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 ?

  • $\begingroup$ Is $R$ fixed? Does the solution need to be finite at the origin? $\endgroup$ – Dylan Mar 10 at 7:58
  • $\begingroup$ 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? $\endgroup$ – never_mind Mar 10 at 8:46
  • $\begingroup$ The solutions are regular Bessel functions, not modified ones. But you won't get any further with these boundary conditions. $\endgroup$ – Dylan Mar 10 at 8:51
  • $\begingroup$ Can you post your work ? $\endgroup$ – 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?

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

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