Large time behavior of solutions to a Stokes-Magneto equations in three dimensions

Abstract

This paper is devoted to the large time decay of solutions of a three-dimensional Stokes-Magneto equations. It is shown that, when initial data belong to $L^2$, weak solutions of the equations decay to zero in $L^{3/2,\infty }\times L^2$ without a uniform rate, and this decay estimate is optimal. Furthermore, the optimal temporal decay estimates for weak solutions are established when initial data belongs to $L^1\cap L^2$.

1 Introduction

In this paper, we study the following equations

$$\begin{aligned} -\nu \Delta u+\nabla p_*&=b\cdot \nabla b \ \ \ \text{ in }\ \ \ \mathbb {R}_+\times \mathbb {R}^3, \end{aligned}$$

(1.1)

$$\begin{aligned} \partial _tb+u\cdot \nabla b-\eta \Delta b&=b\cdot \nabla u \ \ \ \text{ in }\ \ \ \mathbb {R}_+\times \mathbb {R}^3, \end{aligned}$$

(1.2)

$$\begin{aligned} \nabla \cdot u=0,\ \ \nabla \cdot b&=0, \end{aligned}$$

(1.3)

$$\begin{aligned} b|_{t=0}&=b_0. \end{aligned}$$

(1.4)

Here u is the velocity field, b is the magnetic field, $p_*=p+\frac{1}{2}|b|^2$ is the total pressure, p is the pressure, $\nu >0$ is the viscosity coefficient and $\eta >0$ is the magnetic resistivity coefficient.

Equations (1.1)–(1.3) is obtained by removing the advective terms $(\partial _t+u\cdot \nabla )u$ from the u equation of the magnetohydrodynamics (MHD) equations. It is well-known that MHD equations, which was first derived by Alfvén, govern the motion of the electrically conducting fluids arising from plasmas, liquid metals, and electrolytes, etc (see [12]). It is also known that MHD equations are one of the most important equations in the study of phenomena arising from geophysics, astrophysics, cosmology and engineering (see, e.g., [2, 5]).

Equation (1.1)–(1.3) is closely connected with the method of magnetic relaxation (see [14]). When $\eta =0$, Moffatt [16] argued that (1.1)–(1.3) on a smooth bounded domain $\Omega$ should produce a magnetostatic equilibrium $b^E(x)$ that satisfies

$$\begin{aligned} j^E\times b^E=\nabla p^E, \ \ \ j^E=\nabla \times b^E, \ \ \ \nabla \cdot b^E=0 \ \ \text{ in }\ \ \Omega ,\ \ \ \ \ \ \ \ b^E\cdot n=0\ \ \text{ on }\ \partial \Omega , \end{aligned}$$

(1.5)

if the topology of the magnetic field is non-trivial. Note that (1.5) almost shares the same form with the steady Euler equation:

$$\begin{aligned} u^E\times \omega ^E=\nabla h^E, \ \ \ \omega ^E=\nabla \times u^E, \ \ \ \nabla \cdot u^E=0 \ \ \text{ in }\ \ \Omega ,\ \ \ \ \ \ \ \ u^E\cdot n=0\ \ \text{ on }\ \partial \Omega , \end{aligned}$$

(1.6)

if one “identifies” $b^E$ with velocity field $u^E$, here $\nabla h^E=\nabla (p_e+\frac{1}{2}|u^E|^2)$ and $p_e$ denotes the pressure of the Euler equation. This indicates that the study of (1.1)–(1.3) might be helpful to understand the unstable Euler flows. Moffatt also argued that the steady state of some non-resistive MHD equations should also obey (1.5) (see [15]). However, there is no rigorous proof that the magnetic relaxation will yield a steady Euler flow. One of the reasons is that the global well-posedness of 3D MHD equations remains open (see [14] and references therein).

From a limiting state point of view, the dynamical model used to obtain the above steady state is not particularly important (see [8, 14, 16]). In fact, it was argued by Moffatt that dropping the acceleration terms from the u equation and working with a “Stokes” model might prove more mathematically amenable (see [8, 14, 16]). In recent years, the well-posedness of (1.1)–(1.4) and related models have attracted great attention. McCormick et al. proved the existence of weak solutions of the equations in [14], where the uniqueness of weak solutions for two-dimensional case is also shown. Furthermore, they proved that weak solutions of the 2D equations become regular if $b_0$ is smooth (see [14]). We refer readers to [4, 8] for the local-in-time existence of regular solutions of 3D non-resistive MHD equations. Recently, we established an optimal regularity criterion for (1.1)–(1.4), and studied the global-in-time existence of strong solutions when initial data is small in critical Sobolev spaces or critical Besov spaces (see [22]). We also established global-in-time existence of strong solutions of the equations with arbitrary initial data when $-\Delta$ in (1.2) is replaced by $(-\Delta )^\alpha$ with $\alpha \ge 3/2$ (see [10]).

The purpose of this paper is to investigate the decay of weak solutions of (1.1)–(1.4). The analysis of decay of solutions of fluid flow motions originally goes back to Leray [13], in which he asked whether or not weak solutions of 3D Navier–Stokes equations decay to zero in $L^2$ as time tends to infinity. Since then, this kind of problem has been extensively studied, see [3, 17,18,19,20] for Navier–Stokes equations and [1, 6, 7, 21] for MHD equations. In this paper, motivated by the work of [1, 20], we show that the $L^{3/2,\infty }$ norm of velocity u and $L^2$ norm of magnetic field b are decay to zero without a uniform rate when initial data belong to $L^2$. It stated as follows:

Theorem 1.1

Let $b_0\in L^2$ with $\nabla \cdot b_0=0$. Assume that (u, b) a weak solution of the initial value problem (1.1)–(1.4). Then (u, b) satisfies

$$\begin{aligned} \lim _{t\rightarrow \infty }(\Vert u(t)\Vert _{L^{3/2,\infty }}+\Vert b(t)\Vert _{L^2})=0. \end{aligned}$$

(1.7)

For the proof, it should be pointed out that in contrast with Navier–Stokes equations or MHD, there is no a priori bound for u in $L^\infty (\mathbb {R}_+,L^2(\mathbb {R}^3))$ because of the absence of $\partial _tu$ in (1.1). Note that this estimate for u is crucial to study the corresponding part for Navier–Stokes or MHD (see [1, 17, 20]). However, by the energy estimate of solutions, we would overcome this difficulty by some proper interpolation inequalities, see Sect. 3 below.

Furthermore, we show that the decay result obtained in Theorem 1.1 is optimal in the sense that for any sphere with radius $\alpha$ in $L^2(\mathbb {R}^3)$, there exists a $b_0$ on the sphere such that the corresponding solutions should decay arbitrarily slow. The result is stated as follows:

Theorem 1.2

For any $T>0$, $\alpha >0$ and $0<\epsilon <1$, there exists $b_0\in L^2$ with $\nabla \cdot b_0=0$ and $\Vert b_0\Vert _{L^2}=\alpha$, such that if (u, b) is a weak solution of (1.1)–(1.4) corresponding to the initial data $b_0$, then $\frac{\Vert b(T)\Vert _{L^2}}{\Vert b_0\Vert _{L^2}}\ge 1-\epsilon$.

Motivated by [1, 20], we prove this result by choosing a suitable scaling transform $\delta _\lambda$ on $L^2$ that preserves $L^2$-norm. Since this scaling does not preserve the semi-norm in $\dot{H}^1$, we assume without loss of generality that $b_0$ belongs to a more regular space, say $H^1$. Taking $\delta _\lambda b_0$ as the initial data, we establish a global-in-time bound of solutions of (1.1)–(1.4) when $\lambda$ is small, then we show Theorem 1.2, see Sect. 4.

The non-uniform decay of weak solutions derived in Theorem 1.1 can be improved if initial data satisfies some additional assumptions. More precisely, it is shown that when $b_0$ belongs to $L^1\cap L^2$, the $L^2$ norm of b(t) will decay like $O(t^{-3/4})$ as $t\rightarrow \infty$. This indicates that, on the one hand, the temporal decay of $\Vert b(t)\Vert _{L^2}$ can be uniformly dominated by $t^{-3/4}$ (in the sense of ‘$\lesssim$’). On the other hand, this decay rate is optimal in the sense that the lower bound for rate of decay is proportional to $t^{-3/4}$. The two results are stated as follows:

Theorem 1.3

Let $b_0\in L^1\cap L^2$ with $\nabla \cdot b_0=0$. Then there exists a Leray-Hopf weak solution of the initial value problem (1.1)–(1.4), which satisfies

$$\begin{aligned} \Vert u(t)\Vert _{L^{3/2,\infty }}\le c(\nu ,\eta ,\Vert b_0\Vert _{L^1\cap L^2})(1+t)^{-\frac{3}{2}},\ \ \ \ \ \ \Vert b(t)\Vert _{L^2}\le c(\nu ,\eta ,\Vert b_0\Vert _{L^1\cap L^2})(1+t)^{-\frac{3}{4}}. \end{aligned}$$

(1.8)

Theorem 1.4

Let $R_\beta =\{f\in L^1:\inf _{|\xi |\le \beta }|\hat{f}(\xi )|\ge \beta \}$ for $\beta >0$ and let $b_0\in L^1\cap L^2\cap R_\beta$ with $\nabla \cdot b_0=0$. Then there exists a Leray-Hopf weak solution of (1.1)–(1.4) such that

$$\begin{aligned} \Vert b(t)\Vert _{L^2}\ge c(\nu ,\eta ,\Vert b_0\Vert _{L^1\cap L^2},\beta )(1+t)^{-\frac{3}{4}}. \end{aligned}$$

(1.9)

Remark 1.5

The space $R_\beta$ plays a crucial role in the proof of the optimal decay rate of solutions. When $b_0\in L^2\cap R_\beta$, the weak solution of the heat equation

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _tb'-\eta \Delta b'=0 \ \ \ \text{ in } \ \ \ \mathbb {R}^+\times \mathbb {R}^3, \\ b'(0)=b_0, \end{array}\right. } \end{aligned}$$

(1.10)

decays at most as $\Vert b'(t)\Vert _{L^2}\ge c(\eta ,\beta )(1+t)^{-3/4}$ (see Lemma 2.5 below). Compared with the linear equation (1.10), the system (1.1)–(1.4) contains complicated nonlinearities, but they do not make the decay of solutions worse (see Sect. 6). Thus, if $b_0\in L^2\cap R_\beta$, the component b of the weak solutions (u, b) of the system in general cannot decay faster than $(1+t)^{-3/4}$.

Remark 1.6

The space $R_\beta$ is strictly contained in $L^1$. This means that there are functions that contained in $L^1$ but they do not belong to $R_\beta$. In fact, let $\chi :\mathbb {R}^3\rightarrow [0,1]$ be a smooth function that satisfies $\chi (x)=1$ for $|x|\le 1$ and $\chi (x)=0$ for $|x|\ge 2$. Let $\phi (x)=\chi (x)-\chi (2x)$. Then both $\chi$ and $\phi$ are Schwartz functions on $\mathbb {R}^3$. It is clear that $\mathcal {F}^{-1}\chi \in R_\beta$, and $\mathcal {F}^{-1}\phi \in L^1$ whereas $\mathcal {F}^{-1}\phi \notin R_\beta$, here $\mathcal {F}^{-1}$ denotes the Fourier inverse transform.

The proof of Theorems 1.3 and 1.4 are based on the Fourier splitting method [19]. The main task is to estimate $\hat{b}$ for its lower frequency part. We point out that there is a difficulty similar as that has been stated above. That is, the absence of $\partial _tu$ in (1.1) leads to the absence of a priori bound of u in $L^\infty (\mathbb {R}_+,L^2(\mathbb {R}^3))$, and this is much unlike the Navier–Stokes equations [17, 19, 20] or MHD equations [1, 21]. However, we overcome the difficulty by energy estimate of solutions and applications of interpolation inequalities.

2 Preliminaries

Throughout the paper, c represents a positive constant (depending only on $\nu$, $\eta$) whose value may change at each occurrence. $A\lesssim B$ denotes the inequality $A\le cB$. $c(\alpha _1,\alpha _2,\ldots )$ stands for a positive constant that depends on $\alpha _1$, $\alpha _2$, ...etc. We denote by $\hat{f}$ the Fourier transform of f, while the inverse Fourier transform of f is denoted by $\mathcal {F}^{-1}f$. We consider function spaces on $\mathbb {R}^3$, for instance, $C_c^\infty :=C_c^\infty (\mathbb {R}^3)$, $L^p:=L^p(\mathbb {R}^3)$, $H^s:=H^s(\mathbb {R}^3)$. $L^{p,\infty }$ denotes the weak $L^p$ space. We will use $\int :=\int _{\mathbb {R}^3}$, $\Vert \cdot \Vert _p:=\Vert \cdot \Vert _{L^p}$ and $\Vert \cdot \Vert :=\Vert \cdot \Vert _{2}$ for convenient. We define $\mathcal {D}_\sigma =\{\mathbf{f}\in C_c^\infty :\nabla \cdot \mathbf{f}=0\}$. Let $L_\sigma ^2$ and $H_\sigma ^1$ be the closure of $\mathcal {D}_\sigma$ in the $L^2$ and $H^1$ norm, respectively.

Definition 2.1

[14] Let $T>0$ and let $b_0\in L_\sigma ^2$. A function (u, b) is called a weak solution of the equation (1.1)–(1.4) on (0, T), if

• (i) $u\in L^\infty (0,T;L^{3/2,\infty })\cap L^2(0,T;H^1_\sigma )$ and $b\in L^\infty (0,T;L^2)\cap L^2(0,T;H^1_\sigma )$,

• (ii) (u, b) verifies:

  $$\begin{aligned}&\int \nu \nabla u:\nabla \phi _1+(b\cdot \nabla )\phi _1\cdot b\mathrm{d}x=0, \\&\int b_0\cdot \phi _2(0)\mathrm{d}x{-}\int _0^T\int b\cdot \partial _t\phi _2{-}\eta \nabla b:\nabla \phi _2{+}(u\cdot \nabla ) \phi _2\cdot b{-}(b\cdot \nabla )\phi _2\cdot u\mathrm{d}x\mathrm{d}t{=}0, \end{aligned}$$

  for all test functions $\phi _1$, $\phi _2\in C_c^\infty ([0,T);\mathcal {D}_\sigma )$.

Motivated by Ogawa, Rajopadhye and Schonbek [17] about the decay of weak solutions of forced Navier–Stokes equations or Agapito and Schonbek [1] about the analysis of decay of MHD equations, we formulate the following technical lemma:

Lemma 2.2

Let $b_0\in L^2_\sigma$. Assume that (u, b) is a weak solution of (1.1)–(1.4). Then for $E(t)\in C^1(\mathbb {R};\mathbb {R}_+)$ with $E(t)\ge 0$ and $\psi \in C^1(\mathbb {R};C^1\cap L^2)$ such that $\psi (t)$ is radial on $\mathbb {R}^3$, the solution (u, b) satisfies the following equations:

$$\begin{aligned} 0=-2\nu \int _s^tE(\tau )\Vert \nabla \psi *u(\tau )\Vert ^2\mathrm{d}\tau +2\int _s^tE(\tau )\langle b\cdot \nabla b(\tau ), \psi *\psi *u(\tau )\rangle \mathrm{d}\tau , \end{aligned}$$

(2.1)

and

$$\begin{aligned} E(t)\Vert \psi *b(t)\Vert ^2&=E(s)\Vert \psi *b(s)\Vert ^2+\int _s^tE'(\tau )\Vert \psi *b(\tau )\Vert ^2\mathrm{d}\tau \nonumber \\&\quad +2\int _s^tE(\tau )(\langle \psi '*b(\tau ),\psi *b(\tau )\rangle -\eta \Vert \nabla \psi *b(\tau )\Vert ^2)\mathrm{d}\tau \nonumber \\&\quad -2\int _s^tE(\tau )(\langle u\cdot \nabla b(\tau ),\psi *\psi *b(\tau )\rangle \nonumber \\&\quad -\langle b\cdot \nabla u(\tau ),\psi *\psi *b(\tau )\rangle )\mathrm{d}\tau , \end{aligned}$$

(2.2)

for all $0\le s\le t\le \infty$.

Proof

We first give the proof of (2.2). Taking the inner product of (1.2) with $2E(t)\psi *\psi *b(t)$, then integrating in $[s,t]\times \mathbb {R}^3$, we obtain that

$$\begin{aligned}&2\int _s^t\int \partial _\tau b(\tau )\cdot E(\tau )\psi *\psi *b(\tau )\mathrm{d}x\mathrm{d}\tau \nonumber \\&\qquad -2\eta \int _s^t\int \Delta b(\tau )\cdot E(\tau )\psi *\psi *b(\tau )\mathrm{d}x\mathrm{d}\tau \nonumber \\&\quad = -2\int _s^tE(\tau )\int (u\cdot \nabla b(\tau )-b\cdot \nabla u(\tau ))\cdot \psi *\psi *b(\tau )\mathrm{d}x\mathrm{d}\tau . \end{aligned}$$

(2.3)

The main task is to deal with the terms on the left-hand side of (2.3). For the first term, by integration by parts, we have

$$\begin{aligned}&\int \partial _\tau b(\tau )\cdot E(\tau )\psi *\psi *b(\tau )\mathrm{d}x\nonumber \\&\quad =\int \frac{\mathrm{d}}{\mathrm{d}\tau }\Big (b(\tau )\cdot E(\tau )\psi *\psi *b(\tau )\Big )-b(\tau )\cdot \frac{\mathrm{d}}{\mathrm{d}\tau }\Big (E(\tau )\psi *\psi *b(\tau )\Big )\mathrm{d}x \nonumber \\&\quad =\frac{\mathrm{d}}{\mathrm{d}\tau }\int b(\tau )\cdot E(\tau )\psi *\psi *b(\tau )\mathrm{d}x\nonumber \\&\qquad -\int b(\tau )\cdot \Big (E'(\tau ) \psi *\psi *b(\tau )+2E(\tau )\psi '*\psi *b(\tau )\nonumber \\&\qquad +E(\tau )\psi *\psi *\partial _\tau b(\tau )\Big )\mathrm{d}x, \nonumber \\&\quad =\frac{\mathrm{d}}{\mathrm{d}\tau }\int b(\tau )\cdot E(\tau )\psi *\psi *b(\tau )\mathrm{d}x\nonumber \\&\qquad -\int b(\tau )\cdot \Big (E'(\tau ) \psi *\psi *b(\tau )+2E(\tau )\psi '*\psi *b(\tau )\Big )\mathrm{d}x \nonumber \\&\qquad \ \ \ \ -\int \partial _\tau b(\tau )\cdot E(\tau )\psi *\psi *b(\tau )\mathrm{d}x, \end{aligned}$$

(2.4)

here $\psi$ is radial has been used to derive the third equation. Thus, we apply Parseval’s relation to obtain that

$$\begin{aligned} 2\int _s^t&\int \partial _\tau b(\tau )\cdot E(\tau )\psi *\psi *b(\tau )\mathrm{d}x\mathrm{d}\tau \nonumber \\&=E(t)\int b(t)\cdot \psi *\psi *b(t)\mathrm{d}x-E(s)\int b(s)\cdot \psi *\psi *b(s)\mathrm{d}x \nonumber \\&\ \ \ \ -\int _s^t\int b(\tau )\cdot (E'(\tau )\psi *\psi *b(\tau )+2E(\tau )\psi '*\psi *b(\tau ))\mathrm{d}x\mathrm{d}\tau \nonumber \\&=E(t)\int \hat{b}(t)\cdot (\bar{\hat{\psi }})^2\bar{\hat{b}}(t)\mathrm{d}\xi -E(s)\int \hat{b}(s)\cdot (\bar{\hat{\psi }})^2\bar{\hat{b}}(s)\mathrm{d}\xi \nonumber \\&\ \ \ \ -\int _s^t\Big (E'(\tau )\int \hat{b}(\tau )\cdot (\bar{\hat{\psi }})^2\bar{\hat{b}}(\tau )\mathrm{d}\xi +2E(\tau )\int \hat{b}(\tau )\cdot \bar{\hat{\psi '}}\bar{\hat{\psi }}\bar{\hat{b}}(\tau )\mathrm{d}\xi \Big )\mathrm{d}\tau . \end{aligned}$$

(2.5)

Since $\psi$ is radial, it follows that $\bar{\hat{\psi }}=\hat{\psi }$ and $(\bar{\hat{\psi }})^2=|\hat{\psi }|^2$. Hence the previous equation turns to the following:

$$\begin{aligned}&2\int _s^t\int \partial _\tau b(\tau )\cdot E(\tau )\psi *\psi *b(\tau )\mathrm{d}x\mathrm{d}\tau \nonumber \\&\quad =E(t)\int |\hat{\psi }|^2|\hat{b}(t)|^2\mathrm{d}\xi -E(s)\int |\hat{\psi }|^2|\hat{b}(s)|^2\mathrm{d}\xi \nonumber \\&\qquad -\int _s^t\Big (E'(\tau )\int |\hat{\psi }|^2|\hat{b}(\tau )|^2\mathrm{d}\xi +2E(\tau )\int \hat{\psi }\hat{b}(\tau )\cdot \bar{\hat{\psi '}}\bar{\hat{b}}(\tau )\mathrm{d}\xi \Big )\mathrm{d}\tau \nonumber \\&\quad =E(t)\Vert \psi *b(t)\Vert ^2-E(s)\Vert \psi *b(s)\Vert ^2-\int _s^tE'(\tau )\Vert \psi *b(\tau )\Vert ^2\nonumber \\&\qquad +2E(\tau )\langle \psi '*b(\tau ),\psi *b(\tau )\rangle \mathrm{d}\tau , \end{aligned}$$

(2.6)

here Plancherel’s theorem has been used to deduce the second equation.

For the second term, a similar computation gives that

$$\begin{aligned} -\int _s^t\int \Delta b(\tau )\cdot E(\tau )\psi *\psi *b(\tau )\mathrm{d}x \mathrm{d}\tau&=\int _s^tE(\tau )\int |\xi |^2 \hat{b}(\tau )\cdot (\bar{\hat{\psi }})^2\bar{\hat{b}}(\tau )\mathrm{d}\xi \mathrm{d}\tau \nonumber \\&=\int _s^tE(\tau )\int |\xi |^2|\hat{\psi }|^2|\hat{b}(\tau )|^2\mathrm{d}\xi \mathrm{d}\tau \nonumber \\&=\int _s^tE(\tau )\Vert \nabla \psi *b(\tau )\Vert ^2\mathrm{d}\tau . \end{aligned}$$

(2.7)

Substituting (2.6) and (2.7) into (2.3), we conclude that (2.2) holds.

The proof of (2.1) is slightly simpler and can be shown in a way similar to that of (2.2). Taking the inner product of (1.1) with $2E(t)\psi *\psi *u(t)$, then integrating in $[s,t]\times \mathbb {R}^3$, we obtain that

$$\begin{aligned} -2\nu \int _s^t\int \Delta u(\tau )\cdot E(\tau )\psi *\psi *u(\tau )\mathrm{d}x\mathrm{d}\tau =2\int _s^tE(\tau )\int b\cdot \nabla b(\tau )\cdot \psi *\psi *u(\tau )\mathrm{d}x\mathrm{d}\tau . \end{aligned}$$

(2.8)

By repeating the manipulation of the derivation of (2.7), we know that

$$\begin{aligned} -\int _s^t\int \Delta u(\tau )\cdot E(\tau )\psi *\psi *u(\tau )\mathrm{d}x\mathrm{d}\tau =\int _s^tE(\tau )\Vert \nabla \psi *u(\tau )\Vert ^2\mathrm{d}\tau . \end{aligned}$$

(2.9)

Substituting (2.9) into (2.8), it follows that (2.1) holds. The proof of Lemma 2.2 is completed. $\square$

The following result is a straightforward of Lemma 2.2.

Corollary 2.3

Let $b_0\in L^2_\sigma$. Assume that (u, b) is a weak solution of (1.1)–(1.4). Then for a radial function $\varphi \in L^2$, (u, b) satisfies

$$\begin{aligned}&\Vert \mathcal {F}^{-1}\varphi *b(t)\Vert ^2\le \Vert e^{\eta (t-s)\Delta }\mathcal {F}^{-1}\varphi *b(s)\Vert ^2 \nonumber \\&\quad +2\int _s^t|\langle u\cdot \nabla b(\tau ),e^{2\eta (t-\tau )\Delta }\mathcal {F}^{-1}\varphi ^2*b(\tau )\rangle |\nonumber \\&\quad +|\langle b\cdot \nabla u(\tau ),e^{2\eta (t-\tau )\Delta }\mathcal {F}^{-1}\varphi ^2*b(\tau )\rangle |\mathrm{d}\tau , \end{aligned}$$

(2.10)

for all $0\le s\le t\le \infty$.

Proof

For any $\epsilon >0$, set $\psi (\tau )=\mathcal {F}^{-1}(e^{-\eta |\xi |^2(t+\epsilon -\tau )}\varphi (\xi ))$ and $E(t)=1$ in (2.2), we deduce that

$$\begin{aligned}&\Vert e^{\epsilon \eta \Delta }\mathcal {F}^{-1}\varphi *b(t)\Vert ^2=\Vert e^{\eta (t+\epsilon -s)\Delta } \mathcal {F}^{-1}\varphi *b(s)\Vert ^2 \nonumber \\&\quad +2\int _s^t(\langle \eta (-\Delta )e^{\eta (t+\epsilon -\tau )\Delta }\mathcal {F}^{-1}\varphi *b(\tau ),e^{\eta (t+\epsilon -\tau )\Delta }\mathcal {F}^{-1}\varphi *b(\tau )\rangle \nonumber \\&\quad -\eta \Vert \nabla e^{\eta (t+\epsilon -\tau )\Delta }\mathcal {F}^{-1}\varphi *b(\tau )\Vert ^2)\mathrm{d}\tau \nonumber \\&\quad -2\int _s^t(\langle u\cdot \nabla b(\tau ),e^{2\eta (t+\epsilon -\tau )\Delta }\mathcal {F}^{-1}\varphi ^2*b(\tau )\rangle \nonumber \\&\quad -\langle b\cdot \nabla u(\tau ),e^{2\eta (t+\epsilon -\tau )\Delta }\mathcal {F}^{-1}\varphi ^2*b(\tau )\rangle )\mathrm{d}\tau . \end{aligned}$$

(2.11)

By integration by parts, it is seen that

$$\begin{aligned}&\langle (-\Delta )e^{\eta (t+\epsilon -\tau )\Delta }\mathcal {F}^{-1}\varphi *b(\tau ),e^{\eta (t+\epsilon -\tau )\Delta }\mathcal {F}^{-1}\varphi *b(\tau )\rangle \nonumber \\&\quad =\Vert \nabla e^{\eta (t+\epsilon -\tau )\Delta }\mathcal {F}^{-1}\varphi *b(\tau )\Vert ^2. \end{aligned}$$

(2.12)

Substituting (2.12) into (2.11), we have

$$\begin{aligned}&\Vert e^{\epsilon \eta \Delta }\mathcal {F}^{-1}\varphi *b(t)\Vert ^2\le \Vert e^{\eta (t+\epsilon -s)\Delta } \mathcal {F}^{-1}\varphi *b(s)\Vert ^2 \nonumber \\&\quad +2\int _s^t|\langle u\cdot \nabla b(\tau ),e^{2\eta (t+\epsilon -\tau )\Delta }\mathcal {F}^{-1}\varphi ^2*b(\tau )\rangle |\nonumber \\&\quad +|\langle b\cdot \nabla u(\tau ),e^{2\eta (t+\epsilon -\tau )\Delta }\mathcal {F}^{-1}\varphi ^2*b(\tau )\rangle |\mathrm{d}\tau . \end{aligned}$$

(2.13)

By passing to the limit as $\epsilon \rightarrow 0$ in (2.13), we finally conclude that (2.10) holds true. This completes the proof of Corollary 2.3. $\square$

The following $L^p$-$L^q$ estimate for heat operator will be frequently used in the rest of the paper.

Lemma 2.4

[11]. Let $\mu >0$, $1\le p\le q\le \infty$, $f\in L^p$ and let $m\ge 0$. Then the following $L^p$-$L^q$ estimate holds

$$\begin{aligned} \Vert \nabla ^me^{\mu t\Delta }f\Vert _q\le c(\mu )t^{-\frac{m}{2}-\frac{3}{2}(\frac{1}{p}-\frac{1}{q})}\Vert f\Vert _p, \ \ \ \ \text{ for } \text{ any } \ \ \ t>0. \end{aligned}$$

(2.14)

Lemma 2.5

[20]. Let $\mu >0$, $f\in L^2\cap R_\beta$ for some $\beta >0$. Let $e^{\mu t\Delta }f=\int K_t(x-y)f(y)dy$ with $K_t(x)=\frac{1}{(4\pi \mu t)^{3/2}}e^{-\frac{|x|^2}{4\mu t}}$. Then there exists $c(\mu ,\beta )>0$ such that $\Vert e^{\mu t\Delta }f\Vert \ge c(\mu ,\beta )(1+t)^{-\frac{3}{4}}$.

3 Proof of Theorem 1.1

We begin with the $L^{3/2,\infty }\times L^2$ estimate of (u, b). Taking the inner product of (1.1) and (1.2) with u and b, respectively, then integrating in $\mathbb {R}^3$ and summing the resultant equations, we use integration by parts and (1.3) to obtain that

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\Vert b\Vert ^2+2\nu \Vert \nabla u\Vert ^2+2\eta \Vert \nabla b\Vert ^2=0. \end{aligned}$$

(3.1)

Integrating with respect to t, we deduce that for any $t\ge 0$,

$$\begin{aligned} \Vert b(t)\Vert ^2+2\int _0^t\big (\nu \Vert \nabla u(\tau )\Vert ^2+\eta \Vert \nabla b(\tau )\Vert ^2\big )\mathrm{d}\tau \le \Vert b_0\Vert ^2. \end{aligned}$$

(3.2)

Based on this $L^2$ estimate of b, we can deduce that u(t) is bounded in $L^{3/2,\infty }$ (see [10, 14, 22]). In fact, consider the following nonhomogeneous Stokes equation

$$\begin{aligned} {\left\{ \begin{array}{ll} -\nu \Delta u+\nabla p_*=b\cdot \nabla b, \\ \nabla \cdot u=0, \end{array}\right. } \end{aligned}$$

(3.3)

we know that $(u,p_*)$ is solved by

$$u(t,x)=\int \mathbf{U}(x-y)\cdot (b\cdot \nabla b)(t,y)dy\ \ \ \text{ and }\ \ \ p_*(t,x)=\int \mathbf{q}(x-y)\cdot (b\cdot \nabla b)(t,y)dy,$$

here $(\mathbf{U}(\cdot ),\mathbf{q}(\cdot ))$ is the fundamental solution of Stokes equations and $\mathbf{U}(x)=O(|x|^{-1})$ as either $|x|\rightarrow 0$ or $|x|\rightarrow \infty$, see Section IV.2 in [9] for details. Thus, $\nabla \mathbf{U}\in L^{3/2,\infty }$. Moreover, by $\nabla \cdot b=0$ and Young inequality in weak $L^p$ spaces, we deduce that for any $t\ge 0$,

$$\begin{aligned} \Vert u(t)\Vert _{L^{3/2,\infty }}&=\Big \Vert \int \nabla \mathbf{U}(x-y)(b\otimes b)(t,y)dy\Big \Vert _{L^{3/2,\infty }} \nonumber \\&\lesssim \Vert \nabla \mathbf{U}\Vert _{L^{3/2,\infty }}\Vert (b\otimes b)(t)\Vert _1 \nonumber \\&\lesssim \Vert b(t)\Vert ^2. \end{aligned}$$

(3.4)

Let $\varphi :\mathbb {R}^3\rightarrow [0,1]$ be a smooth, radial cutoff function such that $\varphi (\xi )=1$ for $|\xi |\le 1$, $\varphi (\xi )=0$ for $|\xi |\ge 2$. By Plancherel’s theorem, $\Vert b(t)\Vert =\Vert \hat{b}(t)\Vert \le \Vert \varphi \hat{b}(t)\Vert +\Vert (1-\varphi )\hat{b}(t)\Vert$. It suffices to show that $\lim _{t\rightarrow \infty }\Vert \varphi \hat{b}(t)\Vert =0$ and $\lim _{t\rightarrow \infty }\Vert (1-\varphi )\hat{b}(t)\Vert =0$, respectively.

We apply (2.10) and Plancherel’s theorem to obtain

$$\begin{aligned} \Vert \varphi \hat{b}(t)\Vert ^2&\le \Vert e^{-\eta (t-s)|\xi |^2}\varphi \hat{b}(s)\Vert ^2 \nonumber \\&\quad +2\int _s^t|\langle u\cdot \nabla b(\tau ),e^{2\eta (t-\tau )\Delta }\mathcal {F}^{-1}\varphi ^2*b(\tau )\rangle |\nonumber \\&\quad +|\langle b\cdot \nabla u(\tau ),e^{2\eta (t-\tau )\Delta }\mathcal {F}^{-1}\varphi ^2*b(\tau )\rangle |\mathrm{d}\tau . \end{aligned}$$

(3.5)

By Young inequality, the second term on the right-hand side of (3.5) is bounded by (in the sense of ‘$\lesssim$’)

$$\begin{aligned}&\int _s^t(\Vert u\cdot \nabla b(\tau )\cdot e^{2\eta (t-\tau )\Delta }\mathcal {F}^{-1}\varphi ^2*b(\tau )\Vert _1\nonumber \\&\qquad +\Vert b\cdot \nabla u(\tau )\cdot e^{2\eta (t-\tau )\Delta }\mathcal {F}^{-1}\varphi ^2*b(\tau )\Vert _1)\mathrm{d}\tau \nonumber \\&\quad \lesssim \int _s^t(\Vert u\cdot \nabla b(\tau )\Vert _\frac{3}{2}+\Vert b\cdot \nabla u(\tau )\Vert _\frac{3}{2}) \Vert e^{2\eta (t-\tau )\Delta }\mathcal {F}^{-1}\varphi ^2*b(\tau )\Vert _3\mathrm{d}\tau \nonumber \\&\quad \lesssim \int _s^t(\Vert u(\tau )\Vert _6\Vert \nabla b(\tau )\Vert +\Vert b(\tau )\Vert _6\Vert \nabla u(\tau )\Vert )\Vert e^{2\eta (t-\tau )\Delta }\mathcal {F}^{-1}\varphi ^2\Vert _{\frac{6}{5}}\Vert b(\tau )\Vert \mathrm{d}\tau \nonumber \\&\quad \lesssim \Vert b_0\Vert \int _s^t(\Vert \nabla u(\tau )\Vert ^2+\Vert \nabla b(\tau )\Vert ^2)\mathrm{d}\tau . \end{aligned}$$

(3.6)

Combining (3.5) and (3.6), and passing to the limit as $t\rightarrow \infty$, we use Lebesgue’s dominated convergence theorem and (3.2) to deduce that

$$\begin{aligned} \lim _{t\rightarrow \infty }\Vert \varphi \hat{b}(t)\Vert&\lesssim \lim _{t\rightarrow \infty }\Vert e^{-\eta (t-s)|\xi |^2} \varphi \hat{b}(s)\Vert ^2+\Vert b_0\Vert \int _s^\infty (\Vert \nabla u(\tau )\Vert ^2+\Vert \nabla b(\tau )\Vert ^2)\mathrm{d}\tau \nonumber \\&\lesssim \Vert b_0\Vert \int _s^\infty (\Vert \nabla u(\tau )\Vert ^2+\Vert \nabla b(\tau )\Vert ^2)\mathrm{d}\tau \rightarrow 0 \ \ \ \ \ \text{ as } \ \ \ \ s\rightarrow \infty . \end{aligned}$$

(3.7)

This implies that $\lim _{t\rightarrow \infty }\Vert \varphi \hat{b}(t)\Vert =0$.

Next, we prove that $\lim _{t\rightarrow \infty }\Vert (1-\varphi )\hat{b}(t)\Vert =0$. Let $\gamma >0$. Setting $E(t)=(1+t)^\gamma$ and $\psi =\mathcal {F}^{-1}(1-\varphi )$ in (2.1) and (2.2), then summing the resultant equations, we apply Plancherel’s theorem to deduce that

$$\begin{aligned} E(t)\Vert (1-\varphi )\hat{b}(t)\Vert ^2&\le E(s)\Vert (1-\varphi )\hat{b}(s)\Vert ^2+\int _s^tE'(\tau )\Vert (1-\varphi )\hat{b} (\tau )\Vert ^2\mathrm{d}\tau \nonumber \\&\quad -2\eta \int _s^tE(\tau )\Vert \xi \cdot (1-\varphi )\hat{b}(\tau )\Vert ^2\mathrm{d}\tau \nonumber \\&\quad +2\int _s^tE(\tau )\langle \widehat{b\cdot \nabla b}(\tau ),(1-\varphi )^2\hat{u}(\tau )\rangle \mathrm{d}\tau \nonumber \\&\quad -2\int _s^tE(\tau )(\langle \widehat{u\cdot \nabla b}(\tau ),(1-\varphi )^2\hat{b}(\tau )\rangle \mathrm{d}\tau \nonumber \\&\quad +2\int _s^tE(\tau )\langle \widehat{b\cdot \nabla u}(\tau ),(1-\varphi )^2\hat{b}(\tau )\rangle )\mathrm{d}\tau \nonumber \\&:=\sum _{j=1}^6I_j. \end{aligned}$$

(3.8)

The bound of $I_2+I_3$ is obtained by Fourier splitting method (see [19, 20]). In fact, for each $t>0$, we define $r(t)=\sqrt{\frac{\gamma }{2\eta (1+t)}}$, here $\gamma >0$ is given in the previous paragraph. Thus, there holds

$$\begin{aligned} I_2+I_3&=\int _s^tE'(\tau )\Big (\int _{|\xi |<r(\tau )}+\int _{|\xi |\ge r(\tau )}\Big )|(1-\varphi )\hat{b}(\tau ) |^2\mathrm{d}\xi \mathrm{d}\tau \\&\quad -2\eta \int _s^tE(\tau )\int _{|\xi |\ge r(\tau )}|\xi \cdot (1-\varphi )\hat{b}(\tau )|^2\mathrm{d}\xi \mathrm{d}\tau \\&\le \int _s^tE'(\tau )\int _{|\xi |<r(\tau )}|(1-\varphi )\hat{b}(\tau )|^2\mathrm{d}\xi d\tau +\int _s^t(E'(\tau )\\&\quad -2\eta E(\tau )r^2(\tau ))\int _{|\xi |\ge r(\tau )}|(1-\varphi )\hat{b}(\tau )|^2\mathrm{d}\xi \mathrm{d}\tau . \end{aligned}$$

Since $E'(\tau )-2\eta E(\tau )r^2(\tau )=0$ for all $\tau \in [s,t]$, we deduce that

$$\begin{aligned} I_2+I_3&\le \int _s^tE'(\tau )\int _{|\xi |<r(\tau )}|(1-\varphi )\hat{b}(\tau )|^2\mathrm{d}\xi \mathrm{d}\tau \nonumber \\&\le \sup _{\tau \in [s,t]}\Big (\int _{|\xi |<r(\tau )}|(1-\varphi )\hat{b}(\tau )|^2\mathrm{d}\xi \Big )\int _s^tE'(\tau )\mathrm{d}\tau \nonumber \\&\le E(t)\int _{|\xi |<r(s)}|(1-\varphi )\hat{b}(\tau )|^2\mathrm{d}\xi . \end{aligned}$$

(3.9)

Now we estimate $I_4+I_5+I_6$. Let $\zeta =-2\varphi +\varphi ^2$. Then $\zeta \in C^\infty _c$ and $\mathcal {F}^{-1}\zeta$ belongs to Schwartz space. By the divergence free condition (1.3), there hold $\langle \widehat{u\cdot \nabla b},\hat{b}\rangle =0$, $\langle \widehat{b\cdot \nabla b},\hat{u}\rangle +\langle \widehat{b\cdot \nabla u},\hat{b}\rangle =0$. Thus, we use Plancherel’s theorem to deduce that

$$\begin{aligned} |I_4+I_5+I_6|&\lesssim \int _s^tE(\tau )(|\langle b\cdot \nabla b(\tau ),\mathcal {F}^{-1}\zeta *u(\tau )\rangle |+|\langle u\cdot \nabla b(\tau ),\mathcal {F}^{-1}\zeta *b(\tau )\rangle |\nonumber \\&\quad +|\langle b\cdot \nabla u(\tau ),\mathcal {F}^{-1}\zeta *b(\tau )\rangle |)\mathrm{d}\tau \nonumber \\&\lesssim E(t)\int _s^t\Vert b\cdot \nabla b(\tau )\Vert _1\Vert \mathcal {F}^{-1}\zeta *u(\tau )\Vert _\infty \nonumber \\&\quad +(\Vert u\cdot \nabla b(\tau )\Vert _\frac{3}{2}+\Vert b\cdot \nabla u(\tau )\Vert _\frac{3}{2})\Vert \mathcal {F}^{-1}\zeta *b(\tau )\Vert _3\mathrm{d}\tau \nonumber \\&\lesssim E(t)\int _s^t\Vert b\Vert \Vert \nabla b(\tau )\Vert \Vert \mathcal {F}^{-1}\zeta \Vert _\frac{6}{5}\Vert u(\tau )\Vert _6 +(\Vert u(\tau )\Vert _6\Vert \nabla b(\tau )\Vert \nonumber \\&\quad +\Vert b(\tau )\Vert _6\Vert \nabla u(\tau )\Vert )\Vert \mathcal {F}^{-1}\zeta \Vert _\frac{6}{5} \Vert b(\tau )\Vert \mathrm{d}\tau \nonumber \\&\lesssim E(t)\Vert b_0\Vert \int _s^t(\Vert \nabla u(\tau )\Vert ^2+\Vert \nabla b(\tau )\Vert ^2)\mathrm{d}\tau . \end{aligned}$$

(3.10)

Substituting (3.9) and (3.10) into (3.8), then multiplying the resultant equation by $E(t)^{-1}$, we obtain

$$\begin{aligned}&\Vert (1-\varphi )\hat{b}(t)\Vert ^2\lesssim \frac{E(s)}{E(t)}\Vert (1-\varphi )\hat{b}(s)\Vert ^2\nonumber \\&\quad +\int _{|\xi |<r(s)}| (1-\varphi )\hat{b}(\tau )|^2\mathrm{d}\xi +\Vert b_0\Vert \int _s^t(\Vert \nabla u(\tau )\Vert ^2+\Vert \nabla b(\tau )\Vert ^2)\mathrm{d}\tau . \end{aligned}$$

(3.11)

Passing to the limit as t, $s\rightarrow \infty$, we have

$$\begin{aligned}&\lim _{t\rightarrow \infty }\Vert (1-\varphi )\hat{b}(t)\Vert ^2=\lim _{s\rightarrow \infty }\lim _{t\rightarrow \infty } \Vert (1-\varphi )\hat{b}(t)\Vert ^2 \nonumber \\&\quad \lesssim \lim _{s\rightarrow \infty }\int _{|\xi |<r(s)}|(1-\varphi )\hat{b}(\tau )|^2\mathrm{d}\xi +\lim _{s\rightarrow \infty }\Vert b_0\Vert \int _s^\infty (\Vert \nabla u(\tau )\Vert ^2\nonumber \\&\qquad +\Vert \nabla b(\tau )\Vert ^2)\mathrm{d}\tau =0. \end{aligned}$$

(3.12)

This completes the proof of Theorem 1.1.

4 Proof of Theorem 1.2

We assume that $b_0$ belongs to some Sobolev spaces, that is, $b_0\in H^1$ with $\Vert b_0\Vert =\alpha$. For $\lambda >0$, consider the scaling transformation $\delta _\lambda :L^2\rightarrow L^2$ via $f(\cdot )\mapsto \lambda ^{3/2}(\lambda \cdot )$. It’s clear that it preserves the $L^2$ norm, that is, $\Vert \delta _\lambda f\Vert =\Vert f\Vert$. Let $(u^\lambda ,b^\lambda )$ be a solution of (1.1)–(1.4) with the initial data $\delta _\lambda b_0$, then $(u^\lambda ,b^\lambda )$ satisfies following equations:

$$\begin{aligned} -\nu \Delta u^\lambda +\nabla p_*^\lambda&=b^\lambda \cdot \nabla b^\lambda , \end{aligned}$$

(4.1)

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

(4.2)

$$\begin{aligned} \nabla \cdot u^\lambda =0,\ \ \nabla \cdot b^\lambda&=0, \end{aligned}$$

(4.3)

$$\begin{aligned} b^\lambda |_{t=0}&=\delta _\lambda b_0, \end{aligned}$$

(4.4)

here $p_*^\lambda =p^\lambda +\frac{1}{2}|b^\lambda |^2$.

Lemma 4.1

(i) For any $\lambda >0$, there exists a $T_{\lambda }>0$, such that (4.1)–(4.4) admits a unique strong solution $(u^\lambda ,b^\lambda )\in C([0,T_\lambda );H^{1})\cap L^2(0,T_\lambda ;H^2)$. (ii) There exist $\lambda _0=\lambda _0(\nu ,\eta , \Vert b_0\Vert _{H^1})>0$ and $c(\nu ,\eta ,\Vert b_0\Vert )>0$, such that for any $0<\lambda <\lambda _0$, (4.1)–(4.4) admits a unique global solution $(u^\lambda ,b^\lambda )\in C([0,\infty );H^{1})\cap L^2(0,\infty ;H^2)$, and there holds

$$\begin{aligned} \Vert \nabla u^\lambda (t)\Vert \le c(\nu ,\eta ,\Vert b_0\Vert )\Vert \nabla \delta _\lambda b_0\Vert ^\frac{3}{2},\ \ \ \ \Vert \nabla b^\lambda (t)\Vert \le c(\nu ,\eta ,\Vert b_0\Vert )\Vert \nabla \delta _\lambda b_0\Vert , \end{aligned}$$

(4.5)

for all $t\ge 0$.

Proof of Lemma 4.1.

• (i) The local-in-time existence of strong solutions are ensured by the global existence of weak solutions $(u^\lambda ,b^\lambda )$ (see [14, page 521]) together with a priori $H^1$ estimate near the initial time (see (4.7)–(4.9) below). This a priori estimate also leads to the uniqueness of strong solutions, as well as the continuity with respect to t or initial data. The proof is similar to that of Navier–Stokes equations and thus it is omitted here.

• (ii) By repeating the manipulation of (3.1)–(3.4), we have the following bounds:

  $$\begin{aligned} \Vert u^\lambda (t)\Vert _{L^{3/2,\infty }}\lesssim \Vert b^\lambda (t)\Vert ^2,\ \ \ \ \Vert b^\lambda (t)\Vert ^2 +2\int _0^t\big (\nu \Vert \nabla u^\lambda (\tau )\Vert ^2+\eta \Vert \nabla b^\lambda (\tau )\Vert ^2\big )\mathrm{d}\tau \le \Vert b_0\Vert ^2, \end{aligned}$$

  (4.6)

  for all $t\ge 0$.

Taking the inner product of (4.1) with $u^\lambda$ and integrating in $\mathbb {R}^3$, we use interpolation inequality and (4.6) to obtain $\Vert \nabla u^\lambda (t)\Vert \le c(\nu ,\eta ,\Vert b_0\Vert )\Vert \nabla b^\lambda (t)\Vert ^{3/2}$. Thus, it sufficient to show $\Vert \nabla b^\lambda (t)\Vert \le c(\nu ,\eta ,\Vert b_0\Vert )\Vert \nabla \delta _\lambda b_0\Vert$.

Taking the inner product of (4.1) and (4.2) with $-\Delta u^\lambda$ and $-\Delta b^\lambda$ respectively, integrating in $\mathbb {R}^3$ and then summing the resultant equations, we use $\nabla \cdot u=0$ to obtain that

$$\begin{aligned}&\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}\Vert \nabla b^\lambda \Vert ^2+\nu \Vert \Delta u^\lambda \Vert ^2+\eta \Vert \Delta b^\lambda \Vert ^2 \nonumber \\&\quad =\int \nabla (b^\lambda \cdot \nabla b^\lambda ):\nabla u^\lambda \mathrm{d}x{-}\int \nabla (u^\lambda \cdot \nabla b^\lambda ):\nabla b^\lambda \mathrm{d}x{+}\int \nabla (b^\lambda \cdot \nabla u^\lambda ):\nabla b^\lambda \mathrm{d}x .\nonumber \\ \end{aligned}$$

(4.7)

By (4.3) and integration by parts, the absolute value of the right-hand side of (4.7) is bounded (in the sense of ‘$\lesssim$’) by

$$\begin{aligned} \int |\nabla u^\lambda ||\nabla b^\lambda |^2\mathrm{d}x&\le c\Vert \nabla u^\lambda \Vert _{6}\Vert \nabla b^\lambda \Vert _{\frac{12}{5}}^2 \nonumber \\&\le \frac{\nu }{2}\Vert \Delta u^\lambda \Vert ^2+c\Vert \nabla b^\lambda \Vert ^3\Vert \Delta b^\lambda \Vert \nonumber \\&\le \frac{\nu }{2}\Vert \Delta u^\lambda \Vert ^2+\frac{\eta }{2}\Vert \Delta b^\lambda \Vert ^2+c\Vert \nabla b^\lambda \Vert ^6 \nonumber \\&\le \frac{\nu }{2}\Vert \Delta u^\lambda \Vert ^2+\frac{\eta }{2}\Vert \Delta b^\lambda \Vert ^2+c\Vert \nabla b^\lambda \Vert ^2\Vert b^\lambda \Vert ^2\Vert \Delta b^\lambda \Vert ^2, \end{aligned}$$

(4.8)

where we have used interpolation inequality to derive the second and the fourth inequalities. Substituting (4.8) into (4.7), we apply (4.6) to deduce that

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\Vert \nabla b^\lambda \Vert ^2+\nu \Vert \Delta u^\lambda \Vert ^2+\eta \Vert \Delta b^\lambda \Vert ^2\le c(\nu ,\eta ,\Vert b_0\Vert )\Vert \nabla b^\lambda \Vert ^2\Vert \Delta b^\lambda \Vert ^2. \end{aligned}$$

(4.9)

Since $\Vert \nabla \delta _\lambda b_0\Vert =\lambda \Vert \nabla b_0\Vert$, we choose $\lambda _0$ small enough such that $c(\nu ,\eta ,\Vert b_0\Vert )\Vert \nabla \delta _{\lambda _0}b_0\Vert ^2<\eta$. For any $\lambda \in (0,\lambda _0)$, by the fact that $b^\lambda \in C([0,T_\lambda );H^{1})$, we deduce that there exists a $\tilde{T}_\lambda \in (0,T_\lambda )$ such that $c(\nu ,\eta ,\Vert b_0\Vert )\Vert \nabla \delta _\lambda b_0(t)\Vert ^2<\eta$ for $t\in [0,\tilde{T}_\lambda ]$. Thus, it follows from (4.9) that $\Vert \nabla b^\lambda (t)\Vert ^2\le \Vert \nabla \delta _\lambda b_0\Vert ^2$ for $t\in [0,\tilde{T}_\lambda ]$. By induction, we deduce that (4.5) holds for all $t\ge 0$. $\square$

Assume that $\lambda <\lambda _0$. By Lemma 4.1 (ii), (4.1)–(4.4) admits a unique global strong solution, which is denoted by $(u^\lambda ,b^\lambda )$. By Fourier transformation, $b^\lambda$ is solved as follows:

$$\begin{aligned} b^\lambda (t,x)=e^{\eta t\Delta }\delta _\lambda b_0(x)+\int _0^te^{\eta (t-s)\Delta }(-u^\lambda \cdot \nabla b^\lambda (s) +b^\lambda \cdot \nabla u^\lambda (s))\mathrm{d}s. \end{aligned}$$

(4.10)

Hence

$$\begin{aligned} \Vert b^\lambda (t)\Vert \ge \Vert e^{\eta t\Delta }\delta _\lambda b_0\Vert -\Big \Vert \int _0^te^{\eta (t-s)\Delta } (-u^\lambda \cdot \nabla b^\lambda (s)+b^\lambda \cdot \nabla u^\lambda (s))\mathrm{d}s\Big \Vert . \end{aligned}$$

(4.11)

The main task is to calculate the limit (as $\lambda \rightarrow 0^+$) of terms on the right-hand side of (4.11). For the first term, by Plancherel’s theorem and Lebesgue’s dominated convergence theorem, we know that

$$\begin{aligned} \lim _{\lambda \rightarrow 0^+}\Vert e^{\eta t\Delta }\delta _\lambda b_0\Vert ^2&=\lim _{\lambda \rightarrow 0^+}\int e^{-2\eta t|\xi |^2}|\widehat{\delta _\lambda b_0}(\xi )|^2\mathrm{d}x \nonumber \\&=\lim _{\lambda \rightarrow 0^+}\lambda ^3\int e^{-2\eta t|\xi |^2}|\widehat{b_0(\lambda \cdot )}(\xi )|^2\mathrm{d}x \nonumber \\&=\lim _{\lambda \rightarrow 0^+}\lambda ^{-3}\int e^{-2\eta t|\xi |^2}|\widehat{b_0}(\lambda ^{-1}\xi )|^2\mathrm{d}x \nonumber \\&=\lim _{\lambda \rightarrow 0^+}\int e^{-2\eta t\lambda ^2|\xi |^2}|\widehat{b_0}(\xi )|^2\mathrm{d}x \nonumber \\&=\Vert b_0\Vert ^2. \end{aligned}$$

(4.12)

While for the second term, we claim that

$$\begin{aligned} \lim _{\lambda \rightarrow 0^+}\Big \Vert \int _0^te^{\eta (t-s)\Delta }(-u^\lambda \cdot \nabla b^\lambda (s)+b^\lambda \cdot \nabla u^\lambda (s))\mathrm{d}s\Big \Vert =0. \end{aligned}$$

(4.13)

In fact, for small enough $\lambda >0$, one applies Lemma 2.4 and (4.5) to deduce that

$$\begin{aligned}&\Big \Vert \int _0^te^{\eta (t-s)\Delta }(-u^\lambda \cdot \nabla b^\lambda (s)+b^\lambda \cdot \nabla u^\lambda (s))\mathrm{d}s\Big \Vert \nonumber \\&\quad \le \int _0^t\Vert e^{\eta (t-s)\Delta }(-u^\lambda \cdot \nabla b^\lambda (s)+b^\lambda \cdot \nabla u^\lambda (s))\Vert \mathrm{d}s \nonumber \\&\quad \le c(\eta )\int _0^t(t-s)^{-\frac{1}{4}}(\Vert u^\lambda \cdot \nabla b^\lambda (s)\Vert _\frac{3}{2}+\Vert b^\lambda \cdot \nabla u^\lambda (s)\Vert _\frac{3}{2})\mathrm{d}s \nonumber \\&\quad \le c(\eta )\int _0^t(t-s)^{-\frac{1}{4}}\Vert \nabla u^\lambda (s)\Vert \Vert \nabla b^\lambda (s)\Vert \mathrm{d}s \nonumber \\&\quad \le c(\nu ,\eta ,\Vert b_0\Vert )\Vert \nabla \delta _\lambda b_0\Vert ^{\frac{5}{2}}t^\frac{3}{4} \nonumber \\&\quad \le c(\nu ,\eta ,\Vert b_0\Vert _{H^1})\lambda ^\frac{5}{2}t^\frac{3}{4}. \end{aligned}$$

(4.14)

Passing to the limit as $\lambda \rightarrow 0^+$, it follows that (4.13) holds true.

Multiplying (4.11) by $\Vert b^\lambda (0)\Vert ^{-1}$ and passing to the limit as $\lambda \rightarrow 0^+$, we use $\Vert b^\lambda (0)\Vert =\Vert b_0\Vert$, (4.12) and (4.13) to deduce that

$$\begin{aligned} \lim _{\lambda \rightarrow 0^+}\frac{\Vert b^\lambda (t)\Vert }{\Vert b^\lambda (0)\Vert }&=\frac{1}{\Vert b_0\Vert } \lim _{\lambda \rightarrow 0^+}\Vert b^\lambda (t)\Vert \nonumber \\&=\frac{1}{\Vert b_0\Vert }\lim _{\lambda \rightarrow 0^+}\Vert e^{\eta t\Delta }\delta _\lambda b_0\Vert \nonumber \\&\quad -\frac{1}{\Vert b_0\Vert }\lim _{\lambda \rightarrow 0^+}\Big \Vert \int _0^te^{\eta (t-s)\Delta }(-u^\lambda \cdot \nabla b^\lambda (s)+b^\lambda \cdot \nabla u^\lambda (s))\mathrm{d}s\Big \Vert =1. \end{aligned}$$

(4.15)

This completes the proof of Theorem 1.2.

5 Proof of Theorem 1.3

Step 1. A formal proof of $L^{3/2,\infty }\times L^2$ decay. Assume that (u, b) is a smooth solution of (1.1)–(1.4). Removing $2\nu \Vert \nabla u\Vert ^2$ from (3.1) and using Plancherel’s theorem, it follows that

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\Vert \hat{b}\Vert ^2+2\eta \int |\xi |^2|\hat{b}|^2\mathrm{d}\xi \le 0. \end{aligned}$$

(5.1)

Let $r(t)=\sqrt{\frac{3}{2\eta (1+t)}}$, we deduce that

$$\begin{aligned} \int |\xi |^2|\hat{b}|^2\mathrm{d}\xi&=\int _{|\xi |<r(t)}|\xi |^2|\hat{b}|^2\mathrm{d}\xi +\int _{|\xi |\ge r(t)}|\xi |^2|\hat{b}|^2 \mathrm{d}\xi \nonumber \\&\ge \frac{3}{2\eta (1+t)}\int _{|\xi |\ge r(t)}|\hat{b}|^2\mathrm{d}\xi \nonumber \\&\ge \frac{3}{2\eta (1+t)}\int |\hat{b}|^2\mathrm{d}\xi -\frac{3}{2\eta (1+t)}\int _{|\xi |<r(t)}|\hat{b}|^2\mathrm{d}\xi . \end{aligned}$$

(5.2)

Substituting (5.2) into (5.1), and multiplying the resultant equation by $(1+t)^3$, we obtain

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\Big ((1+t)^3\Vert \hat{b}\Vert ^2\Big )\le 3(1+t)^{2}\int _{|\xi |<r(t)}|\hat{b}|^2\mathrm{d}\xi . \end{aligned}$$

(5.3)

Now we prove that for all $|\xi |<r(t)$, there holds

$$\begin{aligned} |\hat{b}(\xi )|\le c(\nu ,\eta ,\Vert b_0\Vert _{L^1\cap L^2})(1+\xi ^{-1}). \end{aligned}$$

(5.4)

In fact, by taking Fourier transformation to (1.2), $\hat{b}$ is solved as

$$\begin{aligned} \hat{b}(\xi )=e^{-\eta t|\xi |^2}\hat{b}(0)+\int _0^te^{-\eta (t-s)|\xi |^2}(\widehat{b\cdot \nabla u}-\widehat{u\cdot \nabla b})(s)\mathrm{d}s. \end{aligned}$$

(5.5)

Since $b_0\in L^1$, we deduce from (5.5) that

$$\begin{aligned} |\hat{b}(\xi )|\le e^{-\eta t|\xi |^2}\Vert b_0\Vert _1+\int _0^te^{-\eta (t-s)|\xi |^2}|\xi |(\Vert b\otimes u\Vert _1+\Vert u\otimes b\Vert _1)(s)\mathrm{d}s. \end{aligned}$$

(5.6)

By interpolation inequality and (3.2), the second term on the right-hand side of (5.6) is bounded by

$$\begin{aligned}&|\xi |\int _0^te^{-\eta (t-s)|\xi |^2}\Vert u(s)\Vert \Vert b(s)\Vert \mathrm{d}s \nonumber \\&\quad \lesssim |\xi |\Vert b_0\Vert \int _0^te^{-\eta (t-s)|\xi |^2}\Vert u(s)\Vert _{L^{3/2,\infty }}^\frac{2}{3}\Vert \nabla u(s)\Vert ^\frac{1}{3}\mathrm{d}s \nonumber \\&\quad \lesssim |\xi |\Vert b_0\Vert \int _0^te^{-\eta (t-s)|\xi |^2}(\Vert u(s)\Vert _{L^{3/2,\infty }}+\Vert \nabla u(s)\Vert )\mathrm{d}s \nonumber \\&\quad \lesssim |\xi |\Vert b_0\Vert \Big [\Vert b_0\Vert ^2\int _0^te^{-\eta (t-s)|\xi |^2}\mathrm{d}s\nonumber \\&\qquad +\Big (\int _0^t e^{-\eta (t-s)|\xi |^2}\mathrm{d}s\Big )^\frac{1}{2}\Big (\int _0^t\Vert \nabla u(s)\Vert ^2\mathrm{d}s\Big )^\frac{1}{2}\Big ] \nonumber \\&\quad \lesssim \Vert b_0\Vert ^3|\xi |^{-1}+\Vert b_0\Vert ^2, \end{aligned}$$

(5.7)

Substituting (5.7) into (5.6), this proves (5.4).

Integrating (5.3) with respect to t and using (5.4), we deduce that

$$\begin{aligned} \Vert \hat{b}(t)\Vert ^2&\le (1+t)^{-3}\Vert b_0\Vert ^2+c(\nu ,\eta ,\Vert b_0\Vert _{L^1\cap L^2})[(1+t)^{-\frac{3}{2}} +(1+t)^{-\frac{1}{2}}] \nonumber \\&\le c(\nu ,\eta ,\Vert b_0\Vert _{L^1\cap L^2})(1+t)^{-\frac{1}{2}}, \end{aligned}$$

(5.8)

for all $t\ge 0$. Hence, we have

$$\begin{aligned} \Vert b(t)\Vert\le & {} c(\nu ,\eta ,\Vert b_0\Vert _{L^1\cap L^2})(1+t)^{-\frac{1}{4}} \ \ \ \ \ \text{ and } \ \ \ \ \Vert u(t)\Vert _{L^{3/2,\infty }}\nonumber \\\le & {} c(\nu ,\eta ,\Vert b_0\Vert _{L^1\cap L^2})(1+t)^{-\frac{1}{2}}. \end{aligned}$$

(5.9)

Now we prove that, by a iteration process, the decay rate for b(t) in (5.9) can be improved to a much faster decay rate, which is proportional to that of solutions of heat equation. More precisely, we show that $\Vert b(t)\Vert \le c(\nu ,\eta ,\Vert b_0\Vert _{L^1\cap L^2})(1+t)^{-3/4}$ for all $t\ge 0$.

By a calculation similar to that of (5.7), we apply (5.9) to deduce that

$$\begin{aligned}&\Big |\int _0^te^{-\eta (t-s)|\xi |^2}(\widehat{b\cdot \nabla u}-\widehat{u\cdot \nabla b})(s)\mathrm{d}s\Big | \nonumber \\&\quad \le c|\xi |\int _0^te^{-\eta (t-s)|\xi |^2}(\Vert u(s)\Vert _{L^{3/2,\infty }}+\Vert \nabla u(s)\Vert )\Vert b(s)\Vert \mathrm{d}s \nonumber \\&\quad \le c(\nu ,\eta ,\Vert b_0\Vert _{L^1\cap L^2})|\xi |\Big \{\int _0^te^{-\eta (t-s)|\xi |^2}(1+s)^{-\frac{3}{4}}\mathrm{d}s \nonumber \\&\qquad +\Big (\int _0^t e^{-2\eta (t-s)|\xi |^2}(1+s)^{-\frac{1}{2}}\mathrm{d}s\Big )^\frac{1}{2}\Big (\int _0^t\Vert \nabla u(s)\Vert ^2\mathrm{d}s\Big )^\frac{1}{2}\Big \} \nonumber \\&\quad \le c(\nu ,\eta ,\Vert b_0\Vert _{L^1\cap L^2})|\xi |\int _0^te^{-\eta (t-s)|\xi |^2}(1+s)^{-\frac{3}{4}}\mathrm{d}s \nonumber \\&\qquad +c(\nu ,\eta ,\Vert b_0\Vert _{L^1\cap L^2})|\xi |\Big (\int _0^te^{-2\eta (t-s)|\xi |^2}(1+s)^{-\frac{1}{2}}\mathrm{d}s\Big )^\frac{1}{2} \nonumber \\&\quad :=I_1+I_2. \end{aligned}$$

(5.10)

The bound of $I_1$ is obtained by dividing the integration interval [0, t] into [0, t/2] and [t/2, t].

$$\begin{aligned} I_1&\le c(\nu ,\eta ,\Vert b_0\Vert _{L^1\cap L^2})|\xi |\Big (\int _0^\frac{t}{2}+\int _\frac{t}{2}^t\Big ) e^{-\eta (t-s)|\xi |^2}(1+s)^{-\frac{3}{4}}\mathrm{d}s \nonumber \\&\le c(\nu ,\eta ,\Vert b_0\Vert _{L^1\cap L^2})|\xi |\nonumber \\&\quad \Big (e^{-\frac{\eta t|\xi |^2}{2}}\int _0^\frac{t}{2}(1+s)^{-\frac{3}{4}}\mathrm{d}s+(1+t)^{-\frac{3}{4}}\int _\frac{t}{2}^te^{-\eta (t-s)|\xi |^2}\mathrm{d}s\Big ) \nonumber \\&\le c(\nu ,\eta ,\Vert b_0\Vert _{L^1\cap L^2})(|\xi |e^{-\frac{\eta t|\xi |^2}{2}}(1+t)^\frac{1}{4} +|\xi |^{-1}(1+t)^{-\frac{3}{4}}). \end{aligned}$$

(5.11)

Similarly, $I_2$ is bounded by

$$\begin{aligned} I_2\le c(\nu ,\eta ,\Vert b_0\Vert _{L^1\cap L^2})(|\xi |e^{-\frac{\eta t|\xi |^2}{2}}(1+t)^\frac{1}{4}+(1+t)^{-\frac{1}{4}}). \end{aligned}$$

(5.12)

Substituting (5.11) and (5.12) into (5.10), we deduce from (5.5) that

$$\begin{aligned} |\hat{b}(\xi )|\le c(\nu ,\eta ,\Vert b_0\Vert _{L^1\cap L^2})(1+|\xi |e^{-\frac{\eta t|\xi |^2}{2}}(1+t)^\frac{1}{4} +(1+t)^{-\frac{1}{4}}+|\xi |^{-1}(1+t)^{-\frac{3}{4}}). \end{aligned}$$

(5.13)

This implies that

$$\begin{aligned} \int _{|\xi |<r(t)}|\hat{b}|^2\mathrm{d}\xi&\le c(\nu ,\eta ,\Vert b_0\Vert _{L^1\cap L^2})\nonumber \\&\quad \int _{|\xi |<r(t)}(1+|\xi |^2e^{-\eta t|\xi |^2}(1+t)^\frac{1}{2}+(1+t)^{-\frac{1}{2}}\nonumber \\&\quad +|\xi |^{-2}(1+t)^{-\frac{3}{2}})\mathrm{d}\xi \nonumber \\&\le c(\nu ,\eta ,\Vert b_0\Vert _{L^1\cap L^2})((1+t)^{-\frac{3}{2}}+(1+t)^{-2}) \nonumber \\&\le c(\nu ,\eta ,\Vert b_0\Vert _{L^1\cap L^2})(1+t)^{-\frac{3}{2}} \end{aligned}$$

(5.14)

Integrating (5.3) with respect to t and using (5.14), we obtain the desired decay estimate of b(t):

$$\begin{aligned} \Vert b(t)\Vert ^2= & {} \Vert \hat{b}(t)\Vert ^2\le (1+t)^{-3}\Vert b_0\Vert ^2+c(\nu ,\eta ,\Vert b_0\Vert _{L^1\cap L^2})(1+t)^{-\frac{3}{2}}\nonumber \\\le & {} c(\nu ,\eta ,\Vert b_0\Vert _{L^1\cap L^2})(1+t)^{-\frac{3}{2}}. \end{aligned}$$

(5.15)

Furthermore, the equation (5.15), together with (3.4), implies that $\Vert u(t)\Vert \le c(\nu ,\eta ,\Vert b_0\Vert _{L^1\cap L^2})(1+t)^{-3/2}$.

Step 2. The decay of weak solutions. Let $\epsilon >0$ be arbitrary. For $1\le p\le \infty$, we define a mollification $\mathcal {J_\epsilon }:L^p\rightarrow L^p$ via $f\mapsto \rho _\epsilon *f$, where $\rho _\epsilon (\cdot )=\frac{1}{\epsilon ^3}\rho (\frac{\cdot }{\epsilon })$ and $\rho \in C_0^\infty$ is a nonnegative, radial function that satisfies $\int \rho (x)\mathrm{d}x=1$.

Consider the following equations:

$$\begin{aligned} -\nu \Delta u^\epsilon +\nabla p_*^\epsilon&=b^\epsilon \cdot \nabla b^\epsilon , \end{aligned}$$

(5.16)

$$\begin{aligned} \partial _tb^\epsilon +u^\epsilon \cdot \nabla b^\epsilon -\eta \Delta b^\epsilon +\epsilon (-\Delta )^{3/2} b^\epsilon&=b^\epsilon \cdot \nabla u^\epsilon , \end{aligned}$$

(5.17)

$$\begin{aligned} \nabla \cdot u^\epsilon =0,\ \ \nabla \cdot b^\epsilon&=0, \end{aligned}$$

(5.18)

$$\begin{aligned} b^\epsilon |_{t=0}&=\mathcal {J}_\epsilon b_0. \end{aligned}$$

(5.19)

This system is obtained by adding an artificial diffusion term $\epsilon (-\Delta )^{3/2}$ to the b-equation of (1.1)–(1.4), then replacing the initial datum $b_0$ by a smooth function $\mathcal {J}_\epsilon b_0$. The fractional Laplacian $(-\Delta )^{3/2}$ is defined by the Fourier transform, namely, $(-\Delta )^{3/2}=\mathcal {F}^{-1}(|\xi |^3\mathcal {F})$.

We say that $(u^\epsilon ,b^\epsilon )$ is a global-in-time strong solution of the equations (5.16)–(5.19) if it is a weak solution of the system (this means that $(u^\epsilon ,b^\epsilon )$ belongs to a proper integrable space and it solves the system in the sense of distribution) and it satisfies

$$\begin{aligned}&u^\epsilon \in C([0,\infty ;H^{3/2}_\sigma )\cap L_{loc}^2(0,\infty ;H^2) \ \ \ \ \text{ and } \nonumber \\&\quad b^\epsilon \in C([0,\infty ;H^1_\sigma )\cap L_{loc}^2(0,\infty ;H^{5/2}). \end{aligned}$$

(5.20)

It can be proved that (5.16)–(5.19) admits a unique global-in-time strong solution $(u^\epsilon ,b^\epsilon )$. In fact, it was recently shown by the authors in [10] that (5.16)–(5.19) admits a unique global-in-time strong solution when $\eta =0$. This global well-posedness result is absolutely true for the same equations when $\eta >0$.

Thus, for (5.16)–(5.19), we repeat the manipulation of Step 1 to deduce that for all $\epsilon >0$ and $t\ge 0$,

$$\begin{aligned} \Vert u^\epsilon (t)\Vert _{L^{3/2,\infty }}&\le c(\nu ,\eta ,\Vert b_0\Vert _{L^1\cap L^2})(1+t)^{-\frac{3}{2}}, \end{aligned}$$

(5.21)

$$\begin{aligned} \Vert b^\epsilon (t)\Vert&\le c(\nu ,\eta ,\Vert b_0\Vert _{L^1\cap L^2})(1+t)^{-\frac{3}{4}} . \end{aligned}$$

(5.22)

Note that the presence of $\epsilon (-\Delta )^{3/2}b^\epsilon$ does not affect the decay of $(u^\epsilon ,b^\epsilon )$ in $L^{3/2,\infty }\times L^2$.

It remains to show that $\{(u^\epsilon ,b^\epsilon )\}_{\epsilon >0}$ (or a subsequence) converges in some sense to (u, b) as $\epsilon$ tends to zero, and (u, b) is a weak solution of (1.1)–(1.4) as well as (u, b) satisfies (1.8).

By the energy method, it is clear that

$$\begin{aligned}&u^\epsilon \ \ \ \ \text{ is } \text{ uniformly } \text{ bounded } \text{ in } \ \ \ L^\infty (0,\infty ;L^{3/2,\infty })\cap L_{loc}^2(0,\infty ;H^1_\sigma ), \end{aligned}$$

(5.23)

$$\begin{aligned}&b^\epsilon \ \ \ \ \ \text{ is } \text{ uniformly } \text{ bounded } \text{ in } \ \ \ L^\infty (0,\infty ;L^2)\cap L_{loc}^2(0,\infty ;H^1_\sigma ), \end{aligned}$$

(5.24)

$$\begin{aligned}&\partial _tb^\epsilon \ \ \text{ is } \text{ uniformly } \text{ bounded } \text{ in } \ \ \ L_{loc}^{24/19}(0,\infty ;H^{-1}), \end{aligned}$$

(5.25)

here $H^{-1}$ denotes the dual space of $H^1_\sigma$. Thus, it follows from Banach-Alaoglu theorem that there exists a subsequence which is still denoted by $\{(u^\epsilon ,b^\epsilon ,\partial _tb^\epsilon )\}_{\epsilon >0}$ and an element $(u,b,\partial _tb)$, such that

$$\begin{aligned}&(u^\epsilon ,b^\epsilon ,\partial _tb^\epsilon )\rightarrow (u,b,\partial _tb) \ \text{ weakly } \text{ star } \text{ in } \ L^\infty (0,\infty ;L^{3/2,\infty })\times L^\infty (0,\infty ;L^2)\nonumber \\&\quad \times L_{loc}^{24/19}(0,\infty ;H^{-1}), \end{aligned}$$

(5.26)

$$\begin{aligned}&(u^\epsilon ,b^\epsilon )\rightarrow (u,b) \ \ \ \ \ \ \ \ \ \ \ \ \ \text{ weakly } \text{ in } \ \ \ L_{loc}^2(0,\infty ;H^1_\sigma )\nonumber \\&\quad \times L_{loc}^2(0,\infty ;H^1_\sigma ) \ \ \ \text{ as } \ \ \ \epsilon \rightarrow 0. \end{aligned}$$

(5.27)

Furthermore, for any bounded Lipschitz domain $\Omega \subset \mathbb {R}^3$, since $H^1(\Omega )\hookrightarrow L^2(\Omega )$ is compact, it follows from Aubin-Lions theorem, (5.26) and (5.27) that there exists a subsequence which is still denoted by $\{b^\epsilon \}$ such that

$$\begin{aligned} b^\epsilon \rightarrow b \ \ \ \text{ strongly } \text{ in } \ \ \ L_{loc}^2(0,\infty ;L^2(\Omega )) \ \ \ \text{ as } \ \ \ \epsilon \rightarrow 0. \end{aligned}$$

(5.28)

This convergence, together with the $L^{3/2,\infty }$-estimate of $u^\epsilon$ (see (3.4)), implies that

$$\begin{aligned} u^\epsilon \rightarrow u \ \ \ \text{ strongly } \text{ in } \ \ \ L_{loc}^2(0,\infty ;L^{3/2,\infty }(\Omega )) \ \ \ \text{ as } \ \ \ \epsilon \rightarrow 0. \end{aligned}$$

(5.29)

Hence, by (5.28), (5.29) and the uniform boundedness of $\{(u^\epsilon ,b^\epsilon )\}_{\epsilon >0}$, we deduce that

$$\begin{aligned}&b^\epsilon \cdot \nabla b^\epsilon \rightarrow b\cdot \nabla b \ \ \ \text{ weakly } \text{ star } \text{ in } \ \ \ L_{loc}^{4/3}(0,\infty ;H^{-1}(\Omega )), \end{aligned}$$

(5.30)

$$\begin{aligned}&u^\epsilon \cdot \nabla b^\epsilon \rightarrow u\cdot \nabla b \ \ \ \text{ weakly } \text{ star } \text{ in } \ \ \ L_{loc}^{24/19}(0,\infty ;H^{-1}(\Omega )), \end{aligned}$$

(5.31)

$$\begin{aligned}&b^\epsilon \cdot \nabla u^\epsilon \rightarrow b\cdot \nabla u \ \ \ \text{ weakly } \text{ star } \text{ in } \ \ \ L_{loc}^{24/19}(0,\infty ;H^{-1}(\Omega )) \ \ \ \text{ as } \ \ \ \epsilon \rightarrow 0. \end{aligned}$$

(5.32)

By passing to the limit as $\epsilon \rightarrow 0$, we conclude that (u, b) is a weak solution of (1.1)–(1.4).

Moreover, by (5.28) and Fatou’s lemma, we deduce from (5.22) that

$$\begin{aligned} \Vert b(t)\Vert \le c(\nu ,\eta ,\Vert b_0\Vert _{L^1\cap L^2})(1+t)^{-\frac{3}{4}} \ \ \ \text{ for } \text{ all } \ \ \ t\ge 0. \end{aligned}$$

(5.33)

This equation, together with (3.4), implies that

$$\begin{aligned} \Vert u(t)\Vert _{L^{3/2,\infty }}\le c(\nu ,\eta ,\Vert b_0\Vert _{L^1\cap L^2})(1+t)^{-\frac{3}{2}} \ \ \ \text{ for } \text{ all } \ \ \ t\ge 0. \end{aligned}$$

(5.34)

This completes the proof of Theorem 1.3.

6 Proof of Theorem 1.4

Let $b_1(t)=e^{\eta t\Delta }b_0$, and let $w=b-b_1$. A direct computation yields $\Vert b(t)\Vert \ge \Vert b_1(t)\Vert -\Vert w(t)\Vert$. The lower bound of decay ratio for b(t) is obtained by establishing the lower bound and upper bound for $b_1(t)$ and w(t), respectively. By Lemma 2.5, it is seen that $\Vert b_1(t)\Vert \ge c(\eta ,\beta )(1+t)^{-3/4}$. Thus, (1.9) holds if one shows that there exists a $\rho >3/4$ such that $\Vert w(t)\Vert \le c(1+t)^{-\rho }$.

Consider the equations that satisfied by (u, w):

$$\begin{aligned} -\nu \Delta u+\nabla p_*&=b\cdot \nabla b, \end{aligned}$$

(6.1)

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

(6.2)

$$\begin{aligned} \nabla \cdot u&=0,\ \ \nabla \cdot w=0, \end{aligned}$$

(6.3)

$$\begin{aligned} w|_{t=0}&=0, \end{aligned}$$

(6.4)

Taking the inner product of (6.1) and (6.2) with u and w, respectively, then integrating in $\mathbb {R}^3$ and summing the resultant equations, we use integration by parts and $\nabla \cdot u=\nabla \cdot b=0$ to obtain that

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\Vert w\Vert ^2+2\nu \Vert \nabla u\Vert ^2+2\eta \Vert \nabla w\Vert ^2&=\int \big (b\cdot \nabla b\big )\cdot u\mathrm{d}x\nonumber \\&\quad -\int \big (u\cdot \nabla b\big )\cdot w\mathrm{d}x+\int \big (b\cdot \nabla u\big )\cdot w\mathrm{d}x \nonumber \\&=\int \big (u\cdot \nabla b\big )\cdot b_1\mathrm{d}x-\int \big (b\cdot \nabla u\big )\cdot b_1\mathrm{d}x \nonumber \\&=\int \big (b\otimes u-u\otimes b\big )\cdot \nabla b_1\mathrm{d}x. \end{aligned}$$

(6.5)

Let $r(t)=\sqrt{\frac{3}{2\eta (1+t)}}$. By repeating the manipulation of the derivation of (5.3), we find that

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\Big ((1+t)^3\Vert \hat{w}\Vert ^2\Big )\le 3(1+t)^{2}\int _{|\xi |<r(t)}|\hat{w}|^2\mathrm{d}\xi +(1+t)^3\Big |\int \big (b\otimes u-u\otimes b\big )\cdot \nabla b_1\mathrm{d}x\Big |. \end{aligned}$$

(6.6)

Applying interpolation inequality, Lemma 2.4 and (1.8), we bound the integration of the second term on the right-hand side of (6.6) as follows:

$$\begin{aligned}&\Big |\int \big (b\otimes u-u\otimes b\big )(t)\cdot \nabla b_1(t)\mathrm{d}x\Big |\le c\Vert \nabla b_1(t)\Vert _\infty \Vert u(t)\Vert \Vert b(t)\Vert \nonumber \\&\quad \le c(\nu ,\eta ,\Vert b_0\Vert _{L^1\cap L^2})t^{-\frac{5}{4}}(1+t)^{-\frac{3}{4}}\Vert u(t)\Vert _{L^{3/2,\infty }}^\frac{2}{3}\Vert \nabla u(t)\Vert ^\frac{1}{3} \nonumber \\&\quad \le c(\nu ,\eta ,\Vert b_0\Vert _{L^1\cap L^2})(1+t)^{-3}\Vert \nabla u(t)\Vert ^\frac{1}{3}. \end{aligned}$$

(6.7)

Substituting (6.7) into (6.6), we know that

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\Big ((1+t)^3\Vert \hat{w}\Vert ^2\Big )\le 3(1+t)^{2}\int _{|\xi |<r(t)}|\hat{w}|^2\mathrm{d}\xi +c(\nu ,\eta , \Vert b_0\Vert _{L^1\cap L^2})\Vert \nabla u(t)\Vert ^\frac{1}{3}. \end{aligned}$$

(6.8)

Now we estimate $|\hat{w}|$ on the ball $\{\xi \in \mathbb {R}^3:|\xi |<r(t)\}$. Taking Fourier transformation to (6.2), we use (6.4) to deduce that

$$\begin{aligned} \hat{w}(\xi )=\int _0^te^{-\eta (t-s)|\xi |^2}(\widehat{b\cdot \nabla u}-\widehat{u\cdot \nabla b})(s)\mathrm{d}s. \end{aligned}$$

(6.9)

Thus, by repeating the manipulation of (5.7) or (5.10), we apply (1.8) to obtain

$$\begin{aligned} |\hat{w}(\xi )|&\le c|\xi |\int _0^te^{-\eta (t-s)|\xi |^2}(\Vert u(s)\Vert _{L^{3/2,\infty }}+\Vert \nabla u(s)\Vert )\Vert b(s)\Vert \mathrm{d}s \nonumber \\&\le c(\nu ,\eta ,\Vert b_0\Vert _{L^1\cap L^2})|\xi |\Big [\int _0^te^{-\eta (t-s)|\xi |^2}(1+s)^{-\frac{9}{4}}\mathrm{d}s\nonumber \\&\quad +\Big (\int _0^t e^{-2\eta (t-s)|\xi |^2}(1+s)^{-\frac{3}{2}}\mathrm{d}s\Big )^\frac{1}{2}\Big (\int _0^t\Vert \nabla u(s)\Vert ^2\mathrm{d}s\Big )^\frac{1}{2}\Big ] \nonumber \\&\le c(\nu ,\eta ,\Vert b_0\Vert _{L^1\cap L^2})|\xi |\Big [\int _0^t(1+s)^{-\frac{9}{4}}\mathrm{d}s+\Big (\int _0^t(1+s)^{-\frac{3}{2}}\mathrm{d}s \Big )^\frac{1}{2}\Big ] \nonumber \\&\le c(\nu ,\eta ,\Vert b_0\Vert _{L^1\cap L^2})|\xi |. \end{aligned}$$

(6.10)

Substituting (6.10) into (6.8) and integrating the resultant equation in [0, t], we see that

$$\begin{aligned} (1+t)^3\Vert \hat{w}(t)\Vert ^2&\le c(\nu ,\eta ,\Vert b_0\Vert _{L^1\cap L^2})\Big [\int _0^t(1+s)^2\int _{|\xi |<r(s)} |\xi |^2\mathrm{d}\xi \mathrm{d}s+\int _0^t\Vert \nabla u(s)\Vert ^\frac{1}{3}\mathrm{d}s\Big ] \nonumber \\&\le c(\nu ,\eta ,\Vert b_0\Vert _{L^1\cap L^2})\Big [\int _0^t(1+s)^{-\frac{1}{2}}\mathrm{d}s\nonumber \\&\quad +t^\frac{5}{6}\Big (\int _0^t\Vert \nabla u(s)\Vert ^2\mathrm{d}s\Big )^\frac{1}{6}\Big ] \nonumber \\&\le c(\nu ,\eta ,\Vert b_0\Vert _{L^1\cap L^2})(1+t)^\frac{5}{6}. \end{aligned}$$

(6.11)

Thus, we finally obtain the desired upper decay rate for w(t):

$$\begin{aligned} \Vert w(t)\Vert \le c(\nu ,\eta ,\Vert b_0\Vert _{L^1\cap L^2})(1+t)^{-\frac{13}{12}}. \end{aligned}$$

(6.12)

This completes the proof of Theorem 1.4.