On non-resistive limit of 1D MHD equations with no vacuum at infinity
aa r X i v : . [ m a t h . A P ] F e b On non-resistive limit of 1D MHD equations with no vacuumat infinity
Zilai Li , Huaqiao Wang , Yulin Ye ∗ February 8, 2021 School of Mathematics and Information Science, Henan Polytechnic University,Jiaozuo 454000, P.R. China.
Email:[email protected] College of Mathematics and Statistics, Chongqing University,Chongqing 401331, P.R. China.
Email:[email protected] School of Mathematics and Statistics, Henan University,Kaifeng 475004, P.R. China.
Email:[email protected]
Abstract:
In this paper, we consider the Cauchy problem for the one-dimensionalcompressible isentropic magnetohydrodynamic (MHD) equations with no vacuum at in-finity, but the initial vacuum can be permitted inside the region. By deriving a priori ν (resistivity coefficient)-independent estimates, we establish the non-resistive limit of theglobal strong solutions with large initial data. Moreover, as a by-product, the globalwell-posedness of strong solutions for both the compressible resistive MHD equations andnon-resistive MHD equations are also established, respectively. Keywords:
1D compressible MHD equations; Cauchy problem; global strong solu-tions; non-resistive limit.
Compressible magnetohydrodynamics (MHD) is used to describe the macroscopic be-havior of the electrically conducting fluid in a magnetic field. The application of MHDhas a very wide of physical objects from liquid metals to cosmic plasmas. The system ofthe resistive MHD equations has the form: ρ t + ( ρu ) x = 0 , ( ρu ) t + ( ρu + P ( ρ ) + b ) x = ( µu x ) x ,b t + ( ub ) x = νb xx , (1.1)in R × [0 , ∞ ). Here, ρ, u, P ( ρ ) and b denote the density, velocity, pressure and magneticfield, respectively. µ > ν > ∗ Corresponding author. n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity
Compressible magnetohydrodynamics (MHD) is used to describe the macroscopic be-havior of the electrically conducting fluid in a magnetic field. The application of MHDhas a very wide of physical objects from liquid metals to cosmic plasmas. The system ofthe resistive MHD equations has the form: ρ t + ( ρu ) x = 0 , ( ρu ) t + ( ρu + P ( ρ ) + b ) x = ( µu x ) x ,b t + ( ub ) x = νb xx , (1.1)in R × [0 , ∞ ). Here, ρ, u, P ( ρ ) and b denote the density, velocity, pressure and magneticfield, respectively. µ > ν > ∗ Corresponding author. n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity P ( ρ ) = Rρ γ , γ > . For simplicity, we set R = 1. We focus on the initial condition:( ρ, u, b ) | t =0 = ( ρ ( x ) , u ( x ) , b ( x )) → (¯ ρ, , ¯ b ) as | x | → + ∞ , (1.2)where ¯ ρ and ¯ b are both non-zero constants. Note that we can always normalize ¯ ρ suchthat ¯ ρ ≥ ν is inversely proportional tothe electrical conductivity, therefore it is more reasonable to ignore the magnetic diffusionwhich means ν = 0, when the conducting fluid considered is of highly conductivity, forexample the ideal conductors. So instead of equations (1.1), when there is no resistivity,the system reduces to the so called compressible, isentropic, viscous and non-resistiveMHD equations: ˜ ρ t + (˜ ρ ˜ u ) x = 0 , (˜ ρ ˜ u ) t + (˜ ρ ˜ u + P (˜ ρ ) + ˜ b ) x = ( µ ˜ u x ) x , ˜ b t + (˜ u ˜ b ) x = 0 , (1.3)in R × [0 , + ∞ ), with the following initial condition:(˜ ρ, ˜ u, ˜ b ) | t =0 = (˜ ρ , ˜ u , ˜ b ) → (¯ ρ, , ¯ b ) as | x | → + ∞ . (1.4)Because of the tight interaction between the dynamic motion and the magnetic field,the presence of strong nonlinearities, rich phenomena and mathematical challenges, manyphysicists and mathematics are attracted to study in this field. Before stating our maintheorems, we briefly recall some previous known results on compressible MHD equations.Firstly, we begin with the MHD equations with magnetic diffusion. For one-dimensionalcase, Vol’pert-Hudjaev [21] proved the local existence and uniqueness of strong solutionsto the Cauchy problem and Kawashima-Okada [13] obtained the global smooth solutionswith small initial data. For large initial data and the density containing vacuum, Ye-Li [26]proved the global existence of strong solutions to the 1D Cauchy problem with vacuum atinfinity. When considering the full MHD equations and the heat conductivity depends onthe temperature θ , Chen-Wang [3] studied the free boundary value problem and establishedthe existence, uniqueness and Lipschitz dependence of strong solutions. Recently, Fan-Huang-Li [6] obtained the global strong solutions to the initial boundary value problemto the planner MHD equations with temperature-dependent heat conductivity. Later,with the effect of self-gravitation as well as the influence of radiation on the dynamics athigh temperature regimes taken into account, Zhang-Xie [28] obtained the global strongsolutions to the initial boundary value problem for the nonlinear planner MHD equations.For multi-dimensional MHD equations, Lv-Shi-Xu [18] considered the 2-D isentropic MHDequations and proved the global existence of classical solutions provided that the initialenergy is small, where the decay rates of the solutions were also obtained. Vol’pert-Hudjaev[21] and Fan-Yu [5] obtained the local classical solution to the 3-D compressible MHDequations with the initial density is strictly positive or could contain vacuum, respectively.Hu-Wang [9] derived the global weak solutions to the 3-D compressible MHD equationswith large initial data. Recently, Li-Xu-Zhang [14] established the global existence of n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity
Compressible magnetohydrodynamics (MHD) is used to describe the macroscopic be-havior of the electrically conducting fluid in a magnetic field. The application of MHDhas a very wide of physical objects from liquid metals to cosmic plasmas. The system ofthe resistive MHD equations has the form: ρ t + ( ρu ) x = 0 , ( ρu ) t + ( ρu + P ( ρ ) + b ) x = ( µu x ) x ,b t + ( ub ) x = νb xx , (1.1)in R × [0 , ∞ ). Here, ρ, u, P ( ρ ) and b denote the density, velocity, pressure and magneticfield, respectively. µ > ν > ∗ Corresponding author. n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity P ( ρ ) = Rρ γ , γ > . For simplicity, we set R = 1. We focus on the initial condition:( ρ, u, b ) | t =0 = ( ρ ( x ) , u ( x ) , b ( x )) → (¯ ρ, , ¯ b ) as | x | → + ∞ , (1.2)where ¯ ρ and ¯ b are both non-zero constants. Note that we can always normalize ¯ ρ suchthat ¯ ρ ≥ ν is inversely proportional tothe electrical conductivity, therefore it is more reasonable to ignore the magnetic diffusionwhich means ν = 0, when the conducting fluid considered is of highly conductivity, forexample the ideal conductors. So instead of equations (1.1), when there is no resistivity,the system reduces to the so called compressible, isentropic, viscous and non-resistiveMHD equations: ˜ ρ t + (˜ ρ ˜ u ) x = 0 , (˜ ρ ˜ u ) t + (˜ ρ ˜ u + P (˜ ρ ) + ˜ b ) x = ( µ ˜ u x ) x , ˜ b t + (˜ u ˜ b ) x = 0 , (1.3)in R × [0 , + ∞ ), with the following initial condition:(˜ ρ, ˜ u, ˜ b ) | t =0 = (˜ ρ , ˜ u , ˜ b ) → (¯ ρ, , ¯ b ) as | x | → + ∞ . (1.4)Because of the tight interaction between the dynamic motion and the magnetic field,the presence of strong nonlinearities, rich phenomena and mathematical challenges, manyphysicists and mathematics are attracted to study in this field. Before stating our maintheorems, we briefly recall some previous known results on compressible MHD equations.Firstly, we begin with the MHD equations with magnetic diffusion. For one-dimensionalcase, Vol’pert-Hudjaev [21] proved the local existence and uniqueness of strong solutionsto the Cauchy problem and Kawashima-Okada [13] obtained the global smooth solutionswith small initial data. For large initial data and the density containing vacuum, Ye-Li [26]proved the global existence of strong solutions to the 1D Cauchy problem with vacuum atinfinity. When considering the full MHD equations and the heat conductivity depends onthe temperature θ , Chen-Wang [3] studied the free boundary value problem and establishedthe existence, uniqueness and Lipschitz dependence of strong solutions. Recently, Fan-Huang-Li [6] obtained the global strong solutions to the initial boundary value problemto the planner MHD equations with temperature-dependent heat conductivity. Later,with the effect of self-gravitation as well as the influence of radiation on the dynamics athigh temperature regimes taken into account, Zhang-Xie [28] obtained the global strongsolutions to the initial boundary value problem for the nonlinear planner MHD equations.For multi-dimensional MHD equations, Lv-Shi-Xu [18] considered the 2-D isentropic MHDequations and proved the global existence of classical solutions provided that the initialenergy is small, where the decay rates of the solutions were also obtained. Vol’pert-Hudjaev[21] and Fan-Yu [5] obtained the local classical solution to the 3-D compressible MHDequations with the initial density is strictly positive or could contain vacuum, respectively.Hu-Wang [9] derived the global weak solutions to the 3-D compressible MHD equationswith large initial data. Recently, Li-Xu-Zhang [14] established the global existence of n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity γ − + ν − ) E is suitably small. When the resistivity is zero, then the magnetic equation is reducedfrom the heat-type equation to the hyperbolic-type equation, the problem becomes morechallenging, hence the results are few. Kawashima [12] obtained the classical solutionsto 3-D MHD equations when the initial data are of small perturbations in H -norm andaway from vacuum. Xu-Zhang [22] proved a blow-up criterion of strong solutions for 3-Disentropic MHD equations with vacuum. Fan-Hu [4] established the global strong solutionsto the initial boundary value problem of 1-D heat-conducting MHD equations. With moregeneral heat-conductivity, Zhang-Zhao [30] established the global strong solutions and alsoobtained the non-resistivity limits of the solutions in L -norm. Jiang-Zhang [10] obtainedthe non-resistive limit of the strong solution and the “magnetic boundary layer” estimatesto the initial boundary value problem of 1-D isentropic MHD equations as the resistivity ν →
0. Yu [27] obtained the global existence of strong solutions to the initial boundaryvalue problem of 1-D isentropic MHD equations. For the Cauchy problem with largeinitial data and vacuum, Li-Wang-Ye [16] established the global well-posedness of strongsolutions to the 1D isentropic MHD equations with vacuum at infinity, that is ¯ ρ = 0.However, for the Cauchy problem with no vacuum at infinity, the global well-posedness ofstrong solutions and the non-resistive limits when the resistivity coefficient ν → ν . Thus, to obtain the apriori ν (resistivity coefficient)-independent estimates, some of the main new difficultieswill be encountered due to the absence of resistivity and the initial density and initialmagnetic which approach non-zero constants at infinity.It turns out that the key issue in this paper is to derive upper bound for the density,magnetic field and the time-dependent higher norm estimates which are independent ofresistivity ν . This is achieved by modifying upper bound estimate for the density developedin [24] and [26] in the theory of Cauchy problem with vacuum and initial magnetic approachzero at infinity. However, in comparison with the Cauchy problem with vacuum at infinityin [24] and [16], some new difficulties will be encountered. The first difficulty lies in nointegrability for the density itself just from the elementary energy estimate (see Lemma3.1), which is required when deriving the upper bound of the density. To overcome thisdifficulty, we use the technique of mathematical frequency decomposition to divide themomentum ξ into two parts: ξ = ˆ ρudx = ˆ (cid:0) √ ρ − √ ¯ ρ (cid:1) √ ρudx + √ ¯ ρ ˆ √ ρudx = ξ + ξ . It is crucial to obtain the upper bound of ξ and ξ . For getting k ξ k L ∞ , we use thetechnique of mathematical frequency decomposition to get the estimate of k√ ρ − √ ¯ ρ k L by the elementary energy estimates, and then using H¨older’s inequality, we can obtainthe upper bound of ξ (see (3.23)). For obtaining k ξ k L ∞ , due to the Sobolev embeddingtheory, we need some L ˜ p integrability of ξ . However, we can not obtain it just from k ξ x k L directly, because the Poinc´are type inequality is no longer valid in the whole space R . To overcome this difficulty, we use the Caffarelli-Kohn-Nirenberg weighted inequalityand the G-N inequality to obtain the upper bound of ξ (see (3.24)). It is worth noting n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity
0. Yu [27] obtained the global existence of strong solutions to the initial boundaryvalue problem of 1-D isentropic MHD equations. For the Cauchy problem with largeinitial data and vacuum, Li-Wang-Ye [16] established the global well-posedness of strongsolutions to the 1D isentropic MHD equations with vacuum at infinity, that is ¯ ρ = 0.However, for the Cauchy problem with no vacuum at infinity, the global well-posedness ofstrong solutions and the non-resistive limits when the resistivity coefficient ν → ν . Thus, to obtain the apriori ν (resistivity coefficient)-independent estimates, some of the main new difficultieswill be encountered due to the absence of resistivity and the initial density and initialmagnetic which approach non-zero constants at infinity.It turns out that the key issue in this paper is to derive upper bound for the density,magnetic field and the time-dependent higher norm estimates which are independent ofresistivity ν . This is achieved by modifying upper bound estimate for the density developedin [24] and [26] in the theory of Cauchy problem with vacuum and initial magnetic approachzero at infinity. However, in comparison with the Cauchy problem with vacuum at infinityin [24] and [16], some new difficulties will be encountered. The first difficulty lies in nointegrability for the density itself just from the elementary energy estimate (see Lemma3.1), which is required when deriving the upper bound of the density. To overcome thisdifficulty, we use the technique of mathematical frequency decomposition to divide themomentum ξ into two parts: ξ = ˆ ρudx = ˆ (cid:0) √ ρ − √ ¯ ρ (cid:1) √ ρudx + √ ¯ ρ ˆ √ ρudx = ξ + ξ . It is crucial to obtain the upper bound of ξ and ξ . For getting k ξ k L ∞ , we use thetechnique of mathematical frequency decomposition to get the estimate of k√ ρ − √ ¯ ρ k L by the elementary energy estimates, and then using H¨older’s inequality, we can obtainthe upper bound of ξ (see (3.23)). For obtaining k ξ k L ∞ , due to the Sobolev embeddingtheory, we need some L ˜ p integrability of ξ . However, we can not obtain it just from k ξ x k L directly, because the Poinc´are type inequality is no longer valid in the whole space R . To overcome this difficulty, we use the Caffarelli-Kohn-Nirenberg weighted inequalityand the G-N inequality to obtain the upper bound of ξ (see (3.24)). It is worth noting n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity ρ, ˜ b ≥ ≤ ˜ b ˜ ρ < ∞ in (1.3), which implies the magnetic field ˜ b is bounded provided thedensity ˜ ρ is bounded. However, this is not physical and realistic in magnetohydrodynamics.Moreover, compared with that for the Cauchy problem with vacuum at infinity in [16],since the magnetic field ˜ b → ¯ b , as | x | → + ∞ , the method of getting the upper boundof magnetic field b in [16] can not be used here anymore. So another difficulty is howto get the uniform (independent of ν ) upper bound of the magnetic field b without theassumption similar as that in two-phase fluids. To overcome this difficulty, we will makefull use of the structure of the momentum equation and effective viscous flux (see Lemma3.4 and Lemma 3.5). Notations : We denote the material derivative of u and effective viscous flux by˙ u , u t + uu x and F , µu x − (cid:18) P ( ρ ) − P (¯ ρ ) + b − ¯ b (cid:19) , and define potential energy byΦ( ρ ) = ρ ˆ ρ ¯ ρ P ( s ) − P (¯ ρ ) s ds = 1 γ − (cid:0) ρ γ − ¯ ρ γ − γ ¯ ρ γ − ( ρ − ¯ ρ ) (cid:1) . Sometimes we write ´ R f ( x ) dx as ´ f ( x ) for simplicity.The main results of this paper can be stated as:As a by-product, the first result is the global existence of strong solutions to 1Dnon-resistive MHD equations (1.3)-(1.4) when the the initial density and initial magneticapproach non-zero constants at infinity. Theorem 1.1.
Suppose that the initial data (˜ ρ , ˜ u , ˜ b )( x ) satisfies˜ ρ − ¯ ρ ∈ H ( R ) , ˜ b − ¯ b ∈ H ( R ) , ˜ u ∈ H ( R ) ,
12 ˜ ρ ˜ u + Φ(˜ ρ ) + (˜ b − ¯ b ) ! | x | α ∈ L ( R ) (1.5)for 1 < α ≤
2, and the compatibility condition (cid:18) µ ˜ u x − P (˜ ρ ) − b (cid:19) x = p ˜ ρ ˜ g ( x ) , x ∈ R (1.6)with some ˜ g satisfying ˜ g ∈ L ( R ). Then for any T >
0, there exists a unique global strongsolution (˜ ρ, ˜ u, ˜ b ) to the Cauchy problem (1.3) and (1.4) such that0 ≤ ˜ ρ ≤ C, (˜ ρ − ¯ ρ, ˜ b − ¯ b ) ∈ L ∞ (0 , T ; H ( R )) , ˜ ρ t ∈ L ∞ (0 , T ; L ( R )) , ˜ b t ∈ L (0 , T ; L ( R )) ,
12 ˜ ρ ˜ u + Φ(˜ ρ ) + (˜ b − ¯ b ) ! | x | α ∈ L ∞ (0 , T ; L ( R )) , ˜ u ∈ L ∞ (0 , T ; H ( R )) , p ˜ ρ ˜ u t ∈ L ∞ (0 , T ; L ( R )) , ˜ u t ∈ L (0 , T ; H ( R )) . The second result is the global existence and non-resistive limits of strong solutions to1D MHD equations (1.1)-(1.2) when the the initial density and initial magnetic approachnon-zero constants at infinity. n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity
12 ˜ ρ ˜ u + Φ(˜ ρ ) + (˜ b − ¯ b ) ! | x | α ∈ L ∞ (0 , T ; L ( R )) , ˜ u ∈ L ∞ (0 , T ; H ( R )) , p ˜ ρ ˜ u t ∈ L ∞ (0 , T ; L ( R )) , ˜ u t ∈ L (0 , T ; H ( R )) . The second result is the global existence and non-resistive limits of strong solutions to1D MHD equations (1.1)-(1.2) when the the initial density and initial magnetic approachnon-zero constants at infinity. n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity Theorem 1.2.
Suppose that the initial data ( ρ , u , b )( x ) satisfies ρ − ¯ ρ ∈ H ( R ) , b − ¯ b ∈ H ( R ) , u ∈ H ( R ) , (cid:18) ρ u + Φ( ρ ) + ( b − ¯ b ) (cid:19) | x | α ∈ L ( R ) (1.7)for 1 < α ≤
2, and the compatibility condition (cid:18) µu x − P ( ρ ) − b (cid:19) x = √ ρ g ( x ) , x ∈ R (1.8)with some g satisfying g ∈ L ( R ). Then for each fixed ν >
0, there exist a positiveconstant C and a unique global strong solution ( ρ, u, b ) to the Cauchy problem (1.1)-(1.2)such that 0 ≤ ρ ( x, t ) ≤ C, ∀ ( x, t ) ∈ R × [0 , T ] , (1.9)sup ≤ t ≤ T (cid:13)(cid:13)(cid:13)(cid:13)(cid:18) ρu + Φ( ρ ) + ( b − ¯ b ) (cid:19) (1 + | x | α ) (cid:13)(cid:13)(cid:13)(cid:13) L ( t )+ ˆ T (cid:18) µ (cid:13)(cid:13)(cid:13) u x (1 + | x | α ) (cid:13)(cid:13)(cid:13) L + ν (cid:13)(cid:13)(cid:13) b x (1 + | x | α ) (cid:13)(cid:13)(cid:13) L (cid:19) dt ≤ C, (1.10)and sup ≤ t ≤ T ( k u k L + k u xx k L + k u x k L + k b x k L + k ρ x k L + k√ ρ ˙ u k L ) ( t )+ ˆ T (cid:0) k u xt k L + ν k b xx k L (cid:1) dt ≤ C. (1.11)Moreover, as ν →
0, we have ( ( ρ, u, b ) → (˜ ρ, ˜ u, ˜ b ) strongly in L ∞ (0 , T ; L ) ,νb x → , u → ˜ u, u x → ˜ u x strongly in L (0 , T ; L ) , (1.12)and sup ≤ t ≤ T (cid:16) k ρ − ˜ ρ k L + k u − ˜ u k L + k b − ˜ b k L (cid:17) + ˆ T µ k ( u − ˜ u ) x k L dt ≤ Cν, (1.13)where C is a positive constant independent of ν . Remark 1.1.
In Theorems 1.1 and 1.2, we do not need the artificial assumption similarlyas that in two-phase fluids. Moreover, if ignoring the magnetic field, then MHD systemreduces to the compressible Navier-Stokes equations. So, Theorems 1.1 and 1.2 can beseen as an extension of that in [24].
Remark 1.2.
In Theorem 1.2, we give that the global strong solution of resistive MHDequation (1.1)-(1.2) converges to that of non-resistive MHD equation (1.3)-(1.4) in L -norm as ν →
0, moreover, the convergence rates are also justified.The rest of the paper is organized as follows. In Section 2, we recall some preliminarylemmas which will be used later. Section 3 is devoted to establishing global ν -independentestimates for (1.1) and (1.2), which will be used to justify the non-resistive limit. Sections4 and 5 are devoted to proving Theorem 1.1 and Theorem 1.2, respectively. n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity
0, moreover, the convergence rates are also justified.The rest of the paper is organized as follows. In Section 2, we recall some preliminarylemmas which will be used later. Section 3 is devoted to establishing global ν -independentestimates for (1.1) and (1.2), which will be used to justify the non-resistive limit. Sections4 and 5 are devoted to proving Theorem 1.1 and Theorem 1.2, respectively. n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity In this section, we will recall some known facts and elementary inequalities that willbe used frequently later.The following well-known inequality will be used frequently later.
Lemma 2.1 ( Gagliardo-Nirenberg inequality [7, 19]).
For any f ∈ W ,m ( R ) ∩ L r ( R ) ,there exists some generic constant C > which may depend on q, r such that k f k L q ≤ C k f k − θL r k∇ f k θL m , (2.1) where θ = ( r − q )( r − m + 1) − , if m = 1 , then q ∈ [ r, ∞ ) , if m > , then q ∈ [ r, ∞ ] . The following Caffarelli-Kohn-Nirenberg weighted inequality is the key to deal withthe Cauchy problem in this paper.
Lemma 2.2 ( Caffarelli-Kohn-Nirenberg weighted inequality [1]). (1) ∀ h ∈ C ∞ ( R ) ,it holds that k| x | κ h k r ≤ C k| x | α | ∂ x h |k θp k| x | β h k − θq (2.2) where ≤ p, q < ∞ , < r < ∞ , ≤ θ ≤ , p + α > , q + β > , r + κ > and satisfy r + κ = θ (cid:18) p + α − (cid:19) + (1 − θ ) (cid:18) q + β (cid:19) , (2.3) and κ = θσ + (1 − θ ) β with ≤ α − σ if θ > and ≤ α − σ ≤ if θ > and p + α − r + κ .Proof. The proof of (1) can be found in [1]. Here, we omit the details.By direct calculation, the potential energy Φ( ρ, ¯ ρ ) has the following properties: Lemma 2.3.
Observing the function of the potential energy Φ( ρ, ¯ ρ ) , we will easily findthe following properties for positive constants c , c , C , C :(1) if ≤ ρ ≤ ρ , c ( ρ − ¯ ρ ) ≤ Φ( ρ, ¯ ρ ) ≤ c ( ρ − ¯ ρ ) ; (2) if ρ > ρ, ρ γ − ¯ ρ γ ≤ C ( ρ − ¯ ρ ) γ ≤ C Φ( ρ, ¯ ρ ) . ν -independent estimates for (1.1) and (1.2) The main purpose of this section is to derive the global ν -independent a priori estimatesof the solutions ( ρ, u, b ) to the system (1.1) and (1.2), which is used to justify the non-resistive limit. Hence, in this section, we assume ( ρ, u, b ) is a smooth solution of the system(1.1) and (1.2) on R × [0 , T ] with 0 < T < ∞ . For the sake of simplicity, we denote by C the generic positive constant, which may depend on γ, µ, T , but is independent of ν .First of all, we can prove the following elementary energy estimates. n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity
Observing the function of the potential energy Φ( ρ, ¯ ρ ) , we will easily findthe following properties for positive constants c , c , C , C :(1) if ≤ ρ ≤ ρ , c ( ρ − ¯ ρ ) ≤ Φ( ρ, ¯ ρ ) ≤ c ( ρ − ¯ ρ ) ; (2) if ρ > ρ, ρ γ − ¯ ρ γ ≤ C ( ρ − ¯ ρ ) γ ≤ C Φ( ρ, ¯ ρ ) . ν -independent estimates for (1.1) and (1.2) The main purpose of this section is to derive the global ν -independent a priori estimatesof the solutions ( ρ, u, b ) to the system (1.1) and (1.2), which is used to justify the non-resistive limit. Hence, in this section, we assume ( ρ, u, b ) is a smooth solution of the system(1.1) and (1.2) on R × [0 , T ] with 0 < T < ∞ . For the sake of simplicity, we denote by C the generic positive constant, which may depend on γ, µ, T , but is independent of ν .First of all, we can prove the following elementary energy estimates. n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity Lemma 3.1.
Let ( ρ, u, b ) be a smooth solution of (1.1) and (1.2) . Then for any T > ,it holds that sup ≤ t ≤ T (cid:13)(cid:13)(cid:13)(cid:13) ρu + Φ( ρ ) + ( b − ¯ b ) (cid:13)(cid:13)(cid:13)(cid:13) L ( R ) + ˆ T (cid:0) µ k u x k L + ν k b x k L (cid:1) dt ≤ C. (3.1) Proof.
By the defination of the potential energy Φ, and using (1.1) , we deduceΦ t + ( u Φ) x + ( ρ γ − ¯ ρ γ ) u x = 0 . (3.2)Adding the equation (1.1) multiplied by u , (1.1) multiplied by b , into (3.2), then inte-grating the resulting equation over R × [0 , T ] with respect to the variables x and t , wehave ˆ (cid:18) ρu + Φ( ρ ) + ( b − ¯ b ) (cid:19) dx + ˆ T (cid:0) µ k u x k L + ν k b x k L (cid:1) dt ≤ ˆ (cid:18) ρ u + Φ( ρ ) + ( b − ¯ b ) (cid:19) dx ≤ k ρ k L ∞ k u k L + C ( k ρ − ¯ ρ k L + k b − ¯ b k L ) ≤ C ( k ρ − ¯ ρ k H + ¯ ρ )( k u k L + 1) + C k b − ¯ b k L ≤ C. Then the proof of Lemma 3.1 is completed.To obtain the upper bound of the density ρ , we need the following weighted energyestimates. Lemma 3.2.
Let ( ρ, u, b ) be a smooth solution of (1.1) and (1.2) . Then for any T > and some index < α ≤ , it holds that sup ≤ t ≤ T (cid:13)(cid:13)(cid:13)(cid:13)(cid:18) ρu + Φ( ρ ) + ( b − ¯ b ) (cid:19) | x | α (cid:13)(cid:13)(cid:13)(cid:13) L ( R ) + ˆ T (cid:16) µ k u x | x | α k L + ν k b x | x | α k L (cid:17) dt ≤ C. (3.3) Proof.
Multiplying the equation (1.1) by u | x | α and integrating the resulting equationover R with respect to x , we have12 ddt ˆ ρu | x | α + µ ˆ u x | x | α = 12 ˆ ρu α | x | α − x − µ ˆ α | x | α − xuu x − ˆ (cid:18) P ( ρ ) + b (cid:19) x u | x | α . (3.4)It follows from the integration by parts that − ˆ (cid:18) P ( ρ ) + b (cid:19) x u | x | α = − ˆ (( P ( ρ ) − P (¯ ρ )) x + bb x ) u | x | α = − ˆ (cid:0) ( P ( ρ ) − P (¯ ρ )) x + ( b − ¯ b )( b − ¯ b ) x + ¯ b ( b − ¯ b ) x (cid:1) u | x | α = ˆ (cid:18) P ( ρ ) − P (¯ ρ ) + ( b − ¯ b ) b ( b − ¯ b ) (cid:19) (cid:0) u x | x | α + uα | x | α − x (cid:1) . (3.5) n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity
Multiplying the equation (1.1) by u | x | α and integrating the resulting equationover R with respect to x , we have12 ddt ˆ ρu | x | α + µ ˆ u x | x | α = 12 ˆ ρu α | x | α − x − µ ˆ α | x | α − xuu x − ˆ (cid:18) P ( ρ ) + b (cid:19) x u | x | α . (3.4)It follows from the integration by parts that − ˆ (cid:18) P ( ρ ) + b (cid:19) x u | x | α = − ˆ (( P ( ρ ) − P (¯ ρ )) x + bb x ) u | x | α = − ˆ (cid:0) ( P ( ρ ) − P (¯ ρ )) x + ( b − ¯ b )( b − ¯ b ) x + ¯ b ( b − ¯ b ) x (cid:1) u | x | α = ˆ (cid:18) P ( ρ ) − P (¯ ρ ) + ( b − ¯ b ) b ( b − ¯ b ) (cid:19) (cid:0) u x | x | α + uα | x | α − x (cid:1) . (3.5) n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity ddt ˆ ρu | x | α + µ ˆ u x | x | α = 12 ˆ ρu α | x | α − x − µ ˆ α | x | α − xuu x + ˆ (cid:18) P ( ρ ) − P (¯ ρ ) + ( b − ¯ b ) b ( b − ¯ b ) (cid:19) (cid:0) u x | x | α + uα | x | α − x (cid:1) . (3.6)To deal with the last term on the right-hand side of (3.6), first, multiplying (3.2) by | x | α and integrating over R with respect to x , yields ddt ˆ Φ( ρ ) | x | α − ˆ u Φ( ρ ) α | x | α − x + ˆ ( ρ γ − ¯ ρ γ ) | x | α u x = 0 , (3.7)and then multiplying the equation (1.1) by ( b − ¯ b ) | x | α and integrating over R with respectto x , we have12 ddt ˆ ( b − ¯ b ) | x | α + 12 ˆ (cid:0) ( b − ¯ b ) (cid:1) x u | x | α + ˆ ( b − ¯ b ) u x | x | α + ¯ b ˆ ( b − ¯ b ) u x | x | α = ν ˆ b xx ( b − ¯ b ) | x | α , which together with the integration by parts implies that12 ddt ˆ ( b − ¯ b ) | x | α + ν ˆ b x | x | α = − ˆ ( b − ¯ b ) u x | x | α − uα | x | α − x ) − ¯ b ˆ ( b − ¯ b ) u x | x | α − ν ˆ b x ( b − ¯ b ) α | x | α − x. (3.8)Now, putting (3.7) and (3.8) into (3.6), we have ddt ˆ (cid:18) ρu + Φ( ρ ) + ( b − ¯ b ) (cid:19) | x | α dx + µ ˆ u x | x | α + ν ˆ b x | x | α = ˆ (cid:18) ρu + Φ( ρ ) + ( b − ¯ b ) (cid:19) uα | x | α − x + ˆ ( ρ γ − ¯ ρ γ ) uα | x | α − x − µ ˆ α | x | α − xuu x + ¯ b ˆ ( b − ¯ b ) uα | x | α − x − ν ˆ ( b − ¯ b ) b x α | x | α − x =: I + I + I + I + I . (3.9)Next, we estimate the terms I - I as follows: I ≤ ˆ (cid:18)(cid:12)(cid:12)(cid:12)(cid:12) ρu + Φ( ρ ) + ( b − ¯ b ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:19) | u || x | α − ≤ ˆ (cid:18)(cid:12)(cid:12)(cid:12)(cid:12) ρu + Φ( ρ ) + ( b − ¯ b ) (cid:12)(cid:12)(cid:12)(cid:12) | x | α (cid:19) α − α (cid:18)(cid:12)(cid:12)(cid:12)(cid:12) ρu + Φ( ρ ) + ( b − ¯ b ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:19) α | u |≤ C k u k L ∞ (cid:13)(cid:13)(cid:13)(cid:13)(cid:18) ρu + Φ( ρ ) + ( b − ¯ b ) (cid:19) | x | α (cid:13)(cid:13)(cid:13)(cid:13) − α L (cid:13)(cid:13)(cid:13)(cid:13) ρu + Φ( ρ ) + ( b − ¯ b ) (cid:13)(cid:13)(cid:13)(cid:13) α L ≤ C (1 + k u x k L ) (cid:13)(cid:13)(cid:13)(cid:13)(cid:18) ρu + Φ( ρ ) + ( b − ¯ b ) (cid:19) | x | α (cid:13)(cid:13)(cid:13)(cid:13) − α L ≤ C (1 + k u x k L ) (cid:18)(cid:13)(cid:13)(cid:13)(cid:13)(cid:18) ρu + Φ( ρ ) + ( b − ¯ b ) (cid:19) | x | α (cid:13)(cid:13)(cid:13)(cid:13) L + 1 (cid:19) , (3.10) n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity
Multiplying the equation (1.1) by u | x | α and integrating the resulting equationover R with respect to x , we have12 ddt ˆ ρu | x | α + µ ˆ u x | x | α = 12 ˆ ρu α | x | α − x − µ ˆ α | x | α − xuu x − ˆ (cid:18) P ( ρ ) + b (cid:19) x u | x | α . (3.4)It follows from the integration by parts that − ˆ (cid:18) P ( ρ ) + b (cid:19) x u | x | α = − ˆ (( P ( ρ ) − P (¯ ρ )) x + bb x ) u | x | α = − ˆ (cid:0) ( P ( ρ ) − P (¯ ρ )) x + ( b − ¯ b )( b − ¯ b ) x + ¯ b ( b − ¯ b ) x (cid:1) u | x | α = ˆ (cid:18) P ( ρ ) − P (¯ ρ ) + ( b − ¯ b ) b ( b − ¯ b ) (cid:19) (cid:0) u x | x | α + uα | x | α − x (cid:1) . (3.5) n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity ddt ˆ ρu | x | α + µ ˆ u x | x | α = 12 ˆ ρu α | x | α − x − µ ˆ α | x | α − xuu x + ˆ (cid:18) P ( ρ ) − P (¯ ρ ) + ( b − ¯ b ) b ( b − ¯ b ) (cid:19) (cid:0) u x | x | α + uα | x | α − x (cid:1) . (3.6)To deal with the last term on the right-hand side of (3.6), first, multiplying (3.2) by | x | α and integrating over R with respect to x , yields ddt ˆ Φ( ρ ) | x | α − ˆ u Φ( ρ ) α | x | α − x + ˆ ( ρ γ − ¯ ρ γ ) | x | α u x = 0 , (3.7)and then multiplying the equation (1.1) by ( b − ¯ b ) | x | α and integrating over R with respectto x , we have12 ddt ˆ ( b − ¯ b ) | x | α + 12 ˆ (cid:0) ( b − ¯ b ) (cid:1) x u | x | α + ˆ ( b − ¯ b ) u x | x | α + ¯ b ˆ ( b − ¯ b ) u x | x | α = ν ˆ b xx ( b − ¯ b ) | x | α , which together with the integration by parts implies that12 ddt ˆ ( b − ¯ b ) | x | α + ν ˆ b x | x | α = − ˆ ( b − ¯ b ) u x | x | α − uα | x | α − x ) − ¯ b ˆ ( b − ¯ b ) u x | x | α − ν ˆ b x ( b − ¯ b ) α | x | α − x. (3.8)Now, putting (3.7) and (3.8) into (3.6), we have ddt ˆ (cid:18) ρu + Φ( ρ ) + ( b − ¯ b ) (cid:19) | x | α dx + µ ˆ u x | x | α + ν ˆ b x | x | α = ˆ (cid:18) ρu + Φ( ρ ) + ( b − ¯ b ) (cid:19) uα | x | α − x + ˆ ( ρ γ − ¯ ρ γ ) uα | x | α − x − µ ˆ α | x | α − xuu x + ¯ b ˆ ( b − ¯ b ) uα | x | α − x − ν ˆ ( b − ¯ b ) b x α | x | α − x =: I + I + I + I + I . (3.9)Next, we estimate the terms I - I as follows: I ≤ ˆ (cid:18)(cid:12)(cid:12)(cid:12)(cid:12) ρu + Φ( ρ ) + ( b − ¯ b ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:19) | u || x | α − ≤ ˆ (cid:18)(cid:12)(cid:12)(cid:12)(cid:12) ρu + Φ( ρ ) + ( b − ¯ b ) (cid:12)(cid:12)(cid:12)(cid:12) | x | α (cid:19) α − α (cid:18)(cid:12)(cid:12)(cid:12)(cid:12) ρu + Φ( ρ ) + ( b − ¯ b ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:19) α | u |≤ C k u k L ∞ (cid:13)(cid:13)(cid:13)(cid:13)(cid:18) ρu + Φ( ρ ) + ( b − ¯ b ) (cid:19) | x | α (cid:13)(cid:13)(cid:13)(cid:13) − α L (cid:13)(cid:13)(cid:13)(cid:13) ρu + Φ( ρ ) + ( b − ¯ b ) (cid:13)(cid:13)(cid:13)(cid:13) α L ≤ C (1 + k u x k L ) (cid:13)(cid:13)(cid:13)(cid:13)(cid:18) ρu + Φ( ρ ) + ( b − ¯ b ) (cid:19) | x | α (cid:13)(cid:13)(cid:13)(cid:13) − α L ≤ C (1 + k u x k L ) (cid:18)(cid:13)(cid:13)(cid:13)(cid:13)(cid:18) ρu + Φ( ρ ) + ( b − ¯ b ) (cid:19) | x | α (cid:13)(cid:13)(cid:13)(cid:13) L + 1 (cid:19) , (3.10) n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity α > ρ ˆ u = ˆ (¯ ρ − ρ + ρ ) u ≤ ˆ (cid:0) (¯ ρ − ρ ) | { ≤ ρ ≤ ρ } + (¯ ρ − ρ ) | { ρ> ρ } (cid:1) u + ˆ ρu ≤ C (cid:18) k ( ρ − ¯ ρ ) | { ≤ ρ ≤ ρ } k L k u k L + k ( ρ − ¯ ρ ) | { ρ> ρ } k L γ k u k L γγ − + 1 (cid:19) ≤ C " k Φ( ρ ) k L (cid:18) k u k L k u x k L (cid:19) + k Φ( ρ ) k γ L (cid:18) k u k − γ L k u x k γ L (cid:19) + 1 ≤ ε k u k L + C (cid:0) k u x k L (cid:1) , where we have used the properties of the potential energy Φ in Lemma 2.3. This togetherwith G-N inequality implies that k u k L ≤ C (1 + k u x k L ) , (3.11)and k u k L ∞ ≤ C k u k L k u x k L ≤ C (1 + k u x k L ) . (3.12)Applying the property of Φ( ρ ), H¨older’s inequality, the C-K-N weighted inequality (2.3),Lemma 3.1, Young’s inequality and (3.12), we obtain I = ˆ (cid:0) ( ρ γ − ¯ ρ γ ) | { ≤ ρ ≤ ρ } + ( ρ γ − ¯ ρ γ ) | { ρ> ρ } (cid:1) uα | x | α − x ≤ C ˆ (cid:0) | ρ − ¯ ρ || { ≤ ρ ≤ ρ } + ( ρ − ¯ ρ ) γ | { ρ> ρ } (cid:1) | u || x | α − ≤ C ˆ (cid:16) Φ ( ρ ) | { ≤ ρ ≤ ρ } + Φ( ρ ) | { ρ> ρ } (cid:17) | u || x | α − ≤ C ˆ (cid:16) Φ ( ρ ) | { ≤ ρ ≤ ρ } | x | α (cid:17) | u || x | α − + C ˆ (cid:0) Φ( ρ ) | { ρ> ρ } | x | α (cid:1) α − α Φ α ( ρ ) | u |≤ C (cid:18) k Φ( ρ ) | x | α k L (cid:13)(cid:13)(cid:13) | x | α − u (cid:13)(cid:13)(cid:13) L + k Φ( ρ ) | x | α k − α L k Φ( ρ ) k α L k u k L ∞ (cid:19) ≤ + C (1 + k u x k L )(1 + k Φ( ρ ) | x | α k L ) , (3.13)where we have used the fact: k| x | α − u k L ≤ C k u k α L k u x k − α L ≤ C ( k u k L + k u x k L ) ≤ C (1 + k u x k L ) , (3.14)here, the index 1 < α ≤ I ≤ C k| x | α u x k L k| x | α − u k L ≤ ε k| x | α u x k L + C (cid:0) k u x k L (cid:1) . (3.15) n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity
Multiplying the equation (1.1) by u | x | α and integrating the resulting equationover R with respect to x , we have12 ddt ˆ ρu | x | α + µ ˆ u x | x | α = 12 ˆ ρu α | x | α − x − µ ˆ α | x | α − xuu x − ˆ (cid:18) P ( ρ ) + b (cid:19) x u | x | α . (3.4)It follows from the integration by parts that − ˆ (cid:18) P ( ρ ) + b (cid:19) x u | x | α = − ˆ (( P ( ρ ) − P (¯ ρ )) x + bb x ) u | x | α = − ˆ (cid:0) ( P ( ρ ) − P (¯ ρ )) x + ( b − ¯ b )( b − ¯ b ) x + ¯ b ( b − ¯ b ) x (cid:1) u | x | α = ˆ (cid:18) P ( ρ ) − P (¯ ρ ) + ( b − ¯ b ) b ( b − ¯ b ) (cid:19) (cid:0) u x | x | α + uα | x | α − x (cid:1) . (3.5) n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity ddt ˆ ρu | x | α + µ ˆ u x | x | α = 12 ˆ ρu α | x | α − x − µ ˆ α | x | α − xuu x + ˆ (cid:18) P ( ρ ) − P (¯ ρ ) + ( b − ¯ b ) b ( b − ¯ b ) (cid:19) (cid:0) u x | x | α + uα | x | α − x (cid:1) . (3.6)To deal with the last term on the right-hand side of (3.6), first, multiplying (3.2) by | x | α and integrating over R with respect to x , yields ddt ˆ Φ( ρ ) | x | α − ˆ u Φ( ρ ) α | x | α − x + ˆ ( ρ γ − ¯ ρ γ ) | x | α u x = 0 , (3.7)and then multiplying the equation (1.1) by ( b − ¯ b ) | x | α and integrating over R with respectto x , we have12 ddt ˆ ( b − ¯ b ) | x | α + 12 ˆ (cid:0) ( b − ¯ b ) (cid:1) x u | x | α + ˆ ( b − ¯ b ) u x | x | α + ¯ b ˆ ( b − ¯ b ) u x | x | α = ν ˆ b xx ( b − ¯ b ) | x | α , which together with the integration by parts implies that12 ddt ˆ ( b − ¯ b ) | x | α + ν ˆ b x | x | α = − ˆ ( b − ¯ b ) u x | x | α − uα | x | α − x ) − ¯ b ˆ ( b − ¯ b ) u x | x | α − ν ˆ b x ( b − ¯ b ) α | x | α − x. (3.8)Now, putting (3.7) and (3.8) into (3.6), we have ddt ˆ (cid:18) ρu + Φ( ρ ) + ( b − ¯ b ) (cid:19) | x | α dx + µ ˆ u x | x | α + ν ˆ b x | x | α = ˆ (cid:18) ρu + Φ( ρ ) + ( b − ¯ b ) (cid:19) uα | x | α − x + ˆ ( ρ γ − ¯ ρ γ ) uα | x | α − x − µ ˆ α | x | α − xuu x + ¯ b ˆ ( b − ¯ b ) uα | x | α − x − ν ˆ ( b − ¯ b ) b x α | x | α − x =: I + I + I + I + I . (3.9)Next, we estimate the terms I - I as follows: I ≤ ˆ (cid:18)(cid:12)(cid:12)(cid:12)(cid:12) ρu + Φ( ρ ) + ( b − ¯ b ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:19) | u || x | α − ≤ ˆ (cid:18)(cid:12)(cid:12)(cid:12)(cid:12) ρu + Φ( ρ ) + ( b − ¯ b ) (cid:12)(cid:12)(cid:12)(cid:12) | x | α (cid:19) α − α (cid:18)(cid:12)(cid:12)(cid:12)(cid:12) ρu + Φ( ρ ) + ( b − ¯ b ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:19) α | u |≤ C k u k L ∞ (cid:13)(cid:13)(cid:13)(cid:13)(cid:18) ρu + Φ( ρ ) + ( b − ¯ b ) (cid:19) | x | α (cid:13)(cid:13)(cid:13)(cid:13) − α L (cid:13)(cid:13)(cid:13)(cid:13) ρu + Φ( ρ ) + ( b − ¯ b ) (cid:13)(cid:13)(cid:13)(cid:13) α L ≤ C (1 + k u x k L ) (cid:13)(cid:13)(cid:13)(cid:13)(cid:18) ρu + Φ( ρ ) + ( b − ¯ b ) (cid:19) | x | α (cid:13)(cid:13)(cid:13)(cid:13) − α L ≤ C (1 + k u x k L ) (cid:18)(cid:13)(cid:13)(cid:13)(cid:13)(cid:18) ρu + Φ( ρ ) + ( b − ¯ b ) (cid:19) | x | α (cid:13)(cid:13)(cid:13)(cid:13) L + 1 (cid:19) , (3.10) n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity α > ρ ˆ u = ˆ (¯ ρ − ρ + ρ ) u ≤ ˆ (cid:0) (¯ ρ − ρ ) | { ≤ ρ ≤ ρ } + (¯ ρ − ρ ) | { ρ> ρ } (cid:1) u + ˆ ρu ≤ C (cid:18) k ( ρ − ¯ ρ ) | { ≤ ρ ≤ ρ } k L k u k L + k ( ρ − ¯ ρ ) | { ρ> ρ } k L γ k u k L γγ − + 1 (cid:19) ≤ C " k Φ( ρ ) k L (cid:18) k u k L k u x k L (cid:19) + k Φ( ρ ) k γ L (cid:18) k u k − γ L k u x k γ L (cid:19) + 1 ≤ ε k u k L + C (cid:0) k u x k L (cid:1) , where we have used the properties of the potential energy Φ in Lemma 2.3. This togetherwith G-N inequality implies that k u k L ≤ C (1 + k u x k L ) , (3.11)and k u k L ∞ ≤ C k u k L k u x k L ≤ C (1 + k u x k L ) . (3.12)Applying the property of Φ( ρ ), H¨older’s inequality, the C-K-N weighted inequality (2.3),Lemma 3.1, Young’s inequality and (3.12), we obtain I = ˆ (cid:0) ( ρ γ − ¯ ρ γ ) | { ≤ ρ ≤ ρ } + ( ρ γ − ¯ ρ γ ) | { ρ> ρ } (cid:1) uα | x | α − x ≤ C ˆ (cid:0) | ρ − ¯ ρ || { ≤ ρ ≤ ρ } + ( ρ − ¯ ρ ) γ | { ρ> ρ } (cid:1) | u || x | α − ≤ C ˆ (cid:16) Φ ( ρ ) | { ≤ ρ ≤ ρ } + Φ( ρ ) | { ρ> ρ } (cid:17) | u || x | α − ≤ C ˆ (cid:16) Φ ( ρ ) | { ≤ ρ ≤ ρ } | x | α (cid:17) | u || x | α − + C ˆ (cid:0) Φ( ρ ) | { ρ> ρ } | x | α (cid:1) α − α Φ α ( ρ ) | u |≤ C (cid:18) k Φ( ρ ) | x | α k L (cid:13)(cid:13)(cid:13) | x | α − u (cid:13)(cid:13)(cid:13) L + k Φ( ρ ) | x | α k − α L k Φ( ρ ) k α L k u k L ∞ (cid:19) ≤ + C (1 + k u x k L )(1 + k Φ( ρ ) | x | α k L ) , (3.13)where we have used the fact: k| x | α − u k L ≤ C k u k α L k u x k − α L ≤ C ( k u k L + k u x k L ) ≤ C (1 + k u x k L ) , (3.14)here, the index 1 < α ≤ I ≤ C k| x | α u x k L k| x | α − u k L ≤ ε k| x | α u x k L + C (cid:0) k u x k L (cid:1) . (3.15) n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity I , it follows from the H¨older inequality and C-K-N weighted inequality that I ≤ C (cid:13)(cid:13)(cid:13) ( b − ¯ b ) | x | α (cid:13)(cid:13)(cid:13) L (cid:13)(cid:13)(cid:13) | x | α − u (cid:13)(cid:13)(cid:13) L ≤ C (cid:18)(cid:13)(cid:13)(cid:13) ( b − ¯ b ) | x | α (cid:13)(cid:13)(cid:13) L + k u x k L + 1 (cid:19) . (3.16)Similarly, using the C-K-N weighted inequality (2.3), one has I ≤ Cν k| x | α b x k L k ( b − ¯ b ) | x | α − k L ≤ Cν k| x | α b x k L (cid:16) k b − ¯ b k α L k b x k − α L (cid:17) ≤ εν k| x | α b x k L + Cν (cid:0) k b x k L (cid:1) , (3.17)where 1 < α ≤ ε > ddt ˆ (cid:18) ρu + Φ( ρ ) + ( b − ¯ b ) (cid:19) | x | α + µ ˆ | x | α u x + ν ˆ | x | α b x ≤ C (1 + k u x k L ) (cid:13)(cid:13)(cid:13)(cid:13)(cid:18) ρu + Φ( ρ ) + ( b − ¯ b ) (cid:19) | x | α (cid:13)(cid:13)(cid:13)(cid:13) L + C (1 + k u x k L + ν k b x k L ) . This together with Gronwall’s inequality, (1.7) and Lemma 3.1 gives ˆ (cid:18) ρu + Φ( ρ ) + ( b − ¯ b ) (cid:19) | x | α + µ ˆ T ˆ | x | α u x + ν ˆ T ˆ | x | α b x ≤ C ( T ) . Then, the proof of Lemma 3.2 is completed.The upper bound of the density ρ can be shown in the similar manner as that in [24].However, for completeness of the paper, we give the details here. Lemma 3.3.
Let ( ρ, u, b ) be a smooth solution of (1.1) and (1.2) . Then for any T > ,it holds that ≤ ρ ( x, t ) ≤ C, ∀ ( x, t ) ∈ R × [0 , T ] , and sup ≤ t ≤ T k ρ − ¯ ρ k L ( R ) ≤ C. Proof.
Let ξ = ´ x −∞ ρudy , then the momentum equation (1.1) can be rewritten as ξ tx + (cid:18) ρu + P ( ρ ) − P (¯ ρ ) + b − ¯ b (cid:19) x = ( µu x ) x . Integrating the above equality with respect to x over ( −∞ , x ) yields that ξ t + ρu + P ( ρ ) − P (¯ ρ ) + b − ¯ b µu x , (3.18) n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity
Let ξ = ´ x −∞ ρudy , then the momentum equation (1.1) can be rewritten as ξ tx + (cid:18) ρu + P ( ρ ) − P (¯ ρ ) + b − ¯ b (cid:19) x = ( µu x ) x . Integrating the above equality with respect to x over ( −∞ , x ) yields that ξ t + ρu + P ( ρ ) − P (¯ ρ ) + b − ¯ b µu x , (3.18) n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity gives ξ t + ρu + P ( ρ ) − P (¯ ρ ) + b − ¯ b µ ρ t + uρ x ρ = 0 . (3.19)Next, we define the particle trajectory X ( x, t ) as follows: ( dX ( x,t ) dt = u ( X ( x, t ) , t ) ,X ( x,
0) = x, (3.20)which implies dξdt = ξ t + uξ x = ξ t + ρu . Then combining the above equality and (3.19), we infer that ddt ( ξ + µ ln ρ ) ( X ( x, t ) , t ) + (cid:18) P ( ρ ) + b (cid:19) ( X ( x, t ) , t ) − (cid:18) P (¯ ρ ) + ¯ b (cid:19) = 0 , which together with (cid:16) P ( ρ ) + b (cid:17) ( X ( x, t ) , t ) ≥ ddt ( ξ + µ ln ρ ) ( X ( x, t ) , t ) ≤ (cid:18) P (¯ ρ ) + ¯ b (cid:19) ( X ( x, t ) , t ) ≤ C. (3.21)Thus, integrating it over [0 , T ] with respect to t , we have( ξ + µ ln ρ ) ( X ( x, t ) , t ) ≤ ( ξ + µ ln ρ )( x,
0) + C. By direct calculation, we obtainln ρ ≤ µ ( ξ + µ ln ρ + C − ξ ) ≤ µ (cid:18) ξ + µ ln ρ + C − ˆ x −∞ ρudy (cid:19) ≤ µ (cid:18) ξ + µ ln ρ + C − ˆ x −∞ √ ρu ( √ ρ − √ ¯ ρ ) dy − √ ¯ ρ ˆ x −∞ √ ρudy (cid:19) ≤ µ ( ξ + ln ρ + C − ξ − ξ ) . (3.22)Firstly, it follows from Lemma 2.3 and Lemma 3.1 that k ξ k L ∞ can be estimated as | ξ | = | ˆ x −∞ √ ρu ( √ ρ − √ ¯ ρ ) dy |≤ k√ ρu k L (cid:0) k ( √ ρ − √ ¯ ρ ) | { ≤ ρ ≤ ρ } k L + k ( √ ρ − √ ¯ ρ ) | { ρ> ρ } k L (cid:1) ≤ C k√ ρu k L (cid:0) k ( ρ − ¯ ρ ) | { ≤ ρ ≤ ρ } k L + k√ ρ − ¯ ρ | { ρ> ρ } k L (cid:1) ≤ C k√ ρu k L (cid:18) k ( ρ − ¯ ρ ) | { ≤ ρ ≤ ρ } k L + k ( ρ − ¯ ρ ) { ρ> ρ } k L (cid:19) ≤ C k√ ρu k L (cid:18) k ( ρ − ¯ ρ ) | { ≤ ρ ≤ ρ } k L + k ( ρ − ¯ ρ ) γ | { ρ> ρ } k L (cid:19) ≤ C k√ ρu k L (cid:18) k Φ( ρ ) k L + k Φ( ρ ) k L (cid:19) ≤ C. (3.23) n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity
0) + C. By direct calculation, we obtainln ρ ≤ µ ( ξ + µ ln ρ + C − ξ ) ≤ µ (cid:18) ξ + µ ln ρ + C − ˆ x −∞ ρudy (cid:19) ≤ µ (cid:18) ξ + µ ln ρ + C − ˆ x −∞ √ ρu ( √ ρ − √ ¯ ρ ) dy − √ ¯ ρ ˆ x −∞ √ ρudy (cid:19) ≤ µ ( ξ + ln ρ + C − ξ − ξ ) . (3.22)Firstly, it follows from Lemma 2.3 and Lemma 3.1 that k ξ k L ∞ can be estimated as | ξ | = | ˆ x −∞ √ ρu ( √ ρ − √ ¯ ρ ) dy |≤ k√ ρu k L (cid:0) k ( √ ρ − √ ¯ ρ ) | { ≤ ρ ≤ ρ } k L + k ( √ ρ − √ ¯ ρ ) | { ρ> ρ } k L (cid:1) ≤ C k√ ρu k L (cid:0) k ( ρ − ¯ ρ ) | { ≤ ρ ≤ ρ } k L + k√ ρ − ¯ ρ | { ρ> ρ } k L (cid:1) ≤ C k√ ρu k L (cid:18) k ( ρ − ¯ ρ ) | { ≤ ρ ≤ ρ } k L + k ( ρ − ¯ ρ ) { ρ> ρ } k L (cid:19) ≤ C k√ ρu k L (cid:18) k ( ρ − ¯ ρ ) | { ≤ ρ ≤ ρ } k L + k ( ρ − ¯ ρ ) γ | { ρ> ρ } k L (cid:19) ≤ C k√ ρu k L (cid:18) k Φ( ρ ) k L + k Φ( ρ ) k L (cid:19) ≤ C. (3.23) n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity k ξ k L ∞ as k ξ k L ∞ ≤ C k ξ k ˜ p ˜ p +2 L ˜ p k ξ x k p +2 L ≤ C ( k ξ x k ηL k| x | κ ξ k − ηL q ) ˜ p ˜ p +2 k ξ x k p +2 L ≤ C ( k ξ x k ηL k| x | α ξ x k − ηL ) ˜ p ˜ p +2 k ξ x k p +2 L ≤ C k ξ x k − (1 − η ) ˜ p ˜ p +2 L k| x | α ξ x k (1 − η ) ˜ p ˜ p +2 L = C k ξ x k − α L k| x | α ξ x k α L ≤ C ( k√ ρu k − α L k| x | α √ ρu k α L ) ≤ C. (3.24)Here the indexes 1 ≤ ˜ p < ∞ , q > , η ∈ (0 ,
1) and satisfy1˜ p = ( 12 − η + (cid:18) q + κ (cid:19) (1 − η ) , q + κ = 12 + α − > , which gives α > , η = ˜ p ( α − − α ˜ p > ⇒ ˜ p > α − . (3.25)Similarly as that for (3.23) and (3.24), we can obtain | ξ | ≤ C. (3.26)Then, substituting (3.23), (3.24) and (3.26) into (3.21), we haveln ρ ≤ C, which gives ρ ≤ C. It follows from Lemma 2.3 and Lemma 3.1 that k ρ − ¯ ρ k L = (cid:13)(cid:13)(cid:13) ( ρ − ¯ ρ ) | { ρ ≤ ρ } (cid:13)(cid:13)(cid:13) L + (cid:13)(cid:13) ( ρ − ¯ ρ ) | { ρ ≥ ρ } (cid:13)(cid:13) L ≤ k Φ( ρ ) k L + (cid:13)(cid:13) ( ρ − ¯ ρ ) | { ρ ≥ ρ } (cid:13)(cid:13) L ≤ C + (cid:13)(cid:13) ( ρ − ¯ ρ ) | { ρ ≥ ρ } (cid:13)(cid:13) L . (3.27)To deal with the last term on the right-hand side of (3.27), we discuss it in the followingtwo cases:Case 1: if 1 < γ ≤
2, by Lemma 2.3, Lemma3.1 and the fact that ρ ≤ C , it holds k ( ρ − ¯ ρ ) | { ρ ≥ ρ } k L ( R ) ≤ k ρ − ¯ ρ k γ L γ ( R ) k ρ − ¯ ρ k − γ L ∞ ( R ) ≤ k Φ( ρ ) k L ( R ) k ρ − ¯ ρ k − γ L ∞ ( R ) ≤ C ( T ) . (3.28) n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity
2, by Lemma 2.3, Lemma3.1 and the fact that ρ ≤ C , it holds k ( ρ − ¯ ρ ) | { ρ ≥ ρ } k L ( R ) ≤ k ρ − ¯ ρ k γ L γ ( R ) k ρ − ¯ ρ k − γ L ∞ ( R ) ≤ k Φ( ρ ) k L ( R ) k ρ − ¯ ρ k − γ L ∞ ( R ) ≤ C ( T ) . (3.28) n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity γ >
2, using the fact that ρ − ¯ ρ > ¯ ρ ⇒ ρ − ¯ ρ ) γ − < ρ γ − and Lemma 3.1, we have k ( ρ − ¯ ρ ) | { ρ ≥ ρ } k L ( R ) ≤ (cid:13)(cid:13) ( ρ − ¯ ρ ) | { ρ ≥ ρ } (cid:13)(cid:13) L ≤ (cid:13)(cid:13)(cid:13)(cid:13) ( ρ − ¯ ρ ) γ | { ρ ≥ ρ } ρ − ¯ ρ ) γ − | { ρ ≥ ρ } (cid:13)(cid:13)(cid:13)(cid:13) L ≤ (cid:13)(cid:13)(cid:13)(cid:13) ( ρ − ¯ ρ ) γ | { ρ ≥ ρ } ρ γ − (cid:13)(cid:13)(cid:13)(cid:13) L ≤ ρ γ − k Φ( ρ ) k L ≤ C. (3.29)Thus, combining (3.28), (3.29) with (3.27), we can obtain k ρ − ¯ ρ k L ≤ C. This completesthe proof of Lemma 3.3.
Lemma 3.4.
Let ( ρ, u, b ) be a smooth solution of (1.1) - (1.2) . Then for any T > , itholds that sup ≤ t ≤ T (cid:16) µ k u x k L + ν k b x k L + k b − ¯ b k L (cid:17) + ˆ T k b − ¯ b k L + µν k ( b − ¯ b ) b x k L dt + ˆ T ν k b xx k L dt + ˆ T k√ ρ ˙ u k L dt ≤ C and sup ≤ t ≤ T ( k u k L + k u k L ∞ ) ≤ C. Proof.
The proof of Lemma 3.4 will be divided into four steps.Step 1. Multiplying the equation (1.1) by ˙ u and integrating the resulting equationover R with respect to x yields µ ddt ˆ u x + ˆ ρ ˙ u = − µ ˆ u x ( uu x ) x − ˆ P ( ρ ) x ( u t + uu x ) − ˆ (cid:18) b (cid:19) x ( u t + uu x )=: J + J + J . (3.30)Firstly, by integration by parts, we find J = − µ ˆ u x − µ ˆ u (cid:18) u x (cid:19) x = − µ ˆ u x + µ ˆ u x · u x = − µ ˆ u x . (3.31) n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity
The proof of Lemma 3.4 will be divided into four steps.Step 1. Multiplying the equation (1.1) by ˙ u and integrating the resulting equationover R with respect to x yields µ ddt ˆ u x + ˆ ρ ˙ u = − µ ˆ u x ( uu x ) x − ˆ P ( ρ ) x ( u t + uu x ) − ˆ (cid:18) b (cid:19) x ( u t + uu x )=: J + J + J . (3.30)Firstly, by integration by parts, we find J = − µ ˆ u x − µ ˆ u (cid:18) u x (cid:19) x = − µ ˆ u x + µ ˆ u x · u x = − µ ˆ u x . (3.31) n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity J = − ˆ ( P ( ρ ) − P (¯ ρ )) x ( u t + uu x )= ddt ˆ ( P ( ρ ) − P (¯ ρ )) u x − ˆ [( P ( ρ ) − P (¯ ρ )) t + ( P ( ρ ) − P (¯ ρ )) x u ] u x = ddt ˆ ( P ( ρ ) − P (¯ ρ )) u x + γ ˆ ρ γ u x , (3.32)and J = − ˆ b ( b − ¯ b ) x ( u t + uu x )= − ˆ (cid:18) ( b − ¯ b ) (cid:19) x ( u t + uu x ) − ¯ b ˆ ( b − ¯ b ) x ( u t + uu x )= ddt ˆ (cid:18) ( b − ¯ b ) b ( b − ¯ b ) (cid:19) u x − ˆ (cid:18)(cid:18) ( b − ¯ b ) (cid:19) t + (cid:18) ( b − ¯ b ) (cid:19) x u (cid:19) u x − ¯ b ˆ (cid:0) ( b − ¯ b ) t + u ( b − ¯ b ) x (cid:1) u x = ddt ˆ (cid:18) ( b − ¯ b ) b ( b − ¯ b ) (cid:19) u x − ˆ (cid:0) ( b − ¯ b ) + ¯ b (cid:1) ( νb xx − bu x ) u x , (3.33)where we have used the following facts:( P ( ρ ) − P (¯ ρ )) t + u ( P ( ρ ) − P (¯ ρ )) x + γρ γ u x = 0 , (cid:18) ( b − ¯ b ) (cid:19) t + u (cid:18) ( b − ¯ b ) (cid:19) x + b ( b − ¯ b ) u x = νb xx ( b − ¯ b ) , and ( b − ¯ b ) t + u ( b − ¯ b ) x + bu x = νb xx . Substituting (3.31)-(3.33) into (3.30), using H¨older’s and Cauchy-Schwarz’s inequalities,we have ddt ˆ µ u x − ( P ( ρ ) − P (¯ ρ )) u x − ( b − ¯ b ) u x − ¯ b ( b − ¯ b ) u x dx + ˆ ρ ˙ u = − µ ˆ u x − γ ˆ ρ γ u x − ν ˆ b xx ( b − ¯ b ) u x + ˆ ( b − ¯ b ) u x + 2¯ b ˆ ( b − ¯ b ) u x − ¯ b ˆ νb xx u x + ¯ b ˆ u x ≤ C (cid:0) k u x k L + k u x k L + ν k b xx k L k b − ¯ b k L k u x k L + k u x k L k b − ¯ b k L + k u x k L k b − ¯ b k L + ν k b xx k L k u x k L (cid:1) ≤ εν k b xx k L + C (cid:18) k u x k L + k b − ¯ b k L + ( k b − ¯ b k L k b − ¯ b k L ) + k u x k L (cid:19) ≤ εν k b xx k L + C (cid:0) k u x k L + k b − ¯ b k L + k u x k L + 1 (cid:1) . (3.34) n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity
The proof of Lemma 3.4 will be divided into four steps.Step 1. Multiplying the equation (1.1) by ˙ u and integrating the resulting equationover R with respect to x yields µ ddt ˆ u x + ˆ ρ ˙ u = − µ ˆ u x ( uu x ) x − ˆ P ( ρ ) x ( u t + uu x ) − ˆ (cid:18) b (cid:19) x ( u t + uu x )=: J + J + J . (3.30)Firstly, by integration by parts, we find J = − µ ˆ u x − µ ˆ u (cid:18) u x (cid:19) x = − µ ˆ u x + µ ˆ u x · u x = − µ ˆ u x . (3.31) n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity J = − ˆ ( P ( ρ ) − P (¯ ρ )) x ( u t + uu x )= ddt ˆ ( P ( ρ ) − P (¯ ρ )) u x − ˆ [( P ( ρ ) − P (¯ ρ )) t + ( P ( ρ ) − P (¯ ρ )) x u ] u x = ddt ˆ ( P ( ρ ) − P (¯ ρ )) u x + γ ˆ ρ γ u x , (3.32)and J = − ˆ b ( b − ¯ b ) x ( u t + uu x )= − ˆ (cid:18) ( b − ¯ b ) (cid:19) x ( u t + uu x ) − ¯ b ˆ ( b − ¯ b ) x ( u t + uu x )= ddt ˆ (cid:18) ( b − ¯ b ) b ( b − ¯ b ) (cid:19) u x − ˆ (cid:18)(cid:18) ( b − ¯ b ) (cid:19) t + (cid:18) ( b − ¯ b ) (cid:19) x u (cid:19) u x − ¯ b ˆ (cid:0) ( b − ¯ b ) t + u ( b − ¯ b ) x (cid:1) u x = ddt ˆ (cid:18) ( b − ¯ b ) b ( b − ¯ b ) (cid:19) u x − ˆ (cid:0) ( b − ¯ b ) + ¯ b (cid:1) ( νb xx − bu x ) u x , (3.33)where we have used the following facts:( P ( ρ ) − P (¯ ρ )) t + u ( P ( ρ ) − P (¯ ρ )) x + γρ γ u x = 0 , (cid:18) ( b − ¯ b ) (cid:19) t + u (cid:18) ( b − ¯ b ) (cid:19) x + b ( b − ¯ b ) u x = νb xx ( b − ¯ b ) , and ( b − ¯ b ) t + u ( b − ¯ b ) x + bu x = νb xx . Substituting (3.31)-(3.33) into (3.30), using H¨older’s and Cauchy-Schwarz’s inequalities,we have ddt ˆ µ u x − ( P ( ρ ) − P (¯ ρ )) u x − ( b − ¯ b ) u x − ¯ b ( b − ¯ b ) u x dx + ˆ ρ ˙ u = − µ ˆ u x − γ ˆ ρ γ u x − ν ˆ b xx ( b − ¯ b ) u x + ˆ ( b − ¯ b ) u x + 2¯ b ˆ ( b − ¯ b ) u x − ¯ b ˆ νb xx u x + ¯ b ˆ u x ≤ C (cid:0) k u x k L + k u x k L + ν k b xx k L k b − ¯ b k L k u x k L + k u x k L k b − ¯ b k L + k u x k L k b − ¯ b k L + ν k b xx k L k u x k L (cid:1) ≤ εν k b xx k L + C (cid:18) k u x k L + k b − ¯ b k L + ( k b − ¯ b k L k b − ¯ b k L ) + k u x k L (cid:19) ≤ εν k b xx k L + C (cid:0) k u x k L + k b − ¯ b k L + k u x k L + 1 (cid:1) . (3.34) n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity k u x k L on the right-hand side of the equation (3.34). Employingthe effective viscous flux, momentum equation, Lemma 3.1 and Lemma 3.3, we have k F k L ≤ k F k L k F x k L ≤ C (cid:13)(cid:13)(cid:13)(cid:13)(cid:20) µu x − ( P ( ρ ) − P (¯ ρ )) − (cid:18) ( b − ¯ b ) b ( b − ¯ b ) (cid:19)(cid:21)(cid:13)(cid:13)(cid:13)(cid:13) L k ρ ˙ u k L ≤ C (cid:0) k u x k L + k ρ − ¯ ρ k L + k b − ¯ b k L + k b − ¯ b k L (cid:1) k ρ ˙ u k L ≤ C (cid:18) k u x k L + k b − ¯ b k L k b − ¯ b k L + 1 (cid:19) k ρ ˙ u k L ≤ C (cid:18) k u x k L + k b − ¯ b k L + 1 (cid:19) k√ ρ ˙ u k L . (3.35)Moreover, we can obtain k u x k L = (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) F + P ( ρ ) − P (¯ ρ ) + b − ¯ b µ (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) L ≤ C (cid:18) k F k L + k ρ − ¯ ρ k L + (cid:13)(cid:13)(cid:13)(cid:13) ( b − ¯ b ) b ( b − ¯ b ) (cid:13)(cid:13)(cid:13)(cid:13) L (cid:19) ≤ C (cid:18) k F k L + k ρ − ¯ ρ k L k ρ − ¯ ρ k L ∞ + k b − ¯ b k L + k b − ¯ b k L (cid:19) ≤ C (cid:18) k u x k L + k b − ¯ b k L + 1 (cid:19) k√ ρ ˙ u k L + k b − ¯ b k L + 1 ! , (3.36)which implies k u x k L ≤ C (cid:18) k u x k L + k b − ¯ b k L + 1 (cid:19) k√ ρ ˙ u k L ! + k b − ¯ b k L + 1 ≤ ε k√ ρ ˙ u k L + C (cid:0) k u x k L + k b − ¯ b k L + 1 (cid:1) . (3.37)Then, substituting (3.37) into (3.34) and choosing ε sufficiently small, we have ddt ˆ µ u x − ( P ( ρ ) − P (¯ ρ )) u x − ( b − ¯ b ) u x − ¯ b ( b − ¯ b ) u x + ˆ ρ ˙ u ≤ εν k b xx k L + ε k√ ρ ˙ u k L + C (cid:0) k u x k L + k b − ¯ b k L + 1 (cid:1) . (3.38)Step 2. To control the term k b − ¯ b k L on the right-hand side of (3.38), we rewrite themagnetic field equation as( b − ¯ b ) t + u ( b − ¯ b ) x + ( b − ¯ b ) u x + ¯ bu x = νb xx . Then multiplying the above equation by ( b − ¯ b ) and integrating it over R with respect to n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity
The proof of Lemma 3.4 will be divided into four steps.Step 1. Multiplying the equation (1.1) by ˙ u and integrating the resulting equationover R with respect to x yields µ ddt ˆ u x + ˆ ρ ˙ u = − µ ˆ u x ( uu x ) x − ˆ P ( ρ ) x ( u t + uu x ) − ˆ (cid:18) b (cid:19) x ( u t + uu x )=: J + J + J . (3.30)Firstly, by integration by parts, we find J = − µ ˆ u x − µ ˆ u (cid:18) u x (cid:19) x = − µ ˆ u x + µ ˆ u x · u x = − µ ˆ u x . (3.31) n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity J = − ˆ ( P ( ρ ) − P (¯ ρ )) x ( u t + uu x )= ddt ˆ ( P ( ρ ) − P (¯ ρ )) u x − ˆ [( P ( ρ ) − P (¯ ρ )) t + ( P ( ρ ) − P (¯ ρ )) x u ] u x = ddt ˆ ( P ( ρ ) − P (¯ ρ )) u x + γ ˆ ρ γ u x , (3.32)and J = − ˆ b ( b − ¯ b ) x ( u t + uu x )= − ˆ (cid:18) ( b − ¯ b ) (cid:19) x ( u t + uu x ) − ¯ b ˆ ( b − ¯ b ) x ( u t + uu x )= ddt ˆ (cid:18) ( b − ¯ b ) b ( b − ¯ b ) (cid:19) u x − ˆ (cid:18)(cid:18) ( b − ¯ b ) (cid:19) t + (cid:18) ( b − ¯ b ) (cid:19) x u (cid:19) u x − ¯ b ˆ (cid:0) ( b − ¯ b ) t + u ( b − ¯ b ) x (cid:1) u x = ddt ˆ (cid:18) ( b − ¯ b ) b ( b − ¯ b ) (cid:19) u x − ˆ (cid:0) ( b − ¯ b ) + ¯ b (cid:1) ( νb xx − bu x ) u x , (3.33)where we have used the following facts:( P ( ρ ) − P (¯ ρ )) t + u ( P ( ρ ) − P (¯ ρ )) x + γρ γ u x = 0 , (cid:18) ( b − ¯ b ) (cid:19) t + u (cid:18) ( b − ¯ b ) (cid:19) x + b ( b − ¯ b ) u x = νb xx ( b − ¯ b ) , and ( b − ¯ b ) t + u ( b − ¯ b ) x + bu x = νb xx . Substituting (3.31)-(3.33) into (3.30), using H¨older’s and Cauchy-Schwarz’s inequalities,we have ddt ˆ µ u x − ( P ( ρ ) − P (¯ ρ )) u x − ( b − ¯ b ) u x − ¯ b ( b − ¯ b ) u x dx + ˆ ρ ˙ u = − µ ˆ u x − γ ˆ ρ γ u x − ν ˆ b xx ( b − ¯ b ) u x + ˆ ( b − ¯ b ) u x + 2¯ b ˆ ( b − ¯ b ) u x − ¯ b ˆ νb xx u x + ¯ b ˆ u x ≤ C (cid:0) k u x k L + k u x k L + ν k b xx k L k b − ¯ b k L k u x k L + k u x k L k b − ¯ b k L + k u x k L k b − ¯ b k L + ν k b xx k L k u x k L (cid:1) ≤ εν k b xx k L + C (cid:18) k u x k L + k b − ¯ b k L + ( k b − ¯ b k L k b − ¯ b k L ) + k u x k L (cid:19) ≤ εν k b xx k L + C (cid:0) k u x k L + k b − ¯ b k L + k u x k L + 1 (cid:1) . (3.34) n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity k u x k L on the right-hand side of the equation (3.34). Employingthe effective viscous flux, momentum equation, Lemma 3.1 and Lemma 3.3, we have k F k L ≤ k F k L k F x k L ≤ C (cid:13)(cid:13)(cid:13)(cid:13)(cid:20) µu x − ( P ( ρ ) − P (¯ ρ )) − (cid:18) ( b − ¯ b ) b ( b − ¯ b ) (cid:19)(cid:21)(cid:13)(cid:13)(cid:13)(cid:13) L k ρ ˙ u k L ≤ C (cid:0) k u x k L + k ρ − ¯ ρ k L + k b − ¯ b k L + k b − ¯ b k L (cid:1) k ρ ˙ u k L ≤ C (cid:18) k u x k L + k b − ¯ b k L k b − ¯ b k L + 1 (cid:19) k ρ ˙ u k L ≤ C (cid:18) k u x k L + k b − ¯ b k L + 1 (cid:19) k√ ρ ˙ u k L . (3.35)Moreover, we can obtain k u x k L = (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) F + P ( ρ ) − P (¯ ρ ) + b − ¯ b µ (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) L ≤ C (cid:18) k F k L + k ρ − ¯ ρ k L + (cid:13)(cid:13)(cid:13)(cid:13) ( b − ¯ b ) b ( b − ¯ b ) (cid:13)(cid:13)(cid:13)(cid:13) L (cid:19) ≤ C (cid:18) k F k L + k ρ − ¯ ρ k L k ρ − ¯ ρ k L ∞ + k b − ¯ b k L + k b − ¯ b k L (cid:19) ≤ C (cid:18) k u x k L + k b − ¯ b k L + 1 (cid:19) k√ ρ ˙ u k L + k b − ¯ b k L + 1 ! , (3.36)which implies k u x k L ≤ C (cid:18) k u x k L + k b − ¯ b k L + 1 (cid:19) k√ ρ ˙ u k L ! + k b − ¯ b k L + 1 ≤ ε k√ ρ ˙ u k L + C (cid:0) k u x k L + k b − ¯ b k L + 1 (cid:1) . (3.37)Then, substituting (3.37) into (3.34) and choosing ε sufficiently small, we have ddt ˆ µ u x − ( P ( ρ ) − P (¯ ρ )) u x − ( b − ¯ b ) u x − ¯ b ( b − ¯ b ) u x + ˆ ρ ˙ u ≤ εν k b xx k L + ε k√ ρ ˙ u k L + C (cid:0) k u x k L + k b − ¯ b k L + 1 (cid:1) . (3.38)Step 2. To control the term k b − ¯ b k L on the right-hand side of (3.38), we rewrite themagnetic field equation as( b − ¯ b ) t + u ( b − ¯ b ) x + ( b − ¯ b ) u x + ¯ bu x = νb xx . Then multiplying the above equation by ( b − ¯ b ) and integrating it over R with respect to n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity x , we have14 ddt ˆ ( b − ¯ b ) + 3 ν ˆ ( b − ¯ b ) b x = − ˆ (cid:18)
34 ( b − ¯ b ) + ¯ b ( b − ¯ b ) (cid:19) u x = − ˆ (cid:18)
34 ( b − ¯ b ) + ¯ b ( b − ¯ b ) (cid:19) F + P ( ρ ) − P (¯ ρ ) + b − ¯ b µ = − ˆ (cid:18)
34 ( b − ¯ b ) + ¯ b ( b − ¯ b ) (cid:19) F + P ( ρ ) − P (¯ ρ ) + ( b − ¯ b ) + ¯ b ( b − ¯ b ) µ = − µ ˆ ( b − ¯ b ) ( F + P ( ρ ) − P (¯ ρ )) − ¯ bµ ˆ ( b − ¯ b ) ( F + P ( ρ ) − P (¯ ρ )) − µ ˆ ( b − ¯ b ) − b µ ˆ ( b − ¯ b ) − ¯ b µ ˆ ( b − ¯ b ) , which gives14 ddt ˆ ( b − ¯ b ) + 38 µ ˆ ( b − ¯ b ) + ¯ b µ ˆ ( b − ¯ b ) + 3 ν ˆ ( b − ¯ b ) b x = − µ ˆ ( b − ¯ b ) ( F + P ( ρ ) − P (¯ ρ )) − ¯ bµ ˆ ( b − ¯ b ) ( F + P ( ρ ) − P (¯ ρ )) − b µ ˆ ( b − ¯ b ) =: K + K + K , (3.39)Next, we estimate K − K term by term. Using (3.35), Lemmas 3.3 and Young’s inequality,we have K ≤ C k b − ¯ b k L ( k F k L + k P ( ρ ) − P (¯ ρ ) k L ) ≤ C k b − ¯ b k L (cid:18) k u x k L + k b − ¯ b k L + 1 (cid:19) k√ ρ ˙ u k L + k ρ − ¯ ρ k L ! ≤ ε k b − ¯ b k L + ε k√ ρ ˙ u k L + C (1 + k u x k L ) . (3.40)Similarly, we can get K ≤ k b − ¯ b k L ( k F k L + k P ( ρ ) − P (¯ ρ ) k L ) ≤ k b − ¯ b k L (cid:0) k u x k L + k ρ − ¯ ρ k L + k b − ¯ b k L + k b − ¯ b k L (cid:1) ≤ k b − ¯ b k L (cid:18) k u x k L + k ρ − ¯ ρ k L + k b − ¯ b k L k b − ¯ b k L + k b − ¯ b k L (cid:19) ≤ ε k b − ¯ b k L + C (cid:0) k u x k L + 1 (cid:1) , (3.41)and K ≤ C k b − ¯ b k L ≤ C k b − ¯ b k L k b − ¯ b k L ≤ ε k b − ¯ b k L + C. (3.42)Then, substituting (3.40)-(3.42) into (3.39) and choosing ε > ddt ˆ ( b − ¯ b ) + 38 µ ˆ ( b − ¯ b ) + ¯ b µ ˆ ( b − ¯ b ) + 3 ν ˆ ( b − ¯ b ) b x ≤ ε k√ ρ ˙ u k L + C (cid:0) k u x k L (cid:1) . (3.43) n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity
34 ( b − ¯ b ) + ¯ b ( b − ¯ b ) (cid:19) F + P ( ρ ) − P (¯ ρ ) + ( b − ¯ b ) + ¯ b ( b − ¯ b ) µ = − µ ˆ ( b − ¯ b ) ( F + P ( ρ ) − P (¯ ρ )) − ¯ bµ ˆ ( b − ¯ b ) ( F + P ( ρ ) − P (¯ ρ )) − µ ˆ ( b − ¯ b ) − b µ ˆ ( b − ¯ b ) − ¯ b µ ˆ ( b − ¯ b ) , which gives14 ddt ˆ ( b − ¯ b ) + 38 µ ˆ ( b − ¯ b ) + ¯ b µ ˆ ( b − ¯ b ) + 3 ν ˆ ( b − ¯ b ) b x = − µ ˆ ( b − ¯ b ) ( F + P ( ρ ) − P (¯ ρ )) − ¯ bµ ˆ ( b − ¯ b ) ( F + P ( ρ ) − P (¯ ρ )) − b µ ˆ ( b − ¯ b ) =: K + K + K , (3.39)Next, we estimate K − K term by term. Using (3.35), Lemmas 3.3 and Young’s inequality,we have K ≤ C k b − ¯ b k L ( k F k L + k P ( ρ ) − P (¯ ρ ) k L ) ≤ C k b − ¯ b k L (cid:18) k u x k L + k b − ¯ b k L + 1 (cid:19) k√ ρ ˙ u k L + k ρ − ¯ ρ k L ! ≤ ε k b − ¯ b k L + ε k√ ρ ˙ u k L + C (1 + k u x k L ) . (3.40)Similarly, we can get K ≤ k b − ¯ b k L ( k F k L + k P ( ρ ) − P (¯ ρ ) k L ) ≤ k b − ¯ b k L (cid:0) k u x k L + k ρ − ¯ ρ k L + k b − ¯ b k L + k b − ¯ b k L (cid:1) ≤ k b − ¯ b k L (cid:18) k u x k L + k ρ − ¯ ρ k L + k b − ¯ b k L k b − ¯ b k L + k b − ¯ b k L (cid:19) ≤ ε k b − ¯ b k L + C (cid:0) k u x k L + 1 (cid:1) , (3.41)and K ≤ C k b − ¯ b k L ≤ C k b − ¯ b k L k b − ¯ b k L ≤ ε k b − ¯ b k L + C. (3.42)Then, substituting (3.40)-(3.42) into (3.39) and choosing ε > ddt ˆ ( b − ¯ b ) + 38 µ ˆ ( b − ¯ b ) + ¯ b µ ˆ ( b − ¯ b ) + 3 ν ˆ ( b − ¯ b ) b x ≤ ε k√ ρ ˙ u k L + C (cid:0) k u x k L (cid:1) . (3.43) n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity ν k b xx k L on the right hand-side of (3.38), we multiplythe equation (1.1) by νb xx and integrate it over R to get ν ddt ˆ b x + ν ˆ b xx = ν ˆ b xx ub x + ν ˆ b xx ( b − ¯ b ) u x + ¯ bν ˆ b xx u x = − ν ˆ b x u x + ν ˆ b xx ( b − ¯ b ) u x + ¯ bν ˆ b xx u x =: H + H + H . (3.44)For the term H , by the effective viscous flux, it shows that H = − ν ˆ b x F + P ( ρ ) − P (¯ ρ ) + b − ¯ b µ ≤ − ν µ ˆ b x F + ν µ ˆ b x (cid:18) P (¯ ρ ) + ¯ b (cid:19) ≤ C (1 + k F k L ∞ ) ν k b x k L ≤ C (cid:18) k√ ρ ˙ u k L + k u x k L + k b − ¯ b k L (cid:19) ν k b x k L ≤ ε k√ ρ ˙ u k L + k b − ¯ b k L + C (cid:0) k u x k L + ν k b x k L (cid:1) ν k b x k L , (3.45)where we have used the following inequality: k F k L ∞ ≤ C k F k L k F x k L ≤ C (cid:0) k u x k L + k ρ − ¯ ρ k L + k b − ¯ b k L + k b − ¯ b k L (cid:1) k√ ρ ˙ u k L ≤ C (cid:18) k√ ρ ˙ u k L + k u x k L + k b − ¯ b k L (cid:19) . (3.46)For the terms H and H , using H¨older’s inequality, Cauchy’s inequality and (3.37), wehave H ≤ ν k b xx k L k b − ¯ b k L k u x k L ≤ εν k b xx k L + C (cid:0) k b − ¯ b k L + k u x k L (cid:1) ≤ εν k b xx k L + ε k√ ρ ˙ u k L + C (cid:0) k u x k L + k b − ¯ b k L (cid:1) , (3.47)and H ≤ Cν k b xx k L k u x k L ≤ εν k b xx k L + C k u x k L . (3.48)Thus, substituting (3.45), (3.47) and (3.48) into (3.44) and choosing ε > ν ddt ˆ b x + ν ˆ b xx ≤ ε k√ ρ ˙ u k L + C (cid:0) k u x k L + ν k b x k L (cid:1) ν k b x k L + C (cid:0) k u x k L + k b − ¯ b k L (cid:1) . (3.49)Step 4. Adding the equation (3.43) multiplied by µ (2 C + 1) and (3.38) into (3.49), wehave ddt ˆ (cid:18) µ u x + ν b x + 2 µ (2 C + 1)3 ( b − ¯ b ) − ψ ( ρ, u, b ) (cid:19) + ˆ ρ ˙ u + ν ˆ b xx + ˆ (cid:18) ( b − ¯ b ) + 8¯ b (2 C + 1)3 ( b − ¯ b ) + 8 µν (2 C + 1)( b − ¯ b ) b x (cid:19) ≤ C + C (cid:0) k u x k L + ν k b x k L (cid:1) (cid:0) k u x k L + ν k b x k L (cid:1) , (3.50) n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity
34 ( b − ¯ b ) + ¯ b ( b − ¯ b ) (cid:19) F + P ( ρ ) − P (¯ ρ ) + ( b − ¯ b ) + ¯ b ( b − ¯ b ) µ = − µ ˆ ( b − ¯ b ) ( F + P ( ρ ) − P (¯ ρ )) − ¯ bµ ˆ ( b − ¯ b ) ( F + P ( ρ ) − P (¯ ρ )) − µ ˆ ( b − ¯ b ) − b µ ˆ ( b − ¯ b ) − ¯ b µ ˆ ( b − ¯ b ) , which gives14 ddt ˆ ( b − ¯ b ) + 38 µ ˆ ( b − ¯ b ) + ¯ b µ ˆ ( b − ¯ b ) + 3 ν ˆ ( b − ¯ b ) b x = − µ ˆ ( b − ¯ b ) ( F + P ( ρ ) − P (¯ ρ )) − ¯ bµ ˆ ( b − ¯ b ) ( F + P ( ρ ) − P (¯ ρ )) − b µ ˆ ( b − ¯ b ) =: K + K + K , (3.39)Next, we estimate K − K term by term. Using (3.35), Lemmas 3.3 and Young’s inequality,we have K ≤ C k b − ¯ b k L ( k F k L + k P ( ρ ) − P (¯ ρ ) k L ) ≤ C k b − ¯ b k L (cid:18) k u x k L + k b − ¯ b k L + 1 (cid:19) k√ ρ ˙ u k L + k ρ − ¯ ρ k L ! ≤ ε k b − ¯ b k L + ε k√ ρ ˙ u k L + C (1 + k u x k L ) . (3.40)Similarly, we can get K ≤ k b − ¯ b k L ( k F k L + k P ( ρ ) − P (¯ ρ ) k L ) ≤ k b − ¯ b k L (cid:0) k u x k L + k ρ − ¯ ρ k L + k b − ¯ b k L + k b − ¯ b k L (cid:1) ≤ k b − ¯ b k L (cid:18) k u x k L + k ρ − ¯ ρ k L + k b − ¯ b k L k b − ¯ b k L + k b − ¯ b k L (cid:19) ≤ ε k b − ¯ b k L + C (cid:0) k u x k L + 1 (cid:1) , (3.41)and K ≤ C k b − ¯ b k L ≤ C k b − ¯ b k L k b − ¯ b k L ≤ ε k b − ¯ b k L + C. (3.42)Then, substituting (3.40)-(3.42) into (3.39) and choosing ε > ddt ˆ ( b − ¯ b ) + 38 µ ˆ ( b − ¯ b ) + ¯ b µ ˆ ( b − ¯ b ) + 3 ν ˆ ( b − ¯ b ) b x ≤ ε k√ ρ ˙ u k L + C (cid:0) k u x k L (cid:1) . (3.43) n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity ν k b xx k L on the right hand-side of (3.38), we multiplythe equation (1.1) by νb xx and integrate it over R to get ν ddt ˆ b x + ν ˆ b xx = ν ˆ b xx ub x + ν ˆ b xx ( b − ¯ b ) u x + ¯ bν ˆ b xx u x = − ν ˆ b x u x + ν ˆ b xx ( b − ¯ b ) u x + ¯ bν ˆ b xx u x =: H + H + H . (3.44)For the term H , by the effective viscous flux, it shows that H = − ν ˆ b x F + P ( ρ ) − P (¯ ρ ) + b − ¯ b µ ≤ − ν µ ˆ b x F + ν µ ˆ b x (cid:18) P (¯ ρ ) + ¯ b (cid:19) ≤ C (1 + k F k L ∞ ) ν k b x k L ≤ C (cid:18) k√ ρ ˙ u k L + k u x k L + k b − ¯ b k L (cid:19) ν k b x k L ≤ ε k√ ρ ˙ u k L + k b − ¯ b k L + C (cid:0) k u x k L + ν k b x k L (cid:1) ν k b x k L , (3.45)where we have used the following inequality: k F k L ∞ ≤ C k F k L k F x k L ≤ C (cid:0) k u x k L + k ρ − ¯ ρ k L + k b − ¯ b k L + k b − ¯ b k L (cid:1) k√ ρ ˙ u k L ≤ C (cid:18) k√ ρ ˙ u k L + k u x k L + k b − ¯ b k L (cid:19) . (3.46)For the terms H and H , using H¨older’s inequality, Cauchy’s inequality and (3.37), wehave H ≤ ν k b xx k L k b − ¯ b k L k u x k L ≤ εν k b xx k L + C (cid:0) k b − ¯ b k L + k u x k L (cid:1) ≤ εν k b xx k L + ε k√ ρ ˙ u k L + C (cid:0) k u x k L + k b − ¯ b k L (cid:1) , (3.47)and H ≤ Cν k b xx k L k u x k L ≤ εν k b xx k L + C k u x k L . (3.48)Thus, substituting (3.45), (3.47) and (3.48) into (3.44) and choosing ε > ν ddt ˆ b x + ν ˆ b xx ≤ ε k√ ρ ˙ u k L + C (cid:0) k u x k L + ν k b x k L (cid:1) ν k b x k L + C (cid:0) k u x k L + k b − ¯ b k L (cid:1) . (3.49)Step 4. Adding the equation (3.43) multiplied by µ (2 C + 1) and (3.38) into (3.49), wehave ddt ˆ (cid:18) µ u x + ν b x + 2 µ (2 C + 1)3 ( b − ¯ b ) − ψ ( ρ, u, b ) (cid:19) + ˆ ρ ˙ u + ν ˆ b xx + ˆ (cid:18) ( b − ¯ b ) + 8¯ b (2 C + 1)3 ( b − ¯ b ) + 8 µν (2 C + 1)( b − ¯ b ) b x (cid:19) ≤ C + C (cid:0) k u x k L + ν k b x k L (cid:1) (cid:0) k u x k L + ν k b x k L (cid:1) , (3.50) n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity ψ ( ρ, u, b ) = ( P ( ρ ) − P (¯ ρ )) u x + ( b − ¯ b ) u x + ¯ b ( b − ¯ b ) u x . Integrating (3.50) over [0 , T ]with respect to t , we obtain ˆ (cid:0) µu x + νb x + ( b − ¯ b ) (cid:1) + ˆ T ˆ (cid:0) ρ ˙ u + ( b − ¯ b ) + µν ( b − ¯ b ) b x + ν b xx (cid:1) ≤ C + C ˆ T (cid:0) k u x k L + ν k b x k L (cid:1) (cid:0) k u x k L + ν k b x k L (cid:1) + ˆ ψ ( ρ, u, b ) − ˆ ψ ( ρ , u , b ) . (3.51)By Lemma 3.1, Lemma 3.3 and Cauchy-Schwarz’s inequality, we can obtain ˆ ψ ( ρ, u, b ) − ˆ ψ ( ρ , u , b ) ≤ C (cid:0) k P ( ρ ) − P (¯ ρ ) k L + k b − ¯ b k L + k b − ¯ b k L (cid:1) k u x k L + C ≤ C (cid:0) k ρ − ¯ ρ k L + k b − ¯ b k L + 1 (cid:1) k u x k L + C ≤ εµ k u x k L + C (1 + k b − ¯ b k L ) , which together with (3.51) gives (choosing ε > ˆ (cid:0) µu x + νb x + ( b − ¯ b ) (cid:1) + ˆ T ˆ (cid:0) ρ ˙ u + ( b − ¯ b ) + µν ( b − ¯ b ) b x + ν b xx (cid:1) ≤ C (1 + k b − ¯ b k L ) + C ˆ T (cid:0) k u x k L + ν k b x k L (cid:1) (cid:0) k u x k L + ν k b x k L (cid:1) . (3.52)Now, integrating (3.43) over [0 , T ] with respect to t , and then adding the resultinginequality multiplied by 4 C into (3.52) show that ˆ (cid:0) µu x + νb x + ( b − ¯ b ) (cid:1) + ˆ T ˆ (cid:0) ρ ˙ u + ( b − ¯ b ) + µν ( b − ¯ b ) b x + ν b xx (cid:1) ≤ C + C ˆ T (cid:0) k u x k L + ν k b x k L (cid:1) (cid:0) k u x k L + ν k b x k L (cid:1) , which together with Gronwall’s inequality and Lemma 3.1 yields ˆ µu x + νb x + ( b − ¯ b ) dx + ˆ T ˆ ( b − ¯ b ) + µν ( b − ¯ b ) b x dxdt + ˆ T ˆ ν b xx dxdt + ˆ T ˆ ρ ˙ u dxdt ≤ C ( T ) . Thus, it follows from (3.11) and (3.12) that k u k L ≤ C (1 + k u x k L ) ≤ C, and k u k L ∞ ≤ C k u k L k u x k L ≤ C (1 + k u x k L ) ≤ C. Then, we complete the proof of Lemma 3.4. n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity
34 ( b − ¯ b ) + ¯ b ( b − ¯ b ) (cid:19) F + P ( ρ ) − P (¯ ρ ) + ( b − ¯ b ) + ¯ b ( b − ¯ b ) µ = − µ ˆ ( b − ¯ b ) ( F + P ( ρ ) − P (¯ ρ )) − ¯ bµ ˆ ( b − ¯ b ) ( F + P ( ρ ) − P (¯ ρ )) − µ ˆ ( b − ¯ b ) − b µ ˆ ( b − ¯ b ) − ¯ b µ ˆ ( b − ¯ b ) , which gives14 ddt ˆ ( b − ¯ b ) + 38 µ ˆ ( b − ¯ b ) + ¯ b µ ˆ ( b − ¯ b ) + 3 ν ˆ ( b − ¯ b ) b x = − µ ˆ ( b − ¯ b ) ( F + P ( ρ ) − P (¯ ρ )) − ¯ bµ ˆ ( b − ¯ b ) ( F + P ( ρ ) − P (¯ ρ )) − b µ ˆ ( b − ¯ b ) =: K + K + K , (3.39)Next, we estimate K − K term by term. Using (3.35), Lemmas 3.3 and Young’s inequality,we have K ≤ C k b − ¯ b k L ( k F k L + k P ( ρ ) − P (¯ ρ ) k L ) ≤ C k b − ¯ b k L (cid:18) k u x k L + k b − ¯ b k L + 1 (cid:19) k√ ρ ˙ u k L + k ρ − ¯ ρ k L ! ≤ ε k b − ¯ b k L + ε k√ ρ ˙ u k L + C (1 + k u x k L ) . (3.40)Similarly, we can get K ≤ k b − ¯ b k L ( k F k L + k P ( ρ ) − P (¯ ρ ) k L ) ≤ k b − ¯ b k L (cid:0) k u x k L + k ρ − ¯ ρ k L + k b − ¯ b k L + k b − ¯ b k L (cid:1) ≤ k b − ¯ b k L (cid:18) k u x k L + k ρ − ¯ ρ k L + k b − ¯ b k L k b − ¯ b k L + k b − ¯ b k L (cid:19) ≤ ε k b − ¯ b k L + C (cid:0) k u x k L + 1 (cid:1) , (3.41)and K ≤ C k b − ¯ b k L ≤ C k b − ¯ b k L k b − ¯ b k L ≤ ε k b − ¯ b k L + C. (3.42)Then, substituting (3.40)-(3.42) into (3.39) and choosing ε > ddt ˆ ( b − ¯ b ) + 38 µ ˆ ( b − ¯ b ) + ¯ b µ ˆ ( b − ¯ b ) + 3 ν ˆ ( b − ¯ b ) b x ≤ ε k√ ρ ˙ u k L + C (cid:0) k u x k L (cid:1) . (3.43) n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity ν k b xx k L on the right hand-side of (3.38), we multiplythe equation (1.1) by νb xx and integrate it over R to get ν ddt ˆ b x + ν ˆ b xx = ν ˆ b xx ub x + ν ˆ b xx ( b − ¯ b ) u x + ¯ bν ˆ b xx u x = − ν ˆ b x u x + ν ˆ b xx ( b − ¯ b ) u x + ¯ bν ˆ b xx u x =: H + H + H . (3.44)For the term H , by the effective viscous flux, it shows that H = − ν ˆ b x F + P ( ρ ) − P (¯ ρ ) + b − ¯ b µ ≤ − ν µ ˆ b x F + ν µ ˆ b x (cid:18) P (¯ ρ ) + ¯ b (cid:19) ≤ C (1 + k F k L ∞ ) ν k b x k L ≤ C (cid:18) k√ ρ ˙ u k L + k u x k L + k b − ¯ b k L (cid:19) ν k b x k L ≤ ε k√ ρ ˙ u k L + k b − ¯ b k L + C (cid:0) k u x k L + ν k b x k L (cid:1) ν k b x k L , (3.45)where we have used the following inequality: k F k L ∞ ≤ C k F k L k F x k L ≤ C (cid:0) k u x k L + k ρ − ¯ ρ k L + k b − ¯ b k L + k b − ¯ b k L (cid:1) k√ ρ ˙ u k L ≤ C (cid:18) k√ ρ ˙ u k L + k u x k L + k b − ¯ b k L (cid:19) . (3.46)For the terms H and H , using H¨older’s inequality, Cauchy’s inequality and (3.37), wehave H ≤ ν k b xx k L k b − ¯ b k L k u x k L ≤ εν k b xx k L + C (cid:0) k b − ¯ b k L + k u x k L (cid:1) ≤ εν k b xx k L + ε k√ ρ ˙ u k L + C (cid:0) k u x k L + k b − ¯ b k L (cid:1) , (3.47)and H ≤ Cν k b xx k L k u x k L ≤ εν k b xx k L + C k u x k L . (3.48)Thus, substituting (3.45), (3.47) and (3.48) into (3.44) and choosing ε > ν ddt ˆ b x + ν ˆ b xx ≤ ε k√ ρ ˙ u k L + C (cid:0) k u x k L + ν k b x k L (cid:1) ν k b x k L + C (cid:0) k u x k L + k b − ¯ b k L (cid:1) . (3.49)Step 4. Adding the equation (3.43) multiplied by µ (2 C + 1) and (3.38) into (3.49), wehave ddt ˆ (cid:18) µ u x + ν b x + 2 µ (2 C + 1)3 ( b − ¯ b ) − ψ ( ρ, u, b ) (cid:19) + ˆ ρ ˙ u + ν ˆ b xx + ˆ (cid:18) ( b − ¯ b ) + 8¯ b (2 C + 1)3 ( b − ¯ b ) + 8 µν (2 C + 1)( b − ¯ b ) b x (cid:19) ≤ C + C (cid:0) k u x k L + ν k b x k L (cid:1) (cid:0) k u x k L + ν k b x k L (cid:1) , (3.50) n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity ψ ( ρ, u, b ) = ( P ( ρ ) − P (¯ ρ )) u x + ( b − ¯ b ) u x + ¯ b ( b − ¯ b ) u x . Integrating (3.50) over [0 , T ]with respect to t , we obtain ˆ (cid:0) µu x + νb x + ( b − ¯ b ) (cid:1) + ˆ T ˆ (cid:0) ρ ˙ u + ( b − ¯ b ) + µν ( b − ¯ b ) b x + ν b xx (cid:1) ≤ C + C ˆ T (cid:0) k u x k L + ν k b x k L (cid:1) (cid:0) k u x k L + ν k b x k L (cid:1) + ˆ ψ ( ρ, u, b ) − ˆ ψ ( ρ , u , b ) . (3.51)By Lemma 3.1, Lemma 3.3 and Cauchy-Schwarz’s inequality, we can obtain ˆ ψ ( ρ, u, b ) − ˆ ψ ( ρ , u , b ) ≤ C (cid:0) k P ( ρ ) − P (¯ ρ ) k L + k b − ¯ b k L + k b − ¯ b k L (cid:1) k u x k L + C ≤ C (cid:0) k ρ − ¯ ρ k L + k b − ¯ b k L + 1 (cid:1) k u x k L + C ≤ εµ k u x k L + C (1 + k b − ¯ b k L ) , which together with (3.51) gives (choosing ε > ˆ (cid:0) µu x + νb x + ( b − ¯ b ) (cid:1) + ˆ T ˆ (cid:0) ρ ˙ u + ( b − ¯ b ) + µν ( b − ¯ b ) b x + ν b xx (cid:1) ≤ C (1 + k b − ¯ b k L ) + C ˆ T (cid:0) k u x k L + ν k b x k L (cid:1) (cid:0) k u x k L + ν k b x k L (cid:1) . (3.52)Now, integrating (3.43) over [0 , T ] with respect to t , and then adding the resultinginequality multiplied by 4 C into (3.52) show that ˆ (cid:0) µu x + νb x + ( b − ¯ b ) (cid:1) + ˆ T ˆ (cid:0) ρ ˙ u + ( b − ¯ b ) + µν ( b − ¯ b ) b x + ν b xx (cid:1) ≤ C + C ˆ T (cid:0) k u x k L + ν k b x k L (cid:1) (cid:0) k u x k L + ν k b x k L (cid:1) , which together with Gronwall’s inequality and Lemma 3.1 yields ˆ µu x + νb x + ( b − ¯ b ) dx + ˆ T ˆ ( b − ¯ b ) + µν ( b − ¯ b ) b x dxdt + ˆ T ˆ ν b xx dxdt + ˆ T ˆ ρ ˙ u dxdt ≤ C ( T ) . Thus, it follows from (3.11) and (3.12) that k u k L ≤ C (1 + k u x k L ) ≤ C, and k u k L ∞ ≤ C k u k L k u x k L ≤ C (1 + k u x k L ) ≤ C. Then, we complete the proof of Lemma 3.4. n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity b , we need to re-estimate the k b x k L independent of ν as follows. Lemma 3.5.
Let ( ρ, u, b ) be a smooth solution of (1.1) - (1.2) . Then for any T > , itholds that sup ≤ t ≤ T (cid:16) k b k L ∞ ( R ) + k ρ x k L ( R ) + k b x k L ( R ) (cid:17) + ˆ T (cid:16) µ k u xx k L ( R ) + ν k b xx k L ( R ) (cid:17) dt ≤ C. Proof.
Differentiating the equality (1.1) with respect to x , then multiplying the resultingequation by b x and integrating by parts over R , we have12 ddt ˆ b x + ν ˆ b xx = − ˆ b x u x − ˆ bb x u xx − ˆ ub x b xx = − ˆ b x u x − ˆ bb x u xx . (3.53)To control the second term on the right-hand side of (3.53), multiplying the momentumequation (1.1) by u xx gives µ ˆ u xx = ˆ ρ ˙ uu xx + γ ˆ ρ γ − ρ x u xx + ˆ bb x u xx . (3.54)Combining (3.53) and (3.54) yields12 ddt ˆ b x + ν ˆ b xx + µ ˆ u xx = − ˆ b x u x + ˆ ρ ˙ uu xx + γ ˆ ρ γ − ρ x u xx = − ˆ b x F + P ( ρ ) − P (¯ ρ ) + b − ¯ b µ + ˆ ρ ˙ uu xx + γ ˆ ρ γ − ρ x u xx ≤ − ˆ b x F − P (¯ ρ ) + − ¯ b µ + ˆ ρ ˙ uu xx + γ ˆ ρ γ − ρ x u xx ≤ C (cid:0) k F k L ∞ k b x k L + k b x k L + k√ ρ ˙ u k L k u xx k L + k ρ x k L k u xx k L (cid:1) ≤ ε k u xx k L + C ( k F k L ∞ + 1) k b x k L + C (cid:0) k ρ x k L + k√ ρ ˙ u k L (cid:1) ≤ ε k u xx k L + C (cid:0) k√ ρ ˙ u k L + k b − ¯ b k L + 1 (cid:1) k b x k L + C (cid:0) k ρ x k L + k√ ρ ˙ u k L (cid:1) , (3.55)where we have used (3.46) and Lemma 3.4.Next, differentiating the density equation (1.1) with respect to x , multiplying theresulting equation by ρ x , and integrating it over R implies that12 ddt ˆ ρ x = − ˆ ρ x u x − ˆ uρ x ρ xx − ˆ ρρ x u xx = − ˆ ρ x u x − ˆ ρρ x u xx = − ˆ ρ x F + P ( ρ ) − P (¯ ρ ) + b − ¯ b µ − ˆ ρρ x u xx ≤ − µ ˆ ρ x F + 32 µ ˆ (cid:18) P (¯ ρ ) + ¯ b (cid:19) ρ x + C k ρ x k L k u xx k L ≤ ε k u xx k L + C ( k F k L ∞ + 1) k ρ x k L ≤ ε k u xx k L + C (cid:0) k√ ρ ˙ u k L + k b − ¯ b k L + 1 (cid:1) k ρ x k L , (3.56) n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity
Differentiating the equality (1.1) with respect to x , then multiplying the resultingequation by b x and integrating by parts over R , we have12 ddt ˆ b x + ν ˆ b xx = − ˆ b x u x − ˆ bb x u xx − ˆ ub x b xx = − ˆ b x u x − ˆ bb x u xx . (3.53)To control the second term on the right-hand side of (3.53), multiplying the momentumequation (1.1) by u xx gives µ ˆ u xx = ˆ ρ ˙ uu xx + γ ˆ ρ γ − ρ x u xx + ˆ bb x u xx . (3.54)Combining (3.53) and (3.54) yields12 ddt ˆ b x + ν ˆ b xx + µ ˆ u xx = − ˆ b x u x + ˆ ρ ˙ uu xx + γ ˆ ρ γ − ρ x u xx = − ˆ b x F + P ( ρ ) − P (¯ ρ ) + b − ¯ b µ + ˆ ρ ˙ uu xx + γ ˆ ρ γ − ρ x u xx ≤ − ˆ b x F − P (¯ ρ ) + − ¯ b µ + ˆ ρ ˙ uu xx + γ ˆ ρ γ − ρ x u xx ≤ C (cid:0) k F k L ∞ k b x k L + k b x k L + k√ ρ ˙ u k L k u xx k L + k ρ x k L k u xx k L (cid:1) ≤ ε k u xx k L + C ( k F k L ∞ + 1) k b x k L + C (cid:0) k ρ x k L + k√ ρ ˙ u k L (cid:1) ≤ ε k u xx k L + C (cid:0) k√ ρ ˙ u k L + k b − ¯ b k L + 1 (cid:1) k b x k L + C (cid:0) k ρ x k L + k√ ρ ˙ u k L (cid:1) , (3.55)where we have used (3.46) and Lemma 3.4.Next, differentiating the density equation (1.1) with respect to x , multiplying theresulting equation by ρ x , and integrating it over R implies that12 ddt ˆ ρ x = − ˆ ρ x u x − ˆ uρ x ρ xx − ˆ ρρ x u xx = − ˆ ρ x u x − ˆ ρρ x u xx = − ˆ ρ x F + P ( ρ ) − P (¯ ρ ) + b − ¯ b µ − ˆ ρρ x u xx ≤ − µ ˆ ρ x F + 32 µ ˆ (cid:18) P (¯ ρ ) + ¯ b (cid:19) ρ x + C k ρ x k L k u xx k L ≤ ε k u xx k L + C ( k F k L ∞ + 1) k ρ x k L ≤ ε k u xx k L + C (cid:0) k√ ρ ˙ u k L + k b − ¯ b k L + 1 (cid:1) k ρ x k L , (3.56) n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity ddt ˆ (cid:0) b x + ρ x (cid:1) + ν ˆ b xx + µ ˆ u xx ≤ C (cid:0) k√ ρ ˙ u k L + k b − ¯ b k L + 1 (cid:1) (cid:0) k ρ x k L + k b x k L (cid:1) , which together Gronwall’s inequality and Lemma 3.4 gives ˆ (cid:0) ρ x + b x (cid:1) + ˆ T ˆ (cid:0) νb xx + µu xx (cid:1) ≤ C ( T ) . Thus, it follows from Lemma 3.1 and the G-N inequality (2.1) that k b k L ∞ ≤ k b − ¯ b k L ∞ + ¯ b ≤ C k b − ¯ b k L k b x k L + ¯ b ≤ C ( T ) . Thus, the proof of Lemma 3.5 is finished.With the help of the Lemma 3.5, we can get the estimates of the first order derivativewith respect to t of the density and the magnetic field, respectively. Lemma 3.6.
Let ( ρ, u, b ) be a smooth solution to (1.1) - (1.2) . Then for any T > , itholds that sup ≤ t ≤ T k ρ t k L ( R ) + k b t k L (0 ,T ; L ( R )) ≤ C. Proof.
By direct calculation, we have k ρ t k L ≤ C ( k ρu x k L + k uρ x k L ) ≤ C ( k ρ k L ∞ k u x k L + k u k L ∞ k ρ x k L ) ∈ L ∞ (0 , T ) , and k b t k L ≤ C ( k bu x k L + k ub x k L + k νb xx k L ) ≤ C ( k b k L ∞ k u x k L + k u k L ∞ k b x k L + k νb xx k L ) ≤ C (1 + k νb xx k L ) ∈ L (0 , T ) , where we have used Lemma 3.4 and Lemma 3.5.The following estimates of the second order derivative of the velocity plays an importantrole in the analysis of the non-resistive limit. Lemma 3.7.
Let ( ρ, u, b ) be a smooth solution to (1.1) - (1.2) . Then for any T > , itholds that sup ≤ t ≤ T k√ ρ ˙ u k L ( R ) + ˆ T µ k u xt k L ( R ) ≤ C, and sup ≤ t ≤ T (cid:0) k u xx k L ( R ) + k√ ρu t k L ( R ) (cid:1) ≤ C. Proof.
Differentiating the momentum equation (1.1) with respect to t , we have ρ t ˙ u + ρ ˙ u t + (cid:18) P ( ρ ) + b (cid:19) xt = µu xxt . n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity
Differentiating the momentum equation (1.1) with respect to t , we have ρ t ˙ u + ρ ˙ u t + (cid:18) P ( ρ ) + b (cid:19) xt = µu xxt . n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity u and integrating the resulting equation over R yields12 ddt ˆ ρ ˙ u + µ ˆ u xt = 12 ˆ ρ t ˙ u − ˆ (cid:18) P ( ρ ) + b (cid:19) xt ˙ u − µ ˆ u xt (cid:0) u x + uu xx (cid:1) =: L + L + L . (3.57)Now, we estimate the terms L − L as L = − ˆ ( ρu ) x ˙ u = ˆ ρu ˙ u ˙ u x ≤ C k√ ρ ˙ u k L k u k L ∞ k ˙ u x k L ≤ C k√ ρ ˙ u k L (cid:0) k u xt k L + k u x k L + k u k L ∞ k u xx k L (cid:1) (3.58) ≤ C k√ ρ ˙ u k L (cid:18) k u xt k L + k u x k L k u xx k L + k u xx k L (cid:19) ≤ ε k u xt k L + C (cid:0) k√ ρ ˙ u k L + k u xx k L + 1 (cid:1) , and similarly, we also have L = ˆ (cid:18) P ( ρ ) + b (cid:19) t ˙ u x ≤ C k ρ γ − ρ t + bb t k L k ˙ u x k L ≤ C ( k ρ t k L + k b t k L ) k u xt + u x + uu xx k L ≤ C (1 + k b t k L ) (cid:18) k u xt k L + k u x k L k u xx k L + k u k L ∞ k u xx k L (cid:19) ≤ ε k u xt k L + C (cid:0) k u xx k L + k b t k L + 1 (cid:1) , (3.59)and L ≤ C k u xt k L (cid:0) k u x k L + k u k L ∞ k u xx k L (cid:1) ≤ ε k u xt k L + C (cid:0) k u xx k L (cid:1) , (3.60)where we have used the Gagliardo-Nirenberg inequality, Lemma 3.4 and Lemma 3.5. Thus,substituting (3.58)-(3.60) into (3.57) and choosing ε > ddt ˆ ρ ˙ u + µ ˆ u xt ≤ C (cid:0) k√ ρ ˙ u k L + k u xx k L + k b t k L (cid:1) , which combining with Gronwall’s inequality, Lemma 3.4 and Lemma 3.5 gives12 ˆ ρ ˙ u + ˆ T ˆ µu xt ≤ C ( T ) . This together with the momentum equation yields that µ k u xx k L ≤ C (cid:18) k ρ ˙ u k L + k (cid:18) P ( ρ ) + b (cid:19) x k L (cid:19) ≤ C ( k√ ρ ˙ u k L + k ρ x k L + k b x k L ) ≤ C, and k√ ρu t k L ≤ C k√ ρ ˙ u − √ ρuu x k L ≤ C k√ ρ ˙ u k L + k√ ρ k L ∞ k u k L ∞ k u x k L ≤ C. We complete the proof of Lemma 3.7. n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity
Differentiating the momentum equation (1.1) with respect to t , we have ρ t ˙ u + ρ ˙ u t + (cid:18) P ( ρ ) + b (cid:19) xt = µu xxt . n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity u and integrating the resulting equation over R yields12 ddt ˆ ρ ˙ u + µ ˆ u xt = 12 ˆ ρ t ˙ u − ˆ (cid:18) P ( ρ ) + b (cid:19) xt ˙ u − µ ˆ u xt (cid:0) u x + uu xx (cid:1) =: L + L + L . (3.57)Now, we estimate the terms L − L as L = − ˆ ( ρu ) x ˙ u = ˆ ρu ˙ u ˙ u x ≤ C k√ ρ ˙ u k L k u k L ∞ k ˙ u x k L ≤ C k√ ρ ˙ u k L (cid:0) k u xt k L + k u x k L + k u k L ∞ k u xx k L (cid:1) (3.58) ≤ C k√ ρ ˙ u k L (cid:18) k u xt k L + k u x k L k u xx k L + k u xx k L (cid:19) ≤ ε k u xt k L + C (cid:0) k√ ρ ˙ u k L + k u xx k L + 1 (cid:1) , and similarly, we also have L = ˆ (cid:18) P ( ρ ) + b (cid:19) t ˙ u x ≤ C k ρ γ − ρ t + bb t k L k ˙ u x k L ≤ C ( k ρ t k L + k b t k L ) k u xt + u x + uu xx k L ≤ C (1 + k b t k L ) (cid:18) k u xt k L + k u x k L k u xx k L + k u k L ∞ k u xx k L (cid:19) ≤ ε k u xt k L + C (cid:0) k u xx k L + k b t k L + 1 (cid:1) , (3.59)and L ≤ C k u xt k L (cid:0) k u x k L + k u k L ∞ k u xx k L (cid:1) ≤ ε k u xt k L + C (cid:0) k u xx k L (cid:1) , (3.60)where we have used the Gagliardo-Nirenberg inequality, Lemma 3.4 and Lemma 3.5. Thus,substituting (3.58)-(3.60) into (3.57) and choosing ε > ddt ˆ ρ ˙ u + µ ˆ u xt ≤ C (cid:0) k√ ρ ˙ u k L + k u xx k L + k b t k L (cid:1) , which combining with Gronwall’s inequality, Lemma 3.4 and Lemma 3.5 gives12 ˆ ρ ˙ u + ˆ T ˆ µu xt ≤ C ( T ) . This together with the momentum equation yields that µ k u xx k L ≤ C (cid:18) k ρ ˙ u k L + k (cid:18) P ( ρ ) + b (cid:19) x k L (cid:19) ≤ C ( k√ ρ ˙ u k L + k ρ x k L + k b x k L ) ≤ C, and k√ ρu t k L ≤ C k√ ρ ˙ u − √ ρuu x k L ≤ C k√ ρ ˙ u k L + k√ ρ k L ∞ k u k L ∞ k u x k L ≤ C. We complete the proof of Lemma 3.7. n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity This section is devoted to the proof of Theorem 1.1. To do this, we also need thefollowing global a priori estimates of the solution (˜ ρ, ˜ u, ˜ b ) to the problem (1.3) and (1.4). Proposition 4.1.
Let (˜ ρ, ˜ u, ˜ b ) be a smooth solution of (1.3) , (1.4) and (1.5) , then forsome index < α ≤ , we have ≤ ˜ ρ ( x, t ) ≤ C, ∀ ( x, t ) ∈ R × [0 , T ] , (4.1)sup ≤ t ≤ T (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)
12 ˜ ρ ˜ u + Φ(˜ ρ ) + (˜ b − ¯ b ) ! (1 + | x | α ) (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) L + ˆ T µ (cid:13)(cid:13)(cid:13) ˜ u x (1 + | x | α ) (cid:13)(cid:13)(cid:13) L dt ≤ C, (4.2) and sup ≤ t ≤ T (cid:16) k ˜ u k H + k ˜ b x k L + k ˜ ρ x k L + k p ˜ ρ ˙˜ u k L (cid:17) + ˆ T µ k ˜ u xt k L ≤ C. (4.3) Proof.
In fact, repeating the arguments in the proofs of Lemma 3.1-Lemma 3.7 step bystep, we can easily obtain (4.1)- (4.3). Here, we omit the details.The existence and uniqueness of global strong solution in Theorem 1.1 can be obtainedfrom the local existence in time and the global (in time) a priori estimates in Proposition4.1 by a standard continuity argument c.f. [16] .
The existence and uniqueness of global strong solution in Theorem 1.2:
The existence and uniqueness of global strong solution in Theorem 1.2 can be obtainedby the local existence in time and the global (in time) a priori estimates in Sections 3 bya standard continuity argument c.f. [26].
The non-resistive limit in Theorem 1.2:
To justify the non-resistive limit as ν →
0, we consider the the difference of these two solutions ( ρ − ˜ ρ, u − ˜ u, b − ˜ b ) whichsatisfy the following: ( ρ − ˜ ρ ) t + ( ρ − ˜ ρ ) u x + ˜ ρ ( u − ˜ u ) x + ( ρ − ˜ ρ ) x u + ˜ ρ x ( u − ˜ u ) = 0 ,ρ ( u − ˜ u ) t + ρu ( u − ˜ u ) x − µ ( u − ˜ u ) xx = − ( ρ − ˜ ρ )(˜ u t + ˜ u ˜ u x ) − ρ ( u − ˜ u )˜ u x − ( P ( ρ ) − P (˜ ρ )) x − (cid:16) b − ˜ b (cid:17) x , ( b − ˜ b ) t + u x ( b − ˜ b ) + ˜ b ( u − ˜ u ) x + u ( b − ˜ b ) x + ( u − ˜ u )˜ b x = νb xx . (5.1)Firstly, multiplying the equation (5.1) by 2( ρ − ˜ ρ ), integrating the resultant over R , and integrating by parts, it follows from H¨older’s inequality, the Gagliardo-Nirenberg n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity
0, we consider the the difference of these two solutions ( ρ − ˜ ρ, u − ˜ u, b − ˜ b ) whichsatisfy the following: ( ρ − ˜ ρ ) t + ( ρ − ˜ ρ ) u x + ˜ ρ ( u − ˜ u ) x + ( ρ − ˜ ρ ) x u + ˜ ρ x ( u − ˜ u ) = 0 ,ρ ( u − ˜ u ) t + ρu ( u − ˜ u ) x − µ ( u − ˜ u ) xx = − ( ρ − ˜ ρ )(˜ u t + ˜ u ˜ u x ) − ρ ( u − ˜ u )˜ u x − ( P ( ρ ) − P (˜ ρ )) x − (cid:16) b − ˜ b (cid:17) x , ( b − ˜ b ) t + u x ( b − ˜ b ) + ˜ b ( u − ˜ u ) x + u ( b − ˜ b ) x + ( u − ˜ u )˜ b x = νb xx . (5.1)Firstly, multiplying the equation (5.1) by 2( ρ − ˜ ρ ), integrating the resultant over R , and integrating by parts, it follows from H¨older’s inequality, the Gagliardo-Nirenberg n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity ddt ˆ ( ρ − ˜ ρ ) = − ˆ ( ρ − ˜ ρ ) u x − ˆ ˜ ρ ( ρ − ˜ ρ )( u − ˜ u ) x − ˆ ˜ ρ x ( ρ − ˜ ρ )( u − ˜ u ) ≤ (cid:0) k u x k L ∞ k ρ − ˜ ρ k L + k ˜ ρ k L ∞ k ρ − ˜ ρ k L k ( u − ˜ u ) x k L + k ˜ ρ x k L k ρ − ˜ ρ k L k u − ˜ u k L ∞ (cid:1) ≤ C (cid:20) k u x k L k u xx k L k ρ − ˜ ρ k L + (cid:18) k ( u − ˜ u ) x k L + k u − ˜ u k L k ( u − ˜ u ) x k L (cid:19) k ρ − ˜ ρ k L (cid:21) ≤ ε k ( u − ˜ u ) x k L + C (1 + k u xx k L ) (cid:0) k ρ − ˜ ρ k L + 1 (cid:1) . (5.2)Secondly, we multiply (5.1) by 2( u − ˜ u ), integrate the resultant over R , integrate byparts, and use H¨older’s inequality, the Gagliardo-Nirenberg inequality, Young’s inequality,the fact that k u − ˜ u k L ≤ C (cid:0) k ( u − ˜ u ) x k L (cid:1) (see (3.11)) and Proposition 4.1 to get ddt ˆ ρ ( u − ˜ u ) + 2 µ ˆ ( u − ˜ u ) x = − ˆ ρ − ˜ ρ )( u − ˜ u )(˜ u t + ˜ u ˜ u x ) − ˆ ρ ( u − ˜ u ) ˜ u x − ˆ P ( ρ ) − P (˜ ρ )) x ( u − ˜ u ) − ˆ ( b − ˜ b ) x ( u − ˜ u ) ≤ C (cid:16) k ˙˜ u k L k ρ − ˜ ρ k L k u − ˜ u k L ∞ + k ˜ u x k L ∞ k√ ρ ( u − ˜ u ) k L + ( k ρ − ˜ ρ k L + k b − ˜ b k L ) k ( u − ˜ u ) x k L (cid:17) ≤ C h k ˙˜ u k L k u − ˜ u k L k ( u − ˜ u ) x k L k ρ − ˜ ρ k L + k ˜ u x k L k ˜ u xx k L k√ ρ ( u − ˜ u ) k L + ( k ρ − ˜ ρ k L + k b − ˜ b k L ) k ( u − ˜ u ) x k L i ≤ ε k ( u − ˜ u ) x k L + C (cid:16) k ρ − ˜ ρ k L + k b − ˜ b k L + k√ ρ ( u − ˜ u ) k L (cid:17)(cid:0) k ˙˜ u k L + k ˜ u xx k L (cid:1) . (5.3)Thirdly, multiplying the equation (5.1) by 2( b − ˜ b ), integrating the resultant over R ,integrating by parts and using Proposition 4.1, the third equation of (1.3), the fact that k ˜ b k L ∞ ≤ C k ˜ b k L k ˜ b x k L , we deduce ddt ˆ ( b − ˜ b ) = − ˆ ( b − ˜ b ) u x − ˆ ˜ b ( b − ˜ b )( u − ˜ u ) x − ˆ ( b − ˜ b )( u − ˜ u )˜ b x + 2 ν ˆ ( b − ˜ b ) b xx ≤ C (cid:16) k u x k L ∞ k b − ˜ b k L + k ˜ b k L ∞ k b − ˜ b k L k ( u − ˜ u ) x k L + k b − ˜ b k L k ˜ b x k L k u − ˜ u k L ∞ + ν k b − ˜ b k L k b xx k L (cid:17) ≤ ε k ( u − ˜ u ) x k L + C (1 + k u xx k L ) k b − ˜ b k L + ν k b xx k L . (5.4)Then combining (5.2)-(5.4) and choosing ε > ddt ˆ (cid:16) ( ρ − ˜ ρ ) + ρ ( u − ˜ u ) + ( b − ˜ b ) (cid:17) + 2 µ ˆ ( u − ˜ u ) x ≤ C (cid:0) k u xx k L + k ˜ u xx k L + k ˙˜ u k L (cid:1) (cid:16) k ρ − ˜ ρ k L + k b − ˜ b k L + k√ ρ ( u − ˜ u ) k L + 1 (cid:17) + ν k b xx k L , n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity
0, we consider the the difference of these two solutions ( ρ − ˜ ρ, u − ˜ u, b − ˜ b ) whichsatisfy the following: ( ρ − ˜ ρ ) t + ( ρ − ˜ ρ ) u x + ˜ ρ ( u − ˜ u ) x + ( ρ − ˜ ρ ) x u + ˜ ρ x ( u − ˜ u ) = 0 ,ρ ( u − ˜ u ) t + ρu ( u − ˜ u ) x − µ ( u − ˜ u ) xx = − ( ρ − ˜ ρ )(˜ u t + ˜ u ˜ u x ) − ρ ( u − ˜ u )˜ u x − ( P ( ρ ) − P (˜ ρ )) x − (cid:16) b − ˜ b (cid:17) x , ( b − ˜ b ) t + u x ( b − ˜ b ) + ˜ b ( u − ˜ u ) x + u ( b − ˜ b ) x + ( u − ˜ u )˜ b x = νb xx . (5.1)Firstly, multiplying the equation (5.1) by 2( ρ − ˜ ρ ), integrating the resultant over R , and integrating by parts, it follows from H¨older’s inequality, the Gagliardo-Nirenberg n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity ddt ˆ ( ρ − ˜ ρ ) = − ˆ ( ρ − ˜ ρ ) u x − ˆ ˜ ρ ( ρ − ˜ ρ )( u − ˜ u ) x − ˆ ˜ ρ x ( ρ − ˜ ρ )( u − ˜ u ) ≤ (cid:0) k u x k L ∞ k ρ − ˜ ρ k L + k ˜ ρ k L ∞ k ρ − ˜ ρ k L k ( u − ˜ u ) x k L + k ˜ ρ x k L k ρ − ˜ ρ k L k u − ˜ u k L ∞ (cid:1) ≤ C (cid:20) k u x k L k u xx k L k ρ − ˜ ρ k L + (cid:18) k ( u − ˜ u ) x k L + k u − ˜ u k L k ( u − ˜ u ) x k L (cid:19) k ρ − ˜ ρ k L (cid:21) ≤ ε k ( u − ˜ u ) x k L + C (1 + k u xx k L ) (cid:0) k ρ − ˜ ρ k L + 1 (cid:1) . (5.2)Secondly, we multiply (5.1) by 2( u − ˜ u ), integrate the resultant over R , integrate byparts, and use H¨older’s inequality, the Gagliardo-Nirenberg inequality, Young’s inequality,the fact that k u − ˜ u k L ≤ C (cid:0) k ( u − ˜ u ) x k L (cid:1) (see (3.11)) and Proposition 4.1 to get ddt ˆ ρ ( u − ˜ u ) + 2 µ ˆ ( u − ˜ u ) x = − ˆ ρ − ˜ ρ )( u − ˜ u )(˜ u t + ˜ u ˜ u x ) − ˆ ρ ( u − ˜ u ) ˜ u x − ˆ P ( ρ ) − P (˜ ρ )) x ( u − ˜ u ) − ˆ ( b − ˜ b ) x ( u − ˜ u ) ≤ C (cid:16) k ˙˜ u k L k ρ − ˜ ρ k L k u − ˜ u k L ∞ + k ˜ u x k L ∞ k√ ρ ( u − ˜ u ) k L + ( k ρ − ˜ ρ k L + k b − ˜ b k L ) k ( u − ˜ u ) x k L (cid:17) ≤ C h k ˙˜ u k L k u − ˜ u k L k ( u − ˜ u ) x k L k ρ − ˜ ρ k L + k ˜ u x k L k ˜ u xx k L k√ ρ ( u − ˜ u ) k L + ( k ρ − ˜ ρ k L + k b − ˜ b k L ) k ( u − ˜ u ) x k L i ≤ ε k ( u − ˜ u ) x k L + C (cid:16) k ρ − ˜ ρ k L + k b − ˜ b k L + k√ ρ ( u − ˜ u ) k L (cid:17)(cid:0) k ˙˜ u k L + k ˜ u xx k L (cid:1) . (5.3)Thirdly, multiplying the equation (5.1) by 2( b − ˜ b ), integrating the resultant over R ,integrating by parts and using Proposition 4.1, the third equation of (1.3), the fact that k ˜ b k L ∞ ≤ C k ˜ b k L k ˜ b x k L , we deduce ddt ˆ ( b − ˜ b ) = − ˆ ( b − ˜ b ) u x − ˆ ˜ b ( b − ˜ b )( u − ˜ u ) x − ˆ ( b − ˜ b )( u − ˜ u )˜ b x + 2 ν ˆ ( b − ˜ b ) b xx ≤ C (cid:16) k u x k L ∞ k b − ˜ b k L + k ˜ b k L ∞ k b − ˜ b k L k ( u − ˜ u ) x k L + k b − ˜ b k L k ˜ b x k L k u − ˜ u k L ∞ + ν k b − ˜ b k L k b xx k L (cid:17) ≤ ε k ( u − ˜ u ) x k L + C (1 + k u xx k L ) k b − ˜ b k L + ν k b xx k L . (5.4)Then combining (5.2)-(5.4) and choosing ε > ddt ˆ (cid:16) ( ρ − ˜ ρ ) + ρ ( u − ˜ u ) + ( b − ˜ b ) (cid:17) + 2 µ ˆ ( u − ˜ u ) x ≤ C (cid:0) k u xx k L + k ˜ u xx k L + k ˙˜ u k L (cid:1) (cid:16) k ρ − ˜ ρ k L + k b − ˜ b k L + k√ ρ ( u − ˜ u ) k L + 1 (cid:17) + ν k b xx k L , n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity ˆ (cid:16) ( ρ − ˜ ρ ) + ρ ( u − ˜ u ) + ( b − ˜ b ) (cid:17) + 2 µ ˆ ( u − ˜ u ) x ≤ Cν ˆ T k b xx k L dt exp (cid:26) C ˆ T (cid:0) k u xx k L + k ˜ u xx k L + k ˙˜ u k L (cid:1) dt (cid:27) ≤ Cν, (5.5)where we have used Lemmas 3.5 and 3.7, Proposition 4.1 and the following fact: k ˙˜ u k L ≤ C k√ ¯ ρ ˙˜ u k L ≤ C (cid:16) k ( √ ¯ ρ − p ˜ ρ )˜ u k L + k p ˜ ρ ˜ u k L (cid:17) ≤ C (cid:0) k ˜ ρ − ¯ ρ k L k ˙˜ u k L ∞ + 1 (cid:1) ≤ ε k ˙˜ u k L + C ( k ˙˜ u x k L + 1) , which gives k ˙˜ u k L ≤ C ( k (˜ u t + ˜ u ˜ u x ) x k L + 1) ≤ C (cid:0) k ˜ u xt k L + k ˜ u x k L + k ˜ u ˜ u xx k L + 1 (cid:1) ≤ C ( k ˜ u xt k L + k ˜ u xx k L + 1) ∈ L (0 , T ) . Moreover, we have ˆ ¯ ρ ( u − ˜ u ) = ˆ ( ρ − ¯ ρ )( u − ˜ u ) + ˆ ρ ( u − ˜ u ) ≤ C k ρ − ¯ ρ k L k u − ˜ u k L + Cν ≤ C k ρ − ¯ ρ k L k u − ˜ u k L k ( u − ˜ u ) x k L + Cν ≤ ε k u − ˜ u k L + C k ρ − ¯ ρ k L k ( u − ˜ u ) x k L + Cν ≤ ε k u − ˜ u k L + Cν + Cν which together with (5.5) yields thatsup ≤ t ≤ T k u − ˜ u k L ≤ Cν. (5.6)Then we have proved the results of Theorem 1.2.
Z. Li is supported by the NSFC (No.11671319, No.11931013), Fund of HPU (No.B2016-57). H. Wang is supported by the National Natural Science Foundation of China (GrantNo.11901066), the Natural Science Foundation of Chongqing (Grant No.cstc2019jcyj-msxmX0167) and Project No.2019CDXYST0015 and No. 2020CDJQY-A040 supportedby the Fundamental Research Funds for the Central Universities. Y. Ye is supported bythe National Natural Science Foundation of China (No.11701145) and China PostdoctoralScience Foundation (No.2020M672196). n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity
Z. Li is supported by the NSFC (No.11671319, No.11931013), Fund of HPU (No.B2016-57). H. Wang is supported by the National Natural Science Foundation of China (GrantNo.11901066), the Natural Science Foundation of Chongqing (Grant No.cstc2019jcyj-msxmX0167) and Project No.2019CDXYST0015 and No. 2020CDJQY-A040 supportedby the Fundamental Research Funds for the Central Universities. Y. Ye is supported bythe National Natural Science Foundation of China (No.11701145) and China PostdoctoralScience Foundation (No.2020M672196). n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity References [1] L. Caffarelli, R. Kohn, L. Nirenberg, First order interpolation inequality with weights.Compositio Mathematica 53 (1984), 259-275.[2] F. Catrina, Z.Q. Wang, On the Caffarelli-Kohn-Nirenberg inequalities: Sharp con-stants, existence (and nonexistence), and symmetry of extremal functions. Comm.Pure Appl. Math. LIV (2001), 229-258.[3] G.Q. Chen, D.H. Wang, Globla solutions of nonlinear magnetohydrodynamics withlarge initial data. J. Differential Equations 182 (2002), 344-376.[4] J.S. Fan, Y.X. Hu, Global strong solutions to the 1-D compressible magnetohydro-dynamic equations with zero resistivity. J. Math. Phys. 56 (2015), no. 2, 023101, 13pp.[5] J.S. Fan, W.H. Yu, Strong solution to the compressible magnetohydrodynamic equa-tions with vacuum. Nonlinear Anal. Real World Appl. 10 (2009), no. 1, 392-409.[6] J.S. Fan, S.X. Huang, F.C. Li, Global strong solutions to the planar compressiblemagnetohydrodynamic equations with large initial data and vacuum. Kinet. Relat.Models 10 (2017), no. 4, 1035-1053.[7] E. Gagliardo, Ulteriori propriet di alcune classi di funzioni in pi`u vari-abili. RicercheMat., 8 (1959), 24-51.[8] G.Y. Hong, X.F. Hou, H.Y. Peng, C.J. Zhu, Global existence for a class of largesolutions to three-dimensional compressible magnetohydrodynamic equations withvacuum. SIAM J. Math. Anal. 49 (2017), no. 4, 2409-2441.[9] X.P. Hu, D.H. Wang, Global existence and large-time behavior of solutions to thethree-dimensional equations of compressible magnetohydrodynamic flows. Arch. Ra-tion. Mech. Anal. 197 (2010), no. 1, 203-238.[10] S. Jiang, J.W. Zhang, On the non-resistive limit and the magnetic boundary-layer forone-dimensional compressible magnetohydrodynamics. Nonlinearity 30 (2017), no. 9,3587-3612.[11] Q.S. Jiu, M.J. Li, Y.L. Ye, Global classical solution of the Cauchy problem to 1Dcompressible Navier-Stokes equations with large initial data. J. Differential Equations257 (2014), no. 2, 311-350.[12] S. Kawashima, Systems of a hyperbolic-parabolic composite type, with applicationsto the equations of magnetohydrodynamics. PhD thesis, Kyoto University, (1983).[13] S. Kawashima, M. Okada, Smooth global solutions for the one-dimensional equationsin magnetohydrodynamics. Proc. Japan Acad. Ser. A Math. Sci. 58 (1982), 384-387.[14] H.L. Li, X.Y. Xu, J.W. Zhang, Global classical solutions to 3D compressible magne-tohydrodynamic equations with large oscillations and vacuum. SIAM J. Math. Anal.45 (2013), no. 3, 1356-1387. n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity
Z. Li is supported by the NSFC (No.11671319, No.11931013), Fund of HPU (No.B2016-57). H. Wang is supported by the National Natural Science Foundation of China (GrantNo.11901066), the Natural Science Foundation of Chongqing (Grant No.cstc2019jcyj-msxmX0167) and Project No.2019CDXYST0015 and No. 2020CDJQY-A040 supportedby the Fundamental Research Funds for the Central Universities. Y. Ye is supported bythe National Natural Science Foundation of China (No.11701145) and China PostdoctoralScience Foundation (No.2020M672196). n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity References [1] L. Caffarelli, R. Kohn, L. Nirenberg, First order interpolation inequality with weights.Compositio Mathematica 53 (1984), 259-275.[2] F. Catrina, Z.Q. Wang, On the Caffarelli-Kohn-Nirenberg inequalities: Sharp con-stants, existence (and nonexistence), and symmetry of extremal functions. Comm.Pure Appl. Math. LIV (2001), 229-258.[3] G.Q. Chen, D.H. Wang, Globla solutions of nonlinear magnetohydrodynamics withlarge initial data. J. Differential Equations 182 (2002), 344-376.[4] J.S. Fan, Y.X. Hu, Global strong solutions to the 1-D compressible magnetohydro-dynamic equations with zero resistivity. J. Math. Phys. 56 (2015), no. 2, 023101, 13pp.[5] J.S. Fan, W.H. Yu, Strong solution to the compressible magnetohydrodynamic equa-tions with vacuum. Nonlinear Anal. Real World Appl. 10 (2009), no. 1, 392-409.[6] J.S. Fan, S.X. Huang, F.C. Li, Global strong solutions to the planar compressiblemagnetohydrodynamic equations with large initial data and vacuum. Kinet. Relat.Models 10 (2017), no. 4, 1035-1053.[7] E. Gagliardo, Ulteriori propriet di alcune classi di funzioni in pi`u vari-abili. RicercheMat., 8 (1959), 24-51.[8] G.Y. Hong, X.F. Hou, H.Y. Peng, C.J. Zhu, Global existence for a class of largesolutions to three-dimensional compressible magnetohydrodynamic equations withvacuum. SIAM J. Math. Anal. 49 (2017), no. 4, 2409-2441.[9] X.P. Hu, D.H. Wang, Global existence and large-time behavior of solutions to thethree-dimensional equations of compressible magnetohydrodynamic flows. Arch. Ra-tion. Mech. Anal. 197 (2010), no. 1, 203-238.[10] S. Jiang, J.W. Zhang, On the non-resistive limit and the magnetic boundary-layer forone-dimensional compressible magnetohydrodynamics. Nonlinearity 30 (2017), no. 9,3587-3612.[11] Q.S. Jiu, M.J. Li, Y.L. Ye, Global classical solution of the Cauchy problem to 1Dcompressible Navier-Stokes equations with large initial data. J. Differential Equations257 (2014), no. 2, 311-350.[12] S. Kawashima, Systems of a hyperbolic-parabolic composite type, with applicationsto the equations of magnetohydrodynamics. PhD thesis, Kyoto University, (1983).[13] S. Kawashima, M. Okada, Smooth global solutions for the one-dimensional equationsin magnetohydrodynamics. Proc. Japan Acad. Ser. A Math. Sci. 58 (1982), 384-387.[14] H.L. Li, X.Y. Xu, J.W. Zhang, Global classical solutions to 3D compressible magne-tohydrodynamic equations with large oscillations and vacuum. SIAM J. Math. Anal.45 (2013), no. 3, 1356-1387. n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity n non-resistive limit of 1D MHD equations with no vacuum at infinityn non-resistive limit of 1D MHD equations with no vacuum at infinity