On existence of global classical solutions to the 3D compressible MHD equations with vacuum

Abstract

In this paper, the existence of global classical solutions is justified for the three-dimensional compressible magnetohydrodynamic (MHD) equations with vacuum. The main goal of this paper is to obtain a unique global classical solution on $\mathbb{R}^{3}\times [0, T]$ with any $T\in (0, \infty )$, provided that the initial magnetic field in the $L^{3}$-norm and the initial density are suitably small. Note that the first result is obtained under the condition of $\rho _{0}\in L^{\gamma }\cap W^{2, q}$ with $q\in (3, 6)$ and $\gamma \in (1, 6)$. It should be noted that the initial total energy can be arbitrarily large, the initial density allowed to vanish, and the system does not satisfy the conservation law of mass (i.e., $\rho _{0} \notin L^{1}$). Thus, the results obtained particularly extend the one due to Li–Xu–Zhang (Li et al. in SIAM J. Math. Anal. 45:1356–1387, 2013), where the global well-posedness of classical solutions with small energy was proved.

1 Introduction

One of the important problems in the theory of magnetohydrodynamics (MHD) is that of existence of global solutions to the equations of motion for a viscous compressible fluid. In this paper, we consider the MHD system of equations for a compressible isentropic MHD flows which in the case of 3D motion has the form (cf. [1, 7]):

$$ \textstyle\begin{cases} \rho _{t} +\operatorname{div}(\rho \mathbf {u})=0, \\ (\rho \mathbf {u})_{t}+\operatorname{div}(\rho \mathbf {u}\otimes \mathbf {u})+\nabla P( \rho )=\mu \Delta \mathbf {u}+(\mu +\lambda )\nabla \operatorname{div} \mathbf {u}+( \nabla \times \mathbf {B})\times \mathbf {B}, \\ \mathbf {B}_{t}+\mathbf {u}\cdot \nabla \mathbf {B}-\mathbf {B}\cdot \nabla \mathbf {u}+\mathbf {B}\operatorname{div} \mathbf {u}=\nu \Delta \mathbf {B}, \\ \operatorname{div} \mathbf {B}=0, \end{cases} $$

(1.1)

where $t\geqslant 0$ is the time, $x\in \mathbb{R}^{3}$ is the spatial coordinate, and $\rho \geqslant 0$, $\mathbf {u}=(u^{1}, u^{2}, u^{3})$, $\mathbf {B}=(B^{1}, B^{2}, B^{3})$ are the fluid density, velocity, and magnetic field, respectively. The pressure $P=P(\rho )$ satisfies the condition

$$ P(\rho )=A\rho ^{\gamma } \quad \text{with } A>0, \gamma >1, $$

(1.2)

where $\gamma >1$ is the adiabatic exponent, and $A>0$ is a physical constant. The viscosity coefficients μ and λ satisfy

$$ \mu >0,\qquad \lambda +\frac {2}{3}\mu \geqslant 0. $$

(1.3)

Positive constant ν is the resistivity coefficient acting as a magnetic diffusivity of magnetic field. In the second equation in (1.1), the circled times ⊗ means matrix multiplication, namely if $\mathbf{ a}=(a^{1}, a^{2}, a^{3})$, $\mathbf{ b}=(b^{1}, b^{2}, b^{3})$, then

$$\mathbf{a}\otimes \mathbf{b}=\left(\begin{array}{ccc}{a}^1{b}^1& a^1{b}^2& a^1{b}^3\\ a^2{b}^1& a^2{b}^2& a^2{b}^3\\ a^3{b}^1& a^3{b}^2& a^3{b}^3\end{array}\right).$$

Now, we consider the Cauchy problems of (1.1)–(1.3) with $(\rho , \mathbf {u}, \mathbf {B})(x, t)$ vanishing at infinity:

$$ (\rho , \mathbf {u}, \mathbf {B}) (x, t)\to 0\quad \text{as } \vert x \vert \to \infty , $$

(1.4)

and the initial conditions:

$$ (\rho , \mathbf {u}, \mathbf {B}) (x, 0)=(\rho _{0}, \mathbf {u}_{0}, \mathbf {B}_{0}) (x)\quad \text{with } x\in \mathbb{R}^{3}. $$

(1.5)

A great number of works have been devoted to the well-posedness theory of the multidimensional compressible MHD equations. The system of equations (1.1) describes the interaction between fluid flow and magnetic field. If we ignore the magnetic effects in (1.1) (i.e., $\mathbf {B}=0$), then the MHD system reduces to the Navier–Stokes system, which has been discussed by many mathematicians (see, for example, [11, 17, 18]). In [5], Huang et al. (2012) established the global existence and uniqueness of classical solutions to the Cauchy problem for the compressible Navier–Stokes equations in 3D with smooth initial data that are of small energy. Then, in [6], Huang et al. (2014) considered the two-dimensional density-dependent Navier–Stokes equations over bounded domains, and they derived a new blow-up criterion for strong solutions with vacuum. In [10], Li et al. (2019) were concerned with the global well-posedness and large time asymptotic behavior of strong solutions to the Cauchy problems of the Navier–Stokes equations for viscous compressible barotropic flows in 2D and 3D. However, if we consider the influence of magnetic field, the physical phenomena and mathematical structure of these equations make it more complex than the Navier–Stokes system, which makes more and more researchers begin to study the equations with magnetic field (see [2, 14, 16] and the references therein). Moreover, in [3], Fan and Li obtained the global strong solutions to the 3D compressible nonisentropic MHD equations with zero resistivity, and the results do not need the positivity of initial density, thus, it may vanish in an open subset of the domain. Hu and Wang in [4] considered the equations of the three-dimensional viscous, compressible, and heat-conducting magnetohydrodynamic flows in a bounded domain, and they obtained a solution to the initial-boundary value problem through an approximation scheme and a weak convergence method, and then, the existence of a global variational weak solution to the three-dimensional full magnetohydrodynamic equations with large data was established. Later, in [4], they got the global existence and large-time behavior of solutions to the three-dimensional equations of compressible magnetohydrodynamic flows. In [15], Zhang et al. (2009) studied the initial boundary value problems of MHD equations in plasma physics, and obtained the global existence of weak solutions with cylindrical symmetry. Recently, for the Cauchy problem, Li et al. (2013) in [9] considered the three-dimensional isentropic compressible magnetohydrodynamic equations, and they proved the global well-posedness of a classical solution with small energy but possibly large oscillations, where the flow density was allowed to contain vacuum states. Later, Si et al. (2018) in [13] improved the result of [17], and excluded the unsatisfactory restriction on the adiabatic exponent (i.e., $\gamma \in (1, 3/2)$), and they obtained the global classical solutions of compressible isentropic Navier–Stokes equations with small density and the adiabatic exponent $\gamma \in (1, 6)$ and $\gamma \in (1, \infty )$, respectively.

The main purpose of this paper is to obtain the global existence and uniqueness of classical solution of the problem (1.1)–(1.5). Before stating the main results, we explain the notation and conventions used throughout this paper. We denote

$$ \int f(x)\,dx= \int _{\mathbb{R}^{3}} f(x)\,dx. $$

For $1< r<\infty$ and $k\in \mathbb{Z}$, we denote the standard homogeneous and inhomogeneous Sobolev spaces:

$$ \textstyle\begin{cases} L^{r}=L^{r}(\mathbb{R}^{3}), \qquad D^{k, r}=\{\mathbf {u}\in L^{1}_{ \mathrm{loc}} | \Vert \nabla ^{k} \mathbf {u}\Vert _{L^{r}}< \infty \},\qquad \Vert \mathbf {u}\Vert _{D^{k, r}}= \Vert \nabla ^{k}\mathbf {u}\Vert _{L^{r}}, \\ W^{k, r}=L^{r} \cap D^{k, r}, \qquad H^{k}=W^{k, 2}, \qquad D^{k}=D^{k, 2}, \qquad D^{1}=\{\mathbf {u}\in L^{6} | \Vert \nabla \mathbf {u}\Vert _{L^{2}}< \infty \}. \end{cases} $$

The total energy is defined as follows:

$$ E(t)= \int \biggl(\frac {1}{2}\rho \vert \mathbf {u}\vert ^{2}+\frac {A}{\gamma -1} \rho ^{\gamma }+\frac {1}{2} \vert \mathbf {B}\vert ^{2} \biggr) (x, t)\,dx, $$

(1.6)

and the initial energy is denoted by $E_{0}$, i.e.,

$$ E_{0}\triangleq E(0)= \int \biggl(\frac {1}{2}\rho _{0} \vert \mathbf {u}_{0} \vert ^{2}+ \frac {A}{\gamma -1}\rho _{0}^{\gamma }+\frac {1}{2} \vert \mathbf {B}_{0} \vert ^{2} \biggr) (x)\,dx. $$

(1.7)

The first result of this paper is formulated in the following theorem.

Theorem 1.1

For any given numbers $M_{0}, M_{1}>0$ and $q\in (3, 6)$, suppose that

$$ \textstyle\begin{cases} \rho _{0} \vert \mathbf{ u}_{0} \vert ^{2}+\rho _{0}^{\gamma }+ \vert \mathbf{ B}_{0} \vert ^{2} \in L^{1}, \qquad (\mathbf{ u}_{0}, \mathbf{ B}_{0})\in D^{1}\cap D^{2}, \\ 0\leqslant \inf \rho _{0}\leqslant \rho _{0}(x)\leqslant \sup \rho _{0} \leqslant M_{0},\qquad (\rho _{0}, P(\rho _{0}))\in H^{2} \cap W^{2, q}, \\ \Vert \nabla \mathbf{ u}_{0} \Vert _{L^{2}}^{2}+ \Vert \nabla \mathbf{ B}_{0} \Vert _{H^{1}}^{2} \leqslant M_{1}, \end{cases} $$

(1.8)

and the compatibility condition holds

$$ \begin{aligned} &-\mu \Delta \mathbf{ u}_{0}-(\lambda +\mu )\nabla \operatorname{div} \mathbf{ u}_{0}+\nabla P(\rho _{0})-(\nabla \times \mathbf{ B}_{0}) \times \mathbf{ B}_{0}=\rho _{0} g \quad \textit{with} \\ &\Vert \sqrt{ \rho _{0}}g \Vert _{L^{2}}^{2}\leqslant M_{2}, \end{aligned}$$

(1.9)

for some $g\in D^{1}$. Then, there exists a positive constant ε depending only on μ, ν, λ, A, γ, $M_{0}$, $M_{1}$, and $M_{2}$ such that if

$$ \Vert \mathbf{ B}_{0} \Vert _{L^{3}}^{\frac {1}{2\gamma }} \leqslant M_{0} \leqslant \varepsilon \quad \textit{and}\quad 1< \gamma < 6, $$

(1.10)

then for any $0< T<\infty$, there exists a unique global classical solution $(\rho , \mathbf{ u}, \mathbf{ B})$ of the problem (1.1)–(1.5) on $\mathbb{ R}^{3} \times [0, T]$, satisfying

$$ 0\leqslant \rho \leqslant 2M_{0} \quad \textit{for all } x\in \mathbb{R}^{3}, t \geqslant 0, $$

(1.11)

and

$$ \textstyle\begin{cases} (\rho , P(\rho ))\in C([0,T]; H^{2} \cap W^{2, q}),\quad \sqrt{\rho }\mathbf{ u}_{t}\in L^{\infty }(0, T; L^{2}), \\ \mathbf{ u}\in C([0,T]; D^{1} \cap D^{2}) \cap L^{\infty }(\tau ,T; D^{3} \cap W^{3, q}), \\ \mathbf{ u}_{t}\in L^{\infty }(\tau ,T; D^{1} \cap D^{2}) \cap H^{1}(\tau ,T; D^{1}), \\ \mathbf{ B}\in C([0,T]; H^{2}) \cap L^{\infty }(\tau ,T; H^{3}), \\ \mathbf{ B}_{t}\in C([0, T]; L^{2}) \cap H^{1}(\tau ,T; H^{1}) \end{cases} $$

(1.12)

for any $0<\tau <T<\infty$.

Remark 1.1

In Theorem 1.1, the classical solution of (1.1)–(1.5) is justified under the condition that the initial density and the $L^{3}$-norm of the initial magnetic field are sufficiently small, and this solution is far away from the initial time. It is worth noting that the total initial energy $E_{0}$ can be arbitrarily large and the vacuum states are allowed.

Remark 1.2

The proof of Theorem 1.1 is based on a new t-weighted estimate of $\|(\nabla \dot{\mathbf {u}}, \nabla \mathbf {B}_{t})\|_{L^{2}}$ (see (3.56)), and the $L^{1}(0, T; L^{\infty })$-estimate of the effective viscous flux F will be achieved by making full use of (3.56). It is worth pointing out that the effective viscous flux F plays an important role in applying the Zlotnik’s inequality (see Lemma 2.3) to finish the proof of the (a priori) upper bound of the density.

Remark 1.3

Indeed, if, in addition, the conservation law of the total mass holds (i.e., $\|\rho (t)\|_{L^{1}}=\|\rho _{0}\|_{L^{1}}$ for all $t>0$), then Theorem 1.1 is similar to the results of [12] for all $\gamma >1$.

The rest of the paper is organized as follows. In Sect. 2, we recall some known facts and elementary inequalities which will be frequently used later. Section 3 is devoted to the global a priori estimates, which are necessary for the proof of Theorem 1.1.

2 Preliminaries

In this section, we will recall some known facts and elementary inequalities which will be used frequently later. We start with the well-known Gagliardo–Nirenberg inequality [8].

Lemma 2.1

For $p\in [2, 6]$, $q\in (1, \infty )$, and $r\in (3, \infty )$, assume that $f\in H^{1}(\mathbb{R}^{3})$ and $g\in L^{q}(\mathbb{R}^{3}) \cap D^{1, r}(\mathbb{R}^{3})$. Then there exists a generic constant $C>0$, depending only on q and r, such that

$$\begin{aligned}& \Vert f \Vert _{L^{p}}\leqslant C \Vert f \Vert _{L^{2}}^{\frac {6-p}{2p}} \Vert \nabla f \Vert _{L^{2}}^{ \frac {3p-6}{2p}}, \end{aligned}$$

(2.1)

$$\begin{aligned}& \Vert g \Vert _{L^{\infty }}\leqslant C \Vert g \Vert _{L^{q}}^{\frac {q(r-3)}{3r+q(r-3)}} \Vert \nabla g \Vert _{L^{r}}^{\frac {3r}{3r+q(r-3)}}. \end{aligned}$$

(2.2)

As in [9], we introduce the effective viscous flux F, the vorticity ω, and the material derivative“⋅”, which are defined as follows:

$$ F \triangleq (2\mu +\lambda )\operatorname{div} \mathbf {u}-P(\rho )-\frac {1}{2} \vert \mathbf {B}\vert ^{2}, \qquad \omega \triangleq \nabla \times \mathbf {u},\qquad \dot{\mathbf {u}} \triangleq \mathbf {u}_{t}+\mathbf {u}\cdot \nabla \dot{\mathbf {u}}. $$

(2.3)

Then it is easily derived from (1.1) that

$$ \Delta F=\operatorname{div}(\rho \dot{\mathbf {u}})- \operatorname{div}\operatorname{div}(\mathbf {B}\otimes \mathbf {B}) \quad \text{and}\quad \mu \Delta \omega =\nabla \times \bigl(\rho \dot{\mathbf {u}}-\operatorname{div}(\mathbf {B}\otimes \mathbf {B}) \bigr). $$

(2.4)

Thus, it follows from Lemma 2.1 and the standard $L^{p}$-estimates of elliptic equations that we have the following lemma.

Lemma 2.2

Let $(\rho , \mathbf{ u}, \mathbf{ B})$ be a smooth solution of (1.1)–(1.4). Then there exists a generic constant $C>0$ such that for any $p\in [2, 6]$,

$$\begin{aligned} & \Vert \nabla F \Vert _{L^{p}}+ \Vert \nabla \omega \Vert _{L^{p}}\leqslant C \bigl( \Vert \rho \dot{\mathbf{ u}} \Vert _{L^{p}}+ \Vert \nabla \mathbf{ B}\cdot \mathbf{ B} \Vert _{L^{p}} \bigr), \end{aligned}$$

(2.5)

$$\begin{aligned} & \Vert F \Vert _{L^{p}}\leqslant C \bigl( \Vert \nabla \mathbf{ u} \Vert _{L^{2}}+ \bigl\Vert P(\rho ) \bigr\Vert _{L^{2}}+ \bigl\Vert |\mathbf{ B}|^{2} \bigr\Vert _{L^{2}} \bigr)^{(6-p)/2p} \\ &\hphantom{\Vert F \Vert _{L^{p}}\leqslant}{}\times \bigl( \Vert \rho \dot{\mathbf{ u}} \Vert _{L^{2}}+ \Vert \nabla \mathbf{ B}\cdot \mathbf{ B} \Vert _{L^{2}} \bigr)^{(3p-6)/2p}, \end{aligned}$$

(2.6)

$$\begin{aligned} & \Vert \omega \Vert _{L^{p}}\leqslant C \Vert \nabla \mathbf{ u} \Vert _{L^{2}}^{(6-p)/2p} \bigl( \Vert \rho \dot{\mathbf{ u}} \Vert _{L^{2}}+ \Vert \nabla \mathbf{ B}\cdot \mathbf{ B} \Vert _{L^{2}} \bigr)^{(3p-6)/2p}, \end{aligned}$$

(2.7)

$$\begin{aligned} & \Vert \nabla \mathbf{ u} \Vert _{L^{p}}\leqslant C \bigl( \Vert F \Vert _{L^{p}}+ \bigl\Vert P(\rho ) \bigr\Vert _{L^{p}}+ \bigl\Vert |\mathbf{ B}|^{2} \bigr\Vert _{L^{p}}+ \Vert \omega \Vert _{L^{p}} \bigr), \end{aligned}$$

(2.8)

$$\begin{aligned} & \Vert \nabla \mathbf{ u} \Vert _{L^{p}}\leqslant C \Vert \nabla \mathbf{ u} \Vert _{L^{2}}^{(6-p)/2p} \bigl( \Vert \rho \dot{\mathbf{ u}} \Vert _{L^{2}}+ \bigl\Vert P(\rho ) \bigr\Vert _{L^{6}}+ \Vert \nabla \mathbf{ B}\cdot \mathbf{ B} \Vert _{L^{2}} \bigr)^{(3p-6)/2p}. \end{aligned}$$

(2.9)

The proof of Lemma 2.2 can be found in [9, 13], hence, we skip it for simplicity.

Lemma 2.3

Let $y\in W^{1,1}(0,T)$ satisfy the ODE system:

$$ y'=g(y)+b'(t) \quad \textit{on } [0,T],\qquad y(0)=y_{0}, $$

where $b\in W^{1,1}(0,T)$, $g\in C(\mathbb{R})$, and $g(+\infty )=-\infty$. Assume that there are two constants $N_{0}\geqslant 0$ and $N_{1}\geqslant 0$ such that for all $0\leqslant t_{1}< t_{2}\leqslant T$,

$$ b(t_{2})-b(t_{1})\leqslant N_{0}+N_{1}(t_{2}-t_{1}). $$

(2.10)

Then

$$ y(t)\leqslant \max \bigl\{ y_{0}, \xi ^{*} \bigr\} +N_{0}< +\infty\quad \textit{on } [0,T], $$

where $\xi ^{*}\in \mathbb{R}$ is a constant such that

$$ g(\xi )\leqslant -N_{1} \quad \textit{for } \xi \geqslant \xi ^{*}. $$

(2.11)

We can use the above Zlotnik’s inequality (see [19]) to prove the t-independent upper bound of the density.

Finally, we state the local existence result of classical solutions to the problem (1.1)–(1.5) with large initial data which may contain vacuum states (see [9]).

Lemma 2.4

Assume that the initial data $(\rho _{0}, \mathbf{ u}_{0}, \mathbf{ B}_{0})$ satisfy the conditions (1.8) and (1.9) of Theorem 1.1. Then there exists a positive time $T_{0}>0$ and a unique classical solution $(\rho , \mathbf{ u}, \mathbf{ B})$ of (1.1)–(1.5) on $\mathbb{R}^{3}\times (0, T_{0}]$, satisfying $\rho \geqslant 0$, and for any $\tau \in (0, T_{0})$,

$$ \textstyle\begin{cases} (\rho , P(\rho ))\in C([0,T_{0}]; H^{1} \cap W^{1, q}) \cap L^{\infty }(0, T_{0}; H^{2} \cap W^{2, q}), \\ \mathbf{ u}\in C([0,T_{0}]; D^{1} \cap D^{2}) \cap L^{\infty }(\tau ,T_{0}; D^{3} \cap W^{3, q}), \\ \mathbf{ u}_{t}\in L^{\infty }(\tau ,T_{0}; D^{1} \cap D^{2}) \cap H^{1}(\tau ,T_{0}; D^{1}), \\ \mathbf{ B}\in C([0,T_{0}]; H^{2}) \cap L^{\infty }(\tau ,T_{0}; H^{3}), \\ \mathbf{ B}_{t}\in C([0, T_{0}]; L^{2}) \cap H^{1}(\tau ,T_{0}; H^{1}). \end{cases} $$

(2.12)

3 Proof of Theorem 1.1

In the section, we will establish the uniform a priori bounds of local solutions $(\rho , \mathbf {u}, \mathbf {B})$ to the Cauchy problems (1.1)–(1.5) whose existence is guaranteed by Lemma 2.4. Thus, let $T> 0$ be a fixed time and $(\rho , \mathbf {u}, \mathbf {B})$ be the smooth solution of (1.1)–(1.5) on $\mathbb{R}^{3}\times [0, T]$ with smooth initial data $(\rho _{0}, \mathbf {u}_{0}, \mathbf {B}_{0})$ satisfying (1.8). To estimate this solution, we define

$$\begin{aligned}& A_{1}(T) \triangleq \sup_{t\in [0,T]} \bigl\Vert (\nabla \mathbf {u}, \nabla \mathbf {B}) \bigr\Vert _{L^{2}}^{2}+ \int _{0}^{T} \bigl\Vert \bigl(\sqrt{\rho }\dot{ \mathbf {u}}, \nabla ^{2} \mathbf {B}, \mathbf {B}_{t} \bigr) \bigr\Vert _{L^{2}}^{2}\,dt, \\& A_{2}(T) \triangleq \sup_{t\in [0,T]} \bigl\Vert \bigl( \sqrt{\rho }\dot{\mathbf {u}}, \nabla ^{2} \mathbf {B}, \mathbf {B}_{t} \bigr) \bigr\Vert _{L^{2}}^{2}+ \int _{0}^{T} \bigl\Vert ( \nabla \dot{\mathbf {u}}, \nabla \mathbf {B}_{t}) \bigr\Vert _{L^{2}}^{2} \,dt. \end{aligned}$$

Here, $\|(f, g)\|_{L^{p}}\triangleq \|f\|_{L^{p}}+\|g\|_{L^{p}}$.

The proof of Theorem 1.1 is based on the following key a priori estimates of $(\rho , \mathbf {u}, \mathbf {B})$.

Proposition 3.1

Let the conditions (1.8) and (1.9) be in force. Assume that $(\rho , \mathbf{ u}, \mathbf{ B})$ is a smooth solution of (1.1)–(1.5) on $\mathbb{R}^{3} \times [0, T]$ with $T>0$. Then there exist positive constants K and ε, depending only on μ, λ, ν, γ, A, $E_{0}$, $M_{1}$, and $M_{2}$, such that if

$$ \textstyle\begin{cases} 0\leqslant \rho (x, t)\leqslant 2M_{0}, \quad \forall (x, t)\in \mathbb{R}^{3}\times [0, T], \\ A_{1}(t)+A_{2}(t)\leqslant 2K, \end{cases} $$

(3.1)

then one has

$$ \textstyle\begin{cases} 0\leqslant \rho (x, t)\leqslant \frac {7}{4}M_{0}, \quad \forall (x, t) \in \mathbb{R}^{3}\times [0, T], \\ A_{1}(t)+A_{2}(t)\leqslant K, \end{cases} $$

(3.2)

provided

$$ \Vert \mathbf{ B}_{0} \Vert _{L^{3}}^{\frac {1}{2\gamma }} \leqslant M_{0} \leqslant \varepsilon \quad \textit{and}\quad \varepsilon \in (1, 6). $$

(3.3)

The proof of Proposition 3.1 will be presented by a series of lemmas below. For simplicity, we will use the conventions that C and $C_{i}$ ($i = 1, 2, \dots$) denote various positive constants, which may depend on μ, λ, ν, γ, A, $E_{0}$, $M_{1}$, and $M_{2}$, but are independent of T and $M_{0}$. Sometimes we also write $C(\alpha )$ to emphasize the dependence on α.

We first begin with the following standard energy estimates, which can be easily deduced from (1.1)–(1.5).

Lemma 3.1

Let $(\rho , \mathbf{ u}, \mathbf{ B})$ be a smooth solution of (1.1)–(1.5) on $\mathbb{R}^{3}\times [0,T]$. Then

$$ E(t)+ \int _{0}^{T} \bigl(\mu \Vert \nabla \mathbf{ u} \Vert _{L^{2}}^{2}+(\mu + \lambda ) \Vert \operatorname{div} \mathbf{ u} \Vert _{L^{2}}^{2}+\nu \Vert \nabla \mathbf{ B} \Vert _{L^{2}}^{2} \bigr)\,dt\leqslant E_{0}, $$

(3.4)

where $E(t)\geqslant 0$ and $E_{0}$ are as given in (1.6) and (1.7).

Proof

Multiplying (1.1)₁, (1.1)₂, and (1.1)₃ by $\frac {\gamma }{\gamma -1}A\rho ^{\gamma -1}$, $\mathbf{ u,}$ and B, respectively, and integrating the resulting equations by parts over $\mathbb{R}^{3}$, we obtain after adding them together that

$$ \frac {d}{dt}E(t)+\mu \Vert \nabla \mathbf{ u} \Vert _{L^{2}}^{2}+(\mu + \lambda )\|\operatorname{div} \mathbf{ u}\|_{L^{2}}^{2}+\nu \|\nabla \mathbf{ B} \|_{L^{2}}^{2}=0, $$

which, integrated over $(0, t)$, $\forall t\in [0, T]$, immediately leads to (3.4). □

By virtue of (3.1) and (3.4), we infer from Lemma 2.1 ($p=6$ in (2.1)) that

$$ \begin{aligned} & \int _{0}^{T} \bigl( \Vert \mathbf{ u} \Vert _{L^{6}}^{4}+ \Vert \nabla \mathbf{ u} \Vert _{L^{2}}^{4}+ \Vert \nabla \mathbf{ B} \Vert _{L^{2}}^{4} \bigr)\,dt \\ &\quad \leqslant C \int _{0}^{T} \bigl( \Vert \nabla \mathbf{ u} \Vert _{L^{2}}^{4}+ \Vert \nabla \mathbf{ B} \Vert _{L^{2}}^{4} \bigr)\,dt \\ &\quad \leqslant C\sup_{t\in [0, T]} \bigl( \Vert \nabla \mathbf{ u} \Vert _{L^{2}}^{2}+ \Vert \nabla \mathbf{ B} \Vert _{L^{2}}^{2} \bigr) \int _{0}^{T} \bigl( \Vert \nabla \mathbf{ u} \Vert _{L^{2}}^{2}+ \Vert \nabla \mathbf{ B} \Vert _{L^{2}}^{2} \bigr)\,dt \leqslant C(E_{0})K. \end{aligned} $$

(3.5)

Here the constant $C>0$ comes from the Gagliardo–Nirenberg–Sobolev inequality in (2.1), and we use the constant $C(E_{0})>0$ to emphasize the dependence on $E_{0}$. Thus, combining (3.4) and (3.5) yields the following lemma:

Lemma 3.2

Let $(\rho , \mathbf{ u}, \mathbf{ B})$ be a smooth solution of (1.1)–(1.5) on $\mathbb{R}^{3}\times [0,T]$ satisfying (3.1). Then

$$ \sup_{t\in [0, T]} \Vert \mathbf{ B} \Vert _{L^{3}}^{3}+ \int _{0}^{T} \Vert \mathbf{ B} \Vert _{L^{9}}^{3}\,dt\leqslant e^{CK} \Vert \mathbf{ B}_{0} \Vert _{L^{3}}^{3}, $$

(3.6)

where the constant $C>0$ depends on ν, $E_{0}$, and the coefficients of the Gagliardo–Nirenberg–Sobolev inequality in Lemma 2.1, but is independent of $M_{0}$.

Proof

Multiplying the third equation of (1.1) by $3|\mathbf {B}|\mathbf {B}$ and integrating by parts over $\mathbb{R}^{3}$, we have

$$\begin{aligned}& \frac {d}{dt} \Vert \mathbf {B}\Vert _{L^{3}}^{3}+3 \nu \int \bigl( \vert \mathbf {B}\vert \vert \nabla \mathbf {B}\vert ^{2}+ \vert \mathbf {B}\vert \bigl\vert \nabla \bigl( \vert \mathbf {B}\vert \bigr) \bigr\vert ^{2} \bigr)\,dx \\& \quad \leqslant \nu \int \vert \mathbf {B}\vert \vert \nabla \mathbf {B}\vert ^{2}\,dx+C \Vert \nabla \mathbf {u}\Vert _{L^{2}}^{2} \Vert \mathbf {B}\Vert _{L^{9/2}}^{3}, \end{aligned}$$

(3.7)

where the last term on the right-hand in (3.7) comes from the following inequality:

$$ \begin{aligned} \int \vert \nabla \mathbf{ u} \vert \vert \mathbf{ B} \vert ^{3}\,dx & \leqslant C \Vert \nabla \mathbf{ u} \Vert _{L^{2}} \Vert \mathbf{ B} \Vert _{L^{9/2}}^{3/2} \Vert \mathbf{ B} \Vert _{L^{9}}^{3/2} \\ &\leqslant C \Vert \nabla \mathbf{ u} \Vert _{L^{2}} \Vert \mathbf{ B} \Vert _{L^{9/2}}^{3/2} \bigl\Vert \vert \mathbf{ B} \vert ^{3/2} \bigr\Vert _{L^{6}} \\ &\leqslant C \bigl( \Vert \nabla \mathbf{ u} \Vert _{L^{2}}^{2} \Vert \mathbf{ B} \Vert _{L^{9/2}}^{3} \bigr)^{1/2} \bigl\Vert \vert \nabla \mathbf{ B} \vert \vert \mathbf{ B} \vert ^{1/2} \bigr\Vert _{L^{2}}. \end{aligned} $$

To deal with the right-hand side of (3.7), we notice that

$$ \Vert \mathbf {B}\Vert _{L^{9}}^{3}\leqslant C \bigl\Vert \vert \mathbf {B}\vert ^{3/2} \bigr\Vert _{L^{6}}^{2} \leqslant C \bigl\Vert \vert \nabla \mathbf {B}\vert \vert \mathbf {B}\vert ^{1/2} \bigr\Vert _{L^{2}}^{2}, $$

then

$$ \Vert \mathbf {B}\Vert _{L^{9/2}}\leqslant C \Vert \mathbf {B}\Vert _{L^{3}}^{1/2} \Vert \mathbf {B}\Vert _{L^{9}}^{1/2} \leqslant C \Vert \mathbf {B}\Vert _{L^{3}}^{1/2} \bigl\Vert \nabla \vert \mathbf {B}\vert ^{3/2} \bigr\Vert _{L^{2}}^{1/3}, $$

which, together with (3.7), yields

$$ \frac {d}{dt} \Vert \mathbf {B}\Vert _{L^{3}}^{3}+ \Vert \mathbf {B}\Vert _{L^{9}}^{3} \leqslant C \Vert \nabla \mathbf {u}\Vert _{L^{2}}^{4} \Vert \mathbf {B}\Vert _{L^{3}}^{3}. $$

As a result, we deduce from (3.5) and the Gronwall’s inequality that (3.6) holds. □

Now, to estimate $A_{1}(T)$ and $A_{2}(T)$, we first prove the following lemma.

Lemma 3.3

Let $(\rho , \mathbf{ u}, \mathbf{ B})$ be a smooth solution of (1.1)–(1.5) on $\mathbb{R}^{3}\times [0,T]$ satisfying (3.1). Then

$$\begin{aligned}& A_{1}(T) \leqslant C \Bigl(\sup_{t\in [0, T]} \Vert \mathbf{ B} \Vert _{L^{3}}^{2} \Bigr)A_{1}(T)+ C \bigl(1+M_{0}^{\gamma } \bigr) +C \int _{0}^{T} \Vert \nabla \mathbf{ u} \Vert _{L^{4}}^{4}\,dt, \end{aligned}$$

(3.8)

$$\begin{aligned}& \begin{aligned} &A_{2}(T) \leqslant C+ \frac {1}{4}A_{1}(T)+C \Bigl[(M_{0}^{\frac {1}{2}}K+M_{0}^{ \frac {5\gamma }{6}}K^{1/2}+ \Bigl(\sup_{t\in [0, T]} \Vert \mathbf{ B} \Vert _{L^{3}} \Bigr)K \Bigr]A_{1}(T) \\ &\hphantom{A_{2}(T) \leqslant}{} +C \Bigl(\sup_{t\in [0, T]} \Vert \mathbf{ B} \Vert _{L^{3}}^{2}+K\sup_{t \in [0, T]} \Vert \mathbf{ B} \Vert _{L^{3}}^{4} \Bigr)A_{2}(T) \\ &\hphantom{A_{2}(T) \leqslant}{}+C \int _{0}^{T} \bigl( \Vert \nabla \mathbf{ u} \Vert _{L^{4}}^{4}+ \bigl\Vert P(\rho ) \bigr\Vert _{L^{4}}^{4} \bigr)\,dt, \end{aligned} \end{aligned}$$

(3.9)

where the constant $C>0$ depends on μ, λ, ν, γ, A, $E_{0}$, $M_{1}$, $M_{2}$, and the coefficients of the Gagliardo–Nirenberg–Sobolev inequality in Lemma 2.1, but is independent of $M_{0}$.

Proof

In order to prove (3.8), using Lemma 2.1 and equation (1.1)₃, we have

$$ \begin{aligned} &\nu \bigl( \Vert \nabla \mathbf {B}\Vert _{L^{2}}^{2} \bigr)_{t}+\nu ^{2} \bigl\Vert \nabla ^{2} \mathbf {B}\bigr\Vert _{L^{2}}^{2}+ \Vert \mathbf {B}_{t} \Vert _{L^{2}}^{2} \\ &\quad = \int (\mathbf {B}_{t}-\nu \Delta \mathbf {B})^{2}\,dx \\ &\quad = \int \vert \mathbf {B}\cdot \nabla \mathbf {u}-\mathbf {u}\cdot \nabla \mathbf {B}- \mathbf {B}\operatorname{div} \mathbf {u}\vert ^{2}\,dx \\ &\quad \leqslant C \bigl( \Vert \mathbf {B}\Vert _{L^{6}}^{2} \Vert \nabla \mathbf {u}\Vert _{L^{3}}^{2}+ \Vert \mathbf {u}\Vert _{L^{\infty }}^{2} \Vert \nabla \mathbf {B}\Vert _{L^{2}}^{2} \bigr) \\ &\quad \leqslant C \Vert \mathbf {B}\Vert _{L^{2}} \bigl\Vert \nabla ^{2} \mathbf {B}\bigr\Vert _{L^{2}} \bigl( \Vert \nabla \mathbf {u}\Vert _{L^{3}}^{2}+ \Vert \mathbf {u}\Vert _{L^{\infty }}^{2} \bigr) \\ &\quad \leqslant \frac {\nu ^{2}}{2} \bigl\Vert \nabla ^{2} \mathbf {B}\bigr\Vert _{L^{2}}^{2}+C(E_{0}) \bigl( \Vert \nabla \mathbf {u}\Vert _{L^{3}}^{4}+ \Vert \mathbf {u}\Vert _{L^{\infty }}^{4} \bigr), \end{aligned} $$

which, together with the Cauchy–Schwarz inequality, one obtains

$$ \bigl(\nu \Vert \nabla \mathbf {B}\Vert _{L^{2}}^{2} \bigr)_{t}+\nu ^{2} \bigl\Vert \nabla ^{2} \mathbf {B}\bigr\Vert _{L^{2}}^{2}+ \Vert \mathbf {B}_{t} \Vert _{L^{2}}^{2} \leqslant C \bigl( \Vert \nabla \mathbf {u}\Vert _{L^{3}}^{4}+ \Vert \mathbf {u}\Vert _{L^{\infty }}^{4} \bigr). $$

(3.10)

Next, multiplying (1.1)₂ by $\dot{\mathbf {u}}$ and integrating by parts, we deduce

$$ \begin{aligned} \int \rho \vert \dot{\mathbf {u}} \vert ^{2}\,dx ={}& \int \biggl(-\nabla P \cdot \dot{\mathbf {u}}+\mu \Delta \mathbf {u}\cdot \dot{ \mathbf {u}}+(\lambda + \mu )\nabla \operatorname{div} \mathbf {u}\cdot \dot{\mathbf {u}} \\ &{} +\mathbf {B}\cdot \nabla \mathbf {B}\cdot \dot{\mathbf {u}}-\frac {1}{2} \dot{ \mathbf {u}}\cdot \nabla \bigl( \vert \mathbf {B}\vert ^{2} \bigr) \biggr) \,dx\triangleq \sum_{i=1}^{5}I_{i}. \end{aligned} $$

(3.11)

The right-hand side of (3.11) can be estimated as follows. It holds by (1.1)₁ that

$$ P(\rho )_{t}+\mathbf {u}\cdot \nabla P(\rho )+\gamma P( \rho )\operatorname{div} \mathbf {u}=0, $$

(3.12)

which, together with (3.1), yields

$$ \begin{aligned} I_{1}&=- \int \nabla P\cdot (\mathbf {u}_{t}+\mathbf {u}\cdot \nabla \mathbf {u})\,dx \\ &= \int \bigl(P(\rho )\operatorname{div} \mathbf {u}_{t}-(\mathbf {u}\cdot \nabla \mathbf {u}) \cdot \nabla P(\rho ) \bigr)\,dx \\ &= \biggl( \int P(\rho )\operatorname{div} \mathbf {u}\,dx \biggr)_{t}- \int \bigl(P(\rho )_{t}\operatorname{div} \mathbf {u}+\mathbf {u}\cdot \nabla \mathbf {u}\cdot \nabla P(\rho ) \bigr)\,dx \\ &= \biggl( \int P(\rho )\operatorname{div} \mathbf {u}\,dx \biggr)_{t}+ \int \bigl((\gamma -1)P( \rho ) (\operatorname{div} \mathbf {u})^{2}+P( \rho ) \vert \nabla \mathbf {u}\vert ^{2} \bigr)\,dx \\ &\leqslant C \biggl( \int P(\rho )\operatorname{div} \mathbf {u}\,dx \biggr)_{t}+C(A, \gamma )M_{0}^{\gamma } \Vert \nabla \mathbf {u}\Vert _{L^{2}}^{2}. \end{aligned} $$

(3.13)

Thanks to (3.4), we find (keep in mind that $0\leqslant \rho \leqslant 2M_{0}$)

$$ \begin{aligned} \int P(\rho )\operatorname{div} \mathbf {u}\,dx &\leqslant \frac {\mu }{8} \Vert \nabla \mathbf {u}\Vert _{L^{2}}^{2}+C \bigl\Vert P( \rho ) \bigr\Vert _{L^{\infty }} \bigl\Vert P(\rho ) \bigr\Vert _{L^{1}} \\ &\leqslant \frac {\mu }{8} \Vert \nabla \mathbf {u}\Vert _{L^{2}}^{2}+C(E_{0})M_{0}^{\gamma }. \end{aligned} $$

(3.14)

Integrating by parts, one has

$$ \begin{aligned} I_{2}&=\mu \int \Delta \mathbf {u}\cdot (\mathbf {u}_{t}+ \mathbf {u}\cdot \nabla \mathbf {u})\,dx=-\frac {\mu }{2} \bigl( \Vert \nabla \mathbf {u}\Vert _{L^{2}}^{2} \bigr)_{t}+ \mu \int \partial _{i} u^{j}\partial _{i} \bigl(u^{k}\partial _{k}u^{j} \bigr)\,dx \\ &\leqslant -\frac {\mu }{2} \bigl( \Vert \nabla \mathbf {u}\Vert _{L^{2}}^{2} \bigr)_{t}+\mu \int \partial _{i} u^{j}\partial _{i} u^{k}\partial _{k}u^{j}\,dx- \frac {1}{2} \int \vert \nabla \mathbf {u}\vert ^{2} \operatorname{div} \mathbf {u}\,dx \\ &\leqslant -\frac {\mu }{2} \bigl( \Vert \nabla \mathbf {u}\Vert _{L^{2}}^{2} \bigr)_{t}+C \Vert \nabla \mathbf {u}\Vert ^{3}_{L^{3}}. \end{aligned} $$

(3.15)

Similarly,

$$ I_{3}=(\lambda +\mu ) \int \dot{\mathbf {u}}\cdot \nabla \operatorname{div} \mathbf {u}\,dx \leqslant - \frac {\lambda +\mu }{2} \bigl( \Vert \operatorname{div} \mathbf {u}\Vert _{L^{2}}^{2} \bigr)_{t}+C \Vert \nabla \mathbf {u}\Vert _{L^{3}}^{3}. $$

(3.16)

Integration by parts also gives

$$ \begin{aligned} I_{4}={}& \int \mathbf {B}\cdot \nabla \mathbf {B}\cdot (\mathbf {u}_{t}+ \mathbf {u}\cdot \nabla \mathbf {u})\,dx \\ ={}&- \int \mathbf {B}\cdot \nabla \mathbf {u}_{t} \cdot \mathbf {B}\,dx+ \int B^{i}\partial _{i}B^{j} u^{k} \partial _{k} u^{j}\,dx \\ ={}& - \biggl( \int \mathbf {B}\cdot \nabla \mathbf {u}\cdot \mathbf {B}\,dx \biggr)_{t}+ \int \bigl(\mathbf {B}_{t} \cdot \nabla \mathbf {u}\cdot \mathbf {B}+ \mathbf {B}\cdot \nabla \mathbf {u}\cdot \mathbf {B}_{t}+B^{i} \partial _{i}B^{j} u^{k} \partial _{k} u^{j} \bigr)\,dx \\ \leqslant{}& - \biggl( \int \mathbf {B}\cdot \nabla \mathbf {u}\cdot \mathbf {B}\,dx \biggr)_{t} \\ &{}+C \Vert \mathbf {B}_{t} \Vert _{L^{2}} \Vert \nabla \mathbf {u}\Vert _{L^{3}} \Vert \mathbf {B}\Vert _{L^{6}}+C \Vert \mathbf {B}\Vert _{L^{2}} \Vert \nabla \mathbf {B}\Vert _{L^{6}} \Vert \mathbf {u}\Vert _{L^{\infty }} \Vert \nabla \mathbf {u}\Vert _{L^{3}} \\ \leqslant{}& - \biggl( \int \mathbf {B}\cdot \nabla \mathbf {u}\cdot \mathbf {B}\,dx \biggr)_{t} \\ &{}+ \frac {1}{4} \bigl( \Vert \mathbf {B}_{t} \Vert _{L^{2}}^{2}+\nu ^{2} \bigl\Vert \nabla ^{2}\mathbf {B}\bigr\Vert _{L^{2}}^{2} \bigr)+C (E_{0}) \bigl( \Vert \nabla \mathbf {u}\Vert _{L^{3}}^{4}+ \Vert \mathbf {u}\Vert _{L^{\infty }}^{4} \bigr), \end{aligned} $$

(3.17)

where we also used (2.1), (3.4), and the simple fact

$$ \Vert \mathbf {B}\Vert _{L^{6}}\leqslant C \Vert \mathbf {B}\Vert _{L^{2}}^{1/2} \Vert \nabla \mathbf {B}\Vert _{L^{6}}^{1/2}\leqslant C \Vert \mathbf {B}\Vert _{L^{2}}^{1/2} \bigl\Vert \nabla ^{2} \mathbf {B}\bigr\Vert _{L^{2}}^{1/2}, $$

which, together with Cauchy–Schwarz inequality, gives

$$ \int \mathbf {B}\cdot \nabla \mathbf {u}\cdot \mathbf {B}\,dx \leqslant \frac { \mu }{8} \Vert \nabla \mathbf {u}\Vert _{L^{2}}^{2}+C \Vert \mathbf {B}\Vert _{L^{3}}^{2} \Vert \nabla \mathbf {B}\Vert _{L^{2}}^{2}. $$

(3.18)

Similarly,

$$ \begin{aligned} I_{5}={}&-\frac {1}{2} \int \nabla \bigl( \vert \mathbf {B}\vert ^{2} \bigr)\cdot ( \mathbf {u}_{t}+\mathbf {u}\cdot \nabla \mathbf {u})\,dx \\ \leqslant{}& \frac {1}{2} \biggl( \int \vert \mathbf {B}\vert ^{2} \operatorname{div} \mathbf {u}\,dx \biggr)_{t} \\ &{}+\frac {1}{4} \bigl( \Vert \mathbf {B}_{t} \Vert _{L^{2}}^{2}+\nu ^{2} \bigl\Vert \nabla ^{2}\mathbf {B}\bigr\Vert _{L^{2}}^{2} \bigr)+C (E_{0}) \bigl( \Vert \nabla \mathbf {u}\Vert _{L^{3}}^{4}+ \Vert \mathbf {u}\Vert _{L^{\infty }}^{4} \bigr) \end{aligned} $$

(3.19)

and

$$ \frac {1}{2} \int \vert \mathbf {B}\vert ^{2} \operatorname{div} \mathbf {u}\,dx \leqslant \frac { \mu }{8} \Vert \nabla \mathbf {u}\Vert _{L^{2}}^{2}+C \Vert \mathbf {B}\Vert _{L^{3}}^{2} \Vert \nabla \mathbf {B}\Vert _{L^{2}}^{2}. $$

(3.20)

Thus, substituting (3.13), (3.15)–(3.17), and (3.19) into (3.11), using (3.10), we obtain

$$ \begin{aligned} & \biggl(\frac {\mu }{2} \Vert \nabla \mathbf {u}\Vert _{L^{2}}^{2}+ \frac {\mu +\lambda }{2} \Vert \operatorname{div} \mathbf {u}\Vert _{L^{2}}^{2}+\nu \Vert \nabla \mathbf {B}\Vert _{L^{2}}^{2} \biggr)_{t} \\ &\qquad {}+ \Vert \sqrt{\rho }\dot{\mathbf {u}} \Vert _{L^{2}}^{2}+ \Vert \mathbf {B}_{t} \Vert _{L^{2}}^{2}+ \nu ^{2} \bigl\Vert \nabla ^{2} \mathbf {B}\bigr\Vert _{L^{2}}^{2} \\ &\quad \leqslant \bigg( \int \biggl[ \biggl(P(\rho )\operatorname{div} \mathbf {u}-\mathbf {B}\cdot \nabla \mathbf {u}\cdot \mathbf {B}+\frac {1}{2} \vert \mathbf {B}\vert ^{2} \operatorname{div} \mathbf {u}\biggr]\,dx \biggr)_{t}+C M_{0}^{\gamma } \Vert \nabla \mathbf {u}\Vert _{L^{2}}^{2} \\ & \qquad {}+C \bigl( \Vert \nabla \mathbf {u}\Vert _{L^{3}}^{3}+ \Vert \nabla \mathbf {u}\Vert _{L^{3}}^{4}+ \Vert \mathbf {u}\Vert _{L^{\infty }}^{4} \bigr). \end{aligned} $$

(3.21)

Thanks to (2.1) and (2.2), we find from (3.21) that

$$ \begin{aligned} & \int _{0}^{T} \bigl( \Vert \nabla \mathbf {u}\Vert _{L^{3}}^{3}+ \Vert \nabla \mathbf {u}\Vert _{L^{3}}^{4}+ \Vert \mathbf {u}\Vert _{L^{\infty }}^{4} \bigr)\,dt \\ &\quad \leqslant C \int _{0}^{T} \bigl( \Vert \nabla \mathbf {u}\Vert _{L^{2}}^{2}+ \Vert \nabla \mathbf {u}\Vert _{L^{4}}^{4}+ \Vert \nabla \mathbf {u}\Vert _{L^{2}}^{4/3} \Vert \nabla \mathbf {u}\Vert _{L^{4}}^{8/3} \bigr)\,dt \\ &\quad \leqslant C+\delta \int _{0}^{T} \Vert \nabla \mathbf {u}\Vert _{L^{2}}^{4}\,dt+C( \delta ) \int _{0}^{T} \Vert \nabla \mathbf {u}\Vert _{L^{4}}^{4}\,dt \\ &\quad \leqslant C+\delta E_{0}/\mu \sup_{t\in [0, T]} \Vert \nabla \mathbf {u}\Vert _{L^{2}}^{2}+C(\delta ) \int _{0}^{T} \Vert \nabla \mathbf {u}\Vert _{L^{4}}^{4}\,dt, \end{aligned} $$

(3.22)

where $\delta >0$ is an undetermined small constant and the constant $C(\delta )>0$ depends on δ and the coefficients in (2.1). Moreover, the derivation of inequality (3.22) also uses the following inequalities:

$$ \Vert \nabla \mathbf {u}\Vert _{L^{3}}\leqslant C \Vert \nabla \mathbf {u}\Vert _{L^{2}}^{1/3} \Vert \nabla \mathbf {u}\Vert _{L^{4}}^{2/3} \quad \text{and}\quad \Vert \mathbf {u}\Vert _{L^{\infty }}\leqslant C \Vert \mathbf {u}\Vert _{L^{6}}^{1/3} \Vert \nabla \mathbf {u}\Vert _{L^{4}}^{2/3}. $$

Combining (3.14), (3.18), and (3.20), then integrating (3.21) over $(0, T)$, and choosing $\delta >0$ suitably small in (3.22), by virtue of the Gronwall’s inequality, we obtain (3.8).

To prove (3.9), applying $\dot{u}^{j}[\partial _{t}+\operatorname{div}(\mathbf {u}\cdot )]$ to $\text{(1.1)}_{2}^{j}$, summing with respect to j, and integrating the resulting equation over $\mathbb{R}^{3}$, we obtain after integration by parts

$$ \begin{aligned} \biggl(\frac {1}{2} \int \rho \vert \dot{\mathbf {u}} \vert ^{2}\,dx \biggr)_{t} ={}&- \int \dot{u}^{j} \bigl[\partial _{j}P_{t}+\operatorname{div}(\mathbf {u}\partial _{j}P) \bigr]\,dx \\ &{}+ \mu \int \dot{u}^{j} \bigl[\Delta {u}^{j}_{t}+\operatorname{div} \bigl(\mathbf {u}\Delta u^{j} \bigr) \bigr]\,dx \\ &{} +(\lambda +\mu ) \int \dot{u}^{j} \bigl[\partial _{j}\partial _{t}(\operatorname{div} \mathbf {u})+\operatorname{div} \bigl(\mathbf {u}\partial _{j}(\operatorname{div} \mathbf {u}) \bigr) \bigr]\,dx \\ & {}+ \int \dot{u}^{j} \bigl[\partial _{t} \bigl(\mathbf {B}\cdot \nabla B^{j} \bigr)+\operatorname{div} \bigl(\mathbf {u}\mathbf {B}\cdot \nabla B^{j} \bigr) \bigr]\,dx \\ &{} -\frac {1}{2} \int \dot{u}^{j} \bigl[\partial _{t}\partial _{j} \bigl( \vert \mathbf {B}\vert ^{2} \bigr)+\operatorname{div} \bigl(\mathbf {u}\partial _{j} \bigl( \vert \mathbf {B}\vert ^{2} \bigr) \bigr) \bigr]\,dx \triangleq \sum _{i=1}^{5}J_{i}, \end{aligned} $$

(3.23)

where the first term on the right-hand side can be estimated as follows, based on integration by parts and (3.12):

$$ \begin{aligned} J_{1} &= \int \bigl(\partial _{j} \dot{u}^{j} P(\rho )_{t}+ \partial _{k} \dot{u}^{j} u^{k} \partial _{j}P(\rho ) \bigr)\,dx \\ &= \int \bigl(-\gamma P(\rho )\operatorname{div} \mathbf {u}\partial _{j} \dot{u}^{j}- \partial _{j}\dot{u}^{j} u^{k} \partial _{k}P(\rho )-\partial _{j} \bigl( \partial _{k}\dot{u}^{j} u^{k} \bigr)P(\rho ) \bigr)\,dx \\ &= \int \bigl(-\gamma P(\rho )\operatorname{div} \mathbf {u}\partial _{j} \dot{u}^{j}+ \partial _{j}\dot{u}^{j} \partial _{k} u^{k} P(\rho )-\partial _{k} \dot{u}^{j} \partial _{j} u^{k} P(\rho ) \bigr) \,dx \\ &\leqslant \frac {\mu }{8} \Vert \nabla \dot{\mathbf {u}} \Vert _{L^{2}}^{2}+C \bigl( \Vert \nabla \mathbf {u}\Vert _{L^{4}}^{4}+ \bigl\Vert P(\rho ) \bigr\Vert _{L^{4}}^{4} \bigr). \end{aligned} $$

(3.24)

Similarly,

$$ J_{2}=\mu \int \dot{u}^{j} \bigl[\Delta {u}^{j}_{t}+\operatorname{div} \bigl(\mathbf {u}\Delta u^{j} \bigr) \bigr]\,dx\leqslant - \frac {3\mu }{4} \Vert \nabla \dot{\mathbf {u}} \Vert _{L^{2}}^{2}+C \Vert \nabla \mathbf {u}\Vert _{L^{4}}^{4} $$

(3.25)

and

$$ J_{3}\leqslant -\frac {\lambda +\mu }{2} \Vert \operatorname{div}\dot{\mathbf {u}} \Vert _{L^{2}}^{2}+ \frac {\mu }{4} \Vert \nabla \dot{\mathbf {u}} \Vert _{L^{2}}^{2}+C \Vert \nabla \mathbf {u}\Vert _{L^{4}}^{4}. $$

(3.26)

Next, integrating by parts, one has (keeping in mind that $\operatorname{div} \mathbf {B}=0$)

$$ \begin{aligned} J_{4} &= \int \dot{u}^{j} \bigl[\partial _{t} \bigl(\mathbf {B}\cdot \nabla B^{j} \bigr)+\operatorname{div} \bigl(\mathbf {u}\mathbf {B}\cdot \nabla B^{j} \bigr) \bigr]\,dx \\ &= \int \bigl(\dot{u}^{j} \bigl(B_{t}^{i} \partial _{i}B^{j}+B^{i}\partial _{i}B_{t}^{j} \bigr)- \partial _{k} \dot{u}^{j}u^{k}B^{i}\partial _{i}B^{j} \bigr)\,dx \\ &\leqslant - \int \bigl(B^{j}\partial _{i}\dot{u}^{j}B_{t}^{i}+B_{t}^{j} \partial _{i}\dot{u}^{j}B^{i}\,dx+ \partial _{k}\dot{u}^{j}u^{k}B^{i} \partial _{i}B^{j} \bigr)\,dx \\ &\leqslant \frac {\mu }{8} \Vert \nabla \dot{\mathbf {u}} \Vert _{L^{2}}^{2}+C \Vert \mathbf {B}\Vert _{L^{3}}^{2} \Vert \nabla \mathbf {B}_{t} \Vert _{L^{2}}^{2}+ C \Vert \mathbf {B}\Vert _{L^{3}}^{4} \bigl\Vert \nabla ^{2}\mathbf {B}\bigr\Vert _{L^{2}}^{4}+ \Vert \mathbf {u}\Vert _{L^{\infty }}^{4}. \end{aligned} $$

(3.27)

Similarly,

$$ J_{5}\leqslant \frac {\mu }{8} \Vert \nabla \dot{ \mathbf {u}} \Vert _{L^{2}}^{2}+C \Vert \mathbf {B}\Vert _{L^{3}}^{2} \Vert \nabla \mathbf {B}_{t} \Vert _{L^{2}}^{2}+ C \Vert \mathbf {B}\Vert _{L^{3}}^{4} \bigl\Vert \nabla ^{2}\mathbf {B}\bigr\Vert _{L^{2}}^{4}+ \Vert \mathbf {u}\Vert _{L^{\infty }}^{4}. $$

(3.28)

Substituting (3.24)–(3.28) into (3.23), we obtain

$$ \begin{aligned} & \bigl( \Vert \sqrt{\rho } \dot{ \mathbf {u}} \Vert _{L^{2}}^{2} \bigr)_{t}+ \Vert \nabla \dot{\mathbf {u}} \Vert _{L^{2}}^{2} \\ &\quad \leqslant C \bigl( \Vert \nabla \mathbf {u}\Vert _{L^{4}}^{4}+ \bigl\Vert P(\rho ) \bigr\Vert _{L^{4}}^{4} \bigr)+C \Vert \mathbf {B}\Vert _{L^{3}}^{2} \Vert \nabla \mathbf {B}_{t} \Vert _{L^{2}}^{2}+ C \Vert \mathbf {B}\Vert _{L^{3}}^{4} \bigl\Vert \nabla ^{2}\mathbf {B}\bigr\Vert _{L^{2}}^{4}+ \Vert \mathbf {u}\Vert _{L^{\infty }}^{4}. \end{aligned} $$

(3.29)

On the other hand, it follows from (1.1)₃ that

$$ \mathbf {B}_{tt}-\nu \Delta \mathbf {B}_{t}= ( \mathbf {B}\cdot \nabla \mathbf {u}- \mathbf {u}\cdot \nabla \mathbf {B}-\mathbf {B}\operatorname{div} \mathbf {u})_{t}. $$

(3.30)

Multiplying (3.30) by $\mathbf {B}_{t}$ and integrating over $\mathbb{R}^{3}$ yields

$$ \begin{aligned} &\biggl(\frac {1}{2} \Vert \mathbf {B}_{t} \Vert _{L^{2}}^{2} \biggr)_{t}+ \nu \Vert \nabla \mathbf {B}_{t} \Vert _{L^{2}}^{2} \\ &\quad = \int ( \mathbf {B}_{t} \cdot \nabla \mathbf {u}-\mathbf {u}\cdot \nabla \mathbf {B}_{t}-\mathbf {B}_{t} \cdot \operatorname{div} \mathbf {u}) \cdot \mathbf {B}_{t}\,dx \\ &\qquad {} + \int \bigl(-\mathbf {B}\cdot \nabla (\mathbf {u}\cdot \nabla \mathbf {u})+( \mathbf {u}\cdot \nabla \mathbf {u})\cdot \nabla \mathbf {B}+\mathbf {B}\operatorname{div}( \mathbf {u}\cdot \nabla \mathbf {u}) \bigr)\cdot \mathbf {B}_{t}\,dx \\ &\qquad {} + \int ( \mathbf {B}\cdot \nabla \dot{\mathbf {u}}-\dot{\mathbf {u}}\cdot \nabla \mathbf {B}-\mathbf {B}\cdot \operatorname{div}\dot{\mathbf {u}} )\cdot \mathbf {B}_{t} \,dx\triangleq \sum_{i=1}^{3}N_{i}. \end{aligned} $$

(3.31)

Now, we estimate $N_{i}$ as follows. By using (2.1), (2.2), and integrating by parts, we obtain

$$ \begin{aligned} N_{1} &= \int ( \mathbf {B}_{t} \cdot \nabla \mathbf {u}- \mathbf {u}\cdot \nabla \mathbf {B}_{t}-\mathbf {B}_{t} \cdot \operatorname{div} \mathbf {u})\cdot \mathbf {B}_{t}\,dx \\ &\leqslant C \Vert \nabla \mathbf {B}_{t} \Vert _{L^{2}} \Vert \mathbf {B}_{t} \Vert _{L^{2}} \bigl( \Vert \nabla \mathbf {u}\Vert _{L^{3}}+ \Vert \mathbf {u}\Vert _{L^{\infty }} \bigr) \\ & \leqslant \frac {\nu }{8} \Vert \nabla \mathbf {B}_{t} \Vert _{L^{2}}^{2}+C \Vert \mathbf {B}_{t} \Vert _{L^{2}}^{2} \Vert \nabla \mathbf {u}\Vert _{L^{2}} \Vert \nabla \mathbf {u}\Vert _{L^{6}}, \end{aligned} $$

(3.32)

and

$$ \begin{aligned} N_{2} &= \int ( \mathbf {B}\cdot \nabla \dot{\mathbf {u}}- \dot{\mathbf {u}}\cdot \nabla \mathbf {B}-\mathbf {B}\cdot \operatorname{div} \dot{\mathbf {u}} )\cdot \mathbf {B}_{t}\,dx\leqslant \frac {1}{4} \Vert \nabla \dot{\mathbf {u}} \Vert _{L^{2}}^{2}+C \Vert \mathbf {B}\Vert _{L^{3}}^{2} \Vert \nabla \mathbf {B}_{t} \Vert _{L^{2}}^{2}, \end{aligned} $$

(3.33)

and

$$ \begin{aligned} N_{3} &= \int \bigl(-\mathbf {B}\cdot \nabla (\mathbf {u}\cdot \nabla \mathbf {u})+(\mathbf {u}\cdot \nabla \mathbf {u})\cdot \nabla \mathbf {B}+ \mathbf {B}\operatorname{div}(\mathbf {u}\cdot \nabla \mathbf {u}) \bigr)\cdot \mathbf {B}_{t}\,dx \\ &= \int \bigl( u^{k} \partial _{k} u^{j} B^{i} \partial _{i}B_{t}^{j} + u^{k} \partial _{k} u^{i} \partial _{i}B^{j} B_{t}^{j}-u^{k} \partial _{k} u^{i} \partial _{i}B^{j} B_{t}^{j} -u^{k} \partial _{k} u^{i} B^{j}\partial _{i}B_{t}^{j} \bigr)\,dx \\ &= \int \bigl( u^{k} \partial _{k} u^{j} B^{i} \partial _{i}B_{t}^{j} -u^{k} \partial _{k} u^{i} B^{j} \partial _{i}B_{t}^{j} \bigr)\,dx \\ &\leqslant \frac {\nu }{8} \Vert \nabla \mathbf {B}_{t} \Vert _{L^{2}}^{2}+C \Vert \nabla \mathbf {B}\Vert _{L^{2}}^{2} \Vert \nabla \mathbf {u}\Vert _{L^{2}}^{2} \Vert \nabla \mathbf {u}\Vert _{L^{6}}^{2}. \end{aligned} $$

(3.34)

Thus, substituting (3.32)–(3.34) into (3.31), we infer that

$$ \begin{aligned} \bigl( \Vert \mathbf {B}_{t} \Vert _{L^{2}}^{2} \bigr)_{t}+ \Vert \nabla \mathbf {B}_{t} \Vert _{L^{2}}^{2}\leqslant{}& \frac {1}{4} \Vert \nabla \dot{\mathbf {u}} \Vert _{L^{2}}^{2}+C \Vert \mathbf {B}\Vert _{L^{3}}^{2} \Vert \nabla \mathbf {B}_{t} \Vert _{L^{2}}^{2} \\ &{} +C \bigl( \Vert \mathbf {B}_{t} \Vert _{L^{2}}^{2} \Vert \nabla \mathbf {u}\Vert _{L^{2}} \Vert \nabla \mathbf {u}\Vert _{L^{6}}+ \Vert \nabla \mathbf {B}\Vert _{L^{2}}^{2} \Vert \nabla \mathbf {u}\Vert _{L^{2}}^{2} \Vert \nabla \mathbf {u}\Vert _{L^{6}}^{2} \bigr). \end{aligned} $$

(3.35)

The combination of (3.29) and (3.35) gives

$$ \begin{aligned} & \bigl( \Vert \sqrt{\rho } \dot{ \mathbf {u}} \Vert _{L^{2}}^{2}+ \Vert \mathbf {B}_{t} \Vert _{L^{2}}^{2} \bigr)_{t}+ \Vert \nabla \dot{ \mathbf {u}} \Vert _{L^{2}}^{2}+ \Vert \nabla \mathbf {B}_{t} \Vert _{L^{2}}^{2} \\ &\quad \leqslant C \bigl( \Vert \nabla \mathbf {u}\Vert _{L^{4}}^{4}+ \bigl\Vert P(\rho ) \bigr\Vert _{L^{4}}^{4} \bigr)+C \Vert \mathbf {B}\Vert _{L^{3}}^{2} \Vert \nabla \mathbf {B}_{t} \Vert _{L^{2}}^{2}+ C \Vert \mathbf {B}\Vert _{L^{3}}^{4} \bigl\Vert \nabla ^{2}\mathbf {B}\bigr\Vert _{L^{2}}^{4}+ \Vert \mathbf {u}\Vert _{L^{\infty }}^{4} \\ &\qquad {} +C \bigl( \Vert \mathbf {B}_{t} \Vert _{L^{2}}^{2} \Vert \nabla \mathbf {u}\Vert _{L^{2}} \Vert \nabla \mathbf {u}\Vert _{L^{6}}+ \Vert \nabla \mathbf {B}\Vert _{L^{2}}^{2} \Vert \nabla \mathbf {u}\Vert _{L^{2}}^{2} \Vert \nabla \mathbf {u}\Vert _{L^{6}}^{2} \bigr). \end{aligned} $$

(3.36)

The right-hand side of (3.36) on $[0, T]$ can be estimated as follows. It holds by (3.1) that

$$ \begin{aligned} \int _{0}^{T} \Vert \mathbf {B}\Vert _{L^{3}}^{4} \bigl\Vert \nabla ^{2} \mathbf {B}\bigr\Vert _{L^{2}}^{4}\,dt &\leqslant C\sup_{t\in [0, T]} \bigl( \Vert \mathbf {B}\Vert _{L^{3}}^{4} \bigl\Vert \nabla ^{2}\mathbf {B}\bigr\Vert _{L^{2}}^{2} \bigr) \int _{0}^{T} \bigl\Vert \nabla ^{2} \mathbf {B}\bigr\Vert _{L^{2}}^{2}\,dt \\ &\leqslant CKA_{2}(T) \Bigl(\sup_{t\in [0, T]} \Vert \mathbf {B}\Vert _{L^{3}}^{4} \Bigr). \end{aligned} $$

(3.37)

An application of the $L^{p}$-theory for the elliptic equation (1.1)₂ leads to

$$ \begin{aligned} \Vert \nabla \mathbf {u}\Vert _{L^{6}} & \leqslant C \bigl( \Vert \rho \dot{\mathbf {u}} \Vert _{L^{2}}+ \bigl\Vert P(\rho ) \bigr\Vert _{L^{6}}+ \Vert \mathbf {B}\Vert _{L^{\infty }} \Vert \nabla \mathbf {B}\Vert _{L^{2}} \bigr) \\ & \leqslant C \bigl(M_{0}^{1/2} \Vert \sqrt{\rho } \dot{ \mathbf {u}} \Vert _{L^{2}}+M_{0}^{5 \gamma /6} \bigl\Vert P(\rho ) \bigr\Vert _{L^{1}}^{1/6}+ \Vert \mathbf {B}\Vert _{L^{3}} \bigl\Vert \nabla ^{2} \mathbf {B}\bigr\Vert _{L^{2}} \bigr) \\ & \leqslant C \bigl(M_{0}^{1/2} \Vert \sqrt{\rho } \dot{ \mathbf {u}} \Vert _{L^{2}}+M_{0}^{5 \gamma /6}E_{0}^{1/6}+ \Vert \mathbf {B}\Vert _{L^{3}} \bigl\Vert \nabla ^{2} \mathbf {B}\bigr\Vert _{L^{2}} \bigr) \\ & \leqslant C \bigl(M_{0}^{1/2}K^{1/2}+M_{0}^{5\gamma /6}+ \Vert \mathbf {B}\Vert _{L^{3}}K^{1/2} \bigr), \end{aligned} $$

(3.38)

where we have used (3.1). Thus

$$ \begin{aligned} & \int _{0}^{T} \bigl( \Vert \mathbf {B}_{t} \Vert _{L^{2}}^{2} \Vert \nabla \mathbf {u}\Vert _{L^{2}} \Vert \nabla \mathbf {u}\Vert _{L^{6}}+ \Vert \nabla \mathbf {B}\Vert _{L^{2}}^{2} \Vert \nabla \mathbf {u}\Vert _{L^{2}}^{2} \Vert \nabla \mathbf {u}\Vert _{L^{6}}^{2} \bigr)\,dt \\ &\quad \leqslant C K^{1/2} \Bigl(M_{0}^{1/2}K^{1/2}+M_{0}^{5\gamma /6}+ \Bigl( \sup_{t\in [0, T]} \Vert \mathbf {B}\Vert _{L^{3}} \Bigr)K^{1/2} \Bigr) \int _{0}^{T} \Vert \mathbf {B}_{t} \Vert _{L^{2}}^{2}\,dt \\ & \qquad {}+ C \Bigl(\sup_{t\in [0, T]} \Vert \nabla \mathbf {u}\Vert _{L^{2}}^{2} \Bigr) \Bigl(M_{0}K+M_{0}^{5\gamma /3}+ \Bigl(\sup_{t\in [0, T]} \Vert \mathbf {B}\Vert _{L^{3}}^{2} \Bigr)K \Bigr) \int _{0}^{T} \Vert \nabla \mathbf {B}\Vert _{L^{2}}^{2}\,dt \\ & \quad \leqslant C \Bigl(M_{0}^{1/2}K+M_{0}^{5\gamma /6}K^{1/2}+ \Bigl( \sup_{t\in [0, T]} \Vert \mathbf {B}\Vert _{L^{3}} \Bigr)K \Bigr)A_{1}(T). \end{aligned} $$

(3.39)

In addition, it holds by (1.1)₃, (2.1), (2.2), and (3.1) that

$$ \bigl\Vert \nabla ^{2}\mathbf {B}\bigr\Vert _{L^{2}}^{2} \leqslant C \bigl( \Vert \mathbf {B}_{t} \Vert _{L^{2}}^{2}+K \Vert \mathbf {B}\Vert _{L^{3}}^{2} \Vert \nabla \mathbf {u}\Vert _{L^{2}}^{2} \bigr), $$

which, together with (3.36)–(3.39), (3.22), and Gronwall’s inequality, gives rise to (3.9). □

We are now in a position of providing the concluding estimates of $A_{1}(T)$ and $A_{2}(T)$.

Lemma 3.4

Let $(\rho , \mathbf{ u}, \mathbf{ B})$ be a smooth solution of (1.1)–(1.5) on $\mathbb{R}^{3}\times [0,T]$, satisfying $0\leqslant \rho \leqslant 2M_{0}$. Then there exist positive numbers $\varepsilon _{1}$ and K, depending only on μ, λ, ν, γ, A, $E_{0}$, $M_{1}$, and $M_{2}$, such that

$$ A_{1}(T)+A_{2}(T)\leqslant K, $$

(3.40)

provided

$$ A_{1}(T)+A_{2}(T)\leqslant 2K\quad \textit{and}\quad \Vert \mathbf{ B}_{0} \Vert _{L^{3}}^{\frac {1}{2\gamma }} \leqslant M_{0}\leqslant \varepsilon _{1}. $$

(3.41)

Proof

Indeed, it follows from (2.8) that

$$ \int _{0}^{T} \Vert \nabla \mathbf {u}\Vert _{L^{4}}^{4}\,dt\leqslant C \int _{0}^{T} \bigl( \Vert F \Vert _{L^{4}}^{4}+ \bigl\Vert P(\rho ) \bigr\Vert _{L^{4}}^{4}+ \bigl\Vert |\mathbf {B}|^{2} \bigr\Vert _{L^{4}}^{4}+ \Vert \omega \Vert _{L^{4}}^{4} \bigr)\,dt. $$

(3.42)

To estimate $\|P(\rho )\|_{L^{4}}$, we multiply (3.12) by $3(P(\rho ))^{2}$, integrate it over $\mathbb{R}^{3}$ and utilize (2.3) to get that

$$ \begin{aligned} & \biggl( \int \bigl(P(\rho ) \bigr)^{3}\,dx \biggr)_{t}+ \frac {3\gamma -1}{2 \mu +\lambda } \bigl\Vert P(\rho ) \bigr\Vert _{L^{4}}^{4}+ \frac {3\gamma -1}{2(2\mu + \lambda )} \int \vert \mathbf {B}\vert ^{2}P^{3}\,dx \\ &\quad =-\frac {3\gamma -1}{2\mu +\lambda } \int F\cdot P^{3}\,dx \leqslant \frac {3\gamma -1}{2(2\mu +\lambda )} \bigl\Vert P(\rho ) \bigr\Vert _{L^{4}}^{4}+C \Vert F \Vert _{L^{4}}^{4}, \end{aligned} $$

(3.43)

thus

$$ \int _{0}^{T} \bigl\Vert P(\rho ) \bigr\Vert _{L^{4}}^{4}\,dt\leqslant C \int _{0}^{T} \Vert F \Vert _{L^{4}}^{4}\,dt+CM_{0}^{2 \gamma }. $$

(3.44)

Using (2.1), (2.2), (2.5)–(2.7), and the fact that $0\leqslant \rho \leqslant 2M_{0}$, we deduce

$$ \begin{aligned} & \int _{0}^{T} \bigl( \Vert F \Vert _{L^{4}}^{4}+ \bigl\Vert |\mathbf {B}|^{2} \bigr\Vert _{L^{4}}^{4}+ \Vert \omega \Vert _{L^{4}}^{4} \bigr)\,dt \\ &\quad \leqslant \int _{0}^{T} \bigl( \Vert F \Vert _{L^{2}}^{2}+ \Vert \omega \Vert _{L^{2}}^{2} \bigr) \bigl( \Vert F \Vert _{L^{\infty }}^{2}+ \Vert \omega \Vert _{L^{\infty }}^{2} \bigr)\,dt+ \int _{0}^{T} \Vert \mathbf {B}\Vert _{L^{8}}^{8}\,dt \\ &\quad \leqslant \int _{0}^{T} \bigl( \Vert \nabla \mathbf {u}\Vert _{L^{2}}^{2}+ \bigl\Vert P(\rho ) \bigr\Vert _{L^{2}}^{2}+ \bigl\Vert |\mathbf {B}|^{2} \bigr\Vert _{L^{2}}^{2} \bigr) \bigl( \Vert \rho \dot{\mathbf {u}} \Vert _{L^{2}}+ \Vert \nabla \mathbf {B}\cdot \mathbf {B}\Vert _{L^{2}} \bigr) \\ & \qquad {}\times \bigl( \Vert \rho \dot{\mathbf {u}} \Vert _{L^{6}}+ \Vert \nabla \mathbf {B}\cdot \mathbf {B}\Vert _{L^{6}} \bigr)\,dt + \int _{0}^{T} \Vert \mathbf {B}\Vert _{L^{8}}^{8}\,dt \end{aligned} $$

(3.45)

Now, we estimate the right-hand side of (3.45) as follows:

$$ \begin{aligned} & \int _{0}^{T} \bigl( \Vert \nabla \mathbf {u}\Vert _{L^{2}}^{2}+ \bigl\Vert P(\rho ) \bigr\Vert _{L^{2}}^{2} \bigr) \Vert \rho \dot{\mathbf {u}} \Vert _{L^{2}} \Vert \rho \dot{\mathbf {u}} \Vert _{L^{6}}\,dt \\ &\quad \leqslant CM_{0}^{3/2} \bigl(A_{1}(T)+M_{0}^{\gamma } \bigr) \biggl( \int _{0}^{T} \Vert \sqrt{ \rho } \dot{\mathbf {u}} \Vert _{L^{2}}^{2}\,dt \biggr)^{1/2} \biggl( \int _{0}^{T} \Vert \nabla \dot{\mathbf {u}} \Vert _{L^{2}}^{2}\,dt \biggr)^{1/2} \\ &\quad \leqslant CM_{0}^{3/2} \bigl(A_{1}(T)+M_{0}^{\gamma } \bigr) A_{1}^{1/2}(T)A_{2}^{1/2}(T). \end{aligned} $$

(3.46)

It follows from (2.1) and (2.2) that

$$ \Vert \nabla \mathbf {B}\cdot \mathbf {B}\Vert _{L^{6}} \leqslant C \bigl\Vert \nabla ( \nabla \mathbf {B}\cdot \mathbf {B}) \bigr\Vert _{L^{2}}\leqslant C \Vert \nabla \mathbf {B}\Vert _{L^{2}}^{1/2} \bigl\Vert \nabla ^{2} \mathbf {B}\bigr\Vert _{L^{2}}^{3/2}. $$

(3.47)

Hence

$$ \begin{aligned} & \int _{0}^{T} \bigl( \Vert \nabla \mathbf {u}\Vert _{L^{2}}^{2}+ \bigl\Vert P(\rho ) \bigr\Vert _{L^{2}}^{2} \bigr) \Vert \rho \dot{\mathbf {u}} \Vert _{L^{2}} \Vert \nabla \mathbf {B}\cdot \mathbf {B}\Vert _{L^{6}} \,dt \\ &\quad \leqslant CM_{0}^{1/2} \bigl(A_{1}(T)+M_{0}^{\gamma } \bigr)A_{2}^{1/2}(T) \biggl( \int _{0}^{T} \Vert \nabla \mathbf {B}\Vert _{L^{2}}^{2}\,dt \biggr)^{1/4} \biggl( \int _{0}^{T} \bigl\Vert \nabla ^{2} \mathbf {B}\bigr\Vert _{L^{2}}^{2}\,dt \biggr)^{3/4} \\ &\quad \leqslant CM_{0}^{1/2} \bigl(A_{1}(T)+M_{0}^{\gamma } \bigr) A_{1}^{3/4}(T)A_{2}^{1/2}(T), \end{aligned} $$

(3.48)

and

$$ \begin{aligned} & \int _{0}^{T} \bigl( \Vert \nabla \mathbf {u}\Vert _{L^{2}}^{2}+ \bigl\Vert P(\rho ) \bigr\Vert _{L^{2}}^{2} \bigr) \Vert \rho \dot{\mathbf {u}} \Vert _{L^{6}} \Vert \nabla \mathbf {B}\cdot \mathbf {B}\Vert _{L^{2}} \,dt \\ &\quad \leqslant CM_{0} \bigl(A_{1}(T)+M_{0}^{\gamma } \bigr) \int _{0}^{T} \bigl\Vert \nabla ^{2} \mathbf {B}\bigr\Vert _{L^{2}} \Vert \mathbf {B}\Vert _{L^{3}} \Vert \nabla \dot{\mathbf {u}} \Vert _{L^{2}} \\ &\quad \leqslant CM_{0} \Bigl(\sup_{t\in [0, T]} \Vert \mathbf {B}\Vert _{L^{3}} \Bigr) \bigl(A_{1}(T)+M_{0}^{\gamma } \bigr) \biggl( \int _{0}^{T} \bigl\Vert \nabla ^{2} \mathbf {B}\bigr\Vert _{L^{2}}^{2}\,dt \biggr)^{1/2} \\ &\qquad {}\times \biggl( \int _{0}^{T} \Vert \nabla \dot{\mathbf {u}} \Vert _{L^{2}}^{2}\,dt \biggr)^{1/2} \\ &\quad \leqslant CM_{0} \Bigl(\sup_{t\in [0, T]} \Vert \mathbf {B}\Vert _{L^{3}} \Bigr) \bigl(A_{1}(T)+M_{0}^{\gamma } \bigr) A_{1}^{1/2}(T)A_{2}^{1/2}(T). \end{aligned} $$

(3.49)

It follows from (3.47) that

$$ \begin{aligned} & \int _{0}^{T} \bigl( \Vert \nabla \mathbf {u}\Vert _{L^{2}}^{2}+ \bigl\Vert P(\rho ) \bigr\Vert _{L^{2}}^{2} \bigr) \Vert \nabla \mathbf {B}\cdot \mathbf {B}\Vert _{L^{6}} \Vert \nabla \mathbf {B}\cdot \mathbf {B}\Vert _{L^{2}}\,dt \\ &\quad \leqslant C \bigl(A_{1}(T)+M_{0}^{\gamma } \bigr) \int _{0}^{T} \bigl\Vert \nabla ^{2} \mathbf {B}\bigr\Vert _{L^{2}}^{5/2} \Vert \mathbf {B}\Vert _{L^{3}} \Vert \nabla \mathbf {B}\Vert _{L^{2}}^{1/2} \\ &\quad \leqslant C \Bigl(\sup_{t\in [0, T]} \Vert \mathbf {B}\Vert _{L^{3}} \Bigr) \bigl(A_{1}(T)+M_{0}^{\gamma } \bigr)A_{2}^{1/2} \biggl( \int _{0}^{T} \bigl\Vert \nabla ^{2} \mathbf {B}\bigr\Vert _{L^{2}}^{2}\,dt \biggr)^{3/4} \biggl( \int _{0}^{T} \Vert \nabla \mathbf {B}\Vert _{L^{2}}^{2}\,dt \biggr)^{1/4} \\ &\quad \leqslant C \Bigl(\sup_{t\in [0, T]} \Vert \mathbf {B}\Vert _{L^{3}} \Bigr) \bigl(A_{1}(T)+M_{0}^{\gamma } \bigr) A_{1}^{3/4}(T)A_{2}^{1/2}(T). \end{aligned} $$

(3.50)

Similarly,

$$ \begin{aligned} & \int _{0}^{T} \bigl\Vert \vert \mathbf {B}\vert ^{2} \bigr\Vert _{L^{2}}^{2} \Vert \rho \dot{ \mathbf {u}} \Vert _{L^{2}} \bigl( \Vert \rho \dot{\mathbf {u}} \Vert _{L^{6}}+ \Vert \nabla \mathbf {B}\cdot \mathbf {B}\Vert _{L^{6}} \bigr)\,dt \\ &\quad \leqslant C \Bigl(\sup_{t\in [0, T]} \Vert \mathbf {B}\Vert _{L^{3}}^{2} \Bigr) \bigl(M_{0}^{3/2}A_{1}(T)+M_{0}^{1/2}A_{1}^{5/4}(T) \bigr) A_{1}^{1/2}(T)A_{2}^{1/2}(T), \end{aligned} $$

(3.51)

and

$$ \begin{aligned} & \int _{0}^{T} \bigl[ \bigl\Vert \vert \mathbf {B}\vert ^{2} \bigr\Vert _{L^{2}}^{2} \Vert \nabla \mathbf {B}\cdot \mathbf {B}\Vert _{L^{2}} \bigl( \Vert \rho \dot{\mathbf {u}} \Vert _{L^{6}}+ \Vert \nabla \mathbf {B}\cdot \mathbf {B}\Vert _{L^{6}} \bigr)+ \Vert \mathbf {B}\Vert _{L^{8}}^{8} \bigr]\,dt \\ &\quad \leqslant C \Bigl(\sup_{t\in [0, T]} \Vert \mathbf {B}\Vert _{L^{3}}^{3} \Bigr) \bigl(M_{0}A_{1}(T)+A_{1}^{3/4}(T)A_{2}^{1/4}(T) \bigr) A_{1}^{1/2}(T)A_{2}^{1/2}(T). \end{aligned} $$

(3.52)

Substituting (3.46)–(3.52) into (3.45), by virtue of (3.42) and (3.44), one obtains

$$\begin{aligned}& \int _{0}^{T} \bigl( \Vert \nabla \mathbf {u}\Vert _{L^{4}}^{4}+ \bigl\Vert P(\rho ) \bigr\Vert _{L^{4}}^{4} \bigr)\,dt \\& \quad \leqslant CM_{0}^{2\gamma }+C \bigl(A_{1}(T)+M_{0}^{\gamma } \bigr) \bigl(M_{0}^{3/2}+ M_{0}^{1/2}A_{1}^{1/4} \bigr)A_{1}^{1/2}(T)A_{2}^{1/2}(T) \\& \qquad {} +C \bigl(A_{1}(T)+M_{0}^{\gamma } \bigr) \Bigl(\sup _{t\in [0, T]} \Vert \mathbf {B}\Vert _{L^{3}} \Bigr) \bigl(M_{0}+A_{1}^{1/4}(T) \bigr)A_{1}^{1/2}(T)A_{2}^{1/2}(T) \\& \qquad {} +C \Bigl(\sup_{t\in [0, T]} \Vert \mathbf {B}\Vert _{L^{3}}^{2} \Bigr) \bigl(M_{0}^{3/2}A_{1}(T)+M_{0}^{1/2}A_{1}^{5/4}(T) \bigr) A_{1}^{1/2}(T)A_{2}^{1/2}(T) \\& \qquad {} +C \Bigl(\sup_{t\in [0, T]} \Vert \mathbf {B}\Vert _{L^{3}}^{3} \Bigr) \bigl(M_{0}A_{1}(T)+A_{1}^{3/4}(T)A_{2}^{1/4}(T) \bigr)A_{1}^{1/2}(T)A_{2}^{1/2}(T). \end{aligned}$$

(3.53)

Collecting (3.6), (3.8), (3.9), and (3.53) together, and choosing $K\geqslant 1$, $M_{0}\leqslant 1/2$, we immediately obtain

$$ \begin{aligned} &A_{1}(T)+A_{2}(T) \\ &\quad \leqslant C \bigl(M_{0}^{1/2}K+M_{0}^{5\gamma /6}K^{1/2}+e^{CK}M_{0}^{2 \gamma +1}+e^{CK}M_{0}^{2\gamma } \bigr)A_{1}(T) \\ &\qquad {} +C \bigl(e^{2CK}M_{0}^{4\gamma }+e^{4CK}M_{0}^{8\gamma } \bigr)A_{2}(T)+C \bigl(1+M_{0}^{\gamma }+M_{0}^{2\gamma } \bigr) \\ &\qquad {} + C \bigl(M_{0}^{1/2}+e^{CK}M_{0}^{2\gamma } \bigr) \bigl(A_{1}(T)+M_{0}^{\gamma } \bigr) \bigl(M_{0}+ A_{1}^{1/4}(T) \bigr)A_{1}^{1/2}(T)A_{2}^{1/2}(T) \\ &\qquad {} +Ce^{2CK}M_{0}^{4\gamma } \bigl(M_{0}^{3/2}A_{1}(T)+M_{0}^{1/2}A_{1}^{5/4}(T) \bigr) A_{1}^{1/2}(T)A_{2}^{1/2}(T) \\ &\qquad {} +Ce^{3CK}M_{0}^{6\gamma } \bigl(M_{0}A_{1}(T)+A_{1}^{3/4}(T)A_{2}^{1/4}(T) \bigr)A_{1}^{1/2}(T)A_{2}^{1/2}(T) \\ &\quad \leqslant C_{1}e^{4C_{0}K}M_{0}^{1/2} \bigl[A_{1}(T)+A_{2}(T) \bigr] +C_{2} \bigl(1+M_{0}^{\gamma }+M_{0}^{2\gamma } \bigr) \\ &\qquad {} +C_{3}e^{3C_{0}K}M_{0}^{1/2} \bigl(A_{1}^{5/4}(T)+A_{2}^{5/8}(T) \bigr)A_{1}^{1/2}(T)A_{2}^{1/2}(T), \end{aligned} $$

(3.54)

where $C_{i}$ $(i=0, 1, 2, 3)>0$ depend on μ, λ, ν, A, $E_{0}$, $M_{1}$, and $M_{2}$, but not on $M_{0}$ and T.

To continue, set

$$ K \triangleq \max \{1, 8C_{2}, M_{1}+M_{2}\}. $$

Thus, if it holds that

$$ M_{0}\leqslant \varepsilon _{1} \triangleq \min \biggl\{ \frac {1}{2}, \biggl(\frac {1}{2C_{1}e^{4C_{0}K}} \biggr)^{2}, \biggl( \frac {1}{16C_{3}e^{3C_{0}K^{5/4}}} \biggr)^{2} \biggr\} , $$

then one infers from (3.54) that

$$ \begin{aligned} A_{1}(T)+A_{2}(T) & \leqslant 4C_{2}+16C_{3}e^{3C_{0}K} M_{0}^{1/2}K^{5/4}A_{1}^{1/2}(T)A_{2}^{1/2}(T) \\ &\leqslant \frac {K}{2}+\frac {1}{2} \bigl[A_{1}(T)+A_{2}(T) \bigr], \end{aligned} $$

(3.55)

provided (3.41) holds. Thus, the desired estimate of (3.40) immediately follows from (3.55) by choosing K and $\varepsilon _{1}$ as above. □

In order to derive a uniform upper bound of the density, we still need the following t-weighted estimate.

Lemma 3.5

Let the conditions of Lemma 3.4be in force. Then there exists a positive constant $\varepsilon _{2}$, depending only on μ, λ, ν, γ, A, $E_{0}$, $M_{1}$, and $M_{2}$, such that if $\|\mathbf{ B}_{0}\|_{L^{3}}^{1/2\gamma }\leqslant M_{0}\leqslant \varepsilon _{2}$, then for any $0\leqslant t_{1}< t_{2}\leqslant T$,

$$ \begin{aligned} &\sup_{t_{1}\leqslant t\leqslant t_{2}} \bigl[(t-t_{1}) \bigl( \Vert \sqrt{\rho }\dot{\mathbf{ u}} \Vert _{L^{2}}^{2}+ \Vert \mathbf{ B}_{t} \Vert _{L^{2}}^{2}+ \bigl\Vert \nabla ^{2} \mathbf{ B} \bigr\Vert _{L^{2}}^{2} \bigr) \bigr] \\ &\quad {} + \int _{t_{1}}^{t_{2}}(t-t_{1}) \bigl( \Vert \nabla \dot{\mathbf{ u}} \Vert _{L^{2}}^{2}+ \Vert \nabla \mathbf{ B}_{t} \Vert _{L^{2}}^{2} \bigr)\,dt \leqslant C \bigl[1+M_{0}^{\frac {5\gamma }{2}}(t_{2}-t_{1}) \bigr], \end{aligned} $$

(3.56)

where the constant $C>0$ depends on μ, λ, ν, γ, A, $E_{0}$, $M_{1}$, $M_{2}$, K, and the coefficients of the Gagliardo–Nirenberg–Sobolev inequality in Lemma 2.1, but is independent of $M_{0}$.

Proof

Indeed, it follows from (3.6), (3.38), and (3.40) that

$$ \Vert \nabla \mathbf {u}\Vert _{L^{6}} \leqslant C \bigl(M_{0}^{1/2} \Vert \sqrt{ \rho } \dot{\mathbf {u}} \Vert _{L^{2}}+M_{0}^{5\gamma /6}+M_{0}^{2\gamma } \bigl\Vert \nabla ^{2} \mathbf {B}\bigr\Vert _{L^{2}} \bigr), $$

(3.57)

thus

$$ \begin{aligned} &\Vert \nabla \mathbf {u}\Vert _{L^{4}}^{4}+ \bigl\Vert P(\rho ) \bigr\Vert _{L^{4}}^{4} \\ &\quad \leqslant C \Vert \nabla \mathbf {u}\Vert _{L^{2}} \Vert \nabla \mathbf {u}\Vert _{L^{6}}^{3}+CM_{0}^{3 \gamma } \\ &\quad \leqslant C \bigl(M_{0}^{5\gamma /2}+M_{0}^{3/2} \Vert \nabla \mathbf {u}\Vert _{L^{2}} \Vert \sqrt{\rho } \dot{\mathbf {u}} \Vert _{L^{2}}^{3}+M_{0}^{6\gamma } \Vert \nabla \mathbf {u}\Vert _{L^{2}} \bigl\Vert \nabla ^{2} \mathbf {B}\bigr\Vert _{L^{2}}^{3} \bigr). \end{aligned} $$

(3.58)

Multiplying (3.36) by $(t-t_{1})$, integrating it over $(t_{1}, t)$ with $t_{1}\leqslant t\leqslant t_{2}$, using (3.6) and (3.40), we get

$$ \begin{aligned} &\sup_{t_{1}\leqslant t \leqslant t_{2}} \bigl[(t-t_{1}) \bigl( \Vert \sqrt{\rho } \dot{\mathbf {u}} \Vert _{L^{2}}^{2}+ \Vert \mathbf {B}_{t} \Vert _{L^{2}}^{2} \bigr) \bigr]+ \int _{t_{1}}^{t_{2}}(t-t_{1}) \bigl( \Vert \nabla \dot{\mathbf {u}} \Vert _{L^{2}}^{2}+ \Vert \nabla \mathbf {B}_{t} \Vert _{L^{2}}^{2} \bigr)\,dt \\ &\quad \leqslant C+ \int _{t_{1}}^{t_{2}}(t-t_{1}) \bigl( \Vert \nabla \mathbf {u}\Vert _{L^{4}}^{4}+ \bigl\Vert P(\rho ) \bigr\Vert _{L^{4}}^{4}+M_{0}^{8\gamma } \bigl\Vert \nabla ^{2} \mathbf {B}\bigr\Vert _{L^{2}}^{4} \bigr)\,dt \\ &\qquad {} +C \int _{t_{1}}^{t_{2}}(t-t_{1}) \bigl( \Vert \mathbf {B}_{t} \Vert _{L^{2}}^{2} \Vert \nabla \mathbf {u}\Vert _{L^{2}} \Vert \nabla \mathbf {u}\Vert _{L^{6}} \bigr)\,dt \\ &\qquad {} +C \int _{t_{1}}^{t_{2}}(t-t_{1}) \bigl( \Vert \nabla \mathbf {B}\Vert _{L^{2}}^{2} \Vert \nabla \mathbf {u}\Vert _{L^{2}}^{2} \Vert \nabla \mathbf {u}\Vert _{L^{6}}^{2}+ \Vert \mathbf {u}\Vert _{L^{\infty }}^{4} \bigr)\,dt \\ &\quad \triangleq C+J_{1}+J_{2}+J_{3}, \end{aligned} $$

(3.59)

where the second term on the right-hand side can be estimated as follows based on (3.58):

$$ \begin{aligned} J_{1}\leqslant{}& C \int _{t_{1}}^{t_{2}}(t-t_{1}) \bigl(M_{0}^{5 \gamma /2}+M_{0}^{3/2} \Vert \nabla \mathbf {u}\Vert _{L^{2}} \Vert \sqrt{\rho } \dot{\mathbf {u}} \Vert _{L^{2}}^{3} +M_{0}^{6\gamma } \Vert \nabla \mathbf {u}\Vert _{L^{2}} \bigl\Vert \nabla ^{2} \mathbf {B}\bigr\Vert _{L^{2}}^{3} \\ &{} +M_{0}^{8\gamma } \bigl\Vert \nabla ^{2} \mathbf {B}\bigr\Vert _{L^{2}}^{4} \bigr)\,dt \\ \leqslant{}& CM_{0}^{3/2}\sup_{t_{1}\leqslant t\leqslant t_{2}} \bigl[(t-t_{1}) \Vert \sqrt{\rho } \dot{\mathbf {u}} \Vert _{L^{2}}^{2} \bigr] \biggl( \int _{t_{1}}^{t_{2}} \Vert \nabla \mathbf {u}\Vert _{L^{2}}^{2}\,dt \biggr)^{1/2} \biggl( \int _{t_{1}}^{t_{2}} \Vert \sqrt{\rho }\dot{\mathbf {u}} \Vert _{L^{2}}^{2}\,dt \biggr)^{1/2} \\ &{} +CM_{0}^{6\gamma }\sup_{t_{1}\leqslant t\leqslant t_{2}} \bigl[(t-t_{1}) \bigl\Vert \nabla ^{2} \mathbf {B}\bigr\Vert _{L^{2}}^{2} \bigr] \biggl( \int _{t_{1}}^{t_{2}} \Vert \nabla \mathbf {u}\Vert _{L^{2}}^{2}\,dt \biggr)^{1/2} \biggl( \int _{t_{1}}^{t_{2}} \bigl\Vert \nabla ^{2} \mathbf {B}\bigr\Vert _{L^{2}}^{2}\,dt \biggr)^{1/2} \\ &{} +CM_{0}^{8\gamma }\sup_{t_{1}\leqslant t\leqslant t_{2}} \bigl[(t-t_{1}) \bigl\Vert \nabla ^{2} \mathbf {B}\bigr\Vert _{L^{2}}^{2} \bigr] \int _{t_{1}}^{t_{2}} \bigl\Vert \nabla ^{2} \mathbf {B}\bigr\Vert _{L^{2}}^{2}\,dt+C M_{0}^{5\gamma /2}(t_{1}-t_{2})^{2} \\ \leqslant{}& C M_{0}^{3/2}\sup_{t_{1}\leqslant t\leqslant t_{2}} \bigl[(t-t_{1}) \Vert \sqrt{\rho } \dot{\mathbf {u}} \Vert _{L^{2}}^{2} \bigr] +CM_{0}^{6\gamma } \sup _{t_{1}\leqslant t\leqslant t_{2}} \bigl[(t-t_{1}) \bigl\Vert \nabla ^{2} \mathbf {B}\bigr\Vert _{L^{2}}^{2} \bigr] \\ &{} + C M_{0}^{5\gamma /2}(t_{1}-t_{2})^{2}. \end{aligned} $$

By virtue of (3.57) and Cauchy–Schwarz inequality, we also have

$$\begin{aligned} J_{2} \leqslant& C \int _{t_{1}}^{t_{2}}(t-t_{1}) \Vert \mathbf {B}_{t} \Vert _{L^{2}}^{2} \Vert \nabla \mathbf {u}\Vert _{L^{2}} \bigl(M_{0}^{5 \gamma /6}+M_{0}^{1/2} \Vert \sqrt{\rho } \dot{\mathbf {u}} \Vert _{L^{2}} +M_{0}^{2 \gamma } \bigl\Vert \nabla ^{2} \mathbf {B}\bigr\Vert _{L^{2}} \bigr) \,dt \\ \leqslant& CM_{0}^{1/2}\sup_{t_{1}\leqslant t\leqslant t_{2}} \bigl[(t-t_{1}) \Vert \mathbf {B}_{t} \Vert _{L^{2}}^{2} \bigr] \biggl( \int _{t_{1}}^{t_{2}} \Vert \nabla \mathbf {u}\Vert _{L^{2}}^{2}\,dt \biggr)^{1/2} \biggl( \int _{t_{1}}^{t_{2}} \Vert \sqrt{\rho }\dot{\mathbf {u}} \Vert _{L^{2}}^{2}\,dt \biggr)^{1/2} \\ &{} +CM_{0}^{5\gamma /6}\sup_{t_{1}\leqslant t\leqslant t_{2}} \bigl[(t-t_{1}) \Vert \mathbf {B}_{t} \Vert _{L^{2}} \bigr] \biggl( \int _{t_{1}}^{t_{2}} \Vert \nabla \mathbf {u}\Vert _{L^{2}}^{2}\,dt \biggr)^{1/2} \biggl( \int _{t_{1}}^{t_{2}} \Vert \mathbf {B}_{t} \Vert _{L^{2}}^{2}\,dt \biggr)^{1/2} \\ &{} +CM_{0}^{2\gamma }\sup_{t_{1}\leqslant t\leqslant t_{2}} \bigl[(t-t_{1}) \Vert \mathbf {B}_{t} \Vert _{L^{2}}^{2} \bigr] \biggl( \int _{t_{1}}^{t_{2}} \bigl\Vert \nabla ^{2} \mathbf {B}\bigr\Vert _{L^{2}}^{2}\,dt \biggr)^{1/2} \biggl( \int _{t_{1}}^{t_{2}} \Vert \nabla \mathbf {u}\Vert _{L^{2}}^{2}\,dt \biggr)^{1/2} \\ \leqslant& C \bigl(M_{0}^{1/2}+M_{0}^{5\gamma /12} \bigr)\sup_{t_{1}\leqslant t \leqslant t_{2}} \bigl[(t-t_{1}) \Vert \mathbf {B}_{t} \Vert _{L^{2}}^{2} \bigr]+ C M_{0}^{5\gamma /2}(t_{1}-t_{2})^{2}+C. \end{aligned}$$

Similarly, we infer from (3.40) that

$$ \begin{aligned} J_{3}\leqslant{}& C \int _{t_{1}}^{t_{2}}(t-t_{1}) \Vert \nabla \mathbf {u}\Vert _{L^{2}}^{2} \bigl(M_{0}^{5\gamma /3}+M_{0} \Vert \sqrt{ \rho } \dot{\mathbf {u}} \Vert _{L^{2}}^{2} +M_{0}^{4\gamma } \bigl\Vert \nabla ^{2} \mathbf {B}\bigr\Vert _{L^{2}}^{2} \bigr)\,dt \\ \leqslant{}& C M_{0}\sup_{t_{1}\leqslant t\leqslant t_{2}} \bigl[(t-t_{1}) \Vert \sqrt{\rho }\dot{\mathbf {u}} \Vert _{L^{2}}^{2} \bigr]+C M_{0}^{4\gamma } \sup _{t_{1}\leqslant t\leqslant t_{2}} \bigl[(t-t_{1}) \bigl\Vert \nabla ^{2} \mathbf {B}\bigr\Vert _{L^{2}}^{2} \bigr] \\ &{} + C M_{0}^{10\gamma /3}(t_{1}-t_{2})^{2}+C. \end{aligned} $$

In addition, it holds by (1.1)₃ and (3.40) that

$$ \begin{aligned} \sup_{t_{1}\leqslant t\leqslant t_{2}} \bigl[(t-t_{1}) \bigl\Vert \nabla ^{2} \mathbf {B}\bigr\Vert _{L^{2}}^{2} \bigr] &\leqslant C \sup_{t_{1} \leqslant t\leqslant t_{2}} \bigl[(t-t_{1}) \bigl( \Vert \mathbf {B}_{t} \Vert _{L^{2}}^{2} + \Vert \mathbf {B}\Vert _{L^{3}}^{2} \Vert \nabla \mathbf {u}\Vert _{L^{2}}^{2} \bigr) \bigr] \\ &\leqslant C \sup_{t_{1}\leqslant t\leqslant t_{2}} \bigl[(t-t_{1}) \Vert \mathbf {B}_{t} \Vert _{L^{2}}^{2} \bigr] + C M_{0}^{8\gamma }(t_{1}-t_{2})^{2}+C. \end{aligned} $$

(3.60)

Substituting $J_{1}$, $J_{2}$, and $J_{3}$ into (3.59) and using (3.60), we obtain

$$ \begin{aligned} &\sup_{t_{1}\leqslant t \leqslant t_{2}} \bigl[(t-t_{1}) \bigl( \Vert \sqrt{\rho } \dot{\mathbf {u}} \Vert _{L^{2}}^{2}+ \Vert \mathbf {B}_{t} \Vert _{L^{2}}^{2}+ \bigl\Vert \nabla ^{2} \mathbf {B}\bigr\Vert _{L^{2}}^{2} \bigr) \bigr] \\ &\qquad {}+ \int _{t_{1}}^{t_{2}}(t-t_{1}) \bigl( \Vert \nabla \dot{\mathbf {u}} \Vert _{L^{2}}^{2}+ \Vert \nabla \mathbf {B}_{t} \Vert _{L^{2}}^{2} \bigr)\,dt \\ & \quad \leqslant C_{4} M_{0}\sup_{t_{1}\leqslant t\leqslant t_{2}} \bigl[(t-t_{1}) \Vert \sqrt{\rho }\dot{\mathbf {u}} \Vert _{L^{2}}^{2} \bigr] \\ &\qquad {}+ C_{5} \bigl(M_{0}^{1/2}+M_{0}^{5 \gamma /12} \bigr)\sup_{t_{1}\leqslant t\leqslant t_{2}} \bigl[(t-t_{1}) \Vert \mathbf {B}_{t} \Vert _{L^{2}}^{2} \bigr] \\ &\qquad {} + C_{6} M_{0}^{4\gamma }\sup_{t_{1}\leqslant t\leqslant t_{2}} \bigl[(t-t_{1}) \bigl\Vert \nabla ^{2} \mathbf {B}\bigr\Vert _{L^{2}}^{2} \bigr] +C M_{0}^{5 \gamma /2}(t_{1}-t_{2})^{2}+C. \end{aligned} $$

(3.61)

Thus, if $M_{0}$ is chosen to be suitably small such that

$$ M_{0}\leqslant \varepsilon _{2} \triangleq \min \biggl\{ \frac {1}{2C_{4}}, \biggl(\frac {1}{2C_{5}} \biggr)^{{\frac {12}{5\gamma +6}}}, \biggl( \frac {1}{2C_{6}} \biggr)^{{\frac {1}{4\gamma }}} \biggr\} , $$

then we immediately obtain (3.56) from (3.61). □

We are now in a position of estimating an upper bound of the density.

Lemma 3.6

Let the conditions of Lemma 3.4and 3.5be in force. Then there exists a positive constant ε, depending only on μ, λ, ν, γ, A, $E_{0}$, $M_{1}$, and $M_{2}$, such that if $\|\mathbf{ B}_{0}\|_{L^{3}}^{1/2\gamma }\leqslant M_{0}\leqslant \varepsilon$, then

$$ \rho (x, t)\leqslant \frac {7M_{0}}{4} \quad \textit{for all } (x, t) \in \mathbb{R}^{3}\times [0, T]. $$

(3.62)

Proof

In view of (2.3)₁, we can rewrite (1.1)₁ in the form:

$$ D_{t} \rho =g(\rho )+b'(t), $$

(3.63)

where

$$ g(\rho )\triangleq -\frac {A \rho }{2\mu +\lambda }\rho ^{\gamma }, \qquad b(t) \triangleq -\frac {1}{2\mu +\lambda } \int _{0}^{t} \biggl( \rho F+\frac {1}{2} \rho \vert \mathbf {B}\vert ^{2} \biggr)\,ds. $$

(3.64)

Obviously, it holds that $g(\infty )=-\infty$. So, to apply Lemma 2.3, we still need to deal with $b(t)$. To do this, we first utilize (2.2), (2.5), (2.6), and the fact that $0\leqslant \rho \leqslant 2M_{0}$ to deduce that for any $0\leqslant t_{1} < t_{2} \leqslant T$,

$$ \begin{aligned} \bigl\vert b(t_{2})-b(t_{1}) \bigr\vert \leqslant{}& CM_{0} \int _{t_{1}}^{t_{2}} \bigl( \Vert F \Vert _{L^{\infty }}+ \Vert \mathbf {B}\Vert _{L^{\infty }}^{2} \bigr) \,dt \\ \leqslant{}& CM_{0} \int _{t_{1}}^{t_{2}} \bigl( \Vert F \Vert _{L^{2}}^{1/4} \Vert \nabla F \Vert _{L^{6}}^{3/4}+ \Vert \mathbf {B}\Vert _{L^{3}}^{2/3} \Vert \nabla \mathbf {B}\Vert _{L^{6}}^{4/3} \bigr)\,dt \\ \leqslant{}& C M_{0} \int _{t_{1}}^{t_{2}} \bigl( \Vert \nabla \mathbf{ u} \Vert _{L^{2}}+M_{0}^{\gamma /2}+ \Vert \mathbf{ B} \Vert _{L^{4}}^{2} \bigr)^{1/4} \\ &{}\times \bigl(M_{0} \Vert \nabla \dot{\mathbf{ u}} \Vert _{L^{2}}+ \Vert \mathbf {B}\Vert _{L^{3}}^{1/3} \Vert \nabla \mathbf{ B} \Vert _{L^{6}}^{5/3} \bigr)^{3/4}\,dt \\ &{} +C M_{0} \int _{t_{1}}^{t_{2}} \Vert \mathbf {B}\Vert _{L^{3}}^{2/3} \Vert \nabla \mathbf {B}\Vert _{L^{6}}^{4/3}\,dt \triangleq \sum_{i=1}^{7}N_{i}. \end{aligned} $$

(3.65)

The right-hand side of (3.65) can be estimated as follows. We deduce from (3.4), (3.56), and Hölder inequality that

$$ \begin{aligned} N_{1} & = C M_{0}^{7/4} \int _{t_{1}}^{t_{2}}(t-t_{1})^{-3/8} \bigl( \Vert \nabla \mathbf{ u} \Vert _{L^{2}}^{2} \bigr)^{1/8} \bigl[(t-t_{1}) \Vert \nabla \dot{\mathbf{ u}} \Vert _{L^{2}}^{2} \bigr]^{3/8}\,dt \\ &\leqslant C M_{0}^{7/4} \biggl( \int _{t_{1}}^{t_{2}}(t-t_{1})^{-3/4} \,dt \biggr)^{1/2} \biggl( \int _{t_{1}}^{t_{2}} \Vert \nabla \mathbf{ u} \Vert _{L^{2}}^{2}\,dt \biggr)^{1/8} \biggl( \int _{t_{1}}^{t_{2}}(t-t_{1}) \Vert \nabla \dot{\mathbf {u}} \Vert _{L^{2}}^{2}\,dt \biggr)^{3/8} \\ &\leqslant C M_{0}^{7/4} (t_{2}-t_{1})^{1/8} \bigl[1+M_{0}^{15\gamma /16}(t_{2}-t_{1})^{3/4} \bigr] \\ &\leqslant \frac {AM_{0}^{\gamma +1}}{8(2\mu +\lambda )}(t_{2}-t_{1})+C \bigl(M_{0}^{\frac {13-\gamma }{7}}+M_{0}^{\frac {14+\gamma }{2}} \bigr), \end{aligned} $$

and

$$\begin{aligned} N_{2} =&C M_{0} \int _{t_{1}}^{t_{2}} \Vert \nabla \mathbf{ u} \Vert _{L^{2}}^{1/4} \Vert \mathbf {B}\Vert _{L^{3}}^{1/4} \bigl\Vert \nabla ^{2} \mathbf{ B} \bigr\Vert _{L^{2}}^{3/4} \bigl[(t-t_{1}) \bigl\Vert \nabla ^{2}\mathbf{ B} \bigr\Vert _{L^{2}}^{2} \bigr]^{1/4}(t-t_{1})^{-1/4} \,dt \\ \leqslant& C M_{0}^{(2+\gamma )/2} \Bigl[\sup_{t_{1}\leqslant t \leqslant t_{2}} \bigl((t-t_{1}) \bigl\Vert \nabla ^{2} \mathbf {B}\bigr\Vert _{L^{2}}^{2} \bigr) \Bigr]^{1/4} \biggl( \int _{t_{1}}^{t_{2}} \Vert \nabla \mathbf{ u} \Vert _{L^{2}}^{2}\,dt \biggr)^{1/8} \\ &{}\times\biggl( \int _{t_{1}}^{t_{2}} \Vert \nabla \mathbf{ B} \Vert _{L^{2}}^{2}\,dt \biggr)^{3/8} \biggl( \int _{t_{1}}^{t_{2}}(t-t_{1})^{-1/2} \,dt \biggr)^{1/2} \\ \leqslant& C M_{0}^{(2+\gamma )/2} (t_{2}-t_{1})^{1/4} \bigl[1+M_{0}^{5 \gamma /8}(t_{2}-t_{1})^{1/2} \bigr] \\ \leqslant& \frac {AM_{0}^{\gamma +1}}{8(2\mu +\lambda )}(t_{2}-t_{1})+C \bigl(M_{0}^{\frac {3+\gamma }{3}}+M_{0}^{\frac {2+3\gamma }{2}} \bigr). \end{aligned}$$

Similarly, using (3.56) and Hölder inequality, we have

$$ \begin{aligned} N_{3} & = C M_{0}^{(14+\gamma )/8} \int _{t_{1}}^{t_{2}}(t-t_{1})^{-3/8} \bigl[(t-t_{1}) \Vert \nabla \dot{\mathbf{ u}} \Vert _{L^{2}}^{2} \bigr]^{3/8}\,dt \\ &\leqslant C M_{0}^{(14+\gamma )/8} \biggl( \int _{t_{1}}^{t_{2}}(t-t_{1})^{-3/5} \,dt \biggr)^{5/8} \biggl( \int _{t_{1}}^{t_{2}}(t-t_{1}) \Vert \nabla \dot{\mathbf {u}} \Vert _{L^{2}}^{2}\,dt \biggr)^{3/8} \\ &\leqslant C M_{0}^{(14+\gamma )/8} (t_{2}-t_{1})^{1/4} \bigl[1+M_{0}^{15 \gamma /16}(t_{2}-t_{1})^{3/4} \bigr] \\ &\leqslant \frac {AM_{0}^{\gamma +1}}{2\mu +\lambda } \biggl(\frac {1}{8}+CM_{0}^{ \frac {12+\gamma }{16}} \biggr) (t_{2}-t_{1})+CM_{0}^{ \frac {12-\gamma }{6}}, \end{aligned} $$

and

$$ \begin{aligned} N_{4} ={}&C M_{0}^{(8+\gamma )/8} \int _{t_{1}}^{t_{2}} \Vert \mathbf {B}\Vert _{L^{3}}^{1/4} \bigl\Vert \nabla ^{2}\mathbf{ B} \bigr\Vert _{L^{2}}^{3/4}\,dt \\ \leqslant{} &C M_{0}^{(8+5\gamma )/8} \Bigl[\sup_{t_{1}\leqslant t \leqslant t_{2}} \bigl((t-t_{1}) \bigl\Vert \nabla ^{2} \mathbf {B}\bigr\Vert _{L^{2}}^{2} \bigr) \Bigr]^{1/4} \biggl( \int _{t_{1}}^{t_{2}} \bigl\Vert \nabla ^{2} \mathbf{ B} \bigr\Vert _{L^{2}}^{2}\,dt \biggr)^{3/8} \\ &{} \times \biggl( \int _{t_{1}}^{t_{2}}(t-t_{1})^{-2/5} \,dt \biggr)^{5/8} \\ \leqslant{} &C M_{0}^{(8+5\gamma )/8} (t_{2}-t_{1})^{3/8} \bigl[1+M_{0}^{5 \gamma /8}(t_{2}-t_{1})^{1/2} \bigr] \\ \leqslant{}& \frac {AM_{0}^{\gamma +1}}{8(2\mu +\lambda )}(t_{2}-t_{1})+C \bigl(M_{0}^{\frac {5+2\gamma }{5}}+M_{0}^{1+3\gamma } \bigr). \end{aligned} $$

By virtue of (3.6) and (3.40), we infer from Cauchy–Schwarz inequality that

$$ \begin{aligned} N_{5} &=C M_{0}^{7/4} \int _{t_{1}}^{t_{2}} \Vert \mathbf {B}\Vert _{L^{3}}^{1/4} \Vert \nabla \mathbf{ B} \Vert _{L^{2}}^{1/4} \Vert \nabla \dot{\mathbf {u}} \Vert _{L^{2}}^{3/4}\,dt \\ & \leqslant C M_{0}^{(7+2\gamma )/4} \biggl( \int _{t_{1}}^{t_{2}}(t-t_{1}) \Vert \nabla \dot{\mathbf {u}} \Vert _{L^{2}}^{2}\,dt \biggr)^{3/8} \biggl( \int _{t_{1}}^{t_{2}}(t-t_{1})^{-3/5} \,dt \biggr)^{5/8} \\ &\leqslant C M_{0}^{(7+2\gamma )/4} (t_{2}-t_{1})^{1/4} \bigl[1+M_{0}^{15 \gamma /16}(t_{2}-t_{1})^{3/4} \bigr] \\ &\leqslant \frac {AM_{0}^{\gamma +1}}{2\mu +\lambda } \biggl(\frac {1}{8}+CM_{0}^{ \frac {12+7\gamma }{16}} \biggr) (t_{2}-t_{1})+CM_{0}^{ \frac {6+\gamma }{3}}, \end{aligned} $$

and

$$\begin{aligned} N_{6} ={}&C M_{0} \int _{t_{1}}^{t_{2}} \Vert \mathbf {B}\Vert _{L^{3}}^{1/2} \Vert \nabla \mathbf{ B} \Vert _{L^{2}}^{1/4} \bigl\Vert \nabla ^{2} \mathbf {B}\bigr\Vert _{L^{2}}^{5/4}\,dt \\ \leqslant{}& C M_{0}^{1+\gamma } \Bigl[\sup_{t_{1}\leqslant t \leqslant t_{2}} \bigl((t-t_{1}) \bigl\Vert \nabla ^{2} \mathbf {B}\bigr\Vert _{L^{2}}^{2} \bigr) \Bigr]^{1/4} ( \int _{t_{1}}^{t_{2}} \bigl( \bigl\Vert \nabla ^{2} \mathbf {B}\bigr\Vert _{L^{2}}^{2}\,dt \bigr)^{3/8} \\ &{} \times \biggl( \int _{t_{1}}^{t_{2}}(t-t_{1})^{-2/5} \,dt \biggr)^{5/8} \\ \leqslant{}& C M_{0}^{(1+\gamma } (t_{2}-t_{1})^{3/8} \bigl[1+M_{0}^{5 \gamma /8}(t_{2}-t_{1})^{1/2} \bigr] \\ \leqslant{}& \frac {AM_{0}^{\gamma +1}}{8(2\mu +\lambda )}(t_{2}-t_{1})+C \bigl(M_{0}^{1+\gamma }+M_{0}^{1+2\gamma } \bigr), \end{aligned}$$

and

$$ \begin{aligned} N_{7} &=C M_{0} \int _{t_{1}}^{t_{2}} \Vert \mathbf {B}\Vert _{L^{3}}^{2/3} \bigl\Vert \nabla ^{2} \mathbf {B}\bigr\Vert _{L^{2}}^{4/3}\,dt \\ & \leqslant C M_{0}^{\frac {3+4\gamma }{3}}\biggl( \int _{t_{1}}^{t_{2}} \bigl( \bigl\Vert \nabla ^{2}\mathbf {B}\bigr\Vert _{L^{2}}^{2}\,dt \bigr)^{2/3} \biggl( \int _{t_{1}}^{t_{2}}\,dt \biggr)^{1/3} \\ &\leqslant C M_{0}^{\frac {3+4\gamma }{3}} (t_{2}-t_{1})^{1/3} \\ &\leqslant \frac {AM_{0}^{\gamma +1}}{8(2\mu +\lambda )}(t_{2}-t_{1})+C M_{0}^{\frac {2+3\gamma }{2}}. \end{aligned} $$

Due to $1<\gamma <6$, it is obvious that

$$ \frac {13-\gamma }{7}< \frac {12-\gamma }{6} $$

and hence, putting $N_{i}$ ($i=1, 2, \dots , 7$) into (3.65), we obtain

$$ \bigl\vert b(t_{2})-b(t_{1}) \bigr\vert \leqslant C_{7} \bigl(M_{0}^{\frac {13-\gamma }{7}}+M_{0}^{ \frac {3+\gamma }{3}} \bigr)+\frac {AM_{0}^{\gamma +1}}{2\mu +\lambda } \biggl(\frac {7}{8}+C_{8}M_{0}^{\frac {12+\gamma }{16}} \biggr) (t_{2}-t_{1}). $$

(3.66)

Now, in view of (2.10) and (2.11), we set

$$ N_{0} \triangleq C_{7} \bigl(M_{0}^{\frac {13-\gamma }{7}}+M_{0}^{ \frac {3+\gamma }{3}} \bigr),\qquad N_{1}\triangleq \frac {AM_{0}^{ \gamma +1}}{2\mu +\lambda } \biggl(\frac {7}{8}+C_{8}M_{0}^{ \frac {12+\gamma }{16}} \biggr), \quad \text{and}\quad \xi ^{*} \triangleq M_{0}. $$

Thus, if $M_{0}$ is chosen to be small enough such that (noting that $1<\gamma <6$)

$$ M_{0} \triangleq \biggl\{ \varepsilon _{2}, \biggl( \frac {1}{4C_{7}} \biggr)^{\frac {7}{6-\gamma }}, \biggl(\frac {1}{2C_{7}} \biggr)^{ \frac {3}{\gamma }}, \biggl(\frac {1}{8C_{8}} \biggr)^{ \frac {16}{12+\gamma }} \biggr\} , $$

then it is easy to check that

$$ g(\xi )=-\frac {A \xi ^{\gamma +1}}{2\mu +\lambda }\leqslant -\frac {A M_{0}^{ \gamma +1}}{2\mu +\lambda }\leqslant -N_{1}, \quad \forall \xi \geqslant \xi ^{*}=M_{0}. $$

As a result, we conclude from (3.66) and Lemma 2.3 that

$$ \sup_{0\leqslant t \leqslant T} \bigl\Vert \rho (t) \bigr\Vert _{L^{\infty }}\leqslant M_{0}+N_{0} \leqslant M_{0}+\frac {3M_{0}}{4}=\frac {7M_{0}}{4}. $$

The proof of Lemma 3.6 is therefore complete. □

Proof of Proposition 3.1

In view of (3.40) and (3.62), we obtain the desired estimate of (3.2) with K and ε being the same as in Lemmas 3.4 and 3.6. □

With the help of Proposition 3.1, the higher-order estimates of the solution $(\rho , \mathbf {u}, \mathbf {B})$ can be shown in a manner similar to that in [9] (see Lemmas 4.1–4.6). For completeness and for convenience, we collect these estimates in the following lemma without proofs.

Proposition 3.2

Let the conditions of Proposition 3.1be in force. Then for any $T > 0$ and $0 < \tau < T$,

$$\begin{aligned} &\sup_{t\in [0, T]} \bigl( \bigl\Vert (\nabla \rho , \nabla P) \bigr\Vert _{H^{1} \cap W^{1, q}}+ \bigl\Vert (\rho _{t}, P_{t}) \bigr\Vert _{H^{1}} \bigr)+ \int _{0}^{T} \bigl\Vert (\rho _{tt}, P_{tt}) \bigr\Vert _{L^{2}}^{2}\,dt\leqslant C(T),\quad q \in (3, 6), \\ &\sup_{t\in [0, T]} \bigl( \bigl\Vert (\sqrt{\rho } \mathbf{ u}_{t}, \mathbf{ B}_{t}) \bigr\Vert _{L^{2}}^{2}+ \bigl\Vert (\nabla \mathbf{ u}, \nabla \mathbf{ B}) \bigr\Vert _{H^{1}}^{2} \bigr)+ \int _{0}^{T} \bigl( \bigl\Vert (\nabla \mathbf{ u}_{t}, \nabla \mathbf{ B}_{t}) \bigr\Vert _{L^{2}}^{2}+ \bigl\Vert \nabla ^{2} \mathbf{ u} \bigr\Vert _{H^{1}}^{2} \bigr)\,dt \\ &\quad \leqslant C(T), \\ &\sup_{t\in [\tau , T]} \bigl( \bigl\Vert \nabla ^{2}\mathbf{ u} \bigr\Vert _{H^{1} \cap W^{1, q}}^{2}+ \bigl\Vert \nabla ^{2} \mathbf{ B} \bigr\Vert _{H^{2}}^{2}+ \bigl\Vert (\nabla \mathbf{ u}_{t}, \nabla \mathbf{ B}_{t} ) \bigr\Vert _{H^{1}}^{2} \bigr)+ \int _{\tau }^{T} \bigl\Vert (\nabla \mathbf{ u}_{tt}, \nabla \mathbf{ B}_{tt}) \bigr\Vert _{L^{2}}^{2}\,dt \\ &\quad \leqslant C(\tau , T), \end{aligned}$$

where $C(\tau , T)$ denotes a positive constant depending on τ and T, in addition to μ, λ, ν, γ, A, $E_{0}$, $M_{1}$, $M_{2}$, K, and the coefficients of the Gagliardo–Nirenberg–Sobolev inequality in Lemma 2.1.

With Propositions 3.1–3.2 at hand, we can extend the local solutions obtained in Lemma 2.4 to be a global one in a similar manner as that in [9].

Proof of Theorem 1.1

By Lemma 2.4, there exists a $T_{*} > 0$ such that the problem (1.1)–(1.5) has a classical solution $(\rho , \mathbf {u}, \mathbf {B})$ on $(0, T_{*}]$. Noting that

$$ A_{1}(0)+A_{2}(0)=M_{1}+M_{2}\leqslant K, \quad 0\leqslant \rho _{0} \leqslant M_{0}, $$

and using the continuity arguments, one infers that there exists a $T_{1}\in (0, T_{*}]$ such that (3.1) holds for $T = T_{1}$. Next, let

$$ T^{*} \triangleq \sup \{T | \text{ (3.1) holds}\}. $$

(3.67)

Then $T^{*}\geqslant T_{1}>0$. We claim that

$$ T^{*} =\infty . $$

(3.68)

Indeed, if we had $T^{*}<\infty$, then it would follow from Proposition 3.1 that (3.2) holds for $T = T^{*}$, provided $\|\mathbf {B}_{0}\|_{L^{3}}^{1/2\gamma }\leqslant M_{0}\leqslant \varepsilon$ and $1<\gamma <6$. It would also follow from Proposition 3.2 that for any $0<\tau < T\leqslant T^{*}$,

$$ \textstyle\begin{cases} (\nabla \mathbf {u}_{t}, \nabla \mathbf {B}_{t}, \nabla ^{2} \mathbf {u}, \nabla ^{3} \mathbf {B})\in C([\tau , T]; L^{2} \cap L^{4}), \\ \nabla ^{3} \mathbf {u}\in C([\tau , T]; L^{p}) \quad \text{with } p \in [2, q), q\in (3, 6), \\ (\nabla \mathbf {u}, \nabla \mathbf {B}, \nabla ^{2} \mathbf {u}, \nabla ^{2} \mathbf {B})\in C([\tau , T]; L^{2} \cap C(\mathbb{R}^{3})), \end{cases} $$

(3.69)

where we have also used the following embedding:

$$ L^{\infty } \bigl(\tau , T; H^{1} \bigr) \cap H^{1} \bigl( \tau , T; H^{-1} \bigr) \hookrightarrow C \bigl(\tau , T; L^{\alpha } \bigr),\quad \forall \alpha \in [2, 6). $$

Moreover, for any $0<\tau <T\leqslant T^{*}$, it then follows from (1.1)₁, Propositions 3.1 and 3.2 that

$$ \int _{\tau }^{T} \int \bigl\vert \partial _{t} \bigl(\rho \vert \mathbf {u}_{t} \vert ^{2} \bigr) \bigr\vert \,dx\,dt+ \int _{\tau }^{T} \int \bigl\vert \partial _{t} \bigl(\rho \vert \mathbf {u}\nabla \mathbf {u}\vert ^{2} \bigr) \bigr\vert \,dx\,dt\leqslant C(\tau , T), $$

which, together Proposition 3.2, yields

$$ \sqrt{\rho }\mathbf {u}_{t}, \sqrt{\rho }\mathbf {u}\cdot \nabla \mathbf {u}\in C \bigl(\tau , T; L^{2} \bigr), $$

and, consequently,

$$ \sqrt{\rho }\dot{\mathbf {u}} \in C \bigl(\tau , T; L^{2} \bigr). $$

In a similar manner, we can also show using (3.69) that

$$ \nabla \dot{\mathbf {u}} \in C \bigl(\tau , T; L^{2} \bigr). $$

This particularly implies that $(\rho , \mathbf {u}, \mathbf {B})(x, T^{*})$ satisfies the compatibility condition (1.9) with $g(x)\triangleq \dot{\mathbf {u}}(x, T^{*})$ at $t=T^{*}$. Thus, using Lemma 2.4, (3.2), and the continuity arguments, we know that there exists a $T^{**}>T^{*}$ such that (3.1) holds for $T = T^{**}$, which contradicts (3.67). Hence, (3.68) holds. This, together with Proposition 3.2 again, shows that the solution $(\rho , \mathbf {u}, \mathbf {B})$ is in fact the unique classical solution on $\mathbb{R}^{3} \times (0, T ]$ for any $0< T< T^{*}=\infty$. The proof of Theorem 1.1 is therefore complete. □