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

2005_ds.pdf
2006_ds.pdf
2007_ds.pdf
2008_ds.pdf
2009_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.

Happy reading :).

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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

(more…)

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