00footnotetext: _2020 AMS Mathematics Subject Classification:_ 76B47, 35Q35

# Weakened vortex stretching effect in three scale hierarchy for the 3D Euler
equations

In-Jee Jeong  Department of Mathematics and RIM, Seoul National University.
<EMAIL_ADDRESS>,  Jungkyoung Na  Department of Mathematics, Brown
University<EMAIL_ADDRESS>and  Tsuyoshi Yoneda  Graduate School of
Economics, Hitotsubashi University<EMAIL_ADDRESS>
###### Abstract.

We consider the 3D incompressible Euler equations under the following three
scale hierarchical situation: large-scale vortex stretching the middle-scale,
and at the same time, the middle-scale stretching the small-scale. In this
situation, we show that, the stretching effect of this middle-scale flow is
weakened by the large-scale. In other words, the vortices being stretched
could have the corresponding stress tensor being weakened.

##  1\. Introduction

Recent direct numerical simulations [2, 3, 12, 13] of the 3D Navier–Stokes
turbulence at high Reynolds numbers have shown that there exists a hierarchy
of scale local vortex stretching dynamics. In particular, Goto–Saito–Kawahara
[3] discovered that turbulence at sufficiently high Reynolds numbers in a
periodic cube is composed of a self-similar hierarchy of anti-parallel pairs
of vortex tubes, which is sustained by creation of smaller-scale vortices due
to stretching in larger-scale strain fields. This discovery has been further
investigated by Y.–Goto–Tsuruhashi [16] (see also [15]). From these previous
results, we could conclude physically that the most important features of the
Navier–Stokes turbulence could be scale local vortex stretching which does not
seem to be random (see also [5, 6, 7] for the related mathematical results).
Therefore as the sequence of these studies, our next study will be clarifying
the locality of vortex stretching dynamics precisely, and in this paper we
consider it with three-scale hierarchical structure (c.f. for a geometric
approach considering this locality, see Shimizu–Y. [14]).

In this paper, we shall consider solutions to the 3D incompressible Euler
equations in $\mathbb{T}^{3}$

(1.1) $\left\\{\begin{aligned} \partial_{t}u+u\cdot\nabla u+\nabla p=0,\\\
\nabla\cdot u=0,\end{aligned}\right.$

in which the velocity has the following hierarchical structure:
$u^{\mathcal{L}}+u^{\mathcal{I}}+u^{\mathcal{S}}$, where the letters
$\mathcal{L},\mathcal{I},\mathcal{S}$ stand for large, intermediate, and
small-scale, respectively. They will be arranged in a way that the
corresponding vorticities, denoted by
$\omega^{\mathcal{L}},\omega^{\mathcal{I}},\omega^{\mathcal{S}}$, are mutually
almost orthogonal. This is natural since the rate of vortex stretching
interaction is maximized when the vortex lines are orthogonal to each other.
Indeed, this orthogonality was confirmed in direct numerical simulations [2,
3, 12, 13] by statistical means. The goal of this paper is to understand the
dynamics of vortex stretching under this hierarchical structure.

Before we describe our results, let us give details of our flow configuration.
We take the length scale $L$ of the torus $\mathbb{T}^{3}:=[-L,L)^{3}$ to be
large ($L=100$ suffices) and fix the large-scale velocity to be the linear
strain

(1.2) $\begin{split}u^{\mathcal{L}}(x):=(Mx_{1},-Mx_{2},0),\end{split}$

in the region $|x|\leq 10$, for some large $M\gg 1$. This velocity field
corresponds to a large-scale antiparallel columnar vortex parallel to the
$x_{3}$-axis supported away from the origin. In the following, we shall
implicitly assume that $u^{\mathcal{L}}$ is a steady solution to 3D Euler with
some smooth forcing $f^{\mathcal{L}}$ which is supported in the region
$|x|>10$.

To motivate our choice of smaller scale vorticities, consider the linearized
Euler dynamics around $u^{\mathcal{L}}$ in vorticity form:

(1.3) $\begin{split}\partial_{t}\omega+u^{\mathcal{L}}\cdot\nabla\omega=\nabla
u^{\mathcal{L}}\omega\end{split}$

and since

$\begin{split}\nabla u^{\mathcal{L}}=\begin{pmatrix}M&0&0\\\ 0&-M&0\\\
0&0&0\end{pmatrix},\end{split}$

we see that, with respect to the $L^{\infty}$-norm, the solution to (1.3)
expands and contracts exponentially in time with rate $Mt$ along the $x_{1}$
and $x_{2}$-axis, respectively.

This raises the following question: if we arrange intermediate and small-scale
vorticities to be respectively parallel to $x_{1}$ and $x_{2}$ axis, is it
possible that the intermediate vortex (being exponentially stretched by the
large-scale) significantly stretches the small-scale, by dominating the decay
effect of the large-scale?

It turns out that, interestingly, the answer is no: the small-scale vortex,
even in the presence of exponentially stretched intermediate-scale, still
decays exponentially in time with the same rate $-Mt$. While this is not too
hard to see for the linearized Euler equations, we establish this in the full
nonlinear Euler equations: consider the system

(1.4) $\left\\{\begin{aligned}
\partial_{t}\omega+(u+u^{\mathcal{L}})\cdot\nabla\omega=\nabla(u+u^{\mathcal{L}})\omega,\\\
u=\nabla\times(-\Delta)^{-1}\omega,\end{aligned}\right.$

where we shall assume that $\omega=\omega^{\mathcal{I}}+\omega^{\mathcal{S}}$
is supported in the region $|x|\leq 10$, so that (1.2) applies. We write
$u^{\mathcal{I}}=\nabla\times(-\Delta)^{-1}\omega^{\mathcal{I}}$ and
$u^{\mathcal{S}}=\nabla\times(-\Delta)^{-1}\omega^{\mathcal{S}}$. More
precisely, we take

(1.5)
$\begin{split}\omega^{\mathcal{I}}_{0}(x):=(\omega^{\mathcal{I}}_{1,0}(x_{2},x_{3}),0,0),\qquad\omega^{\mathcal{I}}_{1,0}(x_{2},x_{3}):=\sum_{i,j\in\\{0,1\\}}(-1)^{i+j}\varphi((-1)^{i}x_{2},(-1)^{j}x_{3}),\end{split}$

where $\varphi\in C^{\infty}_{c}$ is a non-negative function supported in
$\left\\{(x_{2},x_{3})\in\mathbb{T}^{2}:1\leq x_{2},x_{3}\leq 2\right\\}$.
Note that $\omega^{\mathcal{I}}_{0}$ is parallel to the $x_{1}$-axis and has
odd symmetry in both $x_{2}$ and $x_{3}$. These odd symmetries are imposed to
maximize the stretching effects of $\omega^{\mathcal{I}}$ near the origin and
has been inspired by several works [8, 17, 1]. Finally, with some sufficiently
small $0<\varepsilon,\ell\ll 1$, we fix some divergence-free vector field
$\tilde{\psi}\in C^{\infty}_{c}(B_{0}(2\ell))$ and require

(1.6)
$\omega^{\mathcal{S}}_{0}(x)=\varepsilon\tilde{\psi}(x),\qquad\tilde{\psi}(x)=\left(0,\psi\left(\ell^{-1}x\right),0\right),\qquad|x|\leq\ell$

for some smooth bump function $\psi\geq 0$.

We consider the flow map $\Phi$ generated by
$u^{\mathcal{L}}+u^{\mathcal{I}}+u^{\mathcal{S}}$: namely, $\Phi(0,x)=x$ and

(1.7)
$\frac{d\Phi}{dt}(t,x)=(u^{\mathcal{L}}+u^{\mathcal{I}}+u^{\mathcal{S}})(t,\Phi(t,x)).$

The solution to (1.4) satisfies the Cauchy formula
$\omega(t,\Phi)=\nabla\Phi\omega_{0}$, and naturally, we define the evolution
of intermediate and small-scale vortex by
$\omega^{\mathcal{I}}(t,\Phi(t,x))=\nabla\Phi(t,x)\omega^{\mathcal{I}}_{0}(x)$
and
$\omega^{\mathcal{S}}(t,\Phi(t,x))=\nabla\Phi(t,x)\omega^{\mathcal{S}}_{0}(x)$.
As long as the solution remains smooth, the support of $\omega^{\mathcal{I}}$
and $\omega^{\mathcal{S}}$ remains disjoint.

We are now ready to state our main theorem. For $0<r<L$, $B_{0}(r)$ denotes
the ball $\\{|x|<r\\}$.

######  Theorem A.

Consider the solution to (1.4) with initial data
$\omega_{0}=\omega^{\mathcal{I}}_{0}+\omega^{\mathcal{S}}_{0}\in
C^{\infty}(\mathbb{T}^{3})$ with $\mathbb{T}^{3}:=[-L,L)^{3}$. The solution
remains smooth in the time interval $[0,T_{M}]$, where
$T_{M}=M^{-1}\log(1+M)$. On this time interval, we have

(1.8) $\begin{split}\left\|{\nabla
u^{\mathcal{I}}(t,\cdot)}\right\|_{L^{\infty}(B_{0}(\ell))}\leq
C\exp(-C^{-1}Mt){\|\nabla
u^{\mathcal{I}}_{0}\|_{L^{\infty}(\mathbb{T}^{3})}}\end{split}$

and

(1.9)
$\begin{split}\left\|{\omega^{\mathcal{S}}(t,\cdot)}\right\|_{L^{\infty}\left({B_{0}(\tilde{\ell})}\right)}\leq
C\exp(-Mt){\|\omega^{\mathcal{S}}_{0}\|_{L^{\infty}(\mathbb{T}^{3})}},\qquad\tilde{\ell}=(M^{-1}\ell)^{C}\end{split}$

uniformly for all $M\gg 1$, $0<\ell\leq M^{-C}$, and
$0<\varepsilon\leq\ell^{2s-3}\exp{(-M^{C})}$ with $s>5/2$, where $C>1$ is a
constant depending only on $L$, $\psi$ and $\varphi$.

######  Remark 1.1.

While $T_{M}\to 0$ as $M\to\infty$, we have that $MT_{M}\to\infty$, so that
within the timescale of $T_{M}$, the exponential terms in the right hand sides
of (1.8) and (1.9) decays to zero. Furthermore, (1.8) should be contrasted
with exponential growth of $\omega^{\mathcal{I}}$:
$\|\omega^{\mathcal{I}}(t,\cdot)\|_{L^{\infty}(\mathbb{T}^{3})}\gtrsim\exp(Mt)\|\omega^{\mathcal{I}}_{0}\|_{L^{\infty}(\mathbb{T}^{3})}$
in the same time interval. In other words, the vortices being stretched could
have the corresponding stress tensor being weakened.

Interpretation of the result. It is important to compare the above with the
case when the large-scale strain field is absent: in this case, the small-
scale vortex gets stretched at least exponentially in time by the
intermediate-scale (cf. [1]). Therefore, we see that the vortex stretching
effect of the intermediate scale is weakened by the large-scale, and the
resulting small-scale vortex dynamics is not really different from the case
when the intermediate vortex is absent. Our result suggests that the vortex
stretching of “adjacent” scales is the one most likely to occur, and thus
possible blow-up solution to the three dimensional Navier–Stokes and/or Euler
equations may not possess multi-scale vortex stretching motion. This insight
is consistent with the numerical result by Kang–Yun–Protas [9]. Based on
solving a suitable optimization problem numerically, they investigated the
largest possible growth of the vorticity in finite time in three-dimensional
Navier–Stokes flows. Their findings revealed that the flows maximizing the
vortex growth exhibit the form of three perpendicular pairs of anti-parallel
vortex tubes with the same size (see Figure 11 in [9]). Furthermore, the flow
evolution resulting from such an initial vorticity is accompanied by
reconnection events. We can at least see that this flow does not possess
multi-scale vortex stretching motion.

######  Remark 1.2.

One could similarly investigate the situation in which the directions of the
intermediate and small-scale vorticities are switched. In this case, both
large and intermediate-scale vortex stretch the small-scale vortex. However,
in this case, the length scale of the intermediate vortex grows at the same
time, and escapes the $O(1)$-region around the origin.

Ideas of the proof. Let us briefly explain the main steps of the proof. To
begin with, when $\omega^{\mathcal{S}}$ is completely absent, one can simply
study the nonlinear evolution of $\omega^{\mathcal{I}}$ and obtain the bound
(1.8). This is already non-trivial since one needs to understand the nonlinear
self-interaction of $\omega^{\mathcal{I}}$.

Then, one can introduce $\omega^{\mathcal{S}}$ and formally analyze the vortex
stretching equation of $\omega^{\mathcal{S}}$, which is
$D_{t}(\omega^{\mathcal{S}})\simeq(\nabla u^{\mathcal{L}}+\nabla
u^{\mathcal{I}})\omega^{\mathcal{S}}$ if one neglects the nonlinear self-
interaction. Applying the bound (1.8) and integrating, one obtains the desired
estimate (1.9). However, in this case, not only one needs to handle the self-
interaction term, but also needs to deal with the linear feedback of
$\omega^{\mathcal{S}}$ onto the intermediate scale $\omega^{\mathcal{I}}$.
Indeed, as soon as $\omega^{\mathcal{S}}$ is introduced,
$\omega^{\mathcal{S}}$ and $\omega^{\mathcal{I}}$ satisfy a coupled system of
PDEs, even at the linearized level. In particular, it becomes tricky to obtain
the bound (1.8) in the first place.

Therefore, our proof consists of a two-step comparison procedure: we introduce
the “pseudo-solution” pair $(\omega^{\mathcal{I},P},\omega^{\mathcal{S},P})$
where $\omega^{\mathcal{I},P}$ corresponds to the intermediate scale in the
absense of the small scale. Then, we compare $\omega^{\mathcal{S}}$ with
$\omega^{\mathcal{S},P}$ which is then compared in turn with the linearized
dynamics around $\omega^{\mathcal{I},P}$.

### Notation

We employ the letters $C,\,C_{1},\,C_{2},\cdots$ to denote any constants which
may change from line to line in a given computation. In particular, the
constants depend only on $L$, $\psi$ and $\varphi$. We sometimes use $A\approx
B$ and $A\lesssim B$, which mean $A=CB$ and $A\leq CB$, respectively, for some
constant $C$.

##  2\. Preliminaries

The aim of this section is to establish two principles for comparing perturbed
solutions of 3D Euler equations in $\mathbb{T}^{3}$.

Let $\bar{u}\in L^{\infty}([0,\bar{T}];H^{s+2}(\mathbb{T}^{3}))$ with
$s>\frac{5}{2}$ be a solution to the following Cauchy problem for the 3D
incompressible Euler equations:

(2.1) $\left\\{\begin{aligned}
\partial_{t}\bar{u}+\bar{u}\cdot\nabla\bar{u}+\nabla\bar{p}&=0,\\\
\nabla\cdot\bar{u}&=0,\\\ \bar{u}(t=0)&=\bar{u}_{0},\end{aligned}\right.$

where $\bar{u}_{0}$ is a function in $H^{s+2}(\mathbb{T}^{3})$. Then we
consider a perturbation problem of (2.1). To be precise, let $u$ be the
solution of the following problem:

(2.2) $\left\\{\begin{aligned} \partial_{t}u+u\cdot\nabla u+\nabla p&=0,\\\
\nabla\cdot u&=0,\\\
u(t=0)&=\bar{u}_{0}+\varepsilon\tilde{u}_{0},\end{aligned}\right.$

where $\varepsilon>0$ and $\tilde{u}_{0}$ is a function belonging to
$H^{s+1}(\mathbb{T}^{3})$.

Now we introduce and prove our two principles.

###  2.1. Principle 1

Defining $\tilde{u}=u-\bar{u}$ and $\tilde{p}=p-\bar{p}$, we have

(2.3) $\left\\{\begin{aligned}
\partial_{t}\tilde{u}+\bar{u}\cdot\nabla\tilde{u}+\tilde{u}\cdot\nabla\bar{u}+\tilde{u}\cdot\nabla\tilde{u}+\nabla\tilde{p}&=0,\\\
\nabla\cdot\tilde{u}&=0,\\\
\tilde{u}(t=0)&=\varepsilon\tilde{u}_{0}.\end{aligned}\right.$

Next, we consider linearization around $\bar{u}$: writing
$u^{Lin}=\bar{u}+\tilde{u}^{Lin}$, and dropping quadratic terms in the
perturbation, we arrive at

(2.4) $\left\\{\begin{aligned}
\partial_{t}\tilde{u}^{Lin}+\bar{u}\cdot\nabla\tilde{u}^{Lin}+\tilde{u}^{Lin}\cdot\nabla\bar{u}+\nabla\tilde{p}^{Lin}&=0,\\\
\nabla\cdot\tilde{u}^{Lin}&=0,\\\
\tilde{u}^{Lin}(t=0)&=\varepsilon\tilde{u}_{0}.\end{aligned}\right.$

Then we estimate
${\left\|\omega(t,\cdot)-\omega^{Lin}(t,\cdot)\right\|}_{L^{\infty}\left(\mathbb{T}^{3}\right)}$
on $[0,\bar{T}]$ for sufficiently small $\varepsilon>0$, where
$\omega=\nabla\times u$ and $\omega^{Lin}=\nabla\times u^{Lin}$ are
corresponding vortices.:

######  Proposition 2.1.

Under the above setting, there exists a constant $C>0$ such that if
$\varepsilon>0$ satisfies

(2.5)
$\varepsilon\leq\frac{1}{C{\left\|\tilde{u}_{0}\right\|}_{H^{s+1}(\mathbb{T}^{3})}\bar{T}}\exp\left(-C\int_{0}^{\bar{T}}{\left\|\bar{u}(\tau,\cdot)\right\|}_{H^{s+2}(\mathbb{T}^{3})}d\tau\right),$

then on $[0,\bar{T}]$, we have

(2.6)
${\left\|\omega(t,\cdot)-\omega^{Lin}(t,\cdot)\right\|}_{L^{\infty}\left(\mathbb{T}^{3}\right)}\leq\varepsilon^{2}{\left\|\tilde{u}_{0}\right\|}_{H^{s+1}(\mathbb{T}^{3})}^{2}C\bar{T}\exp\left(C\int_{0}^{\bar{T}}{\left\|\bar{u}(\tau,\cdot)\right\|}_{H^{s+2}(\mathbb{T}^{3})}d\tau\right).$

######  Remark 2.2.

The point is that the right hand side of (2.6) is $O(\varepsilon^{2})$, while
a priori the left hand side is only $O(\varepsilon)$ a priori. Furthermore,
the right hand side depends only on $\tilde{u}_{0}$ and $\bar{u}$.

###### Proof.

From an elementary estimate

${\left\|\omega(t,\cdot)-\omega^{Lin}(t,\cdot)\right\|}_{L^{\infty}\left(\mathbb{T}^{3}\right)}\lesssim{\left\|\omega(t,\cdot)-\omega^{Lin}(t,\cdot)\right\|}_{H^{s-1}(\mathbb{T}^{3})}\lesssim{\left\|u(t,\cdot)-{u^{Lin}}(t,\cdot)\right\|}_{H^{s}(\mathbb{T}^{3})},$

it suffices to prove

${\left\|u(t,\cdot)-u^{Lin}(t,\cdot)\right\|}_{H^{s}(\mathbb{T}^{3})}\leq\varepsilon^{2}{\left\|\tilde{u}_{0}\right\|}_{H^{s+1}(\mathbb{T}^{3})}^{2}C\bar{T}\exp\left(C\int_{0}^{\bar{T}}{\left\|\bar{u}(\tau,\cdot)\right\|}_{H^{s+2}(\mathbb{T}^{3})}d\tau\right)$

for $\varepsilon>0$ satisfying (2.5). Our first step is to estimate
${\left\|\tilde{u}(t,\cdot)\right\|}_{H^{s}}$ on $[0,\bar{T}]$. Denoting
$J=(I-\Delta)^{\frac{1}{2}}$, (2.3) gives

$\begin{split}\frac{1}{2}\frac{d}{dt}{\left\|\tilde{u}\right\|}_{H^{s+1}}^{2}&=-\int
J^{s+1}(\bar{u}\cdot\nabla\tilde{u})\cdot J^{s+1}\tilde{u}-\int
J^{s+1}(\tilde{u}\cdot\nabla\bar{u})\cdot J^{s+1}\tilde{u}\\\ &\quad-\int
J^{s+1}(\tilde{u}\cdot\nabla\tilde{u})\cdot J^{s+1}\tilde{u}-\int
J^{s+1}\nabla\tilde{p}\cdot J^{s+1}\tilde{u}\\\
&=\mathrm{I}+\mathrm{II}+\mathrm{III}+\mathrm{IV}.\end{split}$

Since $\tilde{u}$ is divergence-free, $\mathrm{IV}=0$. Using the fact that
$H^{s+1}{(\mathbb{T}^{3})}$ is a Banach algebra, we have

$\mathrm{II}\lesssim{\left\|\tilde{u}\cdot\nabla\bar{u}\right\|}_{H^{s+1}}{\left\|\tilde{u}\right\|}_{H^{s+1}}\lesssim{\left\|\bar{u}\right\|}_{H^{s+2}}{\left\|\tilde{u}\right\|}_{H^{s+1}}^{2}.$

For $\mathrm{I}$, we note that $\nabla\cdot\tilde{u}=0$ yields
$\int\left(\bar{u}\cdot\nabla J^{s+1}\tilde{u}\right)\cdot
J^{s+1}\tilde{u}=0,$ so that

$\mathrm{I}=-\int\left(\left[J^{s+1},\bar{u}\cdot\right]\nabla\tilde{u}\right)\cdot
J^{s+1}\tilde{u}.$

Recalling the Kato-Ponce commutator estimate ([10]):

${\left\|\left[J^{s^{\prime}},f\right]g\right\|}_{L^{2}\left(\mathbb{T}^{3}\right)}\lesssim\left(\|f\|_{H^{\frac{5}{2}+\varepsilon^{\prime}}(\mathbb{T}^{3})}{\left\|J^{s^{\prime}-1}g\right\|}_{L^{2}\left(\mathbb{T}^{3}\right)}+{\left\|J^{s^{\prime}}f\right\|}_{L^{2}\left(\mathbb{T}^{3}\right)}\|g\|_{H^{\frac{3}{2}+\varepsilon^{\prime}}(\mathbb{T}^{3})}\right)$

for any $\varepsilon^{\prime}>0$ and $s^{\prime}>0$, we obtain

$\mathrm{I}\lesssim{\left\|\left[J^{s+1},\bar{u}\cdot\right]\nabla\tilde{u}\right\|}_{L^{2}}{\left\|J^{s+1}\tilde{u}\right\|}_{L^{2}}\lesssim{\left\|\bar{u}\right\|}_{H^{s+1}}{\left\|\tilde{u}\right\|}_{H^{s+1}}^{2}.$

Replacing $\bar{u}$ with $\tilde{u}$ and proceeding in the same way, we can
also estimate $\mathrm{III}\lesssim{\left\|\tilde{u}\right\|}_{H^{s+1}}^{3}.$
Combining all, we arrive at

$\frac{d}{dt}{\left\|\tilde{u}\right\|}_{H^{s+1}}\leq
C\left({\left\|\bar{u}\right\|}_{H^{s+2}}{\left\|\tilde{u}\right\|}_{H^{s+1}}+{\left\|\tilde{u}\right\|}_{H^{s+1}}^{2}\right).$

Introducing the quantity
$y(t)={\left\|\tilde{u}(t)\right\|}_{H^{s+1}}\exp(-C\int_{0}^{t}{\left\|\bar{u}(\tau)\right\|}_{H^{s+2}}d\tau),$
we have

$\left\\{\begin{aligned} &\frac{d}{dt}y(t)\leq
C\exp\left(C\int_{0}^{\bar{T}}{\left\|\bar{u}(\tau)\right\|}_{H^{s+2}}d\tau\right)y^{2}(t),\\\
&{y(0)}=\varepsilon{\left\|\tilde{u}_{0}\right\|}_{H^{s+1}}\end{aligned}\right.$

on $[0,\bar{T}]$. Then for $\varepsilon>0$ satisfying (2.5) (by adjusting $C$
if necessary), we have

$y(t)\leq\frac{\varepsilon{\left\|\tilde{u}_{0}\right\|}_{H^{s+1}}}{1-\varepsilon{\left\|\tilde{u}_{0}\right\|}_{H^{s+1}}Ct\exp\left(C\int_{0}^{\bar{T}}{\left\|\bar{u}(\tau)\right\|}_{H^{s+2}}d\tau\right)}\leq
2\varepsilon{\left\|\tilde{u}_{0}\right\|}_{H^{s+1}},$

and consequently on $[0,\bar{T}]$, we have

(2.7) ${\left\|\tilde{u}(t)\right\|}_{H^{s+1}}\leq
2\varepsilon{\left\|\tilde{u}_{0}\right\|}_{H^{s+1}}\exp\left(C\int_{0}^{\bar{T}}{\left\|\bar{u}(\tau)\right\|}_{H^{s+2}}d\tau\right).$

Next, we consider the equation of $\tilde{u}^{D}=\tilde{u}-\tilde{u}^{Lin}$:

$\left\\{\begin{split}\partial_{t}\tilde{u}^{D}+\bar{u}\cdot\nabla\tilde{u}^{D}+\tilde{u}^{D}\cdot\nabla\bar{u}+\tilde{u}\cdot\nabla\tilde{u}+\nabla(\tilde{p}-\tilde{p}^{Lin})&=0,\\\
\nabla\cdot\tilde{u}^{D}&=0,\\\ \tilde{u}^{D}(t=0)&=0,\end{split}\right.$

which gives

$\begin{split}\frac{1}{2}\frac{d}{dt}{\left\|\tilde{u}^{D}\right\|}_{H^{s}}^{2}&=-\int
J^{s}(\bar{u}\cdot\nabla\tilde{u}^{D})\cdot J^{s}\tilde{u}^{D}-\int
J^{s}(\tilde{u}^{D}\cdot\nabla\bar{u})\cdot J^{s}\tilde{u}^{D}\\\ &\qquad-\int
J^{s}(\tilde{u}\cdot\nabla\tilde{u})\cdot J^{s}\tilde{u}^{D}-\int
J^{s}\nabla(\tilde{p}-\tilde{p}^{Lin})\cdot J^{s}\tilde{u}^{D}\\\
&=\mathrm{V}+\mathrm{VI}+\mathrm{VII}+\mathrm{VIII}.\end{split}$

To estimate $\mathrm{V}$, $\mathrm{VI}$, and $\mathrm{VIII}$, proceeding in
the same way as $\mathrm{I}$, $\mathrm{II}$, and $\mathrm{IV}$, respectively,
we have

$\mathrm{V}\lesssim{\left\|\bar{u}\right\|}_{H^{s}}{\left\|\tilde{u}^{D}\right\|}_{H^{s}}^{2},\qquad\mathrm{VI}\lesssim{\left\|\bar{u}\right\|}_{H^{s+1}}{\left\|\tilde{u}^{D}\right\|}_{H^{s}}^{2},\qquad\mathrm{VIII}=0.$

For $\mathrm{VII}$, the fact that $H^{s}{(\mathbb{T}^{3})}$ is a Banach
algebra implies

$\mathrm{VII}\lesssim{\left\|\tilde{u}\cdot\nabla\tilde{u}\right\|}_{H^{s}}{\left\|\tilde{u}^{D}\right\|}_{H^{s}}\lesssim{\left\|\tilde{u}\right\|}_{H^{s+1}}^{2}{\left\|\tilde{u}^{D}\right\|}_{H^{s}}.$

Combining all, we obtain

$\frac{d}{dt}{\left\|\tilde{u}^{D}\right\|}_{H^{s}}\leq
C\left({\left\|\bar{u}\right\|}_{H^{s+1}}{\left\|\tilde{u}^{D}\right\|}_{H^{s}}+{\left\|\tilde{u}\right\|}_{H^{s+1}}^{2}\right).$

Thus, for $\varepsilon>0$ satisfying (2.5), the Grönwall’s inequality and
(2.7) give us

$\begin{split}{\left\|\tilde{u}^{D}(t)\right\|}_{H^{s}}&\leq\exp\left(C\int_{0}^{t}{\left\|\bar{u}(\tau)\right\|}_{H^{s+1}}d\tau\right)C\int_{0}^{t}{\left\|\tilde{u}(\tau)\right\|}_{H^{s+1}}^{2}d\tau\leq
4\varepsilon^{2}{\left\|\tilde{u}_{0}\right\|}_{H^{s+1}}^{2}C\bar{T}\exp\left(3C\int_{0}^{\bar{T}}{\left\|\bar{u}(\tau)\right\|}_{H^{s+2}}d\tau\right)\end{split}$

on $[0,\bar{T}]$. Since $\tilde{u}^{D}=\tilde{u}-\tilde{u}^{Lin}=u-u^{Lin}$,
we are done. ∎

###  2.2. Principle 2

Let $u$ be the solution of (2.2). We consider the following PDE:

(2.8) $\left\\{\begin{aligned}
\partial_{t}\bar{u}^{*}+u\cdot\nabla\bar{u}^{*}+\nabla\bar{p}^{*}-\nabla\bar{u}^{*}\cdot(u-\bar{u}^{*})&=0,\\\
\nabla\cdot\bar{u}^{*}&=0,\\\
\bar{u}^{*}(t=0)&=\bar{u}_{0},\end{aligned}\right.$

where
$(\nabla\bar{u}^{*}\cdot(u-\bar{u}^{*}))_{i}:=\sum_{j=1}^{3}\partial_{i}\bar{u}^{*}_{j}(u-\bar{u}^{*})_{j}$
with $i=1,2,3$. (This is different from
$(u-\bar{u}^{*})\cdot\nabla\bar{u}^{*}$.) This time, we compare $\bar{u}^{*}$
with $\bar{u}$ of (2.1). Note that (2.1) and (2.8) share the same initial
data. We have the following:

######  Proposition 2.3.

Under the above setting, for $t\in[0,\bar{T}]$ and $\varepsilon>0$ satisfying
(2.5), there exists a constant $C>0$ such that

${\left\|\bar{u}^{*}(t,\cdot)-\bar{u}(t,\cdot)\right\|}_{H^{s}(\mathbb{T}^{3})}\leq\varepsilon{\left\|\tilde{u}_{0}\right\|}_{H^{s+1}}C\bar{T}\left(\sup_{t\in[0,\bar{T}]}{\left\|\bar{u}(t)\right\|}_{H^{s+1}(\mathbb{T}^{3})}\right)\exp\left(C\int_{0}^{\bar{T}}\left(1+{\left\|\bar{u}(\tau)\right\|}_{H^{s+2}(\mathbb{T}^{3})}\right)d\tau\right).$

###### Proof.

Denoting $\tilde{u}=u-\bar{u}$ and $\bar{u}^{D*}=\bar{u}^{*}-\bar{u}$, we
obtain the equation of $\bar{u}^{D*}$:

$\left\\{\begin{split}&\partial_{t}\bar{u}^{D*}+\tilde{u}\cdot\nabla\bar{u}^{D*}+\bar{u}\cdot\nabla\bar{u}^{D*}+\tilde{u}\cdot\nabla\bar{u}-\nabla\bar{u}^{D*}\cdot\tilde{u}+\nabla\bar{u}^{D*}\cdot\bar{u}^{D*}-\nabla\bar{u}\cdot\tilde{u}+\nabla\bar{u}\cdot\bar{u}^{D*}+\nabla(\bar{p}^{*}-\bar{p})=0,\\\
&\nabla\cdot\bar{u}^{D*}=0,\\\ &\bar{u}^{D*}(t=0)=0.\end{split}\right.$

This gives

$\begin{split}\frac{1}{2}\frac{d}{dt}{\left\|\bar{u}^{D*}\right\|}_{H^{s}}^{2}&=-\int
J^{s}(\tilde{u}\cdot\nabla\bar{u}^{D*})\cdot J^{s}\bar{u}^{D*}-\int
J^{s}(\bar{u}\cdot\nabla\bar{u}^{D*})\cdot J^{s}\bar{u}^{D*}-\int
J^{s}(\tilde{u}\cdot\nabla\bar{u})\cdot J^{s}\bar{u}^{D*}\\\ &\quad+\int
J^{s}(\nabla\bar{u}^{D*}\cdot\tilde{u})\cdot J^{s}\bar{u}^{D*}-\int
J^{s}(\nabla\bar{u}^{D*}\cdot\bar{u}^{D*})\cdot J^{s}\bar{u}^{D*}+\int
J^{s}(\nabla\bar{u}\cdot\tilde{u})\cdot J^{s}\bar{u}^{D*}\\\
&\quad-J^{s}(\nabla\bar{u}\cdot\bar{u}^{D*})\cdot J^{s}\bar{u}^{D*}-\int
J^{s}\nabla(\tilde{p}-\tilde{p}^{Lin})\cdot J^{s}\bar{u}^{D*}\\\
&=\mathrm{I}+\mathrm{II}+\mathrm{III}+\mathrm{IV}+\mathrm{V}+\mathrm{VI}+\mathrm{VII}+\mathrm{VIII}.\end{split}$

For $\mathrm{I}$, $\mathrm{II}$, and $\mathrm{VIII}$, we proceed in the same
way as the proof of Proposition 2.1 to obtain

$\mathrm{I}\lesssim{\left\|\tilde{u}\right\|}_{H^{s}}{\left\|\bar{u}^{D*}\right\|}_{H^{s}}^{2},\qquad\mathrm{II}\lesssim{\left\|\bar{u}\right\|}_{H^{s}}{\left\|\bar{u}^{D*}\right\|}_{H^{s}}^{2},\qquad\mathrm{VIII}=0.$

For $\mathrm{III}+\mathrm{VI}+\mathrm{VII}$, we use the fact that
$H^{s+1}{(\mathbb{T}^{3})}$ is a Banach algebra to have

$\begin{split}\mathrm{III}+\mathrm{VI}+\mathrm{VII}\lesssim\left({\left\|\tilde{u}\right\|}_{H^{s}}+{\left\|\bar{u}^{D*}\right\|}_{H^{s}}\right){\left\|\bar{u}\right\|}_{H^{s+1}}{\left\|\bar{u}^{D*}\right\|}_{H^{s}}.\end{split}$

For $\mathrm{IV}$ and $\mathrm{V}$, $\nabla\cdot\bar{u}^{D*}=0$ and the
integration by parts give us

$\mathrm{IV}=-\int
J^{s}(\bar{u}^{D*}_{j}\partial_{i}\tilde{u}_{j})J^{s}\bar{u}^{D*}_{i}\lesssim{\left\|\tilde{u}\right\|}_{H^{s+1}}{\left\|\bar{u}^{D*}\right\|}_{H^{s}}^{2},\qquad\mathrm{V}=-\frac{1}{2}\int
J^{s}\partial_{i}|\bar{u}^{D*}|^{2}J^{s}\bar{u}^{D*}_{i}=0.$

Combining all, we arrive at

$\frac{d}{dt}{\left\|\bar{u}^{D*}\right\|}_{H^{s}}\leq
C\left({\left\|\tilde{u}\right\|}_{H^{s+1}}+{\left\|\bar{u}\right\|}_{H^{s+1}}\right){\left\|\bar{u}^{D*}\right\|}_{H^{s}}+{\left\|\tilde{u}\right\|}_{H^{s+1}}{\left\|\bar{u}\right\|}_{H^{s+1}}.$

Thus, for $\varepsilon>0$ satisfying (2.5), the Grönwall’s inequality and
(2.7) yield

$\begin{split}{\left\|\bar{u}^{D*}(t)\right\|}_{H^{s}}&\leq\exp\left(C\int_{0}^{t}\left({\left\|\tilde{u}(\tau)\right\|}_{H^{s+1}}+{\left\|\bar{u}(\tau)\right\|}_{H^{s+1}}\right)d\tau\right)C\int_{0}^{t}{\left\|\tilde{u}(\tau)\right\|}_{H^{s+1}}{\left\|\bar{u}(\tau)\right\|}_{H^{s+1}}d\tau\\\
&\leq
2\varepsilon{\left\|\tilde{u}_{0}\right\|}_{H^{s+1}}C\bar{T}\left(\sup_{t\in[0,\bar{T}]}{\left\|\bar{u}(t)\right\|}_{H^{s+1}}\right)\exp\left(2C\int_{0}^{\bar{T}}\left(1+{\left\|\bar{u}(\tau)\right\|}_{H^{s+2}}\right)d\tau\right)\end{split}$

on $[0,\bar{T}]$. ∎

##  3\. Proof of the main result

The aim of this section is to show Theorem A. In the proof, we shall always
assume that initial data $\omega^{\mathcal{I}}_{0}$ and
$\omega^{\mathcal{S}}_{0}$ satisfy (1.5) and (1.6), respectively. We note that
by our definitions of $\omega^{\mathcal{I}}$ and $\omega^{\mathcal{S}}$, they
solve the following coupled system:
$(\omega^{\mathcal{I}},\omega^{\mathcal{S}})(t=0)=(\omega^{\mathcal{I}}_{0},\omega^{\mathcal{S}}_{0})$
and

(3.1) $\left\\{\begin{aligned}
\partial_{t}\omega^{\mathcal{I}}+(u^{\mathcal{L}}+u^{\mathcal{I}}+u^{\mathcal{S}})\cdot\nabla\omega^{\mathcal{I}}&=\nabla(u^{\mathcal{L}}+u^{\mathcal{I}}+u^{\mathcal{S}})\omega^{\mathcal{I}},\\\
\partial_{t}\omega^{\mathcal{S}}+(u^{\mathcal{L}}+u^{\mathcal{I}}+u^{\mathcal{S}})\cdot\nabla\omega^{\mathcal{S}}&=\nabla(u^{\mathcal{L}}+u^{\mathcal{I}}+u^{\mathcal{S}})\omega^{\mathcal{S}},\\\
(u^{\mathcal{I}},u^{\mathcal{S}})&=\nabla\times(-\Delta)^{-1}(\omega^{\mathcal{I}},\omega^{\mathcal{S}}).\\\
\end{aligned}\right.$

Recalling the definition of $u^{\mathcal{L}}$ in (1.2), we have
$\omega^{\mathcal{L}}:=\nabla\times u^{\mathcal{L}}=0$ in $|x|\leq 10$. Thus,
setting $\bar{u}^{*}:=u^{\mathcal{L}}+u^{\mathcal{I}}$ and
$u:=u^{\mathcal{L}}+u^{\mathcal{I}}+u^{\mathcal{S}}$, and noticing

$\begin{split}\nabla\times\left(u\cdot\nabla\bar{u}^{*}-\nabla\bar{u}^{*}\cdot(u-\bar{u}^{*})\right)&=\nabla\times\left(\bar{u}^{*}\cdot\nabla\bar{u}^{*}+(u-\bar{u}^{*})\cdot\nabla\bar{u}^{*}-\nabla\bar{u}^{*}\cdot(u-\bar{u}^{*})\right)\\\
&=\nabla\times\left((\bar{\omega}^{*}\times\bar{u}^{*})+\frac{1}{2}\nabla|\bar{u}^{*}|^{2}+\bar{\omega}^{*}\times(u-\bar{u}^{*})\right)\qquad(\bar{\omega}^{*}:=\nabla\times\bar{u}^{*})\\\
&=\bar{u}^{*}\cdot\nabla\bar{\omega}^{*}-\nabla\bar{u}^{*}\bar{\omega}^{*}+(u-\bar{u}^{*})\cdot\nabla\bar{\omega}^{*}-\nabla(u-\bar{u}^{*})\bar{\omega}^{*}=u\cdot\nabla\bar{\omega}^{*}-\nabla
u\,\bar{\omega}^{*},\end{split}$

we can check that $\bar{u}^{*}$ and $u$ solve (2.8) and (2.2), respectively.
On the other hand, we shall introduce pseudo-solutions
$(\omega^{\mathcal{I},P},\omega^{\mathcal{S},P})$ as the solutions of

(3.2) $\left\\{\begin{aligned}
\partial_{t}\omega^{\mathcal{I},P}+(u^{\mathcal{L}}+u^{\mathcal{I},P})\cdot\nabla\omega^{\mathcal{I},P}&=\nabla(u^{\mathcal{L}}+u^{\mathcal{I},P})\omega^{\mathcal{I},P},\\\
u^{\mathcal{I},P}&=\nabla\times(-\Delta)^{-1}\omega^{\mathcal{I},P},\end{aligned}\right.$

and

(3.3) $\left\\{\begin{aligned}
\partial_{t}\omega^{\mathcal{S},P}+(u^{\mathcal{L}}+u^{\mathcal{I},P}+u^{\mathcal{S},P})&\cdot\nabla\omega^{\mathcal{S},P}=\nabla(u^{\mathcal{L}}+u^{\mathcal{I},P}+u^{\mathcal{S},P})\omega^{\mathcal{S},P}+\nabla
u^{\mathcal{S},P}\omega^{\mathcal{I},P}-u^{\mathcal{S},P}\cdot\nabla\omega^{\mathcal{I},P},\\\
u^{\mathcal{S},P}&=\nabla\times(-\Delta)^{-1}\omega^{\mathcal{S},P},\end{aligned}\right.$

with initial data $\omega^{\mathcal{I},P}(t=0)=\omega^{\mathcal{I}}_{0}$ and
$\omega^{\mathcal{S},P}(t=0)=\omega^{\mathcal{S}}_{0}$. Then we can check that
$\bar{u}:=u^{\mathcal{L}}+u^{\mathcal{I},P}$ and
$u:=u^{\mathcal{L}}+u^{\mathcal{I},P}+u^{\mathcal{S},P}$ are solutions of
(2.1) and (2.2), respectively. Note that
$u=u^{\mathcal{L}}+u^{\mathcal{I},P}+u^{\mathcal{S},P}=u^{\mathcal{L}}+u^{\mathcal{I}}+u^{\mathcal{S}}$,
which implies
$u^{\mathcal{I},P}-u^{\mathcal{I}}=u^{\mathcal{S}}-u^{\mathcal{S},P}$. Our
strategy is first to analyze pseudo-solutions
$(\omega^{\mathcal{I},P},\omega^{\mathcal{S},P})$, and then to compare them
with real solutions $(\omega^{\mathcal{I}},\omega^{\mathcal{S}})$ using the
principles introduced in the previous section.

###  3.1. Behavior of $\omega^{\mathcal{I},P}$ and $\nabla u^{\mathcal{I},P}$

Abusing the notation for simplicity, we denote pseudo-solutions
$\omega^{\mathcal{I},P}$ and $u^{\mathcal{I},P}$ by $\omega^{\mathcal{I}}$ and
$u^{\mathcal{I}}$, respectively. We note that the the symmetry
$\omega^{\mathcal{I}}(t,x_{1},x_{2},x_{3})=(\omega^{\mathcal{I}}_{1}(t,x_{2},x_{3}),0,0)$
propagates in time for the solutions to (3.2). Namely, the assumptions

(3.4) $\partial_{1}\omega^{\mathcal{I}}_{1}\equiv
0,\quad\omega^{\mathcal{I}}_{2}\equiv 0,\quad\omega^{\mathcal{I}}_{3}\equiv 0$

hold for all times if they are valid at $t=0$. This is not trivial since
$u^{\mathcal{L}}$ depends on $x_{1}$. To see this, first note that (3.4)
implies

(3.5) $u^{\mathcal{I}}_{1}\equiv 0,\quad\partial_{1}u^{\mathcal{I}}_{2}\equiv
0,\quad\partial_{1}u^{\mathcal{I}}_{3}\equiv 0$

by the Biot-Savart law. Together this implies $\nabla
u^{\mathcal{I}}\omega^{\mathcal{I}}\equiv 0$. Denoting
$D_{t}=\partial_{t}+(u^{\mathcal{L}}+u^{\mathcal{I}})\cdot\nabla$, from

$D_{t}(\omega^{\mathcal{I}}_{2})=(\nabla
u^{\mathcal{I}}\omega^{\mathcal{I}})_{2}-M\omega^{\mathcal{I}}_{2},\quad
D_{t}(\omega^{\mathcal{I}}_{3})=(\nabla
u^{\mathcal{I}}\omega^{\mathcal{I}})_{3},$

we see that $(\nabla u^{\mathcal{I}}\omega^{\mathcal{I}})\equiv 0$ is
consistent with $\omega^{\mathcal{I}}_{2},\omega^{\mathcal{I}}_{3}$ being zero
for all times. Lastly,

$\displaystyle D_{t}($ $\displaystyle\omega^{\mathcal{I}}_{1})=(\nabla
u^{\mathcal{I}}\omega^{\mathcal{I}})_{1}+M\omega^{\mathcal{I}}_{1},\qquad
D_{t}(\partial_{1}\omega^{\mathcal{I}}_{1})=\partial_{1}((\nabla
u^{\mathcal{I}}\omega^{\mathcal{I}})_{1}),$

which shows that $\partial_{1}\omega^{\mathcal{I}}_{1}\equiv 0$ propagates in
time. This shows that the equation for $\omega^{\mathcal{I}}_{1}$ is given by

(3.6)
$\partial_{t}\omega^{\mathcal{I}}_{1}+(u^{\mathcal{I}}_{2}-Mx_{2})\partial_{2}\omega^{\mathcal{I}}_{1}+u^{\mathcal{I}}_{3}\partial_{3}\omega^{\mathcal{I}}_{1}=M\omega^{\mathcal{I}}_{1}.$

Comparing (3.6) with 2D Euler equation which has global well-posedness of
smooth solutions, we can check that there exists the unique global smooth
solution $\omega^{\mathcal{I}}_{1}$ of (3.6) with initial data given in (1.5).
Furthermore, we can also observe that $\omega^{\mathcal{I}}_{1}$ keeps the odd
symmetry in both $x_{2}$ and $x_{3}$.

To begin with, we observe temporal behaviors of
${\left\|\omega^{\mathcal{I}}(t,\cdot)\right\|}_{L^{p}}$ for $p\in[1,\infty]$.

######  Lemma 3.1.

For $t\in[0,\infty)$,
$\omega^{\mathcal{I}}(t,\cdot)=(\omega^{\mathcal{I}}_{1}(t,\cdot),0,0)$
satisfies

(3.7)
${\left\|\omega^{\mathcal{I}}_{1}(t,\cdot)\right\|}_{L^{p}}{={\left\|\omega^{\mathcal{I}}_{0}\right\|}_{L^{p}}}e^{\frac{M(p-1)}{p}t}\;(1\leq
p<\infty)\quad\text{and}\quad{\left\|\omega^{\mathcal{I}}_{1}(t,\cdot)\right\|}_{L^{\infty}}{={\left\|\omega^{\mathcal{I}}_{0}\right\|}_{L^{\infty}}}e^{Mt}.$

###### Proof.

Taking the $L^{2}$ inner product of (3.6) with
$\omega^{\mathcal{I}}_{1}|\omega^{\mathcal{I}}_{1}|^{p-2}$, we have

$\begin{split}\frac{1}{p}\frac{d}{dt}{\left\|\omega^{\mathcal{I}}_{1}\right\|}_{L^{p}}^{p}&=-\int(u^{\mathcal{I}}_{2}\partial_{2}\omega^{\mathcal{I}}_{1}+u^{\mathcal{I}}_{3}\partial_{3}\omega^{\mathcal{I}}_{1})\omega^{\mathcal{I}}_{1}|\omega^{\mathcal{I}}_{1}|^{p-2}+\int
Mx_{2}\partial_{2}\omega^{\mathcal{I}}_{1}\omega^{\mathcal{I}}_{1}|\omega^{\mathcal{I}}_{1}|^{p-2}+M{\left\|\omega^{\mathcal{I}}_{1}\right\|}_{L^{p}}^{p}.\end{split}$

After the integration by parts, the first integral vanishes since
$u^{\mathcal{I}}_{1}=0$ and $\nabla\cdot u^{\mathcal{I}}=0$, and the second
integral is equal to
$-\frac{M}{p}{\left\|\omega^{\mathcal{I}}_{1}\right\|}_{L^{p}}^{p}$. Thus we
have

$\begin{split}\frac{d}{dt}{\left\|\omega^{\mathcal{I}}_{1}\right\|}_{L^{p}}^{p}&=M(p-1){\left\|\omega^{\mathcal{I}}_{1}\right\|}_{L^{p}}^{p},\quad\mbox{
which
implies}\qquad{\left\|\omega^{\mathcal{I}}_{1}(t,\cdot)\right\|}_{L^{p}}=e^{\frac{M(p-1)}{p}t}{\left\|\omega^{\mathcal{I}}_{1,0}\right\|}_{L^{p}}.\end{split}$

Passing $p\rightarrow\infty$, we can also obtain $L^{\infty}$ estimate. ∎

Next, using (3.7), we estimate
$\left\|{u^{\mathcal{I}}(t,\cdot)}\right\|_{H^{s}}$ with $s>\frac{5}{2}$ for
$t\in[0,T_{M}]$ with $T_{M}=\frac{\log(M+1)}{M}$. This estimate will be used
in Section 3.3.

######  Lemma 3.2.

For $s>\frac{5}{2}$ and $t\in[0,T_{M}]$,
${\left\|u^{\mathcal{I}}(t,\cdot)\right\|}_{H^{s}(\mathbb{T}^{3})}\lesssim
e^{{C}Mt}$.

###### Proof.

Fix $s>\frac{5}{2}$. Noticing that $u^{\mathcal{I}}$ solves

$\left\\{\begin{aligned}
\partial_{t}u^{\mathcal{I}}+(u^{\mathcal{L}}+u^{\mathcal{I}})\cdot\nabla
u^{\mathcal{I}}+\nabla p^{\mathcal{I}}=0,\\\ \nabla\cdot
u^{\mathcal{I}}=0.\end{aligned}\right.$

and denoting $J=(I-\Delta)^{\frac{1}{2}}$, we have

$\begin{split}\frac{1}{2}\frac{d}{dt}{\left\|u^{\mathcal{I}}\right\|}_{H^{s}}^{2}&=-\int
J^{s}(u^{\mathcal{L}}\cdot\nabla u^{\mathcal{I}})\cdot
J^{s}u^{\mathcal{I}}-\int J^{s}(u^{\mathcal{I}}\cdot\nabla
u^{\mathcal{I}})\cdot J^{s}u^{\mathcal{I}}-\int J^{s}\nabla
p^{\mathcal{I}}\cdot
J^{s}u^{\mathcal{I}}=\mathrm{I}+\mathrm{II}+\mathrm{III}.\end{split}$

$\mathrm{III}=0$ follows from $\nabla\cdot u^{\mathcal{I}}=0$. Noticing
$\int\left(u^{\mathcal{L}}\cdot\nabla J^{s}u^{\mathcal{I}}\right)\cdot
J^{s}u^{\mathcal{I}}=\int\left(u^{\mathcal{I}}\cdot\nabla
J^{s}u^{\mathcal{I}}\right)\cdot J^{s}u^{\mathcal{I}}=0,$ we obtain

$\mathrm{I}=-\int\left(\left[J^{s},u^{\mathcal{L}}\cdot\right]\nabla
u^{\mathcal{I}}\right)\cdot
J^{s}u^{\mathcal{I}}\quad\text{and}\quad\mathrm{II}=-\int\left(\left[J^{s},u^{\mathcal{I}}\cdot\right]\nabla
u^{\mathcal{I}}\right)\cdot J^{s}u^{\mathcal{I}},$

where we recall that $[\cdot,\cdot]$ denotes the commutator. Using
$\left\|{u^{\mathcal{L}}}\right\|_{H^{s}(\mathbb{T}^{3})}\lesssim M$ and the
Sobolev embedding $H^{s}(\mathbb{T}^{3})\hookrightarrow
W^{1,\infty}(\mathbb{T}^{3})$, we obtain

$\mathrm{I}\lesssim{\left\|\left[J^{s},u^{\mathcal{L}}\cdot\right]\nabla
u^{\mathcal{I}}\right\|}_{L^{2}}{\left\|J^{s}u^{\mathcal{I}}\right\|}_{L^{2}}\lesssim
M{\left\|u^{\mathcal{I}}\right\|}_{H^{s}}^{2},\quad\mathrm{II}\lesssim{\left\|\left[J^{s},u^{\mathcal{I}}\cdot\right]\nabla
u^{\mathcal{I}}\right\|}_{L^{2}}{\left\|J^{s}u^{\mathcal{I}}\right\|}_{L^{2}}\lesssim{\left\|\nabla
u^{\mathcal{I}}\right\|}_{L^{\infty}}{\left\|u^{\mathcal{I}}\right\|}_{H^{s}}^{2},$

which lead to

$\frac{d}{dt}{\left\|u^{\mathcal{I}}\right\|}_{H^{s}}\lesssim\left(M+{\left\|\nabla
u^{\mathcal{I}}\right\|}_{L^{\infty}}\right){\left\|u^{\mathcal{I}}\right\|}_{H^{s}}.$

According to the Calderon-Zygmund theory, we have

${\left\|\nabla
u^{\mathcal{I}}\right\|}_{L^{\infty}\left(\mathbb{T}^{3}\right)}\lesssim{\left\|\omega^{\mathcal{I}}\right\|}_{L^{\infty}\left(\mathbb{T}^{3}\right)}\log\left(10+\frac{\left\|{u^{\mathcal{I}}}\right\|_{H^{s}(\mathbb{T}^{3})}}{{\left\|\omega^{\mathcal{I}}\right\|}_{L^{\infty}\left(\mathbb{T}^{3}\right)}}\right)\lesssim
e^{Mt}\log\left(10+{\left\|u^{\mathcal{I}}\right\|}_{H^{s}}\right),$

where we used (3.7) in the last inequality. This gives

$\frac{d}{dt}\log\left(10+{\left\|u^{\mathcal{I}}\right\|}_{H^{s}}\right)\lesssim
M+e^{Mt}\log\left(10+{\left\|u^{\mathcal{I}}\right\|}_{H^{s}}\right).$

Using Grönwall’s inequality, we obtain the desired estimate on $[0,T_{M}]$. ∎

Our next aim is to estimate the maximum of $\nabla u^{\mathcal{I}}(t,\cdot)$
in a small region near the origin up to time $T_{M}=\frac{\log(M+1)}{M}$. The
corresponding velocity
$u^{\mathcal{I}}=(0,u^{\mathcal{I}}_{2},u^{\mathcal{I}}_{3})$ to
$\omega^{\mathcal{I}}=(\omega^{\mathcal{I}}_{1},0,0)$ has explicit formula:

(3.8)
$\begin{split}u^{\mathcal{I}}_{2}(t,x_{2},x_{3})=\frac{1}{2\pi}\sum_{n=(n_{2},n_{3})\in\mathbb{Z}^{2}}\iint_{[-L,L)^{2}}\frac{-x_{3}+y_{3}+2Ln_{3}}{|x-y-2Ln|^{2}}\omega^{\mathcal{I}}_{1}(t,y_{2},y_{3})dy_{2}dy_{3},\\\
u^{\mathcal{I}}_{3}(t,x_{2},x_{3})=\frac{1}{2\pi}\sum_{n=(n_{2},n_{3})\in\mathbb{Z}^{2}}\iint_{[-L,L)^{2}}\frac{x_{2}-y_{2}-2Ln_{2}}{|x-y-2Ln|^{2}}\omega^{\mathcal{I}}_{1}(t,y_{2},y_{3})dy_{2}dy_{3}.\end{split}$

Using this, we can prove a log-Lipschitz estimate of $u^{\mathcal{I}}$.

######  Lemma 3.3.

Let $x,x^{\prime}\in\mathbb{T}^{2}:=[-L,L)^{2}$. Then we have

(3.9) $\left|u^{\mathcal{I}}(t,x)-u^{\mathcal{I}}(t,x^{\prime})\right|\lesssim
e^{tM}|x-x^{\prime}|\left(1+\log\frac{3L}{|x-x^{\prime}|}\right).$

Note that the argument of the logarithm in (3.9) is always greater than 1
because $|x-x^{\prime}|\leq 2\sqrt{2}L$. We omit the proof since it follows
directly from the standard log-Lipschitz estimate for 2d Euler (see for
instance [11]).

Now we consider a characteristic curve
$A^{\mathcal{I}}(t,a_{2},a_{3})=\left(A^{\mathcal{I}}_{2}(t,a_{2},a_{3}),A^{\mathcal{I}}_{3}(t,a_{2},a_{3})\right):[0,\infty)\times\mathbb{T}^{2}\rightarrow\mathbb{T}^{2}$
of (3.6) defined by $A^{\mathcal{I}}(0,a_{2},a_{3})=(a_{2},a_{3})$ and

(3.10) $\left\\{\begin{aligned}
&\frac{d}{dt}A^{\mathcal{I}}_{2}(t,a_{2},a_{3})=u^{\mathcal{I}}_{2}(t,A^{\mathcal{I}}(t,a_{2},a_{3}))-MA^{\mathcal{I}}_{2}(t,a_{2},a_{3}),\\\
&\frac{d}{dt}A^{\mathcal{I}}_{3}(t,a_{2},a_{3})=u^{\mathcal{I}}_{3}(t,A^{\mathcal{I}}(t,a_{2},a_{3})).\end{aligned}\right.$

Evaluating along this curve, we have from (3.6)

(3.11)
$\omega^{\mathcal{I}}_{1}(t,A^{\mathcal{I}}_{2}(t,a),A^{\mathcal{I}}_{3}(t,a))=e^{Mt}\omega^{\mathcal{I}}_{1,0}(a_{2},a_{3}).$

We need the following two lemmas for $A^{\mathcal{I}}$.

######  Lemma 3.4.

The determinant of $\nabla_{a}A^{\mathcal{I}}$ with $a=(a_{2},a_{3})$
satisfies

(3.12) $\det(\nabla_{a}A^{\mathcal{I}})=e^{-Mt}.$

###### Proof.

Using $u^{\mathcal{I}}_{1}=0$ and $\nabla\cdot u^{\mathcal{I}}=0$, we compute

$\displaystyle\frac{d}{dt}\det(\nabla_{a}A^{\mathcal{I}})(t,a)$
$\displaystyle=(\partial_{2}u^{\mathcal{I}}_{2}(t,A^{\mathcal{I}}(t,a))-M+\partial_{3}u^{\mathcal{I}}_{3}(t,A^{\mathcal{I}}(t,a)))\det(\nabla_{a}A^{\mathcal{I}})(t,a)=-M\det(\nabla_{a}A^{\mathcal{I}})(t,a).$

Since $\det(\nabla_{a}A^{\mathcal{I}})(0,a)=1$, we obtain (3.12). ∎

######  Lemma 3.5.

Let $1\leq a_{2},a_{3}\leq 2$. Then for $0\leq t\leq T_{M}$, there exist
constants $C_{1},\,C_{2}>0$, $C_{3}\geq 1$ independent of $M$ such that

(3.13) $C_{1}e^{-C_{3}Mt}\leq A^{\mathcal{I}}_{2}(t,a_{2},a_{3})\leq
C_{2}e^{-C_{3}^{-1}Mt},$

and

(3.14) $C_{1}\leq A^{\mathcal{I}}_{3}(t,a_{2},a_{3})\leq C_{2}.$

###### Proof.

Note that $u^{\mathcal{I}}_{2}(t,0,A^{\mathcal{I}}_{3}(t,a_{2},a_{3}))=0$ for
all $t$ by odd symmetry of $\omega^{\mathcal{I}}_{1}$, which gives

(3.15)
$\frac{d}{dt}A^{\mathcal{I}}_{2}(t,a_{2},a_{3})=u^{\mathcal{I}}_{2}(t,A^{\mathcal{I}}(t,a_{2},a_{3}))-u^{\mathcal{I}}_{2}(t,0,A_{3}^{\mathcal{I}}(t,a_{2},a_{3}))-MA^{\mathcal{I}}_{2}(t,a_{2},a_{3}).$

To begin with, we show the upper bound of $A^{\mathcal{I}}_{2}(t,a_{2},a_{3})$
in (3.13). Applying (3.9) and noticing $A^{\mathcal{I}}_{2}(t,a_{2},a_{3})>0$,
we have

$\frac{d}{dt}A^{\mathcal{I}}_{2}(t,a_{2},a_{3})\leq
Ce^{tM}A_{2}^{\mathcal{I}}(t,a_{2},a_{3})\left(1+\log\frac{3L}{A_{2}^{\mathcal{I}}(t,a_{2},a_{3})}\right)-MA^{\mathcal{I}}_{2}(t,a_{2},a_{3}),$

which leads to

$\frac{d}{dt}\left(\log A^{\mathcal{I}}_{2}(t,a_{2},a_{3})-(1+\log
3L)\right)\leq-Ce^{tM}\left(\log A^{\mathcal{I}}_{2}(t,a_{2},a_{3})-(1+\log
3L)\right)-M.$

From

$\frac{d}{dt}\left(\left(\log A^{\mathcal{I}}_{2}(t,a_{2},a_{3})-(1+\log
3L)\right)\exp{\left(\int_{0}^{t}Ce^{sM}ds\right)}\right)\leq-M\exp{\left(\int_{0}^{t}Ce^{sM}ds\right)}\leq-M,$

we see that for $0\leq t\leq T_{M}$,

(3.16) $\begin{split}\log A^{\mathcal{I}}_{2}(t,a_{2},a_{3})&\leq 1+\log
3L+\exp\left(-\int_{0}^{t}Ce^{sM}ds\right)\left(\log a_{2}-(1+\log
3L)-Mt\right)\\\ &\leq 1+\log 3L+\exp(-C)\left(\log a_{2}-(1+\log
3L)-Mt\right),\end{split}$

where we used $\log a_{2}-(1+\log 3L)-Mt<0$ and $t\leq
T_{M}=\frac{\log(M+1)}{M}$. Since $a_{2}\leq 2$, we arrive at

$A^{\mathcal{I}}_{2}(t,a_{2},a_{3})\lesssim\exp\left(\exp(-C)(\log 2-(1+\log
3L)-Mt)\right)$

for $t\in[0,T_{M}]$. We take $C_{3}:=\exp(C)$, which satisfies $C_{3}\geq 1$.

Now we prove the lower bound of $A^{\mathcal{I}}_{2}(t,a_{2},a_{3})$ in
(3.13). Recalling (3.15) and (3.9), we have

$\begin{split}-\frac{d}{dt}A^{\mathcal{I}}_{2}(t,a_{2},a_{3})&=-u^{\mathcal{I}}_{2}(t,A^{\mathcal{I}}(t,a_{2},a_{3}))+u^{\mathcal{I}}_{2}(t,0,A_{3}^{\mathcal{I}}(t,a_{2},a_{3}))+MA^{\mathcal{I}}_{2}(t,a_{2},a_{3})\\\
&\leq
Ce^{tM}A_{2}^{\mathcal{I}}(t,a_{2},a_{3})\left(1+\log\frac{3L}{A_{2}^{\mathcal{I}}(t,a_{2},a_{3})}\right)+MA^{\mathcal{I}}_{2}(t,a_{2},a_{3}),\end{split}$

which yields

$\frac{d}{dt}\left(1+\log\frac{3L}{A_{2}^{\mathcal{I}}(t,a_{2},a_{3})}\right)\leq
Ce^{tM}\left(1+\log\frac{3L}{A_{2}^{\mathcal{I}}(t,a_{2},a_{3})}\right)+M.$

Noticing

$\frac{d}{dt}\left(\left(1+\log\frac{3L}{A_{2}^{\mathcal{I}}(t,a_{2},a_{3})}\right)\exp{\left(-\int_{0}^{t}Ce^{sM}ds\right)}\right)\leq
M\exp{\left(-\int_{0}^{t}Ce^{sM}ds\right)}\leq M,$

we see that for $0\leq t\leq T_{M}$,

(3.17)
$\begin{split}\log\frac{3L}{A_{2}^{\mathcal{I}}(t,a_{2},a_{3})}&\leq-1+\exp\left(\int_{0}^{t}Ce^{sM}ds\right)\left(1+\log\frac{3L}{a_{2}}+Mt\right)\leq-1+C_{3}\left(1+\log\frac{3L}{a_{2}}+Mt\right)\end{split}$

where $C_{3}:=\exp(C)$. Since $a_{2}\geq 1$, we have
$A^{\mathcal{I}}_{2}(t,a_{2},a_{3})\geq 3L\exp\left(1-C_{3}(1+\log
3L+Mt)\right).$ To obtain the bounds of $A^{\mathcal{I}}_{3}(t,a_{2},a_{3})$
in (3.14), we note that
$u^{\mathcal{I}}_{3}(t,A^{\mathcal{I}}_{2}(t,a_{2},a_{3}),0)=0$ (by the odd
symmetry of $\omega^{\mathcal{I}}_{1}$) and
$A^{\mathcal{I}}_{3}(t,a_{2},a_{3})>0$ for all $t>0$, and proceed as we have
shown the case of $A^{\mathcal{I}}_{2}(t,a_{2},a_{3})$. ∎

Henceforth, let $C_{i}$ $(i=1,2,3)$ denote the constants in Lemma 3.5. Now we
are ready to estimate ${\left\|\nabla
u^{\mathcal{I}}(t,\cdot)\right\|}_{L^{\infty}}$ near the origin.

######  Lemma 3.6.

Let $\delta:=\frac{C_{1}}{10(M+1)^{C_{3}}}$. Then for $t\in[0,T_{M}]$, we have

(3.18) $\left\|{\nabla
u^{\mathcal{I}}(t,\cdot)}\right\|_{L^{\infty}(B_{0}(\delta))}\lesssim
e^{-C_{3}^{-1}Mt}.$

######  Remark 3.7.

We note that $\delta\leq\frac{C_{1}e^{-C_{3}Mt}}{10}$ for $0\leq t\leq T_{M}$
by the definitions of $\delta$ and $T_{M}$.

###### Proof.

Let $x=(x_{2},x_{3})\in B_{0}(\delta)$. We recall explicit formulas

(3.19)
$\begin{split}&\partial_{2}u^{\mathcal{I}}_{2}(t,x_{2},x_{3})=-\partial_{3}u^{\mathcal{I}}_{3}(t,x_{2},x_{3})\\\
&\qquad=\frac{1}{\pi}\sum_{n=(n_{2},n_{3})\in\mathbb{Z}^{2}}p.v.\iint_{[-L,L)^{2}}\frac{(x_{2}-y_{2}-2Ln_{2})(x_{3}-y_{3}-2Ln_{3})}{|x-y-2Ln|^{4}}\omega^{\mathcal{I}}_{1}(t,y_{2},y_{3})dy_{2}dy_{3}\\\
\end{split}$

Moreover, since $\omega^{\mathcal{I}}_{1}=0$ in $[0,T_{M}]\times
B_{0}(\delta)$ by Lemma 3.5,

(3.20)
$\begin{split}&\partial_{2}u^{\mathcal{I}}_{3}(t,x_{2},x_{3})=-\partial_{3}u^{\mathcal{I}}_{2}(t,x_{2},x_{3})\\\
&\qquad=\frac{1}{2\pi}\sum_{n=(n_{2},n_{3})\in\mathbb{Z}^{2}}p.v.\iint_{[-L,L)^{2}}\frac{(x_{2}-y_{2}-2Ln_{2})^{2}-(x_{3}-y_{3}-2Ln_{3})^{2}}{|x-y-2Ln|^{4}}\omega^{\mathcal{I}}_{1}(t,y_{2},y_{3})dy_{2}dy_{3}\end{split}$

in $[0,T_{M}]\times B_{0}(\delta)$. (See e.g. [11] for derivations of (3.19)
and (3.20).) To begin with, we estimate
$\left\|{\partial_{2}u^{\mathcal{I}}_{2}}\right\|_{L^{\infty}(B_{0}(\delta))}=\left\|{\partial_{3}u^{\mathcal{I}}_{3}}\right\|_{L^{\infty}(B_{0}(\delta))}$.

By the odd symmetry of $\omega^{\mathcal{I}}_{1}$ in both $x_{2}$ and $x_{3}$,
(3.19) yields

$\begin{split}&\partial_{2}u^{\mathcal{I}}_{2}(t,x_{2},x_{3})\\\
&\approx\sum_{n\in\mathbb{Z}^{2}}\iint_{\left\\{0\leq y_{2},y_{3}\leq
L\right\\}}\left[\frac{(x_{2}-y_{2}-2Ln_{2})(x_{3}-y_{3}-2Ln_{3})}{|x-y-2Ln|^{4}}-\frac{(x_{2}+y_{2}-2Ln_{2})(x_{3}-y_{3}-2Ln_{3})}{\left((x_{2}+y_{2}-2Ln_{2})^{2}+(x_{3}-y_{3}-2Ln_{3})^{2}\right)^{2}}\right.\\\
&\left.\quad-\frac{(x_{2}-y_{2}-2Ln_{2})(x_{3}+y_{3}-2Ln_{3})}{\left((x_{2}-y_{2}-2Ln_{2})^{2}+(x_{3}+y_{3}-2Ln_{3})^{2}\right)^{2}}+\frac{(x_{2}+y_{2}-2Ln_{2})(x_{3}+y_{3}-2Ln_{3})}{\left((x_{2}+y_{2}-2Ln_{2})^{2}+(x_{3}+y_{3}-2Ln_{3})^{2}\right)^{2}}\right]\omega^{\mathcal{I}}_{1}(t,y_{2},y_{3})dy_{2}y_{3}\\\
&=\mathrm{I}+\mathrm{II}+\mathrm{III}+\mathrm{IV}.\end{split}$

To estimate $\mathrm{I}$, we divide it into $\mathrm{I}_{0}$ and
$\mathrm{I}_{\neq 0}$ which correspond to the term with $n=(0,0)$ and the sum
of all other terms with $n\neq(0,0)$, respectively. For $\mathrm{I}_{0}$, we
make a change of variables $(y_{2},y_{3})\mapsto(a_{2},a_{3})$ by
$y=A^{\mathcal{I}}(t,a)$ and use (3.12) to have

${\mathrm{I}_{0}}\approx\iint_{\left\\{0\leq a_{2},a_{3}\leq
L\right\\}}\frac{(x_{2}-A^{\mathcal{I}}_{2}(t,a))(x_{3}-A^{\mathcal{I}}_{3}(t,a))}{|x-A^{\mathcal{I}}(t,a)|^{4}}\omega^{\mathcal{I}}_{1}(t,A^{\mathcal{I}}_{2}(t,a),A^{\mathcal{I}}_{3}(t,a))e^{-Mt}da_{2}da_{3}.$

By (3.11) and the assumption that $\omega^{\mathcal{I}}_{1,0}$ is supported on
$\left\\{1\leq a_{2},a_{3}\leq 2\right\\}$ (see (1.5)), we obtain

${\mathrm{I}_{0}}\leq{\left\|\omega^{\mathcal{I}}_{1,0}\right\|}_{L^{\infty}}\iint_{\left\\{1\leq
a_{2},a_{3}\leq
2\right\\}}\frac{\left|(x_{2}-A^{\mathcal{I}}_{2}(t,a))(x_{3}-A^{\mathcal{I}}_{3}(t,a))\right|}{|x-A^{\mathcal{I}}(t,a)|^{4}}da_{2}da_{3}.$

Since $x\in B_{0}(\delta)$, Lemma 3.5 and Remark 3.7 imply

(3.21) $\left\\{\begin{aligned} \frac{9C_{1}e^{-C_{3}Mt}}{10}\leq
A^{\mathcal{I}}_{2}(t,a)-|x_{2}|&\leq A^{\mathcal{I}}_{2}(t,a)-x_{2}\leq
A^{\mathcal{I}}_{2}(t,a)+|x_{2}|\leq\frac{11C_{2}e^{-C_{3}^{-1}Mt}}{10},\\\
\frac{9C_{1}}{10}\leq A^{\mathcal{I}}_{3}(t,a)-|x_{3}|&\leq
A^{\mathcal{I}}_{3}(t,a)-x_{3}\leq
A^{\mathcal{I}}_{3}(t,a)+|x_{3}|\leq\frac{11C_{2}}{10}.\end{aligned}\right.$

Consequently,

$\frac{\left|(x_{2}-A^{\mathcal{I}}_{2}(t,a))(x_{3}-A^{\mathcal{I}}_{3}(t,a))\right|}{|x-A^{\mathcal{I}}(t,a)|^{4}}\leq\frac{\left|(x_{2}-A^{\mathcal{I}}_{2}(t,a))(x_{3}-A^{\mathcal{I}}_{3}(t,a))\right|}{(x_{3}-A^{\mathcal{I}}_{3}(t,a))^{4}}\lesssim
e^{-C_{3}^{-1}Mt},$

which gives ${\mathrm{I}_{0}}\lesssim e^{-C_{3}^{-1}Mt}$. To estimate
$\mathrm{I}_{\neq 0}$, denoting $\tilde{n}:=(-n_{2},n_{3})$, we compute

$\begin{split}&\frac{(x_{2}-y_{2}-2Ln_{2})(x_{3}-y_{3}-2Ln_{3})}{|x-y-2Ln|^{4}}+\frac{(x_{2}-y_{2}+2Ln_{2})(x_{3}-y_{3}-2Ln_{3})}{|x-y-2L\tilde{n}|^{4}}\\\
&=\frac{(x_{3}-y_{3}-2Ln_{3})\left((x_{2}-y_{2})(|x-y-2L\tilde{n}|^{4}+|x-y-2Ln|^{4})-2L_{2}(|x-y-2L\tilde{n}|^{4}-|x-y-2Ln|^{4})\right)}{|x-y-2Ln|^{4}|x-y-2L\tilde{n}|^{4}}.\end{split}$

Since
$|x-y-2L\tilde{n}|^{4}-|x-y-2Ln|^{4}=16Ln_{2}(x_{2}-y_{2})\left((x_{2}-y_{2})^{2}+4L^{2}n_{2}^{2})+(x_{3}-y_{3}-2Ln_{3})^{2}\right)$
and $|x-y|\leq|x|+|y|\leq\delta+L$ implies

(3.22) $|x-y-2Ln|,\,|x-y-2L\tilde{n}|\gtrsim|n|,$

we have

$\left|\frac{(x_{2}-y_{2}-2Ln_{2})(x_{3}-y_{3}-2Ln_{3})}{|x-y-2Ln|^{4}}+\frac{(x_{2}-y_{2}+2Ln_{2})(x_{3}-y_{3}-2Ln_{3})}{|x-y-2L\tilde{n}|^{4}}\right|\lesssim\frac{{|x_{2}-y_{2}|}}{|n|^{3}}.$

Hence, making again the change of variables
$(y_{2},y_{3})\mapsto(a_{2},a_{3})$ by $y=A^{\mathcal{I}}(t,a)$ and using
(3.21), we proceed in the same argument to obtain ${\mathrm{I}_{\neq
0}}\lesssim e^{-C_{3}^{-1}Mt}.$ In a similar manner, we can also show that all
of $\mathrm{II},\,\mathrm{III},$ and $\mathrm{IV}$ have the same upper bound,
which implies
$\left\|{\partial_{2}u^{\mathcal{I}}_{2}}\right\|_{L^{\infty}(B_{0}(\delta))}\lesssim
e^{-C_{3}^{-1}Mt}.$

Next, we estimate
$\left\|{\partial_{3}u^{\mathcal{I}}_{2}}\right\|_{L^{\infty}(B_{0}(\delta))}\left(=\left\|{\partial_{2}u^{\mathcal{I}}_{3}}\right\|_{L^{\infty}(B_{0}(\delta))}\right)$.
By the odd symmetry of $\omega^{\mathcal{I}}_{1}$ in both $x_{2}$ and $x_{3}$,
(3.20) yields

$\begin{split}&\partial_{3}u^{\mathcal{I}}_{2}(t,x_{2},x_{3})\\\
&\approx\sum_{n\in\mathbb{Z}^{2}}\iint_{\left\\{0\leq y_{2},y_{3}\leq
L\right\\}}\left[\frac{(x_{2}-y_{2}-2Ln_{2})^{2}-(x_{3}-y_{3}-2Ln_{3})^{2}}{|x-y-2Ln|^{4}}-\frac{(x_{2}+y_{2}-2Ln_{2})^{2}-(x_{3}-y_{3}-2Ln_{3})^{2}}{\left((x_{2}+y_{2}-2Ln_{2})^{2}+(x_{3}-y_{3}-2Ln_{3})^{2}\right)^{2}}\right.\\\
&\left.\quad-\frac{(x_{2}-y_{2}-2Ln_{2})^{2}-(x_{3}+y_{3}-2Ln_{3})^{2}}{\left((x_{2}-y_{2}-2Ln_{2})^{2}+(x_{3}+y_{3}-2Ln_{3})^{2}\right)^{2}}+\frac{(x_{2}+y_{2}-2Ln_{2})^{2}-(x_{3}+y_{3}-2Ln_{3})^{2}}{\left((x_{2}+y_{2}-2Ln_{2})^{2}+(x_{3}+y_{3}-2Ln_{3})^{2}\right)^{2}}\right]\omega^{\mathcal{I}}_{1}(t,y_{2},y_{3})dy_{2}y_{3}\\\
&=\mathrm{I}+\mathrm{II}+\mathrm{III}+\mathrm{IV}.\end{split}$

for $(t,x)\in[0,T_{M}]\times B_{0}(\delta)$. To bound
$\mathrm{I}+\mathrm{II}$, we again divide it into
$(\mathrm{I}+\mathrm{II})_{0}$ and $(\mathrm{I}+\mathrm{II})_{\neq 0}$ which
correspond to the term with $n=(0,0)$ and the sum of all other terms with
$n\neq(0,0)$, respectively. For $(\mathrm{I}+\mathrm{II})_{0}$, we again make
a change of variables $(y_{2},y_{3})\mapsto(a_{2},a_{3})$ by
$y=A^{\mathcal{I}}(t,a)$ and use (3.12) to have

$\begin{split}{(\mathrm{I}+\mathrm{II})_{0}}=\frac{1}{2\pi}\iint_{\left\\{0\leq
a_{2},a_{3}\leq
L\right\\}}K_{x_{2},x_{3}}(A^{\mathcal{I}}_{2}(t,a),A^{\mathcal{I}}_{3}(t,a))\,\omega^{\mathcal{I}}_{1}(t,A^{\mathcal{I}}_{2}(t,a),A^{\mathcal{I}}_{3}(t,a))e^{-Mt}da_{2}da_{3},\end{split}$

where

$K_{x_{2},x_{3}}(z_{2},z_{3})=\frac{\left((x_{2}-z_{2})^{2}-(x_{3}-z_{3})^{2}\right)\left((x_{2}+z_{2})^{2}+(x_{3}-z_{3})^{2}\right)^{2}-\left((x_{2}+z_{2})^{2}-(x_{3}-z_{3})^{2}\right)|x-z|^{4}}{|x-z|^{4}\left((x_{2}+z_{2})^{2}+(x_{3}-z_{3})^{2}\right)^{2}}$

for $z=(z_{2},z_{3})$. From (3.11) and the assumption that
$\omega^{\mathcal{I}}_{1,0}$ is supported on $\left\\{1\leq a_{2},a_{3}\leq
2\right\\}$, we have

${(\mathrm{I}+\mathrm{II})_{0}}\leq{\left\|\omega^{\mathcal{I}}_{1,0}\right\|}_{L^{\infty}}\iint_{\left\\{1\leq
a_{2},a_{3}\leq
2\right\\}}\left|K_{x_{2},x_{3}}(A^{\mathcal{I}}_{2}(t,a),A^{\mathcal{I}}_{3}(t,a))\right|da_{2}da_{3}.$

Note that the numerator of $K_{x_{2},x_{3}}(z_{2},z_{3})$ can be written as

(3.23) $\sum_{\begin{subarray}{c}\alpha_{1},\alpha_{2},\alpha_{4}\geq
0,\alpha_{3}\geq 1\\\
\alpha_{1}+\alpha_{2}+\alpha_{3}+\alpha_{4}=6,\alpha_{3}\geq
1\end{subarray}}C_{(\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4})}x_{2}^{\alpha_{1}}x_{3}^{\alpha_{2}}z_{2}^{\alpha_{3}}z_{3}^{\alpha_{4}}$

for some constants $C_{(\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4})}$’s.
Moreover, the denominator of $K_{x_{2},x_{3}}(z_{2},z_{3})$ is bounded below
by $(x_{3}-z_{3})^{8}$. Thus, using Lemma 3.5 and Remark 3.7, we have

$\begin{split}\left|K_{x_{2},x_{3}}(A^{\mathcal{I}}_{2}(t,a),A^{\mathcal{I}}_{3}(t,a))\right|&\lesssim\sum_{\begin{subarray}{c}\alpha_{1},\alpha_{2},\alpha_{4}\geq
0,\alpha_{3}\geq 1\\\
\alpha_{1}+\alpha_{2}+\alpha_{3}+\alpha_{4}=6,\alpha_{3}\geq
1\end{subarray}}\frac{\left|x_{2}^{\alpha_{1}}x_{3}^{\alpha_{2}}(A^{\mathcal{I}}_{2})^{\alpha_{3}}(A^{\mathcal{I}}_{3})^{\alpha_{4}}\right|}{(x_{3}-A^{\mathcal{I}}_{3}(t,a))^{8}}\\\
&\lesssim\sum_{\begin{subarray}{c}\alpha_{1},\alpha_{2},\alpha_{4}\geq
0,\alpha_{3}\geq 1\\\
\alpha_{1}+\alpha_{2}+\alpha_{3}+\alpha_{4}=6,\alpha_{3}\geq
1\end{subarray}}\left(e^{-C_{3}Mt}\right)^{\alpha_{1}}\left(e^{-C_{3}Mt}\right)^{\alpha_{2}}\left(e^{-C_{3}^{-1}Mt}\right)^{\alpha_{3}}\lesssim
e^{-C_{3}^{-1}Mt}\end{split}$

for $t\in[0,T_{M}]$. In the last line, we used $\alpha_{3}\geq 1$. This gives
${(\mathrm{I}+\mathrm{II})_{0}}\lesssim e^{-C_{3}^{-1}Mt}$. For
$(\mathrm{I}+\mathrm{II})_{\neq 0}$, we recall (3.23) and (3.22), which give

$\begin{split}\left|K_{x_{2}-2Ln_{2},x_{3}-2Ln_{3}}(y_{2},y_{3})\right|&\lesssim\sum_{\begin{subarray}{c}\alpha_{1},\alpha_{2},\alpha_{4}\geq
0,\alpha_{3}\geq 1\\\
\alpha_{1}+\alpha_{2}+\alpha_{3}+\alpha_{4}=6,\alpha_{3}\geq
1\end{subarray}}\frac{\left|(x_{2}-2Ln_{2})^{\alpha_{1}}(x_{3}-2Ln_{3})^{\alpha_{2}}y_{2}^{\alpha_{3}}y_{3}^{\alpha_{4}}\right|}{|n|^{8}}\lesssim\frac{y_{2}}{|n|^{3}},\end{split}$

where in the last inequality, we used $\alpha_{1}+\alpha_{2}\leq 5$ and
$\alpha_{3}\geq 1$. Hence proceeding as before, we obtain
$(\mathrm{I}+\mathrm{II})_{\neq 0}\lesssim e^{-C_{3}^{-1}Mt}.$ In the same
way, we can also show that $\mathrm{III}+\mathrm{IV}\lesssim
e^{-C_{3}^{-1}Mt}$, and consequently, we obtain
$\left\|{\partial_{3}u^{\mathcal{I}}_{2}}\right\|_{L^{\infty}(B_{0}(\delta))}\lesssim
e^{-C_{3}^{-1}Mt}.$ ∎

###  3.2. Linearization of the equation for $\omega^{\mathcal{S},P}$

Abusing the notation as in the last section, we denote pseudo-solutions
$\omega^{\mathcal{S},P}$ and $u^{\mathcal{S},P}$ by $\omega^{\mathcal{S}}$ and
$u^{\mathcal{S}}$, respectively. Dropping nonlinear terms in (3.3), we obtain
the following linearized equation of $\omega^{\mathcal{S}}$:

(3.24)
$\partial_{t}\omega^{\mathcal{S}}+(u^{\mathcal{L}}+u^{\mathcal{I}})\cdot\nabla\omega^{\mathcal{S}}=\nabla(u^{\mathcal{L}}+u^{\mathcal{I}})\omega^{\mathcal{S}}+\nabla
u^{\mathcal{S}}\omega^{\mathcal{I}}-u^{\mathcal{S}}\cdot\nabla\omega^{\mathcal{I}}.$

Now recalling (3.10) and abusing the notation, we denote a characteristic
curve

$A^{\mathcal{I}}(t,a_{1},a_{2},a_{3})=\left(A^{\mathcal{I}}_{1}(t,a_{1},a_{2},a_{3}),A^{\mathcal{I}}_{2}(t,a_{1},a_{2},a_{3}),A^{\mathcal{I}}_{3}(t,a_{1},a_{2},a_{3})\right):[0,\infty)\times\mathbb{T}^{3}\rightarrow\mathbb{T}^{3}$

defined by $A^{\mathcal{I}}(0,a_{1},a_{2},a_{3})=(a_{1},a_{2},a_{3})$ and

(3.25) $\left\\{\begin{aligned}
&\frac{d}{dt}A^{\mathcal{I}}_{1}(t,a_{1},a_{2},a_{3})=MA^{\mathcal{I}}_{1}(t,a_{1},a_{2},a_{3}),\\\
&\frac{d}{dt}A^{\mathcal{I}}_{2}(t,a_{1},a_{2},a_{3})=u^{\mathcal{I}}_{2}(t,A_{2}^{\mathcal{I}}(t,a_{1},a_{2},a_{3}),A_{3}^{\mathcal{I}}(t,a_{1},a_{2},a_{3}))-MA^{\mathcal{I}}_{2}(t,a_{1},a_{2},a_{3}),\\\
&\frac{d}{dt}A^{\mathcal{I}}_{3}(t,a_{1},a_{2},a_{3})=u^{\mathcal{I}}_{3}(t,A_{2}^{\mathcal{I}}(t,a_{1},a_{2},a_{3}),A_{3}^{\mathcal{I}}(t,a_{1},a_{2},a_{3})).\end{aligned}\right.$

Then we can show the following.

######  Lemma 3.8.

Let $\delta=\frac{C_{1}}{10(M+1)^{C_{3}}}$ as in Lemma 3.6. Then for
$a=(a_{1},a_{2},a_{3})\in\mathbb{T}^{3}$ and
$\ell\leq\min\left\\{\frac{\delta}{M+1},(3Le)^{1-C_{3}}\delta^{C_{3}}\right\\}$,
the following statements hold:

  * •

if $a\in B_{0}(\ell)$, then $A^{\mathcal{I}}(t,a)\in B_{0}(\delta)$ for
$t\in[0,T_{M}]$,

  * •

if $a\in\mathbb{T}^{3}\backslash B_{0}(\ell)$, then
$A^{\mathcal{I}}(t,a)\in\mathbb{T}^{3}\backslash
B_{0}\left(C_{4}\left(\frac{\ell}{M+1}\right)^{C_{3}}\right)$ for
$t\in[0,T_{M}]$, where $C_{4}>0$ is a constant.

###### Proof.

Let us prove the first statement. Suppose that $a\in B_{0}(\ell)$. Then
$|a_{1}|\leq\ell$ implies that
$\left|A^{\mathcal{I}}_{1}(t,a_{1},a_{2},a_{3})\right|=|a_{1}|e^{Mt}\leq\ell(M+1)\leq\delta$
for $t\in[0,T_{M}]$. For $A^{\mathcal{I}}_{2}(t,a)$ and
$A^{\mathcal{I}}_{3}(t,a)$, we only need to consider the case when
$a_{2},a_{3}\geq 0$ by the odd symmetry of $\omega^{\mathcal{I}}$ in both
$x_{2}$ and $x_{3}$. We claim that if $a_{2},a_{3}\in B_{0}(\ell)$ with
$a_{2},a_{3}\geq 0$, then

(3.26) $0\leq A^{\mathcal{I}}_{2}(t,a_{1},a_{2},a_{3})\leq\delta
e^{-C_{3}^{-1}Mt},$

and

(3.27) $0\leq A^{\mathcal{I}}_{3}(t,a_{1},a_{2},a_{3})\leq\delta.$

Recalling (3.15), (3.9), and $A^{\mathcal{I}}_{2}(t,a_{1},a_{2},a_{3})\geq 0$,
we have

$\frac{d}{dt}A^{\mathcal{I}}_{2}(t,a_{1},a_{2},a_{3})\leq
Ce^{tM}A_{2}^{\mathcal{I}}(t,a_{1},a_{2},a_{3})\left(1+\log\frac{3L}{A_{2}^{\mathcal{I}}(t,a_{1},a_{2},a_{3})}\right)-MA^{\mathcal{I}}_{2}(t,a_{1},a_{2},a_{3}),$

Then proceeding as we did to derive (3.16), we see that for $0\leq t\leq
T_{M}$,

$\log A^{\mathcal{I}}_{2}(t,a_{1},a_{2},a_{3})\leq 1+\log
3L+C_{3}^{-1}\left(\log a_{2}-(1+\log 3L)-Mt\right),$

which is equivalent to

$A^{\mathcal{I}}_{2}(t,a_{1},a_{2},a_{3})\leq
3Le\left(\frac{a_{2}}{3Le}\right)^{C_{3}^{-1}}e^{-C_{3}^{-1}Mt}.$

But since we assumed $a_{2}\leq\ell\leq(3Le)^{1-C_{3}}\delta^{C_{3}}$, (3.26)
holds. Then, (3.27) can be handled by a parallel argument.

Next, we prove our second statement. Suppose that
$a\in\mathbb{T}^{3}\backslash B_{0}(\ell)$. Then $|a_{1}|\geq\ell$ implies
that
$\left|A^{\mathcal{I}}_{1}(t,a_{1},a_{2},a_{3})\right|=|a_{1}|e^{Mt}\geq\ell$
for all $t$. For $A^{\mathcal{I}}_{2}(t,a)$ and $A^{\mathcal{I}}_{3}(t,a)$, we
only need to consider the case when $a_{2},a_{3}\geq 0$ by the odd symmetry of
$\omega^{\mathcal{I}}$ in both $x_{2}$ and $x_{3}$. We claim that if
$a_{2},a_{3}\in\mathbb{T}^{3}\backslash B_{0}(\ell)$ with $a_{2},a_{3}\geq 0$,
then

(3.28)
$A^{\mathcal{I}}_{2}(t,a_{1},a_{2},a_{3})\geq(3Le)^{1-C_{3}}\ell^{C_{3}}e^{-C_{3}Mt},$

and

(3.29)
$A^{\mathcal{I}}_{3}(t,a_{1},a_{2},a_{3})\geq(3Le)^{1-C_{3}}\ell^{C_{3}}.$

From (3.15), (3.9), and $A^{\mathcal{I}}_{2}(t,a_{1},a_{2},a_{3})\geq 0$, we
have

$-\frac{d}{dt}A^{\mathcal{I}}_{2}(t,a_{1},a_{2},a_{3})\leq
Ce^{tM}A_{2}^{\mathcal{I}}(t,a_{2},a_{3})\left(1+\log\frac{3L}{A_{2}^{\mathcal{I}}(t,a_{1},a_{2},a_{3})}\right)+MA^{\mathcal{I}}_{2}(t,a_{1},a_{2},a_{3}).$

With the same argument as the derivation of (3.17), we obtain for $0\leq t\leq
T_{M}$,

$\log\frac{3L}{A_{2}^{\mathcal{I}}(t,a_{1},a_{2},a_{3})}\leq-1+C_{3}\left(1+\log\frac{3L}{a_{2}}+Mt\right),$

so that

$A^{\mathcal{I}}_{2}(t,a_{1},a_{2},a_{3})\geq
3Le\left(\frac{a_{2}}{3Le}\right)^{C_{3}}e^{-C_{3}Mt}\geq(3Le)^{1-C_{3}}\ell^{C_{3}}e^{-C_{3}Mt}{,}$

where we used the assumption $a_{2}\geq\ell$ in the last inequality.
Similarly, we can show (3.29). Noticing $T_{M}=\frac{\log(M+1)}{M}$, our
second statement follows. ∎

Now we are ready to estimate
$\left\|{\omega^{\mathcal{S}}(t,\cdot)}\right\|_{L^{\infty}}$ near the origin.

######  Lemma 3.9.

Let $\ell$ in (1.6) satisfy
$\ell\leq\min\left\\{\frac{\delta}{M+1},(3Le)^{1-C_{3}}\delta^{C_{3}}\right\\}$.
Then for $0\leq t\leq T_{M}$, we have

$\left\|{\omega^{\mathcal{S}}(t,\cdot)}\right\|_{L^{\infty}\left(B_{0}\left(C_{4}\left(\frac{\ell}{M+1}\right)^{C_{3}}\right)\right)}\lesssim\varepsilon
e^{-Mt}.$

###### Proof.

Recalling $\omega^{\mathcal{I}}\equiv 0$ in $[0,T_{M}]\times B_{\delta}(0)$,
the previous lemma reduces (3.24) to

(3.30)
$\partial_{t}\omega^{\mathcal{S}}+(u^{\mathcal{L}}+u^{\mathcal{I}})\cdot\nabla\omega^{\mathcal{S}}=\nabla(u^{\mathcal{L}}+u^{\mathcal{I}})\omega^{\mathcal{S}}$

in $[0,T_{M}]\times B_{\delta}(0)$. First of all, we claim that
$\omega^{\mathcal{S}}_{1}(t,\cdot)=0$ in $[0,T_{M}]\times
B_{0}\left(C_{4}\left(\frac{\ell}{M+1}\right)^{C_{3}}\right)$. Indeed,
noticing $u^{\mathcal{I}}_{1}=0$ (see (3.5)), (3.30) gives

$\partial_{t}\omega^{\mathcal{S}}_{1}+Mx_{1}\partial_{1}\omega^{\mathcal{S}}_{1}+(-Mx_{2}+u^{\mathcal{I}}_{2})\partial_{2}\omega^{\mathcal{S}}_{1}+u^{\mathcal{I}}_{3}\partial_{3}\omega^{\mathcal{S}}_{1}=M\omega^{\mathcal{S}}_{1}.$

Evaluating along the characteristic $A^{\mathcal{I}}$, Lemma 3.8 implies
$\omega^{\mathcal{S}}_{1}=0$ in $[0,T_{M}]\times
B_{0}\left(C_{4}\left(\frac{\ell}{M+1}\right)^{C_{3}}\right)$ because
$\omega^{\mathcal{S}}_{1,0}=0$ in $B_{0}(\ell)$. Next, (3.5) reduces the
equations of $\omega^{\mathcal{S}}_{2}$ and $\omega^{\mathcal{S}}_{3}$ as
follows:

(3.31) $\left\\{\begin{aligned}
&\partial_{t}\omega^{\mathcal{S}}_{2}+Mx_{1}\partial_{1}\omega^{\mathcal{S}}_{2}+(-Mx_{2}+u^{\mathcal{I}}_{2})\partial_{2}\omega^{\mathcal{S}}_{2}+u^{\mathcal{I}}_{3}\partial_{3}\omega^{\mathcal{S}}_{2}=(-M+\partial_{2}u^{\mathcal{I}}_{2})\omega^{\mathcal{S}}_{2}+\partial_{3}u^{\mathcal{I}}_{2}\omega^{\mathcal{S}}_{3},\\\
&\partial_{t}\omega^{\mathcal{S}}_{3}+Mx_{1}\partial_{1}\omega^{\mathcal{S}}_{3}+(-Mx_{2}+u^{\mathcal{I}}_{2})\partial_{2}\omega^{\mathcal{S}}_{3}+u^{\mathcal{I}}_{3}\partial_{3}\omega^{\mathcal{S}}_{3}=\partial_{2}u^{\mathcal{I}}_{3}\omega^{\mathcal{S}}_{2}+\partial_{3}u^{\mathcal{I}}_{3}\omega^{\mathcal{S}}_{3},\end{aligned}\right.$

so that in $[0,T_{M}]\times
B_{0}\left(C_{4}\left(\frac{\ell}{M+1}\right)^{C_{3}}\right)$. Hence using
(3.18) and Lemma 3.8, we derive

(3.32)
$\begin{split}\partial_{t}|\omega^{\mathcal{S}}_{2}(t,A^{\mathcal{I}}(t,a))|&=\frac{\omega^{\mathcal{S}}_{2}(t,A^{\mathcal{I}}(t,a))\partial_{t}\left(\omega^{\mathcal{S}}_{2}(t,A^{\mathcal{I}}(t,a))\right)}{|\omega^{\mathcal{S}}_{2}(t,A^{\mathcal{I}}(t,a))|}\\\
&\leq\left(-M+Ce^{-C_{3}^{-1}Mt}\right)|\omega^{\mathcal{S}}_{2}(t,A^{\mathcal{I}}(t,a))|+Ce^{-C_{3}^{-1}Mt}|\omega^{\mathcal{S}}_{3}(t,A^{\mathcal{I}}(t,a))|\end{split}$

and similarly

(3.33) $\partial_{t}|\omega^{\mathcal{S}}_{3}(t,A^{\mathcal{I}}(t,a))|\leq
Ce^{-C_{3}^{-1}Mt}\left(|\omega^{\mathcal{S}}_{2}(t,A^{\mathcal{I}}(t,a))|+|\omega^{\mathcal{S}}_{3}(t,A^{\mathcal{I}}(t,a))|\right).$

Thus, from

$\partial_{t}\left(|\omega^{\mathcal{S}}_{2}(t,A^{\mathcal{I}}(t,a))|+|\omega^{\mathcal{S}}_{3}(t,A^{\mathcal{I}}(t,a))|\right)\lesssim
e^{-C_{3}^{-1}Mt}\left(|\omega^{\mathcal{S}}_{2}(t,A^{\mathcal{I}}(t,a))|+|\omega^{\mathcal{S}}_{3}(t,A^{\mathcal{I}}(t,a))|\right),$

we have

$|\omega^{\mathcal{S}}_{2}(t,A^{\mathcal{I}}(t,a))|+|\omega^{\mathcal{S}}_{3}(t,A^{\mathcal{I}}(t,a))|\lesssim\left|\omega^{\mathcal{S}}_{2,0}(a)+\omega^{\mathcal{S}}_{3,0}(a)\right|\lesssim\varepsilon,$

where we used (1.6) in the last inequality. Inserting
$|\omega^{\mathcal{S}}_{2}(t,A^{\mathcal{I}}(t,a))|\lesssim\varepsilon$ into
(3.33), we obtain

$\partial_{t}|\omega^{\mathcal{S}}_{3}(t,A^{\mathcal{I}}(t,a))|\leq
Ce^{-C_{3}^{-1}Mt}|\omega^{\mathcal{S}}_{3}(t,A^{\mathcal{I}}(t,a))|+C\varepsilon
e^{-C_{3}^{-1}Mt}.$

Noticing $\omega^{\mathcal{S}}_{3,0}(a)=0$ in $B_{0}(\ell)$, Grönwall’s
inequality gives

$\begin{split}|\omega^{\mathcal{S}}_{3}(t,A^{\mathcal{I}}(t,a))|&\leq\exp{\left(\int_{0}^{t}Ce^{-C_{3}^{-1}Ms}ds\right)}\int_{0}^{t}C\varepsilon
e^{-C_{3}^{-1}Ms}ds\\\
&=\exp\left(\frac{C\left(1-e^{-C_{3}^{-1}Mt}\right)}{C_{3}^{-1}M}\right)\frac{C\varepsilon\left(1-e^{-C_{3}^{-1}Mt}\right)}{C_{3}^{-1}M}\lesssim\frac{\varepsilon}{M}\lesssim\varepsilon
e^{-Mt}.\end{split}$

for $t\in[0,T_{M}]$. Inserting this into (3.32), we obtain

$\partial_{t}|\omega^{\mathcal{S}}_{2}(t,A^{\mathcal{I}}(t,a))|\leq\left(-M+Ce^{-C_{3}^{-1}Mt}\right)|\omega^{\mathcal{S}}_{2}(t,A^{\mathcal{I}}(t,a))|+C\varepsilon
e^{-(C_{3}^{-1}+1)Mt},$

which yields

$\begin{split}\partial_{t}\left(|\omega^{\mathcal{S}}_{2}(t,A^{\mathcal{I}}(t,a))|\exp\left(\int_{0}^{t}M-Ce^{-C_{3}^{-1}Ms}ds\right)\right)&\leq
C\varepsilon
e^{-(C_{3}^{-1}+1)Mt}\exp\left(\int_{0}^{t}M-Ce^{-C_{3}^{-1}Ms}ds\right)\lesssim\varepsilon
e^{-C_{3}^{-1}Mt}.\end{split}$

Integrating from $0$ to $t$, we obtain

$\begin{split}|\omega^{\mathcal{S}}_{2}(t,A^{\mathcal{I}}(t,a))|&\leq\exp\left(\int_{0}^{t}-M+Ce^{-C_{3}^{-1}Ms}ds\right)\left(\omega^{\mathcal{S}}_{2,0}(a)+\int_{0}^{t}C\varepsilon
e^{-C_{3}^{-1}Ms}ds\right)\lesssim\varepsilon e^{-Mt}.\qed\end{split}$

###  3.3. Comparison between solutions

In this section, we compare our pseudo-solutions with real solutions as we
mentioned in the beginning of this section. In order to distinguish solutions
of (3.3) and (3.24), we denote the solution of the linearized equation (3.24)
by $(\omega^{\mathcal{S},P,Lin},u^{\mathcal{S},P,Lin})$. We fix
$s>\frac{5}{2}$ and set $\bar{u}=u^{\mathcal{L}}+u^{\mathcal{I},P}$ in (2.1),
$u=u^{\mathcal{L}}+u^{\mathcal{I},P}+u^{\mathcal{S},P}$ in (2.2),
$\tilde{u}=u-\bar{u}$ in (2.3), and $\tilde{u}^{Lin}=u^{\mathcal{S},P,Lin}$ in
(2.4) to employ Proposition 2.1. (1.6) implies that $\tilde{u}_{0}$ in (2.3)
satisfies

(3.34)
${\left\|\tilde{u}_{0}\right\|}_{H^{s+1}(\mathbb{T}^{3})}\lesssim{\left\|\tilde{\psi}\right\|}_{H^{s}(\mathbb{T}^{3})}=\ell^{-s+\frac{3}{2}}{\left\|\psi\right\|}_{H^{s}(\mathbb{T}^{3})},$

and Lemma 3.2 gives

(3.35)
$\int_{0}^{T_{M}}{\left\|(u^{\mathcal{L}}+u^{\mathcal{I},P})(\tau,\cdot)\right\|}_{H^{s+2}}ds\lesssim
MT_{M}+\frac{e^{{C}MT_{M}}-1}{{C}M}\lesssim{M^{C}},$

by adjusting the value of absolute constant $C>1$ from an inequality to
another. Therefore, Proposition 2.1 implies

${\left\|\omega^{\mathcal{S},P}(t,\cdot)-\omega^{\mathcal{S},P,Lin}(t,\cdot)\right\|}_{L^{\infty}}\lesssim\varepsilon^{2}\ell^{-2s+3}e^{CM^{C}}$

on $[0,T_{M}]$, whenever $\varepsilon>0$ satisfies

(3.36) $\varepsilon\leq\ell^{s-\frac{3}{2}}e^{-CM^{C}}.$

Hence, it follows from Lemma 3.9 that

(3.37)
$\left\|{\omega^{\mathcal{S},P}(t,\cdot)}\right\|_{L^{\infty}\left(B_{0}\left(C_{4}\left(\frac{\ell}{M+1}\right)^{C_{3}}\right)\right)}\lesssim\varepsilon
e^{-Mt}$

on $[0,T_{M}]$ if we pick $\ell$ and $\varepsilon$ satisfying

(3.38)
$\ell\leq\min\left\\{\frac{\delta}{M+1},(3Le)^{1-C_{3}}\delta^{C_{3}}\right\\},\qquad\varepsilon\leq\ell^{2s-3}e^{-M^{C}}$

with $C$ adjusted. (Here, we have used that $M\gg 1$.)

Next, we set $\bar{u}^{*}:=u^{\mathcal{L}}+u^{\mathcal{I}}$, so that
$\bar{u}^{*}$ solves (2.8). Thus, using (3.18), (3.34), and (3.35),
Proposition 2.3 give us (1.8) for $\varepsilon,\,\ell$ satisfying (3.38) of
which $C$ is adjusted if necessary. To derive (1.9), noticing
$u=u^{\mathcal{L}}+u^{\mathcal{I},P}+u^{\mathcal{S},P}=u^{\mathcal{L}}+u^{\mathcal{I}}+u^{\mathcal{S}}$
and recalling (2.7), (3.34), and (3.35), there exists a constant $C>0$ such
that

$\frac{d}{dt}\left|\Phi(t,a)-A^{\mathcal{I}}(t,a)\right|\leq{\left\|u^{\mathcal{S},P}\right\|}_{L^{\infty}}{\lesssim{\left\|u^{\mathcal{S},P}\right\|}_{H^{s+1}}}\lesssim\varepsilon\ell^{-s+\frac{3}{2}}e^{CM^{C}}$

for $\varepsilon>0$ satisfying (3.36), where $\Phi,A^{\mathcal{I}}$ are from
(1.7), (3.25), respectively. Thus if $\varepsilon,\,\ell$ satisfy (3.38) of
which $C$ is adjusted if necessary, then the estimate

$\left|\Phi(t,a)-A^{\mathcal{I}}(t,a)\right|\lesssim\varepsilon\ell^{-s+\frac{3}{2}}e^{CM^{C}}t\lesssim\sqrt{\varepsilon}$

on $[0,T_{M}]$ and Lemma 3.8 imply

$\left|\Phi(t,a)\right|\leq\left|A^{\mathcal{I}}(t,a)\right|+\left|\Phi(t,a)-A^{\mathcal{I}}(t,a)\right|\leq\delta+\sqrt{\varepsilon}\leq
2\delta$

for $(t,a)\in[0,T_{M}]\times B_{0}(\ell)$ while

$\left|\Phi(t,a)\right|\geq\left|A^{\mathcal{I}}(t,a)\right|-\left|\Phi(t,a)-A^{\mathcal{I}}(t,a)\right|\geq
C_{4}\left(\frac{\ell}{M+1}\right)^{C_{3}}-\sqrt{\varepsilon}\geq
C_{5}\left(\frac{\ell}{M+1}\right)^{C_{3}}$

for $(t,a)\in[0,T_{M}]\times\mathbb{T}^{3}\backslash B_{0}(\ell)$ and some
$C_{5}>0$. Recall that $\omega^{\mathcal{I},P}=0$ in
$B_{0}(2\delta)\times[0,T_{M}]$ (see Lemma 3.5 and Remark 3.7). Hence by
(3.3), $\omega^{\mathcal{S},P}$ solves

$\partial_{t}\left(\omega^{\mathcal{S},P}(t,\Phi(t,a))\right)=\nabla
u(t,\Phi(t,a))\omega^{\mathcal{S},P}(t,\Phi(t,a))$

for $(t,a)\in[0,T_{M}]\times B_{0}(\ell)$. Since $\omega^{\mathcal{S}}$ also
solves

$\partial_{t}\left(\omega^{\mathcal{S}}(t,\Phi(t,a))\right)=\nabla
u(t,\Phi(t,a))\omega^{\mathcal{S}}(t,\Phi(t,a)),$

we have

$\partial_{t}\left|\omega^{\mathcal{S},P}(t,\Phi(t,a))-\omega^{\mathcal{S}}(t,\Phi(t,a))\right|\leq{\left\|\nabla
u\right\|}_{L^{\infty}}\left|\omega^{\mathcal{S},P}(t,\Phi(t,a))-\omega^{\mathcal{S}}(t,\Phi(t,a))\right|$

for $(t,a)\in[0,T_{M}]\times B_{0}(\ell)$. But ${\left\|\nabla
u\right\|}_{L^{\infty}}<\infty$ up to time $T_{M}$, and
$\omega^{\mathcal{S},P}$ and $\omega^{\mathcal{S}}$ have the same initial data
$\omega^{\mathcal{S}}_{0}$, so that
$\omega^{\mathcal{S},P}(t,\Phi(t,a))=\omega^{\mathcal{S}}(t,\Phi(t,a))$ for
$(t,a)\in[0,T_{M}]\times B_{0}(\ell)$. This implies
$\omega^{\mathcal{S},P}(t,x)=\omega^{\mathcal{S}}(t,x)$ for
$(t,x)\in[0,T_{M}]\times
B_{0}\left(C_{5}\left(\frac{\ell}{M+1}\right)^{C_{3}}\right)$, and therefore
(1.9) follows from (3.37). This completes our proof of Theorem A. $\Box$

Acknowledgments. Research of TY was partially supported by Grant-in-Aid for
Scientific Research B (20H01819), Japan Society for the Promotion of Science
(JSPS). IJ has been supported by the National Research Foundation of
Korea(NRF) grant No. 2022R1C1C1011051.

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