I encountered the following question in my studies:

Let us assume we have a real anlaytic solution to an ODE on $\mathbb{R}$ of Schr\"odinger type

$-\psi''(x)+V(x)\psi(x)=\lambda \psi(x)$

but we assume that $V$ is well-behaved (a bounded smooth function for instance).

The question I have is whether it follows that $\psi$ admits a meromorphic extension to $\mathbb{C}$?

Are there any sufficient conditions from which this would follow?

Thank you

I must admit that this question is somewhat unnatural to the tools I know in analysis and so it may be that this question has a simple answer, but I am not aware of any results in that direction.

  • 3
    $\begingroup$ If $\psi$ has a meromorphic extension, then so does $V = \lambda + \psi''/\psi$. $\endgroup$ – Mateusz Kwaśnicki Sep 17 '17 at 21:53
  • 2
    $\begingroup$ By the analytic existence and uniqueness for ODEs $\psi(x)$ extends to an analytic solution to the domain of analyticity of $V(x)$. If $V(x)$ has isolated singularities, then $\psi(x)$ has a chance to be meromorphic. But the singularities (and even the single-valuedness) of $\psi(x)$ has a complicated relationship with the singularities of $V(x)$. $\endgroup$ – Igor Khavkine Sep 17 '17 at 22:37

As Mateusz remarked, $V$ must be assumed to have a meromorphic extension to $\mathbb C$, and $V = \lambda + \psi''/\psi$. Now if $\psi$ has a pole or zero at $z=p$, so that $\psi(z) \sim c (z-p)^d$ as $z \to p$ (with $d \ne 0$), then $V(z) \sim d (d-1) (z-p)^{-2}$. Thus the only poles that $V$ may have are poles of order $2$ with leading coefficient $d (d-1)$ for an integer $d$, or (corresponding to simple zeros of $\psi$) poles of order $1$.

EDIT: When $V$ has a pole of order $1$ or $2$ at $z=p$, the differential equation has a regular singular point there.

In the case of order $1$ the indicial equation is $r^2 - r = 0$, so there is a solution with a simple zero at $z=p$ (corresponding to indicial root $1$); the other solutions (not constant multiples of this one) will have a logarithmic branch point at $z=p$.

In the case of order $2$ with leading coefficient $d(d-1)$, the indicial equation is $r^2 - r = d^2 - d$, so if $d$ is an integer $\ge 2$ there is a solution with a pole of order $d$ at $z=p$ (corresponding to indicial root $d$); the other solutions (not constant multiples of this one) may or may not have a logarithmic branch point.

If there is more than one pole of $V$, there may be compatibility conditions needed to ensure that a solution that works at one pole corresponds to solutions that work at the other poles, rather than those that have branch points.


As Robert Israel mentioned, it is necessary that $V$ is meromorphic, with all poles of order two. This is far from sufficient. If $V$ has no poles then every solution is entire. In general, solutions undergo transformations when one performs analytic continuation around the poles of $V$, the so-called monodromy transformations. There are two questions here: a) when ALL solutions are meromorphic, that is when monodromy is trivial, and b) when there exists one (and thus a one-dimensional space of) meromorphic slutions. Both questions are difficult when $V$ has more than one pole, they are solved only in special cases, and even in these special cases, the criterion is not easily verifiable.


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