Global Controllability of the Navier–Stokes Equations in the Presence of Curved Boundary with No-Slip Conditions

Abstract

We consider the issue of the small-time global exact null controllability of the axi-symmetric incompressible Navier–Stokes equation in a 3D finite vertical cylinder with circular section. We assume that we are able to act on the fluid flow on the top and on the bottom of the cylinder while no-slip conditions are prescribed on the boundary of the lateral section. We also make use of a distributed control, which can be chosen arbitrarily small for any Sobolev regularity in space. Our work improves earlier results in Guerrero et al. (C R Math Acad Sci Paris 343:573–577, 2006; J Math Pures Appl (9) 98:689–709, 2012) where the distributed force is small only in a negative Sobolev space and the recent work Coron et al. (Ann PDE 5(2):1–49, 2019) where the case of the 2D incompressible Navier–Stokes equation in a rectangle was considered. Our analysis actually follows quite narrowly the one in Coron et al. (2019) by making use of Coron’s return method, of the well-prepared dissipation method and of long-time nonlinear Cauchy–Kovalevskaya estimates. An extra difficulty here is the curvature of the uncontrolled part of the boundary which requires further analysis to apply the well-prepared dissipation method to lower order boundary layer terms.

1 Introduction, Statement of the Main Result and Some Open Problems

1.1 Setting

We consider a finite vertical cylinder $\Omega =D_1\times [0,L] \subset {\mathbb {R}}^3$ where $D_1$ is the unit open disc in ${\mathbb {R}}^2$ and $L>0$. Let us denote by $\Gamma = (D_1\times \{0\}) \cup (D_1\times \{L\})$ the union of the bottom and of the top of the cylinder. This surface $\Gamma$ has to be thought as the controlled part of the boundary while the complementary part $\partial \Omega \backslash \Gamma = (\partial D_1)\times [0,L]$ of the cylinder’s boundary, that is the lateral part of the cylinder’s boundary, has to be thought as the uncontrolled part of the boundary. Let us denote by $L^2_{{\mathrm {div}}\,}(\Omega )$ the space of the axi-symmetric functions u in $L^2(\Omega )$ such that ${\mathrm {div}}\,u=0$ and $u=0$ on $\partial \Omega \backslash \Gamma$.

1.2 Main Result

Our main result is the following theorem.

Theorem 1.1

Let $T>0$ and $u_0\in L^2_{{\mathrm {div}}\,}(\Omega )$. For any $k\in {\mathbb {N}}$ and for any $\eta >0$, there exists a force $f\in L^1((0,T);H^k(\Omega ))$ satisfying $\Vert f\Vert _{L^1((0,T);H^k(\Omega ))}\le \eta$, and a Leray weak solution $u\in C_{\mathrm{w}}([0,T];L^2_{{\mathrm {div}}\,}(\Omega ))\cap L^2((0,T);H^1(\Omega ))$ of

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _t u+u\cdot \nabla u-\Delta u+\nabla p=f \quad \text { and } \quad {\mathrm {div}}\,u=0\qquad \text { in }\Omega ,\\ u=0\qquad \text { on }\partial \Omega \backslash \Gamma , \end{array}\right. } \end{aligned}$$

(1.1)

satisfying $u(0,\cdot )=u_0$ and $u(T,\cdot )=0$.

1.3 Notion of Controlled Weak Leray Solution

Above, the notion of weak Leray solution corresponds to the following weak formulation:

$$\begin{aligned} \begin{aligned}&- \int _0^T\int _\Omega u \cdot \partial _t \varphi + \int _0^T\int _\Omega (u \cdot \nabla ) u \cdot \varphi + 2 \int _0^T\int _\Omega D(u) : D(\varphi ) \\&\qquad \quad = \int _\Omega u_0 \cdot \varphi (0,\cdot ) +\int _0^T\int _\Omega u \cdot f , \end{aligned} \end{aligned}$$

(1.2)

for any test function $\varphi \in C^\infty ([0,T]\times {\bar{\Omega }})$ which is divergence free, tangent to $\partial \Omega \backslash \Gamma$, vanishes at $t = T$ and vanishes on $\Gamma$. This last condition encodes that one controls the part $\Gamma$ of the boundary, so that no boundary condition is there required. Above the notation $D(\cdot )$ stands for the symmetric part of the gradient.

1.4 Comparison with the Literature

Theorem  1.1 establishes the small-time global exact null controllability of the axi-symmetric incompressible Navier–Stokes equation in a 3D finite cylinder with circular section in the case where the no-slip Dirichlet boundary condition is imposed on the lateral boundary of the cylinder, while we are able to act on the fluid flow on the top and on the bottom of the cylinder, as well as in the interior of the cylinder through a distributed force which can be chosen arbitrarily small for any Sobolev regularity in space. This result improves earlier results in [12, 13] where the distributed force is small only in a negative Sobolev space and the recent work [8] where the case of the 2D incompressible Navier–Stokes equation in a rectangle was considered. Let us also mention the paper [6], and its corresponding proceeding [7], on the global null controllability of the Navier–Stokes equation in the case where Navier slip-with-friction boundary conditions are prescribed on the uncontrolled part of the boundary rather than the no-slip conditions. In these references some controlled weak Leray solutions are constructed. This has been improved into smooth solutions (in the case where the initial data is smooth) in the recent paper [15]. The main difference between the case of the Navier conditions and the one of the no-slip conditions, when following the method of proof considered in these references and in the present paper, is the size of the boundary layers which are involved. Indeed in both cases one considers a small-viscosity regime where a controlled auxiliary flow which is built for the inviscid limit of the system has to be corrected by some boundary layer terms on the uncontrolled part of the boundary.

1.5 Difficulty of the Proof

Indeed the proof of Theorem 1.1 below closely follows the analysis developed in [8] to deal with the case of the rectangle by making use of Coron’s return method, of the well-prepared dissipation method and of long-time nonlinear Cauchy–Kovalevskaya estimates. One of the main extra difficulty here is the curvature of the uncontrolled part of the cylinder boundary which requires a more delicate analysis, in particular to apply the well-prepared dissipation method, because of lower order boundary layer profiles whose behavior is forced by the higher order boundary layer profiles. This difficulty already appeared in [15] where the case of the Navier boundary conditions, rather than the no-slip condition, was treated.

1.6 Open Problems

It will be interesting to investigate whether or not it is possible to extend the result of Theorem 1.1 to the case of smooth solutions or to the case of non-axisymmetric initial data. We refer to Remark 2.10 below for a few more technical explanations regarding the difficulties of these investigations. An even more challenging issue, as already mentioned in the rectangle case tackled in [8, 9], is to get rid of the “phantom” force f.

2 Strategy of the Proof of the Main Result: Theorem 1.1

In this section we explain the general strategy of the proof of Theorem  1.1.

2.1 A Rapid Reminder on Coron’s Return Method for the Incompressible Euler Equations

Following Coron’s return method, The starting idea is to follow the strategy used by Coron and Glass, see [11], to prove the small-time global exact null controllability of the incompressible Euler equations. In this case, the idea is to introduce an auxiliary flow which is compactly supported in time, and around which the linearized incompressible Euler equations is controllable. To drive the dynamics despite the presence of a non-zero initial data these auxiliary flows have to be chosen with a strong amplitude, which makes them vary quickly. To encode these features we introduce the rescaling

$$\begin{aligned}&u^{\varepsilon }(t,x):=\varepsilon u(\varepsilon t,x),\quad p^{\varepsilon }(t,x):=\varepsilon ^2 p(\varepsilon t,x), \end{aligned}$$

(2.1)

where $\varepsilon$ is thought as a small positive parameter and we observe that the time interval over which the control is required is then stretched from (0, T) to the large time interval $(0, T/\varepsilon )$.

Some appropriate auxiliary flows can be built for very general geometries of the fluid domain with involved gluing technics, but can be taken in a very simple form in the present case of the finite vertical cylinder $\Omega$ as

$$\begin{aligned} h^0(t) e_3 , \end{aligned}$$

(2.2)

where $\{e_i\}_{i=1}^3$ are the unit vectors of the canonical basis of ${\mathbb {R}}^3.$ Since whatever is the scalar function $h^0$, the flow given by (2.2) unconditionally satisfies the incompressible Euler equations in $\Omega$, associated with the pressure $-(h^0)'(t) x_3$, while being tangent to the lateral, uncontrolled, part $\partial \Omega \backslash \Gamma$ of the cylinder’s boundary. This tangency condition is the natural counterpart of the no-slip condition, actually of any impermeable condition, for the Euler equations (instead of the Navier–Stokes equation).

Moreover the linearization of the incompressible Euler equations around this flow is simply the equation

$$\begin{aligned} \partial _t v+h^0(t) \partial _3 v +\nabla q=0 \quad \text { and } \quad {\mathrm {div}}\,v=0, \end{aligned}$$

(2.3)

where, moreover, the pressure term can be discarded since the divergence free constraint satisfies by the initial data is propagated by transport along the flow (2.2). Together with the tangency condition on the lateral boundary, this leads to the even simpler equation than (2.3), that is the pressure-less equation:

$$\begin{aligned} \partial _t v+h^0(t) \partial _3 v =0 . \end{aligned}$$

(2.4)

A controlled solution, in the rescaled variables given by (2.1), can be then constructed as an asymptotic expansion with principal part $h^0(t) e_3 + \varepsilon v$. It is then only a matter to choose the function $h^0$ in order to flush v outside of $\Omega$ and to bound the effect of the remainder term of the expansion to conclude the controllability of the incompressible Euler equations.

2.2 An Almost Returning Solution to the Navier–Stokes Equations Despite the Boundary Layers

Now, the flow (2.2) also satisfies the Navier–Stokes equations inside the cylinder but does not satisfy the no-slip boundary conditions on the boundary of the cross-section. To adapt to the no-slip condition, some boundary layers appear near the boundary. Indeed, under the rescaling (2.1) the (unforced) incompressible Navier–Stokes equations then read as the following small viscosity problem:

$$\begin{aligned} \partial _t u^{\varepsilon }+u^{\varepsilon }\cdot \nabla u^{\varepsilon }-\varepsilon \Delta u^{\varepsilon }+\nabla p^{\varepsilon }=0 \quad \text { and } \quad {\mathrm {div}}\,u^{\varepsilon }=0 . \end{aligned}$$

Thanks to the peculiar form of the flow (2.2), the dynamics of this boundary layer is simply given by a heat equation, rather than by Prandtl’s equations. Indeed in polar coordinates, for a radial function f, the Laplace operator is given by the formula

$$\begin{aligned} \Delta f&= \frac{\partial ^2 f}{\partial r^2} + \frac{1}{r} \frac{\partial f}{\partial r}, \end{aligned}$$

(2.5)

where $r:= \sqrt{x_1^2 + x_2^2 }$ represents the distance to the origin.

The following key intermediate result proves the existence of axisymmetric solutions to the heat equations in the unit disk $D_1$ with appropriate Dirichlet data which almost return to zero at the final time, while the principal part of the Dirichlet data satisfies a flushing condition at an intermediate time.

Proposition 2.1

There are $h^0,h^1$ and $h^2$ in $C^{\infty }_0(0,T)$, satisfying the flushing condition

$$\begin{aligned} \int _0^{\frac{ T }{3 } } h^0(t) \, dt = 2L , \end{aligned}$$

(2.6)

such that the solutions $(v^\varepsilon := v^\varepsilon (t,r) )_\varepsilon$ to

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _t v^\varepsilon -\varepsilon \Delta v^\varepsilon = 0 \qquad t\ge 0, 0 \le r \le 1 ,\\ v^\varepsilon (t,1)= -\sum _{i=0}^2 \varepsilon ^{\frac{i}{2}} h^i(t) \qquad t\ge 0, 0 \le r \le 1,\\ v^\varepsilon (0,r)=0 \qquad 0 \le r \le 1, \end{array}\right. }\nonumber \\ \end{aligned}$$

(2.7)

verifies

$$\begin{aligned} \Vert v^\varepsilon \left( \frac{T}{\varepsilon },\cdot \right) \Vert _{L^2(D_1)}\le C \varepsilon ^{\frac{7}{4}} . \end{aligned}$$

(2.8)

Moreover the function $h^0$ has to satisfy the zeroth order moment condition:

$$\begin{aligned} \int _0^{T } h^0(t) \, dt = 0. \end{aligned}$$

(2.9)

Let us highlight that despite a right hand side $o(\varepsilon )$ in (2.8) would be enough for the sequel, we choose to keep the explicit exponent which comes for free with our method of proof. The zeroth order moment condition (2.9) comes from the construction of $V^0$ in Sects. 2.5 and 3. Since the source term of the Eq. (2.14) for $V^0$ below is zero, we find that $h^0$ satisfies (2.9) from (3.14) in Sect. 3. Indeed the proof of Proposition 2.1 is based on the well-prepared dissipation method, which was initiated in [17], and adapted in [6, 8, 15]. We give below the core of the proof, postponing to the next sections some intermediate results which are themselves quite intricate.

2.3 A Multi-scale Asymptotic Expansion of $v^\varepsilon$

Let $\chi$ be a cut-off function $\chi (r)=0$ when $r\le \frac{1}{3}$ and $\chi (r)=1$ when $r\ge \frac{2}{3}$. Given $h^0,h^1$ and $h^2$ in $C^{\infty }_0(0,T)$, we seek for a multi-scale asymptotic expansion of the solutions $(v^\varepsilon )_\varepsilon$ of the form

$$\begin{aligned} v^{\varepsilon }(t,r):\,= & {} \, -\chi (r) \sum _{i=0}^2\varepsilon ^{\frac{i}{2}} V^i\left( t,\frac{1-r}{\sqrt{\varepsilon }}\right) +\varepsilon r^{\varepsilon }(t,x_1,x_2), \end{aligned}$$

(2.10)

where $r = (x_1^2 + x_2^2)^{\frac{1}{2}}$ and $V^i(t,z)\rightarrow 0$ as $z\rightarrow 0$ for $0\le i\le 2$ and $r^{\varepsilon }$ is considered as a technical lower-order corrector. To infer which relevant “profiles” $V^i(t,z)$ we should consider, we observe that for a function g(t, z), denoting by $\alpha \in C^{\infty }([\frac{1}{3},1];{\mathbb {R}})$ the function such that

$$\begin{aligned} \frac{1}{r}= 1+ (1-r)+(1-r)^2\alpha (r), \end{aligned}$$

(2.11)

we have:

$$\begin{aligned} \varepsilon \Delta \{g\}_{\varepsilon }(t,x)= \{ \partial _z^2 g \}_{\varepsilon }(t,x) - {\varepsilon }^{\frac{1}{2}}\{ \partial _z g\}_{\varepsilon }(t,x) - \varepsilon \{ z \partial _z g\}_{\varepsilon }(t,x) - \varepsilon ^{\frac{3}{2}} \alpha (r) \{ z^2 \partial _z g\}_{\varepsilon }(t,x) , \end{aligned}$$

(2.12)

where the notation $\{\cdot \}_{\varepsilon }$ stands for

$$\begin{aligned} \{g\}_{\varepsilon }(t,x):=g\left( t,\frac{1-r}{\sqrt{\varepsilon }}\right) , \quad r = (x_1^2 + x_2^2)^{\frac{1}{2}} . \end{aligned}$$

(2.13)

Therefore, by inserting the ansatz (2.10) in the Eq. (2.7), and then using (2.12) and equalling by powers of $\varepsilon$, we are led to consider the following three problems:

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _tV^0-\partial _z^2V^0=0,\quad t\ge 0,z\ge 0,\\ V^0(t,1)=h^0(t),\qquad t\ge 0,\\ V^0(0,z)=0,\qquad \qquad \, z\ge 0, \end{array}\right. } \end{aligned}$$

(2.14)

where we observe that the only non-zero source is on the boundary data,

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _t V^1-\partial _z^2V^1=-\partial _zV^0,\quad t\ge 0,z\ge 0,\\ V^1(t,1)=h^1(t),\qquad \qquad \quad t\ge 0,\\ V^1(0,z)=0,\qquad \qquad \qquad z\ge 0, \end{array}\right. }\nonumber \\ \end{aligned}$$

(2.15)

and

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _t V^2-\partial _z^2V^2=-\partial _zV^1-z\partial _z V^0,\quad t\ge 0,z\ge 0,\\ V^2(t,1)=h^{2}(t),\qquad t\ge 0,\\ V^2(0,z)=0,\qquad z\ge 0. \end{array}\right. } \end{aligned}$$

(2.16)

2.4 Boundary Forcing and Enhanced Time-Decay of the Free and Forced Heat Equation in a Disk

In this subsection we establish that there exists an appropriate choice of the boundary data, to impose a rapid decay in time of the solutions to the forced heat equation, such as (2.14)–(2.16), for which the choice is respectively the one of the function $h^0$, $h^1$ and $h^2$.

Let us introduce the following weighted Sobolev spaces.

Definition 2.2

For $z\in {\mathbb {R}}$, we denote $\langle z\rangle :=\sqrt{1+z^2}$ and for s and $q\in {\mathbb {N}},$ we set

endowed with it natural associated norm. In the same way we define $H^{s}_{q}({\mathbb {R}})$ and the norm

$$\begin{aligned} \Vert f\Vert _{H^{s}_{q}({\mathbb {R}})} := \left( \sum _{j=0}^s \int _{{\mathbb {R}}}\langle z\rangle ^{2q}|\partial _z^jf(z)|^2dz \right) ^{\frac{1}{2}} . \end{aligned}$$

Observe that by the Plancherel theorem, we have the following equivalence of norms:

$$\begin{aligned} \Vert f\Vert _{H^{s}_{q}({\mathbb {R}})} \sim \sum _{j=0}^{q} \left( \int _{{\mathbb {R}}} \langle \zeta \rangle ^{2s} | \partial _{\zeta }^j {\hat{f}}(\zeta )|^2 d\zeta \right) ^{\frac{1}{2}} , \end{aligned}$$

(2.17)

where ${\hat{f}}$ denotes the Fourier transform of f.

Definition 2.3

Let $k\in {\mathbb {N}},\gamma >0$ and X a Banach space with norm $\Vert \cdot \Vert _X$. We define the space $C^k_{\gamma }({\mathbb {R}}_+;X)$ of the functions $f\in C^k({\mathbb {R}}_+;X)$ such that

$$\begin{aligned} \Vert f\Vert _{C^k_{\gamma }({\mathbb {R}}_+;X)} :=\sup _{t\ge 0,0\le j\le k}\big (\Vert \partial _t^jf(t)\Vert _X\langle t\rangle ^{\gamma }\big )<+\infty , \end{aligned}$$

where

$$\begin{aligned} C^k({\mathbb {R}}_+;X) :=\big \{ f:\partial _t^jf\in C({\mathbb {R}}_+;X), \ 0\le j\le k \big \}. \end{aligned}$$

The proof of the next proposition is given in Sect. 3.

Proposition 2.4

Let $\gamma >0$, $k,s,q,n\in {\mathbb {N}}$ satisfy

$$\begin{aligned} n\ge \frac{q}{2}+\gamma -1. \end{aligned}$$

(2.18)

and we define

$$\begin{aligned} {\tilde{\gamma }}:= 2n+3, \quad {\tilde{s}}:= s+2k+2n, \quad {\tilde{q}}:= 2n+3. \end{aligned}$$

(2.19)

Given $f\in C^0_{{\tilde{\gamma }}}({\mathbb {R}}_+;H^{{\tilde{s}}}_{{\tilde{q}}}({\mathbb {R}}_+))$ when $k=0$ and $f \in C^{k-1}_{{\tilde{\gamma }}}({\mathbb {R}}_+;H^{{\tilde{s}}}_{{\tilde{q}} } ({\mathbb {R}}_+))$ when $k\ge 1$, we can find a nonzero function $h\in C^{\infty }_0(0,T)$, supported in $(0,\frac{T}{3}]\cup [\frac{2T}{3},T)$, such that the following system

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _t v-\partial _z^2 v=f, \qquad t\ge 0, z\ge 0,\\ v(t,0)=h(t),\qquad t\ge 0,\\ v(0,z)=0,\qquad \qquad z \ge 0, \end{array}\right. }\nonumber \\ \end{aligned}$$

(2.20)

has a unique solution $v\in C^{k}_{\gamma }({\mathbb {R}}_+;H^{s}_{q}({\mathbb {R}}_+).$ Moreover, if f is supported away from $t=0$ as a function of time t, so does v.

2.5 Proof of Proposition 2.1

Let us now see how to conclude the proof of Proposition 2.1 by applying Proposition 2.4 to the three problems (2.14)–(2.16). Let $k_0=0, \gamma _0=23,s_0=28,q_0=23$.

• By Proposition 2.4, there exists $h^0\in C^{\infty }_0(0,T)$ supported in $(0,\frac{T}{3}]\cup [\frac{2T}{3},T)$ such that system (2.14) has a unique solution $V^0\in C^0_{23}({\mathbb {R}}_+;H^{28}_{23}({\mathbb {R}}_+))$. Moreover $V^0$ is supported away from $t=0$ as a function of time t.

• Using Proposition 2.4 again, there exists $h^1\in C^{\infty }_0(0,T)$ such that system (2.15) has a unique solution $V^1\in C^{0}_{7}({\mathbb {R}}_+;H^{7}_{8}({\mathbb {R}}_+))$. Moreover $V^1$ is supported away from $t=0$ as a function of time t.

• By a final use of Proposition 2.4, there exists $h^{2}\in C^{\infty }_0(0,T)$ and $V^2\in C^{0}_{2}({\mathbb {R}}_+;H^{2}_{2}({\mathbb {R}}_+))$ satisfying (2.16). Moreover $V^2$ is supported away from $t=0$ as a function of time t.

By the definition of $\{\cdot \}_{\varepsilon }$, for a profile $\mathcal {V}$ defined in ${\mathbb {R}}^2_+$ with $\mathcal {V}\in C({\mathbb {R}}_+;L^2({\mathbb {R}}_+))$,

$$\begin{aligned} \Vert \{\mathcal {V}(t,\cdot )\}_{\varepsilon }\Vert _{L^2(D_1)}\le C\varepsilon ^{\frac{1}{4}}\Vert \mathcal {V}(t,\cdot )\Vert _{L^2({\mathbb {R}}_+)} \end{aligned}$$

(2.21)

for a constant $C>0.$ Hence, in view of (2.10),

$$\begin{aligned} \Vert v^{\varepsilon }(\frac{T}{\varepsilon },\cdot )\Vert _{L^2(D_1)}\,\le & {} \, C \varepsilon ^{\frac{1}{4}} (\Vert V^0(\frac{T}{\varepsilon },\cdot )\Vert _{L^2({\mathbb {R}}_+)}+\sqrt{\varepsilon }\Vert V^1( \frac{T}{\varepsilon },\cdot )\Vert _{L^{2}({\mathbb {R}}_+)}+\varepsilon \Vert V^2(\frac{T}{\varepsilon },\cdot )\Vert _{L^2({\mathbb {R}}_+)}) \nonumber \\&\,+\varepsilon \Vert r^{\varepsilon }\Vert _{L^2(D_1)} \nonumber \\\le & {} \,C\varepsilon ^{{\frac{9}{4}}}+\varepsilon \Vert r^{\varepsilon }\Vert _{L^2(D_1)}, \end{aligned}$$

(2.22)

where we use $V^i\in C^0_2({\mathbb {R}}_+;L^2({\mathbb {R}}_+))$ by construction. Then the proof of Proposition 2.1 is complete up to the following result on the remainder which is postponed to Sect. 4.

Lemma 2.5

There exists a constant C such that, for $s=0,1,$

$$\begin{aligned} \Vert r^{\varepsilon }\Vert _{L^{\infty }({\mathbb {R}}_+;L^2(D_1))}\,\le & {} \, C \varepsilon ^{\frac{3}{4}}, \end{aligned}$$

(2.23)

$$\begin{aligned} \Vert r^{\varepsilon }\Vert _{L^2({\mathbb {R}}_+;H^{2s}(D_1))}\,\le & {} \, C \varepsilon ^{-\frac{1}{4}.} \end{aligned}$$

(2.24)

2.6 An Auxiliary Solution of the Navier–Stokes Equations in the Infinite Cylinder

We then consider as an auxiliary solution of the Navier–Stokes equations:

$$\begin{aligned} u^{\text {aux}} = \left( \sum _{i=0}^2\varepsilon ^{\frac{i}{2}} h^i(t) + v^\varepsilon \right) e_3 , \end{aligned}$$

(2.25)

where $r := \sqrt{x_1^2 + x_2^2 }$ and $h^0,h^1$, $h^2$ and $v^\varepsilon$ is the family of functions defined in the proof of Proposition 2.1.

Let us extend the finite vertical cylinder $\Omega$ into a cylinder with infinite height:

$$\begin{aligned} \mathcal {C}:=D_1\times {\mathbb {R}}. \end{aligned}$$

(2.26)

We also needs an extension $u_b$ of the initial data $u_0$. Actually, we will need in the sequel to work with analytic data, due to the difficulty related to a loss of derivatives due to the boundary layers in the equation satisfied by the remainder. More precisely one will use analyticity with respect to the vertical variable $x_3$. Therefore we cannot simply extend the initial data $u_0$ by 0 outside of $\Omega$. The reason why we use the index b is that we will need a first regularization step to get a large enough radius of analyticity to completely overrule the effect of the loss of derivative.

Then, by setting

$$\begin{aligned} p^{\text {aux}} = - \,\sum _{i=0}^2\varepsilon ^{\frac{i}{2}} (h^i)'(t) x_3 , \end{aligned}$$

(2.27)

we obtain that

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _t u^{\text {aux}}+u^{\text {aux}}\cdot \nabla u^{\text {aux}}-\varepsilon \Delta u^{\text {aux}}+\nabla p^{\text {aux}}= 0 \qquad \text { in }(0,\frac{T}{{\varepsilon }})\times \mathcal {C},\\ {\mathrm {div}}\,u^{\text {aux}}=0\qquad \text { in }(0,\frac{T}{{\varepsilon }})\times \mathcal {C},\\ u^{\text {aux}}=0 \qquad \text { on }(0,\frac{T}{{\varepsilon }})\times \partial \mathcal {C},\\ u^{\text {aux}}|_{t=0}=0 \qquad \text { in } \mathcal {C}. \end{array}\right. } \end{aligned}$$

(2.28)

and

$$\begin{aligned} \Vert u^{\text {aux}}\left( \frac{T}{\varepsilon },\cdot \right) \Vert _{L^2(\Omega )}\le C\varepsilon ^{\frac{7}{4}}. \end{aligned}$$

(2.29)

We also define (the index “fl” stands for “flushed”):

$$\begin{aligned} u^{\text {fl}}(t,x):= \mu (t) u_b\left( x-\left( \int _0^th^0(s)ds\right) e_3\right) , \end{aligned}$$

(2.30)

where $\mu (t)\in C^{\infty }({\mathbb {R}})$ is a cut-off function which satisfies $\mu (t)=1$ when $t\le \frac{T}{3}$ and $\mu (t)=0$ when $t\ge \frac{2T}{3}$.

We observe that $u^{\text {fl}}$ satisfies

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _t u^{\text {fl}}+h^0\partial _3 u^{\text {fl}}=\xi ^{\text {fl}}+f^{\text {fl}}\qquad \text { in } (0,T)\times \mathcal {C},\\ {\mathrm {div}}\,u^{\text {fl}}=0\qquad \text { in } (0,T)\times \mathcal {C},\\ u^{\text {fl}}=0\qquad \text { on } (0,T)\times \partial \mathcal {C},\\ u^{\text {fl}}=u_b\qquad \text { on } \{0\}\times \mathcal {C}, \end{array}\right. }\nonumber \\ \end{aligned}$$

(2.31)

where $\partial _3:=\partial _{x_3}$ and by virtue of (2.6),

$$\begin{aligned} \xi ^{\text {fl}}(t,x):\,= & {} \,{\dot{\mu }}(t)u_b(x-2Le_3)|_{\mathcal {C}\backslash \Omega }, \end{aligned}$$

(2.32)

$$\begin{aligned} f^{\text {fl}}(t,x):\,= & {} \,{\dot{\mu }}(t)u_b(x-2Le_3)|_{\Omega }. \end{aligned}$$

(2.33)

We also introduce a domain

$$\begin{aligned} G:=D_1\times [-2L,-L], \end{aligned}$$

(2.34)

and a space

$$\begin{aligned} L^2_{{\mathrm {div}}\,}(\mathcal {C}):=\{v\in L^2(\mathcal {C}):v \text { is axi-symmetric, }{\mathrm {div}}\,v=0, v=0\text { on }\partial \mathcal {C}\}. \end{aligned}$$

(2.35)

It is easy to observe that the control profile $\xi ^{\text {fl}}$ is supported in $\mathcal {C}\backslash \Omega$, the phantom profile $f^{\text {fl}}$ is supported in $\Omega ,$ and for any $k\in {\mathbb {N}}$,

$$\begin{aligned} \Vert f^{\text {fl}}\Vert _{L^1(0,T);H^k(\mathcal {C}))}\le T\Vert u_b\Vert _{H^k(G)}. \end{aligned}$$

(2.36)

Now consider any Leray weak solution

$$\begin{aligned} u^{\varepsilon }\in C_{\mathrm{w}}([0,\frac{T}{\varepsilon }];L^2_{{\mathrm {div}}\,,\text {loc}}(\mathcal {C}))\cap L^2\left( (0,\frac{T}{\varepsilon });H^1(\mathcal {C})\right) , \end{aligned}$$

of the rescaled system

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _t u^{\varepsilon }+u^{\varepsilon }\cdot \nabla u^{\varepsilon }-\varepsilon \Delta u^{\varepsilon }+\nabla p^{\varepsilon }=\xi ^{\varepsilon }+f^{\varepsilon }\qquad \text { in }\left( 0,\frac{T}{{\varepsilon }}\right) \times \mathcal {C},\\ {\mathrm {div}}\,u^{\varepsilon }=0\qquad \text { in }\left( 0,\frac{T}{{\varepsilon }}\right) \times \mathcal {C},\\ u^{\varepsilon }=0 \qquad \quad \text { on }\left( 0,\frac{T}{{\varepsilon }}\right) \times \partial \mathcal {C},\\ u^{\varepsilon }|_{t=0}=\varepsilon u_b \qquad \text { in } \mathcal {C}. \end{array}\right. }\nonumber \\ \end{aligned}$$

(2.37)

We decompose $u^{\varepsilon }$ and $p^{\varepsilon }$ into

$$\begin{aligned} u^{\varepsilon }\,= & {} \, u^{\text {aux}} + \varepsilon u^{\text {fl}} + \varepsilon R^{\varepsilon } , \end{aligned}$$

(2.38)

$$\begin{aligned} p^{\varepsilon }\,= & {} \,p^{\text {aux}}+\varepsilon \pi ^{\varepsilon }, \end{aligned}$$

(2.39)

then we take

$$\begin{aligned} \xi ^{\varepsilon }=\varepsilon \xi ^{\text {fl}}\qquad \text {and} \qquad f^{\varepsilon }=\varepsilon f^{\text {fl}}, \end{aligned}$$

(2.40)

and we observe that the remainder $R^{\varepsilon }$ satisfies

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _tR^{\varepsilon }-\varepsilon \Delta R^{\varepsilon }+u^{\varepsilon }\cdot \nabla R^{\varepsilon }+ R^{\varepsilon }\cdot \nabla (u^{\text {aux}} + \varepsilon u^{\text {fl}})+\nabla \pi ^{\varepsilon }=F^{\varepsilon },\\ {\mathrm {div}}\,R^{\varepsilon }=0,\\ R^{\varepsilon }|_{\partial \mathcal {C}}=0,\\ R^{\varepsilon }|_{t=0}=0, \end{array}\right. }\nonumber \\ \end{aligned}$$

(2.41)

where

$$\begin{aligned} F^{\varepsilon }:\,= & {} \,-(-\chi \{V^0\}_{\varepsilon }+\sqrt{\varepsilon }(h^1-\chi \{V^1\}_{\varepsilon })+\varepsilon (h^2-\chi \{V^2\}_{\varepsilon })+\varepsilon r^{\varepsilon })\partial _3 u^{\text {fl}} \nonumber \\&\,+u^{\text {fl}}_r\chi '\{V^0+\sqrt{\varepsilon }V^1+\varepsilon V^2\}_{\varepsilon }e_3-u^{\text {fl}}_r\frac{\chi }{1-r}\{z\partial _z V^0+\sqrt{\varepsilon }z\partial _z V^1+\varepsilon z\partial _z V^2\}_{\varepsilon }e_3 \nonumber \\&\, -\varepsilon u^{\text {fl}}_r\partial _r r^{\varepsilon }e_3-\varepsilon u^{\text {fl}}\cdot \nabla u^{\text {fl}}+\varepsilon \Delta u^{\text {fl}}. \end{aligned}$$

(2.42)

where $u^{\text {fl}}_r$ is the $e_r$ component of $u^{\text {fl}}$ given by (2.55). By construction, $u^{\text {fl}}$ is supported in [0, T], so does $F^{\varepsilon }$.

Indeed, because of the fast variation due to the boundary layer terms in $u^{\text {aux}}$, the term $R^{\varepsilon }\cdot \nabla u^{\text {aux}}$ in (2.41) is singular with respect to $\varepsilon$. A classical trick here is to treat this singularity in $\varepsilon$ against a loss of derivative, see for example [18, 19], and (5.18)–(5.20) below, which leads to consider analytic regularity in order to bootstrap some estimates.

Proposition 2.6

Let $T>0.$ There exists $\rho _b>0$ such that, for every $u_b\in L^2_{{\mathrm {div}}\,}(\Omega )$ for which there exists $C_b$ such that

$$\begin{aligned}&\forall m\ge 0,\quad \Vert \partial _{x_3}^m u_b\Vert _{H^3(\mathcal {C})}\le \frac{m!}{\rho _b^m}C_b, \end{aligned}$$

(2.43)

$$\begin{aligned}&\Vert u_b\Vert _{L^1_{x_3}(H^2(D_1))}\le C_b. \end{aligned}$$

(2.44)

for every $k\in {\mathbb {N}}$, we have the following estimate for $R^{\varepsilon }$, there exists a constant C such that

$$\begin{aligned} \sup _{t\in [0,T/{\varepsilon }]}\Vert R^{\varepsilon }(t)\Vert _{L^2(\mathcal {C})}+\left( \int _0^{T/{\varepsilon }}\varepsilon \Vert \nabla R^{\varepsilon }\Vert _{L^2(\mathcal {C})}^2\right) ^{\frac{1}{2}}\le C \varepsilon ^{\frac{1}{4}}. \end{aligned}$$

(2.45)

The proof of Proposition 2.6 is postponed to Sect. 5.

Let us now explain how to deduce Theorem 1.1 by following the strategy of [8]. It is a matter to glue together three steps, each of which corresponds to a subpart of the time interval which is imparted to realize the control. To explain this, we go back to the original scaling and first recast the result obtained so far, which will be used as a second step (explaining the notation b as an index for the “initial” data below). Setting, for $\varepsilon \in (0,1)$,

$$\begin{aligned}&u^{\varepsilon }(t,x):=\varepsilon u(\varepsilon t,x),\quad p^{\varepsilon }(t,x):=\varepsilon ^2 p(\varepsilon t,x), \end{aligned}$$

(2.46)

$$\begin{aligned}&\xi ^{\varepsilon }(t,x):=\varepsilon ^2 \xi (\varepsilon t,x)\quad \text { and }\quad f^{\varepsilon }(t,x):=\varepsilon ^2 f(\varepsilon t,x), \end{aligned}$$

(2.47)

we deduce from Lemma 2.5, the definition of $u^{\text {fl}}$ and Proposition 2.6 the following result about the possibility to drive a large analytic initial data with a sufficiently large analyticity radius can be driven approximately into the null equilibrium.

Proposition 2.7

Let $T>0$. There exists $\rho _b>0$ such that, for every $\sigma >0$ and each $u_b\in L^2_{{\mathrm {div}}\,}(\mathcal {C})$ for which there exists $C_b$ such that (2.43) and (2.44) hold, for every $k\in {\mathbb {N}}$, there exist two forces $\xi \in C^{\infty }([0,T]\times \overline{\mathcal {C}\backslash \Omega })$ and $f\in C^{\infty }([0,T]\times {\overline{\Omega }})$ satisfying

$$\begin{aligned}&\Vert f\Vert _{L^1((0,T);H^k(\Omega ))}\le C_k\Vert u_b|_{G}\Vert _{H^k(G)}, \end{aligned}$$

(2.48)

$$\begin{aligned}&\text {supp } \xi \subset (0,T)\times \overline{\mathcal {C}\backslash \Omega }, \end{aligned}$$

(2.49)

where the constant $C_k$ depends only on k, and a Leray weak solution $u\in C_{\mathrm{w}}([0,T];L^2_{{\mathrm {div}}\,}(\mathcal {C}))\cap L^2((0,T);H^1(\mathcal {C}))$ to (2.52) associated with the initial data $u_b$, such that there exists $0<T_c\le T$ such that $u_c:=u(T_c,\cdot )$ satisfies

$$\begin{aligned} \Vert u_c|_{\Omega }\Vert _{L^2(\Omega )}\le \sigma . \end{aligned}$$

(2.50)

Since this result requires some analytic initial data, we use as a prequel the following result regarding the regularization of any finite energy initial data into an analytic function with arbitrary analyticity radius.

Proposition 2.8

Let $T>0$, $\rho _b>0$ and $u_0\in L^2_{{\mathrm {div}}\,}(\Omega )$ with $u_0\cdot {\mathbf {n}}=0\text { on }\Gamma$. For any $k\in {\mathbb {N}}$ and $\eta _b>0,$ there exists an extension $u_a\in L^2_{{\mathrm {div}}\,}(\mathcal {C})$ of $u_0$ to the domain $\mathcal {C}$, a control force $\xi \in C^{\infty }([0,T]\times \overline{\mathcal {C}\backslash \Omega })$, a phantom force $f\in C^{\infty }([0,T]\times {\overline{\Omega }})$ satisfying

$$\begin{aligned} \Vert f\Vert _{L^1(0,T);H^k(\Omega )}\le \eta _b, \end{aligned}$$

(2.51)

a Leray weak solution $u\in C_{\mathrm{w}}([0,T];L^2_{{\mathrm {div}}\,}(\mathcal {C}))\cap L^2((0,T);H^1(\mathcal {C}))$ to

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _t u+u\cdot \nabla u-\Delta u+\nabla p=\xi +f\qquad \text { in }(0,T)\times \mathcal {C},\\ {\mathrm {div}}\,u=0 \qquad \text { in }(0,T)\times \mathcal {C},\\ u=0\qquad \text { on }(0,T)\times \partial \mathcal {C}, \end{array}\right. } \end{aligned}$$

(2.52)

associated with initial data $u_a$, $C_b>0$ and $0<T_b\le T$ such that $u_b:=u(T_b,\cdot )\in L^2_{{\mathrm {div}}\,}(\mathcal {C})$ satisfies

$$\begin{aligned}&\Vert u_b|_{G}\Vert _{H^k(G)}\le \eta _b,\end{aligned}$$

(2.53)

(2.43) and (2.44).

Let us emphasize that for $L^2_{{\mathrm {div}}\,}(\Omega )$ initial data, we can find $T_1\in (0,T)$ such that $u(T_1,\cdot )\in H^1(\Omega )$. Now for this new $H^1$ “initial” data, we can find a small time $T_2\in (T_1, T]$ and a solution u to (2.52) with zero force, such that $u\in C^{\infty }((T_1,T_2)\times \mathcal {C})$, the rest of the proof is almost the same as the proof of Proposition 1.7 of [8] and is therefore left to the reader.

Last, we use that small enough states can be driven exactly to the rest state.

Proposition 2.9

Let $T>0$. There exists $\sigma >0$ such that, for any $u_c\in L^2_{{\mathrm {div}}\,}(\Omega )$ which satisfies

$$\begin{aligned} \Vert u_c\Vert _{L^2(\Omega )}\le \sigma , \end{aligned}$$

(2.54)

there exists a Leray weak solution $u\in C_{\mathrm{w}}([0,T];L^2_{{\mathrm {div}}\,}(\Omega ))\cap L^2((0,T);H^1((\Omega ))$ to (1.1) with the associated initial data $u_c$ and $f=0$, which satisfies $u(T,\cdot )=0.$

We refer to Theorem 2 of [10]. Actually, the original theorem of [10] requires the initial date to be small in $L^4(\Omega )\cap L^2_{{\mathrm {div}}\,}(\Omega )$. While if we only have the smallness of the initial data, by using the energy inequality, we can find a $T_1\in (0,\frac{T}{2})$ such that $u(T_1)$ is small in $L^2(\Omega )\cap H^1(\Omega )$ hence small in $L^4(\Omega )\cap L^2_{{\mathrm {div}}\,}(\Omega ).$ Therefore we only require the initial data small in $L^2_{{\mathrm {div}}\,}(\Omega )$.

Theorem 1.1 is then implied by combining the three propositions above.

Remark 2.10

We observe that in the strategy above we only make use of analyticity in the vertical direction. One can wonder if a use of analyticity in the orthoradial direction could be helpful to extend the the result of Theorem 1.1 to the case of smooth solutions or to the case of non-axisymmetric initial data. Unfortunately this does not seem to be the case. One difficulty is that, to deal with the nonlinear feature in the remainder estimates, see Sect. 5, one typically needs to replace the space $L^2$ by the Sobolev space $H^{\frac{1}{2}}$ to bootstrap some estimates. This requires to construct yet more accurate asymptotic expansion. Then one observes that the lower order boundary layer profiles which would be natural to consider depend on the vertical variable and satisfy some equations which couples the terms of a heat equation with respect to the fast variable z, like the ones considered above, and some transport terms in the $x_3$ direction with a prefactor which corresponds to the principal term of $u^{\text {aux}}$. Unfortunately the well-prepared dissipation method seems to be delicate to adapt to such problems.

2.7 Organization of the Rest of the Paper

The rest of the paper is devoted to the proof of the intermediate results which were admitted above in the course of the proof of Theorem  1.1. The two next sections are devoted to the proofs of the intermediate results used to prove Proposition 2.1, that is Proposition 2.4 in Sect. 3, and Lemma  2.5 in Sect. 4. Next, in Sect. 5 we give the proof of Proposition 2.6. Once again, the proof is quite technical, and we will use two technical intermediate results: Proposition 5.4 and Lemma 5.3 whose proofs are postponed respectively to Sects. 6 and 7.

let $x_1=r\cos \theta ,x_2=r\cos \theta ,$ $n=\frac{1}{r}(x_1, x_2, 0)$. Let $\{e_i\}_{i=1}^3$ be the unit vectors of the canonical basis of ${\mathbb {R}}^3$. Let $e_r=(\cos \theta , \sin \theta , 0), e_{\theta }=(-\sin \theta ,\sin \theta , 0).$ For a vector $a\in {\mathbb {R}}^3,$ we write

$$\begin{aligned} a=a_1e_1+a_2e_2+a_3e_3=a_re_r+a_{\theta }e_{\theta }+a_3e_3. \end{aligned}$$

(2.55)

We always denote $\nabla _\mathrm{h}:=(\partial _{x_1},\partial _{x_2})$, $\partial _i:=\partial _{x_i}$ for $1\le i\le 3,$ $\partial _r:=e_r\cdot \nabla$ and $(a,b)=\int _{\mathcal {C}} a(x) b(x)\,dx$ the $L^2(\mathcal {C})$ inner product of a and b.

3 Proof of Proposition 2.4

This section is devoted to the proof of Proposition 2.4. We will rely on the following result from [15, Lemma 3.4] where, for $n\in {\mathbb {N}}$ and $x\in {\mathbb {R}},$ we set

$$\begin{aligned} s_n(x):=\sum _{k=0}^n \frac{x^k}{k!}. \end{aligned}$$

(3.1)

Lemma 3.1

Let $\gamma >0$ and $k,s,q,n\in {\mathbb {N}}$ satisfying (2.18) and ${\tilde{\gamma }}$, ${\tilde{s}}$, and ${\tilde{q}}$ as in (2.19). Let $v_0 \in H^{s+2k}_{{\tilde{q}} } ({\mathbb {R}})$ and $f\in C^0_{{\tilde{\gamma }}}({\mathbb {R}}_+;H^{{\tilde{s}}}_{{\tilde{q}}}({\mathbb {R}}))$ when $k=0$ and $f \in C^{k-1}_{{\tilde{\gamma }}}({\mathbb {R}}_+;H^{{\tilde{s}}}_{{\tilde{q}} } ({\mathbb {R}}))$ when $k\ge 1$, such that

$$\begin{aligned} \left. \Big (\partial ^j_{\zeta }\big ({\hat{v}}_0(\zeta )+\int _0^{\infty }s_n(t\zeta ^2){\hat{f}}(t,\zeta )dt\big )\Big )\right| _{\zeta =0}=0,\quad \text { for } 0\le j\le 2n+1. \end{aligned}$$

(3.2)

Then the following Cauchy problem

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _t v-\partial _z^2 v=f, \quad t\ge 0, z\in {\mathbb {R}},\\ v|_{t=0}=v_0,\quad z\in {\mathbb {R}}, \end{array}\right. } \end{aligned}$$

(3.3)

has a unique solution $v\in C^{k}_{\gamma }({\mathbb {R}}_+;H^{s}_{q}({\mathbb {R}})).$

We are now ready to start the proof of Proposition 2.4.

Proof of Proposition 2.4

We first observe that it is sufficient to deal with the case where $k=0$, since the general case follows by using that for $0\le i\le k$, for z in ${\mathbb {R}}$ and $t\ge 0$, $\partial _t^{i}v=\partial _z^2\partial _t^{i-1}v+\partial _t^{i-1}f$.

Assume that $h\in C_0^{\infty }(0,T)$. Let $j_0=[\frac{{\tilde{s}}+1}{2}]$,

$$\begin{aligned} A_0(t):=h(t), \quad A_j(t):=\partial _tA_{j-1}(t)-\partial _z^{2j-2}f|_{z=0^+}\text { for }1\le j\le j_0. \end{aligned}$$

(3.4)

For $z\ge 0$, we denote

$$\begin{aligned} A(t,z):=\sum _{j=0}^{j_0}A_j(t)\frac{z^{2j}}{(2j)!}\chi _1(z), \end{aligned}$$

(3.5)

where $\chi _1(z)\in C_{0}^{\infty }(\overline{{\mathbb {R}}_+})$ is a cut-off function which satisfies $\chi _1(z)=1$ for $z\in [0,\frac{1}{2}]$ and $\chi _1(z)=0$ for $z\ge 1$. One can check that

$$\begin{aligned} A\in C_{{\tilde{\gamma }}}^0({\mathbb {R}}_+;C_0^{\infty }(\overline{{\mathbb {R}}_+})). \end{aligned}$$

(3.6)

Then the function $V:=v-A$ satisfies, for $t\ge 0,z\ge 0,$

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _tV-\partial _z^2V=F,\\ V(t,0)=0,\\ V(0,z)=0 \end{array}\right. } \end{aligned}$$

(3.7)

where

$$\begin{aligned} F:=f-\partial _tA+\partial _z^2 A. \end{aligned}$$

(3.8)

It follows from the construction of $A_j(t),$ that $\partial _z^{2j}F|_{z=0^+}=0$ for $0\le j\le j_0-1.$ Thus, extending F by $F(t,z)=-F(t,-z)$ to the whole line $z\in {\mathbb {R}}$, we have

$$\begin{aligned} F\in C^0_{{\tilde{\gamma }}}({\mathbb {R}}_+;H^{{\tilde{s}}}_{{\tilde{q}}}({\mathbb {R}})). \end{aligned}$$

(3.9)

Moreover, if f is supported away from $t=0$ as a function of time t, so does F, and by using the energy method it is easy to find that V is also supported away from $t=0$.

Now, let us observe that, to prove Proposition 2.4, it is sufficient to find a function $h\in C_0^{\infty }(0,T)$, supported in $(0,\frac{T}{3}]\cup [\frac{2T}{3},T)$, such that the associated function F satisfies the condition:

$$\begin{aligned} \left. \Big (\partial ^j_{\zeta }\big (\int _0^{\infty }s_n(t\zeta ^2){\hat{F}}(t,\zeta )dt\big )\Big )\right| _{\zeta =0}=0,\quad \text { for } 0\le j\le 2n+1, \end{aligned}$$

(3.10)

where $s_n$ is defined in (3.1) and ${\hat{F}}$ denotes the Fourier transform of F with respect to z. Indeed, then by the result of Lemma 3.1, we obtain that $V\in C^0_{\gamma }({\mathbb {R}}_+;H^s_q({\mathbb {R}}))$. Moreover, by (3.6), it follows that $v\in C^0_{\gamma }({\mathbb {R}}_+;H^s_1({\mathbb {R}}_+))$. Actually, since F is an odd function with respect to z, it suffices to guarantee (3.10) for odd integer j with $1\le j\le 2n+1$. $\square$

By the definition of $A_j(t)$ in (3.4), for $0\le j\le j_0$,

$$\begin{aligned} A_j(t)=\partial _t^jh(t)-\sum _{i=0}^{j-1}\partial _t^{j-1-i}\partial _z^{2i}f|_{z=0^+}. \end{aligned}$$

(3.11)

Then, by combining (3.5), (3.8) and (3.11), we arrive at

$$\begin{aligned} F\,= & {} \,f-\sum _{j=0}^{j_0}\partial _tA_j(t)\frac{z^{2j}}{(2j)!}\chi _1(z)+\sum _{j=0}^{j_0}A_j(t)\partial _z^2(\frac{z^{2j}}{(2j)!}\chi _1(z))\\= & {} \,f-\sum _{j=0}^{j_0-1}\partial _z^{2j}f|_{z=0^+}\frac{z^{2j}}{(2j)!}\chi _1(z) -\partial _tA_{j_0}(t)\frac{z^{2j_0}}{(2j_0)!}\chi _1(z)+A_0(t)\chi _1''(z)\\&\,+\sum _{j=1}^{j_0}A_j(t)\frac{z^{2j-1}}{(2j-1)!}(2\chi _1'(z)+\frac{z}{2j}\chi _1''(z))\\= & {} \,{\tilde{f}}+\sum _{j=0}^{j_0+1}\partial _t^j h(t)\alpha _j(z), \end{aligned}$$

where

$$\begin{aligned} {\tilde{f}}:\,= & {} \,f-\sum _{j=0}^{j_0-1}\partial _z^{2j}f|_{z=0^+}\frac{z^{2j}}{(2j)!}\chi _1(z) +\sum _{i=0}^{j_0-1}\partial _t^{j_0-i}\partial _z^{2i}f|_{z=0^+} \frac{z^{2j_0}}{(2j_0)!}\chi _1(z)\\&\,-\sum _{j=1}^{j_0}\sum _{i=0}^{j-1}\partial _t^{j-1-i}\partial _z^{2i}f|_{z=0^+}\frac{z^{2j-1}}{(2j-1)!}(2\chi _1'(z)+\frac{z}{2j}\chi _1''(z)), \\ \alpha _0(z):\,= & {} \,\chi _1''(z),\\ \alpha _j(z):\,= & {} \,\frac{z^{2j-1}}{(2j-1)!}(2\chi _1'(z)+\frac{z}{2j}\chi _1''(z)), \quad \text { for }1\le j\le j_0,\\ \alpha _{j_0+1}(z):\,= & {} \,-\frac{z^{2j_0}}{(2j_0)!}\chi _1(z). \end{aligned}$$

Next we odd extend ${\tilde{f}},$ $\chi _1(z)$ and $\chi _1''(z),$ and even extend $\chi _1'(z)$ with respect to z to get a odd extension of F.

When $j=1$, the condition (3.10) becomes

$$\begin{aligned} 0\,= & {} \,\partial _{\zeta }\left( \int _0^{\infty }s_n(t\zeta ^2){\hat{F}}(t,\zeta )\,dt\right) |_{\zeta =0}\nonumber \\= & {} \,\int _0^{\infty }\partial _{\zeta }{\hat{F}}(t,0)\,dt \nonumber \\= & {} \,\int _0^{\infty }\partial _{\zeta }\hat{{\tilde{f}}}(t,0)\,dt+ \left( \sum _{i=0}^{j_0+1} \int _0^{\infty }\partial _t^{i}h(t)\,dt\right) \, \partial _{\zeta }{\widehat{\alpha }}_i(0). \end{aligned}$$

(3.12)

Since $h\in C_0^{\infty }((0,T))$, we have that, for $i\ge 1$,

$$\begin{aligned} \int _0^{\infty }\partial _t^{i}h(t)dt=0. \end{aligned}$$

On the other hand,

$$\begin{aligned} \partial _{\zeta }{\widehat{\alpha }}_0(0)=-\text {i}\int _{-\infty }^{\infty }z\alpha _0(z)\,dz =-2\text {i}\int _0^{\infty }z\chi _1''(z)\,dz=-2\text {i}, \end{aligned}$$

(3.13)

where we denote $\text {i}=\sqrt{-1}$. Thus (3.12) is equivalent to

$$\begin{aligned} \int _0^{\infty }h(t)\,dt=c_0:=\frac{1}{2\text {i}}\int _0^{\infty }\partial _{\zeta }\hat{{\tilde{f}}}(t,0)dt. \end{aligned}$$

(3.14)

When $j=3$, the condition (3.10) becomes

$$\begin{aligned} 0\,= & {} \,\partial _{\zeta }^3\left( \int _0^{\infty }s_n(t\zeta ^2){\hat{F}}(t,\zeta )\,dt\right) |_{\zeta =0}\nonumber \\= & {} \, \int _0^{\infty }(6t\partial _{\zeta }\hat{{\tilde{f}}}(t,0)+\partial _{\zeta }^3\hat{{\tilde{f}}}(t,0))dt\nonumber \\&\,+\int _0^{\infty }\sum _{i=0}^{j_0+1}(6t\partial _t^ih(t)\partial _{\zeta }{\widehat{\alpha }}_i(0) +\partial _t^ih(t)\partial _{\zeta }^3{\widehat{\alpha }}_i(0))dt. \end{aligned}$$

(3.15)

Observe that

$$\begin{aligned} \int _0^{\infty }t\partial _th(t)dt=-\int _0^{\infty }h(t)dt=-c_0, \end{aligned}$$

and for $i\ge 2,$

$$\begin{aligned} \int _0^{\infty }t\partial _t^ih(t)dt=-\int _0^{\infty }\partial _t^{i-1}h(t)dt=0. \end{aligned}$$

We find that (3.15) is equivalent to

$$\begin{aligned} \int _0^{\infty }th(t)dt=c_1, \end{aligned}$$

(3.16)

where

$$\begin{aligned} c_1:=\frac{1}{12\text {i}}\int _0^{\infty }(6t\partial _{\zeta }\hat{{\tilde{f}}}(t,0)+\partial _{\zeta }^3\hat{{\tilde{f}}}(t,0))dt -\frac{1}{2\text {i}}c_0\partial _{\zeta }\hat{\alpha _1}(0)+\frac{1}{12\text {i}}c_0\partial _{\zeta }^3{\widehat{\alpha }}_0(0). \end{aligned}$$

(3.17)

Generally, we can find some constants $c_i, 0\le i\le n,$ such that when $j=2i+1,$ (3.10) is equivalent to

$$\begin{aligned} \int _0^{\infty }t^ih(t)dt=c_i. \end{aligned}$$

(3.18)

We can prove it by induction. We already know it holds for $i=0,1$, assume $2\le i \le n$ and assume that (3.18) holds for $0\le l<i$, then when $j=2i+1,$ the condition (3.10) becomes

$$\begin{aligned} 0\,= & {} \,\partial _{\zeta }^{2i+1}\left( \int _0^{\infty }s_n(t\zeta ^2){\hat{F}}(t,\zeta )\,dt\right) |_{\zeta =0} \nonumber \\= & {} \,\sum _{l=0}^i\int _0^{\infty }C_{2i+1}^{2l}\frac{(2l)!}{l!}t^l\partial _{\zeta }^{2i+1-2l}\hat{{\tilde{f}}}(t,0)dt \nonumber \\&\,+\sum _{l=0}^i\int _0^{\infty } C_{2i+1}^{2l}\frac{(2l)!}{l!}t^{l}\sum _{\nu =0}^{j_0+1}\partial _t^{\nu }h(t)\partial _{\zeta }^{2i+1-2l}\hat{\alpha _{\nu }}(0)dt. \end{aligned}$$

(3.19)

Note that, for $l,\nu \in {\mathbb {N}},$

$$\begin{aligned} \int _0^{\infty }t^{l}\partial _t^{\nu }h(t) dt= {\left\{ \begin{array}{ll} 0,\qquad \text { if } l<\nu ,\\ (-1)^{\nu }\frac{l !}{(l-\nu )!}\int _0^{\infty }t^{l-\nu }h(t)dt, \qquad \text { if }l\ge \nu . \end{array}\right. } \end{aligned}$$

(3.20)

Using (3.20) and the induction assumption that (3.18) holds for $0\le l<i$, (3.19) is equivalent to (3.18) with

$$\begin{aligned} c_i:\,= & {} \,\frac{1}{2\text {i}}\cdot \frac{i!}{(2i+1)!}\partial _{\zeta }^{2i+1}\left( \int _0^{\infty }s_n(t\zeta ^2)\hat{{\tilde{f}}}(t,\zeta )\,dt\right) |_{\zeta =0} \nonumber \\&\,+\frac{1}{2\text {i}}\cdot \frac{i!}{(2i+1)!}\sum _{l=0}^{i-1}\sum _{\nu =0}^{l}(-1)^{\nu }C_{2i+1}^{2l}\frac{(2l)!}{(l-\nu )!} c_{l-\nu }\partial _{\zeta }^{2i+1-2l}\hat{\alpha _{\nu }}(0)\nonumber \\&\,+\frac{1}{2\text {i}}\sum _{\nu =1}^{i}(-1)^{\nu }\frac{i!}{(i-\nu )!}c_{i-\nu }\partial _{\zeta }\hat{\alpha _{\nu }}(0). \end{aligned}$$

(3.21)

Thus the condition (3.10) is equivalent to (3.18) holds for all $0\le i\le n.$ Given function f, we can define constants $c_i$ as above, the rest of our task is therefore to find a nonzero function $h\in C_0^{\infty }(0,T)$, supported in $(0,\frac{T}{3}]\cup [\frac{2T}{3},T)$, such that (3.18) holds for all $0\le i\le n.$

We introduce a nonnegative cut-off function $\chi _2\in C_{0}^{\infty }(0,T)$ such that $\chi _2(t)=0$ for $t\in [\frac{T}{3},\frac{2T}{3}]$ and $\chi _2(t)=1$ for $t\in [\frac{T}{9},\frac{2T}{9}]\cup [\frac{7T}{9},\frac{8T}{9}]$, we seek a function h of form

$$\begin{aligned} h(t)=\sum _{i=0}^{n+1}\beta _it^i\chi _2(t), \end{aligned}$$

(3.22)

where $\beta _i$, $0\le i\le n+1,$ are constants to be determined. Then (3.18) holds for $0\le i\le n$ is equivalent to

$$\begin{aligned} {\mathbf {B}}\mathbf {\beta }={\mathbf {c}}, \end{aligned}$$

(3.23)

where $\mathbf {\beta }=(\beta _0,\dots ,\beta _{n+1})^{Tr}, {\mathbf {c}}=(c_0,\dots ,c_n)^{Tr}$ and

$$\begin{aligned} {\mathbf {B}}= \begin{pmatrix} \int _{0}^{\infty }\chi _2(t)dt &{} \int _{0}^{\infty }t\chi _2(t)dt &{} \cdots &{} \int _{0}^{\infty }t^{n+1}\chi _2(t)dt \\ \int _{0}^{\infty }t\chi _2(t)dt &{} \int _{0}^{\infty }t^2\chi _2(t)dt &{} \cdots &{} \int _{0}^{\infty }t^{n+2}\chi _2(t)dt\\ \cdots &{} \cdots &{} \cdots &{} \cdots \\ \int _{0}^{\infty }t^n\chi _2(t)dt&{} \int _{0}^{\infty }t^{n+1}\chi _2(t)dt &{} \cdots &{} \int _{0}^{\infty }t^{2n+1}\chi _2(t)dt \end{pmatrix}. \end{aligned}$$

(3.24)

We denote $\mathbf {B'}$ the sub-matrix of ${\mathbf {B}}$ made of the first n row, $\mathbf {b'}$ the vector of $(n+1)-th$ row of ${\mathbf {B}}$. It is obvious that $\mathbf {b'}$ is a nonzero vector. In order to find a nonzero vector $\mathbf {\beta }$ such that (3.23), we first show that $\mathbf {B'}$ is a nondegenerate matrix. Otherwise, there exists a vector ${\mathbf {y}}=(y_0,\dots ,y_n)^{Tr}\ne 0$ such that

$$\begin{aligned} {\mathbf {y}}^{Tr}\mathbf {B'}{\mathbf {y}}=\int _0^{\infty }|y_0+y_1t+\cdots +y_nt^n|^2\chi _2(t)dt=0. \end{aligned}$$

(3.25)

Hence $y_0+y_1t+\cdots +y_nt^n\equiv 0$ for $t\in [\frac{T}{9},\frac{2T}{9}]\cup [\frac{7T}{9},\frac{8T}{9}],$ this contradict to ${\mathbf {y}}\ne 0$. So $\mathbf {B'}$ is a nondegenerate matrix. If ${\mathbf {c}}\ne 0,$ we can take $(\beta _0,\dots ,\beta _n)^{Tr}=\mathbf {B'}^{-1}{\mathbf {c}}$, $\beta _{n+1}=0$. If ${\mathbf {c}}=0,$ we can take $\beta _{n+1}=1,$ $(\beta _0,\dots ,\beta _n)^{Tr}=-\mathbf {B'}^{-1}\mathbf {b'}$. Either way, we have found a nonzero vector $\beta$ such that (3.23) holds true. This means we have hound a nonzero function $h\in C_0^{\infty }(0,T)$ supported in $(0,\frac{T}{3}]\cup [\frac{2T}{3},T)$ such that (3.18) holds for $0\le i\le n$. This concludes our proof.

$\square$

4 Proof of Lemma 2.5

This section is devoted to the proof of Lemma  2.5.

4.1 Equation Satisfied by the Remainder

By using the equation for $V^i$, for $0\le i\le 2$, see the systems (2.14)–(2.16), the Eq. (2.7) and the composition rule (2.12), we obtain that the remainder $r^{\varepsilon } := r^{\varepsilon }(t,x_1,x_2)$ satisfies

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _t r^{\varepsilon }-\varepsilon \Delta _{\mathrm{h}} r^{\varepsilon }=q^{\varepsilon } \qquad t\ge 0,(x_1,x_2)\in D_1,\\ r^{\varepsilon }(t,x_1,x_2)=0\qquad t\ge 0,(x_1,x_2)\in \partial D_1,\\ r^{\varepsilon }(0,x_1,x_2)=0\qquad (x_1,x_2)\in D_1, \end{array}\right. }\nonumber \\ \end{aligned}$$

(4.1)

where

$$\begin{aligned} \Delta _{\mathrm{h}}:=\partial _{x_1}^2+\partial _{x_2}^2, \end{aligned}$$

(4.2)

and

$$\begin{aligned} q^{\varepsilon }:\,= & {} \,-(\chi ''+\frac{\chi '}{r})\{V^0+\sqrt{\varepsilon }V^1+\varepsilon V^2\}_{\varepsilon }+\frac{2\chi '}{1-r}\{z\partial _z V^0+\sqrt{\varepsilon }z\partial _z V^1+\varepsilon z\partial _z V^2\}_{\varepsilon }\nonumber \\&\,+\sqrt{\varepsilon }\chi \alpha (r)\{z^2\partial _z V^0\}_{\varepsilon }+\sqrt{\varepsilon }\chi \{z\partial _z V^1\}_{\varepsilon }+\varepsilon \chi \alpha (r)\{z^2\partial _z V^1\}_{\varepsilon }+\frac{\varepsilon \chi }{r}\{\partial _z V^2\}_{\varepsilon }. \end{aligned}$$

(4.3)

Moreover, since $V^i$, for $0\le i\le 2,$ are supported way from $t=0$ as a function of time t, so is the forcing term $q^{\varepsilon }$.

4.2 Estimate Satisfied by the Forcing Term of the Remainder’s Equation

In this subsection we prove that $q^{\varepsilon }$ satisfies, for a constant C,

$$\begin{aligned} \Vert q^{\varepsilon }\Vert _{(L^1\cap L^2)({\mathbb {R}}_+\times D_1)}\le C\varepsilon ^{\frac{3}{4}}. \end{aligned}$$

(4.4)

By definition of the cut-off function $\chi$, see Sect. 2.3, there exists a constant C such that

$$\begin{aligned} |(\chi ''+\frac{\chi '}{r})\frac{1}{1-r}|\le C,\qquad |\frac{\chi '}{(1-r)^2}|\le C. \end{aligned}$$

(4.5)

Thus by virtue of (2.21), we find

$$\begin{aligned} \Vert (\chi ''+\frac{\chi '}{r})\{V^0\}_{\varepsilon }\Vert _{(L^1\cap L^2)({\mathbb {R}}_+;L^2( D_1))}\,= & {} \,\Vert (\chi ''+\frac{\chi '}{r})\frac{\sqrt{\varepsilon }}{1-r}\{zV^0\}_{\varepsilon }\Vert _{(L^1\cap L^2)({\mathbb {R}}_+;L^2( D_1))}\nonumber \\\le & {} \,C\varepsilon ^{\frac{3}{4}}\Vert zV^0\Vert _{(L^1\cap L^2)({\mathbb {R}}_+;L^2({\mathbb {R}}_+))} \nonumber \\\le & {} \,C\varepsilon ^{\frac{3}{4}}\Vert V^0\Vert _{C^0_2({\mathbb {R}}_+;H^0_1({\mathbb {R}}_+))}. \end{aligned}$$

(4.6)

Similarly, one has

$$\begin{aligned}&\Vert (\chi ''+\frac{\chi '}{r})\{\sqrt{\varepsilon }V^1+\varepsilon V^2\}_{\varepsilon }\Vert _{(L^1\cap L^2)({\mathbb {R}}_+;L^2( D_1))}\nonumber \\&\le C\varepsilon ^{\frac{3}{4}}\big (\Vert V^1\Vert _{C^0_2({\mathbb {R}}_+;H^0_0({\mathbb {R}}_+))}+\sqrt{\varepsilon } \Vert V^2\Vert _{C^0_2({\mathbb {R}}_+;H^0_0({\mathbb {R}}_+))}\big ), \end{aligned}$$

(4.7)

and

$$\begin{aligned}&\Vert \frac{2\chi '}{1-r}\{z\partial _z V^0+\sqrt{\varepsilon }z\partial _z V^1+\varepsilon z\partial _zV^2\}_{\varepsilon }\Vert _{(L^1\cap L^2)({\mathbb {R}}_+;L^2( D_1))}\nonumber \\&\le C\varepsilon ^{\frac{3}{4}}\big (\Vert V^0\Vert _{C^0_2({\mathbb {R}}_+;H^1_2({\mathbb {R}}_+))}+ \Vert V^1\Vert _{C^0_2({\mathbb {R}}_+;H^1_1({\mathbb {R}}_+))}+\sqrt{\varepsilon }\Vert V^2\Vert _{C^0_2({\mathbb {R}}_+;H^1_1({\mathbb {R}}_+))}\big ), \end{aligned}$$

(4.8)

and

$$\begin{aligned} \Vert \sqrt{\varepsilon }\chi \{z\partial _z V^1\}_{\varepsilon }\Vert _{(L^1\cap L^2)({\mathbb {R}}_+;L^2( D_1))}\le C\varepsilon ^{\frac{3}{4}}\Vert V^1\Vert _{C^0_2({\mathbb {R}}_+;H^1_1({\mathbb {R}}_+))}, \end{aligned}$$

(4.9)

and

$$\begin{aligned} \Vert \frac{\varepsilon \chi }{r}\{\partial _z V^2\}_{\varepsilon }\Vert _{(L^1\cap L^2)({\mathbb {R}}_+;L^2( D_1))}\le C\varepsilon ^{\frac{5}{4}}\Vert V^2\Vert _{C^0_2({\mathbb {R}}_+;H^1_0({\mathbb {R}}_+))}. \end{aligned}$$

(4.10)

Since $\alpha (r)\in C^{\infty }([\frac{1}{3},1])$, recalling that the definition is in (2.11),

$$\begin{aligned}&\Vert \sqrt{\varepsilon }\chi \alpha (r)\{z^2\partial _z V^0\}_{\varepsilon }\Vert _{(L^1\cap L^2)({\mathbb {R}}_+;L^2( D_1))}\le C\varepsilon ^{\frac{3}{4}}\Vert V^0\Vert _{C^0_2({\mathbb {R}}_+;H^1_2({\mathbb {R}}_+))}, \end{aligned}$$

(4.11)

$$\begin{aligned}&\Vert \varepsilon \chi \alpha (r)\{z^2\partial _z V^1\}_{\varepsilon }\Vert _{(L^1\cap L^2)({\mathbb {R}}_+;L^2( D_1))}\le C\varepsilon ^{\frac{5}{4}}\Vert V^1\Vert _{C^0_2({\mathbb {R}}_+;H^1_2({\mathbb {R}}_+))}. \end{aligned}$$

(4.12)

By summarizing the above inequalities, we conclude the proof of (4.4).

4.3 Parabolic Estimates

Now we are ready to prove (2.23) and (2.24). By using energy method, we find that

$$\begin{aligned} \frac{1}{2}\partial _t\Vert r^{\varepsilon }\Vert ^2_{L^2(D_1)}+\varepsilon \Vert \nabla _{\mathrm{h}} r^{\varepsilon }\Vert _{L^2(D_1)}^2\le \Vert q^{\varepsilon }\Vert _{L^2(D_1)}\Vert r^{\varepsilon }\Vert _{L^2(D_1)}, \end{aligned}$$

(4.13)

where $\nabla _\mathrm{h}=(\partial _{x_1},\partial _{x_2}).$ Thanks to (4.4),

$$\begin{aligned} \Vert r^{\varepsilon }\Vert _{L^{\infty }({\mathbb {R}}_+;L^2(D_1))}\le \Vert q^{\varepsilon }\Vert _{L^1({\mathbb {R}}_+;L^2(D_1))}\le C \varepsilon ^{\frac{3}{4}}. \end{aligned}$$

(4.14)

While by taking $L^2(D_1)$ inner product of (4.1) with $\partial _t r^\varepsilon ,$ we obtain

$$\begin{aligned} \Vert \partial _t r^\varepsilon (t)\Vert _{L^2(D_1)}^2+\frac{\varepsilon }{2}\frac{d}{dt}\Vert \nabla _{h} r^\varepsilon (t)\Vert _{L^2(D_1)}^2 \le&\Vert q^\varepsilon \Vert _{L^2(D_1)}\Vert \partial _t r^\varepsilon \Vert _{L^2(D_1)}\\ \le&{\frac{1}{2}}\Vert q^\varepsilon \Vert _{L^2(D_1)}^2+{\frac{1}{2}}\Vert \partial _t r^\varepsilon \Vert _{L^2(D_1)}^2, \end{aligned}$$

from which and (4.4), we infer

$$\begin{aligned} \varepsilon \Vert \nabla _\mathrm{h}r^\varepsilon \Vert _{L^\infty ({\mathbb {R}}_+; L^2(D_1))}^2+ \Vert \partial _t{r}^{\varepsilon }\Vert _{L^2({\mathbb {R}}_+\times D_1)}^2 \le \Vert {q}^{\varepsilon }\Vert _{L^2({\mathbb {R}}_+\times D_1)}^2\le C \varepsilon ^{\frac{3}{2}}. \end{aligned}$$

(4.15)

Thanks to (4.15), we deduce from the $r^\varepsilon$ equation of (4.1) that

$$\begin{aligned} \varepsilon \Vert \Delta _{\mathrm{h}} r^\varepsilon \Vert _{L^2({\mathbb {R}}_+\times D_1)}\le \Vert \partial _t{r}^{\varepsilon }\Vert _{L^2({\mathbb {R}}_+\times D_1)}+\Vert {q}^{\varepsilon }\Vert _{L^2({\mathbb {R}}_+\times D_1)} \le C \varepsilon ^{\frac{3}{4}}, \end{aligned}$$

which together with the homogeneous boundary condition of $r^\varepsilon$ on $\partial D_1$ ensures that

$$\begin{aligned} \Vert \nabla _\mathrm{h}^2 r^{\varepsilon }\Vert _{L^2({\mathbb {R}}_+\times D_1)}\le C \varepsilon ^{-\frac{1}{4}} \end{aligned}$$

(4.16)

Since $r^{\varepsilon }$ vanishes on the boundary $\partial D_1$ and thanks to Poincaré inequality and interpolation inequality,

$$\begin{aligned} \Vert r^{\varepsilon }\Vert _{L^2({\mathbb {R}}_+\times D_1)}\le C\Vert \nabla _\mathrm{h}r^{\varepsilon }\Vert _{L^2({\mathbb {R}}_+\times D_1)}\le C\Vert r^{\varepsilon }\Vert _{L^2({\mathbb {R}}_+\times D_1)}^{\frac{1}{2}}\Vert \nabla _{\mathrm{h}}^2 r^{\varepsilon }\Vert _{L^2({\mathbb {R}}_+\times D_1)}^{\frac{1}{2}}. \end{aligned}$$

(4.17)

Therefore,

$$\begin{aligned} \Vert r^{\varepsilon }\Vert _{L^2({\mathbb {R}}_+\times D_1)}\le C\Vert \nabla _{\mathrm{h}}^2 r^{\varepsilon }\Vert _{L^2({\mathbb {R}}_+\times D_1)}\le C\varepsilon ^{-\frac{1}{4}}. \end{aligned}$$

(4.18)

By summarizing the estimates (4.14), (4.16), and (4.17) and (4.18), we finish the proof of Lemma 2.5.

5 Proof of Proposition 2.6

The section is devoted to the proof of Proposition 2.6. As already explained, despite we only desire to obtain a $L^2$ estimate of the remainder term $R^{\varepsilon }$, the singular feature of the problem satisfied by $R^{\varepsilon }$, due to the large variations in the boundary layer, combined with the nonlinearity of the Navier–Stokes system, leads us to consider analytic estimates in the spirit of Cauchy–Kowaleskaya estimates. As we need a nonlinear long-time version of such Cauchy–Kowaleskaya estimates, we follow the method initiated by Chemin in [3], see also [4, 19], which makes use of Fourier theory and Besov spaces. We introduce, for $s=0$ and $s=\frac{1}{2}$, the Besov spaces ${\dot{B}}^s$ respectively endowed with the norm

$$\begin{aligned} \Vert a\Vert _{{\dot{B}}^s}:=\sum _{k\in {\mathbb {Z}}}2^{ks}\Vert {\dot{\Delta }}_k a\Vert _{L^2(\mathcal {C})} \end{aligned}$$

(5.1)

where the dyadic operator ${{\dot{\Delta }}}_k$ is defined by

where ${\mathcal {F}}^{-1}_{\xi \rightarrow x_3}a$ denotes the inverse Fourier transform of the distribution a with respect to the third variable, and ${\widehat{a}}(x_1,x_2,\xi )={\mathcal {F}}_{x_3\rightarrow \xi }(a)(x_1,x_2,\xi ).$ One may check more details on Littlewood-Paley theory from [1].

For any locally bounded function $\Phi$ on ${\mathbb {R}}^+\times {\mathbb {R}}$, we define

$$\begin{aligned} v_{\Phi }:= {\mathcal {F}}^{-1}_{\xi \rightarrow x_3}(e^{\Phi (t,\xi )}{\hat{v}}(t,x_1,x_2,\xi )). \end{aligned}$$

(5.2)

Then it follows from (2.41) that, for any function $\Phi$ of the form:

$$\begin{aligned} \Phi (t,\xi ):=\rho (t)|\xi |-\beta (t), \end{aligned}$$

(5.3)

the vector field $R^{\varepsilon }_{\Phi }$ satisfies the Navier–Stokes type system:

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _t R^{\varepsilon }_{\Phi }-{\dot{\rho }}|\partial _3|R^{\varepsilon }_{\Phi }+{\dot{\beta }}R^{\varepsilon }_{\Phi }+(u^{\varepsilon }\cdot \nabla R^{\varepsilon }+R^{\varepsilon }\cdot \nabla (u^{\text {aux}} + \varepsilon u^{\text {fl}}))_{\Phi }\\ \qquad -\varepsilon \Delta R^{\varepsilon }_{\Phi }+\nabla \pi ^{\varepsilon }_{\Phi }=F^{\varepsilon }_{\Phi },\\ {\mathrm {div}}\,R^{\varepsilon }_{\Phi }=0,\\ R^{\varepsilon }_{\Phi }|_{\partial \mathcal {C}}=0,\\ R^{\varepsilon }_{\Phi }|_{t=0}=0. \end{array}\right. } \end{aligned}$$

(5.4)

We recall that the source term $F^{\varepsilon }$ is defined in (2.42). Such a form of the exponential Fourier multiplier make appear two possible gains through the second and third terms in the first equation above. The purpose of these gains is to help to deal with the singular part of the convective term, that is the fourth term of the same equation. The choices of $\beta (t)$ and $\rho (t)$ are therefore crucial. We first define $\beta (t)$, by

$$\begin{aligned} {\left\{ \begin{array}{ll} {\dot{\beta }}(t)=&{}C_*\chi _{[0,T]}(t)+C_{*}(\varepsilon \Vert \nabla _{\mathrm{h}} e^{\rho _0|\partial _3|}u^{\text {fl}}\Vert ^2_{{\dot{B}}^{\frac{1}{2}}}+\Vert h^1-\chi \{V^1\}_{\varepsilon }\Vert ^2_{L^{\infty }(D_1)}\\ &{}+\varepsilon \Vert h^2-\chi \{V^2\}_{\varepsilon }\Vert ^2_{L^{\infty }(D_1)}+\Vert V^0\Vert _{L_z^{\infty }}+\Vert z\partial _z V^0\Vert _{L_z^{\infty }}+\varepsilon \Vert r^{\varepsilon }\Vert _{L^{\infty }(D_1)}), \\ \beta (0)=&{}0, \end{array}\right. } \end{aligned}$$

(5.5)

where $C_*$ is a constant which will be determined later.

Proposition 5.1

If $u_b$ satisfies (2.43) for a constant $C_0>0$ and $\rho _b>\rho _0$, there exists $\beta _*>0$ such that

$$\begin{aligned} \sup _{t\in [0,T/{\varepsilon }]}\beta (t)=\beta (T/{\varepsilon })\le \beta _*. \end{aligned}$$

(5.6)

Proof

Since $h^1$ is compactly supported in (0, T), $\chi$ is a cut-off function which satisfies $0\le \chi \le 1,$

$$\begin{aligned} \int _0^{\frac{T}{\varepsilon }}\Vert h^1-\chi \{V^1\}_{\varepsilon }\Vert _{L^{\infty }(D_1)}^2\,dt\le 2T\Vert h^1\Vert _{L^{\infty }}^2+2\int _0^{\frac{T}{\varepsilon }} \Vert V^1\Vert _{L_z^{\infty }}^2\,dt. \end{aligned}$$

By construction, $V^1\in C_{7}^{0}({\mathbb {R}}_+;H^{7}_{8}({\mathbb {R}}_+))$, and recall Definition 2.3, we find that

$$\begin{aligned} \int _0^{\frac{T}{\varepsilon }}\Vert h^1-\chi \{V^1\}_{\varepsilon }\Vert _{L^{\infty }(D_1)}^2\,dt\le 2T\Vert h^1\Vert ^2_{L^{\infty }}+C\Vert V^1\Vert ^2_{C^0_1({\mathbb {R}}_+;H^1({\mathbb {R}}_+))}, \end{aligned}$$

for a constant C. Similarly

$$\begin{aligned} \int _0^{\frac{T}{\varepsilon }}\Vert h^2-\chi \{V^2\}_{\varepsilon }\Vert _{L^{\infty }(D_1)}^2\,dt\le 2T\Vert h^2\Vert ^2_{L^{\infty }}+C\Vert V^2\Vert ^2_{C^0_1({\mathbb {R}}_+;H^1({\mathbb {R}}_+))}. \end{aligned}$$

and

$$\begin{aligned} \int _0^{\frac{T}{\varepsilon }}(\Vert V^0\Vert _{L^{\infty }_z}+\Vert z\partial _z V^0\Vert _{L^{\infty }_z})\,dt\le C\Vert V^0\Vert _{C^0_2({\mathbb {R}}_+;H^2_1({\mathbb {R}}_+))}. \end{aligned}$$

By Proposition 2.5 and Sobolev imbedding inequality,

$$\begin{aligned} \varepsilon \int _0^{\frac{T}{\varepsilon }}\Vert r^{\varepsilon }\Vert _{L^{\infty }(D_1)}\,dt\,\le & {} \, C\varepsilon \int _0^{\frac{T}{\varepsilon }} \Vert r^{\varepsilon }\Vert _{H^2(D_1)}\,dt\nonumber \\\le & {} \, C\varepsilon \Vert r^{\varepsilon }\Vert ^{\frac{1}{2}}_{L^2({\mathbb {R}}_+;H^2(D_1))}\varepsilon ^{-\frac{1}{2}}T^{\frac{1}{2}}\nonumber \\\le & {} \,C\varepsilon ^{\frac{3}{8}}. \end{aligned}$$

It remains to estimate $\int _0^{\frac{T}{\varepsilon }}\varepsilon \Vert \nabla _{\mathrm{h}} e^{\rho _0 |\partial _3|}u^{\text {fl}}\Vert ^2_{{\dot{B}}^{\frac{1}{2}}}\,dt$. Since $u^{\text {fl}}$ is supported in [0, T] as a function of t we have that

$$\begin{aligned} \int _0^{\frac{T}{\varepsilon }}\varepsilon \Vert \nabla _{\mathrm{h}} e^{\rho _0 |\partial _3|}u^{\text {fl}}\Vert ^2_{{\dot{B}}^{\frac{1}{2}}}\,dt\,\le & {} \, T\Vert \nabla _{\mathrm{h}} e^{\rho _0|\partial _3|}u_b\Vert ^2_{{\dot{B}}^{\frac{1}{2}}}\nonumber \\\le & {} \,T\left( \sum _{k\in \mathbb {Z}} 2^{\frac{k}{2}}\Vert {\dot{\Delta }}_k\nabla _{\mathrm{h}} e^{\rho _0|\partial _3|} u_b\Vert _{L^2(\mathcal {C})}\right) ^2\nonumber \\\le & {} \, C T\big (\Vert \nabla _{\mathrm{h}} e^{\rho _0|\partial _3|} u_b\Vert _{L^2(\mathcal {C})}+\Vert \partial _3\nabla _{\mathrm{h}} e^{\rho _0|\partial _3|} u_b\Vert _{L^2(\mathcal {C})}\big )^2\nonumber \\\le & {} \, C T\Vert e^{\rho _0|\partial _3|}u_b\Vert ^2_{H^2(\mathcal {C})}. \end{aligned}$$

Since $u_b$ satisfies (2.43) for $\rho _b>\rho$,

$$\begin{aligned} \int _0^{\frac{T}{\varepsilon }}\varepsilon \Vert \nabla _{\mathrm{h}} e^{\rho _0 |\partial _3|}u^{\text {fl}}\Vert ^2_{{\dot{B}}^{\frac{1}{2}}}\,dt\le C TC_b^2\left( \frac{\rho _b}{\rho _b-\rho _0}\right) ^2. \end{aligned}$$

Eventually, gathering the above inequality concludes the proof of Proposition 5.1. $\square$

Remark 5.2

From now on, for simplification, we shall denote the norm $\Vert \cdot \Vert _{L^2(\mathcal {C})}$ by $\Vert \cdot \Vert$ if there is no ambiguity.

Now we are in a position to complete the proof of Proposition 2.6.

Proof of Proposition 2.6

We define $\rho (t)$ as the solution of the nonlinear ODE:

$$\begin{aligned} {\left\{ \begin{array}{ll} {\dot{\rho }}(t)=-C_*(\varepsilon \Vert \nabla _{\mathrm{h}} R^{\varepsilon }_{\Phi }\Vert ^2_{{\dot{B}}^0}+\Vert z\partial _z V^0\Vert _{L^{\infty }_z}),\qquad t>0,\\ \rho (0)=\rho _0, \end{array}\right. } \end{aligned}$$

(5.7)

where

$$\begin{aligned} \rho _0:=2+C_*\int _0^{\infty }\Vert z\partial _z V^0(t)\Vert _{L_z^{\infty }} dt. \end{aligned}$$

(5.8)

and $C_*$ is a constant which will be determined later. We set

$$\begin{aligned} T^*:=\sup \{t\in [0,\frac{T}{{\varepsilon }}]:\rho (t)\ge 1\}. \end{aligned}$$

(5.9)

We apply the operator ${\dot{\Delta }}_k$ to (5.4) and we use energy estimates to find that

$$\begin{aligned}&\frac{1}{2}\frac{d}{dt}\Vert {\dot{\Delta }}_k R^{\varepsilon }_{\Phi }\Vert ^2+|{\dot{\rho }}|(|\partial _3|{\dot{\Delta }}_k R^{\varepsilon }_{\Phi },{\dot{\Delta }}_k R^{\varepsilon }_{\Phi })+{\dot{\beta }}\Vert {\dot{\Delta }}_k R^{\varepsilon }_{\Phi }\Vert ^2+\varepsilon \Vert \nabla {\dot{\Delta }}_k R^{\varepsilon }_{\Phi }\Vert ^2\nonumber \\&\qquad +({\dot{\Delta }}_k(u^{\varepsilon }\cdot \nabla R^{\varepsilon }+R^{\varepsilon }\cdot \nabla (u^{\text {aux}} + \varepsilon u^{\text {fl}}))_{\Phi },{\dot{\Delta }}_k R^{\varepsilon }_{\Phi })=({\dot{\Delta }}_k F^{\varepsilon }_{\Phi },{\dot{\Delta }}_k R^{\varepsilon }_{\Phi }), \end{aligned}$$

(5.10)

which implies

$$\begin{aligned}&\sum _{k\in \mathbb {Z}}\Vert {\dot{\Delta }}_k R^{\varepsilon }_{\Phi }(t)\Vert +\sum _{k\in \mathbb {Z}}\Big (\int _0^t 2^k|{\dot{\rho }}|\Vert {\dot{\Delta }}_k R^{\varepsilon }_{\Phi }\Vert ^2\,ds\Big )^{\frac{1}{2}}\nonumber \\&\qquad +\sum _{k\in \mathbb {Z}}\Big (\int _0^t {\dot{\beta }}\Vert {\dot{\Delta }}_k R^{\varepsilon }_{\Phi }\Vert ^2\,ds\Big )^{\frac{1}{2}}+\sum _{k\in \mathbb {Z}} \Big (\int _0^t\varepsilon \Vert \nabla {\dot{\Delta }}_k R^{\varepsilon }_{\Phi }\Vert ^2\,ds\Big )^{\frac{1}{2}}\nonumber \\&\le C \sum _{k\in \mathbb {Z}}\Big (\int _0^t |({\dot{\Delta }}_k F^{\varepsilon }_{\Phi },{\dot{\Delta }}_k R^{\varepsilon }_{\Phi })|\,ds\Big )^{\frac{1}{2}}\nonumber \\&\qquad +C\sum _{k\in \mathbb {Z}}\Big (\int _0^t |({\dot{\Delta }}_k (u^{\varepsilon }\cdot \nabla R^{\varepsilon }+R^{\varepsilon }\cdot \nabla (u^{\text {aux}} + \varepsilon u^{\text {fl}}))_{\Phi },{\dot{\Delta }}_k R^{\varepsilon }_{\Phi })|\,ds\Big )^{\frac{1}{2}}. \end{aligned}$$

(5.11)

Since $F^{\varepsilon }$ is supported in [0, T], the first term on the right hand side can be bounded by

$$\begin{aligned} \sum _{k\in \mathbb {Z}}\left( \int _0^t\chi _{[0,T]}(s)\Vert {\dot{\Delta }}_kR^{\varepsilon }_{\Phi }(s)\Vert ^2ds\right) ^{\frac{1}{2}}+\sum _{k\in \mathbb {Z}}\left( \int _0^T\Vert {\dot{\Delta }}_k F^{\varepsilon }_{\Phi }(s)\Vert ^2ds\right) ^{\frac{1}{2}}. \end{aligned}$$

(5.12)

It remains to estimate the last term in (5.11). It is easy to observe from (2.10) and (2.25) that $u^{\text {aux}}$ is independent of $x_3$ variable. By using integration by parts and ${\mathrm {div}}\,u^{\text {aux}}=0$, we find that

$$\begin{aligned}&({\dot{\Delta }}_k (u^{\text {aux}}\cdot \nabla R^{\varepsilon })_{\Phi },{\dot{\Delta }}_k R^{\varepsilon }_{\Phi })= (u^{\text {aux}}\cdot \nabla {\dot{\Delta }}_kR^{\varepsilon }_{\Phi },{\dot{\Delta }}_k R^{\varepsilon }_{\Phi })=0. \end{aligned}$$

(5.13)

While due to ${\mathrm {div}}\,u^{\text {fl}}={\mathrm {div}}\,R^{\varepsilon }=0,$ one has

$$\begin{aligned} ({\dot{\Delta }}_k((\varepsilon u^{\text {fl}}+\varepsilon R^{\varepsilon })\cdot \nabla R^{\varepsilon })_{\Phi },{\dot{\Delta }}_k R^{\varepsilon }_{\Phi })= -({\dot{\Delta }}_k((\varepsilon u^{\text {fl}}+\varepsilon R^{\varepsilon })\otimes R^{\varepsilon })_{\Phi },\nabla {\dot{\Delta }}_k R^{\varepsilon }_{\Phi }). \end{aligned}$$

(5.14)

Next we use the following lemma: $\square$

Lemma 5.3

For any axi-symmetric functions a, b and c in $\mathcal {C}$, assume that function a vanishes on the boundary $\partial \mathcal {C}$, then for any constant $c_0>0,$ there exists $C>0$, such that

$$\begin{aligned}&\sum _{k\in \mathbb {Z}}\left( \int _0^t|({\dot{\Delta }}_k(ab)_{\Phi },{\dot{\Delta }}_k c_{\Phi }))|\,ds\right) ^{\frac{1}{2}}\le c_0\sum _{k\in \mathbb {Z}} \left( \int _0^t \Vert {\dot{\Delta }}_k c_{\Phi }\Vert ^2\,ds\right) ^{\frac{1}{2}}\\&\quad +C\sum _{k\in \mathbb {Z}} \left( \int _0^t \Vert \nabla _{\mathrm{h}}a_{\Phi }\Vert ^2_{{\dot{B}}^{\frac{1}{2}}}\Vert {\dot{\Delta }}_k b_{\Phi }\Vert ^2\,ds\right) ^{\frac{1}{2}}. \end{aligned}$$

When $a=b$ and both vanish on the boundary, we also have

$$\begin{aligned}&\sum _{k\in \mathbb {Z}}\left( \int _0^t|({\dot{\Delta }}_k(a^2)_{\Phi },{\dot{\Delta }}_k c_{\Phi }))|\,ds\right) ^{\frac{1}{2}}\le c_0\sum _{k\in \mathbb {Z}} \left( \int _0^t \Vert {\dot{\Delta }}_k c_{\Phi }\Vert ^2\,ds\right) ^{\frac{1}{2}}\\&\qquad +C\sum _{k\in \mathbb {Z}} \left( \int _0^t 2^k\Vert \nabla _{\mathrm{h}}a_{\Phi }\Vert ^2_{{\dot{B}}^{0}}\Vert {\dot{\Delta }}_k a_{\Phi }\Vert ^2\,ds\right) ^{\frac{1}{2}}. \end{aligned}$$

The proof of Lemma 5.3 is postponed to Sect. 7. Then by Lemma 5.3, for any $c_0>0,$ there exists a constant $C>0$, such that

$$\begin{aligned}&\displaystyle \sum _{k\in \mathbb {Z}}\left( \int _0^t| ({\dot{\Delta }}_k((\varepsilon u^{\text {fl}}+\varepsilon R^{\varepsilon })\cdot \nabla R^{\varepsilon })_{\Phi },{\dot{\Delta }}_k R^{\varepsilon }_{\Phi })|\,ds\right) ^{\frac{1}{2}}\le c_0\sum _{k\in \mathbb {Z}}\left( \int _0^t \varepsilon \Vert \nabla {\dot{\Delta }}_k R^{\varepsilon }_{\Phi }\Vert ^2\,ds\right) ^{\frac{1}{2}}\nonumber \\&\displaystyle +C\sum _{k\in \mathbb {Z}}\left( \int _0^t \varepsilon (\Vert \nabla _{\mathrm{h}} u^{\text {fl}}_{\Phi }\Vert ^2_{{\dot{B}}^{\frac{1}{2}}}+2^k\Vert \nabla _{\mathrm{h}} R^{\varepsilon }_{\Phi }\Vert ^2_{{\dot{B}}^0})\Vert {\dot{\Delta }}_k R^{\varepsilon }_{\Phi }\Vert ^2\,ds\right) ^{\frac{1}{2}} \end{aligned}$$

(5.15)

Again, by integration by parts, we find

$$\begin{aligned} ({\dot{\Delta }}_k(R^{\varepsilon }\cdot \nabla (u^{\text {aux}} + \varepsilon u^{\text {fl}}))_{\Phi },{\dot{\Delta }}_k R^{\varepsilon }_{\Phi })=-({\dot{\Delta }}_k(R^{\varepsilon }\otimes (u^{\text {aux}} + \varepsilon u^{\text {fl}}))_{\Phi },\nabla {\dot{\Delta }}_k R^{\varepsilon }_{\Phi }). \end{aligned}$$

By Lemma 5.3, for any $c_0>0,$ there exists a constant $C>0$, such that

$$\begin{aligned}&\sum _{k\in \mathbb {Z}}\left( \int _0^t\varepsilon |({\dot{\Delta }}_k(R^{\varepsilon }\otimes u^{\text {fl}})_{\Phi },\nabla {\dot{\Delta }}_k R^{\varepsilon })|\,ds\right) ^{\frac{1}{2}}\nonumber \\&\le c_0\sum _{k\in \mathbb {Z}}\left( \int _0^t\varepsilon \Vert \nabla {\dot{\Delta }}_k R^{\varepsilon }_{\Phi }\Vert ^2\,ds\right) ^{\frac{1}{2}} +C\sum _{k\in \mathbb {Z}}\left( \int _0^t\varepsilon \Vert \nabla _{\mathrm{h}} u^{\text {fl}}_{\Phi }\Vert ^2_{{\dot{B}}^{\frac{1}{2}}}\Vert {\dot{\Delta }}_k R^{\varepsilon }_{\Phi }\Vert ^2\,ds\right) ^{\frac{1}{2}}. \end{aligned}$$

(5.16)

and

$$\begin{aligned}&\sum _{k\in \mathbb {Z}}\left( \int _0^t|({\dot{\Delta }}_k(R^{\varepsilon }\otimes ( u^{\text {aux}}-(h^0-\chi \{V^0\}_{\varepsilon })e_3 ))_{\Phi },\nabla {\dot{\Delta }}_k R^{\varepsilon }_{\Phi })|\,ds\right) ^{\frac{1}{2}}\nonumber \\&\quad \le c_0\sum _{k\in \mathbb {Z}}\left( \int _0^t \varepsilon \Vert \nabla {\dot{\Delta }}_k R^{\varepsilon }_{\Phi }\Vert ^2\,ds\right) ^{\frac{1}{2}}\nonumber \\&\qquad +C\sum _{k\in \mathbb {Z}}\left( \int _0^t\Vert {\dot{\Delta }}_k R^{\varepsilon }_{\Phi }\Vert ^2(\Vert h^1\,-\,\chi \{V^1\}_{\varepsilon }\Vert ^2_{L^{\infty }(D_1)}\,\,+\,\varepsilon \Vert h^2\,-\,\chi \{V^2\}_{\varepsilon }\Vert ^2_{L^{\infty }(D_1)}\,\,+\,\varepsilon \Vert r^{\varepsilon }\Vert ^2_{L^{\infty }})\,ds\right) ^{\frac{1}{2}},\nonumber \\ \end{aligned}$$

(5.17)

where in the last step, we used (2.10) and (2.25).

It remains to estimate $({\dot{\Delta }}_k(R^{\varepsilon }\cdot \nabla (h^0-\chi \{V^0\}_{\varepsilon }) )_{\Phi },{\dot{\Delta }}_k R^{\varepsilon }_{3,\Phi })$. Observing that

$$\begin{aligned}&({\dot{\Delta }}_k(R^{\varepsilon }\cdot \nabla (h^0-\chi \{V^0\}_{\varepsilon }) )_{\Phi },{\dot{\Delta }}_k R^{\varepsilon }_{3,\Phi })\nonumber \\&\quad =-({\dot{\Delta }}_k R^{\varepsilon }_{r,\Phi }(\chi '\{V^0\}_{\varepsilon }+\frac{\chi }{1-r}\{z\partial _z V^0\}_{\varepsilon }),{\dot{\Delta }}_{k} R^{\varepsilon }_{3,\Phi }), \end{aligned}$$

which implies

$$\begin{aligned}&\big |({\dot{\Delta }}_k(R^{\varepsilon }\cdot \nabla (h^0-\chi \{V^0\}_{\varepsilon }) )_{\Phi },{\dot{\Delta }}_k R^{\varepsilon }_{3,\Phi })\big |\nonumber \\&\quad \le \Vert V^0\Vert _{L_{z}^{\infty }}\Vert {\dot{\Delta }}_k R^{\varepsilon }_{\Phi }\Vert ^2-({\dot{\Delta }}_k R^{\varepsilon }_{r,\Phi }\frac{\chi }{1-r}\{z\partial _z V^0\}_{\varepsilon },{\dot{\Delta }}_{k} R^{\varepsilon }_{3,\Phi }) \end{aligned}$$

(5.18)

Since $R^{\varepsilon }$ vanishes on the boundary of $\mathcal {C},$ we have

$$\begin{aligned} \frac{{\dot{\Delta }}_k R^{\varepsilon }_{r,\Phi }}{1-r}=-\int _0^1(\partial _r{\dot{\Delta }}_k R^{\varepsilon }_{r,\Phi })(t,1-(1-r)s,x_3)ds. \end{aligned}$$

(5.19)

By the divergence free condition ${\mathrm {div}}\,R^{\varepsilon }=0,$

$$\begin{aligned} \partial _r {\dot{\Delta }}_k R^{\varepsilon }_{r,\Phi }=-\frac{{\dot{\Delta }}_k R^{\varepsilon }_{r,\Phi }}{r}-\partial _3 {\dot{\Delta }}_k R^{\varepsilon }_{3,\Phi }. \end{aligned}$$

(5.20)

Note that $\chi (r)=0$ when $r\le \frac{1}{3}$, we find

$$\begin{aligned}&|({\dot{\Delta }}_k R^{\varepsilon }_{r,\Phi }\frac{\chi }{1-r}\{z\partial _z V^0\}_{\varepsilon },{\dot{\Delta }}_{k} R^{\varepsilon }_{3,\Phi })|\nonumber \\&\quad =\left| \int _{\mathcal {C}}\int _0^1(\partial _r{\dot{\Delta }}_k R^{\varepsilon }_{r,\Phi })(t,1-(1-r)s,x_3)\chi (r)\{z\partial _z V^0\}_{\varepsilon }(t,r){\dot{\Delta }}_k R^{\varepsilon }_{3,\Phi }(t,x)dsdx\right| \nonumber \\&\quad \le (3\Vert {\dot{\Delta }}_k R^{\varepsilon }_{r,\Phi }\Vert +\Vert \partial _3{\dot{\Delta }}_k R^{\varepsilon }_{3,\Phi }\Vert )\Vert z\partial _z V^0\Vert _{L_z^{\infty }}\Vert {\dot{\Delta }}_k R^{\varepsilon }_{3,\Phi }\Vert \nonumber \\&\quad \le C(1+2^k)\Vert z\partial _z V^0\Vert _{L_z^{\infty }}\Vert {\dot{\Delta }}_k R^{\varepsilon }_{\Phi }\Vert ^2, \end{aligned}$$

which implies

$$\begin{aligned}&\sum _{k\in \mathbb {Z}}\left( \int _0^t|({\dot{\Delta }}_k(R^{\varepsilon }\cdot \nabla (h^0-\chi \{V^0\}_{\varepsilon }) )_{\Phi },{\dot{\Delta }}_k R^{\varepsilon }_{3,\Phi })|\,ds\right) ^{\frac{1}{2}}\nonumber \\&\le C\sum _{k\in \mathbb {Z}}\left( \int _0^t(\Vert V^0\Vert _{L_z^{\infty }}+(1+2^k)\Vert z\partial _z V^0\Vert _{L_z^{\infty }})\Vert {\dot{\Delta }}_k R^{\varepsilon }_{\Phi }\Vert ^2\,ds\right) ^{\frac{1}{2}}. \end{aligned}$$

(5.21)

Finally, by summing up the estimates (5.13), (5.15)–(5.17) and (5.21), for any $c_0>0,$ there exists a constant $C>0$ such that

$$\begin{aligned}&\sum _{k\in \mathbb {Z}}\left( \int _0^t |({\dot{\Delta }}_k (u^{\varepsilon }\cdot \nabla R^{\varepsilon }+R^{\varepsilon }\cdot \nabla (u^{\text {aux}} + \varepsilon u^{\text {fl}}))_{\Phi },{\dot{\Delta }}_k R^{\varepsilon }_{\Phi })|\,ds\right) ^{\frac{1}{2}}\nonumber \\&\quad \le c_0\sum _{k\in \mathbb {Z}}\left( \int _0^t \varepsilon \Vert \nabla {\dot{\Delta }}_k R^{\varepsilon }_{\Phi }\Vert ^2\,ds\right) ^{\frac{1}{2}}\nonumber \\&\qquad +C\sum _{k\in \mathbb {Z}} \left( \int _0^t\left( \varepsilon 2^k\Vert \nabla _{\mathrm{h}} R^{\varepsilon }_{\Phi }\Vert ^2_{{\dot{B}}^0}+2^k\Vert z\partial _z V^0\Vert _{L_z^{\infty }}+{\mathcal {K}}_{\varepsilon }\right) \Vert {\dot{\Delta }}_k R^{\varepsilon }_{\Phi }\Vert ^2\,ds\right) ^{\frac{1}{2}}, \end{aligned}$$

(5.22)

where

$$\begin{aligned} {\mathcal {K}}_{\varepsilon }:\,= & {} \,\varepsilon \Vert \nabla _{\mathrm{h}} u^{\text {fl}}_{\Phi }\Vert ^2_{{\dot{B}}^{\frac{1}{2}}}+\Vert h^1-\chi \{V^1\}_{\varepsilon }\Vert _{L^{\infty }(D_1)}^2+\varepsilon \Vert h^2-\chi \{V^2\}_{\varepsilon }\Vert _{L^{\infty }(D_1)}^2\nonumber \\&\,+\Vert V^0\Vert _{L_z^{\infty }}+\Vert z\partial _z V^0\Vert _{L_z^{\infty }}+\varepsilon \Vert r^{\varepsilon }\Vert ^2_{L^{\infty }(D_1)}. \end{aligned}$$

(5.23)

In view of (5.11) and (5.14), there exists a constant $C_*>0$ such that, for any $t\in [0,T^*]$,

$$\begin{aligned}&\sum _{k\in \mathbb {Z}}\Vert {\dot{\Delta }}_kR^{\varepsilon }_{\Phi }(t)\Vert +\sum _{k\in \mathbb {Z}}\left( \int _0^t2^k|{\dot{\rho }}|\Vert {\dot{\Delta }}_k R^{\varepsilon }_{\Phi }\Vert ^2\,ds\right) ^{\frac{1}{2}}\nonumber \\&\qquad +\sum _{k\in \mathbb {Z}}\left( \int _0^t{\dot{\beta }}\Vert {\dot{\Delta }}_k R^{\varepsilon }_{\Phi }\,ds\right) ^{\frac{1}{2}}+\sum _{k\in \mathbb {Z}}\left( \int _0^t\varepsilon \Vert \nabla {\dot{\Delta }}_kR^{\varepsilon }_{\Phi }\Vert ^2\,ds\right) ^{\frac{1}{2}}\nonumber \\&\quad \le C_*\sum _{k\in \mathbb {Z}}\left( \int _0^T\Vert {\dot{\Delta }}_k F^{\varepsilon }_{\Phi }\Vert ^2\,ds\right) ^{\frac{1}{2}}+\sum _{k\in \mathbb {Z}}\left( \int _0^tC_*(\chi _{[0,T]}+{\mathcal {K}}_{\varepsilon })\Vert {\dot{\Delta }}_k R^{\varepsilon }_{\Phi }\Vert ^2\,ds\right) ^{\frac{1}{2}}\nonumber \\&\qquad +\sum _{k\in \mathbb {Z}}\left( \int _0^tC_*2^k(\varepsilon \Vert \nabla _{\mathrm{h}} R^{\varepsilon }_{\Phi }\Vert ^2_{{\dot{B}}^0}+\Vert z\partial _z V^0\Vert _{L_z^{\infty }})\Vert {\dot{\Delta }}_k R^{\varepsilon }_{\Phi }\Vert ^2\,ds\right) ^{\frac{1}{2}}. \end{aligned}$$

(5.24)

Recall the definition of $\beta (t)$ in (5.5) and $\rho (t)$ in (5.7), we find that, for $t\le T^*,$

$$\begin{aligned} \sum _{k\in \mathbb {Z}}\Vert {\dot{\Delta }}_kR^{\varepsilon }_{\Phi }(t)\Vert +\sum _{k\in \mathbb {Z}}\left( \int _0^t\varepsilon \Vert \nabla {\dot{\Delta }}_kR^{\varepsilon }_{\Phi }\Vert ^2\,ds\right) ^{\frac{1}{2}}\le C_*\sum _{k\in \mathbb {Z}}\left( \int _0^T\Vert {\dot{\Delta }}_k F^{\varepsilon }_{\Phi }\Vert ^2\,ds\right) ^{\frac{1}{2}}. \end{aligned}$$

(5.25)

Let us admit the following proposition for the time being.

Proposition 5.4

If $u_b$ satisfies (2.43) and (2.44) for a constant $C_b>0,$ there is a constant $C_F$ such that for any $\varepsilon \in (0,1),$

$$\begin{aligned} \sum _{k\in \mathbb {Z}}\left( \int _0^T\Vert {\dot{\Delta }}_k F^{\varepsilon }_{\Phi }\Vert ^2\,dt\right) ^{\frac{1}{2}}\le C_F \varepsilon ^{\frac{1}{4}}. \end{aligned}$$

(5.26)

The proof of Proposition 5.4 is postponed to Sect. 6.

We then deduce that for $t\le T^*$,

$$\begin{aligned} \sum _{k\in \mathbb {Z}}\Vert {\dot{\Delta }}_kR^{\varepsilon }_{\Phi }(t)\Vert +\sum _{k\in \mathbb {Z}}\left( \int _0^t\varepsilon \Vert \nabla {\dot{\Delta }}_kR^{\varepsilon }_{\Phi }\Vert ^2\,ds\right) ^{\frac{1}{2}}\le C_*C_F\varepsilon ^{\frac{1}{4}}. \end{aligned}$$

(5.27)

For $t\le T^*,$ by the Minkowski inequality,

$$\begin{aligned} \int _0^t\varepsilon \Vert \nabla R^{\varepsilon }_{\Phi }\Vert ^2_{{\dot{B}}^0}\,ds\,= & {} \,\int _0^t\varepsilon \left( \sum _{k\in \mathbb {Z}}\Vert {\dot{\Delta }}_k \nabla R^{\varepsilon }_{\Phi }\Vert \right) ^2\,ds\nonumber \\\le & {} \,\left( \sum _{k\in \mathbb {Z}}\left( \int _0^t\varepsilon \Vert {\dot{\Delta }}_k\nabla R^{\varepsilon }_{\Phi }\Vert ^2\,ds\right) ^{\frac{1}{2}}\right) ^2\nonumber \\\le & {} \,C_*^2C_F^2\varepsilon ^{\frac{1}{2}}. \end{aligned}$$

(5.28)

So that

$$\begin{aligned} \rho (T^*)\ge 2-C_*^3C_F^2\varepsilon ^{\frac{1}{2}}. \end{aligned}$$

(5.29)

Thus, for $\varepsilon$ small enough, $\rho (T^*)> 1$ and thus $T^*=\frac{T}{{\varepsilon }}$ and we have, for $t\in [0, \frac{T}{{\varepsilon }}]$,

$$\begin{aligned} \Vert R^{\varepsilon }_{\Phi }(t)\Vert +\left( \int _0^t\varepsilon \Vert \nabla R^{\varepsilon }_{\Phi }\Vert ^2\right) ^{\frac{1}{2}}\le C \varepsilon ^{\frac{1}{4}}, \end{aligned}$$

(5.30)

which implies (2.45) since $\beta (t)\le \beta _*$ by Proposition 5.1 and $\rho (t)\ge 1$ for $t\in [0,\frac{T}{{\varepsilon }}]$. This completes the proof of Proposition 2.6. $\square$

6 Proof of Proposition 5.4

This section is devoted to the proof of Proposition 5.4. We start with the following observation.

Lemma 6.1

For any axi-symmetric function f in $D_1$, if f vanishes on the boundary $\partial D_1$ and $\nabla _{\mathrm{h}} f\in L^2(D_1)$, where $\nabla _{\mathrm{h}}=(\partial _{x_1},\partial _{x_2})$, then there exists a constant C such that

$$\begin{aligned} \Vert f\Vert _{L^{\infty }(D_1)}\le C\Vert \nabla _{\mathrm{h}}f\Vert _{L^2(D_1)}. \end{aligned}$$

(6.1)

Proof

For $r\in [0,1]$, since f vanishes on $\partial D_1$,

$$\begin{aligned} |f(r)|^2=-\int _r^1 2f(\tau )\partial _\tau f(\tau )\, d\tau \le 2\left( \int _r^1|\partial _\tau f(\tau )|^2\tau \,d\tau \right) ^{\frac{1}{2}}\left( \int _r^1\frac{|f(\tau )|^2}{\tau }\,d\tau \right) ^{\frac{1}{2}}. \end{aligned}$$

On the other hand, one has

$$\begin{aligned} \Vert \nabla _{\mathrm{h}} f\Vert ^2_{L^2(D_1)}=2\pi \int _0^1 \left( |\partial _r f(r)|^2+\frac{|f(r)|^2}{r^2}\right) r\,dr, \end{aligned}$$

so that

$$\begin{aligned} \Vert f\Vert _{L^{\infty }(D_1)}\le \frac{1}{\sqrt{\pi }}\Vert \nabla _{\mathrm{h}} f\Vert _{L^2(D_1)}, \end{aligned}$$

which leads to (6.1). $\square$

First, for any profile $\mathcal {V}\in L^2((0,T)\times D_1)$, any axi-symmetric function v defined in $(0,T)\times \mathcal {C}$, vanishes on the boundary $\mathcal {C}$, and function $g\in L^2((0,T);L^{\infty }( D_1)),$ we deduce from (2.21) and Lemma 6.1 that

$$\begin{aligned} \sum _{k\in \mathbb {Z}}\left( \int _0^T\Vert {\dot{\Delta }}_k(v\{\mathcal {V}\}_{\varepsilon })_{\Phi }\Vert ^2\,dt\right) ^{\frac{1}{2}} \,\le & {} \, \sum _{k\in \mathbb {Z}}\sup _{t\in [0,T]}\Vert {\dot{\Delta }}_k v_{\Phi }\Vert _{L^{\infty }_{x_{\mathrm{h}}}L^2_{x_3}} \left( \int _0^T\Vert \{\mathcal {V}\}_{\varepsilon }\Vert ^2_{L^2_{x_\mathrm{h}}}\,dt\right) ^{\frac{1}{2}}\nonumber \\\le & {} C\varepsilon ^{\frac{1}{4}}\Vert \mathcal {V}\Vert _{L^2((0,T)\times {\mathbb {R}}_+)}\sum _{k\in \mathbb {Z}}\Vert \nabla _{\mathrm{h}}{\dot{\Delta }}_k v_{\Phi }\Vert _{L^\infty ((0,T);L^2(\mathcal {C}))}, \end{aligned}$$

(6.2)

and

$$\begin{aligned} \sum _{k\in \mathbb {Z}}\left( \int _0^T\Vert {\dot{\Delta }}_k(vg)_{\Phi }\Vert ^2\,dt\right) ^{\frac{1}{2}}\le C\sum _{k\in \mathbb {Z}}\Vert {\dot{\Delta }}_k v_{\Phi }\Vert _{L^\infty ((0,T);L^2(\mathcal {C}))}\Vert g\Vert _{L^2((0,T);L^{\infty }(D_1))}. \end{aligned}$$

(6.3)

We recall from (2.42) that

$$\begin{aligned} F^{\varepsilon }:\,= & {} \,\chi (\{V^0\}_{\varepsilon }+\sqrt{\varepsilon }\{V^1\}_{\varepsilon }+\varepsilon \{V^2\}_{\varepsilon })\partial _3 u^{\text {fl}}-(\sqrt{\varepsilon }h^1+\varepsilon h^2+\varepsilon r^{\varepsilon })\partial _3 u^{\text {fl}}\nonumber \\&\,+u^{\text {fl}}_r\chi '\{V^0+\sqrt{\varepsilon }V^1+\varepsilon V^2\}_{\varepsilon }e_3-u^{\text {fl}}_r\frac{\chi }{1-r}\{z\partial _z V^0+\sqrt{\varepsilon }z\partial _z V^1+\varepsilon z\partial _z V^2\}_{\varepsilon }e_3\nonumber \\&\, -\varepsilon u^{\text {fl}}_r\partial _r r^{\varepsilon }e_3-\varepsilon u^{\text {fl}}\cdot \nabla u^{\text {fl}}+\varepsilon \Delta u^{\text {fl}}. \end{aligned}$$

(6.4)

Now we estimate $\sum _{k\in \mathbb {Z}}\big (\int _0^T\Vert {\dot{\Delta }}_k F^{\varepsilon }_{\Phi }\Vert ^2\,dt\big )^{\frac{1}{2}}$ term by term.

We first get, by applying (6.2) to $\mathcal {V}:=V_0$ and $v:=\chi \partial _3 u^{\text {fl}},$ that

$$\begin{aligned}&\sum _{k\in \mathbb {Z}}\left( \int _0^T\Vert {\dot{\Delta }}_k(\chi \partial _3u^{\text {fl}}\{V^0\}_{\varepsilon })_{\Phi }\Vert ^2\,dt\right) ^{\frac{1}{2}}\nonumber \\&\quad \le C\varepsilon ^{\frac{1}{4}}\Vert V^0\Vert _{L^2((0,T)\times {\mathbb {R}}_+)}\sum _{k\in \mathbb {Z}}\Vert \nabla _{\mathrm{h}}{\dot{\Delta }}_k (\chi \partial _3 u^{\text {fl}}_{\Phi })\Vert _{L^\infty ((0,T);L^2(\mathcal {C}))}. \end{aligned}$$

(6.5)

Note that

$$\begin{aligned} u^{\text {fl}}(t,x)=\mu (t)u_b(x-\int _0^t h^0(s)dse_3). \end{aligned}$$

So that

$$\begin{aligned} \sum _{k\in \mathbb {Z}}\Vert \nabla _{\mathrm{h}}{\dot{\Delta }}_k (\chi \partial _3 u^{\text {fl}}_{\Phi })\Vert _{L^\infty ((0,T); L^2(\mathcal {C})} \le C\sum _{k\in \mathbb {Z}}\Vert {\dot{\Delta }}_ke^{\rho _0|\partial _3|}\partial _3u_b\Vert _{H^1(\mathcal {C})}. \end{aligned}$$

(6.6)

Thanks to Cauchy inequality and the properties of operator ${\dot{\Delta }}_k$,

$$\begin{aligned}&\sum _{k\in \mathbb {Z}}\Vert {\dot{\Delta }}_ke^{\rho _0|\partial _3|}\partial _3 u_b\Vert _{H^1(\mathcal {C})}\nonumber \\&\quad \le C \sum _{k\in \mathbb {Z}} 2^k\Vert {\dot{\Delta }}_k e^{\rho _0|\partial _3|} u_b\Vert _{H^1(\mathcal {C})}\nonumber \\&\quad \le C(\sum _{k\le 0}\Vert {\dot{\Delta }}_k e^{\rho _0|\partial _3|} u_b\Vert _{H^1(\mathcal {C})}^2)^{\frac{1}{2}}+C(\sum _{k>0} 2^{4k}\Vert {\dot{\Delta }}_k e^{\rho _0|\partial _3|} u_b\Vert _{H^1(\mathcal {C})}^2)^{\frac{1}{2}}\nonumber \\&\quad \le C\Vert e^{\rho _0|\partial _3|}u_b\Vert _{H^3(\mathcal {C})}\nonumber \\&\quad \le CC_b, \end{aligned}$$

(6.7)

where we used (2.43) by taking $\rho _b=2\rho _0.$

By inserting the above estimates into (6.5), we obtain

$$\begin{aligned} \sum _{k\in \mathbb {Z}}\left( \int _0^T\Vert {\dot{\Delta }}_k(\chi \partial _3u^{\text {fl}}\{V^0\}_{\varepsilon })_{\Phi }\Vert ^2\,dt\right) ^{\frac{1}{2}}\le C\varepsilon ^{\frac{1}{4}}. \end{aligned}$$

(6.8)

Similarly, one has

$$\begin{aligned} \sum _{k\in \mathbb {Z}}\left( \int _0^T\Vert {\dot{\Delta }}_k(\chi \partial _3u^{\text {fl}}(\sqrt{\varepsilon }\{V^1\}_{\varepsilon }+\varepsilon \{V^2\}_{\varepsilon }))_{\Phi }\Vert ^2\,dt\right) ^{\frac{1}{2}}\le C\varepsilon ^{\frac{3}{4}}. \end{aligned}$$

(6.9)

We apply (6.3) to $v=\partial _3 u^{\text {fl}}$ and $g=\sqrt{\varepsilon }h^1+\varepsilon h^2+\varepsilon r^{\varepsilon }$,

$$\begin{aligned}&\sum _{k\in \mathbb {Z}}\left( \int _0^T\Vert {\dot{\Delta }}_k( (\sqrt{\varepsilon }h^1+\varepsilon h^2+\varepsilon r^{\varepsilon })\partial _3u^{\text {fl}})_{\Phi }\Vert ^2\,dt\right) ^{\frac{1}{2}}\nonumber \\&\quad \le C \sum _{k\in \mathbb {Z}}\Vert {\dot{\Delta }}_k\partial _3u^{\text {fl}}_{\Phi }\Vert _{L^\infty ((0,T); L^2(\mathcal {C}))}\Vert \sqrt{\varepsilon }h^1+\varepsilon h^2+\varepsilon r^{\varepsilon }\Vert _{L^2((0,T);L^{\infty }(D_1))} . \end{aligned}$$

(6.10)

In the same way as (6.6) and (6.7), we obtain that

$$\begin{aligned} \sum _{k\in \mathbb {Z}}\Vert {\dot{\Delta }}_k\partial _3u^{\text {fl}}_{\Phi }\Vert _{L^\infty ((0,T);L^2(\mathcal {C}))}\le CC_b. \end{aligned}$$

(6.11)

While it follows from Sobolev imbedding inequality and Lemma 2.5 that

$$\begin{aligned} \Vert r^{\varepsilon }\Vert _{L^2((0,T);L^{\infty }(D_1))}\le C\Vert r^{\varepsilon }\Vert _{L^2((0,T);H^2(D_1))}\le C\varepsilon ^{-\frac{1}{4}}. \end{aligned}$$

(6.12)

We deduce that

$$\begin{aligned} \sum _{k\in \mathbb {Z}}\left( \int _0^T\Vert {\dot{\Delta }}_k( (\sqrt{\varepsilon }h^1+\varepsilon h^2+\varepsilon r^{\varepsilon })\partial _3u^{\text {fl}})_{\Phi }\Vert ^2\,dt\right) ^{\frac{1}{2}}\le C \varepsilon ^{\frac{1}{2}}. \end{aligned}$$

(6.13)

We apply (6.2) to $v=u^{\text {fl}}_r\chi '$ and $\mathcal {V}=V^0$ to obtain that

$$\begin{aligned}&\sum _{k\in \mathbb {Z}}\left( \int _0^T\Vert {\dot{\Delta }}_k(u^{\text {fl}}_r\chi '\{V^0\}_{\varepsilon })_{\Phi }\Vert ^2\,dt\right) ^{\frac{1}{2}}\nonumber \\&\quad \le C\varepsilon ^{\frac{1}{4}}\Vert V^0\Vert _{L^2((0,T)\times {\mathbb {R}}_+)}\sum _{k\in \mathbb {Z}}\Vert \nabla _{\mathrm{h}}{\dot{\Delta }}_k(u^{\text {fl}}_r\chi ')_{\Phi }\Vert _{L^\infty ((0,T);L^2(\mathcal {C}))}\nonumber \\&\quad \le C\varepsilon ^{\frac{1}{4}}\Vert V^0\Vert _{L^2((0,T)\times {\mathbb {R}}_+)}\sum _{k\in \mathbb {Z}}\Vert {\dot{\Delta }}_ke^{\rho _0|\partial _3|}u_b\Vert _{L^2_{x_3}(H^1(D_1))}. \end{aligned}$$

(6.14)

Thanks to Cauchy inequality and the properties of operator ${\dot{\Delta }}_k$,

$$\begin{aligned} \sum _{k\in {\mathbb {Z}}}\Vert \nabla _\mathrm{h}{\dot{\Delta }}_ke^{\rho _0|\partial _3|}u_b\Vert _{L^2(\mathcal {C})}\le & {} \left( \sum _{k\in {\mathbb {Z}}}2^{-\frac{|k|}{2}}\right) ^{\frac{1}{2}}\left( \sum _{k\in {\mathbb {Z}}}2^{\frac{|k|}{2}}\Vert e^{\rho _0|\partial _3|}{\dot{\Delta }}_k u_b\Vert ^2_{L^2_{x_3}(H^1(D_1))}\right) ^{\frac{1}{2}}\nonumber \\\le & {} C\left( \int _{{\mathbb {R}}}(|\xi |^{\frac{1}{2}}+|\xi |^{-\frac{1}{2}})e^{2\rho _0|\xi |}\Vert {\mathcal {F}}u_b(\xi )\Vert ^2_{H^1(D_1)}d\xi \right) ^{\frac{1}{2}}, \end{aligned}$$

(6.15)

where ${\mathcal {F}}u_b(\xi )$ is the Fourier transform of $u_b$ in the direction of $x_3$.

• For low frequencies, by (2.44),

  $$\begin{aligned}&\int _{|\xi |\le 1}(|\xi |^{\frac{1}{2}}+|\xi |^{-\frac{1}{2}})e^{2\rho _0|\xi |}\Vert {\mathcal {F}}u_b(\xi )\Vert ^2_{H^1(D_1)}d\xi \nonumber \\&\le e^{2\rho _0}\Vert {\mathcal {F}}u_b\Vert ^2_{L^\infty _\xi (H^1(D_1))}\int _{|\xi |\le 1}(|\xi |^{\frac{1}{2}}+|\xi |^{-\frac{1}{2}})\,d\xi \le C\Vert u_b\Vert ^2_{L^1_{x_3}(H^1(D_1))}\le CC_b^2. \end{aligned}$$

  (6.16)

• For high frequencies, by (2.43),

  $$\begin{aligned}&\int _{|\xi |\ge 1}(|\xi |^{\frac{1}{2}}+|\xi |^{-\frac{1}{2}})e^{2\rho _0|\xi |}\Vert {\mathcal {F}}u_b(\xi )\Vert ^2_{H^1(D_1)}d\xi \nonumber \\&\le \int _{|\xi |\ge 1}(|\xi |+1)^2e^{2\rho _0|\xi |}\Vert {\mathcal {F}}u_b(\xi )\Vert ^2_{H^1(D_1)}d\xi \nonumber \\&\le C\Vert e^{\rho _0|\partial _3|}u_b\Vert ^2_{H^2(\mathcal {C})}\le CC_b^2. \end{aligned}$$

  (6.17)

Gathering the estimates (6.14)–(6.17), we get

$$\begin{aligned} \sum _{k\in \mathbb {Z}}\left( \int _0^T\Vert {\dot{\Delta }}_k(u^{\text {fl}}_r\chi '\{V^0\}_{\varepsilon })_{\Phi }\Vert ^2\,dt\right) ^{\frac{1}{2}}\le C\varepsilon ^{\frac{1}{4}}. \end{aligned}$$

(6.18)

Similarly, one has

$$\begin{aligned} \sum _{k\in \mathbb {Z}}\left( \int _0^T\Vert {\dot{\Delta }}_k(u^{\text {fl}}_r\chi '(\sqrt{\varepsilon }\{V^1\}_{\varepsilon }+\varepsilon \{V^2\}_{\varepsilon }))_{\Phi }\Vert ^2\,dt\right) ^{\frac{1}{2}}\le C\varepsilon ^{\frac{3}{4}}. \end{aligned}$$

(6.19)

We apply (6.2) to $v=z\partial _z V^0$ and $v=u^{\text {fl}}_r\frac{\chi }{1-r}$ and use (2.30):

$$\begin{aligned}&\sum _{k\in \mathbb {Z}}\left( \int _0^T\Vert {\dot{\Delta }}_k(u^{\text {fl}}_r\frac{\chi }{1-r}\{z\partial _z V^0\}_{\varepsilon })_{\Phi }\Vert ^2\,dt\right) ^{\frac{1}{2}}\nonumber \\&\le C\varepsilon ^{\frac{1}{4}}\Vert z\partial _z V^0\Vert _{L^2((0,T)\times {\mathbb {R}}_+)}\sum _{k\in \mathbb {Z}}\Vert \nabla _{\mathrm{h}}{\dot{\Delta }}_k(u^{\text {fl}}_r\frac{\chi }{1-r})_{\Phi }\Vert _{L^\infty ((0,T); L^2(\mathcal {C}))}\nonumber \\&\le C\varepsilon ^{\frac{1}{4}}\Vert V^0\Vert _{C^0_0({\mathbb {R}}_+;H^1_1({\mathbb {R}}_+))}\sum _{k\in \mathbb {Z}}\Vert {\dot{\Delta }}_ke^{\rho _0|\partial _3|}\frac{u_b}{1-r}\Vert _{L^2_{x_3}(H^1(D_1))}. \end{aligned}$$

Since $u_b$ vanishes on the boundary of $\mathcal {C}$,

$$\begin{aligned} \frac{u_b}{1-r}=-\int _0^1(\partial _ru_b)(1-(1-r)s)ds. \end{aligned}$$

Proceeding in the same way as for the treatment of (6.15)–(6.17), and by using (2.43) and (2.44), we find that

$$\begin{aligned}&\sum _{k\in \mathbb {Z}}\big \Vert {\dot{\Delta }}_ke^{\rho _0|\partial _3|}\frac{u_b}{1-r}\big \Vert _{L^2_{x_3}(H^1(D_1))}\nonumber \\&\le C\left( \int _{{\mathbb {R}}}(|\xi |^{\frac{1}{2}}+|\xi |^{-\frac{1}{2}}) e^{2\rho _0|\xi |}\left( \int _0^1\Vert {\mathcal {F}}(\partial _ru_b)(1-(1-r)s)\Vert _{H^1(D_1)}\,ds\right) ^2d\xi \right) ^{\frac{1}{2}}\nonumber \\&\le C \left( \int _{{\mathbb {R}}}(|\xi |^{\frac{1}{2}}+|\xi |^{-\frac{1}{2}})e^{2\rho _0|\xi |}\Vert {\mathcal {F}}\partial _ru_b\Vert ^2_{H^1(D_1)}d\xi \right) ^{\frac{1}{2}}\nonumber \\&\le C\Vert u_b\Vert _{L^1_{x_3}(H^2(D_1))}+C\Vert e^{\rho _0|\partial _3|}u_b\Vert _{H^3(\mathcal {C})}\nonumber \\&\le CC_b. \end{aligned}$$

Thus, we obtain

$$\begin{aligned} \sum _{k\in \mathbb {Z}}\left( \int _0^T\Vert {\dot{\Delta }}_k(u^{\text {fl}}_r\frac{\chi }{1-r}\{z\partial _z V^0\}_{\varepsilon })_{\Phi }\Vert ^2\,dt\right) ^{\frac{1}{2}}\le C\varepsilon ^{\frac{1}{4}}. \end{aligned}$$

(6.20)

Similarly, one has

$$\begin{aligned} \sum _{k\in \mathbb {Z}}\left( \int _0^T\Vert {\dot{\Delta }}_k(u^{\text {fl}}_r\frac{\chi }{1-r}(\sqrt{\varepsilon }\{z\partial _z V^1\}_{\varepsilon }+\varepsilon \{z\partial _z V^2\}_{\varepsilon }))_{\Phi }\Vert ^2\,dt\right) ^{\frac{1}{2}}\le C\varepsilon ^{\frac{3}{4}}. \end{aligned}$$

(6.21)

We apply (6.3) to $v=\varepsilon u^{\text {fl}}_r$ and $g=\partial _r r^{\varepsilon }$,

$$\begin{aligned}&\sum _{k\in \mathbb {Z}}\Big (\int _0^T\Vert {\dot{\Delta }}_k(\varepsilon u^{\text {fl}}_r\partial _r r^{\varepsilon })_{\Phi }\Vert ^2\,dt\Big )^{\frac{1}{2}}\nonumber \\&\le C\varepsilon \sum _{k\in \mathbb {Z}}\Vert {\dot{\Delta }}_k u^{\text {fl}}_{r,\Phi }\Vert _{L^\infty ((0,T); L^2(\mathcal {C}))}\Vert \partial _r r^{\varepsilon }\Vert _{L^2((0,T);L^{\infty }(D_1))}. \end{aligned}$$

Notice that

$$\begin{aligned} \sum _{k\in \mathbb {Z}}\Vert {\dot{\Delta }}_k u^{\text {fl}}_{r,\Phi }\Vert _{L^\infty ((0,T);L^2(\mathcal {C}))}\le \sum _{k\in \mathbb {Z}}\Vert {\dot{\Delta }}_k e^{\rho _0|\partial _3|}u_b\Vert _{L^2(\mathcal {C})}. \end{aligned}$$

In the same way as (6.15)–(6.17), we can get

$$\begin{aligned} \sum _{k\in \mathbb {Z}}\Vert {\dot{\Delta }}_k e^{\rho _0|\partial _3|}u_b\Vert _{L^2(\mathcal {C})}\le CC_b. \end{aligned}$$

By using Lemmas 6.1 and 2.5, we obtain that

$$\begin{aligned} \Vert \partial _r r^{\varepsilon }\Vert _{L^2((0,T);L^{\infty }(D_1))}\le C\Vert \nabla _{\mathrm{h}}\partial _r r^{\varepsilon }\Vert _{L^2((0,T)\times D_1)}\le C\varepsilon ^{-\frac{1}{4}}. \end{aligned}$$

Therefore, we achieve

$$\begin{aligned} \sum _{k\in \mathbb {Z}}\left( \int _0^T\Vert {\dot{\Delta }}_k(\varepsilon u^{\text {fl}}_r\partial _r r^{\varepsilon })_{\Phi }\Vert ^2\,dt\right) ^{\frac{1}{2}}\le C\varepsilon ^{\frac{3}{4}}. \end{aligned}$$

(6.22)

In view of (2.30), one has

$$\begin{aligned} \sum _{k\in \mathbb {Z}}\left( \int _0^t \Vert {\dot{\Delta }}_k \Delta u^{\text {fl}}_{\Phi } \Vert ^2\,dt\right) ^{\frac{1}{2}}\le C\sum _{k\in \mathbb {Z}}\Vert {\dot{\Delta }}_k e^{\rho _0|\partial _3|}u^{\text {fl}}\Vert _{H^2(\mathcal {C})}. \end{aligned}$$

In the same way as (6.15)–(6.17), we can get

$$\begin{aligned} \sum _{k\in \mathbb {Z}}\Vert {\dot{\Delta }}_k e^{\rho _0|\partial _3|}u_b\Vert _{H^2(\mathcal {C})}\le C\Vert u_b\Vert _{L^1_{x_3}(H^2(D_1))}+C\Vert \left\langle \partial _3\right\rangle e^{\rho _0|\partial _3|}u_b\Vert _{H^2(\mathcal {C})} \le CC_b. \end{aligned}$$

Thus we obtain

$$\begin{aligned} \sum _{k\in \mathbb {Z}}\left( \int _0^t\Vert {\dot{\Delta }}_k(\varepsilon \Delta u^{\text {fl}})_{\Phi }\Vert ^2\,dt\right) ^{\frac{1}{2}}\le C\varepsilon . \end{aligned}$$

(6.23)

It remains to estimate the last term $\varepsilon u^{\text {fl}}\cdot \nabla u^{\text {fl}}$ of $F^{\varepsilon }$. By using (2.30), we find

$$\begin{aligned} \sum _{k\in \mathbb {Z}}\left( \int _0^T\Vert {\dot{\Delta }}_k(u^{\text {fl}}\cdot \nabla u^{\text {fl}})_{\Phi }\Vert ^2\,dt\right) ^{\frac{1}{2}}\le C\sum _{k\in \mathbb {Z}}\Vert {\dot{\Delta }}_k(u_b\cdot \nabla u_b)_{\Phi }\Vert _{L^2(\mathcal {C})}. \end{aligned}$$

(6.24)

In the same way as (6.15)–(6.17), one has

$$\begin{aligned}&\sum _{k\in \mathbb {Z}}\Vert {\dot{\Delta }}_k(u_b\cdot \nabla u_b)_{\Phi }\Vert _{L^2(\mathcal {C})}\nonumber \\&\quad \le C \Vert u_b\cdot \nabla u_b\Vert _{L^1_{x_3}(L^2(D_1))}+ C\Vert \left\langle \partial _3 \right\rangle e^{\rho _0|\partial _3|}(u_b\cdot \nabla u_b)\Vert _{L^2(\mathcal {C})}. \end{aligned}$$

(6.25)

It follows from Lemma 6.1 that

$$\begin{aligned} \Vert u_b\cdot \nabla u_b\Vert _{L^1_{x_3}(L^2(\mathcal {C}))}\,\le & {} \, \Vert u_b\Vert _{L^2_{x_3}(L^{\infty }(D_1))}\Vert \nabla u_b\Vert _{L^2(\mathcal {C})}\nonumber \\\le & {} \, C\Vert \nabla _{\mathrm{h}} u_b\Vert _{L^2(\mathcal {C})}\Vert \nabla u_b\Vert _{L^2(\mathcal {C})}\nonumber \\\le & {} \, CC_b^2. \end{aligned}$$

(6.26)

By the Plancherel theorem and the inequalities

$$\begin{aligned} |\xi |\le |\eta |+|\xi -\eta | \text { and } (1+|\xi |^2)\le 2(1+|\eta |^2)(1+|\xi -\eta |^2)\qquad \forall \,\xi ,\eta \in {\mathbb {R}}, \end{aligned}$$

we get

$$\begin{aligned}&\Vert \left\langle \partial _3 \right\rangle e^{\rho _0|\partial _3|}(u_b\cdot \nabla u_b)\Vert _{L^2(\mathcal {C})}^2= \int _{{\mathbb {R}}}\left\langle \xi \right\rangle ^2e^{2\rho _0|\xi |}\Vert {\mathcal {F}}(u_b\cdot \nabla u_b)\Vert ^2_{L^2(D_1)}d\xi \\&\quad =\int _{{\mathbb {R}}}\left\langle \xi \right\rangle ^2e^{2\rho _0|\xi |}\Vert \int _{{\mathbb {R}}}{\mathcal {F}}u_b(\eta )\cdot {\mathcal {F}}(\nabla u_b)(\xi -\eta )d\eta \Vert ^2_{L^2(D_1)}d\xi \\&\quad \le C\int _{{\mathbb {R}}}\big \Vert \int _{{\mathbb {R}}}\left\langle \eta \right\rangle e^{\rho _0|\eta |}|{\mathcal {F}}(u_b)(\eta )|\left\langle \xi -\eta \right\rangle e^{\rho _0|\xi -\eta |}|{\mathcal {F}}(\nabla u_b)(\xi -\eta )|d\eta \big \Vert ^2_{L^2(D_1)}d\xi \\&\quad \le C\int _{{\mathbb {R}}}\left( \int _{{\mathbb {R}}}\big \Vert \left\langle \eta \right\rangle e^{\rho _0|\eta |}|{\mathcal {F}}(u_b)(\eta )|\big \Vert _{L^\infty (D_1)}\big \Vert \left\langle \xi -\eta \right\rangle e^{\rho _0|\xi -\eta |}|{\mathcal {F}}(\nabla u_b)(\xi -\eta )|\big \Vert _{L^2(D_1)}\,d\eta \right) ^2\,d\xi \\&\quad \le C\big \Vert \left\langle \xi \right\rangle e^{\rho _0|\xi |}|{\mathcal {F}}(u_b)(\xi )|\big \Vert ^2_{L^1_\xi (L^\infty (D_1))}\big \Vert \left\langle \xi \right\rangle e^{\rho _0|\xi |}|{\mathcal {F}}(\nabla u_b)(\xi )|\big \Vert ^2_{L^2_\xi (L^2(D_1))}. \end{aligned}$$

By using again the Plancherel theorem and Lemma 6.1, one has

$$\begin{aligned} \big \Vert \left\langle \xi \right\rangle e^{\rho _0|\xi |}|{\mathcal {F}}(u_b)(\xi )|\big \Vert _{L^1_\xi (L^\infty (D_1))} \le&C\int _{{\mathbb {R}}}\big \Vert \left\langle \xi \right\rangle e^{\rho _0|\xi |}|{\mathcal {F}}(\nabla _\mathrm{h}u_b)(\xi )|\big \Vert _{L^2(D_1)}\,d\xi \\ \le&C\left( \int _{{\mathbb {R}}}\big \Vert \left\langle \xi \right\rangle ^2 e^{\rho _0|\xi |}|{\mathcal {F}}(\nabla _\mathrm{h}u_b)(\xi )|\big \Vert _{L^2(D_1)}^2\,d\xi \right) ^{{\frac{1}{2}}}\\ \le&C\Vert e^{\rho _0|\partial _3|}u_b\Vert _{H^3(\mathcal {C})}, \end{aligned}$$

so that we deduce from (2.43) that

$$\begin{aligned}&\Vert \left\langle \partial _3 \right\rangle e^{\rho _0|\partial _3|}(u_b\cdot \nabla u_b)\Vert _{L^2(\mathcal {C})}\nonumber \\&\le C\Vert e^{\rho _0|\partial _3|}u_b\Vert _{H^3(\mathcal {C})} \Vert \left\langle \partial _3\right\rangle e^{\rho _0|\partial _3|}\nabla u_b\Vert _{L^2(\mathcal {C})}\nonumber \\&\quad \le C\Vert e^{\rho _0|\partial _3|}u_b\Vert ^2_{H^3(\mathcal {C})}\le CC_b^2. \end{aligned}$$

(6.27)

By combining the estimates (6.24)–(6.27), we arrive at

$$\begin{aligned} \sum _{k\in \mathbb {Z}}\left( \int _0^T\Vert {\dot{\Delta }}_k(\varepsilon u^{\text {fl}}\cdot \nabla u^{\text {fl}})_{\Phi }\Vert ^2\,dt\right) ^{\frac{1}{2}}\le C\varepsilon . \end{aligned}$$

(6.28)

Finally, by gathering (6.4) and inequalities (6.8), (6.9), (6.13), (6.18)–(6.22) and (6.28), we obtain that there exist a constant $C_F$ such that

$$\begin{aligned} \sum _{k\in \mathbb {Z}}\left( \int _0^T\Vert {\dot{\Delta }}_k F^{\varepsilon }_{\Phi }\Vert ^2\,dt\right) ^{\frac{1}{2}}\le C_F \varepsilon ^{\frac{1}{4}}, \end{aligned}$$

(6.29)

which finishes the proof of Proposition 5.4.

7 Proof of Lemma 5.3

This section is devoted to the proof of Lemma 5.3. As in Remark 5.2, we always denote the norm $\Vert \cdot \Vert _{L^2(\mathcal {C})}$ by $\Vert \cdot \Vert$ in this section.

Proof of Lemma 5.3

We first get, by using Bony’s decomposition from [2] in the $x_3$ variable, that

$$\begin{aligned} ab=T_a^{\mathrm{v}}b+R^{\mathrm{v}}(a,b)+T_b^{\mathrm{v}}a, \end{aligned}$$

where

For a function $a\in L^2(\mathcal {C})$, we introduce the notation as in [8],

$$\begin{aligned} a^+:={\mathcal {F}}_{\xi \rightarrow x_3}^{-1}|{\hat{a}}|. \end{aligned}$$

(7.1)

It is easy to observe from Lemma 6.1 and Bernstein’s inequality that

$$\begin{aligned} \Vert {\dot{S}}_{k-1}a_{\Phi }^+\Vert _{L^\infty (\mathcal {C})}\le & {} C\sum _{k'\le k-2}\Vert {\dot{\Delta }}_{k'}a_{\Phi }^+\Vert _{L^{\infty }(\mathcal {C})}\nonumber \\\le & {} C\sum _{k'\le k-2} 2^{\frac{k'}{2}}\Vert \nabla _{\mathrm{h}}{\dot{\Delta }}_{k'}a_{\Phi }\Vert \le C\Vert \nabla _\mathrm{h}a_{\Phi }\Vert _{{\dot{B}}^{\frac{1}{2}}}, \end{aligned}$$

(7.2)

so that we get, by a similar proof of Lemma 5.7 of [8] and Proposition 5.1 that

$$\begin{aligned} |({\dot{\Delta }}_k(T^{\mathrm{v}}_a b)_{\Phi },{\dot{\Delta }}_k c_{\Phi })|\,\le & {} \,C\sum _{|k'- k|\le 1}\Vert {\dot{S}}_{k'-1}a_{\Phi }^+\Vert _{L^{\infty }(\mathcal {C})}\Vert {\dot{\Delta }}_{k'} b_{\Phi }\Vert \Vert {\dot{\Delta }}_k c_{\Phi }\Vert \nonumber \\\le & {} \,C\Vert \nabla _{\mathrm{h}} a_{\Phi }\Vert _{{\dot{B}}^{\frac{1}{2}}}\sum _{|k'- k|\le 1}\Vert {\dot{\Delta }}_{k'}b_{\Phi }\Vert \Vert {\dot{\Delta }}_kc_{\Phi }\Vert . \end{aligned}$$

(7.3)

Hence for any $c_0>0,$ there exists $C>0$ such that

$$\begin{aligned} \sum _{k\in \mathbb {Z}}\left( \int _0^t|({\dot{\Delta }}_k(T^{\mathrm{v}}_a b)_{\Phi },{\dot{\Delta }}_k c_{\Phi })|\,ds\right) ^{\frac{1}{2}}\le & {} c_0\sum _{k\in \mathbb {Z}}\left( \int _0^t\Vert {\dot{\Delta }}_k c_{\Phi }\Vert ^2\,ds\right) ^{\frac{1}{2}}\nonumber \\&+C\sum _{k\in \mathbb {Z}}\left( \int _0^t\Vert \nabla _{\mathrm{h}}a_{\Phi }\Vert ^2_{{\dot{B}}^{\frac{1}{2}}}\Vert {\dot{\Delta }}_k b_{\Phi }\Vert ^2\,ds\right) ^{\frac{1}{2}} . \end{aligned}$$

(7.4)

Similarly, by applying Bernstein’s inequality and Lemma 6.1, we find

$$\begin{aligned} |({\dot{\Delta }}_k (R^{\mathrm{v}}(a,b))_{\Phi },{\dot{\Delta }}_kc_{\Phi })| \,\le & {} \,C2^{\frac{k}{2}}\sum _{k'\ge k-3} \Vert \widetilde{{\dot{\Delta }}}_{k'}a^+_{\Phi }\Vert _{L^2_{x_3}(L^\infty (D_1))}\Vert {{\dot{\Delta }}}_{k'}b_{\Phi }\Vert \Vert {\dot{\Delta }}_kc_{\Phi }\Vert \nonumber \\\le & {} \,C2^{\frac{k}{2}}\sum _{k'\ge k-3} \Vert \nabla _{\mathrm{h}}\widetilde{{\dot{\Delta }}}_{k'}a_{\Phi }\Vert \Vert {\dot{\Delta }}_{k'}b_{\Phi }\Vert \Vert {\dot{\Delta }}_kc_{\Phi }\Vert \nonumber \\\le & {} \,C\sum _{k'\ge k-3} 2^{\frac{k-k'}{2}}\Vert \nabla _{\mathrm{h}} a_{\Phi }\Vert _{{\dot{B}}^{\frac{1}{2}}}\Vert {\dot{\Delta }}_{k'}b_{\Phi }\Vert \Vert {\dot{\Delta }}_kc_{\Phi }\Vert . \end{aligned}$$

(7.5)

Then we use the Minkowski inequality and the Hölder inequality to get

$$\begin{aligned}&\sum _{k\in \mathbb {Z}}\left( \int _0^t|({\dot{\Delta }}_k (R^{\mathrm{v}}(a,b))_{\Phi },{\dot{\Delta }}_kc_{\Phi })|\,ds\right) ^{\frac{1}{2}}\nonumber \\&\le c_0\sum _{k\in \mathbb {Z}}\left( \int _0^t\Vert {\dot{\Delta }}_k c_{\Phi }\Vert ^2\,ds\right) ^{\frac{1}{2}} +C\sum _{k\in \mathbb {Z}}\left( \int _0^t \left( \sum _{k'\ge k-3}2^{\frac{k-k'}{2}}\Vert \nabla _{\mathrm{h}}a_{\Phi }\Vert _{{\dot{B}}^{\frac{1}{2}}}\Vert {\dot{\Delta }}_{k'} b_{\Phi }\Vert \right) ^2\right) ^{\frac{1}{2}}\nonumber \\&\le c_0\sum _{k\in \mathbb {Z}}\left( \int _0^t\Vert {\dot{\Delta }}_k c_{\Phi }\Vert ^2\,ds\right) ^{\frac{1}{2}}+C\sum _{k\in \mathbb {Z}}\left( \int _0^t \Vert \nabla _{\mathrm{h}}a_{\Phi }\Vert ^2_{{\dot{B}}^{\frac{1}{2}}}\Vert {\dot{\Delta }}_{k} b_{\Phi }\Vert ^2_{}\,ds\right) ^{\frac{1}{2}}. \end{aligned}$$

(7.6)

For the last term, we observe that

$$\begin{aligned} |({\dot{\Delta }}_k(T_b^{\mathrm{v}} a)_{\Phi },{\dot{\Delta }}_k c_{\Phi })|\,\le & {} \, C\sum _{|k'-k|\le 1}\Vert {\dot{S}}_{k'-1}b_{\Phi }\Vert _{L^2_{x_\mathrm{h}}(L^{\infty }_{x_3})}\Vert {\dot{\Delta }}_{k'}a^+_{\Phi }\Vert _{L^{\infty }_{x_\mathrm{h}}(L^2_{x_3})}\Vert {\dot{\Delta }}_{k}c_{\Phi }\Vert \nonumber \\\le & {} \,C\sum _{|k'-k|\le 1}\sum _{\ell \le k'-2}2^{\frac{\ell }{2}}\Vert {\dot{\Delta }}_{\ell }b_{\Phi }\Vert \Vert \nabla _{\mathrm{h}}{\dot{\Delta }}_{k'}a_{\Phi }\Vert {\dot{\Delta }}_{k}c_{\Phi }\Vert \nonumber \\\le & {} \,C\sum _{\ell \le k-1} 2^{\frac{\ell -k}{2}}\Vert {\dot{\Delta }}_{\ell }b_{\Phi }\Vert \Vert \nabla _{\mathrm{h}}a_{\Phi }\Vert _{{\dot{B}}^{\frac{1}{2}}}\Vert {\dot{\Delta }}_{k}c_{\Phi }\Vert . \end{aligned}$$

Thus we get

$$\begin{aligned}&\sum _{k\in \mathbb {Z}}\left( \int _0^t |({\dot{\Delta }}_k(T_b^{\mathrm{v}} a)_{\Phi },{\dot{\Delta }}_k c_{\Phi })|\,ds\right) ^{\frac{1}{2}}\nonumber \\&\le c_0\sum _{k\in \mathbb {Z}}\left( \int _0^t\Vert {\dot{\Delta }}_k c_{\Phi }\Vert ^2\,ds\right) ^{\frac{1}{2}}+C\sum _{k\in \mathbb {Z}} \left( \int _0^t \left( \sum _{\ell \le k-1}2^{\frac{\ell -k}{2}}\Vert \nabla _{\mathrm{h}}a_{\Phi }\Vert _{{\dot{B}}^{\frac{1}{2}}}\Vert {\dot{\Delta }}_{l} b_{\Phi }\Vert \right) ^2\,ds\right) ^{\frac{1}{2}}\nonumber \\&\le c_0\sum _{k\in \mathbb {Z}}\left( \int _0^t\Vert {\dot{\Delta }}_k c_{\Phi }\Vert ^2\,ds\right) ^{\frac{1}{2}}+C\sum _{k\in \mathbb {Z}}\left( \int _0^t \Vert \nabla _{\mathrm{h}}a_{\Phi }\Vert ^2_{{\dot{B}}^{\frac{1}{2}}}\Vert {\dot{\Delta }}_{k} b_{\Phi }\Vert ^2\,ds\right) ^{\frac{1}{2}} . \end{aligned}$$

(7.7)

By combining the estimates (7.4),(7.6) and (7.7), we conclude the proof of the first part of Lemma 5.3. $\square$

Let us now turn to the case where $a=b$. First, observing from (7.2) that

$$\begin{aligned} \Vert S_{k-1}a^+_{\Phi }\Vert _{L^\infty (\mathcal {C})}\le C2^{\frac{k}{2}}\Vert \nabla _\mathrm{h}a_{\Phi }\Vert _{{\dot{B}}^{0}}, \end{aligned}$$

we get, by a similar derivation of (7.3), that

$$\begin{aligned} |({\dot{\Delta }}_k(T_a^{\mathrm{v}} a)_{\Phi },{\dot{\Delta }}_k c_{\Phi })|\le & {} \,C2^{\frac{k}{2}}\Vert \nabla _{\mathrm{h}} a_{\Phi }\Vert _{{\dot{B}}^0}\Vert {\dot{\Delta }}_kc_{\Phi }\Vert \sum _{|k'- k|\le 1}\Vert {\dot{\Delta }}_{k'}a_{\Phi }\Vert , \end{aligned}$$

which inplies

$$\begin{aligned}&\sum _{k\in \mathbb {Z}}\left( \int _0^t |({\dot{\Delta }}_k(T_a^{\mathrm{v}} a)_{\Phi },{\dot{\Delta }}_k c_{\Phi })|\,ds\right) ^{\frac{1}{2}}\nonumber \\&\quad \le c_0\sum _{k\in \mathbb {Z}}\left( \int _0^t\Vert {\dot{\Delta }}_k c_{\Phi }\Vert \,ds\right) ^{\frac{1}{2}}+C\sum _{k\in \mathbb {Z}}\left( \int _0^t2^k\Vert \nabla _{\mathrm{h}} a_{\Phi }\Vert _{{\dot{B}}^0}^2\Vert {\dot{\Delta }}_ka_{\Phi }\Vert ^2\,ds\right) ^{\frac{1}{2}}. \end{aligned}$$

(7.8)

We deduce from the third inequality of (7.5) that

$$\begin{aligned} |({\dot{\Delta }}_k(R^{\mathrm{v}}(a,a))_{\Phi },{\dot{\Delta }}_k c_{\Phi })|\le C \sum _{k'\ge k-3}2^{\frac{k}{2}}\Vert \nabla _{\mathrm{h}} a_{\Phi }\Vert _{{\dot{B}}^0}\Vert {\dot{\Delta }}_{k'}a_{\Phi }\Vert \Vert {\dot{\Delta }}_k c_{\Phi }\Vert . \end{aligned}$$

Then we can use the Minkowski inequality, again, to find that

$$\begin{aligned}&\sum _{k\in \mathbb {Z}}\left( \int _0^t |({\dot{\Delta }}_k(R^{\mathrm{v}}(a,a))_{\Phi },{\dot{\Delta }}_k c_{\Phi })|\,ds\right) ^{\frac{1}{2}}\nonumber \\&\quad \le c_0\sum _{k\in \mathbb {Z}}\left( \int _0^t\Vert {\dot{\Delta }}_k c_{\Phi }\Vert \,ds\right) ^{\frac{1}{2}} +C\sum _{k\in \mathbb {Z}}\left( \int _0^t 2^k\Vert \nabla _{\mathrm{h}} a_{\Phi }\Vert _{{\dot{B}}^0}^2\Vert {\dot{\Delta }}_ka_{\Phi }\Vert ^2\,ds\right) ^{\frac{1}{2}}, \end{aligned}$$

(7.9)

which finishes the proof of the second part of Lemma 5.3.$\square$