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

by
Kunlong Shi
1,
Jishan Fan
2,* and
Gen Nakamura
3
1
College of Sciences, Nanjing Forestry University, Nanjing 210037, China
2
Department of Applied Mathematics, Nanjing Forestry University, Nanjing 210037, China
3
Department of Mathematics, Hokkaido University, Sapporo 060-0810, Japan
*
Author to whom correspondence should be addressed.
Mathematics 2024, 12(10), 1510; https://doi.org/10.3390/math12101510
Submission received: 16 April 2024 / Revised: 9 May 2024 / Accepted: 11 May 2024 / Published: 12 May 2024

## 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 $η$ is positive.
MSC:
35Q35; 76D03

## 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:
$∂ t ρ + u · ∇ ρ = 0 ,$
$ρ ∂ t u + ρ ( u · ∇ ) u + ∇ π = rot b × b ,$
$∂ t b + u · ∇ b − b · ∇ u = η Δ b ,$
$div u = 0 , div b = 0 in R 3 × ( 0 , ∞ ) ,$
$lim | x | → ∞ ( ρ , u , b ) = ( 1 , 0 , 0 ) ,$
$( ρ , u , b ) ( · , 0 ) = ( ρ 0 , u 0 , b 0 ) in R 3 .$
The unknowns are the fluid velocity field $u = u ( x , t )$, the pressure $π = π ( x , t )$, the density $ρ = ρ ( x , t )$, and the magnetic field $b = b ( x , t )$. $η > 0$ is the resistivity coefficient. The term $rot b × b$ 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:
$∇ u ∈ L 1 ( 0 , T ; L ∞ ( R 3 ) ) .$
This refined the previous blow-up criteria [5,6,7]:
$ω : = rot u ∈ L 1 ( 0 , T ; B ˙ 2 , 1 d 2 ( R d ) ) ,$
$∇ u ∈ L 1 ( 0 , T ; L ∞ ) and ∇ π ∈ L 1 ( 0 , T ; B ∞ , r s − 1 ) , s ≥ 1 , 1 ≤ r ≤ ∞ .$
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 $0 < inf ρ 0 ≤ ρ 0 ≤ C , ∇ ρ 0 ∈ H 2 , u 0 , b 0 ∈ H 3$ with $div u 0 = div b 0 = 0$ in $R 3$. Let $( ρ , u , b )$ be the unique solution to the problem (1)–(6). If (7) holds true with some $0 < T < ∞$, then the solution $( ρ , u , b )$ 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 $η = 0$, 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]:
$∥ Λ s ( f g ) − f Λ s g ∥ L p ≤ C ( ∥ ∇ f ∥ L p 1 ∥ Λ s − 1 g ∥ L q 1 + ∥ g ∥ L p 2 ∥ Λ s f ∥ L q 2 ) ,$
$∥ Λ s ( f g ) ∥ L p ≤ C ( ∥ f ∥ L p 1 ∥ Λ s g ∥ L q 1 + ∥ Λ s f ∥ L p 2 ∥ g ∥ L q 2 ) ,$
with $s > 0 , Λ : = ( − Δ ) 1 2$ and $1 p = 1 p 1 + 1 q 1 = 1 p 2 + 1 q 2$.

## 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
$1 C ≤ ρ ≤ C .$
We will use the identity
$b · ∇ b + b × rot b = 1 2 ∇ | b | 2 .$
Testing (2) by u, using (1), (4) and (13), we find that
$1 2 d d t ∫ ρ | u | 2 d x = ∫ ( b · ∇ ) b · u d x .$
Testing (3) by b and using (4), we obtain
$1 2 d d t ∫ | b | 2 d x + η ∫ | ∇ b | 2 d x = ∫ ( b · ∇ ) u · b d x .$
Summing up (14) and (15), we have the well-known energy identity
$1 2 d d t ∫ ( ρ | u | 2 + | b | 2 ) d x + η ∫ | ∇ b | 2 d x = 0 ,$
and hence
$∫ ( | u | 2 + | b | 2 ) d x + ∫ 0 T ∫ | ∇ b | 2 d x d t ≤ C .$
It is easy to deduce that
$∥ ∇ ρ ∥ L ∞ ≤ ∥ ∇ ρ 0 ∥ L ∞ exp ∫ 0 t ∥ ∇ u ( s ) ∥ L ∞ d s ≤ C .$
Testing (3) by $| b | q − 2 b ( 2 < q < ∞ )$ and using (4), we derive
$1 q d d t ∥ b ∥ L q q + η ∫ | b | q − 2 | ∇ b | 2 d x + η ∫ 1 2 ∇ | b | 2 · ∇ | b | q − 2 d x = ∫ b · ∇ u · | b | q − 2 b d x ≤ ∥ ∇ u ∥ L ∞ ∥ b ∥ L q q ,$
and therefore
$d d t ∥ b ∥ L q ≤ ∥ ∇ u ∥ L ∞ ∥ b ∥ L q ,$
which gives
$∥ b ∥ L q ≤ ∥ b 0 ∥ L q exp ∫ 0 t ∥ ∇ u ( s ) ∥ L ∞ d s .$
Taking $q → ∞$, one has
$∥ b ∥ L ∞ ≤ C .$
(2) can be rewritten as
$∂ t u + u · ∇ u + 1 ρ ∇ π = 1 ρ rot b × b .$
Testing (20) by $∇ π$ and using (4), it follows that
$∫ 1 ρ | ∇ π | 2 d x = − ∫ u · ∇ u · ∇ π d x + ∫ 1 ρ rot b × b · ∇ π d x ≤ C ( ∥ u · ∇ u ∥ L 2 + ∥ rot b × b ∥ L 2 ) ∥ ∇ π ∥ L 2 ≤ C ( ∥ u ∥ L 2 ∥ ∇ u ∥ L ∞ + ∥ b ∥ L ∞ ∥ ∇ b ∥ L 2 ) ∥ ∇ π ∥ L 2 ,$
whence
$∥ ∇ π ∥ L 2 ≤ C ∥ ∇ u ∥ L ∞ + C ∥ ∇ b ∥ L 2 .$
Taking $rot$ to (20) and denoting the vorticity $ω : = rot u$, we obtain
$∂ t ω + u · ∇ ω = ω · ∇ u − ∇ 1 ρ × ∇ π + rot rot b ρ × b .$
Testing (22) by $ω$ and using (4), (17) and (19), we compute
$1 2 d d t ∫ | ω | 2 d x = ∫ ω · ∇ u · ω d x − ∫ ∇ 1 ρ × ∇ π ω d x + ∫ rot rot b ρ × b · ω d x ≤ ∥ ∇ u ∥ L ∞ ∫ | ω | 2 d x + ∇ 1 ρ ∥ ∇ π ∥ L 2 ∥ ω ∥ L 2 + C ∥ ∇ ρ ∥ L ∞ ∥ b ∥ L ∞ ∥ ∇ b ∥ L 2 ∥ ω ∥ L 2 + C ∥ b ∥ L ∞ ∥ Δ b ∥ L 2 ∥ ω ∥ L 2 + C ∥ ∇ b ∥ L 4 2 ∥ ω ∥ L 2 ≤ C ∥ ∇ u ∥ L ∞ ∫ | ω | 2 d x + C ∥ ∇ π ∥ L 2 ∥ ω ∥ L 2 + C ∥ ∇ b ∥ L 2 ∥ ω ∥ L 2 + η 4 ∥ Δ b ∥ L 2 2 + C ∥ ω ∥ L 2 2 .$
Here, we have used the Gagliardo–Nirenberg inequality
$∥ ∇ b ∥ L 4 2 ≤ C ∥ b ∥ L ∞ ∥ Δ b ∥ L 2 .$
On the other hand, testing (3) by $− Δ b$ and using (4) and (19), we achieve
$1 2 d d t ∫ | ∇ b | 2 d x + η ∫ | Δ b | 2 d x = ∑ j ∫ ( u · ∇ ) b ∂ j 2 b d x − ∫ ( b · ∇ ) u · Δ b d x = − ∑ j ∫ ∂ j u · ∇ b · ∂ j b d x − ∫ b · ∇ u · Δ b d x ≤ C ∥ ∇ u ∥ L ∞ ∥ ∇ b ∥ L 2 2 + ∥ b ∥ L ∞ ∥ ∇ u ∥ L 2 ∥ Δ b ∥ L 2 ≤ η 4 ∥ Δ b ∥ L 2 2 + C ∥ ∇ u ∥ L ∞ ∥ ∇ b ∥ L 2 2 + C ∥ ω ∥ L 2 2 .$
Here, we have used the inequality
$∥ ∇ u ∥ L r ≤ C ∥ ω ∥ L r with 1 < r < ∞ .$
Summing up (23) and (25) and using (21) and the Gronwall inequality, we reach
$∥ ω ∥ L 2 + ∥ ∇ b ∥ L 2 ≤ C .$
Taking $div$ to (20) and using (4), we observe that
$− Δ π = f : = ρ div ( u · ∇ u ) + ρ ∇ 1 ρ · ∇ π − ρ div rot b ρ × b ,$
from which, with (17), (19), (21) and (27), we have
$∥ Δ π ∥ L 4 ≤ ∥ f ∥ L 4 ≤ C ∥ ∇ u ∥ L ∞ ∥ ∇ u ∥ L 4 + C ∥ ∇ ρ ∥ L ∞ ∥ ∇ π ∥ L 4 + C ∥ ∇ ρ ∥ L ∞ ∥ b ∥ L ∞ ∥ rot b ∥ L 4 + C ∥ Δ b ∥ L 4 ∥ b ∥ L ∞ + C ∥ ∇ b ∥ L 8 2 ≤ C ∥ ∇ u ∥ L ∞ ∥ ω ∥ L 4 + C ∥ ∇ π ∥ L 4 + C ∥ rot b ∥ L 4 + C ∥ Δ b ∥ L 4 ≤ C ∥ ∇ u ∥ L ∞ ∥ ω ∥ L 4 + C ∥ ∇ π ∥ L 2 4 7 ∥ Δ π ∥ L 4 3 7 + C ∥ rot b ∥ L 4 + C ∥ Δ b ∥ L 4 ≤ 1 2 ∥ Δ π ∥ L 4 + C ∥ ∇ u ∥ L ∞ ∥ ω ∥ L 4 + C ∥ ∇ u ∥ L ∞ + C + C ∥ rot b ∥ L 4 + C ∥ Δ b ∥ L 4 ,$
which yields
$∥ Δ π ∥ L 4 ≤ C ∥ ∇ u ∥ L ∞ ∥ ω ∥ L 4 + C ∥ ∇ u ∥ L ∞ + C + C ∥ rot b ∥ L 4 + C | Δ b ∥ L 4 .$
Here, we have used the Gagliardo–Nirenberg inequalities
$∥ ∇ b ∥ L 8 2 ≤ C ∥ b ∥ L ∞ ∥ Δ b ∥ L 4 ,$
$∥ ∇ π ∥ L 4 ≤ C ∥ ∇ π ∥ L 2 4 7 ∥ Δ π ∥ L 4 3 7 .$
Testing (22) by $| ω | 2 ω$ and using (4), (17), (19), (29), (30) and (31), we have
$1 4 d d t ∥ ω ∥ L 4 4 ≤ C ∥ ∇ u ∥ L ∞ ∥ ω ∥ L 4 4 + C ∇ 1 ρ L ∞ ∥ ∇ π ∥ L 4 ∥ ω ∥ L 4 3 + C ( ∥ b ∥ L ∞ ∥ Δ b ∥ L 4 + ∥ ∇ b ∥ L 8 2 ) ∥ ω ∥ L 4 3 + C ∇ 1 ρ L ∞ ∥ b ∥ L ∞ ∥ rot b ∥ L 4 ∥ ω ∥ L 4 3 ≤ C ∥ ∇ u ∥ L ∞ ∥ ω ∥ L 4 4 + C ∥ ∇ π ∥ L 4 ∥ ω ∥ L 4 3 + C ( ∥ Δ b ∥ L 4 + ∥ rot b ∥ L 4 ) ∥ ω ∥ L 4 3 ,$
which implies
$d d t ∥ ω ∥ L 4 2 ≤ C ∥ ∇ u ∥ L ∞ ∥ ω ∥ L 4 2 + C ( ∥ ∇ u ∥ L ∞ + 1 + ∥ rot b ∥ L 4 + ∥ Δ b ∥ L 4 ) ∥ ω ∥ L 4 ≤ C ∥ ∇ u ∥ L ∞ ∥ ω ∥ L 4 2 + C ( ∥ ∇ u ∥ L ∞ + 1 + ∥ rot b ∥ L 4 ) ∥ ω ∥ L 4 + C ∥ ω ∥ L 4 2 + C ∥ Δ b ∥ L 4 2 ,$
and thus
$∥ ω ∥ L 4 2 ≤ ∥ ω 0 ∥ L 4 2 + C ∫ 0 t [ ( ∥ ∇ u ∥ L ∞ + 1 ) ∥ ω ∥ L 4 2 + ( ∥ ∇ u ∥ L ∞ + 1 + ∥ rot b ∥ L 4 ) ∥ ω ∥ L 4 ] d s + ∫ 0 t ∥ Δ b ∥ L 4 2 d s .$
On the other hand, using the $L 2 ( 0 , T ; W 2 , 4 )$-theory of the heat equation, it follows that
$∫ 0 t ∥ Δ b ∥ L 4 2 d s ≤ C + C ∫ 0 t ∥ u · ∇ b − b · ∇ u ∥ L 4 2 d s ≤ C + C ∫ 0 t ( ∥ u ∥ L ∞ 2 ∥ ∇ b ∥ L 4 2 + ∥ b ∥ L ∞ 2 ∥ ∇ u ∥ L 4 2 ) d s ≤ C + C ∫ 0 t ( ∥ u ∥ L 6 4 3 ∥ ∇ u ∥ L ∞ 2 3 ∥ rot b ∥ L 4 2 + ∥ ω ∥ L 4 2 ) d s ≤ C + C ∫ 0 t ( ∥ ∇ u ∥ L ∞ 2 3 ∥ rot b ∥ L 4 2 + ∥ ω ∥ L 4 2 ) d s .$
Here, we have used the inequality
$∥ ∇ b ∥ L p ≤ C ∥ rot b ∥ L p with 1 < p < ∞ .$
Inserting (33) into (32), we have
$∥ ω ∥ L 4 2 ≤ C + C ∫ 0 t ( ∥ ∇ u ∥ L ∞ + ) ∥ ω ∥ L 4 2 d s + C ∫ 0 t ( ∥ ∇ u ∥ L ∞ + ∥ rot b ∥ L 4 ) ∥ ω ∥ L 4 d s + C ∫ 0 t ∥ ∇ u ∥ L ∞ 2 3 ∥ rot b ∥ L 4 2 d s .$
Taking $rot$ to (3) and denoting the current $J : = rot b$, we infer that
$∂ t J − η Δ J + rot ( u · ∇ b − b · ∇ u ) = 0 .$
Testing (36) by $| J | 2 J$, using (17), (19) and (34), we derive
$1 4 d d t ∥ J ∥ L 4 4 + η ∫ | J | 2 | ∇ J | 2 d x + η ∫ 1 2 ∇ | J | 2 · ∇ | J | 2 d x = ∫ ( b · ∇ u − u · ∇ b ) rot ( | J | 2 J ) d x = ∫ ( b · ∇ u − u · ∇ b ) ( | J | 2 rot J + ∇ | J | 2 × J ) d x ≤ ∥ b · ∇ u − u · ∇ b ∥ L 4 ∥ J ∥ L 4 ∥ | J | · | ∇ J | ∥ L 2 ≤ C ( ∥ b ∥ L ∞ ∥ ∇ u ∥ L 4 + ∥ u ∥ L ∞ ∥ ∇ b ∥ L 4 ) ∥ J ∥ L 4 ∥ | J | · | ∇ J | ∥ L 2 ≤ η 2 ∫ | J | 2 | ∇ J | 2 d x + C ( ∥ ω ∥ L 4 2 + ∥ u ∥ L ∞ 2 ∥ J ∥ L 4 2 ) ∥ J ∥ L 4 2 ,$
which implies
$d d t ∥ J ∥ L 4 2 ≤ C ∥ ω ∥ L 4 2 + C ∥ ∇ u ∥ L ∞ 2 3 ∥ J ∥ L 4 2 .$
Integrating the above inequality, one has
$∥ rot b ∥ L 4 2 ≤ C + C ∫ 0 t ( ∥ ω ∥ L 4 2 + ∥ ∇ u ∥ L ∞ 2 3 ∥ rot b ∥ L 4 2 ) d s .$
Summing up (35) and (37), using the Gronwall inequality, we arrive at
$∥ ∇ u ∥ L 4 + ∥ ∇ b ∥ L 4 + ∫ 0 T ∥ Δ b ∥ L 4 2 d t ≤ C ,$
$∥ u ∥ L ∞ + ∫ 0 T ∥ ∇ b ∥ L ∞ 2 d t ≤ C .$
Applying $Λ 3$ to (1), testing by $Λ 3 ρ$ and using (4) and (10), we obtain
$1 2 d d t ∫ ( Λ 3 ρ ) 2 d x = − ∫ ( Λ 3 ( u · ∇ ρ ) − u · ∇ Λ 3 ρ ) Λ 3 ρ d x ≤ C ( ∥ ∇ u ∥ L ∞ ∥ Λ 3 ρ ∥ L 2 + ∥ ∇ ρ ∥ L ∞ ∥ Λ 3 u ∥ L 2 ) ∥ Λ 3 ρ ∥ L 2 .$
Applying $Λ 3$ to (2), testing by $Λ 3 u$ and using (1) and (4), we have
$1 2 d d t ∫ ρ | Λ 3 u | 2 d x = ∫ ( Λ 3 ( b · ∇ b ) − b · ∇ Λ 3 b ) Λ 3 u d x + ∫ b · ∇ Λ 3 b · Λ 3 u d x − ∫ ( Λ 3 ( ρ ∂ t u ) − ρ Λ 3 ∂ t u ) Λ 3 u d x − ∫ ( Λ 3 ( ρ u · ∇ u ) − ρ u · ∇ Λ 3 u ) Λ 3 u d x = : I 1 + I 2 + I 3 + I 4 .$
Applying $Λ 3$ to (3), testing by $Λ 3 b$ and using (4), we have
$1 2 d d t ∫ | Λ 3 b | 2 d x + η ∫ | Λ 4 b | 2 d x = ∫ ( Λ 3 ( b · ∇ u ) − b · ∇ Λ 3 u ) Λ 3 b d x + ∫ b · Λ 3 u · Λ 3 b d x − ∫ ( Λ 3 ( u · ∇ b ) − u · ∇ Λ 3 b ) Λ 3 b d x = : I 5 + I 6 + I 7 .$
Summing up (41) and (42) and noting that $I 2 + I 6 = 0$, we have
$1 2 d d t ∫ ( ρ | Λ 3 u | 2 + | Λ 3 b | 2 ) d x + η ∫ | Λ 4 b | 2 d x = I 1 + I 3 + I 4 + I 5 + I 7 .$
Using (10) and (11), we bound $I 1 , I 4 , I 5$ and $I 7$ as follows.
$I 1 ≤ C ∥ ∇ b ∥ L ∞ ( ∥ Λ 3 b ∥ L 2 2 + ∥ Λ 3 u ∥ L 2 2 ) ; I 4 ≤ C ( ∥ ∇ ( ρ u ) ∥ L ∞ ∥ Λ 3 u ∥ L 2 + ∥ ∇ u ∥ L ∞ ∥ Λ 3 ( ρ u ) ∥ L 2 ) ∥ Λ 3 u ∥ L 2 ≤ C ( ∥ ∇ u ∥ L ∞ + 1 ) ∥ Λ 3 u ∥ L 2 2 + C ∥ ∇ u ∥ L ∞ ( ∥ Λ 3 u ∥ L 2 + ∥ u ∥ L ∞ ∥ Λ 3 ρ ∥ L 2 ) ∥ Λ 3 u ∥ L 2 ≤ C ( ∥ ∇ u ∥ L ∞ + 1 ) ∥ Λ 3 u ∥ L 2 2 + C ∥ ∇ u ∥ L ∞ ∥ Λ 3 ρ ∥ L 2 2 ; I 5 + I 7 ≤ C ( ∥ ∇ b ∥ L ∞ ∥ Λ 3 u ∥ L 2 + ∥ ∇ u ∥ L ∞ ∥ Λ 3 b ∥ L 2 ) ∥ Λ 3 b ∥ L 2 .$
To bound $I 3$, we proceed as follows.
$I 3 ≤ C ( ∥ ∂ t u ∥ L ∞ ∥ Λ 3 ρ ∥ L 2 + ∥ ∇ ρ ∥ L ∞ ∥ Λ 2 ∂ t u ∥ L 2 ) ∥ Λ 3 u ∥ L 2 ≤ C u · ∇ u + 1 ρ ∇ π − rot b ρ × b L ∞ ∥ Λ 3 ρ ∥ L 2 ∥ Λ 3 u ∥ L 2 + C Δ u · ∇ u + 1 ρ ∇ π − rot b ρ × b L 2 ∥ Λ 3 u ∥ L 2 ≤ C ( ∥ ∇ u ∥ L ∞ + ∥ ∇ π ∥ L ∞ + ∥ ∇ b ∥ L ∞ ) ∥ Λ 3 ρ ∥ L 2 ∥ Λ 3 u ∥ L 2 + C ∥ u ∥ L ∞ ∥ Λ 3 u ∥ L 2 + ∥ Λ 2 ∇ π ∥ L 2 + ∥ ∇ π ∥ L 3 Λ 2 1 ρ L 6 + Δ 1 ρ L 6 ∥ rot b ∥ L 3 ∥ b ∥ L ∞ + ∥ b ∥ L ∞ ∥ Λ 3 b ∥ L 2 ∥ Λ 3 u ∥ L 2 ≤ C ( ∥ ∇ u ∥ L ∞ + ∥ ∇ π ∥ L ∞ + ∥ ∇ b ∥ L ∞ ) ∥ Λ 3 ρ ∥ L 2 ∥ Λ 3 u ∥ L 2 + C ( ∥ Λ 3 u ∥ L 2 + ∥ Λ 2 ∇ π ∥ L 2 + ∥ ∇ π ∥ L 3 ∥ Δ ρ ∥ L 6 + ∥ Δ ρ ∥ L 6 + ∥ Λ 3 b ∥ L 2 ) ∥ Λ 3 u ∥ L 2 ≤ C ( ∥ ∇ u ∥ L ∞ + ∥ ∇ π ∥ L ∞ + ∥ ∇ b ∥ L ∞ ) ∥ Λ 3 ρ ∥ L 2 ∥ Λ 3 u ∥ L 2 + C ( ∥ Λ 3 u ∥ L 2 + ∥ ∇ f ∥ L 2 + ∥ ∇ π ∥ L 3 ∥ Λ 3 ρ ∥ L 2 + ∥ Λ 3 ρ ∥ L 2 + ∥ Λ 3 b ∥ L 2 ) ∥ Λ 3 u ∥ L 2 .$
On the other hand, we have
$∥ ∇ π ∥ L ∞ ≤ C ∥ ∇ π ∥ L 2 + C ∥ Δ π ∥ L 4 ≤ C ∥ ∇ u ∥ L ∞ + C + C ∥ Δ b ∥ L 4 ;$
$∥ ∇ Δ π ∥ L 2 = ∥ ∇ f ∥ L 2 ≤ C ∥ ∇ ρ ∥ L ∞ ∥ ∇ u ∥ L 4 2 + C ∥ u ∥ L ∞ ∥ Λ 3 u ∥ L 2 + C ∥ ∇ ρ ∥ L ∞ ∥ ∇ 2 π ∥ L 2 + C ∥ Δ ρ ∥ L 6 ∥ ∇ π ∥ L 3 + C ∥ ∇ ρ ∥ L ∞ ( ∥ Δ b ∥ L 2 ∥ b ∥ L ∞ + ∥ ∇ b ∥ L 4 2 + ∥ ∇ ρ ∥ L ∞ ∥ ∇ b ∥ L 2 ) + C ∥ b ∥ L ∞ ∥ Λ 3 b ∥ L 2 + C Δ 1 ρ L 6 ∥ b ∥ L ∞ ∥ ∇ b ∥ L 3 ≤ C + C ∥ Λ 3 u ∥ L 2 + C ∥ ∇ 2 π ∥ L 2 + C ∥ Λ 3 ρ ∥ L 2 ∥ ∇ π ∥ L 3 + C ∥ Λ 3 b ∥ L 2 + C ∥ Λ 3 ρ ∥ L 2 ≤ 1 2 ∥ ∇ f ∥ L 2 + C ∥ ∇ π ∥ L 2 + C + C ∥ Λ 3 u ∥ L 2 + C ∥ Λ 3 ρ ∥ L 2 + C ∥ Λ 3 b ∥ L 2 + C ∥ Λ 3 ρ ∥ L 2 ∥ ∇ π ∥ L 3 ,$
which gives
$∥ ∇ f ∥ L 2 ≤ C + C ∥ ∇ u ∥ L ∞ + C ∥ Λ 3 ρ ∥ L 2 + C ∥ Λ 3 u ∥ L 2 + C ∥ Λ 3 b ∥ L 2 + C ∥ Λ 3 ρ ∥ L 2 ( ∥ ∇ π ∥ L 2 + ∥ Δ π ∥ L 4 ) ≤ C + C ∥ ∇ u ∥ L ∞ + C ∥ Λ 3 ( ρ , u , b ) ∥ L 2 + C ∥ Λ 3 ρ ∥ L 2 ( ∥ ∇ u ∥ L ∞ + 1 + ∥ Δ b ∥ L 4 ) .$
Inserting the above estimates into (44), we obtain
$I 3 ≤ C ( ∥ ∇ u ∥ L ∞ + 1 + ∥ ∇ b ∥ L ∞ + ∥ Δ b ∥ L 4 ) ( ∥ Λ 3 ρ ∥ L 2 2 + ∥ Λ 3 u ∥ L 2 2 ) + C ∥ Λ 3 ( ρ , u , b ) ∥ L 2 2 ≤ C ( ∥ ∇ u ∥ L ∞ + 1 + ∥ ∇ b ∥ L ∞ + ∥ Δ b ∥ L 4 ) ∥ Λ 3 ( ρ , u , b ) ∥ L 2 2 .$
Inserting the above estimates of $I 1 , I 3 , I 4 , I 5$ and $I 7$ into (43), we have
$1 2 d d t ∫ ( ρ | Λ 3 u | 2 + | Λ 3 b | 2 ) d x + η ∫ | Λ 4 b | 2 d x ≤ C ( ∥ ∇ u ∥ L ∞ + 1 + ∥ ∇ b ∥ L ∞ + ∥ Δ b ∥ L 4 ) ∥ Λ 3 ( ρ , u , b ) ∥ L 2 2 .$
Summing up (40) and (46) and using the Gronwall inequality, we conclude that
$∥ Λ 3 ( ρ , u , b ) ∥ L 2 + ∫ 0 T ∫ | Λ 4 b | 2 d x d t ≤ C .$
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 $ρ = 1$ and $η = 0$, Caflisch et al. [12] showed the following regularity criterion:
$rot u , rot b ∈ L 1 ( 0 , T ; L ∞ ) .$
Since the problem is very challenging, we are unable to present further developments.

## Author Contributions

Writing—original draft, K.S. and J.F.; Writing—review & editing, G.N. All authors have read and agreed to the published version of the manuscript.

## Funding

J. Fan is partially supported by the NSFC (No. 11971234). The authors are indebted to the referees for their valuable suggestions.

## Data Availability Statement

The data in this study are available from the corresponding author upon request.

## Conflicts of Interest

The authors declare no conflict of interest.

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MDPI and ACS Style

Shi, K.; Fan, J.; Nakamura, G. A Blow-Up Criterion for the Density-Dependent Incompressible Magnetohydrodynamic System with Zero Viscosity. Mathematics 2024, 12, 1510. https://doi.org/10.3390/math12101510

AMA Style

Shi K, Fan J, Nakamura G. A Blow-Up Criterion for the Density-Dependent Incompressible Magnetohydrodynamic System with Zero Viscosity. Mathematics. 2024; 12(10):1510. https://doi.org/10.3390/math12101510

Chicago/Turabian Style

Shi, Kunlong, Jishan Fan, and Gen Nakamura. 2024. "A Blow-Up Criterion for the Density-Dependent Incompressible Magnetohydrodynamic System with Zero Viscosity" Mathematics 12, no. 10: 1510. https://doi.org/10.3390/math12101510

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