Global Strong Solutions of the 2D Density-Dependent Incompressible Magnetic Bénard Problem

Abstract

In this paper, we prove global existence of strong solutions to the 2D density-dependent incompressible magnetic Bénard problem in a bounded domain.

1 Introduction

Magnetohydrodynamics (MHD) studies the interaction of electromagnetic fields and conducting fluids. In this paper, we consider the following 2D density-dependent incompressible magnetic Bénard system [1]:

$$\begin{aligned}&\partial _t\rho +\mathrm {div}\,(\rho u)=0, \end{aligned}$$

(1.1)

$$\begin{aligned}&\partial _t(\rho u)+\mathrm {div}\,(\rho u\otimes u)+\nabla \left( \pi +\frac{1}{2}|b|^2\right) -\mu \Delta u=(b\cdot \nabla )b+\rho \theta e_2, \end{aligned}$$

(1.2)

$$\begin{aligned}&\partial _tb+u\cdot \nabla b-b\cdot \nabla u=\eta \Delta b, \end{aligned}$$

(1.3)

$$\begin{aligned}&\rho \partial _t\theta +\rho u\cdot \nabla \theta =\rho u e_2+k\Delta \theta , \end{aligned}$$

(1.4)

$$\begin{aligned}&\mathrm {div}\,u=\mathrm {div}\,b=0\ \ \mathrm {in}\ \ \Omega \times (0,\infty ), \end{aligned}$$

(1.5)

$$\begin{aligned}&u=0,b\cdot n=0,\mathrm {rot}\,b=0,\theta =0\ \ \mathrm {on}\ \ \partial \Omega \times (0,\infty ), \end{aligned}$$

(1.6)

$$\begin{aligned}&(\rho ,\rho u,b,\rho \theta )(\cdot ,0)=(\rho _0,\rho _0u_0,b_0,\rho _0\theta _0)(\cdot )\ \ \mathrm {in}\ \ \Omega \subset \mathbb {R}^2. \end{aligned}$$

(1.7)

Here $\rho$ denotes the density, u the velocity field, $\pi$ the pressure, b the magnetic field, and $\theta$ the temperature, respectively. $\mu$ is the viscosity coefficient and $\eta$ is the resistivity coefficient. k is the heat conductivity coefficient. $\Omega$ is a bounded domain in $\mathbb {R}^2$ with smooth boundary $\partial \Omega$, n is the unit outward normal vector to the boundary $\partial \Omega$. $e_2:=(0,1)^t$. In (1.6), we denote $\mathrm {rot}\,u:=\partial _1u_2-\partial _2u_1$ for the 2D vector $u:=(u_1,u_2)^t$ and $\mathrm {rot}\,\phi :=(\partial _2\phi ,-\partial _1\phi )^t$ for scalar $\phi$.

Wu [2] shows the local well-posedness of strong solutions to the problem (1.1)-(1.7) with $\inf \rho _0>0$. When $\eta >0$ and $\theta =0$, Huang and Wang [3] (also see [4]) prove the global well-posedness of the strong solutions with the following compatibility condition: $\exists (\nabla \pi _0,g)\in L^2$ such that

$$\begin{aligned} \nabla \left( \pi _0+\frac{1}{2}|b_0|^2\right) -\mu \Delta u_0-b_0\cdot \nabla b_0=\sqrt{\rho _0}g. \end{aligned}$$

(1.8)

Fan-Li-Nakamura [5] showed a regularity criterion. Fan-Zhou [7] proved the uniform-in-$\mu (\eta )$ local well-posedness of smooth solutions when $\Omega :=\mathbb {R}^d$. When $\rho =1$ and $b=0$, Lai-Pan-Zhao [8, 9] showed the global well-posedness of smooth solutions, Jin-Fan-Nakamura-Zhou [10] studied the partial vanishing viscosity limit.

When $b=0$ and $\theta =0$, (1.1), (1.2) and (1.5) is the density-dependent incompressible Navier–Stokes equations. Li [11] and Danchin-Mucha [12] showed the local/global well-posedness of strong/weak solutions without (1.8). For other studies of 2D magnetic Bénard problem, we refer to [6, 13,14,15,16,17].

The aim of this paper is to prove the global well-posedness of strong solutions to the problem without (1.8). We will prove

Theorem 1.1

Let $0\le \rho _0\in W^{1,q}\ (2<q<\infty ), u_0, \theta _0\in H_0^1, b_0\in H^1$ with $\mathrm {div}\,u_0=\mathrm {div}\,b_0=0$ in $\Omega$ and $b_0\cdot n=\mathrm {rot}\,b_0=0$ on $\partial \Omega$. Then, the problem (1.1)-(1.7) has a unique strong solution $(\rho ,u,b,\theta )$ satisfying

$$\begin{aligned} \begin{array}{l} \nabla \rho \in L^\infty (0,T;L^q), \rho _t\in L^4(0,T;L^q),\\ u\in L^\infty (0,T;H^1)\cap L^2(0,T;H^2), \sqrt{t} u\in L^\infty (0,T;H^2)\cap L^2(0,T;W^{2,6}),\\ \sqrt{t}\sqrt{\rho }u_t\in L^\infty (0,T;L^2), \sqrt{t} u_t\in L^2(0,T;H^1),\\ b\in L^\infty (0,T;H^1)\cap L^2(0,T;H^2), \sqrt{t} b\in L^\infty (0,T;H^2)\cap L^2(0,T;W^{2,6}),\\ \sqrt{t} b_t\in L^\infty (0,T;L^2)\cap L^2(0,T;H^1),\\ \theta \in L^\infty (0,T;H^1)\cap L^2(0,T;H^2), \sqrt{t}\theta \in L^\infty (0,T;H^2)\cap L^2(0,T;W^{2,6}),\\ \sqrt{t}\theta _t\in L^\infty (0,T;L^2)\cap L^2(0,T;H^1) \end{array} \end{aligned}$$

(1.9)

for any given $0<T\le \infty$.

In the following proofs, we will use the following two lemmas.

Lemma 1.2

([3]). Assume $\Omega$ is a bounded smooth domain in $\mathbb {R}^2$ and $f\in L^2(s,t;H^1(\Omega ))\cap L^2(s,t;W^{1,q}(\Omega ))$, with $2<q<\infty$ and $0\le s<t\le \infty$. Then it holds that

$$\begin{aligned} \Vert f\Vert _{L^2(s,t;L^\infty (\Omega ))}^2\le C+C\Vert f\Vert _{L^2(s,t;H^1(\Omega ))}^2\ln ^+\Vert f\Vert _{L^2(s,t;W^{1,q}(\Omega ))}, \end{aligned}$$

(1.10)

with some constant C depending only on q and $\Omega$, and independent of s and t.

Lemma 1.3

([3]). Assume that $b\in H^1$ is a weak solution of the Poisson equations

$$\begin{aligned} \left\{ \begin{array}{l} \Delta b=G\ \ \text{ in }\ \ \Omega ,\\ b\cdot n=0, \mathrm {rot}\,b=0\ \ \text{ on }\ \ \partial \Omega . \end{array} \right. \end{aligned}$$

(1.11)

with $G\in L^q,1<q<\infty$. Then it holds that

$$\begin{aligned} \Vert b\Vert _{W^{2,q}}\le C\Vert G\Vert _{L^q}+C\Vert b\Vert _{H^1} \end{aligned}$$

(1.12)

with some constant C depending on q and $\Omega$.

2 Proof of Theorem 1.1

This section is devoted to the proof of Theorem 1.1. Since the local strong solutions to the problem (1.1)–(1.7) was established in [2], we will give a new proof in Appendix when $\inf \rho _0>0$. We only need to show a priori estimates (1.9). For simplicity, we will take $\mu =\eta =k=1$.

First, it follows from (1.1) and (1.5) that

$$\begin{aligned} 0\le \rho \le C. \end{aligned}$$

(2.1)

Testing (1.2) by u and using (1.1) and (1.5), we see that

$$\begin{aligned} \frac{1}{2}\frac{\mathrm {d}}{\mathrm {d}t}\int \rho |u|^2\mathrm {d}x+\int |\nabla u|^2\mathrm {d}x=\int (b\cdot \nabla )b\cdot u\mathrm {d}x+\int \rho \theta e_2u\mathrm {d}x. \end{aligned}$$

(2.2)

Testing (1.3) by b and using (1.5), we find that

$$\begin{aligned} \frac{1}{2}\frac{\mathrm {d}}{\mathrm {d}t}\int |b|^2\mathrm {d}x+\int (\mathrm {rot}\,b)^2\mathrm {d}x=\int (b\cdot \nabla )u\cdot b\mathrm {d}x. \end{aligned}$$

(2.3)

Testing (1.4) by $\theta$ and using (1.1) and (1.5), we get

$$\begin{aligned} \frac{1}{2}\frac{\mathrm {d}}{\mathrm {d}t}\int \rho \theta ^2\mathrm {d}x+\int |\nabla \theta |^2\mathrm {d}x=\int \rho ue_2\theta \mathrm {d}x. \end{aligned}$$

(2.4)

Summing up (2.2), (2.3) and (2.4), we get

$$\begin{aligned}&\frac{1}{2}\frac{\mathrm {d}}{\mathrm {d}t}\int (\rho |u|^2+|b|^2+\rho \theta ^2)\mathrm {d}x+\int |\nabla u|^2\mathrm {d}x+\int (\mathrm {rot}\,b)^2\mathrm {d}x+\int |\nabla \theta |^2\mathrm {d}x\\&\quad \le C\Vert \sqrt{\rho }\theta \Vert _{L^2}\Vert \sqrt{\rho }u\Vert _{L^2}, \end{aligned}$$

which gives

$$\begin{aligned}&\int (\rho |u|^2+|b|^2+\rho \theta ^2)\mathrm {d}x+\int _0^t\int |\nabla u|^2\mathrm {d}x\mathrm {d}s+\int _0^t\int (\mathrm {rot}\,b)^2\mathrm {d}x\mathrm {d}s\nonumber \\&\quad +\int _0^t\int |\nabla \theta |^2\mathrm {d}x\mathrm {d}s \le C\int (\rho _0|u_0|^2+|b_0|^2+\rho _0\theta _0^2)\mathrm {d}x. \end{aligned}$$

(2.5)

Testing (1.2) by $u_t$, using (1.1), (1.5) and (2.1), we derive that

$$\begin{aligned}&\frac{1}{2}\frac{\mathrm {d}}{\mathrm {d}t}\int |\nabla u|^2\mathrm {d}x+\int \rho |u_t|^2\mathrm {d}x\nonumber \\&\quad =-\int \rho u\cdot \nabla u\cdot u_t\mathrm {d}x+\int b\cdot \nabla b\cdot u_t\mathrm {d}x+\int \rho \theta e_2u_t\mathrm {d}x\nonumber \\&\quad \le \Vert \sqrt{\rho }u_t\Vert _{L^2}\Vert \sqrt{\rho }\Vert _{L^\infty }\Vert u\Vert _{L^\infty }\Vert \nabla u\Vert _{L^2}-\int b\otimes b:\nabla u_t\mathrm {d}x+\int \rho \theta e_2u_t\mathrm {d}x\nonumber \\&\quad \le \frac{1}{8}\int \rho |u_t|^2\mathrm {d}x+C\Vert u\Vert _{L^\infty }^2\Vert \nabla u\Vert _{L^2}^2-\frac{\mathrm {d}}{\mathrm {d}t}\int b\otimes b:\nabla u\mathrm {d}x\nonumber \\&\qquad +\int \partial _t(b\otimes b):\nabla u\mathrm {d}x+C\int \rho \theta ^2\mathrm {d}x\nonumber \\&\quad \le \frac{1}{8}\int \rho |u_t|^2\mathrm {d}x+\frac{1}{8}\Vert b_t\Vert _{L^2}^2+C\Vert u\Vert _{L^\infty }^2\Vert \nabla u\Vert _{L^2}^2-\frac{\mathrm {d}}{\mathrm {d}t}\int b\otimes b:\nabla u\mathrm {d}x\nonumber \\&\qquad +C\Vert b\Vert _{L^\infty }^2\Vert \nabla u\Vert _{L^2}^2+C\int \rho \theta ^2\mathrm {d}x. \end{aligned}$$

(2.6)

Similarly, testing (1.3) by $b_t$, we get

$$\begin{aligned}&\frac{1}{2}\frac{\mathrm {d}}{\mathrm {d}t}\int (\mathrm {rot}\,b)^2\mathrm {d}x+\int |b_t|^2\mathrm {d}x=-\int u\cdot \nabla b\cdot b_t\mathrm {d}x+\int b\cdot \nabla u\cdot b_t\mathrm {d}x\nonumber \\&\quad \le \Vert u\Vert _{L^\infty }\Vert \nabla b\Vert _{L^2}\Vert b_t\Vert _{L^2}+\Vert b\Vert _{L^\infty }\Vert \nabla u\Vert _{L^2}\Vert b_t\Vert _{L^2}\nonumber \\&\quad \le \frac{1}{8}\Vert b_t\Vert _{L^2}^2+C\Vert u\Vert _{L^\infty }^2\Vert \nabla b\Vert _{L^2}^2+C\Vert b\Vert _{L^\infty }^2\Vert \nabla u\Vert _{L^2}^2. \end{aligned}$$

(2.7)

On the other hand, we have

$$\begin{aligned} \left| \int b\otimes b:\nabla u\mathrm {d}x\right|\le & {} \Vert b\Vert _{L^4}^2\Vert \nabla u\Vert _{L^2}\le C\Vert b\Vert _{L^2}\Vert \nabla b\Vert _{L^2}\Vert \nabla u\Vert _{L^2}\nonumber \\\le & {} C\Vert \nabla b\Vert _{L^2}\Vert \nabla u\Vert _{L^2}\le \frac{1}{8}\Vert \nabla u\Vert _{L^2}^2+C_1\Vert \nabla b\Vert _{L^2}^2. \end{aligned}$$

(2.8)

Here, we have used the estimate:

$$\begin{aligned} \Vert b\Vert _{H^1}\le C\Vert \nabla b\Vert _{L^2}\le C\Vert \mathrm {rot}\,b\Vert _{L^2}. \end{aligned}$$

(2.9)

Multiplying (2.7) by $2C_1M+2$ for some positive constant $M>0$, adding it to (2.6) and integrating with respect to time, then for every $0\le s<t<T$,

$$\begin{aligned}&\int (|\nabla u(t)|^2+|\mathrm {rot}\,b(t)|^2)\mathrm {d}x+\int _s^t(\Vert \sqrt{\rho }u_t\Vert _{L^2}^2+\Vert b_t\Vert _{L^2}^2)\mathrm {d}\tau \nonumber \\&\quad \le C\int (|\nabla u(s)|^2+|\mathrm {rot}\,b(s)|^2)\mathrm {d}x\exp \left\{ C\int _s^t(\Vert u\Vert _{L^\infty }^2+\Vert b\Vert _{L^\infty }^2)\mathrm {d}\tau \right\} +C.\nonumber \\ \end{aligned}$$

(2.10)

Denote

$$\begin{aligned} \Psi (t):=e+\sup \limits _{0\le \tau \le t}(\Vert u(\tau )\Vert _{H^1}^2+\Vert b(\tau )\Vert _{H^1}^2)+\int _0^t(\Vert \sqrt{\rho }u_t\Vert _{L^2}^2+\Vert b_t\Vert _{L^2}^2)\mathrm {d}\tau \end{aligned}$$

(2.11)

Then, (2.5) and (2.10) give that

$$\begin{aligned} \Psi (t)\le C\psi (s)\exp \left\{ C\int _s^t(\Vert u\Vert _{L^\infty }^2+\Vert b\Vert _{L^\infty }^2)\mathrm {d}\tau \right\} . \end{aligned}$$

(2.12)

Using Lemma 1.2, we have

$$\begin{aligned}&\Vert u\Vert _{L^2(s,t;L^\infty )}^2{+}\Vert b\Vert _{L^2(s,t;L^\infty )}^2\nonumber \\&\quad \le C{+}C(\Vert u\Vert _{L^2(s,t;H^1)}^2+\Vert b\Vert _{L^2(s,t;H^1)}^2)(\ln ^{+}\Vert u\Vert _{L^2(s,t;W^{1,4})}+\ln ^+\Vert b\Vert _{L^2(s,t;W^{1,4})}).\nonumber \\ \end{aligned}$$

(2.13)

On the other hand, (1.2) can be rewritten as

$$\begin{aligned} -\Delta u+\nabla \left( \pi +\frac{1}{2}|b|^2\right) =f:=b\cdot \nabla b+\rho \theta e_2-\rho u_t-\rho u\cdot \nabla u. \end{aligned}$$

(2.14)

By the $W^{2,q}$-theory of Stokes system, we observe that

$$\begin{aligned} \Vert u\Vert _{W^{1,4}}\le & {} C\Vert u\Vert _{W^{2,\frac{4}{3}}}+\Vert u\Vert _{H^1}\\\le & {} C\Vert f\Vert _{L^\frac{4}{3}}+C\Vert u\Vert _{H^1}\\\le & {} C\Vert b\cdot \nabla b+\rho \theta e_2-\rho u_t-\rho u\cdot \nabla u\Vert _{L^\frac{4}{3}}+C\Vert u\Vert _{H^1}, \end{aligned}$$

which gives

$$\begin{aligned}&\Vert u\Vert _{L^2(s,t;W^{1,4})}\nonumber \\&\quad \le C\Vert u\Vert _{L^2(s,t;H^1)}+C\Vert \sqrt{\rho }u_t\Vert _{L^2(s,t;L^2)}+C\Vert u\Vert _{L^2(s,t;H^1)}\Vert \nabla u\Vert _{L^\infty (s,t;L^2)}\nonumber \\&\qquad +C\Vert b\Vert _{L^2(s,t;H^1)}\Vert \nabla b\Vert _{L^\infty (s,t;L^2)}. \end{aligned}$$

(2.15)

Similarly, we have

$$\begin{aligned} \Vert b\Vert _{L^2(s,t;W^{1,4})}\le & {} C\Vert b\Vert _{L^2(s,t;H^1)}{+}C\Vert b_t\Vert _{L^2(s,t;L^2)}\nonumber \\&{+}C\Vert u\Vert _{L^2(s,t;H^1)}\Vert \nabla b\Vert _{L^\infty (s,t;L^2)}+C\Vert b\Vert _{L^2(s,t;H^1)}\Vert \nabla u\Vert _{L^\infty (s,t;L^2)}.\nonumber \\ \end{aligned}$$

(2.16)

Summing up (2.15) and (2.16), we have

$$\begin{aligned}&\Vert u\Vert _{L^2(s,t;W^{1,4})}^2+\Vert b\Vert _{L^2(s,t;W^{1,4})}^2\nonumber \\&\quad \le C+C(\Vert u\Vert _{L^2(s,t;H^1)}^2+\Vert b\Vert _{L^2(s,t;H^1)}^2)\ln (C(M,T)\psi (t)). \end{aligned}$$

(2.17)

Inserting (2.17) into (2.12), it arrives at

$$\begin{aligned} \Psi (T)\le C\Psi (s)[C(M,T)\Psi (T)]^{C_2(\Vert u\Vert _{L^2(s,t;H^1)}^2+\Vert b\Vert _{L^2(s,t;H^1)}^2)}. \end{aligned}$$

(2.18)

One can choose s chose enough to T, such that

$$\begin{aligned}C_2(\Vert u\Vert _{L^2(s,T;H^1)}^2+\Vert b\Vert _{L^2(s,T;H^1)}^2)\le \frac{1}{2},\end{aligned}$$

then we have

$$\begin{aligned}\Psi (T)\le C\Psi ^2(s)\cdot C(M,T^*)^2,\end{aligned}$$

which proves

$$\begin{aligned} \Psi (T)\le C. \end{aligned}$$

(2.19)

Taking the operator $\partial _t$ to (1.2), testing by $u_t$, using (1.1) and (1.5), we have

$$\begin{aligned}&\frac{1}{2}\frac{\mathrm {d}}{\mathrm {d}t}\int \rho |u_t|^2\mathrm {d}x+\int |\nabla u_t|^2\mathrm {d}t\nonumber \\&\quad =-\int \rho _t|u_t|^2\mathrm {d}x-\int \rho _tu\cdot \nabla u\cdot u_t\mathrm {d}x-\int \rho u_t\cdot \nabla u\cdot u_t\mathrm {d}x\nonumber \\&\qquad -\int \partial _t(b\otimes b):\nabla u_t\mathrm {d}x+\int (\rho \theta )_te_2u_t\mathrm {d}x\nonumber \\&\quad =:\sum \limits _{i=1}^5I_i. \end{aligned}$$

(2.20)

We use (2.1), Gagliardo–Nirenberg inequality and the Hölder inequality to bound $I_i\ (i=1,\cdots ,4)$ as follows:

$$\begin{aligned} I_1= & {} -\int \rho u\cdot \nabla |u_t|^2\mathrm {d}x\le C\Vert u\Vert _{L^6}\Vert \sqrt{\rho }u_t\Vert _{L^3}\Vert \nabla u_t\Vert _{L^2}\nonumber \\\le & {} C\Vert u\Vert _{L^6}\Vert \sqrt{\rho }u_t\Vert _{L^2}^\frac{1}{2}\Vert \sqrt{\rho }u_t\Vert _{L^6}^\frac{1}{2}\Vert \nabla u_t\Vert _{L^2}\nonumber \\\le & {} C\Vert u\Vert _{L^6}\Vert \sqrt{\rho }u_t\Vert _{L^2}^\frac{1}{2}\Vert \nabla u_t\Vert _{L^2}^\frac{3}{2}\nonumber \\\le & {} \frac{1}{16}\Vert \nabla u_t\Vert _{L^2}^2+C\Vert \sqrt{\rho }u_t\Vert _{L^2}^2, \end{aligned}$$

(2.21)

$$\begin{aligned} I_2= & {} -\int \rho u\cdot \nabla (u\cdot \nabla u\cdot u_t)\mathrm {d}x\nonumber \\\le & {} C\Vert \sqrt{\rho }u_t\Vert _{L^6}\Vert u\Vert _{L^6}\Vert \nabla u\Vert _{L^3}^2+C\Vert \sqrt{\rho }u_t\Vert _{L^6}\Vert \Delta u\Vert _{L^2}\Vert u\Vert _{L^6}^2\nonumber \\&\quad +C\Vert \nabla u_t\Vert _{L^2}\Vert \nabla u\Vert _{L^6}\Vert u\Vert _{L^6}^2\nonumber \\\le & {} C\Vert \nabla u_t\Vert _{L^2}\Vert u\Vert _{H^1}^2\Vert u\Vert _{H^2}\nonumber \\\le & {} \frac{1}{16}\Vert \nabla u_t\Vert _{L^2}^2+C\Vert u\Vert _{H^2}^2, \end{aligned}$$

(2.22)

$$\begin{aligned} I_3\le & {} C\Vert \sqrt{\rho }u_t\Vert _{L^4}^2\Vert \nabla u\Vert _{L^2}\nonumber \\\le & {} C\Vert \sqrt{\rho }u_t\Vert _{L^2}^\frac{2}{3}\Vert \sqrt{\rho }u_t\Vert _{L^8}^\frac{4}{3}\Vert \nabla u\Vert _{L^2}\le C\Vert \sqrt{\rho }u_t\Vert _{L^2}^\frac{2}{3}\Vert u_t\Vert _{L^8}^\frac{4}{3}\Vert \nabla u\Vert _{L^2}\nonumber \\\le & {} C\Vert \sqrt{\rho }u_t\Vert _{L^2}^\frac{2}{3}\Vert \nabla u_t\Vert _{L^2}^\frac{4}{3}\Vert \nabla u\Vert _{L^2}\le \frac{1}{16}\Vert \nabla u_t\Vert _{L^2}^2+C\Vert \sqrt{\rho }u_t\Vert _{L^2}^2, \end{aligned}$$

(2.23)

$$\begin{aligned} I_4\le & {} 2\Vert b\Vert _{L^\infty }\Vert b_t\Vert _{L^2}\Vert \nabla u_t\Vert _{L^2}\nonumber \\\le & {} \frac{1}{16}\Vert \nabla u_t\Vert _{L^2}^2+C\Vert b\Vert _{L^\infty }^2\Vert b_t\Vert _{L^2}^2, \end{aligned}$$

(2.24)

$$\begin{aligned} I_5= & {} \int (\rho ue_2)e_2u_t\mathrm {d}x-\int \mathrm {div}\,(\rho u\theta )\cdot e_2u_t\mathrm {d}x+\int \Delta \theta \cdot e_2u_t\mathrm {d}x\nonumber \\= & {} \int (\rho ue_2)e_2u_t\mathrm {d}x+\int \rho u\theta \nabla (e_2u_t)\mathrm {d}x-\int \nabla \theta \cdot \nabla (e_2u_t)\mathrm {d}x\nonumber \\\le & {} C\Vert \sqrt{\rho }u\Vert _{L^2}\Vert \sqrt{\rho }u_t\Vert _{L^2}+C\Vert u\Vert _{L^\infty }\Vert \sqrt{\rho }\theta \Vert _{L^2}\Vert \nabla u_t\Vert _{L^2}+\Vert \nabla \theta \Vert _{L^2}\Vert \nabla u_t\Vert _{L^2}\nonumber \\\le & {} C\Vert \sqrt{\rho }u\Vert _{L^2}\Vert \sqrt{\rho }u_t\Vert _{L^2}+\frac{1}{16}\Vert \nabla u_t\Vert _{L^2}^2+C\Vert u\Vert _{L^\infty }^2+C\Vert \nabla \theta \Vert _{L^2}^2. \end{aligned}$$

(2.25)

Inserting the above estimates into (2.10) and testing by t, we arrive at

$$\begin{aligned}&\frac{\mathrm {d}}{\mathrm {d}t}t\int \rho |u_t|^2\mathrm {d}x+t\int |\nabla u_t|^2\mathrm {d}x\nonumber \\&\quad \le C\int \rho |u_t|^2\mathrm {d}x+Ct\int \rho |u_t|^2\mathrm {d}x+Ct\Vert u\Vert _{H^2}^2\nonumber \\&\qquad +Ct\Vert b\Vert _{L^\infty }^2\Vert b_t\Vert _{L^2}^2+Ct\Vert \nabla \theta \Vert _{L^2}^2+C. \end{aligned}$$

(2.26)

Similarly, applying $\partial _t$ to (1.3), testing by $b_t$, using (1.5), we derive

$$\begin{aligned}&\frac{1}{2}\frac{\mathrm {d}}{\mathrm {d}t}\int |b_t|^2\mathrm {d}x+\int (\mathrm {rot}\,b_t)^2\mathrm {d}x\nonumber \\&\quad =-\int u_t\cdot \nabla b\cdot b_t\mathrm {d}x+\int b_t\cdot \nabla u\cdot b_t\mathrm {d}x+\int b\cdot \nabla u_t\cdot b_t\mathrm {d}x\nonumber \\&\quad \le \Vert u_t\Vert _{L^4}\Vert \nabla b\Vert _{L^4}\Vert b_t\Vert _{L^2}+\Vert \nabla u\Vert _{L^2}\Vert b_t\Vert _{L^4}^2+\Vert \nabla u_t\Vert _{L^2}\Vert b\Vert _{L^\infty }\Vert b_t\Vert _{L^2}\nonumber \\&\quad \le \frac{1}{16}\Vert \nabla u_t\Vert _{L^2}^2+C\Vert b\Vert _{W^{1,4}}^2\Vert b_t\Vert _{L^2}^2+C\Vert b_t\Vert _{L^2}\Vert \nabla b_t\Vert _{L^2}\nonumber \\&\quad \le \frac{1}{16}\Vert \nabla u_t\Vert _{L^2}^2+\frac{1}{16}\Vert \mathrm {rot}\,b_t\Vert _{L^2}^2+C\Vert b\Vert _{W^{1,4}}^2\Vert b_t\Vert _{L^2}^2+C\Vert b_t\Vert _{L^2}^2. \end{aligned}$$

(2.27)

Multiplying (2.27) by t, adding it to (2.26) and using the Gronwall inequality, we arrive at

$$\begin{aligned} t\int \rho |u_t|^2\mathrm {d}x+t\int |b_t|^2\mathrm {d}x+\int _0^ts(\Vert \nabla u_t\Vert _{L^2}^2+\Vert \nabla b_t\Vert _{L^2}^2)\mathrm {d}s\le C. \end{aligned}$$

(2.28)

Here, we have used the following fact: $u\in L^2(0,T;H^2)$, which can be proved by the following calculations (2.29) and (2.30).

Similarly to (2.15),

$$\begin{aligned} \Vert u\Vert _{H^2}\le & {} C\Vert f\Vert _{L^2}+C\Vert u\Vert _{H^1}\\\le & {} C\Vert b\cdot \nabla b\Vert _{L^2}+C\Vert \rho \theta \Vert _{L^2}+C\Vert \rho u_t\Vert _{L^2}+C\Vert \rho u\cdot \nabla u\Vert _{L^2}+C\\\le & {} C\Vert b\Vert _{L^4}\Vert \nabla b\Vert _{L^4}+C\Vert \sqrt{\rho }u_t\Vert _{L^2}+C\Vert u\Vert _{L^4}\Vert \nabla u\Vert _{L^4}+C\\\le & {} C\Vert b\Vert _{H^2}^\frac{1}{2}+C\Vert \sqrt{\rho }u_t\Vert _{L^2}+C\Vert u\Vert _{H^2}^\frac{1}{2}+C, \end{aligned}$$

which gives

$$\begin{aligned} \Vert u\Vert _{H^2}\le C\Vert b\Vert _{H^2}^\frac{1}{2}+C\Vert \sqrt{\rho }u_t\Vert _{L^2}+C. \end{aligned}$$

(2.29)

Similarly to (2.29),

$$\begin{aligned} \Vert b\Vert _{H^2}\le & {} C\Vert b_t+u\cdot \nabla b-b\cdot \nabla u\Vert _{L^2}+C\Vert b\Vert _{H^1}\\\le & {} C\Vert b_t\Vert _{L^2}+C\Vert u\Vert _{L^4}\Vert \nabla b\Vert _{L^4}+C\Vert b\Vert _{L^4}\Vert \nabla u\Vert _{L^4}+C\\\le & {} C\Vert b_t\Vert _{L^2}+C\Vert b\Vert _{H^2}^\frac{1}{2}+C\Vert u\Vert _{H^2}^\frac{1}{2}+C, \end{aligned}$$

which leads to

$$\begin{aligned} \Vert b\Vert _{H^2}\le C\Vert b_t\Vert _{L^2}+C\Vert u\Vert _{H^2}^\frac{1}{2}+C. \end{aligned}$$

(2.30)

Summing up (2.29) and (2.30) and using (2.28), we have

$$\begin{aligned} \sqrt{t}u,\sqrt{t}b\in L^\infty (0,T;H^2). \end{aligned}$$

(2.31)

Similarly to (2.15),

$$\begin{aligned} \Vert u\Vert _{W^{2,6}}\le & {} C\Vert f\Vert _{L^6}+C\Vert u\Vert _{H^1}\\\le & {} C\Vert b\cdot \nabla b\Vert _{L^6}+C\Vert \rho \theta \Vert _{L^6}+C\Vert \rho u_t\Vert _{L^6}+C\Vert \rho u\cdot \nabla u\Vert _{L^6}+C\\\le & {} C\Vert b\Vert _{L^6}\Vert \nabla b\Vert _{L^\infty }+C\Vert \theta \Vert _{H^1}+C\Vert \nabla u_t\Vert _{L^2}+C\Vert u\Vert _{L^6}\Vert \nabla u\Vert _{L^\infty }+C\\\le & {} \Vert b\Vert _{W^{2,6}}^\frac{3}{5}+C\Vert \theta \Vert _{H^1}+C\Vert \nabla u_t\Vert _{L^2}+C\Vert u\Vert _{W^{2,6}}^\frac{3}{5}+C, \end{aligned}$$

which implies

$$\begin{aligned} \Vert u\Vert _{W^{2,6}}\le C\Vert b\Vert _{W^{2,6}}^\frac{3}{5}+C\Vert \theta \Vert _{H^1}+C\Vert \nabla u_t\Vert _{L^2}+C. \end{aligned}$$

(2.32)

Similarly,

$$\begin{aligned} \Vert b\Vert _{W^{2,6}}\le & {} C\Vert b_t+u\cdot \nabla b-b\cdot \nabla u\Vert _{L^6}+C\Vert b\Vert _{H^1}\\\le & {} C\Vert b_t\Vert _{L^6}+C\Vert u\Vert _{L^6}\Vert \nabla b\Vert _{L^\infty }+C\Vert b\Vert _{L^6}\Vert \nabla u\Vert _{L^\infty }+C\\\le & {} C\Vert b\Vert _{W^{2,6}}^\frac{3}{5}+C\Vert u\Vert _{W^{2,6}}^\frac{3}{5}+C\Vert \nabla b_t\Vert _{L^2}+C, \end{aligned}$$

which yields

$$\begin{aligned} \Vert b\Vert _{W^{2,6}}\le C\Vert u\Vert _{W^{2,6}}^\frac{3}{5}+C\Vert \nabla b_t\Vert _{L^2}+C. \end{aligned}$$

(2.33)

Summing up (2.32) and (2.33) and using (2.28), we have

$$\begin{aligned} \sqrt{t}u,\sqrt{t}b\in L^2(0,T;W^{2,6}). \end{aligned}$$

(2.34)

On the other hand,

$$\begin{aligned} \int _0^T\Vert \nabla u\Vert _{L^\infty }\mathrm {d}t\le & {} C\int _0^T\Vert \nabla u\Vert _{L^2}^\frac{2}{5}\Vert u\Vert _{W^{2,6}}^\frac{3}{5}\mathrm {d}t=C\int _0^t(\sqrt{t}\Vert u\Vert _{W^{2,6}})^\frac{3}{5}t^{-\frac{3}{10}}\mathrm {d}t\nonumber \\\le & {} C\left( \int _t^T(\sqrt{t}\Vert u\Vert _{W^{2,6}})^2\mathrm {d}t\right) ^\frac{3}{10}\left( \int _0^Tt^{-\frac{3}{10}\cdot \frac{10}{7}}\mathrm {d}t\right) ^\frac{7}{10}\nonumber \\\le & {} C\int _0^Tt\Vert u\Vert _{W^{2,6}}^2\mathrm {d}t\le C. \end{aligned}$$

(2.35)

Taking $\nabla$ to (1.1), testing by $|\nabla \rho |^{q-2}\nabla \rho$, using (1.5) and (2.35), we compute

$$\begin{aligned}\frac{\mathrm {d}}{\mathrm {d}t}\int |\nabla \rho |^q\mathrm {d}x\le C\Vert \nabla u\Vert _{L^\infty }\int |\nabla \rho |^q\mathrm {d}x,\end{aligned}$$

which leads to

$$\begin{aligned} \Vert \nabla \rho \Vert _{L^q}\le C. \end{aligned}$$

(2.36)

It follows from (1.1) that

$$\begin{aligned} \Vert \rho _t\Vert _{L^4(0,T;L^q)}= & {} \Vert u\cdot \nabla \rho \Vert _{L^4(0,T;L^q)}\nonumber \\\le & {} \Vert u\Vert _{L^4(0,T;L^\infty )}\Vert \nabla \rho \Vert _{L^\infty (0,T;L^q)}\le C\Vert u\Vert _{L^4(0,T;L^\infty )}\le C \end{aligned}$$

(2.37)

by the Gagliardo–Nirenberg inequality

$$\begin{aligned} \Vert u\Vert _{L^\infty }^2\le C\Vert u\Vert _{L^2}\Vert u\Vert _{H^2}. \end{aligned}$$

(2.38)

Testing (1.4) by $\theta _t$, we observe that

$$\begin{aligned}&\frac{1}{2}\frac{\mathrm {d}}{\mathrm {d}t}\int |\nabla \theta |^2\mathrm {d}x+\int \rho \theta _t^2\mathrm {d}x\\&\quad =\int \rho ue_2\theta _t\mathrm {d}x-\int \rho u\cdot \nabla \theta \cdot \theta _t\mathrm {d}x\\&\quad \le \frac{1}{2}\int \rho \theta _t^2\mathrm {d}x+C\int \rho |u|^2\mathrm {d}x+C\Vert u\Vert _{L^\infty }\Vert \nabla \theta \Vert _{L^2}, \end{aligned}$$

which gives

$$\begin{aligned} \int |\nabla \theta |^2\mathrm {d}x+\int _0^T\int \rho \theta _t^2\mathrm {d}x\mathrm {d}t\le C. \end{aligned}$$

(2.39)

Taking $\partial _t$ to (1.4), testing by $\theta _t$, using (1.5), we have

$$\begin{aligned}&\frac{1}{2}\frac{\mathrm {d}}{\mathrm {d}t}\int \rho \theta _t^2\mathrm {d}x+\int |\nabla \theta _t|^2\mathrm {d}x\nonumber \\&\quad =-\int \rho _t\theta _t^2\mathrm {d}x-\int \rho _tu\cdot \nabla \theta \cdot \theta _t\mathrm {d}x-\int \rho u_t\cdot \nabla \theta \cdot \theta _t\mathrm {d}x+\int (\rho _tu+\rho u_t)e_2\theta _t\mathrm {d}x\nonumber \\&\quad \le -\int \rho u\nabla \theta _t^2\mathrm {d}x-\int \rho u\cdot \nabla (u\cdot \nabla \theta \cdot \theta _t)\mathrm {d}x-\int \rho u_t\cdot \nabla \theta \cdot \theta _t\mathrm {d}x\nonumber \\&\qquad +\int \rho u\nabla (ue_2\theta _t)\mathrm {d}x+\int \rho u_te_2\theta _t\mathrm {d}x\nonumber \\&\quad \le \Vert \sqrt{\rho }\theta _t\Vert _{L^2}\Vert \sqrt{\rho }\Vert _{L^\infty }\Vert u\Vert _{L^\infty }\Vert \nabla \theta _t\Vert _{L^2}+\Vert \sqrt{\rho }\theta _t\Vert _{L^6}\Vert u\Vert _{L^6}\Vert \nabla u\Vert _{L^6}\Vert \nabla \theta \Vert _{L^2}\nonumber \\&\qquad +C\Vert \sqrt{\rho }\theta _t\Vert _{L^6}\Vert u\Vert _{L^6}^2\Vert \nabla ^2\theta \Vert _{L^2}+C\Vert \nabla \theta _t\Vert _{L^2}\Vert u\Vert _{L^6}^2\Vert \nabla \theta \Vert _{L^6}\nonumber \\&\quad \le \frac{1}{4}\Vert \nabla \theta _t\Vert _{L^2}^2+C\Vert u\Vert _{L^\infty }^2\Vert \sqrt{\rho }\theta _t\Vert _{L^2}^2+C\Vert \nabla u\Vert _{L^6}^2+C\Vert \nabla ^2\theta \Vert _{L^2}^2+C\Vert \nabla \theta \Vert _{L^6}^2. \end{aligned}$$

(2.40)

Multiplying (2.40) by t, we have

$$\begin{aligned}&\frac{\mathrm {d}}{\mathrm {d}t}t\int \rho \theta _t^2\mathrm {d}x+t\int |\nabla \theta _t|^2\mathrm {d}x\nonumber \\&\quad \le C\int \rho \theta _t^2\mathrm {d}x+C\Vert u\Vert _{L^\infty }^2t\Vert \sqrt{\rho }\theta _t\Vert _{L^2}^2+C\Vert u\Vert _{H^2}^2+C\Vert \theta \Vert _{H^2}^2. \end{aligned}$$

(2.41)

On the other hand,

$$\begin{aligned} \Vert \theta \Vert _{H^2}\le & {} C\Vert \rho \theta _t+\rho u\cdot \nabla \theta -\rho ue_2\Vert _{L^2}+C\Vert \theta \Vert _{H^1}\\\le & {} C+C\Vert \sqrt{\rho }\theta _t\Vert _{L^2}+\Vert u\Vert _{L^\infty }\Vert \nabla \theta \Vert _{L^2}\\\le & {} C+C\Vert \sqrt{\rho }\theta _t\Vert _{L^2}+C\Vert u\Vert _{L^\infty }, \end{aligned}$$

which implies

$$\begin{aligned} \theta \in L^2(0,T;H^2). \end{aligned}$$

(2.42)

It follows from (2.41), (2.42) and the Gronwall inequality that

$$\begin{aligned} t\int \rho \theta _t^2\mathrm {d}x+\int _0^ts\Vert \nabla \theta _t\Vert _{L^2}^2\mathrm {d}s\le C. \end{aligned}$$

(2.43)

Similarly, we have

$$\begin{aligned} \sqrt{t}u\in L^\infty (0,T;H^2)\cap L^2(0,T;W^{2,6}). \end{aligned}$$

(2.44)

This completes the proof. $\square$

3 Appendix

The aim of this section is to prove the local well-posedness of the problem (1.1)-(1.7) when $\inf \rho _0>0$. We will prove

Theorem 3.1

Let $\frac{1}{C}\le \rho _0\le C,\rho _0\in H^2, u_0,\theta _0\in H_0^1\cap H^2, b_0\in H^2$ with $\mathrm {div}\,u_0=\mathrm {div}\,b_0=0$ in $\Omega$ and $b_0\cdot n=0, \mathrm {rot}\,b_0=0$ on $\partial \Omega$. Then, the problem (1.1)–(1.7) has a unique strong solution $(\rho ,u,b,\theta )$ satisfying

$$\begin{aligned} \begin{array}{l} \frac{1}{C}\le \rho \le C,\rho \in C([0,T];H^2),\partial _t\rho \in C([0,T];H^1),\\ u,b,\theta \in C([0,T];H^2)\cap L^2(0,T;W^{2,4}),\partial _t(u,b,\theta )\in C([0,T];L^2)\cap L^2(0,T;H^1) \end{array} \end{aligned}$$

(3.1)

for some $0<T\le \infty$.

We will prove Theorem 3.1 by the Banach fixed-point theorem. We denote the nonempty closed set

$$\begin{aligned} \mathcal {A}:=\{\tilde{u}\in \mathcal {A};\tilde{u}(\cdot ,0)=u_0,\mathrm {div}\,\tilde{u}=0,\Vert \tilde{u}\Vert _\mathcal {A}\le A\}\end{aligned}$$

with the norm

$$\begin{aligned}\Vert \tilde{u}\Vert _\mathcal {A}:=\Vert \tilde{u}\Vert _{L^\infty (0,T;H_0^1\cap H^2)}+\Vert \tilde{u}\Vert _{L^2(0,T;W^{2,4})}+\Vert \partial _t\tilde{u}\Vert _{L^\infty (0,T;L^2)}+\Vert \partial _t\tilde{u}\Vert _{L^2(0,T;H^1)}.\end{aligned}$$

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

$$\begin{aligned}&\partial _t\rho +\tilde{u}\cdot \nabla \rho =0, \end{aligned}$$

(3.2a)

$$\begin{aligned}&\rho (\cdot ,0)=\rho _0; \end{aligned}$$

(3.2b)

$$\begin{aligned}&\partial _tb+\tilde{u}\cdot \nabla b-b\cdot \nabla \tilde{u}=\eta \Delta b, \end{aligned}$$

(3.3a)

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

(3.3b)

$$\begin{aligned}&b\cdot n=0,\mathrm {rot}\,b=0\ \ \text{ on }\ \ \partial \Omega \times (0,\infty ), \end{aligned}$$

(3.3c)

$$\begin{aligned}&b(\cdot ,0)=b_0\ \ \text{ in }\ \ \Omega ; \end{aligned}$$

(3.3d)

$$\begin{aligned}&\rho \partial _t\theta +\rho \tilde{u}\cdot \nabla \theta =\rho \tilde{u}e_2+k\Delta \theta , \end{aligned}$$

(3.4a)

$$\begin{aligned}&\theta =0\ \ \text{ on }\ \ \partial \Omega \times (0,\infty ), \end{aligned}$$

(3.4b)

$$\begin{aligned}&\theta (\cdot ,0)=\theta _0\ \ \text{ in }\ \ \Omega ; \end{aligned}$$

(3.4c)

$$\begin{aligned}&\rho \partial _tu+\rho \tilde{u}\cdot \nabla u+\nabla \left( \pi +\frac{1}{2}|b|^2\right) -\mu \Delta u=b\cdot \nabla b+\rho \theta e_2, \end{aligned}$$

(3.5a)

$$\begin{aligned}&\mathrm {div}\,u=0, \end{aligned}$$

(3.5b)

$$\begin{aligned}&u=0\ \ \text{ on }\ \ \partial \Omega \times (0,\infty ), \end{aligned}$$

(3.5c)

$$\begin{aligned}&u(\cdot ,0)=u_0\ \ \text{ in }\ \ \Omega . \end{aligned}$$

(3.5d)

Let u be the unique strong solution to the above problem, we define the fixed-point map $F:\tilde{u}\in \mathcal {A}\rightarrow u\in \mathcal {A}$. 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 3.1.

Lemma 3.2

Let $\tilde{u}\in \mathcal {A}$ be given and $\frac{1}{C}\le \rho _0\in H^2$. Then, the problem (3.2) has a unique solution $\rho$ satisfying

$$\begin{aligned} \frac{1}{C}\le \rho \le C,\Vert \rho \Vert _{L^\infty (0,T;H^2)}\le C,\Vert \rho _t\Vert _{L^\infty (0,T;H^1)}\le CA \end{aligned}$$

(3.6)

for some $0<T\le 1$.

Here, later on, C will denote a constant independent of A.

Proof

Since Eq. (3.2a) is linear with regular $\tilde{u}$, the existence and uniqueness are well known, we only need to show the a priori estimates.

First, it is obvious that

$$\begin{aligned} \frac{1}{C}\le \rho \le C. \end{aligned}$$

(3.7)

Taking $\nabla ^2$ to (3.2a), testing by $\nabla ^2\rho$, using $\mathrm {div}\,\tilde{u}=0$, we find that

$$\begin{aligned} \frac{1}{2}\frac{\mathrm {d}}{\mathrm {d}t}\int |\nabla ^2\rho |^2\mathrm {d}x= & {} -\int (\nabla ^2(\tilde{u}\cdot \nabla \rho )-(\tilde{u}\cdot \nabla )\nabla ^2\rho )\nabla ^2\rho \mathrm {d}x\\\le & {} \Vert \nabla ^2(\tilde{u}\cdot \nabla \rho )-(\tilde{u}\cdot \nabla )\nabla ^2\rho \Vert _{L^2}\Vert \nabla ^2\rho \Vert _{L^2}\\\le & {} C(\Vert \nabla \tilde{u}\Vert _{L^\infty }\Vert \nabla ^2\rho \Vert _{L^2}+\Vert \nabla \rho \Vert _{L^4}\Vert \nabla ^2\tilde{u}\Vert _{L^4})\Vert \nabla ^2\rho \Vert _{L^2}\\\le & {} C\Vert \tilde{u}\Vert _{W^{2,4}}(\Vert \nabla ^2\rho \Vert _{L^2}^2+1), \end{aligned}$$

which gives

$$\begin{aligned}\Vert \nabla ^2\rho \Vert _{L^2}^2\le \left( \Vert \nabla ^2\rho _0\Vert _{L^2}^2+C\int _0^T\Vert \tilde{u}\Vert _{W^{2,4}}\mathrm {d}t\right) \exp \left( C\int _0^T\Vert \tilde{u}\Vert _{W^{2,4}}\mathrm {d}t\right) \end{aligned}$$

and thus

$$\begin{aligned} \Vert \nabla ^2\rho \Vert _{L^2}\le C \end{aligned}$$

(3.8)

if $A\sqrt{T}\le 1$.

It follows from (3.2a) that

$$\begin{aligned}\partial _t\rho =-\tilde{u}\cdot \nabla \rho \end{aligned}$$

and hence

$$\begin{aligned} \Vert \partial _t\rho \Vert _{H^1}\le \Vert \tilde{u}\Vert _{L^6}\Vert \nabla \rho \Vert _{L^3}+\Vert \nabla \tilde{u}\Vert _{L^4}\Vert \nabla \rho \Vert _{L^4}+\Vert \tilde{u}\Vert _{L^\infty }\Vert \nabla ^2\rho \Vert _{L^2}\le CA. \end{aligned}$$

(3.9)

This completes the proof. $\square$

Lemma 3.3

Let $\tilde{u}\in \mathcal {A}$ be given and $b_0\in H^2$ with $\mathrm {div}\,b_0=0$ in $\Omega$ and $b_0\cdot n=\mathrm {rot}\,b_0=0$ on $\partial \Omega$. Then, the problem (3.3) has a unique solution b satisfying

$$\begin{aligned}&\Vert b\Vert _{L^\infty (0,T;H^2)}+\Vert b\Vert _{L^2(0,T;W^{2,4})}\le C, \end{aligned}$$

(3.10)

$$\begin{aligned}&\Vert \partial _tb\Vert _{L^\infty (0,T;L^2)}+\Vert \partial _tb\Vert _{L^2(0,T;H^1)}\le C \end{aligned}$$

(3.11)

for some $0<T\le 1$.

Proof

Since Eqs. (3.3a) and (3.3b) are linear with regular $\tilde{u}$, the existence and uniqueness are well known; we only need to show the a priori estimates.

Since

$$\begin{aligned}\tilde{u}(x,t)=u_0(x)+\int _0^t\partial _t\tilde{u}\mathrm {d}s,\end{aligned}$$

it follows that

$$\begin{aligned} \Vert \nabla \tilde{u}\Vert _{L^\infty (0,T;L^2)}\le & {} \Vert \nabla u_0\Vert _{L^2}+\int _0^T\Vert \nabla \partial _t\tilde{u}\Vert _{L^2}\mathrm {d}t\nonumber \\\le & {} C+C\sqrt{T}A\le C \end{aligned}$$

(3.12)

if $\sqrt{T}A\le 1$.

Testing (3.3a) by b, we see that

$$\begin{aligned}\frac{1}{2}\frac{\mathrm {d}}{\mathrm {d}t}\int |b|^2\mathrm {d}x +\eta \int |\mathrm {rot}\,b|^2\mathrm {d}x=\int b\cdot \nabla \tilde{u}\cdot b\mathrm {d}x\le \Vert \nabla \tilde{u}\Vert _{L^\infty }\int |b|^2\mathrm {d}x,\end{aligned}$$

which implies

$$\begin{aligned} \Vert b\Vert _{L^\infty (0,T;L^2)}+\Vert b\Vert _{L^2(0,T;H^1)}\le C \end{aligned}$$

(3.13)

if $A\sqrt{T}\le 1$.

Testing (3.3a) by $-\Delta b$, using (3.12), (3.13) and Lemma 1.3, we deduce that

$$\begin{aligned}&\frac{1}{2}\frac{\mathrm {d}}{\mathrm {d}t}\int |\mathrm {rot}\,b|^2\mathrm {d}x+\eta \int |\Delta b|^2\mathrm {d}x=\int (\tilde{u}\cdot \nabla b-b\cdot \nabla \tilde{u})\Delta b\mathrm {d}x\\&\quad \le (\Vert \tilde{u}\Vert _{L^4}\Vert \nabla b\Vert _{L^4}+\Vert b\Vert _{L^\infty }\Vert \nabla \tilde{u}\Vert _{L^2})\Vert \Delta b\Vert _{L^2}\\&\quad \le C(\Vert \nabla b\Vert _{L^2}^\frac{1}{2}\Vert b\Vert _{H^2}^\frac{1}{2}+\Vert b\Vert _{L^2}^\frac{1}{2}\Vert b\Vert _{H^2}^\frac{1}{2})\Vert \Delta b\Vert _{L^2}\\&\quad \le \frac{\eta }{2}\Vert \Delta b\Vert _{L^2}^2+C\Vert \nabla b\Vert _{L^2}^4+C, \end{aligned}$$

which yields

$$\begin{aligned} \Vert b\Vert _{L^\infty (0,T;H^1)}+\Vert b\Vert _{L^2(0,T;H^2)}\le C. \end{aligned}$$

(3.14)

Here, we have used the Gagliardo–Nirenberg inequalities:

$$\begin{aligned}&\Vert \nabla b\Vert _{L^4}\lesssim \Vert \nabla b\Vert _{L^2}^\frac{1}{2}\Vert b\Vert _{H^2}^\frac{1}{2}, \end{aligned}$$

(3.15)

$$\begin{aligned}&\quad \Vert b\Vert _{L^\infty }\lesssim \Vert b\Vert _{L^2}^\frac{1}{2}\Vert b\Vert _{H^2}^\frac{1}{2}, \end{aligned}$$

(3.16)

Taking $\partial _t$ to (3.3a), testing by $\partial _tb$, using (3.14), we derive

$$\begin{aligned}&\frac{1}{2}\frac{\mathrm {d}}{\mathrm {d}t}\int |\partial _tb|^2\mathrm {d}x+\eta \int (\mathrm {rot}\,\partial _tb)^2\mathrm {d}x\\&\quad =-\int \partial _t\tilde{u}\cdot \nabla b\cdot \partial _tb\mathrm {d}x+\int \partial _tb\cdot \nabla \tilde{u}\cdot \partial _tb\mathrm {d}x-\int b\cdot \partial _t\tilde{u}\cdot \nabla \partial _tb\mathrm {d}x\\&\quad \le \Vert \nabla b\Vert _{L^2}\Vert \partial _t\tilde{u}\Vert _{L^4}\Vert \partial _tb\Vert _{L^4}+\Vert \nabla \tilde{u}\Vert _{L^\infty }\Vert \partial _tb\Vert _{L^2}^2+\Vert \partial _t\tilde{u}\Vert _{L^4}\Vert b\Vert _{L^4}\Vert \nabla \partial _tb\Vert _{L^2}\\&\quad \le C\Vert \partial _t\tilde{u}\Vert _{L^4}\Vert \partial _tb\Vert _{L^2}^\frac{1}{2}\cdot \Vert \mathrm {rot}\,\partial _tb\Vert _{L^2}^\frac{1}{2}+\Vert \nabla \tilde{u}\Vert _{L^\infty }\Vert \partial _tb\Vert _{L^2}^2+C\Vert \partial _t\tilde{u}\Vert _{L^4}\Vert \nabla \partial _tb\Vert _{L^2}\\&\quad \le \frac{\eta }{2}\Vert \mathrm {rot}\,\partial _tb\Vert _{L^2}^2+C\Vert \partial _t\tilde{u}\Vert _{L^4}^2+C\Vert \partial _tb\Vert _{L^2}^2+\Vert \nabla \tilde{u}\Vert _{L^\infty }\Vert \partial _tb\Vert _{L^2}^2, \end{aligned}$$

which leads to

$$\begin{aligned} \Vert \partial _tb\Vert _{L^\infty (0,T;L^2)}+\Vert \partial _tb\Vert _{L^2(0,T;H^1)}\le C \end{aligned}$$

(3.17)

if $A\sqrt{T}\le 1$ and $T\le 1$.

Using (3.15), (3.16), Lemma 1.3, (3.14) and (3.12), we have

$$\begin{aligned} \Vert b\Vert _{H^2}\le & {} C\Vert \tilde{u}\cdot \nabla b-b\cdot \nabla \tilde{u}\Vert _{L^2}+C\Vert \partial _tb\Vert _{L^2}+C\Vert b\Vert _{H^1}\\\le & {} C\Vert \tilde{u}\Vert _{L^4}\Vert \nabla b\Vert _{L^4}+C\Vert b\Vert _{L^\infty }\Vert \nabla \tilde{u}\Vert _{L^2}+C\\\le & {} C\Vert \nabla b\Vert _{L^4}+C\Vert b\Vert _{L^\infty }+C\\\le & {} C\Vert b\Vert _{H^2}^\frac{1}{2}+C\le \frac{1}{2}\Vert b\Vert _{H^2}+C, \end{aligned}$$

which implies

$$\begin{aligned} \Vert b\Vert _{L^\infty (0,T;H^2)}\le C. \end{aligned}$$

(3.18)

Similarly,

$$\begin{aligned} \Vert b\Vert _{W^{2,4}}\le & {} C\Vert \partial _tb+\tilde{u}\cdot \nabla b-b\cdot \nabla \tilde{u}\Vert _{L^4}+C\Vert b\Vert _{H^1}\\\le & {} C\Vert \partial _tb\Vert _{L^4}+C\Vert \tilde{u}\Vert _{L^8}\Vert \nabla b\Vert _{L^8}+C\Vert b\Vert _{L^\infty }\Vert \nabla \tilde{u}\Vert _{L^4}+C\\\le & {} C\Vert \partial _tb\Vert _{L^4}+C+C\Vert \nabla \tilde{u}\Vert _{L^4}\\\le & {} C\Vert \partial _tb\Vert _{L^4}+C+CA, \end{aligned}$$

which implies

$$\begin{aligned} \Vert b\Vert _{L^2(0,T;W^{2,4})}\le C \end{aligned}$$

(3.19)

if $A\sqrt{T}\le 1$ and $T\le 1$.

This completes the proof. $\square$

Lemma 3.4

Let $\tilde{u}\in \mathcal {A}$ be given and $\theta _0\in H_0^1\cap H^2$. Then, the problem (3.4) has a unique solution $\theta$ satisfying

$$\begin{aligned}&\Vert \theta \Vert _{L^\infty (0,T;H^2)}+\Vert \theta \Vert _{L^2(0,T;W^{2,4})}\le C, \end{aligned}$$

(3.20)

$$\begin{aligned}&\Vert \partial _t\theta \Vert _{L^\infty (0,T;L^2)}+\Vert \partial _t\theta \Vert _{L^2(0,T;H^1)}\le C \end{aligned}$$

(3.21)

for some $0<T\le 1$.

Proof

Since Eq. (3.4a) is linear with regular $(\rho ,\tilde{u})$, the existence and uniqueness are well known; we only need to prove the a priori estimates.

Testing (3.4a) by $\theta$ and using (3.2a), we know that

$$\begin{aligned}&\frac{1}{2}\frac{\mathrm {d}}{\mathrm {d}t}\int \rho \theta ^2\mathrm {d}x+k\int |\nabla \theta |^2\mathrm {d}x =\int \rho \tilde{u}e_2\theta \mathrm {d}x\\&\quad \le \Vert \sqrt{\rho }\Vert _{L^\infty }\Vert \tilde{u}\Vert _{L^2}\Vert \sqrt{\rho }\theta \Vert _{L^2}\le C\Vert \sqrt{\rho }\theta \Vert _{L^2}, \end{aligned}$$

which gives

$$\begin{aligned} \Vert \theta \Vert _{L^\infty (0,T;L^2)}+\Vert \theta \Vert _{L^2(0,T;H^1)}\le C. \end{aligned}$$

(3.22)

Testing (3.4a) by $\partial _t\theta$, we get

$$\begin{aligned}&\frac{k}{2}\frac{\mathrm {d}}{\mathrm {d}t}\int |\nabla \theta |^2\mathrm {d}x +\int \rho (\partial _t\theta )^2\mathrm {d}x\\&\quad =-\int \rho \tilde{u}\cdot \nabla \theta \cdot \partial _t\theta \mathrm {d}x+\int \rho \tilde{u}e_2\cdot \partial _t\theta \mathrm {d}x\\&\quad \le \frac{1}{2}\int \rho (\partial _t\theta )^2\mathrm {d}x+C\Vert \tilde{u}\Vert _{L^\infty }^2\Vert \nabla \theta \Vert _{L^2}^2+C\Vert \tilde{u}\Vert _{L^2}^2, \end{aligned}$$

which gives

$$\begin{aligned} \Vert \theta \Vert _{L^\infty (0,T;H^1)}\le C \end{aligned}$$

(3.23)

if $AT\le 1$ and $T\le 1$.

Taking $\partial _t$ to (3.4a), testing by $\partial _t\theta$, using Lemma 3.2, we compute

$$\begin{aligned}&\frac{1}{2}\frac{\mathrm {d}}{\mathrm {d}t}\int \rho (\partial _t\theta )^2\mathrm {d}x +k\int |\nabla \partial _t\theta |^2\mathrm {d}x\\&\quad =\int (\partial _t\rho \tilde{u}e_2+\rho \partial _t\tilde{u}e_2)\partial _t\theta \mathrm {d}x -\int (\partial _t\rho \tilde{u}+\rho \partial _t\tilde{u})\cdot \nabla \theta \cdot \partial _t\theta \mathrm {d}x -\int \partial _t\rho (\partial _t\theta )^2\mathrm {d}x\\&\quad \le (\Vert \partial _t\rho \Vert _{L^4}\Vert \tilde{u}\Vert _{L^4}+\Vert \rho \Vert _{L^\infty }\Vert \partial _t\tilde{u}\Vert _{L^2})\Vert \partial _t\theta \Vert _{L^2}\\&\quad +(\Vert \partial _t\rho \Vert _{L^8}\Vert \tilde{u}\Vert _{L^8}+\Vert \rho \Vert _{L^\infty }\Vert \partial _t\tilde{u}\Vert _{L^4})\Vert \nabla \theta \Vert _{L^2}\Vert \partial _t\theta \Vert _{L^4} +\Vert \partial _t\rho \Vert _{L^2}\Vert \partial _t\theta \Vert _{L^4}^2\\&\quad \le CA\Vert \partial _t\theta \Vert _{L^2}+CA\Vert \partial _t\theta \Vert _{L^4}+C\Vert \partial _t\tilde{u}\Vert _{L^4}\Vert \partial _t\theta \Vert _{L^4}+CA\Vert \partial _t\theta \Vert _{L^4}^2\\&\quad \le \frac{k}{2}\Vert \nabla \partial _t\theta \Vert _{L^2}^2+CA^2+C\Vert \partial _t\tilde{u}\Vert _{L^2}\Vert \partial _t\tilde{u}\Vert _{H^1}+CA^2\Vert \partial _t\theta \Vert _{L^2}^2, \end{aligned}$$

which gives

$$\begin{aligned} \Vert \partial _t\theta \Vert _{L^\infty (0,T;L^2)}+\Vert \partial _t\theta \Vert _{L^2(0,T;H^1)}\le C \end{aligned}$$

(3.24)

if $A^2T\le 1$ and $T\le 1$.

On the other hand, using the $H^2$-theory of Poisson equation, it is clear that

$$\begin{aligned} \Vert \theta \Vert _{H^2}\le & {} C\Vert \rho \partial _t\theta +\rho \tilde{u}\cdot \nabla \theta -\rho \tilde{u}e_2\Vert _{L^2}\\\le & {} C+\Vert \tilde{u}\Vert _{L^4}\Vert \nabla \theta \Vert _{L^4}\le C+C\Vert \nabla \theta \Vert _{L^4}\\\le & {} C+C\Vert \nabla \theta \Vert _{L^2}^\frac{1}{2}\Vert \theta \Vert _{H^2}^\frac{1}{2}\le \frac{1}{2}\Vert \theta \Vert _{H^2}+C, \end{aligned}$$

and hence

$$\begin{aligned} \Vert \theta \Vert _{L^\infty (0,T;H^2)}\le C. \end{aligned}$$

(3.25)

Similarly,

$$\begin{aligned} \Vert \theta \Vert _{W^{2,4}}\le & {} C\Vert \rho \partial _t\theta +\rho \tilde{u}\cdot \nabla \theta -\rho \tilde{u}e_2\Vert _{L^4}\\\le & {} C\Vert \partial _t\theta \Vert _{L^4}+C\Vert \tilde{u}\Vert _{L^8}\Vert \nabla \theta \Vert _{L^8}+C\\\le & {} C\Vert \partial _t\theta \Vert _{L^4}+C, \end{aligned}$$

which yields

$$\begin{aligned} \Vert \theta \Vert _{L^2(0,T;W^{2,4})}\le C \end{aligned}$$

(3.26)

if $T\le 1$.

This completes the proof. $\square$

Lemma 3.5

Let $\tilde{u}\in \mathcal {A}$ be given and $u_0\in H_0^1\cap H^2$ and $\mathrm {div}\,u_0=0$ in $\Omega$. Then, the problem (3.5) has a unique solution u satisfying

$$\begin{aligned} \Vert u\Vert _\mathcal {A}\le C_1 \end{aligned}$$

(3.27)

for some $0<T\le 1$.

Proof

Since Eq. (3.5a) is linear with regular $(\rho ,\tilde{u},b,\theta )$, the existence and uniqueness are well known; we only need to prove the a priori estimates.

Testing (3.5a) by u and using (3.5b) and (3.2a), we have

$$\begin{aligned}&\frac{1}{2}\frac{\mathrm {d}}{\mathrm {d}t}\int \rho |u|^2\mathrm {d}x+\mu \int |\nabla u|^2\mathrm {d}x=\int (b\cdot \nabla b+\rho \theta e_2)u\mathrm {d}x\\&\quad \le (\Vert b\Vert _{L^\infty }\Vert \nabla b\Vert _{L^2}+\Vert \rho \Vert _{L^\infty }\Vert \theta \Vert _{L^2})\Vert u\Vert _{L^2}\\&\quad \le C\Vert \sqrt{\rho }u\Vert _{L^2}, \end{aligned}$$

which gives

$$\begin{aligned} \Vert u\Vert _{L^\infty (0,T;L^2)}+\Vert u\Vert _{L^2(0,T;H^1)}\le C_1. \end{aligned}$$

(3.28)

Testing (3.5a) by $\partial _tu$, using Lemmas 3.2-3.4, we achieve

$$\begin{aligned}&\frac{\mu }{2}\frac{\mathrm {d}}{\mathrm {d}t}\int |\nabla u|^2\mathrm {d}x+\int \rho |\partial _tu|^2\mathrm {d}x=\int (b\cdot \nabla b+\rho \theta e_2-\rho \tilde{u}\cdot \nabla u)\partial _tu\mathrm {d}x\\&\quad \le (\Vert b\Vert _{L^\infty }\Vert \nabla b\Vert _{L^2}+\Vert \rho \Vert _{L^\infty }\Vert \theta \Vert _{L^2}+\Vert \rho \Vert _{L^\infty }\Vert \tilde{u}\Vert _{L^\infty }\Vert \nabla u\Vert _{L^2})\Vert \partial _tu\Vert _{L^2}\\&\quad \le \frac{1}{2}\int \rho |\partial _tu|^2\mathrm {d}x+C+C\Vert \tilde{u}\Vert _{L^\infty }^2\Vert \nabla u\Vert _{L^2}^2, \end{aligned}$$

which implies

$$\begin{aligned} \Vert u\Vert _{L^\infty (0,T;H^1)}+\Vert u\Vert _{L^2(0,T;H^2)}\le C_1 \end{aligned}$$

(3.29)

if $AT\le 1$ and $T\le 1$.

Taking $\partial _t$ to (3.5a), testing by $\partial _tu$, using Lemmas 3.2–3.4, we reach

$$\begin{aligned}&\frac{1}{2}\frac{\mathrm {d}}{\mathrm {d}t}\int \rho |\partial _tu|^2\mathrm {d}x +\mu \int |\nabla \partial _tu|^2\mathrm {d}x\\&\quad =-\int (\partial _tb\otimes b+b\otimes \partial _tb):\nabla \partial _tu\mathrm {d}x +\int (\partial _t\rho \theta +\rho \partial _t\theta )e_2\cdot \nabla u\mathrm {d}x\\&\quad -\int \partial _t\rho |\partial _tu|^2\mathrm {d}x-\int (\partial _t\rho \tilde{u}+\rho \partial _t\tilde{u})\cdot \nabla u\cdot \partial _tu\mathrm {d}x\\&\quad \le C\Vert b\Vert _{L^\infty }\Vert \partial _tb\Vert _{L^2}\Vert \nabla \partial _tu\Vert _{L^2} +C(\Vert \partial _t\rho \Vert _{L^2}\Vert \theta \Vert _{L^\infty } +\Vert \rho \Vert _{L^\infty }\Vert \partial _t\theta \Vert _{L^2})\Vert \partial _tu\Vert _{L^2}\\&\quad +\Vert \partial _t\rho \Vert _{L^2}\Vert \partial _tu\Vert _{L^4}^2+C(\Vert \partial _t\rho \Vert _{L^2}\Vert \tilde{u}\Vert _{L^\infty }+\Vert \rho \Vert _{L^\infty }\Vert \partial _t\tilde{u}\Vert _{L^2})\Vert \nabla u\Vert _{L^4}\Vert \partial _tu\Vert _{L^4}\\&\quad \le C\Vert \nabla \partial _tu\Vert _{L^2}+CA\Vert \partial _tu\Vert _{L^2}+C\Vert \partial _tu\Vert _{L^2}\\&\quad CA\Vert \partial _tu\Vert _{L^2}\Vert \nabla \partial _tu\Vert _{L^2}+CA^2\Vert \partial _tu\Vert _{L^4} +CA\Vert \partial _tu\Vert _{L^4}\\&\quad \le \frac{\mu }{2}\Vert \nabla \partial _tu\Vert _{L^2}^2+C+CA^2+CA^2\Vert \partial _tu\Vert _{L^2}^2+CA^4, \end{aligned}$$

which implies

$$\begin{aligned} \Vert \partial _tu\Vert _{L^\infty (0,T;L^2)}+\Vert \partial _tu\Vert _{L^2(0,T;H^1)}\le C_1 \end{aligned}$$

(3.30)

if $A^2T\le 1, A^4T\le 1$ and $T\le 1$.

Using the $H^2$-theory of Stokes system, it follows from (3.5a) and (3.5b) that

$$\begin{aligned} \Vert u\Vert _{H^2}\le & {} C\Vert \rho \partial _tu+\rho \tilde{u}\cdot \nabla u-b\cdot \nabla b-\rho \theta e_2\Vert _{L^2}\\\le & {} C+C\Vert \tilde{u}\Vert _{L^4}\Vert \nabla u\Vert _{L^4}\le C+C\Vert \nabla u\Vert _{L^4}\\\le & {} C+C\Vert \nabla u\Vert _{L^2}^\frac{1}{2}\Vert u\Vert _{H^2}^\frac{1}{2}\le \frac{1}{2}\Vert u\Vert _{H^2}+C, \end{aligned}$$

which implies

$$\begin{aligned} \Vert u\Vert _{L^\infty (0,T;H^2)}\le C_1. \end{aligned}$$

(3.31)

Similarly,

$$\begin{aligned} \Vert u\Vert _{W^{2,4}}\le & {} C\Vert \rho \partial _tu+\rho \tilde{u}\cdot \nabla u-b\cdot \nabla b-\rho \theta e_2\Vert _{L^4}\\\le & {} C\Vert \partial _tu\Vert _{L^4}+\Vert \tilde{u}\Vert _{L^8}\Vert \nabla u\Vert _{L^8}+C\\\le & {} C\Vert \partial _tu\Vert _{L^4}+C, \end{aligned}$$

which implies

$$\begin{aligned} \Vert u\Vert _{L^2(0,T;W^{2,4})}\le C_1. \end{aligned}$$

(3.32)

This completes the proof. $\square$

Due to Lemmas 3.2–3.5, 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 3.6

There is a constant $0<\delta <1$ such that for any $\tilde{u}_i\ \ (i=1,2)$,

$$\begin{aligned} \Vert F(\tilde{u}_1)-F(\tilde{u}_2)\Vert _{L^2(0,T;H^1)}\le \delta \Vert \tilde{u}_1-\tilde{u}_2\Vert _{L^2(0,T;H^1)} \end{aligned}$$

(3.33)

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

Proof

Suppose $(\rho _i,u_i,\pi _i,b_i,\theta _i)\ (i=1,2)$ are the solutions to the problem (3.2)-(3.5) corresponding to $\tilde{u}_i\ (i=1,2)$. Denote

$$\begin{aligned}(\rho ,u,\pi ,b,\theta ,\tilde{u}):=(\rho _1-\rho _2,u_1-u_2,\pi _1-\pi _2,b_1-b_2,\theta _1-\theta _2,\tilde{u}_1-\tilde{u}_2).\end{aligned}$$

Then, we have

$$\begin{aligned}&\partial _t\rho +\tilde{u}_1\cdot \nabla \rho =-\tilde{u}\cdot \nabla \rho _2, \end{aligned}$$

(3.34)

$$\begin{aligned}&\partial _tb-\eta \Delta b=-\tilde{u}_1\cdot \nabla b-\tilde{u}\cdot \nabla \nabla b_2+b_1\cdot \nabla \tilde{u}+b\cdot \nabla \tilde{u}_2, \end{aligned}$$

(3.35)

$$\begin{aligned}&\rho _1\partial _t\theta +\rho _1\tilde{u}_1\cdot \nabla \theta -k\Delta \theta =(\rho _1\tilde{u}_1-\rho _2\tilde{u}_2)e_2-(\rho _1\tilde{u}_1-\rho _2\tilde{u}_2)\nabla \theta _2-\rho \partial _t\theta _2 \end{aligned}$$

(3.36)

$$\begin{aligned}&\rho _1\partial _tu+\rho _1\tilde{u}_1\cdot \nabla u-\mu \Delta u+\nabla (\pi _1-\pi _2)+\rho \partial _tu_2+(\rho _1\tilde{u}_1-\rho _2\tilde{u}_2)\nabla u_2\nonumber \\&=\mathrm {div}\,(b_1\otimes b_1-b_2\otimes b_2)+(\rho _1\theta _1-\rho _2\theta _2)e_2. \end{aligned}$$

(3.37)

Testing (3.34) by $\rho$, we have

$$\begin{aligned} \frac{\mathrm {d}}{\mathrm {d}t}\int \rho ^2\mathrm {d}x\le & {} C\Vert \tilde{u}\Vert _{L^4}\Vert \nabla \rho _2\Vert _{L^4}\Vert \rho \Vert _{L^2}\le C\Vert \tilde{u}\Vert _{L^4}\Vert \rho \Vert _{L^2}\nonumber \\\le & {} C\Vert \nabla \tilde{u}\Vert _{L^2}\Vert \rho \Vert _{L^2}\nonumber \\\le & {} \epsilon _1\Vert \nabla \tilde{u}\Vert _{L^2}^2+C\Vert \rho \Vert _{L^2}^2 \end{aligned}$$

(3.38)

for any $0<\epsilon _1<1$.

Testing (3.35) by b, we obtain

$$\begin{aligned}&\frac{1}{2}\frac{\mathrm {d}}{\mathrm {d}t}\int |b|^2\mathrm {d}x+\eta \int (\mathrm {rot}\,b)^2\mathrm {d}x\nonumber \\&\quad \le \Vert \tilde{u}\Vert _{L^4}\Vert \nabla b_2\Vert _{L^4}\Vert b\Vert _{L^2}+\Vert b_1\Vert _{L^\infty }\Vert \nabla \tilde{u}\Vert _{L^2}\Vert b\Vert _{L^2}+\Vert \nabla \tilde{u}_2\Vert _{L^\infty }\Vert b\Vert _{L^2}^2\nonumber \\&\quad \le C\Vert \nabla \tilde{u}\Vert _{L^2}\Vert b\Vert _{L^2}+\Vert \nabla \tilde{u}_2\Vert _{L^\infty }\Vert b\Vert _{L^2}^2\nonumber \\&\quad \le \epsilon _2\Vert \nabla \tilde{u}\Vert _{L^2}^2+C\Vert b\Vert _{L^2}^2+\Vert \nabla \tilde{u}_2\Vert _{L^\infty }\Vert b\Vert _{L^2}^2 \end{aligned}$$

(3.39)

for any $0<\epsilon _2<1$.

Testing (3.36) by $\theta$, using $\partial _t\rho _1+\mathrm {div}\,(\rho _1\tilde{u}_1)=0$, we deduce that

$$\begin{aligned}&\frac{1}{2}\frac{\mathrm {d}}{\mathrm {d}t}\int \rho _1\theta ^2\mathrm {d}x +k\int |\nabla \theta |^2\mathrm {d}x\nonumber \\&\quad \le C(\Vert \rho \Vert _{L^2}\Vert \tilde{u}_1\Vert _{L^\infty }+\Vert \tilde{u}\Vert _{L^2})\Vert \theta \Vert _{L^2} +C(\Vert \rho \Vert _{L^2}\Vert \tilde{u}_1\Vert _{L^\infty }+\Vert \tilde{u}\Vert _{L^2})\Vert \nabla \theta _2\Vert _{L^4}\Vert \theta \Vert _{L^4}\nonumber \\&\quad +C\Vert \rho \Vert _{L^2}\Vert \partial _t\theta _2\Vert _{L^4}\Vert \theta \Vert _{L^4}\nonumber \\&\quad \le C(\Vert \rho \Vert _{L^2}+\Vert \tilde{u}\Vert _{L^2})\Vert \theta \Vert _{L^2}+C(\Vert \rho \Vert _{L^2}+\Vert \tilde{u}\Vert _{L^2})\Vert \theta \Vert _{L^4}+C\Vert \rho \Vert _{L^2}\Vert \partial _t\theta _2\Vert _{L^4}\Vert \theta \Vert _{L^4}\nonumber \\&\quad \le \frac{k}{2}\Vert \nabla \theta \Vert _{L^2}^2+\epsilon _3\Vert \tilde{u}\Vert _{L^2}^2+C\Vert \rho \Vert _{L^2}^2+C\Vert \theta \Vert _{L^2}^2 +C\Vert \partial _t\theta _2\Vert _{L^4}^2\Vert \rho \Vert _{L^2}^2 \end{aligned}$$

(3.40)

for any $0<\epsilon _3<1$.

Testing (3.37) by u and using $\partial _t\rho _1+\mathrm {div}\,(\rho _1\tilde{u}_1)=0$, we have

$$\begin{aligned}&\frac{1}{2}\frac{\mathrm {d}}{\mathrm {d}t}\int \rho _1|u|^2\mathrm {d}x+\mu \int |\nabla u|^2\mathrm {d}x\nonumber \\&\quad \le C\Vert b\Vert _{L^2}(\Vert b_1\Vert _{L^\infty }+\Vert b_2\Vert _{L^\infty })\Vert \nabla u\Vert _{L^2}+C(\Vert \rho \Vert _{L^2}\Vert \theta _1\Vert _{L^\infty }+\Vert \theta \Vert _{L^2})\Vert u\Vert _{L^2}\nonumber \\&\quad +C(\Vert \rho \Vert _{L^2}\Vert \tilde{u}_1\Vert _{L^\infty }+\Vert \tilde{u}\Vert _{L^2})\Vert \nabla u_2\Vert _{L^4}\Vert u\Vert _{L^4}+C\Vert \rho \Vert _{L^2}\Vert \partial _tu_2\Vert _{L^4}\Vert u\Vert _{L^4}\nonumber \\&\quad \le C\Vert b\Vert _{L^2}\Vert \nabla u\Vert _{L^2}+C(\Vert \rho \Vert _{L^2}+\Vert \theta \Vert _{L^2})\Vert u\Vert _{L^2}\nonumber \\&\quad +C(\Vert \rho \Vert _{L^2}+\Vert \tilde{u}\Vert _{L^2})\Vert u\Vert _{L^4}+C\Vert \rho \Vert _{L^2}\Vert \partial _tu_2\Vert _{L^4}\Vert u\Vert _{L^4}\nonumber \\&\quad \le \frac{\mu }{2}\Vert \nabla u\Vert _{L^2}^2+\epsilon _4\Vert \tilde{u}\Vert _{L^2}^2+C\Vert \rho \Vert _{L^2}^2+C\Vert \theta \Vert _{L^2}^2\nonumber \\&\quad +C\Vert b\Vert _{L^2}^2+C\Vert u\Vert _{L^2}^2+C\Vert \partial _tu_2\Vert _{L^4}^2\Vert \rho \Vert _{L^2}^2 \end{aligned}$$

(3.41)

for any $0<\epsilon _4<1$.

Combining (3.38), (3.39), (3.40) and (3.41) and taking $\epsilon _i\ (i=1,2,3,4)$ small enough, and using the Gronwall inequality, we conclude that (3.33) holds true for small $0<T<<1$.

This completes the proof. $\square$

Proof of Theorem 3.1

By Lemmas 3.2–3.6 and the Banach fixed-point theorem, we finish the proof. $\square$