# (geometric) Does an eigenfunction necessarily tends to infinity in regions where time - invariant potential is greater than energy?

A diatomic-like setup:

There are 3 regions: $$x < x', x \in [x', x''], x > x''$$ with potential being $$V\left(x\right) > E$$, $$V\left(x\right) < E$$ and $$V\left(x\right) > E$$, respectively, and in all regions the eigenfunction $$\psi\left(x\right) > 0$$ where $$x$$ describes the distance between two particles in the diatomic setup.

The time-independent Schrodinger equation is given as

$$\frac{d^{2}\psi}{dx^{2}} = \frac{2m}{\hbar^{2}}\left[V\left(x\right) - E \right] \psi$$

In $x < x''$:

The potential $$V\left(x\right)$$ encodes the distance $$x$$, as a denominator term, between the two particles resulting in the potential $$V\left(x\right)$$ tending to $$\infty$$ as the distance $$x$$ tends to 0. This gives $$\frac{d^{2}\psi}{dx^{2}} > 0$$ so the second spatial derivative of $$\psi$$ is concave upwards. So $$\psi$$ is infinitely large in this region as tends to infinity for $$x$$ closer to 0.

In $x \in [x', x'']$:

In this region, the second spatial derivative of $$\psi$$ would be negative.

In $x > x''$:

But if this were a potential with a limiting value, $$V_{l}$$ then there exists some $$x > x''$$ such that $$V$$ converges to $$V_{l}$$. In this region, just after $$x > x''$$, it is true that $$V\left(x\right) - E > 0$$ and $$V\left(x\right) - E$$ increases the further $$x$$ is from $$x''$$ and only for $$V\left(x\right) - E$$ to be constant at some $$x > x''$$ once $$V\left(x\right)$$ converges to $$V_{l}$$. Depending on the specificity of the potential, in this case depending at where $$V\left(x\right)$$ converges to $$V_{l}$$, the second spatial derivative for $$\psi$$, $$\frac{d^{2}\psi}{dx^{2}}$$, can be infinitely large. However, it would necessarily be positive based on the governing potential outlined here.

Am I right in my understanding of the behaviour for $$\psi$$ in the third region? Most authors seem to conclude that $$\psi$$ would tend to infinity in the third region given this specific potential. My understanding is that it may tend to infinity.

At a region where the second derivative of a function is positive. It doesn't lead to a ascending function. For example $$f(x) = e^{-x}$$ at region $$x>0$$
$$\begin{equation} \frac{d^2 f(x)}{dx^2} = + e^{-x} \gt 0 \end{equation}$$
But $$e^{-x}$$ is descending function for $$x>0$$.