Trouble understanding this proof Let $\Omega \subset \mathbb R^3$ be an open,bounded and connected set with $C^2-$regular boundary $\Gamma=\partial \Omega$. Assume the following estimates for functions $u,v,w$ hold:

where $C_2,M$ are some positive constants and $D \ge 1$. We have also the next Theorem:

However what I try to understand is the proof of the following Theorem:

QUESTIONS:


*

*Is the sequence $D_k$ somehow connected to the sequence $(u_k,v_k,w_k)$ (let's keep the same notation with the subsequence) and if yes how?

*Although I think I understand why we extract the subsequence $(u_k,v_k,w_k)$ and how we deduce the strong convergence in $L^2$, I can not see the need of having such a subsequence in the proof. Why do we need it?

*Since $D_k \to \infty$ what can we say about $\sqrt D_k {\vert \vert \nabla w_k \vert \vert}_{L^2(\Omega_T)}$?

*How do we conclude that $w$ is a non negative constant?


I've been stuck for hours in this proof and I haven't done much progress. I would really appreciate if somebody could explain to me the logic behind this short proof.
Thanks in advance!
 A: 1) Yes.  The idea is that for each $D_k$ you use Theorem 2.4 to solve for the trio $(u_k,v_k,w_k)$ solving the PDE in Theorem 2.4 with $D = D_k.$  Then $(21)$ gives the estimates
$$
\sup_{k} \Vert u_k,v_k \Vert_{H^1} + \sup_k \Vert w_k \Vert_{L^2} +  \sup_k \sqrt{D_k} \Vert \nabla w_k \Vert_{L^2}  <\infty.
$$
2) We now use weak sequential compactness in $H^1(\Gamma)$ and $H^1(\Omega)$ along with the previously stated bound to deduce that there exist subsequences converging weakly.  However, we can then invoke Rellich's compactness theorem to conclude that if we extract a further subsequence then the convergence is actually strong in $L^2(\Gamma)$  and in $L^2(\Omega)$.  Extracting a further subsequence and using a basic result from measure theory, we can further assume that the sequence actually converges almost everywhere.  
The need for subsequences comes from the fact that we're using "soft analysis" tools such as compactness here.  Weak sequential compactness doesn't tell us that bounded sequences converge (because it's not true!) but that bounded sequences have weakly converging subsequences.  Similarly, $L^2$ convergence doesn't imply a.e. convergence (again, it's not true!), but it does imply that we can extract a subsequence along which we get a.e. convergence.  The key point here is that we don't really care about the actual value of $D_k$ in the structure of the argument.  We just want it to be increasing to $\infty$, and extracting any number of subsequences won't change this.
3) The last of the bounds above tells us that 
$$
\Vert \nabla w_k \Vert_{L^2} \le \frac{C}{\sqrt{D_k}} \to 0 \text{ as } k \to \infty.
$$
Thus, along our subsequence we have that $\nabla w_k \to 0$ strongly in $L^2$ as $k \to \infty$, and so the limiting $w$ satisfies $\nabla w =0$, which means that $w$ is a constant, as $\Omega$ is connected (this is a basic result in Sobolev theory).  This seems to be the essential reason for taking $D_k \to \infty$.  Without that here we couldn't conclude that $w$ is constant.
4) Theorem 2.4 asserts that the solution triple $(u_k,v_k,w_k)$ is non-negatives, so $w_k \ge 0$ for each $k$.  When we pass to the limit along the subsequence we will get that $w \ge 0$ a.e. but since $w$ is constant we conclude that $w$ is a non-negative constant.
I hope this helps!  Let me know if you would like any clarification.
