\documentclass[reqno]{amsart}
\usepackage{hyperref}
\usepackage{amssymb,mathrsfs}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 250, pp. 1--9.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/250\hfil axisymmetric Navier-Stokes equations]
{Regularity for the axisymmetric Navier-Stokes equations}

\author[P. Wang \hfil EJDE-2015/250\hfilneg]
{Peng Wang}

\address{Peng Wang \newline
Department of Mathematics, Zhejiang Normal University,
Jinhua 321004, Zhejiang, China}
\email{wpmath2013@gmail.com}


\thanks{Submitted June 11, 2015. Published September 25, 2015.}
\subjclass[2010]{35Q30, 76D03}
\keywords{ Navier-Stokes equations; axi-symmetric flow;
 regularity criterion}

\begin{abstract}
 In this article, we establish a regularity criterion for the Navier-Stokes
 system with axisymmetric initial data. It is proved that if the
 local axisymmetric smooth solution $u$ satisfies
 ${\|u^\theta\|_{L^{\alpha}(0,T; L^{\beta})}}<\infty$ , where
 $\frac{2}{\alpha}+\frac{3}{\beta} \leq 1 $, and
 $3 < \beta \leq \infty$, then the strong solution keeps smoothness up 
 to time $T$.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\allowdisplaybreaks

\section{Introduction}

We study the following classic 3D incompressible Navier-Stokes equations
in the whole space,
\begin{equation} \label{e1.1}
\begin{gathered}
\partial_tu + (u\cdot \nabla u)u + \nabla p=\nu \Delta u, \\
\nabla \cdot u=0,\\
u(x,t=0)=u_0,
\end{gathered}
\end{equation}
where $u(x, t)\in \mathbb{R}^3$ and $p(x, t) \in \mathbb{R}$
denote the unknowns, velocity and pressure respectively, while $\nu$
denotes the viscous coefficient of the system.

A lot of works have been devoted to study the above system, but
global well-posedness for \eqref{e1.1} with arbitrary large initial data
is still a challenging open problem, see \cite{CF, F, MB, YZ, YZ1, YZ2}.

Here, we are concerned with \eqref{e1.1} with axisymmetric initial data.
If $u_0$ is axisymmetric in system \eqref{e1.1}, then the solution
$u(x,t)$ of system \eqref{e1.1}is also axisymmetric~\cite{UY, LMNP}.
So, it is convenient to write $u(x,t)$ as in the form
\[
u(x,t)=u^r(r, z, t)e_r+u^\theta(r, z, t)e_\theta +u^z(r, z, t)e_z,
\]
where $e_r$, $e_\theta$ and $e_z$ are the standard orthonormal unit
vectors in cylindrical coordinate system
\begin{gather*}
e_r=({\frac{x_1}{r}}, {\frac{x_2}{r}}, 0)=(\cos{\theta}, \sin{\theta}, 0),\\
e_\theta=({-\frac{x_2}{r}}, {\frac{x_1}{r}}, 0)=(-\sin{\theta},\cos{\theta}, 0),\\
e_z=(0, 0, 1),
\end{gather*}
with $r=(x_1^2+x_2^2)^{1/2}$.
By direct computations, it is easy to show the following relations.
\begin{gather*}
\nabla =(\partial_{x_1}, \partial_{x_2}, \partial_{z})^{T}
= \partial_r e_r+\frac{\partial_\theta}{r}e_\theta + \partial_ze_z, \\
\Delta =\nabla \cdot \nabla=\frac{1}{r}\partial_r(r\partial_r)
 +\frac{1}{r^2}\frac{\partial^2}{\partial_\theta^2}
 +\frac{\partial^2}{\partial z^2}, \\
\frac{\partial e_r}{\partial \theta}=e_\theta, \quad
 \frac{\partial e_\theta}{\partial \theta}=-e_r.
\end{gather*}
Accordingly, the system \eqref{e1.1} can be rewritten equivalently as
\begin{equation} \label{e1.2}
\begin{gathered}
\frac{\tilde{D}}{Dt}u^r-\nu (\partial_r^2+\partial_z^2
 +\frac{1}{r}\partial_r-\frac{1}{r^2})u^r-\frac{(u^\theta)^2}{r}+\partial_rp=0 , \\
\frac{\tilde{D}}{Dt}u^\theta-\nu (\partial_r^2+\partial_z^2
 +\frac{1}{r}\partial_r-\frac{1}{r^2})u^\theta+\frac{u^ru^\theta}{r}=0,\\
\frac{\tilde{D}}{Dt}u^z-\nu (\partial_r^2+\partial_z^2
 +\frac{1}{r}\partial_r)u^z+\partial_zp=0,\\
u|_{t=0}=u_0^r\cdot e_r + u_0^\theta\cdot e_\theta + u_0^z\cdot e_z,
\end{gathered}
\end{equation}
where ${\frac{\tilde{D}}{Dt}}$ denotes the material derivative
\[
{\frac{\tilde{D}}{Dt}}=\partial_t +u^r\partial_r +u^z\partial_z.
\]

If $u^\theta=0$ (so-called without swirl), Ukhovskii and Yudovich \cite{UY}
(see also \cite{LMNP}) proved the existence of generalized solutions,
uniqueness and regularity. When $u^\theta\neq 0$ (with swirl), it is much
complicated and difficult. For recent progress, one can find results
on regularity criteria or global existence with small initial data
in \cite{CL,JN,LMNP,ZZ}. In particular, very recently in \cite{ZZ},
the following regularity criterion was established:
\begin{equation} \label{e1.3}
\|u^\theta 1_{r\leq\varsigma}\|_{L^\alpha((0, T);L^\beta)}<\infty,\quad
\text{with } \frac{2}{\alpha}+\frac{3}{\beta}<1, \quad
\beta>6, \quad \text{or } (\alpha,\beta)=(4, 6),
\end{equation}
where $\varsigma>0$ is given.

The aim of this paper is to give a regularity criteria in terms of
$u^\theta$. More precisely, we have the following theorem.

\begin{theorem} \label{thm1.1}
Let $u_0 \in H^2$, and
$u\in C([0,T);H^2(\mathbb{R}^3))\cap L^2_{loc}([0,T);\dot{H}^3(\mathbb{R}^3))$
be the solution of \eqref{e1.1}. If it satisfies
\begin{equation} \label{e1.4}
 {\|u^\theta\|_{L^\alpha(0,T;L^\beta)}} <\infty , \quad
\text{where } {\frac{2}{\alpha}+\frac{3}{\beta}=1}, \text{ and } 3< \beta\leq \infty,
\end{equation}
then $u(x,t)$ can be continued beyond $T$.
\end{theorem}

In Section 2 some key lemmas are given. Then Section 3 is devoted to the proof
of the main result.

\section{Key lemmas}

Before going to the details, let us introduce some notation.
$L^{p,q}$ norm be defined by
\begin{equation} \label{e2.1}
\|u\|_{L^{p,q}}
= \begin{cases}
 \big(\int_{0}^{t}{\|u\|_{L^q}^p}{\rm d}\tau\big)^{1/p} &\text{if }
 1 \leq p <\infty,\\[4pt]
 \operatorname{ess\,sup}_{ 0<\tau<t} {\|u\|_{L^q}}
&\text{if } q =\infty.
 \end{cases}
\end{equation}
And we define $\tilde{\nabla}\doteq(\partial_r, \partial_z)$.

Next, let us introduce the vorticity field and the corresponding equation,
\[
 \omega=\nabla\times u=(\frac{1}{r}\frac{\partial u^z}{\partial \theta}
-\frac{\partial u^\theta}{\partial z})e_r +(\frac{\partial u^r}{\partial z}
-\frac{\partial u^z}{\partial r})e_\theta+(\frac{1}{r}
\frac{\partial}{\partial r}(u^\theta r)-\frac{1}{r}
\frac{\partial u^r}{\partial \theta})e_z,
\]
or equivalently,
\[
\omega = \omega^r e_r + \omega^\theta e_\theta + \omega^z e_z
= -\partial_z u^\theta e_r + (\partial_z u^r-\partial_r u^z)e_\theta
+(\partial_r u^\theta + \frac{u^\theta}{r})e_z.
\]
Then we have the  vorticity equation
\begin{equation} \label{e2.2}
\begin{gathered}
\frac{\tilde{D}}{Dt}\omega^r-\nu(\partial_r^2+\partial_z^2
+\frac{1}{r}\partial_r-\frac{1}{r^2})\omega^r-(\omega^r\partial_r
+\omega^z\partial_z)u^r=0,
\\
\frac{\tilde{D}}{Dt}\omega^\theta-\nu(\partial_r^2+\partial_z^2
+\frac{1}{r}\partial_r-\frac{1}{r^2})\omega^\theta
-\frac{2u^\theta \partial_z u^\theta}{r}-\frac{u^r \omega^\theta}{r}=0,
\\
\frac{\tilde{D}}{Dt}\omega^z-\nu(\partial_r^2+\partial_z^2
+\frac{1}{r}\partial_r)\omega^z-(\omega^r\partial_r+\omega^3\partial_3)u^3=0,
\\
\omega|_{t=0} =\omega_0^re_r+\omega_0^\theta e_\theta+\omega_0^z e_z.
\end{gathered}
\end{equation}
If, we set $\tilde{u}\doteq u^re_r+u^z e_z$, then
\[
\nabla\cdot\tilde{u}=0 \quad\text{and} \quad
\nabla\times \tilde{u}=\omega^\theta e_\theta.
\]
To proof Theorem \ref{thm1.1}, we need the following key lemmas.

\begin{lemma}[{\cite[Lemma 2]{CL}}]  \label{lem2.1}
Suppose that $u(x,t)$ is an axisymmetric vector field with $\operatorname{div}u = 0$,
 and $\omega$ = curl $u$ vanishes sufficiently fast near infinity in $\mathbb{R}^3$,
then $\nabla u $ and $\nabla (u^{\theta}e_{\theta})$ can be represented
as the singular integral form
\begin{gather*}
\nabla \tilde{u}(x) = C \omega^\theta e_{\theta} (x)
 +[K\ast (\omega ^ {\theta} e_{\theta} )](x), \\
\nabla (u^{\theta}e_{\theta}(x)) = C \tilde{\omega}(x)
+ {[ H\ast(\tilde{\omega})](x)},
\end{gather*}
where the kernels $K(x)$ and $H(x)$ are matrix valued functions homogeneous
of degree $-3$, defining a singular integral operator by convolution,
and $f\ast g(x)= \int_{\mathbb{R}^3}{f(x-y)g(y)} {\rm d}y $ denotes
the standard convolution operator. The matrices $C$ and $\tilde{C}$ are constant.
\end{lemma}

\begin{lemma} \label{lem2.2}
Based on the above Lemma \ref{lem2.1} and the $L^p$ boundness of Calderon-Zygmund
singular integral operators with $1 < p < \infty$, we can deduce that
\[
{\|\nabla \tilde{u}\|_{L^p}} \lesssim {\|\omega\|_{L^p}},\quad
{\|\nabla (u^\theta e_{\theta})\|_{L^p}} \lesssim {\|\omega\|_{L^p}}.
\]
\end{lemma}

\begin{lemma} \label{lem2.3}
Let $u$ be a sufficiently smooth vector field, then for all $ 1 < p <\infty $,
we have
\[
{\|\nabla u\|_{L^{p}}} \leq C(p) {\|\omega\|_{L^p}}.
\]
\end{lemma}

\begin{lemma}[{\cite[Lemma 3]{JN}}]  \label{lem2.4}
Let $u$ be a sufficiently smooth divergence-free axisymmetric vector field.
Then there exist constants $C_{1}(p) > 0$ and $C_{2} > 0$,
 independent of $u$, such that for $ 1 < p < \infty$, we have
\begin{gather*}
{\|\nabla u^r\|_{L^p}} +{\|\frac{u^r}{r}\|_{L^p}}
 \leq C_{1}(p){\|\omega^\theta\|_{L^p}},\\
{\|\partial_{r}(\frac{u^\theta}{r})\|_{L^p}} \leq C_{2} {\|\nabla^2 u\|_{L^p}}.
\end{gather*}
\end{lemma}

\begin{lemma}[{\cite[Lemma 4]{JN}}] \label{lem2.5}
 Suppose that $u$ is a sufficiently smooth axisymmetric vector field,
then there exists a constant $C > 0$ that is independent of $u$,
such that for all $1 \leq p \leq \infty$, we have
\begin{gather*}
{\|\nabla u^\theta\|_{L^p}} +{\|\frac{u^\theta}{r}\|_{L^p}}
 \leq C{\|\nabla u\|_{L^p}},\\
{\|\partial_{r}(\frac{u^\theta}{r}) \|_{L^p}} \leq C {\|\Delta u\|_{L^p}}.
\end{gather*}
\end{lemma}

\begin{lemma}[{\cite[Lemma 5]{JN}}] \label{lem2.6}
 Let $u$ be the sufficiently smooth and divergence-free axisymmetric
vector field. Then there exist $C_{1}(p)$, $C_{2}$, independent of $u$, such that
for $1 < p < \infty$
\begin{align*}
C_{1}(p){\|\Delta u\|_{L^p}}
&\leq {\|\frac{\omega^r}{r}\|_{L^p}}+{\|\frac{\omega^\theta}{r}\|_{L^p}}
 +{\|\nabla \omega^r\|_{L^p}},
+{\|\nabla \omega^\theta \|_{L^p}}+{\|\nabla \omega^z\|_{L^P}} \\
&\leq C_{2}{\|\Delta u\|_{L^p}}.
\end{align*}
\end{lemma}

\begin{lemma} \label{lem2.7}
Let $u$ be the unique local axisymmetric solution of \eqref{e1.1}, then we have
\begin{align*}
{\|\nabla^2 u\|_{L^2}^2}
&={\|{\nabla}\partial_ru^r\|_{L^2}^2} +
{\|{\nabla}\frac{u^r}{r}\|_{L^2}^2} + {\|{\nabla}\partial_zu^r\|_{L^2}^2}
+{\|\partial_z\frac{u^r}{r}\|_{L^2}^2}\\
&\quad+{\|{\nabla}\partial_ru^\theta\|_{L^2}^2}+{\|{\nabla}
 \frac{u^\theta}{r}\|_{L^2}^2}
+{\|{\nabla}\partial_zu^\theta\|_{L^2}^2}+{\|\partial_z\frac{u^\theta}{r}\|_{L^2}^2}
+{\|\nabla^2u^z\|_{L^2}^2}\\
&\quad +\int_{\mathbb{R}^3}{\frac{2}{r^2}\big\{(\partial_ru^r)^2
 +(\frac{u^r}{r})^2-\partial_ru^r\frac{u^r}{r}\big\}}{\rm d}x\\
&\quad +\int_{\mathbb{R}^3}{\frac{2}{r^2}\big\{(\partial_ru^\theta)^2
 +(\frac{u^\theta}{r})^2-\partial_ru^\theta\frac{u^\theta}{r}\big\}}{\rm d}x
\end{align*}
\end{lemma}

\begin{lemma}[{Proposition 2.5]{MZ}}] \label{lem2.8}
Let $u$ be the sufficiently smooth and divergence-free axisymmetric
vector field, and $\nabla \times u = \omega$, then one can obtain that
\begin{equation} \label{e2.3}
\frac{u^r}{r} = \Delta^{-1}\partial_z(\frac{\omega^\theta}{r})
- 2\frac{\partial_r}{r}\Delta^{-2}{\partial_z}(\frac{\omega^\theta}{r})
\end{equation}
where
\begin{equation} \label{e2.4}
\frac{\partial_r}{r}f(r,z) = \frac{x_2^2}{r^2}R_{11}f
+\frac{x_1^2}{r^2}R_{22}f -2\frac{x_1x_2}{r^2}R_{12}f
\end{equation}
here $R_{ij}= \Delta^{-1}\partial_i\partial_j$.
\end{lemma}

\begin{lemma} \label{lem2.9}
Based on Lemma \ref{lem2.7}, for $ 1 < p < \infty$, one can deduce easily the following 
results
\begin{gather}
{\|\hat{\nabla}\frac{u^r}{r}\|_{L^p}} \leq C(p){\|\frac{\omega^\theta}{r}\|_{L^p}}
\label{e2.5}\\
\|\hat{\nabla}\hat{\nabla}\frac{u^r}{r}\|_{L^p} 
\leq C(p){\|\partial_z(\frac{\omega^\theta}{r})\|_{L^p}} \label{e2.6}
\end{gather}
\end{lemma}

The Lemma below is a general Sobolev-Hardy inequality, which was deduced 
by Hui chen et al \cite[v]{ZC}. About more Sobolev-Hardy inequality one 
can see \cite[Theorem 2.1]{BT}.

\begin{lemma}[{\cite[Lemma 2.4]{ZC}}] \label{lem2.10}
We assume that There exist a positive constant $C(s, q^*)$, $q^* \in [2, 2(2-s)]$ 
with $0 \leq s <2$ and $r = (x_1^2 + x_2^2)^{1/2}$ such that for all
$u \in \mathscr{D}^{1, q}(\mathbb{R}^3)$, one can obtain that
\[
{\|\frac{u}{r^{\frac{s}{q^*}}}\|_{L^{q^*}}} 
\leq C(q^*, s){\|u\|_{L^2}^{\frac{3-s}{q^*}
-\frac{1}{2}}}{\|\nabla u\|_{L^2}^{\frac{3}{2}-\frac{3-s}{q^*}}}.
\]
\end{lemma}

\begin{lemma}[\cite{ZC}] \label{lem2.11}
Let $u$ be the unique axisymmetric solution of \eqref{e1.1}, then we have
\begin{align*}
&{\|\frac{\omega^\theta}{r}\|^2_{L^\infty(0, T; L^2)}} 
+{\|\frac{\omega^r}{r}\|^2_{L^\infty(0,T; L^2)}}
+ {\|\nabla \frac{\omega^\theta}{r}\|_{L^2(0, T;L^2)}^2} 
+{\|\nabla{\frac{\omega^r}{r}}\|^2_{L^2(0,T; L^2)}}\\
& \leq C\Big\{{\|\frac{\omega^r_0}{r}\|_{L^2}}
+{\|\frac{\omega^\theta_0}{r}\|_{L^2}}\Big\}
\exp\Big\{C\int_{0}^{T}{\|u^\theta\|_{L^\beta}^{\frac{2}{1-\frac{3}{\beta}}}}
{\rm d}t\Big\}.
\end{align*}
where $(\alpha, \beta)$ satisfies $\frac{2}{\alpha}+\frac{3}{\beta}\leq 1$ 
with $ 3 < \beta \leq \infty$.
\end{lemma}

\begin{proof}
This proof can be found in \cite{ZC}. For reader's convenience, 
we give it here. Multiplying the $\omega^r$ 
equation of $\eqref{e2.2}$ by $\frac{\omega^r}{r^2}$
and integrating the resulting equation over ${\mathbb{R}^3}$ leads to
\begin{align*}
&{\frac{1}{2}{\frac{d}{dt}}{\|\frac{\omega^r}{r}\|_{L^2}^2}} 
+\nu{\|\hat{\nabla}\frac{\omega^r}{r}\|_{L^2}^2} \\
&= \int_{\mathbb{R}^3}{(\omega^r\partial_r 
 + \omega^z \partial_z){\frac{u^r}{r}}{\frac{\omega^r}{r}}\cdot r} {\rm d}x \\
&= -2\pi \int_{\mathbb{R}}\int_{0}^{\infty} {\partial_z u^\theta
 \cdot\partial_r{\frac{u^r}{r}}\cdot{\frac{\omega^r}{r}}}{\rm d}r {\rm d}z
+ 2\pi \int_{\mathbb{R}}\int_{0}^{\infty} {\frac{\partial_r(ru^\theta)}{r}
 \cdot{\partial_z{\frac{u^r}{r}}\cdot{\frac{\omega^r}{r}}}\cdot r}
 {\rm d}r {\rm d}z\\
& = \int_{\mathbb{R}^3}{u^\theta(\partial_z\partial_r\frac{u^r}{r})
 {\frac{\omega^r}{r}}
 + u^\theta(\partial_r{\frac{u^r}{r}})(\partial_z{\frac{\omega^r}{r}})} {\rm d}x\\
&\quad - \int_{\mathbb{R}^3}{u^\theta\cdot(\partial_r\partial_z{\frac{u^r}{r}})
 (\frac{\omega^r}{r})} {\rm d}x
-\int_{\mathbb{R}^3}{u^\theta\cdot(\partial_z{\frac{u^r}{r}})
 (\partial_r\frac{\omega^r}{r})} {\rm d}x\\
&= \int_{\mathbb{R}^3}{u^\theta\cdot(\partial_r\frac{u^r}{r})\cdot
 (\partial_z\frac{\omega^r}{r})}{\rm d}x
-\int_{\mathbb{R}^3}{u^\theta\cdot(\partial_z\frac{u^r}{r})
 \cdot(\partial_r\frac{\omega^r}{r})}{\rm d}x = H_1 + H_2
\end{align*}
Form Lemma \ref{lem2.9} we  obtain 
\begin{align*}
 |H_1| 
&\leq \int_{\mathbb{R}^3}{|u^\theta\cdot(\partial_r\frac{u^r}{r}
 \partial_z\frac{\omega^r}{r}) |}{\rm d}x \\
&\leq {\|u^\theta\|_{L^\beta}}{\|\partial_r\frac{u^r}{r}
 \|_{L^{\frac{2\beta}{\beta-2}}}{\|\partial_z\frac{\omega^r}{r}\|_{L^2}}} 
 \quad\text{(H\"older inequality)}\\
 & \leq C{\|u^\theta\|_{L^\beta}}{\|\hat{\nabla}
 \partial_r\frac{u^r}{r}\|_{L^2}^{\frac{3}
 {\beta}}{\|\partial_r\frac{u^r}{r}\|_{L^2}^{1-\frac{3}{\beta}}}
 {\|\partial_z\frac{\omega^r}{r}\|_{L^2}}} \\
 &\quad \text{((Lemma \ref{lem2.10} $s=0$, $q^* =\frac{2\beta}{\beta-2} $)}\\
 & \leq C{\|u^\theta\|_{L^\beta}{\|\hat{\nabla}
 \frac{\omega^\theta}{r}\|_{L^2}^{\frac{3}{\beta}}}
 {\|\frac{\omega^\theta}{r}\|_{L^2}^{1-\frac{3}{\beta}}}
 {\|\hat{\nabla}\frac{\omega^r}{r}\|_{L^2}}} \quad \text{(Lemma \ref{lem2.9})}\\
 & \leq C_{\delta}{\|u^\theta\|_{L^\beta}^{\frac{2}{1-\frac{3}{\beta}}}
 {\|\frac{\omega^\theta}{r}\|_{L^2}^2}} 
 +\delta{\|\hat{\nabla}\frac{\omega^r}{r}\|_{L^2}^2} 
+\delta{\|\hat{\nabla}\frac{\omega^\theta}{r}\|_{L^2}^2} \quad 
\text{(Young inequality)}
\end{align*}
The quantity $H_2$ can be estimated  similarly as $H_1$:
\[
|H_2| \leq C_{\delta}{\|u^\theta\|_{L^\beta}^{\frac{2}{1-\frac{3}{\beta}}}}
{\|\frac{\omega^\theta}{r}\|_{L^2}^2} +\delta{\|\hat{\nabla}
\frac{\omega^r}{r}\|_{L^2}^2} +\delta{\|\hat{\nabla}
\frac{\omega^\theta}{r}\|_{L^2}^2}.
\]
Thus we have
\begin{equation} \label{e2.7}
\begin{aligned}
&{\frac{1}{2}{\frac{d}{dt}}{\|\frac{\omega^r}{r}\|_{L^2}^2}}
  +\nu{\|\hat{\nabla}\frac{\omega^r}{r}\|_{L^2}^2}\\
&\leq C_{\delta}{\|u^\theta\|_{L^\beta}^{\frac{2}{1-\frac{3}{\beta}}}
 {\|\frac{\omega^\theta}{r}\|_{L^2}^2}} +2\delta{\|\hat{\nabla}
 \frac{\omega^r}{r}\|_{L^2}^2} +2\delta{\|\hat{\nabla}
 \frac{\omega^\theta}{r}\|_{L^2}^2}
\end{aligned}
\end{equation}
Multiplying $\omega^\theta$ equation of \eqref{e2.2} by 
$\frac{\omega^\theta}{r^2}$, and integrating over $\mathbb{R}^3$, 
after integrating by parts we obtain that
\[
\frac{1}{2}{\frac{d}{dt}{\|\frac{\omega^\theta}{r}\|_{L^2}^2}} 
+\nu{\|\hat{\nabla}\frac{\omega^\theta}{r}\|_{L^2}^2} 
=2\int_{\mathbb{R}^3}{\frac{u^\theta}{r}{\frac{\omega^r}{r}}
{\frac{\omega^\theta}{r}}}{\rm d}x := H_3
\]
\begin{align*}
 |H_3| 
&\leq \int_{\mathbb{R}^3}{|u^\theta\cdot({r^{-\frac{1}{2}}
 \frac{\omega^\theta}{r}})({r^{-\frac{1}{2}}{\frac{\omega^r}{r}}})|}{\rm d}x\\
 &\leq C{\|u^\theta\|_{L^\beta}}{\|{r^{-\frac{1}{2}}
 \frac{\omega^\theta}{r}}\|_{L^{\frac{2\beta}{\beta-1}}}}
 {\|{r^{-\frac{1}{2}}{\frac{\omega^r}{r}}}\|_{L^{\frac{2\beta}{\beta-1}}}} \quad 
 \text{(H\"older inequality)}\\
 & \leq C{\|u^\theta\|_{L^\beta}}{\|\frac{\omega^\theta}{r}\|_{L^2}^{\frac{1}{2}
 -\frac{3}{2\beta}}}
 {\|\hat{\nabla}\frac{\omega^\theta}{r}\|_{L^2}^{\frac{1}{2}
 +\frac{3}{2\beta}}}{\|\frac{\omega^r}{r}\|_{L^2}^{\frac{1}{2}-\frac{3}{2\beta}}}
 {\|\hat{\nabla}\frac{\omega^r}{r}\|_{L^2}^{\frac{1}{2}+\frac{3}{2\beta}}}\\
 &\text{(where we used Lemma \ref{lem2.10} $s=\frac{\beta}{\beta-1}$, 
$q^* =\frac{2\beta}{\beta-1} $)}\\
 &\leq C_{\delta}{\|u^\theta\|_{L^\beta}^{\frac{2}{1-\frac{3}{\beta}}}}
 {\|\frac{\omega^r}{r}\|_{L^2}}{\|\frac{\omega^\theta}{r}\|_{L^2}}
 +\delta{\|\hat{\nabla}\frac{\omega^\theta}{r}\|_{L^2}^2} 
 + \delta{\|\hat{\nabla}\frac{\omega^r}{r}\|_{L^2}^{2}}
 \quad \text{(Young ineq.)}\\
 &\leq C_{\delta}{\|u^\theta\|_{L^\beta}^{\frac{2}{1-\frac{3}{\beta}}}}
({\|\frac{\omega^r}{r}\|_{L^2}^2} + {\|\frac{\omega^\theta}{r}\|_{L^2}^2} ) 
+\delta{\|\hat{\nabla}\frac{\omega^\theta}{r}\|_{L^2}^2} 
+ \delta{\|\hat{\nabla}\frac{\omega^r}{r}\|_{L^2}^{2}}
\end{align*}
Then we obtain 
\begin{equation} \label{e2.8}
\begin{aligned}
&\frac{1}{2}{\frac{d}{dt}{\|\frac{\omega^\theta}{r}\|_{L^2}^2}} 
 + \nu{\|\hat{\nabla}\frac{\omega^\theta}{r}\|_{L^2}^2}\\
&\leq C_{\delta}{\|u^\theta\|_{L^\beta}^{\frac{2}{1-\frac{3}{\beta}}}}
 ({\|\frac{\omega^r}{r}\|_{L^2}^2} + {\|\frac{\omega^\theta}{r}\|_{L^2}^2} ) 
+\delta{\|\hat{\nabla}\frac{\omega^\theta}{r}\|_{L^2}^2}
 + \delta{\|\hat{\nabla}\frac{\omega^r}{r}\|_{L^2}^{2}}
\end{aligned}
\end{equation}
Combining \eqref{e2.7} and \eqref{e2.8} together and let $\delta$ be a 
small enough constant, we have then by Gronwall's inequality that
\begin{align*}
&{{\|\frac{\omega^r}{r}\|^2_{L^\infty(0,T;L^2)}}} 
+{\|\hat{\nabla}\frac{\omega^r}{r}\|^2_{L^2(0,T;L^2)}}
+{{\|\frac{\omega^\theta}{r}\|^2_{L^\infty(0,T;L^2)}}} 
+ {\|\hat{\nabla}\frac{\omega^\theta}{r}\|^2_{L^2(0,T;L^2)}}\\
&\leq C({\|\frac{\omega^r_0}{r}\|_{L^2}} 
 + {\|\frac{\omega^\theta_0}{r}\|_{L^2}})
\exp\Big\{C\int_{0}^{T}{\|u^\theta\|_{L^\beta}^{\frac{2}{1-\frac{3}{\beta}}}}
{\rm d}t\Big\}.
\end{align*}
\end{proof}

\section{Proof of regularity criteria}

\begin{proof}
Let $u$ be an axisymmetric smooth solution of the Navier-Stokes equations. 
Taking curl on the both sides of the Navier-Stokes equations, then we can 
obtain the  equation
\[
\partial_t \omega - \Delta \omega +(u\cdot \nabla)\omega 
= (\omega\cdot\nabla)u.
\]
By multiplying $\omega$ on the both sides of the above equations, 
and integrating over $\mathbb{R}^3$, one obtain:
\begin{align*}
& \frac{1}{2}\frac{d}{dt} \int_{\mathbb{R}^3} {|\omega|}^2 {\rm d}x 
 + \int_{\mathbb{R}^3} {|\nabla\omega|}^2 {\rm d}x\\
&=\int_{\mathbb{R}^3} {(\omega \cdot \nabla)u\cdot \omega} {\rm d}x \\
&= \int_{\mathbb{R}^3} {\omega^r \partial_{r} u^r \omega^r} {\rm d}x 
 -\int_{\mathbb{R}^3} {\frac{\omega^\theta}{r} u^\theta \omega^r} {\rm d}x
 + \int_{\mathbb{R}^3} {\omega^z \partial_{z}u^r \omega^{r}} {\rm d}x 
 +\int_{\mathbb{R}^3} {\omega^r \partial_r u^\theta \omega^\theta} {\rm d}x \\
&\quad + \int_{\mathbb{R}^3}{\frac{\omega^\theta}{r}u^r \omega^\theta} {\rm d}x
 + \int_{\mathbb{R}^3}{\omega^z \partial_{z} u^\theta \omega^\theta} {\rm d}x
 +\int_{\mathbb{R}^3}{\omega^r \partial_{r}u^z \omega^\theta} {\rm d}x
 +\int_{\mathbb{R}^3}{\omega^z \partial_z u^z \omega^z} {\rm d}x\\
&:= I_{1}+I_{2}+I_{3}+I_{4}+I_{5}+I_{6}+I_{7}+I_{8}.
\end{align*}

We will estimate the terms one by one, for the term $I_1$, using integration 
by parts, we have
\begin{align*}
I_{1} 
&= \int_{\mathbb{R}^3} {\omega^r \partial_{r} u^r \omega^r} {\rm d}x \\
&= -{\int_{\mathbb{R}^3} {{\partial_{z}u^\theta}\cdot\partial_{r} 
 u^r\cdot\omega^r} {\rm d}x}\\
&=\int_{\mathbb{R}^3}({u^\theta\cdot{\partial_z\partial_ru^r}\cdot\omega^r 
 +u^\theta\cdot{\partial_ru^r}{\partial_z\omega^r}}){\rm d}x 
\end{align*}
then
\[
|I_{1}| \leq \int_{\mathbb{R}^3}|{u^\theta\cdot{\partial_z\partial_ru^r}
\cdot\omega^r}| {\rm d}x +\int_{\mathbb{R}^3}|{u^\theta
\cdot{\partial_ru^r}{\partial_z\omega^r}}|{\rm d}x \doteq I_1^1+I_1^2
\]
For the term $I_1^1$,
\begin{align*}
I_1^1
& \leq \Big\{ \int_{{R}^3}|{u^\theta\cdot\omega^r}|^2{\rm d}x\Big\}^{1/2}
 \cdot{\|\partial_z\partial_r u^r\|_{L^2}}
\quad \text{(H\"older inequality)}\\
&\leq C_\delta\int_{{R}^3}|{u^\theta\cdot\omega^r}|^2{\rm d}x 
 + \delta {\|\partial_z\partial_r u^r\|^2_{L^2}}
\quad \text{(Young inequality)}\\
&\leq C_\delta\int_{{R}^3}|{u^\theta\cdot\omega^r}|^2{\rm d}x 
 + \delta {\|\nabla\omega\|^2_{L^2}}
\quad \text{(By Lemma \ref{lem2.6})}\\
&\leq C_\delta{\|u^\theta\|^2_{L^\beta}}{\|\omega^r\|^2_{L^{\frac{2\beta}{\beta-2}}}} 
+ \delta {\|\nabla\omega\|^2_{L^2}}
\quad \text{(H\"older inequality)}\\
&\leq C_\delta{\|u^\theta\|^2_{L^\beta}}
 \left\{{\|\omega\|^\theta_{L^2}{\|\nabla\omega\|^{1-\theta}_{L^2}}} \right\}^2 
 +\delta {\|\nabla\omega\|^2_{L^2}}\\
&\quad \text{(Gagliardo-Nirenberg inequlity and $\theta = 1 - \frac{3}{\beta}$)}\\
&\leq C_\delta {\|u^\theta\|^{\frac{2}{\theta}}_{L^\beta}}{\|\omega\|_{L^2}^2} 
+ 2\delta{\|\nabla\omega\|^2_{L^2}} \quad \text{(Young inequality)}
\end{align*}
For the term of $I_1^2$,
\begin{align*}
 I_1^2& = \int_{\mathbb{R}^3}{|u^\theta\cdot{\partial_{r}u^r
 \cdot{\partial_{z}\omega^r}}|}{\rm d}x \\
&\leq \Big\{\int_{\mathbb{R}^3}{|u^\theta\cdot{\partial_{r}u^r}|^2}{\rm d}x
 \Big\}^{1/2}\cdot{\|{\partial_z\omega^r}\|_{L^2}}
 \quad {\text{(H\"older inequality)}}\\
&\leq C_\delta\Big\{\int_{\mathbb{R}^3}{|u^\theta\cdot{\partial_{r}u^r}|^2}
 {\rm d}x\Big\} + \delta{\|{\partial_z\omega^r}\|_{L^2}^2}
\quad \text{(Young inequality)}\\
&\leq C_\delta{\|u^\theta\|_{L^\beta}^2}{\|{\partial_{r}u^r\|_{L^{\frac{2\beta}
 {\beta-2}}}^2}}+\delta{\|{\partial_z\omega^r}\|_{L^2}^2}
\quad {\text{(H\"older inequality)}}\\
&\leq C_\delta{\|u^\theta\|_{L^\beta}^2}
 \Big\{{\|\omega\|_{L^2}^{\theta}}{\|\nabla\omega\|_{L^2}^{1-\theta}}\Big\}^{1/2}
+\delta{\|{\partial_z\omega^r}\|_{L^2}^2}\\
&\quad \text{(Gagliardo-Nirenberg inequlity and $\theta = 1 - \frac{3}{\beta}$)}\\
&\leq C_\delta{\|u^\theta\|_{L^\beta}^{\frac{2}{\theta}}}
 {\|\omega\|_{L^2}^2}+2\delta{\|\nabla\omega\|^2_{L^2}}
\quad \text{(Young inequality and Lemma \ref{lem2.6})}
\end{align*}
Then we obtain 
\[
|I_1| \leq C_\delta{\|u^\theta\|_{L^\beta}^{\frac{2}{1-{\frac{3}{\beta}}}}}
{\|\omega\|_{L^2}^2}+2\delta{\|\nabla\omega\|^2_{L^2}}
\]
Similarly, one has
\begin{align*}
|I_{2}|, |I_{3}|, |I_{4}|, |I_{6}|, |I_{7}| 
\leq C_\delta{\|u^\theta\|_{L^\beta}^{\frac{2}{1-{\frac{3}{\beta}}}}}
{\|\omega\|_{L^2}^2}+2\delta{\|\nabla\omega\|^2_{L^2}}.
\end{align*}
One can see that $I_{5}$ is a difficult term, but we can obtain a term that 
can be estimated by Lemma \ref{lem2.10},
\begin{align*}
|I_{5}| 
&\leq \int_{\mathbb{R}^3}{|\frac{\omega^\theta}{r}\omega^\theta u^r|}{\rm d}x 
\leq C{\|u^r\|_{L^2}}(\int_{\mathbb{R}^3}{\frac{(\omega^\theta)^4}{r^2}}
 {\rm d}x)^{1/2}
\quad {\text{(H\"older inequality)}}\\
&\leq C{\|\frac{\omega^\theta}{r}\|_{L^6}} {\|\omega^\theta\|_{L^3}}
 \leq C{\|\frac{\omega^\theta}{r}\|_{L^6}}{\|\omega^\theta\|_{L^2}^{1/2}}
 {\|\nabla \omega^\theta\|_{L^2}^{1/2}}
\; \text{ (Gagliardo-Nirenberg ineq.)}\\
& \leq C_{\delta}{\|\nabla\frac{\omega^\theta}{r}\|_{L^{2}}^{\frac{4}{3}}}
 {\|\omega^\theta\|_{L^2}^{\frac{2}{3}}}
+\delta{\|\nabla \omega^\theta\|_{L^2}^2}
\quad \text{(Young inequality )}\\
&\leq {\|\nabla\frac{\omega^\theta}{r}\|_{L^{2}}^{2}} 
+ C_{\delta}{\|\omega^\theta\|_{L^2}^{2}}
+\delta{\|\nabla \omega^\theta\|_{L^2}^2}.
\quad \text{(Young inequality )}\\
&\leq\|\nabla\frac{\omega^\theta}{r}\|_{L^{2}}^2 
 +C_{\delta}{\|\omega\|_{L^2}^{2}}+\delta{\|\nabla \omega\|_{L^2}^2}.
\quad \text{(Lemma \ref{lem2.6})}\\
\end{align*}
Using integration by parts, one has
\begin{align*}
I_{8}&=\int_{\mathbb{R}^3}{\omega^z \partial_{z} u^z \omega^z} {\rm d}x 
 = \int_{\mathbb{R}^3} {(\partial_{r}{u^\theta}+\frac{u^\theta}{r})
 {\partial_{z}u^z}{\omega^r}}{\rm d}x\\
&= \int_{\mathbb{R}^3}{{\partial_{r}{u^\theta}}{\partial_z\cdot
  u^z\cdot\omega^r}}{\rm d}x +\int_{\mathbb{R}^3}{{\frac{u^\theta}{r}
 \cdot{\partial_z{u^z}\cdot\omega^r}}}{\rm d}x\\
& = -\int_{\mathbb{R}^3}{u^\theta\cdot{\partial_r\partial_z u^z}\cdot 
 \omega^r}{\rm d}x -\int_{\mathbb{R}^3}{u^\theta \cdot \partial_z u^z 
\cdot\partial_r \omega^r}{\rm d}x +\int_{\mathbb{R}^3}{u^\theta 
 \cdot {\partial_z u^z}\cdot{\frac{\omega^r}{r}}}{\rm d}x\\
&:= I_8^1 +I_8^2 +I_8^3
\end{align*}
For the term $I_{8}^{1}$,
\begin{align*}
|I_{8}^{1}| 
&\leq \int_{\mathbb{R}^3}{|u^\theta\cdot \partial_r \partial_z 
\cdot\omega^r|}{\rm d}x \leq 
\Big\{\int_{\mathbb{R}^3}{|u^\theta\cdot \omega^r|^2}{\rm d}x\Big\}^{1/2}
 {\|\partial_r\partial_z u^z\|_{L^2}}
\quad {\text{(H\"older ineq.)}}\\
&\leq C_\delta\int_{\mathbb{R}^3}{|u^\theta\cdot \omega^r|^2}{\rm d}x 
+\delta{\|\partial_r\partial_z u^z\|_{L^2}}
\quad \text{(Young inequality )}\\
&\leq C_\delta {\|u^\theta\|_{L^\beta}^2}{\|\omega\|_{\frac{2\beta}{\beta-2}}^2}
 +\delta{\|\nabla\omega\|_{L^2}^2}
\quad \text{(H\"older inequality and Lemma \ref{lem2.7})}\\
&\leq C_\delta {\|u^\theta\|_{L^\beta}^2} 
\left\{{\|\omega\|_{L^2}^\theta}{\|\nabla\omega\|_{L^2}^{1-\theta}}\right\}^{2} 
+\delta{\|\nabla\omega\|_{L^2}^2}
\quad \text{(Gagliardo-Nirenberg inequlity )}\\
&\leq C_\delta{\|u^\theta\|_{L^\beta}^{\frac{2}{\theta}}}{\|\omega\|_{L^2}^2}
+2\delta{\|\nabla\omega\|_{L^2}^2}
\quad \text{(Young inequality )}
\end{align*}
The quantities $|I_{8}^{2}|$, $|I_{8}^{3}|$ can be estimated similarly to 
$|I_{8}^{1}|$. Putting together the above estimates, and taking $\delta$ 
small enough, then one have
\[
\frac{d}{dt}\|{\omega}\|_{L^2}^2+ \|{\nabla \omega}\|_{L^2}^2 
\leq C\Big\{1+{\|u^\theta\|_{L^\beta}^{\frac{2}{1-\frac{3}{\beta}}}} \Big\}
{\|\omega\|_{L^2}^2} + C{\|\nabla\frac{\omega^\theta}{r}\|_{L^2}^2}
\]
Using Gronwall's inequality, we have
\begin{align*}
&{\|\omega\|_{L^\infty((0,T);L^2)}^2}+{\|\nabla\omega\|_{L^2((0,T);L^2)}^2} \\
&\leq C\big\{{\|\omega_{0}\|_{L^2}^2}+{\|\nabla\frac{\omega^\theta}{r}\|_{L^2
(0, T; L^{2})}^{2}}\big\}
\exp\Big\{C + C\int_{0}^{T}{\|u^\theta\|_{L^\beta}^{\frac{2\beta}{\beta-3}}}{\rm d}t
\Big\}.
\end{align*}
Therefore, combining with Lemma \ref{lem2.11} then we completes the proof of 
Theorem \ref{thm1.1}.
\end{proof}


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\end{document}