# Global Existence and Asymptotic Behavior of Solutions for Compressible Two-Fluid Euler–Maxwell Equation

Authors: Ismahan Binshati; Harumi Hattori
Journal: International Journal of Differential Equations
 (2020)

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## Abstract

We study the global existence and asymptotic behavior of the solutions for two-fluid compressible isentropic Euler–Maxwell equations by the Fourier transform and energy method. We discuss the case when the pressure for two fluids is not identical, and we also add friction between the two fluids. In addition, we discuss the rates of decay ofLp−Lq norms for a linear system. Moreover, we use the result for Lp−Lq estimates to prove the decay rates for the nonlinear systems.


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## Body

## 1. Introduction

We consider the Cauchy problem for the first-order nonlinear two-fluid compressible isentropic Euler–Maxwell equations in three dimensions. In the following system of equations, the first equation is the conservation of the mass. The second equation is conservation of the momentum, to which we added frictional dampingα±u±−u∓ besides the damping ν±u±. Then, the compressible two-fluid Euler–Maxwell equations can be written as(1)∂tn±+∇.n±u±=0,∂tu±+u±.∇u±+1n±∇p±n±=±E+u±×B−ν±u±−α±u±−u∓,∂tE−∇×B=n−u−−n+u+,∂tB+∇×E=0,∇.E=n+−n−,∇.B=0,where n±=n±t,x∈ℝ3 denotes the density of electrons (n−) and ions (n+), u±=u±t,x∈ℝ3 denotes the velocity of electrons (u−) and ions (u+), E=Et,x∈ℝ3 denotes the electric field, and B=Bt,x∈ℝ3 denotes the magnetic field for t>0,x∈ℝ3. The initial data are given by(2)n±,u±,E,Bt=0=n±0,u±0,E0,B0,x∈ℝ3,with the compatibility conditions(3)∇.E0=n+0−n−0,∇.B0=0,x∈ℝ3.The Euler–Maxwell system (1) is a symmetrizable hyperbolic system for n>0, and the initial value problems (1) and (2) have a local smooth solution when the initial data are smooth. The global existence of smooth solutions to the initial boundary value problem has been given in [1] by the compensated compactness method. The authors in [2, 3] studied the existence of global smooth solutions for the three-dimensional isentropic Euler–Maxwell system with small amplitude, and the periodic problem was discussed by Uedaet al. [4]. For the special case where the solution to the Euler–Maxwell equation has asymptotic limits with small parameters, see [5, 6]. The special case of the diffusive relaxation limit of the three-dimensional nonisentropic Euler–Maxwell equation is considered in [7, 8]. Two hierarchies of models of the ionospheric plasma for two-fluid Euler–Maxwell equations were presented in [9]. The Fourier transform method was considered by Duan [2, 10] and Kawashima and Ueda [11]. Jerome [12] adapted the classical semigroup-resolvent approach of Kato [13] to the Cauchy problem in ℝ3 and established a local smooth solution. In [2], Duan considered the case when the pressure function p±⋅ depend only on density, having the expression p±n±=A±n±γ with constants A+=A−>0 and the adiabatic exponent γ>1.In this paper, we consider the global existence of smooth solutions for the two-fluid compressible isentropic Euler–Maxwell equation extending the results of Duan [2]. In contrast to Duan, we suppose A+≠A−, and we add the friction α±u±−u∓ where α+=α/n+>0,α−=α/n−>0, and α is a constant.Theorem 1.
LetN≥4 and (3) hold. There are δ0>0andC0 such that if(4)n±0−1,u±0,E0,B0N≤δ0,where .N is the HN norm, then the Cauchy problems (1) and (2) of the Euler–Maxwell system admit a unique global solution n±t,x,u±t,x,Et,x,Bt,x with(5)n±t,x−1,u±t,x,Et,x,Bt,x∈C0,∞;HNℝ3∩Lip0,∞;HN−1ℝ3,supt≥0n±t−1,u±t,Et,BtN≤C0n±0−1,u±0,E0,B0N.We obtain the decay rates of smooth solutions by the Fourier transform. The main results are stated as follows.Theorem 2.
There areδ1>0andC1 such that if(6)n±0−1,u±0,E0,B013+n±0−1,u±0,E0,B0L1≤δ1,where .13 is the H13 norm, then the solution n±t,x,u±t,x,Et,x,Bt,x satisfies that for any t≥0,(7)n±tLq≤C11+t−11/4,(8)u±t,EtLq≤C11+t−2+3/2q,(9)BtLq≤C11+t−3/2+3/2q,with 2≤q≤∞. Furthermore,(10)U±t−Gt∗U±0Lq≤Cε13V01+t−2+3/2q,where Gt,x is Green’s matrix for the linearized system.The proof of Theorem1 and Theorem 2 is based on the energy method and the Fourier transform, as in [2]. There are three key steps: the first key step is the a priori estimate to establish the global solution and has the form(11)ΕNVt+∫0tDNVsds≤ΕNV0,where Vt is the perturbation of solution (1) and ΕN.andDN. denote the energy functional and energy dissipation rate functional as in [2]. This differs from [4, 10] because the two-fluid system has a more complex structure than one fluid, so obtaining energy estimates for the density, velocity, and electric magnetic fields for Euler–Maxwell require a different strategy. The time decay property of solutions to the nonlinear system requires the construction of functionals, capturing the optimal energy dissipation rate. The second key step is linearizing the homogeneous form of (1) and using the Fourier transform to obtain the Lp−Lq time decay rate and the explicit representation of the solution. The third step is combining the previous two steps and applying the Fourier transform to obtain the time decay rate of the solution to the reformulated nonlinear system to finish the proof of Theorem 1. Thus, the solutions can be represented by the solution of the linearized system and the refined energy estimates using Duhamel’s principle.We introduce some notations that we will use later in this paper. For any integerN≥0,HNandH˙N denote the Sobolev space HNℝ3 and the Nth-order homogeneous Sobolev space, respectively. Set L2=H0. The norm of HN is denoted by ⋅N with ⋅=⋅0. The inner product in L2ℝ3 is denoted by ⋅,⋅, i.e.,(12)f,g=∫ℝ3fxgxdx,f=fx,g=gx∈L2ℝ3.We denote∂α=∂x1α1∂x2α2∂x3α3=∂1α1∂2α2∂3α3 for the multi-index α=α1,α2,α3, and the length of α is α=α1+α2+α3. In addition, C and λ denote some positive constants, where both C and λ may take different values in different places.We organize this paper as follows. InSection 2, we reformulate the Cauchy problem and consider the proof of global existence and uniqueness of solutions. In Section 3, we discuss the time rate of decay for linearized systems, and we obtain the linearized system for 9×9. Finally, in Section 4, we discuss the time decay rate of solutions of the nonlinear system (15) and complete the proof of Theorems and 2.


## 2. Global Solution for the Nonlinear System

### 2.1. Reformulation of the Problem

Denote byn±,u±,E,B a smooth solution to system (1) with initial data (2) satisfying (3). Let(13)σ±t,x=A±2γ−1n±tγ,xγ−1/2−1,υ±=1γu±tγ,x,E˜=1γEtγ,x,B˜=1γBtγ,x.DefineV≔σ±,υ±,E˜,B˜ and(14)ϕσ±≔γ−12A±σ±+12/γ−1−σ±A±−1.Note thatV satisfies(15)∂tσ±+A±γ−12A±σ±+1∇.υ±+υ±.∇σ±=0,∂tυ±+υ±.∇υ±+A±γ−12A±σ±+1∇σ±=±1γE˜+υ±×B˜−1γν±υ±−1γα±υ±−υ∓,∂tE˜−1γ∇×B˜=1γυ−+1γϕσ−+σ−A−υ−−1γυ+−1γϕσ++σ+A+υ+,∂tB˜−1γ∇×E˜=0,∇.E˜=−1γϕσ−+σ−A−+1γϕσ++σ+A+,∇.B˜=0,t>0,x∈ℝ3,with the initial data(16)V|t=0=V0≔σ±0,υ±0,E0˜,B0˜,x∈ℝ3.Here, we have used the notationV0=σ±0,υ±0,E0˜,B0˜ for the special case where n±0,u±0,E0,B0 is substituted into (13). Note that V0 satisfies(17)∇.E0˜=−1γϕσ−0+σ−0A−+1γϕσ+0+σ+0A+,∇.B0˜=0,x∈ℝ3.SupposeU≔n±,u±,E,B is a smooth solution to the initial value problem of the original Cauchy problems (1) and (2), which satisfy (3). Now, we introduce another transformation by setting ρ±t,x=n±t,x−1, then U≔ρ±,u±,E,B satisfies(18)∂tρ±+∇.u±=−∇.ρ±u±,∂tu±+u±∓E+A±γ∇ρ±=−u±.∇u±−A±γρ±+1γ−2−1∇ρ±±u±×B−α±u±−u∓,∂tE−∇×B−u−+u+=ρ−u−−ρ+u+,∂tB+∇×E=0,∇.E=ρ+−ρ−,∇.B=0,t>0,x∈ℝ3,with the initial data(19)Ut=U0≔ρ±0,u±0,E0,B0,x∈ℝ3,satisfying the compatibility condition(20)∇.E0=ρ+0−ρ−0,∇.B0=0,where ρ±0=n±0−1.We will assumeN≥4 is an integer. In addition to V=σ±,υ±,E˜,B˜, define the full instant energy functional ΕNVt and the high-order instant energy functional ΕNhVt via(21)εNVt=VtN2+k1∑α|≤N−1∂αυ+,∇∂ασ++∂αυ−,∇∂ασ−+k2∑α≤N−1∂αυ+−υ−,∇∂αE˜−k3∑α≤N−2∇×∂αE˜,∂αB˜,(22)εNhVt=∇VtN−12+k1∑1≤α≤N−1∂αυ+,∇∂ασ++∂αυ−,∇∂ασ−+k2∑1≤α≤N−1∂αυ+−υ−,∇∂αE˜−k3∑1≤α≤N−2∇×∂αE˜,∂αB˜,where 0<k3≪k2≪k1≪1 are constants to be chosen later in the proof such that ki,i=1,2,3 are small enough compared to 1 and satisfy(23)ℰNVt∼σ±,υ±,E˜,B˜N2,ℰNhVt∼∇σ±,υ±,E˜,B˜N−12.Define the dissipation ratesDNVtandDNhVt by(24)DNVt=υ+,υ−N2+∇σ+,σ−N−12+∇E˜,B˜N−22+E˜2+σ+−σ−2,(25)DNhVt=∇υ+,υ−N−12+∇2σ+,σ−N−22+∇2E˜,B˜N−32+∇E˜2+∇σ+−σ−2.Proposition 1.
Suppose initial dataV0=σ±0,υ±0,E0˜,B0˜ satisfies (17). Then, there exist ℰN⋅ and DN⋅ having the forms (21) and (24), respectively, such that if ℰNV0>0 is sufficiently small, the Cauchy problems (15) and (16) admit a unique, global, nonzero solution V=σ±,υ±,E˜,B˜, satisfying(26)V∈C0,∞;HNℝ3∩Lip0,∞;HN−1ℝ3,(27)ℰNVt+λ∫0tDNVsds≤ℰNV0,for any t≥0.Remark 1.
The solutions obtained in Proposition1 indeed represent the decay rates in time under some regularity and integrability conditions on initial data V0=σ±0,V±0,E0˜,B0˜ and set(28)ϵmV0=V0m+υ±0,E0˜,B0˜L1,for the integer m≥4.Remark 2.
Note that the existence result in Theorem1 follows from Proposition 1, the derivation of rates of (7) and (9) in Theorem 1, and Proposition 2. The proof of Proposition 2 is analogous to that of Lemma 5.2 in [10].


### 2.2. A Priori Estimates

In this section, we obtain uniform-in-time a priori estimates for smooth solutions to the Cauchy problems (15) and (16) by using the classical energy method.Theorem 3.
Let0<T≤∞ be given. Suppose that V=σ±,υ±,E˜,B˜∈C0,T;HNℝ3 is smooth, that σ satisfies(29)sup0≤t≤Tσ±tN≤1,and that V solves system (15) for t∈0,T. Then, there exist ℰN. and DN. having the forms (21) and (24) such that for all 0≤t≤T,(30)ddtℰNVt+λDNVt≤CℰNVt1/2+ℰNVtDNVt.Proof.
Performing the energy estimate, we obtain the following results:Step 1. We apply ∂α to the first equation of (15) and then multiply that equation by ∂ασ±; also, we apply ∂α to the second equation of (15) and then multiply that equation by ∂αυ±; after many steps, we get(31)12ddt‖V‖N2+1γν±‖υ+,υ−‖N2≤C‖V‖N‖υ+,υ−‖2+‖∇σ+,σ−,υ+,υ−‖N−12.Step 2. We rewrite the first and second equations of (15) by putting the linear terms on the left-hand sides and the nonlinear terms on the right-hand sides:(32)∂tσ++A+∇⋅υ+=−υ+⋅∇σ+−γ−12σ+∇⋅υ+≔f1+,(33)∂tσ−+A−∇⋅υ−=−υ−⋅∇σ−−γ−12σ−∇⋅υ−≔f1−,(34)∂tυ++A+∇σ+−1γE˜+ν+γυ++α+γυ+−υ−=−υ+⋅∇υ+−γ−12σ+∇σ++υ+×B˜≔f2+,(35)∂tυ−+A−∇σ−+1γE˜+ν−γυ−+α−γυ−−υ+=−υ−⋅∇υ−−γ−12σ−∇σ−−υ−×B˜≔f2−.Letα≤N−1. If we apply ∂α to (34), multiply by ∇∂ασ+, integrate in x, and then combine the result with an application of ∂α to (35) after which we multiply by ∇∂ασ− and integrate in x, then we get(36)ddtℰN,1intV+λ∇σ+,σ−N−12+σ+A+−σ−A−N2≤Cυ+,υ−N2+Cσ+,σ−,υ+,υ−,B˜N2∇σ+,σ−,υ+,υ−N−12,where(37)ℰN,1intV=∑α≤N−1∂αυ+,∇∂ασ++∂αυ−,∇∂ασ−.Step 3. We subtract equation (35) from equation (33) to get(38)∂tυ+−υ−+A+∇σ+−A−∇σ−+2γE˜=−∂αυ+⋅υ++∂αυ+⋅υ+−γ−12∂ασ+∇σ++γ−12∂ασ−∇σ−−∂αυ+×B˜−∂αυ−×B˜−1γ∂αν+υ+−ν−υ−−1γ∂αα+υ+−υ−−α−υ−−υ+.Now, we apply∂α to (38), multiply by ∂αE, integrate by parts in x, and replace ∂αE˜ with the third equation of (16). Then, we have(39)ddtℰN,2intV+λE˜N−12≤Cυ+,υ−N2+Cσ+,σ−N−12+Cυ+,υ−N∇B˜N−2+Cσ+,σ−,υ+,υ−,B˜N2∇σ+,σ−,υ+,υ−N−12,where(40)ℰN,2intυ=∑α≤N−1∂αυ+−υ−,∂αE˜forα≤N−1.Step 4. We apply ∂α to the third equation of (15), multiply by ∂α∇×B˜, integrate by parts in x, and use the relation(41)∂α∂xiB˜=∂xiΔ−1∇×∇×∂αB˜≤∇×∂αB˜,for each1≤i≤3. Then, we obtain(42)ddtℰN,3intV+λ∇B˜N−22≤Cυ+,υ−,E˜N−12+Cσ+,σ−N2∇υ+,υ−N−12,where(43)ℰN,3intυ=−∑α≤N−2∇×∂αE˜,∂αB˜for allα≤N−2.Step 5. Utilizing steps (1)–(4) above, we can now prove (30). Define(44)ℰNVt=∇VtN2+∑i=13ℰNintVt,and note that constants 0<k3≪k2≪k1≪1 are to be determined. We observe that if 0<ki≪1 for i=1,2,3, are sufficiently small, then ℰNV∼VN2 holds. Furthermore, by letting 0≪k3≪k2≪k1≪1 be sufficiently small, taking k23/2≪k3, and taking the sum of (31), k1× (36), k2× (42), and k1× (42), we find that there exists λ>0andC>0 such that (30) is satisfied:(45)12ddtVN2+ν±γυ+,υ−N2+k1ddt∑α≤N∂αυ+,∇∂ασ++∂αυ−,∇∂ασ−+k1λ∇σ+,σ−N−12+σ+A+−σ−A−N2+k2ddt∑α≤N−1∂αυ+−υ−,∂αE˜+k2λE˜N−12−k3ddt∑α≤N−2∇×∂αE˜,∂αB˜+k3λ∇B˜N−22≤CVNυ+,υ−2+∇σ+,σ−,υ+,υ−N−12+k1Cυ+,υ−N2+k1Cσ+,σ−,υ+,υ−,B˜N2∇σ+,σ−,υ+,υ−N−12+k2Cυ+,υ−N2+k2Cσ+,σ−N−12+k2Cυ+,υ−N∇B˜N−2+k2Cσ+,σ−,υ+,υ−,B˜N2∇σ+,σ−,υ+,υ−N−12+k3Cυ+,υ−,E˜N−12+k3Cσ+,σ−N2∇υ+,υ−N−12.
If we now let(46)V≤ℰ1/2,σ+,σ−,υ+,υ−,B˜N2=ℰN,∇σ+,σ−,υ+,υ−N−12=DN,then(47)ddtVN2+k1∑α|≤N∂αυ+,∇∂ασ++∂αυ−,∇∂ασ−+k2∑α≤N−1∂αυ+−υ−,∂αE−k3ddt∑α≤N−2∇×∂αE˜,∂αB˜+ν±γυ+,υ−N2+k2λ∇σ+,σ−N−12+k3λ∇B˜N−22+k2λ∇E˜N−22+k1λσ+A+−σ−A−N2≤CVN∇σ+,σ−,υ+,υ−N−12+k1Cσ+,σ−,υ+,υ−,B˜N2∇σ+,σ−,υ+,υ−N−12.
It follows that(48)ddtℰNVt+ν±γυ+,υ−N2+k2λ∇σ+,σ−N−12+k3λ∇B˜N−22+k2λ∇E˜N−22+k2λE˜2+k1λσ+A+−σ−A−N2≤CVN+∇σ+,σ−,υ+,υ−N−12∇σ+,σ−,υ+,υ−N−12.
Thus,(49)ddtℰNVt+λDNVt≤CℰNVt1/2+ℰNVtDNVt,and this concludes the proof.


### 2.3. Proof of Global Existence

We consider the global existence of the smooth solution to the isentropic Euler–Maxwell system for a quasilinear symmetric hyperbolic system (15). Therefore, we combine those a priori estimates with the local existence of solutions to extend the local solution up to infinite time by using the continuity of ℰNVt.Lemma 1 (local existence of smooth solution, see [2, 13, 14]).
AssumeV0∈HNℝ3 satisfies (17). Then, there exists T0>0 such that the Cauchy problems (15) and (16) admit a unique solution on 0,T0 with(50)V∈C0,T0;HNℝ3∩Lip0,T0;HN−1ℝ3.Proof of Proposition.
1. Since (15) is a quasilinear symmetric hyperbolic system, the global existence of smooth solutions follows from the local existence result in Lemma 1 (see also Section 16 of [14]). In addition, the a priori estimate (30) in Theorem 3 and the continuity argument show that ℰNVt is bounded uniformly in time under the assumption that ℰNV0>0 is sufficiently small. Therefore, global solutions satisfying (26) and (27) exist. This concludes the proof of Proposition 1.


## 2.1. Reformulation of the Problem

Denote byn±,u±,E,B a smooth solution to system (1) with initial data (2) satisfying (3). Let(13)σ±t,x=A±2γ−1n±tγ,xγ−1/2−1,υ±=1γu±tγ,x,E˜=1γEtγ,x,B˜=1γBtγ,x.DefineV≔σ±,υ±,E˜,B˜ and(14)ϕσ±≔γ−12A±σ±+12/γ−1−σ±A±−1.Note thatV satisfies(15)∂tσ±+A±γ−12A±σ±+1∇.υ±+υ±.∇σ±=0,∂tυ±+υ±.∇υ±+A±γ−12A±σ±+1∇σ±=±1γE˜+υ±×B˜−1γν±υ±−1γα±υ±−υ∓,∂tE˜−1γ∇×B˜=1γυ−+1γϕσ−+σ−A−υ−−1γυ+−1γϕσ++σ+A+υ+,∂tB˜−1γ∇×E˜=0,∇.E˜=−1γϕσ−+σ−A−+1γϕσ++σ+A+,∇.B˜=0,t>0,x∈ℝ3,with the initial data(16)V|t=0=V0≔σ±0,υ±0,E0˜,B0˜,x∈ℝ3.Here, we have used the notationV0=σ±0,υ±0,E0˜,B0˜ for the special case where n±0,u±0,E0,B0 is substituted into (13). Note that V0 satisfies(17)∇.E0˜=−1γϕσ−0+σ−0A−+1γϕσ+0+σ+0A+,∇.B0˜=0,x∈ℝ3.SupposeU≔n±,u±,E,B is a smooth solution to the initial value problem of the original Cauchy problems (1) and (2), which satisfy (3). Now, we introduce another transformation by setting ρ±t,x=n±t,x−1, then U≔ρ±,u±,E,B satisfies(18)∂tρ±+∇.u±=−∇.ρ±u±,∂tu±+u±∓E+A±γ∇ρ±=−u±.∇u±−A±γρ±+1γ−2−1∇ρ±±u±×B−α±u±−u∓,∂tE−∇×B−u−+u+=ρ−u−−ρ+u+,∂tB+∇×E=0,∇.E=ρ+−ρ−,∇.B=0,t>0,x∈ℝ3,with the initial data(19)Ut=U0≔ρ±0,u±0,E0,B0,x∈ℝ3,satisfying the compatibility condition(20)∇.E0=ρ+0−ρ−0,∇.B0=0,where ρ±0=n±0−1.We will assumeN≥4 is an integer. In addition to V=σ±,υ±,E˜,B˜, define the full instant energy functional ΕNVt and the high-order instant energy functional ΕNhVt via(21)εNVt=VtN2+k1∑α|≤N−1∂αυ+,∇∂ασ++∂αυ−,∇∂ασ−+k2∑α≤N−1∂αυ+−υ−,∇∂αE˜−k3∑α≤N−2∇×∂αE˜,∂αB˜,(22)εNhVt=∇VtN−12+k1∑1≤α≤N−1∂αυ+,∇∂ασ++∂αυ−,∇∂ασ−+k2∑1≤α≤N−1∂αυ+−υ−,∇∂αE˜−k3∑1≤α≤N−2∇×∂αE˜,∂αB˜,where 0<k3≪k2≪k1≪1 are constants to be chosen later in the proof such that ki,i=1,2,3 are small enough compared to 1 and satisfy(23)ℰNVt∼σ±,υ±,E˜,B˜N2,ℰNhVt∼∇σ±,υ±,E˜,B˜N−12.Define the dissipation ratesDNVtandDNhVt by(24)DNVt=υ+,υ−N2+∇σ+,σ−N−12+∇E˜,B˜N−22+E˜2+σ+−σ−2,(25)DNhVt=∇υ+,υ−N−12+∇2σ+,σ−N−22+∇2E˜,B˜N−32+∇E˜2+∇σ+−σ−2.Proposition 1.
Suppose initial dataV0=σ±0,υ±0,E0˜,B0˜ satisfies (17). Then, there exist ℰN⋅ and DN⋅ having the forms (21) and (24), respectively, such that if ℰNV0>0 is sufficiently small, the Cauchy problems (15) and (16) admit a unique, global, nonzero solution V=σ±,υ±,E˜,B˜, satisfying(26)V∈C0,∞;HNℝ3∩Lip0,∞;HN−1ℝ3,(27)ℰNVt+λ∫0tDNVsds≤ℰNV0,for any t≥0.Remark 1.
The solutions obtained in Proposition1 indeed represent the decay rates in time under some regularity and integrability conditions on initial data V0=σ±0,V±0,E0˜,B0˜ and set(28)ϵmV0=V0m+υ±0,E0˜,B0˜L1,for the integer m≥4.Remark 2.
Note that the existence result in Theorem1 follows from Proposition 1, the derivation of rates of (7) and (9) in Theorem 1, and Proposition 2. The proof of Proposition 2 is analogous to that of Lemma 5.2 in [10].


## 2.2. A Priori Estimates

In this section, we obtain uniform-in-time a priori estimates for smooth solutions to the Cauchy problems (15) and (16) by using the classical energy method.Theorem 3.
Let0<T≤∞ be given. Suppose that V=σ±,υ±,E˜,B˜∈C0,T;HNℝ3 is smooth, that σ satisfies(29)sup0≤t≤Tσ±tN≤1,and that V solves system (15) for t∈0,T. Then, there exist ℰN. and DN. having the forms (21) and (24) such that for all 0≤t≤T,(30)ddtℰNVt+λDNVt≤CℰNVt1/2+ℰNVtDNVt.Proof.
Performing the energy estimate, we obtain the following results:Step 1. We apply ∂α to the first equation of (15) and then multiply that equation by ∂ασ±; also, we apply ∂α to the second equation of (15) and then multiply that equation by ∂αυ±; after many steps, we get(31)12ddt‖V‖N2+1γν±‖υ+,υ−‖N2≤C‖V‖N‖υ+,υ−‖2+‖∇σ+,σ−,υ+,υ−‖N−12.Step 2. We rewrite the first and second equations of (15) by putting the linear terms on the left-hand sides and the nonlinear terms on the right-hand sides:(32)∂tσ++A+∇⋅υ+=−υ+⋅∇σ+−γ−12σ+∇⋅υ+≔f1+,(33)∂tσ−+A−∇⋅υ−=−υ−⋅∇σ−−γ−12σ−∇⋅υ−≔f1−,(34)∂tυ++A+∇σ+−1γE˜+ν+γυ++α+γυ+−υ−=−υ+⋅∇υ+−γ−12σ+∇σ++υ+×B˜≔f2+,(35)∂tυ−+A−∇σ−+1γE˜+ν−γυ−+α−γυ−−υ+=−υ−⋅∇υ−−γ−12σ−∇σ−−υ−×B˜≔f2−.Letα≤N−1. If we apply ∂α to (34), multiply by ∇∂ασ+, integrate in x, and then combine the result with an application of ∂α to (35) after which we multiply by ∇∂ασ− and integrate in x, then we get(36)ddtℰN,1intV+λ∇σ+,σ−N−12+σ+A+−σ−A−N2≤Cυ+,υ−N2+Cσ+,σ−,υ+,υ−,B˜N2∇σ+,σ−,υ+,υ−N−12,where(37)ℰN,1intV=∑α≤N−1∂αυ+,∇∂ασ++∂αυ−,∇∂ασ−.Step 3. We subtract equation (35) from equation (33) to get(38)∂tυ+−υ−+A+∇σ+−A−∇σ−+2γE˜=−∂αυ+⋅υ++∂αυ+⋅υ+−γ−12∂ασ+∇σ++γ−12∂ασ−∇σ−−∂αυ+×B˜−∂αυ−×B˜−1γ∂αν+υ+−ν−υ−−1γ∂αα+υ+−υ−−α−υ−−υ+.Now, we apply∂α to (38), multiply by ∂αE, integrate by parts in x, and replace ∂αE˜ with the third equation of (16). Then, we have(39)ddtℰN,2intV+λE˜N−12≤Cυ+,υ−N2+Cσ+,σ−N−12+Cυ+,υ−N∇B˜N−2+Cσ+,σ−,υ+,υ−,B˜N2∇σ+,σ−,υ+,υ−N−12,where(40)ℰN,2intυ=∑α≤N−1∂αυ+−υ−,∂αE˜forα≤N−1.Step 4. We apply ∂α to the third equation of (15), multiply by ∂α∇×B˜, integrate by parts in x, and use the relation(41)∂α∂xiB˜=∂xiΔ−1∇×∇×∂αB˜≤∇×∂αB˜,for each1≤i≤3. Then, we obtain(42)ddtℰN,3intV+λ∇B˜N−22≤Cυ+,υ−,E˜N−12+Cσ+,σ−N2∇υ+,υ−N−12,where(43)ℰN,3intυ=−∑α≤N−2∇×∂αE˜,∂αB˜for allα≤N−2.Step 5. Utilizing steps (1)–(4) above, we can now prove (30). Define(44)ℰNVt=∇VtN2+∑i=13ℰNintVt,and note that constants 0<k3≪k2≪k1≪1 are to be determined. We observe that if 0<ki≪1 for i=1,2,3, are sufficiently small, then ℰNV∼VN2 holds. Furthermore, by letting 0≪k3≪k2≪k1≪1 be sufficiently small, taking k23/2≪k3, and taking the sum of (31), k1× (36), k2× (42), and k1× (42), we find that there exists λ>0andC>0 such that (30) is satisfied:(45)12ddtVN2+ν±γυ+,υ−N2+k1ddt∑α≤N∂αυ+,∇∂ασ++∂αυ−,∇∂ασ−+k1λ∇σ+,σ−N−12+σ+A+−σ−A−N2+k2ddt∑α≤N−1∂αυ+−υ−,∂αE˜+k2λE˜N−12−k3ddt∑α≤N−2∇×∂αE˜,∂αB˜+k3λ∇B˜N−22≤CVNυ+,υ−2+∇σ+,σ−,υ+,υ−N−12+k1Cυ+,υ−N2+k1Cσ+,σ−,υ+,υ−,B˜N2∇σ+,σ−,υ+,υ−N−12+k2Cυ+,υ−N2+k2Cσ+,σ−N−12+k2Cυ+,υ−N∇B˜N−2+k2Cσ+,σ−,υ+,υ−,B˜N2∇σ+,σ−,υ+,υ−N−12+k3Cυ+,υ−,E˜N−12+k3Cσ+,σ−N2∇υ+,υ−N−12.
If we now let(46)V≤ℰ1/2,σ+,σ−,υ+,υ−,B˜N2=ℰN,∇σ+,σ−,υ+,υ−N−12=DN,then(47)ddtVN2+k1∑α|≤N∂αυ+,∇∂ασ++∂αυ−,∇∂ασ−+k2∑α≤N−1∂αυ+−υ−,∂αE−k3ddt∑α≤N−2∇×∂αE˜,∂αB˜+ν±γυ+,υ−N2+k2λ∇σ+,σ−N−12+k3λ∇B˜N−22+k2λ∇E˜N−22+k1λσ+A+−σ−A−N2≤CVN∇σ+,σ−,υ+,υ−N−12+k1Cσ+,σ−,υ+,υ−,B˜N2∇σ+,σ−,υ+,υ−N−12.
It follows that(48)ddtℰNVt+ν±γυ+,υ−N2+k2λ∇σ+,σ−N−12+k3λ∇B˜N−22+k2λ∇E˜N−22+k2λE˜2+k1λσ+A+−σ−A−N2≤CVN+∇σ+,σ−,υ+,υ−N−12∇σ+,σ−,υ+,υ−N−12.
Thus,(49)ddtℰNVt+λDNVt≤CℰNVt1/2+ℰNVtDNVt,and this concludes the proof.


## 2.3. Proof of Global Existence

We consider the global existence of the smooth solution to the isentropic Euler–Maxwell system for a quasilinear symmetric hyperbolic system (15). Therefore, we combine those a priori estimates with the local existence of solutions to extend the local solution up to infinite time by using the continuity of ℰNVt.Lemma 1 (local existence of smooth solution, see [2, 13, 14]).
AssumeV0∈HNℝ3 satisfies (17). Then, there exists T0>0 such that the Cauchy problems (15) and (16) admit a unique solution on 0,T0 with(50)V∈C0,T0;HNℝ3∩Lip0,T0;HN−1ℝ3.Proof of Proposition.
1. Since (15) is a quasilinear symmetric hyperbolic system, the global existence of smooth solutions follows from the local existence result in Lemma 1 (see also Section 16 of [14]). In addition, the a priori estimate (30) in Theorem 3 and the continuity argument show that ℰNVt is bounded uniformly in time under the assumption that ℰNV0>0 is sufficiently small. Therefore, global solutions satisfying (26) and (27) exist. This concludes the proof of Proposition 1.


## 3. Linearized Homogeneous System

### 3.1. Linearized Equations

To obtain the time decay rates of a solution to the nonlinear system (15) or (18), we consider the linearized homogeneous equations of system (18):(51)∂tρ±+∇⋅u±=0,∂tu±+u±+α±u±−α±u∓∓E+A±γ∇ρ±=0,∂tE−∇×B−u−+u+=0,∂tB+∇×E=0,∇.E=ρ+−ρ−,∇.B=0,t>0,x∈ℝ3,with the given initial data(52)Ut=0=U0≔ρ±0,u±0,E0,B0,x∈ℝ3,which satisfies the compatibility conditions(53)∇.E0=ρ+0−ρ−0,∇.B0=0,x∈ℝ3.Throughout this section, we letU=ρ±,u±,E,B be the solution to system (51). Moreover, in this section, we introduce some notation about Fourier transform f:ℝ3⟶ℝ, defined by(54)f^k=∫ℝ3e−ix⋅kfxdx,x⋅k≔∑j=13xjkj,4k∈ℝ3,where i is the complex number, and we use the energy method to the initial value problems (51) and (53) in Fourier space to show that there is a time-frequency Lyapunov inequality, which leads to the pointwise time-frequency upper-bound of the solution.We will use the energy method to the initial value problems (51) and (53) in the Fourier transform to show that there is a time-frequency Lyapunov functional which is equivalent to Ut,k^2 and moreover its dissipation rate can be represented by itself.


### 3.2. Representation of Solution

Denote byU=ρ±,u±,E,B=etLU0 the explicit solution to the Cauchy problems (51) and (52), satisfying (53). In this section, we study the representation of U.First, we take the time derivative for the first equation and the divergence of the second equation of system (51) and substitute ∇⋅E=ρ+−ρ−. So,(55)∂ttρ±+∂t∇⋅u±=0,(56)∇⋅∂tu±+∇⋅u±+α±∇⋅u±−u∓∓∇⋅E+A±γΔρ±=0.By combining the two equations (55) and (56), we have(57)∂ttρ±+1+α±∂tρ±−α±∂tρ∓±ρ+−ρ−−A±γΔρ±=0,with the initial data given by(58)±ρ+−ρ−|t=0=±ρ+0−ρ−0=±∇⋅E0,∂tρ±t=0=−∇⋅u±0.Then, taking the Fourier transform of the second-order ODE (57) with (58), we get(59)∂ttρ^±+1+α±∂tρ^±−α±∂tρ^∓±ρ^+−ρ^−+A±γk2ρ^±=0,±ρ^+−ρ^−t=0=±ρ^+0−ρ^−0=±ik⋅E^0,∂tρ^±t=0=−ik⋅u^±0.Now, set(60)x1=ρ^+,x2=x˙1=∂tρ^+,x3=ρ^−,x4=x˙3=∂tρ^−,x=x1,x2,x3,x4T.Then,(61)xt=Akx,where(62)Ak=0100−1−A+γk2−1+α+1α+00011α−−1−A−γk2−1+α−.Note that the eigenvector of the matrixAk is given by(63)a1λ=1+A−γk2+1+α−λ+λ21+A−γk2+1+α−λ+λ2λ1+α−λ1+α−λλ,a2λ=1+α+λ1+α+λλ1+A+γk2+1+α+λ+λ21+A+γk2+1+α+λ+λ2λ.In the next two sections, we provide an estimate foru^±,ρ^±,E^,andB^. In Section 3.2.1, we estimate for u^±andρ^±, and in Section 3.2.2, we estimate for u^±,E^,andB^. To do so, we set k˜=k/k, and we use the relation u^±=k˜k˜⋅u^±−k˜×k˜×u^± where we refer to k˜k˜⋅u^± as the “parallel part” and k˜×k˜×u^± as the “perpendicular part.”


#### 3.2.1. Parallel Part

We proceed with the asymptotic expansion of eigenvalues: letλjk,j=1,2,3,4, be the eigenvalues of the matrix Ak. Taking the determinant, we see the eigenvalues satisfy(64)detA−λI=λ4+2+α++α−λ3+3+α++α−+A+γk2+A−γk2λ2+2+A+γk2+A−γk2+α−A+γk2+α+A−γk2λ+A+γk2+A−γk2+A−A+γ2k4=0.(i)
First we consider whenk⟶0, since detA−λI=0 satisfies(65)λλ+1λ2+1+α−+α+λ+2=0,λjk has the following asymptotic expansion:(66)λjk=λj0+λj1k+λj2k2+…,where each coefficientλjn is given by direct computation as(67)λ10=λ11=0,λ12=−12A+γ+A−γ,λ20=−1,λ21=0,λ22=−A+γ+A−γ+α−A+γ+α+A−γ2−α+−α−,λ30=−1+α++α−−1+α++α−2−82,λ31=0,λ32=0,λ40=−1+α++α−+1+α++α−2−82,λ41=0,λ42=0.Thus, the approximation of the eigenvalue whenk⟶0 is(68)λ1=−12A+γ+A−γk2,λ2=−1−A+γ+A−γ+α−A+γ+α+A−γ2−α+−α−k2,λ3=−1+α++α−−1+α++α−2−82,λ4=−1+α++α−+1+α++α−2−82.Therefore, if we define(69)Bt=a1λ1eλ1ta1λ2eλ2ta2λ3eλ3ta2λ4eλ4t,then the Green matrixG for (61) is given by Gt=BtB−10, and the solution is(70)x=Gtx0.Thus, after a series of calculation using symbolic manipulator, we have(71)x1x2x3x4=O1e−θt+O1e−θk2tO1e−θt+O1e−θk2tO1e−θt+O1e−θk2tO1e−θt+O1e−θk2tO1e−θt+O1e−θk2tk2O1e−θt+O1e−θk2tk2O1e−θt+O1e−θk2tk2O1e−θt+O1e−θk2tk2O1e−θt+O1e−θk2tO1e−θt+O1e−θk2tO1e−θt+O1e−θk2tO1e−θt+O1e−θk2tO1e−θt+O1e−θk2tk2O1e−θt+O1e−θk2tk2O1e−θt+O1e−θk2tk2O1e−θt+O1e−θk2tk2x1,0x2,0x3,0x4,0,whereθ is the sum of real parts of λj,j=1,2,3,4. Thus,(72)ρ^+t,k=O1e−θt+O1e−θk2tρ^+0−O1e−θt+O1e−θk2tik⋅u^+0+O1e−θt+O1e−θk2tρ^−0−O1e−θt+O1e−θk2tik⋅u^−0,−ik⋅u^+t,k=O1e−θt+O1e−θk2tk2ρ^+0−O1e−θt+O1e−θk2tk2ik⋅u^+0+O1e−θt+O1e−θk2tk2ρ^−0−O1e−θt+O1e−θk2tk2ik⋅u^−0,ρ^−t,k=O1e−θk2t+O1e−θtρ^+0−O1e−θk2t+O1e−θtik⋅u^+0+O1e−θk2t+O1e−θtρ^−0−O1e−θk2t+O1e−θtik⋅u^−0,−ik⋅u^−t,k=O1e−θt+O1e−θk2tk2ρ^+0−O1e−θt+O1e−θk2tk2ik⋅u^+0+O1e−θt+O1e−θk2tk2ρ^−0−O1e−θt+O1e−θk2tk2ik⋅u^−0.In fact, we know that(73)u^±=k˜k˜⋅u^±−k˜×k˜×u^±impliesk˜k˜⋅u^±=u^±+k˜×k˜×u^±.Therefore, plugging (73) in the second and the fourth equations of (72), we obtain(74)u^+t,k=−k˜×k˜×u^++ik˜kO1e−θt+O1e−θk2tk2ρ^+0+O1e−θt+O1e−θk2tk2k˜k˜⋅u^+0+ik˜kO1e−θt+O1e−θk2tk2ρ^−0+O1e−θt+O1e−θk2tk2k˜k˜⋅u^−0,u^−t,k=−k˜×k˜×u^−+ik˜kO1e−θt+O1e−θk2tk2ρ^+0+O1e−θt+O1e−θk2tk2k˜k˜⋅u^+0+ik˜kO1e−θt+O1e−θk2tk2ρ^−0+O1e−θt+O1e−θk2tk2k˜k˜⋅u^−0.(ii)
Whenk⟶∞,λjk has the following asymptotic expansion:(75)λjk=μj1k+μj0+μj−1k−1+μj−2k−2+μj−3k−3+…,where each coefficientμjn is given by direct computation as(76)μ11=iA−γ,μ10=−1+α−2,μ21=−iA−γ,μ20=−1+α−2,μ31=iA+γ,μ30=−1+α+2,μ41=−iA+γ,μ40=−1+α+2.The approximation of the eigenvalue whenk⟶∞ is(77)λ1=−1+α−2+iA−γk,λ2=−1+α−2−iA−γk,λ3=−1+α+2+iA+γk,λ4=−1+α+2−iA+γk.So,G=BtB−10 is the Green matrix for (61), and the solution is(78)x=Gtx0.Thus, after a series of calculation using the symbolic manipulator, we obtain(79)x1x2x3x4=O1e−θtO1e−θtk−1O1e−θtk−1O1e−θtO1e−θtkO1e−θtO1e−θtO1e−θtk−1O1e−θtk−1O1e−θtO1e−θtO1e−θtk−1O1e−θtO1e−θt|k|−1O1e−θtkO1e−θtx1,0x2,0x3,0x4,0,ρ^+t,k=O1e−θtρ^+0−O1e−θtk−1ik⋅u^+0+O1e−θtk−1ρ^−0−O1e−θtik⋅u^−0,−ik⋅u^+t,k=O1e−θtkρ^+0−O1e−θtik⋅u^+0+O1e−θtρ^−0−O1e−θtk−1ik⋅u^−0,ρ^−t,k=O1e−θtk−1ρ^+0−O1e−θtik⋅u^+0+O1e−θtρ^−0−O1e−θtk−1ik⋅u^−0,−ik⋅u^−t,k=O1e−θtρ^+0−O1e−θtk−1ik⋅u^+0+O1e−θtkρ^−0−O1e−θtik⋅u^−0.Therefore, after plugging (72) into the second and fourth equations of (79), we obtain(80)u^+t,k=−k˜×k˜×u^++O1e−θtik˜k˜⋅ρ^+0+O1e−θtk˜k˜⋅u^+0+O1e−θtik˜kρ^−0+O1e−θtk˜k˜k⋅u^−0,u^−t,k=−k˜×k˜×u^−+O1e−θtik˜kρ^+0+O1e−θtk˜k˜k⋅u^+0+O1e−θtik˜k˜ρ^−0+O1e−θtk˜k˜⋅u^−0.(iii)
When0<k<∞, we consider the Routh–Hurwitz stability condition of the characteristic polynomial (64). That is, if we write (64) in the form(81)b0λ4+b1λ3+b2λ2+b3λ+b4=0.The system stability requires(82)b1>0,b1b2−b0b3>0,b1b2−b0b3b3−b12b4>0,b4>0.It can be shown that, in our case, the conditions corresponding to (82) are given, respectively, by(83)2+α++α−>0,2+α++α−3+α++α−+A+γk2+A−γk2−2+A+γk2+A−γk2+α−A+γk2+α+A−γk2>0,4+2α++2α−+A+γk2+A−γk2+3α++α+2+2α+α−+α+A+γk2+3α−+α−2+α−A−γk22+A+γk2+A−γk2+α−A+γk2+α+A−γk2−2+α++α−2A+γk2+A−γk2+A−A+γ2k4>0,A+γk2+A−γk2+A−A+γ2k4>0.It is not difficult to show that the above inequalities are satisfied, and this implies that all roots of the characteristic equation have negative real parts.Although the eigenvalues may coalesce, the computations in (i), (ii), and (iii) show that coalescence occurs when the real parts of the eigenvalues are negative. Therefore, the stability conditions are satisfied.


#### 3.2.2. Perpendicular Part

Now, we consider(84)M1t,k≔−k˜×k˜×u^+t,k,M2t,k≔−k˜×k˜×u^−t,k,M3t,k≔−k˜×k˜×E^t,k,M4t,k≔−k˜×k˜×B^t,k,when t>0 and k≠0. Taking the curl of the second, third, and fourth equations in system (51), we get(85)∂t∇×u++∇×u++α+∇×u+−α+∇×u−−∇×E=0,∂t∇×u−+∇×u−+α−∇×u−−α−∇×u++∇×E=0,∂t∇×E−∇×∇×B+∇×u+−∇×u−=0,∂t∇×B+∇×∇×E=0.Now, taking the Fourier Transform and multiplying by−ik, we obtain(86)∂tM1+M1+α+M1−α+M2−M3=0,∂tM2+M2+α−M2−α−M1+M3=0,∂tM3−ik×M4+M1−M2=0,∂tM4+ik×M3=0.Subtracting the first equation from the second equation in (86), we obtain(87)∂tM2−M1+M2−M1+α−M2−M1+α+M2−M1+2M3=0,and we simplify the above computation by letting M5=M2−M1.So,(88)∂tM5+1+α++α−M5+2M3=0.Thus,(89)∂tM5=−1+α++α−M5−2M3,∂tM3=ik×M4+M5,∂tM4=−ik×M3,where we write the initial data in the form(90)M3,M4,M5t=0=M3,0,M4,0,M5,0,with(91)M5,0=−k˜×k˜×u^−−u^+,M3,0=−k˜×k˜×E^0,M4,0=−k˜×k˜×B^0.Taking the time derivative of∂tM3 and substituting ∂tM5 and ∂tM4 as given in (89), we get that(92)∂ttM3=k×k×M3−1+α++α−M5−2M3.Fromk⋅M2=0, we find k×k×M3=−k2M3; thus,(93)∂ttM3+2+k2M3=−1+α++α−M5.Now, taking the time derivative of (93) and replacing ∂tM5 by the first equation of (89) and taking the sum with (93), we get(94)∂tttM3+1+α++α−∂ttM3+2+k2∂tM3+1+α++α−k2M3=0,with the initial data(95)M3t=0=M3,0,∂tM3t=0=ik×M4,0+M5,0,∂ttM3t=0=−1+α++α−M5,0−2+k2M3,0.Note that the characteristic equation of (94) is(96)Fx≔χ3+1+α++α−χ2+2+k2χ+1+α++α−k2=0.Set(97)x1=M3,x2=x˙1=∂tM3,x3=x˙2=∂ttM3,x=x1,x2,x3T.Then, equation (94) is written as(98)xt=Φkx,where(99)Φk=010001−1+α++α−k2−2+k2−1+α++α−.Asymptotic Expansion of Eigenvalues. Let λjk be the eigenvalues of the matrix Φk. We will find the asymptotic expansion of the eigenvalues λjkfork⟶0andk⟶∞.The eigenvaluesλjk,j=1,2,3, are the solutions of the characteristic equation which can be written as(100)detΦk−λI=λ3+1+α++α−λ2+2+k2λ+1+α++α−k2=0.The eigenvector forΦk is(101)bλ=1λλ2.(i)
Whenk⟶0,λjk has the following asymptotic expansion:(102)λjk=λj0+λj1k+λj2k2+…,where each coefficientλjn is given by the direct computation as(103)λ10=λ11=0,λ12=−121+α++α−,λ20=−1+α++α−+1+α++α−2−82,λ21=0,λ22=−1+α++α−−1+α++α−2−84,λ30=−1+α++α−−1+α++α−2−82,λ31=0,λ32=−1+α++α−+1+α++α−2−84.Thus, the approximation of the eigenvalue whenk⟶0 is(104)λ1=−121+α++α−k2,λ2=−1+α++α−+1+α++α−2−82−1+α++α−+1+α++α−2−84k2,λ3=−1+α++α−−1+α++α−2−82−1+α++α−−1+α++α−2−84k2.Therefore, if we set(105)Dt=bλ1eλ1tbλ2eλ2tbλ3eλ3t,the Green matrixG for (98) is given by Gt=DtD−10, and the solution is represented as(106)x=Gtx0,where each component of this solution is itself a3×3 diagonal matrix. We integrate the first and the third equations of (89), and we get the following expressions:(107)M5t,k=M5,0ke−t1+α++α−+2e−t1+α++α−∫0tc1ke1+α++α−+ηsds+∫0te−t1+α++α−+βsc2kcosωs+c3ksinωsds,M4t,k=M4,0k−ik×∫0tc1keηs+eβsc2kcosωs+c3ksinωsds,where(108)c1=13η2+21+α++α−η+2+k2−ηI3M5,0t,k+η1+α++α−+ηI3M3,0t,k+1+α++α−+ηik×M4,0t,k,c2=13η2+21+α++α−η+2+k2ηI3M5,0t,k+2η+η1+α++α−+η+k2+2I3M3,0t,k−1+α++α−+ηik×M4,0t,k,c3=13η2+21+α++α−η+2+k2−3/2η2−3/2η1+α++α−−2+k2ωI3M5,0t,k+1+α++α−+ηη1+α++α−+2+k22ωI3M3,0t,k+−1+α++α−2/2+3/2η2+2+k2ωik×M4,0t,k.After a series of calculation using the symbolic manipulator and definition (84), the solution can be written as(109)−k˜×k˜×u^−−−k˜×k˜×u^+=O1e−θk2t−k˜×k˜×u^−0−−k˜×k˜×u^+0+O1e−θk2t−O1e−θt−k˜×k˜×E^0+O1e−θk2t−O1e−θt−O1ie−θt−k˜×k˜×B^0,−k˜×k˜×E^=O1e−θt−O1ie−θt−k˜×k˜×E^0+O1ie−θt−k˜×k˜×B^0,−k˜×k˜×B^=O1ie−θt−k˜×k˜×E^0+O1e−θt−O1ie−θt−k˜×k˜×B^0.By using (73) and substituting (74) in the first equation of (109), we obtain(110)u^−−u^+=ik˜kO1e−θt+O1e−θk2tk2ρ^+0+O1e−θt+O1e−θk2tk2k˜k˜⋅u^+0+ik˜kO1e−θt+O1e−θk2tk2ρ^−0+O1e−θt+O1e−θk2tk2k˜k˜⋅u^−0+O1e−θk2t−k˜×k˜×u^−0−−k˜×k˜×u^+,0+O1e−θk2t−O1e−θt−k˜×k˜×E^0+O1e−θk2t−O1e−θt−O1ie−θt−k˜×k˜×B^0.Furthermore, we know from (51) that(111)∇⋅E=ρ+−ρ−,and thus(112)E^=−k˜×k˜×E^−ik˜kρ^+−ρ^−.Multiplying the first and the third equation of (72) by k˜ and substituting the result into equation (112), we obtain(113)E^=−ik˜kO1e−θt+O1e−θk2tρ^+0−O1e−θt+O1e−θk2tik⋅u^+0+O1e−θt+O1e−θk2tρ^−0−O1e−θt+O1e−θk2tik⋅u^−0+O1e−θt−O1ie−θt−k˜×k˜×E^0+O1ie−θt−k˜×k˜×B^0.Now, sincek˜k˜⋅B^=0, we have(114)B^=O1ie−θt−k˜×k˜×E^0+O1e−θt−O1ie−θt−k˜×k˜×B^0.Moreover, taking the sum of the first and the second equations of (86), we obtain(115)∂tM1+M2+M1+M2+M2−M1α−−α+=0.Thus,(116)M1+M2=e−θtM1,0+M2,0+∫0te−t−sM2−M1α−−α+ds.Substituting the first and the second equations of (84) into equation (116), we get that(117)−k˜×k˜×u^++−k˜×k˜×u^−=e−θt−k˜×k˜×u^+0+−k˜×k˜×u^−0+∫0te−t−sO1e−θk2s−k˜×k˜×u^−0−−k˜×k˜×u^+0+O1e−θk2s−O1e−θs−k˜×k˜×E^0+O1e−θk2s−O1e−θs−O1ie−θs−k˜×k˜×B^0ds.Next, substituting (74) into the above computations, we obtain(118)u^+t,k+u^−t,k=ik˜kO1e−θt+O1e−θk2tk2ρ^+0+O1e−θt+O1e−θk2tk2k˜k˜⋅u^+0+ik˜kO1e−θt+O1e−θk2tk2ρ^−0+O1e−θt+O1e−θk2tk2k˜k˜⋅u^−0+e−θt−k˜×k˜×u^+0+−k˜×k˜×u^−0+O1e−θk2t−e−t−k˜×k˜×u^+0−−k˜×k˜×u^−0+O1e−θk2t−e−t−k˜×k˜×E^0+O1e−θt−e−t−k˜×k˜×E^0+O1e−θk2t−e−t−k˜×k˜×B^0−O1e−θt−e−t−k˜×k˜×B^0−O1e−θt−e−t−k˜×ik˜×B^0.Now, taking the sum and difference, respectively, of (110) and (118), we obtain(119)u^±t,k=ik˜kO1e−θt+O1e−θk2tk2ρ^+0+O1e−θt+O1e−θk2tk2k˜k˜⋅u^+0+ik˜kO1e−θt+O1e−θk2tk2ρ^−0+O1e−θt+O1e−θk2tk2k˜k˜⋅u^−0+O1e−θk2t−k˜×k˜×u^−0−−k˜×k˜×u^+0+O1e−θk2t−O1e−θt−k˜×k˜×E^0+O1e−θk2t−O1e−θt−O1ie−θt−k˜×k˜×B^0+ik˜kO1e−θtρ^+0+O1e−θtk˜k˜⋅u^+0−ik˜kO1e−θtρ^−0+O1e−θtk˜k˜⋅u^−0+e−θt−k˜×k˜×u^+0+−k˜×k˜×u^−0+O1e−θk2t+O1e−θk2t−O1e−θt+O1e−θk2t+O1e−θt−O1ie−θt−O1e−t.(ii)
Whenk⟶∞,λjk has the following asymptotic expansion:(120)λjk=μj1k+μj0+μj−1k−1+μj−2k−2+….Each coefficientμjn is given by direct computation as(121)μ11=0,μ10=−1+α++α−,μ21=i,μ20=0,μ2−1=−23i,μ2−2=−49,μ31=−i,μ30=0,μ3−1=23i,μ3−2=−49.The approximation of the eigenvalue whenk⟶∞ is(122)λ1=−1+α++α−,λ2=ik−23ik−1−49k−2,λ3=−ik+23ik−1−49k−2.Hence, the Green matrixG for (98) is given by Gt=DtD−10, and the solution is represented as(123)x=Gtx0,where each component of this solution is itself a 3×3 diagonal matrix. After a series of calculation using the symbolic manipulator and equation (84), the solution can be written as(124)−k˜×k˜×u^−−−k˜×k˜×u^+=O1e−θt−k˜×k˜×u^−,0−−k˜×k˜×u^+0+O1ke−θk−2t−k˜×k˜×E^0+O1k2e−θt−O1k2e−θk−2t−k˜×k˜×B^0,−k˜×k˜×E^=−O1e−θt+O1e−θk−2t−k˜×k˜×u^−0−−k˜×k˜×u^+0+O1−θk−2t−k˜×k˜×E^0+O1ke−θk−2t−k˜×k˜×B^0,−k˜×k˜×B^=O1ike−θk−2t−k˜×k˜×u^−0−−k˜×k˜×u^+0−O1ike−θt−k˜×k˜×E^0+O1e−θk−2t−k˜×k˜×B^0.By using (73) and substituting (80) into the first equation of the above computation, we obtain(125)u^−−u^+=O1e−θtk˜k˜−O1e−θtik˜kiρ^+0+O1e−θt−O1e−θt1kk˜k˜⋅u^+0+O1e−θtik˜k−O1e−θtik˜k˜ρ^−0+O1e−θt1k−O1e−θtk˜k˜⋅u^−0+O1e−θt−k˜×k˜×u^−0−−k˜×k˜×u^+,0+O1ke−θk−2t−k˜×k˜×E^0+O1k2e−θt−O1k2e−θk−2t−k˜×k˜×B^0.Furthermore, we have(126)E^=−k˜×k˜×E^−ik˜kρ^+−ρ^−.Therefore, multiplying the first and the third equation of (79) by k˜ and substituting into the above equation, we obtain(127)E^=−O1−O1k−1e−θtρ^+0−O1k−1−O1e−θtik⋅u^+0+O1k−1−O1e−θtρ^−0−O1+O1k−1e−θtik⋅u^−0+−O1e−θt+O1e−θk−2t−k˜×k˜×u^−0−−k˜×k˜×u^+,0+O1e−θk−2t−k˜×k˜×E^0+O1ke−θk−2t−k˜×k˜×B^0.Sincek˜k˜⋅B^=0, we get(128)B^=O1ike−θk−2t−k˜×k˜×u^−,0−−k˜×k˜×u^+,0−O1ike−θt−k˜×k˜×E^0+O1e−θk−2t−k˜×k˜×B^0.Substituting the first equation of (124) into equation (116) results in(129)−k˜×k˜×u^++−k˜×k˜×u^−=e−θt−k˜×k˜×u^+0+−k˜×k˜×u^−0+∫0te−t−sO1e−θs−k˜×k˜×u^−0−−k˜×k˜×u^+0+O1ke−θk−2s−k˜×k˜×E^0+O1k2e−θs−O1k2e−θk−2s−k˜×k˜×B^0ds.Next, substituting equation (80) into equation (129), we obtain(130)u^+t,k+u^−t,k=O1e−θtk˜k˜−O1e−θtik˜kiρ^+0+O1e−θt−O1e−θt1kk˜k˜⋅u^+0+O1e−θtik˜k−O1e−θtik˜k˜ρ^−0+O1e−θt1k−O1e−θtk˜k˜⋅u^−0+e−θt−k˜×k˜×u^+0+−k˜×k˜×u^−0+O1e−θt−e−t−k˜×k˜×u^+0+−k˜×k˜×u^−0+O1e−θk−2t−e−t−k˜×k˜×E^0+O1e−θt−e−t−k˜×k˜×B^0+O1e−θk−2t−e−t−k˜×k˜×B^0.Now, taking the sum and difference, respectively, of equations (125) and (130), we obtain(131)u^±t,k=O1e−θtk˜k˜−O1e−θtik˜kiρ^+0+O1e−θt−O1e−θt1kk˜k˜⋅u^+0+O1e−θtik˜k−O1e−θtik˜k˜ρ^−0+O1e−θt1k−O1e−θtk˜k˜⋅u^−0x+O1e−θt−k˜×k˜×u^+0+−k˜×k˜×u^−0+O1ke−θk−2t−k˜×k˜×E^0+O1k2e−θt−O1k2e−θk−2t−k˜×k˜×B^0+ik˜k|2O1e−θtρ^+0+O1ke−θtk˜k˜⋅u^+0−ik˜O1e−θtρ^−0+O1e−θtk˜k˜⋅u^−0+e−θt−k˜×k˜×u^+0+−k˜×k˜×u^−0+O1e−θt+O1e−θk−2t+O1e−θt−O1e−θk−2t−O1e−t.Theorem 4.
Let1≤p,r≤2≤q≤∞,andl≥0 and let m≥0 be an integer. Define(132)l+31r−1q+=l+31r−1q−+1,whenr≠2orq≠2orlis not an integer,l,whenr=q=2andlis an integer,where ⋅− denotes the integer part of the argument. Suppose U0 satisfies (53). Then, for any t≥0, ∇metLU0 satisfies the following time decay property:(133)∇metLU0Lq≤C1+t−3/21/p−1/q−m/2U0Lp+C1+t−l/2∇m+l+31/r−1/q+U0Lr,where C=Cp,q,r,l,m.Theorem 5.
Let1≤p,r≤2≤q≤∞,andl≥0 and let m≥0 be an integer. Assume Ut=etLU0 is the solution of the initial value problems (51) and (52) with initial data ρ±0,u±0,E0,B0 which satisfies (53). Then, for any t≥0, U=ρ±,u±,E,B satisfies(134)∇mρ±tLq≤C1+t−3/21/p−1/q−m/2ρ^±0,u^±,0Lp+Cρ±0,u±0Lr,∇mu±tLq≤Ce−t/2ρ±0Lp+∇m+31/r−1/q+ρ±0,u±0Lr(135)+C1+t−3/21/p−1/q−m+1/2u±0,E0,B0Lp+C1+t−l+1/2∇m+l+31/r−1/q+u±0,E0,B0Lr,(136)∇mEtLq≤C1+t−3/21/p−1/q−m+1/2u±0,E0,B0Lp+C1+t−l/2∇m+l+31/r−1/q+u±0,E0,B0Lr,(137)∇mBtLq≤C1+t−3/21/p−1/q−m/2u±0,E0,B0Lp+C1+t−l/2∇m+l+31/r−1/q+u±0,E0,B0Lr,where C=Cp,q,r,l,m and l+31/r−1/q+ is defined in (132).Proof.
Let1≤p,r≤2≤q≤∞, and m be a nonnegative integer. Let q′ satisfy 1/q+1/q=1. Using the Hausdorff–Young inequality, we prove (134) as follows:(138)∇mρ±t,kLxq≤Ckmρ^±Lq′|k|≤1+Ckmρ^±Lq′k≥1.
We estimate the first term of (138) using the Hölder inequality 1/q′=1/p′+p′−q′/p′q′ with 1/p′+1/p=1 and fixing ϵ>0 sufficiently small; that is,(139)kmρ^±0,u^±0e−θt+e−θ|k|2tLq′k≤1=k−p′−q′/p′q′3−εkm+p′−q′/p′q′3−ερ^±0,u^±0e−θt+e−θ|k|2tLq′k≤1≤k−3−ε−p′−q′/p′q′km+p′−q′/p′q′ρ^±0,u^±0e−θt+e−θ|k|2tLp′k≤1≤km+p′−q′/p′q′3−ερ^±0,u^±0e−θt+e−θ|k|2tLp′k≤1≤Ce−θtρ^±0,u^±0Lp′k≤1+Ct+1−3/21/q′−1/p′−m+1/2ρ^±0,u^±0Lp′k≤1≤Ce−θtρ±0,u±0Lpk≤1+Ct+1−3/21/p−1/q−m+1/2ρ±0,u±0Lpk≤1.
Now, we estimate the second term; by takingϵ>0 sufficiently small and using the Hölder inequality 1/q′=1/r′+r′−q′/r′q′ with 1/r′+1/r=1, we find(140)kmρ^±0,u^±0e−θt+e−θk2tLq′k≥1=k−r′−q′/r′q′3+εkm+p′−q′/p′q′3+ερ^±0,u^±0e−θtLq′k≥1≤k−3+ε−r′−q′/r′q′km+r′−q′/r′q′3+ερ^±0,u^±0e−θtLr′k≥1≤Ckm+1/q′−1/r′3+ερ^±0,u^±0e−θtLr′k≥1≤Ce−θt∇m+31/r−1/q+ρ±0,u±0Lr′k≥1.
We prove (135) similarly. That is,(141)∇mu±tLxq≤Ckmu^±Lq′k≤1+Ckmu^±Lq′k≥1,where(142)kmu^±t,kLq′k≤1=kme−t/2ρ^±0,u^±0,E^0Lq′+kme−θtρ^±0,u^±0,E^0,B^0Lq′+kme−θk2tρ^±0,u^±0,E^0,B^0Lq′.
We estimate of the first term of (142) using the Hölder inequality 1/q′=1/p′+p′−q′/p′q′ with 1/p′+1/p=1 and fixing ϵ>0 sufficiently small:(143)kme−t/2ρ^±0,u^±0,E^0Lq′=∫ℝ3kme−t/2ρ^±0,u^±0,E^0q′1/q′=∫ℝ3k−p′−q′/p′q′3−εkm+p′−q′/p′q′3−εe−t/2ρ^±0,u^±0,E^0q′1/q′≤e−t/2k−3−ε−p′−q′/p′q′km+p′−q′/p′q′ρ^±0Lp′k≤1≤e−t/2km+p′−q′/p′q′3−ερ^±0Lp′k≤1≤Ce−t/2ρ^±0Lp′k≤1≤Ce−t/2ρ±0Lpk≤1.
We estimate the second term of (142) by(144)kme−θk2tρ^±0,u^±0,E^0,B^0Lq′=∫ℝ3km+1e−θk2tu^±0,E^0,B^0q′1/q′=∫ℝ3k−p′−q′/p′q′3−εkm+p′−q′/p′q′3−εe−θk2tu^±0,E^0,B^0q′1/q′≤k−3−ε−p′−q′/p′q′km+p′−q′/p′q′e−θk2tu^±0,E^0,B^0Lp′k≤1≤e−t/2km+p′−q′/p′q′3−ερ^±0Lp′k≤1≤Ce−t/2ρ^±0Lp′k≤1≤Ce−t/2ρ±0Lpk≤1.
We estimate the third term of (142) in two parts. First, take ϵ>0 sufficiently small and use the Hölder inequality 1/q′=1/r′+r′−q′/r′q′ with 1/r′+1/r=1:(145)kmu^±t,kLq′k≥1=kme−t/2ρ^±0,u^±0,E^0Lq′+kme−θtρ^±0,u^±0,E^0,B^0Lq′+km1ke−θk2tρ^±0,u^±0,E^0,B^0Lq′,where(146)kme−t/2ρ^±0,u^±0,E^0Lq′=∫ℝ3kme−t/2ρ^±0,u^±0,E^0q′1/q′=∫ℝ3k−r′−q′/r′q′3+εkm+r′−q′/r′q′3+εe−t/2ρ^±0,u^±0,E^0q′1/q′≤e−t/2k−3+ε‖−r′−q′/r′q′‖km+r′−q′/r′q′3+ερ^±0Lp′k≤1≤e−t/2km+1/q′−1/r′3+ερ^±0Lp′k≤1≤Ce−t/2∇m+31/r−1/q+ρ^±0Lr.
To estimate the second part of the third term, we have(147)km1ke−θt/k2ρ^±0,u^±0,E^0,B^0Lq′=∫ℝ3km−1e−θt/k2u^±0,E^0,B^0q′1/q′=∫ℝ3k−r′−q′/r′q′3+ϵkm+r′−q′/r′q′3+ϵe−θt/k2u^±0,E^0,B^0q′1/q′≤k−3+ϵ−r′−q′/r′q′km+r′−q′/r′q′3+ϵe−θt/k2u^±0,E^0,B^0Lr′k≤1≤e−θt/k2km+l+31/q′−1/r′k−l+1u±0,E0,B0Lr′k≤1≤C1+t−l+1/2∇m+l+31/r−1/q+u±0,E0,B0Lr,where we used(148)supk≤11kl+1e−θt+1/k2≤C1+t−l+1/2.
The proofs of (136) and (137) are similar.
Note that we transformed to the Fourier space to find the solution representation. Now, we go back to the physical spacext and obtain the decay rates for the solutions to the linearized system. Fundamental properties of the system are given by the following corollary based on Theorem 5.Corollary 1.
AssumeUt=etLU0 is the solution of the initial value problems (51) and (52) with initial data ρ±0,u±0,E0,B0 which satisfies (53). Then, U=ρ±,u±,E,B satisfies(149)ρ±t≤Ce−t/2ρ±0,u±0,u±t≤Ce−t/2ρ±0+C1+t−5/4u±0,E0,B0L1∩H˙2,Et≤C1+t−5/2u±0,E0,B0L1∩H˙3,Bt≤C1+t−3/4u±0,E0,B0L1∩H˙2,(150)ρ±tL∞≤Ce−t/2ρ±0,u±0L2∩H˙2,u±tL∞≤Ce−t/2ρ±0L1∩H˙2+C1+t−2u±0,E0,B0L1∩H˙5,EtL∞≤C1+t−2u±0,E0,B0L1∩H˙6,BtL∞≤C1+t−3/2u±0,E0,B0L1∩H˙5,(151)Bt≤C1+t−5/4u±0,E0,B0L1∩H˙4,∇NEt,Bt≤C1+t−5/4u±0,E0,B0L1∩H˙N+3.Proof.
The results of Corollary1 are particular cases of Theorem 5.
For example, we explain how we get the decay rates foru±. From (135) for L2, take q=2,p=1,m=0,l=3/2, and r=2:(152)∇mu±tL2≤Ce−t/2ρ±0L1+∇3/2+1ρ±0,u±0L1+C1+t−5/4u±0,E0,B0L1+C1+t−5/4∇1+1/2+1u±0,E0,B0L1.
Thus,(153)u±tL2≤Ce−t/2ρ±0+C1+t−5/4u±0,E0,B0L1∩H˙2.
Now, forL∞, take q=∞,p=1,m=0,l=3, and r=2:(154)∇mu±tL∞≤Ce−t/2ρ±0L1+∇3/2+1ρ±0,u±0L2+C1+t−2u±0,E0,B0L1+C1+t−2∇4+1/2+1u±0,E0,B0L2.
Thus,(155)u±tL∞≤Ce−t/2ρ±0L1∩H˙2+C1+t−2u±0,E0,B0L1∩H˙5.


## 3.1. Linearized Equations

To obtain the time decay rates of a solution to the nonlinear system (15) or (18), we consider the linearized homogeneous equations of system (18):(51)∂tρ±+∇⋅u±=0,∂tu±+u±+α±u±−α±u∓∓E+A±γ∇ρ±=0,∂tE−∇×B−u−+u+=0,∂tB+∇×E=0,∇.E=ρ+−ρ−,∇.B=0,t>0,x∈ℝ3,with the given initial data(52)Ut=0=U0≔ρ±0,u±0,E0,B0,x∈ℝ3,which satisfies the compatibility conditions(53)∇.E0=ρ+0−ρ−0,∇.B0=0,x∈ℝ3.Throughout this section, we letU=ρ±,u±,E,B be the solution to system (51). Moreover, in this section, we introduce some notation about Fourier transform f:ℝ3⟶ℝ, defined by(54)f^k=∫ℝ3e−ix⋅kfxdx,x⋅k≔∑j=13xjkj,4k∈ℝ3,where i is the complex number, and we use the energy method to the initial value problems (51) and (53) in Fourier space to show that there is a time-frequency Lyapunov inequality, which leads to the pointwise time-frequency upper-bound of the solution.We will use the energy method to the initial value problems (51) and (53) in the Fourier transform to show that there is a time-frequency Lyapunov functional which is equivalent to Ut,k^2 and moreover its dissipation rate can be represented by itself.


## 3.2. Representation of Solution

Denote byU=ρ±,u±,E,B=etLU0 the explicit solution to the Cauchy problems (51) and (52), satisfying (53). In this section, we study the representation of U.First, we take the time derivative for the first equation and the divergence of the second equation of system (51) and substitute ∇⋅E=ρ+−ρ−. So,(55)∂ttρ±+∂t∇⋅u±=0,(56)∇⋅∂tu±+∇⋅u±+α±∇⋅u±−u∓∓∇⋅E+A±γΔρ±=0.By combining the two equations (55) and (56), we have(57)∂ttρ±+1+α±∂tρ±−α±∂tρ∓±ρ+−ρ−−A±γΔρ±=0,with the initial data given by(58)±ρ+−ρ−|t=0=±ρ+0−ρ−0=±∇⋅E0,∂tρ±t=0=−∇⋅u±0.Then, taking the Fourier transform of the second-order ODE (57) with (58), we get(59)∂ttρ^±+1+α±∂tρ^±−α±∂tρ^∓±ρ^+−ρ^−+A±γk2ρ^±=0,±ρ^+−ρ^−t=0=±ρ^+0−ρ^−0=±ik⋅E^0,∂tρ^±t=0=−ik⋅u^±0.Now, set(60)x1=ρ^+,x2=x˙1=∂tρ^+,x3=ρ^−,x4=x˙3=∂tρ^−,x=x1,x2,x3,x4T.Then,(61)xt=Akx,where(62)Ak=0100−1−A+γk2−1+α+1α+00011α−−1−A−γk2−1+α−.Note that the eigenvector of the matrixAk is given by(63)a1λ=1+A−γk2+1+α−λ+λ21+A−γk2+1+α−λ+λ2λ1+α−λ1+α−λλ,a2λ=1+α+λ1+α+λλ1+A+γk2+1+α+λ+λ21+A+γk2+1+α+λ+λ2λ.In the next two sections, we provide an estimate foru^±,ρ^±,E^,andB^. In Section 3.2.1, we estimate for u^±andρ^±, and in Section 3.2.2, we estimate for u^±,E^,andB^. To do so, we set k˜=k/k, and we use the relation u^±=k˜k˜⋅u^±−k˜×k˜×u^± where we refer to k˜k˜⋅u^± as the “parallel part” and k˜×k˜×u^± as the “perpendicular part.”


### 3.2.1. Parallel Part

We proceed with the asymptotic expansion of eigenvalues: letλjk,j=1,2,3,4, be the eigenvalues of the matrix Ak. Taking the determinant, we see the eigenvalues satisfy(64)detA−λI=λ4+2+α++α−λ3+3+α++α−+A+γk2+A−γk2λ2+2+A+γk2+A−γk2+α−A+γk2+α+A−γk2λ+A+γk2+A−γk2+A−A+γ2k4=0.(i)
First we consider whenk⟶0, since detA−λI=0 satisfies(65)λλ+1λ2+1+α−+α+λ+2=0,λjk has the following asymptotic expansion:(66)λjk=λj0+λj1k+λj2k2+…,where each coefficientλjn is given by direct computation as(67)λ10=λ11=0,λ12=−12A+γ+A−γ,λ20=−1,λ21=0,λ22=−A+γ+A−γ+α−A+γ+α+A−γ2−α+−α−,λ30=−1+α++α−−1+α++α−2−82,λ31=0,λ32=0,λ40=−1+α++α−+1+α++α−2−82,λ41=0,λ42=0.Thus, the approximation of the eigenvalue whenk⟶0 is(68)λ1=−12A+γ+A−γk2,λ2=−1−A+γ+A−γ+α−A+γ+α+A−γ2−α+−α−k2,λ3=−1+α++α−−1+α++α−2−82,λ4=−1+α++α−+1+α++α−2−82.Therefore, if we define(69)Bt=a1λ1eλ1ta1λ2eλ2ta2λ3eλ3ta2λ4eλ4t,then the Green matrixG for (61) is given by Gt=BtB−10, and the solution is(70)x=Gtx0.Thus, after a series of calculation using symbolic manipulator, we have(71)x1x2x3x4=O1e−θt+O1e−θk2tO1e−θt+O1e−θk2tO1e−θt+O1e−θk2tO1e−θt+O1e−θk2tO1e−θt+O1e−θk2tk2O1e−θt+O1e−θk2tk2O1e−θt+O1e−θk2tk2O1e−θt+O1e−θk2tk2O1e−θt+O1e−θk2tO1e−θt+O1e−θk2tO1e−θt+O1e−θk2tO1e−θt+O1e−θk2tO1e−θt+O1e−θk2tk2O1e−θt+O1e−θk2tk2O1e−θt+O1e−θk2tk2O1e−θt+O1e−θk2tk2x1,0x2,0x3,0x4,0,whereθ is the sum of real parts of λj,j=1,2,3,4. Thus,(72)ρ^+t,k=O1e−θt+O1e−θk2tρ^+0−O1e−θt+O1e−θk2tik⋅u^+0+O1e−θt+O1e−θk2tρ^−0−O1e−θt+O1e−θk2tik⋅u^−0,−ik⋅u^+t,k=O1e−θt+O1e−θk2tk2ρ^+0−O1e−θt+O1e−θk2tk2ik⋅u^+0+O1e−θt+O1e−θk2tk2ρ^−0−O1e−θt+O1e−θk2tk2ik⋅u^−0,ρ^−t,k=O1e−θk2t+O1e−θtρ^+0−O1e−θk2t+O1e−θtik⋅u^+0+O1e−θk2t+O1e−θtρ^−0−O1e−θk2t+O1e−θtik⋅u^−0,−ik⋅u^−t,k=O1e−θt+O1e−θk2tk2ρ^+0−O1e−θt+O1e−θk2tk2ik⋅u^+0+O1e−θt+O1e−θk2tk2ρ^−0−O1e−θt+O1e−θk2tk2ik⋅u^−0.In fact, we know that(73)u^±=k˜k˜⋅u^±−k˜×k˜×u^±impliesk˜k˜⋅u^±=u^±+k˜×k˜×u^±.Therefore, plugging (73) in the second and the fourth equations of (72), we obtain(74)u^+t,k=−k˜×k˜×u^++ik˜kO1e−θt+O1e−θk2tk2ρ^+0+O1e−θt+O1e−θk2tk2k˜k˜⋅u^+0+ik˜kO1e−θt+O1e−θk2tk2ρ^−0+O1e−θt+O1e−θk2tk2k˜k˜⋅u^−0,u^−t,k=−k˜×k˜×u^−+ik˜kO1e−θt+O1e−θk2tk2ρ^+0+O1e−θt+O1e−θk2tk2k˜k˜⋅u^+0+ik˜kO1e−θt+O1e−θk2tk2ρ^−0+O1e−θt+O1e−θk2tk2k˜k˜⋅u^−0.(ii)
Whenk⟶∞,λjk has the following asymptotic expansion:(75)λjk=μj1k+μj0+μj−1k−1+μj−2k−2+μj−3k−3+…,where each coefficientμjn is given by direct computation as(76)μ11=iA−γ,μ10=−1+α−2,μ21=−iA−γ,μ20=−1+α−2,μ31=iA+γ,μ30=−1+α+2,μ41=−iA+γ,μ40=−1+α+2.The approximation of the eigenvalue whenk⟶∞ is(77)λ1=−1+α−2+iA−γk,λ2=−1+α−2−iA−γk,λ3=−1+α+2+iA+γk,λ4=−1+α+2−iA+γk.So,G=BtB−10 is the Green matrix for (61), and the solution is(78)x=Gtx0.Thus, after a series of calculation using the symbolic manipulator, we obtain(79)x1x2x3x4=O1e−θtO1e−θtk−1O1e−θtk−1O1e−θtO1e−θtkO1e−θtO1e−θtO1e−θtk−1O1e−θtk−1O1e−θtO1e−θtO1e−θtk−1O1e−θtO1e−θt|k|−1O1e−θtkO1e−θtx1,0x2,0x3,0x4,0,ρ^+t,k=O1e−θtρ^+0−O1e−θtk−1ik⋅u^+0+O1e−θtk−1ρ^−0−O1e−θtik⋅u^−0,−ik⋅u^+t,k=O1e−θtkρ^+0−O1e−θtik⋅u^+0+O1e−θtρ^−0−O1e−θtk−1ik⋅u^−0,ρ^−t,k=O1e−θtk−1ρ^+0−O1e−θtik⋅u^+0+O1e−θtρ^−0−O1e−θtk−1ik⋅u^−0,−ik⋅u^−t,k=O1e−θtρ^+0−O1e−θtk−1ik⋅u^+0+O1e−θtkρ^−0−O1e−θtik⋅u^−0.Therefore, after plugging (72) into the second and fourth equations of (79), we obtain(80)u^+t,k=−k˜×k˜×u^++O1e−θtik˜k˜⋅ρ^+0+O1e−θtk˜k˜⋅u^+0+O1e−θtik˜kρ^−0+O1e−θtk˜k˜k⋅u^−0,u^−t,k=−k˜×k˜×u^−+O1e−θtik˜kρ^+0+O1e−θtk˜k˜k⋅u^+0+O1e−θtik˜k˜ρ^−0+O1e−θtk˜k˜⋅u^−0.(iii)
When0<k<∞, we consider the Routh–Hurwitz stability condition of the characteristic polynomial (64). That is, if we write (64) in the form(81)b0λ4+b1λ3+b2λ2+b3λ+b4=0.The system stability requires(82)b1>0,b1b2−b0b3>0,b1b2−b0b3b3−b12b4>0,b4>0.It can be shown that, in our case, the conditions corresponding to (82) are given, respectively, by(83)2+α++α−>0,2+α++α−3+α++α−+A+γk2+A−γk2−2+A+γk2+A−γk2+α−A+γk2+α+A−γk2>0,4+2α++2α−+A+γk2+A−γk2+3α++α+2+2α+α−+α+A+γk2+3α−+α−2+α−A−γk22+A+γk2+A−γk2+α−A+γk2+α+A−γk2−2+α++α−2A+γk2+A−γk2+A−A+γ2k4>0,A+γk2+A−γk2+A−A+γ2k4>0.It is not difficult to show that the above inequalities are satisfied, and this implies that all roots of the characteristic equation have negative real parts.Although the eigenvalues may coalesce, the computations in (i), (ii), and (iii) show that coalescence occurs when the real parts of the eigenvalues are negative. Therefore, the stability conditions are satisfied.


### 3.2.2. Perpendicular Part

Now, we consider(84)M1t,k≔−k˜×k˜×u^+t,k,M2t,k≔−k˜×k˜×u^−t,k,M3t,k≔−k˜×k˜×E^t,k,M4t,k≔−k˜×k˜×B^t,k,when t>0 and k≠0. Taking the curl of the second, third, and fourth equations in system (51), we get(85)∂t∇×u++∇×u++α+∇×u+−α+∇×u−−∇×E=0,∂t∇×u−+∇×u−+α−∇×u−−α−∇×u++∇×E=0,∂t∇×E−∇×∇×B+∇×u+−∇×u−=0,∂t∇×B+∇×∇×E=0.Now, taking the Fourier Transform and multiplying by−ik, we obtain(86)∂tM1+M1+α+M1−α+M2−M3=0,∂tM2+M2+α−M2−α−M1+M3=0,∂tM3−ik×M4+M1−M2=0,∂tM4+ik×M3=0.Subtracting the first equation from the second equation in (86), we obtain(87)∂tM2−M1+M2−M1+α−M2−M1+α+M2−M1+2M3=0,and we simplify the above computation by letting M5=M2−M1.So,(88)∂tM5+1+α++α−M5+2M3=0.Thus,(89)∂tM5=−1+α++α−M5−2M3,∂tM3=ik×M4+M5,∂tM4=−ik×M3,where we write the initial data in the form(90)M3,M4,M5t=0=M3,0,M4,0,M5,0,with(91)M5,0=−k˜×k˜×u^−−u^+,M3,0=−k˜×k˜×E^0,M4,0=−k˜×k˜×B^0.Taking the time derivative of∂tM3 and substituting ∂tM5 and ∂tM4 as given in (89), we get that(92)∂ttM3=k×k×M3−1+α++α−M5−2M3.Fromk⋅M2=0, we find k×k×M3=−k2M3; thus,(93)∂ttM3+2+k2M3=−1+α++α−M5.Now, taking the time derivative of (93) and replacing ∂tM5 by the first equation of (89) and taking the sum with (93), we get(94)∂tttM3+1+α++α−∂ttM3+2+k2∂tM3+1+α++α−k2M3=0,with the initial data(95)M3t=0=M3,0,∂tM3t=0=ik×M4,0+M5,0,∂ttM3t=0=−1+α++α−M5,0−2+k2M3,0.Note that the characteristic equation of (94) is(96)Fx≔χ3+1+α++α−χ2+2+k2χ+1+α++α−k2=0.Set(97)x1=M3,x2=x˙1=∂tM3,x3=x˙2=∂ttM3,x=x1,x2,x3T.Then, equation (94) is written as(98)xt=Φkx,where(99)Φk=010001−1+α++α−k2−2+k2−1+α++α−.Asymptotic Expansion of Eigenvalues. Let λjk be the eigenvalues of the matrix Φk. We will find the asymptotic expansion of the eigenvalues λjkfork⟶0andk⟶∞.The eigenvaluesλjk,j=1,2,3, are the solutions of the characteristic equation which can be written as(100)detΦk−λI=λ3+1+α++α−λ2+2+k2λ+1+α++α−k2=0.The eigenvector forΦk is(101)bλ=1λλ2.(i)
Whenk⟶0,λjk has the following asymptotic expansion:(102)λjk=λj0+λj1k+λj2k2+…,where each coefficientλjn is given by the direct computation as(103)λ10=λ11=0,λ12=−121+α++α−,λ20=−1+α++α−+1+α++α−2−82,λ21=0,λ22=−1+α++α−−1+α++α−2−84,λ30=−1+α++α−−1+α++α−2−82,λ31=0,λ32=−1+α++α−+1+α++α−2−84.Thus, the approximation of the eigenvalue whenk⟶0 is(104)λ1=−121+α++α−k2,λ2=−1+α++α−+1+α++α−2−82−1+α++α−+1+α++α−2−84k2,λ3=−1+α++α−−1+α++α−2−82−1+α++α−−1+α++α−2−84k2.Therefore, if we set(105)Dt=bλ1eλ1tbλ2eλ2tbλ3eλ3t,the Green matrixG for (98) is given by Gt=DtD−10, and the solution is represented as(106)x=Gtx0,where each component of this solution is itself a3×3 diagonal matrix. We integrate the first and the third equations of (89), and we get the following expressions:(107)M5t,k=M5,0ke−t1+α++α−+2e−t1+α++α−∫0tc1ke1+α++α−+ηsds+∫0te−t1+α++α−+βsc2kcosωs+c3ksinωsds,M4t,k=M4,0k−ik×∫0tc1keηs+eβsc2kcosωs+c3ksinωsds,where(108)c1=13η2+21+α++α−η+2+k2−ηI3M5,0t,k+η1+α++α−+ηI3M3,0t,k+1+α++α−+ηik×M4,0t,k,c2=13η2+21+α++α−η+2+k2ηI3M5,0t,k+2η+η1+α++α−+η+k2+2I3M3,0t,k−1+α++α−+ηik×M4,0t,k,c3=13η2+21+α++α−η+2+k2−3/2η2−3/2η1+α++α−−2+k2ωI3M5,0t,k+1+α++α−+ηη1+α++α−+2+k22ωI3M3,0t,k+−1+α++α−2/2+3/2η2+2+k2ωik×M4,0t,k.After a series of calculation using the symbolic manipulator and definition (84), the solution can be written as(109)−k˜×k˜×u^−−−k˜×k˜×u^+=O1e−θk2t−k˜×k˜×u^−0−−k˜×k˜×u^+0+O1e−θk2t−O1e−θt−k˜×k˜×E^0+O1e−θk2t−O1e−θt−O1ie−θt−k˜×k˜×B^0,−k˜×k˜×E^=O1e−θt−O1ie−θt−k˜×k˜×E^0+O1ie−θt−k˜×k˜×B^0,−k˜×k˜×B^=O1ie−θt−k˜×k˜×E^0+O1e−θt−O1ie−θt−k˜×k˜×B^0.By using (73) and substituting (74) in the first equation of (109), we obtain(110)u^−−u^+=ik˜kO1e−θt+O1e−θk2tk2ρ^+0+O1e−θt+O1e−θk2tk2k˜k˜⋅u^+0+ik˜kO1e−θt+O1e−θk2tk2ρ^−0+O1e−θt+O1e−θk2tk2k˜k˜⋅u^−0+O1e−θk2t−k˜×k˜×u^−0−−k˜×k˜×u^+,0+O1e−θk2t−O1e−θt−k˜×k˜×E^0+O1e−θk2t−O1e−θt−O1ie−θt−k˜×k˜×B^0.Furthermore, we know from (51) that(111)∇⋅E=ρ+−ρ−,and thus(112)E^=−k˜×k˜×E^−ik˜kρ^+−ρ^−.Multiplying the first and the third equation of (72) by k˜ and substituting the result into equation (112), we obtain(113)E^=−ik˜kO1e−θt+O1e−θk2tρ^+0−O1e−θt+O1e−θk2tik⋅u^+0+O1e−θt+O1e−θk2tρ^−0−O1e−θt+O1e−θk2tik⋅u^−0+O1e−θt−O1ie−θt−k˜×k˜×E^0+O1ie−θt−k˜×k˜×B^0.Now, sincek˜k˜⋅B^=0, we have(114)B^=O1ie−θt−k˜×k˜×E^0+O1e−θt−O1ie−θt−k˜×k˜×B^0.Moreover, taking the sum of the first and the second equations of (86), we obtain(115)∂tM1+M2+M1+M2+M2−M1α−−α+=0.Thus,(116)M1+M2=e−θtM1,0+M2,0+∫0te−t−sM2−M1α−−α+ds.Substituting the first and the second equations of (84) into equation (116), we get that(117)−k˜×k˜×u^++−k˜×k˜×u^−=e−θt−k˜×k˜×u^+0+−k˜×k˜×u^−0+∫0te−t−sO1e−θk2s−k˜×k˜×u^−0−−k˜×k˜×u^+0+O1e−θk2s−O1e−θs−k˜×k˜×E^0+O1e−θk2s−O1e−θs−O1ie−θs−k˜×k˜×B^0ds.Next, substituting (74) into the above computations, we obtain(118)u^+t,k+u^−t,k=ik˜kO1e−θt+O1e−θk2tk2ρ^+0+O1e−θt+O1e−θk2tk2k˜k˜⋅u^+0+ik˜kO1e−θt+O1e−θk2tk2ρ^−0+O1e−θt+O1e−θk2tk2k˜k˜⋅u^−0+e−θt−k˜×k˜×u^+0+−k˜×k˜×u^−0+O1e−θk2t−e−t−k˜×k˜×u^+0−−k˜×k˜×u^−0+O1e−θk2t−e−t−k˜×k˜×E^0+O1e−θt−e−t−k˜×k˜×E^0+O1e−θk2t−e−t−k˜×k˜×B^0−O1e−θt−e−t−k˜×k˜×B^0−O1e−θt−e−t−k˜×ik˜×B^0.Now, taking the sum and difference, respectively, of (110) and (118), we obtain(119)u^±t,k=ik˜kO1e−θt+O1e−θk2tk2ρ^+0+O1e−θt+O1e−θk2tk2k˜k˜⋅u^+0+ik˜kO1e−θt+O1e−θk2tk2ρ^−0+O1e−θt+O1e−θk2tk2k˜k˜⋅u^−0+O1e−θk2t−k˜×k˜×u^−0−−k˜×k˜×u^+0+O1e−θk2t−O1e−θt−k˜×k˜×E^0+O1e−θk2t−O1e−θt−O1ie−θt−k˜×k˜×B^0+ik˜kO1e−θtρ^+0+O1e−θtk˜k˜⋅u^+0−ik˜kO1e−θtρ^−0+O1e−θtk˜k˜⋅u^−0+e−θt−k˜×k˜×u^+0+−k˜×k˜×u^−0+O1e−θk2t+O1e−θk2t−O1e−θt+O1e−θk2t+O1e−θt−O1ie−θt−O1e−t.(ii)
Whenk⟶∞,λjk has the following asymptotic expansion:(120)λjk=μj1k+μj0+μj−1k−1+μj−2k−2+….Each coefficientμjn is given by direct computation as(121)μ11=0,μ10=−1+α++α−,μ21=i,μ20=0,μ2−1=−23i,μ2−2=−49,μ31=−i,μ30=0,μ3−1=23i,μ3−2=−49.The approximation of the eigenvalue whenk⟶∞ is(122)λ1=−1+α++α−,λ2=ik−23ik−1−49k−2,λ3=−ik+23ik−1−49k−2.Hence, the Green matrixG for (98) is given by Gt=DtD−10, and the solution is represented as(123)x=Gtx0,where each component of this solution is itself a 3×3 diagonal matrix. After a series of calculation using the symbolic manipulator and equation (84), the solution can be written as(124)−k˜×k˜×u^−−−k˜×k˜×u^+=O1e−θt−k˜×k˜×u^−,0−−k˜×k˜×u^+0+O1ke−θk−2t−k˜×k˜×E^0+O1k2e−θt−O1k2e−θk−2t−k˜×k˜×B^0,−k˜×k˜×E^=−O1e−θt+O1e−θk−2t−k˜×k˜×u^−0−−k˜×k˜×u^+0+O1−θk−2t−k˜×k˜×E^0+O1ke−θk−2t−k˜×k˜×B^0,−k˜×k˜×B^=O1ike−θk−2t−k˜×k˜×u^−0−−k˜×k˜×u^+0−O1ike−θt−k˜×k˜×E^0+O1e−θk−2t−k˜×k˜×B^0.By using (73) and substituting (80) into the first equation of the above computation, we obtain(125)u^−−u^+=O1e−θtk˜k˜−O1e−θtik˜kiρ^+0+O1e−θt−O1e−θt1kk˜k˜⋅u^+0+O1e−θtik˜k−O1e−θtik˜k˜ρ^−0+O1e−θt1k−O1e−θtk˜k˜⋅u^−0+O1e−θt−k˜×k˜×u^−0−−k˜×k˜×u^+,0+O1ke−θk−2t−k˜×k˜×E^0+O1k2e−θt−O1k2e−θk−2t−k˜×k˜×B^0.Furthermore, we have(126)E^=−k˜×k˜×E^−ik˜kρ^+−ρ^−.Therefore, multiplying the first and the third equation of (79) by k˜ and substituting into the above equation, we obtain(127)E^=−O1−O1k−1e−θtρ^+0−O1k−1−O1e−θtik⋅u^+0+O1k−1−O1e−θtρ^−0−O1+O1k−1e−θtik⋅u^−0+−O1e−θt+O1e−θk−2t−k˜×k˜×u^−0−−k˜×k˜×u^+,0+O1e−θk−2t−k˜×k˜×E^0+O1ke−θk−2t−k˜×k˜×B^0.Sincek˜k˜⋅B^=0, we get(128)B^=O1ike−θk−2t−k˜×k˜×u^−,0−−k˜×k˜×u^+,0−O1ike−θt−k˜×k˜×E^0+O1e−θk−2t−k˜×k˜×B^0.Substituting the first equation of (124) into equation (116) results in(129)−k˜×k˜×u^++−k˜×k˜×u^−=e−θt−k˜×k˜×u^+0+−k˜×k˜×u^−0+∫0te−t−sO1e−θs−k˜×k˜×u^−0−−k˜×k˜×u^+0+O1ke−θk−2s−k˜×k˜×E^0+O1k2e−θs−O1k2e−θk−2s−k˜×k˜×B^0ds.Next, substituting equation (80) into equation (129), we obtain(130)u^+t,k+u^−t,k=O1e−θtk˜k˜−O1e−θtik˜kiρ^+0+O1e−θt−O1e−θt1kk˜k˜⋅u^+0+O1e−θtik˜k−O1e−θtik˜k˜ρ^−0+O1e−θt1k−O1e−θtk˜k˜⋅u^−0+e−θt−k˜×k˜×u^+0+−k˜×k˜×u^−0+O1e−θt−e−t−k˜×k˜×u^+0+−k˜×k˜×u^−0+O1e−θk−2t−e−t−k˜×k˜×E^0+O1e−θt−e−t−k˜×k˜×B^0+O1e−θk−2t−e−t−k˜×k˜×B^0.Now, taking the sum and difference, respectively, of equations (125) and (130), we obtain(131)u^±t,k=O1e−θtk˜k˜−O1e−θtik˜kiρ^+0+O1e−θt−O1e−θt1kk˜k˜⋅u^+0+O1e−θtik˜k−O1e−θtik˜k˜ρ^−0+O1e−θt1k−O1e−θtk˜k˜⋅u^−0x+O1e−θt−k˜×k˜×u^+0+−k˜×k˜×u^−0+O1ke−θk−2t−k˜×k˜×E^0+O1k2e−θt−O1k2e−θk−2t−k˜×k˜×B^0+ik˜k|2O1e−θtρ^+0+O1ke−θtk˜k˜⋅u^+0−ik˜O1e−θtρ^−0+O1e−θtk˜k˜⋅u^−0+e−θt−k˜×k˜×u^+0+−k˜×k˜×u^−0+O1e−θt+O1e−θk−2t+O1e−θt−O1e−θk−2t−O1e−t.Theorem 4.
Let1≤p,r≤2≤q≤∞,andl≥0 and let m≥0 be an integer. Define(132)l+31r−1q+=l+31r−1q−+1,whenr≠2orq≠2orlis not an integer,l,whenr=q=2andlis an integer,where ⋅− denotes the integer part of the argument. Suppose U0 satisfies (53). Then, for any t≥0, ∇metLU0 satisfies the following time decay property:(133)∇metLU0Lq≤C1+t−3/21/p−1/q−m/2U0Lp+C1+t−l/2∇m+l+31/r−1/q+U0Lr,where C=Cp,q,r,l,m.Theorem 5.
Let1≤p,r≤2≤q≤∞,andl≥0 and let m≥0 be an integer. Assume Ut=etLU0 is the solution of the initial value problems (51) and (52) with initial data ρ±0,u±0,E0,B0 which satisfies (53). Then, for any t≥0, U=ρ±,u±,E,B satisfies(134)∇mρ±tLq≤C1+t−3/21/p−1/q−m/2ρ^±0,u^±,0Lp+Cρ±0,u±0Lr,∇mu±tLq≤Ce−t/2ρ±0Lp+∇m+31/r−1/q+ρ±0,u±0Lr(135)+C1+t−3/21/p−1/q−m+1/2u±0,E0,B0Lp+C1+t−l+1/2∇m+l+31/r−1/q+u±0,E0,B0Lr,(136)∇mEtLq≤C1+t−3/21/p−1/q−m+1/2u±0,E0,B0Lp+C1+t−l/2∇m+l+31/r−1/q+u±0,E0,B0Lr,(137)∇mBtLq≤C1+t−3/21/p−1/q−m/2u±0,E0,B0Lp+C1+t−l/2∇m+l+31/r−1/q+u±0,E0,B0Lr,where C=Cp,q,r,l,m and l+31/r−1/q+ is defined in (132).Proof.
Let1≤p,r≤2≤q≤∞, and m be a nonnegative integer. Let q′ satisfy 1/q+1/q=1. Using the Hausdorff–Young inequality, we prove (134) as follows:(138)∇mρ±t,kLxq≤Ckmρ^±Lq′|k|≤1+Ckmρ^±Lq′k≥1.
We estimate the first term of (138) using the Hölder inequality 1/q′=1/p′+p′−q′/p′q′ with 1/p′+1/p=1 and fixing ϵ>0 sufficiently small; that is,(139)kmρ^±0,u^±0e−θt+e−θ|k|2tLq′k≤1=k−p′−q′/p′q′3−εkm+p′−q′/p′q′3−ερ^±0,u^±0e−θt+e−θ|k|2tLq′k≤1≤k−3−ε−p′−q′/p′q′km+p′−q′/p′q′ρ^±0,u^±0e−θt+e−θ|k|2tLp′k≤1≤km+p′−q′/p′q′3−ερ^±0,u^±0e−θt+e−θ|k|2tLp′k≤1≤Ce−θtρ^±0,u^±0Lp′k≤1+Ct+1−3/21/q′−1/p′−m+1/2ρ^±0,u^±0Lp′k≤1≤Ce−θtρ±0,u±0Lpk≤1+Ct+1−3/21/p−1/q−m+1/2ρ±0,u±0Lpk≤1.
Now, we estimate the second term; by takingϵ>0 sufficiently small and using the Hölder inequality 1/q′=1/r′+r′−q′/r′q′ with 1/r′+1/r=1, we find(140)kmρ^±0,u^±0e−θt+e−θk2tLq′k≥1=k−r′−q′/r′q′3+εkm+p′−q′/p′q′3+ερ^±0,u^±0e−θtLq′k≥1≤k−3+ε−r′−q′/r′q′km+r′−q′/r′q′3+ερ^±0,u^±0e−θtLr′k≥1≤Ckm+1/q′−1/r′3+ερ^±0,u^±0e−θtLr′k≥1≤Ce−θt∇m+31/r−1/q+ρ±0,u±0Lr′k≥1.
We prove (135) similarly. That is,(141)∇mu±tLxq≤Ckmu^±Lq′k≤1+Ckmu^±Lq′k≥1,where(142)kmu^±t,kLq′k≤1=kme−t/2ρ^±0,u^±0,E^0Lq′+kme−θtρ^±0,u^±0,E^0,B^0Lq′+kme−θk2tρ^±0,u^±0,E^0,B^0Lq′.
We estimate of the first term of (142) using the Hölder inequality 1/q′=1/p′+p′−q′/p′q′ with 1/p′+1/p=1 and fixing ϵ>0 sufficiently small:(143)kme−t/2ρ^±0,u^±0,E^0Lq′=∫ℝ3kme−t/2ρ^±0,u^±0,E^0q′1/q′=∫ℝ3k−p′−q′/p′q′3−εkm+p′−q′/p′q′3−εe−t/2ρ^±0,u^±0,E^0q′1/q′≤e−t/2k−3−ε−p′−q′/p′q′km+p′−q′/p′q′ρ^±0Lp′k≤1≤e−t/2km+p′−q′/p′q′3−ερ^±0Lp′k≤1≤Ce−t/2ρ^±0Lp′k≤1≤Ce−t/2ρ±0Lpk≤1.
We estimate the second term of (142) by(144)kme−θk2tρ^±0,u^±0,E^0,B^0Lq′=∫ℝ3km+1e−θk2tu^±0,E^0,B^0q′1/q′=∫ℝ3k−p′−q′/p′q′3−εkm+p′−q′/p′q′3−εe−θk2tu^±0,E^0,B^0q′1/q′≤k−3−ε−p′−q′/p′q′km+p′−q′/p′q′e−θk2tu^±0,E^0,B^0Lp′k≤1≤e−t/2km+p′−q′/p′q′3−ερ^±0Lp′k≤1≤Ce−t/2ρ^±0Lp′k≤1≤Ce−t/2ρ±0Lpk≤1.
We estimate the third term of (142) in two parts. First, take ϵ>0 sufficiently small and use the Hölder inequality 1/q′=1/r′+r′−q′/r′q′ with 1/r′+1/r=1:(145)kmu^±t,kLq′k≥1=kme−t/2ρ^±0,u^±0,E^0Lq′+kme−θtρ^±0,u^±0,E^0,B^0Lq′+km1ke−θk2tρ^±0,u^±0,E^0,B^0Lq′,where(146)kme−t/2ρ^±0,u^±0,E^0Lq′=∫ℝ3kme−t/2ρ^±0,u^±0,E^0q′1/q′=∫ℝ3k−r′−q′/r′q′3+εkm+r′−q′/r′q′3+εe−t/2ρ^±0,u^±0,E^0q′1/q′≤e−t/2k−3+ε‖−r′−q′/r′q′‖km+r′−q′/r′q′3+ερ^±0Lp′k≤1≤e−t/2km+1/q′−1/r′3+ερ^±0Lp′k≤1≤Ce−t/2∇m+31/r−1/q+ρ^±0Lr.
To estimate the second part of the third term, we have(147)km1ke−θt/k2ρ^±0,u^±0,E^0,B^0Lq′=∫ℝ3km−1e−θt/k2u^±0,E^0,B^0q′1/q′=∫ℝ3k−r′−q′/r′q′3+ϵkm+r′−q′/r′q′3+ϵe−θt/k2u^±0,E^0,B^0q′1/q′≤k−3+ϵ−r′−q′/r′q′km+r′−q′/r′q′3+ϵe−θt/k2u^±0,E^0,B^0Lr′k≤1≤e−θt/k2km+l+31/q′−1/r′k−l+1u±0,E0,B0Lr′k≤1≤C1+t−l+1/2∇m+l+31/r−1/q+u±0,E0,B0Lr,where we used(148)supk≤11kl+1e−θt+1/k2≤C1+t−l+1/2.
The proofs of (136) and (137) are similar.
Note that we transformed to the Fourier space to find the solution representation. Now, we go back to the physical spacext and obtain the decay rates for the solutions to the linearized system. Fundamental properties of the system are given by the following corollary based on Theorem 5.Corollary 1.
AssumeUt=etLU0 is the solution of the initial value problems (51) and (52) with initial data ρ±0,u±0,E0,B0 which satisfies (53). Then, U=ρ±,u±,E,B satisfies(149)ρ±t≤Ce−t/2ρ±0,u±0,u±t≤Ce−t/2ρ±0+C1+t−5/4u±0,E0,B0L1∩H˙2,Et≤C1+t−5/2u±0,E0,B0L1∩H˙3,Bt≤C1+t−3/4u±0,E0,B0L1∩H˙2,(150)ρ±tL∞≤Ce−t/2ρ±0,u±0L2∩H˙2,u±tL∞≤Ce−t/2ρ±0L1∩H˙2+C1+t−2u±0,E0,B0L1∩H˙5,EtL∞≤C1+t−2u±0,E0,B0L1∩H˙6,BtL∞≤C1+t−3/2u±0,E0,B0L1∩H˙5,(151)Bt≤C1+t−5/4u±0,E0,B0L1∩H˙4,∇NEt,Bt≤C1+t−5/4u±0,E0,B0L1∩H˙N+3.Proof.
The results of Corollary1 are particular cases of Theorem 5.
For example, we explain how we get the decay rates foru±. From (135) for L2, take q=2,p=1,m=0,l=3/2, and r=2:(152)∇mu±tL2≤Ce−t/2ρ±0L1+∇3/2+1ρ±0,u±0L1+C1+t−5/4u±0,E0,B0L1+C1+t−5/4∇1+1/2+1u±0,E0,B0L1.
Thus,(153)u±tL2≤Ce−t/2ρ±0+C1+t−5/4u±0,E0,B0L1∩H˙2.
Now, forL∞, take q=∞,p=1,m=0,l=3, and r=2:(154)∇mu±tL∞≤Ce−t/2ρ±0L1+∇3/2+1ρ±0,u±0L2+C1+t−2u±0,E0,B0L1+C1+t−2∇4+1/2+1u±0,E0,B0L2.
Thus,(155)u±tL∞≤Ce−t/2ρ±0L1∩H˙2+C1+t−2u±0,E0,B0L1∩H˙5.


## 3.2.1. Parallel Part

We proceed with the asymptotic expansion of eigenvalues: letλjk,j=1,2,3,4, be the eigenvalues of the matrix Ak. Taking the determinant, we see the eigenvalues satisfy(64)detA−λI=λ4+2+α++α−λ3+3+α++α−+A+γk2+A−γk2λ2+2+A+γk2+A−γk2+α−A+γk2+α+A−γk2λ+A+γk2+A−γk2+A−A+γ2k4=0.(i)
First we consider whenk⟶0, since detA−λI=0 satisfies(65)λλ+1λ2+1+α−+α+λ+2=0,λjk has the following asymptotic expansion:(66)λjk=λj0+λj1k+λj2k2+…,where each coefficientλjn is given by direct computation as(67)λ10=λ11=0,λ12=−12A+γ+A−γ,λ20=−1,λ21=0,λ22=−A+γ+A−γ+α−A+γ+α+A−γ2−α+−α−,λ30=−1+α++α−−1+α++α−2−82,λ31=0,λ32=0,λ40=−1+α++α−+1+α++α−2−82,λ41=0,λ42=0.Thus, the approximation of the eigenvalue whenk⟶0 is(68)λ1=−12A+γ+A−γk2,λ2=−1−A+γ+A−γ+α−A+γ+α+A−γ2−α+−α−k2,λ3=−1+α++α−−1+α++α−2−82,λ4=−1+α++α−+1+α++α−2−82.Therefore, if we define(69)Bt=a1λ1eλ1ta1λ2eλ2ta2λ3eλ3ta2λ4eλ4t,then the Green matrixG for (61) is given by Gt=BtB−10, and the solution is(70)x=Gtx0.Thus, after a series of calculation using symbolic manipulator, we have(71)x1x2x3x4=O1e−θt+O1e−θk2tO1e−θt+O1e−θk2tO1e−θt+O1e−θk2tO1e−θt+O1e−θk2tO1e−θt+O1e−θk2tk2O1e−θt+O1e−θk2tk2O1e−θt+O1e−θk2tk2O1e−θt+O1e−θk2tk2O1e−θt+O1e−θk2tO1e−θt+O1e−θk2tO1e−θt+O1e−θk2tO1e−θt+O1e−θk2tO1e−θt+O1e−θk2tk2O1e−θt+O1e−θk2tk2O1e−θt+O1e−θk2tk2O1e−θt+O1e−θk2tk2x1,0x2,0x3,0x4,0,whereθ is the sum of real parts of λj,j=1,2,3,4. Thus,(72)ρ^+t,k=O1e−θt+O1e−θk2tρ^+0−O1e−θt+O1e−θk2tik⋅u^+0+O1e−θt+O1e−θk2tρ^−0−O1e−θt+O1e−θk2tik⋅u^−0,−ik⋅u^+t,k=O1e−θt+O1e−θk2tk2ρ^+0−O1e−θt+O1e−θk2tk2ik⋅u^+0+O1e−θt+O1e−θk2tk2ρ^−0−O1e−θt+O1e−θk2tk2ik⋅u^−0,ρ^−t,k=O1e−θk2t+O1e−θtρ^+0−O1e−θk2t+O1e−θtik⋅u^+0+O1e−θk2t+O1e−θtρ^−0−O1e−θk2t+O1e−θtik⋅u^−0,−ik⋅u^−t,k=O1e−θt+O1e−θk2tk2ρ^+0−O1e−θt+O1e−θk2tk2ik⋅u^+0+O1e−θt+O1e−θk2tk2ρ^−0−O1e−θt+O1e−θk2tk2ik⋅u^−0.In fact, we know that(73)u^±=k˜k˜⋅u^±−k˜×k˜×u^±impliesk˜k˜⋅u^±=u^±+k˜×k˜×u^±.Therefore, plugging (73) in the second and the fourth equations of (72), we obtain(74)u^+t,k=−k˜×k˜×u^++ik˜kO1e−θt+O1e−θk2tk2ρ^+0+O1e−θt+O1e−θk2tk2k˜k˜⋅u^+0+ik˜kO1e−θt+O1e−θk2tk2ρ^−0+O1e−θt+O1e−θk2tk2k˜k˜⋅u^−0,u^−t,k=−k˜×k˜×u^−+ik˜kO1e−θt+O1e−θk2tk2ρ^+0+O1e−θt+O1e−θk2tk2k˜k˜⋅u^+0+ik˜kO1e−θt+O1e−θk2tk2ρ^−0+O1e−θt+O1e−θk2tk2k˜k˜⋅u^−0.(ii)
Whenk⟶∞,λjk has the following asymptotic expansion:(75)λjk=μj1k+μj0+μj−1k−1+μj−2k−2+μj−3k−3+…,where each coefficientμjn is given by direct computation as(76)μ11=iA−γ,μ10=−1+α−2,μ21=−iA−γ,μ20=−1+α−2,μ31=iA+γ,μ30=−1+α+2,μ41=−iA+γ,μ40=−1+α+2.The approximation of the eigenvalue whenk⟶∞ is(77)λ1=−1+α−2+iA−γk,λ2=−1+α−2−iA−γk,λ3=−1+α+2+iA+γk,λ4=−1+α+2−iA+γk.So,G=BtB−10 is the Green matrix for (61), and the solution is(78)x=Gtx0.Thus, after a series of calculation using the symbolic manipulator, we obtain(79)x1x2x3x4=O1e−θtO1e−θtk−1O1e−θtk−1O1e−θtO1e−θtkO1e−θtO1e−θtO1e−θtk−1O1e−θtk−1O1e−θtO1e−θtO1e−θtk−1O1e−θtO1e−θt|k|−1O1e−θtkO1e−θtx1,0x2,0x3,0x4,0,ρ^+t,k=O1e−θtρ^+0−O1e−θtk−1ik⋅u^+0+O1e−θtk−1ρ^−0−O1e−θtik⋅u^−0,−ik⋅u^+t,k=O1e−θtkρ^+0−O1e−θtik⋅u^+0+O1e−θtρ^−0−O1e−θtk−1ik⋅u^−0,ρ^−t,k=O1e−θtk−1ρ^+0−O1e−θtik⋅u^+0+O1e−θtρ^−0−O1e−θtk−1ik⋅u^−0,−ik⋅u^−t,k=O1e−θtρ^+0−O1e−θtk−1ik⋅u^+0+O1e−θtkρ^−0−O1e−θtik⋅u^−0.Therefore, after plugging (72) into the second and fourth equations of (79), we obtain(80)u^+t,k=−k˜×k˜×u^++O1e−θtik˜k˜⋅ρ^+0+O1e−θtk˜k˜⋅u^+0+O1e−θtik˜kρ^−0+O1e−θtk˜k˜k⋅u^−0,u^−t,k=−k˜×k˜×u^−+O1e−θtik˜kρ^+0+O1e−θtk˜k˜k⋅u^+0+O1e−θtik˜k˜ρ^−0+O1e−θtk˜k˜⋅u^−0.(iii)
When0<k<∞, we consider the Routh–Hurwitz stability condition of the characteristic polynomial (64). That is, if we write (64) in the form(81)b0λ4+b1λ3+b2λ2+b3λ+b4=0.The system stability requires(82)b1>0,b1b2−b0b3>0,b1b2−b0b3b3−b12b4>0,b4>0.It can be shown that, in our case, the conditions corresponding to (82) are given, respectively, by(83)2+α++α−>0,2+α++α−3+α++α−+A+γk2+A−γk2−2+A+γk2+A−γk2+α−A+γk2+α+A−γk2>0,4+2α++2α−+A+γk2+A−γk2+3α++α+2+2α+α−+α+A+γk2+3α−+α−2+α−A−γk22+A+γk2+A−γk2+α−A+γk2+α+A−γk2−2+α++α−2A+γk2+A−γk2+A−A+γ2k4>0,A+γk2+A−γk2+A−A+γ2k4>0.It is not difficult to show that the above inequalities are satisfied, and this implies that all roots of the characteristic equation have negative real parts.Although the eigenvalues may coalesce, the computations in (i), (ii), and (iii) show that coalescence occurs when the real parts of the eigenvalues are negative. Therefore, the stability conditions are satisfied.


## 3.2.2. Perpendicular Part

Now, we consider(84)M1t,k≔−k˜×k˜×u^+t,k,M2t,k≔−k˜×k˜×u^−t,k,M3t,k≔−k˜×k˜×E^t,k,M4t,k≔−k˜×k˜×B^t,k,when t>0 and k≠0. Taking the curl of the second, third, and fourth equations in system (51), we get(85)∂t∇×u++∇×u++α+∇×u+−α+∇×u−−∇×E=0,∂t∇×u−+∇×u−+α−∇×u−−α−∇×u++∇×E=0,∂t∇×E−∇×∇×B+∇×u+−∇×u−=0,∂t∇×B+∇×∇×E=0.Now, taking the Fourier Transform and multiplying by−ik, we obtain(86)∂tM1+M1+α+M1−α+M2−M3=0,∂tM2+M2+α−M2−α−M1+M3=0,∂tM3−ik×M4+M1−M2=0,∂tM4+ik×M3=0.Subtracting the first equation from the second equation in (86), we obtain(87)∂tM2−M1+M2−M1+α−M2−M1+α+M2−M1+2M3=0,and we simplify the above computation by letting M5=M2−M1.So,(88)∂tM5+1+α++α−M5+2M3=0.Thus,(89)∂tM5=−1+α++α−M5−2M3,∂tM3=ik×M4+M5,∂tM4=−ik×M3,where we write the initial data in the form(90)M3,M4,M5t=0=M3,0,M4,0,M5,0,with(91)M5,0=−k˜×k˜×u^−−u^+,M3,0=−k˜×k˜×E^0,M4,0=−k˜×k˜×B^0.Taking the time derivative of∂tM3 and substituting ∂tM5 and ∂tM4 as given in (89), we get that(92)∂ttM3=k×k×M3−1+α++α−M5−2M3.Fromk⋅M2=0, we find k×k×M3=−k2M3; thus,(93)∂ttM3+2+k2M3=−1+α++α−M5.Now, taking the time derivative of (93) and replacing ∂tM5 by the first equation of (89) and taking the sum with (93), we get(94)∂tttM3+1+α++α−∂ttM3+2+k2∂tM3+1+α++α−k2M3=0,with the initial data(95)M3t=0=M3,0,∂tM3t=0=ik×M4,0+M5,0,∂ttM3t=0=−1+α++α−M5,0−2+k2M3,0.Note that the characteristic equation of (94) is(96)Fx≔χ3+1+α++α−χ2+2+k2χ+1+α++α−k2=0.Set(97)x1=M3,x2=x˙1=∂tM3,x3=x˙2=∂ttM3,x=x1,x2,x3T.Then, equation (94) is written as(98)xt=Φkx,where(99)Φk=010001−1+α++α−k2−2+k2−1+α++α−.Asymptotic Expansion of Eigenvalues. Let λjk be the eigenvalues of the matrix Φk. We will find the asymptotic expansion of the eigenvalues λjkfork⟶0andk⟶∞.The eigenvaluesλjk,j=1,2,3, are the solutions of the characteristic equation which can be written as(100)detΦk−λI=λ3+1+α++α−λ2+2+k2λ+1+α++α−k2=0.The eigenvector forΦk is(101)bλ=1λλ2.(i)
Whenk⟶0,λjk has the following asymptotic expansion:(102)λjk=λj0+λj1k+λj2k2+…,where each coefficientλjn is given by the direct computation as(103)λ10=λ11=0,λ12=−121+α++α−,λ20=−1+α++α−+1+α++α−2−82,λ21=0,λ22=−1+α++α−−1+α++α−2−84,λ30=−1+α++α−−1+α++α−2−82,λ31=0,λ32=−1+α++α−+1+α++α−2−84.Thus, the approximation of the eigenvalue whenk⟶0 is(104)λ1=−121+α++α−k2,λ2=−1+α++α−+1+α++α−2−82−1+α++α−+1+α++α−2−84k2,λ3=−1+α++α−−1+α++α−2−82−1+α++α−−1+α++α−2−84k2.Therefore, if we set(105)Dt=bλ1eλ1tbλ2eλ2tbλ3eλ3t,the Green matrixG for (98) is given by Gt=DtD−10, and the solution is represented as(106)x=Gtx0,where each component of this solution is itself a3×3 diagonal matrix. We integrate the first and the third equations of (89), and we get the following expressions:(107)M5t,k=M5,0ke−t1+α++α−+2e−t1+α++α−∫0tc1ke1+α++α−+ηsds+∫0te−t1+α++α−+βsc2kcosωs+c3ksinωsds,M4t,k=M4,0k−ik×∫0tc1keηs+eβsc2kcosωs+c3ksinωsds,where(108)c1=13η2+21+α++α−η+2+k2−ηI3M5,0t,k+η1+α++α−+ηI3M3,0t,k+1+α++α−+ηik×M4,0t,k,c2=13η2+21+α++α−η+2+k2ηI3M5,0t,k+2η+η1+α++α−+η+k2+2I3M3,0t,k−1+α++α−+ηik×M4,0t,k,c3=13η2+21+α++α−η+2+k2−3/2η2−3/2η1+α++α−−2+k2ωI3M5,0t,k+1+α++α−+ηη1+α++α−+2+k22ωI3M3,0t,k+−1+α++α−2/2+3/2η2+2+k2ωik×M4,0t,k.After a series of calculation using the symbolic manipulator and definition (84), the solution can be written as(109)−k˜×k˜×u^−−−k˜×k˜×u^+=O1e−θk2t−k˜×k˜×u^−0−−k˜×k˜×u^+0+O1e−θk2t−O1e−θt−k˜×k˜×E^0+O1e−θk2t−O1e−θt−O1ie−θt−k˜×k˜×B^0,−k˜×k˜×E^=O1e−θt−O1ie−θt−k˜×k˜×E^0+O1ie−θt−k˜×k˜×B^0,−k˜×k˜×B^=O1ie−θt−k˜×k˜×E^0+O1e−θt−O1ie−θt−k˜×k˜×B^0.By using (73) and substituting (74) in the first equation of (109), we obtain(110)u^−−u^+=ik˜kO1e−θt+O1e−θk2tk2ρ^+0+O1e−θt+O1e−θk2tk2k˜k˜⋅u^+0+ik˜kO1e−θt+O1e−θk2tk2ρ^−0+O1e−θt+O1e−θk2tk2k˜k˜⋅u^−0+O1e−θk2t−k˜×k˜×u^−0−−k˜×k˜×u^+,0+O1e−θk2t−O1e−θt−k˜×k˜×E^0+O1e−θk2t−O1e−θt−O1ie−θt−k˜×k˜×B^0.Furthermore, we know from (51) that(111)∇⋅E=ρ+−ρ−,and thus(112)E^=−k˜×k˜×E^−ik˜kρ^+−ρ^−.Multiplying the first and the third equation of (72) by k˜ and substituting the result into equation (112), we obtain(113)E^=−ik˜kO1e−θt+O1e−θk2tρ^+0−O1e−θt+O1e−θk2tik⋅u^+0+O1e−θt+O1e−θk2tρ^−0−O1e−θt+O1e−θk2tik⋅u^−0+O1e−θt−O1ie−θt−k˜×k˜×E^0+O1ie−θt−k˜×k˜×B^0.Now, sincek˜k˜⋅B^=0, we have(114)B^=O1ie−θt−k˜×k˜×E^0+O1e−θt−O1ie−θt−k˜×k˜×B^0.Moreover, taking the sum of the first and the second equations of (86), we obtain(115)∂tM1+M2+M1+M2+M2−M1α−−α+=0.Thus,(116)M1+M2=e−θtM1,0+M2,0+∫0te−t−sM2−M1α−−α+ds.Substituting the first and the second equations of (84) into equation (116), we get that(117)−k˜×k˜×u^++−k˜×k˜×u^−=e−θt−k˜×k˜×u^+0+−k˜×k˜×u^−0+∫0te−t−sO1e−θk2s−k˜×k˜×u^−0−−k˜×k˜×u^+0+O1e−θk2s−O1e−θs−k˜×k˜×E^0+O1e−θk2s−O1e−θs−O1ie−θs−k˜×k˜×B^0ds.Next, substituting (74) into the above computations, we obtain(118)u^+t,k+u^−t,k=ik˜kO1e−θt+O1e−θk2tk2ρ^+0+O1e−θt+O1e−θk2tk2k˜k˜⋅u^+0+ik˜kO1e−θt+O1e−θk2tk2ρ^−0+O1e−θt+O1e−θk2tk2k˜k˜⋅u^−0+e−θt−k˜×k˜×u^+0+−k˜×k˜×u^−0+O1e−θk2t−e−t−k˜×k˜×u^+0−−k˜×k˜×u^−0+O1e−θk2t−e−t−k˜×k˜×E^0+O1e−θt−e−t−k˜×k˜×E^0+O1e−θk2t−e−t−k˜×k˜×B^0−O1e−θt−e−t−k˜×k˜×B^0−O1e−θt−e−t−k˜×ik˜×B^0.Now, taking the sum and difference, respectively, of (110) and (118), we obtain(119)u^±t,k=ik˜kO1e−θt+O1e−θk2tk2ρ^+0+O1e−θt+O1e−θk2tk2k˜k˜⋅u^+0+ik˜kO1e−θt+O1e−θk2tk2ρ^−0+O1e−θt+O1e−θk2tk2k˜k˜⋅u^−0+O1e−θk2t−k˜×k˜×u^−0−−k˜×k˜×u^+0+O1e−θk2t−O1e−θt−k˜×k˜×E^0+O1e−θk2t−O1e−θt−O1ie−θt−k˜×k˜×B^0+ik˜kO1e−θtρ^+0+O1e−θtk˜k˜⋅u^+0−ik˜kO1e−θtρ^−0+O1e−θtk˜k˜⋅u^−0+e−θt−k˜×k˜×u^+0+−k˜×k˜×u^−0+O1e−θk2t+O1e−θk2t−O1e−θt+O1e−θk2t+O1e−θt−O1ie−θt−O1e−t.(ii)
Whenk⟶∞,λjk has the following asymptotic expansion:(120)λjk=μj1k+μj0+μj−1k−1+μj−2k−2+….Each coefficientμjn is given by direct computation as(121)μ11=0,μ10=−1+α++α−,μ21=i,μ20=0,μ2−1=−23i,μ2−2=−49,μ31=−i,μ30=0,μ3−1=23i,μ3−2=−49.The approximation of the eigenvalue whenk⟶∞ is(122)λ1=−1+α++α−,λ2=ik−23ik−1−49k−2,λ3=−ik+23ik−1−49k−2.Hence, the Green matrixG for (98) is given by Gt=DtD−10, and the solution is represented as(123)x=Gtx0,where each component of this solution is itself a 3×3 diagonal matrix. After a series of calculation using the symbolic manipulator and equation (84), the solution can be written as(124)−k˜×k˜×u^−−−k˜×k˜×u^+=O1e−θt−k˜×k˜×u^−,0−−k˜×k˜×u^+0+O1ke−θk−2t−k˜×k˜×E^0+O1k2e−θt−O1k2e−θk−2t−k˜×k˜×B^0,−k˜×k˜×E^=−O1e−θt+O1e−θk−2t−k˜×k˜×u^−0−−k˜×k˜×u^+0+O1−θk−2t−k˜×k˜×E^0+O1ke−θk−2t−k˜×k˜×B^0,−k˜×k˜×B^=O1ike−θk−2t−k˜×k˜×u^−0−−k˜×k˜×u^+0−O1ike−θt−k˜×k˜×E^0+O1e−θk−2t−k˜×k˜×B^0.By using (73) and substituting (80) into the first equation of the above computation, we obtain(125)u^−−u^+=O1e−θtk˜k˜−O1e−θtik˜kiρ^+0+O1e−θt−O1e−θt1kk˜k˜⋅u^+0+O1e−θtik˜k−O1e−θtik˜k˜ρ^−0+O1e−θt1k−O1e−θtk˜k˜⋅u^−0+O1e−θt−k˜×k˜×u^−0−−k˜×k˜×u^+,0+O1ke−θk−2t−k˜×k˜×E^0+O1k2e−θt−O1k2e−θk−2t−k˜×k˜×B^0.Furthermore, we have(126)E^=−k˜×k˜×E^−ik˜kρ^+−ρ^−.Therefore, multiplying the first and the third equation of (79) by k˜ and substituting into the above equation, we obtain(127)E^=−O1−O1k−1e−θtρ^+0−O1k−1−O1e−θtik⋅u^+0+O1k−1−O1e−θtρ^−0−O1+O1k−1e−θtik⋅u^−0+−O1e−θt+O1e−θk−2t−k˜×k˜×u^−0−−k˜×k˜×u^+,0+O1e−θk−2t−k˜×k˜×E^0+O1ke−θk−2t−k˜×k˜×B^0.Sincek˜k˜⋅B^=0, we get(128)B^=O1ike−θk−2t−k˜×k˜×u^−,0−−k˜×k˜×u^+,0−O1ike−θt−k˜×k˜×E^0+O1e−θk−2t−k˜×k˜×B^0.Substituting the first equation of (124) into equation (116) results in(129)−k˜×k˜×u^++−k˜×k˜×u^−=e−θt−k˜×k˜×u^+0+−k˜×k˜×u^−0+∫0te−t−sO1e−θs−k˜×k˜×u^−0−−k˜×k˜×u^+0+O1ke−θk−2s−k˜×k˜×E^0+O1k2e−θs−O1k2e−θk−2s−k˜×k˜×B^0ds.Next, substituting equation (80) into equation (129), we obtain(130)u^+t,k+u^−t,k=O1e−θtk˜k˜−O1e−θtik˜kiρ^+0+O1e−θt−O1e−θt1kk˜k˜⋅u^+0+O1e−θtik˜k−O1e−θtik˜k˜ρ^−0+O1e−θt1k−O1e−θtk˜k˜⋅u^−0+e−θt−k˜×k˜×u^+0+−k˜×k˜×u^−0+O1e−θt−e−t−k˜×k˜×u^+0+−k˜×k˜×u^−0+O1e−θk−2t−e−t−k˜×k˜×E^0+O1e−θt−e−t−k˜×k˜×B^0+O1e−θk−2t−e−t−k˜×k˜×B^0.Now, taking the sum and difference, respectively, of equations (125) and (130), we obtain(131)u^±t,k=O1e−θtk˜k˜−O1e−θtik˜kiρ^+0+O1e−θt−O1e−θt1kk˜k˜⋅u^+0+O1e−θtik˜k−O1e−θtik˜k˜ρ^−0+O1e−θt1k−O1e−θtk˜k˜⋅u^−0x+O1e−θt−k˜×k˜×u^+0+−k˜×k˜×u^−0+O1ke−θk−2t−k˜×k˜×E^0+O1k2e−θt−O1k2e−θk−2t−k˜×k˜×B^0+ik˜k|2O1e−θtρ^+0+O1ke−θtk˜k˜⋅u^+0−ik˜O1e−θtρ^−0+O1e−θtk˜k˜⋅u^−0+e−θt−k˜×k˜×u^+0+−k˜×k˜×u^−0+O1e−θt+O1e−θk−2t+O1e−θt−O1e−θk−2t−O1e−t.Theorem 4.
Let1≤p,r≤2≤q≤∞,andl≥0 and let m≥0 be an integer. Define(132)l+31r−1q+=l+31r−1q−+1,whenr≠2orq≠2orlis not an integer,l,whenr=q=2andlis an integer,where ⋅− denotes the integer part of the argument. Suppose U0 satisfies (53). Then, for any t≥0, ∇metLU0 satisfies the following time decay property:(133)∇metLU0Lq≤C1+t−3/21/p−1/q−m/2U0Lp+C1+t−l/2∇m+l+31/r−1/q+U0Lr,where C=Cp,q,r,l,m.Theorem 5.
Let1≤p,r≤2≤q≤∞,andl≥0 and let m≥0 be an integer. Assume Ut=etLU0 is the solution of the initial value problems (51) and (52) with initial data ρ±0,u±0,E0,B0 which satisfies (53). Then, for any t≥0, U=ρ±,u±,E,B satisfies(134)∇mρ±tLq≤C1+t−3/21/p−1/q−m/2ρ^±0,u^±,0Lp+Cρ±0,u±0Lr,∇mu±tLq≤Ce−t/2ρ±0Lp+∇m+31/r−1/q+ρ±0,u±0Lr(135)+C1+t−3/21/p−1/q−m+1/2u±0,E0,B0Lp+C1+t−l+1/2∇m+l+31/r−1/q+u±0,E0,B0Lr,(136)∇mEtLq≤C1+t−3/21/p−1/q−m+1/2u±0,E0,B0Lp+C1+t−l/2∇m+l+31/r−1/q+u±0,E0,B0Lr,(137)∇mBtLq≤C1+t−3/21/p−1/q−m/2u±0,E0,B0Lp+C1+t−l/2∇m+l+31/r−1/q+u±0,E0,B0Lr,where C=Cp,q,r,l,m and l+31/r−1/q+ is defined in (132).Proof.
Let1≤p,r≤2≤q≤∞, and m be a nonnegative integer. Let q′ satisfy 1/q+1/q=1. Using the Hausdorff–Young inequality, we prove (134) as follows:(138)∇mρ±t,kLxq≤Ckmρ^±Lq′|k|≤1+Ckmρ^±Lq′k≥1.
We estimate the first term of (138) using the Hölder inequality 1/q′=1/p′+p′−q′/p′q′ with 1/p′+1/p=1 and fixing ϵ>0 sufficiently small; that is,(139)kmρ^±0,u^±0e−θt+e−θ|k|2tLq′k≤1=k−p′−q′/p′q′3−εkm+p′−q′/p′q′3−ερ^±0,u^±0e−θt+e−θ|k|2tLq′k≤1≤k−3−ε−p′−q′/p′q′km+p′−q′/p′q′ρ^±0,u^±0e−θt+e−θ|k|2tLp′k≤1≤km+p′−q′/p′q′3−ερ^±0,u^±0e−θt+e−θ|k|2tLp′k≤1≤Ce−θtρ^±0,u^±0Lp′k≤1+Ct+1−3/21/q′−1/p′−m+1/2ρ^±0,u^±0Lp′k≤1≤Ce−θtρ±0,u±0Lpk≤1+Ct+1−3/21/p−1/q−m+1/2ρ±0,u±0Lpk≤1.
Now, we estimate the second term; by takingϵ>0 sufficiently small and using the Hölder inequality 1/q′=1/r′+r′−q′/r′q′ with 1/r′+1/r=1, we find(140)kmρ^±0,u^±0e−θt+e−θk2tLq′k≥1=k−r′−q′/r′q′3+εkm+p′−q′/p′q′3+ερ^±0,u^±0e−θtLq′k≥1≤k−3+ε−r′−q′/r′q′km+r′−q′/r′q′3+ερ^±0,u^±0e−θtLr′k≥1≤Ckm+1/q′−1/r′3+ερ^±0,u^±0e−θtLr′k≥1≤Ce−θt∇m+31/r−1/q+ρ±0,u±0Lr′k≥1.
We prove (135) similarly. That is,(141)∇mu±tLxq≤Ckmu^±Lq′k≤1+Ckmu^±Lq′k≥1,where(142)kmu^±t,kLq′k≤1=kme−t/2ρ^±0,u^±0,E^0Lq′+kme−θtρ^±0,u^±0,E^0,B^0Lq′+kme−θk2tρ^±0,u^±0,E^0,B^0Lq′.
We estimate of the first term of (142) using the Hölder inequality 1/q′=1/p′+p′−q′/p′q′ with 1/p′+1/p=1 and fixing ϵ>0 sufficiently small:(143)kme−t/2ρ^±0,u^±0,E^0Lq′=∫ℝ3kme−t/2ρ^±0,u^±0,E^0q′1/q′=∫ℝ3k−p′−q′/p′q′3−εkm+p′−q′/p′q′3−εe−t/2ρ^±0,u^±0,E^0q′1/q′≤e−t/2k−3−ε−p′−q′/p′q′km+p′−q′/p′q′ρ^±0Lp′k≤1≤e−t/2km+p′−q′/p′q′3−ερ^±0Lp′k≤1≤Ce−t/2ρ^±0Lp′k≤1≤Ce−t/2ρ±0Lpk≤1.
We estimate the second term of (142) by(144)kme−θk2tρ^±0,u^±0,E^0,B^0Lq′=∫ℝ3km+1e−θk2tu^±0,E^0,B^0q′1/q′=∫ℝ3k−p′−q′/p′q′3−εkm+p′−q′/p′q′3−εe−θk2tu^±0,E^0,B^0q′1/q′≤k−3−ε−p′−q′/p′q′km+p′−q′/p′q′e−θk2tu^±0,E^0,B^0Lp′k≤1≤e−t/2km+p′−q′/p′q′3−ερ^±0Lp′k≤1≤Ce−t/2ρ^±0Lp′k≤1≤Ce−t/2ρ±0Lpk≤1.
We estimate the third term of (142) in two parts. First, take ϵ>0 sufficiently small and use the Hölder inequality 1/q′=1/r′+r′−q′/r′q′ with 1/r′+1/r=1:(145)kmu^±t,kLq′k≥1=kme−t/2ρ^±0,u^±0,E^0Lq′+kme−θtρ^±0,u^±0,E^0,B^0Lq′+km1ke−θk2tρ^±0,u^±0,E^0,B^0Lq′,where(146)kme−t/2ρ^±0,u^±0,E^0Lq′=∫ℝ3kme−t/2ρ^±0,u^±0,E^0q′1/q′=∫ℝ3k−r′−q′/r′q′3+εkm+r′−q′/r′q′3+εe−t/2ρ^±0,u^±0,E^0q′1/q′≤e−t/2k−3+ε‖−r′−q′/r′q′‖km+r′−q′/r′q′3+ερ^±0Lp′k≤1≤e−t/2km+1/q′−1/r′3+ερ^±0Lp′k≤1≤Ce−t/2∇m+31/r−1/q+ρ^±0Lr.
To estimate the second part of the third term, we have(147)km1ke−θt/k2ρ^±0,u^±0,E^0,B^0Lq′=∫ℝ3km−1e−θt/k2u^±0,E^0,B^0q′1/q′=∫ℝ3k−r′−q′/r′q′3+ϵkm+r′−q′/r′q′3+ϵe−θt/k2u^±0,E^0,B^0q′1/q′≤k−3+ϵ−r′−q′/r′q′km+r′−q′/r′q′3+ϵe−θt/k2u^±0,E^0,B^0Lr′k≤1≤e−θt/k2km+l+31/q′−1/r′k−l+1u±0,E0,B0Lr′k≤1≤C1+t−l+1/2∇m+l+31/r−1/q+u±0,E0,B0Lr,where we used(148)supk≤11kl+1e−θt+1/k2≤C1+t−l+1/2.
The proofs of (136) and (137) are similar.
Note that we transformed to the Fourier space to find the solution representation. Now, we go back to the physical spacext and obtain the decay rates for the solutions to the linearized system. Fundamental properties of the system are given by the following corollary based on Theorem 5.Corollary 1.
AssumeUt=etLU0 is the solution of the initial value problems (51) and (52) with initial data ρ±0,u±0,E0,B0 which satisfies (53). Then, U=ρ±,u±,E,B satisfies(149)ρ±t≤Ce−t/2ρ±0,u±0,u±t≤Ce−t/2ρ±0+C1+t−5/4u±0,E0,B0L1∩H˙2,Et≤C1+t−5/2u±0,E0,B0L1∩H˙3,Bt≤C1+t−3/4u±0,E0,B0L1∩H˙2,(150)ρ±tL∞≤Ce−t/2ρ±0,u±0L2∩H˙2,u±tL∞≤Ce−t/2ρ±0L1∩H˙2+C1+t−2u±0,E0,B0L1∩H˙5,EtL∞≤C1+t−2u±0,E0,B0L1∩H˙6,BtL∞≤C1+t−3/2u±0,E0,B0L1∩H˙5,(151)Bt≤C1+t−5/4u±0,E0,B0L1∩H˙4,∇NEt,Bt≤C1+t−5/4u±0,E0,B0L1∩H˙N+3.Proof.
The results of Corollary1 are particular cases of Theorem 5.
For example, we explain how we get the decay rates foru±. From (135) for L2, take q=2,p=1,m=0,l=3/2, and r=2:(152)∇mu±tL2≤Ce−t/2ρ±0L1+∇3/2+1ρ±0,u±0L1+C1+t−5/4u±0,E0,B0L1+C1+t−5/4∇1+1/2+1u±0,E0,B0L1.
Thus,(153)u±tL2≤Ce−t/2ρ±0+C1+t−5/4u±0,E0,B0L1∩H˙2.
Now, forL∞, take q=∞,p=1,m=0,l=3, and r=2:(154)∇mu±tL∞≤Ce−t/2ρ±0L1+∇3/2+1ρ±0,u±0L2+C1+t−2u±0,E0,B0L1+C1+t−2∇4+1/2+1u±0,E0,B0L2.
Thus,(155)u±tL∞≤Ce−t/2ρ±0L1∩H˙2+C1+t−2u±0,E0,B0L1∩H˙5.


## 4. Decay Rates for Nonlinear Systems

### 4.1. Decay Rates for System (15)

In this section, we apply the linearLp−Lq time decay property of the homogeneous system (51) to the nonlinear case. Throughout this section, we suppose that U=ρ±,u±,E,B is the solution to the Cauchy problem (15) satisfying (16).By Duhamel’s principle, the solutionU can be formally written as(156)U=etLU0+∫0tet−sLg1s,g2s,g3s,0ds,where etLU0 is as defined in Section 3.2 and(157)g1=−∇.ρ±u±,g2=−u±.∇u±−A±γρ±+1γ−2−1∇ρ±±u±×B+α±u±−u∓,g3=ρ−u−−ρ+u+.Remark 3.
Note that in (156), g1s,g2s,g3s,0 satisfies the compatibility condition (53). Thus, et−sL acts on g1s,g2s,g3s,0 for all 0≤s≤t.Proposition 2.
Suppose initial dataV0=σ±0,υ±0,E0˜,B0˜ satisfies (17). If εN+2V0>0 is small enough, then the solution V=σ±,υ±,E˜,B˜ satisfies(158)VtN≤CεN+2V01+t−3/4,for any t>0. Moreover, εN+6V0>0 being sufficiently small implies that the solution V=σ±,υ±,E˜,B˜ satisfies(159)∇VtN−1≤CεN+6V01+t−5/4,for any t≥0.Lemma 2.
SupposeV=σ±,υ±,E,B is the solution of the initial value problems (15) and (16) with initial data V0=σ±,υ±,E0˜,B0˜ satisfying (17) as described in Proposition 1. If ℰNV0 is sufficiently small, then(160)ddtℰNVt+λDNVt≤0.Proof.
Letl⩾0. Multiplying equation (160) by 1+tl, we get(161)ddtℰNVt1+tl+λDNVt1+tl≤0.
Integrating over0,1, we find(162)∫0tddtℰNVs1+slds+λ∫0t1+slDNVsds≤0,which implies(163)1+tlℰNVt+λ∫0t1+slDNVsds≤ℰNV0+l∫0t1+sl−1ℰNVsds.
Recalling(164)ℰNVt≤CDNVt+B˜2+σ++σ−2,we obtain(165)1+tlℰNVt+λ∫0t1+slDNVsds≤ℰNV0+Cl∫0t1+sl−1DNVt+B˜2+σ++σ−2ds+Cl∫0t1+sl−1DN+1Vsds.
Similarly,(166)1+tl−1ℰN+1Vt+λ∫0t1+sl−1DN+1Vsds≤ℰN+1V0+Cl−1∫0t1+sl−2DNVt+B˜2+σ++σ−2ds+Cl+1∫0t1+sl−2DN+2Vsds,λ∫0tDN+2Vsds≤ℰN+2Vt+λ∫0tDN+2Vsds≤ℰN+2V0,for 1≤l≤2. Thus,(167)1+tlℰNVt+λ∫0t1+slDNVsds≤ℰNV0+C∫0t1+sl−1B˜2+σ++σ−2.
To estimate the term on the right-hand side of (167), define(168)ℰN,∞≤1+s3/2ℰNVs.
Applying Duhamel’s principle fort≥0 along with the linear estimate on B from (149) to (156), we get(169)Bt≤1+t−3/4‖u±0,E0,B0‖L1∩H˙2+C∫0t1+s−3/4g2s,g3sL1∩H˙2,for all 0≤s≤t.
Since(170)g2sL1=∫−u±.∇u±−A±γρ±+1γ−2−1∇ρ±±u±×B+α±u±−u∓ds≤∫−u±.∇u±ds+∫−u±×Bds+∫γ1+ρ±γ−2−1∇ρ±ds,we can apply the Sobolev imbedding theorem along with the Hölder and Cauchy inequalities to obtain(171)g2sL1≤C∇u±+u±+B+∇ρ±,(172)g3sL1=∫ρ±u±ds.
Thus,(173)g3sL1≤Cρ±+u±.
Therefore, we get(174)g2tH˙2=∑α=12Dα−u±.∇u±−A±γρ±+1γ−2−1∇ρ±±u±×B+α±u±−u∓.
Thus,(175)g2tH˙2≤∇Ut2u±,BtL∞,g3sH˙2=∑α=12∫Dαρ±u±2ds1/2.
Thus,(176)g3tH˙2≤∇Ut2ρ±,u±L∞.
Therefore,(177)g2s,g3sL1∩H˙2≤CℰNUs.
Notice that(178)ℰNUs≤CℰNVγs.
From (168),(179)ℰNVγs≤1+γs−3/2ℰN,∞Vγt,for any 0≤s≤t.
Then,(180)g2s,g3sL1∩H˙2≤C1+γs−3/2ℰN,∞Vγt.
Substituting (180) in (169), we get(181)Bt≤1+t−3/4u±0,E0,B0L1∩H˙2+C1+s−3/2ℰN,∞Vγt≤1+t−3/4u±0,E0,B0L1∩H˙2+ℰN,∞Vγt,which implies(182)B˜t2≤1+t−3/2υ±0,E˜0,B˜0L1∩H˙22+ℰN,∞Vγt2.
Finally, we need to show the uniform-in-time bound ofℰN,∞Vt which implies the decay rates of the energy functional ℰNVt. Using l=3/2+ϵ in (172) with ϵ>0 sufficiently small and using (181), we get(183)1+t3/2+ϵℰNVt+λ∫0t1+s3/2+ϵDNVsds≤CℰN+2V0+C∫0t1+s1/2+ϵB˜2+σ++σ−2≤CℰN+1V0+1+t3/2+ϵC1+t−3/2υ±0,E˜0,B˜0L1∩H˙22+C1+s−3/2ℰN,∞Vt2≤CℰN+2V0+C1+tϵυ±0,E˜0,B˜0L1∩H˙22+ℰN,∞Vt2,which implies(184)1+t3/2ℰNVt≤CℰN+2V0+υ±0,E˜0,B˜0L12+ℰN,∞Vt2,1+t3/2ℰN,∞Vt≤CϵN+2V02+ℰN,∞Vt2.
Now, from the definition (28) and since ϵN+2V0>0 is sufficiently small, ℰN,∞Vt≤ϵN+2V02 holds for any t≥0, and we get(185)VtN≤CℰNVt1/2≤CϵN+2V01+t−3/4,which is the proof of (158) in Proposition 2.


### 4.2. Time Rate for the High-Order Energy Functional

In this section, we determine the time decay estimates of the high-order energy functional∇VN2, that is, the proof (159) of Proposition 2. To do so, we investigate the time decay estimates on ∇B˜ and ∇NE˜,B˜ using the following lemma.Lemma 3.
SupposeV=σ±,υ±,E˜,B˜ is the solution to the Cauchy problems (15) and (16) with initial condition V0=σ0,υ0,E0˜,B0˜ satisfying (17) as obtained in Proposition 1. If ℰNV0 is sufficiently small, then there exists the high-order energy functional ℰNh⋅ and the high-order dissipation rate DNh⋅ such that for all ≥0,(186)ddtℰNhVt+λDNhVt≤C∇B˜2.By comparing the definitions ofℰNhVt (22) and DNhVt (25) and based on Lemma 3 equation (186), we have(187)ddtℰNhVt+λℰNhVt≤C∇B˜2+∇NE˜,B˜+∇σ++σ−2,which implies(188)ℰNhVt≤e−λtℰNhV0+C∫0te−λt−s∇B˜s2+∇NE˜,B˜s+∇σ++σ−s2.Now, to estimate the time integral term on the right-hand side of inequality (189), we have the next lemma.Lemma 4.
SupposeV=σ±,υ±,E˜,B˜ is the solution of the initial value problems (15) and (16) with initial condition V0=σ0,υ0,E0˜,B0˜ satisfying (17) as obtained in Proposition 1. If ϵN+6V0 is sufficiently small, where defined in (28), then for all t≥0,(189)∇B˜t2+∇NE˜t,B˜t2+∇Nσ+t+σ−t2≤εN+6V021+t5/2.The proofs of Lemmas3 and 4 are analogous to those of Lemmas 5.1 and 5.2 in [10].Now, we suppose that the above lemma is true. Then, by putting (188) into (186), we get(190)ℰNhVt≤e−λtℰNhV0+CϵN+6V021+t5/2,in which (159) in Proposition 2 holds.


### 4.3. Decay Rate inLq

In this section, we determine the decay rates inLq, 2≤q≤+∞, of (7)–(9) for solutions U=ρ±,u±,E,B to the Cauchy problems (51) and (52. Suppose that ϵ13V0 is sufficiently small. In addition, for N≥4, Proposition 2 shows that if ϵN+2V0 is sufficiently small, then(191)UtN≤CϵN+2V01+t−3/4,and if ϵN+6V0 is sufficiently small, then(192)∇UtN−1≤CϵN+6V01+t−5/4.We establish the estimates onB,u±,E and ρ± as follows.To estimate theL2 decay rate on BLq, note that from (191), we obtain(193)Bt≤Cϵ6V01+t−3/4.To estimate theL∞ decay rate, note that the L∞ estimates on B in (150) to (156) give(194)BtL∞≤C1+t−3/2u±0,E0,B0L1∩H˙5+C∫0t1+t−s−3/2g2s,g3sL1∩H˙5ds.From (191), we obtain(195)g2s,g3sL1∩H˙5≤CUt62≤Cϵ82V01+t−3/2,and thus(196)BtL∞≤Cϵ8V01+t−3/2.Moreover, byL2−L∞ interpolation,(197)BtLq≤Cϵ8V01+t−3/2+3/2q,for 2≤q≤∞.To estimateu±,ELq, note that for the L2 decay rate, we can utilize the L2 estimate on u± and E in (149) to (156). That is,(198)u±t≤C1+t−54ρ±0+u±0,E0,B0L1∩H˙2+C∫0t1+t−s−5/4g1s+g2s,g3sL1∩H˙2ds,Et≤C1+t−5/4u±0,E0,B0L1∩H˙3+C∫0t1+t−s−5/4g2s,g3sL1∩H˙3ds.From (191),(199)g1t+g2t,g3tL1∩H˙3≤Ut42≤Cϵ62V01+t−3/2,and it therefore holds that(200)u±t,Et≤Cϵ6V01+t−5/4.For theL∞ decay rate, we can utilize the L∞ estimates on u± and E in (150) to (156). That is,(201)u±tL∞≤C1+t−2ρ±0L2∩H˙2+u±0,E0,B0L1∩H˙5+C∫0t1+t−s−2g1sL2∩H˙2+g2s,g3sL1∩H˙5ds,EtL∞≤C1+t−2u±0,E0,B0L1∩H˙6+C∫0t1+t−s−2g2s,g3sL1∩H˙6ds.Since(202)g1tL2∩H˙2+g2t,g3tH˙5∩H˙6≤∇Ut62≤Cϵ132V01+t−5/2,g2t,g3tL1≤CUtu±t+∇Ut≤Cϵ6V01+t−3/4⋅ϵ10V01+t−5/4≤Cϵ102V01+t−2,it holds that(203)u±t,EtL∞≤Cε13V01+t−2.Moreover, byL2−L∞ interpolation,(204)u±t,EtLq≤Cε13V01+t−2+3/2q,for 2≤q≤∞.To estimate theL2 decay rate on ρ±Lq, we utilize the L2 estimate on ρ± in (149) to (156). We get(205)ρ±t≤Ce−t/2ρ±0,u±0+C∫0te−t−s/2g1s,g2sds.Since(206)g1t,g2t≤C∇Ut12+u±t⋅BtL∞≤Cϵ102V01+t−5/2,(207)ρ±t≤Cϵ10V01+t−5/2.(205) implies the slower decay estimate:Moreover, after estimatingg1t,g2t and utilizing the previous slower decay estimate, we obtain(208)g1t,g2t≤Cu±tL∞∇ρ±t+∇u±t+Bt≤Cρ±t∇ρ±t2+∇u±t2≤Cϵ13V021+t−11/4,and it follows from (205) that(209)ρ±t≤Cϵ13V01+t−11/4.For theL∞ decay rate, we utilize the L∞ estimates on ρ± in (150) to (156). We have(210)ρ±tL∞≤Cet/2ρ±0,u±L2∩H˙2+C∫0te−t−s/2g1s,g2sL2∩H˙2ds,and it is simple to check that(211)g1t,g2tL2∩H˙2≤C∇Ut4ρ±t+u±t,BtL∞+∇ρ±t,u±tL∞≤Cϵ13V01+t−11/4.Applying (210), we obtain(212)ρ±tL∞≤Cϵ13V01+t−11/4.Thus, byL2−L∞ interpolation,(213)ρ±tLq≤Cϵ13V01+t−11/4,for 2≤q≤∞. Note that (197), (204), and (213) correspond to (9), (8), and (7), respectively. This completes the proof of Theorem 1.Now, to obtain the decay rates of smooth solutions from (156) let Gt∗u±0=etLU0. For the L2 decay rate, by applying (156) and (157), we have(214)U−etLU02=∫0tet−sLg1s,g2s,g3s,0ds2≤∫0tet−sLg1s,g2s,g3s,02ds,since(215)g1tL22=∫0tet−sL−∇.ρ±u±ds2≤∫0tet−sL∇.ρ±u±2ds≤Cϵ102V01+t−5/2,g2tL22=∫0tet−sL−u±.∇u±−A±γρ±+1γ−2−1∇ρ±±u±×B+α±u±−u∓ds2≤∫0tet−sL−u±.∇u±−A±γρ±+1γ−2−1∇ρ±±u±×B+α±u±−u∓2ds≤Cϵ102V01+t−5/2,g3tL22=∫0tet−sLρ−u−−ρ+u+ds2≤∫0tet−sLρ−u−−ρ+u+2ds≤Cϵ102V01+t−5/2.Therefore,(216)U−etLU0L2≤Cϵ10V01+t−5/2.For theL∞ decay rate, by applying (156) and (157), we get(217)g1tL∞=sup0≤s≤t∫0tet−sL−∇.ρ±u±ds≤sup0≤s≤t∫0tet−sL∇.ρ±u±ds≤Cϵ13V01+t−2,g2tL∞=sup0≤s≤t∫0tet−sL−u±.∇u±−A±γρ±+1γ−2−1∇ρ±±u±×B+α±u±−u∓ds≤sup0≤s≤t∫0tet−sL−u±.∇u±−A±γρ±+1γ−2−1∇ρ±±u±×B+α±u±−u∓ds≤Cϵ13V01+t−2,‖g3t‖L∞=sup0≤s≤t|∫0tet−sLρ−u−−ρ+u+ds|≤sup0≤s≤t∫0tet−sL|ρ−u−−ρ+u+|ds≤Cϵ13V01+t−2.Thus,(218)U−etLU0L∞≤Cϵ13V01+t−2.So, byL2−L∞ interpolation,(219)U−etLU0Lq≤Cϵ13V01+t−2+3/2q,for 2≤q≤∞. This completes the proof of Theorem 2.


## 4.1. Decay Rates for System (15)

In this section, we apply the linearLp−Lq time decay property of the homogeneous system (51) to the nonlinear case. Throughout this section, we suppose that U=ρ±,u±,E,B is the solution to the Cauchy problem (15) satisfying (16).By Duhamel’s principle, the solutionU can be formally written as(156)U=etLU0+∫0tet−sLg1s,g2s,g3s,0ds,where etLU0 is as defined in Section 3.2 and(157)g1=−∇.ρ±u±,g2=−u±.∇u±−A±γρ±+1γ−2−1∇ρ±±u±×B+α±u±−u∓,g3=ρ−u−−ρ+u+.Remark 3.
Note that in (156), g1s,g2s,g3s,0 satisfies the compatibility condition (53). Thus, et−sL acts on g1s,g2s,g3s,0 for all 0≤s≤t.Proposition 2.
Suppose initial dataV0=σ±0,υ±0,E0˜,B0˜ satisfies (17). If εN+2V0>0 is small enough, then the solution V=σ±,υ±,E˜,B˜ satisfies(158)VtN≤CεN+2V01+t−3/4,for any t>0. Moreover, εN+6V0>0 being sufficiently small implies that the solution V=σ±,υ±,E˜,B˜ satisfies(159)∇VtN−1≤CεN+6V01+t−5/4,for any t≥0.Lemma 2.
SupposeV=σ±,υ±,E,B is the solution of the initial value problems (15) and (16) with initial data V0=σ±,υ±,E0˜,B0˜ satisfying (17) as described in Proposition 1. If ℰNV0 is sufficiently small, then(160)ddtℰNVt+λDNVt≤0.Proof.
Letl⩾0. Multiplying equation (160) by 1+tl, we get(161)ddtℰNVt1+tl+λDNVt1+tl≤0.
Integrating over0,1, we find(162)∫0tddtℰNVs1+slds+λ∫0t1+slDNVsds≤0,which implies(163)1+tlℰNVt+λ∫0t1+slDNVsds≤ℰNV0+l∫0t1+sl−1ℰNVsds.
Recalling(164)ℰNVt≤CDNVt+B˜2+σ++σ−2,we obtain(165)1+tlℰNVt+λ∫0t1+slDNVsds≤ℰNV0+Cl∫0t1+sl−1DNVt+B˜2+σ++σ−2ds+Cl∫0t1+sl−1DN+1Vsds.
Similarly,(166)1+tl−1ℰN+1Vt+λ∫0t1+sl−1DN+1Vsds≤ℰN+1V0+Cl−1∫0t1+sl−2DNVt+B˜2+σ++σ−2ds+Cl+1∫0t1+sl−2DN+2Vsds,λ∫0tDN+2Vsds≤ℰN+2Vt+λ∫0tDN+2Vsds≤ℰN+2V0,for 1≤l≤2. Thus,(167)1+tlℰNVt+λ∫0t1+slDNVsds≤ℰNV0+C∫0t1+sl−1B˜2+σ++σ−2.
To estimate the term on the right-hand side of (167), define(168)ℰN,∞≤1+s3/2ℰNVs.
Applying Duhamel’s principle fort≥0 along with the linear estimate on B from (149) to (156), we get(169)Bt≤1+t−3/4‖u±0,E0,B0‖L1∩H˙2+C∫0t1+s−3/4g2s,g3sL1∩H˙2,for all 0≤s≤t.
Since(170)g2sL1=∫−u±.∇u±−A±γρ±+1γ−2−1∇ρ±±u±×B+α±u±−u∓ds≤∫−u±.∇u±ds+∫−u±×Bds+∫γ1+ρ±γ−2−1∇ρ±ds,we can apply the Sobolev imbedding theorem along with the Hölder and Cauchy inequalities to obtain(171)g2sL1≤C∇u±+u±+B+∇ρ±,(172)g3sL1=∫ρ±u±ds.
Thus,(173)g3sL1≤Cρ±+u±.
Therefore, we get(174)g2tH˙2=∑α=12Dα−u±.∇u±−A±γρ±+1γ−2−1∇ρ±±u±×B+α±u±−u∓.
Thus,(175)g2tH˙2≤∇Ut2u±,BtL∞,g3sH˙2=∑α=12∫Dαρ±u±2ds1/2.
Thus,(176)g3tH˙2≤∇Ut2ρ±,u±L∞.
Therefore,(177)g2s,g3sL1∩H˙2≤CℰNUs.
Notice that(178)ℰNUs≤CℰNVγs.
From (168),(179)ℰNVγs≤1+γs−3/2ℰN,∞Vγt,for any 0≤s≤t.
Then,(180)g2s,g3sL1∩H˙2≤C1+γs−3/2ℰN,∞Vγt.
Substituting (180) in (169), we get(181)Bt≤1+t−3/4u±0,E0,B0L1∩H˙2+C1+s−3/2ℰN,∞Vγt≤1+t−3/4u±0,E0,B0L1∩H˙2+ℰN,∞Vγt,which implies(182)B˜t2≤1+t−3/2υ±0,E˜0,B˜0L1∩H˙22+ℰN,∞Vγt2.
Finally, we need to show the uniform-in-time bound ofℰN,∞Vt which implies the decay rates of the energy functional ℰNVt. Using l=3/2+ϵ in (172) with ϵ>0 sufficiently small and using (181), we get(183)1+t3/2+ϵℰNVt+λ∫0t1+s3/2+ϵDNVsds≤CℰN+2V0+C∫0t1+s1/2+ϵB˜2+σ++σ−2≤CℰN+1V0+1+t3/2+ϵC1+t−3/2υ±0,E˜0,B˜0L1∩H˙22+C1+s−3/2ℰN,∞Vt2≤CℰN+2V0+C1+tϵυ±0,E˜0,B˜0L1∩H˙22+ℰN,∞Vt2,which implies(184)1+t3/2ℰNVt≤CℰN+2V0+υ±0,E˜0,B˜0L12+ℰN,∞Vt2,1+t3/2ℰN,∞Vt≤CϵN+2V02+ℰN,∞Vt2.
Now, from the definition (28) and since ϵN+2V0>0 is sufficiently small, ℰN,∞Vt≤ϵN+2V02 holds for any t≥0, and we get(185)VtN≤CℰNVt1/2≤CϵN+2V01+t−3/4,which is the proof of (158) in Proposition 2.


## 4.2. Time Rate for the High-Order Energy Functional

In this section, we determine the time decay estimates of the high-order energy functional∇VN2, that is, the proof (159) of Proposition 2. To do so, we investigate the time decay estimates on ∇B˜ and ∇NE˜,B˜ using the following lemma.Lemma 3.
SupposeV=σ±,υ±,E˜,B˜ is the solution to the Cauchy problems (15) and (16) with initial condition V0=σ0,υ0,E0˜,B0˜ satisfying (17) as obtained in Proposition 1. If ℰNV0 is sufficiently small, then there exists the high-order energy functional ℰNh⋅ and the high-order dissipation rate DNh⋅ such that for all ≥0,(186)ddtℰNhVt+λDNhVt≤C∇B˜2.By comparing the definitions ofℰNhVt (22) and DNhVt (25) and based on Lemma 3 equation (186), we have(187)ddtℰNhVt+λℰNhVt≤C∇B˜2+∇NE˜,B˜+∇σ++σ−2,which implies(188)ℰNhVt≤e−λtℰNhV0+C∫0te−λt−s∇B˜s2+∇NE˜,B˜s+∇σ++σ−s2.Now, to estimate the time integral term on the right-hand side of inequality (189), we have the next lemma.Lemma 4.
SupposeV=σ±,υ±,E˜,B˜ is the solution of the initial value problems (15) and (16) with initial condition V0=σ0,υ0,E0˜,B0˜ satisfying (17) as obtained in Proposition 1. If ϵN+6V0 is sufficiently small, where defined in (28), then for all t≥0,(189)∇B˜t2+∇NE˜t,B˜t2+∇Nσ+t+σ−t2≤εN+6V021+t5/2.The proofs of Lemmas3 and 4 are analogous to those of Lemmas 5.1 and 5.2 in [10].Now, we suppose that the above lemma is true. Then, by putting (188) into (186), we get(190)ℰNhVt≤e−λtℰNhV0+CϵN+6V021+t5/2,in which (159) in Proposition 2 holds.


## 4.3. Decay Rate inLq

In this section, we determine the decay rates inLq, 2≤q≤+∞, of (7)–(9) for solutions U=ρ±,u±,E,B to the Cauchy problems (51) and (52. Suppose that ϵ13V0 is sufficiently small. In addition, for N≥4, Proposition 2 shows that if ϵN+2V0 is sufficiently small, then(191)UtN≤CϵN+2V01+t−3/4,and if ϵN+6V0 is sufficiently small, then(192)∇UtN−1≤CϵN+6V01+t−5/4.We establish the estimates onB,u±,E and ρ± as follows.To estimate theL2 decay rate on BLq, note that from (191), we obtain(193)Bt≤Cϵ6V01+t−3/4.To estimate theL∞ decay rate, note that the L∞ estimates on B in (150) to (156) give(194)BtL∞≤C1+t−3/2u±0,E0,B0L1∩H˙5+C∫0t1+t−s−3/2g2s,g3sL1∩H˙5ds.From (191), we obtain(195)g2s,g3sL1∩H˙5≤CUt62≤Cϵ82V01+t−3/2,and thus(196)BtL∞≤Cϵ8V01+t−3/2.Moreover, byL2−L∞ interpolation,(197)BtLq≤Cϵ8V01+t−3/2+3/2q,for 2≤q≤∞.To estimateu±,ELq, note that for the L2 decay rate, we can utilize the L2 estimate on u± and E in (149) to (156). That is,(198)u±t≤C1+t−54ρ±0+u±0,E0,B0L1∩H˙2+C∫0t1+t−s−5/4g1s+g2s,g3sL1∩H˙2ds,Et≤C1+t−5/4u±0,E0,B0L1∩H˙3+C∫0t1+t−s−5/4g2s,g3sL1∩H˙3ds.From (191),(199)g1t+g2t,g3tL1∩H˙3≤Ut42≤Cϵ62V01+t−3/2,and it therefore holds that(200)u±t,Et≤Cϵ6V01+t−5/4.For theL∞ decay rate, we can utilize the L∞ estimates on u± and E in (150) to (156). That is,(201)u±tL∞≤C1+t−2ρ±0L2∩H˙2+u±0,E0,B0L1∩H˙5+C∫0t1+t−s−2g1sL2∩H˙2+g2s,g3sL1∩H˙5ds,EtL∞≤C1+t−2u±0,E0,B0L1∩H˙6+C∫0t1+t−s−2g2s,g3sL1∩H˙6ds.Since(202)g1tL2∩H˙2+g2t,g3tH˙5∩H˙6≤∇Ut62≤Cϵ132V01+t−5/2,g2t,g3tL1≤CUtu±t+∇Ut≤Cϵ6V01+t−3/4⋅ϵ10V01+t−5/4≤Cϵ102V01+t−2,it holds that(203)u±t,EtL∞≤Cε13V01+t−2.Moreover, byL2−L∞ interpolation,(204)u±t,EtLq≤Cε13V01+t−2+3/2q,for 2≤q≤∞.To estimate theL2 decay rate on ρ±Lq, we utilize the L2 estimate on ρ± in (149) to (156). We get(205)ρ±t≤Ce−t/2ρ±0,u±0+C∫0te−t−s/2g1s,g2sds.Since(206)g1t,g2t≤C∇Ut12+u±t⋅BtL∞≤Cϵ102V01+t−5/2,(207)ρ±t≤Cϵ10V01+t−5/2.(205) implies the slower decay estimate:Moreover, after estimatingg1t,g2t and utilizing the previous slower decay estimate, we obtain(208)g1t,g2t≤Cu±tL∞∇ρ±t+∇u±t+Bt≤Cρ±t∇ρ±t2+∇u±t2≤Cϵ13V021+t−11/4,and it follows from (205) that(209)ρ±t≤Cϵ13V01+t−11/4.For theL∞ decay rate, we utilize the L∞ estimates on ρ± in (150) to (156). We have(210)ρ±tL∞≤Cet/2ρ±0,u±L2∩H˙2+C∫0te−t−s/2g1s,g2sL2∩H˙2ds,and it is simple to check that(211)g1t,g2tL2∩H˙2≤C∇Ut4ρ±t+u±t,BtL∞+∇ρ±t,u±tL∞≤Cϵ13V01+t−11/4.Applying (210), we obtain(212)ρ±tL∞≤Cϵ13V01+t−11/4.Thus, byL2−L∞ interpolation,(213)ρ±tLq≤Cϵ13V01+t−11/4,for 2≤q≤∞. Note that (197), (204), and (213) correspond to (9), (8), and (7), respectively. This completes the proof of Theorem 1.Now, to obtain the decay rates of smooth solutions from (156) let Gt∗u±0=etLU0. For the L2 decay rate, by applying (156) and (157), we have(214)U−etLU02=∫0tet−sLg1s,g2s,g3s,0ds2≤∫0tet−sLg1s,g2s,g3s,02ds,since(215)g1tL22=∫0tet−sL−∇.ρ±u±ds2≤∫0tet−sL∇.ρ±u±2ds≤Cϵ102V01+t−5/2,g2tL22=∫0tet−sL−u±.∇u±−A±γρ±+1γ−2−1∇ρ±±u±×B+α±u±−u∓ds2≤∫0tet−sL−u±.∇u±−A±γρ±+1γ−2−1∇ρ±±u±×B+α±u±−u∓2ds≤Cϵ102V01+t−5/2,g3tL22=∫0tet−sLρ−u−−ρ+u+ds2≤∫0tet−sLρ−u−−ρ+u+2ds≤Cϵ102V01+t−5/2.Therefore,(216)U−etLU0L2≤Cϵ10V01+t−5/2.For theL∞ decay rate, by applying (156) and (157), we get(217)g1tL∞=sup0≤s≤t∫0tet−sL−∇.ρ±u±ds≤sup0≤s≤t∫0tet−sL∇.ρ±u±ds≤Cϵ13V01+t−2,g2tL∞=sup0≤s≤t∫0tet−sL−u±.∇u±−A±γρ±+1γ−2−1∇ρ±±u±×B+α±u±−u∓ds≤sup0≤s≤t∫0tet−sL−u±.∇u±−A±γρ±+1γ−2−1∇ρ±±u±×B+α±u±−u∓ds≤Cϵ13V01+t−2,‖g3t‖L∞=sup0≤s≤t|∫0tet−sLρ−u−−ρ+u+ds|≤sup0≤s≤t∫0tet−sL|ρ−u−−ρ+u+|ds≤Cϵ13V01+t−2.Thus,(218)U−etLU0L∞≤Cϵ13V01+t−2.So, byL2−L∞ interpolation,(219)U−etLU0Lq≤Cϵ13V01+t−2+3/2q,for 2≤q≤∞. This completes the proof of Theorem 2.



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