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)


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 }


\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 }.

Adding both inequalities we obtain

\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)


\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


\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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