# Ngô Quốc Anh

## October 12, 2012

### Đề thi Cao học ĐHKHTN Hà Nội

Filed under: Đề Thi — Ngô Quốc Anh @ 10:20

Đề thi Cao học Đại học Khoa học Tự Nhiên Hà Nội

 __________Môn Đại số__________ __________Môn Giải tích__________ 2000_ds.pdf 2001_ds.pdf 2002_ds.pdf 2003_ds.pdf 2004_ds.pdf 2010_ds.pdf 2011_ds.pdf 2012_ds.pdf 2000_gt.pdf 2001_gt.pdf 2002_gt.pdf 2003_gt.pdf 2004_gt.pdf 2005_gt.pdf 2006_gt.pdf 2007_gt.pdf 2008_gt.pdf 2009_gt.pdf 2010_gt.pdf 2011_gt.pdf 2012_gt.pdf

Vì lý do khách quan nên đề thi trong 1 hoặc 2 năm gần nhất sẽ chưa được cập nhật ở trang web này. Mọi đóng góp về đề thi cũng như những chỉnh sửa về sai sót luôn được chào đón.

## October 4, 2012

### Integral representation of radially symmetric smooth solutions of bilaplacian equations

Filed under: PDEs — Ngô Quốc Anh @ 8:16

This entry came from one of my calculation concerning to my recent work. Let us consider the following PDE

$\displaystyle \Delta^2 u = -u^{-q} \quad \text{ in }\mathbb R^3$

with $q>3$.

We assume throughout this entry that the function $u$ is radially symmetric and also smooth. In this manner, we can write $u=u(r)$ where $r \geqslant 0$ is the radius. Using the calculation the previous entry, we obtain the following ODE

$\displaystyle\frac{1}{{{r^4}}}({r^4}{u^{(3)}})' = - {u^{ - q}}.$

The aim of this entry is to transform the above ODE to its corresponding integral equation. To understand the procedure, we do it step by step.

First, we multiply both sides by $r^4$ and integrate over $[0,r]$ to gain the following

$\displaystyle {r^4}{u^{(3)}}(r) = - \int_0^r {{s^4}u{{(s)}^{ - q}}ds} .$

Next, we divide both sides by $r^4$ and integrate over $[o,r]$ again to get

$\displaystyle u''(r) - u''(0)= - \int_0^r {\left( {\frac{1}{{{t^4}}}\int_0^t {{s^4}u{{(s)}^{ - q}}ds} } \right)dt} = \frac{1}{3}\int_0^r {\left( {\int_0^t {{s^4}u{{(s)}^{ - q}}ds} } \right)d(\frac{1}{{{t^3}}})} .$

By integration by parts, we get