Let $u \in C^{\infty}_c(\mathbb{R})$ and consider the Schrödinger operator $$ L =  \frac{\partial^2}{\partial x^2} + u(x).$$ I am looking for bound state solutions to $L \psi =  \kappa^2 \psi$ as explained below. Since $u(x)$ has compact support, for large $x$ the eigenvalue equation looks like $$  \frac{\partial^2 \psi}{\partial x^2} + \kappa^2 \psi = 0,$$ the solutions to which are linear combinations of $\mathrm{e}^{\pm \kappa x}$. I am interested in the specific solution which is asymptotic to $ \mathrm{e}^{\kappa x}$ as $x \to +\infty$. Now, at the other end of the real line, such a solution should be asymptotic to $$ \alpha(\kappa) \mathrm{e}^{\kappa x} + \beta(\kappa) \mathrm{e}^{\kappa x}, \qquad x \to \infty, $$ and I wish to require that $\beta(\kappa) = 0$. I have a feeling that there should only be finitely many values of $\kappa$ for which this is satisfied, but I don't really have a clue how to prove it. KreinRutmann Theoremlike arguments I think may give you that there are only countably many such eigenvalues, but that's a general theorem and I don't see how it should say anything about the particular solutions I am considering.

$\begingroup$ There are only finitely many negative eigenvalues. This question is particularly discussed in connection with the inverse scattering method as applied to the Kortewegde Vries equation. $\endgroup$ – ifw Jan 29 '17 at 21:05

$\begingroup$ Yes, I am trying to understand this in connection with the inverse scattering transform. Could you point to a reference that explains this issue? $\endgroup$ – onamoonlessnight Jan 29 '17 at 21:40

2$\begingroup$ This is e.g. explained in Chapter 2 of Solitons, Nonlinear Evolution Equations and Inverse Scattering by Ablowitz and Clarkson. See there in particular Lemma 2.2.2 and its proof. $\endgroup$ – ifw Jan 29 '17 at 22:18
The number of bound states is indeed finite. This can be proved as follows. First of all, for a bound state your eigenvalue $k^2$ must be real. This is because your
operator with real $u$ and zero boundary conditions at $\pm\infty$ is selfadjoint.
Next, if real $\kappa^2$ is very large by absolute value, then you cannot have a bound state: if $\kappa^2$ is very negative, $\psi$ oscillates at $\infty$,
if $\kappa^2$ is very positive, $\phi$ cannot be zero on
both infinities by the Sturm comparison theorem. It remains to notice that your function $\beta(\kappa)$ is
analytic, so it cannot have infinitely many zeros on a finite interval.

$\begingroup$ $\beta(\kappa)=W(e^{\kappa x}, \psi)/(2\kappa)$ is actually not holomorphic at $\kappa =0$, which is exactly the point that matters here, but of course there are alternative arguments that show that there are only finitely many eigenvalues (oscillation theory works fine). $\endgroup$ – Christian Remling Jan 30 '17 at 22:17

$\begingroup$ Just a pole at $\kappa=0$ does not alter my argument. $\endgroup$ – Alexandre Eremenko Jan 31 '17 at 7:55