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

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2005(2005), No. 125, pp. 1--13.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2005 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2005/125\hfil Decay of solutions]
{Decay of solutions to equations modelling incompressible bipolar
non-newtonian fluids}
\author[B.-Q. Dong\hfil EJDE-2005/125\hfilneg]
{Bo-Qing Dong}

\address{Bo-Qing Dong \hfill\break
School of  Mathematical Sciences, Nankai University,
Tianjin 300071, China}
\email{bqdong@mail.nankai.edu.cn}

\date{}
\thanks{Submitted March 28, 2005. Published November 7, 2005.}
\subjclass[2000]{35B40, 35Q35, 76A05}
\keywords{Decay; bipolar non-Newtonian fluids; Fourier splitting method}

\begin{abstract}
 This article concerns  systems of equations
 that model incompressible bipolar non-Newtonian fluid motion
 in  the whole space $\mathbb{R}^n$.  Using the  improved Fourier
 splitting method,  we prove that  a  weak solution decays in
 the $L^2$ norm at the same rate as  $(1+t)^{-n/4}$ as the time
 $t$ approaches infinity.
 Also we obtain optimal $L^2$ error-estimates for Newtonian
 and  Non-Newtonian flows.
\end{abstract}

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

\def\re{{\rm Re}}
\def\im{{\rm Im}}
\def\H{{\bf H}}
\def\ep{{ \varepsilon }}

\section{Introduction}
  Consider  the  viscous incompressible fluid motion governed by
  the  momentum and continuity equations
\begin{gather}\label{e1.1}
    \partial_t u+ (u\cdot\nabla)u-\nabla\cdot\tau^v+\nabla\pi =0\quad
     \mbox{in } \mathbb{R}^n\times(0, \infty),   \\
    \nabla\cdot u  =  0
\quad     \mbox{ in }  \mathbb{R}^n\times(0, \infty),\label{e1.2}
\end{gather}
with  the initial condition
\begin{equation}\label{e1.3}
    u(x, 0)=u_0\quad \mbox{in } \mathbb{R}^n.
\end{equation}
 Here $n\geq 2$, the gradient $\nabla =(\partial_{x_1},\dots ,
\partial_{x_n})$, $ u=(u_1 ,\dots , u_n ) $ and $\pi$ denote the
unknown velocity and pressure of the fluid motion.
 $\tau^v=(\tau^v_{ij}) $ is  the stress tensor specified in the
 form
\begin{equation}\label{nonlin}
    \tau_{ij}^v=2\ ( \mu_0+\mu_1
   |e(u)|^{p-2})\ e_{ij}(u)-2\mu_2\Delta e_{ij}(u)
\end{equation}
with the constant viscosities $ \mu_0> 0$, $\mu_1$, $\mu_2\geq 0  $
 and the symmetric deformation velocity tensor  $e(u) =(e_{ij}(u))$,
   \begin{equation}\label{nonlin2}
e_{ij}(u)  =
{\frac{1}{2}}\Big({\frac{\partial u_i}{\partial
x_j}}+{\frac{\partial u_j}{\partial x_i}}\Big),\quad
|e(u)|=(e_{ij}(u)  e_{ij}(u) )^{1/2}.
\end{equation}


When $\mu_1 =\mu_2=0$, the Stokes Law
\begin{equation}\label{lin}
\tau_{ij}^v=2  \mu_0 e_{ij}(u)
\end{equation}
holds. The fluids, such as  water and  alcohol, satisfying
the  linear equation expressed by (\ref{lin}) is said to be
Newtonian, and \eqref{e1.1} turns out to be the famous Navier-Stokes equations \ (refer to
\cite{Lad69,Temam}). However the nonlinear constitutive equation
expressed by (\ref{nonlin}) with $\mu_1, \mu_2>0$ is related to
other non-Newtonian fluids such as the molten plastics, dyes,
adhesives, paints and greases. When $\mu_2=0$, system
\eqref{e1.1}-\eqref{nonlin}
 was first proposed by  Ladyzhenskaya \cite{Lad} and is
known as the Ladyzhenskaya equations.  The fluid is said to be
monopolar because the   only first order derivative of the
velocity field is involved in the stress tensor (see
(\ref{nonlin}) for $\mu_2=0$), whereas the fluid is bipolar if the
second order derivative arises in $\tau^v$ (see (\ref{nonlin}) for
$\mu_1, \mu_2>0$). The theory of bipolar fluids is compatible with
 the basic principles of thermodynamics, including the Clausius-Duhem
 inequality and the material frame indifference (See \cite{BBN}
  for a detailed description of multipolar fluids).  Moreover, the
fluid  is shear thinning if $p<2$ and shear thickening  if $p>2$
(when $p=2$, the system turns out to be
 Navier-Stokes equations).

 There is an extensive literature on the  solutions
of the incompressible non-Newtonian fluids.  Ladyzhenskaya
\cite{Lad} and Lions \cite{Lions}  first discussed the existence
and uniqueness of weak solutions of the sort monopolar model (see
(\ref{nonlin}) for $\mu_2=0$), and more recently, Du and
Gunzguiger \cite{DG}  studied the somewhat more general existence
and uniqueness results in  bounded domains. Pokorny \cite{Pokorny}
investigated the Cauchy problem for both monopolar and bipolar
fluids in whole spaces. As for the decay properties of solutions,
on the one hand, Nec\u{a}sov\'{a} and  Penel \cite{NP} recently
examined the  logarithmic
  decay in  $\mathbb{R}^2$ and
 algebraic decay
in  $\mathbb{R}^3$  with respect to  the monopolar shear thickening fluids
$(p\geq 3)$ by the Schonbek's Fourier splitting method
 \cite{Schonbek}. With the aid of  Wiegner's  method \cite{Wiegner},
  Guo and Zhu  \cite{GZ 2000} improved the algebraic
decay rates. Higher decay rates were recently  proved by Dong
\cite{Dong06} based on  the arguments of Kajikiya and Miyakawa
\cite{KM}. In particular, by improving Schonbek's Fourier
splitting method, the optimal algebraic decay rate in $\mathbb{R}^2$ of
this monopolar model was  obtained by Dong and Li \cite{Dong} in
the following form
\begin{equation}\label{1.7}
  \|u(t)\|_{L^2}\leq C(1+t)^{-1/2},\quad
  \|u(t)-e^{t\Delta}u_0\|_{L^2}\leq C(1+t)^{-3/4},\quad \forall \ t>0.
\end{equation}
On the other hand, Guo and Zhu \cite{GZ 2001} considered also the
decay of the weak solution of the bipolar fluids (see
(\ref{nonlin}) for $\mu_1, \mu_2>0$). Based on the fourth order
linear parabolic equation and the Wiegner's method \cite{Wiegner},
the decay rate of the $L^2$ norm they obtained is only one-half of
the decay rate to the   linear heat equation, that is
\begin{equation}\label {1.8}
\begin{gathered}
  \|u(t)\|_{L^2}\leq C(1+t)^{-\frac{n}{4r}-\frac{n}{8}},\forall \ t>0;\\
  \|u(t)-e^{t\Delta}u_0\|_{L^2}\to 0,\quad \mbox{as }
  t\to  \infty.
\end{gathered}
\end{equation}
assuming $L^r\cap L^2$ integrability of the initial data.
Furthermore, as for   the time decay of Navier-Stokes equations \ in whole spaces, the
sharp decay rates were obtained by Schonbek\cite{Schonbek},
Kajikiya and Miyakawa \cite{KM},  Wiegner\cite{Wiegner} and
references cited therein.

The aim of this paper is investigate the optimal rate of decay
of global solutions to  the Cauchy problem of  the bipolar shear
thinning fluids \eqref{e1.1}-\eqref{nonlin}
 ($p\geq 3,\  \mu_1, \mu_2>0$ in (\ref{nonlin})). We use the
improved Fourier splitting  methods developed by Dong et al
\cite{Dong,Dong05} and  Zhang \cite{Zhang95}, and the  rigorous
analysis of the lower frequency effect of the lower dissipative
term $\Delta u $   which    determines mainly the time decay rates
of the solutions. We obtain the optimal $L^2$-decay rate, which is
the same as that of  the linear heat equation
\begin{equation}\label {1.9}
  \|u(t)\|_{L^2}\leq C(1+t)^{-n/4},\quad \forall \ t>0,
\end{equation}
assuming $L^1\cap L^2$ integrability of the initial data.
Furthermore, since the bipolar non-Newtonian flow is  modified
from  the Newtonian flow, we examine the $L^2$-decay estimates of
the error $u(t)-\tilde u(t)$. Here $u(t)$ denotes weak solution of
the non-Newtonian system \eqref{e1.1}-\eqref{nonlin} with $\mu_1,\mu_2>0 $,
whereas $\tilde u$ denotes the weak solution of the Newtonian
system of \eqref{e1.1}-\eqref{nonlin} with $\mu_1=\mu_2=0$. The optimal error
estimates we obtained are the following
\begin{equation}\label {1.10}
\|u(t)-\tilde{u}(t)\| = o\left((1+t)^{-n/4}
          \right),\quad\mbox{as } t\to \infty.
\end{equation}

This paper is organized as follows. In Sections 2
we  define weak solutions and state some preliminary lemmas.
Decay estimates of the non-Newtonian flow are described in Section
3. Decay estimates of the error between the
non-Newtonian and Newtonian flows $u(t)-\tilde u(t)$ are derived
in Section 4.

\section{Preliminaries}

Let  $\|\cdot\|_q= \|\cdot\|_{L^q}$   ($\|\cdot\|=\|\cdot\|_2$)
be the norm of the usual scalar and vector Lebesgue space
$L^q(\mathbb{R}^n)$ and $\|\cdot\|_{m, p}= \|\cdot\|_{W^{m, p}}$ be the
norm of the Sobolev space $W^{m,p}(\mathbb{R}^n)$. The space $\H$ denotes
the $L^2-$closure of $C_{0,\sigma}^{\infty}(\mathbb{R}^n)$, which is the
set of smooth divergence-free vector fields with compact supports
in $\mathbb{R}^n$. The space $W^{1, q}_{0,\sigma}(\mathbb{R}^n)$ denotes the
closure of $C_{0,\sigma}^{\infty}(\mathbb{R}^n)$ in $W^{1,q}(\mathbb{R}^n)$. The
Fourier transformation of a function $f$ is denoted by $\hat f$ or
$F[f]$. $C>0$, independent of the quantities $t$, $x$, $\rho$, $u$
and $\tilde u $, is a generic constant, which may depend on
  the initial data $u_0$.


Without loss of generality, we assume that $ \mu_0=\mu_1=\mu_2=1$
in \eqref{nonlin}. Substitution of \eqref{nonlin} into \eqref{e1.1} produces
\begin{equation}\label{2.1}
    u_t-\triangle u +\Delta^2u
    +(u\cdot\nabla)u-\nabla\cdot(|\nabla u|^{p-2}\nabla u)
    +\nabla\pi = 0,
\end{equation}
   in $\mathbb{R}^n\times(0, \infty)$.

By a weak solution of the initial value problem \eqref{e1.2}-\eqref{nonlin} and
\eqref{2.1} for $n \geq 2$ and $p\geq1+\frac{2n}{n+2}$ (see
\cite{Lad,Lions,Pokorny}), we mean a vector field
$$
 u\in L^p(0,  T;W^{1,p}_{0,\sigma}(\mathbb{R}^n))\cap L^{\infty}(0,  T;\H)\cap
 L^2(0, T;W^{2, 2}_{0,\sigma}(\mathbb{R}^n)),\quad  \forall \, T>0
$$
 satisfying
\begin{equation}
\begin{aligned}
\int_{\mathbb{R}^n} u(t)\cdot \varphi(t)\,dx
- \int_0^t \int_{\mathbb{R}^n} u\cdot \frac {\partial \varphi}{\partial s}\,
   dx ds&\\
+ \int_0^t\int_{\mathbb{R}^n}u_j\frac{\partial u_i}{\partial x_j}\varphi_i\,
  dx ds
  +  \int_0^t\int_{\mathbb{R}^n}\tau_{ij}(e(u))\cdot e_{ij}(\varphi)
   dx ds
 &=   \int_{\mathbb{R}^n} u_0\cdot \varphi(0)\,dx
 \end{aligned} \label{2.2}
\end{equation}
 a. e. $t\in (0,  T)$ for every $ \varphi\in C^1([0,T),\H)
 \cap C([0,T), W^{2,2}_{0,\sigma}(\mathbb{R}^n)\cap W^{1,p}_{0,\sigma}(\mathbb{R}^n)
  )$ and $\varphi(x,T)=0$.
Moreover, we assume that the weak solution also satisfies the
following energy inequality
\begin{equation} \label {2.3}
   \frac{1}{2}\frac{d}{dt}
   \int_{\mathbb{R}^n}|u|^2 dx
  +\int_{\mathbb{R}^n}|\nabla u|^2\,dx
  +\int_{\mathbb{R}^n}|\Delta u|^2\,dx
  +\int_{\mathbb{R}^n}|\nabla u|^p\,dx
 \leq 0.
\end{equation}

It should be noted that a weak solution can be specified  as a
limit of a sequence of smooth approximate solutions in a local
$L^2$-norm due to the standard  Faedo-Galerkin argument. Thus the
decay estimates with respect to the weak solution become   limits
of those of the smooth approximate solutions (see, for example,
Kajikiya and Miyakawa \cite{KM}). Since we only consider the decay
estimates of the weak solution  in $L^2$-norm, without loss of
generality, we may suppose that the weak solution admit enough
regularity so that we can work on the weak solution directly
rather than on the sequence of smooth approximate solutions.

Let us now recall some preliminary lemmas.

\begin{lemma}[Gronwall Inequality] \label{lem2.1}
Let $f(t), g(t), h(t) $ be nonnegative continuous functions and
satisfying the inequality
  $$
    g(t)\leq f(t)+\int^t_0g(s)h(s)ds,   \quad  \forall \ t>0,
  $$
  where $f'(t)\ge 0$.  Then
 \begin{equation}\label{2.4}
      g(t)
 \leq f(t)\exp \Big(\int_0^{t} h(s)ds \Big), \quad  \forall \ t>0.
 \end{equation}
\end{lemma}

\begin{lemma} \label{lem2.2}
Assume that $u_0\in  \H \cap
 L^1(\mathbb{R}^n) $ and $u$ is a weak solution of \eqref{e1.2}-\eqref{nonlin} and
 \eqref{2.1}.
 Then
 \begin{equation}\label{2.5}
\sup_{0\leq t\leq\infty }\|u(t)\|   \leq \|u_0\|,
\end{equation}
and (i)  $ 2 < p <3,n=2$,
\begin{equation}\label{2.6}
|\hat{u}(\xi, t)|  \leq  C+
    C|\xi|\int^t_0\|u(s)\|^2ds
    + C|\xi|\Big(\int^t_0\|u(s)\|^{\frac{2}{4-p}}ds
            \Big)^{\frac{4-p}{2}},
\end{equation}
(ii) $  1+\frac{2n}{n+2}\leq p < 3,  n\geq 3$,
\begin{equation}\label{2.7}
|\hat{u}(\xi, t)|  \leq
  C+C|\xi|\int^t_0\|u(s)\|^2ds
  +C|\xi|\left(\int^t_0\|u(s)\|
             ^{\frac{2
             \alpha}{2-\beta}}ds\right)^{\frac{2-\beta}{2}},
\end{equation}
where $\alpha=\frac{2n-(n-2)(p-1)}{4}, \ \beta
     =\frac{(n+2)(p-1)-2n}{4}$,
\\
(iii) $   p \geq 3, n\geq 2$,
\begin{equation}\label{2.8}
|\hat{u}(\xi,t)|  \leq  C+ C|\xi|\int^t_0\|u(s)\|^2ds+
                                     C|\xi|.
\end{equation}
\end{lemma}

\begin{proof}
From the energy inequality (\ref{2.3}), it is
easy to get the first inequality (\ref{2.5}). We now prove
(\ref{2.6})-(\ref{2.8}), first, applying  the Fourier
transformation of \eqref{2.1} we have
\begin{equation}\label{2.9}
    \hat{u}_t+(|{\xi}|^2+|{\xi}|^4)\hat{u}
= F[\nabla\cdot(|e(u)|^{p-2}e(u))
   -(u\cdot\nabla)u-\nabla \pi]
=:G(\xi, t).
\end{equation}
Now we estimate $G(\xi, t)$.   Taking divergence in \eqref{2.1} to get,
 $$
 \Delta \pi=\sum_{i, j}{\frac{{\partial}^2}{\partial {x_i}\partial {x_j}}}
          [-u_iu_j+|e(u)|^{p-2}e_{ij}(u)].
 $$
The Fourier transformation yields
$$
|{\xi}|^2F[\pi]=\sum_{i, j}{\xi}_i{\xi}_jF[-u_iu_j
    +|e(u)|^{p-2}e_{ij}(u)],
$$
and thus
\begin{equation}\label{2.10}
       |F[\nabla \pi]|
 =     |{\xi}| F[\pi]
\leq   |F[\nabla\cdot(|e(u)|^{p-2}e(u))]|+|F[(u\cdot\nabla)u]|.
\end{equation}
Furthermore,
\begin{gather}\label{2.11}
 |F[(u\cdot\nabla)u]|
                =  |F[\textrm{div}(u\otimes u)]|
               \leq \sum_{i, j}\int_{\mathbb{R}^n}|u_iu_j|\>|{\xi}_j|dx
                \leq   |\xi|\> \|u\|^2,\\
 |F[\nabla\cdot(|e(u)|^{p-2}e(u))]|\leq |\xi||F[|e(u)|^{p-2}e(u)]|
                  \leq |\xi|\|\nabla u\|_{p-1}^{p-1}.\label{2.12}
\end{gather}
So inserting  (\ref{2.10})-(\ref{2.12})  into $G(\xi, t)$,  we
have
\begin{equation}\label{2.13}
|G(\xi,t)| \leq C|\xi|\|u\|^2+C|\xi|\|\nabla u\|^{p-1}_{p-1}.
\end{equation}
From (\ref{2.9}), it follows easily  that,
$$
\frac{d}{dt}\left(\hat u e^{(|\xi|^2+|\xi|^4)t}\right )\leq
G(\xi,t)e^{(|\xi|^2+|\xi|^4)t}.
$$
Integrating in time gives,
\begin{equation}\label{2.14}
\begin{aligned} |\hat{u}(\xi,t)|
 & \leq \left|e^{-(|\xi|^2+|\xi|^4)t}\hat{u}_0(\xi)
                  +\int_0^tG(\xi,t)e^{-(|\xi|^2+|\xi|^4)(t-s)}ds\right|\\
 & \leq |\hat{u}_0(\xi)|+\int_0^t|G(\xi,t)|ds \\
&\leq  C+|\xi|\int_0^t\|u(s)\|^2ds+C|\xi|\int_0^t\|\nabla
u(s)\|^{p-1}_{p-1}ds.
\end{aligned}
\end{equation}

Now we estimate $ \int_0^t\|\nabla u(s)\|^{p-1}_{p-1}ds $ in three
cases (note that the case $  p =2,  n=2 $ is the  Navier-Stokes equations \
\cite{Lad69,Temam}):
\begin{itemize}
\item[(i)] $ 2 < p <3$, $n=2$;

\item[(ii)]  $  1+\frac{2n}{n+2}\leq p < 3$, $n\geq 3$;

\item[(iii)] $   p \geq 3$, $n\geq 2$.
\end{itemize}
{\bf Case (i):}  $ 2 <p< 3, n=2$.
 By Gagliardo-Nirenberg
inequality (refer to \cite{H}),
\begin{equation}\label{2.15}
\begin{aligned}
   \int_0^t\|\nabla u(s)\|^{p-1}_{p-1}ds
&\leq  \nonumber  \int^t_0|\xi|\|u(s)\|\|D^2u(s)\|^{p-2}ds\\
& \leq  \Big( \int^t_0\|u(s)\|^{\frac{2}{4-p}}ds\Big)^{\frac{4-p}{2}}
 \Big(\int^{\infty}_0\|D^2u(t)\|^2dt\Big)^{\frac{p-2}{2}}\\
& \leq \Big(\int^t_0\|u(s)\|^{\frac{2}{4-p}}ds\Big)^{\frac{4-p}{2}},
\end{aligned}
\end{equation}
noting that $\int^{\infty}_0\|D^2u(t)\|^2dt \leq C $.

\noindent {\bf Case (ii):}  $  1+\frac{2n}{n+2}\leq p < 3,  n\geq 3$.
\begin{equation}\label{2.16}
\begin{aligned}
 \int_0^t\|\nabla u(s)\|^{p-1}_{p-1}ds
         &\leq   \int^t_0\|u(s)\|^{\alpha}\|D^2 u(s)\|^{\beta}ds\\
 & \leq  \Big(\int^t_0\|u(s)\|
             ^{\frac{2 \alpha}{2-\beta}}ds\Big)^{\frac{2-\beta}{2}}
 \Big(\int_0^{\infty}\|D^2 u(t)\|^2dt\Big)^{\frac{\beta}{2}}\\
  & \leq  \Big(\int^t_0\|u(s)\|^{\frac{2\alpha}{2-\beta}}ds
  \Big)^{\frac{2-\beta}{2}},
\end{aligned}
\end{equation}
where $\alpha=\frac{2n-(n-2)(p-1)}{4}, \ \beta
     =\frac{(n+2)(p-1)-2n}{4}$, and $0<\beta<1$.

\noindent{\bf Case (iii):}  $ p \geq 3$, $n\geq 2$.
With the above definition of the weak solution,  we know also that
$\nabla u \in L^2((0,\infty)\times \mathbb{R}^n)\cap L^p((0,\infty)\times
\mathbb{R}^n)$, so by using the interpolation technology, $ \nabla u\in
L^{p-1}((0,\infty)\times \mathbb{R}^n)$, i.e.
\begin{equation}\label{2.17}
\int_0^{\infty}\|\nabla u(s)\|^{p-1}_{p-1}ds  \leq   C .
\end{equation}
Hence (\ref{2.14})-(\ref{2.17}) imply the  assertions of lemma \ref{lem2.2}
and the proof  is complete.
\end{proof}

\section{Decay estimates of the non-Newtonian flows}

 As is well known that the weak solutions
of Navier-Stokes equations have the optimal  decay estimates in
the whole space  \cite{KM,Schonbek,Wiegner}.  In this section,
we show that the non-Newtonian flow has also the same  optimal $L^2$
time  decay estimates. The results read as follows.

\begin{theorem} \label{thm3.1}
Assume that $u_0\in  \H \cap L^1(\mathbb{R}^n)$. Let $u(t)$ be a weak
 solution of  \eqref{e1.2}-\eqref{nonlin} and \eqref{2.1}.
 Then, for $ n=2$, $p>2$ and for  $n\geq3$, $p\geq1+\frac{2n}{n+2}$,
 we have
\begin{equation}\label{3.1}
\|u(t)\|\leq C(1+t)^{-n/4}, \quad    \forall\, t> 0.
\end{equation}
\end{theorem}

\begin{proof} From the energy inequality (\ref{2.3}), it follows
that
\begin{equation}\label{3.2}
\frac{1}{2}\frac{d}{dt}\int_{\mathbb{R}^n}|u|^2 dx+\int_{\mathbb{R}^n}|\nabla
u|^2\leq0.
\end{equation}
Applying  Plancherel's theorem to (\ref{3.2}) yields
\begin{equation}\label{3.3}
\frac{1}{2}\frac{d}{dt}\int_{\mathbb{R}^n}|\hat u(\xi,t)|^2
d\xi+\int_{\mathbb{R}^n}|\xi|^2|\hat u(\xi,t)|^2d\xi\leq 0.
\end{equation}
Let $f(t)$ be a smooth function of t with $ f(0)=1,f(t)>0 $ and
$f'(t)>0 $, then
$$
\frac{d}{dt}\Big(f(t)\int_{\mathbb{R}^n}|\hat u(\xi,t)|^2 d\xi\Big)
+2f(t)\int_{\mathbb{R}^n}|\xi|^2|\hat u(\xi,t)|^2d\xi\leq
f'(t)\int_{\mathbb{R}^n}|\hat u(\xi,t)|^2 d\xi.
$$
Set $  B(t)=\{\xi \in \mathbb{R}^n :2f(t)|\xi|^2\leq f'(t)\}$. Then
\begin{align*}
&2f(t)\int_{\mathbb{R}^n}|\xi|^2|\hat u(\xi,t)|^2d\xi\\
&=2f(t)\int_{B(t)}|\xi|^2|\hat u(\xi,t)|^2d\xi
 +2f(t)\int_{B(t)^c}|\xi|^2|\hat u(\xi,t)|^2d\xi\\
&\geq 2f(t)\int_{B(t)^c}|\xi|^2|\hat u(\xi,t)|^2d\xi\\
&\geq f'(t)\int_{\mathbb{R}^n}|\hat u(\xi,t)|^2 d\xi-f'(t)
\int_{B(t)}|\hat u(\xi,t)|^2 d\xi.
\end{align*}
Therefore,
$$
\frac{d}{dt}\Big(f(t)\int_{\mathbb{R}^n}|\hat u(\xi,t)|^2 d\xi\Big)
\leq f'(t)\int_{B(t)}|\hat u(\xi,t)|^2d\xi.
$$
Integrating in time yields
\begin{equation}\label{3.4}
f(t)\int_{\mathbb{R}^n}|\hat u(\xi,t)|^2 d\xi \leq \int_{\mathbb{R}^n}|\hat
u_0|^2d\xi+ C \int_0^t f'(s)\int_{B(s)}|\hat u(\xi,s)|^2d\xi ds.
\end{equation}
Now we study three cases:
(i) $2<p<3$, $n = 2$;
(ii) $1+\frac{2n}{n+2}\leq p < 3$, $n\geq 3$;
(iii) $p \geq 3$, $n\geq 2$.


\noindent {\bf (i) Case $ 2<p<3$, $n = 2$.}
Let $A^2=\frac{f'(t)}{2f(t)}$, and  $\omega_n $ be  volume
of unit ball in $\mathbb{R}^n $. According to (\ref{2.6}),
\begin{equation}\label{3.5}
\begin{aligned}
 & f(t)\int_{\mathbb{R}^2}|\hat u(\xi,t)|^2 d\xi \\
 & \leq  \int_{\mathbb{R}^2}|\hat u_0|^2d\xi
   +C\omega_n \int_0^t f'(s)\int_0^A\Big\{1\\
 &\quad + \rho\int^s_0\|u(\tau)\|^2d\tau
 +\rho\Big(\int^s_0\|u(\tau)\|^{\frac{2}{4-p}}d\tau\Big)^{\frac{4-p}{2}}
 \Big\}^2\rho\,  d\rho\, ds\\
& \leq  C+C \int_0^t f'(s)\Big\{ \frac{f'(s)}{2f(s)}
 +\Big( \frac{f'(s)}{2f(s)} \Big)^2 \Big(\int^s_0\|u(\tau)\|^2d\tau
  \Big)^2\Big\} ds\\
&\quad+C \int_0^t f'(s)\Big\{\Big( \frac{f'(s)}{2f(s)}\Big)^2
\Big( \int^s_0\|u(\tau)\|^{\frac{2}{4-p}}d\tau\Big)^{4-p} \Big\} ds.
\end{aligned}
\end{equation}
On the one hand, using  (\ref{2.5}) on the right hand side of
(\ref{3.5}), we obtain
\begin{equation}\label{3.6}
 f(t)\int_{\mathbb{R}^2}|\hat u(\xi,t)|^2 d\xi
\leq C+C \int_0^t f'(s)\Big\{ \frac{f'(s)}{2f(s)}
+\big( \frac{f'(s)}{2f(s)} \big)^2(s^2+s^{4-p})\Big\}ds.
\end{equation}
Let $f(t)=(\ln(e+t))^5$. \ Then\
$f'(t)=\frac{5(\ln(e+t))^4}{e+t}$, and $\frac{f'(t)}{
f(t)}=\frac{5}{(e+t)\ln(e+t)}. $  By   (\ref{3.6}) and an
elementary calculation based on  Plancherel's theorem, we have
\begin{align*}
& (\ln(e+t))^5\int_{\mathbb{R}^2}| u( x ,t)|^2 d x\\
&=(\ln(e+t))^5\int_{\mathbb{R}^2}|\hat u(\xi,t)|^2 d\xi\\
&\leq C+C \int_0^t  \Big\{\frac{ (\ln(e+s))^3 }{(e+s)^2}+
\frac{s^2(\ln(e+s))^2 }{(e+s)^3}+\frac{s^{ 4-p }(\ln(e+s))^2 }{(e+s)^3}
 \Big\}ds\\
&\leq C+C \int_0^t   \frac{ (\ln(e+s))^2 }{ e+s }  ds
 \quad \mbox{(because $1< 4-p<2$)} \\
&\leq C (\ln(e+t))^3,
\end{align*}
and so
\begin{equation}\label{3.7}
\|u(t)\|\leq C(\ln(e+t))^{-1}.
\end{equation}
By the inductive argument, we suppose that
\begin{equation}\label{3.8}
 \|u(t)\|\leq C(\ln(e+t))^{-m}\quad
\forall \, m \in {\bf{N}}.
\end{equation}
Inserting  (\ref{3.8}) into the right hand side of (\ref{3.5}),
letting  $f(t)=(\ln(e+t))^{2m+3}$, and using  (\ref{3.5}) and the
inequality $ \int_0^t(\ln (e+s))^{-m}ds \leq C(e+t)\ln
(e+t))^{-m}, $ thus   from (\ref{3.5})
\begin{align*}
& (\ln(e+t))^{2m+3}\int_{\mathbb{R}^2}| u( x ,t)|^2 d x\\
&=(\ln(e+t))^{2m+3}\int_{\mathbb{R}^2}|\hat u(\xi,t)|^2 d\xi\\
&\leq C+C \int_0^t  \Big\{\frac{ (\ln(e+s))^{2m+1} }{(e+s)^2}+
 \frac{1 }{ e+s }+\frac{(\ln(e+s))^{2m-(4-p)m} } { (e+s)(e+s)^{p-2} } \Big\}ds\\
&\leq C+C \int_0^t   \frac{ 1 }{ e+s }  ds\\
&\leq C  \ln(e+t).
\end{align*}
Here we used that $(\ln(e+s))^{k}\leq c(k)(e+s)$, for all $k>0$.
The above inequality implies
\begin{equation}\label{3.9}
\|u(t)\|\leq C(\ln(e+t))^{-m-1}\quad  \forall\,  m \in {\bf{N}}.
\end{equation}
 On the other hand, from (\ref{3.5}) and the H\"{o}lder
 inequality,
\begin{align*}
&f(t)\int_{\mathbb{R}^2}|\hat u(\xi,t)|^2 d\xi \\
&\leq C+C\int_0^tf'(s)\frac{f'(s)}{2f(s)}ds
  +\int_0^t sf'(s)\Big( \frac{f'(s)}{2f(s)} \Big)^2 ds
  \int^t_0\|u(s)\|^4ds\\
 &\quad + C\int_0^t s^{\frac{4-p}{2}}f'(s)\Big( \frac{f'(s)}{2f(s)}\Big)^2ds
 \Big( \int^t_0\|u(s)\|^{\frac{4}{4-p}}ds\Big)^{\frac{4-p}{2}} .
\end{align*}
Let $ f(t)=(1+t)^2 $. Then
\begin{equation}\label{3.10}
\begin{aligned}
&(1+t)^2\int_{\mathbb{R}^2}|\hat u(\xi,t)|^2 d\xi \\
&\leq  C(1+t)+ C(1+t)\int^t_0\|u(s)\|^4ds
 +  C(1+t)^{\frac{4-p}{2}}
 \Big( \int^t_0\|u(s)\|^{\frac{4}{4-p}}ds\Big)^{\frac{4-p}{2}} .
\end{aligned}
\end{equation}
Noting that $\frac{1}{2}<\frac{4-p}{2}< 1$ and applying  the Young
inequality to the last term of (\ref{3.10}), we have
 \begin{equation}\label{3.11}
\Big(\int^t_0\|u(s)\|^{\frac{4}{4-p}}ds\Big)^{\frac{4-p}{2}}
\leq C\int^t_0\|u(s)\|^{\frac{4}{4-p}}ds+C.
\end{equation}
Inserting  (\ref{3.8}) and  (\ref{3.11}) into  (\ref{3.10}),  we
get the following estimate
\begin{align*}
(1+t)\int_{\mathbb{R}^2}|\hat u(\xi,t)|^2 d\xi
&\leq C+C\int^t_0\|u(s)\|^2(1+s)\Big\{
(1+s)^{-1}(\ln(e+s))^{-m}\\
&\quad +(1+s)^{-1}(\ln(e+s))^{-\frac{m(2p-4)}{4-p}}\Big\}ds.
\end{align*}
Let
\begin{gather*}
g(t)=(1+t)\int_{\mathbb{R}^2}|\hat u(\xi,t)|^2 d\xi=(1+t)\int_{\mathbb{R}^2}|
u(x,t)|^2dx,f(t)=C,\\
h(t)=(1+t)^{-1}(\ln(e+t))^{-m}+(1+t)^{-1}(\ln(e+t))^{-\frac{m(2p-4)}{4-p}}.
\end{gather*}
When the integer $m$ is suitable large, it is simple to deduce
that $\int_0^{\infty} h(t) dt < \infty$.
Applying Lemma \ref{lem2.1}, we have
$$
g(t)\leq C\exp \Big(\int_0^{\infty} h(t)dt \Big)\leq C,
 $$
and thus
$$
\|u(t)\| \leq C(1+t)^{-1/2}.
$$

\noindent{\bf (ii)  Case  $1+\frac{2n}{n+2}\leq p < 3$, $n\geq3$.}
Inserting (\ref{2.7}) into the right hand of  (\ref{3.4}) and
using H\"{o}lder inequality and Young inequality,  we have
 \begin{align*}
&f(t)\int_{\mathbb{R}^n}|\hat u(\xi,t)|^2 d\xi  \\
& \leq \int_{\mathbb{R}^n}|\hat u_0|d\xi
  +C\omega_n \int_0^t f'(s)\int_0^A\Big\{1 \\
&\quad + \rho\int^s_0\|u(\tau)\|^2d\tau
+\rho \Big(\int^s_0\|u(\tau)\|^{\frac{2\alpha}{2-\beta}}
 d\tau\Big)^{\frac{2-\beta}{2}} \Big\}^2\rho^{n-1}d\rho ds\\
&\leq  C+C \int_0^t f'(s)\Big\{\Big(\frac{f'(s)}{2f(s)}\Big)^{n/2}
   +\Big( \frac{f'(s)}{2f(s)}\Big)^{\frac{n+2}{2}}
 s\int^s_0\|u(\tau)\|^4d\tau \Big\}ds \\
 &\quad+C \int_0^t f'(s) \Big( \frac{f'(s)}{2f(s)} \Big)^{\frac{n+2}{2}}
 s^{\frac{2-\beta}{2}}ds \Big( \int^t_0\|u(s)\|^{\frac{4\alpha}{2-\beta}}ds
\Big)^ {\frac{2-\beta}{2}}\\
& \leq  C+C \int_0^t f'(s)\Big\{ \Big(\frac{f'(s)}{2f(s)}\Big)^{n/2}
  +\Big( \frac{f'(s)}{2f(s)} \Big)^{\frac{n+2}{2}}
 s \int^t_0\|u(s)\|^4ds \Big\} ds\\
&\quad+C \int_0^t f'(s) \Big( \frac{f'(s)}{2f(s)} \Big)^{\frac{n+2}{2}}
 s^{\frac{2-\beta}{2}}ds
 \Big( \int^t_0\|u(s)\|^{\frac{4\alpha}{2-\beta}}ds  +C\Big).
 \end{align*}
 Let $f(t)=(1+t)^n$. Noting that $\frac{1}{2}<\frac{2-\beta}{2}< 1$,
$\frac{4\alpha}{2-\beta}>2$ and $\|u(t)\|\leq C$, we have
\begin{align*}
&(1+t)^n\int_{\mathbb{R}^n}|\hat u(\xi,t)|^2 d\xi\\
 &\leq C(1+t)^{n/2}+C(1+t)^{n/2}\int^t_0\|u(s)\|^4ds
  +C(1+t)^{n/2} \int^t_0\|u(s)\|^{\frac{4\alpha}{2-\beta}}ds \\
&\leq  C(1+t)^{n/2}+C(1+t)^{n/2}\int^t_0\|u(s)\|^2ds.
\end{align*}
This yields
\begin{align*}
(1+t)^{n/2}\|u(t)\|^2 &=(1+t)^{n/2}\int_{\mathbb{R}^n}|\hat u(\xi,t)|^2 d\xi\\
 &\leq  C+\int^t_0(1+s)^{n/2}\|u(s) \|^2(1+s)^{-\frac{n}{2}}ds.
\end{align*}
Letting
\[
f(t)=C,\quad g(t)=(1+t)^{n/2}\|u(t)\|^2,\quad
h(t)=(1+t)^{-n/2},
\]
applying  Lemma \ref{lem2.1} and the bound  $\int_0^{\infty}h(s)ds \leq
C$, we deduce  readily that $$ \|u(t)\|\leq C(1+t)^{-n/4}.
$$


\noindent {\bf (iii) Case $p\geq 3$, $n \geq2$.}
Inserting (\ref{2.8}) into the right hand of (\ref{3.4}),
\begin{equation}\label{3.12}
\begin{aligned}
&f(t)\int_{\mathbb{R}^n}|\hat u(\xi,t)|^2 d\xi  \\
&\leq   C    +C\omega_n \int_0^t f'(s)\int_0^A
\Big(1+\rho\int^s_0\|u(\tau)\|^2d\tau+\rho
 \Big)^2\rho^{n-1}d\rho \,ds\\
&\leq  C+C \int_0^t f'(s)\left\{\Big(
  \frac{f'(s)}{2f(s)}\Big)^{n/2}
  +\Big(\frac{f'(s)}{2f(s)} \Big)^{\frac{n+2}{2}}
 \Big(\Big(\int^s_0\|u(\tau)\|^2d\tau  \Big)^2+1\Big)\right\}ds.
\end{aligned}
\end{equation}
First, we discuss the case  $n=2$.  It follows from (\ref{3.12})
and the bound $\|u(t)\|\leq C$ that
\[
 f(t)\int_{\mathbb{R}^2}|\hat u(\xi,t)|^2 d\xi\leq  C+C \int_0^t f'(s)
\Big(  \frac{f'(s)}{2f(s)} +\Big( \frac{f'(s)}{2f(s)}
 \Big)^2 ( s^2+1)\Big)ds.
\]
Let $f(t)=(\ln(e+t))^5$. By the same calculation as that  of
(\ref{3.7}),  we have
\begin{equation}\label{3.13}
\|u(t)\|\leq C(\ln(e+t))^{-1}.
\end{equation}
Hence,  letting $ f(t)=(1+t)^2 $ in (\ref{3.10}) and   using the
H\"{o}lder inequality,
\begin{align*}
(1+t)^2\int_{\mathbb{R}^2}|\hat u(\xi,t)|^2 d\xi
&\leq   C(1+t)+
C\int_0^t(1+s)^{-1}\Big(\int^s_0\|u(\tau)\|^2d\tau \Big)^2ds\\
&\leq C(1+t)+C\int_0^t\int^s_0\|u(\tau)\|^4d\tau ds\\
&\leq C(1+t)+C(1+t)\int^t_0\|u(s)\|^4 ds.
\end{align*}
By (\ref{3.11}), we obtain the inequality
\[
(1+t)\int_{\mathbb{R}^2}|\hat u(\xi,t)|^2 d\xi
\leq C +C\int^t_0\|u(s)\|^2(1+s)\left(
(1+s)^{-1}(\ln(e+s))^{-2}\right)ds.
\]
Let
\begin{gather*}
g(t)=(1+t)\int_{\mathbb{R}^2}|\hat u(\xi,t)|^2 d\xi=(1+t)\int_{\mathbb{R}^2}|
u(x,t)|^2 dx,\\
 h(t)=(1+t)^{-1}(\ln(e+t))^{-2},\quad  f(t)=C\,.
\end{gather*}
Applying Lemma \ref{lem2.1}, we have
$$
g(t)\leq C\exp \Big( \int_0^{\infty} h(t)dt\Big)\leq C,
$$
and so
$$
 \|u(t)\| \leq C(1+t)^{-1/2}.
$$
Next, we carry out the proof in  the case $n\geq 3$. Letting
$f(t)=(1+t)^n $ in (\ref{3.10}) and  using the H\"{o}lder
inequality,  we have, similar to the argument in the case of
$n=2$,
 \begin{align*}
&(1+t)^n\int_{\mathbb{R}^n}|\hat u(\xi,t)|^2 d\xi \\
 & \leq   C +C(1+t)^{-\frac{n}{2}} +C(1+t)^{-\frac{n+2}{2}}
       + C(1+t)^{-\frac{n}{2}}\int^t_0\|u(s)\|^4 \, ds.
 \end{align*}
Thus
\[
(1+t)^{n/2}\int_{\mathbb{R}^n}|\hat u(\xi,t)|^2 d\xi   \leq   C
+C\int^t_0\|u(s)\|^2(1+s)^{n/2}(1+s)^{-\frac{n}{2}} ds.
 \]
Since $ \int_0^{\infty}(1+s)^{-\frac{n}{2}} ds\leq C$  when
$n\geq 3 $. Thus  applying  Lemma \ref{lem2.1}, we have
$$
(1+t)^{n/2}\int_{\mathbb{R}^n}|\hat u(\xi,t)|^2 d\xi\leq C\exp
\Big( \int_0^{\infty} (1+t)^{-\frac{n}{2}}dt \Big)\leq C.
$$
Hence $ \|u(t)\| \leq C(1+t)^{-n/4}$. The proof of
Theorem \ref{thm3.1} is complete.
\end{proof}

\section{Error estimates for   Newtonian and non-Newtonian
flows}

\begin{theorem} \label{thm4.1}
In addition to the  assumption of  Theorem \ref{thm3.1}, suppose that
$\tilde u$ denotes the weak solution of the Newtonian system
\eqref{e1.1}-\eqref{nonlin} with $\mu_1=\mu_2=0$. Then
 $$
\|u(t)-\tilde{u}(t)\| = o\left((1+t)^{-n/4}\right),\quad
  \mbox{as } t\to \infty.
 $$
\end{theorem}

 Note that the estimates  of
Theorem \ref{thm4.1} with  $u(t)-\tilde u(t)$  replaced by $e^{t
\Delta}u_0-\tilde u(t)$ also hold  (see Kajikiya and Miyakawa
\cite{KM}). Thus from the inequality
$$
\|u(t)-\tilde u(t)\|\leq
\|e^{t \Delta}u_0- \tilde u(t)\|+\|e^{t \Delta}u_0- u(t)\|,
$$
we need to prove the validity of the estimates  of Theorem \ref{thm4.1}
with $u(t)-\tilde u(t)$  replaced by $e^{t \Delta}u_0- u(t)$.
So we only need the following lemma.

\begin{lemma} \label{lem4.1}
 In addition to the assumption of  Theorem \ref{thm3.1}, let $v(t)=e^{t\Delta}u_0$
 be the solution of  the linear heat equation  with the same initial
data $u_0$, then for $ t\geq 1$,
 $$
\|u(t)-v(t)\|^2\leq C
\begin{cases}
(1+t)^{-p/2}, & 2<p<3,\,n=2 \\
(1+t)^{-\frac{n}{2}-\frac{1}{2}}, &  1+\frac{2n}{n+2} \leq p<3,\,n\geq 3\\
(1+t)^{-\frac{n}{2}-\frac{1}{2}}, & p\geq 3,\  n\geq 2.
\end{cases}
$$
 \end{lemma}

We remark that from Lemma \ref{lem4.1}, it is readily seen  that when
 $ u_0\in \H\cap L^1 $,
$$
\|u(t)-v(t)\| = o\left((1+t)^{-n/4}\right),\quad t\to \infty.
$$

\begin{proof}[Proof of Lemma \ref{lem4.1}]
Denote the difference $w(t)=u(t)-v(t)$. Thus
$w(t) $ satisfies
\begin{equation}\label{4.1}
w_t-\triangle w +\Delta^2 w  = B(u,v),
 \quad w(x,0)=0,
\end{equation}
where
$B(u,v)=-(u\cdot\nabla)u+\nabla\cdot(|e(u)|^{p-2}e(u))-\Delta^2
v-\nabla\pi$. Since $u_0\in \H,$\  $v$ is divergence free, and so
is $w$.

 Multiplying by $w$ and integrating with respect to
$\mathbb{R}^n$, it follows that
\begin{equation}\label{4.2}
 \frac{d}{dt}\|w\|^2+2\|\nabla w\|^2+2\|\Delta w\|^2=:2B(u,v,w),
 \end{equation}
where
\begin{equation}\label{4.3}
\begin{aligned}
B(u,v,w)&= -((u\cdot\nabla)u,w)-((|e(u)|^{p-2}e(u)),\nabla w)-(\Delta^2 v,w)\\
&=((u\cdot\nabla)w,w+v)+(|e(u)|^{p-2}e(u),\nabla v)-\|\nabla u\|^p_p+(\Delta v,\Delta w)\\
&=((u\cdot\nabla)w,v)+(|e(u)|^{p-2}e(u),\nabla
v)+(\Delta v,\Delta w)-\|\nabla u\|^p_p.
\end{aligned}
\end{equation}
Since, for $1\leq q\leq \infty$ and $k\in N $,
\begin{equation}\label{4.4}
\|D^k v(t)\|_q\leq
(1+t)^{-\frac{n}{2}(1-\frac{1}{q})-\frac{k}{2}}\|u_0\|_1 \quad
\forall\,  t\geq 1,
\end{equation}
which  follows from the properties of the heat kernel (see
Kajikiya and Miyakawa \cite{KM}).  Thus we estimate the first three
terms. Noting that $\|u(t)\|\leq C(1+t)^{-n/4}$, we have
\begin{equation}\label{4.5}
\begin{aligned}
& 2|((u\cdot\nabla)w,v)+(|e(u)|^{p-2}e(u),\nabla v)+(\Delta
v,\Delta w)| \\
&\leq  2\|u\|\|\nabla w\|\|v\|_{\infty}+2\|\nabla
           u\|_{p-1}^{p-1}\|\nabla v\|_{\infty}+2\|\Delta v\|\|\Delta w\|\\
&\leq  \|\nabla w\|^2+\|u\|^2\|v\|_{\infty}^2+2\|\nabla
  u\|_{p-1}^{p-1}\|\nabla v\|_{\infty}+2\|\Delta w\|^2+\frac{1}{2}\|\Delta
           v\|^2\\
&\leq  \|\nabla w\|^2+2\|\Delta
       w\|^2+(1+t)^{-\frac{3n}{2}}+2(1+t)^{-\frac{n}{2}-\frac{1}{2}}\|\nabla
       u\|_{p-1}^{p-1}+\frac{1}{2}(1+t)^{-\frac{n}{2}-2}.
\end{aligned}
\end{equation}
Hence (\ref{4.2})-(\ref{4.5}) yield
\begin{equation}\label{4.6}
\frac{d}{dt}\|w\|^2+ \|\nabla w\|^2 \leq
2(1+t)^{-\frac{n}{2}-\frac{1}{2}}\|\nabla
           u\|_{p-1}^{p-1}+\frac{1}{2}(1+t)^{-\frac{n}{2}-2}.
\end{equation}
Similar to (\ref{3.4}) with $f(t)=(1+t)^{2n}$,  the  derivation of
(\ref{4.6})
 implies
\begin{align*}
&(1+t)^{2n}\int_{\mathbb{R}^n}|\hat w(\xi,t)|^2 d\xi \\
&\leq  C(1+t)^{2n }\int_{B(t)}|\hat w(\xi,s)|^2d\xi
+C(1+t)^{\frac{3n}{2}-1}
+C(1+t)^{\frac{3n}{2}-\frac{1}{2}}\int_0^t\|\nabla
           u\|_{p-1}^{p-1} ds.
\end{align*}
Similar to the proof of the Lemma \ref{lem2.2},  we use (\ref{4.1}) and
$\|v(t)\|_1\leq \|u_0\|_1\leq C$ to obtain
\begin{equation}\label{4.7}
\begin{aligned}
|\hat{w}(\xi, t)|  &\leq   C|\xi|\int^t_0\|u(s)\|^2ds
  +C|\xi|\int_0^t\|\nabla
           u\|_{p-1}^{p-1} ds+C|\xi|^4 \int_0^t\|v(s)\|_1ds\\
&\leq   C|\xi|\int^t_0\|u(s)\|^2ds+C|\xi|\int_0^t\|\nabla
           u\|_{p-1}^{p-1} ds+C|\xi|^4 t.
\end{aligned}
\end{equation}
Therefore,  (\ref{4.6}) and (\ref{4.7}) yield
\begin{align*}
 \|w(t)\|^2 &= \int_{\mathbb{R}^n}|\hat w(\xi,t)|^2 d\xi\leq \nonumber
               C(1+t)^{-\frac{n}{2}-1}\Big(\int^t_0\|u(s)\|^2ds\Big)^2
               +C(1+t)^{-\frac{n}{2}-1}\\
 &\quad +C(1+t)^{-\frac{n}{2}-\frac{1}{2}}\int_0^t\|\nabla
           u\|_{p-1}^{p-1} ds +C(1+t)^{-\frac{n}{2}-1}\Big(\int_0^t\|\nabla
           u\|_{p-1}^{p-1} ds\Big)^2.
\end{align*}
When $2<p<3$, $n=2$, it follows from (\ref{2.15}) and
$\|u(t)\|\leq C(1+t)^{1/2}$ that
\begin{align*}
\|w(t)\|^2
&  \leq  C(1+t)^{-2}(\ln(1+t))^2
 +C(1+t)^{-(p-1)}+C(1+t)^{-\frac{p}{2}}\\
          & \leq C(1+t)^{-\frac{p}{2}}.
\end{align*}
When $1+\frac{n+2}{2n}\leq p <3$ and $n\geq 3$, equation
(\ref{2.16}) and  the inequalities
$\|u(t)\|\leq C(1+t)^{-n/4}$ and  $\frac{n\alpha}{4-2\beta}> 1$ imply
$$
\int_0^t\|\nabla  u\|_{p-1}^{p-1} ds
=\int _0^t(1+s)^{-\frac{n\alpha}{4-2\beta}}ds\leq C,
$$
which yields
\[
\|w(t)\|^2
 \leq  C(1+t)^{-\frac{n}{2}-1}+ C(1+t)^{-\frac{n}{2}-\frac{1}{2}}
 \leq C(1+t)^{-\frac{n}{2}-\frac{1}{2}}.
\]
Similarly, for the case of  $p\geq 3$ and $n\geq 2$, we derive
from  (\ref{2.17}) that
\[
\|w(t)\|^2 \leq C(1+t)^{-\frac{n}{2}-\frac{1}{2}}.
\]
 Hence the proof of Lemma \ref{lem4.1} is complete.
\end{proof}


\subsection*{Acknowledgements}
 The author would like to express his gratitude to the
referees for  his/her valuable comments and suggestions. He is
also grateful to  Zhi-Min Chen and Yongsheng Li for many helpful
discussions.

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\end{thebibliography}

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