Assuming that $\Omega $ is Lipschitz- continuous subset of $\mathbb{R}^n$ and making use of the identity I.2.17 and theorem 2.4 and 2.5 in [V. Girault and P. A. Raviart, Finite Element Methods for Navier-Stokes Equations. Theory and Algorithms, Springer Series in Computational Mathematics, 5 Springer, Berlin, 1986.]
Which state that\
- $\mathcal{D}(\bar{\Omega})^n$ is dense in $ H(\operatorname{div},\Omega)$\
and
- the map $\gamma: v \rightarrow v\cdot\eta|_{\Gamma_n}$ defined on $\mathcal{D}(\bar{\Omega})^n$ can be extended by continuity to a linear and continuous mapping from $H(\operatorname{div},\Omega)$ into $H^{-\frac{1}{2}}(\Gamma_n)$\
So, we have the following Green formula
$$(v,\nabla \phi)+(\operatorname{div} v,\phi)=(v\cdot \eta,\phi)_{\Gamma_n} \quad \forall v \in H(\operatorname{div},\Omega), \forall \phi \in H^1(\Omega)$$
Therefore, we have for every $v \in V_0$
\begin{align}
a(u,v)=\int_\Omega a(x)\nabla u(x) \cdot \nabla v(x)+\bigl (b(x) \cdot \nabla u(x) \bigr)v(x) \, \mathrm{dx} \\
= \int_\Omega -\operatorname{div} \big(a(x)\nabla u(x)\big)v(x)+\bigl (b(x) \cdot \nabla u(x) \bigr)v(x) \, \mathrm{dx}\\+\int_{\Gamma_n} a(x)\nabla u(x) \eta v(x) \mathrm{ds}
\end{align}
while $\eta$ is the outward unit normal vector.
Now, taking $v \in \mathcal{D}(\Omega)$ we have
$$ \int_\Omega -\operatorname{div} \big(a(x)\nabla u(x)\big)v(x)+\bigl (b(x) \cdot \nabla u(x) \bigr)v(x) \, \mathrm{dx}=\int_{\Omega} f(x)v(x)\, \mathrm{dx}$$
So,
$$ -\operatorname{div} \big(a(x)\nabla u(x)\big)+\bigl (b(x) \cdot \nabla u(x) \bigr)=f(x) \quad \text{a.e. in } \Omega$$
Which can be extended into all $\Omega$
Making use of this, we find
$$ a(v) \frac{\partial u}{\partial \eta}=g(x), \quad \forall x \in \Gamma_n$$
Finely, the equation is
$$
\left\{\begin{aligned}
-\operatorname{div} \big(a(x)\nabla u(x)\big)+\bigl (b(x) \cdot \nabla u(x) \bigr)=f(x) \quad \forall x \in \Omega\\
a(x) \frac{\partial u}{\partial \eta}=g(x), \quad \forall x \in \Gamma_n
\end{aligned}\right.
$$
The second one can be treated the same way to get
$$
\left\{\begin{aligned}
-\operatorname{div} \big(a(x)\nabla u(x)\big)-\bigl (b(x) \cdot \nabla u(x) \bigr)=f(x) \quad \forall x \in \Omega\\
a(x) \frac{\partial u}{\partial \eta}+b(x)u(x)\eta=g(x), \quad \forall x \in \Gamma_n
\end{aligned}\right.
$$
Best Answer
No, this operator is not lower bounded in your sense. If it were, we would have $$ \gamma \int_\Omega |\nabla u|^2\leq c\gamma\int_\Omega A\nabla u\cdot\nabla u=c\gamma\int_\Omega vu\leq c\gamma\|u\|_2\|v\|_2\leq c\|u\|_2^2 $$ for $u\in C_c^\infty(\Omega)$. To see that this is not true, take any $u\in L^2(\Omega)\setminus H^1(\Omega)$ and a sequence $(u_n)$ in $C_c^\infty(\Omega)$ such that $u_n\to u$ in $L^2$.