0

Let consider the problem $$\inf\left\{I(u)=\int_\Omega f(x,u(x),\nabla u(x))\mathrm d x\mathrm dx\mid u\in u_0+W_0^{1,p}(\Omega ) \right\}\tag{P},$$ where $u_0+W_0^{1,p}(\Omega )=\{u\in W^{1,p}(\Omega )\mid u|_{\partial \Omega }=u_0\}$ and $\Omega \subset \mathbb R^n$ is an open bounded with Lipschitz boundary. Let $m$ s.t. $$\inf\left\{I(u)=\int_\Omega f(x,u(x),\nabla u(x))\mathrm d x\mathrm dx\mid u\in u_0+W_0^{1,p}(\Omega ) \right\}=m.$$ In the proof of the existence of $\bar u\in u_0+W_0^{1,p}(\Omega )$ s.t. $m=I(\bar u)$ they prove that if $(u_n)\subset u_0+W_0^{1,p}(\Omega )$ is s.t. $I(u_n)\to m$, then there is a subsequence (still denoted $u_n$) that converge weakly to $\bar u\in W^{1,p}$ and that $I(u)$ is lower semi-continuous at $\bar u$, i.e. $$u_n\rightharpoonup u\text{ in }W^{1,p}\implies \liminf_{n\to \infty }I(u_n)\geq I(\bar u).$$

I don't really understand why this prove the existence.

Q1) The first part is called compactness... is there a link between the fact that $\{I(u)\mid u\in u_0+W^{1,p}(\Omega )\}$ is compact and that there is a sequence $(u_n)\subset u_0+W^{1,p}_0$ that converge converge weakly in $\bar u$ s.t. $I(u_n)\to m$ ?

Q2) I don't really understand this lower semi-continuity. In what that help us to conclude on the existence ?

user330587
  • 1,624

1 Answers1

1

For Q1, yes, there is a link. You are using the fact that in a reflexive space $W^p_0$, $p>1$ a bounded sequence $\{u_n\}$ has a subsequence $\{u_{n_k}\}$ which converges weakly. This is a compactness result (the equivalent of the Bolzano's theorem in $\mathbb{R}^n$). As for Q2, weak convergence in $W^p_0$, $p>1$, implies strong convergence of traces so $\bar u$ is such that $\bar u\in u_0+W^p_0$. This implies that $I(\bar u)\ge m$. On the other hand, by the lower semicontinuity and the choice of $u_n$, $$m=\lim_{n\to\infty}I(u_n)=\lim_{k\to\infty}I(u_{n_k})\ge I(\bar u)$$ and so $I(\bar u)=m$.

Gio67
  • 20,905
  • thanks. What do you mean by "weak convergence in $W_0^p$ implies strong convergence of trace" ? – user330587 May 29 '17 at 12:15
  • if $u_n$ goes weakly to $\bar u$, then $tr(u_n)$ goes strongly to $tr(\bar u)$ in $L^p(\partial \Omega)$. Since all your functions are equal $u_0$ on $\partial \Omega$ so does $\bar u$. You need to make sure that $\bar u\in u_0+W^p_0$. This does it. – Gio67 May 29 '17 at 12:18
  • It mean that $\int_{\partial \Omega }|tr(\bar u)-tr(u_n)|\to 0$ ? – user330587 May 29 '17 at 12:20
  • yes but with power p – Gio67 May 29 '17 at 13:01
  • By curiosity, do we have something as bounded sequence converge weakly under some condition or not ? By the way, I don't understand why $I(bat u)\geq m$. – user330587 May 29 '17 at 14:14
  • I don't understand your first question. For the second $m$ is the infimum over all $I(u)$ where $u\in u_0+W^{1,p}_0$. Since $\bar u$ belongs to this space, $I(\bar u)$ is one of the admissible numbers and so above the infimum. – Gio67 May 29 '17 at 22:01