(Question) on Time-dependent Sobolev spaces for evolution equations I have got a question on so-called time-dependet Sobolev spaces - in particular as introduced in Evans book on PDE for the treatment of parabolic and hyperbolic PDE.
Let us take a look at a linear hyperbolic PDE in $n$ spatial dimension and $1$ time dimension.
$u_{,tt} - Lu = 0$
where
$Lu = \sum_{i,j} a^{ij} u_{,i} u_{,j} + \sum_{i} b^i u_{,i} + c u$.
Furthermore, we impose zero-boundary conditions on a open, bounded, smooth-boundary set $U \subset \mathbb R^n$, and regard a finite time interval $R := [0,T]$
First of all, it is natural to grant the time variable not special treatment, and regard this PDE as a PDE on $R \times U$. We then apply the test function machinery and obtain the notion of weak solution in a natural way. We might want the solution to be two-times weakly differentiable in time and space directions.
We then assume our function lies within $H^2(R) \otimes H^2_0{U}$ with the topological tensor product. I am not too well-versed with this construction, as it does not belong to my university's canon, but for $u \in H^2(R) \otimes H^2_0{U}$ you would expect $u_{,tt} \in H^0(R) \otimes H^2_0{U}$.
On the other hand, we can regard our hyperbolic equation as an second-order ODE on a Hilbert space. Then we might want our solution to be $u \in H^2( R, H^2_0(U) )$. In that case $u_{,tt} \in H^2( R, H^2_0(U) )$.
In most of the above, we can assume weaker regularity, i.e. replace $H^2$ by $H^1$. This works, as in the weak formulation, the second distributional derivatives may be handed over to a test function. Then we the second-derivatives in any direction can be found in $H^{-1}$.
This makes sense, at least to me. However, the book of Evans treats weak derivatives in the above setting in in a different way, and I do not understand the transition.
For example, he defines the solution of the above hyperbolic equation, assuming zero boundary conditions, to have the properties [1]


*

*$u \in L^2( \mathbb R, H_0^1(U) )$

*$u_{,t} \in L^2( \mathbb R, L^2(U) )$

*$u_{,tt} \in L^2( \mathbb R, H^{-1}(U) )$.


This is clearly the ODE-on-VS approach, but it appears the time-derivative is taken over to the spatial derivatives. Of course, he may do so, as this setting is more general than what proposed as a weak solution in the above paragraphs. But then we might use, say, distributions as well.
Whence I wonder why he does so - whether this is just for the reader convenience for whatever reason, whether it really points to what the solution really behaves like most regularly.
Can somebody explain this to me?
[1] L.C.Evans, Partial Differential Equations, 2nd Edition, p.400.
 A: Firstly, you shouldn't think of the weak solutions as ODE on VS. For ODE on VS (using something like the Hille-Yosida theorem for semigroups of linear operator), the correct regularity should be strong continuity: that $u \in C^1(R,\mathcal{H})$ where $\mathcal{H}$ is the Hilbert space. 
Secondly, an intuitive explanation of why $u_{tt} \in L^2(R, H^{-1}(U))$. Just use the hyperbolic equation: $u_{tt} = L u$. So $u_{tt}$ should belong to the same space as $Lu$. With two (spatial) derivatives acting on $u\in L^2(R,H^1(U))$, it is natural that $Lu \in L^2(R,H^{-1}(U))$, and hence also $u_{tt}$. 
Thirdly: I don't think a naive application of the weak solution idea should give you what you claimed ($H^2(R)\otimes H^2_0(U)$). Compare, say, to the elliptic case of a function in a box. Demanding that a function admits 2 weak derivatives in each direction separately is rather stronger than demanding the function admits 2 weak derivative overall. The former is not isotropic. The latter is. (In particular, the former says that $\partial_x\partial_x\partial_y\partial_y\partial_z\partial_z u \in L^2$, which is much stronger than just $u\in H^2$.) In particular, if you have a function $u \in H^1(R\times U)\cap C^\infty(R\times U)$, you'd see that the $H^1(R\times U)$ norm is comparable to the sum of the $u\in L^2(R, H^1(U))$ and $u_t \in L^2(R, L^2(U)) = L^2(R\times U)$ norms. 
