Ngô Quốc Anh

April 22, 2011

On Costa-Hardy-Rellich inequalities


This note is to concern a recent result by David G. Costa [here]. Here the statement

Theorem 1.1. For all a,b\in \mathbb R and u \in C^\infty_0(\mathbb R^N\backslash\{0\}) one has

\displaystyle\left| {\frac{{N - 2 - \gamma }}{2}\int_{\mathbb R^N} {\frac{{|\nabla u{|^2}}}{{|x{|^\gamma }}}dx} + \gamma \int_{\mathbb R^N} {\frac{{{{(x \cdot \nabla u)}^2}}}{{|x{|^{\gamma + 2}}}}dx} } \right| \leqslant {\left( {\int_{\mathbb R^N} {\frac{{|\Delta u{|^2}}}{{|x{|^{2b}}}}dx} } \right)^{\frac{1}{2}}}{\left( {\int_{\mathbb R^N} {\frac{{|\nabla u{|^2}}}{{|x{|^{2a}}}}dx} } \right)^{\frac{1}{2}}}

where \gamma=a+b+1. In addition, if \gamma \leqslant N-2, then

\displaystyle\widehat C\int_{\mathbb R^N} {\frac{{{{(x \cdot \nabla u)}^2}}}{{|x{|^{\gamma + 2}}}}dx} \leqslant {\left( {\int_{\mathbb R^N} {\frac{{|\Delta u{|^2}}}{{|x{|^{2b}}}}dx} } \right)^{\frac{1}{2}}}{\left( {\int_{\mathbb R^N} {\frac{{|\nabla u{|^2}}}{{|x{|^{2a}}}}dx} } \right)^{\frac{1}{2}}}

where the constant \widehat C=|\frac{N+a+b-1}{2}| is sharp.

Here’s the proof.

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January 6, 2011

The Alexandrov-Bol inequality

Filed under: Giải Tích 6 (MA5205) — Tags: , — Ngô Quốc Anh @ 19:55

In the literature, there is an inequality called the Alexandrov-Bol inequality which is frequently used in partial differential equations. Here we just recall its statement without any proof.

Theorem. Let \Omega be a good domain in \mathbb R^2. Assume p \in C^2(\Omega)\cap C^0(\overline \Omega) be a positive function satisfying the elliptic inequality

\displaystyle -\Delta \log p \leqslant p

in \Omega. Then it holds

\displaystyle l^2(\partial\Omega) \geqslant \frac{1}{2} \big(8\pi-m(\Omega)\big)m(\Omega)

where

\displaystyle l(\partial\Omega)=\int_{\partial\Omega}\sqrt{p}ds

and

\displaystyle m(\Omega)=\int_\Omega pdx.

An analytic proof was given by C. Bandle aroud 1975 when she assumed p to be real analytic. The above version was due to Suzuki in an elegant paper published in the Ann. Inst. H. Poincare in 1992 [here]. The proof is mainly depended on the isoperimetric inequality for the flat Riemannian surfaces. We refer the reader to the paper by Suzuki for the proof.

September 14, 2010

Asympotic behavior of integrals, 2

Filed under: Giải Tích 6 (MA5205), PDEs — Tags: — Ngô Quốc Anh @ 15:07

We now prove the following result

Theorem. Let u and f be two smooth functions on \mathbb R^2 satisfying

\Delta u(x)=f(x), \quad x \in \mathbb R^2.

Suppose that f is bounded and also f \in L^1(\mathbb R^2) and

|u(x)| \leqslant o(|x|), \quad |x| \to \infty.

Then

\displaystyle\mathop {\lim }\limits_{|x| \to \infty } \frac{{u(x)}}{{\log |x|}} = \frac{1}{{2\pi }}\int_{{\mathbb{R}^2}} {f(y)dy} .

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September 7, 2010

Asympotic behavior of integrals

Filed under: Giải Tích 6 (MA5205), PDEs — Tags: — Ngô Quốc Anh @ 10:52

Long time ago, we studied [here] the following fact

Suppose f \in L^1(\mathbb R^n) \cap L_{loc}^\infty (\mathbb R^n) with f \geq 0. Define

\displaystyle Sf\left( x \right) = \int_{\mathbb{R}^n } {\log \frac{{\left| y \right|}}{{\left| {x - y} \right|}}f\left( y \right)dy}.

Show that Sf(x) is finite for all x \in \mathbb R^n and Sf \in L_{loc}^1(\mathbb R^n).

In this entry, from now on we continue to prove several useful results appearing in PDE. We shall prove the following

Theorem. Assume u is a solution to

\displaystyle (-\Delta)^\frac{3}{2} u(x)=-2e^{3u(x)}, \quad x \in \mathbb R^3

with finite energy

\displaystyle \int_{{\mathbb{R}^3}} {{e^{3u(x)}}dx} < \infty.

Then

\displaystyle\mathop {\lim }\limits_{|x| \to \infty } \frac{{u(x)}}{{\log |x|}} = - \frac{1}{{{\pi ^2}}}\int_{{\mathbb{R}^3}} {{e^{3u(y)}}dy} .

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August 29, 2010

Achieving regularity results via bootstrap argument, 4

Filed under: Giải Tích 6 (MA5205), PDEs — Tags: — Ngô Quốc Anh @ 2:36

Let us consider the following equation

\displaystyle u(x) = \int_{{\mathbb{R}^n}} {\frac{{u{{(y)}^{\frac{{n + \alpha }}{{n - \alpha }}}}}}{{{{\left| {x - y} \right|}^{n - \alpha }}}}dy} , \quad x \in {\mathbb{R}^n}

for n\geqslant 1 and 0<\alpha<n. In this entry, by using boothstrap argument, we show that

Theorem. If positive function u \in L_{loc}^\frac{2n}{n-\alpha}(\mathbb R^n) solves the equation, then u \in C^\infty(\mathbb R^n).

In the process of proving the result, we need the following result

Proposition. Let V \in L^\frac{n}{\alpha}(B_3) be a non-negative function and set

\displaystyle \delta(V)=\|V\|_{L^\frac{n}{\alpha}(B_3)}.

For \nu >r>\frac{n}{n-\alpha}, there exist positive constants \overline \delta<1 and C \geqslant 1 depending only on n, \alpha, r and \nu such that for any 0 \leqslant V \in L^\frac{n}{\alpha}(B_3) with \delta(V) \leqslant \overline \delta, h \in L^\nu(B_2) and 0 \leqslant u \in L^r(B_3) satisfying

\displaystyle u(x) \leqslant \int_{{B_3}} {\frac{{V(y)u(y)}}{{{{\left| {x - y} \right|}^{n - \alpha }}}}dy} + h(x), \quad x \in {B_2}

we have

\displaystyle {\left\| u \right\|_{{L^\nu }({B_{1/2}})}} \leqslant C\left( {{{\left\| u \right\|}_{{L^r}({B_3})}} + {{\left\| h \right\|}_{{L^\nu }({B_2})}}} \right).

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July 24, 2010

Regularity theory for integral equations

Filed under: Giải Tích 6 (MA5205), PDEs — Tags: — Ngô Quốc Anh @ 20:54

My purpose is to derive some regularity result concerning the following integral equation

\displaystyle u(x) = \int_\Omega {\frac{{u(y)}}{{{{\left| {x - y} \right|}^{n - \alpha }}}}dy}

where \Omega \subset \mathbb R^n is open and bounded and 0<\alpha<n. To this purpose, in this entry we first consider the equation

\displaystyle u(x) = \int_\Omega {\frac{{f(y)}}{{{{\left| {x - y}  \right|}^{n - \alpha }}}}dy}

for a suitable choice of f.

The case f \in L^\infty(\Omega). We will prove that u \in C^{1,\beta}(\Omega) for any \beta\in (0,1). Indeed, up to a constant factor, the first derivative of u are given by

\displaystyle {D_i}u(x) = \int_\Omega {\frac{{{x_i} - {y_i}}}{{{{\left| {x - y} \right|}^{n + 2 - \alpha }}}}f(y)dy}.

From this formula,

\displaystyle\left| {{D_i}u({x^1}) - {D_i}u({x^2})} \right| = \mathop {\sup }\limits_\Omega |f|\int_\Omega {\left| {\frac{{x_i^1 - {y_i}}}{{{{\left| {{x^1} - y} \right|}^{n + 2 - \alpha }}}} - \frac{{x_i^2 - {y_i}}}{{{{\left| {{x^2} - y} \right|}^{n + 2 - \alpha }}}}} \right|dy} .

By the intermediate value theorem, on the line from x^1 to x^2, there exists some x^3 with

\displaystyle\left| {\frac{{x_i^1 - {y_i}}}{{{{\left| {{x^1} - y} \right|}^{n + 2 - \alpha }}}} - \frac{{x_i^2 - {y_i}}}{{{{\left| {{x^2} - y} \right|}^{n + 2 - \alpha }}}}} \right| \leqslant \frac{C}{{{{\left| {{x^3} - y} \right|}^{n + 2 - \alpha }}}}\left| {{x^1} - {x^2}} \right|.

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June 9, 2010

Invariance under fractional linear transformations

Filed under: Giải Tích 5, Giải Tích 6 (MA5205) — Tags: — Ngô Quốc Anh @ 5:02

I just read the following result due to Loss-Sloane published in J. Funct. Anal. this year [here]. This is just a lemma in their paper that I found very interesting.

Let f be any function in C_0^\infty(\mathbb R \setminus \{0\}). Consider the inversion x \mapsto \frac{1}{x} and set

\displaystyle g(x) = {\left| x \right|^{\alpha - 1}}f\left( {\frac{1}{x}} \right).

Then g \in C_0^\infty(\mathbb R) and

\displaystyle\iint\limits_{{\mathbb{R}^2}} {\frac{{{{\left| {g(x) - g(y)} \right|}^2}}}{{{{\left| {x - y} \right|}^{\alpha + 1}}}}dxdy} = \iint\limits_{{\mathbb{R}^2}} {\frac{{{{\left| {f(x) - f(y)} \right|}^2}}}{{{{\left| {x - y} \right|}^{\alpha + 1}}}}dxdy}.

Proof. For fixed \varepsilon consider the regions

\displaystyle {R_1}: = \left\{ {(x,y) \in {\mathbb{R}^2}:\left| {\frac{x}{y}} \right| > 1 + \varepsilon } \right\}

and likewise,

\displaystyle {R_2}: = \left\{ {(x,y) \in {\mathbb{R}^2}:\left|  {\frac{y}{x}} \right| > 1 + \varepsilon } \right\}.

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May 30, 2010

Co-area formula for gradient, 2

Filed under: Giải Tích 6 (MA5205) — Tags: — Ngô Quốc Anh @ 20:29

Let us recall the following result

Theorem. Let \Omega \subset \mathbb R^n be an open set and u \in \mathcal D(\Omega). If u \geqslant 0 then for any 1 \leqslant p<\infty, we have

\displaystyle\int_\Omega|\nabla u|^p dx =\int_0^M \left(\int_{\{u=t\}}|\nabla u|^{p-1}d\sigma\right)dt

where M=\sup u over \overline \Omega.

The above result has been proven in this entry. If we chose p=0 then we would have

\displaystyle\int_\Omega dx =\int_0^M \left(\int_{\{u=t\}}\frac{1}{|\nabla u|}d\sigma\right)dt.

If we now replace \Omega=\{u>t\}, we get

\displaystyle\int_{\{u>t\}}dx =\int_t^M \left(\int_{\{u=t\}}\frac{1}{|\nabla u|}d\sigma\right)dt.

By differentiating with respect to t, we arrive at

\displaystyle \frac{d}{dt}\int_{\{u>t\}}dx =-\int_{\{u=t\}}\frac{1}{|\nabla u|}d\sigma.

In this entry, we shall prove the foregoing identity is actually true. This is the second interesting formula I have mentioned before. The proof, of course again, is based on a clever choice of test function.

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May 27, 2010

A Simple Approach to the Hardy and Rellich inequalities

Filed under: Giải Tích 6 (MA5205) — Tags: , — Ngô Quốc Anh @ 16:07

The classical Hardy inequality in \mathbb R^n, n \geqslant 3, is stated as follows

Theorem (Hardy’s inequality). Let u \in \mathcal D^{1,2}(\mathbb R^n) with n \geqslant 3. Then

\displaystyle\frac{{{u^2}}}{{{{\left| x \right|}^2}}} \in {L^1}({\mathbb{R}^n})

and

\displaystyle {\left( {\frac{{n - 2}}{2}} \right)^2}\int_{{\mathbb{R}^n}} {\frac{{{u^2}}}{{{{\left| x \right|}^2}}}dx} \leqslant \int_{{\mathbb{R}^n}} {{{\left| {\nabla u} \right|}^2}dx}.

The constant {\left( {\frac{{n - 2}}{2}} \right)^2} is the best possible constant.

I suddenly found a very simple proof due to E. Mitidieri [here].

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May 24, 2010

Co-area formula for gradient

Filed under: Giải Tích 6 (MA5205) — Tags: — Ngô Quốc Anh @ 18:55

I found two interesting formulas related to co-area formula while reading some tricks done by Talenti regarding to the best constant of the Sobolev inequality. The first result is to derive a representation of

\displaystyle \int_\Omega |\nabla u|^pdx

and the second result is to deal with differentiation of level sets. Having all these stuffs, I will derive a very short and beautiful proof concerning the lower bound of \int \exp(u) dx where u, a positive solution to the following PDE

\displaystyle-\Delta u = e^u

This proof I firstly learned from a paper published in Duke Math. J. in 1991 by W. Cheng and C. Li [here].

Co-area formula. Suppose that \Omega is an open set in \mathbb R^n, and u is a real-valued Lipschitz function on \Omega. Then, for an integrable function g

\displaystyle\int_\Omega g(x) |\nabla u(x)| dx = \int_{-\infty}^\infty \left(\int_{\{u=t\}}g(x) dH_{n-1}(x)\right)dt

where H^{n-1} is the (n-1)-dimensional Hausdorff measure.

The Sard theorem. Let f :\mathbb{R}^n \rightarrow \mathbb{R}^m be C^k, k times continuously differentiable, where k \geqslant \max\{n-m+1, 1\}. Let X be the critical set of f, the set of points x in \mathbb R^n at which the Jacobian matrix of f has {\rm rank} < m. Then f(X) has Lebesgue measure 0 in \mathbb R^m.

The Sard theorem has some useful applications. For example, if u \in \mathcal D(\Omega) the space of test functions where \Omega \subset \mathbb R^n, then for almost every t in the range of u, we have that |\nabla u|\ne 0 on the level set \{u=t\}. Thus that level set will be an (n-1)-dimensional surface. Furthermore

\{u=t\}=\partial \{u>t\}

and

|\{u=t\}|=0.

Theorem. Let \Omega \subset \mathbb R^n be an open set and u \in \mathcal D(\Omega). If u \geqslant 0 then for any 1 \leqslant p<\infty, we have

\displaystyle\int_\Omega|\nabla u|^p dx =\int_0^M \left(\int_{\{u=t\}}|\nabla u|^{p-1}d\sigma\right)dt

where M=\sup u over \overline \Omega.

(more…)

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