# (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 

• $u \in L^2( \mathbb R, H_0^1(U) )$
• $u_{,t} \in L^2( \mathbb R, L^2(U) )$
• $u_{,tt} \in L^2( \mathbb R, H^{-1}(U) )$.

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?

 L.C.Evans, Partial Differential Equations, 2nd Edition, p.400.

• A quibble on your title: it really isn't a time-dependent Sobolev space, which I tend to think as the family of Sobolev spaces $\mathcal{H}_t$ depending on a time parameter. What you really described is $u$ being a Sobolev space-valued function $u: R\to \mathcal{H}$. Mar 12, 2011 at 22:50

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.
• One might consider any function that is discontinuous at a point in some directions but continuous in others. For example, $\frac{x}{\sqrt{x^2+y^2}}$. If you take a weak derivative, instead of getting a distribution, you instead get a dipole-like singularity, and the "strength" of the singularity increases by 1 power for each direction the function is discontinuous in. Apr 22, 2011 at 4:05