A Blow-Up Criterion for the Density-Dependent Incompressible Magnetohydrodynamic System with Zero Viscosity

Abstract

In this paper, we provide a blow-up criterion for the density-dependent incompressible magnetohydrodynamic system with zero viscosity. The proof uses the $L^p$-method and the Kato–Ponce inequalities in the harmonic analysis. The novelty of our work lies in the fact that we deal with the case in which the resistivity $\eta$ is positive.

1. Introduction

Magnetohydrodynamics (MHD) is concerned with the study of applications between magnetic fields and fluid conductors of electricity. The application of magnetohydrodynamics covers a very wide range of physical objects, from liquid metals to cosmic plasmas.

We consider the following 3D density-dependent incompressible magnetohydrodynamic system:

$$\begin{array}{ccc}& & {\partial }_t\rho +u·\nabla \rho =0,\hfill \end{array}$$

(1)

$$\begin{array}{ccc}& & \rho {\partial }_{t}u+\rho (u·\nabla )u+\nabla \pi =\mathrm{rot}b\times b,\hfill \end{array}$$

(2)

$$\begin{array}{ccc}& & {\partial }_{t}b+u·\nabla b-b·\nabla u=\eta \Delta b,\hfill \end{array}$$

(3)

$$\begin{array}{ccc}& & \mathrm{div}u=0,\mathrm{div}b=0\mathrm{in}{\mathbb{R}}^3\times (0,\infty ),\hfill \end{array}$$

(4)

$$\begin{array}{ccc}& & \underset{\left|x\right|\to \infty }{lim}(\rho ,u,b)=(1,0,0),\hfill \end{array}$$

(5)

$$\begin{array}{ccc}& & (\rho ,u,b)(·,0)=({\rho }_0,u_0,b_0)\mathrm{in}{\mathbb{R}}^3.\hfill \end{array}$$

(6)

The unknowns are the fluid velocity field $u=u(x,t)$, the pressure $\pi =\pi (x,t)$, the density $\rho =\rho (x,t)$, and the magnetic field $b=b(x,t)$. $\eta >0$ is the resistivity coefficient. The term $\mathrm{rot}b\times b$ in (2) is the Lorentz force with low regularity, and thus it is the difficult term.

For the case of $b=0$, there are many studies. Beirão da Veiga and Valli [1,2] and Valli and Zajaczkowski [3] proved the unique solvability, local in time, in some supercritical Sobolev spaces and Hölder spaces in bounded domains. It is worth pointing out that, in 1995, Berselli [4] discussed the standard ideal flow. Danchin [5] and Danchin and Fanelli [6] (see also [7,8]) proved the unique solvability, local in time, in some critical Besov spaces. Recently, Bae et al [9] showed a regularity criterion:

$$\nabla u\in L^1(0,T;L^{\infty }\left({\mathbb{R}}^3\right)).$$

(7)

This refined the previous blow-up criteria [5,6,7]:

$$\begin{array}{ccc}& & \omega :=\mathrm{rot}u\in L^1(0,T;{\dot{B}}_{2,1}^{\frac{d}{2}}\left({\mathbb{R}}^d\right)),\hfill \end{array}$$

(8)

$$\begin{array}{ccc}& & \nabla u\in L^1(0,T;L^{\infty })\mathrm{and}\nabla \pi \in L^1(0,T;B_{\infty ,r}^{s-1}),s\ge 1,1\le r\le \infty .\hfill \end{array}$$

(9)

In [10], the authors proved the local well-posedness of smooth solutions in Sobolev spaces. The aim of this article is to prove (7) for the system (1)–(6). We will prove the following.

Theorem 1.

Let $0<inf{\rho }_0\le {\rho }_0\le C,\nabla {\rho }_0\in H^2,u_0,b_0\in H^3$ with $\mathrm{div}{u}_0=\mathrm{div}{b}_0=0$ in ${\mathbb{R}}^3$. Let $(\rho ,u,b)$ be the unique solution to the problem (1)–(6). If (7) holds true with some $0<T<\infty$, then the solution $(\rho ,u,b)$ can be extended beyond $T>0$.

Remark 1.

In [8], Zhou, Fan and Xin showed the same blow-up criterion (8), which is refined by (7) for the ideal MHD system.

Remark 2.

When $\eta =0$, we are unable to show a similar result.

In the following proofs, we will use the bilinear commutator and product estimates due to Kato–Ponce [11]:

$$\begin{array}{ccc}& & \parallel {\Lambda }^s\left(fg\right)-f{\Lambda }^s{g\parallel }_{L^p}\le {C(\parallel \nabla f\parallel }_{L^{p_1}}\parallel {\Lambda }^{s-1}{g\parallel }_{L^{q_1}}+{\parallel g\parallel }_{L^{p_2}}\parallel {\Lambda }^{s}f{\parallel }_{L^{q_2}}),\hfill \end{array}$$

(10)

$$\begin{array}{ccc}& & \parallel {\Lambda }^s{\left(fg\right)\parallel }_{L^p}\le {C(\parallel f\parallel }_{L^{p_1}}\parallel {\Lambda }^s{g\parallel }_{L^{q_1}}+\parallel {\Lambda }^s{f\parallel }_{L^{p_2}}{\parallel g\parallel }_{L^{q_2}}),\hfill \end{array}$$

(11)

with $s>0,\Lambda :={(-\Delta )}^{\frac{1}{2}}$ and ${\displaystyle \frac{1}{p}=\frac{1}{p_1}+\frac{1}{q_1}=\frac{1}{p_2}+\frac{1}{q_2}}$.

2. Proof of Theorem 1

We only need to prove a priori estimates.

First, thanks to the maximum principle, it is easy to see that

$$\frac{1}{C}\le \rho \le C.$$

(12)

We will use the identity

$$b·\nabla b+b\times \mathrm{rot}b=\frac{1}{2}\nabla {\left|b\right|}^2.$$

(13)

Testing (2) by u, using (1), (4) and (13), we find that

$$\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}\int \rho {\left|u\right|}^2\mathrm{d}x=\int (b·\nabla )b·u\mathrm{d}x.$$

(14)

Testing (3) by b and using (4), we obtain

$$\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}{\int \left|b\right|}^2\mathrm{d}x+\eta \int {|\nabla b|}^2\mathrm{d}x=\int (b·\nabla )u·b\mathrm{d}x.$$

(15)

Summing up (14) and (15), we have the well-known energy identity

$$\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}{\int \left(\rho \right|u|}^2+{\left|b\right|}^2{)\mathrm{d}x+\eta \int |\nabla b|}^2\mathrm{d}x=0,$$

and hence

$${\int \left(\right|u|}^2+{\left|b\right|}^2)\mathrm{d}x+{\int }_0^T\int {|\nabla b|}^2\mathrm{d}x\mathrm{d}t\le C.$$

(16)

It is easy to deduce that

$${\parallel \nabla \rho \parallel }_{L^{\infty }}\le {\parallel \nabla {\rho }_0\parallel }_{L^{\infty }}exp\left({\int }_0^t{\parallel \nabla u\left(s\right)\parallel }_{L^{\infty }}\mathrm{d}s\right)\le C.$$

(17)

Testing (3) by ${\left|b\right|}^{q-2}b(2<q<\infty )$ and using (4), we derive

$$\begin{array}{ccc}& & \frac{1}{q}\frac{\mathrm{d}}{\mathrm{d}t}{\parallel b\parallel }_{L^q}^q+{\eta \int \left|b\right|}^{q-2}{|\nabla b|}^2\mathrm{d}x+\eta \int \frac{1}{2}{\nabla \left|b\right|}^2·\nabla {\left|b\right|}^{q-2}\mathrm{d}x\hfill \\ & =& \int b·\nabla u·{\left|b\right|}^{q-2}b\mathrm{d}x\le {\parallel \nabla u\parallel }_{L^{\infty }}{\parallel b\parallel }_{L^q}^q,\hfill \end{array}$$

and therefore

$$\frac{\mathrm{d}}{\mathrm{d}t}{\parallel b\parallel }_{L^q}\le {\parallel \nabla u\parallel }_{L^{\infty }}{\parallel b\parallel }_{L^q},$$

which gives

$${\parallel b\parallel }_{L^q}\le {\parallel b_0\parallel }_{L^q}exp\left({\int }_0^t{\parallel \nabla u\left(s\right)\parallel }_{L^{\infty }}\mathrm{d}s\right).$$

(18)

Taking $q\to \infty$, one has

$${\parallel b\parallel }_{L^{\infty }}\le C.$$

(19)

(2) can be rewritten as

$${\partial }_{t}u+u·\nabla u+\frac{1}{\rho }\nabla \pi =\frac{1}{\rho }\mathrm{rot}b\times b.$$

(20)

Testing (20) by $\nabla \pi$ and using (4), it follows that

$$\begin{array}{ccc}\hfill \int \frac{1}{\rho }{|\nabla \pi |}^2\mathrm{d}x& =& -\int u·\nabla u·\nabla \pi \mathrm{d}x+\int \left(\frac{1}{\rho }\mathrm{rot}b\times b\right)·\nabla \pi \mathrm{d}x\hfill \\ & \le & {C(\parallel u·\nabla u\parallel }_{L^2}{+\parallel \mathrm{rot}b\times b\parallel }_{L^2}{)\parallel \nabla \pi \parallel }_{L^2}\hfill \\ & \le & {C(\parallel u\parallel }_{L^2}{\parallel \nabla u\parallel }_{L^{\infty }}+{\parallel b\parallel }_{L^{\infty }}{\parallel \nabla b\parallel }_{L^2}{)\parallel \nabla \pi \parallel }_{L^2},\hfill \end{array}$$

whence

$${\parallel \nabla \pi \parallel }_{L^2}\le {C\parallel \nabla u\parallel }_{L^{\infty }}+C{\parallel \nabla b\parallel }_{L^2}.$$

(21)

Taking $\mathrm{rot}$ to (20) and denoting the vorticity $\omega :=\mathrm{rot}u$, we obtain

$${\partial }_t\omega +u·\nabla \omega =\omega ·\nabla u-\nabla \frac{1}{\rho }\times \nabla \pi +\mathrm{rot}\left(\frac{\mathrm{rot}b}{\rho }\times b\right).$$

(22)

Testing (22) by $\omega$ and using (4), (17) and (19), we compute

$$\begin{array}{ccc}& & \frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}\int {\left|\omega \right|}^2\mathrm{d}x\hfill \\ & =& \int \omega ·\nabla u·\omega \mathrm{d}x-\int \left(\nabla \frac{1}{\rho }\times \nabla \pi \right)\omega \mathrm{d}x+\int \mathrm{rot}\left(\frac{\mathrm{rot}b}{\rho }\times b\right)·\omega \mathrm{d}x\hfill \\ & \le & {\parallel \nabla u\parallel }_{L^{\infty }}{\int \left|\omega \right|}^2\mathrm{d}x+∥\nabla \frac{1}{\rho }∥{\parallel \nabla \pi \parallel }_{L^2}{\parallel \omega \parallel }_{L^2}\hfill \\ & & +{C\parallel \nabla \rho \parallel }_{L^{\infty }}{\parallel b\parallel }_{L^{\infty }}{\parallel \nabla b\parallel }_{L^2}{\parallel \omega \parallel }_{L^2}+{C\parallel b\parallel }_{L^{\infty }}{\parallel \Delta b\parallel }_{L^2}{\parallel \omega \parallel }_{L^2}+{C\parallel \nabla b\parallel }_{L^4}^2{\parallel \omega \parallel }_{L^2}\hfill \\ & \le & {C\parallel \nabla u\parallel }_{L^{\infty }}{\int \left|\omega \right|}^2\mathrm{d}x+{C\parallel \nabla \pi \parallel }_{L^2}{\parallel \omega \parallel }_{L^2}+{C\parallel \nabla b\parallel }_{L^2}{\parallel \omega \parallel }_{L^2}\hfill \\ & & +\frac{\eta }{4}{\parallel \Delta b\parallel }_{L^2}^2+C{\parallel \omega \parallel }_{L^2}^2.\hfill \end{array}$$

(23)

Here, we have used the Gagliardo–Nirenberg inequality

$${\parallel \nabla b\parallel }_{L^4}^2\le {C\parallel b\parallel }_{L^{\infty }}{\parallel \Delta b\parallel }_{L^2}.$$

(24)

On the other hand, testing (3) by $-\Delta b$ and using (4) and (19), we achieve

$$\begin{array}{ccc}& & \frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}{\int |\nabla b|}^2\mathrm{d}x+\eta \int {|\Delta b|}^2\mathrm{d}x\hfill \\ & =& \sum _j\int (u·\nabla )b{\partial }_j^{2}b\mathrm{d}x-\int (b·\nabla )u·\Delta b\mathrm{d}x\hfill \\ & =& -\sum _j\int {\partial }_{j}u·\nabla b·{\partial }_{j}b\mathrm{d}x-\int b·\nabla u·\Delta b\mathrm{d}x\hfill \\ & \le & {C\parallel \nabla u\parallel }_{L^{\infty }}{\parallel \nabla b\parallel }_{L^2}^2+{\parallel b\parallel }_{L^{\infty }}{\parallel \nabla u\parallel }_{L^2}{\parallel \Delta b\parallel }_{L^2}\hfill \\ & \le & \frac{\eta }{4}{\parallel \Delta b\parallel }_{L^2}^2+{C\parallel \nabla u\parallel }_{L^{\infty }}{\parallel \nabla b\parallel }_{L^2}^2+C{\parallel \omega \parallel }_{L^2}^2.\hfill \end{array}$$

(25)

Here, we have used the inequality

$${\parallel \nabla u\parallel }_{L^r}\le C{\parallel \omega \parallel }_{L^r}\mathrm{with}1<r<\infty .$$

(26)

Summing up (23) and (25) and using (21) and the Gronwall inequality, we reach

$${\parallel \omega \parallel }_{L^2}+{\parallel \nabla b\parallel }_{L^2}\le C.$$

(27)

Taking $\mathrm{div}$ to (20) and using (4), we observe that

$$-\Delta \pi =f:=\rho \mathrm{div}(u·\nabla u)+\rho \nabla \frac{1}{\rho }·\nabla \pi -\rho \mathrm{div}\left(\frac{\mathrm{rot}b}{\rho }\times b\right),$$

(28)

from which, with (17), (19), (21) and (27), we have

$$\begin{array}{ccc}\hfill {\parallel \Delta \pi \parallel }_{L^4}& \le & {\parallel f\parallel }_{L^4}\le {C\parallel \nabla u\parallel }_{L^{\infty }}{\parallel \nabla u\parallel }_{L^4}+{C\parallel \nabla \rho \parallel }_{L^{\infty }}{\parallel \nabla \pi \parallel }_{L^4}\hfill \\ & & +{C\parallel \nabla \rho \parallel }_{L^{\infty }}{\parallel b\parallel }_{L^{\infty }}{\parallel \mathrm{rot}b\parallel }_{L^4}+{C\parallel \Delta b\parallel }_{L^4}{\parallel b\parallel }_{L^{\infty }}+C{\parallel \nabla b\parallel }_{L^8}^2\hfill \\ & \le & {C\parallel \nabla u\parallel }_{L^{\infty }}{\parallel \omega \parallel }_{L^4}+{C\parallel \nabla \pi \parallel }_{L^4}+{C\parallel \mathrm{rot}b\parallel }_{L^4}+C{\parallel \Delta b\parallel }_{L^4}\hfill \\ & \le & {C\parallel \nabla u\parallel }_{L^{\infty }}{\parallel \omega \parallel }_{L^4}+{C\parallel \nabla \pi \parallel }_{L^2}^{\frac{4}{7}}{\parallel \Delta \pi \parallel }_{L^4}^{\frac{3}{7}}+{C\parallel \mathrm{rot}b\parallel }_{L^4}+C{\parallel \Delta b\parallel }_{L^4}\hfill \\ & \le & \frac{1}{2}{\parallel \Delta \pi \parallel }_{L^4}+{C\parallel \nabla u\parallel }_{L^{\infty }}{\parallel \omega \parallel }_{L^4}+{C\parallel \nabla u\parallel }_{L^{\infty }}+C+{C\parallel \mathrm{rot}b\parallel }_{L^4}+C{\parallel \Delta b\parallel }_{L^4},\hfill \end{array}$$

which yields

$${\parallel \Delta \pi \parallel }_{L^4}\le {C\parallel \nabla u\parallel }_{L^{\infty }}{\parallel \omega \parallel }_{L^4}+{C\parallel \nabla u\parallel }_{L^{\infty }}+C+{C\parallel \mathrm{rot}b\parallel }_{L^4}+C{|\Delta b\parallel }_{L^4}.$$

(29)

Here, we have used the Gagliardo–Nirenberg inequalities

$$\begin{array}{ccc}\hfill {\parallel \nabla b\parallel }_{L^8}^2& \le & {C\parallel b\parallel }_{L^{\infty }}{\parallel \Delta b\parallel }_{L^4},\hfill \end{array}$$

(30)

$$\begin{array}{ccc}\hfill {\parallel \nabla \pi \parallel }_{L^4}& \le & {C\parallel \nabla \pi \parallel }_{L^2}^{\frac{4}{7}}{\parallel \Delta \pi \parallel }_{L^4}^{\frac{3}{7}}.\hfill \end{array}$$

(31)

Testing (22) by ${\left|\omega \right|}^2\omega$ and using (4), (17), (19), (29), (30) and (31), we have

$$\begin{array}{ccc}\hfill \frac{1}{4}\frac{\mathrm{d}}{\mathrm{d}t}{\parallel \omega \parallel }_{L^4}^4& \le & {C\parallel \nabla u\parallel }_{L^{\infty }}{\parallel \omega \parallel }_{L^4}^4+C{∥\nabla \frac{1}{\rho }∥}_{L^{\infty }}{\parallel \nabla \pi \parallel }_{L^4}{\parallel \omega \parallel }_{L^4}^3\hfill \\ & & +{C(\parallel b\parallel }_{L^{\infty }}{\parallel \Delta b\parallel }_{L^4}+{\parallel \nabla b\parallel }_{L^8}^2{)\parallel \omega \parallel }_{L^4}^3+C{∥\nabla \frac{1}{\rho }∥}_{L^{\infty }}{\parallel b\parallel }_{L^{\infty }}{\parallel \mathrm{rot}b\parallel }_{L^4}{\parallel \omega \parallel }_{L^4}^3\hfill \\ & \le & {C\parallel \nabla u\parallel }_{L^{\infty }}{\parallel \omega \parallel }_{L^4}^4+{C\parallel \nabla \pi \parallel }_{L^4}{\parallel \omega \parallel }_{L^4}^3+{C(\parallel \Delta b\parallel }_{L^4}+{\parallel \mathrm{rot}b\parallel }_{L^4}{)\parallel \omega \parallel }_{L^4}^3,\hfill \end{array}$$

which implies

$$\begin{array}{ccc}\hfill \frac{\mathrm{d}}{\mathrm{d}t}{\parallel \omega \parallel }_{L^4}^2& \le & {C\parallel \nabla u\parallel }_{L^{\infty }}{\parallel \omega \parallel }_{L^4}^2+{C(\parallel \nabla u\parallel }_{L^{\infty }}+1+{\parallel \mathrm{rot}b\parallel }_{L^4}+{\parallel \Delta b\parallel }_{L^4}{)\parallel \omega \parallel }_{L^4}\hfill \\ & \le & {C\parallel \nabla u\parallel }_{L^{\infty }}{\parallel \omega \parallel }_{L^4}^2+{C(\parallel \nabla u\parallel }_{L^{\infty }}+1+{\parallel \mathrm{rot}b\parallel }_{L^4}{)\parallel \omega \parallel }_{L^4}+{C\parallel \omega \parallel }_{L^4}^2+C{\parallel \Delta b\parallel }_{L^4}^2,\hfill \end{array}$$

and thus

$$\begin{array}{ccc}\hfill {\parallel \omega \parallel }_{L^4}^2& \le & {\parallel {\omega }_0\parallel }_{L^4}^2+C{\int }_0^t{\left[\right(\parallel \nabla u\parallel }_{L^{\infty }}+{1)\parallel \omega \parallel }_{L^4}^2\hfill \\ & & +{(\parallel \nabla u\parallel }_{L^{\infty }}+1+{\parallel \mathrm{rot}b\parallel }_{L^4}{)\parallel \omega \parallel }_{L^4}]\mathrm{d}s+{\int }_0^t{\parallel \Delta b\parallel }_{L^4}^2\mathrm{d}s.\hfill \end{array}$$

(32)

On the other hand, using the $L^2(0,T;W^{2,4})$-theory of the heat equation, it follows that

$$\begin{array}{ccc}\hfill {\int }_0^t{\parallel \Delta b\parallel }_{L^4}^2\mathrm{d}s& \le & C+C{\int }_0^t{\parallel u·\nabla b-b·\nabla u\parallel }_{L^4}^2\mathrm{d}s\hfill \\ & \le & C+C{\int }_0^t{(\parallel u\parallel }_{L^{\infty }}^2{\parallel \nabla b\parallel }_{L^4}^2+{\parallel b\parallel }_{L^{\infty }}^2{\parallel \nabla u\parallel }_{L^4}^2)\mathrm{d}s\hfill \\ & \le & C+C{\int }_0^t{(\parallel u\parallel }_{L^6}^{\frac{4}{3}}{\parallel \nabla u\parallel }_{L^{\infty }}^{\frac{2}{3}}{\parallel \mathrm{rot}b\parallel }_{L^4}^2+{\parallel \omega \parallel }_{L^4}^2)\mathrm{d}s\hfill \\ & \le & C+C{\int }_0^t{(\parallel \nabla u\parallel }_{L^{\infty }}^{\frac{2}{3}}{\parallel \mathrm{rot}b\parallel }_{L^4}^2+{\parallel \omega \parallel }_{L^4}^2)\mathrm{d}s.\hfill \end{array}$$

(33)

Here, we have used the inequality

$${\parallel \nabla b\parallel }_{L^p}\le C{\parallel \mathrm{rot}b\parallel }_{L^p}\mathrm{with}1<p<\infty .$$

(34)

Inserting (33) into (32), we have

$$\begin{array}{ccc}\hfill {\parallel \omega \parallel }_{L^4}^2& \le & C+C{\int }_0^t{(\parallel \nabla u\parallel }_{L^{\infty }}+{)\parallel \omega \parallel }_{L^4}^2\mathrm{d}s+C{\int }_0^t{(\parallel \nabla u\parallel }_{L^{\infty }}+{\parallel \mathrm{rot}b\parallel }_{L^4}{)\parallel \omega \parallel }_{L^4}\mathrm{d}s\hfill \\ & & +C{\int }_0^t{\parallel \nabla u\parallel }_{L^{\infty }}^{\frac{2}{3}}{\parallel \mathrm{rot}b\parallel }_{L^4}^2\mathrm{d}s.\hfill \end{array}$$

(35)

Taking $\mathrm{rot}$ to (3) and denoting the current $J:=\mathrm{rot}b$, we infer that

$${\partial }_{t}J-\eta \Delta J+\mathrm{rot}(u·\nabla b-b·\nabla u)=0.$$

(36)

Testing (36) by ${\left|J\right|}^{2}J$, using (17), (19) and (34), we derive

$$\begin{array}{ccc}& & \frac{1}{4}\frac{\mathrm{d}}{\mathrm{d}t}{\parallel J\parallel }_{L^4}^4+{\eta \int \left|J\right|}^2{|\nabla J|}^2\mathrm{d}x+\eta \int \frac{1}{2}{\nabla \left|J\right|}^2·\nabla {\left|J\right|}^2\mathrm{d}x\hfill \\ & =& \int (b·\nabla u-u·\nabla b)\mathrm{rot}\left(\right|J{|}^{2}J)\mathrm{d}x\hfill \\ & =& \int (b·\nabla u-u·\nabla b)\left(\right|J{|}^2\mathrm{rot}J+\nabla |J{|}^2\times J)\mathrm{d}x\hfill \\ & \le & {\parallel b·\nabla u-u·\nabla b\parallel }_{L^4}{\parallel J\parallel }_{L^4}\parallel \left|J\right|·{|\nabla J|\parallel }_{L^2}\hfill \\ & \le & {C(\parallel b\parallel }_{L^{\infty }}{\parallel \nabla u\parallel }_{L^4}+{\parallel u\parallel }_{L^{\infty }}{\parallel \nabla b\parallel }_{L^4}{)\parallel J\parallel }_{L^4}\parallel \left|J\right|·{|\nabla J|\parallel }_{L^2}\hfill \\ & \le & \frac{\eta }{2}{\int \left|J\right|}^2{|\nabla J|}^2\mathrm{d}x+{C(\parallel \omega \parallel }_{L^4}^2+{\parallel u\parallel }_{L^{\infty }}^2{\parallel J\parallel }_{L^4}^2{)\parallel J\parallel }_{L^4}^2,\hfill \end{array}$$

which implies

$$\frac{\mathrm{d}}{\mathrm{d}t}{\parallel J\parallel }_{L^4}^2\le {C\parallel \omega \parallel }_{L^4}^2+{C\parallel \nabla u\parallel }_{L^{\infty }}^{\frac{2}{3}}{\parallel J\parallel }_{L^4}^2.$$

Integrating the above inequality, one has

$${\parallel \mathrm{rot}b\parallel }_{L^4}^2\le C+C{\int }_0^t{(\parallel \omega \parallel }_{L^4}^2+{\parallel \nabla u\parallel }_{L^{\infty }}^{\frac{2}{3}}{\parallel \mathrm{rot}b\parallel }_{L^4}^2)\mathrm{d}s.$$

(37)

Summing up (35) and (37), using the Gronwall inequality, we arrive at

$$\begin{array}{ccc}& & {\parallel \nabla u\parallel }_{L^4}+{\parallel \nabla b\parallel }_{L^4}+{\int }_0^T{\parallel \Delta b\parallel }_{L^4}^2\mathrm{d}t\le C,\hfill \end{array}$$

(38)

$$\begin{array}{ccc}& & {\parallel u\parallel }_{L^{\infty }}+{\int }_0^T{\parallel \nabla b\parallel }_{L^{\infty }}^2\mathrm{d}t\le C.\hfill \end{array}$$

(39)

Applying ${\Lambda }^3$ to (1), testing by ${\Lambda }^3\rho$ and using (4) and (10), we obtain

$$\begin{array}{ccc}& & \frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}\int {\left({\Lambda }^3\rho \right)}^2\mathrm{d}x=-\int ({\Lambda }^3(u·\nabla \rho )-u·\nabla {\Lambda }^3\rho ){\Lambda }^3\rho \mathrm{d}x\hfill \\ & \le & {C(\parallel \nabla u\parallel }_{L^{\infty }}\parallel {\Lambda }^3{\rho \parallel }_{L^2}+{\parallel \nabla \rho \parallel }_{L^{\infty }}\parallel {\Lambda }^3{u\parallel }_{L^2})\parallel {\Lambda }^3{\rho \parallel }_{L^2}.\hfill \end{array}$$

(40)

Applying ${\Lambda }^3$ to (2), testing by ${\Lambda }^{3}u$ and using (1) and (4), we have

$$\begin{array}{ccc}& & \frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}\int \rho {\left|{\Lambda }^{3}u\right|}^2\mathrm{d}x=\int ({\Lambda }^3(b·\nabla b)-b·\nabla {\Lambda }^{3}b){\Lambda }^{3}u\mathrm{d}x+\int b·\nabla {\Lambda }^{3}b·{\Lambda }^{3}u\mathrm{d}x\hfill \\ & & -\int ({\Lambda }^3\left(\rho {\partial }_{t}u\right)-\rho {\Lambda }^3{\partial }_{t}u){\Lambda }^{3}u\mathrm{d}x-\int ({\Lambda }^3(\rho u·\nabla u)-\rho u·\nabla {\Lambda }^{3}u){\Lambda }^{3}u\mathrm{d}x\hfill \\ & =& :I_1+I_2+I_3+I_4.\hfill \end{array}$$

(41)

Applying ${\Lambda }^3$ to (3), testing by ${\Lambda }^{3}b$ and using (4), we have

$$\begin{array}{ccc}& & \frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}\int |{\Lambda }^3{b|}^2\mathrm{d}x+\eta \int {\left|{\Lambda }^{4}b\right|}^2\mathrm{d}x\hfill \\ & =& \int ({\Lambda }^3(b·\nabla u)-b·\nabla {\Lambda }^{3}u){\Lambda }^{3}b\mathrm{d}x+\int b·{\Lambda }^{3}u·{\Lambda }^{3}b\mathrm{d}x\hfill \\ & & -\int ({\Lambda }^3(u·\nabla b)-u·\nabla {\Lambda }^{3}b){\Lambda }^{3}b\mathrm{d}x=:I_5+I_6+I_7.\hfill \end{array}$$

(42)

Summing up (41) and (42) and noting that $I_2+I_6=0$, we have

$$\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}\int (\rho |{\Lambda }^3{u|}^2+|{\Lambda }^3{b|}^2)\mathrm{d}x+\eta \int |{\Lambda }^4{b|}^2\mathrm{d}x=I_1+I_3+I_4+I_5+I_7.$$

(43)

Using (10) and (11), we bound $I_1,I_4,I_5$ and $I_7$ as follows.

$$\begin{array}{ccc}\hfill I_1& \le & {C\parallel \nabla b\parallel }_{L^{\infty }}(\parallel {\Lambda }^3{b\parallel }_{L^2}^2+\parallel {\Lambda }^{3}u{\parallel }_{L^2}^2);\hfill \\ \hfill I_4& \le & {C(\parallel \nabla \left(\rho u\right)\parallel }_{L^{\infty }}\parallel {\Lambda }^3{u\parallel }_{L^2}+{\parallel \nabla u\parallel }_{L^{\infty }}\parallel {\Lambda }^3{\left(\rho u\right)\parallel }_{L^2})\parallel {\Lambda }^3{u\parallel }_{L^2}\hfill \\ & \le & {C(\parallel \nabla u\parallel }_{L^{\infty }}+1)\parallel {\Lambda }^3{u\parallel }_{L^2}^2+{C\parallel \nabla u\parallel }_{L^{\infty }}(\parallel {\Lambda }^3{u\parallel }_{L^2}+{\parallel u\parallel }_{L^{\infty }}\parallel {\Lambda }^3{\rho \parallel }_{L^2})\parallel {\Lambda }^3{u\parallel }_{L^2}\hfill \\ & \le & {C(\parallel \nabla u\parallel }_{L^{\infty }}+1)\parallel {\Lambda }^3{u\parallel }_{L^2}^2+{C\parallel \nabla u\parallel }_{L^{\infty }}{\parallel {\Lambda }^3\rho \parallel }_{L^2}^2;\hfill \\ \hfill I_5+I_7& \le & {C(\parallel \nabla b\parallel }_{L^{\infty }}\parallel {\Lambda }^3{u\parallel }_{L^2}+{\parallel \nabla u\parallel }_{L^{\infty }}\parallel {\Lambda }^3{b\parallel }_{L^2})\parallel {\Lambda }^3{b\parallel }_{L^2}.\hfill \end{array}$$

To bound $I_3$, we proceed as follows.

$$\begin{array}{ccc}\hfill I_3& \le & C(\parallel {\partial }_t{u\parallel }_{L^{\infty }}\parallel {\Lambda }^3{\rho \parallel }_{L^2}+{\parallel \nabla \rho \parallel }_{L^{\infty }}\parallel {\Lambda }^2{\partial }_t{u\parallel }_{L^2})\parallel {\Lambda }^3{u\parallel }_{L^2}\hfill \\ & \le & C{∥u·\nabla u+\frac{1}{\rho }\nabla \pi -\frac{\mathrm{rot}b}{\rho }\times b∥}_{L^{\infty }}\parallel {\Lambda }^3{\rho \parallel }_{L^2}{\parallel {\Lambda }^{3}u\parallel }_{L^2}\hfill \\ & & +C{∥\Delta \left(u·\nabla u+\frac{1}{\rho }\nabla \pi -\frac{\mathrm{rot}b}{\rho }\times b\right)∥}_{L^2}{\parallel {\Lambda }^{3}u\parallel }_{L^2}\hfill \\ & \le & {C(\parallel \nabla u\parallel }_{L^{\infty }}+{\parallel \nabla \pi \parallel }_{L^{\infty }}+{\parallel \nabla b\parallel }_{L^{\infty }})\parallel {\Lambda }^3{\rho \parallel }_{L^2}{\parallel {\Lambda }^{3}u\parallel }_{L^2}\hfill \\ & & +C\left({\parallel u\parallel }_{L^{\infty }}\parallel {\Lambda }^3{u\parallel }_{L^2}+\parallel {\Lambda }^2{\nabla \pi \parallel }_{L^2}+{\parallel \nabla \pi \parallel }_{L^3}{∥{\Lambda }^2\frac{1}{\rho }∥}_{L^6}\right.\hfill \\ & & \left.+{∥\Delta \frac{1}{\rho }∥}_{L^6}{\parallel \mathrm{rot}b\parallel }_{L^3}{\parallel b\parallel }_{L^{\infty }}+{\parallel b\parallel }_{L^{\infty }}{\parallel {\Lambda }^{3}b\parallel }_{L^2}\right){\parallel {\Lambda }^{3}u\parallel }_{L^2}\hfill \\ & \le & {C(\parallel \nabla u\parallel }_{L^{\infty }}+{\parallel \nabla \pi \parallel }_{L^{\infty }}+{\parallel \nabla b\parallel }_{L^{\infty }})\parallel {\Lambda }^3{\rho \parallel }_{L^2}{\parallel {\Lambda }^{3}u\parallel }_{L^2}\hfill \\ & & +C(\parallel {\Lambda }^3{u\parallel }_{L^2}+\parallel {\Lambda }^2{\nabla \pi \parallel }_{L^2}+{\parallel \nabla \pi \parallel }_{L^3}{\parallel \Delta \rho \parallel }_{L^6}+{\parallel \Delta \rho \parallel }_{L^6}+\parallel {\Lambda }^3{b\parallel }_{L^2})\parallel {\Lambda }^3{u\parallel }_{L^2}\hfill \\ & \le & {C(\parallel \nabla u\parallel }_{L^{\infty }}+{\parallel \nabla \pi \parallel }_{L^{\infty }}+{\parallel \nabla b\parallel }_{L^{\infty }})\parallel {\Lambda }^3{\rho \parallel }_{L^2}{\parallel {\Lambda }^{3}u\parallel }_{L^2}\hfill \\ & & +C(\parallel {\Lambda }^3{u\parallel }_{L^2}+{\parallel \nabla f\parallel }_{L^2}+{\parallel \nabla \pi \parallel }_{L^3}\parallel {\Lambda }^3{\rho \parallel }_{L^2}+\parallel {\Lambda }^3{\rho \parallel }_{L^2}+\parallel {\Lambda }^3{b\parallel }_{L^2})\parallel {\Lambda }^3{u\parallel }_{L^2}.\hfill \end{array}$$

(44)

On the other hand, we have

$$\begin{array}{ccc}\hfill {\parallel \nabla \pi \parallel }_{L^{\infty }}& \le & {C\parallel \nabla \pi \parallel }_{L^2}+C{\parallel \Delta \pi \parallel }_{L^4}\hfill \\ & \le & {C\parallel \nabla u\parallel }_{L^{\infty }}+C+C{\parallel \Delta b\parallel }_{L^4};\hfill \end{array}$$

(45)

$$\begin{array}{ccc}\hfill {\parallel \nabla \Delta \pi \parallel }_{L^2}& =& {\parallel \nabla f\parallel }_{L^2}\hfill \\ & \le & {C\parallel \nabla \rho \parallel }_{L^{\infty }}{\parallel \nabla u\parallel }_{L^4}^2+{C\parallel u\parallel }_{L^{\infty }}\parallel {\Lambda }^3{u\parallel }_{L^2}+{C\parallel \nabla \rho \parallel }_{L^{\infty }}{\parallel {\nabla }^2\pi \parallel }_{L^2}\hfill \\ & & +{C\parallel \Delta \rho \parallel }_{L^6}{\parallel \nabla \pi \parallel }_{L^3}+{C\parallel \nabla \rho \parallel }_{L^{\infty }}{(\parallel \Delta b\parallel }_{L^2}{\parallel b\parallel }_{L^{\infty }}+{\parallel \nabla b\parallel }_{L^4}^2+{\parallel \nabla \rho \parallel }_{L^{\infty }}{\parallel \nabla b\parallel }_{L^2})\hfill \\ & & +{C\parallel b\parallel }_{L^{\infty }}\parallel {\Lambda }^3{b\parallel }_{L^2}+C{∥\Delta \frac{1}{\rho }∥}_{L^6}{\parallel b\parallel }_{L^{\infty }}{\parallel \nabla b\parallel }_{L^3}\hfill \\ & \le & C+C\parallel {\Lambda }^3{u\parallel }_{L^2}+C\parallel {\nabla }^2{\pi \parallel }_{L^2}+C\parallel {\Lambda }^3{\rho \parallel }_{L^2}{\parallel \nabla \pi \parallel }_{L^3}+C\parallel {\Lambda }^3{b\parallel }_{L^2}+C{\parallel {\Lambda }^3\rho \parallel }_{L^2}\hfill \\ & \le & \frac{1}{2}{\parallel \nabla f\parallel }_{L^2}+{C\parallel \nabla \pi \parallel }_{L^2}+C+C{\parallel {\Lambda }^{3}u\parallel }_{L^2}\hfill \\ & & +C\parallel {\Lambda }^3{\rho \parallel }_{L^2}+C\parallel {\Lambda }^3{b\parallel }_{L^2}+C\parallel {\Lambda }^3{\rho \parallel }_{L^2}{\parallel \nabla \pi \parallel }_{L^3},\hfill \end{array}$$

which gives

$$\begin{array}{ccc}\hfill {\parallel \nabla f\parallel }_{L^2}& \le & C+{C\parallel \nabla u\parallel }_{L^{\infty }}+C\parallel {\Lambda }^3{\rho \parallel }_{L^2}+C{\parallel {\Lambda }^{3}u\parallel }_{L^2}\hfill \\ & & +C\parallel {\Lambda }^3{b\parallel }_{L^2}+C\parallel {\Lambda }^3{\rho \parallel }_{L^2}{(\parallel \nabla \pi \parallel }_{L^2}+{\parallel \Delta \pi \parallel }_{L^4})\hfill \\ & \le & C+{C\parallel \nabla u\parallel }_{L^{\infty }}+C\parallel {\Lambda }^3{(\rho ,u,b)\parallel }_{L^2}+C\parallel {\Lambda }^3{\rho \parallel }_{L^2}{(\parallel \nabla u\parallel }_{L^{\infty }}+1+{\parallel \Delta b\parallel }_{L^4}).\hfill \end{array}$$

Inserting the above estimates into (44), we obtain

$$\begin{array}{ccc}\hfill I_3& \le & {C(\parallel \nabla u\parallel }_{L^{\infty }}+1+{\parallel \nabla b\parallel }_{L^{\infty }}+{\parallel \Delta b\parallel }_{L^4}\left)\right(\parallel {\Lambda }^3{\rho \parallel }_{L^2}^2+\parallel {\Lambda }^3{u\parallel }_{L^2}^2)+C\parallel {\Lambda }^3{(\rho ,u,b)\parallel }_{L^2}^2\hfill \\ & \le & {C(\parallel \nabla u\parallel }_{L^{\infty }}+1+{\parallel \nabla b\parallel }_{L^{\infty }}+{\parallel \Delta b\parallel }_{L^4})\parallel {\Lambda }^3{(\rho ,u,b)\parallel }_{L^2}^2.\hfill \end{array}$$

Inserting the above estimates of $I_1,I_3,I_4,I_5$ and $I_7$ into (43), we have

$$\begin{array}{ccc}& & \frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}\int (\rho |{\Lambda }^3{u|}^2+|{\Lambda }^3{b|}^2)\mathrm{d}x+\eta \int |{\Lambda }^4{b|}^2\mathrm{d}x\hfill \\ & \le & {C(\parallel \nabla u\parallel }_{L^{\infty }}+1+{\parallel \nabla b\parallel }_{L^{\infty }}+{\parallel \Delta b\parallel }_{L^4})\parallel {\Lambda }^3{(\rho ,u,b)\parallel }_{L^2}^2.\hfill \end{array}$$

(46)

Summing up (40) and (46) and using the Gronwall inequality, we conclude that

$$\parallel {\Lambda }^3{(\rho ,u,b)\parallel }_{L^2}+{\int }_0^T\int {\left|{\Lambda }^{4}b\right|}^2\mathrm{d}x\mathrm{d}t\le C.$$

This completes the proof. □

3. Conclusions

In this paper, we prove a refined blow-up criterion for the inhomogeneous incompressible MHD system with zero viscosity, which is important and can be used for the simulation of MHD. For $\rho =1$ and $\eta =0$, Caflisch et al. [12] showed the following regularity criterion:

$$\mathrm{rot}u,\mathrm{rot}b\in L^1(0,T;L^{\infty }).$$

(47)

Since the problem is very challenging, we are unable to present further developments.