4
$\begingroup$

I'm trying to understand Lemma 3.2 from p. 355 of the paper R. C. Carlson and K. R. Goodearl, Commutants of Ordinary Differential Operators, Journal of Differenial Equations 35 (1980), 339–365.

enter image description here

The authors define $B$ as a matrix coefficient with $C^\infty$ entries, $D$ is the ordinary derivative operator on $\bigoplus^k C^\infty(\mathscr J)$ for some open interval $\mathscr J$, and $L$ as the differential operator given below (from the previous page)

enter image description here

$\textbf{Question}$:

I can't see how this is true that any differential operator can be written in this way. I'm trying to see it in the scalar case, i.e. where $A,B$ are $1\times 1$ matrices but I can't even construct non-trivial examples that can be written in that way. Moreover, I don't even understand what the proof is doing.

$\endgroup$
  • $\begingroup$ @epimorphic Nope! Thanks for your detailed response! $\endgroup$ – user119264 Nov 9 '18 at 3:25
1
$\begingroup$

Writing out the lower-order terms explicitly, we want to have $$C_m D^m = C_m A_n^{-r} D^s L^r + \sum_{j=0}^{m-1} E_j D^j$$ for some matrices $E_j$, so that $\sum_{j=0}^m C_j D^j$ can be written in a form known to satisfy the statement of the lemma: $$\sum_{j=0}^m C_j D^j = C_m A_n^{-r} D^s L^r + \sum_{j=0}^{m-1} (C_j + E_j) D^j.$$ To prove, it suffices to show that $D^s L^r$, when rearranged so that all of the $D$s (including the ones in $L$) are on the right, has the form $$D^s L^r = A_n^r D^m + \sum_{j=0}^{m-1} F_j D^j \tag{*}\label{com}$$ for some matrices $F_j$. This is done through the product rule $$DA = A' + AD,$$ where $A'$ is the component-wise derivative of $A$. Intuitively, the rule says that when trying to "commute" $D$ and $A$, we get a term where $D$ hits the $A$ and disappears, and another where $D$ actually manages to pass through. The only way to get a $D^m$ all the way to the right from $D^s L^r$, therefore, is to take the highest-order term $D^s (A_n D^n)^r$ and pass every $D$ to the right through the $A_n$; every other term will end up with a lower power of $D$. This results in $\eqref{com}$.


For a simple example, let $k=1$ as you said, and let $L = (x^2 + 1) D^2$ and $T = \sin(x) D^3$. Then $r = s = 1$. By the product rule, $DL = 2x D^2 + (x^2 + 1) D^3$, so $$T = \frac{\sin(x)}{x^2 + 1}DL - \frac{2x\sin(x)}{x^2 + 1} D^2 = \frac{\sin(x)}{x^2 + 1}DL - \frac{2x\sin(x)}{(x^2 + 1)^2} L.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.