Ngô Quốc Anh

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}


  1. em không biết điều kiện thứ nhất dùng để làm gì mà vẫn ra đáp án? Bài này e giải bằng biến đổi Furier.nhờ thầy giúp đỡ
    Giải phương trình

    u_{tt}+a^2u_{xxxx}=0, \quad -\infty \leq x<0

    với điều kiện biên và điều kiện ban đầu:

    \begin{array}{lcl} u(0,t) &=& 0,\quad\forall t > 0, \hfill \\ u(x,0) &=& \varphi (x),\quad0 < x < \infty , \hfill \\ {u_t}(x,0) &=& \psi (x),\quad0 < x < \infty .\end{array}

    Comment by TriPhuong — February 5, 2013 @ 18:58

    • Em xem lại đề như trên đã chuẩn chưa nhé, hình như có cái gì đó ko ổn.

      Comment by Ngô Quốc Anh — February 5, 2013 @ 19:14

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

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

Blog at

%d bloggers like this: