# 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

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