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Research article

Theoretical and numerical stability results for a viscoelastic swelling porous-elastic system with past history

  • Received: 08 April 2021 Accepted: 08 August 2021 Published: 16 August 2021
  • MSC : 35B40, 45K05, 74S05, 93D15, 93D20

  • The purpose of this paper is to establish a general stability result for a one-dimensional linear swelling porous-elastic system with past history, irrespective of the wave speeds of the system. First, we establish an explicit and general decay result under a wider class of the relaxation (kernel) functions. The kernel in our memory term is more general and of a broader class. Further, we get a better decay rate without imposing some assumptions on the boundedness of the history data considered in many earlier results in the literature. We also perform several numerical tests to illustrate our theoretical results. Our output extends and improves some of the available results on swelling porous media in the literature.

    Citation: Adel M. Al-Mahdi, Mohammad M. Al-Gharabli, Mohamed Alahyane. Theoretical and numerical stability results for a viscoelastic swelling porous-elastic system with past history[J]. AIMS Mathematics, 2021, 6(11): 11921-11949. doi: 10.3934/math.2021692

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  • The purpose of this paper is to establish a general stability result for a one-dimensional linear swelling porous-elastic system with past history, irrespective of the wave speeds of the system. First, we establish an explicit and general decay result under a wider class of the relaxation (kernel) functions. The kernel in our memory term is more general and of a broader class. Further, we get a better decay rate without imposing some assumptions on the boundedness of the history data considered in many earlier results in the literature. We also perform several numerical tests to illustrate our theoretical results. Our output extends and improves some of the available results on swelling porous media in the literature.



    Over the few last decades, a great number of scientists and mathematicians have got interested in the theory of elastic materials with voids (porous materials), see Figure 1. This theory is considered to be a simple extension of the classical theory of elasticity. See in this regard the concept of granular materials with voids introduced by Goodman and Cowin [1] and the idea of Nuziato and Cowin [2] related to the nonlinear theory of elastic materials with voids. The importance of materials with microstructure has been demonstrated by the very large number of published work and the various applications in the fields of petroleum industry, material science, biomedical, etc. Scientists investigated the coupling of macroscopic and microscopic structures in the porous material and its strength. Moreover, the stability has been studied by adding some dissipation mechanisms at the microscopic and/or macroscopic level. We start with the pioneer contribution of Quintanilla [3] in 2003, where he considered the following problem

    {ρ0utt=μuxx+βϕx,x(0,π),t>0,ρ0κϕtt=αϕxxβuxτϕtξϕ,x(0,π),t>0,u(x,0)=u0(x),ϕ(x,0)=ϕ0(x),x(0,π),ut(x,0)=u1(x),ϕt(x,0)=ϕ1(x),x(0,π),u(0,t)=u(π,t)=ϕ(0,t)=ϕ(π,t)=0,t0, (1.1)
    Figure 1.  Cracking and uneven foundation.

    where u,ϕ, and ρ0,κ>0 are the longitudinal displacement, the volume fraction, the mass density, and the equilibrated inertia respectively; and μ,α,β,τ,ξ are positive constitutive constants which satisfy μξ>β2. He showed that the damping in the porous equation (τϕt) is not strong enough to establish an exponential decay. However, a slow decay was obtained. Casas and Quintanilla [4] looked into a thermo-porous system of the form

    {ρ0utt=μuxx+bϕxβθx,x(0,π),t>0,ρ0κϕtt=αϕxxbuxτϕtξϕ+mθ,x(0,π),t>0,cθt=κθxxβuxtmϕt,x(0,π),t>0,u(x,0)=u0(x),ϕ(x,0)=ϕ0(x),θ(x,0)=θ0(x)x(0,π),ut(x,0)=u1(x),ϕt(x,0)=ϕ1(x),x(0,π),u(0,t)=u(π,t)=ϕ(0,t)=ϕ(π,t)=0,t0, (1.2)

    where θ is the temperature difference and the other parameters are as defined above. Under the same conditions, the authors showed that the presence of the macro-temperature and the porous dissipations acting simultaneously are able to stabilize the system exponentially. In addition, the same authors, in [5], considered (1.2), with τ=0, and proved that heat effect alone is not strong enough to obtain an exponential decay. In this case, a slow decay was established. But, adding a micro-temperature to the system (1.1) with τ=0 leads to an exponential decay. Managa and Quintanilla [6] studied diverse one-dimensional porous systems and established many slow and exponential decay results. The summary of their result is that the exponential decay can only be obtained if two dissipations from the macroscopic and microscopic equations are combined. Otherwise, only the slow decay can be obtained. In [7], Pamplona et al. studied the problem

    {ρ0utt=μuxx+bϕxβθx+γuxxt,x(0,π),t>0,ρ0κϕtt=αϕxxbuxτϕtξϕ+mθ,x(0,π),t>0,cθt=κθxxβuxtmϕt,x(0,π),t>0,u(x,0)=u0(x),ϕ(x,0)=ϕ0(x),θ(x,0)=θ0(x)x(0,π),ut(x,0)=u1(x),ϕt(x,0)=ϕ1(x),x(0,π),u(0,t)=u(π,t)=ϕ(0,t)=ϕ(π,t)=0,t0. (1.3)

    and showed that combining the strong damping in the elastic equation with the macro-temperature effect is not enough to obtain an exponential decay. However, for regular enough solutions, the decay is polynomial. Managa and Quintanilla [8] discussed several systems with quasi-static micro-voids and proved many slow and exponential decay theorems. Rivera and Quintanilla [9] improved the slow-decay results obtained for some systems in [8] and proved several polynomial results. They also showed that the rate of the polynomial decay can be improved for more regular solutions. Soufyane [10] used a viscoelastic damping in the porous equation together with a macro-temperature effect and established an exponential (respect. polynomial) decay for relaxation functions of exponential (respect. polynomial) decay. A similar result was also obtained by Soufyane et al. [11,12], for system (1.2) supplemented with the following boundary conditions:

    u(0,t)=ϕ(0,t)=θ(0,t)=θ(L,t)=0,t0,u(L,t)=t0g1(ts)[μux(L,s)+bϕ(L,s)]ds,t0,ϕ(L,t)=δt0g2(ts)ϕx(L,s)ds,t0,

    where g1 and g2 are relaxation functions of general type. They obtained a general decay result, from which the usual exponential and polynomial decay rates are particular cases. Pamplona et al. [13] treated a one-dimensional porous elastic problem with history and proved, in the absence of the porous or the elastic dissipation, the lack of exponential decay. Messaoudi and Fareh [14,15] considered the following viscoelastic porous system

    {ρ1utt=k(ϕx+ψ)x+θx,in(0,1)×IR+ρ2ψttαψxx+k(ϕx+ψ)θ+t0g(ts)ψxx(x,s)ds=0,in(0,1)×IR+ρ3θtκθxx+ϕxt+ψt=0,in(0,1)×IR+, (1.4)

    together with initial and boundary conditions. Here, ρ1,ρ2,ρ3,k,κ,α are positive constants and g:[0,+)(0,+) is a non-increasing differentiable function satisfying

    g(0)>0,t0g(s)ds=l>0,g(t)γ(t)g(t),t0,

    where γ(t) is a positive nonincreasing differentiable function. They established general decay results for the case of equal and non-equal speed of wave propagations. Also, Messaoudi and Fareh [16] considered the following

    {ρutt=μuxx+bϕxβθx,in(0,1)×IR+Jϕtt=αϕxxbuxξϕ+mθτϕt,in(0,1)×IR+cθt=qxβutxmϕt,in(0,1)×IR+τ0qt+q+κθx=0,in(0,1)×IR+, (1.5)

    together with initial and boundary conditions. They used the multiplier method and established an expoential decay result. See also a similar result by Han and Xu [17], where they established the well posedness and proved an exponential decay result using a very lengthy and detailed spectral theory approach. For the Cauchy problems, Said-Houari and Messaoudi [18] looked into the following

    {ρutt=μuxx+bϕxβθx+γuxxt,Jϕtt=αϕxxbuxaϕ+mθ,cθt=qxβutxmϕt,τ0qt+q+κθx=0,xIR,t>0 (1.6)

    together with initial data. Here ρ,μ,α,J,a are strictly positive constants such that μa>b2 and b, β and m are different from zero. They showed that the decay rate of the system is very slow and of regularity-loss type. Apalara [19] considered

    {ρuttμuxxbϕx=0,in(0,1)×IR+Jϕttδϕxx+bux+ξϕ+t0g(ts)ϕxx(x,s)ds=0,in(0,1)×IR+, (1.7)

    together with initial and boundary conditions where the relaxation function satisfied g(t)η(t)g(t) and established a general decay result under the assumption of equal-speed wave propagations. After that, Feng and Yin [20] extended the result of [19] to the case of non-equal wave speeds. Magana and Quintanilla [6] considered

    {ρuttμuxxγutxx=0,in(0,1)×IR+Jϕttδδϕxx+bux+ξϕ=0,in(0,1)×IR+, (1.8)

    together with initial and boundary conditions and proved that the viscoelasticity damping is not strong enough to bring about exponential stability. However, they showed that the presence of both porous-viscosity and viscoelasticity stabilizes the system exponentially. Some other forms of damping were also considered in the literature. See [21] for thermal damping, [5] for thermal damping with microtemperatures, [4] for porous-elasticity with thermal damping, [22] for porous dissipation with microtemperatures. For more recent stability results related to porous problems, we refer the reader to [23,24] and the references therein.

    Swelling (also called expansive) soils have been characterized under porous media theory. They contain clay minerals that attract and absorb water, which may lead to increase pressure. In architectural and civil engineering, swelling soils are considered to be sources of problems and harms. If the pressure of the soil is higher compared to the main structure, it could result in heaving. The more the initial dry density of a soil, the more its potential to swell due to capillary action that accompanies absorption of underground water or shrinkage due to dryness as a result of changes in weather condition. Swelling soils causes serious engineering problems. Estimates indicate that about 20–25% of land area in the United State is covered with such problematic soils with the accompanied economic loss of 5.5 to 7 billions USD in 2003 [25]. Hence, it is crucial to study the ways to annihilate or at least minimize such damages. Reader is referred to [26,27,28,29,30,31,32] for other details concerning swelling soil. As established by Ieșan [33] and simplified by Quintanilla [34], the basic field equations for the linear theory of swelling porous elastic soils are mathematically given by

    {ρzztt=P1xG1+F1ρuutt=P2x+G2+F1, (1.9)

    where the constituents z and u represent the displacement of the fluid and the elastic solid material, respectively. The positive constant coefficients ρz and ρu are the densities of each constituent. The functions (P1,G1,F1) represent the partial tension, internal body forces, and eternal forces acting on the displacement, respectively. Similar definition holds for (P2,G2,F2) but acting on the elastic solid. In addition, the constitutive equations of partial tensions are given by

    [P1P2]=[a1a2a2a3]A[zxux], (1.10)

    where a1,a3 are positive constants and a20 is a real number. The matrix A is positive definite in the sense that a1a3a22.

    Quintanilla [34] investigated (1.9) by taking

    G1=G2=ξ(ztut),F1=a3zxxt,F2=0,

    where ξ is a positive coefficient, with initial and homogeneous Dirichlet boundary conditions and obtained an exponential stability result. Similarly, Wang and Guo [35] considered (1.9) with initial and some mixed boundary conditions, taking

    G1=G2=0,F1=ρzγ(x)zt,F2=0,

    where γ(x) is an internal viscous damping function with a positive mean. They used the spectral method to establish an exponential stability result. Ramos et al. [36] looked into the following swelling porous elastic soils

    {ρzztta1zxxa2uxx=0,in(0,L)×IR+ρuutta3uxxa2zxx+γ(t)g(ut)=0,in(0,L)×IR+, (1.11)

    and established an exponential decay rate provided that the wave speeds of the system are equal. Recently, Apalara [37] considered the following

    ρzztta1zxxa2uxx=0,in(0,1)×(0,)ρuutta3uxxa2zxx+t0g(ts)uxx(x,s)ds=0,in(0,1)×(0,)z(x,0)=z0(x), zt(x,0)=z1(x), u(x,0)=u0(x), ut(x,0)=u1(x), x[0,1]u(0,t)=u(1,t)=z(0,t)=z(1,t)=0t0, (1.12)

    where the relaxation function satisfies the condition

    g(t)ξ(t)g(t),t0, (1.13)

    and established a general decay result. For more results in porous elasticity system, porous-thermo-elasticity systems, Timoshenko system and other systems, we refer the reader to see [38,39,40,41,42].

    Motivated by all the above works and the other works of some viscoelastic problems with infinite memory [43,44,45,46,47,48,49,50,51,52], we consider the following problem

    ρzztta1zxxa2uxx=0,in(0,1)×(0,)ρuutta3uxxa2zxx+0g(s)uxx(x,ts)ds=0,in(0,1)×(0,)z(x,0)=z0(x), zt(x,0)=z1(x), u(x,t)=u0(x,t), ut(x,0)=u1(x), x[0,1]u(0,t)=u(1,t)=z(0,t)=z(1,t)=0t0. (1.14)

    where the solution is (z,u) such that u and u represent the displacement of the fluid and the elastic solid material. The positive constant coefficients ρu and ρz are the densities of each constituent. The coefficients a1,a2 and a3 are positive constants satisfying specific conditions.

    Our main objectives

    We consider Problem (1.14) and intend to establish a four fold objective:

    ● extend many earlier works such as the ones in [37] from finite memory to infinite memory.

    ● prove a general decay estimate for the solution of Problem (1.14) with a wider class of relaxation functions; that is

    g(t)ξ(t)Ψ(g(t)),t0, (1.15)

    where ξ and Ψ are two functions that satisfy some conditions to be specified later.

    ● get a better decay rate without imposing some assumptions on the boundedness of initial data considered in many papers in the literature such as the ones assumed in [48,53,54].

    ● perform some numerical tests to illustrate our theoretical results.

    We establish our result by using the multiplier method and some convexity properties. In fact, the result in the present paper is essential to engineers and architects in order to plan and to ensure safe construction.

    The rest of this paper is organized as follows. In section 2, we present some assumptions and material needed for our work. Some technical lemmas are presented and proved in section 3. We state and prove our main decay result and provide some conclusions in section 4. The numerical illustrations are presented in section 5.

    In this section, we state some assumptions needed in the proof of our results. Throughout this paper, c is used to denote a generic positive constant. For the relaxation function g, we assume the following:

    (A) g:[0,+)(0,+) is a C1 nonincreasing function satisfying

    g(0)>0and0<:=+0g(s)ds<a3a22a1, (2.1)

    In addition, there exists a C1 function Ψ:R+R+ which is linear or it is strictly increasing and strictly convex C2 function on (0,r] for some r>0 with Ψ(0)=Ψ(0)=0, lims+Ψ(s)=+, ssΨ(s) and ss(Ψ)1(s) are convex on (0,r]. Moreover, there exists a positive nonincreasing differentiable function ξ such that

    g(t)ξ(t)Ψ(g(t)),t0, (2.2)

    Remark 2.1. [55] If Ψ is a strictly increasing, strictly convex C2 function over (0,r] and satisfying Ψ(0)=Ψ(0)=0, then it has an extension ¯Ψ, that is also strictly increasing and strictly convex C2 over (0,). For example, if Ψ(r)=a,Ψ(r)=b,Ψ(r)=c, and for t>r, ¯Ψ can be defined by

    ¯Ψ(t)=c2t2+(bcr)t+(a+c2r2br). (2.3)

    For simplicity, in the rest of this paper, we use Ψ instead of ¯Ψ

    Remark 2.2. Since Ψ is strictly convex on (0,r] and Ψ(0)=0, then

    Ψ(θz)θΨ(z),0θ1andz(0,r]. (2.4)

    In this section, we establish some essential lemmas needed for the proof of our stability result.

    Lemma 3.1. The energy functional E, defined by

    E(t)=1210[ρzz2t+a1z2x+ρuu2t+(a30g(s)ds)u2x+2a2zxux]dx+12(gux)(t), (3.1)

    where

    (gv)(t)=100g(s)|v(t)v(ts)|2 dsdx, (3.2)

    and satisfies

    E(t)=12(gux)(t)0. (3.3)

    Proof. By multiplying the first two equations of (1.14) by zt and ut, respectively, and then, integrating by parts over (0,1), we end up with (3.3).

    Lemma 3.2. ([55]) For all uH10([0,1]),

    10(0g(s)|ux(t)ux(ts)|ds)2dxCα(hux)(t), (3.4)

    for any 0<α<1,

    Cα=(t0g2(s)αg(s)g(s)ds)andh(t)=αg(t)g(t). (3.5)

    Lemma 3.3. There exists a positive constant M1 such that

    10tg(s)(ux(t)ux(ts))2dsdxM1h0(t), (3.6)

    where h0(t)=0g(t+s)(1+||u0x(s)||2)ds.

    Proof.

    10tg(s)||ux(t)ux(ts)||2ds2||ux(t)||2tg(s)ds+2tg(s)||ux(ts)||2ds2sups0||ux(s)||20g(t+s)ds+20g(t+s)||ux(s)||2ds(8a22a1sups0E(s))0g(t+s)ds+20g(t+s)||u0x(s)||2ds(8a22a1E(0))0g(t+s)ds+20g(t+s)||u0x(s)||2dsM10g(t+s)(1+||u0x(s)||2)ds, (3.7)

    where M1=max{2,(8a22a1E(0))}.

    Lemma 3.4. The functional

    F1(t):=ρu10utudxa2a1ρz10ztudx

    satisfies, for any ε1>0 and some constant a0,

    F1(t)a0210u2xdx+ε110z2tdx+(ρu+a22ρ2z4ε1a21)10u2tdx+Cα2a0(hux)(t), (3.8)

    where a0=a3a22a10g(s)ds.

    Proof. Direct computations using integration by parts gives

    F1(t)=(a3a22a1)10u2xdx+ρu10u2tdxa2ρza110ztutdx+10ux0g(s)ux(ts)dsdx. (3.9)

    Applying Young's inequality, we get for ε1>0,

    a2ρza110ztutdxε110z2tdx+a22ρ2z4ε1a2110u2tdx, (3.10)

    and for ε2>0,

    10ux0g(s)ux(ts)dsdx=0g(s)ds10u2xdx10ux0g(s)(ux(t)ux(ts))dsdx(0g(s)ds+ε22)10u2xdx+Cα2ε2(hux)(t). (3.11)

    Combining (3.9)–(3.11), we end up with

    F1(t)(a3a22a10g(s)dsε22)10u2xdx+ε110z2tdx+(ρu+a22ρ2zε1a21)10u2tdx+Cα2ε2(hux)(t). (3.12)

    Using assumption (A), taking ε2=a0, where a0=a3a22a10g(s)ds, we obtain (3.8).

    Lemma 3.5. Assume that (A) holds. Then, the functional

    F2(t):=ρu10ut0g(s)(u(t)u(ts))dsdx

    satisfies, for any δ2,δ3>0,

    F2(t)ρuc0210u2tdx+δ110u2xdx+δ3a2210z2xdx+[cCαδ1+cδ2(1+Cα)+Cαδ3+Cα](hux)(t). (3.13)

    Proof. Differentiating F2, taking into account (1.14), and using integrating by parts, we obtain

    F2(t)=ρu0g(s)ds10u2tdx+a310ux0g(s)(ux(t)ux(ts))dsdx100g(s)ux(s)ds0g(s)(ux(t)ux(ts))dsdxρu10ut0g(s)(u(t)u(ts))dsdx+a210zx0g(s)(ux(t)ux(ts))dsdx=ρu0g(s)ds10u2tdx+10(0g(s)(ux(t)ux(ts)))2dx+(a30g(s)ds)10ux0g(s)(ux(t)ux(ts))dsdx+a210zx0g(s)(ux(t)ux(ts))dsdxρu10ut0g(s)(u(t)u(ts))dsdx. (3.14)

    Using Young's inequality and Lemma 3.2, for any δ1>0, we obtain

    (a30g(s)ds)10ux0g(s)(ux(t)ux(ts))dsdxδ110u2x(t)dx+cCαδ1(hux)(t). (3.15)

    Similarly, we can get for any δ2>0

    ρu10ut(t)0g(s)(z(t)z(ts))dsdx=ρu10ut(t)0h(s)(u(t)u(ts))dsdxρu10ut(t)0αg(s)(u(t)u(ts))dsdxδ210u2t(t)dx+ρ2u(0h(s) ds)2δ2(hu)(t)+ρ2uα22δ210(0g(s)(u(t)u(ts)ds)2dxδ210u2t(t)dx+c2δ2(hux)(t)+α2cCα2δ2(hux)(t)δ210u2t(t)dx+cδ2(1+Cα)(hux)(t).

    Using Young's inequality and performing similar calculations as in (3.15), we obtain for any δ3>0

    a210uxt0g(ts)(ux(t)ux(s))dsdxδ3a2210u2x dx+Cαδ3(hux)(t). (3.16)

    Combining all the above estimates, we get

    F2(t)ρu(0g(s)dsδ2)10u2tdx+δ210u2xdx+δ3a2210z2xdx+[cCαδ1+cδ2(1+Cα)+Cαδ3+Cα](hux)(t). (3.17)

    By taking δ2=ρuc02, we end up with the desired inequality (3.13).

    Lemma 3.6. Assume that (A) holds. Then the functional

    F3(t):=a210(ρzuztρuzut)dx

    satisfies

    F3(t)a22210z2xdx+(a22+32(a21+a23)+32)10u2xdx+32(1)(hux)(t)+2η110u2tdx+a22η1(ρ2u+ρ2z)10z2tdx. (3.18)

    Proof. Direct computations give

    F3(t)=a2210z2xdxa2210u2xdx+a2(a3a1)10uxzxdxa210zx0g(s)ux(ts)dsdx+a2(ρzρu)10utztdx. (3.19)

    Using Young's inequality,

    a2(a1a3)10uxzxdxa22310z2xdx+32(a21+a23)210u2xdx.

    Similarly, we can get for any η1>0,

    a2(ρzρu)10utztdxη1210u2tdx+(a22ρ2zη1+a22ρ2uη1)10z2tdx. (3.20)

    Exploiting Young's inequality again, we obtain

    a210zx0g(s)ux(ts)dsdxa22610z2xdx+3210(0g(s)(|ux(ts)||ux(t)+|ux(t)|)ds)2dx.

    Recalling that 0g(s)ds= and exploiting Young's and Cauchy-Schwarz inequalities and using (a+b)2(1+η)a2+(1+1η)b2, η>0, we obtain

    3210(0g(s)(|ux(ts)|ux(t)|+|ux(t)|)ds)2dx32(1+η2)210u2xdx+32(1+1η2)10(0g(s)|ux(t)ux(ts)|ds)2dx. (3.21)

    By taking η2=1, and aiming to lemma 3.2, we obtain

    a210zx(t)t0g(ts)ux(s)dsdxa22610u2x dx+3210z2xdx+32(1)Cα(hux)(t).

    Estimate (3.18) follows by combining all the above estimates.

    Lemma 3.7. The functional

    F4(t):=ρz10ztzdx

    satisfies, for any ε3>0,

    F4(t)ρz10z2tdx+(a1+a22ε3)10z2xdx+ε310u2xdx. (3.22)

    Proof. It is straightforward to see that

    F4(t)=ρz10z2tdx+a110z2xdx+a210zxuxdx.

    Using Young's inequality and the fact that a1a3>a22, we end up with (3.22).

    Lemma 3.8. ([55]) Under the assumption (A), the functional

    F5(t):=Ωt0r(ts)|ux(s)|2dsdx, (3.23)

    satisfies the estimate

    F5(t)12(gux)(t)+310u2x dx+1210+tg(s)(ux(t)ux(ts)ds)2dsdx, (3.24)

    where r(t)=+tg(s)ds.

    Proof. The proof can be achieved by following the same calculations in [40].

    Lemma 3.9. The functional L defined by

    L(t)=μE(t)+4i=1μiFi(t), (3.25)

    satisfies, for suitable choice of μ,μ1,μ2,μ3,μ4 and for all t0,

    L(t)410u2x dx10z2x dx10u2t dx10z2t dx+14(gux)(t), (3.26)

    and

    L(t)E(t). (3.27)

    Proof. By taking the derivative of the functional L and using the above estimates, we get

    L(t)μα2(gux)(t)(μ3a222μ2δ3a22μ4[a1+a22ϵ3])10z2xdx(μ1a02μ2δ1μ3[32(a21+a23+)+a22]μ4ϵ3)10u2x dx(μ2ρzc02μ1[ρu+a22ρ2z4ϵ1a21]2η1μ3)10u2tdx(μ4ρzμ1ϵ1μ3[a22ρ2zη1+a22ρ2uη1])10z2tdx(μ2μ1Cα2a0μ2[cCαδ1+cδ2(1+Cα)+Cαδ3+Cα]3μ32(1)Cα)(hux)(t).

    Firstly, we select μ4 large enough such that

    β1:=μ4ρzμ3[a22ρ2zη1+a22ρ2uη1]1>0. (3.28)

    Next, we take μ3 large enough such that

    β2:=μ3a222μ4[a3+a22ϵ3]1>0, (3.29)

    and then, we choose μ1 so large that

    β3:=μ1a02μ3[32(a21+a23+)+a22]4>0. (3.30)

    After fixing μ1, we select μ2 large enough such that

    β4:=μ2ρzc02μ1[ρu+a224ϵ1a21ρ2z]1>0. (3.31)

    Now, for any fixed μ1,μ2,μ3,μ4>0, we pick δ3<β22μ2a22, δ1<β34μ2, ϵ3<β34μ4, ϵ1<β12μ1 and η1<β44μ3, so that the following estimates are satisfied

    μ3a222μ2δ3a22μ4[a3+a22ϵ3]>1, (3.32)
    μ1a02μ2δ1μ3[32(a21+a23+)+a22]μ4ϵ3>4, (3.33)
    μ4ρzμ1ϵ1μ3[a22ρ2zη1+a22ρ2uη1]>1. (3.34)

    Since αg2(s)αg(s)g(s)<g(s), using the Lebesgue dominated convergence theorem, we can get

    αCα=0αg2(s)αg(s)g(s) ds0,asα0. (3.35)

    Hence, there exists some 0<α<1, such that if α<α, then

    αCα<18(μ12a0+μ2[cδ1+cδ2+1δ3+1]+μ332(1)). (3.36)

    By putting α=12μ and choosing μ sufficiently large such that

    μ4cμ2δ2>0, (3.37)

    which gives

    μ2cμ2δ2Cα(μ12a0+μ2[cδ1+cδ2+1δ3+1]+3μ32(1))>0. (3.38)

    Hence, we conclude that (3.26) holds. Moreover, we can choose μ even larger (if needed) so that (3.27) is satisfied, which means that, for some constants α1,α2>0,

    α1E(t)L(t)α2E(t).

    Lemma 3.10. Assume that (A) holds. Then, the energy functional satisfies, for all tR+, the following estimate

    t0E(s)ds<˜mh1(t), (3.39)

    where h1(t)=(1+t0h0(s)ds) and h0 is defined in Lemma (3.3).

    Proof. Let F(t)=L(t)+F5(t), then using (3.26) and (3.23), we obtain for all tR+,

    F(t)10u2x dx10z2x dx10u2t dx10z2t dx14(gux)(t)λE(t)+1210+tg(s)(ux(t)ux(ts))2dsdx, (3.40)

    where λ is some positive constant. Therefore,

    λt0E(s)dsF(0)F(t)+M12t0+0g(τ+s)(1+|u0x(s)|ds)2dτdsF(0)+M12t0h0(s)ds. (3.41)

    Therefore, (3.39) is established with ˜m=max{F(0)λ,M12λ}.

    Corollary 3.11. There exists 0<q0<1 such that, for all t0, we have the following estimate:

    t0g(s)10|ux(t)ux(ts)|2dxds1q(t)Ψ1(q(t)μ(t)ξ(t)), (3.42)

    where Ψ is defined in Remark (2.1),

    μ(t):=t0g(s)10|ux(t)ux(ts)|2dxdscE(t), (3.43)

    and

    q(t):=q0h1(t)<1. (3.44)

    Proof. Using (3.1) and (3.39), we have

    q(t)10t0(ux(t)ux(ts))2dsdx2q(t)10t0(|ux(t)|2+|ux(ts)|2)dsdx(4a22q(t)a1)t0(E(t)+E(ts))dsdx(8a22q(t)a1)t0E(s)dsdx(8a22q(t)a1)˜mh1(t),tR+. (3.45)

    Thanks to (3.39), then for all t0 and for 0<q0<min{1,(a18a22˜m)}, we have

    q(t)t010|ux(t)ux(ts)|2dxds<1.

    Then, the rest of the proof of (3.42) is straightforward as the one in [56].

    In this section, we state and prove a new general decay result for our problem (1.14). We introduce the following functions:

    Ψ1(t):=1t1sΨ(s)ds, (4.1)
    Ψ2(t)=tΨ(t),Ψ3(t)=t(Ψ)1(t),Ψ4(t)=Ψ3(t). (4.2)

    Further, we introduce the class S of functions χ:R+R+ satisfying for fixed c1,c2>0 (should be selected carefully in (4.18)):

    χC1(R+),χ1,χ0, (4.3)

    and

    c2Ψ4[cdq(t)h0(t)]c1(Ψ2(Ψ5(t)χ(t))Ψ2(Ψ5(t))χ(t)), (4.4)

    where d>0, c is a generic positive constant which may change from line to line, h0 and q are defined in (3.6) and (3.44) and

    Ψ5(t)=Ψ11(c1t0ξ(s)ds). (4.5)

    Remark 4.1. According to the properties of Ψ introduced in (A), Ψ2 is convex increasing and defines a bijection from R+ to R+, Ψ1 is decreasing defines a bijection from (0,1] to R+, and Ψ3 and Ψ4 are convex and increasing functions on (0,r]. Then the set S is not empty because it contains χ(s)=εΨ5(s) for any 0<ε1 small enough. Indeed, (4.3) is satisfied (since (1.15) and (4.5)). On the other hand, we have q(t)h0(t) is nonincreasing, 0<Ψ51, and Ψ and Ψ4 are increasing, then (4.4) is satisfied if

    c2Ψ4[cdq0h0(0)]c1ε(Ψ(1ε)Ψ(1))

    which holds, for 0<ε1 small enough, since limt+Ψ(t)=+. But with the choice χ=εΨ5, (4.6) (below) does not lead to any stability estimate. The idea is to choose χ satisfy (4.3) and (4.4) such that (4.6) gives the best possible decay rate for E.

    Theorem 4.2. Assume that (A) holds, then there exists a strictly positive constant C such that, for any χ satisfying (4.3) and (4.4), the solution of (1.14) satisfies, for all t0,

    E(t)CΨ5(t)χ(t)q(t). (4.6)

    Proof. We start by combining (3.3), (3.6), (3.26), and (3.42), then, for some m>0 and for any t0, we have

    L(t)mE(t)+cq(t)Ψ1(q(t)μ(t)ξ(t))+ch0(t). (4.7)

    Without loss of generality, one can assume that E(0)>0. For ε0<r, let the functional F defined by

    F(t):=Ψ(ε0E(t)q(t)E(0))L(t),

    which satisfies FE. By noting that Ψ0, q0, E0 and using (4.7), we get

    F(t)=ε0(qE)(t)E(0)Ψ(ε0E(t)q(t)E(0))L(t)+Ψ(ε0E(t)q(t)E(0))L(t)mE(t)Ψ(ε0E(t)q(t)E(0))+cq(t)Ψ(ε0E(t)q(t)E(0))Ψ1(q(t)μ(t)ξ(t))+ch0(t)Ψ(ε0E(t)q(t)E(0)). (4.8)

    Let Ψ be the convex conjugate of Ψ in the sense of Young (see [57]), then

    Ψ(s)=s(Ψ)1(s)Ψ[(Ψ)1(s)],ifs(0,Ψ(r)] (4.9)

    and Ψ satisfies the following generalized Young inequality

    ABΨ(A)+Ψ(B),ifA(0,Ψ(r)],B(0,r]. (4.10)

    So, with A=Ψ(ε0E(t)q(t)E(0)) and B=Ψ1(q(t)μ(t)ξ(t)) and using (3.3) and (4.8)–(4.10), we arrive at

    F(t)mE(t)Ψ(ε0E(t)q(t)E(0))+cq(t)Ψ(Ψ(ε0E(t)q(t)E(0)))+c(μ(t)ξ(t))+ch0(t)Ψ(ε0E(t)q(t)E(0))mE(t)Ψ(ε0E(t)q(t)E(0))+cε0E(t)E(0)Ψ(ε0E(t)q(t)E(0))+c(μ(t)ξ(t))+ch0(t)Ψ(ε0E(t)q(t)E(0)). (4.11)

    So, multiplying (4.11) by ξ(t) and using (3.43) and the fact that ε0E(t)q(t)E(0)<r gives

    ξ(t)F(t)mξ(t)E(t)Ψ(ε0E(t)q(t)E(0))+cξ(t)ε0E(t)E(0)Ψ(ε0E(t)q(t)E(0))+cμ(t)+cξ(t)h0(t)Ψ(ε0E(t)q(t)E(0))ε0(mE(0)ε0c)ξ(t)E(t)E(0)Ψ(ε0E(t)q(t)E(0))cE(t)+cξ(t)h0(t)Ψ(ε0E(t)q(t)E(0)). (4.12)

    Consequently, recalling the definition of Ψ2 and choosing ε0 so that k=(mE(0)ε0c)>0, we obtain, for all tR+,

    F1(t)kε0ξ(t)(E(t)E(0))Ψ(ε0E(t)q(t)E(0))+cξ(t)h0(t)Ψ(ε0E(t)q(t)E(0))=kξ(t)q(t)Ψ2(ε0E(t)q(t)E(0))+cξ(t)h0(t)Ψ(ε0E(t)q(t)E(0)), (4.13)

    where F1=ξF+cEE and satisfies for some α1,α2>0.

    α1F1(t)E(t)α2F1(t). (4.14)

    Since Ψ2(t)=Ψ(t)+tΨ(t), then, using the strict convexity of Ψ on (0,r], we find that Ψ2(t),Ψ2(t)>0 on (0,r].

    Let d>0. Using the general Young's inequality (4.10) on the last term in (4.13) with A=Ψ(ε0E(t)q(t)E(0)) and B=[cdh0(t)], we have

    ch0(t)Ψ(ε0E(t)q(t)E(0))=dq(t)[cdq(t)h0(t)](Ψ(ε0E(t)q(t)E(0)))dq(t)Ψ3(Ψ(ε0E(t)q(t)E(0)))+dq(t)Ψ3[cdq(t)h0(t)]dq(t)(ε0E(t)q(t)E(0))(Ψ(ε0E(t)q(t)E(0)))+dq(t)Ψ4[cdq(t)h0(t)]dq(t)Ψ2(ε0E(t)q(t)E(0))+dq(t)Ψ4[cdq(t)h0(t)]. (4.15)

    Now, combining (4.13) and (4.15) and choosing d small enough so that k1=(kd)>0, we arrive at

    F1(t)kξ(t)q(t)Ψ2(ε0E(t)q(t)E(0))+dξ(t)q(t)Ψ2(ε0E(t)q(t)E(0))+dξ(t)q(t)Ψ4[cdq(t)h0(t)]k1ξ(t)q(t)Ψ2(ε0E(t)q(t)E(0))+dξ(t)q(t)Ψ4[cdq(t)h0(t)]. (4.16)

    Using the equivalent property in (4.14) and the nonincreasing of Ψ2, we have, for some d0=α1E(0)>0,

    Ψ2(ε0E(t)q(t)E(0))Ψ2(d0F1(t)q(t)).

    Letting F2(t):=d0F1(t)q(t) and recalling q0, then we arrive at,

    F2(t)d0q(t)(k1ξ(t)q(t)Ψ2(ε0E(t)q(t)E(0))+dξ(t)q(t)Ψ4[cdq(t)h0(t)]). (4.17)

    Then, (4.17) becomes for some constant c1=d0k1>0 and c2=d0d>0,

    F2(t)c1ξ(t)Ψ2(F2(t))+c2ξ(t)Ψ4[cdq(t)h0(t)]. (4.18)

    Since d0q(t) is is nonincreasing. Using the equivalent property F1E implies that there exists b0>0 such that F2(t)b0E(t)q(t). Let tR+ and χ(t) satisfying (4.3) and (4.4).

    If

    b0q(t)E(t)2Ψ5(t)χ(t), (4.19)

    then, we have

    E(t)2b0Ψ5(t)χ(t)q(t). (4.20)

    If

    b0q(t)E(t)>2Ψ5(t)χ(t), (4.21)

    then, for any 0st, we get

    b0q(s)E(s)>2Ψ5(t)χ(t), (4.22)

    since, q(t)E(t) is nonincreasing function. Therefore, for any 0st, we have

    F2(s)>2Ψ5(t)χ(t). (4.23)

    Using (2.4), 0<χ1 and the fact that Ψ2 is convex, we get, for any 0<ϵ11,

    Ψ2(ϵ1χ(s)F2(s)ϵ1Ψ5(s))=Ψ2(ϵ1χ(s)F2(s)ϵ1χ(s)Ψ5(s)χ(s))ϵ1χ(s)Ψ2(F2(s)Ψ5(s)χ(s)). (4.24)

    Recalling the definition of Ψ2, that is Ψ2(t)=tΨ(t), we obtain

    Ψ2(ϵ1χ(s)F2(s)ϵ1Ψ5(s))ϵ1χ(s)(F2(s)Ψ5(s)χ(s))Ψ(F2(s)Ψ5(s)χ(s))ϵ1χ(s)F2(s)Ψ(F2(s)Ψ5(s)χ(s))ϵ1χ(s)Ψ5(s)χ(s)Ψ(F2(s)Ψ5(s)χ(s)). (4.25)

    Now, using (4.23) and the fact that Ψ is increasing, for any 0st, we arrive at

    Ψ(F2(s)Ψ5(s)χ(s))<Ψ(F2(s)),Ψ(F2(s)Ψ5(s)χ(s))>Ψ(Ψ5(s)χ(s)). (4.26)

    Therefore, we have

    Ψ2(ϵ1χ(s)F2(s)ϵ1Ψ5(s))ϵ1χ(s)F2(s)Ψ(F2(s))ϵ1χ(s)Ψ5(s)χ(s)Ψ(Ψ5(s)χ(s)). (4.27)

    Now, we let

    F3(s)=ϵ1χ(s)F2(s)ϵ1Ψ5(s), (4.28)

    where ϵ1 small enough so that F3(0)1. Then, (4.24) becomes, for any 0st,

    Ψ2(F3(s))ϵ1χ(t)Ψ2(F2(s))ϵ1χ(t)Ψ2(Ψ5(s)χ(s)). (4.29)

    Further, we get

    F3(t)=ϵ1χ(t)F2(t)+ϵ1χ(s)F2(t)ϵ1Ψ5(t). (4.30)

    Since χ0 and using (4.18), then for any 0st, 0<ϵ11, we obtain

    F3(t)ϵ1χ(s)F2(t)ϵ1Ψ5(t)c1ϵ1ξ(t)χ(t)Ψ2(F2(t))+c2ϵ1ξ(t)χ(s)Ψ4[cdq(t)h0(t)]ϵ1Ψ5(t). (4.31)

    Then, using (4.29), we get

    F3(t)c1ξ(t)Ψ2(F3(t))c1ϵ1ξ(t)χ(t)Ψ2(Ψ5(s)χ(s))+c2ϵ1ξ(t)χ(t)Ψ4[cdq(t)h0(t)]ϵ1Ψ5(t). (4.32)

    From the definition of Ψ1 and Ψ5, we have

    Ψ1(Ψ5(s))=c1s0ξ(τ)dτ,

    hence,

    Ψ5(s)=c1ξ(s)Ψ2(Ψ5(s)). (4.33)

    Now, we have

    c2ϵ1ξ(t)χ(t)Ψ4[cdq(t)h0(t)]c1ϵ1ξ(t)χ(t)Ψ2(Ψ5(s)χ(s))ϵ1Ψ5(t)=c2ϵ1ξ(t)χ(t)Ψ4[cdq(t)h0(t)]c1ϵ1ξ(t)χ(t)Ψ2(Ψ5(s)χ(s))+c1ϵ1ξ(t)Ψ2(Ψ5(t))=ϵ1ξ(t)χ(t)(c2Ψ4[cdq(t)h0(t)]c1Ψ2(Ψ5(s)χ(s))+c1Ψ2(Ψ5(t))χ(t)). (4.34)

    Then, according to (4.4), we get

    ϵ1ξ(t)χ(t)(c2Ψ4[cdq(t)h0(t)]c1Ψ2(Ψ5(s)χ(s))c1Ψ2(Ψ5(t))χ(t))0.

    Then (4.28) gives

    F3(t)c1ξ(t)Ψ2(F3(t)). (4.35)

    Thus from (4.35) and the definition of Ψ1 and Ψ2 in (1.15) and (4.2), we obtain

    (Ψ1(F3(t)))c1ξ(t). (4.36)

    Integrating (4.36) over [0,t], we get

    Ψ1(F3(t))c1t0ξ(s)ds+Ψ1(F3(0)). (4.37)

    Since Ψ1 is decreasing, F3(0)1 and Ψ1(1)=0, then

    F3(t)Ψ11(c1t0ξ(s)ds)=Ψ5(t). (4.38)

    Recalling that F3(t)=ϵ1χ(t)F2(t)ϵ1Ψ5(t), we have

    F2(t)(1+ϵ1)ϵ1Ψ5(t)χ(t). (4.39)

    Similarly, recall that F2(t):=d0F1(t)q(t), then

    F1(t)(1+ϵ1)d0ϵ1Ψ5(t)χ(t)q(t). (4.40)

    Since F1E, then for some b>0, we have E(t)bF1; which gives

    E(t)b(1+ϵ1)d0ϵ1Ψ5(t)χ(t)q(t). (4.41)

    From (4.20) and (4.41), we obtain the following estimate

    E(t)c3(Ψ5(t)χ(t)q(t)), (4.42)

    where c3=max{2b0,b(1+ϵ1)d0ϵ1}.

    Example 1: Let g(t)=a(1+t)ν, where ν>1 and 0<a<ν1 so that (A) is satisfied. In this case ξ(t)=νa1ν and Ψ(t)=tν+1ν. Then, there exist positive constants ai(i=0,...,3) depending only on a,ν such that

    Ψ4(t)=a0tν+1ν,Ψ2(t)=a1tν+1ν,Ψ1(t)=a2(t1ν1),Ψ5(t)=(a3t+1)ν. (4.43)

    We will discuss two cases:

    Case 1: if

    m0(1+t)r1+||u0x||2m1(1+t)r, (4.44)

    where 0<r<ν1 and m0,m1>0, then we have, for some positive constants ai(i=4,...,7) depending only on a,ν,m0,m1,r, the following:

    a4(1+t)ν+1+rh0(t)a5(1+t)ν+1+r, (4.45)
    q0q(t)a6{1+ln(1+t),νr=2;2,νr>2;(1+t)ν+r+2,1<νr<2. (4.46)
    q0q(t)a7{1+ln(1+t),νr=2;2,νr>2;(1+t)ν+r+2,1<νr<2. (4.47)

    We notice that condition (4.4) is satisfied if

    (t+1)νq(t)h0(t)χ(t)a8(1(χ)1ν)νν+1, (4.48)

    where a8>0 depending on a,ν,c1 and c2. Choosing χ(t) as the following

    χ(t)=λ{(1+t)p,p=r+1νr2;(1+t)p,p=ν1,1<νr<2. (4.49)

    with 0<λ1, so that (4.3) is valid. Moreover, using (4.45) and (4.46), we see that (4.48) is satisfied if 0<λ1 is small enough, and then, (4.4) is satisfied. Hence (4.6) and (4.47) imply that, for any tR+

    E(t)a9{(1+ln(1+t))(1+t)(νr1),νr=2;(1+t)(νr1),νr>2;(1+t)(νr1),1<νr<2. (4.50)

    Thus, the estimate (4.50) gives limt+E(t)=0.

    Case 2: if m01+||u0x||2m1. That is r=0 in (4.44) (as it was assumed in [48,53,54], then (4.50) holds with r=0.

    In this section, some numerical experiments are performed to illustrate the energy decay results in Theorem 4.2. For this purpose, we use a finite element scheme in space, where for the time discretization, we use the Crank-Nicolson method, which is a second-order method in time that has the property to be unconditionally stable. The spatial interval (0,L)=(0,1) is subdivided into 100 subintervals, where the temporal interval (0,T) is subdivided into N with a time step Δt=T/N. The solution at the time step n+1 is given by

    {(M+γ1R)Zn+1=(2Mγ1R)ZnMZn1β1RUn(M+γ2R)Un+1=(2Mγ2R)UnMUn1β2RZn+1+α2RT0g(s)U(tn+1s)ds, (5.1)

    where

    γ1=a1Δt22ρz,β1=a2Δt2ρz,
    γ2=a3Δt22ρu,β2=a2Δt2ρu,α2=Δt2ρu,

    and the matrices M and R are the mass matrix and the stiffness matrix respectively.

    We run our code for N time steps (N=T/Δt) using the following initial conditions:

    z0(x)=sin(πx),u0(x)=2sin(πx),
    z1(x)=u1(x)=sin(π2x).

    For the numerical experiments, we choose the functional g as in the Example 1 in the previous section

    g(t)=a(1+t)ν,

    where ν>1 and 0<a<ν1.

    Based on the Example 1, we perform three tests in order to the energy satisfies

    E(t)Eh(t)foranytR+,

    where Eh given by (4.50) and the tests done as follows:

    Test 1: For the first numerical test, we choose the following entries:

    ρz=ρu=a1=a2=1,a3=2,

    and

    a=1,ν=4,r=2,

    where

    Eh(t)=a9(1+ln(1+t)(1+t)νr1).

    Test 2: For the second test, we choose the following entries:

    ρz=12,ρu=14,a1=a2=1,a3=2,

    and

    a=1,ν=6,r=3,

    where

    Eh(t)=a9(1(1+t)νr1).

    Test 3: Then, we consider the last case of (4.50) by choosing the following quantities:

    ρz=ρu=a1=a2=1,a3=2,

    and

    a=1,ν=3,r=3/2,

    where

    Eh(t)=a9(1(1+t)νr1).

    Test 4: Finally, we consider the case whene r=0 and keeping the same quantities of Test 3:

    ρz=ρu=a1=a2=1,a3=2,

    and

    a=1,ν=3,r=0,

    where

    Eh(t)=a9(1(1+t)νr1).

    In Figure 2, we show the cross section cuts of the numerical solution (z,u) at x=0.3, x=0.5 and at x=0.9 for Test 1. Figure 3 presents the evolution in time of the system energy E(t) compared with Eh(t) for Test 1, we made a zoom on the first part to show the difference between the curves. Next, by following the same process, we present the rest of numerical results in Figures 4 and 5 for Test 2, in Figures 6 and 7 for Test 3 and in Figures 8 and 9 for Test 4.

    Figure 2.  Test 1: the cross section of the approximate solution (z,u).
    Figure 3.  Test 1: energy decay.
    Figure 4.  Test 2: the cross section of the approximate solution (z,u).
    Figure 5.  Test 2: energy decay.
    Figure 6.  Test 3: the cross section of the approximate solution (z,u).
    Figure 7.  Test 3: energy decay.
    Figure 8.  Test 4: the cross section of the approximate solution (z,u).
    Figure 9.  Test 4: energy decay.

    The computational simulations show the decay behaviour for the solutions and the energy of the system (1.14), we examined four tests based on Example 1 (in both cases r0 and r=0). As a conclusion, we observed that the energy decay uniformly for all tests and satisfies (4.50) in Example 1, and moreover independently of the waves speeds of the system.

    In this work, we considered a viscoelastic swelling porous elastic system with an infinite memory term. We proved a general decay result with a large class of the relaxation functions associated with the memory term. The proof is based on the multiplier method and some properties of convex functions and without imposing some assumptions on the boundedness of the history data considered in many earlier results in the literature. We tested our decay results numerically through several numerical tests and we found the energy decay uniformly for all tests and satisfies our decay theory. Our result generalizes some earlier works on swelling porous media in the literature and it is significant to engineers and architects as it might help to attenuate the harmful effects of swelling soils swiftly.

    The authors thank KFUPM and University of Sharjah for their continuous support. The authors also thank the referee for her/his very careful reading and valuable comments. This work is funded by KFUPM, Grant No. SB201005.

    The authors declare that there is no conflict of interest regarding the publication of this paper.



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