\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 72, pp. 1--14.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2012 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2012/72\hfil Blow-up criterion]
{Blow-up criterion for two-dimensional heat
convection equations with zero heat conductivity}

\author[Y.-Z. Wang, Z. Wei \hfil EJDE-2012/72\hfilneg]
{Yu-Zhu Wang, Zhiqiang Wei}

\address{Yu-Zhu Wang \newline
School of Mathematics and Information Sciences \\
North China University of Water Resources and Electric
Power,  Zhengzhou 450011, China}
\email{yuzhu108@163.com}

\address{Zhiqiang Wei \newline
School of Mathematics and Information Sciences \\
North China University of Water Resources and Electric Power,
Zhengzhou 450011, China}
\email{weizhiqiang@ncwu.edu.cn}

\thanks{Submitted February 28, 2012. Published May 10, 2012.}
\subjclass[2000]{76D03, 35Q35}
\keywords{Heat convection equations; smooth solutions; blow-up criterion}

\begin{abstract}
 In this article we obtain a blow-up criterion of smooth solutions
 to Cauchy problem for  the incompressible  heat convection equations
 with  zero  heat conductivity in $\mathbb{R}^2$.
 Our proof is based on careful H\"oder estimates of heat and transport
 equations and the standard Littlewood-Paley theory.
\end{abstract}

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

\section{Introduction}

 The  incompressible  heat convection equations in two space
dimensions take the form
\begin{equation}
\begin{gathered}
 \partial_t u+u\cdot\nabla u+\nabla \pi =\mu \Delta u+\theta e_2, \\
 \partial_t\theta +u\cdot\nabla \theta-\nu\Delta \theta
=\frac{\mu}{2}\sum^2_{i, j=1}(\partial_iu^j+\partial_ju^i)^2,  \\
\nabla \cdot u=0,
\end{gathered}
\label{e1.1}
\end{equation}
where $u=(u^1, u^2)^t$ is the fluid velocity, $ \pi $ is the
pressure, $\theta$ stands for the  absolute temperature, $\mu$ is
the coefficient of viscosity, $\nu$ is  the coefficient of heat
conductivity and $e_2=(0, 1)$.

 Some problems related to \eqref{e1.1} have been studied in recent
years (see \cite{lamb}, \cite{fo}, \cite{hi}-\cite{rk} and \cite{lk}).
 Fan and Ozawa \cite{fo} obtained some regularity criteria of strong
solutions to the Cauchy problem for the  \eqref{e1.1} in $\mathbb{R}^3$.
Hiroshi \cite{hi} proved the existence of the strong
solutions for the initial boundary value problems for \eqref{e1.1}.
Kagei and Skowron \cite{ks} discussed the existence and uniqueness
of solutions of the initial-boundary value problem for the heat
convection equations \eqref{e1.1} of incompressible asymmetric fluids in
$\mathbb{R}^3$. Moreover, Kagei \cite{kage} considered global
attractors for the initial-boundary value problem for \eqref{e1.1} in
$\mathbb{R}^2$. Lukaszewicz and Krzyzanowski \cite{lk} treated
the initial-boundary value problem for \eqref{e1.1} with moving boundaries
in $\mathbb{R}^3$.  Kakizawa \cite{rk} proved that \eqref{e1.1} has
uniquely a mild solution. Moreover, a mild solution of \eqref{e1.1} can be
a strong or classical solution under appropriate assumptions for
initial data.

 It is well known that the Boussinesq approximation \cite{bj1} is a
simplified model of heat convection of incompressible viscous
fluids. There is no doubt that many investigations on the Boussinesq
approximation have been carried out for a hundred years.
For regularity criteria of weak solutions and blow up
criteria of smooth solutions, we refer to \cite{fz}  and  so on.

Equation \eqref{e1.1} is the Navier-Stokes equations coupled with the heat equation.
Due to its importance in mathematics and physics, there is  lots of literature
devoted to the mathematical theory of the Navier-Stokes equations.
Leray-Hopf weak solution were constructed by Leray \cite{leray} and
Hopf \cite{hopf}, respectively. Later on, much effort has been
devoted to establish the global existence and uniqueness of smooth solutions
 to the Navier-Stokes equations.
Different criteria for regularity of the weak solutions have been proposed and
 many interesting results were established
 (see  \cite{cg},  \cite{fo}-\cite{fjnz},  \cite{gala1}, \cite{he},
\cite{serrin} and \cite{zhou}-\cite{zg}). Serrin-type regularity
criteria of Leray weak solutions  in terms of pressure in Besov space were
 obtained in \cite{gg} and \cite{hg}.

 In this paper, we consider \eqref{e1.1} with the zero heat
conductivity; i.e., $\nu=0$. Without loss of generality, we take
$\mu=1$. The corresponding  heat convection equations thus reads
\begin{equation}
\begin{gathered}
\partial_t u+u\cdot\nabla u+\nabla \pi = \Delta u+\theta e_2, \\
\partial_t\theta +u\cdot\nabla \theta
=\frac{1}{2}\sum^2_{i, j=1}(\partial_iu^j+\partial_ju^i)^2, \\
\nabla \cdot u=0.
\end{gathered}
\label{e1.3}
\end{equation}
Due to the  term $\frac{1}{2}\sum^2_{i,j=1}(\partial_iu^j+\partial_ju^i)^2$,
 it is very difficult to deal with  \eqref{e1.3}.
The local well-posedness of the Cauchy problem for \eqref{e1.3} is rather standard,
 which can be obtained by standard Galerkin
method and energy estimates (for example see \cite{fo}). In the
absence of global well-posedness, the development of blow-up/ non
blow-up theory (see \cite{bkm}) is of major importance for both theoretical and
pratical purposes. In this paper, we obtain a blow-up criterion of
smooth solutions to the Cauchy problem  for
\eqref{e1.3}. Our main theorem is as follows.

\begin{theorem} \label{thm1.1}
Assume that $(u, \theta)$ is a local smooth
solution to the heat convection  equations with  zero  heat
conductivity  \eqref{e1.3} on $[0, T)$ and
$\|u(0)\|_{H^1\cap\dot{C}^{1+\alpha}}+\|\theta(0)\|_{L^2\cap \dot{C}^\alpha}<\infty
$ for some $\alpha \in (0, 1)$.
Then
$$
\|u(t)\|_{\dot{C}^{1+\alpha}}+\|\theta(t)\|_{\dot{C}^\alpha}<\infty
$$
for all $0\leq t\leq T$ provided that
\begin{equation}
\|u\|_{L^2_T(\dot{B}^0_{\infty, \infty})}<\infty, \quad
\|\theta\|_{L^1_T(\dot{B}^0_{\infty, \infty})}<\infty. \label{e1.4}
\end{equation}
\end{theorem}

 This article is organized as follows. We first state some preliminary
on functional settings and some important inequalities in Section 2
and then prove the blow-up criterion of smooth solutions of
 \eqref{e1.3} in Section 3.


\section{Preliminaries}

 Let $\mathcal{S}(\mathbb{R}^2)$ be the Schwartz class of rapidly
decreasing functions. Given $f \in \mathcal{S}(\mathbb{R}^2)$, its
Fourier transform $\mathcal{F}f=\hat{f}$ is defined by
$$
\hat{f}(\xi)=\int_{\mathbb{R}^2} e^{-ix\cdot\xi}f(x)dx
$$
and for any given $g \in \mathcal{S}(\mathbb{R}^2)$, its inverse
Fourier transform $\mathcal{F}^{-1}g=\check{g}$ is defined by
$$
\check{g}(x)=\int_{\mathbb{R}^2} e^{ix\cdot\xi}g(\xi)d\xi.
$$

 Next let us recall the Littlewood-Paley decomposition. Choose
two non-negative radial functions $\chi, \phi \in
\mathcal{S}(\mathbb{R}^2)$, supported respectively in
$\mathbb{B}=\{\xi \in \mathbb{R}^2: |\xi|\leq \frac{4}{3}\}$ and
$\mathcal{C}=\{ \xi \in \mathbb{R}^2: \frac{3}{4}\leq|\xi|\leq
\frac{8}{3}\}$ such that
$$
\chi(\xi)+\sum_{k\geq 0}\phi(2^{-k}\xi)=1, \quad \forall \xi \in \mathbb{R}^2
$$
and
$$
\sum ^{\infty}_{k=-\infty}\phi (2^{-k}\xi)=1, \quad \forall
\xi \in \mathbb{R}^2\backslash\{0\}.
$$
The frequency localization operator is defined by
$$
\Delta_kf =\int_{\mathbb{R}^2}\check{\phi}(y)f(x-2^{-k}y)dy,
\quad S_kf=\sum_{k'\leq k-1}\Delta_{k'}f.
$$

 Let us now recall  homogeneous  Besov spaces (for example,
see \cite{bl} and \cite{ht}). For $(p, q)\in [1, \infty]^2 $ and
$s \in \mathbb{R}$, the homogeneous Besov space $\dot{B}^s_{p, q}$ is
defined as the set of $f$ up to polynomials such that
$$ \|f\|_{\dot{B}^s_{p,q}} = \|2^{ks}
\|\Delta_kf\|_{L^p}\|_{l^q(\mathbb{Z})}< \infty.
$$

Finally, we recall the following space, which is defined
in \cite{cjy2}. Let $ p$  be in $[1, \infty]$  and
$ r \in \mathbb{R}$; the space $\tilde{L}^p_T(C^r)$ is the space of the
distributions $ f $ such that
$$
\|f\|_{\tilde{L}^p_T(C^r)}=\sup_k 2^{kr}
\|\Delta_kf\|_{L^p_T(L^\infty)}<\infty.
$$

The open  ball with radius $R$ centered at $x_0 \in
\mathbb{R}^2$ is denoted by $\textbf{B}(x_0,R)$. The ring $\{\xi\in
\mathbb{R}^2 | R_1 \leq |\xi|\leq R_2\}$ is denoted by
$\textbf{C}(0,R_1,R_2)$.

 In what follows, we  shall use  Bernstein
inequalities, which can be found in \cite{cjy}.

\begin{lemma} \label{lem2.1}
Let $k$ a positive integer and $\sigma$ any smooth
homogeneous function of degree $m\in \mathbb{R}$.
A constant $ C $ exists such that, for any positive real number
 $\lambda$ and any function $f $ in  $L^p(\mathbb{R}^2)$, we have
\begin{gather}
\operatorname{supp} \hat{f}\subset \lambda\textbf{B}\Rightarrow
 \sup_{|\beta|=k}\|\partial^\beta f\|_{L^q}\leq
C\lambda^{k+2(\frac{1}{p}-\frac{1}{q})}\|f\|_{L^p},\label{e2.1}
\\
\operatorname{supp} \hat{f}\subset \lambda \textbf{C}
\Rightarrow C^{-1}\lambda^{k}\|f\|_{L^p}\leq \sup_{|\beta|=k}
\|\partial^\beta f\|_{L^p}\leq C \lambda^{k}\|f\|_{L^p}.
\label{e2.2}
\end{gather}
Moreover, if $\sigma$ is a smooth function on $\mathbb{R}^2$ which
is homogeneous of degree $m$ outside a fixed ball, then we have
\begin{equation}
\operatorname{supp} \hat{f}\subset \lambda\textbf{C} \Rightarrow
\|\sigma(D)f\|_{L^q}\leq C \lambda^{(m+2(\frac{1}{p}-\frac{1}{q}))}\|f\|_{L^p}.
\label{e2.3}
\end{equation}
\end{lemma}

\begin{lemma} \label{lem2.2}
For any $f \in L^p(\mathbb{R}^2)(p>1)$ and any
positive real number $\lambda$,
\begin{equation}
\operatorname{supp} \hat{f}\subset \lambda \textbf{C} \Rightarrow
 \|e^{t\Delta}f\|_{L^p}\leq Ce^{-c \lambda^{2}t}\|f\|_{L^p},
\label{e2.4}
\end{equation}
where  $C$ and $c$ are positive constants.
 See \cite{cjy1} for the proof of
 \eqref{e2.4}.
\end{lemma}

The following lemma plays an important role in the proof of
Theorem \ref{thm1.1} (see also \cite {mzz} and \cite{m} where similar estimate were
established).

\begin{lemma} \label{lem2.3}
Assume that $\gamma>0$, then there exists a positive constant $C>0$ such that
\begin{equation}
\|f\|_{L^\infty}\leq C \left(1+\|f\|_{L^2}+\|f\|_{\dot{B}^0_{\infty, \infty}}\ln(e+\|f\|_{\dot{C}^\gamma})\right)
\label{e2.5}
\end{equation}
and
\begin{equation}
\begin{split}
\int^T_0\|\nabla f(\tau)\|_{L^\infty}d\tau 
&\leq C\Big(1+\int^T_0\| f(\tau) \|_{L^2}d\tau
+  \sup_k\int^T_0\|\Delta_k\nabla 
f(\tau)\|_{L^\infty}d\tau\\
&\quad\times \ln\big(e+\int^T_0\|\nabla
f(\tau)\|_{\dot{C}^\gamma}d\tau\big)\Big).
\end{split}\label{e2.6}
\end{equation}
\end{lemma}

\begin{proof}
If $f\in W^{m,p}$, $m>\frac{2}{p}$,
 $C^\gamma $ in \eqref{e2.5} is replaced by $W^{m, p}$, then \eqref{e2.5}
still holds. For example, see \cite{bkm,kt}. It is not difficult to prove
\eqref{e2.5} (see \cite{ww}). For the reader convenience, we give a detail proof. It
follows from Littlewood-Paley composition that
\begin{equation}
 f=\sum^0_{k=-\infty }\Delta_k
f+\sum^A_{k=1}\Delta_kf+\sum^\infty_{k= A+1}\Delta_k f.
\label{e2.7}
\end{equation}
Using  \eqref{e2.7} and \eqref{e2.3}, we obtain
\begin{align*}
{ \| f \|_{L^\infty}}
&\leq {  \sum^0_{k=-\infty}
\|\Delta_k f\|_{L^\infty}+ A\max_{1\leq k\leq A}\|\Delta_k f\|_{L^\infty}
+\sum^\infty_{k= A+1}\|\Delta_kf\|_{L^\infty}} \\
&\leq  { C\sum^0_{k=-\infty }
2^k\|\Delta_k f\|_{L^2}+A\|f\|_{\dot{B}^0_{\infty, \infty}}+
 \sum^\infty_{k= A+1}2^{-\gamma k}2^{\gamma k}\|\Delta_kf\|_{L^\infty}} \\
&\leq  {  C\|f\|_{L^2}+A\|f\|_{\dot{B}^0_{\infty, \infty}}
 + \sum^\infty_{k= A+1}2^{-\gamma k}\| f\|_{\dot{C}^\gamma}} \\
&\leq {  C\| f\|_{L^2}+A\|f\|_{\dot{B}^0_{\infty, \infty}}+2^{-\gamma A}
 \| f\|_{\dot{C}^\gamma}.}
\end{align*}
Equation \eqref{e2.5} follows immediately by choosing
$$
A=\frac{1}{\gamma}\log_2(e+\| f\|_{\dot{C}^\gamma})
\leq C\ln (e+\|f\|_{\dot{C}^\gamma}).
$$
Similar to the proof of \eqref{e2.5}, we can obtain \eqref{e2.6}
(see also \cite{lmz}). Thus the proof is complete.
\end{proof}

To prove Theorem \ref{thm1.1}, we need the following interpolation
inequalities  in  two space dimensions.

\begin{lemma} \label{lem2.4}
The following inequalities hold
\begin{equation}
\|f\|_{L^p}\leq C\|f\|^{1-\frac{2}{q}
+\frac{2}{p}}_{L^q}\|\nabla f\|^{\frac{2}{q}-\frac{2}{p}}_{L^q},\quad
 -\frac{2}{p}\leq 1-\frac{2}{q}, \quad
 p\geq q.
\label{e2.8}
\end{equation}
\end{lemma}

\begin{proof}
 Noting $ -\frac{2}{p}\leq 1-\frac{2}{q}$,
$p\geq q$ and using the Sobolev embedding theorem, we obtain
\begin{equation}
\|f\|_{L^p}\leq C(\|f\|_{L^q}+\|\nabla f\|_{L^q}). \label{e2.9}
\end{equation}
Let
$f_\lambda(x)=f(\lambda x)$.
From \eqref{e2.9}, we obtain
$$
\|f_\lambda\|_{L^p}\leq C(\|f_\lambda\|_{L^q}+\|\nabla f_\lambda\|_{L^q}),
$$
which implies
\begin{equation}
\|f\|_{L^p}\leq C(\lambda^{\frac{2}{p}-\frac{2}{q}}\|f\|_{L^q}
+\lambda^{1+\frac{2}{p}-\frac{2}{q}}\|\nabla f\|_{L^q}).
\label{e2.10}
\end{equation}
Taking $\lambda=\|f\|_{L^q}\|\nabla f\|^{-1}_{L^q}$, from \eqref{e2.10}, we
immediately obtain \eqref{e2.8}. Thus,   the proof is complete.
\end{proof}


\section{Proof of main results}

 This section is devoted to the proof of Theorem \ref{thm1.1},
for which we need the following Lemma  that is basically established
in \cite{fo}. For completeness, the proof
is also sketched here.

\begin{lemma} \label{lem3.1}
 Assume $\|u(0)\|_{ H^1}+ \|\theta(0)\|_{L^2}<\infty$ and  assume furthermore
 that $ (u, \theta) $ is a smooth solution
to the Cauchy problem for \eqref{e1.3} on $\times [0, T)$. If
\begin{equation}
 u\in L^2\left(0, T; \dot{B}^0_{\infty,
\infty}(\mathbb{R}^2)\right),
\label{e3.1}
\end{equation}
then
\begin{equation}
\begin{aligned}
&\|u(t)\|^2_{L^2}+\|\nabla u(t)\|^2_{L^2}+\|\theta(t)\|^2_{L^2}
+\int^T_0(\|\nabla u(t)\|^2_{L^2}+\|\Delta u(t)\|^2_{L^2})dt\\
&\leq C(\|u(0)\|^2_{H^1}+\|\theta(0)\|^2_{L^2}).
\end{aligned} \label{e3.2}
\end{equation}
\end{lemma}

\begin{proof}
 Multiplying the first equation in \eqref{e1.3} by
$u$ and using Cauchy inequality, we obtain
\begin{equation}
\frac{1}{2}\frac{d}{dt}\|u(t)\|^2_{L^2}+\|\nabla u(t)\|^2_{L^2}
\leq \frac{1}{2}\int_{\mathbb{R}^2}(|\theta|^2+|u|^2)(x, t)dx.
\label{e3.3}
\end{equation}
Multiplying the first equation in \eqref{e1.3} by $ -\Delta u,$ using
integration by parts, we obtain
\begin{equation}
\frac{1}{2}\frac{d}{dt}\|\nabla u(t)\|^2_{L^2}+\|\Delta u(t)\|^2_{L^2}
=-\int_{\mathbb{R}^2}\theta e_2\cdot \Delta u dx
+\int_{\mathbb{R}^2}u\cdot\nabla u\cdot \Delta udx.
\label{e3.4}
\end{equation}
Note that (see \cite{zf})
$$
-\Delta u=\nabla \times (\nabla \times u), \quad
 \nabla \times (u\cdot\nabla u)=u\cdot\nabla (\nabla\times u)
$$
provided that $\nabla\cdot u=0$.

Using integration by parts,  we obtain
\begin{equation}
\begin{aligned}
{ \int_{\mathbb{R}^2}u\cdot\nabla u\cdot \Delta
udx}&=  -\int_{\mathbb{R}^2} (u\cdot \nabla u)\cdot \nabla
\times (\nabla \times u)dx \\
&=  -\int_{\mathbb{R}^2} \nabla \times (u\cdot\nabla u)\cdot\nabla \times udx    \\
&=   -\int_{\mathbb{R}^2}u\cdot\nabla (\nabla \times u)\cdot(\nabla \times u)dx
=    0.
\end{aligned} \label{e3.5}
\end{equation}

It follows from \eqref{e3.4}, \eqref{e3.5} and Young inequality that
\begin{equation}
\frac{1}{2}\frac{d}{dt}\|\nabla u(t)\|^2_{L^2}+\frac{1}{2}\|\Delta
u(t)\|^2_{L^2} \leq
C\|\theta(t)\|^2_{L^2}. \label{e3.6}
\end{equation}
Multiplying the second equation in \eqref{e1.3} by $\theta$, using
H\"{o}lder inequality and Young inequality, it holds that
\begin{equation}
\begin{aligned}
 \frac{1}{2}\frac{d}{dt}\|\theta(t)\|^2_{L^2}
&= -\frac{1}{2}\sum^2_{i, j=1}\int_{\mathbb{R}^2}
 \theta(\partial_iu_j+\partial_ju_i)^2dx   \\
&\leq   C\|\theta(t)\|_{L^2}\|\nabla u(t)\|^2_{L^4}  \\
&\leq  C\|\theta(t)\|_{L^2} \|u(t)\| _{\dot{B}^0_{\infty, \infty}}
 \|\Delta u(t)\|_{L^2} \\
&\leq   \frac{1}{6}\|\Delta u(t)\|^2_{L^2}+C\|u(t)\|^2 _{\dot{B}^0_{\infty, \infty}}
\|\theta(t)\|^2_{L^2},
\end{aligned} \label{e3.7}
\end{equation}
where we have used  the interpolation inequality (see for example \cite{mo})
\begin{equation}
\|\nabla u(t)\|_{L^4} \leq C \|u(t)\|^{1/2}
_{\dot{B}^0_{\infty, \infty}}\|\Delta u(t)\|^{1/2}_{L^2}.
\label{e3.8}
\end{equation}
Collecting \eqref{e3.3}, \eqref{e3.6} and \eqref{e3.7} gives
\begin{equation}
\begin{aligned}
&\frac{d}{dt}(\|u(t)\|^2_{L^2}+\|\nabla
u(t)\|^2_{L^2}+\|\theta(t)\|^2_{L^2})+\|\nabla
u(t)\|^2_{L^2}+\|\Delta u(t)\|^2_{L^2}\\
& \leq C\left(\|u(t)\|^2_{L^2}+\|\theta(t)\|^2_{L^2}+\|u(t)\|^2_{\dot{B}^{0}_{\infty,
\infty}}(\|\nabla u(t)\|^2_{L^2}+\|\theta(t)\|^2_{L^2})\right).
\end{aligned} \label{e3.9}
\end{equation}
Inequality \eqref{e3.2}  follows immediately  from \eqref{e3.1},
\eqref{e3.9} and Gronwall's inequality.
 Thus, the proof complete.
\end{proof}

 We also need the following lemma (see also  \cite{cjy2,lmz}
 where similar estimates were established).

\begin{lemma} \label{lem3.2}
Assume that $F\in \tilde{L}^1_T(C^{-1})\cap L^2_T({L}^{2})$ and
 $u_0\in L^2$. Let $u$ be a solution of the
Navier-Stokes equations
\begin{equation}
\begin{gathered}
\partial_t u+u\cdot\nabla u+\nabla \pi = \Delta u+F, \\
 \nabla \cdot u=0,  \\
t=0:\quad u=u_0(x).
\end{gathered} \label{e3.10}
\end{equation}
Then it holds that
\begin{equation}
\begin{aligned}
 \|u\|_{\tilde{L}^1_T(C^{1})}
&\leq C(\sup_k\|\Delta_ku_0\|_{L^2}(1-\exp\{-c2^{2k}T\})
+(\|u_0\|_{L^2} \\
&\quad +\|F\|_{L^2_T({L}^{2})})\|\nabla u\|^2_{L^2_T(L^2)}
+ \sup_k\int^T_02^{-k}\|\Delta_kF(\tau)\|_{L^\infty}d\tau ).
\end{aligned} \label{e3.11}
\end{equation}
\end{lemma}

\begin{proof}
Applying $\Delta_k$ to \eqref{e3.10},  we obtain
\begin{equation}
\Delta_ku=e^{\Delta t}\Delta_ku_0
 +\int^t_0e^{\Delta(t-\tau)}\Delta_k \mathbb{P}
\big(\nabla \cdot(u\otimes u)+F\big)(\tau)d\tau, \label{e3.12}
\end{equation}
where operator $\mathbb{P}$ satisfies
\[
(\hat{\mathbb{P}u})^i=\sum^2_{j=1}
(\delta_{ij}-\frac{\xi^i\xi^j}{|\xi|^2})\hat{u}^j(\xi).
\]
 It follows from \eqref{e2.3} and \eqref{e2.4} that
\begin{equation}
\begin{aligned}
&\|\Delta_ku(t)\|_{L^\infty}\\
&\leq C \Big(e^{-c2^{2k}t}\|\Delta_ku_0\|_{L^\infty}
+\int^t_0e^{-c2^{2k}(t-\tau)}\|\Delta_k\nabla \cdot(u\otimes u)(\tau)
\|_{L^\infty}d\tau\Big)\\
&\quad + C\int^t_0e^{-c2^{2k}(t-\tau)}\|\Delta_k F(\tau)\|_{L^\infty}d\tau.
\end{aligned} \label{e3.13}
\end{equation}
This implies that
\begin{equation}
\begin{aligned}
&\|u\|_{\tilde{L}^1_T(C^{1})}\\
&\leq   C\sup_k\int^T_0 2^ke^{-c2^{2k}t}\|\Delta_ku_0\|_{L^\infty}dt \\
&\quad + C\sup_k\int^T_0\int^t_02^{2k}e^{-c2^{2k}(t-\tau)}
 \|\Delta_k u\otimes u(\tau)\|_{L^\infty}d\tau dt \\
&\quad + C\sup_k\int^T_0\int^t_02^{k}e^{-c2^{2k}(t-\tau)}
\|\Delta_k F(\tau)\|_{L^\infty}d\tau dt \\
&\leq   C\sup_k\|\Delta_ku_0\|_{L^2}(1-e^{-c2^{2k}T})\\
&\quad + C\sup_k\int^T_0\|\Delta_k(u\otimes u)(\tau)\|_{L^\infty}d\tau
 +\sup_k\int^T_02^{-k}\|\Delta_kF(\tau)\|_{L^\infty}d\tau .
\end{aligned} \label{e3.14}
\end{equation}
It follows from Bony  decomposition that
\begin{align*}
&\|\Delta_k(u\otimes u)(\tau)\|_{L^\infty}\\
&=  \sum_{|m-n|\leq 1}\|\Delta_k(\Delta_mu\otimes\Delta_nu)(\tau)\|_{L^\infty}
+\sum_{m-n \geq  2}\|\Delta_k(\Delta_mu\otimes\Delta_nu)(\tau)\|_{L^\infty}\\
&\quad + \sum_{n-m\geq 2}\|\Delta_k(\Delta_mu\otimes\Delta_nu)(\tau)\|_{L^\infty}
\end{align*}
By  \eqref{e2.1} and \eqref{e2.2}, a straight computation gives
\begin{align*}
& \int^T_0\sum_{|m-n|\leq 1}\|\Delta_k(\Delta_mu\otimes\Delta_nu)
 (\tau)\|_{L^\infty}d\tau \\
&\leq  C\int^T_0\sum_{|m-n|\leq 1}2^k\|\Delta_k(\Delta_mu\otimes\Delta_nu)(\tau)
 \|_{L^2}d\tau \\
&\leq  C\int^t_0\sum_{|m-n|\leq 1, m\geq k-3}2^{k-\frac{m+n}{2}}
 \|2^m\Delta_m u(\tau)\|^{1/2}_{L^\infty}
\|\Delta_n u(\tau)\|^{1/2}_{L^2}\|\Delta_m u(\tau)\|^{1/2}_{L^\infty}\\
&\quad \times \|2^n\Delta_nu(\tau)\|^{1/2}_{L^2}d\tau \\
&\leq  C\int^t_0\sum_{|m-n|\leq 1, m\geq k-3}2^{k-\frac{m+n}{2}}
 \|2^m\Delta_m u(\tau)\|^{1/2}_{L^\infty}
\|\Delta_n u(\tau)\|^{1/2}_{L^2}\|2^m\Delta_m u(\tau)\|^{1/2}_{L^2}\\
&\quad\times \|2^n\Delta_nu(\tau)\|^{1/2}_{L^2}d\tau \\
& \leq  C\|u\|^{1/2}_{L^\infty_T(L^2)}\|\nabla u\|_{L^2_T(L^2)}
 \|u\|^{1/2}_{\tilde{L}^1_T(C^1)}.
\end{align*}
Similarly, we obtain
\begin{align*}
&\int^T_0\Big(\sum_{m-n\geq 2}\|\Delta_k(\Delta_mu\otimes \Delta_nu)(\tau)\|_{L^\infty}
+\sum_{n-m\geq 2}\|\Delta_k(\Delta_mu\otimes \Delta_nu)(\tau)\|_{L^\infty}\Big)d\tau \\
& \leq  C\int^T_0\sum_{m-n\geq 2, |m-k|\leq 2}\|\Delta_m u(\tau)
 \|_{L^\infty}\|\Delta_n u(\tau)\|_{L^\infty}d\tau \\
&\leq  C\sum_{m-n\geq 2, |m-k|\leq 2}\|2^m\Delta_m u(\tau)\|
^{1/2}_{L^\infty}\|2^m\Delta_m u(\tau)\|^{1/2}_{L^2}2^{n-\frac{m}{2}}
\|\Delta_n u(\tau)\|_{L^2}d\tau \\
 &\leq   C\|u\|^{1/2}_{L^\infty_T(L^2)}\|\nabla u\|_{L^2_T(L^2)}
\|u\|^{1/2}_{\tilde{L}^1_T(C^1)}.
\end{align*}
Using the above two estimates, from \eqref{e3.14} and Young inequality, we
obtain
\begin{equation}
\begin{aligned}
 \|u\|_{\tilde{L}^1_T(C^1)}
&\leq C(\sup_k\|\Delta_ku_0\|_{L^2}(1-\exp\{-c2^{2k}T\})
+\|u\|_{L^\infty_T(L^2)}\|\nabla u\|^2_{L^2_T(L^2)} \\
&\quad + \sup_k\int^T_02^{-k}\|\Delta_kF(\tau)\|_{L^\infty}d\tau ).
\end{aligned} \label{e3.15}
\end{equation}
Combining \eqref{e3.15} and the basic energy estimate
\begin{equation}
\|u\|^2_{L^\infty_T(L^2)}+\|\nabla u\|^2_{L^2_T(L^2)}\leq
C(\|u_0\|^2_{L^2}+\|F\|^2_{L^2_T({L}^{2})}) \label{e3.16}
\end{equation}
gives \eqref{e3.11}. Thus, the proof is complete.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm1.1}]
Set $F=u\cdot\nabla u+\theta e_2$.
 It follows from \eqref{e1.4} and \eqref{e3.2} that
$F\in \tilde{L}^1_T(C^{-1})\cap  L^2_T({L}^{2})$. Applying $\Delta_k$ to both
sides of \eqref{e3.10} and using standard energy estimate, \eqref{e2.2} and
Young inequality, we have
\begin{align*}
&\frac{1}{2}\frac{d}{dt}\|\Delta_ku\|^2_{L^2}+c2^{2k}\|\Delta_ku\|^2_{L^2}\\
&\leq \frac{c}{2}2^{2k}\|\Delta_ku\|^2_{L^2}
+C(\|\Delta_ku\|_{L^2}+\|\Delta_kF\|^2_{L^2}+\|\Delta_k(u\otimes
u)\|^2_{L^2}).
\end{align*}
 Integrating  the above inequality with respect to $ t $ and summing over $k$,
 we obtain
\begin{equation}
\begin{aligned}
&\sum_{k}\|\Delta_ku\|^2_{L^\infty_T(L^2)}
 +\sum_{k}\int^t_02^{2k}\|\Delta_ku(\tau)\|^2_{L^2}d\tau \\
&\leq   C(\|u_0\|^2_{L^2}+\|F\|^2_{L^2_T({L}^{2})}+
\|u\|^2_{L^\infty_T(L^2)}\|\nabla u\|^2_{L^2_T(L^2)}),
\end{aligned}\label{e3.17}
\end{equation}
where we used the interpolation inequality (see Lemma \ref{lem2.4})
$$
\|u\|_{L^4}\leq C\|u\|^{1/2}_{L^2}\|\nabla u\|^{1/2}_{L^2}.
$$
It follows from \eqref{e3.16} and \eqref{e3.17} that
\begin{equation}
\begin{aligned}
&\sum_{k}\|\Delta_ku\|^2_{L^\infty_T(L^2)}
+\sum_{k}\int^t_02^{2k}\|\Delta_ku(\tau)\|^2_{L^2}d\tau \\
&\leq C(\|u_0\|^2_{L^2}
+\|F\|^2_{L^2_T({L}^{2})})(1+\|u_0\|^2_{L^2}+\|F\|^2_{L^2_T({L}^{2})}).
\end{aligned}\label{e3.18}
\end{equation}
Using \eqref{e3.18}, for any $ t_0\in [0, T)$, we can choose $k_0>0$ such
that
$$
\sup_{k\geq k_0}\|\Delta_ku\|_{L^\infty_{[t_0, T]}(L^2)}\leq \frac{\varepsilon}{4C}.
$$
By \eqref{e3.16}, we can choose $t_1\in[t_0, T]$ such that
\begin{align*}
&\sup_{t_1\leq t\leq T}\sup_{k\leq
k_0}\|\Delta_ku(t)\|_{L^2}(1-\exp\{-c2^{2k}(T-t)\})\\
&\leq  \sup_{t_1\leq t\leq T}2c2^{2k_0}(T-t_1)\|u(t)\|_{L^2} \\
&\leq   C2^{2k_0}(\|u_0\|_{L^2}+\|F\|_{L^2_T({L}^{2})})(T-t_1) 
\leq  \frac{\varepsilon}{4C}.
\end{align*}
Consequently,
\begin{equation}
\sup_{t_1\leq t\leq T}\sup_k\|\Delta_ku(t)\|_{L^2}(1-\exp\{-c2^{2k}(T-t)\})
\leq \frac{\varepsilon}{2C}.
\label{e3.19}
\end{equation}
On the other hand, we can choose $t_2\in[t_1, T)$ such that
\begin{equation}
\begin{aligned}
&\Big(\sup_{t_2\leq t\leq T}\|u(t)\|_{L^2}+\|F\|_{L^2_{[t_2, T]}({L}^{2})}\Big)
\|\nabla u\|^2_{L^2_{[t_2, T]}(L^2)}\\
&+ \sup_k\int^T_{t_2}2^{-k}\|\Delta_kF(\tau)\|_{L^\infty}d\tau )\\
&\leq \frac{\varepsilon}{2C}.
\end{aligned}\label{e3.20}
\end{equation}
It follows from \eqref{e3.11} that
\begin{equation}
\begin{aligned}
\|u\|_{\tilde{L}^1_{[t_2, T]}(C^{1})}
&\leq
C\Big(\sup_k\|\Delta_ku(t_2)\|_{L^2}(1-\exp\{-c2^{2k}(T-t_2)\})\\
&\quad + \big(\|u(t_2)\|_{L^2}+\|F\|_{L^2_{[t_2, T]}({L}^{2})}\big)\|\nabla
u\|^2_{L^2_{[t_2, T]}(L^2)}\\
&\quad + \sup_k\int^T_{t_2}2^{-k}\|\Delta_kF(\tau)\|_{L^\infty}d\tau \Big).
\end{aligned}\label{e3.21}
\end{equation}
Combining \eqref{e3.19}-\eqref{e3.21} gives
\begin{equation}
 \|u\|_{\tilde{L}^1_{[t_2, T]}(C^{1})}\leq \varepsilon.
\label{e3.22}
\end{equation}
Using \eqref{e3.22} and \eqref{e1.4}, we can
choose $t^\ast\in [t_2, T)$ such that
\begin{equation}
\|u\|_{\tilde{L}^1_{[t^\ast, T]}(C^{1})}\leq \varepsilon, \quad
\|\theta\|_{L^1_{[t^\ast, T]}(\dot{B}^0_{\infty, \infty})}\leq \varepsilon.
\label{e3.23}
\end{equation}
 For $0\leq t< T$, define
$$
M(t)=\sup_{0\leq \tau <t}\|u(\tau)\|_{\dot{C}^{1+\alpha}}, \quad
N(t)=\sup _{0\leq \tau <t}\|\theta(\tau)\|_{\dot{C}^\alpha}.
$$
 In what follows, we estimate $M(t)$ and  $N(t)$ for $0\leq t<T$.
Applying $\Delta_k$ to the first and second equation in \eqref{e1.3}, we
obtain
\begin{equation}
\begin{gathered}
\partial_t \Delta_ku-\Delta \Delta_k u +\nabla \Delta_k\pi
 =-\nabla\cdot\Delta_k(u\otimes u)+\Delta_k(\theta e_2), \\
\partial_t \Delta_k\theta +u\cdot\nabla \Delta_k\theta
=\Delta_k\Big(\frac{1}{2}\sum^2_{i, j=1}(\partial_iu^j+\partial_ju^i)^2\Big)
+[u\cdot\nabla, \Delta_k]\theta.
\end{gathered}
\label{e3.24}
\end{equation}
Firstly, we make estimate $\|u(t)\|_{\dot{C}^{1+\alpha}}$. It follows
from the first equation in \eqref{e3.24} and  \eqref{e2.4} that
\begin{align*}
\|\Delta_ku(t)\|_{L^\infty}
&\leq  Ce^{-c2^{2k}t}\|\Delta_ku(0)\|_{L^\infty}
+C\int^t_0e^{-c2^{2k}(t-\tau)}\|\nabla \cdot \Delta_k(u\otimes u)
(\tau)\|_{L^\infty}d\tau\\
&\quad + C\int^t_0e^{-c2^{2k}(t-\tau)}\|\Delta_k(\theta e_2)(\tau)
\|_{L^\infty}d\tau.
\end{align*}
By the above inequality, \eqref{e2.1}, \eqref{e2.2} and  H\"{o}lder inequality, we obtain
\begin{equation}
\begin{aligned}
\|u(t)\|_{\dot{C}^{1+\alpha}}
&\leq  C\|u(0)\|_{\dot{C}^{1+\alpha}}+
C\int^t_02^{3k/2}e^{-c2^{2k}(t-\tau)}\|u\otimes
 u(\tau)\|_{\dot{C}^{\frac{1}{2}+\alpha}}d\tau   \\
 &\quad+ C\int^t_02^ke^{-c2^{2k}(t-\tau)}\|\theta (\tau)\|_{\dot{C}^\alpha}d\tau  \\
&\leq      C(\|u(0)\|_{\dot{C}^{1+\alpha}}+N(t))
 +  C(\int^t_0\|u\otimes u(\tau)\|^4_{\dot{C}^{\frac{1}{2}
+\alpha}}d\tau )^{1/4}.
\end{aligned}
\label{e3.26}
\end{equation}
By \eqref{e2.8}, we obtain
\begin{equation}
\|u\|_{L^4}\leq C\|u\|^{1/2}_{L^2}\|\nabla
u\|^{1/2}_{L^2}. \label{e3.27}
\end{equation}
Using  this inequality and the fact
$\|u \otimes u\|_{\dot{C}^{\frac{1}{2}+\alpha}}\leq C
\|u\|_{L^4}\|u\|_{\dot{C}^{1+\alpha}}$, we obtain
\begin{equation}
\begin{aligned}
& \|u(t)\|^4_{\dot{C}^{1+\alpha}}\\
&\leq  C(\|u(0)\|_{\dot{C}^{1+\alpha}}+N(t))^4+
C\int^t_0\|u(\tau)\|^2_{L^2}\|\nabla u(\tau)\|^2_{L^2}\|
\|u(\tau)\|^4_{\dot{C}^{1+\alpha}}d\tau \\
&\leq C(\|u(0)\|_{\dot{C}^{1+\alpha}}+N(\tilde{t}))^4+
 C\int^t_0\|u(\tau)\|^2_{L^2}\|\nabla u(\tau)\|^2_{L^2}\|
\|u(\tau)\|^4_{\dot{C}^{1+\alpha}}d\tau,
\end{aligned}\label{e3.28}
\end{equation}
for any fixed $\tilde{t}:  0\leq \tilde{t}\leq T$ and
$t\leq \tilde{t}<T$. Here we have used the fact that $N(t)$ is
nondecreasing.
Consequently, Gronwall's inequality gives
\begin{align*}
 M(\tilde{t})^4
&=  \sup_{0\leq t<
\tilde{t}}\|u(t)\|^4_{\dot{C}^{1+\alpha}}  \\
& \leq C(\|u(0)\|_{\dot{C}^{1+\alpha}}+N(\tilde{t}))^4\exp\{
C\int^t_0\|u(\tau)\|^2_{L^2}\|\nabla u(\tau)\|^2_{L^2}\|d\tau\}.
\end{align*}
Since $\tilde{t}\in [0, T)$ is arbitrary, by \eqref{e3.2}, we obtain
\begin{equation}
M(t)\leq C(\|u(0)\|_{\dot{C}^{1+\alpha}}+N(t)),  \quad \forall
t\in[0, T). \label{e3.29}
\end{equation}
We next continue to estimate $N(t)$. It follows from the second
equation in \eqref{e3.24} that
\begin{equation}
\begin{aligned}
 \|\Delta_k\theta\|_{L^\infty}
&\leq  C\|\Delta_k\theta(0)\|_{L^\infty}
+C\int^t_0\sum_{i,
j=1}^2\|\Delta_k(\partial_iu^j+\partial_ju^i)^2(\tau)\|_{L^\infty}d\tau \\
&\quad +  C\int^t_0\|[u\cdot\nabla, \Delta_k]\theta(\tau)\|_{L^\infty}d\tau.
\end{aligned}
\label{e3.30}
\end{equation}
Using  \eqref{e2.1}, \eqref{e2.2}, \eqref{e3.30} and H\"{o}lder inequality,
we have
\begin{equation}
\begin{aligned}
 \|\theta(t)\|_{\dot{C}^\alpha}
&\leq  C\|\theta(0)\|_{\dot{C}^\alpha}+C\int^t_0\|\nabla
u(\tau)\|_{L^\infty}\|u(\tau)\|_{\dot{C}^{1+\alpha}}d\tau \\
&\quad + C\int^t_02^{k\alpha}\|[u\cdot\nabla,
\Delta_k]\theta(\tau)\|_{L^\infty}d\tau.
\end{aligned}
\label{e3.31}
\end{equation}
It follows from Bony decomposition that
\begin{equation}
\begin{aligned}
[u\cdot\nabla, \Delta_k]\theta
&=  \sum_{|k'-r|\leq 1}[\Delta_{k'}u\cdot\nabla, \Delta_k]\Delta_r\theta
 + \sum_{k'\leq r-2}[\Delta_{k'}u\cdot\nabla, \Delta_k]\Delta_r\theta\\
&\quad  +\sum_{k'\leq r-2}[\Delta_{r}u\cdot\nabla, \Delta_k]\Delta_{k'}\theta\\
&=  \sum_{|k'-r|\leq 1}[\Delta_{k'}u\cdot\nabla, \Delta_k]\Delta_r\theta
+\sum_{|r-k|\leq 2}[S_{r-1}u\cdot\nabla, \Delta_k]\Delta_r\theta\\
&\quad + \sum_{|r-k|\leq 2}[\Delta_{r}u\cdot\nabla, \Delta_k]S_{r-1}\theta.
\end{aligned} \label{e3.32}
\end{equation}
Note that
$$
[S_{r-1}u, \Delta_k]f=\int_{\mathbb{R}^2} h(y)
[S_{r-1}u(x)-S_{r-1}u(x-2^{-k}y)]f(x-2^{-k}y)dy
,$$
we obtain
$$
\|[S_{r-1}u, \Delta_k]f\|_{L^\infty}\leq C2^{-k}
\|\nabla S_{r-1}u\|_{L^\infty}\|f\|_{L^\infty}.
$$
Hence
\begin{equation}
\begin{aligned}
& \sum_{|r-k|\leq 2}\int^t_02^{k\alpha}\|[S_{r-1}u\cdot\nabla,
\Delta_k]\Delta_r\theta(\tau)\|_{L^\infty}d\tau \\
& \leq
  C\sum_{|r-k|\leq 2}   \int^t_0 2^{k(\alpha-1)}\|\nabla S_{r-1}u\|_{L^\infty}
 \|\nabla \Delta_r\theta\|_{L^\infty}(\tau)d\tau \\
&\leq   C\sum_{|r-k|\leq 2}   \int^t_0 \|\nabla S_{r-1}u\|_{L^\infty} 2^{r\alpha}
 \| \Delta_r\theta\|_{L^\infty}(\tau)d\tau  \\
&\leq   C\int^t_0\|\nabla u(\tau)\|_{L^\infty}\|\theta(\tau)\|_{\dot{C}^\alpha} d\tau.
 \end{aligned}\label{e3.33}
\end{equation}
Note that
$$
[\Delta_ru, \Delta_k]f=\int_{\mathbb{R}^2} h(y)[\Delta_ru(x)
-\Delta_ru(x-2^{-k}y)]f(x-2^{-k}y)dy.
$$
Then, we have
$$
\|[\Delta_ru, \Delta_k]f\|\leq C2^{-k}\|\nabla \Delta_ru\|_{L^\infty}\|f\|_{L^\infty}.
$$
It follows from the above inequality and \eqref{e2.1}, \eqref{e2.2} that
\begin{equation}
\begin{aligned}
& \sum_{|r-k|\leq 2} \int^t_0 2^{k\alpha} \|[\Delta_ru\cdot\nabla,
 \Delta_k]S_{r-1}\theta(\tau)\| _{L^\infty}  \\
&\leq  C\sum_{|r-k|\leq 2}\int^t_02^{k(\alpha-1)}\|\nabla
 \Delta_ru\|_{L^\infty}\|\nabla S_{r-1}\theta\|_{L^\infty}(\tau)d\tau  \\
&\leq  C\int^t_0\|\theta(\tau)\|_{L^\infty}\|u(\tau)\|_{\dot{C}^{1+\alpha}}
 (\tau)d\tau.
\end{aligned} \label{e3.34}
\end{equation}
By a straightforward computation, we obtain
\begin{equation}
\begin{aligned}
&\sum_{|k'-r|\leq 1}\int^t_02^{k\alpha}\|[\Delta_{k'}u\cdot\nabla,
\Delta_k]\Delta_r\theta(\tau)\|_{L^\infty}d\tau \\
&\leq
C\int^t_0\|\theta(\tau)\|_{L^\infty}\|u(\tau)\|_{\dot{C}^{1+\alpha}}d\tau.
\end{aligned}\label{e3.35}
\end{equation}
Collecting \eqref{e3.31}-\eqref{e3.35} gives
\begin{equation}
\begin{aligned}
 \|\theta(t)\|_{\dot{C}^\alpha}
&\leq  C\|\theta(0)\|_{\dot{C}^\alpha}
 +C\int^t_0\|\nabla u(\tau)\|_{L^\infty}\|u(\tau)\|_{\dot{C}^{1+\alpha}}d\tau\\
&\quad + C\int^t_0(\|\nabla u(\tau)\|_{L^\infty}\|\theta(\tau)\|_{\dot{C}^\alpha}
+\|\theta(\tau)\|_{L^\infty}\|u(\tau)\|_{\dot{C}^{1+\alpha}})d\tau  \\
&\leq  C\|\theta(0)\|_{\dot{C}^\alpha}+
 C\int^t_0(\|\nabla u(\tau)\|_{L^\infty}
+\|\theta(\tau)\|_{L^\infty})(\|u(\tau)\|_{\dot{C}^{1+\alpha}}\\
&\quad +\|\theta(\tau)\|_{\dot{C}^\alpha})d\tau.
\end{aligned} \label{e3.36}
\end{equation}
From \eqref{e3.29} and \eqref{e3.36}, we obtain
\begin{equation}
 N(t) \leq   C\|\theta(0)\|_{\dot{C}^\alpha}+
 C\int^t_0(\|\nabla u(\tau)\|_{L^\infty}+\|\theta(\tau)\|_{L^\infty})
(\|u(0)\|_{\dot{C}^{1+\alpha}}+N(\tau))d\tau.
\label{e3.37}
\end{equation}
With the help of Lemma \ref{lem2.3} and \eqref{e3.23}, we obtain
\begin{equation}
\begin{aligned}
& C\int^t_0(\|\nabla u(\tau)\|_{L^\infty}+\|\theta(\tau)\|_{L^\infty})d\tau   \\
&\leq  C\int^{t_\star}_0 (\|\nabla u(\tau)\|_{L^\infty}
 +\|\theta(\tau)\|_{L^\infty})d\tau\\
&\quad + C\int^t_{t_\star} (1+\|u(\tau)\|_{L^2}+\|\theta(\tau)\|_{L^2})d\tau
\\
&\quad + C\int^t_{t_{\star}}\|\theta\|_{\dot{B}^0_{\infty, \infty}}
\ln(e+\|\theta(\tau)\|_{\dot{C}^\alpha})d\tau
\\
&\quad +C\sup_k\int^t_{t_\star}
\|\nabla \Delta_ku(\tau)\|_{L^\infty}d\tau
\ln\Big(e+\int^t_0\|u(\tau)\|_{\dot{C}^{1+\alpha}}d\tau\Big) \\
&\leq  C_\star+C\varepsilon \ln\left(e+N(t)\right)
 +C\varepsilon \ln[e+Ct(\|u(0)\|_{\dot{C}^{1+\alpha}}+N(t))] \\
&\leq   C_\star+C\varepsilon \ln(e+\|u(0)\|_{\dot{C}^{1+\alpha}}+N(t)),
\end{aligned}
\label{e3.38}
\end{equation}
where $C_\star$ is a positive constant depending on the solution
$(u, \theta)$ on $[0, t_\star]$.
 It follows from \eqref{e3.37}-\eqref{e3.38} that
 $$
N(t)\leq C_\star(1+\|u(0)\|_{\dot{C}^{1+\alpha}}
+\|\theta(0)\|_{\dot{C}^{\alpha}})+C\int^t_0(\|\nabla u(\tau)\|_{L^\infty}
 +\|\theta(\tau)\|_{L^\infty})N(\tau)d\tau,
$$
 provided that $\varepsilon >0$ is suitably small.
 By Gronwall's inequality and \eqref{e3.38}, we obtain
\begin{align*}
&e+\|u(0)\|_{\dot{C}^{1+\alpha}}+N(t) \\
&\leq   C_\star(e+\|u(0)\|_{\dot{C}^{1+\alpha}}
+\|\theta(0)\|_{\dot{C}^\alpha})
 \exp\{C\int^t_0(\|\nabla u(\tau)\|_{L^\infty}
+\|\theta(\tau)\|_{L^\infty})d\tau\} \\
&\leq  C_\star(e+\|u(0)\|_{\dot{C}^{1+\alpha}}+\|\theta(0)\|_{\dot{C}^\alpha})
 \exp\{C_\star+C\varepsilon \ln(e+\|u(0)\|_{\dot{C}^{1+\alpha}}+N(t))\} \\
&\leq  C_\star(e+\|u(0)\|_{\dot{C}^{1+\alpha}}+\|\theta(0)\|_{\dot{C}^\alpha})
(e+\|u(0)\|_{\dot{C}^{1+\alpha}}+N(t))^{C\varepsilon}.
\end{align*}
 Choosing $\varepsilon>0$ suitably small,  the above inequality and
 \eqref{e3.29} yields
 $$
M(t)+N(t)\leq C_\star(1+\|u(0)\|_{\dot{C}^{1+\alpha}}
+\|\theta(0)\|_{\dot{C}^\alpha})^2.
$$
 The proof is complete.
\end{proof}

\subsection*{Acknowledgements}
The research is supported by grant 11101144 from the  NNSF of China.

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