Assume $v(t)\in H_0^1(U)$ for all $0\leq t\leq T$. The weak derivative $v'$ of $v$ is defined by \begin{equation} \int_0^T \phi'(t)v(t)\,\mathrm{d}t=\int_0^T\phi(t)v'(t)\,\mathrm{d}t\quad\text{for all }\phi\in C_c^\infty([0,T]) \end{equation} (Bochner integral). In Evans' proof of the uniqueness of weak solutions of second order hyperbolic pdes one has to calculate \begin{equation} I=\frac12\frac{\mathrm{d}}{\mathrm{d}t}\int_Ua^{ij}\partial_i v\partial_j v+b^i\partial_i v\cdot v\,\mathrm{d}x. \end{equation} The integrand reads pointwise, of course, thinking of $v$ as $[v(t)](x),x\in U,t\in[0,T]$, $\partial_i$ denotes weak derivatives in space; So this is an ordinary Lebesgue integral. We assume $a^{ij}=a^{ij}(x,t),b^i=b^i(x,t)\in C^1(U\times[0,T])$ and $v'(t)\in H_0^1(U)$ for all $t\in[0,T]$. We can even assume $v\in W^{1,1}(0,T;H_0^1(U))$.
I want to prove, that \begin{equation} I=\int_Ua^{ij}\partial_i(v')\partial_j v+b^i\partial_i(v')v\,\mathrm{d}x+\frac12\int_U\partial_t a^{ij}\partial_i v\partial_j v+\partial_t b^i\partial_i v\cdot v+b^i\partial_i v\cdot v'\,\mathrm{d}x. \end{equation} So usally one would swap integration and derivation in $I$, use product rule and integration by parts. E.g., because of $v(t),v'(t)\in H_0^1(U)$ we got \begin{equation} \int_U b^i\partial_i(v')v\,\mathrm{d}x=-\int_U b^i\partial_i v\cdot v'\,\mathrm{d}x. \end{equation} But what is \begin{equation} \frac{\mathrm{d}}{\mathrm{d}t}\left(a^{ij}\partial_i v\partial_j v+b^i\partial_i v\cdot v\right) \end{equation} ? Can we use product rule or integration by parts in the Bochner Sobolev space?
Addendum
I think given that the product rule holds, the problem would reduce to why \begin{equation} \partial_i(v')=(\partial_i v)' \end{equation} holds, which I should be able to proof using Fubini: We have the formula \begin{equation} v(t)=v(0)+\int_0^tv'(s)\,\mathrm{d}s \end{equation} for all $t\in[0,T]$. Let $\phi\in C_c^\infty(U)$, writing $\partial_i v$ for $t\mapsto \partial_i(v(t))$ we have \begin{equation} \int_0^t\int_U(\partial_i v)'\phi(x)\,\mathrm{d}x\,\mathrm{d}s=\int_U\phi(x)(\partial_i v(t)-\partial_i v(0))\,\mathrm{d}x=-\int_U\partial_i\phi(x)(v(t)-v(0))\,\mathrm{d}x\\=-\int_0^t\int_U\partial_i\phi(x)v'(s)\,\mathrm{d}x\,\mathrm{d}s. \end{equation} So given sufficient continuity of the integrand we got \begin{equation} \int_U(\partial_i v)'\phi(x)\,\mathrm{d}x=-\int_U\partial_i\phi(x)v'(s)\,\mathrm{d}x. \end{equation}