# Ngô Quốc Anh

## October 3, 2010

### An identity of differentiation involving the Kelvin transform, 2

Filed under: Giải Tích 1 — Tags: — Ngô Quốc Anh @ 11:36

I found the following interesting identity which is similar to what I have showed recent days [here]. In that entry, we showed that

$\displaystyle {\nabla _x}\left( {u\left( {\frac{x}{{{{\left| x \right|}^2}}}} \right)} \right) \cdot x = - {\nabla _y}\left( {u\left( y \right)} \right) \cdot y$

where $x$ and $y$ are connected by

$\displaystyle y = \frac{x}{{{{\left| x \right|}^2}}} \in {\mathbb{R}^2}$.

For $\lambda>0$ we denote by $y$ the following

$\displaystyle y = \frac{\lambda^2 x}{{{{\left| x \right|}^2}}} \in {\mathbb{R}^n}$.

Then we show that

Lemma.

$\displaystyle {\left| x \right|^2}{\Delta _x}u\left( {\frac{{{\lambda ^2}x}}{{{{\left| x \right|}^2}}}} \right) - (n - 2)x \cdot {\nabla _x}u\left( {\frac{{{\lambda ^2}x}}{{{{\left| x \right|}^2}}}} \right) = {\left| y \right|^2}{\Delta _y}u\left( y \right) - (n - 2)y \cdot {\nabla _y}u\left( y \right)$.

## September 22, 2010

### An identity of differentiation involving the Kelvin transform

Filed under: Các Bài Tập Nhỏ, Giải Tích 1 — Tags: — Ngô Quốc Anh @ 15:47

This short note is to prove the following

$\displaystyle {\nabla _x}\left( {u\left( {\frac{x}{{{{\left| x \right|}^2}}}} \right)} \right) \cdot x = - {\nabla _y}\left( {u\left( y \right)} \right) \cdot y$

where $x$ and $y$ are connected by

$\displaystyle y = \frac{x}{{{{\left| x \right|}^2}}} \in {\mathbb{R}^2}$.

The proof is straightforward as follows.

• Calculation of $\frac{\partial}{\partial x_1}$.

We see that

$\displaystyle\begin{gathered} \frac{\partial }{{\partial {x_1}}}\left( {u\left( {\frac{x}{{{{\left| x \right|}^2}}}} \right)} \right){x_1} = \frac{\partial }{{\partial {y_1}}}\left( {u\left( y \right)} \right)\frac{\partial }{{\partial {x_1}}}\left( {\frac{{{x_1}}}{{{{\left| x \right|}^2}}}} \right){x_1} + \frac{\partial }{{\partial {y_2}}}\left( {u\left( y \right)} \right)\frac{\partial }{{\partial {x_1}}}\left( {\frac{{{x_2}}}{{{{\left| x \right|}^2}}}} \right){x_1} \hfill \\ \qquad\qquad\qquad= \frac{\partial }{{\partial {y_1}}}\left( {u\left( y \right)} \right)\left( {\frac{1}{{{{\left| x \right|}^2}}} - \frac{{2x_1^2}}{{{{\left| x \right|}^4}}}} \right){x_1} + \frac{\partial }{{\partial {y_2}}}\left( {u\left( y \right)} \right)\left( { - \frac{{2{x_1}{x_2}}}{{{{\left| x \right|}^4}}}} \right){x_1}. \hfill \\ \end{gathered}$

• Calculation of $\frac{\partial}{\partial x_2}$.

Similarly, we get

$\displaystyle\begin{gathered} \frac{\partial }{{\partial {x_2}}}\left( {u\left( {\frac{x}{{{{\left| x \right|}^2}}}} \right)} \right){x_2} = \frac{\partial }{{\partial {y_1}}}\left( {u\left( y \right)} \right)\frac{\partial }{{\partial {x_2}}}\left( {\frac{{{x_1}}}{{{{\left| x \right|}^2}}}} \right){x_2} + \frac{\partial }{{\partial {y_2}}}\left( {u\left( y \right)} \right)\frac{\partial }{{\partial {x_2}}}\left( {\frac{{{x_2}}}{{{{\left| x \right|}^2}}}} \right){x_2} \hfill \\ \qquad\qquad\qquad= \frac{\partial }{{\partial {y_1}}}\left( {u\left( y \right)} \right)\left( { - \frac{{2{x_1}{x_2}}}{{{{\left| x \right|}^4}}}} \right){x_2} + \frac{\partial }{{\partial {y_2}}}\left( {u\left( y \right)} \right)\left( {\frac{1}{{{{\left| x \right|}^2}}} - \frac{{2x_2^2}}{{{{\left| x \right|}^4}}}} \right){x_2}. \hfill \\ \end{gathered}$

## April 16, 2010

### Kelvin transform: Biharmonic

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 0:36

In this topic, we consider the Kelvin transform for Laplaction operators. Precisely, what we get is the following

$\displaystyle\Delta \left( {\frac {1} {{{{\left| x \right|}^{n - 2}}}}u\left( {\frac {x} {{{{\left| x \right|}^2}}}} \right)} \right){\text{ }} = \Delta {u^\sharp }(x) = {| {{x^\sharp }} |^{n + 2}}(\Delta u)\left( {{x^\sharp }} \right) = \frac {1} {{{{\left| x \right|}^{n + 2}}}}(\Delta u)\left( {\frac {x} {{{{\left| x \right|}^2}}}} \right)$.

We now consider a different situation. The detail can be found in the following paper due to X.W. Xu published in Proc. Roy. Soc. Edinburgh Sect. A, 2000.

Theorem. If $u$ is a sufficiently good function then $v$ satisfies the equation

$\displaystyle \Delta^2 v = \frac{1}{|x|^{n+4}}(\Delta^2 u)\left(\frac{x}{|x|^2}\right)$

where $v$ is defined to be

$\displaystyle v(x)=|x|^{4-n}u\left(\frac{x}{|x|^2}\right)$.

## April 9, 2010

### Kelvin transform: Laplacian

Filed under: Các Bài Tập Nhỏ, Linh Tinh, Nghiên Cứu Khoa Học, PDEs — Tags: — Ngô Quốc Anh @ 0:01

For each point $x \ne 0$, denote $x=(x_1,...,x_n)$ and

$\displaystyle \xi = {x^\sharp } = \left( {\frac {{{x_1}}} {{{{\left| x \right|}^2}}},...,\frac {{{x_n}}} {{{{\left| x \right|}^2}}}} \right)$

is the inversion of  $x$ with respect to the unit sphere. We have the following identities

$\displaystyle\frac {{\partial {\xi _j}}} {{\partial {x_k}}} = \frac {1} {{{{\left| x \right|}^2}}}\left( {{\delta _{jk}} - 2\frac {{{x_j}{x_k}}} {{{{\left| x \right|}^2}}}} \right)$

and

$\displaystyle\sum\limits_{l = 1}^n {\frac {{\partial {\xi _l}}} {{\partial {x_j}}}\frac {{\partial {\xi _l}}} {{\partial {x_k}}}} = \frac {1} {{{{\left| x \right|}^4}}}{\delta _{jk}}$.

Thus,

$\displaystyle\sum\limits_{l = 1}^n {\frac {{\partial {x_l}}} {{\partial {\xi _j}}}\frac {{\partial {x_l}}} {{\partial {\xi _k}}}} = \frac {1} {{{{\left| \xi \right|}^4}}}{\delta _{jk}}$.

Next, think of $\xi$ as a system of orthogonal curvilinear coordinates for $x$, we deduce that the metric tensor of the Euclidean space in curvilinear coordinates

$\displaystyle {g_{j,k}}\left( {{\xi _j},{\xi _k}} \right) = \sum\limits_{l = 1}^n {\frac {{\partial {x_l}}} {{\partial {\xi _j}}}\frac {{\partial {x_l}}} {{\partial {\xi _k}}}} = \frac {1} {{{{\left| \xi \right|}^4}}}{\delta _{jk}}$.

This implies that the so-called Lame coefficients is

$\displaystyle {h_j} = \sqrt {{g_{j,j}}\left( {{\xi _j},{\xi _j}} \right)} = \frac {1} {{{{\left| \xi \right|}^2}}}$.

## February 25, 2010

### Double Kelvin transform being the identity map

In this topic, we proved a very interesting property involving the Laplacian of the Kelvin transform of a function. Recall that, for a given function $u$, its Kelvin transform is defined to be

$\displaystyle {u^\sharp }(x) = \frac{1}{{{{\left| x \right|}^{n - 2}}}}u\left( {\frac{x}{{{{\left| x \right|}^2}}}} \right)$.

We then have

$\displaystyle\Delta {u^\sharp }(x) = |{x^\sharp }{|^{n + 2}}\Delta u\left( {{x^\sharp }} \right)$

where the inversion point $x^\sharp$ of $x$ is defined to be

$\displaystyle {x^\sharp } = \frac{x}{{\left| x \right|^2}}$.

Therefore, the Kelvin transform can be defined to be

$\displaystyle {u^\sharp }({x^\sharp }) = \frac{1}{{{{\left| {{x^\sharp }} \right|}^{n - 2}}}}u\left( x \right)$.

We also have another formula

$\displaystyle\Delta {u^\sharp }({x^\sharp }) = |{({x^\sharp })^\sharp }{|^{n + 2}}\Delta u\left( {{{({x^\sharp })}^\sharp }} \right) = {\left| x \right|^{n + 2}}\Delta u(x)$.

The right hand side of the above identity involves $\Delta u(x)$, we can rewrite this one in terms of $(\Delta u)^\sharp$. Actually, we have the following

$\displaystyle {(\Delta u)^\sharp }({x^\sharp }) = \frac{1}{{{{\left| {{x^\sharp }} \right|}^{n -2}}}}\Delta u(x)$

which gives

$\displaystyle\Delta {u^\sharp }({x^\sharp }) = {\left| x \right|^{n + 2}}\Delta u(x) = {\left| x \right|^{n + 2}}{\left| {{x^\sharp }} \right|^{n - 2}}{(\Delta u)^\sharp }({x^\sharp }) =|x|^4 {(\Delta u)^\sharp }({x^\sharp })$.

Today, we study the a more general form of the Kelvin transform. The above definition of the Kelvin transform is with respect to the origin and unit ball in $\mathbb R^n$. We are now interested in the case when the Kelvin transform is defined with respect to a fixed ball. The following result is adapted from a paper due to M.C. Leung published in Math. Ann. in 2003.

Define the reflection on the sphere with center at $\xi$ and radius $a>0$ by

$\displaystyle {R_{\xi ,a}}(x) = \xi + {a^2}\frac{{x - \xi }}{{{{\left| {x - \xi } \right|}^2}}}, \quad x \ne \xi$.

It is direct to check that

$\displaystyle {R_{\xi ,a}}({R_{\xi ,a}}(x)) = x,\quad \forall x \ne \xi$.

We observe that

$\displaystyle \xi + {a^2}\frac{{x - \xi }}{{{{\left| {x - \xi } \right|}^2}}} \ne \xi , \quad \forall x \ne \xi$.

The  Kelvin transform with center at $\xi$ and radius $a>0$ of a function $u$ is given by

$\displaystyle u_{\xi ,a}^\sharp (x) = \frac{{{a^{n - 2}}}}{{{{\left| {x - \xi } \right|}^{n - 2}}}}u\left( {\xi + {a^2}\frac{{x - \xi }}{{{{\left| {x - \xi } \right|}^2}}}} \right), \quad \forall x \ne \xi$.

We remark that if $\xi$ and $a$ are fixed, and if $u(y)\gg 1$, then $u_{\xi ,a}^\sharp (x) \gg 1$ as well, where $y=R_{\xi, a}(x)$. In addition, $R_{\xi, a}$ sends a set of small diameter not too close to $\xi$ to a set of small diameter. We verify that double Kelvin transform is the identity map. We have

$\displaystyle\begin{gathered} (u_{\xi ,a}^\sharp )_{\xi ,a}^\sharp (x) = \frac{{{a^{n - 2}}}}{{{{\left| {x - \xi } \right|}^{n - 2}}}}u_{\xi ,a}^\sharp \left( {\xi + {a^2}\frac{{x - \xi }}{{{{\left| {x - \xi } \right|}^2}}}} \right) \hfill \\ \qquad\qquad= \frac{{{a^{n - 2}}}}{{{{\left| {x - \xi } \right|}^{n - 2}}}}\frac{{{a^{n - 2}}}}{{\frac{{{a^{2(n - 2)}}}}{{{{\left| {x - \xi } \right|}^{n - 2}}}}}}u\left( {\xi + {a^2}\frac{{{a^2}\frac{{x - \xi }}{{{{\left| {x - \xi } \right|}^2}}}}}{{\frac{{{a^4}}}{{{{\left| {x - \xi } \right|}^2}}}}}} \right) \hfill \\ \qquad\qquad= u(x), \quad\forall x \ne \xi . \hfill \\ \end{gathered}$