Ngô Quốc Anh

February 28, 2012

The Rellich embedding theorem on a bounded domain

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 18:18

Today, let us summarize steps during the standard proof of the Rellich embedding theorem on a bounded domain \Omega \subset \mathbb R^n. This theorem says that

Theorem (Rellich). Suppose that \Omega \subset \mathbb R^n is an open, bounded domain with C^1 boundary,  and that 1<p< n. Then W^{1,p}(\Omega) is compactly embedded in L^q(\Omega) for all 1\leqslant q <\frac{np}{n-p}.

In particular, for any sequence \{u_j\}_j \subset W^{1,p}(\Omega) , there exists a subsequence \{u_{j_k}\}_k \subset \{u_j\}_j such that u_{j_k} \to u strongly in L^q(\Omega) for some u \in L^q(\Omega).

In order to prove the Rellich theorem, we need the so-called Arzela-Ascoli theorem.

Lemma (The Arzela-Ascoli theorem). Suppose that u_j \in C^0(\overline\Omega), \|u_j\|_{C^0(\overline\Omega)} \leqslant M < \infty, and \{u_j\}_j is equicontinuous. Then there exists a subsequence u_{j_k} \to u uniformly on \Omega.

The Arzela-Ascoli theorem is well-known. To prove the Rellich theorem, we shall use the standard mollifiers. To do that, we have to extend \Omega a little bit.

Step 1. Assume that \overline\Omega \subset \mathbb R^n is also an open, bounded domain with C^1 boundary with \Omega \subset\subset \overline \Omega. By the Sobolev extension theorem, we can define Eu_j by \overline u_j with

\sup_j \|\overline u_j\|_{W^{1,p}(\mathbb R^n)} < C<\infty.

By the Gagliardo-Nirenberg inequality,

\sup_j \|\overline u_j\|_{L^q(\Omega} < C<\infty.

For \varepsilon>0, let \eta_\varepsilon denote the standard mollifiers and set \overline u_j^\varepsilon = \eta_\varepsilon * \overline u_j. By choosing \varepsilon sufficiently small, \overline u_j^\varepsilon \in C^\infty(\overline \Omega).

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February 24, 2012

The pullback metric under scaling

Filed under: Uncategorized — Ngô Quốc Anh @ 20:11

Today, let us discuss a very simple question. Assume that (M,g) is a Riemannian manifold and \lambda>0 is constant. We study the following so-called pullback metric \lambda^\star g of g under the scaling x \mapsto \lambda x. Precisely, we aim to compare \lambda^\star g and g.

For simplicity, we follow definition of the pullback metric. Besides, by \lambda we mean the map x \mapsto \lambda x.

The pushforward \lambda_\star by \lambda. This is the first step. Suppose X \in T_xM for some x \in M. By definition, we have \lambda_\star X \in T_{\lambda (x)}M which is just

\displaystyle (\lambda_\star X) f = X ( f \circ \lambda), \quad \forall f \in C^\infty(M,\mathbb R).

By the definition of derivation, in local coordinates, we clearly have

\displaystyle X(f \circ \lambda ) = X(f \circ (\lambda {\rm id})) = \lambda X(f \circ {\rm id}) = \lambda X(f).

The pullback metric \lambda^\star g. By definition, we have

\displaystyle \lambda^\star g (X, Y) = g(\lambda_\star X, \lambda_\star Y), \quad \forall X, Y \in T_xM.

Obviously,

\displaystyle g({\lambda _ \star }X,{\lambda _ \star }Y) = {\lambda ^2}g({\rm id}_*X,{\rm id}_ \star Y)= {\lambda ^2}g(X,Y).

Thus, we can conclude that

\displaystyle \lambda^\star g=\lambda^2 g.

February 6, 2012

A note on the maximum principle

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 18:18

Today, let us discuss a very interesting stuff. Let say M is a compact manifold without boundary of dimention n \geqslant 3.

On M, we consider the following simple PDE

\displaystyle -\Delta u + hu = a

where h and a are smooth functions with h>0 and a \geqslant 0. Since h>0, it is well-known that the operator -\Delta +h is coercive, see here. A standard variation method tells us that there exists a weak solution u to the above PDE. By regularity theorem, u is at least a C^2 function, thus, a strong solution (in the classical sense).

Next, we claim that u \geqslant 0. To this purpose, assume that the solution u achieves its minimum at some point x_0 \in M. In particular, there holds

-\Delta u (x_0) \leqslant 0.

This, together with the fact that h>0 and a \geqslant 0, implies that u(x_0) \geqslant 0. Thus, we have shown that u \geqslant 0 in M.

Once we have the non-negativity of u, in view of the strong maximum principle, either u \equiv 0 in M or u>0. In other words, the solution u cannot achieves its minimum inside the manifold. Since the manifold has no boundary, it is natural to think that the solution u cannot achieve its minimum although u>0. This is clear a contradiction to the fact that the manifold is compact and u is of class C^2(M).

So something went wrong but what and why?

In fact, we have made a small mistake. In view of the strong maximum principle, we can only claim that the solution can only achieve its non-positive minimum value on the boundary. Therefore, there are cases so that u may achieve its positive minimum inside M. Thus, there is no contradiction here.

In order to see this, let us go back to a proof of the strong maximum principle. Roughly speaking, it starts with the following simple one.

Lemma 1. If -\Delta u +hu >0 at any point in M, then u cannot have non-positive minimum value in M.

Proof. The proof is standard. Assume that at x_0 \in M, the function u realizes its minimum, besides, u(x_0) \leqslant 0. In particular, -\Delta u(x_0) \leqslant 0 and h(x_0)u(x_0) \leqslant 0. These force -\Delta u + hu \leqslant 0 at x_0. A contradiction.

Form the proof above, if u(x_0)>0, we cannot get any contradiction. This is why we can claim either u \equiv 0 or u>0 since u can achieve its positive minimum in M.

Notice that, if we don’t have any h (in the operator), the non-positivity can be dropped.

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