Ngô Quốc Anh

March 10, 2012

An integral of 1/(1+|x|^2) over the whole Euclidean space

Filed under: Giải Tích 2, Giải Tích 3, Linh Tinh — Ngô Quốc Anh @ 2:46

Sometimes, we need a precise value for following

\displaystyle\int_{{\mathbb{R}^n}} {\frac{{dx}}{{{{(1 + |x{|^2})}^\alpha }}}}, \quad \alpha>\frac{n}{2}.

As such, I am going to calculate it and place the result here for future works.

In order to evaluate the above integral, we need to use the so-called co-area formula. We first write

\displaystyle\int_{{\mathbb{R}^n}} {\frac{{dx}}{{{{(1 + |x{|^2})}^\alpha }}}} = \int_0^{ + \infty } {\left( {\int_{\partial {B_0}(r)} {\frac{{dS}}{{{{(1 + {r^2})}^\alpha }}}} } \right)dr}.

Note that

\displaystyle \begin{gathered} \int_{\partial {B_0}(r)} {\frac{{dS}}{{{{(1 + {r^2})}^\alpha }}}} = \frac{1}{{{{(1 + {r^2})}^\alpha }}}\int_{\partial {B_0}(r)} {dS} \hfill \\ \qquad\qquad\qquad= \frac{1}{{{{(1 + {r^2})}^\alpha }}}\text{Area}({B_0}(r)) = \frac{{2{\pi ^{\frac{n}{2}}}}}{{\Gamma \left( {\frac{n}{2}} \right)}}\frac{{{r^{n - 1}}}}{{{{(1 + {r^2})}^\alpha }}}\end{gathered}.

Therefore,

\displaystyle\int_{{\mathbb{R}^n}} {\frac{{dx}}{{{{(1 + |x{|^2})}^\alpha }}}} = \frac{{2{\pi ^{\frac{n}{2}}}}}{{\Gamma \left( {\frac{n}{2}} \right)}}\int_0^{ + \infty } {\frac{{{r^{n - 1}}}}{{{{(1 + {r^2})}^\alpha }}}} dr = {\pi ^{\frac{n}{2}}}\frac{{\Gamma \left( {\alpha - \frac{n}{2}} \right)}}{{\Gamma (\alpha )}}

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November 5, 2011

MuPad: Heart in 3D

Filed under: Giải Tích 2, Giải Tích 5, Liên Kết — Ngô Quốc Anh @ 0:26

This is not mathematics. I just found an equation so that we can draw a heart in 3D. Indeed, the following equation

\displaystyle {\left( {{x^2} + \frac{9}{4}{y^2} + {z^2} - 1} \right)^3} - {x^2}{z^3} - \frac{9}{{80}}{y^2}{z^3} = 0

will generate a heart. I have tried and the following pictures show that fact.

(more…)

April 22, 2011

On Costa-Hardy-Rellich inequalities

Filed under: Giải Tích 2, Giải Tích 6 (MA5205) — Tags: , , , — Ngô Quốc Anh @ 15:44

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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April 1, 2011

Several interesting limits from a paper by Chang-Qing-Yang

Filed under: Các Bài Tập Nhỏ, Giải Tích 1, Giải Tích 2, Giải Tích Cổ Điển, Linh Tinh — Ngô Quốc Anh @ 16:42

Recently, I have learnt from my friend, ZJ, the following result

Assume that F:\mathbb R \to \mathbb R is absolutely integrable. Then

\displaystyle\begin{gathered} \mathop {\lim }\limits_{t \to \pm \infty } {e^{2t}}\int_t^{ + \infty } {F(x){e^{ - 2x}}dx} = 0, \hfill \\ \mathop {\lim }\limits_{t \to \pm \infty } {e^{ - 2t}}\int_{ - \infty }^t {F(x){e^{ - 2x}}dx} = 0. \hfill \\ \end{gathered}

The result seems reasonable by the following observation, for example, we consider the first identity when t \to +\infty. Then the factor

\displaystyle\int_t^{ + \infty } {F(x){e^{ - 2x}}dx}

decays faster then the exponent function \exp (2t). This may be true, of course we need to prove mathematically, because the integrand contains the term \exp (-2x) which turns out to be a good term since x \geqslant t. So here is the trick in order to solve such a problem.

(more…)

March 1, 2011

The implicit function theorem: A PDE example

Filed under: Giải Tích 3, PDEs — Tags: — Ngô Quốc Anh @ 23:29

This entry devotes an existence result for the following semilinear elliptic equation

-\Delta u + u = u^p+f(x)

in the whole space \mathbb R^n where 0<u \in H^1(\mathbb R^n).

Our aim is to apply the implicit function theorem. It is known in the literature that

Theorem (implicit function theorem). Let X, Y, Z be Banach spaces. Let the mapping f:X\times Y\to Z be continuously Fréchet differentiable.

If

(x_0,y_0)\in X\times Y, \quad F(x_0,y_0) = 0,

and

y\mapsto DF(x_0,y_0)(0,y)

is a Banach space isomorphism from Y onto Z, then there exist neighborhoods U of x_0 and V of y_0 and a Frechet differentiable function g:U\to V such that

F(x,g(x)) = 0

and F(x,y) = 0 if and only if y = g(x), for all (x,y)\in U\times V.

Let us now consider

X=L^2(\mathbb R^n), \quad Y=H_+^2(\mathbb R^n), \quad Z=L^2(\mathbb R^n).

Let us define

F(f,u)=-\Delta u + u - u^p-f(x), \quad f \in X, \quad u \in Y, \quad x \in \mathbb R^n.

It is not hard to see that Fréchet derivative of F at (f,u) with respect to u in the direction v is given by

{D_u}F(f,u)v = - \Delta v + v - p{u^{p - 1}}v.

Since -\Delta +I defines an isomorphism from Y to Z, it is clear to see that our PDE is solvable for f small enough in the X-norm.

January 8, 2011

A funny limit involving sine function

Filed under: Các Bài Tập Nhỏ, Giải Tích 1 — Ngô Quốc Anh @ 2:32

Today, I have been asked to calculate the following limit

\displaystyle \mathop {\lim }\limits_{n \to + \infty } \sin (\sin \overbrace {(...(}^n\sin x)...))

for each fixed x \in [0,2\pi]. From the mathematical point of view, we can assume x \in (-\frac{\pi}{2}, \frac{\pi}{2}) as we just replace x by \sin (\sin x)) if necessary.

There are three possible cases

Case 1. x \in (0, \frac{\pi}{2}). In this case, it is well known that function \frac{\sin x}{x} is monotone decreasing since

\displaystyle {\left( {\frac{{\sin x}}{x}} \right)^\prime } = \frac{{x\cos x - \sin x}}{{{x^2}}} = \frac{{\cos x}}{{{x^2}}}\left( {x - \tan x} \right) \leqslant 0

in its domain. Consequently, it holds

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October 3, 2010

An identity of differentiation involving the Kelvin transform, 2

Filed under: Giải Tích 1 — Tags: — Ngô Quốc Anh @ 11:36

I found the following interesting identity which is similar to what I have showed recent days [here]. In that entry, we showed that

\displaystyle {\nabla _x}\left( {u\left( {\frac{x}{{{{\left| x \right|}^2}}}} \right)} \right) \cdot x = - {\nabla _y}\left( {u\left( y \right)} \right) \cdot y

where x and y are connected by

\displaystyle y = \frac{x}{{{{\left| x \right|}^2}}} \in {\mathbb{R}^2}.

For \lambda>0 we denote by y the following

\displaystyle y = \frac{\lambda^2 x}{{{{\left| x \right|}^2}}} \in {\mathbb{R}^n}.

Then we show that

Lemma.

\displaystyle {\left| x \right|^2}{\Delta _x}u\left( {\frac{{{\lambda ^2}x}}{{{{\left| x \right|}^2}}}} \right) - (n - 2)x \cdot {\nabla _x}u\left( {\frac{{{\lambda ^2}x}}{{{{\left| x \right|}^2}}}} \right) = {\left| y \right|^2}{\Delta _y}u\left( y \right) - (n - 2)y \cdot {\nabla _y}u\left( y \right).

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September 22, 2010

An identity of differentiation involving the Kelvin transform

Filed under: Các Bài Tập Nhỏ, Giải Tích 1 — Tags: — Ngô Quốc Anh @ 15:47

This short note is to prove the following

\displaystyle {\nabla _x}\left( {u\left( {\frac{x}{{{{\left| x \right|}^2}}}} \right)} \right) \cdot x = - {\nabla _y}\left( {u\left( y \right)} \right) \cdot y

where x and y are connected by

\displaystyle y = \frac{x}{{{{\left| x \right|}^2}}} \in {\mathbb{R}^2}.

The proof is straightforward as follows.

  • Calculation of \frac{\partial}{\partial x_1}.

We see that

\displaystyle\begin{gathered} \frac{\partial }{{\partial {x_1}}}\left( {u\left( {\frac{x}{{{{\left| x \right|}^2}}}} \right)} \right){x_1} = \frac{\partial }{{\partial {y_1}}}\left( {u\left( y \right)} \right)\frac{\partial }{{\partial {x_1}}}\left( {\frac{{{x_1}}}{{{{\left| x \right|}^2}}}} \right){x_1} + \frac{\partial }{{\partial {y_2}}}\left( {u\left( y \right)} \right)\frac{\partial }{{\partial {x_1}}}\left( {\frac{{{x_2}}}{{{{\left| x \right|}^2}}}} \right){x_1} \hfill \\ \qquad\qquad\qquad= \frac{\partial }{{\partial {y_1}}}\left( {u\left( y \right)} \right)\left( {\frac{1}{{{{\left| x \right|}^2}}} - \frac{{2x_1^2}}{{{{\left| x \right|}^4}}}} \right){x_1} + \frac{\partial }{{\partial {y_2}}}\left( {u\left( y \right)} \right)\left( { - \frac{{2{x_1}{x_2}}}{{{{\left| x \right|}^4}}}} \right){x_1}. \hfill \\ \end{gathered}

  • Calculation of \frac{\partial}{\partial x_2}.

Similarly, we get

\displaystyle\begin{gathered} \frac{\partial }{{\partial {x_2}}}\left( {u\left( {\frac{x}{{{{\left| x \right|}^2}}}} \right)} \right){x_2} = \frac{\partial }{{\partial {y_1}}}\left( {u\left( y \right)} \right)\frac{\partial }{{\partial {x_2}}}\left( {\frac{{{x_1}}}{{{{\left| x \right|}^2}}}} \right){x_2} + \frac{\partial }{{\partial {y_2}}}\left( {u\left( y \right)} \right)\frac{\partial }{{\partial {x_2}}}\left( {\frac{{{x_2}}}{{{{\left| x \right|}^2}}}} \right){x_2} \hfill \\ \qquad\qquad\qquad= \frac{\partial }{{\partial {y_1}}}\left( {u\left( y \right)} \right)\left( { - \frac{{2{x_1}{x_2}}}{{{{\left| x \right|}^4}}}} \right){x_2} + \frac{\partial }{{\partial {y_2}}}\left( {u\left( y \right)} \right)\left( {\frac{1}{{{{\left| x \right|}^2}}} - \frac{{2x_2^2}}{{{{\left| x \right|}^4}}}} \right){x_2}. \hfill \\ \end{gathered}

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September 4, 2010

Convergence of improper integrals does not imply the integrands go to zero

Filed under: Counter-examples, Giải Tích 3 — Ngô Quốc Anh @ 12:31

This entry devotes a similar question that raises during a course of series. We all know that for a convergent series of (positive) real number

\displaystyle \sum_{n=1}^\infty a_n

it is necessary to have

\displaystyle\mathop {\lim }\limits_{n \to \infty } {a_n} = 0.

This is the so-called n-th term test. A natural extension is the following question

Question. Suppose f(x) is positive on [0,\infty) and

\displaystyle\int_0^{ + \infty } {f(x)dx}

exists. Must f(x) tend to zero as x \to +\infty?

(more…)

July 27, 2010

The implicit function theorem: How to prove a continuously dependence on parameters for solutions of ODEs

Filed under: Giải Tích 3, PDEs — Tags: — Ngô Quốc Anh @ 0:25

It is clear that the implicit function theorem plays an important role in analysis. From now on, I am going to demonstrate this significant matter from the theory of differential equations, both ODE and PDE, point of view.

Let us start with the following ODE

-u''-\alpha^2 u^{-q-1}+\beta^2u^{q-1}=0

on some domain \Omega \subset \mathbb R^n with \alpha \not\equiv 0 and \beta \not\equiv 0. We assume the existence result on W_+^{2,p} is proved for some p>1. We prove the following

Theorem. The solution u \in W_+^{2,p} depends continuously on (\alpha, \beta) \in L^\infty \times L^\infty.

Proof. Consider the map

\mathcal N : W_+^{2,p} \times (L^\infty \times L^\infty) \to L^p

taking

(u,\alpha,\beta) \mapsto -u''-\alpha^2 u^{-q-1}+\beta^2u^{q-1}.

This map is evidently continuous (since W_+^{2,p} is an algebra). One readily shows that its Fréchet derivative at (u, \alpha, \beta) with respect to u in the direction h is

\mathcal N'[u,\alpha ,\beta ]h = - h'' + \left[ {(q + 1){\alpha ^2}{u^{ - q - 2}} + (q - 1){\beta ^2}{u^{q - 2}}} \right]h.

The continuity of the map

(u,\alpha,\beta) \mapsto \mathcal N'[u,\alpha ,\beta ]

follows from the fact that W_+^{2,p} is an algebra continuously embedded in C^0(\Omega).

Since \alpha \not\equiv 0 and \beta \not\equiv 0, the potential

V={(q + 1){\alpha ^2}{u^{ - q - 2}} + (q - 1){\beta ^2}{u^{q - 2}}}

is not identically zero. Thus it is well-known that the map

-\Delta +V : W^{2,p} \to L^p

is an isomorphism.

The implicit function theorem then implies that if u_0 is a solution for data (\alpha_0, \beta_0), there is a continuous map defined near (\alpha_0, \beta_0) taking (\alpha, \beta) to the corresponding solution of the ODE. This establishes the conclusion.

For the more details, we prefer the reader to this preprint.

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