Consider the differential equation $(\nabla^2+\frac{1}{R^2})\psi(\bar{r}) = 0$ in 2 dimensions, with the boundary condition $\partial_r\psi(R)+ \kappa \psi(R) = 0$, on unit disk of radius R. What is the solution of this boundary value problem ?
My work: I expanded $\psi(\bar{r}) =\sum_{p} a_p e^{ip.r}+a_p^\dagger e^{-ip.r}$, then we get $p^2 = m^2 $. Hence we get $\psi$. Using the boundary condition didn't give me anything ?
Not a full answer, but I'll show what I tried
Find a solution of the form
$$ \psi(r,\phi) = \sum_n P_n(r)\Phi_n(\phi) $$
where $\Phi_n(\phi) = A_n\cos(n\phi) + B_n\sin(n\phi)$ can be obtained through separation of variables.
The remaining radial equation is
$$ P_n''(r) + \frac{1}{r}P_n'(r) - \frac{n^2}{r^2}P_n(r) + \frac{1}{R^2}P_n(r) = 0 $$
Rearrange to
$$ r^2 P_n''(r) + r P_n'(r) + \left(\frac{r^2}{R^2}-n^2\right)P_n(r) = 0 $$
The general solution is
$$ P_n(r) = J_n\left(\frac{r}{R}\right) $$
where $J_n$ is the Bessel function of the first kind.
The boundary condition gives
$$ \frac{1}{R}J_n'(1) + \kappa J_n(1) = 0 $$
which is nonsense. Unless $R$ or $\kappa$ is allowed to vary, this problem is over-determined.
Are you sure the problem isn't
$$ \nabla^2 \psi + \frac{1}{\lambda^2}\psi = 0 $$
where $\lambda$ is some unknown constant?