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.