Page 623 - Advanced_Engineering_Mathematics o'neil

16.7 Vibrations in a Circular Membrane I 603

Because the membrane is fixed on a circular frame,
ω ω

z ω (R,t) = a ω J 0 R cos(ωt) + b ω J 0 R sin(ωt) = 0
c c
for t > 0. This equation will be satisfied for all t > 0if J 0 (ωR/c) = 0. Thus, choose ω so that
ωR/c is a positive zero of J 0 (x). Let these zeros be j 1 < j 2 < ···. For each positive integer n,
choose ω n so that ω n R/c = j n . This gives us the eigenvalues
2
j n c

2
λ n = ω =
n
R
and the corresponding eigenfunctions

j n
J 0 r .
R
The functions

j n j n ct j n j n ct
z n (r,t) = a n J 0 r cos + b n J 0 r sin
R R R R
satisfy the wave equation and the boundary condition that z(R,0) = 0, for each positive integer
n. To satisfy the initial condition z(r,0) = f (r), use a superposition
∞
j n j n

j n ct j n ct
z(r,t) = a n J 0 r cos + b n J 0 r sin . (16.18)
R R R R
n=1
The initial condition gives us
∞
j n r

z(r,0) = f (r) = a n J 0 .
R
n=1
This is a Fourier-Bessel expansion, which we developed in Chapter 15 for the interval [0,1].Let
s =r/R to convert this expansion to
∞

f (Rs) = a n J 0 ( j n s)
n=1
on 0 ≤ s ≤ 1. Choose
1
2 sf (Rs)J 0 ( j n s)ds
a n = 0
2
J ( j n )
1
for n = 1,2,···.
Next solve for the b n ’s. We need
∞
∂z
j n c j n r
(r,0) = g(r) = b n J 0 .
∂t R R
n=1
Againwelet s =r/R to normalize the interval to [0,1] and obtain

1
2R sg(Rs)J 0 ( j n s)ds
b n = 0 .
2
j n c J ( j n )
1
With these coefficients, equation (16.18) is the solution for z(r,t).

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October 14, 2010 15:23 THM/NEIL Page-603 27410_16_ch16_p563-610