On the Cauchy problem of compressible full Hall-MHD equations

Abstract

This paper is concerned with an initial value problem of the compressible full Hall-MHD equations in three-dimensional whole space. Both the global existence and the optimal decay rates of solutions are obtained, when the smooth initial data are sufficiently close to the non-vacuum equilibrium in $H^1$. As a by-product of the uniform estimates, the vanishing limit of Hall coefficient is also justified.

1 Introduction

The Hall effect was firstly discovered by Edwin Hall in 1879 (cf. [19]) and is the production of a voltage difference (the Hall voltage) across an electrical conductor, transverse to an electric current in the conductor and to an applied magnetic field perpendicular to the current. It restores the influence of the electric current in the Lorentz force occurring in Ohm’s law. The motion of a conducting fluid with Hall effects in magnetic field is governed by the mathematical model of Hall magnetohydrodynamics (for short, Hall-MHD, cf. [1, 23]):

$$\begin{aligned} \left\{ \begin{aligned}&\rho _{t}+\mathop {\mathrm{div}}\nolimits (\rho u)=0,\\&(\rho u)_{t}+\mathop {\mathrm{div}}\nolimits (\rho u\otimes u)+ \nabla p= \mu \Delta u +(\mu +\lambda )\nabla \mathop {\mathrm{div}}\nolimits u+(\nabla \times b )\times b,\\&\mathrm{c}_{\mathrm{V}}\left[ (\rho \theta )_{t}+ \mathop {\mathrm{div}}\nolimits (\rho u\theta )\right] + p\mathop {\mathrm{div}}\nolimits u=\kappa \Delta \theta +\frac{\mu }{2}\left| \nabla u+(\nabla u)^{\top }\right| ^{2}+\lambda (\mathop {\mathrm{div}}\nolimits u)^{2}+\nu |\nabla \times b|^{2},\\&b_{t}-\nabla \times (u\times b )+\varepsilon \nabla \times \left( \rho ^{-1}(\nabla \times b)\times b\right) =\nu \Delta b,\\&\mathop {\mathrm{div}}\nolimits b=0, \end{aligned}\right. \end{aligned}$$

(1.1)

where the unknown functions $\rho , u\in \mathbb R^3, \theta$ and $b\in \mathbb R^3$ are the density of the fluid, velocity, temperature and magnetic field, respectively. In this paper, we focus our interest on the polytropic ideal fluids which satisfy the equations of states $p=\mathrm{R}\rho \theta$ and $e=\mathrm{c}_{\mathrm{V}}\theta$, where p and e denote the pressure and internal energy of the fluids, respectively. The constants $\mu$ and $\lambda$ are the viscosity coefficients, $\kappa >0$ is the heat conductivity, $\nu >0$ is the resistivity coefficient acting as the diffusion coefficient of the magnetic field, $\varepsilon >0$ is the Hall coefficient, while $\mathrm{R}$ and $\mathrm{c}_{\mathrm{V}}$ are the perfect gas constant and the specific heat at constant volume, respectively. For simplicity, we assume throughout this paper that the physical constants $\kappa =\nu =\mathrm{R}=\mathrm{c}_{\mathrm{V}}\equiv 1$. Moreover, the viscosity coefficients $\mu$ and $\lambda$ are assumed to satisfy the physical conditions:

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

which ensures that (1.1)$_2$ is a L$\mathrm{\acute{a}}$me system.

In this paper, we consider an initial value problem of (1.1) with the far-field behavior

$$\begin{aligned} (\rho , u, \theta , b)(x,t)\rightarrow (\rho _\infty ,0,\theta _\infty ,0)\quad \mathrm{as}\quad |x|\rightarrow \infty , \end{aligned}$$

(1.2)

and the initial conditions

$$\begin{aligned} (\rho , u,\theta ,b)|_{t=0}=(\rho _0, u_{0},\theta _{0},b_{0})(x),\quad x\in \mathbb R^3. \end{aligned}$$

(1.3)

The Hall-MHD system (1.1) can be derived from fluid mechanics with appropriate modifications to account for electrical forces and Hall effects, and is believed to be an essential feature in the problems of magnetic reconnection in space plasmas, star formulation, neutron stars and geo-dynamo (see, for example, [3, 16, 20, 23, 30, 33]). From the mathematical point of view, the Hall term is quadratic in magnetic field and involves second-order derivatives. So, the mathematical analysis of Hall-MHD is more complicated than that of the conventional MHD system, where the Hall effect is small and can be neglected (e.g., in the laminar situation). We refer to [15, 21, 26, 27, 29, 31] and among others for the studies of compressible MHD equations without Hall term. In the past few years, the incompressible Hall-MHD system has been studied by Chae and his coauthors (cf. [4,5,6,7,8,9] and among others), where the global weak/local strong solutions with large data, the global strong solutions with small data, the blow-up criteria and the singularity analysis results were obtained. Compared with the incompressible case, the compressible Hall-MHD system receives less attention from mathematicians. For the compressible isentropic case, the global existence and asymptotic behavior were proved in [13, 18, 34], when the initial data are sufficiently close to the non-vacuum equilibrium. The aim of this paper is to treat the compressible heat-conductive case and to study the large-time behavior of global classical solutions to the initial value problem (1.1)–(1.3).

Without loss of generality, it is assumed that $\rho _\infty =\theta _\infty =1$. Then, the main result of this paper can be stated as follows.

Theorem 1.1

(I. Global Existence) For any given positive number $M_0>0$ (not necessarily small), assume that the initial data $(\rho _0-1, u_0, \theta _0-1, b_0)\in H^3$ satisfy

$$\begin{aligned} \Vert \nabla ^2(\rho _0-1, u_0, \theta _0-1, b_0)\Vert _{H^1}\le M_0, \end{aligned}$$

(1.4)

then there exists a positive constant $\eta _0>0$, depending on $M_0$, such that the problem (1.1)–(1.3) has a unique global solution $(\rho , u, \theta , b)$ on $\mathbb R^3\times (0,\infty )$ satisfying $\inf _{x,t}\rho (x,t)>0$, $\inf _{x,t}\theta (x,t)>0$, and

$$\begin{aligned} \begin{aligned}&\Vert (\rho -1, u, \theta -1, b)(t)\Vert ^2_{H^3}+\int \limits _0^t(\Vert \nabla \rho (\tau )\Vert ^2_{H^2}+ \Vert \nabla (u, \theta ,b)(\tau )\Vert ^2_{H^3})d\tau \\&\quad \le C\Vert (\rho _0-1, u_0, \theta _0-1, b_0)\Vert ^2_{H^3} \end{aligned} \end{aligned}$$

(1.5)

for all $t\ge 0$, provided

$$\begin{aligned} \Vert (\rho _0-1, u_0, \theta _0-1, b_0)\Vert _{H^1}\le \eta _0, \end{aligned}$$

(1.6)

where $C>0$ is a generic positive constant independent of t.

(II. Decay Rates) Assume further that $\Lambda \triangleq \Vert (\rho _0-1, u_0, \theta _0-1, b_0)\Vert _{L^1}$ is bounded. Then, there exists a positive constant $0<\eta _1\le \eta _0$, depending on $M_0$ and $\Lambda$, such that for any $t\ge 0$

$$\begin{aligned} \Vert \nabla ^k(\rho -1, u, \theta -1, b)(t)\Vert _{L^{2}}\le C(1+t)^{-\left( \frac{3}{4}+\frac{k}{2}\right) },\quad k=0,1, \end{aligned}$$

(1.7)

and that for any $t\ge {\bar{T}}$ with ${\bar{T}}>0$ being large enough and depending on $M_0,\Lambda$,

$$\begin{aligned} \Vert \nabla ^2(\rho -1, u, \theta -1)(t)\Vert _{H^1}\le C(1+t)^{-\frac{7}{4}}, \end{aligned}$$

(1.8)

$$\begin{aligned} \Vert \nabla ^k b(t)\Vert _{L^{2}}\le C(1+t)^{-(\frac{3}{4}+\frac{k}{2})},\quad k=2,3, \end{aligned}$$

(1.9)

provided $\Vert (\rho _0-1, u_0, \theta _0-1, b_0)\Vert _{H^1}\le \eta _1.$

Remark 1.1

Compared with the decay estimates of linearized system (cf. Lemma 2.1), the decay rates stated in the second part of Theorem 1.1 are optimal, except the one of $\Vert \nabla ^3(\rho -1,u,\theta -1)\Vert _{L^2}$. Indeed, if the initial data are more regular, say $(\rho _0-1, u_0, \theta _0-1, b_0)\in H^k$ with $k\ge 4$, then one can obtain the optimal decay rates of the solutions up to the $(k-1)$th order derivatives of $(\rho -1,u,\theta -1)$ and the kth order derivatives of b. The lack of the optimal decay estimates of $\Vert \nabla ^k(\rho -1,u,\theta -1)\Vert _{L^2}$ is mainly due to the insufficient dissipation of density and the strong coupling of fluid quantities.

As a by-product, we can study the vanishing limit of Hall coefficient (i.e., $\varepsilon \rightarrow 0$).

Theorem 1.2

Let $(\rho , u, \theta , b)$ be the classical solution of Hall-MHD system (1.1) obtained in Theorem 1.1. Then, as $\varepsilon \rightarrow 0$, it holds that

$$\begin{aligned} (\rho , u, \theta , b)\rightarrow ({{\bar{\rho }}},{\bar{u}}, {\bar{\theta }}, {\bar{b}})\quad {strongly \ \ in} \quad C([0,T]; H^2) \end{aligned}$$

(1.10)

for any $0<T<\infty$, where $({\bar{\rho }},{\bar{u}}, {\bar{\theta }}, {\bar{b}})$ is the classical solution of MHD system (4.1) without Hall effects (cf. Theorem 4.1). Moreover,

$$\begin{aligned} \sup _{0\le t\le T}\Vert (\rho -{\bar{\rho }}, u-{\bar{u}}, \theta -{\bar{\theta }}, b-{\bar{b}})(t)\Vert _{H^2}^2\le C(T)\varepsilon ^2, \end{aligned}$$

(1.11)

where C(T) is a positive constant depending on T.

The proofs of Theorems 1.1 and 1.2 are similar to the ones in [34, 35], based on the standard $L^2$-method and the origin ideas developed by Matsumura-Nishida [28]. It is worth noting that though the $H^1$-perturbation of initial data is small, the higher-order derivatives could be of large oscillations. Compared with the results in [11, 35] where the authors only obtained the optimal decay estimates for the $L^p$-norm ($2\le p\le 6$) of the solution and the $L^2$-norm of its first-order derivative, the decay rates of both $\Vert \nabla ^2(\rho ,u,\theta ,b)(t)\Vert _{L^2}$ and $\Vert \nabla ^3b(t)\Vert _{L^2}$ for large $t>0$ are also optimal in the present paper. This will be achieved by making a full use of the $H^1$-decay estimates and the Sobolev interpolation inequality (see (3.74) and (3.75)). The key point here is that all the estimates are uniform in the Hall coefficient $\varepsilon$.

The rest of this paper is organized as follows. In Sect. 2, we first reformulate the problem and recall some known results about the linear system. Theorems 1.1 and 1.2 will be shown in Sects. 3 and 4, respectively, based on the global uniform-in-$\varepsilon$ estimates.

2 Reformulation

In order to estimate the solutions, we set

$$\begin{aligned} \phi \triangleq \rho -1,\quad \psi \triangleq \theta -1\quad \mathrm{with}\quad \phi _0\triangleq \rho _0-1,\quad \psi _0\triangleq \theta _0-1 . \end{aligned}$$

Then, the system (1.1)–(1.3) can be reformulated in the form:

$$\begin{aligned} \left\{ \begin{aligned}&\phi _{t}+ \mathop {\mathrm{div}}\nolimits u= S_{1},\\&u_{t}- \mu \Delta u -(\mu +\lambda )\nabla \mathop {\mathrm{div}}\nolimits u+\nabla \phi +\nabla \psi =S_{2},\\&\psi _{t}-\Delta \psi + \mathop {\mathrm{div}}\nolimits u=S_{3},\\&b_{t}-\Delta b=S_{4} ,\\&\mathop {\mathrm{div}}\nolimits b=0, \end{aligned}\right. \end{aligned}$$

(2.1)

with the initial data

$$\begin{aligned} (\phi , u, \psi , b)|_{t=0}=(\phi _0, u_{0},\psi _0,b_{0}) \end{aligned}$$

(2.2)

and the far-field boundary conditions

$$\begin{aligned} (\phi , u, \psi , b)\rightarrow (0,0,0,0),\quad \mathrm{as}\quad |x|\rightarrow \infty , \end{aligned}$$

(2.3)

where the nonlinear terms $S_{i}(i=1, 2, 3, 4)$ on the right-hand side are defined as follows:

$$\begin{aligned} \left\{ \begin{aligned}&S_{1}\triangleq -\phi \mathop {\mathrm{div}}\nolimits u-u\cdot \nabla \phi ,\\&S_{2}\triangleq -u\cdot \nabla u-h(\phi )\left[ \mu \Delta u +(\mu +\lambda )\nabla \mathop {\mathrm{div}}\nolimits u\right] +g(\phi )(\nabla \times b )\times b+\left[ h(\phi )-\psi g(\phi )\right] \nabla \phi ,\\&S_{3}\triangleq -u\cdot \nabla \psi -\psi \mathop {\mathrm{div}}\nolimits u -h(\phi )\Delta \psi +g(\phi )\left[ \frac{\mu }{2}\left| \nabla u+(\nabla u)^{\top }\right| ^{2}+\lambda (\mathop {\mathrm{div}}\nolimits u)^{2}+|\nabla \times b|^{2}\right] ,\\&S_{4}\triangleq -u\cdot \nabla b + b\cdot \nabla u-b \mathop {\mathrm{div}}\nolimits u-\varepsilon \nabla \times \left[ g(\phi )((\nabla \times b )\times b)\right] ,\\ \end{aligned}\right. \end{aligned}$$

(2.4)

and $h(\cdot ),g(\cdot )$ are the functions of $\phi$ given by

$$\begin{aligned} h(\phi )=\frac{\phi }{\phi +1} \quad \mathrm{and}\quad g(\phi )=\frac{1}{\phi +1}. \end{aligned}$$

(2.5)

It is clear that for smooth solutions, the problem (1.1)–(1.3) is equivalent to the one (2.1)–(2.3). Moreover, the left-hand side of (2.1) is indeed the linearized Navier–Stokes equations coupled with a heat equation. In fact, let

$$\begin{aligned} \begin{aligned}&A\triangleq \left( \begin{array}{ccc} 0 &{} \mathop {\mathrm{div}}\nolimits &{} 0\\ \nabla &{}-\mu \Delta -(\mu +\lambda )\nabla \mathop {\mathrm{div}}\nolimits &{} \nabla \\ 0&{} \mathop {\mathrm{div}}\nolimits &{} -\Delta \\ \end{array} \right) .\ \ \end{aligned} \end{aligned}$$

(2.6)

Then, it is easy to see that ($U\triangleq (\phi , u, \psi )$)

$$\begin{aligned} U(t)=e^{-tA}U(0)\quad \mathrm{and}\quad B(t)=e^{-t\Delta }B(0) \end{aligned}$$

solve the linearized Navier–Stokes equations $U_t+AU=0$ (cf. [11, 24, 25]) and the heat equations $B_t-\Delta B=0$ (cf. [32]), respectively. Thus, it follows from [11, 24, 25] and [35] that

Lemma 2.1

Assume that $(U_0,B_0)\in L^1\cap H^3$. Let $U\triangleq U(x,t)$ and $B(t)\triangleq B(x,t)$ be the smooth solutions of $U_t+AU=0$ and $B_t-\Delta B=0$, respectively. Then, for any $k\in \{0, 1, 2, 3\}$,

$$\begin{aligned} \Vert \nabla ^k(U,B)(t)\Vert _{L^2}\le C(1+t)^{-(\frac{3}{4}+\frac{k}{2})}\left( \Vert (U_0,B_0)\Vert _{L^1}+\Vert \nabla ^k(U_0,B_0)\Vert _{L^2}\right) , \end{aligned}$$

(2.7)

where C is a generic positive constant depending only on $\mu$ and $\lambda$.

Based on Lemma 2.1, it is easily deduced from the Duhamel’s principle that

Lemma 2.2

Assume that the pair of functions $(\phi ,u,\psi ,b)$ is the smooth solution of (2.1) with initial data $(\phi _0,u_0,\psi _0,b_0)\in L^1\cap H^3$. Then, for any $k\in \{0, 1, 2, 3\}$,

$$\begin{aligned} \begin{aligned} \Vert \nabla ^k(\phi ,u,\psi ,b)(t)\Vert _{L^2}&\le C(1+t)^{-(\frac{3}{4}+\frac{k}{2})}\Vert (\phi _0,u_0,\psi _0,b_0)\Vert _{L^1\cap H^k}\\&\quad +\,C\int \limits _0^t(1+t-\tau )^{-(\frac{3}{4}+\frac{k}{2})}\Vert (S_1,S_2,S_3,S_4)\Vert _{L^1\cap H^k}d\tau , \end{aligned} \end{aligned}$$

(2.8)

where C is a generic positive constant depending only on $\mu$ and $\lambda$.

The following product and commutator estimates [22] will be used repeatedly in the derivations of the global a priori estimates.

Lemma 2.3

Let f, g be the smooth functions in Schwartz class. Then, for any $s>0$ and $p\in (1,+\infty )$, there exists a generic positive constant $C>0$ such that

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

(2.9)

and

$$\begin{aligned} \Vert D^{s}(fg)-fD^{s}g\Vert _{L^{p}}\le C(\Vert \nabla f\Vert _{L^{p_1}}\Vert D^{s-1}g\Vert _{L^{q_1}}+\Vert D^{s}f\Vert _{L^{p_2}}\Vert g\Vert _{L^{q_2}}), \end{aligned}$$

(2.10)

where $p_1, p_2>1$ satisfy

$$\begin{aligned} \frac{1}{p}=\frac{1}{p_1}+\frac{1}{q_1}=\frac{1}{p_2}+\frac{1}{q_2}. \end{aligned}$$

(2.11)

We also need the following simple facts (see, for example, [35]), which will be used to derive the decay rates.

Lemma 2.4

Assume that $a,b,c\in R$ satisfy $0\le a\le b, b>1$ and $c>0$. Then, there exists a positive constant C, depending only on a, b, c, such that for any $t>0$,

$$\begin{aligned} \int \limits _0^t(1+t-\tau )^{-a}(1+\tau )^{-b}d\tau +\int \limits _0^t(1+\tau )^{-a}e^{-c(t-\tau )}d\tau \le C(1+t)^{-a}. \end{aligned}$$

(2.12)

Finally, we recall the local existence theorem of (1.1)–(1.3) [also cf. the problem (2.1)–(2.3)], which can be proved by using the Schauder fixed-point theorem (see, e.g., [12,13,14,15]).

Lemma 2.5

Assume that the initial data satisfy

$$\begin{aligned} (\rho _0-1, u_0, \theta _0-1, b_0) \in H^3,\quad \inf \limits _{x\in \mathbb R^3}\rho _0(x)>0,\quad \mathop {\mathrm{div}}\nolimits b_0=0. \end{aligned}$$

(2.13)

Then, there exists a small positive time $T^*>0$ such that the problem (1.1)–(1.3) has a unique classical solution $(\rho ,u,\theta ,b)$ on $\mathbb R^3\times [0,T^*]$ satisfying

$$\begin{aligned} (\rho -1, u, \theta -1, b)\in C([0,T^*]; H^3)\cap L^2(0,T^*; H^4),\quad \inf \limits _{x\in \mathbb R^3,t\in [0,T^*]}\rho (x,t)>0. \end{aligned}$$

(2.14)

3 Proof of Theorem 1.1

3.1 Global existence of classical solutions

To establish the global well-posedness of classical solutions stated in Theorem 1.1, it suffices to prove Theorem 3.1.

Theorem 3.1

Given any positive number $K>0$ (not necessarily small), assume that

$$\begin{aligned} (\phi _0, u_0, \psi _0, b_0)\in H^3,\quad \Vert \nabla ^2(\phi _0, u_0, \psi _0, b_0)\Vert _{H^1}\le K. \end{aligned}$$

(3.1)

Then, there exists a positive constant $\eta >0$, depending only on $\mu ,\lambda$ and K, such that the problem (2.1)–(2.3) has a unique global classical solution $(\phi , u, \psi , b)$ on $\mathbb R^3\times [0,\infty )$ satisfying

$$\begin{aligned} {\left\{ \begin{array}{ll} \Vert \nabla (\phi ,u,\psi ,b)(t)\Vert _{H^2}^2+\int \limits _0^t\left( \Vert \nabla ^2(u,\psi ,b)\Vert _{H^2}^2+\Vert \nabla ^2\phi \Vert _{H^1}^2\right) {\mathrm{ds}}\le C\Vert \nabla (\phi _0,u_0,\psi _0,b_0)\Vert _{H^2}^2,\\ \Vert (\phi ,u,\psi ,b)(t)\Vert _{H^1}^2+\int \limits _0^t\left( \Vert \nabla (u,\psi ,b)\Vert _{H^1}^2+\Vert \nabla \phi \Vert _{L^2}^2\right) {\mathrm{ds}}\le C\Vert (\phi _0,u_0,\psi _0,b_0)\Vert _{H^1}^2, \end{array}\right. } \end{aligned}$$

(3.2)

for any $t\ge 0$, provided

$$\begin{aligned} \Vert (\phi _0, u_0, \psi _0, b_0)\Vert _{H^1}\le \eta . \end{aligned}$$

(3.3)

In order to prove Theorem 3.1, we make the following a priori assumptions. For any given $M>1$ (not necessarily small), assume that

$$\begin{aligned} \sup _{0\le t\le T}\Vert \nabla ^2(\phi , u, \psi , b)(t)\Vert _{H^1}\le M \end{aligned}$$

(3.4)

and

$$\begin{aligned} \sup _{0\le t\le T}\Vert (\phi , u, \psi , b)(t)\Vert _{H^1}\le \delta , \end{aligned}$$

(3.5)

where $\delta >0$ is a positive constant, depending on M, and satisfies

$$\begin{aligned} 0<\delta \le \delta _0\triangleq (4C^2M)^{-1}, \quad (C ~ \text {is the Sobolev embedding constant of}~ (3.7)) . \end{aligned}$$

(3.6)

Our main purpose is to close the a priori assumptions (3.4) and (3.5) by choosing $\Vert (\phi _0,u_0,\psi _0,b_0)\Vert _{H^1}$ suitably small. To begin, we first observe that due to (3.4)–(3.6) and Sobolev embedding inequality (cf. [2]),

$$\begin{aligned} \Vert \phi (t)\Vert _{L^\infty }\le C\Vert \nabla \phi \Vert ^{\frac{1}{2}}_{L^{2}}\Vert \nabla ^2\phi \Vert ^{\frac{1}{2}}_{L^{2}}\le \frac{1}{2},\quad \forall \ t\in [0,T], \end{aligned}$$

(3.7)

which indicates that

$$\begin{aligned} \frac{1}{2}\le \inf \limits _{x,t}\phi (x,t)+1\le \sup \limits _{x,t}\phi (x,t)+1\le \frac{3}{2}, \end{aligned}$$

(3.8)

and moreover,

$$\begin{aligned} |h(\phi )|\le C|\phi |,\quad |h^{(k)}(\phi )|, \ |g^{(k-1)}(\phi )|\le C, \quad \forall \ 1\le k\in \mathbb {Z}^+. \end{aligned}$$

(3.9)

First of all, we derive the elementary $L^2$-estimates of $(\phi ,u,\psi ,b)$.

Lemma 3.1

Under the assumptions (3.4) and (3.5), one has

$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}\Vert (\phi ,u,\psi ,b)\Vert ^{2}_{L^{2}}+ \Vert \nabla (u,\psi ,b)\Vert ^{2}_{L^{2}}\le C\delta ^{\frac{1}{2}}M^{\frac{1}{2}}\Vert \nabla (\phi ,u,\psi ,b)\Vert ^2_{L^2}. \end{aligned}$$

(3.10)

Proof

Multiplying (2.1)$_{1}$, (2.1)$_{2}$, (2.1)$_{3}$, (2.1)$_{4}$ by $\phi ,u, \psi , b$ in $L^2$, respectively, and integrating by parts, we obtain after adding them together that

$$\begin{aligned} \begin{aligned}&\frac{1}{2}\frac{{\mathrm{d}}}{{\mathrm{d}}t}\Vert (\phi ,u,\psi ,b)\Vert _{L^2}^{2}+\left( \mu \Vert \nabla u\Vert _{L^2}^{2}+(\mu +\lambda )\Vert \mathop {\mathrm{div}}\nolimits u\Vert _{L^2}^{2}+\Vert \nabla \psi \Vert _{L^2}^{2}+\Vert \nabla b\Vert _{L^2}^{2}\right) \\&\quad =\langle S_{1},\phi \rangle +\langle S_{2}, u\rangle +\langle S_{3},\psi \rangle +\langle S_{4}, b\rangle , \end{aligned} \end{aligned}$$

(3.11)

where $\langle \cdot ,\cdot \rangle$ denotes the standard $L^2$-inner product.

We are now in a position of estimating each term on the right-hand side of (3.11). First, integrating by parts and using Sobolev inequality, by (3.5) we easily get

$$\begin{aligned} \langle S_{1},\phi \rangle \le C\Vert \phi \Vert _{L^{3}}\Vert \nabla u\Vert _{L^{2}}\Vert \phi \Vert _{L^{6}} \le C\Vert \phi \Vert _{H^{1}}\Vert \nabla u\Vert _{L^{2}}\Vert \nabla \phi \Vert _{L^{2}} \le C\delta (\Vert \nabla u\Vert ^{2}_{L^{2}}+\Vert \nabla \phi \Vert ^{2}_{L^{2}}). \end{aligned}$$

(3.12)

Integrating by parts and using Sobolev inequality (cf. [2]), we infer from (3.4), (3.5), (3.8) and (3.9) that

$$\begin{aligned} \begin{aligned} \langle S_{2}, u\rangle&\le C\int \left( |u|^2|\nabla u|+|\phi ||\nabla u|^2+|u||\nabla u||\nabla \phi |+|b||\nabla b||u|+|\phi ||\nabla \phi ||u|+|\psi ||\nabla \phi ||u|\right) dx\\&\le C\Vert u\Vert _{L^{3}}\Vert \nabla u\Vert _{L^{2}}\Vert u\Vert _{L^{6}}+C\Vert \nabla u\Vert _{L^{3}}\left( \Vert \phi \Vert _{L^6}\Vert \nabla u\Vert _{L^{2}}+\Vert u\Vert _{L^{6}}\Vert \nabla \phi \Vert _{L^{2}}\right) \\&\quad +\,C\Vert b\Vert _{L^3}\Vert \nabla b\Vert _{L^2}\Vert u\Vert _{L^6}+C\left( \Vert \phi \Vert _{L^{3}} +\Vert \psi \Vert _{L^{3}}\right) \Vert \nabla \phi \Vert _{L^{2}}\Vert u\Vert _{L^{6}}\\&\le C\left( \Vert (\phi ,u,\psi , b)\Vert _{H^{1}}+\Vert \nabla u\Vert _{L^2}^{1/2}\Vert \nabla ^2u\Vert _{L^2}^{1/2}\right) \left( \Vert \nabla \phi \Vert _{L^{2}}^2+\Vert \nabla u\Vert _{L^{2}}^2+\Vert \nabla b\Vert _{L^2}^2\right) \\&\le C\delta ^{\frac{1}{2}}M^{\frac{1}{2}}\left( \Vert \nabla \phi \Vert _{L^{2}}^2+\Vert \nabla u\Vert _{L^{2}}^2+\Vert \nabla b\Vert _{L^2}^2\right) , \end{aligned} \end{aligned}$$

(3.13)

since $\delta \in (0,1)$ and $M>1$. In a similar manner, we also have

$$\begin{aligned} \begin{aligned}&\langle S_{3}, \psi \rangle +\langle S_{4}, b\rangle \\&\quad \le C\int \left( |\psi ||u||\nabla \psi |+|\psi |^2|\nabla u|+|\phi ||\nabla \psi |^2+|\psi ||\nabla \psi ||\nabla \phi |+|\nabla b|^2|\psi |+|\nabla u|^2|\psi |\right) dx\\&\qquad +C\int \left( |b||u||\nabla b|+|b|^2|\nabla u|+|\nabla b|^2|b|\right) dx\\&\quad \le C\left( \Vert (\psi ,b)\Vert _{H^1}+\Vert (\nabla u,\nabla \psi ,\nabla b)\Vert _{L^3}\right) \left( \Vert \nabla \phi \Vert _{L^2}^2+\Vert \nabla u\Vert _{L^2}^2+\Vert \nabla \psi \Vert _{L^2}^2+\Vert \nabla b\Vert _{L^2}^2\right) \\&\quad \le C\delta ^{\frac{1}{2}}M^{\frac{1}{2}}\left( \Vert \nabla \phi \Vert _{L^2}^2+\Vert \nabla u\Vert _{L^2}^2+\Vert \nabla \psi \Vert _{L^2}^2+\Vert \nabla b\Vert _{L^2}^2\right) . \end{aligned} \end{aligned}$$

(3.14)

Thus, substituting (3.12)–(3.14) into (3.11) immediately leads to (3.10). $\square$

Lemma 3.2

Under the assumptions (3.4) and (3.5), one has

$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}\Vert \nabla (\phi ,u,\psi ,b)\Vert ^{2}_{L^{2}}+ \Vert \nabla ^{2}(u,\psi ,b)\Vert ^{2}_{L^{2}}\le C\delta ^{\frac{1}{2}} M^{\frac{3}{2}}\left( \Vert \nabla ^2(u,\psi ,b)\Vert _{L^2}^2+\Vert \nabla \phi \Vert _{L^2}^2\right) . \end{aligned}$$

(3.15)

Proof

Operating $\nabla$ to (2.1)$_{1}$, (2.1)$_{2}$, (2.1)$_{3}$ and (2.1)$_{4}$, multiplying them by $\nabla \phi$, $\nabla u$, $\nabla \psi$ and $\nabla b$ in $L^2$, respectively, and integrating by parts, we have

$$\begin{aligned} \begin{aligned}&\frac{1}{2}\frac{{\mathrm{d}}}{{\mathrm{d}}t}\Vert \nabla (\phi ,u,\psi ,b)\Vert ^{2}_{L^{2}} +\left( \mu \Vert \nabla ^{2} u\Vert _{L^2}^{2}+(\mu +\lambda )\Vert \nabla \mathop {\mathrm{div}}\nolimits u\Vert _{L^2}^{2}+\Vert \nabla ^{2} \psi \Vert _{L^2}^{2}+\Vert \nabla ^{2} b\Vert _{L^2}^{2}\right) \\&\quad =\langle \nabla S_{1}, \nabla \phi \rangle -\langle S_{2}, \nabla ^2 u\rangle -\langle S_{3},\nabla ^2 \psi \rangle -\langle S_{4}, \nabla ^2 b\rangle . \end{aligned} \end{aligned}$$

(3.16)

Upon integration by parts, we infer from (3.4) and (3.5) that

$$\begin{aligned} \begin{aligned} \langle \nabla S_{1}, \nabla \phi \rangle \le&C\int \left( |\nabla u||\nabla \phi |^2+|\phi ||\nabla ^2u||\nabla \phi |\right) dx\\ \le&C\Vert \nabla \phi \Vert _{L^3}\left( \Vert \nabla \phi \Vert _{L^2}\Vert \nabla u\Vert _{L^6}+\Vert \phi \Vert _{L^6}\Vert \nabla ^2 u\Vert _{L^2}\right) \\ \le&C\Vert \nabla \phi \Vert _{L^2}^{1/2}\Vert \nabla ^2\phi \Vert _{L^2}^{1/2}\left( \Vert \nabla \phi \Vert _{L^2}^2+\Vert \nabla ^2 u\Vert _{L^2}^2\right) \\ \le&C\delta ^{\frac{1}{2}}M^{\frac{1}{2}}\Vert \left( \Vert \nabla \phi \Vert _{L^{2}}^2+\Vert \nabla ^2 u\Vert _{L^{2}}^2\right) . \end{aligned} \end{aligned}$$

(3.17)

By direct computations, we easily get that

$$\begin{aligned} \begin{aligned} \Vert (S_2,S_3,S_4)\Vert _{L^2}&\le C\left( \Vert (u,\psi ,b)\Vert _{L^3}\Vert \nabla (u,\psi ,b)\Vert _{L^6}+\Vert (\phi ,b)\Vert _{L^\infty }\Vert \nabla ^2(u,\psi ,b)\Vert _{L^2}+\Vert \nabla (u,b)\Vert _{L^4}^2\right) \\&\quad +C\left( \Vert (\phi ,\psi )\Vert _{L^\infty }\Vert \nabla \phi \Vert _{L^2}+\Vert b\Vert _{L^\infty }\Vert \nabla b\Vert _{L^\infty }\Vert \nabla \phi \Vert _{L^2}\right) \\&\le C\left( \Vert (u,\psi ,b)\Vert _{H^1}+\Vert \nabla (\phi ,u,b)\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla ^2(\phi ,u,b)\Vert _{L^2}^{\frac{1}{2}}\right) \Vert \nabla ^2(u,\psi ,b)\Vert _{L^2}\\&\quad +C\left( \Vert \nabla (\phi ,\psi )\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla ^2(\phi ,\psi )\Vert _{L^2}^{\frac{1}{2}}+\Vert \nabla b\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla ^2 b\Vert _{L^2}\Vert \nabla ^3b\Vert _{L^2}^{\frac{1}{2}}\right) \Vert \nabla \phi \Vert _{L^2}\\&\le C\delta ^{\frac{1}{2}} M^{\frac{3}{2}}\left( \Vert \nabla ^2(u,\psi ,b)\Vert _{L^2}+\Vert \nabla \phi \Vert _{L^2}\right) , \end{aligned} \end{aligned}$$

(3.18)

and consequently,

$$\begin{aligned} \left| \langle S_{2}, \nabla ^2 u\rangle \right| +\left| \langle S_{3},\nabla ^2 \psi \rangle \right| +\left| \langle S_{4}, \nabla ^2 b\rangle \right| \le C\delta ^{\frac{1}{2}} M^{\frac{3}{2}}\left( \Vert \nabla ^2(u,\psi ,b)\Vert _{L^2}^2+\Vert \nabla \phi \Vert _{L^2}^2\right) . \end{aligned}$$

(3.19)

Now, putting (3.17) and (3.19) into (3.16), we arrive at the desired result of (3.15). $\square$

Remark 3.1

In fact, it can be also deduced from (3.17) that

$$\begin{aligned} \begin{aligned} \langle \nabla S_{1}, \nabla \phi \rangle \le&C\int \left( |\nabla u||\nabla \phi |^2+|\phi ||\nabla ^2u||\nabla \phi |\right) dx\\&\le C\left( \Vert \nabla u\Vert _{L^3}\Vert \nabla \phi \Vert _{L^3}^2+\Vert \phi \Vert _{L^3}\Vert \nabla ^2 u\Vert _{L^2}\Vert \nabla \phi \Vert _{L^6}\right) \\&\le C\left( \Vert \nabla u\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla ^2u\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla \phi \Vert _{L^2}\Vert \nabla ^2\phi \Vert _{L^2}+\Vert \phi \Vert _{H^1}\Vert \nabla ^2 u\Vert _{L^2}\Vert \nabla ^2\phi \Vert _{L^2}\right) \\&\le C\left( \Vert \nabla u\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla ^2u\Vert _{L^2}^{\frac{1}{2}}\Vert \phi \Vert _{L^2}^{\frac{1}{2}}\Vert \nabla ^2\phi \Vert _{L^2}^{\frac{3}{2}}+\Vert \phi \Vert _{H^1}\Vert \nabla ^2 u\Vert _{L^2}\Vert \nabla ^2\phi \Vert _{L^2}\right) \\&\le C\delta \left( \Vert \nabla ^2\phi \Vert _{L^2}^{2}+\Vert \nabla ^2 u\Vert _{L^2}^2\right) , \end{aligned} \end{aligned}$$

(3.20)

where we have used the Sobolev interpolation inequality $\Vert \nabla \phi \Vert _{L^2}^2\le C\Vert \phi \Vert _{L^2}\Vert \nabla ^2\phi \Vert _{L^2}$. Similarly,

$$\begin{aligned} \Vert (S_2,S_3,S_4)\Vert _{L^2}\le C\delta ^{\frac{1}{2}} M^{\frac{3}{2}}\left( \Vert \nabla ^2(u,\psi ,b)\Vert _{L^2}+\Vert \nabla ^2\phi \Vert _{L^2}\right) . \end{aligned}$$

This, combined with (3.16) and (3.20), shows

$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}\Vert \nabla (\phi ,u,\psi ,b)\Vert ^{2}_{L^{2}}+ \Vert \nabla ^{2}(u,\psi ,b)\Vert ^{2}_{L^{2}}\le C\delta ^{\frac{1}{2}} M^{\frac{3}{2}} \Vert \nabla ^2(\phi ,u,\psi ,b)\Vert _{L^2}^2, \end{aligned}$$

(3.21)

which will be used to close the estimate of $\Vert \nabla (\phi ,u,\psi ,b)\Vert ^{2}_{L^{2}}$.

Clearly, we need to estimate $\Vert \nabla \phi \Vert _{L^2(0,T;L^2)}$.

Lemma 3.3

Under the assumptions (3.4) and (3.5), one has

$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}\langle u,\nabla \phi \rangle + \Vert \nabla \phi \Vert ^{2}_{L^{2}} \le C\Vert \nabla (u,\psi )\Vert ^{2}_{H^{1}}+C\delta ^{\frac{1}{2}} M^{\frac{3}{2}}\left( \Vert \nabla \phi \Vert _{L^2}^2+\Vert \nabla (u,\psi ,b)\Vert ^{2}_{H^{1}}\right) . \end{aligned}$$

(3.22)

Proof

Multiplying (2.1)$_{2}$ by $\nabla \phi$ in $L^2$, we have

$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}\langle u,\nabla \phi \rangle + \Vert \nabla \phi \Vert _{L^2}^{2}=\langle u,\nabla \phi _t\rangle +\langle \mu \Delta u +(\mu +\lambda )\nabla \mathop {\mathrm{div}}\nolimits u- \nabla \psi ,\nabla \phi \rangle +\langle S_{2},\nabla \phi \rangle . \end{aligned}$$

(3.23)

Upon integration by parts, one infers from (2.1) that

$$\begin{aligned} \begin{aligned} \langle u,\nabla \phi _t\rangle =\,&-\langle u,\nabla (\mathop {\mathrm{div}}\nolimits u+\mathop {\mathrm{div}}\nolimits (\phi u))\rangle =\langle \mathop {\mathrm{div}}\nolimits u,\mathop {\mathrm{div}}\nolimits u+\mathop {\mathrm{div}}\nolimits (\phi u)\rangle \\&\le C\Vert \nabla u\Vert _{L^2}^2+C\Vert \nabla u\Vert _{L^3}\left( \Vert \nabla \phi \Vert _{L^2}\Vert u\Vert _{L^6}+\Vert \phi \Vert _{L^6}\Vert \nabla u\Vert _{L^2}\right) \\&\le C\Vert \nabla u\Vert _{L^2}^2+C\Vert \nabla u\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla ^2 u\Vert _{L^2}^{\frac{1}{2}}\left( \Vert \nabla \phi \Vert _{L^2}^2+\Vert \nabla u\Vert _{L^2}^2\right) \\&\le C\Vert \nabla u\Vert _{L^2}^2+C\delta ^{\frac{1}{2}}M^{\frac{1}{2}}\left( \Vert \nabla \phi \Vert _{L^2}^2+\Vert \nabla u\Vert _{L^2}^2\right) . \end{aligned} \end{aligned}$$

(3.24)

It is easy to see that

$$\begin{aligned} \langle \mu \Delta u +(\mu +\lambda )\nabla \mathop {\mathrm{div}}\nolimits u- \nabla \psi ,\nabla \phi \rangle \le \frac{1}{4}\Vert \nabla \phi \Vert _{L^2}^2+C\left( \Vert \nabla ^2u\Vert _{L^2}^2+\Vert \nabla \psi \Vert _{L^2}^2\right) , \end{aligned}$$

(3.25)

and moreover, it follows from (3.18) that

$$\begin{aligned} \langle S_{2},\nabla \phi \rangle \le C \Vert S_{2}\Vert _{L^2}\Vert \nabla \phi \Vert _{L^2}\le C\delta ^{\frac{1}{2}} M^{\frac{3}{2}}\left( \Vert \nabla ^2(u,\psi ,b)\Vert _{L^2}^2+\Vert \nabla \phi \Vert _{L^2}^2\right) \end{aligned}$$

(3.26)

Hence, substituting (3.24)–(3.26) into (3.23), we obtain (3.22).

We proceed to prove the $L^2$-estimates of the second-order derivatives.

Lemma 3.4

Under the assumptions (3.4) and (3.5), one has

$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}\Vert \nabla ^{2}(\phi ,u,\psi ,b)\Vert ^{2}_{L^{2}}+ \Vert \nabla ^{3}(u,\psi ,b)\Vert ^{2}_{L^{2}}\le C\delta ^{\frac{1}{4}}M^{\frac{3}{2}}\left( \Vert \nabla ^3(u,\psi ,b)\Vert _{L^2}^2+\Vert \nabla ^2\phi \Vert _{L^2}^2\right) . \end{aligned}$$

(3.27)

Proof

Similarly to the proof of Lemma 3.2, we deduce from (2.1) that

$$\begin{aligned} \begin{aligned}&\frac{1}{2}\frac{{\mathrm{d}}}{{\mathrm{d}}t} \Vert \nabla ^{2}(\phi ,u,\psi ,b)\Vert _{L^2}^{2}+\left( \mu \Vert \nabla ^{3} u\Vert _{L^2}^{2}+(\mu +\lambda )\Vert \nabla ^{2}\mathop {\mathrm{div}}\nolimits u\Vert _{L^2}^{2}+\Vert \nabla ^{3} \psi \Vert _{L^2}^{2}+\Vert \nabla ^{3} b\Vert _{L^2}^{2}\right) \\&\quad =\langle \nabla ^{2}S_{1},\nabla ^{2}\phi \rangle +\langle \nabla ^{2}S_{2},\nabla ^{2}u \rangle +\langle \nabla ^{2}S_{3},\nabla ^{2}\psi \rangle +\langle \nabla ^{2}S_{4},\nabla ^{2}b\rangle . \end{aligned} \end{aligned}$$

(3.28)

Using Lemma 2.3 and Gagliardo–Nirenberg inequality (cf. [2]), we find (keeping in mind that $M>1$)

$$\begin{aligned} \begin{aligned} \langle \nabla ^{2}S_{1},\nabla ^{2}\phi \rangle&=-\langle \nabla ^2(\phi \mathop {\mathrm{div}}\nolimits u),\nabla ^2\phi \rangle -\langle \nabla ^2(u\cdot \nabla \phi )-u\cdot \nabla ^2(\nabla \phi ),\nabla ^2\phi \rangle +\frac{1}{2}\langle \mathop {\mathrm{div}}\nolimits u,|\nabla ^2\phi |^2 \rangle \\&\le C\left( \Vert \nabla u\Vert _{L^{\infty }}\Vert \nabla ^{2}\phi \Vert _{L^{2}}+\Vert \phi \Vert _{L^{\infty }}\Vert \nabla ^{3}u\Vert _{L^{2}}+\Vert \nabla ^2 u\Vert _{L^{6}}\Vert \nabla \phi \Vert _{L^{3}}\right) \Vert \nabla ^2 \phi \Vert _{L^{2}}\\&\le C\left( \Vert \nabla u\Vert _{L^{2}}^{\frac{1}{4}}\Vert \nabla ^3 u\Vert _{L^{2}}^{\frac{3}{4}}\Vert \nabla ^2\phi \Vert _{L^{2}}+\Vert \nabla \phi \Vert _{L^{2}}^{\frac{1}{2}}\Vert \nabla ^2 \phi \Vert _{L^{2}}^{\frac{1}{2}}\Vert \nabla ^{3} u\Vert _{L^{2}}\right) \Vert \nabla ^2 \phi \Vert _{L^{2}}\\&\le C\delta ^{\frac{1}{4}}M^{\frac{3}{4}}\Vert \left( \Vert \nabla ^2 \phi \Vert _{L^{2}}^2+\Vert \nabla ^3 u\Vert _{L^{2}}^2\right) . \end{aligned} \end{aligned}$$

(3.29)

Next, we estimate some terms involving in $\nabla S_i$ for $i=2,3,4$. Indeed, by making use of the Sobolev interpolation inequality (cf. [10, 17])

$$\begin{aligned} \Vert D^j f\Vert _{L^2}\le C\Vert f\Vert _{L^2}^{\frac{m-j}{m}}\Vert D^m f\Vert _{L^2}^{\frac{j}{m}},\quad \forall \ 0\le j\le m, \end{aligned}$$

we get that

$$\begin{aligned} \begin{aligned} \Vert \nabla (u\cdot \nabla u)\Vert _{L^2}\le&\Vert u\Vert _{L^3}\Vert \nabla ^2u\Vert _{L^6}+\Vert \nabla u\Vert _{L^3}\Vert \nabla u\Vert _{L^6}\\&\le C\left( \Vert u\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla u\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla ^3u\Vert _{L^2}+\Vert \nabla u\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla ^2 u\Vert _{L^2}^{\frac{3}{2}}\right) \\&\le C\left( \Vert u\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla u\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla ^3u\Vert _{L^2}+\Vert \nabla u\Vert _{L^2}^{\frac{1}{2}}\Vert u\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla ^3u\Vert _{L^2} \right) \\&\le C \delta \Vert \nabla ^3u\Vert _{L^2}. \end{aligned} \end{aligned}$$

(3.30)

Similarly to the derivation of (3.30), we also have

$$\begin{aligned} \begin{aligned}&\Vert \nabla (u\cdot \nabla \psi )\Vert _{L^2}+\Vert \nabla (u\cdot \nabla b)\Vert _{L^2}+\Vert \nabla (b\cdot \nabla u)\Vert _{L^2}+\Vert \nabla (\psi \mathop {\mathrm{div}}\nolimits u)\Vert _{L^2}+\Vert \nabla (b\mathop {\mathrm{div}}\nolimits u)\Vert _{L^2}\\&\quad \le C\Vert (u,\psi ,b)\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla (u,\psi ,b)\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla ^3(u,\psi ,b)\Vert _{L^2}\le C\delta \Vert \nabla ^3(u,\psi ,b)\Vert _{L^2}. \end{aligned} \end{aligned}$$

(3.31)

It is easy to deduce from (3.7)–(3.9) and the interpolation inequality that

$$\begin{aligned}&\Vert \nabla (h(\phi )\nabla \phi )\Vert _{L^2}+\Vert \nabla (\psi g(\phi )\nabla \phi )\Vert _{L^2}+\Vert \nabla (g(\phi )|\nabla (u,b)|^2)\Vert _{L^2}\nonumber \\&\quad \le C\left( \Vert (\phi ,\psi )\Vert _{L^\infty }\Vert \nabla ^2\phi \Vert _{L^2}+\Vert \nabla (\phi ,\psi )\Vert _{L^3}\Vert \nabla \phi \Vert _{L^6} +\Vert \psi \Vert _{L^\infty }\Vert \nabla \phi \Vert _{L^3}\Vert \nabla \phi \Vert _{L^6}\right) \nonumber \\&\qquad +C\left( \Vert \nabla \phi \Vert _{L^6}\Vert \nabla (u,b)\Vert _{L^6}^2+\Vert \nabla (u,b)\Vert _{L^3}\Vert \nabla ^2(u,b)\Vert _{L^6}\right) \nonumber \\&\quad \le C\left( \Vert \nabla (\phi ,\psi )\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla ^2(\phi ,\psi )\Vert _{L^2}^{\frac{1}{2}}+\Vert \nabla \psi \Vert _{L^2}^{\frac{1}{2}}\Vert \nabla ^2\psi \Vert _{L^2}^{\frac{1}{2}} \Vert \nabla \phi \Vert _{L^2}^{\frac{1}{2}}\Vert \nabla ^2\phi \Vert _{L^2}^{\frac{1}{2}}\right) \Vert \nabla ^2\phi \Vert _{L^2}\nonumber \\&\qquad +C\left( \Vert \nabla ^2\phi \Vert _{L^2}\Vert \nabla (u,b)\Vert _{L^2}+ \Vert \nabla (u,b)\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla ^2(u,b)\Vert _{L^2}^{\frac{1}{2}}\right) \Vert \nabla ^3 (u,b)\Vert _{L^2}\nonumber \\&\quad \le C\delta ^{\frac{1}{2}}M \left( \Vert \nabla ^2\phi \Vert _{L^2}+\Vert \nabla ^3 (u,b)\Vert _{L^2}\right) ,\end{aligned}$$

(3.32)

$$\begin{aligned}&\Vert \nabla ( h(\phi )(\Delta u,\nabla \mathop {\mathrm{div}}\nolimits u,\Delta \psi ))\Vert _{L^2}\le C\left( \Vert \phi \Vert _{L^\infty }\Vert \nabla ^3( u,\psi )\Vert _{L^2}+\Vert \nabla \phi \Vert _{L^3}\Vert \nabla ^2(u,\psi )\Vert _{L^6}\right) \nonumber \\&\quad \le C\Vert \nabla \phi \Vert _{L^2}^{\frac{1}{2}}\Vert \nabla ^2\phi \Vert _{L^2}^{\frac{1}{2}}\Vert \nabla ^3( u,\psi )\Vert _{L^2}\le C\delta ^{\frac{1}{2}}M^{\frac{1}{2}}\Vert \nabla ^3(u,\psi )\Vert _{L^2},\end{aligned}$$

(3.33)

$$\begin{aligned}&\Vert \nabla (g(\phi )((\nabla \times b)\times b))\Vert _{L^2}\le C\left( \Vert \nabla \phi \Vert _{L^6}\Vert b\Vert _{L^6}\Vert \nabla b\Vert _{L^6}+\Vert b\Vert _{L^3}\Vert \nabla ^2b\Vert _{L^6}+\Vert \nabla b\Vert _{L^3}\Vert \nabla b\Vert _{L^6}\right) \nonumber \\&\quad \le C\left( \Vert \nabla ^2\phi \Vert _{L^2}\Vert \nabla b\Vert _{L^2}\Vert \nabla ^2 b\Vert _{L^2}+\Vert b\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla b\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla ^3 b\Vert _{L^2}\right) \nonumber \\&\quad \le C\delta ^{\frac{1}{2}}M\left( \Vert \nabla ^2\phi \Vert _{L^2}+\Vert \nabla ^3 b\Vert _{L^2}\right) \end{aligned}$$

(3.34)

and

$$\begin{aligned} \begin{aligned} \Vert \nabla ^2(g(\phi )((\nabla \times b)\times b))\Vert _{L^2}\le&C\left( \Vert \nabla ^2\phi \Vert _{L^2}\Vert b\Vert _{L^\infty }+\Vert \nabla \phi \Vert _{L^6}\Vert \nabla \phi \Vert _{L^\infty }\Vert b\Vert _{L^3}\right) \Vert \nabla b\Vert _{L^\infty }\\&+C\left( \Vert b\Vert _{L^\infty }\Vert \nabla ^3b\Vert _{L^2}+\Vert \nabla b\Vert _{L^3}\Vert \nabla ^2 b\Vert _{L^6}\right) \\&+C\left( \Vert \nabla \phi \Vert _{L^6}\Vert b\Vert _{L^\infty }\Vert \nabla ^2b\Vert _{L^3}+\Vert \nabla \phi \Vert _{L^6}\Vert \nabla b\Vert _{L^6}^2\right) \\ \le&\, C\delta ^{\frac{1}{2}}M^{\frac{3}{2}}\left( \Vert \nabla ^2\phi \Vert _{L^2}+\Vert \nabla ^3 b\Vert _{L^2}\right) . \end{aligned} \end{aligned}$$

(3.35)

Collecting (3.30)–(3.35) together, we arrive at

$$\begin{aligned} \Vert \nabla (S_2,S_3,S_4)\Vert _{L^2}\le C\delta ^{\frac{1}{2}}M^{\frac{3}{2}}\left( \Vert \nabla ^3(u,\psi ,b)\Vert _{L^2}+\Vert \nabla ^2\phi \Vert _{L^2}\right) , \end{aligned}$$

(3.36)

from which and (3.28), (3.29), we obtain (3.27) after integrating by parts and using Cauchy–Schwarz inequality. $\square$

The next lemma is concerned with the estimate of $\Vert \nabla ^2\phi \Vert _{L^2(0,T;L^2)}$, which can be achieved in a similar manner as that used in the proof of Lemma 3.3.

Lemma 3.5

Under the assumptions (3.4) and (3.5), one has

$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}\langle \mathop {\mathrm{div}}\nolimits u,\Delta \phi \rangle + \Vert \nabla ^2\phi \Vert ^{2}_{L^{2}} \le C\Vert \nabla ^2 (u,\psi )\Vert ^{2}_{H^{1}}+C\delta ^{\frac{1}{2}}M^{\frac{3}{2}}\left( \Vert \nabla ^2(u,\psi ,b)\Vert _{H^1}^2+\Vert \nabla ^2 \phi \Vert ^{2}_{L^{2}}\right) . \end{aligned}$$

(3.37)

Proof

Operating $\mathop {\mathrm{div}}\nolimits$ to (2.1)$_{2}$ and multiplying it by $\Delta \phi$ in $L^2$, we obtain after integrating by parts that

$$\begin{aligned} \begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\langle \mathop {\mathrm{div}}\nolimits u,\Delta \phi \rangle + \Vert \nabla ^2 \phi \Vert _{L^2}^{2} =\,&-\langle \nabla \mathop {\mathrm{div}}\nolimits u,\nabla \phi _t\rangle -\langle \Delta \psi ,\Delta \phi \rangle +\langle \mathop {\mathrm{div}}\nolimits S_{2},\Delta \phi \rangle \\&+\langle \mathop {\mathrm{div}}\nolimits [\mu \Delta u +(\mu +\lambda )\nabla \mathop {\mathrm{div}}\nolimits u],\Delta \phi \rangle . \end{aligned} \end{aligned}$$

(3.38)

In view of (2.1)$_1$, we have

$$\begin{aligned} \begin{aligned} \left| \langle \nabla \mathop {\mathrm{div}}\nolimits u,\nabla \phi _t\rangle \right| \le&\, C\left( \Vert \nabla ^2 u\Vert _{L^2}+\Vert (\phi ,u)\Vert _{L^\infty }\Vert \nabla ^2(\phi ,u)\Vert _{L^2}+\Vert \nabla \phi \Vert _{L^3}\Vert \nabla u\Vert _{L^6}\right) \Vert \nabla ^2u\Vert _{L^2}\\ \le&\,C\Vert \nabla ^2u\Vert _{L^2}^2+C\Vert \nabla (\phi ,u)\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla ^2(\phi ,u)\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla ^2(\phi ,u)\Vert _{L^2}^2\\ \le&\,C\Vert \nabla ^2u\Vert _{L^2}^2+C\delta ^{\frac{1}{2}}M^{\frac{1}{2}}\Vert \nabla ^2(\phi ,u)\Vert _{L^2}^2, \end{aligned} \end{aligned}$$

(3.39)

and by virtue of (3.36) we see that

$$\begin{aligned} \begin{aligned}&\left| \langle \Delta \psi ,\Delta \phi \rangle \right| +\left| \langle \mathop {\mathrm{div}}\nolimits S_{2},\Delta \phi \rangle \right| +\left| \langle \mathop {\mathrm{div}}\nolimits [\mu \Delta u +(\mu +\lambda )\nabla \mathop {\mathrm{div}}\nolimits u],\Delta \phi \rangle \right| \\&\quad \le \frac{1}{2}\Vert \nabla ^2\phi \Vert _{L^2}^2+C\left( \Vert \nabla ^3u\Vert ^2_{L^2}+\Vert \nabla ^2\psi \Vert _{L^2}^2\right) +C\delta ^{\frac{1}{2}}M^{\frac{3}{2}}\left( \Vert \nabla ^3(u,\psi ,b)\Vert _{L^2}^2+\Vert \nabla ^2\phi \Vert _{L^2}^2\right) \end{aligned} \end{aligned}$$

which, combined with (3.38) and (3.39), yields (3.37). $\square$

Next, it remains to deal with $H^3$-estimates of the solutions.

Lemma 3.6

Under the assumptions (3.4) and (3.5), one has

$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}\Vert \nabla ^{3}(\phi ,u,\psi ,b)\Vert ^{2}_{L^{2}}+ \Vert \nabla ^{4}(u,\psi ,b)\Vert ^{2}_{L^{2}}\le C\delta ^{\frac{1}{4}}M^{\frac{5}{2}}\left( \Vert \nabla ^3\phi \Vert _{L^2}^2+\Vert \nabla ^4 (u,\psi ,b)\Vert _{L^2}^2\right) . \end{aligned}$$

(3.40)

Proof

It follows from (2.1) that

$$\begin{aligned} \begin{aligned}&\frac{1}{2}\frac{{\mathrm{d}}}{{\mathrm{d}}t}\Vert \nabla ^{3}(\phi ,u,\psi ,b)\Vert _{L^2}^{2}+\left( \mu \Vert \nabla ^{4} u\Vert _{L^2}^{2}+(\mu +\lambda )\Vert \nabla ^{3}\mathop {\mathrm{div}}\nolimits u\Vert _{L^2}^{2}+\Vert \nabla ^{4} \psi \Vert _{L^2}^{2}+\Vert \nabla ^{4} b\Vert _{L^2}^{2}\right) \\&\quad =\langle \nabla ^{3}S_{1},\nabla ^{3}\phi \rangle +\langle \nabla ^{3}S_{2},\nabla ^{3}u\rangle +\langle \nabla ^{3}S_{3},\nabla ^{3}\psi \rangle +\langle \nabla ^{3}S_{4},\nabla ^{3}b \rangle . \end{aligned} \end{aligned}$$

(3.41)

Based on Lemma 2.3 and Sobolev inequality, it is easily seen that

$$\begin{aligned} \begin{aligned} \langle \nabla ^{3}S_{1},\nabla ^{3}\phi \rangle =\,&-\langle \nabla ^3(\phi \mathop {\mathrm{div}}\nolimits u),\nabla ^3\phi \rangle -\langle \nabla ^3(u\cdot \nabla \phi )-u\cdot \nabla ^3(\nabla \phi ),\nabla ^3\phi \rangle -\frac{1}{2}\langle \mathop {\mathrm{div}}\nolimits u,|\nabla ^3\phi |^2 \rangle \\ \le&\, C\left( \Vert \nabla u\Vert _{L^{\infty }}\Vert \nabla ^{3}\phi \Vert _{L^{2}}+\Vert \phi \Vert _{L^{\infty }}\Vert \nabla ^{4}u\Vert _{L^{2}}+\Vert \nabla ^3 u\Vert _{L^{6}}\Vert \nabla \phi \Vert _{L^{3}}\right) \Vert \nabla ^3 \phi \Vert _{L^{2}}\\ \le&C\left( \Vert \nabla u\Vert _{L^{2}}^{\frac{1}{4}}\Vert \nabla ^3 u\Vert _{L^{2}}^{\frac{3}{4}}\Vert \nabla ^3\phi \Vert _{L^{2}}+\Vert \nabla \phi \Vert _{L^{2}}^{\frac{1}{2}}\Vert \nabla ^2 \phi \Vert _{L^{2}}^{\frac{1}{2}}\Vert \nabla ^{4} u\Vert _{L^{2}}\right) \Vert \nabla ^3 \phi \Vert _{L^{2}}\\ \le&C\delta ^{\frac{1}{4}}M^{\frac{3}{4}}\left( \Vert \nabla ^3 \phi \Vert _{L^{2}}^2+\Vert \nabla ^4 u\Vert _{L^{2}}^2\right) . \end{aligned} \end{aligned}$$

(3.42)

In the following, similarly to the proofs of Lemma 3.4, we estimate the terms involved in $\Vert \nabla ^2S_i\Vert _{L^2}$ for $i=2,3,4$. First, by virtue of the interpolation inequality we have

$$\begin{aligned} \begin{aligned} \Vert \nabla ^2(u\cdot \nabla u)\Vert _{L^2}\le&\, C\left( \Vert \nabla u\Vert _{L^3}\Vert \nabla ^2u\Vert _{L^6}+\Vert u\Vert _{L^3}\Vert \nabla ^3 u\Vert _{L^6}\right) \\ \le&\,C\left( \Vert \nabla u\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla ^2 u\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla ^3u\Vert _{L^2}+\Vert u\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla u\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla ^4u\Vert _{L^2}\right) \\ \le&\, C\Vert u\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla u\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla ^4u\Vert _{L^2} \le C \delta \Vert \nabla ^4u\Vert _{L^2} \end{aligned} \end{aligned}$$

(3.43)

and similarly,

$$\begin{aligned} \begin{aligned}&\Vert \nabla ^2(u\cdot \nabla \psi )\Vert _{L^2}+\Vert \nabla ^2(u\cdot \nabla b)\Vert _{L^2}+\Vert \nabla ^2(b\cdot \nabla u)\Vert _{L^2}+\Vert \nabla ^2(\psi \mathop {\mathrm{div}}\nolimits u)\Vert _{L^2}+\Vert \nabla ^2(b\mathop {\mathrm{div}}\nolimits u)\Vert _{L^2}\\&\quad \le C\Vert (u,\psi ,b)\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla (u,\psi ,b)\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla ^4(u,\psi ,b)\Vert _{L^2}\le C\delta \Vert \nabla ^4(u,\psi ,b)\Vert _{L^2}. \end{aligned} \end{aligned}$$

(3.44)

By (3.8) and (3.9), we obtain

$$\begin{aligned} \begin{aligned} \Vert \nabla ^2(h(\phi )\nabla \phi )\Vert _{L^2}\le&\, C\left( \Vert \phi \Vert _{L^\infty }\Vert \nabla ^3\phi \Vert _{L^2}+\Vert \nabla ^2\phi \Vert _{L^6}\Vert \nabla \phi \Vert _{L^3}+\Vert \nabla ^2\phi \Vert _{L^2}^3\right) \\ \le&\, C\left( \Vert \nabla \phi \Vert _{L^2}^{\frac{1}{2}}\Vert \nabla ^2\phi \Vert _{L^2}^{\frac{1}{2}}+\Vert \phi \Vert _{L^2}\Vert \nabla ^3\phi \Vert _{L^2}\right) \Vert \nabla ^3\phi \Vert _{L^2}\\ \le&\, C\delta ^{\frac{1}{2}} M\Vert \nabla ^3\phi \Vert _{L^2}. \end{aligned} \end{aligned}$$

(3.45)

In a similar manner,

$$\begin{aligned}&\Vert \nabla ^2(\psi g(\phi )\nabla \phi )\Vert _{L^2}\le C\left( \Vert \psi \Vert _{L^\infty }\Vert \nabla ^2(g(\phi )\nabla \phi )\Vert _{L^2} +\Vert \nabla ^2\psi \Vert _{L^3}\Vert \nabla \phi \Vert _{L^6}\right) \nonumber \\&\quad \le C \delta ^{\frac{1}{2}} M^{\frac{1}{2}}\Vert \nabla ^3\phi \Vert _{L^2}\left( 1+\delta ^{\frac{1}{2}} M\right) +C\Vert \nabla ^2\phi \Vert _{L^2}\Vert \nabla ^2\psi \Vert _{L^2}^{\frac{1}{2}}\Vert \nabla ^3\psi \Vert _{L^2}^{\frac{1}{2}}\nonumber \\&\quad \le C \delta ^{\frac{1}{2}} M^{\frac{3}{2}}\Vert \nabla ^3\phi \Vert _{L^2}+C\Vert \nabla \phi \Vert _{L^2}^{\frac{1}{2}}\Vert \nabla ^3\phi \Vert _{L^2}^{\frac{1}{2}}\Vert \psi \Vert _{L^2}^{\frac{1}{4}}\Vert \nabla ^2\psi \Vert _{L^2}^{\frac{1}{4}}\Vert \nabla ^4\psi \Vert _{L^2}^{\frac{1}{2}}\nonumber \\&\quad \le C \delta ^{\frac{1}{2}} M^{\frac{3}{2}}\Vert \nabla ^3\phi \Vert _{L^2}+C\delta ^{\frac{3}{4}} M^{\frac{1}{4}} \left( \Vert \nabla ^3\phi \Vert _{L^2}+\Vert \nabla ^4\psi \Vert _{L^2}\right) \nonumber \\&\quad \le C \delta ^{\frac{1}{2}} M^{\frac{3}{2}}\left( \Vert \nabla ^3\phi \Vert _{L^2}+\Vert \nabla ^4\psi \Vert _{L^2}\right) ,\end{aligned}$$

(3.46)

$$\begin{aligned}&\Vert \nabla ^2(g(\phi )|\nabla (u, b)|^2)\Vert _{L^2} \le C\left( \Vert \nabla ^2 \phi \Vert _{L^6} +\Vert \nabla \phi \Vert _{L^\infty }\Vert \nabla \phi \Vert _{L^6}\right) \Vert \nabla (u, b)|\Vert ^2_{L^6}\nonumber \\&\qquad +C\Vert \nabla ^2(u, b)\Vert _{L^3}\Vert \nabla ^2(u, b)\Vert _{L^6}\nonumber \\&\qquad + C\left( \Vert \nabla \phi \Vert _{L^\infty }\Vert \nabla ^2(u, b)\Vert _{L^6}\Vert \nabla (u, b)\Vert _{L^3} +\Vert \nabla (u, b)\Vert _{L^3}\Vert \nabla ^3(u, b)\Vert _{L^6}\right) \nonumber \\&\quad \le C\left( \Vert \nabla ^3\phi \Vert _{L^2}+\Vert \phi \Vert ^{\frac{1}{2}}_{L^2}\Vert \nabla ^3\phi \Vert ^{\frac{3}{2}}_{L^2}\right) \Vert (u,b)\Vert _{L^2}\Vert \nabla ^4(u,b)\Vert _{L^2}\nonumber \\&\qquad +C\left( \Vert \nabla ^2 \phi \Vert _{L^2}\Vert \nabla ^3\phi \Vert _{L^2}\Vert (u,b)\Vert _{H^1}+\Vert \nabla (u,b)\Vert ^{\frac{1}{2}}_{L^2}\Vert \nabla ^2 (u,b)\Vert ^{\frac{1}{2}}_{L^2}\right) \Vert \nabla ^4(u,b)\Vert _{L^2}\nonumber \\&\quad \le C\delta ^{\frac{1}{2}} M^{\frac{3}{2}}\Vert \nabla ^4(u, b)\Vert _{L^2}. \end{aligned}$$

(3.47)

In terms of Lemma 2.3 and the interpolation inequality again, we see that

$$\begin{aligned} \begin{aligned}&\Vert \nabla ^2( h(\phi )\Delta u)\Vert _{L^2}\le C\left[ \Vert \phi \Vert _{L^\infty }\Vert \nabla ^4 u\Vert _{L^2}+\left( \Vert \nabla ^2\phi \Vert _{L^3}+\Vert \nabla \phi \Vert _{L^6}^2\right) \Vert \nabla ^2u\Vert _{L^6}\right] \\&\quad \le C\delta ^{\frac{1}{2}}M^{\frac{1}{2}}\Vert \nabla ^4 u\Vert _{L^2}+C\left( \Vert \nabla ^2\phi \Vert _{L^2}^{\frac{1}{2}}\Vert \nabla ^3\phi \Vert _{L^2}^{\frac{1}{2}}+\Vert \nabla ^2\phi \Vert _{L^2}^2\right) \Vert \nabla ^3u\Vert _{L^2}\\&\quad \le C\delta ^{\frac{1}{2}}M^{\frac{1}{2}}\Vert \nabla ^4u\Vert _{L^2}+C\left( \Vert \nabla \phi \Vert _{L^2}^{\frac{1}{4}}\Vert \nabla ^3\phi \Vert _{L^2}^{\frac{3}{4}}+\Vert \nabla \phi \Vert _{L^2}\Vert \nabla ^3\phi \Vert _{L^2}\right) \Vert \nabla ^2u\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla ^4u\Vert _{L^2}^{\frac{1}{2}}\\&\quad \le C\delta ^{\frac{1}{4}}M\left( \Vert \nabla ^4 u\Vert _{L^2}+\Vert \nabla ^3\phi \Vert _{L^2}\right) , \end{aligned} \end{aligned}$$

(3.48)

and analogously,

$$\begin{aligned} \Vert \nabla ^2( h(\phi )(\nabla \mathop {\mathrm{div}}\nolimits u,\Delta \psi ))\Vert _{L^2}\le C\delta ^{\frac{1}{4}}M\left( \Vert \nabla ^4 (u,\psi )\Vert _{L^2}+\Vert \nabla ^3\phi \Vert _{L^2}\right) . \end{aligned}$$

(3.49)

Since

$$\begin{aligned} |\nabla ^2g(\phi )|\le C\left( |\nabla ^2\phi |+|\nabla \phi |^2\right) ,\quad |\nabla ^3g(\phi )|\le C\left( |\nabla ^3\phi |+|\nabla \phi ||\nabla ^2\phi |+|\nabla \phi |^3\right) . \end{aligned}$$

we deduce from Lemma 2.3 and the interpolation inequality that

$$\begin{aligned} \begin{aligned}&\Vert \nabla ^2(g(\phi )((\nabla \times b)\times b))\Vert _{L^2}\\&\quad \le C\left( \Vert \nabla ^2\phi \Vert _{L^3}\Vert b\Vert _{L^\infty }\Vert \nabla b\Vert _{L^6}+\Vert \nabla \phi \Vert _{L^6}^2\Vert b\Vert _{L^\infty }\Vert \nabla b\Vert _{L^6}+\Vert b\Vert _{L^3}\Vert \nabla ^3b\Vert _{L^6}+\Vert \nabla b\Vert _{L^3}\Vert \nabla ^2b\Vert _{L^6}\right) \\&\quad \le C\left( \Vert \nabla ^2\phi \Vert _{L^2}^{\frac{1}{2}}\Vert \nabla ^3\phi \Vert _{L^2}^{\frac{1}{2}}+\Vert \nabla \phi \Vert _{L^2}\Vert \nabla ^3\phi \Vert _{L^2}\right) \Vert b\Vert _{H^1}^{\frac{1}{2}}\Vert \nabla ^2 b\Vert _{L^2}^{\frac{3}{2}}+C\Vert b\Vert _{H^1}\Vert \nabla ^4b\Vert _{L^2} \\&\quad \le C\left( \Vert \nabla ^2\phi \Vert _{L^2}^{\frac{1}{2}}\Vert \nabla ^3\phi \Vert _{L^2}^{\frac{1}{2}}+\Vert \nabla \phi \Vert _{L^2}\Vert \nabla ^3\phi \Vert _{L^2}\right) \Vert b\Vert _{H^1}^{\frac{5}{4}}\Vert \nabla ^4 b\Vert _{L^2}^{\frac{3}{4}}+C\Vert b\Vert _{H^1}\Vert \nabla ^4b\Vert _{L^2} \\&\quad \le C\delta M^{\frac{3}{4}}\left( \Vert \nabla ^3\phi \Vert _{L^2}+\Vert \nabla ^4 b\Vert _{L^2}\right) \end{aligned} \end{aligned}$$

(3.50)

and

$$\begin{aligned} \begin{aligned}&\Vert \nabla ^3(g(\phi )((\nabla \times b)\times b))\Vert _{L^2}\\&\quad \le C\left( \Vert \nabla ^3\phi \Vert _{L^2}+\Vert \nabla \phi \Vert _{L^3}\Vert \nabla ^2\phi \Vert _{L^6}+\Vert \nabla \phi \Vert _{L^6}^3\right) \Vert b\Vert _{L^\infty }\Vert \nabla b\Vert _{L^\infty }+C\Vert \nabla b\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla ^2 b\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla ^4 b\Vert _{L^2}\\&\quad \le C\delta ^{\frac{1}{2}}M^{\frac{5}{2}}\left( \Vert \nabla ^3\phi \Vert _{L^2}+\Vert \nabla ^4 b\Vert _{L^2}\right) . \end{aligned} \end{aligned}$$

(3.51)

Collecting (3.43)–(3.51) together, we find

$$\begin{aligned} \Vert \nabla ^2(S_2,S_3,S_4)\Vert _{L^2}\le C\delta ^{\frac{1}{4}}M^{\frac{5}{2}}\left( \Vert \nabla ^3\phi \Vert _{L^2}+\Vert \nabla ^4 (u,\psi ,b)\Vert _{L^2}\right) , \end{aligned}$$

(3.52)

which, together (3.42) and (3.41), immediately leads to (3.40), using integration by parts and Cauchy–Schwarz inequality. $\square$

Finally, to close the estimate, we still need to deal with $\Vert \nabla ^3\phi \Vert _{L^2(0,T;L^2)}$.

Lemma 3.7

Under the assumptions (3.4) and (3.5), one has

$$\begin{aligned} \frac{{\mathrm{d}}}{\mathrm{d}t}\langle \nabla \mathop {\mathrm{div}}\nolimits u,\nabla \Delta \phi \rangle + \Vert \nabla ^3\phi \Vert ^{2}_{L^{2}} \le C\Vert \nabla ^3(u,\psi )\Vert _{H^1}^2+C\delta ^{\frac{1}{4}}M^{\frac{5}{2}}\left( \Vert \nabla ^3\phi \Vert _{L^2}^2+\Vert \nabla ^3 (u,\psi ,b)\Vert _{H^1}^2\right) . \end{aligned}$$

(3.53)

Proof

Operating $\nabla \mathop {\mathrm{div}}\nolimits$ to both sides of (2.1)$_{2}$ and multiplying it by $\nabla \Delta \phi$ in $L^2$, we have

$$\begin{aligned} \begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}\langle \nabla \mathop {\mathrm{div}}\nolimits u ,\nabla \Delta \phi \rangle + \Vert \nabla \Delta \phi \Vert _{L^2}^{2}=\,&\langle \nabla \mathop {\mathrm{div}}\nolimits u,\nabla \Delta \phi _t\rangle +\langle \nabla \mathop {\mathrm{div}}\nolimits S_{2},\nabla \Delta \phi \rangle \\&+\langle \nabla \mathop {\mathrm{div}}\nolimits (\mu \Delta u +(\mu +\lambda )\nabla \mathop {\mathrm{div}}\nolimits u)-\nabla \Delta \psi ,\nabla \Delta \phi \rangle . \end{aligned} \end{aligned}$$

(3.54)

In view of (2.1)$_{1}$, we have from Lemma 2.3 that

$$\begin{aligned} \begin{aligned} \langle \nabla \mathop {\mathrm{div}}\nolimits u,\nabla \Delta \phi _t\rangle =\,&-\langle \Delta \mathop {\mathrm{div}}\nolimits u,\Delta \phi _t\rangle \\ \le&\, C\Vert \nabla ^3u\Vert _{L^2}\left( \Vert \nabla ^3u\Vert _{L^2}+\Vert \nabla ^3(\phi u)\Vert _{L^2}\right) \\ \le&\, C\Vert \nabla ^3u\Vert _{L^2}\left( \Vert \nabla ^3u\Vert _{L^2}+\Vert \nabla ^3\phi \Vert _{L^2}\Vert u\Vert _{L^\infty }+\Vert \phi \Vert _{L^\infty }\Vert \nabla ^3u\Vert _{L^2}\right) \\ \le&\, C\Vert \nabla ^3u\Vert _{L^2}^2+C\delta ^{\frac{1}{2}}M^{\frac{1}{2}}\left( \Vert \nabla ^3u\Vert _{L^2}^2+\Vert \nabla ^3\phi \Vert _{L^2}^2\right) . \end{aligned} \end{aligned}$$

(3.55)

For the other terms, we have

$$\begin{aligned} \begin{aligned}&\left| \langle \nabla \mathop {\mathrm{div}}\nolimits S_{2},\nabla \Delta \phi \rangle \right| +\left| \langle \nabla \mathop {\mathrm{div}}\nolimits (\mu \Delta u +(\mu +\lambda )\nabla \mathop {\mathrm{div}}\nolimits u)-\nabla \Delta \psi ,\nabla \Delta \phi \rangle \right| \\&\quad \le \frac{1}{2}\Vert \nabla ^3\phi \Vert _{L^2}^2+C\Vert \nabla ^3(u,\psi )\Vert _{H^1}^2+C\delta ^{\frac{1}{4}}M^{\frac{5}{2}}\left( \Vert \nabla ^3\phi \Vert _{L^2}^2+\Vert \nabla ^4 (u,\psi ,b)\Vert _{L^2}^2\right) . \end{aligned} \end{aligned}$$

(3.56)

Thus, substituting (3.55) and (3.56) into (3.54), we obtain (3.53). $\square$

With Lemmas 3.1–3.7 at hand, we are now ready to prove Theorem 3.1.

Proof of Global Existence

On the one hand, in view of (3.10), (3.15) and (3.22), we have

$$\begin{aligned} \begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}\Vert (\phi ,u,\psi ,b)\Vert _{H^1}^2+\Vert \nabla (u,\psi ,b)\Vert _{H^1}^2\le&\, C_1\delta ^{\frac{1}{2}} M^{\frac{3}{2}}\left( \Vert \nabla (u,\psi ,b)\Vert _{H^1}^2+\Vert \nabla \phi \Vert _{L^2}^2\right) \\ \le&\, \frac{1}{2}\Vert \nabla (u,\psi ,b)\Vert _{H^1}^2+C_1\delta ^{\frac{1}{2}} M^{\frac{3}{2}}\Vert \nabla \phi \Vert _{L^2}^2 \end{aligned} \end{aligned}$$

(3.57)

and

$$\begin{aligned} \begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}\langle u,\nabla \phi \rangle + \Vert \nabla \phi \Vert ^{2}_{L^{2}} \le&\, C_2\Vert \nabla (u,\psi )\Vert ^{2}_{H^{1}}+C_2\delta ^{\frac{1}{2}} M^{\frac{3}{2}}\left( \Vert \nabla \phi \Vert _{L^2}^2+\Vert \nabla (u,\psi ,b)\Vert ^{2}_{H^{1}}\right) \\ \le&\, 2C_2\Vert \nabla (u,\psi , b)\Vert ^{2}_{H^{1}}+C_2\delta ^{\frac{1}{2}} M^{\frac{3}{2}}\Vert \nabla \phi \Vert _{L^2}^2, \end{aligned} \end{aligned}$$

(3.58)

provided $\delta >0$ is chosen to be small enough such that

$$\begin{aligned} 0<\delta \le \delta _1\triangleq \min \left\{ \delta _0,\left( 2C_1M^{\frac{3}{2}}\right) ^{-2},M^{-3}\right\} . \end{aligned}$$

It follows from (3.57) that

$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}\Vert (\phi ,u,\psi ,b)\Vert _{H^1}^2+\frac{1}{2}\Vert \nabla (u,\psi ,b)\Vert _{H^1}^2\le C_1\delta ^{\frac{1}{2}} M^{\frac{3}{2}}\Vert \nabla \phi \Vert _{L^2}^2, \end{aligned}$$

which, multiplied by a suitably large number $K_1=\max \{4,8C_2\}$ and added to (3.58), yields

$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}\left( K_1\Vert (\phi ,u,\psi ,b)\Vert _{H^1}^2+\langle u,\nabla \phi \rangle \right) +2C_2\Vert \nabla (u,\psi ,b)\Vert _{H^1}^2+\Vert \nabla \phi \Vert _{L^2}^2\le C_3(K_1)\delta ^{\frac{1}{2}} M^{\frac{3}{2}}\Vert \nabla \phi \Vert _{L^2}^2, \end{aligned}$$

where $C_3(K_1)$ is a positive constant depending on $K_1$. So, if $\delta >0$ is chosen to be such that

$$\begin{aligned} 0<\delta \le \delta _2\triangleq \min \left\{ \delta _1, \left( 2C_3(K_1)M^{\frac{3}{2}}\right) ^{-2}\right\} , \end{aligned}$$

then one has

$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}\left( K_1\Vert (\phi ,u,\psi ,b)\Vert _{H^1}^2+\langle u,\nabla \phi \rangle \right) +2C_2\Vert \nabla (u,\psi ,b)\Vert _{H^1}^2+\frac{1}{2}\Vert \nabla \phi \Vert _{L^2}^2\le 0, \end{aligned}$$

which, integrated over (0, T), immediately gives

$$\begin{aligned} \sup _{0\le t\le T}\Vert (\phi ,u,\psi ,b)(t)\Vert _{H^1}^2+\int \limits _0^T\left( \Vert \nabla (u,\psi ,b)\Vert _{H^1}^2+\Vert \nabla \phi \Vert _{L^2}^2\right) {\mathrm{d}}t\le {\bar{C}}\Vert (\phi _0,u_0,\psi _0,b_0)\Vert _{H^1}^2, \end{aligned}$$

(3.59)

since the fact $K_1=\max \{4,8C_2\}$ implies

$$\begin{aligned} \left| \langle u,\nabla \phi \rangle \right| \le \frac{K_1}{2} \Vert (\phi ,u)\Vert _{H^1}^2. \end{aligned}$$

On the other hand, one easily concludes from (3.21), Lemmas 3.4 and 3.6 that

$$\begin{aligned} \begin{aligned} \frac{{\mathrm{d}}}{\mathrm{d}t}\Vert \nabla (\phi ,u,\psi ,b)(t)\Vert _{H^2}^2+ \Vert \nabla ^2(u,\psi ,b)\Vert _{H^2}^2\le&\, C_4\delta ^{\frac{1}{4}}M^{\frac{5}{2}}\left( \Vert \nabla ^2(u,\psi ,b)\Vert _{H^2}^2+\Vert \nabla ^2\phi \Vert _{H^1}^2\right) \\ \le&\frac{1}{2}\Vert \nabla ^2(u,\psi ,b)\Vert _{H^2}^2+C_4\delta ^{\frac{1}{4}}M^{\frac{5}{2}} \Vert \nabla ^2\phi \Vert _{H^1}^2, \end{aligned} \end{aligned}$$

and hence,

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\Vert \nabla (\phi ,u,\psi ,b)(t)\Vert _{H^2}^2+\frac{1}{2} \Vert \nabla ^2(u,\psi ,b)\Vert _{H^2}^2\le C_4\delta ^{\frac{1}{4}}M^{\frac{5}{2}} \Vert \nabla ^2\phi \Vert _{H^1}^2, \end{aligned}$$

(3.60)

provided $\delta >0$ is chosen to be such that

$$\begin{aligned} 0<\delta \le \delta _3\triangleq \min \left\{ \delta _2,\left( 2C_4M^{\frac{5}{2}}\right) ^{-4}\right\} . \end{aligned}$$

It follows from (3.37) and (3.53) that

$$\begin{aligned} \begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t}\left( \langle \mathop {\mathrm{div}}\nolimits u,\Delta \phi \rangle +\langle \nabla \mathop {\mathrm{div}}\nolimits u,\nabla \Delta \phi \rangle \right) + \Vert \nabla ^2\phi \Vert ^{2}_{H^{1}}\\&\quad \le C_5\Vert \nabla ^2(u,\psi )\Vert _{H^2}^2+C_5\delta ^{\frac{1}{4}}M^{\frac{5}{2}}\left( \Vert \nabla ^2\phi \Vert _{H^1}^2+\Vert \nabla ^2 (u,\psi ,b)\Vert _{H^2}^2\right) \\&\quad \le 2C_5\Vert \nabla ^2(u,\psi ,b)\Vert _{H^2}^2+\frac{1}{2}\Vert \nabla ^2\phi \Vert _{H^1}^2, \end{aligned} \end{aligned}$$

so that

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\left( \langle \mathop {\mathrm{div}}\nolimits u,\Delta \phi \rangle +\langle \nabla \mathop {\mathrm{div}}\nolimits u,\nabla \Delta \phi \rangle \right) + \frac{1}{2}\Vert \nabla ^2\phi \Vert ^{2}_{H^{1}} \le 2C_5\Vert \nabla ^2(u,\psi ,b)\Vert _{H^2}^2 , \end{aligned}$$

(3.61)

provided

$$\begin{aligned} 0<\delta \le \delta _4\triangleq \min \left\{ \delta _3,\left( 2C_5M^{\frac{5}{2}}\right) ^{-4}, \left( M^{\frac{5}{2}}\right) ^{-4}\right\} . \end{aligned}$$

Similarly to the proof of (3.59), multiplying (3.60) by a suitably large number $K_2=\max \{8,8C_5\}$, adding it to (3.61), and integrating the resulting relation over (0, T), we obtain

$$\begin{aligned} \sup _{0\le t\le T}\Vert \nabla (\phi ,u,\psi ,b)(t)\Vert _{H^2}^2+\int \limits _0^T\left( \Vert \nabla ^2(u,\psi ,b)\Vert _{H^2}^2+\Vert \nabla ^2\phi \Vert _{H^1}^2\right) {\mathrm{d}}t\le {\tilde{C}}\Vert \nabla (\phi _0,u_0,\psi _0,b_0)\Vert _{H^2}^2. \end{aligned}$$

(3.62)

With the help of (3.59) and (3.62), by bootstrap arguments we easily close the a priori assumptions (3.4) and (3.5) by taking $M^2=2{\tilde{C}}\Vert \nabla (\phi _0,u_0,\psi _0,b_0)\Vert _{H^2}^2$ and choosing $\eta >0$ sufficiently small to be such that $2{\bar{C}} \Vert (\phi _0,u_0,\psi _0,b_0)\Vert _{H^1}^2 \le \delta ^2$. This, together with the local existence result (cf. Lemma 2.5), finishes the proof of Theorem 3.1, and thus, the proof of the first part of Theorem 1.1 is complete. $\square$

3.2 Decay rates

This subsection is devoted to the derivations of the decay rates of the solutions. As a first step, we derive the decay estimates of $\Vert \nabla (\phi ,u,\psi ,b)\Vert _{H^2}$.

Lemma 3.8

Assume that the conditions of Theorem 3.1 are satisfied. If $\eta >0$ is small enough and $\Vert (\phi _0, u_0, \psi _0, b_0)\Vert _{L^1}$ is bounded, then for any $t\ge 0$,

$$\begin{aligned} \Vert \nabla (\phi ,u,\psi ,b)(t)\Vert _{H^2}^2\le C(1+t)^{-\frac{5}{2}}\quad {and}\quad \Vert (\phi ,u,\psi ,b)(t)\Vert _{L^2}^2\le C(1+t)^{-\frac{3}{2}}. \end{aligned}$$

(3.63)

Proof

Similarly to the derivation of (3.59), it is easily deduced from (3.21) and Lemmas 3.4–3.7 that there exist positive numbers ${\bar{K}}>0$ and k such that if $\eta >0$ is small enough, then

$$\begin{aligned} E'(t)+k\left( \Vert \nabla ^2(u,\psi ,b)\Vert _{H^2}^2+\Vert \nabla ^2\phi \Vert _{H^1}^2\right) \le 0, \end{aligned}$$

where

$$\begin{aligned} E(t)\triangleq {\bar{K}}\Vert \nabla (\phi ,u,\psi ,b)\Vert _{H^2}^2+\langle \mathop {\mathrm{div}}\nolimits u,\Delta \phi \rangle +\langle \nabla \mathop {\mathrm{div}}\nolimits u,\nabla \Delta \phi \rangle . \end{aligned}$$

Adding $\Vert \nabla (\phi ,u,\psi ,b)\Vert _{L^2}^2$ to both sides of the last inequality, we have

$$\begin{aligned} E'(t)+k\Vert \nabla (\phi ,u,\psi ,b)\Vert _{H^2}^2\le C\Vert \nabla (\phi ,u,\psi ,b)\Vert _{L^2}^2. \end{aligned}$$

(3.64)

Noting that for suitably large ${\bar{K}}>0$,

$$\begin{aligned} E(t)\sim \Vert \nabla (\phi ,u,\psi ,b)\Vert _{H^2}^2, \end{aligned}$$

and thus, it follows from (3.64) that there exists some positive number $c>0$ such that

$$\begin{aligned} E'(t)+cE(t)\le C\Vert \nabla (\phi ,u,\psi ,b)\Vert _{L^2}^2, \end{aligned}$$

so that

$$\begin{aligned} E(t)\le E(0)e^{-ct}+C\int \limits _0^t e^{-c(t-\tau )}\Vert \nabla (\phi ,u,\psi ,b)(\tau )\Vert _{L^2}^2d\tau . \end{aligned}$$

(3.65)

In order to deal with $\Vert \nabla (\phi ,u,\psi ,b)\Vert _{L^2}^2$, we first make use of Lemma 2.2 to see that

$$\begin{aligned} \begin{aligned} \Vert \nabla (\phi ,u,\psi ,b)(t)\Vert _{L^2}\le&\, C(1+t)^{-\frac{5}{4}}+C\int \limits _0^t(1+t-\tau )^{-\frac{5}{4}}\Vert (S_1,S_2,S_3,S_4)\Vert _{L^1\cap H^1}d\tau \\ \le&\, C(1+t)^{-\frac{5}{4}}+C\eta ^{\frac{1}{2}}\int \limits _0^t(1+t-\tau )^{-\frac{5}{4}}\sqrt{E(\tau )}d\tau , \end{aligned} \end{aligned}$$

(3.66)

where Theorem 3.1, (3.18), (3.36) and the following simple facts were also used.

$$\begin{aligned}&\Vert (S_1,S_2,S_3,S_4)\Vert _{L^1}\le C\left( \Vert (\phi ,u,\psi ,b)\Vert _{L^2}\Vert \nabla (\phi ,u,\psi ,b)\Vert _{H^1}+\Vert \nabla (u,b)\Vert _{L^2}^2+\Vert b\Vert _{L^2}\Vert \nabla \phi \Vert _{L^3}\Vert \nabla b\Vert _{L^6}\right) \nonumber \\&\quad \le \eta C(K) \Vert \nabla (\phi ,u,\psi ,b)\Vert _{H^1}\le \eta C(K)\sqrt{E(t)}, \end{aligned}$$

(3.67)

$$\begin{aligned}&\Vert S_1\Vert _{H^1}+\Vert (S_2,S_3,S_4)\Vert _{H^1}\le C\Vert (\phi ,u)\Vert _{H^1}\Vert \nabla (\phi ,u)\Vert _{H^2}+\eta ^{\frac{1}{2}} C(K) \Vert \nabla (\phi ,u,\psi ,b)\Vert _{H^2}\nonumber \\&\quad \le \eta ^{\frac{1}{2}} C(K)\sqrt{E(t)}. \end{aligned}$$

(3.68)

Define

$$\begin{aligned} \mathcal {E}(t)\triangleq \sup _{0\le \tau \le t}(1+\tau )^{\frac{5}{2}}E(\tau ). \end{aligned}$$

Then, it follows from Lemma 2.4 that

$$\begin{aligned} \begin{aligned} \Vert \nabla (\phi ,u,\psi ,b)(t)\Vert _{L^2}\le&\, C(1+t)^{-\frac{5}{4}}+C\eta ^{\frac{1}{2}}\sqrt{\mathcal {E}(t)}\int \limits _0^t(1+t-\tau )^{-\frac{5}{4}}(1+\tau )^{-\frac{5}{4}}d\tau \\ \le&\, C(1+t)^{-\frac{5}{4}}\left( 1+\eta ^{\frac{1}{2}}\sqrt{\mathcal {E}(t)}\right) \end{aligned} \end{aligned}$$

which, inserted into (3.65), shows that for any $t\ge 0$ (noting that $\mathcal {E}(t)$ is non-decreasing)

$$\begin{aligned} \begin{aligned} (1+t)^{\frac{5}{2}}E(t)\le&\, C(1+t)^{\frac{5}{2}}e^{-ct}+C\left( 1+\eta \mathcal {E}(t)\right) (1+t)^{\frac{5}{2}}\int \limits _0^t e^{-c(t-\tau )}(1+\tau )^{-\frac{5}{2}} d\tau \\ \le&\,C+C \eta \mathcal {E}(t). \end{aligned} \end{aligned}$$

As a result, if $\eta >0$ is chosen to be small enough, then one has $\mathcal {E}(t) \le C$. This, together with the fact that $E(t)\sim \Vert \nabla (\phi ,u,\psi ,b)\Vert _{H^2}^2$, proves (3.63)$_1$.

Moreover, using (3.63)$_1$, (3.67) and (3.68), we have from Lemmas 2.2 and 2.4 that

$$\begin{aligned} \begin{aligned} \Vert (\phi ,u,\psi ,b)(t)\Vert _{L^2}\le&\, C(1+t)^{-\frac{3}{4}}+C\int \limits _0^t(1+t-\tau )^{-\frac{3}{4}}\Vert (S_1,S_2,S_3,S_4)\Vert _{L^1\cap L^2}d\tau \\ \le&C(1+t)^{-\frac{3}{4}}+C \int \limits _0^t(1+t-\tau )^{-\frac{3}{4}}\sqrt{E(\tau )}d\tau \\ \le&C(1+t)^{-\frac{3}{4}}+C \mathcal {E}(t)\int \limits _0^t(1+t-\tau )^{-\frac{3}{4}}(1+\tau )^{-\frac{5}{4}}d\tau \\ \le&C(1+t)^{-\frac{3}{4}}, \end{aligned} \end{aligned}$$

(3.69)

which immediately yields the desired estimate of (3.63)$_2$. The proof of (3.63) is thus finished. $\square$

Compared with Lemma 2.1, it is easily seen that the decay rates of $H^1$-norm of solutions stated in (3.63) are optimal. In the next lemma, we aim to improve the decay estimates of higher derivatives, which will be achieved by using Lemma 3.8.

Lemma 3.9

There exists a positive time $T_1>0$ such that if $\eta >0$ is small enough, then

$$\begin{aligned} \Vert \nabla ^2(\phi ,u,\psi ,b)(t)\Vert _{H^1}^2\le C(1+t)^{-\frac{7}{2}},\quad \forall \ t\ge T_1. \end{aligned}$$

(3.70)

Proof

In view of Lemma 2.3, we have

$$\begin{aligned} \begin{aligned} \Vert \nabla S_1\Vert _{L^2}\le&\, \Vert \nabla ^2(\phi u)\Vert _{L^2}\le C\left( \Vert \phi \Vert _{L^3}\Vert \nabla ^2u\Vert _{L^6}+\Vert u\Vert _{L^3}\Vert \nabla ^2\phi \Vert _{L^6}\right) \\ \le&\, C\Vert (\phi ,u)\Vert _{H^1}\Vert \nabla ^3(\phi ,u)\Vert _{L^2}\le C\eta \Vert \nabla ^3(\phi ,u)\Vert _{L^2}. \end{aligned} \end{aligned}$$

(3.71)

From the derivations of (3.30)–(3.35), by Theorem 3.1 and Lemma 3.8 we obtain

$$\begin{aligned} \begin{aligned} \Vert \nabla (S_2,S_3,S_4)\Vert _{L^2}\le&\, C\eta \Vert \nabla ^3(u,\psi ,b)\Vert _{L^2}+C(K)\Vert \nabla (\phi ,u,\psi ,b)\Vert _{H^2}^2\\ \le&\, C\eta \Vert \nabla ^3(u,\psi ,b)\Vert _{L^2}+C(K)(1+t)^{-\frac{5}{2}}. \end{aligned} \end{aligned}$$

(3.72)

Thus, substituting (3.71) and (3.72) into (3.28) and integrating by parts, by Young inequality we see that if $\eta >0$ is small enough, then

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\Vert \nabla ^2(\phi ,u,\psi , b)\Vert _{L^2}^2+\Vert \nabla ^3(u,\psi ,b)\Vert _{L^2}^2\le C\eta \Vert \nabla ^3\phi \Vert _{L^2}^2+C(1+t)^{-5}, \end{aligned}$$

which, combined with (3.40) and (3.54), yields that there exist positive numbers ${\tilde{K}}$ (suitably large) and $k_1$ (suitably small) such that if $\eta >0$ is small enough, then

$$\begin{aligned} E_1'(t)+k_1\left( \Vert \nabla ^3(u,\psi ,b)\Vert _{H^1}^2+\Vert \nabla ^3\phi \Vert _{L^2}^2\right) \le C(1+t)^{-5}, \end{aligned}$$

(3.73)

where

$$\begin{aligned} E_1(t)\triangleq {\tilde{K}}\Vert \nabla ^2(\phi ,u,\psi , b)\Vert _{H^1}^2+\langle \nabla \mathop {\mathrm{div}}\nolimits u,\nabla \Delta \phi \rangle \sim \Vert \nabla ^2(\phi ,u,\psi , b)\Vert _{H^1}^2. \end{aligned}$$

Noting that the Sobolev interpolation inequality, together with the Young inequality, gives

$$\begin{aligned} \Vert \nabla ^2\phi \Vert _{L^2}^2\le C\Vert \nabla \phi \Vert _{L^2}\Vert \nabla ^3\phi \Vert _{L^2}\le \alpha ^{-1}(1+t)\Vert \nabla ^3\phi \Vert _{L^2}^2+C(\alpha )(1+t)^{-1}\Vert \nabla \phi \Vert _{L^2}^2 \end{aligned}$$

where $\alpha >0$ is a positive constant to be chosen later. This particularly implies that

$$\begin{aligned} \Vert \nabla ^3\phi \Vert _{L^2}^2\ge \alpha (1+t)^{-1}\Vert \nabla ^2\phi \Vert _{L^2}^2-C(\alpha )(1+t)^{-2}\Vert \nabla \phi \Vert _{L^2}^2. \end{aligned}$$

(3.74)

In a similar manner,

$$\begin{aligned} \Vert \nabla ^3(u,\psi ,b)\Vert _{H^1}^2\ge \alpha (1+t)^{-1}\Vert \nabla ^2(u,\psi ,b)\Vert _{H^1}^2-C(\alpha )(1+t)^{-2}\Vert \nabla (u,\psi ,b)\Vert _{H^1}^2. \end{aligned}$$

(3.75)

As a result of (3.74) and (3.75), we deduce from (3.73) that

$$\begin{aligned} \begin{aligned}&E_1'(t)+\frac{\alpha k_1}{2}(1+t)^{-1}\Vert \nabla ^2(u,\psi ,b)\Vert _{H^1}^2 +\frac{k_1}{2}\left( \alpha (1+t)^{-1}\Vert \nabla ^2\phi \Vert _{L^2}^2+\Vert \nabla ^3\phi \Vert _{L^2}^2\right) \\&\quad \le C(1+t)^{-5}+C(1+t)^{-2}\left( \Vert \nabla \phi \Vert _{L^2}^2+\Vert \nabla (u,\psi ,b)\Vert _{H^1}^2\right) \le C(\alpha )(1+t)^{-\frac{9}{2}}. \end{aligned} \end{aligned}$$

(3.76)

It is clear that if $t\ge \alpha >0$, then $\Vert \nabla ^3\phi \Vert _{L^2}\ge \alpha (1+t)^{-1}\Vert \nabla ^3\phi \Vert _{L^2}$. So, we have from (3.76) that

$$\begin{aligned} E_1'(t)+\frac{\alpha k_1}{2}(1+t)^{-1} \Vert \nabla ^2(\phi ,u,\psi ,b)\Vert _{H^1}^2\le C(\alpha )(1+t)^{-\frac{9}{2}}. \end{aligned}$$

(3.77)

Moreover, since $E_1(t)\sim \Vert \nabla ^2(\phi ,u,\psi ,b)\Vert _{H^1}^2$ for suitably large ${\tilde{K}}>0$, there exists a positive constant $c_1$, depending only on ${\tilde{K}}$ and $k_1$, such that

$$\begin{aligned} E_1'(t)+\alpha c_1 (1+t)^{-1}E_1(t)\le C(\alpha )(1+t)^{-\frac{9}{2}}. \end{aligned}$$

(3.78)

Thus, if $\alpha >0$ is chosen to be $\alpha =4c_1^{-1}$, then one easily concludes from (3.78) that

$$\begin{aligned} E_1'(t)+4(1+t)^{-1}E_1(t)\le C(1+t)^{-\frac{9}{2}}, \end{aligned}$$

and hence,

$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}\left[ (1+t)^4E_1(t)\right] = (1+t)^4\left( E_1'(t)+4(1+t)^{-1}E_1(t)\right) \le C(1+t)^{-\frac{1}{2}}, \end{aligned}$$

which, integrated over (0, t), results in

$$\begin{aligned} (1+t)^4E_1(t)\le E_1(0)+C_1(1+t)^{\frac{1}{2}}\le 2C_1(1+t)^{\frac{1}{2}}, \end{aligned}$$

(3.79)

provided $t>0$ is large enough such that

$$\begin{aligned} t\ge T_1\triangleq \max \left\{ \alpha ,\left( \frac{E_1(0)}{C_1}\right) ^2-1\right\} . \end{aligned}$$

Now, (3.70) readily follows from (3.79) and the fact that $E_1(t)\sim \Vert \nabla ^2(\phi ,u,\psi ,b)\Vert _{H^1}^2$. The proof of Lemma 3.9 is therefore finished. $\square$

Finally, we improve the decay rates of $\Vert \nabla ^3b\Vert _{L^2}$, based on a full use of Lemmas 3.8 and 3.9.

Lemma 3.10

There exists a positive time $T_2>0$ such that if $\eta >0$ is small enough, then

$$\begin{aligned} \Vert \nabla ^3b(t)\Vert _{H^1}^2\le C(1+t)^{-\frac{9}{2}},\quad \forall \ t\ge T_2. \end{aligned}$$

(3.80)

Proof

Indeed, similarly to the derivation of (3.41), we obtain after integrating by parts and using Cauchy–Schwarz inequality that

$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}\Vert \nabla ^3b\Vert _{L^2}^2+\Vert \nabla ^4b\Vert _{L^2}^2\le C\Vert \nabla ^2S_4\Vert _{L^2}^2. \end{aligned}$$

(3.81)

We are now ready to deal with $\Vert \nabla ^2S_4\Vert _{L^2}$. First, using Lemmas 3.8 and 3.9, we have

$$\begin{aligned} \begin{aligned}&\Vert \nabla ^2(u\cdot \nabla b)\Vert _{L^2}+\Vert \nabla ^2(b\cdot \nabla u)\Vert _{L^2}+\Vert \nabla ^2(b\mathop {\mathrm{div}}\nolimits u)\Vert _{L^2}\\&\quad \le C\left( \Vert (u,b)\Vert _{L^\infty }\Vert \nabla ^3(u,b)\Vert _{L^2}+\Vert \nabla (u,b)\Vert _{L^3}\Vert \nabla ^2(u,b)\Vert _{L^6}\right) \\&\quad \le C \Vert \nabla (u,b)\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla ^2(u,b)\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla ^3(u,b)\Vert _{L^2} \le C(1+t)^{-\frac{26}{8}} \end{aligned} \end{aligned}$$

(3.82)

and moreover, by Lemma 2.3 we obtain in a manner similarly to the derivation of (3.51) that

$$\begin{aligned} \begin{aligned} \Vert \nabla ^3(g(\phi )(\nabla \times b)\times b)\Vert _{L^2} \le&\, C(K)(1+t)^{-\frac{26}{8}}+C\Vert b\Vert _{L^\infty }\Vert \nabla ^4 b\Vert _{L^2}\\ \le&\, C(K)(1+t)^{-\frac{26}{8}}+\eta ^{\frac{1}{2}} C(K)\Vert \nabla ^4 b\Vert _{L^2}. \end{aligned} \end{aligned}$$

(3.83)

Taking (3.82) and (3.83) into account using Cauchy–Schwarz inequality and choosing $\eta >0$ small enough, we infer from (3.81) that

$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}\Vert \nabla ^3b\Vert _{L^2}^2+\Vert \nabla ^4b\Vert _{L^2}^2\le C(1+t)^{-\frac{26}{4}}. \end{aligned}$$

(3.84)

Similarly to (3.74), by Lemma 3.9 we have

$$\begin{aligned} \begin{aligned} \Vert \nabla ^4b\Vert _{L^2}^2\ge&5 (1+t)^{-1}\Vert \nabla ^3b\Vert _{L^2}^2-C(1+t)^{-2}\Vert \nabla ^2b\Vert _{L^2}^2\\ \ge&5 (1+t)^{-1}\Vert \nabla ^3b\Vert _{L^2}^2-C (1+t)^{-\frac{11}{2}}, \end{aligned} \end{aligned}$$

which, inserted into (3.84), gives

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\Vert \nabla ^3b\Vert _{L^2}^2+5(1+t)^{-1}\Vert \nabla ^3b\Vert _{L^2}^2\le C(1+t)^{-\frac{11}{2}}. \end{aligned}$$

(3.85)

Thus, we conclude from (3.85) that

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\left( (1+t)^5\Vert \nabla ^3b\Vert _{L^2}^2\right) =(1+t)^5\left( \frac{{\mathrm{d}}}{{\mathrm{d}}t}\Vert \nabla ^3b\Vert _{L^2}^2+5(1+t)^{-1}\Vert \nabla ^3b\Vert _{L^2}^2\right) \le C(1+t)^{-\frac{1}{2}}, \end{aligned}$$

(3.86)

which immediately shows

$$\begin{aligned} (1+t)^5\Vert \nabla ^3b(t)\Vert _{L^2}^2\le \Vert \nabla ^3b_0\Vert _{L^2}^2+C_2(1+t)^{\frac{1}{2}}\le 2C_2(1+t)^{\frac{1}{2}}, \end{aligned}$$

provided

$$\begin{aligned} t\ge T_2\triangleq \max \left\{ T_1,\left( \frac{\Vert \nabla ^3b_0\Vert _{L^2}^2}{C_2}\right) ^2-1\right\} . \end{aligned}$$

As a result of (3.86), we arrive at the desired decay rate stated in (3.81). $\square$

Proof of Decay Rates

Collecting Lemmas 3.8–3.10 together, one immediately obtains the desired decay rates stated in the second part of Theorem 1.1. $\square$

4 Proof of Theorem 1.2

To prove Theorem 1.2, we first observe that all the global estimates and the decay rates established in Sect. 3 also hold for the standard MHD equations without Hall effects (i.e., $\varepsilon \equiv 0$):

$$\begin{aligned} \left\{ \begin{aligned}&{\bar{\rho }}_{t}+\mathop {\mathrm{div}}\nolimits ({\bar{\rho }} {\bar{u}})=0,\\&({\bar{\rho }} {\bar{u}})_{t}+\mathop {\mathrm{div}}\nolimits ({\bar{\rho }}{\bar{u}}\otimes {\bar{u}})+ \nabla {\bar{p}}= \mu \Delta {\bar{u}} +(\mu +\lambda )\nabla \mathop {\mathrm{div}}\nolimits {\bar{u}}+(\nabla \times {\bar{b}} )\times {\bar{b}},\\&\mathrm{c}_{\mathrm{V}}\left[ ({\bar{\rho }}{\bar{\theta }})_{t}+ \mathop {\mathrm{div}}\nolimits ({\bar{\rho }} {\bar{u}}{\bar{\theta }})\right] + {\bar{p}}\mathop {\mathrm{div}}\nolimits {\bar{u}}=\kappa \Delta {\bar{\theta }}+\frac{\mu }{2}\left| \nabla {\bar{u}}+(\nabla {\bar{u}})^{\top }\right| ^{2}+\lambda (\mathop {\mathrm{div}}\nolimits {\bar{u}})^{2}+\nu |\nabla \times {\bar{b}}|^{2},\\&{\bar{b}}_{t}-\nabla \times ({\bar{u}} \times {\bar{b}} )=\nu \Delta {\bar{b}},\\&\mathop {\mathrm{div}}\nolimits {\bar{b}}=0, \end{aligned}\right. \end{aligned}$$

(4.1)

which are equipped with the far-field boundary conditions and the initial conditions:

$$\begin{aligned} \left\{ \begin{aligned}&({\bar{\rho }},{\bar{u}}, {\bar{\theta }},{\bar{b}})(x,t)\rightarrow (1,0,1,0)\quad \mathrm{as}\quad |x|\rightarrow \infty ,\\&({\bar{\rho }},{\bar{u}},{\bar{\theta }},{\bar{b}})|_{t=0}=(\rho _0, u_{0},\theta _{0},b_{0})(x),\quad x\in \mathbb R^3. \end{aligned}\right. \end{aligned}$$

(4.2)

Here, the pressure ${\bar{p}}=R{\bar{\rho }}{\bar{\theta }}$.

With the help of the global a priori estimates, we can show the following global existence theorem for the Cauchy problem (4.1)–(4.2).

Theorem 4.1

For any given positive number ${\bar{M}}_0>0$ (not necessarily small), assume that the initial data $(\rho _0-1, u_0, \theta _0-1, b_0)\in H^3$ satisfy

$$\begin{aligned} \Vert \nabla ^2(\rho _0-1, u_0, \theta _0-1, b_0)\Vert _{H^1}\le {{\bar{M}}_0}, \end{aligned}$$

(4.3)

then there exists a positive constant ${\bar{\eta }}_0>0$, depending on ${\bar{M}}_0$, such that the problem (4.1)–(4.2) has a unique global classical solution $({\bar{\rho }},{\bar{u}}, {\bar{\theta }}, {\bar{b}})$ on $\mathbb R^3\times (0,\infty )$ satisfying

$$\begin{aligned} \begin{aligned}&\Vert ({\bar{\rho }}-1, {\bar{u}}, {\bar{\theta }}-1,{\bar{b}})(t)\Vert ^2_{H^3}+\int \limits _0^t(\Vert \nabla {\bar{\rho }}(\tau )\Vert ^2_{H^2}+ \Vert \nabla ({\bar{u}}, {\bar{\theta }},{\bar{b}})(\tau )|^2_{H^3})d\tau \\&\quad \le C\Vert (\rho _0-1, u_0, \theta _0-1, b_0)\Vert ^2_{H^3} \end{aligned} \end{aligned}$$

(4.4)

for all $t\ge 0$, provided

$$\begin{aligned} \Vert (\rho _0-1, u_0, \theta _0-1, b_0)\Vert _{H^1}\le {\bar{\eta }}_0, \end{aligned}$$

(4.5)

where $C>0$ is a generic positive constant independent of t.

Let $(\rho ,u,\theta ,b)$ be the global solutions of the problem (1.1)–(1.3) obtained in Theorem 1.1. Indeed, based on the global uniform-in-$\varepsilon$ estimates derived in Sect. 3, it is easily checked that as $\varepsilon \rightarrow 0$, there exists a subsequence of $(\rho ,u,\theta ,b)$ (still denoted by $(\rho ,u,\theta ,b)$) such that

$$\begin{aligned} (\rho ,u,\theta ,b)\rightarrow ({\bar{\rho }},{\bar{u}},{\bar{\theta }},{\bar{b}}) \quad \mathrm{strongly\ \ in}\quad C([0,T];H^2),\quad \forall \ T\in (0,\infty ). \end{aligned}$$

To prove the convergence rates, set

$$\begin{aligned} \Phi =\rho -{\bar{\rho }}, \quad U=u-{\bar{u}},\quad \Psi =\theta -{\bar{\theta }}, \quad B=b-{\bar{b}}. \end{aligned}$$

Then, it can be easily derived from (1.1) and (4.1) that

$$\begin{aligned}&\Phi _t+U\cdot \nabla \rho +{\bar{u}}\cdot \nabla \Phi +\rho \mathop {\mathrm{div}}\nolimits U+\Phi \mathop {\mathrm{div}}\nolimits {\bar{u}}=0, \end{aligned}$$

(4.6)

$$\begin{aligned}&\rho U_t+\rho u\cdot \nabla U-\mu \Delta U-(\mu +\lambda )\nabla \mathop {\mathrm{div}}\nolimits U=-\Phi {\bar{u}}_t-\rho U\cdot \nabla {\bar{u}}-\Phi {\bar{u}}\cdot \nabla {\bar{u}}-\nabla (p-{\bar{p}})\nonumber \\&\quad -\frac{1}{2}\nabla \left( |b|^2-|{\bar{b}}|^2\right) +b\cdot \nabla B+B\cdot \nabla {\bar{b}}, \end{aligned}$$

(4.7)

$$\begin{aligned}&\rho \Psi _t+\rho u\cdot \nabla \Psi -\Delta \Psi =-\Phi {\bar{\theta }}_t-\rho U\cdot \nabla {\bar{\theta }}-\Phi {\bar{u}}\cdot \nabla {\bar{\theta }}-p\mathop {\mathrm{div}}\nolimits U-(p-{\bar{p}})\mathop {\mathrm{div}}\nolimits {\bar{u}}\nonumber \\&\quad +\frac{\mu }{2}\left( |\nabla u+(\nabla u)^\top |^2-|\nabla {\bar{u}}+(\nabla {\bar{u}})^\top |^2\right) \nonumber \\&\quad +\lambda \left( (\mathop {\mathrm{div}}\nolimits u)^2-(\mathop {\mathrm{div}}\nolimits {\bar{u}})^2\right) +\left( |\nabla \times b|^2-|\nabla \times {\bar{b}}|^2\right) \end{aligned}$$

(4.8)

and

$$\begin{aligned} \begin{aligned} B_t-\Delta B=\,&-u\cdot \nabla B-U\cdot \nabla {\bar{b}}+b\cdot \nabla U+B\cdot \nabla {\bar{u}}-b\mathop {\mathrm{div}}\nolimits U-B\mathop {\mathrm{div}}\nolimits {\bar{u}}\\&-\varepsilon \nabla \times \left( \rho ^{-1}(\nabla \times b)\times b\right) . \end{aligned} \end{aligned}$$

(4.9)

Moreover, the pair of functions $(\Phi , U,\Psi , B)$ satisfies the vanishing far-field and initial conditions.

We first prove the $L^2$-convergence rate of the vanishing limit of Hall coefficient.

Lemma 4.1

For any $T\in (0,\infty )$, there exists a positive constant $C=C(T)$, independent of $\varepsilon$, such that

$$\begin{aligned} \sup _{0\le t\le T}\Vert (\Phi ,U,\Psi , B)(t)\Vert _{L^2}^2+\int \limits _0^T\Vert \nabla (U,\Psi ,B)\Vert _{L^2}^2{\mathrm{d}}t\le C\varepsilon ^2. \end{aligned}$$

(4.10)

Proof

The proofs will be carried out by the standard $L^2$-method, based on the global estimates stated in Theorems 1.1 and 4.1. For completeness, we sketch the proofs.

First, multiplying (4.6) by $\Phi$ in $L^2$, integrating parts, using Theorems 1.1, 4.1 and Cauchy–Schwarz inequality, we easily get that

$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}\Vert \Phi \Vert _{L^2}^2\le C\Vert (\nabla \rho ,\nabla {\bar{u}})\Vert _{H^2}\left( \Vert (\Phi ,U)\Vert _{L^2}^2+\Vert \nabla U\Vert _{L^2}^2\right) \le C\left( \Vert (\Phi ,U)\Vert _{L^2}^2+\Vert \nabla U\Vert _{L^2}^2\right) . \end{aligned}$$

(4.11)

Note that

$$\begin{aligned} |p-{\bar{p}}|+|b|^2-|{\bar{b}}|^2=|\Phi \theta +{\bar{\rho }}\Psi |+(b+{\bar{b}})\cdot B\le C\left( |\Phi |+|\Psi |+|B|\right) . \end{aligned}$$

(4.12)

Thus, multiplying (4.7) by U in $L^2$ and integrating by parts, by Theorems 1.1 and 4.1 we deduce

$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}\Vert \sqrt{\rho }U\Vert _{L^2}^2+\Vert \nabla U\Vert _{L^2}^2\le C\Vert (\Phi ,U,\Psi ,B)\Vert _{L^2}^2 \end{aligned}$$

(4.13)

where we have used the following estimates:

$$\begin{aligned} \Vert (u_t,{\bar{u}}_t,\theta _t,{\bar{\theta }}_t,b_t,{\bar{b}}_t)\Vert _{H^1}^2+\int \limits _0^T \Vert (u_t,{\bar{u}}_t,\theta _t,{\bar{\theta }}_t,b_t,{\bar{b}}_t)\Vert _{H^2}^2{\mathrm{d}}t\le C, \end{aligned}$$

(4.14)

due to (1.1), (4.1), Theorems 1.1 and 4.1. Similarly, since

$$\begin{aligned} \begin{aligned}&\left( |\nabla u+(\nabla u)^\top |^2-|\nabla {\bar{u}}+(\nabla {\bar{u}})^\top |^2\right) +\left( (\mathop {\mathrm{div}}\nolimits u)^2-(\mathop {\mathrm{div}}\nolimits {\bar{u}})^2\right) +\left( |\nabla \times b|^2-|\nabla \times {\bar{b}}|^2\right) \\&\quad \le C\Vert \nabla (u,{\bar{u}},b,{\bar{b}})\Vert _{L^\infty }\left( |\nabla U|+|\nabla B|\right) \le C\left( |\nabla U|+|\nabla B|\right) , \end{aligned} \end{aligned}$$

(4.15)

we obtain after multiplying (4.8) by $\Psi$ in $L^2$ that

$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}\Vert \sqrt{\rho }\Psi \Vert _{L^2}^2+\Vert \nabla \Psi \Vert _{L^2}^2\le C\Vert (\Phi ,U,\Psi )\Vert _{L^2}^2+C\Vert \nabla (U,B)\Vert _{L^2}^2. \end{aligned}$$

(4.16)

In a similar manner,

$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}\Vert B\Vert _{L^2}^2+\Vert \nabla B\Vert _{L^2}^2\le C\Vert (U,B)\Vert _{L^2}^2+C\varepsilon ^2, \end{aligned}$$

(4.17)

since Theorem 1.1 implies that

$$\begin{aligned} \varepsilon \Vert \rho ^{-1}(\nabla \times b)\times b\Vert _{H^2}\le C\varepsilon . \end{aligned}$$

(4.18)

Multiplying (4.13) and (4.17) by a suitable large constant, then adding them with (4.11) and (4.16) together, we find

$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}\Vert (\Phi ,\sqrt{\rho }U,\sqrt{\rho }\Psi , B)(t)\Vert _{L^2}^2+\Vert \nabla (U,\Psi ,B)\Vert _{L^2}^2\le C\Vert (\Phi ,U,\Psi ,B)\Vert _{L^2}^2+C\varepsilon ^2, \end{aligned}$$

which, combined with Gronwall inequality and the fact that $\rho$ is strictly positive, leads to the desired result of (4.10). $\square$

The next lemma is concerned with the $H^1$-convergence rate of the vanishing limit of Hall coefficient.

Lemma 4.2

There exists a positive constant $C=C(T)$, independent of $\varepsilon$, such that for any $T>0$,

$$\begin{aligned} \sup _{0\le t\le T}\Vert \nabla (\Phi ,U,\Psi , B)(t)\Vert _{L^2}^2+\int \limits _0^T\Vert \nabla ^2(U,\Psi ,B)\Vert _{L^2}^2{\mathrm{d}}t\le C\varepsilon ^2. \end{aligned}$$

(4.19)

Proof

Operating $\nabla$ to both sides of (4.6) and multiplying it by $\nabla \Phi$ in $L^2$, by Lemma 4.1 and Young inequality we obtain after integrating by parts that

$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}\Vert \nabla \Phi \Vert _{L^2}^2\le C(\delta )\Vert (\Phi ,U)\Vert _{H^1}^2+\delta \Vert \nabla ^2U\Vert _{L^2}^2\le C\varepsilon ^2+C(\delta )\Vert \nabla (\Phi ,U)\Vert _{L^2}^2+\delta \Vert \nabla ^2U\Vert _{L^2}^2. \end{aligned}$$

(4.20)

Multiplying (4.7) by $U_t$ in $L^2$, integrating by parts, using (4.10), (4.14), Theorems 1.1, 4.1, Sobolev inequalities and Cauchy–Schwarz inequality, we find (noting that $\rho$ is strictly lower-bounded)

$$\begin{aligned} \begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}\Vert \nabla U\Vert _{L^2}^2+\Vert \sqrt{\rho }U_t\Vert _{L^2}^2\le&\, C\Vert (\Phi ,U,\Psi ,B)\Vert _{L^2}^2+C\Vert \nabla (\Phi ,U,\Psi ,B)\Vert _{L^2}^2\\ \le&\, C\varepsilon ^2+C\Vert \nabla (\Phi ,U,\Psi ,B)\Vert _{L^2}^2. \end{aligned} \end{aligned}$$

(4.21)

Similarly, using (4.10), (4.14), (4.15), (4.18), Theorems 1.1 and 4.1, we have

$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}\Vert \nabla \Psi \Vert _{L^2}^2+\Vert \sqrt{\rho }\Psi _t\Vert _{L^2}^2\le C\varepsilon ^2+C\Vert \nabla (\Phi ,U,\Psi ,B)\Vert _{L^2}^2, \end{aligned}$$

(4.22)

and

$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}\Vert \nabla B\Vert _{L^2}^2+\Vert B_t\Vert _{L^2}^2\le C\varepsilon ^2+C\Vert \nabla (U,B)\Vert _{L^2}^2. \end{aligned}$$

(4.23)

Noting that $\rho$ is strictly lower-bounded due to (3.8), it follows from (4.20)–(4.23) that

$$\begin{aligned} \begin{aligned}&\frac{{\mathrm{d}}}{{\mathrm{d}}t}\Vert \nabla (\Phi ,U,\Psi , B)\Vert _{L^2}^2+\frac{1}{2}\left( \Vert U_t\Vert _{L^2}^2+\Vert \Psi _t\Vert _{L^2}^2+\Vert B_t\Vert _{L^2}^2\right) \\&\quad \le C\varepsilon ^2+C(\delta )\Vert \nabla (\Phi ,U,\Psi ,B)\Vert _{L^2}^2+\delta \Vert \nabla ^2 U\Vert _{L^2}^2. \end{aligned} \end{aligned}$$

(4.24)

In view of (4.10), (4.14), (4.15), (4.18), Theorems 1.1 and 4.1, we easily have from (4.7)–(4.9) that

$$\begin{aligned} \begin{aligned} \Vert \nabla ^2(U,\Psi , B)\Vert _{L^2}\le&\, C\Vert (U_t,\Psi _t,B_t)\Vert _{L^2}+C\Vert (\Phi ,U,\Psi ,B)\Vert _{H^1}+C\varepsilon \\ \le&\, C\Vert (U_t,\Psi _t,B_t)\Vert _{L^2}+C\Vert \nabla (\Phi ,U,\Psi ,B)\Vert _{L^2}+C\varepsilon , \end{aligned} \end{aligned}$$

(4.25)

Choosing $\delta$ to be suitable small in (4.24), then (4.24) together with (4.25) implies

$$\begin{aligned} \begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}\Vert \nabla (\Phi ,U,\Psi , B)\Vert _{L^2}^2+\tilde{C}\Vert \nabla ^2(U,\Psi , B)\Vert _{L^2}\le C\varepsilon ^2+C\Vert \nabla (\Phi ,U,\Psi ,B)\Vert _{L^2}^2 \end{aligned} \end{aligned}$$

for some constant $\tilde{C}>0$. Consequently, Lemma 4.1 and Gronwall inequality give the desired result of Lemma 4.2. $\square$

Finally, we derive the $H^2$-convergence rate of the vanishing limit of Hall coefficient.

Lemma 4.3

There exists a positive constant $C=C(T)$, independent of $\varepsilon$, such that for any $T>0$,

$$\begin{aligned} \sup _{0\le t\le T}\Vert \nabla ^2(\Phi ,U,\Psi , B)(t)\Vert _{L^2}^2+\int \limits _0^T\Vert \nabla ^3(U,\Psi ,B)\Vert _{L^2}^2{\mathrm{d}}t\le C\varepsilon ^2. \end{aligned}$$

(4.26)

Proof

Using Lemma 2.3, Theorems 1.1, 4.1, (4.10), (4.19), Young inequality and Sobolev inequalities, by direct calculations we have from (4.6) that

$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}\Vert \nabla ^2 \Phi \Vert _{L^2}^2\le C(\delta )\Vert (\Phi ,U)\Vert _{H^2}^2+\delta \Vert \nabla ^3U\Vert _{L^2}^2\le C\varepsilon ^2+C(\delta )\Vert \nabla ^2 (\Phi ,U)\Vert _{L^2}^2+\delta \Vert \nabla ^3U\Vert _{L^2}^2. \end{aligned}$$

(4.27)

Rewriting (4.7) in the form:

$$\begin{aligned} \begin{aligned} U_t+ u\cdot \nabla U-\rho ^{-1}\left( \mu \Delta U+(\mu +\lambda )\nabla \mathop {\mathrm{div}}\nolimits U\right) =\,&-\rho ^{-1}\Big [\Phi {\bar{u}}_t+\rho U\cdot \nabla {\bar{u}}+\Phi {\bar{u}}\cdot \nabla {\bar{u}}+\nabla (p-{\bar{p}}) \\&+\frac{1}{2}\nabla \left( |b|^2-|{\bar{b}}|^2\right) -b\cdot \nabla B-B\cdot \nabla {\bar{b}}\Big ]. \end{aligned} \end{aligned}$$

Operating $\nabla ^2$ to the above equation and multiplying it by $\nabla ^2U$ in $L^2$, by tedious but straightforward computations we obtain after integrating by parts and using Lemma 2.3 that

$$\begin{aligned} \begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}\Vert \nabla ^2U\Vert _{L^2}^2+\Vert \nabla ^3U\Vert _{L^2}^2\le&\,C \Vert (\Phi ,U,\Psi ,B)\Vert _{H^2}^2\le C\varepsilon ^2+C\Vert \nabla ^2(\Phi ,U,\Psi , B)\Vert _{L^2}^2, \end{aligned} \end{aligned}$$

(4.28)

where we have used (4.10), (4.14), (4.19), Theorems 1.1 and 4.1. Similarly,

$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}\Vert \nabla ^2\Psi \Vert _{L^2}^2+\Vert \nabla ^3\Psi \Vert _{L^2}^2\le C\varepsilon ^2+C\Vert \nabla ^2(\Phi ,U,\Psi , B)\Vert _{L^2}^2. \end{aligned}$$

(4.29)

Operating $\nabla ^2$ to both sides of (4.9) and multiplying the resulting equation by $\nabla ^2B$ in $L^2$, by virtue of (4.10), (4.18) and (4.19) we deduce

$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}\Vert \nabla ^2B\Vert _{L^2}^2+\Vert \nabla ^3B\Vert _{L^2}^2\le C\varepsilon ^2+C\Vert \nabla ^2(\Phi ,U,\Psi , B)\Vert _{L^2}^2, \end{aligned}$$

(4.30)

choosing $\delta$ to be suitable small in (4.27), then (4.27)–(4.30) gives

$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}\Vert \nabla ^2(\Phi ,U,\Psi , B)(t)\Vert _{L^2}^2+\Vert \nabla ^3(U,\Psi ,B)\Vert _{L^2}^2\le C\varepsilon ^2+C\Vert \nabla ^2(\Phi ,U,\Psi , B)\Vert _{L^2}^2, \end{aligned}$$

(4.31)

and thus, an application of Gronwall inequality yields (4.26). $\square$

Proof of Theorem 1.2

Now, the convergence rates of vanishing limit of Hall coefficient stated in Theorem 1.2 readily follow from Lemmas 4.1–4.3. $\square$