Ngô Quốc Anh

May 15, 2010

Two fundamental results in the calculus of variation

Filed under: Giải tích 8 (MA5206), PDEs — Ngô Quốc Anh @ 20:28

I suddenly think that I should post this entry ‘cos sometimes I don’t remember these stuffs. These results appear frequently in solving PDEs especially when using the direct method. For example, the simplest case is the following eigenvalue problem

\displaystyle -{\rm div}(|\nabla u|^{p-2}\nabla u)=\lambda |u|^{p-2}u

over a bounded domain \Omega \subset \mathbb R^n with Dirichlet boundary condition. We assume p<n. Our aim is to show the existence of the first eigenvalue \lambda_1>0. Obviously, our problem is to solve the following optimization

\displaystyle\mathop {\inf }\limits_{u \in W^{1,2}_0(\Omega)} \left\{ {\int_\Omega {|\nabla u{|^p}dx} :\int_\Omega {|u{|^p}dx} = 1} \right\}.

The direct method says that we firstly select a minimizing sequence, say \{u_n\}_n, then we need to prove \{u_n\} is convergent. There are two steps in the above argument which lead to this entry. Our first claim is the following.

Boundedness implies weakly convergence. The first result says that

If E is a reflexive Banach space and \{x_n\}_n \subset E is a bounded sequence. Then up to a subsequence x_n converges weakly to some x in X.

The proof of this claim can be found in a book due to Brezis (Theorem III.27). Interestingly, its converse also holds by the Eberlein-Šmulian theorem.

Theorem (Eberlein-Šmulian). Suppose E is a Banach space such that every bounded sequence \{x_n\}_n contains a weakly convergent subsequence. Then E is reflexive.

There was an elementary proof of this theorem. We refer the reader to a paper due to Whitley [here]. Let us get back to our optimization problem. Once we have a minimizing sequence \{u_n\}_n \subset W^{1,p}_0(\Omega) it is clear to see that \{u_n\}_n is bounded in W^{1,p}_0(\Omega) since

\displaystyle\int_\Omega {|{u_n}{|^p}dx} = 1

and

\displaystyle\int_\Omega {|\nabla {u_n}{|^p}dx} \leqslant C, \quad \forall n.

By using the first claim, u_n converges weakly to some u \in W^{1,p}_0(\Omega). It is worth noticing that by saying u_n \rightharpoonup u in W^{1,p}_0(\Omega) we mean u_n \rightharpoonup u in L^p(\Omega) and Du_n \rightharpoonup Du in L^p(\Omega, \mathbb R^n). Now we need further argument

Weakly convergence becomes strongly convergence via compact operator. This second result says that

A compact operator C : E \to X between Banach spaces maps every weakly convergent sequence in E into one that converges strongly in X.

The proof of this relies on the contradiction argument and the fact that once a sequence converges strongly to some limit, this limit is unique. However, the converse is no long true. For example, by the Schur theorem, a sequence \{x_n\} in \ell^1 converges weakly, it also converges strongly. We take C to be the identity I in \ell^1. Since \ell^1 has infinity dimensional, by using the Riesz theorem, I cannot be compact.

Using this claim we deduce that u_n converges strongly to u in L^q for any 1\leqslant q<p^\star=\frac{np}{n-p}. By using the Minkowski and Holder inequalities we can show that u satisfies the constraint. It now follows from the weakly lower semi-continuous of norm that u indeed satisfies the equation. The proof follows.

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