I have the given problem to solve:
Solve the eigenvalue problem for $-\Delta$ on the quarter-circle $x^2+y^2\leq R^2, x\geq 0,\ y\geq 0$ with homogeneous Dirichlet conditions.
This is what I did. Since we have a Dirichlet homogeneous condition, we can prepare the Ansatz :
$$u(r,\theta)=u(r)\sin2n\theta,$$
for the PDE problem
\begin{equation} \Delta u=-\lambda u\\ u(r,0)=0, \ \ \ \ u(r,\pi/2)=0 \ \ \ \ \ \ 0\leq r\leq R\\ u(0,\theta)=0, \ \ \ \ \ u(R,\theta)=0 \ \ \ \ \ \ \ \ 0\leq \theta \leq\frac{\pi}{2} \end{equation}
Since, the operator is $\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r^2}+\frac{\partial}{\partial\theta^2}$ we get:
\begin{equation} u_{rr}+\frac{1}{r}u_r-4n^2\frac{1}{r^2}u(r)=-\lambda u(r) \\ u_{rr}+\frac{1}{r}u_r+\bigg(\lambda-\frac{4n^2}{r^2}\bigg)u(r)= 0\\ \end{equation}
So this is a Bessel equation, and $\lambda>0$, on a bounded domain, so we obtain the general solution, with $v=2n$:
\begin{equation} R(r)=aJ_v(\sqrt{\lambda}r) \end{equation}
We then have the form for $u(r,\theta)$:
\begin{equation} u(r,\theta)=aJ_v(\sqrt{\lambda}r)\sin 2n\theta \end{equation}
But how do I find this on the quarter-circle and with the right coefficient for $R(r)$?
Any hints appreciated.
Thanks
Using BCs as hinted, we get, where $\alpha_{n,k}$ are the Bessel zeros:
\begin{equation} \alpha_{n,k}=R(R)\\ \alpha_{n,k}=aJ_v(\sqrt{\lambda}R)\\ \alpha_{n,k}=\sqrt{\lambda}R\\ \lambda=\bigg(\frac{\alpha_{n,k}}{R}\bigg)^2 \end{equation}
So the overall solution is:
\begin{equation} u(r,\theta)=\sum_{n=1}^\infty\sum_{k=1}^\infty J_{2n}\bigg(\frac{\alpha_{n,k}}{R}r\bigg) \sin2n\theta \end{equation}