Matrix-valued ordinary differential equation with symmetry I am considering the following equation
$$\begin{pmatrix} -\frac{d}{dx} + \lambda \sin(2\pi x) & \lambda - \lambda \cos(2\pi x) \\  -\lambda-\lambda \cos(2\pi x) & -\frac{d}{dx} - \lambda \sin(2\pi x) \end{pmatrix}\varphi(x)=0$$
and I am wondering if there is an explicit characterization of $\lambda \neq 0$ for which there exist $1$-periodic solutions to this ODE.
This ODE, though matrix-valued has a nice symmetry in the sense that the off-diagonal entries have only even Taylor polynomials whereas the diagonal ones have odd ones.
However, it is of course difficult to say based on a Taylor series, if this will yield a periodic function.
Observation:
Close to zero, i.e. $x=0$, the solution should heuristically look like $\varphi(x)=(0,e^{-\lambda \pi x^2}),$ as it solves the equation "to leading-order"
 A: Your equation in fact admits a simple explicit solution. As noted in the previous answer, the general solution is encoded in a matrix ODE,
$$
M'(x) = \lambda A(x)M(x), \qquad M(0)=I\,.
$$
Now, the simplification becomes apparent if change to a new basis, using the $x$-dependent rotation
$$
R(x) = \begin{pmatrix} \cos(\pi x) & -\sin(\pi x) \\ \sin(\pi x) & \cos(\pi x)\end{pmatrix}.
$$
Define $N(x)$ by $M(x)=R(x)N(x)$, and the ODE becomes
$$
N' = \left(\lambda R^{-1}AR-R^{-1}R' \right)N = \begin{pmatrix} 0 & \pi \\ -\pi-2\lambda & 0\end{pmatrix} N.
$$
Writing $\lambda = \frac{\pi}{2}(\mu^2-1)$, the solution is
$$
N(x) = \begin{pmatrix}
 \cos (\pi\mu  x) & \frac{1}{\mu}  \sin (\pi\mu  x) \\
 -\mu \sin (\pi\mu  x) & \cos (\pi\mu  x)
\end{pmatrix} \implies M(1) = \begin{pmatrix}
 -\cos (\pi\mu)  & -\frac{1}{\mu}  \sin(\pi\mu)  \\
 \mu  \sin(\pi\mu) & -\cos(\pi\mu)
\end{pmatrix}.
$$
We get periodic solutions when $\mu=2n+1$, $n=1,2,3,\ldots$, so $\lambda=2\pi n(n+1)$.
A: I doubt that this vector ODE be explicitly integrable. However you can do the following.
Write it as $\frac{d\phi}{dx}=\lambda A(x)\phi$, where
$$A(x)=\begin{pmatrix} \sin2\pi x & 1- \cos2\pi x \\ -1- \cos2\pi x & - \sin2\pi x \end{pmatrix}.$$
Notice ${\rm Tr} A\equiv0$.
The general solution of the Cauchy problem is given in the form $\phi(x)=M(\lambda;x)\phi(0)$. Your question amounts to whether $\mu=1$ is an eigenvalue of $M(\lambda;1)$, or not.
First of all $\frac{d\det M}{dx}=\lambda ({\rm Tr}A(x))\det M\equiv0$, hence $\det M\equiv1$ (because $M(\lambda;0)=I_2$). Thus $1$ is an eigenvalue if and only if the trace of $M(\lambda;1)$ equals $2$.
Thus we are led to the study of $\lambda\mapsto{\rm Tr}M(\lambda;1)$. To this end, we notice that
$$\frac{dM}{dx}=\lambda A(x)M,\qquad M(\lambda;0)=I_2.$$
The map $\lambda\to M$ is analytic. Expanding $M=I_2+\lambda M_1+\lambda^2M_2+\cdots$, we have $M_1'=A$ and $M_2'=AM_1$ with $M_j(\lambda;0)=0_2$. This can be integrated explicitly. Obviously ${\rm Tr}M_1\equiv0$. If my calculation is correct, we have ${\rm Tr}M_2(1)=-1$. Thus $|{\rm Tr}M(\lambda;1)|<2$ for small $\lambda\ne0$.
This is a useful information: when $\lambda\ne0$ is small, then the eigenvalues of $M(\lambda;1)$ are $e^{\pm i\theta}$, where $\theta(\lambda)$ varies analytically, with $\theta'(0)\ne0$. For a dense subset near zero, the frequency has the form $\theta(\lambda)=\frac{2\pi p}q$ where $\frac pq$ is rational (irreducible). Then every solution is $q$-periodic.
To answer properly and positively your question, there remains to check whether $\lambda\to{\rm Tr}M(\lambda;1)$ takes values above $2$. If so, there is a parameter $\lambda$ for which this trace equals $2$, and thus $\mu=1$ is a double eigenvalue. Then the ODE admits a $1$-periodic solution.
