2.1 Elsässer variables under the restriction $\mu =\nu$

Since we focus on the case $\nu =\mu$ in this paper, the symmetry of the system (1.3) enables us to reformulate it by using the Elsässer variables

\[ w^\pm{=}\omega\mp j. \]

Then $w^{\pm }$ solves

\begin{align*} & \partial_tw^\pm+\bar{U}\partial_xw^\pm{\pm}\alpha\partial_xw^{{\pm}}-\nu\Delta w^\pm\mp2\bar{U}'\partial_{xy}{\Delta^{{-}1}(w^{-}-w^+)}-\bar{U}''\partial_x{\Delta^{{-}1}w^{{\mp}}}\nonumber\\ & \quad= - \nabla^\bot\Delta^{{-}1}w^{{\mp}}\cdot\nabla w^{{\pm}}+\partial_{xy}\Delta^{{-}1}w^{{\mp}}(2\partial_{xx}\Delta^{{-}1}w^{{\pm}}-w^{{\pm}})\nonumber\\ & \qquad-\partial_{xy}\Delta^{{-}1}w^{{\pm}}(2\partial_{xx}\Delta^{{-}1}w^{{\mp}}-w^{{\mp}}). \end{align*}

Such kind of variables have played important roles in the study of large time behaviours of solutions to MHD equations in the absence of shear flows [Reference Bardos, Sulem and Sulem2, Reference Cai and Lei9, Reference He, Xu, Li and Yu19, Reference Wei and Zhang43], and the nonlinear stability result [Reference Liss31] where the three dimensional Couette flow $(y,0,0)^{\top }$ is taken into consideration as mentioned in § 1.

Similar to the 3D case, the background magnetic field $(\alpha, 0)^\top$ introduces oscillations (see the $-\alpha \partial _xw^\pm$ terms in (2.1)) that may stabilize the system. Following [Reference Liss31], we define the profiles

\[ z^{{\pm}}=O^t_{{\pm}\alpha}w^\pm \]

to hide the oscillations in the new unknowns $z^{\pm }$, where $O^t_{\pm \alpha }$ is defined in (1.12). Then $z^{\pm }$ solves

(2.1)\begin{equation} \begin{aligned} & \partial_tz^\pm+\bar{U}\partial_xz^\pm - \nu\Delta z^\pm+\bar{U}'\partial_{xy}\Delta^{{-}1}\left(z^\pm - O^t_{\pm2\alpha}z^\mp\right)-\bar{U}''\partial_x\Delta^{{-}1}O^t_{\pm2\alpha}z^{{\mp}}\\ & \quad={-}\nabla^\bot\Delta^{{-}1}O^t_{\pm2\alpha}z^{{\mp}}\cdot\nabla z^{{\pm}}+\partial_{xy}\Delta^{{-}1}O^t_{\pm2\alpha}z^{{\mp}}(2\partial_{xx}\Delta^{{-}1}z^{{\pm}}-z^{{\pm}})\\ & \qquad-\partial_{xy}\Delta^{{-}1}z^{{\pm}}(2\partial_{xx}\Delta^{{-}1}O^t_{\pm2\alpha}z^{{\mp}}-O^t_{\pm2\alpha}z^{{\mp}}). \end{aligned} \end{equation}

2.2 Change of coordinates

In this paper we use the coordinate transform introduced by Bedrossian, Vicol and Wang in [Reference Bedrossian, Vicol and Wang8] to unwind the decaying background shear flow $\bar {U}(t,y)$ in (2.1):

(2.2)\begin{equation} \left\{ \begin{array}{@{}ll} X=x-t\bar{U}(t,y)\\ Y=\bar{U}(t,y). \end{array} \right. \end{equation}

Denote the spatial derivatives of the shear flow in the new coordinates as follows

(2.3)\begin{equation} \begin{cases} a(t,Y(t,y)) & =\partial_y\bar{U}(t,y),\\ b(t,Y(t,y)) & =\partial_{y}^2\bar{U}(t,y),\\ c(t,Y(t,y)) & =\partial_y^3\bar{U}(t,y),\\ d(t,Y(t,y)) & =\partial_y^4\bar{U}(t,y). \end{cases} \end{equation}

Note that by the chain rule, we have

(2.4)\begin{equation} b=a\partial_Y a. \end{equation}

For any function $\tilde {h}$ in the $(x,y)$ coordinates, the corresponding function $h$ in the $(X,Y)$ coordinates is given by

\[ h(t,X,Y)=\tilde{h}(t,x,y). \]

Then $\nabla \tilde h$ and $\Delta \tilde h$ can be rewritten in the new coordinate system (2.2) in terms of $h$ as follows (the notations $\nabla _t$ and $\Delta _t$ are introduced naturally):

(2.5)\begin{equation} \nabla\tilde{h}(t,x,y)=(\partial_{x}\tilde{h}, \partial_{y}\tilde{h})=(\partial_{X}h, a(\partial_{Y}-t\partial_{X})h)=(\partial_{X}^t h, \partial_{Y}^t h)=\nabla _t h, \end{equation}

and

(2.6)\begin{equation} \begin{aligned} \Delta \tilde{h}(t,x,y) & =(\partial_{XX}+a^2\partial_{YY}^L+b\partial_{Y}^L)h=\Delta_t h\\ & =\Delta_L h +\left((a^2-1)\partial_{YY}^L+b\partial_{Y}^L\right)h\\ & =\tilde{\Delta}_t h + b\partial_{Y}^L h, \end{aligned} \end{equation}

where we have used the notations

\[ \partial_Y^L=\partial_Y-t\partial_X,\quad \partial_{YY}^L=(\partial_Y-t\partial_X)^2, \]

and the modified Laplace operator is given by

(2.7)\begin{equation} \tilde{\Delta}_t = \Delta_L +(a^2-1)\partial_{YY}^L,\quad \mathrm{with}\quad \Delta_L=\partial_{XX}+\partial_{YY}^L. \end{equation}

Finally, using the fact $\partial _t \bar {U} =\nu \partial _y^2\bar {U}$ and the definitions of $a$ and $b$, we have

(2.8)\begin{equation} \partial_t{\tilde{h}}=\partial_t h +\partial_X h(-\bar{U}-t\partial_t \bar{U})+\partial_Y h \partial_t \bar{U} =\partial_t h-Y\partial_X h +\nu b\partial_Y^Lh.\end{equation}

In particular,

\[ \partial_t\partial_y\bar{U}=\partial_ta+\nu b\partial_Ya,\quad\mathrm{and}\quad \partial_t\partial_y^2\bar{U}=\partial_tb+\nu b\partial_Yb, \]

which, together with the fact $\partial _t\bar {U}=\nu \partial _y^2\bar {U}$ imply that

(2.9)\begin{equation} \partial_ta=\nu c-\nu b\partial_Ya,\quad\mathrm{and}\quad \partial_tb=\nu d-\nu b\partial_Yb. \end{equation}

2.3 Definition of $\Delta _t^{-1}$ and system (2.1) under new coordinates

In this subsection, we first give the definition of the inverse of $\Delta _t$ in the spirit of Antonelli, Dolce, and Marcati [Reference Antonelli, Dolce and Marcati1]. For the sake of completeness, we sketch the definitions below.

To begin with, assume that $a^2-1=(a^2-1)(t, Y)$ and $b=b(t, Y)$ are given functions in $L^\infty (\mathbb {R}^+; H^1(\mathbb {R}))$, then the fact that $\Delta _L^{-1}$ is well defined for $k\ne 0$ enables us to define an operator on $L^2(\mathbb {T}\times \mathbb {R})$:

(2.10)\begin{equation} \tilde \Lambda=\left((a^2-1)\partial_{YY}^L+b\partial^L_Y\right)(-\Delta_L)^{{-}1}, \end{equation}

with

(2.11)\begin{equation} \|\tilde \Lambda\|_{L^2\rightarrow L^2}\le C_* \left(\|a^2-1\|_{L^\infty H^1}+\|b\|_{L^\infty H^1}\right), \end{equation}

for some constant $C_*\ge 1$, see proposition 4.1 in [Reference Antonelli, Dolce and Marcati1] for more details.

Definition 2.1 Assume that $(\|a^2-1\|_{L^\infty H^1}+\|b\|_{L^\infty H^1})\le \delta$ such that $C_*\delta <1$, then for $k\ne 0$, let us define

(2.12)\begin{equation} \Delta_t^{{-}1}:=\Delta_L^{{-}1}\Lambda, \quad\mathrm{i. e.}\quad \Lambda\Delta_t=\Delta_L, \end{equation}

where

\[ \Lambda:=(I-\tilde \Lambda)^{{-}1}=\sum_{n=0}^\infty\tilde \Lambda^n. \]

Remark 2.2 For the operator $\Lambda$, we also have the following useful identity

(2.13)\begin{equation} \Lambda=I+\tilde \Lambda \Lambda. \end{equation}

Now we are in a position to rewrite (2.1) under the coordinates defined by (2.2). To this end, let us denote

\begin{align*} & Z^\pm(t,X,Y)=z^\pm(t, x,y),\\ & U^1_0(t, Y)=u^1_0(t,y),\\ & B^1_0(t, Y)=b^1_0(t,y). \end{align*}

Note that

(2.14)\begin{equation} \partial_{XY}^L\Delta_t^{{-}1}=\partial_{XY}^L\Delta_L^{{-}1}\Delta_L\Delta_t^{{-}1}=S\Lambda, \quad\mathrm{with}\quad S=\partial^L_{XY}\Delta_L^{{-}1}, \end{equation}

and

\[ \nabla^\bot\Delta^{{-}1}O^t_{\pm2\alpha}z^{{\mp}}_0\cdot\nabla z^{{\pm}}={-}\partial_y\partial_{yy}^{{-}1}z^\mp_0\partial_xz^\pm{=}{-}\partial_y\partial_{yy}^{{-}1}w^\mp_0\partial_xz^\pm{=}\left(u^1_0\pm b^1_0\right)\partial_xz^\pm_\ne. \]

Then it follows from (2.1) that the equations of $Z^\pm$ take the form of

(2.15)\begin{equation} \partial_tZ^\pm - \nu\tilde{\Delta}_t Z^\pm+a^2S\Lambda\left(Z^\pm - O^t_{\pm2\alpha}Z^\mp\right)-b\partial_X\Delta^{{-}1}_tO^t_{\pm2\alpha}Z^{{\mp}}=\mathrm{NL}^\pm, \end{equation}

or

(2.16)\begin{equation} \partial_tZ^\pm - \nu{\Delta}_L Z^\pm+S\Lambda\left(Z^\pm - O^t_{\pm2\alpha}Z^\mp\right)=\mathrm{LP}^\pm+\mathrm{NL}^\pm, \end{equation}

where

(2.17)\begin{equation} \mathrm{NL}^{{\pm}}:=\mathrm{NLT}^{{\pm}}+\mathrm{NLS1}^{{\pm}}+\mathrm{NLS2}^{{\pm}}, \end{equation}

with

\begin{align*} \mathrm{NLT}^{{\pm}}& ={-}\left(U^1_0\pm B^1_0\right)\partial_X Z^{{\pm}}_\ne-\nabla^\bot_t\Delta^{{-}1}_tO^t_{\pm2\alpha}Z^{{\mp}}_\ne\cdot\nabla_t Z^{{\pm}},\\ \mathrm{NLS1}^{{\pm}}& =aS\Lambda O^t_{\pm2\alpha}Z^{{\mp}}_\ne(2\partial_{XX}\Delta^{{-}1}_tZ^{{\pm}}_\ne-Z^{{\pm}}),\\ \mathrm{NLS2}^{{\pm}}& ={-}aS\Lambda Z^{{\pm}}_\ne(2\partial_{XX}\Delta^{{-}1}_tO^t_{\pm2\alpha}Z^{{\mp}}_\ne-O^t_{\pm2\alpha}Z^{{\mp}}), \end{align*}

and

(2.18)\begin{align} \mathrm{LP}^{{\pm}}:={-}(a^2-1)S\Lambda\left(Z^\pm - O^t_{\pm2\alpha}Z^\mp\right)+b\partial_X\Delta^{{-}1}_tO^t_{\pm2\alpha}Z^{{\mp}}+\nu(a^2-1)\partial_{YY}^LZ^\pm. \end{align}

Here ‘$\mathrm {NLT}$’, ‘$\mathrm {NLS}$’ and ‘$\mathrm {LP}$’ stand for ‘nonlinear transport’, ‘nonlinear stretch’ and ‘linear perturbation’, respectively.

2.4 Toy model and key ideas

Compared with the Navier–Stokes equations around 2D shear flows near Couette [Reference Bedrossian, Vicol and Wang8], and the MHD equations around 3D Couette flows [Reference Liss31], we will encounter new difficulties for the MHD equations around 2D shear flows near Couette. In order to track the difficulties precisely, let us consider the toy model

(2.19)\begin{equation} \partial_tf^\pm+S\Lambda f^\pm - \nu\Delta_Lf^\pm = 0, \end{equation}

or equivalently in Fourier variables

(2.20)\begin{equation} \partial_t\widehat{f^\pm}+\frac{k(\eta-kt)}{k^2+(\eta-kt)^2}\widehat{\Lambda f^\pm}-\nu\left(k^2+(\eta-kt)^2\right)\widehat{f^\pm}=0. \end{equation}

It is worth pointing out that if the shear flow is the Couette flow, i.e., $\bar {U}(t,y)=U(y)=y$, the operator $\Lambda$ in (2.20) will not appear. This scenario is reminiscent of the toy model introduced by Bedrossian, Germain and Masmoudi in [Reference Bedrossian, Germain and Masmoudi3] for the 3D Navier–Stokes equations near Couette:

\begin{align*} & \partial_t\hat{g}(t,k,\eta,l)+\frac{2k(\eta-kt)}{k^2+(\eta-kt)^2+l^2}\hat{g}(t, k,\eta,l)\\& \quad -\nu\left(k^2+(\eta-kt)^2+l^2\right)\hat{g}(t,k,\eta,l)=0. \end{align*}

To balance the interaction between $\frac {2k(\eta -kt)}{k^2+(\eta -kt)^2+l^2}\hat {g}$ and $-\nu (k^2+(\eta -kt)^2+l^2)\hat {g}$, the authors in [Reference Bedrossian, Germain and Masmoudi3] constructed a multiplier $m(t,k, \eta,l)$ satisfying

\[ \frac{\partial_tm}{m}=\frac{2k(\eta-kt)}{k^2+(\eta-kt)^2+l^2}\quad \mathrm{for} \quad t\in \left[\frac{\eta}{k}, \frac{\eta}{k}+1000\nu^{-\frac {1}{3}}\right]. \]

More precisely, the two dimensional analogue of the multiplier $m(t,k, \eta,l)$ is given as follows:

1. (1) if $k = 0: m(t,0,\eta ) = 1$;

2. (2) if $k \ne 0, \frac {\eta }{k} < -1000\nu ^{-\frac {1}{3}}: m(t,k,\eta ) = 1$;

3. (3) if $k \ne 0, -1000\nu ^{-\frac {1}{3}} < \frac {\eta }{k} < 0$:

   • • $m(t,k,\eta ) = \frac {k^2+\eta ^2}{k^2+(\eta -kt)^2}$ if $0 < t < \frac {\eta }{k}+1000\nu ^{-\frac {1}{3}}$,

   • • $m(t,k,\eta ) = \frac {k^2+\eta ^2}{k^2+(1000k\nu ^{-\frac {1}{3}})^2}$ if $t > \frac {\eta }{k}+1000\nu ^{-\frac {1}{3}}$;

4. (4) if $k \ne 0, \frac {\eta }{k} > 0$:

   • • $m(t,k,\eta ) = 1$ if $t < \frac {\eta }{k}$,

   • • $m(t,k,\eta ) = \frac {k^2}{k^2+(\eta -kt)^2}$ if $\frac {\eta }{k} < t < \frac {\eta }{k} + 1000\nu ^{-\frac {1}{3}}$,

   • • $m(t,k,\eta ) =\frac {k^2}{k^2+(1000k\nu ^{-\frac {1}{3}})^2}$ if $t > \frac {\eta }{k} + 1000\nu ^{-\frac {1}{3}}$.

Notice, in particular, that

(2.21)\begin{equation} \nu^{\frac{2}{3}}\lesssim m(t, k, \eta)\leq 1, \end{equation}

and

(2.22)\begin{equation} m(t, k, \eta)\gtrsim \frac{k^2}{k^2+(\eta-kt)^2}. \end{equation}

In our case, for the Couette flow, it suffices to use $m^{\frac {1}{2}}(t,k,\eta )$ instead of $m(t,k,\eta )$ to suppress the potential growth caused by the linear stretch term $\frac {k(\eta -kt)}{k^2+(\eta -kt)^2}\widehat {f^{\pm }}$. Nevertheless, for general shear flows the presence of the operator $\Lambda$ may amplify the linear stretch effect since the norm of $\Lambda$ may be larger than 1 even though the shear flow $\bar {U}$ is close to Couette. Roughly speaking, one can regard the linear stretch term in (2.20) as $(1+O(\tilde \delta ))\frac {k(\eta -kt)}{k^2+(\eta -kt)^2}\widehat {f^{\pm }}$ with $\tilde \delta >0$. Then it is natural to modify the multiplier $m^{\frac {1}{2}}(t,k,\eta )$ so as to suppress the extra growth resulting from the operator $\Lambda$. To this end, our strategy is to replace $m^{\frac {1}{2}}(t,k,\eta )$ with $\tilde {m}^{\frac {1}{2}}(t,k,\eta )$, where $\tilde m(t,k,\eta )$ is defined by

(2.23)\begin{equation} \tilde m(t, k, \eta)=m^{1+\frac {3}{2}\tilde\delta}(t, k,\eta), \end{equation}

with arbitrary $\tilde \delta >0$. Note that

(2.24)\begin{equation} \frac{\partial_t\tilde m(t, k,\eta)}{\tilde m(t, k,\eta)}=\left(1+\frac 32\tilde\delta\right)\frac{\partial_tm(t, k,\eta)}{m(t, k,\eta)}. \end{equation}

Then (2.21) and (2.22) reduce to

(2.25)\begin{equation} \nu^{\frac{2}{3}+\tilde\delta}\lesssim \tilde m(t, k, \eta)\leq 1, \end{equation}

and

(2.26)\begin{equation} \tilde m(t, k, \eta)\gtrsim \left(\frac{k^2}{k^2+(\eta-kt)^2}\right)^{1+\frac 32\tilde\delta}. \end{equation}

For $(t,k,\eta )\in [0,\infty )\times \mathbb {Z}\backslash \{0\}\times \mathbb {R}$, let us define the following three disjoint sets

\begin{align*} D_{dam}& =\left\{ (t,k,\eta): t<\frac{\eta}{k}\right\},\\ D_{dis}& =\left\{ (t,k,\eta): \frac{\eta}{k}<{-}1000\nu^{-\frac {1}{3}}\right\}\cup\left\{{-}1000\nu^{-\frac {1}{3}}<\frac{\eta}{k}, t\ge\frac{\eta}{k}+1000\nu^{-\frac {1}{3}}\right\},\\ D_{mul}& =\left\{(t,k,\eta): -1000\nu^{-\frac {1}{3}}<\frac{\eta}{k}\le0, t<\frac{\eta}{k}+1000\nu^{-\frac {1}{3}}\right\}\\& \quad \cup\left\{(t,k,\eta): \frac{\eta}{k}>0, \frac{\eta}{k}\le t<\frac{\eta}{k}+1000\nu^{-\frac {1}{3}}\right\}. \end{align*}

Then the effects of the linear stretch are summarized as follows:

1. (1) if $(t, k, \eta )\in D_{dam}$, then $k(\eta -kt)\ge 0$, and thus the linear stretch term behaves as a damping.

2. (2) if $(t, k, \eta )\in D_{dis}$, we have

   \[ \left|t-\frac{\eta}{k}\right|\ge1000\nu^{-\frac {1}{3}}, \]
   and thus
   (2.27)\begin{equation} \frac{|k(\eta-kt)|}{k^2+(\eta-kt)^2}\le \frac{1}{1000^3} \nu\left(k^2+(\eta-kt)^2\right), \end{equation}
   which means that the linear stretch is dominated by the dissipation.

3. (3) by the definition of $\tilde m$, there holds

   (2.28)\begin{equation} \frac{|k(\eta-kt)|}{k^2+(\eta-kt)^2}{\bf 1}_{D_{mul}}(t, k,\eta)=\frac{1}{1+\frac 32\tilde\delta}\frac{-\partial_t(\tilde m^{1/2})(t, k,\eta)}{\tilde m^{1/2}(t, k, \eta)}, \end{equation}
   that is to say, the effect of the linear stretch is balanced by the evolution of ${\tilde m}^\frac {1}{2}$ on $D_{mul}$.

Remark 2.3 Clearly, the following decomposition of unity holds

\begin{align*} 1={\bf 1}_{D_{dam}}(t, k,\eta)+{\bf 1}_{D_{dis}}(t, k,\eta)+{\bf 1}_{D_{mul}}(t, k,\eta), \quad \mathrm{for \ \ all} \\ (t,k,\eta)\in[0,\infty)\times\mathbb{Z}\backslash\{0\}\times\mathbb{R}. \end{align*}

Then it follows from (2.27) and (2.28) that

(2.29)\begin{align} & \frac{|k(\eta-kt)|}{k^2+(\eta-kt)^2}\\ & \quad=\frac{k(\eta-kt)}{k^2+(\eta-kt)^2}{\bf 1}_{D_{dam}}(t, k,\eta)+\frac{|k(\eta-kt)|}{k^2+(\eta-kt)^2}{\bf 1}_{D_{dis}}(t, k,\eta)\nonumber\\ & \qquad-\frac {1}{2}\frac{1}{1+\frac 32\tilde\delta}\frac{\dot {\tilde m}(t, k,\eta)}{\tilde m(t, k,\eta)}\nonumber\\ & \quad\le\frac{k(\eta-kt)}{k^2+(\eta-kt)^2}{\bf 1}_{D_{dam}}(t, k,\eta)+\frac{1}{1000^3} \nu\left(k^2+(\eta-kt)^2\right)\nonumber\\ & \qquad-\frac{1}{1+\frac 32\tilde\delta}\frac{\partial_t (\tilde m^{1/2})(t, k,\eta)}{\tilde m^{1/2}(t, k, \eta)}. \nonumber \end{align}

The treatment of the general shear flow $(\bar {U},0)^\top$ is much more complicated than that of the Couette flow $(y,0)^\top$. In fact, for the case $\bar {U}(t,y)=U(y)=y$, once the multiplier $m(t,k,\eta )$ is well defined as above, then it is straightforward to deal with the linear stretch term, see (3.24). In particular, the damping effect stemming from the linear stretch term when $(t,k,\eta )\in D_{dam}$ can be ignored. However, for the case that $U(y)$ is close to $y$, the appearance of the operator $\Lambda$ makes the damping effect of the linear stretch $\frac {k(\eta -kt)}{k^2+(\eta -kt)^2}\widehat {\Lambda f^\pm }$ unclear even though $(t,k,\eta )\in D_{dam}$. Our strategy is to isolate the damping effect by using (2.13):

(2.30)\begin{align} {\bf 1}_{D_{dam}}\frac{k(\eta-kt)}{k^2+(\eta-kt)^2}\widehat{\Lambda f^\pm}={\bf 1}_{D_{dam}}\frac{k(\eta-kt)}{k^2+(\eta-kt)^2}\widehat{ f^\pm}+{\bf 1}_{D_{dam}}\frac{k(\eta-kt)}{k^2+(\eta-kt)^2}\widehat{\tilde{\Lambda}\Lambda f^\pm}. \end{align}

The first term on the right hand side of (2.30) is the main part, and the second term is the perturbation. Some delicate commutator estimates will be performed to treat the perturbation. As a result, lots of errors appear. To close the estimates, the damping effect captured by the main part of the linear stretch in (2.30), together with the dissipation and other good terms, will be used to absorb the errors. See Step I of § 4.1 for more details.

In addition, the presence of the operator $\Lambda$ leads us to estimate the composition $B\circ \Lambda$ for some multiplier $B$ involved in this paper. The continuity of $B\circ \Lambda$ depends on the commutator estimates of $B$. That's why we give a collection of commutator estimates in Appendix B. In particular, some extra terms appear in the commutator estimate of $\sqrt {-\frac {\partial _t(\tilde m^\frac {1}{2})}{\tilde m^\frac {1}{2}}}\tilde K$, see lemma B.4. This motivates us to establish a composition inequality for the multipliers whose commutator estimates have extra errors with better commutator estimates, see lemma C.1 for the details.

On the other hand, it is worth pointing out that the above toy model (2.19) ignored the oscillation terms $O^t_{\pm 2\alpha }S\Lambda f^{\mp }$, which should be taken into consideration as well in the energy estimates. In order to take advantage of the oscillator $O^t_\alpha$, noting that

(2.31)\begin{equation} \widehat{O^t_\alpha}=\frac{1}{i\alpha k}\partial_t\widehat{O^t_\alpha}\quad \mathrm{for}\quad k\ne0, \end{equation}

integrating by parts with respect to the time variable will be exploited as that in [Reference Liss31]. In this way, we will inevitably encounter the time derivative of the operator $\Lambda$. To achieve this, using again (2.13), we obtain an important relation

(2.32)\begin{equation} \partial_t\Lambda=\Lambda\partial_t\tilde \Lambda\Lambda, \end{equation}

where $\partial _t\tilde \Lambda$ is derived from (2.10)

(2.33)\begin{align} \partial_t\tilde \Lambda\,{=}\left[2(a^2\,{-}\,1)S+b\partial_X\Delta^{{-}1}_{L}+2\tilde \Lambda S\right]\,{+}\,\left(2a\partial_ta\partial_{YY}^L\,{+}\,\partial_tb\partial^L_Y\right)(-\Delta_L)^{{-}1}=:\tilde \Lambda_t^1+\tilde\Lambda_t^2. \end{align}

See the estimates of $\mathrm {OLS}_5$ in Step II of § 4.1 for more details.

Finally, we would like to remark that, unlike [Reference Antonelli, Dolce and Marcati1], the coefficients $a$ and $b$ hidden in the definition of $\Delta _t^{-1}$ are time dependent (see (2.10) and (2.12)), and $\partial _ta$ and $\partial _tb$ are involved when performing integrating by parts in time (see (2.33)). Accordingly, recalling (2.3) and (2.9), we find that some higher derivatives of $\bar {U}$ are actually involved. This partially explains the extra regularities required in (1.8).

3.1 Proof of theorem 3.1

To prove theorem 3.1, it suffices to establish the following a priori estimates:

(3.19)\begin{equation} \|KZ^\pm\|_{L^\infty L^2}+\nu^\frac {1}{2}\|\nabla_LKZ^\pm\|_{L^2L^2}+\left\|\sqrt{-\frac{\dot{M}}{M}}KZ^\pm\right\|_{L^2L^2}\le8\epsilon, \end{equation}

and

(3.20)\begin{equation} \left\|u^1_0\mp b^1_0\right\|_{L^\infty H^N}+\nu^\frac {1}{2}\left\|\partial_y(u^1_0\mp b^1_0)\right\|_{L^2H^N}\le8\epsilon. \end{equation}

Indeed, (3.4) is a direct consequence of (3.11) and (3.19), and (3.5) is nothing but (3.20). To prove (3.6), by the definition of $M_2$, we have

(3.21)\begin{equation} 1\lesssim \nu^{-\frac 16}\left( \sqrt{-\frac{\dot{M_2}}{M_2}(t,k,\eta)}+\nu^{\frac {1}{2}}\left| k,\eta-kt\right| \right). \end{equation}

It follows that

(3.22)\begin{equation} \|KZ_{\ne} \|_{L^2L^2}\lesssim \nu^{-\frac 16}\left( \nu^{\frac {1}{2}}\left\| \nabla_L KZ_{\ne}\right\|_{L^2L^2}+\left\|\sqrt{-\frac{\dot{M}}{M}}KZ_{{\neq}} \right\|_{L^2L^2} \right). \end{equation}

Thus, (3.6) follows from (3.11), (3.19) and (3.22) immediately.

Next we will prove (3.19) and (3.20) by using the standard continuity method. First of all, the local well-posedness of the 2D MHD equations in $H^N$ ensures that there exists a $T_0>0$, such that

\[ \|KZ^\pm\|_{L^\infty(0,T_0;L^2)}+\nu^\frac {1}{2}\|\nabla_LKZ^\pm\|_{L^2(0,T_0;L^2)}+\left\|\sqrt{-\frac{\dot{M}}{M}}KZ^\pm\right\|_{L^2(0,T_0;L^2)}\leq2\epsilon, \]

and

\[ \left\|u^1_0\mp b^1_0\right\|_{L^\infty (0,T_0;H^N)}+\nu^\frac {1}{2}\left\|\partial_y(u^1_0\mp b^1_0)\right\|_{L^2(0,T_0;H^N)}\leq2\epsilon. \]

Then define $T^* \le \infty$ to be the maximum of all time $T$ such that (3.19) and (3.20) hold on $[0,T]$. By the continuity, $T^* > T_0$.

We are left to prove that the constant 8 on the right of (3.19) and (3.20) can be replaced by 4, which implies that $T^* = \infty$. In fact, we have the following propositon.

The proof of proposition 3.3 will be achieved in the following two subsections.

3.2 Improvement of (3.19)

Recalling the definition of the multiplier $K$ in (3.10), from (3.2), we derive the following energy identity:

(3.23)\begin{align} & \frac{1}{2}\|KZ^+(t)\|^2_{L^2}+\nu\|\nabla_LKZ^+\|^2_{L^2L^2}\nonumber\\ & \qquad+\left\|\sqrt{-\frac{\dot{M}}{M}}KZ^+\right\|^2_{L^2L^2}+\left\|\sqrt{-\frac{\partial_t(m^{1/2})}{m^{1/2}}}KZ^+\right\|^2_{L^2L^2}\nonumber\\ & \quad=\frac {1}{2}\|KZ^+(0)\|^2_{L^2}-\int_0^t\left\langle SKZ^+, KZ^+\right\rangle {\rm d}t'+\int_0^t\left\langle SO^t_{2\alpha}KZ^-, KZ^+\right\rangle {\rm d}t'\nonumber\\ & \qquad+\int_0^t\left\langle KZ^+, K\left(\partial_tZ^+\right)_\mathcal{NL}\right\rangle {\rm d}t'\nonumber\\ & \quad=\frac {1}{2}\|KZ^+(0)\|^2_{L^2}+\mathrm{LS}+\mathrm{OLS}+\mathcal{NL}. \end{align}

By the definition of $m$, we have

(3.24)\begin{equation} \mathrm{LS}\le\left\|\sqrt{-\frac{\partial_t(m^{1/2})}{m^{1/2}}}KZ^+\right\|^2_{L^2L^2}+\frac{\nu}{2}\|\nabla_LKZ^+\|^2_{L^2L^2}. \end{equation}

Thanks to the fact (2.31), one can estimate $\mathrm {OLS}$ by integrating by parts in time:

(3.25)\begin{equation} \mathrm{OLS}=\sum_{i=1}^5\mathrm{OLS}_i, \end{equation}

where

\begin{align*} \mathrm{OLS}_1& =\frac{1}{2\alpha}\left\langle S\partial_X^{{-}1}O^t_{2\alpha}KZ^-_\neq(t), KZ^+(t)\right\rangle, \nonumber\\ \mathrm{OLS}_2& ={-}\frac{1}{2\alpha}\left\langle S\partial_X^{{-}1}O^t_{2\alpha}KZ^-_\neq(0), KZ^+(0)\right\rangle, \nonumber\\ \mathrm{OLS}_3& ={-}\frac{1}{2\alpha}\int_0^t\left\langle \left(S\frac{2\dot{K}}{K}+\dot{S}\right)\partial_X^{{-}1}O^t_{2\alpha}KZ^-_\neq, KZ^+\right\rangle {\rm d}t',\nonumber\\ \mathrm{OLS}_4& ={-}\frac{1}{2\alpha}\int_0^t\left\langle S\partial_X^{{-}1}O^t_{2\alpha}KZ^-_\neq, K\partial_tZ^+\right\rangle {\rm d}t',\nonumber\\ \mathrm{OLS}_5& ={-}\frac{1}{2\alpha}\int_0^t\left\langle S\partial_X^{{-}1}O^t_{2\alpha}K\partial_tZ^-_\neq, KZ^+\right\rangle {\rm d}t'. \end{align*}

Clearly,

(3.26)\begin{align} \mathrm{OLS}_1+\mathrm{OLS}_2\le\frac{1}{2|\alpha|}\left(\|KZ^-_{{\neq}}(t)\|_{L^2}\|KZ^+_{{\neq}}(t)\|_{L^2}+\|KZ^-_{{\neq}}(0)\|_{L^2}\|KZ^+_{{\neq}}(0)\|_{L^2}\right). \end{align}

To bound $\mathrm {OLS}_3$, in Fourier variables, we find that

(3.27)\begin{equation} \frac{\dot{K}}{K}=\frac{\partial_t(m^{1/2})}{m^{1/2}}+\frac{\dot{M}}{M}=\frac {1}{2}\frac{\dot{m}}{m}+\frac{\dot{M}}{M}. \end{equation}

On the support of $\dot {m}$, there holds

(3.28)\begin{equation} \frac{\dot{m}}{m}=\frac{2k(\eta-kt)}{k^2+(\eta-kt)^2}=2S(t, k, \eta). \end{equation}

Thus,

(3.29)\begin{equation} \left|\frac{\partial_t(m^{1/2})}{m^{1/2}}\right|+|S|\le\frac{|k(\eta-kt)|}{k^2+(\eta-kt)^2}\le\min\left\{\frac {1}{2}, \sqrt{-\frac{\dot{M}_1}{M_1}}\right\}. \end{equation}

Moreover,

(3.30)\begin{equation} |\dot{S}|\le\frac{k^2}{k^2+(\eta-kt)^2}={-}\frac{\dot{M}_1}{M_1}\le-\frac{\dot{M}}{M}. \end{equation}

It follows that

(3.31)\begin{equation} \mathrm{OLS}_3\le\frac{2}{\left|\alpha\right|}\left\|\sqrt{-\frac{\dot{M}}{M}}KZ^-_{{\neq}}\right\|_{L^2L^2}\left\|\sqrt{-\frac{\dot{M}}{M}}KZ^+\right\|_{L^2L^2}. \end{equation}

As for $\mathrm {OLS}_4$, in view of (3.2), we have

(3.32)\begin{equation} \mathrm{OLS}_4=\sum_{i=1}^4\mathrm{OLS}_4^{(i)}, \end{equation}

with

\begin{align*} \mathrm{OLS}_4^{(1)}& ={-}\frac{\nu}{2\alpha}\int_0^t\left\langle S\partial_X^{{-}1}O^t_{2\alpha}KZ^-_{{\neq}}, K\Delta_L Z^+\right\rangle {\rm d}t',\\ \mathrm{OLS}_4^{(2)}& =\frac{1}{2\alpha}\int_0^t\left\langle S\partial_X^{{-}1}O^t_{2\alpha}KZ^-_{{\neq}}, KSZ^+\right\rangle {\rm d}t',\\ \mathrm{OLS}_4^{(3)}& ={-}\frac{1}{2\alpha}\int_0^t\left\langle S\partial_X^{{-}1}O^t_{2\alpha}KZ^-_{{\neq}}, KO^t_{2\alpha}Z^{-}_{{\neq}}\right\rangle {\rm d}t',\\ \mathrm{OLS}_4^{(4)}& ={-}\frac{1}{2\alpha}\int_0^t\left\langle S\partial_X^{{-}1}O^t_{2\alpha}KZ^-_{{\neq}}, K\left(\partial_tZ^+\right)_{\mathcal{NL}}\right\rangle {\rm d}t'. \end{align*}

Integrating by parts, and using the fact $|S|\le \frac {1}{2}$, we have

\begin{align*} \mathrm{OLS}_4^{(1)}& =\frac{\nu}{2\alpha}\int_0^t\left\langle S\partial_X^{{-}1}O^t_{2\alpha}\nabla_LKZ^-_{{\neq}}, \nabla_L KZ^+\right\rangle {\rm d}t'\nonumber\\ & \le\frac{\nu}{4\left|\alpha\right|}\|\nabla_LKZ^-_{{\neq}}\|_{L^2L^2}\|\nabla_LKZ^+_{{\neq}}\|_{L^2L^2}. \end{align*}

Thanks to the fact $|S|\le \sqrt {-\frac {\dot {M}_1}{M_1}}$, one deduces that

\[ \mathrm{OLS}_4^{(2)}\le\frac{1}{2\left|\alpha\right|}\left\|\sqrt{-\frac{\dot{M}}{M}}KZ_{\ne}^+\right\|_{L^2L^2}\left\|\sqrt{-\frac{\dot{M}}{M}}KZ_{\ne}^-\right\|_{L^2L^2}. \]

Owing to the periodicity in $X$ variable, we find that

(3.33)\begin{equation} \mathrm{OLS}_4^{(3)}={-}\frac{1}{4\alpha}\int_0^t\int_{\mathbb{T}\times\mathbb{R}} \partial_X\left(S\partial_X^{{-}1}O^t_{2\alpha}KZ^-_\neq\right)^2 dXdYdt'=0.\end{equation}

To bound $\mathrm {OLS}_4^{(4)}$, by (3.2), we get

(3.34)\begin{align} \mathrm{OLS}_4^{(4)}& =\frac{1}{2\alpha}\int_0^t\left\langle S\partial_X^{{-}1}O^t_{2\alpha}KZ^-_{{\neq}}, K\left(\nabla^\bot_L\Delta^{{-}1}_{L}O^t_{2\alpha}Z^-{\cdot}\nabla_LZ^+\right)\right\rangle {\rm d}t'\nonumber\\ & \quad-\frac{1}{2\alpha}\int_0^t\left\langle S\partial_X^{{-}1}O^t_{2\alpha}KZ^-_{{\neq}}, K\left(SO^t_{2\alpha}Z^{-}(2\partial_{XX}\Delta^{{-}1}_LZ^+{-}Z^+)\right)\right\rangle {\rm d}t'\nonumber\\ & \quad+\frac{1}{2\alpha}\int_0^t\left\langle S\partial_X^{{-}1}O^t_{2\alpha}KZ^-_{{\neq}}, K\left(SZ^+(2\partial_{XX}\Delta^{{-}1}_LO^t_{2\alpha}Z^{-}-O^t_{2\alpha}Z^{-})\right)\right\rangle {\rm d}t'\nonumber\\ & =\mathcal{NLT}+\mathcal{NLS}1+\mathcal{NLS}2, \end{align}

with

\begin{align*} \mathcal{NLT}& =\frac{1}{2\alpha}\int_0^t\left\langle S\partial_X^{{-}1}O^t_{2\alpha}KZ^-_\neq, K\left(\nabla^\bot_L\Delta^{{-}1}_{L}O^t_{2\alpha}Z^-_\neq{\cdot}\nabla_LZ^+_\neq\right)\right\rangle {\rm d}t'\nonumber\\ & \quad+\frac{1}{2\alpha}\int_0^t\left\langle S\partial_X^{{-}1}O^t_{2\alpha}KZ^-_\neq, K\left(\nabla^\bot_L\Delta^{{-}1}_{L}O^t_{2\alpha}Z^-_0\cdot\nabla_LZ^+_\neq\right)\right\rangle {\rm d}t'\nonumber\\ & \quad+\frac{1}{2\alpha}\int_0^t\left\langle S\partial_X^{{-}1}O^t_{2\alpha}KZ^-_\neq, K\left(\nabla^\bot_L\Delta^{{-}1}_{L}O^t_{2\alpha}Z^-_\neq{\cdot}\nabla_LZ^+_0\right)\right\rangle {\rm d}t'\nonumber\\ & \quad=\mathcal{NLT}(\ne,\ne)+\mathcal{NLT}(0,\ne)+\mathcal{NLT}(\ne,0). \end{align*}

By using lemma 3.2, we are led to

(3.35)\begin{align} & \mathcal{NLT}(\ne,\ne)\nonumber\\ & \lesssim\frac{1}{\left|\alpha\right|}\|KZ^-_{{\neq}}\|_{L^\infty L^2}\|\nabla^\bot_L\Delta^{{-}1}_{L}O^t_{2\alpha}Z^-_{{\neq}}\cdot\nabla_LZ^+_{{\neq}}\|_{L^1H^N}\nonumber\\ & \lesssim\frac{1}{\left|\alpha\right|}\|KZ^-_{{\neq}}\|_{L^\infty L^2}\left(\|KZ^-_{{\neq}}\|_{L^2L^2}+\nu^{-\frac {1}{3}}\left\|\sqrt{-\frac{\dot{M}}{M}}KZ^-_{{\neq}}\right\|_{L^2L^2}\right)\|\nabla_LKZ^+_{{\neq}}\|_{L^2L^2}\nonumber\\ & \lesssim\nu^{-\frac 56}\epsilon^3. \end{align}

In view of (3.16), one deduces that

(3.36)\begin{align} \mathcal{NLT}(\ne,0)& =\frac{1}{2\alpha}\int_0^t\left\langle S\partial_X^{{-}1}O^t_{2\alpha}KZ^-_{{\neq}}, K\left(\partial_X\Delta^{{-}1}_{L}O^t_{2\alpha}Z^-_{{\neq}}\partial_YZ^+_0\right)\right\rangle {\rm d}t'\nonumber\\ & \lesssim\frac{1}{\left|\alpha\right|}\|KZ^-_{{\neq}}\|_{L^\infty L^2}\left\|\sqrt{-\frac{\dot{M}}{M}}KZ^-_{{\neq}}\right\|_{L^2L^2}\|\partial_YZ^+_0\|_{L^2H^N}\nonumber\\ & \lesssim\nu^{-\frac {1}{2}}\epsilon^3. \end{align}

Recalling that $z^{\pm }=O^t_{\pm \alpha }w^{\pm }$, we then have $z^\pm _0=w^\pm _0=\omega _0\mp j_0$. Consequently,

\[ \partial_yz^\pm_0=\partial_y\omega_0\mp\partial_yj_0={-}\left(\partial_{yy}u^1_0\mp\partial_{yy}b^1_0\right), \]

and hence

(3.37)\begin{equation} \partial_y\partial_{yy}^{{-}1}z^\pm_0={-}\left(u^1_0\mp b^1_0\right). \end{equation}

Using (3.13), we arrive at

(3.38)\begin{align} \mathcal{NLT}(0,\ne)& ={-}\frac{1}{2\alpha}\int_0^t\left\langle S\partial_X^{{-}1}O^t_{2\alpha}KZ^-_{{\neq}}, K\left(\partial_Y\partial_{YY}^{{-}1}O^t_{2\alpha}Z^-_0\partial_XZ^+_{{\neq}}\right)\right\rangle {\rm d}t'\nonumber\\ & \lesssim\frac{1}{\left|\alpha\right|}\|KZ^-_{{\neq}}\|_{L^2L^2}\|\partial_Y\partial_{YY}^{{-}1}Z^-_0\|_{L^\infty H^N}\|\partial_XZ^+_{{\neq}}\|_{L^2H^N}\nonumber\\ & \lesssim\frac{1}{\left|\alpha\right|}\|KZ^-_{{\neq}}\|_{L^2L^2}\|u^1_0+b^1_0\|_{L^\infty H^N}\|\nabla_LKZ^+_{{\neq}}\|_{L^2L^2}\nonumber\\ & \lesssim\nu^{-\frac 23}\epsilon^3. \end{align}

$\mathcal {NLS}1$ and $\mathcal {NLS}2$ can be treated in the same way, we only estimate $\mathcal {NLS}2$ now. To this end, we infer from (2.22) that

(3.39)\begin{equation} |S|\le\frac{|k(\eta-kt)|}{k^2+(\eta-kt)^2}\le\frac{|k|}{\sqrt{k^2+(\eta-kt)^2}}\lesssim m^{1/2}. \end{equation}

This, together with (2.21) and the obvious fact $|\partial _{XX}\Delta _L^{-1}|\le 1$, implies that

(3.40)\begin{align} \mathcal{NLS}2& \lesssim\frac{1}{\left|\alpha\right|}\|KZ^-_{{\neq}}\|_{L^\infty L^2}\|SZ^+(2\partial_{XX}\Delta^{{-}1}_LO^t_{2\alpha}Z^{-}-O^t_{2\alpha}Z^{-})\|_{L^1H^N}\nonumber\\ & \lesssim\frac{1}{\left|\alpha\right|}\|KZ^-_{{\neq}}\|_{L^2 L^2}\|m^{1/2}Z^+\|_{L^2H^N}\left(\nu^{-\frac {1}{3}}\|m^{1/2}Z^-\|_{L^\infty H^N}\right)\nonumber\\ & \lesssim\nu^{-\frac 23}\epsilon^3. \end{align}

Note that $\mathrm {OLS}_5$ can be treated in the same manner as $\mathrm {OLS}_4$, and $\mathcal {NL}$ in (3.23) can be treated in the same way as $\mathrm {OLS}_4^{(4)}$. On the other hand, one can obtain an energy identity for $KZ^-$ similar to (3.23), and estimate the corresponding right hand side terms analogously. Putting the energy estimates for $KZ^+$ and $KZ^-$ together, we find that for sufficiently large $|\alpha |$, $\mathrm {LS}$, $\mathrm {OLS}_1$, $\mathrm {OLS}_3$, $\mathrm {OLS}_4^{(1)}$, $\mathrm {OLS}_4^{(2)}$ can be obsorbed by the left hand. In conclusion, there exist a universal constant $\alpha _0>0$, and a positive constant $C$ depending only on $N$ and $\alpha$, such that if $\left |\alpha \right | > \alpha _0$, we have

(3.41)\begin{align} \|KZ^\pm(t)\|^2_{L^2}\,{+}\,\nu\|\nabla_LKZ^\pm\|^2_{L^2L^2}\,{+}\,\left\|\sqrt{-\frac{\dot{M}}{M}}KZ^\pm\right\|^2_{L^2L^2}\le2\|KZ^\pm(0)\|^2_{L^2}+C\nu^{-\frac 56}\epsilon^3, \end{align}

which suffices to improve (3.19) as long as $\epsilon \ll \nu ^\frac 56$.

3.3 Improvement of (3.20)

Since $u^2_0=b^2_0=0$, we derive from (1.2) for $\bar {U}(t,y)=U(y)=y$ that $(u^1_0, b^1_0)$ solves

\[ \begin{cases} \partial_tu^1_0-\nu\partial_{yy}u^1_0={-}\left(u\cdot\nabla u^1-b\cdot\nabla b^1\right)_0,\\ \partial_tb^1_0-\nu\partial_{yy}b^1_0={-}\left(u\cdot\nabla b^1-b\cdot\nabla u^1\right)_0. \end{cases} \]

Accordingly, in the coordinate system (3.1), we have

(3.42)\begin{equation} \partial_t\left(U^1_0\mp B^1_0\right)-\nu\partial_{YY}\left(U^1_0\mp B^1_0\right)={-}\left((U_\neq{\pm} B_\neq)\cdot\nabla_L (U^1_\neq{\mp} B^1_\neq)\right)_0. \end{equation}

Then

(3.43)\begin{align} & \frac {1}{2}\left\|(U^1_0\mp B^1_0)(t)\right\|_{H^N}^2+\nu\|\partial_Y(U^1_0\mp B^1_0)\|^2_{L^2H^N}\nonumber\\ & \quad =\frac {1}{2}\left\|(U^1_0\mp B^1_0)(0)\right\|_{H^N}^2\nonumber\\& \qquad -\int_0^t\left\langle U^1_0\mp B^1_0, \left((U_\neq{\pm} B_\neq)\cdot\nabla_L (U^1_\neq{\mp} B^1_\neq)\right)_0\right\rangle_{H^N}{\rm d}t'. \end{align}

Thanks to the divergence free condition, we have

(3.44)\begin{align} & -\int_0^t\left\langle U^1_0\mp B^1_0, \left((U_\neq{\pm} B_\neq)\cdot\nabla_L (U^1_\neq{\mp} B^1_\neq)\right)_0\right\rangle_{H^N}{\rm d}t'\nonumber\\ & \quad=\int_0^t\left\langle \partial_Y(U^1_0\mp B^1_0), \left((U_\neq^2\pm B_\neq^2) (U^1_\neq{\mp} B^1_\neq)\right)_0\right\rangle_{H^N}{\rm d}t'\nonumber\\ & \quad\lesssim\left\|\partial_Y(U^1_0\mp B^1_0)\right\|_{L^2H^N}\left\|U_\neq^2\pm B_\neq^2\right\|_{L^2H^N}\left\|U^1_\neq{\mp} B^1_\neq\right\|_{L^\infty H^N}. \end{align}

Note that

\[ u_\neq^2\pm b_\neq^2=\partial_x\Delta^{{-}1}(\omega\pm j)=\partial_x\Delta^{{-}1}w^\mp{=}\partial_x\Delta^{{-}1}O^t_\alpha z^\mp, \]

and

\[ u^1_\neq{\mp} b^1_\neq{=}{-}\partial_y\Delta^{{-}1}(\omega_\ne\mp j_\ne)={-}\partial_y\Delta^{{-}1}w^\pm_\ne={-}\partial_y\Delta^{{-}1}O ^t_{-\alpha}z^\pm_\ne. \]

Then in view of (3.14) and (3.16), we find that

(3.45)\begin{equation} \left\|U_\neq^2\pm B_\neq^2\right\|_{L^2H^N}= \left\|\partial_X\Delta_L^{{-}1}O^t_\alpha Z^\mp \right\|_{L^2H^N}\lesssim\left\|\sqrt{-\frac{\dot{M}}{M}}KZ^\mp\right\|_{L^2L^2}, \end{equation}

and

(3.46)\begin{equation} \|U^1_\neq{\mp} B^1_\neq\|_{L^\infty H^N}= \left\|\partial_Y^L\Delta^{{-}1}_LO^t_{-\alpha}Z^\pm_\ne\right\|_{L^\infty H^N}\lesssim\left\|KZ^\pm_\ne\right\|_{L^\infty L^2}. \end{equation}

Substituting (3.44)–(3.46) into (3.43), noting that the coordinate system (3.1) is the same as the original system in $y$ variable, and using the hypotheses (3.19) and (3.20), we are led to

(3.47)\begin{align} & \frac {1}{2}\left\|(u^1_0\mp b^1_0)(t)\right\|_{H^N}^2+\nu\left\|\partial_y(u^1_0\mp b^1_0)\right\|^2_{L^2H^N}\nonumber\\ & \quad\le\frac {1}{2}\left\|(u^1_0\mp b^1_0)(0)\right\|_{H^N}^2+C\left\|\partial_y(u^1_0\mp b^1_0)\right\|_{L^2H^N}\left\|\sqrt{-\frac{\dot{M}}{M}}KZ^-_{{\neq}}\right\|_{L^2L^2}\left\|KZ^+_{{\neq}}\right\|_{L^\infty L^2}\nonumber\\ & \quad\le\frac {1}{2}\|(u^1_0\mp b^1_0)(0)\|_{H^N}^2+C\nu^{-\frac {1}{2}}\epsilon^3, \end{align}

which is sufficient to improve (3.20) provided $\epsilon \ll \nu ^{\frac {1}{2}}$. Combining (3.41) with (3.47), we complete the proof of proposition 3.3 and hence of theorem 3.1.

4.1 Proof of theorem 4.1

The proof of theorem 4.1 is similar to that of theorem 3.1. By the definition of $M_2$, (3.22) still holds with $K$ replaced by $\tilde K$, it suffices to establish the following a priori estimates:

(4.10)\begin{equation} \|\tilde KZ^\pm\|_{L^\infty L^2}+\nu^\frac {1}{2}\|\nabla_L \tilde KZ^\pm\|_{L^2L^2}+\left\|\sqrt{-\frac{\dot{M}}{M}}\tilde KZ^\pm\right\|_{L^2L^2}\le8\epsilon, \end{equation}

and

(4.11)\begin{equation} \left\|{U}^1_0\mp {B}^1_0\right\|_{L^\infty H^N}+\nu^\frac {1}{2}\left\|\partial_y({U}^1_0\mp {B}^1_0)\right\|_{L^2H^N}\le8\epsilon. \end{equation}

Let us define $T^*$ to be the end point of the largest interval $[0,T]$ such that (4.10) and (4.11) hold for all $0\leq t\leq T$. We are left to establish the following proposition.

Proof. We first improve (4.10). Similar to (3.23), from (2.16) and the definition of $\tilde K$ in (4.2), we have the following energy identity:

(4.12)\begin{align} & \frac {1}{2}\|\tilde KZ^+(t)\|^2_{L^2}+\nu\|\nabla_L\tilde KZ^+\|^2_{L^2L^2}+\left\|\sqrt{-\frac{\dot{M}}{M}}\tilde KZ^+\right\|^2_{L^2L^2}\nonumber\\ & \qquad+\left\|\sqrt{-\frac{\partial_t(\tilde m^{1/2})}{\tilde m^{1/2}}}\tilde KZ^+\right\|^2_{L^2L^2}\nonumber\\ & \quad=\frac {1}{2}\|\tilde KZ^+(0)\|^2_{L^2}-\int_0^t\left\langle \tilde KS\Lambda Z^+, \tilde KZ^+\right\rangle {\rm d}t'+\int_0^t\left\langle \tilde KO^t_{2\alpha}S\Lambda Z^-, \tilde KZ^+\right\rangle {\rm d}t'\nonumber\\ & \qquad+\int_0^t \left\langle \tilde KZ^+, \tilde K\mathrm{LP}^+\right\rangle {\rm d}t'+\int_0^t\left\langle \tilde KZ^+, \tilde K\mathrm{NL}^+\right\rangle {\rm d}t'\nonumber\\ & \quad=\frac {1}{2}\|\tilde KZ^+(0)\|^2_{L^2}+\mathrm{LS}+\mathrm{OLS}+\mathcal{LP}+\mathcal{NL}. \end{align}

The improvement of (4.10) will be achieved by the following four steps.

Step I: estimates of $\mathrm {LS}$. We first split $\mathrm {LS}$ into two parts:

(4.13)\begin{align} \mathrm{LS}& ={-}\frac{1}{4\pi^2}\sum_{k\neq0}\iint\tilde K(k,\eta) \frac{k(\eta-kt)}{k^2+(\eta-kt)^2}\left({\bf 1}_{D_{dam}}(t, k,\eta)+{\bf 1}_{D_{dam}^c}(t, k,\eta)\right)\nonumber\\ & \quad\times\widehat{\Lambda Z^+_{{\neq}}}(k, \eta) \tilde K(k,\eta)\bar{\hat{Z}}^+_{{\neq}}(k,\eta) d\eta {\rm d}t\nonumber\\ & =\mathrm{LS}^{dam}+\mathrm{LS}^{*}. \end{align}

Thanks to (2.13), one can split $\mathrm {LS}^{dam}$ into two parts

\begin{align*} \mathrm{LS}^{dam}& ={-}\left\| \sqrt{\left|S\right|_d}\tilde K Z_{{\neq}}^+\right\|^2_{L^2L^2}\nonumber\\ & \quad-\frac{1}{4\pi^2}\sum_{k\neq0}\iint \tilde K(k,\eta) \frac{|k(\eta-kt)|}{k^2+(\eta-kt)^2}{\bf 1}_{D_{dam}}\nonumber\\ & \quad\times \widehat{\tilde \Lambda \Lambda Z_{{\neq}}^+}(k, \eta)\tilde K(k,\eta)\bar{\hat{Z}}^+_{{\neq}}(k,\eta) d\eta {\rm d}t\nonumber\\ & =:\mathrm{LS}^{dam}_1+\mathrm{LS}_{2}^{dam}. \end{align*}

By the definition of $\tilde \Lambda$ in (2.10), there holds

\begin{align*} & \left|\widehat{\tilde\Lambda\Lambda Z_{{\neq}}^+}\right|(k,\eta)\nonumber\\ & \quad\le\int_\xi\left(|\widehat{a^2-1}|(\eta-\xi)\frac{(\xi-kt)^2}{k^2+(\xi-kt)^2}+|\hat{b}|(\eta-\xi)\frac{|\xi-kt|}{k^2+(\xi-kt)^2}\right)|\widehat{\Lambda Z_{{\neq}}^+}|(k,\xi)d\xi\nonumber\\ & \quad\le\int_\xi\left(|\widehat{a^2-1}|(\eta-\xi)+|\hat{b}|(\eta-\xi)\right)|\widehat{\Lambda Z_{{\neq}}^+}|(k,\xi)d\xi. \end{align*}

Combining this with (B.8), (B.11), and (A.3) yields

(4.14)\begin{align} & \mathrm{LS}_{2}^{dam}\nonumber\\ & \quad\le C\sum_{k\ne0}\iiint\langle\eta-\xi\rangle^{N+(1+\frac {3}{2}\tilde\delta)+\frac {3}{2}}\left(|\widehat{a^2-1}|(\eta-\xi)+|\hat{b}|(\eta-\xi)\right)|\widehat{\tilde K\Lambda Z^+_{{\neq}}}|(k,\xi)\nonumber\\ & \qquad\times\left(\sqrt{\frac{|k(\xi-kt)|}{k^2+(\xi-kt)^2}}+\sqrt{\frac{|k|}{k^2+(\xi-kt)^2}}\right)\nonumber\\ & \qquad\times \sqrt{\frac{k(\eta-kt)}{k^2+(\eta-kt)^2}}{\bf 1}_{D_{dam}}|\widehat{\tilde KZ^+_{{\neq}}}|(k,\eta)d\xi d\eta {\rm d}t\nonumber\\ & \quad\le C\delta\left(\left\|\sqrt{\left|S\right|_d}\tilde K \Lambda Z^+_{{\neq}}\right\|_{L^2L^2}+\left\|\sqrt{-\frac{\dot M_1}{M_1}}\tilde K \Lambda Z^+_{{\neq}}\right\|_{L^2L^2}\right)\left\|\sqrt{\left|S\right|_d}\tilde KZ^+_{{\neq}}\right\|_{L^2L^2}. \end{align}

Using (B.8), (B.9), (B.11), lemma C.1 and (2.29), we find that

(4.15)\begin{align} & \left\|\sqrt{\left|S\right|}\tilde K \Lambda Z^+_{{\neq}}\right\|_{L^2L^2}\nonumber\\ & \quad\le\frac{1}{1-C\delta}\left\|\sqrt{\left|S\right|}\tilde K Z^+_{{\neq}}\right\|_{L^2L^2}+\frac{C\delta}{1-C\delta}\left\|\sqrt{-\frac{\dot M_1}{M_1}}\tilde K \Lambda Z^+_{{\neq}}\right\|_{L^2L^2}\nonumber\\ & \quad\le\frac{1}{1-C\delta}\left(\left\|\sqrt{\left|S\right|_d}\tilde K Z^+_{{\neq}}\right\|_{L^2L^2}+\sqrt{\frac{1}{1000^3}}\nu^\frac {1}{2}\left\|\nabla_L\tilde KZ^+_{{\neq}}\right\|_{L^2L^2}\right.\nonumber\\ & \qquad\left.+\frac{1}{\sqrt{1+\frac {3}{2}\tilde\delta}}\left\|\sqrt{-\frac{\partial_t(\tilde m^{1/2})}{\tilde m^{1/2}}}\tilde KZ^+_{{\neq}}\right\|_{L^2L^2}\right)+\frac{C\delta}{(1-C\delta)^2}\left\|\sqrt{-\frac{\dot M_1}{M_1}}\tilde K Z^+_{{\neq}}\right\|_{L^2L^2}, \end{align}

where $|S|$ denotes the multiplier with symbol $\frac {|k(\eta -kt)|}{k^2+(\eta -kt)^2}$. Substituting this into (4.14), and using (B.8), (B.9) and lemma C.1 again to bound the second term on the right hand side of (4.14), one deduces that

(4.16)\begin{align} \mathrm{LS}_{2}^{dam} & \le C\delta\Bigg\{\frac{1}{1-C\delta}\Bigg(\left\|\sqrt{\left|S\right|_d}\tilde K Z^+_{{\neq}}\right\|_{L^2L^2}+\sqrt{\frac{1}{1000^3}}\nu^\frac {1}{2}\left\|\nabla_L\tilde KZ^+_{{\neq}}\right\|_{L^2L^2}\nonumber\\ & \quad+\frac{1}{\sqrt{1+\frac {3}{2}\tilde\delta}}\left\|\sqrt{-\frac{\partial_t(\tilde m^{1/2})}{\tilde m^{1/2}}}\tilde KZ^+_{{\neq}}\right\|_{L^2L^2}\Bigg)+\frac{C\delta}{(1-C\delta)^2}\left\|\sqrt{-\frac{\dot M_1}{M_1}}\tilde K Z^+_{{\neq}}\right\|_{L^2L^2}\nonumber\\ & \quad+\frac{1}{1-C\delta}\left\|\sqrt{-\frac{\dot M_1}{M_1}}\tilde K Z^+_{{\neq}}\right\|_{L^2L^2}\Bigg\}\left\|\sqrt{\left|S\right|_d}\tilde KZ^+_{{\neq}}\right\|_{L^2L^2}\nonumber\\ & \le \left(\frac{2C\delta}{1-C\delta}+\frac{C\delta}{2(1-C\delta)^2}\right)\left\|\sqrt{\left|S\right|_d}\tilde KZ^+_{{\neq}}\right\|_{L^2L^2}^2\nonumber\\ & \quad+\frac{C\delta}{2(1-C\delta)}\left(\vphantom{\left\|\sqrt{-\frac{\partial_t(\tilde m^{1/2})}{\tilde m^{1/2}}}\tilde KZ^+_{{\neq}}\right\|^2_{L^2L^2}}\frac{1}{1000^3}\nu\left\|\nabla_L\tilde KZ^+_{{\neq}}\right\|^2_{L^2L^2}\right.\nonumber\\ & \left.\quad +\,\frac{1}{{1+\frac {3}{2}\tilde\delta}}\left\|\sqrt{-\frac{\partial_t(\tilde m^{1/2})}{\tilde m^{1/2}}}\tilde KZ^+_{{\neq}}\right\|^2_{L^2L^2}\right)\nonumber\\ & \quad+\frac{C\delta}{2(1-C\delta)^2}\left\|\sqrt{-\frac{\dot M_1}{M_1}}\tilde KZ^+_{{\neq}}\right\|^2_{L^2L^2}. \end{align}

Next we turn to bound $\mathrm {LS}^*$. From (2.27) and (2.28), we infer that

(4.17)\begin{align} \mathrm{LS}^*& \le\frac{1}{4\pi^2}\sum_{k\neq0}\iint \tilde K(k,\eta) \frac{|k(\eta-kt)|}{k^2+(\eta-kt)^2}{\bf 1}_{D_{dam}^c}|\widehat{\Lambda Z^+_{{\neq}}}|(k, \eta)\tilde K(k,\eta)|\hat{Z}^+_{{\neq}}|(k,\eta) d\eta {\rm d}t\nonumber\\ & \le \frac{\nu}{1000^3}\sum_{k\neq0}\iint \tilde K(k,\eta) \left(k^2+(\eta-kt)^2\right)|\widehat{\Lambda Z^+_{{\neq}}}|(k, \eta)\tilde K(k,\eta)|\hat{Z}^+_{{\neq}}|(k,\eta) d\eta {\rm d}t\nonumber\\ & \quad+\frac{1}{1\,{+}\,\frac {3}{2}\tilde\delta}\sum_{k\neq0}\iint\tilde K(k,\eta) \frac{-\partial_t(\tilde m^{1/2})(t, k, \eta)}{\tilde m^{1/2}(t, k, \eta)}|\widehat{\Lambda Z^+_{{\neq}}}|(k, \eta) \tilde K(k,\eta)|\hat{Z}^+_{{\neq}}|(k,\eta) d\eta {\rm d}t\nonumber\\ & \le\frac{\nu}{1000^3}\|\nabla_L\tilde K\Lambda Z^+_{{\neq}}\|_{L^2L^2}\|\nabla_L\tilde KZ^+_{{\neq}}\|_{L^2L^2}\nonumber\\ & \quad+\frac{1}{1+\frac {3}{2}\tilde\delta}\left\|\sqrt{-\frac{\partial_t(\tilde m^{1/2})}{\tilde m^{1/2}}}\tilde K\Lambda Z^+_{{\neq}}\right\|_{L^2L^2}\left\|\sqrt{-\frac{\partial_t(\tilde m^{1/2})}{\tilde m^{1/2}}}\tilde KZ^+_{{\neq}}\right\|_{L^2L^2}. \end{align}

By virtue of (B.8) and lemma C.1, we have

(4.18)\begin{equation} \|\nabla_L\tilde K\Lambda Z^+_{{\neq}}\|_{L^2L^2}\le\frac{1}{1-C\delta}\|\nabla_L\tilde KZ^+_{{\neq}}\|_{L^2L^2}. \end{equation}

Now we are left to bound $\left \|\sqrt {-\frac {\partial _t(\tilde m^{1/2})}{\tilde m^{1/2}}}\tilde K\Lambda Z^+_{\neq }\right \|_{L^2L^2}$. In fact, thanks to (B.8), (B.10), and lemma C.1, we are led to

(4.19)\begin{align} & \left\|\sqrt{-\frac{\partial_t(\tilde m^{1/2})}{\tilde m^{1/2}}}\tilde K\Lambda Z^+_{{\neq}}\right\|_{L^2L^2}\nonumber\\ & \quad\leq\frac{1}{1-{C}\delta}\left\|\sqrt{-\frac{\partial_t(\tilde m^{1/2})}{\tilde m^{1/2}}}\tilde KZ^+_{{\neq}}\right\|_{L^2L^2}+\sqrt{\frac{1+\frac {3}{2}\tilde\delta}{1000^3}}\frac{C\delta}{(1-{C}\delta)^2}\nu^\frac {1}{2}\left\|\nabla_L\tilde KZ^+_{{\neq}}\right\|_{L^2L^2}\nonumber\\ & \quad+\frac{C\delta\sqrt{1+\frac {3}{2}\tilde\delta}}{(1-{C}\delta)^2}\left\|\sqrt{-\frac{\dot{M}_1}{M_1}}\tilde KZ^+_{{\neq}}\right\|_{L^2L^2}+\frac{C\delta\sqrt{1+\frac {3}{2}\tilde\delta}}{1-{C}\delta}\left\|\sqrt{\left|S\right|}\tilde K\Lambda Z^+_{{\neq}}\right\|_{L^2L^2}, \end{align}

where we have used (4.18) to bound $\|\nabla _L\tilde K\Lambda Z^+_{\neq }\|_{L^2L^2}$, and used (B.8), (B.9) and lemma C.1 to bound $\left \|\sqrt {-\frac {\dot {M}_1}{M_1}}\tilde K\Lambda Z^+_{\neq }\right \|_{L^2L^2}$, respectively. It follows from (4.15) and (4.19) that

(4.20)\begin{align} & \left\|\sqrt{-\frac{\partial_t(\tilde m^{1/2})}{\tilde m^{1/2}}}\tilde K\Lambda Z^+_{{\neq}}\right\|_{L^2L^2}\nonumber\\ & \quad\le \frac{1}{(1-C\delta)^2}\left\|\sqrt{-\frac{\partial_t(\tilde m^{1/2})}{\tilde m^{1/2}}}\tilde KZ^+_{{\neq}}\right\|_{L^2L^2}\nonumber\\ & \qquad+\sqrt{\frac{1+\frac {3}{2}\tilde\delta}{1000^3}}\frac{2{C}\delta}{(1-{C}\delta)^2}\nu^\frac {1}{2}\left\|\nabla_L\tilde KZ^+_{{\neq}}\right\|_{L^2L^2}\nonumber\\ & \qquad+\frac{C\delta\sqrt{1+\frac {3}{2}\tilde\delta}}{(1-{C}\delta)^2}\left\|\sqrt{\left|S\right|_d}\tilde K Z^+_{{\neq}}\right\|_{L^2L^2}+\frac{C\delta\sqrt{1+\frac {3}{2}\tilde\delta}}{(1-{C}\delta)^3}\left\|\sqrt{-\frac{\dot{M}_1}{M_1}}\tilde KZ^+_{{\neq}}\right\|_{L^2L^2}. \end{align}

Substituting (4.18) and (4.20) into (4.17), and using Cauchy–Schwarz inequality, we arrive at

\begin{align*} \mathrm{LS}^* & \le\frac{1}{1\,{-}\,C\delta}\frac{1}{1000^3}\nu\|\nabla_L\tilde KZ^+_{{\neq}}\|^2_{L^2L^2}+\frac{1}{1+\frac {3}{2}\tilde\delta}\frac{1}{(1\,{-}\,{C}\delta)^2}\left\|\sqrt{-\frac{\partial_t(\tilde m^{1/2})}{\tilde m^{1/2}}}\tilde KZ^+_{{\neq}}\right\|^2_{L^2L^2}\nonumber\\ & \quad+\frac{1}{1\,{+}\,\frac {3}{2}\tilde\delta}\frac{{C}\delta}{(1\,{-}\,{C}\delta)^2}\left({\frac{1+\frac {3}{2}\tilde\delta}{1000^3}}\nu\left\|\nabla_L\tilde KZ^+_{{\neq}}\right\|_{L^2L^2}^2+\left\|\sqrt{-\frac{\partial_t(\tilde m^{1/2})}{\tilde m^{1/2}}}\tilde KZ^+_{{\neq}}\right\|^2_{L^2L^2}\right)\nonumber\\ & \quad+\frac {1}{2}\frac{1}{1+\frac {3}{2}\tilde\delta}\frac{{C}\delta}{(1-{C}\delta)^2}\left(\vphantom{\left\|\sqrt{-\frac{\partial_t(\tilde m^{1/2})}{\tilde m^{1/2}}}\tilde KZ^+_{{\neq}}\right\|_{L^2L^2}^2}\left(1+\frac {3}{2}\tilde\delta\right)\left\|\sqrt{\left|S\right|_d}\tilde KZ^+_{{\neq}}\right\|_{L^2L^2}^2\right.\nonumber\\ & \left.\quad+\,\left\|\sqrt{-\frac{\partial_t(\tilde m^{1/2})}{\tilde m^{1/2}}}\tilde KZ^+_{{\neq}}\right\|_{L^2L^2}^2\right)\nonumber\\ & \quad+\frac {1}{2}\frac{1}{1+\frac {3}{2}\tilde\delta}\frac{{C}\delta}{(1-{C}\delta)^3}\left(\left(1+\frac {3}{2}\tilde\delta\right)\left\|\sqrt{-\frac{\dot{M}_1}{M_1}}\tilde KZ^+_{{\neq}}\right\|_{L^2L^2}^2\right.\nonumber\\ & \left.\quad+\,\left\|\sqrt{-\frac{\partial_t(\tilde m^{1/2})}{\tilde m^{1/2}}}\tilde KZ^+_{{\neq}}\right\|_{L^2L^2}^2\vphantom{\left\|\sqrt{-\frac{\dot{M}_1}{M_1}}\tilde KZ^+_{{\neq}}\right\|_{L^2L^2}^2}\right)\nonumber\\ & =\frac{C_\delta}{1+\frac {3}{2}\tilde\delta}\left\|\sqrt{-\frac{\partial_t(\tilde m^{1/2})}{\tilde m^{1/2}}}\tilde KZ^+_{{\neq}}\right\|^2_{L^2L^2}+\frac{1}{(1-{C}\delta)^2}\frac{1}{1000^3}\nu\|\nabla_L\tilde KZ^+_{{\neq}}\|^2_{L^2L^2}\nonumber\\ & \quad+\frac{{C}\delta}{2(1-{C}\delta)^2}\left\|\sqrt{\left|S\right|_d}\tilde KZ^+_{{\neq}}\right\|_{L^2L^2}^2+\frac{{C}\delta}{2(1-{C}\delta)^3}\left\|\sqrt{-\frac{\dot{M}_1}{M_1}}\tilde KZ^+_{{\neq}}\right\|_{L^2L^2}^2, \end{align*}

where

(4.21)\begin{equation} C_\delta:=\frac{1}{(1-{C}\delta)^2}+\frac{3{C}\delta}{2(1-{C}\delta)^2}+\frac{{C}\delta}{2(1-{C}\delta)^3},\end{equation}

which is increasing in $\delta \in (0, \frac {1}{3C}]$, and $C_\delta \rightarrow 1$ as $\delta \rightarrow 0+$.

Step II: estimates of $\mathrm {OLS}$. Similar to (3.25), we write

(4.22)\begin{equation} \mathrm{OLS}=\sum_{i=1}^5\mathrm{OLS}_i, \end{equation}

where

\begin{align*} \mathrm{OLS}_1 & =\frac{1}{2\alpha}\left\langle \tilde K\partial_X^{{-}1}O^t_{2\alpha}S\Lambda Z^-_{{\neq}}(t), \tilde KZ^+(t)\right\rangle, \nonumber\\ \mathrm{OLS}_2& ={-}\frac{1}{2\alpha}\left\langle \tilde K\partial_X^{{-}1}O^t_{2\alpha}S\Lambda Z^-_{{\neq}}(0), \tilde KZ^+(0)\right\rangle, \nonumber\\ \mathrm{OLS}_3& ={-}\frac{1}{2\alpha}\int_0^t\left\langle \partial_X^{{-}1}O^t_{2\alpha}\left(S\frac{2\dot{\tilde K}}{\tilde K}+\dot{S}\right)\tilde K\Lambda Z^-_{{\neq}}, \tilde KZ^+\right\rangle {\rm d}t',\nonumber\\ \mathrm{OLS}_4& ={-}\frac{1}{2\alpha}\int_0^t\left\langle \tilde KS\partial_X^{{-}1}O^t_{2\alpha}\Lambda Z^-_{{\neq}}, \tilde K\partial_tZ^+\right\rangle {\rm d}t',\nonumber\\ \mathrm{OLS}_5& ={-}\frac{1}{2\alpha}\int_0^t\left\langle \tilde KS\partial_X^{{-}1}O^t_{2\alpha}\partial_t(\Lambda Z^-_{{\neq}}), \tilde KZ^+\right\rangle {\rm d}t'. \end{align*}

We postpone the treatment of the nonlinear terms in $\mathrm {OLS}_4$ and $\mathrm {OLS}_5$ to Step IV, and focus on the linear terms here. The estimates for $\mathrm {OLS}_1$ and $\mathrm {OLS}_2$ are the same as (3.26), and thus omitted. To estimate $\mathrm {OLS}_3$, similar to (3.27)–(3.29), we have

\begin{align*} \frac{\dot{\tilde K}}{\tilde K}& =\frac{\partial_t(\tilde m^{1/2})}{\tilde m^{1/2}}+\frac{\dot{M}}{M}=\frac {1}{2}\frac{\dot{\tilde m}}{\tilde m}+\frac{\dot{M}}{M}, -\frac{\dot{\tilde m}}{\tilde m}\\& =(1+\frac {3}{2}\tilde\delta)\frac{2k(\eta-kt)}{k^2+(\eta-kt)^2}{\bf 1}_{D_{mul}}(t,k,\eta)=2(1+\frac {3}{2}\tilde\delta)S(t, k, \eta){\bf 1}_{D_{mul}}(t,k,\eta), \end{align*}

and

\[ \left|\frac{\partial_t(\tilde m^{1/2})}{\tilde m^{1/2}}\right|\le(1+\frac {3}{2}\tilde\delta) |S|\le\frac{(1+\frac {3}{2}\tilde\delta)|k(\eta-kt)|}{k^2+(\eta-kt)^2}\le(1+\frac {3}{2}\tilde\delta)\min\left\{\frac {1}{2}, \sqrt{-\frac{\dot{M}_1}{M_1}}\right\}. \]

Combining these calculations with (3.30), and using (B.8), (B.9) and lemma C.1, we are led to

(4.23)\begin{align} \mathrm{OLS}_3& \le\frac{C}{|\alpha|}\left\|\sqrt{-\frac{\dot{M}}{M}}\tilde K\Lambda Z^-_{{\neq}}\right\|_{L^2L^2}\left\|\sqrt{-\frac{\dot{M}}{M}}\tilde KZ^+\right\|_{L^2L^2}\nonumber\\ & \le\frac{C}{|\alpha|}\frac{1}{1-C\delta}\left\|\sqrt{-\frac{\dot{M}}{M}}\tilde KZ^-_{{\neq}}\right\|_{L^2L^2}\left\|\sqrt{-\frac{\dot{M}}{M}}\tilde KZ^+\right\|_{L^2L^2}. \end{align}

To bound $\mathrm {OLS}_4$, by (2.16), we have

(4.24)\begin{equation} \mathrm{OLS}_4=\sum_{i=1}^5\mathrm{OLS}_4^{(i)},\end{equation}

where

\begin{align*} \mathrm{OLS}_4^{(1)}& ={-}\frac{\nu}{2\alpha}\int_0^t\left\langle S\partial_X^{{-}1}O^t_{2\alpha}\tilde K\Lambda Z^-_{{\neq}}, \tilde K\Delta_L Z^+\right\rangle {\rm d}t',\\ \mathrm{OLS}_4^{(2)}& =\frac{1}{2\alpha}\int_0^t\left\langle S\partial_X^{{-}1}O^t_{2\alpha}\tilde K\Lambda Z^-_{{\neq}}, \tilde KS\Lambda Z^+\right\rangle {\rm d}t',\\ \mathrm{OLS}_4^{(3)}& ={-}\frac{1}{2\alpha}\int_0^t\left\langle S\partial_X^{{-}1}O^t_{2\alpha}\tilde K\Lambda Z^-_{{\neq}}, \tilde KO^t_{2\alpha}Z^{-}_{{\neq}}\right\rangle {\rm d}t',\\ \mathrm{OLS}_4^{(4)}& ={-}\frac{1}{2\alpha}\int_0^t\left\langle S\partial_X^{{-}1}O^t_{2\alpha}\tilde K\Lambda Z^-_{{\neq}}, \tilde K\mathrm{LP}^+\right\rangle {\rm d}t',\\ \mathrm{OLS}_4^{(5)}& ={-}\frac{1}{2\alpha}\int_0^t\left\langle S\partial_X^{{-}1}O^t_{2\alpha}\tilde K\Lambda Z^-_{{\neq}}, \tilde K\mathrm{NL}^+\right\rangle {\rm d}t'. \end{align*}

Integrating by parts, using the fact $|S|\le \frac{1}{2}$ and (4.18), we have

(4.25)\begin{align} \mathrm{OLS}_4^{(1)}& =\frac{\nu}{2\alpha}\int_0^t\left\langle S\partial_X^{{-}1}O^t_{2\alpha}\nabla_L\tilde K\Lambda Z^-_{{\neq}}, \nabla_L \tilde KZ^+\right\rangle {\rm d}t'\nonumber\\ & \le \frac{\nu}{4|\alpha|}\|\nabla_L\tilde K\Lambda Z^-_{{\neq}}\|_{L^2L^2}\|\nabla_L\tilde KZ^+_{{\neq}}\|_{L^2L^2}\nonumber\\ & \le\frac{\nu}{4|\alpha|(1-C\delta)}\|\nabla_L\tilde KZ^-_{{\neq}}\|_{L^2L^2}\|\nabla_L\tilde KZ^+_{{\neq}}\|_{L^2L^2}. \end{align}

Thanks to the fact $|S|\le \sqrt {-\frac {\dot {M}_1}{M_1}}$, one deduces that

(4.26)\begin{align} \mathrm{OLS}_4^{(2)}& \le\frac{1}{2|\alpha|}\left\|\sqrt{-\frac{\dot{M}}{M}}\tilde K\Lambda Z^+\right\|_{L^2L^2}\left\|\sqrt{-\frac{\dot{M}}{M}}\tilde K\Lambda Z^-\right\|_{L^2L^2}\nonumber\\ & \le \frac{1}{|\alpha|(1-C\delta)^2}\left\|\sqrt{-\frac{\dot{M}}{M}}\tilde KZ^+\right\|_{L^2L^2}\left\|\sqrt{-\frac{\dot{M}}{M}}\tilde KZ^-\right\|_{L^2L^2}. \end{align}

Owing to the periodicity in $X$ variable, we find that

(4.27)\begin{equation} \mathrm{OLS}_4^{(3)}=\frac{1}{4\alpha}\int_0^t\int_{\mathbb{T}\times\mathbb{R}} \partial_X\left(S\partial_X^{{-}1}O^t_{2\alpha}\tilde K\Lambda Z^-_{{\neq}}\right)^2 dXdYdt'=0.\end{equation}

By (2.18), we deduce that

(4.28)\begin{equation} \mathrm{OLS}_4^{(4)}=\sum_{i=1}^3\mathrm{OLS}_{4}^{(4,i)},\end{equation}

with

\begin{align*} \mathrm{OLS}_4^{(4,1)}& =\frac{1}{2\alpha}\int_0^t\left\langle S\partial_X^{{-}1}O^t_{2\alpha}\tilde K\Lambda Z^-_{{\neq}}, \tilde K\left((a^2-1)S\Lambda \left(Z^+_{{\neq}}-O^t_{2\alpha}Z^-_{{\neq}}\right)\right)\right\rangle {\rm d}t',\nonumber\\ \mathrm{OLS}_4^{(4,2)}& ={-}\frac{1}{2\alpha}\int_0^t\left\langle S\partial_X^{{-}1}O^t_{2\alpha}\tilde K\Lambda Z^-_{{\neq}}, \tilde K\left(b\partial_X\Delta^{{-}1}_tO^t_{2\alpha}Z^{-}_{{\neq}}\right)\right\rangle {\rm d}t',\nonumber\\ \mathrm{OLS}_4^{(4,3)}& ={-}\frac{\nu}{2\alpha}\int_0^t\left\langle S\partial_X^{{-}1}O^t_{2\alpha}\tilde K\Lambda Z^-_{{\neq}}, \tilde K\left((a^2-1)\partial_{YY}^LZ^+_{{\neq}}\right)\right\rangle {\rm d}t'. \end{align*}

Using $|S|\le \sqrt {-\frac {\dot {M}_1}{M_1}}$, (A.3) and lemma C.1, we have

(4.29)\begin{align} \mathrm{OLS}_{4}^{(4,1)}& \le\frac{1}{2|\alpha|}\left\|\tilde KS\Lambda Z^-_{{\neq}}\right\|_{L^2L^2}\left\|\tilde K\left((a^2-1)S\Lambda \left(Z^+_{{\neq}}-O^t_{2\alpha}Z^-_{{\neq}}\right)\right)\right\|_{L^2L^2}\nonumber\\ & \le\frac{C\delta}{|\alpha|}\left\|\sqrt{-\frac{\dot{M}_1}{M_1}}\tilde K\Lambda Z^-_{{\neq}}\right\|_{L^2L^2}\left\|\sqrt{-\frac{\dot{M}_1}{M_1}}\tilde K\Lambda \left(Z^+_{{\neq}}-O^t_{2\alpha}Z^-_{{\neq}}\right)\right\|_{L^2L^2}\nonumber\\ & \le\frac{C\delta}{|\alpha|(1-C\delta)^2}\left\|\sqrt{-\frac{\dot M_1}{M_1}}\tilde KZ^-_{{\neq}}\right\|_{L^2L^2}\nonumber\\ & \quad\times\left(\left\|\sqrt{-\frac{\dot M_1}{M_1}}\tilde KZ^+_{{\neq}}\right\|_{L^2L^2}+\left\|\sqrt{-\frac{\dot M_1}{M_1}}\tilde KZ^-_{{\neq}}\right\|_{L^2L^2}\right). \end{align}

Noting that

\[ \partial_X\Delta_t^{{-}1}=\partial_X\Delta_L^{{-}1}\Lambda, \]

and

(4.30)\begin{align} \left|\mathcal{F}\left(b\partial_X\Delta^{{-}1}_tO^t_{2\alpha}Z^{-}_{{\neq}}\right)\right|(t,k, \eta)& \le \int_\xi|\hat{b}|(\eta-\xi)\frac{|k|}{k^2+(\xi-kt)^2}|\widehat{\Lambda {O^t_{2\alpha}}Z^-_{{\neq}}}|(k,\xi)d\xi\nonumber\\ & \le\left(|\hat{b}|*\left(\sqrt{-\frac{\dot{M}_1}{M_1}}|\widehat{\Lambda {O^t_{2\alpha}}Z^-_{{\neq}}}|\right)\right)(t,k, \eta), \end{align}

then we have

(4.31)\begin{align} \mathrm{OLS}_{4}^{(4,2)}& \le\frac{C\delta}{|\alpha|}\left\|\tilde KS\Lambda Z^-_{{\neq}}\right\|_{L^2L^2}\left\|\tilde K\sqrt{-\frac{\dot{M}_1}{M_1}}\Lambda {O^t_{2\alpha}}Z^-_{{\neq}}\right\|_{L^2L^2}\nonumber\\ & \le\frac{C\delta}{|\alpha|(1-C\delta)^2}\left\|\sqrt{-\frac{\dot{M}_1}{M_1}}\tilde KZ^-_{{\neq}}\right\|_{L^2L^2}^2. \end{align}

Integrating by parts and using (2.4), we find that

(4.32)\begin{align} \mathrm{OLS}_{4}^{(4,3)}& =\frac{\nu}{\alpha}\int_0^t\left\langle S\partial_X^{{-}1}O^t_{2\alpha}\tilde K\Lambda Z^-_{{\neq}}, \tilde K\left(b\partial_{Y}^LZ^+_{{\neq}}\right)\right\rangle {\rm d}t'\nonumber\\ & \quad+\frac{\nu}{2\alpha}\int_0^t\left\langle \partial_Y^LS\partial_X^{{-}1}O^t_{2\alpha}\tilde K\Lambda Z^-_{{\neq}}, \tilde K\left((a^2-1)\partial_{Y}^LZ^+_{{\neq}}\right)\right\rangle {\rm d}t'\nonumber\\ & \le\frac{C\delta\nu}{|\alpha|(1-C\delta)}\left\|\sqrt{-\frac{\dot{M}_1}{M_1}}\tilde KZ^-_{{\neq}}\right\|_{L^2L^2}\left\|\partial_Y^L\tilde KZ^+_{{\neq}}\right\|_{L^2L^2}\nonumber\\ & \quad+\frac{C\delta\nu}{4|\alpha|}\left\|\partial_Y^L\tilde K\Lambda Z^-_{{\neq}}\right\|_{L^2L^2}\left\|\partial_Y^L\tilde KZ^+_{{\neq}}\right\|_{L^2L^2}\nonumber\\ & \le\frac{C\delta\nu}{|\alpha|(1-C\delta)}\left\|\nabla_L\tilde KZ^-_{{\neq}}\right\|_{L^2L^2}\left\|\nabla_L\tilde KZ^+_{{\neq}}\right\|_{L^2L^2}, \end{align}

where we have used (A.3), (A.8) to bound $\left \|\tilde K(b\partial _Y^L)\right \|_{L^2L^2}$ and $\left \|\tilde K((a^2-1)\partial _Y^L)\right \|_{L^2L^2}$. The estimates of $\mathrm {OLS}_{4}^{(5)}$ will be postponed in Step IV.

Now we rewrite $\mathrm {OLS}_5$ as follows:

\begin{align*} \mathrm{OLS}_5& ={-}\frac{1}{2\alpha}\int_0^t\left\langle \tilde KS\partial_X^{{-}1}O^t_{2\alpha}\Lambda \partial_tZ^-_{{\neq}}, \tilde KZ^+_{{\neq}}\right\rangle {\rm d}t'\nonumber\\ & \quad-\frac{1}{2\alpha}\int_0^t\left\langle \tilde KS\partial_X^{{-}1}O^t_{2\alpha}\partial_t\Lambda Z^-_{{\neq}}, \tilde KZ^+_{{\neq}}\right\rangle {\rm d}t'\nonumber\\ & =\mathrm{OLS}_{5,1}+\mathrm{OLS}_{5,2}. \end{align*}

To estimate $\mathrm {OLS}_{5,1}$, instead of using (2.16), up to the nonlinear terms and some linear errors, we write $\partial _tZ^{-}$ in terms of $\nu \Delta _tZ^-$ by virtue of (2.6) and (2.15)

\[ \partial_tZ^- = \nu{\Delta}_t Z^-{-}\nu b\partial_Y^LZ^-{-}a^2S\Lambda \left(Z^-_{{\neq}}-O^t_{{-}2\alpha}Z^+_{{\neq}}\right)+b\partial_X\Delta^{{-}1}_tO^t_{{-}2\alpha}Z^+_{{\neq}}+\mathrm{NL}^-. \]

Then thanks to the relation (2.12), we have

\begin{align*} \Lambda \partial_tZ^-_{{\neq}}& =\nu{\Delta}_L Z^-_{{\neq}}-\nu \Lambda \left(b\partial_Y^LZ^-_{{\neq}}\right) -\Lambda\left(a^2S\Lambda\left(Z^-_{{\neq}}-O^t_{{-}2\alpha}Z^+_{{\neq}}\right)\right)\nonumber\\ & \quad+\Lambda\left(b\partial_X\Delta^{{-}1}_tO^t_{{-}2\alpha}Z^+_{{\neq}}\right)+\Lambda\mathrm{NL}^-_{{\neq}}. \end{align*}

Thus,

(4.33)\begin{equation} \mathrm{OLS}_{5,1}=\sum_{i=1}^5\mathrm{OLS}_{5,1}^{(i)},\end{equation}

where

\begin{align*} \mathrm{OLS}_{5,1}^{(1)}& ={-}\frac{\nu}{2\alpha}\int_0^t\left\langle \tilde K\partial_X^{{-}1}O^t_{2\alpha}\Delta_LZ^-_{{\neq}}, \tilde KSZ^+_{{\neq}}\right\rangle {\rm d}t',\\ \mathrm{OLS}_{5,1}^{(2)}& =\frac{\nu}{2\alpha}\int_0^t\left\langle \tilde K\partial_X^{{-}1}O^t_{2\alpha}\Lambda\left(b\partial_Y^LZ^-_{{\neq}}\right), \tilde KSZ^+_{{\neq}}\right\rangle {\rm d}t',\\ \mathrm{OLS}_{5,1}^{(3)}& =\frac{1}{2\alpha}\int_0^t\left\langle \tilde K\partial_X^{{-}1}O^t_{2\alpha}\Lambda\left(a^2S\Lambda\left(Z^-_{{\neq}}-O^t_{{-}2\alpha}Z^+_{{\neq}}\right)\right), \tilde KSZ^+_{{\neq}}\right\rangle {\rm d}t',\\ \mathrm{OLS}_{5,1}^{(4)}& ={-}\frac{1}{2\alpha}\int_0^t\left\langle \tilde K\partial_X^{{-}1}O^t_{2\alpha}\Lambda\left(b\partial_X\Delta^{{-}1}_tO^t_{{-}2\alpha}Z^+_{{\neq}}\right), \tilde KSZ^+_{{\neq}}\right\rangle {\rm d}t',\\ \mathrm{OLS}_{5,1}^{(5)}& ={-}\frac{1}{2\alpha}\int_0^t\left\langle \tilde K\partial_X^{{-}1}O^t_{2\alpha}\Lambda\mathrm{NL}^-_{{\neq}}, \tilde KSZ^+_{{\neq}}\right\rangle {\rm d}t'. \end{align*}

Similar to (4.32), we obtain

(4.34)\begin{align} \left|\mathrm{OLS}_{5,1}^{(1)}\right|+\left|\mathrm{OLS}_{5,1}^{(2)}\right|\le\left(\frac{\nu}{4|\alpha|}+\frac{C\delta\nu}{|\alpha|(1-C\delta)}\right)\left\|\nabla_L\tilde KZ^-_{{\neq}}\right\|_{L^2L^2}\left\|\nabla_L\tilde KZ^+_{{\neq}}\right\|_{L^2L^2}.\end{align}

Similar to (4.26) and (4.29), we arrive at

(4.35)\begin{align} \mathrm{OLS}_{5,1}^{(3)} & \le\frac{1}{2|\alpha|}\left(\left\|\tilde K\Lambda\left(S\Lambda\left(Z^-_{{\neq}}-O^t_{{-}2\alpha}Z^+_{{\neq}}\right)\right)\right\|_{L^2L^2}\right.\nonumber\\ & \quad\left.+\left\|\tilde K\Lambda\left((a^2-1)S\Lambda\left(Z^-_{{\neq}}-O^t_{{-}2\alpha}Z^+_{{\neq}}\right)\right)\right\|_{L^2L^2}\right)\|\tilde KSZ^+_{{\neq}}\|_{L^2L^2}\nonumber\\ & \le \left(\frac{1}{|\alpha|(1-C\delta)^2}+\frac{C\delta}{|\alpha|(1-C\delta)^2}\right)\left\|\sqrt{-\frac{\dot M_1}{M_1}}\tilde KZ^+_{{\neq}}\right\|_{L^2L^2}\nonumber\\ & \quad\times\left(\left\|\sqrt{-\frac{\dot M_1}{M_1}}\tilde KZ^+_{{\neq}}\right\|_{L^2L^2}+\left\|\sqrt{-\frac{\dot M_1}{M_1}}\tilde KZ^-_{{\neq}}\right\|_{L^2L^2}\right). \end{align}

Clearly, $\mathrm {OLS}_{5,1}^{(4)}$ can be bounded in the same way as (4.31)

(4.36)\begin{equation} \left|\mathrm{OLS}_{5,1}^{(4)}\right|\le\frac{C\delta}{|\alpha|(1-C\delta)^2}\left\|\sqrt{-\frac{\dot{M}_1}{M_1}}\tilde KZ^+_{{\neq}}\right\|_{L^2L^2}^2.\end{equation}

The estimates of $\mathrm {OLS}_{5,1}^{(5)}$ will be postponed in Step IV.

To estimate $\mathrm {OLS}_{5,2}$, we split it into two parts according to (2.32) and (2.33):

\[ \mathrm{OLS}_{5,2}={-}\frac{1}{2\alpha}\int_0^t\left\langle \tilde K\partial_X^{{-}1}O^t_{2\alpha}\partial_t\Lambda Z^-_{{\neq}}, \tilde KSZ^+\right\rangle {\rm d}t'=\mathrm{OLS}_{5,2}^{(1)}+\mathrm{OLS}_{5,2}^{(2)}, \]

where

\begin{align*} \mathrm{OLS}_{5,2}^{(1)}& ={-}\frac{1}{2|\alpha|}\int_0^t\left\langle \tilde K\partial_X^{{-}1}O^t_{2\alpha}\Lambda\tilde \Lambda^1_t\Lambda Z^-_{{\neq}}, \tilde KSZ^+_{\ne}\right\rangle {\rm d}t',\\ \mathrm{OLS}_{5,2}^{(2)}& ={-}\frac{1}{2|\alpha|}\int_0^t\left\langle \tilde K\partial_X^{{-}1}O^t_{2\alpha}\Lambda\tilde \Lambda^2_t\Lambda Z^-_{{\neq}}, \tilde KSZ^+_{\ne}\right\rangle {\rm d}t'. \end{align*}

From (A.8) and (3.29), we obtain

\begin{align*} \mathrm{OLS}_{5,2}^{(1)}& \le\frac{1}{|\alpha|}\left\|\tilde K\Lambda\tilde \Lambda^1_t\Lambda Z^-_{{\neq}}\right\|_{L^2L^2}\left\|\sqrt{-\frac{\dot M_1}{M_1}}\tilde KZ^+_{{\neq}}\right\|_{L^2L^2}\nonumber\\ & \le\frac{1}{|\alpha|}\frac{C\delta}{(1-C\delta)^2}\left\|\sqrt{-\frac{\dot M_1}{M_1}}\tilde K Z^-_{{\neq}}\right\|_{L^2L^2}\left\|\sqrt{-\frac{\dot M_1}{M_1}}\tilde K Z^+_{{\neq}}\right\|_{L^2L^2}. \end{align*}

The rest part $\mathrm {OLS}_{5,2}^{(2)}$ can be bounded as follows. Using the definition of $\tilde \Lambda ^2_t$ in (2.33), lemmas A.2 and C.1, we are led to

\begin{align*} & \left\|\tilde K\tilde\Lambda^2_t\Lambda Z^-_{{\neq}}\right\|_{L^2L^2}\\ & \quad\le{C}\left\|\left(\langle\cdot \rangle^{N+(1+\frac{3}{2}\tilde\delta)}\left(|\widehat{a\partial_ta}|+|\widehat{\partial_tb}|\right)\right)*|\widehat{\tilde K\Lambda Z^-_{{\neq}}}|\right\|_{L^2_tL^2_{k,\eta}}\nonumber\\ & \quad\le{C}\left\|\langle\cdot \rangle^{N+(1+\frac{3}{2}\tilde\delta)}\left(|\widehat{a\partial_ta}|+|\widehat{\partial_tb}|\right)\right\|_{L^2_tL^1_\eta}\|\tilde K\Lambda Z^-_{{\neq}}\|_{L^\infty L^2}\nonumber\\ & \quad\le\frac{C}{1-C\delta}\left(\left\|{a\partial_ta}\right\|_{L^2H^{N+(1+\frac{3}{2}\tilde\delta)+1}}+\left\|{\partial_tb}\right\|_{L^2H^{{N+(1+\frac{3}{2}\tilde\delta)+1}}}\right)\|\tilde KZ^-_{{\neq}}\|_{L^\infty L^2}\nonumber\\ & \quad \le\frac{C\delta\nu^{1/2}}{1-C\delta}\|\tilde KZ^-_{{\neq}}\|_{L^\infty L^2}. \end{align*}

Then it follows this and lemma C.1 that

(4.37)\begin{align} \mathrm{OLS}_{5,2}^{(2)}& \le\frac{1}{|\alpha|(1-C\delta)}\left\|\tilde K\tilde \Lambda^2_t\Lambda Z^-_{{\neq}}\right\|_{L^2L^2}\left\|\tilde KSZ^+_{{\neq}}\right\|_{L^2L^2}\nonumber\\ & \le\frac{C\delta\nu^{1/2}}{|\alpha|(1-C\delta)^2}\left\|\tilde KZ^-_{{\neq}}\right\|_{L^\infty L^2}\left\|\sqrt{-\frac{\dot M_1}{M_1}}\tilde KZ^+_{{\neq}}\right\|_{L^2L^2}. \end{align}

Step III: estimates of $\mathcal {LP}$. By (2.18), we rewrite $\mathcal {LP}$ as

(4.38)\begin{equation} \mathcal{LP}=\int_0^t\left\langle \tilde KZ^+, \tilde K\mathrm{LP}^+\right\rangle {\rm d}t'=\mathcal{LP}_1+\mathcal{LP}_2+\mathcal{LP}_3,\end{equation}

where

\begin{align*} \mathcal{LP}_1& ={-}\int_0^t\left\langle \tilde KZ^+_{{\neq}},\tilde K\left((a^2-1)S\Lambda (Z^+_{{\neq}}-O^t_{2\alpha}Z^{-}_{\ne})\right)\right\rangle {\rm d}t',\\ \mathcal{LP}_2& =\int_0^t\left\langle \tilde KZ^+_{{\neq}},\tilde K\left(b\partial_X\Delta^{{-}1}_L\Lambda O^t_{2\alpha}Z^-_{{\neq}}\right)\right\rangle {\rm d}t',\\ \mathcal{LP}_3& =\nu\int_0^t\left\langle \tilde KZ^+,\tilde K\left((a^2-1)\partial^L_{YY}Z^+\right)\right\rangle {\rm d}t'. \end{align*}

To estimate $\mathcal {LP}_1$, let us denote $Z_\ne :=Z^+_\ne -O^t_{2\alpha }Z^{-}_{\ne }$, and rewrite $\mathcal {LP}_1$ as

\begin{align*} \mathcal{LP}_1& ={-}\frac{1}{4\pi^2}\sum_{k\neq0}\int_0^t\int_\eta\int_\xi \tilde K(k,\eta) \widehat{(a^2-1)}(\eta-\xi)\frac{k(\xi-kt)}{k^2+(\xi-kt)^2}\widehat{\Lambda Z_{{\neq}}}(k, \xi)\nonumber\\ & \quad\times \tilde K(k,\eta)\bar{\hat{Z}}^+_{{\neq}}(k,\eta)d\xi d\eta {\rm d}t'. \end{align*}

Swapping the positions of $\eta$ and $\xi$ in (A.11), then using the resulting inequality, (A.3), lemma C.1, and (4.15), we have

(4.39)\begin{align} \mathcal{LP}_1& \le C\sum_{k\ne0}\int_0^t\int_\eta\int_\xi\langle\eta-\xi\rangle^{N+(1+\frac{3}{2}\tilde\delta)+\frac{3}{2}}|\widehat{a^2-1}|(\eta-\xi)\nonumber\\ & \quad \times\sqrt{\frac{|k(\xi-kt)|}{k^2+(\xi-kt)^2}}|\widehat{\tilde K\Lambda Z_{{\neq}}}|(k,\xi)\nonumber\\ & \quad \times\left(\sqrt{\frac{|k(\eta-kt)|}{k^2+(\eta-kt)^2}}+\sqrt{\frac{|k|}{k^2+(\eta-kt)^2}}\right)|\widehat{\tilde {K}Z^+_{{\neq}}}|(k,\eta)d\xi d\eta {\rm d}t\nonumber\\ & \le C\delta\left\|\sqrt{\left|S\right|} \tilde K\Lambda Z_{{\neq}}\right\|_{L^2L^2}\left(\left\|\sqrt{\left|S\right|}\tilde KZ^+_{{\neq}}\right\|_{L^2L^2}+\left\|\sqrt{-\frac{\dot M_1}{M_1}} \tilde KZ^+_{{\neq}}\right\|_{L^2L^2}\right)\nonumber\\ & \le \frac{C\delta}{1-C\delta}\left(\left\|\sqrt{\left|S\right|}\tilde KZ_{{\neq}}\right\|_{L^2L^2}+\frac{C\delta}{1-C\delta}\left\|\sqrt{-\frac{\dot M_1}{M_1}}\tilde K Z_{{\neq}}\right\|_{L^2L^2}\right)\nonumber\\ & \quad\times\left(\left\|\sqrt{\left|S\right|}\tilde K Z^+_{{\neq}}\right\|_{L^2L^2}+\left\|\sqrt{-\frac{\dot M_1}{M_1}} \tilde KZ^+_{{\neq}}\right\|_{L^2L^2}\right)\nonumber\\ & \le \frac{C\delta}{1-C\delta}\Bigg(\left\|\sqrt{\left|S\right|_d}\tilde K Z^\pm_{{\neq}}\right\|^2_{L^2L^2}+{\frac{1}{1000^3}}\nu\left\|\nabla_L\tilde KZ^\pm_{{\neq}}\right\|^2_{L^2L^2}\nonumber\\ & \quad+\frac{1}{{1+\frac{3}{2}\tilde\delta}}\left\|\sqrt{-\frac{\partial_t(\tilde m^{1/2})}{\tilde m^{1/2}}}\tilde KZ^\pm_{{\neq}}\right\|^2_{L^2L^2}+\left\|\sqrt{-\frac{\dot M_1}{M_1}} \tilde KZ^\pm_{{\neq}}\right\|^2_{L^2L^2}\Bigg). \end{align}

Now we turn to estimate $\mathcal {LP}_2$. Compared with (4.31), we need an extra commutator estimate of $\sqrt {\left |\partial _X\Delta _L^{-1}\right |}$. Indeed, using (A.8), (A.9) and lemma C.1, one easily deduces that

(4.40)\begin{align} \mathcal{LP}_2& \le C\sum_{k\ne0}\int_0^t\int_\eta\int_\xi \tilde K(k,\eta)\left(|\hat{b}|(\eta-\xi)\frac{k^2}{k^2+(\xi-kt)^2}|\widehat{\Lambda Z^-_{{\neq}}}|(k,\xi)\right)\nonumber\\ & \quad \times |\widehat{\tilde KZ^+_{{\neq}}}|(k,\eta)d\xi d\eta {\rm d}t'\nonumber\\ & \le C\sum_{k\ne0}\int_0^t\int_\eta\int_\xi \langle\eta-\xi\rangle^{N+(1+\frac{3}{2}\tilde\delta)+1}|\hat{b}|(\eta-\xi)\frac{|k|}{\sqrt{k^2+(\xi-kt)^2}}|\widehat{\tilde K\Lambda Z^-_{{\neq}}}|(k,\xi)\nonumber\\ & \quad\times \frac{|k|}{\sqrt{k^2+(\eta-kt)^2}}|\widehat{\tilde KZ^+_{{\neq}}}|(k,\eta)d\xi d\eta {\rm d}t' \nonumber\\ & \le C\delta\left\|\sqrt{-\frac{\dot{M}_1}{M_1}}\tilde K\Lambda Z^-_{{\neq}}\right\|_{L^2L^2}\left\|\sqrt{-\frac{\dot{M}_1}{M_1}}\tilde KZ^+_{{\neq}}\right\|_{L^2L^2}\nonumber\\ & \le\frac{C\delta}{1-C\delta}\left\|\sqrt{-\frac{\dot{M}_1}{M_1}}\tilde KZ^-_{{\neq}}\right\|_{L^2L^2}\left\|\sqrt{-\frac{\dot{M}_1}{M_1}}\tilde KZ^+_{{\neq}}\right\|_{L^2L^2}. \end{align}

To bound $\mathcal {LP}_3$, it is natural to divide it into two parts as follows

\begin{align*} \mathcal{LP}_3& =\nu\int_0^t\left\langle \tilde KZ^+_{{\neq}},\tilde K\left((a^2-1)\partial^L_{YY}Z^+_{{\neq}}\right)\right\rangle {\rm d}t'\\& \quad +\nu\int_0^t\left\langle \tilde KZ^+_0,\tilde K\left((a^2-1)\partial_{YY}Z^+_0\right)\right\rangle {\rm d}t'\nonumber\\ & =:\mathcal{LP}_{3,\ne}+\mathcal{LP}_{3,0}. \end{align*}

Similar to (4.32), we obtain

(4.41)\begin{equation} \mathcal{LP}_{3,\ne}\le C\delta\nu\|\nabla_L\tilde KZ^+_\ne\|^2_{L^2L^2}.\end{equation}

For $\mathcal {LP}_{3,0}$, integrating by parts and using (2.4) and (A.3) yields

(4.42)\begin{align} \mathcal{LP}_{3,0}& \le C\nu\|a^2-1\|_{L^\infty H^N}\|\partial_YZ^+_0\|_{L^2H^N}^2+C\nu\|b\|_{L^2H^N}\|Z^+_0\|_{L^\infty H^N}\|\partial_YZ^+_0\|_{L^2H^N}\nonumber\\ & \le C \delta \left( \nu\left\| \nabla_L \tilde K Z_0^+\right\|_{L^2L^2}^2+\left\|\tilde K Z_0^+ \right\|_{L^\infty L^2}^2 \right) . \end{align}

Step IV: nonlinear estimates. We first collect all the above nonlinear terms to be estimated:

(4.43)\begin{equation} \begin{aligned} \mathrm{OLS}_4^{(5)} & ={-}\frac{1}{2\alpha}\int_0^t\left\langle S\partial_X^{{-}1}O^t_{2\alpha}\tilde K\Lambda Z^-_{{\neq}},\tilde K\mathrm{NL}_{{\neq}}^+\right\rangle {\rm d}t',\\ \mathrm{OLS}_{5,1}^{(5)} & ={-}\frac{1}{2\alpha}\int_0^t\left\langle \tilde K\partial_X^{{-}1}O^t_{2\alpha}\Lambda\mathrm{NL}^-_{{\neq}}, \tilde KSZ^+_{{\neq}}\right\rangle {\rm d}t',\\ \mathcal{NL} & =\int_0^t\left\langle \tilde KZ^+, \tilde K\mathrm{NL}^+\right\rangle {\rm d}t'. \end{aligned} \end{equation}

Owing to lemma C.1 and the fact $|S|\le \frac{1}{2}$, we observe that the bounds of the three quantities are essentially the same. To avoid unnecessary repetition, we only sketch the treatment of $\mathcal {NL}$ by modifying the nonlinear estimates in the Couette case, see (3.35)–(3.40). In fact, recalling the definition of $\mathrm {NL}^\pm$ (2.17), we write

\[ \mathcal{NL}=\int_0^t\left\langle \tilde KZ^+, \tilde K\mathrm{NLT}^+\right\rangle {\rm d}t'+\int_0^t\left\langle \tilde KZ^+, \tilde K\mathrm{NLS}^+\right\rangle {\rm d}t'=\mathcal{NLT}+\mathcal{NLS}, \]

here we use the shorthand notation $\mathrm {NLS}^\pm :=\mathrm {NLS}1^\pm +\mathrm {NLS}2^\pm$. Noting that

\[ \left\langle \tilde KZ^+_0, \tilde K \left((U^1_0+B^1_0)\partial_XZ^+\right)\right\rangle=0, \]

by virtue of (4.6), (4.8), lemmas 4.2 and C.1, and the hypotheses (4.10) and (4.11), one deduces that

(4.44)\begin{align} \mathcal{NLT} & \le\|\tilde KZ^+_{{\neq}}\|_{L^2L^2}\|U^1_0+ B^1_0\|_{L^\infty H^N}\|\partial_XZ^+_{{\neq}}\|_{L^2H^N}\nonumber\\ & \quad+\|\tilde KZ^+\|_{L^\infty L^2}\|\nabla^\bot_t\Delta^{{-}1}_{t}O^t_{2\alpha}Z^-_{{\neq}}\cdot\nabla_tZ^+_{{\neq}}\|_{L^1H^N}\nonumber\\ & \quad+\|\tilde KZ^+\|_{L^\infty L^2}\left\|a\partial_X\Delta_L^{{-}1}\Lambda O^t_{2\alpha}Z^-_{{\neq}}\partial_YZ^+_0\right\|_{L^1H^N}\nonumber\\ & \le C\nu^{-\frac{\tilde\delta}{2}}\|\tilde KZ^+_{{\neq}}\|_{L^2L^2}\|U^1_0+B^1_0\|_{L^\infty H^N}\|\nabla_L\tilde KZ^+_{{\neq}}\|_{L^2L^2}\nonumber\\ & \quad+C\nu^{-\tilde\delta}\|\tilde KZ^+\|_{L^\infty L^2}\left(\|\tilde KZ^-_{{\neq}}\|_{L^2L^2}+\nu^{-\frac{1}{3}}\left\|\sqrt{-\frac{\dot{M}_1}{M_1}}\tilde KZ^-_{{\neq}}\right\|_{L^2L^2}\right)\nonumber\\ & \quad\times \|\nabla_L\tilde KZ^+_{{\neq}}\|_{L^2L^2}\nonumber\\ & \quad+C\nu^{-\frac{\tilde\delta}{2}}\|\tilde KZ^+\|_{L^\infty L^2}\left\|\sqrt{-\frac{\dot{M}_1}{M_1}}\tilde KZ^-_{{\neq}}\right\|_{L^2L^2}\|\partial_YZ^+_0\|_{L^2H^N}\nonumber\\ & \le C\nu^{-\frac{5}{6}-\tilde\delta}\epsilon^3. \end{align}

Similar to (3.40), using (2.21), (2.23), (3.39), (A.3), (A.1), and lemma C.1, we arrive at

(4.45)\begin{align} \mathcal{NLS}& \le C \|\tilde KZ^+\|_{L^\infty L^2}\left(\left\| aS\Lambda O^t_{2\alpha}Z^{-}_{{\neq}}(2\partial_{XX}\Delta^{{-}1}_L\Lambda Z^+_{{\neq}}-Z^+)\right\|_{L^1H^N}\right.\nonumber\\ & \quad\left.+\left\|aS\Lambda Z^+_{{\neq}}(2\partial_{XX}\Delta^{{-}1}_L\Lambda O^t_{2\alpha}Z^{-}_{{\neq}}-O^t_{2\alpha}Z^{-})\right\|_{L^1H^N}\right)\nonumber\\ & \le C\|\tilde KZ^+\|_{L^\infty L^2}\left(\nu^{-\frac{\tilde\delta}{2}}\|\tilde m^{1/2}Z^\mp_{{\neq}}\|_{L^2H^N}\right)\left(\nu^{-(\frac{1}{3}+\frac{\tilde\delta}{2})}\|\tilde m^{1/2}Z^\pm_\ne\|_{L^2H^N}\right)\nonumber\\ & \quad+C\|\tilde KZ^+_{{\neq}}\|_{L^2 L^2}\left(\nu^{-\frac{\tilde\delta}{2}}\|\tilde m^{1/2}Z^\mp_{{\neq}}\|_{L^2H^N}\right)\left(\nu^{-(\frac{1}{3}+\frac{\tilde\delta}{2})}\|\tilde m^{1/2}Z^\pm_0\|_{L^\infty H^N}\right)\nonumber\\ & \le C\nu^{-\frac{2}{3}-\tilde\delta}\epsilon^3. \end{align}

Now collecting the above estimates in Step I–IV, we conclude that there exist positive constant $\alpha _0$ (sufficiently large) and $\delta _0$ (sufficiently small) independent of $\nu$, such that if $|\alpha |\ge \alpha _0$ and $\delta \le \delta _0$, there holds

(4.46)\begin{align} & \|\tilde KZ^+(t)\|^2_{L^2}+\nu\|\nabla_L\tilde KZ^+\|^2_{L^2L^2}+\left\|\sqrt{-\frac{\dot{M}}{M}}\tilde KZ^+\right\|^2_{L^2L^2}\nonumber\\& \quad \le3\|\tilde KZ^+(0)\|^2_{L^2}+C\nu^{-\frac{5}{6}-\tilde\delta}\epsilon^3, \end{align}

where the constant $C$ depends only on $N$ and $\alpha$. It suffices to improve (4.10) as long as $\epsilon \ll \nu ^{\frac{5}{6}+\tilde \delta }$.

The improvement of (4.11) is similar to that of (3.20). Firstly, we write the equations of $U_0^1\mp B_0^1$:

\begin{align*} \partial_t\left(U^1_0\mp B^1_0\right)-\nu\partial_{YY}\left(U^1_0\mp B^1_0\right)& =\nu(a^2-1)\partial_{YY}\left(U^1_0\mp B^1_0\right)+\nu b\partial_Y\left(U^1_0\mp B^1_0\right)\\ & \quad-\left((U_\neq{\pm} B_\neq)\cdot\nabla_t (U^1_\neq{\mp} B^1_\neq)\right)_0. \end{align*}

Then we have the energy identity

(4.47)\begin{align} & \frac{1}{2}\left\|(U^1_0\mp B^1_0)(t)\right\|_{H^N}^2+\nu\|\partial_Y(U^1_0\mp B^1_0)\|^2_{L^2H^N}\nonumber\\ & \quad=\frac{1}{2}\left\|(U^1_0\mp B^1_0)(0)\right\|_{H^N}^2\nonumber\\ & \qquad-\int_0^t\left\langle U^1_0\mp B^1_0, \left((U_\neq{\pm} B_\neq)\cdot\nabla_L (U^1_\neq{\mp} B^1_\neq)\right)_0\right\rangle_{H^N}{\rm d}t'\nonumber\\ & \qquad+\nu\int_0^t\left\langle U^1_0\mp B^1_0, (a^2-1)\partial_{YY}\left(U^1_0\mp B^1_0\right) \right\rangle_{H^N} {\rm d}t'\nonumber\\ & \qquad+\nu\int_0^t\left\langle U^1_0\mp B^1_0, b\partial_Y\left(U^1_0\mp B^1_0\right) \right\rangle_{H^N} {\rm d}t'\nonumber\\ & \quad=\frac{1}{2}\left\|(U^1_0\mp B^1_0)(0)\right\|_{H^N}^2+\sum_{q=1}^3I_q. \end{align}

Using the divergence free condition $\nabla _t\cdot (U\pm B)=0$, we find that

\begin{align*} I_1& =\int_0^t\left\langle \partial_Y\left(U^1_0\mp B^1_0\right), a\left((U_{{\neq}}^2\pm B_{{\neq}}^2) (U^1_\neq{\mp} B^1_\neq)\right)_0\right\rangle_{H^N}{\rm d}t'\nonumber\\ & \quad+\int_0^t\left\langle U^1_0\mp B^1_0, \partial_Ya\left((U_{{\neq}}^2\pm B_{{\neq}}^2) (U^1_\neq{\mp} B^1_\neq)\right)_0\right\rangle_{H^N}{\rm d}t'. \end{align*}

Recalling (2.4), it is easy to rewrite $\partial _Ya$ as follows

\[ \partial_Ya=b+\left(\frac{1}{1-(1-a)}-1\right)b=b+b\sum_{n=1}^{+\infty}a^n. \]

Combining this with (A.3) yields $\nu ^{\frac{1}{2}}\|\partial _Ya\|_{L^2H^N}\lesssim \delta$. On the other hand, similar to (3.45) and (3.46), we have

\[ \left\|U_\neq^2\pm B_\neq^2\right\|_{L^2H^N}= \left\|\partial_X\Delta_L^{{-}1}\Lambda O^t_\alpha Z^\mp \right\|_{L^2H^N}\lesssim \nu^{-\frac{\tilde\delta}{2}}\left\|\sqrt{-\frac{\dot{M}}{M}}\tilde KZ^\mp\right\|_{L^2L^2}, \]

and

\[ \|U^1_\neq{\mp} B^1_\neq\|_{L^\infty H^N}= \left\|a\partial_Y^L\Delta^{{-}1}_L\Lambda O^t_{-\alpha}Z^\pm_\ne\right\|_{L^\infty H^N}\lesssim\nu^{-\frac{\tilde\delta}{2}}\left\|\tilde KZ^\pm_\ne\right\|_{L^\infty L^2}. \]

It follows that

(4.48)\begin{align} I_1& \lesssim\nu^{-\tilde\delta}\left(1+\|a-1\|_{L^\infty H^N}\right)\left\|\partial_Y(U_0^1\mp B^1_0) \right\|_{L^2H^N}\nonumber\\ & \quad\times \left\|\sqrt{-\frac{\dot{M}}{M}}\tilde KZ^\mp\right\|_{L^2L^2}\left\|\tilde KZ^\pm_\ne\right\|_{L^\infty L^2}\nonumber\\ & \quad+\nu^{-\tilde\delta}\|\partial_ta\|_{L^2H^N}\left\|U_0^1\mp B^1_0 \right\|_{L^\infty H^N}\left\|\sqrt{-\frac{\dot{M}}{M}}\tilde KZ^\mp\right\|_{L^2L^2}\left\|\tilde KZ^\pm_\ne\right\|_{L^\infty L^2}. \end{align}

Integrating by parts, and using (A.3), it is easy to see that

(4.49)\begin{align} I_2+I_3\lesssim \delta \nu\|\partial_Y(U_0^1\mp B^1_0)\|_{L^2H^N}^2+\delta \nu^\frac{1}{2}\|U_0^1\mp B^1_0\|_{L^\infty H^N}\|\partial_Y(U_0^1\mp B^1_0)\|_{L^2H^N}.\end{align}

Substituting (4.48) and (4.49) into (4.47), and using the hypotheses (4.10) and (4.11), we have

(4.50)\begin{align} \|({U}^1_0\mp{B}^1_0)(t)\|_{H^N}^2+\nu\|\partial_Y({U}^1_0\mp{B}^1_0)\|^2_{L^2H^N}\le 2\|({U}^1_0\mp{B}^1_0)(0)\|_{H^N}^2+C\nu^{-\frac{1}{2}-\tilde\delta}\epsilon^3,\end{align}

for some constant $C$ independent of $\nu$. Combining (4.46) and (4.50), one deduces proposition 4.3 under the hypotheses $\epsilon \ll \nu ^{\frac{5}{6}+\tilde \delta }$ and hence of theorem 4.1.