4
$\begingroup$

Let $\Omega\subset \mathbb R^n$ be a convex subset. All the objects below will be defined on this set.

Let us assume $P(x,D)$ to be a differentiable operator order $m$ and of square size, that is having as many equations as unknowns, with variable smooth coefficients.

Let moreover $Q(D)$ another differentiable operator of order $m$ and of square size (the same of $P$), but this time having constant coefficients.

We also assume that

  1. $P$ is of constant strength in the sense of Hormander, that is for any $x,y\in \Omega$ there exist a constant $C_{xy}$ such that $$|P(x,\xi)|/|P(y,\xi)|\leq C_{xy},$$ for any $\xi\in\mathbb R^n$
  2. The operators $P$ and $Q$ are of equal strength, that is for any $x\in\Omega$ (it is not anymore important who is $x$ by the previous point), we have that there exist a constant $C$ such that $$\frac{1}{C}\leq\frac{|P(x,\xi)|}{|Q(\xi)|}\leq C,\quad\forall \xi\in\mathbb R^n$$

I want to solve now the equation $P(x,D)u=f$, for $f=(f_1,\dotso,f_k)\in (C^\infty_c(\Omega))^k$, and of course deduce something on the regularity of the solution $u$.

If I understand well, the heuristic should be as follows:

  1. The equation $Q(D)u=f$ admits a smooth solution $u$ (Ehrenpreis fundamental principle), this should be the hard part to prove, but I take it as a black box
  2. Since $P$ and $Q$ are of equal strength, by the previous hypotheses, then the solution $u$ to $P(x,D)u=f$ has to be "as regular as" the solution $v$ to $Q(D)v=f$. In particular I expect $u$ to exist and to be smooth, well, even compactly supported since so is $f$. This is in fact the guiding principle of Hormander when he introduce this concept of strength as a preorder relation between operators in Linear partial differential operators, chap. 3.

Is my heuristic correct? How can one be rigorous in proving this? Thank you very much in advance

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ This cannot be correct. First, the solution will not necessarily be compactly supported. That's not even true when $f = 0$. Some indirect evidence is that if this heuristic were correct, then the generic system of PDE's would be solvable on a given (convex) domain. Nothing remotely close to this is known. $\endgroup$
    – Deane Yang
    Commented Sep 23, 2016 at 13:04
  • $\begingroup$ @DeaneYang This is indeed enlightening. So let me refine a bit my question: are there some "known" classes of perturbation of a differential operator with constant coefficients $Q$ so that the perturbed operator $P$ still admits solutions? Whatever sense this question makes. Thank you $\endgroup$
    – user17697
    Commented Sep 23, 2016 at 13:21
  • $\begingroup$ The idea of studying a variable coefficient PDE using a nearby constant coefficient PDE is called "freezing the coefficients" and is commonly used for elliptic PDE's. It could be used for hyperbolic PDE's but it's easier just to prove the estimates directly using the variable coefficient system. There are also results known (but maybe not existence) for a PDE of real principal type. For systems, there is a paper of Nils Dencker. $\endgroup$
    – Deane Yang
    Commented Sep 23, 2016 at 19:34

0

Reset to default

You must log in to answer this question.