 Solution

The Lax-Milgram theorem says that if $H$ is a real Hilbert space, $B:H\times H\to \mathbb{R}$ is a bounded bilinear form satisfyingfor some $\alpha >0$, and $F\in H^{\ast }$, then there is a unique $u\in H$ such that $B(u,v)=F(v)$ for every $v\in H$.

For this problem take $H=H_0^1(U)$ andThe Cauchy-Schwarz inequality and the continuous embedding $H_0^1(U)\hookrightarrow L^2(U)$ make both maps bounded. For smooth zero-boundary functions, integration by parts givesdensity extends this identity to $H_0^1(U)$. Consequentlyso $B$ is coercive. Lax–Milgram supplies the unique weak solution.