We consider $2a$  periodic smooth solutions for \begin{eqnarray*} \Delta u+V(x)\,u=0\qquad\hbox{in}\:[a,a] \end{eqnarray*} We assume that $V$ is smooth and even (i.e. $V(x)=V(x)$). We also assume that (up to multiplication with a real number) there exists only one odd $2a$  periodic solution. Can one say anything about the number of even $2a$  periodic solutions?

$\begingroup$ It seems like this should be derivable as a special case from the usual analysis of Hill's equation. $\endgroup$– BuzzJun 29, 2021 at 1:41
1 Answer
Because of the uniqueness of the initial value problem, there can be at most two solutions, i.e., if we have one odd $2a$periodic solution, then there can be at most one more even $2a$periodic solution. For example, for $V(x)=(\pi /a)^2 $, we have the odd solution $\sin \pi x/a $ and the even solution $\cos \pi x/a $. On the other hand, for generic $V(x)$, the eigenvalues are simple, i.e., if we have an odd $2a$periodic solution, there is no additional even $2a$periodic solution. An example is the Mathieu equation, rescaled to $2a$periodicity, where $V(x)=(\pi^2/(2a)^2 ) (2q\cos (\pi x/a)\lambda (q))$, with nonzero $q$ and associated eigenvalue $\lambda (q)$; cf. Ince's Theorem.
In summary, there are either zero or one even $2a$periodic solutions.

$\begingroup$ Many thanks! That closes the case. In fact I am interested in the nonexistence of an even function. Is there a reference for this generic property? $\endgroup$– guest61Jun 29, 2021 at 15:18

$\begingroup$ For Mathieu's equation, the statement is referred to as Ince's Theorem  I've added a link to the corresponding section in the NIST Handbook. Not sure how this statement is referred to in the more general Hill's equation context. $\endgroup$ Jun 29, 2021 at 15:22
