The Poisson problem $-\Delta u =0$ with $u=g$ on the boundary where $g \in H^{\frac{1}{2}}$

Question

Consider $$ \begin{align} -\Delta u =0 & \text{on $\Omega$} \\ u = g & \text{on $\partial\Omega$} \end{align} $$ where $g \in H^{\frac{1}{2}}(\partial\Omega)$.

It seems there exists a solution $u \in H^1(\Omega)$ to this problem, but how is existence proved? I am also interested in the estimate $$|u| \leq C|g|$$ and what the constant depends on. Thanks for any references.

Answer 1

As $g\in H^{\frac{1}{2}}(\partial \Omega)$,$\exists v\in H^1(\Omega)$ such that $\gamma_0(v)=g$ where $\gamma_0$ is the trace operator. If you define $w=u-v$ you problem is equivalent to $$-\Delta w= \Delta v \ \ \Omega$$ $$w =0 \ \ \ \partial \Omega$$ Notice that $\Delta v\in H^{-1}(\Omega)$ , so using Lax-Milgram theorem (or Riesz in this case) you have that the problem have an unique solution and $$\|w\|_{H^1}\leq C \|\Delta v\|_{H^{-1}}\leq C'\|v\|_{H^1}$$ so $$\|u\|_{H^1}\leq \|w\|_{H^1}+\|v\|_{H^1}\leq (C'+1)\|v\|_{H^1}\leq C'' \|g\|_{H^{\frac 1 2 }} $$ the last estimete because of the trace theorem.

Answer 2

A suggestion:

Say $u$ and $v$ both solve the equation. Then:

$-\Delta(u-v) = 0$ on $\Omega$ and $(u-v) = 0$ on $\partial \Omega$

Can you now show that $u-v$ must be zero?