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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

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    $\begingroup$ Hint: use the B.C. on $r=R$ $\endgroup$
    – user619894
    Aug 22, 2022 at 13:02
  • $\begingroup$ Got it. Thanks. What puzzles me is that $a$ in front of the Bessel function. Is it just $1$? $\endgroup$ Aug 22, 2022 at 13:57
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    $\begingroup$ 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. $\endgroup$
    – user619894
    Aug 22, 2022 at 14:12

1 Answer 1

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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}

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