# A New Result of Stability for Thermoelastic-Bresse System of Second Sound Related with Forcing, Delay, and Past History Terms

Authors: Djamel Ouchenane; Zineb Khalili; Fares Yazid; Mohamed Abdalla; Bahri Belkacem Cherif; Ibrahim Mekawy
Journal: Journal of Function Spaces
 (2021)

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## Abstract

We consider a one-dimensional linear thermoelastic Bresse system with delay term, forcing, and infinity history acting on the shear angle displacement. Under an appropriate assumption between the weight of the delay and the weight of the damping, we prove the well-posedness of the problem using the semigroup method, where an asymptotic stability result of global solution is obtained.


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## Body

## 1. Introduction

In this work, we considered with the following problem:(1)ρ1φtt−kφx+lw+ψx−k0lwx−lφ+μ1φtx,t+μ2φtx,t−τ=0,ρ2ψtt−bψxx+kφx+lw+ψ+∫0∞gsψxxx,t−sds+γθx+fψ=0,ρ1wtt−k0wx−lφx+klφx+lw+ψ=0,ρ3θt+κqx+γψtx=0,αqt+βq+κθx=0,x,t∈0,1×0,∞, with initial-boundary conditions
(2)φ0,t=φx1,t=ψx0,t=ψ1,t=wx0,t=w1,t=θ0,t=q1,t=0,t≥0,(3)φx,0=φ0x,φtx,0=φ1x,x∈0,1,ψx,0=ψ0x,ψtx,0=ψ1x,x∈0,1,wx,0=w0x,wtx,0=w1x,x∈0,1θx,0=θ0x,qx,0=q0x,φtx,t−τ=f0x,t−τ,,with τ>0 is a time delay and μ1 and μ2 are positive real numbers. The function θ is the temperature difference, q is the heat flux, and ρ1,ρ2,ρ3,k,l,k0,b,γ,κ,α,β are positive constants. We use the energy method and assume that the relaxation function g satisfies the following hypotheses:(G1)g:ℝ+⟶ℝ+ is a C1 function such that
(4)g0>0,b−∫0∞gsds=b−g0=L>0.(G2) Letζ be a positive constant with
(5)g′t≤−ζgt,∀t≥0,and we suppose that the forcing term fψx,t satisfies some hypotheses.(A1)f:ℝ⟶ℝ such that
(6)fψ2−fψ1≤k0ψ1θ−ψ2θψ1−ψ2for all ψ1,ψ2∈ℝ,wherek0>0,θ>0.(A2)(7)0≤f^ψ≤fψψforallψ∈ℝ,with
(8)f^z=∫0zfsds.Depending on some of the following parameters, we consider(9)η~=1−αkρ3ρ1ρ1k−ρ2b−γ2αb,k=k0.It is well known that, in the single wave equation, ifμ2=0, that is, in the absence of a delay, the energy of system exponentially decays (see, e.g., [1–22]). On the contrary, if μ1=0, that is, there exists only the delay part in the interior, the system becomes unstable.Bresse system is a mathematical model that describes the vibration of a planar, linear shearable curved beam. The model was first derived by Bresse [23], and it consists of three coupled wave equations given by
(10)ρ1φtt=Qx+lN+F1,ρ2ψtt=Mx−Q+F2,ρ1wtt=Nx−IQ+F3,where
(11)N=k0wx−lφ,Q=kφx+lw+ψ,M=bψx.We use N,Q, and M to denote the axial force, the shear force, and the bending moment. By w,φ, and ψ, we are denoting the longitudinal, vertical, and shear angle displacements. Here, ρ1=ρA,ρ2=ρI, b=EI, k0=EA, k=k′GA, and l=R−1 (see, e.g., [23]).The Bresse system (10) is more general than the well-known Timoshenko system where the longitudinal displacement ω is not considered l=0. The reader may refer to, for example, [24–34].System (10) is an undamped system, and its associated energy remains constant when the time t evolves. To stabilize system (10), many damping terms have been considered by several authors (see, e.g., [1, 35–40]).In the succeeding text, we will present some works, which studied the stability of the dissipatif Bresse system. The paper [41] was concerned with asymptotic stability of a Bresse system with two frictional dissipations.
(12)ρ1φtt−kφx+lw+ψx−k0lwx−lφ=−γ1φt,ρ2ψtt−bψxx+kφx+lw+ψ=−γ2ψt,ρ1wtt−k0wx−lφx+lkφx+lw+ψ=0.Under the condition of equal speeds of wave propagation, the authors proved that the system is exponentially stable. Otherwise, they show that Bresse system is not exponentially stable. Then, they proved that the solution decays polynomially to zero with optimal decay rate, depending on the regularity of initial data.There are several works dedicated to the mathematical analysis of the Bresse system. They are mainly concerned with decay rates of solutions of the linear system. This is done by adding suitable damping effects that can be of thermal, viscous, or viscoelastic nature (see for instance [42–44]), among others.Concerning thermoelastic Bresse system, [37] considered
(13)ρ1φtt−kφx+lw+ψx−k0lwx−lφ+lγθ1=0,ρ2ψtt−bψxx+kφx+lw+ψ+γθx=0,ρ1wtt−k0wx−lφx+klφx+lw+ψ+γθ1x=0,ρ3θt−θxx+γψtx=0,ρ3θ1x−θ1xx+γwtx−lφt=0,together with initial and specific boundary conditions and proved an exponential and only polynomial-type decay stabilities results.


## 2. Preliminaries and Well-Posedness

Firstly, we assume the following hypothesis:(14)μ2<μ1.Using semigroup theory, we will prove that systems (1)–(3) are well posed by introducing the following new variable [17].
(15)zx,ρ,t=φtx,t−τρ,x∈0,1,ρ∈0,1,t>0.Then, we have(16)τztx,ρ,t+zρx,ρ,t=0in0,1×0,1×0,∞.Further, let(17)ηtx,s=ψx,t−ψx,t−s,s≥0.For this reason, we observe that(18)ηttx,s+ηstx,s=ψtx,t.Therefore, problem (1) takes the form
(19)ρ1φtt−kφx+lw+ψx−lk0wx−lφ+μ1φtx,t+μ2zx,1,t=0,τztx,ρ,t+zρx,ρ,t=0,ρ2ψtt−Lψxx+kφx+lw+ψ+∫0∞gsηxxtx,sds+γθx+fψx,t=0,ρ1wtt−k0wx−lφx+lkφx+lw+ψ=0,ρ3θt+qx+γψtx=0,αqt+βq+θx=0,ηttx,s+ηstx,s=ψtx,t.The following are with the boundary conditions:(20)φ0,t=φx1,t=ψx0,t=ψ1,t=wx0,t=w1,t=θ0,t=q1,t=0,t≥0.The initial conditions are as follows:
(21)φx,0=φ0x,φtx,0=φ1x,ψx,0=ψ0x,x∈0,1,ψtx,0=ψ1x,wx,0=w0x,wtx,0=w1x,x∈0,1,θx,0=θ0x,qx,0=q0x,x∈0,1,φtx,−t=f0x,tin0,1×0,τ,zx,1,t=fx,t−τin0,1×0,τ,ηtx,0=0,∀t≥0,ηt0,s=ηt1,s=0∀s≥0,η0x,s=η0s=0∀s≥0.Letξ be positive constants such that
(22)τμ2<ξ<τ2μ1−μ2,where τ is a real number with 0<τ and μ1,μ2 are a positive constants, and the initial data are φ0,φ1,ψ0,ψ1,w0,w1,f,θ0,q0,η0.If we set(23)U=φ,φt,z,ψ,ψt,w,wt,θ,q,ηtT,then
(24)U′=φt,φtt,zt,ψt,ψtt,wt,wtt,θt,qt,ηttT.Therefore, problems (19)–(21) can be written as
(25)U′t=AUt+F,U0=φ0,φ1,f1.−,τ,ψ0,ψ1,w0,w1,θ0,q0,η0,where the operator A is defined by
(26)Aφuzψvwϖθqϕ=ukρ1φx+lw+ψx+k0lρ1wx−lφ−μ1ρ1u−μ2ρ1z.,1−1τzρvLρ2ψxx−kρ2φx+lw+ψ+1ρ2∫0∞gsϕxxsds−γρ2θxϖk0ρ1wx−lφx−klρ1φx+lw+ψ−1ρ3qx−γρ3vx−βαq−kαθx−ϕs+v,(27)F=0000−1ρ2fψ00000.We consider the following spaces:(28)H∗10,1=h∈H10,1:h0=0,H~∗10,1=h∈H10,1:h1=0,H∗20,1=H20,1∩H∗10,1,H~∗20,1=H20,1∩H~∗10,1,H=H∗10,1×L20,1×L20,1,H010,1×H~∗10,1×L20,1×H~∗10,1×L20,1×H~∗10,1×L20,1×L20,1×L20,1×Lg2ℝ+,H010,1,where Lg2ℝ+,H010,1 denotes the Hilbert space of H01−valued functions on ℝ+, endowed with the inner product
(29)V1,V2Lg2ℝ+,H01Ω=∫01∫01gsV1xsV2xsdsdx.We will show under the assumption (22) that A generates a C0 semigroup on H.Now, we consider the vectors(30)U=φ,u,z,ψ,v,w,ϖ,θ,q,ϕT,U¯=φ¯,u¯,z¯,ψ¯,v¯,w¯,ϖ¯,θ¯,q¯,ϕ¯T,and we define the inner product
(31)U,U¯H=k∫01φx+ψ+lwφ¯x+ψ¯+lw¯dx+ρ2∫01vv¯dx+ρ1∫01ϖϖ¯dx+k0∫01wx−lφw¯x−lφ¯dx+l∫01ψxψx¯dx+ρ1∫01uu¯dx+ξ∫01∫01zz¯dρdx+ρ3∫01θθ¯dx+α∫01qq¯dx+∫01∫0∞gsϕxsϕ¯xsdxds,where the domain of A is defined by
(32)DA=U∈H/φ∈H∗20,1;ψ,w∈H~∗20,1,u,θ∈H∗10,1;v,ϖ,q∈H~∗10,1,u=z.,0,zρ∈L20,1;L20,1,φx1=0,wx0=ψx0=0,ϕs∈Lg2ℝ+,H010,1,ϕx,0=0,.Important properties of the above metrics are stated in the following lemmas. Although most of these results are followed straightforwardly from the known results, they are crucial for what follows. So for the convenience of the reader, we give their proofs here.Lemma 1.
The operatorA is dissipative and satisfies, for any U∈DA,(33)AU,UH=−β∫01q2dx+−μ1+μ22+ξ2τ∫01u2dx+μ22−ξ2τ∫01z2x,1dx+12∫01∫0∞g′sϕxx,s2dsdx≤0.Proof.
For anyU∈DA, using the inner product,
(34)AU,UH=ukρ1φx+lw+ψx+k0lρ1wx−lφ−μ1ρ1u−μ2ρ1z.,1−1τzρvLρ2ψxx−kρ2φx+lw+ψ+1ρ2∫0∞gsϕxxsds−γρ2θx−ϖk0ρ1wx−lφx−klρ1φx+lw+ψ−1ρ3qx−γρ3vx−βαq−1αθx−ϕs+v,φuzψvwϖθqϕH.
Then,(35)AU,UH=k∫01ux+v+lϖφx+lw+ψdx+k0∫01ϖx−luwx−lφdx+k∫01φx+lw+ψudx+k0l∫01wx−lφudx−μ1∫01u2dx−μ2∫01zx,1udx+L∫01ψxxvdx−k∫01φx+lw+ψvdx−γ∫01θxvdx+k0∫01wx−lφϖdx−kl∫01φx+lw+ψϖdx+L∫01vxψxdx+∫01∫0∞gsϕxs−ϕs+vdxds−∫01qxθdx−γ∫01uxθdx−β∫01q2dx−∫01θxqdx−ξ∫01∫01zzρdρdx.
By the fact that(36)−β∫01q2dx−μ1∫01u2dx−μ2∫01zx,1udx+∫01∫0∞gsϕxs−ϕs+vdxds−ξτ∫01∫01zx,ρzρx,ρdρdx=−β∫01q2dx−μ1∫01u2dx−μ2∫01zx,1udx+∫01∫0∞gsϕxs−ϕs+vdxds−ξ2τ∫01∫01∂∂ρz2x,ρdρdx=−β∫01q2dx−μ1∫01u2dx−μ2∫01zx,1udx+∫01∫0∞gsϕxs−ϕs+vdxds−ξ2τ∫01z2x,1−z2x,0dx=−β∫01q2dx−μ1∫01u2dx−μ2∫01zx,1udx+∫01∫0∞gsϕxs−ϕs+vdxds−ξ2τ∫01z2x,1dx+ξ2τ∫01u2dx,and using Young’s inequality, we find
(37)AU,UH≤−β∫01q2dx+−μ1+μ22+ξ2τ∫01u2dx+μ22−ξ2τ∫01z2x,1dx+12∫01∫0∞g′sϕxx,s2dsdx.
Keeping in mind condition (22), the desired result yields.Lemma 2.
The operatorI−A is surjective.Proof.
We need to show that for allF=f1,f2,f3,f4,f5,f6,f7f8,f9,f10T∈H, there exists U∈DA such that
(38)U−AU=F,that is,
(39)−u+φ=f1∈H∗10,1,−kφx+lw+ψx−k0lwx−lφ+ρ1u+μ1u+μ2z.,1=ρ1f2∈L20,1,z+τ−1zρ=τf3∈L20,1,H10,1,−v+ψ=f4∈H~∗10,1,−Lψxx+kφx+lw+ψ+ρ2v−∫0∞gsϕxxsds+γθx=ρ2f5∈L20,1,−ϖ+w=f6∈H~∗10,1,−k0wx−lφx+klφx+lw+ψ+ρ1ϖ=ρ1f7∈L20,1,qx+γvx+ρ3θ=ρ3f8∈L20,1,β+αq+θx=αf9∈L20,1,ϕ+ϕs−v=f10∈L20,1.
From (39), we define
(40)θ=αk∫0xf9ydy−αkβ+α∫0xqydy,
soθ0,t=0.
Insertingu=φ−f1,v=ψ−f4,ϖ=w−f6 and (39) into (40), we get
(41)−kφx+lw+ψx−k0lwx−lφ+ρ1φ+μ1u+μ2z.,1=h1∈L20.1,−Lψxx+kφx+lw+ψ+ρ2ψ−∫0∞gsϕxxsds−γβ+αq=h2∈L20.1,−k0wx−lφx+klφx+lw+ψ+ρ1w=h3∈L20.1,qx+β+α∫0xqydy−γψx=h4∈L20.1,z+τ−1zρ=h5∈L20.1,ϕ+ϕs−v=h6∈L20.1,where
(42)h1=ρ1f1+f2,h2=ρ2f4+f5−αkγf9,h3=ρ1f6+f7,h4=γf4x+ρ3f8−αk∫0xf9ydy,h5=z+τ−1zρ,h6=ϕ+ϕs−v.
Furthermore, by (39), we can find as zx,0=ux for x∈0,1. Following the same last approach, we obtain by using equation for z in (39)
(43)zx,ρ=uxe−τρ+τe−τρ∫0ρf3x,seτρsds.
From (39), we obtain
(44)zx,ρ=φxe−τρ−f1e−τρ+τe−τρ∫0ρf3x,seτρsds.Then,
(45)zx,1=φxe−τ+z0x,such that
(46)z0x=−f1e−τ+τe−τ∫0ρf3x,seτsds.
We note that the last equation in (41) with ϕx,0=0 has a unique solution
(47)ϕx,s=∫0xeyf10x,y+vxdye−s=∫0xeyf10x,y+ψx−f4xdye−s.
In order to solve (42), we consider
(48)aφ,ψ,w,q,φ~,ψ~,w~,q~=Lφ~,ψ~,w~,q~,where
(49)a:H∗10,1×H~∗10,1×H~∗10,1×L20,12⟶ℝis the bilinear form given by
(50)aφ,ψ,w,q,φ~,ψ~,w~,q~=k∫01φx+lw+ψφ~x+lw~+ψ~dx+β+α∫01qq~dx+b∫01ψxψ~xdx+ρ2∫01ψψ~dx−γβ+α∫01qψ~dx+ρ1∫01ψψ~dx+γβ+α∫01ψq~dx+ρ1∫01ww~dx+k0∫01wx−lφw~x−lφ~dx+∫01φφ~μ1+μ2e−τdx+ρ3β+α∫01∫0xqydy∫0xq~ydydx.L:H∗10,1×H~∗10,1×H~∗10,1×L20,1⟶ℝis the linear form defined by
(51)Lφ~,ψ~,w~,q~=∫01h1φ~dx+∫01h2ψ~dx+∫01h3w~dx+α+β∫01h4∫0xq~ydydx+∫01μ1f1μ2z0φ~dx.
It is easy to verify thata is continuous and coercive, and L is continuous. So applying the Lax-Milgram theorem, we deduce that for all φ~,ψ~,w~,q~∈H∗10,1×H~∗10,1×H~∗10,1×L20,1, problem (48) admits a unique solution φ,ψ,w,q∈H∗10,1×H~∗10,1×H~∗10,1×L20,1. Since DA is dense in H consequently, using Lemmas 1 and 2, we conclude that A is a maximal monotone operator. Hence, by Hille-Yosida theorem (see [45]), we have the following well-posedness result such that (25) is satisfied.Theorem 3.
LetU0∈H, then there exists a unique weak solution U∈Cℝ+,H of problems (1)–(3). Moreover, if U0∈DA, then U∈Cℝ+,DA∩C1ℝ+,H.Lemma 4.
The operatorF defined in (26) is locally Lipschitz in H.Proof.
LetU=φ,u,z,ψ,v,w,ϖ,θ,q,ϕT,U¯=φ¯,u¯,z¯,ψ¯,v¯,w¯,ϖ¯,θ¯,q¯,ϕ¯T, then we have
(52)FU−FU¯H≤fψ−fψ¯L2.
By using (6), Holder’s and Poincaré’s inequalities, we can obtain
(53)fψ−fψ¯L2≤ψ2θθ+ψ¯2θθψ−ψ¯≤c1ψ−ψ¯,which gives us
(54)FU−FU¯H≤c1ψ−ψ¯H.
Then, the operatorF is locally Lipschitz in H. The proof is hence complete.


## 3. Exponential Stability

Here, we present our stability result for the energy of the solution of systems (1)–(3), by using the multiplier technique. So we define the energy of our system by
(55)Et=12∫01ρ1φt2+ρ2ψt2+ρ1wt2+bψx2+ρ3θ2+αq2+kφx+ψ+lw2+k0wx−lφ2dx+ξ2∫01∫01z2x,ρ,tdρdx+12∫01∫0∞gsηxtx,s2dsdx+∫01f^ψtdx.The proof of the stability for our system is based on the following lemmas:Lemma 5.
Letφ,ψ,w,θ,q,z,ηt be the solution of (19)-(21). Then, the energy functional, defined by (55), satisfies
(56)E′t≤−β∫01q2dx−C∫01ψt2dx−μ1−ξ2τ−μ22φt22−ξ2τ−μ22zx,1,t22+12∫01∫0∞g′sηxtx,s2dsdx,such that C>0.Proof.
Multiplying1.11, 1.12, 1.13, 1.14, and 1.15 by φt,ψt,wt,θ, and q, respectively, and after simplification, we have (56).
With the fact(57)ddtf^ψ=fψψ,it gives us (56).Lemma 6.
Letφ,ψ,w,θ,q,z,ηt be the solution of (19)–(21). We have
(58)F1t≔αρ3∫01θ∫0xqydydxsatisfies, for any ε1>0, the estimate
(59)F1′t≤−ρ32∫01θ2dx+ε1∫01ψt2dx+c1+1ε1∫01q2dx.Proof.
Taking the derivative ofF1, using the fourth and fifth equations in (1) and performing integration by parts, we get
(60)F1′t=−ρ3k∫01θ2dx−αk∫01q2dx−αγ∫01ψtx∫01qydydx−βρ3∫01θ∫0xqydydx.
According to Cauchy–Schwarz and Young’s inequalities withε1>0, we get (59).Lemma 7.
Letφ,ψ,w,θ,q,z,ηt be the solution of (19)–(21). We have
(61)F2t≔−ρ2ρ3γ∫01θ∫0xψtydydxsatisfies, for any ε1,ε2,δ1>0, the estimate
(62)F2′t≤−ρ2γ∫01ψt2dx+ε2∫01φx+ψ+lw2dx+ε3+ρ3γε2b2λ2+b22ε2λ2∫01ψx2dx+c1+1ε2+1ε3∫01θ2dx+c∫01q2dx+g04δ1∫01∫0∞gsηxtx,s2dsdx.+δ1+C1∫01θx2dx.Proof.
For differentiation ofF2, using equations in (1) and integration by parts, we obtain
(63)F2′t=−ρ2∫01ψt2dx−ρ2kγ∫01qψtdx+ρ3∫01θ2dx−bρ3γ∫01θψxdx+kρ3γ∫01φx+ψ+lw∫0xθydydx+ρ3γ∫01∫0∞gsηxtx,sds∫0xθxydydx+ρ3γ∫01θ∫0xfψdydx.
Estimate (62) follows by using Cauchy–Schwarz, Young’s, and Poincaré’s inequalities that
(64)∫01fψθdx≤∫01ψθψθdx≤ψ2θ+1θψ2θ+1θ≤C1∫01θ2dx.Lemma 8.
Letφ,ψ,w,θ,q,z,ηt be the solution of (19)–(21). Then, the energy functional
(65)F3t≔−ρ1∫01φφt+wwtdxsatisfies the estimate
(66)F3′t≤−ρ1−14ε4∫01φt2dx+c∫01ψx2dx+k0∫01wx−lφ2dx+c∫01φx+ψ+lw2dx−ρ1∫01wt2dx+ε5μ2+μ1ε4∫01φ2dx+μ24ε5∫01z2x,1,tdx.Proof.
Using (1)–(3) gives
(67)F3′t=−ρ1∫01φt2dx+k∫01φx+ψ+lw2dx−k∫01φx+ψ+lwψdx−ρ1∫01wt2dx+k0∫01wx−lφ2dx+μ1∫01φφtdx+μ2∫01φzx,1,tdx.
Using Young’s and Poincaré’s inequalities, estimate (66) is established.Lemma 9.
Letφ,ψ,w,θ,q,z,ηt be the solution of (19)–(21). Then, the energy functional
(68)F4t≔ρ2∫01ψψtdxsatisfies for any δ2>0 the estimate
(69)F4′t≤b2+δ2+C2∫01ψx2dx+ρ2∫01ψt2dx+k2b∫01φx+ψ+lw2dx+c∫01θ2dx+g04δ2∫01∫0∞gsηxtx,s2dsdx.Proof.
Taking the derivative ofF4 and using the second equation in (1), it follows that
(70)F4′t=−b∫01ψx2dx+ρ2∫01ψt2dx+γ∫01ψxθdx−k∫01φx+ψ+lwdx+∫01ψxx∫0∞gsηxtx,sdsdx.−∫01ψfψdx,(71)∫01fψψdx≤∫01ψθψψdx≤ψ2θ+1θψ2θ+1ψ≤C2∫01ψx2dx.
Young’s and Poincaré’s inequalities for (70) yield (69).Lemma 10.
Letφ,ψ,w,θ,q,z,ηt be the solution of (19)–(21). Then, the energy functional
(72)F5t≔−ρ1∫01φtwx−lφdx−ρ1∫01wtφx+ψ+lwdxsatisfies the estimate
(73)F5′t≤−lk0−μ14ε6−μ24ε7∫01wx−lφ2dx−lρ12∫01wt2dx+lρ1+ε6μ1∫01φt2dx+c∫01ψt2dx+lk∫01φx+ψ+lw2dx+ε7μ2∫01z2x,1,tdx.Proof.
For differentiation ofF5, using 1.11 and 1.13, we arrive at
(74)F5′t=−lk0∫01wx−lφ2dx−lρ1∫01wt2dx+lρ1∫01φt2dx+lk∫01φx+ψ+lw2dx−ρ1∫01ψtwtdx+μ1∫01φtwx−lφdx+μ2∫01zx,1,twx−lφdx.
Young’s inequality for (74) yields (73).Lemma 11.
Letφ,ψ,w,θ,q,z,ηt be the solution of (19)-(21) and let k=k0. Then, the functional
(75)F6t≔−ρ1∫01wx−lφ∫0xwtydydx−ρ1∫01φt∫0xφx+ψ+lwdydxsatisfies the estimate
(76)F6′t≤−ρ12∫01φt2dx−k0∫01wx−lφ2dx+ρ1∫01wt2dx+k∫01φx+ψ+lw2dx+ρ12∫01ψt2dx.Proof.
A simple differentiation ofF6, using the first and third equations in (1), leads to
(77)F6′t=−ρ1∫01φt2dx−k0∫01wx−lφ2dx+ρ1∫01wt2dx−ρ1∫01φt∫0xψtydy+k∫01φx+ψ+lw2dx+lk−k0∫01wx−lφ∫0xφx+ψ+lwdydx,
and using Young’s and Cauchy-Schwarz inequalities, with the fact thatk=k0, gives (76).Lemma 12.
Letφ,ψ,w,θ,q,z,ηt be the solution of (19)–(21) and let (9) holds, and we have
(78)F7t≔ρ2∫01ψtφx+ψ+lwdx+bρ1k∫01φtψxdx+bρ3γρ1k−ρ2b∫01θφtdx−bγρ1k−ρ2b∫01qφx+ψ+lwdx−bl2ρ2k0∫01ψψtdx+blρ1k0∫01ψwtdxsatisfies, for any ε4,ε5,δ3>0, the estimate
(79)F7′t≤−k2−bηγαε10+γ4ε1+kbρ3γ4ε2ρ1ρ1k−ρ2b+b4ε3∫01φx+ψ+lw2dx+ε8∫01wt2dx+b2l2k+bl2ρ2δ3k0+bε3+bl2k0c1+2εb2λ1+b22ελ1+c2∫01ψx2dx+ε9∫01wx−lφ2dx+c1+1ε8+bρ1ε4k∫01ψt2dx+c1+1ε8∫01q2dx+c1+1ε9+bρ3μ1γρ1ε5ρ1k−ρ2b∫01θ2dx+bηγαε10+γε1+kbρ3γρ1ε2ρ1k−ρ2b∫01θx2dx+bρ1k4ε4+bρ3μ14ε5γρ1ρ1k−ρ2b∫01φt2dx+lε1b2∫01w+ψ2dx+ε2∫01φx+ψ2+g0bl2ρ2k04δ3∫01∫0∞gsηxtx,s2dsdx.Proof.
Taking the deviate ofF7, we obtain
(80)F7′t=ρ2∫01ψttφx+ψ+lwdx+ρ2∫01ψtφx+ψ+lwtdx+bρ1k∫01φttψxdx−bρ1k∫01φtψxtdx+bρ3γρ1k−ρ2b∫01θtφtdx+bρ3γρ1k−ρ2b∫01θφttdx−bγρ1k−ρ2b∫01qtφx+ψ+lwdx−bγρ1k−ρ2b∫01qφx+ψ+lwtdx−bl2ρ2k0∫01ψt2dx−bl2ρ2k0∫01ψttψdx+blρ1k0∫01wttψdx+blρ1k0∫01wtψtdx.
From the RHS of (80) and the relations in (1)–(3), we arrive at
(81)ρ2∫01ψttφx+ψ+lwdx=−k∫01φx+ψ+lw2dx−γ∫01θxφx+ψ+lwdx−b∫01ψxφx+ψ+lwxdx−∫01∫0∞gsψxxx,t−sdsφx+ψ+lwdx−∫01fψφx+ψ+lwdx,(82)ρ1∫01φttψxdx=k∫01ψxφx+ψ+lwxdx+k0l∫01wx−lφdx−μ1∫01φtψxdx,(83)ρ3∫01θtφtdx=k∫01qφxtdx+γ∫01ψtφxtdx,(84)∫01θφttdx=−kρ1∫01θxφx+ψ+lwdx+lk0ρ1∫01θwx−lφdx−μ1ρ1∫01θφtdx,(85)−∫01qtφx+ψ+lwdx=βα∫01qφx+ψ+lwdx+kα∫01θxφx+ψ+lwdx,(86)−ρ2∫01ψttψdx=b∫01ψx2dx+k∫01ψφx+ψ+lwdx−γ∫01θψxdx+∫01∫0∞gsψxxdsψdx+∫01fψψdx,(87)ρ1∫01wttψdx=−k0∫01ψxwx−lφdx−kl∫01ψφx+ψ+lwdx.
Invoking to (81)–(87) into (80), we arrive at
(88)F7′t=−k∫01φx+ψ+lw2dx+ρ2−bl2ρ2k0∫01ψt2dx+lρ2+blρ1k0∫01ψtwtdx+bηαγ∫01θxφx+ψ+lwdx−bγρ1k−ρ2b∫01qψtdx−blγρ1k−ρ2b∫01qwtdx+blk0ρ3γρ1ρ1k−ρ2b∫01θwx−lφdx−γbl2k0∫01θψxdx+bβαγρ1k−ρ2b∫01qφx+ψ+lwdx+b2l2k0∫01ψx2dx−bl∫01ψxwx−lφdx−γ∫01θxφx+ψ+lwdx−kbρ3γρ1ρ1k−ρ2b∫01θxφx+ψ+lwdx−b∫01ψxφx+ψ+lwxdx+bρ1k∫01ψtφxtdx−bρ3μ1γρ1ρ1k−ρ2b∫01θφtdx−∫01fψφx+ψ+lwdx+bl2k0∫01fψψdx+bl2ρ2k0∫01ψxx∫0∞gsηxtx,s.dsdx.
We thus have(89)∫01φxfψdx≤φxψ2θ+1θψ2θ+1≤ε2b2∫01φx2dx+b22ελ1∫01ψx2dx≤εb2∫01φx+ψ2dx+εb2∫01ψ2dx+b22ελ1∫01ψx2dx≤εb2∫01φx+ψ2dx+εb2λ1+b22ελ1∫01ψx2dx.
Estimate (79) follows thanks to Young’s inequality and the fact that k=k0.Lemma 13.
Letφ,ψ,w,θ,q,z,ηt be the solution of (19)–(21). Then, the energy functional
(90)F8t≔∫01ρ1φtφdx+μ12∫01φ2dx.
Then, we have the following estimate, for anyε11>0,(91)F8′t≤−K+ε11K2+μ2c2∫01φx2dx+K2ε11∫01ψx2dx+μ22ε11∫01z2x,1,tdx+ρ1∫01φt2dx,where c=1/π2 is the Poincaré constant.Proof.
Taking the derivative of (90) with respect to t, we have
(92)F8′t=ρ1∫01φttφdx+ρ1∫01φt2dx+μ1∫01φtφdx.
Then, by using the first equation in (1), we find
(93)F8′t=k∫01φx+ψ+lwxφdx−μ2∫01φzx,1,tdx+ρ1∫01φt2dx.
Consequently, we arrive at(94)F8′t=−k∫01φx+ψ+lwφxdx−μ2∫01φzx,1,tdx+ρ1∫01φt2dx.
Applying Young’s inequality and Poincaré’s inequality, we find (90).Lemma 14.
Letφ,ψ,w,θ,q,z,ηt be the solution of (19)–(21). Then, we define the functional
(95)F9t≔∫01∫01e−2τρz2x,ρ,tdρdx.
Then, the following result holds.(96)F9′t≤−F9t−c12τ∫01z2x,1,tdx+12τ∫01ψt2x,tdx,where c is a positive constant.Proof.
Taking the deviate of (95) with respect to t and using the equation (16), we get
(97)ddt∫01∫01e−2τρz2x,ρ,tdρdx=−1τ∫01∫01e−2τρzzρx,ρ,tdρdx=−∫01∫01e−2τρz2x,ρ,tdρdx−12τ∫01∫01∂∂ρe−2τρz2x,ρ,tdρdx.
Making use of the estimate above, implies that there exists a positive constantc1 such that (96) holds.Theorem 15.
Assume thatη=0 and k=k0. Then, φ,ψ,w,θ,q,z,ηt the solution of (19)–(21) satisfies
(98)Et≤c0e−c1t,t≥0,where the positive constant c0 is directly depending on initial data and the uniform constant c1 is depending only on the coefficients of the system. For N,Ni>0,(99)Lt≔NEt+∑i=1i=9NiFit,
Then, from (56), (59), (62), (66), (69), (73), (76), (79), (91), and (96), we have
(100)L′t≤−βN+c11+1ε1+cN2+c1+1ε8N7∫01q2dx−Nμ1−ξ2−μ22φt22−Nξ2−μ22zx,1,t22−N1ρ32−N2C1+1+1ε2+1ε3−cN4−c1+1ε9+bρ3μ1γρ1cε5ρ1k−ρ2bN7∫01θ2dx+ε1N1−CN−N2ρ2γ+ρ2N4+cN5+ρ12N6+c1+1ε8+bρ1ε4kN7+ρ1N8+12τN9∫01ψt2dx+ε2N2+cN3+k2bN4+lkN5+kN6−k2−bη~αγε10+γ4ε1+kbρ3γ4ε2ρ1ρ1k−ρ2b+b4ε3N7∫01φx+lw+ψ2dx+ε3+ρ3γε2b2λ2+b22ε2λ2N2+cN3+δ2+b2+C2N4+bl2ρ2δ3k0+b2l2k+bε3+bl2k0c1+2εb2λ1+b22ελ1+c2N7+k2ε11−k+ε11k2+ε11μ2c2N8∫01ψx2dx+−ρ1N3−lρ12N5+ρ1N6+ε8N7∫01wt2dx+k0N3−lk0−μ14ε6−μ24ε7N5−k0N6+ε9N7∫01wx−lφ2dx+μ24ε5N3+ε7μ2N5+μ22ε11N8−c12τN9∫01z2x,1,tdx+ε5μ2+ε4μ1N3∫01φ2dx+−1−14ε4N3+lρ1+ε6μ1N5−ρ12N6+ρ1N8+bρ1k4ε4+bρ3μ14ε5γρ1ρ1k−ρ2bN7∫01φt2dx+−k+ε11k2+μ2c2N8∫01φx2dx+bη~ε10αγ+γε1+kbρ3γρ1ε2ρ1k−ρ2bN7+N2δ1∫01θx2dx−N9F9t+N2g04δ1+N4g04δ2+N7g0bl2ρ2k04δ3−Nζ2∫01∫0∞gsηxtx,s2dsdx.
At this point, we have to choose our constants very carefully. First, choosingεii=1,...10 small enough such that
(101)ε1≤N2ρ2/γ+ρ2N4+cN5+ρ1/2N6N1.
Moreover, we pickN9 large enough so that
(102)μ24ε5N3+ε7μ2N5+μ22ε11N8−c12τN9≤0,N9≥μ2/4ε5N3+ε7μ2N5+μ2/2ε11N8c1/2τ,and we take ε11 small enough such that
(103)ε11≤kk/2+μ2c/2N8.
Next, choosingN5 large enough such that
(104)N5ρ3κ4≥N4γρ3+ρ32ε4b+2κ.
After that, we can chooseN large enough such that
(105)N≥c11+1/ε1+cN2+c1+1/ε8N7β,N2g04δ1+N4g04δ2+g0bl2ρ2k04δ3−Nζ2≤0.
Thus, the relation (100) becomes
(106)ddtLt≤−η1∫01ψt2+ψx2+φt2+φx+lw+ψ2+θ2+q2dx−η1∫01∫01z2x,ρ,tdρdx,which leads by (55) that there exists also η2, such that
(107)ddtLt≤−η2Et,∀t≥0.Lemma 16.
ForN large enough, there exist two positive constants β1 and β2 depending on Ni,i=1,⋯,9 and εi,i=1,⋯,11 such that
(108)β1Et≤Lt≤β2Et,∀t≥0.Proof.
We consider the functional(109)Ht=∑i=1i=9NiFitand show that
(110)Ht≤CEt,C>0.
From (58), (61), (65), (68), (72), (75), (78), (90), and (95), we obtain
(111)Ht≤N1αρ3∫01θ∫0xqydydx+N2−ρ2ρ3γ∫01θdx∫0xψtydydx+N3ρ1∫01φφt+wwtdx+N4ρ2∫01∫0xψψtt,xdx+N5−ρ1∫01φtwx−lφdx−ρ1∫01wtφx+ψ+lwdx+N6−ρ1∫01wx−lφ∫0xwtydydx−ρ1∫01φt∫0xφx+ψ+lwdydx+N7ρ2∫01φx+ψ+lwdx+bρ1k∫01φtψxdx+N8∫01ρ1φφtdx+μ12∫01φ2dx+N9∫01∫01e−2τρz2x,ρ,tdρdx.
By using, the trivial relation(112)∫01φ+lw2dx≤2c∫01φx+lw+ψ2dx+2c∫01ψx2dx,
Young’s and Poincaré’s inequalities, we get(113)Ht≤α1∫01φt2dx+α2∫01ψt2dx+α3∫01wt2+α4∫01ψx2+α5∫01θ2dx+α6∫01q2dx+α7∫01φx+lw+ψ2+wx−lφ2dx+∫01∫01z2x,ρ,tdρdx,where α1,⋯,α6 are the positive constants as follows:
(114)α1≔12N3ρ1+N8ρ1,α2≔12N4ρ2+N2ρ2ρ3γ,α3=12N3ρ1+N6ρ1,α4≔bρ12k,α5≔12N1ρ3+ρ2ρ3γ,α6≔12N1ρ3+N5τ0ρ3,α7≔12N7ρ2+3ρ1.
From (113), we have
(115)Ht≤C^Et,for
(116)C^=maxα1,α2,α3,α4,α5,α6minρ1,ρ2,ρ3,k,b,κ,γ,δ,τ0.
Therefore, we get(117)Lt−NEt≤C^Et.
Then, we can chooseN large enough so that β1=N−C^>0. Then, (108) holds true for β2=N+C^>0, and this concludes the proof of the Lemma.
Combining now (107) and (108), we conclude that there exists some Λ>0 such that
(118)ddtLt≤−ΛLt,∀t≥0.
Integration of (118) yields
(119)Lt≤L0e−Λt,∀t≥0.
Finally, using (108) and (119), so (98) is satisfied, we thus immediately reach to Theorem 15.


## 4. Conclusion and Perspective

In this current study, a one-dimensional linear thermoelastic Bresse system with delay term, forcing, and infinity history acting on the shear angle displacement is considered. According to an appropriate assumption between the weight of the delay and the weight of the damping, the well-posedness of the problem using the semigroup method is proved, where an asymptotic stability result of global solution is obtained. In next article, we will generalize this result to convex bounded domain with a holomorphic map, and letx and y be two distinct fixed points for our problem. We will suppose there is at least one complex geodesics passing through two distinct variables. We will see that this method of proof cannot be generalized to the case of a bounded domain of a complex Banach space. Also, in the last part of the next article, we will study the fixed points of the analytical automorphisms of the open unit-ball B of a complex Banach space. More precisely, we will assume that B is homogeneous and we will show that, if the right hand side is an analytical automorphism of B, there exists a complex geodesic which we will specify formed of fixed points of the right hand. We will see that the set of fixed points of the right hand can be much larger by using the studied algorithm in ([46–51]).



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