2
$\begingroup$

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[1] + 2 BesselY[0, (2 Sqrt[E^(-x α)])/α] C[2]}}

However, some is strange about this solution. Specifically notice that if we consider the equation:

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

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[1] + 2 BesselY[0, (2 Sqrt[E^(x α)])/α] C[2]}}

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

$\endgroup$
2
  • $\begingroup$ But then, you should be adding initial conditions that express the fact that $\psi(x)$ is regular at the origin, no? $\endgroup$ Commented Aug 3, 2017 at 2:59
  • $\begingroup$ 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? $\endgroup$
    – ace7047
    Commented Aug 3, 2017 at 3:21

1 Answer 1

7
$\begingroup$

They are both correct:

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

(*Out[70]= True*)

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

(*Out[67]= 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?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.