Let $(V,H,V^*)$ be an evolution triple. Let
$$\mathcal{V}=L^2(0,T,V),$$ $$\mathcal{W}=\{v\in \mathcal{V}\,|\,v'\in \mathcal{V}^* \}.$$
Problem $\mathcal{P}$. Find $u\in \mathcal{V}$ such that $u'\in\mathcal{W}$ and $$u''(t)+A(t,u'(t))=f(t)$$ holds with $u(0)=u_0$ and $u'(0)=u_1$.
Problem $\mathcal{Q}$. Find $z\in \mathcal{W}$ such that $$z'(t)+A(t,z(t))=f(t)$$ holds with $z(0)=u_1$.
How to show that $\mathcal{P}$ and $\mathcal{Q}$ are equivalent?
"$\mathcal{P} \Rightarrow \mathcal{Q}$". Since $u'\in \mathcal{W}=W^{1,2}(0,T;V,H)$ then it can be indentyfied with $AC(0,T;V)$ (absolutely continuous functions with values in $V$) and so
$$u'(t)=\int_{0}^{t}u''(s)ds+u'(0).$$ Since $u'$ exists (in what sense actually? Why can I write (1)? $u'\in \mathcal{V}$ is enough?), then $$u(t)=\int_{0}^{t}u'(s)ds+u(0).\qquad (1)$$ Let $z(s)=u'(s)$. Observe that $$z(0)=\int_{0}^{0}u''(s)ds+u'(0),$$ impying $z(0)=u_1$. Also, It follows from the assumptions that $z\in \mathcal{W}$. Hence, every colution of $\mathcal{P}$ is a solution to $\mathcal{Q}$.
I have some troubles with "$\mathcal{Q} \Rightarrow \mathcal{P}$". If $z\in \mathcal{W}$, then $z\in AC(0,T,V)$ and as a result $$z(t)=\int_{0}^{t}z'(s)ds+z(0)=\int_{0}^{t}z'(s)ds+u_1.$$ Where does $z'$ belong? From the very definition I suppose it is from $\mathcal{V}^*$. Correct? Let $u'(t)=z(t)$. Since $z\in \mathcal{W}$, then $u'$ has a generalized weak derivative $u''$. $$u''(t)=(u'(t))'=(\int_{0}^{t}z'(s)ds+z(0))'=z'(t).$$ Why the latter inequality holds?