Integral of a weak derivative

Question

While reading chapter 6 of John Hunter's notes (https://www.math.ucdavis.edu/~hunter/pdes/pde_notes.pdf) I got stuck on some steps. I think they are all based on a similar idea as the following.
Let $u \in L^2(0,T; H_0^1(U))$ and it's weak time derivative $u_t \in L^2(0,T; H^{-1}(U))$ does there hold (for $U$ a bounded set of $\mathbb{R}^N$) $$ u(t) = u(0) + \int_0^t u_t(s) ds \: \: \: ?$$ To write this in a meaningfull manner $u(0)$ has to be defined uniquely, therefore is $u$ continuous?
My questions are

1) Does there hold $$ \int_0^t \frac{d}{dt}||u||_{L^{2}(U)}^2(s) ds = ||u||_{L^{2}(U)}^2(t)-||u||_{L^{2}(U)}^2(0) \: \: \: ?$$ again $||u||_{L^{2}(U)}^2(0)$ has to be well-defined.

2) In proposition 6.5 the author gives a prove for the existence of solution of the differential equation $$ \vec{c_n}_t + A(t) \vec{c_n} = \vec{f}(t), \: \: \: \vec{c_n}(0)= \vec{g} $$ Where $A \in L^{\infty}(0,T; \mathbb{R}^{N \times N}), \: \: \: \vec{f} \in L^2(0,T; \mathbb{R}^N) $
And then he writes the 'equivalent' integral equation $$ \vec{c_n}(t) = \vec{g}- \int_0^t A(s) \vec{c_n}(s) ds + \int_0^t \vec{f}(s)ds $$ Why is this equivalent to the original problem? I know this holds in the classical sence, but here we are considering weak derivatives. I don't see how we get from the weak equation to the integral form and backwards.

3) My last question involves a Gronwall inequality in the context of weak derivatives. If for a.e. $t \in [0,T]: \frac{d}{dt}||u(t)||_{L^2(U)} \leq ||u(t)||_{L^2(U)}$ can we bound $u(t)$?

I searched a lot about this but I couldn't find any answer. The most problems will be solved if $u$ was absolutely continous. But since $u_t(t) \in H^{-1}(U)$ and not per se in $H_0^1(U))$ so we can't use the Sobolev theorem directly? The literature I found didn't consider the case for $u_t(t) \in H^{-1}(U)$.

EDIT: In the notes of John Hunter and in the book of Evans they use $${u_n}_t(u_n) = \int_{U} {u_n}_t u_n dx $$ Where $u_n$ are approximate solutions. So if the action of ${u_n}_t$ is just the inner product of $L^2(\Omega)$ why not take ${u_n}_t(t)$ to be in $H_0^1(\Omega)$? Can't it be bounded in $H_0^1(\Omega)$-norm?

Answer 1

1. If $u \in L^2(0,T;H^1_0)$ with $u' \in L^2(0,T;H^{-1})$ then $u \in C^0([0,T];L^2)$, so $u(t)$ makes sense for all $t$. And yes that identity does hold, since $(u(t), u(t)_{L^2}$ is absolutely continuous wrt. $t$.

2. I think the $c_n$ are actually classically differentiable since if I am right they come from the Galerkin approximations.

3. Applying Gronwall's lemma gives a bound on $\lVert u(t) \rVert_{L^2}$ in terms of the $L^2$ norm of the initial data. Just check wikipedia.

For your edit: you are confusing $u'$ with $u_n'$. The former is a weak time derivative, the latter is a time derivative of the Galerkin approximation to the problem, and $u_n'(t) \in L^2$ is a lot smoother than $u'$, so the action agrees with the inner product.