Quick question, any contribution or hint would be appreciated:
How does it follow that:
$$\lim\limits_{k \rightarrow \infty}\int_{\Omega}a(u_{k})\frac{\partial u_{k}}{\partial x}\frac{\partial v}{\partial x} + \int_{\Omega}b(u_{k})\frac{\partial v}{\partial x} = \int_{\Omega}a(u)\frac{\partial u}{\partial x}\frac{\partial v}{\partial x} + \int_{\Omega}b(u)\frac{\partial v}{\partial x}$$
Where $v \in W^{1,\infty}(\Omega)$ and $u_{k} \rightharpoonup u$ in $W^{1,2}(\Omega) \Subset L^{2-\epsilon}(\Omega)$, so it follows that $u_{k} \rightarrow u$ in $L^{2-\epsilon}(\Omega)$ where $\epsilon > 0$. Also $a(u_{k}) \rightarrow a(u)$ in $L^{2}(\Omega)$ and $b(u_{k}) \rightarrow b(u)$ in $L^{1}(\Omega)$.
$\Omega$ is given as a bounded domain in $\mathbb{R}^{n}$
Let $w=\dfrac{\partial v}{\partial x}$. Since $v\in W^{1,\infty}$, we have $w\in L^\infty$. The space $L^\infty$ is dual to $L^1$. Thus, pairing a fixed $L^\infty$ function with a convergent sequence of $L^1$ functions produces a convergent sequence of numbers. This is why $$\int b(u_k) w\to \int b(u)w$$
For the other integral, note that $a(u_k)w$ converges in $L^2$, since multiplication by a bounded function is a continuous operator. Then use the following
General fact: if $f_k \rightharpoonup f$ in $L^p$ and $g_k\to g$ in $L^q$, where $p^{-1}+q^{-1}=1$, then $\int f_kg_k\to \int fg $. Reason: $$f_kg_k - fg = (f_k - f)g+f_k(g_k - g)$$ where the integral of the first term goes to zero by the definition of weak convergence, and the second by Hölder's inequality.