Ngô Quốc Anh

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}

  • Calculation of \nabla_x \cdot x.

Thus

\displaystyle\begin{gathered} {\nabla _x}\left( {u\left( {\frac{x}{{{{\left| x \right|}^2}}}} \right)} \right) \cdot x = \frac{\partial }{{\partial {y_1}}}\left( {u\left( y \right)} \right){x_1}\left( {\frac{1}{{{{\left| x \right|}^2}}} - \frac{{2x_1^2}}{{{{\left| x \right|}^4}}}} \right) + \frac{\partial }{{\partial {y_2}}}\left( {u\left( y \right)} \right){x_1}\left( { - \frac{{2{x_1}{x_2}}}{{{{\left| x \right|}^4}}}} \right) +\\ \quad\qquad\qquad \frac{\partial }{{\partial {y_1}}}\left( {u\left( y \right)} \right){x_2}\left( { - \frac{{2{x_1}{x_2}}}{{{{\left| x \right|}^4}}}} \right) + \frac{\partial }{{\partial {y_2}}}\left( {u\left( y \right)} \right){x_2}\left( {\frac{1}{{{{\left| x \right|}^2}}} - \frac{{2x_2^2}}{{{{\left| x \right|}^4}}}} \right) \hfill \\ \qquad\qquad= \frac{\partial }{{\partial {y_1}}}\left( {u\left( y \right)} \right)\left[ {\frac{{{x_1}}}{{{{\left| x \right|}^2}}} - \frac{{2x_1^3}}{{{{\left| x \right|}^4}}} - \frac{{2{x_1}x_2^2}}{{{{\left| x \right|}^4}}}} \right] + \frac{\partial }{{\partial {y_2}}}\left( {u\left( y \right)} \right)\left[ {\frac{{{x_2}}}{{{{\left| x \right|}^2}}} - \frac{{2x_2^3}}{{{{\left| x \right|}^4}}} - \frac{{2x_1^2{x_2}}}{{{{\left| x \right|}^4}}}} \right] \hfill \\ \qquad\qquad= \frac{\partial }{{\partial {y_1}}}\left( {u\left( y \right)} \right)\left[ {\frac{{{x_1}}}{{{{\left| x \right|}^2}}} - 2{x_1}\left( {\frac{{x_1^2}}{{{{\left| x \right|}^4}}} + \frac{{x_2^2}}{{{{\left| x \right|}^4}}}} \right)} \right] + \frac{\partial }{{\partial {y_2}}}\left( {u\left( y \right)} \right)\left[ {\frac{{{x_2}}}{{{{\left| x \right|}^2}}} - 2{x_2}\left( {\frac{{x_2^2}}{{{{\left| x \right|}^4}}} + \frac{{x_1^2}}{{{{\left| x \right|}^4}}}} \right)} \right] \hfill \\ \qquad\qquad= \frac{\partial }{{\partial {y_1}}}\left( {u\left( y \right)} \right)\left[ {\frac{{{x_1}}}{{{{\left| x \right|}^2}}} - \frac{{2{x_1}}}{{{{\left| x \right|}^2}}}} \right] + \frac{\partial }{{\partial {y_2}}}\left( {u\left( y \right)} \right)\left[ {\frac{{{x_2}}}{{{{\left| x \right|}^2}}} - \frac{{2{x_2}}}{{{{\left| x \right|}^2}}}} \right]. \hfill \\ \end{gathered}

It is now easy to see 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.

The proof follows.

2 Comments »

  1. QA vẫn viết đều nhỉ:)

    Comment by Đặng Tuấn Hiệp — September 22, 2010 @ 18:16

    • Cảm ơn Hiệp, thay giấy nháp thôi à😉

      Comment by Ngô Quốc Anh — September 23, 2010 @ 0:55


RSS feed for comments on this post. TrackBack URI

Leave a Reply

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

WordPress.com Logo

You are commenting using your WordPress.com 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 )

Google+ photo

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

Connecting to %s

Create a free website or blog at WordPress.com.

%d bloggers like this: