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16.8 Vibrations in a Circular Membrane II 607

are the eigenvalues in the problem for F(r). Corresponding eigenfunctions are

j nk
J n r for n = 0,1,2,··· ,k = 1,2,··· .
R
With these eigenvalues λ, the problem for T is
2

T + c 2 j nk T = 0; T (0) = 0.

R
Solutions are constant multiples of

j nk ct
T nk (r) = cos .
R
We now have functions

j nk j nk
z nk (r,θ,t) =[a nk cos(nθ) + b nk sin(nθ)]J n r cos ct
R R
for n = 0,1,2··· and k = 1,2,3,···. Each of these functions satisfies the wave equation and the
boundary conditions, together with the initial condition of zero velocity. To satisfy the condition
on initial position given by f , we must in general use the superposition
∞ ∞
j nk j nk

z(r,θ,t) = [a nk cos(nθ) + b nk sin(nθ)]J n r cos ct .
R R
n=0 k=1
We must choose the constants a nk and b nk to satisfy
∞ ∞

j nk
z(r,θ,0) = [a nk cos(nθ) + b nk sin(nθ)]J n r = f (r,θ)).
R
n=0 k=1
To see how to choose these coefficients, first write this equation in the more suggestive form
∞
j nk

f (r,θ) = a 0k J 0 r
R
k=1

∞ ∞ ∞ ∞

j nk j nk
+ a nk J n r cos(nθ) + b nk J n r sin(nθ) .
R R
n=1 k=1 n=1 k=1
For a given r, f (r,θ) is a function of θ, and this is the Fourier series for this function of θ on
[−π,π]. In this case, the coefficients are infinite series, but they are also the Fourier coefficients
of f (r,θ) for a fixed r. We know these Fourier coefficients. For a given r,
∞ π
j nk

1
a 0k J 0 r = f (r,θ)dθ = α 0 (r),
R 2π
k=1 −π
and, for n = 1,2,···,
∞ π
j nk

1
a nk J n r = f (r,θ)cos(nθ)dθ = α n (r),
R π
k=1 −π
and
∞ π
j nk

1
b nk J n r = f (r,θ)sin(nθ)dθ = β n (r),
R π
k=1 −π
Now recognize that, for n = 0,1,2,···, each of the last three equations is an expansion of a
function of r in a Fourier-Bessel series, with coefficients a 0k , a nk and b nk , respectively. We know
the coefficients in these expansions. Make the change of variables r/R = ξ so that ξ varies from
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October 14, 2010 15:23 THM/NEIL Page-607 27410_16_ch16_p563-610