Eigenvalues of Sturm–Liouville operator

Question

Can we calculate the eigenvalues and eigenfunctions of the following operator in $W^{1,2}(\mathbb{R})$?

$$-\left(\frac{1}{\cosh^2x}\right)y''-\frac{2}{\cosh^4x}y=\lambda y.$$

Answer 1

For me, the first (and usually only) attempt with such Sturm-Liouville equations is to rewrite them as Schrodinger equations, using this transformation.

If I got everything right, ... EDIT: but I didn't in the first attempt, as hyh pointed out. The correct transformed equation is $$ -u'' + V(t)u = \lambda u , \quad\quad V(t) =\frac{-t^2-6}{4(t^2+1)^2} ; $$ the variables are related to the old ones by $t=\sinh x$, $u=\sqrt{\cosh x} y$.

This is a similar, but slightly more subtle situation than with my original incorrect potential. We now have $V=-1/(4t^2)-1/t^4+O(t^{-6})$. What is clear is that we have purely absolutely continuous spectrum of multiplicity $2$ on $\sigma_{ac}=[0,\infty)$. Below $\lambda =0$, we have purely discrete spectrum.

The question of whether we have finitely or infinitely many eigenvalues below zero is probably answerable, but seems a bit tricky. The potential $W=-1/(4t^2)$ would lead to finite discrete spectrum, but is exactly the borderline case in the sense that $W=-c/t^2$ for $c>1/4$ would give infinitely many eigenvalues.

EDIT 2: I'm reasonably confident now that there are only finitely many eigenvalues, though to show it properly would probably require some work. We could try to argue as follows: on the interval $2^n\le t\le 2^{n+1}$, we can bound $V$ from below by $-c_n/t^2$, $c_n=1/4+2^{-2n}$. For this modified potential, the solutions at $\lambda =0$ look like $t^{1/2}t^{i2^{-n}}=t^{1/2}e^{i2^{-n}\log t}$ (the equation for the modified potential is an Euler equation and can be solved explicitly). So the phase increases by $\approx n2^{-n}$ on the interval under consideration. Since this is summable, we expect (real valued) solutions to have only finitely many zeros, and then the claim follows from oscillation theory (see the comments).

Third and (let's hope) final EDIT: There are indeed only finitely many eigenvalues. A much better argument (than the previous paragraph) is given by Giorgio in the comments.