Is the eigenvalue decomposition of the Sturm-Liouville operator $$ Lf(x)=-f''(x)+h\sin(x)f'(x),\quad h>0, $$ with Neumann boundary conditions $f'(-\pi)=f'(\pi)=0$ on the Hilbert space $L^2([-\pi,\pi],\mu_h)$ known? Here $d\mu_h(x)=Z^{-1}e^{h\cos x}dx$ and $Z>0$ is chosen such that $\mu_h$ is a probability measure. I suspect it involves Bessel functions.

N.B. The Sturm-Liouville problem is unitarily equivalent to the Schrödinger operator $$ Hf(x)=-f''(x)+\frac{h}{4}(h\sin^2(x)-2\cos(x))f(x) $$ on $L^2([-\pi,\pi])$.

Edit: By the hint of Sascha, we transform $H$ into a Schrödinger operator with Whittaker-Hill potential. Set $y=x/2$, then $$ Hf(y)=-\frac14 f''(y)+\left(\frac{h^2}{8}-\frac{h^2}{8}\cos(4y)-\frac{h}{2}\cos(2y)\right)f(y). $$ Hence, $4H-h^2/2$ is a Schrödinger operator with Whittaker-Hill potential with parameters $\alpha=h/2$, $s=1$ (in the convention of the paper by Hemery and Veselov).

Is anything known about the spectral gap of this operator (as noted by Hemery and Veselov, the ground state eigenvalue is (not very surprisingly) $-h^2/2$)?

  • $\begingroup$ The operator $L$ is not self-adjoint, right? Could you thus specify what exactly you mean by "eigenvalue decomposition"? $\endgroup$ – Jochen Glueck Feb 27 '19 at 19:07
  • $\begingroup$ My bad. I missed some crucial information. I edited the question accordingly. $\endgroup$ – julian Feb 27 '19 at 19:16
  • $\begingroup$ I'd be surprised if you could find the eigenvalues explicitly here, but I might be wrong of course. $\endgroup$ – Christian Remling Feb 27 '19 at 19:42
  • $\begingroup$ So what do precise information on the spectral gap do you need? $\endgroup$ – Sascha Feb 28 '19 at 14:51
  • $\begingroup$ I need the second lowest eigenvalue. $\endgroup$ – julian Feb 28 '19 at 14:52

I think this is one of the quasi exactly solvable potentials for the Schrödinger equation, see this paper.

To say a bit more: For particular choices of $h$ certain eigenfunctions are explicit but not all of them.

  • $\begingroup$ Thank you for this hint. It works out in fact (see the edited question). $\endgroup$ – julian Feb 28 '19 at 14:06

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