# A Stability Result for a Swelling Porous System with Nonlinear Boundary Dampings

Authors: Adel M. Al-Mahdi; Salim A. Messaoudi; Mohammad M. Al-Gharabli
Journal: Journal of Function Spaces
 (2022)

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## Abstract

In this work, we consider a swelling porous system where the damping terms are on the boundary. We establish an explicit and general decay result, without imposing restrictive growth assumption near the origin on the damping terms. Our result allows a larger class of damping terms, and the usual exponential and polynomial decay estimates are only special cases. We also give some illustrative examples.


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## Body

## 1. Introduction

Soil swelling is one of the problems which has been studied under the porous media theory. The soils with highly swelling index have high swelling properties. In the swelling soils, clay minerals are commonly found there, and they attract and absorb the water which may lead to increase in pressure. When the swelling soil is exposed to the water, the water molecules are attracted into the interstices of the soil matrix. With the water drawn is getting increasing, the soil plates are forced apart due to the rise in pressure within the soil pores leading to swelling or heaving of the soils. So, swelling soils are considered to be one of the sources of problems in foundation design and construction. For more information in the soil swelling problems, we refer the reader to [1–4] and the references therein. The mathematical model of linear theory of swelling porous elastic soils was established by Ieşan [5] and simplified by Quintanilla [6] as follows:
(1)ρ1φtt=P1x−G1+F1,ρ2ψtt=P2x+G2+F2,where the functions P1,G1,F1 represent the partial tension, internal body forces, and external forces acting on the displacement, respectively. Similar definitions hold for P2,G2,F2 but acting on the elastic solid. The constituent φ represents the displacement of the fluid and ρ1>0 is the coefficient density of φ. Here, ψ is the elastic solid material and its coefficient density is ρ2>0. Moreover, the constitutive equations of partial tensions are given by
(2)P1P2=a1a2a2a3⏟Aφxψx,where a1,a3 are positive constants and a2≠0 is a real number. The matrix A is positive definite in the sense that a1a3≥a22.Quintanilla [6] investigated (1) by taking
(3)G1=G2=βφt−ψt,F1=a3φxxt,F2=0,and established an exponential stability result where β>0. Wang and Guo [7] considered (1) by taking
(4)G1=G2=0,F1=−ρ1ξxφt,F2=0,and they establish an exponential stability result by using spectral method, where ξx is an internal viscous damping function with a positive mean. Ramos et al. [8] studied (1), with a damping acting on the domain; that is,
(5)G1=G2=0,F1=0,F2=−γtgψt.They established an exponential decay rate. Recently, Apalara [9] looked into (1), with viscoelastic damping acting on the domain. So, he took
(6)G1=G2=0,F1=0,F2=−∫0tgt−sψxxsdsand established a general decay rate irrespective of the wave speed of the system. Very recently, Al-Mahdi et al. [10] also considered (1) with
(7)G1=G2=0,F1=0,F2=−∫0∞gsψxxt−sds,and established explicit and general decay results under a wider class of relaxation functions, and they also performed several numerical tests to illustrate their theoretical results. The reader can consult [11–20] and the references therein for some other interesting related results.In this paper, we are concerned with the following swelling system(8)ρ1φtt−a1φxx−a2ψxx=0,in0,1×0,∞,ρ2ψtt−a3ψxx−a2φxx=0,in0,1×0,∞,together with initial and boundary conditions:
(9)ψ0,t=φ0,t=0,a2φx1,t+a3ψx1,t=−h1ψt1,t,a1φx1,t+a2ψx1,t=−h2φt1,t,(10)φx,0=φ0x,φtx,0=φ1x,ψx,0=ψ0x,ψtx,0=ψ1x,where h1 and h2 are specific functions, a1,a3 are positive constants, and a2≠0 is a real number.The use of dampings at the boundary is well known in mechanical structure. It is employed to stabilize motions and absorb shock. This can be achieved by adding some controllers at the boundary (see [21, 22]). In this paper, we aim to establish an explicit and general decay rate result for system (8). We obtain our result without imposing any restrictive growth assumption near the origin on the damping functions. The result in this paper allows a larger class of functions h1 and h2, from which the energy decay rates are not necessarily of exponential or polynomial types (see the examples in Section 3).The proofs of our results are carried out, using the multiplier method and benefit from [22, 23] with necessary modifications dictated by the nature of our problem. To the best of our knowledge, this is the first work of this nature. For more works used the frictional damping acting in a part/whole domain or in the boundary, we point out to the work of [24–27].The rest of the paper is organized as follows: in Section2, we present some hypotheses and material needed for our work. Some essential lemmas and the statement with the proof of the decay result are given in Section 3.


## 2. Preliminaries

In this section, we present some material needed for the proofs of our main results.In the sequel, we assume that system (8) has a unique solution
(11)ψ,φ∈L∞ℝ+;H20,1∩H∩W1,∞ℝ+;H∩W2,∞ℝ+;L20,1,where H=f∈H10,1:f0=0. This result can be proved, for initial data in suitable function spaces, using standard arguments such as the Galerkin method.We consider the following hypotheses:A1.hi:ℝ⟶ℝ (for i=1,2) is a nondecreasing C1 function such that
(12)Λis≤his≤Λi−1sforalls≤r,i=1,2,c1s≤his≤c2sforalls≥r,where Λ1 and Λ2 are strictly increasing C1 functions on 0,+∞,Λ10=Λ20=0, and r,c1,c2 are positive constants.A2. The coefficientsa1,a3 are positive constants and a2≠0 is a real number such that
(13)a1a3−a22>0.Remark 1.
Hypothesis A1 implies thatshis>0, for all s≠0,i=1,2.Remark 2.
Hypothesis A1 was introduced by Lasiecka and Tataru [28].The following lemma will be of essential use in establishing the main result.Lemma 3 (see [23]).
LetF:ℝ+⟶ℝ+ be a nonincreasing function and β:ℝ+⟶ℝ+ be a strictly increasing C1 function, with βt⟶+∞ as t⟶+∞. Assume that there exist m,n≥0 and c>0 such that
(14)∫S∞β′tFt1+mdt≤cFS1+m+cFSβn,1≤S<+∞.
Then, there exist two positive constantsλ1 and λ2 such that
(15)Et≤λ1e−λ2βt∀t≥1,ifm=n=0,Et≤λ1βt1+n/m∀t≥1,ifm>0.
The energy functional associated with problem (8) is
(16)Et=12∫01ρ1φt2+a1φx2+ρ2ψt2+a3ψx2+2a2ψxφxdx.Lemma 4.
The energy functional satisfies, along the solution of (8),
(17)E′t=−ψth1ψt1,t−φth2φt1,t≤0.Proof.
By multiplying the two equations of (8) by φt and ψt, respectively, and then integrating over 0,1 with using both boundary and initial conditions (9) and (10), the estimate (17) is established.Throughout this paper, we will usec to denote a generic positive constant.


## 3. Stability

In this section, we state and prove our main stability result which reads as follows:Theorem 5.
Letψ,φ be the solution of (8) and assume that A1 and A2 hold. Then, there exists a constant c¯>0 such that, for t large, the solution of (8) satisfies
(18)Et≤c¯G−11t2,where
(19)Gs=sΛ1−1+Λ2−1−1s.Proof of Theorem8 will be carried out through several lemmas.Lemma 6.
Letψ,φ be the solution of (8); assume that A1 and A2 hold, and β:ℝ+⟶ℝ+ be a concave nondecreasing C2 function. Then, for T≥S≥0 and some positive constant c, the energy functional satisfies
(20)∫STβ′tE2tdt≤cE2S+c∫STβ′tEth12ψt1,tdt+h22φt1,tdt.Proof.
Multiplying the first equation in (8) by 3xφx−φβ′E and the second by 3xψx+ψβ′E, integrating over 0,1×S,T, and adding the results and recalling (16), we get
(21)∫STβ′tE2tdt=∫STβ′tEt32∫01ρ2ψt2dx−12∫01ρ1φt2dx+a3φx2dx−2∫01a2ψxφxdt−β′tEt∫013ρ1xφxφt+ρ2xψxψt+ρ2ψψt−ρ1φφtdxST+∫STβ′′E+β′E′t∫013ρ1xφxφt+ρ2ψxψt+ρ2ψψt−ρ1φφtdxdt+∫STβ′tEtφ1,th2φt1,tdt−N2∫STβ′tEtψ1,th1ψt1,tdt−3∫STβ′tEtφx1,th2φt1,tdt−N2∫STβ′tEtψx1,th1ψt1,tdt+a22∫STβ′tEtψx1,tφx1,tdt.
Using Young’s inequality, (16), and the properties of E and β′, we have
(22)∫STβ′E32∫01ρ2ψt2dx−12∫01ρ1φt2dx+a3ψx2dx−2∫01a2ψxφxdt≤cβ′SE2S.
Exploiting Young’s and Poincaré’s inequalities and (16), we obtain
(23)∫013ρ1xφxφt+ρ2xψxψt+ρ2ψψt−ρ1φφtdx≤C∫01φx2+φt2+ψx2+ψt2dx≤cEt.
Using the properties ofE and β′, we conclude
(24)−β′E∫013ρ1xφxφt+ρ2xψxψt+N2ρ2ψψt−ρ1φφtdxST≤Cβ′tE2tST≤cβ′SE2S.
As in the above calculations, we handle the third term in the right hand side of (21) as follows:
(25)∫STβ′′E+β′E′∫01N+1ρ1xφxφt+ρ2ψxψt+ρ2ψψt−ρ1φφtdxdt≤c∫STβ′′E2dt+c∫STβ′E′Edt≤CE2S∫STβ′′dt+cβ′S∫STE′Edt≤cβ′SE2S.
Using Cauchy-Schwarz’s inequality, we conclude that(26)ψ1,t2=∫01ψxdx2≤∫01ψx2dx.
Similarly, we have(27)φ1,t2≤∫01φx2dx.
Now, using (26) and Young’s inequality, we find that
(28)∫STβ′Eψ1,th1ψt1,tdt≤cε∫STβ′E∫01ψx2dxdt+cε∫STβ′Eh12ψt1,tdt.
Similarly, using (27) and Young’s inequality, we see that
(29)∫STβ′Eφ1,th2φt1,tdt≤cε∫STβ′E∫01φx2dxdt+cε∫STβ′Eh22φt1,tdt.
From (9), we have
(30)a1a2a2a3φx1,tψx1,t=−h1ψt1,t−h2φt1,t.
Therefore, we get(31)ψx=a2h2φt1,t−a1h1ψt1,ta1a3−a22,φx=a2h1ψt1,t−a3h2φt1,ta1a3−a22,where a1a3−a22>0 (as assumed in A2). Using (31) and Young’s inequality, we have
(32)−3∫STβ′Eφx1,th2φt1,tdt−∫STβ′Eψx1,th1ψt1,tdt+a22∫STβ′Eψx1,tφx1,tdt≤c∫STβ′Eh12ψt1,t+h22φt1,tdt.
We use (16), (28), (29), and (32), to estimate the last five terms in (21) as follows:
(33)∫STβ′Eφ1,th2φt1,tdt−∫STβ′Eψ1,th1ψt1,tdt−3∫STβ′Eφx1,th2φt1,tdt−∫STβ′Eψx1,th1ψt1,tdt+a22∫STβ′Eψx1,tφx1,tdt≤cβ′SE2S+c∫STβ′Eh12ψt1,t+h22φt1,tdt.
Combining (21)–(25) and (28)–(33), then (20) is established.Lemma 7.
Letψ,φ be the solution of (8); assume that A1 and A2 hold. Then, the energy functional satisfies
(34)Et≤cβ0t2∀t≥t1,where β0:R+⟶R+ is a strictly increasing C1 function, with β0t⟶+∞ as t⟶+∞, β0′t=Λ01/β0t, and Λ0≔Λ1−1+Λ2−1−1.Proof.
First, we define the following function:(35)χt≔1+∫1t1Λ01/sds,t≥t0,for some t0>max1,1/r. Then,
(36)χ′t=1Λ01/t>0∀t≥t0,χ′t⟶+∞ast⟶+∞,and χ′t is strictly increasing. Thus, χ is a convex and strictly increasing C2 function, with χt⟶+∞ as t⟶+∞. If we let
(37)β0≔χ−1,t≥t0,then it is easy to check that β0 is strictly increasing and β0′t=Λ01/β0t is strictly decreasing. So, β0 is a concave C2 function, with β0t⟶+∞ as t⟶+∞. We use this particular function β0 and take t1≥t0 such that β0′t1<r, to estimate the last integral in the right hand side of (20), for T≥S≥t1.
To estimate∫STβ′0Eh12ψt1,tdt, we consider the following cases:
(38)C1=t∈S,T:ψt1,t>r,C2=t∈S,T:ψt1,t≤randψt1,t≤β0′t,C3=t∈S,T:ψt1,t≤randψt1,t>β0′t.
Using A1, Remark2, (17), and the first case in (38), we have
(39)h12ψt1,t=h1ψt1,th1ψt1,t≤c2ψt1,th1ψt1,t≤c−E′t,which gives
(40)∫S,T∩C1β0′Eh12ψt1,tdt≤∫S,T∩C1cβ0′E−E′t≤∫STcβ0′E−E′tdt≤cβ0′SE2S.
Now, using A1, the properties ofΛ1−1 and Λ0−1, and the second case in (38), we get
(41)∫S,T∩C2β0′Eh12ψt1,tdt≤c∫S,T∩C2β0′EΛ1−1ψt1,t2dt≤c∫S,T∩C2β0′EΛ1−1β0′t2dt≤c∫S,T∩C2β0′EΛ0−1β0′t2dt≤∫STβ0′EΛ0−1β0′t2dt.
Finally, using A1, (17), the properties of Λ1−1, and the last case in (38), we obtain
(42)∫S,T∩C3β0′Eh12ψt1,tdt=∫S,T∩C3β0′Eh1ψt1,th1ψt1,tdt≤∫S,T∩C3Eψt1,tΛ1−1ψt1,th1ψt1,tdt≤Λ1−1r∫S,T∩C3Eψt1,th1ψt1,tdt≤c∫S,T∩C3Eψt1,th1ψt1,tdt≤c∫S,T∩C3E−E′dt≤∫STE−E′dt≤cE2S−E2T≤cE2S.
Combining (40)–(42), we get
(43)∫STβ0′Eh12ψt1,tdt≤cE2S+c∫STβ0′EH0−1β0′t2dt.
Repeating the same above calculations, we also find(44)∫STβ0′Eh22ψt1,tdt≤cE2S+c∫STβ0′EΛ0−1β0′t2dt.
Then, (20) becomes
(45)∫STβ0′tE2tdt≤cES2+c∫STβ0′EΛ0−1β0′t2dt.
Recallingβ0′t=Λ01/β0t and letting β0t=s, then for T≥S≥t1, we have
(46)∫S∞β0′tE2tdt≤cES2+cES∫S∞β0′tΛ0−1β0′t2dt=cES2+cES∫β0S∞Λ0−1Λ01s2ds=cES2+cESβ0S.
Using (46) and Lemma 3 with m=n=1, F=E, and β=β0, we obtain
(47)Et≤cβ0t2,∀t≥t1.
Hence, (34) is established.Proof of Theorem 8.
To prove (18), let us define Gs=sΛ0s and take s0>t0 such that Λ01/s0≤1.
SinceΛ0 is increasing, then we have Λ01/s≤Λ01/s0≤1,∀s≥s0. Therefore, we obtain
(48)β0−1s≤1+s−11Λ01/s=Λ01/s+s−1Λ01/s≤sΛ01/s=1G1/s,∀s≥s0.
So, lettingt=1/G1/s, we can see that
(49)G1s=1t⇒1s=G−11t⇒s=1G−11/t.
Then, using (48) and (49), we see that
(50)β0−1s≤1GG−11/t=11/t=t.
Hence,(51)s=1G−11/t≤β0t.
Now, it is easy to see that(52)1β0t≤G−11t∀t≥t0.
Therefore, using (34), estimate (18) is established.Example 9.
As in [22], we consider the following examples to illustrate our decay result:
(1)
LetΛ1s=Λ2s=e−lns2 near zero. Then, (18) gives
(53)Et≤ce−2lnt1/2(2)
LetΛ1s=Λ2s=e−1/s near zero. Then, (18) implies
(54)Et≤clnt2(3)
IfΛ1s=Λ2s=e−e1/s near zero, then, using (18), we have
(55)Et≤clnlnt2(4)
IfΛ1s=Λ2s=s, then, using (18), we have
(56)Et≤ct(5)
IfΛ1s=s3/2, Λ2s=s3 near zero. Then,(57)Gs=s−1+1+4s23,then,
(58)Gss1/3=−1+1+4s2
By approximating1+4s near s=0, we have
(59)1+4s≈1+2s.
Hence,(60)Gss1/3≈s,nears=0,which implies that G−1s≈s1/4. Then, using (18), we have
(61)Et≤1t.



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