# Ngô Quốc Anh

## September 20, 2012

### Poly-Laplacian of rotationally symmetric functions in R^3

Filed under: Các Bài Tập Nhỏ, PDEs — Ngô Quốc Anh @ 5:24

In $\mathbb R^3$, it is known that for any rotationally symmetric function $f$, i.e. $f$ depends only on the radius $r$, the following holds

$\displaystyle \Delta f= f'' + \frac{2}{r} f' = \frac{1}{r^2}(r^2 f')' .$

By a simple calculation, it is easy to have

$\displaystyle\begin{gathered} {\Delta ^2}f = \Delta \left( {f'' + \frac{2}{r}f'} \right) \hfill \\ \qquad= {\left( {f'' + \frac{2}{r}f'} \right)^\prime }^\prime + \frac{2}{r}{\left( {f'' + \frac{2}{r}f'} \right)^\prime } \hfill \\ \qquad= {f^{(4)}} + {\left( { - \frac{2}{{{r^2}}}f' + \frac{2}{r}f''} \right)^\prime } + \frac{2}{r}{f^{(3)}} - \frac{4}{{{r^3}}}f' + \frac{4}{{{r^2}}}f'' \hfill \\ \qquad= {f^{(4)}} + \left( {\frac{4}{{{r^3}}}f' - \frac{2}{{{r^2}}}f'' - \frac{2}{{{r^2}}}f'' + \frac{2}{r}{f^{(3)}}} \right) + \frac{2}{r}{f^{(3)}} - \frac{4}{{{r^3}}}f' + \frac{4}{{{r^2}}}f'' \hfill \\ \qquad= {f^{(4)}} + \frac{4}{r}{f^{(3)}}. \hfill \\ \end{gathered}$

In other words, there holds

$\displaystyle {\Delta ^2}f = \frac{1}{{{r^4}}}({r^4}{f^{(3)}})'.$

## September 8, 2012

### CE: Liminf of products

Filed under: Counter-examples, Giải Tích 1 — Ngô Quốc Anh @ 4:46

Today we discuss the inferior limit of the product of two functions. Let us take the following simple question:

Question. Given a function $f:\mathbb R^n \to \mathbb R$ with the following property

$\displaystyle \liminf_{|x| \to +\infty} f(x) \geqslant 0,$

do we always have the following

$\displaystyle \liminf_{|x| \to +\infty} |x|^5f(x) \geqslant 0.$

It turns out that the statement should be hold since $|x|>0$. Unfortunately, since the function $|x|$ blows up of order $5$, the behavior of the product $|x|^5f(x)$ depends on the order of decay of the function $f$. Let take the following counter-example.

We consider the function

$\displaystyle f(x)=-\frac{1}{1+|x|}, \quad x \in \mathbb R^n.$

Although the function $f$ is everywhere negative, there holds

$\displaystyle \liminf_{|x| \to +\infty} f(x) =0.$

Now it is clear that

$\displaystyle \liminf_{|x| \to +\infty} |x|^5f(x)=\liminf_{|x| \to +\infty} \frac{-|x|^5}{1+|x|}=-\infty.$

By definition, one can easily prove that the statement of the question holds if we have

$\displaystyle \liminf_{|x| \to +\infty} f(x)> 0.$