# Ngô Quốc Anh

## October 29, 2010

### Blowup of semilinear heat equation with critical sobolev exponent via the concavity method

Filed under: PDEs — Ngô Quốc Anh @ 21:52

Let us consider the following semilinear heat equation

$u_t - \Delta u =u^{2^\star}, \quad (x,t) \in \Omega \times (0,T)$

together with the following conditions

$u(x,t)=0, \quad (x,t) \in \partial\Omega \times (0,T)$

and

$u(x,0)=u_0(x) \geqslant 0, \not\equiv 0$.

This equation corresponds formally to the $L^2$ gradient flow associated to the energy functional

$\displaystyle E(u)=\frac{1}{2}\int_\Omega |\nabla u|^2dx - \frac{1}{2^\star}\int_\Omega |u|^{2^\star}dx$.

We prove the following

If there exists some $t_0$ such that $E(u(t_0)) \leqslant 0$ then $u(u,t;u_0)$ blows up in finite time.

This result is adapted from a paper by Zhong Tan published in Commun. Partial Differential Equations in 2001 [here].

## 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.

## October 11, 2010

### The Agmon type inequality in 1D

Filed under: Uncategorized — Tags: , , — Ngô Quốc Anh @ 18:14

In this note, we prove a very interesting inequality known as the Agmon type inequality in space dimension 1.

$\displaystyle |g(x){|^2} \leqslant {\left( {\int_\mathbb{R} {|g(\xi ){|^2}d\xi } } \right)^{\frac{1}{2}}}{\left( {\int_\mathbb{R} {|g'(\xi ){|^2}d\xi } } \right)^{\frac{1}{2}}},\quad \forall x \in \mathbb{R}$

for any smooth function $g$ with compact support in $\mathbb R$.

The proof is standard and classical. The trick is to use the integral representation for functions that we have already discussed when we talk about the Poincare inequality.

Proof. Since $g$ has compact support, there exists some $L>0$ sufficiently large such that $g$ vanishes outside of $(-L,L)$. Then we can write

$\displaystyle g{(x)^2} = 2\int_{ - L}^x {g(\xi )g'(\xi )d\xi } \leqslant 2\int_{ - L}^x {|g(\xi )||g'(\xi )|d\xi }$

and

$\displaystyle g{(x)^2} = -2\int_x^{ L}{g(\xi )g'(\xi )d\xi } \leqslant 2\int_x^{L}{|g(\xi )||g'(\xi )|d\xi }$.

$\displaystyle g{(x)^2} \leqslant \int_{ - L}^L {|g(\xi )||g'(\xi )|d\xi } = \int_\mathbb{R} {|g(\xi )||g'(\xi )|d\xi }$

Then using the Cauchy-Schwarz inequality we find the desired inequality.

I will show some application of this equality, precisely, I will derive a proof of the Ladyzhenskaya inequalities

$\displaystyle \int_{{\mathbb{R}^2}} {|u(x){|^4}dx} \leqslant \left( {\int_{{\mathbb{R}^2}} {|u(x){|^2}dx} } \right)\left( {\int_{{\mathbb{R}^2}} {|\nabla u(x){|^2}dx} } \right)$

and

$\displaystyle \int_{{\mathbb{R}^3}} {|u(x){|^4}dx} \leqslant {\left( {\int_{{\mathbb{R}^3}} {|u(x){|^2}dx} } \right)^{\frac{1}{2}}}{\left( {\int_{{\mathbb{R}^3}} {|\nabla u(x){|^2}dx} } \right)^{\frac{3}{2}}}$

which plays an important role in the theory of Navier-Stokes equations.

## 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.

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