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.

Blog at