4
$\begingroup$

Let $\Omega\subset\mathbb{R}^{d}$ be a bounded Lipschitz domain. Let $f\in H^{s}(\Omega)$, for $s\geq -1$. Then Lax-Milgram guarantees a unique solution to the Poisson problem

$$\begin{cases}\Delta u=f & \Omega \\ u=0 & \partial\Omega \end{cases} \tag{Poisson}$$

in $H_{0}^{1}(\Omega)$, where the boundary condition is satisfied in the sense of trace.

It is known that if we impose the stronger assumption that $\Omega$ is a $C^{k,1}$ (i.e. $k$ times continuously differentiable with $k^{th}$ derivative Lipschitz) domain, for some nonnegative integer $k$, and $f\in H^{k}(\Omega)$, then we have the elliptic regularity result $u\in H^{k+2}(\Omega)$ and moreover,

$$\|u\|_{H^{k+2}(\Omega)} \lesssim_{\Omega,k} \|f\|_{H^{k}(\Omega)}.$$

Now let $0<s<1$. If we assume that $\Omega$ is a $C^{2,1}$ domain, then by interpolation between the estimates $$\|u\|_{H^{2}(\Omega)} \leq C(\Omega,0)\|f\|_{L^{2}(\Omega)}$$ and $$\|u\|_{H^{3}(\Omega)} \leq C(\Omega,1)\|f\|_{H^{1}(\Omega)},$$

we have the estimate $$\|u\|_{H^{2+s}(\Omega)} \leq C(\Omega,s)\|f\|_{H^{s}(\Omega)}.$$

In the preceding argument, it seems that we're assuming much more regularity on the domain $\Omega$ then we need to. Heuristically interpolating between $C^{1,1}$ and $C^{2,1}$, my guess is that the natural regularity assumption on the domain $\Omega$ is $C^{2,s}$, but I do not know how to prove this, nor can I find a reference.

Question. Let $0<s<1$. Is it true that if $\Omega$ is $C^{2,s}$ domain and $f\in H^{s}(\Omega)$, then the solution $u$ to (Poisson) is in $H^{2+s}(\Omega)$, and moreover, $u$ satisfies the estimate $$\|u\|_{H^{2+s}(\Omega)} \leq C(\Omega,s) \|f\|_{H^{s}(\Omega)}?$$

$\endgroup$
  • $\begingroup$ Nice question, but you have an answer right where you study: ask Caffarelli (assuming the information in your profile is up to date). If you find the answer, let me know! $\endgroup$ – Piotr Hajlasz May 3 '18 at 21:30
  • $\begingroup$ I looked at your questions. I know it must be quite frustrating about not getting answers, but as I pointed in some other comments your questions are very technical so nobody will read them except those who work exactly in the same area. The above question is simple and very interesting so think about asking questions no more complicated than this one. $\endgroup$ – Piotr Hajlasz May 4 '18 at 2:17
  • $\begingroup$ @PiotrHajlasz: Thanks for your comment. When I saw him two days ago, he indicated that he would be out of office for the remainder of the week. If I or someone else has not come up with answer before then, I will try to ask him next week. $\endgroup$ – Matt Rosenzweig May 4 '18 at 23:48
  • $\begingroup$ Good. If you know the answer should write it here. I am interested in it. $\endgroup$ – Piotr Hajlasz May 5 '18 at 1:02

Your Answer

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

Browse other questions tagged or ask your own question.