1
$\begingroup$

Let $ \Omega $ be a bounded domain in $\mathbb R^3$ with smooth boundary. Consider the Poisson equation $$ -\Delta u=f $$ where $ f\in C_0^{\infty}(\Omega) $ and $f$ is null outside $\Omega$.

I'm not sure I understand the Lax-Milgram theorem. Does it say that we can find the solution in $ H_0^1$ whose support is included inside $\Omega$?

$\endgroup$
2

1 Answer 1

4
$\begingroup$

Consider the continuous bilinear function: \begin{equation} B\colon H_0^1(\Omega)\times H_0^1(\Omega)\rightarrow\mathbb{R}, (u,v)\mapsto B(u,v)=\int_\Omega\nabla u\cdot\nabla v, \end{equation} which is coercive due to the Poincaré inequality. There exists a constant $C>0$ only depending on $\Omega$, so that: \begin{equation} \forall u\in H_0^1(\Omega)\colon B(u,u)=\|\nabla u\|_{L^2(\Omega)}^2 \geq C\|u\|_{L^2(\Omega)}^2. \end{equation} Notice that $H_0^1(\Omega)=W_0^{1,2}(\Omega)\subseteq H^1(\Omega)=W^{1,2}(\Omega)\subseteq L^2(\Omega)$ and since $W_0^{1,p}(\Omega)$ is defined as the closure of $C_\mathrm{c}^\infty(\Omega)$ in $W^{1,p}(\Omega)$, $H_0^1(\Omega)$ is a closed subspace of the Hilbert space $L^2(\Omega)$ and therefore itself a Hilbert space with the restriction of $\langle\cdot,\cdot\rangle_{L^2(\Omega)}$ to $H_0^1(\Omega)$.

The Lax-Milgram theorem (which can be proved using the Riesz representation theorem to which it is similar) can now be applied to $B$. For every $f\in H_0^1(\Omega)$ there exists a $u\in H_0^1(\Omega)$, so that: \begin{equation} \forall v\in H_0^1(\Omega)\colon B(u,v)=\langle f,v\rangle_{L^2(\Omega)} \Leftrightarrow \int_\Omega(\Delta u+f)v=0, \end{equation} which means that $u$ is a weak solution of $\Delta u=-f$.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .