Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Suppose I have an annulus $U\subset \mathbb{C}$ and a single-valued holomorphic function $V:U\to \mathbb{C}$.

I would like to know if there are (tractable) conditions on $V$ that ensure that the second order linear ODE $$ \partial_{zz}^2 f+ V f=0 $$ admits a single-valued holomorphic solution $f: U\to \mathbb{C}$ (here $z$ is the usual complex coordinate).

Such a question is easy to answer for the first order linear ODE $$ \partial_{z} f+ V f=0 $$ Indeed, in this case separation of variables tells us that a necessary and sufficient condition is that the residue of $V$ is an integer.

I should note that taking $V=\alpha z^{-2}$ and $U$ the annulus $\lbrace z: 0<|z|<1\rbrace$ allows one to give explicit necessary and sufficient conditions on $\alpha$.

Where are questions of this sort treated? It seems like a classical problem...

share|improve this question
add comment

1 Answer

up vote 3 down vote accepted

When the annulus is $0<|z|<1$, and $V$ has at most a second order pole at $0$, the answer is known, and is given by Fuchs theory. This is the so-called regular singularity at $0$. The reference is any book on analytic theory of linear differential equations, for example, Ince, Ordinary differential equations. In all other cases, the question is much harder, and there is no simple explicit answer.

Putting $z=\exp(w)$ you obtain a linear differential equation of second order with analytic periodic coefficients, and you are interested in the existence of a periodic solution with the same period. This was subject to much research since the 19 century, and of course a lot is known, but there is no simple answer. The key words are "Hill's equation", "Hill's determinant", "Lyapunov", etc. You can look into Whittaker Watson Chapter 19 "Mathieu functions" for the beginning.

share|improve this answer
    
Its good to know why I wasn't getting anywhere. –  Rbega Sep 27 '12 at 10:12
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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