Ngô Quốc Anh

March 29, 2011

The (original) Picone identity

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 17:02

For differentiable functions v > 0 and u \geqslant 0, the following Picone’s identity is well known

\displaystyle {\left| {\nabla u - \frac{u}{v}\nabla v} \right|^2} = {\left| {\nabla u} \right|^2} - 2\frac{u}{v}\nabla u \cdot \nabla v + \frac{{{u^2}}}{{{v^2}}}{\left| {\nabla v} \right|^2} = {\left| {\nabla u} \right|^2} - \nabla \left( {\frac{{{u^2}}}{v}} \right) \cdot \nabla v \geqslant 0.

The proof is very simple. For each partial derivative \frac{\partial}{\partial x_i} we have

\displaystyle\frac{\partial }{{\partial {x_i}}}\left( {\frac{{{u^2}}}{v}} \right) = \frac{1}{{{v^2}}}\left[ {\frac{{\partial ({u^2})}}{{\partial {x_i}}}v - {u^2}\frac{{\partial v}}{{\partial {x_i}}}} \right] = \frac{1}{{{v^2}}}\left[ {2u\frac{{\partial u}}{{\partial {x_i}}}v - {u^2}\frac{{\partial v}}{{\partial {x_i}}}} \right]

which implies

\displaystyle\nabla \left( {\frac{{{u^2}}}{v}} \right) = \frac{1}{{{v^2}}}\left[ {2uv\nabla u - {u^2}\nabla v} \right] = \frac{{2u}}{v}\nabla u - \frac{{{u^2}}}{{{v^2}}}\nabla v.


\displaystyle - \nabla \left( {\frac{{{u^2}}}{v}} \right) \cdot \nabla v = - \left[ {\frac{{2u}}{v}\nabla u - \frac{{{u^2}}}{{{v^2}}}\nabla v} \right] \cdot \nabla v = - 2\frac{u}{v}\nabla u \cdot \nabla v + \frac{{{u^2}}}{{{v^2}}}{\left| {\nabla v} \right|^2}.

The Picone identity is very useful. We shall address this later on.

Leave a Comment »

No comments yet.

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.

Create a free website or blog at

%d bloggers like this: