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16.9 Vibrations in a Rectangular Membrane 609
The left side depends only on y and t and the right only on x, and these variables are independent,
so both sides must be constant:
T Y X
− = =−λ.
2
c T Y X
Then
T Y
X + λX = 0 and + λ = .
c T Y
2
The equation for T and Y has one side dependent only on t and the other side only on y.
Therefore, for some constant μ,
T Y
+ λ = =−μ.
c T Y
2
Then
2
Y + μY = 0 and T + c (λ + μ)T = 0.
Separation of variables has introduced two separation constants. From the boundary conditions,
X(0) = X(L) = Y(0) = Y(K) = 0.
We have solved these problems for X and Y before, obtaining eigenvalues and eigenfunctions
2
n π 2 nπx
λ n = , X n (x) = sin
L 2 L
and
2
m π 2 mπy
μ m = ,Y m (x) = sin
L 2 K
with n and m varying independently over the positive integers. The problem for T becomes
2 2 2 2
n π m π
2
T + c + T = 0.
L 2 K 2
With zero initial velocity we have T (0) = 0. Therefore, T (t) must be a constant multiple of
cos(α nm πct), where
n 2 m 2
α nm = + .
L 2 K 2
For each positive integer n and m, we now have functions
nπx mπy
z nm (x, y,t) = a nm sin sin cos(α nm πct).
L K
that satisfy the wave equation and the boundary conditions, as well as the condition of zero initial
velocity. To satisfy z(x, y,0) = f (x, y), attempt a superposition, which is now a double sum:
∞ ∞
z(x, y,t) = z nm (x, y,t).
n=1 m=1
We must choose the coefficients so that
∞ ∞
nπx mπy
z(x, y,0) = a nm sin sin = f (x, y).
L K
n=1 m=1
If we think of y as fixed for the moment, then f (x, y) = h y (x) is a function of x.Now
∞ ∞
mπy nπx
f (x, y) = h y (x) = a nm sin sin
K L
n=1 m=1
is the Fourier sine expansion in x of f (x, y) on [0, L]. Therefore, the coefficient of sin(nπx/L),
which is the entire sum in square brackets, is the Fourier sine coefficient of this function. For a
given n,
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October 14, 2010 15:23 THM/NEIL Page-609 27410_16_ch16_p563-610

