# Strange behavior of solution to $- \frac{\partial^2 \psi\left(x\right)}{\partial x^2} - e^{-\alpha x} \psi =0$

Say we have the following equation the following equation (inspired from quantum mechanics with $V\left(x\right) = - e^{- \alpha x}$):

\begin{align} \label{eq:first-eq} - \frac{\partial^2 \psi\left(x\right)}{\partial x^2} - e^{-\alpha x} \psi =0 \tag{1} \end{align}

According to Mathematica the solutions are linear combinations of Bessel functions:

\begin{align} \label{eq:second-eq} \psi\left(x\right) = A J_{0} \left(\frac{2}{\alpha }e^{-\alpha x/2} \right) + B Y_{0} \left(\frac{2}{\alpha }e^{-\alpha x/2} \right) \tag{2} \end{align}

The code:

eq = -D[ψ[x], {x, 2}] - E^(-α*x)*ψ[x] == 0
DSolve[eq, ψ[x], x]
{{ψ[x] ->
BesselJ[0, (2 Sqrt[E^(-x α)])/α] C + 2 BesselY[0, (2 Sqrt[E^(-x α)])/α] C}}


$$- \frac{\partial^2 \psi\left(x\right)}{\partial x^2} - e^{\alpha x} \psi =0$$

Which is Eq. \ref{eq:first-eq} but with $\alpha \rightarrow -\alpha$, then we should get Eq. \ref{eq:second-eq} with the $\alpha \rightarrow -\alpha$. However, instead we get:

$$\psi\left(x\right) = A J_{0} \left(\frac{2}{\alpha }e^{\alpha x/2} \right) + B Y_{0} \left(\frac{2}{\alpha }e^{\alpha x/2} \right)$$

eq = -D[ψ[x], {x, 2}] - E^(α*x)*ψ[x] == 0
DSolve[eq, ψ[x], x]
{{ψ[x] ->
{{ψ[x] -> BesselJ[0, (2 Sqrt[E^(x α)])/α] C + 2 BesselY[0, (2 Sqrt[E^(x α)])/α] C}}


I know that $J_{0}$ is even but $Y_{0}$ is not, what is happening? Also, which of the solutions is actually correct?

• But then, you should be adding initial conditions that express the fact that $\psi(x)$ is regular at the origin, no? – J. M.'s technical difficulties Aug 3 '17 at 2:59
• Sorry I might be misunderstanding @J.M., if I plot this function it looks regular at the origin $x = 0$ ... Or how would you do that? – ace7047 Aug 3 '17 at 3:21

They are both correct:

eq = -D[ψ[x], {x, 2}] - E^(-α*x)*ψ[x] == 0;
DSolve[eq, ψ, x];
FullSimplify[eq /. First[%]]

(*Out= True*)

eq = -D[ψ[x], {x, 2}] - E^(α*x)*ψ[x] == 0;
DSolve[eq, ψ, x];
FullSimplify[eq /. First[%]]

(*Out= True*)


The parity of BesselJ and BesselY is completely irrelevant, given that their arguments are neither even or odd in α. And since you have two free constants any way, what's the contradiction?