Timeline for Weak convergence of 4-th degrees

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Mar 26, 2014 at 23:49	comment	added	jokersobak		@AthanagorWurlitzer We have a priori estimates for the solution $y$: $y$ is bounded in $L^\infty(Q) \cap W$ where $Q = \Omega \times (0,T)$, $W = \{y \in L^2(0,T;V) : y' \in L^2(0,T;V') \}$, $V = H^1(\Omega)$. Thus $y$ is bounded in $L^2(0,T;H^1(\Omega))$. I think that without the term $\mu\\|u\\|^2$ in $J$ proving of the existence may be impossible. I will think further. Thank you!
Mar 26, 2014 at 19:39	comment	added	username		hum, yes, silly me. And no $H^1$ bound comes from $A$? Then I don't see.
Mar 26, 2014 at 18:41	comment	added	jokersobak		@AthanagorWurlitzer It's clear that if $\{u_k^4\}$ is bounded in $L^2$ then (subsequence of) $u_k^4 \to \chi$ in $L^2$ weakly. But we need that $\chi = u^4$ where $u_k \to u$ weakly in $L^2$ (or in $L^8$).
Mar 26, 2014 at 15:00	comment	added	username		What I meant was that you can, by extracting a first subsequence which converges weakly in $L^2$, and from that subsequence, which is still bounded in $L^8$, extract another subsequence so that $u^4$ converges weakly as well in $L^2$.
Mar 26, 2014 at 12:34	comment	added	jokersobak		@AthanagorWurlitzer Thanks a lot for your answer! Yes, $\Omega$ is bounded. Could you explain why the fact that $u_k \to u$ weakly in $L^8$ implies that $u_k^4 \to u^4$ weakly in $L^2$?
Mar 26, 2014 at 8:15	review	First posts
Mar 26, 2014 at 8:21
Mar 26, 2014 at 7:57	history	asked	jokersobak	CC BY-SA 3.0