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.