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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 104, 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}
\thanks{\copyright 2016 Texas State University.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2016/104\hfil Long time decay for 3D Navier-Stokes equations]
{Long time decay for 3D Navier-Stokes equations in Sobolev-Gevrey spaces}

\author[J. Benameur, L. Jlali \hfil EJDE-2016/104\hfilneg]
{Jamel Benameur, Lotfi Jlali}

\address{Jamel Benameur \newline
Institut Sup\'erieur des Sciences Appliqu\'ees et de Technologie de Gab\`es,
Universit\'e de Gab\`es,  Tunisia}
\email{jamelbenameur@gmail.com}

\address{Lotfi Jlali \newline
Facult\'e de Sciences Math\'ematiques,
Physiques et Naturelles de Tunis,
Universit\'e de Tunis El Manar, Tunisia}
\email{lotfihocin@gmail.com}

\thanks{Submitted February 2, 2016. Published April 21, 2016.}
\subjclass[2010]{35Q30, 35D35}
\keywords{Navier-Stokes Equation; critical spaces; long time decay}

\begin{abstract}
 In this article, we  study the long time decay of global solution to
 $3$D incompressible Navier-Stokes equations. We prove
 that if $u\in{\mathcal C}([0,\infty),H^1_{a,\sigma}(\mathbb{R}^3))$ is a
 global solution, where $H^1_{a,\sigma}(\mathbb{R}^3)$ is the Sobolev-Gevrey
 spaces with parameters $a>0$ and $\sigma>1$, then
 $\|u(t)\|_{H^1_{a,\sigma}(\mathbb{R}^3)}$ decays to zero as time approaches
 infinity. Our technique is based on Fourier analysis.
\end{abstract}

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


\section{Introduction}

The $3$D incompressible Navier-Stokes equations are 
\begin{equation} 
 \begin{gathered}
 \partial_t u  -\Delta u+ u\cdot\nabla u
 =-\nabla p \quad \text{in } \mathbb{R}^+\times \mathbb{R}^3\\
    \operatorname{div} u = 0 \quad \text{in } \mathbb{R}^+\times \mathbb{R}^3\\
    u(0,x) =u^0(x) \quad \text{in }\mathbb{R}^3,
  \end{gathered}\label{NSE}
\end{equation}
where, we assume that the fluid viscosity $\nu=1$, and
 $u=u(t,x)=(u_1,u_2,u_3)$ and $p=p(t,x)$ denote respectively the
unknown velocity and the unknown pressure of the fluid at the point
 $(t,x)\in \mathbb{R}^+\times \mathbb{R}^3$,
$(u\cdot\nabla u):=u_1\partial_1 u+u_2\partial_2 u+u_3\partial_3u$, and
$u^0=(u_1^o(x),u_2^o(x),u_3^o(x))$ is a given initial velocity.
If $u^0$ is quite regular, the divergence free condition determines
the pressure $p$.

We define the Sobolev-Gevrey spaces as follows; for $a, s\geq0$, $\sigma>1$ and 
$|D|=(-\Delta)^{1/2}$,
$$
H^s_{a,\sigma}(\mathbb{R}^3)=\{f\in L^2(\mathbb{R}^3) : 
e^{a|D|^{1/\sigma}}f\in H^s(\mathbb{R}^3)\}
$$
which is equipped with the norm
$$
\|f\|_{H^s_{a,\sigma}}=\|e^{a|D|^{1/\sigma}}f\|_{H^s}
$$
and its associated inner product
$$
\langle f\mid g\rangle_{{H}^s_{a,\sigma}}
=\langle e^{a|D|^{1/\sigma}}f\mid e^{a|D|^{1/\sigma}}g\rangle_{H^s}.
$$

There are several authors who have studied the behavior of the norm of the 
solution to infinity in the different Banach spaces. 
Wiegner \cite{MW1} proved  that the $L^2$ norm of the solutions vanishes 
for any square integrable initial data, as time approaches infinity, and gave
a decay rate that seems to be optimal for a class of initial data. 
Schonbek and  Wiegner \cite{MS2,MWS} derived some asymptotic 
properties of the solution and its higher derivatives under additional assumptions 
on the initial data. Benameur and Selmi \cite {BS1} proved that if 
$u$ is a Leray solution of the 2D Navier-Stokes equation, then 
$\lim_{t\to\infty}\|u(t)\|_{L^2(\mathbb{R}^2)}=0$. 
For the critical Sobolev spaces $\dot{H}^{1/2}$,  Gallagher,  Iftimie and Planchon
\cite{GIP}  proved that $\|u(t)\|_{\dot{H}^{1/2}}$ approaches zero at infinity.
Now, we state our main result.

\begin{theorem}\label{thm1}
Let $a>0$ and $\sigma>1$. Let 
$u\in{\mathcal C}([0,\infty),H^1_{a,\sigma}(\mathbb{R}^3))$ 
be a global solution to \eqref{NSE}. Then
\begin{equation}\label{eq1}
\limsup_{t\to\infty}\|u(t)\|_{H^1_{a,\sigma}}=0.
\end{equation}
\end{theorem}

Note that the existence of local solutions to \eqref{NSE} was studied 
recently in \cite{BL1}.

This article is organized as follows: 
In section $2$, we give some notations and important preliminary results. 
Section $3$ is devoted to prove that if 
$u\in{\mathcal C}(\mathbb{R}^+, H^1(\mathbb{R}^3))$ is a global solution 
to \eqref{NSE} then $\|u(t)\|_{H^1}$ decays to zero as time approaches infinity. 
The proof is based on the fact that
\begin{equation}\label{eq2}
\lim_{t\to\infty}\|u(t)\|_{\dot{H}^{1/2}}= 0
\end{equation}
and the energy estimate
\begin{equation}\label{enq1}
\|u(t)\|_{L^2}^2+\int_0^t\|\nabla u(\tau)\|_{L^2}^2d\tau\leq\|u^0\|_{L^2}^2.
\end{equation}
In section $4$, we generalize the results of Foias-Temam \cite{FT} 
to $\mathbb{R}^3$ and in section $5$, we prove the main theorem.

\section{Notation and preliminary results}
\subsection{Notation}
In this section, we collect  notation and definitions that will be used later.
First, the Fourier transformation is normalized as
$$
\mathcal{F}(f)(\xi)=\widehat{f}(\xi)
=\int_{\mathbb{R}^3}\exp(-ix\cdot\xi)f(x)dx,\quad
\xi=(\xi_1,\xi_2,\xi_3)\in\mathbb{R}^3,
$$
the inverse Fourier formula is
$$
\mathcal{F}^{-1}(g)(x)=(2\pi)^{-3}\int_{\mathbb{R}^3}
\exp(i\xi\cdot x)g(\xi)d\xi,\quad x=(x_1,x_2,x_3)\in\mathbb{R}^3,
$$
and the convolution product of a suitable pair of functions $f$ and 
$g$ on $\mathbb{R}^3$ is 
$$
(f\ast g)(x):=\int_{\mathbb{R}^3}f(y)g(x-y)dy.
$$
For $ s\in\mathbb{R} $, $H^s(\mathbb{R}^3)$ denotes the usual non-homogeneous 
Sobolev space on $\mathbb{R}^3$ and  $\langle\cdot\mid\cdot\rangle_{H^s}$ 
denotes the usual  scalar product on $H^s(\mathbb{R}^3)$.
For $ s\in\mathbb{R} $, $\dot{H}^s(\mathbb{R}^3)$ denotes the usual 
homogeneous Sobolev space on $\mathbb{R}^3$ and 
$\langle\cdot\mid\cdot\rangle_{\dot{H}^s}$ denotes the usual scalar product on 
$\dot{H}^s(\mathbb{R}^3)$.
We denote by $\mathbb{P}$ the Leray projection operator defined by the formula
$$
\mathcal{F}(\mathbb{P}f)(\xi)=\widehat{f}(\xi)-\frac{(f(\xi)\cdot\xi)}{|\xi|^2}\xi.
$$
The fractional Laplacian operator $(-\Delta)^{\alpha}$ for a real number 
$\alpha$ is defined through the Fourier transform, namely
$$
\widehat{(-\Delta)^{\alpha}f(\xi)}=|\xi|^{2\alpha}\hat{f}(\xi).
$$
Finally, If $f=(f_1,f_2,f_3)$ and $g=(g_1,g_2,g_3)$ are two vector fields, we set
$$
f\otimes g:=(g_1f,g_2f,g_3f),
$$
and
$$
\operatorname{div}(f\otimes g):=(\operatorname{div}(g_1f),
\operatorname{div}(g_2f),\operatorname{div}(g_3f)).
$$

\subsection{Preliminary results}
In this section, we recall some classical results and we give a new 
technical lemma.

\begin{lemma}[\cite{HB}] \label{lem1}
Let $(s,t)\in{\mathbb{R}^2}$ be such that $s<3/2$ and $s+t>0$. 
Then, there exists a constant $C :=C(s,t)>0$, such that for all 
$u,v\in \dot H^{s}(\mathbb{R}^3)\cap \dot H^{t}(\mathbb{R}^3)$, we have
$$
\|uv\|_{\dot{H}^{{s+t-{\frac{3}{2}}}}(\mathbb{R}^3)}
\leq C(\|u\|_{\dot{H}^s(\mathbb{R}^3)}\|v\|_{\dot{H}^t(\mathbb{R}^3)}
+\|u\|_{\dot{H}^t(\mathbb{R}^3)}\|v\|_{\dot{H}^s(\mathbb{R}^3)}).
$$
If $s<3/2$, $t<3/2$ and $s+t>0$,  then there exists a constant
 $c :=c(s,t)>0$, such that
$$
\|uv\|_{\dot{H}^{{s+t-{\frac{3}{2}}}}(\mathbb{R}^3)}
\leq c \|u\|_{\dot{H}^s(\mathbb{R}^3)}\|v\|_{\dot{H}^t(\mathbb{R}^3)}.
$$
\end{lemma}


\begin{lemma}\label{lem3}
Let $f\in \dot H^{s_1}(\mathbb{R}^3)\cap \dot H^{s_2}(\mathbb{R}^3)$, 
where $s_1 < \frac{3}{2} < s_2 $. Then, there is a constant 
$c=c(s_1,s_2)$ such that
$$
\|f\|_{L^{\infty}(\mathbb{R}^3)}
\leq \|\hat{f}\|_{L^{1}(\mathbb{R}^3)}
\leq c\|f\|^{\frac{s_2-\frac{3}{2}}{s_2-s_1}}_{ \dot H^{s_1}
(\mathbb{R}^3)}\|f\|^{\frac{\frac{3}{2}-s_1}{s_2-s_1}}_{\dot H^{s_2}(\mathbb{R}^3)}.
$$
\end{lemma}

\begin{proof}
 We have
\begin{align*}
\|f\|_{L^{\infty}(\mathbb{R}^3)}
&\leq \|\widehat{f}\|_{L^{1}(\mathbb{R}^3)}\\
&\leq   \int_{\mathbb{R}^3} |\widehat{f(\xi)}|d\xi\\
&\leq  \int_{|\xi|<{\lambda}}|\widehat{f(\xi)}|d\xi
 +\int_{|\xi|>{\lambda}}|\widehat{f(\xi)}|d\xi.
\end{align*}
We take
$$
I_1=\int_{|\xi|<{\lambda}}\frac{1}{|\xi|^{s_1}}|\xi|^{s_1}|\widehat{f(\xi)}|d\xi.
$$
Using the Cauchy-Schwarz inequality, we obtain
\begin{align*}
I_1&\leq  
\Big(\int_{|\xi|<{\lambda}}\frac{1}{|\xi|^{2s_1}}d\xi\Big)^{1/2}
\|f\|_ {\dot H^{s_1}}\\
&\leq  2\sqrt{\pi}
\Big(\int^{\lambda}_{0}\frac{1}{r^{2s_{1}-2}}dr\Big)^{1/2}
\|f\|_ {\dot H^{s_1}}\\
&\leq  c_{s_1}\lambda^{\frac{3}{2}-s_1}\|f\|_ {\dot H^{s_1}}.
\end{align*}
Similarly, take
$$
I_2=\int_{|\xi|>{\lambda}}\frac{1}{|\xi|^{s_2}}|\xi|^{s_2}|\widehat{f(\xi)}|d\xi.
$$
Then we have
\begin{align*}
I_2&\leq   \Big(\int_{|\xi|>{\lambda}}\frac{1}{|\xi|^{2s_2}}d\xi\Big)^{1/2}
\|f\|_ {\dot H^{s_2}}\\
&\leq  2\sqrt{\pi}\Big(\int^{\infty}_{\lambda}\frac{1}{r^{2s_{2}-2}}dr\Big)^{1/2}
\|f\|_ {\dot H^{s_2}}\\&\leq c_{s_2}\lambda^{\frac{3}{2}-s_2}\|f\|_{\dot H^{s_2}}.
\end{align*}
Therefore,
$$
\|f\|_{L^{\infty}}\leq A\lambda^{\frac{3}{2}-s_1}+B\lambda^{\frac{3}{2}-s_2},
$$
with $A=c_{s_1}\|f\|_ {\dot H^{s_1}}$ and $B=c_{s_2}\|f\|_{\dot H^{s_2}}$.

Since the  function
$$
\lambda\mapsto \varphi(\lambda)= A\lambda^{\frac{3}{2}-s_1}
+B\lambda^{\frac{3}{2}-s_2}
$$
attains its minimum at   
$\lambda=\lambda^* =c(s_1,s_2)(B/A)^{\frac{1}{{s_2}-{s_1}}}$.
Then
$$
\|f\|_{L^{\infty}(\mathbb{R}^3)}
\leq c' A^{\frac{{s_2}-\frac{3}{2}}{{s_2}-{s_1}}} 
B^{\frac{{\frac{3}{2}}-{s_1}}{{s_2}-{s_1}}}.
$$
\end{proof}

We remark that, for $s_1=1$ and $s_2=2$, where 
$f\in \dot H^{1}(\mathbb{R}^3)\cap \dot H^{2}(\mathbb{R}^3)$, we obtain
\begin{equation}\label{rem1}
\|f\|_{L^{\infty}(\mathbb{R}^3)}
\leq \|\hat{f}\|_{L^{1}(\mathbb{R}^3)}
\leq c\|f\|^{1/2}_ {\dot H^{1}(\mathbb{R}^3)}\|f\|^{1/2}_ {\dot H^{2}(\mathbb{R}^3)}.
\end{equation}

\section{Long time decay of \eqref{NSE} in $H^1(\mathbb{R}^3)$}

In this section, we prove that if 
$u\in{\mathcal C}(\mathbb{R}^+,H^1(\mathbb{R}^3))$ is a global solution 
of \eqref{NSE}, then
\begin{equation}\label{eq3}
\limsup_{t\to\infty}\|u(t)\|_{H^1}=0.
\end{equation}
This proof is done in two steps.
\smallskip

\noindent\textbf{Step 1:}
 We shall prove that
\begin{equation}\label{eq4}
\limsup_{t\to\infty}\|u(t)\|_{\dot H^1}=0.
\end{equation}
We have
$$
\partial_t u-\Delta u+ u\cdot\nabla u =-\nabla p.
$$
Taking the $\dot {H}^{1/2}(\mathbb{R}^3)$ inner product of the above 
equality with $u$, we obtain
$$
\frac{1}{2}\frac{d}{dt} \|u\|_{\dot {H}^{1/2}}^2+\|\nabla u\|_{\dot {H}^{1/2}}^2
\leq |\langle (u\cdot\nabla u) \mid u \rangle_{\dot {H}^{1/2}}|.
$$
Using the fundamental property $u\cdot\nabla v=\operatorname{div}(u\otimes v)$ 
if $\operatorname{div}v=0$, we obtain
\begin{align*}
\frac{1}{2}\frac{d}{dt}  \|u\|_{\dot {H}^{1/2}}^2+ \|\nabla u\|_{\dot H^{1/2}}^2
&\leq  |\langle (u\cdot\nabla u) \mid u \rangle_{\dot {H}^{1/2}}|\\
&\leq  |\langle \operatorname{div}(u\otimes u) \mid u \rangle_{\dot {H}^{1/2}}|\\
&\leq  |\langle u\otimes u \mid \nabla u \rangle_{\dot {H}^{1/2}}|\\
&\leq  \|u\otimes u\|_{\dot {H}^{1/2}} \|\nabla u\|_{\dot H^{1/2}}\\
&\leq \|u\otimes u\|_{\dot {H}^{1/2}} \|u\|_{\dot H^{3/2}}.
\end{align*}
Hence, from Lemma \eqref{lem1} there would exist a constant $c>0$ such that
$$
\frac{1}{2}\frac{d}{dt}\|u\|_{\dot H^{1/2}}^2+ \|u\|_{\dot H^{3/2}}^2
\leq c \|u\|_{\dot{H}^{1/2}} \|u\|_{\dot {H}^{3/2}}^2.
$$
From the equality \eqref{eq2} there  would exist $t_0>0$  such that, 
for all $t\geq t_0 $,
$$
\|u(t)\|_{\dot H^{1/2}}<\frac{1}{2c}.
$$
Then
$$
\frac{1}{2}\frac{d}{dt} \|u\|_{\dot H^{1/2}}^2
+\frac{1}{2} \|u\|_{\dot H^{3/2}}^2\leq 0,\quad \forall t\geq t_0.
$$
Integrating with respect to time, we obtain
\begin{align*}
\|u(t)\|_{\dot H^{1/2}}^2+\int_{t_0}^t \| u(\tau)\|_{\dot H^{3/2}}^2
d \tau \leq  \|u(t_0)\|_{\dot H^{1/2}}^2,\quad \forall t\geq t_0.
\end{align*}
Let $s>0$ and $c=c_s$. There exists $T_0=T_0(s,u^0)>0$, such that
$$
\|u(T_0)\|_{\dot H^{1/2}}<\frac{1}{2c_s}.
$$
Then
$$
\|u(t)\|_{\dot H^{1/2}}<\frac{1}{2c_s},\quad \forall t\geq T_0.
$$
Now, for $s>0$ we have
$$
\partial_t u-\Delta u+ u\cdot\nabla u =-\nabla p.
$$
Taking the $\dot {H}^s(\mathbb{R}^3)$ inner product of the above equality
 with $u$, we obtain
$$
\frac{1}{2}\frac{d}{dt} \|u\|_{\dot {H}^s}^2+\|\nabla u\|_{\dot {H}^s}^2
\leq|\langle (u\cdot\nabla u) \mid u \rangle_{\dot {H}^s}|.
$$
Using the fundamental property $u\cdot\nabla v=\operatorname{div}(u\otimes v)$ if $\operatorname{div}v=0$, we obtain
\begin{align*}
\frac{1}{2}\frac{d}{dt}  \|u\|_{\dot {H}^s}^2+ \|u\|_{\dot H^{s+1}}^2
&\leq  |\langle (u\cdot\nabla u) \mid u \rangle_{\dot {H}^s}|\\
&\leq  |\langle \operatorname{div}(u\otimes u) /u \rangle_{\dot {H}^s}|\\
&\leq  |\langle u\otimes u \mid \nabla u \rangle_{\dot {H}^s}|\\
&\leq  \|u\otimes u\|_{\dot {H}^s} \|\nabla u\|_{\dot H^s}\\
&\leq \|u\otimes u\|_{\dot {H}^s} \|u\|_{\dot H^{s+1}}\\
&\leq c_s \|u\|_{\dot H^{1/2}}\|u\|_{\dot H^{s+1}}^2.
\end{align*}
Thus
$$
\frac{1}{2}\frac{d}{dt} \|u\|_{\dot {H}^s}^2+\frac{1}{2} \|u(t)\|_{\dot H^{s+1}}^2
\leq 0,\quad  \forall t\geq T_0.
$$
So, for $T_0\leq t'\leq t$,
$$
\|u(t)\|_{\dot {H}^s}^2+\int_{t'}^t \|u(\tau)\|_{\dot H^{s+1}}^2d\tau 
\leq \|u(t')\|_{\dot {H}^s}^2.
$$
In particular, for $s=1$,
\begin{align*}
\|u(t)\|_{\dot {H}^1}^2+\int_{t'}^t \|u(\tau)\|_{\dot H^{2}}^2 d\tau
\leq \|u(t')\|_{\dot {H}^1}^2.
\end{align*}
Then, the map  $t\to\|u(t)\|_{\dot H^1}$ is decreasing on $[T_0,\infty)$ 
and $u\in L^2([0,\infty),\dot H^2(\mathbb{R}^3))$.
Now, let $\varepsilon>0$ be small enough. Then the $L^2$-energy estimate
$$
\|u(t)\|_{L^2}^2+2\int_{T_0}^t\|\nabla u(\tau)\|_{L^2}^2d\tau
\leq\|u(T_0)\|_{L^2}^2,\quad \forall t\geq T_0
$$
implies that $u\in L^2([T_0,\infty),\dot H^1(\mathbb{R}^3))$ 
and there is a time $t_{\varepsilon}\geq T_0$ such that
$$
\|u(t_\varepsilon)\|_{\dot H^1}<\varepsilon.
$$
Since  the map $t\mapsto \|u(t)\|_{\dot H^1}$ is decreasing on 
$[T_0,\infty)$, it follows that
$$
\|u(t)\|_{\dot H^1}<\varepsilon,\quad \forall t\geq t_\varepsilon.
$$
Therefore \eqref{eq4} is proved.
\smallskip

\noindent\textbf{Step 2:}
 In this step, we prove that
\begin{equation}\label{eq5}
\limsup_{t\to\infty}\|u(t)\|_{L^2}= 0.
\end{equation}
This proof is inspired by \cite{BB1} and \cite{BS1}. For $\delta>0$ and 
a given distribution $f$, we define the operators $A_{\delta}(D)$ and 
$B_{\delta}(D)$ as follows
$$
A_{\delta}(D)f=\mathcal{F}^{-1}(\mathbf{1}_{\{|\xi|<\delta\}}
\mathcal{F}(f)),\quad
B_{\delta}(D)f=\mathcal{F}^{-1}(\mathbf{1}_{\{|\xi|\geq\delta\}}\mathcal{F}(f)).
$$
It is clear that when applying $A_{\delta}(D)$ (respectively, $B_{\delta}(D)$) 
to any distribution, we are dealing with its low-frequency part 
(respectively, high-frequency part).

Let $u$ be a solution to \eqref{NSE}. Denote by $\omega_{\delta}$ and 
$\upsilon_{\delta}$, respectively, the low-frequency part and the
 high-frequency part of $u$ and so on ${\omega_{\delta}}^0$ and 
${\upsilon_{\delta}}^0$ for the initial data $u^0$. We have
$$
\partial_t u-\Delta u+ u\cdot\nabla u =-\nabla p.
$$
Then
$$
\partial_t u-\Delta u+ \mathbb{P}(u\cdot\nabla u) =0.
$$
Applying the pseudo-differential operators $A_{\delta}(D)$ to the above equality, 
we obtain
\begin{gather*}
\partial_t A_{\delta}(D)u- \Delta A_{\delta}(D)u+A_{\delta}
(D)\mathbb{P}(u\cdot\nabla u)=0, \\
\partial_t \omega_{\delta}-\Delta\omega_{\delta}
 +A_{\delta}(D)\mathbb{P}(u\cdot\nabla u)=0.
\end{gather*}
Taking the $L^2(\mathbb{R}^3)$ inner product of the above equality 
with $\omega_{\delta}(t)$, we obtain
\begin{align*}
\frac{1}{2}\frac{d}{dt} \|\omega_{\delta}(t)\|_{L^2}^2
 +\|\nabla \omega_{\delta}(t)\|_{L^2}^2
&\leq |\langle A_{\delta}(D)\mathbb{P}(u(t)\cdot\nabla u(t)) \mid \omega_{\delta}(t) \rangle_{L^2}|\\
&\leq  |\langle A_{\delta}(D)\operatorname{div}(u\otimes u)(t)\mid \omega_{\delta}(t) \rangle_{L^2}|\\
&\leq |\langle A_{\delta}(D)(u\otimes u )(t)\mid \nabla\omega_{\delta}(t) \rangle_{L^2}|\\
&\leq  |\langle (u\otimes u)(t)\mid \nabla\omega_{\delta}(t) \rangle_{L^2}|\\
&\leq \|u\otimes u(t)\|_{L^2}\|\nabla\omega_{\delta}(t)\|_{L^2}\\
&\leq \|u\otimes u(t)\|_{L^2}\|\nabla\omega_{\delta}(t)\|_{L^2}.
\end{align*}
Lemma \ref{lem1} gives
\begin{align*}
\frac{1}{2}\frac{d}{dt} \|\omega_{\delta}(t)\|_{L^2}^2
+\|\nabla \omega_{\delta}(t)\|_{L^2}^2
&\leq  C\|u(t)\| _{\dot H^{1/2}} \|\nabla u(t)\|_{L^2}
 \|\nabla\omega_{\delta}(t)\|_{L^2}\\
&\leq  CM\|\nabla u(t)\|_{L^2}\|\nabla\omega_{\delta}(t)\|_{L^2}.
\end{align*}
with $M=\sup_{t\geq0}\|u(t)\| _{\dot H^{1/2}})$. Integrating with respect to $t$, 
we obtain
$$
\|\omega_{\delta}(t)\|_{L^2}^2\leq \|{\omega_{\delta}}^0\|_{L^2}^2
+CM\int_0^t \|\nabla u(\tau)\|_{L^2}\|\nabla\omega_{\delta}(\tau)\|_{L^2}d\tau.
$$
Hence, we have $\|\omega_{\delta}(t)\|_{L^2}^2\leq M_{\delta}$ for all $t\geq0$, 
where
$$
M_{\delta}=\|{\omega_{\delta}}^0\|_{L^2}^2
+CM\int_0^{\infty} \|\nabla u(\tau)\|_{L^2}
\|\nabla\omega_{\delta}(\tau)\|_{L^2}d\tau.
$$
Using the fact that  
$\lim_{\delta\to0}\|{\omega_{\delta}}^0\|_{L^2(\mathbb{R}^3)}^2=0$ 
and thanks to  the Lebesgue-dominated convergence theorem we deduce that
\begin{equation}\label{eq6}
\lim_{\delta\to0}\int_0^{\infty}\|\nabla u(\tau)\|_{L^2}
\|\nabla \omega_{\delta}(\tau)\|_{L^2}d\tau=0.
\end{equation}
Hence $\lim_{\delta\to0} M_{\delta}=0$, and thus
\begin{equation}\label{eq7}
\lim_{\delta\to0}\sup_{t\geq0}\|\omega_{\delta}(t)\|_{L^2}= 0.
\end{equation}
We can take  time equal to $\infty$ in the integral \eqref{eq6} 
because by definition of  $\omega_{\delta}$ we have
\begin{align*}
\|\nabla \omega_{\delta}\|_{L^2}
&= \|\mathcal{F}(\nabla \omega_{\delta})\|_{L^2}\\
&= \|\xi|\mathbf{1}_{\{|\xi|<\delta\}}\mathcal{F}(u)\|_{L^2}\\
&\leq \|\xi|\mathcal{F}(u)\|_{L^2}\\
&\leq \|\nabla u\|_{L^2}.
\end{align*}
Now, using the fact that  $\lim_{\delta\to0}\|\nabla \omega_{\delta}(t)\|_{L^2}=0$ 
almost everywhere. Then, the  sequence
$$
\|\nabla u(t)\|_{L^2}\|\nabla \omega_\delta(t)\|_{L^2}
$$
converges point-wise to zero. Moreover, using the above computations and the 
energy estimate \eqref{enq1}, we obtain
$$
\|\nabla u(t)\|_{L^2}\|\nabla \omega_{\delta}(t)\|_{L^2}
\leq \|\nabla u(t)\|_{L^2}^2\in L^1(\mathbb{R}^+).
$$
Thus, the integral sequence is dominated. Hence, the limiting function is 
integrable and one can take the time $T=\infty $ in \eqref{eq6}.

Now, let us investigate the high-frequency part. For this, we apply the 
pseudo-differential operators $B_{\delta}(D)$ to the \eqref{NSE} to obtain
$$
\partial_t \upsilon_{\delta}-\Delta\upsilon_{\delta}
+B_{\delta}(D)\mathbb{P}(u\cdot\nabla u)=0.
$$
Taking the Fourier transform with respect to the space variable, we obtain
\begin{align*}
\partial_t |\widehat{\upsilon_\delta}(t,\xi)|^2
+2 |\xi|^2 |\widehat{\upsilon_\delta}(t,\xi)|^2
&\leq 2|\mathcal{F}(B_{\delta}(D)\mathbb{P}(u\cdot\nabla u))(t,\xi)
 \|\widehat{\upsilon_\delta}(t,\xi)|\\
&\leq 2|\mathcal{F}(B_{\delta}(D)\mathbb{P}(\operatorname{div}
 (u\otimes u)))(t,\xi)\|\widehat{\upsilon_\delta}(t,\xi)|\\
&\leq 2|\xi\|\mathcal{F}(B_{\delta}(D)\mathbb{P}(u\otimes u))
 (t,\xi)\|\widehat{\upsilon_\delta}(t,\xi)|\\
&\leq 2|\xi\|\mathcal{F}(u\otimes u)(t,\xi)\|\widehat{\upsilon_\delta}(t,\xi)|\\
&\leq 2|\mathcal{F}(u\otimes u)(t,\xi)\|\widehat{\nabla\upsilon_\delta}(t,\xi)|.
\end{align*}
Multiplying the obtained equation by $\exp(2t|\xi|^2)$ and integrating with 
respect to time, we obtain
$$
|\widehat{\upsilon_\delta}(t,\xi)|^2
\leq e^{-2 t|\xi|^2}|\widehat{\upsilon_\delta^0}(\xi)|^2
+2\int_0^t e^{-2(t-\tau)|\xi|^2} |\mathcal{F}(u\otimes u)
(\tau,\xi)\|\widehat{\nabla\upsilon_\delta}(\tau,\xi)|d\tau.
$$
Since $|\xi|>\delta$, we have
$$
|\widehat{\upsilon_\delta}(t,\xi)|^2
\leq e^{-2 t\delta^2}|\widehat{\upsilon_\delta^0}(\xi)|^2
+2\int_0^t e^{-2(t-\tau)\delta^2} |\mathcal{F}(u\otimes u)
(\tau,\xi)\|\widehat{\nabla\upsilon_\delta}(\tau,\xi)|d\tau.
$$
Integrating with respect to the frequency variable $\xi$ and using 
Cauchy-Schwarz inequality, we obtain
$$
\|\upsilon_{\delta}(t)\|_{L^2}^2\leq  e^{-2 t{\delta}^2}\|\upsilon_{{\delta}^0}\|_{L^2}^2+2\int_0^t e^{-2(t-\tau){\delta}^2}
\|u\otimes u(\tau)\|_{L^2}\|\nabla\upsilon_{\delta}(\tau) \|_{L^2}d\tau.
$$
By the definition of $\upsilon_\delta$, we have
$$
\|\upsilon_{\delta}(t)\|_{L^2}^2
\leq  e^{-2 t{\delta}^2}\|u^0\|_{L^2}^2
+2\int_0^t e^{-2(t-\tau){\delta}^2}\|u\otimes u(\tau)\|_{L^2}
\|\nabla u (\tau)\|_{L^2}d\tau.
$$
Lemma \ref{lem1} and the equality \eqref{eq2} yield
\begin{align*}
\|\upsilon_{\delta}(t)\|_{L^2(\mathbb{R}^3)}^2
&\leq  e^{-2 t{\delta}^2}\|u^0\|_{L^2(\mathbb{R}^3)}^2
 +C\int_0^t e^{-2(t-\tau){\delta}^2}
\|u (\tau)\|_{\dot H^{1/2}}\|\nabla u (\tau)\|_{L^2}^2d\tau \\
&\leq  e^{-2t{\delta}^2}\|u^0\|_{L^2}^2+CM\int_0^t e^{-2(t-\tau){\delta}^2}
\|\nabla u(\tau) \|_{L^2}^2d\tau,
\end{align*}
where $M=\sup_{t\geq0}\|u \|_{\dot H^{1/2}}$.
Hence, $\|\upsilon_{\delta}(t)\|_{L^2}^2\leq N_{\delta}(t)$, where
$$
N_{\delta}(t)= e^{-2 t{\delta}^2}\|u^0\|_{L^2}^2
+CM\int_0^t e^{-2(t-\tau){\delta}^2}\|\nabla u (\tau)\|_{L^2}^2d\tau.
$$
Using the energy estimate \eqref{enq1}, we obtain 
$N_{\delta}\in L^1(\mathbb{R}^+)$ and
$$
\int_0^{\infty}N_{\delta}(t)dt
\leq \frac{\|u^0\|_{L^2}^2}{2\delta^2}+\frac{CM\|u^0\|_{L^2}^2}{4\delta^2}.
$$
This leads to the fact that the function $t\to\|\upsilon_{\delta}(t)\|_{L^2}^2$ 
is both continuous and Lebesgue integrable over $\mathbb{R}^+$.

Now, let $\varepsilon>0$. At first, the inequality \eqref{eq7} implies 
that there exists some $\delta_0>0$ such that
\begin{align*}
\|\omega_{\delta_0}(t)\|_{L^2}\leq \varepsilon/2,\,\forall\,t\geq0.
\end{align*}
Let us consider the set $\mathrm{R}_{\delta_0}$ defined by $\mathrm{R}_{\delta_0}:=\{t\geq0,\,\|\upsilon_{\delta}(t)\|_{L^2(\mathbb{R}^3)}>\varepsilon/2\}$. If we denote by $\lambda_1(\mathrm{R}_{\delta_0})$ the Lebesgue measure of $\mathrm{R}_{\delta_0}$, we have
$$
\int_0^{\infty}\|\upsilon_{\delta_0}(t)\|_{L^2(\mathbb{R}^3)}^2dt
\geq\int_{\mathrm{R}_{\delta_0}}\|\upsilon_{\delta}(t)\|_{L^2(\mathbb{R}^3)}^2dt
\geq (\varepsilon/2)^2 \lambda_1(\mathrm{R}_{\delta_0}).
$$
By doing this, we can deduce that 
$\lambda_1(\mathrm{R}_{\delta_0})= T^\varepsilon_{\delta^0}<\infty$,
 and there exists $t^\varepsilon_{\delta^0}>T^\varepsilon_{\delta^0}$ such that
\[
\|\upsilon_{\delta_0}(t^\varepsilon_{\delta^0})\|_{L^2}^2\leq (\varepsilon/2)^2.
\]
So, $\|u(t^\varepsilon_{\delta^0})\|_{L^2}\leq \varepsilon$ and from 
the energy estimate \eqref{enq1} we have
$$
\|u(t)\|_{L^2}\leq \varepsilon,\,\,\,\forall t\geq t^\varepsilon_{\delta^0}.
$$
This completes the proof of \eqref{eq5}. 


\section{Generalization of Foias-Temam result in $H^1(\mathbb{R}^3)$}

 Fioas and Temam \cite{FT} proved an analytic property for the Navier-Stokes
 equations on the torus $\mathbb T^3=\mathbb{R}^3/\mathbb Z^3$. 
Here, we give a similar result on the whole space $\mathbb{R}^3$.
\begin{theorem}\label{thm2}
We assume that $u^0\in H^{1}(\mathbb{R}^3)$.  Then, there exists a 
time $T$ that depends only on the $\|u^0\|_{ H^{1}(\mathbb{R}^3)}$, such that
\begin{itemize}
\item \eqref{NSE} possesses on $(0,T)$ a unique regular solution $u$ such 
that the function  $t\mapsto e^{t|D|}u(t)$ is continuous from $[0,T] $ 
into $H^{1}(\mathbb{R}^3)$.

\item If $u\in{\mathcal C}(\mathbb{R}^+,H^1(\mathbb{R}^3))$ is a global 
and bounded solution to \eqref{NSE}, then there are $M\geq0$ and
 $t_0>0$ such that
$$
\|e^{t_0|D|}u(t)\|_{H^1(\mathbb{R}^3)}\leq M,\quad \forall t\geq t_0.
$$
\end{itemize}
\end{theorem}

Before proving this Theorem, we need the following Lemmas.


\begin{lemma}\label{lem5}
Let $ t\mapsto e^{t|D|}u$ belong to 
$\dot H^{1}(\mathbb{R}^3)\cap\dot H^{2}(\mathbb{R}^3)$. 
Then
$$
\|e^{t|D|}(u\cdot\nabla v)\|_{L^2(\mathbb{R}^3)}
\leq \|e^{t|D|}u\|^{1/2}_{H^1(\mathbb{R}^3)} \|e^{t|D|} u\|^{1/2}_{H^2(\mathbb{R}^3)}
\|e^{t|D|}v\|_{H^1(\mathbb{R}^3)}.
$$
\end{lemma}

\begin{proof}
 We have
\begin{align*}
\|e^{t|D|}(u\cdot\nabla v)\|^2_{L^2}
&= \int_{\mathbb{R}^3} e^{2t|\xi|}|\widehat{u\cdot\nabla v}(\xi)|^2d\xi\\
&\leq  \int_{\mathbb{R}^3} e^{2t|\xi|}
\Big(\int_{\mathbb{R}^3} |\hat{u}(\xi-\eta)\|\widehat{\nabla v}
(\eta)|d\eta\Big)^2d\xi\\
&\leq  \int_{\mathbb{R}^3}
\Big(\int_{\mathbb{R}^3} e^{t|\xi|}|\hat{u}(\xi-\eta)
\|\widehat{\nabla v}(\eta)|d\eta\Big)^2d\xi.
\end{align*}
Using the inequality $e^{|\xi|}\leq e^{|\xi-\eta|}e^{|\eta|}$, we obtain
\begin{align*}
\|e^{t|D|}(u\cdot\nabla v)\|^2_{L^2}
&\leq \int_{\mathbb{R}^3}\Big(\int_{\mathbb{R}^3}e^{t|\xi-\eta|}|\hat{u}(\xi-\eta)|
e^{t|\eta|}|\widehat{\nabla v}(\eta)|d\eta\Big)^2d\xi\\
&\leq \int_{\mathbb{R}^3}\Big(\int_{\mathbb{R}^3}
\big(e^{t|\xi-\eta|}|\hat{u}(\xi-\eta)|\big)
\big(e^{t|\eta|}|\eta\|\hat{v}(\eta)|\big)d\eta\Big)^2d\xi\\
&\leq \Big(\int_{\mathbb{R}^3} e^{t|\xi|}|\hat{u}(\xi)|d\xi\Big)^2
\|e^{t|D|}\nabla v\|^2_{L^2}.
\end{align*}
Hence, for $f={\mathcal F}^{-1}( e^{t|\xi|}|\hat{u}(\xi)|)
\in \dot H^{1}(\mathbb{R}^3)\cap\dot H^{2}(\mathbb{R}^3)$, 
inequality \eqref{rem1} gives
\begin{align*}
\|e^{t|D|}(u\cdot\nabla v)\|_{L^2}
&\leq \|e^{t|D|}u\|^{1/2}_{\dot H^1} \|e^{t|D|} u\|^{1/2}_{\dot H^2}
\|e^{t|D|}\nabla v\|_{L^2}\\
&\leq \|e^{t|D|}u\|^{1/2}_{\dot H^1} \|e^{t|D|} u\|^{1/2}_{\dot H^2}
\|e^{t|D|}v\|_{\dot H^1}\\
&\leq \|e^{t|D|}u\|^{1/2}_{ H^1} \|e^{t|D|} u\|^{1/2}_{ H^2}\|e^{t|D|}v\|_{ H^1}.
\end{align*}
\end{proof}


\begin{lemma}\label{lem6}
Let $t\mapsto e^{t|D|}u \in \dot H^{1}(\mathbb{R}^3)\cap\dot H^{2}(\mathbb{R}^3)$. 
Then
$$
\big|\langle e^{t|D|}(u\cdot\nabla v) \mid e^{t|D|}w \rangle_{{H^1}}\big|
\leq\|e^{t|D|}u\|^{1/2}_{H^1} \|e^{t|D|} u\|^{1/2}_{H^2} 
\|e^{t|D|}v\|_{{H^1}} \|e^{t|D|} w\|_{{H^2}}.
$$
\end{lemma}

\begin{proof}
 We have
\begin{align*}
\langle u\cdot\nabla v \mid w \rangle_{H^1}
&= \sum_{|j|=1}\langle\partial_j ( u\cdot\nabla v)\mid \partial_j w\rangle_{L^2}\\
&= - \sum_{|j|=1}\langle u\cdot\nabla v\mid \partial^{2}_j w\rangle_{L^2}\\
&= -\langle u\cdot\nabla v \mid \Delta w\rangle_{L^2}.
\end{align*}
Then
\begin{align*}
\big|\langle e^{t|D|}(u\cdot\nabla v )\mid e^{t|D|}w \rangle_{H^1}\big|
&= \big|\langle e^{t|D|}(u\cdot\nabla v) \mid e^{t|D|}\Delta w \rangle_{L^2}\big|\\
&\leq \|e^{t|D|}(u\cdot\nabla v)\|_{L^2} \|e^{t|D|}\Delta w\|_{L^2}\\
&\leq \|e^{t|D|}(u\cdot\nabla v)\|_{L^2}\|e^{t|D|} w\|_{ \dot H^2}\\
&\leq \|e^{t|D|}(u\cdot\nabla v)\|_{L^2}\|e^{t|D|} w\|_{H^2}.
\end{align*}
Finally, using Lemma \ref{lem5}, we obtain the desired result.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm2}]
 We have
\[
\partial_{t}u -\Delta u +u\cdot\nabla u =-\nabla p.
\]
Applying  the fourier transform to the last equation  and multiplying by
 $\overline{\widehat{u}}$, we obtain
\begin{align*}
\partial_{t} \widehat{u}\cdot\overline{\widehat{u}} 
+ |\xi|^2 |\widehat{u}|^2 = -(\widehat{u\cdot\nabla u})\cdot \overline{\widehat{u}}.
\end{align*}
Then
\begin{align*}
\partial_{t}|\widehat{u}|^2+ 2 |\xi|^2 |\widehat{u}|^2
 = - 2 \operatorname{Re} ((\widehat{u\cdot\nabla u})\cdot\widehat{u}).
\end{align*}
Multiplying the above equation by $(1+|\xi|^2) e^{2t|\xi|}$, we obtain
$$
(1+|\xi|^2) e^{2t|\xi|}\partial_{t}|\widehat{u}|^2 
 + 2(1+|\xi|^2) |\xi|^2 e^{2t|\xi|} |\widehat{u}|^2 
= - 2 \operatorname{Re} ((\widehat{u\cdot\nabla u})\cdot \widehat{u})
(1+|\xi|^2) e^{2t|\xi|}.
$$
Integrating with respect to  $\xi$, we obtain
\begin{align*}
&\int_{\mathbb{R}^3}(1+|\xi|^2) e^{2t|\xi|}\partial_{t}|\widehat{u}(\xi)|^2 d\xi 
+ 2 \int_{\mathbb{R}^3}(1+|\xi|^2)|\xi|^2 e^{2t|\xi|} |\widehat{u}(\xi)|^2 d\xi \\
&=  - 2 \operatorname{Re}
\int_{\mathbb{R}^3}((\widehat{u\cdot\nabla u})\cdot\widehat{u})
(1+|\xi|^2) e^{2t|\xi|} d\xi.
\end{align*}
Thus
\begin{equation}\label{enq2}
\langle e^{t|D|}\partial_{t} u / e^{t|D|} u \rangle_{H^1} 
+ 2 \|e^{t|D|}\nabla u\|^2_{H^1(\mathbb{R}^3)} 
= - 2 Re \langle e^{t|D|} (u\cdot\nabla u) \mid e^{t|D|}u \rangle_{H^1}.
\end{equation}
Therefore,
\begin{align*}
\langle e^{t|D|} u'(t)\mid  e^{t|D|} u(t) \rangle_{H^1}
&= \langle ( e^{t|D|} u(t))'-|D|e^{t|D|}u(t) \mid e^{t|D|} u(t) \rangle_{H^1}\\
&= \frac{1}{2} \frac{d}{dt} \|e^{t|D|}u\|^2_{H^1}
-\langle e^{t|D|}|D| u(t) \mid e^{t|D|} u(t) \rangle_{H^1}\\
&\geq \frac{1}{2} \frac{d}{dt} \|e^{t|D|}u\|^2_{H^1}
- \|e^{t|D|}u\|_{H^1} \|e^{t|D|}u\|_{H^2}.
\end{align*}
Using the Young inequality, we obtain
\begin{equation}\label{enq3}
\frac{d}{dt} \|e^{t|D|}u\|^2_{H^1} -2 \|e^{t|D|}u\|^2_{H^1} 
-\frac{1}{2}\|e^{t|D|}u\|^2_{H^2}
\leq 2 \langle e^{t|D|} u'(t) \mid e^{t|D|} u(t) \rangle_{H^1}.
\end{equation}
Hence, using  Lemma \ref{lem6} and Young inequality the right hand of 
\eqref{enq2} satisfies
\begin{align*}
|-2 \operatorname{Re}\langle e^{t|D|}( u\cdot\nabla u) \mid  e^{t|D|} 
u \rangle_{H^1}|
&\leq 2 \|e^{t|D|}u\|^{3/2}_{H^1} \|e^{t|D|}u\|^{3/2}_{H^2}\\
&\leq  \frac{3}{4}\|e^{t|D|}u\|^2_{H^2} + \frac{c_1}{2} \|e^{t|D|}u\|^6_{H^1},
\end{align*}
where $c_1$ is a positive  constant. Then,  \eqref{enq2} yields
\begin{equation}\label{enq4}
\langle e^{t|D|}u'(t) \mid e^{t|D|} u(t) \rangle_{H^1}+2 \|e^{t |D|}
 \nabla u\|^2_{H^1} 
\leq  \frac{3}{4}\|e^{t|D|}u\|^2_{H^2}+\frac{c_1}{2}\|e^{t|D|}u\|^6_{H^1}.
\end{equation}
Hence, using  \eqref{enq3}--\eqref{enq4}, we obtain
\[
\frac{d}{dt} \|e^{t|D|}u\|^2_{H^1} -2\|e^{t |D|} u\|^2_{H^1}
-2\|e^{t |D|} u\|^2_{H^2}+ 4 \|e^{t |D|} \nabla u\|^2_{H^1}
\leq c_1\|e^{t|D|}u\|^6_{H^1}.
\]
The equality $\|e^{t |D|} u\|^2_{H^2}
=\|e^{t |D|} u\|^2_{H^1}+\|e^{t |D|} \nabla u\|^2_{H^1}$ yields
\begin{align*}
\frac{d}{dt} \|e^{t|D|}u\|^2_{H^1} + 2 \|e^{t |D|} \nabla u\|^2_{H^1}
&\leq  4 \|e^{t|D|}u\|^2_{H^1} +c_1\|e^{t|D|}u\|^6_{H^1}\\
&\leq c_2 + 2 c_1\|e^{t|D|}u\|^6_{H^1}.
\end{align*}
where  $ c_2$ is a positive constant.
Finally, we obtain
\[
y(t)\leq y(0)+K_1 \int^t_0y^3(s)ds.
\]
where
\begin{equation*}
 y(t)=1 + \|e^{t|D|} u(t) \|^2_{H^1} \quad\mbox{and}\quad K_1 =  2 c_1+c_2.
\end{equation*}
Let
\begin{equation*}
T_1=\frac{2}{K_1 y^2(0)}
\end{equation*}
and $0< T \leq T^*$ be such that
$T =\sup \{t\in[0,T^*)\,|\,\sup_{0\leq s\leq t}y(s)\leq 2y(0)\}$. 
Hence for $0 \leq t \leq\min(T_1,T)$, we have
\begin{align*}
y(t)
&\leq  y(0)+K_1 \int^t_0 y^3(s)ds\\
&\leq  y(0)+K_1 \int^t_0 8 y^3(0)ds\\
&\leq  \left(1+K_1 8 T_1 y^2(0)\right)y(0).
\end{align*}
Taking $1+K_1 8 T_1 y^2(0)<2$, we obtain $ T>T_1$. Then
$y(t)\leq2y(0)$ for all $t \in [0,T_1]$.
This shows that $ t\mapsto e^{t|D|}u(t)\in H^1{(\mathbb{R}^3)}$ for all 
$t \in [0,T_1]$.
In particular
$$
\| e^{T_1|D|}u(T_1)\|^2_{H^1} \leq 2+2\|u_0\|^2_{H^1}.
$$
Now, from the hypothesis, we assume that there exists $M_1>0$ such that
\[
\|u(t)\|_{H^1}\leq M_1 \quad   \text{for all } t \geq 0.
\]
Define the system
\begin{gather*}
\partial_t w  -\Delta w+ w.\nabla w   
= -\nabla p \quad \text{in } \mathbb{R}^+\times \mathbb{R}^3\\
    \operatorname{div} w = 0 \quad \text{in } \mathbb{R}^+\times \mathbb{R}^3\\
    w(0) =u(T) \quad \text{in }\mathbb{R}^3,
  \end{gather*}
where $w(t)=u(T+t)$.
Using a similar technique, we can prove that there exists 
$T_2 =\frac{2}{K_1}(1+M^{2}_1)^{-2}$  such that
$$
y(t)=1+\|e^{t|D|}w(t)\|^2_{H^1} \leq 2(1+M^{2}_1), \quad \forall t \in [0,T_2].
$$
This implies $1+\|e^{t|D|}u(T+t)\|^2_{H^1} \leq 2(1+M^{2}_1)$. 
Hence, for $t=T_2$ we have
$$
\|e^{T_2|D|}u(T+T_2)\|^2_{H^1} \leq 2(1+M^{2}_1).
$$
Since $t=T+T_2\geq T_2$ for all $T\geq 0$, we obtain
$$
\|e^{T_2|D|}u(t)\|^2_{H^1} \leq 2(1+M^{2}_1),\quad \forall t\geq T_2.
$$
Then
$$
\|e^{T_2|D|}u(t)\|^2_{H^1} \leq 2(1+M^{2}_1),\quad \forall t\geq T_2,
$$
where
\begin{equation*}
T_2 = T_2(M_1) =\frac{2}{K_1}(1+M^{2}_1)^{-2}.
\end{equation*}
\end{proof}


\section{Proof of the main result}

In this section, we prove  Theorem \ref{thm1}. This proof uses the results 
of sections $3$ and $4$.

 Let $u\in{\mathcal C}(\mathbb{R}^+, H^1_{a,\sigma}(\mathbb{R}^3) )$.
As $ H^1_{a,\sigma}(\mathbb{R}^3)\hookrightarrow H^1(\mathbb{R}^3)$, then 
$u\in{\mathcal C}(\mathbb{R}^+,H^1(\mathbb{R}^3))$.
Applying Theorem \ref{thm2}, there exist $t_0>$ such that
\begin{equation}\label{enq5}
\|e^{t_0|D|}u(t)\|_{H^1}\leq c_0=2+M_1^2,\quad \forall t\geq t_0,
\end{equation}
where  $t_0=\frac{2}{K_1}(1+M_1^2)^{-2}$.
Let $a>0$, $\beta>0$. Then there exists $c_3\geq0$ such that
\begin{align*}
ax^{1/\sigma}\leq c_3+\beta x,\quad \forall x\geq0.
\end{align*}
Indeed, $\frac{1}{\sigma}+\frac{\sigma-1}{\sigma}=\frac{1}{p}+\frac{1}{q}=1$. 
Using the Young inequality, we obtain
\begin{align*}
ax^{1/\sigma}&= a \beta^{\frac{-1}{\sigma}}(\beta^{1/\sigma}x^{1/\sigma})\\
&\leq \frac{(a \beta^{\frac{-1}{\sigma}})^q}{q}
 +\frac{( \beta^{1/\sigma}x^{1/\sigma})^p}{p}\\
&\leq c_3+\frac{\beta x}{\sigma} \leq c_3+\beta x,
\end{align*}
where $c_3=\frac{\sigma-1}{\sigma}a^{\frac{\sigma}{\sigma-1}}
\beta^{\frac{1}{1-\sigma}}$.

Take $\beta=\frac{t_0}{2}$, using  \eqref{enq5} and the  Cauchy Schwarz inequality, 
we have
\begin{align*}
\|u(t)\|_{ H^1_{a,\sigma}}^2
&= \|e^{a|D|^{1/\sigma}}u(t)\|_{H^1}^2\\
&=  \int(1+|\xi|^2) e^{2a|\xi|^{1/\sigma}}|\widehat{u}(t,\xi)|^2d\xi\\
&\leq \int(1+|\xi|^2) e^{2(c_3+\beta|\xi|})|\widehat{u}(t,\xi)|^2d\xi\\
&\leq \int(1+|\xi|^2) e^{2c_3}e^{t_0 |\xi|}|\widehat{u}(t,\xi)|^2d\xi\\
&\leq e^{2c_3}\Big(\int(1+|\xi|^2)|\widehat{u}(t,\xi)|^2d\xi\Big)^{1/2}
\Big(\int(1+|\xi|^2)e^{2t_0|\xi|}|\widehat{u}(t,\xi)|^2d\xi\Big)^{1/2}\\
&\leq e^{2c_3}\|u\|_{ H^1}^{1/2}\|e^{t_0|D|}u(t)\|_{H^1}^{1/2}\\
&\leq c\|u\|_{ H^1}^{1/2},
\end{align*}
where $c=e^{2c_3}c_0^{1/2}$.
Using the inequality \eqref{eq3}, we obtain
$$
\limsup_{t\to\infty}\|e^{a|D|^{1/\sigma}}u(t)\|_{H^1}=0.
$$


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\end{document}