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

Lemma.

\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).

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September 22, 2010

An identity of differentiation involving the Kelvin transform

Filed under: Các Bài Tập Nhỏ, Giải Tích 1 — Tags: — Ngô Quốc Anh @ 15:47

This short note is to prove the following

\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}.

The proof is straightforward as follows.

  • Calculation of \frac{\partial}{\partial x_1}.

We see that

\displaystyle\begin{gathered} \frac{\partial }{{\partial {x_1}}}\left( {u\left( {\frac{x}{{{{\left| x \right|}^2}}}} \right)} \right){x_1} = \frac{\partial }{{\partial {y_1}}}\left( {u\left( y \right)} \right)\frac{\partial }{{\partial {x_1}}}\left( {\frac{{{x_1}}}{{{{\left| x \right|}^2}}}} \right){x_1} + \frac{\partial }{{\partial {y_2}}}\left( {u\left( y \right)} \right)\frac{\partial }{{\partial {x_1}}}\left( {\frac{{{x_2}}}{{{{\left| x \right|}^2}}}} \right){x_1} \hfill \\ \qquad\qquad\qquad= \frac{\partial }{{\partial {y_1}}}\left( {u\left( y \right)} \right)\left( {\frac{1}{{{{\left| x \right|}^2}}} - \frac{{2x_1^2}}{{{{\left| x \right|}^4}}}} \right){x_1} + \frac{\partial }{{\partial {y_2}}}\left( {u\left( y \right)} \right)\left( { - \frac{{2{x_1}{x_2}}}{{{{\left| x \right|}^4}}}} \right){x_1}. \hfill \\ \end{gathered}

  • Calculation of \frac{\partial}{\partial x_2}.

Similarly, we get

\displaystyle\begin{gathered} \frac{\partial }{{\partial {x_2}}}\left( {u\left( {\frac{x}{{{{\left| x \right|}^2}}}} \right)} \right){x_2} = \frac{\partial }{{\partial {y_1}}}\left( {u\left( y \right)} \right)\frac{\partial }{{\partial {x_2}}}\left( {\frac{{{x_1}}}{{{{\left| x \right|}^2}}}} \right){x_2} + \frac{\partial }{{\partial {y_2}}}\left( {u\left( y \right)} \right)\frac{\partial }{{\partial {x_2}}}\left( {\frac{{{x_2}}}{{{{\left| x \right|}^2}}}} \right){x_2} \hfill \\ \qquad\qquad\qquad= \frac{\partial }{{\partial {y_1}}}\left( {u\left( y \right)} \right)\left( { - \frac{{2{x_1}{x_2}}}{{{{\left| x \right|}^4}}}} \right){x_2} + \frac{\partial }{{\partial {y_2}}}\left( {u\left( y \right)} \right)\left( {\frac{1}{{{{\left| x \right|}^2}}} - \frac{{2x_2^2}}{{{{\left| x \right|}^4}}}} \right){x_2}. \hfill \\ \end{gathered}

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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).

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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}}}.

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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}

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