# Legendre differential equation with additional term

In an application I encountered the ODE $$\left( {x}^{2}-1 \right) {\frac {{\rm d}^{2}}{{\rm d}{x}^{2}}}f \left( x \right) +x \left( {\frac {\rm d}{{\rm d}x}}f \left( x \right) \right) \left( 8\,{x}^{2}-7 \right) -4\, \left( C+1 \right) f \left( x \right) =0.$$ In principle, this is Legendre's differential equation with an additional term $\tilde \quad x^3 \frac{d}{dx}f(x)$ and with a solution that can be expressed by Maple as $f(x) = K_1 HeunC(4, -\frac{1}{2}, -\frac{1}{2}, -2, \frac{3}{8}-C, x^2)$, where $K_1 \in \mathbb{R}$ satisfying the boundary conditions indicated below. Of course, this confluent Heun's function does not tell me anything if I am looking for actual solutions of this Sturm-Liouville problem that can be expressed in terms of analytical functions( which may or may not exist). The main problem here is that the unknown eigenvalues are part of this solution. Especially, it does not help me identifying the eigenvalues ( which should behave asymptotically Schrödinger-operator like $C_n \tilde \quad n^2$). So I was only able to identify one 'easy solution' $f(x) = 1$ and $C=-1$. But this is it. I don't see how I can find any other solution analytically. Does anybody here know whether solutions to this function are known or whether there are any solutions that we can write down? Has this particular ODE ever been studied? I am greatful to every comment.

Edit: I am looking for bounded solutions (eigenfunctions and eigenvalues) on $[-1,1]$ satisfying $f(-1) = f(1)$ and $f'(-1) = f'(1)$. I am interested in finding explicit representations (sure this is a vague term, but probably you know what I am talking about. A function defined by a series or an integral is an explicit representation, but a confluent heun's function that is just defined as a solution to a ODE is not). I am aware of the fact that there is a high chance that things like recurrence relations cannot be found. Despite, you may be able to identify how solutions could look like (e.g. a polynomial of degree n). This would also help very much.

• Could you make it clear, what "solutions" are you looking for? Do you want to find the eigenvalues $C$ (perhaps together with eigenfunctions)? If so, what boundary conditions are you taking? And what limits, if any, do you care about if explicit solutions are not available? – Igor Khavkine Jun 8 '14 at 21:28
• If you ask Maple to expand the Heun function in Series, it will be happy to oblige. So this representation of the solution should be explicit enough by your definition. On the other hand, a series expansion at $x=0$ or $x=1$ or $x=-1$ will not tell you anything about eigenvalues. Your best bet is probably numerics for now values of $C$ and a WKB approximation for large values of $C$. – Igor Khavkine Jun 9 '14 at 0:44
• I think you've just restated the observation that representing the solution as a power series in $x$ will not determine which values of $C$ are eigenvalues. Beyond that, I can only reiterate the simple truth that every student of differential equations should know: it is unrealistic to expect to find solutions of a generic ODE without some kind of approximation. If you insist on no approximation, I don't know what else to tell you. – Igor Khavkine Jun 9 '14 at 1:33
• (1) The coefficients of the powers of $x$ will depend on $C$. (2) CASs know how to expand special functions in series. It is not magic, they apply the method of solution by series to the corresponding ODE. – Igor Khavkine Jun 9 '14 at 1:54
• Downvote. Your question is not a good fit for a research site. What you need some basic textbook knowledge about ODE and Sturm-Liouville theory, like Coddington & Levinson. – Igor Khavkine Jun 9 '14 at 3:06

HeunC is a perfectly good "analytical" function. Note that the symmetry $f(1) = f(-1)$ means you want to take the even solution, so _C2 $= 0$ and _C1 $= 1$ (say). You want $f'(-1) = f'(1)$, but since $f$ is even this means you want $f'(1) = 0$. Unfortunately $1$ is right at the radius of convergence of the HeunC series, which makes things a bit more challenging, but it appears that after $C=-1$ the next eigenvalue is near $C = 0.67$. In fact here is a plot of the curve $f'(x) = 0$ as a function of $x$ and $C$: