# Ngô Quốc Anh

## 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

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

Now Laplacian of a function can be computed as following

$\displaystyle\Delta u = \frac {1} {{\prod\limits_{k = 1}^n {{h_k}} }}\sum\limits_{j = 1}^n {\frac {\partial } {{\partial {x_j}}}\left( {\frac {{\prod\limits_{k = 1}^n {{h_k}} }} {{h_j^2}}\frac {{\partial u}} {{\partial {\xi _j}}}} \right)}$

which implies that

$\displaystyle\Delta u\left( x \right) = {\left| \xi \right|^{2n}}\sum\limits_{j = 1}^n {\frac {\partial } {{\partial {\xi _j}}}} \left( {\frac {1} {{{{\left| \xi \right|}^{2n - 4}}}}\frac {{\partial u}} {{\partial {\xi _j}}}} \right)$.

Next using the rule

$\displaystyle\Delta \left( {uv} \right) = v\Delta u + 2\nabla u \cdot \nabla v + u\Delta v$

we get

$\displaystyle\begin{gathered} \frac{1}{{{{\left| \xi \right|}^{2 - n}}}}\frac{\partial }{{\partial {\xi _j}}}\left( {\frac{1}{{{{\left| \xi \right|}^{2n - 4}}}}\frac{{\partial u}}{{\partial {\xi _j}}}} \right) = 2\frac{{\partial u}}{{\partial {\xi _j}}}\frac{\partial }{{\partial {\xi _j}}}\left( {\frac{1}{{{{\left| \xi \right|}^{n - 2}}}}} \right) + \frac{1}{{{{\left| \xi \right|}^{n - 2}}}}\frac{{{\partial ^2}u}}{{\partial \xi _j^2}} \hfill \\ \qquad= \frac{{{\partial ^2}}}{{\partial \xi _j^2}}\left( {\frac{1}{{{{\left| \xi \right|}^{n - 2}}}}u} \right) - u\frac{{{\partial ^2}}}{{\partial \xi _j^2}}\left( {\frac{1}{{{{\left| \xi \right|}^{n - 2}}}}} \right) \hfill \\ \end{gathered}$

and finally conclude with the help of

$\displaystyle\Delta \left( {\frac {1} {{{{\left| \xi \right|}^{n - 2}}}}} \right) = \sum\limits_{j = 1}^n {\frac {{{\partial ^2}}} {{\partial \xi _j^2}}\left( {\frac {1} {{{{\left| \xi \right|}^{n - 2}}}}} \right)} = 0$

that

$\displaystyle\begin{gathered} \Delta u\left( x \right) = {\left| \xi \right|^{n + 2}}\sum\limits_{j = 1}^n {\frac{{{\partial ^2}}}{{\partial \xi _j^2}}} \left( {\frac{1}{{{{\left| \xi \right|}^{n - 2}}}}u} \right) = \frac{1}{{{{\left| x \right|}^{n + 2}}}}\sum\limits_{j = 1}^n {\frac{{{\partial ^2}{u^\sharp }}}{{\partial \xi _j^2}}} \hfill \\ \qquad= \frac{1}{{{{\left| x \right|}^{n + 2}}}}\Delta {u^\sharp }\left( \xi \right) = \frac{1}{{{{\left| x \right|}^{n + 2}}}}\Delta {u^\sharp }\left( {{x^\sharp }} \right). \hfill \\ \end{gathered}$

Therefore,

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

Finally,

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

This property is very useful. Actually, we can prove a more general formula by replacing $\frac{1}{|x|^{n-2}}$ to a suitable term.

1. Hi Ngo,
Do you know if I can use the Kelvin transform for the $n$-laplacian in $\mathbb R^n$?
in general, for the $p$-laplacian in $\mathbb R^n$ with $p$ not equal to $2$ is not longer available. But I believe that for the $n$-laplacian it holds.
Do you Know more?
FAB

Comment by Fab — March 7, 2013 @ 16:59

• Dear Fab, I am sorry but I don’t have any information about the case $p=n$. In case I have some idea, I will blog it.

Comment by Ngô Quốc Anh — March 27, 2013 @ 10:03

2. Dear Ngo, very nice proof! Do you in which book I could find this proof? Thank you.

Comment by João Henrique Santos de Andrade — May 23, 2020 @ 3:13

• Hi,

thanks for your interest in the post. For the case of $\Delta^2$, you can find a detailed proof at

https://anhngq.wordpress.com/2010/04/16/kelvin-transform-biharmonic/

There is also a general formula for $\Delta^m$. I think the formula for the higher-order case can be proved by induction.

Comment by Ngô Quốc Anh — May 23, 2020 @ 6:54

This site uses Akismet to reduce spam. Learn how your comment data is processed.