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}


\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}}.



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}


\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


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

That means

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


\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.

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

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