I was wondering whether this ODE has been studied yet or whether there is anything we can say about its solutions?

$$(1-t^2)u_{tt}-tu_t+\left[ n \beta (2t^2-1)+ \beta^2 (2t^2-1)^2+C\right]u=0$$

$C$ is a free parameter. So if you know a function that would fulfill this equation only for particular $C$, this would be perfectly fine. I am interested in its solutions on $[-1,1]$.

I should include the motivation/reference here: Physics (Quantum Mechanics) and you might want to see this great answer that gives us first hints about the structure of the solution (due to O.L.) https://math.stackexchange.com/questions/780591/how-to-make-a-smart-guess-for-this-ode

The problem is that the approach taken here does not offer an analytical representation of the solution. I suspect that the solutions form a nice orthogonal basis of $L^2[-1,1]$, but was incapable of constructing them by recursion or explicit representation.

  • $\begingroup$ What is the motivation for this question? $\endgroup$ – Noah Schweber May 4 '14 at 22:08

Your equation has a symmetry which allows you to separately consider even and odd solutions. For either class of solutions you get an equation which has regular singular points at 0 and 1 and an irregular point of rank 1 at infinity. Such an equation is known as a confluent Heun equation. Heun functions are implemented in Maple.

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    $\begingroup$ Specifically, the solution Maple 18 finds is $$ u \left( t \right) ={ a_1}\,{{\rm e}^{\beta\,\sqrt {n}{t}^{2}}}{ \it HeunC} \left( 2\,\beta\,\sqrt {n},-1/2,-1/2,-\beta/2,3/8-1/4\,n{ \beta}^{2}-C/4+\beta/4,{t}^{2} \right) +{a_2}\,{{\rm e}^{\beta\, \sqrt {n}{t}^{2}}}{\it HeunC} \left( 2\,\beta\,\sqrt {n},1/2,-1/2,- \beta/2,3/8-1/4\,n{\beta}^{2}-C/4+\beta/4,{t}^{2} \right) t $$ $\endgroup$ – Robert Israel May 5 '14 at 7:18
  • $\begingroup$ @MichaelRenardy could you give me a hint/reference how such a separation is done? Cause I have difficulties to see the relationship betwen the Heun's function and my equation. How do I use the symmetry? $\endgroup$ – user37929 May 5 '14 at 13:32
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    $\begingroup$ Substitute $s=t^2$ and $u(t)=v(s)$ or, respectively $u(t)=tv(s)$. $\endgroup$ – Michael Renardy May 5 '14 at 13:42
  • $\begingroup$ @MichaelRenardy so I get: $u(t)=v(s) \Rightarrow u′(t)=v′(s)2t \Rightarrow u′′(t)=v′′(s)4s+2v′(s) $This should give me the ODE: $v''(s) + \frac{1}{2} \left(\frac{1}{s} + \frac{1}{s-1} \right)v'(s) + \left( n \beta \frac{2s-1}{s(1-s)} + \beta^2 \frac{2s-1)^2}{s(1-s)} +C \right)v(s)=0$. Now in the equation you mention at dlmf.nist.gov/31.12 there is no term containing $s^2$ in the nominator in the last term. So does this mean, that I have to take $C$ such that the $s^2$ does not appear? How do I see that by doing so, I don't throw away solutions? $\endgroup$ – user37929 May 5 '14 at 21:05
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    $\begingroup$ You can get rid of this term by an elementary transformation like $v(s)=\exp(\alpha s)w(s)$ for an appropriate $\alpha$. There is a book by Ronveaux on the Heun equation, it discusses reduction to standard form, among other things. $\endgroup$ – Michael Renardy May 6 '14 at 9:50

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