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


\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).

Proof. Writing

\displaystyle w(x) = {\left| x \right|^{ - \frac{{n - 2}}{2}}}u\left( {\frac{{{\lambda ^2}x}}{{{{\left| x \right|}^2}}}} \right)

we have

\displaystyle {\left| x \right|^{\frac{{n + 2}}{2}}}{\Delta _x}w(x) = {\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) - \frac{{{{(n - 2)}^2}}}{4}u\left( {\frac{{{\lambda ^2}x}}{{{{\left| x \right|}^2}}}} \right).

Similarly, writing

\displaystyle W(y) = {\left| y \right|^{ - \frac{{n - 2}}{2}}}u\left( y \right)

we have

\displaystyle {\left| y \right|^{\frac{{n + 2}}{2}}}{\Delta _y}W(y) = {\left| y \right|^2}{\Delta _y}u\left( y \right) - (n - 2)y \cdot {\nabla _y}u\left( y \right) - \frac{{{{(n - 2)}^2}}}{4}u\left( y \right).


\displaystyle |y|=\frac{\lambda^2}{|x|}

it follows that

\displaystyle {\left| x \right|^{\frac{{n + 2}}{2}}}{\Delta _x}w(x) = u\left( {\frac{{{\lambda ^2}x}}{{{{\left| x \right|}^2}}}} \right) = u(y) = {\left| y \right|^{\frac{{n + 2}}{2}}}{\Delta _y}W(y) = {\left( {\frac{{{\lambda ^2}}}{{|x|}}} \right)^{\frac{{n - 2}}{2}}}W(y).

Then we obtain

\displaystyle W(y) = {\left( {\frac{{|x|}}{\lambda }} \right)^{n - 2}}w(x).

By the property of the Kelvin transformation, we obtain

\displaystyle {\Delta _y}W(y) = {\left( {\frac{{|x|}}{\lambda }} \right)^{n - 2}}{\Delta _x}w(x).

Then we have

\displaystyle {\left| x \right|^{\frac{{n + 2}}{2}}}{\Delta _x}w(x) = {\left( {\frac{{{\lambda ^2}}}{{|x|}}} \right)^{\frac{{n + 2}}{2}}}{\left( {\frac{{|x|}}{\lambda }} \right)^{n - 2}}{\Delta _x}w(x) = {\left| y \right|^{\frac{{n + 2}}{2}}}{\Delta _y}W(y).

This completes the proof. For more details, we refer the reader to this paper [here].


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