# 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

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

Thus

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

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

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