# Existence of periodic solution to ODE

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?

• The operator is ellptic. So it's resolvent is compact. This implies your claim. – Zero Apr 29 at 4:22
• 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. – Kung Yao Apr 29 at 5:33

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.