6
$\begingroup$

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{equation} \begin{cases} \Delta v =0 & \mbox{on } \Omega \\ v=f|_{\partial\Omega} & \mbox{on } \partial \Omega \end{cases} \end{equation} 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!

$\endgroup$
  • $\begingroup$ @Paul: is there a proof and if yes: how do I have to do it. I'll delete the word "somehow" $\endgroup$ – mjb Jun 20 '13 at 12:30
  • 1
    $\begingroup$ 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 $f$. $\endgroup$ – Shuhao Cao Jun 20 '13 at 12:49
  • $\begingroup$ @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. $\endgroup$ – mjb Jun 20 '13 at 13:02
4
$\begingroup$

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 $F'(v)=0$.

$\endgroup$
  • 1
    $\begingroup$ Dirichlet boundary requires the data to be $H^{1/2}$ regular in order that the solution in $H^1$. $\endgroup$ – Shuhao Cao Jun 20 '13 at 12:41
  • 1
    $\begingroup$ You are right, I misred the question, but anyway, with the new hypothesis $Df\in L^2$, we have that $f\in H^{1/2}(\partial\Omega)$ $\endgroup$ – Tomás Jun 20 '13 at 12:43
  • $\begingroup$ Now I think it is right @ShuhaoCao. What do you think? $\endgroup$ – Tomás Jun 20 '13 at 12:49
  • 1
    $\begingroup$ Hmm...well if $Df\in L^2$ and $v = f|_{\partial \Omega}$ then sure. $\endgroup$ – Shuhao Cao Jun 20 '13 at 12:52
  • 1
    $\begingroup$ 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. $\endgroup$ – Tomás Jun 20 '13 at 13:15

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.