Poisson's Equation in a Disk

Question

How can we solve Poisson's equation in a disk in plane polar coordinates?:

$$ \nabla^2 \phi = u_{rr} + \frac{1}{r}u_r + \frac{1}{r^2} u_{\theta \theta} = f(r, \theta)$$

(My attempt):

We know that the eigenfunctions of the operator $ \frac{\partial^2 \theta}{\partial \theta^2} $ with periodicity $2 \pi$ are a superposition of sins and cosines, with eigenvalues $\lambda_n = n^2 $

So we want an eigenfunction of the form:

$$ \phi_{\lambda}(r, \theta) = \psi_{\lambda}(r) * [A_n \space \mathrm{cos}(n\theta) + B_n \space \mathrm{sin}(n\theta)] $$

Plugging this back into the PDE, we obtain:

$$ [\psi_{\lambda}''(r) + \frac{1}{r} \psi_{\lambda}'(r) - \frac{n^2}{r^2} \psi_{\lambda}(r)] * [A_n \space \mathrm{cos}(n\theta) + B_n \space \mathrm{sin}(n\theta)] + f(r, \theta) = 0$$

What do we do from here - in the Helmholtz equation, $\space f(r, \theta) = \lambda u $ and we obtain an nth order Bessel equation for the radial eigenfunction $\psi(r) $, but I can't figure out how to proceed from here, any gentle hints?

Answer 1

Try to expand $ f(r, \theta)$ using: $$\sum_n\left([\psi_{n}''(r) + \frac{1}{r} \psi_{n}'(r) - \frac{n^2}{r^2} \psi_{n}(r)] [A_n \space \mathrm{cos}(n\theta) + B_n \space \mathrm{sin}(n\theta)]\right) = - f(r, \theta)$$

Answer 2

If you have uniform Dirichlet conditions on the boundary, say $u(R,\theta) = -U$, then $u_{\theta\theta}=0$ and the solution is

$u(r,\theta) = C \ln \frac{r}R + \frac{R^2 - r^2}4 + U$

Assuming $u(0,\theta)$ is finite, C must be zero.

At least that's the solution Maple gave to a lazy old man.