# Solving an eigenvalue problem on a circular domain

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

$$$$\Delta u=-\lambda u\\ u(r,0)=0, \ \ \ \ u(r,\pi/2)=0 \ \ \ \ \ \ 0\leq r\leq R\\ u(0,\theta)=0, \ \ \ \ \ u(R,\theta)=0 \ \ \ \ \ \ \ \ 0\leq \theta \leq\frac{\pi}{2}$$$$

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:

$$$$u_{rr}+\frac{1}{r}u_r-4n^2\frac{1}{r^2}u(r)=-\lambda u(r) \\ u_{rr}+\frac{1}{r}u_r+\bigg(\lambda-\frac{4n^2}{r^2}\bigg)u(r)= 0\\$$$$

So this is a Bessel equation, and $$\lambda>0$$, on a bounded domain, so we obtain the general solution, with $$v=2n$$:

$$$$R(r)=aJ_v(\sqrt{\lambda}r)$$$$

We then have the form for $$u(r,\theta)$$:

$$$$u(r,\theta)=aJ_v(\sqrt{\lambda}r)\sin 2n\theta$$$$

But how do I find this on the quarter-circle and with the right coefficient for $$R(r)$$?

Any hints appreciated.

Thanks

• Hint: use the B.C. on $r=R$ Aug 22, 2022 at 13:02
• Got it. Thanks. What puzzles me is that $a$ in front of the Bessel function. Is it just $1$? Aug 22, 2022 at 13:57
• Since the BC are homogeneous, $a$ is arbitrary. Consider the matrix equation $Ax=\lambda x$. If $U$ is a solution, then $aU$ is also a solution. PS, don't put your solution in the question, just self answer and accept. Aug 22, 2022 at 14:12

Using BCs as hinted, we get, where $$\alpha_{n,k}$$ are the Bessel zeros:
$$$$\alpha_{n,k}=R(R)\\ \alpha_{n,k}=aJ_v(\sqrt{\lambda}R)\\ \alpha_{n,k}=\sqrt{\lambda}R\\ \lambda=\bigg(\frac{\alpha_{n,k}}{R}\bigg)^2$$$$
$$$$u(r,\theta)=\sum_{n=1}^\infty\sum_{k=1}^\infty J_{2n}\bigg(\frac{\alpha_{n,k}}{R}r\bigg) \sin2n\theta$$$$