Consider the problem
$$\left\{\begin{array}{ll} -\Delta u+c\cdot \nabla u=f, & \Omega \\
\partial_{\nu} u=0, & \partial \Omega \end{array} \right.$$
with $\Omega$ smooth, $c\in C^1(\bar{\Omega})$ and $f\in L^2(\Omega)$. Assume further that $div c=0$ and that $c\cdot \nu\geq c_0>0$.
The variational formlation of this problem consists of finding $u\in H^1(\Omega)$ such that $$\int_{\Omega} \nabla u\cdot \nabla v+(c\cdot \nabla u)v dx=\int_{\Omega} fv dx, \qquad \forall v\in H^1(\Omega)$$
Let $V=H^1(\Omega)$ and consider the bilinear form $$B(u,v)=\int_{\Omega} \nabla u\cdot \nabla v+(c\cdot \nabla u)v dx$$
I am wondering whether there is something wrong with the following argument showing existence and uniqueness of a solution to the previous problem using the Lax-Milgram theorem. By direct computation, from the above assumptions we easily conclude that
$$\int_{\Omega} (c\cdot \nabla u)u dx=\frac{1}{2}\int_{\Omega} c\cdot \nabla (u^2) dx=\int_{\partial \Omega} u^2c\cdot \nu d\sigma$$
Recall also that in $H^1(\Omega)$, the norm $$||u||_{1,\partial}=\int_{\partial \Omega} u^2\,d\sigma+\int_{\Omega} |\nabla u|^2\,dx$$ is equivalent to the standard norm $||u||_{1,2}$
Thus, if $c\cdot \nu\geq c_0>0$ then $$B(u,u)=\int_{\Omega} |\nabla u|^2+(c\cdot \nabla u)u \,dx \geq \int_{\Omega} |\nabla u|^2+c_0\int_{\partial \Omega} u^2\,d\sigma\geq C||u||_{1,2}$$
so that $B$ is $V$-coercive. Checking that both B and $Lv=\int_{\Omega} fv\,dx$ are continuous, one concludes that the problem has a unique solution $u\in H^1(\Omega)$.
