Weak solution of PDE

Question

Let $\Omega\subset \mathbb{R}^d$ be a domain. $f\in L^2(\Omega)$ is a known function. There exists a function $q\in L^2(\Omega)$, such that for all $\zeta\in H^1_0(\Omega)$, we have $$ \int_{\Omega}2q\nabla f\cdot\nabla \zeta + q|\zeta(x)|^2dx = \int_{\Omega} r(x)\zeta(x) dx. $$ Here, I would like to show that $q$ is the weak form of a solution of some PDE. I tried reverse engineering by applying the divergence theorem, but the term $|\zeta(x)|^2$ is causing problem. Suppose the problem comes without the term $|\zeta(x)|^2$, then it will be quite straightforward. May I know there is a way to get around with it? Or is there any alternative approach to solve for $q$?

Answer 1

The $|\zeta|^2$ term can actually be removed from your equality. The stated identity holds for all $\zeta \in H^1_0$, so fix $\zeta_0 \in H^1_0$ and consider $\zeta = \lambda \zeta_0 \in H^1_0$ for $\lambda \in \mathbb{R} \backslash \{0\}$. The equality then requires that $$ \lambda \int_\Omega [2q \nabla f \cdot \nabla \zeta_0 - r \zeta_0 ] = -\lambda^2 \int_\Omega q |\zeta_0|^2. $$ Divide by $\lambda$ and then send $\lambda \to 0$. This yields $$ \int_\Omega [2q \nabla f \cdot \nabla \zeta_0 - r \zeta_0 ] =0 $$ for all $\zeta_0 \in H^1_0$. From here you can integrate by parts, as you mention in the post, to find that $$ -\nabla \cdot(2 q \nabla f) = r. $$