A question about the PDE: $u_t-\Delta u=au-bvu$ Consider the following system of partial differential equations (see here for more details):

I just want to know how the author got that expression after multiplying by $u^-$ and integrating over $\Omega$ in the proof of Lemma 2.3. I think the function $u^-=\min(0,u)$ is not differentiable  in time.
 A: $$u_t-\Delta u = au-buv \tag{RD} $$
As written, let's multiply (RD) by $u^{-}:=\text{min} \{0,u \}$ and integrate over $\Omega$ we get
$$ \underbrace{\int \partial_t u \ u^{-}}_{(1)} ~\underbrace{- \int \Delta u \ u^{-}}_{(2)}=\underbrace{\int u(a-bv)u^{-}}_{(3)}$$
Now notice that by Green and by the homogenous Dirichlet boundary we get
$$(2)=- \int \Delta u\ u^{-}=\int \nabla u \ \nabla u^{-}-\int_\Gamma \partial_n u \ u^{-}=\int \nabla u \ \nabla u^{-}$$
and  $$(3)=\int u(a-bv)u^- \leq C \int (u^{-})^2$$
since $u \leq u^-$. Lastly notice that as written in my comment above that the weak derivative (both valid for time and space derivative) of $u^{-}$ is given by
$$(u^{-})'=\begin{cases} u', &u\geq0 \\ 0, &u<0 \end{cases}$$
and therefore by the definition of weak derivatives with an appropriate test function one gets $$\int \partial_t u^ \ u^-=\int \partial_t u^- \ u^-~ \text{ and } ~\int \nabla u \ \nabla u^-=\int \nabla (u^-) \ \nabla u^-$$ (valid for integration on $\Omega$ by extending with $0$) and finally with the chain rule $$(1)=\int u^- \partial_t u^-=\int \frac12 \frac{d}{dt} (u^-)^2$$ we get
$$\frac12 \frac{d}{dt} \int (u^-)^2+\int (\nabla u^-)^2=\int u(a-bv)u^- \leq C \int (u^-)^2 $$
