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610 CHAPTER 16 The Wave Equation

∞ L

mπy 2 nπξ
a nm sin = h y (ξ)sin dξ
K L L
m=1 0
2 L nπξ
= f (ξ, y)sin dξ
L 0 L
The integral on the right is a function of y, and on the left is a series that we can think of as its
Fourier sine expansion on [0, K], with the a nm ’s as coefficients. Therefore,
2 K 2 L nπξ mπη
a nm = f (ξ, y)sin dξ sin dη
K 0 L 0 L K
4 L K nπξ mπη
= f (ξ,η)sin sin dη dξ.
LK 0 0 L K
With this choice of constants we have the solution for the displacement function z(x, y,t).

EXAMPLE 16.15
Suppose the initial position function is
z(x, y,0) = x(L − x)y(K − y)
and the initial velocity is zero. Compute

4 L K nπξ mπη
a nm = ξ(L − ξ)η(K − η)sin sin dη dξ
LK 0 0 L K
2
16L K 2
n
m
= [(−1) − 1][(−1) − 1].
2 3
(nmπ )
The solution is
z(x, y,t) =
∞ ∞ 2 2

16L K nπx mπy
n
m
[(−1) − 1][(−1) − 1]sin sin cos(α nm πct).
(nmπ ) L K
2 3
n=1 m=1
SECTION 16.9 PROBLEMS

2
In each of Problems 1, 2, and 3, solve the problem for the 1. c = 1, L = K = 2π, f (x, y) = x sin(y), g(x, y) = 0
rectangular membrane with the given c, L, K, f (x, y) and 2. c = 3, L = K = π, f (x, y) = 0, g(x, y) = xy
g(x, y). 3. c = 2, L = K = 2π, f (x, y) = 0, g(x, y) = 1

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October 14, 2010 15:23 THM/NEIL Page-610 27410_16_ch16_p563-610