Time dependent Sobolev spaces and parabolic PDE

Question

Let $X$ be a Banach space and $u \in L^p(0,T;X).$ Then $v \in L^p(0,T;X)$ is called weak derivative of $u$ if \begin{eqnarray}\tag{1} \int\limits_0^T u(t)\,\phi_t \,\mathrm dt = -\int\limits_0^T v(t)\,\phi(t)\,\mathrm dt \quad \text{for all } \phi \in C_c^{\infty}((0,T)). \end{eqnarray} The weak derivative $v$ is denoted by $u'$.

While defining the weak formulation of Linear parabolic PDEs (for example, $u_t-\Delta u=0$), the weak formulation is defined for $u \in L^2(0,T;H^1_0(\Omega))$ and $u'\in L^2(0,T;H^{-1}(\Omega)).$

My doubts are the following:

1. Why do we need $u'\in L^2(0,T;H^{-1}(\Omega))?$ and why not $u' \in L^2(0,T;H^1_0(\Omega))$ as in the definition of weak derivative?
2. In view of (1), how to interpret $u'=v \in L^p(0,T;X^*)$ as the weak derivative of $u \in L^p(0,T;X)?$

P.S. : Notations are same as the ones used Chapter 5 and 7 of the book "Partial Differential Equations'' by L.C. Evans.

Answer 1

So the two assumptions are independent assumptions. As a first assumption you want $u∈ L^2_t(H^1_0)$ (shortcut for $L^2(0,T;H^1_0(\Omega))$) because you want to be able to define for example $∫ |∇ u|^2 = -∫ u\,\Delta u$ and also the value of $u$ at the boundary (so you need to have some notion of trace at the boundary, which is ok if you are in $H^1$).

Then, on a completely different and independent hypothesis, you want $u∈ L^2_t(H^{-1})$. I say independent because for different equations, the hypotheses could be for example $u∈ L^2_t(H^1_0)$ and $u'∈ C^0_t(H^{-2})$. Here the fact that you get $H^{-1}$ is not due to the fact that it is the dual of $H^1_0$ but more to your particular equation. If $u∈ H^1$, then it has one derivative in $L^2$. But when taking the Laplacian, you take two derivatives which is why you arrive at $u' = \Delta u ∈ H^{1-2} = H^{-1}$.

And to define the weak derivative, in general you do not need a function and its weak derivative to be in the same space. The only requirement if you want your weak derivatives to remain functions (and not distributions) is that $u∈ L^1_{\mathrm{loc}}$ and $u'∈ L^1_{\mathrm{loc}}$ so that you can make sense of $(1)$. Once this is ok, if you furthermore have $v∈ L^p_t(Y)$ for some space $Y$ then you can say that $u'∈ L^p_t(Y)$.