On the Temporal Decay for the 2D Non-resistive Incompressible MHD Equations

Abstract

Califano–Chiuderi (Phys Rev E 60(Part B): 4701–4707, 1999) gave the numerical observation that the energy of the MHD equations is dissipated at a rate independent of the ohmic resistivity, which was first proved by Ren et al. (J Funct Anal 267: 503–541, 2014) (the initial data near $(0,\vec {e}_1)$, $\vec {e}_1=(1,0)$). Precisely, they showed some explicit decay rates of solutions in $L^2$ norm. So a nature question is whether the obtained decay rates in Ren et al. (J Funct Anal 267: 503–541, 2014) are optimal. In this paper, we aim at giving the explicit decay rates of solutions in both $L^2$ norm and $L^\infty$ norm. In particular, our decay rate in terms of $L^2$ norm improves the previous work Ren et al. (J Funct Anal 267: 503–541, 2014).

Appendix A

In this section, we give the proof of some lemmas.

Proof of Lemma 4.3

Thanks to (3.13), (4.3)$_7$ for $k=0$ and (4.3)$_1$ for $k=1$ and $r=2$, we get

$$\begin{aligned} \begin{aligned} \Vert M_1(\partial ,t)f\Vert _{\mathcal {F}L^2(D_4)} \lesssim&\ \min \{\Vert \mathcal {G}_{1,2}e^{-\mathcal {G}_{2,2}t}\widehat{f}\Vert _{L^2(D_4)}, \Vert \mathcal {G}_{1,1}e^{-\mathcal {G}_{2,2}t}\widehat{|\nabla |^{-1}f}\Vert _{L^2(D_4)}\}\\ \lesssim&\ \langle t\rangle ^{-\frac{1}{2}}\min \{\Vert f\Vert _{L^1\cap L^2},\Vert |\nabla |^{-1}f\Vert _{L^2}\}, \end{aligned} \end{aligned}$$

which yields (4.11)$_1$. It follows (4.11)$_2$ by using (3.13),

$$\begin{aligned} \Vert |\nabla |M_1(\partial ,t)f\Vert _{\mathcal {F}L^r(D_4)} \lesssim \ \Vert \mathcal {G}_{1,1}e^{-\mathcal {G}_{2,2}t}\widehat{f}\Vert _{L^r(D_4)} \end{aligned}$$

and (4.3)$_1$ for $k=1$. (4.11)$_3$ can be obtained by using (3.13),

$$\begin{aligned} \Vert \partial _xM_1(\partial ,t)f\Vert _{\mathcal {F}L^r(D_4)} \lesssim \ \Vert \mathcal {G}_{2,2}e^{-\mathcal {G}_{2,2}t}\widehat{f}\Vert _{L^r(D_4)} \end{aligned}$$

and (4.3)$_1$ for $k=2$. Combining with (3.13),

$$\begin{aligned} \Vert M_1(\partial ,t)f\Vert _{\mathcal {F}L^r(D_4)} \lesssim \ \Vert \mathcal {G}_{1,2}e^{-\mathcal {G}_{2,2}t}\widehat{f}\Vert _{L^r(D_4)}, \end{aligned}$$

(4.3)$_2$ for $k=1$ and $\delta =0.01$ leads (4.11)$_4$. Combining with (3.13),

$$\begin{aligned} \Vert |\nabla |^{-1}M_1(\partial ,t)f\Vert _{\mathcal {F}L^1(D_{42})} \lesssim \ \Vert \mathcal {G}_{1,3}e^{-\mathcal {G}_{2,2}t}\widehat{f}\Vert _{L^1(D_{42})} \end{aligned}$$

and (4.3)$_6$ for $k=0$, $q=p=4/3$, we can get (4.11)$_5$. It follows (4.11)$_6$ by using (3.13),

$$\begin{aligned} \Vert \mathcal {R}_1M_1(\partial ,t)f\Vert _{\mathcal {F}L^1(D_{42})} \lesssim \ \Vert \mathcal {G}_{2,3}e^{-\mathcal {G}_{2,2}t}\widehat{f}\Vert _{L^1(D_{42})}, \end{aligned}$$

and (4.3)$_4$ for $k=1$. It follows (4.11)$_7$ by using (3.13),

$$\begin{aligned} \Vert \mathcal {R}_1M_1(\partial ,t)f\Vert _{\mathcal {F}L^1(D_{4})} \lesssim \ \Vert \mathcal {G}_{2,3}e^{-\mathcal {G}_{2,2}t}\widehat{f}\Vert _{L^1(D_4)}, \end{aligned}$$

and (4.3)$_3$ for $k=1$. Applying

$$\begin{aligned} \Vert |\nabla |\mathcal {R}_1M_1(\partial ,t)f\Vert _{\mathcal {F}L^1(D_{4})} +\Vert \mathcal {R}_{11}M_1(\partial ,t)f\Vert _{\mathcal {F}L^1(D_{4})} \lesssim \ \Vert \mathcal {G}_{2,2}e^{-\mathcal {G}_{2,2}t}\widehat{f}\Vert _{L^1(D_4)}, \end{aligned}$$

and (4.3)$_1$ for $k=2$ and $r=1$ can lead (4.11)$_8$. Similarly, using

$$\begin{aligned} \Vert \partial _x\mathcal {R}_1M_1(\partial ,t)f\Vert _{\mathcal {F}L^1(D_{4})} \!+\!\Vert |\nabla |\mathcal {R}_{11}M_1(\partial ,t)f\Vert _{\mathcal {F}L^1(D_{4})} \!\lesssim \! \Vert \mathcal {G}_{3,3}e^{-\mathcal {G}_{2,2}t}\widehat{f}\Vert _{L^1(D_4)}, \end{aligned}$$

and (4.3)$_1$ for $k=3$ and $r=1$ can lead (4.11)$_9$. $\square$

Proof of Lemma 4.4

Using (3.13) and (4.3)$_1$ for $k=0$, one can easily get (4.12)$_1$. By (3.13),

$$\begin{aligned} \Vert \partial _xM_2(\partial ,t)f\Vert _{\mathcal {F}L^r(D_{4})} \lesssim \ \Vert |\nabla |\mathcal {G}_{1,1}e^{-\mathcal {G}_{2,2}t}\widehat{f}\Vert _{L^r(D_4)}, \end{aligned}$$

and (4.3)$_1$ for $k=1$, we can obtain (4.12)$_2$. Similarly, one can also get (4.12)$_3$. Using a similar way leading (4.5) for $k=0$, we can get (4.12)$_4$. Thanks to

$$\begin{aligned} \Vert \mathcal {R}_1^lM_2(\partial ,t)f\Vert _{\mathcal {F}L^1(D_{4})} \lesssim \ \langle t\rangle ^{-\frac{l}{2}}\Vert M_2(\partial ,t)f\Vert _{\mathcal {F}L^1(D_{4})} \lesssim \ \langle t\rangle ^{-\frac{l}{2}} \Vert \mathcal {G}_{0,0}e^{-\mathcal {G}_{2,2}t}\widehat{f}\Vert _{L^1(D_4)}, \end{aligned}$$

it follows (4.12)$_5$ by using (4.3)$_1$ for $k=0$ and $r=1$. Due to

$$\begin{aligned} \Vert \partial _x\mathcal {R}_1M_2(\partial ,t)f\Vert _{\mathcal {F}L^1(D_{4})} \lesssim \ \Vert |\nabla |\mathcal {R}_1^2M_2(\partial ,t)f\Vert _{\mathcal {F}L^1(D_{4})}, \end{aligned}$$

it follows (4.12)$_6$ by using (4.12)$_5$. $\square$

Proof of Lemma 4.5

Using (3.13), (4.1)$_2$ for $k=1$, (4.3)$_1$ for $k=1$ and $r=2$, we have

$$\begin{aligned} \Vert M_3(\partial ,t)f\Vert _{\mathcal {F}L^2(D_{4})}\lesssim & {} \ \Vert e^{\frac{1}{2}t\Delta }f\Vert _{\mathcal {F}L^2(D_{4})}+ \Vert |\vec {\xi }|^{-1}\mathcal {G}_{1,1}e^{-\mathcal {G}_{2,2}t}\widehat{f}\Vert _{L^2(D_4)}\\\lesssim & {} \ \langle t\rangle ^{-\frac{1}{2}}\Vert |\nabla |^{-1}f\Vert _{H^1}. \end{aligned}$$

Using (3.13), (4.1)$_2$ for $k=0$ and (4.3)$_7$ for $k=0$, we have

$$\begin{aligned} \Vert M_3(\partial ,t)f\Vert _{\mathcal {F}L^2(D_{4})} \lesssim \ \Vert e^{\frac{1}{2}t\Delta }f\Vert _{\mathcal {F}L^2(D_{4})}+ \Vert \mathcal {G}_{1,2}e^{-\mathcal {G}_{2,2}t}\widehat{f}\Vert _{L^2(D_4)} \lesssim \ \langle t\rangle ^{-\frac{1}{2}}\Vert f\Vert _{L^1\cap L^2}. \end{aligned}$$

Thus, we complete the proof of (4.13)$_1$. Using (3.13), one can get the first estimate of (4.13)$_2$ by applying

$$\begin{aligned} \Vert M_3(\partial ,t)f\Vert _{\mathcal {F}L^1(D_{4})} \lesssim \ \Vert e^{\frac{1}{2}t\Delta }f\Vert _{\mathcal {F}L^1(D_{4})}+ \Vert \mathcal {G}_{2,4}e^{-\mathcal {G}_{2,2}t}\widehat{f}\Vert _{L^1(D_4)}, \end{aligned}$$

(4.1)$_4$ for $k=0$, and (4.3)$_5$ for $k=0$. Using (3.13), one can get the second estimate of (4.13)$_2$ by applying

$$\begin{aligned} \Vert M_3(\partial ,t)f\Vert _{\mathcal {F}L^1(D_{4})} \lesssim \ \Vert |\nabla |e^{\frac{1}{2}t\Delta }|\nabla |^{-1}f\Vert _{\mathcal {F}L^1(D_{4})}+ \Vert \mathcal {G}_{2,3}e^{-\mathcal {G}_{2,2}t}|\vec {\xi }|^{-1}\widehat{f}\Vert _{L^1(D_4)}, \end{aligned}$$

(4.1)$_4$ for $k=1$, and (4.3)$_3$ for $k=1$. Using (3.13), one can get the first estimate of (4.13)$_3$ by applying

$$\begin{aligned} \Vert \partial _yM_3(\partial ,t)f\Vert _{\mathcal {F}L^2(D_{4})} \lesssim \ \Vert |\nabla |e^{\frac{1}{2}t\Delta }f\Vert _{\mathcal {F}L^2(D_{4})}+ \Vert \mathcal {G}_{1,1}e^{-\mathcal {G}_{2,2}t}\widehat{f}\Vert _{L^2(D_4)}, \end{aligned}$$

(4.1)$_2$ for $k=1$, and (4.3)$_1$ for $k=1$. Using (3.13), one can get the second estimate of (4.13)$_3$ by applying

$$\begin{aligned} \Vert \partial _yM_3(\partial ,t)f\Vert _{\mathcal {F}L^2(D_{4})} \lesssim \ \Vert |\nabla |^2e^{\frac{1}{2}t\Delta }|\nabla |^{-1}f\Vert _{\mathcal {F}L^2(D_{4})}+ \Vert \mathcal {G}_{2,2}e^{-\mathcal {G}_{2,2}t}|\vec {\xi }|^{-1}\widehat{f}\Vert _{L^2(D_4)}, \end{aligned}$$

(4.1)$_2$ for $k=2$, and (4.3)$_1$ for $k=2$. Using (3.13), one can get the third estimate of (4.13)$_3$ by applying

$$\begin{aligned} \Vert \partial _yM_3(\partial ,t)f\Vert _{\mathcal {F}L^2(D_{4})} \lesssim \ \Vert |\nabla |e^{\frac{1}{2}t\Delta }f\Vert _{\mathcal {F}L^2(D_{4})}+ \Vert \mathcal {G}_{2,3}e^{-\mathcal {G}_{2,2}t}\widehat{f}\Vert _{L^2(D_4)}, \end{aligned}$$

(4.1)$_1$ for $k=1$, and (4.3)$_7$ for $k=1$. So we conclude the proof of (4.13)$_3$. For (4.13)$_4$, we only show the case $r=2$, and other cases can be bounded similarly. Using $|\xi |\lesssim |\vec {\xi }|^2$,

$$\begin{aligned} \Vert \partial _xM_3(\partial ,t)f\Vert _{\mathcal {F}L^2(D_{4})} \lesssim \ \Vert \partial _xe^{\frac{1}{2}t\Delta }f\Vert _{\mathcal {F}L^2(D_{4})}+ \Vert \mathcal {G}_{2,2}e^{-\mathcal {G}_{2,2}t}\widehat{f}\Vert _{L^2(D_4)}, \end{aligned}$$

one can get (4.13)$_4$ for $r=2$ by (4.1)$_2$ for $k=2$ and (4.3)$_1$ for $k=2$. Using $|\xi |\lesssim |\vec {\xi }|^2$,

$$\begin{aligned} \Vert \partial _xM_3(\partial ,t)f\Vert _{\mathcal {F}L^1(D_{4})} \lesssim \ \Vert \partial _xe^{\frac{1}{2}t\Delta }f\Vert _{\mathcal {F}L^1(D_{4})}+ \Vert \mathcal {G}_{2,4}e^{-\mathcal {G}_{2,2}t}\widehat{\partial _xf}\Vert _{L^1(D_4)}, \end{aligned}$$

one can get the first bound of (4.13)$_5$ by (4.1)$_4$ for $k=2$ and (4.3)$_5$ for $k=0$. Using $|\xi |\lesssim |\vec {\xi }|^2$,

$$\begin{aligned} \Vert \partial _xM_3(\partial ,t)f\Vert _{\mathcal {F}L^1(D_{4})} \lesssim \ \Vert \partial _xe^{\frac{1}{2}t\Delta }f\Vert _{\mathcal {F}L^1(D_{4})}+ \Vert \mathcal {G}_{3,4}e^{-\mathcal {G}_{2,2}t}\widehat{f}\Vert _{L^1(D_4)}, \end{aligned}$$

one can get the second bound of (4.13)$_5$ by (4.1)$_4$ for $k=2$ and (4.3)$_3$ for $k=2$. Using

$$\begin{aligned} \Vert e^{ct\Delta }f\Vert _{\mathcal {F}L^1} \lesssim \ \langle t\rangle ^{-\frac{3}{4}}(\Vert f\Vert _{L^1_y(L^2_x)} +\Vert \widehat{f}\Vert _{L^1}), \end{aligned}$$

which can be proved by using the similar arguments yielding (4.2), and

$$\begin{aligned} \Vert \mathcal {G}_{2,4}e^{-\mathcal {G}_{2,2}t}\widehat{f}\Vert _{L^1(D_4)}\le & {} \ \Vert \mathcal {G}_{2,2}e^{-\mathcal {G}_{2,2}t}\widehat{f}\Vert _{L^1(D_{41})} +\Vert \mathcal {G}_{2,4}e^{-\mathcal {G}_{2,2}t}\widehat{f}\Vert _{L^1(D_{42})}\\\le & {} \ \langle t\rangle ^{-\frac{3}{4}}(\Vert \widehat{f}\Vert _{L^1}+\Vert f\Vert _{L^1_y(L^2_x)}), \end{aligned}$$

which can be obtained by using (4.3)$_1$ for $k=2$ and (4.3)$_6$ for $k=1$, $p=\frac{4}{3}$ and $q=2$, we have

$$\begin{aligned} \Vert \partial _xM_3(\partial ,t)f\Vert _{\mathcal {F}L^1(D_{4})}\lesssim & {} \ \Vert e^{\frac{1}{2}t\Delta }\partial _xf\Vert _{\mathcal {F}L^1(D_{4})}+ \Vert \mathcal {G}_{2,4}e^{-\mathcal {G}_{2,2}t}\widehat{\partial _xf}\Vert _{L^1(D_4)}\\\lesssim & {} \ \langle t\rangle ^{-\frac{3}{4}}(\Vert \widehat{\partial _xf}\Vert _{L^1}+\Vert \partial _xf\Vert _{L^1_y(L^2_x)}), \end{aligned}$$

which completes the proof of the third bound of (4.13)$_5$. Using (4.1)$_2$ for $k=2$, (4.3)$_1$ for $k=2$ and

$$\begin{aligned} \Vert \Delta M_3(\partial ,t)f\Vert _{\mathcal {F}L^2(D_{4})} \lesssim \ \Vert \Delta e^{\frac{1}{2}t\Delta }f\Vert _{\mathcal {F}L^2(D_{4})}+ \Vert \mathcal {G}_{2,2}e^{-\mathcal {G}_{2,2}t}\widehat{f}\Vert _{L^2(D_4)}, \end{aligned}$$

we can get (4.13)$_6$. Using $|\xi |\lesssim |\vec {\xi }|^2$, (4.1)$_4$ for $k=5$, (4.3)$_1$ for $k=5$ and

$$\begin{aligned} \Vert \partial _x^2\mathcal {R}_1 M_3(\partial ,t)f\Vert _{\mathcal {F}L^1(D_{4})} \lesssim \ \Vert \partial _x^2\mathcal {R}_1 e^{\frac{1}{2}t\Delta }f\Vert _{\mathcal {F}L^1(D_{4})}+ \Vert \mathcal {G}_{5,5}e^{-\mathcal {G}_{2,2}t}\widehat{f}\Vert _{L^1(D_4)}, \end{aligned}$$

we can get (4.13)$_7$. Using $|\xi |\lesssim |\vec {\xi }|^2$, (4.1)$_4$ for $k=3$, (4.3)$_1$ for $k=3$ and

$$\begin{aligned} \Vert \partial _x|\nabla | M_3(\partial ,t)f\Vert _{\mathcal {F}L^1(D_{4})} \lesssim \ \Vert \partial _x |\nabla | e^{\frac{1}{2}t\Delta }f\Vert _{\mathcal {F}L^1(D_{4})}+ \Vert \mathcal {G}_{3,3}e^{-\mathcal {G}_{2,2}t}\widehat{f}\Vert _{L^1(D_4)}, \end{aligned}$$

we can get (4.13)$_8$ for $r=1$. Other cases $1<r\le 2$ can be bounded similarly. Using $|\xi |\lesssim |\vec {\xi }|^2$, (4.1)$_4$ for $k=4$, (4.3)$_1$ for $k=4$ and

$$\begin{aligned} \Vert \partial _x^2 M_3(\partial ,t)f\Vert _{\mathcal {F}L^1(D_{4})} \lesssim \ \Vert \partial _x^2 e^{\frac{1}{2}t\Delta }f\Vert _{\mathcal {F}L^1(D_{4})}+ \Vert \mathcal {G}_{4,4}e^{-\mathcal {G}_{2,2}t}\widehat{f}\Vert _{L^1(D_4)}, \end{aligned}$$

we can get (4.13)$_9$. Using $|\xi |\lesssim |\vec {\xi }|^2$, (4.1)$_4$ for $k=2$, (4.3)$_1$ for $k=2$ and

$$\begin{aligned} \Vert \mathcal {R}_1 M_3(\partial ,t)f\Vert _{\mathcal {F}L^1(D_{4})} \lesssim \ \Vert \partial _x e^{\frac{1}{2}t\Delta }|\nabla |^{-1}f\Vert _{\mathcal {F}L^1(D_{4})}+ \Vert \mathcal {G}_{2,2}e^{-\mathcal {G}_{2,2}t}|\vec {\xi }|^{-1}\widehat{f}\Vert _{L^1(D_4)}, \end{aligned}$$

we can get (4.13)$_{10}$. Using $|\xi |\lesssim |\vec {\xi }|^2$, (4.1)$_4$ for $k=3$, (4.3)$_1$ for $k=3$ and

$$\begin{aligned} \Vert \partial _x\mathcal {R}_1 M_3(\partial ,t)f\Vert _{\mathcal {F}L^1(D_{4})} \lesssim \ \Vert \partial _x\mathcal {R}_1 e^{\frac{1}{2}t\Delta }f\Vert _{\mathcal {F}L^1(D_{4})}+ \Vert \mathcal {G}_{3,3}e^{-\mathcal {G}_{2,2}t}\widehat{f}\Vert _{L^1(D_4)}, \end{aligned}$$

we can get (4.13)$_{11}$. $\square$

Proof of Lemma 4.6

$\underline{(4.17)_1}$ By Hölder’s inequality, product estimate in one dimension, interpolation inequality and

$$\begin{aligned} \Vert \nabla P_\backsim (\vec {R'}\cdot \vec {b})\Vert _{L^2} \lesssim \ \Vert \vec {b}\Vert _{L^2}\lesssim \ \langle t\rangle ^{-\frac{1}{4}}\Vert V\Vert _3, \end{aligned}$$

(A.1)

we have

$$\begin{aligned}{} & {} \Vert \nabla P_\backsim (\vec {R'}\cdot \vec {b}) b\Vert _{L^1}+\Vert \vec {b}b\Vert _{L^1}\le \ \Vert \vec {b}\Vert _{L^2}\Vert b\Vert _{L^2} \lesssim \ \langle t\rangle ^{-\frac{1}{2}}\Vert V\Vert _3^2, \\{} & {} \Vert |\partial _x|^\beta (bb)\Vert _{L^1_x} \lesssim \ \Vert b\Vert _{L^2_x}\Vert |\partial _x|^\beta b\Vert _{L^2_x} \lesssim \ \Vert b\Vert _{L^2_x}^{2-\beta } \Vert \partial _xb\Vert _{L^2_x}^\beta , \end{aligned}$$

which yields

$$\begin{aligned} \Vert |\partial _x|^\beta (bb)\Vert _{L^1} \lesssim \ \Vert b\Vert _{L^2}^{2-\beta } \Vert \partial _xb\Vert _{L^2}^\beta \lesssim \ \langle t\rangle ^{-\frac{1+\beta }{2}}\Vert V\Vert _3^2. \end{aligned}$$

$\underline{(4.17)_2}$ Using Hölder’s inequality and $\partial _y v=-\partial _xu$, we have

$$\begin{aligned} \Vert \partial _yv_{<\langle t\rangle ^{-8}}b\Vert _{L^1} +\Vert \partial _yv_{>2\langle t\rangle ^{-0.05}}b\Vert _{L^1} \lesssim \ \Vert \partial _xu\Vert _{L^2}\Vert b\Vert _{L^2} \lesssim \ \langle t\rangle ^{-\frac{5}{4}}\Vert V\Vert _3^2. \end{aligned}$$

Thanks to $\Vert \partial _x P_\backsim (\vec {R'}\cdot \vec {b})\Vert _{L^2} \lesssim \ \Vert B\Vert _{L^2}$, we can get

$$\begin{aligned} \Vert \partial _xP_\backsim (\vec {R'}\cdot \vec {b})\partial _yu\Vert _{L^1} \le \ \Vert \partial _xP_\backsim (\vec {R'}\cdot \vec {b})\Vert _{L^2} \Vert \partial _yu\Vert _{L^2} \lesssim \ \Vert B\Vert _{L^2}\Vert \partial _yu\Vert _{L^2} \lesssim \ \langle t\rangle ^{-\frac{5}{4}}\Vert V\Vert _3^2. \end{aligned}$$

$\underline{(4.17)_3}$ Using product estimate and interpolation inequality, we have

$$\begin{aligned} \Vert |\nabla |^\beta (\vec {u}\otimes \vec {u})\Vert _{L^1} \lesssim \ \Vert \vec {u}\Vert _{L^2}\Vert |\nabla |^\beta \vec {u}\Vert _{L^2} \lesssim \ \Vert \vec {u}\Vert _{L^2}^{2-\beta }\Vert \nabla \vec {u}\Vert _{L^2}^\beta \lesssim \ \langle t\rangle ^{-1-\frac{\beta }{4}}\Vert V\Vert _3^2 \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \Vert |\nabla |^{0.99}(\vec {b}B)\Vert _{L^1} \lesssim&\ \Vert |\nabla |^{0.99}\vec {b}\Vert _{L^2} \Vert B\Vert _{L^2} +\Vert |\nabla |^{0.99} B\Vert _{L^2} \Vert \vec {b}\Vert _{L^2}\\ \lesssim&\ \langle t\rangle ^{-0.21-0.5}\Vert V\Vert _3^2 +\langle t\rangle ^{-0.25-0.6}\Vert V\Vert _3^2\\ \lesssim&\ \langle t\rangle ^{-0.7}\Vert V\Vert _3^2, \end{aligned} \end{aligned}$$

where we have used

$$\begin{aligned} \Vert |\nabla |^{0.99}\vec {b}\Vert _{L^2} \le \ \Vert |\nabla |^{0.99}\vec {b}_{\le \langle t\rangle ^{0.03}}\Vert _{L^2} +\Vert |\nabla |^{0.99}\vec {b}_{>\langle t\rangle ^{0.03}}\Vert _{L^2}\lesssim \langle t\rangle ^{-0.21} \Vert V\Vert _3 \end{aligned}$$

and

$$\begin{aligned} \Vert |\nabla |^{0.99}B\Vert _{L^2} \lesssim \ \Vert B\Vert _{L^2}^{0.01}\Vert \nabla B\Vert _{L^2}^{0.99} \lesssim \ \langle t\rangle ^{-0.6}\Vert V\Vert _3. \end{aligned}$$

This concludes the proof of Lemma 4.6. $\square$

Proof of Lemma 4.7

$\underline{(4.18)_1}$ Using interpolation inequality

$$\begin{aligned} \Vert f\Vert _{L^\frac{2p}{2-p}_x}\lesssim \ \Vert f\Vert _{L^2_x}^\frac{1}{p}\Vert \partial _xf\Vert _{L^2_x}^{1-\frac{1}{p}}, \end{aligned}$$

we obtain

$$\begin{aligned} \begin{aligned} \Vert b\partial _xb\Vert _{L^1_y(L^p_x)} \lesssim&\ \big \Vert \Vert b\Vert _{L^\frac{2p}{2-p}_x}\Vert \partial _xb\Vert _{L^2_x} \big \Vert _{L^1_y} \lesssim \ \big \Vert \Vert b\Vert _{L^2_x}^\frac{1}{p}\Vert \partial _xb\Vert _{L^2_x}^{2-\frac{1}{p}} \big \Vert _{L^1_y}\\ \lesssim&\ \Vert b\Vert _{L^2}^\frac{1}{p} \Vert \partial _xb\Vert _{L^2}^{2-\frac{1}{p}} \lesssim \ \langle t\rangle ^{-\frac{3}{2}+\frac{1}{2p}}\Vert V\Vert _3^2. \end{aligned} \end{aligned}$$

$\underline{(4.18)_2}$ We only give the estimate of $\Vert \langle \nabla \rangle ^3(\vec {b}\cdot \nabla u)\Vert _{L^1_x(L^2_y)}$, since other terms can be bounded similarly. Using

$$\begin{aligned} \Vert \langle \nabla \rangle ^k f\Vert _{L^1_x(L^2_y)} \lesssim \ \Vert f\Vert _{L^1_x(L^2_y)} +\Vert \nabla ^k f\Vert _{L^1_x(L^2_y)},\ k\ge 0, \end{aligned}$$

(A.2)

we have

$$\begin{aligned} \begin{aligned}&\Vert \langle \nabla \rangle ^3(\vec {b}\cdot \nabla u)\Vert _{L^1_x(L^2_y)}\\&\quad \lesssim \ \Vert \langle \nabla \rangle ^3(B\partial _y u)\Vert _{L^1_x(L^2_y)} +\Vert \langle \nabla \rangle ^3(b\partial _x u)\Vert _{L^1_x(L^2_y)}\\&\quad \lesssim \ \Vert B\partial _y u\Vert _{L^1_x(L^2_y)} +\Vert B\partial _y \nabla ^3u\Vert _{L^1_x(L^2_y)}+\ other\ similar\ terms. \end{aligned} \end{aligned}$$

By interpolation inequality, we can get

$$\begin{aligned} \Vert B\partial _y u\Vert _{L^1_x(L^2_y)} \lesssim \ \Vert B\Vert _{L^2_x(L^\infty _y)}\Vert u\Vert _{L^2} \lesssim \ \Vert B\Vert _{L^2}^\frac{1}{2} \Vert \partial _xb\Vert _{L^2}^\frac{1}{2}\Vert u\Vert _{L^2} \lesssim \ \langle t\rangle ^{-\frac{9}{8}}\Vert V\Vert _3^2. \end{aligned}$$

Using

$$\begin{aligned} \Vert \nabla \langle \nabla \rangle ^3 u\Vert _{L^2} \le \ \Vert \nabla \langle \nabla \rangle ^3 u_{<\langle t\rangle ^{\frac{3}{28}}}\Vert _{L^2} +\Vert \nabla \langle \nabla \rangle ^3 u_{\ge \langle t\rangle ^{\frac{3}{28}}}\Vert _{L^2} \lesssim \ \langle t\rangle ^{-\frac{3}{7}}\Vert V\Vert _3^2,\nonumber \\ \end{aligned}$$

(A.3)

then

$$\begin{aligned} \begin{aligned} \Vert B\partial _y \nabla ^3u\Vert _{L^1_x(L^2_y)} \lesssim&\ \Vert B\Vert _{L^2_x(L^\infty _y)} \Vert \partial _y \nabla ^3u\Vert _{L^2} \lesssim \ \langle t\rangle ^{-\frac{3}{7}-\frac{5}{8}}\Vert V\Vert _3^2 \lesssim \ \langle t\rangle ^{-1.01}\Vert V\Vert _3^2. \end{aligned} \end{aligned}$$

Thus,

$$\begin{aligned} \Vert \langle \nabla \rangle ^3(\vec {b}\cdot \nabla u)\Vert _{L^1_x(L^2_y)}\lesssim \ \langle t\rangle ^{-1.01}\Vert V\Vert _3^2. \end{aligned}$$

$\underline{(4.18)_3}$ Use (A.2), we only show the estimates of

$$\begin{aligned} \Vert v_{<\langle t\rangle ^{-8}}\partial _y\nabla ^3b\Vert _{L^1_x(L^2_y)},\ \textrm{and}\ \ \Vert v_{>2\langle t\rangle ^{-0.05}}\partial _y\nabla ^3b\Vert _{L^1_x(L^2_y)}, \end{aligned}$$

and other terms can be bounded similarly. Using Hölder’s inequality and interpolation inequality, one can get

$$\begin{aligned} \begin{aligned} \Vert v_{<\langle t\rangle ^{-8}}\partial _y\nabla ^3b\Vert _{L^1_x(L^2_y)} \lesssim&\ \Vert v_{<\langle t\rangle ^{-8}}\Vert _{L^2_x(L^\infty _y)}\Vert \partial _y\nabla ^3 b\Vert _{L^2}\\ \lesssim&\ \Vert v\Vert _{L^2}^\frac{1}{2}\Vert \partial _yv_{<\langle t\rangle ^{-8}}\Vert _{L^2}^\frac{1}{2} \Vert b\Vert _{H^4}\\ \lesssim&\ \langle t\rangle ^{-2}\Vert V\Vert _3^2,\\ \Vert v_{>2\langle t\rangle ^{-0.05}}\partial _y\nabla ^3b\Vert _{L^1_x(L^2_y)} \lesssim&\ \Vert v_{>2\langle t\rangle ^{-0.05}}\Vert _{L^2_x(L^\infty _y)} \Vert \partial _y\nabla ^3b\Vert _{L^2}\\ \lesssim&\ \Vert v_{>2\langle t\rangle ^{-0.05}}\Vert _{L^2}^\frac{1}{2} \Vert \partial _yv_{>2\langle t\rangle ^{-0.05}}\Vert _{L^2}^\frac{1}{2}\\&\ \times (\Vert \partial _y\nabla ^3b_{<\langle t\rangle ^{0.05}}\Vert _{L^2}+ \Vert \partial _y\nabla ^3b_{\ge \langle t\rangle ^{0.05}}\Vert _{L^2})\\ \lesssim&\ \langle t\rangle ^{0.025}\Vert \nabla v\Vert _{L^2} (\Vert \partial _y\nabla ^3b_{<\langle t\rangle ^{0.05}}\Vert _{L^2}+ \Vert \partial _y\nabla ^3b_{\ge \langle t\rangle ^{0.05}}\Vert _{L^2})\\ \lesssim&\ \langle t\rangle ^{0.025-1}(\langle t\rangle ^{0.2-0.25}+\langle t\rangle ^{-0.2})\Vert V\Vert _3^2\\ \lesssim&\ \langle t\rangle ^{-1.01}\Vert V\Vert _3^2. \end{aligned} \end{aligned}$$

$\underline{(4.18)_4}$ Using (A.2), we only estimate $\Vert B\partial _y \nabla ^2b\Vert _{L^1_x(L^2_y)}$, and other terms can be controlled similarly. Using interpolation inequality, and

$$\begin{aligned} \Vert \langle \nabla \rangle ^3b\Vert _{L^2}\lesssim \ \Vert \langle \nabla \rangle ^3b_{<\langle t\rangle ^{\frac{1}{32}}}\Vert _{L^2}+ \Vert \langle \nabla \rangle ^3b_{\ge \langle t\rangle ^{\frac{1}{32}}}\Vert _{L^2} \lesssim \ \langle t\rangle ^{-\frac{5}{32}}\Vert V\Vert _3, \end{aligned}$$

(A.4)

we have

$$\begin{aligned} \Vert B\partial _y \nabla ^2b\Vert _{L^1_x(L^2_y)} \lesssim \ \Vert B\Vert _{L^2_x(L^\infty _y)}\Vert \langle \nabla \rangle ^3b\Vert _{L^2} \lesssim \ \langle t\rangle ^{-\frac{5}{8}-\frac{5}{32}}\Vert V\Vert _3^2 \lesssim \ \langle t\rangle ^{-0.75}\Vert V\Vert _3^2. \end{aligned}$$

$\underline{(4.18)_5}$ Using (A.2) again, we only bound $\Vert \partial _x\langle \nabla \rangle ^2P_\backsim u b \Vert _{L^1_x(L^2_y)}$, while other terms can be bounded similarly. Using interpolation inequality, and

$$\begin{aligned} \Vert \partial _x\langle \nabla \rangle ^3 u\Vert _{L^2} \le \ \Vert \partial _x\langle \nabla \rangle ^3 u_{<\langle t\rangle ^{0.2}}\Vert _{L^2} +\Vert \partial _x\langle \nabla \rangle ^3 u_{\ge \langle t\rangle ^{0.2}}\Vert _{L^2} \lesssim \ \langle t\rangle ^{-0.8}\Vert V\Vert _3, \end{aligned}$$

we have

$$\begin{aligned} \begin{aligned} \Vert \partial _x\langle \nabla \rangle ^2P_\backsim u b \Vert _{L^1_x(L^2_y)} \lesssim&\ \Vert \partial _x\langle \nabla \rangle ^2 u\Vert _{L^2_x(L^\infty _y)} \Vert b\Vert _{L^2}\\ \lesssim&\ \Vert \partial _x\langle \nabla \rangle ^3 u\Vert _{L^2} \Vert b\Vert _{L^2}\\ \lesssim&\ \langle t\rangle ^{-1.05}\Vert V\Vert _3^2. \end{aligned} \end{aligned}$$

$\square$

Proof of Lemma 4.8

By Hölder’s inequality, and (A.1), it is easy to get the estimate of (4.19)$_1$ and (4.19)$_2$. Let us begin with the estimate of (4.19)$_3$.

$\underline{(4.19)_3}$ We only give the estimate of $\Vert \vec {b}\cdot \nabla \vec {b}\Vert _{H^2}$, while one can bound other terms by the similar way. We have

$$\begin{aligned} \Vert \vec {b}\cdot \nabla \vec {b}\Vert _{H^2} \le \ \Vert b\partial _x\nabla ^2b\Vert _{L^2}+\ other\ similar\ terms. \end{aligned}$$

Using

$$\begin{aligned} \Vert \partial _x\nabla ^2b\Vert _{L^2} \lesssim \ \Vert \partial _x\nabla ^2b_{<\langle t\rangle ^{\frac{1}{8}}}\Vert _{L^2} +\Vert \partial _x\nabla ^2b_{\ge \langle t\rangle ^{\frac{1}{8}}}\Vert _{L^2} \lesssim \ \langle t\rangle ^{-\frac{5}{8}}\Vert V\Vert _3^2, \end{aligned}$$

(A.5)

we have

$$\begin{aligned} \Vert b\partial _x\nabla ^2b\Vert _{L^2} \lesssim \ \Vert b\Vert _{L^\infty }\Vert \partial _x\nabla ^2b\Vert _{L^2} \lesssim \ \langle t\rangle ^{-1.1}\Vert V\Vert _3^2. \end{aligned}$$

Thus,

$$\begin{aligned} \Vert \vec {b}\cdot \nabla \vec {b}\Vert _{H^2} \lesssim \ \langle t\rangle ^{-1.1}\Vert V\Vert _3^2. \end{aligned}$$

$\underline{(4.19)_4}$ Using

$$\begin{aligned} \Vert \partial _xP_\backsim (\vec {R'}\cdot \vec {b})\Vert _{L^\infty }\lesssim \ \Vert \widehat{B}\Vert _{L^1}\lesssim \ \langle t\rangle ^{-1}\Vert V\Vert _3 \end{aligned}$$

and (A.4), one can get

$$\begin{aligned} \Vert \partial _xP_\backsim (\vec {R'}\cdot \vec {b})\partial _yb\Vert _{L^2} \lesssim \ \langle t\rangle ^{-1.1}\Vert V\Vert _3^2. \end{aligned}$$

Then we can get the estimates of other terms by the similar way.

$\underline{(4.19)_5}$ Here we only bound $\Vert P_\backsim (\vec {R'}\cdot \vec {b})\partial _x\nabla ^2b\Vert _{L^2}$. Using (A.5), and

$$\begin{aligned} \Vert P_\backsim (\vec {R'}\cdot \vec {b})\Vert _{\mathcal {F}L^1} \lesssim&\ \Vert |\nabla |^{-1}b\Vert _{\mathcal {F}L^1}+\sum _{\langle t\rangle ^{-8}\le M\le 2\langle t\rangle ^{-0.05}} \Vert |\nabla |^{-1}P_M B\Vert _{\mathcal {F}L^1}\nonumber \\ \lesssim&\ \langle t\rangle ^{-0.5}\Vert V\Vert _3+\sum _{\langle t\rangle ^{-8}\le M\le 2\langle t\rangle ^{-0.05}} M^{-1} \Vert P_M B\Vert _{\mathcal {F}L^1}\nonumber \\ \lesssim&\ \langle t\rangle ^{-0.5}\Vert V\Vert _3+\sum _{\langle t\rangle ^{-8}\le M\le 2\langle t\rangle ^{-0.05}} \Vert P_M B\Vert _{ L^2}\nonumber \\ \lesssim&\ \langle t\rangle ^{-0.5}\Vert V\Vert _3+\Vert B\Vert _{L^2}\sum _{\langle t\rangle ^{-8}\le M\le 2\langle t\rangle ^{-0.05}} M^{0.001}M^{-0.001}\nonumber \\ \lesssim&\ \langle t\rangle ^{-0.5}\Vert V\Vert _3+\langle t\rangle ^{0.008-0.5}\Vert V\Vert _3 \lesssim \ \langle t\rangle ^{-0.492}\Vert V\Vert _3. \end{aligned}$$

(A.6)

we have

$$\begin{aligned} \Vert P_\backsim (\vec {R'}\cdot \vec {b})\partial _x\nabla ^2b\Vert _{L^2} \lesssim \Vert P_\backsim (\vec {R'}\cdot \vec {b})\Vert _{\mathcal {F}L^1} \Vert \partial _x\nabla ^2b\Vert _{L^2} \lesssim \ \langle t\rangle ^{-1.1}\Vert V\Vert _3^2. \end{aligned}$$

$\underline{(4.19)_6}$ Here we only show the estimates of $\Vert b\partial _x^2\nabla ^2b\Vert _{L^2}$ and $\Vert P_\backsim (\vec {R'}\cdot \vec {b})\partial _y\nabla ^3u\Vert _{L^2}$. Using

$$\begin{aligned} \Vert \partial _x\langle \nabla \rangle ^3b\Vert _{L^2} \lesssim \ \Vert \partial _x\langle \nabla \rangle ^3b_{<\langle t\rangle ^\frac{1}{8}}\Vert _{L^2} +\Vert \partial _x\langle \nabla \rangle ^3b_{\ge \langle t\rangle ^\frac{1}{8}}\Vert _{L^2} \lesssim \ \langle t\rangle ^{-\frac{1}{2}}\Vert V\Vert _3^2, \end{aligned}$$

we have

$$\begin{aligned} \Vert b\partial _x^2\nabla ^2b\Vert _{L^2} \lesssim \ \Vert b\Vert _{L^\infty }\Vert \partial _x^2\nabla ^2b\Vert _{L^2} \lesssim \ \langle t\rangle ^{-1}\Vert V\Vert _3^2. \end{aligned}$$

Thanks to (A.6) and (A.3), we can get the estimate of $\Vert P_\backsim (\vec {R'}\cdot \vec {b})\partial _y\nabla ^3u\Vert _{L^2}$ by using Hölder’s inequality.

$\underline{(4.19)_7}$ By using (A.6) and (A.4), one can easily get this estimate. $\square$

Proof of Lemma 4.9

$\underline{(4.20)_1}$ The first three terms can be bounded easily. By

$$\begin{aligned} \Vert \nabla P_\backsim (\vec {R'}\cdot \vec {b})\Vert _{\mathcal {F}L^1}\lesssim \ \Vert \vec {b}\Vert _{\mathcal {F}L^1} \lesssim \ \langle t\rangle ^{-\frac{1}{2}}\Vert V\Vert _3, \end{aligned}$$

we can get

$$\begin{aligned} \Vert [\partial _x(\partial _yP_\backsim (\vec {R'}\cdot \vec {b})b),\nabla P_\backsim (\vec {R'}\cdot \vec {b})u]\Vert _{\mathcal {F}L^1}\lesssim & {} \ \Vert \vec {b}\Vert _{\mathcal {F}L^1}(\Vert \partial _xb\Vert _{\mathcal {F}L^1}+\Vert \widehat{u}\Vert _{L^1})\\\lesssim & {} \ \langle t\rangle ^{-\frac{3}{2}}\Vert V\Vert _3^2. \end{aligned}$$

$\underline{(4.20)_2}$ Like the previous procedure, it suffices to estimate $\Vert b\partial _{xx}u\Vert _{\mathcal {F}L^1}$ and $\Vert \partial _yb\partial _{x}v\Vert _{\mathcal {F}L^1}$. Using

$$\begin{aligned}&\Vert \langle \nabla \rangle \partial _{x}u\Vert _{\mathcal {F}L^1} \le \ \Vert \langle \nabla \rangle \partial _{x}u_{\ge \langle t\rangle ^{0.5}}\Vert _{\mathcal {F}L^1} +\Vert \langle \nabla \rangle \partial _{x}u_{< \langle t\rangle ^{0.5}}\Vert _{\mathcal {F}L^1} \lesssim \ \langle t\rangle ^{-0.75}\Vert V\Vert _3,\nonumber \\&\Vert \langle \nabla \rangle b\Vert _{\mathcal {F}L^1}\le \ \Vert \langle \nabla \rangle b_{<\langle t\rangle ^{\frac{5}{69}}}\Vert _{\mathcal {F}L^1}+\Vert \langle \nabla \rangle b_{\ge \langle t\rangle ^\frac{5}{69}}\Vert _{\mathcal {F}L^1} \lesssim \ \langle t\rangle ^{-0.42}\Vert V\Vert _3,\nonumber \\&\Vert \partial _xv\Vert _{\mathcal {F}L^1} \lesssim \ \Vert \partial _xv_{<\langle t\rangle ^{0.15}}\Vert _{\mathcal {F}L^1}+\Vert \partial _xv_{\ge \langle t\rangle ^{0.15}}\Vert _{\mathcal {F}L^1} \lesssim \ \langle t\rangle ^{-0.85}\Vert V\Vert _3, \end{aligned}$$

(A.7)

one gets

$$\begin{aligned} \Vert b\partial _{xx}u\Vert _{\mathcal {F}L^1} \lesssim \ \Vert \widehat{b}\Vert _{L^1}\Vert \partial _{xx}u\Vert _{\mathcal {F}L^1} \lesssim \ \langle t\rangle ^{-\frac{5}{4}}\Vert V\Vert _3^2 \end{aligned}$$

and

$$\begin{aligned} \Vert \partial _yb\partial _xv\Vert _{\mathcal {F}L^1} \lesssim \ \Vert \partial _yb\Vert _{\mathcal {F}L^1}\Vert \partial _x v\Vert _{\mathcal {F}L^1} \lesssim \ \langle t\rangle ^{-\frac{5}{4}}\Vert V\Vert _3^2. \end{aligned}$$

$\underline{(4.20)_3}$ Using (A.6) and the fact that $P_\thickapprox$ can be bounded by the process dealing with $P_\thicksim$, we can get the desired estimate by Hölder’s inequality.

$\underline{(4.20)_4}$ Here we only show the estimate of $\Vert P_\backsim (\vec {R'}\cdot \vec {b})\partial _y\nabla ^2u\Vert _{\mathcal {F}L^1}$. Using

$$\begin{aligned} \Vert \langle \nabla \rangle ^3u\Vert _{\mathcal {F}L^1} \lesssim \ \Vert \langle \nabla \rangle ^3u_{<\langle t\rangle ^\frac{10}{69}}\Vert _{\mathcal {F}L^1} +\Vert \langle \nabla \rangle ^3u_{\ge \langle t\rangle ^\frac{10}{69}}\Vert _{\mathcal {F}L^1} \lesssim \ \langle t\rangle ^{-\frac{39}{69}}\Vert V\Vert _3 \end{aligned}$$

and (A.6), we have

$$\begin{aligned} \Vert P_\backsim (\vec {R'}\cdot \vec {b})\partial _y\nabla ^2u\Vert _{\mathcal {F}L^1} \lesssim \ \Vert P_\backsim (\vec {R'}\cdot \vec {b})\Vert _{\mathcal {F}L^1} \Vert \partial _y\nabla ^2u\Vert _{\mathcal {F}L^1} \lesssim \ \langle t\rangle ^{-1.01}\Vert V\Vert _3^2. \end{aligned}$$

$\underline{(4.20)_5}$ We only give the estimate of $\Vert \partial _x(b\partial _xb)\Vert _{\mathcal {F}L^1}$. Using

$$\begin{aligned} \Vert \partial _x^2b\Vert _{\mathcal {F}L^1} \le \ \Vert \partial _x^2b_{<\langle t\rangle ^\frac{10}{59}}\Vert _{\mathcal {F}L^1} +\Vert \partial _x^2b_{\ge \langle t\rangle ^\frac{10}{59}}\Vert _{\mathcal {F}L^1} \lesssim \ \langle t\rangle ^{-0.83}\Vert V\Vert _3, \end{aligned}$$

we have

$$\begin{aligned} \Vert \partial _x(b\partial _xb)\Vert _{\mathcal {F}L^1} \le \ \Vert \partial _xb\partial _xb\Vert _{\mathcal {F}L^1}+\Vert b\partial _x^2b\Vert _{\mathcal {F}L^1} \lesssim \ \langle t\rangle ^{-1.3}\Vert V\Vert _3^2. \end{aligned}$$

$\underline{(4.20)_6}$ Using the same way yielding (A.7)$_3$, we can get

$$\begin{aligned} \Vert \nabla B\Vert _{\mathcal {F}L^1}\lesssim \ \langle t\rangle ^{-0.85}\Vert V\Vert _3, \end{aligned}$$

which, together with (A.7)$_2$ yields

$$\begin{aligned} \Vert \nabla (B\vec {b})\Vert _{\mathcal {F}L^1} \lesssim \ \langle t\rangle ^{-1.2}\Vert V\Vert _3^2. \end{aligned}$$

Other terms can be bounded similarly.

$\underline{(4.20)_7}$ We only estimate $\Vert \nabla ^2 B\partial _y b\Vert _{\mathcal {F}L^1}$, while other terms can be bounded similarly. Using the same way yielding (A.7)$_3$, one has

$$\begin{aligned} \Vert \nabla ^2 B\Vert _{\mathcal {F}L^1}\lesssim \ \langle t\rangle ^{-0.71}\Vert V\Vert _3, \end{aligned}$$

which, with (A.7)$_2$ leads

$$\begin{aligned} \Vert \nabla ^2 B\partial _y b\Vert _{\mathcal {F}L^1} \lesssim \ \Vert \nabla ^2 B\Vert _{\mathcal {F}L^1} \Vert \partial _y b\Vert _{\mathcal {F}L^1} \lesssim \ \langle t\rangle ^{-1.1}\Vert V\Vert _3^2. \end{aligned}$$

Using (A.7)$_2$ and (A.6), we have

$$\begin{aligned} \Vert P_\backsim (\vec {R'}\cdot \vec {b})\partial _yb\Vert _{\mathcal {F}L^1} \lesssim \ \Vert P_\backsim (\vec {R'}\cdot \vec {b})\Vert _{\mathcal {F}L^1} \Vert \partial _yb\Vert _{\mathcal {F}L^1} \lesssim \ \langle t\rangle ^{-0.9}\Vert V\Vert _3^2. \end{aligned}$$

$\square$