1 Introduction

Magnetohydrodynamics (MHD) studies the electromagnetic fields and conducting fluids. In this paper, we consider the time decay of the following 3D full compressible Hall-MHD system [16]:

$$\begin{aligned}&\partial _t\rho +\mathrm {div}\,(\rho u)=0, \end{aligned}$$
(1.1)
$$\begin{aligned}&\partial _t(\rho u)+\mathrm {div}\,(\rho u\otimes u)+\nabla p-\mu \Delta u-(\lambda +\mu )\nabla \mathrm {div}\,u=\mathrm {rot}\,B\times B, \end{aligned}$$
(1.2)
$$\begin{aligned}&C_V(\partial _t(\rho \theta )+\mathrm {div}\,(\rho u\theta ))-\Delta \theta +p\mathrm {div}\,u=\frac{\mu }{2}|\nabla u+\nabla u^t|^2\nonumber \\&\quad +\;\lambda (\mathrm {div}\,u)^2+|\mathrm {rot}\,B|^2,\qquad \end{aligned}$$
(1.3)
$$\begin{aligned}&\partial _tB+\mathrm {rot}\,(B\times u)+\mathrm {rot}\,\left( \frac{\mathrm {rot}\,B\times B}{\rho }\right) =\Delta B,\ \ \mathrm {div}\,B=0, \end{aligned}$$
(1.4)
$$\begin{aligned}&(\rho ,u,\theta ,B)(\cdot ,0)=(\rho _0,u_0,\theta _0,B_0)(\cdot )\ \ \mathrm {in}\ \ {\mathbb {R}^3}, \end{aligned}$$
(1.5)
$$\begin{aligned}&\lim \limits _{|x|\rightarrow \infty }(\rho ,u,\theta ,B)=(1,0,1,0). \end{aligned}$$
(1.6)

Here, \(\rho , u, \theta \), and B denote the density, velocity, temperature, and the magnetic field, respectively. The pressure \(p:=R\rho \theta \) with physical constant \(R>0\). \(\mu \) and \(\lambda \) are the shear viscosity and bulk viscosity of the flow and satisfy \(\mu >0\) and \(\displaystyle \lambda +\frac{2}{3}\mu \ge 0\). The positive constant \(C_V\) is the specific heat at constant volume. For simplicity, we will take \(R=C_V=1\). \(\nabla u^t\) is the transpose of the \(\nabla u\).

The applications of the Hall-MHD system cover a very wide range of physical objects, for example, magnetic reconnection in space plasmas, star formation, neutron stars, and geo-dynamo.

Continuity equation (1.1) allows the simplification of Eqs. (1.2) and (1.3) in their first two terms as \(\rho \partial _tu+\rho u\cdot \nabla u\) and \(C_V\rho \partial _t\theta +C_V\rho u\cdot \nabla \theta \), respectively.

When the Hall effect term \(\displaystyle {\mathrm {rot}\,\left( \frac{\mathrm {rot}\,B\times B}{\rho }\right) }\) is neglected, the system (1.1)–(1.4) reduces to the well-known full compressible MHD system, which has received many studies [2, 10, 11, 13, 14, 19]. The local strong solution was proved by Fan and Yu [10]. Fan and Yu [11], Ducomet-Feireisl [2], and Hu and Wang [13, 14] established the global weak solutions. Wei et al. [19] showed the time decay of the smooth solutions.

Very recently, in [4, 5, 8, 9], local well-posedness and low Mach number limit and blow-up criteria of smooth solutions to the problem (1.1)–(1.6) were established.

The aim of this paper is to use the method in [19] to prove the time decay rate of the Cauchy problem (1.1)–(1.6). We will prove

Theorem 1.1

Let \(N\ge 3\) and \(\rho _0-1, u_0, \theta _0-1, B_0\in H^N\). Then, there exists a positive constant \(\delta _0\) such that if

$$\begin{aligned} \Vert (\rho _0-1, u_0, \theta _0-1, B_0)\Vert _{H^N}\le \delta _0, \end{aligned}$$
(1.7)

then the Cauchy problem (1.1)–(1.6) has a unique global solution \((\rho , u, \theta , B)\), satisfying that for all \(t\ge 0\),

$$\begin{aligned}&\Vert (\rho -1, u, \theta -1, B)(t)\Vert _{H^N}^2+\Vert (\rho -1, u, \theta -1, B)\Vert _{L^2(0,t;H^{N+1})}^2\nonumber \\&\quad \le C\Vert (\rho _0-1, u_0, \theta _0-1, B_0)\Vert _{H^N}^2. \end{aligned}$$
(1.8)

If further, \((\rho _0-1, u_0, \theta _0-1, B_0)\in \dot{H}^{-s}\) for some \(\displaystyle {s\in \left( 0,\frac{3}{2}\right) }\), then for all \(t\ge 0\),

$$\begin{aligned} \Vert \Lambda ^{-s}(\rho -1, u, \theta -1, B)(t)\Vert _{L^2}\le C_0, \end{aligned}$$
(1.9)

and

$$\begin{aligned} \Vert D^\ell (\rho -1, u, \theta -1, B)\Vert _{H^{N-1}}\le C_0(1+t)^{-\frac{\ell +s}{2}},\ \ for\ \ \ell =0,1,\ldots ,N-1,\nonumber \\ \end{aligned}$$
(1.10)

where \(\Lambda :=(-\Delta )^\frac{1}{2}\). Here, \(C_0\) is a positive constant depending on \(N,s,\lambda ,\mu ,R,C_V\), and the initial data.

Remark 1.1

Wei et al. [19] showed the same time decay rate (1.8) and (1.10) when the Hall effect term is neglected. The new input of this paper is that it includes the Hall effect term.

Remark 1.2

For the corresponding incompressible model (incompressible Hall-MHD), we refer to [1, 3, 6, 7, 12, 17, 18] and references therein.

Notation. Throughout this paper, \(D^k\) with an integer k stands for the usual any spatial derivatives of order k. We define the operator \(\Lambda ^s, s\in \mathbb {R}\) by \(\displaystyle \Lambda ^sf:=\int |\xi |^s{\hat{f}}(\xi )e^{2\pi ix\cdot \xi }\hbox {d}\xi \), where \({\hat{f}}\) is the Fourier transform of f. We define the homogeneous Sobolev space \(\dot{H}^s\) of all f for which \(\Vert f\Vert _{\dot{H}^s}\) is finite, where \(\Vert f\Vert _{\dot{H}^s}:=\Vert \Lambda ^sf\Vert _{L^2}=\Vert |\xi |^s{\hat{f}}\Vert _{L^2}\), and we use \(H^s\) to denote the usual Sobolev spaces with norm \(\Vert \cdot \Vert _{H^s}\). We will employ the notation \(A\lesssim B\) to mean that \(A\le CB\) for a universal constant \(C>0\).

2 Proof of Theorem 1.1

This section is devoted to the proof of Theorem 1.1. We only need to establish a priori estimates.

First, we rewrite (1.1)–(1.3) as

$$\begin{aligned}&\partial _t\rho +\mathrm {div}\,u=-(\rho -1)\mathrm {div}\,u-u\cdot \nabla \rho , \end{aligned}$$
(2.1)
$$\begin{aligned}&\partial _tu-\mu \Delta u-(\lambda +\mu )\nabla \mathrm {div}\,u+\nabla \rho +\nabla \theta =-u\nonumber \\&\quad \cdot \nabla u-\frac{\rho -1}{\rho }(\mu \Delta u+(\lambda +\mu )\nabla \mathrm {div}\,u)\nonumber \\&\quad -\frac{(\theta -1)-(\rho -1)}{\rho }\nabla \rho +\frac{\mathrm {rot}\,B\times B}{\rho }, \end{aligned}$$
(2.2)
$$\begin{aligned}&\partial _t\theta -\Delta \theta +\mathrm {div}\,u=-u\cdot \nabla \theta -\frac{\rho -1}{\rho }\Delta \theta -(\theta -1)\mathrm {div}\,u\nonumber \\&\quad +\frac{1}{\rho }\left( \frac{\mu }{2}|\nabla u+\nabla u^t|^2+\lambda (\mathrm {div}\,u)^2+|\mathrm {rot}\,B|^2\right) . \end{aligned}$$
(2.3)

In the following proofs, we will use the following bilinear product estimate due to Kato and Ponce [15]:

$$\begin{aligned} \Vert D^s(fg)\Vert _{L^p}\le C(\Vert f\Vert _{L^{p_1}}\Vert D^sg\Vert _{L^{q_1}}+\Vert D^sf\Vert _{L^{p_2}}\Vert g\Vert _{L^{q_2}}), \end{aligned}$$
(2.4)

with \(s>0\) and \(\frac{1}{p}=\frac{1}{p_1}+\frac{1}{q_1}=\frac{1}{p_2}+\frac{1}{q_2}\).

We assume a priori that for sufficiently small \(\delta >0\),

$$\begin{aligned} \Vert (\rho -1, u, \theta -1, B)(t)\Vert _{H^3}\le \delta . \end{aligned}$$
(2.5)

By (2.5) and Sobolev’s inequality, we see that

$$\begin{aligned} \frac{1}{2}\le \rho \le \frac{3}{2}. \end{aligned}$$
(2.6)

Lemma 2.1

Let (2.5) hold true, we have

$$\begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t}\Vert D^k(\rho -1, u, \theta -1, B)\Vert _{L^2}^2+C\Vert D^{k+1}(u, \theta -1, B)\Vert _{L^2}^2\nonumber \\&\quad \lesssim \delta \Vert D^{k+1}(\rho -1, u, \theta -1, B)\Vert _{L^2}^2+\delta \Vert D^k(\rho -1, B)\Vert _{L^2}^2 \end{aligned}$$
(2.7)

for \(k=0,1,\ldots ,N-1\).

Proof

Applying \(D^k\) to (2.1), (2.2), (2.3), and (1.4), testing by \(D^k(\rho -1), D^ku, D^k(\theta -1)\), and \(D^kB\), respectively, and summing up them, we find that

$$\begin{aligned}&\frac{1}{2}\frac{\hbox {d}}{\hbox {d}t}\Vert D^k(\rho -1, u, \theta -1, B)\Vert _{L^2}^2+C\Vert D^{k+1}(u, \theta -1, B)\Vert _{L^2}^2\nonumber \\&\quad =I_1+\int D^k\mathrm {rot}\,\left( \frac{B\times \mathrm {rot}\,B}{\rho }\right) D^kB \hbox {d}x\nonumber \\&\quad =:I_1+I_2. \end{aligned}$$
(2.8)

It has been proved in [19] that

$$\begin{aligned} I_1:= & {} -\int D^k[(\rho -1)\mathrm {div}\,u+u\cdot \nabla \rho ]D^k(\rho -1)\hbox {d}x\\&-\int D^k\left[ u\cdot \nabla u+\frac{\rho -1}{\rho }(\mu \Delta u+(\lambda +\mu )\nabla \mathrm {div}\,u)\right. \\&\left. +\frac{(\theta -1) -(\rho -1)}{\rho }\nabla \rho -\frac{\mathrm {rot}\,B\times B}{\rho }\right] D^ku \hbox {d}x\\&+\int D^k\left[ -u\cdot \nabla \theta -\frac{\rho -1}{\rho }\Delta \theta -(\theta -1)\mathrm {div}\,u\right. \\&\left. +\frac{1}{\rho }\left( \frac{\mu }{2}|\nabla u+\nabla u^t|^2+\lambda (\mathrm {div}\,u)^2+|\mathrm {rot}\,B|^2\right) \right] D^k\theta \hbox {d}x\\&+\int D^k\mathrm {rot}\,\left[ u\times B-\frac{\mathrm {rot}\,B\times B}{\rho }\right] D^k B \hbox {d}x\\\lesssim & {} \delta \Vert D^{k+1}(\rho -1, u, \theta -1, B)\Vert _{L^2}^2. \end{aligned}$$

We use (2.4) and (2.5) to bound \(I_2\) as follows:

$$\begin{aligned} I_2= & {} -\int D^k\left( \mathrm {rot}\,B\times \frac{B}{\rho }\right) D^k\mathrm {rot}\,B \hbox {d}x\\\lesssim & {} \left\| D^k\left( \mathrm {rot}\,B\times \frac{B}{\rho }\right) \right\| \Vert D^k\mathrm {rot}\,B\Vert _{L^2}\\\lesssim & {} \left( \Vert D^k\mathrm {rot}\,B\Vert _{L^2}\left\| \frac{B}{\rho }\right\| _{L^\infty }+\Vert \mathrm {rot}\,B\Vert _{L^\infty }\left\| D^k\left( \frac{B}{\rho }\right) \right\| _{L^2}\right) \Vert D^k\mathrm {rot}\,B\Vert _{L^2}\\\lesssim & {} \delta (\Vert D^k\mathrm {rot}\,B\Vert _{L^2}+\Vert D^kB\Vert _{L^2}+\Vert D^k(\rho -1)\Vert _{L^2})\Vert D^k\mathrm {rot}\,B\Vert _{L^2}\\\le & {} \delta (\Vert D^{k+1}B\Vert _{L^2}^2+\Vert D^kB\Vert _{L^2}^2+\Vert D^k(\rho -1)\Vert _{L^2}^2). \end{aligned}$$

Inserting the above estimates into (2.8) gives the lemma. \(\square \)

Lemma 2.2

Let (2.5) hold true, we have

$$\begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t}\Vert D^{k+1}(\rho -1, u, \theta -1, B)\Vert _{L^2}^2+C\Vert D^{k+2}(u, \theta -1, B)\Vert _{L^2}^2\nonumber \\&\quad \lesssim \delta (\Vert D^{k+1}(\rho -1)\Vert _{L^2}^2+\Vert D^{k+1}B\Vert _{L^2}^2+\Vert D^{k+2}(u, \theta -1, B)\Vert _{L^2}^2) \end{aligned}$$
(2.9)

for \(k=0,1,\ldots ,N-1\).

Proof

Applying \(D^{k+1}\) to (2.1), (2.2), (2.3), and (1.4), testing by \(D^{k+1}(\rho -1), D^{k+1}u, D^{k+1}(\theta -1)\), and \(D^{k+1}B\), respectively, and summing up them, we obtain

$$\begin{aligned}&\frac{1}{2}\frac{\hbox {d}}{\hbox {d}t}\Vert D^{k+1}(\rho -1, u, \theta -1, B)\Vert _{L^2}^2+C\Vert D^{k+2}(u, \theta -1, B)\Vert _{L^2}^2\nonumber \\&\quad =I_3+\int D^{k+1}\mathrm {rot}\,\left( \frac{B\times \mathrm {rot}\,B}{\rho }\right) D^{k+1}B \hbox {d}x\nonumber \\&\quad =:I_3+I_4. \end{aligned}$$
(2.10)

It has been proved in [19] that

$$\begin{aligned} I_3:= & {} -\int D^{k+1}[(\rho -1)\mathrm {div}\,u+u\cdot \nabla \rho ]D^{k+1}(\rho -1)\hbox {d}x\\&-\int D^{k+1}\left[ u\cdot \nabla u+\frac{\rho -1}{\rho }(\mu \Delta u+(\lambda +\mu )\nabla \mathrm {div}\,u)\right. \\&\quad \left. +\frac{(\theta -1)-(\rho -1)}{\rho }\nabla \rho -\frac{\mathrm {rot}\,B\times B}{\rho }\right] D^{k+1}u \hbox {d}x\\&+\int D^{k+1}\left[ -u\cdot \nabla \theta -\frac{\rho -1}{\rho }\Delta \theta -(\theta -1)\mathrm {div}\,u\right. \\&\quad \left. +\frac{1}{\rho }\left( \frac{\mu }{2}|\nabla u+\nabla u^t|^2+\lambda (\mathrm {div}\,u)^2+|\mathrm {rot}\,B|^2\right) \right] D^{k+1}\theta \hbox {d}x\\&+\int D^{k+1}\mathrm {rot}\,\left[ u\times B-\frac{\mathrm {rot}\,B\times B}{\rho }\right] D^{k+1} B \hbox {d}x\\\lesssim & {} \delta (\Vert D^{k+1}(\rho -1)\Vert _{L^2}^2+\Vert D^{k+2}(u, \theta -1, B)\Vert _{L^2}^2). \end{aligned}$$

Similarly to \(I_2\), we use (2.4) and (2.5) to bound \(I_4\) as

$$\begin{aligned} I_4\lesssim \delta (\Vert D^{k+2}B\Vert _{L^2}^2+\Vert D^{k+1}B\Vert _{L^2}^2+\Vert D^{k+1}(\rho -1)\Vert _{L^2}^2). \end{aligned}$$

Inserting the above estimates into (2.10) gives the lemma. \(\square \)

In [19], it is proved that

$$\begin{aligned}&\frac{\hbox {d}}{\hbox {d}t}\int D^ku\cdot D^{k+1}(\rho -1)dx+C\Vert D^{k+1}(\rho -1)\Vert _{L^2}^2\nonumber \\&\quad \lesssim \Vert D^{k+1}u\Vert _{L^2}^2+\Vert D^{k+1}(\theta -1)\Vert _{L^2}^2+\Vert D^{k+2}(u, \theta -1, B)\Vert _{L^2}^2 \end{aligned}$$
(2.11)

for \(k=0,1,\ldots ,N-1\).

Lemma 2.3

For \(s\in \left( 0,\frac{1}{2}\right] \), we have

$$\begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t}\Vert \Lambda ^{-s}(\rho -1, u, \theta -1, B)\Vert _{L^2}^2+C\Vert \nabla \Lambda ^{-s}(u, \theta -1, B)\Vert _{L^2}^2\nonumber \\&\quad \lesssim \Vert \nabla (\rho -1, u, \theta -1, B)\Vert _{H^2}^2\Vert \Lambda ^{-s}(\rho -1, u, \theta -1, B)\Vert _{L^2} \end{aligned}$$
(2.12)

and for \(\displaystyle {s\in \left( \frac{1}{2},\frac{3}{2}\right) }\), we have

$$\begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t}\Vert \Lambda ^{-s}(\rho -1, u, \theta -1, B)\Vert _{L^2}^2+C\Vert \nabla \Lambda ^{-s}(u, \theta -1, B)\Vert _{L^2}^2\nonumber \\&\quad \lesssim \Vert (\rho -1, u, \theta -1, B)\Vert _{H^1}^{s-\frac{1}{2}}\Vert \nabla (\rho -1, u, \theta -1, B)\Vert _{H^1}^{\frac{5}{2}-s}\nonumber \\&\qquad \times \Vert \Lambda ^{-s}(\rho -1, u, \theta -1, B)\Vert _{L^2}.\quad \qquad \end{aligned}$$
(2.13)

Proof

Applying \(\Lambda ^{-s}\) to (2.1), (2.2), (2.3), and (1.4), testing by \(\Lambda ^{-s}(\rho -1), \Lambda ^{-s}u, \Lambda ^{-s}(\theta -1)\), and \(\Lambda ^{-s}B\), respectively, and summing up them, we derive

$$\begin{aligned}&\frac{1}{2}\frac{\hbox {d}}{\hbox {d}t}\Vert \Lambda ^{-s}(\rho -1, u, \theta -1, B)\Vert _{L^2}^2\nonumber \\&\qquad +\int (\mu |\nabla \Lambda ^{-s}u|^2+(\lambda +\mu )(\mathrm {div}\,\Lambda ^{-s}u)^2+|\nabla \Lambda ^{-s}(\theta -1)|^2+|\nabla \Lambda ^{-s}B|^2) \hbox {d}x\nonumber \\&\quad =I_5+\int \Lambda ^{-s}\mathrm {rot}\,\left( \frac{B\times \mathrm {rot}\,B}{\rho }\right) \Lambda ^{-s}B \hbox {d}x\nonumber \\&\quad =:I_5+I_6. \end{aligned}$$
(2.14)

It has been proved in [19] that

$$\begin{aligned} I_5:= & {} -\int \Lambda ^{-s}[(\rho -1)\mathrm {div}\,u+u\cdot \nabla \rho ]\Lambda ^{-s}(\rho -1)\hbox {d}x\\&-\int \Lambda ^{-s}\left[ u\cdot \nabla u+\frac{\rho -1}{\rho }(\mu \Delta u+(\lambda +\mu )\nabla \mathrm {div}\,u)\right. \\&\quad \left. +\frac{(\theta -1)-(\rho -1)}{\rho }\nabla \rho -\frac{\mathrm {rot}\,B\times B}{\rho }\right] \Lambda ^{-s}u \hbox {d}x\\&-\int \Lambda ^{-s}\left[ u\cdot \nabla \theta +\frac{\rho -1}{\rho }\Delta \theta +(\theta -1)\mathrm {div}\,u\right. \\&\quad \left. -\frac{1}{\rho }\left( \frac{\mu }{2}|\nabla u+\nabla u^t|^2+\lambda (\mathrm {div}\,u)^2+|\mathrm {rot}\,B|^2\right) \right] \Lambda ^{-s}\theta \hbox {d}x\\&+\int \Lambda ^{-s}\mathrm {rot}\,\left[ u\times B-\frac{\mathrm {rot}\,B\times B}{\rho }\right] \Lambda ^{-s}B \hbox {d}x\\\lesssim & {} \Vert \nabla (\rho -1, u, \theta -1, B)\Vert _{H^2}^2\Vert \Lambda ^{-s}(\rho -1, u, \theta -1, B)\Vert _{L^2}\ \ \mathrm {for}\ \ s\in \left( 0,\frac{1}{2}\right] , \end{aligned}$$

and

$$\begin{aligned}&I_5\lesssim \Vert (\rho -1, u, \theta -1, B)\Vert _{H^1}^{s-\frac{1}{2}}\Vert \nabla (\rho -1, u, \theta -1, B)\Vert _{H^1}^{\frac{5}{2}-s}\Vert \nonumber \\&\quad \times \Lambda ^{-s}(\rho -1, u, \theta -1, B)\Vert _{L^2} \end{aligned}$$

for \(\displaystyle {s\in \left( \frac{1}{2},\frac{3}{2}\right) }\).

In the following proofs, we will use the following inequality [19]:

$$\begin{aligned} \Vert \Lambda ^{-s}w\Vert _{L^2}\lesssim \Vert w\Vert _{L^\frac{3}{s+\frac{3}{2}}}, \end{aligned}$$
(2.15)

and the Gagliardo–Nirenberg inequalities [19]:

$$\begin{aligned}&\Vert B\Vert _{L^2}\lesssim \Vert \Lambda ^{-s}B\Vert _{L^2}^\frac{1}{1+s}\Vert \nabla B\Vert _{L^2}^\frac{s}{1+s}, \end{aligned}$$
(2.16)
$$\begin{aligned}&\Vert \Lambda ^{1-s}B\Vert _{L^2}\lesssim \Vert \Lambda ^{-s}B\Vert _{L^2}^\frac{s}{1+s}\Vert \nabla B\Vert _{L^2}^\frac{1}{1+s}. \end{aligned}$$
(2.17)

Using (2.15), (2.16), and (2.17), we bound \(I_6\) as follows:

$$\begin{aligned} I_6\lesssim & {} \left\| \Lambda ^{-s}\left( \frac{\mathrm {rot}\,B\times B}{\rho }\right) \right\| _{L^2}\Vert \Lambda ^{-s}\mathrm {rot}\,B\Vert _{L^2}\\\lesssim & {} \left\| \frac{\mathrm {rot}\,B\times B}{\rho }\right\| _{L^\frac{3}{s+\frac{3}{2}}}\Vert \Lambda ^{1-s}B\Vert _{L^2}\\\lesssim & {} \Vert \mathrm {rot}\,B\Vert _{L^\frac{3}{s}}\Vert B\Vert _{L^2}\Vert \Lambda ^{1-s}B\Vert _{L^2}\\\lesssim & {} \Vert \mathrm {rot}\,B\Vert _{L^\frac{3}{s}}\Vert \nabla B\Vert _{L^2}\Vert \Lambda ^{-s}B\Vert _{L^2}. \end{aligned}$$

When \(\displaystyle 0<s\le \frac{1}{2}\), we have

$$\begin{aligned} I_6\lesssim \Vert \nabla B\Vert _{H^2}\Vert \nabla B\Vert _{L^2}\Vert \Lambda ^{-s}B\Vert _{L^2}\lesssim \Vert \nabla B\Vert _{H^2}^2\Vert \Lambda ^{-s}B\Vert _{L^2}. \end{aligned}$$
(2.18)

When \(\displaystyle \frac{1}{2}<s<\frac{3}{2}\), using the Gagliardo–Nirenberg inequality

$$\begin{aligned} \Vert \mathrm {rot}\,B\Vert _{L^\frac{3}{s}}\lesssim \Vert \nabla B\Vert _{L^2}^{s-\frac{1}{2}}\Vert \nabla ^2B\Vert _{L^2}^{\frac{3}{2}-s}, \end{aligned}$$

we bound \(I_6\) as

$$\begin{aligned} I_6\lesssim & {} \Vert \nabla B\Vert _{L^2}^{s-\frac{1}{2}}\Vert \nabla ^2B\Vert _{L^2}^{\frac{3}{2}-s}\Vert \nabla B\Vert _{L^2}\Vert \Lambda ^{-s}B\Vert _{L^2}\nonumber \\\lesssim & {} \Vert \nabla B\Vert _{L^2}^{s-\frac{1}{2}}\Vert \nabla B\Vert _{H^1}^{\frac{5}{2}-s}\Vert \Lambda ^{-s}B\Vert _{L^2}. \end{aligned}$$
(2.19)

Inserting the above estimates (2.18) and (2.19) into (2.14) leads to the lemma. \(\square \)

Now, by the same calculations as that in [19], we can finish the proof of Theorem 1.1.

We first close the energy estimates at each \(l\hbox {th}\) level in our weaker sense. Let \(N\ge 3\) and \(0\le l\le m-1\) with \(1\le m\le N\). Summing up the estimates (2.7) of Lemma 2.1 for from \(k=l\) to \(m-1\), we obtain

$$\begin{aligned}&\frac{\hbox {d}}{\hbox {d}t}\sum \limits _{l\le k\le m-1}\Vert D^k(\rho ,u,\theta -1,B)\Vert _{L^2}^2+C\sum \limits _{l+1\le k\le m}\Vert D^k(u,\theta -1,B)\Vert _{L^2}^2\nonumber \\&\quad \lesssim \delta \sum \limits _{l+1\le k\le m}\Vert D^k(\rho -1,u,\theta -1,B)\Vert _{L^2}^2. \end{aligned}$$
(2.20)

Let \(k=m-1\) in the estimates (2.9) of Lemma 2.2, we have

$$\begin{aligned}&\frac{\hbox {d}}{\hbox {d}t}\Vert D^m(\rho -1,u,\theta -1,B\Vert _{L^2}^2+C\Vert D^{m+1}(u,\theta -1,B)\Vert _{L^2}^2\nonumber \\&\quad \lesssim \delta (\Vert D^m(\rho -1)\Vert _{L^2}^2+\Vert D^{m+1}u\Vert _{L^2}^2+\Vert D^{m+1}(\theta -1)\Vert _{L^2}^2+\Vert D^{m+1}B\Vert _{L^2}^2).\nonumber \\ \end{aligned}$$
(2.21)

Adding the inequality (2.20) with (2.21), we get

$$\begin{aligned}&\frac{\hbox {d}}{\hbox {d}t}\sum \limits _{l\le k\le m}\Vert D^k(\rho -1,u,\theta -1,B)\Vert _{L^2}^2+C_1\sum \limits _{l+1\le k\le m+1}\Vert D^k(u,\theta -1,B)\Vert _{L^2}^2\nonumber \\&\quad \le C_2\delta \sum \limits _{l+1\le k\le m}\Vert D^k(\rho -1)\Vert _{L^2}^2. \end{aligned}$$
(2.22)

Summing up the estimates (2.11) for from \(k=l\) to \(m-1\), we have

$$\begin{aligned}&\frac{\hbox {d}}{\hbox {d}t}\sum \limits _{l\le k\le m}\int _{\mathbb {R}^3}D^ku\cdot D^{k+1}(\rho -1) \hbox {d}x+C_3\sum \limits _{l+1\le k\le m}\Vert D^k(\rho -1)\Vert _{L^2}^2\nonumber \\&\quad \le C_4\sum \limits _{l+1\le k\le m+1}(\Vert D^ku\Vert _{L^2}^2+\Vert D^k(\theta -1)\Vert _{L^2}^2)+C_4\sum \limits _{l+2\le k\le m+1}\Vert D^kB\Vert _{L^2}^2.\qquad \qquad \end{aligned}$$
(2.23)

Multiplying (2.23) by \(\displaystyle {\frac{2C_2\delta }{C_3}}\) and adding it with (2.22), since \(\delta >0\) is small, we deduce that there exists a constant \(C_5>0\) such that for \(0\le l\le m-1\)

$$\begin{aligned}&\frac{\hbox {d}}{\hbox {d}t}\left\{ \sum \limits _{l\le k\le m}\Vert D^k(\rho -1,u,\theta -1,B)\Vert _{L^2}^2+\frac{2C_2\delta }{C_3}\sum \limits _{l\le k\le m-1}\int _{\mathbb {R}^3}D^ku\cdot D^{k+1}(\rho -1)\hbox {d}x\right\} \nonumber \\&\quad +C_5\left\{ \sum \limits _{l+1\le k\le m}\Vert D^k(\rho -1)\Vert _{L^2}^2+\sum \limits _{l+1\le k\le m+1}\Vert D^k(u,\theta -1,B)\Vert _{L^2}^2\right\} \le 0. \end{aligned}$$
(2.24)

Next, we define \(\mathcal {E}_l^m(t)\) to be \(C_5^{-1}\) times the expression under the time derivative in (2.24). Observe that since \(\delta \) is small, \(\mathcal {E}_l^m(t)\) is equivalent to \(\Vert D^l(\rho -1,u,\theta -1,B)(t)\Vert _{H^{m-l}}^2\), that is, there exists a constant \(C_6>0\) such that for \(0\le l\le m-1\)

$$\begin{aligned} C_6^{-1}\Vert D^l(\rho -1,u,\theta -1,B)\Vert _{H^{m-l}}^2\le \mathcal {E}_l^m(t)\le C_6\Vert D^l(\rho -1,u,\theta -1,B)\Vert _{H^{m-l}}^2. \end{aligned}$$
(2.25)

Then, we may write (2.24) as that for \(0\le l\le m-1\)

$$\begin{aligned}&\frac{\hbox {d}}{\hbox {d}t}\mathcal {E}_l^m(t)+\Vert D^{l+1}(\rho -1)\Vert _{H^{m-l-1}}^2+\Vert D^{l+1}u\Vert _{H^{m-l}}^2\nonumber \\&\quad +\Vert D^{l+1}(\theta -1)\Vert _{H^{m-l}}^2+\Vert D^{l+1}B\Vert _{m-l}^2\le 0. \end{aligned}$$
(2.26)

Proof of (1.8). Taking \(l=0\) and \(m=3\) in (2.26) and then integrating directly in time, we get

$$\begin{aligned} \Vert (\rho -1,u,\theta -1,B)(t)\Vert _{H^3}^2\lesssim \mathcal {E}_0^3(t)\lesssim \mathcal {E}_0^3(0)\lesssim \Vert (\rho _0-1,u_0,\theta _0-1,B_0)\Vert _{H^3}^2. \end{aligned}$$
(2.27)

By a standard continuity argument, this closes the a priori estimates if at the initial time we assume that \(\Vert (\rho _0-1,u_0,\theta _0-1,B_0)\Vert _{H^3}^2\le \delta _0\) is sufficiently small. This in turn allows us to take \(l=0\) and \(m=N\) in (2.26) and then integrate it directly in time to obtain

$$\begin{aligned}&\Vert (\rho -1,u,\theta -1,B)(t)\Vert _{H^N}^2\\&\quad +\int _0^t\left( \Vert \nabla (\rho -1) (\tau )\Vert _{H^{N-1}}^2+\Vert \nabla u(\tau )\Vert _{H^N}^2+\Vert \nabla (\theta -1)(\tau )\Vert _{H^N}^2+\Vert \nabla B(\tau )\Vert _{H^N}^2\right) \hbox {d}\tau \\&\qquad \le C\Vert (\rho _0-1,u_0,\theta _0-1,B_0)\Vert _{H^N}^2. \end{aligned}$$

This proved (1.8).

Next, we turn to prove (1.10). However, we are not able to prove them for all \(\displaystyle s\in \left[ 0,\frac{3}{2}\right) \), and at this moment, we shall first prove them for \(\displaystyle s\in \left[ 0,\frac{1}{2}\right] \).

Define \(\mathcal {E}_{-s}(t):=\Vert \Lambda ^{-s}(\rho -1,u,\theta -1,B)(t)\Vert _{L^2}^2\). Then, integrating in time (2.12) of Lemma 2.3, by the bound (1.8), we obtain that for \(\displaystyle s\in \left( 0,\frac{1}{2}\right] \),

$$\begin{aligned}&\mathcal {E}_{-s}(t)\le \mathcal {E}_{-s}(0)+C\int _0^t\Vert \nabla (\rho -1,u,\theta -1,B)(\tau )\Vert _{H^2}^2 \sqrt{\mathcal {E}_{-s}(\tau )}\hbox {d}\tau \nonumber \\&\quad \le C_0\left( 1+\sup \limits _{0\le \tau \le t}\sqrt{\mathcal {E}_{-s}(\tau )}\right) . \end{aligned}$$
(2.28)

This implies (1.10) for \(\displaystyle s\in \left[ 0,\frac{1}{2}\right] \), that is,

$$\begin{aligned} \Vert \Lambda ^{-s}(\rho -1,u,\theta -1,B)(t)\Vert _{L^2}^2\le C_0. \end{aligned}$$
(2.29)

If \(l=1,\ldots ,N-1\), we may use the Gagliardo–Nirenberg inequality [19]:

$$\begin{aligned} \Vert \nabla ^{l+1}f\Vert _{L^2}\ge C\Vert \Lambda ^{-s}f\Vert _{L^2}^{-\frac{1}{l+s}}\Vert \nabla ^lf\Vert _{L^2}^{1+\frac{1}{l+s}}. \end{aligned}$$

By this fact and (2.29), we may find

$$\begin{aligned} \Vert D^{l+1}(\rho -1,u,\theta -1,B)\Vert _{L^2}^2\ge & {} C_0\left( \Vert D^l(\rho -1)\Vert _{L^2}^2+\Vert D^lu\Vert _{L^2}^2\right. \nonumber \\&\left. +\Vert D^l(\theta -1)\Vert _{L^2}^2+\Vert D^lB\Vert _{L^2}^2\right) ^{1+\frac{1}{l+s}}.\qquad \end{aligned}$$
(2.30)

This together with (1.8) implies in particular that for \(l=1,\ldots ,N-1\),

$$\begin{aligned}&\Vert D^{l+1}(\rho -1,u,\theta -1,B)\Vert _{H^{N-l-1}}^2\ge C_0\left( \Vert D^l(\rho -1)\Vert _{H^{N-l}}^2\right. \nonumber \\&\quad \left. +\Vert D^lu\Vert _{H^{N-l}}^2+\Vert D^l(\theta -1)\Vert _{H^{N-t}}^2+\Vert D^lB\Vert _{H^{N-l}}\right) ^{1+\frac{1}{l+s}}.\qquad \end{aligned}$$
(2.31)

Thus, we deduce from (2.26) with \(m=N\) the following time differential inequality

$$\begin{aligned} \frac{\hbox {d}}{\hbox {d}t}\mathcal {E}_t^N(t)+C_0(\mathcal {E}_l^N)^{1+\frac{1}{l+s}}\le 0\ \ {\mathrm {for}}\ \ l=1,\ldots ,N-1. \end{aligned}$$

Solving this inequality directly gives

$$\begin{aligned} \mathcal {E}_l^N(t)\le C_0(1+t)^{-(l+s)}\ \ {\mathrm {for}}\ \ l=1,\ldots ,N-1. \end{aligned}$$

This implies that for \(s\in \left[ 0,\frac{1}{2}\right] \),

$$\begin{aligned} \Vert D^l(\rho -1,u,\theta -1,B)(t)\Vert _{H^{N-l}}^2\le C_0(1+t)^{-(l+s)}\ \ {\mathrm {for}}\ \ l=1,\ldots ,N-1. \end{aligned}$$
(2.32)

Hence, by (2.29), (2.32), and the interpolation, we get (1.10) for \(\displaystyle s\in \left[ 0,\frac{1}{2}\right] \).

For \(\displaystyle s\in \left( \frac{1}{2},\frac{3}{2}\right) \), notice that the arguments for the case \(\displaystyle s\in \left[ 0,\frac{1}{2}\right] \) cannot be applied to this case. However, observing that we have \(\rho _0-1,u_0,\theta _0-1,B_0\in \dot{H}^\frac{1}{2}\) since \(\dot{H}^{-s}\cap L^2\subset \dot{H}^{-s'}\) for any \(\displaystyle s'\in \left[ 0,\frac{1}{2}\right] \), we then deduce from what we have proved for (1.10) with \(s=\frac{1}{2}\) that the following decay result holds:

$$\begin{aligned} \Vert D^l(\rho -1,u,\theta -1,B)(t)\Vert _{H^{N-l}}^2\le C_0(1+t)^{-\left( l+\frac{1}{2}\right) }\ \ {\mathrm {for}}\ \ l=0,1,\ldots ,N-1. \end{aligned}$$
(2.33)

Hence, by (2.33), we deduce from (2.13) that for \(\displaystyle s\in \left( \frac{1}{2},\frac{3}{2}\right) \),

$$\begin{aligned} \mathcal {E}_{-s}(t)\le & {} \mathcal {E}_{-s}(0) +C\int _0^t\Vert (\rho -1,u,\theta -1,B)(\tau )\Vert _{H^1}^{s-\frac{1}{2}}\Vert \nonumber \\&\times \nabla (\rho -1,u,\theta -1,B)(\tau )\Vert _{H^1}^{\frac{5}{2}-s}\sqrt{\mathcal {E}_{-s}(\tau )} \hbox {d}\tau \nonumber \\\le & {} C_0+C_0\int _0^t(1+\tau )^{-\left( \frac{7}{4}-\frac{s}{2}\right) }d\tau \sup \limits _{0\le \tau \le t}\sqrt{\mathcal {E}_{-s}(\tau )}\nonumber \\\le & {} C_0(1+\sup \limits _{0\le \tau \le t}\sqrt{\mathcal {E}_{-s}(\tau )}. \end{aligned}$$
(2.34)

This implies (1.10) for \(\displaystyle s\in \left( \frac{1}{2},\frac{3}{2}\right) \), that is,

$$\begin{aligned} \Vert \Lambda ^{-s}(\rho -1,u,\theta -1,B)(t)\Vert _{L^2}^2\le C_0,\ \ \mathrm {for}\ \ s\in \left( \frac{1}{2},\frac{3}{2}\right) . \end{aligned}$$
(2.35)

Now that we have proved (2.35), we may repeat the arguments leading to (1.10) for \(\displaystyle s\in \left[ 0,\frac{1}{2}\right] \) to prove that they hold also for \(\displaystyle s\in \left( \frac{1}{2},\frac{3}{2}\right) \). This completes the whole proof of Theorem 1.1. \(\square \)