Magnetohydrodynamics approximation of the compressible full magneto- micropolar system

Jishan Fan; Tohru Ozawa; Jishan Fan; Tohru Ozawa

doi:10.3934/math.2022878

AIMS Mathematics

2022, Volume 7, Issue 9: 16037-16053. doi: 10.3934/math.2022878

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Research article

Magnetohydrodynamics approximation of the compressible full magneto- micropolar system

Jishan Fan ¹,
Tohru Ozawa ^{2
,
,}

1.
Department of Applied Mathematics, Nanjing Forestry University, Nanjing 210037, China
2.
Department of Applied Physics, Waseda University, Tokyo, 169-8555, Japan

Received: 23 May 2022 Revised: 14 June 2022 Accepted: 20 June 2022 Published: 30 June 2022
MSC : 35B25, 35Q60, 76W05

In this paper, we will use the Banach fixed point theorem to prove the uniform-in- $\epsilon$ existence of the compressible full magneto-micropolar system in a bounded smooth domain, where $\epsilon$ is the dielectric constant. Consequently, the limit as $\epsilon\rightarrow0$ can be established. This approximation is usually referred to as the magnetohydrodynamics approximation and is equivalent to the neglect of the displacement current.

Keywords:

Citation: Jishan Fan, Tohru Ozawa. Magnetohydrodynamics approximation of the compressible full magneto- micropolar system[J]. AIMS Mathematics, 2022, 7(9): 16037-16053. doi: 10.3934/math.2022878

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Abstract

1. Introduction

In this paper, we consider the compressible full magneto-micropolar system ^[1]

$\begin{eqnarray} &&\partial_t\rho+ \mathrm{div}\,(\rho u) = 0, \end{eqnarray}$

(1.1)

$\begin{eqnarray} &&\partial_t(\rho u)+ \mathrm{div}\,(\rho u\otimes u)+\nabla(\rho\theta)-(\mu+\mu_r)\Delta u-(\lambda+\mu-\mu_r)\nabla \mathrm{div}\, u\\ &&\qquad\qquad = 2\mu_r \mathrm{rot}\, w+(E+u\times b)\times b, \end{eqnarray}$

(1.2)

$\begin{eqnarray} &&\partial_t(\rho w)+ \mathrm{div}\,(\rho uw)-(c_a+c_d)\Delta w-(c_0+c_d-c_a)\nabla \mathrm{div}\, w+4\mu_rw = 2\mu_r \mathrm{rot}\, u, \end{eqnarray}$

(1.3)

$\begin{eqnarray} &&\partial_t(\rho\theta)+ \mathrm{div}\,(\rho u\theta)-k\Delta\theta+\rho\theta \mathrm{div}\, u = \frac{\mu}{2}(\nabla u+\nabla u^T):(\nabla u+\nabla u^T)+\lambda( \mathrm{div}\, u)^2\\ &&\qquad\qquad+4\mu_r\left|\frac12 \mathrm{rot}\, u-w\right|^2+c_0( \mathrm{div}\, w)^2+(c_a+c_d)\nabla w:\nabla w\\ &&\qquad\qquad+(c_d-c_a)\nabla w:\nabla w^T+|E+u\times b|^2, \end{eqnarray}$

(1.4)

$\begin{eqnarray} &&\epsilon\partial_tE-\mathrm{rot}b+E+u\times b = 0, \end{eqnarray}$

(1.5)

$\begin{eqnarray} &&\partial_tb+\mathrm{rot}E = 0, \mathrm{div}\, b = 0, \end{eqnarray}$

(1.6)

in $Q_T: = \Omega\times(0, T)$ for any $T > 0$ , with the initial and boundary conditions

$\begin{eqnarray} &&(\rho, u, w, \theta, E, b)(\cdot,0) = (\rho_0, u_0, w_0, \theta_0, E_0,b_0)\ \ \mathrm{in}\ \ \Omega\subseteq {\mathbb{R}^3}, \end{eqnarray}$

(1.7)

$\begin{eqnarray} &&u = 0,\ w = 0,\ \theta = 0,\ E\times n = 0,\ b\cdot n = 0\ \ \mathrm{on}\ \ \partial\Omega\times(0,T). \end{eqnarray}$

(1.8)

Here, $\rho$ is the density of the fluid, $u$ is the fluid velocity field, $w$ is the micro-rotational velocity, $\theta$ is the temperature, $E$ is the electric field, and $b$ is the magnetic field. $\Omega$ is a bounded domain in ${\mathbb{R}^3}$ with smooth boundary $\partial\Omega$ , whose outward normal vector is denoted by $n$ . The positive constant $k$ is the heat conductivity, the physical constants $\mu$ and $\lambda$ are the shear viscosity and bulk viscosity and satisfy $\mu > 0$ and $\lambda+\frac23\mu\geq0$ . $\epsilon > 0$ is the dielectric constant. The positive constant $\mu_r$ represents the dynamic microrotation viscosity. $c_0, c_a, c_d$ are constants called coefficients of angular viscosities, which satisfy $\lambda+\mu-\mu_r > 0, c_0+c_d-c_a > 0$ .

When $w = 0$ , the above system is symmetric hyperbolic-parabolic. Kawashima-Shizuta ^[2,3,4] proved the local existence of smooth solutions for large data and global existence of smooth solutions for small data and studied the limit as $\epsilon\rightarrow0$ when $\Omega: = \mathbb{R}^2$ . Jiang-Li ^[5,6] studied the limit of $\epsilon\rightarrow0$ when $\Omega: = \mathbb{T}^3$ . Similar results have been obtained in ^{[7,8,9,10,11,12,13]}. Li-Mu ^[14] studied the low Mach number limit of the problem (1.1)–(1.6) when $\Omega: = {\mathbb{R}^3}$ .

When $\epsilon = 0$ and the entropy is a constant, Wei-Guo-Li ^[15] and Wu-Wang ^[16] studied the long-time behavior of smooth solutions. Zhang ^[17] showed the local well-posedness (without proof) and a blow-up criterion.

The well-posedness of the problem has been studied in ^{[18,19,20,21]}. The numerical analysis of some related problems has been considered in ^{[22,23,24,25,26,27,28,29]}.

The aim of this paper is to prove the uniform-in- $\epsilon$ existence of unique local strong solutions to the problem (1.1)–(1.8) when $\Omega$ is a bounded domain.

Here, we impose the following regularity conditions on the initial data:

$\begin{equation} \begin{array}{l} \theta_0\geq0,\ 0 < \frac1C\leq\rho_0\leq C,\ \rho_0\in W^{1,6},\ \mathrm{div}\, b_0 = 0\ \ \mathrm{in}\ \ \Omega,\\ E_0\times n = 0,\ b_0\cdot n = 0\ \ \mathrm{on}\ \ \partial\Omega,\ E_0,b_0\in H^2,\ u_0,w_0,\theta_0\in H_0^1\cap H^2. \end{array} \end{equation}$

(1.9)

Theorem 1.1. Let (1.9) hold true and $0 < \epsilon < 1$ , and let $k = 1$ . Then, there exist a small time $\tilde T > 0$ independent of $\epsilon > 0$ and a unique strong solution $(\rho, u, w, \theta, E, b)$ to the initial boundary value problem (1.1)–(1.8) such that

$\begin{equation} \begin{array}{l} \theta\geq0,\ \frac1C\leq\rho\leq C,\ \rho\in L^\infty(0,\tilde T;W^{1,6}),\ \partial_t\rho\in L^\infty(0,\tilde T;L^6),\\ u,w\in L^\infty(0,\tilde T;H^2)\cap L^2(0,\tilde T;W^{2,6}),\ \theta\in L^\infty(0,\tilde T;H^2),\\ u_t,w_t,\theta_t\in L^\infty(0,\tilde T;L^2)\cap L^2(0,\tilde T;H^1),\\ b\in L^\infty(0,\tilde T;H^2), b_t\in L^\infty(0,\tilde T;H^1),\\ E\in L^\infty(0,\tilde T;H^1)\cap L^2(0,\tilde T;H^2),\ E_t\in L^2(0,\tilde T;H^1), \end{array} \end{equation}$

(1.10)

with the corresponding norms that are uniformly bounded with respect to $\epsilon > 0$ .

We will prove Theorem 1.1. by the Banach fixed point theorem. We define the nonempty closed set

$\mathcal{A}: = \{(\tilde u,\tilde w)\in\mathcal{A};\tilde u(\cdot,0) = u_0,\tilde w(\cdot,0) = w_0,\|(\tilde u,\tilde w)\|_{\mathcal{A}}\leq A\}$

with the norm

$\|(\tilde u,\tilde w)\|_{\mathcal{A}}: = \|(\tilde u,\tilde w)\|_{L^\infty(0,T;H^2)}+\|(\tilde u,\tilde w)\|_{L^2(0,T;W^{2,6})}+\|\partial_t(\tilde u,\tilde w)\|_{L^\infty(0,T;L^2)}+\|\partial_t(\tilde u,\tilde w)\|_{L^2(0,T;H^1)}.$

Let $\tilde u\in\mathcal{A}$ be given, and we consider the following linear problems:

$\begin{eqnarray} &&\partial_t\rho+ \mathrm{div}\,(\rho\tilde u) = 0, \end{eqnarray}$

(1.11)

$\begin{eqnarray} &&\rho(\cdot,0) = \rho_0; \end{eqnarray}$

(1.12)

$\begin{eqnarray} &&\epsilon\partial_tE-\mathrm{rot}b+E+\tilde u\times b = 0, \end{eqnarray}$

(1.13)

$\begin{eqnarray} &&\partial_tb+\mathrm{rot}E = 0, \end{eqnarray}$

(1.14)

$\begin{eqnarray} && \mathrm{div}\, b = 0, \end{eqnarray}$

(1.15)

$\begin{eqnarray} &&(E,b)(\cdot,0) = (E_0,b_0), \end{eqnarray}$

(1.16)

$\begin{eqnarray} &&E\times n = 0,\ b\cdot n = 0\ \ \mathrm{on}\ \ \partial\Omega\times(0,T); \end{eqnarray}$

(1.17)

$\begin{eqnarray} &&\rho\partial_t\theta+\rho\tilde u\cdot\nabla\theta-\Delta\theta+\rho\theta \mathrm{div}\,\tilde u = \frac\mu2(\nabla\tilde u+\nabla\tilde u^T):(\nabla\tilde u+\nabla\tilde u^T)+\lambda( \mathrm{div}\,\tilde u)^2\\ &&\qquad+4\mu_r\left|\frac12 \mathrm{rot}\,\tilde u-\tilde w\right|^2+c_0( \mathrm{div}\,\tilde w)^2+(c_a+c_d)\nabla\tilde w:\nabla\tilde w\\ &&\qquad+(c_d-c_a)\nabla\tilde w:\nabla\tilde w^T+|E+\tilde u\times b|^2, \end{eqnarray}$

(1.18)

$\begin{eqnarray} &&\theta(\cdot,0) = \theta_0, \end{eqnarray}$

(1.19)

$\begin{eqnarray} &&\theta = 0\ \ \mathrm{on}\ \ \partial\Omega\times(0,T). \end{eqnarray}$

(1.20)

$\begin{eqnarray} &&\rho\partial_tu+\rho\tilde u\cdot\nabla u+\nabla(\rho\theta)-(\mu+\mu_r)\Delta u-(\lambda+\mu-\mu_r)\nabla \mathrm{div}\, u\\ &&\qquad = 2\mu_r \mathrm{rot}\,\tilde w+(E+u\times b)\times b, \end{eqnarray}$

(1.21)

$\begin{eqnarray} &&u(\cdot,0) = u_0, \end{eqnarray}$

(1.22)

$\begin{eqnarray} &&u = 0\ \ \mathrm{on}\ \ \partial\Omega\times(0,T). \end{eqnarray}$

(1.23)

$\begin{eqnarray} &&\rho\partial_tw+\rho\tilde u\cdot\nabla w-(c_a+c_d)\Delta w-(c_0+c_d-c_a)\nabla \mathrm{div}\, w+4\mu_rw = 2\mu_r \mathrm{rot}\, u, \end{eqnarray}$

(1.24)

$\begin{eqnarray} &&w(\cdot,0) = w_0, \end{eqnarray}$

(1.25)

$\begin{eqnarray} &&w = 0\ \ \mathrm{on}\ \ \partial\Omega\times(0,T). \end{eqnarray}$

(1.26)

Let $(u, w)$ be the unique strong solution to the above problem, and we define the fixed point map $F:(\tilde u, \tilde w)\in\mathcal{A}\rightarrow (u, w)\in\mathcal{A}$ with $\tilde u = \tilde w = 0$ on $\partial\Omega\times(0, T)$ . We will prove that the map $F$ maps $\mathcal{A}$ into $\mathcal{A}$ for suitable constant $A$ and small $T$ , and $F$ is a contraction mapping on $\mathcal{A}$ , and thus $F$ has a unique fixed point in $\mathcal{A}$ . This proves Theorem 1.1.

2. Proof of Theorem 1.1

This section is devoted to the proof of Theorem 1.1.

Lemma 2.1. Let $(\tilde u, \tilde w)\in\mathcal{A}$ be given. Then, the problem (1.11) and (1.12) has a unique solution $\rho$ satisfying

$\frac1C\leq\rho\leq C,\ \|\rho\|_{L^\infty(0,T;W^{1,6})}\leq C,\ \|\rho_t\|_{L^\infty(0,T;L^6)}\leq CA$

for some small $0 < T\leq 1$ .

Here and later on, $C$ will denote a constant independent of $\epsilon$ and $A$ .

Proof. Since Eq (1.11) is linear with regular $\tilde u$ , the existence and uniqueness are well-known, and we only need to establish a priori estimates.

Let

$\frac{dx(X,t)}{dt} = \tilde u(x(X,t),t)\ \ \mathrm{and}\ \ x(X,0) = X,$

and we see that

$\frac{d\rho(x(X,t),t)}{dt} = -\rho \mathrm{div}\,\tilde u,$

whence

$\begin{equation} \rho(x,t) = \rho_0\exp\left(-\int_0^t \mathrm{div}\,\tilde u ds\right). \end{equation}$

(2.1)

It follows from (2.1) that

$\begin{eqnarray*} \rho(x,t)&\leq&\rho\exp\left(\int_0^T\| \mathrm{div}\,\tilde u\|_{L^\infty}dt\right)\\ &\leq&\rho_0\exp\left(\int_0^T C\|\tilde u\|_{W^{2,6}}dt\right)\\ &\leq&\rho_0\exp(CA\sqrt T)\leq C\|\rho_0\|_{L^\infty} \end{eqnarray*}$

if $A\sqrt T\leq 1$ ;

$\begin{eqnarray*} \rho(x,t)&\geq&\rho_0\exp\left(-\int_0^T\| \mathrm{div}\,\tilde u\|_{L^\infty} dt\right)\\ &\geq&\inf\rho_0\exp(-CA\sqrt T)\\ &\geq&C\inf\rho_0 \end{eqnarray*}$

if $A\sqrt T\leq 1$ ;

$\nabla\rho = \nabla\rho_0\exp\left(-\int_0^t \mathrm{div}\,\tilde u ds\right)-\rho_0\exp\left(-\int_0^t \mathrm{div}\,\tilde u ds\right)\int_0^t\nabla \mathrm{div}\,\tilde u ds,$

whence

$\begin{eqnarray*} \|\nabla\rho\|_{L^\infty(0,T;L^6)}&\leq&\exp\left(\int_0^T\| \mathrm{div}\,\tilde u\|_{L^\infty}dt\right)\left(\|\nabla\rho_0\|_{L^6}+\|\rho_0\|_{L^\infty}\int_0^T\|\nabla \mathrm{div}\,\tilde u\|_{L^6}dt\right)\\ &\leq&C\exp(CA\sqrt T)(1+A\sqrt T)\\ &\leq&C \end{eqnarray*}$

if $A\sqrt T\leq 1$ .

It follows from (1.11) that

$\rho_t = -\tilde u\nabla\rho-\rho \mathrm{div}\,\tilde u,$

and

$\begin{eqnarray*} \|\rho_t\|_{L^\infty(0,T;L^6)}&\leq&\|\tilde u\|_{L^\infty(0,T;L^\infty)}\|\nabla\rho\|_{L^\infty(0,T;L^6)}+\|\rho\|_{L^\infty(0,T;L^\infty)}\| \mathrm{div}\,\tilde u\|_{L^\infty(0,T;L^6)}\\ &\leq&CA \end{eqnarray*}$

if $A\sqrt T\leq 1$ .

This completes the proof.

Lemma 2.2. Let $(\tilde u, \tilde w)\in\mathcal{A}$ be given. Then, the problem (1.13)–(1.17) has a unique solution $(E, b)$ satisfying (2.2), (2.4), (2.6), (2.8), (2.9), (2.10), (2.11) and (2.12) for some $0 < T\leq 1$ .

Proof. Since Eqs (1.13)–(1.15) are linear with regular $(\rho, \tilde u)$ , the existence and uniqueness are well- known, and we only need to establish the a priori estimates.

Testing (1.13) and (1.14) by $E$ and $b$ and summing up the result, we see that

$\begin{eqnarray*} &&\frac12\frac{d}{dt}\int(\epsilon E^2+b^2)dx+\int E^2dx\\ & = &-\int(\tilde u\times b)E dx\\ &\leq&\|\tilde u\|_{L^\infty}\|b\|_{L^2}\|E\|_{L^2}\leq CA\|b\|_{L^2}\|E\|_{L^2}\\ &\leq&\frac12\|E\|_{L^2}^2+CA^2\|b\|_{L^2}^2, \end{eqnarray*}$

which gives

$\begin{equation} \int(\epsilon E^2+b^2)dx+\int_0^T\int E^2 dx dt\leq C \end{equation}$

(2.2)

if $A^2T\leq 1$ .

Note that (1.13), (1.17) and (1.23) give the boundary condition

$\begin{equation} \mathrm{rot}b\times n = 0\ \ \mathrm{on}\ \ \partial\Omega\times(0,T). \end{equation}$

(2.3)

Taking $\mathrm{rot}$ to (1.13) and (1.14), testing by $\mathrm{rot}E$ and $\mathrm{rot}b$ and using (2.3), summing up the result and integrating by parts, we find that

$\begin{eqnarray*} &&\frac12\frac{d}{dt}\int(\epsilon|\mathrm{rot}E|^2+|\mathrm{rot}b|^2)dx+\int|\mathrm{rot}E|^2dx\\ & = &-\int\mathrm{rot}(\tilde u\times b)\cdot\mathrm{rot}E dx\\ &\leq&C\|\tilde u\|_{H^2}\|\mathrm{rot}b\|_{L^2}\|\mathrm{rot}E\|_{L^2}\\ &\leq&\frac12\int|\mathrm{rot}E|^2dx+CA^2\|\mathrm{rot}b\|_{L^2}^2, \end{eqnarray*}$

which yields

$\begin{equation} \int(\epsilon|\mathrm{rot}E|^2+|\mathrm{rot}b|^2)dx+\int_0^T\int|\mathrm{rot}E|^2dx dt\leq C \end{equation}$

(2.4)

if $A^2T\leq 1$ .

Here, we have used the Poincaré inequality

$\begin{equation} \|b\|_{L^2}\leq C\|\mathrm{rot}b\|_{L^2}. \end{equation}$

(2.5)

Taking $\partial_t$ to (1.13) and (1.14), testing by $\partial_tE$ and $\partial_tb$ and summing up the result and using (2.4), we infer that

$\begin{eqnarray*} &&\frac12\frac{d}{dt}\int(\epsilon|E_t|^2+|\partial_tb|^2)+\int|E_t|^2dx\\ & = &\int\partial_t(b\times\tilde u)\cdot\partial_tE dx\\ &\leq&(\|\partial_tb\|_{L^2}\|\tilde u\|_{L^\infty}+\|b\|_{L^6}\|\partial_t\tilde u\|_{L^3})\|E_t\|_{L^2}\\ &\leq&\frac12\int|E_t|^2dx+CA^2\|\partial_tb\|_{L^2}^2+C\|\partial_t\tilde u\|_{L^2}\|\nabla\partial_t\tilde u\|_{L^2}, \end{eqnarray*}$

which implies

$\begin{equation} \int(\epsilon|E_t|^2+|b_t|^2)dx+\int_0^T\int|E_t|^2dx dt\leq C \end{equation}$

(2.6)

if $A^2T\leq 1$ .

(1.14) and (2.3) give the boundary condition

$\begin{equation} \mathrm{rot}^2E\times n = 0\ \ \mathrm{on}\ \ \partial\Omega\times(0,T). \end{equation}$

(2.7)

Taking $\mathrm{rot}^2$ to (1.13) and (1.14), testing by $\mathrm{rot}^2E$ and $\mathrm{rot}^2b$ and using (2.7) and summing up the result, we derive

$\begin{eqnarray*} &&\frac12\frac{d}{dt}\int(|\epsilon\mathrm{rot}^2E|^2+|\mathrm{rot}^2b|^2)dx+\int|\mathrm{rot}^2E|^2dx\\ & = &\int \mathrm{rot}\,^2(b\times\tilde u) \mathrm{rot}\,^2E dx\\ &\leq&C\|\tilde u\|_{H^2}\| \mathrm{rot}\,^2b\|_{L^2}\| \mathrm{rot}\,^2E\|_{L^2}\\ &\leq&\frac12\int| \mathrm{rot}\,^2E|^2dx+CA^2\| \mathrm{rot}\,^2b\|_{L^2}^2, \end{eqnarray*}$

which implies

$\begin{equation} \int(\epsilon| \mathrm{rot}\,^2E|^2+| \mathrm{rot}\,^2b|^2)dx+\int_0^T\int| \mathrm{rot}\,^2E|^2dx dt\leq C \end{equation}$

(2.8)

if $A^2T\leq 1$ .

Taking $\nabla \mathrm{div}\,$ to (1.13), testing by $\nabla \mathrm{div}\, E$ and using (2.8), we get

$\begin{eqnarray*} &&\frac{\epsilon}{2}\frac{d}{dt}\int|\nabla \mathrm{div}\, E|^2dx+\int|\nabla \mathrm{div}\, E|^2dx\\ & = &\int\nabla \mathrm{div}\,(b\times\tilde u)\cdot\nabla \mathrm{div}\, E dx\\ &\leq&C\|\tilde u\|_{H^2}\| \mathrm{rot}\,^2b\|_{L^2}\|\nabla \mathrm{div}\, E\|_{L^2}\\ &\leq&\frac12\int|\nabla \mathrm{div}\, E|^2dx+CA^2, \end{eqnarray*}$

which leads to

$\begin{equation} \epsilon\int|\nabla \mathrm{div}\, E|^2dx+\int_0^T\int|\nabla \mathrm{div}\, E|^2dx dt\leq C \end{equation}$

(2.9)

if $A^2T\leq 1$ .

Taking $\partial_t \mathrm{rot}\,$ to (1.13) and (1.14), testing by $\partial_t \mathrm{rot}\, E$ and $\partial_t \mathrm{rot}\, b$ and using (2.3) and (2.8), we obtain

$\begin{eqnarray*} &&\frac12\frac{d}{dt}\int(\epsilon| \mathrm{rot}\, E_t|^2+| \mathrm{rot}\, b_t|^2)dx+\int| \mathrm{rot}\, E_t|^2dx\\ & = &\int \mathrm{rot}\,(b_t\times\tilde u+b\times\tilde u_t) \mathrm{rot}\, E_t dx\\ &\leq&\frac12\int| \mathrm{rot}\, E_t|^2dx+C\|b_t\|_{L^6}^2\|\nabla\tilde u\|_{L^3}^2+C\|\tilde u\|_{L^\infty}^2\| \mathrm{rot}\, b_t\|_{L^2}^2\\ &&+C\|b\|_{L^\infty}^2\|\nabla\tilde u_t\|_{L^2}^2+C\| \mathrm{rot}\, b\|_{L^6}^2\|\tilde u_t\|_{L^3}^2\\ &\leq&\frac12\int| \mathrm{rot}\, E_t|^2dx+CA^2\| \mathrm{rot}\, b_t\|_{L^2}^2+C\|\nabla\tilde u_t\|_{L^2}^2, \end{eqnarray*}$

which yields

$\begin{equation} \int(\epsilon| \mathrm{rot}\, E_t|^2+| \mathrm{rot}\, b_t|^2)dx+\int_0^T\int| \mathrm{rot}\, E_t|^2dx dt\leq CA^2 \end{equation}$

(2.10)

if $A^2T\leq 1$ .

Applying $\partial_t \mathrm{div}\,$ to (1.13), testing by $\partial_t \mathrm{div}\, E$ , and using (2.10) and (2.8), we have

$\begin{eqnarray*} &&\frac{\epsilon}{2}\frac{d}{dt}\int| \mathrm{div}\, E_t|^2dx+\int| \mathrm{div}\, E_t|^2dx\\ & = &\int \mathrm{div}\,(b_t\times\tilde u+b\times\tilde u_t)\partial_t \mathrm{div}\, E dx\\ & = &\int(\tilde u \mathrm{rot}\, b_t-b_t \mathrm{rot}\,\tilde u+\tilde u_t \mathrm{rot}\, b-b \mathrm{rot}\,\tilde u_t)\partial_t \mathrm{div}\, E dx\\ &\leq&(\|\tilde u\|_{L^\infty}\| \mathrm{rot}\, b_t\|_{L^2}+\|b_t\|_{L^6}\| \mathrm{rot}\,\tilde u\|_{L^3}+\|\tilde u_t\|_{L^6}\| \mathrm{rot}\, b\|_{L^3}+\|b\|_{L^\infty}\| \mathrm{rot}\,\tilde u_t\|_{L^2})\| \mathrm{div}\, E_t\|_{L^2}\\ &\leq&C(A^2+\|\nabla\tilde u_t\|_{L^2})\| \mathrm{div}\, E_t\|_{L^2}\\ &\leq&\frac12\| \mathrm{div}\, E_t\|_{L^2}^2+CA^4+C\|\nabla\tilde u_t\|_{L^2}^2, \end{eqnarray*}$

which gives

$\begin{equation} \epsilon\int( \mathrm{div}\, E_t)^2dx+\int_0^T\int( \mathrm{div}\, E_t)^2dx dt\leq CA^2 \end{equation}$

(2.11)

if $A^2T\leq 1$ .

(2.6), (2.10) and (2.11) imply

$\begin{eqnarray*} &&\frac{d}{dt}\int(E^2+( \mathrm{div}\, E)^2+| \mathrm{rot}\, E|^2)dx\\ & = &2\int(E\cdot E_t+ \mathrm{div}\, E \mathrm{div}\, E_t+ \mathrm{rot}\, E \mathrm{rot}\, E_t)dx\\ &\leq&\left(\int(E^2+( \mathrm{div}\, E)^2+| \mathrm{rot}\, E|^2)\right)^\frac12\left(\int(E_t^2+( \mathrm{div}\, E_t)^2+( \mathrm{rot}\, E_t)^2)dx\right)^\frac12, \end{eqnarray*}$

whence

$\frac{dy}{dt}\leq C(\|E_t\|_{L^2}+\| \mathrm{div}\, E_t\|_{L^2}+\| \mathrm{rot}\, E_t\|_{L^2}),$

with

$y(t): = \left(\int(E^2+( \mathrm{div}\, E)^2+( \mathrm{rot}\, E)^2)dx\right)^\frac12.$

Integrating the above inequality, we have

$\begin{equation} y(t)\leq y(0)+CAT\leq C \end{equation}$

(2.12)

if $AT\leq 1$ .

This completes the proof.

Lemma 2.3. Let $(\tilde u, \tilde w)\in\mathcal{A}$ be given. Then, the problem (1.18) and (1.19) has a unique solution $\theta$ satisfying $\theta\geq0$ , (2.14), (2.15) and (2.16).

Proof. Since Eq (1.18) is linear with regular $(\rho, \tilde u, \tilde w, E, b)$ , the existence and uniqueness are well- known, and we only need to establish a priori estimates.

Testing (1.18) by $\theta$ and using (1.11) and Lemmas 2.1–2.2, we infer that

$\begin{eqnarray*} &&\frac12\frac{d}{dt}\int\rho\theta^2dx+\int|\nabla\theta|^2dx\\ & = &-\int\rho\theta^2 \mathrm{div}\,\tilde u dx+\int\Big[\frac\mu2(\nabla\tilde u+\nabla\tilde u^T):(\nabla\tilde u+\nabla\tilde u^T)+\lambda( \mathrm{div}\, \tilde u)^2\\ &&+4\mu_r\left|\frac12 \mathrm{rot}\,\tilde u-\tilde w\right|^2+c_0( \mathrm{div}\,\tilde w)^2+(c_a+c_d)\nabla\tilde w:\nabla\tilde w\\ &&+(c_d-c_a)\nabla\tilde w:\nabla\tilde w^T+|E+\tilde u\times b|^2\Big]\theta dx\\ &\leq&\| \mathrm{div}\,\tilde u\|_{L^\infty}\int\rho\theta^2dx+C\|\nabla\tilde u\|_{L^6}\|\nabla\tilde u\|_{L^3}\|\theta\|_{L^2}+C\|\tilde w\|_{L^6}\|\tilde w\|_{L^3}\|\theta\|_{L^2}\\ &&+C\|\nabla\tilde w\|_{L^6}\|\nabla\tilde w\|_{L^3}\|\theta\|_{L^2}+C\|E\|_{L^3}\|E\|_{L^6}\|\theta\|_{L^2} +C\|\tilde u\|_{L^\infty}^2\|b\|_{L^4}^2\|\theta\|_{L^2}\\ &\leq&\| \mathrm{div}\,\tilde u\|_{L^\infty}\int\rho\theta^2dx+CA^2\|\theta\|_{L^2}+C\|\theta\|_{L^2}, \end{eqnarray*}$

which yields

$\begin{equation} \int\theta^2dx+\int_0^T\int|\nabla\theta|^2dx dt\leq C \end{equation}$

(2.13)

if $A^2T\leq 1$ .

Testing (1.18) by $\theta_t$ and using Lemmas 2.1–2.2, we deduce that

$\begin{eqnarray*} &&\frac12\frac{d}{dt}\int|\nabla\theta|^2dx+\int\rho\theta_t^2dx\\ & = &-\int(\rho\theta \mathrm{div}\,\tilde u+\rho\tilde u\cdot\nabla\theta)\theta_t dx\\ &&+\int\Big[\frac\mu2(\nabla\tilde u+\nabla\tilde u^T):(\nabla\tilde u+\nabla\tilde u^T)+\lambda( \mathrm{div}\,\tilde u)^2\\ &&+4\mu_r\left|\frac12 \mathrm{rot}\,\tilde u-\tilde w\right|^2+c_0( \mathrm{div}\,\tilde w)^2+(c_a+c_d)\nabla\tilde w:\nabla\tilde w\\ &&+(c_d-c_a)\nabla\tilde w:\nabla\tilde w^T+|E+\tilde u\times b|^2\Big]\theta_t dx\\ &\leq&\| \mathrm{div}\,\tilde u\|_{L^6}\|\theta\|_{L^3}\|\sqrt\rho\theta_t\|_{L^2}\|\sqrt\rho\|_{L^\infty}+\|\tilde u\|_{L^\infty}\|\nabla\theta\|_{L^2}\|\sqrt\rho\theta_t\|_{L^2}\|\sqrt\rho\|_{L^\infty}\\ &&+C(\|\nabla\tilde u\|_{L^4}^2+\|\tilde w\|_{L^4}^2+\|\nabla\tilde w\|_{L^4}^2+\|E\|_{L^4}^2+\|\tilde u\|_{L^\infty}^2\|b\|_{L^4}^2)\|\theta_t\|_{L^2}\\ &\leq&\frac12\int\rho\theta_t^2dx+CA^2(\|\theta\|_{L^3}^2+\|\nabla\theta\|_{L^2}^2)+CA^4+C, \end{eqnarray*}$

which gives

$\begin{equation} \int|\nabla\theta|^2dx+\int_0^T\int\theta_t^2dx dt\leq C \end{equation}$

(2.14)

if $A^4T\leq 1$ .

Taking $\partial_t$ to (1.18), testing by $\theta_t$ , and using (1.11) and Lemmas 2.1–2.2, we have

$\begin{eqnarray*} &&\frac12\frac{d}{dt}\int\rho\theta_t^2dx+\int|\nabla\theta_t|^2dx\\ &\leq&-\int\rho_t\theta_t^2dx-\int\partial_t(\rho\tilde u)\cdot\nabla\theta\cdot\theta_tdx-\int\partial_t(\rho\theta \mathrm{div}\,\tilde u)\cdot\theta_tdx\\ &&+C\int|\nabla\tilde u||\nabla\tilde u_t||\theta_t|dx+C\int|\tilde w||\tilde w_t||\theta_t|dx\\ &&+C\int|\nabla\tilde w||\nabla\tilde w_t||\theta_t|dx+C\int|EE_t\theta_t|dx\\ &&+C\left|\int(\tilde u\times b)(\tilde u_t\times b+\tilde u\times b_t)\theta_t dx\right|\\ &\leq&\|\rho_t\|_{L^6}\|\theta_t\|_{L^2}\|\theta_t\|_{L^3}+\|\rho_t\|_{L^6}\|\tilde u\|_{L^\infty}\|\nabla\theta\|_{L^2}\|\theta_t\|_{L^3}\\ &&+C\|\tilde u_t\|_{L^3}\|\nabla\theta\|_{L^2}\|\theta_t\|_{L^6}+C\|\rho_t\|_{L^6}\|\theta\|_{L^6}\| \mathrm{div}\,\tilde u\|_{L^6}\|\theta_t\|_{L^2}\\ &&+C\|\sqrt\rho\theta_t\|_{L^2}^2\| \mathrm{div}\,\tilde u\|_{L^\infty}+C\|\theta\|_{L^6}\|\nabla\tilde u_t\|_{L^2}\|\theta_t\|_{L^3}\\ &&+C\|\nabla\tilde u\|_{L^6}\|\nabla\tilde u_t\|_{L^2}\|\theta_t\|_{L^3}+C\|\tilde w\|_{L^6}\|\tilde w_t\|_{L^2}\|\theta_t\|_{L^3}\\ &&+C\|\nabla\tilde w\|_{L^6}\|\nabla\tilde w_t\|_{L^2}\|\theta_t\|_{L^3}+C\|E\|_{L^6}\|E_t\|_{L^2}\|\theta_t\|_{L^3}\\ &&+C\|\tilde u\|_{L^\infty}\|b\|_{L^\infty}^2\|\tilde u_t\|_{L^2}\|\theta_t\|_{L^2}+C\|\tilde u\|_{L^\infty}^2\|b\|_{L^\infty}\|b_t\|_{L^2}\|\theta_t\|_{L^2}\\ &\leq&CA\|\theta_t\|_{L^2}^\frac32\|\nabla\theta_t\|_{L^2}^\frac12+CA^2\|\theta_t\|_{L^2}^\frac12 \|\nabla\theta_t\|_{L^2}^\frac12\\ &&+C\|\tilde u_t\|_{L^2}^\frac12\|\nabla\tilde u_t\|_{L^2}^\frac12\|\nabla\theta_t\|_{L^2}+CA^2\|\theta_t\|_{L^2}+C\| \mathrm{div}\,\tilde u\|_{L^\infty}\int\rho\theta_t^2 dx\\ &&+CA\|\nabla\tilde u_t\|_{L^2}\|\theta_t\|_{L^2}^\frac12\|\nabla\theta_t\|_{L^2}^\frac12+CA\|\nabla\tilde w_t\|_{L^2}\|\theta_t\|_{L^2}^\frac12\|\nabla\theta_t\|_{L^2}^\frac12+C\|E_t\|_{L^2}\|\theta_t\|_{L^3}, \end{eqnarray*}$

which yields

$\begin{equation} \int\theta_t^2dx+\int_0^T\int|\nabla\theta_t|^2 dx dt\leq C \end{equation}$

(2.15)

if $(A^4+A^3+A^2)T\leq 1$ . Here, we bound

$\begin{eqnarray*} CA\|\nabla\tilde u_t\|_{L^2}\|\theta_t\|_{L^2}^\frac12\|\nabla\theta_t\|_{L^2}^\frac12&\leq&\frac{1}{16}\|\nabla\theta_t\|_{L^2}^2 +CA^\frac43\|\nabla\tilde u_t\|_{L^2}^\frac43\|\theta_t\|_{L^2}^\frac23\\ &\leq&\frac{1}{16}\|\nabla\theta_t\|_{L^2}^2+CA^\frac43\|\nabla\tilde u_t\|_{L^2}^\frac43(1+\|\theta_t\|_{L^2}^2)\\ CA\|\nabla\tilde w_t\|_{L^2}\|\theta_t\|_{L^2}^\frac12\|\nabla\theta_t\|_{L^2}^\frac12&\leq&\frac{1}{16}\|\nabla\theta_t\|_{L^2}^2 +CA^\frac43\|\nabla\tilde w_t\|_{L^2}^\frac43\|\theta_t\|_{L^2}^\frac23\\ &\leq&\frac{1}{16}\|\nabla\theta_t\|_{L^2}^2+CA^\frac43\|\nabla\tilde w_t\|_{L^2}^\frac43(1+\|\theta_t\|_{L^2}^2),\\ C\|E_t\|_{L^2}\|\theta_t\|_{L^3}&\leq&C\|E_t\|_{L^2}\|\theta_t\|_{L^2}^\frac12\|\nabla\theta_t\|_{L^2}^\frac12\\ &\leq&\frac{1}{16}\|\nabla\theta_t\|_{L^2}^2+C\|E_t\|_{L^2}^\frac43(1+\|\theta_t\|_{L^2}^2). \end{eqnarray*}$

It follows from (1.18), (2.15) and (2.14) that

$\begin{equation} \|\theta\|_{L^\infty(0,T;H^2)}\leq CA^2+C. \end{equation}$

(2.16)

This completes the proof.

Lemma 2.4. Let $(\tilde u, \tilde w)\in\mathcal{A}$ be given. Then, the problem (1.21)–(1.23) has a unique solution $u$ satisfying

$\begin{equation} \|u\|_{L^\infty(0,T;H^2)}+\|u\|_{L^2(0,T;W^{2,6})}+\|u_t\|_{L^\infty(0,T;L^2)}+\|u_t\|_{L^2(0,T;H^1)}\leq C_1 \end{equation}$

(2.17)

for some small $0 < T\leq 1$ . Here, $C_1$ is a positive constant independent of $\epsilon$ and $\mathcal{A}$ .

Proof. Since Eq (1.21) is linear with regular $(\rho, \tilde u, \tilde w, \theta, E, b)$ , the existence and uniqueness are well- known, and we only need to establish (2.17).

Testing (1.21) by $u$ and using (1.11) and Lemmas 2.1–2.3, we see that

$\begin{eqnarray*} &&\frac12\frac{d}{dt}\int\rho|u|^2dx+\int((\mu+\mu_r)|\nabla u|^2+(\lambda+\mu-\mu_r)( \mathrm{div}\, u)^2)dx+\int|u\times b|^2dx\\ & = &\int\rho\theta \mathrm{div}\, u dx+2\mu_r\int u \mathrm{rot}\,\tilde wdx+\int(E\times b)u dx\\ &\leq&\|\rho\|_{L^\infty}\|\theta\|_{L^2}\| \mathrm{div}\, u\|_{L^2}+C\|u\|_{L^2}\| \mathrm{rot}\,\tilde w\|_{L^2}+\|E\|_{L^6}\|b\|_{L^3}\|u\|_{L^2}\\ &\leq&C\| \mathrm{div}\, u\|_{L^2}+C\|u\|_{L^2}+CA\|u\|_{L^2}\\ &\leq&\frac{\lambda+\mu-\mu_r}{2}\| \mathrm{div}\, u\|_{L^2}^2+C\|u\|_{L^2}+C+CA\|u\|_{L^2}, \end{eqnarray*}$

which gives

$\begin{equation} \int|u|^2dx+\int_0^T\int|\nabla u|^2dx dt\leq C \end{equation}$

(2.18)

if $A^2T\leq 1$ .

Testing (1.21) by $u_t$ and using Lemmas 2.1–2.3, we find that

$\begin{eqnarray*} &&\frac12\frac{d}{dt}\int[(\mu+\mu_r)|\nabla u|^2+(\lambda+\mu-\mu_r)( \mathrm{div}\, u)^2]dx+\int\rho|u_t|^2dx\\ & = &-\int\rho\tilde u\cdot\nabla u\cdot u_t dx-\int\nabla(\rho\theta)\cdot u_t dx+2\mu_r\int u_t \mathrm{rot}\,\tilde w dx+\int[(E+u\times b)\times b]u_t dx\\ &\leq&\frac12\int\rho|u_t|^2dx+C\|\tilde u\|_{L^\infty}^2\|\nabla u\|_{L^2}^2+C\|\rho\|_{L^\infty}^2\|\nabla\theta\|_{L^2}^2+C\|\theta\|_{L^3}^2\|\nabla\rho\|_{L^6}^2\\ &&+C\| \mathrm{rot}\,\tilde w\|_{L^2}^2+C\|E\|_{L^6}^2\|b\|_{L^3}^2+C\|u\|_{L^\infty}^2\|b\|_{L^4}^2\\ &\leq&\frac12\int\rho|u_t|^2dx+CA^2\|\nabla u\|_{L^2}^2+C+CA^2, \end{eqnarray*}$

which implies

$\begin{equation} \int|\nabla u|^2dx+\int_0^T\int|u_t|^2dx dt\leq C \end{equation}$

(2.19)

if $A^2T\leq 1$ .

Applying $\partial_t$ to (1.21), testing by $u_t$ , and using (1.11), Lemmas 2.1–2.3 and (2.19), we have

$\begin{eqnarray*} &&\frac12\frac{d}{dt}\int\rho|u_t|^2dx+\int[(\mu+\mu_r)|\nabla u_t|^2+(\lambda+\mu-\mu_r)( \mathrm{div}\, u_t)^2]dx+\int|u_t\times b|^2dx\\ & = &-\int\rho_tu_t^2dx-\int(\rho\tilde u)_t\cdot\nabla u\cdot u_t dx+\int(\rho\theta)_t \mathrm{div}\, u_t dx+2\mu_r\int\tilde w_t \mathrm{rot}\, u_t dx\\ &&+\int(E_t\times b+E\times b_t)u_t dx+\int[(u\times b_t)\times b+(u\times b)\times b_t]u_t dx\\ &\leq&\|\rho_t\|_{L^6}\|u_t\|_{L^3}\|u_t\|_{L^2}+\|\rho_t\|_{L^6}\|\tilde u\|_{L^\infty}\|\nabla u\|_{L^2}\|u_t\|_{L^3}\\ &&+\|\rho\|_{L^\infty}\|\tilde u_t\|_{L^3}\|\nabla u\|_{L^2}\|u_t\|_{L^6}+\|\rho_t\|_{L^6}\|\theta\|_{L^3}\| \mathrm{div}\, u_t\|_{L^2}\\ &&+\|\rho\|_{L^\infty}\|\theta_t\|_{L^2}\| \mathrm{div}\, u_t\|_{L^2}+2\mu_r\|\tilde w_t\|_{L^2}\| \mathrm{rot}\, u_t\|_{L^2}+\|E_t\|_{L^2}|\|b\|_{L^\infty}\|u_t\|_{L^2}\\ &&+\|E\|_{L^6}\|b_t\|_{L^3}\|u_t\|_{L^2}+\|u\|_{L^6}\|b_t\|_{L^3}\|b\|_{L^\infty}\|u_t\|_{L^2}\\ &\leq&C\|u_t\|_{L^2}\|u_t\|_{L^3}+CA\|u_t\|_{L^3}+C\|\tilde u_t\|_{L^3}\|u_t\|_{L^6}+C\| \mathrm{div}\, u_t\|_{L^2}\\ &&+CA\| \mathrm{rot}\, u_t\|_{L^2}+C\|E_t\|_{L^2}\|u_t\|_{L^2}+C\|b_t\|_{L^3}\|u_t\|_{L^2}\\ &\leq&\frac\mu2\int|\nabla u_t|^2dx+C\|u_t\|_{L^2}^2+CA^2+C\|\tilde u_t\|_{L^2}\|\nabla\tilde u_t\|_{L^2}+C\\ &&+C\|E_t\|_{L^2}\|u_t\|_{L^2}+C\|b_t\|_{L^3}\|u_t\|_{L^2}, \end{eqnarray*}$

which gives

$\begin{equation} \int|u_t|^2dx+\int_0^T\int|\nabla u_t|^2dx dt\leq C_1 \end{equation}$

(2.20)

if $A^2T\leq 1$ .

Since

$\tilde u(x,t) = u_0(x)+\int_0^t\partial_t\tilde u ds,$

and

$\begin{equation} \|\nabla\tilde u\|_{L^\infty(0,T;L^2)}\leq C+\int_0^T\|\nabla\partial_t\tilde u\|_{L^2}dt\leq C+C\sqrt TA\leq C \end{equation}$

(2.21)

if $A^2T\leq 1$ .

Similarly, we have

$\begin{equation} \|\nabla\tilde w\|_{L^\infty(0,T;L^2)}\leq C \end{equation}$

(2.22)

if $A^2T\leq 1$ .

We rewrite (1.21) as

$-\mu\Delta u-(\lambda+\mu)\nabla \mathrm{div}\, u = f: = 2\mu_r \mathrm{rot}\,\tilde w+(E+u\times b)\times b-\rho\partial_tu-\rho\tilde u\cdot\nabla u-\nabla(\rho\theta).$

By the $H^2$ -theory of the elliptic system, we get

$\begin{eqnarray*} \|u\|_{H^2}&\leq&C\|f\|_{L^2}\\ &\leq&C\|\nabla\tilde w\|_{L^2}+C\|E\|_{L^6}\|b\|_{L^3}+C\|u\|_{L^6}\|b\|_{L^6}^2+C\|\rho\partial_tu\|_{L^2}\\ &&+C\|\tilde u\|_{L^6}\|\nabla u\|_{L^3}+C\|\rho\nabla\theta\|_{L^2}+C\|\theta\|_{L^3}\|\nabla\rho\|_{L^6}\\ &\leq&C+C\|\nabla u\|_{L^3}, \end{eqnarray*}$

which yields

$\begin{equation} \|u\|_{L^\infty(0,T;H^2)}\leq C_1. \end{equation}$

(2.23)

Similarly, by the $W^{2, 6}$ -theory of the elliptic system, we have

$\begin{eqnarray*} \|u\|_{W^{2,6}}&\leq&C\|f\|_{L^6}\\ &\leq&C\| \mathrm{rot}\,\tilde w\|_{L^6}+C\|E\|_{L^6}\|b\|_{L^\infty}+C\|u\|_{L^\infty}\|b\|_{L^\infty}\|b\|_{L^6}\\ &&+C\|u_t\|_{L^6}+C\|\tilde u\|_{L^6}\|\nabla u\|_{L^\infty}+C\|\nabla\theta\|_{L^6}+C\|\nabla\rho\|_{L^6}\\ &\leq&C+C\|\nabla u\|_{L^\infty}+CA^2+C\|u_t\|_{L^6}\\ &\leq&C+C\|\nabla u\|_{L^2}^\frac14\|u\|_{W^{2,6}}^\frac34+CA^2+C\|\nabla u_t\|_{L^2}, \end{eqnarray*}$

whence

$\|u\|_{W^{2,6}}\leq C+CA^2+C\|\nabla u_t\|_{L^2},$

which yields

$\begin{equation} \|u\|_{L^2(0,T;W^{2,6})}\leq C_1 \end{equation}$

(2.24)

if $A^4T\leq 1$ .

This completes the proof.

Similarly to Lemma 2.4, we have the following.

Lemma 2.5. Let $(\tilde u, \tilde w)\in\mathcal{A}$ be given. Then, the problem (1.24)–(1.26) has a unique solution $w$ satisfying (2.17) with $u: = w$ .

Proof. Since the proof is very similar to that of Lemma 2.4, we omit the details here.

Due to the above Lemmas 2.1–2.5, we can take $A: = C_1$ , and thus $F$ maps $\mathcal{A}$ into $\mathcal{A}$ . The following lemma tells us that $F$ is a contraction mapping in the sense of weaker norm.

Lemma 2.6. There is a constant $0 < \delta < 1$ such that for any $\tilde u_i\ (i = 1, 2)$ ,

$\begin{equation} \|F(\tilde u_1,\tilde w_1)-F(\tilde u_2,\tilde w_2)\|_{L^2(0,T;H^1)}\leq\delta\|(\tilde u_1-\tilde u_2,\tilde w_1-\tilde w_2)\|_{L^2(0,T;H^1)} \end{equation}$

(2.25)

for some small $0 < T\leq 1$ .

Proof. Suppose $(\rho_i, u_i, w_i, \theta_i, E_i, b_i)\ (i = 1, 2)$ are the solutions to the problem (1.11)–(1.26) corresponding to $\tilde u_i\ (i = 1, 2)$ . Define

$\begin{eqnarray*} &&\rho: = \rho_1-\rho_2,u: = u_1-u_2,w: = w_1-w_2,\theta: = \theta_1-\theta_2,\\ &&E: = E_1-E_2,b: = b_1-b_2,\tilde u: = \tilde u_1-\tilde u_2,\tilde w: = \tilde w_1-\tilde w_2. \end{eqnarray*}$

Then, we have

$\begin{eqnarray} &&\rho_t+ \mathrm{div}\,(\rho\tilde u_1) = - \mathrm{div}\,(\rho_2\tilde u), \end{eqnarray}$

(2.26)

$\begin{eqnarray} &&\epsilon\partial_tE- \mathrm{rot}\, b+E+\tilde u\times b_1+\tilde u_2\times b = 0, \end{eqnarray}$

(2.27)

$\begin{eqnarray} &&\partial_tb+ \mathrm{rot}\, E = 0, \mathrm{div}\, b = 0, \end{eqnarray}$

(2.28)

$\begin{eqnarray} &&\rho_1\partial_t\theta+\rho_1\tilde u_1\cdot\nabla\theta-\Delta\theta+\rho_1\theta_1 \mathrm{div}\,\tilde u_1-\rho_2\theta_2 \mathrm{div}\,\tilde u_2+\rho\partial_t\theta_2+(\rho_1\tilde u_1-\rho_2\tilde u_2)\nabla\theta_2\\ &&\qquad = Q_1-Q_2, \end{eqnarray}$

(2.29)

$\begin{eqnarray} &&\rho_1\partial_tu+\rho_1\tilde u_1\cdot\nabla u+\nabla(\rho_1\theta_1-\rho_2\theta_2)-(\mu+\mu_r)\Delta u-(\lambda+\mu-\mu_r)\nabla \mathrm{div}\, u\\ &&\qquad+\rho\partial_tu_2+(\rho_1\tilde u_1-\rho_2\tilde u_2)\nabla u_2\\ &&\qquad = 2\mu_r \mathrm{rot}\,\tilde w+(E_1+u_1\times b_1)\times b_1-(E_2+u_2\times b_2)\times b_2, \end{eqnarray}$

(2.30)

with

$\begin{eqnarray*} Q_i:& = &\frac\mu2(\nabla\tilde u_i+\nabla\tilde u_i^T):(\nabla\tilde u_i+\nabla\tilde u_i^T)+\lambda( \mathrm{div}\,\tilde u_i)^2+4\mu_r\left|\frac12 \mathrm{rot}\,\tilde u_i-\tilde w_i\right|^2+c_0( \mathrm{div}\,\tilde w_i)^2\\ &&+(c_a+c_d)\nabla\tilde w_i:\nabla\tilde w_i+(c_d-c_a)\nabla\tilde w_i:\nabla\tilde w_i^T+|E_i+\tilde u_i\times b_i|^2\ \ (i = 1,2) \end{eqnarray*}$

$\begin{eqnarray} &&\rho_1\partial_tw+\rho_1\tilde u_1\cdot\nabla w-(c_a+c_d)\Delta w-(c_0+c_d-c_a)\nabla \mathrm{div}\, w+4\mu_rw\\ & = &2\mu_r \mathrm{rot}\, u-\rho\partial_tw_2-(\rho_1\tilde u_1-\rho_2\tilde u_2)\cdot\nabla w_2. \end{eqnarray}$

(2.31)

Testing (2.26) by $\rho$ , we see that

$\begin{eqnarray} \frac{d}{dt}\int\rho^2dx&\leq&C\|\nabla\tilde u_1\|_{L^\infty}\int\rho^2dx+C(\|\rho_2\|_{L^\infty}\|\nabla\tilde u\|_{L^2}+\|\tilde u\|_{L^6}\|\nabla\rho_2\|_{L^3}\|\rho\|_{L^2}\\ &\leq&C\|\tilde u_1\|_{W^{2,6}}\int\rho^2dx+C\|\nabla\tilde u\|_{L^2}\|\rho\|_{L^2}\\ &\leq&\eta_1\|\nabla\tilde u\|_{L^2}^2+C(1+\|\tilde u_1\|_{W^{2,6}})\|\rho\|_{L^2}^2 \end{eqnarray}$

(2.32)

for any $0 < \eta_1 < 1$ .

Testing (2.27) and (2.28) by $E$ and $b$ and summing up the result, we find that

$\begin{eqnarray} &&\frac12\frac{d}{dt}\int(\epsilon E^2+b^2)dx+\int E^2dx\\ & = &\int(b_1\times\tilde u+b\times\tilde u_2)E dx\\ &\leq&\frac14\int E^2dx+C\|b_1\|_{L^\infty}^2\|\tilde u\|_{L^2}^2+C\|\tilde u_2\|_{L^\infty}^2\|b\|_{L^2}^2\\ &\leq&\frac14\int E^2dx+C\|\tilde u\|_{L^2}^2+C\|b\|_{L^2}^2. \end{eqnarray}$

(2.33)

Testing (2.29) by $\theta$ and using $\partial_t\rho_1+ \mathrm{div}\, (\rho_1\tilde u_1) = 0$ , we infer that

$\begin{eqnarray} &&\frac12\frac{d}{dt}\int\rho_1\theta^2dx+\int|\nabla\theta|^2dx\\ & = &\int[\rho\theta_1 \mathrm{div}\,\tilde u_1+\rho_2\theta \mathrm{div}\,\tilde u_1+\rho_2\theta_2 \mathrm{div}\,\tilde u+\rho\partial_t\theta_2+(\rho\tilde u_1+\rho_2\tilde u)\nabla\theta_2]\theta dx+\int(Q_1-Q_2)\theta dx\\ &\leq&\|\rho\|_{L^2}\|\theta_1\|_{L^\infty}\| \mathrm{div}\,\tilde u_1\|_{L^\infty}\|\theta\|_{L^2}+\|\rho_2\|_{L^\infty}\| \mathrm{div}\,\tilde u_1\|_{L^\infty}\|\theta\|_{L^2}^2\\ &&+\|\rho_2\|_{L^\infty}\|\theta_2\|_{L^\infty}\| \mathrm{div}\,\tilde u\|_{L^2}\|\theta\|_{L^2}+\|\rho\|_{L^2}\|\partial_t\theta_2\|_{L^3}\|\theta\|_{L^6}\\ &&+\|\rho\|_{L^2}\|\tilde u_1\|_{L^\infty}\|\nabla\theta_2\|_{L^6}\|\theta\|_{L^3}+\|\rho_2\|_{L^\infty}\|\tilde u\|_{L^2}\|\nabla\theta_2\|_{L^6}\|\theta\|_{L^3}\\ &&+C(\|\tilde w_1\|_{W^{1,6}}+\|\tilde w_2\|_{W^{1,6}})\|\nabla\tilde w\|_{L^2}\|\theta\|_{L^3}\\ &&+C(\|\nabla\tilde u_1\|_{L^6}+\|\nabla\tilde u_2\|_{L^6})\|\nabla\tilde u\|_{L^2}\|\theta\|_{L^3}+C(\|E_1\|_{L^6}+\|E_2\|_{L^6})\|E\|_{L^2}\|\theta\|_{L^3}\\ &&+C(\|\tilde u_1\|_{L^\infty}\|b_1\|_{L^\infty}+\|\tilde u_2\|_{L^\infty}\|b_2\|_{L^\infty})(\|\tilde u\|_{L^2}\|b_1\|_{L^\infty}+\|\tilde u_2\|_{L^\infty}\|b\|_{L^2})\|\theta\|_{L^2}\\ &\leq&C\| \mathrm{div}\,\tilde u_1\|_{L^\infty}(\|\rho\|_{L^2}^2+\|\theta\|_{L^2}^2)+\eta_1\|\nabla\tilde u\|_{L^2}^2+\eta_1\|\nabla\tilde w\|_{L^2}^2+C\|\theta\|_{L^2}^2+\frac{1}{24}\|\nabla\theta\|_{L^2}^2\\ &&+C\|\partial_t\theta_2\|_{L^3}^2\|\rho\|_{L^2}^2+C\|\rho\|_{L^2}^2+\eta_1\|\tilde u\|_{L^2}^2+\eta_2\|E\|_{L^2}^2+\eta_1\|b\|_{L^2}^2 \end{eqnarray}$

(2.34)

for any $0 < \eta_1, \eta_2 < 1$ .

Testing (2.30) by $u$ and using $\partial_t\rho_1+ \mathrm{div}\, (\rho_1\tilde u_1) = 0$ , we deduce that

$\begin{eqnarray} &&\frac12\frac{d}{dt}\int\rho_1|u|^2dx+\int[(\mu+\mu_r)|\nabla u|^2+(\lambda+\mu-\mu_r)( \mathrm{div}\, u)^2]dx+\int(u\times b_1)^2dx\\ & = &-\int[\rho\partial_tu_2+(\rho\tilde u_1+\rho_2\tilde u)\cdot\nabla u_2]u dx+\int(\rho_1\theta_1-\rho_2\theta_2) \mathrm{div}\, u dx+2\mu_r\int \mathrm{rot}\,\tilde w\cdot u dx\\ &&+\int[(E_1+u_1\times b_1)\times b_1-(E_2+u_2\times b_2)\times b_2]u dx\\ &\leq&\|\rho\|_{L^2}\|\partial_tu_2\|_{L^3}\|u\|_{L^6}+(\|\rho\|_{L^2}\|\tilde u_1\|_{L^\infty}+\|\rho_2\|_{L^\infty}\|\tilde u\|_{L^2})\|\nabla u_2\|_{L^6}\|u\|_{L^3}\\ &&+(\|\rho\|_{L^2}\|\theta_1\|_{L^\infty}+\|\rho_2\|_{L^\infty}\|\theta\|_{L^2})\| \mathrm{div}\, u\|_{L^2}+C\| \mathrm{rot}\,\tilde w\|_{L^2}\|u\|_{L^2}\\ &&+\|E\|_{L^2}\|b_1\|_{L^\infty}\|u\|_{L^2}+\|b\|_{L^2}\|E_2\|_{L^6}\|u\|_{L^3}\\ &&+(\|u_2\|_{L^\infty}\|b\|_{L^2}\|b_1\|_{L^\infty} +\|u_2\|_{L^\infty}\|b_2\|_{L^\infty}\|b\|_{L^2})\|u\|_{L^2}\\ &\leq&\frac{\mu}{16}\|\nabla u\|_{L^2}^2+C\|\partial_tu_2\|_{L^3}^2\|\rho\|_{L^2}^2+C(\|\rho\|_{L^2}+\|\tilde u\|_{L^2})\|u\|_{L^3}\\ &&+C\|\rho\|_{L^2}^2+C\|\theta\|_{L^2}^2+C\|E\|_{L^2}\|u\|_{L^2}+C\|b\|_{L^2}\|u\|_{L^3}+C\|b\|_{L^2}\|u\|_{L^2} +C\| \mathrm{rot}\,\tilde w\|_{L^2}\|u\|_{L^2}\\ &\leq&\frac\mu8\|\nabla u\|_{L^2}^2+C\|\partial_tu_2\|_{L^3}^2\|\rho\|_{L^2}^2+\eta_1\|\tilde u\|_{L^2}^2+C\|u\|_{L^2}^2+C\|\rho\|_{L^2}^2\\ &&+C\|\theta\|_{L^2}^2+C\eta_2\|E\|_{L^2}^2+\eta_1\|b\|_{L^2}^2+\eta_1\| \mathrm{rot}\,\tilde w\|_{L^2}^2 \end{eqnarray}$

(2.35)

for any $0 < \eta_1, \eta_2 < 1$ .

Testing (2.31) by $w$ and using $\partial_t\rho_1+ \mathrm{div}\, (\rho_1\tilde u_1) = 0$ , we compute

$\begin{eqnarray} &&\frac12\frac{d}{dt}\int\rho_1|w|^2dx+(c_a+c_d)\int|\nabla w|^2dx+(c_0+c_d-c_a)\int( \mathrm{div}\, w)^2 dx+4\mu_r\int|w|^2 dx\\ & = &2\mu_r\int w \mathrm{rot}\, u dx-\int[\rho\partial_tw_2+(\rho\tilde u_1+\rho_2\tilde u)\cdot\nabla w_2]w dx\\ &\leq&C\|w\|_{L^2}^2+\frac\mu8\|\nabla u\|_{L^2}^2+\|\rho\|_{L^2}\|\partial_tw_2\|_{L^3}\|w\|_{L^6}\\ &&+(\|\rho\|_{L^2}\|\tilde u_1\|_{L^\infty}+\|\rho_2\|_{L^\infty}\|\tilde u\|_{L^2})\|\nabla w_2\|_{L^6}\|w\|_{L^3}\\ &\leq&C\|w\|_{L^2}^2+\frac\mu8\|\nabla u\|_{L^2}^2+\frac{c_a+c_d}{8}\|\nabla w\|_{L^2}^2+C\|\partial_tw_2\|_{L^3}^2\|\rho\|_{L^2}^2\\ &&+C\|\rho\|_{L^2}^2+\eta_1\|\tilde u\|_{L^2}^2. \end{eqnarray}$

(2.36)

Taking (2.32) $+\eta_1\times$ (2.33) $+$ (2.34) $+$ (2.35) $+$ (2.36), taking $\eta_2 < < \eta_1$ and using the Gronwall inequality, we arrive at (2.25) for small $0 < T\leq 1$ .

This completes the proof.

Proof of Theorem 1.1.

By Lemmas 2.1–2.6 and the Banach fixed point theorem, we finish the proof.

Acknowledgments

J. Fan is partially supported by NSFC (No. 11971234).

Conflict of interest

The authors declare no conflict of interest.

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Magnetohydrodynamics approximation of the compressible full magneto- micropolar system

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Magnetohydrodynamics approximation of the compressible full magneto- micropolar system

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