Is this bootstrap argument correct?

Question

The setup. Assume $\Omega \subset \mathbb{R}^3$ bounded and bilipschitz equivalent to the unit cube and has smooth boundary. Let $v \in W^{1,2}(\Omega,\mathbb{C})$ be a weak solution of $$ \begin{cases} -\Delta v = g & \text{in } \Omega \\ v=0 & \text{on } \partial \Omega\end{cases} \tag{1}$$

Now assume we can show that $g \in L^{12/11}(\Omega)$. From standard elliptic theory (theorem 7.4 in Giaquinta's introduction to elliptic systems) we may infer that $v \in W^{2,12/11}(\Omega)$.

The problem. Now, I even know that $D g (:=\nabla g) \in L^{12/11}(\Omega)$ and want to infer that $v \in W^{3,12/11}$ and I want to use a bootstrap argument to do so.

My attempt. Since $v$ is a weak solution to (1) we know that $$ \int_\Omega Dv Dw = \int_\Omega g w $$ for all $w \in H_0^1(\Omega)$.

We choose such test function $w \in C_c^\infty(\Omega)$ and set $$ u:=-Dw.$$

Inserting $u$ for $w$ in the equation above, we get $$ -\int_\Omega DvD^2w = -\int_\Omega gDw. $$ Integration by parts yields $$ \int_\Omega D^2 v Dw = \int_\Omega Dg w. $$

This means that $\widetilde v = Dv$ is also weak solution to $$ \begin{cases} -\Delta \widetilde v = Dg & \text{in } \Omega \end{cases} $$

The question. How would one then conclude that $\widetilde v = Dv \in W^{2,12/11}(\Omega)$? Note that we don't have boundary conditions anymore. Would it be easier to withdrawing to an open subset $U \subset \Omega$? How would that exact argument go?

Answer 1

As I understand, the question is: If $g\in W^{1,p}(\Omega)$ can we show that $v\in W^{3,p}(\Omega)$, where $1<p<\infty$?

To answer this let me first assume that the boundary of $\Omega$ is smooth. Nonsmooth boundaries will be discussed afterwards.

We start by considering a sequence $\{g_n\}\subset C^\infty(\bar\Omega)$ of smooth functions, converging to $g$ in $W^{1,p}$, and solve $$ \Delta v_n = g_n\quad\textrm{in }\Omega, \qquad v_n=0\quad\textrm{on }\partial\Omega. $$ Because $g_n$ are smooth, by for instance the $L^2$ theory all $v_n$ are smooth. Moreover, as you mentioned, the basic $L^p$ theory gives the estimate $$ \|v_n\|_{W^{2,p}} \leq c\|g_n\|_{L^p}.\qquad\qquad(*) $$ Recall also the estimate $$ \|u\|_{W^{k,p}} \leq c\|\Delta u\|_{W^{k-2,p}} + c\|u\|_{W^{k,p}(\partial\Omega)} + c\|u\|_{L^p},\qquad\qquad(**) $$ that is true for any $u\in W^{k,p}(\Omega)$ with $k\geq2$ and $1<p<\infty$. By applying this estimate to $v_n$, we have $$ \|v_n\|_{W^{3,p}} \leq c\|g_n\|_{W^{1,p}} + c\|v_n\|_{L^p} \leq c\|g_n\|_{W^{1,p}}. $$ So $v_n$ are uniformly bounded in $W^{3,p}$, which means that there is $w\in W^{3,p}(\Omega)$ such that $v_n\to w$ weakly in $W^{3,p}$, and by compactness, strongly in $W^{2,p}$. This implies that $\Delta w = g$ in $\Omega$, because $$ \|\Delta w - g\|_{L^p} \leq \|\Delta(w-v_n)\|_{L^p} + \|g_n-g\|_{L^p} \leq \|w-v_n\|_{W^{2,p}} + \|g_n-g\|_{L^p}. $$ Trace of $w$ on $\partial\Omega$ is zero because the trace is preserved under weak convergence (or if you want, the trace map is continuous from $W^{1,p}(\Omega)$ to $L^p(\partial\Omega)$). To conclude, we have $w=v$, and thus $v\in W^{3,p}(\Omega)$.

Now let me discuss a bit about nonsmooth boundaries. A good (and perhaps primary) reference on this subject is Grisvard's book. Of course, you have the standard interior estimates regardless of the boundary regularity. Only the estimates up-to-boundary depend on how regular the boundary is. In Grisvard's book, the basic $L^p$ theory, including the estimate $(*)$, was proved assuming that the boundary $\partial\Omega$ is $C^{1,1}$, and the estimate $(**)$ was proved for $C^{\,2,1}$ boundaries. I think you get the drift: If you want $v\in W^{k,p}$ then you need a $C^{\,k-1,1}$ boundary. This cannot be improved in general: For instance in polyhedral domains, generally $v\not\in H^2(\Omega)$ even for smooth right hand side $g$. For convex domains though, assuming for instance a Lipschitz condition, at least $(*)$ is true. I expect $(**)$ to be also true. You might be able to prove it if you follow the steps in Grisvard.