Finding an operator C that satisfies AB=CA Let $D=\frac{d}{dx}$ , $A=\sum_{i=-n}^{i=n} a_i(x)D^{i}$ and $B=b(x)D$, where $a_n(x)$ and $b(x)$ are sufficiently smooth functions and $n$ is an arbitrary positive integer. $A$ may not be invertible. Assume that $AB=CA$. What can be said about operator $C$? Is it possible to write $C$ as a function of $A$, $D$ and $B$?
Note that if $A$ is invertible, $C=ABA^{-1}$.
Additionally, if $A,B$ and $C$ are matrices, this problem is similar to the homogeneous Sylvester matrix equation .
 A: The algebra of differential operators is graded by $\deg(A):=n$ for $A=\sum_{i=0}^{i=n}a_n(x)D^n$, if $a_n\not\equiv 0$. It has the property that $\deg(AB)\leq \deg(A)+\deg(B)$, and the equality holds if either $a_n$ or $b$ is not a zero divisor (otherwise it might happen that $a_n(x)b(x)=0$ for all $x\in I$, reducing the degree of $AB$).
Thus in the generality of your question we should assume that $\deg(AB)=\deg(A)+\deg(B)$. Hence the leading coefficient (w.r.t. $D^{n+1})$ of $AB$ is given by $a_n\cdot b$ and the candidate for $C$ is given by $C=bD+c$ for $b$ given and some $c$. On the other hand the lowest coefficient (w.r.t. $D^0$) of $AB$ is $0$, since $B=bD+0$, but for $CA$ it is given by $(bD+c)a_0$, which determines $c$ for most $x\in I$. Yet there are many more constraints given by the coefficients of the other degree terms.
Taking another approach one can directly show, that the equality $AB=CA$ holds for $C=B$, if $a_i,b$ are all constant, since constant functions commute with $D$.
Just another remark: If we exclude the question of the existence of an inverse of a differential operator $A$, its inverse should satisfy $\deg(A^{-1})=\deg(A)^{-1}=-n$, since $0=\deg(I)=\deg(AA^{-1})=\deg(A)+\deg(A^{-1})$ (assuming equality holds). So $A^{-1}$ wouldn't be a (classical) differential operator and neither would $C=ABA^{-1}$ be in general.
