I'm trying to understand the following convergence:
$\int _{[0,T] \times \Omega } (v_nu_n(\phi _t + u_n \cdot \nabla)\phi) dxdt \rightarrow \int _{[0,T] \times \Omega } (vu(\phi _t + u \cdot \nabla)\phi) dxdt$
We know \begin{align} (1)\ \ \ \phi &\in H^1([0,T] \times \Omega) \cap L^2(0,T;H^1(\Omega)) \\ (2)\ \ \ u &\in L^\infty(0,T;L^2(\Omega)) \cap L^2(0,T;H^1(\Omega)) \\ (3)\ \ \ v &\in L^\infty([0,T]\times \Omega) \end{align}
and further
\begin{align} (4)\ \ \ &u_n \rightarrow u \ \ \ \text{in} \ \ \ L^2([0,T] \times \Omega) \\ (5)\ \ \ &u_n \rightharpoonup u \ \ \ \text{in} \ \ \ L^2(0,T; H^1 (\Omega)) \ \ \ \text{weakly}\\ (6)\ \ \ &u_n \overset{\ast}{\rightharpoonup} u \ \ \ \text{in} \ \ \ L^\infty(0,T;L^2(\Omega)) \ \ \ \text{weakly}^* \\ (7)\ \ \ &v_n \rightarrow v \ \ \ \text{in} \ \ \ C(0,T;L^p(\Omega))\ \ \ \forall \ p < \infty \\ (8)\ \ \ &v_n \overset{\ast}{\rightharpoonup} v \ \ \ \text{in} \ \ \ L^\infty([0,T]\times \Omega) \ \ \ \text{weakly}^* \end{align}
For the first term in the integral I can see
$\int_{[0,T]\times \Omega} \! v_n u_n \phi _t - vu\phi_t \, = \int_{[0,T]\times \Omega} \! (v_n - v) u \phi _t \, - \int_{[0,T]\times \Omega} \! (u - u_n) v_n \phi _t \, \rightarrow 0$,
since $u, \phi _t \in L^2([0,T]\times \Omega)$, hence $u \phi _t \in L^1([0,T]\times \Omega)$ and thus the first integral here goes to 0 by $(8)$. For the second integral one can use
$\int_{[0,T]\times \Omega} \! (u - u_n) v_n \phi _t \, \leq \|u - u_n \|_{L^2([0,T]\times \Omega)}\|v_n \phi _t\|_{L^2([0,T]\times \Omega)} \rightarrow 0$,
by $(4)$ and since $v_n$ is bounded in $L^\infty ([0,T]\times \Omega)$
My two questions are: 1.) Is this correct? 2.) I'm looking for a similar approach for the second term, but i couldn't find one yet. Has anybody some hint for me?
Thank you in advance for any answers.