Meromorphic solutions to Legendre's equation

Question

I just saw the following question that was asked yesterday on math overflow on meromorphic solutions to ODEs Although, I understand the answers and comments to the questions, I did not understand how this applies to some of the standard ODEs.

Thus, I would like to devote this question to the question: Can we show that (at least some) solutions to the associated Legendre equation are meromorphic, just to see some theory in practice.

The associated Legendre equation reads for $m \in \left\{-l,..,l\right\}$ and $l \in \mathbb{N}_0$

$$-y''(x)-\cot(x)y'(x)+\frac{m^2}{\sin^2(x)}y(x)=l(l+1) y(x).$$

Its solutions are the associated Legendre polynomials with (even) analytic continuation and so-called associated Legendre functions of the second kind. see here

I ask: Is there a way to show (without explicit construction) that the associated Legendre functions of the first kind, i.e. the eigenfunctions to this equation have a meromorphic continuation (note that the continuation is in fact analytic)?

So to make it clear: The answer to this question is not that we can write them down explicitly and show that they are analytic. The answer should be a proof that any eigenfunction to this equation is meromorphic.

Answer 1

Solutions of the equation you wrote are not polynomials. Legendre polynomials are solutions of the equation $$(1-x^2)y''-2xy'+l(l+1)y=0.$$ This equation is written in the same article you refer to. It has two regular singularities: at $1,-1$, so this is a relatively simple case. The general theory only says that at each regular singularity there is a basis of power series solutions (whose radius of convergence is at least $2$, and whose coefficients can be found explicitly). If there is a meromorphic solution, at least one of these power series must be single-valued (does not involve fractional power). This is the case in our example. If we are searching a polynomial solution, we can impose the condition that the power series of this solution terminates, and this singles out a polynomial solution. That's about all that the general theory tells us in this case. (See also my answer to the question you refer to).

If you consider, for example a more general equation with two finite regular singularities, the so-called oblate/prolate spheroid equation, then the existence of a single valued solution, which is not a polynomial in this case, is a more challenging problem. For the most general equation with two finite regular singularities (and one irregular at infinity), there is no reasonable answer.

So, roughly speaking the answer to your question is this: there is no general theory which will tell you whether a second order linear equation with rational coefficients, and two singularities has a meromorphic solution or not. One has to use the specific form of the equation, and only for simplest forms the answer is known.