# Regularity of a solution of Laplace equation

Assume $\Omega$ is some open, bounded domain with smooth boundary - say $\Omega = B(0,1) \subset \mathbb{R}^3$. Let $v$ be a solution of the Laplace equation $$\begin{cases} \Delta v =0 & \mbox{on } \Omega \\ v=f|_{\partial\Omega} & \mbox{on } \partial \Omega \end{cases}$$ and assume furthermore $f \in L^1_{loc} (\overline{\Omega}) \cap L^4(\overline{\Omega})$ and $Df \in L^2(\overline{\Omega})$.

Can one prove that $v \in L^4$?

What I found so far: this works for $f \in C(\overline{\Omega})$ by the Poisson Integral Formula, but my $f$ is not as smooth (as I can't assert that $f \in H^4$).

Any help is much appreciated!

• @Paul: is there a proof and if yes: how do I have to do it. I'll delete the word "somehow" – mjb Jun 20 '13 at 12:30
• Hi, mjb do you mean by that $f\in L^4(\partial \Omega)$ or the boundary data is the restriction of $f$ on boundary? Talking about $Df$ for boundary data $f$ normally doesn't have full meaning, for if $f$ lives on the boundary, only surface gradient is meaningful for $f$. – Shuhao Cao Jun 20 '13 at 12:49
• @ShuhaoCao I indeed mean $v=f|_{\partial \Omega}$ on $\partial \Omega$, i.e. the boundary data is the restriction of f to the boundary. I will edit my post. – mjb Jun 20 '13 at 13:02

First note that $f\in L^4(\Omega)$ implies $f\in L^1(\Omega)$, hence the hypothesis $f\in L^4(\Omega)\cap L^1_{loc}(\Omega)$ is unnecessary. Now, consider the problem

$$\tag{P} \left\{ \begin{array}{cc} \Delta v=0 &\mbox{ in \Omega} \\ v=f &\mbox{ in \partial\Omega} \end{array} \right.$$

It is well know that if $f\in H^{1/2}(\partial\Omega)$, then problem $(P)$ has a unique solution $v\in H^1(\Omega)$ satisfying $(P)$ in the weak sense. But $H^1(\Omega)$ is contained in $L^{2^\star}$ (Sobolev Embedding), where in our case $$\tag{1}2^\star=\frac{2N}{N-2}=6$$

From $(1)$ we conclude that $v\in L^4(\Omega)$.

Remark 1: $f\in L^4(\Omega)$ with $Df\in L^2(\Omega)$ implies that $f\in H^1(\Omega)$, which implies that $\operatorname{trace}(f)\in H^{1/2}(\partial\Omega)$

Remark 2: To solve problem $(P)$ we procced as follows:

Claim: The solution $v\in H^1$ is characterized by $$\tag{2}\int_\Omega |\nabla v|^2=\min\{\int_\Omega |\nabla u|^2: u\in H^1(\Omega)\ \mbox{and}\ \ \operatorname{trace}u=f\}$$

Denote $\mathcal{K}=\{\int_\Omega |\nabla u|^2: u\in H^1(\Omega)\ \mbox{and}\ \ \operatorname{trace}u=f\}$

First note that $\mathcal{K}$ non empty, because $\operatorname{trace}:H^1(\Omega)\to H^{1/2}(\Omega)$ is surjective, hence, we can take a minimizing sequence in $\mathcal{K}$, i.e. a sequence $u_n\in\mathcal{K}$ satisfying $$\int_\Omega |\nabla u_n|^2\to \inf\mathcal{K}$$

Try to prove that $\|\nabla u_n-\nabla u_m\|_2\to 0$ as $n,m\to\infty$. Note that $u_m-u_n\in H_0^1(\Omega)$, hence, by Poincare inequality we can conclude that $$\|u_n-u_m\|_{1,2}\to 0$$

This implies the existence of some $v\in H^1(\Omega)$ such that $u_n\to v$ in $H^1$. Now you can conclude.

Remark 3: Let me propose you another way to solve this problem.

Let $K=\{\int_\Omega |\nabla u|^2: u\in H^1(\Omega)\ \mbox{and}\ \ \operatorname{trace}u=f\}$ and define $F:K\to\mathbb{R}$ by $$\tag{3}F(u)=\int_\Omega|\nabla u|^2$$

First note that $K$ is closed and convex. Now, try and show:

I - $F$ is coercive, i.e. if $\|u\|_{1,2}\to\infty$ in $K$, then $F(u)\to\infty$,

II - $F$ is weakly lower semicontinuous, i.e. if $u_n\in K$ weakly converges to $u\in K$, then $F(u)\leq\liminf F(u_n)$,

III - $F$ is convex.

I, II and III implies that $F$ is minimized by some $v\in K$ which implies that $F'(v)=0$.

• Dirichlet boundary requires the data to be $H^{1/2}$ regular in order that the solution in $H^1$. – Shuhao Cao Jun 20 '13 at 12:41
• You are right, I misred the question, but anyway, with the new hypothesis $Df\in L^2$, we have that $f\in H^{1/2}(\partial\Omega)$ – Tomás Jun 20 '13 at 12:43
• Now I think it is right @ShuhaoCao. What do you think? – Tomás Jun 20 '13 at 12:49
• Hmm...well if $Df\in L^2$ and $v = f|_{\partial \Omega}$ then sure. – Shuhao Cao Jun 20 '13 at 12:52
• For your first question, the answer is like you've said "the domain is bounded". For you second question $H^{1/2}=W^{1/2,2}$. I'm gonna post the solution of problem $(P)$ to you, wait a moment pkease. – Tomás Jun 20 '13 at 13:15