# 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}|, 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.

## 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 