convergence exercice

Question

I have a question please, thanks to help me. Let $\Omega$ an open bounded, connexe and regular Let $(v_n)$ an sequence in $H^1(\Omega)$ and let $v \in H^1(\Omega)$ such that $v_n$ converge weakly in $L^2(\Omega)$ to $v$. How we can compute the limits $$\int_{\Omega} A |\nabla v_n|^2 dx$$ and $$\int_{\Omega} v_n dx$$ when $n$ tends to $+\infty$

($A$ is such that $\exists \alpha > 0, A(x) \xi \xi \geq \alpha |\xi|^2, \forall \xi \in \mathbb{R}^n$ and $\exists \beta > 0, |A(x) \xi| \leq \beta |\xi|, \forall \xi \in \mathbb{R}^n$) Thanks for the help.

Answer 1

The limit of the second sequence should be obvious from the definition of weak convergence once you notice that $$ \int_\Omega v_n dx = (v_n, 1)_{L^2(\Omega)}$$ where $1$ is the constant function on $\Omega$ with value 1.

Answer 2

Let $\Omega = (-\pi,\pi)$. Let $v_n(x) = \dfrac{\sin(n^2x)}{n}$. Let $A\xi = \xi$. It is clear that $v_n \rightharpoonup 0$ in $L^2(\Omega)$, since in fact $v_n$ converges uniformly. Now, $\nabla v_n(x) = n\cos(n^2 x)$ and thus $(A\nabla v_n) \cdot \nabla v_n = n^2 \cos^2(n^2 x)$. You get $$\int_{-\pi}^\pi (A\nabla v_n) \cdot \nabla v_n \, dx = \int_{-\pi}^\pi n^2\cos^2(n^2 x) \, dx = n^2 \pi$$ for all $n$. In this case the limit is infinite. You can modify this example in a trivial way to get a finite limit or a zero limit, or even a divergent sequence.