#  On The Continuous Periodic Weak Solutions Of Boussinesq Equations

Tao Tao  School of Mathematics Sciences, Peking University, Beijing, China
<EMAIL_ADDRESS>and  Liqun Zhang  Academy of mathematic and system science ,
Beijing , China<EMAIL_ADDRESS>
###### Abstract.

The Boussingesq equations was introduced in understanding the coupling nature
of the thermodynamics and the fluid dynamics. We show the existence of
continuous periodic weak solutions of the Boussinesq equations which satisfies
the prescribed kinetic energy or some other prescribed property. Our results
represent the conversions between internal energy and mechanical energy.

Keywords: Boussinesq equations, Continuous periodic weak solutions, Prescribed
kinetic

energy

AMS Subject Classification (2000): 35Q30, 76D03

##  1\. Introduction

In this paper, we consider the following Boussinesq equations

$\begin{cases}v_{t}+\hbox{div}(v\otimes v)+\nabla p=\theta
e_{3},\quad\quad\mbox{in}\quad T^{3}\times[0,1]\\\ \hbox{div}v=0,\\\
\theta_{t}+\hbox{div}(v\theta)=h,\quad\quad\mbox{in}\quad
T^{3}\times[0,1],\end{cases}$ (1.1)

where $T^{3}$ denote the 3-dimensional torus, i.e. $T^{3}=S^{1}\times
S^{1}\times S^{1}$ and $e_{3}=(0,0,1)^{T}$. Here in our notations, $v$ is the
velocity vector, $p$ is the pressure, $\theta$ denotes the temperature or
density which is a scalar function and $h$ is the heat sources. The Boussinesq
equations model many geophysical flows, such as atmospheric fronts and ocean
circulations (see, for example, [22],[24]).

A recurrent issue in the modern theory of PDEs is that one needs to go beyond
classical solutions. The pair $(v,\theta)$ on $T^{3}\times[0,1]$ is called a
weak solution of (1.1) if they solve (1.1) in the following sense:

$\displaystyle\int_{0}^{1}\int_{T^{3}}(\partial_{t}\varphi\cdot
v+\nabla\varphi:v\otimes v+p\hbox{div}\varphi-\theta
e_{3}\cdot\varphi)dxdt=0,$

for all $\varphi\in C_{c}^{\infty}(T^{3}\times(0,1);R^{3}),$

$\displaystyle\int_{0}^{1}\int_{T^{3}}(\partial_{t}\phi\theta+v\cdot\nabla\phi\theta+h\phi)dxdt=0,$

for all $\phi\in C_{c}^{\infty}(T^{3}\times(0,1);R)$ and

$\displaystyle\int_{0}^{1}\int_{T^{3}}v\cdot\nabla\psi dxdt=0.$

for all $\psi\in C_{c}^{\infty}(T^{3}\times(0,1);R).$

Moreover, the study of weak solutions in fluid dynamics, including those which
fail to conserve energy, is natural in the context of turbulent flows. One of
the famous example is the Onsager conjecture on Euler equation which claims
that the incompressible Euler equation admits Hölder continuous solution which
dissipates kinetic energy. More precisely, the Onsager conjecture on Euler
equation can be stated as following:

  1. (1)

$C^{0,\alpha}$ solution are energy conservative when $\alpha>\frac{1}{3}$.

  2. (2)

For any $\alpha<\frac{1}{3}$, there exist dissipative solutions with
$C^{0,\alpha}$ regularity .

For this conjecture, the part (a) has been proved by P. Constantin, E, Weinan
and E. Titi in [10]. Slightly weak assumption on the solution was subsequently
shown to be sufficient for energy conservation by P. Constantin, etc. in [5,
17], also see [28]. More recently, P. Isett and Sung-jin Oh give a proof to
this part of Onsager’s conjecture for the Euler equations on manifolds by heat
flow method in [18].

The part (b) has been treated by many authors. For weak solutions, there are
some results by V. Scheffer ([25]), A. Shnirelman ([26, 27]) and Camillo De
Lellis, László Székelyhidi ([12, 13]). In recent year, a great progress in the
construction of continuous and Hölder solution was made by Camillo De Lellis,
László Székelyhidi, P. Isett and T. Buckmaster. Camillo De Lellis and László
Székelyhidi developed an iterative scheme (some kind of convex integration) in
[15]. Together with the aid of Beltrami flow on $T^{3}$ and Geometric Lemma,
they constructed a continuous periodic solution on $T^{3}$ which satisfies the
prescribed kinetic energy. The solution is a superposition of infinitely many
(perturbed) and weakly interacting Beltrami flows. Building on the iterative
techniques in [15] and Nash-Moser mollify techniques, they constructed Hölder
continuous periodic solution with exponent $\theta$, for any
$\theta<\frac{1}{10}$, which satisfies the prescribed kinetic energy in [16].
Later, in his PhD Thesis [20], P. Isett introduced some ideas and construct
Hölder continuous periodic solutions with any $\theta<\frac{1}{5}$, and the
solution has compact supports in time. Adhering more to the original scheme
and introducing some new devices in [3], Camillo De Lellis, László Székelyhidi
and T. Buckmaster constructed Hölder continuous weak solutions with any
$\theta<\frac{1}{5}$, which satisfies the prescribed kinetic energy, also see
[2]. Also, in whole space $R^{3}$, P. Isett and Sung-jin Oh in [19]
constructed Hölder continuous solutions with any $\theta<\frac{1}{5}$, which
satisfies the prescribed kinetic energy or is a perturbation of smooth Euler
flow.

For the Onsager critical spatial regularity (Hölder exponent
$\theta=\frac{1}{3}$), there are also some interesting results. In line with
previous scheme, by keeping track of sharper, time localized estimates in [1],
T. Buckmaster constructed Hölder continuous(with exponent
$\theta<\frac{1}{5}-\varepsilon$ in time-space) periodic solutions which for
almost every time belongs to $C_{x}^{\theta}$, for any $\theta<\frac{1}{3}$,
and is compactly temporal supported. By the subdivision of the time interval
and careful smoothing Reynolds stress for some time intervals in [4] , Camillo
De Lellis, László Székelyhidi and T. Buckmaster constructed Hölder continuous
periodic solution which belongs to $L^{1}_{t}C_{x}^{\theta}$, for any
$\theta<\frac{1}{3}$, and has compact support in time. Recently, S. Daneri
consider the Cauchy problem for dissipative Hölder Euler flow in [11].

Motivated by Onsager’s conjecture of Euler equation and the above earlier
works, we consider the Boussinesq equations and want to know if the similar
phenomena can also happen when considering the temperature effects. The
difference is that there are conversions between internal energy and
mechanical energy. What we need is to overcome the difficulty of interactions
between velocity and temperature. To this end, we first consider the existence
of continuous periodic solution of Boussinesq equations which satisfies the
prescribed kinetic energy. We follow the general scheme in the construction of
Euler equations. By establishing the corresponding geometric lemma and
improving the iteration scheme, we obtained some existence results. Besides,
our results represent some conversions between internal energy and mechanical
energy.

Before presenting our theorem, we introduce some notations

$\displaystyle\Theta:=\\{\theta(x)\in C^{\infty}(T^{3}):\theta~{}~{}\mbox{only
dependent on}~{}~{}x_{3},\quad
i.e.\quad\theta(x)=\overline{\theta}(x_{3})\\},$

and

$\displaystyle\Xi:=\\{a(t)b(x_{3}):a\in
C^{\infty}([0,1])~{}~{}\mbox{and}~{}~{}b\in C^{\infty}([0,2\pi])\\}.$

First, for the case without heat source, i.e. $h=0$, we have the following
theorem:

######  Theorem 1.1.

Assume that $e(t):[0,1]\rightarrow R$ is a given positive smooth function and
$\theta_{0}\in\Theta$. Then there exist

$\displaystyle(v,p,\theta)\in C(T^{3}\times[0,1];R^{3}\times R\times R)$

and a positive number $M=M(e)$ such that they solve the system (1.1) with
$h=0$ in the sense of distribution and

$e(t)=\int_{T^{3}}|v|^{2}(x,t)dx,\qquad\|\theta-\theta_{0}\|_{0}<4M,$

where $\|f\|_{0}=\hbox{sup}_{x,t}|f(x,t)|.$

######  Remark 1.1.

In our theorem 1.1, if $\theta=0$, then it’s the continuous Euler flow with
prescribed kinetic energy and have been constructed by Camillo De Lellis and
László Székelyhidi in [15]. In general, for example, if we take
$\theta_{0}=10Mcosx_{3}\in\Theta$, then we must have $\theta\neq 0$.

Next, we consider the Boussinesq equations (1.1) with heat source. For some
kinds of heat sources, we can also construct continuous periodic solution with
prescribed kinetic energy. Specially, we have the following theorem:

######  Theorem 1.2.

Assume that $e(t):[0,1]\rightarrow R$ is a positive smooth function and
$h\in\Xi$. Then there exist

$\displaystyle(v,p,\theta)\in C(T^{3}\times[0,1];R^{3}\times R\times R)$

such that they solve the system (1.1) in the sense of distribution and

$e(t)=\int_{T^{3}}|v|^{2}(x,t)dx.$

Furthermore, to explore the possible effect of temperature on the velocity
field, we can also prove the following theorem for Boussinesq equations
without heat source.

######  Theorem 1.3.

For a given positive constant $M$ and any positive number $\lambda$, there
exist

$\displaystyle(v,p,\theta)\in C(T^{3}\times[0,1];R^{3}\times R\times R)$

such that they solve the system (1.1) in the sense of distribution and

$\displaystyle\|v(x,0)\|_{0}\leq$ $\displaystyle
4M,~{}~{}~{}~{}~{}\int_{0}^{1}\int_{T^{3}}|\theta|^{2}(x,t)dxdt\geq\lambda^{2},$
$\displaystyle{\sup}_{x\in T^{3}}|v(x,t)|\geq$
$\displaystyle\lambda,\qquad{\inf}_{x\in T^{3}}|v(x,t)|\leq 4M,\qquad\forall
t\in[\frac{1}{2},1].$

######  Remark 1.2.

For the Boussinesq system on $T^{3}$, even the initial velocity is small, the
oscillation of velocity after sometime could be as large as possible if we
have enough thermo in the systems.

The result of last theorem coincide with some nature observations and may be
helpful in understanding some nature phenomenon.

######  Remark 1.3.

The above theorems also hold for the two-dimensional Boussinesq system on
$T^{2}$.

Finally, we brief some main ideas in our proof. In [15], they make use of a
special families of stationary, plane wave solutions of Euler
equation–Beltrami flow–which allows for the control of interference terms
between different waves in the construction. For the Boussinesq equations, we
don’t know the existence of the analogous families of stationary, plane wave
solutions. Following an idea in [23, 21], we make use of a multi-steps
iteration scheme and one-dimensional oscillation. That is, in each step, we
add several plane waves which oscillate along the same direction with
different frequency. Thus we can remove one component of stress error in each
step. To reduce the whole stress errors, we then divide into several steps and
finish it one by one. Moreover, in Boussinesq equations, the velocity and
temperature are coupled together, so we need to reduce two stress errors
simultaneously. To achieve this, we first extend the geometric lemma proved in
[15] and add an associated plane wave in the velocity and temperature
simultaneously in each step. The consistence is important for us to reduce the
temperature stress errors in our construction of continuous temperature.

##  2\. Proof of main result and Plan of the paper

As [15], the proof of Theorem 1.1, Theorem 1.2 and Theorem 1.3 will be
achieved through an iteration procedure. In this and subsequent section,
$\mathcal{S}_{0}^{3\times 3}$ denotes the vector space of symmetric trace-free
$3\times 3$ matrices.

######  Definition 2.1.

Assume $v,p,\theta,\mathring{R},f$ are smooth functions on $T^{3}\times[0,1]$
taking values, respectively, in $R^{3},R,R,\mathcal{S}_{0}^{3\times 3},R^{3}$.
We say that they solve the Boussinesq-Reynolds system (with or without heat
source) if

$\begin{cases}\partial_{t}v+{\rm div}(v\otimes v)+\nabla p=\theta e_{3}+{\rm
div}\mathring{R}\\\ {\rm div}v=0\\\ \theta_{t}+{\rm div}(v\theta)=h+{\rm
div}f.\end{cases}$ (2.2)

We now state the first main proposition of the paper, of which Theorem 1.1 and
Theorem 1.2 are implied directly. The proposition is also an extension of the
corresponding result given in [15].

######  Proposition 2.1.

Let $e(t)$ be as in Theorem 1.1 and Theorem 1.2. Then we can choose two
positive constants $\eta$ and $M$ only dependent of $e(t)$, such that the
following properties hold:

For any $0<\delta\leq 1$, if $(v_{,}~{}p,~{}\theta,~{}\mathring{R},~{}f)\in
C^{\infty}([0,1]\times T^{3})$ solve Boussinesq-Reynolds system (2.2) and

$\displaystyle\frac{3\delta}{4}e(t)\leq e(t)-\int_{T^{3}}|v|^{2}(x,t)dx\leq$
$\displaystyle\frac{5\delta}{4}e(t),\quad\forall t\in[0,1],$ (2.3)
$\displaystyle\sup_{x,t}|\mathring{R}(x,t)|\leq$ $\displaystyle\eta\delta,$
(2.4) $\displaystyle\sup_{x,t}|f(x,t)|\leq$ $\displaystyle\eta\delta,$ (2.5)

then we can construct new functions
$(\tilde{v},~{}\tilde{p},~{}\tilde{\theta},~{}\mathring{\tilde{R}},~{}\tilde{f})\in
C^{\infty}([0,1]\times T^{3})$ , they also solve Boussinesq-Reynolds system
(2.2) and satisfy

$\displaystyle\frac{3\delta}{8}e(t)\leq
e(t)-\int_{T^{3}}|\tilde{v}|^{2}(x,t)dx\leq$
$\displaystyle\frac{5\delta}{8}e(t),\quad\forall t\in[0,1],$ (2.6)
$\displaystyle\sup_{x,t}|\mathring{\tilde{R}}(x,t)|\leq$
$\displaystyle\frac{\eta\delta}{2},$ (2.7)
$\displaystyle\sup_{x,t}|\tilde{f}(x,t)|\leq$
$\displaystyle\frac{\eta\delta}{2},$ (2.8)
$\displaystyle\sup_{x,t}|\tilde{v}(x,t)-v(x,t)|\leq$ $\displaystyle
M\sqrt{\delta},$ (2.9)
$\displaystyle\sup_{x,t}|\tilde{\theta}(x,t)-\theta(x,t)|\leq$ $\displaystyle
M\sqrt{\delta},$ (2.10) $\displaystyle\sup_{x,t}|\tilde{p}(x,t)-p(x,t)|\leq$
$\displaystyle M\sqrt{\delta}.$ (2.11)

We will prove Proposition 2.1 in the subsequent sections. First, we give a
proof of Theorem 1.1 and Theorem 1.2 by using this proposition.

###### Proof of Theorem 1.1.

In this case, $h=0$. We first set

$\displaystyle
v_{0}:=0,~{}\theta_{0}\in\Theta,~{}p_{0}:=\int_{0}^{x_{3}}\theta_{0}(y)dy,~{}\mathring{R}_{0}:=0,~{}f_{0}:=0$

and $\delta=1$. Obviously, they solve Boussinesq-Reynolds system (2.2) and
satisfy the following estimates

$\displaystyle\frac{3\delta}{4}e(t)\leq
e(t)-\int_{T^{3}}|v_{0}|^{2}(x,t)dx\leq$
$\displaystyle\frac{5\delta}{4}e(t),\qquad\forall t\in[0,1]$
$\displaystyle\sup_{x,t}|\mathring{R}_{0}(x,t)|=0\quad(\leq$
$\displaystyle\eta\delta),$ $\displaystyle\sup_{x,t}|f_{0}(x,t)|=0\quad(\leq$
$\displaystyle\eta\delta).$

Then using Proposition 2.1 iteratively, we can construct a sequence
$(v_{n},~{}p_{n},~{}\theta_{n},~{}\mathring{R_{n}},~{}f_{n})$, which solve
(2.2) and satisfy

$\displaystyle\frac{3}{4}\frac{e(t)}{2^{n}}\leq
e(t)-\int_{T^{3}}|v_{n}|^{2}(x,t)dx\leq$
$\displaystyle\frac{5}{4}\frac{e(t)}{2^{n}},\quad\forall t\in[0,1]$ (2.12)
$\displaystyle\sup_{x,t}|\mathring{R_{n}}(x,t)|\leq\frac{\eta}{2^{n}},$ (2.13)
$\displaystyle\sup_{x,t}|f_{n}(x,t)|\leq\frac{\eta}{2^{n}},$ (2.14)
$\displaystyle\sup_{x,t}|v_{n+1}(x,t)-v_{n}(x,t)|\leq
M\sqrt{\frac{1}{2^{n}}},$ (2.15)
$\displaystyle\sup_{x,t}|\theta_{n+1}(x,t)-\theta_{n}(x,t)|\leq
M\sqrt{\frac{1}{2^{n}}},$ (2.16)
$\displaystyle\sup_{x,t}|p_{n+1}(x,t)-p_{n}(x,t)|\leq
M\sqrt{\frac{1}{2^{n}}}.$ (2.17)

Therefore from (2.13)-(2.17), we know that
$(v_{n},~{}p_{n},~{}\theta_{n},~{}\mathring{R}_{n},~{}f_{n})$ are Cauchy
sequence in $C(T^{3}\times[0,1])$, therefore there exist

$\displaystyle(v,p,\theta)\in C(T^{3}\times[0,1])$

such that

$\displaystyle v_{n}\rightarrow v,\qquad p_{n}\rightarrow
p,\qquad\theta_{n}\rightarrow\theta,\qquad\mathring{R_{n}}\rightarrow 0,\qquad
f_{n}\rightarrow 0.$

in $C(T^{3}\times[0,1])$ as $n\rightarrow\infty$.

Moreover, by (2.12),

$e(t)=\int_{T^{3}}|v|^{2}(x,t)dx\qquad\forall t\in[0,1].$

By (2.16),

$\|\theta-\theta_{0}\|_{0}\leq
M\sum\limits_{n=0}^{\infty}\sqrt{\frac{1}{2^{n}}}<4M.$

Passing into the limit in (2.2), we conclude that $v,~{}p,~{}\theta$ solve
(1.1) in the sense of distribution. MM $\square$

###### Proof of Theorem 1.2.

For any $h\in\Xi$, there are two functions $a\in C^{\infty}$ and $b\in
C^{\infty}$ such that $h(t,x_{3})=a(t)b(x_{3})$. Thus if we set

$\displaystyle
v_{0}:=0,~{}\quad\theta_{0}:=\int_{0}^{t}a(s)dsb(x_{3}),~{}\quad
p_{0}:=\int_{0}^{t}a(s)ds\int_{0}^{x_{3}}b(y)dy,~{}\quad\mathring{R}_{0}:=0,~{}\quad
f_{0}:=0$

and $\delta=1$. Then they solve (2.2) and satisfy the following estimates

$\displaystyle\frac{3\delta}{4}e(t)\leq
e(t)-\int_{T^{3}}|v_{0}|^{2}(x,t)dx\leq$
$\displaystyle\frac{5\delta}{4}e(t),\qquad\forall t\in[0,1]$
$\displaystyle\sup_{x,t}|\mathring{R}_{0}(x,t)|=0(\leq$
$\displaystyle\eta\delta),$ $\displaystyle\sup_{x,t}|f_{0}(x,t)|=0(\leq$
$\displaystyle\eta\delta).$

Then we can complete our proof by following the proof of Theorem 1.1 line by
line. We omit the detail here. MM $\square$

To prove Theorem 1.3, we again make use of iterative techniques. We state the
following iterative proposition, which implies Theorem 1.3.

######  Proposition 2.2.

There exist two absolute constants $M$ and $\eta$ such that the following
properties hold:

For any $0<\delta\leq 1$, if $(v_{,}~{}p,~{}\theta,~{}\mathring{R},~{}f)\in
C^{\infty}([0,1]\times T^{3})$ solve Boussinesq-Reynolds system (2.2) and

$\displaystyle\sup_{x,t}|\mathring{R}(x,t)|\leq$ $\displaystyle\eta\delta,$
(2.18) $\displaystyle\sup_{x,t}|f(x,t)|\leq$ $\displaystyle\eta\delta,$ (2.19)

then we can construct new functions
$(\tilde{v},~{}\tilde{p},~{}\tilde{\theta},~{}\mathring{\tilde{R}},~{}\tilde{f})\in
C^{\infty}([0,1]\times T^{3})$, they also solve (2.2) and satisfy

$\displaystyle\sup_{x,t}|\mathring{\tilde{R}}(x,t)|\leq$
$\displaystyle\frac{\eta}{2}\delta,$ (2.20)
$\displaystyle\sup_{x,t}|\tilde{f}(x,t)|\leq$
$\displaystyle\frac{\eta}{2}\delta,$ (2.21)
$\displaystyle\sup_{x,t}|\tilde{v}(x,t)-v(x,t)|\leq$ $\displaystyle
M\sqrt{\delta},$ (2.22)
$\displaystyle\sup_{x,t}|\tilde{\theta}(x,t)-\theta(x,t)|\leq$ $\displaystyle
M\sqrt{\delta},$ (2.23) $\displaystyle\sup_{x,t}|\tilde{p}(x,t)-p(x,t)|\leq$
$\displaystyle M\sqrt{\delta}.$ (2.24)

###### Proof of Theorem 1.3.

First, we set

$\displaystyle v_{0}=$
$\displaystyle\left(\begin{array}[]{c}tNsin(N^{2}x_{2})\\\ 0\\\
0\end{array}\right),\quad\mathring{R_{0}}=\left(\begin{array}[]{ccc}0&-\frac{cos(N^{2}x_{2})}{N}&0\\\
-\frac{cos(N^{2}x_{2})}{N}&0&0\\\ 0&0&0\end{array}\right),\quad
f_{0}=\left(\begin{array}[]{c}0\\\ 0\\\
\frac{cos(N^{2}x_{3})}{N}\end{array}\right),$ (2.34) $\displaystyle p_{0}=$
$\displaystyle-(1-t)\frac{cos(N^{2}x_{3})}{N},\qquad\qquad\theta_{0}=(1-t)Nsin(N^{2}x_{3}),$

and $\delta=1$. Then they solve (2.2). If we take $N\geq\frac{2}{\eta}$, then
they satisfy the following estimates

$\displaystyle\sup_{x,t}|\mathring{R}_{0}(x,t)|\leq$
$\displaystyle\eta\delta,$ $\displaystyle\sup_{x,t}|f_{0}(x,t)|\leq$
$\displaystyle\eta\delta.$

By using Proposition 2.2 iteratively, we can construct a sequence
$(v_{n},~{}p_{n},~{}\theta_{n},~{}\mathring{R_{n}},~{}f_{n})$, which solve
(2.2) and satisfy

$\displaystyle\sup_{x,t}|\mathring{R_{n}}(x,t)|\leq$
$\displaystyle\frac{\eta}{2^{n}},$ (2.35)
$\displaystyle\sup_{x,t}|f_{n}(x,t)|\leq$ $\displaystyle\frac{\eta}{2^{n}},$
(2.36) $\displaystyle\sup_{x,t}|v_{n+1}(x,t)-v_{n}(x,t)|\leq$ $\displaystyle
M\sqrt{\frac{1}{2^{n}}},$ (2.37)
$\displaystyle\sup_{x,t}|\theta_{n+1}(x,t)-\theta_{n}(x,t)|\leq$
$\displaystyle M\sqrt{\frac{1}{2^{n}}},$ (2.38)
$\displaystyle\sup_{x,t}|p_{n+1}(x,t)-p_{n}(x,t)|\leq$ $\displaystyle
M\sqrt{\frac{1}{2^{n}}}.$ (2.39)

Then we know that
$(v_{n},~{}p_{n},~{}\theta_{n},~{}\mathring{R}_{n},~{}f_{n})$ are Cauchy
sequence in $C(T^{3}\times[0,1])$, therefore there exist

$\displaystyle(v,p,\theta)\in C(T^{3}\times[0,1])$

such that

$\displaystyle v_{n}\rightarrow v,\qquad p_{n}\rightarrow
p,\qquad\theta_{n}\rightarrow\theta,\qquad\mathring{R_{n}}\rightarrow 0,\qquad
f_{n}\rightarrow 0,$

in $C(T^{3}\times[0,1])$ as $n\rightarrow\infty$.

By (2.37) and (2.38), we have

$\|v-v_{0}\|_{0}\leq M\sum\limits_{n=0}^{\infty}\sqrt{\frac{1}{2^{n}}}<4M,$

and

$\|\theta-\theta_{0}\|_{0}\leq
M\sum\limits_{n=0}^{\infty}\sqrt{\frac{1}{2^{n}}}<4M.$

Finally, let $\lambda$ be as in Theorem 1.3 and take $N={\rm
max}\\{\frac{2}{\eta},4\lambda,16M\\}$, then for $t\in[\frac{1}{2},1]$

$\displaystyle{\sup_{x\in T^{3}}}|v(x,t)|\geq{\sup}_{x\in
T^{3}}|v_{0}(x,t)|-4M\geq\frac{N}{4}\geq\lambda,$ $\displaystyle{\inf}_{x\in
T^{3}}|v(x,t)|\leq{\inf}_{x\in T^{3}}|v_{0}(x,t)|+4M\leq 4M.\qquad$

Moreover, since $v_{0}(x,0)=0$ , we have

$\displaystyle\|v(x,0)\|_{0}\leq 4M.$

A direct calculation gives,

$\displaystyle\int_{0}^{1}\int_{T^{3}}|\theta_{0}|^{2}(x,t)dxdt=\frac{4\pi^{3}}{3}N^{2},$

therefore

$\displaystyle\int_{0}^{1}\int_{T^{3}}|\theta|^{2}(x,t)dxdt$
$\displaystyle\geq$
$\displaystyle\frac{1}{2}\int_{0}^{1}\int_{T^{3}}|\theta_{0}|^{2}(x,t)dxdt-\int_{0}^{1}\int_{T^{3}}|\theta-\theta_{0}|^{2}(x,t)dxdt$
$\displaystyle\geq$ $\displaystyle\frac{2\pi^{3}}{3}N^{2}-(2\pi)^{3}(4M)^{2}$
$\displaystyle\geq$ $\displaystyle\lambda^{2}.$

Passing into the limit in (2.2) we conclude that $v,~{}p,~{}\theta$ solve
(1.1) in the sense of distribution. MM $\square$

###  2.1. Outline of the proof of propositions

The constructions of the functions $\tilde{v},\tilde{\theta}$ consist of
several steps. In the first steps, we add perturbations to $v_{0},\theta_{0}$
and get new functions $v_{01},\theta_{01}$ as

$\displaystyle v_{01}=$ $\displaystyle v_{0}+w_{1o}+w_{1oc}:=v_{0}+w_{1},$
$\displaystyle\theta_{01}=$ $\displaystyle\theta_{0}+\chi_{1}.$

where $w_{1o},w_{1oc},\chi_{1}$ are highly oscillated functions and given by
explicit formulas. We introduce two parameters $\mu_{1},~{}\lambda_{1}$ in the
construction of perturbation, which satisfy
$\mu_{1},~{}\lambda_{1},~{}\frac{\lambda_{1}}{\mu_{1}}\in{\rm N}$. After the
construction of $v_{01},\theta_{01}$, we then focus on constructing functions
$R_{01},p_{01}$ and $f_{01}$ which satisfies the desired estimate and solves
the system (2.2). After the first step, the stress becomes smaller in the
following sense: if

$\displaystyle\rho(t)Id-\mathring{R_{0}}=$
$\displaystyle\sum\limits_{i=1}^{L}a_{i}^{2}\Big{(}Id-\frac{k_{i}}{|k_{i}|}\otimes\frac{k_{i}}{|k_{i}|}\Big{)},$
$\displaystyle f_{0}=$ $\displaystyle\sum\limits_{i=1}^{3}b_{i}A_{k_{i}},$

then

$\displaystyle R_{01}=$
$\displaystyle\sum\limits_{i=2}^{L}a_{i}^{2}\Big{(}Id-\frac{k_{i}}{|k_{i}|}\otimes\frac{k_{i}}{|k_{i}|}\Big{)}+\delta
R_{01},$ $\displaystyle f_{01}=$
$\displaystyle\sum\limits_{i=2}^{3}b_{i}A_{k_{i}}+\delta f_{01}.$

where $\delta R_{01},\delta f_{0}$ can be arbitrary small through the
appropriate choice on $\mu_{1}$ and $\lambda_{1}$.

Repeating the above process, finally we can obtain the needed functions
$(\tilde{v},~{}\tilde{p},~{}\tilde{\theta},~{}\mathring{\tilde{R}},~{}\tilde{f}).$

The rest of paper is organized as follows. In section 3, we do some
preliminaries. We extend the Gemmetric Lemma in [15] and construct a new
operator $\mathcal{G}$ which will be be used in order to deal the stresses
arising from iteration scheme. Then we perform the first step in next several
sections. In section 4, we not only define the perturbations
$w_{1o},w_{1oc},\chi_{1}$ and new stresses $R_{01},f_{01}$, but also determine
the constants $\eta$ and $M$ appeared in the estimates in Proposition 2.1. In
section 5 and section 6, we calculate the main parts of $R_{01},f_{01}$ and
prove the relevant estimates of the various terms involved in the construction
separately. After completing the first step, in section 7, section 8 and
section 9 we then construct $v_{0n},p_{0n},\theta_{0n},R_{0n},f_{0n}$ and
prove relevant estimates by inductions. The construction in section 7 is quite
similar to that of the first step given in section 4. We calculate the main
error parts $R_{0n},f_{0n}$ in section 8 and obtain various error estimates in
section 9 respectively. Those two sections are also similar to that of section
5 and section 6 respectively. And we obtain the energy estimates in section
10. Finally, in section 11, we give a proof of proposition 2.1 by choosing
appropriate parameters $\mu_{n},\lambda_{n}$ for $1\leq n\leq L$. After the
proof of proposition 2.1, we give a proof of proposition 2.2 and this is done
in section 12.

##  3\. Preliminaries

As [15], we introduction some notations: $R^{3\times 3}$ denotes the space of
$3\times 3$ matrices; $\mathcal{S}^{3\times 3}$ denotes the spaces of $3\times
3$ symmetric matrices and $\mathrm{Id}$ denotes $3\times 3$ identity matrix .
The matrix norm $|R|:=\max_{1\leq i,j\leq 3}|R_{ij}|$ if $R=(R_{ij})_{3\times
3}$.

###  3.1. Geometric Lemma

The following lemma is an extension of geometric lemma given in [15] to our
case. Within it, we not only represent a prescribed symmetric matrix $R$, but
also a prescribed vector simultaneously.

######  Lemma 3.1 (Geometric Lemma).

For every $N\in{\mathbf{N}}$, we can choose $r_{0}>0$ and $\bar{\lambda}>1$
such that the following property holds: There exist pairwise disjoint subsets

$\displaystyle\Lambda_{j}\subseteq\\{k\in Z^{3}:|k|=\bar{\lambda}\\},\quad
j\in\\{1,\cdots,N\\},$

smooth positive functions

$\displaystyle\gamma_{k}^{(j)}\in
C^{\infty}(B_{r_{0}}(Id)),j\in\\{1,\cdots,N\\},~{}~{}k\in\Lambda_{j},$

vectors

$\displaystyle A^{j}_{k}\in
R^{3},~{}~{}|A^{j}_{k}|=\frac{1}{\sqrt{2}},~{}~{}k\cdot
A^{j}_{k}=0,~{}~{}j\in\\{1,\cdots,N\\},~{}~{}k\in\Lambda_{j}$

and smooth functions

$\displaystyle g_{k}^{(j)}\in
C^{\infty}(R^{3}),j\in\\{1,\cdots,N\\},~{}~{}k\in\Lambda_{j},$

such that

  1. (1)

$k\in\Lambda_{j}$ implies $-k\in\Lambda_{j}$ and
$\gamma_{k}^{(j)}=\gamma_{-k}^{(j)};$

  2. (2)

for every $R\in B_{r_{0}}(Id)$, the following identity holds:

$R=\frac{1}{2}\sum\limits_{k\in\Lambda_{j}}(\gamma_{k}^{(j)}(R))^{2}\Big{(}Id-\frac{k}{|k|}\otimes\frac{k}{|k|}\Big{)};$

  3. (3)

for every $f\in C^{\infty}(R^{3})$, we have the identity

$f=\sum_{k\in\Lambda_{j}}g_{k}^{(j)}(f)A^{j}_{k}.$

###### Proof.

The proof is essentially an extension of it given in [15]. The choosing of
$r_{0},~{}\lambda_{0},~{}\Lambda_{j},~{}\gamma_{k}^{j}$ and the results (1),
(2) are due to Camillo De Lellis and László Székelyhidi and can be found in
[15]. We only give the proof on existence of $A^{j}_{k}$, $g^{j}_{k}$ and the
related property.

For the given $j\in\\{1,\cdot\cdot\cdot,N\\}$, we can choose
$k^{j}_{1},~{}k^{j}_{2},~{}k^{j}_{3}$ in $\Lambda_{j}$ such that they are
linearly independent, and then we can choose 3-dimensional vectors
$A^{j}_{k_{1}},A^{j}_{k_{2}},A^{j}_{k_{3}}$ such that $A^{j}_{k_{i}}\cdot
k^{j}_{i}=0$ and $|A^{j}_{k_{i}}|=\frac{1}{\sqrt{2}}$, $i=1,2,3$, which are
linear independent.

For $k\in\Lambda_{j}\backslash\\{k^{j}_{1},k^{j}_{2},k^{j}_{3}\\}$, we take
$A^{j}_{k}$ be a 3-dimensional vector such that $A^{j}_{k}\cdot
k=0,|A^{j}_{k}|=\frac{1}{\sqrt{2}}$. Then, we can choose $g_{k}^{j}(f)=0$ for
$k\in\Lambda_{j}\backslash\\{k^{j}_{1},k^{j}_{2},k^{j}_{3}\\}$ and

$\left(\begin{array}[]{cc}g_{k_{1}}^{j}(f)\\\\[12.0pt]
g_{k_{2}}^{j}(f)\\\\[12.0pt]
g_{k_{2}}^{j}(f)\end{array}\right):=(A^{j}_{k_{1}},A^{j}_{k_{2}},A^{j}_{k_{3}})^{-1}f.$

where $f\in C^{\infty}(T^{3})$ and
$(A^{j}_{k_{1}},A^{j}_{k_{2}},A^{j}_{k_{3}})^{-1}$ denote the inverse matrix
of $(A^{j}_{k_{1}},A^{j}_{k_{2}},A^{j}_{k_{3}}).$ Thus, the linear functionals
$g_{k}^{(j)}$ satisfies

$\displaystyle\forall f\in
C^{\infty}(R^{3}),f=\sum_{k\in\Lambda_{j}}g_{k}^{(j)}(f)A^{j}_{k}.$

Then we finished the proof of the geometric lemma.

MM $\square$

###  3.2. The operator $\mathcal{R}$ and $\mathcal{G}$

We introduce some operators in order to deal with the Reynolds stresses. The
operator $\mathcal{R}$ was introduced in [15] and the operator $\mathcal{G}$
is given by us.

####  3.2.1. The operator $\mathcal{R}$

The following lemma is taken from [15], we copy it here for the completeness
of the paper (the proof refers to [15]).

######  Lemma 3.1 ($\mathcal{R}=\textrm{div}^{-1}$).

There exist a linear operator $\mathcal{R}$ from $C^{\infty}(T^{3},R^{3})$ to
$C^{\infty}(T^{3},R^{3\times 3})$ such that the following property hold: for
any $v\in C^{\infty}(T^{3},R^{3})$ we have

  * (a)

$\mathcal{R}v(x)$ is a symmetric trace-free matrix for each $x\in T^{3}$;

  * (b)

${\rm div}\mathcal{R}v=v-\fint_{T^{3}}v$.

Here and subsequent, we use the natation
$\fint_{T^{3}}v(x)dx=\frac{1}{|T^{3}|}\int_{T^{3}}v(x)dx$.

####  3.2.2. The operator $\mathcal{G}$

######  Definition 3.2.

Let $b\in C^{\infty}(T^{3};R)$ be a smooth function. We define a vector-valued
periodic function $\mathcal{G}b$ by

$\displaystyle\mathcal{G}b=\nabla a$

where $a\in C^{\infty}(T^{3};R)$ is the solution of

$\displaystyle\Delta a=b-\fint_{T^{3}}b\textrm{ in }T^{3}$

with $\int_{T^{3}}a(x)dx=0$.

######  Lemma 3.3.

$(\mathcal{G}={\rm div}^{-1})$For any $b\in C^{\infty}(T^{3};R)$, we have

$\displaystyle{\rm div}(\mathcal{G}b)=b-\fint_{T^{3}}b.$

###### Proof.

The proof is elementary. In fact, we calculate directly

$\displaystyle{\rm div}(\mathcal{G}b)=\triangle a=b-\fint_{T^{3}}b.$

MM $\square$

Some estimates related to the operators $\mathcal{R}$ and $\mathcal{G}$ can be
founded in the Appendix A.

##  4\. the constructions of $v_{01},p_{01},\theta_{01},R_{01},f_{01}$

The constructions of
$\tilde{v},~{}\tilde{p},~{}\tilde{\theta},~{}\mathring{\tilde{R}},~{}\tilde{f}$
from $v,~{}p,~{}\theta,~{}\mathring{R},~{}f$ consist of several steps. The
main idea is to decompose the stress error into blocks with the help of
geometric lemma and to remove one block in one step. In this section, we
perform the first step.

For convenience, we set
$v_{0}:=v,~{}p_{0}:=p,~{}\theta_{0}:=\theta,~{}\mathring{R}_{0}:=\mathring{R},~{}f_{0}:=f$.

###  4.1. The main perturbation $w_{1o}$

We first introduce some notations. From [15], we have the following partition
of unity: for two constants $c_{1}$ and $c_{2}$ such that
$\frac{\sqrt{3}}{2}<c_{1}<c_{2}<1$ , we have a family of functions
$\alpha_{l}\in C_{c}^{\infty}(R^{3}):l\in Z^{3}$ such that

$\displaystyle\sum\limits_{l\in
Z^{3}}\alpha_{l}^{2}=1,\qquad\hbox{supp}\alpha_{l}\subseteq B_{c_{2}}(l).$
(4.40)

We apply the geometric Lemma 3.1 with $N=1$ and obtain
$\bar{\lambda}>1,r_{0}>0$ , subset $\Lambda=\\{\pm k_{1},...,\pm k_{L}\\}$ and
vectors $\\{A_{\pm k_{j}},j=1,\cdot\cdot\cdot,L\\}$ together with
corresponding functions

$\displaystyle\gamma_{k_{i}}\in C^{(\infty)}(B_{r_{0}}(Id)),\qquad
g_{k_{i}}\in C^{(\infty)}(R^{3}),\qquad i=1,\cdots,L.$

where $L$ is a fixed integer. Thus the result can be restated as following:
For any $R\in B_{r_{0}}(Id)$, we have the identity

$\displaystyle
R=\sum\limits_{i=1}^{L}\gamma^{2}_{k_{i}}(R)\Big{(}Id-\frac{k_{i}}{|k_{i}|}\otimes\frac{k_{i}}{|k_{i}|}\Big{)}.$
(4.41)

and for any $f\in C^{(\infty)}(R^{3})$, we have

$\displaystyle f=\sum\limits_{i=1}^{L}g_{k_{i}}(f)A_{k_{i}}.$

In fact, from the construction of $g_{k_{i}}$, after possibly relabel the
index, we have

$\displaystyle f=\sum\limits_{i=1}^{3}g_{k_{i}}(f)A_{k_{i}}.$ (4.42)

Next, as in [15], to define the amplitude of the perturbation, we set

$\displaystyle\bar{\rho}(t):=\frac{1}{(2\pi)^{3}}\Big{(}e(t)\Big{(}1-\frac{\delta}{2}\Big{)}-\int_{T^{3}}|v_{0}|^{2}(x,t)dx\Big{)},$
(4.43)

and

$\displaystyle R_{0}(x,t):=\bar{\rho}(t)Id-\mathring{R}_{0}(x,t).$ (4.44)

then for any $l\in Z^{3}$, we denote $b_{1l}$ by

$\displaystyle b_{1l}(x,t):=$
$\displaystyle\sqrt{\bar{\rho}(t)}\alpha_{l}(\mu_{1}v_{0})\gamma_{k_{1}}\Big{(}\frac{R_{0}(x,t)}{\bar{\rho}(t)}\Big{)},$
(4.45)

and define

$\displaystyle B_{k_{1}}:=A_{k_{1}}+i\frac{k_{1}}{|k_{1}|}\times A_{k_{1}}.$
(4.46)

Then we define $l$-perturbation

$\displaystyle
w_{1ol}:=b_{1l}(x,t)\Big{(}B_{k_{1}}e^{i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu_{1}}t)}+B_{-k_{1}}e^{-i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu_{1}}t)}\Big{)}.$
(4.47)

where we set $A_{-k_{1}}=A_{k_{1}}$.

Finally, we define 1-th perturbation

$\displaystyle w_{1o}:=\sum_{l\in Z^{3}}w_{1ol}.$ (4.48)

Obviously, $w_{1ol},w_{1o}$ are all real 3-dimensional vector functions.
According to (4.40), we have
$\hbox{supp}\alpha_{l}\cap\hbox{supp}\alpha_{l^{\prime}}=\emptyset$ if
$|l-l^{\prime}|\geq 2$, therefore the above summation is meaningful.

###  4.2. The constants $\eta$ and $M$

From [15], we known that there exist constant $\eta=\eta(e)$ (independent on
$\delta$) such that

$\displaystyle\Big{\|}\frac{R_{0}}{\bar{\rho}(t)}-Id\Big{\|}_{0}\leq\frac{\|\mathring{R}_{0}\|_{0}}{c\delta
m}\leq\frac{\eta}{cm}\leq\frac{r_{0}}{2},$ (4.49)

so $b_{1l}$ is well-defined, where $\|f\|_{0}=\sup_{x,t}|f(x,t)|.$ Also from
[15], there exists a constant $M=M(e)>1$ (in particular independent of
$\delta$) such that

$\displaystyle\|b_{1l}\|_{0}\leq\frac{M\sqrt{\delta}}{10L},$

where $L$ is fixed constant from geometric lemma.
Moreover, from the property (4.40) of $\alpha_{l}$, we know that

$\displaystyle\|w_{1o}\|_{0}\leq\frac{M\sqrt{\delta}}{2L}.$ (4.50)

###  4.3. The correction $w_{1oc}$ and the perturbations $w_{1}$ and
$\chi_{1}$

We denote the $l$-correction

$\displaystyle w_{1ocl}:=$
$\displaystyle\frac{1}{\lambda_{1}\lambda_{0}}\Big{(}\frac{\nabla
b_{1l}(x,t)\times
B_{k_{1}}}{2^{|l|}}e^{i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu_{1}}t)}+\frac{\nabla
b_{1l}(x,t)\times
B_{-{k_{1}}}}{2^{|l|}}e^{-i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu_{1}}t)}\Big{)},$
(4.51)

then denote 1-th correction

$\displaystyle w_{1oc}:=\sum_{l\in Z^{3}}w_{1ocl}.$ (4.52)

Finally, we denote 1-th perturbation

$\displaystyle w_{1}:=w_{1o}+w_{1oc}.$ (4.53)

Thus, if we denote $w_{1l}$ by

$\displaystyle w_{1l}:=$ $\displaystyle w_{1ol}+w_{1ocl}$ $\displaystyle=$
$\displaystyle\frac{1}{\lambda_{1}\lambda_{0}}\hbox{curl}\Big{(}\frac{b_{1l}(x,t)B_{k_{1}}}{2^{|l|}}e^{i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu_{1}}t)}+\frac{b_{1l}(x,t)B_{-{k_{1}}}}{2^{|l|}}e^{-i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu_{1}}t)}\Big{)},$
(4.54)

then

$\displaystyle w_{1}=\sum\limits_{l\in Z^{3}}w_{1l},\qquad{\rm div}w_{1l}=0,$

and

$\displaystyle\hbox{div}w_{1}=0.$

Moreover, if we set

$\displaystyle B_{1lk_{1}}:=$ $\displaystyle
b_{1l}(x,t)B_{k_{1}}+\frac{1}{\lambda_{1}\lambda_{0}}\frac{\nabla
b_{1l}(x,t)\times B_{k_{1}}}{2^{|l|}},$ $\displaystyle B_{-1lk_{1}}:=$
$\displaystyle
b_{1l}(x,t)B_{-k_{1}}+\frac{1}{\lambda_{1}\lambda_{0}}\frac{\nabla
b_{1l}(x,t)\times B_{-k_{1}}}{2^{|l|}},$

then

$\displaystyle
w_{1l}=B_{1lk_{1}}e^{i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu_{1}}t)}+B_{-1lk_{1}}e^{-i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu_{1}}t)}.$

Thus we complete the construction of perturbation $w_{1}$.

To construct $\chi_{1}$, we first denote $\beta_{1l}$ by

$\displaystyle\beta_{1l}(x,t):=\frac{\alpha_{l}(\mu_{1}v_{0})}{2\sqrt{\bar{\rho}(t)}}\frac{g_{k_{1}}(-f_{0}(x,t))}{\gamma_{k_{1}}\big{(}\frac{R_{0}(x,t)}{\bar{\rho}(t)}\big{)}},$
(4.55)

then denote the $l$-perturbation

$\displaystyle\chi_{1l}(x,t):=\beta_{1l}(x,t)\Big{(}e^{i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu_{1}}t)}+e^{-i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu_{1}}t)}\Big{)}.$
(4.56)

Finally, we define the perturbation

$\displaystyle\chi_{1}(x,t):=\sum_{l\in Z^{3}}\chi_{1l}.$ (4.57)

Thus, $\chi_{1l}$ and $\chi_{1}$ are both real scalar functions, and as the
perturbation of $w_{1}$, the summation in the definition of $\chi_{1}$ is
meaningful.

Moreover, since
$\Big{\|}\frac{R_{0}(x,t)}{\bar{\rho}(t)}\Big{\|}_{0}\leq\frac{r_{0}}{2}$,
there exist two positive constants $c_{11}$ and $c_{21}$ such that

$c_{11}\leq\gamma_{k_{1}}\Big{(}\frac{R(x,t)}{\bar{\rho}(t)}\Big{)}\leq
c_{21}.$

Thus, from (4.55) on $\beta_{1l}$ and the assumption (2.5) on $f_{0}$, after
possibly taking a bigger number $M$ (still is a absolute constant only depend
on $e$), we know that

$\displaystyle\|\beta_{1l}\|_{0}\leq\frac{M\sqrt{\delta}}{10L},$ (4.58)

therefore, we have

$\displaystyle\|\chi_{1}\|_{0}\leq\frac{M\sqrt{\delta}}{2L}.$ (4.59)

###  4.4. The constructions of $v_{01}$, $p_{01}$, $\theta_{01}$, $f_{01}$,
$\mathring{R}_{01}$

.

First, we denote $M_{1}$ by

$\displaystyle M_{1}=$ $\displaystyle\sum\limits_{l\in
Z^{3}}b^{2}_{1l}(x,t)\Big{(}B_{k_{1}}\otimes
B_{k_{1}}e^{2i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu_{1}}t)}+B_{-k_{1}}\otimes
B_{-k_{1}}e^{-2i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu_{1}}t)}\Big{)}$
$\displaystyle+\sum\limits_{l,l^{\prime}\in Z^{3},l\neq
l^{\prime}}w_{1ol}w_{1ol^{\prime}}(x,t),$ (4.60)

and $N_{1},K_{1}$ by

$\displaystyle N_{1}=$ $\displaystyle\sum_{l\in
Z^{3}}\Big{[}w_{1l}\otimes\Big{(}v_{0}-\frac{l}{\mu_{1}}\Big{)}+\Big{(}v_{0}-\frac{l}{\mu_{1}}\Big{)}\otimes
w_{1l}\Big{]}$ $\displaystyle K_{1}=$ $\displaystyle\sum\limits_{l\in
Z^{3}}\beta_{1l}(x,t)b_{1l}(x,t)\Big{(}B_{k_{1}}e^{2i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu_{1}}t)}+B_{-k_{1}}e^{-2i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu_{1}}t)}\Big{)}+\sum\limits_{l,l^{\prime}\in
Z^{3},l\neq l^{\prime}}w_{1ol}\chi_{1l^{\prime}}.$ (4.61)

Notice that $N_{1}$ is a symmetric matrix. Then we define

$\displaystyle v_{01}:=v_{0}+w_{1},$ $\displaystyle
p_{01}:=p_{0}-\frac{2w_{1o}\cdot w_{1oc}+|w_{1oc}|^{2}}{3}-tr(N_{1}),$
$\displaystyle\theta_{01}:=\theta_{0}+\chi_{1}-\fint_{T^{3}}{\chi_{1}}dx,$
$\displaystyle R_{01}:=-R_{0}(x,t)+2\sum\limits_{l\in
Z^{3}}b^{2}_{1l}(x,t)Re(B_{k_{1}}\otimes
B_{-{k_{1}}})+\delta\mathring{R}_{01},$ $\displaystyle
f_{01}:=f_{0}+2\sum\limits_{l\in
Z^{3}}\beta_{1l}(x,t)b_{1l}(x,t)A_{k_{1}}+\delta f_{01},$

where

$\displaystyle\delta\mathring{R}_{01}=$
$\displaystyle\mathcal{R}(\hbox{div}M_{1})+N_{1}-tr(N_{1})Id+\mathcal{R}\Big{\\{}\partial_{t}w_{1}+\hbox{div}\Big{[}\sum_{l\in
Z^{3}}\Big{(}w_{1l}\otimes\frac{l}{\mu_{1}}+\frac{l}{\mu_{1}}\otimes
w_{1l}\Big{)}\Big{]}\Big{\\}}$ $\displaystyle+(w_{1o}\otimes
w_{1oc}+w_{1oc}\otimes w_{1o}+w_{1oc}\otimes w_{1oc})$
$\displaystyle-\frac{2w_{1o}\cdot
w_{1oc}+|w_{1oc}|^{2}}{3}Id-\mathcal{R}\Big{(}\Big{(}\chi_{1}-\fint_{T^{3}}{\chi_{1}}dx\Big{)}e_{3}\Big{)},$
(4.62)

and

$\displaystyle\delta f_{01}=$
$\displaystyle\mathcal{G}(\hbox{div}K_{1})+\mathcal{G}(w_{1}\cdot\nabla\theta_{0})+\mathcal{G}\Big{(}\partial_{t}\chi_{1}+\sum_{l\in
Z^{3}}\Big{(}\frac{l}{\mu_{1}}\cdot\nabla\Big{)}\chi_{1l}\Big{)}+\sum_{l\in
Z^{3}}\Big{(}v_{0}-\frac{l}{\mu_{1}}\Big{)}\chi_{1l}+w_{1oc}\chi_{1}.$ (4.63)

From the property (3.1) of $\mathcal{R}$, we know that
$\delta\mathring{R}_{01}$ is a symmetric and trace-free matrix. Obviously,

$\displaystyle\hbox{div}v_{1}=\hbox{div}v_{0}+\hbox{div}w_{1}=0.$

Moreover, by the definition of $w_{1o},w_{1oc},\delta\mathring{R}_{01}$ as
well as $v_{01},p_{01}$ and notice that
$v_{0},p_{0},\theta_{0},\mathring{R}_{0},f_{0}$ are solutions of the system
(2.2), together with Lemma 3.1, we know that

$\displaystyle\hbox{div}R_{01}=$
$\displaystyle\hbox{div}\mathring{R_{0}}(x,t)+\partial_{t}w_{1}-\nabla(trN_{1})-\nabla\Big{(}\frac{2w_{1o}\cdot
w_{1oc}+|w_{1oc}|^{2}}{3}\Big{)}-\Big{(}\chi_{1}-\fint_{T^{3}}{\chi_{1}}dx\Big{)}e_{3}$
$\displaystyle+\hbox{div}(w_{1o}\otimes w_{1o}+w_{1}\otimes v_{0}+v_{0}\otimes
w_{1}+w_{1o}\otimes w_{1oc}+w_{1oc}\otimes w_{1o}+w_{1oc}\otimes w_{1oc})$
$\displaystyle=$ $\displaystyle\partial_{t}v_{0}+\hbox{div}(v_{0}\otimes
v_{0})+\nabla
p_{0}-\theta_{0}e_{3}+\partial_{t}w_{1}-\nabla(trN_{1})-\nabla\Big{(}\frac{2w_{1o}\cdot
w_{1oc}+|w_{1oc}|^{2}}{3}\Big{)}$
$\displaystyle-\Big{(}\chi_{1}-\fint_{T^{3}}{\chi_{1}}dx\Big{)}e_{3}+\hbox{div}(w_{1o}\otimes
w_{1o}+w_{1}\otimes v_{0}+v_{0}\otimes w_{1}+w_{1o}\otimes w_{1oc}$
$\displaystyle+w_{1oc}\otimes w_{1o}+w_{1oc}\otimes w_{1oc})$ $\displaystyle=$
$\displaystyle\partial_{t}v_{01}+\hbox{div}(v_{01}\otimes v_{01})+\nabla
p_{01}-\theta_{01}e_{3}.$

Where we used

$\fint_{T^{3}}w_{1}(x,t)dx=0,$

and

$\hbox{div}(M_{1})+\hbox{div}\Big{(}2\sum\limits_{l\in
Z^{3}}b^{2}_{1l}(x,t)Re(B_{k_{1}}\otimes
B_{-{k_{1}}})\Big{)}=\hbox{div}(w_{1o}\otimes w_{1o}).$

Furthermore, from the definition of $w_{1o},\chi_{1}$, we know that

$\displaystyle f_{01}=$ $\displaystyle f_{0}+2\sum\limits_{l\in
Z^{3}}\beta_{1l}(x,t)b_{1l}(x,t)A_{k_{1}}+\mathcal{G}(\hbox{div}K_{1})+\mathcal{G}(w_{1}\cdot\nabla\theta_{0})$
$\displaystyle+\mathcal{G}\Big{(}\partial_{t}\chi_{1}+\sum_{l\in
Z^{3}}\Big{(}\frac{l}{\mu_{1}}\cdot\nabla\Big{)}\chi_{1l}\Big{)}+\sum_{l\in
Z^{3}}\Big{(}v_{0}-\frac{l}{\mu_{1}}\Big{)}\chi_{1l}+w_{1oc}\chi_{1}.$

Therefore, by the fact that $v_{0},p_{0},\theta_{0},\mathring{R}_{0},f_{0}$
are solutions of the system (2.2)and by Lemma 3.3

$\displaystyle\hbox{div}f_{01}=$
$\displaystyle\hbox{div}f_{0}+\partial_{t}\Big{(}\chi_{1}-\fint_{T^{3}}\chi_{1}dx\Big{)}+\hbox{div}(w_{1o}\chi_{1}+w_{1oc}\chi_{1}+v_{0}\chi_{1}+w_{1}\theta_{0})$
$\displaystyle=$
$\displaystyle\hbox{div}(v_{0}\theta_{0}+w_{1o}\chi_{1}+w_{1oc}\chi_{1}+v_{0}\chi_{1}+w_{1}\theta_{0})+\partial_{t}\Big{(}\theta_{0}+\chi_{1}-\fint_{T^{3}}\chi_{1}dx\Big{)}-h$
$\displaystyle=$
$\displaystyle\partial_{t}\theta_{01}+\hbox{div}(v_{01}\theta_{01})-h.$

Thus the functions $v_{01},p_{01},\theta_{01},R_{01},f_{01}$ solve the system
(2.2).

##  5\. The representations

In this section, we will calculate the form of

$-R_{0}(x,t)+2\sum\limits_{l\in Z^{3}}b^{2}_{1l}(x,t)Re(B_{k_{1}}\otimes
B_{-{k_{1}}}),$

and

$f_{0}+2\sum\limits_{l\in Z^{3}}\beta_{1l}(x,t)b_{1l}(x,t)A_{k_{1}}.$

We will use the following basic fact.

######  Proposition 5.1.

We have the following identity

$\displaystyle 2Re(B_{k_{1}}\otimes
B_{-{k_{1}}})=Id-\frac{k_{1}}{|k_{1}|}\otimes\frac{k_{1}}{|k_{1}|}.$ (5.64)

###### Proof.

We refer the proof to [15]. MM $\square$

###  5.1. The representation of $-R_{0}(x,t)+2\sum\limits_{l\in
Z^{3}}b^{2}_{1l}(x,t)Re(B_{k_{1}}\otimes B_{-{k_{1}}})$

First, from the definition (4.45) of $b_{1l}(x,t)$ and identity (5.64), we
have

$\displaystyle 2\sum\limits_{l\in Z^{3}}b^{2}_{1l}(x,t)Re(B_{k_{1}}\otimes
B_{-{k_{1}}})=$ $\displaystyle\bar{\rho}(t)\sum\limits_{l\in
Z^{3}}\alpha_{l}^{2}(\mu_{1}v_{0})\gamma_{k_{1}}^{2}\Big{(}\frac{R_{0}(x,t)}{\bar{\rho}(t)}\Big{)}\Big{(}Id-\frac{k_{1}}{|k_{1}|}\otimes\frac{k_{1}}{|k_{1}|}\Big{)}$
$\displaystyle=$
$\displaystyle\bar{\rho}(t)\gamma_{k_{1}}^{2}\Big{(}\frac{R_{0}(x,t)}{\bar{\rho}(t)}\Big{)}\Big{(}Id-\frac{k_{1}}{|k_{1}|}\otimes\frac{k_{1}}{|k_{1}|}\Big{)}.$

where we used $\sum\limits_{l\in Z^{3}}\alpha_{l}^{2}=1$.
Moreover, from the identity (4.41) and estimate (4.49), we can decompose
$R_{0}$

$\displaystyle\frac{R_{0}}{\bar{\rho}(t)}=\sum\limits_{i=1}^{L}\Big{(}\gamma_{k_{i}}\Big{(}\frac{R_{0}}{\bar{\rho}(t)}\Big{)}\Big{)}^{2}\Big{(}Id-\frac{k_{i}}{|k_{i}|}\otimes\frac{k_{i}}{|k_{i}|}\Big{)}.$

Therefore

$\displaystyle-R_{0}(x,t)+2\sum\limits_{l\in
Z^{3}}b^{2}_{1l}(x,t)Re(B_{k_{1}}\otimes
B_{-{k_{1}}})=-\bar{\rho}(t)\sum\limits_{i=2}^{L}\Big{(}\gamma_{k_{i}}\Big{(}\frac{R_{0}}{\bar{\rho}(t)}\Big{)}\Big{)}^{2}\Big{(}Id-\frac{k_{i}}{|k_{i}|}\otimes\frac{k_{i}}{|k_{i}|}\Big{)}.$

Meanwhile, we have

$\displaystyle
R_{01}=-\bar{\rho}(t)\sum\limits_{i=2}^{L}\Big{(}\gamma_{k_{i}}\Big{(}\frac{R_{0}}{\bar{\rho}(t)}\Big{)}\Big{)}^{2}\Big{(}Id-\frac{k_{i}}{|k_{i}|}\otimes\frac{k_{i}}{|k_{i}|}\Big{)}+\delta\mathring{R}_{01}.$
(5.65)

Next section, we will prove that $\delta\mathring{R}_{01}$ is small.

###  5.2. The representation of $f_{0}+2\sum\limits_{l\in
Z^{3}}\beta_{1l}(x,t)b_{1l}(x,t)A_{k_{1}}$

From the definition (4.45) of $b_{1l}(x,t)$ and (4.55) of $\beta_{1l}(x,t)$,
we have

$\displaystyle 2\sum\limits_{l\in
Z^{3}}\beta_{1l}(x,t)b_{1l}(x,t)A_{k_{1}}=\sum\limits_{l\in
Z^{3}}\alpha_{l}^{2}(\mu_{1}v_{0})g_{k_{1}}(-f_{0}(x,t))A_{k_{1}}=-g_{k_{1}}(f_{0}(x,t))A_{k_{1}},$

where we used the fact that $g_{k_{1}}$ is a linear operator.
Moreover, using the identity (4.42), we can decompose $f_{0}$

$\displaystyle
f_{0}(x,t)=\sum\limits_{i=1}^{3}g_{k_{i}}(f_{0}(x,t))A_{k_{i}},$

Therefore,

$\displaystyle f_{0}+2\sum\limits_{l\in
Z^{3}}\beta_{1l}(x,t)b_{1l}(x,t)A_{k_{1}}=\sum\limits_{i=2}^{3}g_{k_{i}}(f_{0}(x,t))A_{k_{i}}$

Meanwhile, we have

$\displaystyle f_{01}=$ $\displaystyle f_{0}+2\sum\limits_{l\in
Z^{3}}\alpha_{l}(\mu_{1}v_{0})\beta_{1}(x,t)b_{l}(x,t)A_{k_{1}}+\delta
f_{01}=\sum\limits_{i=2}^{3}g_{k_{i}}(f_{0}(x,t))A_{k_{i}}+\delta f_{01}$
(5.66)

Again in next section, we will prove that $\delta f_{01}$ is small.

##  6\. Estimates on $\delta\mathring{R}_{01}$ and $\delta f_{01}$

First, we introduce some Hölder (semi)norm:

$\displaystyle\|f\|_{0}:=\hbox{sup}_{T^{3}\times[0,1]}|f|,\qquad[f]_{m}:=\hbox{max}_{|\beta|=m}\|D^{\beta}f\|_{0},$
$\displaystyle[f]_{m+\alpha}:=\hbox{max}_{|\beta|=m}\hbox{sup}_{x\neq
y,t\in[0,1]}\frac{|D^{\beta}f(t,x)-D^{\beta}f(t,y)|}{|x-y|^{\alpha}},$
$\displaystyle\|f\|_{m}:=\sum_{0}^{m}[f]_{j},\|f\|_{m+\alpha}:=\|f\|_{m}+[f]_{m+\alpha}.$

where $m=0,1,2,...$ and $\alpha\in(0,1)$, $\beta$ is a multi index. When
$f(t,x)=f(x)$, the above Hölder(semi) norm denote usual spatial Hölder(semi)
norm. Moreover, in the subsequent estimate, unless otherwise stated,
$\alpha\in(0,1)$ and $C_{1}$ denotes a constant which depends on
$v_{0},\mathring{R_{0}},e(t)$ as well as $\lambda_{0},\alpha,\delta$, but does
not depend on $\mu_{1},\lambda_{1}$, and can change from line to line.
Furthermore, we assume that $1\ll\mu_{1}\ll\lambda_{1}$.

In the following, we frequently use the elementary inequalities

$\displaystyle\|fg\|_{r}\leq$ $\displaystyle
C\bigl{(}\|f\|_{r}\|g\|_{0}+\|g\|_{r}\|f\|_{0}\bigr{)},$ (6.67)
$\displaystyle{[}f{]}_{s}\leq$ $\displaystyle C([f]_{r}+\|f\|_{0}).$ (6.68)

for any $r\geq 0$ and $0\leq s<r$. These inequalities can be found in [15].

We summarize the main properties of $b_{1l}$ and $\beta_{1l}$.

######  Lemma 6.1.

For any $|l|\leq C_{1}\mu_{1}$ and $r\geq 0$, we have

$\displaystyle\|b_{1l}\|_{r}\leq C_{1}\mu_{1}^{r},$
$\displaystyle\|\beta_{1l}\|_{r}\leq C_{1}\mu_{1}^{r},$
$\displaystyle\|(\partial_{t}+\frac{l}{\mu_{1}}\cdot\nabla)b_{1l}\|_{r}\leq
C_{1}\mu_{1}^{r+1},$
$\displaystyle\|(\partial_{t}+\frac{l}{\mu_{1}}\cdot\nabla)\beta_{1l}\|_{r}\leq
C_{1}\mu_{1}^{r+1}.$

###### Proof.

We only need to prove them when $r$ is integer, others can be derived by
inequality (6.68). Recall that

$\displaystyle
b_{1l}(x,t):=\sqrt{\bar{\rho}(t)}\alpha_{l}(\mu_{1}v_{0})\gamma_{k_{1}}\Big{(}\frac{R_{0}(x,t)}{\bar{\rho}(t)}\Big{)},$

Then it is easy to check the first inequality in Lemma 6.1. Other three
estimates are similar. MM $\square$

Then from the definition (4.52) on $w_{10c}$, we have the following estimates

######  Lemma 6.2 (Estimates on corrections).

$\displaystyle\|w_{1oc}\|_{\alpha}\leq
C_{1}\frac{\mu_{1}}{\lambda_{1}^{1-\alpha}},\qquad\forall\alpha\in[0,1)$
(6.69)

###### Proof.

From the definition (4.51), inequality (6.67) and Lemma 6.1

$\displaystyle\|w_{1ocl}\|_{\alpha}\leq C_{1}\Big{(}\frac{\|\nabla
b_{1l}\|_{\alpha}}{\lambda_{1}}+\frac{\|\nabla
b_{1l}\|_{0}}{\lambda_{1}^{1-\alpha}}\Big{)}\leq
C_{1}\frac{\mu_{1}}{\lambda_{1}^{1-\alpha}}.$

Then, from the property (4.40) of $\alpha_{l}$, we have

$\displaystyle\|w_{1oc}\|_{\alpha}\leq
C_{1}\frac{\mu_{1}}{\lambda_{1}^{1-\alpha}}.$

MM $\square$

###  6.1. Estimates on $\delta\mathring{R}_{01}$

As in [15], we also split the stresses into three parts, they are
(1) the oscillation part

$\mathcal{R}(\hbox{div}M_{1})-\mathcal{R}\Big{(}\Big{(}\chi_{1}-\fint_{T^{3}}{\chi_{1}}dx\Big{)}e_{3}\Big{)},$

(2) the transportation part

$\displaystyle\mathcal{R}\Big{\\{}\partial_{t}w_{1oc}+\partial_{t}w_{1o}+\hbox{div}\Big{[}\sum_{l\in
Z^{3}}\Big{(}w_{1l}\otimes\frac{l}{\mu_{1}}+\frac{l}{\mu_{1}}\otimes
w_{1l}\Big{)}\Big{]}\Big{\\}}$ $\displaystyle=$
$\displaystyle\mathcal{R}\Big{(}\partial_{t}w_{1}+\sum_{l\in
Z^{3}}\Big{(}\frac{l}{\mu_{1}}\cdot\nabla\Big{)}w_{1l}\Big{)},$

(3) the error part

$\displaystyle N_{1}-tr(N_{1})Id+(w_{1o}\otimes w_{1oc}+w_{1oc}\otimes
w_{1o}-w_{1oc}\otimes w_{1oc})-\frac{2w_{1o}\cdot
w_{1oc}+|w_{1oc}|^{2}}{3}Id.$

In the following, we will estimate each part separately.

######  Lemma 6.3 (The oscillation part).

$\displaystyle\|\mathcal{R}({\rm div}M_{1})\|_{\alpha}\leq
C_{1}\frac{\mu_{1}^{3}}{\lambda_{1}^{1-\alpha}},$ (6.70)
$\displaystyle\Big{\|}\mathcal{R}\Big{(}\Big{(}\chi_{1}-\fint_{T^{3}}{\chi_{1}}dx\Big{)}e_{3}\Big{)}\Big{\|}_{\alpha}\leq
C_{1}\frac{\mu_{1}}{\lambda_{1}^{1-\alpha}}.$ (6.71)

###### Proof.

We start with the fact that $k_{1}\cdot B_{k_{1}}=0,$ then

$\displaystyle\hbox{div}M_{1}=$ $\displaystyle\sum\limits_{l\in
Z^{3}}\Big{(}B_{k_{1}}\otimes
B_{k_{1}}e^{2i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu}t)}+B_{-k_{1}}\otimes
B_{-k_{1}}e^{-2i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu}t)}\Big{)}\nabla(b^{2}_{1l}(x,t))$
$\displaystyle+\sum\limits_{l,l^{\prime}\in Z^{3},l\neq
l^{\prime}}\hbox{div}(w_{1ol}\otimes w_{1ol^{\prime}})$ $\displaystyle=$
$\displaystyle M_{11}+M_{12}.$

By the Lemma 6.1 and property (A.176) on $\mathcal{R}$,

$\displaystyle\|\mathcal{R}(\hbox{div}M_{11})\|_{\alpha}\leq\sum\limits_{|l|\leq
C_{1}\mu_{1}}C_{1}\frac{\mu_{1}}{\lambda_{1}^{1-\alpha}}\leq
C_{1}\frac{\mu_{1}^{2}}{\lambda_{1}^{1-\alpha}},$

where we used the following inequality which deduced from Lemma 6.1, for
$\mu_{1}\ll\lambda_{1}$ and $m$ a fixed large number,

$\displaystyle\Big{\|}\mathcal{R}\Big{(}\nabla
b_{1l}e^{i\lambda_{1}2^{|l|}k_{1}\cdot x}\Big{)}\Big{\|}_{\alpha}$
$\displaystyle\leq$ $\displaystyle C_{1,m}\Big{(}\frac{\|\nabla
b_{1l}\|_{0}}{\lambda_{1}^{1-\alpha}}+\frac{[\nabla
b_{1l}]_{m}}{\lambda_{1}^{m-\alpha}}+\frac{[\nabla
b_{1l}]_{m+\alpha}}{\lambda_{1}^{m}}\Big{)}$ $\displaystyle\leq$
$\displaystyle C_{1}\frac{\mu_{1}}{\lambda_{1}^{1-\alpha}}$ (6.72)

This fact will be used frequently in the subsequence estimates.

By direct calculations,

$\displaystyle M_{12}=$ $\displaystyle\sum\limits_{l,l^{\prime}\in Z^{3},l\neq
l^{\prime}}b_{1l}b_{1l^{\prime}}\Big{(}B_{k_{1}}\otimes
B_{k_{1}}e^{i\lambda_{1}(2^{|l|}+2^{|l^{\prime}|})k_{1}\cdot
x-ig_{1,l,l^{\prime}}(t)}+B_{k_{1}}\otimes
B_{-k_{1}}e^{i\lambda_{1}(2^{|l|}-2^{|l^{\prime}|})k_{1}\cdot
x-i\overline{g}_{1,l,l^{\prime}}(t)}$ $\displaystyle+B_{-k_{1}}\otimes
B_{k_{1}}e^{i\lambda_{1}(2^{|l^{\prime}|}-2^{|l|})k_{1}\cdot
x+i\overline{g}_{1,l,l^{\prime}}(t)}+B_{-k_{1}}\otimes
B_{-k_{1}}e^{-i\lambda_{1}(2^{|l|}+2^{|l^{\prime}|})k_{1}\cdot
x-ig_{1,l,l^{\prime}}(t)}\Big{)},$

where

$g_{1,l,l^{\prime}}(t)=\lambda_{1}\Big{(}2^{|l|}k_{1}\cdot\frac{l}{\mu_{1}}t+2^{|l^{\prime}|}k_{1}\cdot\frac{l^{\prime}}{\mu_{1}}t\Big{)},$

and

$\overline{g}_{1,l,l^{\prime}}(t)=\lambda_{1}\Big{(}2^{|l|}k_{1}\cdot\frac{l}{\mu_{1}}t-2^{|l^{\prime}|}k_{1}\cdot\frac{l^{\prime}}{\mu_{1}}t\Big{)}.$

Since $k_{1}\cdot B_{k_{1}}=k_{1}\cdot B_{-k_{1}}=0$, we have

$\displaystyle\hbox{div}M_{12}=$ $\displaystyle\sum\limits_{l,l^{\prime}\in
Z^{3},l\neq l^{\prime}}\Big{(}B_{k_{1}}\otimes
B_{k_{1}}e^{i\lambda_{1}(2^{|l|}+2^{|l^{\prime}|})k_{1}\cdot
x-ig_{l,l^{\prime}}(t)}+B_{k_{1}}\otimes
B_{-k_{1}}e^{i\lambda_{1}(2^{|l|}-2^{|l^{\prime}|})k_{1}\cdot
x-i\overline{g}_{l,l^{\prime}}(t)}$ $\displaystyle+B_{-k_{1}}\otimes
B_{k_{1}}e^{i\lambda_{1}(2^{|l^{\prime}|}-2^{|l|})k_{1}\cdot
x+i\overline{g}_{l,l^{\prime}}(t)}+B_{-k_{1}}\otimes
B_{-k_{1}}e^{-i\lambda_{1}(2^{|l|}+2^{|l^{\prime}|})k_{1}\cdot
x-ig_{l,l^{\prime}}(t)}\Big{)}\nabla(b_{1l}b_{1l^{\prime}}),$

then using Lemma 6.1 and property (A.176) on $\mathcal{R}$, we have

$\displaystyle\|\mathcal{R}(\hbox{div}M_{12})\|_{\alpha}\leq\sum\limits_{|l|,|l^{\prime}|\leq
C_{1}\mu_{1}}C_{1}\frac{\mu_{1}}{\lambda_{1}^{1-\alpha}}\leq
C_{1}\frac{\mu_{1}^{3}}{\lambda_{1}^{1-\alpha}}.$

Then we obtain the estimate (6.70).

From Lemma 6.1 and the property (A.176) of $\mathcal{R}$, similarly we obtain

$\displaystyle\Big{\|}\mathcal{R}\Big{(}\Big{(}\chi_{1}-\fint_{T^{3}}{\chi_{1}}dx\Big{)}e_{3}\Big{)}\Big{\|}_{\alpha}\leq\sum\limits_{|l|\leq
C_{1}\mu_{1}}\frac{C_{1}}{\lambda_{1}^{1-\alpha}}\leq
C_{1}\frac{\mu_{1}}{\lambda_{1}^{1-\alpha}}.$

where we used $\mathcal{R}(f(t))=0$ which can be easily proved from the
definition of $\mathcal{R}$ and then we complete the proof of this lemma. MM
$\square$

######  Lemma 6.4 (The transportation part).

$\displaystyle\Big{\|}\mathcal{R}\Big{(}\partial_{t}w_{1}+\sum_{l\in
Z^{3}}\frac{l}{\mu_{1}}\cdot\nabla w_{1l}\Big{)}\Big{\|}_{\alpha}\leq
C_{1}\frac{\mu_{1}^{2}}{\lambda_{1}^{1-\alpha}}$ (6.73)

###### Proof.

Recall that

$\displaystyle
w_{1l}=B_{1lk_{1}}e^{i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu_{1}}t)}+B_{-1lk_{1}}e^{-i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu_{1}}t)}$

and

$w_{1}=\sum_{l\in Z^{3}}w_{1l}.$

From the identity

$\Big{(}\partial_{t}+\frac{l}{\mu_{1}}\cdot\nabla\Big{)}e^{\pm
i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu_{1}}t)}=0,$

we have

$\displaystyle\partial_{t}w_{1}+\sum_{l\in Z^{3}}\frac{l}{\mu_{1}}\cdot\nabla
w_{1l}$ $\displaystyle=$ $\displaystyle\sum_{l\in
Z^{3}}\Big{(}\Big{(}\partial_{t}+\frac{l}{\mu_{1}}\cdot\nabla\Big{)}B_{1lk_{1}}e^{i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu_{1}}t)}+\Big{(}\partial_{t}+\frac{l}{\mu_{1}}\cdot\nabla\Big{)}B_{-1lk_{1}}e^{-i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu_{1}}t)}\Big{)}.$

By the property (A.176) of $\mathcal{R}$ and Lemma 6.1, we have

$\displaystyle\Big{\|}\mathcal{R}\Big{(}\partial_{t}w_{1}+\sum_{l\in
Z^{3}}\frac{l}{\mu_{1}}\cdot\nabla
w_{1l}\Big{)}\Big{\|}_{\alpha}\leq\sum\limits_{|l|\leq
C_{1}\mu_{1}}C_{1}\frac{\mu_{1}}{\lambda_{1}^{1-\alpha}}\leq
C_{1}\frac{\mu_{1}^{2}}{\lambda_{1}^{1-\alpha}}.$

We obtained the proof the Lemma 6.4. MM $\square$

######  Lemma 6.5 (Estimates on error part I).

$\displaystyle\|N_{1}\|_{0}\leq\frac{C_{1}}{\mu_{1}}.$ (6.74)

###### Proof.

From the definition of $N_{1}$, we have

$\displaystyle N_{1}=$ $\displaystyle\sum_{l\in
Z^{3}}\Big{[}w_{1l}\otimes\Big{(}v_{0}-\frac{l}{\mu_{1}}\Big{)}+\Big{(}v_{0}-\frac{l}{\mu_{1}}\Big{)}\otimes
w_{1l}\Big{)}\Big{]}$ $\displaystyle=$ $\displaystyle\sum_{l\in
Z^{3}}\Big{[}w_{1ol}\otimes\Big{(}v_{0}-\frac{l}{\mu_{1}}\Big{)}+\Big{(}v_{0}-\frac{l}{\mu_{1}}\Big{)}\otimes
w_{1ol}\Big{)}\Big{]}$ $\displaystyle+\sum_{l\in
Z^{3}}\Big{[}w_{1ocl}\otimes\Big{(}v_{0}-\frac{l}{\mu_{1}}\Big{)}+\Big{(}v_{0}-\frac{l}{\mu_{1}}\Big{)}\otimes
w_{1ocl}\Big{)}\Big{]}$ $\displaystyle=$ $\displaystyle N_{11}+N_{12}.$

For the term $N_{11}$, using the definition (4.47) of $w_{1ol}$ , we have

$\displaystyle\sum_{l\in
Z^{3}}w_{1ol}\otimes\Big{(}v_{0}-\frac{l}{\mu_{1}}\Big{)}$ $\displaystyle=$
$\displaystyle\sum_{l\in
Z^{3}}b_{1l}(x,t)\Big{(}B_{k_{1}}e^{i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu_{1}}t)}+B_{-{k_{1}}}e^{-i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu_{1}}t)}\Big{)}\otimes\Big{(}v_{0}-\frac{l}{\mu_{1}}\Big{)}.$

Obviously, $b_{1l}(x,t)\neq 0$ if and only if $|\mu_{1}v_{0}-l|\leq 1.$ By
(4.40), it’s easy to check

$\displaystyle\Big{\|}\sum_{l\in
Z^{3}}w_{1ol}\otimes\Big{(}v_{0}-\frac{l}{\mu_{1}}\Big{)}\Big{\|}_{0}\leq\frac{C_{1}}{\mu_{1}}.$

Then, we have

$\displaystyle\|N_{11}\|_{0}\leq\frac{C_{1}}{\mu_{1}}.$

Similarly, by Lemma 6.2, we have

$\displaystyle\|N_{12}\|_{0}\leq\frac{C_{1}}{\lambda_{1}}.$

Then we obtain (6.66), in particular,

$\displaystyle\|trN_{1}\|_{0}\leq\frac{C_{1}}{\mu_{1}},$

and we complete our proof of this lemma. MM $\square$

######  Lemma 6.6 (Estimates on error part II).

$\displaystyle\|w_{1o}\otimes w_{1oc}+w_{1oc}\otimes w_{1o}+w_{1oc}\otimes
w_{1oc}\|_{0}\leq C_{1}\frac{\mu_{1}}{\lambda_{1}}.$ (6.75)

###### Proof.

From the estimate (4.50) and Lemma 6.2, we have

$\displaystyle\|w_{1o}\otimes w_{1oc}+w_{1oc}\otimes w_{1o}+w_{1oc}\otimes
w_{10c}\|_{0}$ $\displaystyle\leq$ $\displaystyle
C_{1}(\|w_{1o}\|_{0}\|w_{1oc}\|_{0}+\|w_{1oc}\|_{0}^{2})$ $\displaystyle\leq$
$\displaystyle C_{1}\frac{\mu_{1}}{\lambda_{1}},$

therefore

$\displaystyle\|2w_{1o}\cdot w_{1oc}+|w_{1oc}|^{2}\|_{0}\leq
C_{1}\frac{\mu_{1}}{\lambda_{1}}.$

MM $\square$

Finally, by Lemma 6.3, Lemma 6.4, Lemma 6.5 and Lemma 6.6, we conclude that

$\displaystyle\|\delta\mathring{R}_{01}\|_{0}\leq
C_{1}\Big{(}\frac{\mu_{1}^{3}}{\lambda_{1}^{1-\alpha}}+\frac{1}{\mu_{1}}\Big{)},\qquad\forall\alpha\in(0,1).$
(6.76)

Furthermore, from the above estimates and constructions of $v_{01},p_{01}$, we
conclude that

$\displaystyle\|v_{01}-v_{0}\|_{0}\leq$
$\displaystyle\frac{M\sqrt{\delta}}{2L}+C_{1}\frac{\mu_{1}}{\lambda_{1}},$
(6.77) $\displaystyle\|p_{01}-p_{0}\|_{0}\leq$ $\displaystyle
C_{1}\Big{(}\frac{1}{\mu_{1}}+\frac{\mu_{1}}{\lambda_{1}}\Big{)}.$ (6.78)

###  6.2. Estimates on $\delta f_{01}$

Recall (4.55), as before, we split $\delta f_{01}$ into three parts:
(1) the oscillation part

$\displaystyle\mathcal{G}(\hbox{div}K_{1})+\mathcal{G}(w_{1}\cdot\nabla\theta_{0}),$

(2) the transportation part

$\displaystyle\mathcal{G}\Big{(}\partial_{t}\chi_{1}+\sum_{l\in
Z^{3}}\Big{(}\frac{l}{\mu_{1}}\cdot\nabla\Big{)}\chi_{1l}\Big{)},$

(3) the error part

$\displaystyle w_{10c}\chi_{1}+\sum\limits_{l\in
Z^{3}}\Big{(}v_{0}-\frac{l}{\mu}\Big{)}\chi_{1l}.$

######  Lemma 6.7 (The Oscillation Part).

$\displaystyle\|\mathcal{G}({\rm
div}K_{1})+\mathcal{G}(w_{1}\cdot\nabla\theta_{0})\|_{\alpha}\leq
C_{1}\frac{\mu_{1}^{3}}{\lambda_{1}^{1-\alpha}}.$ (6.79)

###### Proof.

From (4.53) and the definition of $K_{1}$, we may rewrite

$\displaystyle\hbox{div}K_{1}=$ $\displaystyle\sum\limits_{l\in
Z^{3}}\nabla(\beta_{1l}(x,t)b_{1l}(x,t))\cdot\Big{(}B_{k_{1}}e^{2i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu_{1}}t)}+B_{-k_{1}}e^{-2i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu_{1}}t)}\Big{)}$
$\displaystyle+\sum\limits_{l,l^{\prime}\in Z^{3},l\neq
l^{\prime}}\hbox{div}(w_{1ol}\chi_{1l^{\prime}})$ $\displaystyle=$
$\displaystyle K_{11}+K_{12}.$

By Lemma 6.1 and the property (A.176) of $\mathcal{G}$ , we have

$\displaystyle\|\mathcal{G}(K_{11})\|_{\alpha}\leq\sum\limits_{|l|\leq
C_{1}\mu_{1}}C_{1}\frac{\mu_{1}}{\lambda_{1}^{1-\alpha}}\leq
C_{1}\frac{\mu_{1}^{2}}{\lambda_{1}^{1-\alpha}}.$

Where we used the following inequality which deduced from Lemma 6.1, for
$\mu_{1}\ll\lambda_{1}$ and $m$ a fixed large number,

$\displaystyle\Big{\|}\mathcal{R}\Big{(}\nabla\beta_{1l}e^{i\lambda_{1}2^{|l|}k_{1}\cdot
x}\Big{)}\Big{\|}_{\alpha}$ $\displaystyle\leq$ $\displaystyle
C_{1,m}\Big{(}\frac{\|\nabla\beta_{1l}\|_{0}}{\lambda^{1-\alpha}}+\frac{[\nabla\beta_{1l}]_{m}}{\lambda^{m-\alpha}}+\frac{[\nabla\beta_{1l}]_{m+\alpha}}{\lambda^{m}}\Big{)}$
$\displaystyle\leq$ $\displaystyle
C_{1}\frac{\mu_{1}}{\lambda_{1}^{1-\alpha}}.$ (6.80)

Since $k_{1}\cdot B_{k_{1}}=k_{1}\cdot B_{-k_{1}}=0$, we have

$\displaystyle K_{12}=$
$\displaystyle\hbox{div}\Big{\\{}\sum\limits_{l,l^{\prime}\in Z^{3},l\neq
l^{\prime}}b_{1l}\beta_{1l^{\prime}}(x,t)(x,t)\Big{(}B_{k_{1}}e^{i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu_{1}}t)}$
$\displaystyle+B_{-k_{1}}e^{-i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu_{1}}t)}\Big{)}\Big{(}e^{i\lambda_{1}2^{|l^{\prime}|}k_{1}\cdot(x-\frac{l^{\prime}}{\mu_{1}}t)}+e^{-i\lambda_{1}2^{|l^{\prime}|}k_{1}\cdot(x-\frac{l^{\prime}}{\mu_{1}}t)}\Big{)}\Big{\\}}$
$\displaystyle=$ $\displaystyle\sum\limits_{l,l^{\prime}\in Z^{3},l\neq
l^{\prime}}\Big{(}B_{k_{1}}\cdot\nabla(b_{1l}\beta_{1l^{\prime}})e^{i\lambda_{1}(2^{|l|}+2^{|l^{\prime}|})k_{1}\cdot
x-ig_{1,l,l^{\prime}}(t)}+B_{k_{1}}\cdot\nabla(b_{1l}\beta_{1l^{\prime}})e^{i\lambda_{1}(2^{|l|}-2^{|l^{\prime}|})k_{1}\cdot
x-i\overline{g}_{1,l,l^{\prime}}(t)}$
$\displaystyle+B_{-k_{1}}\cdot\nabla(b_{1l}\beta_{1l^{\prime}})e^{i\lambda_{1}(2^{|l^{\prime}|}-2^{|l|})k_{1}\cdot
x+i\overline{g}_{1,l,l^{\prime}}(t)}+B_{-k_{1}}\cdot\nabla(b_{1l}\beta_{1l^{\prime}})e^{-i\lambda_{1}(2^{|l|}+2^{|l^{\prime}|})k_{1}\cdot
x-ig_{1,l,l^{\prime}}(t)}\Big{)},$

thus from Lemma 6.1 and the property (A.176) of $\mathcal{G}$ , we have

$\displaystyle\|\mathcal{G}(K_{12})\|_{\alpha}\leq\sum\limits_{|l|,|l^{\prime}|\leq
C_{1}\mu_{1}}C_{1}\frac{\mu_{1}}{\lambda_{1}^{1-\alpha}}\leq
C_{1}\frac{\mu_{1}^{3}}{\lambda_{1}^{1-\alpha}}.$

Then we complete the proof of this lemma. MM $\square$

######  Lemma 6.8 (The transportation Part).

$\displaystyle\Big{\|}\mathcal{G}\Big{(}\partial_{t}\chi_{1}+\sum_{l\in
Z^{3}}\Big{(}\frac{l}{\mu_{1}}\cdot\nabla\Big{)}\chi_{1l}\Big{)}\Big{\|}_{\alpha}\leq
C_{1}\frac{\mu_{1}^{2}}{\lambda_{1}^{1-\alpha}}.$ (6.81)

###### Proof.

From (4.48), (4.49) and the definition of $\chi_{1}$, we have

$\displaystyle\partial_{t}\chi_{1}+\sum_{l\in
Z^{3}}\Big{(}\frac{l}{\mu_{1}}\cdot\nabla\Big{)}\chi_{1l}$ $\displaystyle=$
$\displaystyle\sum\limits_{l\in
Z^{3}}\Big{(}\partial_{t}+\frac{l}{\mu_{1}}\cdot\nabla\Big{)}\beta_{1l}\Big{(}e^{i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu_{1}}t)}+e^{-i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu_{1}}t)}\Big{)}.$

By Lemma 6.1 and the property (A.176) of $\mathcal{G}$, we have

$\displaystyle\Big{\|}\mathcal{G}\Big{(}\partial_{t}\chi_{1}+\sum_{l\in
Z^{3}}\Big{(}\frac{l}{\mu_{1}}\cdot\nabla\Big{)}\chi_{1l}\Big{)}\Big{\|}_{\alpha}\leq\sum\limits_{|l|C_{1}\mu_{1}}C_{1}\frac{\mu_{1}}{\lambda_{1}^{1-\alpha}}\leq
C_{1}\frac{\mu_{1}^{2}}{\lambda_{1}^{1-\alpha}}.$

MM $\square$

######  Lemma 6.9 (The error Part).

$\displaystyle\Big{\|}w_{1oc}\chi_{1}+\sum_{l\in
Z^{3}}\Big{(}v_{0}-\frac{l}{\mu_{1}}\Big{)}\chi_{1l}\Big{\|}_{0}\leq
C_{1}\Big{(}\frac{\mu_{1}}{\lambda_{1}}+\frac{1}{\mu_{1}}\Big{)}.$ (6.82)

###### Proof.

First, from Lemma 6.2, we have

$\displaystyle\|w_{1oc}\chi_{1}\|_{0}\leq C_{1}\frac{\mu_{1}}{\lambda_{1}},$

$\displaystyle\sum_{l\in
Z^{3}}\Big{(}v_{0}-\frac{l}{\mu_{1}}\Big{)}\chi_{1l}=\sum\limits_{l\in
Z^{3}}\Big{(}v_{0}-\frac{l}{\mu_{1}}\Big{)}\beta_{1l}\Big{(}e^{i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu_{1}}t)}+e^{-i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu_{1}}t)}\Big{)}.$

Obviously, $\beta_{1l}(x,t)\neq 0$ if and only if $|\mu_{1}v_{0}-l|\leq 1,$
therefore,

$\displaystyle\Big{\|}\sum_{l\in
Z^{3}}\Big{(}v_{0}-\frac{l}{\mu_{1}}\Big{)}\chi_{1l}\Big{\|}_{0}\leq\frac{C_{1}}{\mu_{1}}.$

MM $\square$

Then we conclude that

$\displaystyle\|\delta f_{01}\|_{0}\leq
C_{1}\Big{(}\frac{\mu_{1}^{3}}{\lambda_{1}^{1-\alpha}}+\frac{1}{\mu_{1}}\Big{)}.$
(6.83)

Furthermore, from (4.59), we have

$\displaystyle\|\theta_{01}-\theta_{0}\|_{0}\leq$
$\displaystyle\frac{M\sqrt{\delta}}{2L}+C_{1}\frac{\mu_{1}^{2}}{\lambda_{1}}.$
(6.84)

Where we used the fact

$\displaystyle\sup_{t}\Big{|}\int_{T^{3}}\chi_{1}dx\Big{|}\leq\sum\limits_{|l|\leq
C_{1}\mu_{1}}C_{1}\frac{\mu_{1}}{\lambda_{1}}\leq
C_{1}\frac{\mu_{1}^{2}}{\lambda_{1}}.$

In conclusion, we have

$\displaystyle
R_{01}:=-\rho(t)\sum\limits_{i=2}^{L}\Big{(}\gamma_{k_{i}}\Big{(}\frac{R_{0}}{\rho(t)}\Big{)}\Big{)}^{2}\Big{(}Id-\frac{k_{i}}{|k_{i}|}\otimes\frac{k_{i}}{|k_{i}|}\Big{)}+\delta\mathring{R}_{01},$
$\displaystyle
f_{01}:=\sum\limits_{i=2}^{3}g_{k_{i}}(f_{0}(x,t))A_{k_{i}}+\delta f_{01},$

and the estimates

$\displaystyle\|v_{01}-v_{0}\|_{0}\leq$
$\displaystyle\frac{M\sqrt{\delta}}{2L}+C_{1}\frac{\mu_{1}}{\lambda_{1}},$
$\displaystyle\|p_{01}-p_{0}\|_{0}\leq$ $\displaystyle
C_{1}\Big{(}\frac{1}{\mu_{1}}+\frac{\mu_{1}}{\lambda_{1}}\Big{)},$
$\displaystyle\|\theta_{01}-\theta_{0}\|_{0}\leq$
$\displaystyle\frac{M\sqrt{\delta}}{2L}+C_{1}\frac{\mu_{1}^{2}}{\lambda_{1}},$
$\displaystyle\|\delta\mathring{R_{01}}\|_{0}\leq$ $\displaystyle
C_{1}\Big{(}\frac{\mu_{1}^{3}}{\lambda_{1}^{1-\alpha}}+\frac{1}{\mu_{1}}\Big{)},\qquad\forall\alpha\in(0,1),$
$\displaystyle\|\delta f_{01}\|_{0}\leq$ $\displaystyle
C_{1}\Big{(}\frac{\mu_{1}^{3}}{\lambda_{1}^{1-\alpha}}+\frac{1}{\mu_{1}}\Big{)}.\qquad\forall\alpha\in(0,1).$

Thus we complete the first step.

##  7\. Constructions of $v_{0n},p_{0n},\theta_{0n},R_{0n},f_{0n}$, $2\leq
n\leq L$

In this section, we assume $2\leq n\leq L$ and we will construct
$v_{0n},p_{0n},\theta_{0n},R_{0n},f_{0n}$ by inductions.

Suppose that for $1\leq m<n\leq L$, we have constructed
$v_{0m},~{}p_{0m},~{}\theta_{0m},~{}R_{0m},~{}f_{0m}$ and they are the
solutions of the system (2.2). Furthermore, we have

$\displaystyle
R_{0m}:=-\bar{\rho}(t)\sum\limits_{i=m+1}^{L}\Big{(}\gamma_{k_{i}}\Big{(}\frac{R_{0}}{\bar{\rho}(t)}\Big{)}\Big{)}^{2}\Big{(}Id-\frac{k_{i}}{|k_{i}|}\otimes\frac{k_{i}}{|k_{i}|}\Big{)}+\sum\limits_{i=1}^{m}\delta\mathring{R}_{0i},$
$\displaystyle
f_{0m}:=\left\\{\begin{array}[]{ll}\sum\limits_{i=m+1}^{3}g_{k_{i}}(f_{0}(x,t))A_{k_{i}}+\sum\limits_{i=1}^{m}\delta
f_{0i},\quad m=1,2\\\ \sum\limits_{i=1}^{m}\delta f_{0i},\qquad m\geq
3,\end{array}\right.$ (7.87)

and

$\displaystyle\|v_{0m}-v_{0(m-1)}\|_{0}\leq$
$\displaystyle\frac{M\sqrt{\delta}}{2L}+C_{m}\frac{\mu_{m}}{\lambda_{m}},$
$\displaystyle\|p_{0m}-p_{0(m-1)}\|_{0}\leq$ $\displaystyle
C_{m}\Big{(}\frac{1}{\mu_{m}}+\frac{\mu_{m}}{\lambda_{m}}\Big{)},$
$\displaystyle\|\theta_{0m}-\theta_{0(m-1)}\|_{0}\leq$
$\displaystyle\frac{M\sqrt{\delta}}{2L}+C_{m}\frac{\mu_{m}^{2}}{\lambda_{m}},$
$\displaystyle\|\delta\mathring{R}_{0m}\|_{0}\leq$ $\displaystyle
C_{m}\Big{(}\frac{\mu_{m}^{3}}{\lambda_{m}^{1-\alpha}}+\frac{1}{\mu_{m}}\Big{)},$
$\displaystyle\|\delta f_{0m}\|_{0}\leq$ $\displaystyle
C_{m}\Big{(}\frac{\mu_{m}^{3}}{\lambda_{m}^{1-\alpha}}+\frac{1}{\mu_{m}}\Big{)}.\qquad\forall\alpha\in(0,1).$
(7.88)

where parameters $\mu_{m}$ and $\lambda_{m}$ will be chosen sufficiently
large, which satisfies

$\displaystyle\lambda_{m},\mu_{m},\frac{\lambda_{m}}{\mu_{m}}\in N,$ (7.89)

and the constant $C_{m}$ denotes a constant which depends on
$v_{0(m-1)},\mathring{R_{0}},e(t)$ as well as $\lambda_{0},\alpha,\delta$, but
doesn’t depend on $\mu_{m},\lambda_{m}$, and can change from line to line.

Next, we construct n-th steps by induction.

###  7.1. The perturbations $w_{n}$ and $\chi_{n}$

For $2\leq n\leq L$ and any $l\in Z^{3}$, we denote $b_{nl},B_{k_{n}}$ by

$\displaystyle b_{nl}(x,t):=$
$\displaystyle\sqrt{\rho(t)}\alpha_{l}(\mu_{n}v_{0(n-1)})\gamma_{k_{n}}\Big{(}\frac{R_{0}(x,t)}{\rho(t)}\Big{)},$
$\displaystyle B_{k_{n}}:=$ $\displaystyle
A_{k_{n}}+i\frac{k_{n}}{|k_{n}|}\times A_{k_{n}}.$

And define $l$-perturbations

$\displaystyle
w_{nol}:=b_{nl}(x,t)\Big{(}B_{k_{n}}e^{i\lambda_{n}2^{|l|}k_{n}\cdot(x-\frac{l}{\mu_{n}}t)}+B_{-k_{n}}e^{-i\lambda_{n}2^{|l|}k_{n}\cdot(x-\frac{l}{\mu_{n}}t)}\Big{)},$

where the parameters $\mu_{n}$ and $\lambda_{n}$ will be chosen sufficiently
large, which satisfies the following conditions

$\displaystyle\lambda_{n},\mu_{n},\frac{\lambda_{n}}{\mu_{n}}\in N.$ (7.90)

Then, we define the n-th perturbation

$\displaystyle w_{no}:=\sum_{l\in Z^{3}}w_{nol}.$

Similar as in the first step, the $l$-corrections is denoted by

$\displaystyle w_{nocl}:=$
$\displaystyle\frac{1}{\lambda_{n}\lambda_{0}}\Big{(}\frac{\nabla
b_{nl}(x,t)\times
B_{k_{n}}}{2^{|l|}}e^{i\lambda_{n}2^{|l|}k_{n}\cdot(x-\frac{l}{\mu_{n}}t)}+\frac{\nabla
b_{nl}(x,t)\times
B_{-{k_{n}}}}{2^{|l|}}e^{-i\lambda_{n}2^{|l|}k_{n}\cdot(x-\frac{l}{\mu_{n}}t)}\Big{)},$

and the n-th correction

$\displaystyle w_{noc}:=\sum_{l\in Z^{3}}w_{nocl}.$

Finally, we define n-th perturbation

$\displaystyle w_{n}:=$ $\displaystyle w_{no}+w_{noc}.$

If we denote $w_{nl}$ by

$\displaystyle w_{nl}=$ $\displaystyle\frac{1}{\lambda_{n}\lambda_{0}}{\rm
curl}\Big{(}\frac{b_{nl}(x,t)B_{k_{n}}}{2^{|l|}}e^{i\lambda_{n}10^{|l|}k_{n}\cdot(x-\frac{l}{\mu_{n}}t)}+\frac{b_{nl}(x,t)B_{-{k_{n}}}}{10^{|l|}}e^{-i\lambda_{n}2^{|l|}k_{n}\cdot(x-\frac{l}{\mu_{n}}t)}\Big{)}$

then

$w_{n}=\sum_{l\in
Z^{3}}w_{nl},\qquad\hbox{div}w_{nl}=0,\qquad\hbox{div}w_{n}=0,$

and they are all real vector functions. Moreover,

$\displaystyle\|w_{n0}\|_{0}\leq\frac{M\sqrt{\delta}}{2L}.$ (7.91)

We set

$\displaystyle B_{nlk_{n}}:=$ $\displaystyle
b_{nl}(x,t)B_{k_{n}}+\frac{1}{\lambda_{n}\lambda_{0}}\frac{\nabla
b_{nl}(x,t)\times B_{k_{n}}}{2^{|l|}},$ $\displaystyle B_{-nlk_{n}}:=$
$\displaystyle
b_{nl}(x,t)B_{-k_{n}}+\frac{1}{\lambda_{n}\lambda_{0}}\frac{\nabla
b_{nl}(x,t)\times B_{-k_{n}}}{2^{|l|}},$

then

$\displaystyle
w_{nl}=B_{nlk_{n}}e^{i\lambda_{n}2^{|l|}k_{n}\cdot(x-\frac{l}{\mu_{n}}t)}+B_{-nlk_{n}}e^{-i\lambda_{n}2^{|l|}k_{n}\cdot(x-\frac{l}{\mu_{n}}t)}.$

Then we complete the construction of n-th perturbation $w_{n}$.
To construct $\chi_{n}$, we denote $\beta_{nl}$ by

$\displaystyle\beta_{nl}(x,t):=\left\\{\begin{array}[]{ll}\frac{\alpha_{l}(\mu_{n}v_{0(n-1)})}{2\sqrt{\rho(t)}}\frac{g_{k_{n}}(-f_{0}(x,t))}{\gamma_{k_{n}}\big{(}\frac{R(x,t)}{\rho(t)}\big{)}},\qquad
n=2,3\\\ 0,\qquad n\geq 4,\end{array}\right.$ (7.94)

then define $l$-perturbations

$\displaystyle\chi_{nl}(x,t):=\left\\{\begin{array}[]{ll}\beta_{nl}(x,t)\Big{(}e^{i\lambda_{n}2^{|l|}k_{n}\cdot(x-\frac{l}{\mu_{n}}t)}+e^{-i\lambda_{n}2^{|l|}k_{n}\cdot(x-\frac{l}{\mu_{n}}t)}\Big{)},\quad
n=2,3\\\ 0,\qquad n\geq 4,\end{array}\right.$ (7.97)

and the n-th perturbation

$\displaystyle\chi_{n}(x,t):=\sum\limits_{l\in
Z^{3}}\chi_{nl}(x,t)=\left\\{\begin{array}[]{ll}\sum\limits_{l\in
Z^{3}}\chi_{nl},\quad n=2,3\\\ 0,\qquad n\geq 4.\end{array}\right.$ (7.100)

Moreover, since
$\Big{\|}\frac{R_{0}(x,t)}{\rho(t)}\Big{\|}_{0}\leq\frac{r_{0}}{2}$, there
exist two positive constants $c_{1n}$ and $c_{2n}$ such that

$c_{1n}\leq\gamma_{k_{n}}\Big{(}\frac{R_{0}(x,t)}{\rho(t)}\Big{)}\leq c_{2n}.$

Then, from (7.82) the definition of $\beta_{nl}$, we have

$\displaystyle\|\beta_{nl}\|_{0}\leq\frac{M\sqrt{\delta}}{10L},$ (7.101)
$\displaystyle\|\chi_{n}\|_{0}\leq\frac{M\sqrt{\delta}}{2L}.$ (7.102)

###  7.2. The constructions of $v_{0n}$, $p_{0n}$, $\theta_{0n}$, $f_{0n}$,
$\mathring{R}_{0n}$

First, we denote $M_{n}$ by

$\displaystyle M_{n}=$ $\displaystyle\sum\limits_{l\in
Z^{3}}b^{2}_{nl}(x,t)\Big{(}B_{k_{n}}\otimes
B_{k_{n}}e^{2i\lambda_{n}2^{|l|}k_{n}\cdot(x-\frac{l}{\mu_{n}}t)}+B_{-k_{n}}\otimes
B_{-k_{n}}e^{-2i\lambda_{n}2^{|l|}k_{n}\cdot(x-\frac{l}{\mu_{n}}t)}\Big{)}$
$\displaystyle+\sum\limits_{l,l^{\prime}\in Z^{3},l\neq
l^{\prime}}w_{nol}\otimes w_{nol^{\prime}}(x,t).$ (7.103)

and $N_{n},K_{n}$ by

$\displaystyle N_{n}=$ $\displaystyle\sum_{l\in
Z^{3}}\Big{[}w_{nl}\otimes\Big{(}v_{0(n-1)}-\frac{l}{\mu_{n}}\Big{)}+\Big{(}v_{0(n-1)}-\frac{l}{\mu_{n}}\Big{)}\otimes
w_{nl}\Big{]}$

and

$\displaystyle K_{n}=\left\\{\begin{array}[]{ll}\sum\limits_{l\in
Z^{3}}\beta_{nl}(x,t)b_{nl}(x,t)\Big{(}B_{k_{n}}e^{2i\lambda_{n}2^{|l|}k_{n}\cdot(x-\frac{l}{\mu_{n}}t)}+B_{-k_{n}}e^{-2i\lambda_{n}2^{|l|}k_{n}\cdot(x-\frac{l}{\mu_{n}}t)}\Big{)}\\\
\qquad+\sum\limits_{l,l^{\prime}\in Z^{3},l\neq
l^{\prime}}w_{nol}\chi_{nl^{\prime}},\quad n=2,3\\\ \\\ 0,\quad n\geq
4.\end{array}\right.$ (7.108)

Notice that $N_{n}$ is a symmetric matrix. Then we define

$\displaystyle v_{0n}:=v_{0(n-1)}+w_{n},$ $\displaystyle
p_{0n}:=p_{0(n-1)}-\frac{2w_{no}\cdot w_{noc}+|w_{noc}|^{2}}{3}-tr(N_{n}),$
$\displaystyle\theta_{0n}:=\left\\{\begin{array}[]{ll}\theta_{0(n-1)}+\chi_{n}-\fint_{T^{3}}{\chi_{n}}dx,\quad
n=2,3\\\ \theta_{03},\quad n\geq 4\end{array}\right.$ $\displaystyle
R_{0n}:=R_{0(n-1)}(x,t)+2\sum\limits_{l\in
Z^{3}}b^{2}_{nl}(x,t)Re(B_{k_{n}}\otimes
B_{-{k_{n}}})+\delta\mathring{R}_{0n},$ $\displaystyle
f_{0n}:=\left\\{\begin{array}[]{ll}f_{0(n-1)}+2\sum\limits_{l\in
Z^{3}}\beta_{nl}b_{nl}(x,t)A_{k_{n}}+\delta f_{0n},\quad n=2,3\\\
f_{0(n-1)}+\delta f_{0n},\quad n\geq 4.\end{array}\right.$ (7.111)

Where

$\displaystyle\delta\mathring{R}_{0n}:=$
$\displaystyle\mathcal{R}(\hbox{div}M_{n})+N_{n}-tr(N_{n})Id+\mathcal{R}\Big{\\{}\partial_{t}w_{noc}+\partial_{t}w_{no}$
$\displaystyle+\hbox{div}\Big{[}\sum_{l\in
Z^{3}}\Big{(}w_{nl}\otimes\frac{l}{\mu_{n}}+\frac{l}{\mu_{n}}\otimes
w_{nl}\Big{)}\Big{]}\Big{\\}}-\mathcal{R}\Big{(}\Big{(}\chi_{n}-\fint_{T^{3}}{\chi_{n}}dx\Big{)}e_{3}\Big{)}$
$\displaystyle+(w_{no}\otimes w_{noc}+w_{noc}\otimes w_{no}+w_{noc}\otimes
w_{noc})-\frac{2w_{no}\cdot w_{noc}+|w_{noc}|^{2}}{3}Id,$

and

$\displaystyle\delta
f_{0n}:=\left\\{\begin{array}[]{ll}\mathcal{G}(\hbox{div}K_{n})+\mathcal{G}(w_{n}\cdot\nabla\theta_{0(n-1)})+\mathcal{G}\Big{(}\partial_{t}\chi_{n}+\sum_{l\in
Z^{3}}\Big{(}\frac{l}{\mu_{n}}\cdot\nabla\Big{)}\chi_{nl}\Big{)}\\\
+\sum_{l\in
Z^{3}}\Big{(}v_{0(n-1)}-\frac{l}{\mu_{n}}\Big{)}\chi_{nl}+w_{noc}\chi_{n},\quad
n=2,3\\\ \\\ \mathcal{G}(w_{n}\cdot\nabla\theta_{03}),\quad n\geq
4.\end{array}\right.$ (7.115)

From Lemma 3.1, we know that $\delta\mathring{R}_{0n}$ is a symmetric and
trace-free matrix. And

$\displaystyle\hbox{div}v_{0n}=\hbox{div}v_{0(n-1)}+\hbox{div}w_{n}=0$

By definition, $w_{no}$, $w_{noc}$, $\delta R_{0n}$ together with $v_{0n}$,
$p_{0n}$ and $v_{0(n-1)}$, $p_{0(n-1)}$, $\theta_{0(n-1)}$, $R_{0(n-1)}$,
$f_{0(n-1)}$ are solutions of the system (2.2), we have

$\displaystyle\hbox{div}R_{0n}=$
$\displaystyle\hbox{div}R_{0(n-1)}(x,t)+\partial_{t}w_{n}-\nabla(trN_{n})-\nabla\Big{(}\frac{2w_{no}\cdot
w_{noc}+|w_{noc}|^{2}}{3}\Big{)}-\Big{(}\chi_{n}-\fint_{T^{3}}{\chi_{n}}dx\Big{)}e_{3}$
$\displaystyle+\hbox{div}(w_{no}\otimes w_{no}+w_{n}\otimes
v_{0(n-1)}+v_{0(n-1)}\otimes w_{n}+w_{no}\otimes w_{noc}+w_{noc}\otimes
w_{no}+w_{noc}\otimes w_{noc})$ $\displaystyle=$
$\displaystyle\partial_{t}v_{0(n-1)}+\hbox{div}(v_{0(n-1)}\otimes
v_{0(n-1)})+\nabla
p_{0(n-1)}-\theta_{0(n-1)}e_{3}+\partial_{t}w_{n}-\nabla(trN_{n})$
$\displaystyle-\nabla\Big{(}\frac{2w_{no}\cdot
w_{noc}+|w_{noc}|^{2}}{3}\Big{)}-\Big{(}\chi_{n}-\fint_{T^{3}}{\chi_{n}}dx\Big{)}e_{3}+\hbox{div}(w_{no}\otimes
w_{no}+w_{n}\otimes v_{0(n-1)}$ $\displaystyle+v_{0(n-1)}\otimes
w_{n}+w_{no}\otimes w_{noc}+w_{noc}\otimes w_{no}+w_{noc}\otimes w_{noc})$
$\displaystyle=$ $\displaystyle\partial_{t}v_{0n}+\hbox{div}(v_{0n}\otimes
v_{0n})+\nabla p_{0n}-\theta_{0n}e_{3}.$

Where we used

$\fint_{T^{3}}w_{n}(x,t)dx=0,$

and

$\hbox{div}(M_{n})+\hbox{div}\Big{(}2\sum\limits_{l\in
Z^{3}}b^{2}_{nl}(x,t)Re(B_{k_{n}}\otimes
B_{-{k_{n}}})\Big{)}=\hbox{div}(w_{no}\otimes w_{no}).$

Furthermore, from the definition of $w_{no},\chi_{n}$, we know that

$\displaystyle f_{0n}=\left\\{\begin{array}[]{ll}f_{0(n-1)}+2\sum\limits_{l\in
Z^{3}}\beta_{nl}(x,t)b_{nl}(x,t)A_{k_{n}}+\mathcal{G}(\hbox{div}K_{n})\\\
+\mathcal{G}(w_{n}\cdot\nabla\theta_{0(n-1)})+\mathcal{G}\Big{(}\partial_{t}\chi_{n}+\sum_{l\in
Z^{3}}\Big{(}\frac{l}{\mu_{n}}\cdot\nabla\Big{)}\chi_{nl}\Big{)}\\\
+\sum_{l\in
Z^{3}}\Big{(}v_{0(n-1)}-\frac{l}{\mu_{n}}\Big{)}\chi_{nl}+w_{noc}\chi_{n},\quad
n=2,3\\\ f_{0(n-1)}+\mathcal{G}(w_{n}\cdot\nabla\theta_{3}),\quad n\geq
4\end{array}\right.$

Thus, for $n=2,3$, we have

$\displaystyle\hbox{div}f_{0n}=$
$\displaystyle\hbox{div}f_{0(n-1)}+\partial_{t}\Big{(}\chi_{n}-\fint_{T^{3}}\chi_{n}\Big{)}+\hbox{div}(w_{no}\chi_{n}+w_{noc}\chi_{n}+v_{0(n-1)}\chi_{n}+w_{n}\theta_{0(n-1)})$
$\displaystyle=$
$\displaystyle\hbox{div}(v_{0(n-1)}\theta_{0(n-1)}+w_{no}\chi_{n}+w_{noc}\chi_{n}+v_{0(n-1)}\chi_{n}$
$\displaystyle+w_{n}\theta_{0(n-1)})+\partial_{t}\Big{(}\theta_{0(n-1)}+\chi_{n}-\fint_{T^{3}}\chi_{n}dx\Big{)}-h$
$\displaystyle=$
$\displaystyle\partial_{t}\theta_{0n}+\hbox{div}(v_{0n}\theta_{0n})-h.$

For $n\geq 4$, we have

$\displaystyle\hbox{div}f_{0n}=\hbox{div}f_{0(n-1)}+w_{n}\cdot\nabla\theta_{03}=$
$\displaystyle\partial_{t}\theta_{0(n-1)}+\hbox{div}(v_{0(n-1)}\theta_{0(n-1)}+w_{n}\theta_{0(n-1)})-h$
$\displaystyle=$
$\displaystyle\partial_{t}\theta_{0n}+\hbox{div}(v_{0n}\theta_{0n})-h,$

where $\theta_{0(n-1)}=\theta_{03}$ for $n\geq 4$.

So the functions $v_{0n},~{}p_{0n},~{}\theta_{0n},~{}R_{0n},~{}f_{0n}$ solve
the system (2.2).

##  8\. The representations

In this section, we will calculate the forms of

$R_{0(n-1)}(x,t)+2\sum\limits_{l\in Z^{3}}b^{2}_{nl}(x,t)Re(B_{k_{n}}\otimes
B_{-{k_{n}}}),$

and

$f_{0(n-1)}+2\sum\limits_{l\in
Z^{3}}\beta_{nl}(x,t)(x,t)b_{nl}(x,t)A_{k_{n}},\quad n=2,3.$

As in (5.64), we have the following identity:

$\displaystyle 2Re(B_{k_{n}}\otimes
B_{-{k_{n}}})=Id-\frac{k_{n}}{|k_{n}|}\otimes\frac{k_{n}}{|k_{n}|}.$

###  8.1. The representation of $R_{0(n-1)}(x,t)+2\sum\limits_{l\in
Z^{3}}b^{2}_{nl}(x,t)Re(B_{k_{n}}\otimes B_{-{k_{n}}})$

First, by the notation of $b_{nl}(x,t)$, we have

$\displaystyle 2\sum\limits_{l\in Z^{3}}b^{2}_{nl}(x,t)Re(B_{k_{n}}\otimes
B_{-{k_{n}}})=$ $\displaystyle\bar{\rho}(t)\sum\limits_{l\in
Z^{3}}\alpha_{l}^{2}(\mu_{n}v_{(0(n-1)})\gamma_{k_{n}}^{2}\Big{(}\frac{R_{0}(x,t)}{\bar{\rho}(t)}\Big{)}\Big{(}Id-\frac{k_{n}}{|k_{n}|}\otimes\frac{k_{n}}{|k_{n}|}\Big{)}$
$\displaystyle=$
$\displaystyle\bar{\rho}(t)\gamma_{k_{n}}^{2}\Big{(}\frac{R_{0}(x,t)}{\bar{\rho}(t)}\Big{)}\Big{(}Id-\frac{k_{n}}{|k_{n}|}\otimes\frac{k_{n}}{|k_{n}|}\Big{)}.$

Moreover, by (7),

$\displaystyle
R_{0(n-1)}:=-\bar{\rho}(t)\sum\limits_{i=n}^{L}\Big{(}\gamma_{k_{i}}\Big{(}\frac{R_{0}}{\bar{\rho}(t)}\Big{)}\Big{)}^{2}\Big{(}Id-\frac{k_{i}}{|k_{i}|}\otimes\frac{k_{i}}{|k_{i}|}\Big{)}+\sum\limits_{i=1}^{n-1}\delta\mathring{R}_{0i}.$

Therefore

$\displaystyle R_{0(n-1)}(x,t)+2\sum\limits_{l\in
Z^{3}}b^{2}_{nl}(x,t)Re(B_{k_{n}}\otimes B_{-{k_{n}}})$ $\displaystyle=$
$\displaystyle-\bar{\rho}(t)\sum\limits_{i=n+1}^{L}\Big{(}\gamma_{k_{i}}\Big{(}\frac{R_{0}}{\bar{\rho}(t)}\Big{)}\Big{)}^{2}\Big{(}Id-\frac{k_{i}}{|k_{i}|}\otimes\frac{k_{i}}{|k_{i}|}\Big{)}+\sum\limits_{i=1}^{n-1}\delta\mathring{R}_{0i}.$

Meanwhile, we have

$\displaystyle
R_{0n}=-\bar{\rho}(t)\sum\limits_{i=n+1}^{L}\Big{(}\gamma_{k_{i}}\Big{(}\frac{R_{0}}{\bar{\rho}(t)}\Big{)}\Big{)}^{2}\Big{(}Id-\frac{k_{i}}{|k_{i}|}\otimes\frac{k_{i}}{|k_{i}|}\Big{)}+\sum\limits_{i=1}^{n}\delta\mathring{R}_{0i}.$
(8.116)

In particular,

$\displaystyle R_{0L}=\sum\limits_{i=1}^{L}\delta R_{01}.$

In next section, we will prove that $\delta R_{0n}$ is small.

###  8.2. The representation of $f_{0(n-1)}+2\sum\limits_{l\in
Z^{3}}\beta_{nl}(x,t)b_{nl}(x,t)A_{k_{n}},n=2,3$

From the definition of $b_{nl}(x,t)$ and $\beta_{nl}(x,t)$, we have

$\displaystyle 2\sum\limits_{l\in Z^{3}}\beta_{nl}(x,t)b_{nl}(x,t)A_{k_{n}}=$
$\displaystyle\sum\limits_{l\in
Z^{3}}\alpha_{l}^{2}(\mu_{n}v_{0(n-1)})g_{k_{n}}(-f_{0(n-1)}(x,t))A_{k_{n}}=-g_{k_{n}}(f_{0(n-1)}(x,t))A_{k_{n}}.$

Where we used the fact that $g_{k_{n}}$ is a linear functional. Moreover, by
(7), for $n=2,3$,

$\displaystyle
f_{0(n-1)}:=\sum\limits_{i=n}^{3}g_{k_{i}}(f_{0}(x,t))A_{k_{i}}+\sum\limits_{i=1}^{n-1}\delta
f_{0i}.$

Therefore

$\displaystyle f_{0(n-1)}+2\sum\limits_{l\in
Z^{3}}\beta_{nl}(x,t)b_{nl}(x,t)A_{k_{n}}=$
$\displaystyle\sum\limits_{i=n+1}^{3}g_{k_{i}}(f_{0}(x,t))A_{k_{i}}+\sum\limits_{i=1}^{n-1}\delta
f_{0i}.$

And

$\displaystyle
f_{0n}=\sum\limits_{i=n+1}^{3}g_{k_{i}}(f_{0}(x,t))A_{k_{i}}+\sum\limits_{i=1}^{n}\delta
f_{0i}.$ (8.117)

In particular,

$\displaystyle f_{03}=\sum\limits_{i=1}^{3}\delta f_{0i}.$

And

$\displaystyle f_{0n}=\sum\limits_{i=1}^{n}\delta f_{0i},$

for $n\geq 3$.

##  9\. Estimates on $\delta\mathring{R}_{0n}$ and $\delta f_{0n}$

In this section, as before, $C_{n}$ denotes a constant which depends on
$v_{0(n-1)},~{}\mathring{R_{0}},~{}e(t)$ as well as
$\lambda_{0},~{}\alpha,~{}\delta$, but doesn’t depend on
$\mu_{n},~{}\lambda_{n}$, and can change from line to line. Furthermore, we
will set $1\ll\mu_{n}\ll\lambda_{n}$.

First, we summarize some of the estimates of $b_{nl}$ and $\beta_{nl}$.

######  Lemma 9.1.

For any $|l|\leq C_{n}\mu_{n}$ and $r\geq 0$,

$\displaystyle\|b_{nl}\|_{r}\leq C_{n}\mu_{n}^{r},$
$\displaystyle\|\beta_{nl}\|_{r}\leq C_{n}\mu_{n}^{r},$
$\displaystyle\|(\partial_{t}+\frac{l}{\mu_{n}}\cdot\nabla)b_{nl}\|_{r}\leq
C_{n}\mu_{n}^{r+1},$
$\displaystyle\|(\partial_{t}+\frac{l}{\mu_{n}}\cdot\nabla)b_{nl}\|_{r}\leq
C_{n}\mu_{n}^{r+1}.$

###### Proof.

The proof is similar to that of Lemma 6.1, we omit it here. MM $\square$

From the definition of $w_{noc}$ , we have the following estimates

######  Lemma 9.2 (Estimates on the n-th correction).

$\displaystyle\|w_{noc}\|_{\alpha}\leq
C_{n}\frac{\mu_{n}}{\lambda_{n}^{1-\alpha}},\qquad\forall\alpha\in[0,1)$
(9.118)

###### Proof.

The proof is similar to that of Lemma 6.2, we omit it here. MM $\square$

###  9.1. Estimates on $\delta\mathring{R}_{0n}$

As in (4.54), we have

$\displaystyle\delta R_{0n}=$
$\displaystyle\mathcal{R}(\hbox{div}M_{n})+N_{n}-tr(N_{n})Id+\mathcal{R}\Big{\\{}\partial_{t}w_{n}+\hbox{div}\Big{[}\sum_{l\in
Z^{3}}\Big{(}w_{nl}\otimes\frac{l}{\mu_{n}}+\frac{l}{\mu_{n}}\otimes
w_{nl}\Big{)}\Big{]}\Big{\\}}$ $\displaystyle+N_{n}+(w_{no}\otimes
w_{noc}+w_{noc}\otimes w_{no}+w_{noc}\otimes w_{noc})$
$\displaystyle-\frac{2w_{no}\cdot
w_{noc}+|w_{noc}|^{2}}{3}Id-\mathcal{R}\Big{(}\Big{(}\chi_{n}-\fint_{T^{3}}{\chi_{n}}dx\Big{)}e_{3}\Big{)}.$

We again split the stresses into three parts,
(1) the oscillation part

$\mathcal{R}(\hbox{div}M_{n})-\mathcal{R}\Big{(}\Big{(}\chi_{n}-\fint_{T^{3}}{\chi_{n}}dx\Big{)}e_{3}\Big{)},$

(2) the transportation part

$\displaystyle\mathcal{R}\Big{\\{}\partial_{t}w_{n}+\hbox{div}\Big{[}\sum_{l\in
Z^{3}}\Big{(}w_{nl}\otimes\frac{l}{\mu_{n}}+\frac{l}{\mu_{n}}\otimes
w_{nl}\Big{)}\Big{]}\Big{\\}}=\mathcal{R}\Big{(}\partial_{t}w_{n}+\sum_{l\in
Z^{3}}\Big{(}\frac{l}{\mu_{n}}\cdot\nabla\Big{)}w_{nl}\Big{)},$

(3) the error part

$\displaystyle N_{n}-tr(N_{n})Id+(w_{no}\otimes w_{noc}+w_{noc}\otimes
w_{no}+w_{noc}\otimes w_{noc})-\frac{2w_{no}\cdot
w_{noc}+|w_{noc}|^{2}}{3}Id.$

As before, we will estimate each term separately.

######  Lemma 9.3 (The oscillation part).

$\displaystyle\|\mathcal{R}({\rm div}M_{n})\|_{\alpha}\leq$ $\displaystyle
C_{n}\frac{\mu_{n}^{3}}{\lambda_{n}^{1-\alpha}},\qquad$ (9.119)
$\displaystyle\Big{\|}\mathcal{R}\Big{(}\Big{(}\chi_{n}-\fint_{T^{3}}{\chi_{n}}dx\Big{)}e_{3}\Big{)}\Big{\|}_{\alpha}\leq$
$\displaystyle C_{n}\frac{\mu_{n}}{\lambda_{n}^{1-\alpha}}.$ (9.120)

######  Lemma 9.4 (The transportation part).

$\displaystyle\Big{\|}\mathcal{R}\Big{(}\partial_{t}w_{n}+\sum_{l\in
Z^{3}}\frac{l}{\mu_{n}}\cdot\nabla w_{n}\Big{)}\Big{\|}_{\alpha}\leq
C_{n}\frac{\mu_{n}^{2}}{\lambda_{n}^{1-\alpha}}.$ (9.121)

######  Lemma 9.5 (Estimates on error part I).

$\displaystyle\|N_{n}\|_{0}\leq\frac{C_{n}}{\mu_{n}}.$ (9.122)

######  Lemma 9.6 (Estimates on error part II).

$\displaystyle\|w_{no}\otimes w_{noc}+w_{noc}\otimes w_{no}+w_{noc}\otimes
w_{noc}\|_{0}\leq C_{n}\frac{\mu_{n}}{\lambda_{n}}.$ (9.123)

The proof of the above four lemmas are similar to that of Lemma 6.3, Lemma
6.4, Lemma 6.5 and Lemma 6.6 respectively, we omit it here.

For $n\geq 4$, $\chi_{n}=0$, then

$\mathcal{R}\Big{(}\Big{(}\chi_{n}-\fint_{T^{3}}{\chi_{n}}dx\Big{)}e_{3}\Big{)}\Big{)}=0.$

Finally we obtain

$\displaystyle\|\delta\mathring{R}_{0n}\|_{0}\leq
C_{n}\Big{(}\frac{\mu_{n}^{3}}{\lambda_{n}^{1-\alpha}}+\frac{1}{\mu_{n}}\Big{)},\qquad\forall\alpha\in(0,1).$
(9.124)

From the above estimates and constructions of $v_{0n},p_{0n}$, we also
conclude that

$\displaystyle\|v_{0n}-v_{0(n-1)}\|_{0}\leq$
$\displaystyle\frac{M\sqrt{\delta}}{2L}+C_{n}\frac{\mu_{n}}{\lambda_{n}},$
(9.125) $\displaystyle\|p_{0n}-p_{0(n-1)}\|_{0}\leq$ $\displaystyle
C_{n}\Big{(}\frac{1}{\mu_{n}}+\frac{\mu_{n}}{\lambda_{n}}\Big{)}.$ (9.126)

###  9.2. Estimates on $\delta f_{0n}$

Recall that when $n=2,3$, we have

$\displaystyle\delta f_{0n}=$
$\displaystyle\mathcal{G}(\hbox{div}K_{n})+\mathcal{G}(w_{n}\cdot\nabla\theta_{0(n-1)})+\mathcal{G}\Big{(}\partial_{t}\chi_{n}+\sum_{l\in
Z^{3}}\Big{(}\frac{l}{\mu_{1}}\cdot\nabla\Big{)}\chi_{nl}\Big{)}$
$\displaystyle+\sum_{l\in
Z^{3}}\Big{(}v_{0}-\frac{l}{\mu_{n}}\Big{)}\chi_{nl}+w_{noc}\chi_{n}.$

where we used the fact $\mathcal{G}(f(t))=0.$

As before, we split $\delta f_{0n}$ into three parts:
(1) the Oscillation Part:

$\displaystyle\mathcal{G}(\hbox{div}K_{n})+\mathcal{G}(w_{n}\cdot\nabla\theta_{0(n-1)}).$

(2) the transportation part:

$\displaystyle\mathcal{G}\Big{(}\partial_{t}\chi_{n}+\sum_{l\in
Z^{3}}\Big{(}\frac{l}{\mu_{1}}\cdot\nabla\Big{)}\chi_{nl}\Big{)}.$

(3) the error part:

$\displaystyle w_{noc}\chi_{n}+\sum_{l\in
Z^{3}}\Big{(}v_{0}-\frac{l}{\mu_{n}}\Big{)}\chi_{nl}.$

When $n\geq 4$, we have

$\displaystyle\delta f_{0n}=\mathcal{G}(w_{n}\cdot\nabla\theta_{03}).$

First, we estimate $\delta f_{0n}$ for $n=2,3$.

######  Lemma 9.7 (The Oscillation Part).

$\displaystyle\|\mathcal{G}({\rm
div}K_{n})+\mathcal{G}(w_{n}\cdot\nabla\theta_{0(n-1)})\|_{\alpha}\leq
C_{n}\frac{\mu_{n}^{3}}{\lambda_{n}^{1-\alpha}}.$ (9.127)

######  Lemma 9.8 (The transportation Part).

$\displaystyle\Big{\|}\mathcal{G}\Big{(}\partial_{t}\chi_{n}+\sum_{l\in
Z^{3}}\Big{(}\frac{l}{\mu_{n}}\cdot\nabla\Big{)}\chi_{nl}\Big{)}\Big{\|}_{\alpha}\leq
C_{n}\frac{\mu_{n}^{2}}{\lambda_{n}^{1-\alpha}}.$ (9.128)

######  Lemma 9.9 (The error Part).

$\displaystyle\Big{\|}w_{1oc}\chi_{n}+\sum_{l\in
Z^{3}}\Big{(}v_{0(n-1)}-\frac{l}{\mu_{n}}\Big{)}\chi_{nl}\Big{\|}_{0}\leq
C_{n}\Big{(}\frac{\mu_{n}^{2}}{\lambda_{n}}+\frac{1}{\mu_{n}}\Big{)}.$ (9.129)

The proof of the above four lemmas are similar to that of Lemma 6.7, Lemma
6.8, Lemma 6.9 respectively, we omit it here.
Then we conclude that for $n=2,3$

$\displaystyle\|\delta f_{0n}\|_{0}\leq
C_{n}\Big{(}\frac{\mu_{n}^{3}}{\lambda_{n}^{1-\alpha}}+\frac{1}{\mu_{n}}\Big{)}.$
(9.130)

For $n\geq 4$, since

$\delta f_{0n}=\mathcal{G}(w_{n}\cdot\nabla\theta_{03}),$

then

$\displaystyle\|\delta f_{0n}\|_{\alpha}\leq
C_{n}\frac{\mu_{n}}{\lambda_{n}^{1-\alpha}}.$

Therefore, for any $2\leq n\leq L$, we have

$\displaystyle\|\delta f_{0n}\|_{\alpha}\leq
C_{n}\Big{(}\frac{\mu_{n}^{3}}{\lambda_{n}^{1-\alpha}}+\frac{1}{\mu_{n}}\Big{)}.$
(9.131)

Furthermore, as in (6.84), we have

$\displaystyle\|\theta_{0n}-\theta_{0(n-1)}\|_{0}\leq$
$\displaystyle\frac{M\sqrt{\delta}}{2L}+C_{n}\frac{\mu_{n}^{2}}{\lambda_{n}}.$
(9.132)

Where we used

$\displaystyle\sup_{t}\Big{|}\int_{T^{3}}\chi_{n}dx\Big{|}\leq\sum\limits_{|l|\leq
C_{n}\mu_{n}}C_{n}\frac{\mu_{n}}{\lambda_{n}}\leq
C_{n}\frac{\mu_{n}^{2}}{\lambda_{n}}.$

As before, we have

$\displaystyle
R_{0n}:=\bar{\rho}(t)\sum\limits_{i=n+1}^{L}\Big{(}\gamma_{k_{i}}\Big{(}\frac{R_{0}}{\bar{\rho}(t)}\Big{)}\Big{)}^{2}\Big{(}Id-\frac{k_{i}}{|k_{i}|}\otimes\frac{k_{i}}{|k_{i}|}\Big{)}+\sum\limits_{i=1}^{n}\delta\mathring{R}_{0i},$
(9.133) $\displaystyle
f_{0n}:=\left\\{\begin{array}[]{ll}\sum\limits_{i=n+1}^{3}g_{k_{i}}(f_{0}(x,t))A_{k_{i}}+\sum\limits_{i=1}^{n}\delta
f_{0i},\quad n=2,3\\\ \sum\limits_{i=1}^{n}\delta f_{0i},\quad n\geq
4,\end{array}\right.$ (9.136)

and

$\displaystyle\|v_{0n}-v_{0(n-1)}\|_{0}\leq$
$\displaystyle\frac{M\sqrt{\delta}}{2L}+C_{n}\frac{\mu_{n}}{\lambda_{n}},$
(9.137) $\displaystyle\|p_{0n}-p_{0(n-1)}\|_{0}\leq$ $\displaystyle
C_{n}\Big{(}\frac{1}{\mu_{n}}+\frac{\mu_{n}}{\lambda_{n}}\Big{)},$ (9.138)
$\displaystyle\|v_{0n}-v_{0(n-1)}\|_{0}\leq$
$\displaystyle\frac{M\sqrt{\delta}}{2L}+C_{n}\frac{\mu_{n}^{2}}{\lambda_{n}},$
(9.139) $\displaystyle\|\delta\mathring{R}_{0n}\|_{0}\leq$ $\displaystyle
C_{n}\Big{(}\frac{\mu_{n}^{3}}{\lambda_{n}^{1-\alpha}}+\frac{1}{\mu_{n}}\Big{)},$
(9.140) $\displaystyle\|\delta f_{0n}\|_{0}\leq$ $\displaystyle
C_{n}\Big{(}\frac{\mu_{n}^{3}}{\lambda_{n}^{1-\alpha}}+\frac{1}{\mu_{n}}\Big{)}.\qquad\forall\alpha\in(0,1).$
(9.141)

In particular, we conclude that

$\displaystyle R_{0L}=$
$\displaystyle\sum\limits_{i=1}^{L}\delta\mathring{R}_{01},$ (9.142)
$\displaystyle f_{0L}=$ $\displaystyle\sum\limits_{i=1}^{L}\delta f_{0i}.$
(9.143)

We rewrite $R_{0L}$ as $\mathring{R}_{0L}$, then it’s a symmetric and trace-
free matrix.

##  10\. Energy Estimate

In this section, we assume that $\lambda_{n}\ll\lambda_{n+1}$ and
$C_{n}<C_{n+1}$, then the functions $v_{0(n-1)},~{}w_{n}$ oscillate very
slowly compared with $w_{n+1}$.

######  Lemma 10.1 (Estimates on the energy).

$\displaystyle\Big{|}e(t)\Big{(}1-\frac{\delta}{2}\Big{)}-\int_{T^{3}}|v_{0L}|^{2}(x,t)dx\Big{|}\leq\sum\limits_{i=1}^{L}C_{i}\frac{\mu^{3}_{i}}{\lambda_{i}}.$
(10.144)

###### Proof.

First we have

$\displaystyle\Big{|}e(t)\Big{(}1-\frac{\delta}{2}\Big{)}-\int_{T^{3}}|v_{0L}|^{2}(x,t)dx\Big{|}$
$\displaystyle\leq$
$\displaystyle\Big{|}e(t)\Big{(}1-\frac{\delta}{2}\Big{)}-\int_{T^{3}}\Big{|}v_{0}+\sum\limits_{i=1}^{L}(w_{i0}+w_{i0c})\Big{|}^{2}(x,t)dx\Big{|}$
$\displaystyle\leq$
$\displaystyle\Big{|}e(t)\Big{(}1-\frac{\delta}{2}\Big{)}-\int_{T^{3}}|v_{0}|^{2}(x,t)dx-\sum\limits_{i=1}^{L}\int_{T^{3}}tr(w_{io}\otimes
w_{io})dx\Big{|}$
$\displaystyle+\Big{|}2\sum\limits_{i=1}^{L}\int_{T^{3}}v_{0}\cdot
w_{i}dx+\sum\limits_{1\leq i\neq j\leq L}\int_{T^{3}}tr(w_{i}\otimes
w_{j})dx\Big{|}$
$\displaystyle+\Big{|}\sum\limits_{i=1}^{L}\int_{T^{3}}tr(w_{io}\otimes
w_{ioc}+w_{ioc}\otimes w_{io}+w_{ioc}\otimes w_{ioc})dx\Big{|}$
$\displaystyle\leq$ $\displaystyle I+II+III.$

We estimate each terms separately. From the stationary phase estimates
(A.175), we have

$\displaystyle
II\leq\sum\limits_{i=1}^{L}C_{i}\frac{\mu^{2}_{i}}{\lambda_{i}}.$ (10.145)

where we used the fact that for $i<j$, the oscillation of $w_{j}$ is faster
than that of $w_{i}$ and then,

$\displaystyle\Big{|}\int_{T^{3}}w_{i}\otimes w_{j}(x,t)dx\Big{|}\leq
C_{j}\frac{\mu_{j}^{2}}{\lambda_{j}}.$

From the smallness of $w_{ioc}$, we have

$\displaystyle III\leq\sum\limits_{i=1}^{L}C_{i}\frac{\mu_{i}}{\lambda_{i}}.$
(10.146)

To estimate $I$, we first compute the term
$tr\Big{(}\sum\limits_{i=1}^{L}w_{i0}\otimes w_{i0}\Big{)}$.

$\displaystyle w_{i0}\otimes w_{i0}=$ $\displaystyle 2\sum\limits_{l\in
Z^{3}}b^{2}_{il}Re(B_{k_{i}}\otimes B_{-k_{i}})+\sum\limits_{l\in
Z^{3}}b^{2}_{il}(x,t)\Big{(}B_{k_{i}}\otimes
B_{k_{i}}e^{2i\lambda_{n}2^{|l|}k_{i}\cdot(x-\frac{l}{\mu_{i}}t)}$
$\displaystyle+B_{-k_{i}}\otimes
B_{-k_{i}}e^{-2i\lambda_{i}2^{|l|}k_{i}\cdot(x-\frac{l}{\mu_{i}}t)}\Big{)}+\sum\limits_{l,l^{\prime}\in
Z^{3},l\neq
l^{\prime}}b_{il}b_{il^{\prime}}(x,t)\Big{(}B_{k_{i}}e^{i\lambda_{i}2^{|l|}k_{i}\cdot(x-\frac{l}{\mu_{i}}t)}$
$\displaystyle+B_{-k_{i}}e^{-i\lambda_{i}2^{|l|}k_{i}\cdot(x-\frac{l}{\mu_{i}}t)}\Big{)}\otimes\Big{(}B_{k_{i}}e^{i\lambda_{i}2^{|l^{\prime}|}k_{i}\cdot(x-\frac{l^{\prime}}{\mu_{i}}t)}+B_{-k_{i}}e^{-i\lambda_{i}2^{|l^{\prime}|}k_{i}\cdot(x-\frac{l^{\prime}}{\mu_{i}}t)}\Big{)}$
$\displaystyle=$ $\displaystyle 2\sum\limits_{l\in
Z^{3}}b^{2}_{il}Re(B_{k_{i}}\otimes B_{-k_{i}})+G_{i}$ $\displaystyle=$
$\displaystyle\bar{\rho}(t)\gamma^{2}_{k_{i}}\Big{(}\frac{R_{0}(x,t)}{\bar{\rho}(t)}\Big{)}\Big{(}Id-\frac{k_{i}}{|k_{i}|}\otimes\frac{k_{i}}{|k_{i}|}\Big{)}+G_{i},$

where

$\displaystyle G_{i}=$ $\displaystyle\sum\limits_{l\in
Z^{3}}b^{2}_{il}(x,t)\Big{(}B_{k_{i}}\otimes
B_{k_{i}}e^{2i\lambda_{i}2^{|l|}k_{i}\cdot(x-\frac{l}{\mu_{i}}t)}+B_{-k_{i}}\otimes
B_{-k_{i}}e^{-2i\lambda_{i}2^{|l|}k_{i}\cdot(x-\frac{l}{\mu_{i}}t)}\Big{)}$
$\displaystyle+\sum\limits_{l,l^{\prime}\in Z^{3},l\neq
l^{\prime}}b_{il}b_{il^{\prime}}\Big{(}B_{k_{i}}\otimes
B_{k_{i}}e^{i\lambda_{1}(2^{|l|}+2^{|l^{\prime}|})k_{1}\cdot
x-ig_{1,l,l^{\prime}}(t)}+B_{k_{i}}\otimes
B_{-k_{i}}e^{i\lambda_{1}(2^{|l|}-2^{|l^{\prime}|})k_{i}\cdot
x-i\overline{g}_{1,l,l^{\prime}}(t)}$ $\displaystyle+B_{-k_{i}}\otimes
B_{k_{i}}e^{i\lambda_{1}(2^{|l^{\prime}|}-2^{|l|})k_{i}\cdot
x+i\overline{g}_{1,l,l^{\prime}}(t)}+B_{-k_{i}}\otimes
B_{-k_{i}}e^{-i\lambda_{1}(2^{|l|}+2^{|l^{\prime}|})k_{i}\cdot
x-ig_{1,l,l^{\prime}}(t)}\Big{)},$

is the oscillatory part.

Then, from (4.41) and $tr\mathring{R_{0}}=0$, we have

$\displaystyle tr\Big{(}\sum\limits_{i=1}^{L}w_{i0}\otimes w_{i0}\Big{)}$
$\displaystyle=$ $\displaystyle
tr\Big{(}\sum\limits_{i=1}^{L}\bar{\rho}(t)\gamma^{2}_{k_{i}}\Big{(}\frac{R_{0}(x,t)}{\bar{\rho}(t)}\Big{)}\Big{(}Id-\frac{k_{i}}{|k_{i}|}\otimes\frac{k_{i}}{|k_{i}|}\Big{)}+tr\Big{(}\sum\limits_{i=1}^{L}G_{i}\Big{)}$
$\displaystyle=$ $\displaystyle
3\rho(t)+tr\Big{(}\sum\limits_{i=1}^{L}G_{i}\Big{)}$

From the definition (4.43) of $\bar{\rho}(t)$, we have

$\displaystyle
I=\Big{|}\int_{T^{3}}tr\Big{(}\sum\limits_{i=1}^{L}G_{i}\Big{)}dx\Big{|}.$

By Lemma 6.1 and stationary phase estimate (A.175), we have

$\displaystyle
I\leq\sum\limits_{i=1}^{L}C_{i}\frac{\mu^{3}_{i}}{\lambda_{i}}.$ (10.147)

Finally, from the estimates (10), (10.145), and (10.146), we conclude that

$\displaystyle\Big{|}e(t)\Big{(}1-\frac{\delta}{2}\Big{)}-\int_{T^{3}}|v_{0L}(x,t)|^{2}dx\Big{|}\leq\sum\limits_{i=1}^{L}C_{i}\frac{\mu^{3}_{i}}{\lambda_{i}}.$

MM $\square$

##  11\. Proof of proposition 2.1

In this section, we collect the estimates from the preceding sections. From
these estimates , we can prove Proposition 2.1 by choosing the appropriate
parameters $\mu_{n},\lambda_{n}$ for $1\leq n\leq L$.

###### Proof.

By the definition (9.142), (9.143) and the estimates (9.137)-(9.141),
(10.144), we have
$v_{0N},~{}p_{0N},~{}\theta_{0N},~{}\mathring{R}_{0N},~{}f_{0N}$, which solve
system (2.2) and satisfy

$\displaystyle\|\mathring{R}_{0L}\|_{0}\leq$
$\displaystyle\sum\limits_{n=1}^{L}C_{n}\Big{(}\frac{\mu_{n}^{3}}{\lambda_{n}^{1-\alpha}}+\frac{1}{\mu_{n}}\Big{)},$
$\displaystyle\|f_{0L}\|_{0}\leq$
$\displaystyle\sum\limits_{n=1}^{L}C_{n}\Big{(}\frac{\mu_{n}^{3}}{\lambda_{n}^{1-\alpha}}+\frac{1}{\mu_{n}}\Big{)},\qquad\forall\alpha\in(0,1).$
$\displaystyle\|v_{0L}-v_{0}\|_{0}\leq$
$\displaystyle\frac{M\sqrt{\delta}}{2}+\sum\limits_{n=1}^{L}C_{n}\frac{\mu_{n}}{\lambda_{n}},$
$\displaystyle\|p_{0L}-p_{0}\|_{0}\leq$
$\displaystyle\sum\limits_{n=1}^{L}C_{n}\Big{(}\frac{\mu_{n}}{\lambda_{n}}+\frac{1}{\mu_{n}}\Big{)},$
$\displaystyle\|\theta_{0L}-\theta_{0}\|_{0}\leq$
$\displaystyle\frac{M\sqrt{\delta}}{2}+\sum\limits_{n=1}^{L}C_{n}\frac{\mu_{n}^{2}}{\lambda_{n}},$
$\displaystyle\Big{|}e(t)\Big{(}1-\frac{\delta}{2}\Big{)}-$
$\displaystyle\int_{T^{3}}|v_{0L}(x,t)|^{2}dx\Big{|}\leq\sum\limits_{n=1}^{L}C_{n}\frac{\mu^{3}_{n}}{\lambda_{n}}.$

We let $\mu_{n}=\sqrt[5]{\lambda}_{n}$ , then we have

$\displaystyle\|\mathring{R}_{0L}\|_{0}\leq$
$\displaystyle\sum\limits_{n=1}^{L}C_{n}\Big{(}\frac{1}{\lambda_{n}^{\frac{2}{5}-\alpha}}+\frac{1}{\sqrt[5]{\lambda}_{n}}\Big{)},$
$\displaystyle\|f_{0L}\|_{0}\leq$
$\displaystyle\sum\limits_{n=1}^{L}C_{n}\Big{(}\frac{1}{\lambda_{n}^{\frac{2}{5}-\alpha}}+\frac{1}{\sqrt[5]{\lambda}_{n}}\Big{)},\qquad\forall\alpha\in(0,1).$

and

$\displaystyle\|v_{0L}-v_{0}\|_{0}\leq$
$\displaystyle\frac{M\sqrt{\delta}}{2}+\sum\limits_{n=1}^{L}\frac{C_{n}}{\lambda_{n}^{\frac{4}{5}}},$
$\displaystyle\|p_{0L}-p_{0}\|_{0}\leq$
$\displaystyle\sum\limits_{n=1}^{L}C_{n}\Big{(}\frac{1}{\lambda_{n}^{\frac{4}{5}}}+\frac{1}{\sqrt[5]{\lambda}_{n}}\Big{)},$
$\displaystyle\|\theta_{0L}-\theta_{0}\|_{0}\leq$
$\displaystyle\frac{M\sqrt{\delta}}{2}+\sum\limits_{n=1}^{L}\frac{C_{n}}{\lambda_{n}^{\frac{3}{5}}},$
$\displaystyle\Big{|}e(t)\Big{(}1-\frac{\delta}{2}\Big{)}-$
$\displaystyle\int_{T^{3}}|v_{0L}|^{2}(x,t)dx\Big{|}\leq\sum\limits_{n=1}^{L}\frac{C_{n}}{\lambda_{n}^{\frac{2}{5}}}.$

Let $\lambda_{1}\ll\lambda_{2}\ll\cdot\cdot\cdot\ll\lambda_{N}$ sufficiently
large, such that

$\frac{C_{n}}{\sqrt[5]{\lambda}_{n}}\leq\hbox{min}\Big{(}\frac{\eta\delta}{10L},\frac{M\sqrt{\delta}}{10L},\frac{\delta}{8L}{\rm
min}_{t}e(t)\Big{)}.$

Set $\alpha<\frac{1}{5}$, we conclude that

$\displaystyle\|\mathring{R}_{0L}\|_{0}\leq$
$\displaystyle\frac{\eta\delta}{2},$ $\displaystyle\|f_{0L}\|_{0}\leq$
$\displaystyle\frac{\eta\delta}{2},$ $\displaystyle\|v_{0L}-v_{0}\|_{0}\leq$
$\displaystyle M\sqrt{\delta},$ $\displaystyle\|p_{0L}-p_{0}\|_{0}\leq$
$\displaystyle M\sqrt{\delta},$
$\displaystyle\|\theta_{0L}-\theta_{0}\|_{0}\leq$ $\displaystyle
M\sqrt{\delta},$ $\displaystyle\Big{|}e(t)\Big{(}1-\frac{\delta}{2}\Big{)}-$
$\displaystyle\int_{T^{3}}|v_{0L}|^{2}(x,t)dx\Big{|}\leq\frac{\delta}{8}e(t).$

Finally, we set

$\tilde{v}=v_{0L},~{}\tilde{p}=p_{0L},~{}\tilde{\theta}=\theta_{0L},~{}\mathring{\tilde{R}}=\mathring{R}_{0L},~{}\tilde{f}=f_{0L}.$

then
$\tilde{v},~{}\tilde{p},~{}\tilde{\theta},~{}\mathring{\tilde{R}},~{}\tilde{f}$
are what we need in our Proposition (2.1). MM $\square$

##  12\. Proof of proposition 1.2

In this section, we prove Proposition 2.2. The constructions of
$\tilde{v},~{}\tilde{p},~{}\tilde{\theta},~{}\mathring{\tilde{R}},~{}\tilde{f}$
are similar to the construction in Propositions 2.1. Again, we need multi-
steps iterations. We only give the first step, the others could be constructed
by inductions as in proposition 2.1. For convenience, we use the same
notations as in Proposition 2.1.

###  12.1. The main perturbation $w_{1o}$

The construction is very similar to that of given in subsection 4.1, beside
the choice of amplitude. First, the functions
$\alpha_{l},\gamma_{k_{1}},g_{k_{1}}$ are same as before. We set

$\displaystyle R_{0}(x,t):=\delta Id-\mathring{R}_{0}(x,t).$

For any $l\in Z^{3}$, we set

$\displaystyle b_{1l}(x,t):=$
$\displaystyle\sqrt{\delta}\alpha_{l}(\mu_{1}v_{0})\gamma_{k_{1}}\Big{(}\frac{R(x,t)}{\delta}\Big{)},$
(12.148) $\displaystyle B_{k_{1}}:=$ $\displaystyle
A_{k_{1}}+i\frac{k_{1}}{|k_{1}|}\times A_{k_{1}},$ (12.149)

and $l$-perturbations

$\displaystyle
w_{1ol}:=b_{1l}(x,t)\Big{(}B_{k_{1}}e^{i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu_{1}}t)}+B_{-k_{1}}e^{-i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu_{1}}t)}\Big{)}.$
(12.150)

Finally, we define 1-th perturbation

$\displaystyle w_{1o}:=\sum_{l\in Z^{3}}w_{1ol}.$ (12.151)

Obviously, $w_{1ol},w_{1o}$ are all real 3-dimensional vector function.

We remark that
$\hbox{supp}\alpha_{l}\cap\hbox{supp}\alpha_{l^{\prime}}=\emptyset$ if
$|l-l^{\prime}|\geq 2$, thus the above summation is meaningful.

###  12.2. The constants $\eta$ and $M$

First, $b_{1l}$ is well-defined only if $\frac{R}{\delta}\in B_{r_{0}}(Id)$
where $r_{0}$ is given in geometric Lemma 3.1. However, since

$\displaystyle\Big{\|}\frac{\mathring{R}_{0}}{\delta}-Id\Big{\|}_{0}\leq\frac{\|\mathring{R}_{0}\|_{0}}{\delta}\leq\eta,$
(12.152)

so it suffices to set

$\displaystyle\eta:=\frac{1}{2}r_{0}.$

Next notice that there exist two positive constants $c_{10}$ and $c_{20}$ such
that

$\displaystyle c_{10}\leq\gamma_{k_{1}}\Big{(}\frac{R(x,t)}{\delta}\Big{)}\leq
c_{20}.$

Thus there exists an absolute constant $M>1$ such that

$\displaystyle\|b_{1l}\|_{0}\leq\frac{M\sqrt{\delta}}{10L}.$

Moreover, from the constructions of $\alpha_{l}$, we know that

$\displaystyle\|w_{1o}\|_{0}\leq\frac{M\sqrt{\delta}}{2L}.$ (12.153)

###  12.3. The correction $w_{1oc}$ and the perturbation $w_{1}$ and
$\chi_{1}$

The following constructions are same as in subsection 4.3. We define the
$l$-corrections

$\displaystyle w_{1ocl}:=$
$\displaystyle\frac{1}{\lambda_{1}\lambda_{0}}\Big{(}\frac{\nabla
b_{1l}(x,t)\times
B_{k_{1}}}{2^{|l|}}e^{i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu_{1}}t)}+\frac{\nabla
b_{1l}(x,t)\times
B_{-{k_{1}}}}{2^{|l|}}e^{-i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu_{1}}t)}\Big{)},$
(12.154)

then define 1-th correction

$\displaystyle w_{1oc}:=\sum_{l\in Z^{3}}w_{1ocl},$ (12.155)

Finally, we define 1-th perturbation

$\displaystyle w_{1}:=w_{1o}+w_{1oc}$ (12.156)

Thus, if we denote $w_{1l}$ by

$\displaystyle w_{1l}:=$ $\displaystyle w_{1ol}+w_{1ocl}$ $\displaystyle=$
$\displaystyle\frac{1}{\lambda_{1}\lambda_{0}}\hbox{curl}\Big{(}\frac{b_{1l}(x,t)B_{k_{1}}}{2^{|l|}}e^{i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu_{1}}t)}+\frac{b_{1l}(x,t)B_{-{k_{1}}}}{2^{|l|}}e^{-i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu_{1}}t)}\Big{)},$
(12.157)

then

$\displaystyle w_{1}=\sum\limits_{l\in
Z^{3}}w_{1l},\qquad\hbox{div}w_{1l}=0,\qquad{\rm div}w_{1}=0.$

Moreover, if we set

$\displaystyle B_{1lk_{1}}=$ $\displaystyle
b_{1l}(x,t)B_{k_{1}}+\frac{1}{\lambda_{1}\lambda_{0}}\frac{\nabla
b_{1l}(x,t)\times B_{k_{1}}}{2^{|l|}},$ $\displaystyle B_{-1lk_{1}}=$
$\displaystyle
b_{1l}(x,t)B_{-k_{1}}+\frac{1}{\lambda_{1}\lambda_{0}}\frac{\nabla
b_{1l}(x,t)\times B_{-k_{1}}}{2^{|l|}},$

then

$\displaystyle
w_{1l}=B_{1lk_{1}}e^{i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu_{1}}t)}+B_{-1lk_{1}}e^{-i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu_{1}}t)}.$

Thus, we complete the constructions of perturbation $w_{1}$.

To construct $\chi_{1}$, we first denote $\beta_{1l}$ by

$\displaystyle\beta_{1l}(x,t):=\frac{\alpha_{l}(\mu_{1}v_{0})}{2\sqrt{\delta}}\frac{g_{k_{1}}(-f_{0}(x,t))}{\gamma_{k_{1}}\big{(}\frac{R_{0}(x,t)}{\delta}\big{)}},$
(12.158)

then define $l$-perturbations

$\displaystyle\chi_{1l}(x,t):=\beta_{1l}(x,t)\Big{(}e^{i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu_{1}}t)}+e^{-i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu_{1}}t)}\Big{)}.$
(12.159)

Finally, we define the 1-th perturbation

$\displaystyle\chi_{1}(x,t):=\sum_{l\in Z^{3}}\chi_{1l}.$ (12.160)

Thus, $\chi_{1l}$ and $\chi_{1}$ are both real scalar functions, and as the
perturbation $w_{1}$, the summation in the definition of $\chi_{1}$ is
meaningful.

Thus, from the assumptions of proposition 2.2, we know that

$\|\beta_{1l}\|_{0}\leq\frac{M\sqrt{\delta}}{10L},$

therefore, we have

$\|\chi_{1}\|_{0}\leq\frac{M\sqrt{\delta}}{2L}.$

###  12.4. The constructions of $v_{01}$, $p_{01}$, $\theta_{01}$, $f_{01}$,
$\mathring{R}_{01}$

The constructions in this section are same as in subsection 4.4. First, we
denote $M_{1}$ by

$\displaystyle M_{1}=$ $\displaystyle\sum\limits_{l\in
Z^{3}}b^{2}_{1l}(x,t)\Big{(}B_{k_{1}}\otimes
B_{k_{1}}e^{2i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu_{1}}t)}+B_{-k_{1}}\otimes
B_{-k_{1}}e^{-2i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu_{1}}t)}\Big{)}$
$\displaystyle+\sum\limits_{l,l^{\prime}\in Z^{3},l\neq
l^{\prime}}w_{1ol}w_{1ol^{\prime}}(x,t),$ (12.161)

and $N_{1},K_{1}$ by

$\displaystyle N_{1}=$ $\displaystyle\sum_{l\in
Z^{3}}\Big{[}w_{1l}\otimes\Big{(}v_{0}-\frac{l}{\mu_{1}}\Big{)}+\Big{(}v_{0}-\frac{l}{\mu_{1}}\Big{)}\otimes
w_{1l}\Big{]},$ $\displaystyle K_{1}=$ $\displaystyle\sum\limits_{l\in
Z^{3}}\beta_{1l}(x,t)b_{1l}(x,t)\Big{(}B_{k_{1}}e^{2i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu_{1}}t)}+B_{-k_{1}}e^{-2i\lambda_{1}2^{|l|}k_{1}\cdot(x-\frac{l}{\mu_{1}}t)}\Big{)}+\sum\limits_{l,l^{\prime}\in
Z^{3},l\neq l^{\prime}}w_{1ol}\chi_{1l^{\prime}}.$ (12.162)

Notice that $N_{1}$ is a symmetric matrix. Then we define

$\displaystyle v_{01}:=v_{0}+w_{1},$ $\displaystyle
p_{01}:=p_{0}-\frac{2w_{1o}\cdot w_{1oc}+|w_{1oc}|^{2}}{3}-tr(N_{1}),$
$\displaystyle\theta_{01}:=\theta_{0}+\chi_{1}-\fint_{T^{3}}{\chi_{1}}dx,$
$\displaystyle R_{01}:=-R_{0}(x,t)+2\sum\limits_{l\in
Z^{3}}b^{2}_{1l}(x,t)Re(B_{k_{1}}\otimes
B_{-{k_{1}}})+\delta\mathring{R}_{01},$ $\displaystyle
f_{01}:=f_{0}+2\sum\limits_{l\in
Z^{3}}\beta_{1l}(x,t)b_{1l}(x,t)A_{k_{1}}+\delta f_{01},$

where

$\displaystyle\delta\mathring{R}_{01}=$
$\displaystyle\mathcal{R}(\hbox{div}M_{1})+N_{1}-tr(N_{1})Id+\mathcal{R}\Big{\\{}\partial_{t}w_{1}+\hbox{div}\Big{[}\sum_{l\in
Z^{3}}\Big{(}w_{1l}\otimes\frac{l}{\mu_{1}}+\frac{l}{\mu_{1}}\otimes
w_{1l}\Big{)}\Big{]}\Big{\\}}$ $\displaystyle+(w_{1o}\otimes
w_{1oc}+w_{1oc}\otimes w_{1o}+w_{1oc}\otimes w_{1oc})-\frac{2w_{1o}\cdot
w_{1oc}+|w_{1oc}|^{2}}{3}Id$
$\displaystyle-\mathcal{R}\Big{(}\Big{(}\chi_{1}-\fint_{T^{3}}{\chi_{1}}dx\Big{)}e_{3}\Big{)},$
(12.163)

and

$\displaystyle\delta f_{01}=$
$\displaystyle\mathcal{G}(\hbox{div}K_{1})+\mathcal{G}(w_{1}\cdot\nabla\theta_{0})+\mathcal{G}\Big{(}\partial_{t}\chi_{1}+\sum_{l\in
Z^{3}}\Big{(}\frac{l}{\mu_{1}}\cdot\nabla\Big{)}\chi_{1l}\Big{)}$
$\displaystyle+\sum_{l\in
Z^{3}}\Big{(}v_{0}-\frac{l}{\mu_{1}}\Big{)}\chi_{1l}+w_{1oc}\chi_{1}.$
(12.164)

Obviously,

$\displaystyle\hbox{div}v_{1}=\hbox{div}v_{0}+\hbox{div}w_{1}=0.$

As in subsection 4.4, we have

$\displaystyle\hbox{div}R_{01}=$
$\displaystyle\partial_{t}v_{01}+\hbox{div}(v_{01}\otimes v_{01})+\nabla
p_{01}-\theta_{01}e_{3},$

and

$\displaystyle\hbox{div}f_{01}=$
$\displaystyle\partial_{t}\theta_{01}+\hbox{div}(v_{01}\theta_{01}).$ (12.165)

Thus the functions $v_{01},p_{01},\theta_{01},R_{01},f_{01}$ solve the system
(2.2).

###  12.5. The representation of $-R_{0}(x,t)+2\sum\limits_{l\in
Z^{3}}b^{2}_{1l}(x,t)Re(B_{k_{1}}\otimes B_{-{k_{1}}})$

This subsection is similar to that of subsection 5.1. From the definition of
$b_{1l}(x,t)$ , we have

$\displaystyle 2\sum\limits_{l\in Z^{3}}b^{2}_{1l}(x,t)Re(B_{k_{1}}\otimes
B_{-{k_{1}}})=$ $\displaystyle\delta\sum\limits_{l\in
Z^{3}}\alpha_{l}^{2}(\mu_{1}v_{0})\gamma_{k_{1}}^{2}\Big{(}\frac{R_{0}(x,t)}{\delta}\Big{)}\Big{(}Id-\frac{k_{1}}{|k_{1}|}\otimes\frac{k_{1}}{|k_{1}|}\Big{)}$
$\displaystyle=$
$\displaystyle\delta\gamma_{k_{1}}^{2}\Big{(}\frac{R_{0}(x,t)}{\delta}\Big{)}\Big{(}Id-\frac{k_{1}}{|k_{1}|}\otimes\frac{k_{1}}{|k_{1}|}\Big{)}.$

Where we used $\sum\limits_{l\in Z^{3}}\alpha_{l}^{2}=1$. Moreover, from the
Lemma 3.1 and estimate (12.152), we can decompose $R_{0}$

$\displaystyle\frac{R_{0}}{\delta}=\sum\limits_{i=1}^{N}\Big{(}\gamma_{k_{i}}\Big{(}\frac{R_{0}}{\delta}\Big{)}\Big{)}^{2}\Big{(}Id-\frac{k_{i}}{|k_{i}|}\otimes\frac{k_{i}}{|k_{i}|}\Big{)}.$

Therefore

$\displaystyle-R_{0}(x,t)+2\sum\limits_{l\in
Z^{3}}b^{2}_{l}(x,t)Re(B_{k_{1}}\otimes
B_{-{k_{1}}})=-\delta\sum\limits_{i=2}^{N}\Big{(}\gamma_{k_{i}}\Big{(}\frac{R_{0}}{\delta}\Big{)}\Big{)}^{2}\Big{(}Id-\frac{k_{i}}{|k_{i}|}\otimes\frac{k_{i}}{|k_{i}|}\Big{)}.$

Meanwhile, we have

$\displaystyle
R_{01}=-\delta\sum\limits_{i=2}^{N}\Big{(}\gamma_{k_{i}}\Big{(}\frac{R_{0}}{\delta}\Big{)}\Big{)}^{2}\Big{(}Id-\frac{k_{i}}{|k_{i}|}\otimes\frac{k_{i}}{|k_{i}|}\Big{)}+\delta\mathring{R}_{01}.$

###  12.6. The representation of $f_{0}+2\sum\limits_{l\in
Z^{3}}\beta_{1l}(x,t)b_{1l}(x,t)A_{k_{1}}$

This subsection is similar to that of subsection 5.2, we omit the detailed
proof, only give results. We have

$\displaystyle f_{0}+2\sum\limits_{l\in
Z^{3}}\beta_{1l}(x,t)b_{1l}(x,t)A_{k_{1}}=\sum\limits_{i=2}^{3}g_{k_{i}}(f_{0}(x,t))A_{k_{i}}.$

Meanwhile, we have

$\displaystyle f_{01}=f_{0}+2\sum\limits_{l\in
Z^{3}}\alpha_{l}(\mu_{1}v_{0})\beta_{1}(x,t)b_{l}(x,t)A_{k_{1}}+\delta
f_{01}=\sum\limits_{i=2}^{3}g_{k_{i}}(f_{0}(x,t))A_{k_{i}}+\delta f_{01}.$

###  12.7. Estimates on $\delta\mathring{R}_{01}$ and $\delta f_{01}$

We summarize the main properties of $b_{1l}$ and $\beta_{1l}$.

######  Lemma 12.1.

For any $|l|\leq C_{1}\mu_{1}$ and $r\geq 0$

$\displaystyle\|b_{1l}\|_{r}\leq C_{1}\mu_{1}^{r},$ (12.166)
$\displaystyle\|\beta_{1l}\|_{r}\leq C_{1}\mu_{1}^{r},$ (12.167)
$\displaystyle\|(\partial_{t}+\frac{l}{\mu_{1}}\cdot\nabla)b_{1l}\|_{r}\leq
C_{1}\mu_{1}^{r+1},$ (12.168)
$\displaystyle\|(\partial_{t}+\frac{l}{\mu_{1}}\cdot\nabla)b_{1l}\|_{r}\leq
C_{1}\mu_{1}^{r+1}.$ (12.169)

###### Proof.

The proof is same as in Lemma 6.1, we omit it here. MM $\square$

######  Lemma 12.2 (Estimates on corrections).

For any $\alpha\in[0,1)$,

$\displaystyle\|w_{1oc}\|_{\alpha}\leq
C_{1}\frac{\mu_{1}}{\lambda_{1}^{1-\alpha}},\qquad$ (12.170)

###### Proof.

The proof is same as in Lemma 6.2. We omit it here. MM $\square$

####  12.7.1. Estimates on $\delta\mathring{R}_{01}$

We may rewrite

$\displaystyle\delta\mathring{R}_{01}=$
$\displaystyle\mathcal{R}(\hbox{div}M_{1})+N_{1}-tr(N_{1})Id+\mathcal{R}\Big{\\{}\partial_{t}w_{1}+\hbox{div}\Big{[}\sum_{l\in
Z^{3}}\Big{(}w_{1l}\otimes\frac{l}{\mu_{1}}+\frac{l}{\mu_{1}}\otimes
w_{1l}\Big{)}\Big{]}\Big{\\}}$ $\displaystyle+(w_{1o}\otimes
w_{1oc}+w_{1oc}\otimes w_{1o}+w_{1oc}\otimes w_{1oc})$
$\displaystyle-\frac{2w_{1o}\cdot
w_{1oc}+|w_{1oc}|^{2}}{3}Id-\mathcal{R}\Big{(}\Big{(}\chi_{1}-\fint_{T^{3}}{\chi_{1}}dx\Big{)}e_{3}\Big{)}$

Again, we split the stress into three parts,
(1) the oscillation part

$\mathcal{R}(\hbox{div}M_{1})-\mathcal{R}\Big{(}\Big{(}\chi_{1}-\fint_{T^{3}}{\chi_{1}}dx\Big{)}e_{3}\Big{)},$

(2) the transportation part

$\displaystyle\mathcal{R}\Big{\\{}\partial_{t}w_{1}+\hbox{div}\Big{[}\sum_{l\in
Z^{3}}\Big{(}w_{1l}\otimes\frac{l}{\mu_{1}}+\frac{l}{\mu_{1}}\otimes
w_{1l}\Big{)}\Big{]}\Big{\\}}=\mathcal{R}\Big{(}\partial_{t}w_{1}+\sum_{l\in
Z^{3}}\Big{(}\frac{l}{\mu_{1}}\cdot\nabla\Big{)}w_{1l}\Big{)},$

(3) the error part

$\displaystyle N_{1}-tr(N_{1})Id+(w_{1o}\otimes w_{1oc}+w_{1oc}\otimes
w_{1o}-w_{1oc}\otimes w_{1oc})-\frac{2w_{1o}\cdot
w_{1oc}+|w_{1oc}|^{2}}{3}Id.$

Again, we will estimate each terms separately.

######  Lemma 12.3 (The oscillation part).

$\displaystyle\|\mathcal{R}({\rm div}M_{1})\|_{\alpha}\leq$ $\displaystyle
C_{1}\frac{\mu_{1}^{3}}{\lambda_{1}^{1-\alpha}},\qquad\forall\alpha\in(0,1)$
$\displaystyle\Big{\|}\mathcal{R}\Big{(}\Big{(}\chi_{1}-\fint_{T^{3}}{\chi_{1}}dx\Big{)}e_{3}\Big{)}\Big{\|}_{\alpha}\leq$
$\displaystyle C_{1}\frac{\mu_{1}}{\lambda_{1}^{1-\alpha}}.$

######  Lemma 12.4 (The transportation part).

$\displaystyle\Big{\|}\mathcal{R}\Big{(}\partial_{t}w_{1}+\sum_{l\in
Z^{3}}\frac{l}{\mu_{1}}\cdot\nabla w_{1l}\Big{)}\Big{\|}_{\alpha}\leq$
$\displaystyle C_{1}\frac{\mu_{1}^{2}}{\lambda_{1}^{1-\alpha}}.$

######  Lemma 12.5 (Estimate on error part I).

$\displaystyle\|N_{1}\|_{0}\leq\frac{C_{1}}{\mu_{1}}.$

######  Lemma 12.6 (Estimates on error part II).

$\displaystyle\|w_{1o}\otimes w_{1oc}+w_{1oc}\otimes w_{1o}+w_{1oc}\otimes
w_{1oc}\|_{0}\leq C_{1}\frac{\mu_{1}}{\lambda_{1}}.$

The proof of the above four lemmas are same as in Lemma 6.3, Lemma 6.4, Lemma
6.5 and Lemma 6.6 respectively. We omit it here.

Finally, we conclude that

$\displaystyle\|\delta\mathring{R_{01}}\|_{0}\leq
C_{1}\Big{(}\frac{\mu_{1}^{3}}{\lambda_{1}^{1-\alpha}}+\frac{1}{\mu_{1}}\Big{)},\qquad\forall\alpha\in(0,1).$
(12.171)

Furthermore,

$\displaystyle\|v_{01}-v_{0}\|_{0}\leq$
$\displaystyle\frac{M\sqrt{\delta}}{2L}+C_{1}\frac{\mu_{1}}{\lambda_{1}},$
$\displaystyle\|p_{01}-p_{0}\|_{0}\leq$ $\displaystyle
C_{1}\Big{(}\frac{1}{\mu_{1}}+\frac{\mu_{1}}{\lambda_{1}}\Big{)}.$

####  12.7.2. Estimates on $\delta f_{01}$

Recall that

$\displaystyle\delta f_{01}=$
$\displaystyle\mathcal{G}(\hbox{div}K_{1})+\mathcal{G}(w_{1}\cdot\nabla\theta_{0})+\mathcal{G}\Big{(}\partial_{t}\chi_{1}+\sum_{l\in
Z^{3}}\Big{(}\frac{l}{\mu_{1}}\cdot\nabla\Big{)}\chi_{1l}\Big{)}+\sum_{l\in
Z^{3}}\Big{(}v_{0}-\frac{l}{\mu_{1}}\Big{)}\chi_{1l}+w_{1oc}\chi_{1}.$

As before, we split $\delta f_{01}$ into three parts:
(1) the oscillation part

$\displaystyle\mathcal{G}(\hbox{div}K_{1})+\mathcal{G}(w_{1}\cdot\nabla\theta_{0}).$

(2) the transport part

$\displaystyle\mathcal{G}\Big{(}\partial_{t}\chi_{1}+\sum_{l\in
Z^{3}}\Big{(}\frac{l}{\mu_{1}}\cdot\nabla\Big{)}\chi_{1l}\Big{)}.$

(3) the error part

$\displaystyle w_{10c}\chi_{1}+\sum\limits_{l\in
Z^{3}}\Big{(}v_{0}-\frac{l}{\mu}\Big{)}\chi_{1l}.$

We again estimate each of them. And the proof of the following lemmas are same
in as Lemma 6.7, Lemma 6.8 and Lemma 6.9 respectively. We omit it here.

######  Lemma 12.7 (The Oscillation Part).

$\displaystyle\|\mathcal{G}({\rm
div}K_{1})+\mathcal{G}(w_{1}\cdot\nabla\theta_{0})\|_{\alpha}\leq
C_{1}\frac{\mu_{1}^{3}}{\lambda_{1}^{1-\alpha}}.$ (12.172)

######  Lemma 12.8 (The transportation Part).

$\displaystyle\Big{\|}\mathcal{G}\Big{(}\partial_{t}\chi_{1}+\sum_{l\in
Z^{3}}\Big{(}\frac{l}{\mu_{1}}\cdot\nabla\Big{)}\chi_{1l}\Big{)}\Big{\|}_{\alpha}\leq
C_{1}\frac{\mu_{1}^{2}}{\lambda_{1}^{1-\alpha}}.$ (12.173)

######  Lemma 12.9 (The error Part).

$\displaystyle\Big{\|}w_{1oc}\chi_{1}+\sum_{l\in
Z^{3}}\Big{(}v_{0}-\frac{l}{\mu_{1}}\Big{)}\chi_{1l}\Big{\|}_{0}\leq
C_{1}\Big{(}\frac{\mu_{1}^{2}}{\lambda_{1}}+\frac{1}{\mu_{1}}\Big{)}.$
(12.174)

Finally, we conclude that

$\displaystyle\|\delta f_{01}\|_{0}\leq
C_{1}\Big{(}\frac{\mu_{1}^{3}}{\lambda_{1}^{1-\alpha}}+\frac{1}{\mu_{1}}\Big{)}.$

$\displaystyle\|\theta_{01}-\theta_{0}\|_{0}\leq$
$\displaystyle\frac{\sqrt{M\delta}}{2L}+C_{1}\frac{\mu_{1}^{2}}{\lambda_{1}}.$

And

$\displaystyle
R_{01}:=-\delta\sum\limits_{i=2}^{L}\Big{(}\gamma_{k_{i}}\Big{(}\frac{R_{0}}{\delta}\Big{)}\Big{)}^{2}\Big{(}Id-\frac{k_{i}}{|k_{i}|}\otimes\frac{k_{i}}{|k_{i}|}\Big{)}+\delta\mathring{R}_{01},$
$\displaystyle
f_{01}:=\sum\limits_{i=2}^{3}g_{k_{i}}(f_{0}(x,t))A_{k_{i}}+\delta f_{01},$

with the estimate

$\displaystyle\|v_{01}-v_{0}\|_{0}\leq$
$\displaystyle\frac{M\sqrt{\delta}}{2L}+C_{1}\frac{\mu_{1}}{\lambda_{1}},$
$\displaystyle\|p_{01}-p_{0}\|_{0}\leq$ $\displaystyle
C_{1}\Big{(}\frac{1}{\mu_{1}}+\frac{\mu_{1}}{\lambda_{1}}\Big{)},$
$\displaystyle\|\theta_{01}-\theta_{0}\|_{0}\leq$
$\displaystyle\frac{M\sqrt{\delta}}{2L}+\frac{C_{1}}{\lambda_{1}},$
$\displaystyle\|\delta\mathring{R_{01}}\|_{0}\leq$ $\displaystyle
C_{1}\Big{(}\frac{\mu_{1}^{3}}{\lambda_{1}^{1-\alpha}}+\frac{1}{\mu_{1}}\Big{)},$
$\displaystyle\|\delta f_{01}\|_{0}\leq$ $\displaystyle
C_{1}\Big{(}\frac{\mu_{1}^{3}}{\lambda_{1}^{1-\alpha}}+\frac{1}{\mu_{1}}\Big{)}.\qquad\forall\alpha\in(0,1).$

Then we complete the first step.

As the constructions of
$\tilde{v},~{}\tilde{p},~{}\tilde{\theta},~{}\tilde{\mathring{R}},~{}\tilde{f}$
in Proposition 2.1, we can also construct
$\tilde{v},~{}\tilde{p},~{}\tilde{\theta},~{}\tilde{\mathring{R}},~{}\tilde{f}$
by inductions which satisfies the Proposition 2.2. We omit the detail here.

##  Appendix A Stationary phase lemma

We recall here the following facts about operator $\mathcal{R}$ and
$\mathcal{G}$, where the operator $\mathcal{R}$ is defined in [15] .(for a
proof we refer to [15])

######  Proposition A.1.

(i) Let $k\in Z^{3}\setminus\\{0\\}$ and $\lambda\geq 1$ be fixed. For any
$c\in C^{\infty}(T^{3})$ and $m\in N$ we have

$\left|\int_{T^{3}}c(x)e^{i\lambda k\cdot
x}\,dx\right|\leq\frac{[c]_{m}}{\lambda^{m}}.$ (A.175)

(ii) Let $\varphi_{\lambda}\in C^{\infty}(T^{3})$ be the solution of

$\triangle\varphi_{\lambda}=f_{\lambda}$

in $T^{3}$ with $\int_{T^{3}}\varphi_{\lambda}dx=0$, where

$f_{\lambda}(x):=c(x)e^{i\lambda k\cdot x}-\fint_{T^{3}}c(y)e^{i\lambda k\cdot
y}dy.$

Then for any $\alpha\in(0,1)$ and $m\in N$ we have the estimate

$\displaystyle\|\nabla\phi_{\lambda}\|_{\alpha}\leq
C\Big{(}\frac{\|c\|_{0}}{\lambda^{1-\alpha}}+\frac{[c]_{m}}{\lambda^{m-\alpha}}+\frac{[c]_{m+\alpha}}{\lambda^{m}}\Big{)}.$

where $C=C(\alpha,m)$.

(iii) Let $k\in Z^{3}\setminus\\{0\\}$ be fixed. For a smooth vector field
$c\in C^{\infty}(T^{3};R^{3})$ and a smooth scalar function $d\in
C^{\infty}(T^{3};R)$. Let $F(x):=c(x)e^{i\lambda k\cdot x}$ and
$H(x):=d(x)e^{i\lambda k\cdot x}$. Then we have

$\displaystyle\|\mathcal{R}(F)\|_{\alpha}\leq\frac{C}{\lambda^{1-\alpha}}\|c\|_{0}+\frac{C}{\lambda^{m-\alpha}}[c]_{m}+\frac{C}{\lambda^{m}}[c]_{m+\alpha},$
(A.176)
$\displaystyle\|\mathcal{G}(H)\|_{\alpha}\leq\frac{C}{\lambda^{1-\alpha}}\|d\|_{0}+\frac{C}{\lambda^{m-\alpha}}[d]_{m}+\frac{C}{\lambda^{m}}[d]_{m+\alpha},$
(A.177)

where $C=C(\alpha,m)$.

###### Proof.

(i),(ii) and the estimate of $\mathcal{R}$ in (iii) can be found in [15]. From
the definition of $\mathcal{G}$ and (ii), we directly arrive at the estimate
(A.177). MM $\square$

Acknowledgments. The research is partially supported by the Chinese NSF under
grant 11471320. We thank Tianwen Luo for very valuable discussions.

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