# Ngô Quốc Anh

## January 4, 2014

### A Picone type identity for bi-Laplacian

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 22:22

Simultaneously, I have recently found the following identity in the same fashion of the Picone identity for $\Delta$. It says that

$\displaystyle \left( \Delta u -\frac uv \Delta v\right)^2-\frac {2\Delta v}{v}\left| \nabla u - \frac uv \nabla v\right|^2 = |\Delta u|^2 -\Delta \left( \frac {u^2}v\right)\Delta v$

for any function $v \ne 0$. It is worth noticing that the original Picone identity says that

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

for any function $v \ne 0$. It turns out that a few days ago, this identity appeared in a recent notes by Dwivedi  and Tyagi, see Lemma 2.1 from here. The extra term

$\displaystyle\frac {2\Delta v}{v}\left| \nabla u - \frac uv \nabla v\right|^2$

naturally appears since it only involves up to third order derivatives. However, to compare this term with $0$, we only need to assume that $\Delta v$ has a fixed sign. To see how this identity could be useful, let us consider the following equation

$\displaystyle (-\Delta)^2 u +hu= f u ^\frac {n+4}{n-4}, \quad u>0, h<0$

naturally arises from prescribing $Q$-curvature in Riemannian manifolds of dimension $n \geqslant 5$.

## 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.

## 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.