[Math] Two weak formulations for a time-dependent PDE: are they equivalent

Question

My question is concerned with the way people tend to define weak formulations for time-dependent PDEs. Suppose I have a wave equation in a simplest form
$u_{tt}-a^{2}\Delta u = f$ in a bounded domain in $\mathbb{R}^{2}$ or $\mathbb{R}^{3}$ for $t\in(0,T)$. To make it simple, suppose $u = 0$ on $\partial\Omega$ and initial conditions are homogeneous.
Typical weak formulation is constructed from integration over entire variable space $[0,T]\times\Omega$ and, if we do not use integration by parts in time, has a form
$\int_{0}^{T}\int_{\Omega}(u_{tt}\phi + a^{2}\nabla u\nabla\phi)dxdt = \int_{0}^{T}\int_{\Omega}f\phi dxdt$ for each $\phi$ from the closure of space $\dot{D}((0,T)\times\Omega)$ which consists of all smooth functions with compact support with respect to $\Omega$ ( but not time dimension ). The closure is done with respect to norm $H^{2}((0,T)\times\Omega)$. Function $u$ is sought from the same. A good and consistent theory exists for this formulation, such as existence and uniqueness results, as well as some regularity results. I may not specify the spaces correctly here, but that's not the point right now.
What troubles me is that people don't use this formulation in practical situations, such as proving a priori estimates for Finite Element method. They instead keep using what I call a stronger form:
for a.a. $t\in(0,T)$ we have $\int_{\Omega}(u_{tt}\phi + a^{2}\nabla u\nabla\phi)dx = \int_{\Omega}f\phi dx$ for all twice-differentiable maps $\phi$ belonging to $L^{2}(0,T; H_{0}^{1}(\Omega))$ ( adjust the space in your way, if you like ). So it's the difference in how the time evolution is treated: you integrate over time in the first formulation and
avoid it in the second one.
I've never seen a single proof of existence/uniqueness for the second formulation. It's like it's equivalent to the one above. But if so, can anyone provide at least a sketch of a proof?
I already made a thread concerning the difference between spaces $L^{p}(0,T; L^{p}(\Omega))$ and $L^{p}([0,T]\times\Omega)$: $L^{2}(0,T; L^{2}(\Omega))=L^{2}([0,T]\times\Omega)$?
and it is closesly related to my question here.
Any help will be much appreciated!
By the way, the original weak formulation ( the first one ) implies that you also integrate by parts in time, to reduce regularity assumptions in time. But since you cannot do it in the second formulation, I intentionally skipped it, to give a chance of proving that solutions for both formulations are the same.

Answer 1

For the strong formulation, you only need a test function $\phi \in H_0^1(\Omega)$ (no dependence on time).

Assume that a solution of the weak formulation has the regularity $u_{tt} \in L^2(0,T; H^{-1}(\Omega))$, $u \in L^2(0,T; H_0^1(\Omega))$. Now, take a arbitrary $\phi \in H_0^1(\Omega)$ and some $\psi \in C^\infty(0,T)$. The weak formulation (tested with $\phi(x)\,\psi(t)$) implies $$\int_{0}^{T}\int_{\Omega}(u_{tt}\phi(x) + a^{2}\nabla u\nabla\phi(x) - f \phi(x))dx \, \psi(t) \, dt = 0.$$ Since $\psi$ was arbitrary, the fundamental theorem of calculus of variations implies $$\int_{\Omega}(u_{tt}\phi(x) + a^{2}\nabla u\nabla\phi(x) - f \phi(x))dx = 0$$ a.e. in $(0,T)$.

Hence, both formulations are equivalent if $u$ has the above mentioned regularity.