# Ngô Quốc Anh

## January 31, 2014

### 2014 Vietnamese New Year, aka Tết

Filed under: Linh Tinh — Ngô Quốc Anh @ 1:55

Vietnamese New Year, more commonly known by its shortened name Tết or Tết Nguyên Đán, is the most important and popular holiday and festival in Vietnam. It is the Vietnamese New Year marking the arrival of spring based on the Chinese calendar, a lunisolar calendar. For those who do not know about Tết, please read an article in wikipedia for details.

At the first moment of the new year, I wish you a good health and prosperity all year round and thank you for your interest in my blog.

## September 30, 2013

### The Lichnerowicz equation under some variable changes

Filed under: Linh Tinh, Nghiên Cứu Khoa Học, PDEs — Tags: — Ngô Quốc Anh @ 4:54

Let us consider the so-called Lichnerowicz equation

$-\Delta_g u + hu = fu^{2^\star-1}+au^{-2^\star-1} \quad u>0$

on $(M,g)$, a Riemannian manifold of dimension $n \geq 3$. Here $h$, $f$, and $a$ are smooth function with $a \geq 0$.

• We first use the the following variable change

$\displaystyle v=\log u \quad u=e^ v.$

Clearly,

$\displaystyle\Delta v = \frac{\Delta u}{u} - \frac{|\nabla u|^2}{u^2}$

and

$\displaystyle |\nabla v|^2 = \frac{|\nabla u|^2}{u^2}.$

Therefore, we can write

$\displaystyle -\Delta v =-\frac{\Delta u}{u} +|\nabla v|^2.$

Using this rule, we can rewrite the equation as follows

$\displaystyle \boxed{-\Delta v = -h+fu^{2^\star-2}+au^{-2^\star-2}+|\nabla v|^2=-h+fe^{(2^\star-2)v}+ae^{-(2^\star+2)v}+|\nabla v|^2. }$

## September 15, 2013

### Some integral identities on manifolds with boundary

In this note, I summary several useful integral identities on Riemannian manifolds with boundary.

1. Suppose that $f$ is a function and $X$ is a $1$-form, then

$\displaystyle\boxed{\int_M {f\text{div}Xd{v_g}} = - \int_M {\left\langle {\nabla f,X} \right\rangle_g d{v_g}} + \int_{\partial M} {f\left\langle {X,\nu } \right\rangle_g d{\sigma _g}}.}$

To prove this, we write everything in local coordinates as follows

$\begin{array}{lcl} \displaystyle\int_M {f \text{div} Xd{v_g}} &=& \displaystyle\int_M {f{\nabla _i}{X^i}d{v_g}} \hfill \\ &=& \displaystyle - \int_M {{\nabla _i}f{X^i}d{v_g}} + \int_{\partial M} {f\left\langle {X,\nu } \right\rangle_g d{\sigma _g}} \hfill \\ &=& \displaystyle - \int_M {\left\langle {\nabla f,X} \right\rangle_g d{v_g}} + \int_{\partial M} {f\left\langle {X,\nu } \right\rangle_g d{\sigma _g}}\end{array}$

as claimed.

2. Using the previous identity, we can prove the following

$\displaystyle \boxed{\int_M {{{\left\langle {X,\nabla (\text{div} X)} \right\rangle }_g}d{v_g}} = - \int_M {|\text{div} X|_g^2d{v_g}} + \int_{\partial M} {\text{div} X{{\left\langle {X,\nu } \right\rangle }_g}d{\sigma _g}}}$

where $X$ is again a vector field on $M$. To prove this, we simply apply the previous identity with $f$ replaced by $\text{div}(X)$ to get the desired result.
(more…)

## June 14, 2013

### MuPAD: Drawing asymptotically flat spacetime

Filed under: Linh Tinh — Tags: — Ngô Quốc Anh @ 23:06

Today, I will show how to draw an asymptotically flat spacetime with two ends using MuPad. To draw ends, we use hyperbolic functions.

First, we talk about the end. The function that I am going to use is $h(t)=\frac{1-4t}{t-2}+12$. It is clear that $h$ blow up at $t=2$ and approaches $-4$ at infinity. To use $h$, we use the following

h := proc(t)
begin
(1-4*t)/(t-2)+12
end_proc

I have added $12$ to the function $h$ so that $h$ approaches $8$ at infinity. Then to draw the (upper) end, I use the following function

f := proc(x, y)
begin
if
x^2 + y^2 > 3.4
then
h(x^2+y^2)
else
end_if
end_proc

I have used the number $3.4$ because I do not want my end is too tall, keep in mind that $2$ is the blow-up number. We can calculate

h(3.4)

to see that this number is nothing but $3$ which is close to $1$ as I need. We are now able to draw the first end using the following

plot(
plot::Function3d(f, x = -6 .. 6, y = -6 .. 6, Submesh = [3, 3]),
ViewingBox = [Automatic, Automatic, 3 .. 8],
Scaling = Constrained)

## February 10, 2013

### 2013 Tết Holiday

Filed under: Linh Tinh — Ngô Quốc Anh @ 14:24

Tết Holiday

Vietnamese New Year, more commonly known by its shortened name Tết or Tết Nguyên Đán, is the most important and popular holiday and festival in Vietnam. It is the Vietnamese New Year marking the arrival of spring based on the Chinese calendar, a lunisolar calendar. For those who do not know about Tết, please read an article in wikipedia for details.

At the first moment of the new year, I wish you a good health and prosperity all year round and thank you for your interest in my blog.

## January 29, 2013

### Say HI in MuPAD

Filed under: Linh Tinh — Tags: — Ngô Quốc Anh @ 5:37

I found this interesting note regarding to Matlab. In that note, they plotted the word HI using Matlab. Here I try to use MuPAD in order to get a slightly better picture.

As mentioned in the note, the full function we need to use is

$\displaystyle e^{-x^2-\frac{1}{2}y^2} \cos(4x) + e^{-3\big( (x+\frac{1}{2})^2+\frac{1}{2}y^2 \big)}.$

If you plot that full function, what you are going to have is the following picture

## January 22, 2013

### PhD Thesis Defense

Filed under: Linh Tinh, Luận Văn — Ngô Quốc Anh @ 15:46

I just passed my PhD defense on 18 Jan, 2013. Following is the front page of the slides I used during the defense.

The committee of my defense consists of

Since the thesis contains some unpublished results, I cannot provide the slides here but you can email me if interested.

Title page of the slides used in my PhD thesis defense

## 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,

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

## 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

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.

Older Posts »