Given $\Omega \subset \mathbb{R}^n$ a bounded smooth open set, $T>0$ and $a,b \in L^\infty(\Omega)$ and positive. I'm trying to prove the positiveness of weak solutions to the following PDE
$$ \begin{cases} u_t-\Delta u=a(x)-b(x) u \quad \text{ in } \Omega \times(0,T), \\ u(.,0)\geq 0 \quad \text{ in } \Omega,\\ u=0 \quad\text{ on } \partial \Omega. \end{cases} $$ My attempt: \begin{align*} \dfrac{1}{2}\dfrac{d}{dt} \Vert u^- \Vert_{L^2(\Omega)}^2&=-\int_\Omega u_t u^-\\ &=\int_\Omega -\Delta u u^--\int_\Omega a(x) u^-+\int_\Omega b(x) u u^- \quad \text{from the equation}\\ &=\int_\Omega \nabla u . \nabla u^--\int_\Omega a(x) u^-+\int_\Omega b(x) u u^- \quad \text{from Green's formula}\\ &=-\int_\Omega \vert \nabla u^- \vert^2-\int_\Omega a(x) u^-+\int_\Omega b(x) {u^-}^2 \quad \text{from the property of supports of $u$ and $u^-$}\\ &\leq\Vert b \Vert_{L^\infty(\Omega)} \Vert u^- \Vert_{L^2(\Omega)}^2 \end{align*} By Gronwall's inequality $u^-=0$ and the result follows.
I'm having doubts about the equality $$ \dfrac{1}{2}\dfrac{d}{dt} \Vert u^- \Vert_{L^2(\Omega)}^2=-\int_\Omega u_t u^-. $$