Dear community, 


i have a question regarding differential operators acting on vector valued functions and how to "diagonalize" them.


To explain my question i will use an example:
Let $V^k$ be the space of twice differentiable functions $U:[0,2\pi] \to \mathbb{R}^k$
with periodic boundary conditions.


Consider the differential operator $L:V^2\to V^2$ defined via


$L :=\begin{pmatrix} -\partial^2 & 0 \\ 0 & -\partial^2 \end{pmatrix}$


That means it acts on $U\in V$, $U(t)=(u(t),v(t))^T$, by mapping it to $LU=(-u''(t),-v''(t))^T$.


$L$ is self adjoined with respect to the scalar product
$(U,V) := \int\_0^{2\pi} U^T V \ dt$. So it has real eigenvalues.
It commutes with the operator $J=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$.


In fact, the spectrum of $L$ consists of all $\lambda=k^2$ with $k \in \mathbb{Z}$ with multiplicity 4. The spectrum of $-\partial^2:V^1 \to V^1$ is the same, except that the multiplicities of the eigenvalues are halved. This is evident from the diagonal form of $L$.


Now, what happens for other self-adjoined operators on $V^2$ that commute with $J$?
For example consider the operator $M:V^2 \to V^2$ defined by


$M :=\begin{pmatrix} -\partial^2 & -\partial \\ \partial & -\partial^2 \end{pmatrix}$


It is also self-adjoined with respect to the above mentioned inner product and it also commutes with $J$. Is it possible to "diagonalize" this operator into a form 
$\begin{pmatrix} m & 0 \\ 0 & m \end{pmatrix}$
with a scalar differential operator $m: V^1 \to V^1$ having the same spectrum as $M$ except for the multiplicities?


Any references would be appreciated. Thanks.



Diagonalization of a matrix of differential operators
Another approach, for this particular example is to try to solve the equation $AMA^{-1} - m I = 0$, where $A$ is an invertible $2$-by-$2$ matrix of functions and $m$ is a scalar differential operator. There are a number of solutions to this. For example,
$$
A = \begin{pmatrix}\cos(\tfrac12x) & \sin(\tfrac12x)\cr 
 -\sin(\tfrac12x) & \cos(\tfrac12x) \end{pmatrix}
$$
and $m = -\partial^2 - \tfrac14$. Unfortunately, $A$ is $4\pi$-periodic, not $2\pi$-periodic. In fact, there are no $2\pi$-periodic solutions. 


You can interpret this in two ways. One is that you shouldn't have imposed the $2\pi$-periodicity in the first place, since the conjugacy question for differential operators is really a local one. The other is that $A$ actually represents an isomorphism between two rank $2$-vector bundles over the circle, one with trivial transition, and the other with a twist (so that the sections of the second bundle are represented by functions $U:\mathbb{R}\to \mathbb{R}^2$ such that $U(t+2\pi) = -U(t)$). 


