Wave equation on a disk (circular membrane) Solve wave equation in a disk, axisymmetric case
$$\begin{cases}
\frac{\partial^2u}{\partial t^2}=\frac{c^2}{r}\frac{\partial}{\partial r}(r\frac{\partial u}{\partial r}) \,\,\, \,,0<r<a\quad,t>0\\
u(r,0)=f(r),\quad\frac{\partial u}{\partial t}(r,0)=g(r),\quad u(a,t)=0
\end{cases}$$
My attempt:
Note the function ${\displaystyle u}$  does not depend on the angle ${\displaystyle \theta ,}$ because we have axisymmetric case of a circular membrane.
Let $u(r,t)=R(r)P(t)$ and replacing in the PDE we have:
$$R(r)P''(t)=c^2[\frac{R'(r)}{r}+R''(r)]P(t)\tag1$$
Dividing $(1)$ for $R(r)P(t)$ we have:
$$\frac{P''(t)}{c^2P(t)}=[\frac{1}{r}\frac{R'(r)}{R(r)}+\frac{R''(r)}{R(r)}]$$
The left-hand side of this equality does not depend on ${\displaystyle r,}$ and the right-hand side does not depend on ${\displaystyle t,}$ it follows that both sides must be equal to some constant ${\displaystyle \lambda.}$
Then
$$\lambda=\frac{P''(t)}{c^2P(t)}=[\frac{1}{r}\frac{R'(r)}{R(r)}+\frac{R''(r)}{R(r)}]$$
Of this we have two equations
$$\begin{cases}
P''(t)-c^2\lambda P(t)=0\\
rR''(r)+R'(r)-r\lambda R(r)\tag2
\end{cases}$$
We're going to solve $P''(t)-c^2\lambda P(t)=0$
If $\lambda=0$ then the solution is of the form:
$$P(t)=c_1+c_2t$$
If $\lambda>0$ then the solution is of the form
$$P(t)=c_1e^{ckt}+c_2e^{-ckt}$$
If $\lambda<0$ then the solution is of the form
$$P(t)=c_1\cos(ckt)+c_2\sin(ckt)$$
Here i'm stuck.
 A: The BVP
$$ r^2R'' + rR' - \lambda r^2 R = 0, \quad R(0) < \infty, R(a) = 0 $$
only has a non-trivial solution when $\lambda < 0$. 
You can check that $\lambda = 0$ returns a general solution of $A+B\ln r$, and $\lambda > 0$ returns modified Bessel functions, neither of which will satisfy the boundary conditions.
The substitution $\rho = \sqrt{-\lambda}r$ results in Bessel's equation (check this for yourself), so we have a general solution
$$ R(r) = AJ_0(\sqrt{-\lambda}r) + BY_0(\sqrt{-\lambda}r) $$
Note that $Y_0$ blows up at $r=0$ so we need to set $B=0$. 
This leaves the boundary condition $J_0(\sqrt{-\lambda}a)=0$. Let $\alpha_n$ be the zeroes of $J_0(x)$ with $n=1,2,3,\dots$ then we can rewrite the solution as
$$ R_n(r) = J_0\left(\frac{\alpha_n}{a}r\right) $$
up to a multiplicative constant.
Therefore
$$ u(r,t) = \sum_{n=1}^\infty \left[C_n \cos\left(\frac{\alpha_n}{a}ct\right) + D_n \sin\left(\frac{\alpha_n}{a}ct\right)\right]J_0\left(\frac{\alpha_n}{a}r\right)  $$
The initial conditions give
\begin{align}
u(r,0) &= f(r) = \sum_{n=1}^\infty C_n J_0\left(\frac{\alpha_n}{a}r\right)  \\
u_t(r,0) &= g(r) = \sum_{n=1}^\infty \frac{c\alpha_n}{a}D_n J_0\left(\frac{\alpha_n}{a}r\right) 
\end{align}
To determine the remaining constants, you must find the Fourier-Bessel series of $f(r)$ and $g(r)$. The process is very similar to the Fourier series, since Bessel functions are also mutually orthogonal.
