Obsidian/Zotero/001_artiklid/cuisharpcriticalmassthreshold2025.md

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category: literaturenote citekey: cuisharpcriticalmassthreshold2025 title: On the sharp critical mass threshold for the 3D Patlak-Keller-Segel-Navier-Stokes system via Couette flow authors: "Cui, Shikun; Wang, Lili; Wang, Wendong; Wei, Juncheng" year: 2025 date: 2025-00-00 2025 url: "https://arxiv.org/abs/2506.10578v1" zotero_key: 5SINNICG zotero_storage: JK2JSXJ7 collections: transpordi_planeerimine folder: 001_artiklid firstAuthor: "Cui, Shikun"

status: converted

ON THE SHARP CRITICAL MASS THRESHOLD FOR THE 3D PATLAK-KELLER-SEGEL-NAVIER-STOKES SYSTEM VIA COUETTE FLOW

SHIKUN CUI, LILI WANG, WENDONG WANG, AND JUNCHENG WEI

Abstract. As is well-known, the solution of the Patlak-Keller-Segel system in 3D may blow up in finite time regardless of any initial cell mass. In this paper, we are interested in the suppression of blow-up and the critical mass threshold for the 3D Patlak-Keller-Segel-Navier-Stokes system via the Couette flow (Ay, 0, 0). It is proved that if the Couette flow is sufficiently strong (A is large enough), then the solutions for the system are global in time in the periodic domain (x, y, z) ∈ T 3 as long as the initial cell mass is less than 16π . This result seems to be sharp, since the zero-mode function (the mean value in x−direction) of the three dimensional density is a complication of the two-dimensional Keller-Segel equations, whose critical mass in 2D is 8π. One new observation is the dissipative decay of (ue2,0, ^ue3,0) (see Lemma 4.3 for more details), then we combine the quasilinear method proposed by Wei-Zhang (Comm. Pure Appl. Math., 2021) with the zero-mode estimate of the density by the logarithmic Hardy-Littlewood-Sobolev inequality as Bedrossian-He (SIAM J. Math. Anal., 2017) or He (Nonlinearity, 2025) to obtain the bounded-ness of the density and the velocity.

Keywords: suppression of blow-up; Patlak-Keller-Segel-Navier-Stokes; stability; Couette flow

Contents

1. Introduction	2
2. Key ideas and proof of Theorem 1.1	6
2.1. Fourier analysis	6
2.2. The decomposition of the first component of the velocity	7
2.3. The construction of energy functional	9
2.4. Main steps	9
3. A priori estimates	10
4. Estimates for the zero modes of velocity E1(t): Proof of Proposition 2.1	18
4.1. Energy estimates for u2,0 and u3,0	19
4.2. Heat dissipation estimates for u2,0 and u3,0	21
4.3. Energy estimates for u1,0	25
5. Estimates for the non-zero modes	30
5.1. Energy estimates for E2,1(t)	30
5.2. Energy estimates for E2,2(t)	32
2 6. Estimates for L -norm of the density	35
∞-norm of the density 7. Estimates for L E3(t): Proof of Proposition 2.2	44
8. Energy estimates for E4(t): Proof of Proposition 2.3	46
Appendix A. Some useful estimates and lemmas in the proof	59
A.1. Several useful lemmas	59
A.2. Elliptic estimates	62

Date: June 13, 2025.

A.3. Space-time estimates	63
Acknowledgement	68
References	68

1. Introduction

Consider the following parabolic-elliptic Patlak-Keller-Segel (PKS) system

$$\begin{cases} \partial_t n = \triangle n - \nabla \cdot (n \nabla c), \ \triangle c + n = 0, \end{cases}$$ (1.1)

which is designed to depict the diffusion and chemotactic motion of chemical substances within a population of cells or microorganisms. Patlak made significant contributions in [25], and later, Keller and Segel further advanced it in [21]. Extensive applications have been found across diverse scientific fields, including biology, ecology, and medicine [17]. In the realm of biology, this system provides crucial insights into the complex behavior of cells, such as their migration, aggregation, and diffusion [16].

A well-known characteristic of the PKS system is its critical dependence on spatial dimension. For the one-dimensional PKS system, all its solutions are globally well-posed. When the spacial dimension is higher than one, the solutions of the PKS system (1.1) may blow up in finite time. In the two-dimensional space, the PKS system for both parabolic-elliptic and parabolic-parabolic forms have a critical mass of 8π. In the 2D space, the parabolic-elliptic PKS system is globally wellposed if and only if the total mass M ≤ 8π by Wei [33], see also Blanchet-Dolbeault-Perthame [3] for M < 8π. When M = 8π, the solution may blow up at infinity, see Blanchet-Carrillo-Masmoudi [2] and Davila-del Pino-Dolbeault-Musso-Wei [9] for the blow-up rate at infinity. The parabolicparabolic PKS model (△c is replaced by △c−∂tc in (1.1) 2 ) also has a critical mass of 8π in 2D: if the cell mass M := ||nin||L¹ is less than 8π, the solutions of the system are global in time proved by Calvez-Corrias [4]; if the cell mass is greater than 8π, the solutions will blow up in finite time proved by Schweyer [26].

When the spatial dimension is higher than two, the PKS system (1.1) becomes supercritical. In this condition, regardless of parabolic-elliptic form or parabolic-parabolic form, the finite-time blow-up may occur for arbitrarily small values of the initial mass. In this time, the solution with any initial mass may blow up in finite time. This behavior has been established in various settings: for the parabolic-elliptic case by Nagai [24] and Souplet-Winkler [29], and for the parabolicparabolic case by Winkler [35]. Additional results and further developments on this topic can be found in [2, 9, 31] and the related therein.

Generally, the processes involving chemical attraction take place in fluids. As said in [22]: A natural question is whether the presence of fluid flow can affect singularity formation by mixing the bacteria thus making concentration harder to achieve.

In this paper, we investigate the suppression of blow-up and the critical mass threshold of the following three-dimensional parabolic-elliptic Patlak-Keller-Segel (PKS) system coupled with Navier-Stokes (NS) equations in (x, y, z) ∈ T ³ with T = [0, 2π]:

$$\begin{cases} \partial_t n + v \cdot \nabla n = \triangle n - \nabla \cdot (n \nabla c), \ \Delta c + n - \bar{n} = 0, \ \partial_t v + v \cdot \nabla v + \nabla P = \triangle v + n \nabla \phi, \ \nabla \cdot v = 0, \end{cases}$$ (1.2)

along with initial conditions

$$(n,v)\big|{t=0} = (n{\rm in}, v_{\rm in}),$$

where n represents the cell density, c denotes the chemoattractant density, and v denotes the velocity of fluid. In addition, ¯n = |T| R ^T3 ndxdydz denotes the average of n, P is the pressure and ϕ is the given potential function.

Let us by briefly review some results obtained in the 2D case, which implies the 8π critical mass threshold vanishes by the mixing effect of the fluid. For the parabolic-elliptic PKS system of (1.2)¹ − (1.2)2, Kiselev-Xu [22] showed no critical mass threshold by suppressing the blow-up by stationary relaxation enhancing flows and time-dependent Yao-Zlatos near-optimal mixing flows in T ^d with d = 2, 3. Bedrossian-He [1] studied the suppression of blow-up by non-degenerate shear flows in T 2 for the 2D parabolic-elliptic case. He [12] investigated the suppression of blow-up for the parabolic-parabolic PKS model near the large strictly monotone shear flow in T × R. For the coupled PKS-NS system, Zeng-Zhang-Zi [36] firstly considered the 2D PKS-NS system near the Couette flow in T × R, and proved that if the Couette flow is sufficiently strong, the solution stays globally regular. He [13] considered the blow-up suppression for the parabolic-elliptic PKS-NS system in T × R with the coupling of buoyancy effects for a class of initial data with small vorticity. Wang-Wang-Zhang [32] studied the blow-up suppression of 2D supercritical parabolicelliptic PKS-NS system near the Couette flow in T × R. Li-Xiang-Xu [23] suppressed the blow-up for the PKS-NS system via the Poiseuille flow in T × R, and proved that the solution is global as long as the Poiseuille flow is enough strong. Furthermore, Cui-Wang [6] considered the blow-up suppression for the PKS-NS system in T × I with Navier-slip boundary condition. In addition, Hu [18] proved that sufficiently large buoyancy can also suppress the blow-up of the PKS system, see also the recent results by Hu-Kiselev-Yao [20] and Hu-Kiselev [19]. The blow-up also can be suppressed by adding some logistic terms, see [30] and the references therein.

In 3D, some results have been made in the study of the corresponding problem. Bedrossian-He [1] studied the suppression of blow-up for the parabolic-elliptic PKS system by shear flows in T ³ and T × R . Feng-Shi-Wang [11] suppressed the blow-up for the advective Kuramoto-Sivashinsky and the Keller-Segel equations via the planar helical flows. Shi-Wang [28] considered the suppression effect of the Couette-Poiseuille flow (z, z² , 0) in T ² × R, and Deng-Shi-Wang [10] proved the Couette flow with a sufficiently large amplitude prevents the blow-up of solutions in the whole space for exponential decay data. For the parabolic-parabolic PKS system, He [14] introduced a family of time-dependent alternating shear flows in the domain T 3 , and proved the solution remains globally regular as long as the flow is sufficiently strong. For a time-dependent shear flow, He [15] demonstrated that when the total mass of the cell density is below a specific threshold (8π|T|), the solution remains globally regular in T ³ as long as the flow is sufficiently strong.

For the 3D PKS-NS system, fewer results are obtained. The first three authors [7] considered the PKS system coupled with the linearized NS equations near the Couette flow in T × I × T, and showed that when A is big enough the solutions are global in time as long as M < ^8π 9 and A(∥u2,in∥L² +∥u3,in∥L² ) ≤ C. Then, the first three authors [8] studied the blow-up suppression and the nonlinear stability of the PKS-NS system for the Couette flow (Ay, 0, 0) in T × R × T, and proved that the solutions are global in time as long as the initial cell mass M < ²⁴ 5 π ² and initial velocity A ^{3 +}∥uin∥H² ≤ C.

Our main goal is to investigate the suppression of blow-up and obtain the optimal critical mass for the PKS-NS system (1.2) near the 3D Couette flow (Ay, 0, 0). Introduce a perturbation u = (u1, u2, u3) around the Couette flow (Ay, 0, 0), which u(t, x, y, z) = v(t, x, y, z) − (Ay, 0, 0) satisfying u ^t=0 = uin = (u1,in, u2,in, u3,in). Assume ϕ = x, then we rewrite the system (1.2) into

$$\begin{cases} \partial_t n + Ay \partial_x n + u \cdot \nabla n - \triangle n = -\nabla \cdot (n \nabla c), \ \Delta c + n - \bar{n} = 0, \ \partial_t u + Ay \partial_x u + \begin{pmatrix} Au_2 \ 0 \ 0 \end{pmatrix} - \Delta u + u \cdot \nabla u + \nabla P^{N_1} + \nabla P^{N_2} = \begin{pmatrix} n \ 0 \ 0 \end{pmatrix}, \ \nabla \cdot u = 0, \end{cases} (1.3)$$

where the pressure P ^N1 and P ^N2 are determined by

$$\Delta P^{N_1} = -2A\partial_x u_2 + \partial_x n,$$

$$\Delta P^{N_2} = -\text{div } (u \cdot \nabla u).$$ (1.4)

To estimate non-zero norms more conveniently, we introduce the vorticity ω² = ∂zu¹ − ∂xu³ and △u2, satisfying

$$\partial_t \omega_2 + Ay \partial_x \omega_2 + A \partial_z u_2 - \triangle \omega_2 = \partial_z n - \partial_z (u \cdot \nabla u_1) + \partial_x (u \cdot \nabla u_3)$$

and

$$\partial_t \triangle u_2 + Ay \partial_x \triangle u_2 - \triangle^2 u_2 = -\partial_y \partial_x n - (\partial_x^2 + \partial_z^2)(u \cdot \nabla u_2) + \partial_y [\partial_x (u \cdot \nabla u_1) + \partial_z (u \cdot \nabla u_3)].$$

After the time rescaling $t \mapsto \frac{t}{A}$ , we get

$$\begin{cases} \partial{t}n + y\partial{x}n + \frac{1}{A}u \cdot \nabla n - \frac{1}{A}\triangle n = -\frac{1}{A}\nabla \cdot (n\nabla c), \ \Delta c + n - \bar{n} = 0, \ \partial{t}\omega{2} + y\partial{x}\omega{2} - \frac{1}{A}\triangle\omega{2} + \partial{z}u{2} = -\frac{1}{A}\partial{z}(u \cdot \nabla u{1}) + \frac{1}{A}\partial{x}(u \cdot \nabla u{3}) + \frac{1}{A}\partial{z}n, \ \partial{t}\triangle u{2} + y\partial{x}\triangle u{2} - \frac{1}{A}\triangle(\triangle u{2}) = -\frac{1}{A}\partial{y}\partial{x}n - \frac{1}{A}(\partial{x}^{2} + \partial{z}^{2})(u \cdot \nabla u{2}) \

• \frac{1}{A}\partial{y}\left[\partial{x}(u \cdot \nabla u{1}) + \partial{z}(u \cdot \nabla u_{3})\right], \end{cases}$$

$$(1.5)$$

$$\nabla \cdot u = 0.$$

Before stating the result, we need to define the following modes

Our main result is stated as follows.

$$P_0 f = f0 = \frac{1}{|\mathbb{T}|} \int{\mathbb{T}} f(t, x, y, z) dx$$ , and $P{\neq} f = f{\neq} = f - f_0$ .

Throughout this paper, f⁰ and f̸⁼ respectively represent the zero and non-zero modes of f.

Theorem 1.1. Assume that 0 < nin(x, y, z) ∈ H² (T 3 ) and uin(x, y, z) ∈ H² (T 3 ). There exist a sufficiently small positive constant ϵ depending on ∥nin∥H2(T3) and ∥(uin)̸=∥H2(T3) , and a positive constant A¹ depending on ∥nin∥H2(T3) and ∥uin∥H2(T3) , such that if A ≥ A1,

$$|(u_{2,\text{in}})0|{H^2} + |(u_{3,\text{in}})0|{H^1} \le \epsilon \tag{1.6}$$

and

$$M = \int{\mathbb{T}^3} n{\rm in} dx dy dz < 16\pi^2. \tag{1.7}$$

Then the solution of (1.5) is global in time.

Remark 1.1. For the three-dimensional space, the index 16π 2 in (1.7) seems to be the sharp threshold for initial cell mass. Recall that n⁰ satisfies (see also (6.16))

$$\partial_t n_0 = \frac{1}{A} \Delta n_0 - \frac{1}{A} \nabla \cdot (n_0 \nabla c0) - \frac{1}{A} \nabla \cdot (n{\neq} \nabla c_{\neq})_0 - \frac{1}{A} (u_0 \cdot \nabla n0) - \frac{1}{A} (u{\neq} \cdot \nabla n_{\neq})_0. \tag{1.8}$$

When the velocity u vanishes and n(t, x, y, z) = n(t, y, z), which does not depend on the variable x. Then (1.8) is reduced to

$$\partial_t n_0 = \frac{1}{A} \Delta n_0 - \frac{1}{A} \nabla \cdot (n_0 \nabla c_0) + \text{``good terms''}, \tag{1.9}$$

which is similar to the 2D Keller-Segel equations and the critical mass is R ^T2 n0dydz = 8π = 1 2π R ^T3 ndxdydz. It implies that the critical mass threshold for the initial cell mass to the system (1.8) is just 16π 2 . A new observation is the logarithmic Hardy-Littlewood-Sobolev inequality and the dissipative decay of the (ue²,0, ^ue³,0) (see Lemma 4.3 in Section 4.2 for more details) to estimate the norm of ∥n0∥L² .

Remark 1.2. The small constant ϵ in (1.6) depends on ∥nin∥H2(T3) and ∥(uin)̸=∥H2(T3) , which can be found in (6.8), that is ϵm 1 ³E ³ ≤ C. Moreover, the estimate of E³ is decided by (6.14), (7.4) and Corollary 8.1.

Remark 1.3. In a very recent work [8], the first three authors investigated the suppression of blow-up and the nonlinear stability of the PKS-NS system via the 3D Couette flow in a different domain of T × R × T and ϕ = y. When A is big enough, as long as

$$\begin{cases} A^{\frac{1}{3}+}(|u{\text{in}}|{H^{2}(\mathbb{T}\times\mathbb{R}\times\mathbb{T})} + |(n{\text{in}}){(0,\neq)}|{L^{2}(\mathbb{T}\times\mathbb{R}\times\mathbb{T})}) \leq C, \ M = \int{\mathbb{T}\times\mathbb{R}\times\mathbb{T}} n_{\text{in}} dx dy dz < \frac{24}{5}\pi^{2}, \end{cases}$$

the solutions for the PKS-NS system are global in time, where a new energy estimate for ^√ ∂z ¹−△n(0,̸=) was introduced to control the density. It's still unknown whether the threshold for initial cell mass can be 16π 2 , since it is difficult to apply the logarithmic Hardy-Littlewood-Sobolev inequality without the dissipative decay of the (u2,0, u3,0) at this time.

Remark 1.4. According to the standard arguments, the result of local well-posedness of the system (1.5) exists, which can be refered to [8, 19, 35], and we omitted it.

Here are some notations used in this paper.

Notations:

• For given f(t, x, y, z), the Fourier transform can be defined by

$$f(t,x,y,z) = \sum_{k_1,k_2,k3 \in \mathbb{Z}} \widehat{f}{k_1,k_2,k_3}(t) e^{i(k_1x + k_2y + k_3z)},$$

where ^fb^k1,k2,k³ (t) = ¹ |T| R ^T3 f(t, x, y, z)e ⁻i(k1x+k2y+k3z)dxdydz.

• Especially, we use uj,0, and uj,̸⁼ to represent the zero mode and non-zero mode of the velocity u^j (j = 1, 2, 3), respectively. Similarly, we use ω2,⁰ and ω2,̸⁼ to represent the zero mode and non-zero mode of the vorticity ω2, respectively.
• The norm of the L p space and the time-space norm ∥f∥LqL^p are defined as ∥f∥Lp(T3) = R T3 |f| ^pdxdydz ¹ , and ∥f∥LqL^p = ^∥f∥Lp(T3) Lq(0,t) . Moreover, ⟨·, ·⟩ denotes the standard L 2 scalar product. For simplicity, we write ∥f∥Lp(T3) as ∥f∥L^p .
• For a > 0, we define the norms

$$\begin{split} |f|_{Xa}^2 &= |\mathrm{e}^{aA^{-\frac{1}{3}}t}f|{L^{\infty}L^2}^2 + |\mathrm{e}^{aA^{-\frac{1}{3}}t}\nabla\triangle^{-1}\partialx f|{L^2L^2}^2 + \frac{|\mathrm{e}^{aA^{-\frac{1}{3}}t}f|{L^2L^2}^2}{A^{\frac{1}{3}}} + \frac{|\mathrm{e}^{aA^{-\frac{1}{3}}t}\nabla f|{L^2L^2}^2}{A}, \ |f|_{Y0}^2 &= |f|{L^{\infty}L^2}^2 + \frac{1}{A}|\nabla f|_{L^2L^2}^2. \end{split}$$

• We sometimes denote the partial derivatives ∂x, ∂^y and ∂^z by ∂1, ∂² and ∂3, respectively.
• In this paper, unless otherwise specified, we use the Einstein summation convention.
• The total mass ∥n(t)∥L¹ is denoted by M, and let m := ∥n0∥L¹ . Clearly,

$$M := |n(t)|{L^1} = |n{\text{in}}|_{L^1},$$

$$m := |(n_{\text{in}})0|{L^1} = \frac{|n|_{L^1}}{|\mathbb{T}|} = \frac{M}{|\mathbb{T}|}$$

• Throughout this paper, we denote C by a positive constant independent of A, t and the initial data, and it may be different from line to line.

2. Key ideas and proof of Theorem 1.1

2.1. Fourier analysis. For the given function f, by Fourier series, we get

$$f(t, x, y, z) = \sum_{k_1, k_2, k3 \in \mathbb{Z}} \widehat{f}{k_1, k_2, k_3}(t) e^{i(k_1 x + k_2 y + k_3 z)},$$

where ^fb^k1,k2,k³ (t) = ¹ |T| 3 R ^T3 f(t, x, y, z)e⁻i(k1x+k2y+k3z)dxdydz.

According to the frequency k1, f can be decomposed into

$$f = f_{\neq} + f_0,$$

where

$$f{\neq}(t,y,z) = \sum{k_2,k_3 \in \mathbb{Z}, k1 \neq 0} \widehat{f}{k_1,k_2,k_3}(t) \mathrm{e}^{i(k_1x + k_2y + k_3z)}, \ f0(t,y,z) = \sum{k_2,k3 \in \mathbb{Z}} \widehat{f}{0,k_2,k_3}(t) \mathrm{e}^{i(k_2y + k_3z)}.$$

In fact, f⁰ is the zero mode, and f̸⁼ is the non-zero mode.

Furthermore, according to the frequency k3, the zero mode f⁰ can be decomposed into two parts ^{f0 = f 0 + f}e⁰ satisfying

$$\overline{f}0(t) = \widehat{f}{0,0,0}(t) = \frac{1}{|\mathbb{T}|^3} \int_{\mathbb{T}^3} f(t, x, y, z) dx dy dz, \widetilde{f}0(t, y, z) = \sum{k_2^2 + k3^2 \neq 0} \widehat{f}{0, k_2, k_3}(t) e^{i(k_2 y + k_3 z)},$$ (2.1)

where f ⁰ (t) is called the average of f in the periodic domain T 3 .

Assume that f satisfying

$$\begin{cases} \partial_t f + y \partialx f - \frac{1}{A} \triangle f = 0, \ f|{t=0} = f_{\text{in}}. \end{cases}$$

The function f is divided into three different modes:

$$f(t, x, y, z) = f_{\neq}(t, x, y, z) + \widetilde{f}_0(t, y, z) + \overline{f}_0(t).$$

Then we consider the dynamics for different modes. The non-zero mode f̸⁼ experiences the enhanced dissipation [2]:

$$||f{\neq}(t)||{L^2} \le Ce^{aA^{-\frac{1}{3}}t}||(f{\rm in}){\neq}||_{L^2}.$$

The non-average part ^fe⁰ experiences the heat dissipation with

$$|\widetilde{f}0(t)|{L^2} \le e^{-\frac{t}{2A}} |\widetilde{f}{\text{in}}|{L^2}.$$

The average part f ⁰ (t) is a constant satisfying

$$\overline{f}0(t) = \overline{(f{\rm in})_0}.$$

What we need to emphasize is that the decomposition of different modes and the application of the corresponding dynamics are the keys to research the blow-up suppression problem in the periodic domain T 3 .

2.2. The decomposition of the first component of the velocity. The first component u1,⁰ is affected by the 3D lift-up effect and is the worst part (see Section 2.1 in [5]). Therefore, we must handle it more carefully and meticulously.

Recall that u1,⁰ satisfies

$$\partialt u{1,0} - \frac{1}{A} \triangle u{1,0} + u{2,0} = -\frac{u_{2,0} \partialy u{1,0} + u_{3,0} \partialz u{1,0}}{A} - \frac{(u{\neq} \cdot \nabla u{1,\neq})_0}{A} + \frac{n_0}{A}.$$

First of all, we decompose u1,⁰ into u1,0(t, y, z) = G1(t, y, z) + B1(t, y, z) + B2(t, y, z), satisfying

$$\partial{t}\mathbf{G}{1} - \frac{1}{A}\triangle\mathbf{G}{1} = -\frac{u{2,0}\partial{y}\mathbf{G}{1} + u{3,0}\partial{z}\mathbf{G}{1}}{A} - \frac{(u{\neq} \cdot \nabla u{1,\neq}){0}}{A},$$

$$\partial{t}\mathbf{B}{1} - \frac{1}{A}\triangle\mathbf{B}{1} = -\frac{u{2,0}\partial{y}\mathbf{B}{1} + u{3,0}\partial{z}\mathbf{B}{1}}{A} + \frac{n{0}}{A},$$

$$\partial{t}\mathbf{B}{2} - \frac{1}{A}\triangle\mathbf{B}{2} + u{2,0} = -\frac{u{2,0}\partial{y}\mathbf{B}{2} + u{3,0}\partial{z}\mathbf{B}{2}}{A},$$

$$(2.2)$$

along with the initial conditions

$$\mathbf{G}1|{t=0} = (u_{1,\text{in}})_0, \quad \mathbf{B}1|{t=0} = 0, \quad \mathbf{B}2|{t=0} = 0.$$

In this way, G1(t, y, z) is a good term satisfying the following energy estimate

$$|\mathbf{G}1|{L^{\infty}H^1}^2 + \frac{1}{A} |\nabla \mathbf{G}1|{L^2H^1}^2 \le C(|(u_{1,\text{in}})0|{H^1}^2 + \epsilon^2).$$

In addition, according to the frequency k3, we decompose B1(t, y, z) and B2(t, y, z) into two parts

$$\mathbf{B}{1}(t,y,z) = \overline{\mathbf{B}}{1}(t) + \widetilde{\mathbf{B}}_{1}(t,y,z),$$

$$\mathbf{B}{2}(t,y,z) = \overline{\mathbf{B}}{2}(t) + \widetilde{\mathbf{B}}_{2}(t,y,z),$$

satisfying

$$\begin{cases} \partial_t \widetilde{\mathbf{B}}_1(t, y, z) - \frac{1}{A} \triangle \widetilde{\mathbf{B}}1(t, y, z) = -\frac{1}{A} (u{2,0} \partial_y \widetilde{\mathbf{B}}1 + u{3,0} \partial_z \widetilde{\mathbf{B}}_1) + \frac{1}{A} \widetilde{n}_0(t, y, z), \ \partial_t \overline{\mathbf{B}}_1(t) = \frac{1}{A} \overline{n}_0 = \frac{M}{A|\mathbf{T}|^3}, \end{cases}$$ (2.3)

and

$$\begin{cases} \partial_t \widetilde{\mathbf{B}}_2(t,y,z) - \frac{1}{A} \triangle \widetilde{\mathbf{B}}2(t,y,z) + \widetilde{u}{2,0}(t,y,z) = -\frac{1}{A} (u_{2,0} \partial_y \widetilde{\mathbf{B}}2 + u{3,0} \partial_z \widetilde{\mathbf{B}}_2), \ \partial_t \overline{\mathbf{B}}2(t) + \overline{u}{2,0}(t) = 0. \end{cases}$$ (2.4)

Lastly, we rewrite above decompositions into

$$\mathbf{U}{1}(t,y,z) = \mathbf{G}{1}(t,y,z) + \widetilde{\mathbf{B}}_{1}(t,y,z),$$

$$\mathbf{U}{2}(t,y,z) = \widetilde{\mathbf{B}}{2}(t,y,z) + \overline{\mathbf{B}}{2}(t) + \overline{\mathbf{B}}{1}(t),$$

satisfying U1(t, y, z) + U2(t, y, z) = u1,0(t, y, z).

In this way, we decompose the velocity u1,⁰ into the good part U1(t, y, z) and the bad part U2(t, y, z) satisfying

$$\begin{cases} \partial_t \mathbf{U}_1 - \frac{1}{A} \triangle \mathbf{U}1 = -\frac{1}{A} (u{2,0} \partial_y \mathbf{U}1 + u{3,0} \partial_z \mathbf{U}_1) + \frac{1}{A} \widetilde{n}0(t,y,z) - \frac{(u{\neq} \cdot \nabla u_{1,\neq})_0}{A}, \ \mathbf{U}1|{t=0} = (u_{1,\mathrm{in}})_0, \end{cases}$$

and

$$\begin{cases} \partial_t \mathbf{U}_2 - \frac{1}{A} \triangle \mathbf{U}2 + u{2,0} = -\frac{1}{A} (u_{2,0} \partial_y \mathbf{U}2 + u{3,0} \partial_z \mathbf{U}_2) + \frac{M}{A|\mathbf{T}|^3}, \ \mathbf{U}2|{t=0} = 0. \end{cases}$$

For the good velocity U1, the H¹ norm is enough for us to finish all calculations, and it has a good relation with the cell density n⁰ by

$$|\mathbf{U}1|{L^{\infty}H^1} \le C(|(u_{1,\text{in}})0|{H^1} + |n0|{L^{\infty}L^2} + \epsilon).$$

For the bad velocity U2, when it satisfies ∥△U2∥L∞H^{2 A} + ∥∂tU2∥L∞L^∞ < c for some small c, the operator L^V can be regarded as perturbation of L, which allow us to finish the time-space estimates for the zero modes. Here, L^V and L can be found in (5.10).

2.3. The construction of energy functional. Firstly, we introduce the energy functional

$$\begin{split} E{1,1}(t) &= |u{2,0}|_{Y0} + |u{3,0}|_{Y0} + |\nabla u{2,0}|_{Y0} + |\nabla u{3,0}|_{Y0} + |\Delta u{2,0}|_{Y0} \ &+ |\min{(A^{-\frac{2}{3}} + A^{-1}t)^{\frac{1}{2}}, 1}\Delta u{3,0}|_{Y0}, \ E{1,2}(t) &= A^{-1}(|\Delta \mathbf{U}2|{L^{\infty}H^2} + A^{-\frac{1}{2}}|\nabla\Delta \mathbf{U}2|{L^2H^2}) + |\partial_t \mathbf{U}2|{L^{\infty}H^2}, \ E_{2,1}(t) &= |\partialx^2 n{\neq}|_{Xb}, \ E{2,2}(t) &= |\Delta u{2,\neq}|{X_a} + |\partialx \omega{2,\neq}|_{X_a} + A^{-\frac{1}{3}}\left(|\partialy \omega{2,\neq}|_{X_a} + |\partialz \omega{2,\neq}|_{Xa}\right), \ E{3}(t) &= |n|{L^{\infty}L^{\infty}}, \ E{4}(t) &= |\partialx^2 u{2,\neq}|_{X_b} + |\partialx^2 u{3,\neq}|_{X_b}, \end{split}$$

where a and b are given positive constants with 0 < a < b < 2a. Moreover, we denote

$$E1(t) = E{1,1}(t) + E_{1,2}(t), \quad E2(t) = E{2,1}(t) + E_{2,2}(t).$$

To close the energy, when estimating the nonlinear interactions between the non-zero modes of the velocity of ∥e ²aA^{− 1 3 t}∇(u̸⁼ · ∇u3,̸=)∥L2L² , it is required that the degree of A must be strictly less than ⁵ 3 (see Lemma 3.2). To this end, we introduce the first auxiliary energy

$$E{5,1}(t) = A^{-\frac{2}{3}} | \triangle u{3,\neq} |_{X_b}.$$

Besides, to estimate E⁴ and handle the nonlinear interaction between the bad component U² of u1,⁰ and the non-zero mode u̸=, we also need the second auxiliary energy

$$E{5,2}(t) = \sum{j=2}^{3} (|\partialx^2 u{j,\neq}|_{X_b} + |\partial_x (\partial_z - \kappa \partialy) u{j,\neq}|_{X_b}) + |\partialx \nabla W|{X_b},$$

where the definition of κ and the expression of W can be found in Section 8.

Let us briefly describe the roles of the above-mentioned energy functionals.

• E¹ is introduced to control the zero modes of the velocity u0. E1,¹ is used to control the good velocities u2,⁰ and u3,0. E1,² is used to control the bad part U² of the velocity u1,0.
• E² is introduced to deal with the non-zero mode of the cell density n̸⁼ and the velocity u̸=.
• E³ is introduced to estimate ∥n0(t)∥L² and deal with the nonlinear interaction items in the density n.
• E⁴ is introduced to deal with the nonlinear interaction items with 3D lift-up effect in u̸=.

2.4. Main steps.

Proof of Theorem 1.1. We prove Theorem 1.1 in the following three steps.

Step 1: Let's designate T as the terminal point of the largest range [0, T] such that the following hypothesis hold

$$E_1(t) \le 2E_1, \quad E_2(t) \le 2E_2, \quad E_3(t) \le 2E_3, \quad E_4(t) \le 2E4, \quad E{5,1}(t) \le 2E_5$$ (2.5)

for any t ∈ [0, T], where E1, E2, E3, E⁴ and E⁵ are constants independent of t and A, and will be decided during the calculation.

Step 2: We need to prove the following propositions:

Proposition 2.1. Assume that the initial data (nin, uin) satisfy the assumptions of Theorem 1.1 and (2.5), there exists a positive constant A² independent of A and t, such that if A ≥ A2, there holds

$$E_1(t) \leq E_1$$

for all t ∈ [0, T].

Proposition 2.2. Assume that the initial data (nin, uin) satisfy the assumptions of Theorem 1.1 and (2.5), there exists a positive constant A⁵ independent of A and t, such that if A ≥ A5, there holds

$$E_3(t) \leq E_3$$

for all t ∈ [0, T].

Proposition 2.3. Assume that the initial data (nin, uin) satisfy the assumptions of Theorem 1.1 and (2.5), there exists a positive constant A⁷ independent of A and t, such that if A ≥ A7, there holds

$$E4(t) + E{5,1}(t) \le E_4 + E_5,$$

$E_2(t) \le E_2,$

for all t ∈ [0, T].

Step 3: Combining the above propositions with the well-posedness of system (1.5) given by Remark 1.4, and taking A¹ = max{A2, A5, A7}, we complete the proof.

3. A priori estimates

Firstly, we give some relationships between velocity u̸⁼ and the new vorticity ω2,̸=, which will be frequently used in the later calculations. Since div u̸⁼ = 0, the result follows immediately from Fourier series (see Lemma 3.13 in [8]), we omit the proof.

Lemma 3.1. There holds

$$\begin{split} &|(\partial{x},\partial{z})\partial{x}u{\neq}|{L^{2}} \leq C(|\partial{x}\omega{2,\neq}|{L^{2}} + |\triangle u{2,\neq}|{L^{2}}), \ &|(\partial{x},\partial{z})\partial{y}u{\neq}|{L^{2}} \leq C(|\partial{y}\omega{2,\neq}|{L^{2}} + |\triangle u{2,\neq}|{L^{2}}), \ &|(\partial{x},\partial{z})\partial{z}u{\neq}|{L^{2}} \leq C(|\partial{z}\omega{2,\neq}|{L^{2}} + |\triangle u{2,\neq}|{L^{2}}), \ &|(\partial{x}^{2},\partial{z}^{2})u{3,\neq}|{L^{2}} \leq C(|\partial{x}\omega{2,\neq}|{L^{2}} + |\Delta u{2,\neq}|{L^{2}}), \ &|(\partial{x},\partial{z})\partial{x}\nabla u{\neq}|{L^{2}} \leq C(|\partial{x}\nabla\omega{2,\neq}|{L^{2}} + |\nabla\Delta u{2,\neq}|{L^{2}}), \ &|(\partial{x},\partial{z})\partial{y}\nabla u{\neq}|{L^{2}} \leq C(|\partial{y}\nabla\omega{2,\neq}|{L^{2}} + |\nabla\Delta u{2,\neq}|{L^{2}}), \ &|(\partial{x},\partial{z})\partial{y}\nabla u{\neq}|{L^{2}} \leq C(|\partial{y}\nabla\omega{2,\neq}|{L^{2}} + |\nabla\Delta u{2,\neq}|_{L^{2}}). \end{split}$$

Secondly, we present the nonlinear interactions between the non-zero modes of the velocity, which will often be used to estimate E1, E² and E4.

□

Lemma 3.2. It holds that

$$|e^{2aA^{-\frac{1}{3}}t}|u{\neq}|^{2}|{L^{2}L^{2}}^{2} \leq CA^{\frac{2}{3}}E_{2}^{4},$$

$$|e^{2aA^{-\frac{1}{3}}t}u{\neq} \cdot \nabla u{\neq}|{L^{2}L^{2}}^{2} \leq CA^{\frac{2}{3}}E{2}^{4},$$

$$|e^{2aA^{-\frac{1}{3}}t}\partial{x}(u{\neq} \cdot \nabla u{\neq})|{L^{2}L^{2}}^{2} \leq CA^{\frac{1}{6}+\alpha}E_{2}^{4},$$

$$|e^{2aA^{-\frac{1}{3}}t}\nabla(u{\neq} \cdot \nabla u{2,\neq})|{L^{2}L^{2}}^{2} \leq CA^{\frac{1}{2}+\frac{2}{3}\alpha}E{2}^{4},$$

$$|e^{2aA^{-\frac{1}{3}}t}\partial{z}(u{\neq} \cdot \nabla u{3,\neq})|{L^{2}L^{2}}^{2} \leq CA^{\frac{1}{2}+\frac{2}{3}\alpha}E_{2}^{4},$$

$$|e^{2aA^{-\frac{1}{3}}t}\partial{z}(u{\neq} \cdot \nabla u{1,\neq})|{L^{2}L^{2}}^{2} \leq CA^{\frac{4}{3}}E_{2}^{4},$$

$$|e^{2aA^{-\frac{1}{3}}t}\nabla(u{\neq} \cdot \nabla u{3,\neq})|{L^{2}L^{2}}^{2} \leq C(A^{\frac{7}{6}+\frac{1}{3}\alpha}E{2}^{2}E{5}^{2} + A^{\frac{4}{3}}E{2}^{4}),$$

$$(3.2)$$

where α is a constant with α ∈ ( 1 2 , 3 4 ).

Proof. For convenience, we denote by

$$\Gamma{1} = |\partial{x}\omega{2,\neq}|{L^{2}} + |\triangle u{2,\neq}|{L^{2}}, \Gamma{2} = |\nabla\omega{2,\neq}|{L^{2}} + |\triangle u{2,\neq}|{L^{2}}, \Gamma{3} = |\partial{x}\nabla\omega{2,\neq}|{L^{2}} + |\nabla\triangle u{2,\neq}|{L^{2}}, \Gamma{4} = |\partial{x}\nabla u{2,\neq}|_{L^{2}}.$$

According to the definition of E2(t) and assumptions (2.5), there holds

$$|e^{aA^{-\frac{1}{3}t}}\Gamma1|{L^{\infty}} + \frac{|e^{aA^{-\frac{1}{3}t}}\Gamma1|{L^2}}{A^{\frac{1}{6}}} + \frac{|e^{aA^{-\frac{1}{3}t}}(\Gamma_2, \Gamma3)|{L^2}}{A^{\frac{1}{2}}} + \frac{|e^{aA^{-\frac{1}{3}t}}\Gamma2|{L^{\infty}}}{A^{\frac{1}{3}}} + |e^{aA^{-\frac{1}{3}t}}\Gamma4|{L^2} \le CE_2,$$

which will be frequently used in later calculations.

Estimate of (3.2)1. Using (A.2) in Lemma A.3 and (3.1) 1 in Lemma 3.1, we have

$$|u{\neq}|{L^{\infty}{x,z}L^{2}{y}}^{2} \leq C \left(|\partial{x}u{\neq}|{L^{2}}^{2\alpha}|u{\neq}|{L^{2}}^{2-2\alpha} + |\partial{x}\partial{z}u{\neq}|{L^{2}}^{2\alpha}|u{\neq}|{L^{2}}^{2-2\alpha}\right) \leq C\Gamma{1}^{2}.$$

Therefore, by (A.2) 8 and (3.1) 2 , we arrive

$$|||u{\neq}|^{2}||{L^{2}}^{2} \leq C(||\partial{x}u{\neq}||{L^{2}}^{2\alpha}||u{\neq}||{L^{2}}^{2-2\alpha} + ||\partial{x}\partial{z}u{\neq}||{L^{2}}^{2\alpha}||u{\neq}||{L^{2}}^{2-2\alpha})||\nabla u{\neq}||{L^{2}}||u{\neq}||{L^{2}} \leq C\Gamma{1}^{3}\Gamma_{2},$$

which indicates that

$$|\mathrm{e}^{2aA^{-\frac{1}{3}}t}|u{\neq}|^2|{L^2L^2}^2 \le CA^{\frac{2}{3}}(|\partialx\omega{2,\neq}|_{Xa}^4 + |\triangle u{2,\neq}|_{X_a}^4) \le CA^{\frac{2}{3}}E_2^4.$$

Estimate of (3.2)2. Using Lemma A.3 and Lemma 3.1 again, there hold

$$\begin{split} |u{2,\neq}|{L^{\infty}{x,y}L^{2}{z}}^{2} &\leq C|\partial{x}u{2,\neq}|{L^{2}}|\partial{x}\nabla u{2,\neq}|{L^{2}} \leq C\Gamma{4}^{2}, \ |u{\neq}|{L^{\infty}{x,z}L^{2}{y}}^{2} &\leq C\left(|\partial{x}u{\neq}|{L^{2}}^{2} + |(\partial{x},\partial{z})\partial{x}u{\neq}|{L^{2}}^{2}\right) \leq C\Gamma{1}^{2}, \ |\partial{y}u{\neq}|{L^{\infty}{x}L^{2}{y,z}}^{2} &+ |\partial{y}u{\neq}|{L^{\infty}{z}L^{2}{x,y}}^{2} \leq C\Gamma{2}^{2}, \ |(\partial{x},\partial{z})u{\neq}|{L^{\infty}{y}L^{2}{x,z}}^{2} &\leq C\Gamma{1}\Gamma_{2}. \end{split}$$

Therefore, one gets

$$\begin{split} |u{\neq} \cdot \nabla u{\neq}|{L^{2}}^{2} \leq & |u{1,\neq}\partial{x}u{\neq}|{L^{2}}^{2} + |u{2,\neq}\partial{y}u{\neq}|{L^{2}}^{2} + |u{3,\neq}\partial{z}u{\neq}|{L^{2}}^{2} \ \leq & |u{\neq}|{L^{\infty}{x,z}L^{2}{y}}^{2} |(\partial{x},\partial{z})u{\neq}|{L^{\infty}{y}L^{2}{x,z}}^{2} + |u{2,\neq}|{L^{\infty}{x,y}L^{2}{z}}^{2} |\partial{y}u{\neq}|{L^{\infty}{z}L^{2}{x,y}}^{2} \ \leq & C(\Gamma{1}^{3}\Gamma{2} + \Gamma{2}^{2}\Gamma{4}^{2}), \end{split}$$

which implies that

$$|e^{2aA^{-\frac{1}{3}}t}u{\neq}\cdot\nabla u{\neq}|{L^{2}L^{2}}^{2} \le CA^{\frac{2}{3}}E{2}^{4}$$

Estimate of (3.2)3. For j ∈ {1, 3}, by Lemma A.3 and Lemma 3.1, we have ∥uj,̸=∥ 2 L∞x,zL² + ∥∂xuj,̸=∥ 2 L∞^z L² ≤ C∥∂x(∂x, ∂z)uj,̸=∥ 2 ^L2 ≤ CΓ 2 1

and

$$|\partial_x \partialj u{\neq}|{L^{\infty}{y}L^{2}_{x,z}}^{2} + |\partialj u{\neq}|{L^{\infty}{x,y}L^{2}_{z}}^{2} \le C|\partial_x \partialj \nabla u{\neq}|_{L^{2}} |\partial_x \partialj u{\neq}|_{L^{2}} \le C\Gamma_1 \Gamma_3,$$

x,y

which indicate that

$$|\partialx(u{j,\neq}\partialj u{\neq})|_{L^2}^2 \le C\Gamma_1^3\Gamma_3. \tag{3.3}$$

Using (A.2) 1 , (A.2) 4 and (A.2) 7 , there hold

$$|\partialx u{2,\neq}|{L^\infty{x,y},L^2z}^2 + |u{2,\neq}|_{L^\infty}^2 \leq C |\partialx \nabla u{2,\neq}|{L^2}^{3-2\alpha} |\nabla \triangle u{2,\neq}|_{L^2}^{2\alpha-1} \leq C \Gamma_4^{3-2\alpha} \Gamma_3^{2\alpha-1}$$

and

$$|\partialy u{\neq}|_{Lz^{\infty} L{x,y}^2}^2 \le C |(\partial_x, \partial_z) \partialy u{\neq}|_{L^2}^2 \le C \Gamma_2^2,$$

which show that

$$|\partialx (u{2,\neq} \partialy u{\neq})|_{L^2}^2 \le C \Gamma_2^2 \Gamma_3^{2\alpha - 1} \Gamma_4^{3-2\alpha}. \tag{3.4}$$

Combining (3.3) with (3.4), we conclude that

$$|e^{2aA^{-\frac{1}{3}}t}\partialx(u{\neq}\cdot\nabla u{\neq})|{L^2L^2}^2 \le CA^{\frac{1}{6}+\alpha}E_2^4.$$

Estimate of (3.2)4. By Lemma A.3 and Lemma 3.1, direct calculations show that

$$|u{\neq}|{L^{\infty}}^{2} \leq C(|\partial{x}\omega{2,\neq}|{L^{2}} + |\triangle u{2,\neq}|{L^{2}})(|\nabla\omega{2,\neq}|{L^{2}} + |\triangle u{2,\neq}|{L^{2}}) \leq C\Gamma{1}\Gamma_{2},$$

$$|\nabla u{\neq}|{L^{\infty}{x}L^{2}{x,y}}^{2} + |\nabla u{\neq}|{L^{\infty}{x}L^{2}{y,z}}^{2} \leq C\Gamma_{2}^{2},$$

$$|\nabla u{2,\neq}|{L^{\infty}{x,y}L^{2}{z}}^{2} \leq C|\partial{x}\nabla u{2,\neq}|{L^{2}}|\nabla\triangle u{2,\neq}|{L^{2}}^{2\alpha-1}|\triangle u{2,\neq}|{L^{2}}^{2-2\alpha} \leq C\Gamma{4}\Gamma{3}^{2\alpha-1}\Gamma{1}^{2-2\alpha}.$$ (3.5)

Combining above results with

$$\begin{split} |\nabla(u{\neq} \cdot \nabla u{2,\neq})|{L^{2}}^{2} &\leq C(|\nabla u{\neq}|{L{z}^{\infty}L{x,y}^{2}}^{2} |\nabla u{2,\neq}|{L{x,y}^{\infty}L{z}^{2}}^{2} + |u{\neq}|{L^{\infty}}^{2} |\Delta u{2,\neq}|{L^{2}}^{2}) \ &\leq C(\Gamma{4}\Gamma{3}^{2\alpha-1}\Gamma{1}^{2-2\alpha}\Gamma{2}^{2} + \Gamma{1}^{3}\Gamma_{2}), \end{split}$$

there holds

$$|\mathrm{e}^{2aA^{-\frac{1}{3}}t}\nabla(u{\neq}\cdot\nabla u{2,\neq})|{L^{2}L^{2}}^{2}\leq A^{\frac{1}{2}+\frac{2}{3}\alpha}E{2}^{4}.$$

Estimate of (3.2)5. For j ∈ {1, 3}, using Lemma A.3 and Lemma 3.1, we have ∥∂zuj,̸=∥ 2 L∞^x L² y,z ≤ CΓ 2 1 , ∥∂ju3,̸=∥ 2 L∞y,zL² ≤ CΓ1Γ3,

which along with (3.5) 1 indicates that

$$|\partialz(u{j,\neq}\partialj u{3,\neq})|_{L^2}^2 \le C\Gamma_1^3\Gamma_3. \tag{3.6}$$

For j = 2, by Lemma 3.1 and Lemma A.3, there holds

$$||u{2,\neq}||{L^{\infty}}^{2} \leq C||\partial{x}\nabla u{2,\neq}||{L^{2}}^{3-2\alpha}||\Delta u{2,\neq}||{L^{2}}^{2\alpha-1} \leq C\Gamma{4}^{3-2\alpha}\Gamma_{1}^{2\alpha-1},$$

$$||\partial{y}u{3,\neq}||{L{z}^{\infty}L{x,y}^{2}}^{2} \leq C||(\partial{x},\partial{z})\nabla u{3,\neq}||{L^{2}}^{2} \leq C\Gamma{2}^{2},$$ (3.7)

which along with (3.1) 2 and (3.5) 3 shows that

$$|\partialz(u{2,\neq}\partialy u{3,\neq})|_{L^2}^2 \le C(\Gamma_4 \Gamma_3^{2\alpha-1} \Gamma_1^{2-2\alpha} \Gamma_2^2 + \Gamma_4^{3-2\alpha} \Gamma_1^{2\alpha-1} \Gamma_2^2). \tag{3.8}$$

Combining (3.6) and (3.8), we obtain that

$$|e^{2aA^{-\frac{1}{3}}t}\partialz(u{\neq}\cdot\nabla u{3,\neq})|{L^2L^2}^2 \le A^{\frac{1}{2}+\frac{2}{3}\alpha}E_2^4.$$

Estimate of (3.2)6. According to (3.7) and (3.8), after replacing u3,̸⁼ with u1,̸⁼ , we can prove that

$$|\partialz(u{2,\neq}\partialy u{1,\neq})|_{L^2}^2 \le C(\Gamma_4 \Gamma_3^{2\alpha-1} \Gamma_1^{2-2\alpha} \Gamma_2^2 + \Gamma_4^{3-2\alpha} \Gamma_1^{2\alpha-1} \Gamma_2^2). \tag{3.9}$$

Similar to (3.6), for j ∈ {1, 3}, by using (3.1)3, one deduces

$$|\partialz(u{j,\neq}\partialj u{1,\neq})|_{L^2}^2 \le C\Gamma_1\Gamma_2^2\Gamma_3. \tag{3.10}$$

Collecting (3.9) and (3.10), there holds

$$|e^{2aA^{-\frac{1}{3}}t}\partialz(u{\neq}\cdot\nabla u{1,\neq})|{L^2L^2}^2 \le CA^{\frac{4}{3}}E_2^4.$$

Estimate of (3.2)7. For j ∈ {1, 3}, using Lemma A.3 and Lemma 3.1 again, we have

$$|\partialj u{3,\neq}|{L^\infty{y,z}L^2_x}^2 \le C |\partial_z \partialj \nabla u{3,\neq}|_{L^2} |\partial_z \partialj u{3,\neq}|_{L^2} \le C \Gamma_1 \Gamma_3,$$

which along with (3.5) 1 and (3.5) 2 indicates that

$$|\nabla(u{j,\neq}\partial{j}u{3,\neq})|{L^{2}}^{2} \leq C(|u{j,\neq}|{L^{\infty}}^{2}|\nabla\partial{j}u{3,\neq}|{L^{2}}^{2} + |\nabla u{j,\neq}|{L^{\infty}{x}L^{2}{y,z}}^{2}|\partial{j}u{3,\neq}|{L^{\infty}{y,z}L^{2}{x}}^{2})$$

$$\leq C(\Gamma{1}\Gamma{2}^{3} + \Gamma{2}^{2}\Gamma{1}\Gamma_{3}). \tag{3.11}$$

According to (3.5) 3 and (3.7), one obtains

$$|\nabla(u_{2,\neq}\partialy u{3,\neq})|{L^2}^2 \le C(|u{2,\neq}|_{L^{\infty}}^2 |\partialy \nabla u{3,\neq}|{L^2}^2 + |\nabla u{2,\neq}|{L^{\infty}{x,y}L^2_z}^2 |\partialy u{3,\neq}|_{L^{\infty}z L^2{x,y}}^2)$$

$$\le C(\Gamma_4^{3-2\alpha} \Gamma1^{2\alpha-1} |\Delta u{3,\neq}|_{L^2}^2 + \Gamma_4 \Gamma_3^{2\alpha-1} \Gamma_1^{2-2\alpha} \Gamma_2^2).$$ (3.12)

Therefore, we infer from (3.11) and (3.12) that

$$|e^{2aA^{-\frac{1}{3}}t}\nabla(u{\neq}\cdot\nabla u{3,\neq})|{L^{2}L^{2}}^{2} \le C(A^{\frac{7}{6}+\frac{1}{3}\alpha}E{2}^{2}E{5}^{2}+A^{\frac{4}{3}}E{2}^{4}).$$

We finish the proof. □

Next, we give the nonlinear interaction between the zero mode u1,⁰ and the non-zero mode u̸=. Recall that the zero mode u1,⁰ is decomposed into two components with u1,⁰ = U¹ + U2, where U¹ is the good component and U² is the bad component.

The following result will be used in estimating the energies {∥△u2,̸=∥ 2 Xa , ∥∇ω2,̸=∥ 2 Xa }. Consequently, to close the energy estimates, the degree of A must be strictly less than 1.

Lemma 3.3. It holds that

$$|\mathbf{e}^{aA^{-\frac{1}{3}t}}\mathbf{U}{1}\partial{x}\nabla u{2,\neq}|{L^{2}L^{2}}^{2} \leq CA^{\frac{1}{2}}|\mathbf{U}{1}|{L^{\infty}H^{1}}^{2}|\triangle u{2,\neq}|{X_{a}}^{2},$$

$$|\mathbf{e}^{aA^{-\frac{1}{3}t}}\mathbf{U}{1}\partial{x}(\partial{x},\partial{z})u{3,\neq}|{L^{2}L^{2}}^{2} \leq CA^{\frac{2}{3}}|\mathbf{U}{1}|{L^{\infty}H^{1}}^{2}(|\partial{x}\omega{2,\neq}|{X{a}}^{2} + |\triangle u{2,\neq}|{X_{a}}^{2}),$$

$$|\mathbf{e}^{aA^{-\frac{1}{3}t}}\mathbf{U}{1}\partial{x}^{2}u{1,\neq}|{L^{2}L^{2}}^{2} \leq CA^{\frac{2}{3}}|\mathbf{U}{1}|{L^{\infty}H^{1}}^{2}(|\partial{x}\omega{2,\neq}|{X{a}}^{2} + |\triangle u{2,\neq}|{X_{a}}^{2}),$$

$$|\mathbf{e}^{aA^{-\frac{1}{3}t}}\mathbf{U}{1}\partial{x}\partial{z}u{1,\neq}|{L^{2}L^{2}}^{2} \leq CA^{\frac{2}{3}}|\mathbf{U}{1}|{L^{\infty}H^{1}}^{2}(|\partial{x}\omega{2,\neq}|{X{a}}^{2} + |\triangle u{2,\neq}|{X{a}}^{2}),$$

$$|\mathbf{e}^{aA^{-\frac{1}{3}t}}(\partial{y},\partial{z})\mathbf{U}{1}\partial{x}u{\neq}|{L^{2}L^{2}}^{2} \leq CA^{\frac{2}{3}}|\mathbf{U}{1}|{L^{\infty}H^{1}}^{2}(|\partial{x}\omega{2,\neq}|{X{a}}^{2} + |\triangle u{2,\neq}|{X_{a}}^{2}),$$

$$|\mathbf{e}^{aA^{-\frac{1}{3}t}}\partial{y}\mathbf{U}{1}\partial{y}u{2,\neq}|{L^{2}L^{2}}^{2} \leq CA^{\frac{2}{3}}|\mathbf{U}{1}|{L^{\infty}H^{1}}^{2}|\triangle u{2,\neq}|{X{a}}^{2}.$$

$$(3.13)$$

Proof. Using Lemma A.2 and Lemma A.3, we have

$$\begin{split} |\mathbf{U}{1}\partial{x}\nabla u{2,\neq}|{L^{2}}^{2} &\leq |\mathbf{U}{1}|{L{x}^{\infty}L{y}^{2}}^{2} |\partial{x}\nabla u{2,\neq}|{L{y}^{\infty}L{x,z}^{2}}^{2} \ &\leq C|\mathbf{U}{1}|{H^{1}}^{2} (|\partial{x}\nabla u{2,\neq}|{L^{2}} |\partial{x}\partial{y}\nabla u{2,\neq}|{L^{2}} + |\partial{x}\nabla u{2,\neq}|_{L^{2}}^{2}), \end{split}$$

which implies that

$$|\mathrm{e}^{aA^{-\frac{1}{3}}t}\mathbf{U}{1}\partial{x}\nabla u{2,\neq}|{L^{2}L^{2}}^{2} \leq CA^{\frac{1}{2}}|\mathbf{U}{1}|{L^{\infty}H^{1}}^{2}|\triangle u{2,\neq}|{X_{a}}^{2}$$

For j ∈ {1, 3}, by Lemma A.2, Lemma A.3 and Lemma 3.1, there holds

$$\begin{split} &|\mathbf{U}{1}\partial{x}(\partial{x},\partial{z})u{j,\neq}|{L^{2}}^{2} \ &\leq C|\mathbf{U}{1}|{H^{1}}^{2}|\partial{x}(\partial{x},\partial{z})u{j,\neq}|{L^{2}}|\partial{x}(\partial{x},\partial{z})\nabla u{j,\neq}|{L^{2}} \ &\leq C|\mathbf{U}{1}|{H^{1}}^{2}(|\partial{x}\omega{2,\neq}|{L^{2}} + |\Delta u{2,\neq}|{L^{2}})(|\partial{x}\nabla\omega{2,\neq}|{L^{2}} + |\nabla\Delta u{2,\neq}|{L^{2}}) \end{split}$$

and

$$|\mathbf{e}^{aA^{-\frac{1}{3}}t}\mathbf{U}{1}\partial{x}(\partial{x},\partial{z})u{j,\neq}|{L^{2}L^{2}}^{2} \leq CA^{\frac{2}{3}}|\mathbf{U}{1}|{L^{\infty}H^{1}}^{2}(|\partial{x}\omega{2,\neq}|{X{a}}^{2} + |\triangle u{2,\neq}|{X_{a}}^{2}),$$ (3.13)₂, (3.13)₃ and (3.13)₄.

which give (3.13) 2 , (3.13) 3 and (3.13) 4

Using Lemma A.3, we obtain that

$$\begin{split} |(\partial{y},\partial{z})\mathbf{U}{1}\partial{x}u{\neq}|{L^{2}}^{2} &\leq |\nabla\mathbf{U}{1}|{L^{2}}^{2}|\partial{x}u{\neq}|{L^{2}{y,z}L{x}^{2}}^{2} \ &\leq C|\nabla\mathbf{U}{1}|{L^{2}}^{2}|\partial{x}(\partial{x},\partial{z})u{\neq}|{L^{2}}|\partial{x}(\partial{x},\partial{z})\nabla u{\neq}|{L^{2}} \ &\leq C|\nabla\mathbf{U}{1}|{L^{2}}^{2}(|\partial{x}\omega{2,\neq}|{L^{2}} + |\triangle u{2,\neq}|{L^{2}})(|\partial{x}\nabla\omega{2,\neq}|{L^{2}} + |\nabla\triangle u{2,\neq}|_{L^{2}}), \end{split}$$

which indicates that

$$|\mathbf{e}^{aA^{-\frac{1}{3}}t}\nabla \mathbf{U}{1}\partial{x}u{\neq}|{L^{2}L^{2}}^{2} \leq CA^{\frac{2}{3}}|\mathbf{U}{1}|{L^{\infty}H^{1}}^{2}(|\partial{x}\omega{2,\neq}|{X{a}}^{2} + |\triangle u{2,\neq}|{X_{a}}^{2}).$$

Using Lemma A.3, we obtain that

$$\begin{split} |\partial{y}\mathbf{U}{1}\partial{y}u{2,\neq}|{L^{2}}^{2} &\leq |\nabla\mathbf{U}{1}|{L^{2}}^{2}|\partial{y}u{2,\neq}|{L{y,z}^{\infty}L{x}^{2}}^{2} \ &\leq C|\nabla\mathbf{U}{1}|{L^{2}}^{2}|\triangle u{2,\neq}|{L^{2}}|\nabla\triangle u{2,\neq}|{L^{2}} \end{split}$$

and

$$|\mathrm{e}^{aA^{-\frac{1}{3}}t}\partial_y \mathbf{U}_1 \partialy u{2,\neq}|_{L^2L^2}^2 \leq CA^{\frac{2}{3}} |\mathbf{U}1|{L^\infty H^1}^2 |\triangle u{2,\neq}|{X_a}^2.$$

The following result is only used to estimate ∥(∂y, ∂z)ω2,̸=∥ Xa . Hence, as long as the degree of A is less than ⁵ 3 , the energy estimates become achievable. Sometimes, we need to use the results of Lemma 4.1, and we will prove them in next section.

Lemma 3.4. Under the conditions of Theorem 1.1 and the assumptions (2.5), there exists a constant A³ independent of A and t, such that if A > A3, it holds that

$$|\mathbf{e}^{aA^{-\frac{1}{3}}t}(\partial{y},\partial{z})(u{2,\neq}\nabla\mathbf{U}{1})|{L^{2}L^{2}}^{2} \leq C\left(A^{\frac{2}{3}}|\mathbf{U}{1}|{L^{\infty}H^{1}}^{2}E{2}^{2} + AE{2}^{2\alpha}E{4}^{2-2\alpha}\right),$$

$$|\mathbf{e}^{aA^{-\frac{1}{3}}t}\partial{z}(u{3,\neq}\nabla\mathbf{U}{1})|{L^{2}L^{2}}^{2} \leq C\left(A^{\frac{2}{3}}|\mathbf{U}{1}|{L^{\infty}H^{1}}^{2}E{2}^{2} + A^{\frac{4}{3}}E{2}^{2\alpha}E_{4}^{2-2\alpha}\right),$$ (3.14)

where α is a constant with α ∈ ( 2 4 ).

Proof. First, for j ∈ {2, 3}, direct calculations show that

$$|\partial{j}(u{2,\neq}\nabla\mathbf{U}{1})|{L^{2}}^{2} \leq C\left(|\partial{j}u{2,\neq}\nabla\mathbf{U}{1}|{L^{2}}^{2} + |u{2,\neq}\partial{j}\nabla\mathbf{U}{1}|{L^{2}}^{2}\right), |\partial{z}(u{3,\neq}\nabla\mathbf{U}{1})|{L^{2}}^{2} \leq C\left(|\partial{z}u{3,\neq}\nabla\mathbf{U}{1}|{L^{2}}^{2} + |u{3,\neq}\partial{z}\nabla\mathbf{U}{1}|{L^{2}}^{2}\right).$$ (3.15)

By Lemma A.3, there hold

$$|\partialj u{2,\neq} \nabla \mathbf{U}1|{L^2}^2 \le C |\nabla \mathbf{U}1|{L^2}^2 |\Delta u{2,\neq}|{L^2} |\nabla \Delta u{2,\neq}|{L^2}$$

and

$$\begin{split} |\partialz u{3,\neq} \nabla \mathbf{U}1|{L^2}^2 &\leq C |\nabla \mathbf{U}1|{L^2}^2 |\partial_x (\partial_x, \partialz) u{3,\neq}|_{L^2} |\partial_x (\partial_x, \partial_z) \partialy u{3,\neq}|_{L^2} \ &\leq C |\mathbf{U}1|{H^1}^2 (|\partialx \omega{2,\neq}|{L^2} + |\triangle u{2,\neq}|_{L^2}) (|\partialx \nabla \omega{2,\neq}|{L^2} + |\nabla \triangle u{2,\neq}|_{L^2}), \end{split}$$

which indicate that

$$|\mathbf{e}^{aA^{-\frac{1}{3}t}}\partial{j}u{2,\neq}\nabla\mathbf{U}{1}|{L^{2}L^{2}}^{2} \leq CA^{\frac{2}{3}}|\mathbf{U}{1}|{L^{\infty}H^{1}}^{2}|\Delta u{2,\neq}|{X{a}}^{2} \leq CA^{\frac{2}{3}}|\mathbf{U}{1}|{L^{\infty}H^{1}}^{2}E{2}^{2},$$

$$|\mathbf{e}^{aA^{-\frac{1}{3}t}}\partial{z}u{3,\neq}\nabla\mathbf{U}{1}|{L^{2}L^{2}}^{2} \leq CA^{\frac{2}{3}}|\mathbf{U}{1}|{L^{\infty}H^{1}}^{2}(|\partial{x}\omega{2,\neq}|{X{a}}^{2} + |\Delta u{2,\neq}|{X{a}}^{2}) \leq CA^{\frac{2}{3}}|\mathbf{U}{1}|{L^{\infty}H^{1}}^{2}E{2}^{2}.$$

$$(3.16)$$

Then, we need to deal with ∥e aA^{− 1 3 t}u2,̸=∂j∇U1∥ 2 ^L2L² and ∥e aA^{− 1 3 t}u3,̸=∂z∇U1∥ 2 ^L2L² . Since it is difficult to estimate ∥∂j∇U1∥L∞L² , the traditional Sobolev embedding is insufficient to handle ∥e aA^{− 1 3 t}u2,̸=∂j∇U1∥ 2 ^L2L² and ∥e aA^{− 1 3 t}u3,̸=∂z∇U1∥ 2 ^L2L² .

According to (2.2) and (2.3), U¹ satisfies

$$\begin{cases} \partial_t \mathbf{U}_1 - \frac{1}{A} \triangle \mathbf{U}_1 = \frac{1}{A} \widetilde{n}0 - \frac{1}{A} (u{2,0} \partial_y \mathbf{U}1 + u{3,0} \partial_z \mathbf{U}1) - \frac{1}{A} (u{\neq} \cdot \nabla u_{1,\neq})_0, \ \mathbf{U}1|{t=0} = (u_{1,\text{in}})_0. \end{cases}$$ (3.17)

Therefore, for the given positive constant ϵ1, we have

$$\partial{t}(e^{-\epsilon{1}A^{-\frac{1}{3}}t}\mathbf{U}{1}) + \frac{\epsilon{1}e^{-\epsilon{1}A^{-\frac{1}{3}}t}\mathbf{U}{1}}{A^{\frac{1}{3}}} - \frac{e^{-\epsilon{1}A^{-\frac{1}{3}}t}\triangle\mathbf{U}{1}}{A}$$

$$= \frac{e^{-\epsilon{1}A^{-\frac{1}{3}}t}\widetilde{n}{0}}{A} - \frac{e^{-\epsilon{1}A^{-\frac{1}{3}}t}u{j,0}\partial{j}\mathbf{U}{1}}{A} - \frac{e^{-\epsilon{1}A^{-\frac{1}{3}}t}(u{\neq} \cdot \nabla u{1,\neq}){0}}{A},$$

□

where j ∈ {2, 3}. Multiplying − e−ϵ1^A − 1 ^{3 t}△U¹ 2 on both sides of the above equation, the energy estimate shows that

$$|\mathbf{e}^{-\epsilon{1}A^{-\frac{1}{3}}t}\nabla\mathbf{U}{1}|{L^{\infty}L^{2}}^{2} + \frac{|\mathbf{e}^{-\epsilon{1}A^{-\frac{1}{3}}t}\triangle\mathbf{U}{1}|{L^{2}L^{2}}^{2}}{A}$$

$$\leq |u{1,\text{in}}|{H^{1}}^{2} + \frac{C(|\mathbf{e}^{-\epsilon{1}A^{-\frac{1}{3}}t}\widetilde{n}{0}|{L^{2}L^{2}}^{2} + |\mathbf{e}^{-\epsilon{1}A^{-\frac{1}{3}}t}u{j,0}\partial{j}\mathbf{U}{1}|{L^{2}L^{2}}^{2} + |\mathbf{e}^{-\epsilon{1}A^{-\frac{1}{3}}t}u{\neq} \cdot \nabla u{1,\neq}|{L^{2}L^{2}}^{2})}{A}.$$ (3.18)

When A > A2, by Lemma A.2 and Lemma 4.1, we get

$$||u{2,0}||{L^{\infty}L^{\infty}}^{2} + ||u{3,0}||{L^{\infty}L^{\infty}}^{2} \le C(||u{2,0}||{L^{\infty}H^{2}}^{2} + ||u{3,0}||{L^{\infty}H^{1}}^{2}) \le C\epsilon^{2},$$ (3.19)

which implies that

$$\frac{|\mathbf{e}^{-\epsilon1 A^{-\frac{1}{3}} t} u{j,0} \partial_j \mathbf{U}1|{L^2 L^2}^2}{A} \le \frac{C \epsilon^2 |\mathbf{e}^{-\epsilon_1 A^{-\frac{1}{3}} t} \triangle \mathbf{U}1|{L^2 L^2}^2}{A}.$$ (3.20)

Using (3.2) 2 and (3.20), we infer from (3.18) that

$$|\mathbf{e}^{-\epsilon_1 A^{-\frac{1}{3}}t} \nabla \mathbf{U}1|{L^{\infty}L^2}^2 + \frac{|\mathbf{e}^{-\epsilon_1 A^{-\frac{1}{3}}t} \triangle \mathbf{U}1|{L^2L^2}^2}{A} \leq C \Big( |u{1,\mathrm{in}}|{H^1}^2 + \frac{|\mathbf{e}^{-\epsilon_1 A^{-\frac{1}{3}}t} \widetilde{n}0|{L^2L^2}^2}{A} + \frac{E_2^4}{A^{\frac{1}{3}}} \Big).$$

Due to

$$|\widetilde{n}0|{L^2}^2 \le |n0|{L^2}^2 \le C|n|_{L^\infty L^\infty}|n0|{L^1} \le CE_3M$$

and

$$|e^{-\epsilon1 A^{-\frac{1}{3}}t}|{L^2(0,T)}^2 \le \frac{A^{\frac{1}{3}}}{2\epsilon_1},$$

there holds

$$|\mathbf{e}^{-\epsilon_1 A^{-\frac{1}{3}} t} \nabla \mathbf{U}1|{L^{\infty} L^2}^2 + \frac{|\mathbf{e}^{-\epsilon_1 A^{-\frac{1}{3}} t} \triangle \mathbf{U}1|{L^2 L^2}^2}{A} \le C \left( |u{1,\text{in}}|{H^1}^2 + \frac{E_3 M}{\epsilon_1 A_3^{\frac{2}{3}}} + \frac{E_2^4}{A^{\frac{1}{3}}} \right). \tag{3.21}$$

When A ≥ max{A2, ϵ¹ − ^{3 2}M 3 ²E 3 , E^{12 2} } =: A3, by (A.2) 3 , (3.21) and Lemma 3.1, we have

$$|\mathbf{e}^{aA^{-\frac{1}{3}}t}u{2,\neq}\partial{j}\nabla\mathbf{U}{1}|{L^{2}L^{2}}^{2} \leq C|\Delta u{2,\neq}|{X{a}}^{2\alpha}|u{2,\neq}|{X{b}}^{2-2\alpha}|\mathbf{e}^{-2(1-\alpha)(b-a)A^{-\frac{1}{3}}t}\Delta\mathbf{U}{1}|{L^{2}L^{2}}^{2}$$

$$\leq CAE{2}^{2\alpha}E{4}^{2-2\alpha},$$ (3.22)

where α is a constant satisfying α ∈ 2 , 4 . By (A.2) 3 and Lemma 3.1, there holds

$$\begin{split} |u{3,\neq}|{L^{\infty}{y,z}L^{2}{x}}^{2} &\leq C|\nabla(\partial{x},\partial{z})u{3,\neq}|{L^{2}}|(\partial{x},\partial{z})u{3,\neq}|{L^{2}}^{2\alpha-1}|u{3,\neq}|{L^{2}}^{2-2\alpha} \ &\leq C(|\partial{y}\omega{2,\neq}|{L^{2}} + |\triangle u{2,\neq}|{L^{2}})(|\partial{x}\omega{2,\neq}|{L^{2}} + |\triangle u{2,\neq}|{L^{2}})^{2\alpha-1}|u{3,\neq}|{L^{2}}^{2-2\alpha}, \end{split}$$

which along with (3.21) implies that

$$|\mathbf{e}^{aA^{-\frac{1}{3}t}}u{3,\neq}\partial{z}\nabla\mathbf{U}{1}|{L^{2}L^{2}}^{2} \leq C(|\partial{y}\omega{2,\neq}|{X{a}} + |\triangle u{2,\neq}|{X{a}})(|\partial{x}\omega{2,\neq}|{X{a}} + |\triangle u{2,\neq}|{X{a}})^{2\alpha-1} \ \cdot |\partial{x}^{2}u{3,\neq}|{X{b}}^{2-2\alpha}|\mathbf{e}^{-2(1-\alpha)(b-a)A^{-\frac{1}{3}t}}\Delta\mathbf{U}{1}|{L^{2}L^{2}}^{2} \ \leq CA^{\frac{4}{3}}E{2}^{2\alpha}E{4}^{2-2\alpha}.$$

$$(3.23)$$

Collecting (3.15), (3.16), (3.22) and (3.23), we finish the proof.

Furthermore, we show the the nonlinear interactions between the bad component U² of u1,⁰ and the non-zero mode u̸=.

Lemma 3.5. For j ∈ {2, 3}, there holds

$$|\mathbf{e}^{aA^{-\frac{1}{3}t}}\nabla\mathbf{U}{2}\partial{x}u{j,\neq}|{L^{2}L^{2}}^{2} \leq CAE{1}^{2}E{4}^{\frac{1}{2}}(t)(|\partial{x}\omega{2,\neq}|{X{a}} + |\Delta u{2,\neq}|{X_{a}})^{\frac{3}{2}},$$

$$|\mathbf{e}^{aA^{-\frac{1}{3}t}}\partial{z}\mathbf{U}{2}\partial{x}^{2}u{1,\neq}|{L^{2}L^{2}}^{2} \leq CA^{\frac{17}{12}}E{1}^{2}E{4}^{\frac{1}{4}}(|\partial{x}\omega{2,\neq}|{X{a}} + |\Delta u{2,\neq}|{X{a}})^{\frac{7}{4}},$$

$$|\mathbf{e}^{aA^{-\frac{1}{3}t}}\partial{j}\mathbf{U}{2}\partial{x}\partial{j}u{2,\neq}|{L^{2}L^{2}}^{2} \leq CA^{\frac{4}{3}}E{1}^{2}E{4}^{\frac{1}{2}}|\Delta u{2,\neq}|{X_{a}}^{\frac{3}{2}},$$

$$|\mathbf{e}^{aA^{-\frac{1}{3}t}}\partial{z}(u{2,\neq}\nabla\mathbf{U}{2})|{L^{2}L^{2}}^{2} \leq CAE{1}^{2}E{4}^{\frac{1}{2}}|\Delta u{2,\neq}|{X_{a}}^{\frac{3}{2}},$$

$$|\mathbf{e}^{aA^{-\frac{1}{3}t}}\partial{z}(u{3,\neq}\nabla\mathbf{U}{2})|{L^{2}L^{2}}^{2} \leq CE{1}^{2}A^{\frac{4}{3}}E{4}(|\partial{x}\omega{2,\neq}|{X{a}} + |\Delta u{2,\neq}|{X_{a}}).$$

$$(3.24)$$

Proof. First of all, there holds

$$|\partial_j \mathbf{U}2|{H^1} \leq \int_0^t |\partial_s \partial_j \mathbf{U}2(s)|{H^1.} ds \leq C E_1 t.$$

For the given positive constant ϵ2, since limt→∞ A − ^{1 3} te −ϵ2A − 1 ^{3 t} = 0, there holds

$$|\mathbf{e}^{-\epsilon_2 A^{-\frac{1}{3}} t} |\partial_j \mathbf{U}2|{H^1} |_{L_t^{\infty}} \le C A^{\frac{1}{3}} E_1.$$ (3.25)

By Lemma A.2 and Lemma A.3, we have

∥∇U2∂xuj,̸=∥ ^L2 ≤ ∥∇U2∥ 2 L∞^y L² ∥∂xuj,̸=∥ 2 L∞^z L² x,y ≤ C∥∇U2∥ 2 ^H1 ∥∂xuj,̸=∥ 2 ^L2 ∥∂x∂zuj,̸=∥ 2 L2 , which along with (3.25) and Lemma 3.1 indicates that

$$\begin{split} |\mathrm{e}^{aA^{-\frac{1}{3}t}} \nabla \mathbf{U}_2 \partialx u{j,\neq} |_{L^2 L^2}^2 &\leq C \int_0^t \mathrm{e}^{\frac{a-b}{2}A^{-\frac{1}{3}s}} |\nabla \mathbf{U}2|{H^1}^2 \mathrm{e}^{\frac{b+3a}{2}A^{-\frac{1}{3}s}} |\partialx^2 u{j,\neq}|_{L^2}^{\frac{1}{2}} |\partial_x \partialz u{j,\neq}|_{L^2}^{\frac{3}{2}} ds \ &\leq C A E_1^2 E_4^{\frac{1}{2}}(t) (|\partialx \omega{2,\neq}|_{Xa} + |\triangle u{2,\neq}|_{X_a})^{\frac{3}{2}}. \end{split}$$

Similarly, one obtains that

$$|\partial{z}\mathbf{U}{2}\partial{x}^{2}u{1,\neq}|{L^{2}}^{2} \leq |\partial{z}\mathbf{U}{2}|{L{z}^{\infty}L{y}^{2}}^{2}|\partial{x}^{2}u{1,\neq}|{L{y}^{\infty}L{x,z}^{2}}^{2} \leq C|\partial{z}\mathbf{U}{2}|{H^{1}}^{2}|\partial{x}^{2}u{1,\neq}|{L^{2}}|\partial{x}^{2}\nabla u{1,\neq}|{L^{2}}^{2}$$

$$\leq C|\partial{z}\mathbf{U}{2}|{H^{1}}^{2}|\partial{x}^{2}u{1,\neq}|{L^{2}}^{\frac{1}{4}}(|\partial{x}\omega{2,\neq}|{L^{2}} + |\Delta u{2,\neq}|{L^{2}})^{\frac{3}{4}}(|\partial{x}\nabla\omega{2,\neq}|{L^{2}} + |\nabla\Delta u{2,\neq}|{L^{2}}).$$

By the divergence-free property

$$\partialx^2 u{1,\neq} = -\partial_x (\partialy u{2,\neq} + \partialz u{3,\neq}),$$

we can prove the second result.

Using Lemma A.2 and Lemma A.3 again, by H¨older's inequality, there is

$$\begin{split} |\partial{j}\mathbf{U}{2}\partial{x}\partial{j}u{2,\neq}|{L^{2}}^{2} &\leq C|\partial{j}\mathbf{U}{2}|{H^{1}}^{2}|\partial{x}\partial{j}u{2,\neq}|{L^{2}}|\partial{x}\partial{j}\nabla u{2,\neq}|{L^{2}} \ &\leq C|\partial{j}\mathbf{U}{2}|{H^{1}}^{2}|\partial{x}^{2}u{2,\neq}|{L^{2}}^{\frac{1}{2}}|\Delta u{2,\neq}|{L^{2}}^{\frac{1}{2}}|\partial{x}\partial{j}\nabla u{2,\neq}|_{L^{2}}, \end{split}$$

□

which implies that

$$|\mathbf{e}^{aA^{-\frac{1}{3}}t}\partial{j}\mathbf{U}{2}\partial{x}\partial{j}u{2,\neq}|{L^{2}L^{2}}^{2} \leq CA^{\frac{4}{3}}E{1}^{2}E{4}^{\frac{1}{2}}|\triangle u{2,\neq}|{X_{a}}^{\frac{3}{2}}.$$

According to Lemma A.2 and Lemma A.3, by H¨older's inequality, we get

$$\begin{aligned} |\partialz (u{2,\neq} \nabla \mathbf{U}2)|{L^2}^2 &\leq C |\nabla \mathbf{U}2|{H^1}^2 |\nabla u{2,\neq}|{L^2} |\triangle u{2,\neq}|{L^2} \ &\leq C |\nabla \mathbf{U}2|{H^1}^2 |u{2,\neq}|{L^2}^{\frac{1}{2}} |\triangle u{2,\neq}|{L^2}^{\frac{3}{2}}, \end{aligned}$$

which indicates that

$$\begin{aligned} &|\mathbf{e}^{aA^{-\frac{1}{3}}t}\partial{z}(u{2,\neq}\nabla\mathbf{U}{2})|{L^{2}L^{2}}^{2} \ &\leq C\left|\mathbf{e}^{\frac{a-b}{4}A^{-\frac{1}{3}}t}|\nabla\mathbf{U}{2}|{H^{1}}\right|{L^{\infty}{t}}^{2}|\mathbf{e}^{bA^{-\frac{1}{3}}t}u{2,\neq}|{L^{2}L^{2}}^{\frac{1}{2}}|\mathbf{e}^{aA^{-\frac{1}{3}}t}\triangle u{2,\neq}|{L^{2}L^{2}}^{\frac{3}{2}} \leq CE{1}^{2}E{4}^{\frac{1}{2}}A|\triangle u{2,\neq}|{X_{a}}^{\frac{3}{2}}. \end{aligned}$$

Using Lemma A.3, there holds

$$\begin{split} |\partial{z}u{3,\neq}|{L{y}^{\infty}L{x,z}^{2}}^{2} &\leq C|\partial{z}\partial{y}u{3,\neq}|{L^{2}}|\partial{z}u{3,\neq}|{L^{2}} \ &\leq C|u{3,\neq}|{L^{2}}^{\frac{1}{2}}|\partial{z}^{2}u{3,\neq}|{L^{2}}^{\frac{1}{2}}|\partial{y}u{3,\neq}|{L^{2}}^{\frac{1}{2}}|\partial{y}\partial{z}^{2}u{3,\neq}|{L^{2}}^{\frac{1}{2}} \end{split}$$

and

$$\begin{aligned} |u{3,\neq}|{L^{\infty}{y,z}L^{2}{x}}^{2} &\leq C|(\partial{x},\partial{z})\partial{y}u{3,\neq}|{L^{2}}|(\partial{x},\partial{z})u{3,\neq}|{L^{2}} \ &\leq C|u{3,\neq}|{L^{2}}^{\frac{1}{2}}|(\partial{x}^{2},\partial{z}^{2})u{3,\neq}|{L^{2}}^{\frac{1}{2}}|\partial{y}u{3,\neq}|{L^{2}}^{\frac{1}{2}}|\partial{y}(\partial{x}^{2},\partial{z}^{2})u{3,\neq}|_{L^{2}}^{\frac{1}{2}}. \end{aligned}$$

Using Lemma 3.1, we get

$$|\mathbf{e}^{aA^{-\frac{1}{3}}t}\partialz(u{3,\neq}\nabla\mathbf{U}2)|{L^2L^2}^2 \le CE_1^2A^{\frac{4}{3}}E_4(|\partialx\omega{2,\neq}|_{Xa} + |\Delta u{2,\neq}|_{X_a}).$$

Lastly, we show the the nonlinear interaction between the zero modes {u2,0, u3,0} and the non-zero mode u̸=. The proof can be found in Lemma 3.16 of [8], and we omit it.

Lemma 3.6. For j ∈ {2, 3}, it holds that

$$\begin{split} &|\mathrm{e}^{aA^{-\frac{1}{3}t}}u_{j,0}(\partial_x,\partialz)\nabla u{\neq}|{L^2L^2}^2 \leq CA(|u{2,0}|{L^{\infty}H^2}^2 + |u{3,0}|_{L^{\infty}H^1}^2)(|\partialx\omega{2,\neq}|_{Xa}^2 + |\triangle u{2,\neq}|_{X_a}^2), \ &|\mathrm{e}^{aA^{-\frac{1}{3}t}}\partialz\nabla u{j,0}\cdot u{\neq}|{L^2L^2}^2 \leq CA^{\frac{2}{3}}(|u{2,0}|{L^{\infty}H^2}^2 + |u{3,0}|{L^{\infty}H^1}^2)(|\partialx\omega{2,\neq}|_{Xa}^2 + |\triangle u{2,\neq}|_{X_a}^2), \ &|\mathrm{e}^{aA^{-\frac{1}{3}t}}\partialz u{j,0}\nabla u{\neq}|{L^2L^2}^2 \leq CA(|u{2,0}|{L^{\infty}H^2}^2 + |u{3,0}|{L^{\infty}H^1}^2)(|\partialx\omega{2,\neq}|_{Xa}^2 + |\triangle u{2,\neq}|_{X_a}^2), \ &|\mathrm{e}^{aA^{-\frac{1}{3}t}}\partialy u{3,0}(\partial_x,\partialz)u{2,\neq}|{L^2L^2}^2 \leq CA^{\frac{1}{3}}\left(|u{2,0}|{L^{\infty}H^2}^2 + |u{3,0}|{L^{\infty}H^1}^2\right)|\triangle u{2,\neq}|_{X_a}^2. \end{split}$$

4. Estimates for the zero modes of velocity E1(t): Proof of Proposition 2.1

In this section, we give some estimates for the zero modes of the velocity, which will be used in estimating the zero mode of the density and the non-zero modes.

□

4.1. Energy estimates for u2,⁰ and u3,0. We recall that

$$\begin{cases} \partialt u{2,0} - \frac{1}{A} \triangle u_{2,0} + \frac{1}{A} (u \cdot \nabla u_2)_0 + \frac{1}{A} \partial_y P_0^{N_1} + \frac{1}{A} \partial_y P_0^{N_2} = 0, \ \partialt u{3,0} - \frac{1}{A} \triangle u_{3,0} + \frac{1}{A} (u \cdot \nabla u_3)_0 + \frac{1}{A} \partial_z P_0^{N_1} + \frac{1}{A} \partial_z P_0^{N_2} = 0, \ \partialy u{2,0} + \partialz u{3,0} = 0, \end{cases} (4.1)$$

where

$$\begin{cases} \triangle P^{N_1} = -2A\partial_x u_2 + \partial_x n, \ \triangle P^{N_2} = -\text{div } (u \cdot \nabla u). \end{cases}$$

Lemma 4.1. Under the conditions of Theorem 1.1 and the assumptions (1.6) and (2.5), there exists a positive constant A2,³ independent of A and t, such that if A ≥ A2,3, it holds that

$$||u{2,0}||{Y0} + ||u{3,0}||_{Y0} \le C\epsilon, ||\nabla u{2,0}||_{Y0} + ||\nabla u{3,0}||_{Y0} \le C\epsilon, ||\Delta u{2,0}||_{Y0} \le C\epsilon, ||\min{(A^{-\frac{2}{3}} + A^{-1}t)^{\frac{1}{2}}, 1}\Delta u{3,0}||_{Y_0} \le C\epsilon.$$

$$(4.2)$$

Proof. Estimate of (4.2)1. Due to ∇ · u⁰ = 0, we have

$$<\partial_y P_0^{Nk}, u{2,0}>+<\partial_z P_0^{Nk}, u{3,0}>=-< P_0^{N_k}, \partialy u{2,0}+\partialz u{3,0}>=0, \text{ for } k\in{1,2}$$

$< u0\cdot \nabla u{2,0}, u_{2,0}>+< u0\cdot \nabla u{3,0}, u_{3,0}>=0.$

When A ≥ max{1, ϵ⁻⁶E 12 ² } =: A2,1, using Lemma 3.2 and assumption (1.6), energy estimates of (4.1) show that

$$\begin{aligned} & |u{2,0}|{L^{\infty}L^{2}}^{2} + |u{3,0}|{L^{\infty}L^{2}}^{2} + \frac{1}{A} \left( |\nabla u{2,0}|{L^{2}L^{2}}^{2} + |\nabla u{3,0}|{L^{2}L^{2}}^{2} \right) \ \leq & |(u{2,\text{in}}){0}|{L^{2}}^{2} + |(u{3,\text{in}}){0}|{L^{2}}^{2} + \frac{C||u{\neq}|^{2}|{L^{2}L^{2}}^{2}}{A} \leq |(u{2,\text{in}}){0}|{L^{2}}^{2} + |(u{3,\text{in}}){0}|{L^{2}}^{2} + \frac{CE_{2}^{4}}{A^{\frac{1}{3}}} \leq C\epsilon^{2} \end{aligned}$$

Estimate of (4.2)2. Multiplying 2△u2,⁰ on (4.1)¹ and 2△u3,⁰ on (4.1)2, energy estimates give that

$$\frac{d}{dt} \left( |\nabla u{2,0}|{L^{2}}^{2} + |\nabla u{3,0}|{L^{2}}^{2} \right) + \frac{1}{A} \left( |\Delta u{2,0}|{L^{2}}^{2} + |\Delta u{3,0}|{L^{2}}^{2} \right) \leq C \frac{|u{0} \cdot \nabla u{2,0}|{L^{2}}^{2} + |u{0} \cdot \nabla u{3,0}|{L^{2}}^{2} + |u{\neq} \cdot \nabla u{\neq}|_{L^{2}}^{2}}{A},$$ (4.3)

where we use < ∂yP N^k 0 , △u2,⁰ > + < ∂zP N^k 0 , △u3,⁰ >= 0 for k ∈ {1, 2}.

For j ∈ {2, 3}, by H¨older's inequality and Gagliardo-Nirenberg inequality, there holds

$$|\nabla u{j,0}|{L^{2}}^{2} \leq |u{j,0}|{L^{2}} |\triangle u{j,0}|{L^{2}},$$

$$|u{0} \cdot \nabla u{j,0}|{L^{2}}^{2} \leq C|(u{2}, u{3}){0}|{L^{2}} |\nabla (u{2}, u{3}){0}|{L^{2}} |\nabla u{j,0}|{L^{2}} |\triangle u{j,0}|_{L^{2}}$$

$$+ C|(u{2}, u{3}){0}|{L^{2}}^{2} |\nabla u{j,0}|{L^{2}} |\triangle u{j,0}|{L^{2}}.$$

$$(4.4)$$

According to (4.4) and Young's inequality, we rewrite (4.3) into

$$\frac{d}{dt} |\nabla(u{2}, u{3}){0}|{L^{2}}^{2} \leq -\frac{1}{2A} \left( \frac{|\nabla u{2,0}|{L^{2}}^{4}}{|u{2,0}|{L^{2}}^{2}} + \frac{|\nabla u{3,0}|{L^{2}}^{4}}{|u{3,0}|{L^{2}}^{2}} \right) + \frac{C|u{\neq} \cdot \nabla u{\neq}|{L^{2}}^{2}}{A} + C \frac{|(u{2}, u{3}){0}|{L^{2}}^{2} |\nabla(u{2}, u{3}){0}|{L^{2}}^{4} + |(u{2}, u{3}){0}|{L^{2}}^{4} |\nabla(u{2}, u{3}){0}|_{L^{2}}^{2}}{A}.$$

$$(4.5)$$

Thanks to (4.2)1, by taking ϵ small enough, we infer from (4.5) that

$$\frac{d}{dt} |\nabla(u_2, u_3)0|{L^2}^2 \le C \frac{|u{\neq} \cdot \nabla u{\neq}|_{L^2}^2}{A} + C \frac{|(u_2, u_3)0|{L^2}^4 |\nabla(u_2, u_3)0|{L^2}^2}{A}. \tag{4.6}$$

From this, along with Lemma 3.2, (1.6) and (4.2)1, when A ≥ A2,1, one deduces

$$|\nabla u{2,0}|{L^{\infty}L^{2}}^{2} + |\nabla u{3,0}|{L^{\infty}L^{2}}^{2} \leq |(u{2,\text{in}}){0}|{H^{1}}^{2} + |(u{3,\text{in}}){0}|{H^{1}}^{2} + \frac{C|u{\neq} \cdot \nabla u{\neq}|_{L^{2}L^{2}}^{2}}{A} + C\epsilon^{6}$$

$$\leq |(u{2,\text{in}}){0}|{H^{1}}^{2} + |(u{3,\text{in}}){0}|{H^{1}}^{2} + \frac{CE_{2}^{4}}{A^{\frac{1}{3}}} + C\epsilon^{6} \leq C\epsilon^{2}.$$

$$(4.7)$$

By Lemma A.2, (4.2)¹ and (4.7), we have

$$||u0 \cdot \nabla u{j,0}||_{L^2L^2}^2 \le C||(u_2, u_3)0||{L^{\infty}H^1}^2 ||\Delta u{j,0}||{L^2L^2}^2 \le C\epsilon^2 ||\Delta u{j,0}||{L^2L^2}^2.$$

Integrating in time for (4.3) and taking ϵ small enough, we obtain

$$|\nabla u{2,0}|{L^{\infty}L^{2}}^{2} + |\nabla u{3,0}|{L^{\infty}L^{2}}^{2} + \frac{|\Delta u{2,0}|{L^{2}L^{2}}^{2} + |\Delta u{3,0}|{L^{2}L^{2}}^{2}}{A}$$

$$\leq C \left(|(u{2,\text{in}}){0}|{H^{1}}^{2} + |(u{3,\text{in}}){0}|{H^{1}}^{2} + \frac{|u{\neq} \cdot \nabla u{\neq}|_{L^{2}L^{2}}^{2}}{A}\right) \leq C\epsilon^{2}.$$

$$(4.8)$$

Estimate of (4.2)3. The H² energy estimate for (4.1)¹ shows that

$$|\triangle u{2,0}|{L^{\infty}L^{2}}^{2} + \frac{|\nabla \triangle u{2,0}|{L^{2}L^{2}}^{2}}{A}$$

$$\leq |(u{2,\text{in}}){0}|{H^{2}}^{2} + \frac{C}{A} \left(|\nabla (u{0} \cdot \nabla u{2,0})|{L^{2}L^{2}}^{2} + |\nabla (u{\neq} \cdot \nabla u{2,\neq})|{L^{2}L^{2}}^{2} + |\triangle P{0}^{N{2}}|{L^{2}L^{2}}^{2}\right).$$

$$(4.9)$$

Using Lemma A.2, (4.2)1,² and ∂zu3,⁰ = −∂yu2,0, for j ∈ {2, 3}, there holds

$$\frac{|\nabla(u0 \cdot \nabla u{j,0})|{L^2 L^2}^2}{A} \leq \frac{C}{A} \left( |u{2,0}|{L^{\infty} H^2}^2 + |u{3,0}|{L^{\infty} H^1}^2 \right) \left( |\nabla u{2,0}|{L^2 H^1}^2 + |\nabla u{3,0}|{L^2 H^1}^2 \right) \ \leq C\epsilon^4 + C\epsilon^2 |\Delta u{2,0}|_{L^{\infty} L^2}^2.$$ (4.10)

Moreover, due to div (u · ∇u) = ∂x(u · ∇u1) + ∂y(u · ∇u2) + ∂z(u · ∇u3), there holds

$$|\operatorname{div} (u \cdot \nabla u)0|{L^2}^2 \le |\partial_y (u \cdot \nabla u_2)0|{L^2}^2 + |\partial_z (u \cdot \nabla u_3)0|{L^2}^2$$

$$\leq |\partial_y (u0 \cdot \nabla u{2,0})|_{L^2}^2 + |\partial_z (u0 \cdot \nabla u{3,0})|_{L^2}^2 + |\partialy (u{\neq} \cdot \nabla u{2,\neq})|{L^2}^2 + |\partialz (u{\neq} \cdot \nabla u{3,\neq})|{L^2}^2.$$

From this, along with Lemma 3.2 and (4.10), we have

$$\frac{|\triangle P_0^{N2}|{L^2L^2}^2}{A} = \frac{|\operatorname{div}(u \cdot \nabla u)0|{L^2L^2}^2}{A} \le C\left(\epsilon^4 + \frac{E2^4}{A^{\frac{1}{2} - \frac{2}{3}\alpha}}\right) + C\epsilon^2 |\triangle u{2,0}|_{L^\infty L^2}^2,\tag{4.11}$$

where α ∈ ( 2 , 4 ) is a constant. Then using assumption (1.6), (4.10), (4.11) and Lemma 3.2, we get from (4.9) that

$$|\Delta u{2,0}|{L^{\infty}L^{2}}^{2} + \frac{|\nabla \Delta u{2,0}|{L^{2}L^{2}}^{2}}{A} \le \epsilon^{2} + C\left(\epsilon^{4} + \frac{E{2}^{4}}{A^{\frac{1}{2} - \frac{2}{3}\alpha}}\right) + C\epsilon^{2} |\Delta u{2,0}|_{L^{\infty}L^{2}}^{2}.$$

By taking ϵ small enough, one obtains

$$|\triangle u{2,0}|{Y_0}^2 \le C\epsilon^2$$

provided with A ≥ max{A2,1, E 2 2 ϵ −1 ^{12 3}−4^α } =: A2.2.

Estimate of (4.2)4. Taking H² energy estimate for (4.1)2, there holds

$$\frac{d}{dt} | \triangle u{3,0} |{L^2}^2 + \frac{| \nabla \triangle u{3,0} |{L^2}^2}{A} \le \frac{C}{A} \left( | \nabla (u0 \cdot \nabla u{3,0}) |{L^2}^2 + | \nabla (u{\neq} \cdot \nabla u{3,\neq}) |{L^2}^2 + | \triangle P_0^{N2} |{L^2}^2 \right),$$

which follows that

$$\frac{d}{dt} \left( \min{A^{-\frac{2}{3}} + A^{-1}t, 1} | \Delta u{3,0} |{L^{2}}^{2} \right) + \frac{\min{A^{-\frac{2}{3}} + A^{-1}t, 1}}{A} | \nabla \Delta u{3,0} |{L^{2}}^{2} \ \leq \frac{C}{A} \left( | \nabla (u{0} \cdot \nabla u{3,0}) |{L^{2}}^{2} + (A^{-\frac{2}{3}} + A^{-1}t) | \nabla (u{\neq} \cdot \nabla u{3,\neq}) |{L^{2}}^{2} + | \Delta P{0}^{N{2}} |{L^{2}}^{2} + | \Delta u{3,0} |_{L^{2}}^{2} \right).$$ (4.12)

Thanks to (4.2)3, (4.10) and (4.11), we get

$$\frac{|\nabla(u0 \cdot \nabla u{j,0})|_{L^2 L^2}^2}{A} \le C\epsilon^4, \quad \frac{|\triangle P_0^{N2}|{L^2 L^2}^2}{A} \le C\left(\epsilon^4 + \frac{E_2^4}{A^{\frac{1}{2} - \frac{2}{3}\alpha}}\right). \tag{4.13}$$

Notice that ∥(A − ^{2 3} + A^{−1 t})e⁻aA^{− 1 3 t}∥L∞^{t ≤} CA^{− 2 3} , using (4.2)2, (4.13) and Lemma 3.2, one obtains

$$\begin{split} &|\min{(A^{-\frac{2}{3}}+A^{-1}t)^{\frac{1}{2}},1}\triangle u{3,0}|{Y{0}}^{2}\ \leq &\frac{|(u{3,\mathrm{in}}){0}|{H^{2}}^{2}}{A^{\frac{2}{3}}}+C\left(\epsilon^{2}+\frac{E{2}^{4}}{A^{\frac{1}{2}-\frac{2}{3}\alpha}}\right)+\frac{C}{A}|(A^{-\frac{2}{3}}+A^{-1}t)\mathrm{e}^{-aA^{-\frac{1}{3}}t}|{L{t}^{\infty}}|\mathrm{e}^{2aA^{-\frac{1}{3}}t}\nabla(u{\neq}\cdot\nabla u{3,\neq})|{L^{2}L^{2}}^{2}\ \leq &\frac{|(u{3,\mathrm{in}}){0}|{H^{2}}^{2}}{A^{\frac{2}{3}}}+C\left(\epsilon^{2}+\frac{E{2}^{4}}{A^{\frac{1}{2}-\frac{2}{3}\alpha}}+\frac{E{2}^{2}E{5}^{2}}{A^{\frac{1}{2}-\frac{1}{3}\alpha}}\right),\quad\alpha\in\left(\frac{1}{2},\frac{3}{4}\right). \end{split}$$

When

$$A \ge \max{A{2,2}, (|(u{3,\text{in}})0|{H^2} \epsilon^{-1})^3, (E_2 E5 \epsilon^{-1})^{\frac{12}{3-2\alpha}}} =: A{2,3},$$

we infer from the above inequality that

$$|\min{(A^{-\frac{2}{3}} + A^{-1}t)^{\frac{1}{2}}, 1} \triangle u{3,0}|{Y_0}^2 \le C\epsilon^2.$$

The proof is complete. □

4.2. Heat dissipation estimates for u2,⁰ and u3,0. Next, we consider the standard heat equation in (y, z) ∈ T 2

$$\partial_t f - \frac{1}{A} \triangle f = g, \quad t \in [0, T],$$ (4.14)

where g(t, y, z) is a given function. By performing a decomposition similar to (2.1), we obtain the following heat dissipation estimate for the non-average part ^fe⁼ P k 2 ²+k 2 ̸=0 ^fb^k2,k³ (t)eⁱ(k2y+k3z) .

Lemma 4.2. Let f(t, y, z) be a solution of (4.14), there holds

$$|e^{\frac{t}{2A}}\widetilde{f}|{L^{\infty}L^{2}}^{2} + \frac{1}{A}|e^{\frac{t}{2A}}\nabla\widetilde{f}|{L^{2}L^{2}}^{2} \le C\left(|f{\text{in}}|{L^{2}}^{2} + A|e^{\frac{t}{2A}}(-\Delta)^{-\frac{1}{2}}\widetilde{g}|_{L^{2}L^{2}}^{2}\right),$$

where ^{ge =} P k ²+k ̸=0 ^gb^k2,k³ (t)eⁱ(k2y+k3z) .

Proof. Doing the Fourier transform to (4.14), the Fourier mode ^fe satisfies

$$\partialt \widehat{f}{k_2,k_3} + \frac{k_2^2 + k3^2}{A} \widehat{f}{k_2,k3} = \widehat{g}{k_2,k_3}, \quad k_2^2 + k_3^2 > 0.$$ (4.15)

The solution of (4.15) is given by

$$\widehat{f}_{k_2,k_3} = e^{-\frac{k_2^2 + k3^2}{A}t} (\widehat{f}{in})_{k_2,k_3} + \int_0^t e^{\frac{k_2^2 + k3^2}{A}(s-t)} \widehat{g}{k_2,k3}(s) ds =: F{(1)} + F_{(2)}.$$ (4.16)

For F(1), due to k 2 ² + k 2 ³ > 0, there holds |F(1)| ≤ e − ^{t 2A |}fbin|, which shows that

$$|e^{\frac{t}{2A}}F{(1)}|{L^{\infty}(0,T)}^2 \le |\widehat{f}_{\text{in}}|^2.$$

For F(2), using H¨older's inequality and k ² + k 2 ³ > 0, we have

$$\begin{split} |F_{(2)}| \leq & |\mathrm{e}^{\frac{k_2^2 + k3^2}{2A}(s-t)}|{L^2} |\mathrm{e}^{\frac{k_2^2 + k3^2}{2A}(s-t)} \widehat{g}{k_2,k3}(s)|{L^2} \ \leq & \left(\frac{A}{k_2^2 + k_3^2}\right)^{\frac{1}{2}} |\mathrm{e}^{\frac{k_2^2 + k3^2}{2A}(s-t)} \widehat{g}{k_2,k3}(s)|{L^2} \leq \left(\frac{A}{k_2^2 + k3^2}\right)^{\frac{1}{2}} |\mathrm{e}^{\frac{s-t}{2A}} \widehat{g}{k_2,k3}(s)|{L^2}. \end{split}$$

This implies that

$$|\mathrm{e}^{\frac{t}{2A}}F{(2)}|{L^{\infty}(0,T)}^{2} \leq CA(k{2}^{2}+k{3}^{2})^{-1}|\mathrm{e}^{\frac{t}{2A}}\widehat{g}{k{2},k{3}}|{L^{2}(0,T)}^{2}.$$

Collecting the estimates of F(1), F(2), and using the Plancherel's theorem, we get from (4.16) that

$$\begin{aligned} |\mathbf{e}^{\frac{t}{2A}}\widetilde{f}|{L^{\infty}L^{2}}^{2} &= |\mathbb{T}|^{2} \sum{k{2}^{2}+k{3}^{2}>0} |\mathbf{e}^{\frac{t}{2A}}\widehat{f}{k{2},k{3}}(t)|{L^{\infty}(0,T)}^{2} \ &\leq |\mathbb{T}|^{2} \sum{k{2}^{2}+k{3}^{2}>0} \left( |\mathbf{e}^{\frac{t}{2A}}F{(1)}|{L^{\infty}(0,T)}^{2} + |\mathbf{e}^{\frac{t}{2A}}F{(2)}|{L^{\infty}(0,T)}^{2} \right) \ &\leq C|\mathbb{T}|^{2} \sum{k{2}^{2}+k{3}^{2}>0} \left( |(\widehat{f}{\mathrm{in}}){k{2},k{3}}|^{2} + A(k{2}^{2}+k{3}^{2})^{-1} |\mathbf{e}^{\frac{t}{2A}}\widehat{g}{k{2},k{3}}|{L^{2}(0,T)}^{2} \right) \ &\leq C\left( |f{\mathrm{in}}|{L^{2}}^{2} + A|\mathbf{e}^{\frac{t}{2A}}(-\Delta)^{-\frac{1}{2}}\widetilde{g}|_{L^{2}L^{2}}^{2} \right). \end{aligned} \tag{4.17}$$

We also need to estimate ¹ A ∥e ^{2A ∇}fe^{∥ L}2L² . Multiplying (4.15) by e ^{t 2A} , the energy estimate gives that

$$-\frac{1}{2A}|\mathbf{e}^{\frac{t}{2A}}\widehat{f}_{k_2,k3}|{L^2}^2 + \frac{2(k_2^2 + k3^2)}{3A}|\mathbf{e}^{\frac{t}{2A}}\widehat{f}{k_2,k3}|{L^2}^2 \leq -\frac{1}{2}\left(|\mathbf{e}^{\frac{t}{2A}}\widehat{f}_{k_2,k3}|^2\right)|{t=0}^{t=T} + \frac{CA|\mathbf{e}^{\frac{t}{2A}}\widehat{g}_{k_2,k3}|{L^2}^2}{k_2^2 + k_3^2},$$

which follows that

$$\frac{k_2^2 + k3^2}{A} | \mathbf{e}^{\frac{t}{2A}} \widehat{f}{k_2, k3} |{L^2}^2 \le C \left( |(\widehat{f}{\text{in}}){k_2, k_3}|^2 + A(k_2^2 + k3^2)^{-1} | \mathbf{e}^{\frac{t}{2A}} \widehat{g}{k_2, k3} |{L^2}^2 \right).$$

From this, along with the Plancherel's theorem, we get

$$\frac{1}{A} | \mathbf{e}^{\frac{t}{2A}} \nabla \widetilde{f} |{L^{2}L^{2}}^{2} = \frac{1}{A} |\mathbb{T}|^{2} \sum{k{2}^{2} + k{3}^{2} > 0} | \mathbf{e}^{\frac{t}{2A}} (k{2}^{2} + k{3}^{2})^{\frac{1}{2}} \widehat{f}{k{2}, k{3}} |{L^{2}(0, T)}^{2} \ \leq C |\mathbb{T}|^{2} \sum{k{2}^{2} + k{3}^{2} > 0} \left( |(\widehat{f}{\text{in}}){k{2}, k{3}}|^{2} + A(k{2}^{2} + k{3}^{2})^{-1} | \mathbf{e}^{\frac{t}{2A}} \widehat{g}{k{2}, k{3}} |{L^{2}}^{2} \right) \ \leq C \left( |f{\text{in}}|{L^{2}}^{2} + A | \mathbf{e}^{\frac{t}{2A}} (-\Delta)^{-\frac{1}{2}} \widetilde{g} |{L^{2}L^{2}}^{2} \right). \tag{4.18}$$

Combining (4.17) with (4.18), one obtains

$$|e^{\frac{t}{2A}}\widetilde{f}|{L^{\infty}L^{2}}^{2} + \frac{1}{A}|e^{\frac{t}{2A}}\nabla\widetilde{f}|{L^{2}L^{2}}^{2} \le C\left(|f{\text{in}}|{L^{2}}^{2} + A|e^{\frac{t}{2A}}(-\triangle)^{-\frac{1}{2}}\widetilde{g}|_{L^{2}L^{2}}^{2}\right).$$

The proof is complete. □

Similarly, we decompose the velocity u2,⁰ and u3,⁰ in the same way of (2.1), satisfying

$$u{2,0}(t,y,z) = \overline{u}{2,0}(t) + \widetilde{u}{2,0}(t,y,z), u{3,0}(t,y,z) = \overline{u}{3,0}(t) + \widetilde{u}{3,0}(t,y,z).$$

$$(4.19)$$

In this way, combining with Lemma 4.2, we can prove a typical heat dissipation estimates for ^ue²,0(t, y, z) and ^ue³,0(t, y, z). Furthermore, for ^j ∈ {2, ³}, ^uj,0(t) is a constant satisfying

$$\overline{u}{j,0}(t) \equiv \frac{1}{|\mathbb{T}|^2} \int{|\mathbb{T}|^2} (u_{j,\mathrm{in}})_0 dy dz.$$

Lemma 4.3. Under the assumptions of Lemma 4.1, it holds that

$$|e^{\frac{t}{2A}}\widetilde{u}{2,0}|{L^{\infty}L^{2}}^{2} + |e^{\frac{t}{2A}}\widetilde{u}{3,0}|{L^{\infty}L^{2}}^{2} + \frac{1}{A}\left(|e^{\frac{t}{2A}}\nabla\widetilde{u}{2,0}|{L^{2}L^{2}}^{2} + |e^{\frac{t}{2A}}\nabla\widetilde{u}{3,0}|{L^{2}L^{2}}^{2}\right) \leq C\epsilon^{2}.$$

Proof. Applying Lemma 4.2 to (4.1)1,2, and noting that ∇ · u = 0, we get

$$|\mathbf{e}^{\frac{t}{2A}}\widetilde{u}{2,0}|{L^{\infty}L^{2}}^{2} + |\mathbf{e}^{\frac{t}{2A}}\widetilde{u}{3,0}|{L^{\infty}L^{2}}^{2} + \frac{1}{A}\left(|\mathbf{e}^{\frac{t}{2A}}\nabla\widetilde{u}{2,0}|{L^{2}L^{2}}^{2} + |\mathbf{e}^{\frac{t}{2A}}\nabla\widetilde{u}{3,0}|{L^{2}L^{2}}^{2}\right)$$

$$\leq C\left(|(u{2,\text{in}}){0}|{L^{2}}^{2} + |(u{3,\text{in}}){0}|{L^{2}}^{2} + \frac{1}{A}|\mathbf{e}^{\frac{t}{2A}}\widetilde{(u{2}u{2})}{0}|{L^{2}L^{2}}^{2} + \frac{1}{A}|\mathbf{e}^{\frac{t}{2A}}\widetilde{(u{3}u{2})}{0}|{L^{2}L^{2}}^{2} + \frac{1}{A}|\mathbf{e}^{\frac{t}{2A}}\widetilde{(u{3}u{3})}{0}|{L^{2}L^{2}}^{2} + \frac{1}{A}|\mathbf{e}^{\frac{t}{2A}}\widetilde{P}{0}^{N{1}}|{L^{2}L^{2}}^{2} + \frac{1}{A}|\mathbf{e}^{\frac{t}{2A}}\widetilde{P}{0}^{N{2}}|{L^{2}L^{2}}^{2}\right)$$

$$=:C\left(|(u{2,\text{in}}){0}|{L^{2}}^{2} + |(u{3,\text{in}}){0}|{L^{2}}^{2} + I{1} + \dots + I{5}\right).$$

$$(4.20)$$

For ^I1, as (^u2u2) ⁰ = 2u2,0ue²,^{0 + u}e²,0ue²,^{0 +} (u^2,̸=u2,̸=) 0 , there holds

$$I{1} \leq \frac{C}{A} \left( | e^{\frac{t}{2A}} \overline{u}{2,0} \widetilde{u}{2,0} |{L^{2}L^{2}}^{2} + | e^{\frac{t}{2A}} \widetilde{u}{2,0} \widetilde{u}{2,0} |{L^{2}L^{2}}^{2} + | e^{\frac{t}{2A}} (u{2,\neq} u{2,\neq}){0} |_{L^{2}L^{2}}^{2} \right)$$

$$=: I{11} + I{12} + I_{13}. \tag{4.21}$$

Thanks to ∂tuj,⁰ = 0 for j ∈ {2, 3}, we find

$$|\overline{u}{j,0}| = |(\overline{u}{j,\text{in}})0| \le \frac{1}{|\mathbb{T}|} |(u{j,\text{in}})0|{L^2}.$$ (4.22)

From this and Poincar´e inequality, I¹¹ can be controlled by

$$I{11} \le \frac{C}{A} |(u{2,\text{in}})0|{L^2}^2 |e^{\frac{t}{2A}} \nabla \widetilde{u}{2,0}|{L^2 L^2}^2.$$

For I12, using Lemma 4.1 and Poincar´e inequality, one deduces

$$I{12} \leq \frac{C}{A} |\widetilde{u}{2,0}|{L^{2}L^{\infty}}^{2} |e^{\frac{t}{2A}}\widetilde{u}{2,0}|{L^{\infty}L^{2}}^{2} \leq \frac{C}{A} |\Delta u{2,0}|{L^{2}L^{2}}^{2} |e^{\frac{t}{2A}}\widetilde{u}{2,0}|{L^{\infty}L^{2}}^{2} \leq C\epsilon^{2} |e^{\frac{t}{2A}}\widetilde{u}{2,0}|_{L^{\infty}L^{2}}^{2}.$$

When A is sufficiently large and using Lemma 3.2, we arrive

$$I{13} \le \frac{C}{A} |e^{2aA^{-\frac{1}{3}}t}|u{\neq}|^2|_{L^2L^2}^2 \le \frac{CE_2^4}{A^{\frac{1}{3}}} \le C\epsilon^2.$$

Collecting I¹¹ − I13, (4.21) yields that

$$I{1} \leq \frac{C}{A} |(u{2,\text{in}}){0}|{L^{2}}^{2} |e^{\frac{t}{2A}} \nabla \widetilde{u}{2,0}|{L^{2}L^{2}}^{2} + C\epsilon^{2} |e^{\frac{t}{2A}} \widetilde{u}{2,0}|{L^{\infty}L^{2}} + C\epsilon^{2}.$$

$$(4.23)$$

The estimates of I² and I³ are similar to I1. Note that

$$\widetilde{(u_3 u_2)0} = \overline{u}{3,0} \widetilde{u}{2,0} + \widetilde{u}{3,0} \overline{u}{2,0} + \widetilde{u}{3,0} \widetilde{u}{2,0} + (u{3,\neq} u_{2,\neq})_0$$

$$\widetilde{(u_3 u_3)0} = 2\overline{u}{3,0} \widetilde{u}{3,0} + \widetilde{u}{3,0} \widetilde{u}{3,0} + (u{3,\neq} u_{3,\neq})_0.$$

Using (4.22), Lemma 3.2 and Poincar´e inequality, we have

$$I{2} \leq \frac{C}{A} \left( | e^{\frac{t}{2A}} \overline{u}{3,0} \widetilde{u}{2,0} |{L^{2}L^{2}}^{2} + | e^{\frac{t}{2A}} \widetilde{u}{3,0} \overline{u}{2,0} |{L^{2}L^{2}}^{2} + | e^{\frac{t}{2A}} \widetilde{u}{3,0} \widetilde{u}{2,0} |{L^{2}L^{2}}^{2} + | e^{\frac{t}{2A}} (u{3,\neq} u{2,\neq}){0} |{L^{2}L^{2}}^{2} \right)$$

$$\leq \frac{C}{A} \left( | \overline{u}{3,0}(t) |{L^{\infty}}^{2} | e^{\frac{t}{2A}} \widetilde{u}{2,0} |{L^{2}L^{2}}^{2} + | \overline{u}{2,0}(t) |{L^{\infty}}^{2} | e^{\frac{t}{2A}} \widetilde{u}{3,0} |{L^{2}L^{2}}^{2} \right)$$

$$+ | \widetilde{u}{2,0} |{L^{2}L^{\infty}}^{2} | e^{\frac{t}{2A}} \widetilde{u}{3,0} |{L^{\infty}L^{2}}^{2} + | e^{2aA^{-\frac{1}{3}t}} |u{\neq}|^{2} |{L^{2}L^{2}}^{2} \right)$$

$$\leq \frac{C}{A} \left( | (u{2,\text{in}}){0} |{L^{2}}^{2} + | (u{3,\text{in}}){0} |{L^{2}}^{2} \right) | e^{\frac{t}{2A}} \nabla (\widetilde{u}{2,0}, \widetilde{u}{3,0}) |{L^{2}L^{2}}^{2} + C\epsilon^{2} | e^{\frac{t}{2A}} \widetilde{u}{3,0} |_{L^{\infty}L^{2}}^{2} + C\epsilon^{2}$$

$$(4.24)$$

and

$$I{3} \leq \frac{C}{A} \left( | \mathbf{e}^{\frac{t}{2A}} \overline{u}{3,0} \widetilde{u}{3,0} |{L^{2}L^{2}}^{2} + | \mathbf{e}^{\frac{t}{2A}} \widetilde{u}{3,0} \widetilde{u}{3,0} |{L^{2}L^{2}}^{2} + | \mathbf{e}^{\frac{t}{2A}} (u{3,\neq u{3,\neq 0}}){0} |_{L^{2}L^{2}}^{2} \right)$$

$$\leq \frac{C}{A} \left( | \overline{u}{3,0}(t) |{L^{\infty}}^{2} | \mathbf{e}^{\frac{t}{2A}} \widetilde{u}{3,0} |{L^{2}L^{2}}^{2} + | \widetilde{u}{3,0} |{L^{2}L^{\infty}}^{2} | \mathbf{e}^{\frac{t}{2A}} \widetilde{u}{3,0} |{L^{\infty}L^{2}}^{2} + | \mathbf{e}^{2aA^{-\frac{1}{3}t}} | u{\neq} |^{2} |{L^{2}L^{2}}^{2} \right)$$

$$\leq C\epsilon^{2} | \mathbf{e}^{\frac{t}{2A}} \widetilde{u}{3,0} |{L^{\infty}L^{2}}^{2} + \frac{C}{A} | (u{3,\text{in}}){0} |{L^{2}}^{2} | \mathbf{e}^{\frac{t}{2A}} \nabla \widetilde{u}{3,0} |_{L^{2}L^{2}}^{2} + C\epsilon^{2}.$$

$$(4.25)$$

Taking the Fourier transform for (1.4)1, the Fourier mode ^Pb^N1 k1,k2,k³ follows that

$$-(k_1^2 + k_2^2 + k3^2)\widehat{P}{k_1,k_2,k_3}^{N_1} = ik1(-2A\widehat{u}{2,k_1,k_2,k3} + \widehat{n}{k_1,k_2,k_3}).$$

For the Fourier mode ^Pb^N1 k1,k2,k³ of ^Pe^{N1 0} with k¹ = 0 and k 2 ² + k ³ > 0, we have

$$\widehat{P}_{k_1,k_2,k_3}^{N1}|{k_1=0,k_2^2+k_3^2>0} = \frac{-ik1(-2A\widehat{u}{2,k_1,k_2,k3}+\widehat{n}{k_1,k_2,k_3})}{k_1^2+k_2^2+k3^2}\Big|{k_1=0,k_2^2+k_3^2>0} = 0.$$

This implies that

$$I_4 = 0.$$ (4.26)

It follows from (1.4)² that

$$\Delta P_0^{N_2} = -\text{div } (u \cdot \nabla u)_0 = -\partial_y^2 (u_2 u_2)_0 - 2\partial_y \partial_z (u_2 u_3)_0 - \partial_z^2 (u_3 u_3)_0.$$

Therefore, for I5, one obtains

$$I{5} \leq \frac{C}{A} \left( | e^{\frac{t}{2A}} \widetilde{(u{2}u{2}){0}} |{L^{2}L^{2}}^{2} + | e^{\frac{t}{2A}} \widetilde{(u{2}u{3}){0}} |{L^{2}L^{2}}^{2} + | e^{\frac{t}{2A}} \widetilde{(u{3}u{3}){0}} |{L^{2}L^{2}}^{2} \right) \leq C \left( I{1} + I{2} + I{3} \right). \quad (4.27)$$

Combining (4.23), (4.24), (4.25), (4.26) and (4.27), we get from (4.20) that

$$|\mathbf{e}^{\frac{t}{2A}}\widetilde{u}{2,0}|{L^{\infty}L^{2}}^{2} + |\mathbf{e}^{\frac{t}{2A}}\widetilde{u}{3,0}|{L^{\infty}L^{2}}^{2} + \frac{1}{A}\left(|\mathbf{e}^{\frac{t}{2A}}\nabla\widetilde{u}{2,0}|{L^{2}L^{2}}^{2} + |\mathbf{e}^{\frac{t}{2A}}\nabla\widetilde{u}{3,0}|{L^{2}L^{2}}^{2}\right)$$

$$\leq C\left(|(u{2,\text{in}}){0}|{L^{2}}^{2} + |(u{3,\text{in}}){0}|{L^{2}}^{2}\right) + C\epsilon^{2}\left(|\mathbf{e}^{\frac{t}{2A}}\widetilde{u}{2,0}|{L^{\infty}L^{2}}^{2} + |\mathbf{e}^{\frac{t}{2A}}\widetilde{u}{3,0}|{L^{\infty}L^{2}}^{2}\right)$$

$$+ \frac{C}{A}\left(|(u{2,\text{in}}){0}|{L^{2}}^{2} + |(u{3,\text{in}}){0}|{L^{2}}^{2}\right)\left(|\mathbf{e}^{\frac{t}{2A}}\nabla\widetilde{u}{2,0}|{L^{2}L^{2}}^{2} + |\mathbf{e}^{\frac{t}{2A}}\nabla\widetilde{u}{3,0}|{L^{2}L^{2}}^{2}\right) + C\epsilon^{2}$$

$$\leq C\epsilon^{2}\left(1 + |\mathbf{e}^{\frac{t}{2A}}\widetilde{u}{2,0}|{L^{\infty}L^{2}}^{2} + |\mathbf{e}^{\frac{t}{2A}}\widetilde{u}{3,0}|{L^{\infty}L^{2}}^{2}\right) + \frac{C\epsilon^{2}}{A}\left(|\mathbf{e}^{\frac{t}{2A}}\nabla\widetilde{u}{2,0}|{L^{2}L^{2}}^{2} + |\mathbf{e}^{\frac{t}{2A}}\nabla\widetilde{u}{3,0}|{L^{2}L^{2}}^{2}\right),$$

$$(4.28)$$

where we use assumption (1.6).

Hence, when ϵ is small enough satisfying ϵ ≤ ^√ 2C , we obtain from (4.28) that

$$|\mathrm{e}^{\frac{t}{2A}}\widetilde{u}{2,0}|{L^{\infty}L^{2}}^{2}+|\mathrm{e}^{\frac{t}{2A}}\widetilde{u}{3,0}|{L^{\infty}L^{2}}^{2}+\frac{1}{2A}\left(|\mathrm{e}^{\frac{t}{2A}}\nabla\widetilde{u}{2,0}|{L^{2}L^{2}}^{2}+|\mathrm{e}^{\frac{t}{A}}\nabla\widetilde{u}{3,0}|{L^{2}L^{2}}^{2}\right)\leq C\epsilon^{2}.$$

The proof is complete.

4.3. Energy estimates for u1,0.

Lemma 4.4. Under the assumptions of Lemma 4.1, it holds that

$$|\mathbf{U}1|{L^{\infty}H^1} \le C\left(|(u_{1,\text{in}})0|{H^1} + |n0|{L^{\infty}L^2} + \epsilon\right).$$

Proof. Multiplying △Be¹ on both sides of (2.3)1, and applying H¨older's inequality, the energy estimate shows that

$$\frac{d}{dt} |\nabla \widetilde{\mathbf{B}}{1}|{L^{2}}^{2} + \frac{1}{A} |\Delta \widetilde{\mathbf{B}}{1}|{L^{2}}^{2} \leq \frac{C}{A} \left( |u{2,0}|{L^{\infty}}^{2} + |u{3,0}|{L^{\infty}}^{2} \right) |\nabla \widetilde{\mathbf{B}}{1}|{L^{2}}^{2} + \frac{C}{A} |\widetilde{n}{0}|{L^{2}}^{2}. \tag{4.29}$$

Using (3.19) and Poincar´e inequality

$$|\nabla \widetilde{\mathbf{B}}1|{L^2}^2 \le C |\triangle \widetilde{\mathbf{B}}1|{L^2}^2,$$

□

when choosing ϵ is small enough, we infer from (4.29) that

$$\frac{d}{dt} |\nabla \widetilde{\mathbf{B}}{1}|{L^{2}}^{2} \le -\frac{|\nabla \widetilde{\mathbf{B}}{1}|{L^{2}}^{2}}{2AC} + \frac{C|\widetilde{n}{0}|{L^{2}}^{2}}{A} \le -\frac{|\nabla \widetilde{\mathbf{B}}{1}|{L^{2}}^{2} - C|n{0}|{L^{2}}^{2}}{2AC},\tag{4.30}$$

where we use

$$|\widetilde{n}0|{L^2} = |n_0 - \overline{n}0|{L^2} \le C |n0|{L^2}.$$

According to (4.30), there holds

$$|\nabla \widetilde{\mathbf{B}}1|{L^{\infty}L^2}^2 \le C|n0|{L^{\infty}L^2}^2.$$

Otherwise, there must be a time t = t ∗ , such that

$$\begin{cases} |\nabla \widetilde{\mathbf{B}}{1}|{L^{2}}^{2}|{t=t^{*}} = C|n{0}|{L^{\infty}L^{2}}^{2}, \ \frac{d}{dt}|\nabla \widetilde{\mathbf{B}}{1}|{L^{2}}^{2}\Big|{t=t^{*}} > 0. \end{cases} (4.31)$$

Let t = t ∗ , using (4.31)1, we get from (4.30) that

$$\frac{d}{dt} |\nabla \widetilde{\mathbf{B}}1|{L^2}^2 \Big|_{t=t^} \le -\frac{|\nabla \widetilde{\mathbf{B}}1|{L^2}^2 - C|n0|{L^2}^2}{2AC} \Big|_{t=t^} = -\frac{|n0|{L^\infty L^2}^2 - |n0|{L^2}^2|_{t=t^*}}{2A} \le 0.$$ (4.32)

A contradiction has arisen between (4.31) and (4.32). This shows that

$$|\nabla \widetilde{\mathbf{B}}1|{L^{\infty}L^2}^2 \le C|n0|{L^{\infty}L^2}^2. \tag{4.33}$$

By Poincar´e inequality, there holds

$$|\widetilde{\mathbf{B}}1|{L^2} \le C |\nabla \widetilde{\mathbf{B}}1|{L^2}.$$

Then it follows from (4.33) that

$$|\widetilde{\mathbf{B}}1|{L^{\infty}H^1} \le C |\nabla \widetilde{\mathbf{B}}1|{L^{\infty}L^2} \le C |n0|{L^{\infty}L^2}.$$ (4.34)

Note that the estimate of G¹ is similar to (4.2)1,2, so we omit it. For (2.2)1, energy estimates show that

$$|\mathbf{G}1|{L^{\infty}H^1}^2 + \frac{1}{A}|\nabla\mathbf{G}1|{L^2H^1}^2 \le C|(u_{1,\text{in}})0|{H^1}^2 + C\epsilon^2.$$ (4.35)

Combining (4.34) with (4.35), we conclude that

$$|\mathbf{U}1|{L^{\infty}H^1} \le C(|\widetilde{\mathbf{B}}1|{L^{\infty}H^1} + |\mathbf{G}1|{L^{\infty}H^1}) \le C(|(u_{1,\text{in}})0|{H^1} + |n0|{L^{\infty}L^2} + \epsilon).$$

The proof is complete. □

Lemma 4.5. Under the assumptions of Lemma 4.1, there exists a positive constant A² independent of A and t, such that if A ≥ A2, it holds that

$$\frac{|\triangle \mathbf{U}2|{L^{\infty}H^2}}{A} + \frac{|\nabla \triangle \mathbf{U}2|{L^2H^2}}{A^{\frac{3}{2}}} + |\partial_t \mathbf{U}2|{L^{\infty}H^2} \le C\epsilon.$$

Proof. Estimate of ^Be²(t, y, z). For (2.4)1, the ^H4 energy estimate shows that

$$\frac{|\triangle^2 \widetilde{\mathbf{B}}2|{L^{\infty}L^2}^2}{A^2} + \frac{|\nabla\triangle^2 \widetilde{\mathbf{B}}2|{L^2L^2}^2}{A^3} \le \frac{C}{A} |\nabla\triangle \widetilde{u}{2,0}|{L^2L^2}^2 + \frac{C}{A^3} |\nabla\triangle \left(u_{2,0}\partial_y \widetilde{\mathbf{B}}2 + u{3,0}\partial_z \widetilde{\mathbf{B}}2\right)|{L^2L^2}^2. \tag{4.36}$$

Using Lemma A.2, Lemma 4.1 and Poincar´e inequality, one obtains

$$\frac{|\nabla \triangle (u{2,0}\partial{y}\widetilde{\mathbf{B}}{2})|{L^{2}L^{2}}^{2}}{A^{3}} \leq \frac{C}{A^{3}} \left( |\nabla u{2,0}|{L^{2}H^{2}}^{2} |\partial{y}\widetilde{\mathbf{B}}{2}|{L^{\infty}H^{2}}^{2} + |u{2,0}|{L^{\infty}H^{2}}^{2} |\nabla \partial{y}\widetilde{\mathbf{B}}{2}|{L^{2}H^{2}}^{2} \right) \ \leq C\epsilon^{2} \left( \frac{|\triangle^{2}\widetilde{\mathbf{B}}{2}|{L^{\infty}L^{2}}^{2}}{A^{2}} + \frac{|\nabla \triangle^{2}\widetilde{\mathbf{B}}{2}|{L^{2}L^{2}}^{2}}{A^{3}} \right), \tag{4.37}$$

$$\frac{|u_{3,0}\nabla \triangle \partial_z \widetilde{\mathbf{B}}2 + \nabla u{3,0}\triangle \partial_z \widetilde{\mathbf{B}}2|{L^2L^2}^2}{A^3} \le \frac{C}{A^3} |u{3,0}|{L^{\infty}H^1}^2 |\partial_z \widetilde{\mathbf{B}}2|{L^2H^4}^2 \le C\epsilon^2 \frac{|\nabla \triangle^2 \widetilde{\mathbf{B}}2|{L^2L^2}^2}{A^3}, \quad (4.38)$$

and

$$\frac{|\triangle u_{3,0} \nabla \partial_z \widetilde{\mathbf{B}}2|{L^2 L^2}^2}{A^3} \le \frac{C}{A^3} |\triangle u{3,0}|{L^2 L^2}^2 |\widetilde{\mathbf{B}}2|{L^{\infty} H^4}^2 \le C\epsilon^2 \frac{|\triangle^2 \widetilde{\mathbf{B}}2|{L^{\infty} L^2}^2}{A^2}.$$ (4.39)

Due to ^∥Be^2∥ 2 ^H3 ≤ ∥Be²∥H^{2 ∥}Be²∥H⁴ , and

$$|\widetilde{\mathbf{B}}2|{H^2} \leq \int_0^t |\partial_s \widetilde{\mathbf{B}}2(s)|{H^2} ds \leq t |\partial_t \widetilde{\mathbf{B}}2|{L^\infty H^2},$$

there holds

$$\frac{|\widetilde{\mathbf{B}}{2}|{H^{3}}^{2}}{A^{-1}t} \leq \frac{|\widetilde{\mathbf{B}}|{H^{2}}|\widetilde{\mathbf{B}}|{H^{4}}}{A^{-1}t} \leq A|\partial{t}\widetilde{\mathbf{B}}{2}|{L^{\infty}H^{2}}|\widetilde{\mathbf{B}}{2}|_{H^{4}}.$$ (4.40)

From this, along with ∥∇△u3,0∂zBe^2∥ 2 ^L2 ≤ C∥∇△u3,0∥ 2 ^{L2 ∥}∂zBe^2∥ 2 ^H2 , we get

$$|\nabla \triangle u_{3,0} \partial_z \widetilde{\mathbf{B}}2|{L^2}^2 \le C |\min{(A^{-\frac{2}{3}} + A^{-1}t)^{\frac{1}{2}}, 1} \nabla \triangle u{3,0}|{L^2}^2 \left(\frac{|\widetilde{\mathbf{B}}2|{H^3}^2}{A^{-1}t} + |\widetilde{\mathbf{B}}2|{H^4}^2\right)$$

$$\le C |\min{(A^{-\frac{2}{3}} + A^{-1}t)^{\frac{1}{2}}, 1} \nabla \triangle u{3,0}|{L^2}^2 \left(A^2 |\partial_t \widetilde{\mathbf{B}}2|{L^\infty H^2}^2 + |\widetilde{\mathbf{B}}2|{H^4}^2\right)$$

which implies that

$$\frac{|\nabla \triangle u_{3,0} \partial_z \widetilde{\mathbf{B}}2|{L^2 L^2}^2}{A^3} \le C\epsilon^2 \left( |\partial_t \widetilde{\mathbf{B}}2|{L^\infty H^2}^2 + \frac{|\triangle^2 \widetilde{\mathbf{B}}2|{L^\infty L^2}^2}{A^2} \right),\tag{4.41}$$

where we use Lemma 4.1 and Poincar´e inequality.

Combining (4.37), (4.38), (4.39) and (4.41), we get from (4.36) that

$$\frac{|\Delta^{2}\widetilde{\mathbf{B}}{2}|{L^{\infty}L^{2}}^{2}}{A^{2}} + \frac{|\nabla\Delta^{2}\widetilde{\mathbf{B}}{2}|{L^{2}L^{2}}^{2}}{A^{3}}$$

$$\leq C|\Delta u{2,0}|{Y{0}}^{2} + C\epsilon^{2}|\partial{t}\widetilde{\mathbf{B}}{2}|{L^{\infty}H^{2}}^{2} + C\epsilon^{2}\left(\frac{|\Delta^{2}\widetilde{\mathbf{B}}{2}|{L^{\infty}L^{2}}^{2}}{A^{2}} + \frac{|\nabla\Delta^{2}\widetilde{\mathbf{B}}{2}|{L^{2}L^{2}}^{2}}{A^{3}}\right).$$ (4.42)

When ϵ is small enough satisfying Cϵ² ≤ 2 , using Lemma 4.1, (4.42) yields

$$\frac{|\triangle^2 \widetilde{\mathbf{B}}2|{L^{\infty}L^2}^2}{A^2} + \frac{|\nabla\triangle^2 \widetilde{\mathbf{B}}2|{L^2L^2}^2}{A^3} \le C\epsilon^2 + C\epsilon^2 |\partial_t \widetilde{\mathbf{B}}2|{L^{\infty}H^2}^2. \tag{4.43}$$

From (2.4)¹ and (3.19), direct calculations indicate that

$$|\partial{t}\widetilde{\mathbf{B}}{2}|{L^{\infty}L^{2}} \leq \frac{|\triangle\mathbf{B}{2}|{L^{\infty}L^{2}}}{A} + |\widetilde{u}{2,0}|{L^{\infty}L^{2}} + \frac{1}{A} \left(|u{2,0}|{L^{\infty}L^{\infty}} + |u{3,0}|{L^{\infty}L^{\infty}}\right) |\nabla\widetilde{\mathbf{B}}{2}|{L^{\infty}L^{2}} \ \leq \frac{|\triangle\widetilde{\mathbf{B}}{2}|{L^{\infty}L^{2}}}{A} + |\widetilde{u}{2,0}|{L^{\infty}L^{2}} + C\epsilon \frac{|\nabla\widetilde{\mathbf{B}}{2}|_{L^{\infty}L^{2}}}{A}.$$

$$(4.44)$$

Taking △ for (2.4)1, there holds

$$\partial_t \triangle \widetilde{\mathbf{B}}_2 - \frac{1}{A} \triangle^2 \widetilde{\mathbf{B}}2 = -\frac{1}{A} \triangle \left( u{2,0} \partial_y \widetilde{\mathbf{B}}2 + u{3,0} \partial_z \widetilde{\mathbf{B}}2 \right) - \triangle \widetilde{u}{2,0},$$

which follows that

$$|\partial_t \triangle \widetilde{\mathbf{B}}2|{L^{\infty}L^2}^2 \le \frac{|\triangle^2 \dot{\mathbf{B}}2|{L^{\infty}L^2}^2}{A^2} + \frac{|\triangle(u_{2,0}\partial_y \dot{\mathbf{B}}2 + u{3,0}\partial_z \dot{\mathbf{B}}2)|{L^{\infty}L^2}^2}{A^2} + |\triangle \widetilde{u}{2,0}|{L^{\infty}L^2}^2. \tag{4.45}$$

Using Lemma A.2, Lemma 4.1 and Poincar´e inequality, we have

$$\frac{|\Delta(u_{2,0}\partial_y \widetilde{\mathbf{B}}2)|{L^{\infty}L^2}^2}{A^2} \le \frac{C}{A^2} |u{2,0}|{L^{\infty}H^2}^2 |\partial_y \widetilde{\mathbf{B}}2|{L^{\infty}H^2}^2 \le C\epsilon^2 \frac{|\Delta^2 \widetilde{\mathbf{B}}2|{L^{\infty}L^2}^2}{A^2}$$ (4.46)

and

$$\frac{|u{3,0}\triangle\partial{z}\widetilde{\mathbf{B}}{2} + \nabla u{3,0} \cdot \nabla\partial{z}\widetilde{\mathbf{B}}{2}|{L^{\infty}L^{2}}^{2}}{A^{2}} \leq \frac{C}{A^{2}}|u{3,0}|{L^{\infty}H^{1}}^{2}|\widetilde{\mathbf{B}}{2}|{L^{\infty}H^{4}}^{2} \leq C\epsilon^{2}\frac{|\triangle^{2}\widetilde{\mathbf{B}}{2}|_{L^{\infty}L^{2}}^{2}}{A^{2}}.$$ (4.47)

Note that ∥△u3,0∂zBe^2∥ 2 ^L2 ≤ C∥△u3,0∥ 2 ^{L2 ∥}Be^2∥ 2 ^H3 , by (4.40), and we get

$$\begin{split} |\triangle u_{3,0}\partial_z \widetilde{\mathbf{B}}2|{L^2}^2 \leq & C|\min{(A^{-\frac{2}{3}} + A^{-1}t)^{\frac{1}{2}}, 1} \triangle u{3,0}|{L^2}^2 \left(\frac{|\widetilde{\mathbf{B}}2|{H^3}^2}{A^{-1}t} + |\triangle\partial_z \widetilde{\mathbf{B}}2|{L^2}^2\right) \ \leq & C|\min{(A^{-\frac{2}{3}} + A^{-1}t)^{\frac{1}{2}}, 1} \triangle u{3,0}|{L^2}^2 \left(A^2|\partial_t \widetilde{\mathbf{B}}2|{H^2}^2 + |\triangle^2 \widetilde{\mathbf{B}}2|{L^2}^2\right) \end{split}$$

which along with Lemma 4.1 implies that

$$\frac{|\triangle u_{3,0}\partial_z \widetilde{\mathbf{B}}2|{L^{\infty}L^2}^2}{A^2} \le C\epsilon^2 \left( |\partial_t \widetilde{\mathbf{B}}2|{L^{\infty}H^2}^2 + \frac{|\triangle^2 \widetilde{\mathbf{B}}2|{L^{\infty}L^2}^2}{A^2} \right). \tag{4.48}$$

Collecting (4.46), (4.47) and (4.48), we get from (4.45) that

$$|\partial_t \triangle \widetilde{\mathbf{B}}2|{L^{\infty}L^2}^2 \le C \frac{|\triangle^2 \widetilde{\mathbf{B}}2|{L^{\infty}L^2}^2}{A^2} + C\epsilon^2 |\partial_t \widetilde{\mathbf{B}}2|{L^{\infty}H^2}^2 + |\triangle \widetilde{u}{2,0}|{L^{\infty}L^2}^2.$$

Combining it with (4.44), Lemma 4.1 and Poincar´e inequality, one deduces

$$|\partial_t \widetilde{\mathbf{B}}2|{L^{\infty}H^2}^2 \le C \frac{|\triangle^2 \widetilde{\mathbf{B}}2|{L^{\infty}L^2}^2}{A^2} + C\epsilon^2 |\partial_t \widetilde{\mathbf{B}}2|{L^{\infty}H^2}^2 + C\epsilon^2. \tag{4.49}$$

Substituting (4.43) into (4.49), we have

$$|\partial_t \widetilde{\mathbf{B}}2|{L^{\infty}H^2}^2 \le C\epsilon^2 |\partial_t \widetilde{\mathbf{B}}2|{L^{\infty}H^2}^2 + C\epsilon^2,$$

which indicates that

$$|\partial_t \widetilde{\mathbf{B}}2|{L^\infty H^2}^2 \le C\epsilon^2 \tag{4.50}$$

,

provided with ϵ is sufficiently small satisfying Cϵ² ≤ 2 . By (4.50), we infer from (4.43) that

$$\frac{|\triangle^2 \widetilde{\mathbf{B}}2|{L^{\infty}L^2}^2}{A^2} + \frac{|\nabla \triangle^2 \widetilde{\mathbf{B}}2|{L^2L^2}^2}{A^3} \le C\epsilon^2.$$ (4.51)

For j ∈ {2, 3}, combining (4.50) with (4.51), and using Poincar´e inequality

$$|\nabla^{j}\widetilde{\mathbf{B}}{2}|{L^{2}}^{2} \leq C|\nabla^{j+1}\widetilde{\mathbf{B}}{2}|{L^{2}}^{2},$$

one obtains

$$\frac{|\Delta \widetilde{\mathbf{B}}2|{L^{\infty}H^2}}{A} + \frac{|\nabla \Delta \widetilde{\mathbf{B}}2|{L^2H^2}}{A^{\frac{3}{2}}} + |\partial_t \widetilde{\mathbf{B}}2|{L^{\infty}H^2} \le C\epsilon. \tag{4.52}$$

Estimate of B2(t). According to (4.1) 1 , u2,⁰ satisfies ∂tu2,⁰ = 0, which implies that

$$\overline{u}{2,0}(t) = (\overline{u}{2,\text{in}})_0.$$

Therefore, we rewrite (2.4) 2 into

$$\partial_t \overline{\mathbf{B}}2(t) = -\overline{u}{2,0}(t) = -(\overline{u}_{2,\text{in}})_0.$$

By H¨older's inequality, there holds

$$|\partial_t \overline{\mathbf{B}}2(t)| \le C |(u{2,\text{in}})0|{H^2}.$$ (4.53)

Furthermore, B2(t) only depends on t and does not depend on any spatial variables, and ∂tB² is a constant. Using assumption (1.6), we infer from (4.53) that

$$\frac{|\triangle \overline{\mathbf{B}}{2}|{L^{\infty}H^{2}}}{A} + \frac{|\nabla \triangle \overline{\mathbf{B}}{2}|{L^{2}H^{2}}}{A^{\frac{3}{2}}} + |\partial{t}\overline{\mathbf{B}}{2}|{L^{\infty}H^{2}} \le C|(u{2,\text{in}}){0}|{H^{2}} \le C\epsilon. \tag{4.54}$$

Estimate of B1(t). Next, we rewrite (2.3) into

$$\partial_t \overline{\mathbf{B}}_1(t) = \frac{M}{A|\mathbb{T}|}$$

Moreover, B1(t) only depends on t and does not depend on any spatial variables, and ∂tB¹ is a constant. When A ≥ max{ ϵ , A2,3} =: A2, we obtain that

$$\frac{|\triangle \overline{\mathbf{B}}1|{L^{\infty}H^2}}{A} + \frac{|\nabla \triangle \overline{\mathbf{B}}1|{L^2H^2}}{A^{\frac{3}{2}}} + |\partial_t \overline{\mathbf{B}}1|{L^{\infty}H^2} \le \frac{CM}{A} \le C\epsilon. \tag{4.55}$$

Combining (4.52), (4.54) and (4.55), we conclude that

$$\frac{|\triangle \mathbf{U}2|{L^{\infty}H^2}}{A} + \frac{|\nabla \triangle \mathbf{U}2|{L^2H^2}}{A^{\frac{3}{2}}} + |\partial_t \mathbf{U}2|{L^{\infty}H^2} \le C\epsilon.$$

The proof is complete. □

Corollary 4.1. Under the conditions of Theorem 1.1 and the assumptions (1.6) and (2.5), according to Lemma 4.1 and Lemma 4.5, when A ≥ A2, there holds

$$E_1(t) \leq C\epsilon =: E_1.$$

5. Estimates for the non-zero modes

5.1. Energy estimates for E2,1(t).

Lemma 5.1. Under the conditions of Theorem 1.1, the assumptions (1.6) and (2.5), there exists a positive constant A³ independent of t and A, such that if A ≥ A3, there holds

$$E_{2,1}(t) \le C(|(\partialx^2 n{\rm in}){\ne}|{L^2} + 1).$$

Proof. According to (1.5)1, the non-zero mode ∂ 2 ^xn̸⁼ satisfies

$$\partial_t \partialx^2 n{\neq} + y \partialx^3 n{\neq} - \frac{1}{A} \triangle \partialx^2 n{\neq} = -\frac{1}{A} \nabla \cdot \partialx^2 (un){\neq} - \frac{1}{A} \nabla \cdot \partialx^2 (n \nabla c){\neq}. \tag{5.1}$$

Noting that for given functions f and g, we have

$$(fg)_{\neq} = f0 g{\neq} + f_{\neq} g0 + (f{\neq} g{\neq}){\neq}. \tag{5.2}$$

Therefore, by decomposing u1,⁰ = U¹ + U2, we rewrite (5.1) into

$$\partial_t \partialx^2 n{\neq} + \left( y + \frac{\mathbf{U}_2}{A} \right) \partialx^3 n{\neq} - \frac{\triangle \partialx^2 n{\neq}}{A} = -\frac{1}{A} \nabla \cdot \partialx^2 (u{\neq} n{\neq}){\neq} - \frac{1}{A} \nabla \cdot \partialx^2 (n \nabla c){\neq} \ -\frac{1}{A} \nabla \cdot (\partialx^2 u{\neq} n_0) - \frac{1}{A} \nabla \cdot \partial_x^2 (U0 n{\neq}),$$ (5.3)

where U⁰ = (U1, u2,0, u3,0).

Moreover, a careful deformation of the nonlinear interaction term ∂ 3 x (u1,̸=n̸=)̸⁼ plays a crucial role. Specifically, it can be expanded as follows:

$$\partialx^3(u{1,\neq}n{\neq}){\neq} = (\partialx^3 u{1,\neq}n{\neq}){\neq} + (u_{1,\neq}\partialx^3 n{\neq})_{\neq} + 3\partial_x(\partialx u{1,\neq}\partialx n{\neq})_{\neq}.$$

When A > Cc⁻¹ , it follows from (1.6) and Lemma 4.5 that ∥△U2∥L∞H^{2 A} + ∥∂tU2∥L∞L^∞ < Cϵ. By applying Proposition A.1, we get

$$\begin{split} |\partialx^2 n{\neq}|_{X_b}^2 \leq & C\Big(|(\partialx^2 n{\mathrm{in}}){\neq}|{L^2}^2 + \frac{1}{A}|\mathrm{e}^{bA^{-\frac{1}{3}}t} U_0 \partialx^2 n{\neq}|_{L^2 L^2}^2 + \frac{1}{A^{\frac{5}{3}}}|\mathrm{e}^{bA^{-\frac{1}{3}}t} n_0 \partialx^3 u{1,\neq}|_{L^2 L^2}^2 \ & + \frac{1}{A}|\mathrm{e}^{bA^{-\frac{1}{3}}t} n_0 \partial_x^2 (u_2, u3){\neq}|{L^2 L^2}^2 + \frac{1}{A^{\frac{5}{3}}}|\mathrm{e}^{bA^{-\frac{1}{3}}t} n{\neq} \partialx^3 u{1,\neq}|{L^2 L^2}^2 \ & + \frac{1}{A^{\frac{5}{3}}}|\mathrm{e}^{bA^{-\frac{1}{3}}t} u{1,\neq} \partialx^3 n{\neq}|_{L^2 L^2}^2 + \frac{1}{A}|\mathrm{e}^{bA^{-\frac{1}{3}}t} \partialx u{1,\neq} \partialx n{\neq}|_{L^2 L^2}^2 \ & + \frac{1}{A}|\mathrm{e}^{bA^{-\frac{1}{3}}t} \partial_x^2 \left( (u_2, u3){\neq} n{\neq} \right)|{L^2 L^2}^2 + \frac{1}{A}|\mathrm{e}^{bA^{-\frac{1}{3}}t} \partialx^2 (n \nabla c){\neq}|_{L^2 L^2}^2 \Big) \ = & : C\left( |(\partialx^2 n{\mathrm{in}}){\neq}|{L^2}^2 + T{1,1} + T{1,2} + \dots + T_{1,8} \right). \end{split}$$

Estimate of T1,1. By Lemma A.7, we have

$$|\partialx^2 n{\neq}|_{L^2 L^4} \le C |\partialx^2 n{\neq}|_{L^2 L^2}^{\frac{1}{4}} |\nabla \partialx^2 n{\neq}|_{L^2 L^2}^{\frac{3}{4}}$$

$$(5.4)$$

and

$$||U0||{L^4} \le C||U0||{H^1} \le C(||\mathbf{U}1||{H^1} + ||u{2,0}||{H^1} + ||u{3,0}||{H^1}).$$

.

Then using Lemma 4.1, Lemma 4.4 and ∥n0∥ 2 ^L∞L² ≤ CME3, we obtain

$$\begin{split} |\mathrm{e}^{bA^{-\frac{1}{3}}t}U_0\partialx^2 n{\neq}|_{L^2L^2}^2 \leq & |U0|{L^{\infty}L^4}^2 |\mathrm{e}^{bA^{-\frac{1}{3}}t}\partialx^2 n{\neq}|_{L^2L^4}^2 \ \leq & C|U0|{L^{\infty}H^1}^2 |\mathrm{e}^{bA^{-\frac{1}{3}}t}\partialx^2 n{\neq}|_{L^2L^2}^{\frac{1}{2}} |\mathrm{e}^{bA^{-\frac{1}{3}}t}\nabla\partialx^2 n{\neq}|_{L^2L^2}^{\frac{3}{2}} \leq CA^{\frac{5}{6}}H_2^2E_2^2, \end{split}$$

where H² = ∥(u1,in)0∥H¹ + E³ + M + 1.

Estimates of T1,² and T1,3. Using ∥n0∥L∞L^∞ ≤ ∥n∥L∞L^∞ ≤ 2E³ and ∂xu1,̸⁼ +∂yu2,̸⁼ +∂zu3,̸⁼ = 0, we arrive at

$$|\mathbf{e}^{bA^{-\frac{1}{3}}t}n{0}\partial{x}^{3}u{1,\neq}|{L^{2}L^{2}}^{2} \leq CE{3}^{2}|\mathbf{e}^{bA^{-\frac{1}{3}}t}\partial{x}^{3}u{1,\neq}|{L^{2}L^{2}}^{2} \leq CAE{3}^{2}E{4}^{2},$$

$$|\mathbf{e}^{bA^{-\frac{1}{3}}t}n{0}\partial{x}^{2}(u{2},u{3}){\neq}|{L^{2}L^{2}}^{2} \leq CE{3}^{2}|\mathbf{e}^{bA^{-\frac{1}{3}}t}\partial{x}^{2}(u{2},u{3}){\neq}|{L^{2}L^{2}}^{2} \leq CA^{\frac{1}{3}}E{3}^{2}E{4}^{2}.$$

$$(5.5)$$

Estimate of T1,4. Since ∥n̸=∥L∞L^∞ ≤ 2∥n∥L∞L^∞ ≤ 4E3, by an argument similar to (5.5)1, one obtains

$$|\mathbf{e}^{bA^{-\frac{1}{3}}t}n{\neq}\partial{x}^{3}u{1,\neq}|{L^{2}L^{2}}^{2} \leq C|n|{L^{\infty}L^{\infty}}|\mathbf{e}^{bA^{-\frac{1}{3}}t}\partial{x}^{3}u{1,\neq}|{L^{2}L^{2}}^{2}$$ $$\leq CAE{3}^{2}|\partial{x}^{2}(u{2},u{3}){\neq}|{X{b}}^{2} \leq CAE{3}^{2}E_{4}^{2}.$$

Estimate of T1,5. By (3.5) 1 , we have

$$|\mathrm{e}^{bA^{-\frac{1}{3}}t}u_{1,\neq}\partialx^3n{\neq}|{L^2L^2}^2\leq |u{1,\neq}|_{L^{\infty}L^{\infty}}^2|\mathrm{e}^{bA^{-\frac{1}{3}}t}\partialx^3n{\neq}|_{L^2L^2}^2\leq CA^{\frac{4}{3}}E_2^4.$$

Estimate of T1,6. By Lemma A.3 and Lemma 3.1, one obtains that

$$\begin{split} &| \mathbf{e}^{bA^{-\frac{1}{3}}t} \partialx u{1,\neq} \partialx n{\neq} |_{L^2L^2}^2 \leq | \mathbf{e}^{aA^{-\frac{1}{3}}t} \partialx u{1,\neq} |{L^{\infty}{t,z}L^2_{x,y}}^2 | \mathbf{e}^{aA^{-\frac{1}{3}}t} \partialx n{\neq} |{L^{\infty}{x,y}L^2_{t,z}}^2 \ &\leq C | \mathbf{e}^{aA^{-\frac{1}{3}}t} \partial_x \partialz u{1,\neq} |_{L^{\infty}L^2}^2 | \mathbf{e}^{aA^{-\frac{1}{3}}t} \partialx^2 n{\neq} |_{L^2L^2} | \mathbf{e}^{aA^{-\frac{1}{3}}t} \partial_x^2 \partialy n{\neq} |_{L^2L^2} \leq CA^{\frac{2}{3}} E_2^4 \end{split}$$

Estimate of T1,7. According to (3.5), there holds

$$|\mathbf{e}^{bA^{-\frac{1}{3}}t}(u_2, u3){\neq} \partialx^2 n{\neq}|{L^2L^2}^2 \le |u{\neq}|_{L^{\infty}L^{\infty}}^2 |\mathbf{e}^{bA^{-\frac{1}{3}}t} \partialx^2 n{\neq}|_{L^2L^2}^2 \le CA^{\frac{2}{3}} E_2^4. \tag{5.6}$$

In addition, there also holds

$$|\mathbf{e}^{bA^{-\frac{1}{3}}t}\partial{x}^{2}(u{2},u{3}){\neq}n{\neq}|{L^{2}L^{2}}^{2} \leq |n{\neq}|{L^{\infty}L^{\infty}}^{2}|\mathbf{e}^{bA^{-\frac{1}{3}}t}\partial{x}^{2}(u{2},u{3}){\neq}|{L^{2}L^{2}}^{2} \leq CA^{\frac{1}{3}}E{3}^{2}E_{4}^{2}. \tag{5.7}$$

Using Lemma A.3, we get

$$\begin{split} |\partial{x}(u{2}, u{3}){\neq} \partial{x} n{\neq}|{L^{2}}^{2} \leq & |\partial{x} u{\neq}|{L^{\infty}{y, z} L^{2}{x}}^{2} |\partial{x} n{\neq}|{L^{\infty}{x} L^{2}{y, z}}^{2} \ \leq & C |(\partial{x}, \partial{z}) \partial{x} u{\neq}|{L^{2}} |(\partial{x}, \partial{z}) \partial{x} u{\neq}|{H^{1}} |\partial{x}^{2} n{\neq}|{L^{2}}^{2}, \end{split}$$

which along with Lemma 3.1 shows that

$$|e^{bA^{-\frac{1}{3}}t}\partialx u{\neq}\partialx n{\neq}|_{L^2L^2}^2 \le CA^{\frac{2}{3}}E_2^4.$$ (5.8)

It follows from (5.6), (5.7) and (5.8) that

$$|e^{bA^{-\frac{1}{3}}t}\partialx^2(u{\neq}n{\neq}){\neq}|_{L^2L^2}^2 \le CA^{\frac{2}{3}}(E_2^4 + E_3^2E_4^2).$$

Estimate of T1,8. Due to (5.2), there holds

$$|\mathbf{e}^{bA^{-\frac{1}{3}}t}\partial{x}^{2}(n\nabla c){\neq}|{L^{2}L^{2}}^{2} \leq C\left(|\mathbf{e}^{bA^{-\frac{1}{3}}t}n{0}\partial{x}^{2}\nabla c{\neq}|{L^{2}L^{2}}^{2} + |\mathbf{e}^{bA^{-\frac{1}{3}}t}\partial{x}^{2}n{\neq}\nabla c{0}|{L^{2}L^{2}}^{2} + |\mathbf{e}^{bA^{-\frac{1}{3}}t}\partial{x}^{2}(n{\neq}\nabla c{\neq}){\neq}|{L^{2}L^{2}}^{2}\right).$$

$$(5.9)$$

2 3

Applying Lemma A.8, Lemma A.9 and ∥n0∥L∞L^∞ ≤ ∥n∥L∞L^∞, we get

$$\begin{aligned} |\mathbf{e}^{bA^{-\frac{1}{3}}t} n_0 \partialx^2 \nabla c{\neq} |_{L^2 L^2}^2 &\leq |n0|{L^{\infty} L^{\infty}}^2 |\mathbf{e}^{bA^{-\frac{1}{3}}t} \partialx^2 \nabla c{\neq} |_{L^2 L^2}^2 \ &\leq C E_3^2 |\mathbf{e}^{bA^{-\frac{1}{3}}t} \partialx^2 n{\neq} |_{L^2 L^2}^2 \leq C A^{\frac{1}{3}} E_2^2 E \end{aligned}$$

and

$$\begin{split} |\mathrm{e}^{bA^{-\frac{1}{3}}t}\partialx^2 n{\neq} \nabla c0|{L^2L^2}^2 \leq & |\nabla c0|{L^{\infty}L^{\infty}}^2 |\mathrm{e}^{bA^{-\frac{1}{3}}t}\partialx^2 n{\neq}|_{L^2L^2}^2 \leq CA^{\frac{1}{3}} |n0 - \bar{n}|{L^{\infty}L^3}^2 |\partialx^2 n{\neq}|_{X_b}^2 \ \leq & CA^{\frac{1}{3}} |n0|{L^{\infty}L^{\infty}}^{\frac{4}{3}} |n0|{L^{\infty}L^{\infty}}^{\frac{2}{3}} |\partialx^2 n{\neq}|_{X_b}^2 \leq CA^{\frac{1}{3}} E_2^2 (E_3^2 + M^2). \end{split}$$

Using (5.4), Lemma A.3 and Lemma A.9, one obtains that

$$\begin{split} &| \mathbf{e}^{bA^{-\frac{1}{3}t}} \partialx^2 (n{\neq} \nabla c{\neq}){\neq} |{L^2 L^2}^2 \ \leq & C | \nabla c{\neq} |_{L^{\infty} L^4}^2 | \mathbf{e}^{bA^{-\frac{1}{3}t}} \partialx^2 n{\neq} |{L^2 L^4}^2 + C | n |{L^{\infty} L^{\infty}}^2 | \mathbf{e}^{bA^{-\frac{1}{3}t}} \partialx^2 \nabla c{\neq} |_{L^2 L^2}^2 \ & + C | \partialx n{\neq} |{L^{\infty}{t,x} L^2_{y,z}}^2 | \mathbf{e}^{bA^{-\frac{1}{3}t}} \partialx \nabla c{\neq} |{L^2{t,x} L^{\infty}{y,z}}^2 \ \leq & C | n{\neq} |_{L^{\infty} L^2}^2 | \mathbf{e}^{bA^{-\frac{1}{3}t}} \partialx^2 n{\neq} |_{L^2 L^2}^{\frac{1}{2}} | \mathbf{e}^{bA^{-\frac{1}{3}t}} \nabla \partialx^2 n{\neq} |{L^2 L^2}^{\frac{3}{2}} + C | n |{L^{\infty} L^{\infty}}^2 | \mathbf{e}^{bA^{-\frac{1}{3}t}} \partialx^2 n{\neq} |_{L^2 L^2}^2 \ & + C | \partialx^2 n{\neq} |_{L^{\infty} L^2}^2 | \mathbf{e}^{bA^{-\frac{1}{3}t}} \partialx^2 n{\neq} |_{L^2 L^2} | \mathbf{e}^{bA^{-\frac{1}{3}t}} \nabla \partialx^2 n{\neq} |_{L^2 L^2}^2 \ \leq & C A^{\frac{5}{6}} E_2^2 (E_2^2 + E_3^2), \end{split}$$

where we use ∥n̸=∥L∞L^∞ ≤ 2∥n∥L∞L^∞. Thus, we infer from (5.9) that

$$|\mathrm{e}^{bA^{-\frac{1}{3}}t}\partialx^2(n\nabla c){\neq}|_{L^2L^2}^2 \leq CA^{\frac{5}{6}}(E_2^2+E_3^2+M^2)E_2^2.$$

In conclusion, we obtain that

$$|\partialx^2 n{\neq}|_{X_b}^2 \leq C \left( |(\partialx^2 n{\mathrm{in}}){\neq}|{L^2}^2 + \frac{E_2^4 + E_3^4 + E_4^4 + M^4 + H_2^4}{A^{\frac{1}{6}}} \right).$$

When A ≥ max{A2,(E 4 ² + E 4 ³ + M⁴ + H^{4 2} + E 4 4 ) 6 } =: A3, one deduces

$$E_{2,1}(t) = |\partialx^2 n{\neq}|_{X_b} \le C \left( |(\partialx^2 n{\rm in}){\neq}|{L^2} + 1 \right).$$

The proof is complete. □

5.2. Energy estimates for E2,2(t).

Lemma 5.2. Under the conditions of Theorem 1.1 and the assumptions (1.6) and (2.5), there exists a positive constant A⁴ independent of t and A, such that if A ≥ A4, there holds

$$E{2,2}(t) \le C(|(u{\rm in}){\ne}|{H^2} + E_4 + 1).$$

Proof. It follows from (1.5) that

$$\begin{cases} \partial_t \omega_2 + y \partial_x \omega_2 - \frac{1}{A} \triangle \omega_2 + \partial_z u_2 = -\frac{1}{A} \partial_z (u \cdot \nabla u_1) + \frac{1}{A} \partial_x (u \cdot \nabla u_3) + \frac{1}{A} \partial_z n, \ \partial_t \triangle u_2 + y \partial_x \triangle u_2 - \frac{1}{A} \triangle^2 u_2 = -\frac{1}{A} \partial_y \partial_x n - \frac{1}{A} (\partial_x^2 + \partial_z^2) (u \cdot \nabla u_2) \ + \frac{1}{A} \partial_y \left[ \partial_x (u \cdot \nabla u_1) + \partial_z (u \cdot \nabla u_3) \right]. \end{cases}$$

For convenience, we denote

$$\mathcal{L} = \partial_t + y \partial_x - \frac{1}{A} \triangle,$$

$$\mathcal{L}_V = \partial_t + (y + \frac{\mathbf{U}_2}{A}) \partial_x - \frac{1}{A} \triangle.$$ (5.10)

As previously mentioned, we have decomposed the non-zero mode u1,⁰ into two parts u1,⁰ = U¹ + U2. According to the nonlinear interaction, we will reformulate the coupled system {∂xω2,̸=, △u2,̸=} and (∂y, ∂z)ω2,̸=.

Using u1,⁰ = U¹ + U² and (5.2), the velocity ω2,̸⁼ satisfies

$$\mathcal{L}{V}\omega{2,\neq} + \partial{z}u{2,\neq} = \frac{\partial{x}\left[\left(u{\neq}\cdot\nabla u{3,\neq}\right){\neq} + U{0}\cdot\nabla u{3,\neq} + u{\neq}\cdot\nabla u{3,0}\right]}{A} + \frac{\partial{z}n{\neq}}{A} - \frac{\partial{z}\mathbf{U}{2}\partial{x}u{1,\neq}}{A} - \frac{\partial{z}\left[\left(u{\neq}\cdot\nabla u{1,\neq}\right){\neq} + U{0}\cdot\nabla u{1,\neq} + u{\neq}\cdot\nabla(\mathbf{U}{1} + \mathbf{U}_{2})\right]}{A},$$

where U⁰ = (U1, u2,0, u3,0). Precise calculations show that

$$(\partial_x^2 + \partial_z^2)(\mathbf{U}_2 \partialx u{2,\neq}) = \mathbf{U}_2 \partial_x (\partial_x^2 + \partialz^2) u{2,\neq} + \partial_z (\partial_z \mathbf{U}_2 \partialx u{2,\neq}) + \partial_z \mathbf{U}_2 \partial_x \partialz u{2,\neq}. \tag{5.11}$$

Due to div u̸⁼ = 0, we have

$$\partial{y} \left[ \partial{x} (\mathbf{U}{2} \partial{x} u{1,\neq}) + \partial{z} (\mathbf{U}{2} \partial{x} u{3,\neq}) \right] = \partial{y} \left( -\mathbf{U}{2} \partial{x} \partial{y} u{2,\neq} + \partial{z} \mathbf{U}{2} \partial{x} u{3,\neq} \right) = -\mathbf{U}{2} \partial{x} \partial{y}^{2} u{2,\neq} - \partial{y} \mathbf{U}{2} \partial{x} \partial{y} u{2,\neq} + \partial{y} (\partial{z} \mathbf{U}{2} \partial{x} u{3,\neq}).$$ (5.12)

Using (5.11) and (5.12), △u2,̸⁼ satisfies

$$\begin{split} \mathcal{L}{V} \triangle u{2,\neq} &= -\frac{\partial{x} \partial{y} n{\neq}}{A} - \frac{\left(\partial{x}^{2} + \partial{z}^{2}\right) \left[\left(u{\neq} \cdot \nabla u{2,\neq}\right){\neq} + U{0} \cdot \nabla u{2,\neq} + u{\neq} \cdot \nabla u{2,0}\right]}{A} \ &+ \frac{\partial{y} \partial{x} \left[\left(u{\neq} \cdot \nabla u{1,\neq}\right){\neq} + U{0} \cdot \nabla u{1,\neq} + u{\neq} \cdot \nabla \left(\mathbf{U}{1} + \mathbf{U}{2}\right)\right]}{A} - \frac{\partial{j} \mathbf{U}{2} \partial{x} \partial{j} u{2,\neq}}{A} \ &+ \frac{\partial{y} \partial{z} \left[\left(u{\neq} \cdot \nabla u{3,\neq}\right){\neq} + U{0} \cdot \nabla u{3,\neq} + u{\neq} \cdot \nabla u{3,0}\right]}{A} + \frac{\partial{y} \left(\partial{z} \mathbf{U}{2} \partial{x} u{3,\neq}\right) - \partial{z} \left(\partial{z} \mathbf{U}{2} \partial{x} u{2,\neq}\right)}{A}. \end{split}$$

Therefore, one obtains that

$$\begin{cases} \mathcal{L}{V}\omega{2,\neq} + \partial{z}u{2,\neq} &= \frac{\partial{x}\left[(u{\neq}\cdot\nabla u{3,\neq}){\neq} + U{0}\cdot\nabla u{3,\neq} + u{\neq}\cdot\nabla u{3,0}\right]}{A} + \frac{\partial{z}n{\neq}}{A} - \frac{\partial{z}\mathbf{U}{2}\partial{x}u{1,\neq}}{A} \ &- \frac{\partial{z}\left[(u{\neq}\cdot\nabla u{1,\neq}){\neq} + U{0}\cdot\nabla u{1,\neq} + u{\neq}\cdot\nabla(\mathbf{U}{1} + \mathbf{U}{2})\right]}{A}, \ \mathcal{L}{V}\triangle u{2,\neq} &= -\frac{\partial{x}\partial{y}n{\neq}}{A} - \frac{(\partial{x}^{2} + \partial{z}^{2})\left[(u{\neq}\cdot\nabla u{2,\neq}){\neq} + U{0}\cdot\nabla u{2,\neq} + u{\neq}\cdot\nabla u{2,0}\right]}{A} \ &+ \frac{\partial{y}\partial{x}\left[(u{\neq}\cdot\nabla u{1,\neq}){\neq} + U{0}\cdot\nabla u{1,\neq} + u{\neq}\cdot\nabla(\mathbf{U}{1} + \mathbf{U}{2})\right]}{A} - \frac{\partial{y}\mathbf{U}{2}\partial{x}\partial{y}u{2,\neq}}{A} \ &+ \frac{\partial{y}\partial{z}\left[(u{\neq}\cdot\nabla u{3,\neq}){\neq} + U{0}\cdot\nabla u{3,\neq} + u{\neq}\cdot\nabla u{3,0}\right]}{A} + \frac{\partial{y}(\partial{z}\mathbf{U}{2}\partial{x}u{3,\neq}) - \partial{z}(\partial{z}\mathbf{U}{2}\partial{x}u_{2,\neq})}{A}. \end{cases} (5.13)$$

Taking ∂^x for (5.13) 1 and applying Proposition A.2, we derive

$$|\partial{x}\omega{2,\neq}|{X{a}}^{2} + |\Delta u{2,\neq}|{X{a}}^{2} \leq C\left(|(u{\text{in}}){\neq}|{H^{2}}^{2} + \frac{|\partial{x}n{\neq}|{X{b}}^{2}}{A^{\frac{2}{3}}} + \frac{|e^{aA^{-\frac{1}{3}}t}\partial{z}\mathbf{U}{2}\partial{x}(u{2,\neq},u{3,\neq})|{L^{2}L^{2}}^{2}}{A}\right)$$

$$+ CA^{-1}\left(|e^{aA^{-\frac{1}{3}}t}\partial{x}\left[(u{\neq}\cdot\nabla u{1,\neq}){\neq} + U{0}\cdot\nabla u{1,\neq} + u{\neq}\cdot\nabla(\mathbf{U}{1} + \mathbf{U}{2})\right]|{L^{2}L^{2}}^{2}\right)$$

$$+ |e^{aA^{-\frac{1}{3}}t}(\partial{x},\partial{z})\left[(u{\neq}\cdot\nabla u{2,\neq}){\neq} + U{0}\cdot\nabla u{2,\neq} + u{\neq}\cdot\nabla u{2,0}\right]|{L^{2}L^{2}}^{2}$$

$$+ |e^{aA^{-\frac{1}{3}}t}(\partial{x},\partial{z})\left[(u{\neq}\cdot\nabla u{3,\neq}){\neq} + U{0}\cdot\nabla u{3,\neq} + u{\neq}\cdot\nabla u{3,0}\right]|{L^{2}L^{2}}^{2}\right)$$

$$+ CA^{-\frac{5}{3}}\left(|e^{aA^{-\frac{1}{3}}t}\partial{z}\mathbf{U}{2}\partial{x}^{2}u{1,\neq}|{L^{2}L^{2}}^{2} + |e^{aA^{-\frac{1}{3}}t}\partial{j}\mathbf{U}{2}\partial{x}\partial{j}u{2,\neq}|_{L^{2}L^{2}}^{2}\right)$$

$$=: C\left(|(u{\text{in}}){\neq}|{H^{2}}^{2} + T{2,1} + T{2,2} + \dots + T{2,7}\right). \tag{5.14}$$

Estimates of T2,2, T2,⁶ and T2,7. By Lemma 3.5, Lemma 4.5 and Young's inequality, we have

$$T{2,2} + T{2,6} + T_{2,7} \le C \frac{E_2^2 + E_4^2}{A^{\frac{1}{4}}} + C\epsilon^2 (E_2^2(t) + E_4^2(t)).$$

Estimate of T2,3. Using (3.2) 3 , (3.13) 3,5 , (3.24) 1 , Lemma 3.6, Lemma 4.1, Lemma 4.4, Lemma 6.4 and Lemma 4.5, there holds

$$T_{2,3} \le C \frac{E_2^4 + E_4^4 + H_1^4 + 1}{A^{\frac{5}{6} - \alpha}} + C\epsilon^2 (E_2^2(t) + E_4^2(t)).$$

Estimate of T2,4. Using (3.2) 4 , (3.13) 1,5 , Lemma 3.6, Lemma 4.4 and Lemma 6.4, we get

$$T_{2,4} \leq C \frac{E_2^4 + H_1^4 + 1}{A^{\frac{1}{2} - \frac{2}{3}\alpha}} + C\epsilon^2 E_2^2(t).$$

Estimate of T2,5. Using (3.2) 3,5 , (3.13) 2,5 , Lemma 3.6, Lemma 4.4 and Lemma 6.4, we arrive

$$T_{2,5} \le C \frac{E_2^4 + H_1^4 + 1}{A^{\frac{1}{2} - \frac{2}{3}\alpha}} + C\epsilon^2 E_2^2(t).$$

Therefore, we infer from (5.14) that

$$|\partialx \omega{2,\neq}|_{Xa}^2 + |\Delta u{2,\neq}|_{Xa}^2 \le C\Big(|(u{\rm in}){\neq}|{H^2}^2 + \frac{E_2^4 + E_4^4 + H_1^4 + 1}{A^{\frac{1}{2} - \frac{2}{3}\alpha}} + \epsilon^2(E_2^2(t) + E_4^2(t))\Big). \tag{5.15}$$

For j ∈ {2, 3}, taking ∂^j to (5.13) 1 , one gets

$$\begin{split} \mathcal{L}_V \partialj \omega{2,\neq} + \partial_j \partialz u{2,\neq} &= \frac{\partial_j \partialz n{\neq}}{A} - \frac{\partial_j (\partial_z \mathbf{U}_2 \partialx u{1,\neq})}{A} - \left( \partial_j y + \frac{\partial_j \mathbf{U}_2}{A} \right) \partialx \omega{2,\neq} \ &+ \frac{\partial_j \partialx \left[ (u{\neq} \cdot \nabla u{3,\neq}){\neq} + U0 \cdot \nabla u{3,\neq} + u{\neq} \cdot \nabla u{3,0} \right]}{A} \ &- \frac{\partial_j \partialz \left[ (u{\neq} \cdot \nabla u{1,\neq}){\neq} + U0 \cdot \nabla u{1,\neq} + u_{\neq} \cdot \nabla (\mathbf{U}_1 + \mathbf{U}_2) \right]}{A}. \end{split}$$

Applying Proposition A.1 to it, due to

$$\left|\partial_j y + \frac{\partial_j \mathbf{U}2}{A}\right|{L^{\infty}L^{\infty}} \le C,$$

we obtain that

$$\frac{|\partial{j}\omega{2,\neq}|{X{a}}^{2}}{A^{\frac{2}{3}}} \leq C \left( \frac{|(u{\text{in}}){\neq}|{H^{2}}^{2}}{A^{\frac{2}{3}}} + |\Delta u{2,\neq}|{X{a}}^{2} + |\partial{x}\omega{2,\neq}|{X{a}}^{2} + \frac{|\partial{x}n{\neq}|{X{b}}^{2}}{A^{\frac{2}{3}}} + T{3,1} + T{3,2} \right) + CA^{-\frac{5}{3}} |e^{aA^{-\frac{1}{3}}t}\partial{z}\mathbf{U}{2}\partial{x}u{1,\neq}|_{L^{2}L^{2}}^{2},$$ (5.16)

where

$$\begin{split} T_{3,1} &= A^{-\frac{5}{3}} | \mathrm{e}^{aA^{-\frac{1}{3}}t} \partialx \left[ (u{\neq} \cdot \nabla u{3,\neq}){\neq} + U0 \cdot \nabla u{3,\neq} + u{\neq} \cdot \nabla u{3,0} \right] |{L^2 L^2}^2, \ T{3,2} &= A^{-\frac{5}{3}} | \mathrm{e}^{aA^{-\frac{1}{3}}t} \partialz \left[ (u{\neq} \cdot \nabla u{1,\neq}){\neq} + U0 \cdot \nabla u{1,\neq} + u_{\neq} \cdot \nabla (\mathbf{U}_1 + \mathbf{U}2) \right] |{L^2 L^2}^2. \end{split}$$

It is obvious that

$$T{3,1} \le T{2,5} \le C \frac{E_2^4 + H_1^4 + 1}{A^{\frac{1}{2} - \frac{2}{3}\alpha}} + C\epsilon^2 E_2^2(t)$$

Using (3.2) 6 , (3.13) 5 , (3.14), Lemma 3.6, Lemma 4.4 and Lemma 6.4, we get

$$T_{3,2} \le C \frac{E_2^4 + E_4^4 + H_1^4 + 1}{A^{\frac{1}{2} - \frac{2}{3}\alpha}} + C\epsilon^2 E_2^2(t).$$

By using (5.15), it learns from (5.16) that

$$\frac{|(\partial{y}, \partial{z})\omega{2,\neq}|{X{a}}^{2}}{A^{\frac{2}{3}}} \leq C\left(|(u{\rm in}){\neq}|{H^{2}}^{2} + \frac{E{2}^{4} + E{4}^{4} + H{1}^{4} + 1}{A^{\frac{1}{2} - \frac{2}{3}\alpha}} + \epsilon^{2}E{2}^{2}(t) + \epsilon^{2}E_{4}^{2}(t)\right). \tag{5.17}$$

When ϵ is small satisfying Cϵ² ≤ and

$$A \ge \left(E_2^4 + E_4^4 + H_1^4 + 1\right)^{\frac{6}{3-4\alpha}} =: A_4,$$

we infer from (5.15) and (5.17) that

$$E{2,2}^2(t) \le C\left(|(u{\rm in}){\ne}|{H^2}^2 + E_4^2(t) + 1\right).$$

1. Estimates for L 2 -norm of the density

First, we derive a lower bound for n that decreases exponentially.

Lemma 6.1. For all t ∈ [0, T], there holds

$$\left| \frac{1}{n0(t)} \right|{L^{\infty}(\mathbb{T}^2)} \le \left| \frac{1}{n(t)} \right|_{L^{\infty}(\mathbb{T}^3)} \le \delta^{-1} e^{\frac{\pi}{A}t}, \tag{6.1}$$

where δ > 0 is a constant.

□

Proof. Substituting the minimum point (xmin(t), ymin(t), zmin(t)) into (1.5)¹ and using (1.5)2, we obtain

$$\begin{split} &(\partialt n)(x{\min},y{\min},z{\min}) \ &= \frac{1}{A}(\triangle n)(x{\min},y{\min},z_{\min}) - y(\partialx n)(x{\min},y{\min},z{\min}) - \frac{1}{A}(u\cdot\nabla n)(x{\min},y{\min},z{\min}) \ &- \frac{1}{A}(\nabla n\cdot\nabla c)(x{\min},y{\min},z{\min}) - \frac{1}{A}(n\triangle c)(x{\min},y{\min},z{\min}) \ &\geq - \frac{1}{A}(n\triangle c)(x{\min},y{\min},z{\min}) \geq - \frac{1}{A}\overline{n}n(x{\min},y{\min},z_{\min}), \end{split}$$

which follows that

$$\frac{d}{dt}n{\min}(t) \ge -\frac{1}{A}\overline{n}n{\min}(t). \tag{6.2}$$

Due to nin > 0 in (x, y, z) ∈ T 3 , there must be a δ > 0, such that nin > δ > 0. Therefore, we get by (6.2) that

$$n{\min}(t) \ge n{\inf} e^{\int_0^t -\frac{1}{A}\overline{n}dt} \ge \delta e^{-\frac{\overline{n}}{A}t}.$$

The proof is complete. □

Motivated by [1], we next consider the following 2D free energy of n⁰ on T 2

$$\mathcal{L}[n0] = \int{\mathbb{T}^2} \left[ n_0 \log n_0 - \frac{1}{2} (n_0 - \overline{n}) c_0 \right] dy dz.$$

Lemma 6.2. Under the conditions of Theorem 1.1 and the assumptions (1.6) and (2.5), there exists a positive constant A⁵ independent of t and A, such that if A ≥ A4, there holds

$$\mathcal{L}[n0(t)] \le \mathcal{L}[(n{\rm in})_0] + C, \quad \text{for} \quad t \in [0, T].$$ (6.3)

Proof. Using (1.5), direct calculation shows that

$$\frac{d}{dt}\mathcal{L}[n0] = -\frac{1}{A} \int{\mathbb{T}^2} n_0 |\nabla \log n_0 - \nabla c0|^2 dy dz - \frac{1}{A} \int{\mathbb{T}^2} n_0 u_0 \cdot \nabla c_0 dy dz

• \frac{1}{A} \int{\mathbb{T}^2} \left[ (n{\neq} \nabla c_{\neq})0 + (n{\neq} u_{\neq})_0 \right] \cdot (\nabla \log n_0 - \nabla c0) dy dz =: -\frac{1}{A} \int{\mathbb{T}^2} n_0 |\nabla \log n_0 - \nabla c_0|^2 dy dz + J_1 + J_2.$$ (6.4)

For ^j ∈ {2, ³}, by (4.19), we decompose ^uj,0(t, y, z) = ^uj,0(t) + ^uej,0(t, y, z), therefore

$$J{1} = -\frac{1}{A} \int{\mathbb{T}^{2}} n{0} u{j,0} \partial{j} c{0} dy dz$$

$$= -\frac{1}{A} \int{\mathbb{T}^{2}} n{0} \overline{u}{j,0} \partial{j} c{0} dy dz - \frac{1}{A} \int{\mathbb{T}^{2}} n{0} \widetilde{u}{j,0} \partial{j} c{0} dy dz =: J{11} + J{12}.$$ (6.5)

For J11, using (1.5)² and integration by parts, we have

$$J{11} = \frac{1}{A}\overline{u}{j,0}(t)\int_{\mathbb{T}^2} \left(\partial_y^2 + \partial_z^2\right) c_0 \partial_j c0 dy dz - \frac{1}{A}\overline{u}{j,0}(t)\overline{n}(t)\int_{\mathbb{T}^2} \partial_j c_0 dy dz = 0.$$

For J12, due to ∥n0∥L³ ≤ ∥n0∥ 1 ^L1 ∥n0∥ ^L∞, by Lemma A.8 and H¨older's inequality, there holds

$$J_{12} \leq \frac{|\mathbb{T}|}{A} |n0|{L^{\infty}} |\nabla c0|{L^{\infty}} |\widetilde{u}{j,0}|{L^2} \leq \frac{C}{A} |n0|{L^{\infty}} |n0 - \overline{n}|{L^3} |\widetilde{u}{j,0}|{L^2}$$ $$\leq \frac{C}{A} |n0|{L^{\infty}} |n0|{L^1}^{\frac{1}{3}} |n0|{L^{\infty}}^{\frac{2}{3}} |\widetilde{u}{j,0}|{L^2} \leq \frac{C}{A} m^{\frac{1}{3}} |n|{L^{\infty}}^{\frac{5}{3}} |\widetilde{u}{j,0}|_{L^2}.$$

Combining J¹¹ and J¹² via (6.5), we obtain that

$$J1 \le \frac{C}{A} m^{\frac{1}{3}} ||n||{L^{\infty}}^{\frac{5}{3}} ||\widetilde{u}{j,0}||{L^2}.$$

For J2, using Lemma A.9, (6.1), H¨older's and Young's inequalities, one obtains

$$\begin{split} J2 \leq & \frac{1}{2A} \int{\mathbb{T}^2} \frac{\left| (n{\neq} \nabla c{\neq})0 + (n{\neq} u_{\neq})_0 \right|^2}{n0} dy dz + \frac{1}{2A} \int{\mathbb{T}^2} n_0 \left| \nabla \log n_0 - \nabla c_0 \right|^2 dy dz \ \leq & \frac{1}{2A} \left| \frac{1}{n0} \right|{L^{\infty}} \left| n{\neq} \right|{L^{\infty}}^2 \left( \left| \nabla c{\neq} \right|{L^2}^2 + \left| u{\neq} \right|{L^2}^2 \right) + \frac{1}{2A} \int_{\mathbb{T}^2} n_0 \left| \nabla \log n_0 - \nabla c0 \right|^2 dy dz \ \leq & \frac{1}{2A\delta} e^{\frac{\pi}{A}t} \left| n{\neq} \right|{L^{\infty}}^2 \left( \left| n{\neq} \right|{L^2}^2 + \left| u{\neq} \right|{L^2}^2 \right) + \frac{1}{2A} \int{\mathbb{T}^2} n_0 \left| \nabla \log n_0 - \nabla c_0 \right|^2 dy dz. \end{split}$$

Combining J¹ and J2, we get by (6.4) that

$$\begin{split} \frac{d}{dt}\mathcal{L}[n0] &\leq -\frac{1}{2A} \int{\mathbb{T}^2} n_0 \left| \nabla \log n_0 - \nabla c0 \right|^2 dy dz + \frac{C}{A} m^{\frac{1}{3}} |n|{L^{\infty}}^{\frac{5}{3}} |\widetilde{u}{j,0}|{L^2} \ &+ \frac{1}{2A\delta} \mathrm{e}^{\frac{\overline{n}}{A}t} |n{\neq}|{L^{\infty}}^2 \left( |n{\neq}|{L^2}^2 + |u{\neq}|{L^2}^2 \right), \end{split}$$

which follows that

$$\mathcal{L}[n{0}] - \mathcal{L}[(n{\text{in}}){0}] \leq -\frac{1}{2A} \int{0}^{t} \int{\mathbb{T}^{2}} n{0} |\nabla \log n{0} - \nabla c{0}|^{2} dy dz ds + \frac{C}{A} m^{\frac{1}{3}} ||n||{L^{\infty}L^{\infty}}^{\frac{5}{3}} \int{0}^{t} ||\widetilde{u}{j,0}||{L^{2}} ds

• \frac{1}{2A\delta} ||n{\neq}||{L^{\infty}L^{\infty}}^{2} \int{0}^{t} e^{\frac{\pi}{A}s} \left( ||n{\neq}||{L^{2}}^{2} + ||u{\neq}||_{L^{2}}^{2} \right) ds.$$ (6.6)

Using Lemma 4.3, we have

$$\frac{\int0^t |\widetilde{u}{j,0}|_{L^2} ds}{A} \le \frac{C\epsilon \int_0^t e^{-\frac{s}{2A}} ds}{A} \le C\epsilon. \tag{6.7}$$

Moreover, by Lemma A.1, Lemma 3.1 and assumptions (2.5), one deduces

$$e^{2aA^{-\frac{1}{3}}t}|u{\neq}|{L^{2}}^{2} \le Ce^{2aA^{-\frac{1}{3}}t}\left(|\partial{x}\omega{2,\neq}|{L^{2}}^{2} + |\Delta u{2,\neq}|{L^{2}}^{2}\right) \le CE{2}^{2},$$

and e²aA^{− 1 3 t}∥n̸=∥ 2 ^L2 ≤ C∥∂ ^xn̸=∥ ^Xb ≤ CE² 2 . They imply that

$$||u{\neq}||{L^2}^2 \le CE2^2 e^{-2aA^{-\frac{1}{3}}t}, \quad ||n{\neq}||_{L^2}^2 \le CE_2^2 e^{-2aA^{-\frac{1}{3}}t}.$$

Therefore, when A is sufficiently large satisfying A ≥ n a 3 , we get

$$\int0^t \mathrm{e}^{\frac{\overline{n}}{A}s} \left( |n{\neq}|{L^2}^2 + |u{\neq}|_{L^2}^2 \right) ds \le C E_2^2 \int_0^\infty \mathrm{e}^{\frac{\overline{n}}{A}s - 2aA^{-\frac{1}{3}}s} ds \le C E_2^2 \int_0^\infty \mathrm{e}^{-aA^{-\frac{1}{3}}s} ds = \frac{C E_2^2 A^{\frac{1}{3}}}{a}.$$

Combining it with (2.5) and (6.7), (6.6) yields

$$\mathcal{L}[n0] - \mathcal{L}[(n{\rm in})_0] \le -\frac{1}{2A} \int0^t \int{\mathbb{T}^2} n_0 |\nabla \log n_0 - \nabla c_0|^2 dy dz ds + C\epsilon m^{\frac{1}{3}} E_3^{\frac{5}{3}} + \frac{C E_2^2 E_3^2}{A^{\frac{2}{3}} a \delta}.$$

Hence, when

$$A \ge \max \left{ A_2, A_3, A_4, \left( \frac{E_2^2 E_3^2}{a \delta} \right)^{\frac{3}{2}}, \left( \frac{\overline{n}}{a} \right)^{\frac{3}{2}}, \left( \frac{1}{4a} \right)^{\frac{3}{2}} \right} =: A_5,$$

as long as ϵ is enough small satisfying

$$\epsilon m^{\frac{1}{3}} E_3^{\frac{5}{3}} \le C,$$ (6.8)

we obtain that

$$\mathcal{L}[n0] - \mathcal{L}[(n{\rm in})_0] \le C.$$

Next, as in [1] or [15], we use (6.3) and (A.6) to get a bound on ∥n⁰ log⁺ n0∥L¹ .

Lemma 6.3. Under the assumptions of Lemma 6.2 and m = ∥(nin)0∥L¹ < 8π, there exists a constant C^L log ^L(nin) such that

$$\int_{\mathbb{T}^2} n_0 \log^+ n0 dy dz \le C{L \log L}(n_{\rm in}). \tag{6.9}$$

Proof. Let Y = (y, z) ∈ T ² be fixed. Define the cut-off function φ(τ ) ∈ C ^∞ such that

$$\sup_{\varphi}(\varphi) = B(Y, 1/4),$$

$$\varphi(\tau) = 1, \quad \forall \tau \in B(Y, 1/8),$$

$$\sup_{\varphi}(\nabla \varphi(\tau)) \subset \overline{B}(Y, 1/4) \setminus B(Y, 1/8).$$ (6.10)

By periodically extending n0(τ ) and c0(τ ) to R 2 , we can rewrite the equation −△c⁰ = n⁰ − n that holds on T ² as the following equation that holds on R

$$-\Delta{\tau} (\varphi(\tau)c{0}(\tau)) = -2\nabla{\tau}\varphi(\tau) \cdot \nabla{\tau}c{0}(\tau) - c{0}(\tau)\Delta{\tau}\varphi(\tau) - \varphi(\tau)\Delta{\tau}c_{0}(\tau)$$

$$= -2\nabla{\tau}\varphi(\tau) \cdot \nabla{\tau}c{0}(\tau) - c{0}(\tau)\Delta{\tau}\varphi(\tau) + (n{0}(\tau) - \overline{n})\varphi(\tau).$$ (6.11)

□

Since φ(Y ) = 1 as Y ∈ B(Y, 1/8) and supp(φ) = B(Y, 1/4), using (6.11) and the fundamental solution of the Laplacian on R 2 , we get

$$c{0}(Y) = c{0}(Y)\varphi(Y)$$

$$= -\frac{1}{2\pi} \int{\mathbb{R}^{2}} \log(|Y - \tau|) \left[ (n{0}(\tau) - \overline{n})\varphi(\tau) - 2\nabla{\tau}\varphi(\tau) \cdot \nabla{\tau}c{0}(\tau) - c{0}(\tau)\triangle_{\tau}\varphi(\tau) \right] d\tau$$

$$= -\frac{1}{2\pi} \int{|Y - \tau| \leq \frac{1}{4}} \log(|Y - \tau|) (n{0}(\tau) - \overline{n})\varphi(\tau) d\tau - \frac{1}{\pi} \int{|Y - \tau| \leq \frac{1}{4}} \nabla{\tau} \cdot \left[ \log(|Y - \tau|)\nabla{\tau}\varphi(\tau) \right] c{0}(\tau) d\tau$$

$$+ \frac{1}{2\pi} \int{|Y - \tau| \leq \frac{1}{4}} \log(|Y - \tau|)\triangle{\tau}\varphi(\tau) c_{0}(\tau) d\tau.$$ (6.12)

Due to the support of φ, we can identify the above with an analogous integral on T ² with |Y − τ | replaced by d(Y, τ ). Multiplying (6.12) by − (n0(Y ) − n) and integrating with respect to Y over T 2 , we have

$$-\frac{1}{2}\int{\mathbb{T}^{2}}\left(n{0}(Y) - \overline{n}\right)c_{0}(Y)dY$$

$$=\frac{1}{4\pi}\int\int{\mathbb{T}^{2}\times\mathbb{T}^{2},d(Y,\tau)\leq\frac{1}{4}}\log d(Y,\tau)\left(n{0}(Y) - \overline{n}\right)\left(n_{0}(\tau) - \overline{n}\right)\varphi(\tau)d\tau dY$$

$$+\frac{1}{2\pi}\int\int{\mathbb{T}^{2}\times\mathbb{T}^{2},\frac{1}{8}\leq d(Y,\tau)\leq\frac{1}{4}}\left(n{0}(Y) - \overline{n}\right)\nabla{\tau}\cdot\left(\log d(Y,\tau)\nabla{\tau}\varphi(\tau)\right)c_{0}(\tau)d\tau dY$$

$$-\frac{1}{4\pi}\int\int{\mathbb{T}^{2}\times\mathbb{T}^{2},\frac{1}{8}\leq d(Y,\tau)\leq\frac{1}{4}}\left(n{0}(Y) - \overline{n}\right)\log d(Y,\tau)\triangle{\tau}\varphi(\tau)c{0}(\tau)d\tau dY.$$ (6.13)

By using (6.10)2, one deduces that

$$\begin{split} &\frac{1}{4\pi}\int\int_{d(Y,\tau)\leq\frac{1}{4}}\log d(Y,\tau)\left(n_0(Y)-\overline{n}\right)\left(n0(\tau)-\overline{n}\right)\varphi(\tau)d\tau dY\ =&\frac{1}{4\pi}\int\int{\mathbb{T}^2\times\mathbb{T}^2}\log d(Y,\tau)n_0(Y)n0(\tau)d\tau dY-\frac{1}{4\pi}\int\int{d(Y,\tau)>\frac{1}{8}}\log d(Y,\tau)n_0(Y)n0(\tau)d\tau dY\ &-\frac{1}{2\pi}\overline{n}\int\int{d(Y,\tau)\leq\frac{1}{8}}\log d(Y,\tau)n0(Y)d\tau dY+\frac{1}{4\pi}(\overline{n})^2\int\int{d(Y,\tau)\leq\frac{1}{8}}\log d(Y,\tau)d\tau dY\ &+\frac{1}{4\pi}\int\int_{\frac{1}{8}\leq d(Y,\tau)\leq\frac{1}{4}}\log d(Y,\tau)\left(n_0(Y)-\overline{n}\right)\left(n_0(\tau)-\overline{n}\right)\varphi(\tau)d\tau dY. \end{split}$$

Combining it with (6.13), we have

$$\begin{split} &-\frac{1}{2}\int_{\mathbb{T}^2} \left(n_0(Y) - \overline{n}\right) c0(Y) dY \ = &\frac{1}{4\pi} \int \int{\mathbb{T}^2 \times \mathbb{T}^2} \log d(Y,\tau) n_0(Y) n0(\tau) d\tau dY - \frac{1}{4\pi} \int \int{d(Y,\tau) > \frac{1}{8}} \log d(Y,\tau) n_0(Y) n0(\tau) d\tau dY \ &- \frac{1}{2\pi} \overline{n} \int \int{d(Y,\tau) \le \frac{1}{8}} \log d(Y,\tau) n0(Y) d\tau dY + \frac{1}{4\pi} (\overline{n})^2 \int \int{d(Y,\tau) \le \frac{1}{8}} \log d(Y,\tau) d\tau dY \ &+ \frac{1}{4\pi} \int \int_{\frac{1}{8} \le d(Y,\tau) \le \frac{1}{4}} \log d(Y,\tau) \left(n_0(Y) - \overline{n}\right) \left(n0(\tau) - \overline{n}\right) \varphi(\tau) d\tau dY \ &+ \frac{1}{2\pi} \int \int{\frac{1}{8} \le d(Y,\tau) \le \frac{1}{4}} \left(n0(Y) - \overline{n}\right) \nabla{\tau} \cdot \left(\log d(Y,\tau) \nabla_{\tau} \varphi(\tau)\right) c0(\tau) d\tau dY \ &- \frac{1}{4\pi} \int \int{\frac{1}{8} \le d(Y,\tau) \le \frac{1}{4}} \left(n0(Y) - \overline{n}\right) \log d(Y,\tau) \triangle{\tau} \varphi(\tau) c0(\tau) d\tau dY =: I{1,1} + I{1,2} + \dots + I{1,7}. \end{split}$$

First of all, direct calculations show that

$$I{1,2} + I{1,3} + I{1,4} + I{1,5} \ge -Cm^2$$ .

Moreover, in the region ^{1 8} ≤ |Y − τ | ≤ ¹ 4 , note that

$$|\nabla{\tau} \cdot [\log(|Y - \tau|)\nabla{\tau}\varphi(\tau)]| \le 8|\nabla{\tau}\varphi(\tau)| + \log 8|\triangle{\tau}\varphi(\tau)| \le C,$$

$$|\log(|Y - \tau|)\triangle{\tau}\varphi(\tau)| \le \log 8|\triangle{\tau}\varphi(\tau)| \le C,$$

we have

$$|I{1,6}| + |I{1,7}| \leq C \int_{\mathbb{T}^2} n0(Y) dY \int{\mathbb{T}^2} c0(\tau) d\tau + C \overline{n} \int{\mathbb{T}^2} c_0(\tau) d\tau \leq C m |c0|{L^1(\mathbb{T}^2)}$$

.

Denoting K(τ ) = − 1 2π log |τ | to be the fundamental solution of the Laplacian on T 2 , as −△c⁰ = n⁰ − n, by Young's inequality, we obtain

$$||c0||{L^1(\mathbb{T}^2)} = ||K * (n0 - \overline{n})||{L^1(\mathbb{T}^2)} \le ||K||_{L^1(\mathbb{T}^2)} ||n0 - \overline{n}||{L^1(\mathbb{T}^2)} \le Cm.$$

This follows that

$$I{1,6} + I{1,7} \ge -Cm^2$$ .

Combining the estimates of I1,² − I1,7, we conclude that

$$-\frac{1}{2}\int_{\mathbb{T}^2}(n_0-\overline{n})c0dY \ge \frac{1}{4\pi}\int\int{\mathbb{T}^2\times\mathbb{T}^2}\log d(Y,\tau)n_0(Y)n_0(\tau)d\tau dY - Cm^2.$$

From which along with (6.3), we arrive at

$$\mathcal{L}[(n{\text{in}}){0}] \geq \int{\mathbb{T}^{2}} n{0} \log n{0} dY - \int{\mathbb{T}^{2}} \frac{1}{2} (n{0} - \overline{n}) c{0} dY - C$$

$$\geq \int{\mathbb{T}^{2}} n{0} \log n{0} dY + \frac{1}{4\pi} \int{\mathbb{T}^{2} \times \mathbb{T}^{2}} \log d(Y, \tau) n{0}(Y) n{0}(\tau) d\tau dY - Cm^{2} - C$$

$$= \left(1 - \frac{m}{8\pi}\right) \int{\mathbb{T}^{2}} n{0} \log n_{0} dY$$

$$+ \frac{m}{8\pi} \left(\int{\mathbb{T}^{2}} n{0} \log n{0} dY + \frac{2}{m} \int \int{\mathbb{T}^{2} \times \mathbb{T}^{2}} \log d(Y, \tau) n{0}(Y) n{0}(\tau) d\tau dY\right) - Cm^{2} - C.$$

Applying (A.6) to it, one obtains

$$\mathcal{L}[(n_{\rm in})0] \ge \left(1 - \frac{m}{8\pi}\right) \int{\mathbb{T}^2} n_0 \log n_0 dY - C(m) - Cm^2,$$

which implies that

$$\int_{\mathbb{T}^2} n_0 \log n0 dY \le \frac{\mathcal{L}[(n{\rm in})0] + C(m) + Cm^2}{1 - \frac{m}{8\pi}} \le C{L \log L}(n_{\rm in}) < \infty$$

under the condition of m < 8π. Due to

$$\log^{-} n_0 = -\min{0, \log n_0} = \begin{cases} -\log n_0, & 0 < n_0 < 1, \ 0, & n_0 \ge 1, \end{cases}$$

there holds

$$\int_{\mathbb{T}^2} n_0 \log^+ n_0 dY = \begin{cases} 0, & 0 < n0 < 1, \ \int{\mathbb{T}^2} n_0 \log n_0 dY, & n_0 \ge 1. \end{cases}$$

This shows that

$$\int_{\mathbb{T}^2} n_0 \log^+ n0 dY \le C{L \log L}(n_{\rm in}) < \infty.$$

The proof is complete. □

The following lemma gives a uniform in time L ² bound of n0.

Lemma 6.4. Under the assumptions of Lemma 6.2, there holds

$$||n0||{L^2} \le C(||(n_{\rm in})0||{L^2} + m + 1) =: H_1.$$ (6.14)

Proof. Let Q > max{1, n} be a constant, to be chosen later. Noting that (n0−Q)⁺ = max{0, n0− Q}, and using (6.9), we have

$$\int_{\mathbb{T}^2} (n0 - Q)+ dy dz = \int_{n_0 > Q} (n0 - Q) dy dz \le \int{n_0 > Q} n0 dy dz = \int{n_0 > Q} \frac{1}{\log^+ n_0} n_0 \log^+ n0 dy dz \le \frac{1}{\log Q} \int{n_0 > Q} n_0 \log^+ n0 dy dz \le \frac{C{L \log L}}{\log Q}.$$ (6.15)

Recall that n⁰ satisfies

$$\partial_t n_0 = \frac{1}{A} \triangle n_0 - \frac{1}{A} \nabla \cdot (n_0 \nabla c0) - \frac{1}{A} \nabla \cdot (n{\neq} \nabla c_{\neq})_0 - \frac{1}{A} (u_0 \cdot \nabla n0) - \frac{1}{A} (u{\neq} \cdot \nabla n_{\neq})_0. \tag{6.16}$$

As (n⁰ −Q)+(n⁰ −Q)⁻ = 0, multiplying (6.16) by (n⁰ −Q)⁺ and integrating with respect to (y, z) over T 2 , we get

$$\frac{1}{2} \frac{d}{dt} | (n0 - Q)+ |{L^2(\mathbb{T}^2)}^2 = \frac{1}{A} \Big( \int{\mathbb{T}^2} (n0 - Q)+ \triangle n0 dy dz - \int{\mathbb{T}^2} (n0 - Q)+ \nabla \cdot (n_0 \nabla c_0) dy dz \

• \int_{\mathbb{T}^2} (n0 - Q)+ \nabla \cdot (n{\neq} \nabla c{\neq})0 dy dz - \int{\mathbb{T}^2} (n0 - Q)+ (u_0 \cdot \nabla n_0) dy dz \quad (6.17) \
• \int_{\mathbb{T}^2} (n0 - Q)+ (u{\neq} \cdot \nabla n{\neq})_0 dy dz \Big) =: J_1 + \dots + J_5.$$

For J1, integration by parts shows that

$$J1 = -\frac{1}{A} \int{\mathbb{T}^2} |\nabla ((n0 - Q)+)|^2 dy dz.$$

For J2, using (1.5) and integration by parts, we have

$$J{2} = \frac{1}{2A} \int{\mathbb{T}^{2}} ((n{0} - Q){+})^{2} (n{0} - \overline{n}) dy dz + \frac{Q}{A} \int{\mathbb{T}^{2}} (n{0} - Q){+} (n_{0} - \overline{n}) dy dz$$

$$= \frac{1}{2A} \int{\mathbb{T}^{2}} ((n{0} - Q){+})^{3} dy dz + \frac{3Q - \overline{n}}{2A} \int{\mathbb{T}^{2}} ((n{0} - Q){+})^{2} dy dz + \frac{Q^{2} - Q\overline{n}}{A} \int{\mathbb{T}^{2}} (n{0} - Q)_{+} dy dz$$

$$\leq \frac{1}{2A} \int{\mathbb{T}^{2}} ((n{0} - Q){+})^{3} dy dz + \frac{3Q}{2A} \int{\mathbb{T}^{2}} ((n{0} - Q){+})^{2} dy dz + \frac{Q^{2}m}{A}.$$

For J³ and J5, by integration by parts, one deduces

$$J_3 + J5 \le \frac{1}{16A} \int{\mathbb{T}^2} |\nabla \left( (n0 - Q)+ \right)|^2 dy dz + \frac{C(|(n{\ne} \nabla c{\ne})0|{L^2}^2 + |(n{\ne} u{\ne})0|{L^2}^2)}{A}.$$

For J4, using ∇ · u⁰ = 0, for j ∈ {2, 3}, we have

$$J4 = -\frac{1}{A} \int{\mathbb{T}^2} (n0 - Q)+ u_{j,0} \partial_j n0 dy dz = -\frac{1}{A} \int{\mathbb{T}^2} (n0 - Q)+ u_{j,0} \partial_j (n0 - Q)+ dy dz$$ $$= -\frac{1}{2A} \int{\mathbb{T}^2} u{j,0} \partial_j ((n0 - Q)+)^2 dy dz = 0.$$

Collecting J¹ − J5, we get by (6.17) that

$$\frac{1}{2} \frac{d}{dt} | (n0 - Q)+ |_{L^2}^2 \le -\frac{7 | \nabla (n0 - Q)+ |_{L^2}^2}{8A} + \frac{| (n0 - Q)+ |_{L^3}^3}{2A} + \frac{3Q | (n0 - Q)+ |{L^2}^2}{2A} + \frac{Q^2 m}{A} + \frac{C}{A} \left( | (n{\neq} \nabla c_{\neq})0 |{L^2}^2 + | (u{\neq} n{\neq})0 |{L^2}^2 \right).$$ (6.18)

Using Lemma A.7 and (6.15), one obtains

$$\begin{aligned} |(n0 - Q)+|_{L^3(\mathbb{T}^2)}^3 &\leq C |(n0 - Q)+|_{L^1(\mathbb{T}^2)} |\nabla \left((n0 - Q)+\right)|{L^2(\mathbb{T}^2)}^2 \ &\leq \frac{C{L \log L}}{\log Q} |\nabla \left((n0 - Q)+\right)|_{L^2(\mathbb{T}^2)}^2. \end{aligned}$$

Thus, we choose Q depending only on C^L log ^L such that

$$-\frac{7}{8A} |\nabla \left( (n0 - Q)+ \right)|_{L^2(\mathbb{T}^2)}^2 + \frac{1}{2A} |(n0 - Q)+|_{L^3(\mathbb{T}^2)}^3 \le -\frac{1}{2A} |\nabla \left( (n0 - Q)+ \right)|_{L^2(\mathbb{T}^2)}^2. \tag{6.19}$$

Similarly, it follows from Lemma A.7 and (6.15) that

$$-|\nabla\left((n0 - Q)+\right)|_{L^2(\mathbb{T}^2)}^2 \le -\frac{|(n0 - Q)+|_{L^2(\mathbb{T}^2)}^4}{C|(n0 - Q)+|_{L^1(\mathbb{T}^2)}^2} \le -\frac{1}{Cm^2}|(n0 - Q)+|_{L^2(\mathbb{T}^2)}^4. \tag{6.20}$$

Substituting (6.19) and (6.20) into (6.18), we get

$$\frac{d}{dt} | (n0 - Q)+ |_{L^2(\mathbb{T}^2)}^2 \le -\frac{1}{ACm^2} | (n0 - Q)+ |_{L^2(\mathbb{T}^2)}^4 + \frac{3Q}{A} | (n0 - Q)+ |_{L^2(\mathbb{T}^2)}^2

• \frac{2Q^2m}{A} + \frac{C}{A} \left( | (n{\neq} \nabla c{\neq})0 |{L^2(\mathbb{T}^2)}^2 + | (u{\neq} n{\neq})0 |{L^2(\mathbb{T}^2)}^2 \right).$$ (6.21)

We denote G(t) by

$$G(t) := \frac{C}{A} \int0^t \left( | (n{\neq} \nabla c_{\neq})0 |{L^2(\mathbb{T}^2)}^2 + | (u{\neq} n{\neq})0 |{L^2(\mathbb{T}^2)}^2 \right) ds, \quad \text{for} \quad t \geq 0.$$

Then using Lemma A.1, Lemma A.9, Lemma 3.1 and assumption (2.5), direct calculations indicate that

$$G(t) \leq \frac{C}{A} \left( |n{\neq}|{L^{\infty}L^{\infty}}^{2} |\nabla c{\neq}|{L^{2}L^{2}}^{2} + |n{\neq}|{L^{\infty}L^{\infty}}^{2} |u{\neq}|{L^{2}L^{2}}^{2} \right)$$

$$\leq \frac{C}{A^{\frac{2}{3}}} |n|{L^{\infty}L^{\infty}}^{2} \left( |\partial{x}n{\neq}|{X{a}}^{2} + |\partial{x}\omega{2,\neq}|{X{a}}^{2} + |\triangle u{2,\neq}|{X{a}}^{2} \right) \leq \frac{CE{2}^{2}E{3}^{2}}{A^{\frac{2}{3}}} \leq C$$

$$(6.22)$$

provided with A ≥ A5. Moreover, using Young's inequality, we rewrite (6.21) as

$$\frac{d}{dt} \left( | (n0 - Q)+ |_{L^2}^2 - G(t) \right) \le -\frac{\left[ | (n0 - Q)+ |_{L^2}^2 - G(t) - (9C^2 m^4 Q^2 + 4Cm^3 Q^2)^{\frac{1}{2}} \right]}{2ACm^2} \times \left[ | (n0 - Q)+ |_{L^2}^2 + \left( 9C^2 m^4 Q^2 + 4Cm^3 Q^2 \right)^{\frac{1}{2}} \right],$$

which implies that

$$||(n0 - Q)+||{L^2}^2 - G(t) \le ||(n{\rm in})0||{L^2}^2 + 2\left(9C^2m^4Q^2 + 4Cm^3Q^2\right)^{\frac{1}{2}}.$$

Combining it with (6.22), one deduces

$$|(n0 - Q)+|{L^2} \le C (|(n{\text{in}})0|{L^2} + m + 1).$$ (6.23)

By decomposing n⁰ = (n⁰ − Q)⁺ + min{n0, Q} and using (6.23), we get

$$||n0||{L^2} \le ||(n0 - Q)+||_{L^2} + ||\min{n0, Q}||{L^2}$$

$$\le ||(n0 - Q)+||{L^2} + Q^{\frac{1}{2}}m^{\frac{1}{2}} \le C(||(n{\text{in}})0||{L^2} + m + 1),$$

which gives the result.

The proof is complete. □

7. Estimates for L ^∞-norm of the density E3(t): Proof of Proposition 2.2

Proof of Proposition 2.2. For p = 2^j with j ≥ 1, multiplying (1.5)¹ by 2pn²p−¹ , and integrating by parts the resulting equation over T 3 , one deduces

$$\frac{d}{dt} |n^p|{L^2}^2 + \frac{2(2p-1)}{Ap} |\nabla n^p|{L^2}^2 = \frac{2(2p-1)}{A} \int_{\mathbb{T}^3} n^p \nabla c \cdot \nabla n^p dx dy dz$$

$$\leq \frac{2(2p-1)}{A} |n^p \nabla c|{L^2} |\nabla n^p|{L^2} \leq \frac{2p-1}{Ap} |\nabla n^p|{L^2}^2 + \frac{(2p-1)p}{A} |n^p \nabla c|{L^2}^2.$$

Using H¨older's inequality and Gagliardo-Nirenberg inequality, we get

$$||n^{p}\nabla c||{L^{2}}^{2} \leq ||n^{p}||{L^{4}}^{2}||\nabla c||{L^{4}}^{2} \leq C||n^{p}||{L^{2}}^{\frac{1}{2}}||\nabla n^{p}||{L^{2}}^{\frac{3}{2}}||\nabla c||{L^{4}}^{2} + C||n^{p}||{L^{2}}^{2}||\nabla c||{L^{4}}^{2},$$

which follows that

$$\begin{split} &\frac{d}{dt}|n^p|{L^2}^2 + \frac{2(2p-1)}{Ap}|\nabla n^p|{L^2}^2 \ &\leq &\frac{2p-1}{Ap}|\nabla n^p|{L^2}^2 + \frac{C(2p-1)p}{A}|n^p|{L^2}^{\frac{1}{2}}|\nabla n^p|{L^2}^{\frac{3}{2}}|\nabla c|{L^4}^2 + \frac{C(2p-1)p}{A}|n^p|{L^2}^2|\nabla c|{L^4}^2 \ &\leq &\frac{5(2p-1)}{4Ap}|\nabla n^p|{L^2}^2 + \frac{C(2p-1)p^7}{A}|n^p|{L^2}^2|\nabla c|{L^4}^8 + \frac{C(2p-1)p}{A}|n^p|{L^2}^2|\nabla c|_{L^4}^8. \end{split}$$

Consequently, there holds

$$\frac{d}{dt}|n^p|{L^2}^2 + \frac{1}{2A}|\nabla n^p|{L^2}^2 \le \frac{Cp^8}{A}|n^p|{L^2}^2 \left(1 + |\nabla c|{L^4}^8\right). \tag{7.1}$$

Using Gagliardo-Nirenberg inequality

$$||n^p||{L^2} \le C \left( ||n^p||{L^1}^{\frac{2}{5}} ||\nabla n^p||{L^2}^{\frac{3}{5}} + ||n^p||{L^1} \right),$$

we infer from (7.1) that

$$\frac{d}{dt}|n^p|{L^2}^2 \leq -\frac{|n^p|{L^2}^{\frac{10}{3}}}{2AC|n^p|{L^1}^{\frac{4}{3}}} + \frac{Cp^8}{A}|n^p|{L^2}^2 \left(1 + |\nabla c|_{L^\infty L^4}^8\right).$$

Applying Lemma A.8, Lemma A.9, Lemma 6.4 and the assumption (1.6), there holds

$$\begin{split} |\nabla c|{L^{\infty}L^{4}} &\leq |\nabla c{\neq}|{L^{\infty}L^{4}} + |\nabla c{0}|{L^{\infty}L^{4}} \ &\leq C\left(|\partial{x}^{2}n{\neq}|{L^{\infty}L^{2}} + |n{0}|{L^{\infty}L^{2}}\right) \leq C\left(E{2} + H{1}\right). \end{split}$$

Therefore

$$\frac{d}{dt}|n^p|{L^2}^2 \le -\frac{|n^p|{L^2}^{\frac{10}{3}}}{2CA|n^p|{L^1}^{\frac{4}{3}}} + \frac{Cp^8}{A}|n^p|{L^2}^2(1 + E_2^8 + H_1^8),$$

which indicates that

$$\sup{t\geq 0} |n^p|{L^2}^2 \leq \max\left{8C^3(1+E_2^8+H1^8)^{\frac{3}{2}}p^{12}\sup{t\geq 0} |n^p|{L^1}^2, 2|n{\text{in}}^p|_{L^2}^2\right}. \tag{7.2}$$

Next, the Moser-Alikakos iteration is used to determine E3. Recall p = 2^j with j ≥ 1, and we rewrite (7.2) into

$$\sup{t \ge 0} \int{\mathbb{T}^3} |n(t)|^{2^{j+1}} dx dy dz \le \max \left{ C1 p^{12} \left( \sup{t \ge 0} \int{\mathbb{T}^3} |n(t)|^{2^j} dx dy dz \right)^2, 2 \int{\mathbb{T}^3} |n_{\rm in}|^{2^{j+1}} dx dy dz \right}, \quad (7.3)$$

where C¹ = 8C 3 (1 + E 8 ² + H⁸ 1 ) ² . From Lemma 6.4, we note that

$$||n0||{L^\infty L^2} \le H_1.$$

Hence

$$\sup{t\geq 0} |n(t)|{L^2} \leq |\mathbb{T}||n0|{L^{\infty}L^2} + |n{\neq}|{L^{\infty}L^2} \leq |\mathbb{T}|H_1 + E_2.$$

By interpolation inequality, for 0 < θ < 1, we have

$$|n{\rm in}|{L^{2^j}} \le |n{\rm in}|{L^2}^{\theta} |n{\rm in}|{L^\infty}^{1-\theta} \le |n{\rm in}|{L^2} + |n{\rm in}|{L^\infty} \le |\mathbb{T}|H_1 + E2 + |n{\rm in}|_{L^\infty}$$

for j ≥ 1. This yields that

$$2\int{\mathbb{T}^3} |n{\rm in}|^{2^{j+1}} dx dy dz \le 2 \left( |\mathbb{T}|H_1 + E2 + ||n{\rm in}||_{L^{\infty}} \right)^{2^{j+1}} \le K^{2^{j+1}},$$

where K = 2(|T|H¹ + E² + ∥nin∥L^∞).

Now, we rewrite (7.3) as

$$\sup{t\geq 0}\int{\mathbb{T}^3}|n(t)|^{2^{j+1}}dxdydz\leq \max\left{C14096^j\left(\sup{t\geq 0}\int_{\mathbb{T}^3}|n(t)|^{2^j}dxdydz\right)^2,K^{2^{j+1}}\right}.$$

For j = k, we get

$$\sup{t \ge 0} \int{\mathbb{T}^3} |n(t)|^{2^{k+1}} dx dy dz \le C_1^{a_k} 4096^{b_k} K^{2^{k+1}},$$

where a^k = 1 + 2ak−¹ and b^k = k + 2bk−1.

Generally, one can obtain the following formulas

$$a_k = 2^k - 1$$ , and $b_k = 2^{k+1} - k - 2$ .

Therefore, one deduces

$$\sup{t \geq 0} \left( \int{\mathbb{T}^3} |n(t)|^{2^{k+1}} dx dy dz \right)^{\frac{1}{2^{k+1}}} \leq C_1^{\frac{2^k-1}{2^{k+1}}} 4096^{\frac{2^{k+1}-k-2}{2^{k+1}}} K.$$

Letting k → ∞, there holds

$$\sup{t>0} ||n(t)||{L^{\infty}} \le C(1 + E_2^8 + H_1^8)^{\frac{3}{4}} (|\mathbb{T}|H_1 + E2 + ||n{\rm in}||_{L^{\infty}}) =: E_3.$$ (7.4)

□

8. Energy estimates for E4(t): Proof of Proposition 2.3

To estimate ∥∂ 2 ^xu2,̸=∥^Xb and ∥∂ 2 ^xu3,̸=∥^Xb , it is important to introduce the new quantity W defined by

$$W = u{2,\neq} + \kappa u{3,\neq},$$

where

$$V = y + \frac{\mathbf{U}_2}{A}, \quad \text{and } \kappa = \frac{\partial_z V}{\partial_y V}.$$ (8.1)

The similar quality was first proposed by Wei-Zhang in [34] and further applied in [5] and [8].

For j ∈ {2, 3}, there holds (u · ∇u^j )̸⁼ = u⁰ · ∇uj,̸⁼ + u̸⁼ · ∇uj,⁰ + (u̸⁼ · ∇uj,̸=)̸=. Then we infer from (1.3) that

$$\begin{cases} \mathcal{L}{V} u{2,\neq} + \frac{\mathbf{U}{1} \partial{x} u{2,\neq}}{A} + \frac{g{2,1} + g{2,2} + G{2,3}}{A} + \frac{\partial{y} (P{\neq}^{N{1}} + P{\neq}^{N{2}})}{A} = 0, \ \mathcal{L}{V} u{3,\neq} + \frac{\mathbf{U}{1} \partial{x} u{3,\neq}}{A} + \frac{g{3,1} + g{3,2} + G{3,3}}{A} + \frac{\partial{z} (P{\neq}^{N{1}} + P{\neq}^{N{2}})}{A} = 0, \end{cases}$$ (8.2)

where L^V can be found in (5.10) and

$$g{j,1} = u{2,0} \partialy u{j,\neq} + u_{3,0} \partialz u{j,\neq}, \quad g{j,2} = u{\neq} \cdot \nabla u{j,0}, \quad G{j,3} = (u{\neq} \cdot \nabla u{j,\neq})_{\neq}.$$ (8.3)

Due to div u = 0, we have

$$\operatorname{div} (u \cdot \nabla u)_{\neq} = \partial_x (u \cdot \nabla u1){\neq} + \partial_y (u \cdot \nabla u2){\neq} + \partial_z (u \cdot \nabla u3){\neq}$$

= $$\operatorname{div} (u{\neq} \cdot \nabla u{\neq})_{\neq} + 2(\partialy u{1,0} \partialx u{2,\neq} + \partialz u{1,0} \partialx u{3,\neq}) + 2\partialy g{2,2} + 2\partialz g{3,2},$$

which along with ∂yV = 1 + ^∂yU² A implies that

$$\begin{split} \frac{P_{\neq}^{N1} + P{\neq}^{N_2}}{A} &= -2\triangle^{-1} \left( \partialx u{2,\neq} + \frac{\operatorname{div} \ (u \cdot \nabla u)_{\neq}}{2A} - \frac{\partialx n{\neq}}{2A} \right) \ &= -2\triangle^{-1} \Bigg( \left( 1 + \frac{\partial_y \mathbf{U}_2}{A} \right) \partialx u{2,\neq} + \frac{\partial_z \mathbf{U}_2}{A} \partialx u{3,\neq} + \frac{\operatorname{div} \ (u{\neq} \cdot \nabla u{\neq})_{\neq}}{2A} \ &\quad + \frac{\partial_y \mathbf{U}_1 \partialx u{2,\neq} + \partial_z \mathbf{U}_1 \partialx u{3,\neq}}{A} + \frac{\partialy g{2,2} + \partialz g{3,2}}{A} - \frac{\partialx n{\neq}}{2A} \Bigg) \ &= -2\triangle^{-1} \left( \partial_y V \partial_x W + \frac{\partialy g{2,2} + \partialz g{3,2}}{A} + \frac{P{1,1} + P{1,2} + P_{1,3}}{A} \right), \end{split}$$

where

$$P{1,1} = \frac{\operatorname{div} (u{\neq} \cdot \nabla u{\neq}){\neq}}{2}, \quad P_{1,2} = \partial_y \mathbf{U}_1 \partialx u{2,\neq} + \partial_z \mathbf{U}_1 \partialx u{3,\neq} \quad P_{1,3} = -\frac{\partialx n{\neq}}{2}. \tag{8.4}$$

Using the above decomposition, we rewrite (8.2) into

$$\begin{cases} \mathcal{L}{V}u{2,\neq} - 2\partial{y}\triangle^{-1}(\partial{y}V\partial{x}W) + \frac{\mathbf{U}{1}\partial{x}u{2,\neq}}{A} + \frac{g{2,1} + g{2,2} + G{2,3}}{A} = \frac{2\partial{y}\triangle^{-1}(\partial{y}g{2,2} + \partial{z}g{3,2})}{A} \

• \frac{2\partial{y}\triangle^{-1}(P{1,1} + P{1,2} + P{1,3})}{A}, \ \mathcal{L}{V}u{3,\neq} - 2\partial{z}\triangle^{-1}(\partial{y}V\partial{x}W) + \frac{\mathbf{U}{1}\partial{x}u{3,\neq}}{A} + \frac{g{3,1} + g{3,2} + G{3,3}}{A} = \frac{2\partial{z}\triangle^{-1}(\partial{y}g{2,2} + \partial{z}g{3,2})}{A} \
• \frac{2\partial{z}\triangle^{-1}(P{1,1} + P{1,2} + P{1,3})}{A}. \end{cases} (8.5)$$

Therefore W = u2,̸⁼ + κu3,̸⁼ satisfies

$$\widetilde{\mathcal{L}_V}W + \frac{\mathbf{U}_1 \partial_x W}{A} + \frac{G^{(1)} + G^{(2)}}{A} = \left(\partialt \kappa - \frac{\triangle \kappa}{A}\right) u{3,\neq} - \frac{2\nabla \kappa \cdot \nabla u_{3,\neq}}{A},\tag{8.6}$$

where

$$\widetilde{\mathcal{L}_V}W = \mathcal{L}_V W - 2(\partial_y + \kappa \partial_z) \triangle^{-1}(\partial_y V \partial_x W)$$

and

$$G^{(1)} = G{2,3} + \kappa G{3,3} - 2(\partial_y + \kappa \partialz) \triangle^{-1} (P{1,1} + P{1,2} + P{1,3}),$$

$$G^{(2)} = g{2,1} + g{2,2} + \kappa (g{3,1} + g{3,2}) - 2(\partial_y + \kappa \partial_z) \triangle^{-1} (\partialy g{2,2} + \partialz g{3,2}).$$

In addition, △W satisfies

$$\mathcal{L}V \triangle W = \triangle \left( -\frac{2\nabla \kappa \cdot \nabla u{3,\neq}}{A} \right) + \text{good terms.}$$

To remove the singular term △ − 2∇κ·∇u3,̸⁼ A , motivated by the quasi-linear method in [34], we introduce the following decomposition

$$\nabla \kappa \cdot \nabla u_{3,\neq} = \rho1 \nabla V \cdot \nabla u{3,\neq} + \rho_2 (\partial_z - \kappa \partialy) u{3,\neq}, \tag{8.7}$$

where

$$\rho_1 = \frac{\partial_y \kappa + \kappa \partial_z \kappa}{\partial_y V(1 + \kappa^2)}, \quad \rho_2 = \frac{\partial_z \kappa - \kappa \partial_y \kappa}{1 + \kappa^2}.$$ (8.8)

As (∂^z − κ∂y) has a good commutative relation with L^V , it is a good derivative. Thus the second term in (8.7) is good. To handle the first term in (8.7) and obtain a sharp threshold of velocity, we need to make a further decomposition for W as follows

$$W = W^{(1)} + \frac{1}{A}W^{(2)},$$

where W(1) and W(2) solve

$$\begin{cases} \mathcal{L}_V \triangle W^{(1)} = \text{good terms,} \ \mathcal{L}_V W^{(2)} = -\rho1 \nabla V \cdot \nabla u{3,\neq}, \ W^{(1)}(0) = W(0), \ W^{(2)}(0) = 0. \end{cases}$$

Furthermore, an additional quantity W(3) is also needed satisfying

$$\mathcal{L}V W^{(3)} = -\nabla V \cdot \nabla u{3,\neq}, \quad W^{(3)}(0) = 0.$$

We denote

$$\Delta u{3,\neq} = \left(\Delta u{3,\neq} - 2\partial_x W^{(3)}\right) + 2\partial_x W^{(3)},$$

which satisfies

$$\mathcal{L}{V}\left(\triangle u{3,\neq} - 2\partial{x}W^{(3)}\right) = 2\partial{z}\left(\partial{y}V\partial{x}W\right) - \frac{\triangle(\mathbf{U}{1}\partial{x}u{3,\neq})}{A} - \frac{\triangle(g{3,1} + g{3,2} + G{3,3})}{A} + \frac{2\partial{z}\left(\partial{y}g{2,2} + \partial{z}g{3,2}\right)}{A} + \frac{2\partial{z}\left(P{1,1} + P{1,2} + P{1,3}\right)}{A} - \triangle V\partial{x}u_{3,\neq}$$ (8.9)

and

$$\mathcal{L}V W^{(3)} = -\nabla V \cdot \nabla u{3,\neq}. \tag{8.10}$$

In this way, by space-time estimates, we will prove that

$$E4^2(t) \le E{5,1}^2(t) + E{5,2}^2(t) \le C(|u{\text{in}}|_{H^2}^2 + 1),$$

where E 5,1 (t) and E 5,2 (t) are the auxiliary norms defined by

$$E{5,1}(t) = A^{-\frac{2}{3}} | \triangle u{3,\neq} |_{X_b},$$

$$E{5,2}(t) = \sum{j=2}^{3} (|\partialx^2 u{j,\neq}|_{X_b} + |\partial_x (\partial_z - \kappa \partialy) u{j,\neq}|_{X_b}) + |\partialx \nabla W|{X_b}$$

Lemma 8.1. Under the result of Lemma 4.5, it holds that

$$\begin{split} |\kappa|{H^{1}} &\leq CA^{-1} |\triangle \mathbf{U}{2}|{L^{2}} \leq C\epsilon, \quad |\kappa|{H^{3}} \leq CA^{-1} |\triangle \mathbf{U}{2}|{H^{2}} \leq C\epsilon, \ |\partial{t}\kappa|{H^{1}} &\leq CA^{-1} |\partial{t}\mathbf{U}{2}|{H^{2}} \leq CA^{-1}\epsilon, \quad |\partial{t}\rho{1}|{L^{2}} \leq CA^{-1} |\partial{t}\mathbf{U}{2}|{H^{2}} \leq CA^{-1}\epsilon, \ |\rho{1}|{H^{2}} + |\rho{2}|{H^{2}} &\leq CA^{-1} |\triangle \mathbf{U}{2}|_{H^{2}} \leq C\epsilon, \end{split}$$

where the definitions of κ, ρ¹ and ρ² are given by (8.1) and (8.8).

The following lemma gives the estimates of W(2) and W(3) associated with good derivatives.

Lemma 8.2. Under the conditions of Theorem 1.1 and the assumptions (2.5), it holds that

• (i) ∥∂ 2 ^xW(3)∥ ^Xb + ∥∂x(∂^z − κ∂y)W(3)∥ 2 ^{Xb ≤} CA^{4 3}E 2 5,2 (t),
• (ii) ∥∂ 2 ^xW(2)∥ ^Xb ≤ Cϵ²A ³E 5,2 (t), ∥∂x∇W(2)∥ ^Xb ≤ Cϵ²A²E 2 5,2 (t),
• (iii) ∥∂x(W(2) − ρ1W(3))∥ 2 ^{Xb ≤} CA² 3 ϵ ²E 5,2 (t).

Proof. Estimate (i). Applying Proposition A.3 to (8.10), we obtain

$$|\partial{x}^{2}W^{(3)}|{X{b}}^{2} + |\partial{x}(\partial{z} - \kappa\partial{y})W^{(3)}|{X{b}}^{2} \leq CA^{\frac{1}{3}} (|e^{bA^{-\frac{1}{3}}t}\partial{x}^{2}(\nabla V \cdot \nabla u{3,\neq})|{L^{2}L^{2}}^{2} + |e^{bA^{-\frac{1}{3}}t}\partial{x}(\partial{z} - \kappa\partial{y})(\nabla V \cdot \nabla u{3,\neq})|{L^{2}L^{2}}^{2}) =: CA^{\frac{1}{3}} (I{1} + I{2}).$$

$$(8.11)$$

Recalling that V = y + U2(t,y,z) A , by Lemma 4.5 and Lemma 8.1, we have

$$|\nabla V|_{L^{\infty}} \le 1 + A^{-1} |\nabla \mathbf{U}2|{L^{\infty}} \le C \left(1 + A^{-1} |\Delta \mathbf{U}2|{H^2}\right) \le C,$$ (8.12)

$$|(\partial_z - \kappa \partialy)\nabla V|{L^{\infty}} \le CA^{-1} (1 + |\kappa|_{H^3}) |\Delta \mathbf{U}2|{H^2} \le C.$$ (8.13)

Then for I1, using (8.12), we have

$$I1 \le |\nabla V|{L^{\infty}L^{\infty}}^2 |e^{bA^{-\frac{1}{3}}t} \nabla \partialx^2 u{3,\neq}|_{L^2L^2}^2 \le CA |\partialx^2 u{3,\neq}|_{Xb}^2 \le CA E{5,2}^2(t).$$

For I2, using (8.12) and (8.13), we deduce that

$$|\partial_x(\partial_z - \kappa \partialy) \left(\nabla V \cdot \nabla u{3,\neq}\right)|_{L^2} \leq C \left(|\nabla \partial_x(\partial_z - \kappa \partialy) u{3,\neq}|_{L^2} + |\nabla \partialx^2 u{3,\neq}|_{L^2}\right),$$

which implies

$$I_2 \leq CA \left( |\partial_x (\partial_z - \kappa \partialy) u{3,\neq}|_{X_b}^2 + |\partialx^2 u{3,\neq}|_{Xb}^2 \right) \leq CAE{5,2}^2(t).$$

Combining the estimates of I¹ and I2, (8.11) gives the result of (i).

Estimate (ii). Notice that

$$\mathcal{L}_V \partial_x^2 W^{(2)} = \partial_x^2 \mathcal{L}_V W^{(2)} = -\rho_1 \nabla V \cdot \nabla \partialx^2 u{3,\neq}$$

and ∂ 2 ^xW(2)(0) = 0. Then applying Proposition A.1, there holds

$$|\partialx^2 W^{(2)}|{X_b}^2 \leq C A^{\frac{1}{3}} |\mathrm{e}^{bA^{-\frac{1}{3}}t} \rho_1 \nabla V \cdot \nabla \partialx^2 u{3,\neq}|_{L^2 L^2}^2.$$

Using Lemma 8.1 and (8.12), we obtain

$$|\rho_1 \nabla V \cdot \nabla \partialx^2 u{3,\neq}|_{L^2} \le C |\rho1|{H^2} |\nabla V|_{L^{\infty}} |\nabla \partialx^2 u{3,\neq}|_{L^2} \le C \epsilon |\nabla \partialx^2 u{3,\neq}|_{L^2}.$$

This indicates that

$$|\partialx^2 W^{(2)}|{X_b}^2 \le C\epsilon^2 A^{\frac{1}{3}} |e^{bA^{-\frac{1}{3}}t} \nabla \partialx^2 u{3,\neq}|_{L^2L^2}^2 \le C\epsilon^2 A^{\frac{4}{3}} |\partialx^2 u{3,\neq}|_{Xb}^2 \le C\epsilon^2 A^{\frac{4}{3}} E{5,2}^2(t). \tag{8.14}$$

For j ∈ {1, 2, 3}, we get

$$\mathcal{L}{V}\partial{x}\partial{j}W^{(2)} = \partial{x}\partial{j}\mathcal{L}{V}W^{(2)} - \partial{j}V\partial{x}^{2}W^{(2)} = -\partial{j}\left(\rho{1}\nabla V \cdot \nabla\partial{x}u{3,\neq}\right) - \partial{j}V\partial{x}^{2}W^{(2)}$$

satisfying ∂x∂jW(2)(0) = 0. Applying Proposition A.1 to it, one deduces

$$|\partial_x \partialj W^{(2)}|{X_b}^2 \le CA^{\frac{1}{3}} |e^{bA^{-\frac{1}{3}}t} \partial_j V \partialx^2 W^{(2)}|{L^2L^2}^2 + CA |e^{bA^{-\frac{1}{3}}t} \rho_1 \nabla V \cdot \nabla \partialx u{3,\neq}|_{L^2L^2}^2. \tag{8.15}$$

Due to (8.12) and Lemma 8.1, there holds

$$|\partial_j V \partialx^2 W^{(2)}|{L^2} \le |\partialj V|{L^{\infty}} |\partialx^2 W^{(2)}|{L^2} \le C |\partialx^2 W^{(2)}|{L^2},$$

$$|\rho_1 \nabla V \cdot \nabla \partialx u{3,\neq}|_{L^2} \le C |\rho1|{H^2} |\nabla V|_{L^{\infty}} |\nabla \partialx u{3,\neq}|_{L^2} \le C \epsilon |\nabla \partialx^2 u{3,\neq}|_{L^2},$$

which implies that

$$\begin{split} &| \mathbf{e}^{bA^{-\frac{1}{3}}t} \partial_j V \partialx^2 W^{(2)} |{L^2 L^2}^2 \leq C | \mathbf{e}^{bA^{-\frac{1}{3}}t} \partialx^2 W^{(2)} |{L^2 L^2}^2 \leq C A^{\frac{1}{3}} | \partialx^2 W^{(2)} |{X_b}^2, \ &| \mathbf{e}^{bA^{-\frac{1}{3}}t} \rho_1 \nabla V \cdot \nabla \partialx u{3,\neq} |_{L^2 L^2}^2 \leq C \epsilon^2 | \mathbf{e}^{bA^{-\frac{1}{3}}t} \nabla \partialx^2 u{3,\neq} |_{L^2 L^2}^2 \leq C \epsilon^2 A | \partialx^2 u{3,\neq} |_{X_b}^2. \end{split}$$

Substituting the above estimations into (8.15) and using (8.14), we arrive at

$$|\partialx \nabla W^{(2)}|{X_b}^2 \leq CA^{\frac{2}{3}} |\partialx^2 W^{(2)}|{X_b}^2 + C\epsilon^2 A^2 |\partialx^2 u{3,\neq}|_{Xb}^2 \leq C\epsilon^2 A^2 E{5,2}^2(t).$$

Estimate (iii). Due to

$$\mathcal{L}{V}(\rho{1}f) - \rho{1}\mathcal{L}{V}f = (\partial{t}\rho{1} - A^{-1}\triangle\rho{1})f - 2A^{-1}\nabla\rho{1} \cdot \nabla f$$ $$= (\partial{t}\rho{1} + A^{-1}\triangle\rho{1})f - 2A^{-1}\nabla \cdot (f\nabla\rho{1})$$

and L^V W(2) = ρ1L^V W(3) , there holds

$$\mathcal{L}{V}\partial{x}\left(W^{(2)} - \rho{1}W^{(3)}\right) = \partial{x}\mathcal{L}{V}\left(W^{(2)} - \rho{1}W^{(3)}\right) = \partial{x}\left(\rho{1}\mathcal{L}{V}W^{(3)} - \mathcal{L}{V}(\rho_{1}W^{(3)})\right)$$

$$= -\partial{x}\left[\left(\partial{t}\rho{1} + A^{-1}\triangle\rho{1}\right)W^{(3)} - 2A^{-1}\nabla\cdot\left(W^{(3)}\nabla\rho_{1}\right)\right]$$

$$= -\Delta Q + 2A^{-1}\nabla\cdot\left(\partial{x}W^{(3)}\nabla\rho{1}\right),$$

where △Q = (∂tρ¹ + A⁻¹△ρ1) ∂xW(3) .

Applying Proposition A.1, we get

$$|\partial_x (W^{(2)} - \rho1 W^{(3)})|{Xb}^2 \le C \left( A |e^{bA^{-\frac{1}{3}}t} \nabla Q|{L^2 L^2}^2 + A^{-1} |e^{bA^{-\frac{1}{3}}t} \partial_x W^{(3)} \nabla \rho1|{L^2 L^2}^2 \right). \tag{8.16}$$

It follows from Lemma 8.1 that ∥∇ρ1∥H¹ ≤ ∥ρ1∥H² ≤ Cϵ and

$$|\partial_t \rho_1 + A^{-1} \triangle \rho1|{L^2} \le |\partial_t \rho1|{L^2} + A^{-1} |\rho1|{H^2} \le CA^{-1}\epsilon.$$

Combining them with Lemma A.5, one obtains

$$|\nabla Q|{L^{2}} = |\nabla \triangle^{-1} \triangle Q|{L^{2}} = |\nabla \triangle^{-1} [(\partial{t} \rho{1} + A^{-1} \triangle \rho{1}) \partial{x} W^{(3)}]|_{L^{2}}$$

$$\leq C |\partial{t} \rho{1} + A^{-1} \triangle \rho{1}|{L^{2}} (|\partial{x} W^{(3)}|{L^{2}} + |\partial{x} (\partial{z} - \kappa \partial{y}) W^{(3)}|{L^{2}})$$

$$\leq CA^{-1} \epsilon (|\partial{x}^{2} W^{(3)}|{L^{2}} + |\partial{x} (\partial{z} - \kappa \partial{y}) W^{(3)}|{L^{2}})$$ (8.17)

and

$$|\partial_x W^{(3)} \nabla \rho1|{L^2} \le C |\nabla \rho1|{H^1} \left( |\partialx W^{(3)}|{L^2} + |\partial_x (\partial_z - \kappa \partialy) W^{(3)}|{L^2} \right)$$

$$\le C\epsilon \left( |\partialx^2 W^{(3)}|{L^2} + |\partial_x (\partial_z - \kappa \partialy) W^{(3)}|{L^2} \right).$$ (8.18)

Substituting (8.17) and (8.18) into (8.16), and using the result of (i), we have

$$\begin{split} |\partial_x (W^{(2)} - \rho1 W^{(3)})|{X_b}^2 \leq & CA^{-1} \epsilon^2 \left( | \mathrm{e}^{bA^{-\frac{1}{3}t}} \partialx^2 W^{(3)} |{L^2 L^2}^2 + | \mathrm{e}^{bA^{-\frac{1}{3}t}} \partial_x (\partial_z - \kappa \partialy) W^{(3)} |{L^2 L^2}^2 \right) \ \leq & CA^{-1} \epsilon^2 A^{\frac{1}{3}} \left( |\partialx^2 W^{(3)}|{X_b}^2 + |\partial_x (\partial_z - \kappa \partialy) W^{(3)}|{Xb}^2 \right) \leq CA^{\frac{2}{3}} \epsilon^2 E{5,2}^2(t). \end{split}$$

Lemma 8.3. Under the assumptions of Theorem 1.1 and (2.5), it holds that

$$|\mathbf{e}^{bA^{-\frac{1}{3}t}}\nabla(\mathbf{U}{1}\partial{x}W)|{L^{2}L^{2}}^{2} \leq C\left(|(u{1,\text{in}}){0}|{H^{1}}^{2} + H{1}^{2}\right)\left(A^{\frac{1}{2}}|\Delta W^{(1)}|{X{b}}^{2} + \epsilon^{2}A^{\frac{2}{3}}E{5,2}^{2}(t)\right),$$

$$|\mathbf{e}^{bA^{-\frac{1}{3}t}}\nabla(\mathbf{U}{1}\partial{x}u{3,\neq})|{L^{2}L^{2}}^{2} \leq C\left(|(u{1,\text{in}}){0}|{H^{1}}^{2} + H{1}^{2}\right)AE_{5,2}^{2}(t),$$

$$|\mathbf{e}^{bA^{-\frac{1}{3}t}}\mathbf{U}{1}\partial{x}^{2}(u{2,\neq},u{3,\neq})|{L^{2}L^{2}}^{2} \leq C\left(|(u{1,\text{in}}){0}|{H^{1}}^{2} + H{1}^{2}\right)A^{\frac{1}{3}(2\alpha+1)}E{5,2}^{2}(t),$$

$$|\mathbf{e}^{bA^{-\frac{1}{3}t}}P{1,2}|{L^{2}L^{2}}^{2} \leq C\left(|(u{1,\text{in}}){0}|{H^{1}}^{2} + H{1}^{2}\right)A^{\frac{2}{3}}E_{5,2}^{2}(t),$$

$$(8.19)$$

where α ∈ ( 1 2 , 4 ) and the definition of H¹ is the same as in Lemma 6.4.

Proof. Estimate (8.19)1. Recalling that W = W(1) + 1 AW(2) , we have

$$|\mathbf{e}^{bA^{-\frac{1}{3}t}}\nabla(\mathbf{U}{1}\partial{x}W)|{L^{2}L^{2}}^{2} \leq |\mathbf{e}^{bA^{-\frac{1}{3}t}}\nabla(\mathbf{U}{1}\partial{x}W^{(1)})|{L^{2}L^{2}}^{2} + \frac{1}{A^{2}}|\mathbf{e}^{bA^{-\frac{1}{3}t}}\nabla(\mathbf{U}{1}\partial{x}W^{(2)})|_{L^{2}L^{2}}^{2}.$$

Using Lemma 4.4, Lemma 6.4, Lemma A.2, Lemma A.3 and

$$|e^{bA^{-\frac{1}{3}}t}\partialx \nabla f|{L^2L^2} = |e^{bA^{-\frac{1}{3}}t}\nabla \triangle^{-1}\partialx(\triangle f)|{L^2L^2} \le |\triangle f|_{X_b},$$

one obtains

$$|\mathbf{e}^{bA^{-\frac{1}{3}}t}\nabla(\mathbf{U}{1}\partial{x}W^{(1)})|_{L^{2}L^{2}}^{2}$$

$$\leq |\mathbf{U}{1}|{L^{\infty}{t,z}L^{2}{y}}^{2}|\mathbf{e}^{bA^{-\frac{1}{3}}t}\nabla\partial{x}W^{(1)}|{L^{2}{t,z,z}L^{\infty}{y}}^{2} + |\nabla\mathbf{U}{1}|{L^{\infty}{t}L^{2}{y,z}}^{2}|\mathbf{e}^{bA^{-\frac{1}{3}}t}\partial{x}W^{(1)}|{L^{2}{t,x}L^{\infty}{y,z}}^{2}$$

$$\leq C|\mathbf{U}{1}|{L^{\infty}H^{1}}^{2}(|\mathbf{e}^{bA^{-\frac{1}{3}}t}\partial{x}\nabla W^{(1)}|{L^{2}L^{2}}|\mathbf{e}^{bA^{-\frac{1}{3}}t}\partial{x}\Delta W^{(1)}|{L^{2}L^{2}}^{2}$$

$$+ |\mathbf{e}^{bA^{-\frac{1}{3}}t}\partial{x}\nabla W^{(1)}|{L^{2}L^{2}}^{\frac{3}{2}-\alpha}|\mathbf{e}^{bA^{-\frac{1}{3}}t}\partial{x}\Delta W^{(1)}|{L^{2}L^{2}}^{\alpha-\frac{1}{2}}$$

$$\leq C(|(u{1,\mathrm{in}}){0}|{H^{1}}^{2} + H{1}^{2})A^{\frac{1}{2}}|\Delta W^{(1)}|{X{b}}^{2},$$

$$(8.20)$$

□

where α ∈ ( 1 , 3 4 ). Similarly, by Lemma 4.4, Lemma 6.4, Lemma A.2, Lemma A.3 and (ii) of Lemma 8.2, there holds

$$|\mathbf{e}^{bA^{-\frac{1}{3}}t}\nabla(\mathbf{U}{1}\partial{x}W^{(2)})|_{L^{2}L^{2}}^{2}$$

$$\leq C|\mathbf{U}{1}|{L^{\infty}H^{1}}^{2}(|\mathbf{e}^{bA^{-\frac{1}{3}}t}\partial{x}\nabla W^{(2)}|{L^{2}L^{2}}|\mathbf{e}^{bA^{-\frac{1}{3}}t}\partial{x}\Delta W^{(2)}|{L^{2}L^{2}}$$

$$+|\mathbf{e}^{bA^{-\frac{1}{3}}t}\partial{x}\nabla W^{(2)}|{L^{2}L^{2}}^{3}|\mathbf{e}^{bA^{-\frac{1}{3}}t}\partial{x}\Delta W^{(2)}|{L^{2}L^{2}}^{2})$$

$$\leq C(|(u{1,\text{in}}){0}|{H^{1}}^{2}+H{1}^{2})A^{\frac{2}{3}}|\partial{x}\nabla W^{(2)}|{X{b}}^{2}\leq C(|(u{1,\text{in}}){0}|{H^{1}}^{2}+H{1}^{2})\epsilon^{2}A^{\frac{8}{3}}E{5,2}^{2}(t).$$

$$(8.21)$$

Collecting (8.20) and (8.21), we get the first result.

Estimate (8.19)2. By Lemma 4.4, Lemma 6.4, Lemma A.2 and Lemma A.4, we arrive

$$\begin{split} &|\mathbf{e}^{bA^{-\frac{1}{3}t}}\nabla(\mathbf{U}{1}\partial{x}u{3,\neq})|{L^{2}L^{2}}^{2} \ \leq &|\nabla\mathbf{U}{1}|{L^{\infty}L^{2}}^{2}|\mathbf{e}^{bA^{-\frac{1}{3}t}}\partial{x}u{3,\neq}|{L{t,x}L{y,z}^{\infty}}^{2} + |\mathbf{U}{1}|{L{t,y}L{z}^{2}}^{2}|\mathbf{e}^{bA^{-\frac{1}{3}t}}\nabla\partial{x}u{3,\neq}|{L{t,x,y}L{z}^{\infty}}^{2} \ \leq &C|\mathbf{U}{1}|{L^{\infty}H^{1}}^{2}\left(|\mathbf{e}^{bA^{-\frac{1}{2}t}}\nabla(\partial{x},\partial{z}-\kappa\partial{y})\partial{x}u{3,\neq}|{L^{2}L^{2}}^{2} + |\mathbf{e}^{bA^{-\frac{1}{3}t}}(\partial{x},\partial{z}-\kappa\partial{y})\nabla\partial{x}u{3,\neq}|{L^{2}L^{2}}^{2}\right) \ \leq &C|\mathbf{U}{1}|{L^{\infty}H^{1}}^{2}\left(|\mathbf{e}^{bA^{-\frac{1}{3}t}}\nabla(\partial{x},\partial{z}-\kappa\partial{y})\partial{x}u{3,\neq}|{L^{2}L^{2}}^{2} + |\kappa|{H^{3}}^{2}|\mathbf{e}^{bA^{-\frac{1}{3}t}}\partial{x}\partial{y}u{3,\neq}|{L^{2}L^{2}}^{2}\right) \ \leq &C\left(|(u{1,\mathrm{in}}){0}|{H^{1}}^{2} + H{1}^{2}\right)AE{5,2}^{2}(t), \end{split}$$

where we use Lemma 8.1 and (∂^z − κ∂y)∇∂xu3,̸⁼ = ∇(∂^z − κ∂y)∂xu3,̸⁼ + ∇κ∂y∂xu3,̸=.

Estimate (8.19)3. Due to Lemma 4.4, Lemma 6.4, Lemma A.2 and Lemma A.3, for j ∈ {2, 3} and α ∈ ( , 4 ), there holds

$$\begin{split} |\mathbf{e}^{bA^{-\frac{1}{3}t}}\mathbf{U}{1}\partial{x}^{2}u{j,\neq}|{L^{2}L^{2}}^{2} \leq &|\mathbf{U}{1}|{L{t}^{\infty}L{y}^{\infty}L{z}^{2}}^{2}|\mathbf{e}^{bA^{-\frac{1}{3}t}}\partial{x}^{2}u{j,\neq}|{L{t}^{2}L{z}^{\infty}L{x,y}^{2}}^{2} \ \leq &C|\mathbf{U}{1}|{L^{\infty}H^{1}}^{2}|\mathbf{e}^{bA^{-\frac{1}{3}t}}\partial{x}^{2}u{j,\neq}|{L^{2}L^{2}}^{2-2\alpha}|\mathbf{e}^{bA^{-\frac{1}{3}t}}\nabla\partial{x}^{2}u{j,\neq}|{L^{2}L^{2}}^{2\alpha} \ \leq &C\left(|(u{1,\mathrm{in}}){0}|{H^{1}}^{2}+H{1}^{2}\right)A^{\frac{1}{3}(2\alpha+1)}E{5,2}^{2}(t). \end{split}$$

Estimate (8.19)4. According to P1,² = ∂yU1∂xu2,̸⁼ + ∂zU1∂xu3,̸⁼ in (8.4), using Lemma 4.4, Lemma 6.4 and Lemma A.4, for j ∈ {2, 3}, we get

$$\begin{split} &|\mathbf{e}^{bA^{-\frac{1}{3}}t}P{1,2}|{L^{2}L^{2}}^{2} \leq |\mathbf{e}^{bA^{-\frac{1}{3}}t}\partial{j}\mathbf{U}{1}\partial{x}u{j,\neq}|{L^{2}L^{2}}^{2} \leq |\partial{j}\mathbf{U}{1}|{L{t}^{\infty}L{y,z}^{2}}^{2}|\mathbf{e}^{bA^{-\frac{1}{3}}t}\partial{x}u{j,\neq}|{L{t}^{2}L{y,z}^{\infty}L{x}^{2}}^{2} \ \leq &C|\mathbf{U}{1}|{L^{\infty}H^{1}}^{2}|\mathbf{e}^{bA^{-\frac{1}{3}}t}(\partial{x},\partial{z}-\kappa\partial{y})\partial{x}u{j,\neq}|{L^{2}L^{2}}|\mathbf{e}^{bA^{-\frac{1}{3}}t}\nabla(\partial{x},\partial{z}-\kappa\partial{y})\partial{x}u{j,\neq}|{L^{2}L^{2}} \ \leq &C\left(|(u{1,\mathrm{in}}){0}|{H^{1}}^{2}+H{1}^{2}\right)A^{\frac{2}{3}}E_{5,2}^{2}(t). \end{split}$$

□

Lemma 8.4. Under the assumptions of Theorem 1.1 and (2.5), it holds that

$$|\mathbf{e}^{bA^{-\frac{1}{3}}t}\nabla g{3,1}|{L^{2}L^{2}}^{2} \leq CA^{\frac{5}{3}}\epsilon^{2}\left(E{5,1}^{2}(t) + E{5,2}^{2}(t)\right),$$

$$|\mathbf{e}^{bA^{-\frac{1}{3}}t}\nabla (g{2,1} + \kappa g{3,1})|{L^{2}L^{2}}^{2} + |\mathbf{e}^{bA^{-\frac{1}{3}}t}\partial{x}g{3,1}|{L^{2}L^{2}}^{2}$$

$$+ |\mathbf{e}^{bA^{-\frac{1}{3}}t}\nabla g{2,2}|{L^{2}L^{2}}^{2} + |\mathbf{e}^{bA^{-\frac{1}{3}}t}\nabla g{3,2}|{L^{2}L^{2}}^{2} \leq C\epsilon^{2}AE_{5,2}^{2}(t).$$

$$(8.22)$$

Proof. Estimate (8.22)1. By Lemma 4.1, Lemma A.2 and ∂zu3,⁰ = −∂yu2,0, we get

$$|\nabla(u{2,0}, u{3,0})|_{L^2L^{\infty}}^2 \le CA\epsilon^2, \tag{8.23}$$

which along with (3.19) and

$$|\nabla u{3,\neq}|{L^{2}}^{2} \leq |u{3,\neq}|{L^{2}} |\Delta u{3,\neq}|{L^{2}} \leq A^{\frac{2}{3}} |\partial{x}^{2} u{3,\neq}|{L^{2}}^{2} + A^{-\frac{2}{3}} |\Delta u{3,\neq}|_{L^{2}}^{2}$$

shows that

$$\begin{split} &|\mathbf{e}^{bA^{-\frac{1}{3}}t}\nabla g{3,1}|{L^{2}L^{2}}^{2} = |\mathbf{e}^{bA^{-\frac{1}{3}}t}\nabla\left(u{2,0}\partial{y}u{3,\neq} + u{3,0}\partial{z}u{3,\neq}\right)|{L^{2}L^{2}}^{2} \ \leq &|\nabla(u{2,0},u{3,0})|{L^{2}L^{\infty}}^{2}|\mathbf{e}^{bA^{-\frac{1}{3}}t}\nabla u{3,\neq}|{L^{\infty}L^{2}}^{2} + |(u{2,0},u{3,0})|{L^{\infty}L^{\infty}}^{2}|\mathbf{e}^{bA^{-\frac{1}{3}}t}\triangle u{3,\neq}|{L^{2}L^{2}}^{2} \ \leq &CA\epsilon^{2}|\nabla u{3,\neq}|{X{b}}^{2} \leq CA^{\frac{5}{3}}\epsilon^{2}\left(|\partial{x}^{2}u{3,\neq}|{X{b}}^{2} + A^{-\frac{4}{3}}|\triangle u{3,\neq}|{X{b}}^{2}\right) \leq CA^{\frac{5}{3}}\epsilon^{2}\left(E{5,2}^{2}(t) + E_{5,1}^{2}(t)\right). \end{split}$$

Estimate (8.22)2. As g2,¹ + κg3,¹ = (u2,0∂^y + u3,0∂z) W − u3,̸=(u2,0∂^y + u3,0∂z)κ, by (3.19) and (8.23), we get

$$|e^{bA^{-\frac{1}{3}}t}\nabla(g{2,1}+\kappa g{3,1})|{L^{2}L^{2}}^{2} \leq CA\epsilon^{2}\left(|e^{bA^{-\frac{1}{3}}t}\nabla W|{Y{0}}^{2} + |e^{bA^{-\frac{1}{3}}t}u{3,\neq}\nabla\kappa|{Y{0}}^{2}\right).$$

Direct calculations show that

$$|\mathbf{e}^{bA^{-\frac{1}{3}}t}\nabla W|_{Y_0}^2 \le |\mathbf{e}^{bA^{-\frac{1}{3}}t}\partialx\nabla W|{Y_0}^2 \le |\partialx\nabla W|{Xb}^2 \le E{5,2}^2(t).$$

On the other hand, using Lemma 8.1, one obtains

$$\begin{aligned} &|u{3,\neq}\nabla\kappa|{L^{2}} \leq |u{3,\neq}|{L^{2}} |\nabla\kappa|{L^{\infty}} \leq C |\partial{x}^{2}u{3,\neq}|{L^{2}}, \ &|\nabla(u{3,\neq}\nabla\kappa)|{L^{2}} \leq |u{3,\neq}\nabla\kappa|{H^{1}} \leq C |u{3,\neq}|{H^{1}} |\nabla\kappa|{H^{2}} \leq C |\nabla\partial{x}^{2}u{3,\neq}|{L^{2}}, \end{aligned}$$

which indicates that

$$\begin{split} | \mathrm{e}^{bA^{-\frac{1}{3}t}} u{3,\neq} \nabla \kappa |{Y_0}^2 & \leq C \left( | \mathrm{e}^{bA^{-\frac{1}{3}t}} \partialx^2 u{3,\neq} |_{L^{\infty}L^2}^2 + \frac{1}{A} | \mathrm{e}^{bA^{-\frac{1}{3}t}} \nabla \partialx^2 u{3,\neq} |_{L^2L^2}^2 \right) \ & \leq C | \partialx^2 u{3,\neq} |_{Xb}^2 \leq C E{5,2}^2(t). \end{split}$$

This gives

$$|e^{bA^{-\frac{1}{3}}t}\nabla(g{2,1}+\kappa g{3,1})|{L^{2}L^{2}}^{2} \le C\epsilon^{2}AE{5,2}^{2}(t).$$ (8.24)

Recall that ∂xg3,¹ = (u2,0∂^y + u3,0∂z)∂xu3,̸=. By (3.19), we have

$$|\mathbf{e}^{bA^{-\frac{1}{3}}t}\partialx g{3,1}|{L^2L^2}^2 \le |(u{2,0}, u{3,0})|{L^{\infty}L^{\infty}}^2 |\mathbf{e}^{bA^{-\frac{1}{3}}t}\nabla \partialx u{3,\neq}|{L^2L^2}^2 \le C\epsilon^2 A E{5,2}^2(t). \tag{8.25}$$

For g2,² = u̸⁼ · ∇u2,⁰ and j ∈ {2, 3}, it follows from Lemma A.5 and Lemma 4.1 that

$$\begin{split} |\nabla g{2,2}|{L^{2}} = &|\nabla \left(u{j,\neq}\partial{j}u{2,0}\right)|{L^{2}} \leq C|\partial{j}u{2,0}|{H^{1}}\left(|u{j,\neq}|{H^{1}} + |(\partial{z} - \kappa\partial{y})u{j,\neq}|{H^{1}}\right) \ \leq &C|\nabla u{2,0}|{H^{1}}\left(|\nabla\partial{x}^{2}u{j,\neq}|{L^{2}} + |\nabla\partial{x}(\partial{z} - \kappa\partial{y})u{j,\neq}|_{L^{2}}\right) \end{split}$$

and

$$|\mathbf{e}^{bA^{-\frac{1}{3}}t}\nabla g{2,2}|{L^{2}L^{2}}^{2} \leq C\epsilon^{2} \left(|\mathbf{e}^{bA^{-\frac{1}{3}}t}\nabla\partial{x}^{2}u{j,\neq}|{L^{2}L^{2}}^{2} + |\mathbf{e}^{bA^{-\frac{1}{3}}t}\nabla\partial{x}(\partial{z} - \kappa\partial{y})u{j,\neq}|{L^{2}L^{2}}^{2}\right) \ \leq C\epsilon^{2}A \left(|\partial{x}^{2}u{j,\neq}|{X{b}}^{2} + |\partial{x}(\partial{z} - \kappa\partial{y})u{j,\neq}|{X{b}}^{2}\right) \leq C\epsilon^{2}AE_{5,2}^{2}(t).$$

$$(8.26)$$

For g3,² = u̸⁼ · ∇u3,0, we rewrite it into

$$g{3,2} = (u{2,\neq}\partialy + u{3,\neq}\partialz) u{3,0} = W \partialy u{3,0} + u_{3,\neq} (\partialz u{3,0} - \kappa \partialy u{3,0}),$$

which implies that

$$|\mathbf{e}^{bA^{-\frac{1}{3}}t}\nabla g{3,2}|{L^{2}L^{2}}^{2} \leq |\mathbf{e}^{bA^{-\frac{1}{3}}t}\nabla (W\partial{y}u{3,0})|{L^{2}L^{2}}^{2} + |\mathbf{e}^{bA^{-\frac{1}{3}}t}\nabla \left(u{3,\neq}(\partial{z}u{3,0} - \kappa\partial{y}u{3,0})\right)|_{L^{2}L^{2}}^{2}. \tag{8.27}$$

By Lemma A.2, Lemma A.3, Lemma 4.1 and ∂zu3,⁰ = −∂yu2,0, we get

$$|e^{bA^{-\frac{1}{3}t}}\nabla(W\partial{y}u{3,0})|_{L^{2}L^{2}}^{2}$$

$$\leq |e^{bA^{-\frac{1}{3}t}}\nabla W|{L^{\infty}L^{2}}^{2}|\partial{y}u{3,0}|{L^{2}L^{\infty}}^{2} + |e^{bA^{-\frac{1}{3}t}}W|{L^{\infty}{t,y}L^{2}{x,z}}^{2}|\nabla\partial{y}u{3,0}|{L^{2}{t,y}L^{\infty}{z}}^{2}$$

$$\leq C\left(|u{3,0}|{L^{2}H^{2}}^{2} + |u{2,0}|{L^{2}H^{3}}^{2}\right)|e^{bA^{-\frac{1}{3}t}}\nabla\partial{x}W|{L^{\infty}L^{2}}^{2} \leq C\epsilon^{2}AE_{5,2}^{2}(t).$$ (8.28)

As U2(0) = 0, using Lemma 4.5, there holds

$$|\Delta \mathbf{U}2|{L^2} \le |\mathbf{U}2(t)|{H^2} \le \int_0^t |\partial_t \mathbf{U}2(s)|{H^2} ds \le C\epsilon t.$$

On the other hand, ∥△U2(t)∥H² ≤ CAϵ. Therefore

$$|\triangle \mathbf{U}2|{L^2} \le C\epsilon A \min{A^{-1}t, 1},\,$$

which implies that

$$|\kappa|_{H^1} \le CA^{-1} |\Delta \mathbf{U}2|{L^2} \le C\epsilon \min{A^{-1}t, 1}.$$ (8.29)

From this, along with Lemma 4.1 and Lemma 8.1, one obtains

$$|\kappa \nabla u{3,0}|{H^{1}} \leq C|\kappa|{H^{2}} (|\nabla u{3,0}|{L^{2}} + |\Delta u{3,0}|_{L^{2}})$$

$$\leq C|\kappa|{H^{1}}^{\frac{1}{2}} |\kappa|{H^{3}}^{\frac{1}{2}} (|\nabla u{3,0}|{L^{2}} + |\Delta u{3,0}|{L^{2}})$$

$$\leq C\epsilon \min{A^{-1}t, 1}^{\frac{1}{2}} (|\nabla u{3,0}|{L^{2}} + |\Delta u{3,0}|{L^{2}})$$

$$\leq C\epsilon \left(|\nabla u{3,0}|{L^{2}} + |\min\left(A^{-\frac{2}{3}} + A^{-1}t, 1\right)^{\frac{1}{2}} \Delta u{3,0}|{L^{2}}\right) \leq C\epsilon^{2}.$$ (8.30)

Due to Lemma 4.1, Lemma A.5, (8.30) and ∂zu3,⁰ = −∂yu2,0, we arrive

$$|\mathbf{e}^{bA^{-\frac{1}{3}}t}\nabla\left(u{3,\neq}(\partial{z}u{3,0}-\kappa\partial{y}u{3,0})\right)|{L^{2}L^{2}}^{2}$$

$$\leq C\left(|\partial{z}u{3,0}|{L^{\infty}H^{1}}^{2}+|\kappa\nabla u{3,0}|{L^{\infty}H^{1}}^{2}\right)\left(|\mathbf{e}^{bA^{-\frac{1}{3}}t}u{3,\neq}|{L^{2}H^{1}}^{2}+|\mathbf{e}^{bA^{-\frac{1}{3}}t}(\partial{z}-\kappa\partial{y})u{3,\neq}|_{L^{2}H^{1}}^{2}\right)$$

$$\leq C\epsilon^{2}\left(|\mathbf{e}^{bA^{-\frac{1}{3}}t}\nabla\partial{x}^{2}u{3,\neq}|{L^{2}L^{2}}^{2}+|\mathbf{e}^{bA^{-\frac{1}{3}}t}\nabla(\partial{z}-\kappa\partial{y})u{3,\neq}|{L^{2}L^{2}}^{2}\right) \leq C\epsilon^{2}AE{5,2}^{2}(t).$$

$$(8.31)$$

Combining (8.28) with (8.31), we get from (8.27) that

$$|e^{bA^{-\frac{1}{3}}t}\nabla g{3,2}|{L^{2}L^{2}}^{2} \le C\epsilon^{2}AE_{5,2}^{2}(t).$$ (8.32)

Then (8.22)² follows from (8.24), (8.25), (8.26) and (8.32).

Lemma 8.5. Under the assumptions of Theorem 1.1 and (2.5), it holds that

$$\begin{split} |\mathrm{e}^{bA^{-\frac{1}{3}}t}\nabla G^{(1)}|{L^{2}L^{2}}^{2} \leq & CA^{\frac{1}{6}+\alpha}E{2}^{4} + CA^{\frac{1}{2}+\frac{2}{3}\alpha}E{2}^{4} + CA^{\frac{\alpha}{3}+\frac{1}{2}}E{2}^{2}E{5}^{2} \ & + C\left(|(u{1,\mathrm{in}}){0}|{H^{1}}^{2} + H{1}^{2}\right)A^{\frac{2}{3}}E{5,2}^{2}(t) + CA^{\frac{1}{3}}E{2}^{2}, \ |\mathrm{e}^{bA^{-\frac{1}{3}}t}\nabla G^{(2)}|{L^{2}L^{2}}^{2} \leq & C\epsilon^{2}AE_{5,2}^{2}(t), \end{split}$$

where α ∈ ( 1 , 4 ).

Proof. Due to (8.3) and Lemma 3.2, there holds

$$|\mathbf{e}^{bA^{-\frac{1}{3}}t}\nabla G{2,3}|{L^{2}L^{2}}^{2} \leq |\mathbf{e}^{bA^{-\frac{1}{3}}t}\nabla (u{\neq}\cdot\nabla u{2,\neq}){\neq}|{L^{2}L^{2}}^{2} \leq CA^{\frac{1}{2}+\frac{2}{3}\alpha}E_{2}^{4}.$$

For G3,³ = (u̸⁼ · ∇u3,̸=)̸⁼ in (8.3), using Lemma 3.2 and Lemma 8.1, we have

$$|\mathbf{e}^{bA^{-\frac{1}{3}}t}\nabla\kappa G{3,3}|{L^{2}L^{2}}^{2} \le C|\kappa|{L^{\infty}H^{3}}^{2}|\mathbf{e}^{bA^{-\frac{1}{3}}t}u{\neq} \cdot \nabla u{3,\neq}|{L^{2}L^{2}}^{2} \le CA^{\frac{2}{3}}E_{2}^{4}.$$ (8.33)

On the other hand, by Lemma 8.1 and (8.29), one deduces

$$|\kappa \nabla G{3,3}|{L^{2}} \leq C |\kappa|{H^{2}} |\nabla (u{\neq} \cdot \nabla u{3,\neq})|{L^{2}} \leq C |\kappa|{H^{1}}^{\frac{1}{2}} |\kappa|{H^{3}}^{\frac{1}{2}} |\nabla (u{\neq} \cdot \nabla u{3,\neq})|{L^{2}}$$ $$\leq C (A^{-1}t)^{\frac{1}{2}} |\nabla (u{\neq} \cdot \nabla u{3,\neq})|{L^{2}},$$

which along with Lemma 3.2 and (A − ^{1 3} t) ² ≤ Ce (2a−b)A − 1 3 t implies that

$$|\mathbf{e}^{bA^{-\frac{1}{3}}t}\kappa\nabla G{3,3}|{L^{2}L^{2}}^{2} \leq CA^{-\frac{2}{3}}|(A^{-\frac{1}{3}}t)^{\frac{1}{2}}\mathbf{e}^{bA^{-\frac{1}{3}}t}\nabla(u{\neq}\cdot\nabla u{3,\neq})|{L^{2}L^{2}}^{2} \ \leq CA^{-\frac{2}{3}}|\mathbf{e}^{2aA^{-\frac{1}{3}}t}\nabla(u{\neq}\cdot\nabla u{3,\neq})|{L^{2}L^{2}}^{2} \leq CA^{\frac{\alpha}{3}+\frac{1}{2}}E{2}^{2}E{5}^{2} + CA^{\frac{2}{3}}E_{2}^{4}.$$

$$(8.34)$$

It follows from (8.33) and (8.34) that

$$|e^{bA^{-\frac{1}{3}}t}\nabla(\kappa G{3,3})|{L^{2}L^{2}}^{2} \le CA^{\frac{2}{3}}E{2}^{4} + CA^{\frac{\alpha}{3} + \frac{1}{2}}E{2}^{2}E_{5}^{2}.$$

□

Using Lemma 3.2, we note that

$$|\mathbf{e}^{bA^{-\frac{1}{3}t}}P{1,1}|{L^{2}L^{2}}^{2} \leq C(|\mathbf{e}^{bA^{-\frac{1}{3}t}}\partial{x}(u{\neq}\cdot\nabla u{1,\neq})|{L^{2}L^{2}}^{2} + |\mathbf{e}^{bA^{-\frac{1}{3}t}}\partial{y}(u{\neq}\cdot\nabla u{2,\neq})|{L^{2}L^{2}}^{2} + |\mathbf{e}^{bA^{-\frac{1}{3}t}}\partial{z}(u{\neq}\cdot\nabla u{3,\neq})|{L^{2}L^{2}}^{2}) \leq CA^{\frac{1}{6}+\alpha}E{2}^{4} + CA^{\frac{1}{2}+\frac{2}{3}\alpha}E{2}^{4}.$$

From this, along with (8.4), (8.19)⁴ and Lemma 8.1, one obtains

$$|e^{bA^{-\frac{1}{3}}t}\nabla\left((\partial{y}+\kappa\partial{z})\triangle^{-1}(P{1,1}+P{1,2}+P{1,3})\right)|{L^{2}L^{2}}^{2}$$

$$\leq C\left(|e^{bA^{-\frac{1}{3}}t}P{1,1}|{L^{2}L^{2}}^{2}+|e^{bA^{-\frac{1}{3}}t}P{1,2}|{L^{2}L^{2}}^{2}+|e^{bA^{-\frac{1}{3}}t}P{1,3}|{L^{2}L^{2}}^{2}\right)$$

$$\leq CA^{\frac{1}{6}+\alpha}E{2}^{4}+CA^{\frac{1}{2}+\frac{2}{3}\alpha}E{2}^{4}+C(|(u{1,\text{in}}){0}|{H^{1}}^{2}+H{1}^{2})A^{\frac{2}{3}}E{5,2}^{2}(t)+CA^{\frac{1}{3}}E{2}^{2}.$$

$$(8.35)$$

Based on the above estimates, we conclude that

$$|\mathbf{e}^{bA^{-\frac{1}{3}}t}\nabla G^{(1)}|{L^{2}L^{2}}^{2} \leq CA^{\frac{1}{6}+\alpha}E{2}^{4} + CA^{\frac{1}{2}+\frac{2}{3}\alpha}E{2}^{4} + CA^{\frac{\alpha}{3}+\frac{1}{2}}E{2}^{2}E{5}^{2} + C\left(|(u{1,\mathrm{in}}){0}|{H^{1}}^{2} + H{1}^{2}\right)A^{\frac{2}{3}}E{5,2}^{2}(t) + CA^{\frac{1}{3}}E_{2}^{2}.$$ (8.36)

For G(2) = g2,¹ + g2,² + κ(g3,¹ + g3,2) − 2(∂^y + κ∂z)△⁻¹ (∂yg2,² + ∂zg3,2), by Lemma 8.1 and Lemma 8.4, there holds

$$|\mathbf{e}^{bA^{-\frac{1}{3}t}}\nabla G^{(2)}|{L^{2}L^{2}}^{2} \leq C\left(|\mathbf{e}^{bA^{-\frac{1}{3}t}}\nabla (g{2,1} + \kappa g{3,1})|{L^{2}L^{2}}^{2} + |\mathbf{e}^{bA^{-\frac{1}{3}t}}\nabla g{2,2}|{L^{2}L^{2}}^{2} + |\mathbf{e}^{bA^{-\frac{1}{3}t}}\nabla g{3,2}|{L^{2}L^{2}}^{2}\right) \leq C\epsilon^{2}AE_{5,2}^{2}(t).$$

$$(8.37)$$

Combining (8.36) with (8.37), we finish the proof.

Lemma 8.6. Under the assumptions of Theorem 1.1 and (2.5), there exists a positive constant A⁶ independent of t and A, such that if A ≥ A6, then there holds

$$E4(t) + E{5,1}(t) \le C(|(u{\rm in}){\neq}|_{H^2} + 1) =: E_4 + E_5.$$

Proof. Step I. Estimate ∥△u3,̸⁼ − 2∂xW(3)∥^Xb . Applying Proposition A.1 to (8.9), we obtain

$$|\Delta u{3,\neq} - 2\partial{x}W^{(3)}|{X{b}}^{2}$$

$$\leq C(|(\Delta u{3,\text{in}}){\neq}|{L^{2}}^{2} + A^{\frac{1}{3}}|e^{bA^{-\frac{1}{3}}t}\Delta V\partial{x}u{3,\neq}|{L^{2}L^{2}}^{2}$$

$$+ A^{\frac{1}{3}}|e^{bA^{-\frac{1}{3}}t}\partial{z}(\partial{y}V\partial{x}W)|{L^{2}L^{2}}^{2} + \frac{1}{A}|e^{bA^{-\frac{1}{3}}t}\nabla(\mathbf{U}{1}\partial{x}u{3,\neq})|{L^{2}L^{2}}^{2}$$

$$+ \frac{1}{A}|e^{bA^{-\frac{1}{3}}t}\nabla(g{2,2} + g{3,1} + g{3,2} + G{3,3})|{L^{2}L^{2}}^{2} + \frac{1}{A}|e^{bA^{-\frac{1}{3}}t}(P{1,1} + P{1,2} + P{1,3})|_{L^{2}L^{2}}^{2}),$$

$$(8.38)$$

where we use ∂xW(3)(0) = 0. Due to ∥△V ∥L^∞ ≤ CA⁻¹∥△U2∥H² ≤ C, there holds

$$|\mathbf{e}^{bA^{-\frac{1}{3}}t} \triangle V \partialx u{3,\neq}|{L^2L^2}^2 \leq |\triangle V|{L^{\infty}L^{\infty}}^2 |\mathbf{e}^{bA^{-\frac{1}{3}}t} \partialx u{3,\neq}|_{L^2L^2}^2 \ \leq C |\mathbf{e}^{bA^{-\frac{1}{3}}t} \partialx^2 u{3,\neq}|_{L^2L^2}^2 \leq C A^{\frac{1}{3}} |\partialx^2 u{3,\neq}|_{Xb}^2 \leq C A^{\frac{1}{3}} E{5,2}^2(t).$$

$$(8.39)$$

□

Moreover, for ∥∂yV ∥H³ ≤ C (1 + A⁻¹∥△U2∥H² ) ≤ C, one deduces

$$|\partial{z}(\partial{y}V\partial{x}W)|{L^{2}} \leq |\partial{y}V|{H^{3}}|\partial{x}W|{H^{1}} \leq C|\nabla\partial{x}W|{L^{2}},$$

$$|e^{bA^{-\frac{1}{3}}t}\partial{z}(\partial{y}V\partial{x}W)|{L^{2}L^{2}}^{2} \leq C|e^{bA^{-\frac{1}{3}}t}\nabla\partial{x}W|{L^{2}L^{2}}^{2} \leq CA^{\frac{1}{3}}E_{5,2}^{2}(t).$$ (8.40)

Recall that G3,³ = (u̸⁼ · ∇u3,̸=)̸⁼ in (8.3), by Lemma 3.2, and we get

$$|\mathbf{e}^{bA^{-\frac{1}{3}}t}\nabla G{3,3}|{L^{2}L^{2}}^{2} \le C\left(A^{\frac{\alpha}{3}+\frac{7}{6}}E{2}^{2}E{5}^{2}+A^{\frac{4}{3}}E_{2}^{4}\right),\tag{8.41}$$

where α ∈ ( 2 , 4 ).

Collecting (8.39)-(8.41), and using (8.35), Lemma 8.3 and Lemma 8.4, we get from (8.38) that

$$|\Delta u_{3,\neq} - 2\partialx W^{(3)}|{Xb}^2 \le C\left(|(u{\rm in}){\neq}|{H^2}^2 + A^{\frac{2}{3}}E_{5,2}^2(t) + A^{\frac{\alpha}{3} + \frac{1}{6}}E_2^2 E_5^2 + A^{\frac{1}{3}}E2^4 + \epsilon^2 A^{\frac{2}{3}}E{5,1}^2(t) + 1\right)$$

$$(8.42)$$

provided with A ≥ max{A5, E 24 5−6α 2 , E 24 3−4α 2 ,(∥(u1,in)0∥ 2 ^H1 + H² 1 ) ² } =: A6,1.

Step II. Estimate ∥△W(1)∥^Xb . By (8.6), there holds

$$\widetilde{\mathcal{L}_V}W + \frac{\mathbf{U}_1\partial_x W}{A} + \frac{G^{(1)} + G^{(2)}}{A} = \left(\partialt \kappa - \frac{\triangle \kappa}{A}\right)u{3,\neq} - \frac{2\nabla \kappa \cdot \nabla u_{3,\neq}}{A}.$$

Using the following decomposition

$$\nabla \kappa \cdot \nabla u_{3,\neq} = \rho1 \nabla V \cdot \nabla u{3,\neq} + \rho_2 (\partial_z - \kappa \partialy) u{3,\neq}$$

and W = W(1) + 1 AW(2) , we get

$$\widetilde{\mathcal{L}}{V}W^{(1)} + \frac{\mathbf{U}{1}\partial{x}W}{A} + \frac{G^{(1)} + G^{(2)}}{A} - \left(\partial{t}\kappa - \frac{\Delta\kappa}{A}\right)u_{3,\neq}$$

$$= -\frac{2}{A}\left(\rho{1}\nabla V \cdot \nabla u{3,\neq} + \rho{2}(\partial{z} - \kappa\partial{y})u{3,\neq}\right) - \frac{1}{A}\widetilde{\mathcal{L}}{V}W^{(2)} = -\frac{2}{A}\rho{2}(\partial{z} - \kappa\partial{y})u{3,\neq} - \frac{1}{A}J{11}, \tag{8.43}$$

where

$$J_{11} = \widetilde{\mathcal{L}_V} W^{(2)} + 2\rho1 \nabla V \cdot \nabla u{3, \neq}. \tag{8.44}$$

Applying Proposition A.4 to (8.43), one deduces

$$\begin{split} |\triangle W^{(1)}|{X{b}}^{2} \leq & C\left(|\triangle W^{(1)}(0)|{L^{2}}^{2} + A^{-1}|e^{bA^{-\frac{1}{3}}t}\nabla\left(\rho{2}(\partial{z} - \kappa\partial{y})u{3,\neq}\right)|{L^{2}L^{2}}^{2} \ & + A^{-1}|e^{bA^{-\frac{1}{3}}t}\nabla J{11}|{L^{2}L^{2}}^{2} + A|e^{bA^{-\frac{1}{3}}t}\nabla\left((\partial{t}\kappa - A^{-1}\triangle\kappa)u{3,\neq}\right)|{L^{2}L^{2}}^{2} \ & + A^{-1}|e^{bA^{-\frac{1}{3}}t}\nabla(\mathbf{U}{1}\partial{x}W)|{L^{2}L^{2}}^{2} + A^{-1}|e^{bA^{-\frac{1}{3}}t}\nabla(G^{(1)} + G^{(2)})|_{L^{2}L^{2}}^{2}\right). \end{split}$$ (8.45)

It follows from Lemma 8.1 that ∥ρ2∥H² ≤ Cϵ and

$$|\partialt \kappa - A^{-1} \triangle \kappa|{H^1} \le |\partialt \kappa|{H^1} + CA^{-1} |\kappa|_{H^3} \le CA^{-1} \epsilon. \tag{8.46}$$

Hence we have

$$|\nabla \left(\rho_2(\partial_z - \kappa \partialy)u{3,\neq}\right)|_{L^2} \le C|\rho2|{H^2} |\nabla (\partial_z - \kappa \partialy)u{3,\neq}|_{L^2} \le C\epsilon |\nabla \partial_x(\partial_z - \kappa \partialy)u{3,\neq}|_{L^2}. \tag{8.47}$$ Besides, using (8.46) and Lemma A.5, there holds

$$|\nabla\left((\partialt \kappa - A^{-1} \triangle \kappa) u{3,\neq}\right)|_{L^2} \le C A^{-1} \epsilon \left(|\nabla \partialx^2 u{3,\neq}|_{L^2} + |\nabla \partial_x (\partial_z - \kappa \partialy) u{3,\neq}|_{L^2}\right). \tag{8.48}$$

For J¹¹ in (8.44), we rewrite it as follows:

$$J{11} = \mathcal{L}{V} W^{(2)} - 2(\partial{y} + \kappa \partial{z}) \triangle^{-1}(\partial{y} V \partial{x} W^{(2)}) + 2\rho{1} \nabla V \cdot \nabla u{3,\neq}$$

$$= -2(\partial{y} + \kappa \partial{z}) \triangle^{-1}(\partial{y} V \partial{x} W^{(2)}) + \rho{1} \nabla V \cdot \nabla u{3,\neq}$$

$$= -(\partial{y} + \kappa \partial{z}) \triangle^{-1}(\partial{y} V J{12}) - (\partial{y} + \kappa \partial{z}) \triangle^{-1}(\partial{y} V \rho{1} \triangle u{3,\neq}) + \rho{1} \nabla V \cdot \nabla u_{3,\neq},$$

$$(8.49)$$

where J¹² = 2∂xW(2) − ρ1△u3,̸⁼ = 2∂x(W(2) − ρ1W(3)) − ρ¹ △u3,̸⁼ − 2∂xW(3) . Using Lemma 8.1 and (8.12), we arrive

$$J_{12} \le C \left( |\partial_x (W^{(2)} - \rho1 W^{(3)})|{L^2} + \epsilon |\triangle u_{3,\neq} - 2\partialx W^{(3)}|{L^2} \right)$$

and

$$|(\partial_y + \kappa \partial_z) \triangle^{-1}(\partialy V J{12})|_{H^1} \le C \left( |\partial_y \triangle^{-1}(\partialy V J{12})|{H^1} + |\kappa|{H^3} |\partial_z \triangle^{-1}(\partialy V J{12})|_{H^1} \right)$$

$$\le C |J{12}|{L^2} \le C \left( |\partial_x (W^{(2)} - \rho1 W^{(3)})|{L^2} + \epsilon |\triangle u_{3,\neq} - 2\partialx W^{(3)}|{L^2} \right).$$

$$(8.50)$$

Thanks to ∇V · ∇u3,̸⁼ = ∂yV (∂^y + κ∂z)u3,̸=, there holds

$$\begin{aligned} &-(\partial_y + \kappa \partial_z) \triangle^{-1}(\partial_y V \rho1 \triangle u{3,\neq}) + \rho1 \nabla V \cdot \nabla u{3,\neq} \ &= -\left[ (\partial_y + \kappa \partial_z) \triangle^{-1}, \partial_y V \rho1 \right] \triangle u{3,\neq} \ &= -\left[ \partial_y \triangle^{-1}, \partial_y V \rho1 \right] \triangle u{3,\neq} - \kappa \left[ \partial_z \triangle^{-1}, \partial_y V \rho1 \right] \triangle u{3,\neq}. \end{aligned}$$

From this, along with (8.12), Lemma 8.1 and Lemma A.5, we get

$$| - (\partial{y} + \kappa \partial{z}) \triangle^{-1} (\partial{y} V \rho{1} \triangle u{3,\neq}) + \rho{1} \nabla V \cdot \nabla u{3,\neq} |{H^{1}}$$

$$\leq | \left[ \partial{y} \triangle^{-1}, \partial{y} V \rho{1} \right] \triangle u{3,\neq} |{H^{1}} + | \kappa |{L^{\infty}} | \left[ \partial{z} \triangle^{-1}, \partial{y} V \rho{1} \right] \triangle u{3,\neq} |_{H^{1}}$$

$$\leq C\epsilon \left( | \nabla \partial{x}^{2} u{3,\neq} |{L^{2}} + | \partial{x} (\partial{z} - \kappa \partial{y}) \nabla u{3,\neq} |{L^{2}} \right).$$

$$(8.51)$$

Then we conclude from (8.49), (8.50) and (8.51) that

$$|\nabla J{11}|{L^{2}} \leq C \left( |\partial{x}(W^{(2)} - \rho{1}W^{(3)})|{L^{2}} + \epsilon |\Delta u{3,\neq} - 2\partial{x}W^{(3)}|{L^{2}} \right) + C\epsilon \left( |\nabla \partial{x}^{2}u{3,\neq}|{L^{2}} + |\nabla \partial{x}(\partial{z} - \kappa \partial{y})u{3,\neq}|{L^{2}} \right).$$

$$(8.52)$$

Collecting (8.42), (8.47), (8.48), (8.52) and (iii) of Lemma 8.2, when A ≥ A6,1, one obtains

$$A^{-1} | e^{bA^{-\frac{1}{3}}t} \nabla J{11} |{L^{2}L^{2}}^{2} + A^{-1} | e^{bA^{-\frac{1}{3}}t} \nabla \left( \rho{2}(\partial{z} - \kappa \partial{y}) u{3,\neq} \right) |_{L^{2}L^{2}}^{2}$$

$$+ A | e^{bA^{-\frac{1}{3}}t} \nabla \left( (\partial{t} \kappa - \nu \triangle \kappa) u{3,\neq} \right) |_{L^{2}L^{2}}^{2}$$

$$\leq CA^{-\frac{2}{3}} \left( | \partial{x} \left( W^{(2)} - \rho{1} W^{(3)} \right) |{X{b}}^{2} + \epsilon^{2} | \triangle u{3,\neq} - 2 \partial{x} W^{(3)} |{X{b}}^{2} \right)$$

$$+ C\epsilon^{2} \left( | \partial{x}^{2} u{3,\neq} |{X{b}}^{2} + | \partial{x} (\partial{z} - \kappa \partial{y}) u{3,\neq} |{X{b}}^{2} \right)$$

$$\leq C\epsilon^{2} \left( | (u{\text{in}}){\neq} |{H^{2}}^{2} + E{5,1}^{2}(t) + E_{5,2}^{2}(t) + 1 \right),$$

Combining above with Lemma 8.3 and Lemma 8.5, when

$$A \ge C \max{\epsilon^{-6}(|(u_{1,\text{in}})0|{H^1}^2 + H_1^2)^3, (\epsilon^{-1}E_2)^{\frac{12}{3-2\alpha}}, (E_2E5)^{\frac{12}{3-2\alpha}}, A{6,1}} =: A_{6,2},$$

we get from (8.45) that

$$|\triangle W^{(1)}|_{Xb}^2 \leq C\left(|(u{\rm in}){\neq}|{H^2}^2 + \epsilon^2 E{5,1}^2(t) + \epsilon^2 E{5,2}^2(t) + 1\right) + CA^{-\frac{1}{2}}\left(|(u_{1,\rm in})0|{H^1}^2 + H1^2\right) |\triangle W^{(1)}|{X_b}^2.$$

This implies that

$$|\Delta W^{(1)}|_{Xb}^2 \le C\left(|(u{\rm in}){\neq}|{H^2}^2 + \epsilon^2 E{5,1}^2(t) + \epsilon^2 E{5,2}^2(t) + 1\right). \tag{8.53}$$

Step III. Estimate ∥∂ 2 ^xuj,̸=∥^Xb +∥∂x(∂^z −κ∂y)uj,̸=∥^Xb . As W = W(1) + 1 AW(2) , for j ∈ {2, 3}, we rewrite (8.5) into

$$\mathcal{L}{V}u{j,\neq} = 2\partial{j}\triangle^{-1}(\partial{y}V\partial{x}W^{(1)}) + \frac{2}{A}\partial{j}\triangle^{-1}(\partial{y}V\partial{x}W^{(2)}) - \frac{\mathbf{U}{1}\partial{x}u{j,\neq}}{A} - \frac{g{j,1} + g{j,2} + G{j,3}}{A} + \frac{2\partial{j}\triangle^{-1}(\partial{y}g{2,2} + \partial{z}g{3,2})}{A} + \frac{2\partial{j}\triangle^{-1}(P{1,1} + P{1,2} + P_{1,3})}{A}.$$

Applying Proposition A.3 to it, there holds

$$|\partial{x}u{j,\neq}|{X{b}}^{2} + |\partial{x}(\partial{z} - \kappa\partial{y})u{j,\neq}|{X{b}}^{2}$$

$$\leq C(|(u{\text{in}}){\neq}|{H^{2}}^{2} + |e^{bA^{-\frac{1}{3}}t}\partial{j}(\partial{y}V\partial{x}W^{(1)})|{L^{2}L^{2}}^{2} + A^{-\frac{5}{3}}|e^{bA^{-\frac{1}{3}}t}\partial{y}V\partial{x}^{2}W^{(2)}|{L^{2}L^{2}}^{2}$$

$$+ A^{-1}|e^{bA^{-\frac{1}{3}}t}\mathbf{U}{1}\partial{x}^{2}u{j,\neq}|{L^{2}L^{2}}^{2} + A^{-1}|e^{bA^{-\frac{1}{3}}t}\partial{x}(g{j,1} + G{j,3})|{L^{2}L^{2}}^{2}$$

$$+ A^{-1}|e^{bA^{-\frac{1}{3}}t}\nabla(g{2,2} + g{3,2})|{L^{2}L^{2}}^{2} + A^{-1}|e^{bA^{-\frac{1}{3}}t}(P{1,1} + P{1,2} + P{1,3})|_{L^{2}L^{2}}^{2}),$$

$$(8.54)$$

where we use ∥∂ x f∥L² + ∥∂x(∂^z − κ∂y)f∥L² ≤ C(1 + ∥κ∥L^∞)∥∂x∇f∥L² ≤ C∥∂x∇f∥L² .

By Lemma 4.5, we arrive

$$\begin{split} | \mathrm{e}^{bA^{-\frac{1}{3}}t} \partial_j (\partial_y V \partialx W^{(1)}) |{L^2 L^2}^2 &\leq C \left( | \nabla V |{L^\infty L^\infty}^2 + | \nabla^2 V |{L^\infty L^\infty}^2 \right) | \mathrm{e}^{bA^{-\frac{1}{3}}t} \nabla \partialx W^{(1)} |{L^2 L^2}^2 \ &\leq C \left( 1 + \frac{| \triangle \mathbf{U}2 |{L^\infty H^2}^2}{A} \right) | \triangle W^{(1)} |_{Xb}^2 \leq C | \triangle W^{(1)} |{X_b}^2 \end{split}$$

and

$$A^{-\frac{5}{3}} | \mathrm{e}^{bA^{-\frac{1}{3}}t} \partial_y V \partialx^2 W^{(2)} |{L^2 L^2}^2 \leq A^{-\frac{5}{3}} | \partialy V |{L^\infty L^\infty}^2 | \mathrm{e}^{bA^{-\frac{1}{3}}t} \partialx^2 W^{(2)} |{L^2 L^2}^2 \leq C A^{-\frac{4}{3}} | \partialx^2 W^{(2)} |{X_b}^2.$$

Due to Lemma 8.1 and Lemma 8.4, one obtains

$$|\mathbf{e}^{bA^{-\frac{1}{3}}t}\partialx g{j,1}|{L^2L^2}^2 \le C\left(|\mathbf{e}^{bA^{-\frac{1}{3}}t}\nabla(g{2,1}+\kappa g{3,1})|{L^2L^2}^2 + |\mathbf{e}^{bA^{-\frac{1}{3}}t}\partialx g{3,1}|{L^2L^2}^2\right) \le C\epsilon^2 A E{5,2}^2(t).$$

Moreover, it follows from Lemma 3.2 that

$$|e^{bA^{-\frac{1}{3}}t}\partialx G{j,3}|_{L^2L^2}^2 \le C|e^{2aA^{-\frac{1}{3}}t}\partialx (u{\neq}\cdot\nabla u{\neq})|{L^2L^2}^2 \le CA^{\frac{1}{6}+\alpha}E_2^4.$$

By combining the above estimates and using Lemma 8.3, Lemma 8.4 and (8.35), when

$$A \geq C \max{\epsilon^{\frac{-3}{1-\alpha}} \left( |(u_{1,\mathrm{in}})0|{H^1}^2 + H1^2 \right)^{\frac{3}{2(1-\alpha)}}, A{6,2} } =: A_6.$$

we get from (8.54) that

$$|\partialx u{j,\neq}|_{X_b}^2 + |\partial_x (\partial_z - \kappa \partialy) u{j,\neq}|_{X_b}^2$$

$$\leq C \left( |(u{\text{in}}){\neq}|{H^2}^2 + |\Delta W^{(1)}|{X_b}^2 + A^{-\frac{4}{3}} |\partialx^2 W^{(2)}|{Xb}^2 + \epsilon^2 E{5,2}^2(t) + 1 \right).$$ (8.55)

Step IV. Estimate E 2 5,1 (t) + E 2 5,2 (t) and E 2 4 (t). For W = W(1) + AW(2) , there holds

$$|\partialx \nabla W|{X_b} \leq \frac{1}{A} |\partialx \nabla W^{(2)}|{X_b} + |\partialx \nabla W^{(1)}|{X_b} \leq \frac{1}{A} |\partialx \nabla W^{(2)}|{Xb} + |\triangle W^{(1)}|{X_b},$$

which along with (8.55) gives

$$E{5,2}^2(t) \le C \left( |(u{\rm in}){\neq}|{H^2}^2 + |\Delta W^{(1)}|_{X_b}^2 + A^{-\frac{4}{3}} |\partialx^2 W^{(2)}|{Xb}^2 + \epsilon^2 E{5,2}^2(t) + A^{-2} |\partialx \nabla W^{(2)}|{X_b}^2 + 1 \right).$$

By (8.53) and (ii) of Lemma 8.2, the above inequality indicates that

$$E{5,2}^2(t) \leq C \left( |(u{\rm in}){\neq}|{H^2}^2 + \epsilon^2 E{5,1}^2(t) + 1 \right) + C \epsilon^2 E{5,2}^2(t).$$

Taking ϵ small enough satisfying Cϵ² ≤ 2 , we have

$$E{5,2}^2(t) \le C\left(|(u{\rm in}){\ne}|{H^2}^2 + \epsilon^2 E_{5,1}^2(t) + 1\right).$$ (8.56)

Using (8.42) and (i) of Lemma 8.2, one deduces

$$\begin{split} |\triangle u{3,\neq}|{Xb}^2 &\leq C\left(|\triangle u{3,\neq} - 2\partialx W^{(3)}|{X_b}^2 + |\partialx W^{(3)}|{Xb}^2\right) \ &\leq C\left(|(u{\rm in}){\neq}|{H^2}^2 + A^{\frac{4}{3}} E_{5,2}^2(t) + A^{\frac{\alpha}{3} + \frac{1}{6}} E_2^2 E_5^2 + A^{\frac{1}{3}} E2^4 + \epsilon^2 A^{\frac{2}{3}} E{5,1}^2(t) + 1\right). \end{split}$$ (8.57)

Combining (8.56) and (8.57), when A ≥ A6, there holds

$$E{5,1}^2(t) + E{5,2}^2(t) = A^{-\frac{4}{3}} | \triangle u{3,\neq} |{Xb}^2 + E{5,2}^2(t) \leq C \left( | (u{\rm in}){\neq} |{H^2}^2 + 1 \right) + C \epsilon^2 E{5,1}^2(t).$$

Choosing ϵ small enough satisfying Cϵ² ≤ 2 , we conclude that

$$E4^2(t) \le C(E{5,1}^2(t) + E{5,2}^2(t)) \le C(|(u{\rm in}){\ne}|{H^2}^2 + 1).$$

The proof is complete. □

Corollary 8.1. Under the assumptions of Theorem 1.1 and (2.5), according to Lemma 5.1, Lemma 5.2 and Lemma 8.6, there exists a positive constant A ≥ max{A3, A4, A6} =: A7, such that if A ≥ A7, there holds

$$E_2(t) \le C\left(|(\partialx^2 n{\rm in}){\ne}|{L^2} + |(u{\rm in}){\ne}|_{H^2} + 1\right) =: E_2.$$

Appendix A. Some useful estimates and lemmas in the proof

A.1. Several useful lemmas. We first prove an embedding inequality for non-zero modulus functions.

Lemma A.1. Let f be a function such that f̸⁼ ∈ H¹ (T 3 ), there holds

$$||f{\neq}||{L^{2}(\mathbb{T}^{3})} \le C||\partial{x}f{\neq}||{L^{2}(\mathbb{T}^{3})} \le C||\nabla f{\neq}||_{L^{2}(\mathbb{T}^{3})}.$$

Proof. It follows from Poincar´e inequality immediately and we omit it. □

The following lemma can be used to estimate the L ^∞ norm for the zero mode. Lemma A.2 (Lemma 3.1 in [8]). For a given function f(x, y, z) and f⁰ = |T| R T f(t, x, y, z)dx, we have

$$\begin{split} |f0|{L^{\infty}} &\leq C \left( |\partial_y f0|{L^2}^{\frac{1}{2}} |f0|{L^2}^{\frac{1}{2}} + |\partial_z \nabla f0|{L^2}^{\frac{1}{2}} |\partial_z f0|{L^2}^{\alpha - \frac{1}{2}} |f0|{L^2}^{1-\alpha} + |f0|{L^2} \right), \ |f0|{L^{\infty}} &\leq C \left( |\partial_y f0|{L^2}^{\frac{1}{2}} |f0|{L^2}^{\frac{1}{2}} + |\partial_z \nabla f0|{L^2}^{\alpha - \frac{1}{2}} |\partial_z f0|{L^2}^{\frac{1}{2}} |\partial_y f0|{L^2}^{1-\alpha} + |f0|{L^2} \right), \ |f0|{L^{\infty}{z}L^2{y}} &\leq C \left( |f0|{L^2} + |\partial_z f0|{L^2}^{\alpha} |f0|{L^2}^{1-\alpha} \right), \ |f0|{L^{\infty}{y}L^2{z}} &\leq C \left( |\partial_y f0|{L^2}^{\frac{1}{2}} |f0|{L^2}^{\frac{1}{2}} + |f0|{L^2} \right), \end{split}$$

$$(A.1)$$

where α is a constant with α ∈ 2 , 4 .

The following lemma can be used to estimate the L ^∞ norm for the non-zero mode. We only prove (A.2) 1,2 , and the remaining results are similar to Lemma 3.2 in [8]. The proof is omitted.

Lemma A.3. For a given function g = g(x, y, z) and α ∈ ( 1 2 , 3 4 ), if g⁰ = 1 |T| R g(t, x, y, z)dx = 0, then we have

$$\begin{split} |g|_{L^{\infty}} &\leq C \left( |\partialx \nabla g|{L^2}^{\frac{1}{2}} |\partial_x \partialz g|{L^2}^{1-\alpha} |\partialz^2 g|{L^2}^{\alpha-\frac{1}{2}} + |\partialx \nabla g|{L^2}^{\frac{1}{2}} |\partialx g|{L^2}^{\frac{1}{2}} \right), \ |g|_{L^{\infty}} &\leq C \left( |\partialz \nabla g|{L^2}^{\frac{1}{2}} |\partial_x \partialz g|{L^2}^{1-\alpha} |\partialx^2 g|{L^2}^{\alpha-\frac{1}{2}} + |\partialx \nabla g|{L^2}^{\frac{1}{2}} |\partialx g|{L^2}^{\frac{1}{2}} \right), \ |g|{L^{\infty}{y,z}L^2_x} &\leq C \min \left{ |\partialy g|{L^2}^{\frac{1}{2}} |g|_{L^2}^{\frac{1}{2}} + |\partialz \nabla g|{L^2}^{\frac{1}{2}} |\partialz g|{L^2}^{\alpha-\frac{1}{2}} |g|{L^2}^{1-\alpha} + |g|{L^2}, \ |\partialy g|{L^2}^{\frac{1}{2}} |g|_{L^2}^{\frac{1}{2}} + |\partialz g|{L^2}^{\frac{1}{2}} |\partialz \nabla g|{L^2}^{\alpha-\frac{1}{2}} |\partialy g|{L^2}^{1-\alpha} + |g|{L^2} \right}, \ |g|{L^{\infty}_{x,y}L^2_z} &\leq C \left( |\partialx g|{L^2}^{\frac{1}{2}} |g|_{L^2}^{1-\alpha} + |\partial_x \partialz g|{L^2}^{\alpha-\frac{1}{2}} |g|{L^2}^{1-\alpha} \right), \ |g|{L^{\infty}_{x,z}L^2_y} &\leq C \left( |\partialx g|{L^2}^{\alpha} |g|_{L^2}^{1-\alpha} + |\partial_x \partialz g|{L^2}^{\alpha} |g|{L^2}^{1-\alpha} \right), \ |g|{L^{\infty}xL^2{y,x}} &\leq C \left( |g|_{L^2} + |\partialz g|{L^2}^{\alpha} |g|{L^2}^{1-\alpha} \right), \ |g|{L^{\infty}xL^2{y,x}} &\leq C \left( |\partialy g|{L^2}^{\frac{1}{2}} |g|{L^2}^{\frac{1}{2}} + |g|{L^2} \right), \ |g|_{L^{\infty}xL^2{x,z}} &\leq C \left( |\partialx g|{L^2}^{\alpha} |g|{L^2}^{\frac{1}{2}} + |g|{L^2} \right), \ |g|_{L^{\infty}xL^2{x,z}} &\leq C \left( |\partialx g|{L^2}^{\alpha} |g|{L^2}^{\frac{1}{2}} + |g|{L^2} \right), \ |g|_{L^{\infty}xL^2{x,z}} &\leq C \left( |\partialx g|{L^2}^{\alpha} |g|_{L^2}^{\frac{1}{2}} + |\partial_x \partialz g|{L^2}^{\frac{1}{2}} |\partialx g|{L^2}^{\alpha-\frac{1}{2}} |\partialz g|{L^2}^{\alpha-\frac{1}{2}} |g|_{L^2}^{\frac{3}{2}-2\alpha} \right). \end{aligned}$$

Proof. Due to g⁰ = 0, we denote g(x, y, z) by g = P ^k1,k3∈Z,k1̸=0 ^gb^k1,k³ (y)eⁱ(k1x+k3z) , then

$$||g||{L^{\infty}} \leq \sum{k_1 \neq 0, k3 \in \mathbb{Z}} ||\widehat{g}{k_1, k3}||{Ly^{\infty}} \leq C \sum{k_1 \neq 0, k3 \in \mathbb{Z}} (||\widehat{g}{k_1, k3}||{L_y^2}^{\frac{1}{2}} ||\partialy \widehat{g}{k_1, k3}||{Ly^2}^{\frac{1}{2}} + ||\widehat{g}{k_1, k3}||{L_y^2}). \tag{A.3}$$

First, there holds

$$\begin{split} & \sum_{k_1 \neq 0, k3 \in \mathbb{Z}} |\widehat{g}{k_1, k3}|{L_y^2}^{\frac{1}{2}} |\partialy \widehat{g}{k_1, k3}|{Ly^2}^{\frac{1}{2}} \ &= \sum{k_1 \neq 0, k_3 \in \mathbb{Z}} \frac{|k_1 \partialy \widehat{g}{k_1, k3}|{L_y^2}^{\frac{1}{2}} |k3 \widehat{g}{k_1, k3}|{L_y^2}^{\frac{1}{2}} |k_1 k_3|^{\alpha - \frac{1}{2}} + |k_1 \partialy \widehat{g}{k_1, k3}|{Ly^2}^{\frac{1}{2}} |\widehat{g}{k_1, k3}|{L_y^2}^{\frac{1}{2}} |k_1|^{\alpha - \frac{1}{2}}}{|k_1|^{\alpha} (1 + |k3|^{\alpha})} \ &\leq \sum{k_1 \neq 0, k_3 \in \mathbb{Z}} \frac{|k_1 \partialy \widehat{g}{k_1, k3}|{L_y^2}^{\frac{1}{2}} |k_1 k3 \widehat{g}{k_1, k3}|{L_y^2}^{1 - \alpha} |k3^2 \widehat{g}{k_1, k3}|{L_y^2}^{\alpha - \frac{1}{2}} + |k_1 \partialy \widehat{g}{k_1, k3}|{Ly^2}^{\frac{1}{2}} |\widehat{g}{k_1, k3}|{L_y^2}^{\frac{1}{2}} |k_1|^{\alpha - \frac{1}{2}}}{|k_1|^{\alpha} (1 + |k_3|^{\alpha})} \end{split}$$

Using H¨older's inequality, we get

$$\sum_{k_1 \neq 0, k3 \in \mathbb{Z}} |\widehat{g}{k_1, k3}|{L_y^2}^{\frac{1}{2}} |\partialy \widehat{g}{k_1, k3}|{L_y^2}^{\frac{1}{2}} \leq C (|\partial_x \partialy g|{L^2}^{\frac{1}{2}} |\partial_x \partialz g|{L^2}^{1-\alpha} |\partialz^2 g|{L^2}^{\alpha - \frac{1}{2}} + |\partial_x \partialy g|{L^2}^{\frac{1}{2}} |\partialx g|{L^2}^{\frac{1}{2}}). \tag{A.4}$$

Thus, one deduces

$$\sum_{k_1 \neq 0, k3 \in \mathbb{Z}} |\widehat{g}{k_1, k3}|{L^2_y} \leq \sum \frac{|k1 \widehat{g}{k_1, k3}|{L^2_y}^{\frac{1}{2}} |k_1 k3 \widehat{g}{k_1, k3}|{L^2_y}^{1-\alpha} |k3^2 \widehat{g}{k_1, k3}|{L^2_y}^{\alpha - \frac{1}{2}} + |k1 \widehat{g}{k_1, k3}|{L^2y}^{\alpha} |\widehat{g}{k_1, k3}|{L^2_y}^{1-\alpha}}{|k_1|^{\alpha} (1 + |k_3|^{\alpha})}$$

$$\leq C \left( |\partialx g|{L^2}^{\frac{1}{2}} |\partial_x \partialz g|{L^2}^{1-\alpha} |\partialz^2 g|{L^2}^{\alpha - \frac{1}{2}} + |\partialx g|{L^2} \right).$$ (A.5)

Combining (A.3), (A.4) and (A.5), we finish the proof of (A.1) 1

Similarly, we can prove $$(A.1)_2$$ .

Lemma A.4. For a given function g = g(x, y, z), if g⁰ = 1 |T| R T g(t, x, y, z)dx = 0, there holds

$$\begin{split} |g|{L^\infty{y,z}L^2_x} &\leq C |(\partial_x,\partial_z - \kappa \partialy)g|{L^2}^{\frac{1}{2}} |\nabla(\partial_x,\partial_z - \kappa \partialy)g|{L^2}^{\frac{1}{2}}, \ |g|_{L^\inftyz L^2{y,x}} &\leq C |(\partial_x,\partial_z - \kappa \partialy)g|{L^2}. \end{split}$$

Proof. Let G(X, Y, Z) such that G(x, V (t, y, z), z) = g(x, y, z). Note that (∂^z − κ∂y)g(x, y, z) = ∂ZG(x, V (t, y, z), z) and ∥g∥L² ≤ ∥∂xg∥L² . Then the results follow from (A.2) 3 and (A.2) . □

Lemma A.5 (Lemma 5.5 in [34]). Under the conditions of Lemma 8.1, if ∂xf¹ = 0, P0f² = 0, it holds that

$$\begin{aligned} & |f_1 f2|{L^2} \leq C |f1|{H^1} \left( |f2|{L^2} + |(\partial_z - \kappa \partial_y) f2|{L^2} \right), \ & |\nabla \triangle^{-1} (f_1 f2)|{L^2} \leq C |f1|{L^2} \left( |f2|{L^2} + |(\partial_z - \kappa \partial_y) f2|{L^2} \right), \ & |\nabla (f_1 f2)|{L^2} \leq C |f1|{H^1} \left( |f2|{H^1} + |(\partial_z - \kappa \partial_y) f2|{H^1} \right), \end{aligned}$$

and for j ∈ {2, 3},

$$|[\partial_j \triangle^{-1}, f_1] \triangle f2|{H^1} \le C |f1|{H^2} (|\nabla f2|{L^2} + |(\partial_z - \kappa \partial_y) \nabla f2|{L^2}).$$

Next, we introduce a logarithmic Hardy-Littlewood-Sobolev inequality that plays an important role in estimation ∥n0∥L² and can be found in [27].

Lemma A.6. Let M be a 2D Riemannian compact manifold. For all m > 0, there exists a constant C(m) such that for all nonnegative functions f ∈ L 1 (M) such that f log f ∈ L 1 , if R ^M f dx = m, then

$$\int{\mathcal{M}} f \log f dx + \frac{2}{m} \int{\mathcal{M}} \int_{\mathcal{M}} f(x) f(y) \log d(x, y) \ge -C(m), \tag{A.6}$$

where d(x, y) is the distance on the Riemannian manifold.

The following Gagliardo-Nirenberg-Sobolev inequality on T n is frequently used in the proof.

Lemma A.7 (Lemma 9.2 in [22]). Suppose f ∈ C ^∞(T n ), n ≥ 2, and the set where f vanishes is nonempty. Assume that q, r > 0,∞ > q > r, and ^{1 n} − 1 ² + 1 ^r > 0. Then

$$||f||{L^q} \le C(n,q) ||\nabla f||{L^2}^{\theta} ||f||_{L^r}^{1-\theta}, \quad \theta = \frac{\frac{1}{r} - \frac{1}{q}}{\frac{1}{n} - \frac{1}{2} + \frac{1}{r}}.$$

For a fixed n, the constant C(n, q) is bounded uniformly when q varies in any compact set in (0,∞).

A.2. Elliptic estimates. The following elliptic estimates are necessary.

Lemma A.8. Let c⁰ and n⁰ be the zero mode of c and n, respectively, satisfying

$$-\triangle c_0 + \bar{n} = n_0,$$

then there hold

$$|\Delta c0(t)|{L^2} \le |n0(t)|{L^2},$$

$$|\nabla c0(t)|{L^{\infty}} \le C|n0(t) - \bar{n}|{L^3}$$

$$|\nabla c0(t)|{L^4} \le C|n0(t)|{L^2},$$ (A.7)

for any t ≥ 0.

Proof. The basic energy estimates yield

$$|\triangle c0(t)|{L^2}^2 + |\bar{n}|^2 |\mathbb{T}|^2 = |n0(t)|{L^2}^2,$$

which implies (A.7) 1 . Using Gagliardo-Nirenberg inequality and H¨older's inequality, we have

$$|\nabla c0(t)|{L^{\infty}} \leq C \left( |\nabla c0(t)|{L^2}^{\frac{1}{4}} |\triangle c0(t)|{L^3}^{\frac{3}{4}} + |\nabla c0(t)|{L^2} \right)$$

$$\leq C |\triangle c0(t)|{L^3} \leq C |n0(t) - \bar{n}|{L^3},$$

which gives (A.7) 2 .

Moreover, it follows from Lemma A.7 and (A.7) 1 that

$$|\nabla c0(t)|{L^4} \le C|\nabla c0(t)|{L^2}^{\frac{1}{2}}|\Delta c0(t)|{L^2}^{\frac{1}{2}} \le C|\Delta c0(t)|{L^2} \le C|n0(t)|{L^2},$$

which gives (A.7) 3 . □ Lemma A.9. There holds

$$\begin{aligned} |\partialx^j \triangle c{\neq}(t)|_{L^2} &\leq |\partialx^j n{\neq}(t)|_{L^2}, \ |\partialx^j \nabla c{\neq}(t)|_{L^4} &\leq C |\partialx^j n{\neq}(t)|_{L^2}, \end{aligned}$$

for any t ≥ 0 and j ≥ 0.

Proof. By integration by parts, we have

$$|\triangle c{\neq}(t)|{L^{2}}^{2} = |n{\neq}(t)|{L^{2}}^{2}.$$

Using Gagliardo-Nirenberg inequality and Lemma A.1, we obtain

$$|\nabla c{\neq}(t)|{L^{4}} \leq C\left(|c{\neq}(t)|{L^{2}}^{\frac{1}{8}}|\triangle c{\neq}(t)|{L^{2}}^{\frac{7}{8}} + |c{\neq}(t)|{L^{2}}\right) \leq C|n{\neq}(t)|{L^{2}}.$$

A.3. Space-time estimates. First, we need to prove the following space-time estimate, and this result is also an improvement on previous time-space estimates [7, 34], which allows us to estimate the non-zero mode n̸⁼ in the periodic domain T 3

Proposition A.1. Assume that f satisfies

$$\partial_t f - \frac{1}{A} \triangle f + \left( y + \frac{\mathbf{U}_2(t, y, z)}{A} \right) \partial_x f = \partial_x f_1 + f_2 + \nabla \cdot f_3$$

for t ∈ [0, T], where f, f1, f² and f³ are given functions and P0f = P0f¹ = P0f² = P0f³ = 0. As long as

$$\frac{|\triangle \mathbf{U}2|{L^{\infty}H^2}}{A} + |\partial_t \mathbf{U}2|{L^{\infty}L^{\infty}} < c, \tag{A.8}$$

for some small c independent of A and t, then there holds

$$||f{\neq}||{Xa}^2 \le C \left( ||(f{\text{in}}){\neq}||{L^2}^2 + ||e^{aA^{-\frac{1}{3}}t} \nabla f{1,\neq}||{L^2L^2}^2 + A^{\frac{1}{3}} ||e^{aA^{-\frac{1}{3}}t} f{2,\neq}||{L^2L^2}^2 + A ||e^{aA^{-\frac{1}{3}}t} f{3,\neq}||{L^2L^2}^2 \right),$$

where "a" is a positive constant.

Proof. Our proof was inspired the coordinate transform method of [34]. First, we state the following result, which is classical.

Lemma A.10. Let g : T ³ → R 3 be a C ¹ map such that ∥∇g∥L^∞ < 2 . Then it holds that

• (i) ∥f (Id + g)∥L^p ∼ ∥f∥L^p , ∥∇ (f (Id + g)) ∥L^p ∼ ∥∇f∥L^p for every 1 ≤ p ≤ +∞ and f ∈ W¹,p . Here A ∼ B means C ⁻¹A ≤ B ≤ CA for some absolute constant C.
• (ii) There exists a unique C 1 solution h to h(x, y, z) = g ((x, y, x) + h(x, y, z)) satisfying ∥∇h∥L^∞ ≤ C∥∇g∥L^∞.

Then, we define the map (Id + g) : T ³ → R ³ by

$$(x, y, z) \mapsto (X, Y, Z) = (x, V(t, y, z), z), \quad V = y + \frac{\mathbf{U}_2(t, y, z)}{A}.$$

□

Denote

$$F(t, x, V(t, y, z), z) = f(t, x, y, z),$$

$$F{j}(t, x, V(t, y, z), z) = f{j}(t, x, y, z), \text{ for } j = 1, 2,$$

$$F{3}(t, x, V(t, y, z), z) = \text{div } f{3}(t, x, y, z).$$ (A.9)

According to (A.8), by choosing c small enough, there holds

$$\frac{1}{A} |\nabla \mathbf{U}2|{L^{\infty}} \le \frac{C}{A} |\nabla \mathbf{U}2|{H^2} \le \frac{C}{A} |\triangle \mathbf{U}2|{H^2} < \frac{1}{2},$$

which along with Lemma A.10 implies that F(t, X, Y, Z) ∈ C 1 is well-defined.

Let wt(t, Y, Z), wy(t, Y, Z) and wz(t, Y, Z) be such that

$$w_t(t, V(t, y, z), z) = \frac{\partial_t \mathbf{U}_2(t, y, z)}{A} = \partial_t V(t, y, z),$$

$$w_y(t, V(t, y, z), z) = \frac{\partial_y \mathbf{U}_2(t, y, z)}{A} = \partial_y V(t, y, z) - 1,$$

$$w_z(t, V(t, y, z), z) = \frac{\partial_z \mathbf{U}_2(t, y, z)}{A} = \partial_z V(t, y, z).$$ (A.10)

Direct calculations indicate that

$$\partial_t f = (\partial_t + w_t \partial_Y) F, \quad \partial_x f = \partial_X F, \partial_y f = (1 + w_y) \partial_Y F, \quad \partial_z f = (\partial_Z + w_z \partial_Y) F.$$ (A.11)

Therefore, we obtain that V ∂xf = Y ∂XF, and

$$\Delta f = (\partial_x^2 + \partial_y^2 + \partial_z^2) f$$

$$= \partial_X^2 F + \partial_y \left[ (1 + w_y) \partial_Y F \right] + \partial_z \left[ (\partial_Z + w_z \partial_Y) F \right]$$

$$= (\partial_X^2 + \partial_Y^2 + \partial_Z^2) F + \left[ (1 + w_y)^2 + w_z^2 - 1 \right] \partial_Y^2 F + 2w_z \partial_Z \partialY F + (w{yy} + w_{zz}) \partial_Y F$$

$$=: \Delta F + G \partial_Y^2 F + 2w_z \partial_Z \partial_Y F + H \partial_Y F,$$ (A.12)

where

$$G = (1 + w_y)^2 + w_z^2 - 1, \ H(t, V(t, y, z), z) = \frac{\triangle \mathbf{U}_2(t, y, z)}{A}.$$

Using (A.11)-(A.12), we write the operator L^V into

$$\mathcal{L}{V}f = \partial{t}f - \frac{1}{A}\Delta f + V\partial_{x}f$$

$$= (\partial{t} + w{t}\partial{Y})F - \frac{1}{A}\left(\Delta F + G\partial{Y}^{2}F + 2w{z}\partial{Z}\partial{Y}F + H\partial{Y}F\right) + Y\partial_{X}F$$

$$= \mathcal{L}F - \frac{1}{A}\left(G\partial{Y}^{2}F + 2w{z}\partial{Z}\partial{Y}F\right) + \left(w{t} - \frac{H}{A}\right)\partial{Y}F,$$ (A.13)

where L = ∂^t − 1 ^A△ + Y ∂X. Due to (A.10) and (A.11), there hold

$$\begin{split} \partial_Y F &= \frac{\partial_y f}{\partial_y V}, \quad \partial_Z F = \partial_z f - \frac{\partial_z V \partial_y f}{\partial_y V}, \ G(t, V(t, y, z), z) &= \left(1 + \frac{\partial_y \mathbf{U}_2}{A}\right)^2 + \left(\frac{\partial_z \mathbf{U}_2}{A}\right)^2 - 1 \ &= (\partial_y V)^2 + (\partial_z V)^2 - 1, \end{split}$$

which imply that

$$\begin{split} \partial{Y}G + 2\partial{Z}w{z} &= \frac{\partial{y}\left((\partial{y}V)^{2} + (\partial{z}V)^{2} - 1\right)}{\partial{y}V} + 2\left[\partial{z} - \left(\frac{\partial{z}V}{\partial{y}V}\right)\partial{y}\right]\partial{z}V \ &= 2\partial{y}^{2}V + \frac{2\partial{z}V\partial{y}\partial{z}V}{\partial{y}V} + 2\partial{z}^{2}V - 2\frac{\partial{z}V}{\partial{y}V}\partial{y}\partial{z}V \ &= 2\triangle V = \frac{2}{A}\triangle\mathbf{U}_{2} = 2H. \end{split} \tag{A.14}$$

Combining (A.13) and (A.14), one deduces that

$$\mathcal{L}{V}f = \mathcal{L}F - \frac{1}{A}\partial{Y}(G\partial{Y}F) - \frac{2}{A}\partial{Z}(w{z}\partial{Y}F) + \left(w{t} + \frac{H}{A}\right)\partial{Y}F. \tag{A.15}$$

From (A.9), we get L^V f = ∂XF¹ + F² + F3. Then (A.15) yields

$$\mathcal{L}F = \partial_X F_1 + F_2 + F_3 + \frac{1}{A} \partial_Y (G \partial_Y F) + \frac{2}{A} \partial_Z (w_z \partial_Y F) - \left( w_t + \frac{H}{A} \right) \partial_Y F.$$

This implies that

$$\begin{split} |F|{X{a}}^{2} \leq & C\bigg(|F(0)|{L^{2}}^{2} + |e^{aA^{-\frac{1}{3}}t}\nabla F{1}|{L^{2}L^{2}}^{2} + A^{\frac{1}{3}}|e^{aA^{-\frac{1}{3}}t}F{2}|{L^{2}L^{2}}^{2} + A|e^{aA^{-\frac{1}{3}}t}\nabla\Delta^{-1}F{3}|{L^{2}L^{2}}^{2} \ & + \frac{1}{A}\left|e^{aA^{-\frac{1}{3}}t}\left(G, 2w{z}\right)\partial{Y}F\right|{L^{2}L^{2}}^{2} + A^{\frac{1}{3}}\left|e^{aA^{-\frac{1}{3}}t}\left(w{t} + \frac{H}{A}\right)\partial{Y}F\right|{L^{2}L^{2}}^{2}\bigg) \ := & C\bigg(|F(0)|{L^{2}}^{2} + |e^{aA^{-\frac{1}{3}}t}\nabla F{1}|{L^{2}L^{2}}^{2} + A^{\frac{1}{3}}|e^{aA^{-\frac{1}{3}}t}F{2}|{L^{2}L^{2}}^{2} \ & + A|e^{aA^{-\frac{1}{3}}t}\nabla\Delta^{-1}F{3}|{L^{2}L^{2}}^{2} + K{1} + K{2}\bigg). \end{split} \tag{A.16}$$

For K1, recall that w^z = ∂zU² A and

$$G = (\partial_y V)^2 + (\partial_z V)^2 - 1 = \left(\frac{\partial_y \mathbf{U}_2}{A} + 1\right)^2 + \left(\frac{\partial_z \mathbf{U}_2}{A}\right)^2 - 1 = \frac{|\nabla \mathbf{U}_2|^2}{A^2} + \frac{2\partial_y \mathbf{U}_2}{A}.$$

Then using (A.8), one deduces

$$K{1} \leq \left( |G|{L^{\infty}L^{\infty}}^{2} + |2w{z}|{L^{\infty}L^{\infty}}^{2} \right) \frac{|e^{aA^{-\frac{1}{3}}t}\partial{Y}F|{L^{2}L^{2}}^{2}}{A}$$

$$\leq C(c^{4} + c^{2})|F|{X{a}}^{2},$$ (A.17)

where we use the Poincar´e inequality ∥∇U2∥L² ≤ C∥△U2∥L² .

For K2, as w^t = ∂tU² A and H = △U² A , using (A.8), we have

$$K_2 \le A^{\frac{1}{3}} \left( |wt|{L^{\infty}L^{\infty}}^2 + \left| \frac{H}{A} \right|_{L^{\infty}L^{\infty}}^2 \right) |e^{aA^{-\frac{1}{3}}t} \partialY F|{L^2L^2}^2 \le \frac{Cc^2}{A^{\frac{2}{3}}} |F|_{X_a}^2. \tag{A.18}$$

Combining (A.17) with (A.18), then choosing c small enough, we get by (A.16) that

$$||F||{X{a}}^{2} \leq C \left( ||F(0)||{L^{2}}^{2} + ||e^{aA^{-\frac{1}{3}}t}\nabla F{1}||{L^{2}L^{2}}^{2} + A^{\frac{1}{3}}||e^{aA^{-\frac{1}{3}}t}F{2}||{L^{2}L^{2}}^{2} + A||e^{aA^{-\frac{1}{3}}t}\nabla\Delta^{-1}F{3}||_{L^{2}L^{2}}^{2} \right).$$ (A.19)

It follows from Lemma A.10 that

$$|e^{aA^{-\frac{1}{3}}t}\nabla F|{L^{2}L^{2}}^{2} \le C|e^{aA^{-\frac{1}{3}}t}\nabla f|{L^{2}L^{2}}^{2}, \ |e^{aA^{-\frac{1}{3}}t}F|{L^{2}L^{2}}^{2} \le C|e^{aA^{-\frac{1}{3}}t}f|{L^{2}L^{2}}^{2}$$ (A.20)

and

$$|e^{aA^{-\frac{1}{3}}t}\nabla F1|{L^2L^2}^2 \le C|e^{aA^{-\frac{1}{3}}t}\nabla f1|{L^2L^2}^2, \ |e^{aA^{-\frac{1}{3}}t}F2|{L^2L^2}^2 \le C|e^{aA^{-\frac{1}{3}}t}f2|{L^2L^2}^2. \tag{A.21}$$

Next we claim

$$|\nabla \triangle^{-1} F|{L^2} \sim |\nabla \triangle^{-1} f|{L^2}.$$ (A.22)

As V = y + U2(t,y,z) A and (A.8), we get

$$|\nabla V|{L^{\infty}} + |\nabla^{2}V|{L^{\infty}} \le 1 + \frac{|\nabla \mathbf{U}{2}|{L^{\infty}}}{A} + \frac{|\nabla^{2}\mathbf{U}{2}|{L^{\infty}}}{A} \le 1 + C\frac{|\triangle\mathbf{U}{2}|{H^{2}}}{A} \le C. \tag{A.23}$$

Denote

$$\check{F}_1 = \triangle^{-1}F, \quad \check{f}_1(t, x, y, z) = \check{F}_1(t, x, V(t, y, z), z), \check{f}_2 = \triangle^{-1}f, \quad \check{f}_2(t, x, y, z) = \check{F}_2(t, x, V(t, y, z), z).$$

Then using (A.20) and (A.23), one deduces

$$|\nabla \triangle^{-1} F|{L^{2}}^{2} = |\nabla \check{F}{1}|{L^{2}}^{2} = -\langle F, \check{F}{1} \rangle = -\langle f, \partial{y} V \check{f}{1} \rangle$$

$$= \langle \nabla \check{f}{2}, \nabla (\partial{y} V \check{f}{1}) \rangle \leq |\nabla \check{f}{2}|{L^{2}} |\nabla (\partial{y} V \check{f}{1})|{L^{2}}$$

$$\leq C |\nabla \check{f}{2}|{L^{2}} (|\nabla V|{L^{\infty}} |\nabla \check{f}{1}|{L^{2}} + |\nabla^{2} V|{L^{\infty}} |\check{f}{1}|{L^{2}})$$

$$\leq C |\nabla \check{f}{2}|{L^{2}} |\nabla \check{f}{1}|{L^{2}} \leq C |\nabla \check{f}{2}|{L^{2}} |\nabla \check{f}{1}|{L^{2}},$$ (A.24)

which implies

$$|\nabla \check{F}1|{L^2} \le C |\nabla \check{f}2|{L^2}. \tag{A.25}$$

On the other hand, there holds

$$\begin{split} |\nabla \triangle^{-1} f|{L^{2}}^{2} = &|\nabla \check{f}{2}|{L^{2}}^{2} = - < f, \, \check{f}{2}> = - < F, \, \check{F}{2}/\partial{y}V> \ = &<\nabla \check{F}{1}, \, \nabla (\check{F}{2}/\partial{y}V) > \leq |\nabla \check{F}{1}|{L^{2}} |\nabla (\check{F}{2}/\partial{y}V)|{L^{2}}. \end{split}$$

Notice that ∂yV = A ∂yU² +1 and ¹ A ∥∇U2∥L^∞ < 2 , we have ^{1 2} < ∂yV < ³ 2 . Along this with (A.23), we get ∥∇(1/∂yV )∥L^∞ ≤ C, and

$$|\nabla \Delta^{-1} f|{L^{2}} = |\nabla \check{f}{2}|{L^{2}}^{2} \le C |\nabla \check{F}{1}|{L^{2}} \left(|\nabla \check{F}{2}|{L^{2}} + |\check{F}{2}|_{L^{2}}\right)$$

$$\le C |\nabla \check{F}{1}|{L^{2}} |\nabla \check{F}{2}|{L^{2}} \le C |\nabla \check{F}{1}|{L^{2}} |\nabla \check{f}{2}|{L^{2}},$$ (A.26)

which gives

$$|\nabla \check{f}2|{L^2} \le C |\nabla \check{F}1|{L^2}. \tag{A.27}$$

Combining with (A.24)-(A.27), we conclude that

$$|\nabla \triangle^{-1} F|_{L^2} = |\nabla \check{F}1|{L^2} \sim |\nabla \check{f}2|{L^2} = |\nabla \triangle^{-1} f|_{L^2},$$

which indicates (A.22) holds. Similarly, we obtain

$$|e^{aA^{-\frac{1}{3}}t}\nabla\Delta^{-1}F3|{L^2L^2}^2 \le C|e^{aA^{-\frac{1}{3}}t}f3|{L^2L^2}^2.$$ (A.28)

From (A.19), (A.21) and (A.28), we complete the proof. □

To estimate the coupled terms (△u2,̸=, ∂xω2,̸=), we also need the following proposition.

Proposition A.2. Let (h1, h2) satisfy

$$\begin{cases} \mathcal{L}_V h_1 = \nabla \cdot g_1, \ \mathcal{L}_V h_2 + \partial_x \partial_z \triangle^{-1} h_1 = \nabla \cdot g_2, \end{cases}$$

for t ∈ [0, T]. Assume that

$$\frac{|\triangle \mathbf{U}2|{L^{\infty}H^2}}{A} + |\partial_t \mathbf{U}2|{L^{\infty}L^{\infty}} \le c \tag{A.29}$$

for some small constant c independent of A and T. If P0h¹ = P0h² = P0g¹ = P0g² = 0, then for a ≥ 0, it holds that

$$|h{1,\neq}|{Xa}^2 + |h{2,\neq}|_{Xa}^2 \le C\left(|(h{1,\text{in}}){\neq}|{L^2}^2 + |(h{2,\text{in}}){\neq}|{L^2}^2 + A|e^{aA^{-\frac{1}{3}}t}g{1,\neq}|{L^2L^2}^2 + A|e^{aA^{-\frac{1}{3}}t}g{2,\neq}|_{L^2L^2}^2\right).$$

Proof. From Proposition A.2 in [8], we find that if (h1, h2) satisfy

$$\begin{cases} \partial_t h_1 - \frac{1}{A} \triangle h_1 + y \partial_x h_1 = \nabla \cdot g_1, \ \ \partial_t h_2 - \frac{1}{A} \triangle h_2 + y \partial_x h_2 + \partial_x \partial_z \triangle^{-1} h_2 = \nabla \cdot g_2, \end{cases}$$

for t ∈ [0, T], then it holds that

$$|h{1,\neq}|{Xa}^2 + |h{2,\neq}|_{Xa}^2 \le C\left(|(h{1,\mathrm{in}}){\neq}|{L^2}^2 + |(h{2,\mathrm{in}}){\neq}|{L^2}^2 + A|e^{aA^{-\frac{1}{3}}t}g{1,\neq}|{L^2L^2}^2 + A|e^{aA^{-\frac{1}{3}}t}g{2,\neq}|_{L^2L^2}^2\right)$$

Next following the same route as in Proposition A.1, we complete the proof. □

Proposition A.3 (Proposition 4.7 in [34]). Let f satisfy

$$\mathcal{L}_V f = f_1 + f_2 + f_3$$

for t ∈ [0, T]. Moreover, U² satisfies (A.29) and P0f = P0f¹ = P0f² = P0f³ = 0, then for a ≥ 0, it holds that

$$\begin{split} |\partialx^2 f{\neq}|_{X_a}^2 + |\partial_x (\partial_z - \kappa \partialy) f{\neq}|_{Xa}^2 &\leq C \big( |(f{\rm in}){\neq}|{H^2}^2 + |\mathrm{e}^{aA^{-\frac{1}{3}}t} \triangle f{1,\neq}|{L^2L^2}^2 + A^{\frac{1}{3}} |\mathrm{e}^{aA^{-\frac{1}{3}}t} \partialx^2 f{2,\neq}|_{L^2L^2}^2 \ &\quad + A^{\frac{1}{3}} |\mathrm{e}^{aA^{-\frac{1}{3}}t} \partial_x (\partial_z - \kappa \partialy) f{2,\neq}|_{L^2L^2}^2 + A |\mathrm{e}^{aA^{-\frac{1}{3}}t} \partialx f{3,\neq}|_{L^2L^2}^2 \big). \end{split}$$

Proposition A.4 (Proposition 4.9 in [34]). Let ^f satisfy ^Lf^{V f = f1} for ^{t ∈} [0, T]. Moreover, U² satisfies (A.29) and P0f = P0f¹ = 0, then for a ≥ 0, it holds that

$$|\triangle f{\neq}|{Xa}^2 \le C\left(|(\triangle f{\text{in}}){\neq}|{L^2}^2 + A|e^{aA^{-\frac{1}{3}}t}\nabla f{1,\neq}|{L^2L^2}^2\right).$$

Acknowledgement

W. Wang was supported by National Key R&D Program of China (No.2023YFA1009200) and NSFC under grant 12471219 and 12071054. The research of J. Wei is partially supported by GRF from RGC of Hong Kong entitled "New frontiers in singularity formations in nonlinear partial differential equations".

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(Shikun Cui) School of Mathematical Sciences, Dalian University of Technology, Dalian, 116024, China

Email address: cskmath@163.com

(Lili Wang) School of Mathematical Sciences, Dalian University of Technology, Dalian, 116024, China

Email address: wllmath@163.com

(Wendong Wang) School of Mathematical Sciences, Dalian University of Technology, Dalian, 116024, China

Email address: wendong@dlut.edu.cn

(Juncheng Wei) Department of Mathematics, Chinese University of Hong Kong, Shatin, NT, Hong Kong

Email address: wei@math.cuhk.edu.hk