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We shall consider the matrix-valued differential operator

$$(L u)(x) :=u'(x) - \begin{pmatrix} 0 & \sin(2\pi x-\frac{\pi}{6})\\ - 2\sin(2\pi x+\frac{\pi}{6}) & 0 \end{pmatrix} u(x).$$

This is a $1$-periodic operator. Thus, does there exist a $\lambda \in \mathbb C$ and a $1$-periodic solution to this ODE such that

$$ (L - \lambda)u = 0.$$

Probably there is no explicit solution, but can we show the existence of such a solution?

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  • $\begingroup$ The operator is ellptic. So it's resolvent is compact. This implies your claim. $\endgroup$
    – user130903
    Commented Apr 29, 2021 at 4:22
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    $\begingroup$ Well, the resolvent could just have 0 in its spectrum. In other words, the spectrum of $L$ could be empty. Note that $L$ is not normal. $\endgroup$
    – Kung Yao
    Commented Apr 29, 2021 at 5:33

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The solutions of $(L-\lambda)u=0$ are the functions $u(x)=e^{i\lambda x}v(x)$, where $v$ satisfies $Lv=0$. The periodicity amounts to $e^{i\lambda}v(1)=v(0)$. Thus your problem does admit infinitely many solutions. Just consider the monodromy matrix $M:v(0)\mapsto v(1)$, whose determinant equals $1$ (by the Wronskian). Take an eigenvalue $\mu$ of $M$, which is therefore non-zero. Then any $\lambda\in{\mathbb C}$ such that $e^{i\lambda}=\mu$ solves your problem.

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  • $\begingroup$ thank you for this nice insight. I understand Floquet theory is a very one-dimensional method and so it is taylored to my problem. I wonder whether in higher dimensions, there would have also been a way out?- The comment by Zero was actually a reasonable one, and would apply to higher dimensions, but there could be pathological examples where the spectrum is empty. $\endgroup$
    – Kung Yao
    Commented Apr 29, 2021 at 14:41

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