Global strong solution to 3D full compressible magnetohydrodynamic flows with vacuum at infinity

Abstract

In this paper, we consider the Cauchy problem of the full compressible magnetohydrodynamic equations in ${\mathbb {R}}^3$. When $\Vert \rho _0\Vert _{L^1}+\Vert H_0\Vert _{L^2}$ is suitably small, we establish the global existence of the strong solution, where $\rho _0$ and $H_0$ represent the initial density and magnetic field respectively. Our result shows that the strong solution may have large oscillations and can contain vacuum states.

1 Introduction

The equations of the three-dimensional compressible, viscous and heat-conducting magnetohydrodynamic flows can be written as ([20])

$$\begin{aligned} {\left\{ \begin{array}{ll} \rho _t+\mathrm{div}(\rho u)=0, \\ (\rho u)_t+\mathrm {div}(\rho u\otimes u)-\mu \Delta u-(\mu +\lambda )\nabla \mathrm {div}u+\nabla P =(\mathrm {curl} \ H )\times H,\\ c_v\left[ (\rho \theta )_t+\mathrm {div}(\rho u \theta )\right] -\kappa \Delta \theta +P\mathrm {div}u=2\mu |{\mathfrak {D}}(u)|^2 +\lambda (\mathrm {div}u)^2+\nu |\mathrm {curl}\ H|^2,\\ H_t-\mathrm {curl}\ (u\times H)=\nu \Delta H ,\ \ \ \mathrm {div}H=0. \end{array}\right. } \end{aligned}$$

(1.1)

Here ${\mathfrak {D}}(u)$ is the deformation tensor given by

$$\begin{aligned} {\mathfrak {D}}(u)=\frac{1}{2}\left( \nabla u+(\nabla u)^\prime \right) , \end{aligned}$$

$\rho =\rho (x,t)$, $u=u(x, t)=(u_1, u_2, u_3)(x,t)$, $\theta =\theta (x,t),\ P=R\rho \theta (R>0)$ and $H=(H_1,H_2,H_3)$, are unknown functions denoting the density, velocity, absolute temperature, pressure and magnetic field respectively; $\mu$ and $\lambda$ are coefficients of viscosity, satisfying the following physical restrictions

$$\begin{aligned} \mu >0,\quad \lambda +\frac{2\mu }{3}\ge 0. \end{aligned}$$

(1.2)

The positive constants $\kappa$ and $\nu$ are respectively the heat conduction coefficient and the magnetic diffusion coefficient.

In this paper, we are interested in the global existence of strong solutions to the Cauchy problem for (1.1) with the following initial conditions:

$$\begin{aligned} (\rho ,\ u,\ \theta ,\ H )|_{t=0}=(\rho _0,\ u_0,\ \theta _0,\ H_0 )(x),\ \ x\in {\mathbb {R}}^3 \end{aligned}$$

(1.3)

and the far field behavior:

$$\begin{aligned} \rho (x,t)\rightarrow 0,\ u(x,t)\rightarrow 0,\ \theta (x,t)\rightarrow 0,\ H(x,t)\rightarrow 0, as\ \ |x|\rightarrow \infty ,\ \ for\ \ t\ge 0. \end{aligned}$$

(1.4)

The compressible MHD system (1.1) is obtained by combining the Navier–Stokes equations of fluid dynamics and Maxwell equations of electromagnetism. There are lots of works on the well-posedness of the Cauchy problem and the initial-boundary value problem for Magnetohydrodynamics. The one-dimension problem has been studied in many papers, for example, [3, 4, 8, 9, 18, 32] and the references cited therein. For the multi-dimensional case, see [6, 10, 12,13,14,15,16,17, 22,23,24,25, 27, 29] and so on. To be more specific, for the 2D case, Cao and Wu [2] obtained the global existence, conditional regularity and uniqueness of a weak solution with only magnetic diffusion. Ren et al. [28] proved the global existence and the decay estimates of small smooth solution without magnetic diffusion. Yang and Sun [21] got the global existence of weak solutions for any adiabatic exponent $\gamma \ge 1$. Zhong [35] established the global existence and exponential decay of strong solutions to the initial boundary value problem of two-dimensional nonhomogeneous magnetohydrodynamic (MHD) equations with non-negative density.

For the 3D case, when the initial density is strictly positive, Vol’pert et al. [31] investigated the local strong solutions of the compressible MHD with large initial data. This result was extended by Fan et al. [7] to the case that the initial density could contain vacuum. Suen and Hoff [30] proved the global-in-time existence of weak solutions to the 3D compressible MHD equations with initial data small in $L^2$ and initial density positive and essentially bounded. In [34], they studied the Cauchy problem for the multi-dimensional ($N\ge 3$) non-isentropic full compressible MHD equations.They got the existence and uniqueness of a global strong solution to the system for the initial data close to a stable equilibrium state in critical Besov spaces. For more information on the MHD equations, we refer to [1, 26] and the references therein. Hu and Wang [13] studied the existence of a global variational weak solution to the full MHD equations with large data. Li et al. [22] obtained the global classical solution with small initial energy but large oscillations for the isentropic flows. Later on, Hong et al. [10] got the global classical solution with large initial energy but the adiabatic exponent $\gamma$ is close to 1 and the resistivity coefficient $\nu$ is suitably large. Recently, Chen et al. [5] considered the 3D MHD equations with slip boundary condition and vacuum, and obtained the global classical solutions with small energy but large oscillations.

It is worth noting that, Wen and Zhu [33] got the strong solution with vacuum and large initial data for the three-dimensional full compressible Navier–Stokes equations under the condition that the initial mass is small. One natural question is: does there exist some global solution to (1.1) in some classes of large initial data under the condition that the initial mass is small? We will answer the question in this paper. More precisely, we want to find a global strong solution to the 3D full compressible MHD equations under the small initial mass.

Before we state our main results, we would like to introduce some notations which will be used throughout this paper.

1.1 Notations

• (i) $\displaystyle \int \limits _{{\mathbb {R}}^3} f =\int \limits _{{\mathbb {R}}^3} f \,\mathrm{d}x$.

• (ii) For $1\le l\le \infty$, denote the standard homogeneous and inhomogeneous Sobolev spaces as follows:

  $$\begin{aligned}&L^l=L^l({\mathbb {R}}^3), \ D^{k,l}=\left\{ u\in L^1_{\mathrm{{loc}}}({\mathbb {R}}^3): \Vert \nabla ^k u \Vert _{L^l}<\infty \right\} ,\ \Vert u\Vert _{D^{k,l}}=\Vert \nabla ^k u\Vert _{L^l}, \\&W^{k,l}=L^l\cap D^{k,l}, \ D^k=D^{k,2},\ \ D_0^1=\Big \{u\in L^6: \ \Vert \nabla u\Vert _{L^2}<\infty \Big \},\ \ H^k=W^{k,2}. \end{aligned}$$

• (iii) $G=(2\mu +\lambda )\mathrm {div}u-P$ is the effective viscous flux.

• (iv) $\dot{h}=h_t+u\cdot \nabla h$ denotes the material derivative.

• (v) $\displaystyle m_0=\int \limits _{{\mathbb {R}}^3}\rho _0$ is the initial mass.

The rest of the paper is organized as follows. In Sect. 2, we present our main results. Section 3 focuses on obtaining the necessary a priori estimates for the strong solution to extend the local solution to all time. Finally, we give the proof of the main results.

2 Main results

Before stating the main results, let us make some preliminaries. Assume that $\mu$, $\lambda$, $\kappa$ and $\nu$ are constants. We assume $R=C_\nu =1$ henceforth, since the constants R in the pressure function and $C_\nu$ in the internal energy play no role in the analysis. In this case, if the solutions are regular enough (such as strong solutions and classical solutions), (1.1) is equivalent to the following system

$$\begin{aligned} {\left\{ \begin{array}{ll} \rho _t+\nabla \cdot (\rho u)=0, \\ \rho u_t+\rho u\cdot \nabla u+\nabla P(\rho ,\theta )=\mu \Delta u+(\mu +\lambda )\nabla \mathrm {div}u+(\mathrm {curl} \ H )\times H,\\ \rho \theta _t+\rho u\cdot \nabla \theta +\rho \theta \mathrm {div}u=\frac{\mu }{2}\left| \nabla u+(\nabla u)^\prime \right| ^2+\lambda (\mathrm {div}u)^2+\kappa \Delta \theta +\nu |\mathrm {curl}\ H|^2,\\ H_t-\mathrm {curl}\ (u\times H)=\nu \Delta H,\ \ \ \mathrm {div}H=0\ \ \mathrm {in}\ {\mathbb {R}}^3\times (0,\infty ). \end{array}\right. } \end{aligned}$$

(2.1)

System (2.1) is supplemented with initial conditions

$$\begin{aligned} (\rho , u, \theta , H)|_{t=0}=(\rho _0, u_0, \theta _0, H_0)(x),\ x\in {\mathbb {R}}^3, \end{aligned}$$

(2.2)

and the far-field conditions

$$\begin{aligned} \rho (x,t)\rightarrow 0,\ u(x,t)\rightarrow 0,\ \theta (x,t)\rightarrow 0,\ H(x,t)\rightarrow 0, \mathrm {as}\ |x|\rightarrow \infty ,\ \mathrm {for}\ t\ge 0. \end{aligned}$$

(2.3)

First, the well-known Gagliardo–Nirenberg inequality will be used frequently later (see [19]).

Lemma 2.1

For any $p \in [2,6], q\in (1,\infty )$ and $r\in (3,\infty )$, there exists some generic constant $C>0$ that may depend on q and r such that for $f\in H^1({\mathbb {R}}^3)$ and $g\in L^q({\mathbb {R}}^3)\cap D^{1,r}({\mathbb {R}}^3)$, we have

$$\begin{aligned} \Vert f\Vert ^p_{L^p({\mathbb {R}}^3)}\le C\Vert f\Vert ^{\frac{6-p}{2}}_{L^{2}({\mathbb {R}}^3)}\Vert \nabla f\Vert ^{\frac{3p-6}{2}}_{L^{2}({\mathbb {R}}^3)}, \end{aligned}$$

(2.4)

$$\begin{aligned} \Vert g\Vert _{C({\mathbb {R}}^3)}\le C\Vert g\Vert ^{\frac{q(r-3)}{3r+q(r-3)}}_{L^{q}({\mathbb {R}}^3)}\Vert \nabla g\Vert ^{\frac{3r}{3r+q(r-3)}}_{L^{r}({\mathbb {R}}^3)}. \end{aligned}$$

(2.5)

Next, we give the definition of the strong solution to (2.1)–(2.3) throughout this paper, which is similar to [11].

Definition 2.2

(Strong solution) For $T>0$, $(\rho , u, \theta ,H)$ is called a strong solution to the compressible Magnetohydrodynamic flows (2.1)–(2.3) in ${\mathbb {R}}^3\times [0,T]$, if for some $q\in (3, 6)$,

$$\begin{aligned} \begin{aligned}&0\le \rho \in C([0,T];W^{1,q}\cap H^1),\ \rho _t\in C([0,T];L^2\cap L^q),\\&(u, \theta ,H)\in C([0,T];D^2\cap D^1_0)\cap L^2([0,T];D^{2,q}), \ \theta \ge 0,\\&(u_t, \theta _t,H_t)\in L^2([0,T];D^1_0),\ (\sqrt{\rho } u_t, \sqrt{\rho } \theta _t,H_t)\in L^\infty ([0,T];L^2),\ \end{aligned} \end{aligned}$$

(2.6)

and $(\rho ,u,\theta ,H)$ satisfies (2.1) a.e. in ${\mathbb {R}}^3\times (0,T]$. In particular, the strong solution $(\rho , u, \theta ,H)$ of (2.1)–(2.3) is called a global strong solution, if the strong solution satisfies (2.6) for any $T>0$, and satisfies (2.1) a.e. in ${\mathbb {R}}^3\times (0,\infty )$.

Then the main results in this paper can be stated as follows:

Theorem 2.3

(Global strong solution) Assume that the initial data $(\rho _0,u_0,\theta _0, H_0)$ satisfies

$$\begin{aligned} \rho _0\ge 0,\ \theta _0\ge 0,\ in\ {\mathbb {R}}^3,\ \rho _0\in H^1\cap W^{1,q}\cap L^1,\ (u_0, \theta _0, H_0)\in D^2\cap D_0^1 \end{aligned}$$

(2.7)

and

$$\begin{aligned} {\left\{ \begin{array}{ll} \displaystyle 0\le \rho _0\le {\bar{\rho }},\ \ m_0=\int \limits _{{\mathbb {R}}^3}\rho _0, \ \ \Vert H_0\Vert ^2_{L^2}\le m_0^\frac{1}{2},\\ \displaystyle (1+\nu )\Vert \nabla H_0\Vert ^2_{L^2}+\Vert \sqrt{\rho _0}\theta _0\Vert _{L^2}^2+\mu \Vert \nabla u_0\Vert _{L^2}+(\mu +\lambda )\Vert \mathrm {div} u_0\Vert _{L^2}+ \frac{1}{2(2\mu +\lambda )}\Vert \rho _0\theta _0\Vert _{L^2}^2\\ \quad \displaystyle +\int \limits _{{\mathbb {R}}^3}|\rho _0\theta _0\mathrm {div} u_0|+\int \limits _{{\mathbb {R}}^3}\left( |H_0|^2|\mathrm {div}u_0|+2|H_0\cdot \nabla u_0\cdot H_0|\right) \le K \end{array}\right. } \end{aligned}$$

(2.8)

for some constants $K>1$, ${\bar{\rho }}>0$ and $q\in (3,6)$, and that the following compatibility conditions are satisfied:

$$\begin{aligned} {\left\{ \begin{array}{ll} \mu \Delta u_0+(\mu +\lambda )\nabla \mathrm {div} u_0-\nabla P(\rho _0, \theta _0)-(\mathrm {curl} H_0 )\times H_0=\sqrt{\rho }_0g_1,\\ \kappa \Delta \theta _0+\frac{\mu }{2}\left| \nabla u_0+(\nabla u_0)^\prime \right| ^2+\lambda (\mathrm {div}u_0)^2+\nu |\mathrm {curl} H_0|^2=\sqrt{\rho _0}g_2,\ x\in {\mathbb {R}}^3, \end{array}\right. } \end{aligned}$$

(2.9)

with $g_i\in L^2$, $i=1,2$. Then there exists a unique global strong solution $(\rho ,u,\theta ,H)$ in ${\mathbb {R}}^3\times [0,T]$ for any $T>0$, provided

$$\begin{aligned} m_0\le \varepsilon , \end{aligned}$$

where $\varepsilon$ is a positive constant depending on ${\bar{\rho }}$, K, $\mu$ and $\lambda$ but independent of t.

3 Proof of Theorem 2.3

In this section, we will prove the global existence and uniqueness of the strong solution to the problem (2.1)–(2.3). The local existence and uniqueness of the strong solution has been obtained in [11] under the conditions of Theorem 2.3. In order to get the global solution, we denote

$$\begin{aligned} A(T)=\sup \limits _{0\le t\le T}\int \limits _{{\mathbb {R}}^3}|\nabla u|^2+\int \limits _0^T\int \limits _{{\mathbb {R}}^3}{\rho |\dot{u}|^2} \end{aligned}$$

and

$$\begin{aligned} B(T)=\sup \limits _{0\le t\le T}\int \limits _{{\mathbb {R}}^3}\rho \theta ^2+\int \limits _0^T\int \limits _{{\mathbb {R}}^3}|\nabla \theta |^2. \end{aligned}$$

The following proposition plays a crucial role in the proof.

Proposition 3.1

Assume that the initial data satisfies (1.3), (2.8) and (2.9). If the solution $(\rho , u, \theta , H)$ of (2.1)–(2.3) satisfies that for $(x,t)\in {\mathbb {R}}^3\times [0,T],$

$$\begin{aligned} \begin{aligned}&\Vert H\Vert _{L^2}^2\le 2m_0^{\frac{1}{2}},\quad \Vert \nabla H\Vert _{L^2}^2\le 2K,\quad A(T)\le 2{\tilde{E}} K,\\&B(T)\le 2K,\quad 0\le \rho \le 2{\bar{\rho }}, \end{aligned} \end{aligned}$$

(3.1)

then

$$\begin{aligned} \begin{aligned} \Vert H\Vert _{L^2}^2\le \frac{3m_0^{\frac{1}{2}}}{2},\quad \Vert \nabla H\Vert _{L^2}^2\le \frac{3K}{2},\quad A(T)\le \frac{3{\tilde{E}} K}{2},\\[3mm] B(T)\le \frac{3K}{2},\quad 0\le \rho \le \frac{3{\bar{\rho }}}{2},\quad (x,t)\in {\mathbb {R}}^3\times [0,T],\end{aligned} \end{aligned}$$

(3.2)

provided

$$\begin{aligned}&m_0\le \min \left\{ \frac{1}{CK^6},\quad \frac{1}{C\left( {\tilde{E}}^2K\right) ^2},\quad \frac{1}{C\left( K^\frac{1}{2}+{\tilde{E}}Km_0^\frac{1}{12}\right) ^4},\right. \\&\left. \qquad \qquad \quad \frac{1}{C \left( {\tilde{E}}^{\frac{3}{2}}K+{\tilde{E}}K^{\frac{5}{4}}\right) ^6 } ,\quad \frac{1}{C({\tilde{E}}K)^{\frac{144}{45}}},\quad \frac{1}{C\left( {\tilde{E}}K+m_0^\frac{1}{4}K^\frac{3}{2} \right) ^{48}} \right\} , \end{aligned}$$

where

$$\begin{aligned} {\tilde{E}}=1+\frac{4}{\mu }+\frac{2{\bar{\rho }}}{\mu (\mu +\lambda )}+\frac{6K}{\mu }. \end{aligned}$$

Proof of Proposition 3.1

The Proposition 3.1 can be proved by the Lemmas 3.2–3.7 below.

Throughout the rest of the paper, we denote by C or $C_i, \ (i=1,2,\ldots )$ the generic positive constants which may depend on $\mu ,\lambda ,\kappa , \nu , {\bar{\rho }}$ and K but independent of time T. $\square$

Lemma 3.2

Under the assumptions of Proposition 3.1, it holds that

$$\begin{aligned} \int \limits _{{\mathbb {R}}^3} \rho =\int \limits _{{\mathbb {R}}^3}\rho _0, \end{aligned}$$

(3.3)

for any $t\in [0,T]$.

The proof of Lemma 3.2 can be found in [33], and we omit it here.

Lemma 3.3

Under the assumptions of Proposition 3.1, it holds that

$$\begin{aligned} \int \limits _{{\mathbb {R}}^3}(\rho |u|^2+|H|^2)+\mu \int \limits _0^T\int \limits _{{\mathbb {R}}^3}|\nabla u|^2\mathrm{d}s+\nu \int \limits _0^T\int \limits _{{\mathbb {R}}^3}|\nabla H|^2\mathrm{d}s\le \frac{3m_0^{\frac{1}{2}}}{2}, \end{aligned}$$

(3.4)

provided

$$\begin{aligned} m_0\le \frac{1}{CK^6}. \end{aligned}$$

Proof

Multiplying (2.1)$_2$ and (2.1)$_4$ by u and H respectively, integrating by parts over ${\mathbb {R}}^3$ and summing up the resulting equations, one gets that

(3.5)

Then we estimate the terms on the right-hand side. For $I_1$, it follows that by using Cauchy inequality,

$$\begin{aligned} \begin{aligned} I_1\le (\mu +\lambda )\int \limits _{{\mathbb {R}}^3}|\mathrm {div}u|^2+\frac{1}{4(\mu +\lambda )}\int \limits _{{\mathbb {R}}^3}\rho ^2\theta ^2. \end{aligned} \end{aligned}$$

(3.6)

Next, to estimate $I_2$ and $I_3$, note that

$$\begin{aligned} \begin{aligned} (\mathrm {curl}\ H)\times H=-\frac{1}{2}\nabla |H|^2+H\cdot \nabla H=-\frac{1}{2}\nabla |H|^2+\mathrm {div}(H\otimes H) \end{aligned} \end{aligned}$$

(3.7)

and

$$\begin{aligned} \begin{aligned} \mathrm {curl}\ (u\times H)=H\cdot \nabla u-u\cdot \nabla H-\mathrm {div}u\cdot H, \end{aligned} \end{aligned}$$

(3.8)

where we have used the fact $\mathrm {div} H=0$. This, together with Hölder inequality, Sobolev inequality, Cauchy inequality and (3.1), deduces

(3.9)

and

$$\begin{aligned} I_3&=\int \limits _{{\mathbb {R}}^3}H\cdot \nabla u\cdot H-\int \limits _{{\mathbb {R}}^3}u\cdot \nabla H\cdot H+ \int \limits _{{\mathbb {R}}^3}u\cdot \nabla |H|^2\nonumber \\&\le C \Vert H\Vert _{L^6}\Vert \nabla u\Vert _{L^2}\Vert H\Vert _{L^3}\nonumber \\&\le \frac{\mu }{4}\Vert \nabla u\Vert _{L^2}^2+C{K}^{\frac{1}{2}}m_0^{\frac{1}{4}} \Vert \nabla H\Vert _{L^2}^2. \end{aligned}$$

(3.10)

Substituting (3.6) and (3.9)–(3.10) into (3.5) and setting $\displaystyle m_0\le \frac{\nu ^4}{C^4K^2}$, one may arrive at

(3.11)

Integrating (3.11) over [0, T], and using (3.1) again, we have

$$\begin{aligned} \begin{aligned}&\int \limits _{{\mathbb {R}}^3}(\rho |u|^2+|H|^2)+\mu \int \limits _0^T\int \limits _{{\mathbb {R}}^3}|\nabla u|^2+\nu \int \limits _0^T\int \limits _{{\mathbb {R}}^3}|\nabla H|^2 \\&\quad \le \int \limits _{{\mathbb {R}}^3}(\rho _0|u_0|^2+|H_0|^2)+Cm_0^\frac{2}{3}\int \limits _0^T\Vert \nabla \theta \Vert _{L^2}^2\\&\quad \le C\Vert \rho _0\Vert _{L^\frac{3}{2}}\Vert \nabla u_0\Vert _{L^2}^2+m_0^{\frac{1}{2}}+Cm_0^\frac{2}{3}K\le m_0^{\frac{1}{2}}\left( 1+Cm_0^{\frac{1}{6}}K\right) \le \frac{3m_0^{\frac{1}{2}}}{2}, \end{aligned} \end{aligned}$$

provided

$$\begin{aligned} m_0\le \frac{1}{CK^6}. \end{aligned}$$

$\square$

Lemma 3.4

Under the conditions of Proposition 3.1, it holds that

$$\begin{aligned} \Vert \nabla H\Vert _{L^2}^2 +\int \limits _{0}^{T}\left( \Vert H_t\Vert _{L^2}^2+\Vert \Delta H\Vert _{L^2}^2\right) \mathrm{d}s\le \frac{3K}{2}, \end{aligned}$$

(3.12)

provided

$$\begin{aligned} m_0\le \frac{1}{C\left( {\tilde{E}}^2K\right) ^2}. \end{aligned}$$

Proof

Applying $\partial _j, \ j=1,2,3$ to (2.1)$_4$, multiplying the resulting equations by $\partial _j H$, summing with respect to j, then integrating by parts over ${\mathbb {R}}^3\times [0,T]$ and using Cauchy inequality, one has

(3.13)

We estimate the last term on the right-hand side of (3.13 ) as follows:

$$\begin{aligned} \Vert \mathrm {curl}\ (u\times H)\Vert ^2&\le \Vert u\cdot \nabla H\Vert _{L^2}^2+\Vert H\cdot \nabla u\Vert _{L^2}^2+\Vert \mathrm {div}u H\Vert _{L^2}^2\nonumber \\&\le C \Vert \nabla u\Vert _{L^2}^2\Vert \nabla H\Vert _{L^3}^2+C\Vert H\Vert _{L^{\infty }}^2\Vert \nabla u\Vert _{L^2}^2\nonumber \\&\le C \Vert \nabla u\Vert _{L^2}^2\Vert \nabla H\Vert _{L^2}\Vert \nabla ^2 H\Vert _{L^2}\nonumber \\&\le \frac{\nu ^2}{4}\Vert \Delta H\Vert _{L^2}^2+\frac{C}{\nu ^2} \Vert \nabla u\Vert _{L^2}^4\Vert \nabla H\Vert _{L^2}^2. \end{aligned}$$

(3.14)

On the other hand, multiplying (2.1)$_4$ by $H_t$, integrating by parts over ${\mathbb {R}}^3\times [0,T]$ and using Cauchy inequality, one has

(3.15)

Combing (3.13) with (3.15) implies

$$\begin{aligned} \int \limits _{{\mathbb {R}}^3}|\nabla H|^2 +\int \limits _{0}^{T}\left( \Vert H_t\Vert _{L^2}^2+\Vert \Delta H\Vert _{L^2}^2\right) \mathrm{d}s&\le \frac{1+\nu }{2}\Vert \nabla H_0\Vert _{L^2}^2+\frac{C}{\nu } \int \limits _{0}^{T}\Vert \nabla u\Vert _{L^2}^4\Vert \nabla H\Vert _{L^2}^2\mathrm{d}s\\&\le K+C\sup _{0\le t\le T}\Vert \nabla u\Vert _{L^2}^4\int \limits _{0}^{T}\Vert \nabla H\Vert _{L^2}^2\mathrm{d}s\\&\le K+C{\tilde{E}}^2K^2m_0^\frac{1}{2}\le \frac{3}{2}K, \end{aligned}$$

provided

$$\begin{aligned} m_0\le \frac{1}{C\left( {\tilde{E}}^2K\right) ^2}. \end{aligned}$$

$\square$

Lemma 3.5

Under the conditions of Proposition 3.1, it holds that

$$\begin{aligned} A(T)\le \frac{3 {\tilde{E}}K}{2}, \end{aligned}$$

(3.16)

provided

$$\begin{aligned} m_0\le \min \left\{ \frac{1}{CK^6},\quad \frac{1}{C\left( {\tilde{E}}^2K\right) ^2},\quad \frac{1}{C\left( K^\frac{1}{2}+{\tilde{E}}Km_0^\frac{1}{12}\right) ^4}\right\} . \end{aligned}$$

Proof

Multiplying (2.1)$_2$ by $u_t$ and integrating by parts over ${\mathbb {R}}^3$, we have

(3.17)

where $G=(2\mu +\lambda )\mathrm {div}u-P$. Recalling $P=\rho \theta$, we obtain from (2.1)$_1$ and (2.1)$_3$

$$\begin{aligned} \begin{aligned} P_t=-\mathrm {div}(\rho \theta u)-\rho \theta \mathrm {div}u+\frac{\mu }{2}\left| \nabla u+(\nabla u)^\prime \right| ^2+\kappa \Delta \theta +\nu |\mathrm {curl}\ H|^2. \end{aligned} \end{aligned}$$

(3.18)

Substituting (3.18) into $II_1$, and by integration by parts, Hölder inequality and Sobolev inequality, we have

$$\begin{aligned} II_1&\le C\Vert \rho \theta \Vert _{L^3}\Vert u\Vert _{L^6}\Vert \nabla G\Vert _{L^2}+C\Vert \rho \theta \Vert _{L^3}\Vert \mathrm {div}u\Vert _{L^2} \Vert G\Vert _{L^6}+C\Vert G\Vert _{L^6}\Vert \nabla u\Vert _{L^2}\Vert \nabla u\Vert _{L^3}\\ {}&\quad +\frac{\kappa }{2\mu +\lambda }\Vert \nabla G\Vert _{L^2}\Vert \nabla \theta \Vert _{L^2}+C\Vert G\Vert _{L^6}\Vert \mathrm {curl}\ H\Vert _{L^2}\Vert \mathrm {curl}\ H\Vert _{L^3}\\&\le C\Vert \rho \Vert _{L^6}\Vert \theta \Vert _{L^6}\Vert \nabla u\Vert _{L^2}\Vert \nabla G\Vert _{L^2}+C\Vert \nabla G\Vert _{L^2}\Vert \nabla u\Vert _{L^2}\Vert \nabla u\Vert _{L^3}\\&\quad +\frac{\kappa }{2\mu +\lambda }\Vert \nabla G\Vert _{L^2}\Vert \nabla \theta \Vert _{L^2}+C\Vert \nabla G\Vert _{L^2}\Vert \nabla H\Vert ^\frac{3}{2}_{L^2}\Vert \nabla ^2H\Vert ^\frac{1}{2}_{L^2}, \end{aligned}$$

and

$$\begin{aligned} II_2&\le C\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}\Vert u\Vert _{L^6}\Vert \nabla u\Vert _{L^3} \le C\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}\Vert \nabla u\Vert _{L^2}\Vert \nabla u\Vert _{L^3} . \end{aligned}$$

Taking $\mathrm {div}$ and $\mathrm {curl}$, respectively, on both side of (2.1)$_2$, we get

$$\begin{aligned}&\Delta G= \mathrm {div}(\rho \dot{u})-\mathrm {div}[(\mathrm {curl}\ H) \times H] , \end{aligned}$$

(3.19)

$$\begin{aligned}&{\mu }\Delta (\mathrm {curl}u) =\mathrm {curl}(\rho \dot{u}) -\mathrm {curl}[(\mathrm {curl}\ H)\times H]. \end{aligned}$$

(3.20)

By (3.19), (3.20), the standard $L^2$-estimates, and (3.1), we get

$$\begin{aligned} \begin{aligned} \Vert \nabla G\Vert _{L^{2}}\le \Vert \rho \dot{u}\Vert _{L^{2}}+\Vert (\mathrm {curl}\ H)\times H\Vert _{L^{2}}\le \sqrt{2{\bar{\rho }}}\Vert \sqrt{\rho } \dot{u}\Vert _{L^{2}}+\Vert (\mathrm {curl}\ H)\times H\Vert _{L^{2}} \end{aligned} \end{aligned}$$

(3.21)

and

$$\begin{aligned} \begin{aligned} \Vert \nabla \mathrm {curl}u\Vert _{L^{2}} \le \Vert \rho \dot{u}\Vert _{L^{2}}+ \Vert (\mathrm {curl}\ H)\times H\Vert _{L^{2}}\le \sqrt{2{\bar{\rho }}}\Vert \sqrt{\rho } \dot{u}\Vert _{L^{2}}+\Vert (\mathrm {curl}\ H)\times H\Vert _{L^{2}}. \end{aligned} \end{aligned}$$

(3.22)

Since $\nabla u=\nabla \Delta ^{-1}\left( \nabla \mathrm {div}u-\nabla \times \mathrm {curl}u\right)$, we apply the Calderon-Zygmund inequality to get

$$\begin{aligned} \begin{aligned} \Vert \nabla u\Vert _{L^3}&\le C\big (\Vert \mathrm {curl}u\Vert _{L^3}+\Vert \mathrm {div}u\Vert _{L^3}\big ). \end{aligned} \end{aligned}$$

(3.23)

Using (3.21)–(3.23), Sobolev inequality, Hölder inequality and Gagliardo-Nirenberg inequality, we have

$$\begin{aligned} \Vert \nabla u\Vert _{L^3}&\le C\Vert \mathrm {curl}u\Vert _{L^3} +C\Vert G\Vert _{L^3}+C\Vert \rho \theta \Vert _{L^3}\nonumber \\&\le C\Vert \mathrm {curl}u\Vert _{L^2}^\frac{1}{2}\Vert \nabla \mathrm {curl}u\Vert _{L^2}^\frac{1}{2} +C\Vert G \Vert _{L^2}^\frac{1}{2} \Vert \nabla G\Vert _{L^2}^\frac{1}{2}+C\Vert \rho \Vert _{L^6}\Vert \theta \Vert _{L^6}\nonumber \\&\le C\Vert \mathrm {curl}u\Vert _{L^2}^\frac{1}{2} \left( \Vert \sqrt{\rho }\dot{u}\Vert _{L^2}+C\Vert |H||\nabla H|\Vert _{L^2}\right) ^{\frac{1}{2}}\nonumber \\&\quad +C\Vert G \Vert _{L^2}^\frac{1}{2} \left( \Vert \sqrt{\rho }\dot{u}\Vert _{L^2}+C\Vert |H||\nabla H|\Vert _{L^2}\right) ^{\frac{1}{2}}+Cm_0^{\frac{1}{6}}\Vert \nabla \theta \Vert _{L^2}. \end{aligned}$$

(3.24)

It follows from (3.21),(3.24) that

$$\begin{aligned} II_1+II_2&\le C\left( m_0^\frac{1}{6}\Vert \nabla \theta \Vert _{L^2}\Vert \nabla u\Vert _{L^2} +\Vert \nabla u\Vert _{L^2}\Vert \nabla u\Vert _{L^3}+\kappa \Vert \nabla \theta \Vert _{L^2} +\Vert \nabla H\Vert _{L^2}\Vert \nabla H\Vert _{L^3} \right) \left( \Vert \sqrt{\rho } \dot{u}\Vert _{L^2}\right. \nonumber \\&\left. \quad +\Vert |H||\nabla H|\Vert _{L^2}\right) \nonumber \\&\le C \Vert \nabla u\Vert _{L^2}\left( \Vert \nabla u\Vert _{L^2}+\Vert P\Vert _{L^2} \right) ^{\frac{1}{2}}\left( \Vert \sqrt{\rho }\dot{u}\Vert _{L^2}+\Vert |H||\nabla H|\Vert _{L^2}\right) ^{\frac{3}{2}} \nonumber \\&\quad +C\left( m_0^{\frac{1}{6}}\Vert \nabla u\Vert _{L^2}\right) \Vert \nabla \theta \Vert _{L^2}\left( \Vert \sqrt{\rho }\dot{u}\Vert _{L^2}+\Vert |H||\nabla H|\Vert _{L^2}\right) \nonumber \\&\quad +\frac{\kappa }{2\mu +\lambda }\Vert \nabla \theta \Vert _{L^2}\left( \sqrt{2{\bar{\rho }}}\Vert \sqrt{\rho } \dot{u}\Vert _{L^{2}}+\Vert (\mathrm {curl}\ H)\times H\Vert _{L^{2}}\right) \nonumber \\&\quad +C\Vert \nabla H\Vert ^3_{L^2}\Vert \nabla ^2 H\Vert _{L^2} \left( \Vert \sqrt{\rho }\dot{u}\Vert _{L^2}+\Vert |H||\nabla H|\Vert _{L^2}\right) \nonumber \\&\le C\Vert \nabla u\Vert ^6_{L^2}+Cm_0\Vert \nabla u\Vert _{L^2}^4+\frac{1}{2}\Vert \sqrt{\rho } \dot{u}\Vert ^2_{L^2}+C\Vert |H||\nabla H|\Vert _{L^2}^2+Cm_0^{\frac{1}{3}}\Vert \nabla u\Vert ^2_{L^2}\Vert \nabla \theta \Vert ^2_{L^2}\nonumber \\&\quad +\frac{\kappa ^2(2{\bar{\rho }}+1)}{(2\mu +\lambda )^2}\Vert \nabla \theta \Vert ^2_{L^2}+C\Vert \nabla H\Vert ^6_{L^2}+\Vert \nabla ^2 H\Vert ^2_{L^2}. \end{aligned}$$

(3.25)

For $II_3$, by using (3.1), (3.7), Hölder inequality and Sobolev inequality, we have

(3.26)

By (2.4), Lemmas 3.3 and 3.4 we have

$$\begin{aligned} \Vert |H||\nabla H|\Vert _{L^2}^2 \le C\Vert H\Vert _{L^2}\Vert \nabla H\Vert _{L^2}\Vert \nabla ^2 H\Vert ^2_{L^2}\le C m_0^\frac{1}{4}K^{\frac{1}{2}}\Vert \nabla ^2 H\Vert ^2_{L^2}. \end{aligned}$$

(3.27)

Substituting (3.25), (3.26) into (3.17), and integrating the resulting equation over [0, T], using Cauchy inequality, we have

(3.28)

It follows by Hölder inequality and (2.4) that

$$\begin{aligned}&\int \limits _{{\mathbb {R}}^3}P\mathrm {div}u \le \Vert \rho \theta \Vert _{L^2}\Vert \text{ div }u\Vert _{L^2}\le \frac{{\bar{\rho }}}{4(\mu +\lambda )}\Vert \sqrt{\rho }\theta \Vert ^2_{L^2}+ (\mu +\lambda )\Vert \text{ div }u\Vert ^2_{L^2}, \end{aligned}$$

(3.29)

$$\begin{aligned}&\quad \int \limits _{{\mathbb {R}}^3}\left( |H|^2\mathrm {div}u-2H\cdot \nabla u\cdot H\right) \le C\Vert H\Vert _{L^4}^2\Vert \nabla u\Vert _{L^2}\nonumber \\&\quad \le C\Vert H\Vert _{L^2}^\frac{1}{2}\Vert \nabla H\Vert _{L^2}^{\frac{3}{2}}\Vert \nabla u\Vert _{L^2} \le C\Vert H\Vert _{L^2}\Vert \nabla H\Vert _{L^2}^{3}+\frac{\mu }{2}\Vert \nabla u\Vert ^2_{L^2}. \end{aligned}$$

(3.30)

Then (3.28) together with (3.29), (3.30) leads to

$$\begin{aligned}&\int \limits _0^T\int \limits _{{\mathbb {R}}^3}\rho |\dot{u}|^2\mathrm{d}s+\int \limits _{{\mathbb {R}}^3}|\nabla u|^2\\&\quad \le \frac{2K}{\mu }+\frac{2{\bar{\rho }}}{\mu (\mu +\lambda )}\Vert \sqrt{\rho }\theta \Vert ^2_{L^2}+C\Vert H\Vert _{L^2}\Vert \nabla H\Vert _{L^2}^{3} + C\sup _{0\le t\le T}\left( \Vert \nabla u\Vert _{L^2}^4+m_0\Vert \nabla u\Vert _{L^2}^2 +1\right) \int \limits _{0}^{T} \Vert \nabla u\Vert _{L^2}^2 \mathrm{d}s \\&\qquad +\sup _{0\le t\le T}\left( m_0^\frac{1}{3}\Vert \nabla u\Vert _{L^2}^2 +m_0^\frac{2}{3}+\frac{\kappa ^2(2{\bar{\rho }}+1)}{(2\mu +\lambda )^2}\right) \int \limits _{0}^{T} \Vert \nabla \theta \Vert _{L^2}^2\mathrm{d}s +\frac{2}{\mu }\sup _{0\le t\le T}\Vert \nabla H\Vert _{L^2}^2\int \limits _{0}^{T}\Vert H_t\Vert _{L^2}^2\mathrm{d}s\\&\qquad +\left( Cm_0^\frac{1}{4}K_0^{\frac{1}{2}}+\frac{2}{\mu }\right) \int \limits _0^T\Vert \nabla ^2 H\Vert _{L^2}^2\mathrm{d}s+C\sup _{0\le t\le T}\Vert \nabla H\Vert _{L^2}^4\int \limits _{0}^{T}\Vert \nabla H\Vert _{L^2}^2\mathrm{d}s\\&\quad \le \left( \frac{2}{\mu }+\frac{2{\bar{\rho }}}{\mu (\mu +\lambda )}\right) K+Cm_0^{\frac{1}{4}}K^{\frac{3}{2}} +C\left( 4{\tilde{E}}^2K^2+m_0{\tilde{E}}K+1\right) m_0^\frac{1}{2}+C\left( 2m_0^\frac{1}{3}{\tilde{E}}K+m_0^\frac{2}{3} \right) K\\&\qquad +4K^2 +\left( C m_0^\frac{1}{4}K^\frac{1}{2}+C_1\right) K +CK^2m_0^\frac{1}{2}\\&\quad \le \left( \frac{4}{\mu }+\frac{2{\bar{\rho }}}{\mu (\mu +\lambda )}+\frac{6K}{\mu }\right) K +C\left( K^\frac{1}{2}+{\tilde{E}}Km_0^\frac{1}{4}+{\tilde{E}}m_0^{\frac{5}{4}}+\frac{m_0^{\frac{1}{4}}}{K}+Km_0^{\frac{1}{12}}+m_0^{\frac{5}{12}}+Km_0^\frac{1}{4} \right) Km_0^{\frac{1}{4}}\\&\quad \le {\tilde{E}}K+C\left( K^\frac{1}{2}+{\tilde{E}}Km_0^\frac{1}{12} \right) \le \frac{3{\tilde{E}}K}{2}, \end{aligned}$$

provided

$$\begin{aligned} m_0\le \frac{1}{C\left( K^\frac{1}{2}+{\tilde{E}}Km_0^\frac{1}{12}\right) ^4} \end{aligned}$$

with

$$\begin{aligned} {\tilde{E}}=1+\frac{4}{\mu }+\frac{2{\bar{\rho }}}{\mu (\mu +\lambda )}+\frac{6K}{\mu }. \end{aligned}$$

Thus, the proof of Lemma 3.5 is completed. $\square$

Lemma 3.6

Under the conditions of Proposition 3.1, it holds that

$$\begin{aligned}B(T)\le \frac{3K}{2},\end{aligned}$$

provided

Proof

Multiplying (2.1)$_3$ by $\theta$, and integrating by parts over ${\mathbb {R}}^3$, we obtain

(3.31)

For $III_1$, it follows by Hölder inequality and Sobolev inequality that

$$\begin{aligned} \begin{aligned} III_1\le C\Vert \mathrm {div}u\Vert _{L^2}\Vert \theta \Vert _{L^6}^2\Vert \rho \Vert _{L^6}\le Cm_0^\frac{1}{6} \Vert \nabla u\Vert _{L^2}\Vert \nabla \theta \Vert _{L^2}^2. \end{aligned} \end{aligned}$$

(3.32)

For $III_2$ and $III_3$, by virtue of Hölder inequality and Sobolev inequality, together with (3.24), one has

$$\begin{aligned} III_2+III_3&\le C\Vert \nabla u\Vert _{L^2}\Vert \nabla u\Vert _{L^3}\Vert \theta \Vert _{L^6}\nonumber \\&\le C\Vert \nabla u\Vert _{L^2}\Vert \nabla u\Vert _{L^3}\Vert \nabla \theta \Vert _{L^2}\nonumber \\&\le C\Vert \nabla u\Vert _{L^2}^\frac{3}{2}\Vert \nabla \theta \Vert _{L^2}\left( \Vert \sqrt{\rho }\dot{u}\Vert _{L^2}+\Vert |H||\nabla H|\Vert _{L^2}\right) ^\frac{1}{2}\nonumber \\&\quad +C\Vert \nabla u\Vert _{L^2}\Vert \nabla \theta \Vert _{L^2}\Vert \rho \theta \Vert _{L^2}^\frac{1}{2}\left( \Vert \sqrt{\rho }\dot{u}\Vert _{L^2}+\Vert |H||\nabla H|\Vert _{L^2}\right) ^\frac{1}{2} \nonumber \\&\quad +Cm_0^\frac{1}{6}\Vert \nabla u\Vert _{L^2}\Vert \nabla \theta \Vert _{L^2}^2. \end{aligned}$$

(3.33)

Then we estimate $III_4$ as

$$\begin{aligned} III_4\le C\Vert \nabla H\Vert _{L^2}\Vert \theta \Vert _{L^6}\Vert \nabla H\Vert _{L^3}\le C\Vert \nabla \theta \Vert _{L^2}\Vert \nabla H\Vert _{L^2}^{\frac{3}{2}}\Vert \nabla ^2 H\Vert _{L^2}^{\frac{1}{2}}. \end{aligned}$$

(3.34)

We substitute (3.32)–(3.33) into (3.31) to obtain

(3.35)

Integrating (3.35) over [0, T], and using (3.1), (3.3), (3.4) and (3.27), we have

$$\begin{aligned} B(T)&\le \int \limits _{{\mathbb {R}}^3}\rho _0|\theta _0|^2+C\int \limits _0^T\Vert \nabla u\Vert _{L^2}^3\left( \Vert \sqrt{\rho }\dot{u}\Vert _{L^2}+\Vert |H||\nabla H|\Vert _{L^2}\right) \mathrm{d}s+C\int \limits _0^T\Vert \nabla H\Vert _{L^2}^3\Vert \nabla ^2 H\Vert _{L^2}\mathrm{d}s\nonumber \\ {}&\quad +C\int \limits _0^T\Vert \nabla u\Vert _{L^2}^2\Vert \rho \theta \Vert _{L^2}\left( \Vert \sqrt{\rho }\dot{u}\Vert _{L^2}+\Vert |H||\nabla H|\Vert _{L^2}\right) \mathrm{d}s+C\int \limits _0^T\Vert \nabla u\Vert _{L^2}\Vert \nabla \theta \Vert _{L^2}^2\Vert \rho \Vert _{L^6}\mathrm{d}s\nonumber \\&\le K+C\sup _{0\le t\le T}\Vert \nabla u\Vert _{L^2}^2\left( \int \limits _{0}^{T}\Vert \nabla u\Vert ^2_{L^2}\mathrm{d}s\right) ^\frac{1}{2}\left( \left( \int \limits _{0}^{T}\Vert \sqrt{\rho }\dot{u}\Vert ^2_{L^2}\mathrm{d}s\right) ^\frac{1}{2} +K^{\frac{1}{4}}m_0^{\frac{1}{8}}\left( \int \limits _0^T\Vert \nabla ^2 H\Vert _{L^2}^2\mathrm{d}s\right) ^\frac{1}{2}\right) \nonumber \\&\quad +C\sup _{0\le t\le T}\Vert \nabla H\Vert _{L^2}^2 \left( \int \limits _0^T\Vert \nabla H\Vert _{L^2}^2\mathrm{d}s\right) ^\frac{1}{2} \left( \int \limits _0^T\Vert \nabla ^2 H\Vert _{L^2}^2\mathrm{d}s\right) ^\frac{1}{2}\nonumber \\&\quad +Cm_0^{\frac{1}{3}}\sup _{0\le t\le T}\Vert \nabla u\Vert _{L^2}^2 \left( \int \limits _0^T\Vert \nabla \theta \Vert _{L^2}^2\mathrm{d}s\right) ^\frac{1}{2} \left( \left( \int \limits _0^T\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2\mathrm{d}s\right) ^\frac{1}{2} +m_0^\frac{1}{8}K^\frac{1}{4}\left( \int \limits _0^T\Vert \nabla H^2\Vert _{L^2}^2\mathrm{d}s\right) ^\frac{1}{2} \right) \nonumber \\&\quad +Cm_0^\frac{1}{6}\left( \sup _{0\le t\le T}\Vert \nabla u\Vert _{L^2}^2\right) ^\frac{1}{2}\int \limits _0^T\Vert \nabla \theta \Vert _{L^2}^2\mathrm{d}s \nonumber \\&\le K+Cm_0^\frac{1}{6}K\left( {\tilde{E}}^{\frac{3}{2}}K+{\tilde{E}}K^{\frac{5}{4}}+{\tilde{E}}^{\frac{1}{2}}K^\frac{1}{2} \right) \le \frac{3K}{2}, \end{aligned}$$

(3.36)

provided

$$\begin{aligned}\begin{aligned} m_0 \le \frac{1}{C \left( {\tilde{E}}^{\frac{3}{2}}K+{\tilde{E}}K^{\frac{5}{4}}\right) ^6 }. \end{aligned} \end{aligned}$$

$\square$

Lemma 3.7

Under the assumptions of Proposition 3.1, it holds that

$$\begin{aligned} 0\le \rho \le \frac{3{\bar{\rho }}}{2}, \end{aligned}$$

(3.37)

for any $(x,t)\in {\mathbb {R}}^3\times [0,T]$, provided

$$\begin{aligned} m_0\le&\min \left\{ \frac{1}{CK^6},\quad \frac{1}{C\left( {\tilde{E}}^2K\right) ^2},\quad \frac{1}{C\left( K^\frac{1}{2}+{\tilde{E}}Km_0^\frac{1}{12}\right) ^4},\right. \\&\qquad \quad \left. \frac{1}{C \left( {\tilde{E}}^{\frac{3}{2}}K+{\tilde{E}}K^{\frac{5}{4}} \right) ^6 } ,\quad \frac{1}{C({\tilde{E}}K)^{\frac{144}{45}}},\quad \frac{1}{C\left( {\tilde{E}}K+K^\frac{3}{2}m_0^\frac{1}{4} \right) ^{48}} \right\} .\end{aligned}$$

Proof

It follows from (2.1)$_1$ formally that

$$\begin{aligned} \begin{aligned} Y^\prime (s)=g(Y)+b^\prime (s), \end{aligned} \end{aligned}$$

(3.38)

where

$$\begin{aligned} Y(s)=\log \rho \left( X(t;x,s),s\right) ,\ g(Y)=-\frac{P\left( X(t;x,s),s\right) }{2\mu +\lambda },\ b(s)=-\frac{1}{2\mu +\lambda }\int \limits _0^sG\left( X(t;x,\tau ),\tau \right) \,\mathrm{d}\tau , \end{aligned}$$

and X(t; x, s) is given by

By (2.1)$_1$, we have

This deduces

For $IV_1$, using (2.5), Sobolev inequality, Calderon–Zygmund inequality, Hölder inequality, (3.1) and (3.3), we get

$$\begin{aligned} IV_1&\le C\Vert \Delta ^{-1} \mathrm {div}(\rho u)\Vert _{L^6}^\frac{1}{3}\Vert \nabla \Delta ^{-1} \mathrm {div}(\rho u)\Vert _{L^4}^\frac{2}{3} \le C\Vert \rho u\Vert _{L^2}^\frac{1}{3}\Vert \rho u\Vert _{L^4}^\frac{2}{3} \le C\Vert \rho \Vert _{L^3}^\frac{1}{3} \Vert u\Vert _{L^6}^\frac{1}{3}\Vert \rho \Vert _{L^{12}}^\frac{2}{3}\Vert u\Vert _{L^6}^\frac{2}{3} \le C m_0^{\frac{7}{18}}{\tilde{E}}K. \end{aligned}$$

Similarly, it holds

$$\begin{aligned} IV_2&\le C m_0^{\frac{7}{18}}K, \end{aligned}$$

and then it implies

$$\begin{aligned} IV_1+IV_2\le Cm_0^{\frac{7}{18}}{\tilde{E}}K\le m_0^{\frac{1}{16}}, \end{aligned}$$

(3.39)

provided

$$\begin{aligned} m_0\le \frac{1}{C({\tilde{E}}K)^{\frac{144}{45}}}. \end{aligned}$$

(2.4) and (2.5) give

$$\begin{aligned} IV_3&\le C\int \limits _0^t\Vert \Delta ^{-1} \mathrm {div}\left( \rho u\mathrm {div} u\right) \Vert _{L^6}^\frac{1}{3}\Vert \nabla \Delta ^{-1} \mathrm {div}\left( \rho u\mathrm {div} u\right) \Vert _{L^4}^\frac{2}{3}\,\mathrm{d}\tau \nonumber \\&\le C\int \limits _0^t\Vert \rho u\mathrm {div} u\Vert _{L^2}^\frac{1}{3}\Vert \rho u\mathrm {div} u\Vert _{L^4}^\frac{2}{3}\,\mathrm{d}\tau \nonumber \\&\le C\int \limits _0^t\Vert \rho u\mathrm {div} u\Vert _{L^2}\,\mathrm{d}\tau +C\int \limits _0^t\Vert \rho u\mathrm {div} u\Vert _{L^4}\,\mathrm{d}\tau . \end{aligned}$$

(3.40)

By the definition of G, it reads

$$\begin{aligned} \begin{aligned} \rho u\mathrm {div} u=\frac{1}{2\mu +\lambda }\rho uG+\frac{1}{2\mu +\lambda }\rho ^2 u\theta . \end{aligned} \end{aligned}$$

This, together with Hölder inequality, Sobolev inequality, (3.22), (3.21), (3.3) and (3.1) deduces

$$\begin{aligned} \Vert \rho u\mathrm {div} u\Vert _{L^2}&\le C\Vert \rho uG\Vert _{L^2}+C\Vert \rho ^2 u\theta \Vert _{L^2}\nonumber \\&\le C\Vert \rho \Vert _{L^6} \Vert u\Vert _{L^6}\Vert G\Vert _{L^6}+\Vert \rho \Vert _{L^{12}}^2 \Vert u\Vert _{L^6}\Vert \theta \Vert _{L^6}\nonumber \\&\le C\Vert \rho \Vert _{L^6} \Vert \nabla u\Vert _{L^2}\Vert \nabla G\Vert _{L^2}+C\Vert \rho \Vert _{L^{12}}^2 \Vert \nabla u\Vert _{L^2}\Vert \nabla \theta \Vert _{L^2}\nonumber \\&\le m_0^\frac{1}{6}\Vert \nabla u\Vert _{L^2}\left( \Vert \sqrt{\rho }\dot{u}\Vert _{L^2}+\Vert |H||\nabla H|\Vert _{L^2}+ \Vert \nabla \theta \Vert _{L^2}\right) \end{aligned}$$

(3.41)

and

$$\begin{aligned}&\Vert \rho u\mathrm {div} u\Vert _{L^4}\nonumber \\&\quad \le C\Vert \rho uG\Vert _{L^4}+C\Vert \rho ^2 u\theta \Vert _{L^4}\nonumber \\&\quad \le \Vert u\Vert _{L^\infty }\Big (\Vert \rho \Vert _{L^{12}} \Vert G\Vert _{L^6}+\Vert \rho \Vert _{L^{24}}^2\Vert \theta \Vert _{L^6}\Big )\nonumber \\&\quad \le C(\Vert \nabla u\Vert _{L^2}+\Vert \nabla u\Vert _{L^6})\Big (\Vert \rho \Vert _{L^{12}} \left( \Vert \sqrt{\rho }\dot{u}\Vert _{L^2}+\Vert |H||\nabla H|\Vert _{L^2}\right) +\Vert \rho \Vert _{L^{24}}^2\Vert \nabla \theta \Vert _{L^2}\Big )\nonumber \\&\quad \le Cm_0^\frac{1}{12}(\Vert \nabla u\Vert _{L^2}+\Vert \nabla \times u\Vert _{L^6}+\Vert \mathrm {div} u\Vert _{L^6})\Big ( \Vert \sqrt{\rho }\dot{u}\Vert _{L^2}+\Vert |H||\nabla H|\Vert _{L^2}+\Vert \nabla \theta \Vert _{L^2}\Big )\nonumber \\&\quad \le Cm_0^\frac{1}{12}\Big (\Vert \nabla u\Vert _{L^2}+(\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}+\Vert |H||\nabla H|\Vert _{L^2}+\Vert \nabla \theta \Vert _{L^2})\Big )\Big (\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}+\Vert |H||\nabla H|\Vert _{L^2}+\Vert \nabla \theta \Vert _{L^2}\Big ). \end{aligned}$$

(3.42)

Putting (3.41) and (3.42) into (3.40), and using Cauchy inequality, (3.1) and (3.4), we obtain

$$\begin{aligned} IV_3&\le m_0^\frac{1}{6}\int \limits _0^t\Vert \nabla u\Vert _{L^2}\left( \Vert \sqrt{\rho }\dot{u}\Vert _{L^2}+\Vert |H||\nabla H|\Vert _{L^2}+ \Vert \nabla \theta \Vert _{L^2}\right) \,\mathrm{d}\tau \\&\quad +m_0^\frac{1}{12}\int \limits _0^t\Big (\Vert \nabla u\Vert _{L^2}+(\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}+\Vert |H||\nabla H|\Vert _{L^2}+\Vert \nabla \theta \Vert _{L^2})\Big )\\&\quad \times \Big ( \Vert \sqrt{\rho }\dot{u}\Vert _{L^2}+\Vert |H||\nabla H|\Vert _{L^2}+\Vert \nabla \theta \Vert _{L^2}\Big )\,\mathrm{d}\tau \\&\le C(m_0^\frac{1}{6}+m_0^\frac{1}{12})\int \limits _0^t\Vert \nabla u\Vert _{L^2}^2\,\mathrm{d}\tau +(m_0^\frac{1}{6} +m_0^\frac{1}{12})\int \limits _0^t(\Vert \sqrt{\rho }\dot{u}\Vert _{L^2}^2+\Vert |H||\nabla H|\Vert ^2_{L^2}+\Vert \nabla \theta \Vert _{L^2}^2)\,\mathrm{d}\tau \\&\le C(m_0^\frac{1}{6}+m_0^\frac{1}{12}) m_0^\frac{1}{2}+(m_0^\frac{1}{6}+m_0^\frac{1}{12})\left( {\tilde{E}}K+m_0^\frac{1}{4}K^{\frac{3}{2}}+K\right) \le m_0^{\frac{1}{16}}, \end{aligned}$$

provided

$$\begin{aligned} m_0\le \frac{1}{C\left( {\tilde{E}}K+K^\frac{3}{2}m_0^\frac{1}{4} \right) ^{48}}. \end{aligned}$$

For $IV_4$, it follows that by using the above inequality,

$$\begin{aligned} \begin{aligned} IV_4&\le C\int \limits _0^t\Vert \Delta ^{-1}\mathrm {div}[(\mathrm {curl}\ H)\times H]\Vert _{L^6}^\frac{1}{3}\Vert \nabla \Delta ^{-1}\mathrm {div}[(\mathrm {curl}\ H)\times H]\Vert _{L^4}^\frac{2}{3}\,\mathrm{d}\tau \\&\le C\int \limits _0^t\Vert (\mathrm {curl}\ H)\times H\Vert _{L^2}^\frac{1}{3}\Vert (\mathrm {curl}\ H)\times H\Vert _{L^4}^\frac{2}{3}\,\mathrm{d}\tau \\&\le C\int \limits _0^t\Vert (\mathrm {curl}\ H)\times H\Vert _{L^2}\,\mathrm{d}\tau +\int \limits _0^T\Vert (\mathrm {curl}\ H)\times H\Vert _{L^4}\,\mathrm{d}\tau \\&\le C\int \limits _0^t\Vert H\Vert _{L^6}\Vert \nabla H\Vert _{L^3}\,\mathrm{d}\tau +\int \limits _0^t\Vert H\Vert _{L^\infty }\Vert \nabla H\Vert _{L^4}\,\mathrm{d}\tau \\&\le C\left( \int \limits _0^T\Vert \nabla H\Vert _{L^2}^2\,\mathrm{d}\tau \right) ^\frac{3}{4}\left( \int \limits _0^t\Vert \nabla ^2 H\Vert _{L^2}^2\,\mathrm{d}\tau \right) ^\frac{1}{4} +\left( \int \limits _0^t\Vert \nabla H\Vert _{L^2}^2\,\mathrm{d}\tau \right) ^\frac{3}{8}\left( \int \limits _0^t\Vert \nabla ^2 H\Vert _{L^2}^2\,\mathrm{d}\tau \right) ^\frac{5}{8}\\&\le Cm_0^\frac{3}{8}K^{\frac{1}{4}}+Cm_0^\frac{3}{16}K^{\frac{5}{8}}\le m_0^{\frac{1}{16}}, \end{aligned} \end{aligned}$$

provided

$$\begin{aligned} m_0\le \frac{1}{CK^{5}}. \end{aligned}$$

By the estimates of $IV_i,i=1,2,3,4$, it yields

$$\begin{aligned} b(t)-b(0) \le&m_0^{\frac{1}{16}}\log \frac{3}{2} \le \log \frac{3}{2}, \end{aligned}$$

provided

$$\begin{aligned} m_0\le \min \left\{ \frac{1}{C({\tilde{E}}K)^{\frac{144}{45}}},\quad \frac{1}{C\left( {\tilde{E}}K+K^\frac{3}{2}m_0^\frac{1}{4} \right) ^{48}},\quad \frac{1}{CK^{5}} \right\} . \end{aligned}$$

Integrating (3.38) with respect to s over [0, t], we get

$$\begin{aligned}\begin{aligned} \log \rho (x, t)=&\log \rho _0\left( X(t;x,0)\right) +\int \limits _0^Tg(Y)\,\mathrm{d}t+b(t)-b(0) \le \log {\bar{\rho }}+\log \frac{3}{2}, \end{aligned} \end{aligned}$$

which deduces

$$\begin{aligned} \rho \le \frac{3{\bar{\rho }}}{2}. \end{aligned}$$

$\square$

Remark 3.8

The proof of the upper bound of $\rho$ is not rigorous, since $\rho$ may vanish. In fact, this could be handled by constructing an approximate solution $\rho ^\delta >0$ to (2.1)$_1$ with initial data $\rho _0+\delta >0$ for the constant $\delta >0$. Then one can replace $\log \rho$ by $\log \rho ^\delta$ in the above process and finally pass to the limits $\delta \rightarrow 0^+$ to obtain the desire estimate.

Now we are in the position to prove Theorem 2.3. Recall that in [11], the Serrin-Type blowup criterion for the strong solution to the problem (2.1)–(2.3) is obtained, which is

$$\begin{aligned} \limsup \limits _{T\nearrow T^*}\left( \Vert \rho \Vert _{L^\infty (0,T; L^\infty )}+\Vert u\Vert _{L^s(0,T; L^r)}\right) =\infty ,\ \ \frac{2}{s}+\frac{3}{r}\le 1. \end{aligned}$$

(3.43)

Then combining Proposition 3.1 with (3.43), we can get the global existence and uniqueness of the strong solution to (2.1)–(2.3). Indeed, the boundedness of $\rho$ is given in Lemma 3.7. On the other hand, by choosing $s=6,\ r=6$ in (3.43), and using Lemmas 3.3 and 3.5, one has

$$\begin{aligned} \int \limits _0^T\Vert u\Vert _{L^6}^6\le \sup _{0\le t\le T}\Vert \nabla u\Vert _{L^2}^4\int \limits _0^T\Vert \nabla u\Vert _{L^2}^2\le C. \end{aligned}$$

Thus, we obtain that $T^*=\infty$ and complete the proof of Theorem 2.3.