FOURIER SERIES METHODS AND PARTIAL DIFFERENTIAL …cbafaculty.org/Analysis_Math/DE_C_and_M_Chapter_9_sc.pdf · The basic trigonometric functions cos(t) and sin(t) have period P =

527 Copyright © 2015 Pearson Education, Inc.

CHAPTER 9

FOURIER SERIES METHODS AND PARTIAL DIFFERENTIAL EQUATIONS

SECTION 9.1 PERIODIC FUNCTIONS AND TRIGONOMETRIC SERIES The basic trigonometric functions cos(t) and sin(t) have period P = 2, so the sine or cosine of t (as in Problems 1–4) completes its first period when 2 ;t hence 2 / .P 1. Smallest period P = 2/3 (left-hand figure below) 2. Smallest period P = 1 (right-hand figure above) 3. Smallest period P = 4/3 (left-hand figure below) 4. Smallest period P = 6 (right-hand figure above) However, the basic tangent and cotangent functions have period (instead of 2), so /P in Problems 5 and 6. 5. Smallest period P = ; see the left-hand figure at the top of the next page. 6. Smallest period P = 1/2; see the right-hand figure at the top of the next page.

�Π Πt

�1

1

�1 1 2t

�1

1

�2 Π 2 Π 4 Πt

�1

1

�6 6 12t

�1

1

528 PERIODIC FUNCTIONS AND TRIGONOMETRIC SERIES

Copyright © 2015 Pearson Education, Inc.

The hyperbolic sine and cosine functions of Problems 7 and 8 are steadily increasing (for t > 0), and hence are not periodic. 7. Not periodic (left-hand figure below) 8. Not periodic (right-hand figure above) 9. Smallest period P = (left-hand figure below) 10. Smallest period P = /3 (right-hand figure above) 11. With f(t) = 1 the integral formulas of Eqs. (16) and (17) in the text give a0 = 2 and

an = bn = 0 for n 0. Thus the Fourier series of f is the single term series f(t) = 1.

-1 1 2t

-5

5

-3 3t

1000

-2 2t

-300

300

�3 Π

2�Π

2

Π

2

3 Π

2

5 Π

2

t

�5

5

�Π Π 2 Π 3 Πt

1

�Π Πt

1

�2 Π 2 Π 4 Πt

1

Section 9.1 529


In Problems 12–13, 18–19, 23, and 26 the function ( )f t is defined by one formula on the

interval (–, 0) and by another formula on (0, ). The coefficient integrals must therefore be split accordingly, and the appropriate formula substituted in each integral:

0

0 0

1 1( ) ( ) ,a f t dt f t dt

0

0

1 1( ) cos ( ) cos ,na f t nt dt f t nt dt

(n > 0)

0

0

1 1( ) sin ( ) sin .nb f t nt dt f t nt dt

12. 0

0 0

1 1( 3) ( 3) 0a dt dt

0

0

1 1( 3) cos ( 3) cos 0na nt dt nt dt

0

0

1 1( 3) sin ( 3) sinnb nt dt nt dt

0 for even6cos 1

12 / for odd

nn

n nn

12 sin sin 3 sin5 sin 7

( )1 3 5 7

t t t tf t

�Π Π 3 Π 5 Πt

�3

3

13. 0

0 0

1 1(0) (1) 1a dt dt

0

0

1 1 sin(0) cos (1) cos 0n

na nt dt nt dt

n

0

0

1 1(0) sin (1) sinnb nt dt nt dt

0 for even1 cos

2 / for odd

nn

n nn



1 2 sin sin 3 sin5 sin 7

( )2 1 3 5 7

t t t tf t

(figure below)

�Π Π 3 Π 5 Πt

1

14. 0

0 0

1 1(3) ( 2) 1a dt dt

0

0

1 1 sin(3) cos ( 2) cos 0n

na nt dt nt dt

n

0

0

1 1(3) sin ( 2) sinnb nt dt nt dt

0 for even5(cos 1)

10 / for odd

nn

n nn

1 10 sin sin 3 sin5 sin 7

( )2 1 3 5 7

t t t tf t

(figure below)

�Π Π 3 Π 5 Πt

�2

3

15. 0

10,a t dt

1cos 0na t nt dt

1

sinnb t nt dt

2

2 / for even2sin 2 cos

2 / for odd

n nn n n

n nn

sin sin 2 sin 3 sin 4

( ) 21 2 3 4

t t t tf t

(figure at top of next page)

Section 9.1 531


�Π Π 3 Π 5 Πt

�Π

Π

16. 2

0 0

12a t dt

2

20

1 cos2 2 sin 2 1cos 0n

n n na t nt dt

n

2

0

1sinnb t nt dt

2

2sin 2 2 cos2 2n n n

n n


( ) 21 2 3 4

t t t tf t

(figure below)

�Π Π 3 Π 5 Πt

2 Π

17. 0

0 0

1 1( ) ( )a t dt t dt

0

0

1 1( ) cos ( ) cosna t nt dt t nt dt

22

0 for even2(cos sin 1)

4 / for odd

nn n n

n nn

0

0

1 1( ) sin ( ) sin 0nb t nt dt t nt dt

4 cos cos3 cos5 cos7

( )2 1 9 25 49

t t t tf t

See the figure at the top of the next page.



�Π Π 3 Π 5 Πt

Π

18. 0

0 0

1 1( ) ( )a t dt t dt

0

0

1 1( ) cos ( ) cosna t nt dt t nt dt

22

0 for even2(1 cos )

4 / for odd

nn

n nn

0

0

1 1( ) sin ( ) sin 0nb t nt dt t nt dt


( )2 1 9 25 49

t t t tf t

(figure below)

�Π Π 3 Π 5 Πt

Π

19. 0

0 0

1 1( ) (0)

2a t dt dt

0

0

1 1( ) cos (0) cosna t nt dt nt dt

22

0 for even1 cos

2 / for odd

nn

n nn

0

20

1 1 sin 1( ) sin (0) sinn

n nb t nt dt nt dt

n n

Section 9.1 533


2 cos cos3 cos5 cos7 sin sin 2 sin 3 sin 4

( )1 9 25 49 1 2 3 44

t t t t t t t tf t

�2 Π 2 Π 4 Πt

Π

20. / 2

0 / 2

11 1a dt

/ 2

/ 2

0 for even1 2

cos sin 1 fo 1, 5,2

1 fo 3, 7,n

nn

a nt dt r nn

r n

/ 2

/ 2

1sin 0nb nt dt

1 2 cos cos3 cos5 cos7

( )2 1 3 5 7

t t t tf t

(figure below)

�Π Π 2 Π 3 Π 4 Πt

1

21. 2

20

1 2

3a t dt

22 2

23 2

4 / for even1 4 cos 2( 2)sincos

4 / for evenn

n nn n n na t nt dt

n n n

21sin 0nb t nt dt




( ) 41 4 9 163

t t t tf t

(figure below)

2 Π 4 Π 6 Πt

10

22. 22 2

0 0

1 8

3a t dt

2 22 2

3 20

1 4 cos2 2( 1)sin 2 4cosn

n n n na t nt dt

n n

2 22 2

30

1 (2 4 )cos2 4 sin 2 2 4sinn

n n n nb t nt dt

n n

24 cos cos2 cos3 cos44( )

3 1 4 9 16


1 2 3 4

t t t tf t

t t t t

(figure below)

�Π Π 3 Π 5 Πt

40

23. 2

20 0

1

3a t dt

2 2

23 20

1 2 cos ( 2)sin 2( 1)cos

n

n

n n n na t nt dt

n n

Section 9.1 535


2 2

0

1sinnb t nt dt

2 2

2 2 33

/ for even(2 )cos 2 sin 2

( 4) / for odd

n nn n n n

n n nn

2 cos cos2 cos3 cos4( ) 2 (figure below)

6 1 4 9 16

sin sin 2 sin 3 sin 4 4 sin sin 3 sin5 sin 7

1 2 3 4 1 27 125 343

t t t tf t

t t t t t t t t

2 Π 4 Πt

10

The trigonometric identities 2 cos A cos B = cos(A + B) + cos(A B) 2 sin A cos B = sin(A + B) + sin(A B) 2 sin A sin B = cos(A B) cos(A + B) are needed to evaluate the integrals that appear in Problems 2426.

24. 0

0 0

1 1 4( sin ) (sin )a t dt t dt

0

0

2

2

1 1( sin ) cos (sin ) cos

4 / ( 1) for even2(1 cos )

(1 ) 0 for odd

na t nt dt t nt dt

n nn

n n

0

0

1 1( sin ) sin (sin ) sin 0nb t nt dt t nt dt

2 4 cos2 cos4 cos6 cos8

( )1 15 35 63

t t t tf t

(figure at top of next page)



�Π Π 3 Πt

1

25. In order to evaluate the coefficient integrals in Eqs. (16) and (17) of the text we would need the trigonometric identity

2 1cos 2 1 cos4

2t t

which, however, tells us in advance that the coefficients in the Fourier series of f(t) = cos22t are given by a0 = 1, a4 = 1/2, an = 0 otherwise, and bn = 0 for all n 1.

�Π Π 3 Πt

1

26. 0 0

1 2(sin )a t dt

1 0

1sin cos 0a t t dt

2

20

2 / ( 1) for even1 1 cos(sin ) cos

(1 ) 0 for 1 oddn

n nna t nt dt

n n

21 0

1 1sin

2b t dt

20

1 sin(sin ) sin 0 for 1

(1 )n

nb t nt dt n

n

1 1 2 cos2 cos4 cos6 cos8

( ) sin2 1 15 35 63

t t t tf t t

Section 9.1 537


Note that f(t) = (sin t + |sin t|)/2, so this answer agrees with the answer to Problem 24.

�Π Π 3 Π 5 Πt

SECTION 9.2 GENERAL FOURIER SERIES AND CONVERGENCE

1. 0 3

0 3 0

1 1( 2) (2) 0

3 3a dt dt

0 3

3 0

1 1( 2) cos (2) cos 0

3 3 3 3n

n t n ta dt dt

0 3

3 0

1 1 4(1 cos ) 4( 2) sin (2) sin 1 ( 1)

3 3 3 3n

n

n t n t nb dt dt

n n

8 1 3 1 5 1 7

( ) sin sin sin sin3 3 3 5 3 7 3

t t t tf t

(figure below)

-3 3 6 9 12 15t

-2

2

2. 5

0 0

1(1) 1

5a dt ,

5

0

1 sin(1) cos 0

5 5n

n t na dt

n

5

0

1 1 cos 1 ( 1)(1) sin

5 5

n

n

n t nb dt

n n

538 GENERAL FOURIER SERIES AND CONVERGENCE


1 2 1 3 1 5 1 7

( ) sin sin sin sin2 5 3 5 5 5 7 5

t t t tf t

(figure below)

-5 5 10 15 20 25t

3. 0 2

0 2 0

1 1(2) ( 1) 1

2 2a dt dt

0 2

2 0

1 1 sin(2) cos ( 1) cos 0

2 2 2 2n

nt nt na dt dt

n

0 2

2 0

1 1 3(cos 1) 3(2) sin ( 1) sin ( 1) 1

2 2 2 2n

n

nt nt nb dt dt

n n

1 6 1 3 1 5 1 7

( ) sin sin sin sin2 2 3 2 5 2 7 2

t t t tf t

(figure below)

�2 Π 2 Π 6 Π 10 Πt

4. 2

0 2

10

2a t dt

,

2

2

1cos 0

2 2n

n ta t dt

2 1

2 22

1 4(sin cos ) 4( 1)sin

2 2

n

n

n t n n nb t dt

n n

4 1 2 1 3 1 4

( ) sin sin sin sin2 2 2 3 2 4 2

t t t tf t

See figure at top of next page.

Section 9.2 539


-2 2 6 10t

-2

2

5. 2

0 2

10

2a t dt

,

2

2

1cos 0

2 2n

nta t dt

2 1

22

1 4(sin cos ) 4( 1)sin

2 2

n

n

nt n n nb t dt

n n

1 2 1 3 1 4

( ) 4 sin sin sin sin2 2 2 3 2 4 2

t t t tf t

(figure below)

�2 Π 2 Π 6 Π 10 Πt

�2 Π

2 Π

6. 3

0 0

23

3a t dt

3

2 20

3 cos2 2 sin 2 12 2cos 0

3 3 2n

n n nn ta t dt

n

3 1

2 20

2 2 3sin 2 6 cos2 3( 1)sin

3 3 2

n

n

n t n n nb t dt

n n

3 3 2 1 4 1 6 1 8

( ) sin sin sin sin2 3 2 3 3 3 4 3

t t t tf t

(figure below)

-3 3 6 9 12 15

t

3



7. 0

0 1 0( ) ( ) 1a t dt t dt

0 1

2 2 2 21 0

2 cos sin 1 2( ) cos cos ( 1) 1n

n

n n na t n t dt t n t dt

n n

0 1

1 0( ) sin sin 0nb t n t dt t n t dt

2

1 4 1 1 1( ) cos cos3 cos5 cos7

2 9 25 49f t t t t t

(figure below)

-1 1 2 3 4 5t

1

8. 2

0 1

2 21

3 3a dt

2

1

2 2 1 4 2cos sin sin

3 3 3 3n

n t n na dt

n

2

1

2 2 1 2 4sin cos cos

3 3 3 3n

n t n nb dt

n

Analyzing separately the cases n = 3k, n = 3k + 1, and n = 3k + 2, we find that

a3k = 0, a3k+1 = 3 /n, a3k+2 = + 3 /n, and that bn = 0 for all n. Hence the Fourier series of f(t) is

1 3 2 1 4 1 8 1 10 1 14

( ) cos cos cos cos cos3 3 2 3 4 3 5 3 7 3

t t t t tf t

.

3 6 9t

1

Section 9.2 541


9. 1 2

0 1

2

3a t dt

2 21 2

3 3 2 21

4 cos 2( 2)sin 4( 1)cos

n

n

n n n na t n t dt

n n

1 2

1sin 0nb t n t dt

2

1 4 1 1 1( ) cos cos2 cos3 cos4

3 4 9 16f t t t t t

(figure below)

-1 1 2 3 4 5t

1

10. 2 2

0 0

1 4

2 3a t dt

2 2 2

23 3 2 2

0

1 8 cos ( 2)sin 8( 1)cos

2 2

n

n

n t n n n na t dt

n n

2

2

0

1sin

2 2n

n tb t dt

2 2

3 33 3

4 ( 2)cos 2 sin 2 4 / for even

4 / 16 / for odd

n n n n n n

n n nn

2

2 8 1 2 1 3 1 4( ) cos cos cos cos

3 2 4 2 9 2 16 2

t t t tf t

4 1 2 1 3

sin sin sin2 2 2 3 2

t t t

3

16 1 3 1 5sin sin sin

2 27 2 125 2

t t t




-2 2 4 6 8 10t

2

4

To calculate the Fourier coefficients in Problems 11–14 we use the trigonometric identities for sin A cos B and sin A sin B that are listed above in Section 9.1 (prior to Problems 24–26 there).

11. 1

01

4cos

2

ta dt

1 1

2 21

4cos 4( 1)cos cos

2 (4 1) (4 1)

n

n

t na n t dt

n n

1

1

cos sin 02n

tb n t dt

2 4 1 1 1 1

( ) cos cos2 cos3 cos43 15 35 63

f t t t t t

(figure below)

-1 1 2 3 4 5t

1

12. 1

00

42 sina t dt

21

2 20

4cos 42 sin cos2

(4 1) (4 1)n

na t n t dt

n n

1

20

4cos sin2 sin sin 2 0

(4 1)n

n nb t n t dt

n

Section 9.2 543


2 4 1 1 1 1

( ) cos2 cos4 cos6 cos83 15 35 63

f t t t t t

-1 1 2t

1

13. 1

00

2sina t dt

1

2 20

1 cos 1 ( 1)sin cos for 1

( 1) ( 1)

n

n

na t n t dt n

n n

1

1 0sin cos 0a t t dt

1

20

sinsin sin 0 for 1

( 1)n

nb t n t dt n

n

1 2

1 0

1sin

2b t dt

1 1 2 1 1 1 1

( ) sin cos2 cos4 cos6 cos82 3 15 35 63

f t t t t t t

-1 1 2 3 4 5t

1

14. 2

0 0

1sin 0

2a t dt

2

2 20

2 ( 1) 11 2(cos 1)sin cos for 2

2 2 ( 4) ( 4)

n

n

nt na t dt n

n n

2

2 0

1sin cos 0

2a t t dt



2

20

1 2sinsin sin 0 for 2

2 2 ( 4)n

nt nb t dt n

n

2 2

1 0

1 1sin

2 2b t dt

1 4 1 1 3 1 5 1 7

( ) sin cos cos cos cos2 3 2 5 2 21 2 45 2

t t t tf t t

(figure below)

�2 Π 4 Π 8 Πt

�1

1

15. (a) 2 2

20

0

1 8

3a t dt

2 22 2

3 20

1 4 cos2 2(2 1)sin 2 4cosn

n n n na t nt dt

n n

2 22 2

30

1 (2 4 )cos2 4 sin 2 2 4sinn

n n n nb t nt dt

n n

2

21 1

4 cos sin( ) 4 4

3 n n

nt ntf t

n n

(figure below)

2 Π 4 Π 6 Πt

2 Π2

4 Π2

(b) If we substitute t = 0 in the Fourier series of part (a) and note that

2 2 21 12 2(0) (0 ) (0 ) (2 ) (0) 2 ,f f f we get

Section 9.2 545


2 2

22 2

1 1

4 1 12 4 , so .

3 6n nn n

When we substitute t = and f() = 2 in the series of part (a) we get

2 1 2

22 2

1 1

4 ( 1) ( 1)4 , so .

3 12

n n

n nn n

16. (a) 1

00

1

2a t dt

1

2 2 2 20

cos sin 1 ( 1) 1cos

n

n

n n na t n t dt

n n

11

2 20

sin cos ( 1)sin

n

n

n n nb t n t dt

n n

1

2 2odd 1

1 2 cos 1 ( 1) sin( )

4

n

n n

n t n tf t

n n

(figure below)

1 2 3 4 5t

1

(b) Substitution of t = 0, ( ) 0f t in this series immediately gives 2

2odd

1.

8n n

17. (a) 2

0 02a t dt

2

2 20

cos2 2 sin 2 1cos 0n

n n na t n t dt

n

2

2 20

sin 2 2 cos2 2sinn

n n nb t n t dt

n n

1

2 sin( ) 1

n

n tf t

n

(see figure on next page)

(b) Substitution of t = 1/2, ( ) 1/ 2f t in this series gives



1 2 1 1 1 1 1 1

1 1 , so 1 .2 3 5 7 3 5 7 4

2 4 6t

1

2

The most efficient approach to Problems 18 and 20 is to derive first the expansions

sin 2 sin 3 sin 4

2 sin2 3 4

t t tt t

,

22 4 cos2 cos3 cos4

4 cos3 4 9 16

sin 2 sin 3 sin 44 sin .

2 3 4

t t tt t

t t tt

for 0 < t < 2, as the Fourier series of the functions f(t) and g(t) of period 2 defined for 0 < t < 2 by f(t) = t and g(t) = t2. The first series above yields the series in Problem 18, and a combination of the two yields the series in Problem 20. The expansions in Problems 19 and 21 are valid on the interval < t < rather than the interval 0 < t < 2. When we calculate the Fourier series of the functions f(t) and g(t) of period 2 defined for < t < by f(t) = t and g(t) = t2, we find that

sin 2 sin 3 sin 4

2 sin ,2 3 4

t t tt t

2

2 cos2 cos3 cos44 cos

3 4 9 16

t t tt t

if < t < .

18. First we derive the the Fourier series of the function f(t) of period 2 defined for 0 < t < 2 by f(t) = t.

Section 9.2 547


2

0 0

12a t dt

2

20

1 cos2 2 sin 2 1cos 0n

n n na t nt dt

n

2

20

1 sin 2 2 cos2 2sinn

n n nb t nt dt

n n

1

sin( ) 2

n

ntf t

n

Hence 1

1 sin( )

2 2 n

t ntf t

n

for 0 2t (figure below).

2 Π 4 Π 6 Πt

�Π

2

Π

2

19. 0

10

2

ta dt

,

1cos 0

2n

ta nt dt

1

2

1 sin cos ( 1)sin

2

n

n

t n n nb nt dt

n n

1

1

( 1) sin( )

2

n

n

t ntt

n

(figure below)

�Π Π 3 Π 5 Πt

�Π

2

Π

2

20. First we derive the the Fourier series of the function g(t) of period 2 defined for 0 < t < 2 by g(t) = t2.



2 2

20

0

1 8

3a t dt

2 22 2

3 20

1 4 cos2 2(2 1)sin 2 4cosn

n n n na t nt dt

n n

2 22 2

30

1 (2 4 )cos2 4 sin 2 2 4sinn

n n n nb t nt dt

n n

2

21 1

4 cos sin( ) 4 4

3 n n

nt ntg t

n n

If ( )f t is the function of Problem 18, then for 0 2t we have

2 2 23 6 2 1

( ) ( )12 4 2 6

t tg t f t

2 2

2 21 1 1 1

1 4 cos sin sin cos4 4 2 .

4 3 2 6n n n n

nt nt nt nt

n n n n

2 Π 4 Π 6 Πt

�Π2

12

Π2

6

21. 2

20

1

3a t dt

,

2 2

23 2

1 4 cos 2( 1)sin 4( 1)cos

n

n

n n n na t nt dt

n n

21sin 0nb t nt dt

2

22

1

( 1) cos4 ( )

3

n

n

ntt t

n

2 2 2 2 1

2 21 1

3 1 ( 1) cos ( 1) cos4

12 12 4 3

n n

n n

t nt nt

n n

(figure on next page)

Section 9.2 549


�Π Π 3 Π 5 Πt

�Π2

6

Π2

12

24. 2 4

40

0

1 32

5a t dt

2 4

0

1cosna t nt dt

2 2 4 4 2 2 2

5 2 4

8 2 2 3 cos2 (2 6 3)sin 2 2 316

n n n n n n

n n n

2 4

0

1sinnb t nt dt

4 4 2 2 3 3 2

5 2

8 (2 6 3)cos2 (6 4 )sin 2 3 316

n n n n n n

n n n

4 2 2

2 4 21 1

16 2 3 3( ) 16 cos 16 sin

5 n n

f t nt ntn n n n

(figure below)

2 Π 4 Π 6 Πt

8 Π4

16 Π4

(b) When we substitute t = 0, 4(0) 8f in the series of part (a) we get

4 4 2

4 2 22 4 4

1 1 1

16 1 1 16 18 32 48 32 48 .

5 5 6n n nn n n



We now solve readily for 4 4

11/ / 90.

nn

Similarly, we find that

1 4 4

1( 1) / 7 / 720n

nn

by substituting 4, ( )t f in the series of part (a).

Finally, addition of the first two series stated in part (b) yields the third one.

25. Now we want to sum the alternating series

3 3 3 3

1 1 1 11

3 5 7 9

of reciprocals of odd cubes. Having used a Fourier series of t4 in Problem 24 to evaluate (1/n4), it is natural to look at a Fourier series of t3. Let f(t) be the period 2 function with f(t) = t3 if < t < . We calculate the Fourier coefficients of f(t), and get

30

10,a t dt

31

cos 0na t nt dt

2 2 2 2 2

34 3

2 6 cos 6( 2)sin1 6sin 2n

n n n n nb t nt dt

n n n

3 2 1 1

31 1

sin sin2 ( 1) 12 ( 1) .n n

n n

nt ntt

n n

(figure below)

�Π Π 3 Π 5 Πt

�Π3

Π3

If we substitute t = /2 and use Leibniz's series (1)n+1/n = /4 of Problem 17 we find that

3

3 3 3 3

1 1 1 11 .

3 5 7 9 32

There is no value of t whose substitution in the Fourier series of f(t) = t3 yields the

series (1/n3) containing the reciprocal cubes of both the odd and even integers. Indeed, the summation in "closed form" of the series

3 3 3 3 3

1

1 1 1 1 11

2 3 4 5n n

Section 9.2 551


is a problem that has challenged many fine mathematicians since the time of Euler. Only in modern times (by R. Apery in 1978) has it been shown that this sum is an irrational number. For a delightful account of this work, see the article "A Proof that Euler Missed . . . An Informal Report" by Alfred van der Poorten in the The Mathematical Intelligencer, Volume 1 (1979), pages 195203.

SECTION 9.3 FOURIER SINE AND COSINE SERIES

1. 00

21 2a dt

, 0

2 2sincos 0n

na nt dt

n

Cosine series: ( ) 1f t

�Π Π 2 Π 3 Π 4 Πt

1

0

2 1 cos2 2sin 1 ( 1)n

n

nb nt dt

n n

Sine series: 4 1 1 1

( ) sin sin 3 sin5 sin 73 5 7

f t t t t t

�Π Π 3 Π 5 Πt

�1

1

552 FOURIER SINE AND COSINE SERIES


2. 1

0 02 (1 ) 1a t dt

1 2

2 2 2 20

2(1 cos ) 22 (1 )cos 1 ( 1)n

na t n t dt

n n

Cosine series: 2 2 2 2

1 4 cos3 cos5 cos7( ) cos

2 3 5 7

t t tf t t

-1 1 3 5t

1

1

2 20

2 sin 22 (1 )sinn

n nb t n t dt

n n

Sine series: 2 sin 2 sin 3 sin 4

( ) sin2 3 4

t t tf t t

-1 1 3 5t

1

3. 2

0 0(1 ) 0a t dt

2

22 2 2 2

0

4 4cos 2 sin 4(1 )cos 1 ( 1)

2n

n t n n na t dt

n n

Cosine series: 2 2 2 2

8 1 3 1 5 1 7( ) cos cos cos cos

2 3 2 5 2 7 2

t t t tf t

See figure at top of next page.

Section 9.3 553


-2 2 4 6 8t

-1

1

2

2 20

2 (1 cos ) 2sin 2(1 )sin 1 ( 1)

2n

n

n t n n nb t dt

n n

Sine series: 4 sin sin 2 sin 3 sin 4

( )2 4 6 8

t t t tf t

-2 2 4 6 8t

-1

1

4. 1 2

0 0 1(2 ) 1a t dt t dt

1 2

0 1

cos (2 )cos2 2n

n t n ta t dt t dt

22 2

2 2

0 for odd16

cos sin 0 if 4, 8, 12,2 4

16 / if 2, 6, 10,

nn n

nn

n n

Cosine series: 2 2 2 2 2

16 cos cos3 cos5 cos7( ) 1

2 6 10 14

t t t tf t

-2 2 4 6 8t

1



1 2

0 1

sin (2 )sin2 2n

n t n tb t dt t dt

3 2 22 2

2 2

0 for even32

cos sin 8 / if 1, 5, 9,4 4

8 / if 3, 7, 11,

nn n

n nn

n n


8 1 3 1 5 1 7( ) sin sin sin sin

2 3 2 5 2 7 2

t t t tf t

-1 1 3 5 7 9t

-1

1

5. 2

01

2 21

3 3a dt

2

1

2 3 / if 2, 8,14,2 2 2

cos sin sin 2 3 / if 4,10,16,3 3 3 3

0 otherwisen

n nn t n n

a dt n nn

Cosine series: 1 2 3 1 2 1 4 1 8 1 10

( ) cos cos cos cos3 2 3 4 3 8 3 10 3

t t t tf t

-3 3 6 9 12 15t

1

2

1

0 for even

2 / if 1, 7,13,2 2 2sin cos cos

4 / if 3, 9,15,3 3 3 3

2 / if 5,11,17,

n

n

n nn t n nb dt

n nn

n n

Section 9.3 555


Sine series:

2 22 3 1 5 1 7 9 1 11

sin sin sin sin sin sin( )3 5 7 9 13 3 3 3 3 1 3

t t t t t tf t

-3 3 6 9 12 15t

-1

1

6. 2

20

0

2 2

3a t dt

,

2 2

23 2

0

2 2 cos ( 2)sin2 4( 1)cos

n

n

n n n na t nt dt

n n

Cosine series: 2

2 2 2

1 1 1( ) 4 cos cos2 cos3 cos4

3 2 3 4f t t t t t

�Π Π 3 Π 5 Πt

Π2

2

0

2sinnb t nt dt

2 2 3

3

2 ( 2)cos 2 sin 2 2 / 8 / for odd

2 / for even

n n n n n n n

n n n

Sine series: 1 1 1

( ) 2 sin sin 2 sin 3 sin 42 3 4

f t t t t t

3 3 3

8 1 1 1sin sin 3 sin5 sin 7

3 5 7t t t t



�Π Π 3 Π 5 Πt

Π2

7. 2

00

2( )

3a t t dt

,

3 20

2 cos 2sin2 2( )cos 1 ( 1)n

n

n n n na t t nt dt

n n

Cosine series: 2

2 2 2 2

cos2 cos4 cos6 cos8( ) 4

6 2 4 6 8

t t t tf t

�Π Π 3 Π 5 Πt

Π2

4

3 30

2 2 2cos 2 sin2 4( )sin 1 ( 1)n

n

n n nb t t nt dt

n n

Sine series: 3 3 3

8 sin 3 sin5 sin 7( ) sin

3 5 7

t t tf t t

�Π Π 3 Π 5 Πt

Π2

4

Section 9.3 557


8. 1

20

0

12 ( )

3a t t dt

,

1 2

3 3 2 20

2 cos 2sin 22 ( )cos 1 ( 1)n

n

n n n na t t n t dt

n n

Cosine series: 2 2 2 2 2

1 4 cos2 cos4 cos6 cos8( )

6 2 4 6 8

t t t tf t

�1 1 3 5t

1

4

1 2

3 3 3 30

2 2 2cos sin 42 ( )sin 1 ( 1)n

n

n n nb t t n t dt

n n


8 sin 3 sin5 sin 7( ) sin

3 5 7

t t tf t t

for 0 < t < 1. Note that t = 1/2 yields the summation of Problem 25 in Section 9.2.

�1 1 3 5t

1

4

9. 00

2 4sina t dt

,

2 20

2 1 ( 1)2 1 cos2sin cos if 1

(1 ) (1 )

n

n

na t nt dt n

n n

1 0

2sin cos 0a t t dt

Cosine series: 2

2 4 cos2 cos4 cos6 cos8( )

3 15 35 63

t t t tf t



�Π Π 3 Π 5 Πt

1

20

2 2sinsin sin 0 if 1

(1 )n

nb t nt dt n

n

21 0

2sin 1b t dt

Sine series: ( ) sinf t t

�Π Π 3 Π 5 Πt

1

10. 00

1 2sina t dt

,

2

22

0

4 / ( 4) for odd4 1 cos1 2sin cos 8 / ( 4) if 4, 8, 12,

2 ( 4)0 if 6, 10, 14,

n

n n nnt

a t dt n nn

n

2 0

1sin cos 0a t t dt

Cosine series: 1 4 1 1 3 2 4

( ) cos cos cos3 2 5 2 12 2

t t tf t

1 5 1 7 2 8

cos cos cos21 2 45 2 60 2

t t t


Section 9.3 559


�2 Π 2 Π 4 Π 6 Π 8 Π 10 Πt

1

22

0 2

0 for 2 even4sin1 2sin sin 4 / ( 4) if 1, 5, 9,2 ( 4)

4 / ( 4) if 3, 7,11,n

n nnt

b t dt n nn

n n

22 0

1 1sin

2b t dt

21 0

2sin 1b t dt

Sine series: 1 4 1 1 3 1 5 1 7 1 9

( ) sin sin sin sin sin sin2 3 2 5 2 21 2 45 2 77 2

t t t t tf t t

�2 Π 2 Π 4 Π 6 Π 8 Π 10 Πt

�1

1

11. In order to satisfy the endpoint conditions (0) ( ) 0x x we substitute the sine series

1

( ) sinnn

x t b nt

and odd

4 sin1

n

nt

n (from Example 1 in Section 9.1) into the

differential equation 2 1x x . This gives

2

1 1 odd

4 sinsin 2 sinn n

n n n

ntn b nt b nt

n

.

We therefore choose 24 / (2 )nb n n for n odd, 0nb for n even. This gives the

formal series solution



2odd

4 sin 4 sin 3 sin5 sin 7( ) sin .

(2 ) 21 115 329n

nt t t tx t t

n n

12. In order to satisfy the endpoint conditions (0) ( ) 0x x we substitute the sine series

1

( ) sinnn

x t b nt

and odd

4 sin1

n

nt

n (from Example 1 in Section 9.1) into the

differential equation 4 1x x . This gives

2

1 1 odd

4 sinsin 4 sinn n

n n n

ntn b nt b nt

n

.

We therefore choose 24 / ( 4)nb n n for n odd, 0nb for n even. This gives the

formal series solution

4 sin sin 3 sin5 sin 7

( )5 39 145 371

t t t tx t

.

13. In order to satisfy the endpoint conditions (0) (1) 0x x we substitute the sine series

1

( ) sinnn

x t b n t

and 1

1

2 ( 1) sinn

n

n tt

n

(from Example 1 in Section 9.3, with

L = 1) into the differential equation x x t . This gives

1

2 2

1 1 1

2 ( 1) sinsin sin

n

n nn n n

n tn b n t b n t

n

.

We therefore choose 1 2 22( 1) / (1 )nnb n n . This gives the formal series solution

2 21

( 1) sin2( )

( 1)

n

n

n tx t

n n

of our endpoint value problem.

14. In order to satisfy the endpoint conditions (0) (2) 0x x we substitute the sine series

1

( ) sin2n

n

n tx t b

and 1

1

4 ( 1)sin

2

n

n

n tt

n

(from Example 1 in Section 9.3 ,with

L = 2) into the differential equation 2x x t . This gives

2 2 1

1 1 1

4 ( 1)sin 2 sin sin

4 2

n

n nn n n

n n tb n t b n t

n

.

We therefore choose 1 1

2 2 2 2

4( 1) / 16( 1)

2 / 4 (8 )

n n

n

nb

n n n

Section 9.3 561


for n odd, 0nb for n even. This gives the formal series solution

2 21

16 ( 1) sin ( / 2)( )

( 8)

n

n

n tx t

n n


15. In order to satisfy the endpoint conditions (0) (2) 0x x we substitute the cosine

series 0

1

( ) cos2 n

n

ax t a nt

and 2

odd

4 cos

2 n

ntt

n

(from Example 1 in Section

9.3, with L = ) into the differential equation 2x x t . This gives

20 2

1 1 odd

4 coscos 2 cos

2n nn n n

ntn a nt a a nt

n

.

We therefore choose 0 / 2a , 0na for n > 0 even, and 2 24 / ( 2)na n n for

n odd. This gives the formal series solution

2 2

odd

4 cos 4 cos3 cos5 cos7( ) cos

4 ( 2) 4 63 575 2303n

nt t t tx t t

n n


16. (a) Obviously ( )px t t is a particular solution of 4 4 ,x x t so a general

solution is given by cos2 sin 2 .x t A t B t t We satisfy the endpoint conditions

(0) (2) 0x x by choosing 0 and 1/ sin 2.A B (b) The point is simply that the series in (31) is the Fourier sine series of the period 2 function defined by f(t) = t (sin 2t)/(sin 2) for 0 < t < 1:

1

2 20

2 sin cos 2( 1)2 sin

nn n nt n t dt

n n

1

2 2 2 20

2 2(cos2)sin (sin 2)cos 2 ( 1) (sin 2)2 sin 2 sin

4 4

nn n n nt n t dt

n n

2 2 2 2

2( 1) 2 ( 1) 8( 1)

4 ( 4)

n n n

n

nb

n n n n

2 2

1

sin 2 8 ( 1) sin

sin 2 ( 4)

n

n

t n tt

n n

for 0 < t < 1

17. Suggestion: Substitute u = t in the left-hand integral. 18. The termwise derivative of the given Fourier series is



(4/) (sin nt)/n 4 cos nt. But the series cos nt diverges at t = 0 (for instance). Hence the derived series does

not converge to any function at all, let alone to f'(t).

19. The first termwise integration yields

2

121

( 1) cos2

2

n

n

t ntC

n

,

and substitution of t = 0 gives 1 2 21

1

2 ( 1) / / 6,n

n

C n

so

2 2

21

( 1) cos2 .

2 6

n

n

t nt

n

A second termwise integration gives

3 2

231

( 1) sin2 ,

6 6

n

n

t nt tC

n

and substitution of t = 0 gives C2 = 0. The final termwise integration gives

4 2 2

341

( 1) cos2 ,

24 12

n

n

t nt tC

n

and substitution of t = 0 yields 43

1

2 ( 1) / .n

n

C n

20. Substitution of t = in the formula of Problem 19 above gives

4 4

4 41 1

4 1

4 4 41 1 odd

1 ( 1)2 2 ,

24 12

1 ( 1) 12 2 4

24

n

n n

n

n n n

n n

n n n

which gives 4

4 4 4

1 1 11 .

3 5 7 96

Then

4 4 4 4 4 4 4 4

1

4 4 4 4 4 4 4

1 1 1 1 1 1 1 11

2 3 4 5 6 7 8

1 1 1 1 1 1 11

3 5 7 2 4 6 8

n

Sn

Section 9.3 563


4 4 4 4 4 4 4

4 4 4

1 1 1 1 1 1 11 1

3 5 7 2 2 3 4

1 1 1 11 .

3 5 7 15S S

Solution of this last equation for S now gives

4 4

4 4 4 41

1 16 1 1 1 161 .

15 3 5 7 15 96 90n n

21. We want to calculate the coefficients in the period 4L Fourier sine series

1

( ) sin2n

n

n ttF b

L

which agrees with f(t) if 0 < t < L. Then

2

0

2 2( ) (2 ) .sin sin

2 2 2 2

L L

nL

n t n tf t dt f L t dtb

L L L L

The substitution u = 2L t yields

0

0

1 1 (2 )( ) ( )sin sin

2 2

L

nL

n t n L uf t dt f u dub

L L L L

0 0

1 ( 1)( ) ( ) .sin sin

2 2

L Lnn t n uf t dt f u du

L L L L

Now it is clear that

0

2( )cos

2

L

n

n tf t dtb

L L

if n is odd, whereas bn = 0 if n is even.

22. We want to calculate the coefficients in the period 4L Fourier cosine series

0

1

( ) cos2 2n

n

a n tG t a

L

which agrees with f(t) if 0 < t < L. Then

2

0

2 2( )cos (2 )cos .

2 2 2 2

L L

nL

n t n ta f t dt f L t dt

L L L L

The substitution u = 2L t yields



0

0

1 1 (2 )( )cos ( )cos

2 2

L

nL

n t n L ua f t dt f u du

L L L L

0 0

1 ( 1)( )cos ( )cos .

2 2

L Lnn t n uf t dt f u du

L L L L

Now it is clear that

0

2( )cos

2

L

n

n ta f t dt

L L

if n is odd, whereas an = 0 if n is even (including n = 0).

23. ( 1) / 2

2 20

2 4 8( 1)sin 2sin cos for odd

2 2 2

n

n

nt n nb t dt n n

n n

2 2 2 2

8 1 3 1 5 1 7( ) sin sin sin sin

2 3 2 5 2 7 2

t t t tf t

24. In order to satisfy the endpoint conditions (0) ( ) 0x x we substitute the odd half-

multiple sine series odd

( ) sin2n

n

ntx t b and

( 1) / 2

2odd

8 ( 1)sin

2

n

n

ntt

n

(from Problem

21) into the differential equation x x t . This gives

2 ( 1) / 2

2odd odd odd

8 ( 1)sin sin sin

4 2 2 2

nn

nn n n

n b nt nt ntb

n

.

We therefore choose

( 1) / 2 2 ( 1) / 2

2 2 2

8( 1) / 32( 1)

1 / 4 ( 4)

n n

n

nb

n n n

for n odd. This gives the formal series solution

( 1) / 2

2 2odd

32 ( 1)( ) sin

( 4) 2

n

n

ntx t

n n

32 1 1 3 1 5 1 7

sin sin sin sin3 2 45 2 525 2 2205 2

nt nt nt nt


Section 9.4 565


SECTION 9.4 APPLICATIONS OF FOURIER SERIES

1. We substitute the sine series 1

( ) sinnn

x t b nt

and odd

12 sin( )

n

ntF t

n (from

Example 1 in Section 9.1) into the differential equation 5 ( )x x F t . This gives

2

1 1 odd

12 sinsin 5 sinn n

n n n

ntn b nt b nt

n

.

We therefore choose 0nb for n > 0 even, and 212 / (5 )nb n n for n odd. This

gives the formal series solution

sp 2odd

12 sin 12 sin sin 3 sin5 sin 7( ) .

(5 ) 4 12 100 308n

nt t t t tx t

n n

2 Π 4 Π 6 Πt

�1

1

2. We substitute the cosine series 0

1

( ) cos2 2n

n

a n tx t a

and

( 1) / 2

odd

12 ( 1)( ) cos

n

n

n tF t

n

into the differential equation 10 ( )x x F t . This

gives

2 ( 1) / 2

01 1 odd

12 ( 1)cos 5 10 cos cos

2 2

n

n nn n n

n n t n t n ta a a

n

.

We therefore choose 0 0a and 0na for n > 0 even, and

( 1) / 2 ( 1) / 2

2 2 2 2

12( 1) / 48( 1)

10 / 4 (40 )

n n

n

na

n n n

566 APPLICATIONS OF FOURIER SERIES


for n odd. This gives the formal series solution ( 1) / 2

sp 2 2odd

48 ( 1)( ) cos .

(40 )

n

n

n tx t

n n

5 10 15t

�1

1


( ) sinnn

x t b nt

and 1

1

( 1) sin( ) 2

n

n

ntF t

n

(from

Example 1 in Section 9.3, with L = ) into the differential equation 3 ( )x x F t . This gives

1

2

1 1 1

( 1) sinsin 3 sin 4

n

n nn n n

ntn b nt b nt

n

.

We therefore choose 1 24( 1) / (3 )nnb n n . This gives the formal series solution

1

sp 21

( 1) sin sin sin 2 sin 3 sin 4( ) 4 4 .

(3 ) 2 2 18 52

n

n

nt t t t tx t

n n

2 Π 4 Π 6 Πt

�4

4


1

( ) cos2 2n

n

a n tx t a

and

2 2odd

16 1( ) 2 cos

n

n tF t

n

(from Example 1 in Section 9.3, with L = 2) into the

differential equation 4 ( )x x F t . This gives

Section 9.4 567


2

0 2 21 1 odd

16 1cos 2 4 cos 2 cos

2 2n nn n n

n n t n t n ta a a

n

.

We therefore choose 0 1a and 0na for n > 0 even, and

2 2

2 2 2 2 2 2

16 / 64

4 / 4 (16 )n

na

n n n

for n odd. This gives the formal series solution sp 2 2 2 2 2odd

1 64 cos / 2( ) .

2 (16 )n

n tx t

n n

4 8 12t

1

2


( ) sinnn

x t b n t

and 3 3

odd

8 sin( )

n

n tF t

n

into the

differential equation 10 ( )x x F t . This gives

2 23 3

1 1 odd

8 sinsin 10 sin .n n

n n n

n tn b n t b n t

n

We therefore choose 3 3 2 28 / (10 )nb n n . This gives the formal series solution

sp 3 3 2 2odd

8 sin( )

(10 )n

n tx t

n n

.

2 4 6t

-2

2




1

( ) cos2 n

n

ax t a nt

and 2

even

4 4 cos( )

1n

ntF t

n

into the differential equation 2 ( )x x F t . This gives

20 2

1 1 even

4 4 coscos 2 cos

1n nn n n

ntn a nt a a nt

n

.

We therefore choose 0 4 /a and 0na for n odd, and

2

2 2 2

4 / ( 1) 4

2 ( 1)( 2)n

na

n n n

for n even. This gives the formal series solution

sp 2 2even

4 4 cos 4 cos2 cos4 cos6( ) 1 .

( 1)( 2) 6 210 1190n

nt t t tx t

n n

Π 2 Π 3 Πt

4

Π

In Problems 7–12 we are dealing with the equation mx + kx = F(t) where F(t) is the external

periodic force. The natural frequency is 0 = /k m . If the Fourier series of F(t) contains a term of the form cos(Nt /L) or sin(Nt /L) with 0 = N /L, then pure resonance occurs. Otherwise, it does not. 7. The natural frequency is 0 = 3, and

4 sin 3 sin5 sin 7

( ) sin .3 5 7

t t tF t t

Thus the Fourier series of F(t) contains a sin 3t term, so resonance does occur.

8. The natural frequency is 0 = 5 , and F(t) = bnsin nt. Since n 5 for any integer n, pure resonance does not occur.

Section 9.4 569


9. The natural frequency is 0 = 2, and

4 sin 3 sin5 sin 7

( ) sin .3 5 7

t t tF t t

Because the sin 2t term is missing from the Fourier series of F(t), resonance will not occur.

10. The natural frequency is 0 = 2. From Equation (16) in Section 9.3 of the text we see that the Fourier series of F(t) contains a sin 2t term. Hence pure resonance occurs.

11. The natural frequency is 0 = 4. From Equation (15) in Section 9.3 we see that

2 2

4 cos3 cos5( ) cos .

2 3 5

t tF t t

Because the cos 4t term is missing, we see that resonance will not occur. 12. The natural frequency is 0 = 5, and the Fourier series of F(t) is of the form F(t) = bnsin nt. We calculate b5, and find that

25 0

2 4 4cos5 10 sin5 8( )sin5 0,

125 125b t t t dt

Thus the term sin 5t is present in F(t), and so pure resonance occurs. Problems 13–18 are based on Equations (14)–(16) in the text, according to which the steady periodic solution of

1

sinnn

n tmx cx kx B

L

is given by

sp1

( ) sin( ),n n nn

x t b t

where

,n

n

L

12

tan nn

n

c

k m

in the interval [0, ],

2 22

.nn

n n

Bb

k m c



This calculation is readily automated. The following MATLAB script was written to calculate the coefficients {bn} for Problem 13. Only the values of m, c, k, L and the calculation of the force function coefficients { }nB need to be changed for Problems 14–18.

m = 1; c = 0.1; k = 4; L = pi; results = ones(0,4); for n = 1:9 w = n*pi/L; alpha = atan(c*w/(k-m*w^2)); if k-m*w^2<0 alpha = pi + alpha; end B = 12/(pi*n); % force function coeffs if floor(n/2)==n/2 % are nonzero if n is odd, B = 0; % zero if n is even end b = B/sqrt((k-m*w^2)^2+(c*w)^2); results = [results; n, b, w, alpha]; end results

13. Bn = 12/n for n odd, Bn = 0 for n even xsp(t) 1.2725 sin(t 0.0333) + 0.2542 sin(3t 3.0817) + 0.0364 sin(5t 3.1178) +

14. Bn = 4(1)n+1/n for n = 1, 2, 3, xsp(t) 0.2500 sin(t 0.0063) 0.2000 sin(2t 0.0200) + 4.444 sin(3t 1.5708) 0.0714 sin(4t 3.1130) + Note the dominance of the n = 3 term.

15. Bn = 8/n33 for n odd, Bn = 0 for n even xsp(t) 0.08150 sin(t 1.44692) + 0.00004 sin(3t 3.10176) +

16. F(t) = A0 + Ancos(nt /2) where A0 = 2, An = 16/2n2 for n odd, An = 0 for n even and positive. xsp(t) 0.5000 + 1.0577 cos(t/2 0.0103) 0.0099 cos(3t/2 3.1390) 0.0011 cos(5t/2 3.1402)

Section 9.4 571


17. Bn = 60/n for n odd, Bn = 0 for n even xsp(t) 0.5687 sin(t 0.0562) + 0.4271 sin(3t 0.3891) + 0.1396 sin(5t 2.7899) + 0.0318 sin(7t 2.9874) + xsp(5) 0.248 ft 2.98 in.

18. Bn = (4/n2)sin(n/2) xsp(t) 0.0531 sin(t 0.0004) 0.0088 sin(3t 0.0019) + 1.0186 sin(5t 1.5708) 0.0011 sin(7t 3.1387) + Note the dominance of the n = 5 term. 19. We suppose that ( ) ( ) and ( ) ( )f t P f t g t Q g t for all t. If / /P Q m n where m and n are integers, let .R nP mQ Then ( ) ( ) ( ) ( ) ( ) ( )f t R g t R f t nP g t mQ f t g t for all t, so we see that the sum ( ) ( )f t g t is periodic with period R. 20. Suppose that the function ( ) cos cosf t pt qt is periodic and has period L > 0, so ( ) ( )f t L f t for all t. That is, cos ( ) cos ( ) cos cosp t L q t L pt qt . Substitution of 0t then gives cos cos 2cos0 2pL qL , which implies that cos cos 1pL qL (why?). It follows that andpL qL must both be nonzero integral multiples of 2 . Then

2

2 and 22

pL m p mpL m qL n

qL n q n

with m and n integers. Thus /p q must be rational if the sum ( ) cos cosf t pt qt is to be periodic. SECTION 9.5 HEAT CONDUCTION AND SEPARATION OF VARIABLES 1. From Equation (31) in the text, with L = and k = 3, we get

572 HEAT CONDUCTION AND SEPARATION OF VARIABLES


2

1

( , ) exp( 3 )sinnn

u x t b n t nx

.

With b2 = 4 and bn = 0 otherwise we get the solution u(x, t) = 4e12tsin 2x.

2. From Equation (40) in the text, with k = 10 and L = 5 we get,

2 2

0

1

( , ) exp cos .2 25 5n

n

a n t n xu x t a

With a0 = 14 and an = 0 for n > 0 we get the solution u(x, t) = 7 (constant).

3. With L = 1 and k = 2 in Equation (31), we take b1 = 5, b3 = 1/5, and bn = 0 otherwise. The result is the solution

2 22 181

( , ) 5 sin sin 3 .5

t tu x t e x e x

4. From Equation (31) with k = 1 and L = we get

2

1

( , ) exp( )sinnn

u x t b n t nx

.

But the sin A cos B identity yields 4 sin 4x cos 2x = 2 sin 2x + 2 sin 6x. Hence we choose b2 = b6 = 2 and bn = 0 for n 2, 6. Thus u(x, t) = 2e4tsin 2x + 2e36tsin 6x.

5. From Equation (40) in the text, with k = 2 and L = 3 we get

2 2

0

1

2( , ) exp cos .

2 9 3nn

a n t n xu x t a

With a0 = 0, 2 44, 2,a a and 0na otherwise, and an = 0 we get the solution

2 28 2 32 4

( , ) 4exp cos 2exp cos .9 3 9 3

t x t xu x t

Section 9.5 573


6. From Equation (31) with k = 1/2 and L = 1 we get

2 2

1

( , ) exp sin .2n

n

n tu x t b n x

Trigonometric identities yield 4 sin x cos3x = (2 sin x cos x)(2 cos2x)

= (sin 2x)(1 + cos 2x) = sin 2x + (1/2)sin 4x. Hence we choose b2 = 1, b4 = 1/2, and bn = 0 otherwise to get u(x, t) = exp(22 t) sin 2x + (1/2)exp(82 t) sin 4x.

7. From Equation (40) in the text, with k = 1/3 and L = 2 we get,

2 2

0

1

( , ) exp cos .2 12 2n

n

a n t n xu x t a

Because of the identity cos22x = (1 + cos 4x)/2 , we choose 0 81, 1/ 2,a a and

0na otherwise. This gives the solution

21 1 16

( , ) exp cos4 .2 2 3

tu x t x

8. From Equation (40) with k = 1 and L = 2 we get

2 2

0

1

( , ) exp cos .2 4 2n

n

a n t n xu x t a

But 10 cos x cos 3x = 5 cos 2x + 5 cos 4x. Hence we choose b4 = b8 = 5 and bn = 0 otherwise to get the solution u(x, t) = 5 exp(42 t) cos 2x + 5 exp(162 t) cos 4x.

9. Because of the zero endpoint conditions (0, ) (5, ) 0,u t u t we use the Fourier sine series expansion

odd

100 1( ,0) sin

5n

n xu x

n

of ( ,0) 25u x on the interval 0 < x < 5. When we supply the exponential factors in Eq. (31) with k = 1/10 and L = 5, we get the solution



2 2

odd

100 1( , ) exp sin

250 5n

n t n xu x t

n

10. Because of the zero endpoint conditions (0, ) (10, ) 0,u t u t we use the Fourier sine series expansion

1

1

80 ( 1)( ,0) sin

10

n

n

n xu x

n

of ( ,0) 4u x x on the interval 0 < x < 10 (from Eq. (16) in Section 9.3). When we supply the exponential factors in Eq. (31) here with k = 1/5 and L = 10, we get

1 2 2

1

80 ( 1)( , ) exp sin .

500 10

n

n

n t n xu x t

n

11. Because of the zero-derivative endpoint conditions (0, ) (10, ) 0,x xu t u t we use the

Fourier cosine series expansion

2 2

odd

160 1( ,0) 20 cos

10n

n xu x

n

of ( ,0) 4u x x on the interval 0 < x < 10 (from Eq. (15) in Section 9.3). When we supply the exponential factors in Eq. (40) here with k = 1/5 and L = 10, we get

2 2

2 2odd

160 1( , ) 20 exp cos

500 10n

n t n xu x t

n

12. From Equation (31) with k = 1 and L = 100 we get

2 2

1

( , ) exp sin .10000 100n

n

n t n xu x t b

Because of the zero endpoint conditions (0, ) (100, ) 0,u t u t the {bn} should be the

Fourier sine coefficients of f(x) = x(100 x) on [0, 100], given by

100

0

3 3

3 3

2100 sin

100 100

20000 2 2cos sin 80000 / for odd,

0 for odd.

n

n xb x x dx

n n n n n

n n

This gives the solution

2 2

3 3odd

80000 1( , ) exp sin .

10000 100n

n t n xu x t

n

Section 9.5 575


13. (a) The boundary value problem is ut = kuxx (0 < x < 40),

ux(0, t) = ux(40, t) = 0,

u(x, 0) = 100.

By Equation (31) in the text (with L = 40) the solution is of the form

2 2

1

( , ) exp sin .1600 40n

n

n kt n xu x t b

We use the Fourier sine coefficients 400 /nb n for n odd, bn = 0 otherwise, of the

initial value function ( ) 100f x on the interval 0 < x < 100. This gives

2 2

odd

400 1( , ) exp sin

1600 40n

n kt n xu x t

n

.

(b) With k = 1.15 for copper we find that (20,300) 15.1591 0.000000204 15.16u C . (c) With k = 0.005 for concrete, the first term of the series gives

2400 0.00

(20, ) exp 15,1600

tu t

and we solve for 66,342 sec 19 hr 15 min 42sec.t As a check that the first term suffices for this computation, we find that the next term in the series is then approximately 0.00000019.

14. (a) The boundary value problem is ut = kuxx (0 < x < 50)

ux(0, t) = ux(50, t) = 0

u(x, 0) = 2x

with k = 1.15 cm2/sec for copper. By Equation (40) in the text (with L = 50) the solution is of the form

2 2

0

1

( , ) exp cos .2 2500 50n

n

a n kt n xu x t a

Consulting the Fourier series given in Equation (15) of Section 9.2, we satisfy the initial condition u(x, 0) = 2x by choosing a0 = 100, an = 400/n22 for n odd, and



an = 0 for n even. Thus

2 2

2 2odd

400 1( , ) 50 exp cos .

2500 50n

n kt n xu x t

n

(b) With k = 1.15 for copper we find that (10,60) 50 24.9698 0.1199 0.0018 0.0000 25.15u C . (c) To find out how long it takes the temperature to reach 45C at the point x = 10, we solve the equation

2

2

400 1.1550 exp cos 45

2500 5

t

that we get upon retaining only the first two terms of the series above (with x = 10). Using logarithms (for instance) we find that 414.23 sec 6 min 54 sec.t To

confirm that two terms suffice for this calculation, we retain 3 terms and use a computer or calculator to solve the equation

2 2

2 2

400 1.15 400 9 1.15 350 exp cos exp cos 45

2500 5 9 2500 5

t t

for (again) 414.23 sec.t

15. We need only calculate the coefficients in the usual zero-endpoint series

2 2

21

( , ) exp sin .nn

n kt n xu x t b

L L

For the function ( ) for 0 / 2,f x A x L ( ) 0 for / 2f x L x L we calculate the Fourier sine coefficient

/ 2

2

0

1/ 2 for odd,2 4 4

sin sin 1 for 2,6,10, ,4

0 for 4,8,12, .

L

n

nn x A n A

b A dx nL L n n

n

16. (a) Summing numerically the series in Problem 15 with the values k = 0.15 for

iron, L = 50, A = 100, x = 25, and t = 1800, we find that (25,1800) 21.9259 0.0014 0.0000 21.9245 22 C.u

Section 9.5 577


(b) Because for x fixed the temperature is a function of the product kt, in the case of concrete slabs with k = 0.005 the same temperature will be attained when

(0.005)(t) = (0.15)(1800), that is, when t = 54000 sec = 15 hr. SECTION 9.6

VIBRATING STRINGS AND THE ONE-DIMENSIONAL WAVE EQUATION

In Problems 1–10 we use the general solution

1

( , ) cos sin sinn nn

n at n at n xy x t A B

L L L

(*)

of the string equation ytt = a2yxx with endpoint conditions y(0, t) = y(L, t) = 0. This form of the solution is obtained by superposition of the solutions in Equations (23) and (33) of Problems A and B in this section. It remains only to choose the coefficients {An} and {Bn} so as to satisfy given initial conditions

1

( ,0) sin ( ),nn

n xy x A f x

L

thus, 0

2( )sin ;

L

n

n xA f x dx

L L

and

1

( ,0) sin ( ),t nn

n a n xy x B g x

L L

thus, 0

2( )sin .

L

n

n xB g x dx

n a L

1. Here a = 2 and L = . To satisfy the condition y(x, 0) = (1/10)sin 2x we choose A2 = 1/10 in Eq. (*) above, and An = 0 otherwise. To satisfy the condition yt(x, 0) = 0 we choose Bn = 0 for all n. Thus

y(x, t) = 1

10cos 4t sin 2x.

2. Here a = L = 1. To satisfy the condition

y(x, 0) = 1

10sin x

1

20sin 3x

we choose A1 = 1/10 and A3 = 1/20 in Eq. (*) above, and An = 0 otherwise. To satisfy the condition yt(x, 0) = 0 we choose Bn = 0 for all n. Thus

578 VIBRATING STRINGS AND THE ONE-DIMENSIONAL WAVE EQUATION


y(x, t) = 1

10cos t sin x

1

20cos 3t sin 3x.

3. Here a = 1/2 and L = . Choosing A1 = 1/10 and An = 0 otherwise, B1 = 1/5 and Bn = 0 otherwise, we get

y(x, t) = 1

cos 2sin sin .10 2 2

t tx

4. Here a = 1/2 and L = 2, so nx /L = nx /2 and nat /L = nt /4. To satisfy the condition

y(x, 0) = 1

5sin x cos x =

1

10sin 2x =

1 4sin ,

10 2

x

we choose A4 = 1/10 and An = 0 for n 4. To satisfy the condition yt(x, 0) = 0

we choose Bn = 0 for all n. Thus

y(x, t) = 1

10cos t sin 2x.

5. Here a = 5 and L = 3. Choosing A3 = 1/4 and An = 0 for n 3, B6 = 1/ and Bn = 0 for n 6, we get

y(x, t) = 1

4cos 5t sin x +

1

sin 10t sin 2x.

6. Here a = 10 and L = . To satisfy the condition yt(x, 0) = 0 we choose Bn = 0 for all n, so

y(x, t) = 1

cos10 sin .nn

A nt nx

To satisfy the condition y(x, 0) = x( x) we choose

3

30

8 / for odd,2 4 4cos 2 sinsin

0 for odd.n

n nn n nA x x nx dx

n n

This gives the solution

3

odd

8 cos10 sin( , ) .

n

nt nxy x t

n

7. Here a = 10 and L = 1. To satisfy the condition y(x, 0) = 0 we choose An = 0 for all n, so

Section 9.6 579


y(x, t) = 1

sin10 sin .nn

B n t n x

To satisfy the condition yt(x, 0) = x we choose

1 1

2 2

1 2( 1) ( 1)

10 5

n n

nBn n n

for n 1 (see Equation (16) in Section 9.3). This gives

y(x, t) = 1

2 21

1 ( 1)sin10 sin .

5

n

n

n t n xn

8. Here a = 2 and L = . To satisfy the condition y(x, 0) = sin x we choose A1 = 1 and An = 0 for n > 1, so

y(x, t) = cos 2t sin x + 1

sin 2 sin ,nn

B nt nx

so

yt(x, t) = 2 sin 2t sin x + 1

2 cos2 sin .nn

nB nt nx

The condition yt(x, 0) = 1 will be satisfied if 2nBn = 4/n for n odd and bn = 0 for n even. We therefore choose Bn = 2/n2 for n odd and Bn = 0 for n even, so

odd

4 sin 2 sin( , ) cos2 sin .

n

nt nxy x t t x

n

9. Here a = 2 and L = 1. To satisfy the condition y(x, 0) = 0 we choose An = 0 for

all n, so

y(x, t) = 1

sin 2 sin .nn

B n t n x

To satisfy the condition yt(x, 0) = x(1 x) we choose

1

4 40

1 2 2cos sin(1 )sin .n

n n nB x x n x dx

n n

Hence

4 4

odd

4 sin 2 sin( , ) .

n

n t n xy x t

n

10. Here a = 5 and L = so

y(x, t) = 1

cos5 sin5 sin .n nn

A nt B nt nx



We first compute the Fourier sine series sin2x = 1

sinnn

b nx

and find that bn = 0

if n is even whereas

22 20

2 4(cos 1) 8sin sin

( 4) (4 )n

nb x nx dx

n n n n

if n is odd. To satisfy the condition y(x, t) = sin2x we choose An = bn, and to satisfy the condition yt(x, t) = sin2x we choose Bn = bn /5n. Then

2 2odd

5 cos5 sin5 sin8( , )

5 (4 )n

n nt nt nxy x t

n n

.

11. Substitution of L = 2 ft, T = 32 lb, and the linear density

3

1/ 32 oz 1oz 1lb 1slug 1 slug

2 ft 64ft 16oz 32lb 32 ft

in Eqs. (2) and (26) in the text yields the velocity 4/ 32 1024 ft / seca T with

which waves move along the string, and its fundamental frequency

1

1256 Hz,

2 2

T a

L L

which is approximately middle C. 12. The value of

02 2

odd

4 1( , ) sin sin

n

v L n at n xy x t

a n L L

is maximal when each of the sine products is 1. This happens when / 2, / 2 :x L t L a

2

20 0 0 02 2 2 2 2

odd odd

4 1 4 1 4, sin .

2 2 2 8 2n n

L L v L n v L v L v Ly

a a n a n a a

Using fps units with the string of Problem 11 where L = 2 ft, a = 1024 ft/sec, and v0 = 60 mph = 88 ft/sec, we get

max

88 20.0859ft 1 inch.

2 1024y

13. If ( , ) ( ) ( )y x t F x at F u with ,u x at then the chain rule gives

Section 9.6 581


( ) 1 ( );y dF u

F u F x atx du x

( ) ( ) ;y dF u y

F u a a F x at at du t x

2

2( ) 1 ( );

y dF uF u F x at

x du x

2 2

2 22 2

( ) ( ) .y dF u y

a a F u a a F x at at du t x

14. 1 12 2(0, ) [ ( ) ( )] [ ( ) ( )] 0y t F at F at F at F at

12

1 12 2

( , ) [ ( ) ( )]

[ ( ) ( )] [ (2 ( )) ( )] 0

y L t F L at F L at

F L at F L at F L L at F L at

12( ,0) [ ( ) ( )] ( )y x F x F x F x

12( , ) [ ( ) ( )]ty x t aF x at aF x at

12( ,0) [ ( ) ( )] 0ty x aF x aF x

15. If ( ,0) 0y x then the fundamental theorem of calculus gives

0

0

1( , ) ( , ) ( ,0) ( , ) ( ) ( ) .

2

tt

ty x t y x t y x y x d G x a G x a d

16. If ,u x at v x at then we solve readily for 1 1

2 2( ), ( ).ax u v t u v Hence

1 12 2

1 1( ), ( ) ;

2 2a

y y x y t y yy u v u v

u u x u t u x a t

1 12 2

1 1( ), ( ) ;

2 2a

y y x y t y yy u v u v

v v x v t v x a t

2 1 1

2 2

y y y

v u v x a t

1 1 1 1 1 1

2 2 2 2 2 2

y y y y

x x a t a t x a t

2 2 2 2 2 2

22 2 2 2 2 2

1 1 1 1 10.

4 4 4 4 4

y y y y y ya

x a x t a t x a t a x t

Now if 2

0y y

v u v u

then antidifferentiation with respect to v gives



/ ( ),y u G v an arbitrary function of v. Finally, antidifferentiation with respect to u gives ( ) ( ) ( ) ( ).y F u G v F x at G x at

18. When we separate variables as in Equations (8)(12) in this section, we find that X(x) must satisfy the eigenvalue problem

X'' + X = 0, X(0) = X'(L) = 0. In Example 4 of Section 3.8 we found that the eigenvalues and eigenfunctions of this problem are

2 2

2

(2 1),

4n

n

L

(2 1)

( ) sin2n

n xX x

L

for n = 1, 2, 3, . The function Tn(t) must satisfy the conditions 2 0,n n nT a T (0) 0,nT

so it follows that

(2 1)

( ) cos .2n

n atT t

L

Thus the form of y(t) is

1

(2 1) (2 1)( , ) cos sin .

2 2nn

n at n xy x t A

L L

Finally, in order to satisfy the initial condition y(x, 0) = f(x) we use the odd half- multiple sine series

1

(2 1)( ) sin

2nn

n xf x A

L

discussed in Problem 21 of Section 9.3.

19. The general solution of the second-order ordinary differential equation 2a y g is a

second-order polynomial in x with leading coefficient 2/ 2 .g a But the polynomial 2( ) ( ) / 2x gx x L a has this leading coefficient and satisfies the endpoint conditions

(0) ( ) 0.y y L

20. If ( , ) ( , ) ( ),y x t v x t x then tt tty v and

2 2 2 2 2

2 22 2 2 2 2 2

( ) , so .y v v g y v

x a g ax x x a x x

The transformation of the boundary conditions is straightforward.

Section 9.6 583


22. The eigenvalue problem

X'' + X = 0, X(0) = X(L) = 0 has the usual eigenvalues and eigenfunctions

2 2

2,n

n

L

( ) sinn

n xX x

L

for n = 1, 2, 3, . The function Tn(t) satisfies the equation

2 0,n n nT RT 2 2 2

2 22

0,n

n ah

L

so Tn(t) = Ancos nt + Bnsin nt.

Thus

1

( , ) cos sin sin .n n n nn

n xv x t A t B t

L

To satisfy the conditions v(x, 0) = f(x) and vt(x, 0) = h f(x) we choose An = bn and Bn = hbn/n where

1

( ) sin .nn

n xf x b

L

Then

1

( , ) cos sin sinnn n n

n n

b n xv x t t h t

L

1

cos( )sin ,n n nn

n xc t

L

where

1and tan .cos

nn n

n n

b hc

Finally,

1

( , ) ( , ) cos( )sin .ht htn n n

n

n xy x t e v x t e c t

L

23. First, if 04

x , then

4 4 2x

and 04 4

x , so that



1,

4 2 4 4

1 1 cos2 cos2 1

2 4 4

11

2

y x F x F x

x x

cos 2 cos 2 12 2

x x

1 sin 2 sin 22sin 2 .

x x

x

Next, if 3

4 4x

, then 2 4

x and 0

4 2x

, so that

1,

4 2 4 4

1 1 cos2 1 cos2

2 4 4

11 cos 2 1 cos 2

2 2 2

1 2 sin 22

y x F x F x

x x

x x

x

sin 2x 1.

Finally, if 3

4x

, then 5

4 4x

and 3

2 4 4x

, so that

1,

4 2 4 4

1 cos2 1

2 4

y x F x F x

x

1

cos24

1cos 2 cos 2

2 2 2

1 sin 2 sin 22

sin 2 .

x

x x

x x

x

24. (a) ( )f x = 4 cos 2x = 0 if x = /4 or x = 3/4. (b) If 0 t /4 then 0 /4 ± t /2 so

Section 9.6 585


y(/4, t) = 1

2[F(/4 + t) + F(/4 t)]

= 1

2[1 cos 2(/4 + t) + 1 cos 2(/4 t)]

= 1

2[1 cos (/2 + 2t) + 1 cos (/2 2t)]

= 1

2[2 + sin 2t sin 2t]

y(/4, t) = 1 SECTION 9.7

STEADY-STATE TEMPERATURE AND LAPLACES EQUATION 1. Because Y(0) = Y(b) = 0 we take our separation of variables in the form X'' X = 0 = Y'' + Y with > 0. Then it follows that

( ) sin ,n

n yY y

b

2 2

2n

n

b

and thence that

( ) cosh sinh .n n n

n x n xX x A B

b b

The condition that X(0) = 0 implies that An = 0 so ( ) cosh /n nX x B n x b , and

hence

1

( , ) sinh sin .nn

n x n yu x y C

b b

Finally we satisfy the condition u(a, y) = g(y) by choosing /(sinh / ),n nC b n a b

where the {bn} are the Fourier sine coefficients of g(y) on 0 y b.

2. Because Y(0) = Y(b) = 0 we take our separation of variables in the form X'' X = 0 = Y'' + Y with > 0. Then it follows that

( ) sin ,n

n yY y

b

2 2

2n

n

b

586 STEADY-STATE TEMPERATURE AND LAPLACE’S EQUATION


and thence that


n x n xX x A B

b b

The condition X(a) = 0 implies that

cosh /

.sinh /n

n

A n a bB

n a b

It now follows as in Equation (12) in the text that

( ) sinh ,n n

n a xX x C

b

so

1

( )( , ) sinh sin .n

n

n a x n yu x y C

b b

Finally we satisfy the condition u(0, y) = g(y) by choosing /(sinh / ),n nC b n a b

where the {bn} are the Fourier sine coefficients of g(y) on 0 y b.

3. Just as in Example 1 of Section 9.7 we have Xn(x) = sin nx /a and


n y n yY y A B

a a

The condition Y(0) = 0 now yields An = 0 so ( ) sinh /n nY y B n y a , and hence

1

( , ) sin sinh .nn

n x n yu x y C

a a

Finally we satisfy the condition u(x, b) = f(x) by choosing /(sinh / ),n nC b n b a

where the {bn} are the Fourier sine coefficients of f(x) on 0 x a.

4. Because (0) ( ) 0,X X a we work with the separation of variables X'' + X = 0 = Y'' Y. The eigenvalue problem X'' + X = 0, (0) ( ) 0X X a has eigenvalues and eigenfunctions 0 = 0, X0(x) = 1 and

2 2

2,n

n

a

( ) cosn

n xX x

a

for n = 1, 2, 3, . When n = 0, Y0 = 0 yields Y0(y) = Ay + B. Then Y0(0) = 0

Section 9.7 587


gives B = 0, so we take Y0(y) = y. For n > 0 we have

( ) cosh sinh ,n n n

n y n yY y A B

a a

and Yn(0) = 0 gives An = 0, so

01

( , ) cos sinh .nn

n x n yu x y B y B

a a

Finally

01

( , ) cos sinh ,nn

n x n bu x b B b B

a a

so we satisfy the condition u(x, b) = f(x) by taking B0 = a0/2b and

/ sinh ,n n

n bB a

a

where 0

1

( ) cos .2 n

n

a n xf x a

a

5. Now (0) ( ) 0,Y Y b so we work with the separation of variables X'' X = 0 = Y'' + Y. The eigenvalue problem Y'' + Y = 0, (0) ( ) 0,Y Y b has eigenvalues and eigenfunctions 0 = 0, Y0(y) = 1 and

2 2

2,n

n

b

( ) cosn

n yY y

b

for n = 1, 2, 3, . When n = 0, 0( ) 0X x yields X0(x) = Ax + B. Then

X0(a) = 0 is satisfied by X0(x) = a x. For n > 0 we have

( ) cosh sinh ,n n n

n x n xX x A B

b b

and Xn(a) = 0 is satisfied by the particular linear combination

( ) sinh ,n n

n a xX x C

b

of cosh nx/b and sinh nx/b. Therefore

01

( )( , ) ( ) sinh cos .n

n

n a x n yu x y C a x C

b b

Finally we satisfy the condition u(0, y) = g(y) by choosing



00 2

aC

a and ,

sinh /n

n

bC

n a b

where the {an} are the Fourier cosine coefficients of g(y) on 0 y b.

6. This is the same as Problem 4 except that Y'(0) = 0 instead of Y(0) = 0, so Y0(y) = 1 and Yn(y) = Ancosh ny /a for n > 0. Then

01

( , ) cos cosh ,nn

n x n yu x y A A

a a

so we satisfy the condition u(x, b) = f(x) by choosing A0 = a0/2 and An = an /(cosh nb /a), where {an} are the Fourier cosine coefficients of f(x) on [0, a]. 7. The eigenvalue problem X'' + X = 0, X(0) = X(a) = 0 yields the eigenvalues and eigenfunctions

2 2

2,n

n

a

( ) sinn

n xX x

a

for n = 1, 2, 3, . Then 0n n nY Y

yields / /( ) .n y a n y a

n n nY y A e B e

In order that Y(y) 0 as y we take An = 0, so

/

1

( , ) sin .n y an

n

n xu x y B e

a

Finally we satisfy the condition u(x,0) = f(x) by choosing the constants {Bn} as the Fourier sine coefficients of f(x) on 0 x a.

8. The eigenvalue problem X'' + X = 0, X'(0) = X'(a) = 0 yields 0 = 0, X0(x) = 1 and

2 2

2,n

n

a

( ) cosn

n xX x

a

Section 9.7 589


for n > 0. Then 0n n nY Y

yields Y0(y) = A0y + B0 and / /( ) .n y a n y a

n n nY y A e B e

In order that Y(y) be bounded as y , we take A0 = 0 and An = 0 for n > 0, so

/0

1

( , ) cos .n y an

n

n xu x y B B e

a

Finally we satisfy the condition u(x, 0) = f(x) by choosing B0 = a0/2 and Bn = an where the {an} are the Fourier cosine coefficients of f(x) on [0, a].

9. If in Problem 8 we have ( ) 10f x x on 0 < x < 10, then

10

0 0

10

2 20

210 100,

102 200(cos 1 sin )

10 cos ,10 10n

a x dx

n x n na x dx

n

so

/102 2

odd

400 1( , ) 50 cos .

10n y

n

n xu x y e

n

Then (0,5) 50 8.4250 0.0405 0.0006 0.0000 41.53,u

(5,5) 50 0 0 0 0 50,u

(0,5) 50 8.4250 0.0405 0.0006 0.0000 58.47.u

10. The boundary value problem is uxx + uyy = 0 (0 < x < a, 0 < y < b)

u(0, y) = ux(a, y) = u(x, 0) = 0,

u(x, b) = f(x). The eigenvalue problem X'' + X = 0, X(0) = X'(a) = 0 yields (by Example 4 in Section 3.8)

2 2

2

(2 1),

4n

n

a

(2 1)

( ) sin2n

n xX x

a



for n = 1, 2, 3, . Then 0n n nY Y

yields

(2 1) (2 1)

( ) cosh sinh .2 2n n n

n y n yY y A B

a a

Because Y(0) = 0, we choose An = 0, so

1

(2 1) (2 1)( , ) sin sinh .

2 2nn

n x n yu x y B

a a

Finally we satisfy the condition u(x, b) = f(x) by choosing

2 1 ,sinh[(2 1) / 2 ]

nn

bB

n b a

where the {b2n1} are the odd half-multiple sine coefficients of f(x) on [0, a], as given by Problem 21 in Section 9.3.

11. Now the boundary value problem is uxx + uyy = 0 (0 < x < a, 0 < y < b)

u(a, y) = uy(x, 0) = u(x, b) = 0,

u(0, y) = g(y). The eigenvalue problem Y'' + Y = 0, Y'(0) = Y(b) = 0 yields (similar to Example 4 in Section 3.8)

2 2

2

(2 1),

4n

n

b

(2 1)

( ) cos2n

n yY y

b

for n = 1, 2, 3, . Then 0n n nX X

yields

(2 1) (2 1)

( ) cosh sin .2 2n n n

n x n xX x A B

b b

Now Xn(a) = 0 is satisfied by the particular linear combination

(2 1)

( ) sinh2n n

n a xX x C

b

of cosh (2n–1)x/2b and sinh (2n–1)x/2b. Hence

Section 9.7 591


1

odd

(2 1) ( ) (2 1)( , ) sinh cos

2 2

( )sinh cos .

2 2

nn

nn

n a x n yu x y C

b b

n a x n yA

b b

Finally we satisfy the condition u(0, y) = g(y) by choosing

,sinh / 2

nn

aA

n a b

where the {an} are the odd half-multiple cosine coefficients of g(y) on [0, b], as given by Problem 22 in Section 9.3.

12. The boundary value problem is uxx + uyy = 0 (0 < x < 30, y > 0)

u(0, y) = ux(30, y) = 0

u(x, y) bounded as y

u(x, 0) = 25 The eigenvalue problem X'' + X = 0, X(0) = X'(30) = 0 yields (by Example 4 in Section 3.8)

2 2(2 1)

,3600n

n (2 1)

( ) sin60n

n xX x

for n = 1, 2, 3, . Then

0n n nY Y

yields

(2 1) (2 1)

( ) exp exp .60 60n n n

n y n yY y A B

and we take An = 0 in order that Yn(y) be bounded as y . Hence

/ 60

odd

( , ) sin .60

n yn

n

n xu x y b e

Finally, by Problem 21 in Section 9.3, the odd half-multiple Fourier sine coefficients of u(x, 0) = 25 on [0, 30] are given by



30

2

0

2 200 10025sin sin

30 60 4n

n x nb dx

n n

for n odd. Thus

/ 60

odd

100 1( , ) sin .

60n y

n

n xu x y e

n

13. We start with the periodic polar-coordinate solution

0

1

( , ) ( cos sin )2

nn n

n

au r r a n b n

and choose 0na in order to satisfy the conditions ( ,0) ( , ) 0.u r u r Then

1

( , ) sinnn

n

u r r c n

satisfies the nonhomogeneous boundary condition ( , ) ( )u a f provided that nna c is

is the nth Fourier sine coefficient of ( )f on the interval 0 < < , that is,

0

2( )sin .n n

c f n da

14. We start with the periodic polar-coordinate solution

0

1

( , ) ( cos sin )2

nn n

n

au r r a n b n

and choose 0nb in order to satisfy the conditions ( ,0) ( , ) 0.u r u r Then

0

1

( , ) cos2

nn

n

cu r r c n

satisfies the nonhomogeneous boundary condition ( , ) ( )u a f provided that nna c is

is the nth Fourier sine coefficient of ( )f on the interval 0 < < , that is,

0

2( )cos .n n

c f n da

15. As in the textbook discussion of the polar-coordinate Dirichlet problem, the substitution

( , ) ( ) ( )u r R r in Laplace's equation yields the separated ordinary differential equations

2 0r R rR R (25) and 0. (26)

With 2 the general solution of (26) is

Section 9.7 593


( ) cos sin ,A B and the endpoint condition (0) (0) 0 yields A = 0 and (2 1) / 2,n so the nth eigenvalue and eigenfunction are given by

2(2 1)

,4n

n (2 1)

( ) sin .2n

n

As in the discussion of Eqs. (29) and (30) in the text, the bounded solution of

2

2 (2 1)0

4n n n

nr R rR R

is (2 1) / 2( ) n

nR r r

for 1, 2, 3, .n We thereby obtain the formal series solution

/ 2

odd

( , ) sin .2

nn

n

nu r c r

It remains only to satisfy the nonhomogeneous boundary condition ( , ) ( )u a f by choosing

/ 2

0

2( ) sin ,

2n n

nc f d

a

so that (for n odd) / 2nnc a equals the nth odd half-multiple sine coefficient of ( ).f

16. The only difference between the exterior problem here and the interior problem in the text is that in

Rn(r) = Cnrn + Dnr

n we must choose Cn = 0 in order that Rn(r) be bounded as r .

17. The substitution ( , ) ( ) ( )u r R r in Laplace's equation yields the same separated solution functions

0( ) 1, 0 0 0( ) lnR r C D r

and

( ) cos sin ,n n nA n B n ( ) n nn n n

DR r C r

r

as in Eqs. (28)-(30) in the text. We choose 0nB to satisfy the boundary condition

( , ) ( , )u r u r , and n = 1 with 1 0C U to satisfy the given limit condition as



.r Then the condition that ( , ) 0ru a requires that 21 0 ,D U a so

2 20( , ) cos .U

u r r ar

20. When we substitute v(r, t) = r u(r, t) we get the boundary value problem vt = kvrr (r < a, t > 0)

v(0, t) = v(a, t) = 0

v(r, 0) = T0r that corresponds to a heated rod along the interval 0 r a. It therefore follows from Equation (31) in Section 9.5 that

2 2

21

( , ) exp sin .nn

n kt n xv r t b

a a

To get the formula given in the text it remains only to calculate the Fourier sine coefficients {bn} of f(r) = T0r on 0 < r < a, and finally to divide v(r, t) by r to

get u(r, t).

21. (a) Since we cannot simply substitute r = 0, we apply continuity of u(r, t) at r = 0 and calculate

0

(0, ) lim ,r

u t u r t

noting that

0 0 0

sin / sin / sinlim lim lim

/r r r

n r a n n r a n n

r a n r a a a

by the elementary fact that sin / 1 as 0.

(b) With a = 30 and 0 100T we have

2 2

1

1

(0, ) 200 ( 1) exp .900

n

n

n ktu t

If k = 0.15 for iron then after 15 minutes = 900 seconds the center temperature is (0,900) 45.5075 0.5361 0.0003 0.0000 44.97.u If k = 0.005 for iron then after 15 minutes the center temperature is

Section 9.7 595


(0,900) 190.37 164.174 128.276 90.8081 58.2426

33.8449 17.8190 8.4998 3.6734 1.4384

0.5103 0.1640 0.0478 0.0126 0.0030

0.0007 0.0001 0.00002 0.00000

(0,900) 100.00

u

u

Thus the center of the ball has not yet begun to cool. For the center of this concrete ball to reach 45 (as with the iron ball after 15 minutes) would require (0.15/ 0.005) 15 = 450 minutes, that is, seven and a half hours!

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FOURIER SERIES METHODS AND PARTIAL DIFFERENTIAL …cbafaculty.org/Analysis_Math/DE_C_and_M_Chapter_9_sc.pdf · The basic trigonometric functions cos(t) and sin(t) have period P =