# Ngô Quốc Anh

## April 7, 2011

### Weak comparison principle: p-Laplacian with Neumann boundary condition

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 20:25

I am going to summarize some known comparison principles. I will start with a weak comparison principle for $p$-Laplacian with Neumann boundary condition. This is adapted from a recent paper by J. Giacomoni et al. published in Differential Integral Equations [here]. I will keep the numbering for the convenience.

Lemma 3.1. Let $u,v \in W^{1,N}(\Omega)$ be non-negative functions satisfying $\displaystyle -\Delta_N u + u^{N-1} \geqslant -\Delta_N v + v^{N-1}$

in $\Omega$ and $\displaystyle {\left| {\nabla u} \right|^{N - 2}}\frac{{\partial u}}{{\partial \nu }} \geqslant {\left| {\nabla v} \right|^{N - 2}}\frac{{\partial v}}{{\partial \nu }}$

on the boundary $\partial \Omega$ in the strong sense. Then $u\geqslant v$ in $\overline\Omega$.

Proof. The trick is that if we want to prove $u\geqslant v$ in $\overline\Omega$ then we just show that $(u-v)^- \equiv 0$

in $\overline\Omega$. To this purpose, we use $(u-v)^-$ as a test function in the equation $\displaystyle - ({\Delta _N}u - {\Delta _N}v) + ({u^{N - 1}} - {v^{N - 1}}) \geqslant 0$

and integrate to achive $\displaystyle - \int_\Omega {({\Delta _N}u - {\Delta _N}v){{(u - v)}^ - }} + \int_\Omega {({u^{N - 1}} - {v^{N - 1}}){{(u - v)}^ - }} \leqslant 0$

and use the integration by parts to arrive at $\displaystyle\begin{gathered} \int_\Omega {({{\left| {\nabla u} \right|}^{N - 2}}\nabla u - {{\left| {\nabla v} \right|}^{N - 2}}\nabla v) \cdot \nabla {{(u - v)}^ - }} + \int_\Omega {({u^{N - 1}} - {v^{N - 1}}){{(u - v)}^ - }} \hfill \\ \qquad- \int_{\partial \Omega } {\left( {{{\left| {\nabla u} \right|}^{N - 2}}\frac{{\partial u}}{{\partial \nu }} - {{\left| {\nabla v} \right|}^{N - 2}}\frac{{\partial v}}{{\partial \nu }}} \right){{(u - v)}^ - }}\leqslant 0 . \hfill \\ \end{gathered}$

Keep in mind that $\displaystyle\begin{gathered} \int_\Omega {({{\left| {\nabla u} \right|}^{N - 2}}\nabla u - {{\left| {\nabla v} \right|}^{N - 2}}\nabla v) \cdot \nabla {{(u - v)}^ - }} \hfill \\ \qquad= \int_{\Omega \cap \left\{ {u \leqslant v} \right\}} {({{\left| {\nabla u} \right|}^{N - 2}}\nabla u - {{\left| {\nabla v} \right|}^{N - 2}}\nabla v) \cdot (\nabla u - \nabla v)} \hfill \\\qquad = \int_{\Omega \cap \left\{ {u \leqslant v} \right\}} {{{\left| {\nabla u} \right|}^{N - 2}}\nabla u \cdot (\nabla u - \nabla v)} - \int_{\Omega \cap \left\{ {u \leqslant v} \right\}} {{{\left| {\nabla v} \right|}^{N - 2}}\nabla v \cdot (\nabla u - \nabla v)} .\hfill \\ \end{gathered}$

We now use the following inequality $\displaystyle |{x_2}{|^p} - |{x_1}{|^p} \geqslant p|{x_1}{|^{p - 2}}{x_1} \cdot ({x_2} - {x_1}) + \frac{{|{x_2} - {x_1}{|^p}}}{{{2^{p - 1}} - 1}}$

that proof can be found in this topic. To be precise, we get $\displaystyle {\left| {\nabla u} \right|^{N - 2}}\nabla u \cdot (\nabla u - \nabla v) \geqslant \frac{1}{p}\left[ {{{\left| {\nabla u} \right|}^N} - {{\left| {\nabla v} \right|}^N} + \frac{{|\nabla u - \nabla v{|^p}}}{{{2^{p - 1}} - 1}}} \right]$

and $\displaystyle - {\left| {\nabla v} \right|^{N - 2}}\nabla v \cdot (\nabla u - \nabla v) \geqslant \frac{1}{p}\left[ {{{\left| {\nabla v} \right|}^N} - {{\left| {\nabla u} \right|}^N} + \frac{{|\nabla u - \nabla v{|^p}}}{{{2^{p - 1}} - 1}}} \right].$

Hence, $\displaystyle ({\left| {\nabla u} \right|^{N - 2}}\nabla u - {\left| {\nabla v} \right|^{N - 2}}\nabla v) \cdot (\nabla u - \nabla v) \geqslant \frac{2}{p}\frac{{|\nabla u - \nabla v{|^p}}}{{{2^{p - 1}} - 1}} > 0.$

This proves the positivity of the first term. For the second term, it is clear to see that $\displaystyle\int_\Omega {({u^{N - 1}} - {v^{N - 1}}){{(u - v)}^ - }} = \int_{\Omega \cap \left\{ {u \leqslant v} \right\}} {({u^{N - 1}} - {v^{N - 1}})(u - v)} \geqslant 0.$

Lastly, we know that the integral on $\partial\Omega$ is non-positive since $(u-v)^- \leqslant 0$, thus the left hand side is non-negative. This gives the fact that $(u-v)^- \equiv 0$.

## 4 Comments »

1. […] Quốc Anh: Weak comparison principle: p-Laplacian with Neumann boundary condition, Several interesting limits from a paper by Chang-Qing-Yang, The (original) Picone […]

Pingback by Third Xamuel.com Linkfest — April 10, 2011 @ 23:57

2. Hi there,

thanks for the many interesting articles in your website. Nevertheless, I am slightly confused: in my terminology the negative part of a function, is a nonnegative function, What is the definition of $(u-v)^{-}$ in your context? I ask, since I have no access to the article itself.

Thanks in advance,

ZNS

Comment by ZNS — September 28, 2015 @ 23:07

• Thanks for your interest in my post.

I prefer to the following definition: $u^- = \min \{ u, 0\} \leqslant 0$.

Comment by Ngô Quốc Anh — September 28, 2015 @ 23:13

• Thanks for that. I thought about incorporating a minus sign in front of the usual definition, but wasn’t sure at all. In any case, thanks again, keep up the good work.

Comment by ZNS — September 28, 2015 @ 23:18

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