 Solution

The real Lax-Milgram theorem states that if $V$ is a Hilbert space, $B:V\times V\to \mathbb{R}$ is a continuous bilinear form with $B(v,v)\ge \alpha \Vert v{\Vert }_V^2$ for some $\alpha >0$, and $\ell \in V^{\ast }$ is continuous, then there is a unique $u\in V$ such that $B(u,v)=\ell (v)$ for all $v\in V$. Moreover $\Vert u{\Vert }_V\le \Vert \ell \Vert /\alpha$.

Take the Sobolev space $V=H_0^1(\Omega )$ with norm $\Vert v{\Vert }_V=\Vert \nabla v{\Vert }_2$. Since $\Omega$ is bounded, the Poincaré inequality gives $\Vert v{\Vert }_2\le C_P\Vert \nabla v{\Vert }_2$, making this norm equivalent to the full Sobolev norm. For the given bounded measurable coefficient defineThen $|B(u,v)|\le (1+\bar{a}{C}_P^2)\Vert u{\Vert }_V\Vert v{\Vert }_V$, $B(v,v)\ge \Vert v{\Vert }_V^2$ since $a\ge 0$, and $|\ell (v)|\le C_P\Vert f{\Vert }_2\Vert v{\Vert }_V$. The Lax-Milgram theorem therefore gives a unique $u\in H_0^1(\Omega )$ satisfying this identity, which is exactly the required weak solution. Strict positivity of the lower coefficient bound is unnecessary: the zero boundary condition and boundedness supply coercivity through the gradient term.