We will derive eigenfunctions and eigenvalues on a Pacman domain, which in polar coordinates is $\Omega = \{(r, \theta) : r \in [0, 1], \theta \in [0, 3\pi/2]\}$.

The problem is

\begin{align*}

-\Delta u &= \lambda u \qquad \Omega\\

u &= 0 \qquad \partial \Omega

\end{align*}

In polar coordinates, the Laplacian is

\begin{align}

\Delta = \frac{\partial^2 }{\partial r^2} + \frac{1}{r} \frac{\partial}{\partial r} + \frac{1}{r^2} \frac{\partial^2}{\partial \theta^2}.

\end{align}

Thus, using separation of variables $u(r, \theta) = R(r) \Theta(\theta)$ where $R(1) = 0, \Theta(0) = \Theta(3\pi/2) = 0$, we have

\begin{align*}

\Delta u &= \Theta R” + \frac{1}{r} R’ \Theta + \frac{1}{r^2} R \Theta ” = -\lambda R \Theta.

\end{align*}

Simplifying, we have

\begin{align}\label{eqn:sum0}

\frac{r^2 R” + r R’ + \lambda r^2 R}{R} + \frac{\Theta ”}{\Theta} = 0.

\end{align}

In order for the above to be satisfied, we need each term to be constant, so assume that

\begin{align*}

\frac{\Theta”}{\Theta} = -\lambda_\theta

\end{align*}

where $-\lambda_\theta$ is a constant.

Taking into account the boundary condition, we know that

\begin{align*}

\Theta(\theta) = \sin\left(\frac{2}{3}n \theta \right)

\end{align*}

and $\lambda_\theta = \frac{4}{9}n^2$ for $n \in \mathbb{Z}$.

Now, using (2), we have the corresponding ODE for the $R$ variable

\begin{align*}

r^2 R” + r R’ + (\lambda r^2 – \frac{4}{9}n^2) R = 0.

\end{align*}

Let $\rho = \sqrt\lambda r$, then $R_r = R_\rho \frac{d\rho}{dr} = \sqrt\lambda R_\rho$ and hence $R_{rr} = \lambda R_{\rho\rho}$, hence

\begin{align*}

\rho^2 R” + \rho R’ + (\rho^2 – \frac{4}{9} n^2) R = 0.

\end{align*}

By the change of variables, we know that $R(\rho) = J_{2/3 n}(\rho)$ where $J$ is the Bessel function.

It remains to impose the boundary condition $R = 0$ at $r = 1$, so

\begin{align*}

R(\sqrt\lambda r) = J_{2/3 n}(\sqrt \lambda r) \qquad J_{2/3 n}(\sqrt{\lambda}) = 0.

\end{align*}

meaning that $\lambda = \alpha_{2/3 n, k}^2$ for $k \ge 1$, which are the eigenvalues.