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?
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 $\phi_k = R_k \Theta_k$.