I have the following problem: $$ \begin{cases} -\varepsilon\Delta u +b\nabla u=0 & x \in \Omega\\ u=g & x \in \partial\Omega \end{cases} $$ where $\Omega = (0, 1)^2$, and where
$$ g(x, y) = \begin{cases} 0.5-\lvert 0.5-x\rvert & \text{if } y = 0 \\ 0 & \text{if } y \neq0\end{cases} $$ and $$b=(-y,x).$$
I have to derive the weak formulation and show that there exists a unique solution. I would like to apply the Lax-Milgram theorem so I have to show that it is bounded and coercive. I multiplied by a function $v\in H^{1}_{0}(\Omega)$ and the boundedness is ok. But for the coercivity, I have a problem for the boundary datum.
Here is what I have: $$a(u,u)=\varepsilon\int_{\Omega}(\nabla u)^{2} + \int_{\Omega} b\nabla u\cdot u.$$ For the second integral: $$\int_{\Omega} b\nabla u\cdot u=\frac{1}{2}\int_{\Omega}b \nabla (u^{2})=\frac{1}{2}\int_{\Omega}\text{div}(b\cdot u^{2})-\frac{1}{2}\int_{\Omega}u^{2}\text{div}(b)=\frac{1}{2}\int_{\Omega}\text{div}(b\cdot u^{2})$$ since $\text{div}(b)=0$. Using the divergence theorem, we can say that this last integral equals $\int_{\partial\Omega}b\cdot g^{2}\cdot n$ where $n$ is the unitary vector pointing outwards.
However with this term I cannot show the coercivity. Do you have any suggestion? I tried to consider the part of the boundary where $g\neq 0$ but it doesn't work. At this point I don't even know if the function space for $v$ is correct. I would be very grateful for any help. Thank you in advance!