Let $V \subset H \subset V^*$ be a Gelfand triple (eg. $H^1 \subset L^2 \subset H^{-1}$).
Let $u \in L^2(0,T;V)$ have a distributional derivative $u' \in L^2(0,T;V^*)$. So $\int_0^T u(t)\varphi'(t) = -\int_0^T u'(t)\varphi(t)$ holds for all $\varphi \in C_c^\infty(0,T)$ as an equality in $V^*$.
Let $f:\mathbb{R} \to \mathbb{R}$ be a differentiable function with $f'$ bounded.
Suppose $u' \in L^2(0,T;L^2)$; then the chain rule formula for the weak derivative $(f(u))' = f'(u)u' \in L^2(0,T;L^2)$ makes sense.
But if $u' \in L^2(0,T;H^{-1})$ only, how to make sense of $(f(u))'$? Is it right to define the derivative as $$\langle (f(u))', v \rangle := \langle u', f'(u)v \rangle$$
if for all $v \in L^2(0,T;H^1)$, $f'(u)v \in L^2(0,T;H^1)$?
Where may I find more details about this sort of thing? I'd like to avoid BV spaces and measures because this is simpler. Thank you.
Best Answer
For the $H^1 \subset L^2 \subset H^{-1}$ case.
Let $K_\epsilon$ be a mollifier then for $u\in L^2(0,T;L^2)$ we have that $u_\epsilon = K_\epsilon\ast u$ is actually smooth. In fact, $\lim_{\epsilon \rightarrow 0} u_\epsilon$ converges to $u$ in $L^p_{loc}$ for all $p\in [1,\infty)$. Since $f$ is differentiable, then it is continuous. So, we have the $\lim_{\epsilon \rightarrow 0} f(u_\epsilon) = f(u)$ in the same spaces as before.
Consider the following, \begin{align*} \langle v',f(u_\epsilon)\rangle &= -\langle v, f'(u_\epsilon) u_\epsilon' \rangle \\ &= -\langle v,f'(u) u_\epsilon' \rangle + \langle v,\left[f'(u) - f'(u_\epsilon)\right] u_\epsilon'\rangle \end{align*} A useful fact to use is that since $f$ has bounded derivative then $f$ is Lipshitz. Using this fact and taking $\epsilon$ to zero gives you what you want.