# Helmholtz equation with robin boundary condition

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 ?

• Is $R$ fixed? Does the solution need to be finite at the origin? – Dylan Mar 10 at 7:58
• R is fixed and solution needs to be finite at the origin. I tried it and got solutions to be related to modified bessel functions (two bessel functions) with one of them blowing up at the origin. I don’t know whether it’s right or wrong? – never_mind Mar 10 at 8:46
• The solutions are regular Bessel functions, not modified ones. But you won't get any further with these boundary conditions. – Dylan Mar 10 at 8:51
• Can you post your work ? – never_mind Mar 10 at 9:02

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?

• Ok now I see why lamda should be unknown. Thanks for the answer. – never_mind Mar 10 at 10:21