Ngô Quốc Anh

May 1, 2010

A useful identity in a book due to L. Ahlfors

Filed under: Các Bài Tập Nhỏ, Giải Tích 5, Nghiên Cứu Khoa Học — Tags: — Ngô Quốc Anh @ 3:12

Let \mathbf{x},\mathbf{y} be points in \mathbb R^n. If we denote by \mathbf{x}^\sharp the reflection point of \mathbf{x} with respect to the unit ball, i.e.

\displaystyle \mathbf{x}^\sharp = \frac{\mathbf{x}}{|\mathbf{x}|^2}

we then have the following well-known identity

\displaystyle |\mathbf{x}|\left| {{\mathbf{x}^\sharp } - \mathbf{y}} \right| = |\mathbf{y}|\left| {{\mathbf{y}^\sharp } - \mathbf{x}} \right|.

The proof of the above identity comes from the fact that

\displaystyle |\mathbf{x}|\left| {\frac{\mathbf{x}}{{|\mathbf{x}{|^2}}} - \mathbf{y}} \right| = \sqrt {1 + |\mathbf{x}{|^2}|\mathbf{y}|^2 - 2\mathbf{x} \cdot \mathbf{y}} = |\mathbf{y}|\left| {\frac{\mathbf{y}}{{|\mathbf{y}|^2}} - \mathbf{x}} \right|.

Indeed, by squaring both sides of

\displaystyle |\mathbf{x}|\left| {\frac{\mathbf{x}}{{|\mathbf{x}{|^2}}} - \mathbf{y}} \right| = \sqrt  {1 + |\mathbf{x}|^2|\mathbf{y}|^2 - 2\mathbf{x} \cdot \mathbf{y}}

we arrive at

\displaystyle |\mathbf{x}|^2\left( {\frac{{|\mathbf{x}|^2}}{{|\mathbf{x}|^4}} - 2\frac{{\mathbf{x} \cdot \mathbf{y}}}{{|\mathbf{x}|^2}} + |\mathbf{y}|^2} \right) = 1 + |\mathbf{x}|^2|\mathbf{y}|^2 - 2\mathbf{x} \cdot \mathbf{y}

which is obviously true. Similarly, the last identity also holds. If we replace \mathbf{y} by -\mathbf{y} we also have

\displaystyle |\mathbf{x}|\left| {{\mathbf{x}^\sharp }+ \mathbf{y}} \right| = |\mathbf{y}|\left|  {{\mathbf{y}^\sharp } + \mathbf{x}} \right|.

Generally, if we consider the reflection point of \mathbf{x} over a ball B_r(0), i.e.

\displaystyle \mathbf{x}^\sharp = \frac{r^2\mathbf{x}}{|\mathbf{x}|^2}

we still have the fact

\displaystyle |\mathbf{x}|\left| {{\mathbf{x}^\sharp } - \mathbf{y}} \right| = |\mathbf{y}|\left|  {{\mathbf{y}^\sharp } - \mathbf{x}} \right|.

Indeed, one gets

\displaystyle |\mathbf{x}|\left| {\frac{{{r^2}\mathbf{x}}}{{|\mathbf{x}{|^2}}} - \mathbf{y}} \right| = {r^2}|\mathbf{x}|\left| {\frac{\mathbf{x}}{{|\mathbf{x}{|^2}}} - \frac{\mathbf{y}}{{{r^2}}}} \right| = {r^2}\left| {\frac{\mathbf{y}}{{{r^2}}}} \right|\left| {\frac{{\frac{\mathbf{y}}{{{r^2}}}}}{{{{\left| {\frac{\mathbf{y}}{{{r^2}}}} \right|}^2}}} - \mathbf{x}} \right| = \left| \mathbf{y} \right|\left| {\frac{{{r^2}\mathbf{y}}}{{|\mathbf{y}{|^2}}} - \mathbf{x}} \right|.

Similarly,

\displaystyle |\mathbf{x}|\left| {{\mathbf{x}^\sharp } + y} \right| = |y|\left|   {{y^\sharp } + \mathbf{x}} \right|.

Such identity is very useful. For example, in \mathbb R^n (n\geqslant 3) the following holds

\displaystyle\iint\limits_{{\mathbf{x}} = r} {\frac{{d{\sigma _{\mathbf{x}}}}}{{{{\left| {{\mathbf{x}} - {\mathbf{y}}} \right|}^{n - 2}}}}} = \min \left\{ {\frac{1}{{|{\mathbf{y}}|^{n - 2}}},\frac{1}{r^{n - 2}}} \right\}.

This type of formula has been considered before when n=3 here. For a general case, Lieb and Loss introduced another method in their book published by AMS in 2001. Here we introduce a completely new proof. At first, if |\mathbf{y}|>r by the potential theory, one easily gets

\displaystyle\iint\limits_{{\mathbf{x}} = r} {\frac{{d{\sigma _{\mathbf{x}}}}}{{{{\left| {{\mathbf{x}} - {\mathbf{y}}} \right|}^{n - 2}}}}} = \frac{1}{{|{\mathbf{y}}|^{n - 2}}}.

If |\mathbf{y}|<r, one needs to make use of the reflection point of \mathbf{y} and the above identity to go back to the first case. The point here is |\mathbf{y}^\sharp|>r. The integral is obviously continuous as a function of \mathbf{y}. The above argument is due to professor X.X.W.

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April 29, 2010

Surface integrals over a sphere when its radius varies

Filed under: Các Bài Tập Nhỏ, Giải Tích 5, Giải Tích Cổ Điển — Tags: — Ngô Quốc Anh @ 1:53

Let us consider the following integral in \mathbb R^3

\displaystyle\iint\limits_{|{\mathbf{x}}| = t} {f({\mathbf{x}})d{\sigma _{\mathbf{x}}}}

when t varies.

Obviously, the sphere |{\mathbf{x}}| = t can be parametrized as the following

\displaystyle {\mathbf{x}} = t\left( {\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta } \right)

so

\displaystyle\begin{gathered} \iint\limits_{|{\mathbf{x}}| = t} {f({\mathbf{x}})d{\sigma _{\mathbf{x}}}} = \iint\limits_{\begin{subarray}{c} 0 \leqslant \theta \leqslant \pi \\ 0 \leqslant \varphi \leqslant 2\pi \end{subarray}} {f({\mathbf{x}}(\theta ,\varphi )){t^2}\sin \theta d\theta d\varphi } \hfill \\ \qquad= {t^2}\iint\limits_{\begin{subarray}{c} 0 \leqslant \theta \leqslant \pi \\ 0 \leqslant \varphi \leqslant 2\pi \end{subarray}} {f({\mathbf{x}}(\theta ,\varphi ))\sin \theta d\theta d\varphi } \hfill \\ \qquad= {t^2}\iint\limits_{|{\mathbf{x}}| = 1} {f \left(t\mathbf{x} \right)d{\sigma _{\mathbf{x}}}}. \hfill \\ \end{gathered}

If we wish to work on the average, the formula is much simpler than that, precisely

\displaystyle\frac{1}{{4\pi {t^2}}}\iint\limits_{|{\mathbf{x}}| = t} {f({\mathbf{x}})d{\sigma _{\mathbf{x}}}} = \frac{1}{{4\pi }}\iint\limits_{|{\mathbf{x}}| = 1} {f(t{\mathbf{x}})d{\sigma _{\mathbf{x}}}}

that means

\displaystyle \overline {\iint\limits_{|{\mathbf{x}}| = t} } f({\mathbf{x}})d{\sigma  _{\mathbf{x}}}=\overline {\iint\limits_{|{\mathbf{x}}| = 1} } f({t\mathbf{x}})d{\sigma  _{\mathbf{x}}}

where the bar means the average.

More general, we get

\displaystyle\iint\limits_{|{\mathbf{x}}| = {t_1}} {f({\mathbf{x}})d{\sigma _{\mathbf{x}}}} = {\left( {\frac{{{t_1}}}{{{t_2}}}} \right)^2}\iint\limits_{|{\mathbf{x}}| = {t_2}} {f\left( {\frac{{{t_1}}}{{{t_2}}}{\mathbf{x}}} \right)d{\sigma _{\mathbf{x}}}}.

April 27, 2010

Surface integral: The symmetric property, 2

Filed under: Các Bài Tập Nhỏ, Giải Tích 5 — Tags: — Ngô Quốc Anh @ 16:07

Today, we try to evaluate the following surface integral in \mathbb R^3. I found this result in a paper published in Math. Z. 198, 277-289 (1988). This topic can be considered as a continued part to the following topic.

Proposition. Let f(x) be a continuous function. Then we have

\displaystyle\frac{1}{t}\iint\limits_{|{\mathbf{y}} -  {\mathbf{x}}| = t} {f(|{\mathbf{y}}|)d{\sigma _{\mathbf{y}}}} =  \frac{{2\pi }}{r}\int\limits_{|r - t|}^{r + t} {sf(s)ds}

here we denote r=|\mathbf{x}|.

Proof. Remark that the integral only depends on |\mathbf{x}|, so that we may assume \mathbf{x} = (0, 0, r) and introduce the following spherical coordinate

\mathbf{y}=\mathbf{x}+t\omega

where

\displaystyle \omega = (\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta ).

This system of coordinates can be seen via the picture below

(more…)

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