It seems that the following assertion is widely accepted:

For $k\in\mathbb N$, $p\geq 2$, $\Omega \subset \mathbb R^n$ bounded with $\partial\Omega\in C^{k+2}$ and $f\in W^{k,p}(\Omega)$, the weak solution $u\in H^1_0(\Omega)$ of the problem $$ \begin{equation} \left\{ \begin{aligned} -\Delta u =f \text{ in } \Omega\\ u=0 \text{ on } \partial\Omega \end{aligned} \right. \end{equation} $$ satisfies $u\in W^{k+2,p}(\Omega)$ and $\|u\|_{W^{k+2,p}}\leq C_{\Omega,k,p}\|f\|_{W^{k,p}}$ for some $C_{\Omega,k,p}>0$.

The above is proved in Evans using difference quotients for $p=2$. For $k=0$ it appears to be true due to an interpolation argument (Theorem 7.1 of Giaquinta's and Martinazzi's book on regularity theory). For Hölder continuous domains one can use the classical Schauder theory. But is there a reference for the complete result?

  • $\begingroup$ I am not sure if you would find it there, but Adams: Sobolev Spaces is a good reference for Sobolev spaces' results. $\endgroup$ – Alan Oct 18 '15 at 18:24
  • $\begingroup$ Nope, there is nothing in Adams about regularity theory... $\endgroup$ – Thanasis Stylianou Oct 18 '15 at 18:27
  • $\begingroup$ you can llok to Gilbard Trudinger or the courant lecture notes by Han Lin $\endgroup$ – Paul Oct 18 '15 at 19:15
  • $\begingroup$ Gilbarg and Trudinger provide the result as a product of the Schauder theory which assumes Hölder regularity of the boundary. It seems that this is also the case with Han and Lin $\endgroup$ – Thanasis Stylianou Oct 18 '15 at 20:02

P. Grisvard, Elliptic Problems in Nonsmooth Domains, 1985: Thm. (you even need to impose less regularity on $\partial\Omega$).

  • $\begingroup$ It is not the complete reference but together with Theorem 7.1 of Giaquinta's book provides us with what we need. Thanks, I was well aware of Grisvard's books but I somehow "missed" the section on smoother domains :-) $\endgroup$ – Thanasis Stylianou Oct 19 '15 at 6:50
  • $\begingroup$ Are you missing the explanation of why the solution is $W^{2,p}$ in the very first place? Well, it also follows directly from Thm. from the same book. $\endgroup$ – Delio Mugnolo Oct 19 '15 at 8:57

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