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. $\endgroup$
    – Buzz
    Jun 29, 2021 at 1:41

1 Answer 1


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? $\endgroup$
    – guest61
    Jun 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
  • $\begingroup$ Thanks, I'll check that. $\endgroup$
    – guest61
    Jun 29, 2021 at 15:25

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