Page 370 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 13] FOURIER SERIES 361

13.40. By differentiating the result of Problem 13.35(b), prove that for 0 @ x @ ,
4 cos x cos 3x cos 5x

2 1 3 5
x ¼ 2 þ 2 þ 2 þ
PARSEVAL’S IDENTITY
13.41. By using Problem 13.35 and Parseval’s identity, show that

1 4 1 6
X 1 X 1
n 90 n 945
ðaÞ 4 ¼ ðbÞ 6 ¼
n¼1 n¼1
2
1 1 1 8
13.42. Show that þ þ þ ¼ . [Hint: Use Problem 13.11.]
2
2
2
1 3 2 3 5 2 5 7 2 16
1 1
1 4 1 6
X X
13.43. Show that (a) ; .
4 ¼ 96 ðbÞ 6 ¼ 960
n¼1 ð2n 1Þ n¼1 ð2n 1Þ
2
1 1 1 4 39
13.44. Show that þ þ þ ¼ .
2
2
2
2
2
2
1 2 3 2 2 3 4 2 3 þ 4 þ 5 2 16
BOUNDARY-VALUE PROBLEMS
2
@U @ U
13.45. (a) Solve ¼ 2 subject to the conditions Uð0; tÞ¼ 0; Uð4; tÞ¼ 0; Uðx; 0Þ¼ 3 sin x 2 sin 5 x, where
@t @x 2
0 < x < 4; t > 0.
(b)Give a possible physical interpretation of the problem and solution.
2
Ans: ðaÞ Uðx; tÞ¼ 3e 2 t sin x 2e 50 t 2 sin 5 x.
2
@U @ U 10 < x < 3
13.46. Solve and interpret
@t ¼ @x 2 subject to the conditions Uð0; tÞ¼ 0; Uð6; tÞ¼ 0; Uðx; 0Þ¼ 03 < x < 6
physically.

1
X 2 2 m x
Ans: 2 1 cosðm =3Þ e m t=36 sin
m 6
Uðx; tÞ¼
m¼1
13.47. Show that if each side of equation (3), Page 356, is a constant c where c A 0, then there is no solution
satisfying the boundary-value problem.
13.48. Aflexible string of length is tightly stretched between points x ¼ 0 and x ¼ on the x-axis, its ends are
fixed at these points. When set into small transverse vibration, the displacement Yðx; tÞ from the x-axis of
2
2
@ Y 2 @ Y
2
any point x at time t is given by ¼ a , where a ¼ T= ; T ¼ tension, ¼ mass per unit length.
@t 2 @x 2
2
(a)Find a solution of this equation (sometimes called the wave equation)with a ¼ 4 which satisfies the
conditions Yð0; tÞ¼ 0; Yð ; tÞ¼ 0; Yðx; 0Þ¼ 0:1 sin x þ 0:01 sin 4x; Y t ðx; 0Þ¼ 0for 0 < x < ; t > 0.
(b)Interpret physically the boundary conditions in (a) and the solution.
Ans. (a) Yðx; tÞ¼ 0:1 sin x cos 2t þ 0:01 sin 4x cos 8t
2
2
@ Y @ Y
13.49. (a) Solve the boundary-value problem ¼ 9 subject to the conditions Yð0; tÞ¼ 0; Yð2; tÞ¼ 0,
@t 2 @x 2
Yðx; 0Þ¼ 0:05xð2 xÞ; Y t ðx; 0Þ¼ 0, where 0 < x < 2; t > 0. (b)Interpret physically.
1:6 X 1 ð2n 1Þ x 3ð2n 1Þ t
1
Ans: ðaÞ Yðx; tÞ¼ 3 3 sin cos
2 2
n¼1 ð2n 1Þ
2
@U @ U
13.50. Solve the boundary-value problem ¼ ; Uð0; tÞ¼ 1; Uð ; tÞ¼ 3; Uðx; 0Þ¼ 2.
@t @x 2
[Hint: Let Uðx; tÞ¼ Vðx; tÞþ FðxÞ and choose FðxÞ so as to simplify the differential equation and boundary
conditions for Vðx; tÞ:Š