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?



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


