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Consider $\Omega$ an open, bounded, and convex domain in $\mathbb{R}^n$. Let $g \in L^{2}(\partial \Omega)$ such that the problem

$$ \left\{ \begin{array}{ccccccc} \Delta u = 0, \ \text{in} \ \Omega, \\ u = g, \ \text{on} \ \partial \Omega \\ \end{array} \right. $$

has a unique solution $u \in H^{1}(\Omega).$

My questions are

  1. Let $x_0 \in \partial \Omega$ with $g$ continuous at this point. Is it true that $$ \lim_{y \rightarrow x_0} u(y) = g(x_0) ? $$

  2. Supoose that I have found a function $v \in H^{1}(\Omega)$ with $$ \lim_{y \rightarrow x} v(y) = g(x)$$ for all $x \in \partial \Omega.$ Then $v$ is the unique solution (in $ H^{1}(\Omega)$) of the problem that I said above?

I don't know the answer of these questions, but if they are true then I could understand a passage of an article that I am studying.

Could someone please point to me a reference that can help me with the questions above?

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  • $\begingroup$ The answer is yes to both. To properly support this, I need some context: when you say at the beginning "$u=g$ on $\partial \Omega$"... what exactly does this mean? Also, can you give a reference to the article? $\endgroup$
    – user147263
    Jul 4 '14 at 1:29
  • $\begingroup$ i said "u=g" on $\partial \Omega$ in the trace sense. the article is this hal.archives-ouvertes.fr/docs/00/12/87/60/PDF/fbpLaplacian.pdf specifically in the page two section 1.3 "outline of the proof". My question is about the function $u_{\Omega}$ in the begin of this section. Disconsider the smoothness hipothesis on $\Omega$ (it is a typo from the authors) $\endgroup$ Jul 4 '14 at 1:38
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Quote from the article:

Let $C$ be the class of smooth, bounded and convex domains in $\mathbb R^n$ such that $K$ belongs to the boundary of the domain. Let $\Omega \in C$, we denote furthermore by $u_\Omega$ the function fulfilling $$\begin{align} \Delta u_\Omega &= 0, \\ u_\Omega &= 0 \text{ in }\partial \Omega \setminus K \\ u_\Omega &= 1 \text{ in } K \end{align}$$

First, we cannot expect $u_\Omega$ to be in $H^1(\Omega)$; the jump between $0$ and $1$ will have "ripple effect" inside the domain, making $|\nabla u|^2$ just large enough for the integral to diverge. A typical example of this behavior, in the plane, is $u(z) = \frac{1}{\pi}\arg z$ on the upper half-plane. Along the real axis, this function jumps from $0$ to $1$ at the origin. (I only consider the local behavior near $0$, which is representative of what happens in your case.) Since $|\nabla u(z)| \approx |z|^{-1}$, the $L^2$ norm of $|\nabla u|$ is infinite.

So, $u_\Omega$ is not a variational (Dirichlet-energy-minimizing) solution, since it has infinite energy. Sobolev spaces don't play a role in its existence. The existence and uniqueness are established with the help of potential theory. Key words: Perron solution, harmonic measure, Poisson kernel, Green function. One reference is section 2.8 of Gilbarg & Trudinger. The Perron solution is uniquely defined for every bounded function on the boundary. It satisfies $\lim_{y \rightarrow x_0} u(y) = g(x_0) $ whenever $g$ is continuous at $x_0$ and $x_0$ is a regular boundary point. In a convex domain, every boundary point is regular.

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  • $\begingroup$ Thanks for your answer! for example if i have $\Omega_1 \supset \Omega_2$ elements in $C$, how can i prove that $u_{\Omega_1} \geq u_{\Omega_2}$ in $\Omega_2$ . $\endgroup$ Jul 4 '14 at 2:07
  • $\begingroup$ @mathstudent Applying the maximum principle to $u_{\Omega_2}-u_{\Omega_1}$ in $\Omega_2$ should work. $\endgroup$
    – user147263
    Jul 4 '14 at 2:08
  • $\begingroup$ i will write better : Thanks for your answer! for example if i have $\Omega_1 \supset \Omega_2$ elements in $C$, how can i prove that $u_{\Omega_1} \geq u_{\Omega_2}$ in $\Omega_2$ . if I try to prove $lim sup_{y \rightarrow x} u_{\Omega_2} \leq lim inf_{y \rightarrow x} u_{\Omega_1} (y)$ , for $x \in \partial \Omega$, i will have problem to prove this in the set $\partial K$. Do you know how to prove this ? The author use $u_1 \geq u_2$ in $\Omega_2$, but he dont prove this inequality ... =\ $\endgroup$ Jul 4 '14 at 2:15
  • $\begingroup$ i will write better : Thanks for your answer! for example if i have $\Omega_1 \supset \Omega_2$ elements in $C$, how can i prove that $u_{\Omega_1} \geq u_{\Omega_2}$ in $\Omega_2$ . if I try to prove $lim sup_{y \rightarrow x} u_{\Omega_2} \leq lim inf_{y \rightarrow x} u_{\Omega_1} (y)$ , for $x \in \partial \Omega$, i will have problem to prove this in the set $\partial K$. Do you know how to prove this ? The author use $u_1 \geq u_2$ in $\Omega_2$, but he dont prove this inequality ... =\ $\endgroup$ Jul 4 '14 at 2:20
  • $\begingroup$ @mathstudent You should read up on Perron's method. I suggest Classical Potential Theory by Armitage and Gardner. Write $u_{\Omega_1}$ as the infimum of a family of superharmonic functions; $u_{\Omega_2}$ as the supremum of a family of subharmonic function, and observe that by the maximum principle, every member of the latter family is $\le$ every member of the former family. $\endgroup$
    – user147263
    Jul 4 '14 at 2:57

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