Global Well-Posedness and Large-Time Behavior to the 3D Two-Fluid Model with Degenerate Viscosities

Hongxia Ji, Wenjing Song

School of Science, Xi’an Polytechnic University, Xi’an, China.

DOI: 10.4236/jamp.2025.131010                

Abstract

In the paper, we study a compressible two-fluid model in ${ℝ}^3$ , where ${\gamma }^{\pm }>1$ . The pressure of the two fluids is equal. Different from previous research, we consider that viscosity coefficient both $\mu$ and $\lambda$ are functions of density. The global well-posedness of the three-dimensional compressible two-phase flow model is an open problem due to its dissipative, nonlinear structure. In the paper, setting $m^{\pm }=M^{\pm }$ and $Z=P-\overline{P}$ , by exploiting the dissipation structure, we obtain energy estimates for $\left(Z\mathrm{,}w\mathrm{,}n\right)$ and its derivatives, then we obtain the time decay rates for $\left(Z\mathrm{,}w\mathrm{,}n\right)$ . So we derive global well-posedness and large time behavior to the three dimensional compressible two-fluid model.

Keywords

Two-Fluid Model, Degenerate Viscosities, Large-Time Behavior, Global Well-Posedness

1. Introduction

Two-phase flow refers to a flow system consisting of two substances in different phases (at least one of which is a fluid). The flow system may be in different combinations such as gas-liquid and liquid-solid. Two-phase flow is also widely present in nature and other engineering fields, such as the drift of rain, snow, clouds and fog [1] [2], blood circulation in living organisms [3] [4], sediment movement [5] [6], high-speed aerated water flow in hydraulic engineering, and air pollution caused by smoke and dust in environmental engineering. With the passage of time and the development of science, the problem of two-phase flow has received increasing attention. Two-phase flow plays an important role in aerospace [7] [8], biomedical engineering [9]-[13], chemical engineering [14]-[16] and other fields. However, the analysis of two-phase flow problem is relatively complex. It involves knowledge from multiple fields such as fluid mechanics, thermodynamics, and heat transfer. People use different models to study two-phase flow problems. The most common is the compressible two-fluid model in [17] established using the Navier-Stokes equations:

$\{\begin{array}{l}{\partial }_t\left({\sigma }^{+}\rho \right)+\text{div}\left({\sigma }^{+}\rho w\right)=\mathrm{0,}\hfill \\ {\partial }_t\left({\sigma }^{+}\rho w\right)+\text{div}\left({\sigma }^{+}\rho w\otimes w\right)+{\sigma }^{+}\nabla P^{+}\left(\rho \right)=\text{div}\left({\sigma }^{+}{\tau }_1\right)\hfill \\ {\partial }_t\left({\sigma }^{-}\eta \right)+\text{div}\left({\sigma }^{-}\eta n\right)=\mathrm{0,}\hfill \\ {\partial }_t\left({\sigma }^{-}\eta n\right)+\text{div}\left({\sigma }^{-}\eta n\otimes n\right)+{\sigma }^{-}\nabla P^{-}\left(\eta \right)=\text{div}\left({\sigma }^{-}{\tau }_2\right)\hfill \\ P=P^{+}\left(\rho \right)=P^{-}\left(\eta \right)=a^{+}{\left(\rho \right)}^{{\overline{\gamma }}^{+}}=a^{-}{\left(\eta \right)}^{{\overline{\gamma }}^{-}}\hfill \end{array}$ (1.1)

in which $\rho$ , $\eta$ are no less than 0, ${\overline{\gamma }}^{\pm }\ge 1$ , ${\sigma }^{+}$ is the volume fraction of liquid, while ${\sigma }^{-}$ is that of gas, both ${\sigma }^{+}$ and ${\sigma }^{-}$ are positive constants. $w\left(x\mathrm{,}t\right)$ , $n\left(x\mathrm{,}t\right)$ represent the velocities of each phase. Let $a^{+}=a^{-}=1$ . Moreover, ${\tau }_1$ , ${\tau }_2$ are the viscous stress tensors:

$\{\begin{array}{l}{\tau }_1\mathrm{<mo>:</mo>}={\mu }_1\left(\nabla w+{\nabla }^{t}w\right)+{\lambda }_1\text{div}w\text{Id}\mathrm{,}\hfill \\ {\tau }_2\mathrm{<mo>:</mo>}={\mu }_2\left(\nabla n+{\nabla }^{t}n\right)+{\lambda }_2\text{div}n\text{Id}\mathrm{,}\hfill \end{array}$ (1.2)

where ${\mu }_1$ , ${\mu }_2$ and ${\lambda }_1$ , ${\lambda }_2$ are shear and bulk viscosity coefficients: ${\mu }_1$ , ${\mu }_2\mathrm{<mo>></mo>0}$ , $2{\mu }_1+3{\lambda }_1\ge 0$ , $2{\mu }_2+3{\lambda }_2\ge 0$ , then ${\mu }_1+{\lambda }_1\mathrm{<mo>></mo>0}$ , ${\mu }_2+{\lambda }_2>0$ .

When ${\mu }_1$ , ${\mu }_2$ , ${\lambda }_1$ and ${\lambda }_2$ are constants, there are many studies on the solutions of (1.1). Bresch, Huang and Li obtained global weak solutions of a non-conservative viscous compressible two-phase model in one space dimension in [18]. For the multi-dimensional case, Yao, Zhang and Zhu obtained the existence and asymptotic behavior of global weak solutions for a two-phase model of two-dimensional viscous liquid gas flow by using the idea of studying single-phase Navier-Stokes equations in [19]. Vasseur, Wen and Yu in [20] got the existence of global weak solutions for the model like (1.1), where the pressure contingent upon two variables. Cui, Wang, Yao and Zhu in [21] studied the time-decay rate of a non conservative compressible model by analysing systems and estimating energy. When $P^{+}=P^{-}$ , Wu and Yao obtained the properties of the classical solutions to the three dimensional space in [22].

With the intensive study of the single phase Navier-Stokes equations, researchers have obtained some important conclusions, one of which is particularly noteworthy: The dynamic viscosity coefficient and volumetric viscosity coefficient both depend on density, see [23]-[27] and the references therein. For the two-fluid model, we make the following assumptions:

$\{\begin{array}{l}{\tau }_1\mathrm{<mo>:</mo>}={\rho }^{{\mu }_1}\left(\nabla w+{\nabla }^{t}w\right)+{\rho }^{{\lambda }_1}\text{div}w\text{Id}\mathrm{,}\hfill \\ {\tau }_2\mathrm{<mo>:</mo>}={\eta }^{{\mu }_2}\left(\nabla n+{\nabla }^{t}n\right)+{\eta }^{{\lambda }_2}\text{div}n\text{Id}\mathrm{,.}\hfill \end{array}$ (1.3)

where ${\mu }_1$ , ${\lambda }_1\mathrm{,}$ ${\mu }_2$ , ${\lambda }_2\ge 1$ . The biggest difference between this article and previous research is the viscosity term, in which both $\mu$ and $\lambda$ are functions of density.

When the velocities of the two fluids are the same, Zodji in [28] proved the existence of a unique local-time solution when density and dynamic viscosity coefficients have an infimum. Li, Wang and Zhang in [29] researched a model analogous to (1.1), where the dynamic viscosity coefficient and volumetric viscosity coefficient are functions of density. There are no finite-energy classical solutions to this system for any small time. The assumption made by Bresch in [30] is that the viscosity coefficients of two-phase motion are ${\mu }_1\rho$ and ${\mu }_2\eta$ respectively, and the volume viscosity coefficients are both 0, Bresch obtained the global weak solutions, while $1<{\gamma }^{\pm }<6$ . Evje, Wang and Wen in [31] supposed:

$P^{+}\left(\rho \right)-P^{-}\left(\eta \right)={\rho }^{{\overline{\gamma }}^{+}}-{\eta }^{{\overline{\gamma }}^{-}}=f\left({\sigma }^{-}\eta \right),$ (1.4)

where $f$ represents capillary pressure, $f\in C^3\left(\left[\mathrm{0,}\infty \right)\right)$ . If f satisfies

$-\frac{r_{-}^2\left(1,1\right)}{{\sigma }^{-}\left(1,1\right)}<f^{\prime }\left(1\right)<0,$ (1.5)

then $f$ decreases strictly monotonically in the vicinity of the equilibrium point.

Where $r_{+}^2\mathrm{<mo>:</mo>}=\frac{\text{d}{P}^{+}}{\text{d}\rho }\left(\rho \right)={\overline{\gamma }}^{+}\frac{P^{+}\left(\rho \right)}{\rho }$, $r_{-}^2\mathrm{<mo>:</mo>}=\frac{\text{d}{P}^{-}}{\text{d}\eta }\left(\eta \right)={\overline{\gamma }}^{-}\frac{P^{-}\left(\eta \right)}{\eta }$ represent the

sound speed respectively. They achieved global existence and decay rates of the solutions in [31].

There are currently few research results on viscous stress tensors (1.3). Therefore, we study the properties of solutions when both $\mu$ and $\lambda$ are functions of density.

The research methods and techniques used in this article are derived from [22], which obtains the global goodness and large time behavior of the classical solution of a three-dimensional compressible two-phase flow model with common pressure through energy estimation and continuity induction.

We observed (1.1)₅ and obtained the following relationship

$\text{d}P=r_{+}^2\text{d}\rho =r_{-}^2\text{d}\eta \mathrm{,}$ (1.6)

We introduce fractional density like in [30]

$M^{+}={\sigma }^{+}\rho \mathrm{,}{M}^{-}={\sigma }^{-}\eta \mathrm{,}$ (1.7)

Combined with ${\sigma }^{+}+{\sigma }^{-}=1$ , it gives

$\text{d}\rho =\frac{1}{{\sigma }^{+}}\left(\text{d}{M}^{+}-\rho \text{d}{\sigma }^{+}\right)\mathrm{,}\text{d}\eta =\frac{1}{{\sigma }^{-}}\left(\text{d}{M}^{-}+\eta \text{d}{\sigma }^{-}\right)\mathrm{.}$ (1.8)

From (1.6) and (1.7), we get

$\text{d}{\sigma }^{+}=\frac{{\sigma }^{-}{r}_{+}^2}{{\sigma }^{-}\rho r_{+}^2+{\sigma }^{+}\eta r_{-}^2}\text{d}{M}^{+}-\frac{{\sigma }^{+}{r}_{-}^2}{{\sigma }^{-}\rho r_{+}^2+{\sigma }^{+}\eta r_{-}^2}\text{d}{M}^{-}\mathrm{.}$ (1.9)

Putting (1.9) into (1.8), we derive

$\text{d}\rho =\frac{\rho \eta r_{-}^2}{M^{-}{\left(\rho \right)}^2{r}_{+}^2+M^{+}{\left(\eta \right)}^2{r}_{-}^2}\left(\eta \text{d}{M}^{+}+\rho \text{d}{M}^{-}\right)\mathrm{,}$

and

$\text{d}\eta =\frac{\rho \eta r_{+}^2}{M^{-}{\left(\rho \right)}^2{r}_{+}^2+M^{+}{\left(\eta \right)}^2{r}_{-}^2}\left(\eta \text{d}{M}^{+}+\rho \text{d}{M}^{-}\right)\mathrm{,}$

We get $\text{d}P$ :

$\text{d}P=ℒ\left(\eta \text{d}{M}^{+}+\rho \text{d}{M}^{-}\right)$

where

$ℒ\mathrm{<mo>:</mo><mo>=</mo>}\frac{r_{-}^2{r}_{+}^2}{{\sigma }^{-}\rho r_{+}^2+{\sigma }^{+}\eta r_{-}^2}\mathrm{.}$

From ${\sigma }^{+}+{\sigma }^{-}=1$ , we get

$\frac{M^{+}}{\rho }+\frac{M^{-}}{\eta }=\mathrm{1,<mi>\; </mi><mi>\; </mi>}\text{thus}\mathrm{<mi>\; </mi><mi>\; </mi>}{M}^{-}=\frac{\eta \left(\rho -M^{+}\right)}{\rho }=\frac{P^{1/{\gamma }^{-}}\left(P^{1/{\gamma }^{+}}-M^{+}\right)}{P^{1/{\gamma }^{+}}}\mathrm{.}$ (1.10)

With (1.1)₅, (1.7) and (1.10), we define ${\sigma }^{\pm }$

$\{\begin{array}{l}{\sigma }^{+}\left(P\mathrm{,}{M}^{+}\right)=\frac{M^{+}}{P^{1/{\gamma }^{+}}}\mathrm{,}\hfill \\ {\sigma }^{-}\left(P\mathrm{,}{M}^{+}\right)=1-\frac{M^{+}}{P^{1/{\gamma }^{+}}}\mathrm{.}\hfill \end{array}$ (1.11)

Rewrite (1.1) using $\left(M^{\pm }\mathrm{,}P\mathrm{,}u\mathrm{,}v\right)$ :

$\{\begin{array}{l}{\partial }_t{M}^{+}+\text{div}\left(M^{+}w\right)=\mathrm{0,}\\ {\partial }_t{M}^{-}+\text{div}\left(M^{-}n\right)=\mathrm{0,}\\ {\partial }_{t}P+ℒ\eta \text{div}\left(M^{+}w\right)+ℒ\rho \text{div}\left(M^{-}n\right)=\mathrm{0,}\\ {\partial }_t\left(M^{+}w\right)+\text{div}\left(M^{+}w\otimes w\right)+{\sigma }^{+}\nabla P=\text{div}\left\{{\sigma }^{+}\left[{\rho }^{{\mu }_1}\left(\nabla w+{\nabla }^{t}w\right)+{\rho }^{{\lambda }_1}\text{div}w\text{Id}\right]\right\}\mathrm{,}\\ {\partial }_t\left(M^{-}n\right)+\text{div}\left(M^{-}n\otimes n\right)+{\sigma }^{-}\nabla P=\text{div}\left\{{\sigma }^{-}\left[{\eta }^{{\mu }_2}\left(\nabla n+{\nabla }^{t}n\right)+{\eta }^{{\lambda }_2}\text{div}n\text{Id}\right]\right\}\mathrm{.}\end{array}$ (1.12)

This paper considers (1.12) in 3D space with initial data.

$\left(M^{\pm },P,u,v\right)\left(x,0\right)=\left(M_0^{\pm },P_0,u_0,v_0\right)\left(x\right)\to \left({\overline{M}}^{\pm },\overline{P},\overrightarrow{0},\overrightarrow{0}\right)\text{as}\left|x\right|\to \infty \in {ℝ}^3,$ (1.13)

${\overline{M}}^{\pm }$ and $\overline{P}$ are constants for the background doping profile. Let ${\overline{M}}^{\pm }=1$ , thus $\overline{P}$ depends on $M^{-}\left(\overline{P}\mathrm{,1}\right)-1=0$ .

Moreover, setting

$Z=P-\overline{P},m^{\pm }=M^{\pm }-1,$

We restate (1.12)-(1.13) based on the above equation:

$\{\begin{array}{l}{\partial }_t{m}^{+}+\text{div}w=-\text{div}\left(m^{+}w\right)\mathrm{,}\hfill \\ {\partial }_t{m}^{-}+\text{div}n=-\text{div}\left(m^{-}n\right)\mathrm{,}\hfill \\ {\partial }_{t}Z+{\beta }_1\text{div}w+{\beta }_2\text{div}n=G_1\mathrm{,}\hfill \\ {\partial }_{t}w+{\beta }_3\nabla Z-{\kappa }_1\Delta w-{\kappa }_2\nabla \text{div}w=H_1+R_1\mathrm{,}\hfill \\ {\partial }_{t}n+Z_4\nabla Z-{\varkappa }_1\Delta n-{\varkappa }_2\nabla \text{div}n=H_2+R_2\mathrm{,}\hfill \end{array}$ (1.14)

with initial data

$\left(m^{\pm }\mathrm{,}Z\mathrm{,}w\mathrm{,}n\right)\left(x\mathrm{,0}\right)=\left(m_0^{\pm }\mathrm{,}{Z}_0\mathrm{,}{u}_0\mathrm{,}{v}_0\right)\left(x\right)\to \left(\mathrm{0,0,}\overrightarrow{0}\mathrm{,}\overrightarrow{0}\right)\mathrm{,}\text{as}\left|x\right|\to +\infty \mathrm{.}$ (1.15)

where ${\beta }_1=ℒ\left(\mathrm{1,}\overline{P}\right)\eta \left(\overline{P}\right)$ , ${\beta }_2=ℒ\left(\mathrm{1,}\overline{P}\right)\rho \left(\overline{P}\right)$ , ${\beta }_3=\frac{1}{\rho \left(\overline{P}\right)}$ , ${\beta }_4=\frac{1}{\eta \left(\overline{P}\right)}$ , ${\kappa }_1=\frac{\rho {\left(\overline{P}\right)}^{{\mu }_1}}{\rho \left(\overline{P}\right)}$ , ${\varkappa }_1=\frac{\eta {\left(\overline{P}\right)}^{{\mu }_2}}{\eta \left(\overline{P}\right)}$ , ${\kappa }_2=\frac{\rho {\left(\overline{P}\right)}^{{\mu }_1}+\rho {\left(\overline{P}\right)}^{{\lambda }_1}}{\rho \left(\overline{P}\right)}$ , ${\varkappa }_2=\frac{\eta {\left(\overline{P}\right)}^{{\mu }_2}+\eta {\left(\overline{P}\right)}^{{\lambda }_2}}{\eta \left(\overline{P}\right)}$ and the nonlinear terms are given by

$\begin{array}{c}{G}_1=-h_1\left(m^{+}\mathrm{,}Z\right)\text{div}w-h_2\left(m^{+}\mathrm{,}Z\right)\text{div}n-ℒ\left(m^{+}+\mathrm{1,}Z+\overline{P}\right)\eta u\cdot \nabla m^{+}\\ -ℒ\left(m^{+}+\mathrm{1,}Z+\overline{P}\right)\rho n\cdot \nabla m^{-}\mathrm{,}\end{array}$ (1.16)

$\begin{array}{c}{H}_1=-w\cdot \nabla w-\left(\frac{1}{\rho }-{\beta }_3\right)\nabla Z+\left(\frac{{\rho }^{{\mu }_1}}{\rho }-{\kappa }_1\right)\Delta w\\ +\left(\frac{{\rho }^{{\mu }_1}+{\rho }^{{\lambda }_1}}{\rho }-{\kappa }_2\right)\nabla \text{div}w\mathrm{,}\end{array}$ (1.17)

and

$\begin{array}{c}{R}_1=\frac{{\rho }^{{\mu }_1}\left(\nabla w+{\nabla }^{t}w\right)\nabla {\sigma }^{+}}{M^{+}}+\frac{{\rho }^{{\lambda }_1}\text{div}w\nabla {\sigma }^{+}}{M^{+}}\\ +\frac{\nabla \left({\rho }^{{\mu }_1}\right)\left(\nabla w+{\nabla }^{t}w\right)}{\rho }+\frac{\nabla \left({\rho }^{{\lambda }_1}\right)\text{div}w}{\rho }\mathrm{,}\end{array}$ (1.18)

$\begin{array}{c}{H}_2=-n\cdot \nabla n-\left(\frac{1}{\eta }-{\beta }_4\right)\nabla Z+\left(\frac{{\eta }^{{\mu }_2}}{\eta }-{\varkappa }_1\right)\Delta n\\ +\left(\frac{{\eta }^{{\mu }_2}+{\eta }^{{\lambda }_2}}{\eta }-{\varkappa }_2\right)\nabla \text{div}n\mathrm{,}\end{array}$ (1.19)

$\begin{array}{c}{R}_2=\frac{{\eta }^{{\mu }_2}\left(\nabla n+{\nabla }^{t}n\right)\nabla {\sigma }^{-}}{M^{-}}+\frac{{\eta }^{{\lambda }_2}\text{div}n\nabla {\sigma }^{-}}{M^{-}}\\ +\frac{\nabla \left({\eta }^{{\mu }_2}\right)\left(\nabla n+{\nabla }^{t}n\right)}{\eta }+\frac{\nabla \left({\eta }^{{\lambda }_2}\right)\text{div}n}{\eta }\mathrm{,}\end{array}$ (1.20)

where $h_i\left(i=\mathrm{1,2}\right)$ are given by

$\{\begin{array}{l}{h}_1\left(m^{+}\mathrm{,}Z\right)=ℒ\left(m^{+}+\mathrm{1,}Z+\overline{P}\right)\eta M^{+}-{\beta }_1\mathrm{,}\\ h_2\left(m^{+}\mathrm{,}Z\right)=ℒ\left(m^{+}+\mathrm{1,}Z+\overline{P}\right)\rho M^{-}-{\beta }_2\mathrm{.}\end{array}$ (1.21)

Now, we will show the difficulties in the proof and how to fix them.

First of all, we rewrite (1.12) as (1.14) by taking $m^{\pm }=M^{\pm }-1$ , $Z=P-\overline{P}$ . We obtain the corresponding linear system (1.26) for (1.14). To study time-decay estimates of the linear system (1.26), we must study the properties of the semigroup in detail. We use the Hodge decomposition technique [32] to decompose the linear system into three equations. We can get linear $L^2$ estimates using the Fourier transform.

Second, similar to [22], we multiply (1.14)₃, (1.14)₄, (1.14)₅ by $\frac{{\beta }_3}{{\beta }_1}{\nabla }^{k}Z$ , ${\nabla }^{k}w$ ,

and ${\nabla }^{k}n$ $\left(k=\mathrm{0,1,2}\right)$ , respectively, and then integrate them. The difference between this paper and [22] is that when estimating the term on the right, we need to estimate the norms of ${\rho }^{{\mu }_1}$ , ${\rho }^{{\lambda }_1}$ , ${\eta }^{{\mu }_2}$ and ${\eta }^{{\lambda }_2}$ . We noticed that $\rho$ and $\eta$ are related to $P$ , making $Z=P-\overline{P}$ . Therefore, norm estimates related to ${\rho }^{{\mu }_1}$ , ${\rho }^{{\lambda }_1}$ , ${\eta }^{{\mu }_2}$ and ${\eta }^{{\lambda }_2}$ can all be calculated using norm estimates related to $Z$ . We deduce the energy of $\left(Z\mathrm{,}w\mathrm{,}n\right)$ and its derivatives.

In the last, we need handle $ℒ\eta w\cdot \nabla m^{+}$ and $ℒ\rho n\cdot \nabla m^{-}$ in (1.16). because we have chosen $P$ as our research variable. Dealing with items like ${\displaystyle {\int }_{{ℝ}^3}}ℒ\eta w\cdot \nabla {\nabla }^3{m}^{+}{\nabla }^{3}Z\text{d}x$ and ${\displaystyle {\int }_{{ℝ}^3}}ℒ\rho n\cdot \nabla {\nabla }^3{m}^{-}{\nabla }^{3}Z\text{d}x$ can be more difficult for us, so we have rewritten it as (5.11). At the same time, applying related inequalities, We deduced the time decay rates for $\left(Z\mathrm{,}n\mathrm{,}w\right)$ . We have proven Theorem 1.2.

Main results.

Theorem 1.1. If $M_0^{\pm }-1\in H^3\left({ℝ}^3\right)$ , $P_0-\overline{P}\mathrm{,}{w}_0\mathrm{,}{n}_0\in H^3\left({ℝ}^3\right)\cap L^1\left({ℝ}^3\right)$ , then there is a constant ${\delta }_0$ such that if

$N_0\mathrm{<mo>:</mo>}={\Vert \left(M_0^{\pm }-1\right)\Vert }_{H^3}+{\Vert \left(P_0-\overline{P}\mathrm{,}{w}_0\mathrm{,}{n}_0\right)\Vert }_{H^3\cap L^1}\le {\delta }_0\mathrm{,}$ (1.22)

then the Cauchy problem (1.9)-(1.10) admits a unique global solution $\left(M^{\pm }\mathrm{,}P\mathrm{,}w\mathrm{,}n\right)$ in time, satisfying

$\begin{array}{l}{M}^{\pm }-\mathrm{1,}P-\overline{P}\in C^0\left(\left[\mathrm{0,}\infty \right)\mathrm{<mo>;</mo>}{H}^3\left({ℝ}^3\right)\right)\cap C^1\left(\left[\mathrm{0,}\infty \right)\mathrm{<mo>;</mo>}{H}^2\left({ℝ}^3\right)\right)\mathrm{,}\\ w\mathrm{,}n\in C^0\left(\left[\mathrm{0,}\infty \right)\mathrm{<mo>;</mo>}{H}^3\left({ℝ}^3\right)\right)\cap C^1\left(\left[\mathrm{0,}\infty \right)\mathrm{<mo>;</mo>}{H}^1\left({ℝ}^3\right)\right)\mathrm{.}\end{array}$

After obtaining the global fitness of the model solutions, we further investigate the properties of the solution and study the time decay rate of the solutions.

Theorem 1.2. If $\left(M^{\pm }\mathrm{,}P\mathrm{,}w\mathrm{,}n\right)$ satisfies the conditions in Theorem 1.1, for any $t\ge 0$ , constant $C_0>0$ exists and is independent of $t$ , we can get estimates of the solution $\left(M^{\pm }\mathrm{,}P\mathrm{,}w\mathrm{,}n\right)$ :

${\Vert \left(P-\overline{P}\mathrm{,}w\mathrm{,}n\right)\left(t\right)\Vert }_{L^2}\le C_0{N}_0{\left(1+t\right)}_{}^{-\frac{3}{4}}\mathrm{,}$ (1.23)

${\Vert \nabla P\left(t\right)\Vert }_{H^1}+{\Vert \nabla \left(w\mathrm{,}n\right)\left(t\right)\Vert }_{H^2}\le C_0{N}_0{\left(1+t\right)}_{}^{-\frac{5}{4}}\mathrm{.}$ (1.24)

and

${\Vert {\nabla }^{3}P\left(t\right)\Vert }_{L^2}+{\Vert \left(M^{+}-\mathrm{1,}{M}^{-}-1\right)\left(t\right)\Vert }_{H^3}\le C_0{N}_0\mathrm{.}$ (1.25)

2. Preliminaries

In this paper, ${\Vert \cdot \Vert }_{H^k}$ is expressed as the norm in Sobolev space $H^k\left({ℝ}^3\right)$ . $L^p\left({ℝ}^3\right)$ is expressed as the norm in $L^p$ space, $1\le p\le \infty$ . The norm ${\Vert \left(g\mathrm{,}h\right)\Vert }_X$ is short for ${\Vert g\Vert }_X+{\Vert h\Vert }_X$ . $x\lesssim y$ means that $x\le Ky$ , where positive constant $K>0$ . We represent $\nabla ={\partial }_x=\left({\partial }_1\mathrm{,}{\partial }_2\mathrm{,}{\partial }_3\right)$ like [33], let ${\partial }_x^{\ell }g={\nabla }^{\ell }g=\nabla \left({\nabla }^{\ell -1}g\right)$ . ${\Lambda }^r$ is a pseudo differential operator, defined as

${\Lambda }^{r}g={\mathfrak{G}}^{-1}\left({\left|\xi \right|}^r\widehat{g}\right)\mathrm{,<mi>\; </mi>}\text{for}\mathrm{<mi>\; </mi>}r\in ℝ\mathrm{,}$

where $\widehat{g}$ and $\mathfrak{G}\left(g\right)$ are the Fourier transform of $g$ . ${\Vert g\Vert }_{{\dot{H}}^r}\triangleq {\Vert {\Lambda }^{r}g\Vert }_{L^2}$ represents the norm of homogenous Sobolev space ${\dot{H}}^r\left({ℝ}^3\right)$ . The radial function $\Psi$ satisfies

$\{\begin{array}{l}\Psi \left(\varsigma \right)=\mathrm{1,}\left|\varsigma \right|\le \frac{\zeta }{2}\hfill \\ \Psi \left(\varsigma \right)=\mathrm{0,}\left|\varsigma \right|\ge \zeta \hfill \end{array}$

see Lemma 3.1 for the definition of $\zeta$ .

We define

$g^l={\mathfrak{G}}^{-1}\left[\Psi \left(\varsigma \right)\widehat{g}\right],g^h={\mathfrak{G}}^{-1}\left[\left(1-\Psi \left(\varsigma \right)\right)\widehat{g}\right]\mathrm{.}$

where $g^l$ and $g^h$ respectively represent the high-frequency and low-frequency parts of $g$ . If the Fourier transform of $g$ exists, then $g=g^l+g^h$ .

Moreover, for the convenience of the reader, here is a summary of the Gagliardo-Nirenberg inequality.

Lemma 2.1. If $0\le l\mathrm{,}k\le s$ , then

${\Vert {\nabla }^{l}g\Vert }_{L^{\alpha }}\lesssim {\Vert {\nabla }^{k}g\Vert }_{L^{\theta }}^{1-b}{\Vert {\nabla }^{s}g\Vert }_{L^r}^b$

where $b$ belongs to $\left[\frac{l}{s}\mathrm{,1}\right]$ and satisfies

$\frac{l}{3}-\frac{1}{\alpha }=\left(\frac{k}{3}-\frac{1}{\theta }\right)\left(1-b\right)+\left(\frac{s}{3}-\frac{1}{r}\right)b\mathrm{.}$

Particularly, if $p$ , $q$ and $r$ are all equal to 2, we get

${\Vert {\nabla }^{l}g\Vert }_{L^2}\lesssim {\Vert {\nabla }^{k}g\Vert }_{L^2}^{\frac{s-l}{s-k}}{\Vert {\nabla }^{k}g\Vert }_{L^2}^{\frac{l-k}{s-k}}\mathrm{.}$

Proof. The proof is shown on page 125 of [34]. □

Lemma 2.2. For $s\ge 1$ , we have

${\Vert {\nabla }^s\left(gh\right)\Vert }_{L^r}\lesssim {\Vert g\Vert }_{L^{r_1}}{\Vert {\nabla }^{s}h\Vert }_{L^{r_2}}+{\Vert {\nabla }^{s}g\Vert }_{L^{r_3}}{\Vert h\Vert }_{L^{r_4}}\mathrm{,}$

and

${\Vert {\nabla }^s\left(gh\right)-f{\nabla }^{s}h\Vert }_{L^r}\lesssim {\Vert \nabla g\Vert }_{L^{r_1}}{\Vert {\nabla }^{s-1}h\Vert }_{L^{r_2}}+{\Vert {\nabla }^{s}g\Vert }_{L^{r_3}}{\Vert h\Vert }_{L^{r_4}}\mathrm{,}$

where $r\mathrm{,}{r}_1\mathrm{,}{r}_2\mathrm{,}{r}_3\mathrm{,}{r}_4\in \left[\mathrm{1,}\infty \right]$ and

$\frac{1}{r}=\frac{1}{r_1}+\frac{1}{r_2}=\frac{1}{r_3}+\frac{1}{r_4}\mathrm{.}$

Proof. The lemma has been proved in [35] and [36]. □

3. Spectral Analysis and Linear L² Estimates

3.1. Reformulation

The linearized system for Cauchy problem (1.14)-(1.15) is

$\{\begin{array}{l}{\partial }_t{m}^{+}+\text{div}w=\mathrm{0,}\hfill \\ {\partial }_t{m}^{-}+\text{div}n=\mathrm{0,}\hfill \\ {\partial }_{t}Z+{\beta }_1\text{div}w+{\beta }_2\text{div}n=\mathrm{0,}\hfill \\ {\partial }_{t}w+{\beta }_3\nabla Z-{\kappa }_1\Delta w-{\kappa }_2\nabla \text{div}w=\mathrm{0,}\hfill \\ {\partial }_{t}n+{\beta }_4\nabla Z-{\varkappa }_1\Delta n-{\varkappa }_2\nabla \text{div}n=\mathrm{0,}\hfill \\ \left(m^{+}\mathrm{,}Z\mathrm{,}w\mathrm{,}n\right)\left(x\mathrm{,0}\right)=\left(m_0^{+}\mathrm{,}{Z}_0\mathrm{,}{w}_0\mathrm{,}{n}_0\right)\left(x\right)\hfill \end{array}$ (3.1)

It is straightforward to verify the presence of a zero eigenvalue in the linearized problem (3.1), it adds to the difficulty and complexity of the issue. In order to eliminate the zero eigenvalues, we study the Cauchy problem of $\left(Z\mathrm{,}w\mathrm{,}n\right)$ :

$\{\begin{array}{l}{\partial }_{t}Z+{\beta }_1\text{div}w+{\beta }_2\text{div}n=\mathrm{0,}\hfill \\ {\partial }_{t}w+{\beta }_3\nabla Z-{\kappa }_1\Delta w-{\kappa }_2\nabla \text{div}w=\mathrm{0,}\hfill \\ {\partial }_{t}n+{\beta }_4\nabla Z-{\varkappa }_1\Delta n-{\varkappa }_2\nabla \text{div}n=\mathrm{0,}\hfill \\ \left(Z\mathrm{,}w\mathrm{,}n\right)\left(x\mathrm{,0}\right)=\left(Z_0\mathrm{,}{w}_0\mathrm{,}{n}_0\right)\left(x\right)\mathrm{.}\hfill \end{array}$ (3.2)

According to semigroup theory, (3.2) can be expressed by $\chi ={\left(Z\mathrm{,}w\mathrm{,}n\right)}^t$ as

$\{\begin{array}{l}{\chi }_t=\mathcal{T}\chi \mathrm{,}\\ {\chi |}_{t=0}={\chi }_0\mathrm{,}\end{array}$ (3.3)

where the operator $\mathcal{T}$ is given by

$\mathcal{T}=\left(\begin{array}{ccc}0& -{\beta }_1\text{div}& -{\beta }_2\text{div}\\ -{\beta }_3\nabla & {\kappa }_1\Delta +{\kappa }_2\nabla \text{div}& 0\\ -{\beta }_4\nabla & 0& {\varkappa }_1\Delta +{\varkappa }_2\nabla \text{div}\end{array}\right)\mathrm{.}$

By Fourier transform to the system (3.1), we have

$\{\begin{array}{l}{\widehat{\chi }}_t=\mathcal{K}\left(\varsigma \right)\widehat{\chi },\\ {\widehat{\chi }|}_{t=0}={\widehat{\chi }}_0,\end{array}$ (3.4)

where $\widehat{\chi }\left(\varsigma \mathrm{,}t\right)=\mathfrak{G}\left(\chi \left(x\mathrm{,}t\right)\right)$ , $\varsigma ={\left({\varsigma }^1\mathrm{,}{\varsigma }^2\mathrm{,}{\varsigma }^3\right)}^t$ and $\mathcal{K}\left(\xi \right)$ is defined by

$\mathcal{K}\left(\varsigma \right)=\left(\begin{array}{ccc}0& -i{\beta }_1{\varsigma }^t& -i{\beta }_2{\varsigma }^t\\ -i{\beta }_3\varsigma & -{\kappa }_1{\left|\varsigma \right|}^2{\text{I}}_{3\times 3}-{\kappa }_2\varsigma \otimes \varsigma & 0\\ -i{\beta }_4\varsigma & 0& -{\varkappa }_1{\left|\varsigma \right|}^2{\text{I}}_{3\times 3}-{\varkappa }_2\varsigma \otimes \varsigma \end{array}\right)\mathrm{.}$

We study the properties of hemigroup employing the actual methods in [37]-[39].

Let ${\Psi }^{+}={\Lambda }^{-1}\text{div}w$ and ${\Psi }^{-}={\Lambda }^{-1}\text{div}n$ , they are “compressible part” of $w$ and $n$ , and denote ${\Phi }^{+}={\Lambda }^{-1}\text{curl}w$ and ${\Phi }^{-}={\Lambda }^{-1}\text{curl}n$ (${\left(\text{curl}z\right)}_i^j={\partial }_{x_j}{z}^i-{\partial }_{x_i}{z}^j$ ) by the “incompressible part” of $w$ and $n$ . Rewrite (3.2) as follows:

$\{\begin{array}{l}{\partial }_{t}Z+{\beta }_1\Lambda {\Psi }^{+}+{\beta }_2\Lambda {\Psi }^{-}=\mathrm{0,}\\ {\partial }_t{\Psi }^{+}-{\beta }_3\Lambda Z+\kappa {\Lambda }^2{\Psi }^{+}=\mathrm{0,}\\ {\partial }_t{\Psi }^{-}-{\beta }_4\Lambda Z+\varkappa {\Lambda }^2{\Psi }^{-}=\mathrm{0,}\\ {\left(Z\mathrm{,}{\Psi }^{+}\mathrm{,}{\Psi }^{-}\right)|}_{t=0}=\left({\theta }_0\mathrm{,}{\Lambda }^{-1}\text{div}{w}_0\mathrm{,}{\Lambda }^{-1}\text{div}{n}_0\right)\left(x\right)\mathrm{,}\end{array}$ (3.5)

and

$\{\begin{array}{l}{\partial }_t{\Phi }^{+}+{\kappa }_1{\Lambda }^2{\Phi }^{+}=\mathrm{0,}\\ {\partial }_t{\Phi }^{-}+{\varkappa }_1{\Lambda }^2{\Phi }^{-}=\mathrm{0,}\\ {{\Phi }^{+}|}_{t=0}={\Lambda }^{-1}\text{curl}{w}_0\left(x\right)\mathrm{,}\\ {{\Phi }^{-}|}_{t=0}={\Lambda }^{-1}\text{curl}{n}_0\left(x\right)\mathrm{,}\end{array}$ (3.6)

where $\kappa ={\kappa }_1+{\kappa }_2$ , $\varkappa ={\varkappa }_1+{\varkappa }_2$ .

3.2. Spectral Analysis for IVP (3.5)

The IVP (3.3) for ${\chi }^{\prime }={\left(Z\mathrm{,}{\Psi }^{+}\mathrm{,}{\Psi }^{-}\right)}^t$ can be represented as

$\{\begin{array}{l}{{\chi }^{\prime }}_t={\mathcal{T}}_1{\chi }^{\prime },\\ {{\chi }^{\prime }|}_{t=0}={{\chi }^{\prime }}_0,\end{array}$ (3.7)

where ${\mathcal{T}}_1$ is the operator,

${\mathcal{T}}_1=\left(\begin{array}{ccc}0& -{\beta }_1\Lambda & -{\beta }_2\Lambda \\ {\beta }_3\Lambda & -\kappa {\Lambda }^2& 0\\ {\beta }_4\Lambda & 0& -\varkappa {\Lambda }^2\end{array}\right)\mathrm{.}$

The Fourier transform of (3.5) gives

$\{\begin{array}{l}{\widehat{{\chi }^{\prime }}}_t={\mathcal{K}}_1\left(\varsigma \right)\widehat{{\chi }^{\prime }},\\ {\widehat{{\chi }^{\prime }}|}_{t=0}={\widehat{{\chi }^{\prime }}}_0,\end{array}$ (3.8)

where $\widehat{{\chi }^{\prime }}\left(\varsigma \mathrm{,}t\right)=\mathfrak{G}\left({\chi }^{\prime }\left(x\mathrm{,}t\right)\right)$ and ${\mathcal{K}}_1\left(\varsigma \right)$ is given by

${\mathcal{K}}_1\left(\varsigma \right)=\left(\begin{array}{ccc}0& -{\beta }_1\left|\varsigma \right|& -{\beta }_2\left|\varsigma \right|\\ {\beta }_3\left|\varsigma \right|& -\kappa {\left|\varsigma \right|}^2& 0\\ {\beta }_4\left|\varsigma \right|& 0& -\varkappa {\left|\varsigma \right|}^2\end{array}\right)\mathrm{.}$

We calculate

$\begin{array}{l}\det \left(P\text{I}-{\mathcal{K}}_1\left(\varsigma \right)\right)\\ =P^3+\left(\kappa +\varkappa \right){\left|\varsigma \right|}^2{P}^2+\left({\beta }_1{\beta }_3+{\beta }_2{\beta }_4+\kappa \varkappa {\left|\varsigma \right|}^2\right){\left|\varsigma \right|}^{2}P\\ +\left({\beta }_1{\beta }_3\varkappa +{\beta }_2{\beta }_4\kappa \right){\left|\varsigma \right|}^4\mathrm{,}\end{array}$ (3.9)

so we obtain three different eigenvalues:

$P_1=P_1\left(\left|\varsigma \right|\right),P_2=P_2\left(\left|\varsigma \right|\right),P_3=P\left(\left|\varsigma \right|\right).$

We have

${\text{e}}^{t{\mathcal{K}}_1\left(\varsigma \right)}={\displaystyle \sum _{k=1}^3}{\text{e}}^{P_{k}t}{Q}_k\left(\varsigma \right)\mathrm{,}$ (3.10)

where

$Q_k\left(\varsigma \right)={\displaystyle \prod _{j\ne k}}\frac{{\mathcal{K}}_1\left(\varsigma \right)-P_{j}I}{P_k-P_j},k,j=1,2,3.$ (3.11)

So we get the solution of (3.8):

$\widehat{{\chi }^{\prime }}\left(\varsigma \mathrm{,}t\right)={\text{e}}^{t{\mathcal{K}}_1\left(\varsigma \right)}{\widehat{{\chi }^{\prime }}}_0\left(\varsigma \right)=\left({\displaystyle \sum _{k=1}^3}{\text{e}}^{P_{k}t}{Q}_k\left(\varsigma \right)\right){\widehat{{\chi }^{\prime }}}_0\left(\varsigma \right)\mathrm{.}$ (3.12)

Using comparable methods in [37]-[39], we get

Lemma 3.1. If $\zeta$ is a positive constant and $\zeta \ll 1$ , the following Taylor series expansion exists for $\left|\xi \right|\le \zeta$

$\{\begin{array}{l}{P}_1=-\frac{{\beta }_1{\beta }_3\kappa +{\beta }_2{\beta }_4\varkappa }{2\left({\beta }_1{\beta }_3+{\beta }_2{\beta }_4\right)}{\left|\varsigma \right|}^2+\sqrt{{\beta }_1{\beta }_3+{\beta }_2{\beta }_4}\text{i}\left|\varsigma \right|+\mathcal{O}\left({\left|\varsigma \right|}^3\right)\mathrm{,}\\ P_2=-\frac{{\beta }_1{\beta }_3\kappa +{\beta }_2{\beta }_4\varkappa }{2\left({\beta }_1{\beta }_3+{\beta }_2{\beta }_4\right)}{\left|\varsigma \right|}^2-\sqrt{{\beta }_1{\beta }_3+{\beta }_2{\beta }_4}\text{i}\left|\varsigma \right|+\mathcal{O}\left({\left|\varsigma \right|}^3\right)\mathrm{,}\\ P_3=-\frac{{\beta }_1{\beta }_3\varkappa +{\beta }_2{\beta }_4\kappa }{{\beta }_1{\beta }_3+{\beta }_2{\beta }_4}{\left|\varsigma \right|}^2+\mathcal{O}\left({\left|\varsigma \right|}^3\right)\mathrm{.}\end{array}$ (3.13)

Lemma 3.2. Let $\overline{\nu }=\min \left\{\frac{{\beta }_1{\beta }_3\kappa +{\beta }_2{\beta }_4\varkappa }{2\left({\beta }_1{\beta }_3+{\beta }_2{\beta }_4\right)}\mathrm{,}\frac{{\beta }_1{\beta }_3\varkappa +{\beta }_2{\beta }_4\kappa }{{\beta }_1{\beta }_3+{\beta }_2{\beta }_4}\right\}>0$ , for any $\left|\varsigma \right|\le \eta$ , then

$\left|\widehat{Z}\right|\mathrm{,<mi>\; </mi>}\left|\widehat{{\Psi }^{+}}\right|\mathrm{,<mi>\; </mi>}\left|\widehat{{\Psi }^{-}}\right|\lesssim {\text{e}}^{-\overline{\nu }{\left|\varsigma \right|}^{2}t}\left(\left|\widehat{Z_0}\right|+\left|\widehat{{\Psi }_0^{+}}\right|+\left|\widehat{{\Psi }_0^{-}}\right|\right)\mathrm{,}$ (3.14)

Proof. Owing to (3.11) and $P_k$ $\left(k=\mathrm{1,2,3}\right)$ in (3.13), we can obtain $Q_k$ $\left(k=\mathrm{1,2,3}\right)$ :

$Q_1\left(\varsigma \right)=\left(\begin{array}{ccc}\frac{1}{2}& \frac{{\beta }_1}{2\sqrt{{\beta }_1{\beta }_3+{\beta }_2{\beta }_4}}\text{i}& \frac{{\beta }_2}{2\sqrt{{\beta }_1{\beta }_3+{\beta }_2{\beta }_4}}\text{i}\\ -\frac{{\beta }_3}{2\sqrt{{\beta }_1{\beta }_3+{\beta }_2{\beta }_4}}\text{i}& \frac{{\beta }_1{\beta }_3}{2\left({\beta }_1{\beta }_3+{\beta }_2{\beta }_4\right)}& \frac{{\beta }_2{\beta }_3}{2\left({\beta }_1{\beta }_3+{\beta }_2{\beta }_4\right)}\\ -\frac{{\beta }_4}{2\sqrt{{\beta }_1{\beta }_3+{\beta }_2{\beta }_4}}\text{i}& \frac{{\beta }_1{\beta }_4}{2\left({\beta }_1{\beta }_3+{\beta }_2{\beta }_4\right)}& \frac{{\beta }_2{\beta }_4}{2\left({\beta }_1{\beta }_3+{\beta }_2{\beta }_4\right)}\end{array}\right)+\mathcal{O}\left(\left|\varsigma \right|\right)\mathrm{,}$ (3.15)

$Q_2\left(\varsigma \right)=\left(\begin{array}{ccc}\frac{1}{2}& -\frac{{\beta }_1}{2\sqrt{{\beta }_1{\beta }_3+{\beta }_2{\beta }_4}}\text{i}& -\frac{{\beta }_2}{2\sqrt{{\beta }_1{\beta }_3+{\beta }_2{\beta }_4}}\text{i}\\ \frac{{\beta }_3}{2\sqrt{{\beta }_1{\beta }_3+{\beta }_2{\beta }_4}}\text{i}& \frac{{\beta }_1{\beta }_3}{2\left({\beta }_1{\beta }_3+{\beta }_2{\beta }_4\right)}& \frac{{\beta }_2{\beta }_3}{2\left({\beta }_1{\beta }_3+{\beta }_2{\beta }_4\right)}\\ \frac{{\beta }_4}{2\sqrt{{\beta }_1{\beta }_3+{\beta }_2{\beta }_4}}\text{i}& \frac{{\beta }_1{\beta }_4}{2\left({\beta }_1{\beta }_3+{\beta }_2{\beta }_4\right)}& \frac{{\beta }_2{\beta }_4}{2\left({\beta }_1{\beta }_3+{\beta }_2{\beta }_4\right)}\end{array}\right)+\mathcal{O}\left(\left|\varsigma \right|\right)\mathrm{,}$ (3.16)

and

$Q_3\left(\varsigma \right)=\left(\begin{array}{ccc}0& 0& 0\\ 0& \frac{{\beta }_2{\beta }_4}{{\beta }_1{\beta }_3+{\beta }_2{\beta }_4}& -\frac{{\beta }_2{\beta }_3}{{\beta }_1{\beta }_3+{\beta }_2{\beta }_4}\\ 0& -\frac{{\beta }_1{\beta }_4}{{\beta }_1{\beta }_3+{\beta }_2{\beta }_4}& \frac{{\beta }_1{\beta }_3}{{\beta }_1{\beta }_3+{\beta }_2{\beta }_4}\end{array}\right)+\mathcal{O}\left(\left|\varsigma \right|\right)\mathrm{,}$ (3.17)

for any $\left|\varsigma \right|\le \zeta$ . Therefore, (3.14) follows from (3.12)-(3.13) and (3.15)-(3.17) immediately. □

From (3.14), we can get Proposition 3.3.

Proposition 3.3 (L²-theory). If $k\ge 0$ , $\forall$ $t\ge 0$ , then

${\Vert {\nabla }^k{\text{e}}^{t{\mathcal{K}}_1}{{\chi }^{\prime }}^l\left(0\right)\Vert }_{L^2}\lesssim {\left(1+t\right)}_{}^{\frac{3}{4}-\frac{1}{2}\left(k+\frac{3}{p}\right)}{\Vert {\chi }^{\prime }\left(0\right)\Vert }_{L^p}\mathrm{,}$ (3.18)

where $1\le p\le 2$ .

Proof. Using Plancherel theorem, we have

$\begin{array}{c}{\Vert {\nabla }^k{\text{e}}^{t{\mathcal{T}}_1}{\chi }^l\left(0\right)\Vert }_{L^2}^2=\mathrm{<mi>\; </mi>}{\Vert {\left|\varsigma \right|}^k{\text{e}}^{t{\mathcal{K}}_1\left(\varsigma \right)}{{\widehat{\chi }}^{\prime }}^l\left(0\right)\Vert }_{L^2}^2\\ \lesssim {\displaystyle {\int }_{\left|\varsigma \right|\le \zeta }}{\text{e}}^{-2\overline{\nu }{\left|\varsigma \right|}^{2}t}{\left|\varsigma \right|}^{2k}{\left|{{\widehat{\chi }}^{\prime }}^l\left(0\right)\right|}^2\text{d}\varsigma \\ \lesssim {\left(1+t\right)}_{}^{\frac{3}{2}-\left(\frac{3}{p}+k\right)}{\Vert {{\widehat{\chi }}^{\prime }}^l\left(0\right)\Vert }_{L^q}^2\\ \lesssim {\left(1+t\right)}_{}^{\frac{3}{2}-\left(\frac{3}{p}+k\right)}{\Vert {\chi }^{\prime }\left(0\right)\Vert }_{L^p}^2\mathrm{,}\end{array}$

where $\frac{1}{p}+\frac{1}{q}=1$ .

Proposition 3.3. is proved. □

The $L^2$ -convergence rates obtained above are optimum. In fact, there is a lower bound on the rate of convergence, see Proposition 3.3.

Proposition 3.4. Provided that $\left(\widehat{Z}\mathrm{,}\widehat{{\Psi }^{+}}\mathrm{,}\widehat{{\Psi }^{-}}\right)\in L^1$ , if

$\widehat{{\Psi }_0^{+}}\left(\xi \right)=\widehat{{\Psi }_0^{-}}\left(\xi \right)=\mathrm{0,}\left|\widehat{{\varsigma }_0}\left(\varsigma \right)\right|\ge c_0\mathrm{,}$ (3.19)

for $\forall$ $\left|\xi \right|\le \zeta$ , $t$ is large enough, then the global solution $\left(Z\mathrm{,}{\Psi }^{+}\mathrm{,}{\Psi }^{-}\right)$ of (3.7) satisfies

$\min \left\{{\Vert Z^l\Vert }_{L^2}\mathrm{,}{\Vert {\Psi }^{+\mathrm{,}l}\Vert }_{L^2}\mathrm{,}{\Vert {\Psi }^{-\mathrm{,}l}\Vert }_{L^2}\right\}\gtrsim c_0{\left(1+t\right)}_{}^{-\frac{3}{4}}\mathrm{.}$ (3.20)

Proof. Let ${\overline{\nu }}_1=\frac{{\beta }_1{\beta }_3\kappa +{\beta }_2{\beta }_4\varkappa }{2\left({\beta }_1{\beta }_3+{\beta }_2{\beta }_4\right)}>0$ . Due to (3.19), it follows from (3.12)-(3.13) and (3.15)-(3.17) that

${\widehat{Z}}^l~{\text{e}}^{-{\overline{\nu }}_1{\left|\varsigma \right|}^{2}t}\cos \left(\sqrt{{\beta }_1{\beta }_3+{\beta }_2{\beta }_4}\left|\varsigma \right|t+\mathcal{O}\left({\left|\varsigma \right|}^3\right)t\right){\widehat{Z}}_0^l$

Combining the double angle formula and Plancherel theorem, we get

$\begin{array}{c}{\Vert Z^l\Vert }_{L^2}^2=\mathrm{<mi>\; </mi>}{\Vert {\widehat{Z}}^l\Vert }_{L^2}^2\\ \ge \frac{c_0^2}{2}{\displaystyle {\int }_{\left|\varsigma \right|\le \zeta }}{\text{e}}^{-2{\overline{\nu }}_1{\left|\varsigma \right|}^{2}t}{\cos }^2\left(\sqrt{{\beta }_1{\beta }_3+{\beta }_2{\beta }_4}\left|\varsigma \right|t+\mathcal{O}\left({\left|\varsigma \right|}^3\right)t\right)\text{d}\varsigma \\ \ge \frac{c_0^2}{4}{\left(1+t\right)}_{}^{-\frac{3}{2}}-C{c}_0^2{\left(1+t\right)}_{}^{-\frac{3}{2}}{\left(1+t\right)}_{}^{-\frac{1}{2}}\\ \mathrm{<mi>\; </mi>}\gtrsim c_0^2{\left(1+t\right)}_{}^{-\frac{3}{2}}\mathrm{,}\end{array}$ (3.21)

if $t$ is large enough. (3.20) is derived from (3.21) using the same method. Therefore, we proved Proposition 3.4. □

3.3. Spectral Analysis for IVP (3.6)

Proposition 3.5 (L²-theory). If $k\ge 0$ , $\forall$ $t\ge 0$ , $1\le p\le 2$ , then

$\begin{array}{l}{\Vert {\nabla }^k{\text{e}}^{-{\kappa }_1{\Lambda }^{2}t}{\Psi }^{+\mathrm{,}l}\left(0\right)\Vert }_{L^2}\lesssim {\left(1+t\right)}_{}^{-\frac{1}{2}\left(k+\frac{3}{p}\right)+\frac{3}{4}}{\Vert {\Psi }^{+}\left(0\right)\Vert }_{L^p}\mathrm{,}\\ {\Vert {\nabla }^k{\text{e}}^{-{\varkappa }_1{\Lambda }^{2}t}{\Psi }^{-\mathrm{,}l}\left(0\right)\Vert }_{L^2}\lesssim {\left(1+t\right)}_{}^{-\frac{1}{2}\left(k+\frac{3}{p}\right)+\frac{3}{4}}{\Vert {\Psi }^{-}\left(0\right)\Vert }_{L^p}\mathrm{,}\end{array}$ (3.22)

3.4. L² Decay Estimates for IVP (3.3)

By the definition of ${\Psi }^{\pm }$ , ${\Phi }^{\pm }$ , and

$w=-{\wedge }^{-1}\nabla {\Psi }^{+}-{\wedge }^{-1}\text{div}{\Phi }^{+}$

$n=-{\wedge }^{-1}\nabla {\Psi }^{-}-{\wedge }^{-1}\text{div}{\Phi }^{-}$

By integrating Propositions 3.3, 3.4 and 3.5, we obtain semigroup ${\text{e}}^{-t\mathcal{K}}$ .

Proposition 3.6. If the initial data ${\chi }_0\in L^p\left({ℝ}^3\right)$ . Assume that $k\ge 0$ and $1\le p\le 2$ , $\forall$ $t\ge 0$ , the global solution $\chi ={\left(Z\mathrm{,}w\mathrm{,}n\right)}^t$ of the IVP (3.2) satisfies

${\Vert {\nabla }^k{\text{e}}^{t\mathcal{K}}{\chi }^l\left(0\right)\Vert }_{L^2}\le C{\left(1+t\right)}_{}^{-\frac{1}{2}\left(k+\frac{3}{p}\right)+\frac{3}{4}}{\Vert \chi \left(0\right)\Vert }_{L^p}\mathrm{.}$ (3.23)

If $t$ is large enough and the initial data also satisfy (3.19), then

$\min \left\{{\Vert Z^l\left(t\right)\Vert }_{L^2}\mathrm{,}{\Vert w^l\left(t\right)\Vert }_{L^2}\mathrm{,}{\Vert n^l\left(t\right)\Vert }_{L^2}\right\}\ge C_1{c}_0{\left(1+t\right)}_{}^{-\frac{3}{4}}\mathrm{.}$ (3.24)

4. The Energy Estimate of (Z, w, n) and Its Derivatives

Due to the fact that local classical solutions can be proven in [37] [38], which omit details. Combining the results of local existence with prior estimates through classical argumentation, the global existence of the solution is obtained. So we suppose a priori that

${\Vert \left(m^{+},Z,w,m^{-},n\right)\Vert }_{H^3}\le \delta \ll 1,$ (4.1)

where $\delta$ is little sufficient, $\delta ~{\delta }_0$ . It is in conjunction with Sobolev inequality particularly indicates that

${\Vert \left(m^{+}\mathrm{,}Z\mathrm{,}w\mathrm{,}{m}^{-}\mathrm{,}n\right)\Vert }_{W^{\mathrm{1,}\infty }}\lesssim \delta \mathrm{.}$ (4.2)

The following are a series of lemmas for energy estimation. First, we perform zero order energy estimation.

Lemma 4.1. Under the suppositions of Theorem 1.1 meanwhile (4.1) are valid, then

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}\left\{\frac{{\beta }_3}{{\beta }_1}{\Vert Z\Vert }_{L^2}^2+{\Vert w\Vert }_{L^2}^2+{\Vert n\Vert }_{L^2}^2\right\}+\left({\kappa }_1{\Vert \nabla w\Vert }_{L^2}^2+{\kappa }_2{\Vert \text{div}w\Vert }_{L^2}^2\right)\\ +\left({\varkappa }_1{\Vert \nabla n\Vert }_{L^2}^2+{\varkappa }_2{\Vert \text{div}n\Vert }_{L^2}^2\right)\le C\delta \left({\Vert \nabla w\Vert }_{L^2}^2+{\Vert \nabla Z\Vert }_{L^2}^2+{\Vert \nabla n\Vert }_{L^2}^2\right)\mathrm{.}\end{array}$ (4.3)

Proof. Multiplying (1.14)₃, (1.14)₄, (1.14)₅ by $\frac{{\beta }_3}{{\beta }_1}Z$ , $w$ , $n$ , respectively, we

integrate and sum the resulting equation over ${ℝ}^3$ , then using (1.14)₁ and (1.14)₂, we have

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}\left\{\frac{{\beta }_3}{{\beta }_1}{\Vert Z\Vert }_{L^2}^2+{\Vert w\Vert }_{L^2}^2+{\Vert n\Vert }_{L^2}^2\right\}+\left({\kappa }_1{\Vert \nabla w\Vert }_{L^2}^2+{\kappa }_2{\Vert \text{div}w\Vert }_{L^2}^2\right)\\ +\left({\varkappa }_1{\Vert \nabla n\Vert }_{L^2}^2+{\varkappa }_2{\Vert \text{div}n\Vert }_{L^2}^2\right)\\ ={\displaystyle {\int }_{{ℝ}^3}}\frac{{\beta }_3}{{\beta }_1}{G}_{1}Z\text{d}x+{\displaystyle {\int }_{{ℝ}^3}}w\left(H_1+R_1\right)\text{d}x+{\displaystyle {\int }_{{ℝ}^3}}n\left(H_2+R_2\right)\text{d}x=:{\displaystyle \sum _{i=1}^3}{I}_3.\end{array}$ (4.4)

By employing integration by parts, (4.1), Lemma 2.1, and related Sobolev inequality, then

$\begin{array}{c}\left|I_1\right|\le C{\Vert \left(m^{+}\mathrm{,}Z\right)\Vert }_{L^3}{\Vert \theta \Vert }_{L^6}{\Vert \nabla w\Vert }_{L^2}+{\Vert \left(m^{+}\mathrm{,}Z\right)\Vert }_{L^3}{\Vert Z\Vert }_{L^6}{\Vert \nabla n\Vert }_{L^2}+{\Vert \nabla Z\Vert }_{L^2}{\Vert w\Vert }_{L^6}{\Vert m^{+}\Vert }_{L^3}\\ +{\left|\left|\nabla Z\right|\right|}_{L^2}{\Vert n\Vert }_{L^6}{\Vert m^{-}\Vert }_{L^3}+{\Vert Z\Vert }_{L^6}{\Vert \nabla \left(m^{+}\mathrm{,}Z\right)\Vert }_{L^3}{\Vert w\Vert }_{L^6}{\Vert m^{+}\Vert }_{L^3}\\ +{\Vert \theta \Vert }_{L^6}{\Vert \nabla \left(m^{+}\mathrm{,}Z\right)\Vert }_{L^3}{\Vert n\Vert }_{L^6}{\Vert m^{-}\Vert }_{L^3}+{\Vert Z\Vert }_{L^6}{\left|\left|\nabla w\right|\right|}_{L^2}{\Vert m^{+}\Vert }_{L^3}+{\Vert Z\Vert }_{L^6}{\Vert \nabla n\Vert }_{L^2}{\Vert m^{-}\Vert }_{L^3}\\ \le C\left(\delta {\Vert \nabla Z\Vert }_{L^2}{\Vert \nabla w\Vert }_{L^2}+\delta {\Vert \nabla Z\Vert }_{L^2}{\Vert \nabla n\Vert }_{L^2}+\delta {\Vert \nabla Z\Vert }_{L^2}{\Vert \nabla w\Vert }_{L^2}+\delta {\Vert \nabla Z\Vert }_{L^2}{\Vert \nabla n\Vert }_{L^2}\right)\\ \le C\delta \left({\Vert \nabla w\Vert }_{L^2}^2+{\Vert \nabla Z\Vert }_{L^2}^2+{\Vert \nabla n\Vert }_{L^2}^2\right)\mathrm{.}\end{array}$ (4.5)

For the term $I_2$ , in the same way, we obtain

$\begin{array}{c}\left|I_2\right|\le {\displaystyle {\int }_{{ℝ}^3}}(\left|w\right|\left|w\right|\left|\nabla w\right|+\left|Z\right|\left|\nabla Z\right|\left|w\right|+\left|\nabla Z\right|\left|\nabla w\right|\left|w\right|+\left|Z\right|{\left|\nabla w\right|}^2+\left|\nabla Z\right|\left|\text{div}w\right|\left|w\right|\\ +\left|Z\right|{\left|\text{div}w\right|}^2+\left|Z\right|\left|\nabla m^{+}\right|\left|\nabla w\right|\left|w\right|+\left|\nabla Z\right|\left|m^{+}\right|\left|\nabla w\right|\left|w\right|)\text{d}x\\ \le C({\Vert \nabla w\Vert }_{L^2}{\Vert w\Vert }_{L^3}{\Vert w\Vert }_{L^6}+{\Vert Z\Vert }_{L^6}{\Vert \nabla Z\Vert }_{L^2}{\Vert w\Vert }_{L^3}+{\Vert \nabla Z\Vert }_{L^2}{\Vert \nabla w\Vert }_{L^2}{\Vert w\Vert }_{L^{\infty }}\\ +{\Vert Z\Vert }_{L^{\infty }}{\Vert \nabla w\Vert }_{L^2}^2+{\Vert Z\Vert }_{L^6}{\Vert \nabla m^{+}\Vert }_{L^3}{\Vert \nabla w\Vert }_{L^2}{\Vert w\Vert }_{L^{\infty }}\\ +{\Vert Z\Vert }_{L^2}{\Vert \nabla w\Vert }_{L^2}{\Vert w\Vert }_{L^{\infty }}{\Vert m^{+}\Vert }_{L^{\infty }})\\ \le C\delta \left({\Vert \nabla w\Vert }_{L^2}^2+{\Vert \nabla Z\Vert }_{L^2}^2\right)\mathrm{.}\end{array}$ (4.6)

We also obtain

$\left|I_3\right|\le C\delta \left({\Vert \nabla n\Vert }_{L^2}^2+{\Vert \nabla Z\Vert }_{L^2}^2\right)\mathrm{.}$ (4.7)

Substituting (4.5)-(4.7) into (4.4), when $\delta$ is small enough, we can obtain (4.3) immediately. Lemma 4.1 is proved. □

Lemma 4.2. Under the suppositions of Theorem 1.1 meanwhile (4.1) are valid, then

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}\left\{\frac{{\beta }_3}{{\beta }_1}{\Vert \nabla Z\Vert }_{H^1}^2+{\Vert \nabla w\Vert }_{H^1}^2+{\Vert \nabla n\Vert }_{H^1}^2\right\}+\left({\kappa }_1{\Vert {\nabla }^{2}w\Vert }_{H^1}^2+{\kappa }_2{\Vert \nabla \text{div}w\Vert }_{H^1}^2\right)\\ +\left({\varkappa }_1{\Vert {\nabla }^{2}n\Vert }_{H^1}^2+{\varkappa }_2{\Vert \nabla \text{div}n\Vert }_{H^1}^2\right)\le C\delta \left({\Vert \nabla Z\Vert }_{H^1}^2+{\Vert \nabla w\Vert }_{H^2}^2+{\Vert \nabla n\Vert }_{H^2}^2\right)\mathrm{.}\end{array}$ (4.8)

Proof. First, we consider $k=1$ ,

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}\left\{\frac{{\beta }_3}{{\beta }_1}{\Vert \nabla Z\Vert }_{L^2}^2+{\Vert \nabla w\Vert }_{L^2}^2+{\Vert \nabla n\Vert }_{L^2}^2\right\}+\left({\kappa }_1{\Vert {\nabla }^{2}w\Vert }_{L^2}^2+{\kappa }_2{\Vert \nabla \text{div}w\Vert }_{L^2}^2\right)\\ +\left({\varkappa }_1{\Vert {\nabla }^{2}n\Vert }_{L^2}^2+{\varkappa }_2{\Vert \nabla \text{div}n\Vert }_{L^2}^2\right)\le C\delta \left({\Vert \nabla Z\Vert }_{L^2}^2+{\Vert \nabla w\Vert }_{H^1}^2+{\Vert \nabla n\Vert }_{H^1}^2\right)\mathrm{.}\end{array}$ (4.9)

Multiplying $\nabla$ (1.14)₃, $\nabla$ (1.14)₄, $\nabla$ (1.14)₅ by $\frac{{\beta }_3}{{\beta }_1}\nabla Z$ , $\nabla w$ , $\nabla n$ , respectively, we integrate and sum then the resulting equation over ${ℝ}^3$ , we have

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}\left\{\frac{{\beta }_3}{{\beta }_1}{\Vert \nabla Z\Vert }_{L^2}^2+{\Vert \nabla w\Vert }_{L^2}^2+{\Vert \nabla n\Vert }_{L^2}^2\right\}+\left({\kappa }_1{\Vert {\nabla }^{2}w\Vert }_{L^2}^2+{\kappa }_2{\Vert \nabla \text{div}w\Vert }_{L^2}^2\right)\\ +\left({\varkappa }_1{\Vert {\nabla }^{2}n\Vert }_{L^2}^2+{\varkappa }_2{\Vert \nabla \text{div}n\Vert }_{L^2}^2\right)\end{array}$

$\begin{array}{l}={\displaystyle {\int }_{{ℝ}^3}}\frac{{\beta }_3}{{\beta }_1}\nabla G_1\cdot \nabla Z\text{d}x+{\displaystyle {\int }_{{ℝ}^3}}\left(\nabla H_1+\nabla R_1\right)\cdot \nabla w\text{d}x+{\displaystyle {\int }_{{ℝ}^3}}\left(\nabla H_2+\nabla R_2\right)\cdot \nabla n\text{d}x\\ =:{\displaystyle \sum _{i=1}^3}{J}_i.\end{array}$ (4.10)

Employing relevant inequalities, we obtain

$\left|J_1\right|\le C\delta \left({\Vert \nabla Z\Vert }_{L^2}^2+{\Vert \nabla n\Vert }_{H^1}^2+{\Vert \nabla w\Vert }_{H^1}^2\right)\mathrm{.}$ (4.11)

We calculate

$\begin{array}{c}\left|J_2\right|\le C({\Vert w\Vert }_{L^{\infty }}{\Vert \nabla w\Vert }_{L^2}{\Vert {\nabla }^{2}w\Vert }_{L^2}+{\Vert Z\Vert }_{L^{\infty }}{\Vert \nabla Z\Vert }_{L^2}{\Vert {\nabla }^{2}w\Vert }_{L^2}\\ +{\Vert Z\Vert }_{L^{\infty }}{\Vert {\nabla }^{2}w\Vert }_{L^2}{\Vert {\nabla }^{2}w\Vert }_{L^2}+{\Vert \nabla w\Vert }_{L^2}{\Vert \nabla \left(m^{+},Z\right)\Vert }_{L^{\infty }}{\Vert {\nabla }^{2}w\Vert }_{L^2}{\Vert Z\Vert }_{L^{\infty }}\\ +{\Vert \nabla w\Vert }_{L^2}{\Vert \nabla Z\Vert }_{L^{\infty }}{\Vert \left(m^{+},Z\right)\Vert }_{L^{\infty }}{\Vert {\nabla }^{2}w\Vert }_{L^2})\\ \le C\delta \left({\Vert \nabla w\Vert }_{L^2}^2+{\Vert \nabla Z\Vert }_{L^2}^2+{\Vert {\nabla }^{2}w\Vert }_{L^2}^2\right)\\ \le C\delta \left({\Vert \nabla Z\Vert }_{L^2}^2+{\Vert \nabla w\Vert }_{H^1}^2\right).\end{array}$ (4.12)

Similarly, we obtain

$\left|J_3\right|\le C\delta \left({\Vert \nabla Z\Vert }_{L^2}^2+{\Vert \nabla n\Vert }_{H^1}^2\right)\mathrm{.}$ (4.13)

Substituting (4.11)-(4.13), into (4.10), then, we obtain (4.9) if $\delta$ is small enough.

Next we consider $k=2$ , we have

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}\left\{\frac{{\beta }_3}{{\beta }_1}{\Vert {\nabla }^{2}Z\Vert }_{L^2}^2+{\Vert {\nabla }^{2}w\Vert }_{L^2}^2+{\Vert {\nabla }^{2}n\Vert }_{L^2}^2\right\}+\left({\kappa }_1{\Vert {\nabla }^{3}w\Vert }_{L^2}^2+{\kappa }_2{\Vert {\nabla }^2\text{div}w\Vert }_{L^2}^2\right)\\ +\left({\varkappa }_1{\Vert {\nabla }^{3}n\Vert }_{L^2}^2+{\varkappa }_2{\Vert {\nabla }^2\text{div}n\Vert }_{L^2}^2\right)\le C\delta \left({\Vert \nabla Z\Vert }_{H^1}^2+{\Vert \nabla n\Vert }_{H^2}^2+{\Vert \nabla w\Vert }_{H^2}^2\right)\mathrm{.}\end{array}$ (4.14)

Multiplying ${\nabla }^2$ (1.14)₃, ${\nabla }^2$ (1.14)₃, ${\nabla }^2$ (1.14)₄ by $\frac{{\beta }_3}{{\beta }_1}{\nabla }^{2}Z$ , ${\nabla }^{2}w$ , ${\nabla }^{2}n$ , respectively, we integrate and sum up the resulting equation over ${ℝ}^3$ ,

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}\left\{\frac{{\beta }_3}{{\beta }_1}{\Vert {\nabla }^{2}Z\Vert }_{L^2}^2+{\Vert {\nabla }^{2}w\Vert }_{L^2}^2+{\Vert {\nabla }^{2}n\Vert }_{L^2}^2\right\}+\left({\kappa }_1{\Vert {\nabla }^{3}w\Vert }_{L^2}^2+{\kappa }_2{\Vert {\nabla }^2\text{div}n\Vert }_{L^2}^2\right)\\ +\left({\varkappa }_1{\Vert {\nabla }^{3}n\Vert }_{L^2}^2+{\varkappa }_2{\Vert {\nabla }^2\text{div}n\Vert }_{L^2}^2\right)\\ ={\displaystyle {\int }_{{ℝ}^3}}\frac{{\beta }_3}{{\beta }_1}{\nabla }^2{G}_1\cdot {\nabla }^{2}Z\text{d}x+{\displaystyle {\int }_{{ℝ}^3}}\left({\nabla }^2{H}_1+{\nabla }^2{R}_1\right)\cdot {\nabla }^{2}w\text{d}x\\ +{\displaystyle {\int }_{{ℝ}^3}}\left({\nabla }^2{H}_2+{\nabla }^2{R}_2\right)\cdot {\nabla }^{2}n\text{d}x\\ =:J_4+J_5+J_6.\end{array}$ (4.15)

Similar to (4.11), we obtain

$\left|J_4\right|\le C\delta \left({\Vert {\nabla }^{2}Z\Vert }_{L^2}^2+{\Vert \nabla w\Vert }_{H^2}^2+{\Vert \nabla n\Vert }_{H^2}^2\right)\mathrm{.}$ (4.16)

Similar to (4.12), we obtain

$\left|J_5\right|\le C\delta \left({\Vert \nabla w\Vert }_{L^2}^2+{\Vert {\nabla }^{2}w\Vert }_{L^2}^2+{\Vert {\nabla }^{3}w\Vert }_{L^2}^2+{\Vert \nabla Z\Vert }_{L^2}^2+{\Vert {\nabla }^{2}Z\Vert }_{L^2}^2\right)$

$\le C\delta \left({\Vert \nabla Z\Vert }_{H^1}^2+{\Vert \nabla w\Vert }_{H^2}^2\right)\mathrm{.}$ (4.17)

We also have

$\left|J_6\right|\le C\delta \left({\Vert \nabla Z\Vert }_{H^1}^2+{\Vert \nabla n\Vert }_{H^2}^2\right)\mathrm{.}$ (4.18)

Substituting (4.16)-(4.18) into (4.15), we obtain (4.14). From (4.9) and (4.14), if $\delta$ is little sufficiently, we acquire (4.8). The proof is accomplished. □

Lemma 4.3. Under the suppositions of Theorem 1.1 and (4.1) is valid, then

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}\left\{\frac{{\beta }_3}{{\beta }_1}{\Vert Z_t\Vert }_{L^2}^2+{\Vert w_t\Vert }_{L^2}^2+{\Vert n_t\Vert }_{L^2}^2\right\}+\left({\kappa }_1{\Vert \nabla w_t\Vert }_{L^2}^2+{\kappa }_2{\Vert \text{div}{w}_t\Vert }_{L^2}^2\right)\\ +\left({\varkappa }_1{\Vert \nabla n_t\Vert }_{L^2}^2+{\varkappa }_2{\Vert \text{div}{n}_t\Vert }_{L^2}^2\right)\\ \le C\delta \left({\Vert \nabla w\left(t\right)\Vert }_{H^1}^2+{\Vert \nabla n\left(t\right)\Vert }_{H^1}^2+{\Vert \nabla w_t\left(t\right)\Vert }_{L^2}^2+{\Vert \nabla n_t\left(t\right)\Vert }_{L^2}^2\right)\mathrm{,}\end{array}$ (4.19)

and

$\begin{array}{l}\frac{\text{d}}{\text{d}t}\{{\Vert \sqrt{{\rho }^{{\mu }_1}}\nabla w_t\Vert }_{L^2}^2+{\Vert \sqrt{{\rho }^{{\mu }_1}+{\rho }^{{\lambda }_1}}\text{div}{w}_t\Vert }_{L^2}^2+{\Vert \sqrt{{\eta }^{{\mu }_2}}\nabla n_t\Vert }_{L^2}^2\\ +{\Vert \sqrt{{\eta }^{{\mu }_2}+{\eta }^{{\lambda }_2}}\text{div}{n}_t\Vert }_{L^2}^2\}+\frac{\text{d}}{\text{d}t}\left({\Vert \sqrt{\rho }{w}_{tt}\Vert }_{L^2}^2+{\Vert \sqrt{\eta }{n}_{tt}\Vert }_{L^2}^2\right)\\ \mathrm{<mi>\; </mi>}\le C\delta \left({\Vert \nabla w\Vert }_{H^1}^2+{\Vert \nabla w_t\Vert }_{L^2}^2+{\Vert \nabla n\Vert }_{H^1}^2+{\Vert \nabla n_t\Vert }_{L^2}^2\right)\mathrm{.}\end{array}$ (4.20)

Proof. Calculating the derivative of (1.14)_2,3,4 in terms of $t$ , then

$\{\begin{array}{l}{\partial }_{tt}Z+{\beta }_1\text{div}{w}_t+{\beta }_2\text{div}{n}_t={\left(G_1\right)}_t\mathrm{,}\\ {\partial }_{tt}w+{\beta }_3\nabla Z_t-{\kappa }_1\Delta w_t-{\kappa }_2\nabla \text{div}{w}_t={\left(H_1\right)}_t+{\left(R_1\right)}_t\mathrm{,}\\ {\partial }_{tt}n+{\beta }_4\nabla Z_t-{\varkappa }_1\Delta n_t-{\varkappa }_2\nabla \text{div}{n}_t={\left(H_1\right)}_t+{\left(R_2\right)}_t\mathrm{.}\end{array}$ (4.21)

We multiply (4.21)₁, (4.21)₂, (4.21)₃, by $\frac{{\beta }_3}{{\beta }_1}{Z}_t\mathrm{,}{w}_t\mathrm{,}{n}_t$ respectively, integrate and sum the equality over ${ℝ}^3$ , we have

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{{ℝ}^3}}\frac{{\beta }_3}{{\beta }_1}{\left|Z_t\right|}^2+{\left|w_t\right|}^2+{\left|n_t\right|}^2\text{d}x+{\displaystyle {\int }_{{ℝ}^3}}{\kappa }_1{\left|\nabla w_t\right|}^2+{\kappa }_2{\left|\text{div}{w}_t\right|}^2\text{d}x\\ +{\displaystyle {\int }_{{ℝ}^3}}{\varkappa }_1{\left|\nabla n_t\right|}^2+{\varkappa }_2{\left|\text{div}{n}_t\right|}^2\text{d}x\\ ={\displaystyle {\int }_{{ℝ}^3}}\frac{{\beta }_3}{{\beta }_1}{\left(G_1\right)}_t{Z}_t+\left[{\left(H_1\right)}_t+{\left(R_1\right)}_t\right]w_t+\left[{\left(H_2\right)}_t+{\left(R_2\right)}_t\right]n_t\text{d}x\mathrm{,}\end{array}$ (4.22)

from (1.14),

$\begin{array}{c}{\Vert Z_t\Vert }_{L^2}\le {\Vert G_1\Vert }_{L^2}+{\Vert {\beta }_1\text{div}w\Vert }_{L^2}+{\Vert {\beta }_1\text{div}n\Vert }_{L^2}\\ \le C\left({\Vert \nabla w\Vert }_{L^2}+{\Vert \nabla n\Vert }_{L^2}\right)\mathrm{.}\end{array}$ (4.23)

Similar to the previous process, we have

$\begin{array}{c}{\displaystyle {\int }_{{ℝ}^3}}\frac{{\beta }_3}{{\beta }_1}{\left(G_1\right)}_t{Z}_t\text{d}x\le C{\Vert {\left(G_1\right)}_t\Vert }_{L^2}{\Vert Z_t\Vert }_{L^2}\\ \le C\mathrm{<mi>\; </mi>}({\Vert m_t^{+}\Vert }_{L^3}{\Vert \nabla w\Vert }_{L^6}+{\Vert m^{+}\Vert }_{L^{\infty }}{\Vert \nabla w_t\Vert }_{L^2}+{\Vert \nabla m^{+}\Vert }_{L^3}{\Vert w_t\Vert }_{L^6}\end{array}$

$\begin{array}{c}+{\Vert \nabla m_t^{+}\Vert }_{L^2}{\Vert w\Vert }_{L^{\infty }}+{\Vert Z_t\Vert }_{L^2}{\Vert \nabla w\Vert }_{L^{\infty }}+{\Vert Z\Vert }_{L^{\infty }}{\Vert \nabla w_t\Vert }_{L^2}\\ +{\Vert m_t^{-}\Vert }_{L^3}{\Vert \nabla n\Vert }_{L^6}+{\Vert m^{-}\Vert }_{L^{\infty }}{\Vert \nabla n_t\Vert }_{L^2}+{\Vert \nabla m^{-}\Vert }_{L^3}{\Vert n_t\Vert }_{L^6}\\ +{\Vert \nabla m_t^{-}\Vert }_{L^2}{\Vert n\Vert }_{L^{\infty }}+{\Vert Z_t\Vert }_{L^2}{\Vert \nabla n\Vert }_{L^{\infty }}+{\Vert Z\Vert }_{L^{\infty }}{\Vert \nabla n_t\Vert }_{L^2}){\Vert Z_t\Vert }_{L^2}\\ \le C\delta \left({\Vert \nabla w\Vert }_{H^1}^2+{\Vert \nabla n\Vert }_{H^1}^2+{\Vert \nabla w_t\Vert }_{L^2}^2+{\Vert \nabla n_t\Vert }_{L^2}^2\right)\mathrm{,}\end{array}$ (4.24)

and

${\displaystyle {\int }_{{ℝ}^3}}\left[{\left(H_1\right)}_t+{\left(R_1\right)}_t\right]w_t\text{d}x\le C\delta \left({\Vert \nabla w\Vert }_{H^1}^2+{\Vert \nabla w_t\Vert }_{L^2}^2+{\Vert \nabla n\Vert }_{H^1}^2+{\Vert \nabla n_t\Vert }_{L^2}^2\right)\mathrm{.}$ (4.25)

Similarly, we have

${\displaystyle {\int }_{{ℝ}^3}}\left[{\left(H_2\right)}_t+{\left(R_2\right)}_t\right]n_t\text{d}x\le C\delta \left({\Vert \nabla w\Vert }_{H^1}^2+{\Vert \nabla w_t\Vert }_{L^2}^2+{\Vert \nabla n\Vert }_{H^1}^2+{\Vert \nabla n_t\Vert }_{L^2}^2\right)\mathrm{.}$ (4.26)

Substituting (4.24)-(4.26) into (4.22), we obtain (4.19). We restate the equation (4.21)₂, (4.21)₃, and get the following form:

$\{\begin{array}{l}\rho {\partial }_{tt}w-{\rho }^{{\mu }_1}\Delta w_t-\left({\rho }^{{\mu }_1}+{\rho }^{{\lambda }_1}\right)\nabla \text{div}{w}_t\\ =-{\rho }_t{w}_t-\nabla Z_t-[\rho w\cdot \nabla w+\nabla \left({\rho }^{{\mu }_1}\right)\left(\nabla w+{\nabla }^{t}w\right)+\nabla \left({\rho }^{{\lambda }_1}\right)\text{div}w\begin{array}{c}\stackrel{}{}\\ \end{array}\\ +\frac{\nabla \left({\rho }^{{\mu }_1}\right)\left(\nabla w+{\nabla }^{t}w\right)\nabla {\sigma }^{+}}{{\sigma }^{+}}+{\frac{\nabla \left({\rho }^{{\lambda }_1}\right)\text{div}w\nabla {\sigma }^{+}}{{\sigma }^{+}}]}_t\\ \eta {\partial }_{tt}n-{\eta }^{{\mu }_2}\Delta n_t-\left({\eta }^{{\mu }_2}+{\eta }^{{\lambda }_2}\right)\nabla \text{div}{n}_t\\ =-{\eta }_t{n}_t-\nabla Z_t-[\eta n\cdot \nabla n+\nabla \left({\eta }^{{\mu }_2}\right)\left(\nabla n+{\nabla }^{t}n\right)+\nabla \left({\eta }^{{\lambda }_2}\right)\text{div}n\begin{array}{c}\stackrel{}{}\\ \end{array}\\ +\frac{\nabla \left({\eta }^{{\mu }_2}\right)\left(\nabla n+{\nabla }^{t}n\right)\nabla {\sigma }^{-}}{{\sigma }^{-}}+{\frac{\nabla \left({\eta }^{{\lambda }_2}\right)\text{div}n\nabla {\sigma }^{-}}{{\sigma }^{-}}]}_t\end{array}$ (4.27)

Multiplying (4.27)₁ by $w_{tt}$ respectively, integrating over ${ℝ}^3$ , then

$\begin{array}{l}{\displaystyle {\int }_{{ℝ}^3}}\rho {\left|w_{tt}\right|}^2\text{d}x+\frac{1}{2}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{{ℝ}^3}}{\rho }^{{\mu }_1}{\left|\nabla w_t\right|}^2+\left({\rho }^{{\mu }_1}+{\rho }^{{\lambda }_1}\right){\left|\text{div}{w}_t\right|}^2\text{d}x\\ \le C({\Vert {\rho }_t\Vert }_{L^3}{\Vert w_t\Vert }_{L^6}{\Vert w_{tt}\Vert }_{L^2}+{\Vert \nabla Z_t\Vert }_{L^2}{\Vert w_{tt}\Vert }_{L^2}+{\Vert \nabla Z\Vert }_{L^3}{\Vert \nabla w_t\Vert }_{L^2}{\Vert w_{tt}\Vert }_{L^6}\\ +{\Vert \nabla Z_t\Vert }_{L^{\infty }}{\Vert \nabla w_t\Vert }_{L^2}^2+{\Vert w\Vert }_{L^{\infty }}{\Vert \nabla w\Vert }_{L^2}{\Vert w_{tt}\Vert }_{L^2}+{\Vert {\rho }_t\Vert }_{L^2}{\Vert w\Vert }_{L^{\infty }}{\Vert \nabla w\Vert }_{L^{\infty }}{\Vert w_{tt}\Vert }_{L^2}\\ +{\Vert w_t\Vert }_{L^6}{\Vert \nabla w\Vert }_{L^3}{\Vert w_t\Vert }_{L^2}+{\Vert w\Vert }_{L^{\infty }}{\Vert \nabla w_t\Vert }_{L^2}{\Vert w_{tt}\Vert }_{L^2}+{\Vert \nabla Z_t\Vert }_{L^2}{\Vert \nabla w\Vert }_{L^{\infty }}{\Vert Z_{tt}\Vert }_{L^2}\\ +{\Vert Z_t\Vert }_{L^2}{\Vert \nabla w\Vert }_{L^{\infty }}{\Vert \nabla {\sigma }^{+}\Vert }_{L^{\infty }}{\Vert w_{tt}\Vert }_{L^2}+{\Vert Z\Vert }_{L^{\infty }}{\Vert \nabla w_t\Vert }_{L^2}{\Vert \nabla {\sigma }^{+}\Vert }_{L^{\infty }}{\Vert w_{tt}\Vert }_{L^2}\\ +{\Vert Z\Vert }_{L^{\infty }}{\Vert \nabla w\Vert }_{L^{\infty }}{\Vert \nabla {\sigma }_t^{+}\Vert }_{L^2}{\Vert w_{tt}\Vert }_{L^2}+{\Vert \nabla {\sigma }^{+}\Vert }_{L^6}{\Vert {\sigma }_t\Vert }_{L^6}{\Vert \nabla w\Vert }_{L^6}{\Vert w_{tt}\Vert }_{L^2}\\ +{\Vert {\rho }_t\Vert }_{L^2}{\Vert \nabla {\sigma }^{+}\Vert }_{L^{\infty }}{\Vert \nabla u\Vert }_{L^{\infty }}{\Vert w_{tt}\Vert }_{L^2}\\ \le C\delta \left({\Vert \nabla w\Vert }_{H^1}^2+{\Vert \nabla w_t\Vert }_{L^2}^2\right)+\frac{1}{2}{\displaystyle {\int }_{{ℝ}^3}}\rho {\left|w_{tt}\left(t\right)\right|}^2\text{d}x\mathrm{.}\end{array}$ (4.28)

We multiply (4.27)₂ by $n_{tt}$ respectively, and integrate the resulting equality over ${ℝ}^3$ ,

$\begin{array}{l}{\displaystyle {\int }_{{ℝ}^3}}\eta {\left|n_{tt}\right|}^2\text{d}x+\frac{1}{2}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{{ℝ}^3}}{\eta }^{{\mu }_2}{\left|\nabla n_t\right|}^2+\left({\eta }^{{\mu }_2}+{\eta }^{{\lambda }_2}\right){\left|\text{div}{n}_t\right|}^2\text{d}x\\ \le C\delta \left({\Vert \nabla n\Vert }_{H^1}^2+{\Vert \nabla n_t\Vert }_{L^2}^2\right)+\frac{1}{2}{\displaystyle {\int }_{{ℝ}^3}}\eta {\left|n_{tt}\left(t\right)\right|}^2\text{d}x\mathrm{.}\end{array}$ (4.29)

By using (4.28) and (4.29), we obtain (4.20). □

5. The Time Decay Rates for $\left(Z,w,n\right)$

Theorem 5.1. If $C>0$ , $C$ is not less than 0. Under the conditions of Theorem 1.1 and (4.1), then

${\Vert \left(Z\mathrm{,}w\mathrm{,}n\right)\left(t\right)\Vert }_{L^2}\le C{N}_0{\left(1+t\right)}_{}^{-\frac{3}{4}}\mathrm{,}$ (5.1)

$\begin{array}{l}{\Vert \nabla Z\left(t\right)\Vert }_{H^1}+{\Vert \nabla w\left(t\right)\Vert }_{H^2}+{\Vert \nabla n\left(t\right)\Vert }_{H^2}\\ +{\Vert \left(Z_t\mathrm{,}{w}_t\mathrm{,}{n}_t\mathrm{,}\nabla w_t\mathrm{,}\nabla n_t\right)\Vert }_{L^2}^2\le C{N}_0{\left(1+t\right)}_{}^{-\frac{5}{4}}\end{array}$ (5.2)

and

${\displaystyle {\int }_0^t}{\left(1+\tau \right)}_{}^{\frac{9}{8}}\left({\Vert w_{tt}\left(\tau \right)\Vert }_{L^2}^2+{\Vert n_{tt}\left(\tau \right)\Vert }_{L^2}^2\right)\text{d}\tau \le C{N}_0\mathrm{.}$ (5.3)

Proof. From (1.14), we see that

${\Vert w_t\Vert }_{L^2}\le C\left({\Vert \left({\nabla }^{2}w\mathrm{,}\nabla Z\right)\Vert }_{L^2}+\delta {\Vert \nabla w\Vert }_{H^1}+\delta {\Vert {\nabla }^{2}Z\Vert }_{L^2}\right)\mathrm{,}$ (5.4)

${\Vert n_t\Vert }_{L^2}\le C\left({\Vert \left({\nabla }^{2}n\mathrm{,}\nabla Z\right)\Vert }_{L^2}+\delta {\Vert \nabla n\Vert }_{H^1}+\delta {\Vert {\nabla }^{2}Z\Vert }_{L^2}\right)\mathrm{.}$ (5.5)

From (1.14), (4.19) and (4.20), we have

$\begin{array}{c}{\Vert {\nabla }^{2}Z\Vert }_{L^2}\le \mathrm{<mi>\; </mi>}C{\Vert \nabla \left(\left(1+m^{+}\right)w_t\right)\Vert }_{L^2}+C{\Vert {\nabla }^{3}w\Vert }_{L^2}+C{\Vert \nabla A_1+\nabla B_1\Vert }_{L^2}\\ \le C\left({\Vert \left(\nabla w_t,{\nabla }^{3}w\right)\Vert }_{L^2}+\delta {\Vert {\nabla }^{2}w\Vert }_{L^2}\right).\end{array}$ (5.6)

Similarly, we also have

${\Vert {\nabla }^{3}w\Vert }_{L^2}\le C\left({\Vert \left(\nabla w_t\mathrm{,}{\nabla }^{2}Z\right)\Vert }_{L^2}+\delta {\Vert {\nabla }^{2}w\Vert }_{L^2}\right)\mathrm{,}$

${\Vert {\nabla }^{3}n\Vert }_{L^2}\le C\left({\Vert \left(\nabla n_t\mathrm{,}{\nabla }^{2}Z\right)\Vert }_{L^2}+\delta {\Vert {\nabla }^{2}n\Vert }_{L^2}\right)\mathrm{.}$ (5.7)

Assuming that the constant $D>0$ . From (5.4)-(5.7), summing up $D$ × ((4.8) + (4.19)) + (4.20), then there exists an energy functional, designated as $W\left(w\mathrm{,}Z\mathrm{,}n\right)$ , which is equivalent to ${\Vert \nabla Z\Vert }_{H^1}^2+{\Vert \nabla w\Vert }_{H^2}^2+{\Vert \nabla n\Vert }_{H^2}^2+{\Vert \left(Z_t\mathrm{,}{w}_t\mathrm{,}{n}_t\mathrm{,}\nabla w_t\mathrm{,}\nabla n_t\right)\Vert }_{L^2}^2$ , then for some positive constant $C_1$ , we can get

$\frac{\text{d}}{\text{d}t}W\left(t\right)+C_1\left(W\left(t\right)+{\Vert w_{tt}\Vert }_{L^2}^2+{\Vert n_{tt}\Vert }_{L^2}^2\right)\le C{\Vert \nabla \left(Z^l\mathrm{,}{w}^l\mathrm{,}{n}^l\right)\Vert }_{L^2}^2\mathrm{.}$ (5.8)

In addition, Define the time-weighted energy functional:

$ℐ\left(t\right)=\underset{0\le \tau \le t}{\sup }\left\{{\left(1+\tau \right)}_{}^{\frac{5}{2}}W\left(\tau \right)\right\}\mathrm{,}$ (5.9)

Denoting

$\mathcal{G}=\left(\widehat{G_1}\mathrm{,}\widehat{H_1+R_1}\mathrm{,}\widehat{H_2+R_2}\right)\mathrm{,}$

by Duhamel principle, we have

${\chi }^l={\text{e}}^{t\mathcal{K}}{\chi }^l\left(0\right)+{\displaystyle {\int }_0^t}{\text{e}}^{\left(t-\tau \right)\mathcal{K}}{\mathcal{G}}^l\left(\tau \right)\text{d}\tau \mathrm{,}$

So we can derive

${\Vert \nabla \left(Z^l\mathrm{,}{w}^l\mathrm{,}{n}^l\right)\left(t\right)\Vert }_{L^2}\lesssim {\left(1+t\right)}_{}^{-\frac{5}{4}}{\Vert \left(Z\mathrm{,}w\mathrm{,}n\right)\left(0\right)\Vert }_{L^1}+{\displaystyle {\int }_0^t}{\Vert \nabla {\text{e}}^{\left(t-\tau \right)\mathcal{K}}\mathcal{G}\left(\tau \right)\Vert }_{L^2}\text{d}\tau \mathrm{.}$ (5.10)

Because of the non-dissipative nature of $m^{\pm }$ , we need to devise innovative approaches to address the final two terms of $G_1$ in (1.14)₃. The core concept is to treat $-ℒ\eta w\cdot \nabla m^{+}-ℒ\rho n\cdot \nabla m^{-}$ as a whole, and then restate cleverly. If we notice that $\eta r_{-}^2-\rho r_{+}^2\ne 0$ , and use the subtle relation between the variables, it is somewhat unexpected that $-ℒ\eta u\cdot \nabla m^{+}-ℒ\rho v\cdot \nabla m^{-}$ can be rewritten as

$\begin{array}{l}-ℒ\eta w\cdot \nabla m^{+}-ℒ\rho n\cdot \nabla m^{-}\\ =-ℒ\eta u\cdot \nabla m^{+}-ℒ\rho v\cdot \left(\frac{\nabla Z}{ℒ\rho }-\frac{\eta \nabla m^{+}}{\rho }\right)\\ =-n\cdot \nabla Z+ℒ\eta \left(n-w\right)\cdot \nabla m^{+}\\ =-n\cdot \nabla Z+\frac{r_{+}^2{r}_{-}^2\rho \eta \left(n-w\right)}{{\left(\rho r_{+}\right)}^2+\left(\eta r_{-}^2-\rho r_{+}^2\right)M^{+}}\cdot \nabla m^{+}\\ =-n\cdot \nabla Z+\frac{r_{+}^2{r}_{-}^2\rho \eta \left(n-w\right)}{\eta r_{-}^2-\rho r_{+}^2}\cdot \nabla \ln \left({\left(\rho r_{+}\right)}^2+\left(\eta r_{-}^2-\rho r_{+}^2\right)M^{+}\right)\\ +\frac{r_{+}^2{r}_{-}^2\rho \eta \left(n-w\right)}{\eta r_{-}^2-\rho r_{+}^2}\cdot \left(\nabla {\left(\rho r_{+}\right)}^2+M^{+}\nabla \left(\eta r_{-}^2-\rho r_{+}^2\right)\right)\\ =-n\cdot \nabla Z+\text{div}\left[\frac{r_{+}^2{r}_{-}^2\rho \eta \left(n-w\right)}{\eta r_{-}^2-\rho r_{+}^2}\ln \left(\frac{{\left(\rho r_{+}\right)}^2+\left(\eta r_{-}^2-\rho r_{+}^2\right)M^{+}}{{\left(\overline{\rho }{\overline{r}}_{+}\right)}^2+\left(\overline{\eta }{\overline{r}}_{-}^2-\overline{\rho }{\overline{r}}_{+}^2\right){\overline{M}}^{+}}\right)\right]\\ +\ln \left(\frac{{\left(\rho r_{+}\right)}^2+\left(\eta r_{-}^2-\rho r_{+}^2\right)M^{+}}{{\left(\overline{\rho }{\overline{r}}_{+}\right)}^2+\left(\overline{\eta }{\overline{r}}_{-}^2-\overline{\rho }{\overline{r}}_{+}^2\right){\overline{M}}^{+}}\right)\text{div}\left[\frac{r_{+}^2{r}_{-}^2\rho \eta \left(n-w\right)}{\eta r_{-}^2-\rho r_{+}^2}\right]\\ +\frac{r_{+}^2{r}_{-}^2\rho \eta \left(w-n\right)}{\eta r_{-}^2-\rho r_{+}^2}\cdot \left(\nabla {\left(\rho r_{+}\right)}^2+M^{+}\nabla \left(\eta r_{-}^2-\rho r_{+}^2\right)\right)\mathrm{.}\end{array}$(5.11)

From Proposition 3.6, (5.9) and (5.10), we have

$\begin{array}{l}{\Vert \nabla \left(Z^l\mathrm{,}{w}^l\mathrm{,}{n}^l\right)\left(t\right)\Vert }_{L^2}\\ \lesssim \mathrm{<mi>\; </mi>}{\left(1+t\right)}_{}^{-\frac{5}{4}}{N}_0+{\displaystyle {\int }_0^t}{\left(1+t-\tau \right)}_{}^{-\frac{5}{4}}{\Vert \left(m^{\pm }\mathrm{,}Z\right)\left(\tau \right)\Vert }_{L^{\infty }}{\Vert \left(\nabla w\mathrm{,}\nabla n\right)\left(\tau \right)\Vert }_{L^2}\text{d}\tau \\ +{\displaystyle {\int }_0^t}{\left(1+t-\tau \right)}_{}^{-\frac{5}{4}}{\Vert \left(m^{+}\mathrm{,}Z\mathrm{,}w\mathrm{,}\nabla m^{+}\mathrm{,}\nabla Z\right)\left(\tau \right)\Vert }_{L^{\infty }}{\Vert \left(\nabla Z\mathrm{,}\nabla w\mathrm{,}{\nabla }^{2}w\right)\left(\tau \right)\Vert }_{L^2}\text{d}\tau \\ +{\displaystyle {\int }_0^t}{\left(1+t-\tau \right)}_{}^{-\frac{5}{4}}{\Vert \left(m^{-}\mathrm{,}Z\mathrm{,}n\mathrm{,}\nabla m^{-}\mathrm{,}\nabla Z\right)\left(\tau \right)\Vert }_{L^{\infty }}{\Vert \left(\nabla Z\mathrm{,}\nabla n\mathrm{,}{\nabla }^{2}n\right)\left(\tau \right)\Vert }_{L^2}\text{d}\tau \\ \lesssim {\left(1+t\right)}_{}^{-\frac{5}{4}}{N}_0+\delta {\displaystyle {\int }_0^t}{\left(1+t-\tau \right)}_{}^{-\frac{5}{4}}{\left(1+\tau \right)}_{}^{-\frac{5}{4}}\sqrt{ℐ\left(t\right)}\text{d}\tau \\ \lesssim \mathrm{<mi>\; </mi>}{\left(1+t\right)}_{}^{-\frac{5}{4}}{N}_0+\delta {\left(1+t\right)}_{}^{-\frac{5}{4}}\sqrt{ℐ\left(t\right)}\mathrm{,}\end{array}$ (5.12)

so we can obtain

$\frac{\text{d}}{\text{d}t}W\left(t\right)+C_1\left(W\left(t\right)+{\Vert w_{tt}\Vert }_{L^2}^2+{\Vert n_{tt}\Vert }_{L^2}^2\right)\le C{\left(1+t\right)}_{}^{-\frac{5}{2}}{N}_0^2+C{\delta }^2{\left(1+t\right)}_{}^{-\frac{5}{2}}ℐ\left(t\right)\mathrm{.}$ (5.13)

By Gronwall inequality, we obtain

$W\left(t\right)\le C{\left(1+t\right)}_{}^{-\frac{5}{2}}{N}_0^2+C{\delta }^2{\left(1+t\right)}_{}^{-\frac{5}{2}}ℐ\left(t\right)\mathrm{,}$

that is,

$ℐ\left(t\right)\le C{N}_0^2+C{\delta }^2ℐ\left(t\right)\mathrm{,}$

when $\delta$ is small enough, (5.2) holds.

Employing Parseval’s theorem, we have

$\begin{array}{c}{\Vert \left(Z^l\mathrm{,}{w}^l\mathrm{,}{n}^l\right)\left(t\right)\Vert }_{L^2}\lesssim \mathrm{<mi>\; </mi>}{\left(1+t\right)}_{}^{-\frac{3}{4}}{N}_0+{\delta }^2{\displaystyle {\int }_0^t}{\left(1+t-\tau \right)}_{}^{-\frac{3}{4}}{\left(1+\tau \right)}_{}^{-\frac{5}{4}}\text{d}\tau \\ \lesssim {\left(1+t\right)}_{}^{-\frac{3}{4}}{N}_0+{\delta }^2{\left(1+t\right)}_{}^{-\frac{3}{4}}\mathrm{.}\end{array}$ (5.14)

Substituting (5.14) into (4.3), if $C_2>0$ , $C_2$ a constant, then

$\frac{\text{d}}{\text{d}t}\left\{\frac{{\beta }_3}{{\beta }_1}{\Vert Z\Vert }_{L^2}^2+{\Vert w\Vert }_{L^2}^2+{\Vert n\Vert }_{L^2}^2\right\}+C_2\left(\frac{{\beta }_3}{{\beta }_1}{\Vert Z\Vert }_{L^2}^2+{\Vert w\Vert }_{L^2}^2+{\Vert n\Vert }_{L^2}^2\right)\lesssim {\left(1+t\right)}_{}^{-\frac{3}{2}}{N}_0^2\mathrm{.}$

From the aforesaid inequality, we acquire (5.1). Multiplying (5.13) by ${\left(1+t\right)}_{}^{\frac{9}{8}}$ , we get

$\begin{array}{l}\frac{\text{d}}{\text{d}t}{\left(1+t\right)}_{}^{\frac{9}{8}}W\left(t\right)+C_1{\left(1+t\right)}_{}^{\frac{9}{8}}\left(W\left(t\right)+{\Vert w_{tt}\Vert }_{L^2}^2+{\Vert n_{tt}\Vert }_{L^2}^2\right)\\ \le C\left({\left(1+t\right)}_{}^{\frac{9}{8}}{\left(1+t\right)}_{}^{-\frac{5}{2}}{N}_0^2+{\left(1+t\right)}_{}^{\frac{1}{8}}W\left(t\right)\right)\\ \le C\left({\left(1+t\right)}_{}^{\frac{9}{8}}{\left(1+t\right)}_{}^{-\frac{5}{2}}{N}_0^2+{\left(1+t\right)}_{}^{\frac{1}{8}}{\left(1+t\right)}_{}^{-\frac{5}{2}}{N}_0^2\right)\\ \le C{\left(1+t\right)}_{}^{-\frac{11}{8}}{N}_0^2\mathrm{.}\end{array}$ (5.15)

(5.14) is integrated from 0 to $t$ , we obtain (5.3). The proof is completed. □

Lemma 5.2. Under the assumptions of Theorem 1.1 meanwhile (4.1) are valid, then

${\Vert \left(m^{+}\mathrm{,}{m}^{-}\right)\Vert }_{H^3}\le C{N}_0\mathrm{.}$ (5.16)

Proof. The new semi-linearized system of (1.1) is as follows:

$\{\begin{array}{l}{\partial }_t{m}^{+}+\left(1+m^{+}\right)\text{div}w+w\cdot \nabla m^{+}=\mathrm{0,}\\ \rho {\partial }_{t}w+ℒ\left(\eta \nabla m^{+}+\rho \nabla m^{-}\right)-{\rho }^{{\mu }_1}\Delta w-\left({\rho }^{{\mu }_1}+{\rho }^{{\lambda }_1}\right)\nabla \text{div}w=E_1\mathrm{,}\\ {\partial }_t{m}^{-}+\left(1+m^{-}\right)\text{div}n+n\cdot \nabla m^{-}=\mathrm{0,}\\ \eta {\partial }_{t}n+ℒ\left(\eta \nabla m^{+}+\rho \nabla m^{-}\right)-{\eta }^{{\mu }_2}\Delta n-\left({\eta }^{{\mu }_2}+{\eta }^{{\lambda }_2}\right)\nabla \text{div}n=E_2\mathrm{,}\end{array}$ (5.17)

where

$\begin{array}{c}{E}_1=-\rho w\cdot \nabla w+\frac{{\rho }^{{\mu }_1}\left(\nabla w+{\nabla }^{t}w\right)\nabla {\sigma }^{+}}{{\sigma }^{+}}+\nabla \left({\rho }^{{\mu }_1}\right)\left(\nabla w+{\nabla }^{t}w\right)\\ +\frac{\nabla {\sigma }^{+}{\rho }^{{\lambda }_1}\text{div}w}{{\sigma }^{+}}+\nabla \left({\rho }^{{\lambda }_1}\right)\text{div}w\mathrm{,}\end{array}$ (5.18)

$\begin{array}{c}{E}_2=-\eta v\cdot \nabla n+\frac{{\eta }^{{\mu }_2}\left(\nabla n+{\nabla }^{t}n\right)\nabla {\sigma }^{-}}{{\sigma }^{-}}+\nabla \left({\eta }^{{\mu }_2}\right)\left(\nabla n+{\nabla }^{t}n\right)\\ +\frac{\nabla {\sigma }^{-}{\eta }^{{\lambda }_2}\text{div}n}{{\sigma }^{-}}+\nabla \left({\eta }^{{\lambda }_2}\right)\text{div}n\mathrm{.}\end{array}$ (5.19)

Multiplying ${\nabla }^{\ell }$ (5.17)₁, ${\nabla }^{\ell }$ (5.17)₃ by ${\nabla }^{\ell }{m}^{+}$ , ${\nabla }^{\ell }{m}^{-}$ separately, for $0\le \ell \le 2$ , integrating over ${ℝ}^3$ , then

$\begin{array}{c}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{{ℝ}^3}}{\left|{\nabla }^{\ell }{m}^{\pm }\right|}^2\text{d}x=-{\displaystyle {\int }_{{ℝ}^3}}{\nabla }^{\ell }\left(\text{div}w+m^{+}\text{div}w+w\cdot \nabla m^{+}\right){\nabla }^{\ell }{m}^{+}\text{d}x\\ -{\displaystyle {\int }_{{ℝ}^3}}{\nabla }^{\ell }\left(\text{div}n+m^{-}\text{div}n+n\cdot \nabla m^{-}\right){\nabla }^{\ell }{m}^{-}\text{d}x\mathrm{.}\end{array}$ (5.20)

Integrating by parts, we have

${\displaystyle {\int }_{{ℝ}^3}}w\cdot \nabla {\nabla }^{\ell }{m}^{+}{\nabla }^{\ell }{m}^{+}\text{d}x=-\frac{1}{2}{\displaystyle {\int }_{{ℝ}^3}}\text{div}w{\left|{\nabla }^{\ell }{m}^{+}\right|}^2\text{d}x\mathrm{.}$

Similarly, we have

${\displaystyle {\int }_{{ℝ}^3}}n\cdot \nabla {\nabla }^{\ell }{m}^{-}{\nabla }^{\ell }{m}^{-}\text{d}x=-\frac{1}{2}{\displaystyle {\int }_{{ℝ}^3}}\text{div}n{\left|{\nabla }^{\ell }{m}^{-}\right|}^2\text{d}x\mathrm{.}$

Then we deal with (5.20)

$\begin{array}{l}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{{ℝ}^3}}{\left|{\nabla }^{\ell }{m}^{+}\right|}^2\text{d}x+\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{{ℝ}^3}}{\left|{\nabla }^{\ell }{m}^{-}\right|}^2\text{d}x\\ \le C{\Vert \nabla w\Vert }_{H^2}{\Vert m^{+}\Vert }_{H^2}+C{\Vert \nabla n\Vert }_{H^2}{\Vert m^{-}\Vert }_{H^2}\mathrm{.}\end{array}$

Therefore, we have

$\frac{\text{d}}{\text{d}t}{\Vert m^{+}\Vert }_{H^2}+\frac{\text{d}}{\text{d}t}{\Vert m^{-}\Vert }_{H^2}\le C\left({\Vert \nabla w\Vert }_{H^2}+{\Vert \nabla n\Vert }_{H^2}\right)\mathrm{.}$ (5.21)

By Theorem 5.1, we conclude that ${\displaystyle {\int }_0^t}{\Vert \nabla w\left(\tau \right)\Vert }_{H^2}\text{d}\tau \le C{N}_0$ , ${\displaystyle {\int }_0^t}{\Vert \nabla n\left(\tau \right)\Vert }_{H^2}\text{d}\tau \le C{N}_0$ . We integrate inequality from 0 to $t$ , so ${\Vert m^{\pm }\Vert }_{H^2}\le C{N}_0$ .

However, the uniform boundedness of the $L^1\left(\mathrm{0,}t\right)$ -norm of $\Vert {\nabla }^{4}w\Vert$ and $\Vert {\nabla }^{4}n\Vert$ is uncertain. Therefore, we need another way. The classic $L^p$ -estimate of elliptic systems can be applied to (4.27), which yields

$\begin{array}{c}{\Vert {\nabla }^2{w}_t\Vert }_{L^2}\lesssim {\Vert m_t^{+}\Vert }_{L^3}{\Vert w_t\Vert }_{L^6}+{\Vert {\sigma }^{+}\Vert }_{L^6}{\Vert \nabla Z_t\Vert }_{L^3}+{\Vert {\sigma }_t^{+}\Vert }_{L^2}{\Vert \nabla Z\Vert }_{L^{\infty }}+{\Vert w\Vert }_{L^{\infty }}{\Vert \nabla w\Vert }_{L^2}\\ +{\Vert \nabla w\Vert }_{L^3}{\Vert w_t\Vert }_{L^6}+{\Vert \nabla {\sigma }^{+}\Vert }_{L^{\infty }}{\Vert Z_t\Vert }_{L^2}{\Vert \nabla w\Vert }_{L^{\infty }}+{\Vert {\sigma }_t^{+}\Vert }_{L^6}{\Vert \sigma \Vert }_{L^{\infty }}{\Vert \nabla w\Vert }_{L^3}\\ +{\Vert {\sigma }_t^{+}\Vert }_{L^6}{\Vert \nabla Z\Vert }_{L^{\infty }}{\Vert \nabla w\Vert }_{L^3}+{\Vert {\sigma }^{+}\Vert }_{L^{\infty }}{\Vert \nabla Z\Vert }_{L^{\infty }}{\Vert \nabla w_t\Vert }_{L^2}\\ +{\Vert {\sigma }^{+}\Vert }_{L^{\infty }}{\Vert \nabla {\sigma }_t\Vert }_{L^2}{\Vert \nabla w\Vert }_{L^{\infty }}+{\Vert {\sigma }_t^{+}\Vert }_{L^{\infty }}{\Vert Z\Vert }_{L^{\infty }}{\Vert {\nabla }^{2}w\Vert }_{L^2}\\ +{\Vert {\sigma }^{+}\Vert }_{L^{\infty }}{\Vert Z_t\Vert }_{L^{\infty }}{\Vert {\nabla }^{2}w\Vert }_{L^2}+{\Vert \nabla {\sigma }^{+}\Vert }_{L^{\infty }}{\Vert Z\Vert }_{L^{\infty }}{\Vert \nabla w_t\Vert }_{L^2}\\ +{\Vert {\sigma }^{+}\Vert }_{L^{\infty }}{\Vert \nabla Z\Vert }_{L^{\infty }}{\Vert \nabla w_t\Vert }_{L^2}+\left({\Vert {\sigma }^{+}\Vert }_{L^{\infty }}{\Vert Z_t\Vert }_{L^{\infty }}+{\Vert {\sigma }_t^{+}\Vert }_{L^{\infty }}{\Vert Z\Vert }_{L^{\infty }}\right){\Vert \nabla w_t\Vert }_{L^2}^2\\ \lesssim {\Vert \left(w_{tt}\mathrm{,}{w}_t\mathrm{,}\nabla w_t\right)\Vert }_{L^2}+{\Vert \nabla w\Vert }_{H^1}\mathrm{.}\end{array}$ (5.22)

Similarly,

${\Vert {\nabla }^2{n}_t\Vert }_{L^2}\lesssim {\Vert \left(n_{tt}\mathrm{,}{n}_t\mathrm{,}\nabla n_t\right)\Vert }_{L^2}+{\Vert \nabla n\Vert }_{H^1}\mathrm{.}$ (5.23)

Multiplying ${\nabla }^3$ (5.17)₂ and ${\nabla }^3$ (5.17)₄ by ${\nabla }^{3}w$ and ${\nabla }^{3}n$ , integrating over ${ℝ}^3$ , then

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{{ℝ}^3}}\rho {\left|{\nabla }^{3}w\right|}^2\text{d}x+{\displaystyle {\int }_{{ℝ}^3}}{\rho }^{{\mu }_1}{\left|{\nabla }^{4}w\right|}^2\text{d}x+{\displaystyle {\int }_{{ℝ}^3}}\left({\rho }^{{\mu }_1}+{\rho }^{{\lambda }_1}\right){\left|{\nabla }^3\text{div}w\right|}^2\text{d}x\\ +\frac{1}{2}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{{ℝ}^3}}\eta {\left|{\nabla }^{3}n\right|}^2\text{d}x+{\displaystyle {\int }_{{ℝ}^3}}{\eta }^{{\mu }_2}{\left|{\nabla }^{4}n\right|}^2\text{d}x+{\displaystyle {\int }_{{ℝ}^3}}\left({\eta }^{{\mu }_2}+{\eta }^{{\lambda }_2}\right){\left|{\nabla }^3\text{div}n\right|}^2\text{d}x\\ =\frac{1}{2}{\displaystyle {\int }_{{ℝ}^3}}{\rho }_t{\left|{\nabla }^{3}w\right|}^2\text{d}x+{\eta }_t{\left|{\nabla }^{3}n\right|}^2\text{d}x-{\displaystyle \sum _{z\mathrm{<mo>=</mo>0}}^2}{\displaystyle {\int }_{{ℝ}^3}}{\nabla }^{3-z}\rho {\nabla }^z{w}_t{\nabla }^{3}w\text{d}x\\ -{\displaystyle \sum _{z=0}^2}{\displaystyle {\int }_{{ℝ}^3}}{\nabla }^{3-z}\eta {\nabla }^z{n}_t{\nabla }^{3}n\text{d}x+{\displaystyle \sum _{z\mathrm{<mo>=</mo>0}}^2}{\displaystyle {\int }_{{ℝ}^3}}{\nabla }^{3-z}{\rho }^{{\mu }_1}\cdot {\nabla }^z\Delta w\cdot {\nabla }^{3}w\text{d}x\\ +{\displaystyle \sum _{z=0}^2}{\displaystyle {\int }_{{ℝ}^3}}{\nabla }^{3-z}{\eta }^{{\mu }_2}\cdot {\nabla }^z\Delta n\cdot {\nabla }^{3}n\text{d}x\\ -{\displaystyle {\int }_{{ℝ}^3}}\nabla \left({\rho }^{{\mu }_1}\right)\left|{\nabla }^{4}w\right|\left|{\nabla }^{3}w\right|+\nabla \left({\eta }^{{\mu }_2}\right)\left|{\nabla }^{4}n\right|\left|{\nabla }^{3}n\right|\text{d}x\\ -{\displaystyle {\int }_{{ℝ}^3}}{\displaystyle \sum _{z\mathrm{<mo>=</mo>0}}^2}{\nabla }^{3-z}\left({\rho }^{{\mu }_1}+{\rho }^{{\lambda }_1}\right){\nabla }^z\nabla \text{div}w{\nabla }^{3}w\text{d}x\\ -{\displaystyle {\int }_{{ℝ}^3}}{\displaystyle \sum _{z\mathrm{<mo>=</mo>0}}^2}{\nabla }^{3-z}\left({\eta }^{{\mu }_2}+{\eta }^{{\lambda }_2}\right){\nabla }^z\nabla \text{div}n{\nabla }^{3}n\text{d}x-{\displaystyle {\int }_{{ℝ}^3}}\nabla \left({\rho }^{{\mu }_1}+{\rho }^{{\lambda }_1}\right)\left|{\nabla }^{3}w\right|\left|{\nabla }^3\text{div}w\right|\text{d}x\\ -{\displaystyle {\int }_{{ℝ}^3}}\nabla \left({\eta }^{{\mu }_2}+{\eta }^{{\lambda }_2}\right)\left|{\nabla }^{3}n\right|\left|{\nabla }^3\text{div}n\right|\text{d}x-{\displaystyle {\int }_{{ℝ}^3}}{\nabla }^3\left[ℒ\left(\eta \nabla m^{+}+\rho \nabla m\right)\right]{\nabla }^{3}w\text{d}x\\ -{\displaystyle {\int }_{{ℝ}^3}}{\nabla }^3\left[ℒ\left(\eta \nabla m^{+}+\rho \nabla m\right)\right]{\nabla }^{3}n\text{d}x+{\displaystyle {\int }_{{ℝ}^3}}{\nabla }^3{E}_1{\nabla }^{3}w\text{d}x+{\displaystyle {\int }_{{ℝ}^3}}{\nabla }^3{E}_2{\nabla }^{3}n\text{d}x\\ ={\displaystyle \sum _{i=1}^{14}}{I}_i.\end{array}$ (5.24)

For (1.14)₁, we have

$\begin{array}{c}\left|I_1\right|=\frac{1}{2}{\displaystyle {\int }_{{ℝ}^3}}{\rho }_t{\left|{\nabla }^{3}w\right|}^2\text{d}x+{\eta }_t{\left|{\nabla }^{3}n\right|}^2\text{d}x\\ \lesssim {\Vert {\rho }_t\Vert }_{L^{\infty }}{\Vert {\nabla }^{3}w\Vert }_{L^2}^2+{\Vert {\eta }_t\Vert }_{L^{\infty }}{\Vert {\nabla }^{3}n\Vert }_{L^2}^2\\ \lesssim {\Vert w\Vert }_{W^{\mathrm{1,}\infty }}{\Vert {\nabla }^{3}w\Vert }_{L^2}^2+{\Vert n\Vert }_{W^{\mathrm{1,}\infty }}{\Vert {\nabla }^{3}n\Vert }_{L^2}^2\\ \lesssim \delta \left({\Vert {\nabla }^{3}u\Vert }_{L^2}^2+{\Vert {\nabla }^{3}v\Vert }_{L^2}^2\right)\mathrm{.}\end{array}$ (5.25)

From (5.22), we have

$\begin{array}{c}\left|I_2\right|=-{\displaystyle \sum _{z=0}^2}{\displaystyle {\int }_{{ℝ}^3}}{\nabla }^{3-z}\rho {\nabla }^z{w}_t{\nabla }^{3}w\text{d}x\\ \lesssim \left({\Vert {\nabla }^3\rho \Vert }_{L^2}{\Vert w_t\Vert }_{L^{\infty }}+{\Vert {\nabla }^2\rho \Vert }_{L^3}{\Vert \nabla w_t\Vert }_{L^6}+{\left[\nabla \rho \right]}_{L^{\infty }}{\Vert {\nabla }^2{w}_t\Vert }_{L^2}\right){\Vert {\nabla }^{3}w\Vert }_{L^2}\\ \lesssim \delta \left({\Vert \left(w_{tt}\mathrm{,}{w}_t\mathrm{,}\nabla w_t\right)\Vert }_{L^2}^2+{\Vert \nabla w\Vert }_{H^2}\right)\mathrm{.}\end{array}$ (5.26)

Similarly, we have

$I_3\lesssim \delta \left({\Vert \left(n_{tt}\mathrm{,}{v}_t\mathrm{,}\nabla n_t\right)\Vert }_{L^2}^2+{\Vert \nabla n\Vert }_{H^2}\right)\mathrm{,}$ (5.27)

$\begin{array}{c}\left|I_4\right|\lesssim {\Vert {\nabla }^{2}Z\Vert }_{L^2}{\Vert {\nabla }^{2}w\Vert }_{L^3}{\Vert {\nabla }^{3}w\Vert }_{L^6}+{\Vert {\nabla }^{2}Z\Vert }_{L^2}{\Vert {\nabla }^{3}w\Vert }_{L^6}{\Vert {\nabla }^{3}u\Vert }_{L^3}\\ +{\Vert \nabla Z\Vert }_{L^{\infty }}{\Vert {\nabla }^{4}w\Vert }_{L^2}{\Vert {\nabla }^{3}w\Vert }_{L^2}\\ \lesssim \delta \left({\Vert \nabla w_t\Vert }_{L^2}+{\Vert \nabla w\Vert }_{H^2}+{\Vert {\nabla }^{4}w\Vert }_{L^2}\right)\mathrm{,}\end{array}$ (5.28)

$\begin{array}{c}\left|I_5\right|={\displaystyle \sum _{z=0}^2}{\displaystyle {\int }_{{ℝ}^3}}{\nabla }^{3-z}{\eta }^{{\mu }_2}\cdot {\nabla }^z\Delta n\cdot {\nabla }^{3}n\text{d}x\\ \lesssim \delta \left({\Vert \nabla n_t\Vert }_{L^2}+{\Vert \nabla n\Vert }_{H^2}+{\Vert {\nabla }^{4}n\Vert }_{L^2}\right)\mathrm{,}\end{array}$ (5.29)

$\begin{array}{c}{I}_6=-{\displaystyle {\int }_{{ℝ}^3}}\nabla {\rho }^{{\mu }_1}\left|{\nabla }^{4}w\right|\left|{\nabla }^{3}w\right|+\nabla \left({\eta }^{{\mu }_2}\right)\left|{\nabla }^{4}n\right|\left|{\nabla }^{3}n\right|\text{d}x\\ \lesssim {\Vert \nabla Z\Vert }_{L^{\infty }}{\Vert {\nabla }^{4}w\Vert }_{L^2}{\Vert {\nabla }^{3}w\Vert }_{L^2}+{\Vert \nabla Z\Vert }_{L^{\infty }}{\Vert {\nabla }^{4}n\Vert }_{L^2}{\Vert {\nabla }^{3}n\Vert }_{L^2}\\ \lesssim \delta \left({\Vert {\nabla }^{4}w\Vert }_{L^2}^2+{\Vert {\nabla }^{3}w\Vert }_{L^2}^2+{\Vert {\nabla }^{4}n\Vert }_{L^2}^2+{\Vert {\nabla }^{3}n\Vert }_{L^2}^2\right)\mathrm{.}\end{array}$ (5.30)

Using the similar process to (5.28) and (5.29), we have

$\begin{array}{c}{I}_7=-{\displaystyle {\int }_{{ℝ}^3}}{\displaystyle \sum _{z=0}^2}{\nabla }^{3-z}\left({\rho }^{{\mu }_1}+{\rho }^{{\lambda }_1}\right){\nabla }^z\nabla \text{div}w{\nabla }^{3}w\text{d}x\\ \lesssim \delta \left({\Vert \nabla w_t\Vert }_{L^2}+{\Vert \nabla w\Vert }_{H^2}+{\Vert {\nabla }^{4}w\Vert }_{L^2}\right)\mathrm{,}\end{array}$ (5.31)

$\begin{array}{c}{I}_8=-{\displaystyle {\int }_{{ℝ}^3}}{\displaystyle \sum _{z=0}^2}{\nabla }^{3-z}\left({\eta }^{{\mu }_2}+{\eta }^{{\lambda }_2}\right){\nabla }^z\nabla \text{div}n{\nabla }^{3}n\text{d}x\\ \lesssim \delta \left({\Vert \nabla n_t\Vert }_{L^2}+{\Vert \nabla n\Vert }_{H^2}+{\Vert {\nabla }^{4}n\Vert }_{L^2}\right)\mathrm{,}\end{array}$ (5.32)

$I_9+I_{10}$ is similar to $I_6$ , we have

$I_9+I_{10}\lesssim \delta \left({\Vert {\nabla }^{4}w\Vert }_{L^2}^2+{\Vert {\nabla }^{3}w\Vert }_{L^2}^2+{\Vert {\nabla }^{4}n\Vert }_{L^2}^2+{\Vert {\nabla }^{3}n\Vert }_{L^2}^2\right)\mathrm{,}$ (5.33)

$\begin{array}{c}{I}_{11}=-{\displaystyle {\int }_{{ℝ}^3}}{\nabla }^3\left[ℒ\left(\eta \nabla m^{+}+\rho \nabla m^{-}\right)\right]{\nabla }^{3}w\text{d}x\\ =-{\displaystyle {\int }_{{ℝ}^3}}ℒ\eta {\nabla }^3{m}^{+}{\nabla }^3\left(\frac{m_t^{+}+w\cdot \nabla m^{+}}{1+m^{+}}\right)\text{d}x\\ +{\displaystyle {\int }_{{ℝ}^3}}ℒ\rho {\nabla }^3{m}^{-}{\nabla }^3\left(\frac{m_t^{-}+n\cdot \nabla m^{-}}{1+m^{-}}\right)\text{d}x\\ +{\displaystyle {\int }_{{ℝ}^3}}\left[\nabla \left(ℒ\eta \right){\nabla }^3{m}^{+}+\nabla \left(\mathcal{C}\rho \right){\nabla }^3{m}^{-}\right]{\nabla }^{3}w\text{d}x\\ -{\displaystyle \sum _{z=0}^2}{\displaystyle {\int }_{{ℝ}^3}}{\nabla }^{3-z}\left(ℒ\eta \right)\nabla {\nabla }^z{m}^{+}{\nabla }^{3}w\text{d}x\\ -{\displaystyle \sum _{z=0}^2}{\displaystyle {\int }_{{ℝ}^3}}{\nabla }^{3-z}\left(ℒ\rho \right)\nabla {\nabla }^z{m}^{-}{\nabla }^{3}w\text{d}x\\ \le -\frac{1}{2}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{{ℝ}^3}}\frac{ℒ\eta {\left|{\nabla }^3{m}^{+}\right|}^2}{1+m^{+}}\text{d}x-\frac{1}{2}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{{ℝ}^3}}\frac{ℒ\rho {\left|{\nabla }^3{m}^{-}\right|}^2}{1+m^{-}}\text{d}x\\ +C\delta \left({\Vert \left(w_{tt}\mathrm{,}{n}_{tt}\mathrm{,}{w}_t\mathrm{,}{n}_t\mathrm{,}\nabla w_t\mathrm{,}\nabla n_t\right)\Vert }_{L^2}+{\Vert \left(\nabla w\mathrm{,}\nabla n\right)\Vert }_{H^2}\right)\mathrm{.}\end{array}$ (5.34)

Using the similar process to (5.34), we obtain

$\begin{array}{c}{I}_{12}\le -\frac{1}{2}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{{ℝ}^3}}\frac{ℒ\rho {\left|{\nabla }^3{m}^{-}\right|}^2}{1+m^{-}}\text{d}x-\frac{1}{2}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{{ℝ}^3}}\frac{ℒ\eta {\left|{\nabla }^3{m}^{+}\right|}^2}{1+m^{+}}\text{d}x\\ +C\delta \left({\Vert \left(w_{tt}\mathrm{,}{n}_{tt}\mathrm{,}{w}_t\mathrm{,}{n}_t\mathrm{,}\nabla w_t\mathrm{,}\nabla n_t\right)\Vert }_{L^2}+{\Vert \left(\nabla w\mathrm{,}\nabla n\right)\Vert }_{H^2}\right)\mathrm{.}\end{array}$ (5.35)

By applying integration by parts, we have

$\begin{array}{c}\left|I_{13}\right|\le C{\Vert {\nabla }^2{E}_1\Vert }_{L^2}{\Vert {\nabla }^{4}w\Vert }_{L^2}\\ \le C({\Vert {\nabla }^{2}w\Vert }_{L^2}{\Vert \nabla w\Vert }_{L^{\infty }}+{\Vert {\nabla }^{3}w\Vert }_{L^2}{\Vert \nabla w\Vert }_{L^{\infty }}+{\Vert {\nabla }^{2}Z\Vert }_{L^2}{\Vert {\nabla }^{2}w\Vert }_{L^{\infty }}{\Vert \nabla m^{+}\Vert }_{L^{\infty }}\\ +{\Vert {\nabla }^{2}Z\Vert }_{L^2}{\Vert \nabla w\Vert }_{L^{\infty }}{\Vert {\nabla }^2{m}^{+}\Vert }_{L^{\infty }}+{\Vert Z\Vert }_{L^{\infty }}{\Vert {\nabla }^{3}w\Vert }_{L^2}{\Vert {\nabla }^2{m}^{+}\Vert }_{L^{\infty }}\\ +{\Vert Z\Vert }_{L^{\infty }}{\Vert \nabla w\Vert }_{L^2}{\Vert \nabla m^{+}\Vert }_{L^{\infty }}){\Vert {\nabla }^{4}w\Vert }_{L^2}\\ \le C\delta \left({\Vert \nabla w_t\Vert }_{L^2}^2+{\Vert \nabla w\Vert }_{H^2}^2+{\Vert {\nabla }^{4}w\Vert }_{L^2}^2\right).\end{array}$ (5.36)

Similarly, for $I_{14}$ , we have

$I_{14}\le C\delta \left({\Vert \nabla n_t\Vert }_{L^2}^2+{\Vert \nabla n\Vert }_{H^2}^2+{\Vert {\nabla }^{4}n\Vert }_{L^2}^2\right)\mathrm{.}$ (5.37)

Substituting (5.25)-(5.37) into (5.24) yields,

$\begin{array}{l}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{{ℝ}^3}}\rho {\left|{\nabla }^{3}w\right|}^2+\eta {\left|{\nabla }^{3}n\right|}^2+\frac{ℒ\eta {\left|{\nabla }^3{m}^{+}\right|}^2}{1+m^{+}}+{\displaystyle {\int }_{{ℝ}^3}}\frac{ℒ\rho {\left|{\nabla }^3{m}^{-}\right|}^2}{1+m^{-}}+{\displaystyle {\int }_{{ℝ}^3}}\frac{ℒ\rho {\left|{\nabla }^3{m}^{-}\right|}^2}{1+m^{-}}\\ +{\displaystyle {\int }_{{ℝ}^3}}\frac{ℒ\eta {\left|{\nabla }^3{m}^{+}\right|}^2}{1+m^{+}}\text{d}x+{\displaystyle {\int }_{{ℝ}^3}}{\rho }^{{\mu }_1}{\left|{\nabla }^{4}w\right|}^2\text{d}x+{\displaystyle {\int }_{{ℝ}^3}}\left({\rho }^{{\mu }_1}+{\rho }^{{\lambda }_1}\right){\left|{\nabla }^3\text{div}w\right|}^2\text{d}x\\ +{\displaystyle {\int }_{{ℝ}^3}}{\eta }^{{\mu }_2}{\left|{\nabla }^{4}n\right|}^2\text{d}x+{\displaystyle {\int }_{{ℝ}^3}}\left({\eta }^{{\mu }_2}+{\eta }^{{\lambda }_2}\right){\left|{\nabla }^3\text{div}n\right|}^2\text{d}x\\ \le C\delta \left({\Vert \left(w_{tt}\mathrm{,}{w}_t\mathrm{,}\nabla w_t\right)\Vert }_{L^2}+{\Vert \nabla w\Vert }_{H^2}+{\Vert \left(n_{tt}\mathrm{,}{n}_t\mathrm{,}\nabla n_t\right)\Vert }_{L^2}+{\Vert \nabla n\Vert }_{H^2}\right)\mathrm{.}\end{array}$ (5.38)

In virtue of (5.2) and (5.3), for any $t\ge 0$ , then

${\displaystyle {\int }_0^t}\left({\Vert \left(w_{tt}\mathrm{,}{n}_{tt}\mathrm{,}{w}_t\mathrm{,}{n}_t\mathrm{,}\nabla w_t\mathrm{,}\nabla n_t\right)\left(\tau \right)\Vert }_{L^2}+{\Vert \left(\nabla w\mathrm{,}\nabla n\right)\left(\tau \right)\Vert }_{H^2}\right)\text{d}\tau \le C{N}_0\mathrm{.}$

we deduce that

${\Vert {\nabla }^3{m}^{\pm }\Vert }_{L^2}\le C{N}_0\mathrm{.}$ (5.39)

The proof is completed. □

Proof of Theorem 1.1 and 1.2. We can immediately derive Theorem 1.1 and Theorem 1.2 by Theorem 5.1 and Lemma 5.2.

Funding

This project is supported by National Natural Science Foundation of China 12001415.

Conflicts of Interest

All authors declare no competing interests.

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