Getting an a priori bound on a nonlinear gradient term in PDE; how to adapt trick from $L^2$ case to $H^{-1}$ case?

Question

I have the PDE $$u_t(t) - \Delta f(u(t)) = 0$$ in $H^{-1}(\Omega)$ where $f$ is a nonlinear function.

Define $F(s) = \int_0^s f(s)$. Note that if $u_t(t) \in L^2(\Omega)$, $$\frac{d}{dt}F(u(t)) = f(u(t))u_t(t).$$

We could test the PDE with $f(u(t))$ (i.e multiply by it and integrate) and if $u_t \in L^2(\Omega)$, we can use the above identity to get $$\int_{\Omega}F(u(T)) - \int_{\Omega}F(u_0) + \int_0^T \int_{\Omega} |\nabla f(u(t))|^2 =0$$ and then we can throw away the first term on the LHS since $F$ is positive (a fact for the particular problem). This gives us a bound on $ \nabla f(u)$ in $L^2(0,T;L^2)$.

This is a standard trick. What can I do if I only have $u_t \in H^{-1}(\Omega)$? What is the trick in this case?

Answer 1

Is f continuous with respect to someoing appropriate norm? If so, then I think the argument goes something like the following. Since $$u_t - \Delta f(u(t)) = 0$$ in the $H^{-1}$ sense then $$\langle u_t , v \rangle = \langle \Delta f(u(t)), v \rangle \; \forall v\in H_0^1$$ Take $v = f_\epsilon(u(t)) = \phi_\epsilon \ast f(u(t))$ where $\phi_\epsilon$ is a smooth mollifier. Then we have that \begin{eqnarray} \langle u_t, f_\epsilon(u(t)) \rangle &=& \langle \Delta f(u(t)), f_\epsilon(u(t)) \rangle \nonumber \\ &=& - \langle \nabla f(u(t)), \nabla f_\epsilon(u(t)) \rangle \nonumber \end{eqnarray} Furthermore, you have \begin{eqnarray} \langle u_t, f_\epsilon(u(t)) \rangle &=& \langle (u_\epsilon)_t, f(u(t)) \rangle \nonumber \\ &=& \int \frac{d}{dt} F(u_\epsilon) + \langle (u_\epsilon)_t, f(u(t)) - f(u_\epsilon(t)) \rangle \nonumber \end{eqnarray} Integrate in time and take the limit as $\epsilon$ goes to zero to conclude.

I hope I am not missing anything subtle or doing something silly.