# On a Thermoelastic Laminated Timoshenko Beam: Well Posedness and Stability

Authors: Baowei Feng
Journal: Complexity
 (2020)

---

## Abstract

In this paper, we are concerned with a linear thermoelastic laminated Timoshenko beam, where the heat conduction is given by Cattaneo’s law. We firstly prove the global well posedness of the system. For stability results, we establish exponential and polynomial stabilities by introducing a stability numberχ.


---

## Body

## 1. Introduction

In this paper, we address the following thermoelastic laminated Timoshenko beam in0,1×0,∞:(1)ρωtt+Gψ−ωxx+δθx=0,Iρ3s−ψtt−D3s−ψxx−Gψ−ωx=0,Iρstt−Dsxx+Gψ−ωx+43γs+43βst=0,ρ3θt+qx+δωxt=0,τqt+αq+θx=0,which subject to the following boundary conditions:(2)ωx0,t=ψ0,t=s0,t=θ0,t=0,t∈0,∞,ωx1,t=ψ1,t=s1,t=θ1,t=0,t∈0,∞,and initial conditions(3)ωx,0=ω0x,ψx,0=ψ0x,sx,0=s0x,θx,0=θ0x,x∈0,1,qx,0=q0x,ωtx,0=ω1x,ψtx,0=ψ1x,stx,0=s1x,x∈0,1,where ρ,G,Iρ,D,γ,β,ρ3,δ,τ, and α are positive constants. θx,t represents the difference temperature and qx,t is the heat flux.Laminated beam, which is a relevant research subject due to the high applicability of such materials in the industry, was firstly introduced by Hansen and Spies, see, for instance [1, 2]. They introduced a mathematical model for two-layered beams with structural damping due to the interfacial slip which is given by(4)ρωtt+Gψ−ωxx=0,Iρ3stt−ψtt−Gψ−ωx−D3sxx−ψxx=0,Iρstt+Gψ−ωx+43γs+43βst−Dsxx=0,where the coefficients ρ,G,Iρ,D,γ,andβ are positive constants and represent density, shear stiffness, mass moment of inertia, flexural rigidity, adhesive stiffness, and adhesive damping parameter, respectively. The function ωx,t denotes the transversal displacement, ψx,t represents the rotational displacement, and sx,t is proportional to the amount of slip along the interface at time t and longitudinal spatial variable x. The third equation describes the dynamics of the slip.Up till now, there are some results concerning laminated beam equations, which are mainly concerned with global existence and stability of the related system. By adding suitable damping effects, such as internal damping, (boundary) frictional damping, and viscoelastic damping, it was shown that if the linear damping terms are added in two of the three equations, system (4) is exponentially stable under the “equal wave speeds” assumption ρ/Iρ=G/D. But if the damping terms are added in the three equations, then the system decays exponentially without the equal wave speeds assumption, see, for example, [3–17]. For thermoelastic laminated Timoshenko beam, there are few published works, we can mention the results due to Liu and Zhao [18] and Apalara [19]. In [18], the authors considered the following laminated beams with past history(5)ρφtt+Gψ−φxx+θx=0,Iρ3ω−ψtt−D3ω−ψxx+∫0∞gs3ω−ψxxt−sds−Gψ−φx−θ=0,Iρωtt−Dωxx+Gψ−φx+43γω+43βωt=0,kθt−τθxx+φxt+3ω−ψt=0,together with the following boundary conditions:(6)φx0,t=ψ0,t=ω0,t=θ0,t=0,t∈0,∞,φ1,t=ψx1,t=ωx1,t=θx1,t=0,t∈0,∞.They firstly proved the global well posedness of solutions to the system. The main results are the stability of the system. Ifβ≠0, they proved the exponential and polynomial stabilities depending on the behavior of the kernel function g only. If β=0, they established exponential stability in case of equal wave speeds assumption and lack of exponential stability in case of nonequal wave speeds assumption. Apalara [19] considered a laminated beam with second sound of the form(7)ρωtt+Gψ−ωxx=0,Iρ3s−ψtt−D3s−ψxx−Gψ−ωx+δθx=0,Iρstt−Dsxx+Gψ−ωx+43γs+43βst=0,ρ3θt+qx+δ3s−ψxt=0,τqt+αq+θx=0,together with the following boundary conditions:(8)ωx0,t=ψ0,t=s0,t=q0,t=0,t∈0,∞,ω1,t=ψx1,t=sx1,t=θ1,t=0,t∈0,∞,and proved the global well posedness and established exponential and polynomial stabilities depending on the parameter(9)χτ=1−τρ3GρDIρ−Gρ−τGδ2ρIρ.One can also refer to two recent results of laminated beams with thermal damping in [20, 21], and a result of a coupled hyperbolic equations with a heat equation of second sound in [22].Whens=0, system (4) reduces to the well-known Timoshenko system, which have been widely studied. There are so many papers on the Timoshenko system in the literature, most of those results recover the global well posedness, stability, and long-time dynamics by adding some kinds of damping. Here, we recall some works on the thermoelastic Timoshenko system. Muñoz Rivera and Racke [23] considered a Timoshenko system with thermoelastic dissipation and established exponential stability in case of equal wave speed assumption and polynomial stability if wave speeds are nonequal. Almeida Júnior et al. [24] studied a thermoelastic Timoshenko beam acting on shear force. They obtained the same stability results as in [23]. In addition, they proved that the polynomial decay is optimal. Fernández Sare and Racke [25] considered a Timoshenko system with second sound. They proved that the system is not exponentially stable even if the propagation speeds are equal. The results were generalized by Guesmia et al. [26]. Recently, Santos et al. [27] introduced a stability number χr for the system in [25] and established the exponential decay result for χr=0 and polynomial decay for χr≠0 by using the semigroup method. One can also find a stability result for the Timoshenko system with second sound in Apalara et al. [28]. Feng [29] considered a Timoshenko-Coleman-Gurtin system and studied the long-time dynamics of the system. We at last mention the contribution of Hamadouche and Messaoudi [30] and Aouadi and Boulehmi [31], where the authors considered two classes of nonuniform thermoelastic Timoshenko systems and proved global well posedness and established some stability results.Our goals in the present work are to study the global well posedness and stability of systems (1)–(3). The main points are summarized as follows:(i)
We prove the global well posedness of systems (1)–(3) by using Lumer–Philips theorem. The main result is presented in Theorem 1.(ii)
We introduce a new stability number denoted by(10)χ=τδ2D−Dρ−GIρτρ3DIρ−1,and we show that the system is exponential stable whenχ=0 and polynomial stable when χ≠0. The main results are presented in Theorems 1 and 2.(iii)
The proof of stability results is based on the multiplier method. Since the boundary conditions here we considered are different from those in Apalara [19], so the multipliers we will define are greatly different from the multipliers in Apalara [19].It follows, from (1), that(11)d2dt2∫01ωx,tdx=0,τddt∫01qx,tdx+α∫01qx,tdx=0.If we denote(12)ω¯x,t=ωx,t−∫01ω0x−t∫01ω1xdx,q¯x,t=qx,t−e−α/τt∫01q0xdx,we easily verify that ω¯,ψ,s,θ,q¯ satisfies (1) and in addition,(13)∫01ω¯x,tdx=0,∫01q¯x,tdx=0,∀t≥0.Hence, Poincaré’s inequality holds forω¯. In the following, we work with ω¯ and q¯ but write ω and q for convenience.The remaining paper is planned as follows. In Section2, we study the well posedness of the system. In Section 3, we establish the stability results. Throughout this paper, c>0 is a generic constant that changes from one inequality to another.


## 2. Well Posedness

We start by denoting the vector-valued function byU:(14)U=ω,Φ,3s−ψ,3Λ−Ψ,s,Λ,θ,qT,withΦ=ωt,Ψ=ψt,andΛ=st.Then, systems (1)–(3) can be written as(15)ddtUt=AU,t>0,U0=U0=ω0,ω1,3s0−ψ0,3s1−ψ1,s0,s1,θ0,q0T,where the operator A is defined by(16)AU=Φ−Gρψ−ωxx−δρθx3Λ−ΨDIρ3s−ψxx+GIρψ−ωxΛDIρsxx−GIρψ−ωx−4γ3Iρs−4β3IρΛ−1ρ3qx−δρ3Φx−ατq−1τθx.We consider the following spaces:(17)L∗20,1=v∈L20,1:∫01vxdx=0,H∗10,1=H10,1∩L∗20,1,H∗20,1=v∈H20,1:vx0=vx1=0.Let(18)ℋ=H∗10,1×L∗20,1×H010,1×L20,1×H010,1×L20,1×L20,1×L∗20,1be the Hilbert space equipped with the inner product(19)U,U˜ℋ=ρ∫01ΦΦ˜dx+Iρ∫013Λ−Ψ3Λ˜−Ψ˜dx+3Iρ∫01ΛΛ˜dx+ρ3∫01θθ˜dx+τ∫01qq˜dx+4γ∫01ss˜dx+D∫013s−ψx3s˜−ψ˜xdx+G∫01ψ−ωxψ˜−ω˜xdx+3D∫01sxs˜xdx.The domain ofA is given by(20)DA=U∈Hω∈H∗20,1∩H∗10,1,3s-ψ,s∈H20,1∩H010,1,Φ,q∈H∗10,1,3Λ-Ψ,Λ,θ∈H010,1.The well posedness result can be stated in the following theorem.Theorem 1.
LetU0∈ℋ, then problems (1)–(3) admit a unique weak solution U∈Cℝ+,ℋ. In addition, if U0∈DA, then U∈Cℝ+,DA∩C1ℝ+,ℋ.Proof.
It is easy to obtain that, for anyU=ω,Φ,3s−ψ,3Λ−Ψ,s,Λ,θ,qT∈DA,(21)AU,Uℋ=−4β∫01Λ2dx−α∫01q2dx≤0,which implies the operator A is a dissipative operator.
In what follows, we shall show the operatorId−A is surjective. In other words, given F=f1,f2,f3,f4,f5,f6,f7,f8∈ℋ, we will seek a solution V=v1,v2,v3,v4,v5,v6,v7,v8∈DA of(22)Id−AV=F.
We rewrite (21) as(23)v1−v2=f1,ρv2−Gv1xx−Gv3x+3Gv5x+δv7x=ρf2,v3−v4=f3,Iρv4−Dv3xx−3Gv5+Gv3+Gv1x=Iρf4,v5−v6=f5,Iρ+43βv6−Dv5xx−Gv3−Gv1x+3G+43γv5=Iρf6,ρ3v7+v8x+δv2x=ρ3f7,τ+αv8+v7x=τf8,which implies that(24)v2=v1−f1,(25)v4=v3−f3,(26)v6=v5−f5,(27)v7x=−τ+αv8+τf8.
We infer from (27) that(28)v7=−τ+α∫0xv8ydy+τ∫0xf8ydy.
Replacing (24)–(26) and (28) in (23), we see that(29)ρv1−Gv1xx−Gv3x+3Gv5x−δτ+αv8=ρf1+f2−τf8,Iρv3−Dv3xx−3Gv5+Gv3+Gv1x=Iρf3+f4,Iρ+3G+43β+43γv5−Dv5xx−Gv3−Gv1x=Iρf5+f6+43βf5,−ρ3τ+α∫0xv8ydy+v8x+δv1x=ρ3f7−ρ3τ∫0xf8ydy+δf1x.
We multiply (29) by v˜1, v˜3, v˜5, and τ+α∫0xv˜8ydy, respectively, and integrate their sum over 0,1 to get the following variational formulation:(30)ℬv1,v3,v5,v8,v˜1,v˜3,v˜5,v˜8=ℒv˜1,v˜3,v˜5,v˜8,where the bilinear form ℬ:H∗10,1×H010,1×H010,1×L∗20,12⟶ℝ is given by(31)ℬv1,v3,v5,v8,v˜1,v˜3,v˜5,v˜8=G∫01−v1x−v3+3v5−v˜1x−v˜3+3v˜5dx+ρ∫01v1v˜1dx+Iρ∫01v3v˜3dx+3Iρ+4β+4γ∫01v5v˜5dx+τ+α∫01v8v˜8dx+D∫01v3xv˜3xdx+3D∫01v5xv˜5xdx+ρ3τ+α2∫01∫0xv8ydy∫0xv˜8ydydx,and the linear form ℒ:H∗10,1×H010,1×H010,1×L∗20,1⟶ℝ is defined by(32)ℒv˜1,v˜3,v˜5,v˜8=∫01ρf1+ρf2−τδf8v˜1dx+Iρ∫01f3+f4v˜3dx+∫013Iρ+4βf5+3Iρf6v˜5dx+δτ+α∫01f1v˜8dx+∫01ρ3ττ+α∫0xf8ydy−ρ3τ+αf7∫0xv˜8ydydx.
We denote the Hilbert spaceV by(33)V=H∗10,1×H010,1×H010,1×L∗20,1,equipped with the norm(34)v1,v3,v5,v8V2=−v1x−v3+3v522+v122+v822+v3x22+v5x22.
It is easy to get thatℬ⋅,⋅ and ℒ⋅ are bounded. Moreover there exists a positive constant m such that(35)ℬv1,v3,v5,v8,v1,v3,v5,v8=G∫01−v1x−v3+3v52dx+ρ∫01v12dx+Iρ∫01v32dx+3Iρ+4β+4γ∫01v52dx+τ+α∫01v82dx+D∫01v3x2dx+3D∫01v5x2dx+ρ3τ+α2∫01∫0xv8ydy2dx≥G∫01−v1x−v3+3v52dx+ρ∫01v12dx+τ+α∫01v82dx+D∫01v3x2dx+3D∫01v5x2dx≥mv1,v3,v5,v8V2.
Thus,ℬ is coercive on V×V. Consequently, using Lax–Milgram theorem, we conclude that (30) has a unique solution:(36)v1∈H∗10,1,v3,v5∈H010,1,v8∈L∗20,1.
Substitutingv1,v3,v5, and v8 into (24)–(26) and (28), respectively, we have(37)v2∈H∗10,1,v4,v6∈H010,1,v7∈H010,1.
Letv˜1∈H010,1 and denote(38)v˜^1x=v˜1x−∫01v˜1sds,which gives us v˜^1∈H∗10,1. Now we replace v˜1,v˜3,v˜5,v˜8 by v˜^1,0,0,0 in (30) to obtain(39)G∫01−v1x−v3+3v5−v˜^1xdx+ρ∫01v1v˜^1dx=∫01ρf1+ρf2−τδf8v˜^1dx,i.e.,(40)G∫01v1xxv˜^1dx=ρ∫01v1v˜^1dx−G∫01v3xv˜^1dx+3G∫01v5xv˜^1dx−∫01ρf1+ρf2−τδf8v˜^1dx,∀v˜1∈H010,1,which yields(41)Gv1xx=ρv1−Gv3x+3Gv5x−ρf1+ρf2−τδf8∈L20,1.
Thus,(42)v1∈H20,1.
Moreover, (39) also holds for any ϕ∈C10,1. Then, by using integration by parts, we obtain(43)Gv1x1ϕ1−Gv1x0ϕ0−G∫01v1xxϕdx+ρ∫01v1ϕdx−G∫01v3xϕdx+3G∫01v5xϕdx−∫01ρf1+ρf2−τδf8ϕdx=0.
Then, we get for anyϕ∈C10,1,(44)Gv1x1ϕ1−Gv1x0ϕ0=0.
From (28), we obtain(45)v70=v71=0.
Since ϕ is arbitrary, we get thatv1x0=v1x1=0. Hence, v1∈H∗20,1. Using similar arguments as above, we can obtain(46)v3,v5∈H20,1∩H010,1,v7∈H010,1,v8∈H∗10,1.
Thus,V=v1,v2,v3,v4,v5,v6,v7,v8∈DA and A is maximal. By using Lumer–Philips theorem, see, for example, Liu and Zheng [32] and Pazy [33], we end the proof of the theorem.


## 3. Stability

In this section, we study the stability of systems (1)–(3). More precisely, we establish exponential and polynomial decay results depending on χ defined by(47)χ=τδ2D−Dρ−GIρτρ3DIρ−1.The energy functional of systems (1)–(3) is defined by(48)Et=Eω,ψ,s,θ,q=12∫01ρωt2+Iρ3s−ψt2+3Iρst2+ρ3θ2+τq2+4γs2+D3s−ψx2+Gψ−ωx2+3Dsx2dx.Now we give our stability results.Theorem 2 (exponential decay).
Suppose thatχ=0. For any initial data U0∈ℋ, there exist two positive constants μ and η such that the energy functional (48) satisfies(49)Et≤μe−ηt,∀t≥0.Theorem 3 (polynomial decay).
Suppose thatχ≠0. For any initial data U0∈DA, there exists positive constant μ0 such that the energy functional (48) satisfies(50)Et≤μ0t,∀t>0.
To prove Theorems1 and 2, we need the following technical lemmas.


### 3.1. Technical Lemmas

Lemma 1.
It holds that the energy functionalEt is nonincreasing and satisfies(51)E′t=−4β∫01st2dx−α∫01q2dx≤0.Proof.
Multiplying (1) by ωt, 3s−ψt, st, θ, and q, respectively, integrating the results by parts and using boundary condition (1), we easily get (51).Lemma 2.
Define the functionalF1t by(52)F1t=Iρ∫013s−ψt3s−ψdx−ρ∫01ωt∫0x3s−ψydydx.Then, we have for anyε1>0,(53)F1′t≤−D2∫013s−ψx2dx+ε1∫01ωt2dx+c1+1ε1∫013s−ψt2dx+δ2c∗22D∫01θ2dx,where c∗>0 is the Poincaré constant.Proof.
It follows from (1) that(54)F1′t=D∫013s−ψxx3s−ψdx+G∫01ψ−ωx3s−ψdx+Iρ∫013s−ψt2dx+G∫01ψ−ωx∫0x3s−ψydydx+δ∫01θx∫0x3s−ψydydx−ρ∫01ωt∫0x3s−ψtydydx.
Using integration by parts and boundary condition (1), we arrive at(55)F1′t=−D∫013s−ψx2dx+Iρ∫013s−ψt2dx−δ∫01θ3s−ψdx−ρ∫01ωt∫0x3s−ψtydydx.
Then, by using Hölder’s, Young’s, and Poincaré’s inequalities, we can get (53) from (57).Lemma 3.
The functionalF2t defined by(56)F2t=ρ∫01ψ−ωx∫0xωtydydx,satisfies for any ε2>0,(57)F2′t≤−G2∫01ψ−ωx2dx+ε2∫01ψt2dx+c1+1ε2∫01ωt2dx+δ22G∫01θ2dx.Proof.
DifferentiatingF2t with respect to t and using (1), we see that(58)F2′t=ρ∫01ψt∫0xωtydydx−ρ∫01ωxt∫0xωtydydx−G∫01ψ−ωx∫0xψ−ωyydydx−δ∫01ψ−ωx∫0xθydydx.
Using integration by parts, we obtain(59)F2′t=ρ∫01ψt∫0xωtydydx+ρ∫01ωt2dx−G∫01ψ−ωx2dx−δ∫01θψ−ωxdx.
Then, by using Young’s inequality and Hölder’s inequality, we can get (57).Lemma 4.
Define the functionalF3t by(60)F3t=τρ3∫01θ∫0xqydydx.Then, we can get for anyε3>0,(61)F3′t≤−ρ32∫01θ2dx+ε3∫01ωt2dx+c1+1ε3∫01q2dx.Proof.
DifferentiatingF3 with respect to t and using (1), we obtain(62)F3′t=−τ∫01qx∫0xqydydx−τδ∫01ωxt∫0xqydydx−ρ3α∫01θ∫0xqdydx−ρ3∫01θ∫0xθydydx.
Integration by parts gives us(63)F3′t=τ∫01q2dy+τδ∫01ωtqdx−ρ3α∫01θ∫0xqdydx−ρ3∫01θ2dx.
By using Young’s inequality and Hölder’s inequality, we can get (61).Lemma 5.
The functionalF4t defined by(64)F4t=−ρρ3∫01θ∫0xωtydydx,satisfies for any ε4>0,(65)F4′t≤−ρδ2∫01ωt2dx+ε4∫01ψ−ωx2dx+c1+1ε4∫01θ2dx+ρ2δ∫01q2dx.Proof.
We take the derivative ofF4 and use (1) and integrate by parts to obtain(66)F4′t=ρ∫01qx∫0xωtydydx+ρδ∫01ωxt∫0xωtydydx+ρ3G∫01θ∫0xψ−ωyyydydx+ρ3δ∫01θ∫0xθyy=−ρ∫01qωtdx−ρδ∫01ωt2dx+ρ3G∫01θψ−ωxdx+ρ3δ∫01θ2dx.
Then, using Young’s inequality, we can get (65).Lemma 6.
Define the functionalF5t by(67)F5t=τGδIρ∫013s−ψtψ−ωxdx−τδDρ∫01ωt3s−ψxdx+τρ3Dρ−GIρ∫01θ3s−ψtdx−τDρ−GIρ∫01q3s−ψxdx.Then, we have for anyε5>0,(68)F5′t≤−τGδIρ2∫013s−ψt2dx+c1∫01st2dx+c2∫01θ2dx+c3∫01ψ−ωx2dx+ε5∫013s−ψx2dx+Cε5∫01q2dx+χ∫01θx3s−ψxdx,where ci i=1,2,3 are positive constants.Proof.
By differentiatingF5 with respect to t, we have(69)F5′t=τGδIρ∫013s−ψttψ−ωxdx︸:=I1+τGδIρ∫013s−ψtψ−ωxtdx−τδDρ∫01ωtt3s−ψxdx︸:=I2−τδDρ∫01ωt3s−ψxtdx+τρ3Dρ−GIρ∫01θt3s−ψtdx︸:=I3+τρ3Dρ−GIρ∫01θ3s−ψttdx︸:=I4−τDρ−GIρ∫01qt3s−ψxdx︸:=I5−τDρ−GIρ∫01q3s−ψxtdx︸:=I6.
Using equation (1) and integrating by parts, we see that(70)I1=−τGδD∫013s−ψxψ−ωxxdx+τG2δ∫01ψ−ωx2dx,(71)I2=τδDG∫01ψ−ωxx3s−ψxdx+τδ2D∫01θx3s−ψxdx,(72)I3=−τDρ−GIρ∫01qx3s−ψtdx−τδDρ−GIρ∫01ωxt3s−ψtdx,(73)I4=−τρ3DIρDρ−GIρ∫01θx3s−ψxdx+τρ3GIρDρ−GIρ∫01θψ−ωxdx,(74)I5=αDρ−GIρ∫01q3s−ψxdx+Dρ−GIρ∫01θx3s−ψxdx,(75)I6=τDρ−GIρ∫01qx3s−ψtdx.
Inserting (70)–(75) into (69), we can obtain(76)F5′t=τG2δ∫01ψ−ωx2dx+τGδIρ∫01ψt3s−ψtdx+τρ3GIρDρ−GIρ∫01θψ−ωxdx+αDρ−GIρ∫01q3s−ψxdx+χ∫01θx3s−ψxdx.
Recallingψ=ψ−3s+3s and using Young’s inequality, we conclude that(77)τGδIρ∫01ψt3s−ψtdx=−τGδIρ∫013s−ψt2dx+3τGδIρ∫01st3s−ψtdx≤−τGδIρ2∫013s−ψt2dx+c1∫01st2dx,(78)τρ3GIρDρ−GIρ∫01θψ−ωxdx≤c2∫01θ2dx+c3∫01ψ−ωx2dx,and for any ε5>0,(79)αDρ−GIρ∫01q3s−ψxdx≤ε5∫013s−ψx2dx+Cε5∫01q2dx,which, together with (76)–(78), gives us (68).Lemma 7.
The functionalF6t defined by(80)F6t=3Iρ∫01stsdx+2β∫01s2dx,satisfies(81)F6′t≤−3γ∫01s2dx−3D∫01sx2dx+c4∫01ψ−ωx2dx+3Iρ∫01st2dx,where c4 is a positive constant.Proof.
follows from (1) that(82)F6′t=−3D∫01sx2dx−3G∫01sψ−ωxdx−4γ∫01s2dx+3Iρ∫01st2dx.
Young’s inequality gives us (82).


### 3.2. Exponential Stability: Proof of Theorem1

Proof.
We define the functionalℒt by(83)ℒt=NEt+F1t+N2F2t+N3F3t+N4F4t+N5F5t+F6t,where N and Ni i=2,3,4,5 are positive constants that will be chosen later.
Note that(84)∫01ψt2dx=∫013s−ψ−3st2dx≤2∫013s−ψt2dx+18∫01st2dx.
Replacing (84) in (57) and then combining (51)–(53), (57)–(68), and (82), we obtain(85)ℒ′t≤−4βN−c1N5−18ε2N2−3Iρ∫01st2dx−D2−ε5N5∫013s−ψx2dx−αN−cN31+1ε3−ρ2δN4−Cε5N5∫01q2dx−ρδ2N4−ε1−ε3N3−cN21+1ε2∫01ωt2dx−τGδIρ2N5−2ε2N2−c1+1ε1∫013s−ψt2dx−G2N2−ε4N4−c3N5−c4∫01ψ−ωx2dx−ρ32N3−δ2c∗22D−δ22GN2−cN41+1ε4−c2N5∫01θ2dx−3D∫01sx2dx−3γ∫01s2dx.
Taking(86)ε1=1,ε2=τGδIρ8N2N5,ε3=1N3,ε4=G4N4N2,ε5=D4N5,
we obtain(87)ℒ′t≤−4βN−c1N5−94τGδIρN5−3Iρ∫01st2dx−D4∫013s−ψx2dx−αN−cN31+N3−ρ2δN4−Cε5N5∫01q2dx−ρδ2N4−2−cN21+8N2N5τGδIρ∫01ωt2dx−τGδIρ4N5−2c∫013s−ψt2dx−G4N2−c3N5−c4∫01ψ−ωx2dx−ρ32N3−δ2c∗22D−δ22GN2−cN41+4N4GN2−c2N5∫01θ2dx−3D∫01sx2dx−3γ∫01s2dx.
At this point, we first chooseN5>0 large enough such that(88)τGδIρ4N5−2c>0.
For fixedN5, we take N2>0 so large that(89)G4N2−c3N5−c4>0.
Then, we pickN4>0 large so that(90)ρδ2N4−2−cN21+8N2N5τGδIρ>0.And then we choose N3 so large that(91)ρ32N3−δ2c∗22D−δ22GN2−cN41+4N4GN2−c2N5>0.
At last, we takeN>0 large enough so that the functional ℒt is equivalent to the energy functional Et, i.e., there exist two positive constants:(92)β1Et≤ℒt≤β2Et,and further so that(93)αN−cN31+N3−ρ2δN4−Cε5N5>0.4βN−c1N5−94τGδIρN5−3Iρ>0,
Recalling (48), we infer that there exists a positive constant β3 such that, for any t>0,(94)ℒ′t≤−β3Et,which, along with (92), implies(95)ℒ′t≤−β3β2ℒt.
Integrating (95) over 0,t, we have, for any t>0,(96)ℒt≤ℒ0e−β3/β2=t,which, using (95) again, gives us (49). The proof of Theorem 1 is done.


### 3.3. Polynomial Stability: Proof of Theorem2

In this section, we consider the caseχ≠0 to prove Theorem 2.Differentiating system (1) with respect to time, we obtain the following system:(97)ρωttt+Gψ−ωxxt+δθxt=0,Iρ3s−ψttt−D3s−ψxxt−Gψ−ωxt=0,Iρsttt−Dsxxt+Gψ−ωxt+43γst+43βstt=0,ρ3θtt+qxt+δωxtt=0,τqtt+αqt+θxt=0,which subject to the following boundary conditions:(98)ωxt0,t=ψt0,t=st0,t=θt0,t=0,t∈0,∞,ωxt1,t=ψt1,t=st1,t=θt1,t=0,t∈0,∞.For any initial dataU0∈DA, system (97) is well posed. Next, we introduce second-order energy functional E˜t by(99)E˜t=Eωt,ψt,st,θt,qt=12∫01ρωtt2+Iρ3s−ψtt2+3Iρstt2+ρ3θt2+τqt2+4γst2+D3s−ψxt2+Gψ−ωxt2+3Dsxt2dx.By using the same arguments as in Lemma3, we can get the second-order energy E˜t defined by (99) is nonincreasing and satisfies(100)E˜′t=−4β∫01stt2dx−α∫01qt2dx≤0.In Lemma6, we have proved that, for any ε5>0,(101)F5′t≤−τGδIρ2∫013s−ψt2dx+c1∫01st2dx+c2∫01θ2dx+c3∫01ψ−ωx2dx+ε5∫013s−ψx2dx+Cε5∫01q2dx+χ∫01θx3s−ψxdx.Thanks to (1) and Young’s inequality, we derive that(102)∫01θx2dx≤c∫01q2dx+c∫01qt2dx.Then, for anyε5>0,(103)χ∫01θx3s−ψxdx≤ε5∫013s−ψx2dx+Cε5∫01θx2dx≤ε5∫013s−ψx2dx+Cε5∫01q2dx+Cε5∫01qt2dx.Therefore, the derivative ofF5 satisfies(104)F5′t≤−τGδIρ2∫013s−ψt2dx+c1∫01st2dx+c2∫01θ2dx+c3∫01ψ−ωx2dx+2ε5∫013s−ψx2dx+Cε5∫01q2dx+Cε5∫01qt2dx.Proof.
We define the functionalℒ˜t by(105)ℒ˜t=NEt+E˜t+F1t+N2F2t+N3F3t+N4F4t+N5F5t+F6t.
It follows from (51)–(53), (57)–(65), and (100)–(104) that(106)ℒ˜′t≤−4βN−c1N5−18ε2N2−3Iρ∫01st2dx−D2−2ε5N5∫013s−ψx2dx−αN−cN31+1ε3−ρ2δN4−Cε5N5∫01q2dx−ρδ2N4−ε1−ε3N3−cN21+1ε2∫01ωt2dx−τGδIρ2N5−2ε2N2−c1+1ε1∫013s−ψt2dx−G2N2−ε4N4−c3N5−c4∫01ψ−ωx2dx−ρ32N3−δ2c∗22D−δ22GN2−cN41+1ε4−c2N5∫01θ2dx−3D∫01sx2dx−3γ∫01s2dx−αN−Cε5∫01qt2dx.With the same choice of constants as in Section 3.2, we further take N>0 so large that(107)αN−Cε5>0.
Noting that (48), we know that there exists a positive constant μ1 such that, for any t>0,(108)ℒ˜′t≤−μ1Et.
Since the energy functionalEt is positive and nonincreasing, we infer (108) that, for any t>0,(109)tEt≤∫0tEsds≤1μ1ℒ˜0−ℒ˜t≤ℒ˜0μ1,which gives us(110)Et≤μ0t,∀t>0.
Here,μ0=ℒ˜0/μ1=E0+E˜0/μ1. The proof is complete.Remark 1.
We point out that the functionalℒ˜t is inequivalent to the energy functional Et. That is to say, (92) does not hold true.


## 3.1. Technical Lemmas

Lemma 1.
It holds that the energy functionalEt is nonincreasing and satisfies(51)E′t=−4β∫01st2dx−α∫01q2dx≤0.Proof.
Multiplying (1) by ωt, 3s−ψt, st, θ, and q, respectively, integrating the results by parts and using boundary condition (1), we easily get (51).Lemma 2.
Define the functionalF1t by(52)F1t=Iρ∫013s−ψt3s−ψdx−ρ∫01ωt∫0x3s−ψydydx.Then, we have for anyε1>0,(53)F1′t≤−D2∫013s−ψx2dx+ε1∫01ωt2dx+c1+1ε1∫013s−ψt2dx+δ2c∗22D∫01θ2dx,where c∗>0 is the Poincaré constant.Proof.
It follows from (1) that(54)F1′t=D∫013s−ψxx3s−ψdx+G∫01ψ−ωx3s−ψdx+Iρ∫013s−ψt2dx+G∫01ψ−ωx∫0x3s−ψydydx+δ∫01θx∫0x3s−ψydydx−ρ∫01ωt∫0x3s−ψtydydx.
Using integration by parts and boundary condition (1), we arrive at(55)F1′t=−D∫013s−ψx2dx+Iρ∫013s−ψt2dx−δ∫01θ3s−ψdx−ρ∫01ωt∫0x3s−ψtydydx.
Then, by using Hölder’s, Young’s, and Poincaré’s inequalities, we can get (53) from (57).Lemma 3.
The functionalF2t defined by(56)F2t=ρ∫01ψ−ωx∫0xωtydydx,satisfies for any ε2>0,(57)F2′t≤−G2∫01ψ−ωx2dx+ε2∫01ψt2dx+c1+1ε2∫01ωt2dx+δ22G∫01θ2dx.Proof.
DifferentiatingF2t with respect to t and using (1), we see that(58)F2′t=ρ∫01ψt∫0xωtydydx−ρ∫01ωxt∫0xωtydydx−G∫01ψ−ωx∫0xψ−ωyydydx−δ∫01ψ−ωx∫0xθydydx.
Using integration by parts, we obtain(59)F2′t=ρ∫01ψt∫0xωtydydx+ρ∫01ωt2dx−G∫01ψ−ωx2dx−δ∫01θψ−ωxdx.
Then, by using Young’s inequality and Hölder’s inequality, we can get (57).Lemma 4.
Define the functionalF3t by(60)F3t=τρ3∫01θ∫0xqydydx.Then, we can get for anyε3>0,(61)F3′t≤−ρ32∫01θ2dx+ε3∫01ωt2dx+c1+1ε3∫01q2dx.Proof.
DifferentiatingF3 with respect to t and using (1), we obtain(62)F3′t=−τ∫01qx∫0xqydydx−τδ∫01ωxt∫0xqydydx−ρ3α∫01θ∫0xqdydx−ρ3∫01θ∫0xθydydx.
Integration by parts gives us(63)F3′t=τ∫01q2dy+τδ∫01ωtqdx−ρ3α∫01θ∫0xqdydx−ρ3∫01θ2dx.
By using Young’s inequality and Hölder’s inequality, we can get (61).Lemma 5.
The functionalF4t defined by(64)F4t=−ρρ3∫01θ∫0xωtydydx,satisfies for any ε4>0,(65)F4′t≤−ρδ2∫01ωt2dx+ε4∫01ψ−ωx2dx+c1+1ε4∫01θ2dx+ρ2δ∫01q2dx.Proof.
We take the derivative ofF4 and use (1) and integrate by parts to obtain(66)F4′t=ρ∫01qx∫0xωtydydx+ρδ∫01ωxt∫0xωtydydx+ρ3G∫01θ∫0xψ−ωyyydydx+ρ3δ∫01θ∫0xθyy=−ρ∫01qωtdx−ρδ∫01ωt2dx+ρ3G∫01θψ−ωxdx+ρ3δ∫01θ2dx.
Then, using Young’s inequality, we can get (65).Lemma 6.
Define the functionalF5t by(67)F5t=τGδIρ∫013s−ψtψ−ωxdx−τδDρ∫01ωt3s−ψxdx+τρ3Dρ−GIρ∫01θ3s−ψtdx−τDρ−GIρ∫01q3s−ψxdx.Then, we have for anyε5>0,(68)F5′t≤−τGδIρ2∫013s−ψt2dx+c1∫01st2dx+c2∫01θ2dx+c3∫01ψ−ωx2dx+ε5∫013s−ψx2dx+Cε5∫01q2dx+χ∫01θx3s−ψxdx,where ci i=1,2,3 are positive constants.Proof.
By differentiatingF5 with respect to t, we have(69)F5′t=τGδIρ∫013s−ψttψ−ωxdx︸:=I1+τGδIρ∫013s−ψtψ−ωxtdx−τδDρ∫01ωtt3s−ψxdx︸:=I2−τδDρ∫01ωt3s−ψxtdx+τρ3Dρ−GIρ∫01θt3s−ψtdx︸:=I3+τρ3Dρ−GIρ∫01θ3s−ψttdx︸:=I4−τDρ−GIρ∫01qt3s−ψxdx︸:=I5−τDρ−GIρ∫01q3s−ψxtdx︸:=I6.
Using equation (1) and integrating by parts, we see that(70)I1=−τGδD∫013s−ψxψ−ωxxdx+τG2δ∫01ψ−ωx2dx,(71)I2=τδDG∫01ψ−ωxx3s−ψxdx+τδ2D∫01θx3s−ψxdx,(72)I3=−τDρ−GIρ∫01qx3s−ψtdx−τδDρ−GIρ∫01ωxt3s−ψtdx,(73)I4=−τρ3DIρDρ−GIρ∫01θx3s−ψxdx+τρ3GIρDρ−GIρ∫01θψ−ωxdx,(74)I5=αDρ−GIρ∫01q3s−ψxdx+Dρ−GIρ∫01θx3s−ψxdx,(75)I6=τDρ−GIρ∫01qx3s−ψtdx.
Inserting (70)–(75) into (69), we can obtain(76)F5′t=τG2δ∫01ψ−ωx2dx+τGδIρ∫01ψt3s−ψtdx+τρ3GIρDρ−GIρ∫01θψ−ωxdx+αDρ−GIρ∫01q3s−ψxdx+χ∫01θx3s−ψxdx.
Recallingψ=ψ−3s+3s and using Young’s inequality, we conclude that(77)τGδIρ∫01ψt3s−ψtdx=−τGδIρ∫013s−ψt2dx+3τGδIρ∫01st3s−ψtdx≤−τGδIρ2∫013s−ψt2dx+c1∫01st2dx,(78)τρ3GIρDρ−GIρ∫01θψ−ωxdx≤c2∫01θ2dx+c3∫01ψ−ωx2dx,and for any ε5>0,(79)αDρ−GIρ∫01q3s−ψxdx≤ε5∫013s−ψx2dx+Cε5∫01q2dx,which, together with (76)–(78), gives us (68).Lemma 7.
The functionalF6t defined by(80)F6t=3Iρ∫01stsdx+2β∫01s2dx,satisfies(81)F6′t≤−3γ∫01s2dx−3D∫01sx2dx+c4∫01ψ−ωx2dx+3Iρ∫01st2dx,where c4 is a positive constant.Proof.
follows from (1) that(82)F6′t=−3D∫01sx2dx−3G∫01sψ−ωxdx−4γ∫01s2dx+3Iρ∫01st2dx.
Young’s inequality gives us (82).


## 3.2. Exponential Stability: Proof of Theorem1

Proof.
We define the functionalℒt by(83)ℒt=NEt+F1t+N2F2t+N3F3t+N4F4t+N5F5t+F6t,where N and Ni i=2,3,4,5 are positive constants that will be chosen later.
Note that(84)∫01ψt2dx=∫013s−ψ−3st2dx≤2∫013s−ψt2dx+18∫01st2dx.
Replacing (84) in (57) and then combining (51)–(53), (57)–(68), and (82), we obtain(85)ℒ′t≤−4βN−c1N5−18ε2N2−3Iρ∫01st2dx−D2−ε5N5∫013s−ψx2dx−αN−cN31+1ε3−ρ2δN4−Cε5N5∫01q2dx−ρδ2N4−ε1−ε3N3−cN21+1ε2∫01ωt2dx−τGδIρ2N5−2ε2N2−c1+1ε1∫013s−ψt2dx−G2N2−ε4N4−c3N5−c4∫01ψ−ωx2dx−ρ32N3−δ2c∗22D−δ22GN2−cN41+1ε4−c2N5∫01θ2dx−3D∫01sx2dx−3γ∫01s2dx.
Taking(86)ε1=1,ε2=τGδIρ8N2N5,ε3=1N3,ε4=G4N4N2,ε5=D4N5,
we obtain(87)ℒ′t≤−4βN−c1N5−94τGδIρN5−3Iρ∫01st2dx−D4∫013s−ψx2dx−αN−cN31+N3−ρ2δN4−Cε5N5∫01q2dx−ρδ2N4−2−cN21+8N2N5τGδIρ∫01ωt2dx−τGδIρ4N5−2c∫013s−ψt2dx−G4N2−c3N5−c4∫01ψ−ωx2dx−ρ32N3−δ2c∗22D−δ22GN2−cN41+4N4GN2−c2N5∫01θ2dx−3D∫01sx2dx−3γ∫01s2dx.
At this point, we first chooseN5>0 large enough such that(88)τGδIρ4N5−2c>0.
For fixedN5, we take N2>0 so large that(89)G4N2−c3N5−c4>0.
Then, we pickN4>0 large so that(90)ρδ2N4−2−cN21+8N2N5τGδIρ>0.And then we choose N3 so large that(91)ρ32N3−δ2c∗22D−δ22GN2−cN41+4N4GN2−c2N5>0.
At last, we takeN>0 large enough so that the functional ℒt is equivalent to the energy functional Et, i.e., there exist two positive constants:(92)β1Et≤ℒt≤β2Et,and further so that(93)αN−cN31+N3−ρ2δN4−Cε5N5>0.4βN−c1N5−94τGδIρN5−3Iρ>0,
Recalling (48), we infer that there exists a positive constant β3 such that, for any t>0,(94)ℒ′t≤−β3Et,which, along with (92), implies(95)ℒ′t≤−β3β2ℒt.
Integrating (95) over 0,t, we have, for any t>0,(96)ℒt≤ℒ0e−β3/β2=t,which, using (95) again, gives us (49). The proof of Theorem 1 is done.


## 3.3. Polynomial Stability: Proof of Theorem2

In this section, we consider the caseχ≠0 to prove Theorem 2.Differentiating system (1) with respect to time, we obtain the following system:(97)ρωttt+Gψ−ωxxt+δθxt=0,Iρ3s−ψttt−D3s−ψxxt−Gψ−ωxt=0,Iρsttt−Dsxxt+Gψ−ωxt+43γst+43βstt=0,ρ3θtt+qxt+δωxtt=0,τqtt+αqt+θxt=0,which subject to the following boundary conditions:(98)ωxt0,t=ψt0,t=st0,t=θt0,t=0,t∈0,∞,ωxt1,t=ψt1,t=st1,t=θt1,t=0,t∈0,∞.For any initial dataU0∈DA, system (97) is well posed. Next, we introduce second-order energy functional E˜t by(99)E˜t=Eωt,ψt,st,θt,qt=12∫01ρωtt2+Iρ3s−ψtt2+3Iρstt2+ρ3θt2+τqt2+4γst2+D3s−ψxt2+Gψ−ωxt2+3Dsxt2dx.By using the same arguments as in Lemma3, we can get the second-order energy E˜t defined by (99) is nonincreasing and satisfies(100)E˜′t=−4β∫01stt2dx−α∫01qt2dx≤0.In Lemma6, we have proved that, for any ε5>0,(101)F5′t≤−τGδIρ2∫013s−ψt2dx+c1∫01st2dx+c2∫01θ2dx+c3∫01ψ−ωx2dx+ε5∫013s−ψx2dx+Cε5∫01q2dx+χ∫01θx3s−ψxdx.Thanks to (1) and Young’s inequality, we derive that(102)∫01θx2dx≤c∫01q2dx+c∫01qt2dx.Then, for anyε5>0,(103)χ∫01θx3s−ψxdx≤ε5∫013s−ψx2dx+Cε5∫01θx2dx≤ε5∫013s−ψx2dx+Cε5∫01q2dx+Cε5∫01qt2dx.Therefore, the derivative ofF5 satisfies(104)F5′t≤−τGδIρ2∫013s−ψt2dx+c1∫01st2dx+c2∫01θ2dx+c3∫01ψ−ωx2dx+2ε5∫013s−ψx2dx+Cε5∫01q2dx+Cε5∫01qt2dx.Proof.
We define the functionalℒ˜t by(105)ℒ˜t=NEt+E˜t+F1t+N2F2t+N3F3t+N4F4t+N5F5t+F6t.
It follows from (51)–(53), (57)–(65), and (100)–(104) that(106)ℒ˜′t≤−4βN−c1N5−18ε2N2−3Iρ∫01st2dx−D2−2ε5N5∫013s−ψx2dx−αN−cN31+1ε3−ρ2δN4−Cε5N5∫01q2dx−ρδ2N4−ε1−ε3N3−cN21+1ε2∫01ωt2dx−τGδIρ2N5−2ε2N2−c1+1ε1∫013s−ψt2dx−G2N2−ε4N4−c3N5−c4∫01ψ−ωx2dx−ρ32N3−δ2c∗22D−δ22GN2−cN41+1ε4−c2N5∫01θ2dx−3D∫01sx2dx−3γ∫01s2dx−αN−Cε5∫01qt2dx.With the same choice of constants as in Section 3.2, we further take N>0 so large that(107)αN−Cε5>0.
Noting that (48), we know that there exists a positive constant μ1 such that, for any t>0,(108)ℒ˜′t≤−μ1Et.
Since the energy functionalEt is positive and nonincreasing, we infer (108) that, for any t>0,(109)tEt≤∫0tEsds≤1μ1ℒ˜0−ℒ˜t≤ℒ˜0μ1,which gives us(110)Et≤μ0t,∀t>0.
Here,μ0=ℒ˜0/μ1=E0+E˜0/μ1. The proof is complete.Remark 1.
We point out that the functionalℒ˜t is inequivalent to the energy functional Et. That is to say, (92) does not hold true.



---

*Source: 5139419-2020-02-28.xml*