We consider the boundary value problem on a bounded, open domain $\Omega \subset \mathbb R^n$ determining $u : \Omega \rightarrow \mathbb R$ such that $$-\Delta u(x)=f(x), \qquad u|_{\partial\Omega}=0,$$ for a given function $f : \Omega \rightarrow \mathbb R$. The variational formulation reads: find $u \in H^1_0 (\Omega)$ such that $$a(u, v) := (\nabla u,\nabla v)_0 = (f, v)_0 \ \text{for all}\ v \in H^1_0 (\Omega) \quad (\ast)$$ for a given function $f \in H^{−1}(\Omega)$, where $(\cdot, \cdot)_0$ denotes the standard $L_2$-inner product on $\Omega$. Introducing the differential operator $$A:H^1_0 (\Omega)\rightarrow H^{−1}(\Omega), \quad \langle Au,v\rangle:=a(u,v), \quad u,v\in H^1_0 (\Omega),$$ we can rewrite $(\ast)$ as an operator equation $$Au = f$$ in the Sobolev space $H^1_0 (\Omega)$. Maybe my question may be trivial and sorry if it will be, but I'm not very knowledgeable about this subject: who is the operator? Perhaps it is $-\Delta$, i.e. $A=-\Delta$?
Thanks in advance.
Yes, $A=-\Delta$. You have $$ \begin{split} \langle Au,v\rangle &= a(u,v) \\&= (\nabla u,\nabla v)_{L^2} \\&= \langle-\Delta u,v\rangle. \end{split} $$ Here $\langle\cdot,\cdot\rangle$ denotes the duality product between $H^{-1}$ and $H^1_0$. Integration by parts gives no boundary terms since $v\in H^1_0$. Since the above calculation holds for all $v\in H^1_0$ and $H^{-1}$ is defined to be the dual of this space, we have indeed $Au=-\Delta u$ for any $u\in H^1_0$.