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 )}}

(more…)

April 8, 2011

100k hits

Filed under: Linh Tinh — Ngô Quốc Anh @ 20:52

Hello everyone,

This is a great news and I should share this immediately :D !

My blog has just gone through 100,000 hits at this moment :) . I take this opportunity to thank all of you for your interest in my blog, for your suggestion, for your beautiful comments, and for everything…

As can be seen, I have started using this wordpress blog since May, 2007. However, there was not much information here until the time I came to Singapore to pursue my PhD degree. That was August, 2008. During this period of time, I have learnt much and of course I have some spare time to write down things that I think it may be useful.

A bit for the future, I will keep writing things that I usually face during my mathematics career. Further than that, it totally depends on how much time I will have. Anyway, please do enjoy what we have here.

Again, thank you and have a nice day :) ,

Ngo Quoc Anh.

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…)

October 18, 2010

1/infinity = 0 is equivalent to 1/0=infinity?

Filed under: Các Bài Tập Nhỏ, Linh Tinh — Ngô Quốc Anh @ 12:05

It is now the time to discuss some funny thing. I just learn from GR class this morning a proof of the following statement

\displaystyle \frac{1}{\infty}=0 \quad \Longleftrightarrow \quad \frac{1}{0}=\infty.

Okay, let us start with the left hand side. By rotating 90 degrees counter-clockwise both sides of

\displaystyle \frac{1}{\infty}=0

we get

\displaystyle -18=0.

Now adding both sides by 8 we arrive at

\displaystyle -10=8.

Again, rotating 90 degrees clockwise both sides we reach to

\displaystyle \frac{1}{0}=\infty.

The reverse case can be treated similarly.

April 9, 2010

Kelvin transform: Laplacian

Filed under: Các Bài Tập Nhỏ, Linh Tinh, Nghiên Cứu Khoa Học, PDEs — Tags: — Ngô Quốc Anh @ 0:01

For each point x \ne 0, denote x=(x_1,...,x_n) and

\displaystyle \xi = {x^\sharp } = \left( {\frac {{{x_1}}} {{{{\left| x \right|}^2}}},...,\frac {{{x_n}}} {{{{\left| x \right|}^2}}}} \right)

is the inversion of  x with respect to the unit sphere. We have the following identities

\displaystyle\frac {{\partial {\xi _j}}} {{\partial {x_k}}} = \frac {1} {{{{\left| x \right|}^2}}}\left( {{\delta _{jk}} - 2\frac {{{x_j}{x_k}}} {{{{\left| x \right|}^2}}}} \right)

and

\displaystyle\sum\limits_{l = 1}^n {\frac {{\partial {\xi _l}}} {{\partial {x_j}}}\frac {{\partial {\xi _l}}} {{\partial {x_k}}}} = \frac {1} {{{{\left| x \right|}^4}}}{\delta _{jk}}.

Thus,

\displaystyle\sum\limits_{l = 1}^n {\frac {{\partial {x_l}}} {{\partial {\xi _j}}}\frac {{\partial {x_l}}} {{\partial {\xi _k}}}} = \frac {1} {{{{\left| \xi \right|}^4}}}{\delta _{jk}}.

Next, think of \xi as a system of orthogonal curvilinear coordinates for x, we deduce that the metric tensor of the Euclidean space in curvilinear coordinates

\displaystyle {g_{j,k}}\left( {{\xi _j},{\xi _k}} \right) = \sum\limits_{l = 1}^n {\frac {{\partial {x_l}}} {{\partial {\xi _j}}}\frac {{\partial {x_l}}} {{\partial {\xi _k}}}} = \frac {1} {{{{\left| \xi \right|}^4}}}{\delta _{jk}}.

This implies that the so-called Lame coefficients is

\displaystyle {h_j} = \sqrt {{g_{j,j}}\left( {{\xi _j},{\xi _j}} \right)} = \frac {1} {{{{\left| \xi \right|}^2}}}.

(more…)

January 13, 2010

R-G: A few words about writing covariant derivative of tensors in a short form

Filed under: Linh Tinh, Nghiên Cứu Khoa Học, Riemannian geometry — Ngô Quốc Anh @ 14:21

Before going further, I would like to mention a convention in writing covariant derivative of tensors in a short form. What I mean is how to understand the following notation \nabla_i g_{\alpha\beta} where g is a Riemannian metric.

We all know that we can apply covariant derivative to a scalar function, for example, \nabla_i f is nothing but the partial derivative with respect to x^i. However, the notation \nabla_i g_{\alpha\beta} is a little bit different. For g, a Riemannian metric, which is also a (0,2) tensor, in the full form, we can write

\displaystyle g=g_{\alpha\beta}dx^\alpha \otimes dx^\beta.

The sub indexes \alpha\beta in the notation \nabla_i g_{\alpha\beta} tell us that we are talking about the \alpha\beta-component of the covariant derivative of (0,2) tensor g with respect to the vector field \frac{\partial}{\partial x^i}.

If you have \nabla_i h^{\alpha\beta}, then you are working on some (2,0)-tensor h of the form

\displaystyle h=h^{\alpha\beta} \frac{\partial}{\partial x^\alpha} \otimes \frac{\partial}{\partial x^\beta}.

We now go back to the case \nabla_i g_{\alpha\beta}. Precisely, one should write

\displaystyle {\nabla _i}{g^{\alpha \beta }} = ({\nabla _{\frac{\partial }{{\partial {x^i}}}}}g)\left( {\frac{\partial }{{\partial {x^\alpha }}},\frac{\partial }{{\partial {x^\beta }}}} \right).

Since \{dx^\alpha\} is a basis for the dual space of a space spanned by the basis \{\frac{\partial}{\partial x^\beta}\}, it is clear to see that the right hand side of the above convention is just the \alpha\beta-component, that is, the coefficient of the term

\displaystyle {\nabla _i}{g^{\alpha \beta }}d{x^\alpha } \otimes d{x^\beta }.

Now we show that \nabla_i g^{\alpha\beta}=0 whenever g is a Riemannian metric. This is equivalent to show that every coefficients of \nabla_i g equal to zero, in other word, \nabla_i g=0. Since g is a (0,2)-tensor, then we can compute covariant derivative of g as follows

\displaystyle\begin{gathered}{\nabla _i}g = {\nabla _i}\left( {{g_{\alpha \beta }}d{x^\alpha } \otimes d{x^\beta }} \right) \hfill \\\qquad = \left( {\frac{\partial }{{\partial {x^i}}}{g_{\alpha \beta }}} \right)d{x^\alpha } \otimes d{x^\beta } + {g_{\alpha \beta }}{\nabla _i}\left( {d{x^\alpha } \otimes d{x^\beta }} \right) \hfill \\ \qquad= \left( {\frac{\partial }{{\partial {x^i}}}{g_{\alpha \beta }}} \right)d{x^\alpha } \otimes d{x^\beta } + {g_{\alpha \beta }}{\nabla _i}\left( {d{x^\alpha }} \right) \otimes d{x^\beta } + {g_{\alpha \beta }}d{x^\alpha } \otimes {\nabla _i}\left( {d{x^\beta }} \right) \hfill \\ \qquad= \left( {\frac{\partial }{{\partial {x^i}}}{g_{\alpha \beta }}} \right)d{x^\alpha } \otimes d{x^\beta } - \left( {{g_{\alpha \beta }}\Gamma _{ik}^\alpha+ {g_{\alpha \beta }}\Gamma _{ik}^\beta } \right)d{x^k} \otimes d{x^\beta } \hfill \\ \qquad= \left( {\frac{\partial }{{\partial {x^i}}}{g_{\alpha \beta }} - \Gamma _{i\alpha }^k{g_{k\beta }} - \Gamma _{i\beta }^k{g_{k\alpha }}} \right)d{x^\alpha } \otimes d{x^\beta }. \hfill \\ \end{gathered}

Since g is a metric connection, i.e.

\displaystyle Xg\left( {Y,Z} \right) = g\left( {{\nabla _X}Y,Z} \right) + g\left( {Y,{\nabla _X}Z} \right)

one has

\displaystyle \frac{\partial }{{\partial {x^i}}}g\left( {\frac{\partial }{{\partial {x^\alpha }}},\frac{\partial }{{\partial {x^\beta }}}} \right) = g\left( {{\nabla _i}\frac{\partial }{{\partial {x^\alpha }}},\frac{\partial }{{\partial {x^\beta }}}} \right) + g\left( {\frac{\partial }{{\partial {x^\alpha }}},{\nabla _i}\frac{\partial }{{\partial {x^\beta }}}} \right).

Thus

\displaystyle\frac{\partial }{{\partial {x^i}}}{g_{\alpha \beta }} = g\left( {\Gamma _{i\alpha }^k\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^\beta }}}} \right) + g\left( {\frac{\partial }{{\partial {x^\alpha }}},\Gamma _{i\beta }^k\frac{\partial }{{\partial {x^k}}}} \right) = \Gamma _{i\alpha }^k{g_{k\beta }} + \Gamma _{i\beta }^k{g_{k\alpha }}.

In other word the \alpha\beta-component of \nabla_i g equals to zero. At the last word, in order to calculate covariant derivative of some (p,q)-tensors, you need to use the following three properties

  1. {\nabla _X}\left( {fT} \right) = X(f)T + f{\nabla _X}T.
  2. {\nabla _X}\left( {T \otimes Q} \right) = {\nabla _X}T \otimes Q + T \otimes {\nabla _X}Q.
  3. For (1,0)- or (0,1)-tensors:

    \displaystyle {\nabla _{\frac{\partial }{{\partial {x^i}}}}}\frac{\partial }{{\partial {x^j}}} = \Gamma _{ij}^k\frac{\partial }{{\partial {x^k}}}

    and

    \displaystyle {\nabla _{\frac{\partial }{{\partial {x^i}}}}}d{x^j}=-\Gamma _{ik}^jd{x^k}.

December 2, 2009

MuPAD

Filed under: Linh Tinh — Tags: — Ngô Quốc Anh @ 22:21

MuPAD was a Computer algebra system (CAS). Originally developed by the MuPAD research group at the University of Paderborn, Germany, it was developed by the company SciFace Software GmbH & Co. KG in cooperation with the MuPAD research group and partners from some other universities since 1997.

I found MuPAD a very cool tool for plotting graph of functions. I will show you in details some examples.

1. We start with the following function

f(x,y)=-(x^2+y^2)(\cos (3y) + \sin (7y)).

To plot its graph, I use the following

f := plot::Function3d(exp(-x^2-y^2)*(cos(3*y)+sin(7*x)), x=-2..2, y=-2..2, Submesh=[1,1], FillColorFunction=((x, y, z) -> [(z+2)/4, 0.5, (2-z)/4])): plot(f)

and this is what we get

exp(-x^2-y^2)*(cos(3*y)+sin(7*x))

(more…)

October 30, 2009

A characteristic of essentially bounded functions

Filed under: Các Bài Tập Nhỏ, Giải Tích 6 (MA5205), Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 11:53

In this topic, we prove the following statement

Statement: Let (X,\mathcal B, m) be a probability space. Let h \in L^2(m). Then h is essentially bounded iff h \cdot f \in L^2(m) for all f \in L^2(m).

Proof. If h is bounded, then by using the Holder inequality one has

\displaystyle\int_X {{{\left| {h \cdot f} \right|}^2}dm}\leq \underbrace {\sqrt {\int_X {{{\left| h \right|}^2}dm} } }_{ \leqslant c}\sqrt {\int_X {{{\left| f \right|}^2}dm} }<+\infty

for all f \in L^2(m). Conversely, we suppose h is such that h \cdot f \in L^2(m) whenever f \in L^2(m). Let

\displaystyle X_n = \{ x \in X : n-1 \leq |h(x)| < n\}, \quad \forall n>0.

Then \{X_n\}_1^\infty partitions X. Let

\displaystyle f\left( x \right) =\sum\limits_{n = 1}^\infty{\frac{1}{{n\sqrt {m\left( {{X_n}} \right)} }}{\chi_{{X_n}}}\left( x \right)} ,

where it is understood that the n-term is omiited if m(X_n)=0. Then

\displaystyle\int_X {{{\left| f \right|}^2}dm}=\int_X {{{\left({\sum\limits_{n = 1}^\infty {\frac{1}{{n\sqrt {m\left( {{X_n}}\right)} }}{\chi _{{X_n}}}\left( x \right)} } \right)}^2}dm}\leq \sum\limits_{n = 1}^\infty{\frac{1}{{{n^2}}}}<\infty

which implies f \in L^2(m). Since

\displaystyle\int_X {{{\left| {hf} \right|}^2}dm}=\sum\limits_{n \in F} {\int_{{X_n}} {{{\left| {hf} \right|}^2}dm}}\geq\sum\limits_{n \in F}{{{\left( {\frac{{n - 1}}{n}}\right)}^2}}

where F = \left\{ {n:m\left( {{X_n}} \right) \ne 0} \right\}. Sincc h \cdot f \in L^2(m) we have that F is finite and therefore h is essentially bounded.

October 28, 2009

The weak and weak* topologies: A few words

Filed under: Giải tích 8 (MA5206), Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 2:48

The weak and weak* topologies are the weakest in which certain linear functionals are continuous.

We start with a normed linear space X. The dual space of X, denoted by X', is the collection of all continuous linear functionals, i.e., the set of all mapping \ell : X \to \mathbb R satisfying

\ell(ax)=a \ell (x), \ell(x+y)=\ell(x)+\ell(y)

and

\displaystyle\lim_{n \to \infty} \ell(x_n) = \ell(x) when \displaystyle\lim_{n \to \infty} \|x_n - x\|=0.

Definition 1. In X, the strong topology is the norm topology, i.e., we can talk about an open set of X, for example U in the following sense: U \subset X is said to be open if and only if for each x_0 \in U, there exists \varepsilon>0 such that \{ x \in X: \|x-x_0\|<\varepsilon\} \subset U.

Claim 1. Bounded linear functionals are continuous in the strong topology.

Proof. We first recall that a linear functional \ell is said to be bounded if there is a positive number c such that |\ell (x)| \leq c\|x\| for all x \in X.

Now we assume \ell is continuous but not bounded; then for any choice of c=n, one has \ell(x_n) > n \|x_n\|. Clearly, x_n can be replaced by any multiple of x_n; if we normalize x_n so that

\displaystyle \|x_n\|=\frac{1}{\sqrt{n}}

then x_n \to 0 but \ell (x_n) \to \infty. This shows the lack of boundedness implies the lack of continuity.

Now we assume \ell is bounded. For arbitray x_n and x, one gets

|\ell(x_n)-\ell (x)| = |\ell (x_n-x)| \leq c\|x_n-x\|;

this shows that boundedness implies continuity.

Definition 2. In X, the weak topology is the weakest topology in which all bounded linear functionals are continuous.

The open sets in the weak topology are unions of finite intersections of sets of the form

\{ x : a< \ell(x) < b\}.

Clearly, in an infinite-dimensional space the intersection of a finite number of sets of the above form is unbounded. This shows that every set that is open in the weak topology is unbounded. In particular, the balls

\{ x : \|x\|<R\}

opens in the strong topology, are not open is the weak topology.

Definition 3. In X' the dual space of X, the weak* topology is the crudest topology in which all linear functionals

x: X' \to \mathbb R, x(\ell) := \ell(x)

are continuous.

If X' is nonreflexive, the weak* topology is genuinely coarser than the weak topology, as will be clear from the following theorem due to Alaoglu

Theorem (Alaoglu). The closed unit ball in X' is compact in the weak* topology.

We end this topic by the following theorem

Theorem. The closed unit ball in X is compact in the weak topology if and only if X is reflexive.

September 18, 2009

The method of moving planes: An example inspired by Gidas, Ni, and Nirenberg

Filed under: Linh Tinh, Nghiên Cứu Khoa Học, PDEs — Tags: — Ngô Quốc Anh @ 1:41

In this section we will use the moving plane method to discuss the symmetry of solutions. The following result was first proved by Gidas, Ni and Nirenberg.

Theorem. Suppose u \in C(\bar B_1) \cap C^2(B_1) is a positive solution of

-\Delta u = f(u) in B_1 and u=0 on \partial B_1

where f is locally Lipschitz in \mathbb R. Then u is radialy symmetric in B_1 and \frac{\partial u}{\partial r}(x)<0 for x \ne 0.

The original proof requires that solutions be C^2 up to the boundary. Here we give a method which does not depend on the smoothness of domains nor the smoothness of solutions up to the boundary.

Statement. Suppose that ­ is a bounded domain which is convex in x_1 direction and symmetric with respect to the plane \{x_1 = 0\}. Suppose u \in C^2(\Omega)\cap C(\bar\Omega) is a positive solution of

-\Delta u = f(u) in \Omega and u=0 on \partial \Omega

where f is locally Lipschitz in \mathbb R. Then u is radialy symmetric in x_1 and D_{x_1}u(x)<0 for x \ne 0.

Idea of proof.  Write x=(x_1, y)\in\Omega for y \in \mathbb R^{n-1}. We will prove that

u(x_1,y)<u(x_1^\star,y) for any x_1>0 and x_1^\star<x_1 with x_1^\star+x_1>0.

Then by letting x_1^\star \to -x_1, we get u(x_1,y)\leq u(-x_1,y) for any x_1. Then by changing the direction x_1 \to -x_1 we get the symmetry.

Proof. Let a=\sup x_1 for (x_1, y)\in\Omega. For 0<\lambda<a, define

\Sigma_\lambda=\{x \in \Omega: x_1>\lambda\}

T_\lambda = \{x_1=\lambda\}

\Sigma'_\lambda= the reflection of \Sigma_\lambda with the respect to T_\lambda

x_\lambda=(2\lambda-x_1,x_2,...,x_n) for x=(x_1,x_2,...,x_n).

In \Sigma_\lambda we define

w_\lambda(x)=u(x)-u(x_\lambda) for x\in \Sigma_\lambda.

Then we have by Mean Value Theorem

\Delta w_\lambda+c(x,\lambda)w_\lambda=0 in \Sigma_\lambda

w_\lambda \leq 0 and w_\lambda \not \equiv 0 on \partial \Sigma_\lambda.

where c(x,\lambda) is a bounded function in \Sigma_\lambda.

We need to show w_\lambda<0 in \Sigma_\lambda for any \lambda \in (0,a). This implies in particular that w_\lambda assumes along \partial \Sigma_\lambda \cap \Omega its maximum in \Sigma_\lambda. By Hopf Lemma we have for any such \lambda \in (0,a)

D_{x_1}w_\lambda\bigg|_{x_1=\lambda} = 2D_{x_1}u\bigg|_{x_1=\lambda}<0.

For any \lambda close to a, we have w_\lambda<0 by the maximum principle for narrow domain. Let (\lambda_0, a) be the largest interval of values of \lambda such that w_\lambda<0 in \Sigma_\lambda. We want to show that \lambda_0=0. If \lambda_0>0, by continuity, w_\lambda \leq 0 in \Sigma_{\lambda_0} and w_{\lambda_0} \not \equiv 0 on \partial \Sigma_{\lambda_0}. Then the Strong Maximum Principle implies w_{\lambda_0}<0 in \Sigma_{\lambda_0}. We will show that for any small \varepsilon>0

w_{\lambda_0-\varepsilon} <0 in \Sigma_{\lambda_0-\varepsilon}.

Fix \delta>0 (to be determined). Let K be a closed subset in \Sigma_{\lambda_0} such that |\Sigma_{\lambda_0} - K| <\frac{\delta}{2}. The fact that w_{\lambda_0}<0 in \Sigma_{\lambda_0} implies

w_{\lambda_0}(x)\leq -\eta <0 for any x \in K.

By continuity we have

w_{\lambda_0-\varepsilon}<0 in K.

For \varepsilon>0 small, |\Sigma_{\lambda_0-\varepsilon}-K|<\delta. We choose \delta in such a way that we may apply the Maximum Principle for domain with small volume to w_{\lambda_0-\varepsilon} in \Sigma_{\lambda_0-\varepsilon}-K. Hence we get

w_{\lambda_0-\varepsilon} \leq 0 in \Sigma_{\lambda_0-\varepsilon}-K

and then

w_{\lambda_0-\varepsilon}(x) <0 in \Sigma_{\lambda_0-\varepsilon} -K.

Therefore we obtain for any small \varepsilon>0

w_{\lambda_0-\varepsilon}(x) <0 in \Sigma_{\lambda_0-\varepsilon}.

This contradicts the choice of \lambda_0.

latex
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