Consider the standard uniformly elliptic equation on a domain $\Omega \subset \mathbb{R}^d$:
$$ \mathrm{div}(A(x)\nabla u) = f $$ for $u \in H^{1}(\Omega)$ , $f\in H^{1} (\Omega)$, $a_{ij}(x)$ measurable functions, and the solution being in the sense of distributions. Show that if we have a sequence $u_n,A_n, f_n$ such that: $$ \mathrm{div}(A_n(x)\nabla u_n) = f_n $$ and:
- $a_{ij}^n(x) \to a_{ij}(x)$ pointwise a.e.
- $u_n \to u$ in $H^{1}$
- $f_n \to f $ in $H^{-1}$
Then in the limit we have: $$ \mathrm{div}(A(x)\nabla u) = f $$ with all the above being in the sense of distributions of course.
Does anyone have any ideas? The right hand side should definitely converge to $f$ thanks to the strong convergence of $f$, but i am unsure on how one should handle the matrix terms.
I suppose that $\Omega$ is a bounded open set and the costants $\Lambda$ not depends by $n$, that is $$ a^{ij}_n\le \Lambda \qquad \forall n $$ Then thesis it follows by \begin{equation} \begin{split} \int_\Omega (a^{ij}_n\partial_ju_n-a^{ij}\partial_ju)&= \int_\Omega (a^{ij}_n\partial_ju_n-a^{ij}_n\partial_ju) +\int_\Omega (a^{ij}_n\partial_ju-a^{ij}\partial_ju)\\ & =\int_\Omega a^{ij}_n(\partial_ju_n-\partial_ju) + \int_\Omega (a^{ij}_n-a^{ij})\partial_ju \\ &\le \Lambda||\partial_ju_n-\partial_ju||_{L^1}+ ||\partial_ju||_{L^2}||a^{ij}_n-a^{ij}||_{L^2}\to_{n\to\infty}0 \end{split} \end{equation} by hypothesis and by dominated convergence theorem.