the definition of weak formulation on wiki is: Let $V$ a Banach space. We want to find the solution $u \in V$ of the equation: $$Au=f$$ where $A:V \rightarrow V'$ and $f \in V'$. This is equivalent to finding $u\in V$ such that for all $v\in V$ holds:
$$ [Au](v)=f(v)$$ Here, we call $v$ a test vector or test function.
We bring this into the generic form of a weak formulation, namely, find $ u\in V$ such that,
$$ a(u,v)=f(v)\quad \forall v\in V$$ by defining the bilinear form
$a(u,v):=[Au](v)$.
In poisson's equation we use the bilinear form ( $V=H^1_0(\Omega)$) : $$ a(u,v)=\int _{\Omega }\nabla u\cdot \nabla v\,dx$$ I don't understand how the equation: $$a(u,v)= (Au,v)_{H^1_0(\Omega)} \longrightarrow \int_{\Omega}\nabla u \nabla v \; dx = -\int_{\Omega}(\nabla^2 u) v \; dx$$ is correct when $V=H^1_0(\Omega)$ (i should use Green's identity but $\nabla^2u$ is not defined in $H^1$). I know that: $$\int _{\Omega }\nabla u\cdot \nabla v\,dx=\int _{\Omega }fv\,dx.$$ is well defined but i don't understand how $$a(u,v)= (Au,v)_{H^1_0(\Omega)}$$ is well defined in $H^1$
In the case of the Poisson equation, Lax-Milgram's lemma, tells you that the problem of finding $u\in H_0^1(\Omega)$ such that $$ -\int_{\Omega} \nabla u \cdot \nabla v = \int_{\Omega} fv, \quad \forall v \in H_0^1(\Omega) $$
has one and only one solution. This is the weak solution. A different question, that you can pose after you have guaranteed the existence of a weak solution, is wether that weak solution is actually regular, for instance if $\nabla^2 u \in H^1$, or even if $u \in C^2(\Omega)$. In a way, you derive the weak form from the classical equation but then they become separate equations (until you eventually prove that both problems have the same solution).