Posted by: Shuanglin Shao | November 14, 2009

## Reading: GWP for critical 2D dissipative SQG

Last weekend I attended a wonderful conference SCAPDE at UC Irvine, where I learned an interesting theorem from Kiselev: solutions to critical 2D dissipative quasi-geostrophic equations (SQG) are globally wellposed

$\theta_t=u\cdot \nabla \theta-(-\Delta)^\alpha\theta, u=(u_1,u_2)=(-R_2\theta, R_1\theta),$

where $\theta:\mathbb{R}^2\to \mathbb{R}$ is a scalar function, $R_1, R_2$ are the usual Riesz transform in $\mathbb{R}^2$ defined via $\widehat{R_i(f)}(\xi)=\frac {i\xi_i}{|\xi|}\widehat{f}(\xi)$ and $\alpha\geq 0$.

This is his joint work with Nazarov and Volberg. I took a look at this paper in the past few days. The proof makes good use of classical tools in Fourier analysis such as singular integrals and modulus of continuity, which are familiar topics in Stein’s book, Singular integrals and differentiability properties of functions. It is a place where I do not see Strichartz estimates or Littlewood Paley decompositions and see the power of classical Fourier analysis.

The monotonicity formula they have is

$\|\nabla \theta\|_\infty\le C\|\nabla \theta_0\|_\infty \exp \exp\{C\|\theta_0\|_\infty\}, (1)$

for periodic smooth initial data $\theta_0$. I am new to this field of fluid dynamics but I would still like to say a few words on this estimate. The proof is to find an “upper bound “, modulus of continuity $\omega$, for solutions $\theta$; then they show that a family of modulus of continuity is preserved under evolution of the equation, which is strong enough to control $\|\nabla \theta\|_\infty$. Quoted from the paper, the idea is to show that the critical SQG possesses a stronger “nonlocal” maximum principle than $L^\infty$ control.

Recall a modulus of continuity $\omega: [0,\infty)\mapsto [0,\infty)$ is just an arbitrarily increasing continuous and concave function such that $\omega(0)=0$. A function $f: \mathbb{R}^n\mapsto \mathbb{R}^m$ has modulus of continuity $\omega$ if $|f(x)-f(y)|\le \omega (|x-y|)$ for all $x,y\in \mathbb{R}^n$.

We choose not to report the crucial/essential part of finding the continuity of modulus (maybe later). Instead we assume that $f$ has modolus of continuity $\omega$, which is unbounded and $\omega^\prime (0)<\infty$ and $\lim_{\xi\to 0+}\omega^{\prime\prime}=-\infty$. Then there holds

$\|\nabla f\|_\infty< \omega^\prime (0), (2)$.

The proof is actually very simple. The explicit form of $\omega$ will take care of the implication of (1) from (2).

Assume that $\|\nabla f\|_\infty=|\nabla f(x)|$ for some $x$. We consider the point $y=x+\xi e$ for $e=\frac {\nabla f}{|\nabla f|}$. On the one hand, we have

$f(y)-f(x)\le \omega (\xi), (3)$

for all $\xi\geq 0$. On the other hand, the left hand side of (3) is at least $|\nabla f(x)|\xi -C\xi^2$ where $C=\frac 12 \|\nabla^2 f\|_\infty$ while its right hand side can be represented as $\omega^\prime (0)-\rho(\xi)\xi^2$ with $\rho(\xi)\to\infty$ as $\xi\to 0+$.

Then

$|\nabla f(x)| \le \omega^\prime (0)-(\rho(\xi)-C)\xi$

for all sufficiently small $\xi>0$, and it remains to choose some $\xi>0$ satisfying $\rho(\xi)>C$.