Global classical solution for one-dimensional nonlinear thermoelastiticity with second sound on the semi-axis

Universität Konstanz

Global classical solution for one-dimensional nonlinear thermoelastiticity with second sound on the semi-axis

Yuxi Hu

Konstanzer Schriften in Mathematik Nr. 280, Juli 2011

ISSN 1430-3558

© Fachbereich Mathematik und Statistik Universität Konstanz

Konstanzer Online-Publikations-System (KOPS) URL: http://nbn-resolving.de/urn:nbn:de:bsz:352-140667

Global classical solution for one-dimensional

nonlinear thermoelasticity with second sound on the semi-axis

Yuxi HU

Department of Mathematics Shanghai Jiao Tong University

200240 Shanghai, P.R.China

Abstract

In this paper, we will give a global existence theorem in one-dimensional thermoelasticity with second sound inR⁺. For this purpose, we first give the decay rates of the linearized equations with the help of Fourier sine and cosine transformation and the local existence theorem using theorems for quasi-linear symmetric hyperbolic systems. Then we establish some estimates inL², L¹ andL^∞ norms to get a uniform a priori estimate. Finally, we use the usual continuation argument to get global solution.

Keywords: second sound, linear decay rates, semi-axis, global solution

1 Introduction

The equations of thermoelasticity describe the elastic and the thermal behavior of elastic, heat con- ductive media, in particular the reciprocal actions between elastic stresses and temperature differences.

The common modeling of heat conduction using Fourier law, essentially saying that the heat flux is a certain function of the gradient of the temperature, leads to the well known paradox of infinite propaga- tion of signals, in particular of heat signals. Many efforts are made to remove this paradox (see [3] [4]

[23]). The common feature of these efforts is that all lead to hyperbolic equation rather than parabolic equation and thus has limit speed of propagation. In our case, we consider Cattaneo’s law instead of Fourier’s law and thus get a hyperbolic system.

The general equations for nonlinear thermoelasticity described by the displacement vectoru=u(t, x), the temperatureT =T(t, x) and the heat fluxq=q(t, x), are as follows

ρutt− ∇^′S=ρb, (1.1)

ε_t−tr{SF_t}+∇^′q=r, (1.2)

where ρ is the material density in a domain Ω ofRⁿ, n = 1,2,3, S is the Piola-Kirchhoff stress tensor and b is the specific external body force, ε is the internal energy, q is the heat flux, r is the external heat supply, while∇^′ denotes the divergence operator, trB denotes the trace of a matrixBandF is the deformation gradient

F = 1 +∇u.

The two basic nonlinear differential equations arise from the balance of linear momentum and the balance of energy. For the derivation of the above equations, see Racke and Jiang [10] and Carlson [2].

∗Exchange student from Shanghai Jiaotong University to Konstanz University; Email:huyuxi@sjtu.edu.cn.

We denote byη the entropy and by

ψ:=ε−T η

the Helmholtz free energy. The constitutive assumptions in thermoelasticity with second sound are that S, ε, η, ψare functions of (∇u, θ,∇θ, q) whereθ=T−T0denotes the temperature difference,T0>0 the constant reference temperature. It is always assumed that these functions are smooth and that

detF ̸= 0.

With the help of the second law of thermodynamics, it turns out ψ=ψ(∇u, θ, q) is independent of∇θ and

η=η(∇u, θ, q) =−ψθ(∇u, θ, q), S=S(∇u, θ, q) =ψ_∇u(∇u, θ, q).

So, we can rewrite (1.2) as

(θ+T₀)(tr{−ψ_θ∇u∇u_t} −ψ_θθθ_t) +∇^′q= (θ+T₀)ψ_θqq_t−ψ_qq_t. (1.3) If the thermal behavior is described by Fourier’s law, that is,

q=−κθ_x, (1.4)

then (1.1) and (1.2) constitute the system of classical thermoelasticity. While, the heat flux replacing Fourier’s law by Cattaneo’s law, is given by the following equation

τ(∇u, θ)qt+q+κ(∇u, θ)∇θ= 0, (1.5) where τ is the relaxation time describing the time lag in the response of the heat flux to a gradient in the temperature andκdenotes the heat conductivity tensor. The equations (1.1), (1.3), (1.5) constitute the system of nonlinear thermoelasticity with second sound.

For classical thermoelasticity, many results have been obtained. See [19], [20] [5], [6], [21]. For thermoelasticity with second sound, we already have some results. In one dimension case, Tarabek [24]

had proved a global well posedness for cauchy problem and decay to an equilibrium. Wang and Racke [17]

gave the details for local existence and decay rates. The exponential stability was obtained by Racke [18]

for bounded domain. Extensions were given by Messaoudi and Said-Houari [16]. In three dimension case, the global well-posedness and the exponential stability were given by Irmscher [8] for radially symmetric domain. However, the fully nonlinear Cauchy problem and exterior domains have not yet been treated.

My work here is to be able to prove a global existence theorem for an initial boundary value problem in R⁺ for nonlinear thermoelasticity with second sound.

In one space dimension, (1.1), (1.3), (1.5) turn into







ρu_tt−au_xx+bθ_x=−S_qq_x+ρb, (θ+T0)(˜aθt+buxt) +qx=dqt+r, τ q_t+q+κθ_x= 0,

(1.6)

with

a(∇u, θ, q) =ψu_xu_x, b=−ψu_xθ,˜a=−ψθθ, d= (θ+T0)ψθq−ψq.

In our case, we assume there’s no external energy term, that is,b =r= 0. Without loss of generality, we assumeρ= 1. And for simplicity, we also assumeSq =d= 0 and that Cattaneo’s law is linear, that is,τ, κare positive constants. So the above equations become







utt−a(∇u, θ, q)uxx+b(∇u, θ, q)θx= 0,

(θ+T0)(˜a(∇u, θ, q)θt+b(∇u, θ, q)uxt) +qx= 0, τ qt+q+κθx= 0,

(1.7)

with initial and boundary conditions {

u(0,·) =u0, ut(0,·) =u1, θ(0,·) =θ0, q(0,·) =q0, ∀x∈Ω = [0,¯ +∞) ux|∂Ω=θ|∂Ω= 0.

Letω=u_x, v=u_t, then system (1.7) becomes

ω_t−v_x= 0, (1.8)

vt−a(ω, θ, q)ωx+b(ω, θ, q)θx= 0, (1.9)

˜

a(ω, θ, q)θ_t+b(ω, θ, q)v_x+c(θ)q_x= 0, (1.10)

τ qt+q+κθx= 0, (1.11)

wherec(θ) = _θ+T¹

0(|θ| ≤K < T0will be a posterior estimate justified by the global small solution). With initial and boundary conditions

ω(0,·) = (u0)x=:ω0, v(0,·) =u1=:v0, θ(0,·) =θ0, q(0,·) =q0, (1.12)

ω|∂Ω=θ|∂Ω= 0. (1.13)

Some smooth assumptions on the coefficients of equations (1.8)-(1.11) are needed.

Assumption 1.1. a , b ,˜a , careC³−functions of their arguments and there exist positive constantsγ₀, γ₁ andK < T₀ such that if |ω|,|θ|,|q| ≤K, then

γ₀≤a(ω, θ, q),|b(ω, θ, q)|,˜a(ω, θ, q), c(θ)≤γ₁. (1.14) We introduce some notations which will be frequently used throughout the paper. For a non-negative integer N, let

D^Nu= ∑

l+m=N

∂_t^l∂^m_xu.

We denote by W^m,p(Ω),0 ≤ m ≤ ∞,1 ≤p ≤ ∞, the usual Sobolev space with the norm ∥ · ∥W^m,p. For convenience, H^m(Ω) and L^p(Ω) stand for W^m,2(Ω) and W^0,p(Ω) respectively. Let X be a Banach space. We denote byL^p([α, β], X) (1≤p≤ ∞) and∥ · ∥L^p([α,β,X])the space of all measurablep-th power functions from [α, β] to X and its norm respectively. ForT >0, we use the notation

∥u∥p,T ,k= sup

0≤t≤T

(1 +t)^k∥u(t)∥L^p,1≤p≤ ∞, k≥0.

Throughout this paperCorC^′ will denote a general positive constant which is not necessarily the same in any two places. ForU = (ω, v, θ, q) a function oft andx, we denote

K(U) =







0 −∂_x 0 0

−a∂x 0 b∂x 0 0 _˜a^b∂x 0 _a^c_˜∂x

0 0 ^κ_τ∂x 1 τ





U, E0=

( 1 0 0 0 0 0 1 0

) .

Define (∂_t^kU(0, x) fort≥0 recursively by

(∂_t^kU)(0, x) =∂_t^k−1(−K(U)U)(0, x), k≥1, U(0, x) =U⁰(x) = (ω⁰, v⁰, θ⁰, q⁰). (1.15) Furthermore, we need the smooth assumptions on initial data to ensure global solution.

Assumption 1.2. (1) U_k⁰∈H^3−k(Ω), (regularity on initial data) (2)E0·U_k⁰|∂Ω= 0, (compatibility condition)

whereU_k⁰=∂^k_tU(0, x)andk= 0,1,2,3.

This paper is mainly motivated by Jiang’s paper [9]. In that paper, he was able to prove a global solution for equations of classical one-dimensional thermoelasticity inR⁺for small smooth data. It seems that many results in classical thermoelasticity can be extended to thermoelasticity with second sound, see [17], [18] and [20]. However, it is not true, for example, for Timoshenko-type thermoelastic systems, where a system can be or remain exponentially stable under Fourier’s law, while it loses this property under Cattaneo’s law, see [7]. Our question is that whether a weak damping effect given by equation (1.5) is still predominating to ensure decay rates and global solution compared with a strong impact of dissipation induced by equation (1.4).

We arrange our paper as follows. In section 2, we give a brief introduction to our problem and then study the behavior of the linearized system. Some comparison with Jiang’s paper[9] will be given there.

In section 3, we study local existence theorem based on a local existence theorem for hyperbolic systems.

In section 4, we use energy methods to studyL² estimates for the local solution. In section 5, we obtain L¹ andL^∞estimates based on decay behavior of the linearized system and finally get global solution by usual continuation arguments.

2 Linear Decay Rates

The methods used for obtaining global existence theorems for small data consist of proving suitable a priori estimate, where one often exploits the decay of solutions to the linearized equations. This requires a precise analysis of the asymptotic behavior of such solutions as time tend to infinity, which will finally allow us to describe the asymptotic behavior of solutions to the nonlinear systems as well. In this section, we will give the decay rates of the corresponding linear equations to our system (1.8)-(1.11). Denote

a₀=a(0,0,0), b₀=b(0,0,0), c₀=c(0),˜a₀= ˜a(0,0,0),

where a, b, c,a˜ are defined in section 1. So, by Assumption 1.1, we knowa0 >0, b0 ̸= 0, c0 >0,a˜0 >0.

Then we rewrite equations (1.8)-(1.11) as follows

7.7











ω_t−v_x= 0,

vt−a0ωx+b0θx= (a−a0)ωx+ (b0−b)θx,

˜

a0θt+b0vx+c0qx= (˜a0−˜a)θt+ (b0−b)vx+ (c0−c)qx, τ qt+q+κθx= 0.

(2.16)

Take transformation byU^∗= (ω^∗, v^∗, θ^∗, q^∗) = (√a₀ω, v,√

˜

a₀θ,√_{τ c0}

κ q), system (??) becomes















ω^∗_t − √a0v^∗_x= 0, v^∗_t − √a0ω^∗_x+√^b0

˜

a0θ_x^∗=f(U^∗), θ^∗_t+√^b0

˜

a0v_x^∗+√

κc₀

τ0˜a0q^∗_x=g(U^∗), q_t^∗+¹_τq^∗+√

κc0

τ₀˜a₀θ^∗_x= 0,

where 





f(U^∗) =^(a√⁻a^a₀⁰⁾ω_x^∗+^(b√^0−b)

˜ a₀ θ_x^∗, g(U^∗) =

[(˜a₀−˜a)

√˜a0 θ_t^∗+ (b₀−b)v^∗_x+ (c₀−c)√

κ τ c0q_x^∗

]√

˜ a₀⁻¹. Let α = √

a0, β = √^b0

˜ a₀, γ =

√κc0

τ˜a0. We still denote (ω^∗, v^∗, θ^∗, q^∗) by (ω, v, θ, q). So, the linearized equations wheref(U^∗) =g(U^∗) = 0 are as follows:

wt−αvx= 0, (2.1)

vt−αωx+βθx= 0, (2.2)

θ_t+βv_x+γq_x= 0, (2.3)

qt+1

τq+γθx= 0, (2.4)

with boundary and initial conditions

ω|∂Ω=θ|∂Ω= 0, t≥0 (2.5)

(ω(0, x), v(0, x), θ(0, x), q(0, x))^′ = (ω⁰, v⁰, θ⁰, q⁰)^′≡U⁰, x∈Ω (2.6) where Ω = (0,∞) andα, γ, τ >0, β̸= 0 are constants.

We denote byFs(f(t, x)),Fc(f(t, x)) the Fourier sine and cosine transform of f(t, x) with respect to x, i.e.

Fs(f(t, x))(ξ) =

√2 π

∫ _∞

0

f(t, x) sinξxdx, Fc(f(t, x))(ξ) =

√2 π

∫ _∞

0

f(t, x) cosξxdx.

Taking Fourier sine transform on both sides of (2.1) and (2.3) and Fourier cosine transform on both sides of (2.2) and (2.4), then we get

Uˆ_t+A(ξ) ˆU = 0, t >0, (2.7) Uˆ(0) = (Fs(ω⁰),Fc(v⁰),Fs(θ⁰),Fc(q⁰))^′, (2.8) where ˆU = ˆU(t, ξ) = (Fs(ω(x, t))(ξ),Fc(v(x, t))(ξ),Fs(θ(x, t))(ξ),Fc(q(x, t))(ξ))^′ and

A(ξ) =







0 αξ 0 0

−αξ 0 βξ 0 0 −βξ 0 −γξ

0 0 γξ _τ¹





.

Let det(λI+A(ξ)) be zero, then we get the following characteristic equation λ⁴+1

τλ³+ (α²+β²+γ²)ξ²λ²+α²+β²

τ ξ²λ+α²γ²ξ⁴= 0. (2.9) Using formulae solving quartic equations, we have























 λ1=

√a₁+2y+

√

−3a₁−2y−√_a^2a2

1 +2y

2 −_4τ¹,

λ2=

√a1+2y−√

−3a1−2y−√^2a2

a1 +2y

2 −_4τ¹,

λ₃=

−√ a1+2y+

√−3a1−2y+√_a^2a2

1 +2y

2 −_4τ¹,

λ₄=

−√a₁+2y−√

−3a₁−2y+√_a^2a2

1 +2y

2 −_4τ¹,

(2.10)

where 





a1= (α²+β²+γ²)ξ²−_8τ³2, a₂= ^α2+β_2τ^2−γ2ξ²+_8τ¹₃,

a3=α²γ²ξ⁴+^γ2−3α_16τ²⁻2^3β2ξ²−_256τ³ 4, and













y=−⁵₆a1+ (−^Q₂ +√

∆)¹³ + (−^Q₂ −√

∆)¹³,

P = [−₁₂¹(α²+β²+γ²)²−α²γ²]ξ⁴+^α2_4τ^+β₂²ξ²=:P₁ξ⁴+P₂ξ²,

Q= [−₁₀₈¹ (α²+β²+γ²)³+¹₃(α²+β²+γ²)α²γ²]ξ⁶+^{−2α4−2β4−4α}_24τ^2β22^{+β2γ2−2α2γ2}ξ⁴=:Q1ξ⁶+Q2ξ²

∆ = ^Q₄² +^P₂₇³.

In order to getL¹ decay estimates, we first investigate the low and high frequency expansions of eigen- values in (2.10) in terms ofξ, which are expressed in the following two Lemmas.

Lemma 2.1. (i)λ1(0) =−¹_τ, λj(0) = 0 (j= 2,3,4), and asξ→0, we have













λ1(ξ) =−¹_τ+O(ξ), λ₂(ξ) =−_α^{τ α}2+β^2γ22ξ²+O(ξ³), λ3(ξ) =−_2(α^{τ β}2²+β^γ22)ξ²+i√

α²+β²ξ+O(ξ³), λ₄(ξ) =−_2(α^{τ β}2²+β^γ22)ξ²−i√

α²+β²ξ+O(ξ³),

(2.11)

and 





dλ2

dξ +^{2τ α}_α2+β^2γ22ξ=O(ξ²),

dλ₃ dξ −i√

α²+β²=O(ξ),

dλ₄ dξ +i√

α²+β²=O(ξ),

(2.12)

and asξ→ ∞, we have











λ₁=−_2τ¹ −a−bξ⁻²+O(ξ⁻⁴) +i(eξ+f ξ⁻¹+O(ξ⁻³)), λ2=−_2τ¹ −a−bξ⁻²+O(ξ⁻⁴)−i(eξ+f ξ⁻¹+O(ξ⁻³)), λ₃=a+bξ⁻²+O(ξ⁻⁴) +i(cξ+dξ⁻¹+O(ξ⁻³)), λ4=a+bξ⁻²+O(ξ⁻⁴)−i(cξ+dξ⁻¹+O(ξ⁻³)),

(2.13)

and 













dλ₁

dξ =ie+O(ξ⁻²),

dλ₂

dξ =−ie+O(ξ⁻²),

dλ3

dξ =ic+O(ξ⁻²),

dλ₄

dξ =−ic+O(ξ⁻²),

(2.14)

where

a= 1 4τ

[

2γ² α²+β²+γ²+√

(α²+β²+γ²)−4α²γ² −1 ]

,

c=

√

α²+β²+γ²−√

(α²+β²+γ²)−4α²γ²

2 ,

e=

√

α²+β²+γ²+√

(α²+β²+γ²)−4α²γ²

2 .

(ii)Except at most finite values ofξ >0, λj̸=λi(i̸=j, i, j= 1,2,3,4).

Proof. Ifξ→0,

∆ = Q² 4 +P³

27 =1

4(Q1ξ⁶+Q2ξ⁴)²+ 1

27(P1ξ⁴+P2ξ²)³

= 1

4(Q²₁ξ¹²+ 2Q₁Q₂ξ¹⁰+Q²₂ξ⁸) + 1

27(P₁³ξ¹²+ 3P₁²P₂ξ¹⁰+ 3P₁P₂²ξ⁸+P₂³ξ⁶)

= (Q²₁ 4 +P₁³

27)ξ¹²+ (Q1Q2

2 +P₁²P2

9 )ξ¹⁰+ (Q²₂

4 +P1P₂²

9 )ξ⁸+ 1 27P₂³ξ⁶. Using the formulae (1 +x)^α= 1 +αx+O(x²) asx→0, we get√

∆ =

√P₂³

27ξ³+O(ξ⁵). Then we have

−^Q₂ +√

∆ =

√P₂³

27ξ³−^Q₂²ξ⁴+O(ξ⁵) and (−^Q₂ +√

∆)¹³ =

√P₂

3ξ− _2P^Q2₂ξ²+O(ξ³). According to the definition ofy, we gety=−⁵₆a₁−^2Q_P2²ξ² and therefore get

−3a1−2y−2a2(a1+ 2y)⁻¹² =−4(α²+β²)ξ²+O(ξ³),

and

−3a1−2y+ 2a2(a1+ 2y)⁻¹² = 1

τ²+ ( 2β²γ²

α²+β² −4γ²)ξ²+O(ξ³).

So, we conclude that asξ→0,













λ₁(ξ) =−¹_τ +O(ξ), λ2(ξ) =−_α^{τ α}2+β^2γ22ξ²+O(ξ³), λ3(ξ) =−_2(α^{τ β}2²+β^γ22)ξ²+i√

α²+β²ξ+O(ξ³), λ4(ξ) =−_2(α^{τ β}2²+β^γ22)ξ²−i√

α²+β²ξ+O(ξ³).

Asξ→ ∞, we get as before

∆ = Q² 4 +P³

27 = (Q²₁ 4 +P₁³

27)ξ¹²+ (Q1Q2

2 +P₁²P2

9 )ξ¹⁰+ (Q²₂

4 +P1P₂²

9 )ξ⁸+ 1 27P₂³ξ⁶. Since ^Q₄²¹ +^P₂₇¹³ =−^α2γ2[^{(α2+β2+γ2)2}−4α²γ²]²

432 <0, we get asξ→ ∞ (−Q

2 +

√Q² 4 +P³

27)¹³ = (−Q 2 +i

√

−(Q² 4 +P³

27))¹³ = [Q²

4 −(Q² 4 +P³

27) ]¹₆

e^iθ31 =

√

−P 3e^iθ31, whereθ1 satisfies tanθ1=

√

−(^Q₄²+^P₂₇³)

−^Q₂ .Then after some calculations, we get cosθ1= 3√

3Q1

2P1

√−P1

(1 +1 2(2Q2

Q1 −3P2

P1

)ξ⁻²+O(ξ⁻⁴), sinθ1=

√

4P₁³+ 27Q²₁

4P₁³ +27Q²₁

8P₁³ (4P₁³+ 27Q²

4P₁³ )⁻¹²(2Q2

Q1 −3P2

P1

)ξ⁻²+O(ξ⁻⁴).

Let

θ₁₁= 3√ 3Q1

2P₁√

−P₁, θ₁₂== 3√ 3Q1

4P₁√

−P₁(2Q2

Q₁ −3P2

P₁ ), θ₁₃=

√

4P₁³+ 27Q²₁

4P₁³ , θ₁₄=27Q²₁

8P₁³ (4P₁³+ 27Q²

4P₁³ )⁻¹²(2Q₂ Q1 −3P₂

P1

).

We calculate cos^θ₃¹ in the following formula:

cosθ₁ 3 = 1

2[(cosθ₁+isinθ₁)¹³+ (cosθ₁−isinθ₁)¹³] =m+nξ⁻²+O(ξ⁻⁴),

where {

m=¹₂ [

(θ₁₁+iθ₁₃)¹³ + (θ₁₁−iθ₁₃)¹³ ]

,

n= ¹₆[(θ11+iθ13)⁻²³(θ12+iθ14) + (θ11−iθ13)⁻²³(θ12−θ14)].

Therefore, we geta1+ 2y=c1ξ²+c2+O(ξ⁻²) where





 c1=

[

−²₃(α²+β²+γ²) + 4

√−^P₃¹m ]

<0, c2=_4τ¹2 + 2

√−^P₃^1P_P²₁m+ 4n

√−^P₃¹.

So, we get√

a1+ 2y=i[g1ξ+g2ξ⁻¹+O(ξ⁻³)] and (a1+ 2y)⁻¹² =−i[f1ξ⁻¹+f2ξ⁻³+O(ξ⁻⁵)] where























 g1=√

−c1= [

2

3(α²+β²+γ²)−4

√−^P₃¹m ]¹₂

, g2=−₂^√c₋²_c

1 =−¹₂ [

2

3(α²+β²+γ²)−4

√−^P₃¹m ]₋¹₂

(_4τ¹2 + 2

√−^P₃^1P_P²₁m+ 4n

√−^P₃¹),

f1= ²₃(α²+β²+γ²)−4

√−^P₃¹m,

f2= ¹₂ [

2

3(α²+β²+γ²)−4

√−^P₃¹m ]₋³₂

(_4τ¹2 + 2

√−^P₃^1P_P²₁m+ 4n

√−^P₃¹).

Based on above calculations, we have

−3a1−2y− 2a₂

√a1+ 2y =−2 [

(α²+β²+γ²)ξ²− 3 8τ²

]

−(c1ξ²+c2+O(ξ⁻²)) + 2i( 1

8τ³+α²+β²−γ²

2τ ξ²)(f1ξ⁻¹+f2ξ⁻³+O(ξ⁻⁵)

=r₁ξ²+s₁+O(ξ⁻²) +i(t₁ξ+u₁ξ+O(ξ⁻³)), where

r1=−2(α²+β²+γ²)−c1, r2= 3

4τ²−c2, t1= α²+β²−γ²

τ f1, u1= α²+β²−γ²

τ + 1

8τ³f1. So, we have

(−3a1−2y− 2a2

√a₁+ 2y)¹² =[

(r1ξ²+s1+O(ξ⁻²))²+ (t1ξ+u1ξ⁻¹+O(ξ⁻³))²]¹₄ e^iθ2/2,

= (√

−r₁ξ+2r₁s₁+t₁ 2r₁³²

ξ⁻¹+O(ξ⁻³)) cosθ₂ 2 +i(√

−r₁ξ+2r₁s₁+t₁ 2r₁³²

ξ⁻¹+O(ξ⁻³)) sinθ₂ 2 , whereθ₂ satisfies tanθ₂=^t1_r^{ξ+u1ξ−1+O(ξ−3)}

1ξ²+s1+O(ξ⁻²) . Sincer₁<0 andt₁>0, we get ^π₂ < θ₂< π and cosθ2

2 = t1

2r1

ξ⁻¹−1 2(7t⁴₁

8r⁴₁ +u1r1−t1s1

r²₁ )ξ⁻³,sinθ2

2 = 1−1 8

t²₁

r₁²ξ⁻²+O(ξ⁻⁴).

As a result, we have

(−3a1−2y− 2a2

a₁+ 2y)¹² =v1+w1ξ⁻²+O(ξ⁻⁴) +i(v2ξ+w2ξ⁻¹+O(ξ⁻³)), where

v1= t1

2r

3 2

1

, w1= −7t⁴₁+ 4(2r1s1+t1)t1

16(−r1)⁷² −r1u1−t1s1

2(−r1)³² , v2=√

−r1, w2=− t²₁ 8r

3 2

1

+2r1s1+t1

2r

3 2

1

. Finally we derive the formulae forλ_i asξ→ ∞











λ1= (^v₂¹ −_4τ¹) +^w₂¹ξ⁻²+O(ξ⁻⁴) +i(^v2+g₂ ¹ξ+^w2+g₂ ²ξ⁻¹+O(ξ⁻³)), λ₂= ¯λ₁,

λ3= (−^v₂¹ −_4τ¹)−^w₂¹ξ⁻²+O(ξ⁻⁴) +i(^g1−₂^v2ξ+^g2−₂^w2ξ⁻¹+O(ξ⁻³)), λ₄= ¯λ₃.

For simplicity, we rewrite the above form as follows:











λ1=−_2τ¹ −a−bξ⁻²+O(ξ⁻⁴) +i(eξ+f ξ⁻¹+O(ξ⁻³)), λ₂=−_2τ¹ −a−bξ⁻²+O(ξ⁻⁴)−i(eξ+f ξ⁻¹+O(ξ⁻³)), λ3=a+bξ⁻²+O(ξ⁻⁴) +i(cξ+dξ⁻¹+O(ξ⁻³)), λ4=a+bξ⁻²+O(ξ⁻⁴)−i(cξ+dξ⁻¹+O(ξ⁻³)).

asξ→ ∞ (2.15)

The formulae for a, b, c, c, e, f are very complicated in our case, we use the formulae giving roots to calculate some of them, though these formulae are not crucial in the proof of Theorem 2.7 below. We know that the roots satisfy the following equation:











λ1+λ2+λ3+λ4=−_τ¹,

λ₁λ₂+λ₁λ₃+λ₁λ₄+λ₂λ₃+λ₂λ₄+λ₃λ₄= (α²+β²+γ²)ξ², λ1λ2λ3+λ1λ2λ4+λ1λ3λ4+λ2λ3λ4=−^α2+β_τ ²ξ²,

λ₁λ₂λ₃λ₄=α²γ²ξ⁴.

So, the coefficientsa, c, esatisfy the following equation e²c²=α²γ², e²(2a) +c²(−1

τ −2a) =−α²+β²

τ , e²+c²= (α²+β²+γ²).

Therefore, we have

a= 1 4τ

[

2γ² α²+β²+γ²+√

(α²+β²+γ²)−4α²γ² −1 ]

,

c=

√

α²+β²+γ²−√

(α²+β²+γ²)−4α²γ²

2 ,

e=

√

α²+β²+γ²+√

(α²+β²+γ²)−4α²γ²

2 .

Remark 2.1. We note that the characteristic matrixA(ξ)is a4×4order matrix, which causes technical complications in the discussion of the eigenvalues. However, it is important to note that when τ goes to zero, the three eigenvalues λj(j= 2,3,4)are exactly the same with these in Jiang’s paper [9]. This fact is reasonable since when the parameter τ goes to zero, our system are essentially reduced to the classical system of thermoelasticity.

Lemma 2.2. (i) For any values ofξ withξ >0,Reλ_j(ξ)<0. (j= 1,2,3,4)

(ii) There exists constantsr1, r2 andCn(n= 1,2,3,4) depending onr1, r2 such that for j= 2,3,4







−C2|ξ|²≤Reλj(ξ)≤ −C1|ξ|², 0≤ξ≤r1, Reλ_j(ξ)≤ −C₃, r₁≤ξ≤r₂, Reλj(ξ)≤ −C4, ξ≥r2.

(2.16)

Proof. (i)Since v^′(−A(ξ))v = −¹_τv₄² ≤ 0 for any v = (v1, v2, v3, v4) ∈ R⁴, we know that −A(ξ) is dissipative for anyξ >0. Therefore, we derive that

Reλj(ξ)≤0, ∀ξ >0, j= 1,2,3,4.

Second we claim that there is no purely imaginary root for the characteristic equation (2.9). Otherwise, there exist someξ₀>0 andj such thatReλ_j(ξ₀) = 0. Then take λ_j into (2.9), we get

a⁴−1

τa³i−(α²+β²+γ²)ξ₀²a²+α²+β²

τ ξ₀²ai+α²γ²ξ₀⁴= 0.