Forced vibrations in an annulus / annular membrane. I am trying to find out how to solve the following problem: $$ \frac{ \partial^2 u }{\partial t^2 } = c^2 \nabla^2 + Q(x,y,t) , $$ in which we have the initial conditions $u(x,y,0) = f(x,y)$ and $\frac{ \partial u }{ \partial t } (x,y,0) = 0 $. We look at a membrane that is a circular annulus: $ a < r < b$ (with $u$ fixed on the boundaries: $u = 0 $ there).
The general solution to this problem is that $$u(x,y,t) = \sum_{i} A_i (t) \phi_i (x,y) .$$
Here, $\phi_i$ are the eigenfunctions that solve $$ \nabla^2 \phi = - \lambda \phi . $$
I know these eigenfunctions when the membrane is a rectangle 
( $ \phi_{nm} (x,y) = \sin (n \
pi x / L ) \cdot \sin (m \pi y / H) $ , in which $n , m \in \mathbb{Z}_{>0} $) and when the membrane is a circle with radius $a$ (now the eigenfunctions are : $\phi_m = J_m (z_{n m } r / a) \sin (m \theta) $, where $J$ is the Bessel-function. 
But I don't now how to find the eigenfunctions for this particular membrane. Can you help me out? 
 A: Sorry for putting up a late promise. I am rewriting your $\nabla^2 = \Delta$ here.
So now we wanna solve the eigenvalue problem on this annulus:
$$ \Delta w + \lambda w = 0 . $$
Suppose $w$ is separable in polar coordinates: 
$$w(x,y) = W(r,\theta) = R(r)\Theta(\theta).$$
Up until this step it is pretty standard, being the same with the disk (in your question the second case, it should be a disk, if it is a circle $\mathbb{S}^1$ then the eigenfunction would be simply $e^{i m\pi\theta}$). Now for the annulus the boundary condition should be
$$
R(a) = R(b) = 0,
$$
together with the periodical condition:
$$
\Theta(\theta+2\pi) = \Theta(\theta).
$$
Laplacian operator in 2 dimensional polar coordinate system is:
$$
\Delta w = \frac{\partial^2 w}{\partial r^2} + 
\frac{1}{r}\frac{\partial  w}{\partial r } +
\frac{1}{r^2}\frac{\partial^2 w}{\partial \theta^2} .
$$
Hence: 
$$
\Big(r^2 R'' + rR' + (\lambda r^2-k^2)R\Big)\Theta + (\Theta''+k^2\Theta)R = 0.
$$
Notice we wanna make positive $k^2$ term for $\Theta$ because its periodicity property. The solution to 
$ r^2 R'' + rR' + (\lambda r^2-k^2)R = 0 $ 
is the superposition of the first and second kind of Bessel function:
$$
R_k = \alpha_k J_k(\sqrt{\lambda} r) + \beta_k Y_k(\sqrt{\lambda} r).
$$
The solution to 
$\Theta''+k^2\Theta = 0$ is just 
$$
\Theta_k = a_k \sin(k\theta)+b_k \cos(k\theta).
$$ 
Setting the boundary condition for $R_k$ leads to:
$$
\alpha_k J_k(\sqrt{\lambda} a) + \beta_k Y_k(\sqrt{\lambda} a) = 0,
\\
\alpha_k J_k(\sqrt{\lambda} b) + \beta_k Y_k(\sqrt{\lambda} b) = 0.
$$
Eliminating the coefficients gives us an equation the eigenvalue $\lambda$ must satisfy:
$$
\color{blue}{J_k(\sqrt{\lambda} a)Y_k(\sqrt{\lambda} b) - J_k(\sqrt{\lambda} b)Y_k(\sqrt{\lambda} a)  =0 }.
$$
Then looking for a closed form for $\lambda$ is almost impossible, people normally start using numerical procedure from this point on. Lastly the eigenfunction will be given by $\phi_k = R_k \Theta_k$.
