Global Strong Solutions to the Compressible Nematic Liquid Crystal Flows with Large Oscillations and Vacuum in 2D Bounded Domains

Abstract

We investigate a simplified compressible nematic liquid crystal flow in two-dimensional (2D) bounded domains with Navier slip boundary conditions for velocity and Neumann boundary condition for orientation field. Based on delicate energy method and the structure of the model under consideration, we show the global existence and uniqueness of strong solutions when the initial total energy is suitably small. Our result may be regarded as an extension of the 2D Cauchy problem due to Wang (J Math Fluid Mech 18(3):539–569, 2016).

1 Introduction

The continuum theory of the nematic liquid crystals was developed by Ericksen [3] and Leslie [11] during the period of 1958 through 1968. Since then, there have been remarkable research developments in liquid crystals based on Ericksen–Leslie theory. Especially, the results of Ericksen–Leslie model were extended by Lin and Liu, and a series of great works were produced, which inspired people to think and explore. They not only added a penalization term to the Oseen-Frank energy functional to relax the nonlinear constraint of unit vector length in [14,15,16], but also they made significant results in the development of nematic liquid crystals, for example, they made the rigorous mathematical analysis of a simplified version of the Ericksen–Leslie model in [17]. Afterward, some useful developments concerning the Ericksen–Leslie model describing nematic liquid crystals are available in the survey papers [6, 18, 29]. Let $\Omega \subset \mathbb {R}^2$ be a bounded domain. We consider a simplified version of the Ericksen–Leslie system modeling the hydrodynamic flow of compressible nematic liquid crystals in $\Omega \times (0,T)$:

$$\begin{aligned} \rho _t+{{\,\textrm{div}\,}}(\rho u)&=0, \end{aligned}$$

(1.1)

$$\begin{aligned} \rho u_t+\rho u\cdot \nabla u+\nabla P&=\mathcal Lu -\Delta d\cdot \nabla d, \end{aligned}$$

(1.2)

$$\begin{aligned} d_t+u\cdot \nabla d&=\Delta d+|\nabla d|^2d, \end{aligned}$$

(1.3)

where $\rho$ is the density, $u=(u^1,u^2)$ is the velocity field, $d\in \mathbb {S}^1\triangleq \{d\in \mathbb {R}^2:|d|=1\}$ represents the macroscopic average of the nematic liquid crystal orientation field, $P(\rho )=a\rho ^\gamma \ (a>0, \gamma >1)$ is the pressure, while ${\mathcal {L}}$ denotes the Lamé operator given by

$$\begin{aligned} \mathcal Lu=(\lambda +\mu )\nabla {{\,\textrm{div}\,}}u+\mu \Delta u, \end{aligned}$$

where $\mu$ and $\lambda$ are the shear viscosity and the bulk viscosity coefficients of the fluid, respectively, which satisfy the physical restrictions

$$\begin{aligned} \mu >0, \quad \lambda +\mu \ge 0. \end{aligned}$$

The system (1.1)–(1.3) is supplemented with the initial condition

$$\begin{aligned} (\rho , \rho u, d)(x, 0)=(\rho _0, \rho _0u_0, d_0)(x), \quad x\in \Omega , \end{aligned}$$

(1.4)

and Navier-slip and Neumann boundary conditions for (u, d):

$$\begin{aligned} u\cdot n=0,\quad {{\,\textrm{curl}\,}}u=0, \quad \frac{\partial d}{\partial n}=0, \quad \textrm{on}\ \ \partial \Omega \times (0,T), \end{aligned}$$

(1.5)

where ${{\,\textrm{curl}\,}}u\triangleq \partial _1 u^2-\partial _2 u^1$ and $n=(n^1, n^2)$ is the unit outward normal vector to $\partial \Omega$.

In the last ten years, important progress has been made on the well-posedness for compressible nematic liquid crystal flows. Hu and Wu [7] derived global strong solutions to the 3D Cauchy problem in critical Besov spaces when the initial data are close to an equilibrium state (1,0,$\hat{d}$) with a constant vector $\hat{d}$ satisfying $|\hat{d}|=1$. Wu and Tan [28] proved the global existence and large-time behavior of weak solutions with the initial data of small energy and the initial density being positive and essentially bounded. Moreover, there are several exciting results devoting to the initial density allowing vacuum states. Adopting a celebrated three-level approximation scheme (see [24, Chapter 7] for more details) which consists of the Galerkin approximation, artificial viscosity, an artificial pressure and the weak continuity of the effective viscous flux, Wang and Yu [26] obtained the global existence and large-time behavior of weak solutions to the three-dimensional initial-boundary value problem (IBVP) for compressible liquid crystal flows with the Ginzburg-Landau penalty function under suitable condition on directions. Then Jiang–Jiang–Wang [9] extended the Ginzburg-Landau approximation model considered in [26] to (1.1)–(1.3) and showed global weak solutions with the Neumann boundary condition $\frac{\partial d}{\partial n}|_{\partial \Omega }=0$ in a bounded domain $\Omega \subset \mathbb {R}^N\ (N=2,3)$ provided that the N-th component of initial direction field $d_0$ satisfies a smallness condition. Later on, based on some new estimates to deal with the direction field and its coupling/interaction with the fluid variables, Jiang–Jiang–Wang [10] derived the global existence of weak solutions under restrictions imposed on the small initial energy for the 2D IBVP and some geometric angle condition of the direction field for the 2D Cauchy problem, respectively. Lin–Lai–Wang [19] generalized the main results in [10] to the 3D IBVP under the condition that the initial orientational director field $d_0$ lies in the hemisphere $\mathbb {S}^2_+$. It is worth mentioning that the regularity requirement $d_0\in H^2(\Omega )$ in [9] is weakened to $d_0\in H^1(\Omega )$ in [10, 19] and the Neumann boundary condition $\frac{\partial d}{\partial n}|_{\partial \Omega }=0$ is replaced by the non-homogeneous Dirichlet boundary condition $d|_{\partial \Omega }=d_0$. However, the uniqueness of such weak solutions has not be solved. Huang–Wang–Wen [8] first addressed the question of unique solvability of (1.1)–(1.3). More precisely, they studied the system (1.1)–(1.3) in a domain $\Omega \subset \mathbb {R}^3$ with one of three types of boundary conditions. Under the compatibility condition

$$\begin{aligned} \mathcal Lu_0-\Delta d_0\cdot \nabla d_0-\nabla P(\rho _0) =\sqrt{\rho _0}g\ \ \text {for some}\ \ g\in L^2(\Omega ), \end{aligned}$$

(1.6)

they [8] established the local existence of a unique strong solution. This local-in-time existence result on the Cauchy problem was later generalized to be a global one by Li–Xu–Zhang in [12] with small initial energy. In 2020, Liu [20] removed the additional assumption $(u_0,\nabla d_0)\in \dot{H}^{\beta }(\mathbb {R}^3)$ with $\beta \in (\frac{1}{2},1]$ in [12] and proved global classical solutions by requiring that $\rho _0\in L^1(\mathbb {R}^3)$ and $\Vert \rho _0\Vert _{L^\infty }+\Vert \nabla d_0\Vert _{L^3}$ is properly small. Very recently, Liu and Zhong [22] improved the local existence result on Navier-slip and Neumann boundary conditions for (u, d) in [8] to be a global one provided that the initial energy is small enough. When we consider the well-posedness of strong solutions to the 2D Cauchy problem, it is quite different from the 3D case if the far field condition is vacuum. The main difference lies in the fact that if $\lim \limits _{|x|\rightarrow \infty }u=0$ and $\nabla u\in L^2(\mathbb {R}^3)$, then $u\in L^6(\mathbb {R}^3)$, while it is impossible to obtain that $u\in L^p(\mathbb {R}^2)\ (p>1)$ for u satisfying $\lim \limits _{|x|\rightarrow \infty }u=0$ and $\nabla u\in L^2(\mathbb {R}^2)$. By spatially weighted energy method, Liu–Zheng–Li–Liu [21] and Wang [27] proved the local well-posedness and global well-posedness of strong solutions to the 2D Cauchy problem of (1.1)–(1.3), respectively.

In summary, all the global well-posedness results (even weak solutions) of (1.1)–(1.3) obtained so far are under the condition that the initial data satisfy some additional assumption. Moreover, it should be noted that different techniques were taken to get global (weak or strong) solutions for 3D and 2D problems. Thus, it is natural to ask whether similar result as that in [22] for the 2D initial-boundary value problem (1.1)–(1.5) holds true. Indeed, this is the main aim of the present paper.

Before stating our main result, we first explain the notation and conventions used throughout the paper. We denote the initial total energy of (1.1) by

$$\begin{aligned} C_0\triangleq \int \Big (\frac{1}{2}\rho _0|u_0|^2+\frac{1}{2}|\nabla d_0|^2+G(\rho _0)\Big )dx, \end{aligned}$$

(1.7)

where

$$\begin{aligned} G(\rho )\triangleq \rho \int _{\bar{\rho }}^\rho \frac{P(\xi )-\bar{P}}{\xi ^2}d\xi , \quad \bar{\rho }\triangleq \frac{1}{|\Omega |}\int \rho _0dx, \quad \bar{P}\triangleq P(\bar{\rho }). \end{aligned}$$

(1.8)

Moreover, we write

$$\begin{aligned}&H^1_{\omega } \triangleq \left\{ v\in H^{1}(\Omega ): v\cdot n=0\ \text {and}\ {{\,\textrm{curl}\,}}v=0\ \text {on}\ \partial \Omega \right\} ,\\&\quad H_{n}^{2}\triangleq \{v\in H^{2}(\Omega ): \nabla v\cdot n=0\ \text {on}\ \partial \Omega \}. \end{aligned}$$

When f is a scalar function, we set $\nabla ^{\perp }f\triangleq (\partial _2 f, -\partial _1 f)$ and $\nabla ^{\perp }_jf\triangleq (\nabla ^{\perp }f)^j$. For vector function $v=(v^1,v^2)$, we write $\nabla ^{\perp }v\triangleq (\nabla ^{\perp } v^1, \nabla ^{\perp } v^2)$ and $\nabla ^{\perp }_jv\triangleq (\nabla ^{\perp }_jv^1,\nabla ^{\perp }_jv^2)$. The material derivative of v is denoted by $\dot{v}\triangleq v_t+u\cdot \nabla v$.

Now we state our main result for the problem (1.1)–(1.5).

Theorem 1.1

Let $\Omega$ be a bounded simply connected smooth domain in $\mathbb {R}^2$. For given numbers $M_1$, $M_2>0$ (not necessarily small) and $\hat{\rho }\ge \bar{\rho }+1$, suppose that the initial data $(\rho _0, u_0, d_0)$ satisfies, for $q\in (3, 6)$,

$$\begin{aligned} {\left\{ \begin{array}{ll} 0\le \rho _0\le \hat{\rho }, \ (\rho _0, P(\rho _0))\in W^{1, q},\ u_0\in H^1_{\omega },\ d_0\in H_n^2, \\[3pt] \Vert \nabla u_0\Vert _{L^2}^2\le M_1, \ \Vert \Delta d_0\Vert _{L^2}^2\le M_2,\ |d_0|=1.\\ \end{array}\right. } \end{aligned}$$

(1.9)

There exists a positive constant $\varepsilon$ depending only on $\mu$, $\lambda$, $\gamma$, a, $\hat{\rho }$, $\Omega$, $M_1$, and $M_2$ such that if

$$\begin{aligned} C_0\le \varepsilon , \end{aligned}$$

(1.10)

the problem (1.1)–(1.5) has a unique global strong solution $(\rho , u, d)$ in $\Omega \times (0, \infty )$ satisfying

$$\begin{aligned} 0\le \rho (x, t)\le 2\hat{\rho }, \quad (x, t)\in \Omega \times (0, \infty ), \end{aligned}$$

(1.11)

and for any $0<\tau<T<\infty$,

$$\begin{aligned} {\left\{ \begin{array}{ll} (\rho , P)\in C([0, T]; W^{1, q}),\\ \nabla u\in C([0, T]; H^1)\cap L^2(\tau , T; W^{2, q}),\\ u_t\in L^2(\tau , T; H^1), \sqrt{\rho }u_t\in L^\infty (\tau , T; L^2),\\ \nabla d\in C([0, T]; H^2)\cap L^2(\tau , T; H^3),|d|=1,\\ d_t\in C([0, T];L^2)\cap L^2(\tau , T; H^2). \end{array}\right. } \end{aligned}$$

(1.12)

Remark 1.1

Theorem 1.1 extends the 2D local strong solutions with Navier-slip and Neumann boundary condition for (u, d) obtained by Huang–Wang–Wen [8] to be a global one. It is an interesting question whether 2D strong solutions with homogeneous Dirichlet and Neumann boundary condition for (u, d) in [8] can exist globally. Some new ideas are needed to handle this case, and it will be left for future studies.

Remark 1.2

Our result may be regarded as a generalization of the 2D Cauchy problem [27]. However, this is a non-trivial extension since we need to deal with many surface integrals caused by the boundary condition (1.5).

Remark 1.3

Compared with the results for the Cauchy problem [12, 27], here we cannot obtain large-time behavior of solutions. This reveals some difference between the whole space case and bounded domains case.

The proof of Theorem 1.1 is based on delicate energy estimates. Similarly to previous related results on global strong solutions for the compressible nematic liquid crystal flows [12, 20, 22, 27], the key issue is to derive the uniform-in-time lower-order estimates and uniform upper bound of the density. However, there are some additional difficulties in our analysis. On the one hand, compared with the 2D Cauchy problem [27], the main difficulty lies in treating many surface integrals (see (3.20), (3.21), and (3.32) for example) caused by the boundary condition (1.5). To overcome this obstacle, inspired by [2], we can parameterize $\partial {\Omega }$ by arc length

$$\begin{aligned} \frac{\partial n}{\partial \omega }=\kappa \omega , \end{aligned}$$

where $\omega =(-n^2,n^1)$ and $\kappa$ is the curvature of $\partial \Omega$. Then we obtain from (1.5) that

$$\begin{aligned} (u\cdot \nabla u\cdot n)|_{\partial \Omega }=-(u\cdot \nabla n\cdot u)|_{\partial \Omega }=-\kappa |u|^2, \end{aligned}$$

hence we can transform surface integrals into estimable terms with the help of the trace theorem (see Lemma 2.3) and Gagliardo-Nirenberg inequality (see Lemma 2.2) as well as $L^p$-estimates based on the effective viscous flux $F\triangleq (\lambda +2\mu ){{\,\textrm{div}\,}}u-(P-\bar{P})$. On the other hand, compared with the 3D situation [22], ${{\,\textrm{curl}\,}}u$ is a scalar function in 2D case, thus the term ${{\,\textrm{div}\,}}({{\,\textrm{curl}\,}}u\times \dot{u})$ does not exist and the equalities

$$\begin{aligned} {{\,\textrm{div}\,}}({{\,\textrm{curl}\,}}u\times \dot{u}) ={{\,\textrm{curl}\,}}{{\,\textrm{curl}\,}}u\cdot \dot{u}-{{\,\textrm{curl}\,}}u\cdot {{\,\textrm{curl}\,}}\dot{u}, \end{aligned}$$

and

$$\begin{aligned} \int {{\,\textrm{div}\,}}({{\,\textrm{curl}\,}}u\times \dot{u})dx=\int _{\partial \Omega }{{\,\textrm{curl}\,}}u\times \dot{u}\cdot n dx, \end{aligned}$$

cannot be used in our case. To solve this difficulty, we see that

$$\begin{aligned} {{\,\textrm{curl}\,}}({{\,\textrm{curl}\,}}u \dot{u})={{\,\textrm{curl}\,}}u{{\,\textrm{curl}\,}}\dot{u}-\nabla ^{\perp }{{\,\textrm{curl}\,}}u\cdot \dot{u}, \end{aligned}$$

and

$$\begin{aligned} \int {{\,\textrm{curl}\,}}({{\,\textrm{curl}\,}}u \dot{u})dx=\int _{\partial \Omega } ({{\,\textrm{curl}\,}}u \dot{u})\cdot \omega dx, \end{aligned}$$

which play significant roles in obtaining lower-order a priori estimates of solutions.

The rest of the paper is arranged as follows. In Sect. 2, we collect some elementary facts and give crucial $L^p$-estimates involving the effective viscous flux and the vorticity. In Sect. 3.1, we make some a priori assumptions and devoted to the uniformly a priori estimates of local strong solutions independent of the time, while the energy estimates for the higher-order derivatives are derived in Sect. 3.2. Finally, we give the proof of Theorem 1.1 in Sect. 4.

2 Preliminaries

In this section, we recall some known facts and elementary inequalities which will be used frequently later.

First of all, by time-weighted techniques used in [5] and arguments as in [8], we can obtain the following local existence theorem of strong solutions of (1.1)–(1.5). Here we omit the details for simplicity.

Lemma 2.1

Assume that the initial data $(\rho _0, u_0, d_0)$ satisfy the condition (1.9). Then there exists a positive time $T_0>0$ and a unique strong solution $(\rho , u, d)$ of the problem (1.1)–(1.5) in $\Omega \times (0, T_0]$.

For any smooth vector field v satisfying $(v\cdot n)|_{\partial \Omega }=0$, a simple computation shows that

$$\begin{aligned} (v\cdot \nabla )v\cdot n=-(v\cdot \nabla )n\cdot v\ \ \text{ on }\ \ \partial \Omega . \end{aligned}$$

(2.1)

Set $\omega =(-n^2,n^1)$-the unit tangent vector to $\partial \Omega$. Since $(u\cdot n)|_{\partial \Omega }=0$, we see that $\omega$ is parallel to u. Thus we obtain that

$$\begin{aligned} u=(u\cdot \omega )\omega . \end{aligned}$$

(2.2)

We write

$$\begin{aligned} \kappa =\kappa (x)\triangleq \omega \cdot \nabla n\cdot \omega , \end{aligned}$$

(2.3)

which is the curvature at the point $x\in \partial \Omega$. Then we derive from (2.1)–(2.3) that

$$\begin{aligned} (u\cdot \nabla )u\cdot n=-\kappa (u\cdot \omega )^2=-\kappa |u|^2\ \ \text{ on }\ \ \partial \Omega . \end{aligned}$$

(2.4)

Next, the well-known Gagliardo-Nirenberg inequality (see [23]) will be used frequently.

Lemma 2.2

Assume that $\Omega$ is a bounded Lipschitz domain in $\mathbb {R}^2$. For $p\in [2, \infty )$, $q\in (1, \infty )$, and $r\in (3, \infty )$, there exist two generic constants C, $C_1$, $C_2>0$, which may depend on p, q, r, and $\Omega$ such that, for any $f\in H^1(\Omega )$ and $g\in L^q(\Omega )\cap W^{1, r}(\Omega )$,

$$\begin{aligned}&\Vert f\Vert _{L^p}\le C\Vert f\Vert _{L^2}^\frac{2}{p}\Vert \nabla f\Vert _{L^2}^\frac{p-2}{p}+C_1\Vert f\Vert _{L^2}, \end{aligned}$$

(2.5)

$$\begin{aligned}&\Vert g\Vert _{L^\infty }\le C\Vert g\Vert _{L^q}^\frac{q(r-2)}{2r+q(r-2)}\Vert \nabla g\Vert _{L^r}^\frac{2r}{2r+q(r-2)}+C_2\Vert g\Vert _{L^2}. \end{aligned}$$

(2.6)

Moreover, if $(f\cdot n)|_{\partial \Omega }=0$ or $\bar{f}=0$, then $C_1=0$; when $(g\cdot n)|_{\partial \Omega }=0$ or $\bar{g}=0$, then $C_2=0$.

Next, the following trace theorem (see [4, p. 272]) plays a significant role in dealing with the boundary integral in the next section.

Lemma 2.3

Assume that $\Omega$ is a bounded domain and $\partial \Omega$ is $C^1$. Then there exists a bounded linear operator

$$\begin{aligned} T:W^{1,p}(\Omega )\rightarrow L^p(\partial \Omega ),\ 1\le p<\infty \end{aligned}$$

such that

$$\begin{aligned} Tv=v|_{\partial \Omega }\ \ \text {for}\ \ v\in W^{1,p}(\Omega )\cap C(\overline{\Omega }), \end{aligned}$$

and

$$\begin{aligned} \Vert Tv\Vert _{L^p(\partial \Omega )}\le C\Vert v\Vert _{W^{1,p}(\Omega )}\ \ \text {for}\ \ v\in W^{1,p}(\Omega ), \end{aligned}$$

with the constant C depending only on p and $\Omega$.

The following two lemmas are given in [1, 25].

Lemma 2.4

Let u be a solution of the following Lamé system

$$\begin{aligned} {\left\{ \begin{array}{ll} -\mu \Delta u-(\lambda +\mu )\nabla \textrm{div}\, u=f,\quad &{}x\in \Omega ,\\ u\cdot n=0, \ \textrm{curl}\, u=0, \quad &{}x\in \partial \Omega . \end{array}\right. } \end{aligned}$$

(2.7)

There exists a positive constant C depending only on q, k, $\mu$, $\lambda$, and $\Omega$ such that

(1) If $f\in W^{k, p}$ for some $q\in (1, \infty )$, $k\ge 0$, then we have $u\in W^{k+2,p}$ and

$$\begin{aligned} \Vert u\Vert _{W^{k+2, q}}\le C(\Vert f\Vert _{W^{k, q}}+\Vert u\Vert _{L^q}). \end{aligned}$$

(2) If $f=\nabla g$ and $g\in W^{k,q}$, for some $q\in (1,\infty )$, $k\ge 0$, then we have $u\in W^{k+1,q}$ and

$$\begin{aligned} \Vert u\Vert _{W^{k+1,q}}\le C(\Vert g\Vert _{W^{k, q}}+\Vert u\Vert _{L^q}). \end{aligned}$$

(2.8)

(3) If $f={{\,\textrm{curl}\,}}g$ and $g\in W^{k,q}$, for some $q\in (1,\infty )$ and $k\ge 0$, the conclusion (2) also holds.

Lemma 2.5

Let $k\ge 0$ be an integer and $1<q<+\infty$. Assume that $\Omega$ is a simply connected bounded domain in $\mathbb {R}^2$ with $C^{k+1, 1}$ boundary $\partial \Omega$. Then, for $v\in W^{1, q}(\Omega )$ with $(v\cdot n)|_{\partial \Omega }=0$, it holds that

$$\begin{aligned} \Vert v\Vert _{W^{k+1, q}}\le C\big (\Vert \textrm{div}\,v\Vert _{W^{k, q}}+\Vert \textrm{curl}\,v\Vert _{W^{k, q}}\big ). \end{aligned}$$

In particular, for $k=0$, we have

$$\begin{aligned} \Vert \nabla v\Vert _{L^q}\le C\big (\Vert \textrm{div}\,v\Vert _{L^q}+\Vert \textrm{curl}\,v\Vert _{L^q}\big ). \end{aligned}$$

The following estimates (see [2, Lemma 3.1]) on the material derivative of u will be useful.

Lemma 2.6

If $(\rho , u, d)$ is a smooth solution of (1.1)–(1.5). Then there exists a positive constant C depending only on p and $\Omega$ such that

$$\begin{aligned} \Vert \dot{u}\Vert _{L^p}&\le C\big (\Vert \nabla \dot{u}\Vert _{L^2}+\Vert \nabla u\Vert _{L^2}^2\big ), \end{aligned}$$

(2.9)

$$\begin{aligned} \Vert \nabla \dot{u}\Vert _{L^2}&\le C\big (\Vert \textrm{div}\,\dot{u}\Vert _{L^2}+\Vert \textrm{curl}\,\dot{u}\Vert _{L^2}+\Vert \nabla u\Vert _{L^4}^2\big ), \end{aligned}$$

(2.10)

for any $p\ge 2$.

Next, we introduce the effective viscous flux of the system (1.1)–(1.3) as the following

$$\begin{aligned} F=(2\mu +\lambda ){{\,\textrm{div}\,}}u-(P-\bar{P}). \end{aligned}$$

(2.11)

Lemma 2.7

Let $(\rho , u, d)$ be a strong solution of (1.1)–(1.5) in $\Omega \times (0, T]$. Then, for any $p\in [2, +\infty )$, there exists a positive constant C depending only on p, r, $\mu$, $\lambda$, and $\Omega$ such that

$$\begin{aligned}&\Vert \nabla u\Vert _{L^r}\le C\big (\Vert \textrm{div}\,u\Vert _{L^r}+\Vert \textrm{curl}\,u\Vert _{L^r}\big ), \end{aligned}$$

(2.12)

$$\begin{aligned}&\quad \Vert {{\,\textrm{curl}\,}}u\Vert _{L^p}\le C\big (\Vert \rho \dot{u}\Vert _{L^2}+\Vert |\nabla d||\nabla ^2 d|\Vert _{L^2}+\Vert \nabla u\Vert _{L^2}\big ) \end{aligned}$$

(2.13)

$$\begin{aligned}&\quad \Vert \nabla F\Vert _{L^p}+\Vert \nabla {{\,\textrm{curl}\,}}u\Vert _{L^p}\le C\big (\Vert \rho \dot{u}\Vert _{L^p}+\Vert |\nabla d||\nabla ^2d|\Vert _{L^p}+\Vert \nabla u\Vert _{L^p}\big ), \end{aligned}$$

(2.14)

$$\begin{aligned}&\quad \Vert F\Vert _{L^p}\le C\big (\Vert \rho \dot{u}\Vert _{L^2}+\Vert |\nabla d||\nabla ^2 d|\Vert _{L^2}\big )^{1-\frac{2}{p}} \big (\Vert \nabla u\Vert _{L^2}+\Vert P-\bar{P}\Vert _{L^2}\big )^\frac{2}{p}\nonumber \\&\quad +C\big (\Vert \nabla u\Vert _{L^2}+\Vert P-\bar{P}\Vert _{L^2}\big ), \end{aligned}$$

(2.15)

$$\begin{aligned}&\quad \Vert \nabla u\Vert _{L^p}\le C\big (\Vert \rho \dot{u}\Vert _{L^2}+\Vert |\nabla d||\nabla ^2d|\Vert _{L^2}\big )^{1-\frac{2}{p}}\big (\Vert \nabla u\Vert _{L^2}+\Vert P-\bar{P}\Vert _{L^2}\big )^\frac{2}{p}\nonumber \\&\quad +\big (\Vert \nabla u\Vert _{L^2}+\Vert P-\bar{P}\Vert _{L^2}\big ). \end{aligned}$$

(2.16)

Proof

Since $u\cdot n=0$ on $\partial \Omega$ and $\Omega$ is simply connected, so we immediately get (2.12) from Lemma 2.5. Applying ${{\,\textrm{curl}\,}}$ to (1.2), we see that

$$\begin{aligned} {\left\{ \begin{array}{ll} \mu \Delta {{\,\textrm{curl}\,}}u={{\,\textrm{curl}\,}}\big (\rho \dot{u}+{{\,\textrm{div}\,}}(M(d))\big ),\quad &{}x\in \Omega , \\ {{\,\textrm{curl}\,}}u=0,\quad &{}x\in {\partial {\Omega }}. \end{array}\right. } \end{aligned}$$

(2.17)

This combined with the standard $L^p$-estimate of elliptic equations and Lemma 2.4 shows that

$$\begin{aligned} \Vert {{\,\textrm{curl}\,}}u\Vert _{W^{1,p}}\le C(\Vert \rho \dot{u}\Vert _{L^p}+\Vert |\nabla d||\nabla ^2 d|\Vert _{L^p}+\Vert \nabla u\Vert _{L^p}). \end{aligned}$$

(2.18)

By Gagliardo-Nirenberg inequality and (2.18), we get that, for $p\in [2, +\infty )$,

$$\begin{aligned} \Vert \textrm{curl}\,u\Vert _{L^p}&\le C\Vert \textrm{curl}\,u\Vert _{L^2}^\frac{2}{p}\Vert \nabla \textrm{curl}\,u\Vert _{L^2}^{1-\frac{2}{p}}\nonumber \\&\le C\Vert \nabla u\Vert _{L^2}^\frac{2}{p}(\Vert \rho \dot{u}\Vert _{L^2}+\Vert |\nabla d||\nabla ^2 d|\Vert _{L^2})^{1-\frac{2}{p}}+C\Vert \nabla u\Vert _{L^2}\nonumber \\&\le C\big (\Vert \rho \dot{u}\Vert _{L^2}+\Vert |\nabla d||\nabla ^2 d|\Vert _{L^2}+\Vert \nabla u\Vert _{L^2}\big ), \end{aligned}$$

(2.19)

which leads to (2.13).

Since $\nabla F=\mu \nabla ^{\perp }{{\,\textrm{curl}\,}}u+\rho \dot{u}$, we derive from (2.18) that

$$\begin{aligned}&\Vert \nabla F\Vert _{L^p}\le C(\Vert \nabla ^{\perp }{{\,\textrm{curl}\,}}u\Vert _{L^p}+\Vert \rho \dot{u}\Vert _{L^p}\nonumber \\&\quad +\Vert \Delta d\cdot \nabla d\Vert _{L^p}) \le C(\Vert \rho \dot{u}\Vert _{L^p}+\Vert \nabla u\Vert _{L^p})+\Vert |\nabla d||\nabla ^2 d|\Vert _{L^p}, \end{aligned}$$

(2.20)

from which we deduce (2.14).

By (2.5) of Lemma 2.2, Young’s inequality and the fact $\bar{F}=0$, we obtain that

$$\begin{aligned} \Vert F\Vert _{L^p}\le&C\Vert F\Vert _{L^2}^\frac{2}{p} \Vert \nabla F\Vert _{L^2}^{1-\frac{2}{p}}\\ \le&C(\Vert \rho \dot{u}\Vert _{L^2}+\Vert \nabla u\Vert _{L^2}+\Vert |\nabla d||\nabla ^2 d|\Vert _{L^2})^{1-\frac{2}{p}}(\Vert \nabla u\Vert _{L^2}+\Vert P-{\bar{P}}\Vert _{L^2})^\frac{2}{p}\\ \le&C(\Vert \rho \dot{u}\Vert _{L^2}+\Vert \nabla u\Vert _{L^2}+\Vert |\nabla d||\nabla ^2 d|\Vert _{L^2})^{1-\frac{2}{p}}(\Vert \nabla u\Vert _{L^2}+\Vert P-{\bar{P}}\Vert _{L^2})^\frac{2}{p}\\&+\Vert \nabla u\Vert _{L^2}^{1-\frac{2}{p}} (|\nabla u\Vert _{L^2}+\Vert P-{\bar{P}}\Vert _{L^2})^\frac{2}{p}\\ \le&C(\Vert \rho \dot{u}\Vert _{L^2}+\Vert \nabla u\Vert _{L^2}+\Vert |\nabla d||\nabla ^2 d|\Vert _{L^2})^{1-\frac{2}{p}}(\Vert \nabla u\Vert _{L^2}\\&+\Vert P-{\bar{P}}\Vert _{L^2})^\frac{2}{p} +C(\Vert \nabla u\Vert _{L^2}+\Vert P-{\bar{P}}\Vert _{L^2}), \end{aligned}$$

from which we have (2.15).

From (2.12) and the definition of F, we deduce that

$$\begin{aligned}&\Vert \nabla u\Vert _{L^p}\le C\big (\Vert \textrm{div}\,u\Vert _{L^p}+\Vert \textrm{curl}\,u\Vert _{L^p}\big ) \\&\quad \le C(\Vert F\Vert _{L^p}+\Vert P-\bar{P}\Vert _{L^p}+\Vert \textrm{curl}\,u\Vert _{L^p}) \\&\quad \le C(\Vert \rho \dot{u}\Vert _{L^2}+\Vert \nabla u\Vert _{L^2}+\Vert |\nabla d||\nabla ^2 d|\Vert _{L^2})^{1-\frac{2}{p}}(\Vert \nabla u\Vert _{L^2}+\Vert P-{\bar{P}}\Vert _{L^2})^\frac{2}{p}\\&\quad +C(\Vert \nabla u\Vert _{L^2}+\Vert P-{\bar{P}}\Vert _{L^2}) \\&\quad \le C(\Vert \rho \dot{u}\Vert _{L^2}+\Vert |\nabla d||\nabla ^2 d|\Vert _{L^2})^{1-\frac{2}{p}}(\Vert \nabla u\Vert _{L^2}\\&\quad +\Vert P-{\bar{P}}\Vert _{L^2})^\frac{2}{p}+C(\Vert \nabla u\Vert _{L^2}+\Vert P-{\bar{P}}\Vert _{L^2}), \end{aligned}$$

from which, (2.16) follows. $\square$

The following Beale-Kato-Majda type inequality (see [2, Lemma 2.6]) will be used to estimate $\Vert \nabla u\Vert _{L^\infty }$.

Lemma 2.8

Let $\Omega$ be a bounded simply connected domain in $\mathbb {R}^2$ with smooth boundary. Assume that $v\in W^{2, q}(\Omega )\ (2<q<\infty )$ satisfying $v\cdot n=0$ and ${{\,\textrm{curl}\,}}v=0$ on $\partial \Omega$, then there exists a constant $C=C(q, \Omega )$ such that

$$\begin{aligned} \Vert \nabla v\Vert _{L^\infty } \le C\big (\Vert {{\,\textrm{div}\,}}v\Vert _{L^\infty }+\Vert {{\,\textrm{curl}\,}}v\Vert _{L^\infty }\big )\ln \big (e+\Vert \nabla ^2 v\Vert _{L^q}\big )+C\Vert \nabla v\Vert _{L^2}+C. \end{aligned}$$

Finally, the following Zlotnik inequality (see [30, Lemma 1.3]) will play a key role to obtain the uniform-in-time upper bound of the density.

Lemma 2.9

Suppose the function y satisfies

$$\begin{aligned} y'(t)=g(y)+b'(t)~on~[0, T], \quad y(0)=y^0, \end{aligned}$$

with $g\in C(R)$ and $y, b\in W^{1, 1}(0, T)$. If $g(\infty )=-\infty$ and

$$\begin{aligned} b(t_2)-b(t_1)\le N_0+N_1(t_2-t_1), \end{aligned}$$

for all $0\le t<t_2\le T$ with some $N_0\ge 0$ and $N_1\ge 0$, then

$$\begin{aligned} y(t)\le \max \{y^0, \xi _0\}+N_0<\infty ~on~[0, T], \end{aligned}$$

where $\xi _0$ is a constant such that

$$\begin{aligned} g(\xi )\le -N_1, \quad for\quad \xi \ge \xi _0. \end{aligned}$$

3 A Priori Estimates

In this section, we will establish some necessary a priori bounds for strong solutions to the problem (1.1)–(1.5) in order to extend the local strong solutions guaranteed by Lemma 2.1. This section is divided into two subsections. Section 3.1 aims at deriving the uniform-in-time lower-order estimates and uniform upper bound of the density. To this end, we first assume the a priori hypothesis (3.2) and then we close the a priori hypothesis through Lemmas 3.1–3.8, that is, we should show (3.3). In Sect. 3.2, we establish the time-dependent higher-order estimates of solutions. In what follows, let $T>0$ be a fixed time and $(\rho , u, d)$ be a strong solution to (1.1)–(1.5) in $\Omega \times (0, T]$ with initial data $(\rho _0, u_0, d_0)$ satisfying (1.9).

3.1 Lower-order estimates

Throughout this subsection, we will use C or $C_i\ (i=1, 2, \ldots )$ to denote the generic positive constants, which may depend on $\mu$, $\lambda$, $\gamma$, a, $\hat{\rho }$, $\Omega$, $M_1$, $M_2$, and $\bar{\rho }$. Especially, they are independent of T. $C(\alpha )$ is used to emphasize the dependence of C on $\alpha$.

Set $\sigma =\sigma (t)\triangleq \min \{1, t\}$, we define

$$\begin{aligned} {\left\{ \begin{array}{ll} A_1(T)\triangleq \sup \limits _{0\le t\le T}\big [\sigma \big (\Vert \nabla u\Vert _{L^2}^2+\Vert \Delta d\Vert _{L^2}^2\big )\big ]+\int _{0}^{T}\sigma \Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2dt,\\ A_2(T)\triangleq \sup \limits _{0\le t\le T}\big (\sigma ^2\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2\big )+\int _{0}^{T}\sigma ^2\Vert \nabla \dot{u}\Vert _{L^2}^2dt,\\ A_3(T)\triangleq \sup \limits _{0\le t\le T}\big (\Vert \nabla u\Vert _{L^2}^2+\Vert \Delta d\Vert _{L^2}^2\big ). \end{array}\right. } \end{aligned}$$

(3.1)

We intend to obtain the following key a priori estimates in this subsection, which implies the uniform upper bound of the density.

Proposition 3.1

Under the conditions of Theorem 1.1, there exist positive constants $\varepsilon$ and K both depending on $\mu$, $\lambda$, $\gamma$, a, $\bar{\rho }$, $\hat{\rho }$, $\Omega$, $M_1$, and $M_2$ such that if $(\rho , u, d)$ is a strong solution of (1.1)–(1.5) in $\Omega \times (0, T]$ satisfying

$$\begin{aligned} \sup _{\Omega \times [0, T]}\rho \le 2\hat{\rho }, \quad A_1(T)\le 2C_0^\frac{1}{2}, \quad A_2(T)\le 2C_0^\frac{1}{2}, \quad A_3(\sigma (T))\le 4K, \end{aligned}$$

(3.2)

then the following estimates hold

$$\begin{aligned} \sup _{\Omega \times [0, T]}\rho \le \frac{3}{2}\hat{\rho }, \quad A_1(T)\le C_0^\frac{1}{2}, \quad A_2(T)\le C_0^\frac{1}{2}, \quad A_3(\sigma (T))\le 3K, \end{aligned}$$

(3.3)

provided that $C_0\le \varepsilon$.

Remark 3.1

Recalling the definition of $\sigma (t)$, we then obtain from (3.2) that

$$\begin{aligned} \sup _{0\le t\le T}\big (\Vert \nabla u\Vert _{L^2}^2+\Vert \Delta d\Vert _{L^2}^2\big )\le C. \end{aligned}$$

(3.4)

Before proving Proposition 3.1, we show some necessary a priori estimates, see Lemmas 3.1–3.9 below.

Lemma 3.1

Let the assumptions of Proposition 3.1 be satisfied, then it holds that

$$\begin{aligned}&\sup _{0\le t\le T}\Big (\frac{1}{2}\Vert \sqrt{\rho }u\Vert _{L^2}^2+\frac{1}{2}\Vert \nabla d\Vert _{L^2}^2+\Vert G(\rho )\Vert _{L^1}\Big )\nonumber \\&\quad +\int _0^T\big [\mu \Vert \nabla u\Vert _{L^2}^2+(\mu +\lambda )\Vert {{\,\textrm{div}\,}}u\Vert _{L^2}^2+\Vert \Delta d+|\nabla d|^2d\Vert _{L^2}^2\big ]dt\le C_0, \end{aligned}$$

(3.5)

$$\begin{aligned}&\quad \sup _{0\le t\le T}\big (\Vert d_t\Vert _{L^2}^2+\Vert \nabla ^2d\Vert _{L^2}^2\big )\le C. \end{aligned}$$

(3.6)

Moreover, for any integer $1\le i\le [T]-1$, one has

$$\begin{aligned} \sup _{0\le t\le T}\Vert \rho -\bar{\rho }\Vert _{L^2}^2 +\int _{i-1}^{i+1}\big (\Vert d_t\Vert _{L^2}^2+\Vert \nabla ^2d\Vert _{L^2}^2\big )dt\le CC_0^{\frac{1}{2}}, \end{aligned}$$

(3.7)

provided that $C_0\le 1$.

Proof

1. Due to

$$\begin{aligned} -\Delta u=-\nabla \textrm{div}\,u+\nabla ^{\perp }{{\,\textrm{curl}\,}}u, \end{aligned}$$

(3.8)

we rewrite (1.2) as

$$\begin{aligned} \rho u_t+\rho u\cdot \nabla u-(2\mu +\lambda )\nabla {{\,\textrm{div}\,}}u+\mu \nabla ^{\perp }{{\,\textrm{curl}\,}}u +\nabla (P-\bar{P})+\Delta d\cdot \nabla d=0. \end{aligned}$$

(3.9)

Multiplying (3.9) by u and (1.1) by $G'(\rho )$, respectively, summing up, and integrating the resulting equality over $\Omega$, we get that

$$\begin{aligned}&\frac{d}{dt}\int \Big (\frac{1}{2}\rho |u|^2 +G(\rho )\Big )dx+(2\mu +\lambda )\int ({{\,\textrm{div}\,}}u)^2dx+\mu \int ({{\,\textrm{curl}\,}}u)^2dx\nonumber \\&\quad =-\int {{\,\textrm{div}\,}}(\rho u)G'(\rho )dx-\int u\cdot \nabla (P-\bar{P})dx-\int u \cdot \nabla d\cdot \Delta ddx\nonumber \\&\quad =\int \rho u\cdot \nabla Q(\rho )dx-\int u\cdot \nabla Pdx-\int u \cdot \nabla d\cdot \Delta d dx\nonumber \\&\quad =-\int u \cdot \nabla d\cdot \Delta d dx, \end{aligned}$$

(3.10)

where we have used (3.8) and (1.5) to obtain

$$\begin{aligned} \mathcal Lu\cdot udx&=\int \big [(2\mu +\lambda )\nabla {{\,\textrm{div}\,}}u\cdot u-\mu \nabla ^{\perp }{{\,\textrm{curl}\,}}u\cdot u\big ]dx\\&=-\int \big [(2\mu +\lambda )({{\,\textrm{div}\,}}u)^2+\mu ({{\,\textrm{curl}\,}}u)^2\big ]dx, \end{aligned}$$

and

$$\begin{aligned} G'(\rho )=Q(\rho )-Q(\bar{\rho }),\quad Q'(\rho )=P'(\rho )/\rho . \end{aligned}$$

Multiplying (1.3) by $\Delta d+|\nabla d|^2d$ and integrating over $\Omega$, we derive after using $|d|=1$ and $\frac{\partial d}{\partial n}|_{\partial \Omega }=0$ that

$$\begin{aligned}&\frac{1}{2}\frac{d}{dt}\int |\nabla d|^2dx+\int |\Delta d+|\nabla d|^2d|^2dx\\&\quad =\int u\cdot \nabla d\cdot \Delta ddx +\int \big (|\nabla d|^2 d\cdot d_t+|\nabla d|^2(u\cdot \nabla ) d\cdot d\big )dx\\&\quad =\int u\cdot \nabla d\cdot \Delta ddx +\frac{1}{2}\int \big (|\nabla d|^2 \partial _t|d|^2+|\nabla d|^2u\cdot \nabla |d|^2\big )dx\\&\quad =\int u\cdot \nabla d\cdot \Delta ddx, \end{aligned}$$

which together with (3.10) yields (3.5).

2. Integration by parts, we deduce from (1.5), (2.1), and Lemma 2.3 that

$$\begin{aligned} \Vert \Delta d\Vert _{L^2}^2&=\sum _{i, j=1}^2\int \partial _{ii}d\cdot \partial _{jj}ddx= -\sum _{i, j=1}^2\int \partial _id\cdot \partial _i\partial _{jj}ddx\nonumber \\&=\sum _{i, j=1}^2\int |\partial _{ij}d|^2dx -\sum _{i, j=1}^2\int _{\partial \Omega }\partial _id\cdot \partial _{ij}dn^jdS\nonumber \\&=\sum _{i, j=1}^2\int |\partial _{ij}d|^2dx+\sum _{i, j=1}^2\int _{\partial \Omega }\partial _id\partial _in^j\partial _jddS\nonumber \\&\ge \Vert \nabla ^2 d\Vert _{L^2}^2-C\Vert |\nabla d|^2\Vert _{W^{1, 1}}\nonumber \\&\ge \Vert \nabla ^2 d\Vert _{L^2}^2-C\Vert \nabla d\Vert _{L^2}\Vert \nabla ^2d\Vert _{L^2}-C\Vert \nabla d\Vert _{L^2}^2\nonumber \\&\ge \frac{1}{2}\Vert \nabla ^2 d\Vert _{L^2}^2-C\Vert \nabla d\Vert _{L^2}^2, \end{aligned}$$

(3.11)

which along with (3.4) and (3.5) gives that

$$\begin{aligned} \sup _{0\le t\le T}\Vert \nabla ^2 d\Vert _{L^2}^2\le C. \end{aligned}$$

(3.12)

It holds from (1.3), (2.5), and (1.5) that

$$\begin{aligned} \Vert d_t\Vert _{L^2}^2&\le C\big (\Vert |u||\nabla d|\Vert _{L^2}^2+\Vert |\nabla d|^2\Vert _{L^2}^2+\Vert \Delta d\Vert _{L^2}^2\big )\\&\le C\big (\Vert u\Vert _{L^6}^2\Vert \nabla d\Vert _{L^3}^2+\Vert \nabla d\Vert _{L^4}^4+\Vert \Delta d\Vert _{L^2}^2\big )\\&\le C\Vert \nabla u\Vert _{L^2}^2\Vert \nabla d\Vert _{L^2}^\frac{4}{3}\Vert \nabla ^2d\Vert _{L^2}^\frac{4}{3} +C\Vert \nabla d\Vert _{L^2}^2\Vert \nabla ^2d\Vert _{L^2}^2+C\Vert \Delta d\Vert _{L^2}^2, \end{aligned}$$

which combined with (3.5) and (3.12) implies (3.6).

3. From (1.8), we see that there exists a positive constant C depending only on a, $\gamma$, and $\hat{\rho }$ such that

$$\begin{aligned}&|P-\bar{P}|\le C|\rho -\bar{\rho }|, \quad C^{-1}(\rho -\bar{\rho })^2\le G(\rho )\le C(\rho -\bar{\rho })^2, \end{aligned}$$

which along with (3.5) leads to

$$\begin{aligned} \sup _{0\le t\le T}\Vert \rho -\bar{\rho }\Vert _{L^2}^2\le CC_0. \end{aligned}$$

(3.13)

We derive from (1.3), (1.5), (2.5), (3.4), and (3.12) that

$$\begin{aligned} \frac{d}{dt}\Vert \nabla d\Vert _{L^2}^2+\Vert d_t\Vert _{L^2}^2+\Vert \Delta d\Vert _{L^2}^2&=\int |d_t-\Delta d|^2dx \\&=\int |u\cdot \nabla d-|\nabla d|^2|^2dx\\&\le C\Vert u\Vert _{L^6}^2\Vert \nabla d\Vert _{L^3}^2+C\Vert \nabla d\Vert _{L^4}^4\\&\le C\Vert \nabla u\Vert _{L^2}^2\Vert \nabla d\Vert _{L^2}^\frac{4}{3}\Vert \nabla ^2d\Vert _{L^2}^3 +C\Vert \nabla d\Vert _{L^2}^2\Vert \nabla ^2d\Vert _{L^2}^2\\&\le CC_0^\frac{1}{2}, \end{aligned}$$

which together with (3.11) and (3.5) yields that

$$\begin{aligned} \frac{d}{dt}\Vert \nabla d\Vert _{L^2}^2+\Vert d_t\Vert _{L^2}^2+\Vert \nabla ^2 d\Vert _{L^2}^2 \le CC_0^{\frac{1}{2}}, \end{aligned}$$

(3.14)

provided that $C_0\le 1$. Integrating (3.14) over $[0, \sigma (T)]$ and combining the resultant with (3.5) lead to

$$\begin{aligned} \int _0^{\sigma (T)}\big (\Vert d_t\Vert _{L^2}^2+\Vert \nabla ^2d\Vert _{L^2}^2\big )dt\le CC_0^{\frac{1}{2}}. \end{aligned}$$

Denote $\sigma _i\triangleq \sigma (t+1-i)$. For any integer $1\le i\le [T]-1$, multiplying (3.14) by $\sigma _i$, we get that

$$\begin{aligned} \frac{d}{dt}\big (\sigma _i\Vert \nabla d\Vert _{L^2}^2\big )+\sigma _i\big (\Vert d_t\Vert _{L^2}^2+\Vert \nabla ^2d\Vert _{L^2}^2\big ) \le \sigma _i'\Vert \nabla d\Vert _{L^2}^2+CC_0^{\frac{1}{2}}\sigma _i \le \Vert \nabla d\Vert _{L^2}^2+CC_0^{\frac{1}{2}}. \end{aligned}$$

(3.15)

Integrating (3.15) over $(i-1, i+1]$, we derive (3.7) from (3.5) and (3.13). $\square$

Lemma 3.2

Let $(\rho , u, d)$ be a strong solution of (1.1)–(1.5) satisfying (3.2). Assume that $\eta (t)\ge 0$ is a piecewise differentiable function, then it holds that

$$\begin{aligned}&\frac{d}{dt}\Big (\frac{2\mu +\lambda }{2}\eta (t)\Vert \textrm{div}\,u\Vert _{L^2}^2 +\frac{\mu }{2}\eta (t)\Vert \textrm{curl}\,u\Vert _{L^2}^2+\eta (t)\Vert \Delta d\Vert _{L^2}^2\Big )\nonumber \\&\quad +\frac{1}{2}\eta (t)\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2 +\frac{1}{2}\eta (t)\Vert \nabla d_t\Vert _{L^2}^2\nonumber \\&\quad \le \frac{d}{dt}\int \eta (t)(P-\bar{P})\textrm{div}\,udx+\frac{d}{dt}\int \eta (t)M(d):\nabla udx +C\big (\eta (t)+|\eta '(t)|\big )\Vert \nabla u\Vert _{L^2}^2 \nonumber \\&\quad +C\eta (t)\big (\Vert \nabla u\Vert _{L^3}^3+\Vert \nabla u\Vert _{L^2}^4 d+\Vert \nabla u\Vert _{L^2}^2+\Vert \nabla \Vert _{H^1}^2+\Vert \nabla d\Vert _{H^1}^6 +\Vert \nabla u\Vert _{L^2}^4\Vert \nabla d\Vert _{H^1}^2\big )\nonumber \\&\quad +C|\eta '(t)|\big (\Vert \nabla d\Vert _{L^2}\Vert \nabla d\Vert _{H^1}^3+\Vert \nabla u\Vert _{L^2}^2\big )\nonumber \\&\quad +\eta '(t)\Vert \Delta d\Vert _{L^2}^2 +C\eta (t)\Vert \nabla u\Vert _{L^2}^2\Vert \nabla d\Vert _{H^1}^2+C|\eta '(t)|C_0, \end{aligned}$$

(3.16)

provided that $C_0\le \varepsilon _2$.

Proof

1. Multiplying (3.9) by $\eta (t)\dot{u}$ and integrating the resulting equality over $\Omega$, we get that

$$\begin{aligned} \int \eta (t)\rho |\dot{u}|^2dx&=-\int \eta (t)\dot{u}\cdot \nabla (P-\bar{P})dx+(2\mu +\lambda )\int \eta (t)\nabla {{\,\textrm{div}\,}}u\cdot \dot{u}dx\nonumber \\&\quad -\mu \int \eta (t)\nabla ^{\perp }{{\,\textrm{curl}\,}}u\cdot \dot{u}dx -\int \eta (t)\dot{u}\cdot \Delta d\cdot \nabla ddx \triangleq \sum _{i=1}^4I_i. \end{aligned}$$

(3.17)

By (1.1)$_1$ and $P=a\rho ^\gamma$, we have

$$\begin{aligned} P_t+{{\,\textrm{div}\,}}(Pu)+(\gamma -1)P{{\,\textrm{div}\,}}u=0, \end{aligned}$$

(3.18)

which together with integration by parts and (3.2) shows that

$$\begin{aligned} I_1&=-\int \eta (t)u_t\cdot \nabla (P-\bar{P})dx-\int \eta (t)u\cdot \nabla u\cdot \nabla Pdx\nonumber \\&=\frac{d}{dt}\int \eta (t)(P-\bar{P})\textrm{div}\,udx-\eta '(t)\int (P-\bar{P})\textrm{div}\,udx\nonumber \\&\quad -\int \eta (t)\textrm{div}\,uP_tdx -\int \eta (t)u\cdot \nabla u\cdot \nabla Pdx\nonumber \\&=\frac{d}{dt}\int \eta (t)(P-\bar{P})\textrm{div}\,udx-\eta '(t)\int (P-\bar{P})\textrm{div}\,udx+\int \eta (t)\textrm{div}\,u\textrm{div}\,(Pu)dx\nonumber \\&\quad +(\gamma -1)\int \eta (t)P(\textrm{div}\,u)^2dx-\int \eta (t)u\cdot \nabla u\cdot \nabla Pdx\nonumber \\&=\frac{d}{dt}\int \eta (t)(P-\bar{P})\textrm{div}\,udx-\eta '(t)\int (P-\bar{P})\textrm{div}\,udx +\int \eta (t)P\nabla u:\nabla udx\nonumber \\&\quad +(\gamma -1)\int \eta (t)P(\textrm{div}\,u)^2dx-\int _{\partial \Omega }\eta (t)Pu\cdot \nabla u\cdot ndS \nonumber \\&\le -\frac{d}{dt}\int \eta (t)(P-\bar{P})\textrm{div}\,udx+C\eta (t)\Vert \nabla u\Vert _{L^2}^2 +|\eta '(t)|\Vert P-\bar{P}\Vert _{L^2}\Vert \nabla u\Vert _{L^2},\nonumber \\&\le -\frac{d}{dt}\int \eta (t)(P-\bar{P})\textrm{div}\,udx+C(\eta (t)+|\eta '(t)|)\Vert \nabla u\Vert _{L^2}^2+C|\eta '(t)|\Vert P-\bar{P}\Vert _{L^2}^2,\nonumber \\&\le -\frac{d}{dt}\int \eta (t)(P-\bar{P})\textrm{div}\,udx+C(\eta (t)+|\eta '(t)|)\Vert \nabla u\Vert _{L^2}^2+C|\eta '(t)|C_0, \end{aligned}$$

(3.19)

where we have used

$$\begin{aligned}{} & {} \int \eta (t)\textrm{div}\,(Pu)\textrm{div}\,udx =-\int \eta (t) Pu^j\partial _{ji}u^idx\\{} & {} \quad =-\int _{\partial \Omega }\eta (t)Pu\cdot \nabla u\cdot ndS+\int \eta (t)\partial _i(Pu^j)\partial _ju^idx\\{} & {} \quad =-\int _{\partial \Omega }\eta (t)Pu\cdot \nabla u\cdot ndS +\int \eta (t)\partial _iPu^j\partial _ju^idx\\{} & {} \qquad +\int \eta (t)P\partial _iu^j\partial _ju^idx\\{} & {} \quad =-\int _{\partial \Omega }\eta (t)Pu\cdot \nabla u\cdot ndS\\{} & {} \quad +\int \eta (t) u\cdot \nabla u\cdot \nabla Pdx+\int \eta (t) P\nabla u:\nabla udx, \end{aligned}$$

and

$$\begin{aligned} -\int _{\partial \Omega }\eta (t)Pu\cdot \nabla u\cdot ndS&\le C\int _{\partial \Omega }\eta (t)\kappa |u|^2 dS \le C\eta (t)\int _{\partial \Omega }|u|^2dS\le C\eta (t)\Vert \nabla u\Vert _{L^2}^2, \end{aligned}$$

(3.20)

due to (2.1), (3.2), Lemma 2.3, and (2.5).

2. By (1.5) and (2.1), we derive from integration by parts that

$$\begin{aligned} I_2&=(2\mu +\lambda )\int _{\partial \Omega }\eta (t)\textrm{div}\,u(\dot{u}\cdot n)dS-(2\mu +\lambda )\int \eta (t)\textrm{div}\,u\textrm{div}\,\dot{u}dx\\&=(2\mu +\lambda )\int _{\partial \Omega }\eta (t)\textrm{div}\,u(u\cdot \nabla u\cdot n)dS-\frac{2\mu +\lambda }{2}\frac{d}{dt}\int \eta (t)(\textrm{div}\,u)^2dx\\&\quad -(2\mu +\lambda )\int \eta (t)\textrm{div}\,u\textrm{div}\,(u\cdot \nabla u)dx+\frac{2\mu +\lambda }{2}\eta '(t)\int (\textrm{div}u)^2dx\\&=-\frac{2\mu +\lambda }{2}\frac{d}{dt}\int \eta (t)({{\,\textrm{div}\,}}u)^2dx -(2\mu +\lambda )\int _{\partial \Omega }\eta (t){{\,\textrm{div}\,}}u(u\cdot \nabla n\cdot u)dS\\&\quad -(2\mu +\lambda )\int \eta (t)\textrm{div}\,u\partial _i(u^j\partial _ju^i)dx+\frac{2\mu +\lambda }{2}\eta '(t)\int (\textrm{div}\,u)^2dx\\&=-\frac{2\mu +\lambda }{2}\frac{d}{dt}\int \eta (t)(\textrm{div}\,u)^2dx -(2\mu +\lambda )\int _{\partial \Omega }\eta (t)\textrm{div}\,u(u\cdot \nabla n\cdot u)dS\\&\quad -(2\mu +\lambda )\int \eta (t)\textrm{div}\,u\nabla u:\nabla udx-(2\mu +\lambda )\\&\quad \int \eta (t)\textrm{div}\,uu^j\partial _{ji}u^idx+\frac{2\mu +\lambda }{2}\eta '(t)\int (\textrm{div}\,u)^2dx\\&=-\frac{2\mu +\lambda }{2}\frac{d}{dt}\int \eta (t)(\textrm{div}\,u)^2dx -(2\mu +\lambda )\int _{\partial \Omega }\eta (t)\textrm{div}\,u(u\cdot \nabla n\cdot u)dS\\&\quad -(2\mu +\lambda )\int \eta (t)\textrm{div}\,u\nabla u:\nabla udx+\frac{2\mu +\lambda }{2}\\&\quad \int \eta (t)(\textrm{div}\,u)^3dx +\frac{2\mu +\lambda }{2}\eta '(t)\int (\textrm{div}\,u)^2dx\\&\le -\frac{2\mu +\lambda }{2}\frac{d}{dt}\int \eta (t)(\textrm{div}\,u)^2dx\\&\quad +\frac{1}{2}\eta (t)\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2+\delta \eta (t)\Vert \nabla ^3d\Vert _{L^2}^2 +C|\eta '(t)|\Vert \nabla u\Vert _{L^2}^2\\&\quad +C\eta (t)\big (\Vert \nabla u\Vert _{L^3}^3+\Vert \nabla u\Vert _{L^2}^4 +\Vert \nabla u\Vert _{L^2}^2+\Vert \nabla d\Vert _{H^1}^2\big ), \end{aligned}$$

where we have used

$$\begin{aligned} \int \eta (t)\textrm{div}\,uu^j\partial _{ji}u^idx&=-\int \eta (t)\partial _j(\partial _ku^ku^j)\partial _iu^idx\\&=-\int \eta (t)\partial _{jk}u^ku^j\partial _iu^idx -\int \eta (t){{\,\textrm{div}\,}}u\partial _ju^j\partial _iu^idx\\&=-\int \eta (t)\partial _{ji}u^iu^j{{\,\textrm{div}\,}}udx -\int \eta (t)(\textrm{div}\,u)^3dx, \end{aligned}$$

and

$$\begin{aligned}&\Big |-(2\mu +\lambda )\int _{\partial \Omega }\textrm{div}\,u(u\cdot \nabla n\cdot u)dS\Big |\nonumber \\&\quad =\Big |-\int _{\partial \Omega }\big (F+(P-\bar{P})\big )(u\cdot \nabla n\cdot u)dS\Big |\nonumber \\&\quad \le \Big |\int _{\partial \Omega }F(u\cdot \nabla n\cdot u)dS\Big |+\Big |\int _{\partial \Omega }(P-\bar{P})(u\cdot \nabla n\cdot u)dS\Big |, \nonumber \\&\quad \le C\int _{\partial \Omega }\big |F|u|^2\big |dS+C\int _{\partial \Omega }|u|^2dS\nonumber \\&\quad \le C\Vert F\Vert _{H^1}\Vert u\Vert _{H^1}^2+C\Vert \nabla u\Vert _{L^2}^2\nonumber \\&\quad \le \frac{1}{2}\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2+\delta \Vert \nabla ^3 d\Vert _{L^2}^2 +C\big (\Vert \nabla u\Vert _{L^2}^4+\Vert \nabla u\Vert _{L^2}^2+\Vert \nabla d\Vert _{H^1}^2\big ), \end{aligned}$$

(3.21)

due to Lemma 2.3, (2.5), Lemma 2.7,

$$\begin{aligned} \int _{\partial \Omega }|F||u|^2dS&\le C\int _{\Omega } |F| |u|^2dx+C\int _{\Omega } |\nabla F| |u|^2dx+C\int _{\Omega }|F||\nabla |u|^2|dx \\&\le C\Vert F\Vert _{L^6}\Vert u\Vert _{L^3}\Vert u\Vert _{L^2}+C\Vert \nabla F\Vert _{L^2}\Vert u\Vert ^2_{L^4}+C\Vert F\Vert _{L^6}\Vert u\Vert _{L^3}\Vert \nabla u\Vert _{L^2}\\&\le C\Vert F\Vert _{H^1}\Vert u\Vert _{H^1}^2, \end{aligned}$$

and

$$\begin{aligned} \Vert |\nabla d||\nabla ^2 d|\Vert _{L^2}&\le C\Vert \nabla d\Vert _{L^6}\Vert \nabla ^2d\Vert _{L^3} \\&\le C\Vert \nabla d\Vert _{H^1}\Vert \nabla ^2d\Vert _{L^2}^\frac{1}{2}\Vert \nabla ^2d\Vert _{L^6}^\frac{1}{2} \\&\le C\Vert \nabla d\Vert _{H^1}^\frac{3}{2}\Vert \nabla ^2 d\Vert _{L^6}^\frac{1}{2}\ \\&\le C\Vert \nabla d\Vert _{H^1}^\frac{3}{2}\big (\Vert \nabla ^3d\Vert _{L^2}^\frac{2}{3}+\Vert \nabla ^2d\Vert _{L^2}\big )^\frac{1}{2}\ \ \\&\le C\Vert \nabla d\Vert _{H^1}^\frac{3}{2}\Vert \nabla ^3d\Vert _{L^2}^\frac{1}{3}+\Vert \nabla d\Vert _{H^1}^2. \end{aligned}$$

3. Noticing that

$$\begin{aligned} \int {{\,\textrm{div}\,}}(({{\,\textrm{curl}\,}}u)^2 u) dx=\int ({{\,\textrm{curl}\,}}u)^2\textrm{div}\,udx +2\int {{\,\textrm{curl}\,}}u\nabla {{\,\textrm{curl}\,}}u\cdot u dx. \end{aligned}$$

Thus, we have

$$\begin{aligned} \int {{\,\textrm{curl}\,}}u(u^i\partial _i\textrm{curl}\,u)dx =-\frac{1}{2}\int ({{\,\textrm{curl}\,}}u)^2\textrm{div}\,udx. \end{aligned}$$

This gives that

$$\begin{aligned} \mu \int {{\,\textrm{curl}\,}}u {{\,\textrm{curl}\,}}(u\cdot \nabla u)dx&=\mu \int {{\,\textrm{curl}\,}}u {{\,\textrm{curl}\,}}(u^i\partial _i u)dx\\&=\mu \int {{\,\textrm{curl}\,}}u\big (u^i\textrm{curl}\,\partial _iu-\partial _iu\cdot \nabla ^{\perp }u^i\big )dx\\&=-\frac{\mu }{2}\int ({{\,\textrm{curl}\,}}u)^2{{\,\textrm{div}\,}}udx -\mu \int \big (\partial _iu\cdot \nabla ^{\perp }u^i\big ){{\,\textrm{curl}\,}}u dx, \end{aligned}$$

which combined with (1.5) and integration by parts leads to

$$\begin{aligned} I_3&=-\mu \int \eta (t){{\,\textrm{curl}\,}}u {{\,\textrm{curl}\,}}\dot{u}dx\nonumber \\&=-\frac{\mu }{2}\frac{d}{dt}\int \eta (t)({{\,\textrm{curl}\,}}u)^2dx +\frac{\mu }{2}\eta '(t)\int ({{\,\textrm{curl}\,}}u)^2dx -\mu \int \eta (t){{\,\textrm{curl}\,}}u {{\,\textrm{curl}\,}}(u\cdot \nabla u)dx\nonumber \\&=-\frac{\mu }{2}\frac{d}{dt}\int \eta (t)({{\,\textrm{curl}\,}}u)^2dx +\frac{\mu }{2}\eta '(t)\int ({{\,\textrm{curl}\,}}u)^2dx\nonumber \\&\quad +\mu \int \eta (t)(\partial _iu\cdot \nabla ^{\perp }u^i) {{\,\textrm{curl}\,}}udx +\frac{\mu }{2}\int \eta (t)({{\,\textrm{curl}\,}}u)^2\textrm{div}\,udx\nonumber \\&\le -\frac{\mu }{2}\frac{d}{dt}\int \eta (t)({{\,\textrm{curl}\,}}u)^2dx +C|\eta '(t)|\Vert \nabla u\Vert _{L^2}^2+C\eta (t)\Vert \nabla u\Vert _{L^3}^3. \end{aligned}$$

(3.22)

Moreover, in (3.22), we have also used

$$\begin{aligned} {{\,\textrm{curl}\,}}({{\,\textrm{curl}\,}}u \dot{u})={{\,\textrm{curl}\,}}u {{\,\textrm{curl}\,}}\dot{u}-\nabla ^{\perp }{{\,\textrm{curl}\,}}u\cdot \dot{u} \end{aligned}$$

and

$$\begin{aligned} \int {{\,\textrm{curl}\,}}({{\,\textrm{curl}\,}}u \dot{u})dx =\int _{\partial \Omega } {{\,\textrm{curl}\,}}u \dot{u}\cdot \omega dx, \end{aligned}$$

which combined with the boundary condition (1.5) yields surface integral vanishing.

4. Noting that

$$\begin{aligned} I_4&=-\int \eta (t)u_t\cdot \Delta d\cdot \nabla ddx-\int \eta (t) u\cdot \nabla u\cdot \Delta d\cdot \nabla ddx \triangleq I_{41}+I_{42}. \end{aligned}$$

By (1.5), Hölder’s inequality, Sobolev’s inequality, and (2.5), we deduce that

$$\begin{aligned} I_{41}&=-\int \eta (t)u_t^j\partial _{ii}d\partial _{j}ddx\nonumber \\&=\int \eta (t)\partial _{i}u_t^j\partial _i d\partial _{j}d dx+\int \eta (t)u_t^j\partial _i d\partial _{ij}d dx\nonumber \\&=\int \eta (t)M(d):\nabla u_tdx\nonumber \\&=\frac{d}{dt}\int \eta (t)M(d):\nabla udx-\eta '(t)\int M(d):\nabla udx-\int \eta (t)M(d)_t:\nabla udx\nonumber \\&\le \frac{d}{dt}\int \eta (t)M(d):\nabla udx+C|\eta '(t)|\Vert \nabla d\Vert _{L^4}^2\Vert \nabla u\Vert _{L^2}+C\eta (t)\Vert \nabla u\Vert _{L^3}\Vert \nabla d_t\Vert _{L^2}\Vert \nabla d\Vert _{L^6}\nonumber \\&\le \frac{d}{dt}\int \eta (t)M(d):\nabla udx+C|\eta '(t)|\big (\Vert \nabla d\Vert _{L^2}\Vert \nabla d\Vert _{H^1}^3+\Vert \nabla u\Vert _{L^2}^2\big )\nonumber \\&\quad +\delta \eta (t)\Vert \nabla d_t\Vert _{L^2}^2+C\eta (t)\big (\Vert \nabla u\Vert _{L^3}^3+\Vert \nabla d\Vert _{H^1}^6\big ), \end{aligned}$$

(3.23)

and

$$\begin{aligned} I_{42}&\le C\eta (t)\Vert u\Vert _{L^6}\Vert \nabla u\Vert _{L^2}\Vert \Delta d\Vert _{L^6}\Vert \nabla d\Vert _{L^6}\nonumber \\&\le C\eta (t)\Vert \nabla u\Vert _{L^2}^2\big (\Vert \nabla ^3 d\Vert _{L^2}+\Vert \nabla ^2d\Vert _{L^2}\big )\Vert \nabla d\Vert _{H^1}\nonumber \\&\le \delta \eta (t)\Vert \nabla ^3d\Vert _{L^2}^2+C\eta (t)\Vert \nabla u\Vert _{L^2}^4\Vert \nabla d\Vert _{H^1}^2+C\eta (t)\Vert \nabla u\Vert _{L^2}^2\Vert \nabla d\Vert _{H^1}^2, \end{aligned}$$

(3.24)

due to

$$\begin{aligned} \int \eta (t)u_t^j\partial _i d\partial _{ij}d dx=-\int \eta (t)\partial _{j}u_t^j\partial _{i}d\cdot \partial _{i}ddx-\int \eta (t)u_t^j\partial _{ij}d\partial _{i}ddx, \end{aligned}$$

so that

$$\begin{aligned} \int \eta (t)u_t^j\partial _{i}d\partial _{ij}ddx=-\frac{1}{2}\int \eta (t)|\nabla d|^2\mathbb {I}_2:\nabla u_tdx. \end{aligned}$$

Combining (3.23) and (3.24), we deduce that

$$\begin{aligned} I_4&\le \frac{d}{dt}\int \eta (t)M(d):\nabla udx+\delta \eta (t)\big (\Vert \nabla d_t\Vert _{L^2}^2+\Vert \nabla ^3d\Vert _{L^2}^2\big ) \\&\quad +C|\eta '(t)|\big (\Vert \nabla d\Vert _{L^2}^2\Vert \nabla d\Vert _{H^1}^2+\Vert \nabla u\Vert _{L^2}^2\big )\\&\quad +C\eta (t)\big (\Vert \nabla u\Vert _{L^3}^3+\Vert \nabla d\Vert _{H^1}^6 +\Vert \nabla u\Vert _{L^2}^4\Vert \nabla d\Vert _{H^1}^2+\Vert \nabla u\Vert _{L^2}^2\Vert \nabla d\Vert _{H^1}^2\big ). \end{aligned}$$

Plugging the above estimates on $I_i\ (i=1, 2, 3, 4)$ into (3.17), one arrives at

$$\begin{aligned}&\frac{d}{dt}\Big (\frac{2\mu +\lambda }{2}\eta (t)\Vert \textrm{div}\,u\Vert _{L^2}^2 +\frac{\mu }{2}\eta (t)\Vert \textrm{curl}\,u\Vert _{L^2}^2\Big )+\frac{1}{2}\eta (t)\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2\nonumber \\&\le \frac{d}{dt}\int \eta (t)(P-\bar{P})\textrm{div}\,udx+\frac{d}{dt}\int \eta (t)M(d):\nabla udx +\delta \eta (t)\big (\Vert \nabla ^3d\Vert _{L^2}^2+\Vert \nabla d_t\Vert _{L^2}^2\big )\nonumber \\&\quad +C\big (\eta (t)+|\eta '(t)|\big )\Vert \nabla u\Vert _{L^2}^2+C|\eta '(t)|\big (\Vert \nabla d\Vert _{L^2}\Vert \nabla d\Vert _{H^1}^3+\Vert \nabla u\Vert _{L^2}^2\big )\nonumber \\&\quad +C\eta (t)\big (\Vert \nabla u\Vert _{L^3}^3+\Vert \nabla u\Vert _{L^2}^4 +\Vert \nabla u\Vert _{L^2}^2+\Vert \nabla d\Vert _{H^1}^2+\Vert \nabla d\Vert _{H^1}^6 +\Vert \nabla u\Vert _{L^2}^4\Vert \nabla d\Vert _{H^1}^2\big )\nonumber \\&\quad +C\eta (t)\Vert \nabla u\Vert _{L^2}^2\Vert \nabla d\Vert _{H^1}^2+C|\eta '(t)|C_0. \end{aligned}$$

(3.25)

5. $\Vert \nabla d_t\Vert _{L^2}^2$ remains to be estimated. To this end, applying the operator $\nabla$ to (1.3) gives that

$$\begin{aligned} \nabla d_t-\nabla \Delta d=-\nabla (u\cdot \nabla d)+\nabla (|\nabla d|^2d). \end{aligned}$$

(3.26)

Multiplying (3.26) by $\nabla d_t$ and integrating the resultant by parts, we find that

$$\begin{aligned}&\frac{1}{2}\frac{d}{dt}\int |\Delta d|^2dx+\int |\nabla d_t|^2dx\\&=\int \big (\nabla (|\nabla d|^2d)-\nabla (u\cdot \nabla d)\big )\nabla d_tdx\\&\le \frac{1}{4}\Vert \nabla d_t\Vert _{L^2}^2+C\int \big (|\nabla d|^2|\nabla ^2d|^2+|\nabla d|^6+|\nabla u|^2|\nabla d|^2+|u|^2|\nabla ^2d|^2\big )dx\\&\le \delta \Vert \nabla d_t\Vert _{L^2}^2+C\Vert \nabla d\Vert _{L^3}^2\Vert \nabla ^2d\Vert _{L^6}^2+C\Vert \nabla d\Vert _{L^6}^6 +C\Vert \nabla u\Vert _{L^3}^2\Vert \nabla d\Vert _{L^6}^2 +C\Vert u\Vert _{L^6}^2\Vert \nabla ^2d\Vert _{L^3}^2\\&\le \frac{1}{4}\Vert \nabla d_t\Vert _{L^2}^2+C\Vert \nabla d\Vert _{L^2}\Vert \nabla d\Vert _{L^6}\big (\Vert \nabla ^3d\Vert _{L^2}^2+\Vert \nabla ^2d\Vert _{L^2}^2\big ) +C\Vert \nabla d\Vert _{H^1}^6\\&\quad +C\Vert \nabla u\Vert _{L^3}^3+C\Vert \nabla u\Vert _{L^2}^2\Vert \nabla ^2d\Vert _{L^2}\big (\Vert \nabla ^3d\Vert _{L^2}+\Vert \nabla ^2d\Vert _{L^2}\big )\\&\le \frac{1}{4}\Vert \nabla d_t\Vert _{L^2}^2+\Big (CC_0^\frac{1}{2}+\delta \Big )\Vert \nabla ^3d\Vert _{L^2}^2 +C\Vert \nabla u\Vert _{L^2}^2\Vert \nabla ^2d\Vert _{L^2}^2 +C\Vert \nabla u\Vert _{L^2}^4\Vert \nabla ^2d\Vert _{L^2}^2 \\&\quad +C\Vert \nabla u\Vert _{L^3}^3+C\Vert \nabla d\Vert _{H^1}^2+C\Vert \nabla d\Vert _{H^1}^6, \end{aligned}$$

which implies that

$$\begin{aligned}&\frac{d}{dt}(\eta (t)\Vert \Delta d\Vert _{L^2}^2)+\eta (t)\Vert \nabla d_t\Vert _{L^2}^2\nonumber \\&\le \eta '(t)\Vert \Delta d\Vert _{L^2}^2+\frac{1}{4}\eta (t)\Vert \nabla d_t\Vert _{L^2}^2+\Big (CC_0^\frac{1}{2}+\delta \Big )\eta (t)\Vert \nabla ^3d\Vert _{L^2}^2 +C\eta (t)\Vert \nabla d\Vert _{H^1}^4\nonumber \\&\quad +C\eta (t)\Vert \nabla u\Vert _{L^2}^2\Vert \nabla ^2d\Vert _{L^2}^2+C\eta (t)\Vert \nabla u\Vert _{L^2}^4\Vert \nabla ^2d\Vert _{L^2}^2\nonumber \\&\quad +C\eta (t)\Vert \nabla u\Vert _{L^3}^3+C\eta (t)\Vert \nabla d\Vert _{H^1}^2+C\eta (t)\Vert \nabla d\Vert _{H^1}^6. \end{aligned}$$

(3.27)

Adopting the $L^2$-theory to the Neumann boundary value problem of elliptic equations (see [13]), we infer from (3.26), (3.5), and (3.6) that

$$\begin{aligned} \Vert \nabla ^3d\Vert _{L^2}^2&\le C\Vert \nabla \Delta d\Vert _{L^2}^2+\Vert \nabla d\Vert _{H^1}^2\\&\le C\Vert \nabla d_t\Vert _{L^2}^2+C\Vert \nabla (u\cdot \nabla d)\Vert _{L^2}^2 +C\Vert \nabla (|\nabla d|^2d)\Vert _{L^2}^2+C\Vert \nabla d\Vert _{H^1}^2\\&\le C\Vert \nabla d_t\Vert _{L^2}^2+\Big (CC_0^\frac{1}{2}+\frac{1}{4}\Big )\Vert \nabla ^3d\Vert _{L^2}^2\\&\quad +C\Vert \nabla d\Vert _{H^1}^2+C\Vert \nabla u\Vert _{L^2}^2\Vert \nabla ^2d\Vert _{L^2}^2 +C\Vert \nabla u\Vert _{L^2}^4\Vert \nabla ^2d\Vert _{L^2}^2, \end{aligned}$$

which leads to

$$\begin{aligned} \Vert \nabla ^3d\Vert _{L^2}^2\le C\Vert \nabla d_t\Vert _{L^2}^2+C\Vert \nabla d\Vert _{H^1}^2+C\Vert \nabla u\Vert _{L^2}^2\Vert \nabla ^2d\Vert _{L^2}^2 +C\Vert \nabla u\Vert _{L^2}^4\Vert \nabla ^2d\Vert _{L^2}^2, \end{aligned}$$

(3.28)

if $C_0\le \varepsilon _1$ is suitably small. Putting (3.28) into (3.27), one obtains that

$$\begin{aligned}&\frac{d}{dt}(\eta (t)\Vert \Delta d\Vert _{L^2}^2)+\frac{1}{2}\eta (t)\Vert \nabla d_t\Vert _{L^2}^2\\&\le \eta '(t)\Vert \Delta d\Vert _{L^2}^2 +C\eta (t)\Vert \nabla u\Vert _{L^2}^2\Vert \nabla ^2d\Vert _{L^2}^2+C\eta (t)\Vert \nabla u\Vert _{L^2}^4\Vert \nabla ^2d\Vert _{L^2}^2 \\&\quad +C\eta (t)\Vert \nabla u\Vert _{L^3}^3+C\eta (t)\Vert \nabla d\Vert _{H^1}^2+C\eta (t)\Vert \nabla d\Vert _{H^1}^6. \end{aligned}$$

This combined with (3.25) and (3.28) gives (3.16) after choosing $C_0\le \varepsilon _2\le \varepsilon _1$ and $\delta$ sufficiently small. $\square$

Lemma 3.3

Let $(\rho , u, d)$ be a strong solution of (1.1)–(1.5) satisfying (3.2) and $\eta (t)$ be as in Lemma 3.2, then it holds that

$$\begin{aligned}&\frac{d}{dt}\Big (\frac{\eta (t)}{2}\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2+\frac{\eta (t)}{2}\Vert \nabla d_t\Vert _{L^2}^2\Big ) +(2\mu +\lambda )\eta (t)\Vert \textrm{div}\,\dot{u}\Vert _{L^2}^2\nonumber \\&\quad +\mu \eta (t)\Vert \textrm{curl}\,\dot{u}\Vert _{L^2}^2+\eta (t)\Vert d_{tt}\Vert _{L^2}^2\nonumber \\&\le -C\frac{d}{dt}\int _{\partial \Omega }\eta (t)(u\cdot \nabla n\cdot u)FdS+C|\eta '(t)|\big (\Vert \nabla u\Vert _{L^2}^4+\Vert \nabla ^2d\Vert _{L^2}^4+\Vert \nabla ^2d\Vert _{L^2}^5+C_0\big )\nonumber \\&\quad +C|\eta '(t)|\big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2+\Vert \nabla u\Vert _{L^2}^2+\Vert \nabla ^3d\Vert _{L^2}^2 +CC_0^2\Vert \nabla ^2d\Vert _{L^2}^2+\Vert \nabla d_t\Vert _{L^2}^2\big )\nonumber \\&\quad +C\eta (t)\big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2\Vert \nabla u\Vert _{L^2}^2 +\Vert \nabla u\Vert _{L^2}^4\Vert \nabla ^3d\Vert _{L^2}^\frac{5}{3}+\Vert \nabla u\Vert _{L^2}^2\Vert \nabla ^3d\Vert _{L^2}^2 \big )\nonumber \\&\quad +C\eta (t)\big (\Vert \nabla u\Vert _{L^2}^4+C_0^2\Vert \nabla u\Vert _{L^2}^2 +\Vert \nabla u\Vert _{L^4}^4+\Vert \nabla u\Vert _{L^2}^6+\Vert \nabla ^2d\Vert _{L^2}^4\Vert \nabla u\Vert _{L^2}^2\big )\nonumber \\&\quad +C\delta \eta (t)\big (\Vert \nabla ^2d\Vert _{L^2}\Vert \nabla ^3d\Vert _{L^2}^3+\Vert \nabla ^2d\Vert _{L^2}^2\Vert \nabla ^3d\Vert _{L^2}^2 +\Vert \nabla ^2d\Vert _{L^2}^4 +\Vert \nabla ^2d\Vert _{L^2}^3\Vert \nabla ^3d\Vert _{L^2}\big )\nonumber \\&\quad +C\eta (t)\big (\Vert \nabla u\Vert _{L^2}^2\Vert \nabla d\Vert _{H^1}^3 +\Vert \nabla u\Vert _{L^2}^2\Vert \nabla d\Vert _{H^1}^2\Vert \nabla ^3d\Vert _{L^2}^2 +\Vert \nabla d\Vert _{H^1}^2\Vert \nabla d_t\Vert _{L^2}^2\big )\nonumber \\&\quad +C\eta (t)\big (\Vert \nabla u\Vert _{L^2}^4\Vert \nabla d_t\Vert _{L^2}^2+\Vert \nabla u\Vert _{L^2}^2\Vert \nabla d_t\Vert _{L^2}^2\big ) +C\eta (t)\Vert \nabla d\Vert _{H^1}^4\big (\Vert d_t\Vert _{L^2}^2\nonumber \\&\quad +\Vert \nabla d_t\Vert _{L^2}^2+\Vert \nabla u\Vert _{L^3}^2\big ), \end{aligned}$$

(3.29)

provided that $C_0\le \varepsilon _3$.

Proof

1. By (2.11) and (3.8), we rewrite (1.2) as

$$\begin{aligned} \rho \dot{u}=\nabla F-\mu \nabla ^{\perp }\textrm{curl}\,u-\Delta d\cdot \nabla d. \end{aligned}$$

(3.30)

Applying $\eta (t)\dot{u}^j[\partial /\partial t+{{\,\textrm{div}\,}}(u\cdot )]$ to the jth-component of (3.30), and then integrating the results over $\Omega$, we derive that

$$\begin{aligned}&\frac{1}{2}\left( \frac{d}{dt}\int \eta (t)\rho |\dot{u}|^2dx -\eta '(t)\int \rho |\dot{u}|^2dx\right) \nonumber \\&\quad =\int \eta (t)\big (\dot{u}\cdot \nabla F_t+\dot{u}^j\textrm{div}\,(u\partial _jF)\big )dx\nonumber \\&\quad -\mu \int \eta (t)\big (\dot{u}^j\cdot (\nabla ^{\perp }\textrm{curl}\,u_t)^j+\dot{u}^j\textrm{div}\,((\nabla ^{\perp }\textrm{curl}\,u)^ju)\big )dx\nonumber \\&\quad +\int \eta (t)\big (M(d)_t:\nabla u+{{\,\textrm{div}\,}}M(d)\cdot (u\cdot \nabla \dot{u})\big )dx \triangleq \sum _{i=1}^3J_i. \end{aligned}$$

(3.31)

We denote by $h\triangleq u\cdot (\nabla n+(\nabla n)^{tr})$, we infer from (2.4) and Lemma 2.3 that

$$\begin{aligned}&-\int _{\partial \Omega }\eta (t)F_t(u\cdot \nabla n\cdot u)dS\nonumber \\&\quad =-\frac{d}{dt}\int _{\partial \Omega }\eta (t)(u\cdot \nabla n\cdot u)FdS +\int _{\partial \Omega }\eta (t)Fh\cdot \dot{u}dS-\int _{\partial \Omega }\eta (t) Fh\cdot (u\cdot \nabla u)dS\nonumber \\&\quad +\eta '(t)\int _{\partial \Omega }(u\cdot \nabla n\cdot u)FdS\nonumber \\&=-\frac{d}{dt}\int _{\partial \Omega }\eta (t)(u\cdot \nabla n\cdot u)FdS +\int _{\partial \Omega }\eta (t)Fh\cdot \dot{u}dS+\eta '(t)\int _{\partial \Omega }(\kappa |u|^2)FdS\nonumber \\&\quad +\eta (t)\int _{\partial \Omega }Fh\cdot \kappa |u|^2\cdot ndS\nonumber \\&\quad =-\frac{d}{dt}\int _{\partial \Omega }\eta (t)(u\cdot \nabla n\cdot u)FdS +\int _{\partial \Omega }\eta (t)Fh\cdot \dot{u}dS +\eta '(t)\int _{\partial \Omega }(\kappa |u|^2)FdS\nonumber \\&\quad -\kappa \int \eta (t){{\,\textrm{div}\,}}(Fh|u|^2)dx\nonumber \\&\quad =-\frac{d}{dt}\int _{\partial \Omega }\eta (t)(u\cdot \nabla n\cdot u)FdS +\int _{\partial \Omega }\eta (t)Fh\cdot \dot{u}dS +\eta '(t)\int _{\partial \Omega }(\kappa |u|^2)FdS \nonumber \\&\quad +\kappa \int \eta (t)F|u|^2{{\,\textrm{div}\,}}hdx\nonumber \\&\quad +\kappa \int \eta (t)h\cdot \nabla (F|u|^2)dx\nonumber \\&\quad \le -\frac{d}{dt}\int _{\partial \Omega }\eta (t)(u\cdot \nabla n\cdot u)FdS +C\eta (t)\Vert \nabla F\Vert _{L^2}\Vert u\Vert _{L^3}\Vert \dot{u}\Vert _{L^6}\nonumber \\&\quad +C\eta (t)\big (\Vert F\Vert _{L^3}\Vert u\Vert _{L^6}\Vert \nabla \dot{u}\Vert _{L^2} +\Vert F\Vert _{L^3}\Vert u\Vert _{L^6}\Vert \dot{u}\Vert _{L^2} +\Vert F\Vert _{L^3}\Vert \nabla u\Vert _{L^2}\Vert \dot{u}\Vert _{L^6}\big )\nonumber \\&\quad +C\eta (t)\big (\Vert u\Vert _{L^4}^2\Vert F\Vert _{L^6}\Vert \nabla u\Vert _{L^3} +\Vert \nabla F\Vert _{L^6}\Vert u\Vert _{L^6}^2\Vert \nabla u\Vert _{L^2}\big )\nonumber \\&\quad +C|\eta '(t)|\big (\Vert \nabla u\Vert _{L^2}\Vert u\Vert _{L^6}\Vert F\Vert _{L^3}+\Vert u\Vert _{L^4}^2\Vert F\Vert _{L^2} +\Vert \nabla F\Vert _{L^2}\Vert u\Vert _{L^4}^2\big )\nonumber \\&\quad \le -\frac{d}{dt}\int _{\partial \Omega }\eta (t)(u\cdot \nabla n\cdot u)FdS +C|\eta '(t)|\Vert \nabla u\Vert _{L^2}^2\big (\Vert \rho \dot{u}\Vert _{L^2}\nonumber \\&\quad +\Vert \nabla d\Vert _{L^6}\Vert \nabla ^2d\Vert _{L^3} +\Vert \nabla u\Vert _{L^2}+\Vert P-\bar{P}\Vert _{L^2}\big )\nonumber \\&\quad +C\eta (t)\big (\Vert \rho \dot{u}\Vert _{L^2}+\Vert \nabla d\Vert _{L^6}\Vert \nabla ^2d\Vert _{L^3} +\Vert \nabla u\Vert _{L^2}\nonumber \\&\quad +\Vert P-\bar{P}\Vert _{L^2}\big )\Vert \nabla u\Vert _{L^2}\big (\Vert \nabla \dot{u}\Vert _{L^2} +\Vert \nabla u\Vert _{L^2}^2+\Vert \nabla u\Vert _{L^3}\big )\nonumber \\&\quad +C\eta (t)\Vert \nabla u\Vert _{L^2}^3\Vert \nabla F\Vert _{L^6}\nonumber \\&\quad \le -\frac{d}{dt}\int _{\partial \Omega }\eta (t)(u\cdot \nabla n\cdot u)FdS +C|\eta '(t)|\big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2\nonumber \\&\quad +\Vert \nabla u\Vert _{L^2}^2+\Vert \nabla ^3d\Vert _{L^2}^2+CC_0^\frac{3}{2}\Vert \nabla ^2d\Vert _{L^2}^2\big )\nonumber \\&\quad +C|\eta '(t)|(\Vert \nabla u\Vert _{L^2}^4+\Vert \nabla ^2d\Vert _{L^2}^4+\Vert \nabla ^2d\Vert _{L^2}^5+C_0)+\delta \eta (t) \big (\Vert \nabla \dot{u}\Vert _{L^2}^2+\Vert \nabla F\Vert _{L^6}^2\big )\nonumber \\&\quad +C\eta (t)\big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2\Vert \nabla u\Vert _{L^2}^2 +\Vert \nabla u\Vert _{L^2}^4\Vert \nabla ^3d\Vert _{L^2}^2\nonumber \\&\quad +\Vert \nabla u\Vert _{L^2}^2\Vert \nabla ^3d\Vert _{L^2}^2+\Vert \nabla ^2d\Vert _{L^2}^4\Vert \nabla u\Vert _{L^2}^2 \big )\nonumber \\&\quad +C\eta (t)\big (\Vert \nabla u\Vert _{L^2}^4+C_0^\frac{3}{2}\Vert \nabla u\Vert _{L^2}^2 +\Vert \nabla u\Vert _{L^3}^2+\Vert \nabla u\Vert _{L^2}^6\big ), \end{aligned}$$

(3.32)

due to

$$\begin{aligned} -\int _{\partial \Omega }\eta (t)Fh\cdot (u\cdot \nabla u)dS&=-\int _{\partial \Omega }\eta (t)Fh\cdot (u\cdot \nabla u)|n|^2dS\\&=-\int _{\partial \Omega }\eta (t)Fh\cdot (u\cdot \nabla u)\cdot n\cdot ndS\\&=\int _{\partial \Omega }\eta (t)Fh\cdot \kappa |u|^2\cdot ndS,\\&\Vert \dot{u}\Vert _{L^6}\le C\big (\Vert \nabla \dot{u}\Vert _{L^2}+\Vert \nabla u\Vert _{L^2}^2\big )\ \ (\text {see}\ (2.9)), \end{aligned}$$

and

$$\begin{aligned}&\Vert \nabla d\Vert _{L^6}\Vert \nabla ^2d\Vert _{L^3}\le C(\Vert \nabla d\Vert _{L^2}+\Vert \nabla ^2d\Vert _{L^2})(\Vert \nabla ^2d\Vert _{L^2}^\frac{2}{3} \Vert \nabla ^3d\Vert _{L^2}^\frac{1}{3}+\Vert \nabla ^2d\Vert _{L^2})\\&\qquad \le CC_0^\frac{1}{2}\Vert \nabla ^2d\Vert _{L^2}^\frac{2}{3}\Vert \nabla ^3d\Vert _{L^2}^\frac{1}{2} +C\Vert \nabla ^2d\Vert _{L^2}^2+C\Vert \nabla ^2d\Vert _{L^2}^\frac{5}{3}\Vert \nabla ^3d\Vert _{L^2}^\frac{1}{3} +CC_0. \end{aligned}$$

Thus, it follows from integration by parts, (1.5), (3.2), Hölder’s inequality, (2.5), (2.15), and (3.32) that

$$\begin{aligned} J_1&=\int _{\partial \Omega }\eta (t)F_t\dot{u}\cdot ndS-\int \eta (t)F_t\textrm{div}\,\dot{u}dx -\int \eta (t)u\cdot \nabla \dot{u}\cdot \nabla Fdx\nonumber \\&=-\int _{\partial \Omega }\eta (t)F_t(u\cdot \nabla n\cdot u)dS-(2\mu +\lambda )\int \eta (t)(\textrm{div}\,\dot{u})^2dx +(2\mu +\lambda )\nonumber \\&\int \eta (t)\textrm{div}\,\dot{u}\nabla u:\nabla udx\nonumber \\&\quad -\gamma \int \eta (t) P\textrm{div}\,\dot{u}\textrm{div}\,udx +\int \eta (t)\textrm{div}\,\dot{u}u\cdot \nabla Fdx-\int \eta (t)u\cdot \nabla \dot{u}\cdot \nabla Fdx\nonumber \\&\le -\int _{\partial \Omega }\eta (t)F_t(u\cdot \nabla n\cdot u)dS -(2\mu +\lambda )\int \eta (t)(\textrm{div}\,\dot{u})^2dx+\delta \eta (t)\Vert \nabla \dot{u}\Vert _{L^2}^2\nonumber \\&\quad +C(\delta )\eta (t)\big (\Vert \nabla u\Vert _{L^2}^2\Vert \nabla F\Vert _{L^3}^2 +\Vert \nabla u\Vert _{L^4}^4+\Vert \nabla u\Vert _{L^2}^2\big )\nonumber \\&\le -\int _{\partial \Omega }\eta (t)F_t(u\cdot \nabla n\cdot u)dS -(2\mu +\lambda )\int \eta (t)(\textrm{div}\,\dot{u})^2dx+\delta \eta (t)\Vert \nabla \dot{u}\Vert _{L^2}^2\nonumber \\&\quad +C\eta (t)\big (\Vert \nabla u\Vert _{L^2}^2\Vert \nabla F\Vert _{L^2}\Vert \nabla F\Vert _{L^6}+\Vert \nabla u\Vert _{L^4}^4+\Vert \nabla u\Vert _{L^2}^2 \big )\nonumber \\&\le -\int _{\partial \Omega }\eta (t)F_t(u\cdot \nabla n\cdot u)dS -(2\mu +\lambda )\int \eta (t)(\textrm{div}\,\dot{u})^2dx+\delta \eta (t)\big (\Vert \nabla \dot{u}\Vert _{L^2}^2+\Vert \nabla F\Vert _{L^6}^2\big )\nonumber \\&\quad +C\eta (t)\big (\Vert \nabla u\Vert _{L^2}^4\Vert \nabla F\Vert _{L^2}^2+\Vert \nabla u\Vert _{L^4}^4+\Vert \nabla u\Vert _{L^2}^2 \big )\nonumber \\&\le -\frac{d}{dt}\int _{\partial \Omega }(u\cdot \nabla n\cdot u)FdS-(2\mu +\lambda )\int \eta (t)(\textrm{div}\,\dot{u})^2dx+\delta \eta (t)\big (\Vert \nabla \dot{u}\Vert _{L^2}^2+\Vert \nabla F\Vert _{L^6}^2\big )\nonumber \\&\quad +C|\eta '(t)|\big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2+\Vert \nabla u\Vert _{L^2}^2+\Vert \nabla ^3d\Vert _{L^2}^2 +CC_0^\frac{3}{2}\Vert \nabla ^2d\Vert _{L^2}^2\big )\nonumber \\&\quad +C|\eta '(t)|\big (\Vert \nabla u\Vert _{L^2}^4+\Vert \nabla ^2d\Vert _{L^2}^4+\Vert \nabla ^2d\Vert _{L^2}^5+C_0\big )\nonumber \\&\quad +C\eta (t)\big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2\Vert \nabla u\Vert _{L^2}^2 +\Vert \nabla u\Vert _{L^2}^4\Vert \nabla ^3d\Vert _{L^2}^2+\Vert \nabla u\Vert _{L^2}^2\Vert \nabla ^3d\Vert _{L^2}^2 \big )\nonumber \\&\quad +C\eta (t)\big (\Vert \nabla u\Vert _{L^2}^4+C_0^\frac{3}{2}\Vert \nabla u\Vert _{L^2}^2 +\Vert \nabla u\Vert _{L^3}^2+\Vert \nabla u\Vert _{L^2}^6+\Vert \nabla ^2d\Vert _{L^2}^4\Vert \nabla u\Vert _{L^2}^2\big )\nonumber \\&\le -\frac{d}{dt}\int _{\partial \Omega }(u\cdot \nabla n\cdot u)FdS-(2\mu +\lambda )\int \eta (t)(\textrm{div}\,\dot{u})^2dx+\delta \eta (t)\Vert \nabla \dot{u}\Vert _{L^2}^2\nonumber \\&\quad +C|\eta '(t)|\big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2+\Vert \nabla u\Vert _{L^2}^2+\Vert \nabla ^3d\Vert _{L^2}^2 +CC_0^\frac{3}{2}\Vert \nabla ^2d\Vert _{L^2}^2\big )\nonumber \\&\quad +C|\eta '(t)|\big (\Vert \nabla u\Vert _{L^2}^4+\Vert \nabla ^2d\Vert _{L^2}^4+\Vert \nabla ^2d\Vert _{L^2}^5+C_0\big )\nonumber \\&\quad +C\eta (t)\big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2\Vert \nabla u\Vert _{L^2}^2 +\Vert \nabla u\Vert _{L^2}^4\Vert \nabla ^3d\Vert _{L^2}^2+\Vert \nabla u\Vert _{L^2}^2\Vert \nabla ^3d\Vert _{L^2}^2 \big )\nonumber \\&\quad +C\eta (t)\big (\Vert \nabla u\Vert _{L^2}^4+C_0^\frac{3}{2}\Vert \nabla u\Vert _{L^2}^2 +\Vert \nabla u\Vert _{L^4}^4+\Vert \nabla u\Vert _{L^2}^6+\Vert \nabla ^2d\Vert _{L^2}^4\Vert \nabla u\Vert _{L^2}^2\big )\nonumber \\&\quad +C\delta \eta (t)\big (\Vert \nabla ^2d\Vert _{L^2}\Vert \nabla ^3d\Vert _{L^2}^3+\Vert \nabla ^2d\Vert _{L^2}^2\Vert \nabla ^3d\Vert _{L^2}^2 +\Vert \nabla ^2d\Vert _{L^2}^4 +\Vert \nabla ^2d\Vert _{L^2}^3\Vert \nabla ^3d\Vert _{L^2}\big ), \end{aligned}$$

(3.33)

where we have used

$$\begin{aligned} F_t&=(2\mu +\lambda )\textrm{div}\,u_t-(P-\bar{P})_t\\&=(2\mu +\lambda )\textrm{div}\,\dot{u}-(2\mu +\lambda )\textrm{div}\,(u\cdot \nabla u)+u\cdot \nabla P+\gamma P{{\,\textrm{div}\,}}u \\&=(2\mu +\lambda )\textrm{div}\,\dot{u}-(2\mu +\lambda )u\cdot \nabla \textrm{div}\,u -(2\mu +\lambda )\nabla u:\nabla u+u\cdot \nabla P +\gamma P{{\,\textrm{div}\,}}u\\&=(2\mu +\lambda )\textrm{div}\,\dot{u} -(2\mu +\lambda )\nabla u:\nabla u+\gamma P\textrm{div}\,u-u\cdot \nabla F, \end{aligned}$$

and

$$\begin{aligned} \Vert \nabla F\Vert _{L^6}^2&\le C\big (\Vert \rho \dot{u}\Vert _{L^6}+\Vert |\nabla d||\nabla ^2d|\Vert _{L^6}\big )^2\\&\le C(\hat{\rho })\big (\Vert \dot{u}\Vert _{L^6}^2\\&\quad +\Vert \nabla d|_{L^{12}}^2\Vert \nabla ^2d\Vert _{L^{12}}^2\big )\\&\le C\big (\Vert \nabla d\Vert _{L^2}^\frac{1}{6}\Vert \nabla ^2 d\Vert _{L^2}^\frac{5}{6}\big )^2\big (|\nabla ^2 d\Vert _{L^2}^\frac{1}{6}\Vert \nabla ^3 d\Vert _{L^2}^\frac{5}{6}\\&\quad +C\Vert \nabla ^2 d\Vert _{L^2}\big )^2 +C(\hat{\rho })\big (\Vert \nabla \dot{u}\Vert _{L^2}+\Vert \nabla u\Vert _{L^2}^2\big )\\&\le C\Vert \nabla d\Vert _{L^2}^\frac{1}{3}\Vert \nabla ^2 d\Vert _{L^2}^2\Vert \nabla ^3 d\Vert _{L^2}^ \frac{5}{3}\\&\quad +C\Vert \nabla d\Vert _{L^2}\Vert \nabla ^2 d\Vert _{L^2}^\frac{11}{3}\\&\quad +C\Vert \nabla d\Vert _{L^2}^2\Vert \nabla ^2 d\Vert _{L^2}^\frac{1}{3}\Vert \nabla ^3 d\Vert _{L^2}^\frac{5}{3}\\&\quad +C(\hat{\rho })\big (\Vert \nabla \dot{u}\Vert _{L^2}^2+\Vert \nabla u\Vert _{L^2}^4)\\&\le C\Vert \nabla ^2 d\Vert _{L^2}^2\Vert \nabla ^3 d\Vert _{L^2}^2+C\Vert \nabla ^2 d\Vert _{L^2}\Vert \nabla ^3 d\Vert _{L^3}^3+C\Vert \nabla ^2 d\Vert _{L^2}^4\\&\quad +C(\hat{\rho })\big (\Vert \nabla \dot{u}\Vert _{L^2}+\Vert \nabla u\Vert _{L^2}^2\big ). \end{aligned}$$

2. By a direct calculation, one obtains that

$$\begin{aligned} J_2&=-\mu \int \eta (t)\dot{u}\cdot (\nabla ^{\perp }\textrm{curl}\,u_t)dx\nonumber \\&\quad -\mu \int \eta (t)\dot{u}^j\textrm{div}\,((\nabla ^{\perp }{{\,\textrm{curl}\,}}u)^ju)dx\nonumber \\&=-\mu \int \eta (t)(\textrm{curl}\,\dot{u})^2dx +\mu \int \eta (t){{\,\textrm{curl}\,}}\dot{u}{{\,\textrm{curl}\,}}(u\cdot \nabla u)dx\nonumber \\&\quad +\mu \int \eta (t)\nabla \dot{u}^j(\nabla ^{\perp }{{\,\textrm{curl}\,}}u)^judx+\int _{\partial {\Omega }}\dot{u}^j(\nabla ^{\perp }{{\,\textrm{curl}\,}}u)^ju\cdot ndx\nonumber \\&=-\mu \int \eta (t)(\textrm{curl}\,\dot{u})^2dx \nonumber \\&\quad +\mu \int \eta (t)\textrm{curl}\,\dot{u} \textrm{curl}\,(u\cdot \nabla u)dx+\mu \int \eta (t)u\cdot \nabla \dot{u}\big (\nabla ^{\perp }{{\,\textrm{curl}\,}}u\big )dx\nonumber \\&=-\mu \int \eta (t)(\textrm{curl}\,\dot{u})^2dx+\mu \int \eta (t)\textrm{curl}\,\dot{u}\partial _iu\nabla ^\perp u^idx\nonumber \\&\quad -\mu \int \eta (t)\textrm{curl}\,(u\cdot \nabla \dot{u}{{\,\textrm{curl}\,}}u)dx+\mu \int \eta (t){{\,\textrm{curl}\,}}(u\cdot \nabla \dot{u}){{\,\textrm{curl}\,}}udx\nonumber \\&=-\mu \int \eta (t)(\textrm{curl}\,\dot{u})^2dx+\mu \int \eta (t)\textrm{curl}\,\dot{u}\partial _iu\nabla ^{\perp } u^idx\nonumber \\&\quad -\mu \int \eta (t)\textrm{curl}\,(\dot{u}\cdot \nabla u){{\,\textrm{curl}\,}}udx\nonumber \\&=-\mu \int \eta (t)(\textrm{curl}\,\dot{u})^2dx+\mu \int \eta (t)\textrm{curl}\,\dot{u}\partial _iu\nabla ^{\perp } u^idx \nonumber \\&\quad +\mu \int \eta (t)\textrm{curl}\,u\nabla ^{\perp }u^i\cdot \partial _i\dot{u}dx\nonumber \\&\le \delta \eta (t)\Vert \nabla \dot{u}\Vert _{L^2}^2+C\eta (t)\Vert \nabla u\Vert _{L^4}^4-\mu \eta (t)\Vert \textrm{curl}\,\dot{u}\Vert _{L^2}^2, \end{aligned}$$

(3.34)

due to

$$\begin{aligned} -\int \eta (t)\dot{u}(\nabla ^{\perp }{{\,\textrm{curl}\,}}u_t) dx&=\int \eta (t){{\,\textrm{curl}\,}}\dot{u}{{\,\textrm{curl}\,}}u_tdx+\int {{\,\textrm{curl}\,}}(\dot{u}{{\,\textrm{curl}\,}}u_t)dx\\&=-\int {{\,\textrm{curl}\,}}\dot{u}{{\,\textrm{curl}\,}}u_tdx+\int _{\partial \Omega }\dot{u}\cdot {{\,\textrm{curl}\,}}u_t\cdot \omega dx\\&=-\int ({{\,\textrm{curl}\,}}\dot{u})^2dx+\int {{\,\textrm{curl}\,}}\dot{u}{{\,\textrm{curl}\,}}(u\cdot \nabla u)dx \end{aligned}$$

and

$$\begin{aligned}&\int \textrm{curl}\,\dot{u}\cdot \textrm{curl}\,(u\cdot \nabla u)dx\\&\quad =\int \textrm{curl}\,\dot{u}\cdot \textrm{curl}\,(u^i\partial _iu)dx \\&\quad =\int \textrm{curl}\,\dot{u}\big (u^i\textrm{curl}\,\partial _iu+\partial _iu\cdot \nabla ^{\perp } u^i\big )dx\\&\quad =\int \textrm{curl}\,\dot{u}\partial _iu\nabla ^{\perp } u^idx. \end{aligned}$$

By Hölder’s inequality, (2.9), Sobolev’s inequality, (2.5), and (3.5), we derive that

$$\begin{aligned} J_3&\le C\eta (t)\Vert \nabla \dot{u}\Vert _{L^2}\big (\Vert \nabla d\Vert _{L^6}\Vert \nabla d_t\Vert _{L^3} +\Vert \nabla d\Vert _{L^6}\Vert \nabla ^2d\Vert _{L^6}\Vert u\Vert _{L^6}\big )\nonumber \\&\le C\eta (t)\Vert \nabla \dot{u}\Vert _{L^2}\Vert \nabla d\Vert _{H^1}\big (\Vert \nabla d_t\Vert _{L^2}^\frac{2}{3}\Vert \nabla ^2d_t\Vert _{L^2}^\frac{1}{3}+\Vert \nabla d_t\Vert _{L^2}\big )\nonumber \\&\quad +C\eta (t)\Vert \nabla \dot{u}\Vert _{L^2}\Vert \nabla u\Vert _{L^2}\big (\Vert \nabla d\Vert _{L^2}+\Vert \nabla ^2d\Vert _{L^2}\big ) \big (\Vert \nabla ^2d\Vert _{L^2}+\Vert \nabla ^3d\Vert _{L^2}\big )\nonumber \\&\le \delta \eta (t)\big (\Vert \nabla \dot{u}\Vert _{L^2}^2+\Vert \nabla ^2d_t\Vert _{L^2}^2\big ) +C\eta (t)\Vert \nabla d\Vert _{H^1}^2\Vert \nabla d_t\Vert _{L^2}^2\nonumber \\&\quad +C\eta (t)\big (\Vert \nabla u\Vert _{L^2}^2\Vert \nabla d\Vert _{H^1}^4 +\Vert \nabla u\Vert _{L^2}^2\Vert \nabla d\Vert _{H^1}^2\Vert \nabla ^3d\Vert _{L^2}^2\big ). \end{aligned}$$

(3.35)

Putting above estimates on $J_1$, $J_2$, and $J_3$ into (3.31), we obtain after choosing $\delta$ suitably small that

$$\begin{aligned}&\frac{d}{dt}\Big (\frac{\eta (t)}{2}\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2\Big ) +(2\mu +\lambda )\eta (t)\Vert \textrm{div}\,\dot{u}\Vert _{L^2}^2+\mu \eta (t)\Vert \textrm{curl}\,\dot{u}\Vert _{L^2}^2\nonumber \\&\le -\frac{d}{dt}\int _{\partial \Omega }(u\cdot \nabla n\cdot u)FdS+C|\eta '(t)|\big (\Vert \nabla u\Vert _{L^2}^4+\Vert \nabla ^2d\Vert _{L^2}^4+\Vert \nabla ^2d\Vert _{L^2}^5+C_0\big )\nonumber \\&\quad +C|\eta '(t)|\big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2+\Vert \nabla u\Vert _{L^2}^2+\Vert \nabla ^3d\Vert _{L^2}^2 +CC_0^\frac{3}{2}\Vert \nabla ^2d\Vert _{L^2}^2\big )\nonumber \\&\quad +C\eta (t)\big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2\Vert \nabla u\Vert _{L^2}^2 +\Vert \nabla u\Vert _{L^2}^4\Vert \nabla ^3d\Vert _{L^2}^2+\Vert \nabla u\Vert _{L^2}^2\Vert \nabla ^3d\Vert _{L^2}^2\big )\nonumber \\&\quad +C\eta (t)\big (\Vert \nabla u\Vert _{L^2}^4+C_0^2\Vert \nabla u\Vert _{L^2}^2 +\Vert \nabla u\Vert _{L^3}^2+\Vert \nabla u\Vert _{L^2}^6+\Vert \nabla ^2d\Vert _{L^2}^4\Vert \nabla u\Vert _{L^2}^2\big )\nonumber \\&\quad +C\delta \eta (t)\big (\Vert \nabla ^2d\Vert _{L^2}\Vert \nabla ^3d\Vert _{L^2}^3+\Vert \nabla ^2d\Vert _{L^2}^2\Vert \nabla ^3d\Vert _{L^2}^2 +\Vert \nabla ^2d\Vert _{L^2}^4 +\Vert \nabla ^2d\Vert _{L^2}^3\Vert \nabla ^3d\Vert _{L^2}\big )\nonumber \\&\quad +C\eta (t)\big (\Vert \nabla u\Vert _{L^2}^2\Vert \nabla d\Vert _{H^1}^4 +\Vert \nabla u\Vert _{L^2}^2\Vert \nabla d\Vert _{H^1}^2\Vert \nabla ^3d\Vert _{L^2}^2 +\Vert \nabla d\Vert _{H^1}^2\Vert \nabla d_t\Vert _{L^2}^2\big ).\nonumber \\ \end{aligned}$$

(3.36)

3. Differentiating (1.3) with respect to t and multiplying the results by $d_{tt}$, we obtain from integration by parts, $\frac{\partial d_t}{\partial n}|_{\partial \Omega }=0$, Sobolev’s inequality, and Hölder’s inequality that

$$\begin{aligned}&\frac{1}{2}\frac{d}{dt}\int |\nabla d_t|^2dx+\int |d_{tt}|^2dx\nonumber \\&=\int \langle \partial _t\big (|\nabla d|^2d-u\cdot \nabla d\big ), d_{tt}\rangle dx\nonumber \\&\le C\int |d_{tt}||u_t||\nabla d|dx+C\int |d_{tt}||u||\nabla d_t|dx\nonumber \\&\quad +C\int |d_{tt}||d_t||\nabla d|^2dx +C\int |d_{tt}||\nabla d_t||\nabla d|dx\nonumber \\&\le C\int |d_{tt}||\dot{u}||\nabla d|dx+C\int |d_{tt}||u||\nabla d_t|dx+C\int |d_{tt}||d_t||\nabla d|^2dx\nonumber \\&\quad +C\int |d_{tt}||\nabla d_t||\nabla d|dx\nonumber \\&\quad +C\int |d_{tt}||u||\nabla u||\nabla d|dx\triangleq \sum _{i=1}^5U_i. \end{aligned}$$

(3.37)

By a direct computation, one has

$$\begin{aligned} U_1&\le \delta \Vert d_{tt}\Vert _{L^2}^2+C\Vert \dot{u}\Vert _{L^6}^2\Vert \nabla d\Vert _{L^3}^2\\&\le \delta \Vert d_{tt}\Vert _{L^2}^2+C\big (\Vert \nabla \dot{u}\Vert _{L^2}^2+\Vert \nabla u\Vert _{L^2}^4\big ) \big (\Vert \nabla d\Vert _{L^2}^\frac{4}{3}\Vert \nabla ^2d\Vert _{L^2}^\frac{2}{3}+\Vert \nabla d\Vert _{L^2}^2\big ),\\&\le \delta \Vert d_{tt}\Vert _{L^2}^2+CC_0^\frac{1}{2}\Vert \nabla \dot{u}\Vert _{L^2}^2+C\Vert \nabla u\Vert _{L^2}^4,\\ U_2&\le \delta \Vert d_{tt}\Vert _{L^2}^2+C\Vert u\Vert _{L^6}^2\Vert \nabla d_t\Vert _{L^3}^2\\&\le \delta \Vert d_{tt}\Vert _{L^2}^2+C\Vert \nabla u\Vert _{L^2}^2\big (\Vert \nabla d_t\Vert _{L^2}^ \frac{4}{3}\Vert \nabla ^2d_t\Vert _{L^2}^\frac{2}{3}\\&\quad +\Vert \nabla d_t\Vert _{L^2}^2\big )\\&\le \delta \big (\Vert d_{tt}\Vert _{L^2}^2+\Vert \nabla ^2d_t\Vert _{L^2}^2\big ) +C\Vert \nabla u\Vert _{L^2}^3\Vert \nabla d_t\Vert _{L^2}^2\\&\quad +C\Vert \nabla u\Vert _{L^2}^2\Vert \nabla d_t\Vert _{L^2}^2,\\ U_3&\le \delta \Vert d_{tt}\Vert _{L^2}^2+C\Vert d_t\Vert _{L^6}^2\Vert \nabla d\Vert _{L^6}^4\\&\le \delta \Vert d_{tt}\Vert _{L^2}^2+C\Vert \nabla d\Vert _{H^1}^4\big (\Vert d_t\Vert _{L^2}^2+\Vert \nabla d_t\Vert _{L^2}^2\big ),\\ U_4&\le \delta \Vert d_{tt}\Vert _{L^2}^2+C\Vert \nabla d_t\Vert _{L^3}^2\Vert \nabla d\Vert _{L^6}^2\\&\le \delta \Vert d_{tt}\Vert _{L^2}^2+C\big (\Vert \nabla d_t\Vert _{L^2}^\frac{4}{3}\Vert \nabla ^2d_t\Vert _{L^2}^\frac{2}{3}\\&\quad +\Vert \nabla d_t\Vert _{L^2}^2\big )\Vert \nabla d\Vert _{H^1}^2\\&\le \delta \big (\Vert d_{tt}\Vert _{L^2}^2+\Vert \nabla ^2d_t\Vert _{L^2}^2\big ) +C\Vert \nabla d\Vert _{H^1}^4\Vert \nabla d_t\Vert _{L^2}^2+C\Vert \nabla d\Vert _{H^1}^2\Vert \nabla d_t\Vert _{L^2}^2,\\ U_5&\le \delta \Vert d_{tt}\Vert _{L^2}^2+C\Vert u\Vert _{L^6}^2\Vert \nabla u\Vert _{L^4}^2\Vert \nabla d\Vert _{L^{12}}^2\\&\le \delta \Vert d_{tt}\Vert _{L^2}^2+C\Vert \nabla u\Vert _{L^4}^4 +C\Vert \nabla u\Vert _{L^2}^4\Vert \nabla d\Vert _{L^2}^\frac{1}{6}\Vert \nabla ^2d\Vert _{L^2}^\frac{5}{6}\\&\le \delta \Vert d_{tt}\Vert _{L^2}^2+C\Vert \nabla u\Vert _{L^4}^4+C\Vert \nabla u\Vert _{L^2}^4, \end{aligned}$$

where we have used (2.1), (3.5), and (3.4). Thus, substituting the above estimates on $U_i\ (i=1, 2, \ldots , 5)$ into (3.37) shows that

$$\begin{aligned}&\frac{d}{dt}\Big (\frac{\eta (t)}{2}\Vert \nabla d_t\Vert _{L^2}^2\Big ) +\eta (t)\Vert d_{tt}\Vert _{L^2}^2-\frac{1}{2}\eta '(t)\Vert \nabla d_t\Vert _{L^2}^2\nonumber \\&\le C\delta \eta (t)(\Vert d_{tt}\Vert _{L^2}^2+\Vert \nabla ^2d_t\Vert _{L^2}^2)+CC_0^\frac{1}{2}\eta (t)\Vert \nabla \dot{u}\Vert _{L^2}^2 +C\eta (t)\Vert \nabla u\Vert _{L^4}^4+C\eta (t)\Vert \nabla u\Vert _{L^2}^4\nonumber \\&\quad +C\eta (t)\Vert \nabla u\Vert _{L^2}^3\Vert \nabla d_t\Vert _{L^2}^2+C\eta (t)\Vert \nabla u\Vert _{L^2}^2\Vert \nabla d_t\Vert _{L^2}^2 +C\eta (t)\Vert \nabla d\Vert _{H^1}^4(\Vert d_t\Vert _{L^2}^2+\Vert \nabla d_t\Vert _{L^2}^2)\nonumber \\&\quad +C\eta (t)\Vert \nabla d\Vert _{H^1}^2\Vert \nabla d_t\Vert _{L^2}^2+C\eta (t)\Vert \nabla u\Vert _{L^2}^4\Vert \nabla ^3d\Vert _{L^2}^2. \end{aligned}$$

(3.38)

4. It remains to estimate $\Vert \nabla ^2d_t\Vert _{L^2}$. Taking $v=\nabla d_t$, $k=0$, $q=2$ in Lemma 2.5, we obtain from (2.5), Hölder’s inequality, interpolation inequality, Young’s inequality, (2.5), and Lemma 3.1 that

$$\begin{aligned} \Vert \nabla ^2d_t\Vert _{L^2}^2&\le \Vert \Delta d_t\Vert _{L^2}\nonumber \\&\le C(\Vert d_{tt}\Vert _{L^2}^2+\Vert \partial _t(u\cdot \nabla d)\Vert _{L^2}^2 +\Vert \partial _t(|\nabla d|^2d)\Vert _{L^2}^2)\nonumber \\&\le \Vert u\Vert _{L^6}^2\Vert \nabla d_t\Vert _{L^3}^2+C\Vert d_{tt}\Vert _{L^2}^2+C\Vert \dot{u}\Vert _{L^6}^2\Vert \nabla d\Vert _{L^3}^2+C\Vert d_t\Vert _{L^6}^2\Vert \nabla d\Vert _{L^6}^4\nonumber \\&\quad +C\Vert \nabla d_t\Vert _{L^3}^2\Vert \nabla d\Vert _{L^6}^2+C\Vert u\Vert _{L^6}^2\Vert \nabla u\Vert _{L^4}^2\Vert \nabla d\Vert _{L^{12}}^2\nonumber \\&\le \frac{1}{2}\Vert \nabla ^2d_t\Vert _{L^2}^2+C\Vert d_{tt}\Vert _{L^2}^2+CC_0^\frac{1}{2}\Vert \nabla \dot{u}\Vert _{L^2}^2 +C\Vert \nabla u\Vert _{L^4}^4+C\Vert \nabla u\Vert _{L^2}^4\nonumber \\&\quad +C\Vert \nabla u\Vert _{L^2}^4\Vert \nabla d_t\Vert _{L^2}^2+C\Vert \nabla u\Vert _{L^2}^2\Vert \nabla d_t\Vert _{L^2}^2 +C\Vert \nabla d\Vert _{H^1}^4(\Vert d_t\Vert _{L^2}^2+\Vert \nabla d_t\Vert _{L^2}^2)\nonumber \\&\quad +C\Vert \nabla d\Vert _{H^1}^2\Vert \nabla d_t\Vert _{L^2}^2+C\Vert \nabla u\Vert _{L^2}^4\Vert \nabla ^3d\Vert _{L^2}^2. \end{aligned}$$

(3.39)

This together with (3.38) gives that

$$\begin{aligned}&\frac{d}{dt}\Big (\frac{\eta (t)}{2}\Vert \nabla d_t\Vert _{L^2}^2\Big ) +\eta (t)\Vert d_{tt}\Vert _{L^2}^2-\frac{1}{2}\eta '(t)\Vert \nabla d_t\Vert _{L^2}^2\nonumber \\&\le C\delta \eta (t)\Vert d_{tt}\Vert _{L^2}^2+CC_0^\frac{1}{2}\eta (t)\Vert \nabla \dot{u}\Vert _{L^2}^2 +C\eta (t)\Vert \nabla u\Vert _{L^4}^4+C\eta (t)\Vert \nabla u\Vert _{L^2}^4\nonumber \\&\quad +C\eta (t)\Vert \nabla u\Vert _{L^2}^3\Vert \nabla d_t\Vert _{L^2}^2+C\eta (t)\Vert \nabla u\Vert _{L^2}^2\Vert \nabla d_t\Vert _{L^2}^2 +C\eta (t)\Vert \nabla d\Vert _{H^1}^4(\Vert d_t\Vert _{L^2}^2+\Vert \nabla d_t\Vert _{L^2}^2)\nonumber \\&\quad +C\eta (t)\Vert \nabla d\Vert _{H^1}^2\Vert \nabla d_t\Vert _{L^2}^2+C\eta (t)\Vert \nabla u\Vert _{L^2}^4\Vert \nabla ^3d\Vert _{L^2}^\frac{5}{3}, \end{aligned}$$

(3.40)

which combined with (3.36) leads to (3.29) after choosing $C_0\le \varepsilon _3$ and $\delta$ sufficiently small. $\square$

Lemma 3.4

Let the assumptions of Proposition 3.1 be satisfied. Then there exists a positive constant $\varepsilon _4$ such that

$$\begin{aligned} A_3(\sigma (T))+\int _0^{\sigma (T)}\big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2+\Vert \nabla d_t\Vert _{L^2}^2\big )dt\le 3K, \end{aligned}$$

(3.41)

provided that $C_0\le \varepsilon _4$.

Proof (Proof)

In view of (2.16) and (3.5), we have

$$\begin{aligned} \Vert \nabla u\Vert _{L^3}^3&\le C\big (\Vert \rho \dot{u}\Vert _{L^2}+\Vert |\nabla d||\nabla ^2d|\Vert _{L^2}\big )^3\big (\Vert \nabla u\Vert _{L^2}+\Vert P-\bar{P}\Vert _{L^2}\big )^2 +C\big (\Vert \nabla u\Vert _{L^2}^3+\Vert P-\bar{P}\Vert _{L^3}^3\big )\nonumber \\&\le \delta \Vert \rho \dot{u}\Vert _{L^2}^3+C\Vert \nabla d\Vert _{L^6}^3\Vert \nabla ^2d\Vert _{L^3}^3+C\Vert \nabla u\Vert _{L^2}^4+C\Vert \nabla u\Vert _{L^2}^2+CC_0\nonumber \\&\le \delta \Vert \rho \dot{u}\Vert _{L^2}^3+C\Vert \nabla d\Vert _{H^1}^2(\Vert \nabla ^2d\Vert _{L^2}^2\Vert \nabla ^3d\Vert _{L^2} +\Vert \nabla ^2d\Vert _{L^2}^3)\nonumber \\&\quad +C\Vert \nabla u\Vert _{L^2}^4+C\Vert \nabla u\Vert _{L^2}^2+CC_0\nonumber \\&\le \delta \Vert \rho \dot{u}\Vert _{L^2}^3+\delta \Vert \nabla d\Vert _{H^1}^2\Vert \nabla ^3d\Vert _{L^2}+C\Vert \nabla d\Vert _{H^1}^4\nonumber \\&\quad +C\Vert \nabla u\Vert _{L^2}^4+C\Vert \nabla u\Vert _{L^2}^2+CC_0\nonumber \\&\le \delta \Vert \rho \dot{u}\Vert _{L^2}^3+\delta \Vert \nabla d\Vert _{H^1}^2\Vert \nabla d_t\Vert _{L^2}+C\Vert \nabla d\Vert _{H^1}^4+C\Vert \nabla d\Vert _{H^1}^6 \nonumber \\&\quad +C\Vert \nabla u\Vert _{L^2}^4+C\Vert \nabla u\Vert _{L^2}^2+CC_0, \end{aligned}$$

(3.42)

due to

$$\begin{aligned} \Vert P-\bar{P}\Vert _{L^3}\le C\Vert P-\bar{P}\Vert _{L^\infty }^\frac{1}{3}\Big (\int |P-\bar{P}|^2dx\Big )^\frac{1}{3}\le C(\hat{\rho })\Vert \rho -\bar{\rho }\Vert _{L^2}^\frac{2}{3}. \end{aligned}$$

By (3.42), (3.5), (3.4), and Lemma 3.2, one can check that

$$\begin{aligned} \int _0^{\sigma (T)}\Vert \nabla u\Vert _{L^3}^3dt&\le \delta C(\hat{\rho })\int _0^{\sigma (T)}\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^3dt+C\delta \int _0^{\sigma (T)}\Vert \nabla d_t\Vert _{L^2}\nonumber \\&\quad +C\int _0^{\sigma (T)}\big (\Vert \nabla d\Vert _{H^1}^4+\Vert \nabla u\Vert _{L^2}^4+\Vert \nabla d\Vert _{H^1}^6\big )dt+CC_0. \end{aligned}$$

(3.43)

Taking $\eta (t)=1$ and integrating (3.16) over [0, t] for $0<t\le \sigma (T)$, we deduce from (3.1), (3.2), and (3.43) that

$$\begin{aligned}&\frac{2\mu +\lambda }{2}\Vert \textrm{div}\,u\Vert _{L^2}^2+\frac{\mu }{2}\Vert \textrm{curl}\,u\Vert _{L^2}^2+\Vert \Delta d\Vert _{L^2}^2+\frac{1}{2}\int _{0}^{\sigma (T)}\left( \Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2+\Vert \nabla d_t\Vert _{L^2}^2 \right) dt\\&\le \frac{2\mu +\lambda }{2}\Vert \textrm{div}\,u_0\Vert _{L^2}^2+\frac{\mu }{2}\Vert \textrm{curl}\,u_0\Vert _{L^2}^2+\Vert \Delta d_0\Vert _{L^2}^2+\int (P-P(\rho _{\infty }))\textrm{div}\,udx\Big |_0^{\sigma (T)}\\&\quad +\int M(d):\nabla udx\Big |_0^{\sigma (T)}+C\int _{0}^{\sigma (T)}\Vert \nabla u\Vert _{L^2}^2dt\\&\quad +C\int _{0}^{\sigma (T)}\left( \Vert \nabla u\Vert _{L^3}^3+\Vert \nabla u\Vert _{L^2}^4+\Vert \nabla u\Vert _{L^2}^2+\Vert \nabla d\Vert _{H^1}^2+\Vert \nabla d\Vert _{H^1}^6+\Vert \nabla u\Vert _{L^2}^4\Vert \nabla d\Vert _{H^1}^2\right) dt\\&\le \int (P-P(\rho _{\infty }))\textrm{div}\,udx\Big |_0^{\sigma (T)}+\int M(d):\nabla udx\Big |_0^{\sigma (T)}+C\Vert \nabla u_0\Vert _{L^2}^2+\Vert \Delta d_0\Vert _{L^2}^2\\&\quad +C(\hat{\rho })\delta \int _{0}^{\sigma (T)}\big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2+\Vert \nabla d_t\Vert _{L^2}^2\big )dt+CC_0^{\frac{1}{2}}, \end{aligned}$$

which implies that

$$\begin{aligned}&\frac{2\mu +\lambda }{2}\Vert \textrm{div}\,u\Vert _{L^2}^2+\frac{\mu }{2}\Vert \textrm{curl}\,u\Vert _{L^2}^2+\Vert \Delta d\Vert _{L^2}^2+(\frac{1}{2}-C(\hat{\rho })\delta )\int _{0}^{\sigma (T)}\left( \Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2+\Vert \nabla d_t\Vert _{L^2}^2 \right) dt\\&\le \int (P-P(\rho _{\infty }))\textrm{div}\,udx\Big |_0^{\sigma (T)}+\int M(d):\nabla udx\Big |_0^{\sigma (T)}+C\Vert \nabla u_0\Vert _{L^2}^2+\Vert \Delta d_0\Vert _{L^2}^2+CC_0^{\frac{1}{2}}. \end{aligned}$$

Therefore, choosing $\delta$ sufficiently small, we get from Hölder’s and Young’s inequalities that

$$\begin{aligned}&\Vert \nabla u\Vert _{L^2}^2+\Vert \Delta d\Vert _{L^2}^2+\int _{0}^{\sigma (T)}\left( \Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2+\Vert \nabla d_t\Vert _{L^2}^2 \right) dt\nonumber \\&\le \frac{2C}{2\mu +\lambda }\cdot \frac{2\mu +\lambda }{2}\Vert \textrm{div}\,u\Vert _{L^2}^2+\frac{2C}{\mu }\cdot \frac{\mu }{2}\Vert \textrm{curl}\,u\Vert _{L^2}^2\nonumber \\&\quad +\Vert \Delta d\Vert _{L^2}^2+\frac{1}{\frac{1}{2}-C(\hat{\rho })\delta }\left( \frac{1}{2}-C(\hat{\rho })\delta \right) \int _{0}^{\sigma (T)}\left( \Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2+\Vert \nabla d_t\Vert _{L^2}^2 \right) dt\nonumber \\&\le C_3\left( \frac{2\mu +\lambda }{2}\Vert \textrm{div}\,u\Vert _{L^2}^2+\frac{\mu }{2}\Vert \textrm{curl}\,u\Vert _{L^2}^2+\Vert \Delta d\Vert _{L^2}^2+(\frac{1}{2}-C_3\delta )\int _{0}^{\sigma (T)}\left( \Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2+\Vert \nabla d_t\Vert _{L^2}^2 \right) dt\right) \nonumber \\&\le C_4C\Vert \nabla u_0\Vert _{L^2}^2+C_4C\Vert \Delta d_0\Vert _{L^2}^2\nonumber \\&\quad +C_4\left( \int (P-P(\rho _{\infty }))\textrm{div}\,udx\Big |_0^{\sigma (T)}+\int M(d):\nabla udx\Big |_0^{\sigma (T)}\right) +CC_0^{\frac{1}{2}}\nonumber \\&\le 2K+CC_0^{\frac{1}{2}}\le 3K, \end{aligned}$$

(3.44)

provided that $C_0\le \varepsilon _4$ is suitably small and $C_3\triangleq max\left\{ \frac{2C}{2\mu +\lambda },\frac{2C}{\mu },1,\frac{1}{\frac{1}{2}-C(\hat{\rho })\delta }\right\}$. We immediately obtain (3.41) from (3.44). $\square$

Lemma 3.5

Let the assumptions of Proposition 3.1 be satisfied. For $\sigma _i\triangleq \sigma (t+1-i)$ with i being an integer satisfying $1\le i\le [T]-1$, then there exists a positive constant $\varepsilon _5$ such that

$$\begin{aligned} A_1(T)+\int _{i-1}^{i+1}\sigma _i\big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2+\Vert \nabla d_t\Vert _{L^2}^2\big )dt\le C_0^\frac{1}{2}, \end{aligned}$$

(3.45)

provided that $C_0\le \varepsilon _5$.

Proof

For simplicity, we only prove the case $T>2$. Otherwise, the same thing can be done by choosing a suitably small step size. For integer $i\ (1\le i\le [T]-1)$, taking $\eta (t)=\sigma _i$ and integrating (3.16) over $(i-1, i+1]$, we derive

$$\begin{aligned}&\sup _{i-1\le t\le i+1}\big [\sigma _i\big (\Vert \nabla u\Vert _{L^2}^2 +\Vert \Delta d\Vert _{L^2}^2\big )\big ]+\int _{i-1}^{i+1}\sigma _i\big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2+\Vert \nabla d_t\Vert _{L^2}^2\big )dt\nonumber \\&\le \sigma _i\int (P-\bar{P})\textrm{div}\,udx+\sigma _i\int M(d):\nabla udx+C\int _{i-1}^{i+1}\big (\sigma _i+\sigma _i'\big )\Vert \nabla u\Vert _{L^2}^2dt\nonumber \\&\quad +C\int _{i-1}^{i+1}\sigma _i'\big (\Vert \nabla ^2 d\Vert _{L^2}^4+\Vert \nabla u\Vert _{L^2}^4+C_0^2\big )dt +C\int _{i-1}^{i+1}\sigma _i\Vert \nabla u\Vert _{L^3}^3dt\nonumber \\&\quad +C\int _{i-1}^{i+1}\sigma _i\big (\Vert \nabla u\Vert _{L^2}^4+\Vert \nabla ^2d\Vert _{L^2}^4+\Vert \nabla u\Vert _{L^2}^6+\Vert \nabla ^2d\Vert _{L^2}^6\big )dt +CC_0\nonumber \\&\le \frac{1}{2}\sup _{i-1\le t\le i+1}\big (\sigma _i\Vert \nabla u\Vert _{L^2}^2\big )+C\sup _{i-1\le t\le i+1}\big (\sigma _i\Vert \nabla d\Vert _{L^4}^4\big ) +C\int _{i-1}^{i+1}\Vert \nabla u\Vert _{L^2}^2dt\nonumber \\&\quad +C\sup _{i-1\le t\le i+1}\big (\Vert \nabla u\Vert _{L^2}^2 +\Vert \Delta d\Vert _{L^2}^2+\Vert \nabla d\Vert _{L^2}^2\big )\int _{i-1}^{i+1}\big (\Vert \nabla u\Vert _{L^2}^2+\Vert \nabla ^2d\Vert _{L^2}^2\big )dt\nonumber \\&\quad +C\sup _{i-1\le t\le i+1}\big (\Vert \nabla u\Vert _{L^2}^4 +\Vert \Delta d\Vert _{L^2}^4+\Vert \nabla d\Vert _{L^2}^4\big )\int _{i-1}^{i+1}\big (\Vert \nabla u\Vert _{L^2}^2+\Vert \nabla ^2d\Vert _{L^2}^2\big )dt\nonumber \\&\quad +C\sup _{i-1\le t\le i}\Vert \nabla u\Vert _{L^2}^2\int _{i-1}^i\Vert \nabla u\Vert _{L^2}^2dt+CC_0+\delta \int _{i-1}^{i+1}\sigma _i\big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2 +\Vert \nabla d_t\Vert _{L^2}^2\big )dt\nonumber \\&\le \frac{1}{2}\sup _{i-1\le t\le i+1}\big (\sigma _i\Vert \nabla u\Vert _{L^2}^2\big )+CC_0^\frac{1}{2}\sup _{i-1\le t\le i+1}\big (\sigma _i\Vert \Delta d\Vert _{L^2}^2\big )\nonumber \\&\quad +CC_0\big (A_1(T)+A_3(\sigma (T)+C_0)\big )+CC_0\big (A_1^2(T)+A_3^2(\sigma (T))+C_0^2\big )\nonumber \\&\quad +\delta \int _{i-1}^{i+1}\sigma _i\big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2} +\Vert \nabla d_t\Vert _{L^2}^2\big )dt+CC_0\nonumber \\&\le \frac{1}{2}\sup _{i-1\le t\le i+1}\big (\sigma _i\Vert \nabla u\Vert _{L^2}^2\big )+CC_0^\frac{1}{2}\sup _{i-1\le t\le i+1}\big (\sigma _i\Vert \Delta d\Vert _{L^2}^2\big )\nonumber \\&\quad +\delta \int _{i-1}^{i+1}\sigma _i\big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2} +\Vert \nabla d_t\Vert _{L^2}^2\big )dt+CC_0, \end{aligned}$$

(3.46)

where we have used (3.4), (3.5), and (3.12). Choosing $\delta$ and $C_0$ suitably small, we deduce from (3.46) that

$$\begin{aligned} \sup _{0\le t\le \sigma (T)}\big [\sigma \big (\Vert \nabla u\Vert _{L^2}^2 +\Vert \Delta d\Vert _{L^2}^2\big )\big ]\le CC_0\le C_0^\frac{1}{2}, \end{aligned}$$

(3.47)

due to $\sigma _1(t)=\sigma (t)$ and

$$\begin{aligned} \sup _{i\le t\le i+1}\big [\sigma _i\big (\Vert \nabla u\Vert _{L^2}^2 +\Vert \Delta d\Vert _{L^2}^2\big )\big ] +\int _{i-1}^{i+1}\sigma _i\big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2 +\Vert \nabla d_t\Vert _{L^2}^2\big )dt\le CC_0\le C_0^\frac{1}{2}, \end{aligned}$$

(3.48)

provided that $C_0\le \varepsilon _5$ is properly small. Note that the constant $C_3$ is independent of i. Thus, (3.45) follows from (3.47) and (3.48). $\square$

Lemma 3.6

Assume that Proposition 3.1 can be satisfied. There exists a positive constant $\varepsilon _6$ such that, for $\sigma (T)\le t_1<t_2\le T$,

$$\begin{aligned}&A_2(T)\le C_0^\frac{1}{2}, \end{aligned}$$

(3.49)

$$\begin{aligned} \sup _{0\le t\le T}\big [\sigma ^2\big (&\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2 +\Vert \nabla d_t\Vert _{L^2}^2\big )\big ]\le C_0^\frac{1}{2}, \end{aligned}$$

(3.50)

and

$$\begin{aligned} \int _{t_1}^{t_2}\sigma ^2\big (\Vert \nabla \dot{u}\Vert _{L^2}^2+\Vert d_{tt}\Vert _{L^2}^2\big )dt \le CC_0^\frac{1}{2}+CC_0(t_2-t_1), \end{aligned}$$

(3.51)

provided that $C_0\le \varepsilon _6$.

Proof

1. We obtain from Lemma 2.5 that

$$\begin{aligned}&\Vert \nabla u\Vert _{L^4}^4\le C\big (\Vert \rho \dot{u}\Vert _{L^2}+\Vert |\nabla d||\nabla ^2d|\Vert _{L^2}\big )^2\big (\Vert \nabla u\Vert _{L^2}\\&\quad +\Vert P-\bar{P}\Vert _{L^2}\big )^2+C\big (\Vert \nabla u\Vert _{L^2}^4+\Vert P-\bar{P}\Vert _{L^4}^4\big ), \end{aligned}$$

from which we arrive at

$$\begin{aligned} \int _{i-1}^{i+1}\sigma _i^2\Vert \nabla u\Vert _{L^4}^4dt&\le C\int _{i-1}^{i+1}\sigma _i^2\Big (\Vert \nabla u\Vert _{L^2}+C_0^\frac{1}{2}\Big )\big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2 +\Vert \nabla d\Vert _{L^3}^2\Vert \nabla ^2d\Vert _{L^6}^2\big )dt\\&\quad +C\int _{i-1}^{i+1}\sigma _i^2\big (\Vert \nabla u\Vert _{L^2}^4\big )dt+CC_0\\&\le C\sup _{i-1\le t\le i+1}\Big (\Vert \nabla u\Vert _{L^2}+C_0^\frac{1}{2}\Big )\int _{i-1}^{i+1}\sigma _i^2\big ( \Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2 +\Vert \nabla ^3d\Vert _{L^2}^2\big )dt\\&\quad +C\int _{i-1}^{i+1}\sigma _i^2(\Vert \nabla u\Vert _{L^2}^4+\Vert \nabla d\Vert _{H^1}^4)dt+CC_0\\&\le C\sup _{i-1\le t\le i+1}\Big (C+C_0^\frac{1}{2}\Big )\int _{i-1}^{i+1}\sigma _i^2\big ( \Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2 +\Vert \nabla d_t\Vert _{L^2}^2\big )dt\\&\quad +C\int _{i-1}^{i+1}\sigma _i^2\big (\Vert \nabla u\Vert _{L^2}^4+\Vert \Delta d\Vert _{L^2}^2+\Vert \nabla d\Vert _{L^2}^4\big )dt+CC_0\\&\le C\sup _{i-1\le t\le i+1}\Big (C+C_0^\frac{1}{2}\Big )\int _{i-1}^{i+1}\sigma _i^2\big ( \Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2 +\Vert \nabla d_t\Vert _{L^2}^2\big )dt+CC_0, \end{aligned}$$

and

$$\begin{aligned} \Vert P-\bar{P}\Vert _{L^4}\le C\Vert P-\bar{P}\Vert _{L^\infty }^\frac{1}{2}\Big (\int |P-\bar{P}|^2dx\Big )^\frac{1}{4}\le C(\hat{\rho })\Vert \rho -\bar{\rho }\Vert _{L^2}^\frac{1}{2}. \end{aligned}$$

(3.52)

From Lemma 2.7, (3.1) and Young’s inequality, one obtains that

$$\begin{aligned} \int _{\partial \Omega }\sigma _i^2(u\cdot \nabla n\cdot u)FdS&\le C\sigma _i^2\Vert |u|^2|F|\Vert _{W^{1,1}}\le C\sigma _i^2\Vert \nabla u\Vert _{L^2}^2\Vert F\Vert _{H^1}\\&\le \frac{1}{4}\sigma _i^2(\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2+\Vert \nabla d_t\Vert _{L^2}^2) +C\sigma _i^2\big (\Vert \nabla u\Vert _{L^2}^2+\Vert \nabla d\Vert _{H^1}^2\big )\\&\le \frac{1}{4}\sigma _i^2(\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2+\Vert \nabla d_t\Vert _{L^2}^2) +C\sigma _i^2\Vert \nabla ^2 d\Vert _{L^2}^2+C_0\\&\le \frac{1}{4}\sigma _i^2(\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2+\Vert \nabla d_t\Vert _{L^2}^2) +C\sigma _i^2\big (\Vert \Delta d\Vert _{L^2}^2+\Vert \nabla d\Vert _{L^2}^2\big )+C_0\\&\le \frac{1}{4}\sigma _i^2(\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2+\Vert \nabla d_t\Vert _{L^2}^2)+C_0. \end{aligned}$$

For any integer $1\le i\le [T]-1$, integrating (3.29) with $\eta (t)=\sigma _i^2$ over $(i-1, i+1]$, we get from (3.2), (3.7), (3.41) and Young’s inequality that

$$\begin{aligned}{} & {} \sup _{i-1\le t\le i+1}\big [\sigma _i^2\big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2 +\Vert \nabla d_t\Vert _{L^2}^2\big )\big ]+\int _{i-1}^{i+1}\sigma _i^2\big (\Vert \nabla \dot{u}\Vert _{L^2}^2 +\Vert d_{tt}\Vert _{L^2}^2\big )dt\nonumber \\{} & {} \le -\int _{\partial \Omega }\sigma _i^2(u\cdot \nabla n\cdot u)FdS\Big |_{i-1}^{i+1} +C\int _{i-1}^{i+1}\sigma _i^2\Vert \nabla u\Vert _{L^4}^4dt\nonumber \\{} & {} \quad +C\int _{i-1}^{i+1}\sigma _i\sigma _i'\big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2 +\Vert \nabla d_t\Vert _{L^2}^2+\Vert \nabla u\Vert _{L^2}^2 +\Vert \nabla ^2d\Vert _{L^2}^2+\Vert \nabla u\Vert _{L^2}^4+\Vert \nabla d\Vert _{H^1}^4\big )dt\nonumber \\{} & {} \quad +C\int _{i-1}^{i+1}\sigma _i^2\big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2\Vert \nabla u\Vert _{L^2}^2 +\Vert \nabla u\Vert _{L^2}^6+\Vert \nabla d\Vert _{H^1}^6\big )dt\nonumber \\{} & {} \quad +C\int _{i-1}^{i+1}\sigma _i^2\big (\Vert \nabla u\Vert _{L^2}^3+\Vert \nabla d\Vert _{H^1}^4+\Vert \nabla u\Vert _{L^2}^2 +\delta \Vert \nabla ^2d\Vert _{L^2}\Vert \nabla d_t\Vert _{L^2}^3\big )dt\nonumber \\{} & {} \quad +C\int _{i-1}^{i+1}\sigma _i^2\big (\Vert \nabla d\Vert _{H^1}^2\Vert \nabla d_t\Vert _{L^2}^2\nonumber \\{} & {} \quad +\Vert \nabla d\Vert _{H^1}^4\Vert d_t\Vert _{L^2}^2+\Vert \nabla u\Vert _{L^2}^2\Vert \nabla d_t\Vert _{L^2}^2\big )dt+CC_0\nonumber \\{} & {} \le \frac{1}{4}\sup _{i-1\le t\le i+1}\big [\sigma _i^2\big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2 +\Vert \nabla d_t\Vert _{L^2}^2\big )\big ]+CC_0^\frac{1}{2}+(C+C_0)\nonumber \\{} & {} \quad \int _{i-1}^{i+1}\sigma _i^2\big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2 +\Vert \nabla d_t\Vert _{L^2}^2\big )dt+CC_0\nonumber \\{} & {} \quad +\int _{i-1}^{i+1}\sigma _i\big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2 +\Vert \nabla d_t\Vert _{L^2}^2\big )dt+ \int _{i-1}^{i+1}\sigma _i\Vert \nabla d\Vert _{H^1}^4 dt\nonumber \\{} & {} \quad +C\int _{i-1}^{i+1}\sigma _i^2 (\Vert \nabla d\Vert _{H^2}^6+\Vert \nabla d\Vert _{H^2}^4)dt\nonumber \\{} & {} \quad +C\sup _{i-1\le t\le i+1}\sigma _i\Vert \nabla d_t\Vert _{L^2}\int _{i-1}^{i+1} \sigma _i \Vert \nabla d_t\Vert _{L^2}^2dt\nonumber \\{} & {} \le \frac{1}{4}\sup _{i-1\le t\le i+1}\big [\sigma _i^2\big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2 +\Vert \nabla d_t\Vert _{L^2}^2\big )\big ] \nonumber \\{} & {} \quad +C\sup _{i-1\le t\le i+1}\big [\sigma _i\big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2} +\Vert \nabla d_t\Vert _{L^2}\big )\big ] \int _{i-1}^{i+1}\sigma _i\big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2 +\Vert \nabla d_t\Vert _{L^2}^2\big )dt \nonumber \\{} & {} \quad +CC_0\sup _{i-1\le t\le i+1}\big [\sigma _i^2\big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2 +\Vert \nabla d_t\Vert _{L^2}^2\big )\big ]+CC_0^\frac{1}{2}\nonumber \\{} & {} \le \frac{1}{4}\sup _{i-1\le t\le i+1}\big [\sigma _i^2\big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2 +\Vert \nabla d_t\Vert _{L^2}^2\big )\big ]+CC_0^\frac{1}{2}\sup _{i-1\le t\le i+1}\big [\sigma _i\big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2} +\Vert \nabla d_t\Vert _{L^2}\big )\big ]\nonumber \\{} & {} \quad +CC_0\sup _{i-1\le t\le i+1}\big [\sigma _i^2\big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2 +\Vert \nabla d_t\Vert _{L^2}^2\big )\big ]+CC_0^\frac{1}{2}\nonumber \\{} & {} \le (\frac{1}{4}+C_0)\sup _{i-1\le t\le i+1}\big [\sigma _i^2\big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2 +\Vert \nabla d_t\Vert _{L^2}^2\big )\big ]+CC_0^\frac{1}{2}. \end{aligned}$$

(3.53)

According to the above inequality, we get that

$$\begin{aligned}{} & {} \sup _{i-1\le t\le i+1}\big [\sigma _i^2\big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2 +\Vert \nabla d_t\Vert _{L^2}^2\big )\big ]\nonumber \\{} & {} \quad +\int _{i-1}^{i+1}\sigma _i^2\big (\Vert \nabla \dot{u}\Vert _{L^2}^2 +\Vert d_{tt}\Vert _{L^2}^2\big )dt\le CC_0^\frac{1}{2}, \end{aligned}$$

(3.54)

so we have

$$\begin{aligned} A_2(T)+\sup _{i-1\le t\le i+1}\Vert \nabla d_t\Vert _{L^2}^2+\int _{i-1}^{i+1}\sigma _i^2\Vert d_{tt}\Vert _{L^2}^2\le CC_0^\frac{2}{3}\le C_0^\frac{1}{2}. \end{aligned}$$

(3.55)

Hence

$$\begin{aligned} A_2(T)\le C_0^\frac{1}{2}, \end{aligned}$$

(3.56)

which combined with (3.2) and (3.54) leads to (3.49). (3.50) can be derived by (3.53) directly.

2. From (3.2) and (3.7), we integrate (3.16) over $[t_1, t_2]\subseteq [\sigma (T), T]$ and take $\eta (t)=\sigma$ to obtain that

$$\begin{aligned}&\int _{t_1}^{t_2}\sigma \big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2+\Vert \nabla d_t\Vert _{L^2}^2\big )dt \nonumber \\&\quad \le C(C_0+A_1(T))+CC_0(t_2-t_1)+C\int _{t_1}^{t_2}\sigma \big (\Vert \nabla u\Vert _{L^2}^6\nonumber \\&\quad +\Vert \nabla d\Vert _{H^1}^6\big )dt +C\int _{t_1}^{t_2}\sigma \big (\Vert \nabla u\Vert _{L^2}^4+\Vert \nabla d\Vert _{H^1}^4\big )dt\nonumber \\&\quad \le CC_0^\frac{1}{2}+CC_0(t_2-t_1)+C\int _{t_1}^{t_2}\big (\Vert \nabla u\Vert _{L^2}^2 +\Vert \nabla d\Vert _{H^1}^2\big )dt\nonumber \\&\quad \le CC_0^\frac{1}{2}+CC_0(t_2-t_1). \end{aligned}$$

(3.57)

Similarly to (3.53), integrating (3.29) over $[t_1, t_2]$ and taking $\eta =\sigma ^2$, we find that

$$\begin{aligned}&\int _{t_1}^{t_2}\sigma ^2\big (\Vert \nabla \dot{u}\Vert _{L^2}^2+\Vert d_{tt}\Vert _{L^2}^2\big )dt\nonumber \\&\le C\sup _{t_1\le t\le t_2}\big [\sigma \big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2} +\Vert \nabla d_t\Vert _{L^2}\big )\big ] \int _{t_1}^{t_2}\sigma \big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2 +\Vert \nabla d_t\Vert _{L^2}^2\big )dt +CC_0^\frac{1}{2}\nonumber \\&\le CC_0^\frac{1}{2}+CC_0^\frac{1}{4}\int _{t_1}^{t_2} \sigma \big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2+\Vert \nabla d_t\Vert _{L^2}^2\big )dt\nonumber \\&\le CC_0^\frac{1}{2}+CC_0(t_2-t_1), \end{aligned}$$

(3.58)

owing to (3.2), (3.7), (3.49), and (3.57). The conclusion follows. $\square$

Consider the problem

$$\begin{aligned} {\left\{ \begin{array}{ll} {{\,\textrm{div}\,}}v=g,\quad &{}x\in \Omega ,\\ v=0,\quad &{}x\in \partial \Omega , \end{array}\right. } \end{aligned}$$

(3.59)

where $\Omega$ is a bounded domain in $\mathbb {R}^2$. Define linear operator $\mathcal {B} =[\mathcal {B}_1, \mathcal {B}_2]$, a bounded operator

$$\begin{aligned} \mathcal {B}:\left\{ f\in L^p(\Omega ):\bar{f}=0\right\} \longmapsto \big (W_0^{1,p}(\Omega )\big )^2, \end{aligned}$$

satisfying the properties: (1) For any $p\in (1,\infty )$, there holds

$$\begin{aligned} \Vert \mathcal {B}[f]\Vert _{W_0^{1,p}(\Omega )}\le C(p)\Vert f\Vert _{L^p(\Omega )}. \end{aligned}$$

(2) The function $v=\mathcal {B}[f]$ is the solution of the problem (3.59).

(3) if f can be written as $f={{\,\textrm{div}\,}}g$ with $g\in L^p(\Omega )$ satisfying $(g\cdot n)|_{\partial \Omega }=0$, then

$$\begin{aligned} \Vert \mathcal {B}[f]\Vert _{L^p(\Omega )}\le C(p)\Vert g\Vert _{L^p(\Omega )}, \end{aligned}$$

(3.60)

for any $p\in (1,\infty )$.

Lemma 3.7

Under the assumptions of Proposition 3.1, there exists a positive constant $C=C(\hat{\rho })$ such that

$$\begin{aligned} \int _{0}^{T}\int |P-{\bar{P}}|^2dxdt\le CC_0. \end{aligned}$$

(3.61)

Proof

Multiplying (1.2) by $\mathcal {B}[P-\bar{P}]$ and integrating the resultant over $\Omega$, we obtain that

$$\begin{aligned}&\int |P-\bar{P}|^2dx\nonumber \\&\quad =\big (\int \rho u\cdot \mathcal {B}[P-\bar{P}]dx\big )_t-\int \rho u\nonumber \\&\quad \cdot \nabla \mathcal {B}[P-\bar{P}] dx-\int \rho u\cdot \mathcal {B}[P_t-\bar{P}_t] dx\nonumber \\&\quad +\mu \int \nabla u\dot{\nabla }\mathcal {B}[P-\bar{P}] dx+(\mu +\lambda )\int |P-{\bar{P}}| {{\,\textrm{div}\,}}udx\nonumber \\&\quad -\int M(d)\cdot \nabla \mathcal {B}[P-\bar{P}]dx\nonumber \\&\quad \le \big (\int \rho u\cdot \mathcal {B}[P-\bar{P}]dx\big )_t\nonumber \\&\quad +C\Vert u\Vert _{L^4}^2\Vert P-\bar{P}\Vert _{L^2}+C\Vert u\Vert _{L^2}\Vert \nabla u\Vert _{L^2}\nonumber \\&\quad +C\Vert P-\bar{P}\Vert _{L^2}\Vert \nabla u\Vert _{L^2}+C\Vert M(d)\Vert _{L^2}\Vert P-\bar{P}\Vert _{L^2}\nonumber \\&\quad \le \big (\int \rho u\cdot \mathcal {B}[P-\bar{P}]dx\big )_t+\delta \Vert P-\bar{P}\Vert _{L^2}^2+C\Vert \nabla u\Vert _{L^2}^2+CC_0^\frac{1}{2}\Vert M(d)\Vert _{L^2}, \end{aligned}$$

(3.62)

where in the last inequality we have used (3.5) and

$$\begin{aligned} \Vert \mathcal {B}[P_t-\bar{P}_t]\Vert _{L^2}=\Vert \mathcal {B}[{{\,\textrm{div}\,}}(Pu)]+(\gamma -1)\mathcal {B}[P{{\,\textrm{div}\,}}u-\overline{P{{\,\textrm{div}\,}}u}]\Vert _{L^2}\le C\Vert \nabla u\Vert _{L^2}. \end{aligned}$$

Thus, (3.61) can be deduced by Lemma 3.1 and the properties of $\mathcal {B}$. $\square$

With above Lemmas at hand, the uniform upper bound of the density can be derived, which plays crucial roles to get all the higher-order estimates and thus to extend the local solution globally in time.

Lemma 3.8

There exists a positive constant $\varepsilon$ as in Theorem 1.1 such that if $(\rho , u, d)$ is a strong solution of (1.1)–(1.5) in $\Omega \times (0, T]$ satisfying (3.2), then

$$\begin{aligned} \sup _{0\le t\le T}\Vert \rho (t)\Vert _{L^\infty }\le \frac{3}{2}\hat{\rho }, \end{aligned}$$

(3.63)

provided that $C_0\le \varepsilon \triangleq \min \{1, \varepsilon _2, \varepsilon _3, \varepsilon _4, \varepsilon _5, \varepsilon _6, \varepsilon _7\}$.

Proof

1. We rewrite (1.1)$_1$ as

$$\begin{aligned} D_t\rho =g(\rho )+b'(t), \end{aligned}$$

(3.64)

where

$$\begin{aligned} D_t\rho =\rho _t+u\cdot \nabla \rho , \quad g(\rho )=-\frac{\rho P}{2\mu +\lambda }, \quad b(t)=-\frac{1}{2\mu +\lambda }\int _0^t(\rho \bar{P}-\rho F)d\tau . \end{aligned}$$

For $t\in [0, \sigma (T)]$, by defining $r\ge 4$, we deduce from Hölder’s inequality, Lemma 2.7, and (3.41) that, for $0\le t_1<t_2\le \sigma (T)$,

$$\begin{aligned} |b(t_2)-b(t_1)|&=\frac{1}{\lambda +2\mu }\Big |\int _0^{\sigma (T)}(\rho \bar{P}-\rho F)dt\Big |\\&\le C(\hat{\rho })\int _0^{\sigma (T)}\Vert F\Vert _{L^\infty }dt+CC_0\\&\le C\int _0^{\sigma (T)}\Vert F\Vert _{L^2}^\frac{r-2}{2(r-1)}\Vert \nabla F\Vert _{L^r}^\frac{r}{2(r-1)}dt +CC_0\\&\le C\int _{0}^{\sigma (T)}(\Vert \nabla u\Vert _{L^2}^2+\Vert P-\bar{P}\Vert _{L^2}^2)^\frac{r-2}{4(r-1)}\Vert \rho \dot{u}\Vert _{L^r}^\frac{r}{2(r-1)}dt\\&\quad +C\int _{0}^{\sigma (T)}(\Vert \nabla u\Vert _{L^2}^2+\Vert P-\bar{P}\Vert _{L^2}^2)^\frac{r-2}{4(r-1)}\Vert |\nabla d||\nabla ^2 d|\Vert _{L^r}^\frac{r}{2(r-1)}dt+CC_0, \end{aligned}$$

from which, due to $\frac{2r}{3r-4}\in (0,1)$ and (3.2), one has that

$$\begin{aligned}&\int _{0}^{\sigma (T)}(\Vert \nabla u\Vert _{L^2}^2+\Vert P-\bar{P}\Vert _{L^2}^2)^\frac{r-2}{4(r-1)}\Vert \rho \dot{u}\Vert _{L^r}^\frac{r}{2(r-1)}dt\\&\le C\int _{0}^{\sigma (T)} \Vert \dot{u}\Vert ^\frac{r}{2(r-1)}_{L^r}dt\\&\le C\int _{0}^{\sigma (T)}(\Vert \nabla \dot{u}\Vert _{L^2}^2+\Vert \nabla u\Vert _{L^2}^4)^\frac{r}{4(r-1)} dt\\&\le C\int _{0}^{\sigma (T)}\big (t^2(\Vert \nabla \dot{u}\Vert ^2_{L^2}+\Vert \nabla u\Vert ^4_{L^2})\big )^\frac{r}{4(r-1)}\cdot t^{-\frac{r}{2(r-1)}}dt\\&\le C\left( \int _{0}^{\sigma (T)}\sigma ^2\Vert \nabla \dot{u}\Vert ^2_{L^2}+\sigma ^2\Vert \nabla u\Vert ^4_{L^2}dt\right) ^\frac{r}{4(r-1)}\left( \int _{0}^{\sigma (T)} \sigma ^\frac{-2r}{3r-4}dt\right) ^\frac{3r-4}{4(r-1)}\\&\le CC_0^\frac{r}{16(r-1)}. \end{aligned}$$

Taking the advantage of Lemma 2.7, (2.5), Lemma 3.1, Lemma 3.4, (3.61), Hölder’s inequality, Young’s inequality, and Lemma 2.10, we find that

$$\begin{aligned}&\int _{0}^{\sigma (T)}(\Vert \nabla u\Vert _{L^2}^2+\Vert P-\bar{P}\Vert _{L^2}^2)^\frac{r-2}{4(r-1)}\Vert |\nabla d||\nabla ^2 d|\Vert _{L^r}^\frac{r}{2(r-1)}dt\\&\quad \le C\int _{0}^{\sigma (T)}\big (\Vert \nabla u\Vert _{L^2}^2+\Vert P-\bar{P}\Vert _{L^2}^2\big )^ \frac{r-2}{4(r-1)}\\&\quad \Big (\Vert \nabla d\Vert _{L^2}^\frac{2}{r-1}\Vert \nabla ^2 d\Vert _{L^2}^\frac{5r-4}{2(r-1)}\Vert \nabla ^3 d\Vert _{L^2}^\frac{r-4}{2(r-1)}+\Vert \nabla d\Vert _{L^2}^\frac{2}{r-1}\Vert \nabla ^2 d\Vert _{L^2}^\frac{r-2}{r-1}\Big )dt\\&\quad \le C\int _{0}^{\sigma (T)}\big (\Vert \nabla u\Vert _{L^2}^2+\Vert P-\bar{P}\Vert _{L^2}^2\big )^ \frac{r-2}{4(r-1)}\big (\Vert \nabla d\Vert _{L^2}^\frac{2}{r-1}\Vert \nabla ^2 d\Vert _{L^2}^\frac{5r-4}{2(r-1)}\big )\\&\quad +C\big (\Vert \nabla u\Vert _{L^2}^2+\Vert P-\bar{P}\Vert _{L^2}^2\big )^\frac{r-2}{4(r-1)}\Vert \nabla d\Vert _{L^2}^\frac{2}{r-1}\Vert \nabla ^2 d\Vert _{L^2}^\frac{r}{2(r-1)}\Vert \nabla d_t\Vert _{L^2}^\frac{r-4}{2(r-1)}\\&\quad +C\big (\Vert \nabla u\Vert _{L^2}^2+\Vert P-\bar{P}\Vert _{L^2}^2\big )^\frac{r-2}{4(r-1)}\Vert \nabla d\Vert _{L^2}^\frac{2}{r-1}\Vert \nabla ^2 d\Vert _{L^2}^\frac{r-2}{r-1}dt\\&\quad \le C\int _{0}^{\sigma (T)}\Big (\Vert \nabla u\Vert _{L^2}^\frac{r-2}{2(r-1)}+\Vert P-\bar{P}\Vert _{L^2}^\frac{r-2}{2(r-1)}\Big )C_0^\frac{1}{r-1}dt\\&\quad +C\int _{0}^{\sigma (T)}\big (\Vert \nabla u\Vert _{L^2}^{2}+\Vert P-\bar{P}\Vert _{L^2}^2\big )^\frac{r-2}{4(r-1)}\Vert \nabla d\Vert _{L^2}^ \frac{2}{r-1}\Vert \nabla d_t\Vert _{L^2}^\frac{r-4}{2(r-1)} dt\\&\quad \le C\int _{0}^{\sigma (T)}\Big (\Vert \nabla u\Vert _{L^2}^\frac{r-2}{2(r-1)} C_0^\frac{1}{r-1}+\Vert P-\bar{P}\Vert _{L^2}^\frac{r-2}{2(r-1)}C_0^\frac{1}{r-1}\Big )dt\\&\quad +\left( \int _{0}^{\sigma (T)}\Vert \nabla d_t\Vert _{L^2}^2dt\right) ^\frac{r-4}{4(r-1)}\left( \int _{0}^{\sigma (T)} \Big [(\Vert \nabla u\Vert _{L^2}^{2}\right. \\&\left. \quad +\Vert P-\bar{P}\Vert _{L^2}^2)^\frac{r-2}{4(r-1)}\Vert \nabla d\Vert _{L^2}^\frac{2}{r-2}\Big ]^\frac{4(r-1)}{3r}dt\right) ^\frac{3r}{4(r-1)}\\&\quad \le C\left( \int _{0}^{\sigma (T)}\big (\Vert \nabla u\Vert _{L^2}^2+\Vert P-\bar{P}\Vert _{L^2}^2\big )dt\right) ^\frac{r-2}{4(r-1)}\left( \int _{0}^{\sigma (T)}C_0^\frac{4}{3r-2}dt\right) ^\frac{3r-2}{4(r-1)}\\&\quad +K\left( \int _{0}^{\sigma (T)}\Big [(\Vert \nabla u\Vert _{L^2}^2+\Vert P-\bar{P}\Vert _{L^2}^2)^\frac{r-2}{4(r-1)}\Vert \nabla d\Vert _{L^2}^\frac{2}{r-2}\Big ]^\frac{4(r-1)}{3r}dt\right) ^\frac{3r}{4(r-1)}\\&\quad \le CC_0^\frac{r+2}{4(r-1)}+K\left( \int _{0}^{\sigma (T)}\big (\Vert \nabla u\Vert _{L^2}^2+\Vert P-\bar{P}\Vert _{L^2}^2\big )^\frac{r-2}{3r}\cdot \Vert \nabla d\Vert _{L^2} ^\frac{8(r-1)}{3r(r-2)}dt\right) ^\frac{3r}{4(r-1)}\\&\quad \le CC_0^\frac{r+2}{4(r-1)} +\left( \int _{0}^{\sigma (T)}\big (\Vert \nabla u\Vert _{L^2}^2+\Vert P-\bar{P}\Vert _{L^2}^2\big )dt\right) ^ {\frac{r-2}{3r}\cdot \frac{3r}{4(r-1)}}\\&\quad \cdot \left( \int _{0}^{\sigma (T)}\Vert \nabla d\Vert _{L^2}^\frac{4(r-1)}{(r-2)(r+1)}dt\right) ^{\frac{2(r+1)}{3r}\cdot \frac{3r}{4(r-1)}}\\&\quad \le CC_0^\frac{r+2}{4(r-1)}+CC_0^\frac{r^2}{4(r-1)(r-2)}\\&\quad \le CC_0^\frac{1}{4}, \end{aligned}$$

owing to

$$\begin{aligned}&\Vert |\nabla d||\nabla ^2 d|\Vert _{L^r}^\frac{r}{2(r-1)}\le \big [\Vert \nabla d\Vert _{L^{\frac{r}{2}}}\Vert \nabla ^2 d\Vert _{L^{\frac{r}{2}}}\big ]^\frac{r}{2(r-1)}\\&\quad \le C\big (\Vert \nabla d\Vert _{L^2}^\frac{4}{r}\Vert \nabla ^2 d\Vert _{L^2}^\frac{r-4}{r}\big )^\frac{r}{2(r-1)}\big (\Vert \nabla ^2 d\Vert _{L^2}^\frac{4}{r}\Vert \nabla ^3 d\Vert _{L^2}^\frac{r-4}{r}+\Vert \nabla ^2 d\Vert _{L^2}\big )^\frac{r}{2(r-1)}\\&\quad \le \Vert \nabla d\Vert _{L^2}^\frac{2}{r-1}\Vert \nabla ^2 d\Vert _{L^2}^\frac{5r-4}{2(r-1)}\Vert \nabla ^3 d\Vert _{L^2}^\frac{r-4}{2(r-1)}+C\Vert \nabla d\Vert _{L^2}^\frac{2}{r-1}\Vert \nabla ^2 d\Vert _{L^2}^\frac{r-2}{r-1}. \end{aligned}$$

Hence, we have $|b(t_2)-b(t_1)|\le CC_0^\frac{1}{16}$, so we can choose $N_1=0$, $N_0=CC_0^\frac{1}{16}$, and $\xi _0=\hat{\rho }$. Thus, for all $\xi \ge \xi _0=\hat{\rho }$, we have that

$$\begin{aligned} g(\xi )=-\frac{a\xi ^{\gamma +1}}{2\mu +\lambda }\le -N_1=0. \end{aligned}$$

Combining the above results with (2.9) gives that

$$\begin{aligned} \sup _{0\le t\le \sigma (T)}\Vert \rho \Vert _{L^\infty }\le \max \{\hat{\rho },\xi _0\}+N_0\le \hat{\rho }+CC_0^\frac{1}{7}\le \frac{3}{2}\hat{\rho }. \end{aligned}$$

2. For $t\in [\sigma (T),T]$, we infer from Lemma 2.10, Lemma 2.7, (3.45), Lemma 3.6, and (3.61) that

$$\begin{aligned}&|b(t_2)-b(t_1)|\le C(\hat{\rho })\int _{t_1}^{t_2}\Vert F\Vert _{L^\infty }dt+\int _{t_1}^{t_2}\rho \bar{P}dt\le \frac{(a+C(\hat{\rho })C_0)\hat{\rho }^{\gamma +1}}{2(\lambda +2\mu )}(t_2-t_1)\\&\qquad +C(\hat{\rho })\int _{t_1}^{t_2}\Vert F\Vert _{L^\infty }^3dt\le \frac{(a+C(\hat{\rho })C_0)\hat{\rho }^{\gamma +1}}{2(\lambda +2\mu )}(t_2-t_1)+C(\hat{\rho })\int _{t_1}^{t_2}\Vert F\Vert _{L^2}\Vert \nabla F\Vert _{L^4}^2dt\\&\quad \le \frac{(a+C(\hat{\rho })C_0)\hat{\rho }^{\gamma +1}}{2(\lambda +2\mu )}(t_2-t_1)+C\int _{t_1}^{t_2}(\Vert \nabla u\Vert _{L^2}+\Vert P-\bar{P}\Vert _{L^2})(\Vert \sqrt{\rho }\dot{u}\Vert _{L^4} +\Vert |\nabla d||\nabla ^2 d|\Vert _{L^4})^2dt\\&\quad \le \frac{(a+C(\hat{\rho })C_0)\hat{\rho }^{\gamma +1}}{2(\lambda +2\mu )}(t_2-t_1)+C\int _{t_1}^{t_2} \big (\Vert \nabla u\Vert _{L^2}+\Vert P-\bar{P}\Vert _{L^2}\big ) \big (\Vert \nabla \dot{u}\Vert _{L^2}^2+\Vert \nabla u\Vert _{L^2}^4+\Vert \nabla ^2 d\Vert _{L^2}^4\\&\quad \quad +\Vert \nabla ^2 d\Vert _{L^2}^\frac{7}{3}\Vert \nabla ^3 d\Vert _{L^2}^\frac{5}{3}\big )dt \le \frac{(a+C(\hat{\rho })C_0)\hat{\rho }^{\gamma +1}}{2(\lambda +2\mu )}(t_2-t_1)\\&\quad \quad +C\int _{t_1}^{t_2}\sigma ^2\Vert \nabla \dot{u}\Vert _{L^2}^2dt+C\int _{t_1}^{t_2}\Vert \nabla u\Vert _{L^2}^2dt+\int _{t_1}^{t_2}\Vert P-\bar{P}\Vert _{L^2}^2dt+\int _{t_1}^{t_2}\Big (\Vert \nabla d_t\Vert _{L^2}^\frac{5}{3}\\&\quad \quad +C\Vert \nabla d\Vert _{H^1}^\frac{5}{3}+C\Vert \nabla u\Vert _{L^2}^\frac{5}{3}\Vert \nabla ^2d\Vert _{L^2}^\frac{5}{3}+\Vert \nabla u\Vert _{L^2}^\frac{10}{3}\Vert \nabla ^2d\Vert _{L^2}^\frac{5}{3}\Big )\cdot \big (\Vert \nabla u\Vert _{L^2}+\Vert P-\bar{P}\Vert _{L^2}\big ) dt\\&\quad \quad \le \frac{(a+C(\hat{\rho })C_0)\hat{\rho }^{\gamma +1}}{2(\lambda +2\mu )}(t_2-t_1) +C\int _{t_1}^{t_2}\sigma ^2\Vert \nabla \dot{u}\Vert _{L^2}^2dt+C\int _{t_1}^{t_2}\Vert \nabla u\Vert _{L^2}^2dt\\&\quad \quad +C\int _{t_1}^{t_2}\Vert P-\bar{P}\Vert _{L^2}^2dt+C\left( \int _{t_1}^{t_2}\Vert \nabla d\Vert _{H^1}^\frac{10}{3}dt\right) ^\frac{1}{2}\left( \int _{t_1}^{t_2}\big (\Vert \nabla u\Vert _{L^2}^2+\Vert P-\bar{P}\Vert _{L^2}^2\big )dt\right) ^\frac{1}{2}\\&\quad \quad +\left( \int _{t_1}^{t_2}\Vert \nabla ^2 d\Vert _{L^2}^\frac{10}{3}dt\right) ^\frac{1}{2}\left( \int _{t_1}^{t_2}\big (\Vert \nabla u\Vert _{L^2}^2+\Vert P-\bar{P}\Vert _{L^2}^2\big )dt\right) ^\frac{1}{2}\\&\quad \quad +C\left( \int _{t_1}^{t_2}\big (\Vert \nabla u\Vert _{L^2}^6+\Vert P-\bar{P}\Vert _{L^2}^6\big )dt\right) ^\frac{1}{6} \left( \int _{t_1}^{t_2}\sigma ^2\Vert \nabla d_t\Vert _{L^2}^2dt\right) ^\frac{5}{6}\\&\quad \le \big (\frac{(a+C(\hat{\rho })C_0)\hat{\rho }^{\gamma +1}}{(\lambda +2\mu )}\big )(t_2-t_1)+ CC_0^\frac{1}{2}, \end{aligned}$$

where we have used

$$\begin{aligned} \Vert |\nabla d||\nabla ^2 d|\Vert _{L^4}&\le C\Vert \nabla d\Vert _{L^6}\Vert \nabla ^2 d\Vert _{L^{12}}\\&\le C\Vert \nabla ^2 d\Vert _{L^2}(\Vert \nabla ^2 d\Vert _{L^2}^\frac{1}{6}\Vert \nabla ^3 d\Vert _{L^2}^\frac{5}{6}+\Vert \nabla ^2 d\Vert _{L^2})\\&\le C\Vert \nabla ^2 d\Vert _{L^2}^\frac{7}{6}\Vert \nabla ^3 d\Vert _{L^2}^\frac{5}{6}+\Vert \nabla ^2 d\Vert _{L^2}^2, \end{aligned}$$

and

$$\begin{aligned}&\Vert \nabla ^2 d\Vert _{H^1}\le C\Vert \nabla ^3 d\Vert _{L^2}^2 \le C\Vert \nabla d_t\Vert _{L^2}^2+C\Vert \nabla d\Vert _{H^1}^2\nonumber \\&\quad +C\Vert \nabla u\Vert _{L^2}^2\Vert \nabla ^2d\Vert _{L^2}^2 +C\Vert \nabla u\Vert _{L^2}^4\Vert \nabla ^2d\Vert _{L^2}^2. \end{aligned}$$

(3.65)

Hence, we can view $CC_0^\frac{1}{2}$ as $N_0$ and $\frac{(a+C(\hat{\rho })C_0)\hat{\rho }^{\gamma +1}}{(\lambda +2\mu )}$ as $N_1$, and we choose $\xi _0=\hat{\rho }$.

For all $\xi \ge \xi _0=\hat{\rho }$, we have

$$\begin{aligned} g(\xi )=-\frac{a\xi ^{1+\gamma }}{2\mu +\lambda }\le -\frac{(a+CC_0)\hat{\rho }^{\gamma +1}}{\lambda +2\mu }=-N_1, \end{aligned}$$

where $C_0$ is suitably small satisfying

$$\begin{aligned} (a+CC_0)\hat{\rho }^{\gamma +1}\le a\xi ^{\gamma +1}. \end{aligned}$$

(3.66)

By Zlotnik’s inequality, it gives that

$$\begin{aligned} \sup _{\sigma (T)\le t\le T}\Vert \rho \Vert _{L^\infty }\le \max \{\hat{\rho }, \xi _0\}+N_0\le \hat{\rho }+CC_0^\frac{1}{16}\le \frac{3\hat{\rho }}{2}. \end{aligned}$$

This finishes the proof of Lemma 3.8. $\square$

Lemma 3.9

Let the assumptions of Proposition 3.1 be satisfied. Then there exists a positive constant $\varepsilon _7$ such that

$$\begin{aligned} \sup _{0\le t\le \sigma (T)}\big [\sigma \big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2+ \Vert \nabla d_t\Vert _{L^2}^2\big )\big ]+\int _0^{\sigma (T)}\sigma \big (\Vert \nabla \dot{u}\Vert _{L^2}^2+\Vert d_{tt}\Vert _{L^2}^2\big )dt\le C, \end{aligned}$$

(3.67)

provided that $E_0\le \varepsilon _7$.

Proof

Taking $\eta (t)=\sigma$ and integrating (3.29) over $[0, \sigma (T)]$, we get from (3.2), (3.7), (3.41), and Young’s inequality that

$$\begin{aligned}&\sup _{0\le t\le \sigma (T)}\big [\sigma \big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2 +\Vert \nabla d_t\Vert _{L^2}^2\big )\big ] +\int _0^{\sigma (T)}\sigma (\Vert \nabla \dot{u}\Vert _{L^2}^2+\Vert d_{tt}\Vert _{L^2}^2)dt\\&\le C\int _0^{\sigma (T)}\sigma '\big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2+\Vert \nabla d_t\Vert _{L^2}^2\big )dt+CC_0^\frac{1}{2}\\&\quad +C\int _0^{\sigma (T)}\sigma \big (\Vert \nabla u\Vert _{L^2}+\Vert \Delta d\Vert _{L^2}+E_0^\frac{1}{2}+\delta \big ) \big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2 +\Vert \nabla d_t\Vert _{L^2}^2\big )dt\\&\le \sup _{0\le t\le \sigma (T)}\big [\sigma \big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2 +\Vert \nabla d_t\Vert _{L^2}^2\big )\big ]+C, \end{aligned}$$

from which, the conclusion follows. $\square$

Now we are ready to prove Proposition 3.1.

Proof of Proposition 3.1

Proposition 3.1 follows from Lemma 3.4, Lemma 3.5, Lemma 3.6, and Lemma 3.8. $\square$

3.2 Higher-Order Estimates

In this subsection, we will establish the time-dependent higher-order estimates of solutions. It is denoted by C or $C_i\ (i=1, 2,\ldots )$ the various positive constants, which may depend on the initial data, $\mu$, $\lambda$, $\gamma$, a, $\hat{\rho }$, $\Omega$, $M_1$, $M_2$, $\bar{\rho }$, and T as well.

Lemma 3.10

Under the conditions of Theorem 1.1, it holds that

$$\begin{aligned}&\sup _{0\le t\le T}\Vert \nabla \rho \Vert _{L^q}+\int _0^T\Vert \nabla u\Vert _{L^\infty } dt\le C, \end{aligned}$$

(3.68)

$$\begin{aligned}&\sup _{0\le t\le T}\big [t\big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2+ \Vert \nabla d_t\Vert _{L^2}^2\big )\big ]+\int _0^Tt\big (\Vert \nabla \dot{u}\Vert _{L^2}^2+\Vert d_{tt}\Vert _{L^2}^2\big )dt\le C, \end{aligned}$$

(3.69)

$$\begin{aligned}&\sup _{0\le t\le T}\big [t\big (\Vert u\Vert _{H^2}^2+\Vert \nabla ^3d\Vert _{L^2}^2\big )\big ] +\int _0^Tt\big (\Vert \nabla ^2d_t\Vert _{L^2}^2+\Vert \nabla ^4d\Vert _{L^2}^2\big )dt\le C. \end{aligned}$$

(3.70)

Proof

First, based on Lemma 2.8, (3.68) can be derived by the standard estimates.

Taking the advantage of (3.9), it is not hard to get that

$$\begin{aligned} \sup _{0\le t\le \sigma (T)}\big [t\big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2+\Vert \nabla d_t\Vert _{L^2}^2\big )\big ]+\int _{0}^{\sigma (T)}t\big (\Vert \nabla \dot{u}\Vert _{L^2}^2dx+\Vert d_{tt}\Vert _{L^2}^2\big )dt\le CC_0^\frac{2}{3}, \end{aligned}$$

which together with (3.51) leads to (3.69).

Next, it follows from Lemma 2.4, Lemma 2.7, and (2.11) that

$$\begin{aligned} \Vert \nabla ^2u\Vert _{L^2}&\le C\big (\Vert {{\,\textrm{div}\,}}u\Vert _{H^1}+\Vert {{\,\textrm{curl}\,}}u\Vert _{H^1}\big )\\&\le C\big (\Vert F+P-\bar{P}\Vert _{H^1}+\Vert {{\,\textrm{curl}\,}}u\Vert _{H^1}\big )\\&\le C\big (\Vert \rho \dot{u}\Vert _{L^2}+\Vert |\nabla d||\nabla ^2d|\Vert _{L^2}+\Vert \nabla P\Vert _{L^2}+\Vert P-\bar{P}\Vert _{L^2}+\Vert \nabla u\Vert _{L^2}\big ), \end{aligned}$$

which combined with (3.69), (3.28), (3.5), (3.6), and (3.2) gives that

$$\begin{aligned} \sup _{0\le t\le T}\big [t\big (\Vert u\Vert _{H^2}^2+\Vert \nabla ^3d\Vert _{L^2}^2\big )\big ] \le C. \end{aligned}$$

(3.71)

Furthermore, taking the operator $\nabla$ to (3.26) yields that

$$\begin{aligned} -\nabla ^2\Delta d=\nabla ^2(|\nabla d|^2d-u\cdot \nabla d-d_t). \end{aligned}$$

(3.72)

Applying the standard $L^2$-estimate to (3.72), we derive that

$$\begin{aligned}&\Vert \nabla ^4d\Vert _{L^2}^2 \le C\big (\Vert \nabla ^2d_t\Vert _{L^2}^2+\Vert \nabla ^2(u\cdot \nabla d)\Vert _{L^2}^2 +\Vert \nabla ^2(|\nabla d|^2d)\Vert _{L^2}^2\big )+C\Vert \nabla d\Vert _{H^1}^2\\&\quad \le C\Vert \nabla ^2d_t\Vert _{L^2}^2+C\Vert u\Vert _{L^\infty }^2\Vert \nabla ^3d\Vert _{L^2}^2 +C\Vert \nabla d\Vert _{L^\infty }^2\Vert \nabla ^2u\Vert _{L^2}^2+C\Vert \nabla u\Vert _{L^6}^2\Vert \nabla ^2d\Vert _{L^3}^2\\&\qquad +C\Vert \nabla ^2d\Vert _{L^4}^4+C\Vert \nabla d\Vert _{L^6}^2\Vert \nabla ^3d\Vert _{L^3}^2 +C\Vert \nabla d\Vert _{L^2}^2\Vert \nabla ^2d\Vert _{L^2}^2\Vert \nabla ^2d\Vert _{L^6}^2+C\\&\quad \le C\Vert \nabla ^2d_t\Vert _{L^2}^2+C\Vert \nabla u\Vert _{H^1}^2\Vert \nabla ^2d\Vert _{H^1}^2+C\Vert \nabla ^2d\Vert _{L^2}\Vert \nabla ^3d\Vert _{L^2}^3 +C\Vert \nabla ^3d\Vert _{L^2}^2\\&\qquad +C\Vert \nabla d\Vert _{H^1}^2\big (\Vert \nabla ^3d\Vert _{L^2}\Vert \nabla ^4d\Vert _{L^2}+\Vert \nabla ^3d\Vert _{L^2}^2\big ) +C\\&\quad \le \frac{1}{2}\Vert \nabla ^4d\Vert _{L^2}^2+C\Vert \nabla d\Vert _{H^1}^2\Vert \nabla ^3d\Vert _{L^2}^2+C\Vert \nabla u\Vert _{H^1}^2\Vert \nabla ^2d\Vert _{H^1}^2 +C\Vert \nabla ^3d\Vert _{L^2}^2\\&\quad +C\Vert \nabla ^2d\Vert _{L^2}\Vert \nabla ^3d\Vert _{L^2}^3+C, \end{aligned}$$

which leads to

$$\begin{aligned}&`\Vert \nabla ^4d\Vert _{L^2}^2\le C\Vert \nabla d\Vert _{H^1}^2\Vert \nabla ^3d\Vert _{L^2}^2+C\Vert \nabla u\Vert _{H^1}^2\Vert \nabla ^2d\Vert _{H^1}^2 +C\Vert \nabla ^3d\Vert _{L^2}^2\\&\quad +C\Vert \nabla ^2d\Vert _{L^2}\Vert \nabla ^3d\Vert _{L^2}^3+C. \end{aligned}$$

This along with (3.71), (3.69), (3.28), (3.2), (3.5), (3.41), and (3.31) implies (3.70). $\square$

Lemma 3.11

Under the assumptions of Theorem 1.1, it holds that

$$\begin{aligned} \sup _{0\le t\le T}\big (t\Vert \sqrt{\rho }u_t\Vert _{L^2}^2\big )+\int _0^Tt\Vert \nabla u_t\Vert _{L^2}^2dt\le C. \end{aligned}$$

Proof

By Lemma 3.10, Proposition 3.1, (2.5), and Sobolev’s inequality, we deduce that

$$\begin{aligned} t\Vert \sqrt{\rho }u_t\Vert _{L^2}^2&\le t\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2+t\Vert \sqrt{\rho }u\cdot \nabla u\Vert _{L^2}^2\\&\le C(T)+Ct\Vert u\Vert _{L^6}^2\Vert \nabla u\Vert _{L^4}^2\\&\le C(T)+C\Vert \nabla u\Vert _{L^2}^2\big (t\Vert u\Vert _{H^2}^2\big )\\&\le C, \end{aligned}$$

and

$$\begin{aligned} \int _0^Tt\Vert \nabla u_t\Vert _{L^2}^2dt&\le \int _0^Tt\Vert \nabla \dot{u}\Vert _{L^2}^2dt +\int _0^Tt\Vert \nabla (u\cdot \nabla u)\Vert _{L^2}^2dt\\&\le C(T)+\int _0^Tt\big (\Vert \nabla u\Vert _{L^4}^4+\Vert u\Vert _{L^\infty }^2\Vert \nabla ^2u\Vert _{L^2}^2\big )dt\\&\le C(T)+C\int _0^Tt\Vert \nabla u\Vert _{H^1}^2\Vert u\Vert _{H^2}^2dt\\&\le C. \end{aligned}$$

The conclusion of Lemma 3.11 follows. $\square$

4 Proof of Theorem 1.1

With all the a priori estimates in Sect. 3 at hand, we now prove Theorem 1.1.

Proof of Theorem 1.1

By Lemma 2.1, there exists a $T_*>0$ such that the system (1.1)–(1.5) has a unique classical solution $(\rho , u, d)$ in $\Omega \times (0, T_*]$. According to the definition of (3.1) and (1.9), we obtain that

$$\begin{aligned} 0\le \rho _0\le \hat{\rho },\quad A_1(0)=0, \quad A_2(0)=0, \quad A_3(0)\le K. \end{aligned}$$

To this end, there exists a $T_1\in (0, T_*]$ such that

$$\begin{aligned} 0\le \rho _0\le 2\hat{\rho },\quad A_1(T_1)\le 2C_0^\frac{1}{2},\quad A_2(T_1)\le 2C_0^\frac{1}{2}, \quad A_2(\sigma (T_1))\le 4K. \end{aligned}$$

(4.1)

Next, we define

$$\begin{aligned} T^*=\sup \{T|(4.1) ~\text {holds}\}. \end{aligned}$$

(4.2)

Then $T^*\ge T_1>0$. Hence, for any $0<\tau <T\le T^*$ with T finite, it holds from Lemmas 3.10 and 3.11 that

$$\begin{aligned} \rho \in C([0, T]; W^{1, q}),\ (\nabla u, \, \nabla ^2d) \in C(\tau , T; L^q), \end{aligned}$$

(4.3)

where we have used the standard embedding

$$\begin{aligned} L^\infty (\tau , T; H^1)\cap H^1(\tau , T; H^{-1})\hookrightarrow C([\tau , T]; L^q), \quad \mathrm{for~any}~q\in (3, 6). \end{aligned}$$

In the end, we claim that

$$\begin{aligned} T^*=\infty . \end{aligned}$$

(4.4)

Otherwise, $T^*<\infty$. Then, by Proposition 3.1, (3.3) follows when $T=T^*$. It holds from (3.5) and (4.3) that $(\rho (x, T^*), u(x, T^*), d(x, T^*))$ satisfies (1.9). Thus, Lemma 2.1 shows that there exists some $T^{**}>T^*$ such that (4.1) holds when $T=T^{**}$, which contradicts the definition of $T^*$. As a result, (4.4) follows. By Lemma 2.1, Lemma 3.10, and Lemma 3.11, it shows that $(\rho , u, d)$ is indeed the unique strong solution defined in $\Omega \times (0, T]$ for any $0<T<T^*=\infty$. $\square$