I was solving the Schrödinger equation with $V(r)=-\frac{a}{r}+br$, which, reduced to the non-dimensional form only for the radial part reads $$\left(-\frac{\mathrm{d}^2}{\mathrm{d} z^2}+\frac{\ell(\ell+1)}{z^2}-\frac{2}{z}+\lambda z\right)u(z)=\varepsilon u (z)$$
Here, $\lambda$ is parameter (basically tweaking the strength of the conical part of the potential) and $\varepsilon$ is the energy and $u(z)$ is a function s.t. $u (0) = 0$ and $u (z) \sim z^{l+1}$ when $z$ is small. Ideally, if $\varepsilon$ is an eigenvalue, then the solution also satisfies $\displaystyle\lim_{z\to\infty}u(z)=0$.
In practice, if you implement a stepping scheme from $z=0$ to larger values of $z$, the solution will eventually diverge (or oscillate like crazy if $\varepsilon$ is far away from any eigenvalue). I noticed, that numerically, whenever I just cross the "correct" eigenvalue $\varepsilon$, the direction of divergence changes sign. For example, if $\lambda = 0$ (hydrogen atom), the first eigenvalue is $\varepsilon=-1$. If $\varepsilon$ is set to $-1.001$ and the solution $u (z)\to -\infty$ for large $z$, then for $\varepsilon = -0.999$ the solution $u(z)\to+ \infty$ instead. Same is true near every higher energy and for $\lambda\ne 0$.
I naturally expect the solution to diverge if $\varepsilon$ does not match eigenvalues, but I find it interesting that the sign of this divergence always changes when $\varepsilon$ crosses over an eigenvalue.
My question:
Can I gain some insight into this sign changing from the differential equation?
Is this true for a wide variety of potentials when solving for bound states by the shooting method, or is there something special about this class of potentials I chose?
What exactly is the main reason for this sign flipping?
Thank you in advance!
If you trace the function $ε\mapsto u_ε(T)$ for some large value of $T$, then the graph will be continuous and its roots simple (if the eigenvalues are simple, which they usually are). The roots are close to the eigenvalues, and the values close-by change sign at the root. The larger $T$, the steeper the slope, so that without any renormalization it might look like a jump from $-\infty$ to $+\infty$ or the other way around.