# Why does the “shooting” or “wag the dog” method give a bound state?

I am using numerical methods to solve Schrodingers equation. I have identified an interval for E (energy) in which one solution tends to infinity, but on the other side the other solution tends to negative infinity. (see https://ocw.mit.edu/courses/physics/8-04-quantum-physics-i-spring-2016/lecture-notes/MIT8_04S16_LecNotes12.pdf page 12, for example).

But I don't understand why there must be a solution in this interval? Why can't all values of E give solutions which rise to positive/negative infinity?

• Some reformulation of the claim: In the regular situation, the function $x(T;E)$, where $t\mapsto x(t;E)$ is the solution for energy $E$ is differentiable and has a regular $E$-derivative, so that the equation $x(T;E)$ can be solved for $E(T)$, at least for $T$ large enough. Then the claim is that this implicitly defined function $E(T)$ is asymptotically constant, with the asymptotic value an energy eigenvalue of the original equation. – LutzL Apr 26 at 17:12