Evans defines a function $u=[u(t)](x):=u(t,x)$, \begin{equation} u\in L^2(0,T;H_0^1(U))\quad\text{with}\quad u'\in L^2(0,T;H^{-1}(U)) \end{equation} as a weak solution of a certain parabolic initial/boundary-value problem (i.e. if \begin{equation} \langle u',v\rangle+B[u,v;t]=(f,v)\quad\text{for all }v\in H_0^1(U) \end{equation} and $u(0)=g$).
My question: The above time derivative $u'$ is understood in a distributional/weak sense, alright, but how do we know that $u'$ exists, or is this an implicit assumption? (This is from §7 in Evans)
Yes，it is just the definition of the weak solution. You can use many methods to show the existence of such weak solutions, say Galerkin method, which is exactly what Evans used in his book. For a vector-valued function $u\in L^p(0,T;X)$, the weak derivative $u'$ is defined as the distribution $T(u)$ such that $$ \int_0^T T(u)vdt=-\int_0^T u v' dt\quad \forall v\in C_c^\infty((0,T)). $$ Then we define $W^{1,p}(0,T;X)=\{u\in L^p(0,T;X); u'\in L^p(0,T;X)\}$. $ u'\in L^p(0,T;X)$ if and only if there exits a $L^p$ function $u'$ such that for any $f\in X'$, the function $t\to \langle f, u(t)\rangle$ is absolutely continuous, and $\langle f, u'\rangle=\frac{d}{dt}\langle f, u(t)\rangle$.