

,

and uxx =

2 x3 − 3 x y 2 (x2 + y2 )

(d) We have u =

3

,

y x2 +y 2 .

uyy =

−2 x3 − 3 xy2 (x2 + y2 )

⇒

3

∆u = uxx + uyy = 0.

Switching x and y in (c), it follows immediately that

uy =

−y2 + x2 (x2 + y2 )

2,

ux =

−2 x y (x2 + y2 )

2,

and uyy =

2 y3 − 3 y x2 (x2 + y2 )

3

,

uxx =

−2 y3 − 3 x2y (x2 + y2 )

⇒

3

∆u = uxx + uyy = 0.

(e) We have u = ln(x2 + y2 ), so ux =

2x , x2 + y 2

uy =

2y , x2 + y 2

and uxx =

−2 x2 − y2 (x2 + y2 )2

,

uyy =

2 x2 − y 2

⇒

(x2 + y2 )2

∆u = uxx + uyy = 0.

(f) We have u = ey cos x, so ux = −ey sin(x),

uy = ey cos(x),

and uxx = −ey cos(x),

uyy = ey cos(x)

⇒

∆u = uxx + uyy = 0.

(g) We have u = ln(x2 + y2 ) + ey cos x. Since u is the sum of two solutions of Laplace’s equation (by (e) and (f)), it is itself a solution of Laplace’s equation.

Section 3.1

83

Partial Differential Equations in Physics and Engineering

8. We note that the function is symmetric with respect to x, y, and z, so we only need to compute the derivatives with respect to x (or any other one of the variables) and then replace x by the other variables to get the other derivatives. We have x y z uy = − uz = − ux = − 3 , 3 , 3 , 2 2 2 2 2 2 2 2 2 (x + y + z ) (x + y + z ) (x + y2 + z 2 ) 2 and uxx =

2 x2 − y 2 − z 2

5 ,

(x2 + y2 + z 2 ) 2

uyy =

−x2 + 2 y2 − z 2 5

,

(x2 + y2 + z 2 ) 2

uzz =

−x2 − y2 + 2 z 2 5

;

(x2 + y2 + z 2 ) 2

so ∆u = uxx + uyy + uzz = 0. 9. (a) Let u(x, y) = eax eby . Then ux uy

= =

aeax eby beax eby

uxx uyy uxy

= = =

a2eax eby b2eax eby abeax eby .

So Auxx + 2Buxy + Cuyy + Dux + Euy + F u = 0 ⇔ Aa2eax eby + 2Babeax eby + Cb2eax eby +Daeax eby + Ebeax eby + F eax eby = 0 ⇔ eax eby Aa2 + 2Bab + Cb2 + Da + Eb + F = 0 ⇔ Aa2 + 2Bab + Cb2 + Da + Eb + F = 0, because eax eby 6= 0 for all x and y. (b) By (a), in order to solve uxx + 2uxy + uyy + 2 ux + 2 uy + u = 0, we can try u(x, y) = eax eby , where a and b are solutions of a2 + 2ab + b2 + 2a + 2b + 1 = 0. But a2 + 2ab + b2 + 2a + 2b + 1 = (a + b + 1)2. So a + b + 1 = 0. Clearly, this equation admits infinitely many pairs of solutions (a, b). Here are four possible solutions of the partial differential equation: a = 1, b = −2 a = 0, b = −1 a = −1/2, b = −1/2 a = −3/2, b = 1/2

⇒ ⇒ ⇒ ⇒

u(x, u(x, u(x, u(x,

y) = ex e−2y y) = e−y y) = e−x/2 e−y/2 y) = e−3x/2 ey/2

10. According to Exercise 9(a), to find solutions of utt + 2But − c2 uxx + Au = 0 (c > 0, A > 0) we can we can try u(x, y) = eax eby , where a and b are solutions of a2 − c2b2 + 2Ba + A = 0 or

a(a + 2B) − c2b2 + A = 0.

84 Chapter 3 Partial Differential Equations in Rectangular Coordinates This equation admits infinitely many pairs of solutions (a, b). Here are two possible solutions of the partial differential equation: √ a = 0, b =

√

A/c ⇒ u(t, x) = e

Ax/c

√ √ a = −2B, b = c A/c ⇒ u(t, x) = e−2Bt e Ax/c .

12. Consider the nonlinear equation (1)

ut + A(u)ux = 0, with initial condition u(x, 0) = φ(x),

(∗)

where A(u) is a function of u. Let x(t) denote a solution of the first order differential equation dx = A(u(x(t), t)). (∗∗) dt

(2)

This is a characteristic curve of (*). Restrict a solution u of (*) to a characteristic curve and consider the function of t given by t 7→ u((x(t), t)). The derivative with respect to t of this function is d u((x(t), t)) dt

d x(t) + ut ((x(t), t)) dt = ux ((x(t), t))A(u(x(t), t)) + ut((x(t), t)) = 0, = ux ((x(t), t))

since u is a solution of (*). Thus, t 7→ u((x(t), t)) is constant on the characteristic curves. But this implies that t 7→ A(u((x(t), t))) is constant on the characteristic curves, and hence from (**) it follows that the characteristic curves have constant derivatives and so the characteristic curves are straight lines with slopes A(u(x(t), t))). Setting t = 0. We conclude that the slope of a characteristic line is A(u(x(0), 0))) = A(φ(x(0))). We write these lines in the form x = tA(φ(x(0))) + x(0) . We complete the solution of (*) by solving for x(0) in the preceding equation to get an implicit relation for the characteristic lines, of the form L(x, t) = x(0). The final solution of (*) will be of the form u(x, t) = f(L(x, t)), where f is a function chosen so as to satisfy the initial condition in (*). That is f(L(x, 0)) = φ(x). 13. We follow the outlined solution in Exercise 12. We have A(u) = ln(u), φ(x) = ex ,

⇒

A(u(x(t)), t)) = A(φ(x(0))) = ln(ex(0) ) = x(0).

So the characteristic lines are x = tx(0) + x(0) So u(x, t) = f(L(x, t)) = f ex and so

x t+1

⇒

x(0) = L(x t) =

x . t+1

. The condition u(x, 0) = ex implies that f(x) = x

u(x, t) = e t+1 . x

x

x t+1 Check: ut = −e t+1 (t+1) 2 , ux = e x

ut + ln(u)ux = −e t+1

1 t+1 , x x 1 x t+1 + e = 0. 2 (t + 1) t+1 t+1

14. We follow the outlined solution in Exercise 12. We have A(u) = u + 1, φ(x) = x2 ,

⇒

A(u(x(t)), t)) = A(φ(x(0))) = x(0)2 + 1.

Section 3.1

Partial Differential Equations in Physics and Engineering

85

So the characteristic lines are x = t(x(0)2 + 1) + x(0)

tx(0)2 + x(0) + t − x = 0.

⇒

Solving this quadratic equation in x(0), we find two solutions p −1 ± 1 − 4t(t − x) x(0) = , 2t and so −1 ±

u(x, t) = f

! p 1 − 4t(t − x) . 2t

We now use the initial condition u(x, 0) = x2 to determine the sign in front of the radical. Since we cannot simply set t = 0 in the formula, because of the denominator, we will take the limit as t → 0 and write: ! p 1 − 4t(t − x) −1 ± lim u(x, t) = x2 ⇒ lim f = x2 . t→0 t→0 2t Now

p 1 − 4t(t − x) t→0 2t does not exist, but using l’Hospital’s rule, we find that p −1 + 1 − 4t(t − x) (1 − 4t(t − x))−1/2(−8t + 4x) lim = lim = x. t→0 t→0 2t 4 −1 −

lim

So assuming that f is continuous, we get ! p −1 + 1 − 4t(t − x) 2 x = f lim = f (x) , t→0 2t and hence −1 +

u(x, t) =

!2 p 1 − 4t(t − x) . 2t

15. We will just modify the solution from the previous exercise. We have A(u) = u + 2, φ(x) = x2 ,

A(u(x(t)), t)) = A(φ(x(0))) = x(0)2 + 2.

⇒

So the characteristic lines are x = t(x(0)2 + 2) + x(0)

⇒

tx(0)2 + x(0) + 2t − x = 0.

Solving for x(0), we find x(0) =

−1 ±

p 1 − 4t(2t − x) , 2t

and so u(x, t) = f

−1 ±

Now

! p 1 − 4t(2t − x) . 2t

p 1 − 4t(2t − x) lim t→0 2t does not exist, but using l’Hospital’s rule, we find that p −1 + 1 − 4t(2t − x) (1 − 4t(2t − x))−1/2(−16t + 4x) lim = lim = x. t→0 t→0 2t 4 −1 −

86 Chapter 3 Partial Differential Equations in Rectangular Coordinates So u(x, t) =

−1 +

!2 p 1 − 4t(2t − x) . 2t

16. We have A(u) = u2, φ(x) = x,

A(u(x(t)), t)) = A(φ(x(0))) = x(0)2.

⇒

So the characteristic lines are x = tx(0)2 + x(0)

⇒

tx(0)2 + x(0) − x = 0.

−1 ±

√ 1 + 4tx , 2t

Solving for x(0), we find x(0) = and so u(x, t) = f

−1 ±

√ 1 + 4tx . 2t

Now

√ 1 + 4tx t→0 2t does not exist, but using l’Hospital’s rule, we find that √ −1 + 1 + 4tx (1 + 4tx)−1/2(4x) lim = lim = x. t→0 t→0 2t 4 lim

−1 −

So u(x, t) =

√

−1 +

1 + 4tx . 2t

17. We have A(u) = u2, φ(x) =

√ x,

⇒

A(u(x(t)), t)) = A(φ(x(0))) = x(0).

So the characteristic lines are x = tx(0) + x(0)

⇒

Solving for x(0), we find x(0) = and so u(x, t) = f

x(0)(t + 1) − x = 0. x , t+1

x t+1

Now u(x, 0) = f(x) = So u(x, t) =

r

√

.

x.

x . t+1

Section 3.3

Wave Equation, the Method of Separation of Variables

87

Solutions to Exercises 3.3 1. The solution is u(x, t) =

∞ X n=1

nπx sin L

nπt nπt bn cos c + b∗n sin c L L

,

L where bn are the Fourier sine coefficients of f and b∗n are cnπ times the ∗ Fourier coefficients of g. In this exercise, bn = 0, since g = 0, b1 = 0.05; and bn = 0 for all n > 1, because f is already given by its Fourier sine series (period 2). So u(x, t) = 0.05 sin πx cos t.

2. The solution is u(x, t) =

∞ X

sin nπx (bn cos nt + b∗n sin nt) ,

n=1

where bn are the Fourier sine coefficients of f and b∗n are n1 times the Fourier coefficients of g. In this exercise, b∗n = 0 since g = 0. To get the Fourier coefficients of f , we note that f (x) = sin πx cos πx = 12 sin(2πx). So b2 = 12 , and all other bn = 0. So u(x, t) = 0.5 sin(2πx) cos(2t). 3. The solution is ∞ X

u(x, t) =

sin(nπx) (bn cos(nπt) + b∗n sin(nπt)) ,

n=1

where b∗n = 0 since g = 0. The Fourier coefficients of f are b1 = 1, b2 = 3, b5 = −1 and all other bn = 0. So u(x, t) = sin πx cos(πt) + 3 sin(2πx) cos(2πt) − sin(5πx) cos(5πt). 4. The solution is ∞ X

u(x, t) =

sin(nπx) (bn cos(nπt) + b∗n sin(nπt)) ,

n=1

where b∗2 = So

1 2π ,

all other b∗n = 0, b1 = 1, b3 = 12 , b7 = 3 and all other bn = 0.

u(x, t) = sin πx cos(πt) +

1 sin(2πx) sin(2πt) 2π

1 + sin(3πx) cos(3πt) + 3 sin(7πx) cos(7πt). 2 5. (a) The solution is u(x, t) =

∞ X

sin(nπx) (bn cos(4nπt) + b∗n sin(4nπt)) ,

n=1

where bn is the nth sine Fourier coefficient of f and b∗n is L/(cn) times the Fourier coefficient of g, where L = 1 and c = 4. Since g = 0, we have b∗n = 0

88 Chapter 3 Partial Differential Equations in Rectangular Coordinates for all n. As for the Fourier coefficients of f , we can get them by using Exercise 17, Section 2.4, with p = 1, h = 1, and a = 1/2. We get bn =

nπ 8 2 sin . π2 n2

Thus u(x, t) =

∞ 8 X sin nπ 2 sin(nπx) cos(4nπt) π2 n2 n=1

=

∞ 8 X (−1)k sin((2k + 1)πx) cos(4(2k + 1)πt). π2 (2k + 1)2 k=0

(b) Here is the initial shape of the string. Note the new Mathematica command that we used to define piecewise a function. (Previously, we used the If command.) Clear f f x_ : 2 x ;0 x 1 2 ;1 2 x f x_ : 2 1 x Plot f x , x, 0, 1

1

Because the period of cos(4(2k + 1)πt) is 1/2, the motion is periodic in t with period 1/2. This is illustrated by the following graphs. We use two different ways to plot the graphs: The first uses simple Mathematica commands; the second one is more involved and is intended to display the graphs in a convenient array. Clear partsum partsum x_, t_ : 8 Pi ^ 2 Sum Sin 1 ^ k 2 k 1 Pi x Cos 4 2 k Plot Evaluate partsum x, 0 , f x , x, 0, 1

Here is the motion in an array.

1 Pi t

2k

1 ^ 2, k, 0, 10

Section 3.3

Wave Equation, the Method of Separation of Variables

Table Plot Evaluate partsum x, t , x, 0, 1 , PlotRange 0, 1 , 1, 1 , .5 , 1, .5, .5, 1 , DisplayFunction Identity , t, 0, 1, 1 20 Ticks Show GraphicsArray Partition tt, 4

89

tt

1 0.5

1 0.5

1 0.5

1 0.5

-0.5 -1

-0.5 -1

-0.5 -1

-0.5 -1

1 0.5

1 0.5

1 0.5

1 0.5

-0.5 -1

-0.5 -1

-0.5 -1

-0.5 -1

1 0.5

1 0.5

1 0.5

1 0.5

-0.5 -1

-0.5 -1

-0.5 -1

-0.5 -1

1 0.5

1 0.5

1 0.5

1 0.5

-0.5 -1

-0.5 -1

-0.5 -1

-0.5 -1

1 0.5

1 0.5

1 0.5

1 0.5

-0.5 -1

-0.5 -1

-0.5 -1

-0.5 -1

;

6. (a) Using the formula from the text with c = 1/π and L = 1, we find the solution u(x, t) =

∞ X

sin nπx (bn cos nt + b∗n sin nt) ,

n=1

where bn is the nth sine Fourier coefficient of f and Z 2 1 b∗n = g(x) sin nπx dx n 0 1 4 4 n = − 2 cos nπx = 1 − (−1) . πn πn2 0 We now compute the Fourier coefficients of f . We have Z 2/3 Z 1 2 2 bn = (x − 1/3) sin nπx dx + (1 − x) sin nπx dx. 30 1/3 30 2/3 We evaluate the integral with the help of the following identity, that can be derived using integration by parts: Z a + bx b (a + bx) sin cx dx = − cos cx + 2 sin cx + C (c 6= 0). c c

90 Chapter 3 Partial Differential Equations in Rectangular Coordinates Thus

bn =

=

= =

1 x − 1/3 1 2/3 − cos nπx + 2 2 sin nπx 15 nπ n π 1/3 1 x−1 1 1 + cos nπx − 2 2 sin nπx 15 nπ n π 2/3 1 2nπ nπ 1/3 2nπ 1 1 − cos + 2 2 sin − 2 2 sin 15 nπ 3 n π 3 n π 3 1 1/3 2nπ 2nπ 1 + cos + 2 2 sin 15 nπ 3 n π 3 1 2nπ nπ 2 sin − sin 2 2 15 n π 3 3 i nπ 1 nπ h sin 4 cos − 1 15 n2π 2 3 3

Thus u(x, t) = ∞ i X sin nπx 1 nπ h nπ n sin nt . sin 4 cos − 1 cos nt + 4 1 − (−1) πn2 15 π 3 3 n=1

(b) Here is the initial shape of the string approximated using a partial sum of the series solution at time t = 0. Clear partsum, n, t, f Clear f f x_ : 0 ;0 x 1 3 ;1 3 x 2 3 f x_ : 1 30 x 1 3 ; 2 3 x 1 f x_ : 1 30 1 x partsum x_, t_ Sum Sin n Pi x 1 n ^ 2 Sin n Pi 3 15 Pi ^ 2 4 n ^ 2 Pi 1 1 ^ n Sin n t , n, 1, 10 ; , x, 0, 1 Plot Evaluate partsum x, 0 , f x

4 Cos n Pi 3

1 Cos n t

0.01 0.008 0.006 0.004 0.002 0.2

0.4

0.6

0.8

1

Because the period of cos nt and sin nt is 2π, the motion is 2π-periodic in t. This is illustrated by the following graphs.

Section 3.3

Wave Equation, the Method of Separation of Variables

Plot Evaluate Table partsum x, t ,

t, 0, 2 Pi, 2 Pi 20

,

91

x, 0, 1

3 2 1 0.2

0.4

0.6

0.8

1

-1 -2 -3

It is interesting to note that soon after the string is released, the initial velocity affects the motion in such a way that the initial shape does not appear to have a significant impact on the subsequent motion. Here are snapshots of the string very soon after it is released. Plot Evaluate

partsum x, 0 , partsum x, .01 , partsum x, .1

,

x, 0, 1

0.2

0.1

0.2

0.4

0.6

0.8

1

7. (a) Using the formula from the text with c = 4 and L = 1, we find the solution ∞ X u(x, t) = sin nπx (bn cos 4nπt + b∗n sin 4nπt) , n=1

where bn is the nth sine Fourier coefficient of f and b∗n =

2 4nπ

Z

1

g(x) sin nπx dx 1 1 1 n = − 2 2 cos nπx = 1 − (−1) . 2π n 2π 2n2 0 0

The Fourier coefficients of f . We have bn = 8

Z

1/4

x sin nπx dx + 2 0

Z

3/4

1/4

sin nπx dx + 8

Z

1

(1 − x) sin nπx dx. 3/4

92 Chapter 3 Partial Differential Equations in Rectangular Coordinates We evaluate the integral with the help of the identity

Z

(a + bx) sin cx dx = −

b a + bx cos cx + 2 sin cx + C c c

(c 6= 0).

We obtain

bn

x 1 1/4 = 8 − cos nπx + 2 2 sin nπx nπ n π 0 3/4 2 − cos nπx 1/4 nπ x−1 1 1 +8 cos nπx − 2 2 sin nπx nπ n π 3/4 1/4 nπ nπ 1 = 8 − cos + 2 2 sin nπ 4 n π 4 2 3nπ 2 nπ − cos + cos nπ 4 nπ 4 1/4 3nπ 3nπ 1 +8 cos + 2 2 sin nπ 4 n π 4 8 nπ 3nπ 8 = sin + 2 2 sin 2 2 n π 4 n π 4 8 nπ 3nπ = sin + sin . n2 π 2 4 4

Thus

u(x, t) =

∞ X n=1

+

1 2π 2n2

nπ 3nπ + sin 4 4 1 − (−1)n sin 4nπt .

sin nπx

8 n2 π 2

sin

cos 4nπt

(b) Here is the initial shape of the string approximated using a partial sum of the series solution at time t = 0.

Section 3.3

Wave Equation, the Method of Separation of Variables

Clear partsum, n, t, f Clear f f x_ : 4 x ;0 x 1 4 ;1 4 x 3 4 f x_ : 1 f x_ : 4 1 x ;3 4 x 1 Sum Sin n Pi x partsum x_, t_ 8 Pi ^ 2 n ^ 2 Sin n Pi 4 Sin 3 n Pi 4 1 2 n ^ 2 Pi ^ 2 1 1 ^ n Sin 4 n Pi t , n, 1, 10 Plot Evaluate

; partsum x, 0 , f x

93

Cos 4 n Pi t

, x, 0, 1 , PlotRange

All

1

1

Because the period of cos 4nπt and sin 4nπt is 1/2, the motion is 1/2periodic in t. This is illustrated by the following graphs. Clear tt, t Plot Evaluate Table partsum x, t ,

t, 0, 1 2 , .04

,

x, 0, 1

1 0.5

0.2

0.4

0.6

0.8

1

-0.5 -1

8. (a) The solution is

u(x, t) =

∞ X

bn sin(nπx) cos(nt),

n=1

where bn is the nth sine Fourier coefficient of f . We will get bn from the

94 Chapter 3 Partial Differential Equations in Rectangular Coordinates series that we found in Exercise 8, Section 2.4. We have, for 0 < x < π, x sin x =

∞ π n 4X [1 + (−1)n ] 2 sin nx. sin x − 2 π (n − 1)2 n=2

Let x = πt. Then, for 0 < t < 1, ∞ π n 4X πt sin πt = sin πt − [1 + (−1)n ] 2 sin nπt. 2 π (n − 1)2 n=2

Equivalently, for 0 < x < 1, ∞ 1 n 4 X x sin πx = sin πx − 2 [1 + (−1)n ] 2 sin nπx. 2 π n=2 (n − 1)2

So b1 =

1 2

and

bn = [1 + (−1)n]

−4n (n ≥ 2). π 2(n2 − 1)2

Thus u(x, t) = =

∞ sin πx cos t n 4 X [1 + (−1)n ] 2 sin(nπx) cos(nt) − 2 2 π (n − 1)2 n=2 ∞ X

sin πx cos t 16 n sin(2nπx) cos(2nt). − 2 2 2 π n=1 (4n − 1)2

(b) Here is the initial shape of the string. Clear partsum, n, t, f Clear f f x_ x Sin Pi x partsum x_, t_ Sin Pi x Cos t 2 16 Pi ^ 2 Sum Sin 2 n Pi x Cos 2 n t n 4 n^2 , n, 1, 10 ; , x, 0, 1 , PlotRange All Plot Evaluate partsum x, 0 , f x x Sin

x

0.5 0.4 0.3 0.2 0.1 0.2

0.4

0.6

0.8

1

9. The solution is u(x, t) =

∞ X n=1

sin(nπx) (bn cos(nπt) + b∗n sin(nπt)) ,

1 ^2

Section 3.3

where b∗1 =

1 π

Wave Equation, the Method of Separation of Variables

95

and all other b∗n = 0. The Fourier coefficients of f are Z

bn = 2

1

x(1 − x) sin(nπx) dx. 0

To evaluate this integral, we will use integration by parts to derive first the formula: for a 6= 0, Z x cos(a x) sin(a x) + C, x sin(ax) dx = − + a a2 and Z

x2 sin(ax) dx =

2 cos(a x) x2 cos(a x) 2 x sin(a x) − + C; + a3 a a2

thus Z x(1 − x) sin(ax) dx =

−2 cos(a x) x cos(a x) x2 cos(a x) sin(a x) 2 x sin(a x) − − + C. + + a3 a a a2 a2

Applying the formula with a = nπ, we get Z

1

x(1 − x) sin(nπx) dx

0

−2 cos(nπ x) x cos(nπ x) x2 cos(nπ x) sin(nπ x) 2 x sin(nπ x) 1 = − − + + (nπ)3 nπ nπ (nπ)2 (nπ)2 0 −2 ((−1)n − 1) (−1)n (−1)n −2 ((−1)n − 1) = − + = (nπ)3 nπ nπ (nπ)3 ( 4 if n is odd, (nπ)3 = 0 if n is even.

Thus

(

bn =

8 (nπ)3

0

if n is odd, if n is even,

and so ∞ 8 X sin((2k + 1)πx) cos((2k + 1)πt) 1 u(x, t) = 3 + sin(πx) sin(πt). π (2k + 1)3 π k=0

10. (a) Using the formula from the text with c = 1 and L = 1, we find the solution ∞ X u(x, t) = sin nπxbn cos nπt, n=1

where bn is the nth sine Fourier coefficient of f , bn = 8

Z 0

1/4

x sin nπx dx−8

Z

3/4

1/4

(x−1/2) sin nπx dx+8

Z

1

3/4

(x−1) sin nπx dx.

96 Chapter 3 Partial Differential Equations in Rectangular Coordinates We evaluate the integral as we did in Exercise 7:

bn

1 x 1/4 cos nπx + 2 2 sin nπx = 8 − nπ n π 0 (x − 1/2) 1 3/4 −8 − cos nπx + 2 2 sin nπx nπ n π 1/4 1−x 1 1 +8 cos nπx + 2 2 sin nπx 3/4 nπ n π 1/4 nπ nπ 1 = 8 − cos + 2 2 sin nπ 4 n π 4 1/4 3nπ 1/4 nπ 3nπ 1 nπ 1 −8 − cos + 2 2 sin − cos − 2 2 sin nπ 4 n π 4 nπ 4 n π 4 −1/4 3nπ 3nπ 1 +8 cos − 2 2 sin nπ 4 n π 4 16 nπ 3nπ = sin − sin . 2 2 n π 4 4

Thus u(x, t) =

∞ 16 X sin nπx cos nπt nπ 3nπ sin − sin . π2 n2 4 4 n=1

11. (a) From (11),

un (x, t) = sin

nπx L

bn cos

cnπt cnπt + b∗n sin L L

,

hence

2L nπx un (x, t + ) = sin nc L

cnπ(t + bn cos L

2L nc )

cnπ(t + + b∗n sin L

2L nc )

!

nπx cnπt cnπt ∗ = sin bn cos( + 2π) + bn sin( + 2π) L L L nπx cnπt cnπt ∗ = sin bn cos + bn sin = un (x, t), L L L

so un (x, t) has time period 2L nc . Since any positive integer multiple of a period is also a period, we conclude that un (x, t) has time period 2L c , and since this period is independent of n, it follows that any linear combination of un ’s has period 2L c . This applies to an infinite sum also. Hence the series solution (8) is time periodic with period 2L c , which proves (b). To prove (c),

Section 3.3

Wave Equation, the Method of Separation of Variables

note that cnπ(t + Lc ) cnπ(t + Lc ) bn cos + b∗n sin L L

L nπx un (x, t + ) = sin c L

= = =

!

cnπt cnπt ∗ bn cos( + nπ) + bn sin( + nπ) L L nπx cnπt cnπt (−1)n sin bn cos + b∗n sin L L L nπx cnπt cnπt ∗ sin( − nπ) bn cos + bn sin L L L nπ(L − x) cnπt cnπt − sin( ) bn cos + b∗n sin L L L −un (L − x, t).

nπx = sin L =

97

Since this relation holds for every n, summing over n, we obtain u(x, t +

∞ ∞ X X L L u(x, t + ) = − u(L − x, t) = −u(L − x, t). )= c c n=1 n=1

Think of the string at time t as part of an odd 2L-periodic graph, so that we can talk about u(x, t) for x outside the interval [0, L]. Moreover, we have u(−x, t) = −u(x, t) and u(x + 2L, t) = u(x, t). This graph moves as t varies and returns to its original shape at t + 2L c . The shape of the string at time t + Lc (half a period) is obtained by translating the portion over the interval [−L, 0] to the interval [0, L]. Thus, u(x, t+ Lc ) = u(x−L, t) = −u(L−x, t), since u is odd. 12. In this exercise, we solve the wave equation 2 ∂ 2u ∂u 2∂ u + 2k , = c ∂t2 ∂t ∂x2

where 0 < x < L, t > 0, subject to the conditions u(0, t) = 0,

u(L, t) = 0,

and

∂u (x, 0) = g(x) 0 < x < L. ∂t (a) To find the product solutions, plug u(x, t) = X(x)T (t) into the equation and get u(x, 0) = f (x),

XT 00 + 2kXT 0 = c2X 00T ; T 00 T0 X 00 + 2k = (Divide both sides by c2XT ). c2 T c2 T X Since the left side is a function of t only and the right side is a function of x only, the variables are separated and the equality holds only if T 00 T0 + 2k = C T T

and

X 00 = C, X

where C is a separation constant. Equivalently, we have the following equations in X and T : T 00 + 2kT 0 = CT ⇒ T 00 + 2kT 0 − CT = 0

98 Chapter 3 Partial Differential Equations in Rectangular Coordinates and X 00 − CX = 0. Plugging u = XT into the boundary conditions, we arrive at X(0) = 0 and X(L) = 0, as we did in the text when solving equation (1), subject to (2) and (3). Also, as in the text, the nontrivial solutions of the initial value problem X 00 − CX = 0, X(0) = 0 and X(L) = 0, occur when C = −µ2 < 0. Otherwise, if C ≥ 0, then the only solution of the initial value problem is X = 0. Thus, we must solve the equations X 00 + µ2 X = 0,

X(0) = 0, X(L) = 0,

T 00 + 2kT 0 + (µc)2 T = 0, where now µ is the separation constant. (b) As in the text, for the solution of equation (1), the nontrivial solutions in X are nπ X = Xn = sin x, n = 1, 2, . . . , L and these correspond to the separation constant µ = µn =

nπ . L

2 (c) To determine the solutions in T we have to solve T 00+2k T 0 +( nπ L c) T = 0. This is a second order linear differential equation with constant coefficients (Appendix A.2). Its general solution has three different cases depending on the roots of the characteristic equation nπ 2 λ2 + 2kλ + c = 0. L

By the quadratic formula, the roots of this equation are r nπ 2 λ = −k ± k2 − c . L nπ 2 nπ Lk Case I: k2 − c > 0; equivalently, k > c > 0 or > n > 0. For L L πc these positive integers n, we have two characteristic roots r r nπ 2 nπ 2 λ1 = −k + k2 − and λ2 = −k − k2 − c c . L L To these roots correspond the following solutions: √

√

f1 (t) = e(−k+ ∆)t and f2 (t) = e(−k− ∆)t , 2 where ∆ = k2 − nπ L c . By taking linear combinations of these solutions, we obtain new solutions of the equation. Consider the following linear combinations: f1 + f2 f 1 − f2 and . 2 2 We have √ f1 + f2 1 (−k+√ ∆)t + e(−k− ∆)t = e 2 2 √ √ ∆t + e− ∆t √ e = e−kt = e−kt cosh( ∆t). 2

Section 3.3

Wave Equation, the Method of Separation of Variables

99

Similarly, we have

√ f1 + f2 = e−kt sinh( ∆t). 2 It is not difficult to show that these solutions are linearly independent. Thus the general solution in this case is Tn = e−kt (an cosh λn t + bn sinh λn t), where

r

nπ 2 c . L nπ 2 nπ Lk Case II: k2 − c = 0; equivalently, k = c > 0 or n = . In this L L πc case, we have a double characteristic root, λ = −k, and the corresponding solutions are f1 (t) = e−kt and f2 (t) = t e−kt . λn =

k2 −

The general solution in this case is Tn = e−kt (an + bn t)

n=

Lk . πc

nπ 2 nπ Lk Case III: k2 − c < 0; equivalently, k < c > 0 or < n. For these L L πc positive integers n, we have two complex conjugate characteristic roots r r nπ 2 nπ 2 2 2 λ1 = −k + i k − and λ2 = −k − i k − c c . L L Thus the general solution in this case is Tn = e−kt (an cos λn t + bn sin λn t), where λn =

r

−k2 +

nπ 2 c . L

(d) When kL πc is not a positive integer, Case II does not occur. So only Cases I and II are to be considered. These yield the product solutions sin

nπx Tn (t), L

where Tn is as described in (c). Adding all the product solutions, we get solution is X nπ u(x, t) = e−kt sin x(an cosh λn t + bn sinh λn t) L kL 1≤n< πc

+e−kt

X

kL 1, using the formula sin x sin nx = 12 (cos((n − 1)x) − cos(n + 1)x) and integration by parts, we first derive the formula Z x sin x sin nx dx Z 1 = x(cos((n − 1)x) − cos(n + 1)x) dx 2 cos((−1 + n) x) x sin((−1 + n) x) cos((1 + n) x) x sin((1 + n) x) = + − − +C 2(−1 + n) 2(1 + n)2 2(1 + n) 2(−1 + n)2 b1 =

Thus, for n ≥ 2, 2 (−1)n−1 − 1 2 (−1)n+1 − 1 − π 2(−1 + n)2 π 2(1 + n)2 ( −2 2 if n is even, π(−1+n)2 + π(1+n)2 = 0 if n is odd, ( −8n if n is even, π(n2 −1)2 = 0 if n is odd.

an =

Consequently, a2k =

1 −16k −16k √ , a2k+1 = 0, b2k = , b2k+1 = 0, 2 2 2 2 π(4k − 1) π(4k − 1) 16k2 − 1

and so

√ √ ! π −.5t 3 3 1 u(x, t) = e sin x cos t + √ sin t 2 2 2 3 " ! !# r r ∞ 1 1 1 16 −.5t X k sin(2kx) cos 4k2 − t + √ 4k2 − t . sin − e 2−1 π (4k2 − 1)2 4 4 16k k=1 15. (a) We use Exercise 12 to solve ∂ 2u ∂u ∂ 2u + 3 , = ∂t2 ∂t ∂x2 u(0, t) = u(π, t) = 0, ∂u u(x, 0) = 0, (x, 0) = 10. ∂t

Section 3.3

Wave Equation, the Method of Separation of Variables

103

We have c = 1, k = 3/2, L = π, f (x) = 0, and g(x) = 10. So kL cπ = 3/2, which is not an integer. We distinguish two cases: n 3/2 (that is, n > 1). For n = 1, we have λ1 = 5/2, a1 = 0 and Z π 2 80 b1 = 10 sin x dx = √ . λ1 π 0 π 5 The corresponding solution in T is √ 80 −1.5t 5 √ e sinh t. 2 π 5 For n > 1, we have λn =

p

and bn = = =

2 πλn

|(3/2)2 − n2 | =

Z

1p 2 4n − 9 2

π

sin nx dx 0

40 √ 1 − (−1)n nπ 4n2 − 9 ( √80 if n = 2k + 1 2 (2k+1)π

4(2k+1) −9

0 Thus

otherwise.

√ 80 −1.5t 5 √ e u(x, t) = sin x sinh t 2 π 5 p ∞ 80 −1.5t X sin[(2k + 1)x] sin[ (2k + 1)2 − 9/4 t] p + e . π (2k + 1)π 4(2k + 1)2 − 9 k=1

(b) The following figure shows that the string rises and reaches a maximum height of approximately 3, and then starts to fall back down. As it falls down, it never rises up again. That is the string does not oscillate, but simply falls back to its initial rest position. u x_, t_ 80 Pi Sqrt 5 E ^ 3 2 t Sin x Sinh Sqrt 5 2t 40 Pi E ^ 3 2 t Sum Sin 2 j 1 x Sin Sqrt 2 j 1 ^ 2 9 4 t 2 j 1 Sqrt 2 j 1 ^ 2 9 4 tt

Table u x, t , t, .1, 8, .6

;

Plot Evaluate tt , x, 0, Pi , PlotRange

3 2.5 2 1.5 1 0.5 0.5

1

1.5

2

2.5

3

All

, j, 1, 7

;

104 Chapter 3 Partial Differential Equations in Rectangular Coordinates

Solutions to Exercises 3.4 1. We will use (5), since g ∗ = 0. The odd extension of period 2 of f (x) = sin πx is f ∗ (x) = sin πx. So u(x, t) =

1 t t 1 sin(π(x + )) + sin(π(x − )) = sin(πx + t) + sin(πx − t) . 2 π π 2

2. Reasoning as in Exercise 1, we find u(x, t) =

1 sin(πx + t) cos(πx + t) + sin(πx − t) cos(πx − t) . 2

3. We have f ∗ (x) = sin πx + 3 sin 2πx, g ∗(x) = sin πx. Both f ∗ and g ∗ are periodic with period R 1 2 and they are odd functions. So, the integral of g ∗ over one period, −1 g ∗(x) dx = 0. This implies that an antiderivative of g ∗ is also 2-periodic. Indeed, let Z x x 1 1 G(x) = sin πs ds = − cos πs = (1 − cos πx). π π 0 0 This is clearly 2-periodic. By (6) (or (4)), we have 1 sin(π(x + t)) + 3 sin(2π(x + t)) + sin(π(x − t)) + 3 sin(2π(x − t)) 2 i 1 h + 1 − cos π(x + t) − 1 − cos π(x − t) 2π i 1h = sin(π(x + t)) + 3 sin(2π(x + t)) + sin(π(x − t)) + 3 sin(2π(x − t)) 2 1 + cos π(x − t) − cos π(x + t) . 2π

u(x, t) =

4. For −1 < x < 1, we have ∗

g (x) =

1 if 0 < x < 1, −1 if − 1 < x < 0.

An antiderivative of this function on the interval [−1, 1] is given by x if 0 < x < 1, G(x) = = −x if − 1 < x < 0. We define G outside the interval [−1, 1] by extending it periodically. Thus G(x + 2) = G(x) for all x. By (6), u(x, t) =

1 G(x + t) − G(x − t) . 2

5. The solution is of the form 1 1 ∗ u(x, t) = f (x − t) + f (x + t) + G(x + t) − G(x − t) 2 2 1 ∗ = f (x − t) − G(x − t) + f ∗ (x + t) + G(x + t) , 2 where f ∗ is the odd extension of f and G is as in Example 3. In the second equality, we expressed u as the average of two traveling waves: one wave

Section 3.4

D’Alembert’s Method

105

traveling to the right and one to the left. Note that the waves are not the same, because of the G term. We enter the formulas in Mathematica and illustrate the motion of the string. The difficult part in illustrating this example is to define periodic functions with Mathematica. This can be done by appealing to results from Section 2.1. We start by defining the odd extensions of f and G (called big g) on the interval [-1, 1]. Clear f, bigg f x_ : 2 x ; 1 2 x 1 2 ;1 2 x 1 f x_ : 2 1 x 2 1 x ; 1 x 1 2 f x_ : 1 2 x^2 1 2 bigg x_ Plot

f x , bigg x

, x,

1, 1

1

0.5

-1

-0.5

0.5

1

-0.5

-1

Here is a tricky Mathematica construction. (Review Section 2.1.) extend x_ : x 2 Floor x 1 2 periodicf x_ : f extend x periodicbigg x_ : bigg extend x Plot periodicf x , periodicbigg x

, x,

3, 3

1

0.5

-3

-2

-1

1

2

3

-0.5

-1

Because f ∗ and G are 2-periodic, it follows immediately that f ∗ (x ± ct) and G(x ± ct) are 2/c-periodic in t. Since c = 1, u is 2-periodic in t. The following is an array of snapshots of u. You can also illustrate the

106 Chapter 3 Partial Differential Equations in Rectangular Coordinates motion of the string using Mathematica (see the Mathematica notebooks). Note that in this array we have graphed the exact solution and not just an approximation using a Fourier series. This is a big advantage of the d’Alembert’s solution over the Fourier series solution. u x_, t_ : 1 2 periodicf x t periodicf x t 1 2 periodicbigg x t periodicbigg x t tt Table Plot Evaluate u x, t , x, 0, 1 , PlotRange 0, 1 , 1, 1 , .5 , 1, .5, .5, 1 , DisplayFunction Identity , t, 0, 2.3, 1 5 Ticks Show GraphicsArray Partition tt, 4

;

6. The solution is of the form 1 G(x + t) − G(x − t) , 2

u(x, t) =

where G(x) is a continuous antiderivative of the odd extension of g (denoted g also). Since g(−x) = −g(x), we have ( Rx if − 1 < x < 0 −1 − cos πt dt G(x) = R0 Rx −1 − cos πt dt + 0 cos πt dt if 0 < x < 1. But

Z

0

− cos πt dt = 0, −1

so G(x) =

(

− π1 sin πx if − 1 < x < 0 1 π

sin πx

if 0 < x < 1.

From this we conclude that, for all x, G(x) =

1 sin πx . π

Section 3.4

D’Alembert’s Method

107

7. Very much like Exercise 4. 8. Very much like Exercise 8 but even easier. We have, for all x, G(x) =

1 cos πx. π

9. You can use Exercise 11, Section 3.3, which tells us that the time period of motion is T = 2L c . So, in the case of Exercise 1, T = 2π, and in the case of Exercise 5, T = 2. You can also obtain these results directly by considering the formula for u(x, t). In the case of Exercise 1, u(x, t) = 12 sin(πx + t) + 1 sin(πx − t) so u(x, t + 2π) = 2 sin(πx + t2π) + sin(πx − t2π) = u(x, t). In the case of Exercise 5, use the fact that f ∗ and G are both 2-periodic. 10. Here is a plot with Mathematica. Clear c 1 f x_ u x_, Plot

f, u, c Pi Sin Pi x t_ : 1 2 f x c t f x ct f x , f x c 2 , f x c 2 , u x, 1 2

, x, 0, 1

The string at time t = 1: Plot

f x

1 Pi , f x

1 Pi , u x, 1

, x, 0, 1

1

1

-0.75

108 Chapter 3 Partial Differential Equations in Rectangular Coordinates 12. Let us recall some facts about periodic functions. (a) If f is T -periodic, then any translate of f , say f (x+τ ) is also T -periodic. (b) If f is T -periodic, then any dilate of f , say f (±αx) is T /|α|-periodic (α 6= 0). (c) Linear combinations of T -periodic functions are also T -periodic. Properties (a)–(c) are straightforward to prove. (See Exercise 9, Section 2.1.) In d’Alembert’s solution, f ∗ and G are 2L-periodic. So, the functions f ∗ (ct) and G(ct) are 2L/c-periodic. Thus, for fixed x, the translated functions f ∗(x ± ct) and G(±cx) are 2L/c-periodic in the t-variable. Hence u(x, t) is 2L/c-periodic in t, being a linear combination of 2L/c-periodic functions. 13. We have 1 1 u(x, t) = [f ∗ (x + ct) + f ∗ (x − ct)] + 2 2c ∗

Z

x+ct

g ∗(s) ds, x−ct

∗

where f and g are odd and 2L-periodic. So L 1 1 u(x, t + ) = [f ∗ (x + ct + L) + f ∗ (x − ct − L)] + c 2 2c

Z

x+ct+L

g ∗(s) ds. x−ct−L

Using the fact that f ∗ is odd, 2L-period, and satisfies f ∗ (L − x) = f ∗ (x) (this property is given for f but it extends to f ∗ ), we obtain f ∗ (x + ct + L) = f ∗ (x + ct + L − 2L) = f ∗ (x + ct − L) = −f ∗ (L − x − ct) = −f ∗ (L − (x + ct)) = −f ∗ (x + ct). Similarly f ∗ (x − ct − L) = −f ∗ (L − x + ct) = −f ∗ (L − x + ct) = −f ∗ (L − (x − ct)) = −f ∗ (x − ct). Also g ∗(s + L) = −g ∗ (−s − L) = −g ∗ (−s − L + 2L) = −g ∗(L − s) = −g ∗(s), by the given symmetry property of g. So, using a change of variables, we have Z Z Z 1 x+ct+L ∗ 1 x+ct ∗ 1 x+ct ∗ g (s) ds = g (s + L) ds = − g (s) ds. 2c x−ct−L 2c x−ct 2c x−ct Putting these identities together, it follows that u(x, t + Lc ) = −u(x, t). 14. We have u(x, t) =

1 ∗ 1 [f (x + ct) + f ∗ (x − ct)] + [G(x + ct) − G(x − ct)], 2 2c

where f ∗ is odd and 2L-periodic and G is 2L-periodic. In fact, G is even (see Exercise 17 or see below). For any 2L-periodic even function φ, we have φ(L + a) = φ(L + a − 2L) = φ(a − L) = φ(L − a). And for any 2L-periodic odd function ψ, we have ψ(L + a) = ψ(L + a − 2L) = ψ(a − L) = −ψ(L − a). Using these two facts and that f ∗ is odd and G is even, we get Then u(x, L) = =

1 ∗ 1 [f (L + ct) + f ∗ (L − ct)] + [G(L + ct) − G(L − ct)] 2 2c 1 1 [−f ∗ (L − ct) + f ∗ (L − ct)] + [G(L − ct) − G(L − ct)] = 0. 2 2c

Section 3.4

D’Alembert’s Method

109

To have a complete proof, let us show that G is even. We know that g ∗ is 2L-periodic and odd. For any x, we have

G(−x) =

Z

Z

−x ∗

g (t) dt = a

=

zZ

=0 a

}| ∗

{

g (t) dt + −a

zZ

x ∗

g (−t)(−1)dt = −a G(x)

x

}| ∗

Z

x

g ∗(t) dt −a

{

g (t) dt

a

= G(x), where we have used that g ∗ is odd to infer that its integral over a symmetric interval is 0. (b) Checking the initial condition is easier. We have

u(x, 0) =

1 ∗ 1 [f (x) + f ∗ (x)] + [G(x) − G(x)] = f ∗ (x), 2 2c

and ut(x, 0) =

1 1 [cf ∗(x) − cf ∗ (x)] + [cg(x) + cg(x)] = g(x). 2 2c

Here we have used the fact that G0 = g. You can check that this equality is true at the points of continuity of g. 15. Let us define the function f (x) on (0, 1) and then plot it. Clear f f x_ : 4 x ; 1 4 x 1 4 ;1 4 x 1 2 f x_ : 4 1 2 x ;1 2 x 1 f x_ : 0 1 2 x ; 1 2 x 1 4 f x_ : 4 ; 1 x 1 2 f x_ : 0 Plot f x , x, 1, 1

1

0.5

-1

-0.5

0.5 -0.5

-1

Let us now plot the periodic extension of f

1

110 Chapter 3 Partial Differential Equations in Rectangular Coordinates

extend x_ : x 2 Floor x 1 periodicf x_ : f extend x Plot periodicf x , x, 3 , 3

2

1

0.5

-3

-2

-1

1

2

3

-0.5

-1

We are now ready to define u and plot it for any value of t. We plot u(x, 1/4) along with f (x − 1/4) and f (x + 1/4) to illustrate the fact that u is the average of these two traveling waves. u x_, t_ : 1 2 periodicf x t periodicf x t Plot Evaluate periodicf x , periodicf x 1 4 , periodicf x x, 0, 1 , PlotRange All

1 4 , u x, 1 4

1 0.5

0.5 -0.5 -1

1

,

Section 3.4

111

D’Alembert’s Method

Next, we show the string at t = 1/2.

u x_, t_ : 1 2 periodicf x t periodicf x t Plot Evaluate periodicf x , periodicf x 1 2 , periodicf x x, 0, 1 , PlotRange All

1 2 , u x, 1 2

,

1

1

0.5

-1

(b) It is clear from the graph of u(x, 1/4) that the points .75 < x < 1 are still at rest when t = 1/4. (c) Because f ∗ (x) = 0 for .5 < x < 1.5 and u(x, t) = 12 (f (x + t) + f (x − t)), a point x0 in the interval (1/2, 1) will feel a vibration as soon as the right traveling wave reaches it. Since the wave is traveling at speed c = 1 and is supported on the interval (0, 1/2), the vibration will reach the point x0 after traveling a distance x0 − 1/2. Since the wave is traveling at speed c = 1, it will take x0 − 1/2 time for the point x0 to start move from rest. For all values of t < x0 − 1/2, the point x0 will remain at rest. For example, if 3/4 < x0 < 1, then this point will remain at rest for all time t < 1/4, as we saw in the previously. (d) For an arbitrary value of c > 0, the wave will travel at the speed of c. Thus it will take (x0 − 1/2)/c time to reach the point x0 located in the interval 1/2 < x < 1. 16. (a) From (8), Section 3.3, with b∗n = 0 (since g = 0), we have

u(x, t) = =

∞ X

bn sin

n=1 ∞ X

1 2

nπ nπ xbn cos c t L L

bn sin

n=1

nπ nπ (x − ct) + sin (x + ct) , L L

where we have used the formula sin a cos b = (b) Since, for all x, f ∗ (x) =

∞ X n=1

bn sin

1 2

sin(a − b) + sin(a + b) .

nπ x, L

112 Chapter 3 Partial Differential Equations in Rectangular Coordinates putting x + ct and x − ct in place of x, we get ∞

u(x, t) = =

∞

1X nπ nπ 1X bn sin[ (x − ct)] + bn sin[ (x + ct)] 2 L 2 L n=1 n=1 1 ∗ f (x − ct) + f ∗(x + ct) . 2

17. (a) To prove that G is even, see Exercise 14(a). That G is 2L-periodic follows from the fact that g is 2L-periodic and its integral over one period is 0, because it is odd (see Section 2.1, Exercise 15). Since G is an antiderivative of g ∗, to obtain its Fourier series, we apply Exercise 33, Section 3.3, and get ∞

L X bn (g) nπ G(x) = A0 − cos x, π n L n=1

where bn (g) is the nth Fourier sine coefficient of g ∗, Z 2 L nπ bn (g) = g(x) sin x dx L 0 L and

∞

L X bn (g) . π n

A0 =

n=1

In terms of b∗n , we have L bn (g) 2 = π n nπ

Z

L

g(x) sin 0

nπ x dx = cb∗n , L

and so ∞

∞

L X bn (g) L X bn (g) nπ − cos x π n π n L

G(x) =

n=1

∞ X

=

n=1

cb∗n 1 − cos(

n=1

nπ x) . L

(b) From (a), it follows that G(x + ct) − G(x − ct) =

∞ X

=

n=1 ∞ X n=1

cb∗n

h

1 − cos(

i nπ nπ (x + ct)) − 1 − cos( (x − ct)) L L

h i nπ nπ −cb∗n cos( (x + ct)) − cos( (x − ct)) L L

(c) Continuing from (b) and using the notation in the text, we obtain Z 1 x+ct ∗ 1 g (s) ds = [G(x + ct) − G(x − ct)] 2c x−ct 2c ∞ i X 1h nπ nπ = −b∗n cos( (x + ct)) − cos( (x − ct)) 2 L L =

n=1 ∞ X

b∗n sin(

=

n=1 ∞ X

nπ nπ x) sin( ct) L L

b∗n sin(

nπ x) sin(λn t). L

n=1

Section 3.4

D’Alembert’s Method

113

(d) To derive d’Alembert’s solution from (8), Section 3.3, proceed as follows: ∞ X

∞

X nπ nπ b∗n sin( x) sin(λnt) x) cos(λn t) + L L n=1 n=1 1 ∗ 1 = f (x − ct) + f ∗ (x + ct) + [G(x + ct) − G(x − ct)] , 2 2c

u(x, t) =

bn sin(

where in the last equality we used Exercise 16 and part (c). 18. (a) Recall that u and its partial derivatives are functions of x and t. By differentiating under the integral sign, we obtain Z 1 L d 2 d E(t) = ut + c2 u2x dx dt 2 0 dt Z 1 L = 2ut utt + 2c2ux uxt dx. 2 0 (b) But utt = c2 uxx, so Z L 2 d c ut uxx + c2ux uxt dx E(t) = dt 0 Z L 2 = c [ut uxx + ux uxt ] dx 0 Z L 2 = c ut ux x dx 0 L = c2ut ux = c2[ut(L, t) ux (L, t) − ut(0, t) ux(0, t)]. 0

(c) Since u is a solution of the wave equation representing a string with fixed ends, we have that the velocity at the endpoints is 0. That is, ut (0, t) = ut (L, t) = 0. (d) From (c) and (d), we conclude that d E(t) = 0. dt Thus E(t) is constant for all t. Setting t = 0 in the formula for E(t) and using the initial conditions, we obtain Z Z 1 L 2 1 L 2 2 2 E(0) = ut (x, 0) + c ux (x, 0) dx = g (x) + c2[f 0 (x)]2 dx. 2 0 2 0 Since E is constant in time, this gives the energy of the string for all t > 0. 19. (a) The characteristic lines are lines with slope ±1/c = ±1 in the xt-plane. (b) The interval of dependence of a point (x0 , t0 ) is the interval on the xaxis, centered at x0, with radius ct0 . For the point (.5, .2) this interval is .3 ≤ x ≤ .7. For the point (.3, 2) this interval is −1.8 ≤ x ≤ 2.3. (c) The horizontal, semi-infinite, strip S is bounded by the line x = 0 and x = 1 and lies in the upper half-plane. The region I is the triangular region with base on the interval (0, 1), bounded by the lines through (0, 0) at slope 1, and through (1, 0) at slope −1. The three vertices of I are (0, 0), (1/2, 1/2), and (0, 1). The point (.5, .2) is in I but the point (.3, 2) is not

114 Chapter 3 Partial Differential Equations in Rectangular Coordinates in I. (d) For the points (x, t) in I, u(x, t) depends only on f and g and not their periodic extension. We have Z 1 x+t 1 u(x, t) = = f (x − t) + f (x + t) + g(s) ds 2 2 x−t Z 1 x+t = s ds 2 x−t 1 2 x+t 1 = = (x + t)2 − (x − t)2 = xt. s 4 x−t 4 20. Let P1 = (x0, t0 ) be an arbitrary point in the region II. Form a characteristic parallelogram with vertices P1 , P2 , Q1 , Q2 , as shown in Figure 7 in Section 3.4. The vertices P2 and Q2 are on the characteristic line x−t = 0 and the vertex Q1 is on the boundary line x = 0. From Proposition 1, we have u(P1) = u(Q1 ) + u(Q2) − u(P2). We will find u(P2) and u(Q2) by using the formula u(x, t) = xt from Exercise 19(d), because P2 and Q2 are in the region I. Also, u(Q1) = 0, because of the boundary condition u(0, t) = 0 for all t > 0. The point Q2 is the intersection point of the characteristic lines x + t = x0 + t0 and x − t = 0. Adding the equations, we get 2x = x0 + t0 or x0 +t0 x0 +t0 x0 +t0 0 x = x0 +t 2 . From x = t, we obtain t = 2 , and so Q2 = ( 2 , 2 ). The coordinates of Q1 and P2 are computed similarly. The point Q1 is the intersection point of the characteristic line x − t = x0 − t0 and the boundary x = 0. Thus t = t0 − x0 and so Q1 = (0, t0 − x0). Similarly, the point P2 is the intersection point of the characteristic lines x + t = t0 − x0 and 0 x − t = 0. Adding the equations, we get 2x = t0 − x0 or x = t0 −x 2 . Thus −x0 +t0 −x0 +t0 P2 = ( 2 , ). Let us simplify the notation and write P1 = (x, t) 2 instead of (x0, t0 ). Then, for P1 = (x, t) in the region II, u(x, t) = u(Q2) − u(P2) x+t x+t −x + t −x + t = u( , ) − u( , ) 2 2 2 2 x + t x + t −x + t −x + t = · − · 2 2 2 2 1 2 2 = (x + t) − (−x + t) 4 = tx. Interestingly, the formula for u(x, t) for (x, t) in region II is the same as the one for (x, t) in the region I. In particular, this formula for u(x, t) satisfies the wave equation and the boundary condition at x = 0. The other conditions in the wave problem do not concern the points in the region II and thus should not be checked. 21. Follow the labeling of Figure 8 in Section 3.4. Let P1 = (x0 , t0 ) be an arbitrary point in the region II. Form a characteristic parallelogram with vertices P1 , P2 , Q1 , Q2 , as shown in Figure 8 in Section 3.4. The vertices P2 and Q1 are on the characteristic line x + 2t = 1 and the vertex Q2 is on the boundary line x = 1. From Proposition 1, we have u(P1 ) = u(Q1) + u(Q2 ) − u(P2 ) = u(Q1) − u(P2),

Section 3.4

D’Alembert’s Method

115

because u(Q2) = 0. We will find u(P2 ) and u(Q1) by using the formula u(x, t) = −4t2 + x − x2 + 8tx from Example 4, because P2 and Q1 are in the region I. The point Q1 is the intersection point of the characteristic lines x − 2t = x0 − 2t0 and x + 2t = 1. Adding the equations and then solving for x, we get x0 + 1 − 2t0 x= . 2 The second coordinate of Q1 is then t=

1 − x0 + 2t0 . 4

The point Q2 is the intersection point of the characteristic line x + 2t = x0 + 2t0 and x = 1. Thus t=

x0 + 2t0 − 1 . 2

The point P2 is the intersection point of the characteristic lines x+2t = 1 and x − 2t = 1 − (x0 + 2t0 − 1). Solving for x and t, we find the coordinates of P2 to be 3 − x0 − 2t0 −1 + x0 + 2t0 x= and t = . 2 4 To simplify the notation, replace x0 and t0 by x and y in the coordinates of the points Q1 and P2 and let φ(x, t) = −4t2 + x − x2 + 8tx. We have u(x, t) = u(Q1) − u(P2 ) x + 1 − 2t 1 − x + 2t 3 − x − 2t −1 + x + 2t = φ , −φ , 2 4 2 4 = 5 − 12t − 5x + 12tx, where the last expression was derived after a few simplifications that we omit. It is interesting to note that the formula satisfies the wave equation and the boundary condition u(1, t) = 0 for all t > 0. Its restriction to the line x + 2t = 1 (part of the boundary of region I) reduces to the formula for u(x, t) for (x, t) in region I. This is to be expected since u is continuous in (x, t).

116 Chapter 3 Partial Differential Equations in Rectangular Coordinates 22. Label the problem as in the following figure:

Let P1 = (x0, t0 ) be an arbitrary point in the region IV . Form a characteristic parallelogram with vertices P1 , P2 , Q1 , Q2 , as shown in the figure. The point P2 is on the intersection of x − 2t = 0 and x + 2t = 1. Its coordinates are found to be (.5, .25). Since P2 belongs to three regions, we can compute u(P2) from one of three formulas that we have thus far. Let’s use the formula for the points in region I. We have (from Example 4) u(P2 ) = u(.5, .25) = −4(.25)2 + .5 − (.5)2 + 8(.25)(.5) = 1. The point Q1 is on the intersection of the lines x − 2t = x0 − 2t0 and x + 2t = 1. Its coordinates are

x0 + 1 − 2t0 x0 , t0 − 2 2

.

Since Q1 is in region II, we compute u(Q1) using the formula from Example 5. We find u

x0 + 1 − 2t0 x0 , t0 − 2 2

x0 + 1 − 2t0 2 x0 + 1 − 2t0 x0 +4 · t0 − 2 2 1 x 2 = + t − 4 t − + 4 t x − x2 . 2 2

=

We now consider the point Q2 on the intersection of x − 2t = 0 and x + 2t = x0 + 2t0 . We have x0 + 2t0 x0 + 2t0 Q2 = , . 2 4 Since Q2 is in region III, we compute u(Q2) using the formula from Exer-

Section 3.4

D’Alembert’s Method

117

cise 21. We find x0 + 2t0 x0 + 2t0 x0 + 2t0 x0 + 2t0 u , = 5 − 12 −5 2 4 4 2 x0 + 2t0 x0 + 2t0 +12 4 2 11 x = 5 − 11 t + 8 t2 − + 8 t x + 2 x2 . 2 From Proposition 1, we have u(P1 ) = u(Q1) + u(Q2 ) − u(P2 ) 1 x = + t − 4 t2 − + 4 t x − x2 2 2 11 x 2 +5 − 11 t + 8 t − + 8 t x + 2 x2 − 1 2 9 = − 10 t + 4 t2 − 6 x + 12 t x + x2 2 It is not difficult to show that this function satisfies the wave equation. As further verification, you can check hat this formula yields 1 when evaluated at (.5, .25), which are the coordinates of the points P2 .

118 Chapter 3 Partial Differential Equations in Rectangular Coordinates

Solutions to Exercises 3.5 78 100

1. Multiply the solution in Example 1 by ∞

u(x, t) =

to obtain

312 X e−(2k+1) t sin(2k + 1)x. π 2k + 1 2

k=0

2. The function f is given by its Fourier series. So u(x, t) = 30 sin x e−t . 3. We have

∞ X

u(x, t) =

2

bn e−n t sin(nx),

n=1

where bn =

66 π

Z

π/2

x sin nx dx +

66 π

Z

π

(π − x) sin nx dx 66 x cos(n x) sin(n x) π/2 = − + π n n2 0 66 −(π − x) cos(n x) sin(n x) π + − π n n2 π/2 π π π π π sin(n 2 ) sin(n π2 ) 66 2 cos(n 2 ) 2 cos(n 2 ) = + − + + π n n2 n n2 π 132 sin(n 2 ) = n2 (π 0

(−1)k

=

Thus

π/2

if n = 2k + 1, if n = 2k.

132 π(2k+1)2 0

∞

132 X e−(2k+1) t sin((2k + 1)x) u(x, t) = . π (2k + 1)2 2

k=0

4. We have u(x, t) =

∞ X

2

bn e−n t sin nx,

n=1

where bn is the nth sine Fourier coefficient of the function f (x) = 100 if 0 < x < π/2 and 0 if π/2 < x < π. Instead of computing bn , we will appeal to Exercise 29, Section 3.3, with d = π/2 and p = π. We get 200 nπ bn = 1 − cos . πn 2 Thus

200 X nπ −n2 t sin nx u(x, t) = 1 − cos e . π 2 n ∞

n=1

5. We have u(x, t) =

∞ X n=1

2

bn e−(nπ) t sin(nπx),

Section 3.5

The One Dimensional Heat Equation

119

where bn

x cos(n π x) sin(n π x) 1 + = 2 x sin(nπx) dx = 2 − nπ n2 π 2 0 0 n+1 cos n π (−1) = −2 =2 . nπ nπ Z

1

So u(x, t) =

2 ∞ 2 X (−1)n+1 e−(nπ) t sin(nπx) . π n

n=1

6. We have u(x, t) =

∞ X

2

bn e−(nπ) t sin(nπx),

n=1

where bn = 2

Z

1

e−x sin(nπx) dx 0

= = = So u(x, t) = 2π

1 2e−x − sin nπx − nπ cos nπx 1 + (nπ)2 0 −1 2 2e − nπ(−1)n − − nπ 2 2 1 + (nπ) 1 + (nπ) 2nπ 1 + (−1)n+1 e−1 . 2 1 + (nπ) ∞ X n=1

n 2 1 + (−1)n+1 e−1 e−(nπ) t sin(nπx). 2 1 + (nπ)

Putting t = 0, we obtain the sine series expansion of e−x for 0 < x < 1. Let’s check this result with a graph. Clear f, partsum f x_ : E ^ x partsum x_, t_, n_ : 2 Pi Sum k 1 k ^ 2 Pi ^ 2 1 1 ^ k 1 E^ 1 E^ k ^ 2 Pi ^ 2 t Sin k Pi x , Plot Evaluate f x , partsum x, 0, 20 , x, 0, 1

1

k, 0, n

;

120 Chapter 3 Partial Differential Equations in Rectangular Coordinates 7. (a) From the figures in Example 1, we can see that when t is equal to 1 the temperature is already below 50 degrees. We will give a more accurate approximation. Let us first check the solution by plotting an approximation of the initial temperature distribution.

Clear f, partsum f x_ : 100 partsum x_, t_, n_ : 400 Pi Sum E ^ 2 k 1 ^ 2 t Sin 2 k 1 x 2 k 1 , k, 0, n ; , x, 0, Pi , PlotRange All Plot Evaluate f x , 50, partsum x, 0, 20

120 100 80 60 40 20 0.5

1

1.5

2

2.5

3

Let us now examine the temperature distribution, starting with t = .5 to t = 1.2 with increments of t = .1.

tt

Table Plot Evaluate partsum x, t, 10 , 50 , x, 0, Pi , PlotRange All, DisplayFunction Identity , t, 0.5, 1.2, 1 10 Show GraphicsArray Partition tt, 4

60 40

60 40

50

50

50

50

50

50 30

;

It appears that the maximum temperature dips under 50 degree between t = .9 and t = 1. You can repeat the plot for t = .9 to t = 1, using increments of .01 and get a more accurate value of t. (b)

Section 3.5

The One Dimensional Heat Equation

Plot3D partsum x, t, 20 , x, 0, Pi , t, 0, 5 , PlotRange All, ViewPoint

121

3, 3, .4

100 50 0 0

1

2

3

4

2

5

1

0

To prove that the temperature tends uniformly to 0 in the bar, you can proceed as follows (this is more than what the question is asking for): for t > 0 and all x,

∞ 400 X 2 t sin(2k + 1)x −(2k+1) u(x, t)| = e π 2k + 1 k=0

= =

∞ 400 X

∞

400 X −t k e e = π π k=1 k=1 400 1 . 1− π 1 − e−t −kt

As t → ∞, the last displayed expression tends to 0, which shows that u(x, t) tends to 0 uniformly in x. 8. (a) Applying (4), we see that, for an arbitrary c > 0, the solution becomes

∞

u(x, t) =

400 X −c2 (2k+1)2t sin(2k + 1)x e . π 2k + 1 k=0

(b) As c increases, the rate of heat transfer increases as illustrated by the following figures. This is also

122 Chapter 3 Partial Differential Equations in Rectangular Coordinates

Clear f, partsum f x_ : 100 partsum x_, t_, n_, c_ 400 Pi Sum E ^ 2k Plot Evaluate 50, partsum x, 0, Pi , PlotRange

: 1 ^ 2 t c Sin

2k

1

x

2k

1 ,

k, 0, n

;

x, 1, 20, 1 , partsum x, 1, 20, 2 , partsum x, 1, 20, 3 All

,

50 40 30 20 10

9. (a) The steady-state solution is a linear function that passes through the points (0, 0) and (1, 100). Thus, u(x) = 100x. (b) The steady-state solution is a linear function that passes through the points (0, 100) and (1, 100). Thus, u(x) = 100. This is also obvious: If you keep both ends of the insulated bar at 100 degrees, the steady-state temperature will be 100 degrees. 10. (a) Recall that u1 satisfies (u1)t = c2 (u1)xx , and u2 satisfies (u2 )t = 0 and (u2 )xx = 0. So ut = (u1 + u2 )t = (u1 )t + (u2)t = (u1 )t = c2(u1 )xx = c2 (u1)xx + c2 (u2)xx = c2 [(u1)xx + (u2 )xx] = c2 uxx .

Thus u satisfies (6). Now recall that u1(0) = T1, u1 (L) = T2, and u2 (0, t) = 0 = u2 (L, t). Thus

u(0, t) = u1(0) + u2 (0, t) = T1 and u(L, t) = u1 (L) + u2 (L, t) = T2.

Thus u satisfies (7). Now u1(x) = So

T2 −T1 L x+

T1 and u2 (x, 0) = f (x) − u1 (t).

u(x, 0) = u1 (x) + u2 (x, 0) = f (x), showing that u satisfies (8). (b) According to (14), we have (see the integral formula in Exercise 7, Sec-

Section 3.5

The One Dimensional Heat Equation

123

tion 3.3) bn = = =

=

T 2 − T1 nπ f (x) − x + T1 sin x dx L L 0 Z Z nπ 2 L nπ 2 L T2 − T1 f (x) sin x dx − x + T1 sin x dx L 0 L L 0 L L Z nπ 2 L nπ 2 L T 2 − T1 f (x) sin x dx − − x + T1 cos x L 0 L L nπ L L T 2 − T1 L 2 nπ L + sin x L nπ L 0 Z L 2 nπ 2 f (x) sin x dx − T1 − (−1)n T2 . L 0 L nπ 2 L

Z

L

11. We use the solution outlined in the text. We have the steady-state solution 0 − 100 u1 (x) = x + 100 = 100(1 − x). 1 Then ∞ X 2 u2 (x, t) = bn e−(nπ) t sin(nπx), n=1

where bn = 2

Z

1

30 sin(πx) − 100(1 − x) sin(nπx) dx

0 = 200 =30 if n=1, 0 if n6=1 nπ z Z }| { z Z }| { 1 1 = 60 sin(πx) sin(nπx) dx − 200 (1 − x) sin(nπx) dx 0 0 200 30 − π if n = 1, = − 200 if n > 1. nπ

Thus So ∞

u(x, t) = u1 (x)+u2(x, t) = 100(1−x)+30 e

−π 2 t

200 X e−(nπ) t sin(nπx) sin πx− . π n n=1

12. Let us use (13) and the formula from Exercise 10. We have u(x, t) = u1 (x) +

∞ X

bn e−n

2 π2 t

sin nπx,

n=1

where u1 (x) = 100 and bn = 100

Z

1

x(1 − x) sin nπx dx − 200

0

(1 − (−1)n ) 1 − (−1)n = 200 − 200 (nπ)3 nπ (1 − (−1)n ) 1 − (−1)n = 200 − 200 . (nπ)3 nπ

1 − (−1)n nπ

2

124 Chapter 3 Partial Differential Equations in Rectangular Coordinates Thus ∞ X (1 − (−1)n )

1 − (−1)n −n2 π2 t u(x, t) = 100 + 200 − sin nπx e (nπ)3 nπ n=1 ∞ X 1 2 2 1 = 100 + 400 − e−(2k+1) π t sin[(2k + 1)πx]. 3 ((2k + 1)π) (2k + 1)π k=0

13. We have u1(x) = − 50 π x + 100. We use (13) and the formula from Exercise 10, and get (recall the Fourier coefficients of f from Exercise 3) 50 x + 100 π ∞ X 132 sin(n π2 ) 2 2 − (−1)n + − 100 e−n t sin nx. 2 π n nπ

u(x, t) = −

n=1

14. We have u1 (x) = 100 π x. We use (13) and the formula from Exercise 10, and get (recall the Fourier coefficients of f from Exercise 4) u(x, t) =

100 x π ∞ X 200 nπ (−1)n −n2 t + sin nx 1 − cos + 200 e πn 2 nπ n=1

=

100 nπ i e−n t sin nx 200 X h 1 + (−1)n − cos x+ π π n=1 2 n

=

100 e−(2n) t sin(2nx) 100 X [2 − (−1)n ] x+ . π π n

∞

∞

2

2

n=1

It is good at this point to confirm our results with some graphs. We will plot a partial sum of the series solution at t = 0 (this should be close to the initial temperature distribution). We also plot the partial sum of the series solution for large t (the graphs should be close to the steady-state solution u1 (x)). The graphs suggest that u(1/2, t) = 50 for all t. Can you explain this result using the formula for u and on physical grounds?

Section 3.5

The One Dimensional Heat Equation

Clear f, partsum, n , k, sstate f x_ : 100 ; 0 x Pi 2 ; Pi 2 x Pi f x_ : 0 100 Pi x sstate x_ partsum x_, t_, n_ : sstate x 100 Pi Sum 2 1 ^k E^ 2 k ^ 2 t Sin 2 k x k, , x, 0, Pi Plot Evaluate f x , partsum x, 0, 20 Plot Evaluate sstate x , partsum x, .1, 20 , partsum x, .5, 20 , partsum x, 1, 20 , partsum x, 4, 20

k, 1, n

125

;

, x, 0, Pi

120 100 80 60 40 20 0.5

1

1.5

2

2.5

3

0.5

1

1.5

2

2.5

3

-20

100 80 60 40 20

Graphics

15. From (14), we have bn = 0, since f (x) − u1(x) = 0. Hence u2 (x, t) = 0 and so u(x, t) = u1 (x) = steady-state solution. This is obvious on physical grounds: If the initial temperature distribution is already in equilibrium, then it will not change. 16. This problem is very much like Exercise 10(a). We sketch the details. Let u1 (x, t) = u1 (x) = B−A L x + A. Then u1 satisfies the wave equation utt = c2 uxx , the boundary conditions u1(0, t) = A

and

u1(L, t) = B,

126 Chapter 3 Partial Differential Equations in Rectangular Coordinates and the initial conditions u1 (x, 0) = u1(x) =

B−A x+A L

and (u1 )t(x, 0) = 0.

Let u2 (x, t) denote the solution of the wave equation with 0 boundary data (that is u2 (0, t) = 0 and u2(L, t) = 0) and satisfying the initial conditions u(x, 0) = f (x) − u1(x) and

ut (x, 0) = g(x).

It is straightforward to show that u(x, t) = u1 (x) + u2 (x, t) is a solution of the nonhomogeneous wave problem utt = c2 uxx ; u(0, t) = A, u(x, 0) = f (x),

u(L, t) = B; ut (x, 0) = g(x).

We can appeal to the solution of the wave equation from Sections 3.3 or 3.4 to express the solution u2 in terms of the initial conditions. For example, using the solution from Section 3.3, we obtain u2 (x, t) =

∞ X

sin

n=1

where λn =

nπ x (bn cos λn t + b∗n sin λn t) , L

cnπ L ,

2 bn = L

Z

L

(f (x) − u1 (x)) sin 0

and b∗n

2 = cnπ

Z

L

g(x) sin 0

nπ x dx, L

nπ x dx. L

17. Fix t0 > 0 and consider the solution at time t = t0 : u(x, t0 ) =

∞ X n=1

bn sin

nπ −λ2n t0 . xe L

We will show that this series converges uniformly for all x (not just 0 ≤ x ≤ L) by appealing to the Weierstrass M -test. For this purpose, it suffices to establish the following two inequalities: −λ2n t0 ≤ M for all x; and (a) P bn sin nπ xe n L (b) ∞ M < ∞. n n=1 To establish (a), note that Z L Z 2 nπ nπ 2 L |bn | = f (x) sin f (x) sin x dx ≤ x dx L 0 L L 0 L (The absolute value of the integral is ≤ the integral of the absolute value.) Z 2 L ≤ |f (x)| dx = A (because | sin u| ≤ 1 for all u). L 0

Section 3.5

The One Dimensional Heat Equation

127

Note that A is a finite number because f is bounded, so its absolute value is bounded and hence its integral is finite on [0, L]. We have c2 π 2 t 2 bn sin nπ xe−λ2n t0 ≤ Ae−λ2n t0 = Ae− L2 0 n L 2 2 n c π t0 ≤ A e− L 2 = Arn ,

c2 π 2 t0

where r = e− L2 < 1. Take Mn = Arn . Then a holds and convergent because it is a geometric series with ratio r < 1.

P

Mn is

128 Chapter 3 Partial Differential Equations in Rectangular Coordinates

Solutions to Exercises 3.6 1. Since the bar is insulated and the temperature inside is constant, there is no exchange of heat, and so the temperature remains constant for all t > 0. Thus u(x, t) = 100 for all t > 0. This is also a consequence of (2), since in this case all the an ’s are 0 except a0 = 100. 2. The initial temperature distribution is already given by its Fourier cosine series. We have a1 = 1 and an = 0 for all n 6= 1. Thus 2

u(x, t) = e−(nπ) cos πx. 3. We apply (4): u(x, t) = a0 +

∞ X

2

an e−n t cos nx,

n=1

where a0 = =

Z Z 1 π/2 1 π/2 100 x dx + 100 (π − x) dx π 0 π 0 π 50 2 π/2 50 25 25 − = x (π − x)2 π + π = 25π, π π 2 2 0 π/2

and an = =

= = = So

Z Z 2 π/2 2 π/2 100 x cos nx dx + 100 (π − x) cos nx dx π 0 π 0 200 cos n x x sin n x π/2 + π n2 n 0 200 cos n x (π − x) sin n x π + + − π n2 n π/2 π π π π π π sin n cos n cos n 200 cos n π 1 2 2 2 2 2 sin n 2 + − − + − π n2 n n2 n n2 n2 200 π n 2 cos n ) − (1 + (−1) πn2 2 if n = 4k + 1 or n = 4k + 3, 0 0 if n = 4k or n = 4k + 3, 800 − πn2 if n = 4k + 2 = 2(2k + 1). ∞

200 X 1 2 u(x, t) = 25π − e−4(2k+1) t cos(2(2k + 1)x), 2 π (2k + 1) k=0

4. Refer to Example 1. We need to expand f (x) in a cosine series. The series can be derived from Exercise 6, Section 2.3, with d = 1/2, p = 1, and c = 100. We have, for 0 < x < 1, ∞

f (x) =

100 200 X sin nπ 2 + cos nπx. 2 π n n=1

But

sin nπ 2 =

0 if n = 2k, k (−1) if n = 2k + 1,

Section 3.6

so

Heat Conduction in Bars: Varying the Boundary Conditions

129

∞

200 X (−1)k cos[(2k + 1)πx], f (x) = 50 + π 2k + 1 k=0

and the solution of the heat problem becomes ∞

u(x, t) = 50+

200 X (−1)k 2 2 cos[(2k+1)πx]e−(2k+1) π t π 2k + 1

(t > 0, 0 < x < 1).

k=0

5. Apply the separation of variables method as in Example 1; you will arrive at the following equations in X and T : X 00 − kX = 0,

X(0) = 0, X 0(L) = 0

T 0 − kc2T = 0

.

We now show that the separation constant k has to be negative by ruling out the possibilities k = 0 and k > 0. If k = 0 then X 00 = 0 ⇒ X = ax + b. Use the initial conditions X(0) = 0 implies that b = 0, X 0(L) = 0 implies that a = 0. So X = 0 if k = 0. If k > 0, say k = µ2 , where µ > 0, then X 00 − µ2 X = 0 ⇒ X = c1 cosh µx + c2 sinh µx; X(0) = 0 ⇒ 0 = c1 ; X = c2 sinh µx; X 0(L) = 0 ⇒ 0 = c2 µ cosh(µL) ⇒ c2 = 0, because µ 6= 0 and cosh(µL) 6= 0. So X = 0 if k > 0. This leaves the case k = −µ2 , where µ > 0. In this case X 00 + µ2 X = 0 ⇒ X = c1 cos µx + c2 sin µx; X(0) = 0 ⇒ 0 = c1; X = c2 sin µx; X 0(L) = 0 ⇒ 0 = c2µ cos(µL) ⇒ c2 = 0 or cos(µL) = 0. To avoid the trivial solution, we set cos(µL) = 0, which implies that µ = (2k + 1)

π , 2L

k = 0, 1, . . . .

Plugging this value of k in the equation for T , we find T 0 + µ 2 c2 T = 0

T (t) = Bk e−µ

⇒

2 c2 t

π

= Bk e−((2k+1) 2L )

2 2 c t

.

Forming the product solutions and superposing them, we find that u(x, t) =

∞ X k=0

Bk e−µ

2 c2 t

=

∞ X k=0

π

Bk e−((2k+1) 2L )

2 2 c t

h π i sin (2k + 1) x . 2L

130 Chapter 3 Partial Differential Equations in Rectangular Coordinates To determine the coefficients Bk , we use the initial condition and proceed as in Example 1:

u(x, 0) = f (x) ⇒ f (x) =

∞ X k=0

h π i Bk sin (2k + 1) x ; 2L

h π i ⇒ f (x) sin (2n + 1) x 2L ∞ h h X π i π i = Bk sin (2k + 1) x sin (2n + 1) x 2L 2L k=0

⇒

Z

h π i f (x) sin (2n + 1) x dx 2L 0 Z L ∞ h h X π i π i = Bk sin (2k + 1) x sin (2n + 1) x dx 2L 2L 0 L

k=0

⇒

Z

h π i f (x) sin (2n + 1) x dx 2L 0 Z L h π i = Bn sin2 (2n + 1) x dx, 2L 0 L

where we have integrated the series term by term and used the orthogonality π of the functions sin (2k + 1) 2L x on the interval [0, L]. The orthogonality can be checked directly by verifying that Z 0

L

h h π i π i sin (2k + 1) x sin (2n + 1) x dx = 0 2L 2L

if n 6= k. Solving for Bn and using that Z 0

L

h π i L sin2 (2n + 1) x dx = 2L 2

(check this using a half-angle formula), we find that 2 Bn = L

Z

L 0

h π i f (x) sin (2n + 1) x dx. 2L

6. In order to apply the result of Exercise 5, we must compute the Fourier coefficients of h i the series expansion of f (x) = sin x in terms of the functions (2n+1) sin x (n = 0, 1, 2, . . .) on the interval [0, π]. That is, we must find 2 Bn in the series expansion

sin x =

∞ X n=0

(2n + 1) Bn sin x . 2

Section 3.6

Heat Conduction in Bars: Varying the Boundary Conditions

131

According to Exercise 5, we have Z 2 π (2n + 1) Bn = sin x sin x dx π 0 2 Z 1 π 1 3 = cos[(n − )x] − cos[(n + )x] dx π 0 2 2 " # π 1 1 3 1 1 = sin[(n − sin[(n + )x] − )x] 1 3 π n− 2 2 2 0 n+ 2 3π 2 1 π 1 sin[(nπ + = sin[(nπ − )] − )] π 2n − 1 2 2n + 3 2 2 −(−1)n (−1)n = + π 2n − 1 2n + 3 n+1 8 (−1) = . π (2n − 1)(2n + 3) We can now apply Exercise 5 and get the solution u(x, t) =

∞ 8X (−1)n+1 2 (2n + 1) e−(2n+1) t/4 sin x . π (2n − 1)(2n + 3) 2 n=0

7. In the first part of this solution, we will justify term-by-term integration of the series solution. Recall that the solution is given by u(x, t) = a0 +

∞ X

2

an e−λn t cos

n=1

nπx , L

where an is the cosine Fourier coefficient of f . Since f is a temperature distribution, we can suppose that it is bounded; hence |f (x)| ≤ M for all x, where M is a constant. Hence ≤1

≤M z Z }| { Z nπx 2 L nπ 2 L z }| { |f (x)| cos f (x) cos dx ≤ dx L 0 L L 0 L Z 2 L 2 M dx = L M = 2M, L 0 L

|an | = ≤ and so, for t > 0,

|u(x, t)| ≤ |a0 | +

∞ ∞ X X 2 2 nπx e−λn t < ∞. an e−λn t cos ≤ |a0 | + 2M L n=1

To see that the series e

−λ2n t

tc2 π 2

n=1

P∞

2

n=1

=

e

e−λn t is convergent, write 2 π2 L2

− tc

n 2

0, the series a0 + ∞ a e n=1 n L

132 Chapter 3 Partial Differential Equations in Rectangular Coordinates by the Weierstrass M -test. Thus we can integrate the series term by term (Theorem 3, Section 2.7) and get =a0

=0

z Z }| { z }| { Z ∞ L X 1 L 1 L 2 1 nπx u(x, t) dx = a0 dx + an e−λn t cos dx L 0 L 0 L 0 L n=1 Z 1 L = a0 = f (x) dx, L 0 Z

where the last equality follows from the definition of a0. Thus at any time t > 0, the average of the temperature distribution inside the bar is constant and equals to the average of the initial temperature distribution f (x). This makes sens on physical grounds since the bar is insulated and there is no exchange of heat with the surrounding, the temperature may vary but the total heat inside the bar remains the same, and so the average temperature remains the same at any time t > 0. 8. By Exercise 7, the average temperature in Example 1 is a0 . We now show that the steady-state temperature is also a0. Take the limit as t → ∞, assuming that we can interchange the limit and the summation sign:

lim u(x, t) =

t→∞

lim

t→∞

a0 +

∞ X

2

an e

−( cnπ t L )

n=1

!

nπ cos x L =0

= a0 +

z }| { 2 i nπ −( cnπ ) an cos x lim e L t L t→∞

∞ h X n=1

= a0 . If you really want to justify the interchange of the limit and the summation sign, you can use the estimate that we found in Exercise 7, as follows. Let r = e−

tc2 π 2 L2

. Then ∞ ∞ X X 2 cnπ nπ r an e−( L ) t cos rn = x ≤ . L 1−r n=1 n=1

As t → ∞, r → 0, showing that

lim

t→∞

∞ X n=1

an e− (

) cos nπ x = 0. L

cnπ 2 t L

9. This is a straightforward application of Exercise 7. For Exercise 1 the average is 100. For Exercise 2 the average is a0 = 0. 10. Using the relation sin ax sin bx = 12 cos(a − b)x − cos(a + b)x , we

Section 3.6

Heat Conduction in Bars: Varying the Boundary Conditions

133

obtain, for m 6= n, Z L sin µm x sin µn x dx 0 Z 1 L = cos(µm − µn )x − cos(µm + µn )x dx 2 0 1 1 = sin(µm − µn )L − sin(µm + µn )L 2(µm − µn ) 2(µm + µn ) 1 = sin µm L cos µn L − cos µm L sin µn L] 2(µm − µn ) 1 − sin µm L cos µn L + cos µm L sin µn L 2(µm + µn ) = [µm sin µn L cos µm L − µn cos µn L sin µm L] (µ2n − µ2m ) . We now show that µm sin µn L cos µm L − µn cos µn L sin µm L = 0, by using (7), which states that tan(µL) = − µκ for all µ = µj . In particular, cos µL 6= 0. Dividing through by cos µm L cos µn L 6= 0, we find µm sin µn L cos µm L − µn cos µn L sin µm L = 0 ⇔ µm tan µn L − µn tan µm L = 0 µn µm ⇔ −µm + µn = 0, κ κ which establishes the desired orthogonality relation.

134 Chapter 3 Partial Differential Equations in Rectangular Coordinates 11. To do this problem, as well as Exercises 12-14, we will need the numerical values from Example 3. This will be done entirely on Mathematica. In the first window, we plot f (x), tan x, and −x, and find numerical values for the roots of tan x = −x. f[x_]:=x Plot[f[x],{x,0,1}] 1

0.2

0.4

0.6

0.8

1

Plot[{Tan[x], -x}, {x,0,20}]

approxsol={3,5,8,11,14.3,17.5}; sol=Table[ FindRoot[Tan[x]==-x,{x,approxsol[[j]]}], {j,1,6}] x

2.02876 , x

4.91318 , x

7.97867 , x

11.0855 , x

14.2074 , x

17.3364

In the second window, we form a partial sum of the series solution. This requires computing numerically the generalized Fourier coefficients of f (x). We then evaluate the partial sum solution at time t = 0 (this should approximate f (x)), and plot it against f (x).

Section 3.6

Heat Conduction in Bars: Varying the Boundary Conditions

135

u6[x_,t_]= Sum[c[j] Exp[- sol[[j,1,2]]^2 t] Sin[sol[[j,1,2]] x], {j,1,6}] 4.11586 t

c 1 Sin 2.02876 x c 4 Sin 11.0855 x

122.889 t

24.1393 t

c 2 Sin 4.91318 x c 5 Sin 14.2074 x

201.851 t

63.6591 t 300.55 t

c 3 Sin 7.97867 x c 6 Sin 17.3364 x

c[j_]:=1/Integrate[Sin[sol[[j,1,2]] x]^2,{x,0,1}] *\ Integrate[f[x] Sin[sol[[j,1,2]] x] ,{x,0,1}] tableofcoeff=Table[Chop[c[j]],{j,1,6}] 0.729175,

0.156164, 0.0613973,

0.0321584, 0.0196707,

0.0132429

Plot[Evaluate[{f[x],u6[x,0]}],{x,0,1}, AspectRatio->1.7/2.7,Ticks->None]

In the third window, we plot the series solution at various values of t.

tabplot=Table[ u6[x,t], {t,0,1,.2}]; Plot[Evaluate[{tabplot,.1}],{x,0,1},PlotRange->{0,1}, AspectRatio->1.7/2.7,Ticks->None]

12., 13., and 14. follow by repeating the same commands as in Exercise 11 for the appropriate function f (x). 15. Following the solution in Example 2, we arrive at the following equa-

136 Chapter 3 Partial Differential Equations in Rectangular Coordinates tions T 0 − kc2T = 0, X 00 − kX = 0,

X 0(0) = 0, X 0(L) = −κX(L),

where k is a separation constant. If k = 0 then X 00 = 0 ⇒ X = ax + b, X 0(0) = 0 ⇒ a = 0 X 0(L) = −κX(L) ⇒ b = 0; so k = 0 leads to trivial solutions. If k = α2 > 0, then, as in the text, we conclude that X = 0. This leaves the case k = −µ2 < 0. In this case X(x) = c1 cos µx + c2 sin µx. From X 0(0) = 0 it follows that c2 = 0 or X = c2 cos µx. Set c2 = 1 to simplify the presentation. From X 0(L) = −κX(L), we find that −µ sin µL = −κ cos µL. Hence µ is a solution of the equation µ cos µL = κ sin µL

or

cot µL =

1 µ. κ

Let µn denote the nth positive solution of this equation, and let Xn (x) = cos µn x. Solving the equation for T for each µn , we find 2 µ2 t n

T = Tn = cn e−c

and hence, by superposing the product solutions, we arrive at u(x, t) =

∞ X

2 µ2 t n

cn cos µn xe−c

.

n=1

To prove that the Xn ’s are orthogonal, see Exercise 19. The proof involves ideas that will be studied in greater detail in Chapter 6, in the context of Sturm-Liouville theory. Using the initial condition, it follows that u(x, 0) = f (x) =

∞ X

cn Xn (x).

n=1

Multiplying both sides by Xm and integrating term by term, it follows from the orthogonality of the Xn that Z

∞ X

L

f (x)Xm (x) dx = 0

Hence

n=1

Z

cn

=0

z Z

L

f (x)Xm(x) dx = cn

Z

L

Xn2(x) dx 0

or 1 cn = κn where κn =

Z

L

f (x)Xn (x) dx 0

Z

L

Xn2(x) dx . 0

{

Xn (x)Xm(x) dx . 0

L 0

if m6=n }|

Section 3.6

Heat Conduction in Bars: Varying the Boundary Conditions

137

Let us evaluate κn . Recall that µn is a solution of cot Lµ = κ1 µ. We have κn =

Z

Z 1 L cos µn x dx = (1 + cos 2µn x) dx 2 0 0 L L 1 L 1 sin 2µn x = sin 2µn L. 0 2 4µn 2 4µn

=

L

2

In order to use what we know about µn , namely, that it is a solution of cot Lµ = κ1 µ, we express the sine in terms of the cotangent, as follows. We have sin 2α = 2 sin α cos α = = 2

2 sin α cos α sin2 α + cos2 α

cot α , cot2 α + 1

where the last identity follows upon dividing by cos2 α. Using this in the expression for κn , we find κn = = =

L cot µn L 1 ·2 + 2 4µn 1 + cot2 µn L L (µn /κ) 1 + 2 2µn 1 + (µn /κ)2 L κ + . 2 2 2(κ + µ2n )

16. (a) Take a product solution, u(x, t) = X(x)T (t), plug it into the heat equation ut = c2uxx , proceed formally by using the typical argument of the separation of variables method, and you will get T 0X = c2X 00T T0 c2 T 0

T0 c2 T 2

=k

X 00 X

=

and

T − kc T = 0 and

X 00 X 00

=k

X − kX = 0,

where k is a separation constant. Doing the same with the periodic boundary conditions, you will find X(−π)T (t) = X(π)T (t) and

X 0(−π)T (t) = X 0(π)T (t)

X(−π) = X(π) and

X 0(−π) = X 0(π).

(b) If k = α2 > 0, then the general solution of the X 00 − α2 X = 0 is X = c1 cosh αx + c2 sinh αx (see Appendix A.2). But X(π) = X(−π) ⇒ c1 cosh απ + c2 sinh απ = c1 cosh απ − c2 sinh απ ⇒ c2 sinh απ = 0 ⇒ c2 = 0, because sinh απ 6= 0. Using the second boundary condition, we obtain αc1 sinh απ = −αc1 sinh απ

⇒

c1 = 0.

138 Chapter 3 Partial Differential Equations in Rectangular Coordinates So the case k = α2 > 0 leads only to trivial solutions. (c) If k = 0, then the solution of X 00 = 0 is X = ax + b. The boundary condition X 0(−π) = −X 0(π) implies that a = −a or a = 0. Hence X is constant, and we denote it by a0 . Note that this is an acceptable solution of the differential equation and the periodic boundary conditions. (d) If k = −µ2 < 0, the solution in X is X = c1 cos µx + c2 sin µx. Using the boundary conditions, we have X(−π) = X(π) ⇒ c1 cos µπ − c2 sin µπ = c1 cos µπ + c2 sin µπ ⇒ c2 sin µπ = 0 ⇒ c2 = 0 or

sin µπ = 0.

If c2 = 0, then the condition X 0(−π) = X 0(π) implies that −c2 µ sin µπ = c2µ sin µπ, which in turn implies that sin µπ = 0. So, either way, we have the condition sin µπ = 0, which implies that µ = ±m, where m = 1, 2, 3, . . .. In order not to duplicate solutions, we take µ = m, where m = 1, 2, 3, . . ., and obtain the solutions Xm = am cos mx + bm sin mx. The corresponding solutions in T are T 0 − m 2 c2 T = 0

⇒

T = e−m

2 c2 t

.

Therefore, The product solutions are 2 2 Xm (x)Tm(t) = am cos mx + bm sin mx e−m c t

(m ≥ 1).

Superposing the product solutions, we obtain u(x, t) = a0 +

∞ X

(am cos mx + bm sin mx)e−m

2 c2 t

.

m=1

Setting t = 0 and using the boundary conditions, it follows that u(x, 0) = f (x) = a0 +

∞ X

(am cos mx + bm sin mx).

m=1

Thinking of this as the Fourier series of f , it follows that the an and bn are the Fourier coefficients of f and are given by the Euler formula in Section 2.2. Here we are considering f as a 2π-periodic function of the angle x. 17. Part (a) is straightforward as in Example 2. We omit the details that lead to the separated equations: T 0 − kT = 0, X 00 − kX = 0,

X 0(0) = −X(0), X 0(1) = −X(1),

where k is a separation constant. (b) If k = 0 then X 00 = 0 ⇒ X = ax + b, X 0(0) = −X(0) ⇒ a = −b X 0(1) = −X(1) ⇒ a = −(a + b) ⇒ 2a = −b; ⇒ a = b = 0.

Section 3.6

Heat Conduction in Bars: Varying the Boundary Conditions

139

So k = 0 leads to trivial solutions. (c) If k = α2 > 0, then X 00 − µ2 X = 0 ⇒ X = c1 cosh µx + c2 sinh µx; X 0(0) = −X(0) ⇒ µc2 = −c1 X 0(1) = −X(1) ⇒ µc1 sinh µ + µc2 cosh µ = −c1 cosh µ − c2 sinh µ ⇒ µc1 sinh µ − c1 cosh µ = −c1 cosh µ − c2 sinh µ ⇒ µc1 sinh µ = −c2 sinh µ c1 ⇒ µc1 sinh µ = sinh µ. µ Since µ 6= 0, sinh µ 6= 0. Take c1 6= 0 and divide by sinh µ and get µc1 =

c1 ⇒ µ2 = 1 ⇒ k = 1. µ

So X = c1 cosh x + c2 sinh x. But c1 = −c2 , so X = c1 cosh x + c2 sinh x = c1 cosh x − c1 sinh x = c1 e−x . Solving the equation for T , we find T (t) = et ; thus we have the product solution c0e−x et , where, for convenience, we have used c0 as an arbitrary constant. (d) If k = −α2 < 0, then X 00 + µ2 X = 0 ⇒ X = c1 cos µx + c2 sin µx; X 0(0) = −X(0) ⇒ µc2 = −c1 X 0(1) = −X(1) ⇒ −µc1 sin µ + µc2 cos µ = −c1 cos µ − c2 sin µ ⇒ −µc1 sin µ − c1 cos µ = −c1 cos µ − c2 sin µ ⇒ −µc1 sin µ = −c2 sin µ c1 ⇒ −µc1 sin µ = sin µ. µ Since µ 6= 0, take c1 6= 0 (otherwise you will get a trivial solution) and divide by c1 and get µ2 sin µ = − sin µ ⇒ sin µ = 0 ⇒ µ = nπ, where n is an integer. So X = c1 cos nπx + c2 sin nπx. But c1 = −c2µ, so X = −c1 nπ cos nπx − sin nπx . Call Xn = nπ cos nπx − sin nπx. (e) To establish the orthogonality of the Xn ’s, treat the case k = 1 separately. For k = −µ2 , we refer to the boundary value problem X 00 + µ2n X = 0, X(0) = −X 0 (0), X(1) = −X 0(1), that is satisfied by the Xn ’s, where µn = nπ. We establish orthogonality using a trick from Sturm-Liouville theory (Chapter 6, Section 6.2). Since 00 Xm = µ2m Xm and Xn00 = µ2n Xn ,

140 Chapter 3 Partial Differential Equations in Rectangular Coordinates multiplying the first equation by Xn and the second by Xm and then subtracting the resulting equations, we obtain 00 Xn Xm = µ2m Xm Xn and Xm Xn00 = µ2n Xn Xm 00 Xn Xm − Xm Xn00 = (µ2n − µ2m )XmXn 0 0 X n Xm − Xm Xn0 = (µ2n − µ2m )XmXn

where the last equation follows by simply 0 checking the validity of the identity 00 00 0 0 Xn Xm − Xm Xn = Xn Xm − Xm Xn . So (µ2n

−

µ2m )

Z

Z

0 0 Xn (x)Xm (x) − Xm (x)Xn0 (x) dx 0 1 0 = Xn (x)Xm (x) − Xm (x)Xn0 (x) ,

1

Xm (x)Xn (x) dx = 0

1

0

because the integral of the derivative of a function is the function itself. Now we use the boundary conditions to conclude that 1 0 Xn (x)Xm (x) − Xm (x)Xn0 (x) 0

0 0 = Xn (1)Xm (1) − Xm (1)Xn0 (1) − Xn (0)Xm (0) + Xm (0)Xn0 (0)

= −Xn (1)Xm(1) + Xm (1)Xn(1) + Xn (0)Xm(0) − Xm (0)Xn(0) = 0. Thus the functions are orthogonal. We still have to verify the orthogonality when one of the Xn ’s is equal to e−x . This can be done by modifying the argument that we just gave. (f) Superposing the product solutions, we find that u(x, t) = c0 e−x et +

∞ X

cn Tn (t)Xn (x).

n=1

Using the initial condition, it follows that u(x, 0) = f (x) = c0e

−x

+

∞ X

cn Xn (x).

n=1

The coefficients in this series expansion are determined by using the orthogonality of the Xn ’s in the usual way. Let us determine c0 . Multiplying both sides by e−x and integrating term by term, it follows from the orthogonality of the Xn that Z

1

f (x)e−x dx = c0 0

Hence

Z

1

e−2x dx + 0

Z

1

f (x)e

−x

dx = c0

0

∞ X n=1

Z

cn

=0

z Z

1

c0 =

2e2 e2 − 1

1

e−2x dx = c0

Z

1

f (x)e−x dx. 0

{

Xn (x)e−x dx . 0

0

Thus

}|

1 − e−2 . 2

Section 3.6

Heat Conduction in Bars: Varying the Boundary Conditions

141

In a similar way, we prove that cn =

1 κn

where κn =

Z

1

f (x)Xn (x) dx 0

Z

1

Xn2(x) dx . 0

This integral can be evaluated as we did in Exercise 15 or by straightforward computations, using the explicit formula for the Xn ’s, as follows: Z

1

Xn2(x) dx

=

0

Z

2

1

nπ cos nπx − sin nπx

dx

0

=

Z

n2 π 2 cos2 nπx + sin2 nπx − 2nπ cos(nπx) sin(nπx) dx

1

0

=

=(n2 π 2 )/2

z Z

}|

1

{

n2 π 2 cos2 nπx dx +

=1/2

z Z

0

}|

1

{

sin2 nπx dx 0

−2nπ

=0

zZ

}|

1

{

cos(nπx) sin(nπx) dx 0

=

n2 π 2 + 1 . 2

19. (a) Following the solution in Exercise 16, we arrive at the following equations T 0 − kT = 0, X 00 − kX = 0,

X 0(0) = X(0), X 0(1) = −X(1),

where k is a separation constant. We will not repeat the details. (b) If k = 0 then X 00 = 0 ⇒ X = ax + b, X 0(1) = −X(1) ⇒ a = −a

⇒

a=0

0

X (0) = X(0) ⇒ a = b = 0; so k = 0 leads to trivial solutions. If k = α2 > 0, then X 00 − α2 X = 0

⇒

X = c1 cosh αx + c2 sinh αx,

X(0) = c1 , X 0(0) = αc2 , X(1) = c1 cosh α + c2 sinh α, X 0(1) = αc1 sinh α + αc2 cosh α, X 0(0) = X(0) 0

X (1) = −X(1)

⇒

⇒

c1 = αc2

−(c1 cosh α + c2 sinh α) = αc1 sinh α + αc2 cosh α.

Using αc2 = c1, we obtain from the last equation −(c1 cosh α +

c1 sinh α) = αc1 sinh α + c1 cosh α. α

142 Chapter 3 Partial Differential Equations in Rectangular Coordinates Either c1 = 0, in which case we have a trivial solution, or −(α cosh α + sinh α) = α(α sinh α + cosh α) −2α cosh α = (α2 + 1) sinh α. If α > 0, then the left side is < 0 while the right side is > 0. So there are no positive solutions. Also, if α < 0, then the left side is > 0 while the right side is < 0. So there are no negative solutions. Hence the only solution that satisfies the equation and the boundary conditions is the trivial solution in the case k = α2 > 0. (c) The separation constant must be negative: k = −µ2 < 0. In this case X(x) = c1 cos µx + c2 sin µx. From X 0(0) = X(0) it follows that c1 = µc2 . Hence X(x) = c2µ cos µx + c2 sin µx, where c2 6= 0. From X 0(1) = −X(1), we find that −c2 µ2 sin µ + c2µ cos µ = −c2 µ cos µ − c2 sin µ. Hence µ is a solution of the equation −µ2 sin µ + µ cos µ = −µ cos µ − sin µ

⇒

2µ cos µ = (µ2 − 1) sin µ

Note that µ = ±1 s not a solution and cos µ = 0 is not a possibility, since this would imply sin µ = 0 and the two equations have no common solutions. So we can divide by cos µ(µ2 − 1) and conclude that µ satisfies the equation tan µ = µ22µ . This equation has infinitely many solutions µ = ±µn , that −1 can be seen on a graph. We take only the positive µn since we do not gain any new solutions in X for −µn . Write the corresponding solutions X = Xn = cn (µn cos µn x + sin µn ),

(n = 1, 2, , . . .)

The corresponding solutions in T are 2

Tn (t) = e−µn t . (e) We can prove that the Xn are orthogonal on the interval [0, 1] as follows. (The ideas in the proof will be studied in greater detail in Chapter 6, in the context of Sturm-Liouville theory. They will also appear when we establish the orthogonality of Bessel functions (Chapter 4) and Legendre polynomials (Chapter 5).) We will not need the explicit formulas for Xn and Xm . We will work with the fact hat these are both solutions of the same boundary value problem X 00 + µ2 X = 0, X(0) = X 0(0), X(1) = −X 0 (1), corresponding to different values of µ = µm and µ = µn , with µm 6= µn . So 00 Xm = µ2m Xm and Xn00 = µ2n Xn .

Multiplying the first equation by Xn and the second by Xm and then subtracting the resulting equations, we obtain 00 Xn Xm = µ2m Xm Xn and Xm Xn00 = µ2n Xn Xm 00 Xn Xm − Xm Xn00 = (µ2n − µ2m )XmXn 0 − X X 0 0 = (µ2 − µ2 )X X X n Xm m n m n n m

Section 3.6

Heat Conduction in Bars: Varying the Boundary Conditions

143

where the last equation follows by simply 0 checking the validity of the identity 00 00 0 0 Xn Xm − Xm Xn = Xn Xm − Xm Xn . So (µ2n

−

µ2m )

Z

Z

0 0 Xn (x)Xm (x) − Xm (x)Xn0 (x) dx 0 1 0 (x) − Xm (x)Xn0 (x) , = Xn (x)Xm

1

Xm (x)Xn (x) dx = 0

1

0

because the integral of the derivative of a function is the function itself. Now we use the boundary conditions to conclude that 1 0 Xn (x)Xm (x) − Xm (x)Xn0 (x) 0

0 0 = Xn (1)Xm (1) − Xm (1)Xn0 (1) − Xn (0)Xm (0) + Xm (0)Xn0 (0)

= −Xn (1)Xm(1) + Xm (1)Xn(1) − Xn (0)Xm(0) + Xm (0)Xn(0) = 0. Thus the functions are orthogonal. (f) Superposing the product solutions, we find that u(x, t) =

∞ X

cn Tn (t)Xn (x).

n=1

Using the initial condition, it follows that u(x, 0) = f (x) =

∞ X

cn Xn (x).

n=1

Multiplying both sides by Xm and integrating term by term, it follows from the orthogonality of the Xn that Z

1

f (x)Xm (x) dx = 0

Hence

∞ X n=1

Z

cn

=0

z Z

1

f (x)Xm(x) dx = cn

Z

1

Xn2(x) dx 0

or 1 cn = κn where κn =

Z

1

f (x)Xn (x) dx 0

Z

1

Xn2(x) dx . 0

{

Xn (x)Xm(x) dx . 0

1

0

if m6=n }|

144 Chapter 3 Partial Differential Equations in Rectangular Coordinates

Solutions to Exercises 3.7 3. We use the double sine series solution that is outlined in the text. ∞ ∞ X X mπx nπy ∗ u(x, y, t) = (Bmn cos λmn t + Bmn sin λmn t) sin sin , a b n=1 m=1

where a and b are the dimensions of the membrane, r m2 n 2 λmn = cπ + 2, a2 b Z bZ a 4 mπx nπy Bmn = f (x, y) sin sin dx dy, ab 0 0 a b Z bZ a mπx 4 nπy ∗ Bmn = g(x, y) sin sin dx dy. abλmn 0 0 a b Applying these formulas in our case, we find p λmn = m2 + n2 , ( 2 √2 ∗ λ1,2 = 5 if (m, , n) = (1, 2), Bmn = 0 otherwise, because of the orthogonality of the functions sin mπx sin nπy and the fact that g contains only the function sin πx sin 2ıy. We also have Z 1Z 1 Bmn = 4 x(1 − x)y(1 − y) sin mπx sin nπy dx dy 0 0 Z 1 Z 1 = 4 x(1 − x) sin mπx dx y(1 − y) sin nπy dy 0 0 ( 64 if m and n are both odd, (mnπ 2 )3 = 0 otherwise. Refer to Example 1 of this section or Exercise 9, Section 3.3, for the computation of the definite integrals. Thus u(x, y, t) = ∞ X ∞ X p 64 sin((2k + 1)πx) sin((2l + 1)πy) cos (2k + 1)2 + (2l + 1)2t 2 3 ((2k + 1)(2l + 1)π ) k=0 l=0 √ 2 + √ sin πx sin(2πy) sin( 5 t). 5 5. We proceed as in Exercise 3. We have u(x, y, t) = √ where λmn = ∗ Bmn

∞ ∞ X X

∗ (Bmn cos λmn t + Bmn sin λmn t) sin mπx sin nπy,

n=1 m=1

m2 + n2 , Bmn = 0, and Z 1Z 1 4 = √ sin mπx sin nπy dx dy m2 + n 2 0 0 Z 1 Z 1 4 = √ sin mπx dx sin nπy dy m2 + n 2 0 0 ( 16 √ if m and n are both odd, m2 +n2 (mn)π 2 = 0 otherwise.

Section 3.7

145

The Two Dimensional Wave and Heat Equations

Thus u(x, y, t) = ∞ X ∞ X k=0 l=0

p

p 16 sin((2k + 1)πx) sin((2l + 1)πy) sin (2k + 1)2 + (2l + 1)2t (2k + 1)2 + (2l + 1)2 (2k + 1)(2l + 1)π 2

12. We use the double sine series solution that is outlined in the text. u(x, y, t) =

∞ ∞ X X n=1 m=1

Amn sin

mπx nπy −λ2mn t , sin e a b

where a and b are the dimensions of the membrane, r m2 n 2 λmn = cπ + 2, a2 b Z Z 4 b a mπx nπy Amn = f (x, y) sin sin dx dy. ab 0 0 a b Applying these formulas in our case, we find p λmn = π m2 + n2 , Z 1Z 1 Amn = 4 x(1 − x)y(1 − y) sin mπx sin nπy dx dy 0 0 ( 64 (mnπ 2 )3 if m and n are both odd, = 0 otherwise, We have used the result of Exercise 3. Thus u(x, y, t) = ∞ X ∞ X 64 sin((2k + 1)πx) sin((2l + 1)πy) − e ((2k + 1)(2l + 1)π 2)3 k=0 l=0

(2k+1)2 +(2l+1)2 π 2 t

.

146 Chapter 3 Partial Differential Equations in Rectangular Coordinates

Solutions to Exercises 3.8 1. The solution is given by u(x, y) =

∞ X

Bn sin(nπx) sinh(nπy),

n=1

where Bn = = =

Z 1 2 x sin(nπx) dx sinh(2nπ) 0 2 x cos(n π x) sin(n π x) 1 − + sinh(2nπ) nπ n2 π 2 0 n n+1 2 −(−1) 2 (−1) = . sinh(2nπ) n π sinh(2nπ) n π

Thus, u(x, y) =

∞ 2 X (−1)n+1 sin(nπx) sinh(nπy). π n sinh(2nπ) n=1

3. The solution is given by u(x, y) =

∞ X

∞

An sin(

n=1

X nπx nπ Dn sin(nπy) sinh(nπx), ) sinh( (1 − y)) + 2 2 n=1

where Z 2 2 nπx 100 sin( ) dx nπ sinh( 2 ) 0 2 2 200 nπx 2 = − cos sinh( nπ nπ 2 0 2 ) 200 = (1 − cos(−1)n ) nπ 400 if n is odd, nπ = 0 if n is even;

An =

and Dn = = =

Z 1 2 100(1 − y) sin(nπy) dy sinh(2nπ) 0 200 (−1 + y) cos(n π y) sin(n π y) 1 − sinh(2nπ) nπ n2 π 2 0 200 nπ sinh(2nπ)

Thus, u(x, y) =

∞ ∞ (2k+1)πx (2k+1)π ) sinh( 2 (1 − y)) 200 X sin(nπy) sinh(nπx) 400 X sin( 2 . + π π nπ sinh(2nπ) (2k + 1) sinh( (2k+1)π ) n=1 k=0 2

7. (a) From Example 1, the solution is u(x, y) =

∞ X n=1

Bn sin

nπx nπy sinh , a a

Section 3.8

Laplace’s Equation in Rectangular Coordinates

147

where Bn =

2 a sinh( nπb a )

Z

a

f2 (x) sin 0

nπx dx a

bn

= =

z Z }| { nπx 2 a 1 f2 (x) sin dx a 0 a sinh( nπb a ) 1 bn , sinh( nπb a )

where bn is the sine Fourier coefficient of f2 (x), as given by (3), Section 2.4. Replacing Bn by its value in terms of bn in the series solution, we obtain (a). In the general case, we refer the series solution (9). It is easy to see, as we just did, that bn (f1 )

An =

z Z }| { 2 a 1 nπx 1 f1 (x) sin dx = bn (f1 ), nπb a a sinh( a ) sinh( nπb 0 a ) bn (g1 )

Cn =

z }| { Z b 2 1 nπy 1 g1 (y) sin dy = nπa bn (g1 ), sinh( nπa ) b b sinh( 0 b b ) bn (g2 )

Dn =

z Z }| { 2 b 1 nπy 1 g2 (y) sin dy = bn (g2 ), nπa sinh( b ) b 0 b sinh( nπa b )

where, bn (f ) denotes the nth sine Fourier coefficient of f . So, in terms of the Fourier coefficients of the boundary functions, the general solution (9) becomes u(x, y) =

∞ X n=1 ∞ X n=1

∞

bn (f1 ) sin

X nπ sinh nπ nπ sinh nπ a (b − y) a y + b (f ) sin x x n 2 nπb nπb a a sinh a sinh a

bn (g1 ) sin

nπ sinh nπ b (a − y b sinh nπa b

x)

+

n=1 ∞ X n=1

bn (g2) sin

nπ sinh nπ b x y nπa . b sinh b

148 Chapter 3 Partial Differential Equations in Rectangular Coordinates

Solutions to Exercises 3.9 1. We apply (2), with a = b = 1: u(x, y) =

∞ ∞ X X

Emn sin mπx sin nπy,

n=1 m=1

where Emn =

−4 2 π (m2 + n2 )

1Z 1

Z 0

x sin mπx sin nπy dx dy

0 =

= = =

1−(−1)n nπ

zZ }| { Z 1 1 −4 x sin mπx dx sin nπy dy π 2 (m2 + n2 ) 0 0 −4 1 − (−1)n x cos(m π x) sin(m x) 1 − + π 4 (m2 + n2 ) n m m2 π 0 n m 4 1 − (−1) (−1) . 4 2 2 π (m + n ) n m

Thus u(x, y) =

∞ ∞ 8 XX (−1)m sin mπx sin((2k + 1)πy). π4 (m2 + (2k + 1)2)m(2k + 1) k=0 m=1

3. We use the solution in the text: u(x, y) = u1(x, y) + u2 (x, y) =

∞ ∞ X X

Emn sin mπx sin nπy + u2 (x, y),

n=1 m=1

where u1 is the solution of an associated Poisson problem with zero boundary data, and u2 is the solution of the Dirichlet problem with the given boundary data. We have Z 1Z 1 −4 Emn = sin πx sin mπx sin nπy dx dy π 2(m2 + n2 ) 0 0 0

if

m6=1, 1/2

if

m=1

=

1−(−1)n nπ

z }| {Z z }| { Z 1 1 −4 = sin πx sin mπx dx sin nπy dy π 2(m2 + n2 ) 0 0 ( 0 if m 6= 1, = . (1−(−1)n ) −2 if m = 1 n π 3 (1+n2 ) Thus ∞

X 1 −4 u1(x, y) = 3 sin(πx) sin((2k + 1)πy). π (1 + (2k + 1)2)(2k + 1) k=0

To find u2 , we apply the results of Section 3.8. We have ∞ X n=1

Bn sin nπx sinh nπy,

Section 3.9

Poisson’s Equation: The Method of Eigenfunction Expansions

149

where Bn = = Thus

Z 1 2 x sin nπx dx sinh(nπ) 0 2 (−1)n+1 . sinh(nπ) nπ

∞ 2 X (−1)n+1 sin nπx sinh nπy, π n=1 n sinh(nπ)

and so ∞

u(x, y) =

X −4 1 sin(πx) sin((2k + 1)πy) 3 π (1 + (2k + 1)2)(2k + 1) k=0

∞ 2 X (−1)n+1 + sin nπx sinh nπy. π n=1 n sinh(nπ)

5. We will use an eigenfunction expansion based on the eigenfunctions 2 2 2 φ(x, y) = sin mπx sin nπy, where ∆π(x, y) = −π m +n sin mπx sin nπy. So plug ∞ ∞ X X u(x, y) = Emn sin mπx sin nπy n=1 m=1

into the equation ∆u = 3u − 1, proceed formally, and get P P∞ P∞ P ∞ ∆( ∞ n=1 m=1 Emn sin mπx sin nπy) = 3 n=1 m=1 Emn sin mπx sin nπy − 1 P ∞ P∞ P∞ P∞ Emn sin mπx sin nπy − 1 n=1 m=1 Emn ∆ (sin mπx sin nπy) = 3 n=1 m=1 P ∞ P∞ 2 2 2 −Emn π m + n sin mπx sin nπy n=1 Pm=1 ∞ P∞ = 3 n=1 m=1 Emn sin mπx sin nπy − 1 P ∞ P∞ 2 2 2 Emn sin mπx sin nπy = 1. n=1 m=1 3 + π m + n Thinking of this as the double sine series expansion of the function identically 1, it follows that 3 + π 2 m2 + n2 Emn are double Fourier sine coefficients, given by (see (8), Section 3.7) Z 1Z 1 3 + π 2 m2 + n2 Emn = 4 sin mπx sin nπy dx dy 0

0

1 − (−1)m 1 − (−1)n = 4 mπ nπ 0 if either m or n is even, = . 16 if both m and n are even. π2 m n Thus Emn =

(

0 16 π 2 m n 3+π 2 m2 +n2

if either m or n is even, if both m and n are even,

and so ∞

u(x, y) =

∞

16 X X sin((2k + 1)πx) sin((2l + 1)πy) . 2 π (2k + 1) (2l + 1) 3 + π 2 (2k + 1)2 + (2l + 1)2 k=0 l=0

150 Chapter 3 Partial Differential Equations in Rectangular Coordinates 8. Following the outline in the text, leading to the solution (21), we first compute Z 1 2(−1)m+1 bm(y) = 2 x sin mπx dx = ; mπ 0 then u(x, y) =

∞ X

Em (y) sin mπx,

m=1

where Em (y) =

Z y h −1 2(−1)m+1 sinh(mπs) sinh(mπ(1 − y)) ds + mπ sinh(mπ) mπ 0 Z 1 2(−1)m+1 i sinh(mπy) sinh(mπ(1 − s)) ds mπ y 1 (cosh(mπy)−1) mπ

=

h 2(−1)m sinh(mπ(1 − y)) m2 π 2 sinh(mπ)

+ sinh(mπy)

zZ

1 (cosh(mπ(1−y))−1) mπ

1

}|

zZ

y

}|

{

sinh(mπs) ds

0

{

sinh(mπ(1 − s)) ds

i

y

h 2(−1)m = sinh(mπ(1 − y))(cosh(mπy) − 1) m3 π 3 sinh(mπ) i + sinh(mπy)(cosh(mπ(1 − y)) − 1) h i 2(−1)m = sinh(mπ) − sinh(mπ(1 − y)) − sinh(mπy) , m3 π 3 sinh(mπ) where in the last step we used the identity sinh(a + b) = sinh a cosh b + cosh a sinh b. So ∞ 2 X (−1)m+1 u(x, y) = 3 sinh(mπ)−sinh(mπ(1−y))−sinh(mπy) sin mπx. 3 π m=1 m sinh(mπ)

Section 3.10

Neumann and Robin Conditions

151

Solutions to Exercises 3.10 1. We use a combination of solutions from (2) and (3) and try a solution of the form u(x, y) =

∞ X

sin mx Am cosh m(1 − y) + Bm sinh my .

n=1

(If you have tried a different form of the solution, you can still do the problem, but your answer may look different from the one derived here. The reason for our choice is to simplify the computations that follow.) The boundary conditions on the vertical sides are clearly satisfied. We now determine Am and Bm so as to satisfy the conditions on the other sides. Starting with u(1, 0) = 100, we find that ∞ X

100 =

Am cosh m sin mx.

m=1

Thus Am cosh m is the sine Fourier coefficient of the function f (x) = 100. Hence Z 2 π 200 Am cosh m = 100 sin mx dx ⇒ Am = [1 − (−1)m ] . π 0 πm cosh m Using the boundary condition uy (x, 1) = 0, we find 0=

∞ X m=1

sin mx Am (−m) sinh[m(1 − y)] + mBm cosh my

Thus 0=

∞ X

y=1

.

mBm sin mx cosh m.

m=1

By the uniqueness of Fourier series, we conclude that mBm cosh m = 0 for all m. Since m cosh m 6= 0, we conclude that Bm = 0 and hence u(x, y) = =

∞ 200 X [1 − (−1)m ] sin mx cosh[m(1 − y)] π m cosh m

400 π

m=1 ∞ X k=0

sin[(2k + 1)x] cosh[(2k + 1)(1 − y)]. (2k + 1) cosh(2k + 1)

2. Try a solution of the form u(x, y) =

∞ X

sin mx Am cosh m(1 − y) + Bm sinh my .

n=1

Using the boundary conditions on the horizontal sides, starting with u(x, 0) = 100, we find that ∞ X 100 = Am cosh m sin mx. n=1

152 Chapter 3 Partial Differential Equations in Rectangular Coordinates Thus Am cosh m is the sine Fourier coefficient of the function f (x) = 100 (as in the previous exercise). Hence Am =

200 [1 − (−1)m ] . πm cosh m

Using the boundary condition uy (x, 1) = sin 2x, we find sin 2x =

∞ X

mBm sin mx cosh m.

n=1

By the uniqueness of Fourier series, we conclude that mBm cosh m = 0 for 1 all m 6= 2 and 2B2 cosh 2 = 1 or B2 = 2 cosh 2 . Thus u(x, y) =

1 sin(2x) sinh(2y) 2 cosh 2 ∞ 400 X sin[(2k + 1)x] + cosh[(2k + 1)(1 − y)]. π (2k + 1) cosh(2k + 1) k=0

3. Try a solution of the form u(x, y) =

∞ X

sin mx Am cosh m(1 − y) + Bm sinh my .

n=1

Using the boundary conditions on the horizontal sides, starting with uy (x, 1) = 0, we find ∞ X 0= mBm sin mx cosh m. n=1

Thus Bm = 0 for all m and so u(x, y) =

∞ X

sin mxAm cosh m(1 − y).

n=1

We now work with the Robin condition u(x, 0) + 2uy (x, 0) = g(x) and get ∞ X

sin mxAm cosh m − 2

n=1

∞ X

sin mxAm m sinh m = g(x)

n=1

⇒ g(x) =

∞ X

Am sin mx cosh m − 2m sinh m

n=1

⇒ Am ⇒ Am

Z 2 π cosh m − 2m sinh m = g(x) sin mx dx π 0 Z π 2 = g(x) sin mx dx, π[cosh m − 2m sinh m] 0

which determines the solution. 5. We combine solutions of different types from Exercise 4 and try a solution of the form u(x, y) = A0 + B0 y +

∞ X m=1

cos

mπ mπ mπ x Am cosh[ (b − y)] + Bm sinh[ y] . a a a

Section 3.10

Neumann and Robin Conditions

153

Using the boundary conditions on the horizontal sides, starting with uy (x, b) = 0, we find that 0 = B0 +

∞ X mπ mπ mπ Bm cos x cosh b . a a a

m=1

Thus B0 = 0 and Bm = 0 for all m ≥ 1 and so ∞ X

A0 +

Am cos

m=1

mπ mπ x cosh[ (b − y)]. a a

Now, using u(x, 0) = g(x), we find g(x) = A0 +

∞ X

Am cosh[

m=1

mπ mπ b] cos x. a a

Recognizing this as a cosine series, we conclude that Z 1 a A0 = g(x) dx a 0 and

mπ 2 Am cosh[ b] = a a

Z

a

g(x) cos

mπ x dx; a

g(x) cos

mπ x dx. a

0

equivalently, for m ≥ 1, Am

2 = a cosh[ mπ a b]

Z

a 0

7. The solution is immediate form (5) and (6). We have u(x, y) =

∞ X

Bm sin mx sinh my,

m=1

where Bm are given by (6) with f (x) = sin 2x. By the orthogonality of the 1 trigonometric system, we have Bm = 0 for all m 6= 2 and B2 = 2 cosh(2π) . Thus 1 u(x, y) = sin 2x sinh 2y. 2 cosh(2π) 9. We follow the solution in Example 3. We have u(x, y) = u1 (x, y) + u2 (x, y), where

∞ X

u1 (x, y) =

Bm sin mx sinh my,

m=1

with Bm =

2 πm cosh(mπ)

Z

π

sin mx dx = 0

and u2(x, y) =

∞ X m=1

2 (1 − (−1)m ); πm2 cosh(mπ)

Am sin mx cosh[m(π − y)],

154 Chapter 3 Partial Differential Equations in Rectangular Coordinates with Am

2 = π cosh(mπ)

Z

π

sin mx dx = 0

2 (1 − (−1)m ). πm cosh(mπ)

Hence ∞ 2 X (1 − (−1)m ) sinh my u(x, y) = sin mx + cosh[m(π − y)] π m cosh(mπ) m m=1 ∞ 4X sin(2k + 1)x sinh[(2k + 1)y] = + cosh[(2k + 1)(π − y)] . π (2k + 1) cosh[(2k + 1)π] (2k + 1) k=0

Section 4.1

The Laplacian in Various Coordinate Systems

155

Solutions to Exercises 4.1 1. We could use Cartesian coordinates and compute ux , uy , uxx , and uyy directly from the definition of u. Instead, we will use polar coordinates, because the expression x2 + y 2 = r2, simplifies the denominator, and thus it is easier to take derivatives. In polar coordinates, u(x, y) =

x2

x r cos θ cos θ = = = r−1 cos θ . 2 2 +y r r

So ur = −r−2 cos θ,

urr = 2r−3 cos θ,

uθ = −r−1 sin θ,

uθθ = −r−1 cos θ.

Plugging into (1), we find 1 1 2 cos θ cos θ cos θ ∇2 u = urr + ur + 2 uθθ = − 3 − 3 = 0 (if r 6= 0). r r r3 r r If you used Cartesian coordinates, you should get uxx =

2x(x2 − 3y 2 ) (x2 + y 2 )2

and

uyy = −

2x(x2 − 3y 2 ) . (x2 + y 2)2

2. Reasoning as in Exercises 1, we will use polar coordinates: u(x, y) =

y r sin θ sin θ = = = r−1 sin θ . x2 + y 2 r2 r

So ur = −r−2 sin θ,

urr = 2r−3 sin θ,

uθ = r−1 cos θ,

uθθ = −r−1 sin θ.

Plugging into (1), we find 1 1 2 sin θ sin θ sin θ ∇2u = urr + ur + 2 uθθ = − 3 − 3 = 0 (if r 6= 0). r r r3 r r 3. In polar coordinates: u(r, θ) =

1 1 2 ⇒ ur = − 2 , urr = 3 , uθ = 0, uθθ = 0. r r r

Plugging into (1), we find 1 1 2 1 1 ∇2 u = urr + ur + 2 uθθ = 3 − 3 = 3 r r r r r

(for r 6= 0).

Thus u does not satisfy Laplace’s equation. 4. In cylindrical coordinates: u(ρ, φ, z) =

z 2z ⇒ uρρ = 3 , uφφ = 0, uzz = 0. ρ r

Plugging into (2), we find

∇2 u =

∂ 2 u 1 ∂u 2z z z 1 ∂ 2u ∂ 2u + + 2 2 + 2 = 3 − 3 +0+0= 3 2 ∂ρ ρ ∂ρ ρ ∂φ ∂z r r r

(for ρ 6= 0).

156

Chapter 4

Partial Differential Equations in Polar and Cylindrical Coordinates

5. In spherical coordinates: u(r, θ, φ) = r3 ⇒ urr = 6r, uθ = 0, uθθ = 0, uφφ = 0. Plugging into (3), we find ∇2 u =

2 1 ∂ 2u ∂ 2u 2 ∂u ∂u 2 ∂ u + + csc = 6r + 6r = 12r. + + cot θ θ ∂r2 r ∂r r2 ∂θ2 ∂θ ∂φ2

6. In polar coordinates: u(x, y) = ln(x2 + y 2 ) = 2 ln r; 1 1 ∇2 u = urr + ur + 2 uθθ r r 2 2 = − 2 + 2 + 0 = 0 (for r 6= 0). r r 7. In spherical coordinates: u(r, θ, φ) = r−1 ⇒ ur = −

1 2 , urr = 3 , uθ = 0, uθθ = 0, uφφ = 0. 2 r r

Plugging into (3), we find ∇2 u =

2 2 2 ∂ 2 u 2 ∂u 1 ∂ 2 u ∂u 2 ∂ u = 3 − 3 = 0 (for r 6= 0). + +cot θ θ + +csc ∂r2 r ∂r r2 ∂θ2 ∂θ ∂φ2 r r

8. In polar coordinates: u(r, θ) =

θ sin θ r

θ sin θ 2θ sin θ , urr = , 2 r r3 sin θ + θ cos θ 2 cos θ − θ sin θ uθ = , uθθ = . r r

⇒ ur = −

Plugging into (1), we find ∇2 u =

2θ sin θ θ sin θ 2 cos θ − θ sin θ 2 cos θ − + = r3 r3 r3 r3

(for r 6= 0).

Thus u does not satisfy Laplace’s equation. 9. (a) If u(r, θ, φ) depends only on r, then all partial derivatives of u with respect to θ and φ are 0. So (3) becomes ∇2 u =

2 ∂ 2u 2 ∂u ∂ 2u 2 ∂u ∂u 1 ∂ 2u 2 ∂ u = + + cot θ θ + + + csc . ∂r2 r ∂r r2 ∂θ2 ∂θ ∂φ2 ∂r2 r ∂r

(b) If u(r, θ, φ) depends only on r and θ, then all partial derivatives of u with respect to φ are 0. So (3) becomes ∇2 u =

∂ 2 u 2 ∂u ∂u 1 ∂ 2u + + cot θ + . ∂r2 r ∂r r2 ∂θ2 ∂θ

10. From (7), we have 1 1 1 1 ∇2 u = urr + ur + 2 uθθ + uρ + 2 uφφ . r r ρ ρ

Section 4.1

The Laplacian in Various Coordinate Systems

157

1

Now ρ = r sin θ, r = (ρ2 + z 2 ) 2 , z = r cos θ, θ = tan−1 zρ . Hence, by the chain rule in two dimensions, u ρ = u r rρ + u θ θρ =

ρ z cos θ uθ = sin θur + ur + 2 uθ . 2 r ρ +z r

So 1 ∇ u = urr + ur + r 2 = urr + ur + r 2

1 1 1 cos θ uθθ + sin θur + uθ + 2 2 uφφ r2 r sin θ r r sin θ 1 uθθ + cot θ uθ + csc2 θ uφφ , r2

which is (8). 11. (a) This part follows from the fact that differentiation is a linear operation. Hence ∇2(αu + βv) = α∇2 (u) + β∇2 (v). So if ∇2 u = 0 and ∇2v = 0, then ∇2 (αu + βv) = 0, implying that αu + βv is harmonic. (b) Take u(x, y) = x and v(x, y) = x. Clearly, uxx = uyy = 0 so u and hence v are harmonic. But uv = x2 ; so (uv)xx = 2 and (uv)yy = 0. Hence ∇2 (uv) = 2, implying that uv is not harmonic. (c) If u is harmonic then uxx + uyy = 0. We have (u2)xx = (2uux )x = 2(ux)2 +2uuxx

(u2)yy = (2uuy )y = 2(uy )2 +2uuyy .

and

If u2 is also harmonic, then 0 = (u2 )xx + (u2)yy

= 2(ux )2 + 2uuxx + 2(uy )2 + 2uuyy =0

z }| { = 2((ux ) + (uy ) ) + 2u (uxx + uyy ) 2

2

= 2((ux )2 + (uy )2), which implies that both ux and uy are identically 0. This latter condition implies that u is constant on its domain of definition, if the domain is connected. This result is not true if the domain of the function u is not connected. For example, suppose that u(x, y) = 1 if x > 1 and u(x, y) = −1 if x < −1. Clearly, u is not constant and both u and u2 are harmonic. (For the definition of connected set, see my book “Applied Complex Analysis and Partial Differential Equations.”) Examples of connected sets are any disk, the plane, the upper half-plane, any triangular region any annulus. (d) This part is similar to (c). We suppose that the domain of definition of the functions is connected. (Here again, the result is not true if the domain is not connected.) Suppose that u, v, and u2 + v 2 are harmonic. Then uxx + uyy = 0; =0

vxx + vyy = 0; =0

z }| { z }| { 2((ux) + (uy ) ) + 2u (uxx + uyy ) +2((vx )2 + (vy )2 ) + 2v (vxx + vyy ) . 2

2

The last equation implies that (ux )2 + (uy )2 + (vx )2 + (vy )2 = 0, which, in turn, implies that ux = uy = vx = vy = 0. Hence u and v are constant.

158

Chapter 4

Partial Differential Equations in Polar and Cylindrical Coordinates

Solutions to Exercises 4.2 1. We appeal to the solution (5) with the coefficients (6). Since f(r) = 0, then An = 0 for all n. We have Z 2 1 αn r J0 ( )r dr Bn = 2 αnJ1(αn ) 0 2 Z αn 4 αn = J0(s)s ds (let s = r) α3nJ1(αn )2 0 2 αn 4 = [sJ1 (s)] α3nJ1(αn )2 0 4 = for all n ≥ 1. α2nJ1(αn ) Thus u(r, t) = 4

∞ X J0 ( α2n r ) αn t sin( ). 2 α J (αn ) 2 n=1 n 1

2. We appeal to the solution (5) with the coefficients (6). Since f(r) = 1 − r2, we can use the result of Example 2 and get An =

8 α3nJ1 (αn)

.

For the Bn ’s, we can compute as in Exercise 1 and get Z 1 1 Bn = J0(αn r)r dr 5αnJ1 (αn)2 0 Z αn 1 = J0 (s)s ds (let s = αn r) 5α3nJ1 (αn)2 0 αn 1 = [sJ (s)] 1 3 2 5αnJ1 (αn) 0 1 = for all n ≥ 1. 5α2nJ1 (αn) Thus u(r, t) =

∞ X J0 (αnr) [40 cos(10αnt) + αn sin(10αnt)] . 3 J (α ) 5α n n 1 n=1

3. As in Exercise 1, f(r) = 0 ⇒ An = 0 for all n. We have Z 12 2 Bn = J0 (αnr)r dr αnJ1(αn )2 0 Z αn /2 2 = J0 (s)s ds α3nJ1(αn )2 0 αn /2 2 = [sJ (s)] 1 3 2 αnJ1(αn ) 0 J1 ( α2n ) = for all n ≥ 1. α2nJ1(αn )2 Thus u(r, t) =

∞ X

J1 ( α2n ) J0(αn r) sin(αn t). α2 J (αn)2 n=1 n 1

4. As in Exercise 1, f(r) = 0 ⇒ An = 0 for all n. We have Z 1 2 Bn = J0 (α3r)J0 (αnr)r dr = 0 for n 6= 3 by orthogonality. αnJ1(αn )2 0

Section 4.2

Vibrations of a Circular Membrane: Symmetric Case

159

For n = 3, 2 B3 = α3J1 (α3)2

Z

1

J0 (α3r)2 r dr = 0

1 2 1 , J1 (α3)2 = 2 α3J1 (α3) 2 α3

where we have used the orthogonality relation (12), Section 4.8, with p = 0. Thus u(r, t) =

1 J0 (α3r) sin(α3t). α3

5. Since g(r) = 0, we have Bn = 0 for all n. We have Z 1 2 An = J0(α1 r)J0(αn r)r dr = 0 for n = 6 1 by orthogonality. J1 (αn)2 0 For n = 1, 2 A1 = J1 (α1)2

Z

1

J0(α1 r)2r dr = 1, 0

where we have used the orthogonality relation (12), Section 4.8, with p = 0. Thus u(r, t) = J0 (α1r) cos(α1t). 6. a = 2, c = 1, f(r) = 1 − r, and g(r) = 0. We have g(r) = 0

⇒

Bn = 0 for all n;

Also, An

= = = = = = = =

1 2J1 (αn)2

Z

2

(1 − r)J0 (αnr/2)r dr 0

Z αn 2 2s 1− J0 (s)s ds (let s = αn r/2) α2nJ1 (αn)2 0 αn Z αn αn 2 2 2 sJ (s) − s J (s) ds 1 1 α2nJ1 (αn)2 αn 0 0 αn Z αn 2 4 2 s J (s) − sJ (s) ds − 3 1 1 αnJ1 (αn) αnJ1 (αn)2 0 0 Z αn −2 4 sJ1 (s) ds + αnJ1 (αn) α3nJ1 (αn)2 0 Z αn −2 4 sJ00 (s) ds − 3 αnJ1 (αn) αnJ1 (αn)2 0 αn Z αn h i −2 4 sJ (s) − J (s) ds − 3 0 0 αnJ1 (αn) αnJ1 (αn)2 0 0 Z αn −2 4 J0 (s) ds + αnJ1 (αn) α3nJ1 (αn)2 0

Thus u(r, t) = 2

∞ X 2 n=1

R αn 0

J0 (s) ds − α2n J1(αn) αnr J0 cos α3nJ1 (αn)2 2

αn t 2

.

7. Using orthogonality as in Exercise 5, since f(r) = J0(α3 r), we find that An = 0 for all n 6= 3 and A3 = 1. For the Bn ’s, we have Z 1 2 Bn = (1 − r2)J0 (αnr)r dr αnJ1 (αn)2 0 Z αn 2 = (α2n − s2 )J0(s)s ds (let s = αnr). α5nJ1 (αn)2 0

160

Chapter 4

Partial Differential Equations in Polar and Cylindrical Coordinates

The last integral can be evaluated by parts: u = α2n − s2 , du = −2s, J0 (s)s ds = dv, v = sJ1 (s). So Z αn αn 2 2 2 2 Bn = (αn − s )sJ1 (s) 0 + 2 s J1 (s) ds α5nJ1(αn )2 0 Z αn 4 s2 J1(s) ds = α5nJ1(αn )2 0 αn 4 2 = s J2 (s) (by (7), Section 4.8) α5nJ1(αn )2 0 4J2(αn ) = α3nJ1(αn )2 8 2 = J1(αn)). (using J2 (αn) = α4nJ1(αn ) αn Thus u(r, t) = J0 (α3r) cos(α3 t) + 8

∞ X J0 (αnr) sin(αnt). α4 J (αn) n=1 n 1

8. Since g(r) = 0, we have Bn = 0 for all n. We have Z 1 1 2 An = (3 − 4r2 + r4)J0 (αn r)r dr 2 J1 (αn) 0 128 Z αn 1 = (3α4n − 4α2ns2 + s4 )J0 (s)s ds 64α6nJ1 (αn)2 0

(let s = αnr).

We now integrate by parts: u = 3α4n −4α2n s2 +s4 , du = 4(s3 −2α2ns) ds, J0(s)s ds = dv, v = sJ1 (s). Then Z αn αn 1 4 2 2 4 2 2 2 An = (3αn − 4αns + s )sJ1 (s) 0 − 4 (s − 2αn)s J1(s) ds 64α6nJ1(αn )2 0 Z αn 1 = − (s2 − 2α2n)s2 J1 (s) ds 16α6nJ1 (αn)2 0 Integrate by parts again: u = s2 − 2α2n, du = 2s ds, s2 J1(s) ds = dv, v = s2 J2 (s). Then Z αn αn 1 2 2 2 3 An = − (s − 2αn)s J2 (s) 0 − 2 s J2 (s) ds 16α6nJ1(αn )2 0 h αn i 1 4 3 = − −α J (α ) − 2 s J (s) ds 2 n 3 n 0 16α6nJ1(αn )2 1 = [αnJ2(αn ) + 2J3(αn)] . 16α3nJ1 (αn)2 Now recall that Jp−1 (x) + Jp+1 (x) =

2p Jp (x). x

Take p = 1 and x = αn, and get J3 (αn) = Hence αn J2(αn) + 2J3 (αn) =

4 J2(αn ) − J1(αn ). αn

8 αn J2 (αn )

An =

=

16 α2n J1 (αn ).

Therefore,

1 α5nJ1(αn )

and the solution becomes u(r, t) =

∞ X J0 (αnr) cos(αnt). 5 J (α ) α n n=1 n 1

Section 4.2

Vibrations of a Circular Membrane: Symmetric Case

161

9. (a) Modifying the solution of Exercise 3, we obtain u(r, t) =

∞ X J1 (αn/2) J0(αn r) sin(αn ct). α2 cJ (αn)2 n=1 n 1

(b) Under suitable conditions that allow us to interchange the limit and the summation sign (for example, if the series is absolutely convergent), we have, for a given (r, t), lim u(r, t) =

c→∞

=

∞ X J1(αn /2) J (αnr) sin(αnct) 2 cJ (α )2 0 c→∞ α n n=1 n 1

lim

∞ X n=1

=

lim

c→∞

J1(αn /2) J0 (αnr) sin(αnct) α2n cJ1(αn )2

0,

because limc→∞ αJ2 1cJ(α1n(α/2) 2 = 0 and sin(αn ct) is bounded. If we let u1 (r, t) denote n) n the solution corresponding to c = 1 and uc(r, t) denote the solution for arbitrary c > 0. Then, it is easy t check that uc(r, t) =

1 u1(r, ct). c

This shows that if c increases, the time scale speeds proportionally to c, while the displacement decreases by a factor of 1c . 10. Separating variables, we look for solutions of the form u(r, t) = R(r)T (t). Plugging this into the heat equation, we find 1 T0 R00 1 R0 RT 0 = c2 R00T + R0 T ⇒ = + = −λ2 , 2 r c T R R R where λ2 is a separation constant. (The choice of a nonnegative sign for the separation constant will be justified momentarily.) Hence we arrive at two ordinary differential equations: T 0 + λ2 c2T = 0; r2R00 + rR0 + λ2 r2R = 0. Note that the equation in T is first order. Its solution is an exponential function T (t) = Ce−c

2

λ2 t

.

If we had chosen a negative separation constant −λ2 < 0, the solution in T becomes T (t) = Cec

2

λ2 t

,

which blows up exponentially with t > 0. This does not make sense on physical grounds and so we discard this solution and the corresponding separation constant, and take the separation to be nonnegative. The boundary conditions on R are R(0) is finite and R(a) = 0. The solution of the parametric form of Bessel’s equation r2 R00 + rR0 + λ2 r2R = 0 is R(r) = c1J0 (λr) + c2 Y0(λr).

162

Chapter 4

Partial Differential Equations in Polar and Cylindrical Coordinates

In order for R(0) to be finite, we take c2 = 0, because Y0 is not bounded near zero and J0(0) = 1. The boundary condition R(a) = c1 J0 (λa) = 0 implies that λa = αn

⇒ λ = λn =

αn , n = 1, 2, . . . , a

where αn is the nth positive zero of J0. Solving the corresponding equation in T , we find Tn (t) = An ec

2

λ2n t

.

Thus the product solutions are AnJ0 (λn r)ec

2

λ2n t

.

These satisfy the heat equation and the boundary conditions. In order to satisfy the given initial condition, we form a series solution by superposing the product solutions and take ∞ X 2 2 u(r, t) = An e−c λn tJ0 (λn r). n=1

Setting t = 0, we find that f(r) = u(r, 0) =

∞ X

AnJ0 (λn r).

n=1

Hence the An ’s are the Bessel Fourier coefficients of f and are given by An =

2 2 2 a J1 (αn)

Z

a

f(r)J0 (λn r)r dr. 0

11. (a) With f(r) = 100, we have from Exercise 10, An

= = =

Z a 200 αn r J0 ( r)r dr a2 J1 (αn) 0 a Z αn 200 αn r J0(s)s ds (let =s α2n J1(αn )2 0 a αn 200 200 = J (s)s . 1 2 2 αn J1(αn ) α J n 1 (αn ) 0

Hence u(r, t) = 200

∞ X J0 ( αan r ) − c2 α22n t e a . α J (α ) n 1 n n=1

The solution represents the time evolution of the temperature of the disk, starting with an initial uniform temperature distribution of 100◦ and with its boundary held at 0◦ . The top and bottom faces of the disk are assumed to be completely insulated so that no heat flows through them. (b) The maximum value of the temperature should occur at the center, since heat is being lost only along the circumference of the disk. Of all the points in the disk, the center is least affected by this loss of heat. If we take a = 1, c = 1, and t = 3 in the solution to part (a), we obtain u(r, 3) = 200

∞ X J0 (αnr) −3α2n . e α J (αn) n=1 n 1

Section 4.2

Vibrations of a Circular Membrane: Symmetric Case

163

To plot the solution, we need the zeros of J0 x). You can find these zeros by using the FindRoot command to solve J0 x =0 or, better yet, you can use the built-in values, as we now show. First, load the following package 1, we can divide by α20,n − 1 and get Z

1

J0(r)J0 (α0,nr)r dr = 0

So a∗0,n =

(α20,n

α0,nJ1 (α0,n)J0 (1) . α20,n − 1

2J0(1) − 1)J1 (α0,n)

and the solution is u(r, θ, t) = 8

∞ ∞ X X J0 (α0,nr) 2J0 (1) J0(α0,nr) t) + cos(α sin(α0,nt). 0,n 3 α J (α0,n) (α20,n − 1)J1 (α0,n) n=1 0,n 1 n=1

7. The partial differential equation is 1 1 utt = c2 urr + ur + 2 uθθ , r r

for 0 < r < a, 0 ≤ θ < 2π, t > 0.

The boundary condition is u(a, θ, t) = 0,

for 0 ≤ θ < 2π, t > 0.

The initial conditions are u(r, θ, 0) = 0 and ut(r, θ, 0) = g(r, θ),

for 0 < r < a, 0 ≤ θ < 2π.

Section 4.3

Vibrations of a Circular Membrane: General Case

169

(b) Assume that u(r, θ, t) = R(r)Θ(θ)T (t). Since u(r, θ, 0) = 0 for all 0 < r < a and 0 ≤ θ < 2π, we set T (t) = 0 to avoid having to take R(r) = 0 or Θ(θ) = 0, which would lead to trivial solutions. Separating variables, as in the text, we arrive at ∞ ∞ X α r X αmn ct mn ∗ ∗ amn cos mθ + bmn sin mθ sin . Jm u(r, θ, t) = a a m=0 n=1 (c) From (b), ut(r, θ, t) =

∞ ∞ X α r X αmnc αmn ct mn Jm a∗mn cos mθ + b∗mn sin mθ cos a a a m=0 n=1

and hence g(r, θ) = ut(r, θ, 0) ∞ X ∞ α r X αmn c mn = Jm a∗mn cos mθ + b∗mn sin mθ a a m=0 n=1 =a∗ 0 (r)

z }| { ∞ α r X α0nc ∗ 0n = a0nJ0 a a n=1 =a∗ m (r)

=b∗ m (r)

z }| { z }| { ∞ nX ∞ ∞ o X X αmnr αmn r αmn c ∗ αmnc ∗ + amn Jm cos mθ + bmn Jm sin mθ . a a a a m=1 n=1 n=1 (d) Thinking of the result of (c) as a Fourier series (for fixed r and 0 ≤ θ < 2π) g(r, θ) = a∗0 (r) +

∞ X

a∗m (r) cos mθ + b∗m (r) sin mθ ,

m=1

we see that we must have a∗0 (r)

a∗m (r) =

1 = 2π

1 2π

and b∗m (r) =

1 2π

Z

Z

2π

g(r, θ) dθ = 0

∞ α r X α0nc ∗ 0n a0nJ0 ; a a n=1

g(r, θ) cos mθ dθ =

∞ α r X αmn c ∗ mn amn Jm ; a a n=1

g(r, θ) sin mθ dθ =

∞ α r X αmnc ∗ mn bmn Jm , a a n=1

π 0

Z

π 0

for m = 1, 2, . . . . (e) We now let r vary in [0, a] and view the expansions in (d) as Bessel series. Then, by Theorem 2, Section 4.8, we must have Z a α r a 2 0n ∗ ∗ a0n = a (r)J r dr 0 0 α0nc a2J1 (α0n)2 0 a Z a Z 2π α r 1 0n = g(r, θ)J r dθ dr, 0 πα0ncaJ1 (α0n)2 0 0 a which implies (17). Also, for m = 1, 2, . . ., Z a α r a 2 mn ∗ a∗mn = a (r)J r dr m αmn c a2Jm+1 (αmn )2 0 m a Z a Z 2π α r 2 mn = g(r, θ)Jm cos(mθ) r dθ dr, 2 παmn caJm+1 (αmn) 0 0 a

170

Chapter 4

Partial Differential Equations in Polar and Cylindrical Coordinates

which implies (18), and b∗mn

= =

a

2 αmnc a2 Jm+1 (αmn )2 2

Z Z

παmncaJm+1 (αmn )2

a

b∗m (r)Jm 0 a

0

Z

α

mn r

a

2π

g(r, θ)Jm 0

α

r dr

mn r

a

sin(mθ) r dθ dr,

which implies (19). 8. (a) Subproblem #1 (u1 )tt

= c2 ∇2u1 ,

0 < r < a, 0 ≤ θ < 2π, t > 0,

u1 (a, θ, t)

= 0,

u1(r, θ, 0)

= f(r, θ),

0 < r < a, 0 ≤ θ < 2π,

= 0,

0 < r < a, 0 ≤ θ < 2π.

(u1)t (r, θ, 0)

0 ≤ θ < 2π, t > 0,

Subproblem #2 (u2)tt

=

c2∇2u2 , 0 < r < a, 0 ≤ θ < 2π, t > 0,

u2 (a, θ, t)

=

0,

0 ≤ θ < 2π, t > 0,

u2(r, θ, 0)

=

0,

0 < r < a, 0 ≤ θ < 2π,

(u2 )t(r, θ, 0)

=

g(r, θ),

0 < r < a, 0 ≤ θ < 2π.

By linearity, or superposition, it follows that u(r, θ, t) = u1(r, θ, t) + u2 (r, θ, t) is a solution of utt

=

c2 ∇2u,

0 < r < a, 0 ≤ θ < 2π, t > 0,

u(a, θ, t) =

0,

0 ≤ θ < 2π, t > 0,

u(r, θ, 0) =

f(r, θ),

0 < r < a, 0 ≤ θ < 2π,

ut (r, θ, 0) =

g(r, θ),

0 < r < a, 0 ≤ θ < 2π.

(b) By Example 1, u1(r, θ, t) =

∞ ∞ X X

Jm

α

mn r

m=0 n=1

a

amn cos mθ + bmn sin mθ cos

αmnct a

,

with the amn and bmn as given by (12)–(14). By Exercise 7, u2 (r, θ, t) =

∞ ∞ X X m=0 n=1

Jm

α

mn r

a

αmnct a∗mn cos mθ + b∗mn sin mθ sin , a

with the a∗mn and b∗mn as given by (17)–(19). This implies the solution (16). 9. (a) For l = 0 and all k ≥ 0, the formula follows from (7), Section 4.8, with p = k: Z rk+1 Jk (r) dr = rk+1Jk+1 (r) + C. (b) Assume that the formula is true for l (and all k ≥ 0). Integrate by parts, using

Section 4.3

Vibrations of a Circular Membrane: General Case

171

u = r2l , dv = rk+1 Jk+1(r) dr, and hence du = 2lr2l−1dr and v = rk+1 Jk+1(r): Z Z rk+1+2l Jk (r) dr = r2l [rk+1Jk (r)] dr = r2l rk+1Jk+1(r) − 2l = rk+1+2l Jk+1(r) − 2l = rk+1+2l Jk+1(r) − 2l

Z Z Z

r2l−1rk+1 Jk+1(r) dr rk+2l Jk+1 (r) dr r(k+1)+1+2(l−1) Jk+1(r) dr

and so, by the induction hypothesis, we get Z

rk+1+2l Jk (r) dr

=

rk+1+2l Jk+1(r) − 2l

=

rk+1+2l Jk+1(r) +

l−1 X (−1)n2n (l − 1)! k+2l−n Jk+n+2(r) + C r (l − 1 − n)! n=0

l−1 X (−1)n+1 2n+1 l! k+1+2l−(n+1) Jk+(n+1)+1(r) + C r (l − (n + 1))! n=0

k+1+2l

l X (−1)m 2m l! k+1+2l−m Jk+1(r) + Jk+m+1 (r) + C r (l − m)! m=1

=

r

=

l X (−1)m 2m l! k+1+2l−m Jk+m+1 (r) + C, r (l − m)! m=0

which completes the proof by induction for all integers k ≥ 0 and all l ≥ 0. 10. a = c = 1, f(r, θ) = (1 − r2)( 14 − r2 )r3 sin θ, g(r, θ) = 0. Since g(r, θ) = 0, then a∗mn and b∗mn are 0. Since f(r, θ) is proportional to sin 3θ, by orthogonality, only b3n can possibly be nonzero. All other coefficients amn and bmn must be 0. We have Z 1 2 1 b3n = (1 − r2 )( − r2 )r3J3 (α3nr)r dr J4 (α3n)2 0 4 Z α3n 2 1 = (α23n − s2 )( α23n − s2 )J3 (s)s4 ds (where s = α3nr) α93nJ4(α3n)2 0 4 Z α3n 1 = 4s2 − 5α23ns2 + α43n s4 J3 (s) ds 9 2 2α3nJ4(α3n) 0 =

1 2α93nJ4(α3n)2

4I2 − 5α23nI1 + α43nI0 ,

where, using the result of the previous exercise, Z α3n Il ≡ s4+2l J3 (s) ds 0

=

l α3n X (−1)m 2m l! 4+2l−m J4+m (s) s (l − m)! 0 m=0

=

l!

l X (−1)m 2m l! 4+2l−m J4+m (α3n). α (l − m)! 3n m=0

172

Chapter 4

Partial Differential Equations in Polar and Cylindrical Coordinates

Thus I0

=

α43nJ4(α3n)

I1

=

α63nJ4(α3n) − 2α53nJ5(α3n),

I2

=

α83nJ4(α3n) − 4α73nJ5(α3n) + 8α63nJ6 (α3n).

Hence 4I2 − 5α23nI1 + α43nI0

=

4α83nJ4 (α3n) − 16α73nJ5(α3n) + 32α63nJ6 (α3n)

=

−5α83nJ4 (α3n) + 10α73nJ5 (α3n) + α83nJ4 (α3n) −6α73nJ5 (α3n) + 32α63nJ6 (α3n).

Thus b3n =

−6α73nJ5(α3n) + 32α63nJ6 (α3n) −3α3nJ5(α3n) + 16J6(α3n) = , 2α93nJ4 (α3n)2 α33nJ4 (α3n)2

and the solution is u(r, θ, t) =

∞ X −3α3nJ5 (α3n) + 16J6(α3n) J3 (α3nr) sin 3θ cos(α3nt). α33nJ4 (α3n)2 n=1

Note: Using the identity Jp−1 (x) + Jp+1 (x) =

2p Jp (x), x

we can simplify the b3n into b3n =

−40(α23n − 32) . α53nJ4(α3n)

11. The solution is similar to the solution of the wave equation. The only difference is the equation in T , which is in this case T 0 = −c2 λT with solution T (t) = Ce−c

2

λt

= Ce−c

2

α2mn t/a

.

The other parts of the product solutions are as in the wave equation α r mn Jm amn cos mθ + bmn sin mθ . a Thus the solution is u(r, θ, t) ==

∞ X ∞ X m=0 n=1

Jm

α

mn r

a

2 2 2 amn cos mθ + bmn sin mθ e−c αmn t/a .

Using the initial condition u(r, θ, 0) = f(r, θ), we see that the coefficients amn and bmn are determined by (12)–(14), as in the case of the wave equation. 12. a = c = 1, f(r, θ) = (1 − r2)r sin θ. Using Example 2, we see that the solution is given by ∞ X J1 (α1nr) −α21n t u(r, θ, t) == 16 sin θ . e 3 α J (α1n) n=1 1n 2 (b) We have u(r, θ, 0) = (1 − r2 )r sin θ. Since (1 − r2 )r ≥ 0, the maximum occurs when θ = π/2 (sin θ = 1) at the value of r in [0, 1] that maximizes the function

Section 4.3

Vibrations of a Circular Membrane: General Case

173

2 3 0 2 0 h(r) √ = r(1 − r ) = r − r . We have h (r) = 1 − 3r √ . So h (r) = 0 ⇒ r = 1/ 3. So the maximum occurs at θ = π/2 and r = 1/ 3. The maximum value is √ 2 f 1/ 3, π/2 = 3√ . 3 (c) We have ∞ X J1 (α1nr) −α21n . e u(r, θ, 1) == 16 sin θ α3 J (α1n) n=1 1n 2

The maximum will occur when θ = π/2. Also ∞ X J1 (α1nr) −2α21n u(r, θ, 2) == 16 sin θ . e 3 J (α ) α 1n n=1 1n 2

The maximum will occur when θ = π/2. 13. The proper place for this problem is in the next section, since its solution invovles solving a Dirichlet problem on the unit disk. The initial steps are similar to the solution of the heat problem on a rectangle with nonzero boundary data (Exercise 11, Section 3.8). In order to solve the problem, we consider the following two subproblems: Subproblem #1 (Dirichlet problem) 1 1 (u1)rr + (u1)r + 2 (u1)θθ r r u1(1, θ)

= 0,

0 < r < 1, 0 ≤ θ < 2π,

= sin 3θ,

0 ≤ θ < 2π.

Subproblem #2 (to be solved after finding u1(r, θ) from Subproblem #1) (u2)t

= (u2 )rr + 1r (u2 )r +

u2 (1, θ, t)

= 0,

u2(r, θ, 0)

= −u1 (r, θ),

1 r 2 (u2 )θθ ,

0 < r < 1, 0 ≤ θ < 2π, t > 0, 0 ≤ θ < 2π, t > 0, 0 < r < 1, 0 ≤ θ < 2π.

You can check, using linearity (or superposition), that u(r, θ, t) = u1(r, θ) + u2(r, θ, t) is a solution of the given problem. The solution of subproblem #1 follows immediately from the method of Section 4.5. We have u2 (r, θ) = r3 sin 3θ. We now solve subproblem #2, which is a heat problem with 0 boundary data and initial temperature distribution given by −u2(r, θ) = −r3 sin 3θ. reasoning as in Exercise 10, we find that the solution is u2(r, θ, t) =

∞ X

2

b3nJ3 (α3nr) sin(3θ)e−α3n t,

n=1

where b3n

= = = = =

Z 1 Z 2π −2 r3 sin2 3θJ3 (α3nr)r dθ dr πJ4 (α3n)2 0 0 Z 1 −2 r4J3 (α3nr) dr J4 (α3n)2 0 Z α3n −2 1 s4 J3(s) ds (where α3nr = s) J4 (α3n)2 α53n 0 α3n −2 1 4 s J (s) 4 J4 (α3n)2 α53n 0 −2 . α3nJ4 (α3n)

174

Chapter 4

Partial Differential Equations in Polar and Cylindrical Coordinates

Hence u(r, θ, t) = r3 sin 3θ − 2 sin(3θ)

∞ X J3 (α3nr) −α23n t . e α J (α3n) n=1 3n 4

Section 4.4

Steady-State Temperature in a Disk

175

Exercises 4.4 1. Since f is already given by its Fourier series, we have from (4) u(r, θ) = r cos θ = x. 2. Since f is already given by its Fourier series, we have from (4) u(r, θ) = r2 sin 2θ = 2r2 sin θ cos θ = 2xy. 3. Recall the Fourier series of the sawtooth function f(θ) =

∞ X sin nθ n n=1

(see Example 1, Section 2.2). From (4) u(r, θ) =

∞ X

rn

n=1

sin nθ . n

4. Let us compute the Fourier coefficients of f (or use Example 4, Section 2.2). We have π Z π 1 1 π a0 = (π − θ) dθ = − (π − θ)2 = ; 2π 0 4π 4 0 an

Z 1 π (π − θ) cos nθ dθ π 0 π Z 1 1 1 π sin nθ dθ (π − θ) sin nθ + π n n 0 0 π n −1 = 1 − (−1) ; cos nθ πn2 πn2

= = =

0

bn

Z 1 π (π − θ) sin nθ dθ π 0 π Z 1 −1 1 π cos nθ dθ (π − θ) cos nθ − π n n 0 π 0 1 π 1 1 − 2 sin nθ = . π n n n 0

= = =

Hence f(θ) =

π X + 4 n=1

1 − (−1)n 1 cos nθ + sin nθ ; πn2 n

u(r, θ) =

π X + 4 n=1

1 − (−1)n 1 cos nθ + sin nθ rn. πn2 n

∞

and

∞

5. Let us compute the Fourier coefficients of f. We have a0 =

an =

100 π

Z

50 π

Z

π/4

dθ = 0

π/4

cos nθ dθ = 0

25 ; 2

π 100 100 nπ sin nθ = sin ; nπ nπ 4 0

176

Chapter 4

Partial Differential Equations in Polar and Cylindrical Coordinates

bn =

100 π

Z

π/4

sin nθ dθ = − 0

π 100 100 nπ cos nθ = (1 − cos ). nπ nπ 4 0

Hence ∞ 25 100 X 1 nπ nπ f(θ) = + sin cos nθ + (1 − cos ) sin nθ ; 2 π n=1 n 4 4

and u(r, θ) =

∞ 25 100 X 1 nπ nπ + sin cos nθ + (1 − cos ) sin nθ rn . 2 π n=1 n 4 4

6. In Cartesian coordinates, −1

u(x, y) = tan

r sin θ 1 − r cos θ

= tan

−1

y 1−x

.

Let U (x, y) = u(−x + 1, y). Note that U is a solution of Laplace’s equation if and only if u is a solution of Laplace’s equation. But y y −1 U (x, y) = u(−x + 1, y) = tan = tan−1 = θ + 2kπ. 1 − (−x + 1) x Back to polar coordinates, it is clear that U is a solution of Laplace’s equation. 7. Using (6), we have u(r, θ) =

∞ X

r

n sin nθ

n=1

Let − π2 < T

a.

Setting r = a and using the boundary condition, we obtain f(θ) = a0 +

∞ X

(an cos nθ + bn sin nθ) ,

n=1

which implies that the an and bn are the Fourier coefficients of f and hence are given by (5). 22. The Fourier coefficients of the boundary function are computed in Example 1. Applying the result of the previous exercise, we find 100 X 1 − (−1)n sin nθ , π n=1 n rn ∞

u(r, θ) = 50 +

r > 1.

This is precisely the solution in Example 1, except that r is replaced by 1r . So to find the isotherms, we can proceed exactly as in Example 2, change r to 1r , and get: for 0 < T < 100, πT 2 2 u(x, y) = T ⇔ x + y − 1 = −2 tan y. 100 If T = 50, we find that y = 0 is the corresponding isotherm. If T 6= 50, obtain 2 πT πT x2 + y + tan = sec2 . 100 100

184

Chapter 4

Partial Differential Equations in Polar and Cylindrical Coordinates

πT πT Thus the isotherms lie on circles with centers at (0, − tan 100 ) and radii sec 100 . Note that the isotherms here are the complementary portions of the circles found in Example 3, and these correspond to the complementary temperature T − 100. 23. Using Exercise 5 and formula (6), we find ∞ 25 100 X 1 sin nπ 4 cos nθ + 1 − cos u(r, θ) = + 2 π n=1 n rn =

=

=

∞ 25 100 X + 2 π n=1

sin nθ n θ − π4 − nrn nrn

nπ 4

" 25 100 sin θ + tan−1 − tan−1 2 π r − cos θ

!# r−1 sin θ − π4 1 − r−1 cos θ − π4

!# sin θ − π4 . r − cos θ − π4

To find the isotherms, let 0 < T < 100, then " 25 100 sin θ u(r, θ) = T ⇔ T = + tan−1 − tan−1 2 π r − cos θ

Thus

!# sin θ − π4 r − cos θ − π4

π sin θ 25 T− = tan−1 − tan−1 100 2 r − cos θ

! sin θ − π4 . r − cos θ − π4

π 25 T− 100 2 " sin θ = tan tan−1 − tan−1 r − cos θ

tan

!

!

" −1 25 100 r sin θ + tan−1 − tan−1 2 π 1 − r−1 cos θ

⇔

sin nθ

sin θ r−cos θ

−

= 1+

=

=

=

=

sin θ r−cos θ

!# sin θ − π4 r − cos θ − π4

sin (θ− π 4) r−cos(θ− π 4)

·

sin( θ− π 4) r−cos( θ− π 4)

r sin θ − sin θ − π4 − sin θ cos θ − π4 − cos θ sin θ − π4 r2 − r cos θ + cos θ − π4 + cos θ cos θ − π4 + sin θ sin θ − π4 r sin θ − sin θ − π4 − sin π4 r2 − r cos θ + cos θ − π4 + cos π4 √ r 2 sin θ − sin θ + cos θ − 1 √ √ 2 r2 − r 2 cos θ + cos θ + sin θ + 1 √ x+ 2−1 y−1 √ √ . 2 (x2 + y2 ) − 2+1 x−y+1

Let K = cot

π 25 T− . 100 2

Section 4.4

Steady-State Temperature in a Disk

185

The equation becomes of the isotherms becomes √ √ √ 2 (x2 + y2 ) − 2 + 1 x − y + 1 = Kx + K( 2 − 1)y − K or 1 1 K 1 1 x2 − 1 + √ + √ x + y2 − √ + 1 − √ K y = − √ (1 + K) . 2 2 2 2 2 Hence 2 2 1 1 1 1 K 1 √ + 1− √ K + y− x− 1+ √ + √ 2 2 2 2 2 2 1 = − √ (1 + K) 2 2 2 1 1 1 1 K 1 √ + 1− √ K + + 1+ √ + √ 4 4 2 2 2 2 2 √ √ 1 1 1 1 1 1 K = 1+ 2+ + −√ + + 1− 2+ K2 4 2 8 8 4 2 2 1 1 = 1− √ 1 − K2 . 2 2 Thus the isotherms are circles with centers at 1 1 1 K 1 1 √ + 1− √ 1+ √ + √ , K 2 2 2 2 2 2 and radii

s 1 1 √ 1− (1 − K 2 ). 2 2

It is now straightforward to check that all these circles pass through the points (1, 0) and √12 , √12 . 24. (a) As in the text, we have Rn(r) = c1(

r n r ) + c2 ( )−n R2 R2

for n = 1, 2, . . . ,

R0(r) = c1 + c2 ln(

r ). R2

and

The condition R(R2) = 0 implies that 0 = Rn (R2) = c1 (

R2 n R2 ) + c2( )−n R2 R2

and 0 = R0 (R2) = c1 + c2 ln( Hence

R2 ) R2

r r Rn(r) = c2 ( )−n − ( )n R2 R2

and R0 (r) = c2 ln(

⇒

⇒

c1 = −c2 ;

c1 = 0.

for n = 1, 2, . . . ,

r ). R2

Hence the general solution is ∞ X r r −n r n u(r, θ) = A0 ln( ) + [An cos nθ + Bn sin nθ] ( ) − ( ) . R2 R2 R2 n=1

186

Chapter 4

Partial Differential Equations in Polar and Cylindrical Coordinates

Using the boundary condition, we obtain f1 (θ) = u(R1, θ) = A0 ln(

∞ X R1 R1 R1 )+ [An cos nθ + Bn sin nθ] ( )−n − ( )n , R2 R2 R2 n=1

which is the Fourier series of f1 (θ). Thus Z 2π R1 1 ) = a0 = f1(θ) dθ; R2 2π 0 Z R2 n R1 n 1 2π An ( ) − ( ) = an = f1 (θ) cos nθ dθ; R1 R2 π 0 A0 ln(

and

Z R2 R1 1 2π Bn ( )n − ( )n = bn = f1 (θ) sin nθ dθ. R1 R2 π 0

Solving for An , Bn in terms of an and bn and plugging back into the solution u, we obtain u(r, θ) =

a0

ln( Rr2 ) 1 ln( R R2 )

∞ X

+

[an cos nθ + bn sin nθ]

n=1

( Rr2 )−n − ( Rr2 )n R1 n 2 n (R R1 ) − ( R2 )

∞

2n X ln r − ln R2 Rn R2n 2 −r a0 + [an cos nθ + bn sin nθ] n1 2n . ln R1 − ln R2 n=1 r R2 − R2n 1

=

(b) Using the condition R(R − 1) = 0 and proceeding as in (a), we find r r Rn(r) = c2 ( )n − ( )−n for n = 1, 2, . . . , R1 R1 and R0 (r) = c2 ln(

r ). R1

Hence the general solution is ∗

u (r, θ) =

A∗0

∞ X r r n r −n ∗ ∗ . ln( ) + [An cos nθ + Bn sin nθ] ( ) − ( ) R1 R1 R1 n=1

Using the boundary condition, we obtain f2 (θ) = u∗ (R2, θ) = A∗0 ln(

∞ X R2 R2 R1 )+ [A∗n cos nθ + Bn∗ sin nθ] ( )n − ( )n , R1 R1 R2 n=1

which is the Fourier series of f2 (θ). Thus Z 2π R2 1 ) = a∗0 = f2 (θ) dθ; R1 2π 0 Z R2 n R1 n 1 2π ∗ ∗ An ( ) − ( ) = an = f2 (θ) cos nθ dθ; R1 R2 π 0 A∗0 ln(

and

Z R2 R1 1 2π Bn∗ ( )n − ( )n = b∗n = f2 (θ) sin nθ dθ. R1 R2 π 0

Solving for A∗n , Bn∗ in terms of a∗n and b∗n and plugging back into the solution u∗ , we obtain u∗ (r, θ)

= a∗0

ln( Rr1 ) 2 ln( R ) R1

+

∞ X

[a∗n cos nθ + b∗n sin nθ]

n=1

2 n 1 n (R ) − (R ) R1 R2

X ln r − ln R1 Rn r2n − R2n 1 + [a∗n cos nθ + b∗n sin nθ] n2 2n , ln R2 − ln R1 n=1 r R2 − R2n 1 ∞

= a∗0

( Rr1 )n − ( Rr1 )−n

Section 4.4

Steady-State Temperature in a Disk

187

where a∗0 , a∗n , and b∗n are the Fourier coefficients of f2 (θ). (c) Just add the solutions in (a) and (b). 25. The hint does it. 26. (a) The function is already given by its Fourier series. We have f(θ) = 50 − 50 cos θ. (b) The steady-state solution follows by usng the Fourier series of f. We have u(r, θ) = 50 − 50 r cos θ. (d) To find the isotherms, we must solve u(r, θ) = T , where T is any value between 0 and 100. Using x = r cos θ, we find 50 − 50 r cos θ = T ⇒ 50 − 50 x = T. Thus the isotherms are portions of the vertical lines x=

50 − T 50

inside the unit disk. (e) We follow the steps outlined in Exercise 25. First subproblem: ∇2u1 = 0, u1(1, θ) = 50 − 50 cos θ. Its solution is u1(r, θ) = 50 − 50 r cos θ. Second subproblem: vt = ∇2v, v(1, θ, t) = 0, v(r, θ, 0) = −50 + 50 r cos θ. We next find the Bessel series expanson of v(r, θ, 0). It is clear that only a0n and a1n are nonzero and that all the remaining coefficients amn and bmn are 0. From Example 1, Section 4.8, we find that, for 0 < r < 1, 1=2

∞ X J0 (α0,nr) . α J (α0,n) n=1 0,n 1

Thus, for 0 < r < 1, −50 = −100

∞ X J0(α0,nr) . α J (α0,n) n=1 0,n 1

Also, from Execrise 20, Section 4.8, we have, for 0 < r < 1, r=2

∞ X J1 (α1,nr) . α J (α1,n) n=1 1,n 2

So, for 0 < r < 1, 50 r = 100

∞ X J1 (α1,nr) . α J (α1,n) n=1 1,n 2

Thus the solution is u(r, θ, t)

=

50 − 50 r cos θ − 100

+100 cos θ

=

100

∞ X J0 (α0,nr) −α20,nt e α J (α0,n) n=1 0,n 1

∞ X J1 (α1,nr) −α21,nt . e α J (α1,n) n=1 1,n 2

∞ X 2 J0 (α0,nr) 1 − e−α0,n t α J (α0,n) n=1 0,n 1

−100 cos θ

∞ X 2 J1 (α1,nr) 1 − e−α1,n t . α J (α ) 1,n n=1 1,n 2

188

Chapter 4

Partial Differential Equations in Polar and Cylindrical Coordinates

27. (a) a = c = 1, F (r, θ) = (1 − r2)r sin θ, G(θ) = sin 2θ. Using the notation of Exercise 25, we have u1 (r, θ) = r2 sin 2θ. Also by Example 2, Section 4.3, we have F (r, θ) = (1 − r2 )r sin θ = 16 sin θ

∞ X J1(α1,nr) . 3 α J (α1,n) n=1 1,n 2

We now find the Bessel series expansion of u1. Since u1 is proportional to sin 2θ, we only only need b2n (all other amn and bmn are 0). We have b2,n

2 J3(α2,n)2

=

Z

1

r2J2 (α2,nr)r dr 0

2 α42,nJ3 (α2,n)2

=

Z

α2,n

s3 J2 (s) ds

(s = α2,n r)

0

=

3 α2,n 2 s J (s) 3 α42,nJ3 (α2,n)2 0

=

2 . α2,nJ3 (α2,n)

Thus u1 (r, θ) = r2 sin 2θ = 2 sin 2θ

∞ X J2 (α2,nr) α J (α2,n) n=1 2,n 3

and v(r, θ, t) = 16 sin θ

∞ ∞ X X J1 (α1,nr) −α21,n t J2(α2,nr) −α22,n t − 2 sin 2θ ; e e 3 α J (α1,n) α J (α2,n) n=1 1,n 2 n=1 2,n 3

and so u(r, θ, t)

= r2 sin 2θ + 16 sin θ

−2 sin 2θ

= 16 sin θ

∞ X J1 (α1,nr) −α21,n t e 3 α J (α1,n) n=1 1,n 2

∞ X J2(α2,nr) −α22,n t e α J (α2,n) n=1 2,n 3

∞ ∞ X X 2 J1 (α1,nr) −α21,nt J2 (α2,nr) + 2 sin 2θ e 1 − e−α2,n t 3 α J (α ) α J (α ) 1,n 2,n n=1 1,n 2 n=1 2,n 3

(b) The steady-state solution is u1 (r, θ) = r2 sin 2θ = 2 sin 2θ

∞ X J2 (α2,nr) . α J (α2,n) n=1 2,n 3

28. (a) For z = reiθ with 0 ≤ r = |z| < 1, the geometric series is ∞ ∞ X X 1 zn = rneinθ . = 1 − z n=0 n=0

Also z = re−iθ , z = r. So if |z| < 1 then z < 1 and we have ∞ ∞ X n X 1 z = rn e−inθ . = 1 − z n=0 n=0

Section 4.4

Steady-State Temperature in a Disk

189

For |z| < 1, we have 1 1 + −1 = 1−z 1−z =

∞ X

rneinθ +

n=0

1+

∞ X

rne−inθ − 1

n=0 ∞ X

rn einθ + e−inθ

n=0

=

1+2

∞ X

rn cos nθ,

n=0

where we have used einθ + e−inθ = 2 cos nθ. Using properties of the complex conjugate, we see that 1 1 = = 1−z 1−z

1 . 1−z

So, because for any complex number w, w + w = 2 Re w, we infer that 1 1 + = 2 Re 1−z 1−z

1 1−z

.

Comparing with the previous equations, we find that

2 Re

1 1−z

−1= 1+2

∞ X

rn cos nθ;

n=0

equivalently, 1 + 2 Re

∞ X 1 rn cos nθ, −1 = 1+2 1−z n=0

for all 0 ≤ r < 1 and all θ. (b) From (a) we have

1+2

∞ X

rn cos nθ

=

1 1 + −1 1−z 1−z

=

2−z−z −1 (1 − z)(1 − z)

=

2 − (z + z) − (1 − (z + z) + |z|2) 1 − (z + z + |z|2

=

1 − r2 , 1 − 2r cos θ + r2

n=0

where we hae used z · z = |z|2 = r2 and z + z = 2 Re z = 2x = 2r cos θ.

190

Chapter 4

Partial Differential Equations in Polar and Cylindrical Coordinates

29. (a) Recalling the Euler formulas for the Fourier coefficients, we have u(r, θ) = a0 +

∞ X r n n=1

=

1 2π +

Z

f(φ) dφ 0

∞ Z X r n 1

1 2π +

Z

=

π

f(φ) cos nφ dφ cos nθ + 0

1 2π

1 π

Z

2π

f(φ) sin nφ dφ sin nθ 0

2π

f(φ) dφ 0

n=1

=

a

2π

∞ n Z X 1 r

1 2π

[an cos nθ + bn sin nθ]

2π

n=1

=

a

Z

a

2π 0

Z

π

2π

f(φ) cos nφ cos nθ + sin nφ sin nθ dφ

0

Z ∞ X r n 1 2π f(φ) dφ + f(φ) cos n(θ − φ) dφ a π 0 n=1

2π

"

f(φ) 1 + 2 0

∞ X r n

a

n=1

#

cos n(θ − φ) dφ.

(b) Continuing from (a) and using Exercise 28, we obtain u(r, θ)

=

=

=

1 2π 1 2π 1 2π

Z

2π 0

Z

2 1 − ar f(φ) 1 − 2 ar cos θ +

2π

f(φ) 0

Z

2π 0

a2

r 2 a

a2 − r2 dφ − 2ar cos(θ − φ) + r2

f(φ)P r/a, θ − φ dφ.

Section 4.5

Steady-State Temperature in a Cylinder

191

Solutions to Exercises 4.5 1. Using (2) and (3), we have that u(ρ, z) =

∞ X

An J0 (λn ρ) sinh(λn z),

λn =

n=1

αn , a

where αn = α0,n is the nth positive zero of J0 , and An = = = =

Z a 2 f (ρ)J0(λnρ)ρ dρ sinh(λnh)a2 J1 (αn )2 0 Z 1 200 J0 (αn ρ)ρ dρ sinh(2αn )J1(αn )2 0 Z αn 200 J0 (s)s ds (let s = αn ρ) sinh(2αn )α2n J1 (αn )2 0 αn 200 [J (s)s] 1 sinh(2αn )α2 J1 (αn )2 n

= So

0

200 . sinh(2αn )αn J1 (αn )

∞ X J0 (αn ρ) sinh(αn z) u(ρ, z) = 200 . sinh(2αn)αn J1 (αn ) n=1

2. Using (2) and (3), we have that u(ρ, z) =

∞ X

An J0 (λn ρ) sinh(λn z),

n=1

λn =

αn , a

where αn = α0,n is the nth positive zero of J0 , and An = = = = = = =

Z a 2 f (ρ)J0(λn ρ)ρ dρ sinh(λn h)a2J1 (αn )2 0 Z 1 2 (100 − ρ2)J0 (αn ρ)ρ dρ sinh(2αn )J1(αn )2 0 Z αn 2 (100α2n − s2 )sJ0 (s) ds (let s = αn ρ) sinh(2αn )α4n J1 (αn )2 0 Z αn αn 2 2 2 2 (100αn − s )sJ1 (s) + 2 s J1 (s) ds sinh(2αn )α4n J1 (αn )2 0 0 αn i h 2 3 2 (99α J (α ) + 2[s J (s)] 2 n 1 n sinh(2αn )α4n J1 (αn )2 0 2 99α3n J1 (αn ) + 2α2n J2 (αn ) sinh(2αn )α4n J1 (αn )2 2 2 J1 (αn )) (99α2n + 4) (because J2 (αn ) = sinh(2αn )α3n J1 (αn ) αn

So u(ρ, z) = 2

∞ X n=1

99α2n + 4 J0 (αn ρ) sinh(αn z). sinh(2αn )α3n J1 (αn )

192

Chapter 4

Partial Differential Equations in Polar and Cylindrical Coordinates

3. We have An = = = =

200 sinh(2αn )J1(αn )2

Z

1 2

J0 (αn ρ)ρ dρ 0

200 sinh(2αn )α2n J1 (αn )2

Z

αn 2

J0 (s)s ds (let s = αn ρ) 0

αn 2 200 [J1(s)s] 2 2 sinh(2αn )αn J1 (αn ) 0 100J1( α2n ) . sinh(2αn )αn J1 (αn )2

So u(ρ, z) = 100

∞ X J1 ( αn )J0 (αn ρ) sinh(αn z) 2

n=1

sinh(2αn )αn J1 (αn )2

.

4. We have An = =

Z a 2 f (ρ)J0(λn ρ)ρ dρ sinh(λn h)a2J1 (αn )2 0 Z 1 140 J0(ρ)J0(αn ρ)ρ dρ. sinh(2αn )J1 (αn )2 0

In evaluating the last integral, we can proceed as we did in Exercise 6, Section 4.3. The details are very similar and so will be omitted. The result is Z 1 αn J0 (1)J1(αn ) J0 (ρ)J0(αn ρ)ρ dρ. = . α2n − 1 0 So u(ρ, z) = 140J0(1)

∞ X n=1

αn J0 (αn ρ) sinh(αn z) . sinh(2αn )(α2n − 1)J1(αn )

5. (a) We proceed exactly as in the text and arrive at the condition Z(h) = 0 which leads us to the solutions Z(z) = Zn (z) = sinh(λn (h − z)),

where λn =

αn . a

So the solution of the problem is u(ρ, z) =

∞ X

Cn J0 (λn ρ) sinh(λn(h − z)),

n=1

where

2 Cn = 2 2 a J1 (αn ) sinh(λn h)

Z

a

f (ρ)J0(λn ρ)ρ dρ. 0

(b) The problem can be decomposed into the sum of two subproblems, one treated in the text and one treated in part (a). The solution of the problem is the sum of the solutions of the subproblems: u(ρ, z) =

∞ X

An J0 (λnρ) sinh(λn z) + Cn J0 (λn ρ) sinh(λn(h − z)) ,

n=1

Section 4.5

where An = and

Steady-State Temperature in a Cylinder

2 2 2 a J1 (αn ) sinh(λnh)

2 Cn = 2 a J1(αn )2 sinh(λnh)

Z

a

f2 (ρ)J0(λn ρ)ρ dρ, 0

Z

a

f1 (ρ)J0(λn ρ)ρ dρ. 0

6. Apply the result of Exercise 5. We have Z 1 200 J0 (αn ρ)ρ dρ J1 (αn )2 sinh(2αn ) 0 Z αn 200 J0 (s)s ds α2n J1 (αn )2 sinh(2αn ) 0 αn 200 J1 (s)s 2 2 αn J1 (αn ) sinh(2αn ) 0 200 . αn J1 (αn ) sinh(2αn )

A n = Cn = = = = Hence the solution u(ρ, z) = 200

∞ X n=1

J0 (αn ρ) [sinh(αn z) + sinh(αn (2 − z))] . αn J1 (αn ) sinh(2αn )

Using identities for the hyperbolic functions, you can check that sinh(αn z) + sinh(αn (2 − z)) = 2 sinh αn cosh(αn (z − 1)) and sinh(2αn ) = 2 sinh αn cosh αn . So u(ρ, z) = 400

∞ X

= 200

∞ X

J0 (αn ρ) sinh αn cosh(αn (z − 1)) α J (α ) sinh(2αn ) n=1 n 1 n

n=1

J0 (αn ρ) cosh(αn (z − 1)). αn J1 (αn ) cosh(αn )

7. Let x = λρ, then d d =λ dρ dx and the equation ρ2R00 + ρR0 − λ2 ρ2R = 0, which is the same as ρ2

d2 R dR +ρ − λ2 ρ2R = 0, 2 dρ dρ

becomes λ2 ρ2 or x2

d2 R dR + λρ − λ 2 ρ2 R = 0 dx2 dx d2 R dR +x − x2R = 0, 2 dx dx

193

194

Chapter 4

Partial Differential Equations in Polar and Cylindrical Coordinates

which is the modified Bessel equation of order 0. Its general solution is R(x) = c1I0 (x) + c2K0 (x) or R(ρ) = c1I0 (λρ) + c2K0(λρ). 8. (a) To separate variables in the problem uρρ + 1ρ uρ + uzz

=

0 for 0 < ρ < a, 0 < z < h,

u(ρ, 0) = u(ρ, h) =

0 for 0 < ρ < a,

u(a, z) = f (z) for 0 < z < h, let u(ρ, z) = R(ρ)Z(z). We obtain 1 R00 Z + R0Z + RZ 00 = 0 ρ or

R00 1 R0 Z 00 + =− = λ. R ρR Z

Hence Z 00 + λZ = 0, ρR00 + R0 − λρR = = 0. Separating variables in the boundary conditions, we have u(ρ, 0) = 0 = u(ρ, h)

⇒ Z(0) = 0 = Z(h).

(b) The boundary value problem Z 00 + λZ = 0 Z(0) = 0 = Z(h) has nontrivial solutions given by Zn (z) = sin

nπz (n = 1, 2, . . .). h

These correspond to the separation constant nπ 2 λ = λn = . h (c) The equation in R becomes ρR00 + R0 −

nπ 2 h

ρR = 0

and, by Exercise 7, this has nontrivial solutions nπ nπ R(ρ) = c1I0 ρ + c 2 K0 ρ . h h For R(ρ) to remain bounded as ρ → 0+ , we must take c2 = 0 and hence we conclude that nπ Rn (ρ) = Bn I0 ρ , h

Section 4.5

Steady-State Temperature in a Cylinder

195

where Bn is a constant. We can now write the series solution of the problem as ∞ nπ X nπz B n I0 u(ρ, z) = ρ sin . h h n=1

To determine Bn , we use the boundary condition f (z) = u(a, z) =

∞ X

B n I0

n=1

nπ nπz a sin . h h

Thinking of this as a sine series of f , we conclude that nπ 2 Z h nπz B n I0 f (z) sin a = dz h h 0 h or Bn =

2

Z

nπ h a

h I0

h

f (z) sin 0

nπz dz. h

9. We use the solution in Exercise 8 with a = 1, h = 2, f (z) = 10z. Then 1

Bn =

I0

2

Z

2

0z sin 0

nπz dz 2

40 (−1)n+1 . nπ I0 nπ 2

= Thus

nπ

40 X (−1)n+1 nπ nπz ρ sin . nπ I0 π 2 2 n I0 2 ∞

u(ρ, z) =

n=1

10. Adding the solutions to Exercises 6 and 9, we obtain, for 0 < ρ < 1 and 0 < z < 2, u(ρ, z) = 200 +

∞ X

J0 (αn ρ) cosh(αn (z − 1)) αn J1 (αn ) cosh(αn )

n=1 ∞ X

40 (−1)n+1 nπ nπz I0 ρ sin . nπ π n=1 n I0 2 2 2

11. We have a = 10, h = 6, u(ρ, 0) = 56, for 0 < ρ < 10, u(ρ, 6) = 78, for 0 < ρ < 10, 56 for 0 < z < 4, u(10, z) = 78 for 4 < z < 6. Hence the solution is ∞ nρ nz X J0 ( α10 ) sinh( α10 ) u(ρ, z) = 156 3 α J (α ) sinh( 5 αn ) n=1 n 1 n

+112

∞ ∞ nρ X J0 ( α10 ) sinh( αn(6−z) ) X nπρ nπz 10 B n I0 ( ) sin , + 3 αn J1 (αn ) sinh( 5 αn ) 6 6 n=1 n=1

196

Chapter 4

Partial Differential Equations in Polar and Cylindrical Coordinates

where Bn = = =

1 3I0 ( 10nπ 6 )

Z

6

u(10, z) sin 0

nπz dz 6

Z 4 Z 6 1 nπz nπz sin sin dz + 78 dz 56 6 6 3I0 ( 5nπ 0 4 3 ) 2 2nπ n − 78(−1) . 56 + 22 cos 3 nπI0 ( 5nπ 3 )

Section 4.6

The Helmholtz and Poisson Equations

197

Solutions to Exercises 4.6 1. Write (1) in polar coordinates: 1 1 φrr + φr + 2 φθθ = −kφ r r

φ(a, θ) = 0.

Consider a product solution φ(r, θ) = R(r)Θ(θ). Since θ is a polar angle, it follows that Θ(θ + 2π) = Θ(θ); in other words, Θ is 2π-periodic. Plugging the product solution into the equation and simplifying, we find R00Θ + 1r R0 Θ + r12 RΘ00 = −kRΘ; R00 + 1r R0 + kR Θ = − r12 RΘ00; 00

0

00

r2 RR + r RR + kr2 = − ΘΘ ; hence r2

R00 R0 +r + kr2 = λ, R R

and

Θ00 = λ ⇒ Θ00 + λΘ = 0, Θ where λ is a separation constant. Our knowledge of solutions of second order linear ode’s tells us that the last equation has 2π-periodic solutions if and only if −

λ = m2 ,

m = 0, ±1, ±2, . . . .

This leads to the equations Θ00 + m2 Θ = 0, and r2

R00 R0 +r + kr2 = m2 R R

⇒

r2 R00 + rR0 + (kr2 − m2 )R = 0.

These are equations (3) and (4). Note that the condition R(a) = 0 follows from φ(a, θ) = 0 ⇒ R(a)Θ(θ) = 0 ⇒ R(a) = 0 in order to avoid the constant 0 solution. 2. For n = 1, 2, . . ., let φ0,n(r, θ) = J0 (λ0,nr) and for m = 1, 2, . . . and n = 1, 2, . . ., let φm,n (r, θ) = Jm (λmn r) cos mθ and ψm,n (r, θ) = Jm (λmn r) sin mθ. The orthogonality relations for these functions state that if (m n) 6= (m0 , n0 ) then Z

2π

Z

0

Z

a

φmn (r, θ)φm0 n0 (r, θ)r dr dθ = 0, 0

2π 0

Z

a

ψmn (r, θ)ψm0 n0 (r, θ)r dr dθ = 0, 0

and for all (m n) and (m0 , n0) Z

2π 0

Z

a

φmn(r, θ)ψm0 n0 (r, θ)r dr dθ = 0. 0

The proofs follow from the orthogonality of the trigonometric functions and the Bessel functions. For example, to prove the first identity, write Z

2π 0

Z

=

Z

=

I1I2 .

a

φmn (r, θ)φm0 n0 (r, θ)r dr dθ 0

2π

cos mθ cos m0 θdθ 0

Z

a 0 Jm (λn r)Jm (λn0 r)r dr 0

198

Chapter 4

Partial Differential Equations in Polar and Cylindrical Coordinates

If m 6= m0 , then I1 = 0 by the orthogonality of the 2π-periodic cosine functions. If m = m0 , then necessarily n 6= n0 (otherwise (m, n) = (m0 , n0 )) and then Z

I2 =

a

Jm (λn r)Jm (λn0 r)r dr = 0, 0

by the orthogonality relations for Bessel functions ((11), Section 4.8). Either way, we have I = I1I2 = 0. The other relations are established similarly. We omit the proof. 3. Let φmn and ψmn be as in Exercise 2. We will evaluate I

=

Z

=

Z

2π

Z

0

a

φ2mn(r, θ)r dr dθ 0

Z

2π 2

cos mθdθ 0

=

a

Jm (λn r)2r dr 0

I1 I2,

where m 6= 0. We have Z

I1 =

2π 2

cos mθ =

Z

0

and I2 =

Z

2π 0

a

Jm (λn r)2 r dr = 0

1 + cos 2θ = π, 2

a2 2 (αmn ), J 2 m+1

by (12), Section 4.8. Thus πa2 2 (αmn). J 2 m+1

I= If m = 0, then I

= =

Z Z

2π 0

Z

a

φ20n(r, θ)r dr dθ 0

2π

dθ 0

=

Z

a

J0 (λ0 r)2r dr 0

2 πa2 Jm+1 (αmn ).

Finally, I

= =

Z Z

2π 0

Z

a 2 ψmn (r, θ)r dr dθ 0

2π 2

sin mθdθ 0

=

Z

a

Jm (λn r)2 r dr 0

πa2 2 (αmn ). J 2 m+1

5. We proceed as in Example 1 and try u(r, θ) =

∞ ∞ X X m=0 n=1

Jm (λmn r)(Amn cos mθ + Bmn sin mθ) =

∞ ∞ X X

φmn(r, θ),

m=0 n=1

where φmn (r, θ) = Jm (λmn r)(Amn cos mθ + Bmn sin mθ). We plug this solution

Section 4.6

The Helmholtz and Poisson Equations

199

into the equation, use the fact that ∇2(φmn ) = −λ2mn φmn , and get ! ∞ ∞ ∞ X ∞ X X X 2 = 1− ∇ φmn (r, θ) φmn (r, θ) m=0 n=1

m=0 n=1

⇒

∞ ∞ X X

⇒

m=0 n=1 ∞ ∞ X X

⇒

m=0 n=1 ∞ X ∞ X

2

∇ (φmn(r, θ)) = 1 −

∞ ∞ X X

φmn (r, θ)

m=0 n=1 ∞ ∞ X X

−λ2mn φmn (r, θ) = 1 −

φmn (r, θ)

m=0 n=1

(1 − α2mn )φmn (r, θ) = 1.

m=0 n=1

We recognize this expansion as the expansion of the function 1 in terms of the functions φmn . Because the right side is independent of θ, it follows that all Amn and Bmn are zero, except A0,n. So ∞ X

(1 − α2mn )A0,nJ0 (α0,n)r) = 1,

n=1

which shows that (1 − α2mn)A0,n = a0,n is the nth Bessel coefficient of the Bessel series expansion of order 0 of the function 1. This series is computed in Example 1, Section 4.8. We have ∞ X 2 1= J0 (α0,nr) 0 < r < 1. α J (α0,n) 0,n 1 n=1 Hence (1 − α2mn )A0,n =

2 α0,nJ1 (α0,n)

and so u(r, θ) =

⇒

A0,n =

2 ; (1 − α2mn )α0,nJ1(α0,n)

∞ X

2 J0(α0,nr). 2 )α (1 − α 0,nJ1 (α0,n ) mn n=1

6. We proceed as in the previous exercise and try u(r, θ) =

∞ ∞ X X

Jm (λmn r)(Amn cos mθ + Bmn sin mθ) =

m=0 n=1

∞ ∞ X X

φmn(r, θ),

m=0 n=1

where φmn (r, θ) = Jm (λmn r)(Amn cos mθ + Bmn sin mθ). We plug this solution into the equation, use the fact that ∇2(φmn ) = −λ2mn φmn = −α2mn φmn , and get ! ∞ ∞ ∞ X ∞ X X X 2 ∇ φmn (r, θ) = 3 φmn (r, θ) + r sin θ m=0 n=1

m=0 n=1

⇒

∞ ∞ X X

⇒

m=0 n=1 ∞ X ∞ X

⇒

m=0 n=1 ∞ ∞ X X

∇2 (φmn (r, θ)) = 3

∞ ∞ X X

φmn (r, θ) + r sin θ

m=0 n=1 ∞ X ∞ X

−α2mn φmn(r, θ) = 3

φmn (r, θ) + r sin θ

m=0 n=1

(−3 − α2mn)φmn (r, θ) = r sin θ.

m=0 n=1

We recognize this expansion as the expansion of the function r sin θ in terms of the functions φmn . Because the right side is proportional to sin θ, it follows that all Amn and Bmn are zero, except B1,n. So sin θ

∞ X

(−3 − α21n)B1,nJ1 (α1nr) = r sin θ,

n=1

200

Chapter 4

Partial Differential Equations in Polar and Cylindrical Coordinates

which shows that (−3 − α21n)B1,n = b1,n is the nth Bessel coefficient of the Bessel series expansion of order 1 of the function r. This series is computed in Exercise 20, Section 4.8. We have r=

∞ X

2 J1(α1,nr) α J (α1,n) n=1 1,n 2

0 < r < 1.

Hence (−3 − α21n)B1,n =

2 α1,nJ2 (α1,n)

and so u(r, θ) = sin θ

⇒

B1,n =

−2

(3 +

; α21,n)α1,nJ2 (α1,n)

∞ X

−2 J1 (α1,nr). 2 )α (3 + α 1,nJ2 (α1,n ) 1,n n=1

7. We proceed as in the previous exercise and try u(r, θ) =

∞ X ∞ X

∞ X ∞ X

Jm (λmn r)(Amn cos mθ + Bmn sin mθ) =

m=0 n=1

φmn(r, θ),

m=0 n=1

where φmn (r, θ) = Jm (λmn r)(Amn cos mθ + Bmn sin mθ). We plug this solution into the equation, use the fact that ∇2(φmn ) = −λ2mn φmn = −α2mn φmn , and get ! ∞ ∞ X X ∇2 φmn (r, θ) = 2 + r3 cos 3θ m=0 n=1

⇒

∞ X ∞ X

∇2 (φmn (r, θ)) = 2 + r3 cos 3θ

⇒

m=0 n=1 ∞ X ∞ X

−α2mn φmn (r, θ) = 2 + r3 cos 3θ.

m=0 n=1

We recognize this expansion as the expansion of the function 2 + r3 cos 3θ in terms of the functions φmn. Because of the special form of the right side, it follows that all Amn and Bmn are zero, except A0,n and A3,n. So ∞ X

−α20nA0,nJ0 (α0nr) = 2

n=1

and cos 3θ

∞ X

−α23nA3,n J3(α3nr) = r3 cos 3θ.

n=1

Using the series in Example 1, Section 4.8: 1= we find that 2= and so

∞ X

∞ X

2 J0 (α0,nr) 0 < r < 1, α J (α0,n) n=1 0,n 1 ∞ X

4 J0 (α0,nr) 0 < r < 1, α J (α0,n) 0,n 1 n=1

−α20nA0,nJ0 (α0nr) =

n=1

∞ X

4 J0 (α0,nr), α J (α0,n) n=1 0,n 1

from which we conclude that −α20nA0,n =

4 α0,nJ1 (α0,n)

Section 4.6

or A0,n =

The Helmholtz and Poisson Equations

201

−4

. α30,nJ1 (α0,n)

From the series in Exercise 20, Section 4.8 (with m = 3): We have 3

r =

∞ X

2 J3 (α3,nr) α J (α3,n) n=1 3,n 4

0 < r < 1,

we conclude that ∞ X

∞ X 2 J3(α3,nr) = −α23nA3,nJ3(α3nr) α J (α ) 3,n 4 3,n n=1 n=1

and so A3,n =

−2 . α33,nJ4 (α3,n)

Hence ∞ X

∞ X −4 −2 u(r, θ) = J0(α0,nr) + cos 3θ J3 (α3,nr). 3 J (α 3 J (α α ) α 0,n 3,n ) n=1 0,n 1 n=1 3,n 4

8. We proceed as in Exercise 5. Let u(r, θ) =

∞ ∞ X X

φmn(r, θ),

m=0 n=1

where φmn(r, θ) = Jm (λmn r)(Amn cos mθ + Bmn sin mθ). Plug this solution into the equation, use the fact that ∇2(φmn ) = −λ2mn φmn , and get ! ∞ ∞ X X 2 ∇ φmn (r, θ) = r2 m=0 n=1

⇒

∞ ∞ X X

−α2mn φmn (r, θ) = r2 .

m=0 n=1

We recognize this expansion as the expansion of the function 1 in terms of the functions φmn . Because the right side is independent of θ, it follows that all Amn and Bmn are zero, except A0,n. So ∞ X

−α2mn A0,nJ0(r) = r2 ,

n=1

which shows that −α2mn A0,n = a0,n is the nth Bessel coefficient of the Bessel series expansion of order 0 of the function r2. By Theorem 2, Section 4.8, we have Z 1 2 a0,n = r2 J0(α0,n)r dr J1 (α0,n)2 0 Z α0,n 2 = s3 J0 (s) ds (let α0,n)r = s) α40,nJ1(α0,n)2 0

= = =

u = s2 , du = 2s ds, dv = sJ0 (s) ds, v = sJ1 (s) Z α0,n α0,n 2 3 2 s J (s) − 2 s J (s) ds 1 1 α40,nJ1(α0,n)2 0 0 α0,n i h 2 3 2 α J (α ) − 2s J (s) 1 0,n 2 α40,nJ1(α0,n)2 0,n 0 3 2 α0,nJ1(α0,n) − 2α20,nJ2 (α0,n) . 4 2 α0,nJ1(α0,n)

202

Chapter 4

Partial Differential Equations in Polar and Cylindrical Coordinates

Hence A0,n =

−2 α40,nJ1 (α0,n)2

[α0,nJ1(α0,n) − 2J2(α0,n)] ,

and so u(r, θ) =

∞ X

−2 [α0,nJ1 (α0,n) − 2J2(α0,n)] J0 (α0,nr). 4 J (α 2 α 0,n ) n=1 0,n 1

9. Let h(r) =

r 0

if 0 < r < 1/2, if 1/2 < r < 1.

Then the equation becomes ∇2u = f(r, θ), where f(r, θ) = h(r) sin θ. We proceed as in the previous exercise and try u(r, θ) =

∞ ∞ X X

Jm (λmn r)(Amn cos mθ + Bmn sin mθ) =

m=0 n=1

∞ ∞ X X

φmn(r, θ),

m=0 n=1

where φmn (r, θ) = Jm (λmn r)(Amn cos mθ + Bmn sin mθ). We plug this solution into the equation, use the fact that ∇2(φmn ) = −λ2mn φmn = −α2mn φmn , and get ! ∞ ∞ X X 2 ∇ φmn(r, θ) = h(r) sin θ m=0 n=1

⇒

∞ ∞ X X

∇2 (φmn (r, θ)) = h(r) sin θ

⇒

m=0 n=1 ∞ ∞ X X

−α2mn φmn(r, θ) = h(r) sin θ.

m=0 n=1

We recognize this expansion as the expansion of the function h(r) sin θ in terms of the functions φmn . Because the right side is proportional to sin θ, it follows that all Amn and Bmn are zero, except B1,n. So sin θ

∞ X

−α21nB1,n J1 (α1nr) = h(r) sin θ,

n=1

which shows that −α21nB1,n is the nth Bessel coefficient of the Bessel series expansion of order 1 of the function h(r): −α21nB1,n

=

=

2 J2 (α1,n)2

Z

1/2

r2J1 (α1,nr) dr 0

2 α31,nJ2(α1,n)2

Z

α1,n /2

s2 J1(s) ds 0

=

α1,n /2 2 2 s J (s) 2 α31,nJ2(α1,n)2 0

=

J2 (α1,n/2) . 2α1,nJ2(α1,n)2

Thus u(r, θ) = sin θ

∞ X n=1

−

J2 (α1,n/2) J1 (α1,nr). 2α31,nJ2 (α1,n)2

Section 4.6

The Helmholtz and Poisson Equations

203

11. We have u(r, θ) = u1(r, θ)+u2 (r, θ), where u1(r, θ) is the solution of ∇2 u = 1 and u(1, θ) = 0, and u2(r, θ) is the solution of ∇2u = 0 and u(1, θ) = sin 2θ. By Example 1, we have ∞ X −2 J (αn r); u1 (r, θ) = 3 J (α ) 0 α n n=1 n 1 while u2 is easily found by the methods of Section 4.4: u2 (r θ) = r2 sin 2θ. Thus, u(r, θ) = r2 sin 2θ +

∞ X

−2 J (αnr). 3 J (α ) 0 α n n=1 n 1

204

Chapter 4

Partial Differential Equations in Polar and Cylindrical Coordinates

Solutions to Exercises 4.7 1. Bessel equation of order 3. Using (7), the first series solution is J3 (x) =

∞ X k=0

(−1)k x 2k+3 1 x3 1 x5 1 x7 = − + +··· . k!(k + 3)! 2 1·6 8 1 · 24 32 2 · 120 128

5. Bessel equation of order 32 . The general solution is y(x)

= =

c1 J 32 + c2J− 32 x 32 x 72 1 1 c1 − + · · · 1 · Γ( 52 ) 2 1 · Γ( 72 ) 2 x − 32 x 12 1 1 +c2 − +··· . 1 · Γ( −1 ) 2 1 · Γ( 12 ) 2 2

Using the basic property of the gamma function and (15), we have √ Γ( 52 ) = 32 Γ( 32 ) = 32 12 Γ( 12 ) = 34 π √ Γ( 72 ) = 52 Γ( 52 ) = 15 π 8 √ √ 1 1 1 − 2 Γ(− 2 ) = Γ( 2 ) = π ⇒ Γ(− 12 ) = −2 π. So y(x)

=

=

r

4 x2 8 x4 c1 − +··· 3 4 15 16 r 2 1 2 x c2 (−1) − − −··· πx 2 x 2 r r 2 2 x 2 x4 1 x c1 − + · · · + c2 + −··· πx 3 30 πx x 2 2 πx

9. Divide the equation through by x2 and put it in the form y00 +

1 0 x2 − 9 y = 0 for x > 0. y + x x2

Now refer to Appendix A.6 for terminology and for the method of Frobenius that we are about to use in this exercise. Let p(x) =

1 x

for q(x) =

x2 − 9 . x2

The point x = 0 is a singular point of the equation. But since x p(x) = 1 and x2 q(x) = x2 − 9 have power series expansions about 0 (in fact, they are already given by their power series expansions), it follows that x = 0 is a regular singular point. Hence we may apply the Frobenius method. We have already found one series solution in Exercise 1. To determine the second series solution, we consider the indicial equation r(r − 1) + p0r + q0 = 0, where p0 = 1 and q0 = −9 (respectively, these are the constant terms in the series expansions of xp(x) and x2q(x)). Thus the indicial equation is r−9=0

⇒

r1 = 3, r2 = −3.

The indicial roots differ by an integer. So, according to Theorem 2, Appendix A.6, the second solution y2 may or may not contain a logarithmic term. We have, for x > 0, ∞ ∞ X X −3 y2 = ky1 ln x + x bm xm = ky1 ln x + bm xm−3 , m=0

m=0

Section 4.7

Bessel’s Equation and Bessel Functions

205

where a0 6= 0 and b0 6= 0, and k may or may not be 0. Plugging this into the differential equation x2y00 + xy0 + (x2 − 9)y = 0 and using the fact that y1 is a solution, we have y2

=

ky1 ln x +

∞ X

bm xm−3

m=0

y20

=

ky10 ln x + k

∞ X y1 (m − 3)bm xm−4 ; + x m=0

y200

=

ky100 ln x + k

∞ X y10 xy0 − y1 + (m − 3)(m − 4)bm xm−5 +k 1 2 x x m=0

=

ky100 ln x + 2k

∞ X y10 y1 (m − 3)(m − 4)bm xm−5 ; −k 2 + x x m=0

x2y200 + xy20 + (x2 − 9)y2 =

kx2y100 ln x + 2kxy10 − ky1 +

∞ X

(m − 3)(m − 4)bm xm−3

m=0 ∞ X

+kxy10 ln x + ky1 + 2

(m − 3)bm xm−3

m=0 ∞ X

"

+(x − 9) ky1 ln x +

bm x

m−3

#

m=0 =0

=

}| { z k ln x x2y100 + xy10 + (x2 − 9)y1 ∞ X 0 +2kxy1 + (m − 3)(m − 4)bm + (m − 3)bm − 9bm xm−3 m=0

+x2

∞ X

bm xm−3

m=0

=

2kxy10 +

∞ X

(m − 6)mbm xm−3 +

m=0

∞ X

bm xm−1.

m=0

To combine the last two series, we use reindexing as follows ∞ X

(m − 6)mbm xm−3 +

m=0

=

bm xm−1

m=0

−5b1x−2 +

∞ X

−5b1x−2 +

∞ X

(m − 6)mbm xm−3 +

m=2

=

∞ X

bm−2xm−3

m=2

∞ X (m − 6)mbm + bm−2 xm−3. m=2

Thus the equation x2y200 + xy20 + (x2 − 9)y2 = 0 implies that 2kxy10 − 5b1x−2 +

∞ X (m − 6)mbm + bm−2 xm−3 = 0. m=2

206

Chapter 4

Partial Differential Equations in Polar and Cylindrical Coordinates

This equation determines the coefficients bm (m ≥ 1) in terms of the coefficients of y1 . Furthermore, it will become apparent that k cannot be 0. Also, b0 is arbitrary but by assumption b0 6= 0. Let’s take b0 = 1 and determine the the first five bm ’s. Recall from Exercise 1 y1 =

1 x5 1 x7 1 x3 − + + ··· . 1·6 8 1 · 24 32 2 · 120 128

So

3 x2 5 x4 7 x6 − + + ··· 1·6 8 1 · 24 32 2 · 120 128 and hence (taking k = 1) y10 =

2kxy10 =

6k x3 10k x5 14k x7 − + +··· . 1·6 8 1 · 24 32 2 · 120 128

The lowest exponent of x in 2kxy10 − 5b1x−2 +

∞ X (m − 6)mbm + bm−2 xm−3 m=2

is x−2. Since its coefficient is −5b1, we get b1 = 0 and the equation becomes 2xy10 +

∞ X

(m − 6)mbm + bm−2 xm−3 .

m=2

Next, we consider the coefficient of x−1. It is (−4)2b2 + b0 . Setting it equal to 0, we find b0 1 b2 = = . 8 8 Next, we consider the constant term, which is the m = 3 term in the series. Setting its coefficient equal to 0, we obtain (−3)3b3 + b1 = 0

⇒

b3 = 0

because b1 = 0. Next, we consider the term in x, which is the m = 4 term in the series. Setting its coefficient equal to 0, we obtain (−2)4b4 + b2 = 0

⇒

b4 =

1 1 b2 = . 8 64

Next, we consider the term in x2, which is the m = 5 term in the series. Setting its coefficient equal to 0, we obtain b5 = 0. Next, we consider the term in x3, which is the m = 6 term in the series plus the first term in 2kxy10 . Setting its coefficient equal to 0, we obtain 0 + b4 +

k =0 8

⇒

1 k = −8b4 = − . 8

Next, we consider the term in x4 , which is the m = 7 term in the series. Setting its coefficient equal to 0, we find that b7 = 0. It is clear that b2m+1 = 0 and that 1 1 1 1 y2 ≈ − y1 ln x + 3 + + x+ ··· 8 x 8 x 64 Any nonzero constant multiple of y2 is also a second linearly independent solution of y1 . In particular, 384 y2 is an alternative answer (which is the answer given in the text). 10. Linear independence for two solutions is equivalent to one not being a constant multiple of the other. Since Jp is not a constant multiple of Yp , it follows that

Section 4.7

Bessel’s Equation and Bessel Functions

207

xp Jp are not a constant multiple of xp Yp and hence xp Jp and xp Yp are linearly independent. We now verify that they are solutions of xy00 + (1 − 2p)y0 + xy = 0. We have y

=

xpJp (x),

y0

=

pxp−1Jp (x) + xp Jp0 ,

y00

=

xy00 + (1 − 2p)y0 + xy

=

=

p(p − 1)xp−2Jp (x) + 2pxp−1Jp0 (x) + xp Jp00 (x), x p(p − 1)xp−2 Jp (x) + 2pxp−1Jp0 (x) + xp Jp00 (x) +(1 − 2p) pxp−1Jp (x) + xp Jp0 (x) + x xpJp (x) xp−1 x2 Jp00(x) + [2px + (1 − 2p)x]Jp0 (x) +[p(p − 1) + (1 − 2p)p + x2]Jp (x) xp−1 x2 Jp00(x) + xJp0 (x) + (x2 − p2 )Jp (x)

=

0

=

because Jp satisfies x2Jp00 (x) + xJp0 (x) + (x2 − p2)Jp (x) = 0. A similar proof shows that xp Yp is also a solution of xy00 + (1 − 2p)y0 + xy = 0. 11. The equation is of the form given in Exercise 10 with p = 0. Thus its general solution is y = c1J0 (x) + c2Y0 (x). This is, of course, the general solution of Bessel’s equation of order 0. Indeed, the given equation is equivalent to Bessel’s equation of order 0. 12. The equation is of the form given in Exercise 10 with p = 1/2. Thus its general solution is y = c1 x1/2J1/2(x) + c2x1/2Y1/2 (x). But it is also clear that y = a cos x + b sin x is the general solution of y00 + y = 0. In fact, the two solutions are equivalent since x1/2 J1/2(x) is a constant multiple of sin x (see Example 1) and x1/2 Y1/2(x) is a constant multiple of cos x. (From Example 1, it follows that x1/2 J−1/2 (x) is a constant multiple of cos x. But, by (10), Y1/2 (x) = −J−1/2(x). So x1/2 Y1/2(x) is a constant multiple of cos x as claimed.) 13. The equation is of the form given in Exercise 10 with p = 3/2. Thus its general solution is y = c1 x3/2J3/2(x) + c2x3/2Y3/2 (x). Using Exercise 22 and (1), you can also write this general solution in the form h cos x i sin x y = c1x − cos x + c2x − − sin x x x =

c1 [sin x − x cos x] + c2 [− cos x − x sin x] .

In particular, two linearly independent solution are y1 = sin x − x cos x and y2 = cos x + x sin x. This can be verified directly by using the differential equation (try it!).

208

Chapter 4

Partial Differential Equations in Polar and Cylindrical Coordinates

14. The equation is of the form given in Exercise 10 with p = 2. Thus its general solution is y = c1x2J2 (x) + c2x2 Y2 (x). 15. All three parts, (a)–(c), follow directly from the series form of Jp (x) (see (7)). 17. We have y

=

x−pu,

y0

=

−px−p−1u + x−pu0 ,

y00

=

xy00 + (1 + 2p)y0 + xy

=

p(p + 1)x−p−2u + 2(−p)x−p−1u0 + x−pu00, x p(p + 1)x−p−2u − 2px−p−1u0 + x−p u00 +(1 + 2p) − px−p−1u + x−p u0 + x x−p u x−p−1 x2 u00 + [−2px + (1 + 2p)x]u0 +[p(p + 1) − (1 + 2p)p + x2]u x−p−1 x2 u00 + xu0 + (x2 − p2 )u .

= =

Thus, by letting y = x−p u, we transform the equation xy00 + (1 + 2p)y0 + xy = 0 into the equation

x−p−1 x2u00 + xu0 + (x2 − p2)u = 0,

which, for x > 0, is equivalent to x2u00 + xu0 + (x2 − p2 )u = 0, a Bessel equation of ordr p > 0 in u. The general solution of the last equation is u = c1 Jp (x) + c2Yp (x). Thus the general solution of the original equation is Y = c1 x−p Jp (x) + c2x−p Yp (x). 18. For x > 0, let z = chain rule,

√

dy dx

=

d2y dx2

= =

= =

x and φ(z) = y(x). Then

dz dx

=

2

1 √

x

=

d dφ dz dφ 1 dφ 1 √ = φ(z) = = , dx dz dx dz 2 x dz 2 z d dy d dφ 1 √ = dx dx dx dz 2 x d dφ 1 d 1 dφ √ + √ dx dz 2 x dx 2 x dz d2 φ dz 1 dφ 1 − x−3/2 dz 2 dx 2 z 4 dz 2 1 d φ −2 dφ z − z −3 . 2 4 dz dz

1 2z

and, by the

Section 4.7

Bessel’s Equation and Bessel Functions

209

So 1 d2φ −2 dφ 1 1 1 −3 dφ + + φ=0 xy00 + y0 + y = 0 ⇒ z 2 z − z 4 4 dz 2 dz dz 2 z 4 ⇒

d2 φ dφ + z −1 +φ=0 2 dz dz

⇒ z2

d2φ dφ + z 2 φ = 0. +z 2 dz dz

This is a Bessel equation of order 0 in φ. Its general solution is φ(z) = c1J0 (z) + c2 Y0(z). Thus the solution of the original equation is √ √ y(x) = c1 J0 ( x) + c2Y0 ( x). 21. Using (7), J− 12 (x)

= = = =

x 2k− 2 (−1)k k!Γ(k − 12 + 1) 2 k=0 r ∞ 2 X (−1)k x2k x k!Γ(k + 12 ) 22k k=0 r ∞ 2 X (−1)k 22k k! x2k √ (by Exercise 44(a)) x k! (2k)! π 22k k=0 r r ∞ 2 X (−1)k x2k 2 = cos x. πx k! (2k)! πx ∞ X

1

k=0

22. (a) Using (7), J 3 (x) = 2

= =

x 2k+ 32 (−1)k k!Γ(k + 32 + 1) 2 k=0 r ∞ 2X (−1)k x2k+2 1 x k!Γ(k + 2 + 2 ) 22k+2 k=0 r ∞ 2 X (−1)k 2 22k+1k! x2k+2 πx k! (2k + 3)(2k + 1)! 22k+2 ∞ X

k=0

= = = = =

1 1 1 (Γ(k + 2 + ) = Γ(k + 1 + )Γ(k + 1 + ) then use Exercise 44(b)) 2 2 2 r ∞ 2 X (−1)k (2k + 2) 2k+2 (multiply and divide by (2k + 2)) x πx (2k + 3)! k=0 r ∞ 2 X (−1)k−1(2k) 2k (change k to k − 1) x πx (2k + 1)! k=1 r ∞ 2 X (−1)k−1[(2k + 1) − 1] 2k x πx (2k + 1)! k=1 r r ∞ ∞ 2 X (−1)k−1 2k 2 X (−1)k−1 2k x − x πx (2k)! πx (2k + 1)! k=1 k=1 r 2 sin x − cos x + . πx x

210

Chapter 4

Partial Differential Equations in Polar and Cylindrical Coordinates

25.

(a) Let u =

2 a

1

e− 2 (at−b), Y (u) = y(t), e−at+b =

a2 4

u2 ; then

dy 1 1 a 1 d2y d 0 dY du = = Y 0 (−e− 2 (at−b)); 2 = Y (−e− 2 (at−b)) = Y 00e−at+b +Y 0 e− 2 (at−b). dt du dt dt du 2 So a 1 Y 00e−at+b + Y 0 e− 2 (at−b) + Y e−at+b = 0 2

⇒

1 a Y 00 + Y 0e− 2 (at−b) + Y = 0, 2

1

upon multiplying by eat−b. Using u = a2 e− 2 (at−b), we get Y 00 +

1 0 Y +Y =0 u

⇒

u2Y 00 + uY 0 + u2Y = 0,

which is Bessel’s equation of order 0. (b) The general solution of u2Y 00 + uY 0 + u2Y = 0 is Y (u) = c1 J0 (u) + c2Y0 (u). 1 But Y (u) = y(t) and u = a2 e− 2 (at−b), so 2 1 2 1 y(t) = c1J0 ( e− 2 (at−b)) + c2 Y0 ( e− 2 (at−b)). a a (c) (i) If c1 = 0 and c2 6= 0, then 2 1 y(t) = c2Y0 ( e− 2 (at−b) ). a As t → ∞, u → 0, and Y0(u) → −∞. In this case, y(t) could approach either +∞ or −∞ depending on the sign would approach infinity linearly as near of 1c2 . y(t) 2 − 2 (at−b) 0, Y0 (x) ≈ ln x so y(t) ≈ ln a e ≈ At. (ii) If c1 6= 0 and c2 = 0, then 2 1 y(t) = c1 J0( e− 2 (at−b)). a As t → ∞, u(t) → 0, J0 (u) → 1, and y(t) → c1. In this case the solution is bounded. (ii) If c1 6= 0 and c2 6= 0, as t → ∞, u(t) → 0, J0 (u) → 1, Y0 (u) → −∞. Since Y0 will dominate, the solution will behave like case (i). It makes sense to have unbounded solutions because eventually the spring wears out and does not affect the motion. Newton’s laws tell us the mass will continue with unperturbed momentum, i.e., as t → ∞, y00 = 0 and so y(t) = c1 t + c2 , a linear function, which is unbounded if c1 6= 0. R∞ 31. (a) From Γ(x) = 0 tx−1 e−t dt, we obtain ∞ Z ∞ −t −t Γ(1) = e dt = −e = 1; 0

Γ(2) =

0

=0

Z

∞ 0

z }| { Z ∞ −t te dt = −te−t + 0

∞

e−t dt = 1. 0

(b) Using (15) and the basic property of the gamma function √ Γ( 52 ) = 32 Γ( 32 ) = 32 12 Γ( 12 ) = 34 π √ √ − 32 Γ(− 32 ) = Γ(− 12 ) = −2 π (see Exercise 5) ⇒ Γ(− 32 ) = 43 π. 33. (a) In (13), let u2 = t, 2u du = dt, then Z ∞ Z ∞ Z 2 Γ(x) = tx−1e−t dt = u2(x−1)e−u (2u)du = 2 0

0

0

∞

2

u2x−1e−u du.

Section 4.7

Bessel’s Equation and Bessel Functions

211

(b) Using (a) Γ(x)Γ(y)

Z ∞ 2 2 u2x−1e−u du2 v2y−1e−v dv 0 Z0 ∞ Z ∞ 2 2 4 e−(u +v ) u2x−1v2y−1 du dv.

=

2

=

Z

∞

0

0

(c) Switching to polar coordinates: u = r cos θ, v = r sin θ, u2 + v2 = r2 , dudv = rdrdθ; for (u, v) varying in the first quadrant (0 ≤ u < ∞ and 0 ≤ v < ∞), we have 0 ≤ θ ≤ π2 , and 0 ≤ r < ∞, and the double integral in (b) becomes Γ(x)Γ(y)

=

4

Z

∞

Z

=

2

2

e−r (r cos θ)2x−1 (r sin θ)2y−1 rdrdθ

0

0

Z

π 2

π 2

(cos θ)

2x−1

(sin θ)

2y−1

∞

dθ 2

0

=

=Γ(x+y)

zZ

r

}|

2(x+y)−1 −r 2

e

{

dr

0

(use (a) with x + y in place of x) Z π2 2Γ(x + y) (cos θ)2x−1 (sin θ)2y−1 dθ, 0

implying (c). 34. Using Exercise 33 with x = y = 12 , 1 [Γ( )]2 = 2Γ(1) 2 √ so Γ( 12 ) = π.

Z

π 2

cos1−1 θ sin1−1 θdθ = 2

0

Z

π 2

dθ = 2 0

π = π, 2

35. From Exercise 33(a) and Exercise 34, √ 1 π = Γ( ) = 2 2 so

√ Z ∞ 2 π e−u du = 2 0

⇒

Z

∞

2

e−u du, 0

√

π=

Z

∞

2

e−u du, −∞

2

because e−u is an even function. R π/2 38. Let I = 0 cos5 θ sin6 θ dθ. Applying Exercise 33, we take 2x − 1 = 5 and 2y − 1 = 6, so x = 3 and y = 72 . Then Γ(3) Γ( 72 ) = 2I. Γ(3 + 72 ) Now, Γ(3) = 2! = 2, and 7 7 7 4+7 2+7 7 11 9 7 7 693 7 Γ(3 + ) = (2 + )Γ(2 + ) = ( )( )Γ(1 + ) = Γ( ) = Γ( ). 2 2 2 2 2 2 2 2 2 2 8 2 So I=

1 Γ(3) Γ( 72 ) = 2 Γ(3 + 72 )

Γ( 72 ) 8 693 7 = 693 . Γ( ) 8 2

R π/2 40. Let I = 0 sin2k θ dθ. Applying Exercise 33, we take 2x − 1 = 0 and 2y − 1 = 2k, so x = 12 and y = k + 12 . Then 2I =

Γ( 12 ) Γ(k + 12 ) Γ( 12 ) Γ(k + 12 ) . = Γ(k + 1) Γ( 12 + k + 12 )

212

Chapter 4

Partial Differential Equations in Polar and Cylindrical Coordinates

Now, Γ( 12 ) =

√

π, Γ(k + 1) = k!, and Γ(k + 12 ) = I=

(2k)! √ π 22k k!

(see Exercise 44). So

√ π (2k)! √ π (2k)! π= . 2(k!) 22k k! 2 22k(k!)2

R π/2 41. Let I = 0 sin2k+1 θ dθ. Applying Exercise 33, we take 2x − 1 = 0 and 2y − 1 = 2k + 1, so x = 12 and y = k + 1. Then √ √ Γ( 12 ) Γ(k + 1) π k! 2 π k! 2I = = = . Γ(k + 1 + 12 ) (k + 12 )Γ(k + 12 ) (2k + 1)Γ(k + 12 ) As in (a), we now use Γ(k + 12 ) =

(2k)! √ 22k k! π,

I=

simplify, and get

22k (k!)2 . (2k + 1)!

44. (a) √ Proof by induction. If n = 0, the formula clearly holds since by (15) Γ(1/2) = π. Now suppose that he formula holds for n and let us establish it for n + 1. That is, suppose that 1 (2n)! √ Γ(n + ) = 2n π 2 2 n! (the induction hypothesis) and let us prove that 1 (2(n + 1))! √ Γ(n + 1 + ) = 2(n+1) π. 2 2 (n + 1)! Starting with the right side, we have (2(n + 1))! √ π + 1)!

22(n+1) (n

(2n + 2)(2n + 1)((2n)!) √ π 2222n (n + 1)(n!) 1 (2n)! √ (n + ) 2n π 2 2 n! 1 1 (n + )Γ(n + ) (by the induction hypothesis) 2 2 1 Γ(n + + 1), 2

= = = =

by the basic property of the gamma function. Thus the formula is true for n + 1 and hence for all n ≥ 0. (b) We use (a) and the basic property of the gamma function: Γ(n +

1 + 1) = 2 = =

1 1 (n + )Γ(n + ) 2 2 1 (2n)! √ (2n + 1) (2n)! √ (n + ) 2n π= π 2 2 n! 2 22n n! (2n + 1)! √ π. 22n+1 n!

Section 4.8

Bessel Series Expansions

213

Solutions to Exercises 4.8 1. (a) Using the series definition of the Bessel function, (7), Section 4.7, we have d −p [x Jp (x)] = dx

x 2k d X (−1)k dx 2p k!Γ(k + p + 1) 2 ∞

k=0

(−1)k (−1)k 2k d x 2k X 1 x 2k−1 = 2p k!Γ(k + p + 1) dx 2 2p k!Γ(k + p + 1) 2 2

=

∞ X

=

k=0 ∞ X

=

k=0 ∞ X

−

=

−x−p

∞

k=0

k

(−1) 2p (k − 1)!Γ(k + p + 1)

x 2k−1 2

x 2m+1 (−1)m 2p m!Γ(m + p + 2) 2 m=0

(set m = k − 1)

x 2m+p+1 (−1)m = −x−p Jp+1 (x). m!Γ(m + p + 2) 2 m=0 ∞ X

To prove (7), use (1): d p [x Jp (x)] = xp Jp−1 (x) dx

⇒

Z

xp Jp−1 (x) dx = xp Jp (x) + C.

Now replace p by p + 1 and get Z xp+1 Jp (x) dx = xp+1 Jp+1 (x) + C, which is (7). Similarly, starting with (2), d −p [x Jp (x)] = −x−p Jp+1 (x) dx

Z

⇒ − x−pJp+1 (x) dx = x−p Jp (x) + C Z ⇒ x−p Jp+1 (x) dx = −x−p Jp (x) + C.

Now replace p by p − 1 and get Z x−p+1 Jp (x) dx = −x−p+1 Jp−1(x) + C, which is (8). (b) To prove (4), carry out the differentiation in (2) to obtain x−p Jp0 (x) − px−p−1Jp (x) = −x−p Jp+1 (x)

⇒

xJp0 (x) − pJp (x) = −xJp+1 (x),

upon multiplying through by xp+1 . To prove (5), add (3) and (4) and then divide by x to obtain Jp−1 (x) − Jp+1 (x) = 2Jp0 (x). To prove (6), subtract (4) from (3) then divide by x. d 2. (a) Since dx J0 (x) = −J1 (x), the critical points (maxima or minima) of J0 occur where its derivative vanishes, hence at the zeros of J1(x). (b) At the maxima or minima of Jp , we have Jp0 (x) = 0. From (5), this implies that, at a critical point, Jp−1 (x) = Jp+1 (x). Also, from (3) and (4) it follows that

x=p

3.

Z

Jp (x) Jp (x) =p . Jp−1 (x) Jp+1 (x)

xJ0(x) dx = xJ1(x) + C, by (7) with p = 0.

214

Chapter 4

Partial Differential Equations in Polar and Cylindrical Coordinates

4.

Z

5.

Z

6.

Z

7.

Z

8.

Z

x4 J3 (x) dx = x4 J4(x) + C, by (7) with p = 3. J1(x) dx = −J0 (x) + C, by (8) with p = 1. x−2J3 (x) dx = −x−2 J2 (x) + C, by (8) with p = 3. x3J2 (x) dx = x3 J3(x) + C, by (7) with p = 2.

x3J0 (x) dx =

= = = 9.

Z

J3 (x) dx =

Z

x2[xJ0(x)] dx

x2 = u, xJ0(x) dx = dv, 2x dx = du, v = xJ1(x) Z x3J1 (x) − 2 x2 J1(x) dx = x3J1 (x) − 2x2J2(x) + C 2 x3J1 (x) − 2x2 J1(x) − J0 (x) + C (use (6) with p = 1) x (x3 − 4x)J1(x) + 2x2J0 (x) + C. Z

x2[x−2J3 (x)] dx

x2 = u, x−2J3(x) dx = dv, 2x dx = du, v = −x−2 J2 (x) Z = −J2 (x) + 2 x−1 J2(x) dx = −J2 (x) − 2x−1 J1(x) + C 2 2 J1(x) − J1(x) + C(use (6) with p = 1) x x 4 = J0 (x) − J1(x) + C. x = J0 (x) −

10. We assume throughout this exercise that n is a positive integer. Let u = J0 (x), then du = −J1 (x). If n 6= −1, then ∞ Z ∞ Z ∞ 1 −1 1 n n n+1 J1(x)[J0(x)] dx − u du = − = [J0 (x)] (0−1) = , n+1 n+1 n+1 0 x=0 0 where we have used limx→∞ J0 (x) = 0 and J0(0) = 1. 11. By (2) with p = 12 , J3/2 (x) = =

r d −1 1 2 d sin x −x x 2 J 12 (x) = −x 2 dx π dx x r r 2 1 cos x sin x 2 sin x 2 − = x − 2 − cos x . π x x πx x 1 2

12. By (2) with p = 32 , J5/2(x)

= = = =

r d −3 3 2 d sin x cos x −x − 2 x 2 J 32 (x) = −x 2 dx π dx x3 x r 2 3 cos x sin x sin x cos x − −3 4 + 2 +2 3 x2 π x3 x x x r cos x 2 3 sin x sin x − x2 3 3 − 3 4 + 2 π x x x r 2 3 3 − 1 sin x − cos x . πx x2 x 3 2

Section 4.8

Bessel Series Expansions

215

13. Use (6) with p = 4. Then J5 (x)

= =

=

=

=

=

=

8 J4 (x) − J3 (x) x 8 6 J3(x) − J2 (x) − J3 (x) (by (6) with p = 3) x x 48 8 − 1 J3 (x) − J2 (x) x2 x 48 4 8 −1 J2 (x) − J1(x) − J2 (x) (by (6) with p = 2) x2 x x 192 12 48 − (x) − − 1 J1 (x) J 2 x3 x x2 12 16 2 48 −1 − 1 J1(x) J1 (x) − J0(x) − x x2 x x2 (by (6) with p = 1) 12 16 384 72 − − 1 J0 (x) + − 2 + 1 J1 (x). x x2 x4 x

q 14. From the formula J1/2(x) = π2x sin x, it follows that the zeros of J1/2(x) are located precisely at the zeros of sin x. The latter are located at nπ, n = 0, 1, 2, . . .. 15. (a) and (b) We will do this problem on Mathematica. Let us make some comments about the equation sinx x = cos x. This equation is equivalent to the equation tan x = x, which we have encountered in Section 3.6. It is clear from the graph that this equation has one root xk in each interval − π2 + kπ, π2 + kπ , k = 0, 1, 2, . . .. We have x0 = 0 and the remaininng roots can be obtained by iteration as follows. Take the case k = 1. Start at z1 = π and go up to the line y = x then over to the right to the graph of tan x. You will intersect this graph at the point z2 = π + tan−1 π. Repeat hese steps by going up to the line y = x and then over to the right to the graph of y = tan x. You will intersect the graph at z2 = π + tan−1 z1 . As you continue this process, you will approach the value of x1. To find xk , start at z1 = kπ and construct the sequence zn = kπ + tan−1 zn−1. Let us compute some numerical values and then compare them with the roots that we find by using built-in commands from Mathematica.

216

Chapter 4

Partial Differential Equations in Polar and Cylindrical Coordinates

grph1 Plot PlotRange

x, Tan x 2, 5

, x, 0, 3 Pi 2 , AspectRatio

1.5,

Section 4.8

z 1 Pi z k_ : N ArcTan z k Table z k , k, 1, 10

1

Bessel Series Expansions

Iterative process to approximate the first root of tan x = x

Pi

, 4.40422, 4.48912, 4.49321, 4.4934, 4.49341, 4.49341, 4.49341, 4.49341, 4.49341 pts

Table

z k , Tan z k

, k, 1, 5

, 0 , 4.40422, 3.14159 , 4.48912, 4.40422 , 4.49321, 4.48912 , 4.4934, 4.49321 grph2

ListPlot pts, PlotStyle

PointSize 0.02

4 3 2 1

3.2

3.4

3.6

3.8

4.2

Show grph1, grph2 , AspectRatio 2, 5 PlotRange

4.4

1.5,

5

4

tan zk , zk

217

218

Chapter 4

Partial Differential Equations in Polar and Cylindrical Coordinates

If you want to use a built-in command to find the root, you can do the following (compare the result with the iterative process above). FindRoot Tan x x

x, x, 4

4.49341

16. (a) There are j − 1 roots in (0, a). They are xk =

αpk a αpj

for k = 1, 2, . . . , j − 1,

which are just the j − 1 positive zeros of Jp (x), rescaled by the factor α reciprocal of the scale factor occuring in the argument of Jp apj x .

a αpj

(the

17. (a) From (17),

Aj

= = =

Z 1 Z c 2 2 f(x)J (α x)x dx = J0 (αj x)x dx 0 j J1(αj )2 0 J1 (αj )2 0 Z cαj 2 J0 (s)s ds (let αj x = s) α2j J1 (αj )2 0 cαj 2 2cJ1(αj ) J (s)s = . 1 α2j J1 (αj )2 αj J1(αj )2 0

Thus, for 0 < x < 1, f(x) =

∞ X 2cJ1(αj ) J0 (αj x). αj J1(αj )2 j=1

(b) The function f is piecewise smooth, so by Theorem 2 the series in (a) converges to f(x) for all 0 < x < 1, except at x = c, where the series converges to the average f (c+)+f (c−) value = 12 . 2 18. Done on Mathematica f x_ : 1 f x_ : 0 Plot f x ,

;0 x 1 2 ;1 2 x 1 x, 0, 1

1 0.8 0.6 0.4 0.2 0.2

0.4

0.6

0.8

1

Section 4.8

Bessel Series Expansions

To describe the Bessel series, we need the zeros of J0 x . These are built-in the program. To recall them, we need a package. 0. This will come out of the formula for the Bessel coefficients, using orthogonality. 31. p = 1, y = c1J1 (λx) + c2Y1 (λx). For y to be bounded near 0, we must take c2 = 0. For y(1) = 0, we must take λ = λj = α1,j , j = 1, 2, . . .; and so y = yi = c1,j J1(α1,j x). 33.

p =

1 2,

y = c1 J 12 (λx) + c2Y 12 (λx). For y to be bounded near 0, we must α 1 ,j

take c2 = 0. For y(π) = 0, we must take λ = λj = π2 = j, j = 1, 2, . . . (see Exercises 21); and so r α 1 ,j 2 2 y = yi = c1,j J1 ( x) = c1,j sin(jx) π πx (see Example 1, Section 4.7). 1

35. (b) Bessel’s equation of order 0 is xy00 + y0 + xy = 0. Take y = x− 2 u. Then 1 1 y0 = x− 2 [u0 − 2x u] and y00

= =

1 0 1 1 0 1 u + 2u − u − u ] 2x 2x 2x 2x 1 1 3 x− 2 [u00 − u0 + 2 u]. x 4x 1

x− 2 [u00 −

Substituting into Bessel’s equation and simplifying, we find 3 0 1 1 xu00 − u0 + u + u − u + xu = 0 ⇒ u00 + 1 + 2 u = 0 (∗). 4x 2x 4x √ Since J0(x) satisfies Bessel’s equation, it follows that u = xJ0(x) is a solution of the last displayed equation (∗). (c) Let v = sin x. Then v00 + v = 0. Multiplying this equation by u and (∗) by v, and then subtracting yields uv00 − vu00 =

1 uv 4x2

uv0 − u0v

⇒

0

=

1 uv, 4x2

which implies (c), since 4x1 2 uv = −(u00 + u)v from (b) (multiply the displayed equation in (b) by v). (d) is clear from (b). (e) Integrating both sides of the equation uv0 − u0v from (c), we obtain

Z

0

=

1 uv 4x2

1 uv dx = uv0 − u0 v + C. 4x2

So for any positive integer k, Z (2k+1)π (2k+1)π (2k+1)π 1 0 0 0 u(x) sin x dx = uv − u v = u cos x − u sin x 4x2 2kπ 2kπ 2kπ = − u(2kπ) + u((2k + 1)π) .

Section 4.8

Bessel Series Expansions

225

sin x (f) Consider the integrand in (e): u(x) . This is a continuous function on 4x2 sin x [2kπ, (2k+1)π]. The function 4x2 does not change signs in the interval (2kπ, (2k+ 1)π) (it is either always positive or always negative depending on the sign of sin x). x Let us consider the case when k is even. Then sin ≥ 0 for all x in (2kπ, (2k +1)π). 4x2 Assume that u(x) does not vanish in (2kπ, (2k + 1)π). So u(x) > 0 or u(x) < 0 for all x in (2kπ, (2k + 1)π). Take the case u(x) > 0 for all x in (2kπ, (2k + 1)π). sin x > 0 for all x in (2kπ, (2k + 1)π), which implies that Then u(x) 4x2 Z (2k+1)π u(x) sin x dx > 0. 4x2 2kπ

By continuity of u, it follows that u(2kπ) ≥ 0 and u((2k + 1)π) ≥ 0, and so (e) implies that Z (2k+1)π u(x) sin x dx ≤ 0, 4x2 2kπ which is a contradiction. Thus u(x) is not positive for all x in (2kπ, (2k + 1)π) and so it must vanish at some point inside this interval. A similar argument shows that u(x) cannot be negative for all x in (2kπ, (2k + 1)π). A similar argument works when k is odd and shows that in all situations, u must vanish at some point in (2kπ, (2k + 1)π). (g) From (f) it follows that u has infinitely many zeros on the positive real line; at least one in each interval of the form (2kπ, (2k + 1)π). Since these intervals are disjoint, these zeros are distinct. Now J0 and u have the same zeros, and so the same applies to the zeros of J0 . Note: You can repeat the above proof with sin x replaced by sin(x−a) and show that u (and hence J0) has at least one zero in every interval of the form (a, a + π) for any a > 0. α 36. (a) We know that the function Jp apk x is a solution of the parametric form of Bessel’s equation of order p > 0, x2y00 + xy0 + (λ2k x2 − p2)y = 0, α

where λk = apk and αk is the kth positive zero of Jp . Rewrite the left side of the equation as follows: for x > 0, x2y00 + xy0 + (λ2k x2 − p2)y

= = = =

Hence Jp

αpk x a

λ2k x2 − p2 y x 2 0 p xy0 + λ2k xy − y x 0 (xy0 ) xy0 + λ2k x2 yy0 − p2yy0 1 0 1 0 (xy0 )2 + (λ2k x2 − p2 ) y2 2 2

xy00 + y0 +

(Divide by x.) (Use product rule.) (Multiply by xy0 .) (Chain rule.)

is a solution of 1 0 1 0 (xy0 )2 + (λ2k x2 − p2 ) y2 = 0 2 2

or

0 2 0 0 (xy ) + (λ2k x2 − p2) y2 = 0.

(b) and (c) Integrate from 0 to a and use that y(a) = 0 (because y(a) = Jp (αpk ) = 0) a Z a 0 0 2 (xy ) + (λ2k x2 − p2 ) y2 dx = 0 (The integral cancels the 0 0 first derivative.) Z a 0 2 a 2 2 2 2 2 2 (xy ) + (λk x − p )y − 2λk y x dx = 0 (Integrate by parts.) 0 Z0 a (ay0 (a))2 − 2λ2k y2 x dx = 0 y(a) = 0, y(0) = 0 if p > 0.) 0

226

Chapter 4

Partial Differential Equations in Polar and Cylindrical Coordinates

Thus, with y = Jp

αpk a x ,

we obtain

[y0 (a)]2 =

2λ2k a2

Z

a

Jp 0

α

pk

a

x

2

x dx.

(d) For y = Jp (λk x), by (4), we have λk xJp0 (λk x) − pJp (λk x) = −λk xJp+1 (λk x) or xy0 (x) − py(x) = −λk xJp+1 (λk x) . Substituting x = a and using y(a) = 0, we get ay0 (a) = −λk aJp+1 (αk ). Thus from (c) Z a αpk 2 a2 a2 Jp x x dx = 2 λ2k Jp+1 (αk )2 = [Jp+1 (αk )]2. a 2λk 2 0 One more formula. To complement the integral formulas from this section, consider the following interesting formula. Let a, b, c, and p be positive real numbers with a 6= b. Then Z c c aJp (bc)Jp−1 (ac) − bJp (ac)Jp−1 (bc) . Jp (ax) Jp (bx)x dx = 2 2 b −a 0 To prove this formula, we note that y1 = Jp (ax) satisfies x2y100 + xy10 + (a2x2 − p2)y1 = 0 and y2 = Jp (bx) satisfies x2 y200 + xy20 + (b2 x2 − p2)y2 = 0. Write these equations in the form 0 a2x2 − p2 xy10 + y10 + y1 = 0 x and

0

b2x2 − p2 y1 = 0. x Multiply the first by y2 and the second by y − 1, subtract, simplify, and get 0 0 y2 xy10 − y1 xy20 = y1 y2 (b2 − a2 )x. xy20

Note that

+ y20 +

0 0 d y2 xy10 − y1 xy20 = y2 (xy10 ) − y1 (xy20 ) . dx

So (b2 − a2 )y1 y2 x =

d y2 (xy10 ) − y1 (xy20 ) , dx

and, after integrating, Z c c c 2 2 (b − a ) y1 (x)y2 (x)x dx = y2 (xy10 ) − y1 (xy20 ) = x y2 y10 − y1 y20 . 0

0

0

On the left, we have the desired integral times (b2 − a2 ) and, on the right, we have c Jp (bc)aJp0 (ac) − bJp (ac)Jp0 (bc) − c aJp (0)Jp0 (0) − bJp (0)Jp0 (0) . Since Jp (0) = 0 if p > 0 and J00 (x) = −J1 (x), it follows that Jp (0)Jp0 (0) − Jp (0)Jp0 (0) = 0 for all p > 0. Hence the integral is equal to Z c c I= aJp (bc)Jp0 (ac) − bJp (ac)Jp0 (bc) . Jp (ax) Jp (bx)x dx = 2 2 b −a 0

Section 4.8

Bessel Series Expansions

227

Now using the formula Jp0 (x) =

1 Jp−1(x) − Jp+1 (x) , 2

we obtain I=

2(b2

c aJp (bc) Jp−1 (ac) − Jp+1 (ac) − bJp (ac) Jp−1 (bc) − Jp+1 (bc) . 2 −a )

Simplify with the help of the formula Jp+1 (x) =

2p Jp (x) − Jp−1 (x) x

and you get I

=

=

h c 2p aJp (bc) Jp−1 (ac) − ( Jp (ac) − Jp−1(ac)) 2 −a ) ac i 2p −bJp (ac) Jp−1 (bc) − ( Jp (bc) − Jp−1 (bc)) bc c aJp (bc)Jp−1(ac) − bJp (ac)Jp−1 (bc) , 2 2 b −a 2(b2

as claimed. Note that this formula implies the orthogonality of Bessel functions. In fact its proof mirrors the proof of orthogonality from Section 4.8.

228

Chapter 4

Partial Differential Equations in Polar and Cylindrical Coordinates

Solutions to Exercises 4.9 1. We have J0 (x) =

1 π

Z

π 0

1 cos − x sin θ dθ = π

So

1 π

J0 (0) = For n 6= 0, Jn(x) = so

1 π

Z

π

|Jn (x)| =

≤

≤

π

cos x sin θ dθ.

0

π

dθ = 1. 0

cos nθ − x sin θ dθ;

0

1 Jn (0) = π

2. We have

Z

Z

Z

π

cos nθ dθ = 0. 0

Z π 1 cos nθ − x sin θ dθ π 0 1 π 1 π

≤1

Z

π 0

Z

z }| { cos nθ − x sin θ dθ

π

dθ = 1. 0

3. taking imaginary parts from teh series in Exampe 2, we obtain ∞ X

sin x =

Jn (x) sin

n=−∞

=

∞ X

Jn(x) sin

n=−∞ X nπ nπ Jn(x) sin + 2 2 −1

Jn(x) sin

nπ nπ X J−n (x) sin − 2 2 n=1

n=1

=

∞ X

∞

n=1

=

∞ X

nπ 2

J2k+1(x)(−1)k −

k=0

=

2

∞ X

∞ X

J−(2k+1)(x)

k=0

(−1)k J2k+1(x).

k=0

4. (a) For any real number w, we have p iw e = |cos w + i sin w| = cos2 + sin2 w = 1. If x and θ are real numbers then so is x sin θ, and hence ix sin θ e = 1. (b) For fixed x, apply Parseval’s identity to the Fourier series eix sin θ =

∞ X n=−∞

Jn(x)einθ ,

Section 4.9

Integral Formulas and Asymptotics for Bessel Functions

229

whose Fourier coefficients are Jn(x), and you get 1 2π or 1 2π

Z

π −π

Z

∞ X ix sin θ 2 2 e dθ = |Jn (x)| n=−∞

π

1 dθ = J0 (x)2 + 2 −π

∞ X

Jn(x)2 ,

n=1

hence 1 = J0 (x)2 + 2

∞ X

Jn (x)2 .

n=1

(c) Since the series converges (for fxed x), its nth term goes to 0. Hence limn→∞ Jn (x) = 0. 5. All the terms in the series 1 = J0 (x)2 + 2

∞ X

Jn (x)2

n=1

are nonnegative. Since they all add-up to 1, each must be less than or equal to 1. Hence J0 (x)2 ≤ 1 ⇒ |J0(x)| ≤ 1 and, for n ≥ 2, 1 2Jn(x)2 ≤ 1 ⇒ |Jn(x)| ≤ √ . 2 6. (4) implies that ∞ X cos x sin θ + i sin x sin θ = Jn (x) cos nθ + i sin nθ . n=−∞

taking real parts, we get cos x sin θ

=

∞ X

Jn(x) cos nθ

n=−∞

=

=

= =

J0(x) +

J0(x) +

−∞ X

Jn (x) cos nθ +

∞ X

n=−1

n=1

∞ X

∞ X

J−n (x) cos nθ +

n=1

n=1

J0(x) +

∞ X

(−1)nJn (x) cos nθ +

J0(x) +

n=1 ∞ h X

Jn (x) cos nθ

Jn (x) cos nθ ∞ X

Jn (x) cos nθ

n=1

i 1 + (−1)n Jn(x) cos nθ

n=1

=

J0(x) + 2

∞ X

J2n(x) cos 2nθ .

n=1

The second identity follows similarly by taking imaginary parts.

230

Chapter 4

Partial Differential Equations in Polar and Cylindrical Coordinates

7. We have f(t) = f(π − t) and g(t) = −g(π − t). So Z

π

f(t)g(t) dt

=

Z

0

π/2

f(t)g(t) dt +

Z

0

=

Z

π/2

f(t)g(t) dt −

Z

Z

0

f(π − x)g(π − x) dx (let t = π − x) π/2

π/2

f(t)g(t) dt − 0

f(t)g(t) dt π/2

0

=

π

Z

π/2

f(x)g(x) dx = 0. 0

8. (a) Apply Exercise 7 with f(t) = cos(x sin t) and g(t) = cos t. Part (b) is true for even n and not true if n is odd. If n is even, apply Exercise 7 with f(t) = sin(x sin t) and g(t) = sin(nt).

Section 5.1

Preview of Problems and Methods

231

Solutions to Exercises 5.1 1. Start with Laplace’s equation in spherical coordinates (1)

2 ∂ 2 u 2 ∂u ∂u 1 ∂2u 2 ∂ u = 0, + + cot θ θ + + csc ∂r2 r ∂r r2 ∂θ2 ∂θ ∂φ2

where 0 < r < a, 0 < φ < 2π, and 0 < θ < π. To separate variable, take a product solution of the form u(r, θ, φ) = R(r)Θ(θ)Φ(φ) = RΘΦ, and plug it into (1). We get 2 1 R00ΘΦ + R0ΘΦ + 2 RΘ00Φ + cot θRΘ0 Φ + csc2 θ RΘΦ00 = 0. r r Divide by RΘΦ and multiply by r2 : r2

R00 Φ00 R0 Θ00 Θ0 + 2r + + cot θ + csc2 θ = 0. R R Θ Θ Φ

Now proceed to separate the variables: 00 R00 Φ00 R0 Θ Θ0 r2 + 2r =− + cot θ + csc2 θ . R R Θ Θ Φ Since the left side is a function of r and the right side is a function of φ and θ, each side must be constant and the constants must be equal. So r2

R00 R0 + 2r =µ R R

and

Θ00 Φ00 Θ0 + cot θ + csc2 θ = −µ. Θ Θ Φ The equation in R is equivalent to (3). Write the second equation in the form Θ00 Φ00 Θ0 + cot θ + µ = − csc 2 θ ; Θ Θ Φ 00 Θ Θ0 Φ00 2 sin θ + cot θ +µ =− . Θ Θ Φ This separates the variables θ and φ, so each side must be constant and the constant must be equal. Hence 00 Θ Θ0 sin2 θ + cot θ +µ = ν Θ Θ

and

Φ00 ⇒ Φ00 + νΦ = 0. Φ We expect 2π-periodic solutions in Φ, because φ is and azimuthal angle. The only way for the last equation to have 2π-periodic solutions that are essentially different is to set ν = m2 , where m = 0, 1, 2, . . .. This gives the two equations ν=−

Φ00 + m2 Φ = 0 (equation (5)) and sin2 θ which is equivalent to (6).

Θ00 Θ0 + cot θ +µ Θ Θ

= m2 ,

232 Chapter 5 Partial Differential Equations in Spherical Coordinates 3. From Section 5.5, we have P0 (x) = 1,

P1(x) = x,

P2(x) =

1 (3x2 − 1). 2

So

1 (3 cos2 θ − 1). 2 Next, we verify that these functions are solutions of (7) with µ = n(n + 1), with n = 0, 1, 2, respectively, or µ = 0, 2, 6. (Recall the values of µ from (14).) For µ = 0, (7) becomes Θ00 + cot θΘ0 = 0, P0 (cos θ) = 1,

P1 (cos θ) = cos θ,

P2 (cos θ) =

and clearly the constant function P0(cos θ) = 1 is a solution. For µ = 2, (7) becomes Θ00 + cot θΘ0 + 2Θ = 0. Taking Θ = P1(cos θ) = cos θ, we have Θ0 = − sin θ, and Θ00 = − cos θ. Plugging into the equation, we find Θ00 + cot θΘ0 + 2Θ = − cos θ +

cos θ (− sin θ) + 2 cos θ = 0. sin θ

Finally, for µ = 6, (7) becomes Θ00 + cot θΘ0 + 6Θ = 0. Taking Θ = P2(cos θ) = 12 (3 cos2 θ − 1), we have Θ0 = −3 sin θ cos θ, Θ00 = −3 cos2 θ − sin2 θ , Θ00 + cot θΘ0 + 6Θ =

=

cos θ −3 cos2 θ − sin2 θ + (−3 sin θ cos θ) sin θ +3(3 cos2 θ − 1) 3(cos2 θ + sin2 θ) − 3 = 0

4. When m = 0, P10(cos θ) =√P1 (cos θ) was treated in Exercise 3. When m = 1, √ P11(cos θ) = − 1 − cos2 θ = sin2 θ = sin θ. You should note that 0 ≤ θ ≤ π, so sin θ ≥ 0, thus the positive sign in front of the square root. With m = 1 and µ = 1, equation (6) becomes Θ00 + cot θΘ0 + (2 − csc2 θ)Θ = 0. Taking Θ = P11(cos θ) = sin θ, we have Θ0 = cos θ, Θ00 = − sin θ. So cos θ 1 00 0 2 Θ + cot θΘ + (2 − csc θ)Θ = − sin θ + cos θ + 2 − sin θ sin θ sin2 θ =− sin 2 θ

=

{ 1 z 2}| sin θ + cos θ − 1 = 0. sin θ

Section 5.2

Dirichlet Problems with Symmetry

233

Solutions to Exercises 5.2 1. This problem is similar to Example 2. Note that f is its own Legendre series: f(θ) = 20 (P1(cos θ) + P0(cos θ)). So really there is no need to compute the Legendre coefficients using integrals. We simply have A0 = 20 and A1 = 20, and the solution is u(r, θ) = 20 + 20 r cos θ. 3. We have u(r, θ) =

∞ X

An rnPn (cos θ),

n=0

with An

= =

2n + 1 2 2n + 1 2

Z Z

π

f(θ)Pn (cos θ) sin θ dθ 0 π 2

0

2n + 1 100 Pn(cos θ) sin θ dθ + 2

Z

π

20 Pn(cos θ) sin θ dθ. π 2

Let x = cos θ, dx = − sin θ dθ. Then Z 1 Z An = 50(2n + 1) Pn(x) dx + 10 (2n + 1) 0

0

Pn (x) dx. −1

The case n = 0 is immediate by using P0(x) = 1, Z 1 Z 0 A0 = 50 dx + 10 dx = 60. 0

−1

For n > 0, the integrals are not straightforward and you need to refer to Exercise 10, Section 5.6, where they are evaluated. Quoting from this exercise, we have Z 1 P2n(x)dx = 0, n = 1, 2, . . . , 0

and

Z

1

P2n+1(x) dx = 0

(−1)n (2n)! , 22n+1(n!)2(n + 1)

n = 0, 1, 2, . . . .

Since P2n(x) is an even function, then, for n > 0, Z 0 Z 1 P2n(x) dx = P2n(x) dx = 0. −1

0

Hence for n > 0, A2n = 0. Now P2n+1(x) is an odd function, so Z 0 Z 1 P2n+1(x) dx = − P2n+1(x) dx = − −1

0

(−1)n (2n)! . 22n+1(n!)2 (n + 1)

Hence for n = 0, 1, 2, . . ., A2n+1

=

50(4n + 3)

Z

1

P2n+1(x) dx + 10 (4n + 3) 0

= = =

Z

0

P2n+1(x) dx −1

(−1)n(2n)! (−1)n (2n)! − 10 (4n + 3) 22n+1(n!)2 (n + 1) 22n+1(n!)2(n + 1) n (−1) (2n)! 40 (4n + 3) 2n+1 2 (n!)2(n + 1) (−1)n (2n)! 20 (4n + 3) 2n . 2 (n!)2(n + 1) 50(4n + 3)

234

Chapter 5

Partial Differential Equations in Spherical Coordinates

So

∞ X

u(r, θ) = 60 + 20

(4n + 3)

n=0

5. Solution We have u(r, θ) =

(−1)n (2n)! r2n+1P2n+1(cos θ). 22n(n!)2(n + 1)

∞ X

An rnPn (cos θ),

n=0

with An

= = =

2n + 1 2 2n + 1 2 2n + 1 2

Z Z Z

π

f(θ)Pn (cos θ) sin θ dθ 0 π 2

cos θ Pn(cos θ) sin θ dθ 0 1

x Pn(x) dx, 0

where, as in Exercise 3, we made the change of variables x = cos θ. At this point, we have to appeal to Exercise 11, Section 5.6, for the evaluation of this integral. (The cases n = 0 and 1 can be done by referring to the explicit formulas for the Pn, but we may as well at this point use the full result of Exercise 11, Section 5.6.) We have Z 1 Z 1 1 1 xP0(x) dx = ; xP1(x) dx = ; 2 0 3 0 Z 1 (−1)n+1 (2n − 2)! x P2n(x) dx = 2n ; n = 1, 2, . . . ; 2 ((n − 1)!)2n(n + 1) 0 and

Z

1

x P2n+1(x) dx = 0;

n = 1, 2, . . . .

0

Thus, A0 =

1 1 1 = ; 2 2 4

31 1 = ; 23 2

A1 =

A2n+1 = 0,

n = 1, 2, 3, . . . ;

and for n = 1, 2, . . ., A2n =

2(2n) + 1 (−1)n+1 (2n − 2)! (−1)n+1 (2n − 2)! = (4n + 1) . 2 22n((n − 1)!)2n(n + 1) 22n+1((n − 1)!)2n(n + 1)

So u(r, θ) =

∞ X 1 1 (−1)n+1 (4n + 1)(2n − 2)! 2n + r cos θ + r P2n(cos θ). 4 2 22n+1((n − 1)!)2 n(n + 1) n=1

7. (a) From (8) u(r, θ) = 50 + 25

∞ X

(4n + 3)

n=0

Setting θ = u(r,

π 2,

(−1)n (2n)! r2n+1P2n+1(cos θ) . + 1)

22n(n!)2(n

we get

π ) = 2

50 + 25

∞ X

(4n + 3)

n=0

(−1)n (2n)! π r2n+1P2n+1(cos ) 2n 2 2 (n!) (n + 1) 2 =0

=

z }| { (−1)n (2n)! 2n+1 50 + 25 P2n+1(0) (4n + 3) 2n r 2 (n!)2(n + 1) n=0

=

50

∞ X

Section 5.2

Dirichlet Problems with Symmetry

235

This is expected, since the points with θ = π2 are located halfway between the boundary with The temperature at these points is the average value 100+0 or 50. 2 (d) For θ = 0, we have u(r, 0) + u(r, π) 2 =0

=

z }| { (−1)n (2n)! 2n+1 50 + 25 (P (4n + 3) 2n (cos 0) + P (cos π)) r 2n+1 2n+1 2 (n!)2(n + 1) n=0

=

50,

∞ X

because P2n+1(cos 0) = P2n+1(1) = 1 and P2n+1(cos π) = P2n+1(−1) = −1. This makes sense because the average of temperatures equidistant from u = 50 in a ±z symmetrical environment ought to be 50 itself. 9. We first describe the boundary function. From the figure, we see that there is an angle θ0 , with cos θ0 = − 13 , and such that 70 if 0 ≤ θ ≤ θ0 , f(θ) = 55 if θ0 ≤ θ ≤ π. Appealing to (5), we have An

2n + 1 2

=

Z

π

f(θ)Pn (cos θ) sin θ dθ 0

Z 2n + 1 θ0 = 70 cos θ Pn (cos θ) sin θ dθ 2 0 Z π 2n + 1 + 55 cos θ Pn(cos θ) sin θ dθ 2 θ0 Z 1 Z 2n + 1 − 3 2n + 1 1 = 55 Pn(t) dt + 70 Pn(t) dt, 2 2 −1 − 13 where we have used the substitution cos θ = t, dt = − sin θdθ, and cos θ0 = − 13 . For n = 0, we have P0(x) = 1 and so 1 2 4 A0 = (55)( + (70)( ) = 65. 2 3 3 For n ≥ 1, we appeal to Exercise 9(a) and (c), Section 5.6. The results that we need state that for n = 1, 2, . . ., Z 1 1 Pn (t) dt = [Pn−1(x) − Pn+1(x)], 2n + 1 x and

Z

x

Pn(t) dt = −1

1 [Pn+1(x) − Pn−1(x)] . 2n + 1

Thus, for n ≥ 1, An

=

= So u(r, θ) = 65 +

2n + 1 1 1 55 Pn+1(− ) − Pn−1(− ) 2 2n + 1 3 3 70 1 1 + Pn−1(− ) − Pn+1(− ) 2n + 1 3 3 15 1 1 Pn−1(− ) − Pn+1(− ) . 2 3 3 ∞ 15 X 1 1 r n Pn−1(− ) − Pn+1(− ) Pn(cos θ). 2 n=1 3 3 3

236

Chapter 5

Partial Differential Equations in Spherical Coordinates

Solutions to Exercises 5.3 1. (c) Starting with (4) with n = 2, we have s 5 (2 − m)! m Y2,m (θ, φ) = P (cos θ)eimφ , 4π (2 + m)! 2 where m = −2, −1, 0, 1, 2. To compute the spherical harmonics explicitely, we will need the explicit formula for the associated Legendre functions from Example 1, Section 5.7. We have √ 2 P2−2(x) = 18 (1 − x2 ); P2−1(x) = 12 x 1 − x2; P20(x) = P2(x) = 3x 2−1 ; √ P21(x) = −3x 1 − x2; P22(x) = 3(1 − x2). So Y2,−2(θ, φ) =

= =

Y2,−1 (θ, φ) =

s

s

5 (2 + 2)! −2 P (cos θ)e−2iφ 4π (2 − 2)! 2

r

=sin 2 θ

}| { 5 4! 1 z (1 − cos2 θ) e−2iφ 4π 1 8 r r 30 1 5 3 2 −2iφ = sin θe sin2 θe−2iφ ; π 8 4 6π

5 (2 + 1)! −1 P (cos θ)e−iφ 4π (2 − 1)! 2

r

=sin θ

zp }| { 5 3! 1 = cos θ 1 − cos2 θ e−iφ 4π 1! 2 r r 15 1 5 3 = cos θ sin θe−iφ = cos θ sin θe−iφ . 2π 2 2 6π √ Note that since 0 ≤ θ ≤ π, we have sin θ ≥ 0, and so the equality 1 − cos2 θ = sin θ that we used above does hold. Continuing the list of spherical harmonics, we have s 5 (2 + 0)! Y2,0(θ, φ) = P2(cos θ)e−iφ 4π (2 − 0)! r r 5 3 cos2 θ − 1 1 5 = = (3 cos2 θ − 1). 4π 2 4 π The other spherical harmonics are computed similarly; or you can use the identity in Exercise 4. We have r 5 3 2 Y2,2 = (−1) Y2,−2 = Y2,−2 = sin2 θe−2iφ 4 6π r 5 3 = sin2 θe2iφ . 4 6π In the preceding computation, we used two basic properties of the operation of complex conjugation: az = az

if a is a real number;

and eia = e−ia

if a is a real number.

Section 5.3

Spherical Harmonics and the General Dirichlet Problem

237

Finally, 1

Y2,1

= (−1) Y2,−1 = −Y2,−1 r 5 3 cos θ sin eiφ . = − 2 6π

3 =− 2

r

5 cos θ sin e−iφ 6π

3. To prove (5), we use the definition of the spherical harmonics, (4), and write Z 2π Z π Yn,m (θ, φ)Y n0,m0 (θ, φ) sin θ dθ dφ 0

0

=

C

=I

z Z

2π

=II

}|

{z Z 0 eimφ eim φ dφ

0

π

}|

0 Pnm (cos θ) Pnm0 (cos θ)

{

sin θdθ,

0

where C is a constant that depends on m, n, m0 , and n0 . If m 6= m0 , then I = 0, by the orthogonality of the complex exponential functions (see (11), Section 2.6, with p = π). If m = m0 but n 6= n0 , then II = 0 by the orthogonality of the associate Legendre functions. (Make the change of variable cos θ = x and then use (6), Section 5.7.) Thus if (m, n) 6= (m0 , n0 ), then the above double integral is 0. Next we prove (6), but first note that the left side of (6) is what you would get if you take (m, n) = (m0 , n0 ) in (5). This is because 2

Ym,n (θ, φ)Ym,n (θ, φ) = |Ym,n (θ, φ)| . Here again we are using a property of complex numbers, which states that z z = |z|2

for any complex number z.

Using (4), we have Z π Yn,m (θ, φ)Y n,m(θ, φ) sin θ dθ dφ 0 =I

=II z }| {z { Z 2π Z }| 1 (2n + 1)(n − m)! π m 2 imφ −imφ = e e dφ [Pn (cos θ)] sin θdθ . 2π 0 2(n + m)! 0

We have I=

1 2π

Z

2π imφ −imφ

e

e

0

1 dφ = 2π

Z

2π

dφ = 1, 0

because eimφ e−imφ = 1. Also, II = 1 by (7), Section 5.7. (You need to make a change of variables in the integral in II, x = cos θ, dx = − sin θ dθ, before applying the result of Section 5.7). 5.

(a) If m = 0, the integral becomes Z 2π 1 2π φ dφ = φ2 = 2π2. 2 0 0

Now suppose that m 6= 0. Using integration by parts, with u = φ, du = dφ, dv = 1 e−imφ , v = −im e−imφ , we obtain: =dv

Z

2π 0

We have

=u z}|{ z }| { φ e−imφ dφ =

φ

2π Z 2π 1 −imφ 1 + e−imφ dφ e −im im 0 0

h e−imφ φ=2π = cos(mφ) − i sin(mφ)

= 1, φ=2π

238

Chapter 5

Partial Differential Equations in Spherical Coordinates

and Z

Z

2π

e−imφ dφ = 0

0

=

cos mφ − i sin mφ dφ

2π

0 2π

if m 6= 0, if m = 0

So if m 6= 0, Z

2π

φ e−imφ dφ = 0

2π 2π = i. −im m

Putting both results together, we obtain Z

2π −imφ

φe

dφ =

2π mi 2

if m 6= 0, if m = 0.

2π

0

(b) Using n = 0 and m = 0 in (9), we get

A0,0

=

1 2π

r

1 0! 4π 0! Z π

1 = π √ 2 π 1 = π √ 2 π

0

=2π 2

z Z

}|

φ dφ

π

P0(cos θ) sin θ dθ 0

1 √ sin θ dθ 2 }|

π

Z

0

=2

zZ

{

2π

{

√

φ dφ

Z

sin θ dθ =

π

0

Using n = 1 and m = 0 in (9), we get

A1,0

=

= =

1 2π 1 4π 0,

r r

Z

3 1 4π 1

2π 0

3 (2π2 ) π

π

P1(cos θ) sin θ dθ

0 =0

z Z

}|

1

{

P1 (x) dx −1

R1 where we used −1 P1 (x) dx = 0, because P1(x) = x is odd. Using n = 1 and m = −1 in (9), and appealing to the formulas for the associated Legendre functions from Section 5.7, we get

A1,−1

=

= =

1 2π

r

r

3 2! 4π 0!

3 1 −i 2π 2 r i 3π − . 4 2

2π = (−1) i

z Z

2π

}|

{

iφ

φ e dφ 0

π

}|

2

{

sin θ dθ 0

π

P1−1(cos θ) sin θ dθ 0

=π 2

zZ

Z

(P1−1(cos θ) =

1 sin θ) 2

Using n = 1 and m = 1 in (9), and appealing to the formulas for the associated

Section 5.3

Spherical Harmonics and the General Dirichlet Problem

239

Legendre functions from Section 5.7, we get

A1,1

1 2π

=

r

r

=

i

=

i − 4

3 0! 4π 2!

3 8π r

zZ

= 2π 1 i

z Z

{

φe

dφ

−iφ

0 =− π 2

}|

π

}|

2π

π

P11(cos θ) sin θ dθ 0

{

2

(P11(cos θ) = − sin θ)

− sin θ dθ 0

3π . 2

(c) The formula for An,0 contains the integral the nth Legendre polynomial; so Z

Z

=

Z

=

Z

π

Pn0(cos θ) sin θ dθ 0

Rπ 0

Pn0(cos θ) sin θ dθ. But Pn0 = Pn,

π

Pn(cos θ) sin θ dθ 0 1

Pn (x) dx −1

=

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

where the last equality follows from the orthogonality of Legendre polynomials (take R1 m = 0 in Theorem 1, Section 5.6, and note that P0(x) = 1, so −1(1)Pn (x) dx = 0, as desired.)

7. (a) The boundary function is given by

f(θ, φ) =

100 if − π4 φ ≤ 0 otherwise.

π 4;

The solution u(r, θ, φ) is given by (11), where the coefficients Anm are given by (8). Since the integrand in (8) is 2π-periodic in φ, the outer limits in the integral in (8) can be changed to −π to π without affecting the value of the integral. So we have

Anm

=

Z

π −π

=

100

Z Z

π

f(θ, φ)Y n,m (θ, φ) sin θ dθ dφ 0 π/4

−π/4

Z

π

Y n,m (θ, φ) sin θ dθ dφ . 0

(b) Using the explicit formulas for the spherical harmonics from Exercise 1, obtain the coefficients given in the table. If m = n = 0, then

A0,0

100 = √ 2 π

Z

π/4 −π/4

=2

zZ

π

}|

{

√ sin θ dθ dφ = 50 π.

0

R1 If m = 0 but n 6= 0, then Anm contains the integral −1 Pn(x) dx (n = 1, 2, . . .), which is equal to 0 by the orthogonality of the Legendre polynomials; and so Amn = 0 if m = 0 and n 6= 0. For the other coefficients, let us compute using the formula

240

Chapter 5

Partial Differential Equations in Spherical Coordinates

for the spherical harmonics:

Anm

=

=

=

s

100

2n + 1 (n − m)! 4π (n + m)!

Z

π/4

e−imφ dφ

Z

Pnm (cos θ) sin θ dθ

0

−π/4

z Z

π

2 =m sin

}|

mπ 4

{ Z 1 2n + 1 (n − m)! π/4 100 (cos mφ − i sin mφ) dφ Pnm (s) ds 4π (n + m)! −π/4 −1 s Z 1 200 2n + 1 (n − m)! mπ P m (s) ds. sin m 4π (n + m)! 4 −1 n s

R1 The integral Inm = −1 Pnm (s) ds can be evaluated for particular values of m and n (or you can use the general formula from Exercises 6). For example, if m = n = 1, then =half area of a disk of radius 1 z }| { Z 1 Z 1p π P11(s) ds = − 1 − s2 ds =− . 2 −1 −1 So A1,1

200 = 1

r

√ 3 0! π π sin (− ) = −25 3π. 4π 2! 4 2

The other coefficients are computed similarly by appealing to the explicit formulas for the associated Legendre functions from Section 5.7 and using the preceding formulas. (c) Using an approximation of the solution with n running from 0 to 2 and m running from −n to n, we get

u(r, θ, φ) =

∞ X n X

Anm rnYn,m (θ, φ) ≈

n=0 m=−n

2 n X X

Anm rnYn,m (θ, φ).

n=0 m=−n

Now substituting the explicit formulas and values, we obtain u(r, θ, φ) ≈ r0 Y0,0A0,0 + r1Y1,−1A1,−1 + r1Y1,1 A1,1 + r2 Y2,−2A2,−2 + r2Y2,2 A2,2 " r # r √ √ √ 3 3 1 1 1 −iφ iφ √ 50 π + r = 25 3π + sin θ e sin θ e 25 3π 2 π 2 2π 2 2π " r r r # r 5 5 5 5 3 2 2 2 3 −2iφ 2iφ +r 100 sin θe + sin θe 100 4 6π 6π 4 6π 6π =cos φ

= =

=cos 2φ

z }| z }| √ { { eiφ + e−iφ e2iφ + e−2iφ 2 21 2 25 + 75 r sin θ +125 r sin θ 2 2 π 2 √ 2 1 25 + 75 r sin θ cos φ + 125 r2 sin2 θ cos 2φ . 2 π

9. We apply (11). Since f is its own spherical harmonics series, we have u(r, θ, φ) = Y0,0(θ, φ) =

2

1 √ . π

Section 5.3

Spherical Harmonics and the General Dirichlet Problem

241

11. The solution is very much like the solution of Exercise 7. In our case, we have s Z π Z 2n + 1 (n − m)! π/3 −imφ Anm = 50 e dφ Pnm (cos θ) sin θ dθ 4π (n + m)! −π/3 0

=

=

=

z Z

2 =m sin

}|

mπ 3

{ Z 1 2n + 1 (n − m)! π/3 50 (cos mφ − i sin mφ) dφ Pnm (s) ds 4π (n + m)! −π/3 −1 s Z 1 100 2n + 1 (n − m)! mπ P m (s) ds sin m 4π (n + m)! 3 −1 n s 100 2n + 1 (n − m)! mπ sin Inm , m 4π (n + m)! 3 s

R1 where Inm = −1 Pnm (s) ds. Several values of Inm were found previously. You can also use the result of Exercise 6 to evaluate it.

242

Chapter 5

Partial Differential Equations in Spherical Coordinates

Solutions to Exercises 5.4 5. We apply Theorem 3 and note that since f depends only on r and not on θ or φ, the series expansion should also not depend on θ or φ. So all the coefficients in the series are 0 except for the coefficients Aj,0,0, which we will write as Aj for 1 simplicity. Using (16) with m = n = 0, a = 1, f(r, θ, φ) = 1, and Y0,0(θ, φ) = 2√ , π we get Aj

=

2 j12 (α 12 , j )

Z

1 0

Z

2π

Z

0

π 0

1 j0 (λ0, j r) √ r2 sin θ dθ dφ dr 2 π

=2π

= =

=2

z }| { z Z 2π Z dφ

}| { Z 1 1 √ 2 sin θ dθ j0 (λ0, j r)r2 dr π j1 (α 21 , j ) 0 0 0 √ Z 1 4 π j0 (λ0, j r)r2 dr, j12 (α 12 , j ) 0 π

where λ0, j = α 1 , j , the jth zero of the Bessel function of order 12 . Now 2

J1/2 (x) =

r

2 sin x πx

(see Example 1, Section 4.7), so the zeros of J1/2 are precisely the zeros of sin x, which are jπ. Hence λ0, j = α 1 , j = jπ. 2

Also, recall that j0 (x) =

sin x x

(Exercises 38, Section 4.8), so Z

1

j0 (λ0, j r)r2 dr

=

Z

0

1

j0 (jπ r)r2 dr =

Z

0

1 0

sin(jπr) 2 r dr jπ r

j+1

= =

= (−1) jπ

z Z

}|

{ 1 1 sin(jπ r)r dr jπ 0 (−1)j+1 , (jπ)2

where the last integral follows by integration by parts. So, Aj =

√ 4 π (−1)j+1 . j12 (jπ) (jπ)2

This can be simplified by using a formula for j1 . Recall from Exercise 38, Section 4.7, sin x − x cos x j1 (x) = . x2 Hence j12 (jπ) and

sin(jπ) − jπ cos(jπ) = (jπ)2

2

− cos(jπ) = jπ

√ Aj = 4(−1)j+1 π,

2

(−1)j+1 = jπ

2

=

1 , (jπ)2

Section 5.4

The Helmholtz, Poisson, Heat, and Wave Equations

243

and so the series expansion becomes: for 0 < r < 1, 1 =

∞ X

=

j=1 ∞ X

=

∞ X

j=1

√ sin jπr 4(−1)j+1 π Y0,0 (θ, φ) jπr √ sin jπr 1 √ 4(−1)j+1 π jπr 2 π 2(−1)j+1

j=1

sin jπr . jπr

It is interesting to note that this series is in fact a half range sine series expansion. Indeed, multiplying both sides by r, we get ∞ 2X sin jπr r= (−1)j+1 π j=1 j

(0 < r < 1),

which is a familiar sines series expansion (compare with Example 1, Section 2.4). 7. Reasoning as we did in Exercise 5, we infer that the series representation is of the form ∞ X Aj Y0,0(θ, φ)j0(α 12 , j r), j=1

where Aj

=

2 2 j1 (α 12 , j )

Z

1 0

Z

2π

Z

0

π 0

1 j0 (λ0, j r) √ r4 sin θ dθ dφ dr 2 π

=2π

= =

zZ }| { zZ 2π dφ

=2

}| { Z 1 1 √ 2 sin θ dθ j0 (λ0, j r)r4 dr π j1 (α 21 , j ) 0 0 0 √ Z 1 4 π j0 (λ0, j r)r4 dr, j12 (α 12 , j ) 0 π

where λ0, j = α 12 , j , the jth zero of the Bessel function of order Exercise 5: r 2 J1/2(x) = sin x, πx λ0, j = α 12 , j = jπ, and j0 (x) =

1 2.

Again, from

sin x . x

So Z

1

j0 (λ0, j r)r4 dr

=

0

Z

1

j0 (jπ r)r4 dr =

Z

0

0 2

=(−6+j π

= =

1

z Z

2

sin(jπr) 4 r dr jπ r

(−1)j+1 ) (jπ)3

}| { 1 1 sin(jπ r)r3 dr jπ 0 (−6 + j 2 π2 )(−1)j+1 , (jπ)4

where the last integral follows by three integrations by parts. So, √ 4 π (−1)j+1(−6 + j 2 π2 ) Aj = 2 . j1 (jπ) (jπ)4

244

Chapter 5

Partial Differential Equations in Spherical Coordinates

From Exercise 5, j14 (jπ)

sin(jπ) − jπ cos(jπ) = (jπ)2

so

4

− cos(jπ) = jπ

4

(−1)j+1 = jπ

4

=

1 , (jπ)4

√ (−6 + j 2 π2 ) Aj = 4(−1)j+1 π , (jπ)3

and so the series expansion becomes: for 0 < r < 1, r2

=

∞ X

=

∞ X

√ sin jπr 1 4(−1)j+1 π (−6 + j 2 π2 ) √ 3 (jπ) r 2 π j=1 2(−1)j+1 (−6 + j 2 π2 )

j=1

sin jπr . (jπ)3 r

Here again, it is interesting to note that this series is a half range sine series expansion. Indeed, multiplying both sides by r, we get r3 =

∞ X j=1

2(−1)j+1 (−6 + j 2 π2 )

sin jπr (jπ)3

(0 < r < 1),

which is a sines series expansion of the function f(r) = r3 , 0 < r < 1.

Section 5.5

Legendre’s Differential Equation

245

Solutions to Exercises 5.5 1. Putting n = 0 in (9), we obtain

P0(x) =

0 (0 − 2m)! 1 X (−1)m x0−2m . 0 2 m=0 m!(0 − m)!(0 − 2m)!

The sum contains only one term corresponding to m = 0. Thus P0(x) = (−1)0

0! 0 x = 1, 0!0!0!

because 0! = 1. For n = 1, formula (9) becomes

P1(x) =

where M = and so

1−1 2

M (2 − 2m)! 1 X (−1)m x1−2m, 21 m=0 m!(1 − m)!(1 − 2m)!

= 0. Thus the sum contains only one term corresponding to m = 0

P1 (x) = For n = 2, we have M =

2 2

=

3−1 2

=

=

= 1 and (9) becomes

6! 3 4! 1 1 1 (−1)0 x + 3 (−1)1 x 23 0!3!3! 2 1!2!1! 5 3 3 x − x. 2 2

=

=

m=1

1 (6 − 2m)! 1 X (−1)m x3−2m 3 2 m=0 m!(3 − m)!(3 − 2m)!

P3 (x) =

P4 (x) =

m=0

z }| { }| { 4! (4 − 2)! 1 1 (−1)0 x2 + 2 (−1)1 x0 22 0!2!2! 2 1!1!0! 1 2 1 1 3 6x + (−1)2 = x2 − . 4 4 2 2 z

=

For n = 4, we have M =

= 1 and (9) becomes 1 (4 − 2m)! 1 X (−1)m x2−2m 22 m=0 m!(2 − m)!(2 − 2m)!

P2 (x) =

For n = 3, we have M =

2! 1 1 (−1)0 x = x. 21 0!1!1!

4 2

= 2 and (9) becomes 2 (8 − 2m)! 1 X (−1)m x4−2m 24 m=0 m!(4 − m)!(4 − 2m)!

1 8! 4 1 6! 2 1 4! 0 x − 4 x + 4 x 4 2 0!4!4! 2 1!3!2! 2 2!2!0! 1 (35x4 − 30x2 + 3) 8

246

Chapter 5

Partial Differential Equations in Spherical Coordinates

3. By the first formula in Exercise 2(b) (applied with n and n + 1), we have P2n(0) − P2n+2(0) =

=

P2n(0) − P2(n+1)(0)

= = = = = =

(2n)! (2(n + 1))! − (−1)n+1 2(n+1) 22n(n!)2 2 ((n + 1)!)2 (2n)! (2n + 2)(2n + 1)((2n)!) (−1)n 2n + (−1)n 2 (n!)2 4 22n(n + 1)2 (n!)2 (2n + 2)(2n + 1) (2n)! 1 + (−1)n 2n 2 (n!)2 4 (n + 1)2 2(n + 1)(2n + 1) (2n)! (−1)n 2n 1 + 2 (n!)2 4 (n + 1)2 (2n + 1) (2n)! (−1)n 2n 1 + 2 (n!)2 2 (n + 1) (2n)!(4n + 3) (−1)n 2n+1 2 (n!)2(n + 1) (−1)n

5. Using the explicit formulas for the Legendre polynomials, we find Z

1

P3(x) dx = −1

=

Z

1

5 3 ( x3 − x) dx 2 −1 2 1 5 4 3 2 ( x − x ) = 0 8 4 −1

Another faster way to see the answer is to simply note that P3 is an odd function, so its integral over any symmetric interval is 0. There is yet another more important reason for this integral to equal 0. In fact, Z

1

Pn(x) dx = 0

for all n 6= 0.

−1

This is a consequence of orthogonality that you will study in Section 5.6. 7. Using the explicit formulas for the Legendre polynomials, we find Z

1

P2 (x) dx = 0

=

Z

1

3 1 ( x2 − ) dx 2 2 0 1 1 3 1 ( x − x) = 0 2 2 0

This integral is a special case of a more general formula presented in Exercise 10, Section 5.6. 9. This is Legendre’s equation with n(n + 1) = 30 so n = 5. Its general solution is of the form y

= = =

c1 P5(x) + c2Q5 (x) 1 c1 (63x5 − 70x3 + 15x) + c2 1 − 15x2 + 30x4 + · · · 8 c1 (63x5 − 70x3 + 15x) + c2 1 − 15x2 + 30x4 + · · ·

In finding P5(x), we used the given formulas in the text. In finding the first few terms of Q5 (x), we used (3) with n = 5. (If you are comparing with the answers in your textbook, just remember that c1 and c2 are arbitrary constants.) 11. This is Legendre’s equation with n(n + 1) = 0 so n = 0. Its general solution

Section 5.5

Legendre’s Differential Equation

247

is of the form y

= = =

c1 P0(x) + c2Q0 (x) −2 3 24 5 c1 + c2 x − x + x + ··· 6 5! 3 5 x x c1 + c2 x + + +··· 3 5

where we used P0(x) = 1 and (4) with n = 0. 13. This is Legendre’s equation with n(n + 1) = 6 or n = 2. Its general solution is y = c1P2 (x) + c2 Q2(x). The solution will be bounded on [−1, 1] if and only if c2 = 0; that’s because P2 is bounded in [−1, 1] but Q2 is not. Now, using P2(x) = 12 (3x2 − 1), we find y(0) = c1 P2(0) + c2Q2 (0) = −

c1 + c2Q2 (0) 2

If c2 = 0, then c1 = 0 and we obtain the zero solution, which is not possible (since we are given y0 (0) = 1, the solution is not identically 0). Hence c2 6= 0 and the solutions is not bounded. 15. This is Legendre’s equation with n(n + 1) = 56 or n = 7. Its general solution is y = c1P7 (x) + c2 Q2(x). The solution will be bounded on [−1, 1] if and only if c2 = 0. Since P7(x) is odd, we have P7(0) = 0 and so 1 = y(0) = c1 P7(0) + c2Q7 (0) = c2 Q2(0). This shows that c2 6= 0 and the solutions is not bounded. 17. (To do this problem we can use the recurrence relation for the coefficients, as we have done below in the solution of Exercise 19. Instead, we offer a different solution based on an interesting observation.) This is Legendre’s equation with n(n + 1) = 34 or n = 12 . Its general solution is still given by (3) and (4), with n = 12 : y = c1 y1 + c2 y2 , where y1(x)

= =

+ 1) 2 ( 12 − 2) 12 ( 12 + 1)( 12 + 3) 4 x + x + ··· 2! 4! 3 21 4 1 − x2 − x +··· 8 128

1−

1 1 ( 2 2

and y2 (x)

= =

( 12 − 1)( 12 + 2) 3 ( 12 − 3)( 12 − 1)( 12 + 2)( 12 + 4) 5 x + x + ··· 3! 5! 5 15 5 x + x3 + x + ··· 24 128

x−

Since y1 (0) = 1 and y2 (0) = 0, y10 (0) = 0 and y20 (0) = 1 (differentiate the series term by term, then evaluate at x = 0), it follows that the solution is y = y1 (x) + y2 (x), where y1 and y2 are as describe above. 19. P This is Legendre’s equation with µ = 3. Its solutions have series expansions m y= ∞ for −1 < x < 1, where m=0 am x am+2 =

m(m + 1) − µ am . (m + 2)(m + 1)

Since y(0) = 0 and y0 (0) = 1, we find that a0 = 0 and a1 = 1. Now because the recurrence relation is a two-step recurrence, we obtain 0 = a0 = a2 = a4 = · · ·

248

Chapter 5

Partial Differential Equations in Spherical Coordinates

The odd-indexed coefficients are determined from a1 . Taking µ = 3 and m = 1 in the recurrence relation, we find: a3 =

2−3 1 a1 = − . (3)(2) 6

Now using m = 3, we find a5 =

1 3 3(4) − 3 9 a3 = (− ) = − , (5)(4) 20 6 40

and so forth. Thus the solution is y = x−

x3 3 − x5 + · · · . 6 40

23. Using the reduction of order formula in Exercise 22 (with n = 0) and the explicit formula for P0 (x) = 1, we find for −1 < x < 1, Z 1 Q0(x) = P0(x) dx [P0(x)]2(1 − x2) Z 1 = dx 1 − x2 Z 1 1 1 = − dx 2 1 + x −1 + x 1 1 |1 + x| = (ln |1 + x| − ln |x − 1|) = ln 2 2 |x − 1| 1 1 + x = ln . 2 1−x In evaluating the integral, we use the partial fraction decomposition 1 1 1 1 1 = = − . 1 − x2 (1 − x)(1 + x) 2 1 + x −1 + x You should verify this identity or derive it using the standard methods from calculus for partial fraction decomposition. The second comment concerns the absolute value inside the logarithm. Because −1 < x < 1, we have |1 + x| = 1 + x and | − 1 + x| = 1 − x. So there is no need to keep the absolute values. 29. (a) Since n p p (x + i 1 − x2 cos θ)n = x + i 1 − x2 cos θ , it suffices to prove the inequality p x + i 1 − x2 cos θ ≤ 1, which in turn will follow from 2 p x + i 1 − x2 cos θ ≤ 1. For any complex number α + iβ, we have |α + iβ|2 = α2 + β 2 . So 2 p x + i 1 − x2 cos θ

= x2 + (

p 1 − x2 cos θ)2

= x2 + (1 − x2) cos2 θ ≤ x2 + (1 − x2) = 1,

Section 5.5

Legendre’s Differential Equation

249

which proves the desired inequality. (b) Using Laplace’s formula, we have, for −1 ≤ x ≤ 1, 1 π 1 π 1 π 1

|Pn (x)| = ≤ ≤ =

Z π p n 2 (x + i 1 − x cos θ) dθ 0 Z π p (x + i 1 − x2 cos θ)n dθ Z0 π dθ (by (a)) 0

31. Recall that if α is any real number and k is a nonnegative integer, the kth binomial coefficient is

and

α 0

α k

=

α(α − 1)(α − 2) . . . (α − k + 1) k!

for k ≥ 1

= 1. With this notation, the binomial theorem asserts that

(1 + x)α =

∞ X α xk , k

−1 < x < 1 .

k=0

(a) Set α = − 12 , and obtain that X 1 (2k)! k √ (−1)k 2k v , = 2 (k!)2 1 + v k=0 ∞

|v| < 1 .

Solution We have

−1/2 k

= = = = = =

k factors z }| { 1 −1 −1 −1 − 1 ··· − (k − 1) k! 2 2 2 1 −1 −3 −(2k − 1) ··· k! 2 2 2

1 (−1)k k! 2k 1 (−1)k k! 2k 1 (−1)k k! 2k 1 (−1)k k! 2k

[1 · 3 · · · (2k − 1)] [1 · 2 · 3 · 4 · · · (2k − 1) · (2k)] 2 · 4 · · · (2k) (2k)! (2 · 1) · (2 · 2) · · · (2 · k) (2k)! (−1)k (2k)! = k 2 (k!) 22k(k!)2

So, for |v| < 1, the binomial series gives ∞

√

X 1 = (1 + v)−1/2 = 1+v k=0

−1/2 k

vk =

∞ X

(−1)k

k=0

(2k)! k v 22k (k!)2

250

Chapter 5

Partial Differential Equations in Spherical Coordinates

(b) We have 1 p 1 + (−2xu + u2 )

=

=

∞ X

(2k)! (−2xu + u2)k 22k (k!)2 k=0 k ∞ X X k k (2k)! (−2xu)k−j u2j (−1) 2k j 2 (k!)2 (−1)k

j=0

k=0

=

k ∞ X X

(−1)k

(2k)! k! (−1)k−j 2k−j xk−j uk−j u2j 22k (k!)2 j!(k − j)!

(−1)j

(2k)! xk−j uk+j 2k+j j!k!(k − j)!

k=0 j=0

=

k ∞ X X k=0 j=0

Let n = k + j, so k = n − j and j cannot exceed n/2, because 0 ≤ j ≤ k. Hence 1 p 1 + (−2xu + u2)

=

∞ X

=

n=0 ∞ X n=0

un

n/2 X

(−1)j

j=0

unPn (x)

(2(n − j))! xn−2j − j)!(n − 2j)!

2n(n

Section 5.6

Legendre Polynomials and Legendre Series Expansions

251

Solutions to Exercises 5.6 1. Bonnet’s relation says: For n = 1, 2, . . . , (n + 1)Pn+1(x) + nPn−1(x) = (2n + 1)xPn(x). We have P0(x) = 1 and P1(x) = x. Take n = 1, then 2P2(x) + P0(x)

= 3xP1(x),

2P2(x)

= 3x · x − 1,

P2(x)

=

1 (3x2 − 1). 2

Take n = 2 in Bonnet’s relation, then 3P3(x) + 2P1(x)

= 5xP2(x),

3P3(x)

= 5x

P3(x)

1 (3x2 − 1) − 2x, 2

5 3 3 x − x. 2 2

=

Take n = 3 in Bonnet’s relation, then 4P4(x) + 3P2(x) =

7xP3(x), 5 3 3 3 2 x − x − x −1 , 2 2 2 1 35 4 3 x − 15x2 + . 4 2 2

4P4(x) =

7x

P4 (x) = 3.

Z

Z

1

x P7(x) dx = −1

1

P1(x) P7(x) dx = 0, −1

by Theorem 1(i). 5. By Bonnet’s relation with n = 3, 7xP3(x)

=

4P4(x) + 3P2(x),

xP3(x)

=

4 3 P4 (x) + P2(x). 7 7

So Z

1

x P2(x)P3 (x) dx = −1

Z

1 −1

= =

4 7

Z

4 3 P4(x) + P2(x) P2(x) dx 7 7

1

3 P4 (x) P2(x) dx + 7 −1

0+

Z

1

[P2(x)]2 dx −1

32 6 = , 75 35

where we have used Theorem 1(i) and (ii) to evaluate the last two integrals. Z 1 7. x2 P7(x) dx = 0, because x2 is even and P7(x) is odd. So their product is −1

odd, and the integral of any odd function over a symmetric interval is 0.

252

Chapter 5

Partial Differential Equations in Spherical Coordinates

9. (a) Write (4) in the form 0 0 (t) − Pn−1 (t). (2n + 1)Pn(t) = Pn+1

Integrate from x to 1, Z 1 (2n + 1) Pn(t) dt

=

x

Z

1 0 Pn+1 (t) dt

−

Z

x

=

0 Pn−1 (t) dt x

1 1 Pn+1(t) − Pn−1(t) x

=

1

x

Pn−1(x) − Pn+1(x) + Pn+1(1) − Pn−1(1) .

By Example 1, we have Pn+1 (1) − Pn−1(1) = 0. So for n = 1, 2, . . . Z 1 1 Pn(t) dt = [Pn−1(x) − Pn+1 (x)] . 2n +1 x (b) First let us note that because Pn is even when is even and odd when is odd, it follows that Pn(−1) = (−1)n Pn(1) = (−1)n . Taking x = −1 in (a), we get Z 1 1 Pn(t) dt = [Pn−1(−1) − Pn+1(−1)] = 0, 2n + 1 −1 because n−1 and n+1 are either both even or both odd, so Pn−1(−1) = Pn+1 (−1). (c) We have Z 1 Z x Z 1 0= Pn(t) dt = Pn(t) dt + Pn(t) dt. −1

−1

x

So Z

x

Pn(t) dt =

−

−1

Z

1

Pn (t) dt x

1 [Pn−1(x) − Pn+1 (x)] 2n + 1

=

−

=

1 [Pn+1(x) − Pn−1(x)] 2n + 1

10. (a) Replace n by 2n in the formula in Exercise 9(a), then take x = 0, and get Z 1 1 P2n(t) dt = [P2n−1(0) − P2n+1(0)] = 0, 4n +1 0 because P2n−1(x) and P2n+1(x) are odd functions, so P2n−1(0) = 0 = P2n+1(0). (b) Applying the result of Exercise 9(a) with x = 0, we find Z 1 1 1 P2n+1(t) dt = [P2n(0) − P2n+2(0)] = [P2n(0) − P2n+2(0)] 2(2n + 1) + 1 4n + 3 0 =

1 (2n)!(4n + 3) (2n)! (−1)n 2n+1 = (−1)n 2n+1 , 4n + 3 2 (n!)2(n + 1) 2 (n!)2(n + 1)

by the result of Exercise 3, Section 5.5. 11. (a)

Z Z

1

xP0(x) dx = 0 1

xP1(x) dx = 0

Z Z

1

x dx =

1 ; 2

x2 dx =

1 3

0 1 0

Section 5.6

Legendre Polynomials and Legendre Series Expansions

253

(b) Using Bonnet’s relation (with 2n in place of n): xP2n =

1 [(2n + 1)P2n+1(x) + 2nP2n−1(x)] . 4n + 1

So, using Exercise 10 (with n and n − 1), we find Z

1

x P2n(x) dx = 0

=

=

Z 1 Z 1 1 P2n+1(x) dx + 2n P2n−1(x) dx (2n + 1) 4n + 1 0 0 1 (−1)n (2n)! (−1)n−1 (2(n − 1))! (2n + 1) 2n+1 + 2n 4n + 1 2 (n!)2 (n + 1) 22(n−1)+1((n − 1)!)2n (−1)n (2n − 2)! (2n)(2n − 1) 2 (2n + 1) − 4(n + 1)n 4n + 1 22n(n!)2(n + 1) 2

=

(−1)n (2n − 2)! [−n(4n + 1)] 2n 4n + 1 2 (n!)2(n + 1)

=

(−1)n+1 (2n − 2)! 22n((n − 1)!)2n(n + 1)

Using Bonnet’s relation (with 2n + 1 in place of n): xP2n+1 =

1 [(2n + 2)P2n+2(x) + (2n + 1)P2n(x)] . 4n + 3

Now using Exercise 10, it follows that

R1 0

x P2n+1(x) dx = 0.

13. We will use Dn f to denote the nth derivative of f. Using Exercise 12, Z

1

(1 − x2) P13(x) dx = −1

(−1)13 213(13)!

Z

1

D13 [(1 − x2 )] (x2 − 1)13 dx = 0 −1

because D13[(1 − x2 )] = 0. 15. Using Exercise 12, Z

1

xn Pn(x) dx = −1

= =

= = =

=

(−1)n 2nn!

Z

1

Dn [xn] (x2 − 1)n dx

−1 n Z 1

(−1) (x2 − 1)n dx 2n −1 Z Z π (−1)n π 1 2 n (cos θ − 1) sin θdθ = n sin2n+1 θdθ 2n 2 0 0 (Let x = cos θ, dx = − sin θdθ.) Z π/2 Z π 1 1 2n+1 sin θdθ + sin2n+1 θdθ 2n 0 2n π/2 Z π/2 Z π/2 1 1 π 2n+1 sin θdθ + sin2n+1(θ + )dθ n n 2 0 2 0 2 Z π/2 Z π/2 1 1 π sin2n+1 θdθ + n cos2n+1 θdθ (sin(θ + ) = cos θ) 2n 0 2 0 2 Z π/2 2 2 22n(n!)2 2n+1 (n!)2 sin2n+1 θdθ = n = , n 2 0 2 (2n + 1)! (2n + 1)!

where we have used the results of Exercises 41 and 42, Section 4.7.

254

Chapter 5

Partial Differential Equations in Spherical Coordinates

17. Using Exercise 12, Z 1 ln(1 − x) P2(x) dx =

Z (−1)2 1 2 D [ln(1 − x)] (x2 − 1)2 dx 22 2! −1 Z 1 1 −1 (x − 1)2(x + 1)2 dx 8 −1 (1 − x)2 Z 1 −1 1 −1 −1 (x + 1)2 dx = (x + 1)3 = . −1 8 −1 24 3

−1

= = 21. For n > 0, we have

Dn [ln(1 − x)] = . So Z

1

ln(1 − x) Pn(x) dx = −1

= = =

(−1)n 2nn!

Z

−(n − 1)! (1 − x)n

1

Dn [ln(1 − x)] (x2 − 1)n dx −1

Z (−1)n+1 (n − 1)! 1 (x + 1)n (x − 1)n dx 2nn! (1 − x)n −1 Z 1 −1 1 −1 n n+1 (x + 1) dx = (x + 1) −1 2nn −1 2n n(n + 1) −2 n(n + 1)

For n = 0, we use integration by parts. The integral is a convergent improper integral (the integrand has a problem at 1) Z 1 Z 1 ln(1 − x) P0(x) dx = ln(1 − x) dx −1

−1

=

1 −(1 − x) ln(1 − x) − x = −2 + 2 ln 2. −1

To evaluate the integral at x = 1, we used limx→1(1 − x) ln(1 − x) = 0. 22. The integral can be reduced to the one in Exercise 21, as follows: Let t = −x, then Z 1 Z 1 ln(1 + x)Pn(x) dx = ln(1 − t)Pn (−t) dt −1

−1

=

(−1)

n

Z

1

ln(1 − t)Pn t dt −1

=

(−1)n+1

2 n(n + 1)

if n > 0, and −2 + 2 ln 2 if n = 0. 23. Using Bonnet’s relation and then the result of Exercise 21: for n > 1, Z 1 Z 1 1 ln(1 − x)xPn(x) dx = ln(1 − x) (n + 1)Pn+1(x) + nPn−1 dx 2n + 1 −1 −1 Z 1 n+1 = ln(1 − x)Pn+1(x) dx 2n + 1 −1 Z 1 n + ln(1 − x)Pn−1 dx 2n + 1 −1 n+1 −2 n −2 = + 2n + 1 (n + 1)(n + 2) 2n + 1 (n − 1)n −2 h 1 1 −2 2n + 1 = + = 2n + 1 n + 2 n − 1 2n + 1 (n + 2)(n − 1) −2 = (n + 2)(n − 1)

Section 5.6

Legendre Polynomials and Legendre Series Expansions

255

The cases n = 0 and n = 1 have to be computed separately. For n = 0, we have Z Z 1 1 1 2 1 1 (x − 1) ln(1 − x) − ln(1 − x)x dx = (1 + x) dx 2 2 −1 −1 −1 = −1. −1 For the integration by parts, we used u = ln(1 − x), dv = x dx, du = 1−x dx, 1 2 v = 2 (x − 1). For n = 1, applying Bonnet’s relation as described previously, we obtain Z 1 Z 1 1 ln(1 − x)xP1(x) dx = ln(1 − x) 2P2(x) + 1P0 dx 3 −1 −1 Z Z 2 1 1 1 = ln(1 − x)P2(x) dx + ln(1 − x)P0 dx 3 −1 3 −1

Now use the results of Exercise 21: Z 1 ln(1 − x)xP1(x) dx =

2 −2 1 + (−2 + 2 ln 2) 3 2(2 + 1) 3 −8 2 + ln 2 9 3

−1

=

26. (a) Using the result of Exercise 10, we have Z 2k + 1 1 Ak = f(x)Pk (x) dx 2 0 Z 2k + 1 1 = Pk (x) dx 2 0 0 if k = 2n, n = 1, 2, . . . , 2k + 1 = (2n)! 2 (−1)n 22n+1 (n!)2 (n+1) if k = 2n + 1, n = 0, 1, . . . , =

0

if k = 2n, n = 1, 2, . . . ,

(4n+3)(2n)! (−1)n 22n+1 (n!)2 (n+1)

Also, A0 = Hence

1 2

Z

if k = 2n + 1, n = 0, 1, . . . .

1

P0 (x) dx = 0

1 2

Z

1

dx = 0

1 . 2

4n + 3 (2n)! 1 X (−1)n P2n+1(x) . + 2 n=0 4n + 4 22n(n!)2 ∞

f(x) =

27. Let f(x) = |x|, −1 ≤ x ≤ 1. (a) We have Z 1 1 A0 = |x|P0(x) dx 2 −1 =

Z

1

x dx = 0

A1

=

A2

=

=

1 2

Z 2+1 1 |x| x dx = 0 2 −1 (Because the integrand is odd.) Z 1 5 1 |x| (3x2 − 1) dx 2 −1 2 5 2

Z

1

x (3x2 − 1) dx = 0

5 . 4

256

Chapter 5

Partial Differential Equations in Spherical Coordinates

(b) In general A2k+1 = 0, because f is even, and A2n

4n + 1 2

=

=

Z

(4n + 1)

1

|x|P2n(x) dx −1

Z

1

x P2n(x) dx 0

=

(4n + 1)

(−1)n+1 (2n − 2)! . 22n((n − 1)!)2 n(n + 1)

by Exercise 11 29. Call the function in Exercise 28 g(x). Then g(x) =

1 1 (|x| + x) = (|x| + P1 (x)) . 2 2

Let Bk denote the Legendre coefficient of g and Ak denote the Legendre coefficient of f(x) = |x|, for −1 < x < 1. Then, because P1 (x) is its own Legendre series, we have ( 1 A if k 6= 1 2 k Bk = 1 (Ak + 1) if k = 1 2 Using Exercise 27 to compute Ak , we find B0 =

1 1 1 1 1 1 A0 = , B1 = + A1 = + 0 = , B2n+1 = 0, n = 1, 2, . . . , 2 4 2 2 2 2

and B2n =

1 (−1)n+1 (2n − 2)! A2n = 2n+1 2 2 ((n − 1)!)2 n

4n + 1 n+1

.

30. (a) With the help of Exercise 21, we find A0

An

=

=

=

1 2

Z

1

ln(1 − x)P0(x) dx = ln 2 − 1; −1

2n + 1 2

Z

1

ln(1 − x)Pn(x) dx (n = 1, 2, . . .) −1

2n + 1 (−2) −(2n + 1) = . 2 n(n + 1) n(n + 1)

31. (a) For −1 < x < 1, we use the result of Exercise 30 and get and the Legendre series ln(1 + x)

= ln(1 − (−x)) ∞ X 2n + 1 = ln 2 − 1 − Pn(−x) n(n + 1) n=1

= ln 2 − 1 −

∞ X

(−1)n

n=1

because Pn(−x) = (−1)n Pn (x).

2n + 1 Pn(x), n(n + 1)

Section 6.1

Orthogonal Functions

257

Solutions to Exercises 6.1 1. Let fj (x) = cos(jπx), j = 0, 1, 2, 3, and gj (x) = sin(jπx), j = 1, 2, 3. We R2 have to show that 0 fj (x)gk (x) dx = 0 for all possible choices of j and k. If j = 0, then Z 2 Z 2 2 −1 fj (x)gk (x) dx = sin kπx dx = cos(kπx) = 0. kπ 0 0 0 If j 6= 0, and j = k, then using the identity sin α cos α = Z

Z

2

fj (x)gj (x) dx = 0

1 2

sin 2α,

2

cos(jπx) sin(jπx) dx 0

Z

1 2 sin(2jπx) dx 2 0 2 −1 cos(2jπx) = 0. 0 4jπ

= =

If j 6= 0, and j 6= k, then using the identity sin α cos β = we obtain Z 2 fj (x)gk (x) dx =

Z

0

1 sin(α + β) + sin(α − β) , 2

2

sin(kπx) cos(jπx) dx 0

Z 1 2 sin(k + j)πx + sin(k − j)πx dx 2 0 2 −1 1 1 cos(k + j)πx + cos(k − j)πx = 0. 2π k + j k−j 0

= =

5. Let f(x) = 1, g(x) = 2x, and h(x) = −1 + 4x. We have to show that Z

1

f(x)g(x)w(x) dx = 0, −1

Z

1

f(x)h(x)w(x) dx = 0,

Z

−1

1

g(x)h(x)w(x) dx = 0. −1

Let’s compute: Z

1

f(x)g(x)w(x) dx = −1

Z

1

p 2x 1 − x2 dx = 0,

−1

because we are integrating an odd function over a symmetric interval. For the second integral, we have Z 1 Z 1 p f(x)h(x)w(x) dx = (−1 + 4x2) 1 − x2 dx −1 −1 Z π = (−1 + 4 cos2 θ) sin2 θ dθ 0

(x = cos θ, dx = − sin θ dθ, sin θ ≥ 0 for 0 ≤ θ ≤ π.) Z π Z π 2 = − sin θ dθ + 4 (cos θ sin θ)2 dθ 0

= −

0 =π 2

z Z

π 0

π = − + 2

}| { Z π 2 1 1 − cos 2θ dθ +4 sin(2θ) dθ 2 2 0 =π 2

zZ

π 0

}| { 1 − cos(4θ) dθ = 0 2

258

Chapter 6

Sturm-Liouville Theory with Engineering Applications

For the third integral, we have Z

1

g(x)h(x)w(x) dx = −1

Z

1

p 2x(−1 + 4x2) 1 − x2 dx = 0,

−1

because we are integrating an odd function over a symmetric interval. 9. In order for the functions 1 and a + bx + x2 to be orthogonal, we must have Z

1

1 · (a + bx + x2) dx = 0 −1

Evaluating the integral, we find b 1 1 ax + x2 + x3 2 3 −1

=

2a +

a

=

1 − . 3

In order for the functions x and Z

1 3

2 =0 3

+ bx + x2 to be orthogonal, we must have

1

1 1 · ( + bx + x2 )x dx = 0 3 −1

Evaluating the integral, we find 1 2 b 3 1 4 1 = x + x + x 6 3 4 −1 b =

b =0 3 0.

13. Using Theorem 1, Section 5.6, we find the norm of Pn(x) to be kPnk =

Z

1 2

Pn(x) dx

12

=

−1

2 2n + 1

12

√ = √

2 . 2n + 1

Thus the orthonormal set of functions obtained from the Legendre polynomials is √ 2 √ Pn(x), n = 0, 2, . . . . 2n + 1 17. For Legendre series expansions, the inner product is defined in terms of integration against the Legendre polynomials. That is, Z 1 2 (f, Pj ) = f(x)Pj (x) dx = Aj 2j +1 −1 where Aj is the Legendre coefficient of f (see (7), Section 5.6). According to the generalized Parseval’s identity, we have Z

1

f 2 (x) dx = −1

=

∞ X |(f, Pj )|2 kPj k2 j=0 2 ∞ X 2 2 Aj 2j + 1 2j + 1 j=0

=

∞ X j=0

2 A2 . 2j + 1 j

(The norm kPj k is computed in Exercise 13.)

Section 6.2

Sturm-Liouville Theory

259

Solutions to Exercises 6.2 0 1. Sturm-Liouville form: xy0 + λy = 0, p(x) = x, q(x) = 0, r(x) = 1. Singular problem because p(x) = 0 at x = 0. 00

0

5. Divide the equation through by x2 and get yx − xy2 + λ yx = 0. Sturm-Liouville 0 form: x1 y0 + λ yx = 0, p(x) = x1 , q(x) = 0, r(x) = x1 . Singular problem because p(x) and r(x) are not continuous at x = 0. 0 9. Sturm-Liouville form: (1 − x2 ) y0 + λy = 0, p(x) = 1 − x2 , q(x) = 0, r(x) = 1. Singular problem because p(±1) = 0. 13. Before we proceed with the solution, we can use our knowledge of Fourier series to guess a family of orthogonal functions that satisfy the Sturm-Liouville problem: yk (x) = sin 2k+1 x, k = 0, 1, 2, . . .. It is straightforward to check the 2 validity of our guess. Let us instead proceed to derive these solutions. We organize our solution after Example 2. The differential equation fits the form of (1) with p(x) = 1, q(x) = 0, and r(x) = 1. In the boundary conditions, a = 0 and b = π, with c1 = d2 = 1 and c2 = d1 = 0, so this is a regular Sturm-Liouville problem. We consider three cases. CASE 1: λ < 0. Let us write λ = −α2 , where α > 0. Then the equation becomes y00 − α2 y = 0, and its general solution is y = c1 sinh αx + c2 cosh αx. We need y(0) = 0, so substituting into the general solution gives c2 = 0. Now using the condition y0 (π) = 0, we get 0 = c1 α cosh απ, and since cosh x 6= 0 for all x, we infer that c1 = 0. Thus there are no nonzero solutions in this case. CASE 2: λ = 0. Here the general solution of the differential equation is y = c1 x + c2 , and as in Case 1 the boundary conditions force c1 and c2 to be 0. Thus again there is no nonzero solution. CASE 3: λ > 0. In this case we can write λ = α2 with α > 0, and so the equation becomes y00 + α2y = 0. The general solution is y = c1 cos αx + c2 sin αx. From y(0) = 0 we get 0 = c1 cos 0 + c2 sin 0 or 0 = c1 . Thus y = c2 sin αx. Now we substitute the other boundary condition to get 0 = c2 α cos απ. Since we are seeking nonzero solutions, we take c2 6= 0. Thus we must have cos απ = 0, and 2 hence α = 2k+1 2 . Since λ = α , the problem has eigenvalues 2 2k + 1 λk = , 2 and corresponding eigenfunctions yk = sin

2k + 1 x, k = 0, 1, 2, . . . . 2

17. Case I If λ = 0, the general solution of the differential equation is X = ax + b. As in Exercise 13, check that the only way to satisfy the boundary conditions is to take a = b = 0. Thus λ = 0 is not an eigenvalue since no nontrivial solutions exist. Case II If λ = −α2 < 0, then the general solution of the differential equation is X = c1 cosh αx + c2 sinh αx. We have X 0 = c1 α sinh x + c2α cosh αx. In order to have nonzero solutions, we suppose throughout the solution that c1 or c2 is nonzero. The first boundary condition implies c1 + αc2 = 0 c1 = −αc2. Hence both c1 and c2 are nonzero. The second boundary condition implies that c1(cosh α + α sinh α) + c2(sinh α + α cosh α) = 0. Using c1 = −αc2, we obtain −αc2 (cosh α + α sinh α) + c2 (sinh α + α cosh α) sinh α(1 − α2) sinh α = 0 or 1 − α2

= = =

0 (divide byc2 6= 0) 0 0

260

Chapter 6

Sturm-Liouville Theory with Engineering Applications

Since α 6= 0, it follows that sinh α 6= 0 and this implies that 1 − α2 = 0 or α = ±1. We take α = 1, because the value −1 does not yield any new eigenfunctions. For α = 1, the corresponding solution is X = c1 cosh x + c2 sinh x = −c2 cosh x + c2 sinh x, because c1 = −αc2 = −c2 . So in this case we have one negative eigenvalue λ = −α2 = −1 with corresponding eigenfunction X = cosh x − sinh x. Case III If λ = α2 > 0, then the general solution of the differential equation is X = c1 cos αx + c2 sin αx . We have X 0 = −c1α sin αx + c2α cos αx. In order to have nonzero solutions, one of the coefficients c1 or c2 must be 6= 0. Using the boundary conditions, we obtain c1 + αc2 = c1(cos α − α sin α) + c2(sin α + α cos α) =

0 0

The first equation implies that c1 = −αc2 and so both c1 and c2 are neq0. From the second equation, we obtain −αc2 (cos α − α sin α) + c2 (sin α + α cos α) −α(cos α − α sin α) + (sin α + α cos α) sin α(α2 + 1)

= 0 = 0 = 0

Since α2 + 1 6= 0, then sin α = 0, and so α = nπ, where n = 1, 2, . . .. Thus the eigenvalues are λn = (nπ)2 with corresponding eigenfunctions yn = −nπ cos nπx + sin nπx,

n = 1, 2, . . . .

21. If λ = α2, then the solutions are of the form c1 cosh αx + c2 sinh αx. Using the boundary conditions, we find y(0) = 0 y(1) = 0

⇒ c1 = 0 ⇒ c2 sinh α = 0.

But α 6= 0, hence sinh α 6= 0, and so c2 = 0. There are no nonzero solutions if λ > 0 and so the problem as no positive eigenvalues. This does not contradict Theorem 1 because if we consider the equation y00 − λy = 0 as being in the form (1), then r(x) = −1 < 0 and so the problem is a singular Sturm-Liouville problem to which Theorem 1 does not apply. 25. The eigenfunctions in Example 2 are yj (x) = sin jx, j = 1, 2, . . .. Since f is one of these eigenfunctions, it is equal to its own eigenfunction expansion. 29. You can verify the orthogonality directly by checking that Z 2π nx mx sin sin dx = 0 if m 6= n (m, n integers). 2 2 0 You can also quote Theorem 2(a) because the problem is a regular Sturm-Liouville problem. 33. (a) From Exercise 36(b), Section 4.8, with y = J0 (λr) and p = 0, we have Z a 2λ2 [y(r)]2 r dr = [ay0(a)]2 + λ2a2 [y(a)]2 . 0

Section 6.2

Sturm-Liouville Theory

261

But y satisfies the boundary condition y0 (a) = −κy(a), so Z a 2 [y(r)]2r dr = a2 κ2[y(a)]2 + λ2a2 [y(a)]2 ; 2λ 0

Z

=

a2 2 [J0(λk a)]2 2 κ + [J (λ a)] 0 k 2 λ2k

=

a2 [J0(λk a)]2 + [J1(λk a)]2 , 2

a 2

[y(r)] r dr 0

2 because, by (7), [J0(λk a)]2 = λκk J1(λk a) . (b) Reproduce the sketch of proof of Theorem 1. The given formula for the coefficients is precisely formula (5) in this case. 34. (a) We apply the result of Exercise 33. For 0 < r < 1, we have f(r) = 100 =

∞ X

Ak J0(λk r),

k=1

where λk (k = 1, 2, . . .) are the positive roots of (7), with a = κ = 1, and Z 1 2 Ak = 100J0(λk r)r dr [J0(λk )]2 + [J1(λk )]2 0 Z λk 200 = J0 (s)s ds (Let λk r = s.) [J0(λk )]2 + [λ2k J1 (λk )]2 0 =

=

λ2k

λk 200 sJ1 (s) 0 [J0(λk )]2 + [J1(λk )]2

200 J1(λk ) . λk [J0(λk )]2 + [J1(λk )]2

(b) By the definition of λk , we know that it is a positive root of (7), which, in this case, becomes λ J1 (λ) = J0(λ). Thus, for all k, λk J1 (λk ) = J0(λk ) or J1 (λk ) = J0 (λk ) λk , and so, for 0 < r < 1, 100 =

200

∞ X

=

200

∞ X

J0 (λk ) J0 (λk r) 2 [J (λ )]2 + [J (λ )]2 λ 0 k 1 k k k=1

k=1

=

200

J1 (λk ) J0 (λk r) λ2k [J0(λk )]2 + [J0(λk )]2

∞ X

J1 (λk ) J0 (λk r). [J (λ )]2 1 + λ2k 0 k k=1

35. Because the initial and boundary values do not depend on θ, it follows that the problem has a radially symmetric solution that does not depend on θ. Following the steps in the solution of Exercise 10, Section 4.2, we find that u(r, t) =

∞ X

2

Ak e−λk tJ0 (λk r) (t > 0, 0 < r < 1),

k=1

where, due to the Robin boundary condition, the radial part R in the product solutions must satisfy the boundary condition R0(1) = −R(1). (If you separate variables, you will arrive at the equation in R that is treated in Example 6.) Considering

262

Chapter 6

Sturm-Liouville Theory with Engineering Applications

that R(r) = J0(λr), this condition gives λJ00 (λ) = −J0 (λ). Using J00 (r) = −J1 (r), we see that λ is a solution of equation (7): λJ1 (λ) = J0 (λ). Setting t = 0 and using u(r, 0) = 100, we find that 100 =

∞ X

Ak J0 (λk r),

k=1

which is the series expansion in Exercise 34. Thus u(r, t) = 200

∞ X

λ2 k=1 k

2 J0(λk ) e−λk tJ0 (λk r) (t > 0, 0 < r < 1). 2 2 [J0(λk )] + [J1(λk )]

Section 6.3

The Hanging Chain

263

Solutions to Exercises 6.3 1. (a) The initial shape of the chain is given by the function f(x) = −.01(x − .5),

0 < x < .5,

and the initial velocity of the chain is zero. So the solution is given by (10), with L = .5 and Bj = 0 for all j. Thus √ p α j Aj J0 αj 2x cos 2g t . 2 j=1

∞ X

u(x, t) =

To compute Aj , we use (11), and get Aj

=

2 J12 (αj )

=

−.02 J12 (αj )

Z Z

.5

√ (−.01)(x − .5)J0 αj 2x dx

.5

√ (x − .5)J0 αj 2x dx

0

0

√ Make the change of variables s = αj 2x, or s2 = 2α2j x, so 2s ds = 2α2j dx or dx = αs2 ds. Thus j

Aj

Z

αj

=

−.02 J12 (αj )

=

−.01 α4j J12(αj )

( 0

Z

.5 2 s s − .5)J0(s) 2 ds α2j αj

αj

(s2 − α2j )J0 (s)s ds 0

=

# " α4j .01 2 J2(αj ) α4j J12(αj ) α2j

=

J2(αj ), α2j J12(αj )

.02

where we have used the integral formula (15), Section 4.3, with a = α = αj . We can give our answer in terms of J1 by using formula (6), Section 4.8, with p = 1, and x = αj . Since αj is a zero of J0 , we obtain 2 J1(αj ) = J0 (αj ) + J2(αj ) = J2 (αj ). αj So Aj =

.02 .04 2 J1(αj ) = 3 . α2j J12(αj ) αj αj J1 (αj )

Thus the solution is u(x, t) =

∞ X j=1

√ p α .04 j α 2x cos 2g J t , 0 j α3j J1(αj ) 2

where g ≈ 9.8 m/sec2 . Going back to the questions, to answer (a), we have the normal modes uj (x, t) =

√ p α .04 j α 2x cos 2g J t . 0 j 3 αj J1(αj ) 2

The frequency of the jth normal mode is νj =

p αj 2g . 4π

264

Chapter 6

Sturm-Liouville Theory with Engineering Applications

A six-term approximation of the solution is u(x, t) ≈

6 X j=1

√ p α .04 j α 2x cos 2g J t . 0 j α3j J1(αj ) 2

At this point, we use Mathematica (or your favorite computer system) to approximate the numerical values of the coefficients. Here is a table of relevant numerical data. j αj νj Aj

1 2.40483 .847231 .005540

2 3 4 5 5.52008 8.65373 11.7915 14.9309 1.94475 3.04875 4.15421 5.26023 −.000699 .000227 −.000105 .000058 Table 1 Numerical data for Exercise 1.

6 18.0711 6.36652 −.000036

Section 6.4

Fourth Order Sturm-Liouville Theory

265

Exercises 6.4 1. This is a special case of Example 1 with L = 2 and λ = α4 . The values of α are the positive roots of the equation cos 2α =

1 . cosh 2α

There are infinitely many roots, αn (n = 1, 2, . . .), that can be approximated with the help of a computer (see Figure 1). To each αn corresponds one eigenfunction Xn (x) = cosh αnx − cos αnx −

cosh 2αn − cos 2αn (sinh αnx − sin αnx) . sinh 2αn − sin 2αn

5. There are infinitely many eigenvalues λ = α4, where α is a positive root of the equation 1 cos α = . cosh α As in Example 1, the roots of this equation, αn (n = 1, 2, . . .), can be approximated with the help of a computer (see Figure 1). To each αn corresponds one eigenfunction cosh αn − cos αn (sinh αnx − sin αnx) . sinh αn − sin αn

Xn (x) = cosh αnx − cos αn x −

The eigenfunction expansion of f(x) = x(1 − x), 0 < x < 1, is f(x) =

∞ X

An Xn (x),

n=1

where An ==

R1 0

x(1 − x)Xn (x) dx . R1 2 (x) dx X n 0

After computing several of these coefficients, it was observed that: Z 1 Xn2 (x) dx = 1 for all n = 1, 2, . . . , 0

A2n = 0 for all n = 1, 2, . . . . The first three nonzero coefficients are A1 = .1788,

A3 = .0331,

A5 = .0134.

So f(x) ≈ .1788 X1(x) + .0331 X3(x) + .0134 X5(x), where Xn described explicitely in Example 1. We have X1 (x)

= cosh(4.7300 x) − cos(4.7300 x) + .9825 (sin(4.7300 x) − sinh(4.7300 x)),

X2 (x) X3 (x) X4 (x) X5 (x)

= = = =

cosh(1.0008 x) − cos(1.0008 x) + 1.0008 (sin(1.0008 x) − sinh(1.0008 x)), cosh(10.9956 x) − cos(10.9956 x) + sin(10.9956 x) − sinh(10.9956 x), cosh(14.1372 x) − cos(14.1372 x) + sin(14.1372 x) − sinh(14.1372 x), cosh(17.2788 x) − cos(17.2788 x) + sin(17.2788 x) − sinh(17.2788 x).

9. Assume that µ and X are an eigenvalue and a corresponding eigenfunction of the Sturm-Liouville problem X 00 + µX = 0,

X(0) = 0, X(L) = 0 .

266

Chapter 6

Sturm-Liouville Theory with Engineering Applications

Differentiate twice to see that X also satisfies the fourth order Sturm-Liouville problem X (4) − λX = 0, 00

X(0) = 0, X (0) = 0, X(L) = 0, X 00 (L) = 0 . If α and X are an eigenvalue and a corresponding eigenfunction of X 00 + µX = 0,

X(0) = 0, X(L) = 0,

then differentiating twice the equation, we find X (4) + µX 00 = 0,

X(0) = 0, X(L) = 0.

But X 00 = −µX, so X (4) −µ2X = 0 and hence X satisfies the equation X (4) −λX = 0 with λ = µ2 . Also, from X(0) = 0, X(L) = 0 and the fact that X 00 = −µX, it follows that X 00 (0) = 0 and X 00 (L) = 0.

Section 6.6

The Biharmonic Operator

267

Exercises 6.6 1. uxxyy = 0, uxxxx = 4!, uyyyy = −4!, ∇4u = 0. x 3. Let u(x, y) = x2 +y 2 and v(x, y) = x · u(x, y). We know that u is harmonic (Exercise 1, Section 4.1), and so v is biharmonic by Example 1, with A = 0, B = 1, C = D = 0.

5. Express v in Cartesian coordinates as follows: v

= = =

r2 cos(2θ)(1 − r2 ) r2[cos2 θ − sin2 θ](1 − r2) (x2 − y2 ) 1 − (x2 + y2 ) .

Let u = x2 − y2 . Then u is harmonic and so v is biharmonic by Example 1, with A = 1, D = 1, B = C = 0. 7. Write v = r2 · rn cos nθ and let u = rn cos nθ. Then u is harmonic (use the Laplacian in polar coordinates to check this last assertion) and so v is biharmonic, by Example 1 with A = 1 and B = C = D = 0. 9. Write v = ar2 ln r + br2 + c ln r + d = φ + ψ, where φ = [ar2 + c] ln r and ψ = br2 +d. From Example 1, it follows that ψ is biharmonic. Also, ln r is harmonic (check the Laplacian in polar coordinates) and so, by Example 1, φ is biharmonic. Consequently, v is biharmonic, being the sum of two biharmonic functions. 11. Since the boundary values are independent of θ, try for a solution a function independent of θ and of the form u = ar2 ln r + br2 + c ln r + d. We have ur = 2ar ln r + ar + 2br +

c c = 2ar ln r + r(a + 2b) + . r r

Now use the boundary conditions to solve for the unknown coefficients as follows: u(1, θ) = 0 ⇒ b + d = 0, u(2, θ) = 0 ⇒ 4a ln 2 + 4b + c ln 2 + d = 0, ur (1, θ) = 1 ⇒ a + 2b + c = 1, ur (2, θ) = 1 ⇒ 4a ln 2 + 2a + 4b +

c = 1. 2

Solving this system of four equations in four unknowns, we find a=

1 , 3 − 4 ln 2

b=

−2 ln 2 , 3 − 4 ln 2

c=

2 , 3 − 4 ln 2

d=

2 ln 2 , 3 − 4 ln 2

and hence the solution u(r, θ) =

1 r2 ln r − 2 ln(2)r2 + 2 ln r + 2 ln 2 . 3 − 4 ln 2

13. We follow the method of Theorem 1, as illustrated by Example 2. First, solve the Dirichlet problem ∇2 w = 0, w(1, θ) = cos 2θ, for 0 ≤ r < 1, 0 ≤ θ ≤ 2π. The solution in this case is w(r, θ) = r2 cos 2θ. (This is a simple application of the method of Section 4.4, since the boundary function is already given by its Fourier series.) We now consider a second Dirichlet problem on the unit disk with boundary values v(1, θ) = 12 wr (1, θ) − g(θ) . Since g(θ) = 0 and wr (r, θ) = 2r cos 2θ, it follows that v(1, θ) = cos 2θ. The solution of th Dirichlet problem in v is v(r, θ) = r2 cos 2θ. Thus the solution of biharmonic problem is u(r, θ) = (1 − r2 )r2 cos 2θ + r2 cos 2θ = 2r2 cos 2θ − r4 cos 2θ.

268

Chapter 6

Sturm-Liouville Theory with Engineering Applications

This can be verified directly by plugging into the biharmonic equation and the boundary conditions. 15. We just give the answers details: w(r, θ) = 1, ∇2v = 0, v(1, θ) = 12 (1 − cos θ), v(r, θ) = − 12 r cos θ; u(r, θ) = − 12 r cos θ(1 − r2 ) + 1. 17. u(1, 0) = 0 implies that w = 0 and so v(1, θ) = − g(θ) . So 2 v(r, θ) = −

∞ X 1 rn(cos nθ + bn sin nθ) , a0 + 2 n=1

where an and bn are the Fourier coefficients of g. Finally, ∞ h i X 1 u(r, θ) = (1 − r2 )v(r, θ) = − (1 − r2 ) a0 + rn (cos nθ + bn sin nθ) . 2 n=1

Section 6.7

Vibrations of Circular Plates

269

Exercises 6.7 12. Correction to the suggested proof: y2 = I0 and not J1 . 15. Repeat Steps 1–3 of Example 2 without any change to arrive at the solution u(r, t) =

∞ X

αn cos(kn2 ct) + βn sin(kn2 ct) φn(r),

n=1

P∞ where the kn s are the positive roots of (16). Putting t = 0 we get 0 = n=1 αnφn (r), which implies that αn = 0 for all n. Differentiating with respect to t then setting t = 0, we obtain ∞ X g(r) = kn2 cβn φn(r), n=1

which implies that Z

1

g(r)φn (r)r dr = kn2 cβn 0

βn = R1 0

1

φ2n (r)r dr 0

or

The integral

Z

R1

g(r)φn (r)r dr 0 . R1 2 kncβn 0 φ2n (r)r dr

φ2n(r)r dr is given by (23).

17. Let u1 (r, t) denote the solution of the problem in Example 3 and let u2(r, t) denote the solution in Example 3. Then, by linearity or superposition, u(r, t) = u1(r, t) + u2 (r, t) is the desired solution. 19. Sketch of the solution: u(r, t)

∞ X

=

An (t)φn (r),

n=1

ut(r, t)

∞ X

=

A0n (t)φn (r),

n=1

∇4u(r, t)

∞ X

=

kn4 An(t)φn (r)

n=1 2

4

ut = −c ∇ u ⇒ A0n (t) = −c2 kn4 ⇒ An (t) = An e−c where An is a constant. So u(r, t) =

∞ X

Ane−c

2

4 kn t

φn (r).

n=1

Setting t = 0, we get, for 0 ≤ r < 1, 100 =

∞ X n=1

Thus An is as in Example 1.

An φn(r).

2

4 kn t

,

270

Chapter 6

Sturm-Liouville Theory with Engineering Applications

20. Sketch of the solution: u(r, t)

=

∞ X

αnφn(r),

n=1

∇4u(r, t)

=

∞ X

kn4 αnφn(r)

n=1

100 = ∇4u → 100

∞ X

kn4 αnφn (r)

n=1

→ kn4 αn = An or αn = where An is as in Example 1.

An , kn

Section 7.1

The Fourier Integral Representation

271

Solutions to Exercises 7.1 1. We have f(x) =

1 0

if −a < x < a, (a > 0) otherwise,

This problem is very similar to Example 1. From (3), if ω 6= 0, then a Z Z 1 a sin ωt 2 sin aω 1 ∞ . f(t) cos ωt dt = cos ωt dt = = A(ω) = π −∞ π −a πω −a πω If ω = 0, then A(0) =

1 π

Z

∞

f(t) dt = −∞

1 π

Z

a

dt = −a

2a . π

Since f(x) is even, B(ω) = 0. For |x| 6= a the function is continuous and Theorem 1 gives Z 2a ∞ sin aω cos ωx f(x) = dω . π 0 ω For x = ±a, points of discontinuity of f, Theorem 1 yields the value 1/2 for the last integral. Thus we have the Fourier integral representation of f Z if |x| < a, 1 2a ∞ sin aω cos ωx 1/2 if |x| = a, dω = π 0 ω 0 if |x| > a. 3. We have f(x) =

1 − cos x if −π/2 < x < π/2, 0 otherwise.

The solution follows by combining Examples 1 and 2 and using the linearity of the Fourier integral, or we can compute directly and basically repeat the computation in these examples. Since f is even, B(w) = 0 for all w, and Z 2 π/2 A(w) = 1 − cos t cos wt dt π 0 Z 2 sin(πw/2) 2 π/2 1 = − [cos(w + 1)t + cos(w − 1)t] dt πw π 0 2 π/2 2 sin(πw/2) 1 sin(w + 1)t sin(w − 1)t = (w 6= 1) − + πw π w+1 w−1 0 sin(w − 1) π2 2 sin(πw/2) 1 sin(w + 1) π2 = − + (w 6= ±1) πw π w+1 w−1 cos wπ 2 sin(πw/2) 1 cos wπ 2 2 = − − πw π w+1 w−1 2 sin(πw/2) wπ 2 1 = cos − , πw π 1 − w2 2 If w = ±1, A(w)

= =

Z 2 sin(±π/2) 2 π/2 cos2 t dt − ±π π 0 Z 2 2 π/2 1 + cos 2t 2 1 − dt = − . π π 0 2 π 2

(Another way to get this answer is to use the fact that A(w) is continuous (whenever f is integrable) and take the limit of A(w) as w tends to ±1. You will get 12 . Applying the Fourier integral representation, we obtain: Z 2 ∞ sin(πw/2) cos πw 1 − cos x if |x| ≤ π2 , 2 cos wxdw = f(x) = − 2 0 if |x| > π2 . π 0 w 1−w

272 Chapter 7 The Fourier Transform and its Applications 5. since f(x) = e−|x| is even, B(w) = 0 for all w, and Z 2 ∞ −t A(w) = e cos wt dt π 0 ∞ 2 e−t = [− cos wt + w sin wt] 2 π1+w 0 2 1 = , π 1 + w2 where we have used the result of Exercise 17, Sec. 2.6, to evaluate the integral. Applying the Fourier integral representation, we obtain: Z 1 2 ∞ e−|x| = cos wx dw. π 0 1 + w2 7. The function f(x) =

1 0

if 0 < x < 1, otherwise,

is neither even nor odd. We will need to compute both A(w) and B(w). For w 6= 0, Z 1 1 1 sin w A(w) = cos wt dt = ; π 0 π w Z 1 1 1 1 − cos w B(w) = sin wt dt = . π 0 π w If w = 0, just take limits of A(w) and B(w) as w tends to 0. You will find: A(0) = and B(0) = 0. Thus we have the Fourier integral representation of f Z if x < 0 or x > 1, 0 1 ∞ 1/2 if x = 0 or x = 1, sin w cos wx + (1 − cos w) sin wx , dw = π 0 1 if 0 < x < 1. 9. The function

x 2−x f(x) = −2 − x 0

if −1 < x < 1, if 1 < x < 2, if −2 < x < −1, otherwise,

is odd, as can be seen from its graph. Hence A(w) = 0 and Z 2 ∞ B(w) = f(t) sin wt dt π 0 Z Z 2 1 2 2 = t sin wt dt + (2 − t) sin wt dt π 0 π 1 " # 1 Z 1 2 −t cos wt = cos wt + dt π w w 0 0 " # 2 Z 2 2 cos wt 2−t + − cos wt + dt π w w 1 1 =

2 [2 sin w − sin 2w] . πw2

Since f is continuous, we obtain the Fourier integral representation of Z 2 ∞ 1 f(x) = [2 sin w − sin 2w] sin wx, dw. π 0 w2 11. For the function f(x) =

sin x if 0 < x < π, 0 otherwise,

1 π

Section 7.1

The Fourier Integral Representation

273

we have A(w)

Z 1 π sin t cos wt dt π 0 Z π 1 [sin(1 + w)t + sin(1 − w)t] dt 2π 0 π 1 cos(1 + w)t cos(1 − w)t − (w 6= 1) + 2π 1+w 1−w 0 1 1 1 cos(w + 1)π cos(1 − w)π + − − − 2π 1+w 1−w 1+w 1−w 1 cos wπ cos wπ 2 + + 2π 1 + w 1−w 1 − w2 1 [cos wπ + 1] . π(1 − w2 )

= = = = = =

B(w)

1 π

=

Z

π

sin t sin wt dt Z π 1 − [cos(1 + w)t − cos(w − 1)t] dt 2π 0 π 1 sin(1 + w)t sin(1 − w)t − (w 6= 1) − 2π 1+w 1−w 0 1 sin(w + 1)π sin(1 − w)π − − 2π 1+w 1−w sin wπ . π(1 − w2 )

= = = =

0

Thus the Fourier integral representation of f is Z 1 ∞ 1 + cos wπ sin wπ f(x) = cos wx + sin wx dx. π 0 1 − w2 1 − w2 13. (a) Take x = 1 in the Fourier integral representation of Example 1: Z Z ∞ 2 ∞ sin w cos w sin w cos w 1 π dw = ⇒ dw = . π 0 w 2 w 4 0 (b) Integrate by parts: u = sin2 w, du = 2 sin w cos w dw, dv =

1 w 2 dw,

v = − w1 :

=0

Z

z }| { Z ∞ ∞ sin2 w sin ω sin w cos w π dω = +2 dw = , 2 2 ω w 0 w 2 0

∞

2

0

by (a). Z ∞ sin ω cos 2ω 15. dω = 0. Solution Just take x = 2 in Example 1. ω 0 17.

Z

∞ 0

Solution. Define

0 cos xω + ω sin xω dω = π/2 1 + ω2 πe−x 0 f(x) = π/2 πe−x

if x < 0, if x = 0, if x > 0.

if x < 0, if x = 0, if x > 0.

274 Chapter 7 The Fourier Transform and its Applications Let us find the Fourier integral representation of f: Z 1 ∞ −x 1 A(w) = πe cos wx dx = π 0 1 + w2 (see Exercise 5); Z

1 B(w) = π

∞

πe−x sin wx dx = 0

w , 1 + w2

(see Exercise 17, Sec. 2.6). So f(x) =

Z

∞ 0

cos wx + w sin wx dw, 1 + w2

which yields the desired formula. 19. Let f(x) = e−a|x| , where a > 0. Then f, satisfies the conditions of Theorem 1, f is continuous and even. So it has a Fourier integral representation of the form Z ∞ e−a |x| = A(w) cos wx dx, 0

where A(w)

2 π

=

Z

∞

e−ax cos wx dx 0

2a , π(a2 + w2)

=

where the integral is evaluated with the help of Exercise 17, Sec. 2.6 (see Exercise 5 of this section). For x ≥ 0, the Fourier integral representation becomes −ax

e

2a = π

Z

∞ 0

cos wx dw a2 + w2

(x ≥ 0).

23. (a) Si(−x)

= = =

Z

−x

sin t dt t 0 Z x sin(−u) (−1) du (Let u = −t.) (−u) 0 Z x sin u − du = −Si(x). u 0

Thus Si(x) is an odd function. (b) limx→−∞ Si(x) = limx→∞ −Si(x) − π2 , by (8). (c) We have sin t = sin t t

=

∞ X

(−1)n

−∞ 0. For the case a < 0, use sin(−ax) = − sin ax and linearity of the Fourier transform. Let f(x) = 1 if |x| < a and 0 otherwise. Recall from Example 1 that r r 2 sin aw 2 F(f(x))(w) = = g(w). π w π p Multiplying both sides by π2 and using the linearity of the Fourier transform, it follows that r π F( f(x))(w) = g(w). 2 So Fg = FF

r

π f(x) 2

=

r

π f(x), 2

Section 7.2

The Fourier Transform

279

by the reciprocity relation. Using the symbol w as a variable, we get ( pπ r if |w| < a, sin ax π 2 F = f(w) = x 2 0 otherwise. 15. Consider the function in Example 2, ( −x e f(x) = 0

if x > 0, if x < 0.

If a > 0, then (see Exercise 17) F(f(ax))(w) =

1 1 − iw/a a − iw √ =√ . a 2π(1 + (w/a)2 ) 2π(a2 + w2 )

But

(

f(ax) = Hence g(w) =

e−ax

if x > 0,

0

if x < 0.

√ a − iw = 2πF(f(ax))(w); 2 2 a +w

and so √ F(g) =

2πf(−ax) =

( √ 0

2πeax

if x < 0, if x > 0.

Using the variable w, F(g)(w) =

( √ 0

2πeaw

if w < 0, if w > 0.

If a < 0, use the identity F(f(−x))(w) = F(f(x))(−w) (see Exercise 17 with a = −1) to conclude that ( √ 2πeaw if w > 0, F(g)(w) = 0 if w < 0. 17. (a) Consider first the case a > 0. Using the definition of the Fourier transform and a change of variables Z ∞ 1 F(f(ax))(w) = √ f(ax)e−iωx dx 2π −∞ Z ∞ 1 1 ω 1 √ = f(x)e−i a x dx (ax = X, dx = dX) a 2π −∞ a 1 w = F(f)( ). a a If a < 0, then F(f(ax))(w)

Z ∞ 1 √ f(ax)e−iωx dx 2π −∞ Z −∞ ω 1 1 √ = f(x)e−i a x dx a 2π ∞ 1 w = − F(f)( ). a a

=

280

Chapter 7

The Fourier Transform and its Applications

Hence for all a 6= 0, we can write ω 1 F(f) . |a| a

= (b) We have −|x|

F(e By (a), −2|x|

F(e

1 )(w) = 2

)(w) =

r

r

2 1 . π 1 + w2

2 1 = π 1 + (w/2)2

r

2 2 . π 4 + w2

(c) Let f(x) denote the function in Example 2. Then g(x) = f(−x). So 1 + iw b g(w) = f(−w) b = √ . 2π(1 + w2 ) Let h(x) = e−|x| . Then h(x) = f(x) + g(x). So 1 − iw 1 + iw b h(w) = fb(w) + b g(w) = √ +√ = 2π(1 + w2 ) 2π(1 + w2)

r

2 1 . π 1 + w2

19. Using the definition of the Fourier transform Z ∞ 1 F(f(x − a))(w) = √ f(x − a)e−iωx dx 2π −∞ Z ∞ 1 = √ f(x)e−iω(x+a) dx (x − a = X, dx = dX) 2π −∞ Z ∞ 1 = e−iωa √ f(x)e−iωx dx = e−iaω F(f)(ω). 2π −∞ 2

21. We have F(e−x = F(

2 √1 e−w /4 , 2

cos x ) ex2

1 23. We have F( 1+x 2) =

F(

cos x + cos 2x ) 1 + x2

pπ 2

by Theorem 5. Using Exercise 20, we have 2

= F(cos xe−x ) (w+a)2 1 (w−a)2 √ e− 4 + e− 4 = 2 2 e−|w| . Using Exercise 20, we have

cos x cos 2x = F( ) + F( ) 1 + x2 1 + x2 r 1 π −|w−1| = + e−|w+1| + e−|w−2| + e−|w+2| . e 2 2

25 Let

1 if |x| < 1, 0 otherwise, q and note that f(x) = cos(x) g(x). Now F(g(x)) = π2 sinww . Using Exercise 20, we have g(x) =

F(f(x))

= =

F(cos xg(x)) r 1 2 sin(w − 1) sin(w + 1) + . 2 π w−1 w+1

Section 7.2

The Fourier Transform

281

27. Call g(x) the function in the figure. The support of g is the interval (a, b), that in order to center its support at has length b − a. Translate the function by b+a 2 b−a b−a the origin, over the interval with endpoints at − 2 , 2 . Denote the translated function by f(x). Thus, g(x) = f x − b+a , where f(x) = h if |x| < b−a 2 2 and 0 otherwise. By Example 1 and Exercise 19, gb(w) = =

=

e−i

b+a 2 w

fb(w) r

w 2 sin b−a 2 e h· π w r 2 −i b+a w sin b−a 2 w 2 he . π w −i b+a 2 w

29. Take a = 0 and relabel b = a in Exercise 27, you will get the function f(x) = h if 0 < x < a. Its Fourier transform is r 2 −i a w sin a2w b f (w) == he 2 . π w Let g(x) denote the function in the figure. Then g(x) = a1 x f(x) and so, by Theorem 3, g(w) b

1 d b i f (w) a dw " r # sin a2w 2h d −i a w = i e 2 π a dw w

=

= i

= i

r r

" a cos 2 h e−i 2 w a πa w " 2 h −i a w −2 sin e 2 πa

aw 2

− i sin 2

aw 2

−

sin

aw 2

#

w

# a + a w e−i 2 w . 2w2

aw 2

31. Let f(x) = 1 if −a < x < a and 0 otherwise (Example 1), and let g(x) denote the function in the figure. Then g(x) = x2f(x) and, by Theorem 3, g(w) = b =

=

d2 b f(w) dw2 r 2 d2 sin aw − π dw2 w r 2 2aw cos aw + a2w2 sin aw − 2 sin aw − . π w3 −

33. Let g(x) denote the function in this exercise. By the reciprocity relation, since the function is even, we have F(F(g)) = g(−x) = g(x). Taking inverse Fourier transforms, we obtain F −1(g) = F(g). Hence it is enough to compute the Fourier transform. We use the notation and the result of Exercise 34. We have g(x) = 2f2a (x) − fa (x). Verify this identity by drawing the graphs of f2a and fa and then drawing the graph of f2a (x) − fa (x). With the help of this identity and the result of Exercise 34, we

282

Chapter 7

The Fourier Transform and its Applications

have g(w) b

b = 2fc 2a (w) − fa (w)

8a sin2 (aw) 4a sin2 aw 2 = 2√ −√ 2π 4(aw)2 2π (aw)2 " # 4 sin2 (aw) sin2 aw 2 √ . = − w2 w2 a 2π

34. Let f(x) denote the function inExercise 5 and fa (x) the function in the figure of this exercise. Then fa (x) = f xa and so x \ b f (w) = af(aw) a r 2 1 − cos(aw) a (Exercise 5) π (aw)2 r 2 2 sin2 aw 2 a π (aw)2 4a sin2 aw 2 √ . 2π (aw)2

fba (w) = =

= =

35. (a) Apply Theorem 5 with a = 2, then 2 2 1 F(e−x ) = √ e−w /4 . 2

36. (a) Theorem 5 with a = 1, 1

2

F(e− 2 x

+2

1

2

1

2

) = e2 F(e− 2 x ) = e2 e− 2 w .

37. By Exercise 27, 2 2 1 F(e−x ) = √ e−w /4 . 2

By Theorem 3(i) 2

F(xe−x ) = =

39. Let

d 1 −w2 /4 √ e dw 2 i −w2 /4 √ e . 2 2

i

1 if |x| < 1, 0 otherwise, q and note that f(x) = xg(x). Now F(g(x)) = π2 sinww . By Theorem 3(i) g(x) =

F(f(x))

= = =

F(xg(x)) r d 2 sin w i dw π w r 2 w cos w − sin w i . π w2

Section 7.2 1 41. We have F( 1+x 2) =

F(

pπ 2

e−|w| . So if w > 0

x )(w) = i 1 + x2

r

If w < 0 x F( )(w) = i 1 + x2

π d −w = −i e 2 dw

r

π d w e =i 2 dw

If w = 0, F(

The Fourier Transform

x 1 )(0) = √ 2 1+x 2π

Z

∞ −∞

r

r

π −w e . 2

π w e . 2

x dx = 0 1 + x2

(odd integrand). We can combine these answers into one formula r x π F( )(w) = −i sgn (w)e−|w| . 1 + x2 2 43. Using linearity, 2 2 2 F (1 − x2)e−x = F e−x − F x2e−x . By Theorem 3(ii) and Exercise 27, 2 F (1 − x2)e−x = =

2 1 1 d2 −w2 /4 √ e−w /4 + √ e 2 2 dw2 2 1 √ e−w /4 2 + w2 . 4 2

45. Theorem 3 (i) and Exercise 19: 2

1

F(xe− 2 (x−1) )

=

2 1 d F(e− 2 (x−1) ) dw d −iw 1 2 i F(e− 2 x ) e dw d −iw − 1 w2 d − 1 w2 −iw i e 2 ) =i ) e e 2 dw dw 1 2 i(−w − i)e− 2 w −iw

=

(1 − iw)e− 2 w

= = =

49. 2 b h(ω) = e−ω ·

i

1

2

−iw

.

2 1 1 = F √ e−x /4 · F 1 + ω2 2

r

π −|x| . e 2

Hence h(x) = =

r Z ∞ (x−t)2 1 1 π −|t| √ e− 4 f ∗ g(x) = √ e dt 2 2π −∞ 2 Z ∞ (x−t)2 1 √ e− 4 e−|t|dt. 2 2 −∞

51. Let fb(ω) =

1 if |ω| < 1, 0 otherwise.

Then 1 2 b h(ω) = fb(ω)e− 2 ω .

283

284

Chapter 7

The Fourier Transform and its Applications

1

2

1

2

We have F −1 (e− 2 ω ) = e− 2 x and f(x) =

= 2

Thus

2

/2

and g(x) = e−x .

2 − w2

(a) F(f)(w) = −iwe (b)

2 sin x π x .

r Z ∞ 2 1 2 sin t − (x−t) 2 √ f ∗ g(x) = e dt π t 2π −∞ Z ∞ (x−t)2 sin t 1 e− 2 dt. π −∞ t

h(x) =

53. Let f(x) = xe−x

q

2

, and F(g)(w) =

{∗} = = =

w √1 e− 4 2

.

{·} w w2 w2 −i √ e− 4 e− 2 2 w −3 w2 −i √ e 4 . 2

(c) With the help of Theorem 3 w w2 f ∗ g = F −1 − i √ e−3 4 2 1 4 d w2 −1 = F i√ e−3 4 2 6 dw 1 2 −1 d −3 w2 1 2 w2 = √ F i e 4 = √ xF −1 e−3 4 dw 23 23 2 1 2 √ xe− 3 x . = 3 3 w2 In computing F −1 e−3 4 , use Exercise 10(a) and (5) to obtain (−x)2 2 1 x2 1 F −1 e−aw = √ e− 4a = √ e− 4a . 2a 2a 55. We use the uniqueness of the Fourier transform, which states that if b h=b k then h = k. (a) Let h = f ∗ g and k = g ∗ f and use Theorem 4. Then F(h) = F(f) · F(g) = F(g) · F(f) = F(g ∗ f). hence f ∗ g = g ∗ f. (b) Let u = f ∗ (g ∗ h) and v = (f ∗ g) ∗ h. Then F(u) = = =

F(f) · F(g ∗ h) = F(f) · (F(g) · F(h)) (F(f) · F(g)) · F(h) = F(f ∗ g) · F(h) F((f ∗ g) ∗ h) = F(v).

Hence u = v or f ∗ (g ∗ h) = (f ∗ g) ∗ h). (c) Use Exercise19: F(fa ) = F(f(x − a)) = e−iaw F(f). So F((f ∗ g)a )

= F(f ∗ g(x − a)) = e−iaw F(f ∗ g) F (fa )

z }| { = e−iaw F(f) F(g) = F(fa )F(g) = F(fa ∗ g).

Section 7.2

The Fourier Transform

285

Thus (f ∗ g)a = fa ∗ g. Similarly, (f ∗ g)a = f ∗ (ga ). 57. Recall that f is integrable means that Z ∞ |f(x)| dx < ∞. −∞

If f and g are integrable, then 1 f ∗ g(x) = √ 2π

Z

∞

f(x − t)g(t) dt. −∞

So, using properties of the integral: Z ∞ Z ∞ Z ∞ 1 dx √ |f ∗ g(x)|dx = f(x − t)g(t) dt 2π −∞ −∞ −∞ Z ∞ Z ∞ 1 ≤ √ |f(x − t)g(t)| dx dt 2π −∞ −∞ (Interchange order of integration. =

R∞

−∞

|f (x)|dx

{ ∞ ∞ 1 ≤ √ |f(x − t)| dx |g(t)| dt 2π −∞ −∞ (Change variables in the inner integral X = x − t.) Z ∞ Z ∞ 1 ≤ √ |f(x)| dx |g(t)| dt < ∞; 2π −∞ −∞ Z

zZ

}|

thus f ∗ g is integrable. b 60. Suppose that f(w) = 0 for all w not in [a, b]. This fact is expressed by saying that the support of fb is contained in [a, b]. Suppose also that support of b g is contained in [c, d], where [a, b] is disjoint from [c, d]. Consider the function h = f ∗g. b Then for all w, b h(w) = f(w) · gb(w). In order for the product fb(w) · b g(w) to be nonzero, w has to be simultaneously in [a, b] and [c, d]. But this is impossible, since [a, b] and [c, d] are disjoint. So b h(w) = fb(w) · b g(w) = 0 for w, and hence f ∗ g = 0. To complete the solution of this exercise, we must construct nonzero functions f and g with the stipulated Fourier transform properties. Let r π sin2 x f(x) = . 2 x2 From Example 1 (and Exercise 10(b)), we have 1 − |w| if |w| < 2, b 2 f (w) = 0 othewise. The function fb has support contained in [−2, 2], because it is zero outside |w| < 2. To construct a function whose Fourier transform is supported outside the support of [−2, 2], it suffices to take f and translate its Fourier transform by, say, 5. This is done by multiplying f by e5ix (see Exercise 20(a). Let g(x) = e5ixf(x). Then 1 − |w−5| if |w − 5| < 2, 2 g(w) = b 0 othewise. This Fourier transform is just the translate of fb. It is supported in the interval [3, 7]. By the first part of the solution, f ∗ g = 0.

286

Chapter 7

The Fourier Transform and its Applications

Solutions to Exercises 7.3 1. ∂2u ∂2u = , 2 ∂t ∂x2 u(x, 0) =

1 , 1 + x2

∂u (x, 0) = 0. ∂t

Follow the solution of Example 1. Fix t and Fourier transform the problem with respect to the variable x: d2 u b(w, t) = −w2 u b(w, t), dt2 r 1 π −|w| u b(w, 0) = F = , e 1 + x2 2

d u b(w, 0) = 0. dt

Solve the second order differential equation in u b(w, t): u b(w, t) = A(w) cos wt + B(w) sin wt. Using

d b(w, dt u

0) = 0, we get

−A(w)w sin wt + B(w)w cos wt

= 0 ⇒ B(w)w = 0 ⇒ B(w) = 0.

t=0

Hence Using u b(w, 0) =

pπ 2

−|w|

e

u b(w, t) = A(w) cos wt. p , we see that A(w) = π2 e−|w| and so u b(w, t) =

r

π −|w| cos wt. e 2

Taking inverse Fourier transforms, we get Z ∞ u(x, t) = e−|w| cos wt eixw dw. −∞

3. ∂u 1 ∂2u , = ∂t 4 ∂x2 2 u(x, 0) = e−x . Fix t and Fourier transform the problem with respect to the variable x: d w2 u b(w, t) = − u b(w, t), dt 4 2 w2 1 u b(w, 0) = F e−x = √ e− 4 . 2 Solve the first order differential equation in u b(w, t): u b(w, t) = A(w)e− Using u b(w, 0) =

2

w √1 e− 4 2

w2 4

.

2

, we get A(w) =

w √1 e− 4 2

. Hence

w2 w2 w2 1 1 u b(w, t) = √ e− 4 e− 4 t = √ e− 4 (1+t) . 2 2

Section 7.3

The Fourier Transform Method

287

Taking inverse Fourier transforms, we get Z ∞ 1 w2 u(x, t) = √ e− 4 (1+t) eixw dw. 2 π −∞ 5. ∂ 2u ∂2u = c2 2 , 2 ∂t ∂x r 2 sin x u(x, 0) = , π x

∂u (x, 0) = 0. ∂t

Fix t and Fourier transform the problem with respect to the variable x: d2 u b(w, t) = −c2 w2u b(w, t), dt2 r sin x 1 if |w| < 1 b u b(w, 0) = F (w) = f (w) = 0 if |w| > 1, x d u b(w, 0) = 0. dt Solve the second order differential equation in u b(w, t): u b(w, t) = A(w) cos cwt + B(w) sin cwt. Using

d b(w, dt u

0) = 0, we get u b(w, t) = A(w) cos cwt.

Using u b(w, 0) = fb(w), we see that u b(w, t) = fb(w) cos wt. Taking inverse Fourier transforms, we get Z ∞ Z 1 1 1 ixw b u(x, t) = √ f (w) cos cwt e dw = √ cos cwt eixw dw. 2π −∞ 2π −1 7. ∂u ∂u +3 = 0, ∂x ∂t u(x, 0) = f(x) . Fourier transform the problem with respect to the variable x: d iw u b(w, t) = − u b(w, t), dt 3 u b(w, 0) = F u(x, 0) (w) = fb(w). Solve the first order differential equation in u b(w, t): w

u b(w, t) = A(w)e−i 3 t. Using the transformed initial condition, we get w u b(w, t) = fb(w)e−i 3 t.

Taking inverse Fourier transforms, we get Z ∞ Z ∞ 1 1 t −i w iw(x− 3t ) b b 3 t eixw dw = √ u(x, t) = √ f(w)e f(w)e dw = f(x − ). 3 2π −∞ 2π −∞

288

Chapter 7

The Fourier Transform and its Applications

9. ∂u ∂u − = 0, ∂x ∂t u(x, 0) = 3 cos x .

t2

The solution of this problem is very much like the solution of Exercise 7. However, there is a difficulty in computing the Fourier transform of cos x, because cos x is not integrable on the real line. One can make sense of the Fourier transform by treating cos x as a generalized function, but there is no need for this in this solution, since we do not need the exact formula of the Fourier transform, as you will see shortly. Let f(x) = 3 cos x and Fourier transform the problem with respect to the variable x: d u b(w, t) = 0, dt u b(w, 0) = F 3 cos x) (w) = fb(w). u(w, t) − t2iwb

Solve the first order differential equation in u b(w, t): w 3

u b(w, t) = A(w)e−i 3 t . Using the transformed initial condition, we get w 3

u b(w, t) = fb(w)e−i 3 t . Taking inverse Fourier transforms, we get Z ∞ 1 w 3 u(x, t) = √ fb(w)e−i 3 t eixw dw 2π −∞ Z ∞ t3 1 = √ fb(w)eiw(x+ 3 ) dw 2π −∞ t3 t3 = f(x + ) = 3 cos(x + ). 3 3 11. ∂u ∂u = , ∂x ∂t u(x, 0) = f(x) . Fourier transform the problem with respect to the variable x: d u b(w, t) − iwb u(w, t) = 0, dt u b(w, 0) = fb(w). Solve the first order differential equation in u b(w, t): u b(w, t) = A(w)eiwt .

Using the transformed initial condition, we get iwt b u b(w, t) = f(w)e .

Taking inverse Fourier transforms, we get Z ∞ 1 u(x, t) = √ fb(w)e−iwt eixw dw 2π −∞ Z ∞ 1 = √ fb(w)eiw(x+t) dw 2π −∞ = f(x + t).

Section 7.3

The Fourier Transform Method

289

13. ∂u ∂2u = t 2, ∂t ∂x u(x, 0) = f(x) . Fix t and Fourier transform the problem with respect to the variable x: d b(w, t) = 0, u b(w, t) + tw2u dt b u b(w, 0) = f(w). Solve the first order differential equation in u b(w, t): u b(w, t) = A(w)e−

t2 w 2 2

.

Use the initial condition: A(w) = fb(w). Hence

t u b(w, t) = fb(w)e−

2 w2 2

.

Taking inverse Fourier transforms, we get Z ∞ 1 t2 w 2 u(x, t) = √ fb(w)e− 2 eixw dw. 2π −∞ 15. ∂2u ∂u +2 = −u, 2 ∂t ∂t u(x, 0) = f(x), ut(x, 0) = g(x) . Fourier transform the problem with respect to the variable x: d2 d u b(w, t) + 2 u b(w, t) + u b(w, t) = 0, dt2 dt d u b(w, 0) = fb(w), . u b(w, t) = gb(w). dt Solve the second order differential equation in u b(w, t). The characteristic equation is λ2 + 2λ + 1 = 0. Since we have a repeated root λ = −1, the general solution is of the form: u b(w, t) = A(w)e−t + B(w)te−t . The initial condition u b(w, 0) = fb(w) implies that fb(w) = A(w). So u b(w, t) = d −t −t fb(w)e +B(w)te . The condition dt u b(w, t) = b g(w) implies that −fb(w)+B(w) = g(w) or B(w) = fb(w) + b b g(w). So u b(w, t) = fb(w)e−t + (fb(w) + b g(w))te−t .

Hence Taking inverse Fourier transforms, we get Z ∞ 1 −t b u(x, t) = √ f(w)e + (fb(w) + b g(w))te−t eixw dw 2π −∞ Z ∞ Z ∞ 1 1 b = e−t √ f(w) +b g (w) eixw dw fb(w) eixw dw + te−t √ 2π −∞ 2π −∞ −t −t = e f(x) + te (f(x) + g(x). 17. ∂2u ∂4u = ∂t2 ∂x4 100 if |x| < 2, u(x, 0) = 0 otherwise.

290

Chapter 7

The Fourier Transform and its Applications

Fourier transform the problem with respect to the variable x: d2 u b(w, t) = w4u b(w, t), dt2 r 2 sin 2w b . u b(w, 0) = f (w) = 100 π w Solve the second order differential equation in u b(w, t): 2

2

u b(w, t) = A(w)e−w t + B(w)ew t.

Because a Fourier transform is expected to tend to 0 as w → ±∞, if we fix t > 0 and let w → ∞ or w → −∞, we see that one way to make u b(w, t) → 0 is to take 2 B(w) = 0. Then u b(w, t) = A(w)e−w t, and from the initial condition we obtain B(w) = fb(w). So r 2 2 sin 2w −w2 t −w t u b(w, t) = fb(w)e = 100 . e π w Taking inverse Fourier transforms, we get r Z ∞ 1 2 sin 2w −w2 t ixw u(x, t) = √ 100 e dw e π w 2π −∞ Z 100 ∞ sin 2w −w2 t ixw = e dw. e π −∞ w 19. ∂2u ∂3u = , 2 ∂t ∂t∂x2 u(x, 0) = f(x),

ut(x, 0) = g(x) .

Fourier transform the problem with respect to the variable x: d2 d u b(w, t) = −w2 u b(w, t), dt2 dt d b u b(w, 0) = f(w), u b(w, 0) = gb(w). dt Solve the second order differential equation in u b(w, t). The characteristic equation is λ2 + w2λ = 0, with roots λ = 0 or λ = −w2 . The general solution is of the form: 2

u b(w, t) = A(w) + B(w)e−w t .

b The initial condition u b(w, 0) = fb(w) implies that f(w) = A(w) + B(w) or A(w) = b f (w) − B(w). So 2 u b(w, t) = fb(w) + B(w)(e−w t − 1).

The condition

d b(w, dt u

0) = b g(w)

g(w) = −B(w)w2 ⇒ B(w) = − b

gb(w) . w2

So 2 gb(w) −w2 t gb(w) b b u b(w, t) = u b(w, t) = f(w) − (e − 1) = f(w) + (1 − e−w t). 2 2 w w Taking inverse Fourier transforms, we get Z ∞ 2 1 g(w) b b u(x, t) = √ f(w) + (1 − e−w t ) eixw dw 2 w 2π −∞ Z ∞ Z ∞ 2 1 g(w) b 1 b = √ f(w) eixw dw + √ (1 − e−w t ) eixw dw 2 2π −∞ 2π −∞ w Z ∞ 2 1 g(w) b = f(x) + √ (1 − e−w t ) eixw dw. 2 2π −∞ w

Section 7.3

The Fourier Transform Method

291

21. (a) To verify that 1 1 u(x, t) = [f(x − ct) + f(x + ct)] + 2 2c

Z

x+ct

g(s) ds x−ct

is a solution of the boundary value problem of Example 1 is straightforward. You just have to plug the solution into the equation and the initial and boundary conditions and see that the equations are verified. The details are sketched in Section 3.4, following Example 1 of that section. (b) In Example 1, we derived the solution as an inverse Fourier transform: Z ∞ 1 1 fb(w) cos cwt + g(w) sin cwt eiwx dx. b u(x, t) = √ cw 2π −∞ Using properties of the Fourier transform, we will show that Z ∞ 1 1 b √ f(w) cos cwteiwx dw = (1) [f(x − ct) + f(x + ct)]; 2 2π −∞ Z Z ∞ 1 1 1 x+ct √ (2) g(s) ds. g(w) sin cwt eiwx dw = b 2 x−ct 2π −∞ w To prove (1), note that cos cwt =

eicwt + e−icwt , 2

so 1 √ 2π = = =

Z

∞ −∞

fb(w) cos cwteiwx dw

Z ∞ 1 eicwt + e−icwt iwx b √ f(w) ( dw )e 2 2π −∞ Z ∞ Z ∞ 1 1 1 iw(x+ct) iw(x−ct) b b √ f (w) e f (w) e dw √ dw 2 2π −∞ 2π −∞ 1 [f(x + ct) + f(x − ct)]; 2

because the first integral is simply the inverse Fourier transform of fb evaluated at x + ct, and the second integral is the inverse Fourier transform of fb evaluated at x − ct. This proves (1). To prove (2), we note that the left side of (2) is an inverse Fourier transform. So (2) will follow if we can show that (3)

F

Z

x+ct x−ct

2 g(s) ds = gb(w) sin cwt. w

Let G denote an antiderivative of g. Then (3) is equivalent to F (G(x + ct) − G(x − ct)) (w) =

2 c0 G (w) sin cwt. w

c0 = iwG, b the last equation is equivalent to Since G (4)

b F (G(x + ct)) (w) − F (G(x − ct)) (w) = 2iG(w) sin cwt.

Using Exercise 19, Sec. 7.2, we have F (G(x + ct)) (w) − F (G(x − ct)) (w) = = =

eictw F(G)(w) − e−ictw F(G)(w) F(G)(w) eictw − e−ictw b 2iG(w) sin cwt,

292

Chapter 7

The Fourier Transform and its Applications

where we have applied the formula sin ctw =

eictw − e−ictw . 2i

This proves (4) and completes the solution. 23. Fourier transform the boundary value problem ∂u = ∂t u(x, 0) =

∂2u ∂u +k , ∂x2 ∂x f(x) ,

c2

−∞ < x < ∞,

t > 0,

and get d u b(w, t) = dt u b(w, 0) =

b(w, t) + ikwb u(w, t) = u b(w, t)(−c2 w2 + ikw), −c2 w2 u fb(w).

Solve the first order differential equation in u b(w, t) and get u b(w, t) = A(w)e(−c

2

w 2 +ikw)t

.

The initial condition u b(w, 0) = fb(w) implies (−c b u b(w, t) = f(w)e

2

w 2 +ikw)t

.

Taking inverse Fourier transforms, we get Z ∞ 1 (−c2 w 2 +ikw)t ixw b u(x, t) = √ f(w)e e dw 2π −∞ Z ∞ 1 −c2 w 2 t iw(x+kt) b = √ f(w)e e dw. 2π −∞ 25. Using the Fourier transform, we obtain d2 u b(w, t) dt2 u b(w, 0)

= c2 w4u b(w, t),

d u b(w, 0) = b g(w) . dt

= fb(w),

Thus

2

2

u b(w, t) = A(w)e−cw t + B(w)ecw t. Using the initial conditions: b u b(w, 0) = fb(w) ⇒ A(w) + B(w) = f(w) ⇒ A(w) = fb(w) − B(w); and d u b(w, 0) = b g(w) dt

g(w) ⇒ −cw2 A(w) + cw2B(w) = b gb(w) cw2 g(w) b ⇒ −fb(w) + 2B(w) = cw2 1 b g(w) b ; ⇒ B(w) = f (w) + 2 cw2 1 b gb(w) ⇒ A(w) = . f(w) − 2 cw2 ⇒ −A(w) + B(w) =

Section 7.3

The Fourier Transform Method

Hence u b(w, t)

=

1 2

gb(w) −cw2 t 1 b g(w) cw2 t b e e + . fb(w) − f (w) + cw2 2 cw2 cw 2 t

= =

2

2

2

+ e−cw t) b g(w) (ecw t − e−cw t ) + 2 cw2 2 g b (w) 2 2 fb(w) cosh(cw t) + sinh(cw t) cw2 (e fb(w)

Taking inverse Fourier transforms, we get Z ∞ 1 g(w) b 2 u(x, t) = √ fb(w) cosh(cw2 t) + sinh(cw t) eixw dw. cw2 2π −∞ 27. Fourier transform the boundary value problem: d u b(w, t) = dt u b(w, 0) =

b(w, t), −ic2 w3 u fb(w).

Solve the equation in u b(w, t): u b(w, t) = A(w)e−ic

2

w3 t

.

Apply the initial condition: u b(w, t) = fb(w)e−ic

2

w3 t

.

Inverse Fourier transform: 1 u(x, t) = √ 2π

Z

∞ −∞

2 3 fb(w)e−ic w t eixw dw.

293

294

Chapter 7

The Fourier Transform and its Applications

Solutions to Exercises 7.4 1. Repeat the solution of Example 1 making some adjustments: c = √ x2 √2 e− t t

1 2,

gt(x) =

, u(x, t)

=

f ∗ gt(x) √ Z ∞ 1 2 (x−s)2 √ f(s) √ e− t ds t 2π −∞ Z 1 (x−s)2 x−s 20 1 √ e− t ds (v = √ , dv = − √ ds) tπ −1 t t Z x+1 √ t 2 20 √ e−v ds π x−1 √ t x+1 x−1 10 erf( √ ) − erf( √ ) . t t

= = = =

3. Let us use an approach similar to Example 2. Fourier transform the boundary value problem and get: d u b(w, t) = dt u b(w, 0) = Solve the equation in u b:

b(w, t) −w2 u F(70e−

x2 2

) = 70e−

w2 2

.

2

u b(w, t) = A(w)e−w t.

Apply the boundary condition:

u b(w, t) = 70e−

w2 2

)e−w

2

t

= 70e−w

2

(t+ 12 )

.

Inverse Fourier transform: u(x, t)

= = =

2 1 1 1 F −1 70e−w (t+ 2 ) ( = t+ ) 2a 2 √ 2 70 1 −1 −w 2a √ (a = 2t + 1e ) F 2t + 1 2t + 1 x2 70 √ e− 2(2t+1) , 2t + 1

where we have used Theorem 5, Sec. 7.2. 5. Apply (4) with f(s) = s2 : u(x, t)

= =

f ∗ gt(x) Z ∞ (x−s)2 1 1 √ √ s2 e− t ds. 2t 2π −∞

You can evaluate this integral by using integration by parts twice and then appealing to Theorem 5, Section 7.2. However, we will use a different technique based on the operational properties of the Fourier transform that enables us to evaluate a much more general integral. Let n be a nonnegative integer and suppose that f and sn f(s) are integrable and tend to 0 at ±∞. Then n Z ∞ 1 d √ sn f(s) ds = (i)n F(f)(w) . dwn 2π −∞ w=0 This formula is immediate if we recall Theorem 3(ii), Section 7.2, and that Z ∞ b = √1 φ(0) φ(s) ds. 2π −∞

Section 7.4

The Heat Equation and Gauss’s Kernel

295

We will apply this formula with (x−s)2 1 f(s) = √ e− t . 2t

We have 1 − (x−s)2 t F √ e (w) 2t

−iwx

= e

F

= e−iwx e−w So u(x, t) =

=

=

= =

s2 1 √ e− t 2t

2

t

(w) (by Exercise 19, Sec. 7.2)

= e−(iwx+w

2

t)

(See the proof of Th. 1.)

Z ∞ (x−s)2 1 1 √ √ s2 e− t ds 2t 2π −∞ 2 1 − (x−s)2 d t √ F − (w) e dw2 2t w=0 2 d −(iwx+w2 t) − e dw2 w=0 d −(iwx+w 2 t) − (ix + 2wt) −e dw w=0 i h 2 2 −e−(iwx+w t) (ix + 2wt)2 + 2te−(iwx+w t) w=0

=

x2 + 2t.

You can check the validity of this answer by plugging it back into the heat equation. The initial condition is also obviously met: u(x, 0) = x2 . The approach that we took can be used to solve the boundary value problem with f(x)xn as initial temperature distribution. See the end of this section for interesting applications. 7. Proceed as in Example 2. Fourier transform the problem: du b(w, t), u b(w, t) = −t2 w2u dt Solve for u b(w, t):

u b(w, 0) = fb(w). 3

−w 2 t3 b u b(w, t) = f(w)e .

Inverse Fourier transform and note that

2 t3 u(x, t) = f ∗ F −1 e−w 3 . 3

With the help of Theorem 5, Sec. 7.2 (take a = t3 ), we find √ 2 3 2 t3 3 F −1 e−w 3 = √ e−x 4t3 . 3 2t Thus u(x, t)

= =

√ Z ∞ 3(x−s)2 1 3 √ f(s) √ e− 4t3 ds 2π −∞ 2t3 √ Z ∞ 3(x−s)2 3 √ f(s)e− 4t3 ds. 3 2 πt −∞

9. Fourier transform the problem: du b(w, t), u b(w, t) = −e−t w2u dt

u b(w, 0) = fb(w).

296

Chapter 7

The Fourier Transform and its Applications

Solve for u b(w, t):

2 −t u b(w, t) = fb(w)e−w (1−e ) .

Inverse Fourier transform and note that 2 −t u(x, t) = f ∗ F −1 e−w (1−e ) . With the help of Theorem 5, Sec. 7.2 (take a = 1 − e−t ), we find x2 2 −t 1 − F −1 e−w (1−e ) = √ √ e 4(1−e−t ) . 2 1 − e−t Thus 1 u(x, t) = √ √ 2 π 1 − e−t

Z

∞

−

f(s)e

(x−s)2 4(1−e−t )

ds.

−∞

11. This is a generalization of Exercise 9 that can be solved by similar methods. Fourier transform the problem: du b(w, t), u b(w, t) = −a(t)w2 u dt

b u b(w, 0) = f(w).

Let B(t) denote an antiderivative of a(t), hence B(t) =

Rt 0

a(τ ) dτ . Then u b(w, t):

2 u b(w, t) = fb(w)e−w B(t) .

Inverse Fourier transform and note that 2 u(x, t) = f ∗ F −1 e−w B(t) . With the help of Theorem 5, Sec. 7.2 (take a = B(t)), we find x2 2 1 F −1 e−w B(t) = p e− 4B(t) . 2 B(t) Thus 1 u(x, t) = p 2 πB(t)

Z

∞

f(s)e−

(x−s)2 4B(t)

ds.

−∞

13. If in Exercise 9 we take f(x) =

100 if |x| < 1, 0 otherwise,

then the solution becomes 50 u(x, t) = √ √ π 1 − e−t Let z =

√x−s , 2 1−e−t

dz =

√−ds . 2 1−e−t

u(x, t) =

Z

1

(x−s)2

− 4(1−e−t )

e

Then

Z √ 50 −t 2 1−e √ √ π 1 − e−t

√x+1

= =

√x+1

1−e−t

2

2

e−z dz

x−1

√ 2

Z

ds.

−1

1−e−t

100 2 1−e−t −z2 √ e dz π √x−1 2 1−e−t x+1 x−1 √ √ 50 erf − erf . 2 1 − e−t 2 1 − e−t

Section 7.4

The Heat Equation and Gauss’s Kernel

297

√x−1 −erf approaches very quickly As t increases, the expression erf 2√x+1 1−e−t 2 1−e−t x+1 x−1 erf 2 − erf 2 , which tells us that the temperature approaches the limiting distribution x+1 x−1 50 erf − erf . 2 2 You can verify this assertion using graphs. 15. (a) From the definition erf(−x)

Z −x 2 2 = √ e−z dz π 0 Z x 2 2 = −√ e−(−z) dz π 0 = −erf(x)

(z = −Z, dz = −dZ)

Thus erf is an odd function. (b) From the definition erf(0)

=

erf(∞)

=

Z 0 2 2 √ e−z dz = 0, π 0 Z ∞ 2 2 √ e−z dz = 1, π 0

by (4)m Sec. 4.2. (c) By the fundamental theorem of calculus, d erf(x) dx

= =

Z x 2 2 d √ e−z dz π dx 0 2 2 √ e−x > 0 π

for all x. Thus erf is strictly increasing. (d) We have 2 d 1 (xerf(x) + √ e−x dx π

= = =

Thus

Z

2 d 2x erf(x) − √ e−x dx π 2 2 2x 2x erf(x) + √ e−x − √ e−x π π erf(x).

erf(x) + x

(by (c))

2 1 erf(x) dx = (xerf(x) + √ e−x + C. π

(e) From the power series expansion of ez =

∞ X zn n! n=0

(−∞ < z < ∞),

we obtain the power series 2

e−z =

∞ ∞ X X (−z 2 )n z 2n (−1)n = n! n! n=0 n=0 2

(−∞ < z < ∞).

Since the power series expansion for e−z converges for all z, we can integrate it

298

Chapter 7

The Fourier Transform and its Applications

term by term over any interval. So we have Z x 2 2 erf(x) = √ e−z dz π 0 ! Z x X ∞ 2n 2 nz = √ (−1) dz π 0 n=0 n! x Z ∞ ∞ 2 X (−1)n x 2n 2 X (−1)n 2n+1 = √ z dz = √ z π n=0 n! π n=0 n!(2n + 1) 0 0

∞ 2 X (−1)n √ x2n+1. π n=0 n!(2n + 1)

=

17. (a) If f(x) =

T0 0

if a < x < b, otherwise,

then u(x, t) = (b) Let z =

x−s √ , 2c t

dz =

−ds √ . 2c t

T0 √ 2c πt

Z

b

e−

(x−s)2 4c2 t

ds.

a

Then √ Z 2c√t −z2 T0 √ 2c t e dz x−b 2c πt √ 2c t T0 x−a x−b √ √ erf − erf . 2 2c t 2c t x−a

u(x, t)

= =

19. Using (4) with f(x) = T0, then T0 u(x, t) = √ 2c πt Let z =

x−s √ , 2c t

dz =

−ds √ . 2c t

Z

∞

e−

(x−s)2 4c2 t

ds.

−∞

Then

u(x, t) =

√ T0 √ 2c t 2c πt

Z

∞

2

e−z dz = T0 , −∞

by (4), Sec. 7.2. 21. 23. The solution of the heat problem with initial data f(x) is given by (4), which we recall as follows u(x, t) = gt ∗ f(x), where gt(x) is Gauss’s kernel, 2 2 1 gt(x) = √ e−x /(4c t). c 2t

Let u2(x, t) denote the solution of the heat problem with initial data f(x − a). Applying (4) with f replaced by (f)a (x) = f(x − a), we obtain u2(x, t) = gt ∗ (f)a (x). By Exercise 47(c), Sec. 7.2, we know that the convolution operation commutes with translation. So u2(x, t) = gt ∗ (f)a (x) = (gt ∗ f)a (x) = u(x − a, t).

Section 7.4

The Heat Equation and Gauss’s Kernel

299

25. Let u2 (x, t) denote the solution of the heat problem with initial temperature 2 distribution f(x) = e−(x−1) . Let u(x, t) denote the solution of the problem with 2 initial distribution e−x . Then, by Exercise 23, u2(x, t) = u(x − 1, t) By (4), we have 2 2 2 1 u(x, t) = √ e−x /(4c t) ∗ e−x . c 2t We will apply Exercise 24 with a = 4c12t and b = 1. We have ab a+b

= =

1 p 2(a + b)

=

=

1 1 × 1 2 4c t +1 4c2 t 1 1 + 4c2t 1 q 2( 4c12 t + 1) √ c 2t √ . 4c2t + 1

So u(x, t)

= = =

and hence

2 2 2 1 √ e−x /(4c t) ∗ e−x c 2t √ 2 1 c 2t − x √ ·√ e 1+4c2 t c 2t 4c2 t + 1 2 1 − x √ e 1+4c2 t , 4c2t + 1

(x−1)2 1 − u2(x, t) = √ e 1+4c2 t . 4c2 t + 1

27. We proceed as in Exercise 25. Let u2(x, t) denote the solution of the heat 2 problem with initial temperature distribution f(x) = e−(x−2) /2 . Let u(x, t) denote 2 the solution of the problem with initial distribution e−x /2. Then, by Exercise 23, u2(x, t) = u(x − 2, t) By (4), we have 2 2 2 1 u(x, t) = √ e−x /(4c t) ∗ e−x /2 . c 2t

We will apply Exercise 24 with a =

1 4c2 t

ab a+b

= =

1 p 2(a + b)

=

=

and b = 1/2. We have 1 1 × 1 2 8c t + 4c2 t 1 2 + 4c2t 1 q 1 2( 4c2t + 12 ) √ 2c t √ . 4c2t + 2

1 2

So u(x, t)

= = =

2 2 2 1 √ e−x /(4c t) ∗ e−x /2 c 2t √ 2 1 2c t − x √ ·√ e 2+4c2 t c 2t 4c2 t + 2 2 1 − x √ e 2+4c2 t , 2c2t + 1

300

Chapter 7

The Fourier Transform and its Applications

and hence

(x−2)2 1 − u2(x, t) = √ e 2+4c2 t . 2c2 t + 1

29. Parts (a)-(c) are obvious from the definition of gt(x). (d) The total area under the graph of gt(x) and above the x-axis is Z ∞ Z ∞ 2 2 1 √ gt (x) dx = e−x /(4c t) dx c 2t −∞ −∞ √ Z √ 2c t ∞ −z2 x √ = e dz (z = √ , dx = 2c t dz) 2c t c 2t −∞ √ Z ∞ −z2 √ 2 e dz = 2π, −∞

by (4), Sec. 7.2. (e) To find the Fourier transform of gt(x), apply (5), Sec. 7.2, with a=

√ 1 √ = 2c 2t, 2a

1 , 4c2 t

1 = c2 t. 4a

We get 2 2 1 √ F e−x /(4c t) dx c 2t √ 2 2 1 √ × 2c 2te−c tω c 2t

gt (w) = b =

e−c

=

2

tω 2

.

(f) If f is an integrable and piecewise smooth function, then at its points of continuity, we have lim gt ∗ f(x) = f(x). t→0

This is a true fact that can be proved by using properties of Gauss’s kernel. If we interpret f(x) as an initial temperature distribution in a heat problem, then the solution of this heat problem is given by u(x, t) = gt ∗ f(x). If t → 0, the temperature u(x, t) should approach the initial temperature distribution f(x). Thus limt→0 gt ∗ f(x) = f(x). Alternatively, we can use part (e) and argue as follows. Since lim F (gt ) (ω) = lim e−c

t→0

t→0

2

tω 2

= 1,

So lim F (gt ∗ f) = lim F (gt) F (f) = F (f) .

t→0

t→0

You would expect that the limit of the Fourier transform be the transform of the limit function. So taking inverse Fourier transforms, we get limt→0 gt ∗f(x) = f(x). (Neither one of the arguments that we gave is rigorous.) A generalization of Exercise 5 Suppose that you want to solve the heat equation ut = uxx subject to the initial condition u(x, 0) = xn where n is a nonnegative integer. We have already done the case n = 0 (in Exercise 19) and n = 2 (in Exercise 5). For the general case, proceed as in Exercise 5 and apply (4) with f(s) = sn : u(x, t)

= =

f ∗ gt(x) Z ∞ (x−s)2 1 1 √ √ sn e− t ds. 2t 2π −∞

Section 7.4

The Heat Equation and Gauss’s Kernel

Use the formula 1 √ 2π

Z

∞

sn f(s) ds = (i)n

−∞

with

dn F(f)(w) dwn

301

, w=0

(x−s)2 1 f(s) = √ e− t 2t

and

(x−s)2 2 1 √ e− t (w) = e−(iwx+w t) 2t (see the solution of Exercise 5). So n d −(iwx+w2 t) n u(x, t) = (i) e . dwn w=0 F

To compute this last derivative, recall the Taylor series formula f(x) =

∞ X f (n) (a) (x − a)n . n! n=0

So knowledge of the Taylor series gives immediately the values of the derivatives at a. Since ∞ X (aw)n aw e = , n! n=0 we get

Similarly,

dk bw2 e dwk

dj aw e dwj

w=0

=

= aj .

w=0

0

if k is odd,

j

b (2j)! j!

if k = 2j.

Returning to u(x, t), we compute the nth derivative of e−(iwx+w rule and use the what we just found and get n d −w2 t −iwx n u(x, t) = (i) e e dwn w=0

2

t)

using the Leibniz

=

j n X d −w2 t dn−j −iwx n (i) e · e j dwj dwn−j w=0 j=0

=

[ n2 ] 2j X d dn−2j −iwx n n −w 2 t (i) e · e 2j dw2j dwn−2j w=0 j=0

=

[ n2 ] X (−t)j (2j)! n (i) · (−ix)n−2j 2j j! j=0

=

j [ n2 ] X t (2j)! n−2j n . ·x 2j j!

n

n

j=0

For example, if n = 2, u(x, t)

=

j 1 X t (2j)! 2−2j 2 = x2 + 2t, ·x 2j j! j=0

302

Chapter 7

The Fourier Transform and its Applications

which agrees with the result of Exercise 5. If n = 3, u(x, t)

=

j 1 X t (2j)! 3−2j 3 = x3 + 6tx. ·x 2j j! j=0

You can easily check that this solution verifies the heat equation and u(x, 0) = x3. If n = 4, u(x, t)

=

j 2 X t (2j)! 4−2j 4 = x4 + 12tx2 + 12t2. ·x 2j j! j=0

Here too, you can check that this solution verifies the heat equation and u(x, 0) = x4 . We now derive a recurrence relation that relates the solutions corresponding to n − 1, n, and n + 1. Let un = un(x, t) denote the solution with initial temperature distribution un (x, 0) = xn . We have the following recurrence relation un+1 = xun + 2ntun−1. The proof of this formula is very much like the proof of Bonnet’s recurrence formula for the Legendre polynomials (Section 5.6). Before we give the proof, let us verify the formula with n = 3. The formula states that u4 = 4u3 + 6tu2. Sine u4 = x4 + 12tx2 + 12t2, u3 = x3 + 6tx, and u2 = x2 + 2t, we see that the formula is true for n = 3. We now prove the formula using Leibniz rule of differentiation. As in Section 5.6, let us use the symbol Dn to denote the nth derivative, We have n+1 d n+1 −(iwx+w 2 t) un+1(x, t) = (i) e dwn+1 w=0 h i 2 = (i)n+1 Dn+1 e−(iwx+w t) w=0

=

h i 2 (i)n+1 Dn De−(iwx+w t)

=

h i 2 (i)n+1 Dn −(ix + 2wt)e−(iwx+w t)

=

h i 2 (i)n+1 Dn −(ix + 2wt)e−(iwx+w t)

=

h i 2 2 (i)n+1 −Dn e−(iwx+w t) (ix + 2wt) − 2ntDn−1 e−(iwx+w t)

w=0

w=0

w=0

w=0

= =

x(i)n Dn e−(iwx+w xun + 2ntun−1.

2

t)

+ 2nt (i)n−1 Dn−1 e−(iwx+w w=0

2

t)

w=0

Section 7.5

A Dirichlet Problem and the Poisson Integral Formula

303

Solutions to Exercises 7.5 1. To solve the Dirichlet problem in the upper half-plane with the given boundary function, we use formula (5). The solution is given by

u(x, y)

= = =

y π

Z

∞ −∞

f(s) ds (x − s)2 + y2

Z 50y 1 ds π −1 (x − s)2 + y2 1+x 1−x 50 −1 −1 tan + tan , π y y

where we have used Example 1 to evaluate the definite integral. 3. We appeal to the Poisson integral formula (5). In evaluating the integral, we will come across the improper integral Z

y π

∞ −∞

cos s ds, s2 + y2

which is a nontrivial integral. In order to evaluate it, we will appeal to results about Fourier transforms. Fix y > 0 and consider √ y 2 h(s) = √ 2 π(s + y2 )

(−∞ < s < ∞).

The Fourier transform of h(s) is

b h(w)

= = =

Z y ∞ π −∞ Z y ∞ π −∞ Z y ∞ π −∞

e−isw ds s2 + y2 cos sw − i sin sw ds s2 + y2 cos sw ds, s2 + y2

because the sine integral has an odd integrand, so its value is 0. Let us now use a table of Fourier transforms to evaluate the Fourier transform of h (see also Exercise 12, Section 7.2). From entry 7, form the table of Fourier transforms, Appendix B.1, we have b h(w) = e−y|w|

(y > 0, −∞ < w < ∞).

So

(1)

y π

Z

∞ −∞

cos sw ds = e−y|w| s2 + y2

(y > 0, −∞ < w < ∞).

We are now ready to evaluate the Poisson integral for the given boundary function

304 Chapter 7 The Fourier Transform and its Applications f(x): u(x, y)

= = = = =

=

= =

y π y π y π y π y π

Z

∞

Z

−∞ ∞

Z

−∞ ∞

Z

−∞ ∞

Z

−∞ ∞ −∞

f(s) ds (x − s)2 + y2 f(s) ds (x − s)2 + y2 f(x − s) ds (Change variables s = x − S.) s2 + y2 cos(x − s) ds s2 + y2 cos x cos s + sin x sin s ds s2 + y2 zZ

=0

}| { y cos x cos s sin s y sin x ds + ds 2 2 2 2 π π −∞ s + y −∞ s + y (The integrand is odd in the second integral.) Z y ∞ cos s cos x ds π −∞ s2 + y2 Z

∞

∞

cos x e−y ,

where we have used (1) with w = 1. It is instructive t check that we have indeed found the (bounded) solution to our Dirichlet problem: u(x, y) = cos x e−y . We have uxx = −cosx e−y and uyy = cosxe−y . So uxx + uyy = 0. This shows that u is a solution of Laplace’s equation. Does it satisfy the boundary condition? Let’s see: u(x 0) = cos x = f(x). Yes, it does. Can you guess the solution of the Dirichlet problem with boundary condition f(x) = sin x? 5. Appealing to (4) in Section 7.5, with y = y1 , y2 , y1 + y2 , we find F(Py1 )(w) = e−y1 |w|,

F(Py2 )(w) = e−y2 |w| ,

F(Py1+y2 )(w) = e−(y1 +y2 )|w| .

Hence F(Py1 )(w) · F(Py2 )(w) = e−y1 |w|e−y2 |w| = e−(y1 +y2 )|w| = F(Py1 +y2 )(w). But F(Py1 )(w) · F(Py2 )(w) = F(Py1 ∗ Py2 )(w), Hence F(Py1 +y2 )(w) = F(Py1 ∗ Py2 )(w); and so Py1+y2 = Py1 ∗ Py2 . 7. (a) If f(x) = 7.5,

1 4+x2 ,

then in terms of the Poisson kernel, we have from (3), Sec. 1 f(x) = 2

r

π P2(x). 2

The solution of the Dirichlet problem with boundary value f(x) is then 1 u(x, y) = Py ∗ f(x) = 2

r

π 1 Py ∗ P2(x) = 2 2

r

π P2+y (x), 2

by Exercise 5. More explicitely, using (3), Sec. 7.5, with y + 2 in place of y, we obtain r r 1 π 2 1 y+2 y+2 u(x, y) = = 2 2 π x2 + (y + 2)2 2 x2 + (y + 2)2

Section 7.5

A Dirichlet Problem and the Poisson Integral Formula

305

1 1 (b) The boundary function, 4+x 2 , takes values between 0 and 1/4. For 0 < T ≤ 4 , the points in the upper half-plane with u(x, y) = T satisfy the equation

x2 + (y + 2)2 − x2 + (y + 2 − These points lie on the circle with radius

1 2T

(y + 2) = 0

1 2 4T ) 1 4|T |

=

1 16T 2 .

and center at (0,

1 4T

− 2).

9. Modify the solution of Example 1(a) to obtain that, in the present case, the solution is T0 a+x a−x −1 −1 u(x, y) = tan + tan . π y y To find the isotherms, we must determine the points (x, y) such that u(x, y) = T . As in the solution of Example 1(b), these points satisfy 2 2 πT πT x2 + y − a cot( ) = a csc( ) . T0 T0 Hence the points belong to the arc in the upper half-plane of the circle with center T0 πT (0, a cot( πT T0 )) and radius a csc( T0 ). The isotherm corresponding to T = 2 is the arc of the circle π 2 π 2 x2 + y − a cot( ) = a csc( ) , 2 2 or x2 + y2 = a2. Thus the isotherm in this case is the upper semi-circle of radius a and center at the origin. 13. Parts (a)-(c) are clear. Part (e) fellows from a table. For (d), you can use (e) and the fact that the total area under the graph of Py (x) and above the x-axis is √ √ √ cy (0) = 2πe−0 = 2π. 2πP (f) If f is an integrable function and piecewise smooth, consider the Dirichlet problem with boundary values f(x). Then we know that the solution is u(x, y) = Py ∗ f(x). In particular, the solution tends to the boundary function as y → 0. But this means that limy→0 Py ∗ f(x) = f(x). The proof of this fact is beyond the level of the text. Another way to justify the convergence is to take Fourier transforms. We have F(Py ∗ f)(w) = F(Py )(w) · F(f)(w) = e−y|w| F(f)(w). Since limy→0 e−y|w| = 1, it follows that lim F(Py ∗ f)(w) = lim e−y|w| F(f)(w) = F(f)(w).

y→0

y→0

Taking inverse Fourier transforms, we see that limy→0 Py ∗ f(x) = f(x). The argument that we gave is not rigorous, since we did not justify that the inverse Fourier transform of a limit of functions is the limit of the inverse Fourier transforms.

306

Chapter 7

The Fourier Transform and its Applications

Solutions to Exercises 7.6 1. The even extension of f(x) is fe (x) =

1 if −1 < x < 1, 0 otherwise.

The Fourier transform of fe (x) is computed in Example 1, Sec. 7.2 (with a = 1). We have, for w ≥ 0, r 2 sin w Fc (f)(w) = F(fe )(w) = . π w To write f as an inverse Fourier cosine transform, we appeal to (6). We have, for x > 0, r Z ∞ 2 f(x) = Fc (f)(w) cos wx dw, π 0 or Z ∞ 1 if 0 < x < 1, 2 sin w 0 if x > 1, cos wx dw = 1 π 0 w if x = 1. 2 Note that at the point x = 1, a point of discontinuity of f, the inverse Fourier transform is equal to (f(x+) + f(x−))/2. 3. The even extension of f(x) is fe (x) = 3 e−2|x|. We compute the Fourier transform of fe (x) by using Exercises 5 and 17(a), Sec. 7.2 (with a = 2). We have r 2 2 F(fe )(w) = 3 . π 4 + w2 So, for w ≥ 0, r 2 6 Fc (f)(w) = F(fe )(w) = . π 4 + w2 To write f as an inverse Fourier cosine transform, we appeal to (6). We have, for x > 0, r Z ∞ 2 f(x) = Fc (f)(w) cos wx dw, π 0 or Z 12 ∞ cos wx −2x 3e = dw. π 0 4 + w2 5. The even extension of f(x) is cos x if −2π < x < 2π, fe (x) = 0 otherwise. Let’s compute the Fourier cosine transform using definition (5), Sec. 7.6: r Z 2π 2 Fc(f)(w) = = cos x cos wx dx π 0 r Z 2π 2 1 = [cos(w + 1)x + cos(w − 1)x] dx π 2 r 0 2π 1 2 sin(w + 1)x sin(w − 1)x = (w 6= 1) + 2 π w+1 w−1 0 r 1 2 sin 2(w + 1)π sin 2(w − 1)π = + (w 6= 1) 2 π w+1 w−1 r 1 2 sin 2πw sin 2πw = + (w 6= 1) 2 π w+1 w−1 r 2 w = sin 2πw 2 (w 6= 1). π w −1

Section 7.6

The Fourier Cosine and Sine Transforms

307

Also, by l’Hospital’s rule, we have r √ w 2 sin 2πw 2 = 2π, lim w→0 π w −1 which is the value of the cosine transform at w = 1. To write f as an inverse Fourier cosine transform, we appeal to (6). We have, for x > 0, Z 2 ∞ w cos x if 0 < x < 2π, sin 2πw cos wx dw = 0 if x > 2π. π 0 w2 − 1 For x = 2π, the integral converges to 1/2. So Z 2 ∞ w 1 sin 2πw cos 2πw dw = . 2 π 0 w −1 2 7. The odd extension of f is if 0 < x < 1, 1 fo (x) = −1 if − 1 < x < 0 0 otherwise. To avoid computing the Fourier transform of fo from scratch, let us introduce the function ( 1 if − 12 < x < 12 , g(x) = 0 otherwise. From Example 1, Sec. 7.2, we have F(g)(w) =

r

2 sin w2 . π w

Note that fo (x) = g(x − 12 ) + g(x − 12 ). So F(fo )(w)

= =

= =

1 1 F(g(x − ))(w) + F(g(x − ))(w) 2 2 1 1 e− 2 iw F(g)(w) + e 2 iw F(g)(w) (Exercise 20, Sec. 7.2) z

−2i sin

w 2

}| { 1 − 12 iw iw F(g)(w) e + +e 2 r 2 sin2 w2 −2i . π w

Applying (10), Sec. 7.6, we find Fs (f)(w) = iF(fo )(w) = 2

r

2 sin2 w2 = π w

r

2 1 − cosw . π w

The inverse sine transform becomes Z 2 ∞ 1 − cosw f(x) = sin wx dw. π 0 w 9. Applying the definition of the transform and using Exercise 17, Sec. 2.6 to evaluate the integral, r Z ∞ 2 −2x Fs (e )(w) = e−2x sin wx dx π 0 r ∞ 2 e−2x = [−w cos wx − 2 sin wx] 2 π 4+w x=0 r 2 w = . π 4 + w2

308

Chapter 7

The Fourier Transform and its Applications

The inverse sine transform becomes Z 2 ∞ w sin wx dw. f(x) = π 0 4 + w2 11. The odd extension is fo (x) =

(

sin 2x if − π < x < π, 0

otherwise.

To evaluate the Fourier transform of fo , write fo (x) = sin 2x · g(x), where ( 1 if − π < x < π, g(x) = 0 otherwise. Then by Exercise 20(b), Sec. 7.2, we have F(fo ) = F(sin 2xg(x)) =

1 (F(g)(w − 2) − F(g)(w + 2)) . 2i

By Example 1, Sec. 7.2, F(g)(w) =

r

2 sin πw . π w

So r

2 1 π 2i

sin π(w − 2) sin π(w + 2) − w−2 w+2 r 2 1 sin πw sin πw − π 2i w − 2 w+2 r 2 2 sin πw . π i w2 − 4

F(fo ) = = =

Applying (10), Sec. 7.6, we find Fs(f)(w) = iF(fo )(w) = 2

r

2 sin πw . π w2 − 4

The inverse sine transform becomes Z 4 ∞ sin πw f(x) = sin wx dw. π 0 w2 − 4 13. We have fe (x) = Fc

1 1+x2 .

1 1 + x2

So

=F

1 1 + x2

=

=

r

π −w e 2

(w > 0),

by Exercise 11, Sec. 7.2. 15. We have fo (x) = Fs

x . 1+x2

1 1 + x2

So = iF

x 1 + x2

r

π −w e 2

(w > 0),

by Exercise 35, Sec. 7.2. 17. We have fe (x) = Fc

cos x 1 + x2

cos x 1+x2 .

=F

So

cos x 1 + x2

=

r

π −|w−1| + e−(w+1) e 2

(w > 0),

Section 7.6

The Fourier Cosine and Sine Transforms

309

by Exercises 11 and 20(b), Sec. 7.2. 19. Let fe denote the even extension of f so that fe (x) = f(x) for all x > 0. Then, for w > 0, r Z ∞ r Z ∞ 2 2 f(x) cos wx dx = fe (x) cos wx dx Fc (f)(w) = π 0 π 0 r Z 21 ∞ = fe (x) cos wx dx π 2 −∞ (The integrand is even.) Z ∞ 1 = √ fe (x)(cos wx − i sin wx) dx 2π −∞ (The integral of fe (x) sin wx is 0 because the integrand is odd.) Z ∞ 1 = √ fe (x)e−iwx dx 2π −∞ = F(fe )(w). We proceed in a similar way when dealing with the sine transform. Let fo denote the odd extension of f. The f(x) = fo (x) for x > 0 and r Z ∞ r Z ∞ 2 2 Fs (f)(w) = f(x) sin wx dx = fo (x) sin wx dx π 0 π 0 r Z 21 ∞ = fo (x) sin wx dx π 2 −∞ (The integrand is even.) Z ∞ 1 = i√ fo (x)(cos wx − i sin wx) dx 2π −∞ (The integral of fo (x) cos wx is 0 because the integrand is odd.) Z ∞ i = √ fo (x)e−iwx dx 2π −∞ = iF(fo )(w). 21. From the definition of the inverse transform, we have Fc f = Fc−1f. So Fc Fc f = Fc Fc−1f = f. Similarly, FsFs f = Fs Fs−1f = f.

310

Chapter 7

The Fourier Transform and its Applications

Solutions to Exercises 7.7 1. Fourier sine transform with respect to x: d u b (w, dt s

2

t) = −w u bs (w, t) + d u b (w, dt s

q

=0

z }| { u(0, t)

2 w π

t) = −w2 u bs(w, t).

Solve the first-order differential equation in u bs(w, t) and get 2

u bs(w, t) = A(w)e−w t. Fourier sine transform the initial condition u bs(w, 0) = A(w) = Fs(f(x))(w) = T0 Hence u bs (w, t) =

r

r

2 1 − cos bw . π w

2 1 − cos bw −w2 t . e π w

Taking inverse Fourier sine transform: Z 2 ∞ 1 − cos bw −w2 t u(x, t) = sin wx dw. e π 0 w 3. As in Exercise 1, Fourier sine transform with respect to x: d bs (w, dt u

2

t) = −w u bs (w, t) + d bs(w, dt u

q

=0

2 πw

z }| { u(0, t)

t) = −w2 u bs(w, t).

Solve the first-order differential equation in u bs(w, t) and get 2

u bs(w, t) = A(w)e−w t. Fourier sine transform the initial condition u bs(w, 0) = A(w) = Fs (f(x))(w) = Hence

r

2 −w e . π

r

r 2 −w −w2 t 2 −(w2 t+w) u bs(w, t) = = . e e e π π Taking inverse Fourier sine transform: Z 2 ∞ −(w2 t+w) u(x, t) = e sin wx dw. π 0 5. If you Fourier cosine the equations (1) and (2), using the Neumann type condition ∂u (0, t) = 0, ∂x you will get =0

d bc (w, dt u

h

t) = c2 − w2u bc (w, t) − d u b (w, dt c

q

2 π

z }| { i d u(0, t) dx

t) = −c2 w2u bc (w, t).

Section 7.7

Problems Involving Semi-Infinite Intervals

311

Solve the first-order differential equation in u bc(w, t) and get u bc(w, t) = A(w)e−c

2

w2 t

.

Fourier cosine transform the initial condition u bc(w, 0) = A(w) = Fc (f)(w). Hence u bs(w, t) = Fc (f)(w)e−c

2

w2 t

.

Taking inverse Fourier cosine transform: r Z ∞ 2 2 2 u(x, t) = Fc(f)(w)e−c w t cos wx dw. π 0 7. Solution (4) of Example 1 reads r Z ∞ 2 2 2 u(x, t) = fbs (w)e−c w t sin wt dw, π 0 where fbs (w) =

r

2 π

Z

∞

f(s) sin ws ds. s=0

So u(x, t) = =

=

Z ∞ Z 2 2 2 ∞ f(s) sin ws dse−c w t sin wt dw π w=0 s=0 Z ∞ Z 2 2 2 ∞ f(s) sin ws sin wt e−c w t dwds π s=0 w=0 1 Use sin A sin B = cos(A − B) − cos(A + B) .) r Z 2∞ Z ∞ 2 2 1 2 √ cos(w(x − s)) − cos w(x + s) e−c w t dwds. f(s) π w=0 2π s=0

Evaluate the inner integral by appealing to a result of Example 4, Sec. 7.2, which we recall in the following convenient notation: r Z ∞ 2 x2 2 1 e−aw cos xw dw = √ e− 4a . π w=0 2a Hence r Z ∞ 2 2 2 cos(w(x − s)) − cos w(x + s) e−c w tdw π w=0 r Z ∞ r Z ∞ 2 2 2 2 −c2 w 2 t = cos(w(x − s))e dw − cos w(x + s)e−c w t dw π w=0 π w=0 i (x+s)2 1 h (x−s)2 = √ √ e− 4c2 t − e− 4c2 t ; 2c t and so

1 u(x, t) = √ 2c πt

Z

∞

2

− (x−s) 2

f(s) e

4c t

2

− (x+s) 2

−e

4c t

ds .

0

9. (a) Taking the sine transform of the heat equation (1) and using u(0, t) = T0 for t > 0, we get r h i d 2 2 2 bs(w, t) + u bs(w, t) = c − w u wu(0, t) ; dt π

312

Chapter 7

The Fourier Transform and its Applications

or

r d 2 2 2 2 bs(w, t) = c u bs (w, t) + c ω u wT0 . dt π Taking the Fourier sine transform of the boundary condition u(x, 0) = 0 for x > 0, we get u bs (w, 0) = 0. (b) A particular solution of the differential equation can be guessed easily: u bs (w, t) = q 2 T0 π w .

The general solution of the homogeneous differential equation: d bs(w, t) = 0 u bs (w, t) + c2 ω2 u dt

is u bs (w, t) = A(w)e−c ential equation is

2

w2 t

. So the general solution of the nonhomogeneous differ-

Using u bs(w, 0) = A(w)

q

u bs(w, t) = A(w)e 2 T0 π w

r

2 T0 . π w q = 0, we find A(w) = − π2 Tw0 . So −c2 w 2 t

u bs(w t) =

r

2 T0 − π w

r

2 T0 −c2 w2 t . e π w

Taking inverse sine transforms, we find Z 2 ∞ T0 T0 −c2 w2 t u(x, t) = sin wx dw − e π 0 w w =sgn(x)=1 z Z }| { Z 2 ∞ sin wx 2T0 ∞ sin wx −c2 w2 t = T0 dw dw − e π 0 w π 0 w Z 2T0 ∞ sin wx −c2 w2 t = T0 − dw e π 0 w 11. We will solve the Dirichlet problem uxx + uyy = 0 u(0, y) = 0,

(0 < x < 1, 0 < y)

∂u (x, ∂y

0) = 0, u(1, y) = e−y .

Cosine transform with respect to y: =0

d2 bc (x, dx2 u

w) − w2u bc (x, w) − d2 bc (x, dx2 u

q

2 π

z }| { ∂u (x, 0) = 0 ∂y

w) = w2u bc (x, w).

The general solution is u bc(x, w) = A(w) cosh wx + B(w) sinh wx. Using −y

u bc (0, w) = 0 and u bc (1, w) = Fc (e we get A(w) = 0

and B(w) =

Hence u bc(x, w) =

r

r

)=

r

2 1 , π 1 + w2

2 1 1 · . 2 π 1 + w sinh w

2 1 sinh wx . π 1 + w2 sinh w

Section 7.7

Problems Involving Semi-Infinite Intervals

313

Taking inverse cosine transforms: Z 2 ∞ 1 sinh wx u(x, y) = cos wy dw. π 0 1 + w2 sinh w 13. Proceed as in Exercise 11 using the Fourier sine transform instead of the cosine transform and the condition u(x 0) = 0 instead of uy (x, 0) = 0. This yields =0

d2 bs(x, dx2 u

q z }| { w) − w2 u bs(x, w) + π2 u(x, 0) = 0 d2 bs (x, dx2 u

w) = w2u bs(x, w).

The general solution is u bs (x, w) = A(w) cosh wx + B(w) sinh wx. Using −y

u bs (0, w) = 0 and

u bs(1, w) = Fs(e

we get A(w) = 0

and B(w) =

Hence u bs (x, w) =

r

r

)=

r

2 w , π 1 + w2

2 w 1 · . π 1 + w2 sinh w

2 w sinh wx . π 1 + w2 sinh w

Taking inverse sine transforms: 2 u(x, y) = π

Z

∞ 0

w sinh wx sin wy dw. 1 + w2 sinh w

15. We will solve the Dirichlet problem uxx + uyy = 0 u(x, 0) = 0, where f(y) =

(0 < x, 0 < y) u(0, y) = f(y),

1 if 0 < y < 1, 0 if 1 < y.

Sine transform with respect to y: d2 bs (x, dx2 u

2

w) − w u bs (x, w) + d2 bs (x, dx2 u

q

=0

z }| { 0) = 0

2 π w u(x,

w) = w2u bs(x, w).

The general solution is u bs (x, w) = A(w)e−wx + B(w)ewx . Since we expect a Fourier transform to be bounded and since w and x are > 0, we discard the term in ewx and take u bs (x, w) = A(w)e−wx . Transforming the boundary condition: u bs (0, w) = A(w) = Fs (f(y))(w) =

r

2 1 − cos w . π w

314

Chapter 7

The Fourier Transform and its Applications

Hence u bs(x, w) =

r

2 1 − cos w −wx . e π w

Taking inverse sine transforms: 2 u(x, y) = π

Z

∞ 0

1 − cos w −wx sin wy dw. e w

Section 7.8

Generalized Functions

315

Solutions to Exercises 7.8 1. As we move from left to right at a point x0, if the graph jumps by c units, then we must add the scaled Dirac delta function by cδx0 (x). If the jump is upward, c is positive; and if the jump is downward, c is negative. With this in mind, by looking at the graph, we see that f 0 (x) =

1 1 1 1 1 δ−2 (x)+ δ−1 (x)− δ1 (x)− δ2 (x) = (δ−2 (x) + δ−1 (x) − δ1 (x) − δ2 (x)) . 2 2 2 2 2

3. We reason as in Exercise 1; furthermore, here we have to add the nonzero part of the derivative. We have 0 if x < −2, −1 if − 2 < x < −1, 0 f (x) = δ−1 (x) − 2 δ2 (x) + 1 if − 1 < x < 1, if x > 1, 0 Using unit step functions, we can write f 0 (x) = − U−2(x) − U−1(x) + U−1 (x) − U1(x) + δ−1 (x) − 2 δ2(x). Exercises 5, 7, 9, and 11 can be done by reasoning as in Exercises 1 and 3. See the Answers in the back of the text. 13. We do this problem by reversing the steps in the solutions of the previous exercises. Since f(x) has zero derivative for x < −2 or x > 3, it is therefore constant on these intervals. But since f(x) tends to zero as x → ±∞, we conclude that f(x) = 0 for x < −2 or x > 3. At x = −2, we have a jump upward by one unit, then the function stays constant for −2 < x < −1. At x = −1, we have another jump upward by one unit, then the function stays constant for −1 < x < −1. At x = 1, we have another jump upward by one unit, then the function stays constant for 1 < x < 3. At x = 3, we have a jump downward by three units, then the function stays constant for x > 3. Summing up, we have 0 if x < −2, 1 if − 2 < x < −1, 2 if − 1 < x < 1, f(x) = 3 if 1 < x < 3, 0 if 3 < x. 15. We reason as in the previous exercise and find that f(x) = 0 for x < −1. At x = −1, we have a jump downward by one unit, then the function has derivative f 0 (x) = 2x for −1 < x < 1, implying that the function is f(x) = x2 + c on this interval. Since the graph falls by 1 unit at x = −1, we see that f(x) = x2 − 2 for −1 < x < 1. At x = 1, we have a jump upward by one unit, then the function stays constant and equals 0 for x > 1. Summing up, we have 0 if x < −1, 2 x − 2 if − 1 < x < 1, f(x) = if 1 < x. 0 17. We use the definition (7) of the derivative of a generalized function and the fact that the integral against a delta function δa picks up the value of the function

316

Chapter 7

The Fourier Transform and its Applications

at a. Thus hφ0 (x), f(x)i

=

hφ(x), −f 0 (x)i = −hφ(x), f 0 (x)i

=

−hδ0 (x) − δ1 (x), f 0 (x)i = −f 0 (0) + f 0 (1).

19. From Exercise 7, we have φ0(x) = a1 U−2a (x) − U−a (x) − Using the definition of the unit step function, we find hφ0(x), f(x)i

= =

=

Ua (x) − U2a (x) .

1 1 U−2a (x) − U−a (x) − Ua (x) − U2a(x) , f(x)i a a Z ∞ h 1 i 1 U−2a(x) − U−a (x) − Ua (x) − U2a (x) f(x) dx a −∞ a

h

1 a

Z

−a

1 f(x) dx − a −2a

Z

2a

f(x) dx. a

21. From Exercise 7, we have φ0(x) = a1 U−2a (x) − U−a (x) − Using (9) (or arguing using jumps on the graph), we find

φ00(x) =

1 a

1 a

Ua (x) − U2a (x) .

1 1 1 δ−2a (x)−δ−a (x) − δa (x)−δ2a (x) = δ−2a (x)−δ−a (x)−δa (x)+δ2a (x) . a a a

23. From Figure 19 we see that φ has no jumps on the graph. So, as a generalized function, the derivative has no deltas in it and 0 −1 1 φ0(x) = −1 0

if x < −2, if − 2 < x < −1, if − 1 < x < 1, if 1 < x < 2, if x > 2.

In terms of unit step functions, we have φ0(x) = − U−2(x) − U−1 (x) + U−1 (x) − U1(x) − U1 (x) − U2 (x) . Thus, using (9), φ00(x) = − δ−2 (x)−δ−1 (x) + δ−1 (x)−δ1 (x) − δ1 (x)−δ2 (x) = −δ−2 +2δ−1 −2δ1 +δ2 .

25. Using the definition of φ and the definition of a derivative of a generalized

Section 7.8

317

Generalized Functions

function, and integrating by parts, we find Z ∞ 0 0 hφ (x), f(x)i = −hφ(x), f (x)i = − φ(x)f 0 (x) dx −∞

=

−

Z

0 0

2(x + 1)f (x) dx −

Z

−1

−1

=

−2(x − 1)f 0 (x) dx 0

Z 0 −2(x + 1)f(x) + 2

=

1

−2f(0) + 2

Z

0

−1

Z 1 f(x) dx + 2(x − 1)f(x) − 2 0

0

f(x) dx + 2f(0) − 2

Z

−1

f(x) dx 0

1

f(x) dx 0

h2 U−1 (x) − U0 (x) , f(x)i − h2 U0 (x) − U1 (x) , f(x)i. h2 U−1 (x) − U0 (x) − 2 U0(x) − U1(x) , f(x)i.

= =

Thus

1

φ0 (x) = 2 U−1 (x) − U0 (x) − 2 U0 (x) − U1(x) .

Reasoning similarly, we find hφ00(x), f(x)i

=

−hφ0 (x), f 0 (x)i = −

Z

∞

φ0 (x)f 0 (x) dx −∞

=

−2

Z

0

f 0 (x) dx + 2 −1

Z

1

f 0 (x) dx

0

=

−2 f(0) − f(−1) + 2 f(1) − f(0) = 2f(−1) − 4f(0) + 2f(1)

=

h2δ−1 − 4δ0 + 2δ1 , f(x)i.

Thus φ00(x) = 2δ−1 − 4δ0 + 2δ1 . 27. We use definition (7) of the derivative of a generalized function and the fact that the integral against a delta function δa picks up the value of the function at a: hδ00 (x), f(x)i = −hδ0 (x), f 0 (x)i = −f 0 (0); similarly, hδα0 (x), f(x)i = −hδα (x), f 0 (x)i = −f 0 (α) and hδα(n) (x), f(x)i = (−1)n hφ(x), f (n) (x)i = −f (n) (α). 29. We use (13) and the linearity of the Fourier transform: 1 F 3δ0 − 2δ−2 = √ 3 − 2e2iw . 2π 31. We have ( sgn x)0 = 2δ0(x). This is clear if you draw the graph of sgn x, you will see a jump of 2 units at x = 0, otherwise it is constant. Hence 2 F ( sgn x)0 = F 2δ0 (x) = √ . 2π But

F ( sgn x)0 = i w F sgn x .

318

Chapter 7

The Fourier Transform and its Applications

So

i i 2 F sgn x = − F ( sgn x)0 = − · √ = −i w w 2π

r

2 1 . π w

33. Using the operational property in Theorem 3(i), Section 7.2, we find F x (U−1 − U1)

d F U−1 − U1 dw i d i iw −iw i −√ e +√ e dw 2π w 2π w −i(i) d eiw − e−iw √ w 2π dw 1 d 2i sin w √ (Recall eiu − e−iu = 2i sin u) w 2π dw 2i w cos w − sin w √ w2 2π r 2 w cos w − sin w i . π w2

=

i

=

=

=

=

=

The formula is good at w = 0 if we take the limit as w → 0. You will get r r 2 w cos w − sin w 2 −w sin w F x (U−1 − U1 ) = lim i =i lim = 0. w→0 π w2 π w→0 2w (Use l’Hospital’s rue.) Unlike the Fourier transform in Exercise 31, the transform here is a nice continuous function. There is a major difference between the transforms of the two exercises. In Exercise 31, the function is not integrable and its Fourier transform exists only as a generalized function. In Exercise 33, the function is integrable and its Fourier transform exists in the usual sense of Section 7.2. In fact, look at the transform in Exercise 31, it is not even defined at w = 0. An alternative way to do this problem is to realize that φ0(x) = −δ−1 − δ1 + U−1 − U1. So F φ0(x)

= =

But

F − δ−1 − δ1 + U−1 − U1 1 eiw e−iw iw −iw √ −i +i . −e − e w w 2π F φ0(x) = i w F φ(x) .

So F φ(x)

= =

=

i eiw e−iw F eiw + e−iw + i −i w w w i i √ 2 cos w + (2i sin w) w 2π w r 2 w cos w − sin w i . π w2

Section 7.8

Generalized Functions

319

35. Using linearity and (15), we find 3 3 X X i j (Uj − Uj+1) = − √ j e−ijw − e−i(j+1)w F 2π w j=−3 j=−3 =

−√

3 X i je−ijw 1 − e−iw 2π w j=−3

=

−√

3 X i je−ijw 1 − e−iw 2π w j=−3

=

−√

3 X i −iw j e−ijw − eijw 1−e 2π w j=1

=

−√

X i j(−2i) sin(jw) 1 − e−iw 2π w j=1

3

=

r

3 2 e−iw − 1 X j sin(jw). π w j=1

37. Write τ 2 (x) = x2U0 (x), then use the operational properties F τ 2 (x) = F x2U0 (x) = =

=

39. We have f 0 (x) =

d2 F (U0 (x)) dw2 −i d2 1 −√ 2π dw2 w r 2i 1 2 1 √ =i . 3 π w3 2π w

−

1 (δ−2 (x) + δ−1 (x) − δ1 (x) − δ2 (x)) . 2

So 0

F (f (x)) = = =

F

1 (δ−2 (x) + δ−1 (x) − δ1 (x) − δ2 (x)) 2

1 2iw √ + eiw − e−iw − e−2iw e 2 2π i √ [sin(2w) + sin w] . 2π

Hence F (f(x)) =

−i −i i 1 sin(2w) + sin w √ [sin(2w) + sin w] = √ F (f 0 (x)) = . w w 2π w 2π

41. We have f 0 (x) = −U−2(x) + 2U−1(x) − U1 (x) + δ−1 (x) − 2 δ2 (x).

320

Chapter 7

The Fourier Transform and its Applications

So F (f 0 (x))

= F (−U−2 (x) + 2U−1(x) − U1 (x) + δ−1 (x) − 2 δ2 (x)) −i eiw −i e2iw 2 −i e−iw 1 = −√ + 2√ −√ + √ eiw − √ e−2iw . 2π w 2π w 2π w 2π 2π

Hence F (f(x))

−i F (f 0 (x)) w 2iw 1 eiw e−iw e √ −2 + − ieiw + 2ie−2iw . w w 2π w w

= =

43. Given f(x) = e−x if x > 0 and 0 otherwise, we see that f 0 (x) = δ0 (x) − e−x if x ≥ 0 and 0 otherwise. Hence f 0 (x) = δ0 (x) − f(x) and so F (f 0 (x))

=

F (δ0 (x) − f(x)) ;

iwF (f(x))

=

1 √ − F (f(x)) ; 2π

1 1 ⇒ F (f(x)) = √ 1 + iw 2π 1 1 − iw ⇒ F (f(x)) = √ . 2π 1 + w2

45. You may want to draw a graph to help you visualize the derivatives. For f(x) = sin x if |x| < π and 0 otherwise, we have f”(x) = cos x if |x| < π and 0 otherwise. Note that since f is continuous, we do not add delta functions at the endpoints x = ±π when computing f 0 . For f 00 , the graph is discontinuous at x = ±π and we have f 00 (x) = −δ−π + δπ − sin x if |x| ≤ π and 0 otherwise. Thus f 00 (x) = −δ−π + δπ − f(x)

for all x.

Taking the Fourier transform, we obtain F (f 00 (x))

=

F (−δ−π + δπ − f(x)) ;

−w2 F (f(x))

=

1 1 − √ eiπw + √ e−iπw − F (f(x)) ; 2π 2π =2 cos(πw)

z }| { 1 ⇒ (1 + w )F (f(x)) = √ eiπw + e−iπw 2π 2

1 cos(πw) ⇒ F (f(x)) = √ . 2π 1 + w2

47. In computing the Fourier transform F (f ∗ f), it is definitely q better to use the a formula F (f ∗ f) = F (f) · F (f), since we already have F (f) = π2 a2 +w 2 . So F (f ∗ f) =

2 a2 . 2 π (a + w2 )2

Section 7.8

Generalized Functions

321

The second method is used here only to practice this method, which is a lot more useful in other situations. From Example 8, we have f 00 = a2 f − 2aδ0 . So from

d2f d2 f ∗f = ∗f 2 dx dx2

we have F

=

F

−w2 F (f ∗ f)

=

F

d2 f ∗f 2 dx

d2 f ∗f dx2

a2 f − 2aδ0 ∗ f ;

=F

a2f ∗ f − 2aδ0 ∗ f

√ f / 2π

= =

z }| { a2 F (f ∗ f) − 2aF (δ0 ∗ f) 2a a2 F (f ∗ f) − √ F (f) . 2π

Thus

r

2 a F (f) . π a2 + w2 Writing F (f ∗ f) = F (f) · F (f), we conclude that r 2 a F (f) = π a2 + w2 F (f ∗ f) =

and that F (f ∗ f) =

2 a2 . π (a2 + w2 )2

49. From Example 9, we have f 0 = δ−1 − δ1 . So from

d df f ∗f = ∗f dx dx

we have d f ∗f dx

=

df ∗ f = δ−1 − δ1 ∗ f; dx

=

1 δ−1 ∗ f − δ1 ∗ f; = √ f(x + 1) − f(x − 1) . 2π

Using the explicit formula for f, we find 1 √2π d − √12π f ∗f = dx 0

if − 2 < x < 0, if 0 < x < 2, otherwise,

as can be verified directly from the graph of f ∗ f in Figure 18. 51. Using (20), we have φ∗ψ

= (3δ−1) ∗ (δ2 − δ1 ) = 3δ−1 ∗ δ2 − 3δ−1 ∗ δ1 =

1 3 √ [3δ−1+2 − 3δ−1+1 ] = √ [δ1 − δ0 ] 2π 2π

322

Chapter 7

The Fourier Transform and its Applications

53. Using (20), we have φ∗ψ

=

(δ−1 + 2δ2 ) ∗ (δ−1 + 2δ2 )

=

δ−1 ∗ δ−1 + 2δ−1 ∗ δ2 + 2δ2 ∗ δ−1 + 4δ2 ∗ δ2

=

1 √ [δ−2 + 4δ1 + 4δ4 ] . 2π

55. Following the method of Example 9, we have d (φ ∗ ψ) dx

=

dφ ∗ψ dx

=

(δ−1 − δ1 ) ∗ ((U−1 − U1)x) .

Using (19) and the explicit formula for ψ, it follows that d (φ ∗ ψ) dx

1 √ (left translate by 1 unit of ψ) − (right translate by 1 unit of ψ) 2π 1 √ (x + 1) if − 2 < x < 0 2π √1 (−x + 1) if 0 < x < 2 2π 0 otherwise

=

=

Integrating

d dx

(φ ∗ ψ) and using the fact that φ ∗ ψ equal 0 for large |x|, we find

φ ∗ ψ(x)

=

=

2 √1 ( x 2π 2

+ x) 2

√1 (− x 2 2π

0

+ x)

if − 2 < x < 0 if 0 < x < 2 otherwise

1 x2 1 x2 √ ( + x) U−2 − U0 + √ (− + x) U0 − U2 . 2 2π 2 2π

57. Following the method of Example 9, we have d (φ ∗ ψ) dx

=

dψ ∗φ dx

=

(δ−1 − δ1 ) ∗ (U−1 − U1 + U2 − U3 )

=

δ−1 ∗ U−1 − δ−1 ∗ U1 + δ−1 ∗ U2 − δ−1 ∗ U3 − δ1 ∗ U−1 +δ1 ∗ U1 − δ1 ∗ U2 + δ1 ∗ U3

Integrating

d dx

=

1 √ (U−2 − U0 + U1 − U2 − U0 + U2 − U3 + U4) 2π

=

1 √ ((U−2 − U0) − (U0 − U1) − (U3 − U4)) . 2π

(φ ∗ ψ) and using the fact that φ ∗ ψ equal 0 for large |x| and that

Section 7.8

Generalized Functions

323

there are no discontinuities on the graph, we find

φ ∗ ψ(x)

=

=

√1 (x + 2) 2π √1 (−x + 2) 2π √1 2π √1 (−x + 4) 2π

0

if − 2 < x < 0 if 0 < x < 1 if 1 < x < 3 if 3 < x < 4 otherwise

1 √ (x + 2) U−2 − U0 + (−x + 2) U0 − U1 2π + U1 − U3 + (−x + 4) U3 − U4 .

59. We have φ=

1 1 (U−2 − U−1) + (U−1 − U1) + (U1 − U2 ) 2 2

and ψ = U−1 − U1. So d (φ ∗ ψ) dx

=

dψ ∗φ dx

=

(δ−1 − δ1 ) ∗

=

=

= =

=

Integrating

d dx

1 1 (U−2 − U−1 ) + (U−1 − U1 ) + (U1 − U2) 2 2

1 1 1 1 δ−1 ∗ U−2 − δ−1 ∗ U−1 + δ−1 ∗ U−1 − δ−1 ∗ U1 + δ−1 ∗ U1 − δ−1 ∗ U2 2 2 2 2 1 1 1 1 − δ1 ∗ U−2 + δ1 ∗ U−1 − δ1 ∗ U−1 + δ1 ∗ U1 ) − δ1 ∗ U1 + δ1 ∗ U2 2 2 2 2 1 1 1 1 1 √ U−3 − U−2 + U−2 − U0 ) + U0 − U1 2 2 2 2 2π 1 1 1 1 − U−1 + U0 − U0 + U2 ) − U2 + U3 2 2 2 2 1 1 1 1 1 1 1 √ U−3 + U−2 − U−1 − U0 − U1 + U2 + U3 2 2 2 2 2 2π 2 1 1 √ U−3 + U−2 − U−1 − 2U0 − U1 + U2 + U3 2 2π 1 if − 3 < x < −2 2 if − 2 < x < −1 1 if − 1 < x < 0 1 1 √ × −1 if 0 < x < 1 2 2π −2 if 1 < x < 2 −1 if 2 < x < 3 0 otherwise

(φ ∗ ψ) and using the fact that φ ∗ ψ equal 0 for large |x| and that

324

Chapter 7

The Fourier Transform and its Applications

there are no discontinuities on the graph, we find (x + 3) if − 3 < x < −2 (2x + 5) if − 2 < x < −1 (x + 4) if − 1 < x < 0 1 1 √ × φ ∗ ψ(x) = (−x + 4) if 0 < x < 1 2 2π (−2x + 5) if 1 < x < 2 (−x + 3) if 2 < x < 3 0 otherwise =

1 1 √ (x + 3) U−3 − U−2 + (2x + 5) U−2 − U−1 + (x + 4) U−1 − U0 2 2π +(−x + 4) U0 − U1 + (−2x + 5) U1 − U2 + (−x + 3) U2 − U3 .

Section 7.9

The Nonhomogeneous Heat Equation

325

Solutions to Exercises 7.9 1. Proceed as in Example 1 with c = 1/2. Equation (3) becomes in this case u(x, t)

=

2 2 √ e−x /t ∗ δ1 (x) 2t

=

2 1 √ e−(x−1) /t , πt

since the effect of convolution by δ1 is to shift the function by 1 unit to the right and multiply by √12π . 3. We use the superposition principle (see the discussion preceeding Example 4). If φ is the solution of ut = 14 uxx + δ0 , u(x, 0) = 0 and ψ is the solution of ut = 1 1 4 uxx +δ1 , u(x, 0) = 0, then φ+ψ is the solution of ut = 4 uxx +δ0 +δ1 , u(x, 0) = 0. Applying Examples 2 and 4, we find √ 2 2 t 2|x| 1 x2 φ(x, t) = √ e−x /t − √ Γ , π π 2 t and

√ 2 2 t 2|x − 1| 1 (x − 1)2 ψ = √ e−(x−1) /t − √ Γ , . π π 2 t

5. We use the superposition principle (see the discussion preceeding Example 4). If φ is the solution of ut = 14 uxx + δ0 , u(x, 0) = 0 and ψ is the solution of ut = 14 uxx , u(x, 0) = U0 (x), then you can check that φ + ψ is the solution of ut = 14 uxx + δ0 , u(x, 0) = U0 (x). By Examples 1, √ 2 2 t 2|x| 1 x2 φ(x, t) = √ e−x /t − √ Γ , π π 2 t and by Exercise 20, Section 7.4, 1 ψ(x, t) = erf 2

x √ t

.

7. See the end of Section 7.4 for related topics. 9. Apply Theorem 2 with c = 1 and f(x, t) = cos ax; then Z tZ ∞ 2 1 p u(x, t) = e−(x−y) /(4(t−s)) cos(ay) dyds 0 −∞ 2 π(t − s) Z t Z ∞ 2 1 1 √ p = e−y /(4(t−s)) cos(a(x − y)) dyds 2π −∞ 2(t − s) 0 (Change variables x − y ↔ y) Z t Z ∞ 2 1 1 √ p = cos(ax) e−y /(4(t−s)) cos(ay) dyds 2π 2(t − s) 0 −∞ (Integral of odd function is 0. Z t 2 = cos(ax) e−a (t−s)ds (Fourier transform of a Guaussian.) 0

=

2 1 cos(ax) 1 − e−a t . 2 a

10. Very similar to Exercise 9. The answer is u(x, t) =

2 1 sin(ax) 1 − e−a t , 2 a

326

Chapter 7

The Fourier Transform and its Applications

as you can check by using the equations. 11. Write the Fourier series of f(x) in the form f(x) = a0 +

∞ X

an cos nx + bn sin nx .

n=1

It is not hard to derive the solution of ut = uxx + a0 ,

u(x, 0) = 0,

as u0 (x, t) = a0 t. Now use superposition and the results of Exercises 9 and 10 to conclude that the solution of ut = uxx + a0 +

∞ X

an cos nx + bn sin nx

n=1

is

2 ∞ X 1 − e−n t u(x, t) = a0t + (an cos nx + bn sin nx) . n2 n=1

Section 7.10

Duhamel’s Principle

327

Solutions to Exercises 7.10 2

1. Apply Proposition 1 with f(x, s) = e−(x+s) , then dU dx

=

f(x, x) +

=

Z

Z

x 0

2

e−4x −

∂ f(x, s) ds ∂x

x

2

2(x + s)e−(x+s) ds 0

=

2

e−4x −

Z

2x

2

2ve−v dv

((v = x + s)

x

=

2 2 2x 2 2 e−4x + e−v = 2e−4x − e−x . x

3. Use the product rule for differentiation and Proposition 1 and get Z x Z x Z x d 2 2 d f(x, s) ds = 2x f(x, s) ds + x f(x, s) ds x dx dx 0 0 0 Z x Z x ∂ = 2x f(x, s) ds + x2 f(x, x) + f(x, s) ds . 0 0 ∂x

5. Following Theorem 1, we first solve φt = φxx, φ(x, 0, s) = e−s x2 , where s > 0 is fixed. The solution is φ(x, t, s) = e−s (2t + x2 ) (see the solution of Exercise 5, Section 7.4). The the desired solution is given by u(x, t)

Z

=

t

φ(x, t − s, s)ds 0

Z

=

t

e−s (2(t − s) + x2)ds 0

t −2te−s + 2se−s + 2e−s − x2e−s

=

0

−2 + 2t + x2 + e−t (2 − x2).

=

7. Following Theorem 2, we first solve φtt = φxx, φ(x, 0, s) = 0, φt(x, 0, s) = cos x where s > 0 is fixed. By d’Alembert’s method, the solution is φ(x, t, s) =

1 2

Z

x+t

cos y dy = x−t

1 sin(x + t) − sin(x − t) . 2

The the desired solution is given by u(x, t) =

Z

t

φ(x, t − s, s)ds 0

=

1 2

Z

t

sin(x + t − s) − sin(x − t + s) ds

0

=

t 1 cos(x + t − s) + cos(x − t + s) 0 2

=

1 2 cos x − cos(x + t) − cos(x − t) . 2

328

Chapter 7

The Fourier Transform and its Applications

9. Following Theorem 2, we first solve φtt = φxx , φ(x, 0, s) = 0, φt (x, 0, s) = cos(s + x) where s > 0 is fixed. By d’Alembert’s method, the solution is 1 φ(x, t, s) = 2

Z

x+t

cos(s + y) dy = x−t

1 sin(s + x + t) − sin(s + x − t) . 2

The the desired solution is given by u(x, t)

Z

=

t

φ(x, t − s, s)ds 0

1 2

=

Z

t

sin(x + t) − sin(x − t + 2s) ds

0

=

t 1 1 s sin(x + t) + cos(x − t + 2s) 2 2 0

=

1 1 1 t sin(x + t) + cos(x + t) − cos(x − t) . 2 2 2

11. We use superposition and start by solving φtt = φxx, φ(x, 0, s) = 0, φt (x, 0, s) = x cos s where s > 0 is fixed. By d’Alembert’s method, the solution is 1 φ(x, t, s) = 2

Z

x+t

y cos s dy = x−t

cos s (x + t)2 − (x − t)2 = xt cos s. 4

By Theorem 2, the solution of φtt = φxx + x cos t, φ(x, 0) = 0, φt(x, 0) = 0 is given by φ(x, t)

Z

=

t

φ(x, t − s, s)ds 0

Z

=

t

u

dv

z }| { z }| { x (t − s) cos s ds (Integrate by parts.)

0

t Z t x (t − s) sin s + sin s ds

=

0

=

0

x(1 − cos t).

Next, we solve ψtt = ψxx , ψ(x, 0) = 0, ψt (x, 0) = x. D’Alembert’s method implies that Z 1 x+t 1 x+t ψ(x, t) = y dy = y2 = xt. 2 x−t 4 x−t Thus the solution of the desired problem is u(x, t) = φ(x, t) + ψ(x, t) = x(1 − cos t) + xt = x(1 + t − cos t). The validity of this solution can be checked by plugging it back into the equation and the initial conditions. 13. start by solving φtt = φxx, φ(x, 0, s) = 0, φt(x, 0, s) = δ0 (x) where s > 0 is fixed. By d’Alembert’s method, the solution is φ(x, t, s) =

1 2

Z

x+t

δ0 (y) dy = x−t

1 U0(x + t) − U0(x − t) . 2

By Theorem 2, the solution of utt = uxx + δ0 (x), u(x, 0) = 0, φt(x, 0) = 0 is given

Section 7.10

Duhamel’s Principle

by φ(x, t)

=

Z

t

φ(x, t − s, s)ds 0

= =

1 2

Z

t

U0 (x + t − s) − U0(x − t + s) ds

0

t 1 − τ (x + t − s) − τ (x − t + s) 0 2

= −τ (x) +

1 τ (x + t) + τ (x − t) , 2

where τ = τ0 is the antiderivative of U0 described in Example 2, Section 7.8,

329

A372

Appendix A

Ordinary Differential Equations: Review of Concepts and Methods

Solutions to Exercises A.1 1. We solve the equation y0 +y = 1 in two different ways. The first method basically rederives formula (2) instead of just appealing to it. Using an integrating factor. In the notation R of Theorem 1, we have p(x) = 1 and q(x) = 1. An antiderivative of p(x) is thus 1 · dx = x. The integrating factor is R µ(x) = e p(x) dx = ex . Multiplying both sides of the equation by the integrating factor, we obtain the equivalent equation ex [y0 + y] = ex ; d x [e y] = ex , dx d where we have used the product rule for differentiation to set dx [exy] = ex [y0 + y]. Integrating both sides of theRequation gets rid of the derivative on the left side, and on the right side we obtain ex dx = ex + C. Thus,

ex y = ex + C

⇒

y = 1 + Ce−x,

where the last equality follows by multiplying by e−x the previous equality. This gives the solution y = 1 + Ce−x up to one arbitrary constant, as expected from the solution of a first order differential equation. R Using formula (2). We have, with p(x) = 1, p(x) dx = x (note how we took the constant of integration equal 0): Z y = e−x C + 1 · ex dx = e−x [C + ex ] = 1 + Ce−x .

2. We apply (2) directly. We have p(x) = 2x, an integrating factor is R

µ(x) = e

2x dx

2

= ex ,

and the solution becomes y

=

=

2

e−x

−x2

e

Z 2 C + xex 2 1 x2 1 C+ e = Ce−x + . 2 2

3. This is immediate: y0 = −.5 y

⇒

y = Ce−.5x .

4. Put the equation in the form y0 − 2y = x. Then from (2) Z R R − −2dx −2 dx y = e C + xe dx

=

=

Z e2x C + xe−2x dx 2x

e

1 −2x 1 −2x 1 1 C − xe = Ce2x − x − , − e 2 4 2 4

where we used integration by parts to evaluate the last integral: u = x, dv = e−2x dx, du = dx, v = − 12 e−2x.

Section A.1

5. According to (2),

Linear Ordinary Differential Equations

A373

Z −x y = e C + sin xe dx . x

To evaluate the integral, use integration by parts twice Z Z −x −x sin xe dx = − sin xe + e−x cos x dx −x

= 2

So

− sin xe

−x

+ cos x(−e

Z

sin xe−x dx =

−e−x sin x + cos x

Z

sin xe−x dx =

1 − e−x sin x + cos x . 2

)−

Z

e−x sin x dx;

1 −x 1 y=e C− e sin x + cos x = Cex − sin x + cos x . 2 2 x

6. We have

R

p(x) dx = −x2 , so according to (2), y

x2

= e

Z 3 −x2 C+ x e dx

Z 1 2 −x2 1 −x2 = e C− x e + 2x dx e 2 2 2 2 2 1 1 = ex C − x2e−x − e−x 2 2 x2

2 1 1 = Cex − x2 − . 2 2

7. Put the equation in standard form: y0 +

1 cos x y= . x x

An integrating factor is R

e

p(x) dx

R

=e

1 x

dx ln x

e

= x.

Multiply by the integrating factor (you will see that we just took an unnecessary detour): xy0 + y

=

cos x;

d xy dx

=

cos x;

xy

=

Z

y

=

cos x dx = sin x + C;

sin x C + . x x

8. We use an integrating factor R

e

p(x) dx

R

=e

−2/x dx

= e−2 ln x =

1 . x2

A374

Appendix A

Ordinary Differential Equations: Review of Concepts and Methods

Then 1 0 2 y − 3y 2 x x d 1 y dx x2

=

1;

=

1

1 y x2

=

x+C

y

=

x3 + Cx2.

9. We use an integrating factor R

e

p(x) dx

R

=e

tan x dx

= e− ln(cos x) =

1 = sec x. cos x

Then sec xy0 − sec x tan x y

=

sec x cos x;

d [y sec x] = dx

1

y sec x = y

x+C =

=

x cos x + C cos x.

10. Same integrating factor as in Exercise 9. The equation becomes sec xy0 − sec x tan x y

=

sec3 x;

d [y sec x] = dx

sec3 x Z

y sec x =

sec3 x dx.

To evaluate the last integral, you can use integration by parts: Z

2

Z

sec3 x dx =

Z

du

v z }| { z 2}| { sec x sec x dx

=

(u = tan x, dv = tan x sec x) Z tan x sec x − tan2 x sec x dx

=

(tan2 = sec2 −1) Z Z tan x sec x − sec 3 x dx + sec x dx;

sec3 x dx =

tan x sec x +

Z

sec x dx.

Now we use a known trick to evaluate the last integral: Z Z sec x(tan x + sec x) sec x dx = dx tan x + sec x = ln | tan x + sec x| + C,

Section A.1

Linear Ordinary Differential Equations

A375

where the last integral follows by setting u = tan x+sec x, du = sec x(tan x+sec x). Hence 1 1 tan x sec x + ln | tan x + sec x| + C + C; y sec x = 2 2 y

=

1 cos x tan x + ln | tan x + sec x| + C cos x. 2 2

11. y0 = y

⇒ y = Cex ;

y(0) = 1

⇒ 1=C ⇒ y = ex .

12. An integrating factor is e2x, so y0 + 2 y = 1

⇒ e2x [y0 + 2 y] = e2x ; d 2x e y = e2x dx Z 1 ⇒ ye2x = e2x dx = e2x + C 2 ⇒

⇒ y=

1 + Ce−2x . 2

We now use the initial condition: y(0) = 2 ⇒ 2 =

13. An integrating factor is e e

x2 2

y0 + xe

x2 2

x2 2

y = xe

1 +C 2

⇒ C=

3 2

⇒ y=

1 3 −2x . + e 2 2

, so x2 2

x2 d h x2 i e2 y =e2 x dx Z x2 x2 x2 ⇒ ye 2 = xe 2 dx = e 2 + C

⇒

⇒ y = 1 + Ce−

x2 2

.

We now use the initial condition: y(0) = 0 ⇒ 0 = 1 + C ⇒ C = −1 ⇒ y = 1 − e−

x2 2

.

14. An integrating factor is x, so d [xy] = sin x dx Z ⇒ xy = sin x dx = − cos x + C

xy0 + y = sin x ⇒

⇒ y=−

cos x C + . x x

A376

Appendix A

Ordinary Differential Equations: Review of Concepts and Methods

We now use the initial condition: y(π) = 1 ⇒ 1 =

1 C + π π

⇒ C =π−1 ⇒ y=−

cos x π − 1 π − 1 − cos x + = . x x x

15. An integrating factor is x, so d 2 x y =x dx Z 1 2 ⇒ x y = x dx = x2 + C 2

x2y0 + 2xy = x

⇒

⇒ y=

1 C + . 2 x2

We now use the initial condition: y(−1) = −2 ⇒ −2 =

1 +C 2

⇒ C=− ⇒ y=

5 2

1 5 . − 2 2 x2

16. An integrating factor is 1/x3, so y0 2 1 − y= 4 x2 x3 x

⇒ ⇒

d 1 1 y = 4 dx x2 x Z 1 1 1 y = dx = − x−3 + C x2 x4 3

1 ⇒ y = − x−1 + Cx2. 3 We now use the initial condition: 1 y(1) = 0 ⇒ 0 = − + C 3 ⇒ C=

1 3

⇒ y=−

1 x2 + . 3x 3

17. An integrating factor is sec x (see Exercise 9), so sec xy0 + y tan x sec x = tan x sec x

d [y sec x] = sec x tan x dx Z ⇒ y sec x = tan x sec x dx = sec x + C

⇒

⇒ y = 1 + C cos x. We now use the initial condition: y(0) = 1 ⇒ 1 = 1 + C ⇒ C=0 ⇒ y = 1.

Section A.1

Linear Ordinary Differential Equations

A377

18. Use the solution of the previous exercise, y = 1 + C cos x. The initial condition implies: y(0) = 2

⇒ 2= 1+C ⇒ C=1 ⇒ y = 1 + cos x.

19. An integrating factor is

R

e

ex dx

x

= ee .

The equation becomes x

x

x

ee y0 + ee ex y = ex ee

x d h ex i e y = ex ee dx Z x x x ⇒ ee y = ee ex dx = ee + C

⇒

x

⇒ y = 1 + Ce−e . We now use the initial condition: y(0) = 2

⇒ 2 = 1 + Ce−1

1 = Ce−1

⇒

⇒ C=e x

x

⇒ y = 1 + ee−e = 1 + e1−e . 20. An integrating factor is ex . The equation becomes ex y0 + ex y = ex ex

d x [e y] = e2x dx Z 1 ⇒ ex y = e2x dx = e2x + C 2 ⇒

⇒ y=

1 x e + Ce−x . 2

We now use the initial condition: y(3) = 0

⇒ 0=

1 3 e + Ce−3 2

⇒

1 − e3 = Ce−3 2

1 ⇒ C = − e6 2 ⇒ y=

1 x 1 6−x 1 x = e − e e − e6−x . 2 2 2

21. (a) Clear. (b) ex as a linear combination of the functions cosh x, sinh x: ex = cosh x+sinh x. (c) Let a, b, c d be any real numbers such that ad−bc 6= 0. Let y1 = aex +be−x and y2 = cex + de−x . Then y1 and y2 are solutions, since they are linear combinations of two solutions. We now check that y1 and y2 are linearly independent: y1 y2 W (y1 , y2 ) = y10 y20 =

x ae + be−x x ae − be−x

=

−ad + bc − (ad − bc) = −2(ad − bc) 6= 0.

cex + de−x cex − de−x

A378

Appendix A

Ordinary Differential Equations: Review of Concepts and Methods

Hence y1 and y2 are linearly independent by Theorem 7. 22. (a) Clear. (b) In computing the determinant, we use an expansion along the first row: x e e−x cosh x −e−x sinh x W (ex , e−x , cosh x) = ex x e e−x cosh x −e−x −ex sinh x sinh x = ex − e−x x e−x e cosh x cosh x x e −e−x + cosh x ex e−x = ex −e−x cosh x − e−x sinh x − e−x (ex cosh x − ex sinh x) cosh x ex e−x + ex e−x ≡ 0. (c) Clearly the function 1 is a solution of y000 − y0 = 0. The set of solutions {1, ex , e−x } is a fundamental set if and only if it is linearly independent (Theorem 7). Let us compute the Wronskian. Expand along the 3rd column:

W (ex , e−x , 1) =

=

=

x e e−x 1 x e −e−x 0 x e e−x 0 x x e e −e−x 1· −0· −x ex ex e x e e−x +0 · ex −e−x

e−x e−x

2.

23. (a) Clear (b) 2

W (x, x )

=

x 1

=

x2

x2 2x

(c) Note that the Wronskian vanishes at x = 0. This does not contradict Theorem 2, because if you put the equation in standard form it becomes y00 −

2 0 2 y + 2 y = 0. x x

The coefficients functions are not continuous at x = 0. So we cannot apply Theorem 2 on any interval that that contains 0. Note that W (x, x2) 6= 0 if x 6= 0. 24. (a) and first part of (b) are clear. In standard form, the equation becomes y00 −

1+x 0 1 y + y = 0. x x

Section A.1

Linear Ordinary Differential Equations

A379

Since the coefficient functions are not continuous at 0, Theorem 6 will not apply if we use an initial condition at x = 0. So there is no contradiction with Theorem 6. (c) x e 1 + x W (ex , 1 + x) = ex 1 = ex − ex − xex = −xex . The Wronskian is clearly nonzero on (0, ∞) and hence the solutions are linearly independent on (0, ∞). 25. The general solution is y = c1 ex + c2e2x + 2x + 3. Let’s use the initial conditions: y(0) = 0

⇒ c1 + c2 + 3 = 0

(∗)

y0 (0) = 0

⇒ c1 + 2c2 + 2 = 0

(∗∗)

Subtract (*) from (**) ⇒ c2 − 1 = 0; c2 = 1 Substitute into (*)

⇒ c1 + 4 = 0; c1 = −4.

Thus, y = −4ex + e2x + 2x + 3. 26. The general solution is y = c1 ex + c2e2x + 2x + 3. From the initial conditions, y(0) = 1

⇒ c1 + c2 + 3 = 1

0

y (0) = −1

⇒ c1 + 2c2 + 2 = −1

Subtract (*) from (**)

⇒ c2 − 1 = −2; c2 = −1

Substitute into (*)

⇒ c1 + 2 = 1; c1 = −1.

(∗) (∗∗)

Thus, y = −ex − e2x + 2x + 3. 27. The function ex−1 (= e−1 ex ) is a constant multiple of the ex . Hence it is itself a solution of the homogeneous equation. Similarly, e2(x−1) is a solution of the homogeneous equation. Using the same particular solution as in Example 3, we obtain the general solution y = c1ex−1 + c2e2(x−1) + 2x + 3. From the initial conditions, y(1) = 0

⇒ c1 + c2 + 5 = 0

(∗)

y0 (1) = 2

⇒ c1 + 2c2 + 2 = 2

(∗∗)

Subtract (*) from (**) ⇒ c2 − 3 = 2; c2 = 5 Substitute into (*)

⇒ c1 + 10 = 0; c1 = −10.

Thus, y = −10ex−1 + 5e2(x−1) + 2x + 3. 28. As in the previous exercise, the general solution y = c1ex−1 + c2e2(x−1) + 2x + 3.

A380

Appendix A

Ordinary Differential Equations: Review of Concepts and Methods

From the initial conditions, y(1) = 1

⇒ c1 + c2 + 5 = 1

y0 (1) = −1

⇒ c1 + 2c2 + 2 = −1

(∗) (∗∗)

Subtract (*) from (**) ⇒ c2 − 3 = −2; c2 = 1 Substitute into (*)

⇒ c1 + 6 = 1; c1 = −5.

Thus, y = −5ex−1 + e2(x−1) + 2x + 3. 29. As in the previous exercise, here it is easier to start with the general solution y = c1ex−2 + c2e2(x−2) + 2x + 3. From the initial conditions, y(2) = 0 0

y (1) = 1

⇒ c1 + c2 + 7 = 0

(∗)

⇒ c1 + 2c2 + 2 = 1

(∗∗)

Subtract (*) from (**) ⇒ c2 − 5 = 1; c2 = 6 Substitute into (*)

⇒ c1 + 13 = 0; c1 = −13.

Thus, y = −13ex−2 + 6e2(x−2) + 2x + 3. 30. Start with the general solution y = c1ex−3 + c2e2(x−3) + 2x + 3. From the initial conditions, y(3) = 3

⇒ c1 + c2 + 9 = 3

(∗)

y0 (1) = 1

⇒ c1 + 2c2 + 2 = 3

(∗∗)

Subtract (*) from (**) ⇒ c2 − 7 = 0; c2 = 7 Substitute into (*)

⇒ c1 + 16 = 3; c1 = −13.

Thus, y = −13ex−3 + 7e2(x−3) + 2x + 3. 31. Project Problem: Abel’s formula for n = 3. (a) Let y1 , y2 , y3 be any three solutions of the third order equation y000 + p2(x)y00 + p1 (x)y0 + p0(x)y = 0. Then the Wronskian is (expand along the 3rd row): y1 y2 y3 W (y1 , y2 , y3 ) = y10 y20 y30 00 y y100 y300 1 y2 y1 y 3 = y100 − y200 0 y20 y10 y3 y1 y 2 +y300 0 y10 y2

y3 y30

= (y2 y30 − y20 y3 )y100 − (y1 y30 − y10 y3 )y200 +(y1 y20 − y10 y2 )y300 .

Section A.1

Linear Ordinary Differential Equations

A381

Using the product rule to differentiate W , we find W 0 (y1 , y2 , y3 )

=

(y2 y30 − y20 y3 )y1000 + (y20 y30 + y2 y300 − y200 y3 − y20 y30 )y100 −(y1 y30 − y10 y3 )y2000 − (y10 y30 + y1 y300 − y100 y3 − y10 y30 )y200 +(y1 y20 − y10 y2 )y3000 + (y10 y20 + y1 y200 − y100 y2 − y10 y20 )y300

=

(y2 y30 − y20 y3 )y1000 − (y1 y30 − y10 y3 )y2000 + (y1 y20 − y10 y2 )y3000 .

(b) For each j = 1, 2, 3, we have, from the differential equation, yj000 = −p2 (x)yj00 − p1 (x)yj0 − p0(x)yj . Substituting this into the formula for W 0 , we obtain W0

=

=

−(y2 y30 − y20 y3 ) p2 (x)y100 + p1(x)y10 + p0(x)y1

+(y1 y30 − y10 y3 ) p2 (x)y200 + p1(x)y20 + p0(x)y2

−(y1 y20 − y10 y2 ) p2 (x)y300 + p1(x)y30 + p0(x)y3

−p2 (x)(y2 y30 − y20 y3 )y100 + p2(x)(y1 y30 − y10 y3 )y200 −p2 (x)(y1 y20 − y10 y2 )y300

=

−p2 (x)W.

(c) From (b) W satisfies a the first order differential equation W 0 = −p2W , whose solution is R W = Ce− p2 (x) dx . 32. (a) Let f(x) = |x3|, 3 x 0 f(x) = −x3

if x > 0 if x = 0 if x < 0

Clearly f 0 (x) exists if x 6=. For x = 0, we have f 0 (0)

=

lim

f(x + h) − f(x) h

=

lim

|h|3 h2|h| = lim h→ h h

=

lim h|h| = 0.

h→

h→

h→

Hence the derivative exists for all x and is given by if x > 0 3x2 0 if x = 0 f 0 (x) = −3x2 if x < 0 Another way of writing the derivative is f 0 (x) = 3x2 sgn x, where sgn x = −1, 0, or 1, according as x < 0, x = 0, or x > 0. We have

W (x3 , |x3|) = =

3 x 3x2

|x3| 3x2 sgn x

3x5 sgn x − 3x2|x3|.

A382

Appendix A

Ordinary Differential Equations: Review of Concepts and Methods

You can easily verify that the last expression is identically 0 by considering the cases x > 0, x = 0, and x < 0 separately. (b) The functions x3 and |x3| are linearly independent on the real line because one is not a constant multiple of the other. (c) This does not contradict Theorem 7 because x3 and |x3| are not solution of a differential equation of the form (3).

Section A.1

Using

Linear Ordinary Differential Equations

A383

to solve ODE

Let us start with the simplest command that you can use to solve an ode. It is the DSolve command. We illustrate by examples the different applications of this command. The simplest case is to solve y' = y DSolve y ’ x y x

x

y x ,y x ,x

C 1

The answer is y= C ex as you expect. Note how Mathematica denoted the constant by C[1]. The enxt example is a 2nd order ode DSolve y ’’ x y x

x

y x ,y x ,x x

C 1

C 2

Here we need two arbitrary constants C[1] and C[2]. Let's do an intial value problem.

Solving an Initial Value Problem Here is how you would solve y'' = y, y(0)=0, y'(0)= 1 DSolve y x

y ’’ x 1 2

x

y x ,y 0 1

0, y ’ 0

1 , y x ,x

2x

As you see, the initial value problem has a unique solution (there are no arbitrary constants in the answer.

A384

Appendix A

Ordinary Differential Equations: Review of Concepts and Methods

Plotting the Solution First we need to learn to extract the solution from the output. Here is how it is done. First, solve the problem and call the output solution: DSolve

solution 1 2

y x

x

1

y ’’ x

y x ,y 0

0, y ’ 0

1 , y x ,x

2x

Extract the solution y(x) as follows: y x_ 1 2

y x x

1

. solution

1

2x

Now you can plot the solution: Plot y x , x, 0, 2 3.5 3 2.5 2 1.5 1 0.5 0.5

1

1.5

2

Note the intial conditions on the graph: y(0)=0 and y'(0)=1. To confirm that y'(0)=1 (the slope of the graph at x=0 is 1), plot the tangent line (line with slope 1) Plot

y x , x , x, 0, 2

3.5 3 2.5 2 1.5 1 0.5 0.5

1

1.5

2

Section A.1

Linear Ordinary Differential Equations

The Wronskian The Wronskian is a determinat, so we can compute it using the Det command. Here is an illustration. Clear y sol1 DSolve y ’’ x y x

C 1 Cos x

y x

0, y x , x

C 2 Sin x

Two solutions of the differential equation are obtained different values to the constants c1 and c2. For Clear y1, y2 y1 x_ c1 Cos x ; y2 x_ c2 Sin x ; Their Wronskian is w x_

Det

c1 c2 Cos x

y1 x , y2 x 2

c1 c2 Sin x

, y1 ’ x , y2 ’ x 2

Let's simplify using the trig identity cos^2 x + sin^2 x =1 Simplify w x c1 c2 The Wronskian is nonzero if c1\= 0 and c2 \= 0. Let us try a different problem with a nonhomogeneous Clear y sol2 DSolve y ’’ x y x

1

y x

C 1 Cos x

1, y x , x

C 2 Sin x

Two solutions of the differential equation are obtained different values to the constants c1 and c2. For Clear y1, y2 y1 x_ 1 Cos x ; y2 x_ 1 Sin x ; These solutions are clearly linearly independent (one is not a multiple of the other). Their Wronskian is Clear w w x_ Det Cos x

y1 x , y2 x

Cos x

2

Sin x

, y1 ’ x , y2 ’ x Sin x

2

Let's simplify using the trig identity cos^2 x + sin^2 x =1 Simplify w x 1

Cos x

Let's plot w(x):

Sin x

A385

A386

Appendix A

Ordinary Differential Equations: Review of Concepts and Methods

Plot w x , x, 0, 2 Pi

2 1.5 1 0.5 1

2

3

4

5

6

The Wronskian does vanish at some values of x without being identically 0. Does this contradict Theorem 7 of Appendix A.1?

Section A.2

Linear Ordinary Differential Equations with Constant Coefficients

Solutions to Exercises A.2 1. y00 − 4 y0 + 3 y

Equation: Characteristic equation:

=

0;

λ −4λ +3 =

0

(λ − 1)(λ − 3) =

0;

2

Characteristic roots:

λ1 = 1;

General solution:

y = c1ex + c2 e3x

λ2 = 3

2. y00 − y0 − 6 y

Equation:

= 0;

2

Characteristic equation:

λ − λ−6

= 0

(λ + 2)(λ − 3)

= 0;

Characteristic roots:

λ1 = −2;

General solution:

y = c1e−2x + c2e3x

λ2 = 3

3. Equation: Characteristic equation:

Characteristic roots:

y00 − 5 y0 + 6 y

= 0;

λ2 − 5 λ + 6

= 0

(λ − 2)(λ − 3)

= 0;

λ1 = 2;

λ2 = 3

2x

General solution:

y = c1e

+ c2e3x

4. 2 y00 − 3 y0 + y

Equation: Characteristic equation:

=

0;

2λ2 − 3 λ + 1 =

0

(λ − 1)(2λ − 1)

0;

1 2

Characteristic roots:

λ1 = 1;

General solution:

y = c1 ex + c2 ex/2

λ2 =

=

5 Equation: Characteristic equation:

Characteristic roots: General solution:

y00 + 2 y0 + y

=

0;

λ +2λ +1 =

0

(λ + 1)2

0;

2

=

λ1 = −1 (double root) y = c1 e−x + c2 x e−x

6. Equation: Characteristic equation:

4 y00 − 13 y0 + 9 y

=

0;

4λ2 − 13 λ + 9 =

0

(λ − 1)(4λ − 9) Characteristic roots: General solution:

λ1 = 1;

λ2 =

9 4

y = c1 ex + c2 e−9 x/4

=

0;

A387

A388

Appendix A

Ordinary Differential Equations: Review of Concepts and Methods

7. Equation: Characteristic equation:

4 y00 − 4 y0 + y

=

0;

4λ2 − 4 λ + 1

=

0

2

=

0;

=

0;

4λ2 − 12 λ + 9 =

0

(2λ − 3)2

0;

(2λ − 1) Characteristic roots:

λ1 = 1/2 (double root); y = c1 ex/2 + c2 x ex/2

General solution: 8.

4 y00 − 12 y0 + 9 y

Equation: Characteristic equation:

=

Characteristic roots:

λ1 = 3/2 (double root);

General solution:

y = c1 e3 x/2 + c2 x e3 x/2

9.

y00 + y

Equation: Characteristic equation: Characteristic roots:

λ +1 λ1 = i

Case III:

α = 0,

General solution:

= 0;

2

= 0

λ2 = −i; β = 1;

y = c1 cos x + c2 sin x

10.

9 y00 + 4 y

Equation: Characteristic equation:

=

0;

9λ2 + 4 =

0

λ

2

=

− 49

4 4 i λ2 = − i; 9 9 4 α = 0, β = ; 9 y = c1 cos 49 x + c2 sin 49 x

Characteristic roots:

λ1 =

Case III: General solution: 11.

y00 − 4 y

Equation: Characteristic equation:

Characteristic roots: General solution:

=

0;

λ −4 =

0

(λ − 2)(λ + 2) =

0

2

λ1 = 2 λ2 = −2; y = c1e2x + c2 e−2x.

12. y00 + 3 y0 + 3 y = 0. Equation: Characteristic equation: Characteristic roots:

Case III: General solution:

y00 + 3 y0 + 3 y

=

0;

λ2 + 3 λ + 3 = √ √ −3 ± 9 − 12 −3 ± i 3 λ= = ; 2 √ √2 −3 + i 3 −3 − i 3 λ1 = , λ2 = ; 2 2 √ 3 3 α=− , β = ; 2 √ √2 3 3 y = c1 e− 2 x cos 23 x + c2 e− 2 x sin 23 x

0;

Section A.2

A389

Linear Ordinary Differential Equations with Constant Coefficients

13. y00 + 4 y0 + 5 y

Equation: Characteristic equation: Characteristic roots:

=

0;

λ2 + 4 λ + 5 = √ −4 ± 16 − 20 λ= = −2 ± i; 2 λ1 = −2 + i, λ2 = −2 − i;

0;

Case III:

α = −2, −2x

General solution:

y = c1e

β = 1; −2x

cos x + c2 e

sin x

14. y00 − 2 y0 + 5 y

Equation:

=

0;

λ − 2λ+5 =

0;

2

Characteristic equation: Characteristic roots:

λ=

2±

√

2 λ1 = 1 + 2i,

Case III:

−16

= 1 ± 2i;

λ2 = 1 − 2i;

α = 1,

β = 2; y = c1ex cos 2x + c2 ex sin 2x .

General solution: 15.

y00 + 6 y0 + 13 y

Equation:

=

0;

λ + 6 λ + 13 = √ λ = −3 ± −4 = −3 ± 2i;

0;

2

Characteristic equation: Characteristic roots:

λ1 = −3 + 2i, Case III: General solution:

λ2 = −3 − 2i;

α = −3,

β = 2;

y = c1e−3x cos(2x) + c2 e−3x sin(2x)

16. 2 y00 − 6 y0 + 5 y

Equation: Characteristic equation: Characteristic roots:

Case III: General solution:

=

0;

2λ − 6 λ + 5 = √ 3 ± −1 3±i λ= = ; 2 2 3 3 i i λ1 = + , λ 2 = − ; 2 2 2 2 3 1 α= , β= ; 2 2 y = c1 e3x/2 cos(x/2) + c2 e3x/2 sin(x/2)

0;

2

17. Equation: Characteristic equation:

Characteristic roots: General solution:

λ1 = 0;

y000 − 2 y00 + y0

= 0;

λ3 − 2λ2 + λ

= 0

λ(λ − 1)2

= 0;

λ2 = 1 (double root) y = c1 + c2 ex + c3 x ex

A390

Appendix A

Ordinary Differential Equations: Review of Concepts and Methods

18. y000 − 3 y00 + 2 y

Equation: Characteristic equation:

=

0;

λ3 − 3λ2 + 2 =

0

2

(λ − 1)(λ − 2λ − 2) = √ λ1 = 1, λ = 1 ± −1

Characteristic roots:

0;

λ1 = 1, λ2 = 1 + i, λ3 = 1 − i; y = c1 ex + c2 ex cos x + c3 ex sin x

General solution: 19.

y(4) − 2 y00 + y

Equation: Characteristic equation:

Characteristic roots:

=

0;

λ − 2λ + 1 = 2 λ2 − 1 =

0

4

2

0;

two double roots, λ1 = 1, λ2 = −1; y = c1 ex + c2 e−x + c3 x ex + c4 x e−x

General solution: 20.

y(4) − y

Equation:

4

Characteristic equation:

Characteristic roots:

= 0;

λ −1

= 0;

(λ2 − 1)(λ2 + 1)

= 0

λ1 = 1, λ2 = −1, λ3 = i, λ4 = −i; y = c1 ex + c2 e−x + c3 cos x + c4 sin x

General solution: 21. Equation:

Characteristic equation:

Characteristic roots: General solution:

y000 − 3 y00 + 3y0 − y

= 0;

λ3 − 3λ2 + 3λ − 1

= 0

(λ − 1)3

= 0;

λ1 = 1 (multiplicity 3); y = c1 ex + c2 x ex + c3 x2ex

22. Equation:

y000 + y

=

0;

Characteristic equation:

λ3 + 1 =

0

(λ − 1)(λ − λ + 1) = √ 1 ± −3 λ1 = 1, λ = 2√ √ 1+i 3 1−i 3 λ1 = 1, λ2 = , λ3 = ; 2 2√ √ y = c1 ex + c2 ex/2 cos 23 x + c3 ex/2 sin 23 x

0;

2

Characteristic roots:

General solution:

Section A.2

Linear Ordinary Differential Equations with Constant Coefficients

A391

23. y(4) − 6 y00 + 8y0 − 3y

Equation: Characteristic equation:

=

0;

λ4 − 6λ2 + 8λ − 3 =

0

3

(λ − 1) (λ + 3) = Characteristic roots:

0;

λ1 = 1 (multiplicity 3), λ4 = −3; y = c1 ex + c2 x ex + c3 x2 ex + c4 e−3x.

General solution: 24.

y(4) + 4 y000 + 6 y00 + 4y0 + y

Equation: Characteristic equation:

=

0;

λ4 + 4λ2 + 6λ2 + 4λ + 1 =

0

(λ + 1) Characteristic roots: General solution:

4

=

0;

λ1 = −1 (multiplicity 4); −x

y = c1 e

−x

+ c3 x2 e−x + c4 x3 e−x .

+ c2 x e

25. Equation:

y00 − 4 y0 + 3 y

=

e2x ;

Homogeneous equation:

y00 − 4 y0 + 3 y

=

0;

λ2 − 4λ + 3 =

0

(λ − 1)(λ − 3) =

0;

Characteristic equation:

Characteristic roots:

λ1 = 1, λ2 = 3;

Solution of homogeneous equation: yh = c1 ex + c2 e3x. To find a particular solution, we apply the method of undetermined coefficients. Accordingly, we try yp

=

Ae2x ;

yp0

=

2Ae2x;

yp00

=

4Ae2x.

Plug into the equation y00 − 4 y0 + 3 y = e2x : 4Ae2x − 4(2Ae2x ) + 3Ae2x

2x

Hence yp = −e

= e2x

−Ae2x

= e2x ;

A

= −1.

and so the general solution yg = c1 ex + c2 e3x − e2x .

26. Equation:

y00 − y0 − 6 y

=

ex ;

Homogeneous equation:

y00 − y0 − 6 y

=

0;

λ2 − λ − 6 =

0

(λ + 2)(λ − 3) =

0;

Characteristic equation:

Characteristic roots:

λ1 = −2, λ2 = 3;

Solution of homogeneous equation: yh = c1 e−2x + c2 e3x .

A392

Appendix A

Ordinary Differential Equations: Review of Concepts and Methods

To find a particular solution, we apply the method of undetermined coefficients. Accordingly, we try yp

= Aex ;

yp0

= Aex ;

yp00

= Aex .

Plug into the equation y00 − y0 − 6 y = ex : Aex − Aex − 6Aex

=

ex

−6Aex

=

ex ;

A =

−1/6.

Hence yp = −ex /6 and so the general solution yg = c1 e−2x + c2 e3x −

ex . 6

27. Equation:

y00 − 5 y0 + 6 y

=

ex + x;

Homogeneous equation:

y00 − 5 y0 + 6 y

=

0;

λ2 − 5 λ + 6 =

0

Characteristic equation:

(λ − 2)(λ − 3) Characteristic roots:

=

0;

λ1 = 2, λ2 = 3;

Solution of homogeneous equation: yh = c1 e2x + c2 e3x . To find a particular solution, we apply the method of undetermined coefficients. Accordingly, we try yp

=

Aex + Bx + C;

yp0

=

Aex + B;

yp00

=

Aex .

Plug into the equation y00 − 5 y0 + 6 y = ex + x: Aex − 5 (Aex + B) + 6 (Aex + Bx + C) = 2Aex

=

ex ;

6B

=

1

6C − 5B

=

0.

Hence A = 1/2,

B = 1/6,

C = 5/36;

yp =

ex x 5 + + ; 2 6 36

and so the general solution yg = c1 e2x + c2 e3x +

ex + x

ex x 5 + + . 2 6 36

Section A.2

Linear Ordinary Differential Equations with Constant Coefficients

A393

28. Equation:

2 y00 − 3 y0 + y

=

e2x + sin x;

Homogeneous equation:

2 y00 − 3 y0 + y

=

0;

2λ − 3λ+ 1 =

0

2

Characteristic equation:

(2 λ − 1)(λ − 1) Characteristic roots:

=

0;

λ1 = 1/2, λ2 = 1;

Solution of homogeneous equation: yh = c1 ex/2 + c2 ex . To find a particular solution, we apply the method of undetermined coefficients. Accordingly, we try yp

=

A cos x + B sin x + Ce2x;

yp0

=

−A sin x + B cos x + 2Ce2x;

yp00

=

−A cos x − B sin x + 4Ce2x.

Plug into the equation 2 y00 − 3 y0 + y = e2x + sin x: −2A cos x − 2B sin x + 8Ce2x + 3A sin x − 3B cos x − 6Ce2x +A cos x + B sin x + Ce2x

= e2x + sin x

−A − 3B

= 0;

3A − B

= 1

3C

= 1.

Hence A = 3/10,

B = −1/10,

C = 1/3;

yp =

3 1 1 cos x − sin x + e2x ; 10 10 3

and so the general solution yg = c1 ex/2 + c2 ex +

3 1 1 cos x − sin x + e2x . 10 10 3

29. Equation:

y00 − 4 y0 + 3 y

=

x e−x ;

Homogeneous equation:

y00 − 4 y0 + 3 y

=

0;

λ − 4λ+3 =

0

(λ − 1)(λ − 3) =

0;

2

Characteristic equation:

Characteristic roots:

λ1 = 1, λ2 = 3;

Solution of homogeneous equation: yh = c1 ex + c2 e3x. To find a particular solution, we apply the method of undetermined coefficients. Accordingly, we try yp

= (Ax + B)e−x ;

yp0

= e−x (Ax − B + A);

yp00

= e−x (Ax − B − 2A).

A394

Appendix A

Ordinary Differential Equations: Review of Concepts and Methods

Plug into the equation y00 − 4 y0 + 3 y = e−x : e−x (Ax − B − 2A) − 4e−x (Ax − B + A) + 3(Ax + B)e−x

=

8A = −6A + 8B

Hence A = 1/8,

B = 3/32;

yp = (

=

e−x 1; 0.

3 x + )e−x ; 8 32

and so the general solution yg = c1 ex + c2 e3x + (

x 3 + )e−x . 8 32

30. Equation:

y00 − 4 y

=

cosh x;

Homogeneous equation:

y00 − 4 y

=

0;

λ −4 =

0

2

Characteristic equation: Characteristic roots:

λ1 = −2, λ2 = 2;

Solution of homogeneous equation: yh = c1 cosh 2x + c2 sinh 2x. To find a particular solution, we apply the method of undetermined coefficients. Accordingly, we try yp

=

A cosh x + B sinh x;

yp0

=

A sinh x + B cosh x;

yp00

=

A cosh x + B sinh x.

Plug into the equation y00 − 4 y = cosh x: A cosh x + B sinh x − 4A cosh x − 4B sinh x; = cosh x

Hence A = −1/3,

B = 0;

−3A

= 1;

−3B

= 0.

1 yp = − cosh x; 3

and so the general solution yg = c1 cosh 2x + c2 sinh 2x −

1 cosh x. 3

31. Using the half-angle formula, 1 2

Equation:

y00 + 4 y

=

Homogeneous equation:

y00 + 4 y

= 0;

Characteristic equation:

λ2 + 4

= 0

Characteristic roots:

λ1 = −2i, λ2 = 2i;

Solution of homogeneous equation: yh = c1 cos 2x + c2 sin 2x.

−

1 2

cos 2x;

Section A.2

Linear Ordinary Differential Equations with Constant Coefficients

A395

To find a particular solution, we apply the method of undetermined coefficients. We also notice that 1/8 is a solution of y00 + 4y = 1/8, so we really do not need to look for the solution corresponding to the term 1/8 of the nonhomogeneous part. We try yp

=

1 + Ax cos 2x + Bx sin 2x; 8

yp0

=

A cos 2x − 2Ax sin 2x + B sin 2x + 2Bx cos 2x;

yp00

=

−4A sin 2x − 4Ax cos 2x + 4B cos 2x − 4Bx sin 2x.

Plug into the equation y00 + 4 y =

1 2

−

1 2

cos 2x:

−4A sin 2x − 4Ax cos 2x + 4B cos 2x − 4Bx sin 2x 1 +4 + Ax cos 2x + Bx sin 2x = 8 −4A sin 2x + 4B cos 2x = 4A = 0; 4B = Hence A = 0,

1 B =− ; 8

yp =

1 1 − cos 2x; 2 2 1 − cos 2x; 2

1 . 2

1 x − sin 2x; 8 8

and so the general solution yg = c1 cos 2x + c2 sin 2x +

1 x − sin 2x. 8 8

32. Equation:

y00 + 4 y

=

x sin 2x;

Homogeneous equation:

y00 + 4 y

=

0;

λ2 + 4 =

0

Characteristic equation: Characteristic roots:

λ1 = −2i, λ2 = 2i;

Solution of homogeneous equation: yh = c1 cos 2x + c2 sin 2x. To find a particular solution, we apply the method of undetermined coefficients: yp

= x(Ax + B) cos 2x + x(Cx + D) sin 2x;

yp0

= (2Ax + B) cos 2 − 2(Ax2 + Bx) sin 2x + (2Cx + D) sin 2x + 2(Cx2 + Dx) cos 2x;

yp00

= 2Ax cos 2x − 4(2Ax + B) sin 2x − 4(Ax2 + Bx) cos 2x +2Cx sin 2x + 4(2Cx + D) cos 2x − 4(Cx2 + Dx) sin 2x.

Plug into the equation y00 + 4 y = x sin 2x: 2Ax cos 2x − 4(2Ax + B) sin 2x + 2Cx sin 2x + 4(2Cx + D) cos 2x =

x sin 2x;

(2A + 4(2Cx + D))) cos 2x + (2C − 4(2Ax + B))) sin 2x =

x sin 2x;

−8A = 1; 8C = 0; −4B + 2C = 0 2A + 4D = 0.

A396

Appendix A

Ordinary Differential Equations: Review of Concepts and Methods

Hence 1 1 A = − , C = 0, B = 0, D = ; 8 16

1 1 yp = − x2 cos 2x + x sin 2x; 8 16

and so the general solution 1 1 yg = c1 cos 2x + c2 sin 2x − x2 cos 2x + x sin 2x. 8 16 33. Equation:

y00 + y

=

1 2

Homogeneous equation:

y00 + y

=

0;

λ +1 =

0

2

Characteristic equation: Characteristic roots:

+

1 2

cos 2x;

λ1 = −i, λ2 = i;

Solution of homogeneous equation: yh = c1 cos x + c2 sin x. To find a particular solution, we apply the method of undetermined coefficients. We also use our experience and simplify the solution by trying yp

=

1 + A cos 2x; 2

yp0

=

−2A sin 2x;

yp00

=

−4A cos 2x.

Plug into the equation y00 + y =

1 2

1 2

cos 2x:

1 2

=

1 1 + cos 2x; 2 2

−3A

=

1 ; 2

+

−3A cos 2x +

1 A=− . 6 Hence 1 1 yp = − cos 2x + ; 6 2 and so the general solution yg = c1 cos 2x + c2 sin 2x −

1 1 cos 2x + . 6 2

34. Equation:

y00 + 2 y0 + 2 y

=

e−x cos 2x;

Homogeneous equation:

y00 + 2 y0 + 2y

=

0;

λ2 + 2λ + 2 =

0

Characteristic equation: Characteristic roots:

λ1 = −1 − i, λ2 = −1 + i;

Solution of homogeneous equation: yh = c1 e−x cos x + c2 e−x sin x.

Section A.2

Linear Ordinary Differential Equations with Constant Coefficients

A397

To find a particular solution, we apply the method of undetermined coefficients: yp = e−x x A cos x + B sin x ; yp0 = −e−x x A cos x + B sin x +e−x [ A cos x + B sin x + x − A sin x + B cos x ]; yp00 = e−x x A cos x + B sin x −e−x [ A cos x + B sin x + x − A sin x + B cos x ] −e−x [ A cos x + B sin x + x − A sin x + B cos x ] e−x [ − A sin x + B cos x + − A sin x + B cos x + x − A cos x − B sin x ] = −2e−x [ A cos x + B sin x + x − A sin x + B cos x ] +2e−x − A sin x + B cos x . Plug into the equation y00 + 2 y0 + 2 y = e−x cos 2x: −2e−x [ A cos x + B sin x + x − A sin x + B cos x ] +2e−x − A sin x + B cos x −2e−x x A cos x + B sin x +2e−x [ A cos x + B sin x + x − A sin x + B cos x ] 2e−x x A cos x + B sin x = 2e−x − A sin x + B cos x = Hence A = 0, 2B = 1 ⇒ B =

1 ; 2

yp =

e−x cos 2x; e−x cos 2x.

1 −x x e sin x; 2

and so the general solution 1 yg = c1 e−x cos x + c2 e−x sin x + xe−x sin x. 2 35. Equation:

y00 + 2 y0 + y

=

e−x ;

Homogeneous equation:

y00 + 2 y0 + y

=

0;

Characteristic equation:

λ2 + 2 λ + 1 =

0

Characteristic roots:

λ1 = −1 (double root);

Solution of homogeneous equation:

yh = c1 e−x + c2 x e−x .

To find a particular solution, we apply the method of undetermined coefficients: yp

=

A x2 e−x ;

yp0

=

−A x2 e−x + 2A x e−x ;

yp00

=

A x2 e−x − 2A x e−x + 2A e−x − 2A x e−x .

Plug into the equation y00 + 2 y0 + y = e−x : A x2 e−x − 2A x e−x + 2A e−x − 2A x e−x +2 − A x2 e−x + 2A x e−x + A x2 e−x

=

e−x ;

2Ae−x

=

e−x ;

A=

1 . 2

A398

Appendix A

Ordinary Differential Equations: Review of Concepts and Methods

Hence yp =

1 2 −x x e ; 2

and so the general solution 1 yg = c1 e−x + c2 xe−x + x2 e−x . 2 36. Equation:

y00 + 2y0 + y

=

x e−x + 6;

Homogeneous equation:

y00 + 2 y0 + y

=

0;

λ + 2λ+ 1 =

0

2

Characteristic equation: Characteristic roots:

λ1 = −1 (double root);

Solution of homogeneous equation:

yh = c1 e−x + c2 x e−x .

For a particular solution, try yp

=

6 + A x3 e−x ;

yp0

=

−A x3 e−x + 3A x2 e−x ;

yp00

=

A x3 e−x − 3A x2 e−x + 6A x e−x − 3A x2 e−x .

Plug into the equation y00 + 2 y0 + y = 6 + x e−x : A x3 e−x − 3A x2 e−x + 6A x e−x − 3A x2 e−x −A x3 e−x + 3A x2 e−x + 6 + A x3 e−x 6A x e−x A= Hence

= 6 + x e−x ; = x e−x ;

1 . 6

1 yp = 6 + x3e−x ; 6

and so the general solution 1 yg = c1 e−x + c2 xe−x + x3e−x + 6. 6 37. Equation:

y00 − y0 − 2 y

=

x2 − 4;

Homogeneous equation:

y00 − y0 − 2 y

=

0;

λ −λ−2 =

0

2

Characteristic equation: Characteristic roots:

λ1 = −1,

λ2 = 2;

Solution of homogeneous equation: yh = c1 e−x + c2 e2x . For a particular solution, try yp

=

A x2 + B x + C;

yp0

=

2A x + B;

yp00

=

2A.

Section A.2

Linear Ordinary Differential Equations with Constant Coefficients

A399

Plug into the equation y00 − y0 − 2 y = x2 − 4: 2A − 2A x − B − 2A x2 − 2B x − 2C

=

x2 − 4;

−2A

=

1;

1 A=− ; 2 −2A − 2B = 0

⇒ 1 − 2B = 0; B=

2A − 2C − B = 4

1 1 ⇒ − − − 2C = 4 2 2 C=

Hence

1 ; 2

5 . 4

1 1 5 yp = − x2 + x + ; 2 2 4

and so the general solution 1 1 5 yg = c1 e−x + c2 e2x − x2 + x + . 2 2 4 38. Equation:

y00 − y0 − 2 y

=

2 x2 + x ex;

Homogeneous equation:

y00 − y0 − 2 y

=

0;

λ2 − λ − 2 =

0

Characteristic equation: Characteristic roots:

λ1 = −1,

λ2 = 2;

−x

Solution of homogeneous equation: yh = c1 e

+ c2 e2x .

For a particular solution, try yp

= A x2 + B x + C + (Dx + E)ex ;

yp0

= 2A x + B + Dex + (Dx + E)ex ;

yp00

= 2A + 2Dex + (Dx + E)ex .

Plug into the equation y00 − y0 − 2 y = 2 x2 + x ex : 2A + 2Dex + (Dx + E)ex − 2A x − B − Dex − (Dx + E)ex −2A x2 − 2B x − 2C − 2(Dx + E)ex

=

2 x2 + x ex ;

−2A x2 + x(−2B − 2A) + 2A − B − 2C + (D − 2E)ex − 2Dxex

=

2 x2 + x ex ;

−2A = 2 ⇒ A = −1; 1 −2D = 1 ⇒ D = − ; 2 −2B − 2A = 0 ⇒ B = −A = 1; 2A − 2C − B = 0 ⇒ −2 − 1 − 2C = 0 ⇒ C=− D − 2E = 0 ⇒ E =

3 2

D 1 =− . 2 4

A400

Appendix A

Ordinary Differential Equations: Review of Concepts and Methods

Hence

3 1 1 yp = −x2 + x − x + (− x − )ex ; 2 2 4 and so the general solution 3 1 1 yg = c1 e−x + c2 e2x − x2 + x − x + (− x − )ex . 2 2 4 39. Even though the equation is first order and we can use the integration factor method of the previous section, we will use the method of this section to illustrate another interesting approach. Equation:

y0 + 2 y

=

2 x + sin x;

Homogeneous equation:

y0 + 2 y

=

0;

Characteristic equation:

λ+2

=

0

=

2x + sin x;

Characteristic root:

λ1 = −2;

Solution of homogeneous equation: yh = c1 e−2x . For a particular solution, try yp

=

A x + B + C cos x + D sin x;

yp0

=

A − C sin x + D cos x;

Plug into the equation y0 + 2 y = 2x + sin x: A − C sin x + D cos x + 2A x + 2B + 2C cos x + 2D sin x

2A = 2 ⇒ A = 1; 1 1 A + 2B = 0 ⇒ B = − A = − ; 2 2 2D − C = 1 and D + 2C = 0 ⇒ D = Hence yp = x −

2 1 , C=− . 5 5

1 1 2 − cos x + sin x; 2 5 5

and so the general solution yg = c1 e−2x + x −

1 1 2 − cos x + sin x. 2 5 5

40. Use preceding exercise: Equation: Solution of homogeneous equation:

y0 + 2 y

=

sin x;

yh

=

c1 e−2x .

For a particular solution, try yp = C cos x+D sin x, and as in the previous solution. it follows that 1 2 yp = − cos x + sin x; 5 5 and so the general solution yg = c1 e−2x −

1 2 cos x + sin x. 5 5

Section A.2

Linear Ordinary Differential Equations with Constant Coefficients

A401

41. 2 y0 − y = e2x . Equation:

2 y0 − y

=

e2x;

Homogeneous equation:

2 y0 − y

=

0;

Characteristic equation:

2λ−1

=

0

1 ; 2 Solution of homogeneous equation: yh = c1 ex/2 .

Characteristic root:

λ1 =

For a particular solution, try yp

=

Ae2x ;

yp0

=

2Ae2x ;

Plug into the equation 2y0 − y = e2x : 4Ae2x − Ae2x 3A = 1 Hence yp =

e2x ;

=

⇒ A=

1 . 3

1 2x e ; 3

and so the general solution 1 yg = c1ex/2 + e2x . 3 42. y0 − 7y = ex + cos x. Equation:

y0 − 7y

=

ex + cos x;

Homogeneous equation:

y0 − 7y

=

0;

λ−7 =

0

Characteristic equation: Characteristic root:

λ1 = 7;

Solution of homogeneous equation: yh = c1 e7x . For a particular solution, try yp

=

Aex + B cos x + C sin x;

yp0

=

Aex − B sin x + C cos x;

Plug into the equation y0 − 7y = ex + cos x: Aex − B sin x + C cos x − 7Aex − 7B cos x − 7C sin x −6A = 1 B + 7C = 0 and 7B + 49C = 0 Hence

=

ex + cos x;

1 ⇒ A=− ; 6 ⇒ C=

1 7 1 yp = − ex − cos x + sin x; 6 50 50

1 7 , B =− . 50 50

A402

Appendix A

Ordinary Differential Equations: Review of Concepts and Methods

and so the general solution 1 7 1 yg = c1 e7x + − ex − cos x + sin x. 6 50 50 43. Equation:

y00 + 9y

=

Homogeneous equation:

y00 + 9y

=

Characteristic roots:

λ1 = 3i,

Solution of homogeneous equation:

6 X sin nx ; n n=1 0;

λ2 = −3i;

yh

=

c1 cos 3x + c2 sin 3x.

To find a particular solution, we note that the term sin33x , which appears in the nonhomogeneous term, is part of the homogeneous solution. So for a particular solution that corresponds to this term, we try y3 = A3 x sin 3x + B3 x cos 3x. For the remaining terms of the nonhomogeneous part, use 6 X

Bn sin nx.

n=1, n6=3

Thus we try yp

=

A3 x sin 3x + B3 x cos 3x +

6 X

Bn sin nx;

n=1, n6=3

yp0

=

6 X

A3 sin 3x + 3A3x cos 3x + B3 cos 3x − 3B3 x sin 3x +

nBn cos nx;

n=1, n6=3

yp00

=

3A3 cos 3x + 3A3 cos 3x − 9A3 x sin 3x − 3B3 sin 3x − 3B3 sin 3x −9B3 x cos 3x −

6 X

n2 Bn sin nx;

n=1, n6=3

sin 3x 3

Plug into the equation y00 + 9y =

6A3 cos 3x − 6B3 sin 3x +

6 X

+

P6

n=1, n6=3

(9 − n2)Bn sin nx

sin nx n :

=

n=1, n6=3

6A3 cos 3x − 6B3 sin 3x

=

6 X

=

(9 − n2)Bn sin nx

n=1, n6=3

1 3

1 (n 6= 3) n

6 X

1 x cos 3x + 18

6 X n=1, n6=3

n=1, n6=3

sin nx n

sin nx ; n

⇒ A3 = 0, B3 = − ⇒ Bn =

Hence yp = −

6 X

sin 3x ; 3

n=1, n6=3

6A3 = 0, −6B3 = (n2 − 9)Bn =

sin 3x + 3

sin nx ; n(9 − n2 )

1 ; 18

1 . n(9 − n2)

Section A.2

Linear Ordinary Differential Equations with Constant Coefficients

and so the general solution yg = c1 cos 3x + c2 sin 3x −

1 x cos 3x + 18

6 X n=1, n6=3

sin nx . n(9 − n2)

44. Equation:

y00 + y

=

Homogeneous equation:

y00 + y

=

Characteristic roots:

λ1 = i,

Solution of homogeneous equation:

yh

6 X sin 2nx ; n n=1 0;

λ2 = −i; =

c1 cos x + c2 sin x.

To find a particular solution, try yp

=

6 X

Bn sin 2nx;

n=1

yp00

=

−

6 X

4n2Bn sin 2nx;

n=1

P6

Plug into the equation y00 + y = 6 X

n=1

sin 2nx : n

2

(1 − 4n )Bn sin 2nx =

n=1

(1 − 4n2)Bn Bn =

=

6 X sin 2nx n n=1

1 ; n

1 . n(1 − 4n2)

Hence yp =

6 X

sin nx ; n(1 − 4n2) n=1

and so the general solution yg = c1 cos x + c2 sin x +

6 X

sin nx . n(1 − 4n2 ) n=1

45. Write the equation in the form y00 − 4 y0 + 3 y

=

e2x sinh x

=

e2x

=

1 x e − e−x 2 1 3x e − ex . 2

From Exercise 25, yh = c1ex + c2 e3x . For a particular solution try yp = Axe3x + Bxex .

A403

A404

Appendix A

Ordinary Differential Equations: Review of Concepts and Methods

46. y00 − 4 y0 + 3 y = ex sinh 2x. Write the equation in the form y00 − 4 y0 + 3 y

=

ex sinh 2x

=

ex

=

1 2x e − e−2x 2 1 3x e − e−x . 2

From Exercise 25, yh = c1ex + c2 e3x . For a particular solution try yp = Axe3x + Be−x . 47. y00 + 2y0 + 2 y = cos x + 6x2 − e−x sin x. Characteristic equation λ2 + 2 λ + 2 = 0 ⇒ λ = −1 ± i. So yh = c1e−x cos x + c2 e−x sin x. For a particular solution, try yp = A cos x + B sin x + Cx2 + Dx + E + e−x F x sin x + Gx cos x . 48. y(4) − y = cosh x + cosh 2x. The equation is not of the type that is covered by the method of undetermined coefficients, but we can take a hint from this method. Characteristic equation λ4 − 1 = 0 ⇒ λ = ±1, ±i. So yh = c1e−x + c2 e−x + c3 cos x + c4 sin x. For a particular solution, try yp = Aex + Be−x + Ce2x + De−2x . 49. y00 − 3 y0 + 2 y = 3x4ex + x e−2x cos 3x. Characteristic equation λ2 − 3 λ + 2 = 0 ⇒ λ1 = 1, λ2 = 2. So yh = c1ex + c2 e2x . For a particular solution, try yp = x(Ax4 +Bx3 +Cx2 +Dx+E)ex +(Gx+H) e−2x cos 3x+(Kx+L) e−2x sin 3x. 50. y00 − 3 y0 + 2 y = x4e−x + 7 x2 e−2x . From Exercise 49, yh = c1ex + c2e2x . For a particular solution, try yp = x(Ax4 + Bx3 + Cx2 + Dx + E)e−x + (Gx2 + Hx + K) e−2x . 51. y00 + 4 y = e2x(sin 2x + 2 cos 2x). We have yh = c1 cos 2x + c2 sin 2x. For a particular solution, try yp = e2x (A sin 2x + B cos 2x). 52. y00 + 4 y = x sin 2x + 2 e2x cos 2x. We have yh = c1 cos 2x + c2 sin 2x. For a particular solution, try yp = x(Ax + B) sin 2x + x(Cx + D) cos 2x + e2x (E sin 2x + F cos 2x). 53. y00 − 2 y0 + y = 6x − ex . Characteristic equation λ2 − 2 λ + 1 = 0 ⇒ λ1 = 1(double root). So yh = c1ex + c2 xex . For a particular solution, try yp = Ax + B + Cx2ex .

Section A.2

Linear Ordinary Differential Equations with Constant Coefficients

A405

54. y00 − 3 y0 + 2 y = ex + 3 e−2x . From Exercise 49, yh = c1 ex + c2 e2x . For a particular solution, try yp = Axex + B e−2x. 55. From Exercise 25, we have yh = c1ex + c2 e3x. If α 6= 1 or 3 (the characteristic roots), a particular solution of y00 − 4y0 + 3y = eαx is yp = Aeαx . Plugging into the equation, we find Aα2eαx − 4αAeαx + 3Aeαx

=

A(α2 − 4α + 3) = A

=

eαx ; 1 (divide by eαx 6= 0); 1 . α2 − 4α + 3

Note that α2 − 4α + 3 6= 0, because α 6= 1 or 3. So the general solution in this case is of the form eαx yg = c1ex + c2 e3x + 2 . α − 4α + 3 If α = λ is one of the characteristic roots 1 or 3, then we modify the particular solution and use yp = Axeλx . Then yp0

=

Aeλx + Axλeλx ,

yp00

=

Aλeλx + Aλeλx + Axλ2eλx

=

2Aλeλx + Axλ2 eλx .

Plug into the left side of the equation 2Aλeλx + Axλ2eλx − 4 Aeλx + Axλeλx + 3Axλeλx =0

λx

= Axe

z }| { λ2 − 4λ + 3 −2Aλeλx

= −2Aλeλx . This should equal eλx . So −2A = 1 or A = −1/2 and hence 1 yg = c1 ex + c2 e3x − xeλx . 2 56. The characteristic equation of y00 − 4y0 + 4y = 0 is λ2 − 4λ + 4 = 0, with a double characteristic root λ = 2. Thus yh = c1 e2x + c2xe2x . If α 6= 2, a particular solution of y00 − 4y0 + 4y = eαx is yp = Aeαx . Plugging into the equation, we find Aα2eαx − 4αAeαx + 4Aeαx

=

A(α2 − 4α + 4) = A

=

eαx ; 1 (divide by eαx 6= 0); α2

1 . − 4α + 4

A406

Appendix A

Ordinary Differential Equations: Review of Concepts and Methods

Note that α2 − 4α + 4 6= 0, because α 6= 2. So the general solution in this case is of the form eαx yg = c1 e2x + c2xe2x + 2 . α − 4α + 4 If α = 2 is a characteristic root, then we modify the particular solution and use yp = Ax2e2x , since it is a double root. Then yp0

=

2Axe2x + 2Ax2e2x ,

yp00

=

2Ae2x + 4Axe2x + 4Axe2x + 4Ax2e2x

=

2Ae2x + 8Axe2x + 4Ax2e2x .

Plug into the left side of the equation y00 − 4y0 + 4y = e2x : 2Ae2x + 8Axe2x + 4Ax2 e2x − 4 2Axe2x + 2Ax2e2x + 4Ax2e2x = 2Ae2x . This should equal e2x . So 2A = 1 or A = 1/2 and hence 1 yg = c1 e2x + c2 xe2x + x2e2x . 2 57. y00 + 4y = cos ωx. We have yh = c1 cos 2x + c2 sin 2x. If ω 6= ±2, a particular solution of y00 + 4y = cos ωx is yp = A cos ωx. So yp00 = −Aω2 cos ωx. Plugging into the equation, we find A cos ωx(4 − ω2 ) = A(4 − ω2 ) = A =

cos ωx; 1; 1 . 4 − ω2

Note that 4 − ω2 6= 0, because ω 6= ±2. So the general solution in this case is of the form cos ωx yg = c1 cos 2x + c2 sin 2x + . 4 − ω2 If ω = ±2, then we modify the particular solution and use yp = x A cos ωx + B sin ωx . Then yp0 = A cos ωx + B sin ωx + xω − A sin ωx + B cos ωx , yp00 = xω2 − A cos ωx − B sin ωx + 2ω − A sin ωx + B cos ωx . Plug into the left side of the equation xω2 − A cos ωx − B sin ωx + 2ω − A sin ωx + B cos ωx + 4x A cos ωx + B sin ωx . Using ω2 = 4, this becomes 2ω − A sin ωx + B cos ωx . This should equal cos ωx. So A = 0 and 2ωB = 1 or B = 1/(2ω). yg = c1 cos 2x + c2 sin 2x +

1 x sin ωx (ω = ±2). 2ω

Section A.2

Linear Ordinary Differential Equations with Constant Coefficients

A407

Note that if ω = 2 or ω = −2, the solution is 1 yg = c1 cos 2x + c2 sin 2x + x sin 2x. 4 58. We have yh = c1 cos 3x + c2 sin 3x. If ω 6= ±3, a particular solution of y00 + 9y = sin ωx is yp = A sin ωx. So yp00 = −Aω2 sin ωx. Plugging into the equation, we find A sin ωx(9 − ω2 )

=

sin ωx;

A(9 − ω2 )

=

1;

A

=

1 . 9 − ω2

Note that 9 − ω2 6= 0, because ω 6= ±3. So the general solution in this case is of the form sin ωx yg = c1 cos 3x + c2 sin 3x + . 9 − ω2 If ω = ±3, then we modify the particular solution and use yp = Ax cos ωx. Then yp0

=

A cos ωx − Aωx sin ωx,

yp00

=

−2Aω sin ωx − Aω2 x cos ωx.

Plug into the equation and get −2Aω sin ωx = sin ωx. So −2ωA = 1 or A = −1/(2ω). The general solution is yg = c1 cos 3x + c2 sin 3x −

1 x cos ωx 2ω

(ω = ±3).

If ω = 3 the general solution is 1 yg = c1 cos 3x + c2 sin 3x − x cos 3x. 6 If ω = −3 the general solution is 1 yg = c1 cos 3x + c2 sin 3x + x cos 3x. 6 59. We have yh = c1 cos ωx + c2 sin ωx. If ω 6= ±2, a particular solution of y00 + ω2 y = sin 2x is yp = A sin 2x. So yp00 = −4A sin 2x. Plugging into the equation, we find A sin 2x(−4 + ω2 )

=

sin 2;

A(ω2 − 4)

=

1;

A

=

ω2

1 . −4

A408

Appendix A

Ordinary Differential Equations: Review of Concepts and Methods

Note that ω2 − 4 6= 0, because ω 6= ±2. So the general solution in this case is of the form sin 2x yg = c1 cos ωx + c2 sin ωx + 2 . ω −4 If ω = ±2, then we modify the particular solution and use yp = Ax cos 2x. Then yp0

=

A cos 2x − 2Ax sin 2x,

yp00

=

−4A sin 2x − 4Ax cos 2x.

Plug into the equation and get −4A sin 2x = sin 2x. So −4A = 1 or A = −1/4. The general solution is 1 yg = c1 cos 3x + c2 sin 3x − x cos 2x. 4 60. The characteristic equation of y00 + by0 + y = 0 is λ2 + bλ + 1 = 0. Case 1: b2 − 4 > 0. In this case, we have two distinct roots, √ √ −b + b2 − 4 −b − b2 − 4 λ1 = and λ2 = . 2 2 Then yh = c1 eλ1 x + c2 eλ2 x . For a particular solution, we try yp = A cos x + B sin x. Plugging into the equation, we find −A cos x − B sin x + b(−A sin x + B cos x) + A cos x + B sin x = b(−A sin x + B cos x)

=

sin x; sin x;

1 B = 0, −Ab = 1, A = − . b Note that b 6= 0. So the general solution in this case is of the form yg = c1eλ1 x + c2 eλ2 x −

1 cos x. b

Case 2: b2 − 4 = 0. In this case, we have one double root, λ = b

−b 2 .

Then

b

yh = c1 e− 2 x + c2xe− 2 x . A particular solution has the same form as the previous case. So yg = c1eλ1 x + c2 eλ2 x −

1 cos x. b

Case 3: b2 − 4 < 0. In this case, we have two distinct roots, √ √ −b + i −b2 + 4 −b − i −b2 + 4 λ1 = and λ2 = . 2 2 Then

√

√ −b −b2 + 4 −b2 + 4 x yh = c1 e cos( x) + c2e 2 sin( x). 2 2 Case 3 (a) If b 6= 0. Then we use the same form of the particular solution as in Case 1 and get √ √ −b −b −b2 + 4 −b2 + 4 1 x x yg = c1e 2 cos( x) + c2 e 2 sin( x) − cos x. 2 2 b −b 2 x

Section A.2

Linear Ordinary Differential Equations with Constant Coefficients

A409

Case 3 (b) If b = 0, the equation becomes y00 + y = sin x. Then yh = c1 cos x + c2 sin x. For a particular solution, we try yp = Ax cos x. Then yp0

=

A cos x − Ax sin x

yp00

=

−2A sin x − Ax cos x.

Plugging into the equation, we find −2A sin x =

sin x;

1 − . 2 So the general solution in this case is of the form A =

1 yg = c1 cos x + c2 sin x − x cos x. 2 61. To solve y00 − 4 y = 0, y(0) = 0, y0 (0) = 3, start with the general solution y(x) = c1 cosh 2x + c2 sinh 2x. Then y(0) = 0 ⇒ c1 cosh 0 + c2 sinh 0 = 0 ⇒ c1 = 0; so y(x) = c2 sinh 2x. y0 (0) = 3 ⇒ 2c2 cosh 0 = 3 ⇒ c2 =

3 2

⇒ y(x) =

3 sinh 2x. 2

62. To solve y00 + 2 y0 + y = 0, y(0) = 2, y0 (0) = −1, recall the solution of the homogeneous equation from Exercise 35: y = c1e−x + c2 x e−x. Then y(0) = 2 ⇒ c1 = 2; so y(x) = 2e−x + c2x e−x . y0 (0) = −1

⇒ −2 + c2 = −1 ⇒ c2 = 1 ⇒ y(x) = 2e−x + x e−x .

63. To solve 4y00 − 4 y0 + y = 0, y(0) = −1, y0 (0) = 1, recall the solution of the homogeneous equation from Exercise 7: y = c1 ex/2 + c2 x ex/2. Then y(0) = −1 y0 (0) = 1

⇒ c1 = −1; so y(x) = −ex/2 + c2 x ex/2. 1 ⇒ − + c2 = 1 2 ⇒ c2 =

3 2

3 ⇒ y(x) = −ex/2 + x ex/2 . 2

A410

Appendix A

Ordinary Differential Equations: Review of Concepts and Methods

64. To solve y00 + y = 0,

y(π) = 1, y0 (π) = 0, use the general solution in the form y = c1 cos(x − π) + c2 sin(x − π).

Then y(π) = 1 ⇒ c1 = 1; so y(x) = cos(x − π) + c2 sin(x − π). 1 y0 (π) = 0 ⇒ − + c2 = 1 2 ⇒ c2 = 0 ⇒ y(x) = cos(x − π) = − cos x. 65. To solve y00 − 5 y0 + 6 y = ex , y(0) = 0, y0 (0) = 0, use the general solution from Exercise 27 (modify it slightly): 1 y = c1 e2x + c2 e3x + ex . 2 Then 1 y(0) = 0 ⇒ c1 + c2 = − ; 2 y0 (0) = 0 ⇒ 2c1 + 3c2 = − ⇒ c2 =

1 2

1 ; c1 = −1 2

⇒ y(x) = − e2x +

1 3x 1 x e + e . 2 2

66. To solve 2 y00 − 3 y0 + y = sin x, y(0) = 0, y0 (0) = 0, we first need the general solution of the differential equation. The solution of the homogeneous equation is yh = c1ex + c2 ex/2 (see Exercise 4). For a particular solution, try y = A cos x + B sin x. Then y00 = −A cos x − B sin x, y0 = −A sin x + B cos x. Plug into the equation −2A cos x − 2B sin x + 3A sin x − 3B cos x + A cos x + B sin x =

sin x

(−A − 3B) cos x + (3A − B) sin x =

sin x;

Solving for A and B, we find A =

3 10

−A − 3B

=

0

3A − B

=

1.

1 and B = − 10 . Thus the general solution is

y = c1 ex + c2ex/2 +

3 1 cos x − sin x. 10 10

Using the initial conditions 3 10

y(0) = 0

⇒ c1 + c2 = −

y0 (0) = 0

1 1 ⇒ c1 + c2 = − ; 2 10 4 1 ⇒ c2 = − , c1 = 5 2

Section A.2

Linear Ordinary Differential Equations with Constant Coefficients

y=

A411

1 x 4 x/2 3 1 + e − e cos x − sin x. 2 5 10 10

67. Start with the homogeneous equation y00 − 4 y0 + 3 y = 0. Its characteristic equation is λ2 − 4λ + 3 = 0 ⇒ λ1 = 1, λ2 = 3. Thus yh = c1ex + c2 e3x. For a particular solution of y00 − 4 y0 + 3 y = x e−x, try y = (Ax + b)e−x , y0 = Ae−x − (Ax + b)e−x , y00 = −2Ae−x + (Ax + b)e−x . Plug into the equation and get −2Ae−x + (Ax + b)e−x − 4(Ae−x − (Ax + b)e−x ) + 3(Ax + b)e−x e−x − 2A + b − 4A + 4b + 3b + xe−x A + 4A + 3A

=

x e−x

=

xe−x ;

−6A + 8b = 8A A=

=

0 1;

1 6 3 , b= = . 8 64 32

So the general solution is 1 3 y = c1ex + c2 e3x + ( x + )e−x . 8 32 We can now solve y00 − 4 y0 + 3 y = x e−x, conditions imply that 3 32

=

0

1 3 − 8 32

=

0;

c1 + c2

=

−

3 32

c1 + 3c2

=

−

1 ; 32

c1 + c2 + c1 + 3c2 +

c2 = Hence

y(0) = 0, y0 (0) = 1. The boundary

1 1 ; c1 = − 32 8

1 1 1 3 y = − ex + e3x + ( x + )e−x . 8 32 8 32

68. The general solution of y00 − 4 y = 0 is y = c1 cosh 2x + c2 sinh 2x. For a particular solution of y00 − 4 y = cosh x, we try y = A cosh x. Then y00 = A cosh x. Plugging into the equation: cosh x = y00 − 4y = A cosh x − 4 cosh x = (A − 4) cosh x ⇒ 1 = A − 4, A = 5. So y = c1 cosh 2x + c2 sinh 2x + 5 cosh x. Using the initial conditions y(0) = 1 ⇒ c1 + 5 = 1 ⇒ c1 = −4 y0 (0) = 0 ⇒ c2 = 0;

A412

Appendix A

Ordinary Differential Equations: Review of Concepts and Methods

and so y = −4 cosh 2x + 5 cosh x. 69. Because of the initial conditions, it is more convenient to take π π y = c1 cos[2(x − )] + c2 sin[2(x − )] 2 2 as a general solution of y00 + 4y = 0. For a particular solution of y00 + 4y = cos 2x, we try y = Ax sin 2x, y0 = A sin 2x + 2Ax cos 2x, y00 = 4A cos 2x − 4Ax sin 2x. Plug into the equation, 4A cos 2x = cos 2x ⇒ A =

1 . 4

So the general solution is y = c1 cos[2(x −

π π 1 )] + c2 sin[2(x − )] + x sin 2x. 2 2 4

Using the initial conditions y(π/2) = 1 ⇒ c1 = 1 π cos π = 0 4 π ⇒ c2 = ; 8

y0 (π/2) = 0 ⇒ 2c2 +

and so

π π π 1 )] + sin[2(x − )] + x sin 2x. 2 8 2 4 using the addition formulas for the cosine and sine, we can write π π cos[2(x − )] = − cos 2x and sin[2(x − )] = − sin 2x, 2 2 y = cos[2(x −

and so π 1 π 1 sin 2x + x sin 2x = − cos 2x + − + x sin 2x. 8 4 8 4 P 6 70. From Exercise 43, the general solution of y00 + 9 y = n=1 sinnnx is y = − cos 2x −

y = c1 cos 3x + c2 sin 3x −

1 x cos 3x + 18

6 X n=1, n6=3

sin nx . n(9 − n2)

Using the initial condition y(0) = 0, we find c1 = 0 and so y

y0

=

=

c2 sin 3x −

3c2 cos 3x −

y0 (0) = 2 ⇒ 3c2 −

1 + 18

c2

=

=

1 x cos 3x + 18

n=1, n6=3

sin nx n(9 − n2)

1 1 cos 3x + x sin 3x + 18 6 6 X

n=1, n6=3

1 1 2+ − 3 18 29831 , 45360

6 X

1 =2 9 − n2

6 X n=1, n6=3

1 9 − n2

6 X n=1, n6=3

cos nx 9 − n2

Section A.2

Linear Ordinary Differential Equations with Constant Coefficients

and so y=

6 X

1 29831 sin 3x − x cos 3x + 45360 18

n=1, n6=3

A413

sin nx . n(9 − n2 )

71. An antiderivative of g(x) = ex sin x is a solution of the differential equation y0 = ex sin x. To solve this equation we used the method of undetermined coefficients. The solution of homogeneous equation y0 = 0 is clearly y = c. To find a particular solution of y0 = ex sin x we try y

=

ex (A sin x + B cos x)

y0

=

ex (A sin x + B cos x) + ex (A cos x − B sin x)

=

ex (A − B) sin x + ex (A + B);

Plugging into the equation, we find ex (A − B) sin x + ex (A + B) cos x A − B = 1, A + B = 0

so

Z

ex sin x dx =

=

ex sin x

⇒ A=

1 1 , B=− ; 2 2

1 x 1 ex e sin x − ex cos x + C = sin x − cos x + C. 2 2 2

72. See Exercise 73. 73. An antiderivative of g(x) = ea x cos bx is a solution of the differential equation y0 = ea x cos bx. We assume throughout this exercise that a 6= 0 and b 6= 0. For these special cases the integral is clear. To solve the differential equation we used the method of undetermined coefficients. The solution of homogeneous equation y0 = 0 is y = C. To find a particular solution of y0 = eax cos bx we try y

=

eax (A cos bx + B sin bx)

y0

=

aeax (A cos bx + B sin bx) + eax (−bA sin bx + bB cos bx)

=

eax (Aa + bB) cos bx + eax (aB − bA) sin bx.

Plugging into the equation, we find eax (Aa + bB) cos bx + eax (aB − bA) sin bx

=

eax cos bx

Aa + bB = 1, aB − bA = 0 ⇒ A =

so

Z

eax cos bx dx =

a b , B= 2 ; a2 + b2 a + b2

eax a cos x + b sin bx + C. 2 +b

a2

74. Repeat the steps in the previous exercise. To solve y0 = eax sin bx, try y

=

eax (A cos bx + B sin bx)

y0

=

eax (Aa + bB) cos bx + eax (aB − bA) sin bx.

A414

Appendix A

Ordinary Differential Equations: Review of Concepts and Methods

Plugging into the equation, we find eax (Aa + bB) cos bx + eax (aB − bA) sin bx

=

eax sin bx

Aa + bB = 0, aB − bA = 1 ⇒ A = −

so

Z

eax cos bx dx =

a2

b a , B= 2 ; 2 +b a + b2

eax − b cos x + a sin bx + C. 2 +b

a2

Concerning Exercises 75–78 and for information about the Vandermonde determinant, please refer to the book “Applied Linear Algebra,” by P. Olver and C. Shakiban, Prentice Hall, 2006.

Section A.3

Linear Ordinary Differential Equations with Nonconstant Coefficients

A415

Solutions to Exercises A.3 1. We apply the reduction of order formula and take all constants of integration equal to 0. y00 + 2 y0 − 3 y = 0, Z p(x) = 2, p(x) dx = 2x,

y2 = y1

Z

e−

R

y1 = ex ; e−

p(x) dx

dx = ex

y12

x

= e

R

Z Z

p(x) dx

= e−2x;

e−2x dx e2x −4x

e

1 −4x 1 = − e−3x . dx = e − e 4 4 x

Thus the general solution is y = c1ex + c2 e−3x. 2. We apply the reduction of order formula and take all constants of integration equal to 0. y00 − 5 y0 + 6 y = 0, Z p(x) = −5, p(x) dx = −5x,

y2 = y1

Z

e−

R

y1 = e3x ; e−

R

dx =

e3x

Z

=

e3x

Z

p(x) dx

y12

p(x) dx

= e5x ;

e5x dx e6x e−x dx = e3x −e−x = −e2x .

Thus the general solution is y = c1 e3x + c2 e2x . 3. x y00 − (3 + x)y0 + 3y = 0, y00 −

y1 = ex . We first put the equation in standard form.

3+x 0 3 y + y = 0, x x Z 3 (− − 1) dx = x e−

y2 = y1

Z

e−

R

R

p(x) dx

3+x , x

−3 ln x − x,

=

e3 ln x+x = x3 ex ; Z 3 x x e x e dx e2x Z x e x3 e−x dx

=

ex −x3 e−x − 3x2e−x − 6xe−x − 6e−x

=

−x3 − 3x2 − 6x − 6.

=

p(x) dx

y12

y1 = ex ; p(x) = −

dx =

A416

Appendix A

Ordinary Differential Equations: Review of Concepts and Methods

Thus the general solution is y = c1 ex + c2 (x3 + 3x2 + 6x + 6).

4. xy00 − (2 − x)y0 − 2 y = 0,

y1 = e−x .

We first put the equation in standard form.

y00 −

2−x 0 2 −2 + x y − y = 0, y1 = e−x ; p(x) = , x x x Z 2 (− + 1) dx = −2 ln x + x, x e−

y2 = y1

Z

e−

R

R

p(x) dx

p(x) dx

y12

= e2 ln x−x = x2 e−x ; Z 2 −x x e −x dx = e dx e−2x Z −x = e x2ex dx = e−x x2ex − 2xex + 2ex = x2 − 2x + 2.

Thus the general solution is y = c1 e−x + c2(x2 − 2x + 2).

5. y00 + 4 y = 0,

y1 = cos 2x. y00 + 4 y = 0,

p(x) = 0,

y2 = y1

Z

Z

p(x) dx = 0,

e−

R

y1 = cos 2x; e−

R

p(x) dx

dx = cos 2x

Z

= cos 2x

Z

p(x) dx

y12

= 1;

1 dx cos2 2x sec2 2x dx

1 1 = cos 2x tan 2x = sin 2x. 2 2

Thus the general solution is y = c1 cos 2x + c2 sin 2x.

Section A.3

Linear Ordinary Differential Equations with Nonconstant Coefficients

6. y00 + 9 y = 0, p(x) = 0,

y2 = y1

Z

Z

y1 = sin 3x; e−

p(x) dx = 0,

e−

R

R

p(x) dx

dx = sin 3x

Z

= sin 3x

Z

p(x) dx

y12

= 1;

1 dx sin2 3x csc2 3x dx

1 1 = sin 3x − cot 3x = − cos 3x. 3 3

Thus the general solution is y = c1 cos 3x + c2 sin 3x. 7. y00 − y = 0, p(x) = 0,

y2 = y1

Z

Z

y1 = cosh x; e−

p(x) dx = 0, R

e−

R

p(x) dx

dx =

cosh x

Z

=

cosh x

Z

=

cosh x [tanh x] = sinh x.

p(x) dx

y12

= 1;

1 dx cosh2 x sech 2x dx

Thus the general solution is y = c1 cosh x + c2 sinh x. 8. y00 + 2 y0 + y = 0, Z p(x) = 2, p(x) dx = 2x, −x

y2 = e

Z

y1 = e−x ; e−

R

e−2x dx = e−x e−2x

Thus the general solution is y = c1e−x + c2 xe−x.

p(x) dx

Z

= e−2x ;

dx = e−x x.

A417

A418

Appendix A

Ordinary Differential Equations: Review of Concepts and Methods

9. Put the equation in standard form: (1 − x2 )y00 − 2xy0 + 2y = 0, y00 −

y1 = x;

2x 0 2 y + y = 0, 2 1−x 1 − x2 Z p(x) dx = e−

R

p(x) dx

y2

= =

p(x) = − Z

−

2x dx = ln(1 − x2 ) 1 − x2 2

e− ln(1−x x

Z

2x ; 1 − x2

)

=

1 ; 1 − x2

1 dx. (1 − x2)x2

To evaluate the last integral, we use the partial fractions decomposition 1 (1 − x2 )x2

1

=

1 (1 − x)(1 + x)x2

=

A B C D + + + 2; (1 − x) (1 − x) x x

=

A(1 + x)x2 + B(1 − x)x2 + C(1 − x2)x + D(1 − x2) (1 − x)(1 + x)x2

=

A(1 + x)x2 + B(1 − x)x2 + C(1 − x2)x + D(1 − x2).

Take x = 0

⇒ D = 1.

Take x = 1

⇒ 1 = 2A, A =

1 . 2

Take x = −1

⇒ 1 = 2B, B =

1 . 2

Checking the coefficient of x3 , we find C = 0. Thus 1 (1 − x2)x2 Z

1 1 1 + ++ 2 2(1 − x) 2(1 + x) x

=

1 dx = (1 − x2)x2

1 1 1 − ln(1 − x) + ln(1 + x) − 2 2 x 1 1+x 1 ln − . 2 1−x x

=

So

1 y2 = x ln 2

1+x 1−x

1 x 1+x − = ln − 1. x 2 1−x

Hence the general solution

y = c1 x + c2

x ln 2

1+x 1−x

−1 .

Section A.3

Linear Ordinary Differential Equations with Nonconstant Coefficients

10. Put the equation in standard form: (1 − x2)y00 − 2xy0 = 0, y00 −

y1 = 1;

2x 0 y = 0, 1 − x2 Z p(x) dx = e−

R

p(x) = − Z

−

2x dx = ln(1 − x2) 1 − x2 2

p(x) dx

=

e− ln(1−x

y2

=

Z

= = =

2x ; 1 − x2

)

=

1 ; 1 − x2

1 dx 1 − x2 Z 1 1 1 + dx 2 1−x 1+x −1 1 ln(1 − x) + ln(1 + x) 2 2 1 1+x ln . 2 1−x

Hence the general solution

y = c1 + c2 ln

1+x 1−x

.

11. Put the equation in standard form: x2 y00 + xy0 − y = 0, y00 +

y1 = x;

1 0 1 y − 2 y = 0, x x Z p(x) dx = e−

R

p(x) dx

y2

=

p(x) = Z

1 ; x

1 dx = ln x x

e− ln x = Z

1 ; x

1 dx x3

=

x

=

1 −2 x − x 2

=

−1 . 2x

Hence the general solution

y = c1 x +

c2 . x

A419

A420

Appendix A

Ordinary Differential Equations: Review of Concepts and Methods

12. Put the equation in standard form: x2y00 − xy0 + y = 0, y00 −

y1 = x;

1 0 1 1 p(x) = − ; y + 2 y = 0, x x x Z Z 1 p(x) dx = − dx = − ln x x e−

R

p(x) dx

= eln x = x; Z x = x dx x2 Z 1 = x dx x

y2

= x ln x. Hence the general solution y = c1 x + c2 x ln x. 13. Put the equation in standard form: x2 y00 + xy0 + y = 0, y00 +

y1 = cos(ln x);

1 0 1 y + 2 y = 0, x x Z p(x) dx = e−

R

p(x) = Z

1 ; x

1 dx = ln x x

p(x) dx

=

e− ln x =

y2

=

cos(ln x)

1 ; x Z

=

cos(ln x)

Z

=

cos u tan u = sin u = sin(ln u).

1 dx x cos2(ln x) 1 1 du (u = ln x, du = dx) 2 cos u x

Hence the general solution y = c1 cos(ln x) + c2 sin(ln u). 14. Put the equation in standard form: x2y00 + 2 xy0 + y00 +

1 y = 0, 4

1 y1 = √ ; x

2 0 1 y + 2 y = 0, x 4x Z p(x) dx = e−

R

p(x) = Z

2 ; x

2 dx = 2 ln x x

p(x) dx

=

e−2 ln x =

y2

=

1 √ x

Z

1 ; x2

ln x x dx = √ . x2 x

Section A.3

Linear Ordinary Differential Equations with Nonconstant Coefficients

Hence the general solution 1 ln x y = c1 √ + c2 √ . x x 15. Put the equation in standard form: x y00 + 2 y0 + 4x y = 0, y00 +

y1 =

sin 2x ; x

2 0 2 p(x) = ; y + 4y = 0, x x Z Z 2 p(x) dx = dx = 2 ln x x e−

R

p(x) dx

y2

= e−2 ln x =

1 ; x2

Z

=

sin 2x x

=

sin 2x x

=

cos 2x . −2x

x2 x2 sin2 (2x)

Z

dx

1 sin 2x 1 cot(2x) dx = x −2 sin2 (2x)

Hence the general solution y = c1

sin 2x cos 2x + c2 . x x

16. Put the equation in standard form: x y00 + 2 y0 − x y = 0, y00 +

y1 =

2 0 y − y = 0, x Z p(x) dx = e−

R

ex ; x

p(x) = Z

2 ; x

2 dx = 2 ln x x

p(x) dx

=

e−2 ln x =

y2

=

ex x

=

ex x

=

−

Z Z

x2 −2x e dx x2 e−2x dx =

e−x . 2

Hence the general solution y = c1

1 ; x2

ex e−x + c2 . x x

ex 1 −2x e x −2

A421

A422

Appendix A

Ordinary Differential Equations: Review of Concepts and Methods

17. Put the equation in standard form: x y00 + 2 (1 − x)y0 + (x − 2) y = 0, y00 +

y1 = ex ;

2(1 − x) 0 x − 2 y + y = 0, x x Z p(x) dx = e−

R

p(x) dx

y2

= = =

Hence the general solution y = c1ex + c2

p(x) =

2 − 2; x

2 ln x − 2x e−2 ln x+2x = x

Z

x

Z

e

e

e2x ; x2

e2x dx x2e2x 1 ex dx = − . x2 x

ex . x

18. Put the equation in standard form: (x − 1)2 y00 − 3(x − 1) y0 + 4 y = 0, y00 −

y1 = (x − 1)2 ;

3 4 y = 0, y0 + x−1 (x − 1)2 Z p(x) dx = e−

R

p(x) dx

=

y2

= =

p(x) = −

3 ; x−1

−3 ln(x − 1) e3 ln(x−1) = (x − 1)3; Z (x − 1)3 2 (x − 1) dx (x − 1)4 Z 1 (x − 1)2 dx = (x − 1)2 ln(x − 1). x−1

Hence the general solution y = c1 (x − 1)2 + c2(x − 1)2 ln(x − 1). 19. Put the equation in standard form: x2y00 − 2x y0 + 2 y = 0, y00 −

y 1 = x2 ;

2 0 2 y + 2 y = 0, x x Z p(x) dx = e−

R

p(x) dx

=

y2

= =

2 p(x) = − ; x −2 ln x e2 ln x = x2; Z 2 x x2 dx x4 Z 1 x2 dx = −x. x2

Hence the general solution y = c1 x2 + c2x.

Section A.3

Linear Ordinary Differential Equations with Nonconstant Coefficients

A423

20. Put the equation in standard form: (x2 − 2x)y00 − (x2 − 2) y0 + 2(x − 1)y = 0,

y1 = ex ;

x2 − 2 0 2(x − 1) y − 2 y + 2 y = 0, x − 2x x − 2x Z p(x) dx = 00

e−

R

p(x) dx

=

y2

=

x2 − 2 2x − 2 p(x) = − 2 =− 1+ 2 ; x − 2x x − 2x −x − ln(x2 − 2x) 2

=

ex+ln(x −2x) = (x2 − 2x)ex ; Z (x2 − 2x)ex ex dx e2x Z ex (x2 − 2x)e−x dx

=

ex [−e−x x2 ] = −x2 .

Hence the general solution y = c1ex + c2 x2. 21. y00 − 4y0 + 3 y = e−x . λ2 − 4λ + 3 = 0 ⇒ (λ − 1)(λ − 3) = 0 ⇒ λ = 1 or λ = 3. Linearly independent solutions of the homogeneous equation: y1 = ex Wronskian:

x e W (x) = ex

and y2 = e3x .

e3x = 3e4x − e4x = 2e4x. 3x 3e

We now apply the variation of parameters formula with g(x) = yp

=

= =

e−x ; Z Z −y2 g(x) y1 g(x) y1 dx + y2 dx W (x) W (x) Z Z x −x −e3x e−x e e 3x ex dx + e dx 2e4x 2e4x Z Z ex e3x −2x − dx + e e−4x dx 2 2

=

ex −2x e3x −4x e−x e−x − = e e − 4 8 4 8

=

e−x . 8

Thus the general solution is y = c1 ex + c2e3x +

e−x . 8

22. y00 − 15 y0 + 56 y = e7x + 12 x. λ2 − 15λ + 56 = 0 ⇒ (λ − 7)(λ − 8) = 0 ⇒ λ = 7 or λ = 8.

A424

Appendix A

Ordinary Differential Equations: Review of Concepts and Methods

Linearly independent solutions of the homogeneous equation: y1 = e7x

and y2 = e8x.

Wronskian: 7x e W (x) = 7ex

e8x = 8e15x − 7e15x = e15x . 8e8x

We now apply the variation of parameters formula with g(x) yp

= e7x + 12 x; Z Z −y2 g(x) y1 g(x) = y1 dx + y2 dx W (x) W (x) Z Z 7x 7x −e8x (e7x + 12 x) e (e + 12 x) 8x = e7x dx + e dx e15x e15x Z Z = −e7x (1 + 12xe−x ) dx + e8x e−x + 12xe−8x) dx 7x

= −e

Z Z −8x 8x −x −8x x + 12 xe −e + 12 xe dx + e dx

12 12 12 12 = −e7x x − xe−7x − e−7x + e8x −e−x − xe−8x − e−8x 7 49 8 64 = −xe7x − e7x +

12 12 12 12 x+ − x− 7 49 8 64

= −xe7x − e7x +

3 45 x+ . 14 784

Thus the general solution is y = c1 e7x + c2e8x − xe7x − e7x +

3 45 x+ . 14 784

23. 3 y00 + 13 y0 + 10 y = sin x. 3λ2 + 13λ + 10 = 0 ⇒ λ =

−13 ±

⇒ λ=−

√

169 − 120 −13 ± 7 = 6 6

20 10 =− or λ = −1. 6 3

Two linearly independent solutions of the homogeneous equation: y1 = e−

10 3 x

and y2 = e−x .

Wronskian: e− 10 3 x W (x) = − 10 e− 10 3 x 3

e−x 13 10 13 7 13 = −e− 3 x + e− 3 x = e− 3 x . −x 3 3 −e

Section A.3

Linear Ordinary Differential Equations with Nonconstant Coefficients

A425

We now apply the variation of parameters formula with

g(x)

=

yp

=

=

=

sin x ; 3 Z Z −y2 g(x) y1 g(x) y1 dx + y2 dx W (x) W (x) 10

e− 3 x

Z

−e−x sin x

dx + e−x

Z

10

e− 3 x sin x

dx 13 13 3 73 e− 3 x 3 37 e− 3 x Z Z 1 10 10 1 − e− 3 x e 3 x sin x dx + e−x ex sin x dx. 7 7

Evaluate the integrals with the help of the formulas in the table at the end of the book and get

yp

=

=

= =

10 10 e3x 1 10 1 −x ex − e− 3 x 10 2 (sin x − cos x) sin x − cos x + e 7 7 12 + 12 ( 3 ) + 12 3 1 1 11 10 − 100 sin x − cos x + (sin x − cos x) 7 9 +1 3 72 9 10 1 − sin x − cos x + (sin x − cos x) 763 3 14 7 13 sin x − cos x 218 218

Thus the general solution is

10

y = c1 e− 3 x + c2e−x +

7 13 sin x − cos x. 218 218

24. y00 + 3 y = x. Two linearly independent solutions of the homogeneous equation are:

y1 = cos

√ 3x and

√ y2 = sin 3x.

Wronskian: √ √ cos 3x sin 3x W (x) = √ √ √ √ − 3 sin 3x 3 cos 3x

√ = 3.

A426

Appendix A

Ordinary Differential Equations: Review of Concepts and Methods

We now apply the variation of parameters formula with g(x) yp

= =

=

=

=

=

x; Z

Z −y2 g(x) y1 g(x) y1 dx + y2 dx W (x) W (x) √ √ √ Z x sin 3x √ Z x cos 3x √ √ − cos 3x dx + sin 3x dx 3 3 √ Z √ √ cos 3x 1 x − √ − √ cos 3x + √ cos 3x dx 3 3 3 √ Z √ √ sin 3x x 1 √ sin 3x − √ + √ sin 3x dx 3 3 3 √ √ √ cos 3x 1 − −x cos 3x + √ sin 3x 3 3 √ √ √ sin 3x 1 + x sin 3x + √ cos 3x 3 3 x . 3

Thus the general solution is y = c1 cos

√ √ x 3x + c2 sin 3x + . 3

25. y00 + y = sec x. Two linearly independent solutions of the homogeneous equation are:

Wronskian:

y1 = cos x and

y2 = sin x.

cos x W (x) = − sin x

sin x = 1. cos x

We now apply the variation of parameters formula with g(x) = yp

= = =

sec x =

1 ; cos x

Z

Z −y2 g(x) y1 g(x) y1 dx + y2 dx W (x) W (x) Z Z sin x − cos x dx + sin x dx cos x cos x · ln (| cos x|) + x sin x.

Thus the general solution is y = c1 cos x + c2 sin x + cos x · ln (| cos x|) + x sin x. 26. y00 + y = sin x + cos x. Two linearly independent solutions of the homogeneous equation are: y1 = cos x

and y2 = sin x,

W (x) = 1

Section A.3

A427

Linear Ordinary Differential Equations with Nonconstant Coefficients

(as in previous exercise). We now apply the variation of parameters formula with g(x)

=

=

sin x + cos x; Z Z −y2 g(x) y1 g(x) y1 dx + y2 dx W (x) W (x) Z Z − cos x sin x(sin x + cos x) dx + sin x cos x(sin x + cos x) dx

yp

=

=

− cos x

=

=

=

Z

(sin2 x + sin x cos x) dx + sin x

Z

(sin x cos x + cos2 x) dx

Z

1 − cos 2x 1 dx + sin2 x 2 2 Z 1 1 + cos 2x 2 + sin x sin x + dx 2 2 x sin 2x 1 1 − cos x − + sin2 x + sin x sin2 x + 2 4 2 2 x sin 2x 1 1 2 − cos x − + sin x + sin x sin2 x + 2 4 2 2 − cos x

x sin 2x − 2 4 x sin 2x − 2 4

Thus the general solution is x sin 2x 1 2 1 2 x sin 2x y = c1 cos x+c2 sin x−cos x − + sin x +sin x sin x + − . 2 4 2 2 2 4 27. x y00 − (1 + x)y0 + y = x3 . Two linearly independent solutions of the homogeneous equation are given: y1 = 1 + x and Wronskian:

1 + x ex W (x) = 1 ex x

y2 = ex .

= ex + xex − xex = xex .

We now apply the variation of parameters formula with g(x) = yp

= = =

x3 = x2 ; x Z Z −y2 g(x) y1 g(x) y1 dx + y2 dx W (x) W (x) Z x Z e 2 x −(1 + x) (1 + x)xe−x dx x dx + e xex Z x2 x −(1 + x) + e (xe−x + x2e−x ) dx 2

=

−

x3 x2 − + ex −xe−x − e−x − x2e−x − 2xe−x − 2e−x 2 2

=

−

x3 3 − x2 − 3x − 3. 2 2

Thus the general solution is y = c1 (1 + x) + c2 ex −

x3 3 2 − x − 3x − 3. 2 2

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Appendix A

Ordinary Differential Equations: Review of Concepts and Methods

This can also be written as y = c1(1 + x) + c2ex −

x3 3 2 − x . 2 2

28. x y00 − (1 + x)y0 + y = x4 ex . Two linearly independent solutions of the homogeneous equation are given: y2 = ex .

y1 = 1 + x and Wronskian:

1 + x ex W (x) = 1 ex x

= ex + xex − xex = xex .

We now apply the variation of parameters formula with

=

x3ex ; Z Z −y2 g(x) y1 g(x) y1 dx + y2 dx W (x) W (x) Z x Z e 3 x (1 + x) 3 x x −(1 + x) x e dx + e x e dx xex xex Z Z −(1 + x) x2ex dx + ex (1 + x)x2 dx

=

−(1 + x)

g(x)

=

yp

= =

=

=

=

Z

x2ex dx + ex

Z

(1 + x)x2 dx

3 2 x x4 x x x x −(1 + x) x e − 2xe + 2e + e + 3 4 x3 x4 ex −(1 + x)x2 + 2x(1 + x) − 2(1 + x) + + 3 4 4 x 2 ex − x3 + x2 − 2 . 4 3

Thus the general solution is y = c1(1 + x) + c2 ex + ex

29. x2y00 + 3 xy0 + y = indicial equation is

x4 2 3 − x + x2 − 2 . 4 3

√ x. The homogeneous equation is an Euler equation. The r2 + 2r + 1 = 0

⇒ (r + 1)2 = 0.

We have one double indicial root r = −1. Hence the solutions of the homogenous equation y1 = x−1 Wronskian:

W (x) =

1 x −1 x2

and y2 = x−1 ln x.

ln x 1 − ln x ln x 1 + 3 = 3. = 3 1−ln x x x x 2 1 x

x

Section A.3

Linear Ordinary Differential Equations with Nonconstant Coefficients

A429

We now apply the variation of parameters formula with √ 3 x g(x) = = x− 2 ; x2 Z Z −y2 g(x) y1 g(x) dx + y2 dx yp = y1 W (x) W (x) Z Z 1 ln x ln x 3 − 3 1 3 −3 = − x x 2 dx + x x 2 dx x x x x dv u Z z}|{ z√ }| { ln x Z 1 ln x x dx + x 2 dx x Z 1 2 3/2 2 3/2 1 ln x 2 3/2 − x ln x − x dx + x x 3 3 x x 3

1 − x

= =

1 2 2 3/2 4 1/2 = x . x x 33 9

=

Thus the general solution is 4 y = c1 x−1 + c2x−1 ln x + x1/2. 9 30. The homogeneous equation, x2y00 + x y0 + y = 0, is an Euler equation. The indicial equation is r2 + 1 = 0 ⇒ r = ±i. Hence the solutions of the homogenous equation y1 = cos(ln x) and

y2 = sin(ln x).

Wronskian: cos(ln x) W (x) = − sin(ln x) x

sin(ln x) ln x 1 1 cos2(ln x) + sin2(ln x) = . = cos(ln x) x x x

We now apply the variation of parameters formula with

Z

g(x)

=

1 ; x

yp

=

y1

=

u dv Z z }| Z { z}|{ − cos(ln x) sin(ln x) dx + sin(ln x) cos(ln x) dx

sin(ln x) dx = =

2

Z

−y2 g(x) dx + y2 W (x)

Z

y1 g(x) dx W (x)

u dv Z z }| { z}|{ x sin(ln x) − cos(ln x) dx

x sin(ln x) − x cos(ln x) −

Z

sin(ln x) dx =

x sin(ln x) − x cos(ln x);

Z

sin(ln x) dx =

x [sin(ln x) − cos(ln x)] . 2

Z

sin(ln x) dx;

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Appendix A

Ordinary Differential Equations: Review of Concepts and Methods

Similarly,

Z

cos(ln x) dx =

x [sin(ln x) + cos(ln x)] . 2

So yp

x x = − cos(ln x) [sin(ln x) − cos(ln x)] + sin(ln x) [sin(ln x) + cos(ln x)] 2 2 x = . 2

Thus the general solution is y = c1 cos(ln x) + c2 sin(ln x) +

x . 2

31. x2y00 + 4xy0 + 2 y = 0. Euler equation with α = 4, β = 2, indicial equation r2 + 3r + 2 = 0

(r + 1)(r + 2) = 0;

indicial roots: r1 = −1, r2 = −2; hence the general solution y = c1x−1 + c2 x−2. 32. x2y00 + x y0 − 4y = 0. Euler equation with α = 1, β = −4, indicial equation r2 − 4 = 0

(r + 2)(r − 2) = 0;

indicial roots: r1 = −2, r2 = 2; hence the general solution y = c1x−2 + c2 x2. 33. x2y00 + 3xy0 + y = 0. See Exercise 29. 34. 4 x2y00 + 8 x y0 + y = 0 or x2 y00 + 2 x y0 + 14 y = 0 . Euler equation with α = 2, β = 14 , indicial equation r2 + r +

1 =0 4

1 (r + )2 = 0; 2

one double indicial root: r = − 12 ; hence the general solution y = c1 x−1/2 + c2 x−1/2 ln x. 35. x2y00 + xy0 + 4y = 0. Euler equation with α = 1, β = 4, indicial equation r2 + 4 = 0; indicial roots: r1 = −2i, r2 = 2i; hence the general solution y = c1 cos(2 ln x) + c2 sin(2 ln x). 36. 4 x2y00 + 4x y0 + y = 0 or x2 y00 + x y0 + 14 y = 0. Euler equation with α = 1, β = 14 , indicial equation r2 + 14 = 0; indicial roots: r1 = −i2, r2 = i2; hence the general solution ln x ln x y = c1 cos( ) + c2 sin( ). 2 2 37. x2y00 + 7xy0 + 13 y = 0. Euler equation with α = 7, β = 13, indicial equation r2 + 6 r + 13 = 0; indicial roots: √ r = −3 ± −4; r1 = −3 − 2i, r2 = −3 + 2i. Hence the general solution y = x−3 [c1 cos(2 ln x) + c2 sin(2 ln x)] .

Section A.3

Linear Ordinary Differential Equations with Nonconstant Coefficients

A431

38. x2y00 − x y0 + 5 y = 0. Euler equation with α = −1, β = 5, indicial equation r2 − 2 r + 5 = 0; indicial roots: √ r = 1 ± −4; r1 = 1 − 2i, r2 = 1 + 2i. Hence the general solution y = x [c1 cos(2 ln x) + c2 sin(2 ln x)] . 39. (x − 2)2y00 + 3(x − 2)y0 + y = 0, (x > 2). Let t = x − 2, x = t + 2, y(x) = y(t + 2) = Y (t), dy dY dY dt dY y0 = = = = = Y 0; dx dx dt dx dt d dy dY 0 dt d 0 d2Y y00 = = Y 00. = Y == = dx dx dx dt dx dt2 The equation becomes t2Y 00 + 3tY 0 + Y = 0, t > 0. Euler equation with α = 3, β = 1, indicial equation r2 + 2 r + 1 = 0

(r + 1)2 = 0;

one double indicial root r = −1. Hence the general solution Y = c1t−1 + c2 t−1 ln t, and so y = c1 (x − 2)−1 + c2(x − 2)−1 ln(x − 2). 40. (x + 1)2 y00 + (x + 1) y0 + y = 0, (x y(t − 1) = Y (t), dy dY y0 = = = dx dx d dy d 0 y00 = = Y dx dx dx

> −1). Let t = x + 1, x = t − 1, y(x) = dY dt dY = = Y 0; dt dx dt ==

dY 0 dt d2Y = Y 00. = dt dx dt2

The equation becomes t2 Y 00 + tY 0 + Y = 0, t > 0. Euler equation with α = 1, β = 1, indicial equation r2 + 1 = 0; indicial roots r = ±i. Hence the general solution Y = c1 cos(ln t) + c2 sin(ln t)

(t > 0),

and so y = c1 cos(ln(x + 1)) + c2 sin(ln(x + 1)) 41. We have y2 = y1

Z

e−

R

(x > −1).

p(x) dx

y12

dx.

Using the product rule for differentiation y20

=

y10

=

y10

Z

e−

R

p(x) dx

y12 Z

e−

R

dx + y1

p(x) dx

y12

dx +

e−

e−

R

p(x) dx

y12 R

p(x) dx

y1

.

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Appendix A

Ordinary Differential Equations: Review of Concepts and Methods

So W (y1 , y2 )

y 1 0 y1

=

=

y1 y10

=

e−

R

y1 y10 Z

R

e−

e−

R

R

R

R

p(x) dx

dx

y12

p(x) dx

y12

dx +

e−

p(x) dx −

y12 p(x) dx

e−

dx + e

R

p(x) dx

y1 R

p(x) dx

−

y10 y1

Z

e−

R

p(x) dx

y12

dx

> 0.

This also follows from Abel’s formula, (4), Section A.1. 42. We have

So W (y1 , y2 )

y1

=

xa cos(b ln x)

y10

=

axa−1 cos(b ln x) − bxa−1 sin(b ln x);

y2

=

xa sin(b ln x)

y20

=

axa−1 sin(b ln x) + bxa−1 cos(b ln x).

=

xa sin(b ln x) xa cos(b ln x) axa−1 cos(b ln x) − bxa−1 sin(b ln x) axa−1 sin(b ln x) + bxa−1 cos(b ln x)

=

ax2a−1 cos(b ln x) sin(b ln x) + bx2a−1a cos2 (b ln x) −ax2a−1 sin(b ln x) cos(b ln x) + bx2a−1 sin2 (b ln x)

=

bx2a−1 > 0.

This also follows from Abel’s formula, (4), Section A.1. 43 In the equation x2y00 + αy0 + βy = 0, let t = ln x, x = et , y(x) = y(et ) = Y (t). Then, dt dx

=

1 = e−t ; x

y0

=

dy 1 dY dY dt = = = Y 0 · = Y 0 · e−t ; dx dx dt dx x

y00

=

d d d 0 −t = e−t [Y 0 ] + Y 0 [e−t] Y ·e dx dx dx

= e−t

dY 0 dt dt + Y 0 [−e−t ] dt dx dx

= e−2t Y 00 − e−2tY 0. Substituting into the equation and simplifying and using x = et , x2 = e2t, we obtain d2Y dY + (α − 1) + βY = 0, dt2 dt which is a second order linear differential equation in Y with constant coefficients. 44. Using the change of variables t = ln x, we transformed Euler’s equation into Y 00 + (α − 1)Y 0 + βY = 0. Three cases can occur depending on the characteristic equation r2 + (1 − α)r + β = 0

Section A.3

Linear Ordinary Differential Equations with Nonconstant Coefficients

A433

(we are using r instead of the traditional λ to denote the characteristic roots. Case I: If (1 − α)2 − 4β > 0, then we have two distinct roots p p −1 + α + (1 − α)2 − 4 −1 + α − (1 − α)2 − 4 r1 = r2 = . 2 2 The general solution in this case is Y (t) = c1er1 t + c2er2 t . Using y(x) = Y (ln x), we get y(x) = Y (t) = c1 er1 ln x + c2 er2 ln x = c1 xr1 + c2 xr2 . Case II: If (1 − α)2 − 4β = 0, then we have one double root r=

−1 + α . 2

The general solution in this case is Y (t) = c1ert + c2 tert . Using y(x) = Y (ln x), we get y(x) = Y (t) = c1er ln x + c2(ln x)er ln x = c1xr + c2 xr ln x. Case III: If (1 − α)2 − 4β < 0, then we have two complex conjugate roots p p −1 + α + i −(1 − α)2 + 4 −1 + α − i −(1 − α)2 + 4 r1 = = a+ib r2 = = a−ib. 2 2 The general solution in this case is Y (t) = eat [c1 cos bt + c2 sin bt] . Using y(x) = Y (ln x), we get y(x) = Y (t) = ea ln x [c1 cos(b ln x) + c2 sin(b ln x)] = xa [c1 cos(b ln x) + c2 sin(b ln x)] . 45. (a) From Abel’s formula (Theorem 2, Section A.1), the Wronskian is y1 y20 − y10 y2 = Ce− ∫ p(x) dx , where y1 and y2 are any two solution of (2). (b) Given y1 , set C = 1 in (a) y1 y20 − y10 y2 = e− ∫ p(x) dx . This is a first-order differential equation in y2 that we rewrite as y10 y2 = e− ∫ p(x) dx . y1

y20 − The integrating factor is R

e

y0

− y1 dx 1

= e− ln y1 =

1 . y1

Multiply by the integrating factor: y20 y0 1 − 21 y2 = e− ∫ p(x) dx y1 y1 y1

A434

Appendix A

Ordinary Differential Equations: Review of Concepts and Methods

or

d y2 1 = e− ∫ p(x) dx dx y1 y1

Integrating both sides, we get y2 y1

=

y2

=

Z

1 − ∫ p(x) dx e dx; y1 Z 1 − ∫ p(x) dx dx, y1 e y1

which implies (3). 46. (a) Following the proof of (3), put y = vy1 . Then, y1 v00 + [2y10 + p(x)y1 ]v0 [2y10 + p(x)y1 ] 0 v y1 0 y1 00 v + 2 + p(x) v0 y1

v00 +

=

g(x)

=

g(x) y1

=

g(x) . y1

(b) To solve the first order differential equation in v0 (Theorem 1, Section A.1), we use the integrating factor R

e

y0 2 y1 +p(x) dx 1

R

= y12 e

p(x) dx

.

The equation becomes R y12 e p(x) dx v00

+

R y12 e p(x) dx

0 y1 2 + p(x) v0 y1

R

p(x) dx

g(x);

R

p(x) dx

g(x);

=

y1 e

d h 2 R p(x) dx 0 i = v y e dx 1

y1 e

R

y12 e

p(x) dx 0

v

=

v0

=

Z h R y1 e

p(x) dx

1 − R p(x) dx e y12 +c1

1 −R e y12

i g(x) dx + C1;

Z h R y1 e

p(x) dx

p(x) dx

i g(x) dx

.

Integrating once more, we get Z Z i 1 − R p(x) dx h R p(x) dx v = y1 e e g(x) dx y12 Z 1 − R p(x) dx +c1 dx; e y12 Z − ∫ p(x) dx e y2 = c1 y1 dx y12 Z Z − ∫ p(x) dx e ∫ p(x) dx +y1 dx y1 e g(x) dx dx y12 R where the last two occurrences of p(x) dx represent the same antiderivative of p(x).

Section A.3

Linear Ordinary Differential Equations with Nonconstant Coefficients

47. In solving the equation y00 − 4y0 + 3y = ex , the proof of the previous exercise. Let

A435

y1 = ex , we follow the steps in

y1 = ex , y2 = vex , y20 = v0 ex + vex , y200 = v00 ex + 2v0 ex + vex . Then y00 − 4y0 + 3y = ex

⇒ v00 ex + 2v0 ex + vex − 4(v0 ex + vex ) + 3vex = ex ⇒ v00 + 2v0 + v − 4(v0 + v) + 3v = 1 (divide by ex ) ⇒ v00 − 2v0 = 1.

As expected, we arrive at a first order o.d.e. in v0 . Its solution (using an integrating factor) is e−2x v00 − 2e−2x v0

= e−2x

d −2x 0 v e dx

= e−2x

e−2x v0

=

v0

=

Z

e−2x dx =

−1 −2x +C e 2

−1 + C e2x . 2

Integrating once more, v

−x + C1 e2x + C2 2

=

−x x e + C1 e3x + C2ex . 2 Note that the term C2ex is corresponds to the solution y1 . √ 48. x2y00 + 3xy0 + y = x, y1 = x1 . As in the previous exercise, let y2

y1 =

=

vy1 =

1 v0 x − v v0 v00 v 1 v v0 00 = , y = + 2 3. , y = v , y0 = − − 2 2 2 2 x x x x x x x x

Then x2y00 + 3xy0 + y =

√ √ v v v x ⇒ xv00 − 2v0 + 2 + 3v0 − 3 + = x x x x √ 00 0 ⇒ xv + v = x.

We arrive at a first order o.d.e. in v0 . Its solution (using an integrating factor) is d [xv0 ] = dx 0

=

v0

=

xv

√ x Z

√

x dx =

2 3/2 x +C 3

2 1/2 C x + . 3 x

Integrating once more, 4 3/2 x + C ln x + D 9

v

=

y

= vy1 =

4 1/2 ln x D x +C + . 9 x x

A436

Appendix A

Ordinary Differential Equations: Review of Concepts and Methods

49. 3 y00 + 13 y0 + 10 y = sin x,

y1 = e−x .

As in the previous exercise, let y1 = e−x , y = ve−x , y0 = v0 e−x − ve−x , y00 = v00 e−x − 2v0 e−x + ve−x .

Then 3 y00 + 13 y0 + 10 y = sin x

⇒ 3(v00e−x − 2v0 e−x + ve−x ) +13(v0 e−x − ve−x ) + 10ve−x = sin x ⇒ 3v00 + 7v0 = ex sin x 7 1 ⇒ v00 + v0 = ex sin x. 3 3

We now solve the first order o.d.e. in v0 : 7 e7x/3 v00 + e7x/3v0 = 3 h d 7x/3 0 i = v e dx e7x/3v0

v0

=

1 e7x/3 ex sin x 3 1 10x/3 sin x e 3 Z 1 1 10x/3 sin x dx e 3 3

=

1 e10x/3 10 ( sin x − cos) + C 3 ( 10 )2 + 1 3 3

=

ex 9 (10 sin x − cos) + C. 109 3

(We used the table of integrals to evaluate the preceding integral. We will use it again below.) Integrating once more,

v

y

Z

Z

=

10 109

=

10 ex 9 ex (sin x − cos x) − (cos x + sin x) + C 109 2 327 2

=

vy1 =

=

−

ex sin x dx −

9 327

ex cos x dx

10 9 (sin x − cos x) − (cos x + sin x) + Ce−x 218 654

13 7 cos x + sin x + Ce−x . 218 218

50. x y00 − (1 + x)y0 + y = x3 ,

y1 = ex . We have

y1 = ex , y = vex , y0 = v0 ex + vex , y00 = v00 ex + 2v0 ex + vex .

Section A.3

Linear Ordinary Differential Equations with Nonconstant Coefficients

A437

Then x (v00 ex + 2v0 ex + vex ) − (1 + x)(v0 ex + vex ) + vex = x3

⇒ xex v00 + xex v0 − ex v0 = x3 1 ex 00 1 ⇒ v + ex − 2 + v0 = x x x x

d ex 0 v dx x

=

x

ex 0 v x

=

Z

v0

=

x dx =

1 2 x +C 2

1 3 −x x e + Cxe−x. 2

Integrating once more, Z Z 1 v = x3 e−x dx + +C xe−x dx 2 = y

=

1 3 −x −x e − 3x2e−x − 6xe−x − 6e−x + C −xe−x − e−x 2 1 3 −x − 3x2 − 6x − 6 + C [−x − 1] . 2

A438

Appendix A

Ordinary Differential Equations: Review of Concepts and Methods

Solutions to Exercises A.4 1. Using the ratio test, we have that the series ∞ X

xm 5m + 1 m=0 converges whenever the limit , xm+1 5m + 1 xm lim |x| = |x| = lim m→∞ 5(m + 1) + 1 5m + 1 m→∞ 5m + 6 is less than 1. That is, |x| < 1. Thus, the interval of convergence is |x| < 1 or (−1, 1). Since the series is centered at 0, the radius of convergence is 1. 2. Using the ratio test, we have that the series ∞ X 3m xm m! m=0

converges whenever the following limit is less than 1: , 3m+1 xm+1 m! 3m xm lim = lim 3 |x| m→∞ (m + 1)! m→∞ m! (m + 1)! =

3 |x| lim

m→∞

1 = 3 |x| · 0 = 0. m+1

Since the limit is always less than 1, the series converges for all x. It is cantered at 0 and has radius of convergence R = ∞. 3. Using the ratio test, we have that the series ∞ X (−1)m xm+2 2m m=0

converges whenever the limit , (−1)m+1 x(m+1)+2 (−1)m xm+2 |x| lim = m→∞ 2m+1 2m 2 is less than 1. That is, 12 |x| < 1 or |x| < 2. Thus, the interval of convergence is (−2, 2). Since the series is centered at 0, the radius of convergence is 2. 4. Using the ratio test, we have that the series ∞ X

(x − 2)m

m=0

converges whenever the limit , m+1 m lim (x − 2) (x − 2) = |x − 2| m→∞ is less than 1. That is, |x − 2| < 1. This is an interval centered at 2 with radius 1. Thus interval of convergence is (−1, 3). 5. Using the ratio test, we have that the series ∞ X m m xm m! m=1

Section A.4

The Power Series Method, Part I

A439

converges whenever the following limit is < 1: , (m + 1)m+1 xm+1 (m + 1)m (m + 1) mm xm m! lim · m m · |x| = lim m→∞ m→∞ (m + 1)! m! (m + 1)! m x =

|x| lim

m→∞

m+1 m

m

= e|x|.

We have used the limit lim

m→∞

m+1 m

m

m 1 1+ =e m→∞ m

= lim

(see the remark at the end of the solution). From |x|e < 1 we get |x| < 1/e. Hence the interval of convergence is (−1/e, 1/e). It is centered at 0 and has radius 1/e. One way to show m m+1 =e lim m→∞ m is to show that the natural logarithm of the limit is 1: m m+1 m+1 ln = m ln = m [ln(m + 1) − ln m] . m m By the mean value theorem (applied to the function f(x) = ln x on the interval [m, m + 1]), there is a real number cm in [m, m + 1] such that ln(m + 1) − ln m = f 0 (cm ) = Note that

1 1 1 ≤ . ≤ m+1 cm m

So As m → ∞,

1 cm

m m+1

m ≤ m [ln(m + 1) − ln m] ≤ 1. m+1 → 1, and so by the sandwich theorem, m [ln(m + 1) − ln m] → 1.

Taking the exponential, we derive the desired limit. 6. Using the ratio test, we have that the series ∞ X (m − 1)(m + 3) xm m m=1

converges whenever the limit , (m + 1 − 1)(m + 1 + 3) xm+1 (m − 1)(m + 3) xm lim m→∞ m+1 m =

|x| lim

m→∞

m(m + 4) m · = |x| m+1 (m − 1)(m + 3)

is less than 1. That is, |x| < 1. Thus the interval convergence is (−1, 1). It is centered at 0 and has radius 1. 7. Using the ratio test, we have that the series ∞ X xm ln m m=2

A440

Appendix A

Ordinary Differential Equations: Review of Concepts and Methods

converges whenever the limit , xm+1 xm lim = m→∞ ln(m + 1) ln m

|x| lim

m→∞

ln m = |x| ln(m + 1)

is less than 1 (see the note at the end of the solution). That is, |x| < 1. Thus the interval convergence is (−1, 1). It is centered at 0 and has radius 1. To show that ln m = 1, m→∞ ln(m + 1) lim

you can use l’Hospital rule. Differentiating numerator and denominator, we obtain lim

m→∞

ln m 1/m m+1 = lim = lim = 1. m→∞ m→∞ ln(m + 1) 1/(m + 1) m

8. Using the ratio test, we have that the series ∞ X (x + 1)m m2 m=1

converges whenever the limit , (x + 1)m+1 (x + 1)m lim m→∞ (m + 1)2 m2

m2 = |x + 1| m→∞ (m + 1)2

= |x + 1| lim

is less than 1. That is, |x − (−1)| < 1. Thus the interval convergence is (−2, 0). It is centered at −1 and has radius 1. 9. Using the ratio test, we have that the series ∞ X [10(x + 1)]2m (m!)2 m=1

converges whenever the following limit is < 1: , [10(x + 1)]2(m+1) [10(x + 1)]2m (m!)2 2 2 lim = 10 |x + 1| lim m→∞ m→∞ ((m + 1)!)2 ((m + 1)!)2 (m!)2 = 102 |x + 1|2 lim

(m!)2 (m + 1)2 (m!)2

= 102 |x + 1|2 lim

1 = 0. (m + 1)2

m→∞

m→∞

Thus the series converges for all x, R = ∞. 10. Using the ratio test, we have that the series ∞ X (9 x)2m (m!)2 m=0

converges whenever the following limit is < 1: , (9 x)2(m+1) (9 x)2m (m!)2 2 2 lim = 9 |x| lim m→∞ ((m + 1)!)2 m→∞ ((m + 1)!)2 (m!)2 =

92|x|2 lim

(m!)2 (m + 1)2 (m!)2

=

92|x|2 lim

1 = 0. (m + 1)2

m→∞

m→∞

Section A.4

The Power Series Method, Part I

A441

Thus the series converges for all x, R = ∞. 11. Using the ratio test, we have that the series ∞ X 2m (x − 2)m 10m m=0

converges whenever the following limit is < 1: , 2m+1 (x − 2)m+1 2m (x − 2)m lim = m→∞ 10m+1 10m

2 |x − 2|. 10

Thus |x − 2| < 5. Hence the interval of convergence is centered at 2 and has radius 5; that is, (−3, 7) 12. Using the ratio test, we have that the series ∞ X m! (x − 2)m 10m m=0

converges whenever the following limit is < 1: , (m + 1)! (x − 2)m+1 m! (x − 2)m lim = m→∞ 10m+1 10m

|x − 2| lim

m→∞

m+1 =∞ 10

if x 6= 2. Hence the series converges only at its center 2 and diverges otherwise. Its radius of convergence is 0. 13. We use the geometric series. For |x| < 1, 3−x 1+x

1+x 4 + 1+x 1+x

=

−

=

−1 +

=

4 1 − (−x)

−1 + 4

∞ X

(−1)nxn .

n=0

14. Differentiate the geometric series ∞ X 1 (−1)n xn = 1 − (−x) n=0

(|x| < 1)

term-by-term and you get ∞ X −1 = (−1)n nxn (1 + x)2 n=0

(|x| < 1).

Multiply by −x both sides, ∞ X 1 = (−1)n+1 nxn+1 (1 + x)2 n=0

(|x| < 1).

Replace n by n − 1 and shift the index up by 1: ∞ X 1 = (−1)n (n − 1)xn (1 + x)2 n=1

(|x| < 1).

15. Start with the geometric series ∞ X 1 (−1)n xn = 1 − (−x) n=0

(|x| < 1).

A442

Appendix A

Ordinary Differential Equations: Review of Concepts and Methods

Replace x by x2 : ∞ X 1 = (−1)n x2n 1 + x2 n=0

(|x2| < 1 or |x| < 1).

Differentiate term-by-term: ∞ X −2x = (−1)n (2n)x2n−1 (1 + x2)2 n=1

(|x| < 1).

Divide by −2 both sides, ∞ X x = (−1)n+1nx2n−1 (1 + x2)2 n=1

(|x| < 1).

16. Using a geometric series, ∞ X 1 = x2n 1 − x2 n=0

(|x| < 1).

So for |x| < 1, x+2 1 − x2

=

x 2 + 1 − x2 1 − x2

=

x

∞ X

x2n + 2

n=0 ∞ X

=

∞ X

x2n

n=0

x2n+1 + 2

n=0

=

∞ X

∞ X

x2n

n=0

ak xk ,

k=0

where a0 = 3, a2k = 2, a2k+1 = 1. 17. Use the Taylor series ex = Then 2

eu =

∞ X xn n! n=0

− ∞ < x < ∞.

∞ X u2n n! n=0

− ∞ < u < ∞.

Hence, for all x, 2

e3x

√ +1

3x)2

=

e · e(

=

e

∞ √ X ( 3x)2n n! n=0

=

e

∞ X 3nx2n . n! n=0

18. For |x| < 1, ln(1 + x) =

∞ X (−1)m+1 xm ; m m=1

Section A.4

The Power Series Method, Part I

A443

so ∞ ∞ X X (−1)m+1 xm+1 (−1)m xm x ln(1 + x) = = . m m−1 m=1 m=2

19. Using the sine series expansion, we have for all x x sin x cos x =

∞ ∞ X X x (2x)2n+1 22n+1 2n+1 (−1)n (−1)n . sin(2x) = = x 2 (2n + 1)! n=0 (2n + 1)! n=0

20. Using the sine and cosine series expansions, we have for all x x sin(x + 1)

= x [sin x cos 1 + cos x sin 1] = x cos 1

∞ X

(−1)n

n=0

= cos 1

∞ X

(−1)n

n=0

=

∞ X

∞ X x2n+1 x2n (−1)n + x sin 1 (2n + 1)! (2n)! n=0

∞ X x2n+2 x2n+1 (−1)n + sin 1 (2n + 1)! (2n)! n=0

(−1)k ak xk ,

k=0

where, for k ≥ 0, a2k+1 =

sin 1 (2k)!

and

a2k+2 =

cos 1 . (2k + 1)!

21.We have 1 1 1 = , = 3x 2 + 3x 2(1 − u) 2(1 − (− 2 )) where u = − 3x 2 . So n X ∞ ∞ ∞ 1 3x 3n 1X n 1X − u = = (−1)n n+1 xn. = 2 + 3x 2 n=0 2 n=0 2 2 n=0 The series converges if |u| < 1; that is |x| < 23 . 22. We have 1+x 2 − 3x

where u =

3x 2 .

1 2 − 3x 1 5 + 3 2 − 3x 3 2 − 3x

=

−

=

1 1 5 − + 3 3 2(1 −

=

1 5 1 − + , 3 61−u

3x ) 2

So n ∞ ∞ 1 1 5 X 3x 1 5X n u =− + . =− + 2 + 3x 3 6 n=0 3 6 n=0 2

The series converges if |u| < 1; that is |x| < 23 .

A444

Appendix A

Ordinary Differential Equations: Review of Concepts and Methods

23. We have 3

x + x3 2 + 3x2

(3x2 + 2 + 1) 3x2 + 2 x = x+ 2 3x + 2

= x

= x+

x 2(1 − (− 32 x2 )

= x+

x , 2(1 − u)

where u = − 32 x2. So 3

x + x3 2 + 3x2

∞ xX n u 2 n=0

=

x+

=

n ∞ 3 2 xX − x x+ 2 n=0 2

=

x+

∞ X

(−1)n

n=0

The series converges if |u| < 1; that is x2