# Exponential Stability of Swelling Porous Elastic with a Viscoelastic Damping and Distributed Delay Term

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

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## Abstract

In this paper, we consider a swelling porous elastic system with a viscoelastic damping and distributed delay terms in the second equation. The coupling gives new contributions to the theory associated with asymptotic behaviors of swelling porous elastic soils. The general decay result is established by the multiplier method.


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## Body

## 1. Introduction and Preliminaries

In the late 19th century, Eringen [1] proposed a theory in which he presented a mixture of viscous liquids and elastic solids in addition to gas. And he also studied the equilibrium laws for all components of this mixture, and finally, you get the field equations for a heat conductive mixture (for more details, see [2]). In [3], the author has classified expansive (swelling) soils under the classification of porous media theory.On the other hand, it contains clay minerals that attract and absorb water, which leads to an increase in pressure [4], and this is considered a harmful and dangerous problem in architecture and civil engineering in most countries of the world, especially in foundations, which leads to cracks in buildings and ripples in sidewalks and roads (see [5–8]). From there, studies began to eliminate or reduce the damage, as in ([9–13]), where the basic field equations of the linear theory of swelling porous elastic soils were presented by
(1)ρuutt=P1x+G1+H1,(2)ρϕϕtt=P2x+G2+H2,where u,ϕ are the displacement of the fluid and the elastic solid material. And ρu,ρϕ>0 are the densities of each constituent. The functions (P1,G1,H1) represent the partial tension, internal body forces, and eternal forces acting on the displacement, respectively. Similarly (P2,G2,H2), it works on the elastic solid. In addition, the constitutive equations of partial tensions are given by
(3)P1P1=a1,a2a2,a3⏟A.uxϕx,where a1,a3>0 and a2≠0 is a real number. A is a matrix positive definite in the sense that a1a3>a22.Quintanilla [10] investigated (1) by taking
(4)G1=G2=ξut−ϕt,H1=a3uxxt,H2=0,where ξ>0; they obtained that the stability is exponential. Similarly, in [14], the authors considered (1) with different conditions
(5)G1=G2=0,H1=−ρuγxut,H2=0,where γx is an internal viscous damping function with a positive mean. They established the exponential stability result (see ([10–20]) for some other interesting results on the swelling porous system).Time delays arise in many applications because most phenomena naturally depend not only on the present state but also on some past occurrences.In recent years, the control of PDEs with time delay effects has become an active area of research (see, for example, [15, 20–27]). In many cases, it was shown that delay is a source of instability unless additional condition or control terms are used; the stability issue of systems with delay is of theoretical and practical great importance.A complement to these works, and by introducing the terms of memory and distributed delay, forms a new problem different from previous studies. Under appropriate assumptions and by using the energy method, we prove the stability results.In this paper, we are interested in problem (1) with null internal body forces, but the eternal force acting only on the elastic solid is in the form of viscoelastic damping and distributed delay terms, that is,
(6)G1=G2=H1=0,H2=−∫0tgt−sϕxxx,sds−β1ϕt−∫τ1τ2∣β2σ∣ϕtx,t−σdσ.Remark 1.
Regarding the problems of swelling porous elastic, we believe that there are no studies of viscoelasticity (the memory) and the distributed delay conditions that act as a simultaneous dissipation mechanism, and hence, our coupling constitutes a new contribution.
Thus, we are interested in the following problem:(7)ρuutt−a1uxx−a2ϕxx=0,ρϕϕtt−a3ϕxx−a2uxx+∫0tgt−sϕxxx,sds+β1ϕt+∫τ1τ2∣β2σ∣ϕtx,t−σdσ=0,where
(8)x,σ,t∈H=0,1×τ1,τ2×0,∞,under the initial and boundary conditions
(9)ux,0=u0x,utx,0=u1x,ϕx,0=ϕ0x,ϕtx,0=ϕ1x,x∈0,1,ϕtx,−t=f0x,t,x∈0,1×0,τ2,u0,t=u1,t=ϕ0,t=ϕ1,t=0,t≥0.
First, as in [27], taking the following new variable
(10)Yx,ρ,σ,t=ϕtx,t−σρ,then we obtain
(11)σYtx,ρ,σ,t+Yρx,ρ,σ,t=0,Yx,0,σ,t=ϕtx,t.
Consequently, the problem is equivalent to(12)ρuutt−a1uxx−a2ϕxx=0,ρϕϕtt−a3ϕxx−a2uxx+∫0tgt−sϕxxx,sds+β1ϕt+∫τ1τ2β2σYx,1,σ,tdσ=0,σYtx,ρ,σ,t+Yρx,ρ,σ,t=0,where
(13)x,ρ,σ,t∈0,1×H,with the initial data
(14)ux,0=u0x,utx,0=u1x,ϕx,0=ϕ0x,ϕtx,0=ϕ1x,x∈0,1,Yx,ρ,σ,0=f0x,ρσ,x,ρ,σ∈0,1×0,1×0,τ2,and the boundary conditions
(15)u0,t=u1,t=ϕ0,t=ϕ1,t=0,t≥0.
Here,ρu,ρϕ,a1,a3,β1 are positive constants and a2 is a real number, with a1,a2,a3 satisfying a=a3−a22/a1>0. The integrals represent the memory and the distributed delay terms with τ1,τ2>0 are a time delay, β2 is an L∞ function, and the kernel g is the relaxation function, under the following assumptions.
(H1)g∈C1ℝ+,ℝ+ is a nonincreasing function satisfying
(16)g0>0,a−∫0∞gsds=l>0,where a=a3−a22/a1>0.
(H2) There exists aϑ∈ℝ+,ℝ+ positive nonincreasing differentiable function, such that
(17)g′t≤−ϑtgt,t≥0.
(H3)β2:τ1,τ2→ℝ is a bounded function satisfying
(18)∫τ1τ2∣β2σ∣dσ<β1.Remark 2.
The results that we obtained in this work are also correct with other conditions, including(19)ux0,t=ux1,t=ϕx0,t=ϕx1,t=0,t≥0,u0,t=ux1,t=ϕ0,t=ϕx1,t=0,t≥0,ux0,t=u1,t=ϕx0,t=ϕ1,t=0,t≥0.
Of course, there can be some difficulties with regard to the following boundary conditions:(20)ux0,t=ux1,t=ϕ0,t=ϕ1,t=0,t≥0,u0,t=u1,t=ϕx0,t=ϕx1,t=0,t≥0,unless we assume
(21)∫01u0xdx=0,∫01ϕ0xdx=0,respectively.In this paper, we consideru,ϕ,Y to be a solution of system (12)–(15) with the regularity needed to justify the calculations. In Section 2, we proved our decay result. And we symbolize that c is a positive constant.


## 2. Main Result

In this section, we prove our stability result for the energy of system (12)–(15).We need the following lemmas.Lemma 3.
The energy functionalE, defined by
(22)Et=12∫01ρuut2+a1ux2+ρϕϕt2+a3−∫0tgsdsϕx2+2a2uxϕxdx+12g∘ϕx+12∫01∫01∫τ1τ2σ∣β2σ∣Y2x,ρ,σ,tdσdρdx,satisfies
(23)E′t≤12g′∘ϕx−12gt∫01ϕx2dx−β1−∫τ1τ2∣β2σ∣dσ∫01ϕt2dx≤12g′∘ϕx−η0∫01ϕt2dx≤0,where η0=β1−∫τ1τ2β2σdσ>0 and
(24)g∘vxt=∫01∫0tgt−svxt−vxs2dsdx.Proof.
Multiplying (12)1,2 by ut and ϕt, then integration by parts over 0,1, with (15), gives
(25)12ddt∫01ρuut2+a1ux2+ρϕϕt2+a3ϕx2+2a2uxϕxdx+β1∫01ϕt2dx+∫01ϕt∫τ1τ2∣β2σ∣Yx,1,σ,tdσdx−∫01ϕxt∫0tgt−sϕxsdsdx=0.
The estimate of the last term in the LHS of (25) is as follows:
(26)−∫01ϕxt∫0tgt−sϕxsdsdx=∫01ϕxt∫0tgt−sϕxt−ϕxsdsdx−∫0tgsds∫01ϕxtϕxdx=12ddtg∘ϕx−12ddt∫0tgsds∫01ϕx2dx−12g′∘ϕx+12gt∫01ϕx2dx.
Now, multiplying ((12))3 by Y∣β2σ∣, and by integration over 0,1×0,1×τ1,τ2(27)ddt12∫01∫01∫τ1τ2σ∣β2σ∣Y2x,ρ,σ,tdσdρdx=−∫01∫01∫τ1τ2∣β2σ∣YYρx,ρ,σ,tdσdρdx=−12∫01∫01∫τ1τ2∣β2σ∣ddρY2x,ρ,σ,tdσdρdx=12∫01∫τ1τ2∣β2σ∣Y2x,0,σ,t−Y2x,1,σ,tdσdx=12∫τ1τ2∣β2σ∣dσ∫01ϕt2dx−12∫01∫τ1τ2∣σ2σ∣Y2x,1,σ,tdσdx.
Now, by substituting (26) into (25), and using Young’s inequality, we have
(28)E′t≤12g′∘ϕx−12gt∫01ϕx2dx−β1−∫τ1τ2∣β2σ∣dσ∫01ϕt2d≤12g′∘ϕx−β1−∫τ1τ2∣β2σ∣dσ∫01ϕt2dx,then, by (18), there exists η0>0 so that
(29)E′t≤12g′∘ϕx−η0∫01ϕt2dx,then we obtain (22) and (23) (E is a nonincreasing function).Lemma 4.
The functional(30)D1t≔ρϕ∫01ϕtϕdx−a2a1ρu∫01ϕutdx+β12∫01ϕ2dxsatisfies
(31)D1′t≤−a02∫01ϕx2dx+ε1∫01ut2dx+c1+1ε1∫01ϕt2dx+cg∘ϕx+c∫01∫τ1τ2∣β2σ∣Y2x,1,σ,tdσdx.Proof.
Direct computation using integration by parts and Young’s inequality, forε1>0, yields
(32)D1′t=−a3∫01ϕx2dx+ρϕ∫01ϕt2dx+a22a1∫01ϕx2dx+a2a1ρu∫01ϕtutdx+∫01ϕx∫0tgt−sϕxsdsdx−∫01ϕ∫τ1τ2∣β2σ∣Yx,1,σ,tdσdx≤−a3−a22a1∫01ϕx2dx+ρϕ∫01ϕt2dx+a2a1ρu∫01ϕtutd−∫01ϕ∫τ1τ2∣β2σ∣Yx,1,σ,tdσdx+∫01ϕx∫0tgt−sϕxsdsdx.
The estimate of the two last terms in the RHS of (32) is as follows:
(33)∫01ϕx∫0tgt−sϕxsdsdx=∫0tgsds∫01ϕx2dx−∫01ϕx∫0tgt−sϕxt−ϕxsdsdx≤δ1+∫0tgsds∫01ϕx2dx+14δ1∫0tgsdsg∘ϕx,(34)−∫01ϕ∫τ1τ2∣β2σ∣Yx,1,σ,tdσdx≤cδ2∫τ1τ2∣β2σ∣dσ∫01ϕx2dx+14δ2∫0t∫τ1τ2∣β2σ∣Y2x,1,σ,tdσdx,where we have used Cauchy-Schwartz, Young’s, and Poincaré’s inequalities, for δ1,δ2>0, and (18).
By substituting (33) and (34) into (32), we find
(35)D1′t≤−a3−a22a1−∫0tgsds−δ1−β1cδ2∫01ϕx2dx+ε1∫01ut2dx+c1+1ε1∫01ϕt2dx+14δ1∫0tgsdsg∘ϕx+14δ2∫0t∫τ1τ2∣β2σ∣Y2x,1,σ,tdσdx.
Bearing in mind thata=a3−a22/a1>0 and using (16), we get
(36)∫0tgsds<∫0∞gsds<a,let a0=a3−a22/a1−∫0tgsds>0, and letting δ1=a0/4, δ2=a0/4cμ1, gives (31).Lemma 5.
Assume that (16) hold. Then, the functional
(37)D2t≔a2∫01ϕtudx−∫01ϕutdxsatisfies,
(38)D2′t≤−a222ρϕ∫01ux2dx+c∫01ϕx2dx+c∫01ϕt2dx+cg∘ϕx+c∫01∫τ1τ2∣β2σ∣Y2x,1,σ,tdσdx.Proof.
By differentiatingD2, then using (12), integration by parts, and (15), we find
(39)D2′t=−a22ρϕ∫01ux2dx+a22ρu∫01ϕx2dx−a2a3ρϕ−a1a2ρu∫01ϕxuxdx−a2β1ρϕ∫01uϕtdx+a2ρϕ∫01ux∫0tgt−sϕxsdsd−a2ρϕ∫01u∫τ1τ2∣β2σ∣Yx,1,σ,tdσdx.
In what follows, we estimate the different terms in the RHS of (39); we use Young’s, Cauchy-Schwartz, and Poincaré’s inequalities. For δ3,δ4,δ5>0, we have
(40)−a2a3ρϕ−a1a2ρu∫01ϕxuxdx≤δ3∫01ux2dx+a2a3ρϕ−a1a2ρu214δ3∫01ϕ2dx,a2ρϕ∫01ux∫0tgt−sϕxsdsdx≤2δ4∫01ux2dx+c4δ4∫01ϕx2dx+cδ4g∘ϕx,−a2β1ρϕ∫01uϕtdx≤cδ5∫01ux2dx+c4δ5∫01ϕt2dx,−a2ρϕ∫01u∫τ1τ2∣β2σ∣Yx,1,σ,tdσdx≤cδ6∫01ux2dx−c4δ6∫01∫τ1τ2∣β2σ∣Y2x,1,σ,tdσdx.
By lettingδ3=a2/8ρϕ,δ4=a2/16ρϕ,δ5=δ6=a2/8cρϕ and substituting into (39), we get (38).Lemma 6.
The functional(41)D3t≔−ρu∫01utudxsatisfies
(42)D3′t≤−ρu∫01ut2dx+2a1∫01ux2dx+a34∫01ϕx2dx.Proof.
Direct computations give(43)D3′t=−ρu∫01ut2dx+a1∫01ux2dx+a2∫01uxϕxdx.
Estimate (42) easily follows by using Young’s inequality and a1a3>a22.Now, let us introduce the following functional used.Lemma 7.
The functional(44)D4t≔∫01∫01∫τ1τ2σe−σρ∣β2σ∣Y2x,ρ,σ,tdσdρdxsatisfies
(45)D4′t≤−η1∫01∫01∫τ1τ2σ∣β2σ∣Y2x,ρ,σ,tdσdρdx+β1∫01ϕt2dx−η1∫01∫τ1τ2∣β2σ∣Y2x,1,σ,tdσdx,where η1>0.Proof.
By differentiatingD4, with respect to t and using the last equation in (12), we have
(46)D4′t=−2∫01∫01∫τ1τ2e−σρ∣β2σ∣YYρx,ρ,σ,tdσdρdx=−∫01∫01∫τ1τ2σe−σρ∣β2σ∣Y2x,ρ,σ,tdσdρdx−∫01∫τ1τ2∣β2σ∣e−σY2x,1,σ,t−Y2x,0,σ,tdσdx.
By usingYx,0,σ,t=ϕtx,t, and e−σ≤e−σρ≤1, for all 0<ρ<1, we find
(47)D4′t=−η1∫01∫01∫τ1τ2σ∣β2σ∣Y2x,ρ,σ,tdσdρdx−∫01∫τ1τ2e−σ∣β2σ∣Y2x,1,σ,tdσdx+∫τ1τ2∣β2σ∣dσ∫01ϕt2dx.Because−e−σ is an increasing function, we have −e−σ≤−e−τ2, for all σ∈τ1,τ2.Finally, settingη1=e−τ2 and recalling (18) give (45). We are now ready to prove the main result.Theorem 8.
Assume (16)–(18) hold.
Then,∀t0>0, there exist λ1,λ2>0 such that the energy functional given by (22) satisfies
(48)Et≤λ1e−λ2∫t0tϑsds,∀t≥t0.Proof.
We define the functional of Lyapunov(49)Lt≔NEt+N1D1t+N2D2t+D3t+N4D4t,where N,N1,N2,N4>0 to be selected later.
By differentiating (49) and using (22), (31), (38), (42), and (45), we have
(50)L′t≤−a0N12−cN2−a34∫01ϕx2dx−ρu−N1ε1∫01ut2dx−a22N22ρϕ−2a1∫01ux2dx+cN1+N2g∘ϕx+N2g′∘ϕx−η0N−cN11+1ε1−N2c−β1N4∫01ϕt2dx−N4η1−cN1−cN2∫01∫τ1τ2∣β2σ∣Y2x,1,σ,tdσdx−N4η1∫01∫01∫τ1τ2σ∣β2σ∣Y2x,ρ,σ,tdσdρdx.
By setting(51)ε1=ρu2N1,we obtain
(52)L′t≤−a0N12−cN2−a34∫01ϕx2dx−ρu2∫01ut2dx−a22N22ρϕ−2a1∫01ux2dx+cN1+N2g∘ϕx+N2g′∘ϕx−η0N−cN1−N2c−β1N4∫01ϕt2dx−N4η1−cN1−cN2∫01∫τ1τ2∣β2σ∣Y2x,1,σ,tdσdx−N4η1∫01∫01∫τ1τ2σ∣β2σ∣Y2x,ρ,σ,tdσdρdx.
At this stage, we choose our different constants.
First, choosingN2 large enough such that
(53)α1=a22N22ρϕ−2a1>0,then we pick N1 large enough such that
(54)α2=a0N12−cN2−a34>0,then we select N4 large enough such that
(55)α3=N4η1−cN1−cN2>0.
Thus, we arrive at(56)L′t≤−α2∫01ϕx2dx−ρu2∫01ut2dx−α1∫01ux2dx−η0N−c∫01ϕt2dx+α6g′∘ϕx+α7g∘ϕx−α3∫01∫τ1τ2∣β2σ∣Y2x,1,σ,tdσd−α5∫01∫01∫τ1τ2σ∣β2σ∣Y2x,ρ,σ,tdσdρdx,where α5=η1N4,α6=N/2,α7=cN1+N2.
On the other hand, if we let(57)Lt=N1D1t+N2D2t+D3t+N4D4t,then
(58)Lt≤N1ρϕ∫01ϕϕtdx+N1a2a1ρu∫01ϕutdx+N1μ12∫01ϕ2dx+N2a2∫01ϕut−uϕtdx+ρu∫01utudx+N4∫01∫01∫τ1τ2σe−σρ∣β2σ∣Y2x,ρ,σ,tdσdρdx.
Exploiting Young’s, Cauchy-Schwartz, and Poincaré inequalities, we obtain(59)Lt≤c∫01ut2+ϕt2+ϕx2+ux2dx+cg∘ϕx+c∫01∫01∫τ1τ2σ∣β2σ∣Y2x,ρ,σ,tdσdρ.
On the other hand, from (22), we can write
(60)a1ux2+2a2ϕxux+a4ϕx2=12a1ux+a2a1ϕx2+a4ϕx+a2a4ux2+a1−a22a4ux2+a4−a22a1ϕx2,where
(61)a4=a3−∫0tgsds.
Sincea1a3>a22 and (16), we deduce that
(62)a1ux2+2a2ϕxux+a4ϕx2>12a1−a22a4ux2+a4−a22a1ϕx2.
Consequently, we find(63)Lt=Lt−NEt≤cEt,that is,
(64)N−cEt≤Lt≤N+cEt.
At this point, we chooseN large enough such that
(65)N−c>0,Nη0−c>0,and exploiting (22), estimates (56) and (64), respectively, leads to
(66)c2Et≤Lt≤c3Et,∀t≥0,(67)L′t≤−k1Et+k2g∘ϕx,∀t≥t0,for some k1,k2,c2,c3>0.
By multiplying (67) by ϑt, we get
(68)ϑtL′t≤−k1ϑtEt+k2ϑtg∘ϕx,∀t≥t0.
Now, by using (17), we have the following estimate:
(69)ϑtg∘ϕx=ϑt∫01∫0tgt−sϕxt−ϕxs2dsdx≤∫01∫0tϑt−sgt−sϕxt−ϕxs2dsd≤−∫01∫0tg′t−sϕxt−ϕxs2dsdx=−g′∘ϕx≤−2E′t.
Thus, (68) becomes
(70)ϑtL′t≤−k1ϑtEt−2k2E′t,∀t≥t0,which can be rewritten as
(71)ϑtLt+2k2Et′−ϑ′tLt≤−k1ϑtEt,∀t≥t0.
By usingϑ′t≤0,∀t≥0, we have
(72)ϑtLt+2k2Et′≤−k1ϑtEt,∀t≥t0.
By exploiting (66), we notice that
(73)Kt=ϑtLt+2k2Et~Et.
Consequently, forκ>0, we get
(74)K′t≤−κKtϑt,∀t≥t0.
Integrating (74) over t0,t gives
(75)Kt≤Kt0e−κ∫t0tϑsds,∀t≥t0.
Consequently, (48) is established by virtue of (66) and (75).Remark 9.
The estimate (48) also remains valid for t∈0,t0, thanks to the boundedness and continuity of E and ϑ.



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