# On the Porous-Elastic System with Thermoelasticity of Type III and Distributed Delay: Well-Posedness and Stability

Authors: Djamel Ouchenane; Abdelbaki Choucha; Mohamed Abdalla; Salah Mahmoud Boulaaras; Bahri Belkacem Cherif
Journal: Journal of Function Spaces
 (2021)

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## Abstract

The paper deals with a one-dimensional porous-elastic system with thermoelasticity of type III and distributed delay term. This model is dealing with dynamics of engineering structures and nonclassical problems of mathematical physics. We establish the well posedness of the system, and by the energy method combined with Lyapunov functions, we discuss the stability of system for both cases of equal and nonequal speeds of wave propagation.


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## Body

## 1. Introduction

LetH=0,1×τ1,τ2×0,∞,τ1,τ2>0. For x,s,t∈H, we consider the following porous-elastic system:
(1)ρ1utt=μuxx+bϕx,ρ2ϕtt=δϕxx−bux−ξϕ−βθx−μ1ϕt−∫τ1τ2μ2sϕtx,t−sds,ρ3θtt=lθxx−γϕttx+kθtxx,with the initial data
(2)ux,0=u0x,utx,0=u1x,ϕx,0=ϕ0x,ϕtx,0=ϕ1x,ϕtx,−t=f0x,t,θx,0=θ0x,θtx,0=θ1x,x∈0,1,t>0and boundary conditions
(3)ux0,t=ux1,t=ϕ0,t=ϕ1,t=θx0,t=θx1,t=0,t≥0.Here,ϕ is the volume fraction of the solid elastic material, u is the longitudinal displacement, and θ is the difference in temperatures. The parameters ρ1,ρ2,ρ3,μ,b,δ, ξ,l,γ,β,k are positive constants with μξ>b2. The integral represents the distributed delay term withτ1,τ2which are time delays, μ1 is positive constant, and μ2 is an L∞ function such that(Hyp1)μ2:τ1,τ2⟶ℝ is a bounded function satisfying
(4)∫τ1τ2∣μ2s∣ds<μ1.This type of problem was mainly based on the following equations for one-dimensional theories of porous materials with temperature(5)ρ1utt−Tx=0,ρ2ϕtt−Hx−G=0,ρ3θt+qx+γϕtx=0,where x,t∈0,L×0,∞.According to Green and Naghdis theory, the constitutive equations of system (5) are given by
(6)T=μux+bϕ,(7)G=−bux−ξϕ−μ1ϕt−∫τ1τ2∣μ2s∣ϕtx,t−sds,(8)H=δϕx−βθ,(9)q=−lΦx−kΦtx,where l,k>0 are the thermal conductivity and Φ is the thermal displacement whose time derivative is the empirical temperature θ, that is Φt=θ.We substitute (9) in (5) with the condition b≠0, which results in
(10)ρ1utt=μuxx+bϕx,ρ2ϕtt=δϕxx−bux−ξϕ−μ1ϕt−∫τ1τ2μ2sϕtx,t−sds−βθx,ρ3θt=lΦxx−γϕtx+kΦtxx.By usingΦt=θ in the system (10), we find directly our system (1).By using the multiplier techniques, the exponential decay results have been established. Next, in [1–3], the authors considered three types of thermoelastic theories based on an entropy equality instead of the usual entropy inequality (see [1–21] for more details).According to the distributed delay, we mention, as a matter of course, the work by Nicaise and Pignotti in [16], where the authors studied the following system with distributed delay:
(11)utt−Δu=0,u=0,dudvt+∫τ1τ2μsutt−sds+μ0ut=0,u.,0=u0,ut.,0=u1,utx,−t=f0x,t,and proved the exponential stability result with condition
(12)∫τ1τ2μsds<μ0.See for example [8, 22, 23]. Hao and Wei [24] considered the following problem:
(13)ρ1ϕtt−Kϕx+ψxx=0,ρ2ψtt−bψxx+Kϕx+ψ+βθtx+μ1ψt+μ2ψtt−s+fψt=0,ρ3θtt−δθxx+γφttx−kθtxx=0,and obtained the well-posedness and stability of system.There are many other works done by the authors in this context; our work differs from all of them, since we took the delay in the second equation to make the distributed delay in the rotation angle of the filament, which makes the contributions clear and important. In addition, we established the well-posedness of the system, and we obtain the exponential decay rate whenδ/ρ2=μ/ρ1 and the energy takes the algebraic rate for the case δ/ρ2≠μ/ρ1; these results are mainly stated in Theorem 8.In order to show the dissipativity of systems (1)–(3), we introduce the new variables φ=ut and ψ=ϕt. So, problems (1)–(3) take the form
(14)ρ1φtt=μφxx+bψx,ρ2ψtt=δψxx−bφx−ξψ−μ1ψt−∫τ1τ2μ2sψtx,t−sds−βθtx,ρ3θtt=lθxx−γψtx+kθtxx,with the initial data
(15)φx,0=φ0x,φtx,0=φ1x,ψx,0=ψ0x,ψtx,0=ψ1x,θx,0=θ0x,θtx,0=θ1x,ψtx,−t=−f0x,t,x∈0,1and boundary conditions
(16)φx0,t=φx1,t=ψ0,t=ψ1,t=θx0,t=θx1,t=0,t≥0.First, as in [16], taking the following new variable:
(17)zx,ρ,s,t=ψtx,t−sρ,then we obtain
(18)sztx,ρ,s,t+zρx,ρ,s,t=0,zx,0,s,t=ψtx,t.Consequently, the problem was rewritten as(19)ρ1φtt=μφxx+bψx,ρ2ψtt=δψxx−bφx−ξψ−μ1ψt−∫τ1τ2μ2szx,1,s,tds−βθtx,ρ3θtt=lθxx−γψtx+kθtxx,sztx,ρ,s,t+zρx,ρ,s,t=0,where
(20)x,ρ,s,t∈0,1×H,with the boundary and the initial conditions
(21)φx0,t=φx1,t=ψ0,t=ψ1,t=θx0,t=θx1,t=0,t≥0.(22)φx,0=φ0x,φtx,0=φ1x,ψx,0=ψ0x,(23)ψtx,0=ψ1x,θx,0=θ0x,θtx,0=θ1x,x∈0,1,(24)zx,ρ,s,0=−f0x,ρs=h0x,ρs,x∈0.1,ρ∈0.1,s∈0,τ2.Meanwhile, from (19) and (24), it follows that
(25)d2dt2∫01φx,tdx=0.So, by solving (25) and using (24), we get
(26)∫01φx,tdx=t∫01φ1xdx+∫01φ0xdx.Consequently, if we let(27)φ−x,t=φx,t−t∫01φ1xdx−∫01φ0xdx,we get
(28)∫01φ−x,tdx=0,∀t≥0,and from (19), we have
(29)d2dt2∫01θx,tdx=0.So, by solving (29) and using (24), we get
(30)∫01θx,tdx=t∫01θ1xdx+∫01θ0xdx.Consequently, if we let(31)θ−x,t=θx,t−t∫01θ1xdx−∫01θ0xdx,we get
(32)∫01θ−x,tdx=0,∀t≥0.Then, the Poincaré’s inequality was used forφ− and θ– which are justified. A simple substitution shows that φ−,ψ,θ− satisfies system (19) with initial data for φ− and θ− given as
(33)φ0−x=φ0x−∫01φ0xdx,φ1−x=φ1x−∫01φ1xdx,θ0−x=θ0x−∫01θ0xdx,θ1−x=θ1x−∫01θ1xdx.Now, we useφ−,θ− instead of φ,θ and writing φ,θ for simplicity.


## 2. Well-Posedness

In this section, we give the existence and uniqueness result of the system (19)–(24) using the semigroup theory.First, we introduce the vector function(34)U=φ,φt,ψ,ψt,θ,θt,zT,and the new dependent variables u=φt,v=ψt,w=θt; then the system (19) can be written as follows:
(35)Ut=AU,U0=U0=φ0,φ1,ψ0,ψ1,θ0,θ1,h0T,where A:DA⊂H:⟶H is the linear operator defined by
(36)AU=u1ρ1μφxx+bψxv1ρ2δψxx−bφx−ξψ−βwx−μ1ψt−−∫τ1τ2∣μ2s∣zx,1,s,tdsw1ρ3lθxx−γvx+kwxx−1szρ,and H is the energy space given by
(37)H=H∗1×L∗20,1×H01×L20,1×H∗1×L20,1×L20,1×0,1×τ1,τ2,where
(38)L∗20,1=ϕ∈L20,1∫01ϕxdx=0,H∗10,1=H10,1∩L∗20,1,H∗20,1=ϕ∈H20,1ϕx1=ϕx0=0.For every(39)U=φ,u,ψ,v,θ,w,zT∈H,U^=φ∧,u∧,ψ∧,v∧,θ∧,w∧,z∧T∈H,we equip H with the inner product defined by
(40)<U,U^>H=γρ1∫01uu^dx+γρ2∫01vv^dx+γξ∫01ψψ^dx+βρ3∫01ww^dx+γμ∫01φxφ^xdx+γδ∫01ψxψ^xdx+γb∫01φxψ^+ψφ^dx+lβ∫01θxθ^xdx+γ∫01∫01∫τ1τ2s∣μ2s∣zz^dsdρdx.The domain ofA is given by
(41)DA=U∈H/φ,θ∈H∗20,1∩H∗10,1,ψ∈H20,1∩H010,1u,w∈H∗10,1,v∈H010,1,zx,0,s,t=vz,zρ∈L20,1×0,1×τ1,τ2Clearly,DA is dense in H. Now, we can give the following existence result.Theorem 1.
LetU0∈H and assume that (4) holds. Then, there exists a unique solution U∈Cℝ+,H of problem (19).
Moreover, ifU0∈DA, then
(42)U∈Cℝ+,DA∩C1ℝ+,H.Proof.
First, we prove that the operatorA is dissipative. For any U0∈DA and by using (40), we have
(43)<AU,U>H=−γμ1∫01v2dx−γ∫01∫τ1τ2∣μ2s∣vzx,1,s,tdsdx−γ∫01∫01∫τ1τ2∣μ2s∣zρzdsdρdx−βk∫01wx2dx.For the third term of the right-hand side of (43), we have
(44)−∫01∫01∫τ1τ2∣μ2s∣zρzdsdρdx=−12∫01∫τ1τ2∫01∣μ2s∣ddρz2dρdsdx=−12∫01∫τ1τ2∣μ2s∣z2x,1,s,tdsdx+12∫01∫τ1τ2∣μ2s∣z2x,0,s,tdsdx.By using Young’s inequality, we get(45)−∫01∫τ1τ2∣μ2s∣vzx,1,s,tdsdx≤12∫τ1τ2∣μ2s∣ds∫01v2dx+12∫01∫τ1τ2∣μ2s∣z2x,1,s,tdsdx.Substituting (44) and (45) into (43), using the fact that zx,0,s,t=vx,t and (4), we obtained
(46)<AU,U>H≤−γμ1−∫τ1τ2∣μ2s∣ds∫01v2dx−βk∫01wx2dx≤0.Hence, the operatorA is dissipative.Next, we prove the operatorA is maximal. It is sufficient to show that the operator Id−A is surjective.Indeed, for anyF=f1,f2,f3,f4,f5,f6,f7T∈H, we prove that there exists a unique V=φ,u,ψ,v,θ,w,z∈DA such that
(47)Id−AV=F.That is(48)φ−u=f1,ρ1u−μφxx−bψx=ρ1f2,ψ−v=f3,ρ2v−δψxx+bφx+ξψ+βwx+μ1v+∫τ1τ2μ2szx,1,s,tds=ρ2f4,θ−w=f5,ρ3w−lθxx+γvx−kwxx=ρ3f6,sztx,ρ,s,t+zρx,ρ,s,t=sf7,We note that the last equation in (48) with zx,0,s,t=vx,t has a unique solution given by
(49)zx,ρ,s,t=e−ρsv+sesρ∫0ρesσf7x,σ,s,tdσ,then
(50)zx,1,s,t=e−sv+ses∫01esσf7x,σ,s,tdσ,we have
(51)u=φ−f1,v=ψ−f3,w=θ−f5.Inserting (50) and (51) into (48), (48), and (48), we get
(52)ρ1φ−μφxx−bψx=h1,μ4ψ−δψxx+bφx+βθx=h2,rho3θ−l+kθxx+γψx=h3,where
(53)μ4=ρ2+ξ+μ1+43γ+∫τ1τ2μ2se−sds,h1=ρ1f1+f2,h2=ρ2f3+f4+μ1−∫τ1τ2μ2se−sdsf3ds−∫τ1τ2sμ2ses∫01esσf7x,σ,s,tdσds+βf5xh3=ρ3f5+f6+γf3x−kf5xx.We multiply (52) by φ^,ψ^,θ^, respectively, and integrate their sum over 0,1 to get the following variational formulation:
(54)Bφ,ψ,θ,φ^,ψ^,θ^=Γφ^,ψ^,θ^,where
(55)B:H∗10,1×H010,1×H∗10,12⟶ℝis the bilinear form defined by
(56)Bφ,ψ,θ,φ^,ψ^,θ^=γρ1∫01φφ^dx+γμ∫01φxφ^xdx+γb∫01ψφ^x+φψ^xdx+γμ4∫01ψψ^dx+γδ∫01ψxψ^xdx+γβ∫01θxψ^dx+βγ∫01ψxθ^dx+βρ3∫01θθ^dx+βl+k2∫01θxθ^xdx,Γ:H∗10,1×H010,1×H∗10,1⟶ℝis the linear functional given by
(57)Γφ^,ψ^,θ^=∫01h1φ^dx+∫01h2ψ^dx+∫01h3θ^dx.Now, forV=H∗10,L×H010,L×H∗10,L, equipped with the norm
(58)∥φ,ψ,θ∥V2=∥φ∥22+∥φx∥22+∥ψ∥22+∥ψx∥22+∥θ∥22+∥θx∥22,then, we have
(59)Bφ,ψ,θ,φ,ψ,θ=γρ1∫01φ2dx+γμ∫01φx2dx+γμ4∫01ψ2dx+γδ∫01ψx2dx+ρ3β∫01θ2dx+βl+k∫01θx2dx+2γb∫01φxψdx,we have
(60)μφx2+μ4ψ2+2bφxψ=12μφx+bμψ2+μ4ψ+bμ4φx2+μ−b2μ4φx2+μ4−b2μψ2>12μ−b2μ4φx2+μ4−b2μψ2,by assuming μξ−b2>0, we get
(61)μ−b2μ4>0,μ4−b2μ>0,then, for some M0>0,
(62)Bφ,ψ,θ,φ,ψ,θ≥M0∥φ,ψ,θ∥V2.Thus,B is coercive. Consequently, using the Lax-Milgram theorem, we conclude that the existence of a unique solution φ,ψ,θ in V satisfies
(63)u=φ−f1∈H∗10,1,v=ψ−f3∈H010,1,w=θ−f5∈H∗10,1.Substitutingφ,ψ,θ into (50) and (51), respectively, we have
(64)u,θ∈H∗10,1,ψ∈H010,1,z,zρ∈L20,1×0,1×τ1,τ2.Letφ^∈H010,1 and denote
(65)φ^^=φ^x−∫01φ^ξdξ,which gives us φ^^∈H∗10,1. Now, we replace φ^,ψ^,θ^ by φ^^,0,0 in (54) to obtain
(66)γρ1∫01φφ^^dx+γμ∫01φxφ^^xdx+γb∫01ψxφ^^dx=∫01h1φ^^dx.We get(67)γμ∫01φxφ^^xdx=∫01h1−γρ1φ−γbψxφ^^dx,which yields
(68)γμφxx=γρ1φ−γbψx−h1∈L20,1.Thus,(69)φ∈H20,1.Moreover, (52) also holds for any every φ^∈C10,1. Then, by using integration by parts, we obtain
(70)γμ∫01φxφ^xdx=∫01h1−γρ1φ−γbψxφ^dx.Then, we get for anyφ^∈C10,1(71)φx1φ^1−φx0φ^0=0.Sinceφ^ is arbitrary, we get that φx0=φx1=0. Hence, φ∈H∗20,1. Using similar arguments as above, we can obtain
(72)ψ∈H20,1∩H010,1,θ∈H∗20,1.Finally, the application of regularity theory for the linear elliptic equations guarantees the existence of uniqueU∈DA such that (47) is satisfied.Consequently, we conclude thatA is a maximal dissipative operator. Hence, by Lumer-Philips theorem (see [25, 26]), we have the well-posedness result. This completes the proof.


## 3. Stability Results

We prepare the next lemmas (Lemmas2–7) which will be useful to introduce the Lyapunov function in (104).Lemma 2.
The energy functionalEassociated with our problem defined by(73)Et=γ2∫01ρ1φt2+μφx2+ρ2ψt2+δψx2+ξψ2+2bφxψdx+β2∫01lθx2+ρ3θt2dx+γ2∫01∫01∫τ1τ2s∣μ2s∣z2x,ρ,s,tdsdρdxsatisfies
(74)E′t≤−kβ∫01θtx2dx−γη0∫01ψt2dx≤0,where η0=μ1−∫τ1τ2∣μ2s∣ds≥0.Proof.
Multiplying (19) by γφt, (19) by γψt, and (19) by βθt then integration by parts over 0,1, we get
(75)γ2ddt∫01ρ1φt2+μφx2+ρ2ψt2+δψx2+ξψ2+2bφxψdx+γμ1∫01ψt2dx+β2ddt∫01lθx2+ρ3θt2dx+γ∫01ψt∫τ1τ2∣μ2s∣zx,1,s,tdsdx=0.Now, multiplying (19) by z∣μ2s∣ and integrating the result over 0,1×0,1×τ1,τ2, we get
(76)ddtγ2∫01∫01∫τ1τ2s∣μ2s∣z2x,ρ,s,tdsdρdx=−γ∫01∫01∫τ1τ2∣μ2s∣zzρx,ρ,s,tdsdρdx=−γ2∫01∫01∫τ1τ2∣μ2s∣ddρz2x,ρ,s,tdsdρdx=γ2∫01∫τ1τ2∣μ2s∣z2x,0,s,t−z2x,1,s,tdsdx=γ2∫τ1τ2∣μ2s∣ds∫01ψt2dx−γ2∫01∫τ1τ2∣μ2s∣z2x,1,s,tdsdx.From (75) and (76), we get (73) and (74).Now, using Young’s inequality, (74) can be written as
(77)E′t≤−kβ∫01θtx2dx−γμ1−∫τ1τ2∣μ2s∣ds∫01ψt2dx.Then, by (4), there exists a positive constant η0 such that
(78)E′t≤−kβ∫01θtx2dx−γη0∫01ψt2dx.Thus, the functionalE is nonincreasing.Lemma 3.
The function(79)F1t≔ρ2∫01ψtψdx+bρ1μ∫01ψ∫0xφtydydx+μ12∫01ψ2dxsatisfies
(80)F1′t≤−δ2∫01ψx2dx−μ3∫01ψ2dx+ε1∫01φt2dx+c1+1ε1∫01ψt2dx+c∫01θtx2dx+c∫01∫τ1τ2∣μ2s∣z2x,1,s,tdsdx,where μ3=ξ−b2/μ>0.Proof.
Direct computation, using integration by parts and Young’s inequality, forε1>0, yields
(81)F1′t=−δ∫01ψx2dx−ξ−b2μ∫01ψ2dx+ρ2∫01ψt2dx+bρ1μ∫01ψt∫0xφtydydx−β∫01ψθtxdx−∫01ψ∫τ1τ2∣μ2s∣zx,1,s,tdsdx≤−δ∫01ψx2dx−ξ−b2μ∫01ψ2dx+c1+1ε1∫01ψt2dx+ε1∫01∫0xφtydy2dx−β∫01ψθtxdx−∫01ψ∫τ1τ2∣μ2s∣zx,1,s,tdsdx.By Cauchy-Schwartz’s inequality, it is clear that(82)∫01∫0xφtydy2dx≤∫01∫01φtdx2dx≤∫01φt2dx.So, estimate (81) becomes
(83)F1′t≤−δ∫01ψx2dx−ξ−b2μ∫01ψ2dx+c1+1ε1∫01ψt2dx+ε1∫01φt2dx−β∫01ψθtxdx−∫01ψ∫τ1τ2∣μ2s∣zx,1,s,tdsdx,where the Cauchy-Schwartz, Young, and Poincaré’s inequalities have been used, for ε1>0.By the fact thatμξ>b2, we get the desired result (80).Lemma 4.
Assume that ((4)) holds. Then, the function(84)F2t≔∫01ψxφtdx+∫01ψtφxdxsatisfies
(85)F2′t≤−b2ρ2∫01φx2dx+c∫01ψx2dx+c∫01ψt2+c∫01θtx2+c∫01∫τ1τ2∣μ2s∣z2x,1,s,tdx+δρ2−μρ1∫01φxψxxdx.Proof.
By differentiatingF2, then using (19), integration by parts gives
(86)F2′t=−bρ2∫01φx2dx+δρ2−μρ1∫01φxψxxdx+bρ1∫01ψx2dx−ξρ2∫01φxψdx−μ1ρ2∫01ψtφxdx−βρ2∫01θtxφxdx−1ρ2∫01φx∫τ1τ2∣μ2s∣z2x,1,s,tdsdx.Thanks to Young, Cauchy-Schwartz, and Poincaré’s inequalities to estimate terms in RHS of (86). For δ1,δ2,δ3,δ4>0, we have
(87)−ξρ2∫01φxψdx≤δ1∫01φx2dx+c4δ1∫01ψ2dx,(88)−μ1ρ2∫01ψtφxdx≤δ2∫01φx2dx+c4δ2∫01ψt2dx,(89)−βρ2∫01θtxφxdx≤δ3∫01φx2dx+c4δ3∫01θtx2dx,(90)−1ρ2∫01φx∫τ1τ2∣μ2s∣zx,1,s,tdsdx≤δ4∫01φx2dx+c4δ4∫01∫τ1τ2∣μ2s∣z2x,1,s,tds.The replacement of (87)–(90) into (86) and settingδ1=δ2=δ3=δ4=b/8ρ2helps to obtain (85).Lemma 5.
The function(91)F3t≔−ρ1∫01φtφdxsatisfies
(92)F3′t≤−ρ1∫01φt2dx+3μ2∫01φx2dx+c∫01ψx2dx.Proof.
Direct computations give(93)F3′t=−ρ1∫01φt2dx+μ∫01φx2dx+b∫01φxψdx.Estimate (92) easily follows by using Young’s and Poincaré’s inequalities
(94)F3′t≤−ρ1∫01φt2dx+μ∫01φx2dx+δ5∫01φx2dx+c4δ5∫01ψx2dx,setting δ5=μ/2 to obtain (92).Lemma 6.
The function(95)F4t≔−ρ3∫01θtθdxsatisfies
(96)F4′t≤−l2∫01θx2dx+c∫01ψt2dx+c∫01θtx2dx.Proof.
Direct computations give(97)F4′t=−l∫01θx2dx+γ∫01θxψtdx−k∫01θxθtxdx+ρ3∫01θt2dx.By using Young and Poincaré’s inequalities, we get (96).Lemma 7.
The function(98)F5t≔∫01∫01∫τ1τ2se−sρ∣μ2s∣z2x,ρ,s,tdsdρdxsatisfies
(99)F5′t≤−η1∫01∫01∫τ1τ2s∣μ2s∣z2x,ρ,s,tdsdρdx+μ1∫01ψt2dx−η1∫01∫τ1τ2∣μ2s∣z2x,1,s,tdsdx,where η1 is a given positive constant.Proof.
By differentiatingF5 with respect to t and using the last equation in Hyp1, we have
(100)F5′t=−2∫01∫01∫τ1τ2e−sρ∣μ2s∣zzρx,ρ,s,tdsdρdx=−ddρ∫01∫01∫τ1τ2se−sρ∣μ2s∣z2x,ρ,s,tdsdρdx−∫01∫τ1τ2∣μ2s∣e−sz2x,1,s,t−z2x,0,s,tdsdx.Using the fact thatzx,0,s,t=ψtx,t−s and e−s≤e−sρ≤1, for all 0<ρ<1, we obtain
(101)F5′t=−η1∫01∫01∫τ1τ2s∣μ2s∣z2x,ρ,s,tdsdρdx−∫01∫τ1τ2e−s∣μ2s∣z2x,1,s,tdsdx+∫τ1τ2∣μ2s∣ds∫01ψt2dx.We have−e−s≤−e−τ2∀s∈τ1,τ2. Set η1=e−τ2, and by (4), we get (99).We state and prove the decay result in Theorem8.Theorem 8.
Let ((4)) hold. Then, there exist positive constantsλ1andλ2such that the function ((73)) satisfies, for anyt>0(102)Et≤λ2e−λ1t,ifδρ2=μρ1,(103)Et≤CE10+E20t−1,ifδρ2≠μρ1.Proof.
We define a class of an appropriate Lyapunov function as(104)Lt≔NEt+N1F1t+N2F2t+F3t+F4t+N5F5t,where N, N1, N2, and N5 are positive constants to be selected later.Differentiating (104) and by (74), (80), (85), (92), (96), and (99), we have
(105)L′t≤−δN12−cN2−c∫01ψx2dx−ρ1−N1ε1∫01φt2dx−γη0N−cN11+1ε1−N2c−μ1N5−c∫01ψt2dx−bN22ρ2−3μ2∫01φx2dx−N1μ3∫01ψ2dx−N5η1−cN1−cN2∫01∫τ1τ2∣μ2s∣z2x,1,s,tdsdx−l2∫01θx2dx−N5η1∫01∫01∫τ1τ2s∣μ2s∣z2x,ρ,s,tdsdρdx−Nkβ−cN1−cN2−c∫01θtx2dx+N2δρ2−μρ1∫01φxψxxdx.By settingε1=ρ1/2N1, we obtain
(106)L′t≤−δN12−cN2−c∫01ψx2dx−ρ12∫01φt2dx−bN22ρ2−3μ2∫01φx2dx−γη0N−cN11+N1−cN2−μ1N5−c∫01ψt2dx−N5η1−cN1−cN2∫01∫τ1τ2∣μ2s∣z2x,1,s,tdsdx−N1μ3∫01ψ2dx−Nkβ−cN1−cN2−c∫01θtx2dx−l2∫01θx2dx−N5η1∫01∫01∫τ1τ2s∣μ2s∣z2x,ρ,s,tdsdρdx+N2δρ2−μρ1∫01φxψxxdx.Next, we carefully choose the constants, starting byN2 to be large enough such that
(107)α1=bN22J−3μ2>0,and N1 so that
(108)α2=δN12−cN2−c>0,and N5 large enough such that
(109)α3=N5η1−cN1−cN2>0.We arrive at(110)L′t≤−α2∫01ψx2dx−α0∫01ψ2dx−ρ2∫01φt2dx−α1∫01φx2dx−γη0N−c∫01ψt2dx−kβN−c∫01θtx2dx−l2∫01θx2dx(111)−α3∫01∫τ1τ2∣μ2s∣z2x,1,s,tdsdx+α5∫01φxψxxdx−α4∫01∫01∫τ1τ2s∣μ2s∣z2x,ρ,s,tdsdρdx,where α0=μ3N1=ξ−b2/μN1,α4=N5η1,α5=N2k0=N2δ/ρ2−μ/ρ1.Now, let us define the related function(112)Lt=N1F1t+N2F2t+F3t+F4t+N5F5t,then
(113)Lt≤JN1∫01ψψtdx+bρ1N1μ∫01ψ∫0xφtydydx+μ1N12∫01ψ2dx+N2∫01ψxφt+φxψtdx+ρ1∫01φtφdx+ρ3∫01θtθdx+N5∫01∫01∫τ1τ2se−sρ∣μ2s∣z2x,ρ,s,tdsdρdx.Thanks to Young, Cauchy-Schwartz, and Poincaré’s inequalities, we get(114)Lt≤c∫01φt2+ψt2+ψx2+φx2+ψ2+θt2+θx2dx+c∫01∫01∫τ1τ2s∣μ2s∣z2x,ρ,s,tdsdρ≤cEt.Then,(115)Lt=Lt−NEt≤cEt.Thus,(116)N−cEt≤Lt≤N+cEt.One can nowNlarge enough such that
(117)N−c>0,kβN−c>0,Nγη0−c>0.We get(118)c2Et≤Lt≤c3Et,∀t≥0,and using (73), (110), and (116), and the fact that
(119)∫01θt2dx≤∫01θtx2dx,which gives
(120)L′t≤−k1Et+α5∫01φxψxxdx,∀t≥0.for some k1,c2,c3>0.Case 1.
Ifk0=δ/ρ2−μ/ρ1=0, in this case, ((120)) takes the form(121)L′t≤−k1Et,∀t≥0.The combination of (118) and (121) gives
(122)L′t≤−λ1Lt,∀t≥0,λ1=k1c2.Finally, by integrating (122) and recalling (118), we obtain the first result of (103).Case 2.
Ifk0=δ/ρ2−μ/ρ1≠0, then(123)k0<k1μ2γδ2N2ρ1+b,ifk0>0,k0<k1μ2γ2N2ρ1,ifk0<0.Let(124)Et=Eφ,ψ,θ,z=E1t,be denoted by
(125)E2t=Eφt,ψt,θt,zt.Then, we have(126)E2′t≤−kβ∫01θttx2dx−γη0∫01ψtt2dx.The last term in (120), by using (19), and Young’s inequality, and by setting K=−ρ1α5/μ, we have
(127)α5∫01φxψxxdx=−α5ρ1μ∫01ψxφttdx+bα5μ∫01ψx2dx=−Kddt∫01ψφxtdx−∫01ψtφxdx−K∫01φxψtt2dx+bα5μ∫01ψx2dx≤−Kddt∫01ψφxtdx−∫01ψtφxdx+bα5μ∫01ψx2dx+∣K∣4∫01ψtt2dx+∣K∣∫01φx2dx.Let(128)Nt=∫01ψφxtdx−∫01ψtφxdx,then (120)
(129)L′t+KN′t≤−k1E1′t+bα5μ∫01ψx2dx+∣K∣4∫01ψtt2dx+∣K∣∫01φx2dx≤−k2E1′t+∣K∣4∫01ψtt2dx,where
(130)k2=k1−2μγ∣K∣+bα5δ.Let(131)Gt=Lt+KNt+N3E1t+E2t.IfN3>maxC0∣K∣−c1,∣K∣,∣K∣/4C, indeed,
(132)∣Nt∣=∣∫01ψφxtdx∣+∣∫01ψtφxdx∣≤12∫01φtx2dx+12∫01ψt2dx+12∫01ψ2dx+12∫01φx2dx≤E2t+C0E1t,where C0=max2/γξ,2/γμ,2/γρ2. By (118), we obtain
(133)Gt≤c1E1t−∣K∣E2t+C0E1t+N3E1t+E2t≤N3+c1−C0∣K∣E1t+N3−∣K∣E2t.It is not hard to prove(134)m1E1t+E2t≤Gt≤m2E1t+E2t,where m1,m2>0. By using (129) and (128), we obtain
(135)G′t=L′t+KN′t+N3E1′t+E2′t≤−k2E1t+−CN3+∣K∣4∫01ψtt2dx.ChoosingN3 such that
(136)CN3−∣K∣4>0,we have
(137)G′t≤−k2E1t.Integrating (137), we get
(138)∫0tE1ydy≤1k2G0−G1≤1k2G0≤m2k2E10+E20,using the fact that
(139)tE1t′=tE1′t+E1t≤E1t.We get that(140)tE1t≤m2C2E10+E20,which is desired to be the second result of (103). This completes the proof.


## 4. Conclusion

This paper studied the asymptotic behavior of a one-dimensional thermoelastic system with distributed time delay; namely, an integral damping term on a time intervalt−τ2,t−τ1 is taken into account. Beside the distributed delay term, a standard undelayed damping is included in the model −μ1ϕt. We established the well-posedness of the system, and we proved stability estimates by means of appropriate Lyapunov functions. Exponential decay estimates are proved by nonclassical condition between the delay damping coefficient and the coefficient of the undelayed one which is satisfied. Several papers have been proposed for models including both undelayed and delayed damping of the same form, and exponential stability results have been obtained if the coefficient of the delay is smaller than the one of the undelayed term. This analysis has been extended to the case of a distributed delay in [16]. Also in this case, there are now a few literature, dealing with different PDE models, including thermoelastic systems. Typically, under the assumption (4), the system keeps the same properties, the one without delay but only with a standard frictional damping cϕt, for some coefficient c. Then, this paper introduced a considerable novelties different from those of [15].



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