Maybe you can help me solve this (simple?) problem I'm too stupid to tackle :-( I want to find the eigenfunctions to the Operator $$ \widehat{O} = -\partial_z^2 - \frac{1}{r}\partial_r r\partial_r $$ that is $$ \widehat{O} \, \psi_n = \lambda_n \psi_n \, . $$ So it's basically the cylindrical laplacian without $\varphi$-part. Since the Operator $\partial_z^2$ commutes with $\widehat{O}$ both can be diagonalized simultaneously, right? Call this Eigenvalue $k^2$. Then $$ \left(-k^2 - \frac{1}{r}\partial_r r\partial_r \right) \psi_n = \lambda_n \psi_n $$ is solved by $$ \psi_n = c(z) \, J_0\left(\sqrt{k^2+\lambda} \, r\right) $$ where $c(z)$ is the $z$-dependent constant with respect to $r$ which solves $$ \partial_z^2 \, c(z) = k^2 c(z) \, . $$ We therefore have the solution $$ \psi_n = e^{-kz} \, J_0\left(\sqrt{k^2+\lambda} \, r\right) \, . $$ In choosing this I assumed the solution be regular at $r=0$ and I restrict myself to $z\geq 0$.

So far so simple...Now comes the tricky part.

Let us suppose $\psi_n$ vanishes at some boundary $r=r_0$ which leads to $$ \sqrt{k^2 + \lambda} \, r_0 = j_n \qquad \Longrightarrow \qquad \lambda_n = \left(\frac{j_n}{r_0}\right)^2 - k^2 $$ where $j_n$ are the zeros of the Bessel-function.

Now here is my first question. In memory I tought that any such solution obeys some completeness-relation. However how does it look like? I know that $$ \sum_n {\cal N}_n^{-2} \, J_0\left(\frac{j_n r}{r_0}\right) \, J_0\left(\frac{j_n r'}{r_0}\right) = \frac{\delta(r-r')}{r} $$ where ${\cal N}_n$ is some normalization constant. But how does the corresponding part look for the $z-$ dependence? The problem is that it is not some kind of orthogonal function. When going from $k^2 \rightarrow -k^2$ the problem turns around.

I presume the general solution reads $$ \psi(r,z) = \int_0^\infty {\rm d} k \sum_n c_n(k) \, e^{-kz} \, J_0\left(\frac{j_n r}{r_0}\right) $$ but does this also mean I can expand any function in terms of these eigenfunctions?

  • $\begingroup$ The $z$ dependent part by itself has continuous spectrum, but because the $r$ dependent part has discrete spectrum, doing separation of variables selects only a discrete subset of the spectrum of the $z$ part. In any case completeness of the eigenfunctions is generally nontrivial... $\endgroup$ – Ian Jun 19 '18 at 21:14
  • $\begingroup$ So does this operator have a completeness relation of the form $$ \sum_{n,k} \psi_{n,k}(r,z) \, \psi_{n,k}(r',z') = \frac{\delta(z-z') \delta(r-r')}{r} \, ? $$ $\endgroup$ – Diger Jun 19 '18 at 21:21
  • $\begingroup$ Or must I indeed (for whatever reason) consider $-k^2$ instead of $k^2$ and then use $\int_{-\infty}^{\infty} e^{ikz} \, {\rm d}k = 2\pi \, \delta(z)$ ? $\endgroup$ – Diger Jun 19 '18 at 21:35

Your Answer

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

Browse other questions tagged or ask your own question.