I am working with the following PDE:
\begin{align*} -\nabla \cdot (a(x) \nabla u) + b(x) u = f(x) &, \hspace{.5cm} x = (x_1,x_2) \in \Omega, \\ u = 0&, \hspace{.5cm} x \in \partial \Omega_1 \\ \frac{\partial u}{\partial x_1} + \frac{\partial u}{\partial x_2} + u = 2&, \hspace{.5cm} x \in \partial \Omega_2, \end{align*} where \begin{align*} &\Omega = \left\{ x \vert x_1 > 0, x_2 > 0, x_1 + x_2 < 1 \right\} , \\ &\partial \Omega_1 = \left\{x \vert x_1 = 0, 0 \leq x \leq 1 \right\} \cup \left\{ x \vert x_2 = 0, 0 \leq x_1 \leq 1 \right\} \\ &\partial \Omega_2 = \left\{x \vert x_1 > 0, x_2 > 0, x_1+x_2 = 1 \right\} \\ &0 < a \leq a(x) \leq A, \text{ } 0 < b \leq b(x) \leq B \end{align*}
Here's what I have so far:
Let $H = \left\{ v \in H^1(\Omega) \vert v = 0 \text{ on } \partial \Omega_1 \right\}$. Then multiplying the PDE by $v$ and integrating by parts, we have \begin{align*} \int_{\Omega} fvdx &= - \int_{\partial \Omega} a(x) v \nabla u \cdot \nu ds + \int_{\Omega} (a(x) \nabla u \cdot \nabla v + b(x)uv )dx \\ &= - \int_{\partial \Omega_2} a(x) v \nabla u \cdot \nu ds + \int_{\Omega} (a(x) \nabla u \cdot \nabla v + b(x)uv )dx \\ &= - \frac{\sqrt{2}}{2}\int_{\partial \Omega_2} a(x) v (2-u) ds + \int_{\Omega} (a(x) \nabla u \cdot \nabla v + b(x)uv )dx \\ \end{align*} Let \begin{align*} B(u,v) &= \int_{\Omega} (a(x) \nabla u \cdot \nabla v + b(x)uv) dx + \frac{\sqrt{2}}{2}\int_{\partial \Omega_2} a(x) u v \text{ } ds\\ F(v) &= \int_{\Omega} fv \text{ } dx + \sqrt{2} \int_{\partial \Omega_2} a(x)v \text{ } ds \end{align*} Then the weak variational form of the problem would be
Find $u \in H$ such that $B(u,v) = F(v)$ for all $v \in H$.
I'm fairly confident the weak variational formulation is correct, but I'm having trouble applying Lax-Milgram, particularly with how I apply the trace theorem. When I try to prove that $F$ is bounded, I obtain \begin{align*} \left| F(v) \right| &\leq \int_\Omega \left| f v \right|dx + \sqrt{2} \int_{\partial \Omega_2} \left| a(x) v \right|ds \\ &\leq \| f \|_{L^2(\Omega)}\| v \|_{L^2(\Omega)} + \sqrt{2} \| a(x) \|_{L^2(\partial \Omega_2)} \| v \|_{L^2(\Omega_2)} \\ &\leq \| f \|_{L^2(\Omega)}\| v \|_H + 2^{\frac{3}{4}}A \| v \|_{\partial \Omega_2 }. \end{align*} I know that the trace theorem must be used to complete the proof, but I don't really understand the details, particular when we are only dealing with a subset of the boundary. If anyone could let me know how to proceed from here, I'd really appreciate it!