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606 CHAPTER 16 The Wave Equation

and
F + (1/r)F 2

+ λr =− .
F
Because the left side depends only on r and the right side only on θ, there is some constant μ
such that
F + (1/r)F

2
+ λr =− = μ.
F
Then

+ μ = 0
and
2
2

r F +rF + (λr − μ)F = 0.
These differential equations for F(r), T (t), and (θ) come with the conditions. First, there is
the periodicity condition
(−π) = (π) and (−π) = (π).

Because of the fixed frame,
F(R) = 0.
Finally, if the membrane is released from rest,

T (0) = 0.
The problem for (θ) is a periodic Sturm-Liouville problem and was solved in Example 15.2
(here L = π). The eigenvalues are
2
μ n = n for n = 1,2,···
and the eigenfunctions are
n (θ) = a n cos(nθ) + b n sin(nθ).
2
Now we have μ n = n , so the problem for F(r) is
2
2
2
r F (r) +rF (r) + (kr − n )F(r) = 0; F(R) = 0.

This is a Bessel equation with general solution
√ √
F(r) = αJ n ( λr) + βY n ( λr),
√
with α and β as yet arbitrary constants. Because Y n ( λr) is unbounded as r → 0+ (the center
of the membrane), we must choose β = 0 to have a bounded solution. We expect oscillations of a
√
vibrating membrane to be finite in magnitude. This leaves F(r) = αJ n ( λr). To find admissable
values of λ, use the boundary condition to require that
√
F(R) = αJ n ( λR) = 0.
√
We must have α = 0 for a nontrivial solution. Thus, choose λR to be a positive zero of J n .Let
these positive zeros be
j n1 < j n2 < j n3 < ··· ,
2
which are double indexed because this derivation depends on the eigenvalue μ = n . Then
j 2
λ nk = nk
R 2

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October 14, 2010 15:23 THM/NEIL Page-606 27410_16_ch16_p563-610