# Lax-Milgran theorem with the Poisson equation

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 $$\Omega$$?

• What is your formulation of Lax Milgram? Commented May 11, 2022 at 13:26
• Not very pretentious: en.wikipedia.org/wiki/Weak_formulation. Commented May 11, 2022 at 13:30

Consider the continuous bilinear function: $$$$B\colon H_0^1(\Omega)\times H_0^1(\Omega)\rightarrow\mathbb{R}, (u,v)\mapsto B(u,v)=\int_\Omega\nabla u\cdot\nabla v,$$$$ which is coercive due to the Poincaré inequality. There exists a constant $$C>0$$ only depending on $$\Omega$$, so that: $$$$\forall u\in H_0^1(\Omega)\colon B(u,u)=\|\nabla u\|_{L^2(\Omega)}^2 \geq C\|u\|_{L^2(\Omega)}^2.$$$$ Notice that $$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)$$, so that: $$$$\forall v\in H_0^1(\Omega)\colon B(u,v)=\langle f,v\rangle_{L^2(\Omega)} \Leftrightarrow \int_\Omega(\Delta u+f)v=0,$$$$ which means that $$u$$ is a weak solution of $$\Delta u=-f$$.