This is from Chapter 6 -- Second-Order Elliptic Equations -- of PDE Evans, 2nd edition. See the very bottom of my post for my question.
Pages 311-312:
$\quad$We will in this chapter study the boundary-value problem \begin{cases}\tag{1} Lu=f & \text{in }U \\ u = 0 & \text{on } \partial U, \end{cases} where $U$ is an open, bounded subset of $\mathbb{R}^n$ and $u : \bar{U} \rightarrow \mathbb{R}$ is the unknown, $u=u(x)$. Here $f : U \rightarrow \mathbb{R}$ is given, and $L$ denotes a second-order partial differential operator having either the form $$Lu = -\sum_{i,j=1}^n (a^{ij}(x)u_{x_i})_{x_j} + \sum_{i=1}^n b^i(x) u_{x_i} + c(x)u \tag{2}$$ or else $$Lu=-\sum_{i,j=1}^n a^{ij}(x)u_{x_i x_j} + \sum_{i=1}^n b^i(x) u_{x_i} + c(x)u, \tag{3}$$ for given coefficient fuctions $a^{ij},b^i,c (i,j=1,\ldots,n)$.
Page 314:
DEFINITIONS. (i) The bilinear form $B[\, \, , \, \, ]$ associated with the divergence form elliptic operator $L$ defined by $\text{(2)}$ is $$B[u,v] := \int_u \sum_{i,j=1}^n a^{ij} u_{x_i} v_{x_j} + \sum_{i=1}^n b^i u_{x_i} v + cuv \, dx\tag{8}$$ for all $u,v \in H_0^1(U)$.
Page 318:
THEOREM 2 (Energy estimates). There exist constants $\alpha, \beta > 0$ and $\gamma \ge 0$ such that $$|B[u,v]| \le \alpha \|u \|_{H_0^1(U)} \|v\|_{H_0^1(U)} \tag{i}$$ and $$\beta \|u\|_{H_0^1(U)}^2 \le B[u,u] + \gamma \|u\|_{L^2(U)}^2 \tag{ii}$$ for all $u,v \in H_0^1(U)$.
Page 320:
Examples. In the case $Lu=-\Delta u$, so that $B[u,v] = \int_u Du \cdot Dv \, dx$, we easily check using Poincaré's inequality that Theorem 2 (page 318 of book) holds with $\gamma = 0$. A similar asssertion holds for the general operator $Lu=-\sum_{i=1}^n (a^{ij}u_{x_i})_{x_j} + cu$, provided $c \ge 0$ in $U$.
My question concerns with page 320 (last gray box I listed). When we declare $Lu=-\Delta u$, that is, the operator $L$ is specified to be the negative Laplacian $-\Delta$, how exactly do these following expressions result? $$B[u,v] = \int_u Du \cdot Dv \, dx$$ $$Lu = -\sum_{i,j=1}^n (a^{ij} u_{x_i})_{x_j}+cu(x)$$
Consider $Lu=0$, where $L=-\Delta$ then we have
$Lu=-\Delta u=-\sum_{i,j=1}^n(\delta^{i,j}u_{x_i})_{x_j}=-\sum_{i=1}^nu_{x_ix_i}+0\cdot u(x)$, (since in this case $c=0$).
We find the bilinear form, by testing with a test function $v$ in a suitable space, say $v\in H^1(U)$, which gives us:
$0=\int_UvLu\,dx=-\int_Uv\Delta u\,dx=\int_U\nabla u\cdot\nabla v\,dx=:B[u,v]$. Since now the problem can be set in a weak form, find $u\in H^1_0(U)$ such that $B[u,v]=0,\,\forall v\in H^1(U)$.