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Low Mach number and non-resistive limit of magnetohydrodynamic equations with large temperature variations in general bounded domains

  • Min Liang and Yaobin Ou EMAIL logo
Published/Copyright: August 6, 2025
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Advances in Nonlinear Analysis
From the journal Advances in Nonlinear Analysis Volume 14 Issue 1

Abstract

This article verifies the low Mach number and non-resistive limit of local strong solutions to non-isentropic compressible magnetohydrodynamic (MHD) equations in general three-dimensional bounded domains when the temperature variation is large but finite. The uniform estimates of strong solutions are obtained in a short time interval independent of the Mach number and the magnetic resistivity coefficient, provided that the initial data are well prepared. Previous results on this multi-scale singular limit of compressible MHD equations are either for the cases of the weak solutions or for the cases of small temperature variations or for the cases of flat boundary where the mechanisms are essentially different. The new ingredient of this article is that we develop a new approach of weighted energy to establish the uniform estimates for high-order spatial derivatives of solutions.

Keywords: low Mach number limit; non-resistive limit; large temperature variations; magnetohydrodynamic equations
MSC 2010: 35B40; 35M33; 35Q35; 76W05

1 Introduction

This article is concerned with the low Mach number and non-resistive limit of the following non-dimensional magnetohydrodynamic (MHD) equations for compressible viscous heat-conducting fluids in Ω × ( 0 , T ] (see [17], for instance):

(1.1) ρ t ε + div ( ρ ε , ν u ε , ν ) = 0 , ( ρ ε , ν u ε , ν ) t + div ( ρ ε , ν u ε , ν ⊗ u ε , ν ) − div ( 2 μ D ( u ε , ν ) + λ div u ε , ν I ) + 1 ε 2 ∇ p ε , ν − 1 4 π ( curl B ε , ν ) × B ε , ν = 0 , ( ρ ε , ν e ε , ν ) t + div ( ρ ε , ν u ε , ν e ε , ν ) + p ε , ν div u ε , ν − div ( κ ∇ T ε , ν ) = ε 2 ( 2 μ ∣ D ( u ε , ν ) ∣ 2 + λ ( div u ε , ν ) 2 ) + ν ε 2 4 π ∣ curl B ε , ν ∣ 2 , B t ε , ν − curl ( u ε , ν × B ε , ν ) = − curl ( ν curl B ε , ν ) , div B ε , ν = 0 ,

twhere Ω is a general bounded domain in R 3 , and ρ ε , ν , u ε , ν , B ε , ν , p ε , ν , e ε , ν , and T ε , ν signify the density, velocity field, magnetic field, pressure, internal energy, and temperature, respectively, and D ( u ε , ν ) = 1 2 ( ∇ u ε , ν + ( ∇ u ε , ν ) t ) . The constants μ and λ are the shear and bulk viscosity coefficients, respectively, which satisfies μ > 0 and μ + 3 2 λ ≥ 0 . The constant ε ∈ ( 0 , 1 ] is the Mach number, while κ > 0 is the heat conductivity coefficient, and ν = O ( ε ς ) is the magnetic resistivity coefficient, where ς is a positive constant. Moreover, we suppose that the fluid is an ideal polytropic fluid, which satisfies

(1.2) p ε , ν = R ρ ε , ν T ε , ν , e ε , ν = C v T ε , ν ,

where R > 0 and C v > 0 are the generic gas constant and the specific heat at constant volume, respectively. The ratio of specific heat is γ = 1 + R ⁄ C v . By a formal asymptotic expansion (see [30], for example), as both the Mach number ε and the magnetic resistivity coefficient ν tend to zero, we have

ρ ε , ν → r , p ε , ν → 1 , u ε , ν → ω , B ε , ν → B , ε − 2 ∇ p ε , ν → ∇ π ,

where ( r , ω , B , π ) is a solution to the following problem:

(1.3) r t + div ( r ω ) = 0 , ( r ω ) t + div ( r ω ⊗ ω ) − div ( 2 μ D ( ω ) + λ div ω I ) + ∇ π − 1 4 π ( curl B ) × B = 0 , r div ω = div [ κ R ∇ 1 r ] , B t − curl ( ω × B ) = 0 , div B = 0 .

In particular, if the fluid is isentropic or the temperature has small variations, the limiting velocity is divergence-free when only the Mach number tends to zero. In this case, the low Mach number limit is known as the incompressible limit. However, in this article, we consider the general case where temperature T ε , ν has a large but finite variation, i.e., ∇ T ε , ν = O ( 1 ) , and the divergence of the limiting velocity is nonzero but depends on the diffusion of the temperature. Verifying the low Mach number limit rigorously from a mathematical perspective is crucial in physical applications. According to the basic form of the momentum equation, the nonlinear term with large parameter 1 ⁄ ε 2 leads to essential mathematical difficulties. Klainerman and Majda established the general framework for the low Mach number limit of local smooth solutions to hydrodynamic equations in their seminal works [24,25]. Then, there has been a significant amount of research on the low Mach number limit of hydrodynamic equations. Based on the relation between the pressure term ε − 2 ∇ p ε , ν and the temperature, we introduce some previous results in two aspects.

In the isentropic case, there have already been many results about the low Mach number limit under different assumptions. Lions and Masmoudi [31] established the convergence of global weak solutions of isentropic compressible Navier-Stokes equations ( B ε , ν ≡ 0 ) in various domains using the group method. The study on the compressible MHD equations presents more challenges due to the strong coupling between magnetic field and velocity. For the various boundary conditions, Hu and Wang [14] justified the incompressible limit for global weak solutions of the compressible MHD equations. For the case of the periodic domain, Jiang et al. [15] proved the incompressible limit of the compressible MHD equations with well-prepared initial data and obtained the convergence rates of solutions via the modulated energy method, and the case of general initial data was proved by Li and Mu [27]. For bounded domains, the low Mach number limit of the global smooth solutions to compressible MHD equations with the Navier-slip boundary condition and well-prepared initial data was verified by Dou et al. [6]. And the results were extended to the case with non-slip boundary condition by Dou and Ju [7].

In the non-isentropic case, due to the thermal effect, it is difficult to establish uniform estimates, and the low Mach number limit problem becomes more complex. For small temperature variations, Jiang et al. [17] rigorously justified the low Mach number limit of local smooth solutions to the Cauchy problem of full MHD equations with positive magnetic diffusivity coefficient ( ν > 0 ) and well-prepared initial data by energy estimates and convergence stability lemma, and the case of non-isentropic ideal MHD equations with general initial data was studied in [18]. In a three-dimensional bounded domain, Cui et al. [5] established the uniform estimates and verified the incompressible limit of local strong solutions to the compressible MHD equations, and the case of zero magnetic diffusivity was verified by Ou and Yang [32]. For large temperature variations, in the whole space, the first rigorous analysis of the low Mach number limit for classical solutions of the full Navier-Stokes equations with general initial data was carried out by Alazard [1], while Jiang et al. [19] investigated the case of fully compressible MHD equations with general initial data via the weighted energy method. In bounded domains, Jiang and Ou [20], Ju and Ou [22] verified the low Mach number limit of the Navier-Stokes equations with well-prepared initial data for zero thermal conductivity and positive thermal conductivity, respectively. Recently, as an extension of [22], the low Mach number limit of local strong solutions to the initial boundary value problem of (1.1) with large temperature variations was justified in [29], given a fixed positive magnetic resistivity coefficient or zero magnetic resistivity coefficient, respectively. It is worth noting that the uniform estimates with respect to the magnetic resistivity coefficient were not established in [29]. Therefore, the purpose of the present work is to continue the investigation in [29], establish the uniform estimates in both the Mach number and the magnetic resistivity coefficient, and achieve the asymptotic limit that these two physical parameters tend to zero simultaneously. For more results on the low Mach number limit of the compressible MHD equations, one may refer to [9,26,28,34,36].

When the magnetic resistivity coefficient is small, the fluid is highly conductive. Moreover, in the presence of physical boundaries, the non-resistive limit for the compressible resistive MHD equations is important in application, and it is also a challenging mathematical topic due to the effect of the magnetic boundary layer. For the one-dimensional compressible isentropic viscous MHD equations, Jiang and Zhang [21] justified the vanishing resistivity limit and the global well-posedness of strong solutions by using the effective viscous flux and the structure of the equations. For the multi-dimensional full MHD equations, as the Mach number, the shear viscosity coefficient, and the magnetic resistivity coefficient go to zero at the same time, Jiang et al. [16] studied the incompressible limit with general initial data in the whole space. In a bound domain Ω ⊂ R 3 , as the viscosity and resistivity coefficients go to zero simultaneously, Zhang [39] established the convergence of the global weak solutions for the nonhomogeneous incompressible MHD equations with the following Navier slip boundary conditions:

u ⋅ n = 0 , τ ⋅ S ( u ) ⋅ n + α 1 u ⋅ τ = 0 , on ∂ Ω × ( 0 , T ] , B ⋅ n = 0 , τ ⋅ S ( B ) ⋅ n + α 2 B ⋅ τ = 0 , on ∂ Ω × ( 0 , T ] ,

while Duan et al. [8] discussed the case of the strong solutions in T 2 × ( 0 , 1 ) with T 2 being the two-dimensional torus where the magnetic field satisfies the insulating boundary condition ( B × n = 0 ). Note that the uniform H 3 -estimates of strong solutions and the asymptotic convergence were only investigated for a flat domain in [8]. More relevant literature on the vanishing dissipation limit with respect to more than one parameter (such as the Mach number, viscosity coefficient, and magnetic resistivity coefficient) can be shown in earlier studies [4,15,23,38]. Guo et al. [12] investigated the inviscid and incompressible limit of global weak solutions to isentropic compressible viscous MHD equations with well-prepared initial data in a three-dimensional bounded domain. Recently, Gu and Ou [11] derived the incompressible and non-resistive limit of local strong solutions to isentropic compressible resistive MHD equations with ill-prepared initial data in three-dimensional bounded domains with flat boundaries. However, the case of non-isentropic compressible MHD equations in general bounded domains is still a challenging problem.

In this article, we address on the low Mach number and non-resistive limit of the non-isentropic compressible MHD equations with large temperature variations and well-prepared initial data in general bounded domains. We establish the uniform estimates for the Mach number ε and the magnetic resistivity coefficient ν of the local strong solutions in a short time interval. Moreover, we prove that, as both ε and ν tend to zero, the strong solution of (1.9) will converge to that of (1.19). In comparison with [12] on the case of weak solutions of the isentropic compressible MHD equations, the features of present problem are the effects of large temperature changes and the establishment of the higher-order uniform estimates of local strong solution with respect to both the Mach number and the magnetic resistivity coefficient. Considering the large temperature changes, it is difficult to obtain the uniform estimates of the local strong solutions, because the structure of singular terms of O ( 1 ε ) in (1.4) violates the standard ones in [5,17]. To overcome this difficulty, we introduce the variable v ε , ν = u ε , ν − κ 2 ∇ θ ε , ν (cf. [1]) and then change the original system (1.4) into (1.9), where the resulting singular differential operator is anti-symmetric. Furthermore, we need to additionally handle the high-order terms − β ∇ Δ θ ε , ν introduced in the new system. Upon comparing to [11] for the incompressible and non-resistive limit of isentropic MHD equations in T 2 × ( 0 , 1 ) , where the estimates of the magnetic field were established through several higher-order boundary conditions arising from the insulating boundary condition B ε , ν × n = 0 and the flat boundary assumption Γ = { x 3 = 0 , 1 } , the method therein is ineffective for general bounded domains. In contrast with [29] on the low Mach number limit, this article studies a different kind of singular limit that both the Mach number and the magnetic resistivity coefficient tend to zero, which makes high-order temporal and spatial estimates more subtle and requires balance between the Mach number and the magnetic resistivity coefficient. In particular, the high-order estimates in [29] are not uniform in the magnetic resistivity coefficient ν . Thus, we develop new estimates to overcome this obstacle.

We consider the case when the pressure p ε , ν has a small disturbance near a constant state, while the temperature T ε , ν has a finite variation. Thus, we introduce the change of variables:

p ε , ν = 1 + ε σ ε , ν , T ε , ν = 1 + θ ε , ν .

Then, we convert (1.1)–(1.2) into an “overdetermined” system

(1.4) ρ t ε , ν + div ( ρ ε , ν u ε , ν ) = 0 , ρ ε , ν ( u t ε , ν + u ε , ν ⋅ ∇ u ε , ν ) − μ △ u ε , ν − ( μ + λ ) ∇ div u ε , ν + 1 ε ∇ σ ε , ν − 1 4 π ( curl B ε , ν ) × B ε , ν = 0 , C v [ ρ ε , ν ( θ t ε , ν + u ε , ν ⋅ ∇ θ ε , ν ) ] + ( 1 + ε σ ε , ν ) div u ε , ν = κ Δ θ ε , ν + ε 2 ( 2 μ ∣ D ( u ε , ν ) ∣ 2 + λ ( div u ε , ν ) 2 ) + ν ε 2 4 π ∣ curl B ε , ν ∣ 2 , C v ⁄ R [ σ t ε , ν + u ε , ν ⋅ ∇ σ ε , ν ] + ( 1 + C v ⁄ R ) σ ε , ν div u ε , ν + ( 1 + C v ⁄ R ) 1 ε div u ε , ν − κ ε Δ θ ε , ν = ε ( 2 μ ∣ D ( u ε , ν ) ∣ 2 + λ ( div u ε , ν ) 2 ) + ν ε 4 π ∣ curl B ε , ν ∣ 2 , B t ε , ν − curl ( u ε , ν × B ε , ν ) = − curl ( ν curl B ε , ν ) , div B ε , ν = 0 ,

which coincides (1.1) when the solutions are smooth enough and the initial data are compatible. Suppose that ( ρ ε , ν , u ε , ν , σ ε , ν , θ ε , ν , B ε , ν ) satisfies the following initial conditions:

(1.5) ( ρ ε , ν , u ε , ν , σ ε , ν , θ ε , ν , B ε , ν ) ∣ t = 0 = ( ρ 0 ε , ν , u 0 ε , ν , σ 0 ε , ν , θ 0 ε , ν , B 0 ε , ν ) ,

and the boundary conditions:

(1.6) u ε , ν ⋅ n = 0 , n × curl u ε , ν = 0 , on ∂ Ω × ( 0 , T ] ,

(1.7) ∂ θ ε , ν ∂ n = 0 , B ε , ν ⋅ n = 0 , n × curl B ε , ν = 0 , on ∂ Ω × ( 0 , T ] ,

where ρ 0 ε , ν ≔ ( 1 + ε σ 0 ε , ν ) ⁄ ( 1 + θ 0 ε , ν ) and n is the unit outward normal vector to the boundary. In addition, assume that R = C v = 1 without the loss of generality. To symmetrize the singular operator, we set

(1.8) v ε , ν = u ε , ν − κ 2 ∇ θ ε , ν .

Then, ( ρ ε , ν , u ε , ν , σ ε , ν , θ ε , ν , B ε , ν ) satisfies the following initial boundary value problem:

(1.9) ρ t ε , ν + div ( ρ u ε , ν ) = 0 , in Ω × ( 0 , T ] , ρ ε , ν ( v t ε , ν + u ε , ν ⋅ ∇ v ε , ν ) + 1 ε ∇ σ ε , ν − μ Δ v ε , ν − ξ ∇ div v ε , ν − β ∇ Δ θ ε , ν − 1 4 π ( curl B ε , ν ) × B ε , ν = f ε , ν , in Ω × ( 0 , T ] , 1 2 ( σ t ε , ν + u ε , ν ⋅ ∇ σ ε , ν ) + 1 ε div v ε , ν = g ε , ν , in Ω × ( 0 , T ] , ρ ε , ν ( θ t ε , ν + u ε , ν ⋅ ∇ θ ε , ν ) − κ 2 Δ θ ε , ν + div v ε , ν = h ε , ν , in Ω × ( 0 , T ] , B t ε , ν − curl ( u ε , ν × B ε , ν ) = − curl ( ν curl B ε , ν ) , div B ε , ν = 0 , in Ω × ( 0 , T ] , u ε , ν ⋅ n = B ε , ν ⋅ n = 0 , ∂ θ ε , ν ∂ n = 0 , curl u ε , ν × n = curl B ε , ν × n = 0 , on ∂ Ω × ( 0 , T ] , ( ρ ε , ν , v ε , ν , σ ε , ν , θ ε , ν , B ε , ν ) ∣ t = 0 = ( ρ 0 ε , ν , u 0 ε , ν − κ 2 ∇ θ 0 ε , ν , σ 0 ε , ν , θ 0 ε , ν , B 0 ε , ν ) ( x ) , x ∈ Ω ,

where the constants ξ = μ + λ + κ 2 , β = κ 2 ( 2 μ + λ − κ 2 ) , and the functions

(1.10) f ε , ν = κ 2 [ ∇ ρ ε , ν ( θ t ε , ν + u ε , ν ⋅ ∇ θ ε , ν ) + ρ ε , ν ⋅ ∇ u ε , ν ⋅ ∇ θ ε , ν − ∇ h ε , ν ] , g ε , ν = − σ ε , ν div u ε , ν + ε 2 [ 2 μ ∣ D ( u ε , ν ) ∣ 2 + λ ( div u ε , ν ) 2 ] + ν ε 8 π ∣ curl B ε , ν ∣ 2 , h ε , ν = − ε σ ε , ν div u ε , ν + ε 2 [ 2 μ ∣ D ( u ε , ν ) ∣ 2 + λ ( div u ε , ν ) 2 ] + ν ε 2 4 π ∣ curl B ε , ν ∣ 2 .

The local existence theorem of (1.9)–(1.10) for fixed ε and ν can be proved by the standard fixed point theory [11,13,33]. Thus, we only state the conclusion, but omit the details of the proof here.

Theorem 1.1

(Local existence) Let ε ∈ ( 0 , 1 ] , ν ∈ ( 0 , 1 ] be fixed numbers and Ω ⊂ R 3 be a simply connected, bounded domain with smooth boundary ∂ Ω . Suppose that the initial datum ( ρ 0 ε , ν , v 0 ε , ν , σ 0 ε , ν , B 0 ε , ν ) ∈ H 3 ( Ω ) , θ 0 ε , ν ∈ H 4 ( Ω ) , satisfies ρ 0 ε , ν ≥ m ε , ν and 1 + θ 0 ε , ν ≥ m ε , ν for some positive constant m ε , ν , and

( ∂ t k ρ ε , ν ( 0 ) , ∂ t k v ε , ν ( 0 ) , ∂ t k σ ε , ν ( 0 ) , ∂ t k B ε , ν ( 0 ) ) ∈ H 2 − k ( Ω ) , ∂ t k θ ε , ν ( 0 ) ∈ H 3 − k ( Ω ) , k = 0 , 1 .

Moreover, we assume that the following compatibility conditions are satisfied:

(1.11) v 0 ε , ν ⋅ n = v t ε , ν ( 0 ) ⋅ n = 0 , curl v 0 ε , ν × n = curl v t ε , ν ( 0 ) × n = 0 , on ∂ Ω

and

(1.12) B 0 ε , ν ⋅ n = B t ε , ν ( 0 ) ⋅ n = 0 , curl B 0 ε , ν × n = curl B t ε , ν ( 0 ) × n = 0 , ∂ θ 0 ε , ν ∂ n = ∂ θ t ε , ν ( 0 ) ∂ n = 0 , on ∂ Ω .

Then, there exists a positive constant T ε , ν = T ε , ν ( ρ 0 ε , ν , v 0 ε , ν , σ 0 ε , ν , B 0 ε , ν , θ 0 ε , ν , m ε , ν , ε , ν ) such that the initial boundary value problems (1.9)–(1.10) admits a unique strong solution ( ρ ε , ν , v ε , ν , σ ε , ν , B ε , ν , θ ε , ν ) , satisfying ρ ε , ν > 0 and 1 + θ ε , ν > 0 in Ω × ( 0 , T ε , ν ) and

( ρ ε , ν , σ ε , ν , v ε , ν , B ε , ν ) ∈ C ( [ 0 , T ε , ν ] , H 2 ( Ω ) ) , ( v ε , ν , B ε , ν ) ∈ L 2 ( 0 , T ε , ν ; H 3 ( Ω ) ) , ( ρ t ε , ν , σ t ε , ν , v t ε , ν , B t ε , ν ) ∈ C ( [ 0 , T ε , ν ] , H 1 ( Ω ) ) , ( v t ε , ν , B t ε , ν ) ∈ L 2 ( 0 , T ε , ν ; H 2 ( Ω ) ) , θ ε , ν ∈ C ( [ 0 , T ε , ν ] , H 3 ( Ω ) ) ∩ L 2 ( 0 , T ε , ν ; H 4 ( Ω ) ) , θ t ε , ν ∈ C ( [ 0 , T ε , ν ] , H 2 ( Ω ) ) ∩ L 2 ( 0 , T ε , ν ; H 3 ( Ω ) ) .

Remark 1.1

The notations v t ε , ν ( 0 ) and σ t ε , ν ( 0 ) are defined by taking t = 0 in the system (1.9) 2,3 , respectively:

u t ε , ν ( 0 ) ≔ − u 0 ε , ν ⋅ ∇ u 0 ε , ν + ( ρ 0 ε , ν ) − 1 [ − ε − 1 ∇ σ 0 ε , ν + μ Δ u 0 ε , ν − ( μ + λ ) ∇ div u 0 ε , ν + 1 4 π ( curl B 0 ε , ν ) × B 0 ε , ν ] , σ t ε , ν ( 0 ) ≔ − u 0 ε , ν ⋅ ∇ σ 0 ε , ν − 2 σ 0 ε , ν div u 0 ε , ν − ε − 1 ( 2 div u 0 ε , ν − κ ε Δ θ 0 ε , ν ) + ε ( 2 μ ∣ D ( u 0 ε , ν ) ∣ 2 + λ ( div u 0 ε , ν ) 2 + ν 4 π ∣ curl B 0 ε , ν ∣ 2 ) .

ρ t ε , ν ( 0 ) , B t ε , ν ( 0 ) , and θ t ε , ν ( 0 ) can be defined in the same way. Therefore, to ensure M 0 ε , ν ≤ D 0 , we require

div u 0 ε , ν − κ 2 Δ θ 0 ε , ν L 2 ( Ω ) + ‖ ∇ σ 0 ε , ν ‖ L 2 ( Ω ) ≤ O ( ε ) ,

which means the initial data of (1.9) are “well prepared” in the sense of L 2 -norm.

Based on the local existence theorem (Theorem 1.1), the main results of the article read as follows.

Definition 1.1

Suppose that ε , ν ∈ ( 0 , 1 ] are arbitrary constants with 0 ≤ ν ≤ O ( ε ς ) , 0 < ς < 2 . We define the uniform energy

(1.13) M ε , ν ( t ) ≔ ‖ ( ρ ε , ν , v ε , ν , σ ε , ν , B ε , ν ) ‖ L ∞ ( 0 , t ; H 2 ( Ω ) ) 2 + ‖ θ ε , ν ‖ L ∞ ( 0 , t ; H 3 ( Ω ) ) 2 + ‖ ( ρ ε , ν ) − 1 ‖ L ∞ ( ( 0 , t ) × Ω ) 2 + ‖ ( ρ t ε , ν , v t ε , ν , σ t ε , ν , B t ε , ν ) ‖ L ∞ ( 0 , t ; L 2 ( Ω ) ) 2 + ‖ θ t ε , ν ‖ L ∞ ( 0 , t ; H 1 ( Ω ) ) 2 + ‖ ( v ε , ν , ∇ θ ε , ν , ν B ε , ν ) ‖ L 2 ( 0 , t ; H 3 ( Ω ) ) 2 + ‖ ( v t ε , ν , ∇ θ t ε , ν , ν B t ε , ν , σ t ε , ν ) ‖ L 2 ( 0 , t ; H 1 ( Ω ) ) 2 + ‖ ε ( ρ t ε , ν , v t ε , ν , σ t ε , ν , ∇ θ t ε , ν , B t ε , ν ) ‖ L ∞ ( 0 , t ; H 1 ( Ω ) ) 2 + ‖ ε ( v t ε , ν , ∇ θ t ε , ν , ν B t ε , ν ) ‖ L 2 ( 0 , t ; H 2 ( Ω ) ) 2 .

Theorem 1.2

(Uniform estimates) Let β > 0 and Ω ⊂ R 3 be a simply connected, bounded domain with smooth boundary ∂ Ω . Suppose that ε , ν ∈ ( 0 , 1 ] are arbitrary constants with 0 ≤ ν ≤ O ( ε ς ) , 0 < ς < 2 . For any positive constants A and D 0 , independent of ε and ν , if the initial datum ( ρ 0 ε , ν , v 0 ε , ν , σ 0 ε , ν , B 0 ε , ν , θ 0 ε , ν , m ε , ν ) of (1.9)–(1.10) satisfies

(1.14) 1 + θ 0 ε , ν ≥ A

and

(1.15) ‖ ( ρ 0 ε , ν , v 0 ε , ν , σ 0 ε , ν , B 0 ε , ν ) ‖ H 2 ( Ω ) 2 + ‖ θ 0 ε , ν ‖ H 3 ( Ω ) 2 + ‖ ( ρ 0 ε , ν ) − 1 ‖ L ∞ ( Ω ) 2 + ‖ ( ρ t ε , ν , v t ε , ν , σ t ε , ν , B t ε , ν ) ( 0 ) ‖ L 2 ( Ω ) 2 + ‖ θ t ε , ν ( 0 ) ‖ H 1 ( Ω ) 2 + ‖ ε θ t ε , ν ( 0 ) ‖ H 2 ( Ω ) 2 + ‖ ε ( ρ t ε , ν , v t ε , ν , σ t ε , ν , B t ε , ν ) ( 0 ) ‖ H 1 ( Ω ) 2 ≤ D 0 ,

and the compatibility conditions (1.11)–(1.12) hold. Then, there exist positive constants T and D, independent of ε and ν , such that the unique strong solution ( ρ ε , ν , v ε , ν , σ ε , ν , B ε , ν , θ ε , ν ) to (1.9)–(1.10), which is obtained in Theorem 1.1, satisfies the following uniform estimates:

(1.16) M ε , ν ( t ) ≤ D ,

and

(1.17) 1 + θ ε , ν ( x , t ) ≥ D − 1 ,

where ( x , t ) ∈ Ω × [ 0 , T ] .

Similar to [1], it suffices to show the following proposition to obtain the uniform estimates in (1.16).

Proposition 1.1

Suppose that ε , ν ∈ ( 0 , 1 ] are arbitrary constants with 0 ≤ 0 ≤ ν ≤ O ( ε ς ) , 0 < ς < 2 . Let T ε , ν be the maximal time of existence for the initial boundary value problems (1.9)–(1.10) in the sense of Theorem 1.1. Then, for any 0 ≤ t ≤ min { T ε , ν , 1 } , we have

M ε , ν ( t ) ≤ C 0 ( M 0 ε , ν ) exp { ( t 1 ⁄ 4 + ε 1 − ς ⁄ 2 ) C ( M ε , ν ( t ) ) } ,

for some given nondecreasing and continuous functions C 0 ( ⋅ ) and C ( ⋅ ) .

Remark 1.2

From (1.4), (1.8), and (1.16), we obtain the following estimates:

(1.18) ‖ u ε , ν ‖ L ∞ ( 0 , t ; H 2 ( Ω ) ) 2 + ‖ u ε , ν ‖ L 2 ( 0 , t ; H 3 ( Ω ) ) 2 + ‖ u t ε , ν ‖ L ∞ ( 0 , t ; L 2 ( Ω ) ) 2 + ‖ u t ε , ν ‖ L 2 ( 0 , t ; H 1 ( Ω ) ) 2 + ‖ ε u t ε , ν ‖ L ∞ ( 0 , t ; H 1 ( Ω ) ) 2 + ‖ ε u t ε , ν ‖ L 2 ( 0 , t ; H 2 ( Ω ) ) 2 ≤ D , t ∈ ( 0 , T ] .

With the uniform bounds in (1.16), we derive the singular limits when the Mach number and the magnetic resistivity coefficient tend to zero.

Theorem 1.3

(Singular limit) Let ( ρ ε , ν , v ε , ν , σ ε , ν , B ε , ν , θ ε , ν ) be the local strong solution derived from Theorem 1.1 and satisfy the uniform estimates established in Theorem 1.2. Then, as ε , ν → 0 , for any δ > 0 , we have

( ρ ε , ν , u ε , ν , B ε , ν ) → ( r , ω , B ) , in C ( [ 0 , T ] , H 2 − δ ( Ω ) ) , θ ε , ν → θ , in C ( [ 0 , T ] , H 3 − δ ( Ω ) ) , ( ρ t ε , ν , u t ε , ν , B t ε , ν ) ⇀ ( r t , ω t , B t ) weakly - * , in L ∞ ( 0 , T ; L 2 ( Ω ) ) , θ t ε , ν ⇀ θ t weakly - * , in L ∞ ( 0 , T ; H 1 ( Ω ) ) ,

with r ( 1 + θ ) = 1 . Moreover, there exists a function π ( x , t ) such that ( θ , ω , B , π ) is the unique smooth solution of the following initial boundary value problem of the incompressible MHD equations:

(1.19) ( 1 + θ ) − 1 ( ω t + ω ⋅ ∇ ω ) + ∇ π = μ Δ ω + ( μ + λ ) ∇ div ω + 1 4 π ( curl B ) × B , in Ω × ( 0 , T ] , ( 1 + θ ) − 1 ( θ t + ω ⋅ ∇ θ ) = κ 2 Δ θ , in Ω × ( 0 , T ] , div ω = κ 2 Δ θ , in Ω × ( 0 , T ] , B t − curl ( ω × B ) = 0 , div B = 0 , in Ω × ( 0 , T ] , ω ⋅ n = 0 , curl ω × n = 0 , ∂ θ ∂ n = 0 , B ⋅ n = 0 , curl B × n = 0 , on ∂ Ω × [ 0 , T ] , ( r , ω , B ) ∣ t = 0 = ( r 0 , ω 0 , B 0 ) ( x ) , x ∈ Ω ,

where ( θ 0 , ω 0 , B 0 ) is the weak limit of ( θ 0 ε , ν , u 0 ε , ν , B 0 ε , ν ) in H 2 ( Ω ) .

Remark 1.3

(i) It is worth noting that system (1.3) is equivalent to the incompressible MHD equations (1.19) due to r ( 1 + θ ) = 1 and the regularity of ( θ , ω , B , π ) .

(ii) From Theorems 1.2 and 1.3, the regularity of (1.19) can be obtained that

( ω , B ) ∈ L ∞ ( 0 , T ; H 2 ( Ω ) ) , ω ∈ L 2 ( 0 , T ; H 3 ( Ω ) ) , r ∈ L ∞ ( 0 , T ; H 2 ( Ω ) ) , ( ω t , B t ) ∈ L ∞ ( 0 , T ; L 2 ( Ω ) ) , ω t ∈ L 2 ( 0 , T ; H 1 ( Ω ) ) , r t ∈ L ∞ ( 0 , T ; H 1 ( Ω ) ) , ∇ π ∈ L ∞ ( 0 , T ; L 2 ( Ω ) ) ∩ L 2 ( 0 , T ; H 1 ( Ω ) ) , r ≥ C − 1 > 0 .

The key ingredient for the proof of our main results is to choose a new type of weighted energy functional. First, we derive the L 2 -estimates of ( v ε , ν , ∇ θ ε , ν , σ ε , ν , B ε , ν ) , which is fundamental to gain the estimates for the derivatives of higher order. Second, due to Lemmas 2.3–2.4 and div B ε , ν = 0 , we need to obtain estimates of ( div v ε , ν , ∇ σ ε , ν ) and ( curl v ε , ν , curl B ε , ν ) , respectively. Note that the boundary conditions of velocity and magnetic field have the same structure so that one can apply Lemma 2.4 to derive the estimates for the derivatives of higher order. Compared to the study in [29] for fixed magnetic resistivity coefficient, the main difference in this article is that the L 2 ( 0 , t ; H 3 ( Ω ) ) -norm of B ε , ν and the L 2 ( 0 , t ; H 1 ( Ω ) ) -norm of B t ε , ν are weighted. As a result, one cannot expect to estimate ‖ ∇ div v ε , ν ‖ L 2 ( Ω ) by ‖ B t ε , ν ‖ L t 2 ( H 1 ) as in [29]. Thus, we choose a new approach to estimate this term through the equation of magnetic field. It deserves to note that we have to balance the weighted coefficients of ε and ν for the estimates of ‖ ε div v t ε , ν ‖ L 2 ( Ω ) and other higher-order norms owing to the structure of momentum equation; thus, the assumption 0 ≤ ν ≤ O ( ε ς ) with 0 < ς < 2 is crucial. Due to the structure of the magnetic field equation, the method of estimating ‖ curlcurl B ε , ν ‖ L 2 ( Ω ) in [29] is not applicable when the magnetic resistivity coefficient is small. Therefore, we use a new method to obtain this estimate through Δ B ε , ν = − curlcurl B ε , ν .

The rest of this article is organized as follows. In Section 2, we introduce some preliminary knowledge used in this article, including some inequalities and compactness results. In Section 3, we prove Proposition 1.1 by using Lemmas 2.3–2.5, where the uniform estimates are shown by a composite analysis on both temporal and spatial derivatives. In Section 4, we give the proofs of Theorems 1.2 and 1.3.

2 Preliminaries

In this section, we present some basic lemmas, inequalities, and equalities, which will be used frequently throughout this article.

Lemma 2.1

(Interpolation inequality, cf. Friedman [10], Part 1, Theorem 10.1) Let Ω ⊂ R N be a bounded domain with a C κ -boundary and u be any function in W κ , r ( Ω ) ∩ L q ( Ω ) with 1 ≤ r , q ≤ ∞ . For any integer j with 0 ≤ j < κ , and for any number a in the interval [ j ⁄ κ , 1 ] , set

1 p = j N + a ( 1 r − κ N ) + ( 1 − a ) 1 q .

If κ − j − N ⁄ r is not a nonnegative integer, then

(2.1) ‖ D j u ‖ L p ( Ω ) ≤ C ‖ u ‖ W κ , r ( Ω ) a ‖ u ‖ L q ( Ω ) 1 − a .

If κ − j − N ⁄ r is a nonnegative integer, then (2.1) only holds for a = j ⁄ κ . The constant C depends only on Ω , r , q , κ , j , and a .

For convenience, we introduce some special cases of interpolation inequality, which will be applied frequently in the following sections.

Lemma 2.2

Let U be a bounded, open domain of R 3 with a C 3 boundary. Then, the following inequalities are satisfied:

‖ u ‖ L 3 ≤ C ‖ u ‖ H 1 1 2 ‖ u ‖ L 2 1 2 , ‖ ∇ 2 u ‖ L 2 ≤ C ‖ u ‖ H 3 1 2 ‖ ∇ u ‖ L 2 1 2 , ‖ ∇ 2 u ‖ L 3 ≤ C ‖ u ‖ H 3 3 4 ‖ u ‖ H 1 1 4 , ‖ ∇ 2 u ‖ L 4 ≤ C ‖ u ‖ H 3 7 8 ‖ u ‖ H 1 1 8 , ‖ ∇ u ‖ L ∞ ≤ C ‖ u ‖ W 2,4 ≤ C ‖ u ‖ H 3 7 8 ‖ u ‖ H 1 1 8 .

Moreover, by the Sobolev embedding theorem, we have

‖ u ‖ L ∞ ≤ C ‖ u ‖ W 1,4 ≤ C ‖ u ‖ H 2 3 4 ‖ u ‖ L 2 1 4 ≤ C ‖ u ‖ H 2 7 8 ‖ u ‖ L 2 1 8

and

‖ u ‖ L t 2 ( L 3 ) ≤ C t 1 4 ‖ u ‖ L t ∞ ( L 2 ) 1 2 ‖ u ‖ L t 2 ( H 1 ) 1 2 , ‖ u ‖ L t 2 ( L ∞ ) ≤ C t 1 8 ‖ u ‖ L t 2 ( H 2 ) 3 4 ‖ u ‖ L t ∞ ( H 1 ) 1 4 .

Lemma 2.3

(See [3]). Let Ω be a bounded domain in R N with a smooth boundary ∂ Ω and outward normal n . Then, there exists a constant C > 0 independent of u , such that

‖ u ‖ H s ≤ C ( ‖ div u ‖ H s − 1 ( Ω ) + ‖ curl u ‖ H s − 1 ( Ω ) + ‖ u ⋅ n ‖ H s − 1 ⁄ 2 ( ∂ Ω ) + ‖ u ‖ H s − 1 ( Ω ) ) ,

for any u ∈ [ H s ( Ω ) ] N , s ≥ 1 .

Lemma 2.4

(See [37]). Let Ω be a bounded domain in R N with a smooth boundary ∂ Ω and outward normal n . Then, there exists a constant C > 0 independent of u , such that

‖ u ‖ H s ≤ C ( ‖ div u ‖ H s − 1 ( Ω ) + ‖ curl u ‖ H s − 1 ( Ω ) + ‖ u × n ‖ H s − 1 ⁄ 2 ( ∂ Ω ) + ‖ u ‖ H s − 1 ( Ω ) ) ,

for any u ∈ [ H s ( Ω ) ] N , s ≥ 1 .

Lemma 2.5

(see [2]) Let v be a smooth function in H 1 ( Ω ) such that v × n ∣ ∂ Ω = 0 . Then,

curl v ⋅ n ∣ ∂ Ω = 0 .

Lemma 2.6

(Aubin-Lions-Simon Lemma, see [35]) Let X 0 , X , and X 1 be three Banach spaces with X 0 ⊆ X ⊆ X 1 . Suppose that X 0 is compactly embedded in X and that X is continuously embedded in X 1 . For 1 ≤ p , q ≤ + ∞ , let

W = { u ∈ L p ( 0 , T ; X 0 ) ∣ u t ∈ L q ( 0 , T ; X 1 ) } .

  1. If p < + ∞ , then the embedding of W into L p ( 0 , T ; X ) is compact.

  2. If p = + ∞ and q > 1 , then the embedding of W into C ( [ 0 , T ] ; X ) is compact.

Finally, we list some equalities that are often used in the estimates related to the magnetic field. For any vector fields a and b , we have

(2.2) curl ( a × b ) = ( b ⋅ ∇ ) a − ( a ⋅ ∇ ) b + a ( div b ) − b ( div a ) .

Before ending this section, we give the notations used throughout this article.

Notation.

(i) We denote the usual Sobolev spaces W κ , p ( Ω ) and H κ ( Ω ) by the norms ‖ ⋅ ‖ W k , p and ‖ ⋅ ‖ H k , respectively. We shall also use the following abbreviations for the Sobolev spaces involving time:

L t p ( H κ ) ≡ L p ( 0 , t ; H κ ( Ω ) ) , C t ( H κ ) ≡ C ( [ 0 , t ] , H κ ( Ω ) ) ,

with

‖ ⋅ ‖ L t p ( H κ ) ≡ ‖ ⋅ ‖ L p ( 0 , t ; H κ ) , ‖ ⋅ ‖ C t ( H κ ) ≡ ‖ ⋅ ‖ C ( [ 0 , t ] , H κ ) .

(ii) We will utilize the notation C to signify a generic positive constant independent of ε and ν , and C ( δ ) to denote a uniform constant, which may depend on δ but not on ε and ν , where δ is any of various small parameters introduced later.

(iii) The relation “ f ≲ g ” signifies that “ f ≤ C g .”

3 Uniform energy estimates

In this section, we shall establish the uniform estimates with respect to both the Mach number ε and the magnetic diffusivity coefficient ν in the bounded domain Ω ⊂ R 3 . T ε , ν is defined in Proposition 1.1. We will show that, for 0 ≤ t ≤ min { T ε , ν , 1 } , ε ∈ ( 0 , 1 ] , and 0 ≤ ν ≤ O ( ε ς ) with 0 < ς < 2 ,

M ε , ν ( t ) ≤ C 0 ( M 0 ε , ν ) exp t 1 ⁄ 4 + ε 1 − ς 2 C ( M ε , ν ( t ) ) ,

which can give the uniform bound of M ε , ν ( t ) with respect to ε and ν in this section. For the convenience of description, we will remove the superscript ε and ν in this section.

3.1 Estimates for the density ρ

Since ρ serves as the coefficient of v t in the momentum equation, it is convenient to estimate it first. The following estimates are standard; however, for the sake of complicity, we give a sketch of proof here.

Lemma 3.1

Suppose that the assumptions in Theorem 1.2 are satisfied. For any 0 < ε ≤ 1 and 0 ≤ t ≤ min { T ε , ν , 1 } , we have

(3.1) ‖ ( ρ , ρ − 1 ) ‖ L x , t ∞ + ‖ ρ ‖ L t ∞ ( H 2 ) + ‖ ρ t ‖ L t ∞ ( L 2 ) + ‖ ε ∇ ρ t ‖ L t ∞ ( L 2 ) ≤ C 0 ( M 0 ) exp t 1 2 C 1 ( M ( t ) ) ,

and

(3.2) ‖ ρ t ‖ L t ∞ ( H 1 ) ≤ M ( t ) .

Proof

Define the particle path χ ( x , s ; t ) through ( x , s ) to be the solution of

d d t χ ( x , s ; t ) = u ( χ ( x , s ; t ) , t ) , t , s ∈ [ 0 , T ] , x ∈ Ω , χ ( x , s ; s ) = x , x ∈ Ω .

Then, ρ can be expressed explicitly as

(3.3) ρ ( x , t ) = ρ 0 ( χ ( x , t ; 0 ) ) exp − ∫ 0 t div u ( χ ( x , t ; τ ) , τ ) d τ ,

on the particle path χ . From the explicit expression of ρ in (3.3), we obtain

sup 0 ≤ s ≤ t ‖ ρ − 1 ‖ L ∞ ( s ) ≤ ‖ ρ 0 − 1 ‖ L ∞ exp ∫ 0 t ‖ u ‖ H 3 d s ≤ M 0 1 2 exp t 1 2 C 1 ( M ( t ) ) .

Similarly, we have

sup 0 ≤ s ≤ t ‖ ρ ‖ L ∞ ( s ) ≤ C 0 ( M 0 ) exp t 1 2 C 1 ( M ( t ) ) .

Let α be a multi-index of spatial derivatives with 0 ≤ ∣ α ∣ ≤ 2 . Next, we apply D α to (1.1) 1 , then multiply the resulting equality by D α ρ , and integrate to obtain

‖ D α ρ ‖ L 2 ( t ) ≤ ‖ D α ρ 0 ‖ L 2 exp t 1 2 ‖ u ‖ L t 2 ( H 3 ) .

So we find

‖ ρ ‖ L t ∞ ( H 2 ) ≤ ‖ ρ 0 ‖ H 2 exp C t 1 2 ‖ u ‖ L t 2 ( H 3 ) ≤ C 0 ( M 0 ) exp t 1 2 C 1 ( M ( t ) ) .

Moreover, we differentiate (1.1) 1 with respect to the temporal variable once, multiply the result by ρ t , integrate over Ω × [ 0 , T ] , and use Grönwall’s inequality to discover that

‖ ρ t ‖ L t ∞ ( L 2 ) ≤ ‖ ρ t ( 0 ) ‖ L t ∞ ( L 2 ) + ∫ 0 t ‖ ρ ‖ H 2 ‖ u t ‖ H 1 d s exp { C ∫ 0 t ‖ u ‖ H 3 d s } ≤ C 0 ( M 0 ) exp t 1 2 C 1 ( M ( t ) ) .

Finally, we apply ∂ t ∇ to (1.1) 1 , then multiply ε ∇ ρ t , and integrate the result over Ω × [ 0 , T ] to obtain

‖ ε ∇ ρ t ‖ L t ∞ ( L 2 ) ≤ ∫ 0 t ( ‖ ρ ‖ H 2 ‖ ε u t ‖ H 2 + ‖ ρ t ‖ L 2 ‖ u ‖ H 3 ) d s ⋅ exp C ∫ 0 t ‖ u ‖ H 3 d s ≤ C 0 ( M 0 ) exp t 1 2 C 1 ( M ( t ) ) .

The estimate (3.2) of ρ t follows from (1.1) 1 directly.□

3.2 Estimates of f , g , and h and their derivatives

To obtain the uniform estimates in Proposition 1.1, we first give the estimates for f , g , h , g t , h t , and f t as follows.

First, we will provide estimates of ‖ h ‖ L t 2 ( H 2 ) and ‖ h t ‖ L t ∞ ( L 2 ) . Before that, using (1.4) 2 and the definition of M ( t ) , we derive

(3.4) ‖ ε u ‖ L t ∞ ( H 3 ) 2 ≤ C [ ( M ( t ) ) 2 + ( M ( t ) ) 3 ] .

In view of the boundary conditions curl B × n = 0 , we apply Lemmas 2.3 and 2.4 to (1.9) 5 to obtain

(3.5) ‖ ε ν curl B ‖ L t ∞ ( H 2 ) 2 ≲ ‖ ε ν curlcurlcurl B ‖ L t ∞ ( L 2 ) 2 + ‖ ε ν curl B ‖ L t ∞ ( H 1 ) 2 ≲ ess sup 0 ≤ s ≤ t ∫ Ω ε 2 ∣ curl [ − B t + curl ( u × B ) ] ∣ 2 d x ( s ) + ε 2 ‖ B ‖ L t ∞ ( H 2 ) 2 ≲ ‖ ε curl B t ‖ L t ∞ ( L 2 ) 2 + ε 2 ‖ u ‖ L t ∞ ( H 2 ) 2 ‖ B ‖ L t ∞ ( H 2 ) 2 + ε 2 C 1 ( M ( t ) ) ≲ C 1 ( M ( t ) ) ,

and thus,

(3.6) ‖ ε 2 ν curl B ‖ L t ∞ ( H 2 ) ≲ ε 1 − ς 2 ‖ ε ν curl B ‖ L t ∞ ( H 2 ) ≲ ε 1 − ς 2 C 1 ( M ( t ) ) .

Therefore, according to (1.10) and (3.4)–(3.6), the estimates of ‖ ∇ h ‖ L t ∞ ( L 2 ) , ‖ h ‖ L t 2 ( H 2 ) and ‖ g ‖ L t ∞ ( H 1 ) can be deduced as follows:

(3.7) ‖ ∇ h ‖ L t ∞ ( L 2 ) ≲ ε ‖ σ ‖ L t ∞ ( H 2 ) ‖ u ‖ L t ∞ ( H 2 ) + ε ‖ u ‖ L t ∞ ( H 2 ) ‖ ε u ‖ L t ∞ ( H 3 ) + ε ‖ ε ν curl B ‖ L t ∞ ( H 2 ) ‖ B ‖ L t ∞ ( H 2 ) ≲ ε C 1 ( M ( t ) ) ,

(3.8) ‖ h ‖ L t 2 ( H 2 ) ≲ ε ‖ σ ‖ L t ∞ ( H 2 ) ‖ div u ‖ L t 2 ( H 2 ) + ε ‖ ∇ u ‖ L t 2 ( H 2 ) ‖ ε ∇ u ‖ L t ∞ ( H 2 ) + ‖ ν B ‖ L t 2 ( H 3 ) ‖ ε 2 ν curl B ‖ L t ∞ ( H 2 ) ≲ ε 1 − ς 2 C 1 ( M ( t ) ) ,

and

(3.9) ‖ g ‖ L t ∞ ( H 1 ) ≲ ‖ σ ‖ L t ∞ ( H 2 ) ‖ u ‖ L t ∞ ( H 2 ) + ‖ u ‖ L t ∞ ( H 2 ) ‖ ε u ‖ L t ∞ ( H 3 ) + ‖ ε ν curl B ‖ L t ∞ ( H 2 ) ‖ B ‖ L t ∞ ( H 2 ) ≲ C 1 ( M ( t ) ) .

Furthermore, we have

(3.10) ‖ f ‖ L t ∞ ( L 2 ) = κ 2 [ ∇ ρ ( θ t + u ⋅ ∇ θ ) + ρ ∇ u ⋅ ∇ θ − ∇ h ] L t ∞ ( L 2 ) ≲ ‖ ∇ ρ ‖ L t ∞ ( H 1 ) ( ‖ θ t ‖ L t ∞ ( L 3 ) + ‖ u ‖ L t ∞ ( H 2 ) ‖ ∇ θ ‖ L t ∞ ( L 3 ) ) + ‖ ρ ‖ L t ∞ ( H 2 ) ‖ ∇ u ‖ L t ∞ ( H 1 ) ‖ ∇ θ ‖ L t ∞ ( L 3 ) + ‖ ∇ h ‖ L t ∞ ( L 2 ) ≲ C 1 ( M ( t ) )

and

(3.11) ‖ ∂ i f ‖ L t 2 ( L 2 ) = κ 2 ∂ i [ ∇ ρ ( θ t + u ⋅ ∇ θ ) + ρ ⋅ ∇ u ⋅ ∇ θ − ∇ h ] L t 2 ( L 2 ) ≲ ‖ ρ ‖ L t ∞ ( H 2 ) ( ‖ θ t ‖ L t 2 ( L ∞ ) + ‖ u ‖ L t ∞ ( H 2 ) ‖ ∇ θ ‖ L t 2 ( L ∞ ) ) + ‖ ∂ i ∇ h ‖ L t 2 ( L 2 ) + ‖ ∇ ρ ‖ L t ∞ ( H 1 ) ( ‖ ∇ θ t ‖ L t 2 ( L 3 ) + ‖ ∇ u ‖ L t ∞ ( H 1 ) ‖ ∇ θ ‖ L t 2 ( L ∞ ) + ‖ u ‖ L t ∞ ( H 2 ) ‖ ∂ i ∇ θ ‖ L t 2 ( L 3 ) ) + ‖ ρ ‖ L t ∞ ( H 2 ) ( ‖ ∂ i ∇ u ‖ L t ∞ ( L 2 ) ‖ ∇ θ ‖ L t 2 ( L ∞ ) + ‖ ∇ u ‖ L t ∞ ( H 1 ) ‖ ∂ i ∇ θ ‖ L t 2 ( L 3 ) ) ≲ ( t 1 8 + ε 1 − ς 2 ) C 1 ( M ( t ) ) .

Moreover, we have

(3.12) ‖ ∂ i ∇ g ‖ L t 1 ( L 2 ) = ∂ i ∇ ( − σ div u + ε 2 ( 2 μ ∣ D ( u ) ∣ 2 + λ ( div u ) 2 ) + ν ε 8 π ∣ curl B ∣ 2 ) L t 1 ( L 2 ) ≲ t 1 2 ‖ σ ‖ L t ∞ ( H 2 ) ‖ u ‖ L t 2 ( H 3 ) + ε ‖ u ‖ L t 2 ( H 3 ) 2 + t 1 2 ‖ ε u ‖ L t ∞ ( H 3 ) ‖ u ‖ L t 2 ( H 3 ) + ε ‖ ν B ‖ L t 2 ( H 3 ) 2 ≲ t 1 2 + ε C 1 ( M ( t ) ) .

The temporal derivatives of f , g , and h are expressed as follows:

(3.13) f t = κ 2 [ ∇ ρ t ( θ t + u ⋅ ∇ θ ) + ∇ ρ ( θ t t + u t ⋅ ∇ θ + u ⋅ ∇ θ t ) + ρ t ∇ u ⋅ ∇ θ + ρ ∇ u t ⋅ ∇ θ + ρ ∇ u ⋅ ∇ θ t − ∇ h t ] , g t = − σ t div u − σ div u t + ε 2 [ 2 μ ∣ D ( u ) ∣ 2 + λ ( div u ) 2 ] t + ν ε 4 π curl B ⋅ curl B t , h t = − ε σ t div u − ε σ div u t + ε 2 [ 2 μ ∣ D ( u ) ∣ 2 + λ ( div u ) 2 ] t + ν ε 2 2 π curl B ⋅ curl B t .

Note that, using (3.4) and (3.5), we have

(3.14) ‖ h t ‖ L t ∞ ( L 2 ) ≲ ‖ ε σ t ‖ L t ∞ ( H 1 ) ‖ div u ‖ L t ∞ ( H 1 ) + ‖ σ ‖ L t ∞ ( H 2 ) ‖ ε div u t ‖ L t ∞ ( L 2 ) + ‖ ε ∇ u ‖ L t ∞ ( H 2 ) ‖ ε u t ‖ L t ∞ ( H 1 ) + ‖ ε ν curl B ‖ L t ∞ ( H 2 ) ‖ ε curl B t ‖ L t ∞ ( L 2 ) ≲ C 1 ( M ( t ) ) .

Next, we give the estimates of ‖ ε 2 ∇ g t ‖ L t 2 ( L 2 ) and ‖ ε f t ‖ L t 2 ( L 2 ) from the temporal derivatives of f , g , and h , respectively. From (3.6), we obtain

(3.15) ‖ ε 2 ∇ g t ‖ L t 2 ( L 2 ) = ε 2 ∇ − σ t div u − σ div u t + ε 2 [ 2 μ ∣ D ( u ) ∣ 2 + λ ( div u ) 2 ] t + ν ε 4 π ∣ curl B ∣ curl B t ∣ L t 2 ( L 2 ) ≲ ε ‖ ε σ t ‖ L t ∞ ( H 1 ) ‖ div u ‖ L t 2 ( H 2 ) + ε ‖ σ ‖ L t ∞ ( H 2 ) ‖ ε div u t ‖ L t 2 ( H 1 ) + ε ‖ ε ∇ u t ‖ L t 2 ( H 1 ) ‖ ε ∇ u ‖ L t ∞ ( H 2 ) + ‖ ε ν curl B t ‖ L t 2 ( H 1 ) ‖ ε 2 ν curl B ‖ L t ∞ ( H 2 ) ≲ ε 1 − ς 2 C 1 ( M ( t ) ) ,

which gives that

(3.16) ‖ ε ∇ h t ‖ L t 2 ( L 2 ) ≲ ε 1 − ς 2 C 1 ( M ( t ) ) .

Furthermore, by applying the temporal derivative to (1.9) 4 , we have

(3.17) ‖ ε f t ‖ L t 2 ( L 2 ) = ε κ 2 [ ∇ ρ t ( θ t + u ⋅ ∇ θ ) + ∇ ρ ( θ t t + u t ⋅ ∇ θ + u ⋅ ∇ θ t ) + ρ t ∇ u ⋅ ∇ θ + ρ ∇ u t ⋅ ∇ θ + ρ ∇ u ⋅ ∇ θ t − ∇ h t ] L t 2 ( L 2 ) = ε κ 2 ∇ ρ t ( θ t + u ⋅ ∇ θ ) + ρ − 1 ∇ ρ κ 2 Δ θ t − div v t − ρ t ( θ t + u ⋅ ∇ θ ) + h t + ρ t ∇ u ⋅ ∇ θ + ρ ∇ u t ⋅ ∇ θ + ρ ∇ u ⋅ ∇ θ t − ∇ h t } ∥ L t 2 ( L 2 ) ≲ ε ‖ ρ t ‖ L t ∞ ( H 1 ) ( ‖ θ t ‖ L t 2 ( L ∞ ) + t 1 2 ‖ u ‖ L t ∞ ( H 2 ) ‖ ∇ θ ‖ L t ∞ ( H 2 ) ) + ‖ ∇ ρ ‖ L t ∞ ( H 1 ) ‖ ρ − 1 ‖ L x , t ∞ ( ‖ ε Δ θ t ‖ L t 2 ( L 3 ) + ‖ ε div v t ‖ L t 2 ( L 3 ) + ε ‖ h t ‖ L t 2 ( L 3 ) ) + ε ‖ ∇ ρ ‖ L t ∞ ( H 1 ) ‖ ρ − 1 ‖ L x , t ∞ ‖ ρ t ‖ L t ∞ ( H 1 ) t 1 2 ‖ θ t ‖ L t ∞ ( H 1 ) + t 1 2 ‖ u ‖ L t ∞ ( H 2 ) ‖ ∇ θ ‖ L t ∞ ( H 2 ) + ε ‖ ρ t ‖ L t ∞ ( H 1 ) ‖ ∇ u ‖ L t 2 ( H 2 ) ‖ ∇ θ ‖ L t ∞ ( L 3 ) + t 1 2 ‖ ρ ‖ L t ∞ ( H 2 ) ( ‖ ε ∇ u t ‖ L t ∞ ( L 2 ) ‖ ∇ θ ‖ L t ∞ ( H 2 ) + ‖ ∇ u ‖ L t ∞ ( H 1 ) ‖ ε ∇ θ t ‖ L t ∞ ( H 1 ) ) + ε ‖ ∇ h t ‖ L t 2 ( L 2 ) ≲ t 1 4 + ε 1 − ς 2 C 1 ( M ( t ) ) .

From (3.1) and (3.14), we have

(3.18) ‖ f t ‖ L t 2 L 6 5 = κ 2 [ ∇ ρ t ( θ t + u ⋅ ∇ θ ) + ∇ ρ ( θ t t + u t ⋅ ∇ θ + u ⋅ ∇ θ t ) + ρ t ∇ u ⋅ ∇ θ + ρ ∇ u t ⋅ ∇ θ + ρ ∇ u ⋅ ∇ θ t − ∇ h t ] ∥ L t 2 L 6 5 = κ 2 ∇ ρ t ( θ t + u ⋅ ∇ θ ) + ρ − 1 ∇ ρ κ 2 Δ θ t − div v t − ρ t ( θ t + u ⋅ ∇ θ ) + h t + ρ t ∇ u ⋅ ∇ θ + ρ ∇ u t ⋅ ∇ θ + ρ ∇ u ⋅ ∇ θ t − ∇ h t } ∥ L t 2 L 6 5 ≲ ‖ ρ t ‖ L t ∞ ( H 1 ) ( ‖ θ t ‖ L t 2 ( L 3 ) + t 1 2 ‖ u ‖ L t ∞ ( H 2 ) ‖ ∇ θ ‖ L t ∞ ( H 2 ) ) + ‖ ∇ ρ ‖ L t ∞ ( H 1 ) ‖ ρ − 1 ‖ L x , t ∞ ( ‖ Δ θ t ‖ L t 2 ( L 2 ) ) + ‖ div v t ‖ L t 2 ( L 2 ) + t 1 2 ‖ h t ‖ L t ∞ ( L 2 ) ) + t 1 2 ‖ ∇ ρ ‖ L t ∞ ( H 1 ) ‖ ρ − 1 ‖ L x , t ∞ ‖ ρ t ‖ L t ∞ ( H 1 ) ( ‖ θ t ‖ L t ∞ ( H 1 ) + ‖ u ‖ L t ∞ ( H 2 ) ‖ ∇ θ ‖ L t ∞ ( H 2 ) ) + t 1 2 ‖ ρ t ‖ L t ∞ ( H 1 ) ‖ ∇ u ‖ L t ∞ ( H 1 ) ‖ ∇ θ ‖ L t ∞ ( H 1 ) + ‖ ρ ‖ L t ∞ ( H 2 ) ( ‖ ∇ u t ‖ L t 2 ( L 2 ) ‖ ∇ θ ‖ L t ∞ ( H 2 ) + ‖ ∇ u ‖ L t ∞ ( H 1 ) ‖ ∇ θ t ‖ L t ∞ ( L 2 ) ) + t 1 2 ‖ ∇ h t ‖ L t ∞ ( L 2 ) ≲ t 1 4 C 1 ( M ( t ) ) .

3.3 L 2 -estimates for the solution ( v , ∇ θ , σ , B )

We derive first the L 2 -estimates for ( v , θ , ∇ θ , σ , B ) , which are the basis for obtaining the estimates of higher-order derivatives in view of Lemmas 2.3–2.4.

Lemma 3.2

Suppose that the assumptions in Theorem 1.2 are satisfied. For any 0 < ε ≤ 1 , 0 ≤ ν ≤ O ( ε ς ) with 0 < ς < 2 and 0 ≤ t ≤ min { T ε , ν , 1 } , the solution to system (1.9) satisfies the following estimate:

(3.19) ‖ ( v , θ , ∇ θ , σ , B ) ‖ L 2 2 ( t ) + ‖ ( div u , curl u , Δ θ , ν curl B ) ‖ L t 2 ( L 2 ) 2 ≤ C 0 ( M 0 ) exp { ( t 1 4 + ε ) C 1 ( M ( t ) ) } .

Proof

First, multiplying (1.9) 2 and (1.9) 5 with v and 1 4 π B , respectively, then summing up, and integrating by parts over Ω × ( 0 , t ) , we derive

(3.20) 1 2 ∫ Ω ( ρ ∣ v ∣ 2 + 1 4 π ∣ B ∣ 2 ) d x ( t ) + ∫ 0 t ∫ Ω μ ∣ curl v ∣ 2 + ( μ + ξ ) ∣ div v ∣ 2 + ν 4 π ∣ curl B ∣ 2 d x d s + ∫ 0 t ∫ Ω 1 ε ∇ σ ⋅ v d x d s = 1 2 ∫ Ω ρ 0 ∣ v 0 ∣ 2 + 1 4 π ∣ B 0 ∣ 2 d x + ∫ 0 t ∫ Ω β ∇ Δ θ ⋅ v d x d s + ∫ 0 t ∫ Ω 1 4 π ( curl B ) × B ⋅ v + curl ( u × B ) ⋅ 1 4 π B d x d s + ∫ 0 t ∫ Ω f ⋅ v d x d s ≔ 1 2 ∫ Ω ρ 0 ∣ v 0 ∣ 2 + 1 4 π ∣ B 0 ∣ 2 d x + U 1 + U 2 + U 3 ,

where U 1 , U 2 , and U 3 are estimated as follows:

(3.21) U 1 ≲ t 1 2 ‖ ∇ θ ‖ L t ∞ ( H 2 ) ‖ v ‖ L t 2 ( L 2 ) ≲ t 1 2 C 1 ( M ( t ) ) ,

(3.22) U 2 = ∫ 0 t ∫ Ω 1 4 π ( curl B ) × B ⋅ v + curl ( u × B ) ⋅ 1 4 π B d x d s = ∫ 0 t ∫ Ω 1 4 π ( B ⋅ ∇ ) B − 1 2 ∇ ( ∣ B ∣ 2 ) ⋅ v + B ⋅ [ ( B ⋅ ∇ ) u − ( u ⋅ ∇ ) B − B div u ] d x d s = ∫ 0 t ∫ Ω κ 8 π 1 2 ∇ ( ∣ B ∣ 2 ) ⋅ ∇ θ − ( B ⋅ ∇ ) B ⋅ ∇ θ d x d s ≲ t 1 2 ‖ B ‖ L t ∞ ( H 2 ) 2 ‖ ∇ θ ‖ L t 2 ( L 2 ) ≲ t 1 2 C 1 ( M ( t ) ) ,

by using (1.8) and (2.2), and

(3.23) U 3 = ∫ 0 t ∫ Ω f ⋅ v d x d s ≲ t 1 2 ‖ f ‖ L t ∞ ( L 2 ) ‖ v ‖ L t 2 ( L 2 ) ≲ t 1 2 C 1 ( M ( t ) ) ,

by using (3.10).

In the same way, multiplying (1.9) 3 with σ and integrating over Ω × ( 0 , t ) , we obtain

(3.24) 1 4 ∫ Ω ∣ σ ∣ 2 d x ( t ) + ∫ 0 t ∫ Ω 1 ε σ div v d x d s = 1 4 ∫ Ω ∣ σ 0 ∣ 2 d x + ∫ 0 t ∫ Ω 1 4 σ 2 ∇ u − σ 2 div u + ε 2 σ 2 μ ∣ D ( u ) ∣ 2 + λ ( d i v u ) 2 + ν 8 π ∣ curl B ∣ 2 d x d s ≲ C 0 ( M 0 ) + t 1 2 ‖ u ‖ L t 2 ( H 1 ) ‖ σ ‖ L t ∞ ( H 1 ) 2 + ε ( ‖ σ ‖ L t ∞ ( H 2 ) ‖ u ‖ L t 2 ( H 1 ) 2 + ‖ σ ‖ L t ∞ ( H 2 ) ‖ ν curl B ‖ L t 2 ( H 1 ) 2 ) ≲ C 0 ( M 0 ) exp t 1 2 + ε C 1 ( M ( t ) ) .

Next, multiplying (1.9) 4 with θ and integrating over Ω × ( 0 , t ) , we have that

(3.25) 1 2 ∫ Ω ρ ∣ θ ∣ 2 d x ( t ) + ∫ 0 t ∫ Ω κ 2 ∣ ∇ θ ∣ 2 d x d s = 1 2 ∫ Ω ρ 0 ∣ θ 0 ∣ 2 d x − ∫ 0 t ∫ Ω θ ( div v − h ) d x d s ≲ C 0 ( M 0 ) + t 1 2 ‖ div v ‖ L t 2 ( L 2 ) ‖ θ ‖ L t ∞ ( L 2 ) + ε t 1 2 ‖ σ ‖ L t ∞ ( H 2 ) ‖ div u ‖ L t 2 ( H 1 ) ‖ θ ‖ L t ∞ ( L 2 ) + ε 2 ( ‖ u ‖ L t 2 ( H 2 ) 2 ‖ θ ‖ L t ∞ ( L 2 ) + ‖ ν curl B ‖ L t 2 ( H 1 ) 2 ‖ θ ‖ L t ∞ ( L 2 ) ) ≲ C 0 ( M 0 ) exp ( t 1 2 + ε 2 ) C 1 ( M ( t ) ) .

Subsequently, we apply the gradient operator to (1.9) 4 once and multiply the resulting equality by ∇ θ . Then, integrating the results over Ω × ( 0 , t ) , we discover that

(3.26) 1 2 ∫ Ω ρ ∣ ∇ θ ∣ 2 d x ( t ) + ∫ 0 t ∫ Ω κ 2 ∣ Δ θ ∣ 2 d x d s ≲ C 0 ( M 0 ) + t 1 2 ‖ div v ‖ L t 2 ( L 2 ) ‖ θ ‖ L t ∞ ( H 2 ) + ‖ ∇ ρ ‖ L t ∞ ( H 1 ) ‖ ∇ θ ‖ L t ∞ ( H 1 ) ( t ‖ θ t ‖ L t ∞ ( H 1 ) + t 1 2 ‖ u ‖ L t 2 ( L 2 ) ‖ ∇ θ ‖ L t ∞ ( H 1 ) ) + t 1 2 ‖ ρ ‖ L t ∞ ( H 2 ) ‖ ∇ u ‖ L t 2 ( L 2 ) ‖ ∇ θ ‖ L t ∞ ( L 2 ) + ε ‖ σ ‖ L t 2 ( H 1 ) ‖ u ‖ L t 2 ( H 2 ) ‖ ∇ θ ‖ L t ∞ ( H 1 ) + ε 2 ‖ ∇ θ ‖ L t ∞ ( H 2 ) ( ‖ u ‖ L t 2 ( H 2 ) 2 + ‖ ν curl B ‖ L t 2 ( H 1 ) 2 ) ≲ C 0 ( M 0 ) exp ( t 1 2 + ε 2 ) C 1 ( M ( t ) ) .

Hence, (3.19) can be obtained by summing up (3.20)–(3.23), (3.24) and (3.26).□

3.4 Estimates of σ t , v t , B t , and their spatial derivatives

In order to estimate the temporal derivatives, we can obtain the following results by differential equation (1.9) with respect to t :

(3.27) ρ ( v t t + u ⋅ ∇ v t ) + 1 ε ∇ σ t − μ Δ v t − ξ ∇ div v t − β ∇ Δ θ t = − ρ t ( v t + u ⋅ ∇ v ) + ρ u t ⋅ ∇ v + 1 4 π ( curl B t ) × B + 1 4 π ( curl B ) × B t + f t , in Ω × ( 0 , T ] , 1 2 ( σ t t + u ⋅ ∇ σ t + u t ⋅ ∇ σ ) + 1 ε div v t = g t , in Ω × ( 0 , T ] , ρ ( θ t t + u ⋅ ∇ θ t ) − κ 2 Δ θ t + div v t = − ρ t ( θ t + u ⋅ ∇ θ ) − ρ u t ⋅ ∇ θ + h t , in Ω × ( 0 , T ] , B t t − curl ( u t × B + u × B t ) = − curl ( ν curl B t ) , div B t = 0 , in Ω × ( 0 , T ] , v t ⋅ n = 0 , n × curl v t = 0 , ∂ θ t ∂ n = 0 , B t ⋅ n = 0 , n × curl B t = 0 , on ∂ Ω × ( 0 , T ] , ( v t , σ t , θ t , B t ) ∣ t = 0 = ( v t ( 0 ) , σ t ( 0 ) , θ t ( 0 ) , B t ( 0 ) ) ( x ) , x ∈ Ω ,

where the constants ξ = μ + λ + κ 2 , β = κ 2 ( 2 μ + λ − κ 2 ) and the functions f t , g t , and h t are already expressed by (3.13). And the “initial data” in (3.27) 6 have already defined in Remark 1.1.

From (1.9), we see that the uniform boundedness of ∇ σ and div v can be obtained by establishing the weighted estimates of temporal derivatives of solutions σ and v . Therefore, we show the estimates of σ t , v t , and B t first.

Lemma 3.3

Suppose that the assumptions in Theorem 1.2 are satisfied. For any 0 < ε ≤ 1 , 0 ≤ ν ≤ O ( ε ς ) with 0 < ς < 2 and 0 ≤ t ≤ min { T ε , ν , 1 } , we have

(3.28) ‖ ( ρ v t , B t , σ t ) ‖ L 2 2 ( t ) + ‖ ( v t , ν B t ) ‖ L t 2 ( H 1 ) 2 ≲ C 0 ( M 0 ) exp t 1 4 C 1 ( M ( t ) ) − β ∫ 0 t ∫ Ω div v t Δ θ t d x d s .

Proof

First, multiply (3.27) 1 , (3.27) 2 , and (3.27) 4 by v t , σ t and 1 4 π B t , respectively, then add up the results, and integrate over Ω × ( 0 , t ) . Thus, we can obtain the following result by integration by parts:

(3.29) 1 2 ∫ Ω ρ ∣ v t ∣ 2 + 1 2 σ t 2 + 1 4 π ∣ B t ∣ 2 d x ( t ) + ∫ 0 t ∫ Ω μ ∣ curl v t ∣ 2 + ( μ + ξ ) ( div v t ) 2 + ν 4 π ∣ curl B t ∣ 2 d x d s = 1 2 ∫ Ω ( ρ 0 ∣ v t ( 0 ) ∣ 2 + 1 2 σ t 2 ( 0 ) + 1 4 π ∣ B t ( 0 ) ∣ 2 ) d x + ∫ 0 t ∫ Ω β ∇ Δ θ t ⋅ v t d x d s + H 1 + H 2 + H 3 + H 4 ,

where the terms H 1 through H 4 are estimated as follows:

(3.30) H 1 = − ∫ 0 t ∫ Ω [ ρ t ( u ⋅ ∇ v ) v t + ρ ( u ⋅ ∇ v ) t v t − 1 2 ρ t v t 2 ] d x d s ≲ t 1 2 ‖ v t ‖ L t ∞ ( L 2 ) [ ‖ ρ t ‖ L t ∞ ( L 2 ) ‖ u ‖ L t ∞ ( H 2 ) ‖ ∇ v ‖ L t 2 ( H 2 ) + ‖ ρ ‖ L t ∞ ( H 2 ) ‖ u ‖ L t ∞ ( H 2 ) ‖ ∇ v t ‖ L t 2 ( L 2 ) + ‖ ρ ‖ L t ∞ ( H 2 ) ‖ u t ‖ L t 2 ( H 1 ) ‖ ∇ v ‖ L t ∞ ( H 1 ) + ‖ ρ t ‖ L t ∞ ( H 1 ) ‖ v t ‖ L t 2 ( H 1 ) ] ≲ t 1 2 C 1 ( M ( t ) ) ,

(3.31) H 2 = ∫ 0 t ∫ Ω 1 4 π { ( curl B × B ) t ⋅ v t + ( curl u × B ) t ⋅ B t } d x d s = ∫ 0 t ∫ Ω 1 4 π ( B ⋅ ∇ ) B − 1 2 ∇ ( ∣ B ∣ 2 ) t ⋅ v t + B t ⋅ [ ( B ⋅ ∇ ) u − ( u ⋅ ∇ ) B − B div u ] t d x d s = ∫ 0 t ∫ Ω 1 4 π ( B t ⋅ ∇ ) B ⋅ v t + κ 2 ( B ⋅ ∇ ) ∇ θ t ⋅ B t + B t ⋅ [ ( B t ⋅ ∇ ) u − ( u t ⋅ ∇ ) B ] − 1 2 ( div u ) ∣ B t ∣ 2 + κ 4 ( ∣ B ∣ 2 ) t ( div ∇ θ t ) d x d s ≲ t 1 2 ‖ B t ‖ L t ∞ ( L 2 ) ‖ B ‖ L t ∞ ( H 2 ) ( ‖ v t ‖ L t 2 ( H 1 ) + ‖ ∇ θ t ‖ L t 2 ( H 1 ) ) + t 1 2 ‖ u ‖ L t 2 ( H 3 ) ‖ B t ‖ L t ∞ ( L 2 ) 2 ≲ t 1 2 C 1 ( M ( t ) ) ,

(3.32) H 3 = ∫ 0 t ∫ Ω − 1 2 ( u t ⋅ ∇ σ σ t + u ⋅ ∇ σ t σ t ) − σ t 2 div u − σ σ t div u t + ε 2 [ 2 μ ∣ D ( u ) ∣ 2 + λ ( div u ) 2 ] t σ t + ν ε 4 π σ t curl B ⋅ curl B t d x d s ≲ t 1 2 ‖ u t ‖ L t 2 ( H 1 ) ∥ ∇ σ ∥ L t ∞ ( H 1 ) ‖ σ t ‖ L t ∞ ( L 2 ) + t 1 2 ‖ u ‖ L t ∞ ( H 2 ) ‖ ∇ σ t ‖ L t 2 ( L 2 ) ‖ σ t ‖ L t ∞ ( L 2 ) + t 1 2 ‖ σ t ‖ L t ∞ ( L 2 ) 2 ‖ div u ‖ L t 2 ( H 2 ) + t 1 2 ‖ σ ‖ L t ∞ ( H 2 ) ‖ σ t ‖ L t ∞ ( L 2 ) ‖ div u t ‖ L t 2 ( L 2 ) + t 1 2 ‖ σ t ‖ L t ∞ ( L 2 ) ‖ ∇ u ‖ L t ∞ ( H 1 ) ‖ ε ∇ u t ‖ L t 2 ( H 1 ) + ν t ‖ σ t ‖ L t ∞ ( L 2 ) ‖ ν curl B ‖ L t 2 ( H 2 ) ‖ ε curl B t ‖ L t ∞ ( L 2 ) ≲ t 1 2 C 1 ( M ( t ) ) ,

and by using (3.18),

(3.33) H 4 = ∫ 0 t ∫ Ω f t ⋅ v t d x d s ≲ ‖ f t ‖ L t 2 L 6 5 ‖ v t ‖ L t 2 ( L 6 ) ≲ t 1 4 C 1 ( M ( t ) ) .

In addition, applying Lemma 2.4 to (3.29), we obtain

□ ‖ ν B t ‖ L t 2 ( H 1 ) 2 ≤ C ( ‖ ν curl B t ‖ L t 2 ( L 2 ) 2 + ‖ ν B t ‖ L t 2 ( L 2 ) 2 ) ≤ C 0 ( M 0 ( t ) ) exp t 1 2 C 1 ( M ( t ) ) − β ∫ 0 t ∫ Ω div v t Δ θ t d x d s .

Next, we show the L 2 -estimates of ε ∇ σ t and ε div v t .

Lemma 3.4

Suppose that the assumptions in Theorem 1.2 are satisfied. For any 0 < ε ≤ 1 , 0 ≤ ν ≤ O ( ε ς ) with 0 < ς < 2 and 0 ≤ t ≤ min { T ε , ν , 1 } , we have

(3.34) ‖ ( ε ρ − 1 ∇ σ t , ε div v t ) ‖ L 2 2 ( t ) + ‖ ε div v t ‖ L t 2 ( H 1 ) 2 ≲ C 0 ( M 0 ) exp t 1 4 + ε 1 − ς 2 C 1 ( M ( t ) ) − β ε 2 ∫ 0 t ∫ Ω ρ − 1 ∇ div v t ⋅ Δ ∇ θ t d x d s .

Proof

From (1.9) 1 and (1.1) 1 we derive

1 2 ∫ Ω ( ∇ σ t t + u ⋅ ∇ ( ∇ σ t ) ) ⋅ ε 2 ρ − 1 ∇ σ t d x = d d t ∫ Ω ε 2 4 ρ − 1 ∣ ∇ σ t ∣ 2 d x − ∫ Ω ε 2 4 ( ρ − 1 ) t ∣ ∇ σ t ∣ 2 d x + ∫ Ω ε 2 4 ρ − 1 u ⋅ ∇ ( ∣ ∇ σ t ∣ 2 ) d x = d d t ∫ Ω ε 2 4 ρ − 1 ∣ ∇ σ t ∣ 2 d x + ∫ Ω ε 2 4 ρ − 2 ρ t ∣ ∇ σ t ∣ 2 d x − ∫ Ω ε 2 4 ρ − 1 div u ∣ ∇ σ t ∣ 2 d x − ∫ Ω ε 2 4 ∇ ( ρ − 1 ) ⋅ u ∣ ∇ σ t ∣ 2 d x = d d t ∫ Ω ε 2 4 ρ − 1 ∣ ∇ σ t ∣ 2 d x − ∫ Ω ε 2 2 ρ − 1 div u ∣ ∇ σ t ∣ 2 d x .

Multiply (3.27) 1 and ∇ (3.27) 2 by − ε 2 ρ − 1 ∇ div v t and ε 2 ρ − 1 ∇ σ t , respectively. Summing the results and integrating over Ω × ( 0 , t ) , we obtain

(3.35) 1 2 ∫ Ω ∣ ε div v t ∣ 2 + 1 2 ρ − 1 ∣ ε ∇ σ t ∣ 2 d x ( t ) + ( μ + ξ ) ∫ 0 t ∫ Ω ρ − 1 ∣ ε ∇ div v t ∣ 2 d x d s = 1 2 ∫ Ω ∣ ε div v t ( 0 ) ∣ 2 + 1 2 ρ 0 − 1 ∣ ε ∇ σ t ( 0 ) ∣ 2 d x − β ∫ 0 t ∫ Ω ρ − 1 ε 2 ∇ Δ θ t ⋅ ∇ div v t d x d s + J 1 + J 2 + J 3 + J 4 ,

where

J 1 = μ ∫ 0 t ∫ Ω ρ − 1 ε 2 curlcurl v t ⋅ ∇ div v t d x d s = μ ε 2 ∫ 0 t ∫ Ω − ρ − 2 ∇ ρ × ∇ div v t ⋅ curl v t d x d s ≲ ‖ ρ − 1 ‖ L x , t ∞ 2 ‖ ∇ ρ ‖ L t ∞ ( H 1 ) ‖ ε ∇ div v t ‖ L t 2 ( L 2 ) ‖ ε curl v t ‖ L t 2 ( L 3 ) ≲ t 1 4 C 1 ( M ( t ) ) ,

J 2 = ∫ 0 t ∫ Ω ρ − 1 ε 2 ∇ div v t ⋅ [ − f t + ρ t ( v t + u ⋅ ∇ v ) + ρ ( u ⋅ ∇ v ) t ] d x d s ≲ ‖ ε ∇ div v t ‖ L t 2 ( L 2 ) ‖ ρ − 1 ‖ L x , t ∞ ‖ ε f t ‖ L t 2 ( L 2 ) + t 1 2 ‖ ρ t ‖ L t ∞ ( H 1 ) ( ‖ ε v t ‖ L t ∞ ( H 1 ) + ‖ u ‖ L t ∞ ( H 2 ) ‖ ∇ v ‖ L t ∞ ( H 1 ) ) + t 1 2 ( ‖ ε u t ‖ L t ∞ ( H 1 ) ‖ ∇ v ‖ L t ∞ ( H 1 ) + ‖ u ‖ L t ∞ ( H 2 ) ‖ ε ∇ v t ‖ L t ∞ ( L 2 ) ) ≲ t 1 4 + ε 1 − ς 2 C 1 ( M ( t ) ) ,

J 3 = 1 4 π ∫ 0 t ∫ Ω − ρ − 1 ε 2 ∇ div v t ⋅ ( curl B × B ) t d x d s ≲ t 1 2 ‖ ρ − 1 ‖ L x , t ∞ ‖ ε ∇ div v t ‖ L t 2 ( L 2 ) ‖ ε B t ‖ L t ∞ ( H 1 ) ‖ B ‖ L t ∞ ( H 2 ) ≲ t 1 2 C 1 ( M ( t ) )

and

J 4 = ∫ 0 t ∫ Ω ρ − 1 ε 2 ∇ σ t ⋅ 1 2 ∇ σ t div u + σ t div u − 1 2 ∇ ( ∇ σ ⋅ u t ) + ∇ g t d x d s ≲ t 1 2 ‖ ρ − 1 ‖ L x , t ∞ ‖ ε σ t ‖ L t ∞ ( H 1 ) [ ‖ ε ∇ σ t ‖ L t ∞ ( L 2 ) ‖ div u ‖ L t 2 ( H 2 ) + ‖ σ ‖ L t ∞ ( H 2 ) ‖ ε u t ‖ L t 2 ( H 2 ) ] + ‖ ρ − 1 ‖ L x , t ∞ ‖ σ t ‖ L t 2 ( H 1 ) ‖ ε 2 ∇ g t ‖ L t 2 ( L 2 ) ≲ t 1 4 + ε 1 − ς 2 C 1 ( M ( t ) ) .

Note that the estimates (3.15) and (3.16) of ‖ ε f t ‖ L t 2 ( L 2 ) and ‖ ε 2 ∇ g t ‖ L t 2 ( L 2 ) were used in J 2 and J 4 , respectively.□

In what follows, we will derive the estimates of ‖ ∇ σ t ‖ L t 2 ( L 2 ) and ‖ Δ θ t ‖ L t 2 ( L 2 ) , which give the estimates of ‖ ∇ div v ‖ L ∞ ( L 2 ( Ω ) ) and ‖ ∇ Δ θ ‖ L ∞ ( L 2 ( Ω ) ) simultaneously.

Lemma 3.5

Suppose that the assumptions in Theorem 1.2 are satisfied. For any 0 < ε ≤ 1 , 0 ≤ ν ≤ O ( ε ς ) with 0 < ς < 2 and 0 ≤ t ≤ min { T ε , ν , 1 } , we have

(3.36) ‖ ρ − 1 ∇ div v ‖ L 2 2 ( t ) + ‖ ρ − 1 ∇ Δ θ ‖ L 2 2 ( t ) + ‖ ∇ σ t ‖ L t 2 ( L 2 ) 2 + ‖ Δ θ t ‖ L t 2 ( L 2 ) 2 ≲ C 0 ( M 0 ) exp t 1 4 + ε 2 C 1 ( M ( t ) ) − β ∫ 0 t ∫ Ω div v t Δ θ t d x d s .

Proof

First, we apply the gradient operator ∇ to (1.9) 3 . Multiplying the result by ρ − 1 ∇ σ t and then integrating over Ω × ( 0 , t ) , we obtain

(3.37) 1 2 ∫ 0 t ∫ Ω ρ − 1 ∣ ∇ σ t ∣ 2 d x d s + 1 ε ∫ 0 t ∫ Ω ρ − 1 ∇ div v ⋅ ∇ σ t d x d s = A 1 + A 2 ,

where

A 1 = − 1 2 ∫ 0 t ∫ Ω ( ρ − 1 ∇ σ t ⋅ ∇ u ⋅ ∇ σ + ρ − 1 ∇ σ t ⋅ u ⋅ ∇ 2 σ ) d x d s ≲ δ ∫ 0 t ∫ Ω ρ − 1 ∣ ∇ σ t ∣ 2 d x d s + C ( δ ) ∫ 0 t ∫ Ω ρ − 1 ( ∣ ∇ u ⋅ ∇ σ ∣ 2 + ∣ u ⋅ ∇ 2 σ ∣ 2 ) d x d s ≲ δ ∫ 0 t ∫ Ω ρ − 1 ∣ ∇ σ t ∣ 2 d x d s + C ( δ ) t 1 2 ‖ ρ − 1 ‖ L x , t ∞ [ ‖ ∇ u ‖ L t ∞ ( H 1 ) ‖ ∇ u ‖ L t 2 ( H 2 ) ‖ ∇ σ ‖ L t ∞ ( H 1 ) 2 + ‖ u ‖ L t 2 ( H 3 ) ‖ u ‖ L t ∞ ( H 2 ) ‖ ∇ 2 σ ‖ L t ∞ ( L 2 ) 2 ] ≲ δ ∫ 0 t ∫ Ω ρ − 1 ∣ ∇ σ t ∣ 2 d x d s + C ( δ ) t 1 2 C 1 ( M ( t ) ) ,

and

A 2 = ∫ 0 t ∫ Ω ρ − 1 ∇ σ t ⋅ { − ∇ ( σ div u ) + ε 2 ∇ [ 2 μ ∣ D ( u ) ∣ 2 + λ ( div u ) 2 ] + ν ε 4 π curl B ⋅ ∇ curl B } d x d s ≲ δ ∫ 0 t ∫ Ω ρ − 1 ∣ ∇ σ t ∣ 2 d x d s + C ( δ ) ∫ 0 t ∫ Ω ρ − 1 ∣ ∇ ( σ div u ) ∣ 2 + ε 2 ∇ [ 2 μ ∣ D ( u ) ∣ 2 + λ ( div u ) 2 ] 2 + ν ε 4 π curl B ⋅ ∇ curl B 2 d x d s ≲ δ ∫ 0 t ∫ Ω ρ − 1 ∣ ∇ σ t ∣ 2 d x d s + C ( δ ) t 1 2 ‖ ρ − 1 ‖ L x , t ∞ [ ‖ ∇ σ ‖ L t ∞ ( H 1 ) 2 ‖ div u ‖ L t 2 ( H 2 ) ‖ div u ‖ L t ∞ ( H 1 ) + ‖ σ ‖ L t ∞ ( H 2 ) 2 ‖ ∇ div u ‖ L t 2 ( H 1 ) ‖ ∇ div u ‖ L t ∞ ( L 2 ) ] + C ( δ ) ε 2 ‖ ρ − 1 ‖ L x , t ∞ [ ‖ u ‖ L t 2 ( H 3 ) 2 ‖ u ‖ L t ∞ ( H 2 ) 2 + ‖ ν B ‖ L t 2 ( H 3 ) 2 ‖ B ‖ L t ∞ ( H 2 ) 2 ] ≲ δ ∫ 0 t ∫ Ω ρ − 1 ∣ ∇ σ t ∣ 2 d x d s + C ( δ ) t 1 2 + ε 2 C 1 ( M ( t ) ) .

Hence, we obtain that

(3.38) ∫ 0 t ∫ Ω ρ − 1 ∣ ∇ σ t ∣ 2 d x d s + 1 ε ∫ 0 t ∫ Ω ρ − 1 ∇ div v ⋅ ∇ σ t d x d s ≲ ( t 1 2 + ε 2 ) C 1 ( M ( t ) ) .

Next, multiplying (3.27) 1 by − ρ − 1 ∇ div v and then integrating over Ω × ( 0 , t ) , we can similarly show that

(3.39) μ + ξ 2 ∫ Ω ρ − 1 ∣ ∇ div v ∣ 2 d x ( t ) − 1 ε ∫ 0 t ∫ Ω ρ − 1 ∇ div v ⋅ ∇ σ t d x d s + β ∫ 0 t ∫ Ω ρ − 1 ∇ div v ⋅ ∇ Δ θ t d x d s = μ + ξ 2 ∫ Ω ρ 0 − 1 ∣ ∇ div v 0 ∣ 2 d x + C 1 + C 2 + C 3 + C 4 + C 5 ,

where

C 1 = μ + ξ 2 ∫ 0 t ∫ Ω ( ρ − 1 ) t ∣ ∇ div v ∣ 2 d x d s + μ ∫ 0 t ∫ Ω ρ − 1 ∇ div v ⋅ curlcurl v t d x d s , C 2 = ∫ 0 t ∫ Ω ρ − 1 ∇ div v ⋅ ( ρ v t ) t d x d s , C 3 = ∫ 0 t ∫ Ω ρ − 1 ∇ div v ⋅ [ ρ t u ⋅ ∇ v + ρ ( u ⋅ ∇ v ) t ] d x d s , C 4 = ∫ 0 t ∫ Ω ρ − 1 ∇ div v ⋅ 1 4 π ( curl B × B ) t d x d s , C 5 = ∫ 0 t ∫ Ω − ρ − 1 ∇ div v ⋅ f t d x d s .

Note that

(3.40) ‖ ∇ div v ‖ L t 2 ( L 3 ) ≲ t 1 4 ‖ ∇ div v ‖ L t ∞ ( L 2 ) 1 2 ‖ ∇ div v ‖ L t 2 ( H 1 ) 1 2 ≲ t 1 4 C 1 ( M ( t ) ) .

To show the estimate of ‖ ∇ div v ‖ L t ∞ ( L 2 ) , we have to estimate the reminders C i , i = 1 … 5 . In view of integrating by parts and the boundary condition curl v t × n = 0 , we obtain

C 1 = μ + ξ 2 ∫ 0 t ∫ Ω ( ρ − 1 ) t ∣ ∇ div v ∣ 2 d x d s − μ ∫ 0 t ∫ Ω ρ − 2 ∇ ρ × ∇ div v ⋅ curl v t d x d s ≲ ‖ ρ − 1 ‖ L x , t ∞ 2 ‖ ρ t ‖ L t ∞ ( H 1 ) ‖ ∇ div v ‖ L t 2 ( L 3 ) 2 + ‖ curl v t ‖ L t 2 ( L 2 ) ‖ ρ − 1 ‖ L x , t ∞ 2 ‖ ∇ ρ ‖ L t ∞ ( H 1 ) ‖ ∇ div v ‖ L t 2 ( L 3 ) ≲ t 1 4 C 1 ( M ( t ) ) .

By applying (3.28), one obtains

C 2 = − ∫ 0 t ∫ Ω ( ρ − 1 ∇ div v ) t ⋅ ρ v t d x d s + ∫ Ω ρ − 1 ∇ div v ⋅ ρ v t d x 0 t ≲ t 1 2 ‖ ρ − 1 ‖ L x , t ∞ ‖ ρ t ‖ L t ∞ ( H 1 ) ‖ ∇ div v ‖ L t 2 ( H 1 ) ‖ v t ‖ L t ∞ ( L 2 ) + ∫ 0 t ∫ Ω ∣ div v t ∣ 2 d x d s + C 0 ( M 0 ) + δ ∫ Ω ρ − 1 ∣ ∇ div v ∣ 2 d x + C ( δ ) ∫ Ω ρ ∣ v t ∣ 2 d x ≲ C 0 ( M 0 ) + t 1 4 C 1 ( M ( t ) ) − β ∫ 0 t ∫ Ω div v t Δ θ t d x d s + δ ∫ Ω ρ − 1 ∣ ∇ div v ∣ 2 d x + C ( δ ) ∫ Ω ρ ∣ v t ∣ 2 d x .

Similarly,

C 3 ≲ ‖ ρ − 1 ‖ L x , t ∞ t 1 2 ‖ ρ t ‖ L t ∞ ( H 1 ) ‖ ∇ div v ‖ L t 2 ( H 1 ) ‖ u ‖ L t ∞ ( H 2 ) ‖ ∇ v ‖ L t ∞ ( H 1 ) + ‖ ∇ div v ‖ L t 2 ( L 3 ) ( ‖ u t ‖ L t ∞ ( L 2 ) ‖ v ‖ L t 2 ( H 2 ) + ‖ u ‖ L t ∞ ( H 2 ) ‖ v t ‖ L t 2 ( H 1 ) ) ≲ t 1 4 C 1 ( M ( t ) ) .

Using (3.40), we have

C 4 = ∫ 0 t ∫ Ω ρ − 1 ∇ div v ⋅ [ ( B t ⋅ ∇ ) B − ∇ B ⋅ B t + ( B ⋅ ∇ ) B t − ∇ B t ⋅ B ] d x d s ≔ ℳ 1 ˜ + ℳ 2 ˜ + ℳ 3 ˜ ,

where

ℳ 1 ˜ = ∫ 0 t ∫ Ω ρ − 1 ∇ div v ⋅ [ ( B t ⋅ ∇ ) B − ∇ B ⋅ B t ] d x d s ≲ t 1 2 ‖ ρ − 1 ‖ L x , t ∞ ‖ ∇ div v ‖ L t 2 ( H 1 ) ‖ B t ‖ L t ∞ ( L 2 ) ‖ B ‖ L t ∞ ( H 2 ) ≲ t 1 2 C 1 ( M ( t ) ) ,

ℳ 2 ˜ = ∫ 0 t ∫ Ω ρ − 1 ∇ div v ⋅ ( B ⋅ ∇ ) B t d x d s = ∫ 0 t ∫ Ω − [ ∇ ( ρ − 1 ) ∇ div v + ρ − 1 div ∇ div v ] ⋅ B ⋅ B t d x d s ≲ t 1 2 ( ‖ ρ − 1 ‖ L x , t ∞ 2 ‖ ∇ ρ ‖ L t ∞ ( H 1 ) + ‖ ρ − 1 ‖ L x , t ∞ ) ‖ ∇ div v ‖ L t 2 ( H 1 ) ‖ B t ‖ L t ∞ ( L 2 ) ‖ B ‖ L t ∞ ( H 2 ) ≲ t 1 2 C 1 ( M ( t ) ) ,

and by using (1.9) 5 ,

ℳ 3 ˜ = − ∫ 0 t ∫ Ω ρ − 1 ∇ div v ⋅ ∇ B t ⋅ B d x d s = − ∫ 0 t ∫ Ω ρ − 1 ∇ div v ⋅ B ⋅ ( ∇ curl ( u × B ) − ν ∇ curl curl B ) d x d s ≲ t 1 2 ‖ ρ − 1 ‖ L x , t ∞ ( ‖ ∇ div v ‖ L t 2 ( H 1 ) ‖ B ‖ L t ∞ ( H 2 ) 2 ‖ u ‖ L t ∞ ( H 2 ) + ν ‖ ∇ div v ‖ L t ∞ ( L 2 ) ‖ ν B ‖ L t 2 ( H 3 ) ‖ B ‖ L t ∞ ( H 2 ) ) ≲ t 1 2 C 1 ( M ( t ) ) ,

and

C 5 = ∫ 0 t ∫ Ω ρ − 1 ∇ div v ⋅ κ 2 ∇ ρ t ( θ t + u ⋅ ∇ θ ) + ∇ ρ ( ρ − 1 ( h + κ 2 Δ θ − div v ) ) t + ( ρ ∇ u ⋅ ∇ θ − ∇ h ) t d x d s = ℳ 1 + ℳ 2 + ℳ 3 ,

where, with the help of (3.7), (3.8), and (3.14),

ℳ 1 = ∫ 0 t ∫ Ω ρ − 1 ∇ div v ⋅ κ 2 ∇ ρ t ( θ t + u ⋅ ∇ θ ) + ∇ ρ ρ − 1 h + κ 2 Δ θ − div v t d x d s ≲ ‖ ρ − 1 ‖ L x , t ∞ ‖ ∇ div v ‖ L t 2 ( L 3 ) [ ‖ ∇ ρ t ‖ L t ∞ ( L 2 ) ( ‖ θ t ‖ L t ∞ ( H 1 ) + ‖ u ‖ L t 2 ( H 2 ) ‖ ∇ θ ‖ L t ∞ ( H 2 ) ) + ‖ ∇ ρ ‖ L t ∞ ( H 1 ) ‖ ρ − 1 ‖ L x , t ∞ 2 ‖ ρ t ‖ L t ∞ ( H 1 ) ( ‖ h ‖ L t 2 ( H 2 ) + ‖ κ Δ θ ‖ L t 2 ( H 2 ) + ‖ div v ‖ L t 2 ( H 2 ) ) + ‖ ∇ ρ ‖ L t ∞ ( H 1 ) ‖ ρ − 1 ‖ L x , t ∞ ( ‖ h t ‖ L t 2 ( L 2 ) + ‖ κ Δ θ t ‖ L t 2 ( L 2 ) + ‖ div v t ‖ L t 2 ( L 2 ) ) ] ≲ t 1 4 C 1 ( M ( t ) ) ,

ℳ 2 = ∫ 0 t ∫ Ω ρ − 1 ∇ div v ⋅ κ 2 ( ρ ∇ u ⋅ ∇ θ ) t d x d s ≲ ‖ ρ − 1 ‖ L x , t ∞ ‖ ∇ div v ‖ L t 2 ( L 3 ) ( ‖ ρ t ‖ L t ∞ ( H 1 ) ‖ u ‖ L t 2 ( H 3 ) ‖ ∇ θ ‖ L t ∞ ( H 2 ) + ‖ ρ ‖ L t ∞ ( H 2 ) ‖ ∇ u t ‖ L t 2 ( L 2 ) ‖ ∇ θ ‖ L t ∞ ( H 2 ) + ‖ ρ ‖ L t ∞ ( H 2 ) ‖ ∇ u ‖ L t 2 ( H 2 ) ‖ ∇ θ t ‖ L t ∞ ( L 2 ) ) ≲ t 1 4 C 1 ( M ( t ) )

and

ℳ 3 = − ∫ 0 t ∫ Ω ρ − 1 ∇ div v ⋅ κ 2 ∇ h t d x d s = − ∫ Ω ρ − 1 ∇ div v ⋅ κ 2 ∇ h d x ∣ 0 t + ∫ 0 t ∫ Ω ( ρ − 1 ∇ div v ) t ⋅ κ 2 ∇ h d x d s ≲ C 0 ( M 0 ) + ‖ ρ − 1 ‖ L x , t ∞ ‖ ∇ div v ‖ L t ∞ ( L 2 ) ‖ ∇ h ‖ L t ∞ ( L 2 ) + t 1 2 ( ‖ ρ − 1 ‖ L x , t ∞ 2 ‖ ρ t ‖ L t ∞ ( H 1 ) ‖ ∇ div v ‖ L t ∞ ( L 2 ) ‖ ∇ h ‖ L t 2 ( H 1 ) + ‖ ρ − 1 ‖ L x , t ∞ ‖ ε ∇ div v t ‖ L t 2 ( L 2 ) 1 ε ∇ h L t ∞ ( L 2 ) ≲ C 0 ( M 0 ) exp t 1 2 + ε C 1 ( M ( t ) ) .

Hence, inserting the estimates C 1 – C 5 into (3.38) and choosing δ small enough, we obtain that

(3.41) ∫ Ω ρ − 1 ∣ ∇ div v ∣ 2 d x ( t ) − 1 ε ∫ 0 t ∫ Ω ρ − 1 ∇ div v ⋅ ∇ σ t d x d s + β ∫ 0 t ∫ Ω ρ − 1 ∇ div v ⋅ ∇ Δ θ t d x d s ≲ C 0 ( M 0 ) exp { t 1 4 C 1 ( M ( t ) ) } − β ∫ 0 t ∫ Ω div v t Δ θ t d x d s .

Finally, applying ∇ to (1.9) 4 and multiplying the result by − ρ − 1 κ ∇ Δ θ t , and then integrating over Ω × ( 0 , t ) , we derive that

(3.42) κ 2 4 ∫ Ω ρ − 1 ∣ ∇ Δ θ ∣ 2 d x ( t ) − κ ∫ 0 t ∫ Ω ρ − 1 ∇ div v ⋅ ∇ Δ θ t d x d s + ∫ 0 t ∫ Ω ∣ κ Δ θ t ∣ 2 d x d s = κ 2 4 ∫ Ω ρ 0 − 1 ∣ ∇ Δ θ 0 ∣ 2 d x + D 1 + D 2 ,

where

D 1 = ∫ 0 t ∫ Ω ( ρ − 1 ∇ h ⋅ κ ∇ Δ θ t − ρ − 2 ρ t ∣ κ ∇ Δ θ ∣ 2 ) d x d s ≲ t 1 2 ‖ ρ − 1 ‖ L x , t ∞ 1 ε ∇ h L t ∞ ( L 2 ) ‖ ε ∇ Δ θ t ‖ L t 2 ( L 2 ) + ‖ ρ − 1 ‖ L x , t ∞ 2 ‖ ρ t ‖ L t ∞ ( H 1 ) ‖ ∇ Δ θ ‖ L t 2 ( L 3 ) 2 ≲ C 0 ( M 0 ) exp t 1 2 C 1 ( M ( t ) ) ,

and

D 2 = ∫ Ω [ ρ − 1 ∇ ρ ( θ t + u ⋅ ∇ θ ) + ∇ ( u ⋅ ∇ θ ) ] ︸ ℰ ⋅ κ ∇ Δ θ d x 0 t − ∫ 0 t ∫ Ω [ ρ − 1 ∇ ρ ( θ t + u ⋅ ∇ θ ) + ∇ ( u ⋅ ∇ θ ) ] t ⋅ κ ∇ Δ θ d x d s ≲ C 0 ( M 0 ) + t 1 2 ‖ ℰ ‖ L t ∞ ( L 2 ) ‖ ∇ θ ‖ L t 2 ( H 2 ) + ‖ ℰ t ‖ L t 2 ( L 3 2 ) ‖ ∇ Δ θ ‖ L t 2 ( L 3 ) ≲ C 0 ( M 0 ) exp t 1 4 C 1 ( M ( t ) ) ,

since

‖ ∇ Δ θ ‖ L t 2 ( L 3 ) ≲ t 1 4 ‖ ∇ Δ θ ‖ L t 2 ( L 2 ) 1 2 ‖ ∇ Δ θ ‖ L t 2 ( H 1 ) 1 2 ≲ t 1 4 M ( t ) .

Note that, to estimate the term D 2 , we have to derive ‖ ℰ ‖ L t ∞ ( L 2 ) and ‖ ℰ t ‖ L t 2 ( L 3 2 ) . And using (3.1) and (3.14), we obtain

‖ ℰ t ‖ L t 2 ( L 3 2 ) ≲ ( ‖ ρ − 1 ‖ L t , x ∞ 2 ‖ ρ t ‖ L t ∞ ( H 1 ) ‖ ∇ ρ ‖ L t ∞ ( H 1 ) + ‖ ρ − 1 ‖ L t , x ∞ ‖ ∇ ρ t ‖ L t ∞ ( L 2 ) ) ⋅ × ( ‖ θ t ‖ L t 2 ( H 2 ) + ‖ u ‖ L t ∞ ( H 2 ) ‖ ∇ θ ‖ L t 2 ( H 2 ) ) + ‖ ρ − 1 ‖ L t , x ∞ 2 ‖ ∇ ρ t ‖ L t ∞ ( L 2 ) [ ‖ h t ‖ L t 2 ( L 2 ) + ‖ Δ θ t ‖ L t 2 ( L 2 ) + ‖ div v t ‖ L t 2 ( L 2 ) ] + ‖ u t ‖ L t 2 ( H 1 ) ‖ ∇ θ ‖ L t ∞ ( H 2 ) + + ‖ u ‖ L t ∞ ( H 2 ) ‖ ∇ θ t ‖ L t 2 ( H 1 ) ≲ C 1 ( M ( t ) )

and

‖ ℰ ‖ L t ∞ ( L 2 ) ≲ ‖ ρ − 1 ‖ L t , x ∞ ‖ ∇ ρ ‖ L t ∞ ( H 1 ) ( ‖ θ t ‖ L t ∞ ( H 1 ) + ‖ u ‖ L t ∞ ( H 2 ) ‖ ∇ θ ‖ L t ∞ ( H 1 ) ) + ‖ u ‖ L t ∞ ( H 2 ) ‖ ∇ θ ‖ L t ∞ ( H 1 ) ≲ C 1 ( M ( t ) ) .

Hence, we can obtain (3.36) from (3.38), (3.41), and (3.42).□

3.5 Curl estimates of u , v , and B

Owing to the equation v = u − κ 2 ∇ θ , we obtain that curl v = curl u . Then, we will give the curl estimates of u , v , and B to obtain higher-order derivative estimates based on Lemmas 2.3–2.4.

Lemma 3.6

Suppose that the assumptions in Theorem 1.2 are satisfied. For any 0 < ε ≤ 1 , 0 ≤ ν ≤ O ( ε ς ) with 0 < ς < 2 and 0 ≤ t ≤ min { T ε , ν , 1 } , we have

(3.43) ‖ ( curl u , curl B ) ‖ H 1 2 ( t ) + ‖ ( ε curl u t , ε curl B t ) ‖ L 2 2 ( t ) + ‖ ( curl u , ν curl B ) ‖ L t 2 ( H 2 ) 2 + ‖ ( ε curl u t , ε ν curl B t ) ‖ L t 2 ( H 1 ) 2 ≲ C 0 ( M 0 ) exp t 1 4 + ε C 1 ( M ( t ) ) .

Proof

First, applying the curl operator to (1.4) 2 , we obtain

(3.44) ρ ( curl u t + u ⋅ ∇ curl u ) − μ Δ curl u = − Φ + 1 4 π ( B ⋅ ∇ curl B − curl B ⋅ ∇ B ) ,

where Φ = ( ∂ j ρ u i t − ∂ i ρ u j t ) + [ ∂ j ( ρ u k ) ∂ k u i − ∂ i ( ρ u k ) ∂ k u j ] . Note that we can deduce

(3.45) ‖ Φ ‖ L t 2 ( L 2 ) ≲ ‖ ∇ ρ ‖ L t ∞ ( H 1 ) ‖ u t ‖ L t 2 ( L 3 ) + t 1 2 ‖ ρ ‖ L t ∞ ( H 2 ) ‖ u ‖ L t ∞ ( H 2 ) 2 ≲ t 1 4 M ( t ) .

In view of the boundary condition curl u × n = 0 on ∂ Ω , we multiply (3.44) by curl u and integrate the result by parts over Ω × ( 0 , t ) to obtain

1 2 ∫ Ω ρ ∣ curl u ∣ 2 d x ( t ) + ∫ 0 t ∫ Ω μ ∣ curlcurl u ∣ 2 d x d s = − 1 2 ∫ Ω ρ 0 ∣ curl u 0 ∣ 2 d x + ∫ 0 t ∫ Ω − Φ ⋅ curl u + 1 4 π ( B ⋅ ∇ curl B − curl B ⋅ ∇ B ) ⋅ curl u d x d s + ∫ 0 t ∫ ∂ Ω μ curlcurl u ⋅ ( n × curl u ) d S d s ≲ C 0 ( M 0 ) + δ ‖ curl u ‖ L t ∞ ( L 2 ) 2 + C ( δ ) [ t ‖ Φ ‖ L t 2 ( L 2 ) 2 + t 2 ‖ B ‖ L t ∞ ( H 2 ) 4 ] ,

using of Cauchy’s inequality. Choosing δ small enough and employing (3.45), we have

(3.46) 1 2 ∫ Ω ρ ∣ curl u ∣ 2 d x ( t ) + ∫ 0 t ∫ Ω μ ∣ curlcurl u ∣ 2 d x d s ≲ C 0 ( M 0 ) exp { t C 1 ( M ( t ) ) } .

Multiplying (3.44) by − ρ − 1 Δ curl u and integrating the result over Ω × ( 0 , t ) , we obtain

(3.47) 1 2 ∫ Ω ∣ curl curl u ∣ 2 d x ( t ) + μ ∫ 0 t ∫ Ω ρ − 1 ∣ Δ curl u ∣ 2 d x d s = 1 2 ∫ Ω ∣ curlcurl u 0 ∣ 2 d x + ∫ 0 t ∫ Ω Δ curl u ⋅ u ⋅ ∇ curl u − ρ − 1 Δ curl u ⋅ Φ d x d s − 1 4 π ∫ 0 t ∫ Ω ρ − 1 Δ curl u ( B ⋅ ∇ curl B − curl B ⋅ ∇ B ) d x d s + ∫ 0 t ∫ ∂ Ω curl curl u ⋅ ( n × curl u t ) d S d s ≲ C 0 ( M 0 ) + ‖ Δ curl u ‖ L t 2 ( L 2 ) t 1 2 ‖ u ‖ L t ∞ ( H 2 ) ‖ ∇ curl u ‖ L t ∞ ( L 2 ) + ‖ ρ − 1 ‖ L x , t ∞ ‖ Φ ‖ L t 2 ( L 2 ) + t 1 2 ‖ ρ − 1 ‖ L x , t ∞ ‖ Δ curl u ‖ L t 2 ( L 2 ) ( ‖ B ‖ L t ∞ ( H 2 ) ‖ ∇ curl B ‖ L t ∞ ( L 2 ) + ‖ ∇ B ‖ L t ∞ ( H 1 ) ‖ curl B ‖ L t ∞ ( H 1 ) ) ≲ C 0 ( M 0 ) exp t 1 4 C 1 ( M ( t ) ) ,

where the boundary condition curl u t × n = 0 on ∂ Ω was used.

In the same way, applying the curl operator to (1.9) 5 , multiplying the resulting equality by curl B , and then integrating over Ω × ( 0 , t ) , we have

(3.48) 1 2 ∫ Ω ∣ curl B ∣ 2 d x ( t ) + ∫ 0 t ∫ Ω ν ∣ curlcurl B ∣ 2 d x d s = 1 2 ∫ Ω ∣ curl B 0 ∣ 2 d x + ∫ 0 t ∫ Ω curlcurl B ⋅ curl ( u × B ) d x d s + ν ∫ 0 t ∫ ∂ Ω ( curl B × n ) ⋅ curlcurl B d S d s ≲ C 0 ( M 0 ) + t 1 2 ‖ curlcurl B ‖ L t ∞ ( L 2 ) ‖ B ‖ L t ∞ ( H 2 ) ‖ u ‖ L t 2 ( H 3 ) ≲ C 0 ( M 0 ) + t 1 2 C 1 ( M ( t ) ) .

Next, we apply the operator “ Δ ” to (1.9) 5 to obtain

(3.49) Δ B t + ν Δ curlcurl B + u ⋅ ∇ Δ B = [ Δ , B ⋅ ∇ ] u + B ⋅ ∇ Δ u − [ Δ , u ⋅ ∇ ] B − ( Δ div u ) B − ( div B ) ∇ div u − div u ⋅ Δ B − ∇ div u ⋅ ∇ B .

Note that Δ B = − curlcurl B . In order to obtain the estimates of curl B of higher order, we integrate the product of (3.49) and Δ B . Thus,

(3.50) 1 2 ∫ Ω ∣ curlcurl B ∣ 2 d x ( t ) + ∫ 0 t ∫ Ω ν ∣ curlcurlcurl B ∣ 2 d x d s = 1 2 ∫ Ω ∣ curlcurl B 0 ∣ 2 d x + ∫ 0 t ∫ Ω { Δ B ⋅ { [ Δ , B ⋅ ∇ ] u + B ⋅ ∇ Δ u − [ Δ , u ⋅ ∇ ] B − ( Δ div u ) B − ( div B ) ∇ div u − div u ⋅ Δ B − ∇ div u ⋅ ∇ B } + 1 2 div u ∣ Δ B ∣ 2 d x d s ≲ C 0 ( M 0 ) + t 1 2 ‖ u ‖ L t 2 ( H 3 ) ‖ B ‖ L t ∞ ( H 2 ) 2 ≲ C 0 ( M 0 ) exp t 1 2 C 1 ( M ( t ) ) .

By employing Lemmas 2.3–2.5, (3.46), and (3.47), we have

(3.51) ‖ curl u ‖ L t ∞ ( H 1 ) 2 ≲ ‖ curl u ‖ L t ∞ ( L 2 ) 2 + ‖ curlcurl u ‖ L t ∞ ( L 2 ) 2 + ‖ curl u × n ‖ L t ∞ ( H 1 2 ( ∂ Ω ) ) 2 ≲ C 0 ( M 0 ) exp t 1 4 + ε C 1 ( M ( t ) )

and

(3.52) ‖ curl u ‖ L t 2 ( H 2 ) 2 ≲ ‖ curl u ‖ L t 2 ( H 1 ) 2 + ‖ curlcurl u ‖ L t 2 ( H 1 ) 2 + ‖ curl u × n ‖ L t 2 ( H 3 2 ( ∂ Ω ) ) 2 ≲ ‖ curl u ‖ L t 2 ( L 2 ) 2 + ‖ curlcurl u ‖ L t 2 ( L 2 ) 2 + ‖ curlcurlcurl u ‖ L t 2 ( L 2 ) 2 + ‖ curlcurl u ⋅ n ‖ L t 2 H 1 2 ( ∂ Ω ) 2 ≲ C 0 ( M 0 ) exp t 1 4 + ε C 1 ( M ( t ) ) .

Similarly, by virtue of Lemmas 2.3–2.4, (3.48), and (3.50), we obtain

(3.53) ‖ curl B ‖ L t ∞ ( H 1 ) 2 ≲ ‖ curl B ‖ L t ∞ ( L 2 ) 2 + ‖ curlcurl B ‖ L t ∞ ( L 2 ) 2 + ‖ curl B × n ‖ L t ∞ H 1 2 ( ∂ Ω ) ≲ C 0 ( M 0 ) exp t 1 4 + ε C 1 ( M ( t ) )

and

(3.54) ‖ ν curl B ‖ L t 2 ( H 2 ) 2 ≲ ‖ ν curl B ‖ L t 2 ( H 1 ) 2 + ‖ ν curlcurl B ‖ L t 2 ( H 1 ) 2 + ‖ ν curl B × n ‖ L t 2 H 3 2 ( ∂ Ω ) 2 ≲ ‖ ν curl B ‖ L t 2 ( L 2 ) 2 + ‖ ν curlcurl B ‖ L t 2 ( L 2 ) 2 + ‖ ν curlcurlcurl B ‖ L t 2 ( L 2 ) 2 ≲ C 0 ( M 0 ) exp t 1 4 + ε C 1 ( M ( t ) ) ,

where the boundary conditions curl u × n = curl B × n = 0 and curlcurl u ⋅ n = curlcurl B ⋅ n = 0 on ∂ Ω were used.

In order to obtain the curl estimates of the time derivatives of u and B , we first apply ε ρ ( curl ( ρ − 1 ( ⋅ ) ) ) t to (1.4) 2 . Then, we can similarly show that

(3.55) ρ ( ε curl u t ) t + ρ u ⋅ ∇ ( ε curl u t ) − μ Δ ( ε curl u t ) = − ε { ρ ( [ curl , u ⋅ ∇ ] u ) t + ρ ( u t ⋅ ∇ curl u ) + μ ρ − 1 ρ t Δ curl u } + ρ [ ρ − 2 ∇ ρ × ( ε ( μ Δ u + ( μ + λ ) ∇ div u ) − ∇ σ ) ] t + ε 4 π { [ − ρ − 1 ρ t curl ( curl B × B ) + curl ( curl B × B ) ] t + ρ [ ρ − 2 ∇ ρ × ( curl B × B ) ] t } .

Multiplying (3.55) by ε curl u t and integrating the result over Ω × ( 0 , t ) , we integrate by parts to find that

(3.56) 1 2 ∫ Ω ρ ∣ ε curl u t ∣ 2 d x ( t ) + ∫ 0 t ∫ Ω μ ∣ ε curl curl u t ∣ 2 d x d s ≲ 1 2 ∫ Ω ρ 0 ∣ ε curl u t ( 0 ) ∣ 2 d x + ℱ 1 + ℱ 2 + ℱ 3 ,

where

ℱ 1 ≔ ∫ 0 t ∫ Ω { − ρ ( ε [ curl , u ⋅ ∇ ] u ) t − ρ ( ε u t ⋅ ∇ curl u ) − ε μ ρ − 1 ρ t Δ curl u } ⋅ ε curl u t d x d s , ℱ 2 ≔ ∫ 0 t ∫ Ω ε ρ curl u t ⋅ [ ρ − 2 ∇ ρ × ( ε ( μ Δ u + ( μ + λ ) ∇ div u ) − ∇ σ ) ] t d x d s , ℱ 3 ≔ ∫ 0 t ∫ Ω { ρ − 1 ε ρ t curl ( curl B × B ) + ( ∣ ρ − 1 ∣ 2 ε ρ t ∇ ρ + ρ − 1 ε ∇ ρ t ) × ( curl B × B ) + ε curl ( curl B × B ) t + ρ − 1 ∇ ρ ( ε curl B t × B + ε curl B × B t ) } ⋅ ε curl u t d x d s .

Now, the terms of right-hand side of (3.56) are estimated as follows. By integrating by parts and observing that

‖ ε curl u t ‖ L t 2 ( L 3 ) ≲ t 1 4 ‖ ε curl u t ‖ L t ∞ ( L 2 ) 1 2 ‖ ε curl u t ‖ L t 2 ( H 1 ) 1 2 ≲ t 1 4 M ( t ) ,

we deduce that

ℱ 1 ≲ ( ‖ ρ ‖ L t ∞ ( H 2 ) ‖ u ‖ L t 2 ( H 3 ) ‖ ε u t ‖ L t ∞ ( H 1 ) + ‖ ρ ‖ L t ∞ ( H 2 ) ‖ ε u t ‖ L t ∞ ( H 1 ) ‖ ∇ curl u ‖ L t 2 ( L 2 ) + ‖ ρ − 1 ‖ L x , t ∞ ‖ ε ρ t ‖ L t ∞ ( H 1 ) ‖ Δ curl u ‖ L t 2 ( L 2 ) ) ‖ ε curl u t ‖ L t 2 ( L 3 ) ≲ t 1 4 C 1 ( M ( t ) ) ,

ℱ 2 ≲ ∫ 0 t ∫ Ω ρ − 1 [ ( ρ − 1 ∣ ε ρ t ‖ ∇ ρ ∣ + ∣ ε ∇ ρ t ∣ ) ∣ D 2 u ∣ + ( ρ − 1 ∣ ρ t ‖ ∇ ρ ∣ + ∣ ∇ ρ t ∣ ) ∣ ∇ σ ∣ + ∣ ∇ ρ ∣ ( ∣ ε D 2 u t ∣ + ∣ ∇ σ t ∣ ) ] ⋅ ε curl u t d x d s ≲ ‖ ρ − 1 ‖ L x , t ∞ [ ( ‖ ε ρ t ‖ L t ∞ ( H 1 ) ‖ ∇ ρ ‖ L t ∞ ( H 1 ) ‖ ρ − 1 ‖ L x , t ∞ + ‖ ε ∇ ρ t ‖ L t ∞ ( L 2 ) ) ‖ D 2 u ‖ L t 2 ( H 1 ) + t 1 2 ‖ ∇ σ ‖ L t ∞ ( H 1 ) ( ‖ ρ − 1 ‖ L x , t ∞ ‖ ρ t ‖ L t ∞ ( H 1 ) ‖ ∇ ρ ‖ L t ∞ ( H 1 ) + ‖ ∇ ρ t ‖ L t ∞ ( L 2 ) ) + ‖ ∇ ρ ‖ L t ∞ ( H 1 ) ( ‖ ε D 2 u t ‖ L t 2 ( L 2 ) + ‖ ∇ σ t ‖ L t 2 ( L 2 ) ) ] ‖ ε curl u t ‖ L t 2 ( L 3 ) ≲ t 1 4 C 1 ( M ( t ) ) ,

and

ℱ 3 ≲ t 1 2 ‖ ε curl u t ‖ L t 2 ( H 1 ) { ‖ ρ − 1 ‖ L x , t ∞ ‖ ε ρ t ‖ L t ∞ ( H 1 ) ‖ B ‖ L t ∞ ( H 2 ) ‖ curl B ‖ L t ∞ ( H 1 ) + ( ‖ ρ − 1 ‖ L x , t ∞ 2 ‖ ∇ ρ ‖ L t ∞ ( H 1 ) ‖ ε ρ t ‖ L t ∞ ( H 1 ) + ‖ ρ − 1 ‖ L x , t ∞ ‖ ε ∇ ρ t ‖ L t ∞ ( L 2 ) ) ‖ curl B ‖ L t ∞ ( H 1 ) ‖ B ‖ L t ∞ ( H 2 ) + ‖ ε curl B t ‖ L t ∞ ( L 2 ) ‖ B ‖ L t ∞ ( H 2 ) + ‖ ε B t ‖ L t ∞ ( H 1 ) curl B ‖ L t ∞ ( H 1 ) + ‖ ρ − 1 ‖ L x , t ∞ ‖ ∇ ρ ‖ L t ∞ ( H 1 ) ‖ ε B t ‖ L t ∞ ( H 1 ) ‖ B ‖ L t ∞ ( H 2 ) } ≲ t 1 2 C 1 ( M ( t ) ) .

Concluding the aforementioned estimates, we have

(3.57) 1 2 ∫ Ω ρ ∣ ε curl u t ∣ 2 d x ( t ) + ∫ 0 t ∫ Ω μ ∣ ε curl curl u t ∣ 2 d x d s ≲ C 0 ( M 0 ) exp t 1 4 C 1 ( M ( t ) ) .

Next, using the same method for (3.27) 4 , i.e., multiplying ε curl ((3.27) 4 ) by ε curl B t , and integrating the result over Ω × ( 0 , t ) , we obtain

1 2 ∫ Ω ∣ ε curl B t ∣ 2 d x ( t ) + ∫ 0 t ∫ Ω ν ∣ ε curl curl B t ∣ 2 d x d s = 1 2 ∫ Ω ∣ ε curl B t ( 0 ) ∣ 2 d x + ∫ 0 t ∫ Ω ε curl B t ⋅ ε curlcurl ( u × B ) t d x d s ,

where

∫ 0 t ∫ Ω ε curl B t ⋅ ε curlcurl ( u × B ) t d x d s = ∫ 0 t ∫ Ω ε curl B t ⋅ ε curl [ ( B ⋅ ∇ ) u − ( u ⋅ ∇ ) B − B div u ] t d x d s = ∫ 0 t ∫ Ω ε curl B t ⋅ ε [ ( curl B ⋅ ∇ ) u + ( B ⋅ ∇ ) curl u − ( curl u ⋅ ∇ ) B − ( u ⋅ ∇ ) curl B − ∇ div u × B − ( div u ) curl B ] t d x d s ≲ t 1 2 ‖ ε curl B t ‖ L t ∞ ( L 2 ) ( ‖ ε B t ‖ L t ∞ ( H 1 ) ‖ u ‖ L t 2 ( H 3 ) + ‖ ε u t ‖ L t 2 ( H 2 ) ‖ B ‖ L t ∞ ( H 2 ) ) ≲ t 1 2 C 1 ( M ( t ) ) .

Hence, we find that

(3.58) 1 2 ∫ Ω ∣ ε curl B t ∣ 2 d x ( t ) + ∫ 0 t ∫ Ω ν ∣ ε curl curl B t ∣ 2 d x d s ≲ C 0 ( M 0 ) exp t 1 2 C 1 ( M ( t ) ) .

Moreover, by (3.57) and (3.58), we find

(3.59) ‖ ε curl u t ‖ L t 2 ( H 1 ) 2 ≲ ‖ ε curl u t ‖ L t 2 ( L 2 ) 2 + ‖ ε curlcurl u t ‖ L t 2 ( L 2 ) 2 + ‖ ε curl u t × n ‖ L t 2 H 1 2 ( ∂ Ω ) 2 ≲ C 0 ( M 0 ) exp t 1 4 C 1 ( M ( t ) )

and

(3.60) ‖ ε ν curl B t ‖ L t 2 ( H 1 ) 2 ≲ ‖ ε ν curl B t ‖ L t 2 ( L 2 ) 2 + ‖ ε ν curlcurl B t ‖ L t 2 ( L 2 ) 2 + ‖ ε ν curl B t × n ‖ L t 2 H 1 2 ( ∂ Ω ) 2 ≲ C 0 ( M 0 ) exp t 1 2 C 1 ( M ( t ) ) .

From (3.51)–(3.54) and (3.57)–(3.60), we conclude (3.43) immediately.□

3.6 Estimates of high-order spatial derivatives of θ , θ t , v , and σ

To obtain the estimates of θ t and ∇ θ t and deal with − β ∫ 0 t ∫ Ω div v t Δ θ t d x d s on the right-hand side of (3.28), we need to show the following lemma.

Lemma 3.7

Suppose that the assumptions in Theorem 1.2 are satisfied. For any 0 < ε ≤ 1 , 0 ≤ ν ≤ O ( ε ς ) with 0 < ς < 2 and 0 ≤ t ≤ min { T ε , ν , 1 } , we have

(3.61) ‖ ( θ t , ∇ θ t ) ‖ L 2 2 ( t ) + ‖ ( ∇ θ t , Δ θ t ) ‖ L t 2 ( L 2 ) 2 ≲ C 0 ( M 0 ) exp t 1 4 C 1 ( M ( t ) ) + ∫ 0 t ∫ Ω div v t Δ θ t d x d s .

Proof

First, by differentiating equation (1.9) 4 we observe

(3.62) ρ ( θ t t + u ⋅ ∇ θ t ) − κ 2 Δ θ t = h t − ρ t ( θ t + u ⋅ ∇ θ ) − ρ u t ⋅ ∇ θ − div v t .

Multiplying (3.62) by θ t and integrating the result over Ω × [ 0 , t ] , we integrate by parts to obtain

(3.63) 1 2 ∫ Ω ρ ∣ θ t ∣ 2 d x ( t ) + κ 2 ∫ 0 t ∫ Ω ∣ ∇ θ t ∣ 2 d x d s = 1 2 ∫ Ω ρ 0 ∣ θ t ( 0 ) ∣ 2 d x + ∫ 0 t ∫ Ω [ h t − ρ t ( θ t + u ⋅ ∇ θ ) − ρ u t ⋅ ∇ θ − div v t ] θ t d x d s ≲ C 0 ( M 0 ) + t 1 2 ‖ θ t ‖ L t ∞ ( H 1 ) [ ‖ h t ‖ L t 2 ( L 2 ) + ‖ ρ t ‖ L t ∞ ( H 1 ) ( ‖ θ t ‖ L t 2 ( H 2 ) + ‖ ∇ θ ‖ L t 2 ( H 2 ) ‖ u ‖ L t ∞ ( H 2 ) ) + ‖ ρ ‖ L t ∞ ( H 2 ) ‖ ∇ θ ‖ L t 2 ( H 2 ) ‖ u t ‖ L t ∞ ( L 2 ) + ‖ div v t ‖ L t 2 ( L 2 ) ] ≲ C 0 ( M 0 ) exp t 1 2 C 1 ( M ( t ) ) ,

where (3.14) is used. In order to obtain higher-order estimates, applying ∇ ( ρ − 1 ( ⋅ ) ) to (3.62), multiplying ρ ∇ θ t , and then integrating over Ω × [ 0 , t ] , we have

(3.64) 1 2 ∫ Ω ρ ∣ ∇ θ t ∣ 2 d x + κ 2 ∫ 0 t ∫ Ω ∣ Δ θ t ∣ 2 d x d s = 1 2 ∫ Ω ρ 0 ∣ ∇ θ t ( 0 ) ∣ 2 d x + ∫ 0 t ∫ Ω div v t Δ θ t d x d s + K 1 + K 2 + K 3 + K 4 ,

where the estimates of K 1 , K 2 , K 3 , and K 4 are as follows:

K 1 ≔ − ∫ 0 t ∫ Ω ( − ρ − 1 ρ t ∇ ρ ⋅ ∇ θ t ( θ t + u ⋅ ∇ θ ) + ∇ ρ t ∇ θ t ( θ t + u ⋅ ∇ θ ) ) d x d s ≲ t 1 2 ‖ ρ − 1 ‖ L x , t ∞ ‖ ∇ ρ ‖ L t ∞ ( H 1 ) ‖ ρ t ‖ L t ∞ ( H 1 ) ‖ ∇ θ t ‖ L t ∞ ( L 2 ) ( ‖ θ t ‖ L t 2 ( H 2 ) + ‖ u ‖ L t 2 ( H 2 ) ‖ ∇ θ ‖ L t ∞ ( H 2 ) ) + t 1 2 ‖ ∇ ρ t ‖ L t ∞ ( L 2 ) ‖ ∇ θ t ‖ L t ∞ ( L 2 ) ( ‖ θ t ‖ L t 2 ( H 2 ) + ‖ u ‖ L t ∞ ( H 2 ) ‖ ∇ θ ‖ L t 2 ( H 2 ) ) ≲ t 1 2 C 1 ( M ( t ) ) ,

K 2 ≔ − ∫ 0 t ∫ Ω ρ t ∇ θ t ( ∇ θ t + ∇ u ⋅ ∇ θ + u ⋅ ∇ 2 θ ) d x d s ≲ ‖ ρ t ‖ L t ∞ ( H 1 ) ‖ ∇ θ t ‖ L t 2 ( L 3 ) ( ‖ ∇ θ t ‖ L t 2 ( H 1 ) + t 1 2 ‖ u ‖ L t ∞ ( H 2 ) ‖ ∇ θ ‖ L t ∞ ( H 2 ) + ‖ u ‖ L t ∞ ( H 2 ) ‖ ∇ 2 θ ‖ L t 2 ( H 1 ) ) ≲ t 1 4 C 1 ( M ( t ) ) ,

K 3 ≔ − ∫ 0 t ∫ Ω ρ ∇ u t ⋅ ∇ θ ∇ θ t + ρ u t ⋅ ∇ 2 θ ∇ θ t + ρ ∇ u ∣ ∇ θ t ∣ 2 + κ 2 ρ − 1 ∇ ρ ⋅ Δ θ t ∇ θ t d x d s ≲ t 1 2 ‖ ρ ‖ L t ∞ ( H 2 ) ( ‖ ∇ u t ‖ L t 2 ( L 2 ) ‖ ∇ θ ‖ L t ∞ ( H 2 ) ‖ ∇ θ t ‖ L t ∞ ( L 2 ) + ‖ u t ‖ L t ∞ ( L 2 ) ‖ ∇ 2 θ ‖ L t ∞ ( H 1 ) ‖ ∇ θ t ‖ L t 2 ( H 1 ) + ‖ ∇ u ‖ L t 2 ( H 2 ) ‖ ∇ θ t ‖ L t ∞ ( L 2 ) 2 ) + ‖ ρ − 1 ‖ L x , t ∞ ‖ ∇ ρ ‖ L t ∞ ( H 1 ) ‖ Δ θ t ‖ L t 2 ( L 2 ) ‖ ∇ θ t ‖ L t 2 ( L 3 ) ≲ t 1 4 C 1 ( M ( t ) ) ,

and

K 4 ≔ − ∫ 0 t ∫ Ω ( − ρ − 1 ∇ ρ ⋅ div v t ∇ θ t + ρ − 1 ∇ ρ ⋅ ∇ θ t ⋅ h t − ∇ θ t ∇ h t ) d x d s ≲ ‖ ρ − 1 ‖ L x , t ∞ ‖ ∇ ρ ‖ L t ∞ ( H 1 ) ‖ div v t ‖ L t 2 ( L 2 ) ‖ ∇ θ t ‖ L t 2 ( L 3 ) + ‖ ρ − 1 ‖ L x , t ∞ ‖ ∇ ρ ‖ L t ∞ ( H 1 ) ‖ h t ‖ L t 2 ( L 2 ) ‖ ∇ θ t ‖ L t 2 ( L 3 ) + t 1 2 ‖ h t ‖ L t ∞ ( L 2 ) ‖ ∇ θ t ‖ L t 2 ( H 1 ) ≲ t 1 4 C 1 ( M ( t ) ) ,

by observing that

□ ‖ ∇ θ t ‖ L t 2 ( L 3 ) ≲ t 1 4 ‖ ∇ θ t ‖ L t ∞ ( L 2 ) 1 2 ‖ ∇ θ t ‖ L t 2 ( H 1 ) 1 2 ≲ t 1 4 M ( t ) .

Next, we provide the following lemma in order to eliminate the term − β ε 2 ∫ 0 t ∫ Ω ρ − 1 ∇ div v t ⋅ Δ ∇ θ t d x d s on the right-hand side of (3.34).

Lemma 3.8

Suppose that the assumptions in Theorem 1.2 are satisfied. For any 0 < ε ≤ 1 , 0 ≤ ν ≤ O ( ε ς ) with 0 < ς < 2 and 0 ≤ t ≤ min { T ε , ν , 1 } , we have

(3.65) ‖ ε Δ θ t ‖ L 2 2 ( t ) + ‖ ε ∇ Δ θ t ‖ L t 2 ( L 2 ) 2 ≲ C 0 ( M 0 ) exp t 1 4 + ε 1 − ς 2 C 1 ( M ( t ) ) + ε 2 ∫ 0 t ∫ Ω ρ − 1 ∇ Δ θ t ⋅ ∇ div v t d x d s .

Proof

First, applying ∇ ( ρ − 1 ( ⋅ ) ) to (3.27) 3 , multiplying − ε 2 ∇ Δ θ t , and then integrating the result over Ω × ( 0 , t ) , we verify

(3.66) 1 2 ∫ Ω ∣ ε Δ θ t ∣ 2 d x ( t ) + κ 2 ∫ 0 t ∫ Ω ρ − 1 ∣ ε ∇ Δ θ t ∣ 2 d x d s ≲ 1 2 ∫ Ω ∣ ε Δ θ t ( 0 ) ∣ 2 d x + ε 2 ∫ 0 t ∫ Ω ρ − 1 ∇ div v t ⋅ ∇ Δ θ t d x d s + T 1 + T 2 + T 3 ,

where the terms T i for i = 1, 2, 3 are expressed and estimated as follows:

T 1 ≔ ∫ 0 t ∫ Ω − ε 2 ( ∇ Δ θ t ) ⋅ [ ∇ ( ρ − 1 ) ρ t ( θ t + u ⋅ ∇ θ ) + ρ − 1 ∇ ρ t ( θ t + u ⋅ ∇ θ ) + ρ − 1 ρ t ( ∇ θ t + ∇ u ⋅ ∇ θ + u ⋅ ∇ 2 θ ) ] d x d s ≲ ‖ ε ∇ Δ θ t ‖ L t 2 ( L 2 ) t 1 2 ‖ ρ − 1 ‖ L x , t ∞ 2 ‖ ∇ ρ ‖ L t ∞ ( H 1 ) ‖ ε ρ t ‖ L t ∞ ( H 1 ) ( ‖ θ t ‖ L t ∞ ( H 1 ) + ‖ u ‖ L t ∞ ( H 2 ) ‖ ∇ θ ‖ L t ∞ ( H 1 ) ) t 1 2 ‖ ρ − 1 ‖ L x , t ∞ ‖ ρ t ‖ L t ∞ ( H 1 ) ( ‖ ε θ t ‖ L t ∞ ( H 2 ) + ‖ u ‖ L t ∞ ( H 2 ) ‖ ∇ θ ‖ L t ∞ ( H 2 ) ) + ‖ ε ∇ Δ θ t ‖ L t 2 ( L 2 ) ‖ ρ − 1 ‖ L x , t ∞ ‖ ε ρ t ‖ L t ∞ ( H 1 ) ( ‖ ∇ θ t ‖ L t 2 ( L 3 ) + t 1 2 ‖ u ‖ L t ∞ ( H 2 ) ‖ ∇ θ ‖ L t ∞ ( H 2 ) + t 1 2 ‖ u ‖ L t ∞ ( H 2 ) ‖ ∇ 2 θ ‖ L t ∞ ( H 1 ) ) ≲ t 1 4 C 1 ( M ( t ) ) ,

T 2 ≔ ∫ 0 t ∫ Ω − ε 2 ( ∇ Δ θ t ) [ ∇ u t ⋅ ∇ θ + u t ⋅ ∇ 2 θ + ∇ u ⋅ ∇ θ t + u ⋅ ∇ 2 θ t ] d x d s ≲ ‖ ε ∇ Δ θ t ‖ L t 2 ( L 2 ) ( ε ‖ ∇ u t ‖ L t 2 ( L 2 ) ‖ ∇ θ ‖ L t ∞ ( H 2 ) + ε ‖ u t ‖ L t 2 ( H 1 ) ‖ ∇ 2 θ ‖ L t ∞ ( H 1 ) + t 1 2 ‖ u ‖ L t ∞ ( H 2 ) ‖ ε ∇ θ t ‖ L t ∞ ( H 1 ) + ε ‖ u ‖ L t ∞ ( H 2 ) ‖ ∇ 2 θ t ‖ L t 2 ( L 2 ) ≲ t 1 2 + ε C 1 ( M ( t ) ) ,

and employing (3.14) and (3.16), we derive

T 3 ≔ ∫ 0 t ∫ Ω − ε 2 ( ∇ Δ θ t ) ⋅ − κ 2 ∇ ( ρ − 1 ) Δ θ t + ∇ ( ρ − 1 ) div v t − ∇ ( ρ − 1 ) h t − ρ − 1 ∇ h t d x d s ≲ ‖ ε ∇ Δ θ t ‖ L t 2 ( L 2 ) ‖ ρ − 1 ‖ L x , t ∞ 2 ‖ ∇ ρ ‖ L t ∞ ( H 1 ) ( ‖ ε Δ θ t ‖ L t 2 ( L 3 ) + ‖ ε div v t ‖ L t 2 ( L 3 ) + ‖ ε h t ‖ L t 2 ( L 3 ) ) + ‖ ε ∇ Δ θ t ‖ L t 2 ( L 2 ) ‖ ρ − 1 ‖ L x , t ∞ ‖ ε ∇ h t ‖ L t 2 ( L 2 ) ≲ t 1 4 + ε 1 − ς 2 C 1 ( M ( t ) ) .

Hence, we can obtain

(3.67)□ 1 2 ∫ Ω ∣ ε Δ θ t ∣ 2 d x ( t ) + κ 2 ∫ 0 t ∫ Ω ρ − 1 ∣ ε ∇ Δ θ t ∣ 2 d x d s ≲ C 0 ( M 0 ) exp t 1 4 + ε 1 − ς 2 C 1 ( M ( t ) ) + ε 2 ∫ 0 t ∫ Ω ρ − 1 ∇ div v t ⋅ ∇ Δ θ t d x d s .

According to Lemma 2.4, in order to derive the uniform bounds of ‖ v ‖ L t ∞ ( H 2 ) and ‖ v ‖ L t 2 ( H 3 ) , we need to obtain the estimates of ‖ div v ‖ L t ∞ ( H 1 ) and ‖ ∇ 2 div v ‖ L t 2 ( L 2 ) .

Lemma 3.9

Suppose that the assumptions in Theorem 1.2 are satisfied. For any 0 < ε ≤ 1 , 0 ≤ ν ≤ O ( ε ς ) with 0 < ς < 2 and 0 ≤ t ≤ min { T ε , ν , 1 } , we have

(3.68) ‖ div v ‖ L t ∞ ( H 1 ) 2 + ‖ ε σ t ‖ L t ∞ ( H 1 ) 2 ≲ C 0 ( M 0 ) exp t 1 2 + ε 1 − ς 2 C 1 ( M ( t ) )

and

(3.69) ‖ ∇ 2 σ ‖ L 2 2 ( t ) + ‖ ∇ 2 div v ‖ L t 2 ( L 2 ) 2 ≲ C 0 ( M 0 ) exp ( t 1 4 + ε 2 − ς ) C 1 ( M ( t ) ) − β ∫ 0 t ∫ Ω ∂ i ∇ Δ θ ⋅ ∂ i ∇ div v d x d s .

Proof

First, multiplying ∇ (3.27) 2 by ε 2 ∇ σ t , then integrating the result over Ω × ( 0 , t ) , we deduce that

(3.70) 1 4 ∫ Ω ∣ ε ∇ σ t ∣ 2 d x = 1 4 ∫ Ω ∣ ε ∇ σ t ( 0 ) ∣ 2 d x + ∫ 0 t ∫ Ω ε 2 ∇ σ t ⋅ 1 4 div u ∇ σ t − 1 2 ∇ u ⋅ ∇ σ t − 1 2 ∇ ( u t ⋅ ∇ σ ) − ∇ div v t + ∇ g t d x d s ≲ C 0 ( M 0 ) + t 1 2 ‖ ε ∇ σ t ‖ L t ∞ ( L 2 ) ( ‖ u ‖ L t 2 ( H 3 ) ‖ ε ∇ σ t ‖ L t ∞ ( L 2 ) + ‖ ε u t ‖ L t 2 ( H 2 ) ‖ σ ‖ L t ∞ ( H 2 ) + ‖ ε v t ‖ L t 2 ( H 2 ) ) + ‖ ∇ σ t ‖ L t 2 ( L 2 ) ‖ ε 2 ∇ g t ‖ L t 2 ( L 2 ) ≲ C 0 ( M 0 ) exp t 1 2 + ε 1 − ς 2 C 1 ( M ( t ) ) ,

by using (3.15). Furthermore, from (1.9) 3 and (3.9), we have

(3.71) ‖ div v ‖ L t ∞ ( H 1 ) 2 ≲ ε 2 ‖ u ‖ L t ∞ ( H 2 ) 2 ∥ ∇ σ ∥ L t ∞ ( H 1 ) 2 + ε 2 ‖ g ‖ L t ∞ ( H 1 ) 2 + ‖ ε σ t ‖ L t ∞ ( H 1 ) 2 ≲ C 0 ( M 0 ) exp t 1 2 + ε 1 − ς 2 C 1 ( M ( t ) ) .

Next, multiplying ∂ i [(1.9) 2 ] and ∂ i ∇ [(1.9) 3 ] by − ∂ i ∇ div v and ∂ i ∇ σ for i = 1, 2, 3, respectively, then integrating the sum of the results over Ω × ( 0 , t ) , we have

(3.72) ∫ 0 t ∫ Ω − ∂ i ∇ div v ⋅ [ ∂ i ρ ⋅ ( v t + u ⋅ ∇ v ) + ρ ( ∂ i v t + ∂ i u ⋅ ∇ v + u ⋅ ∂ i ∇ v ) + 1 ε ∂ i ∇ σ − μ ∂ i △ v − ξ ∂ i ∇ div v − β ∂ i ∇ Δ θ + 1 4 π ∂ i ( ( curl B ) × B ) × ∂ i ∇ σ ⋅ 1 2 ∂ i ∇ σ t + 1 2 ( ∂ i ∇ u ⋅ ∇ σ + ∇ u ⋅ ∂ i ∇ σ + ∂ i u ⋅ ∇ 2 σ + u ⋅ ∂ i ∇ 2 σ ) + 1 ε ∂ i ∇ div v d x d s = ∫ 0 t ∫ Ω − ∂ i ∇ div v ⋅ ∂ i f + ∂ i ∇ σ ⋅ ∂ i ∇ g d x d s .

By integrating by parts, we obtain

(3.73) 1 4 ∫ Ω ∣ ∂ i ∇ σ ∣ 2 d x + ( μ + ξ ) ∫ 0 t ∫ Ω ∣ ∂ i ∇ div v ∣ 2 d x d s ≲ 1 4 ∫ Ω ∣ ∂ i ∇ σ 0 ∣ 2 d x − β ∫ 0 t ∫ Ω ∂ i ∇ Δ θ ⋅ ∂ i ∇ div v d x d s + S + ‖ ∂ i ∇ g ‖ L t 1 ( L 2 ) ‖ ∂ i ∇ σ ‖ L t ∞ ( L 2 ) ,

where

S = ∫ 0 t ∫ Ω ∂ i ∇ div v ⋅ ∂ i ρ ⋅ ( v t + u ⋅ ∇ v ) + ρ ( ∂ i v t + ∂ i u ⋅ ∇ v + u ⋅ ∂ i ∇ v ) + 1 4 π ∂ i ( curl B × B ) + μ ∂ i curlcurl v − ∂ i f − ∂ i ∇ σ ⋅ 1 2 ( ∂ i ∇ u ⋅ ∇ σ + ∇ u ⋅ ∂ i ∇ σ + ∂ i u ⋅ ∇ 2 σ ) + 1 4 div u ∣ ∂ i ∇ σ ∣ 2 d x d s ≲ ‖ ρ ‖ L t ∞ ( H 2 ) ‖ ∂ i ∇ div v ‖ L t 2 ( L 2 ) ( ‖ v t ‖ L t 2 ( L 3 ) + t 1 2 ‖ u ‖ L t ∞ ( H 2 ) ‖ ∇ v ‖ L t ∞ ( H 1 ) ) + t 1 2 ‖ ∂ i ∇ div v ‖ L t 2 ( L 2 ) ‖ B ‖ L t ∞ ( H 2 ) 2 + t 1 2 ‖ ∂ i ∇ σ ‖ L t ∞ ( L 2 ) ‖ u ‖ L t 2 ( H 3 ) ‖ σ ‖ L t ∞ ( H 2 ) + ( ‖ curl v ‖ L t 2 ( H 2 ) + ‖ ∂ i f ‖ L t 2 ( L 2 ) ) ‖ ∂ i ∇ div v ‖ L t 2 ( L 2 ) ≲ δ ‖ ∂ i ∇ div v ‖ L t 2 ( L 2 ) 2 + C ( δ ) ‖ ρ ‖ L t ∞ ( H 2 ) ( ‖ v t ‖ L t 2 ( L 3 ) + t 1 2 ‖ u ‖ L t ∞ ( H 2 ) ‖ ∇ v ‖ L t ∞ ( H 1 ) ) + t 1 2 ‖ B ‖ L t ∞ ( H 2 ) + ‖ curl v ‖ L t 2 ( H 2 ) + ‖ ∂ i f ‖ L t 2 ( L 2 ) 2 + δ ‖ ∂ i ∇ σ ‖ L t ∞ ( L 2 ) 2 + C ( δ ) t 1 2 ‖ u ‖ L t 2 ( H 3 ) ‖ σ ‖ L t ∞ ( H 2 ) 2 ≲ δ ( ‖ ∂ i ∇ div v ‖ L t 2 ( L 2 ) 2 + ‖ ∂ i ∇ σ ‖ L t ∞ ( L 2 ) 2 ) + C ( δ ) C 0 ( M 0 ) exp t 1 4 + ε 2 − ς C 1 ( M ( t ) ) .

by employing (3.1), (3.11), (3.12), (3.28), and (3.52). Concluding the aforementioned estimates, we obtain (3.69).□

Finally, we calculate the estimate of ‖ ∇ θ ‖ L t 2 ( H 3 ) to eliminate the integral − β ∫ 0 t ∫ Ω ∂ i ∇ Δ θ ⋅ ∂ i ∇ div v d x d s on the right-hand side of (3.69).

Lemma 3.10

Suppose that the assumptions in Theorem 1.2 are satisfied. For any 0 < ε ≤ 1 , 0 ≤ ν ≤ O ( ε ς ) with 0 < ς < 2 and 0 ≤ t ≤ min { T ε , ν , 1 } , we have

(3.74) ‖ ∂ i ∇ Δ θ ‖ L t 2 ( L 2 ) 2 ≲ C 0 ( M 0 ) exp t 1 4 + ε 2 − ς C 1 ( M ( t ) ) + ∫ 0 t ∫ Ω ∂ i ∇ Δ θ ⋅ ∂ i ∇ div v d x d s .

Proof

Multiply ∂ i ∇ Δ θ by ∂ i ∇ [(1.9) 4 ], and then integrate the result over Ω × ( 0 , t ) to obtain

(3.75) κ 2 ∫ 0 t ∫ Ω ∣ ∂ i ∇ Δ θ ∣ 2 d x d s − ∫ 0 t ∫ Ω ∂ i ∇ Δ θ ⋅ ∂ i ∇ div v d x d s = ∫ 0 t ∫ Ω ∂ i ∇ Δ θ ⋅ ( ∂ i ∇ [ ρ ( θ t + u ⋅ ∇ θ ) ] − ∂ i ∇ h ) d x d s ≲ δ ‖ ∂ i ∇ Δ θ ‖ L t 2 ( L 2 ) 2 + C ( δ ) ( ‖ ρ ‖ L t ∞ ( H 2 ) 2 ‖ θ t ‖ L t 2 ( H 2 ) 2 + t ‖ ρ ‖ L t ∞ ( H 2 ) 2 ‖ u ‖ L t ∞ ( H 2 ) 2 ‖ ∇ θ ‖ L t ∞ ( H 2 ) 2 + ‖ ∂ i ∇ h ‖ L t 2 ( L 2 ) 2 ) ≲ δ ‖ ∂ i ∇ Δ θ ‖ L t 2 ( L 2 ) 2 + C ( δ ) C 0 ( M 0 ) exp t 1 4 + ε 2 − ς C 1 ( M ( t ) ) .

Note that the last inequality holds by using (3.1), (3.28), (3.8), (3.36), and (3.61). Choosing δ small enough, we obtain (3.74).□

3.7 Proof of Proposition 1.1

Because div B = 0 , we can obtain the higher-order estimates of B from the lower-order estimates of curl B . Therefore, from Lemmas 3.2, 3.3, 3.6, and 3.7, we can see that, for any 0 < ε ≤ 1 , 0 ≤ ν ≤ O ( ε ς ) with 0 < ς < 2 and 0 ≤ t ≤ min { T ε , ν , 1 } ,

(3.76) ‖ B ‖ L t ∞ ( H 2 ) 2 + ‖ ν B ‖ L t 2 ( H 3 ) 2 + ‖ B t ‖ L t ∞ ( L 2 ) 2 + ‖ ε B t ‖ L t ∞ ( H 1 ) 2 + ‖ ν B t ‖ L t 2 ( H 1 ) 2 + ‖ ε ν B t ‖ L t 2 ( H 2 ) 2 ≲ C 0 ( M 0 ) exp t 1 4 + ε C 1 ( M ( t ) ) .

Similarly, we can obtain the high-order estimates of the solution of (1.9) via the low-order estimates by Lemmas 2.3 and 2.4. Therefore, from (3.19), (3.28), (3.36), and (3.61), we have, for any 0 < ε ≤ 1 , 0 ≤ ν ≤ O ( ε ς ) with 0 < ς < 2 and 0 ≤ t ≤ min { T ε , ν , 1 } ,

(3.77) ‖ ( v t , σ t , θ t , ∇ θ t , ∇ div v ) ‖ L t ∞ ( L 2 ) 2 + ‖ θ ‖ L t ∞ ( H 3 ) 2 + ‖ ( v t , σ t , ∇ θ t ) ‖ L t 2 ( H 1 ) 2 ≲ C 0 ( M 0 ) exp t 1 4 + ε C 1 ( M ( t ) ) .

By virtue of (3.34), (3.65), and (3.68), we obtain

(3.78) ‖ ε div v t ‖ L t ∞ ( L 2 ) 2 + ‖ ε ( σ t , ∇ θ t ) ‖ L t ∞ ( H 1 ) 2 + ‖ ε div v t ‖ L t 2 ( H 1 ) 2 + ‖ ε ∇ θ t ‖ L t 2 ( H 2 ) 2 ≲ C 0 ( M 0 ) exp t 1 4 + ε 1 − ς 2 C 1 ( M ( t ) ) .

From (3.69) and (3.74), we find

(3.79) ‖ σ ‖ L t ∞ ( H 2 ) 2 + ‖ ∇ 2 div v ‖ L t 2 ( L 2 ) 2 + ‖ ∇ θ ‖ L t 2 ( H 3 ) 2 ≲ C 0 ( M 0 ) exp t 1 4 + ε 1 − ς 2 C 1 ( M ( t ) ) .

Combining the estimates of (3.19), (3.43), (3.68), (3.77)–(3.79), and using Lemmas 2.3–2.5, we obtain that

(3.80) ‖ v ‖ L t ∞ ( H 2 ) 2 + ‖ v ‖ L t 2 ( H 3 ) 2 + ‖ ε v t ‖ L t ∞ ( H 1 ) 2 + ‖ ε v t ‖ L t 2 ( H 2 ) 2 ≲ C 0 ( M 0 ) exp { ( t 1 4 + ε 1 − ς 2 ) C 1 ( M ( t ) ) } .

Hence, summing up (3.77)–(3.80) and (3.76), we obtain Proposition 1.1 immediately.

4 Proofs of main theorems

4.1 Proof of Theorem 1.2

Since

(4.1) 1 + θ ε , ν = 1 + θ 0 ε , ν + ∫ 0 t θ s ε , ν d s ≥ A − t 1 2 ‖ θ t ε , ν ‖ L t 2 ( H 2 ) 2 ≥ A − t 1 2 M ε , ν ( t ) ,

we only need to show that there exists a T ∈ ( 0 , 1 ] , such that (1.16) holds (see Alazard [1]). Assume that Proposition 1.1 holds and T ε , ν < + ∞ is the maximal lifetime of existence for the solution obtained in Theorem 1.1. Then, we have

(4.2) M ε , ν ( t ) ≤ C 0 ( M 0 ε , ν ) exp t 1 4 + ε 1 − ς 2 C 1 ( M ε , ν ( t ) ) , 0 ≤ t ≤ min { T ε , ν , 1 } ,

where M 0 ε , ν ≤ D 0 for 0 < ε ≤ 1 and 0 ≤ ν ≤ O ( ε ς ) with 0 < ς < 2 . In the sequel, we choose D > C 0 ( D 0 ) , ε 1 ∈ ( 0 , 1 ] , ς 1 ∈ ( 0 , 2 ) and T 1 ∈ ( 0 , 1 ] one by one such that

(4.3) C 0 ( D 0 ) exp { ( T 1 1 4 + ε 1 1 − ς 1 2 ) C 1 ( D ) } < D .

Let t < min { T ε , ν , T 1 } . By combining inequalities (4.2) and (4.3) with the assumption M ε , ν ( 0 ) = M 0 ε , ν , we have that M ε , ν ( t ) ≠ D . Otherwise, we find

(4.4) D = M ε , ν ( t ) ≤ C 0 ( M 0 ε , ν ) exp { ( t 1 4 + ε 1 − ς 2 ) C 1 ( M ε , ν ( t ) ) } < D ,

which leads to a contradiction. Besides, we can assume without restriction that D 0 < D , so that M ε , ν ( 0 ) < D . From the continuity of the function M ε , ν ( t ) , we obtain

(4.5) M ε , ν ( t ) < D ,

for t < min { T ε , ν , T 1 } , 0 < ε ≤ ε 1 , and 0 ≤ ν ≤ O ( ε ς ) with 0 < ς ≤ ς 1 . Consequently, we obtain T ε , ν > T 1 for 0 < ε ≤ ε 1 and 0 ≤ ν ≤ O ( ε ς ) with 0 < ς ≤ ς 1 . Otherwise, by using the uniform estimates in (4.5) and applying Theorem 1.1 repeatedly, the time interval of existence can be extended to [ 0 , T 1 ] , which contradicts to the maximality of T ε , ν . Therefore, M ε , ν ( t ) < D for any t ∈ [ 0 , T 1 ] , where T 1 is independent of ε ∈ ( 0 , 1 ] and 0 ≤ ν ≤ O ( ε ς ) with 0 < ς < 2 . Clearly, the conclusion is also true for T ε , ν = + ∞ due to the same argument. This completes the proof of Theorem 1.2.

4.2 Proof of Theorem 1.3.

According to the uniform estimates in (1.16), we obtain

(4.6) u ε ⇀ ω , in L 2 ( 0 , T ; H 3 ( Ω ) ) , ( u ε , B ε ) ⇀ ( ω , B ) weakly- * , in L ∞ ( 0 , T ; H 2 ( Ω ) ) , θ ε , ν ⇀ θ in L 2 ( 0 , T ; H 4 ( Ω ) ) and weakly- * , in L ∞ ( 0 , T ; H 3 ( Ω ) ) , ρ ε ⇀ r weakly- * , in L ∞ ( 0 , T ; H 2 ( Ω ) ) , ( u t ε , B t ε ) ⇀ ( ω t , B t ) in L 2 ( 0 , T ; H 1 ( Ω ) ) and weakly- * , in L ∞ ( 0 , T ; L 2 ( Ω ) ) , θ t ε , ν ⇀ θ t in L 2 ( 0 , T ; H 2 ( Ω ) ) and weakly- * , in L ∞ ( 0 , T ; H 1 ( Ω ) ) , ρ t ε ⇀ r t weakly- * , in L ∞ ( 0 , T ; L 2 ( Ω ) ) ,

which gives

(4.7) ( ρ ε , u ε , B ε ) → ( r , ω , B ) in C ( [ 0 , T ] , H 2 − δ ( Ω ) ) , θ ε , ν → θ in C ( [ 0 , T ] , H 3 − δ ( Ω ) ) ,

for any δ > 0 , by Lemma 2.6. Recall that

(4.8) p ε , ν = ρ ε ( 1 + θ ε , ν ) = 1 + ε σ ε , ν .

Thus, taking ε , ν → 0 , one derives r ( 1 + θ ) = 1 . Multiplying (1.4) 4 by ε and passing to the limit, we obtain (1.19) 3 . And by taking ε , ν → 0 in (1.4) 1 , (1.4) 3 , and (1.4) 5 , we have (1.19) 1 , (1.19) 3 , and (1.19) 5 . Next, apply curl to (1.4) 2 and take the convergence ε , ν → 0 in L 2 ( 0 , T ; L 2 ( Ω ) ) to obtain that

curl { r ( ω t + ω ⋅ ∇ ω ) − ( μ Δ ω + ( μ + λ ) ∇ div ω ) − 1 4 π ( curl B ) × B } = 0 .

Thus, there exists a function π ( x , t ) , such that ∇ π ∈ L 2 ( 0 , T ; H 1 ( Ω ) ) and (1.19) 2 is satisfied. Hence, (1.4) converges to (1.19) in L 2 ( 0 , T ; L 2 ( Ω ) ) as the Mach number and the magnetic resistivity coefficient vanish.

  1. Funding information: The research of this work was supported in part by National Natural Science Foundation of China under Grant Nos 12471216 and 12131007.

  2. Author contributions: All authors have contributed equally.

  3. Conflict of interest: The authors state no conflict of interest.

  4. Data availability statement: Data sharing is not applicable to this article as no datasets were generated or analyzed during the current study.

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Received: 2024-09-21
Revised: 2025-02-02
Accepted: 2025-04-30
Published Online: 2025-08-06

© 2025 the author(s), published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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https://doi.org/10.1515/anona-2025-0087
Keywords for this article
low Mach number limit; non-resistive limit; large temperature variations; magnetohydrodynamic equations
Creative Commons
BY 4.0

Articles in the same Issue

  1. Research Articles
  2. Incompressible limit for the compressible viscoelastic fluids in critical space
  3. Concentrating solutions for double critical fractional Schrödinger-Poisson system with p-Laplacian in ℝ3
  4. Intervals of bifurcation points for semilinear elliptic problems
  5. On multiplicity of solutions to nonlinear Dirac equation with local super-quadratic growth
  6. Entire radial bounded solutions for Leray-Lions equations of (p, q)-type
  7. Combinatorial pth Calabi flows for total geodesic curvatures in spherical background geometry
  8. Sharp upper bounds for the capacity in the hyperbolic and Euclidean spaces
  9. Positive solutions for asymptotically linear Schrödinger equation on hyperbolic space
  10. Existence and non-degeneracy of the normalized spike solutions to the fractional Schrödinger equations
  11. Existence results for non-coercive problems
  12. Ground states for Schrödinger-Poisson system with zero mass and the Coulomb critical exponent
  13. Geometric characterization of generalized Hajłasz-Sobolev embedding domains
  14. Subharmonic solutions of first-order Hamiltonian systems
  15. Sharp asymptotic expansions of entire large solutions to a class of k-Hessian equations with weights
  16. Stability of pyramidal traveling fronts in time-periodic reaction–diffusion equations with degenerate monostable and ignition nonlinearities
  17. Ground-state solutions for fractional Kirchhoff-Choquard equations with critical growth
  18. Existence results of variable exponent double-phase multivalued elliptic inequalities with logarithmic perturbation and convections
  19. Homoclinic solutions in periodic partial difference equations
  20. Classical solution for compressible Navier-Stokes-Korteweg equations with zero sound speed
  21. Properties of minimizers for L2-subcritical Kirchhoff energy functionals
  22. Asymptotic behavior of global mild solutions to the Keller-Segel-Navier-Stokes system in Lorentz spaces
  23. Blow-up solutions to the Hartree-Fock type Schrödinger equation with the critical rotational speed
  24. Qualitative properties of solutions for dual fractional parabolic equations involving nonlocal Monge-Ampère operator
  25. Regularity for double-phase functionals with nearly linear growth and two modulating coefficients
  26. Uniform boundedness and compactness for the commutator of an extension of Riesz transform on stratified Lie groups
  27. Normalized solutions to nonlinear Schrödinger equations with mixed fractional Laplacians
  28. Existence of positive radial solutions of general quasilinear elliptic systems
  29. Low Mach number and non-resistive limit of magnetohydrodynamic equations with large temperature variations in general bounded domains
  30. Sharp viscous shock waves for relaxation model with degeneracy
  31. Bifurcation and multiplicity results for critical problems involving the p-Grushin operator
  32. Asymptotic behavior of solutions of a free boundary model with seasonal succession and impulsive harvesting
  33. Blowing-up solutions concentrated along minimal submanifolds for some supercritical Hamiltonian systems on Riemannian manifolds
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  49. Nonsmooth analysis of doubly nonlinear second-order evolution equations with nonconvex energy functionals
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  57. Local well-posedness for the two-component Benjamin-Ono equation
  58. Self-similar blow-up solutions of the four-dimensional Schrödinger-Wave system
  59. Existence and stability of traveling waves in a Keller-Segel system with nonlinear stimulation
  60. Existence and multiplicity of solutions for a class of superlinear p-Laplacian equations
  61. On the global large regular solutions of the 1D degenerate compressible Navier-Stokes equations
  62. Normal forms of piecewise-smooth monodromic systems
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  65. Global boundedness and stability of a quasilinear two-species chemotaxis-competition model with nonlinear sensitivities
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  70. Hopf's lemma, asymptotic radial symmetry, and monotonicity of solutions to the logarithmic Laplacian parabolic system
  71. Generalized quasi-linear fractional Wentzell problems
  72. Existence, symmetry, and regularity of ground states of a nonlinear Choquard equation in the hyperbolic space
  73. Normalized solutions for NLS equations with general nonlinearity on compact metric graphs
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  75. Review Article
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