\left\{
    \begin{aligned}
        &\text{Find } u \in V, \text{ s.t. } \forall v \in V \\
        &\int_\Omega A \nabla u_\epsilon \cdot \nabla v + \int_{\Gamma_C} \mathrm{B}{\left( u \right)} v = {\left\langle f,v \right\rangle}+\int_{\Gamma_N} gv 
    \end{aligned} ~~,   
\right.$$ if $\kappa_1-L_\mathrm{B} c_j^2 > 0$.

We can reprove this proposition by directly utilizing strongly monotone operator theory (see [@Zeidler1995] sect. 2.14). Moreover, in a special case— Robin problems, we can obtain $L^2(\Omega)$ estimation by utilizing the bilinearity of its variational form.

Let $A^\epsilon$ define as previous and satisfy the assumption \[ass:A\], for simplicity, we state Robin problem on the whole boundary $\Gamma$: $$\left\{
\begin{aligned}
-{\mathrm{div}}(A^\epsilon \nabla u_\epsilon) = f ~~~~ &\text{ in } L^2(\Omega) \\
\bm{n}\cdot A^\epsilon \nabla u_\epsilon + \alpha(\bm{x})u_\epsilon = g ~~~~ &\text{ in } L^2(\Gamma)
\end{aligned} .
\right.$$ And the corresponding variational form is: $$\label{eq:robin variation}
\left\{
\begin{aligned}
&\text{Find } u_\epsilon \in H^1(\Omega), \text{ s.t. } \forall v \in H^1(\Omega) \\
&\int_{\Omega} A^\epsilon \nabla u_\epsilon \cdot \nabla v + \int_{\Gamma} \alpha u_\epsilon v = \int_{\Omega} fv + \int_{\Gamma} gv
\end{aligned} .
\right.$$ We need assumption $$0 < \alpha_1 \leq \alpha(\bm{x}) \leq \alpha_2 < \infty ~~ \forall \bm{x} \in \Gamma$$ to prove this bilinear form is coercive, then Lax-Milgram theorem asserts the 