0
$\begingroup$

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 ?

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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?

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.