0
$\begingroup$

Let

  • $d\in\mathbb N$
  • $\Lambda\subseteq\mathbb R^d$ be bounded and open
  • $f\in L^2(\Lambda)$
  • $p\in L_{\text{loc}}^2(\Lambda)$ admit a weak gradient $\nabla p\in L^2(\Lambda,\mathbb R^d)$

Assuming $$\Delta p=f\;,\tag 1$$ i.e. $$\langle\nabla\phi,\nabla p\rangle_{L^2(\Lambda,\:\mathbb R^d)}=-\langle\phi,f\rangle_{L^2(\Lambda)}\;\;\;\text{for all }\phi\in C_c^\infty(\Lambda)\;,\tag 2$$ are we able to conclude $p\in H^2(\Lambda)$?

I know that this true under some regularity assumptions on $\partial\Lambda$. Since I'm particularly interested in a cube, the only regularity of $\partial\Lambda$ I'm willing to assume is being Lipschitz.

$\endgroup$
  • $\begingroup$ I think you need to specify some boundary conditions in order to get a global estimate. $\endgroup$ – Jose27 Dec 14 '16 at 20:45
  • $\begingroup$ @Jose27 Is a Lipschitz boundary sufficient? $\endgroup$ – 0xbadf00d Dec 14 '16 at 22:04
  • $\begingroup$ I meant boundary conditions on $p $, not restrictions on the domain ( although these play a role too). $\endgroup$ – Jose27 Dec 14 '16 at 22:13
  • $\begingroup$ @Jose27 It would be fine for me, if we assume Dirichlet boundary conditions, $p\in H_0^1(\Lambda)$. Can we conclude $p\in H^2(\Lambda)$ with that assumption, for the general $\Lambda$ of the question? $\endgroup$ – 0xbadf00d Dec 14 '16 at 22:41
4
$\begingroup$

$H^2$-regularity of $p$ is valid if $\Lambda$ is a bounded, polyhedral set, see chapter 4 of Grisvard's book "Elliptic problems in nonsmooth domains".

$\endgroup$
  • $\begingroup$ I guess you've got Theorem 4.3.1.4 in mind. However, it's only stated for a polygon in $\mathbb R^2$. Is there a similar result for polyhedra in higher dimensions (in particular, in $\mathbb R^3$)? $\endgroup$ – 0xbadf00d Feb 2 '17 at 10:14
  • $\begingroup$ What about section 8.1? On p. 371, it is stated "This shows again the regularity of $u$ in $H^2(\Omega)$ when $\Omega$ is a convex polyhedron." $\endgroup$ – gerw Feb 2 '17 at 11:45
  • $\begingroup$ It's rather vague, isn't it? $\endgroup$ – 0xbadf00d Feb 2 '17 at 13:46
  • $\begingroup$ Maybe. I havn't read the preceding theorems. $\endgroup$ – gerw Feb 2 '17 at 17:56

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.