1 Introduction

In this paper, we investigate the global well-posedness of the following 3-D anisotropic Navier–Stokes system:

$$\begin{aligned} (ANS)\quad \left\{ \begin{array}{l} \displaystyle \partial _t u +u\cdot \nabla u-\Delta _\mathrm{h}u=-\nabla p, \qquad (t,x)\in {\mathbb {R}}^+\times {\mathbb {R}}^3, \\ \displaystyle \mathrm{div}\,u = 0, \\ \displaystyle u|_{t=0}=u_0, \end{array}\right. \end{aligned}$$

where \(\Delta _\mathrm{h}\buildrel {\mathrm{def}}\over =\partial _1^2+\partial _2^2,\)u designates the velocity of the fluid and p the scalar pressure function which guarantees the divergence free condition of the velocity field.

Systems of this type appear in geophysical fluid dynamics (see for instance [5, 18]). In fact, meteorologists often model turbulent diffusion by using a viscosity of the form \(-\mu _\mathrm{h}\Delta _\mathrm{h}-\mu _3\partial _3^2\), where \(\mu _\mathrm{h}\) and \(\mu _3\) are empirical constants, and \(\mu _3\) is usually much smaller than \(\mu _\mathrm{h}\). We refer to the book of Pedlovsky [18, Chap. 4], for a complete discussion about this model.

Considering that system (ANS) has only horizontal dissipation, it is reasonable to use functional spaces which distinguish horizontal derivatives from the vertical one, for instance, the anisotropic Sobolev space defined as follows:

Definition 1.1

For any \((s,s')\) in \({\mathbb {R}}^2\), the anisotropic Sobolev space \(H^{s,s'}({\mathbb {R}}^3)\) denotes the space of homogeneous tempered distribution a such that

$$\begin{aligned} \Vert a\Vert ^2_{H^{s,s'}} \buildrel {\mathrm{def}}\over =\int \nolimits _{{\mathbb {R}}^3} |\xi _\mathrm{h}|^{2s}|\xi _3|^{2s'} |\widehat{a} (\xi )|^2\,\mathrm{d}\xi <\infty \quad \hbox {with}\quad \xi _\mathrm{h}=(\xi _1,\xi _2). \end{aligned}$$

Mathematically, Chemin et al. [4] first studied the system (ANS). In particular, Chemin et al. [4] and Iftimie [13] proved that (ANS) is locally well-posed with initial data in \(L^2\cap H^{0,\frac{1}{2}+\varepsilon }\) for some \(\varepsilon >0\), and is globally well-posed if, in addition,

$$\begin{aligned} \Vert u_0\Vert _{L^2}^\varepsilon \Vert u_0\Vert _{H^{0,\frac{1}{2}+\varepsilon }}^{1-\varepsilon }\leqq c \end{aligned}$$
(1.1)

for some sufficiently small constant c.

Notice that just as the classical Navier–Stokes system

$$\begin{aligned} (NS)\quad \left\{ \begin{array}{l} \displaystyle \partial _t u +u\cdot \nabla u-\Delta u=-\nabla p, \qquad (t,x)\in {\mathbb {R}}^+\times {\mathbb {R}}^3, \\ \displaystyle \mathrm{div}\,u = 0, \\ \displaystyle u|_{t=0}=u_0, \end{array}\right. \end{aligned}$$

the system (ANS) has the following scaling invariant property:

$$\begin{aligned} u_\lambda (t,x)\buildrel {\mathrm{def}}\over =\lambda u(\lambda ^2 t,\lambda x) \quad \hbox {and}\quad u_{0,\lambda }(x)\buildrel {\mathrm{def}}\over =\lambda u_0(\lambda x), \end{aligned}$$
(1.2)

which means that if u is a solution of (ANS) with initial data \(u_0\) on [0, T], \(u_\lambda \) determined by (1.2) is also a solution of (ANS) with initial data \(u_{0,\lambda }\) on \([0,T/\lambda ^2]\).

It is easy to observe that the smallness condition (1.1) in [4] is scaling invariant under the scaling transformation (1.2), nevertheless, the norm of the space \(H^{0,\frac{1}{2}+\varepsilon }\) is not. To work (ANS) with initial data in the critical spaces, we first recall the following anisotropic dyadic operators from [2]:

$$\begin{aligned} \begin{aligned}&\Delta _k^\mathrm{h}a\buildrel {\mathrm{def}}\over =\mathcal {F}^{-1}(\varphi (2^{-k}|\xi _\mathrm{h}|)\widehat{a}), \quad \Delta _\ell ^\mathrm{v}a \buildrel {\mathrm{def}}\over =\mathcal {F}^{-1}(\varphi (2^{-\ell }|\xi _3|)\widehat{a}),\\&S^\mathrm{h}_ka\buildrel {\mathrm{def}}\over =\mathcal {F}^{-1}(\chi (2^{-k}|\xi _\mathrm{h}|)\widehat{a}), \quad \ S^\mathrm{v}_\ell a \buildrel {\mathrm{def}}\over =\mathcal {F}^{-1}(\chi (2^{-\ell }|\xi _3|)\widehat{a}), \end{aligned} \end{aligned}$$
(1.3)

where \(\xi _\mathrm{h}=(\xi _1,\xi _2),\)\(\mathcal {F}a\) or \(\widehat{a}\) denotes the Fourier transform of a, while \(\mathcal {F}^{-1} a\) designates the inverse Fourier transform of a, \(\chi (\tau )\) and \(\varphi (\tau )\) are smooth functions such that

$$\begin{aligned}&{\mathrm{Supp}\,}\varphi \subset \Bigl \{\tau \in {\mathbb {R}}\,: \, \frac{3}{4} \leqq |\tau | \leqq \frac{8}{3} \Bigr \}\quad \text{ and }\quad \forall \tau >0\,,\ \sum _{j\in {\mathbb {Z}}}\varphi (2^{-j}\tau )=1;\\&{\mathrm{Supp}\,}\chi \subset \Bigl \{\tau \in {\mathbb {R}}\,: \, |\tau | \leqq \frac{4}{3} \Bigr \}\quad \text{ and }\quad \forall \tau \in {\mathbb {R}}\,,\ \chi (\tau )+ \sum _{j\geqq 0}\varphi (2^{-j}\tau )=1. \end{aligned}$$

Definition 1.2

We define \(\mathcal {B}^{0,\frac{1}{2}}({\mathbb {R}}^3)\) to be the set of homogenous tempered distribution a so that

$$\begin{aligned} \Vert a\Vert _{\mathcal {B}^{0,\frac{1}{2}}}\buildrel {\mathrm{def}}\over =\sum _{\ell \in {\mathbb {Z}}}2^{\frac{\ell }{2}} \Vert \Delta _{\ell }^{\mathrm{v}}a\Vert _{L^2({\mathbb {R}}^3)}<\infty . \end{aligned}$$

The above space was first introduced by Iftimie [12] to study the global well-posedness of the classical 3-D Navier–Stokes system with initial data in the anisotropic functional space. The second author [16] proved the local well-posedness of (ANS) with any solenoidal vector field \(u_0\in \mathcal {B}^{0,\frac{1}{2}}\) and also the global well-posedness with small initial data in \(\mathcal {B}^{0,\frac{1}{2}}.\) This result corresponds to Fujita–Kato’s theorem [11] for the classical Navier–Stokes system. Moreover, the authors [17, 19] proved the global well-posedness of (ANS) with initial data \(u_0=(u_0^\mathrm{h},u_0^3)\) satisfying that

$$\begin{aligned} \Vert u^{\mathrm{h}}_0\Vert _{\mathcal {B}^{0,\frac{1}{2}}}\exp \bigl (C\Vert u^3_0\Vert _{\mathcal {B}^{0,\frac{1}{2}}}^4\bigr )\leqq c_0 \end{aligned}$$
(1.4)

for some \(c_0\) sufficiently small.

Although the norm of \(\mathcal {B}^{0,\frac{1}{2}}\) is scaling invariant under the the scaling transformation (1.2), yet we observe that the solenoidal vector field

$$\begin{aligned} u_0^\varepsilon (x) = \sin \Bigl (\frac{x_{1}}{\varepsilon }\Bigr )\left( 0, - \partial _{3}\varphi , \partial _{2}\varphi \right) \end{aligned}$$
(1.5)

is not small in the space \(\mathcal {B}^{0,\frac{1}{2}}\) no matter how small \(\varepsilon \) is. In order to find a space so that the norm of \(u_0^\varepsilon (x)\) given by (1.5) is small in this space for small \(\varepsilon ,\) Chemin and the third author [8] introduced the following Besov–Sobolev type space with negative index:

Definition 1.3

We define the space \(\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4\) to be the set of a homogenous tempered distribution a so that

$$\begin{aligned} \Vert a\Vert _{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4}\buildrel {\mathrm{def}}\over =\sum _{\ell \in {\mathbb {Z}}} 2^{\frac{\ell }{2}}\Bigl (\bigl (\sum _{k=\ell -1}^\infty 2^{-k} \Vert \Delta ^\mathrm{h}_k \Delta ^\mathrm{v}_\ell a\Vert _{L^4_\mathrm{h}(L^2_\mathrm{v})}^2\bigr )^{\frac{1}{2}} +\Vert S^\mathrm{h}_{\ell -1}\Delta ^\mathrm{v}_\ell a\Vert _{L^2}\Bigr )<\infty . \end{aligned}$$

Chemin and the third author [8] proved the global well-posedness of (ANS) with initial data being small in the space \(\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4.\) In particular, this result ensures the global well-posedness of (ANS) with initial data \(u_0^\varepsilon (x)\) given by (1.5) as long as \(\varepsilon \) is sufficiently small. Furthermore the second and third authors [17] proved the global well-posedness of (ANS) provided that the initial data \(u_0=(u_0^\mathrm{h},u_0^3)\) satisfies that

$$\begin{aligned} \Vert u^{\mathrm{h}}_0\Vert _{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4}\exp \bigl (C\Vert u^3_0\Vert _{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4}^4\bigr )\leqq c_0 \end{aligned}$$
(1.6)

for some \(c_0\) sufficiently small. We remark that this result corresponds to Cannone, Meyer and Planchon’s result in [3] for the classical Navier–Stokes system, where the authors proved that if the initial data satisfies that

$$\begin{aligned} \Vert u_{0}\Vert _{ \dot{B}^{-1+\frac{3}{p}}_{p,\infty } }\leqq c\nu \end{aligned}$$

for some p greater than 3 and some constant c small enough, then (NS) is globally well-posed. The end-point result in this direction is due to Koch and Tataru [14] for initial data in the space of  \(\partial BMO.\)

On the other hand, motivated by the study of the global well-posedness of the classical Navier–Stokes system with slowly varying initial data [6, 7, 9], the first and third authors proved the following theorem for (NS) in [15]:

Theorem 1.1

Let \(\delta \in ]0,1[, u_0=(u_0^\mathrm{h},u_0^3)\in H^{\frac{1}{2}}({\mathbb {R}}^3)\cap B^{0,\frac{1}{2}}_{2,1}({\mathbb {R}}^3)\) with  \(u^{\mathrm{h}}_0\) belonging to  \(L^2({\mathbb {R}}^3)\cap L^\infty ({\mathbb {R}}_\mathrm{v}; H^{-\delta }({\mathbb {R}}^2_\mathrm{h}))\cap L^\infty ({\mathbb {R}}_\mathrm{v}; H^3({\mathbb {R}}^2_\mathrm{h})).\) If we assume in addition that \(\partial _3u_0\in H^{-\frac{1}{2},0},\) then there exists a small enough positive constant \(\varepsilon _0\) such that if

$$\begin{aligned} \Vert \partial _3u_0\Vert _{H^{-\frac{1}{2},0}}^2\exp \left( C\bigl (A_\delta (u^{\mathrm{h}}_0)+B_\delta (u_0)\bigr )\right) \leqq \varepsilon _0, \end{aligned}$$
(1.7)

(NS) has a unique global solution \(u\in C\bigl ({\mathbb {R}}^+;H ^{\frac{1}{2}}\bigr )\cap L^2\bigl ({\mathbb {R}}^+;H ^{\frac{3}{2}}\bigr ),\) where

$$\begin{aligned} \begin{aligned}&A_\delta (u_0^\mathrm{h})\buildrel {\mathrm{def}}\over =\biggl (\frac{\Vert \nabla _\mathrm{h}u^{\mathrm{h}}_0\Vert _{L^\infty _{\mathrm{v}} (L^2_{\mathrm{h}})}^2 \Vert u^{\mathrm{h}}_0\Vert _{L^\infty _\mathrm{v} (B^{-\delta }_{2,\infty })_\mathrm{h}} ^{\frac{2}{\delta }}}{\Vert u^{\mathrm{h}}_0\Vert _{L^\infty _{\mathrm{v}} (L^2_{\mathrm{h}})}^{\frac{2}{\delta }}} +\Vert u^{\mathrm{h}}_0\Vert _{L^\infty _{\mathrm{v}} (L^2_{\mathrm{h}})}^{2}\biggr ) \exp \Big (C_\delta (1+\Vert u^{\mathrm{h}}_0\Vert _{L^\infty _{\mathrm{v}} (L^2_{\mathrm{h}})}^4)\Bigr ),\\&\quad \mathfrak {A}_\delta (u^{\mathrm{h}}_0)\buildrel {\mathrm{def}}\over =\frac{\Vert u^{\mathrm{h}}_0\Vert _{L^\infty _\mathrm{v}(L^2_\mathrm{h})}^3\Vert \nabla _\mathrm{h}^3 u^{\mathrm{h}}_0\Vert _{L^\infty _\mathrm{v}(L^2_\mathrm{h})}^{\frac{1}{2}}}{\Vert \nabla _\mathrm{h}u^{\mathrm{h}}_0\Vert _{L^\infty _\mathrm{v}(L^2_\mathrm{h})}^{\frac{3}{2}}}+A_\delta (u^{\mathrm{h}}_0), \end{aligned} \end{aligned}$$
(1.8)

and

$$\begin{aligned} B_\delta (u_0)\buildrel {\mathrm{def}}\over =\Vert u^{\mathrm{h}}_0\Vert _{B^{0,\frac{1}{2}}_{2,1}} \exp \bigl (C\mathfrak {A}_\delta (u^{\mathrm{h}}_0)\bigr )+ \Vert u_0\Vert _{B^{0,\frac{1}{2}}_{2,1}} \exp \Bigl (\Vert u^{\mathrm{h}}_0\Vert _{B^{0,\frac{1}{2}}_{2,1}} \exp \bigl (C\mathfrak {A}_\delta (u^{\mathrm{h}}_0)\bigr )\Bigr ) \end{aligned}$$
(1.9)

are scaling invariant under the scaling transformation (1.2).

We remark that Theorem 1.1 ensures the global well-posedness of (NS) with initial data

$$\begin{aligned} u_0^\varepsilon (x)=(v^{\mathrm{h}}_0 +\varepsilon w^\mathrm{h}_0,w^3_0)(x_\mathrm{h},\varepsilon x_3) \quad \hbox {with}\quad {{\mathrm{div}}_{\mathrm{h}}}\,v^{\mathrm{h}}_0=0=\mathrm{div}\,w_0 \end{aligned}$$
(1.10)

for \(\varepsilon \leqq \varepsilon _0,\) which was first proved in [6]. We mention that the proof of Theorem 1.1 requires a regularity criteria in [10], which can only be proved for the classical Navier–Stokes system so far.

Motivated by [15, 17, 19], here we are going to study the global well-posedness of (ANS) with initial data \(u_0\) satisfying \(\partial _3u_0\) being sufficiently small in some critical spaces.

The main result of this paper is as follows:

Theorem 1.2

Let \(\Lambda _\mathrm{h}^{-1}\) be a Fourier multiplier with symbol \(|\xi _\mathrm{h}|^{-1},\) let \(u_0\in \mathcal {B}^{0,\frac{1}{2}}\) be a solenoidal vector field with \(\Lambda _\mathrm{h}^{-1}\partial _3 u_0\in {\mathcal {B}^{0,\frac{1}{2}}}.\) Then there exist some sufficiently small positive constant \(\varepsilon _0\) and some universal positive constants LMN so that for \(\mathfrak {A}_N\bigl (\Vert u^\mathrm{h}_0\Vert _{\mathcal {B}^{0,\frac{1}{2}}}\bigr )\) given by (3.5) if

$$\begin{aligned} \begin{aligned} \Vert \Lambda ^{-1}_\mathrm{h}\partial _3 u_0\Vert _{\mathcal {B}^{0,\frac{1}{2}}} \exp \Bigl (L\bigl (1+\Vert u_0^3\Vert _{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4}^4\bigr ) \exp \bigl (M\mathfrak {A}_N^4\bigl (\Vert u^\mathrm{h}_0\Vert _{\mathcal {B}^{0,\frac{1}{2}}}\bigr )\bigr )\Bigr ) \leqq \varepsilon _0, \end{aligned} \end{aligned}$$
(1.11)

(ANS) has a unique global solution \(u=\mathfrak {v}+e^{t\Delta _h} \begin{pmatrix} 0\\ u^3_{0,\mathrm{hh}} \end{pmatrix}\) with \(\mathfrak {v}\in C([0,\infty [\,;\mathcal {B}^{0,\frac{1}{2}})\) and \(\nabla _{\mathrm{h}}\mathfrak {v}\in L^2([0,\infty [\,;\mathcal {B}^{0,\frac{1}{2}}),\) where \(u_{0,\mathrm{hh}}^3\buildrel {\mathrm{def}}\over =\sum _{k\geqq \ell -1}\Delta _k^\mathrm{h}\Delta _\ell ^\mathrm{v}u_0^3.\)

We remark that all the norms of \(u^0\) in (1.11) is scaling invariant under the scaling transformation (1.2). Especially for the term \(\Vert \Lambda ^{-1}_\mathrm{h}\partial _3 u_0\Vert _{\mathcal {B}^{0,\frac{1}{2}}},\) we do not know how to propagate this regularity for the solutions of 3-D Navier–Stokes system. In the sequel, we shall only propagate this regularity for the solutions of 2-D Navier–Stokes system with a parameter [(see (3.4) and (3.7)]. With regular initial data, we may write explicitly the constant \(\mathfrak {A}_N\bigl (\Vert u^\mathrm{h}_0\Vert _{\mathcal {B}^{0,\frac{1}{2}}}\bigr ).\) For instance, we have

Corollary 1.1

Let \(u_0\in L^2\) be a solenoidal vector field with \(\partial _3u_0\in L^2\) and \(\Lambda _\mathrm{h}^{-1}\partial _3 u_0\in {\mathcal {B}^{0,\frac{1}{2}}}.\) Then there exist some sufficiently small positive constant \(\varepsilon _0\) and some universal positive constants LM so that if

$$\begin{aligned} \Vert \Lambda ^{-1}_\mathrm{h}\partial _3 u_0\Vert _{\mathcal {B}^{0,\frac{1}{2}}} \exp \Bigl (L\bigl (1+\Vert u_0^3\Vert _{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4}^4\bigr ) \exp \left( \exp \bigl (M\Vert u^{\mathrm{h}}_0\Vert _{L^2}\Vert \partial _3u^{\mathrm{h}}_0\Vert _{L^2}\bigr )\right) \Bigr ) \leqq \varepsilon _0, \end{aligned}$$
(1.12)

(ANS) has a unique global solution u as in Theorem 1.2.

Remark 1.1

Several remarks are in order about Theorem 1.2:

  1. (a)

    It follows from [8] that

    $$ \Vert u_0^3\Vert _{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4}\lesssim \Vert u_0^3\Vert _{\mathcal {B}^{0,\frac{1}{2}}},$$

    so that the smallness condition (1.11) and (1.12) can also be formulated as

    $$\begin{aligned} \begin{aligned} \Vert \Lambda ^{-1}_\mathrm{h}\partial _3 u_0\Vert _{\mathcal {B}^{0,\frac{1}{2}}} \exp \Bigl (L\bigl (1+\Vert u_0^3\Vert _{\mathcal {B}^{0,\frac{1}{2}}}^4\bigr ) \exp \bigl (M\mathfrak {A}_N^4\bigl (\Vert u^\mathrm{h}_0\Vert _{\mathcal {B}^{0,\frac{1}{2}}}\bigr )\bigr )\Bigr ) \leqq \varepsilon _0, \end{aligned} \end{aligned}$$
    (1.13)

    and

    $$\begin{aligned} \Vert \Lambda ^{-1}_\mathrm{h}\partial _3 u_0\Vert _{\mathcal {B}^{0,\frac{1}{2}}} \exp \Bigl (L\bigl (1+\Vert u_0^3\Vert _{\mathcal {B}^{0,\frac{1}{2}}}^4\bigr ) \exp \left( \exp \bigl (M\Vert u^{\mathrm{h}}_0\Vert _{L^2}\Vert \partial _3u^{\mathrm{h}}_0\Vert _{L^2}\bigr )\right) \Bigr ) \leqq \varepsilon _0. \end{aligned}$$
    (1.14)
  2. (b)

    Due to \(\mathrm{div}\,u_0=0,\) we find

    $$\begin{aligned} \Vert \Lambda ^{-1}_\mathrm{h}\partial _3 u_0\Vert _{\mathcal {B}^{0,\frac{1}{2}}}= \Vert (\Lambda ^{-1}_\mathrm{h}\partial _3u^{\mathrm{h}}_0,-\Lambda ^{-1}_\mathrm{h}{{\mathrm{div}}_{\mathrm{h}}}\,u^{\mathrm{h}}_0)\Vert _{\mathcal {B}^{0,\frac{1}{2}}}. \end{aligned}$$

    Therefore the smallness condition (1.11) is of a similar type as (1.4). Yet roughly speaking, (1.11) requires only \(\partial _3 u^{\mathrm{h}}_0\) and \({{\mathrm{div}}_{\mathrm{h}}}\,u^{\mathrm{h}}_0\) to be small in some scaling invariant space, but without any restriction on \({{\mathrm{curl}}_{\mathrm{h}}}\,u^{\mathrm{h}}_0\). Thus the smallness condition (1.11) is weaker than (1.4).

  3. (c)

    Let \(w_0\) be a smooth solenoidal vector field, we observe that the data

    $$\begin{aligned} u_0^\varepsilon (x)=\bigl (\varepsilon (-\ln \varepsilon )^\delta w^\mathrm{h}_0, (-\ln \varepsilon )^\delta w^3_0\bigr )(x_\mathrm{h},\varepsilon x_3) \quad \hbox {with}\quad \delta \in ]0,1/4[ \end{aligned}$$

    satisies (1.4) for \(\varepsilon \) sufficiently small.

    While since our smallness condition (1.14) does not have any restriction on \({\mathrm{curl}\,}u_0^\mathrm{h},\) for any smooth vector field \(v^\mathrm{h}_0\) satisfying \({{\mathrm{div}}_{\mathrm{h}}}\,v^h_0=0,\) we find

    $$\begin{aligned} u_0^\varepsilon (x)=\bigl (v^{\mathrm{h}}_0+\varepsilon (-\ln \varepsilon )^\delta w^\mathrm{h}_0, (-\ln \varepsilon )^\delta w^3_0\bigr )(x_\mathrm{h},\varepsilon x_3) \quad \hbox {with}\quad \delta \in ]0,1/4[ \end{aligned}$$
    (1.15)

    satisfies (1.14) for any \(\varepsilon \) sufficiently small. Therefore Theorem 1.2 ensures the global well-posedness of (ANS) with initial data given by (1.15). Compared with (1.10), which corresponds to \(\delta =0\) in (1.15), this type of result is new even for the classical Navier–Stokes system.

  4. (d)

    Given \(\phi \in \mathcal {S}({\mathbb {R}}^3)\), we deduce from Proposition 1.1 in [8] that

    $$\begin{aligned} \Vert e^{ix_1/\varepsilon }\phi (x)\Vert _{{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4}} \leqq C\varepsilon ^{\frac{1}{2}}. \end{aligned}$$

    As a result, we find that for any \(\delta \in ]0,1/4[\), the following class of initial data:

    $$\begin{aligned}&u_0^\varepsilon (x)=(v^\mathrm{h}, 0)(x_\mathrm{h},\varepsilon x_3) +(-\ln \varepsilon )^\delta \sin (x_1/\varepsilon )\nonumber \\&\bigl (0,-\varepsilon ^{\frac{1}{2}}\partial _3\phi (x_\mathrm{h},\varepsilon x_3), \varepsilon ^{-\frac{1}{2}}\partial _2\phi (x_\mathrm{h},\varepsilon x_3)\bigr ), \end{aligned}$$
    (1.16)

    satisfies the smallness condition (1.13) for small enough \(\varepsilon \), and hence the data given by (1.16) can also generate unique global solution of (ANS).

  5. (e)

    Since all the results that work for the anisotropic Navier–Stokes system (ANS) should automatically do for the classical Navier–Stokes system (NS),  Theorem 1.2 holds also for (NS).

Let us end this section with some notations that will be used throughout this paper.

Notations: Let AB be two operators, we denote \([A;B]=AB-BA,\) the commutator between A and B,  for \(a\lesssim b\), we means that there is a uniform constant C,  which may be different in each occurrence, such that \(a\leqq Cb\). We shall denote by \((a|b)_{L^2}\) the \(L^2({\mathbb {R}}^3)\) inner product of a and b. \(\left( d_j\right) _{j\in {\mathbb {Z}}}\) designates a generic elements on the unit sphere of \(\ell ^1({\mathbb {Z}})\), i.e. \(\sum _{j\in {\mathbb {Z}}}d_j=1\). Finally, we denote \(L^r_T(L^p_\mathrm{h}(L^q_\mathrm{v}))\) the space \(L^r([0,T]; L^p({\mathbb {R}}_{x_1}\times {\mathbb {R}}_{x_2}; L^q({\mathbb {R}}_{x_3}))),\) and \(\nabla _\mathrm{h}\buildrel {\mathrm{def}}\over =(\partial _{x_1},\partial _{x_2}),\)\(\mathrm{div}\,_\mathrm{h}= \partial _{x_1}+\partial _{x_2}\).

2 Littlewood–Paley Theory

In this section, we shall collect some basic facts on anisotropic Littlewood–Paley theory. We first recall the following anisotropic Bernstein inequalities from [8, 16]:

Lemma 2.1

Let \(\mathbf{B}_{\mathrm{h}}\) (resp. \(\mathbf{B}_\mathrm{v}\)) a ball of \({\mathbb {R}}^2_{\mathrm{h}}\) (resp. \({\mathbb {R}}_\mathrm{v}\)), and \(\mathcal {C}_{\mathrm{h}}\) (resp. \(\mathcal {C}_\mathrm{v}\)) a ring of \({\mathbb {R}}^2_{\mathrm{h}}\) (resp. \({\mathbb {R}}_\mathrm{v}\)); let \(1\leqq p_2\leqq p_1\leqq \infty \) and  \(1\leqq q_2\leqq q_1\leqq \infty .\) Then it holds that

$$\begin{aligned} \begin{aligned} \text{ if }\ \ {\mathrm{Supp}\,}\widehat{a}\subset 2^k\mathbf{B}_{\mathrm{h}}&\Rightarrow \Vert \partial _{x_\mathrm{h}}^\alpha a\Vert _{L^{p_1}_\mathrm{h}(L^{q_1}_\mathrm{v})} \lesssim 2^{k\left( |\alpha |+\frac{2}{p_2}-\frac{2}{p_1}\right) } \Vert a\Vert _{L^{p_2}_\mathrm{h}(L^{q_1}_\mathrm{v})};\\ \text{ if }\ \ {\mathrm{Supp}\,}\widehat{a}\subset 2^\ell \mathbf{B}_\mathrm{v}&\Rightarrow \Vert \partial _{x_3}^\beta a\Vert _{L^{p_1}_\mathrm{h}(L^{q_1}_\mathrm{v})} \lesssim 2^{\ell \left( \beta +\frac{1}{q_2}-\frac{1}{q_1}\right) } \Vert a\Vert _{L^{p_1}_\mathrm{h}(L^{q_2}_\mathrm{v})};\\ \text{ if }\ \ {\mathrm{Supp}\,}\widehat{a}\subset 2^k\mathcal {C}_{\mathrm{h}}&\Rightarrow \Vert a\Vert _{L^{p_1}_\mathrm{h}(L^{q_1}_\mathrm{v})} \lesssim 2^{-kN}\sup _{|\alpha |=N} \Vert \partial _{x_\mathrm{h}}^\alpha a\Vert _{L^{p_1}_\mathrm{h}(L^{q_1}_\mathrm{v})};\\ \text{ if }\ \ {\mathrm{Supp}\,}\widehat{a}\subset 2^\ell \mathcal {C}_\mathrm{v}&\Rightarrow \Vert a\Vert _{L^{p_1}_\mathrm{h}(L^{q_1}_\mathrm{v})} \lesssim 2^{-\ell N} \Vert \partial _{x_3}^N a\Vert _{L^{p_1}_\mathrm{h}(L^{q_1}_\mathrm{v})}. \end{aligned} \end{aligned}$$

Definition 2.1

For any \(p\in [1,\infty ],\), let us define the Chemin–Lerner type norm

$$\begin{aligned} \Vert a\Vert _{\widetilde{L}^p_T(\mathcal {B}^{0,\frac{1}{2}})}\buildrel {\mathrm{def}}\over =\sum _{\ell \in {\mathbb {Z}}}2^{\frac{\ell }{2}} \Vert \Delta _{\ell }^{\mathrm{v}}a\Vert _{L^p_T(L^2({\mathbb {R}}^3))}. \end{aligned}$$

In particular, we denote

$$\begin{aligned} \Vert a\Vert _{\mathcal {B}^{0,\frac{1}{2}}(T)}\buildrel {\mathrm{def}}\over =\Vert a\Vert _{\widetilde{L}^\infty _T(\mathcal {B}^{0,\frac{1}{2}})} +\Vert \nabla _{\mathrm{h}}a\Vert _{\widetilde{L}^2_T(\mathcal {B}^{0,\frac{1}{2}})}. \end{aligned}$$

We remark that the inhomogeneous version of the anisotropic Sobolev space \(H^{0,1}\) can be continuously imbedded into \(\mathcal {B}^{0,\frac{1}{2}}.\) Indeed for any integer N, we deduce from Lemma 2.1 that

$$\begin{aligned} \begin{aligned} \Vert a\Vert _{\mathcal {B}^{0,\frac{1}{2}}}&=\sum _{\ell \leqq N}2^{\frac{\ell }{2}}\Vert \Delta _{\ell }^{\mathrm{v}}a\Vert _{L^2} +\sum _{\ell>N}2^{\frac{\ell }{2}}\Vert \Delta _{\ell }^{\mathrm{v}}a\Vert _{L^2}\\&\leqq \sum _{\ell \leqq N} 2^{\frac{\ell }{2}}\Vert \Delta _{\ell }^{\mathrm{v}}a\Vert _{L^2} +\sum _{\ell >N}2^{-\frac{\ell }{2}}\Vert \partial _3\Delta _{\ell }^{\mathrm{v}}a\Vert _{L^2}\\&\lesssim 2^{\frac{N}{2}}\Vert a\Vert _{L^2}+2^{-\frac{N}{2}}\Vert \partial _3 a\Vert _{L^2}. \end{aligned} \end{aligned}$$

Taking the integer N so that \(2^{N}\sim \Vert \partial _3 a\Vert _{L^2}\Vert a\Vert _{L^2}^{-1}\) in the above inequality leads to

$$\begin{aligned} \Vert a\Vert _{\mathcal {B}^{0,\frac{1}{2}}}\lesssim \Vert a\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3 a\Vert _{L^2}^{\frac{1}{2}}. \end{aligned}$$
(2.1)

Along the same lines, we have

$$\begin{aligned} \Vert a\Vert _{\widetilde{L}^p_T(\mathcal {B}^{0,\frac{1}{2}})}\lesssim \Vert a\Vert _{L^p_T(L^2)}^{\frac{1}{2}} \Vert \partial _3 a\Vert _{L^p_T(L^2)}^{\frac{1}{2}}\quad \forall \ p \in [1,\infty ]. \end{aligned}$$
(2.2)

To overcome the difficulty that one can not use Gronwall’s inequality in the Chemin–Lerner type norms, we recall the following time-weighted Chemin–Lerner norm from [17]:

Definition 2.2

Let \(f(t)\in L^1_\mathrm{{loc}}({\mathbb {R}}_+)\), \(f(t)\geqq 0\). We define

$$\begin{aligned} \Vert a\Vert _{\widetilde{L}^2_{T,f}(\mathcal {B}^{0,\frac{1}{2}})}\buildrel {\mathrm{def}}\over =\sum _{\ell \in {\mathbb {Z}}} 2^{\frac{\ell }{2}} \Bigl (\int \nolimits _0^Tf(t)\Vert \Delta _{\ell }^{\mathrm{v}}a(t)\Vert _{L^2}^2\,\hbox {d}t\Bigr )^{\frac{1}{2}}. \end{aligned}$$

In order to take into account functions with oscillations in the horizontal variables, we recall the following anisotropic Besov type space with negative indices from [8]:

Definition 2.3

For any \( p\in [1,\infty ],\) we define

$$\Vert a\Vert _{\widetilde{L}^p(\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4)}\buildrel {\mathrm{def}}\over =\sum _{\ell \in {\mathbb {Z}}} 2^{\frac{\ell }{2}}\biggl (\Bigl (\sum _{k=\ell -1}^\infty 2^{-k} \Vert \Delta ^\mathrm{h}_k \Delta ^\mathrm{v}_\ell a\Vert _{L^p_T(L^4_\mathrm{h}(L^2_\mathrm{v}))}^2\Bigr )^{\frac{1}{2}} +\Vert S^\mathrm{h}_{\ell -1}\Delta ^\mathrm{v}_\ell a\Vert _{L^p_T(L^2)}\biggr ).$$

In particular, we denote

$$\begin{aligned} \Vert a\Vert _{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4(T)}\buildrel {\mathrm{def}}\over =\Vert a\Vert _{\widetilde{L}^\infty _T({\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4})} +\Vert \nabla _{\mathrm{h}}a\Vert _{\widetilde{L}^2_T({\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4})}. \end{aligned}$$

In the sequel, for \(a\in \mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4,\) we shall frequently use the following decomposition:

$$\begin{aligned} a=a_\mathrm{lh}+a_\mathrm{hh} \quad \hbox {with}\quad a_\mathrm{lh}\buildrel {\mathrm{def}}\over =\sum _{\ell \in {\mathbb {Z}}}S_{\ell -1}^\mathrm{h}\Delta _\ell ^\mathrm{v}a \quad \hbox {and}\quad a_\mathrm{hh}\buildrel {\mathrm{def}}\over =\sum _{k\geqq \ell -1}\Delta _k^\mathrm{h}\Delta _\ell ^\mathrm{v}a. \end{aligned}$$
(2.3)

Lemma 2.2

(Lemma 2.5 in [8]) For any \(a\in {\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4},\) it holds that

$$\begin{aligned} \Vert e^{t\Delta _\mathrm{h}}a_{\mathrm{h}\mathrm{h}}\Vert _{{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4}(\infty )}\lesssim \Vert a\Vert _{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4}. \end{aligned}$$

Definition 2.4

Let us define

$$\begin{aligned} \Vert a\Vert _{\mathcal {B}_4^{0,\frac{1}{2}}}\buildrel {\mathrm{def}}\over =\sum _{\ell \in {\mathbb {Z}}}2^{\frac{\ell }{2}} \Vert \Delta _{\ell }^{\mathrm{v}}a\Vert _{L^4_\mathrm{h}(L^2_\mathrm{v})}\quad \text{ and }\quad \Vert a\Vert _{\widetilde{L}^4_t(\mathcal {B}_4^{0,\frac{1}{2}})}\buildrel {\mathrm{def}}\over =\sum _{\ell \in {\mathbb {Z}}}2^{\frac{\ell }{2}} \Vert \Delta _{\ell }^{\mathrm{v}}a\Vert _{L^4_t(L^4_\mathrm{h}(L^2_\mathrm{v}))}. \end{aligned}$$

In view of the 2-D interpolation inequality that

$$\begin{aligned} \Vert a\Vert _{L^4({\mathbb {R}}^2)}\lesssim \Vert a\Vert _{L^2({\mathbb {R}}^2)}^{\frac{1}{2}}\Vert \nabla _\mathrm{h}a\Vert _{L^2({\mathbb {R}}^2)}^{\frac{1}{2}}, \end{aligned}$$
(2.4)

we find

$$\begin{aligned} \begin{aligned} \Vert a\Vert _{\mathcal {B}_4^{0,\frac{1}{2}}}^2&\lesssim \Bigl (\sum _{\ell \in {\mathbb {Z}}}2^{\frac{\ell }{2}} \Vert \Delta _{\ell }^{\mathrm{v}}a\Vert _{L^2}^{\frac{1}{2}}\Vert \Delta _{\ell }^{\mathrm{v}}\nabla _{\mathrm{h}}a\Vert _{L^2}^{\frac{1}{2}}\Bigr )^2\\&\leqq \Bigl (\sum _{\ell \in {\mathbb {Z}}}2^{\frac{\ell }{2}} \Vert \Delta _{\ell }^{\mathrm{v}}a\Vert _{L^2}\Bigr )\Bigl (\sum _{\ell \in {\mathbb {Z}}}2^{\frac{\ell }{2}}\Vert \Delta _{\ell }^{\mathrm{v}}\nabla _{\mathrm{h}}a\Vert _{L^2}\Bigr ) =\Vert a\Vert _{\mathcal {B}^{0,\frac{1}{2}}}\Vert \nabla _{\mathrm{h}}a\Vert _{\mathcal {B}^{0,\frac{1}{2}}}. \end{aligned} \end{aligned}$$
(2.5)

Similarly, we have

$$\begin{aligned} \Vert a\Vert _{\widetilde{L}^4_t(\mathcal {B}_4^{0,\frac{1}{2}})}^2 \lesssim \Vert a\Vert _{\widetilde{L}^\infty _t(\mathcal {B}^{0,\frac{1}{2}})}\Vert \nabla _{\mathrm{h}}a\Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}. \end{aligned}$$
(2.6)

Before preceding, let us recall Bony’s decomposition for the vertical variable from [1]:

$$\begin{aligned} \begin{aligned} ab=T^\mathrm{v}_a b+ R^\mathrm{v}(a,b)\quad \text{ with }\quad T^\mathrm{v}_a b=\sum _{\ell \in {\mathbb {Z}}}S^\mathrm{v}_{\ell -1}a\Delta ^\mathrm{v}_\ell b,\quad R^\mathrm{v}(a,b)=\sum _{\ell \in {\mathbb {Z}}}\Delta ^\mathrm{v}_\ell a S^\mathrm{v}_{\ell +2}b. \end{aligned} \end{aligned}$$
(2.7)

Sometimes we shall also use Bony’s decomposition for the horizontal variables.

Let us now apply the above basic facts on Littlewood–Paley theory to prove the following proposition:

Proposition 2.1

For any \(a\in {\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4}(T),\) it holds that

$$\begin{aligned} \Vert a\Vert _{\widetilde{L}_T^4(\mathcal {B}_4^{0,\frac{1}{2}})}\lesssim \Vert a\Vert _{{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4}(T)}. \end{aligned}$$
(2.8)

Proof

In view of (2.3) and Definition 2.3, we get, by applying (2.6), that

$$\begin{aligned} \begin{aligned} \Vert a_\mathrm{lh}\Vert _{\widetilde{L}_T^4(\mathcal {B}_4^{0,\frac{1}{2}})}&\lesssim \Vert a_\mathrm{lh}\Vert _{\widetilde{L}_T^\infty (\mathcal {B}^{0,\frac{1}{2}})}^{\frac{1}{2}} \Vert \nabla _\mathrm{h}a_\mathrm{lh}\Vert _{\widetilde{L}_T^2(\mathcal {B}^{0,\frac{1}{2}})}^{\frac{1}{2}}\\&\lesssim \Vert a\Vert _{\widetilde{L}^\infty _T\left( {\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4}\right) }^{\frac{1}{2}} \Vert \nabla _{\mathrm{h}}a\Vert _{\widetilde{L}^2_T\left( {\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4}\right) }^{\frac{1}{2}}. \end{aligned} \end{aligned}$$

Then it remains to prove (2.8) for \(a_\mathrm{hh}.\) Indeed in view of Definition 2.4, we write

$$\begin{aligned} \Vert a_\mathrm{hh}\Vert _{\widetilde{L}_T^4\left( \mathcal {B}_4^{0,\frac{1}{2}}\right) } =\sum _{\ell \in {\mathbb {Z}}}2^{\frac{\ell }{2}} \Vert (\Delta _{\ell }^{\mathrm{v}}a_\mathrm{hh})^2\Vert _{L_T^2(L^2_\mathrm{h}(L^1_\mathrm{v}))}^{\frac{1}{2}}. \end{aligned}$$

Applying Bony’s decomposition for the horizontal variables yields

$$\begin{aligned} (\Delta _{\ell }^{\mathrm{v}}a_\mathrm{hh})^2=\sum _{k\in {\mathbb {Z}}}S^\mathrm{h}_{k-1}\Delta _{\ell }^{\mathrm{v}}a_\mathrm{hh}\Delta _k^{\mathrm{h}}\Delta _{\ell }^{\mathrm{v}}a_\mathrm{hh} +\sum _{k\in {\mathbb {Z}}}S^\mathrm{h}_{k+2}\Delta _{\ell }^{\mathrm{v}}a_\mathrm{hh}\Delta _k^{\mathrm{h}}\Delta _{\ell }^{\mathrm{v}}a_\mathrm{hh}. \end{aligned}$$
(2.9)

We observe that

$$\begin{aligned} \begin{aligned}&\sum _{\ell \in {\mathbb {Z}}}2^{\frac{\ell }{2}}\Bigl (\sum _{k\in {\mathbb {Z}}} \Vert S^\mathrm{h}_{k-1}\Delta _{\ell }^{\mathrm{v}}a_\mathrm{hh}\Delta _k^{\mathrm{h}}\Delta _{\ell }^{\mathrm{v}}a_\mathrm{hh}\Vert _{L_T^2(L^2_\mathrm{h}(L^1_\mathrm{v}))}\Bigr )^{\frac{1}{2}}\\&\quad \leqq \biggl (\sum _{\ell \in {\mathbb {Z}}}2^{\frac{\ell }{2}}\Bigl (\sum _{k\in {\mathbb {Z}}} 2^{-k}\Vert S^\mathrm{h}_{k-1}\Delta _{\ell }^{\mathrm{v}}a_\mathrm{hh}\Vert _{L_T^\infty (L^4_\mathrm{h}(L^2_\mathrm{v}))}^2\Bigr )^{\frac{1}{2}} \biggr )^{\frac{1}{2}}\\&\quad \quad \times \biggl (\sum _{\ell \in {\mathbb {Z}}}2^{\frac{\ell }{2}}\Bigl (\sum _{k\in {\mathbb {Z}}} 2^{k}\Vert \Delta _k^{\mathrm{h}}\Delta _{\ell }^{\mathrm{v}}a_\mathrm{hh}\Vert _{L_T^2(L^4_\mathrm{h}(L^2_\mathrm{v}))}^2\Bigr )^{\frac{1}{2}}\biggr )^{\frac{1}{2}}\\&\quad \leqq \biggl (\sum _{\ell \in {\mathbb {Z}}}2^{\frac{\ell }{2}}\Bigl (\sum _{k\in {\mathbb {Z}}} 2^{-k}\Vert S^\mathrm{h}_{k-1}\Delta _{\ell }^{\mathrm{v}}a_\mathrm{hh}\Vert _{L_T^\infty (L^4_\mathrm{h}(L^2_\mathrm{v}))}^2\Bigr )^{\frac{1}{2}} \biggr )^{\frac{1}{2}}\Vert \nabla _{\mathrm{h}}a_\mathrm{hh}\Vert _{\widetilde{L}^2_T\left( {\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4}\right) }^{\frac{1}{2}}. \end{aligned} \end{aligned}$$

Whereas we get, by using Young’s inequality, that

$$\begin{aligned} \sum _{k\in {\mathbb {Z}}} 2^{-k}\Vert S^\mathrm{h}_{k-1}\Delta _{\ell }^{\mathrm{v}}a_\mathrm{hh}\Vert _{L_T^\infty (L^4_\mathrm{h}(L^2_\mathrm{v}))}^2&=\sum _{k\in {\mathbb {Z}}} \Bigl (\sum _{k'\leqq k-2}2^{-\frac{k-k'}{2}}2^{-\frac{k'}{2}}\Vert \Delta _{k'}^{\mathrm{h}}\Delta _{\ell }^{\mathrm{v}}a_\mathrm{hh}\Vert _{L_T^\infty (L^4_\mathrm{h}(L^2_\mathrm{v}))}\Bigr )^2\\&\leqq \sum _{k\in {\mathbb {Z}}}2^{-k}\Vert \Delta _k^{\mathrm{h}}\Delta _{\ell }^{\mathrm{v}}a_\mathrm{hh}\Vert _{L_T^\infty (L^4_\mathrm{h}(L^2_\mathrm{v}))}^2. \end{aligned}$$

As a result, it turns out that

$$\begin{aligned} \sum _{\ell \in {\mathbb {Z}}}2^{\frac{\ell }{2}}\Bigl (\sum _{k\in {\mathbb {Z}}} 2^{-k}\Vert S^\mathrm{h}_{k-1}\Delta _{\ell }^{\mathrm{v}}a_\mathrm{hh}\Vert _{L_T^\infty (L^4_\mathrm{h}(L^2_\mathrm{v}))}^2\Bigr )^{\frac{1}{2}} \leqq \Vert a\Vert _{\widetilde{L}^\infty _T\left( {\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4}\right) }, \end{aligned}$$

and

$$\begin{aligned} \sum _{\ell \in {\mathbb {Z}}}2^{\frac{\ell }{2}}\Bigl (\sum _{k\in {\mathbb {Z}}} \Vert S^\mathrm{h}_{k-1}\Delta _{\ell }^{\mathrm{v}}a_\mathrm{hh}\Delta _k^{\mathrm{h}}\Delta _{\ell }^{\mathrm{v}}a_\mathrm{hh}\Vert _{L_T^2(L^2_\mathrm{h}(L^1_\mathrm{v}))}\Bigr )^{\frac{1}{2}}\lesssim \Vert a\Vert _{{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4}(T)}. \end{aligned}$$

Along the same lines, we can prove that the second term in (2.9) shares the same estimate. This ensures that (2.8) holds for \(a_\mathrm{hh}.\) We thus complete the proof of the proposition. \(\quad \square \)

3 Sketch of the Proof

Motivated by the study of the global large solutions to the classical 3-D Navier–Stokes system with slowly varying initial data in one direction [6, 7, 9, 15], here we are going to decompose the solution of (ANS) as a sum of a solution to the two-dimensional Navier–Stokes system with a parameter and a solution to the three-dimensional perturbed anisotropic Navier–Stokes system. We point out that compared with the references [6, 7, 9, 15], here the 3-D solution to the perturbed anisotropic Navier–Stokes system will not be small. Indeed only its vertical component is not small. In order to deal with this part, we are going to appeal to the observation from [17, 19], where the authors proved the global well-posedness to 3-D anisotropic Navier–Stokes system with the horizontal components of the initial data being small [see the smallness conditions (1.4) and (1.6)].

For \(u^\mathrm{h}=(u^1,u^2),\) we first recall the two-dimensional Biot–Savart’s law:

$$\begin{aligned} u^\mathrm{h}=u^\mathrm{h}_{{\mathrm{curl}}} +u^\mathrm{h}_{{\mathrm{div}}}\quad \hbox {with}\quad u^\mathrm{h}_{{\mathrm{curl}}}\buildrel {\mathrm{def}}\over =\nabla _{\mathrm{h}}^\perp \Delta _{\mathrm{h}}^{-1}({{\mathrm{curl}}_{\mathrm{h}}}\,u^\mathrm{h}) \quad \hbox {and}\quad ~u^\mathrm{h}_{{\mathrm{div}}}\buildrel {\mathrm{def}}\over =\nabla _{\mathrm{h}}\Delta _{\mathrm{h}}^{-1}({{\mathrm{div}}_{\mathrm{h}}}\,u^{\mathrm{h}}), \end{aligned}$$
(3.1)

where \({{\mathrm{curl}}_{\mathrm{h}}}\,u^\mathrm{h}\buildrel {\mathrm{def}}\over =\partial _1u^2-\partial _2u^1\) and \({{\mathrm{div}}_{\mathrm{h}}}\,u^{\mathrm{h}}\buildrel {\mathrm{def}}\over =\partial _1u^1+\partial _2u^2.\)

In particular, let us decompose the horizontal components \(u_0^\mathrm{h}\) of the initial velocity \(u_0\) of (ANS) as the sum of \(u^\mathrm{h}_{0,{\mathrm{curl}}}\) and \( u^\mathrm{h}_{0,{\mathrm{div}}}\), and let us consider the following 2-D Navier–Stokes system with a parameter:

$$\begin{aligned} \left\{ \begin{array}{l} \displaystyle \partial _t \bar{u}^\mathrm{h}+\bar{u}^\mathrm{h}\cdot \nabla _{\mathrm{h}}\bar{u}^\mathrm{h}-\Delta _\mathrm{h}\bar{u}^\mathrm{h}=-\nabla _{\mathrm{h}}\bar{p}, \qquad (t,x)\in {\mathbb {R}}^+\times {\mathbb {R}}^3, \\ \displaystyle {{\mathrm{div}}_{\mathrm{h}}}\,\bar{u}^\mathrm{h}= 0, \\ \displaystyle \bar{u}^\mathrm{h}|_{t=0}=\bar{u}^\mathrm{h}_0=u^\mathrm{h}_{0,{\mathrm{curl}}}. \end{array}\right. \end{aligned}$$
(3.2)

Concerning the system (3.2), we have the following a priori estimates:

Proposition 3.1

Let \(\bar{u}^\mathrm{h}_0\in \mathcal {B}^{0,\frac{1}{2}} \) with \(\Lambda _\mathrm{h}^{-1}\partial _3\bar{u}^\mathrm{h}_0\in \mathcal {B}^{0,\frac{1}{2}}.\) Then (3.2) has a unique global solution so that for any time \(t>0\), it holds that

$$\begin{aligned} \Vert \bar{u}^\mathrm{h}\Vert _{L^\infty _t\big (\mathcal {B}^{0,\frac{1}{2}}\big )} +\Vert \nabla _{\mathrm{h}}\bar{u}^\mathrm{h}\Vert _{L^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )} \leqq C\mathfrak {A}_N\bigl (\Vert \bar{u}^\mathrm{h}_0\Vert _{\mathcal {B}^{0,\frac{1}{2}}}\bigr ), \end{aligned}$$
(3.3)

and

$$\begin{aligned} \begin{aligned} \Vert \Lambda _\mathrm{h}^{-1}\partial _3\bar{u}^\mathrm{h}\Vert _{\widetilde{L}^\infty _t(\mathcal {B}^{0,\frac{1}{2}})} +\Vert \partial _3\bar{u}^\mathrm{h}\Vert _{\widetilde{L}^2_t(\mathcal {B}^{0,\frac{1}{2}})} \leqq C\Vert \Lambda _\mathrm{h}^{-1}\partial _3\bar{u}^\mathrm{h}_0\Vert _{\mathcal {B}^{0,\frac{1}{2}}} \exp \left( C\mathfrak {A}_N^4\bigl (\Vert \bar{u}^\mathrm{h}_0\Vert _{\mathcal {B}^{0,\frac{1}{2}}}\bigr )\right) , \end{aligned} \end{aligned}$$
(3.4)

where

$$\begin{aligned} \begin{aligned} \bar{u}^\mathrm{h}_{0,N}&\buildrel {\mathrm{def}}\over =\mathcal {F}^{-1}\bigl (\mathbf{1}_{|\xi _3|\leqq \frac{1}{N} \text{ or } |\xi _3|\geqq N}\mathcal {F}({\bar{u}^\mathrm{h}_0})\bigr ) \quad \hbox {and}\quad \\ \mathfrak {A}_N\bigl (\Vert \bar{u}^\mathrm{h}_0\Vert _{\mathcal {B}^{0,\frac{1}{2}}}\bigr )&\buildrel {\mathrm{def}}\over =N^{\frac{1}{2}}\Vert \bar{u}^\mathrm{h}_0\Vert _{\mathcal {B}^{0,\frac{1}{2}}} \exp \bigl (C\Vert \bar{u}^\mathrm{h}_0\Vert _{\mathcal {B}^{0,\frac{1}{2}}}^2\bigr )\\&\quad +\bigl \Vert \bar{u}^\mathrm{h}_{0,N}\bigr \Vert _{\mathcal {B}^{0,\frac{1}{2}}}\exp \left( N^2\exp \bigl (C\Vert \bar{u}^\mathrm{h}_0\Vert _{\mathcal {B}^{0,\frac{1}{2}}}^2\bigr )\right) , \end{aligned} \end{aligned}$$
(3.5)

and N is taken so large that \(\bigl \Vert \bar{u}^\mathrm{h}_{0,N}\bigr \Vert _{\mathcal {B}^{0,\frac{1}{2}}}\) is sufficiently small.

The proof of Proposition 3.1 will be presented in Section 4.

Remark 3.1

Under the assumptions that \(\bar{u}^\mathrm{h}_0\in L^2\) with \(~\partial _3\bar{u}^\mathrm{h}_0\in L^2\) and \(\Lambda _\mathrm{h}^{-1}\partial _3\bar{u}^\mathrm{h}_0\in \mathcal {B}^{0,\frac{1}{2}},\) we have the following alternative estimates for (3.3) and (3.4):

$$\begin{aligned} \Vert \bar{u}^\mathrm{h}\Vert _{\widetilde{L}^\infty _t(\mathcal {B}^{0,\frac{1}{2}})} +\Vert \nabla _{\mathrm{h}}\bar{u}^\mathrm{h}\Vert _{\widetilde{L}^2_t(\mathcal {B}^{0,\frac{1}{2}})} \leqq \Vert \bar{u}^\mathrm{h}_0\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _3\bar{u}^\mathrm{h}_0\Vert _{L^2}^{\frac{1}{2}} \exp \bigl (C\Vert \bar{u}^\mathrm{h}_0\Vert _{L^2}\Vert \partial _3\bar{u}^\mathrm{h}_0\Vert _{L^2}\bigr ), \end{aligned}$$
(3.6)

and

$$\begin{aligned} \Vert \Lambda _\mathrm{h}^{-1}\partial _3\bar{u}^\mathrm{h}\Vert _{\widetilde{L}^\infty _t(\mathcal {B}^{0,\frac{1}{2}})} +\Vert \partial _3\bar{u}^\mathrm{h}\Vert _{\widetilde{L}^2_t(\mathcal {B}^{0,\frac{1}{2}})} \leqq \Vert \Lambda _\mathrm{h}^{-1}\partial _3\bar{u}^\mathrm{h}_0\Vert _{\mathcal {B}^{0,\frac{1}{2}}} \exp \left( \exp \bigl (C\Vert \bar{u}^\mathrm{h}_0\Vert _{L^2}\Vert \partial _3\bar{u}^\mathrm{h}_0\Vert _{L^2}\bigr )\right) . \end{aligned}$$
(3.7)

We shall present the proof right after (4.7).

We notice that

$$\begin{aligned} v_0\buildrel {\mathrm{def}}\over =u_0-\bigl (u^\mathrm{h}_{0,{\mathrm{curl}}},0\bigr )=\bigl (u^\mathrm{h}_{0,{\mathrm{div}}}, u_0^3\bigr ), \end{aligned}$$
(3.8)

which satisfies \(\mathrm{div}\,v_0=0,\) and yet \(v_0\) is not small according to our smallness condition (1.11).

Before proceeding, let us recall the main idea of the proof to Theorem 1.1 in [15]. The authors [15] first constructed \((\bar{u}^\mathrm{h}, \bar{p})\) via the system (3.2). Then in order to get rid of the large part of the initial data \(v_0,\) given by (3.8), the authors introduced a correction velocity, \(\widetilde{u},\) through the system

$$\begin{aligned} \left\{ \begin{array}{l} \displaystyle \partial _t \widetilde{u} +\bar{u}^\mathrm{h}\cdot \nabla _{\mathrm{h}}\widetilde{u} -\Delta \widetilde{u}=-\nabla \widetilde{p},\\ \displaystyle \mathrm{div}\,\widetilde{u} = 0, \\ \displaystyle \widetilde{u}^\mathrm{h}|_{t=0}=\widetilde{u}_0^\mathrm{h}=-\nabla _{\mathrm{h}}\Delta _{\mathrm{h}}^{-1}(\partial _3 u^3_0),\quad \widetilde{u}^3|_{t=0} =\widetilde{u}^3_0=u^3_0. \end{array}\right. \end{aligned}$$
(3.9)

With \(\bar{u}^\mathrm{h}\) and \(\widetilde{u}\) being determined respectively by the systems (3.2) and (3.9), the authors [15] decompose the solution (up) to the classical Navier–Stokes system (NS) as

$$\begin{aligned} u= \begin{pmatrix} \bar{u}^\mathrm{h}\\ 0\end{pmatrix}+\widetilde{u}+v,\quad p=\bar{p}+\widetilde{p}+q. \end{aligned}$$
(3.10)

The key estimate for v is as follows:

Proposition 3.2

Let \(u=(u^\mathrm{h},u^3)\in C([0,T^*[; H^{\frac{1}{2}})\cap L^2(]0,T^*[; H^{\frac{3}{2}})\) be a Fujita–Kato solution of (NS). We denote \(\omega \buildrel {\mathrm{def}}\over =\partial _1 v^2-\partial _2v^1\) and

$$\begin{aligned} \begin{aligned}&M(t)\buildrel {\mathrm{def}}\over =\Vert \nabla v^3(t)\Vert _{H^{-\frac{1}{2},0}}^2 +\Vert \omega (t)\Vert _{H^{-\frac{1}{2},0}}^2,\quad N(t)\buildrel {\mathrm{def}}\over =\Vert \nabla ^2 v^3(t)\Vert _{H^{-\frac{1}{2},0}}^2 +\Vert \nabla \omega (t)\Vert _{H^{-\frac{1}{2},0}}^2. \end{aligned} \end{aligned}$$
(3.11)

Then under the assumption (1.7), there exists some positive constant \(\eta \) such that

$$\begin{aligned} \sup _{t\in [0,T^*[}\Bigl (M(t)+\int \nolimits _0^tN(t')\,\mathrm{d}t'\Bigr ) \leqq \eta . \end{aligned}$$
(3.12)

Then in order to complete the proof of Theorem 1.1, the authors [15] invoked the following regularity criteria for the classical Navier–Stokes system:

Theorem 3.1

(Theorem 1.5 of [10]) Let \(u\in C([0,T^*[; H^{\frac{1}{2}})\cap L^2(]0,T^*[; H^{\frac{3}{2}})\) be a solution of (NS). If the maximal existence time \(T^*\) is finite, then for any \((p_{i,j})\) in \(]1,\infty [^9\), one has

$$\begin{aligned} \sum _{1\leqq i,j\leqq 3} \int \nolimits _0^{T^*} \Vert \partial _{i} u^{j}(t)\Vert ^{p_{i,j}}_{B_{\infty ,\infty } ^{-2+\frac{2}{p_{i,j}}}} \,\mathrm{d}t=\infty . \end{aligned}$$
(3.13)

We remark that Theorem 3.1 only works for the classical 3-D Navier–Stokes system. Therefore the above procedure to prove Theorem 1.1 cannot be applied to construct the global solutions to the 3-D anisotropic Navier–Stokes system.

On the other hand, we remark that the main observation in [17, 19] is that: by using \(\mathrm{div}\,u=0,\) (ANS) can be equivalently reformulated as

$$\begin{aligned} (ANS)\quad \left\{ \begin{array}{l} \displaystyle \partial _t u^\mathrm{h}+u^\mathrm{h}\cdot \nabla _\mathrm{h}u^\mathrm{h}+u^3\partial _3u^\mathrm{h}-\Delta _\mathrm{h}u^\mathrm{h}=-\nabla _\mathrm{h}p, \qquad (t,x)\in {\mathbb {R}}^+\times {\mathbb {R}}^3, \\ \displaystyle \partial _t u^3 +u^\mathrm{h}\cdot \nabla _3 u^\mathrm{h}-u^3{{\mathrm{div}}_{\mathrm{h}}}\,u^\mathrm{h}-\Delta _\mathrm{h}u^3=-\partial _3 p,\\ \displaystyle \mathrm{div}\,u = 0, \\ \displaystyle u|_{t=0}=(u_0^\mathrm{h},u_0^3), \end{array}\right. \end{aligned}$$

so that at least, seemingly, the \(u^3\) equation is a linear one; this explains why there is no size restriction for \(u_0^3\) in (1.4) and (1.6).

Motivated by [17, 19], for \(\bar{u}^\mathrm{h}\) being determined by the systems (3.2), we decompose the solution u of (ANS) as \(u=\begin{pmatrix} \bar{u}^\mathrm{h}\\ 0 \end{pmatrix}+v\). It is easy to verify that the remainder term v satisfies

$$\begin{aligned} \left\{ \begin{array}{l} \displaystyle \partial _t v^\mathrm{h}+v\cdot \nabla v^\mathrm{h}+\bar{u}^\mathrm{h}\cdot \nabla _{\mathrm{h}}v^\mathrm{h}+v\cdot \nabla \bar{u}^\mathrm{h}-\Delta _\mathrm{h}v^\mathrm{h}=-\nabla _{\mathrm{h}}p+\nabla _{\mathrm{h}}\bar{p},\\ \displaystyle \partial _t v^3 +v^\mathrm{h}\cdot \nabla _\mathrm{h}v^3-v^3{{\mathrm{div}}_{\mathrm{h}}}\,v^\mathrm{h}+\bar{u}^\mathrm{h}\cdot \nabla _{\mathrm{h}}v^3 -\Delta _\mathrm{h}v^3=-\partial _3 p,\\ \displaystyle \mathrm{div}\,v=0, \\ \displaystyle v|_{t=0}=v_0=\bigl (-\nabla _{\mathrm{h}}\Delta _{\mathrm{h}}^{-1}(\partial _3 u^3_0),u_0^3\bigr ). \end{array}\right. \end{aligned}$$
(3.14)

We notice that under the smallness condition (1.11), the horizontal components, \(v_0^\mathrm{h},\) are small in the critical space \(\mathcal {B}^{0,\frac{1}{2}}.\) Then the crucial ingredient used in the proof of Theorem 1.2 is that the horizontal components \(v^\mathrm{h}\) of the remainder velocity keeps small for any positive time.

Due to the additional difficulty caused by the fact that \(u_0^3\) belongs to the Sobolev–Besov type space with negative index, as in [8], we further decompose \(v^3\) as

$$\begin{aligned} v^3=v_F+w,\quad \text{ where }\quad v_F(t)\buildrel {\mathrm{def}}\over =e^{t\Delta _\mathrm{h}}u^3_{0,\mathrm{h}\mathrm{h}}\quad \text{ and }\quad u^3_{0,\mathrm{h}\mathrm{h}}\buildrel {\mathrm{def}}\over =\sum _{k\geqq \ell -1} \Delta ^\mathrm{h}_k\Delta _{\ell }^{\mathrm{v}}u^3_0. \end{aligned}$$
(3.15)

Then w solves

$$\begin{aligned} \left\{ \begin{array}{l} \displaystyle \partial _t w-\Delta _\mathrm{h}w +v\cdot \nabla v^3+\bar{u}^\mathrm{h}\cdot \nabla _{\mathrm{h}}v^3=-\partial _3 p,\\ \displaystyle w|_{t=0}=u^3_{0,\mathrm{l}\mathrm{h}}\buildrel {\mathrm{def}}\over =\sum _{\ell \in {\mathbb {Z}}} S^\mathrm{h}_{\ell -1}\Delta _{\ell }^{\mathrm{v}}u^3_0. \end{array}\right. \end{aligned}$$
(3.16)

Proposition 3.3

Let v be a smooth enough solution of (3.14) on \([0,T^*[\). Then there exists some positive constant C so that for any \(t\in ]0,T^*[,\) we have

$$\begin{aligned} \begin{aligned}&\Vert v^{\mathrm{h}}\Vert _{{\widetilde{L}^\infty _t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}} +\bigl (\frac{5}{4}-C\Vert v^{\mathrm{h}}\Vert _{{\widetilde{L}^\infty _t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}}^{\frac{1}{2}}\bigr )\Vert \nabla _{\mathrm{h}}v^{\mathrm{h}}\Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}\leqq \bigl (\Vert v^{\mathrm{h}}_0\Vert _{\mathcal {B}^{0,\frac{1}{2}}}+\Vert \partial _3\bar{u}^\mathrm{h}\Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}\bigr )\\&\quad \ \times \exp \Bigl (C\int \nolimits _0^t\bigl (\Vert w(t')\Vert _{\mathcal {B}^{0,\frac{1}{2}}}^2\Vert \nabla _{\mathrm{h}}w(t')\Vert _{\mathcal {B}^{0,\frac{1}{2}}}^2 +\Vert \bar{u}^\mathrm{h}(t')\Vert _{\mathcal {B}_4^{0,\frac{1}{2}}}^4+\Vert v_F(t')\Vert _{\mathcal {B}_4^{0,\frac{1}{2}}}^4\bigr )\,\mathrm{d}t'\Bigr ), \end{aligned} \end{aligned}$$
(3.17)

and

$$\begin{aligned} \begin{aligned}&\Bigl (\frac{5}{6}-C\bigl (\Vert v^{\mathrm{h}}\Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}^{\frac{1}{2}} +\Vert \partial _3\bar{u}^\mathrm{h}\Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}\bigr )\Bigr )\Vert w\Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}\\&\quad \leqq \Vert u^3_0\Vert _{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4} +C\Bigl (\Vert v^{\mathrm{h}}\Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}+\Vert \partial _3\bar{u}^\mathrm{h}\Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}+\Vert v^{\mathrm{h}}\Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}^2\\&\quad \quad +\bigl (1+ \Vert v^{\mathrm{h}}\Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}+\Vert \partial _3\bar{u}^\mathrm{h}\Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}\bigr ) \Vert v_F\Vert _{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4(t)}\Bigr )\exp \Bigl (C\Vert \bar{u}^\mathrm{h}\Vert _{L^4_t(\mathcal {B}_4^{0,\frac{1}{2}})}^4\Bigr ). \end{aligned} \end{aligned}$$
(3.18)

The proof of the estimates (3.17) and (3.18) will be presented respectively in Sections 5 and 6. Now let us admit the above Propositions 3.1 and 3.3 temporarily, and continue our proof of Theorem 1.2.

Proof of Theorem 1.2

It is well-known that the existence of global solutions to a nonlinear partial differential equations can be obtained by first constructing the approximate solutions, and then performing uniform estimates and finally passing to the limit to such approximate solutions. For simplicity, here we just present the a priori estimates for smooth enough solutions of (ANS).

Let u be a smooth enough solution of (ANS) on \([0, T^*[\) with \(T^*\) being the maximal time of existence. Let \(\bar{u}^\mathrm{h}\) and v be determined by (3.2) and (3.14), respectively. Thanks to (3.1) and Proposition 3.1, we first take LMN large enough and \(\varepsilon _0\) small enough in (1.11) so that

$$\begin{aligned} \begin{aligned} \Vert \Lambda _\mathrm{h}^{-1}\partial _3\bar{u}^\mathrm{h}\Vert _{\widetilde{L}^\infty _t(\mathcal {B}^{0,\frac{1}{2}})} +\Vert \partial _3\bar{u}^\mathrm{h}\Vert _{\widetilde{L}^2_t(\mathcal {B}^{0,\frac{1}{2}})}&\leqq C\Vert \Lambda _\mathrm{h}^{-1}\partial _3u^{\mathrm{h}}_0\Vert _{\mathcal {B}^{0,\frac{1}{2}}} \exp \Bigl (C\mathfrak {A}_N^4\bigl (\Vert u^\mathrm{h}_0\Vert _{\mathcal {B}^{0,\frac{1}{2}}}\bigr ) \Bigr )\\&\leqq \frac{1}{16}\quad \text{ for } \text{ any } \ \ t>0. \end{aligned} \end{aligned}$$
(3.19)

We now define

$$\begin{aligned} T^\star \buildrel {\mathrm{def}}\over =\sup \Bigl \{\ t< T^*,\ \ C\Vert v^\mathrm{h}\Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}\leqq \frac{1}{16}\ \Bigr \}. \end{aligned}$$
(3.20)

Then, thanks to (3.19) and Proposition 3.3, for \(t\leqq T^\star ,\) we find

$$\begin{aligned} \begin{aligned}&\Vert v^{\mathrm{h}}\Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)} \leqq \bigl (\Vert \Lambda _\mathrm{h}^{-1}\partial _3 u^3_0\Vert _{\mathcal {B}^{0,\frac{1}{2}}}+\Vert \partial _3\bar{u}^\mathrm{h}\Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}\bigr )\\&\quad \times \exp \Bigl (C\int \nolimits _0^t \bigl (\Vert w(t')\Vert _{\mathcal {B}^{0,\frac{1}{2}}}^2\Vert \nabla _{\mathrm{h}}w(t')\Vert _{\mathcal {B}^{0,\frac{1}{2}}}^2 +\Vert \bar{u}^\mathrm{h}(t')\Vert _{\mathcal {B}_4^{0,\frac{1}{2}}}^4+\Vert v_F(t')\Vert _{\mathcal {B}_4^{0,\frac{1}{2}}}^4\bigr )\,\mathrm{d}t'\Bigr ), \end{aligned} \end{aligned}$$
(3.21)

and

$$\begin{aligned} \begin{aligned}&\frac{1}{3}\Vert w\Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)} \leqq \Vert u^3_0\Vert _{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4}+C\bigl (1+\Vert v_F\Vert _{{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4}(t)}\bigr ) \exp \Bigl (C\Vert \bar{u}^\mathrm{h}\Vert _{L^4_t(\mathcal {B}_4^{0,\frac{1}{2}})}^4\Bigr ). \end{aligned} \end{aligned}$$
(3.22)

It follows from Lemma 2.2 and Proposition 2.1 that

$$\begin{aligned} \Vert v_F\Vert _{L^4_t(\mathcal {B}_4^{0,\frac{1}{2}})}\lesssim \Vert v_F\Vert _{{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4}(t)} \lesssim \Vert u^3_0\Vert _{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4}, \end{aligned}$$

whereas we deduce from (2.6) and Proposition 3.1 that

$$\begin{aligned} \begin{aligned} \Vert \bar{u}^\mathrm{h}\Vert _{\widetilde{L}^4_t(\mathcal {B}_4^{0,\frac{1}{2}})}^4&\leqq C\Vert \bar{u}^\mathrm{h}\Vert _{\widetilde{L}^\infty _t(\mathcal {B}^{0,\frac{1}{2}})}^2\Vert \nabla _\mathrm{h}\bar{u}^\mathrm{h}\Vert _{\widetilde{L}^2_t(\mathcal {B}^{0,\frac{1}{2}})}^2\\&\leqq C\mathfrak {A}_N^4\bigl (\Vert u^\mathrm{h}_0\Vert _{\mathcal {B}^{0,\frac{1}{2}}}\bigr ). \end{aligned} \end{aligned}$$

By inserting the above two inequalities to (3.22) and using (3.3), we obtain that, for \(t\leqq T^\star \),

$$\begin{aligned} \begin{aligned} \frac{1}{3}\Vert w\Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}\leqq C\bigl (1+\Vert u_0^3\Vert _{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4}\bigr )\exp \Bigl ( C\mathfrak {A}_N^4\bigl (\Vert u^\mathrm{h}_0\Vert _{\mathcal {B}^{0,\frac{1}{2}}}\bigr )\Bigr ). \end{aligned} \end{aligned}$$
(3.23)

Then we deduce that for \(t\leqq T^\star ,\)

$$\begin{aligned}&\int \nolimits _0^t \bigl (\Vert w(t')\Vert _{\mathcal {B}^{0,\frac{1}{2}}}^2\Vert \nabla _{\mathrm{h}}w(t')\Vert _{\mathcal {B}^{0,\frac{1}{2}}}^2 +\Vert \bar{u}^\mathrm{h}(t')\Vert _{\mathcal {B}_4^{0,\frac{1}{2}}}^4+\Vert v_F(t')\Vert _{\mathcal {B}_4^{0,\frac{1}{2}}}^4\bigr )\,\mathrm{d}t'\\&\quad \leqq \Vert w\Vert _{L^\infty _t(\mathcal {B}^{0,\frac{1}{2}})}^2 \Vert \nabla _{\mathrm{h}}w\Vert _{L^2_t(\mathcal {B}^{0,\frac{1}{2}})}^2 +\Vert \bar{u}^\mathrm{h}\Vert _{L^4_t(\mathcal {B}_4^{0,\frac{1}{2}})}^4+\Vert v_F\Vert _{L^4_t(\mathcal {B}_4^{0,\frac{1}{2}})}^4\\&\quad \leqq C\bigl (1+\Vert u_0^3\Vert _{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4}^4\bigr )\exp \Bigl ( C\mathfrak {A}_N^4\bigl (\Vert u^\mathrm{h}_0\Vert _{\mathcal {B}^{0,\frac{1}{2}}}\bigr )\Bigr ). \end{aligned}$$

Inserting the above estimates into (3.21) gives

$$\begin{aligned} \Vert v^\mathrm{h}\Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}\leqq \Vert \Lambda _\mathrm{h}^{-1}\partial _3u_0\Vert _{\mathcal {B}^{0,\frac{1}{2}}}\exp \Bigl ( C\bigl (1+\Vert u_0^3\Vert _{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4}^4\bigr )\exp \bigl ( C\mathfrak {A}_N^4\bigl (\Vert u^\mathrm{h}_0\Vert _{\mathcal {B}^{0,\frac{1}{2}}}\bigr )\bigr )\Bigr ) \end{aligned}$$
(3.24)

for \(t\leqq T^\star .\) Therefore, if we take LMN large enough and \(\varepsilon _0\) small enough in (1.11), we deduce from (3.24) that

$$\begin{aligned} C\Vert v^\mathrm{h}\Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}\leqq \frac{1}{32}\quad \text{ for } \ \ t\leqq T^\star . \end{aligned}$$
(3.25)

(3.25) contradicts (3.20). This in turn shows that \(T^\star =T^*.\) (3.23) along with (3.25) shows that \(T^*=\infty .\) Moreover, thanks to (3.15), we have \(\mathfrak {v}\buildrel {\mathrm{def}}\over =u-e^{t\Delta _h} \begin{pmatrix} 0\\ u^3_{0,\mathrm{hh}} \end{pmatrix}\in C([0,\infty [\,;\mathcal {B}^{0,\frac{1}{2}})\) with \(\nabla _{\mathrm{h}}\mathfrak {v}\in L^2([0,\infty [\,;\mathcal {B}^{0,\frac{1}{2}}).\) This completes the proof of our Theorem 1.2. \(\quad \square \)

Proof of Corollary 1.1

Under the assumptions that \(u^{\mathrm{h}}_0\in L^2\) with \(~\partial _3u^{\mathrm{h}}_0\in L^2\) and \(\Lambda _\mathrm{h}^{-1}\partial _3u^{\mathrm{h}}_0\in \mathcal {B}^{0,\frac{1}{2}},\) we deduce from (3.1), (3.4) and (3.7) that

$$\begin{aligned} \begin{aligned}&\Vert \bar{u}^\mathrm{h}\Vert _{\widetilde{L}^\infty _t(\mathcal {B}^{0,\frac{1}{2}})} +\Vert \nabla _{\mathrm{h}}\bar{u}^\mathrm{h}\Vert _{\widetilde{L}^2_t(\mathcal {B}^{0,\frac{1}{2}})} \leqq \Vert u^{\mathrm{h}}_0\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _3u^{\mathrm{h}}_0\Vert _{L^2}^{\frac{1}{2}} \exp \bigl (C\Vert u^{\mathrm{h}}_0\Vert _{L^2}\Vert \partial _3u^{\mathrm{h}}_0\Vert _{L^2}\bigr ),\\&\quad \Vert \Lambda _\mathrm{h}^{-1}\partial _3\bar{u}^\mathrm{h}\Vert _{\widetilde{L}^\infty _t(\mathcal {B}^{0,\frac{1}{2}})} +\Vert \partial _3\bar{u}^\mathrm{h}\Vert _{\widetilde{L}^2_t(\mathcal {B}^{0,\frac{1}{2}})} \leqq \Vert \Lambda _\mathrm{h}^{-1}\partial _3u^{\mathrm{h}}_0\Vert _{\mathcal {B}^{0,\frac{1}{2}}} \exp \left( \exp \bigl (C\Vert u^{\mathrm{h}}_0\Vert _{L^2}\Vert \partial _3u^{\mathrm{h}}_0\Vert _{L^2}\bigr )\right) . \end{aligned} \end{aligned}$$

Then by repeating the argument from (3.19) to (3.24), we conclude the proof of Corollary 1.1. \(\quad \square \)

4 Estimates of the 2-D Solution \(\pmb {\bar{u}^\mathrm{h}}\)

The goal of this section is to present the proof of Proposition 3.1. Let us start the proof with the following lemma, which is in the spirit of Lemma 3.1 of [6]:

Lemma 4.1

Let \(a^\mathrm{h}=(a^1,a^2)\) be a smooth enough solution of

$$\begin{aligned} \left\{ \begin{array}{l} \displaystyle \partial _t a^\mathrm{h}+a^\mathrm{h}\cdot \nabla _{\mathrm{h}}a^\mathrm{h}-\Delta _\mathrm{h}a^\mathrm{h}=-\nabla _{\mathrm{h}}\pi , \qquad (t,x)\in {\mathbb {R}}^+\times {\mathbb {R}}^3, \\ \displaystyle {{\mathrm{div}}_{\mathrm{h}}}\,a^\mathrm{h}= 0,\\ \displaystyle a^\mathrm{h}|_{t=0}=a^\mathrm{h}_0. \end{array}\right. \end{aligned}$$
(4.1)

Then for any \(t>0\) and any fixed \(x_3\in {\mathbb {R}},\) it holds that

$$\begin{aligned} \Vert a^\mathrm{h}(t,\cdot ,x_3)\Vert _{L^2_\mathrm{h}}^2 +2\int \nolimits _0^t\Vert \nabla _{\mathrm{h}}a^\mathrm{h}(t',\cdot ,x_3)\Vert _{L^2_\mathrm{h}}^2\mathrm{d}t' =\Vert a^\mathrm{h}_0(\cdot ,x_3)\Vert _{L^2_\mathrm{h}}^2, \end{aligned}$$
(4.2)

and

$$\begin{aligned}&\Vert \partial _3a^\mathrm{h}(t,\cdot ,x_3)\Vert _{L^2_\mathrm{h}}^2 +\int \nolimits _0^t\Vert \nabla _{\mathrm{h}}\partial _3a^\mathrm{h}(t',\cdot ,x_3)\Vert _{L^2_\mathrm{h}}^2\,\mathrm{d}t'\nonumber \\&\quad \leqq \Vert \partial _3a^\mathrm{h}_0(\cdot ,x_3)\Vert _{L^2_\mathrm{h}}^2 \exp \bigl (C\Vert a^\mathrm{h}_0\Vert _{L^\infty _\mathrm{v}(L^2_\mathrm{h})}^2\bigr ). \end{aligned}$$
(4.3)

Proof

By taking \(L^2_\mathrm{h}\) inn-product of (4.1) with \(a^\mathrm{h}\) and using \({{\mathrm{div}}_{\mathrm{h}}}\,a^\mathrm{h}=0,\) we obtain (4.2).

While by applying \(\partial _3\) to (4.1) and then taking \(L^2_\mathrm{h}\) inner product of the resulting equation with \(\partial _3a^\mathrm{h},\) we find

$$\begin{aligned} \begin{aligned}&\frac{1}{2}\frac{d}{\mathrm{d}t}\Vert \partial _3a^\mathrm{h}(t,\cdot ,x_3)\Vert _{L^2_\mathrm{h}}^2 +\Vert \nabla _{\mathrm{h}}\partial _3a^\mathrm{h}(t,\cdot ,x_3)\Vert _{L^2_\mathrm{h}}^2\\&\quad =-\bigl (\partial _3(a^\mathrm{h}\cdot \nabla _{\mathrm{h}}a^\mathrm{h})(t,\cdot ,x_3) \big | \partial _3a^\mathrm{h}(t,\cdot ,x_3)\bigr )_{L^2_\mathrm{h}}. \end{aligned} \end{aligned}$$
(4.4)

Due to \({{\mathrm{div}}_{\mathrm{h}}}\,a^\mathrm{h}=0,\) we get, by applying (2.4), that

$$\begin{aligned}&\bigl |\bigl (\partial _3(a^\mathrm{h}\cdot \nabla _{\mathrm{h}}a^\mathrm{h})(t,\cdot ,x_3) | \partial _3a^\mathrm{h}(t,\cdot ,x_3)\bigr )_{L^2_\mathrm{h}}\bigr |\\&\quad =\bigl |\bigl ((\partial _3a^\mathrm{h}\cdot \nabla _\mathrm{h}a^\mathrm{h})(t,\cdot ,x_3) | \partial _3a^\mathrm{h}(t,\cdot ,x_3)\bigr )_{L^2_\mathrm{h}}\bigr |\\&\quad \leqq \Vert \nabla _{\mathrm{h}}a^\mathrm{h}(t,\cdot ,x_3)\Vert _{L^2_\mathrm{h}} \Vert \partial _3a^\mathrm{h}(t,\cdot ,x_3)\Vert _{L^4_\mathrm{h}}^2\\&\quad \leqq C\Vert \nabla _{\mathrm{h}}a^\mathrm{h}(t,\cdot ,x_3)\Vert _{L^2_\mathrm{h}} \Vert \partial _3a^\mathrm{h}(t,\cdot ,x_3)\Vert _{L^2_\mathrm{h}}\Vert \nabla _\mathrm{h}\partial _3a^\mathrm{h}(t,\cdot ,x_3)\Vert _{L^2_\mathrm{h}}. \end{aligned}$$

Applying Young’s inequality yields

$$\begin{aligned}&\bigl |\bigl (\partial _3(a^\mathrm{h}\cdot \nabla _{\mathrm{h}}a^\mathrm{h})(t,\cdot ,x_3) | \partial _3a^\mathrm{h}(t,\cdot ,x_3)\bigr )_{L^2_\mathrm{h}}\bigr |\\&\quad \leqq \frac{1}{2}\Vert \nabla _{\mathrm{h}}\partial _3a^\mathrm{h}(t,\cdot ,x_3)\Vert _{L^2_\mathrm{h}}^2 +C\Vert \nabla _{\mathrm{h}}a^\mathrm{h}(t,\cdot ,x_3)\Vert _{L^2_\mathrm{h}}^2\Vert \partial _3a^\mathrm{h}(t,\cdot ,x_3)\Vert _{L^2_\mathrm{h}}^2. \end{aligned}$$

Inserting the above estimate into (4.4) gives

$$\begin{aligned}&\frac{d}{\mathrm{d}t}\Vert \partial _3a^\mathrm{h}(t,\cdot ,x_3)\Vert _{L^2_\mathrm{h}}^2 +\Vert \nabla _{\mathrm{h}}\partial _3a^\mathrm{h}(t,\cdot ,x_3)\Vert _{L^2_\mathrm{h}}^2\\&\quad \leqq C\Vert \nabla _{\mathrm{h}}a^\mathrm{h}(t,\cdot ,x_3)\Vert _{L^2_\mathrm{h}}^2\Vert \partial _3a^\mathrm{h}(t,\cdot ,x_3)\Vert _{L^2_\mathrm{h}}^2. \end{aligned}$$

Applying Gronwall’s inequality and using (4.2), we achieve

$$\begin{aligned}&\Vert \partial _3a^\mathrm{h}(t,\cdot ,x_3)\Vert _{L^2_\mathrm{h}}^2 +\int \nolimits _0^t\Vert \nabla _{\mathrm{h}}\partial _3a^\mathrm{h}(t',\cdot ,x_3)\Vert _{L^2_\mathrm{h}}^2\,\mathrm{d}t'\\&\quad \leqq \Vert \partial _3a^\mathrm{h}_0(\cdot ,x_3)\Vert _{L^2_\mathrm{h}}^2 \exp \Bigl (C\int \nolimits _0^t \Vert \nabla _{\mathrm{h}}a^\mathrm{h}(t',\cdot ,x_3)\Vert _{L^2_\mathrm{h}}^2\,\mathrm{d}t'\Bigr )\\&\quad \leqq \Vert \partial _3a^\mathrm{h}_0(\cdot ,x_3)\Vert _{L^2_\mathrm{h}}^2 \exp \bigl (C\Vert a^\mathrm{h}_0(\cdot ,x_3)\Vert _{L^2_\mathrm{h}}^2\bigr ), \end{aligned}$$

which leads to (4.3). This completes the proof of this lemma. \(\quad \square \)

Let us now present the proof of Proposition 3.1.

Proof of Proposition 3.1

For any positive integer N,  and \(\bar{u}^\mathrm{h}_{0,N}\) being given by (3.5), we split the solution \(\bar{u}^\mathrm{h}\) to (3.2) as

$$\begin{aligned} \bar{u}^\mathrm{h}=\bar{u}^\mathrm{h}_1+\bar{u}^\mathrm{h}_2, \end{aligned}$$
(4.5)

with \(\bar{u}^\mathrm{h}_1\) and \(\bar{u}^\mathrm{h}_2\) being determined, respectively, by

$$\begin{aligned} \left\{ \begin{array}{l} \displaystyle \partial _t \bar{u}^\mathrm{h}_1 +\bar{u}^\mathrm{h}_1\cdot \nabla _{\mathrm{h}}\bar{u}^\mathrm{h}_1 -\Delta _\mathrm{h}\bar{u}^\mathrm{h}_1=-\nabla _{\mathrm{h}}\bar{p}^{(1)}, \qquad (t,x)\in {\mathbb {R}}^+\times {\mathbb {R}}^3, \\ \displaystyle {{\mathrm{div}}_{\mathrm{h}}}\,\bar{u}^\mathrm{h}_1 = 0,\\ \displaystyle \bar{u}^\mathrm{h}_1|_{t=0}=\bar{u}^\mathrm{h}_{1,0}\buildrel {\mathrm{def}}\over =\bar{u}^\mathrm{h}_0-\bar{u}^\mathrm{h}_{0,N}, \end{array}\right. \end{aligned}$$
(4.6)

and

$$\begin{aligned} \left\{ \begin{array}{l} \displaystyle \partial _t \bar{u}^\mathrm{h}_2 +{{\mathrm{div}}_{\mathrm{h}}}\,\bigl (\bar{u}^\mathrm{h}_2\otimes \bar{u}^\mathrm{h}_2 +\bar{u}^\mathrm{h}_1\otimes \bar{u}^\mathrm{h}_2+\bar{u}^\mathrm{h}_2\otimes \bar{u}^\mathrm{h}_1\bigr ) -\Delta _\mathrm{h}\bar{u}^\mathrm{h}_2=-\nabla _{\mathrm{h}}\bar{p}^{(2)},\\ \displaystyle {{\mathrm{div}}_{\mathrm{h}}}\,\bar{u}^\mathrm{h}_2 = 0,\\ \displaystyle \bar{u}^\mathrm{h}_2|_{t=0}=\bar{u}^\mathrm{h}_{2,0}=\bar{u}^\mathrm{h}_{0,N}. \end{array}\right. \end{aligned}$$
(4.7)

Indeed for smoother initial data \(\bar{u}^\mathrm{h}_0,\) we may write explicitly the constant \(\mathfrak {A}_N\bigl (\Vert \bar{u}^\mathrm{h}_0\Vert _{\mathcal {B}^{0,\frac{1}{2}}}\bigr )\) in (3.3). For instance, if \(\bar{u}^\mathrm{h}_0\in L^2\) with \(~\partial _3\bar{u}^\mathrm{h}_0\in L^2\) and \(\Lambda _\mathrm{h}^{-1}\partial _3\bar{u}^\mathrm{h}_0\in \mathcal {B}^{0,\frac{1}{2}}\), we deduce from Lemma 4.1 that

$$\begin{aligned} \begin{aligned}&\Vert \bar{u}^\mathrm{h}(t)\Vert _{L^2}^2 +2\int \nolimits _0^t\Vert \nabla _{\mathrm{h}}\bar{u}^\mathrm{h}(t')\Vert _{L^2}^2\mathrm{d}t' =\Vert \bar{u}^\mathrm{h}_0\Vert _{L^2}^2,\quad \text{ and }\\&\Vert \partial _3\bar{u}^\mathrm{h}(t)\Vert _{L^2}^2 +\int \nolimits _0^t\Vert \nabla _{\mathrm{h}}\partial _3\bar{u}^\mathrm{h}(t')\Vert _{L^2}^2\,\mathrm{d}t' \leqq \Vert \partial _3\bar{u}^\mathrm{h}_0\Vert _{L^2}^2 \exp \bigl (C\Vert \bar{u}^\mathrm{h}_0\Vert _{L^\infty _\mathrm{v}(L^2)}^2\bigr ), \end{aligned} \end{aligned}$$

which, together with (2.2) and

$$\begin{aligned} \Vert \bar{u}^\mathrm{h}_0\Vert _{L^\infty _\mathrm{v}(L^2_\mathrm{h})}^2\leqq \Vert \bar{u}^\mathrm{h}_0\Vert _{L^2_\mathrm{h}(L^\infty _\mathrm{v})}^2\leqq \Vert u_0^\mathrm{h}\Vert _{\mathcal {B}^{0,\frac{1}{2}}}^2 \leqq \Vert u^\mathrm{h}_0\Vert _{L^2}\Vert \partial _3u^\mathrm{h}_0\Vert _{L^2}, \end{aligned}$$

ensures (3.6). By virtue of (3.6) and (4.22), we deduce (3.7).

In general, we first deduce from Lemma 4.1 that

$$\begin{aligned} \begin{aligned}&\Vert \bar{u}^\mathrm{h}_1(t)\Vert _{L^2}^2 +2\int \nolimits _0^t\Vert \nabla _{\mathrm{h}}\bar{u}^\mathrm{h}_1(t')\Vert _{L^2}^2\mathrm{d}t' =\Vert \bar{u}^\mathrm{h}_{1,0}\Vert _{L^2}^2\lesssim N \Vert \bar{u}^\mathrm{h}_0\Vert _{\mathcal {B}^{0,\frac{1}{2}}}^2,\quad \text{ and }\\&\Vert \partial _3\bar{u}^\mathrm{h}_1(t)\Vert _{L^2}^2 +\int \nolimits _0^t\Vert \nabla _{\mathrm{h}}\partial _3\bar{u}^\mathrm{h}_1(t')\Vert _{L^2}^2\,\mathrm{d}t' \leqq \Vert \partial _3\bar{u}^\mathrm{h}_{1,0}\Vert _{L^2}^2 \exp \bigl (C\Vert \bar{u}^\mathrm{h}_{1,0}\Vert _{L^\infty _\mathrm{v}(L^2)}^2\bigr )\\&\quad \lesssim N\Vert \bar{u}^\mathrm{h}_0\Vert _{\mathcal {B}^{0,\frac{1}{2}}}^2 \exp \bigl (C\Vert \bar{u}^\mathrm{h}_0\Vert _{\mathcal {B}^{0,\frac{1}{2}}}^2\bigr ), \end{aligned} \end{aligned}$$

which, together with (2.2), ensures that

$$\begin{aligned} \Vert \bar{u}^\mathrm{h}_1\Vert _{\widetilde{L}^\infty _t(\mathcal {B}^{0,\frac{1}{2}})} +\Vert \nabla _{\mathrm{h}}\bar{u}^\mathrm{h}_1\Vert _{\widetilde{L}^2_t(\mathcal {B}^{0,\frac{1}{2}})} \leqq C N^{\frac{1}{2}}\Vert \bar{u}^\mathrm{h}_0\Vert _{\mathcal {B}^{0,\frac{1}{2}}} \exp \bigl (C\Vert \bar{u}^\mathrm{h}_0\Vert _{\mathcal {B}^{0,\frac{1}{2}}}^2\bigr ). \end{aligned}$$
(4.8)

Next we handle the estimate of \(\bar{u}^\mathrm{h}_2\). To do this, for any \(\kappa >0,\) we denote

$$\begin{aligned} f^\mathrm{h}(t)\buildrel {\mathrm{def}}\over =\Vert \bar{u}^\mathrm{h}_1(t)\Vert _{\mathcal {B}^{0,\frac{1}{2}}}^2\Vert \nabla _\mathrm{h}\bar{u}^\mathrm{h}_1(t)\Vert _{\mathcal {B}^{0,\frac{1}{2}}}^2 \quad \hbox {and}\quad \bar{u}^\mathrm{h}_{2,\kappa }(t)\buildrel {\mathrm{def}}\over =\bar{u}^\mathrm{h}_2(t) \exp \Bigl (-\kappa \int \nolimits _0^t f^\mathrm{h}(t')\,\mathrm{d}t'\Bigr ). \end{aligned}$$
(4.9)

Then by multiplying \(\exp \Bigl (-\kappa \int \nolimits _0^t f^\mathrm{h}(t')\,\mathrm{d}t'\Bigr )\) to the \(\bar{u}^\mathrm{h}_2\) equation in (4.7), we write

$$\begin{aligned}&\partial _t\bar{u}^\mathrm{h}_{2,\kappa }+\kappa f^\mathrm{h}(t)\bar{u}^\mathrm{h}_{2,\kappa }-\Delta _\mathrm{h}\bar{u}^\mathrm{h}_{2,\kappa } +{{\mathrm{div}}_{\mathrm{h}}}\,(\bar{u}^\mathrm{h}_2\otimes \bar{u}^\mathrm{h}_{2,\kappa } +\bar{u}^\mathrm{h}_1\otimes \bar{u}^\mathrm{h}_{2,\kappa }\\&\quad +\bar{u}^\mathrm{h}_{2,\kappa }\otimes \bar{u}^\mathrm{h}_1)=-\nabla _\mathrm{h}\bar{p}^{(2)}_\kappa . \end{aligned}$$

Applying the operator \(\Delta _\ell ^\mathrm{v}\) to the above equation and taking \(L^2\) inner product of the resulting equation with \(\Delta _\ell ^\mathrm{v}\bar{u}^\mathrm{h}_{2,\kappa },\) and then using integration by parts, we get

$$\begin{aligned} \begin{aligned}&\frac{1}{2}\frac{d}{\mathrm{d}t}\Vert \Delta _\ell ^\mathrm{v}\bar{u}^\mathrm{h}_{2,\kappa }(t)\Vert _{L^2}^2+\kappa f^\mathrm{h}(t)\Vert \Delta _\ell ^\mathrm{v}\bar{u}^\mathrm{h}_{2,\kappa }(t)\Vert _{L^2}^2 +\Vert \Delta _\ell ^\mathrm{v}\nabla _\mathrm{h}\bar{u}^\mathrm{h}_{2,\kappa }\Vert _{L^2}^2\\&\quad =-\bigl (\Delta _\ell ^\mathrm{v}(\bar{u}^\mathrm{h}_2\cdot \nabla _{\mathrm{h}}\bar{u}^\mathrm{h}_{2,\kappa }) \big | \Delta _\ell ^\mathrm{v}\bar{u}^\mathrm{h}_{2,\kappa }\bigr )_{L^2} +\bigl (\Delta _\ell ^\mathrm{v}(\bar{u}^\mathrm{h}_1\otimes \bar{u}^\mathrm{h}_{2,\kappa }+\bar{u}^\mathrm{h}_{2,\kappa }\otimes \bar{u}^\mathrm{h}_1) \big | \Delta _\ell ^\mathrm{v}\nabla _{\mathrm{h}}\bar{u}^\mathrm{h}_{2,\kappa }\bigr )_{L^2}. \end{aligned} \end{aligned}$$
(4.10)

\(\square \)

The estimate of the second line of (4.10) relies on the following lemma, whose proof will be postponed until the “Appendix A”:

Lemma 4.2

Let \(a,b,c\in \mathcal {B}^{0,\frac{1}{2}}(T)\) and \(\mathfrak {f}(t)\buildrel {\mathrm{def}}\over =\Vert a(t)\Vert _{\mathcal {B}_4^{0,\frac{1}{2}}}^4\). Then for any smooth homogeneous Fourier multiplier, A(D),  of degree zero and any \(\ell \in {\mathbb {Z}}\), it holds that

$$\begin{aligned}&\int \nolimits _0^T\bigl |\bigl (\Delta _{\ell }^{\mathrm{v}}A(D)(a\otimes b)\big |\Delta _{\ell }^{\mathrm{v}}c\bigr )_{L^2}\bigr |\,\mathrm{d}t \lesssim d_{\ell }^2 2^{-\ell } \Vert b\Vert _{\widetilde{L}^2_T\big (\mathcal {B}^{0,\frac{1}{2}}\big )}\Vert c\Vert _{\widetilde{L}^2_{T,\mathfrak {f}}\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}\Vert \nabla _{\mathrm{h}}c\Vert _{\widetilde{L}^2_T\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}, \nonumber \\\end{aligned}$$
(4.11)
$$\begin{aligned}&\int \nolimits _0^T\bigl |\bigl (\Delta _{\ell }^{\mathrm{v}}A(D)(a\otimes b)\big |\Delta _{\ell }^{\mathrm{v}}c\bigr )_{L^2}\bigr |\,\mathrm{d}t \lesssim d_{\ell }^2 2^{-\ell } \Vert b\Vert _{\widetilde{L}^2_{T,\mathfrak {f}}\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}\Vert \nabla _{\mathrm{h}}b\Vert _{\widetilde{L}^2_T\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}\Vert c\Vert _{\widetilde{L}^2_T\big (\mathcal {B}^{0,\frac{1}{2}}\big )}.\nonumber \\ \end{aligned}$$
(4.12)

Moreover, for non-negative function \(\mathfrak {g}\in L^\infty (0,T),\) one has

$$\begin{aligned} \begin{aligned}&\int \nolimits _0^T\bigl |\bigl (\Delta _{\ell }^{\mathrm{v}}A(D)(a\otimes b)\big |\Delta _{\ell }^{\mathrm{v}}c\bigr )_{L^2}\bigr | \cdot \mathfrak {g}^2\,\mathrm{d}t \lesssim d_{\ell }^2 2^{-\ell }\Vert a\Vert _{\widetilde{L}^\infty _T\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}\Vert \mathfrak {g}\nabla _{\mathrm{h}}a\Vert _{\widetilde{L}^2_T\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}\\&\quad \times \Vert \mathfrak {g}b\Vert _{\widetilde{L}^2_T\big (\mathcal {B}^{0,\frac{1}{2}}\big )}\Vert c\Vert _{\widetilde{L}^\infty _T\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}\Vert \mathfrak {g}\nabla _{\mathrm{h}}c\Vert _{\widetilde{L}^2_T\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}. \end{aligned} \end{aligned}$$
(4.13)

By applying (4.13) with \(a=c=\bar{u}^\mathrm{h}_2,\)\(b=\nabla _h\bar{u}^\mathrm{h}_{2}\) and \(\mathfrak {g}= \exp \Bigl (-\kappa \int \nolimits _0^t f^\mathrm{h}(t')\,\mathrm{d}t'\Bigr )\), we get

$$\begin{aligned} \int \nolimits _0^t\bigl |\bigl (\Delta _\ell ^\mathrm{v}(\bar{u}^\mathrm{h}_2\cdot \nabla _{\mathrm{h}}\bar{u}^\mathrm{h}_{2,\kappa }) \big | \Delta _\ell ^\mathrm{v}\bar{u}^\mathrm{h}_{2,\kappa }\bigr )_{L^2}\bigr |\,\mathrm{d}t' \lesssim d_\ell ^22^{-\ell }\Vert \bar{u}^\mathrm{h}_2\Vert _{\widetilde{L}^\infty _t(\mathcal {B}^{0,\frac{1}{2}})} \Vert \nabla _{\mathrm{h}}\bar{u}^\mathrm{h}_{2,\kappa }\Vert _{\widetilde{L}^2_{t}(\mathcal {B}^{0,\frac{1}{2}})}^2. \end{aligned}$$
(4.14)

Whereas due to (2.5), one has

$$\begin{aligned} \Vert \bar{u}^\mathrm{h}_1(t)\Vert _{\mathcal {B}_4^{0,\frac{1}{2}}}^4\lesssim \Vert \bar{u}^\mathrm{h}_1(t)\Vert _{\mathcal {B}^{0,\frac{1}{2}}}^2\Vert \nabla _{\mathrm{h}}\bar{u}^\mathrm{h}_1(t)\Vert _{\mathcal {B}^{0,\frac{1}{2}}}^2. \end{aligned}$$

By applying (4.12) with \(a=\bar{u}^\mathrm{h}_1,~b=\bar{u}^\mathrm{h}_{2,\kappa },~ c=\nabla _{\mathrm{h}}\bar{u}^\mathrm{h}_{2,\kappa }\), we infer

$$\begin{aligned} \begin{aligned}&\int \nolimits _0^t\bigl |\bigl (\Delta _\ell ^\mathrm{v}(\bar{u}^\mathrm{h}_1\otimes \bar{u}^\mathrm{h}_{2,\kappa }+\bar{u}^\mathrm{h}_{2,\kappa }\otimes \bar{u}^\mathrm{h}_1) \big |\Delta _\ell ^\mathrm{v}\nabla _{\mathrm{h}}\bar{u}^\mathrm{h}_{2,\kappa }\bigr )_{L^2}\bigr |\,\mathrm{d}t'\\&\quad \lesssim d_\ell ^22^{-\ell }\Vert \bar{u}^\mathrm{h}_{2,\kappa }\Vert _{\widetilde{L}^2_{t,f^\mathrm{h}}(\mathcal {B}^{0,\frac{1}{2}})}^{\frac{1}{2}} \Vert \nabla _{\mathrm{h}}\bar{u}^\mathrm{h}_{2,\kappa }\Vert _{\widetilde{L}^2_{t}(\mathcal {B}^{0,\frac{1}{2}})}^{\frac{3}{2}}. \end{aligned} \end{aligned}$$
(4.15)

Then we get, by first integrating (4.10) over [0, t] and inserting (4.14) and (4.15) into the resulting inequality, that

$$\begin{aligned}&\Vert \Delta _\ell ^\mathrm{v}\bar{u}^\mathrm{h}_{2,\kappa }(t)\Vert _{L^2}^2+2\kappa \int \nolimits _0^tf^\mathrm{h}(t')\Vert \Delta _\ell ^\mathrm{v}\bar{u}^\mathrm{h}_{2,\kappa }(t')\Vert _{L^2}^2\,\mathrm{d}t' +2\Vert \Delta _\ell ^\mathrm{v}\nabla _{\mathrm{h}}\bar{u}^\mathrm{h}_{2,\kappa }\Vert _{L^2_t(L^2)}^2\\&\quad \leqq \Vert \Delta _\ell ^\mathrm{v}\bar{u}^\mathrm{h}_{0,N}\Vert _{L^2}^2 +C d_\ell ^22^{-\ell }\Bigl (\Vert \bar{u}^\mathrm{h}_{2}\Vert _{\widetilde{L}^\infty _t(\mathcal {B}^{0,\frac{1}{2}})} \Vert \nabla _{\mathrm{h}}\bar{u}^\mathrm{h}_{2,\kappa }\Vert _{\widetilde{L}^2_{t}(\mathcal {B}^{0,\frac{1}{2}})}^2\\&\quad \quad +\Vert \bar{u}^\mathrm{h}_{2,\kappa }\Vert _{\widetilde{L}^2_{t,f^\mathrm{h}}(\mathcal {B}^{0,\frac{1}{2}})}^{\frac{1}{2}} \Vert \nabla _{\mathrm{h}}\bar{u}^\mathrm{h}_{2,\kappa }\Vert _{\widetilde{L}^2_t(\mathcal {B}^{0,\frac{1}{2}})}^{\frac{3}{2}}\Bigr ). \end{aligned}$$

Multiplying the above inequality by \(2^\ell \) and taking square root of the resulting inequality, and then summing up the inequalities for \(\ell \in {\mathbb {Z}},\) we arrive at

$$\begin{aligned}&\Vert \bar{u}^\mathrm{h}_{2,\kappa }\Vert _{\widetilde{L}^\infty _t(\mathcal {B}^{0,\frac{1}{2}})} +\sqrt{2\kappa }\Vert \bar{u}^\mathrm{h}_{2,\kappa }\Vert _{\widetilde{L}^2_{t,f^\mathrm{h}}(\mathcal {B}^{0,\frac{1}{2}})} +\sqrt{2}\Vert \nabla _{\mathrm{h}}\bar{u}^\mathrm{h}_{2,\kappa }\Vert _{\widetilde{L}^2_t(\mathcal {B}^{0,\frac{1}{2}})}\\&\quad \leqq \Vert \bar{u}^\mathrm{h}_{0,N}\Vert _{\mathcal {B}^{0,\frac{1}{2}}} +C\Bigl (\Vert \bar{u}^\mathrm{h}_{2}\Vert _{\widetilde{L}^\infty _t(\mathcal {B}^{0,\frac{1}{2}})}^{\frac{1}{2}} \Vert \nabla _{\mathrm{h}}\bar{u}^\mathrm{h}_{2,\kappa }\Vert _{\widetilde{L}^2_{t}(\mathcal {B}^{0,\frac{1}{2}})} +\Vert \bar{u}^\mathrm{h}_{2,\kappa }\Vert _{\widetilde{L}^2_{t,f^\mathrm{h}}(\mathcal {B}^{0,\frac{1}{2}})}^{\frac{1}{4}} \Vert \nabla _{\mathrm{h}}\bar{u}^\mathrm{h}_{2,\kappa }\Vert _{\widetilde{L}^2_t(\mathcal {B}^{0,\frac{1}{2}})}^{\frac{3}{4}}\Bigr )\\&\quad \leqq \Vert \bar{u}^\mathrm{h}_{0,N}\Vert _{\mathcal {B}^{0,\frac{1}{2}}} +\bigl (\sqrt{2}-1+C\Vert \bar{u}^\mathrm{h}_{2}\Vert _{\widetilde{L}^\infty _t(\mathcal {B}^{0,\frac{1}{2}})}^{\frac{1}{2}}\bigr ) \Vert \nabla _{\mathrm{h}}\bar{u}^\mathrm{h}_{2,\kappa }\Vert _{\widetilde{L}^2_{t}(\mathcal {B}^{0,\frac{1}{2}})} +C\Vert \bar{u}^\mathrm{h}_{2,\kappa }\Vert _{\widetilde{L}^2_{t,f^\mathrm{h}}(\mathcal {B}^{0,\frac{1}{2}})}. \end{aligned}$$

In particular, taking \(2\kappa =C^2\) in the above inequality gives

$$\begin{aligned} \Vert \bar{u}^\mathrm{h}_{2,\kappa }\Vert _{\widetilde{L}^\infty _t(\mathcal {B}^{0,\frac{1}{2}})} +\bigl (1-C\Vert \bar{u}^\mathrm{h}_{2}\Vert _{\widetilde{L}^\infty _t(\mathcal {B}^{0,\frac{1}{2}})}^{\frac{1}{2}}\bigr ) \Vert \nabla _{\mathrm{h}}\bar{u}^\mathrm{h}_{2,\kappa }\Vert _{\widetilde{L}^2_t(\mathcal {B}^{0,\frac{1}{2}})} \leqq \Vert \bar{u}^\mathrm{h}_{0,N}\Vert _{\mathcal {B}^{0,\frac{1}{2}}}. \end{aligned}$$
(4.16)

On the other hand, in view of (3.5), we can take N so large that

$$\begin{aligned} C\Vert \bar{u}^\mathrm{h}_{0, N}\Vert _{\mathcal {B}^{0,\frac{1}{2}}}^{\frac{1}{2}} \leqq \frac{1}{2}. \end{aligned}$$
(4.17)

Then a standard continuity argument shows that, for any time \(t>0\), it holds that

$$\begin{aligned} \Vert \bar{u}^\mathrm{h}_{2,\kappa }\Vert _{\widetilde{L}^\infty _t(\mathcal {B}^{0,\frac{1}{2}})} +\frac{1}{2}\Vert \nabla _{\mathrm{h}}\bar{u}^\mathrm{h}_{2,\kappa }\Vert _{\widetilde{L}^2_t(\mathcal {B}^{0,\frac{1}{2}})} \leqq \Vert \bar{u}^\mathrm{h}_{0,N}\Vert _{\mathcal {B}^{0,\frac{1}{2}}}. \end{aligned}$$
(4.18)

Due to the definition of \(\bar{u}^\mathrm{h}_{2,\lambda }\) given by (4.9), one has

$$\begin{aligned} \begin{aligned}&\bigl (\Vert \bar{u}^\mathrm{h}_{2}\Vert _{\widetilde{L}^\infty _t(\mathcal {B}^{0,\frac{1}{2}})} +\Vert \nabla _{\mathrm{h}}\bar{u}^\mathrm{h}_{2}\Vert _{\widetilde{L}^2_t(\mathcal {B}^{0,\frac{1}{2}})}\bigr ) \exp \Bigl (-\kappa \int \nolimits _0^tf^\mathrm{h}(t')\,\mathrm{d}t'\Bigr )\\&\quad \leqq \Vert \bar{u}^\mathrm{h}_{2,\kappa }\Vert _{\widetilde{L}^\infty _t(\mathcal {B}^{0,\frac{1}{2}})} +\Vert \nabla _{\mathrm{h}}\bar{u}^\mathrm{h}_{2,\kappa }\Vert _{\widetilde{L}^2_t(\mathcal {B}^{0,\frac{1}{2}})}, \end{aligned} \end{aligned}$$

which, together with (4.8) and (4.18), implies that

$$\begin{aligned} \begin{aligned} \Vert \bar{u}^\mathrm{h}_2\Vert _{\widetilde{L}^\infty _t(\mathcal {B}^{0,\frac{1}{2}})} +\Vert \nabla _{\mathrm{h}}\bar{u}^\mathrm{h}_2\Vert _{\widetilde{L}^2_t(\mathcal {B}^{0,\frac{1}{2}})}&\leqq 2\Vert \bar{u}^\mathrm{h}_{0,N}\Vert _{\mathcal {B}^{0,\frac{1}{2}}} \exp \Bigl (\kappa \int \nolimits _0^tf^\mathrm{h}(t')\,\mathrm{d}t'\Bigr )\\&\leqq 2\Vert \bar{u}^\mathrm{h}_{0,N}\Vert _{\mathcal {B}^{0,\frac{1}{2}}}\exp \left( N^2\exp \bigl (C\Vert \bar{u}^\mathrm{h}_0\Vert _{\mathcal {B}^{0,\frac{1}{2}}}^2\bigr )\right) , \end{aligned} \end{aligned}$$
(4.19)

By combining (4.8) with (4.19), we obtain (3.3).

It remains to prove (3.4). In order to do, this for any \(\gamma >0,\) we denote

$$\begin{aligned} g^\mathrm{h}(t)\buildrel {\mathrm{def}}\over =\Vert \bar{u}^\mathrm{h}(t)\Vert _{\mathcal {B}^{0,\frac{1}{2}}}^2\Vert \nabla _\mathrm{h}\bar{u}^\mathrm{h}(t)\Vert _{\mathcal {B}^{0,\frac{1}{2}}}^2 \quad \hbox {and}\quad \bar{u}^\mathrm{h}_\gamma (t)\buildrel {\mathrm{def}}\over =\bar{u}^\mathrm{h}(t) \exp \Bigl (-\gamma \int \nolimits _0^t g^\mathrm{h}(t')\,\mathrm{d}t'\Bigr ). \end{aligned}$$
(4.20)

Then, by multiplying \(\exp \left( -\gamma \int \nolimits _0^t g^\mathrm{h}(t')\,\mathrm{d}t'\right) \) to the \(\bar{u}^\mathrm{h}\) equation in (3.2), we write

$$\begin{aligned} \partial _t\bar{u}^\mathrm{h}_\gamma +\gamma g^\mathrm{h}(t)\bar{u}^\mathrm{h}_\gamma -\Delta _\mathrm{h}\bar{u}^\mathrm{h}_\gamma +\bar{u}^\mathrm{h}\cdot \nabla _{\mathrm{h}}\bar{u}^\mathrm{h}_\gamma =-\nabla _\mathrm{h}\bar{p}_\gamma . \end{aligned}$$

Applying the operator \(\Delta _\ell ^\mathrm{v}\Lambda _\mathrm{h}^{-1}\partial _3\) to the above equation and then taking \(L^2\) inner product of the resulting equation with \(\Delta _\ell ^\mathrm{v}\Lambda _\mathrm{h}^{-1}\partial _3\bar{u}^\mathrm{h}_\gamma ,\) we get

$$\begin{aligned} \begin{aligned}&\frac{1}{2}\frac{d}{\mathrm{d}t}\Vert \Delta _\ell ^\mathrm{v}\Lambda _\mathrm{h}^{-1}\partial _3\bar{u}^\mathrm{h}_\gamma (t)\Vert _{L^2}^2 +\gamma g^\mathrm{h}(t)\Vert \Delta _\ell ^\mathrm{v}\Lambda _\mathrm{h}^{-1}\partial _3\bar{u}^\mathrm{h}_\gamma (t)\Vert _{L^2}^2 +\Vert \Delta _\ell ^\mathrm{v}\nabla _\mathrm{h}\Lambda _\mathrm{h}^{-1}\partial _3\bar{u}^\mathrm{h}_\gamma \Vert _{L^2}^2\\&\quad =-\bigl (\Delta _\ell ^\mathrm{v}\Lambda _\mathrm{h}^{-1}\partial _3(\bar{u}^\mathrm{h}\cdot \nabla _{\mathrm{h}}\bar{u}^\mathrm{h}_\gamma ) \big | \Delta _\ell ^\mathrm{v}\Lambda _\mathrm{h}^{-1}\partial _3\bar{u}^\mathrm{h}_\gamma \bigr )_{L^2}\\&\quad =-\bigl (\Delta _\ell ^\mathrm{v}\Lambda _\mathrm{h}^{-1}{{\mathrm{div}}_{\mathrm{h}}}\,(\bar{u}^\mathrm{h}\otimes \partial _3\bar{u}^\mathrm{h}_\gamma +\partial _3\bar{u}^\mathrm{h}_\gamma \otimes \bar{u}^\mathrm{h}) \big | \Delta _\ell ^\mathrm{v}\Lambda _\mathrm{h}^{-1}\partial _3\bar{u}^\mathrm{h}_\gamma \bigr )_{L^2}. \end{aligned} \end{aligned}$$
(4.21)

Noting that \(\Lambda _\mathrm{h}^{-1}{{\mathrm{div}}_{\mathrm{h}}}\,\) is a bounded Fourier multiplier, we get, by using (4.11) with \(a=\bar{u}^\mathrm{h},~b= \partial _3\bar{u}^\mathrm{h}_\gamma \) and \(c=\Lambda _\mathrm{h}^{-1}\partial _3\bar{u}^\mathrm{h}_\gamma ,\) that

$$\begin{aligned}&\int \nolimits _0^t\bigl | \bigl (\Delta _\ell ^\mathrm{v}\Lambda _\mathrm{h}^{-1}{{\mathrm{div}}_{\mathrm{h}}}\,(\bar{u}^\mathrm{h}\otimes \partial _3\bar{u}^\mathrm{h}_\gamma +\partial _3\bar{u}^\mathrm{h}_\gamma \otimes \bar{u}^\mathrm{h}) \big | \Delta _\ell ^\mathrm{v}\Lambda _\mathrm{h}^{-1}\partial _3\bar{u}^\mathrm{h}_\gamma \bigr )_{L^2}\bigr |\,\mathrm{d}t'\\&\quad \lesssim d_\ell ^22^{-\ell }\Vert \partial _3\bar{u}^\mathrm{h}_\gamma \Vert _{\widetilde{L}^2_t(\mathcal {B}^{0,\frac{1}{2}})}^{\frac{3}{2}} \Vert \Lambda _\mathrm{h}^{-1}\partial _3\bar{u}^\mathrm{h}_\gamma \Vert _{\widetilde{L}^2_{t,g^\mathrm{h}}(\mathcal {B}^{0,\frac{1}{2}})}^{\frac{1}{2}}. \end{aligned}$$

By integrating (4.21) over [0, t] and then inserting the above estimate into the resulting inequality, we find

$$\begin{aligned} \begin{aligned}&\Vert \Delta _\ell ^\mathrm{v}\Lambda _\mathrm{h}^{-1}\partial _3\bar{u}^\mathrm{h}_\gamma (t)\Vert _{L^2}^2+2\gamma \int \nolimits _0^tg^\mathrm{h}(t')\Vert \Delta _\ell ^\mathrm{v}\Lambda _\mathrm{h}^{-1}\partial _3\bar{u}^\mathrm{h}_\gamma (t')\Vert _{L^2}^2\,\mathrm{d}t' +2\Vert \Delta _\ell ^\mathrm{v}\partial _3\bar{u}^\mathrm{h}_\gamma \Vert _{L^2_t(L^2)}^2\\&\quad \leqq \Vert \Delta _\ell ^\mathrm{v}\Lambda _\mathrm{h}^{-1}\partial _3\bar{u}^\mathrm{h}_0\Vert _{L^2}^2 +C d_\ell ^22^{-\ell }\Vert \partial _3\bar{u}^\mathrm{h}_\gamma \Vert _{\widetilde{L}^2_t(\mathcal {B}^{0,\frac{1}{2}})}^{\frac{3}{2}} \Vert \Lambda _\mathrm{h}^{-1}\partial _3\bar{u}^\mathrm{h}_\gamma \Vert _{\widetilde{L}^2_{t,g^\mathrm{h}}(\mathcal {B}^{0,\frac{1}{2}})}^{\frac{1}{2}}. \end{aligned} \end{aligned}$$

Multiplying the above inequality by \(2^\ell \) and taking square root of the resulting inequality, and then summing up the inequalities for \(\ell \in {\mathbb {Z}},\) we arrive at

$$\begin{aligned} \begin{aligned}&\Vert \Lambda _\mathrm{h}^{-1}\partial _3\bar{u}^\mathrm{h}_\gamma \Vert _{\widetilde{L}^\infty _t(\mathcal {B}^{0,\frac{1}{2}})} +\sqrt{2\gamma }\Vert \Lambda _\mathrm{h}^{-1}\partial _3\bar{u}^\mathrm{h}_\lambda \Vert _{\widetilde{L}^2_{t,f^\mathrm{h}}(\mathcal {B}^{0,\frac{1}{2}})} +\sqrt{2}\Vert \partial _3\bar{u}^\mathrm{h}_\gamma \Vert _{\widetilde{L}^2_t(\mathcal {B}^{0,\frac{1}{2}})}\\&\quad \leqq \Vert \Lambda _\mathrm{h}^{-1}\partial _3\bar{u}^\mathrm{h}_0\Vert _{\mathcal {B}^{0,\frac{1}{2}}}+C \Vert \partial _3\bar{u}^\mathrm{h}_\gamma \Vert _{\widetilde{L}^2_t(\mathcal {B}^{0,\frac{1}{2}})}^{\frac{3}{4}} \Vert \Lambda _\mathrm{h}^{-1}\partial _3\bar{u}^\mathrm{h}_\gamma \Vert _{\widetilde{L}^2_{t,g^\mathrm{h}}(\mathcal {B}^{0,\frac{1}{2}})}^{\frac{1}{4}}\\&\quad \leqq \Vert \Lambda _\mathrm{h}^{-1}\partial _3\bar{u}^\mathrm{h}_0\Vert _{\mathcal {B}^{0,\frac{1}{2}}}+(\sqrt{2}-1) \Vert \partial _3\bar{u}^\mathrm{h}_\gamma \Vert _{\widetilde{L}^2_t(\mathcal {B}^{0,\frac{1}{2}})}+C \Vert \Lambda _\mathrm{h}^{-1}\partial _3\bar{u}^\mathrm{h}_\gamma \Vert _{\widetilde{L}^2_{t,g^\mathrm{h}}(\mathcal {B}^{0,\frac{1}{2}})}. \end{aligned} \end{aligned}$$

In particular, taking \(2\gamma =C^2\) in the above inequality gives

$$\begin{aligned} \Vert \Lambda _\mathrm{h}^{-1}\partial _3\bar{u}^\mathrm{h}_\gamma \Vert _{\widetilde{L}^\infty _t(\mathcal {B}^{0,\frac{1}{2}})} +\Vert \partial _3\bar{u}^\mathrm{h}_\gamma \Vert _{\widetilde{L}^2_t(\mathcal {B}^{0,\frac{1}{2}})} \leqq \Vert \Lambda _\mathrm{h}^{-1}\partial _3\bar{u}^\mathrm{h}_0\Vert _{\mathcal {B}^{0,\frac{1}{2}}}. \end{aligned}$$

Then a similar derivation from (4.18) to (4.19) leads to

$$\begin{aligned} \Vert \Lambda _\mathrm{h}^{-1}\partial _3\bar{u}^\mathrm{h}\Vert _{\widetilde{L}^\infty _t(\mathcal {B}^{0,\frac{1}{2}})} +\Vert \partial _3\bar{u}^\mathrm{h}\Vert _{\widetilde{L}^2_t(\mathcal {B}^{0,\frac{1}{2}})} \leqq \Vert \Lambda _\mathrm{h}^{-1}\partial _3\bar{u}^\mathrm{h}_0\Vert _{\mathcal {B}^{0,\frac{1}{2}}} \exp \Bigl (\gamma \int \nolimits _0^tg^\mathrm{h}(t')\,\mathrm{d}t'\Bigr ), \end{aligned}$$
(4.22)

which together with (3.3), ensures (3.4). This completes the proof of this proposition. \(\quad \square \)

5 The Estimate of the Horizontal Components \(\pmb {v^\mathrm{h}}\)

The goal of this section is to present the proof of (3.17), namely, we are going to deal with the estimate to the horizontal components of the remainder velocity determined by (3.14).

In order to do this, let u be a smooth enough solution of (ANS) on \([0,T^*[,\) let \(\bar{u}^\mathrm{h}, v_F\) and w be determined respectively by (3.2), (3.15) and (3.16), for any constant \(\lambda >0\), we denote

$$\begin{aligned} \begin{aligned}&v^{\mathrm{h}}_{\lambda }(t)\buildrel {\mathrm{def}}\over =v^{\mathrm{h}}(t)\exp \Bigl (-\lambda \int \nolimits _0^tf(t')\,\mathrm{d}t'\Bigr )\quad \hbox {with}\quad \\&f(t)\buildrel {\mathrm{def}}\over =\Vert w(t)\Vert _{\mathcal {B}^{0,\frac{1}{2}}}^2\Vert \nabla _{\mathrm{h}}w(t)\Vert _{\mathcal {B}^{0,\frac{1}{2}}}^2 +\Vert \bar{u}^\mathrm{h}(t)\Vert _{\mathcal {B}_4^{0,\frac{1}{2}}}^4+\Vert v_F(t)\Vert _{\mathcal {B}_4^{0,\frac{1}{2}}}^4, \end{aligned} \end{aligned}$$
(5.1)

and similar notations for \(\bar{u}^\mathrm{h}_\lambda ,~p_\lambda ,~\bar{p}_\lambda \) and \(v^{\mathrm{h}}_{\lambda /2}\).

By multiplying \(\exp \Bigl (-\lambda \int \nolimits _0^tf(t')\,\mathrm{d}t'\Bigr )\) to the \(v^{\mathrm{h}}\) equation of (3.14), we get

$$\begin{aligned} \partial _t v^{\mathrm{h}}_\lambda +\lambda f(t)v^{\mathrm{h}}_\lambda +v\cdot \nabla v^{\mathrm{h}}_\lambda +\bar{u}^\mathrm{h}\cdot \nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda +v_\lambda \cdot \nabla \bar{u}^\mathrm{h}-\Delta _\mathrm{h}v^{\mathrm{h}}_\lambda =-\nabla _{\mathrm{h}}p_\lambda +\nabla _{\mathrm{h}}\bar{p}_\lambda . \end{aligned}$$

Applying \(\Delta _{\ell }^{\mathrm{v}}\) to the above equation and taking \(L^2\) inner product of the resulting equation with \(\Delta _{\ell }^{\mathrm{v}}v^{\mathrm{h}}_\lambda \), and then integrating the equality over [0, t],  we obtain

$$\begin{aligned}&\frac{1}{2}\Vert \Delta _{\ell }^{\mathrm{v}}v^{\mathrm{h}}_\lambda (t)\Vert _{L^2}^2+\lambda \int \nolimits _0^t f(t')\Vert \Delta _{\ell }^{\mathrm{v}}v^{\mathrm{h}}_\lambda \Vert _{L^2}^2\,\mathrm{d}t' \nonumber \\&\quad +\int \nolimits _0^t\Vert \nabla _{\mathrm{h}}\Delta _{\ell }^{\mathrm{v}}v^{\mathrm{h}}_\lambda \Vert _{L^2}^2\,\mathrm{d}t' =\frac{1}{2}\Vert \Delta _{\ell }^{\mathrm{v}}v^{\mathrm{h}}_0\Vert _{L^2}^2 -\sum _{i=1}^6\mathrm{I}_i, \end{aligned}$$
(5.2)

where

$$\begin{aligned}&\mathrm{I}_1\buildrel {\mathrm{def}}\over =\int \nolimits _0^t\bigl (\Delta _{\ell }^{\mathrm{v}}(\bar{u}^\mathrm{h}\cdot \nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda ) \,\big |\,\Delta _{\ell }^{\mathrm{v}}v^{\mathrm{h}}_\lambda \bigr )_{L^2}\,\mathrm{d}t',\quad \mathrm{I}_2\buildrel {\mathrm{def}}\over =\int \nolimits _0^t\bigl (\Delta _{\ell }^{\mathrm{v}}(v^\mathrm{h}\cdot \nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda ) \,\big |\,\Delta _{\ell }^{\mathrm{v}}v^{\mathrm{h}}_\lambda \bigr )_{L^2}\,\mathrm{d}t',\\&\mathrm{I}_3\buildrel {\mathrm{def}}\over =\int \nolimits _0^t\bigl (\Delta _{\ell }^{\mathrm{v}}(v^{\mathrm{h}}_\lambda \cdot \nabla _{\mathrm{h}}\bar{u}^\mathrm{h}) \,\big |\,\Delta _{\ell }^{\mathrm{v}}v^{\mathrm{h}}_\lambda \bigr )_{L^2}\,\mathrm{d}t',\quad \mathrm{I}_4\buildrel {\mathrm{def}}\over =\int \nolimits _0^t\bigl (\Delta _{\ell }^{\mathrm{v}}(v^3\partial _3 \bar{u}^\mathrm{h}_\lambda ) \,\big |\,\Delta _{\ell }^{\mathrm{v}}v^{\mathrm{h}}_\lambda \bigr )_{L^2}\,\mathrm{d}t',\\&\mathrm{I}_5\buildrel {\mathrm{def}}\over =\int \nolimits _0^t\bigl (\Delta _{\ell }^{\mathrm{v}}(v^3\partial _3 v^{\mathrm{h}}_\lambda ) \,\big |\,\Delta _{\ell }^{\mathrm{v}}v^{\mathrm{h}}_\lambda \bigr )_{L^2}\,\mathrm{d}t',\qquad \mathrm{I}_6\buildrel {\mathrm{def}}\over =\int \nolimits _0^t\bigl (\Delta _{\ell }^{\mathrm{v}}\nabla _{\mathrm{h}}(p_\lambda -\bar{p}_\lambda ) \,\big |\,\Delta _{\ell }^{\mathrm{v}}v^{\mathrm{h}}_\lambda \bigr )_{L^2}\,\mathrm{d}t'. \end{aligned}$$

We mention that since our system (3.14) has only horizontal dissipation, it is reasonable to distinguish the terms above with horizontal derivatives from the ones with vertical derivative. Next let us handle the above term by term.

\(\bullet \) The estimates of \(\underline{\mathrm{I}_1}\) to \(\underline{\mathrm{I}_4}.\)

We first get, by using (4.11) with \(a=\bar{u}^\mathrm{h},~b=\nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda \) and \(c=v^{\mathrm{h}}_\lambda ,\) that

$$\begin{aligned} |\mathrm{I}_1|\lesssim d_{\ell }^2 2^{-\ell }\Vert v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_{t,f}\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}} \Vert \nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{3}{2}}. \end{aligned}$$
(5.3)

Applying (4.13) with \(a=v^{\mathrm{h}},~b=\nabla _{\mathrm{h}}v^{\mathrm{h}},~c=v^{\mathrm{h}}\) and \(\mathfrak {g}(t)=\exp \bigl (-\lambda \int \nolimits _0^tf(t')\,\mathrm{d}t'\bigr )\) yields

$$\begin{aligned} |\mathrm{I}_2| \lesssim d_\ell ^2 2^{-\ell }\Vert v^{\mathrm{h}}\Vert _{{\widetilde{L}^\infty _t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}}\Vert \nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^2. \end{aligned}$$
(5.4)

To handle \(\mathrm{I}_3,\) by using integration by parts, we write

$$\begin{aligned} \mathrm{I}_3=-\int \nolimits _0^t\bigl (\Delta _{\ell }^{\mathrm{v}}({{\mathrm{div}}_{\mathrm{h}}}\,v^{\mathrm{h}}_\lambda \cdot \bar{u}^\mathrm{h}) \big |\Delta _{\ell }^{\mathrm{v}}v^{\mathrm{h}}_\lambda \bigr )_{L^2}\,\mathrm{d}t' -\int \nolimits _0^t\bigl (\Delta _{\ell }^{\mathrm{v}}(\bar{u}^\mathrm{h}\otimes v^{\mathrm{h}}_\lambda ) \big |\Delta _{\ell }^{\mathrm{v}}\nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda \bigr )_{L^2}\,\mathrm{d}t'. \end{aligned}$$

Applying (4.11) with \(a=\bar{u}^\mathrm{h},~b={{\mathrm{div}}_{\mathrm{h}}}\,v^{\mathrm{h}}_\lambda \) and \(c=v^{\mathrm{h}}_\lambda \) gives

$$\begin{aligned} \Bigl |\int \nolimits _0^t\bigl (\Delta _{\ell }^{\mathrm{v}}({{\mathrm{div}}_{\mathrm{h}}}\,v^{\mathrm{h}}_\lambda \cdot \bar{u}^\mathrm{h}) \big |\Delta _{\ell }^{\mathrm{v}}v^{\mathrm{h}}_\lambda \bigr )_{L^2}\,\mathrm{d}t'\Bigr | \lesssim d_{\ell }^2 2^{-\ell }\Vert v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_{t,f}\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}} \Vert \nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{3}{2}}. \end{aligned}$$

Whereas applying (4.12) with \(a=\bar{u}^\mathrm{h},~b=v^{\mathrm{h}}_\lambda \) and \(c=\nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda \) yields

$$\begin{aligned} \Bigl |\int \nolimits _0^t\bigl (\Delta _{\ell }^{\mathrm{v}}(\bar{u}^\mathrm{h}\otimes v^{\mathrm{h}}_\lambda ) \big |\Delta _{\ell }^{\mathrm{v}}\nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda \bigr )_{L^2}\,\mathrm{d}t'\Bigr | \lesssim d_{\ell }^2 2^{-\ell }\Vert v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_{t,f}\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}} \Vert \nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{3}{2}}. \end{aligned}$$

As a result, it turns out that

$$\begin{aligned} |\mathrm{I}_3|\lesssim d_{\ell }^2 2^{-\ell }\Vert v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_{t,f}\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}} \Vert \nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{3}{2}}. \end{aligned}$$
(5.5)

While by applying (4.11) with \(a=v^3,~b=\partial _3 \bar{u}^\mathrm{h}_\lambda ,~c=v^{\mathrm{h}}_\lambda \), and using the fact that

$$\begin{aligned} \Vert v^3(t)\Vert _{\mathcal {B}^{0,\frac{1}{2}}_4}\leqq \Vert v_F(t)\Vert _{\mathcal {B}^{0,\frac{1}{2}}_4}+\Vert w(t)\Vert _{\mathcal {B}^{0,\frac{1}{2}}}^{\frac{1}{2}}\Vert \nabla _\mathrm{h}w(t)\Vert _{\mathcal {B}^{0,\frac{1}{2}}}^{\frac{1}{2}}, \end{aligned}$$

we find

$$\begin{aligned} |\mathrm{I}_4|\lesssim d_{\ell }^2 2^{-\ell }\Vert v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_{t,f}\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}} \Vert \nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}\Vert \partial _3\bar{u}^\mathrm{h}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}. \end{aligned}$$
(5.6)

\(\bullet \)The estimates of \(\underline{\mathrm{I}_5}\).

The estimate of \(\mathrm{I}_5\) is much more complicated, since there is no vertical dissipation in (ANS). To overcome this difficulty, we first use Bony’s decomposition in vertical variable (2.7) to write

$$\begin{aligned} \mathrm{I}_5=\int \nolimits _0^t\bigl (\Delta _\ell ^\mathrm{v}\bigl (T^\mathrm{v}_{v^3}\partial _3 v^{\mathrm{h}}_\lambda + R^\mathrm{v}({v^3},\partial _3 v^{\mathrm{h}}_\lambda ) \bigr )\,\big |\,\Delta _{\ell }^{\mathrm{v}}v^{\mathrm{h}}_\lambda \bigr )_{L^2} \,\mathrm{d}t'\buildrel {\mathrm{def}}\over =\mathrm{I}_5^{T}+\mathrm{I}_5^{R}. \end{aligned}$$

Following [8, 16], we get, by using a standard commutator’s process, that

$$\begin{aligned} \mathrm{I}_5^{T}&=\sum _{|\ell '-\ell |\leqq 5}\Bigl (\int \nolimits _0^t\bigl ([\Delta _{\ell }^{\mathrm{v}}; S^\mathrm{v}_{\ell '-1}v^3]\Delta _{\ell '}^{\mathrm{v}}\partial _3v^{\mathrm{h}}_\lambda \,\big |\,\Delta _{\ell }^{\mathrm{v}}v^{\mathrm{h}}_\lambda \bigr )_{L^2}\,\mathrm{d}t'\\&\quad +\int \nolimits _0^t\bigl ((S^\mathrm{v}_{\ell '-1}v^3-S^\mathrm{v}_{\ell -1}v^3) \Delta _\ell ^\mathrm{v}\Delta _{\ell '}^{\mathrm{v}}\partial _3v^{\mathrm{h}}_\lambda \,\big |\,\Delta _{\ell }^{\mathrm{v}}v^{\mathrm{h}}_\lambda \bigr )_{L^2}\,\mathrm{d}t'\Bigr )\\&\quad +\int \nolimits _0^t\bigl (S^\mathrm{v}_{\ell -1}v^3\Delta _{\ell }^{\mathrm{v}}\partial _3v^{\mathrm{h}}_\lambda \,\big |\,\Delta _{\ell }^{\mathrm{v}}v^{\mathrm{h}}_\lambda \bigr )_{L^2}\,\mathrm{d}t' \buildrel {\mathrm{def}}\over =\mathrm{I}_5^{T,1}+\mathrm{I}_5^{T,2}+\mathrm{I}_5^{T,3}. \end{aligned}$$

By applying the commutator’s estimate (see Lemma 2.97 in [2]), we find

$$\begin{aligned} \bigl |\mathrm{I}_5^{T,1}\bigr |&\leqq \sum _{|\ell '-\ell |\leqq 5}\Vert [\Delta _{\ell }^{\mathrm{v}}; S^\mathrm{v}_{\ell '-1}v^3_\lambda ] \Delta _{\ell '}^{\mathrm{v}}\partial _3 v^{\mathrm{h}}_{\lambda /2}\Vert _{L^{\frac{4}{3}}_t(L^{\frac{4}{3}}_\mathrm{h}(L^2_\mathrm{v}))} \Vert \Delta _{\ell }^{\mathrm{v}}v^{\mathrm{h}}_{\lambda /2}\Vert _{L^4_t(L^4_\mathrm{h}(L^2_\mathrm{v}))}\\&\lesssim \sum _{|\ell '-\ell |\leqq 5}2^{-\ell } \Vert \partial _3S^\mathrm{v}_{\ell '-1}v^3_\lambda \Vert _{L^2_t(L^2_\mathrm{h}(L^\infty _\mathrm{v}))} \Vert \Delta _{\ell '}^{\mathrm{v}}\partial _3 v^{\mathrm{h}}_{\lambda /2}\Vert _{L^4_t(L^4_\mathrm{h}(L^2_\mathrm{v}))} \Vert \Delta _{\ell }^{\mathrm{v}}v^{\mathrm{h}}_{\lambda /2}\Vert _{L^4_t(L^4_\mathrm{h}(L^2_\mathrm{v}))}. \end{aligned}$$

Due to \(\partial _3v^3=-{{\mathrm{div}}_{\mathrm{h}}}\,v^\mathrm{h},\) we get, by applying (2.4), that

$$\begin{aligned} \begin{aligned} \bigl |\mathrm{I}_5^{T,1}\bigr |&\lesssim \sum _{|\ell '-\ell |\leqq 5}2^{-\ell } \Vert S^\mathrm{v}_{\ell '-1}{{\mathrm{div}}_{\mathrm{h}}}\,v^{\mathrm{h}}_\lambda \Vert _{L^2_t(L^2_\mathrm{h}(L^\infty _\mathrm{v}))} 2^{\ell '}\Vert \Delta _{\ell '}^{\mathrm{v}}v^{\mathrm{h}}_{\lambda /2}\Vert _{L^4_t(L^4_\mathrm{h}(L^2_\mathrm{v}))} \Vert \Delta _{\ell }^{\mathrm{v}}v^{\mathrm{h}}_{\lambda /2}\Vert _{L^4_t(L^4_\mathrm{h}(L^2_\mathrm{v}))}\\&\lesssim \sum _{|\ell '-\ell |\leqq 5} \Vert \nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}\Vert \Delta _{\ell '}^{\mathrm{v}}v^{\mathrm{h}}\Vert _{L^\infty _t(L^2)}^{\frac{1}{2}}\Vert \nabla _\mathrm{h}\Delta _{\ell '}^{\mathrm{v}}v^{\mathrm{h}}_\lambda \Vert _{L^2_t(L^2)}^{\frac{1}{2}}\\&\quad \times \Vert \Delta _\ell ^\mathrm{v}v^{\mathrm{h}}\Vert _{L^\infty _t(L^2)}^{\frac{1}{2}} \Vert \nabla _\mathrm{h}\Delta _\ell ^\mathrm{v}v^{\mathrm{h}}_\lambda \Vert _{L^2_t(L^2)}^{\frac{1}{2}}\\&\lesssim d_\ell ^2 2^{-\ell }\Vert v^{\mathrm{h}}\Vert _{\widetilde{L}^\infty _t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}\Vert \nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^2. \end{aligned} \end{aligned}$$

Next, since the support to the Fourier transform of \(\sum _{|\ell '-\ell |\leqq 5} (S^\mathrm{v}_{\ell '-1}v^3-S^\mathrm{v}_{\ell -1}v^3)\) is contained in \({\mathbb {R}}^2\times \cup _{|\ell '-\ell |\leqq 5}2^{\ell '}\mathcal {C}_\mathrm{v},\) we get, by applying Lemma 2.1, that

$$\begin{aligned} \bigl |\mathrm{I}_5^{T,2}\bigr |\lesssim & {} \sum _{|\ell '-\ell |\leqq 5}2^{-\ell }\Vert \partial _3(S^\mathrm{v}_{\ell '-1}v^3_\lambda -S^\mathrm{v}_{\ell -1}v^3_\lambda )\Vert _{L^2_t(L^2_\mathrm{h}(L^\infty _\mathrm{v}))}\\&\Vert \Delta _{\ell '}^{\mathrm{v}}\partial _3v^{\mathrm{h}}_{\lambda /2}\Vert _{L^4_t(L^4_\mathrm{h}(L^2_\mathrm{v}))}\Vert \Delta _{\ell }^{\mathrm{v}}v^{\mathrm{h}}_{\lambda /2}\Vert _{L^4_t(L^4_\mathrm{h}(L^2_\mathrm{v}))}, \end{aligned}$$

from which we infer

$$\begin{aligned} \begin{aligned} \bigl |\mathrm{I}_5^{T,2}\bigr |&\lesssim d_\ell ^2 2^{-\ell }\Vert v^{\mathrm{h}}\Vert _{\widetilde{L}^\infty _t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}\Vert \nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^2. \end{aligned} \end{aligned}$$

Finally, by using integration by parts and \(\partial _3v^3=-{{\mathrm{div}}_{\mathrm{h}}}\,v^\mathrm{h}\) again, we find that

$$\begin{aligned}\begin{aligned} \bigl |\mathrm{I}_5^{T,3}\bigr |&=\frac{1}{2}\Bigl |\int \nolimits _0^t\int \nolimits _{{\mathbb {R}}^3} S^\mathrm{v}_{\ell -1}\partial _3 v^3_\lambda \cdot \bigl |\Delta _{\ell }^{\mathrm{v}}v^{\mathrm{h}}_{\lambda /2}\bigr |^2\,dxdt'\Bigr |\\&\lesssim \Vert S^\mathrm{v}_{\ell -1}{{\mathrm{div}}_{\mathrm{h}}}\,v^{\mathrm{h}}_\lambda \Vert _{L^2_t(L^2_\mathrm{h}(L^\infty _\mathrm{v}))} \Vert \Delta _{\ell }^{\mathrm{v}}v^{\mathrm{h}}_{\lambda /2}\Vert _{L^4_t(L^4_\mathrm{h}(L^2_\mathrm{v}))}^2\\&\lesssim d_\ell ^2 2^{-\ell }\Vert v^{\mathrm{h}}\Vert _{\widetilde{L}^\infty _t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}\Vert \nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^2. \end{aligned} \end{aligned}$$

As a result, it turns out that

$$\begin{aligned} \bigl |\mathrm{I}_5^{T}\bigr |\lesssim d_\ell ^2 2^{-\ell }\Vert v^{\mathrm{h}}\Vert _{\widetilde{L}^\infty _t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}\Vert \nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^2. \end{aligned}$$
(5.7)

On the other hand, by applying Lemma 2.1 once again, we find that

$$\begin{aligned} \bigl |\mathrm{I}_5^{R}\bigr |&\lesssim \sum _{\ell '\geqq \ell -4} \Vert \Delta _{\ell '}^{\mathrm{v}}v^3_\lambda \Vert _{L^2_t(L^2)} 2^{\ell '}\Vert S^\mathrm{v}_{\ell '+2} v^{\mathrm{h}}_{\lambda /2}\Vert _{L^4_t(L^4_\mathrm{h}(L^\infty _\mathrm{v}))} \Vert \Delta _{\ell }^{\mathrm{v}}v^{\mathrm{h}}_{\lambda /2}\Vert _{L^4_t(L^4_\mathrm{h}(L^2_\mathrm{v}))}\\&\lesssim \sum _{\ell '\geqq \ell -4} \Vert \partial _3\Delta _{\ell '}^{\mathrm{v}}v^3_\lambda \Vert _{L^2_t(L^2)} \Vert S^\mathrm{v}_{\ell '+2}v^{\mathrm{h}}_{\lambda /2}\Vert _{L^4_t(L^4_\mathrm{h}(L^\infty _\mathrm{v}))} \Vert \Delta _{\ell }^{\mathrm{v}}v^{\mathrm{h}}_{\lambda /2}\Vert _{L^4_t(L^4_\mathrm{h}(L^2_\mathrm{v}))}. \end{aligned}$$

Observing that

$$\begin{aligned} \begin{aligned} \Vert \partial _3\Delta _{\ell '}^{\mathrm{v}}v^3_\lambda \Vert _{L^2_t(L^2)}&\lesssim d_{\ell '}2^{-\frac{\ell '}{2}}\Vert {{\mathrm{div}}_{\mathrm{h}}}\,v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )},\\ \Vert S^\mathrm{v}_{\ell '+2}v^{\mathrm{h}}_{\lambda /2}\Vert _{L^4_t(L^4_\mathrm{h}(L^\infty _\mathrm{v}))}&\lesssim \Vert v^{\mathrm{h}}\Vert _{\widetilde{L}^\infty _t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}\Vert \nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}, \end{aligned} \end{aligned}$$

we infer

$$\begin{aligned} \bigl |\mathrm{I}_5^{R}\bigr |\lesssim d_\ell ^2 2^{-\ell }\Vert v^{\mathrm{h}}\Vert _{\widetilde{L}^\infty _t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}\Vert \nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^2, \end{aligned}$$

which, together with (5.7), ensures that

$$\begin{aligned} |\mathrm{I}_{5}|\lesssim d_\ell ^2 2^{-\ell }\Vert v^{\mathrm{h}}\Vert _{\widetilde{L}^\infty _t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}\Vert \nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^2. \end{aligned}$$
(5.8)

\(\bullet \)The estimates of \(\underline{\mathrm{I}_6.}\)

We first get, by taking the space divergence operators, \(\mathrm{div}\,\) and \({{\mathrm{div}}_{\mathrm{h}}}\,,\) to (ANS) and (3.2) respectively, that

$$\begin{aligned} -\Delta p=\mathrm{div}\,(u\cdot \nabla u)\quad \text{ and }\quad -\Delta _\mathrm{h}\bar{p}={{\mathrm{div}}_{\mathrm{h}}}\,(\bar{u}\cdot \nabla _{\mathrm{h}}\bar{u}), \end{aligned}$$
(5.9)

so that thanks to the fact that

$$ u=(u^\mathrm{h},u^3)=(\bar{u}^\mathrm{h},0)+(v^\mathrm{h},v^3),$$

we write

$$\begin{aligned} \begin{aligned} \nabla _\mathrm{h}p-\nabla _\mathrm{h}\bar{p}&=\nabla _\mathrm{h}(-\Delta )^{-1}{{\mathrm{div}}_{\mathrm{h}}}\,(v\cdot \nabla u^\mathrm{h}+\bar{u}^\mathrm{h}\cdot \nabla _\mathrm{h}v^\mathrm{h})\\&\quad +\nabla _{\mathrm{h}}(-\Delta )^{-1}\partial _3(u\cdot \nabla v^3)\\&\quad +\nabla _\mathrm{h}\bigl ((-\Delta )^{-1}-(-\Delta _\mathrm{h})^{-1}\bigr ){{\mathrm{div}}_{\mathrm{h}}}\,{{\mathrm{div}}_{\mathrm{h}}}\,\bigl (\bar{u}^\mathrm{h}\otimes \bar{u}^\mathrm{h}\bigr ). \end{aligned} \end{aligned}$$

Accordingly, we decompose \(\mathrm{I}_6\) as

$$\begin{aligned} \mathrm{I}_6= \mathrm{I}_{6,1}+\mathrm{I}_{6,2}+\mathrm{I}_{6,3}+\mathrm{I}_{6,4}, \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} \mathrm{I}_{6,1}&=\int \nolimits _0^t\bigl (\Delta _{\ell }^{\mathrm{v}}\nabla _{\mathrm{h}}(-\Delta )^{-1}{{\mathrm{div}}_{\mathrm{h}}}\,\bigl (\bar{u}^\mathrm{h}\cdot \nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda +v^\mathrm{h}\cdot \nabla _\mathrm{h}v^{\mathrm{h}}_\lambda +v_\lambda \cdot \nabla \bar{u}^\mathrm{h}\bigr )\,\big |\,\Delta _{\ell }^{\mathrm{v}}v^{\mathrm{h}}_\lambda \bigr )_{L^2}\,\mathrm{d}t',\\ \mathrm{I}_{6,2}&=\int \nolimits _0^t\bigl (\Delta _{\ell }^{\mathrm{v}}\nabla _{\mathrm{h}}(-\Delta )^{-1}{{\mathrm{div}}_{\mathrm{h}}}\,(v^3\partial _3 v^\mathrm{h}_\lambda )\,\big |\,\Delta _{\ell }^{\mathrm{v}}v^{\mathrm{h}}_\lambda \bigr )_{L^2}\,\mathrm{d}t',\\ \mathrm{I}_{6,3}&=\int \nolimits _0^t\bigl (\Delta _{\ell }^{\mathrm{v}}\nabla _{\mathrm{h}}(-\Delta )^{-1}\partial _3 \bigl (v_\lambda \cdot \nabla v^3+\bar{u}^\mathrm{h}\cdot \nabla _{\mathrm{h}}v^3_\lambda \bigr )\,\big |\,\Delta _{\ell }^{\mathrm{v}}v^{\mathrm{h}}_\lambda \bigr )_{L^2}\,\mathrm{d}t',\\ \mathrm{I}_{6,4}&=\sum _{i=1}^2\sum _{j=1}^2\int \nolimits _0^t\bigl (\Delta _{\ell }^{\mathrm{v}}\nabla _{\mathrm{h}}\bigl ((-\Delta )^{-1}-(-\Delta _\mathrm{h})^{-1}\bigr ) \partial _i\partial _j(\bar{u}^i \bar{u}^j_\lambda )\,\big |\,\Delta _{\ell }^{\mathrm{v}}v^{\mathrm{h}}_\lambda \bigr )_{L^2}\,\mathrm{d}t'. \end{aligned} \end{aligned}$$

Noticing that \(\nabla _{\mathrm{h}}(-\Delta )^{-1}{{\mathrm{div}}_{\mathrm{h}}}\,\) is a bounded Fourier multiplier. Then along the same line to the estimate of \(\mathrm{I}_1\) to \(\mathrm{I}_4,\) we achieve

$$\begin{aligned} \begin{aligned} |\mathrm{I}_{6,1}|&\lesssim d_{\ell }^2 2^{-\ell } \Bigl (\Vert v^{\mathrm{h}}\Vert _{{\widetilde{L}^\infty _t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}}\Vert \nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^2\\&\quad +\Vert v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_{t,f}\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}\Vert \nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}} \bigl (\Vert \nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}+\Vert \partial _3\bar{u}^\mathrm{h}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}\bigr )\Bigr ). \end{aligned} \end{aligned}$$
(5.10)

However, \(\mathrm{I}_{6,2}\) can not be handled along the same line to that of \(\mathrm{I}_5\), since the symbol of the operator \(\nabla _{\mathrm{h}}(-\Delta )^{-1}{{\mathrm{div}}_{\mathrm{h}}}\,\) depends not only on \(\xi _3\), but also on \(\xi _\mathrm{h},\) which makes it impossible for us to deal with the commutator’s estimate. Fortunately, the appearance of the operator \((-\Delta )^{-1}\) can absorb the vertical derivative. Indeed, by using integration by parts, and the divergence-free condition of v,  we write

$$\begin{aligned} \mathrm{I}_{6,2}&=\int \nolimits _0^t\Bigl (\Delta _{\ell }^{\mathrm{v}}\nabla _{\mathrm{h}}(-\Delta )^{-1}{{\mathrm{div}}_{\mathrm{h}}}\,\bigl (\partial _3(v^3 v^\mathrm{h}_\lambda )-\partial _3 v^3\cdot v^{\mathrm{h}}_\lambda \bigr ) \,\Big |\,\Delta _{\ell }^{\mathrm{v}}v^{\mathrm{h}}_\lambda \Bigr )_{L^2}\,\mathrm{d}t'\\&=-\int \nolimits _0^t\bigl (\Delta _{\ell }^{\mathrm{v}}\nabla _{\mathrm{h}}(-\Delta )^{-1}\partial _3 ( v^3 v^\mathrm{h}_\lambda )\,\big |\,\Delta _{\ell }^{\mathrm{v}}\nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda \bigr )_{L^2}\,\mathrm{d}t'\\&\quad +\int \nolimits _0^t\bigl (\Delta _{\ell }^{\mathrm{v}}\nabla _{\mathrm{h}}(-\Delta )^{-1}{{\mathrm{div}}_{\mathrm{h}}}\,({{\mathrm{div}}_{\mathrm{h}}}\,v^{\mathrm{h}}\cdot v^{\mathrm{h}}_\lambda )\,\big |\,\Delta _{\ell }^{\mathrm{v}}v^{\mathrm{h}}_\lambda \bigr )_{L^2}\,\mathrm{d}t'. \end{aligned}$$

Since both \(\nabla _{\mathrm{h}}(-\Delta )^{-1}\partial _3\) and \(\nabla _{\mathrm{h}}(-\Delta )^{-1}{{\mathrm{div}}_{\mathrm{h}}}\,\) are bounded Fourier multiplier, we get, by applying Lemma 4.2, that

$$\begin{aligned} |\mathrm{I}_{6,2}|\lesssim d_{\ell }^2 2^{-\ell } \bigl (\Vert v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_{t,f}\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}\Vert \nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{3}{2}} +\Vert v^{\mathrm{h}}\Vert _{{\widetilde{L}^\infty _t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}}\Vert \nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^2\bigr ). \end{aligned}$$
(5.11)

To handle \(\mathrm{I}_{6,3}\), we use \(\mathrm{div}\,v={{\mathrm{div}}_{\mathrm{h}}}\,\bar{u}^\mathrm{h}=0\) to write

$$\begin{aligned} \mathrm{I}_{6,3}&=\int \nolimits _0^t\bigl (\Delta _{\ell }^{\mathrm{v}}\nabla _{\mathrm{h}}(-\Delta )^{-1}\partial _3\mathrm{div}\,(v_\lambda v^3)+\Delta _{\ell }^{\mathrm{v}}\nabla _{\mathrm{h}}(-\Delta )^{-1}\partial _3{{\mathrm{div}}_{\mathrm{h}}}\,(\bar{u}^\mathrm{h}v^3_\lambda )\,\big |\,\Delta _{\ell }^{\mathrm{v}}v^{\mathrm{h}}_\lambda \bigr )_{L^2}\,\mathrm{d}t'\\&=\int \nolimits _0^t\left( \nabla _{\mathrm{h}}(-\Delta )^{-1}\Delta _{\ell }^{\mathrm{v}}\bigl ({{\mathrm{div}}_{\mathrm{h}}}\,(v^3\partial _3v^{\mathrm{h}}_\lambda +v^{\mathrm{h}}\partial _3 v^3_\lambda )+2\partial _3(v^3\partial _3v^3_\lambda )\bigr ) \,\big |\,\Delta _{\ell }^{\mathrm{v}}v^{\mathrm{h}}_\lambda \right) _{L^2}\,\mathrm{d}t' \\&\quad +\int \nolimits _0^t\left( \nabla _{\mathrm{h}}(-\Delta )^{-1}{{\mathrm{div}}_{\mathrm{h}}}\,\Delta _{\ell }^{\mathrm{v}}\bigl (v^3\partial _3\bar{u}^\mathrm{h}_\lambda +\bar{u}^\mathrm{h}\partial _3 v^3_\lambda \bigr )\,\big |\,\Delta _{\ell }^{\mathrm{v}}v^{\mathrm{h}}_\lambda \right) _{L^2}\,\mathrm{d}t'\\&=\int \nolimits _0^t\Bigl (\nabla _{\mathrm{h}}(-\Delta )^{-1}\Delta _{\ell }^{\mathrm{v}}\bigl ({{\mathrm{div}}_{\mathrm{h}}}\,(v^3\partial _3v^{\mathrm{h}}_\lambda -v^{\mathrm{h}}{{\mathrm{div}}_{\mathrm{h}}}\,v^{\mathrm{h}}_\lambda )-2\partial _3(v^3{{\mathrm{div}}_{\mathrm{h}}}\,v^{\mathrm{h}}_\lambda )\bigr )\,\big |\,\Delta _{\ell }^{\mathrm{v}}v^{\mathrm{h}}_\lambda \Bigr )_{L^2}\,\mathrm{d}t'\\&\quad +\int \nolimits _0^t\Bigl (\nabla _{\mathrm{h}}(-\Delta )^{-1}{{\mathrm{div}}_{\mathrm{h}}}\,\Delta _{\ell }^{\mathrm{v}}\bigl (v^3\partial _3\bar{u}^\mathrm{h}_\lambda -\bar{u}^\mathrm{h}{{\mathrm{div}}_{\mathrm{h}}}\,v^{\mathrm{h}}_\lambda \bigr )\,\big |\,\Delta _{\ell }^{\mathrm{v}}v^{\mathrm{h}}_\lambda \Bigr )_{L^2}\,\mathrm{d}t'. \end{aligned}$$

Applying (4.11) with \(A(D)=\nabla _{\mathrm{h}}(-\Delta )^{-1}\partial _3, a=v^3, b={{\mathrm{div}}_{\mathrm{h}}}\,v^\mathrm{h}_\lambda \) and \(c=v^\mathrm{h}_\lambda \) yields

$$\begin{aligned}&\int \nolimits _0^t\bigl |\bigl (\nabla _{\mathrm{h}}(-\Delta )^{-1}\partial _3\Delta _{\ell }^{\mathrm{v}}(v^3{{\mathrm{div}}_{\mathrm{h}}}\,v^{\mathrm{h}}_\lambda )\,\big |\,\Delta _{\ell }^{\mathrm{v}}v^{\mathrm{h}}_\lambda \bigr )_{L^2}\bigr |\,\mathrm{d}t'\\&\quad \lesssim d_{\ell }^2 2^{-\ell } \Vert v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_{t,f}\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}\Vert \nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{3}{2}}. \end{aligned}$$

The remaining terms in \(\mathrm{I}_{6,3}\) can be handled along the same lines as to those of \(\mathrm{I}_{6,1}\) and \(\mathrm{I}_{6,2}.\) As a consequence, we obtain

$$\begin{aligned} \begin{aligned} |\mathrm{I}_{6,3}|&\lesssim d_{\ell }^2 2^{-\ell } \Bigl (\Vert v^{\mathrm{h}}\Vert _{{\widetilde{L}^\infty _t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}}\Vert \nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^2\\&\quad +\Vert v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_{t,f}\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}\Vert \nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}} \bigl (\Vert \nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}+\Vert \partial _3\bar{u}^\mathrm{h}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}\bigr )\Bigr ). \end{aligned} \end{aligned}$$
(5.12)

To deal with \(\mathrm{I}_{6,4}\), it is crucial to observe that

$$\begin{aligned} \Delta _{\ell }^{\mathrm{v}}\nabla _{\mathrm{h}}\bigl ((-\Delta )^{-1}-(-\Delta _\mathrm{h})^{-1}\bigr ) \partial _i\partial _j(\bar{u}^i \bar{u}^j_\lambda ) =\Delta _{\ell }^{\mathrm{v}}\nabla _{\mathrm{h}}\partial _3^2(-\Delta )^{-1}(-\Delta _\mathrm{h})^{-1} \partial _i\partial _j(\bar{u}^i \bar{u}^j_\lambda ). \end{aligned}$$

Then due to the fact that \(\sum _{i,j=1}^2\nabla _{\mathrm{h}}\partial _3(-\Delta )^{-1}(-\Delta _\mathrm{h})^{-1} \partial _i\partial _j\) is a bounded Fourier multiplier, we get, by applying (4.11) with \(a=\bar{u}^\mathrm{h},b=\partial _3\bar{u}^\mathrm{h}_\lambda ,c=v^{\mathrm{h}}_\lambda ,\) that

$$\begin{aligned} \begin{aligned} |\mathrm{I}_{6,4}|&\leqq 2\sum _{i=1}^2\sum _{j=1}^3\int \nolimits _0^t\bigl |\bigl (\Delta _{\ell }^{\mathrm{v}}\nabla _{\mathrm{h}}\partial _3(-\Delta )^{-1}(-\Delta _\mathrm{h})^{-1} \partial _i\partial _j(\bar{u}^i\partial _3 \bar{u}^j_\lambda )\,\big |\,\Delta _{\ell }^{\mathrm{v}}v^{\mathrm{h}}_\lambda \bigr )_{L^2}\bigr |\,\mathrm{d}t'\\&\lesssim d_{\ell }^2 2^{-\ell }\Vert v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_{t,f}\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}} \Vert \nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}\Vert \partial _3\bar{u}^\mathrm{h}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}. \end{aligned} \end{aligned}$$
(5.13)

By summing up (5.105.13), we arrive at

$$\begin{aligned} \begin{aligned} |\mathrm{I}_{6}|&\lesssim d_{\ell }^2 2^{-\ell } \Bigl (\Vert v^{\mathrm{h}}\Vert _{{\widetilde{L}^\infty _t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}}\Vert \nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^2\\&\quad +\Vert v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_{t,f}\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}\Vert \nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}} \bigl (\Vert \nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}+\Vert \partial _3\bar{u}^\mathrm{h}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}\bigr )\Bigr ). \end{aligned} \end{aligned}$$
(5.14)

Now we are in a position to complete the proof of (3.17).

Proof of (3.17)

By inserting the estimates (5.35.6), (5.8) and (5.14) into (5.2), we achieve

$$\begin{aligned} \begin{aligned}&\frac{1}{2}\Vert \Delta _{\ell }^{\mathrm{v}}v^{\mathrm{h}}_\lambda (t)\Vert _{L^2}^2+\lambda \int \nolimits _0^t f(t')\Vert \Delta _{\ell }^{\mathrm{v}}v^{\mathrm{h}}_\lambda (t')\Vert _{L^2}^2\,\mathrm{d}t' +\int \nolimits _0^t\Vert \nabla _{\mathrm{h}}\Delta _{\ell }^{\mathrm{v}}v^{\mathrm{h}}_\lambda (t')\Vert _{L^2}^2\,\mathrm{d}t'\\&\quad \leqq \frac{1}{2}\Vert \Delta _{\ell }^{\mathrm{v}}v^{\mathrm{h}}_0\Vert _{L^2}^2+C d_{\ell }^2 2^{-\ell } \Bigl (\Vert v^{\mathrm{h}}\Vert _{{\widetilde{L}^\infty _t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}}\Vert \nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^2\\&\quad \quad +\Vert v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_{t,f}\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}\bigl (\Vert \nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{3}{2}} +\Vert \partial _3\bar{u}^\mathrm{h}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{3}{2}}\bigr )\Bigr ). \end{aligned} \end{aligned}$$

Multiplying the above inequality by \(2^{\ell +1} \) and taking square root of the resulting inequality, and then summing up the inequalities over \({\mathbb {Z}},\) we find that

$$\begin{aligned} \begin{aligned}&\Vert v^{\mathrm{h}}_\lambda \Vert _{{\widetilde{L}^\infty _t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}}+\sqrt{2\lambda }\Vert v^{\mathrm{h}}_\lambda \Vert _{{\widetilde{L}^2_{t,f}\big (\mathcal {B}^{0,\frac{1}{2}}\big )}} +\sqrt{2}\Vert \nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}\\&\quad \leqq \Vert v^{\mathrm{h}}_0\Vert _{\mathcal {B}^{0,\frac{1}{2}}}+C\Vert v^{\mathrm{h}}\Vert _{{\widetilde{L}^\infty _t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}}^{\frac{1}{2}}\Vert \nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}\\&\quad \quad +C \Vert v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_{t,f}\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{4}}\bigl (\Vert \nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{3}{4}} +\Vert \partial _3\bar{u}^\mathrm{h}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{3}{4}}\bigr ). \end{aligned} \end{aligned}$$
(5.15)

It follows from Young’s inequality that

$$\begin{aligned}&C \Vert v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_{t,f}\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{4}}\bigl (\Vert \nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{3}{4}} +\Vert \partial _3\bar{u}^\mathrm{h}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{3}{4}}\bigr )\\&\quad \leqq \frac{1}{10}\Vert \nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}+\Vert \partial _3\bar{u}^\mathrm{h}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}+C\Vert v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_{t,f}\big (\mathcal {B}^{0,\frac{1}{2}}\big )}. \end{aligned}$$

Inserting the above inequality into (5.15) and taking \(\lambda \) so that \(\sqrt{2\lambda }=C\), we obtain

$$\begin{aligned} \begin{aligned} \Vert v^{\mathrm{h}}_\lambda \Vert _{{\widetilde{L}^\infty _t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}}+ \frac{5}{4}\Vert \nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}&\leqq \Vert v^{\mathrm{h}}_0\Vert _{\mathcal {B}^{0,\frac{1}{2}}}+\Vert \partial _3\bar{u}^\mathrm{h}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}\\&\quad +C\Vert v^{\mathrm{h}}\Vert _{{\widetilde{L}^\infty _t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}}^{\frac{1}{2}}\Vert \nabla _{\mathrm{h}}v^{\mathrm{h}}_\lambda \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}, \end{aligned} \end{aligned}$$

which, together with the following consequence of (5.1):

$$\begin{aligned} \Vert a\Vert _{\widetilde{L}^p_t(\mathcal {B}^{0,\frac{1}{2}})}\exp \Bigl (-\lambda \int \nolimits _0^tf(t')\,\mathrm{d}t'\Bigr ) \leqq \Vert a_\lambda \Vert _{\widetilde{L}^p_t(\mathcal {B}^{0,\frac{1}{2}})}\quad \text{ for } \quad p=2\ \text{ or }\ \infty , \end{aligned}$$

gives rise to (3.17). \(\quad \square \)

6 The Estimate of the Vertical Component \(\pmb {v^3}\)

The purpose of this section is to present the proof of (3.18). Compared with [17], where the third component of the velocity field can be estimated in the standard Besov spaces, here, due to the additional terms like \(\bar{u}^\mathrm{h}\cdot \nabla _\mathrm{h}v\) that appears in (3.14), we will have to use the weighted Chemin–Lerner norms once again. Indeed for any constant \(\mu >0\), we denote

$$\begin{aligned} w_{\mu }(t)\buildrel {\mathrm{def}}\over =w(t)\bar{\mathfrak {g}}(t) \quad \text{ with }\quad \bar{\mathfrak {g}}(t)\buildrel {\mathrm{def}}\over =\exp \Bigl (-\mu \int \nolimits _0^t\hbar (t')\,\mathrm{d}t'\Bigr )\quad \hbox {and}\quad \hbar (t)\buildrel {\mathrm{def}}\over =\Vert \bar{u}^\mathrm{h}(t)\Vert _{\mathcal {B}_4^{0,\frac{1}{2}}}^4, \end{aligned}$$
(6.1)

and similar notations for \(v_\mu ,~\bar{u}^\mathrm{h}_\mu ,\) and \(~p_\mu \).

By multiplying \(\bar{\mathfrak {g}}(t)\) to (3.16), we write

$$\begin{aligned} \partial _t w_\mu +\mu \hbar (t)w_\mu -\Delta _\mathrm{h}w_\mu +v\cdot \nabla v^3_\mu +\bar{u}^\mathrm{h}\cdot \nabla _{\mathrm{h}}v^3_\mu =-\partial _3 p_\mu . \end{aligned}$$

By applying \(\Delta _{\ell }^{\mathrm{v}}\) to the above equation and taking \(L^2\) inner product of the resulting equation with \(\Delta _{\ell }^{\mathrm{v}}w_\mu \), and then integrating the equality over [0, t],  we obtain

$$\begin{aligned} \frac{1}{2}\Vert \Delta _{\ell }^{\mathrm{v}}w_\mu (t)\Vert _{L^2}^2+\mu \Vert \sqrt{\hbar }\Delta _{\ell }^{\mathrm{v}}w_\mu \Vert _{L^2_t(L^2)}^2 +\Vert \nabla _{\mathrm{h}}\Delta _{\ell }^{\mathrm{v}}w_\mu \Vert _{L^2_t(L^2)}^2 =\frac{1}{2}\Vert \Delta _{\ell }^{\mathrm{v}}u^3_{0,\mathrm{l}\mathrm{h}}\Vert _{L^2}^2 -\sum _{i=1}^6\mathrm{II}_i, \end{aligned}$$
(6.2)

where

$$\begin{aligned}&\mathrm{II}_1\buildrel {\mathrm{def}}\over =\int \nolimits _0^t\bigl (\Delta _{\ell }^{\mathrm{v}}(\bar{u}^\mathrm{h}\cdot \nabla _{\mathrm{h}}w_\mu ) \,\big |\,\Delta _{\ell }^{\mathrm{v}}w_\mu \bigr )_{L^2}\,\mathrm{d}t',\quad \mathrm{II}_2\buildrel {\mathrm{def}}\over =\int \nolimits _0^t\bigl (\Delta _{\ell }^{\mathrm{v}}(v^{\mathrm{h}}\cdot \nabla _{\mathrm{h}}w_\mu ) \,\big |\,\Delta _{\ell }^{\mathrm{v}}w_\mu \bigr )_{L^2}\,\mathrm{d}t',\\&\mathrm{II}_3\buildrel {\mathrm{def}}\over =\int \nolimits _0^t\bigl (\Delta _{\ell }^{\mathrm{v}}(v^\mathrm{h}_\mu \cdot \nabla _{\mathrm{h}}v_F) \,\big |\,\Delta _{\ell }^{\mathrm{v}}w_\mu \bigr )_{L^2}\,\mathrm{d}t',\quad \mathrm{II}_4\buildrel {\mathrm{def}}\over =\int \nolimits _0^t\bigl (\Delta _{\ell }^{\mathrm{v}}(\bar{u}^\mathrm{h}_\mu \cdot \nabla _{\mathrm{h}}v_F) \,\big |\,\Delta _{\ell }^{\mathrm{v}}w_\mu \bigr )_{L^2}\,\mathrm{d}t',\\&\mathrm{II}_5\buildrel {\mathrm{def}}\over =\int \nolimits _0^t\bigl (\Delta _{\ell }^{\mathrm{v}}(v^3\partial _3 v^3_\mu ) \,\big |\,\Delta _{\ell }^{\mathrm{v}}w_\mu \bigr )_{L^2}\,\mathrm{d}t',\qquad \ \mathrm{II}_6\buildrel {\mathrm{def}}\over =\int \nolimits _0^t\bigl (\Delta _{\ell }^{\mathrm{v}}\partial _3 p_\mu \,\big |\,\Delta _{\ell }^{\mathrm{v}}w_\mu \bigr )_{L^2}\,\mathrm{d}t'. \end{aligned}$$

Let us handle the above term by term.

\(\bullet \) The estimates of \(\underline{\mathrm{II}_1}\) and \(\underline{\mathrm{II}_2}\)

We first get, by applying (4.11) with \(a=\bar{u}^\mathrm{h},~b=\nabla _{\mathrm{h}}w_\mu \) and \(c=w_\mu ,\) that

$$\begin{aligned} |\mathrm{II}_1|\lesssim d_{\ell }^2 2^{-\ell }\Vert w_\mu \Vert _{\widetilde{L}^2_{t,\hbar }\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}} \Vert \nabla _{\mathrm{h}}w_\mu \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{3}{2}}, \end{aligned}$$
(6.3)

whereas by applying a modified version of (4.13) with \(a=v^{\mathrm{h}},~b=\nabla _{\mathrm{h}}w_\mu ,~c=w_\mu \) and \(\mathfrak {g}(t)=\exp \bigl (-\mu \int \nolimits _0^t\hbar (t')\,\mathrm{d}t'\bigr ),\) we find

$$\begin{aligned} |\mathrm{II}_2| \lesssim d_\ell ^2 2^{-\ell }\Vert v^{\mathrm{h}}\Vert _{{\widetilde{L}^\infty _t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}}^{\frac{1}{2}}\Vert \nabla _{\mathrm{h}}v^{\mathrm{h}}\Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}} \Vert w_\mu \Vert _{{\widetilde{L}^\infty _t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}}^{\frac{1}{2}}\Vert \nabla _{\mathrm{h}}w_\mu \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{3}{2}}. \end{aligned}$$
(6.4)

\(\bullet \)The estimate of \(\underline{\mathrm{II}_3}\)

The estimate of \(\mathrm{II}_3\) relies on the following lemma, the proof of which will be postponed until the “Appendix A”:

Lemma 6.1

Let \(a,~c\in \mathcal {B}^{0,\frac{1}{2}}(T) \) and \(b\in {\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4}(T).\) Then for any smooth homogeneous Fourier multiplier, A(D),  of degree zero and any \(\ell \in {\mathbb {Z}}\), it holds that

$$\begin{aligned} \int \nolimits _0^T\bigl |\bigl (A(D)\Delta _{\ell }^{\mathrm{v}}(a\otimes b)\big |\Delta _{\ell }^{\mathrm{v}}c\bigr )_{L^2}\bigr | \,\mathrm{d}t' \lesssim d_{\ell }^2 2^{-\ell }\Vert a\Vert _{\widetilde{L}^4_T(\mathcal {B}_4^{0,\frac{1}{2}})} \Vert b\Vert _{{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4}(T)}\Vert c\Vert _{\widetilde{L}^2_T\big (\mathcal {B}^{0,\frac{1}{2}}\big )}, \end{aligned}$$
(6.5)

and

$$\begin{aligned} \int \nolimits _0^T\bigl |\bigl (A(D)\Delta _{\ell }^{\mathrm{v}}(a\otimes b)\big |\Delta _{\ell }^{\mathrm{v}}c\bigr )_{L^2} \bigr |\,\mathrm{d}t' \lesssim d_{\ell }^2 2^{-\ell }\Vert a\Vert _{\widetilde{L}^2_T\big (\mathcal {B}^{0,\frac{1}{2}}\big )}\Vert b\Vert _{{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4}(T)}\Vert c\Vert _{\mathcal {B}^{0,\frac{1}{2}}(T)}. \end{aligned}$$
(6.6)

Remark 6.1

Indeed the proof of Lemma 6.1 shows that \(\Vert b\Vert _{{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4}(T)}\) in (6.5) and (6.6) can be replaced by \(\Vert b\Vert _{\mathcal {B}^{0,\frac{1}{2}}(T)}.\)

Let us admit this lemma temporarily, and continue our estimate of \(\mathrm{II}_3\). By using integration by parts, we write

$$\begin{aligned} \mathrm{II}_3=-\int \nolimits _0^t\bigl (\Delta _{\ell }^{\mathrm{v}}({{\mathrm{div}}_{\mathrm{h}}}\,v^\mathrm{h}_\mu \cdot v_F) \big |\Delta _{\ell }^{\mathrm{v}}w_\mu \bigr )_{L^2}\,\mathrm{d}t' -\int \nolimits _0^t\bigl (\Delta _{\ell }^{\mathrm{v}}(v^\mathrm{h}_\mu \otimes v_F) \big |\Delta _{\ell }^{\mathrm{v}}\nabla _{\mathrm{h}}w_\mu \bigr )_{L^2}\,\mathrm{d}t'. \end{aligned}$$
(6.7)

Applying (6.6) with \(a={{\mathrm{div}}_{\mathrm{h}}}\,v^\mathrm{h}_\mu ,~b=v_F\) and \(c=w_\mu \) yields

$$\begin{aligned}&\Bigl |\int \nolimits _0^t\bigl (\Delta _{\ell }^{\mathrm{v}}({{\mathrm{div}}_{\mathrm{h}}}\,v^\mathrm{h}_\mu \cdot v_F) \big |\Delta _{\ell }^{\mathrm{v}}w_\mu \bigr )_{L^2}\,\mathrm{d}t'\Bigr |\nonumber \\&\quad \lesssim d_{\ell }^2 2^{-\ell }\Vert \nabla _{\mathrm{h}}v^\mathrm{h}_\mu \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}\Vert v_F\Vert _{{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4}(t)}\Vert w_\mu \Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}, \end{aligned}$$
(6.8)

whereas by applying (6.5) with \(a=v^\mathrm{h}_\mu ,~b=v_F\) and \(c=\nabla _{\mathrm{h}}w_\mu ,\) we obtain

$$\begin{aligned}&\Bigl |\int \nolimits _0^t\bigl (\Delta _{\ell }^{\mathrm{v}}(v^\mathrm{h}_\mu \otimes v_F) \big |\Delta _{\ell }^{\mathrm{v}}\nabla _{\mathrm{h}}w_\mu \bigr )_{L^2}\,\mathrm{d}t'\Bigr |\\&\quad \lesssim d_{\ell }^2 2^{-\ell }\Vert v^\mathrm{h}_\mu \Vert _{\widetilde{L}^4_t(\mathcal {B}_4^{0,\frac{1}{2}})} \Vert v_F\Vert _{{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4}(t)}\Vert \nabla _{\mathrm{h}}w_\mu \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}. \end{aligned}$$

Inserting the above two estimates into (6.7) and using (2.6), we achieve

$$\begin{aligned} \begin{aligned} |\mathrm{II}_3|\lesssim d_{\ell }^2 2^{-\ell }\Vert v_F\Vert _{{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4}(t)} \Vert w_\mu \Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}\Vert v^\mathrm{h}_\mu \Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}. \end{aligned} \end{aligned}$$
(6.9)

\(\bullet \)The estimate of \(\underline{\mathrm{II}_4}\)

Due to \({{\mathrm{div}}_{\mathrm{h}}}\,\bar{u}^\mathrm{h}=0,\) by using integration by parts, we write

$$\begin{aligned} \mathrm{II}_4= & {} \int \nolimits _0^t\bigl (\Delta _{\ell }^{\mathrm{v}}{{\mathrm{div}}_{\mathrm{h}}}\,(\bar{u}^\mathrm{h}v_F) \,\big |\,\Delta _{\ell }^{\mathrm{v}}w_\mu \bigr )_{L^2}\bar{\mathfrak {g}}(t')\,\mathrm{d}t'\\= & {} -\int \nolimits _0^t\bigl (\Delta _{\ell }^{\mathrm{v}}(\bar{u}^\mathrm{h}v_F) \,\big |\,\Delta _{\ell }^{\mathrm{v}}\nabla _{\mathrm{h}}w_\mu \bigr )_{L^2} \bar{\mathfrak {g}}(t')\,\mathrm{d}t'. \end{aligned}$$

By applying Bony’s decomposition (2.7), we get

$$\begin{aligned} \mathrm{II}_4 =-\int \nolimits _0^t\bigl (\Delta _{\ell }^{\mathrm{v}}(T^\mathrm{v}_{\bar{u}^\mathrm{h}} v_F+R^\mathrm{v}(\bar{u}^\mathrm{h}, v_F)) \,\big |\,\Delta _{\ell }^{\mathrm{v}}\nabla _{\mathrm{h}}w_\mu \bigr )_{L^2}\bar{\mathfrak {g}}(t')\,\mathrm{d}t'. \end{aligned}$$

We first observe that

$$\begin{aligned}&\int \nolimits _0^t\bigl |\bigl (\Delta _{\ell }^{\mathrm{v}}(R^\mathrm{v}(\bar{u}^\mathrm{h}, v_F)) \,\big |\,\Delta _{\ell }^{\mathrm{v}}\nabla _{\mathrm{h}}w_\mu \bigr )_{L^2}\bigr |\bar{\mathfrak {g}}(t')\,\mathrm{d}t'\\&\quad \lesssim \sum _{\ell '\geqq \ell -N_0}\int \nolimits _0^t\bar{\mathfrak {g}}(t')\Vert \Delta _{\ell '}^\mathrm{v}\bar{u}^\mathrm{h}(t')\Vert _{L^4_\mathrm{h}(L^2_\mathrm{v})}\Vert S^\mathrm{v}_{\ell +2}v_F(t')\Vert _{L^4_\mathrm{h}(L^\infty _\mathrm{v})} \Vert \Delta _\ell ^\mathrm{v}\nabla _\mathrm{h}w_\mu (t')\Vert _{L^2}\,\mathrm{d}t'\\&\quad \lesssim \sum _{\ell '\geqq \ell -N_0}2^{-\frac{\ell '}{2}}\int \nolimits _0^t d_{\ell '}(t')\bar{\mathfrak {g}}(t')\Vert \bar{u}^\mathrm{h}(t')\Vert _{\mathcal {B}^{0,\frac{1}{2}}_4}\Vert v_F(t')\Vert _{\mathcal {B}^{0,\frac{1}{2}}_4} \Vert \Delta _\ell ^\mathrm{v}\nabla _\mathrm{h}w_\mu (t')\Vert _{L^2}\,\mathrm{d}t'\\&\quad \lesssim \sum _{\ell '\geqq \ell -N_0}d_{\ell '} 2^{-\frac{\ell '}{2}}\int \nolimits _0^t\bar{\mathfrak {g}}(t') \Vert \bar{u}^\mathrm{h}(t')\Vert _{\mathcal {B}^{0,\frac{1}{2}}_4}\Vert v_F(t')\Vert _{\mathcal {B}^{0,\frac{1}{2}}_4} \Vert \Delta _\ell ^\mathrm{v}\nabla _\mathrm{h}w_\mu (t')\Vert _{L^2}\,\mathrm{d}t', \end{aligned}$$

and applying Hölder’s inequality and Proposition 2.1 gives

$$\begin{aligned}&\int \nolimits _0^t\bigl |\bigl (\Delta _{\ell }^{\mathrm{v}}(R^\mathrm{v}(\bar{u}^\mathrm{h}, v_F)) \,\big |\,\Delta _{\ell }^{\mathrm{v}}\nabla _{\mathrm{h}}w_\mu \bigr )_{L^2}\bigr |\bar{\mathfrak {g}}(t')\,\mathrm{d}t'\\&\quad \lesssim \sum _{\ell '\geqq \ell -N_0}d_{\ell '} 2^{-\frac{\ell '}{2}}\Bigl (\int \nolimits _0^t\bar{\mathfrak {g}}^4(t') \Vert \bar{u}^\mathrm{h}(t')\Vert _{\mathcal {B}^{0,\frac{1}{2}}_4}^4\,\mathrm{d}t'\Bigr )^{\frac{1}{4}} \Vert v_F\Vert _{L^4_t(\mathcal {B}^{0,\frac{1}{2}}_4)} \Vert \Delta _\ell ^\mathrm{v}\nabla _\mathrm{h}w_\mu \Vert _{L^2_t(L^2)}\\&\quad \lesssim \mu ^{-\frac{1}{4}}d_{\ell }^2 2^{-\ell }\Vert v_F\Vert _{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4(t)}\Vert \nabla _\mathrm{h}w_\mu \Vert _{L^2_t(\mathcal {B}^{0,\frac{1}{2}})}. \end{aligned}$$

Along the same lines, we find

$$\begin{aligned}&\int \nolimits _0^t\bigl |\bigl (\Delta _{\ell }^{\mathrm{v}}(T^\mathrm{v}_{\bar{u}^\mathrm{h}}v_F) \,\big |\,\Delta _{\ell }^{\mathrm{v}}\nabla _{\mathrm{h}}w_\mu \bigr )_{L^2}\bigr |\bar{\mathfrak {g}}(t')\,\mathrm{d}t'\\&\quad \lesssim \sum _{|\ell '-\ell |\leqq 5}\int \nolimits _0^t\bar{\mathfrak {g}}(t')\Vert S_{\ell '-1}^\mathrm{v}\bar{u}^\mathrm{h}(t') \Vert _{L^4_\mathrm{h}(L^\infty _\mathrm{v})}\Vert \Delta ^\mathrm{v}_{\ell }v_F(t')\Vert _{L^4_\mathrm{h}(L^2_\mathrm{v})} \Vert \Delta _\ell ^\mathrm{v}\nabla _\mathrm{h}w_\mu (t')\Vert _{L^2}\,\mathrm{d}t'\\&\quad \lesssim \sum _{|\ell '-\ell |\leqq 5} \int \nolimits _0^t\bar{\mathfrak {g}}(t') \Vert \bar{u}^\mathrm{h}(t')\Vert _{\mathcal {B}^{0,\frac{1}{2}}_4}\Vert \Delta _{\ell '}^\mathrm{v}v_F(t')\Vert _{L^4_\mathrm{h}(L^2_\mathrm{v})} \Vert \Delta _\ell ^\mathrm{v}\nabla _\mathrm{h}w_\mu (t')\Vert _{L^2}\,\mathrm{d}t'\\&\quad \lesssim \sum _{|\ell '-\ell |\leqq 5}\Bigl (\int \nolimits _0^t\bar{\mathfrak {g}}^4(t') \Vert \bar{u}^\mathrm{h}(t')\Vert _{\mathcal {B}^{0,\frac{1}{2}}_4}^4\,\mathrm{d}t'\Bigr )^{\frac{1}{4}}\Vert \Delta _{\ell '}^\mathrm{v}v_F\Vert _{L^4_t(L^4_\mathrm{h}(L^2_\mathrm{v}))} \Vert \Delta _\ell ^\mathrm{v}\nabla _\mathrm{h}w_\mu \Vert _{L^2_t(L^2)}\\&\quad \lesssim \mu ^{-\frac{1}{4}}d_{\ell }^2 2^{-\ell }\Vert v_F\Vert _{\widetilde{L}^4_t(\mathcal {B}^{0,\frac{1}{2}}_4)}\Vert \nabla _\mathrm{h}w_\mu \Vert _{L^2_t(\mathcal {B}^{0,\frac{1}{2}})}. \end{aligned}$$

As a result, it turns out that

$$\begin{aligned} |\mathrm{II}_4|\lesssim \mu ^{-\frac{1}{4}}d_{\ell }^2 2^{-\ell } \Vert v_F\Vert _{{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4}(t)}\Vert \nabla _{\mathrm{h}}w_\mu \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}. \end{aligned}$$
(6.10)

\(\bullet \)The estimates of \(\underline{\mathrm{II}_5}\)

Due to \(\partial _3v^3=-{{\mathrm{div}}_{\mathrm{h}}}\,v^\mathrm{h}\) and \(v^3=w+v_F,\) we write

$$\begin{aligned} \mathrm{II}_5&=\int \nolimits _0^t\bigl (\Delta _{\ell }^{\mathrm{v}}(-v^3{{\mathrm{div}}_{\mathrm{h}}}\,v^\mathrm{h}_\mu ) \,\big |\,\Delta _{\ell }^{\mathrm{v}}w_\mu \bigr )_{L^2}\,\mathrm{d}t'\\&=-\int \nolimits _0^t\bigl (\Delta _{\ell }^{\mathrm{v}}(v_F{{\mathrm{div}}_{\mathrm{h}}}\,v^\mathrm{h}_\mu +w_\mu {{\mathrm{div}}_{\mathrm{h}}}\,v^\mathrm{h}) \,\big |\,\Delta _{\ell }^{\mathrm{v}}w_\mu \bigr )_{L^2}. \end{aligned}$$

Then applying (6.6) gives rise to

$$\begin{aligned} \begin{aligned} |\mathrm{II}_5|&\lesssim d_\ell ^2 2^{-\ell }\Vert \nabla _{\mathrm{h}}v^{\mathrm{h}}\Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}\bigl (\Vert v_F\Vert _{{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4}(t)}+\Vert w_\mu \Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}\bigr ) \Vert w_\mu \Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}\\&\lesssim d_{\ell }^2 2^{-\ell }\Vert v^{\mathrm{h}}\Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)} \bigl (\Vert v_F\Vert _{{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4}(t)} +\Vert w_\mu \Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}\bigr )\Vert w_\mu \Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}. \end{aligned} \end{aligned}$$
(6.11)

\(\bullet \)The estimates of \(\underline{\mathrm{II}_6}\)

The estimate of \(\mathrm{II}_6\) can be handled similarly as \(\mathrm{I}_6\). Indeed in view of (5.9), we write

$$\begin{aligned} \begin{aligned} \partial _3p&=\partial _3(-\Delta )^{-1}{{\mathrm{div}}_{\mathrm{h}}}\,\bigl (v^\mathrm{h}\cdot \nabla _\mathrm{h}v^\mathrm{h}+\bar{u}^\mathrm{h}\cdot \nabla _\mathrm{h}v^\mathrm{h}+v^\mathrm{h}\cdot \nabla _\mathrm{h}\bar{u}^\mathrm{h}+\bar{u}^\mathrm{h}\cdot \nabla _\mathrm{h}\bar{u}^\mathrm{h}\\&\quad +v^3\partial _3\bar{u}^\mathrm{h}+v^3\partial _3v^\mathrm{h}\bigr )+ \partial _3^2(-\Delta )^{-1}\bigl (v\cdot \nabla v^3+\bar{u}^\mathrm{h}\cdot \nabla _\mathrm{h}v^3\bigr ). \end{aligned} \end{aligned}$$

Accordingly, we have the decomposition \(\mathrm{II}_6=\sum _{i=1}^5\mathrm{II}_{6,i}\) with

$$\begin{aligned} \mathrm{II}_{6,1}&=\int \nolimits _0^t\bigl (\Delta _{\ell }^{\mathrm{v}}\partial _3(-\Delta )^{-1}{{\mathrm{div}}_{\mathrm{h}}}\,\bigl (v^\mathrm{h}\cdot \nabla _\mathrm{h}v^\mathrm{h}_\mu +\bar{u}^\mathrm{h}\cdot \nabla _{\mathrm{h}}v^\mathrm{h}_\mu +v^{\mathrm{h}}_\mu \cdot \nabla _{\mathrm{h}}\bar{u}^\mathrm{h}\bigr )\,\big |\,\Delta _{\ell }^{\mathrm{v}}w_\mu \bigr )_{L^2}\,\mathrm{d}t',\\ \mathrm{II}_{6,2}&=\int \nolimits _0^t\bigl (\Delta _{\ell }^{\mathrm{v}}\partial _3(-\Delta )^{-1}{{\mathrm{div}}_{\mathrm{h}}}\,(v^3\partial _3v^\mathrm{h}_\mu ) \,\big |\,\Delta _{\ell }^{\mathrm{v}}w_\mu \bigr )_{L^2}\,\mathrm{d}t',\\ \mathrm{II}_{6,3}&=\int \nolimits _0^t\bigl (\Delta _{\ell }^{\mathrm{v}}\partial _3(-\Delta )^{-1}{{\mathrm{div}}_{\mathrm{h}}}\,(v^3_\mu \partial _3\bar{u}^\mathrm{h})\,\big |\,\Delta _{\ell }^{\mathrm{v}}w_\mu \bigr )_{L^2}\,\mathrm{d}t',\\ \mathrm{II}_{6,4}&=\int \nolimits _0^t\bigl (\Delta _{\ell }^{\mathrm{v}}\partial _3^2(-\Delta )^{-1} (v\cdot \nabla v^3_\mu +\bar{u}^\mathrm{h}\cdot \nabla _{\mathrm{h}}v^3_\mu )\,\big |\,\Delta _{\ell }^{\mathrm{v}}w_\mu \bigr )_{L^2}\,\mathrm{d}t',\\ \mathrm{II}_{6,5}&=\sum _{i=1}^2\sum _{j=1}^2\int \nolimits _0^t \bigl (2\Delta _{\ell }^{\mathrm{v}}(-\Delta )^{-1}\partial _i\partial _j(\bar{u}^i \partial _3\bar{u}^j_\mu ) \,\big |\,\Delta _{\ell }^{\mathrm{v}}w_\mu \bigr )_{L^2}\,\mathrm{d}t'. \end{aligned}$$

It is easy to observe from the estimate of \(\mathrm{I}_{6,1}\) that

$$\begin{aligned} \begin{aligned} |\mathrm{II}_{6,1}|&\lesssim d_{\ell }^2 2^{-\ell } \Bigl (\Vert v^{\mathrm{h}}\Vert _{{\widetilde{L}^\infty _t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}}^{\frac{1}{2}}\Vert \nabla _{\mathrm{h}}v^{\mathrm{h}}\Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{3}{2}} \Vert w_\mu \Vert _{{\widetilde{L}^\infty _t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}}^{\frac{1}{2}}\Vert \nabla _{\mathrm{h}}w_\mu \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}\\&\quad +\Vert w_\mu \Vert _{\widetilde{L}^2_{t,\hbar }\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}\Vert \nabla _{\mathrm{h}}w_\mu \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}} \Vert \nabla _{\mathrm{h}}v^\mathrm{h}_\mu \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}\Bigr ). \end{aligned} \end{aligned}$$
(6.12)

Mean while, by using \(\partial _3v^3=-{{\mathrm{div}}_{\mathrm{h}}}\,v^\mathrm{h}\) and integration by parts, we write

$$\begin{aligned} \begin{aligned} \mathrm{II}_{6,2}&=\int \nolimits _0^t\bigl (\Delta _{\ell }^{\mathrm{v}}\partial _3(-\Delta )^{-1}{{\mathrm{div}}_{\mathrm{h}}}\,[\partial _3(v^3v^\mathrm{h}_\mu )-v^\mathrm{h}_\mu \partial _3v^3] \,\big |\,\Delta _{\ell }^{\mathrm{v}}w_\mu \bigr )_{L^2}\,\mathrm{d}t'\\&=-\int \nolimits _0^t\bigl (\Delta _{\ell }^{\mathrm{v}}(-\Delta )^{-1}\partial _3^2 (v^3v^\mathrm{h}_\mu ) \,\big |\,\Delta _{\ell }^{\mathrm{v}}\nabla _{\mathrm{h}}w_\mu \bigr )_{L^2}\,\mathrm{d}t'\\&\quad +\int \nolimits _0^t\bigl (\Delta _{\ell }^{\mathrm{v}}\partial _3(-\Delta )^{-1}{{\mathrm{div}}_{\mathrm{h}}}\,(v^{\mathrm{h}}_\mu {{\mathrm{div}}_{\mathrm{h}}}\,v^{\mathrm{h}}) \,\big |\,\Delta _{\ell }^{\mathrm{v}}w_\mu \bigr )_{L^2}\,\mathrm{d}t' \buildrel {\mathrm{def}}\over =\mathrm{II}_{6,2}^a+\mathrm{II}_{6,2}^b. \end{aligned} \end{aligned}$$

It follows from (6.5) and \(v^3=v_F+w\) that

$$\begin{aligned} \bigl |\mathrm{II}_{6,2}^a\bigr |&\lesssim d_{\ell }^2 2^{-\ell }\Vert v^{\mathrm{h}}\Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)} \bigl (\Vert v_F\Vert _{{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4}(t)} +\Vert w_\mu \Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}\bigr )\Vert w_\mu \Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}, \end{aligned}$$

whereas by using a modified version of (4.13), we infer

$$\begin{aligned} \bigl |\mathrm{II}_{6,2}^b\bigr |\lesssim d_{\ell }^2 2^{-\ell } \Vert v^{\mathrm{h}}\Vert _{{\widetilde{L}^\infty _t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}}^{\frac{1}{2}}\Vert \nabla _{\mathrm{h}}v^{\mathrm{h}}\Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{3}{2}} \Vert w_\mu \Vert _{{\widetilde{L}^\infty _t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}}^{\frac{1}{2}}\Vert \nabla _{\mathrm{h}}w_\mu \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}. \end{aligned}$$

Therefore, we obtain

$$\begin{aligned} |\mathrm{II}_{6,2}|\lesssim d_{\ell }^2 2^{-\ell } \Vert v^{\mathrm{h}}\Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)} \bigl (\Vert v_F\Vert _{{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4}(t)}+\Vert v^{\mathrm{h}}\Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)} +\Vert w_\mu \Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}\bigr )\Vert w_\mu \Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)},\nonumber \\ \end{aligned}$$
(6.13)

whereas applying (6.6) with \(a=\partial _3\bar{u}^\mathrm{h},~b=v^3_\mu \) and \(c=w_\mu \) leads to

$$\begin{aligned} |\mathrm{II}_{6,3}|\lesssim d_{\ell }^2 2^{-\ell }\Vert \partial _3\bar{u}^\mathrm{h}\Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}\bigl (\Vert v_F\Vert _{{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4}(t)}+\Vert w_\mu \Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}\bigr )\Vert w_\mu \Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}. \end{aligned}$$
(6.14)

On the other hand, again due to \(\mathrm{div}\,v=0,\) we write

$$\begin{aligned} \mathrm{II}_{6,4}&=\int \nolimits _0^t\bigl (\Delta _{\ell }^{\mathrm{v}}\partial _3^2(-\Delta )^{-1} \bigl (v^\mathrm{h}\cdot \nabla _\mathrm{h}w_\mu +v^\mathrm{h}_\mu \cdot \nabla _\mathrm{h}v_F+v^3\partial _3v^3_\mu \\&\quad +\bar{u}^\mathrm{h}\cdot \nabla _{\mathrm{h}}w_\mu +\bar{u}^\mathrm{h}\cdot \nabla _{\mathrm{h}}v_F \bigr )\,\big |\,\Delta _{\ell }^{\mathrm{v}}w_\mu \bigr )_{L^2}\,\mathrm{d}t'. \end{aligned}$$

Noticing that \((-\Delta )^{-1}\partial _3^2\) is a bounded Fourier operator, we observe that \(\mathrm{II}_{6,4}\) shares the same estimate as \(\sum _{i=1}^5\mathrm{II}_{i}\) given before, that is,

$$\begin{aligned} \begin{aligned} |\mathrm{II}_{6,4}|&\lesssim d_{\ell }^2 2^{-\ell } \Bigl (\Vert v^{\mathrm{h}}\Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)} \Vert w_\mu \Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}^2 +\Vert w_\mu \Vert _{\widetilde{L}^2_{t,\hbar }\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}\Vert w_\mu \Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}^{\frac{3}{2}}\\&\quad +\Vert v_F\Vert _{{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4}(t)}\bigl (\mu ^{-\frac{1}{4}}+\Vert v^\mathrm{h}_\mu \Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}\bigr )\Vert w_\mu \Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}\Bigr ). \end{aligned} \end{aligned}$$
(6.15)

Finally since \((-\Delta )^{-1}\partial _i\partial _j\) is a bounded Fourier operator, we get, by applying (4.11) with \(a=\bar{u}^\mathrm{h},~b=\partial _3\bar{u}^\mathrm{h}_\mu ,~c=w_\mu \), that

$$\begin{aligned} |\mathrm{II}_{6,5}|\lesssim d_{\ell }^2 2^{-\ell }\Vert w_\mu \Vert _{\widetilde{L}^2_{t,\hbar }\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}} \Vert \nabla _{\mathrm{h}}w_\mu \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}\Vert \partial _3\bar{u}^\mathrm{h}_\mu \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}. \end{aligned}$$
(6.16)

By summing (6.126.16), we arrive at

$$\begin{aligned} \begin{aligned} |\mathrm{II}_{6}|&\lesssim d_{\ell }^2 2^{-\ell }\Bigl (\bigl (\Vert v^{\mathrm{h}}\Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)} +\Vert \partial _3\bar{u}^\mathrm{h}\Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}+\Vert w_\mu \Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}\bigr )\Vert w_\mu \Vert _{\widetilde{L}^2_{t,\hbar }\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}} \Vert w_\mu \Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}^{\frac{1}{2}}\\&\quad +\bigl (\mu ^{-\frac{1}{4}}+\Vert \partial _3\bar{u}^\mathrm{h}_\mu \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}+\Vert v^\mathrm{h}\Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}\bigr )\Vert v_F\Vert _{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4}\Vert w_\mu \Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}\\&\quad +\bigl (\Vert v^{\mathrm{h}}\Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}+\Vert \partial _3\bar{u}^\mathrm{h}\Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}\bigr )\Vert w_\mu \Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}^2+\Vert v^\mathrm{h}\Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}^2 \Vert w_\mu \Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}\Bigr ). \end{aligned} \end{aligned}$$
(6.17)

Let us now complete the proof of (3.18).

Proof of (3.18)

By inserting the estimates (6.3), (6.4), (6.96.11) and (6.17) into (6.2), and then multiplying \(2^{\ell +1}\) to the resulting inequality, and finally taking square root and then summing up the resulting inequalities over \({\mathbb {Z}},\) we obtain

$$\begin{aligned} \begin{aligned}&\Vert w_\mu \Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}+\sqrt{2\mu }\Vert w_\mu \Vert _{{\widetilde{L}^2_{t,\hbar }\big (\mathcal {B}^{0,\frac{1}{2}}\big )}}\\&\quad \leqq \Vert u^3_{0,\mathrm{l}\mathrm{h}}\Vert _{\mathcal {B}^{0,\frac{1}{2}}}+C\bigl (\Vert v^{\mathrm{h}}\Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}^{\frac{1}{2}} +\Vert \partial _3\bar{u}^\mathrm{h}_\mu \Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}\bigr )\Vert w_\mu \Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}\\&\quad \quad +C\Bigl (\Vert v^\mathrm{h}\Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}+\bigl (\mu ^{-\frac{1}{8}}+\Vert v^{\mathrm{h}}\Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}^{\frac{1}{2}}+\Vert \partial _3\bar{u}^\mathrm{h}\Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}\bigr ) \Vert v_F\Vert _{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4}^{\frac{1}{2}}\Bigr )\Vert w_\mu \Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}^{\frac{1}{2}}\\&\quad \quad +C\bigl (\Vert v^{\mathrm{h}}\Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}^{\frac{1}{2}} +\Vert \partial _3\bar{u}^\mathrm{h}\Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}+\Vert w_\mu \Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}^{\frac{1}{2}}\bigr )\Vert w_\mu \Vert _{\widetilde{L}^2_{t,\hbar }\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{4}} \Vert \nabla _{\mathrm{h}}w_\mu \Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}^{\frac{1}{4}}. \end{aligned} \end{aligned}$$

Applying Young’s inequality gives

$$\begin{aligned}&C\Bigl (\Vert v^\mathrm{h}\Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}+\bigl (\mu ^{-\frac{1}{8}}+\Vert v^{\mathrm{h}}\Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}^{\frac{1}{2}}+\Vert \partial _3\bar{u}^\mathrm{h}\Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}\bigr ) \Vert v_F\Vert _{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4}^{\frac{1}{2}}\Bigr )\Vert w_\mu \Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}^{\frac{1}{2}}\\&\quad \leqq \frac{1}{12}\Vert w_\mu \Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}+C\bigl (\mu ^{-\frac{1}{4}}+\Vert v^{\mathrm{h}}\Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}+\Vert \partial _3\bar{u}^\mathrm{h}\Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}\bigr )\Vert v_F\Vert _{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4(t)} +C\Vert v^\mathrm{h}\Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}^2, \end{aligned}$$

and

$$\begin{aligned}&C\bigl (\Vert v^{\mathrm{h}}\Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}^{\frac{1}{2}}+\Vert \partial _3\bar{u}^\mathrm{h}\Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}+\Vert w_\mu \Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}^{\frac{1}{2}}\bigr )\Vert w_\mu \Vert _{\widetilde{L}^2_{t,\hbar }\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{4}} \Vert w_\mu \Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}^{\frac{1}{4}}\\&\quad \leqq \frac{1}{12}\Vert w_\mu \Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}+C\Vert w_\mu \Vert _{\widetilde{L}^2_{t,\hbar }\big (\mathcal {B}^{0,\frac{1}{2}}\big )}+C\bigl (\Vert v^{\mathrm{h}}\Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}+\Vert \partial _3\bar{u}^\mathrm{h}\Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}\bigr ). \end{aligned}$$

As a result, we have

$$\begin{aligned}&\Vert w_\mu \Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}+\sqrt{2\mu }\Vert w_\mu \Vert _{{\widetilde{L}^2_{t,\hbar }\big (\mathcal {B}^{0,\frac{1}{2}}\big )}} \leqq \Vert u^3_{0,\mathrm{l}\mathrm{h}}\Vert _{\mathcal {B}^{0,\frac{1}{2}}}+C\Vert w_\mu \Vert _{\widetilde{L}^2_{t,\hbar }\big (\mathcal {B}^{0,\frac{1}{2}}\big )}\\&\quad +\Bigl (\frac{1}{6}+C\bigl (\Vert v^{\mathrm{h}}\Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}^{\frac{1}{2}} +\Vert \partial _3\bar{u}^\mathrm{h}\Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}\bigr )\Bigr )\Vert w_\mu \Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}+C\Bigl (\Vert v^{\mathrm{h}}\Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}\\&\quad +\Vert \partial _3\bar{u}^\mathrm{h}\Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}+\Vert v^{\mathrm{h}}\Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}^2+\bigl (\mu ^{-\frac{1}{4}}+ \Vert v^{\mathrm{h}}\Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}+\Vert \partial _3\bar{u}^\mathrm{h}\Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}\bigr )\Vert v_F\Vert _{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4(t)}\Bigr ). \end{aligned}$$

Taking \(\mu \) in the above inequality so that \(\sqrt{2\mu }=C\) gives rise to

$$\begin{aligned} \begin{aligned}&\Bigl (\frac{5}{6}-C\bigl (\Vert v^{\mathrm{h}}\Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}^{\frac{1}{2}} +\Vert \partial _3\bar{u}^\mathrm{h}\Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}\bigr )\Bigr )\Vert w_\mu \Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}\\&\quad \leqq \Vert u^3_{0,\mathrm{l}\mathrm{h}}\Vert _{\mathcal {B}^{0,\frac{1}{2}}}+C\Bigl (\Vert v^{\mathrm{h}}\Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}+\Vert \partial _3\bar{u}^\mathrm{h}\Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}+\Vert v^{\mathrm{h}}\Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}^2\\&\quad \quad +\bigl (1+ \Vert v^{\mathrm{h}}\Vert _{\mathcal {B}^{0,\frac{1}{2}}(t)}+\Vert \partial _3\bar{u}^\mathrm{h}\Vert _{\widetilde{L}^2_t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}\bigr )\Vert v_F\Vert _{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4(t)}\Bigr ). \end{aligned} \end{aligned}$$
(6.18)

On the other hand, in view of the definition of \(u^3_{0,\mathrm{l}\mathrm{h}}\), it holds for any \(\ell \in {\mathbb {Z}}\) that

$$\begin{aligned} \Vert \Delta _{\ell }^{\mathrm{v}}u^3_{0,\mathrm{l}\mathrm{h}}\Vert _{L^2} \lesssim \sum _{|j-\ell |\leqq 1}\Vert S^\mathrm{h}_{j-1}\Delta _j^{\mathrm{v}}u^3_0\Vert _{L^2} \lesssim d_\ell 2^{-\frac{\ell }{2}}\Vert u^3_0\Vert _{{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4}}, \end{aligned}$$

which indicates that

$$\begin{aligned} \Vert u^3_{0,\mathrm{l}\mathrm{h}}\Vert _{\mathcal {B}^{0,\frac{1}{2}}}\lesssim \Vert u^3_0\Vert _{{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4}}. \end{aligned}$$

Inserting the above estimate into (6.18) and repeating the argument from (4.18) to (4.19), we conclude the proof of (3.18). \(\quad \square \)