Ngô Quốc Anh

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

Proof. Since $\displaystyle v(x) = |x{|^2}\left[ {\frac{1}{{|x{|^{n - 2}}}}u\left( {\frac{x}{{|x{|^2}}}} \right)} \right]$

by using the Kelvin transformation rule above together with the fact that $\displaystyle \Delta (fg) = g\Delta f + 2\nabla f \cdot \nabla g + f\Delta g$

we have $\displaystyle\begin{gathered} \Delta v = \left( {\Delta |x{|^2}} \right)\left( {\frac{1}{{|x{|^{n - 2}}}}u\left( {\frac{x}{{|x{|^2}}}} \right)} \right) + 2\nabla \left( {|x{|^2}} \right) \cdot \nabla \left( {\frac{1}{{|x{|^{n - 2}}}}u\left( {\frac{x}{{|x{|^2}}}} \right)} \right) + |x{|^2}\Delta \left( {\frac{1}{{|x{|^{n - 2}}}}u\left( {\frac{x}{{|x{|^2}}}} \right)} \right) \hfill \\ \qquad= 2n\frac{1}{{|x{|^{n - 2}}}}u\left( {\frac{x}{{|x{|^2}}}} \right) + 2\nabla \left( {|x{|^2}} \right) \cdot \nabla \left( {\frac{1}{{|x{|^{n - 2}}}}u\left( {\frac{x}{{|x{|^2}}}} \right)} \right) + \frac{1}{{|x{|^n}}}(\Delta u)\left( {\frac{x}{{|x{|^2}}}} \right). \hfill \\ \end{gathered}$

We now take $\Delta$ to each term on the right hand side. We first have $\displaystyle\Delta \left( {2n\frac{1}{{|x{|^{n - 2}}}}u\left( {\frac{x}{{|x{|^2}}}} \right)} \right) = \boxed{\frac{2n}{{|x{|^{n + 2}}}}(\Delta u)\left( {\frac{x}{{|x{|^2}}}} \right)}$.

Next $\displaystyle\begin{gathered} \Delta \left( {\frac{1}{{|x{|^n}}}(\Delta u)\left( {\frac{x}{{|x{|^2}}}} \right)} \right) = \Delta \left( {\frac{1}{{|x{|^2}}}\frac{1}{{|x{|^{n - 2}}}}(\Delta u)\left( {\frac{x}{{|x{|^2}}}} \right)} \right) \hfill \\ \qquad= \Delta \left( {\frac{1}{{|x{|^2}}}} \right)\frac{1}{{|x{|^{n - 2}}}}(\Delta u)\left( {\frac{x}{{|x{|^2}}}} \right) + 2\nabla \left( {\frac{1}{{|x{|^2}}}} \right) \cdot \nabla \left( {\frac{1}{{|x{|^{n - 2}}}}(\Delta u)\left( {\frac{x}{{|x{|^2}}}} \right)} \right) + \frac{1}{{|x{|^2}}}\Delta \left( {\frac{1}{{|x{|^{n - 2}}}}(\Delta u)\left( {\frac{x}{{|x{|^2}}}} \right)} \right). \hfill \\ \end{gathered}$

Note that $\displaystyle\Delta \left( {\frac{1}{{|x{|^2}}}} \right) = \frac{8}{{|x{|^4}}} - \frac{{2n}}{{|x{|^4}}}$

and $\displaystyle\nabla \left( {\frac{1}{{|x{|^2}}}} \right) = \frac{{ - 2x}}{{|x{|^4}}}$

so $\displaystyle\begin{gathered} \Delta \left( {\frac{1}{{|x{|^n}}}(\Delta u)\left( {\frac{x}{{|x{|^2}}}} \right)} \right) \hfill \\ \qquad= \left( {\frac{8}{{|x{|^4}}} - \frac{{2n}}{{|x{|^4}}}} \right)\frac{1}{{|x{|^{n - 2}}}}(\Delta u)\left( {\frac{x}{{|x{|^2}}}} \right) - \frac{4}{{|x{|^4}}}x \cdot \nabla \left( {\frac{1}{{|x{|^{n - 2}}}}(\Delta u)\left( {\frac{x}{{|x{|^2}}}} \right)} \right) + \frac{1}{{|x{|^2}}}\frac{1}{{|x{|^{n + 2}}}}({\Delta ^2}u)\left( {\frac{x}{{|x{|^2}}}} \right) \hfill \\\qquad = \boxed{\frac{8}{{|x{|^{n + 2}}}}(\Delta u)\left( {\frac{x}{{|x{|^2}}}} \right) - \frac{{2n}}{{|x{|^{n + 2}}}}(\Delta u)\left( {\frac{x}{{|x{|^2}}}} \right) - \frac{4}{{|x{|^4}}}x \cdot \nabla \left( {\frac{1}{{|x{|^{n - 2}}}}(\Delta u)\left( {\frac{x}{{|x{|^2}}}} \right)} \right) + \frac{1}{{|x{|^{n + 4}}}}({\Delta ^2}u)\left( {\frac{x}{{|x{|^2}}}} \right)}. \hfill \\ \end{gathered}$

Lastly, $\displaystyle\begin{gathered} \Delta \left( {2\nabla \left( {|x{|^2}} \right) \cdot \nabla \left( {\frac{1}{{|x{|^{n - 2}}}}u\left( {\frac{x}{{|x{|^2}}}} \right)} \right)} \right) \hfill \\ \qquad= \underbrace {2\nabla \left( {\Delta \left( {|x{|^2}} \right)} \right)}_0 \cdot \nabla \left( {\frac{1}{{|x{|^{n - 2}}}}u\left( {\frac{x}{{|x{|^2}}}} \right)} \right) + 4\nabla \nabla \left( {|x{|^2}} \right) \cdot \nabla \nabla \left( {\frac{1}{{|x{|^{n - 2}}}}u\left( {\frac{x}{{|x{|^2}}}} \right)} \right) + 2\nabla \left( {|x{|^2}} \right) \cdot \nabla \left( {\Delta \left( {\frac{1}{{|x{|^{n - 2}}}}u\left( {\frac{x}{{|x{|^2}}}} \right)} \right)} \right) \hfill \\\qquad = 8\nabla x \cdot \nabla \left[ {\nabla \left( {\frac{1}{{|x{|^{n - 2}}}}u\left( {\frac{x}{{|x{|^2}}}} \right)} \right)} \right] + 4x \cdot \nabla \left( {\frac{1}{{|x{|^{n + 2}}}}(\Delta u)\left( {\frac{x}{{|x{|^2}}}} \right)} \right). \hfill \\ \end{gathered}$

Note that, $\displaystyle\nabla x \cdot \nabla \left[ {\nabla \left( {\frac{1}{{|x{|^{n - 2}}}}u\left( {\frac{x}{{|x{|^2}}}} \right)} \right)} \right] = \Delta \left( {\frac{1}{{|x{|^{n - 2}}}}u\left( {\frac{x}{{|x{|^2}}}} \right)} \right) = \frac{1}{{|x{|^{n + 2}}}}(\Delta u)\left( {\frac{x}{{|x{|^2}}}} \right)$

and $\displaystyle\begin{gathered} \nabla \left( {\frac{1}{{|x{|^{n + 2}}}}(\Delta u)\left( {\frac{x}{{|x{|^2}}}} \right)} \right) = \nabla \left( {\frac{1}{{|x{|^4}}}\frac{1}{{|x{|^{n - 2}}}}(\Delta u)\left( {\frac{x}{{|x{|^2}}}} \right)} \right) \hfill \\ \qquad= \nabla \left( {\frac{1}{{|x{|^4}}}} \right)\frac{1}{{|x{|^{n - 2}}}}(\Delta u)\left( {\frac{x}{{|x{|^2}}}} \right) + \frac{1}{{|x{|^4}}}\nabla \left( {\frac{1}{{|x{|^{n - 2}}}}(\Delta u)\left( {\frac{x}{{|x{|^2}}}} \right)} \right) \hfill \\ \qquad= \frac{{ - 4x}}{{|x{|^6}}}\frac{1}{{|x{|^{n - 2}}}}(\Delta u)\left( {\frac{x}{{|x{|^2}}}} \right) + \frac{1}{{|x{|^4}}}\nabla \left( {\frac{1}{{|x{|^{n - 2}}}}(\Delta u)\left( {\frac{x}{{|x{|^2}}}} \right)} \right) \hfill \\ = - 4x\frac{1}{{|x{|^{n + 4}}}}(\Delta u)\left( {\frac{x}{{|x{|^2}}}} \right) + \frac{1}{{|x{|^4}}}\nabla \left( {\frac{1}{{|x{|^{n - 2}}}}(\Delta u)\left( {\frac{x}{{|x{|^2}}}} \right)} \right). \hfill \\ \end{gathered}$

Thus $\displaystyle\begin{gathered} \Delta \left( {2\nabla \left( {|x{|^2}} \right) \cdot \nabla \left( {\frac{1}{{|x{|^{n - 2}}}}u\left( {\frac{x}{{|x{|^2}}}} \right)} \right)} \right) \hfill \\ \qquad= \frac{8}{{|x{|^{n + 2}}}}(\Delta u)\left( {\frac{x}{{|x{|^2}}}} \right) + 4x \cdot \left[ { - 4x\frac{1}{{|x{|^{n + 4}}}}(\Delta u)\left( {\frac{x}{{|x{|^2}}}} \right) + \frac{1}{{|x{|^4}}}\nabla \left( {\frac{1}{{|x{|^{n - 2}}}}(\Delta u)\left( {\frac{x}{{|x{|^2}}}} \right)} \right)} \right] \hfill \\ \qquad= \boxed{- \frac{8}{{|x{|^{n + 2}}}}(\Delta u)\left( {\frac{x}{{|x{|^2}}}} \right) + \frac{4}{{|x{|^4}}}x \cdot \nabla \left( {\frac{1}{{|x{|^{n - 2}}}}(\Delta u)\left( {\frac{x}{{|x{|^2}}}} \right)} \right)}. \hfill \\ \end{gathered}$ $\displaystyle {\Delta ^2}v = \frac{1}{{|x{|^{n + 4}}}}({\Delta ^2}u)\left( {\frac{x}{{|x{|^2}}}} \right)$.

In general we have the following

Theorem. If $u$ is a sufficiently good function then $v$ satisfies the equation $\displaystyle \Delta^p v = \frac{1}{|x|^{n+2p}}(\Delta^p u)\left(\frac{x}{|x|^2}\right)$

where $v$ is defined to be $\displaystyle v(x)=\frac{1}{|x|^{n-2p}}u\left(\frac{x}{|x|^2}\right)$.