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For $w \in L^2(0,T;H^1)$, consider the PDE $$\int u'(t)v(t) + \int g(w(t))\nabla u(t) \nabla v(t) = \int f(t) v(t)\quad \forall v \in L^2(0,T;H^1)$$ where $u \in H^1(0,T;L^2)\cap L^2(0,T;H^1)$, and $f$ is given and $g$ is a nonlinear function from $\mathbb{R}$ to $\mathbb{R}$.

Define $T(w) = u$, the solution operator. Suppose we have $w_n \rightharpoonup w$ in $W:=L^2(0,T;H^1)\cap H^1(0,T;L^2)$ (weakly). $$\int Tw_n'(t)v(t) + \int g(w_n(t))\nabla Tw_n(t) \nabla v(t) = \int f(t) v(t)\quad \forall v \in L^2(0,T;H^1)$$ So $w_n \to w$ in $L^2(0,T;L^2)$ by compact embedding, so $w_{n_m} \to w$ a.e. for a subsequence of $w_n$.

I know that $Tw_n$ is bounded in $W$. Thus $Tw_{n_l} \to \eta$ because bounded sequences have a weakly convergent subsequence, and we have a compact embedding as I wrote above.

Assume that $g$ is nice enough so that we can use DCT to deduce that $g(w_{n_m}) \to g(w)$ in $L^2(0,T;L^2)$.

Using this information, how do I pass to the limit in the equation above to deduce that $Tw = \eta$? My only problem is that I have two different subsequences $n_m$ and $n_l$ and I don't know how to "put them together" in the equation to pass to the limit properly.

(I did post this on StackExchange here https://math.stackexchange.com/questions/718864/passing-to-the-limit-in-a-pde-subsequence-problems but got no replies. I apologise if this question is too easy but I don't see anywhere where this is addressed).

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  • $\begingroup$ @AthanagorWurlitzer I don't know if the limit equation has a unique solution. I didn't understand your first line, please can you elaborate? $\endgroup$
    – TheBook
    Commented Mar 21, 2014 at 8:15
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    $\begingroup$ On your question: you have sequence $w_n\to w $. Extract a subsequence $w_{f(n)}$ so that $Tw_{f(n)}\to \eta$. From that sequence $g(w_{f(n)}))$, extract a subsequence so that $g(w_{k(f(n))})\to g(w)$. Then for the sub sub sequence $n^*=k(f(n))$ all convergences hold: $w_{n^*}\to w$, $g(w_{n^*})\to g(w)$ and $Tw_{n^*} \to \eta$. But I don't think that answers your problem. $\endgroup$
    – username
    Commented Mar 21, 2014 at 8:26
  • $\begingroup$ @AthanagorWurlitzer I think your method works: as in the OP, we have $Tw_{n_l} \to \eta$ in $L^2(L^2)$. Since we know $w_n \to w$ in $L^2(L^2)$, we have $w_{n_l} \to w$ also, and therefore $w_{n_{l_k}} \to w$ a.e. for another subsquence. DCT implies $g(w_{n_{l_k}}) \to g(w)$. Since $Tw_{n_l} \to \eta$, we must also have $Tw_{n_{l_k}} \to \eta$. So we have a subsequence $n_{l_k}$ that does the job. Do you think this is right? $\endgroup$
    – TheBook
    Commented Mar 21, 2014 at 10:18
  • $\begingroup$ You need $g(w_n)\rightharpoonup g(w)$ weak in $H^1$, or $\nabla Tw_n\rightharpoonup \nabla \eta$ weak in $L^2$ to pass to the limit. $\endgroup$
    – username
    Commented Mar 21, 2014 at 20:01
  • $\begingroup$ @AthanagorWurlitzer You're right. But since we have $Tw_n$ bounded in $W$, there's a weak convergent subsequence $Tw_{n_l}$ in $W$. So if we replace in my last comment the convergence of $Tw_{n_l} \to \eta$ in $L^2(L^2)$ with $Tw_{n_l} \rightharpoonup \eta$ in $W$, and change the convergence to weak in the 3rd last sentence it would work. $\endgroup$
    – TheBook
    Commented Mar 23, 2014 at 10:34

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