# Ngô Quốc Anh

## April 27, 2011

### The Picone identity for p-Laplacian

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 19:37

Last time we discussed the Picone identity for general purpose [here]. Now we present a generalization of this for $p$-Laplacian operator. This can be seen from the a paper by Walter Allegretto and Yin Xi Huang [here].

Let $v > 0$, $u \geqslant 0$ be differentiable over a domain $\Omega$. Denote

$\displaystyle L(u,v) = {\left| {\nabla u} \right|^p} - p{\left( {\frac{u}{v}} \right)^{p = 1}}\nabla u \cdot \nabla v{\left| {\nabla v} \right|^{p - 2}} + (p - 1){\left( {\frac{u}{v}} \right)^p}{\left| {\nabla v} \right|^p}$

and

$\displaystyle R(u,v) = {\left| {\nabla u} \right|^p} - \nabla \left( {\frac{{{u^p}}}{{{v^{p - 1}}}}} \right) \cdot \nabla v{\left| {\nabla v} \right|^{p - 2}}.$

Then

$L(u,v)=R(u,v).$

Moreover, $L(u, v) \geqslant 0$, and $L(u, v) = 0$ a.e. if and only if $\nabla \left(\frac{u}{v}\right) = 0$ a.e. , i.e. $u = kv$ for some constant $k$ in each component of the domain.

Proof. For each partial derivative, it holds

$\displaystyle\begin{gathered} \frac{\partial }{{\partial {x_i}}}\left( {\frac{{{u^p}}}{{{v^{p - 1}}}}} \right) = \frac{1}{{{v^{2p - 2}}}}\left[ {{v^{p - 1}}\frac{{\partial ({u^p})}}{{\partial {x_i}}} - {u^p}\frac{{\partial ({v^{p - 1}})}}{{\partial {x_i}}}} \right] \hfill \\ \qquad= \frac{1}{{{v^{2p - 2}}}}\left[ {p{u^{p - 1}}{v^{p - 1}}\frac{{\partial u}}{{\partial {x_i}}} - (p - 1){u^p}{v^{p - 2}}\frac{{\partial v}}{{\partial {x_i}}}} \right] \hfill \\ \qquad= p\frac{{{u^{p - 1}}}}{{{v^{p - 1}}}}\frac{{\partial u}}{{\partial {x_i}}} - (p - 1)\frac{{{u^p}}}{{{v^p}}}\frac{{\partial v}}{{\partial {x_i}}}. \hfill \\ \end{gathered}$

Therefore,

$\displaystyle\nabla \left( {\frac{{{u^p}}}{{{v^{p - 1}}}}} \right) = p\frac{{{u^{p - 1}}}}{{{v^{p - 1}}}}\nabla u - (p - 1)\frac{{{u^p}}}{{{v^p}}}\nabla v$

which implies

$\displaystyle\nabla \left( {\frac{{{u^p}}}{{{v^{p - 1}}}}} \right) \cdot \nabla v{\left| {\nabla v} \right|^{p - 2}} = p\frac{{{u^{p - 1}}}}{{{v^{p - 1}}}}\nabla u \cdot \nabla v{\left| {\nabla v} \right|^{p - 2}} - (p - 1)\frac{{{u^p}}}{{{v^p}}}|\nabla v{|^p}.$

Thus $L(u,v)=R(u,v)$. The rest of the proof follows from the Young inequality. Precisely, we first write

$\displaystyle\begin{gathered} L(u,v) = {\left| {\nabla u} \right|^p} - p{\left( {\frac{u}{v}} \right)^{p = 1}}\nabla u \cdot \nabla v{\left| {\nabla v} \right|^{p - 2}} + (p - 1){\left( {\frac{u}{v}} \right)^p}{\left| {\nabla v} \right|^p} \hfill \\ \qquad= {\left| {\nabla u} \right|^p} - p{\left( {\frac{u}{v}} \right)^{p = 1}}\left| {\nabla u} \right|{\left| {\nabla v} \right|^{p - 1}} + (p - 1){\left( {\frac{u}{v}} \right)^p}{\left| {\nabla v} \right|^p} \hfill \\ \qquad\qquad+ p{\left( {\frac{u}{v}} \right)^{p = 1}}{\left| {\nabla v} \right|^{p - 2}}(\left| {\nabla u} \right| \cdot \left| {\nabla v} \right| - \nabla u \cdot \nabla v). \hfill \\ \end{gathered}$

We then have

$\displaystyle\frac{1}{p}{\left| {\nabla u} \right|^p} + \frac{{p - 1}}{p}{\left( {\frac{u}{v}} \right)^p}{\left| {\nabla v} \right|^p} \geqslant {\left( {{{\left| {\nabla u} \right|}^p}} \right)^{\frac{1}{p}}}{\left( {{{\left( {\frac{u}{v}} \right)}^p}{{\left| {\nabla v} \right|}^p}} \right)^{\frac{{p - 1}}{p}}} = {\left( {\frac{u}{v}} \right)^{p = 1}}\left| {\nabla u} \right|{\left| {\nabla v} \right|^{p - 1}}.$

This together the fact that

$\displaystyle\left| {\nabla u} \right| \cdot \left| {\nabla v} \right| - \nabla u \cdot \nabla v \geqslant 0$

prove the non-negativity of $L$.

Lastly, if $L(u,v)(x_0)=0$ with $u(x_0)\ne 0$, we must then have

$\displaystyle\left| {\nabla u} \right| \cdot \left| {\nabla v} \right| - \nabla u \cdot \nabla v = 0$

and

$\displaystyle\left| {\nabla u} \right| = \frac{u}{v}\left| {\nabla v} \right|.$

That means

$\displaystyle\nabla u = \frac{u}{v}\nabla v$

or

$\displaystyle \nabla \left(\frac{u}{v}\right)(x_0)=0.$

On the other hand, if

$S=\{x\in \Omega: u(x)=0\}$

then $\nabla u=0$ a.e. in $S$ and thus $\nabla \left(\frac{u}{v}\right)=0$ a.e. in $S$. We conclude that $\nabla \left(\frac{u}{v}\right)=0$ a.e. in $\Omega$.

## 1 Comment »

1. My sincere thanks for this blog.
It’s clear as crystal.
I try to express $t^p \Delta u(t,x)$ in terms of the $p$-Laplacian.
Sincerly.

Comment by alabair — July 7, 2011 @ 6:32

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