# Exponential Stabilization of a Swelling Porous-Elastic System with Microtemperature Effect and Distributed Delay

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

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## Abstract

The swelling porous thermoelastic system with the presence of temperatures, microtemperature effect, and distributed delay terms is considered. We will establish the well posedness of the system, and we prove the exponential stability result.


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## Body

## 1. Introduction and Preliminaries

Eringen was the first to present a theory in which a mixture of viscous liquid and solids mixed with gas [1]. Then, after studying this heat-resistant mixture, you get to the field equations [2].Expansive (swelling) soils have also been classified under porous media theory which studies this type of problem. This is why this field is considered fertile for study, as there are many studies to reduce the damage caused by swelling soil, especially in civil engineering and architecture, for more depth (see [3–8]).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. And (P1,G1,H1) are the partial tension, internal body forces, and eternal forces acting on the displacement, respectively, similarly (P2,G2,H2), but acting on the elastic solid. In addition, the constitutive equations of partial tensions are given by
(3)P1P2=a1,a2a2,a3⏟A.uxϕx,where a1,a3>0 and a2≠0 are real numbers. A is matrix positive definite with a1a3>a22.Quintanilla [8] investigated (2) by taking
(4)G1=G2=ξut−ϕt,H1=a3uxxt,H2=0,where ξ>0; they obtained the stability is exponentially. Similarly, in [9], the authors are considered (2) with a different conditions
(5)G1=G2=0,H1=−ρuγxut,H2=0,where γx is an internal viscous damping function with positive mean. By the spectral method, they obtained the exponential stability result. For more details, see [8–15].Time delays are very important in most natural phenomena and industrial devices, where the time lag is a source of instability, and it is a problem worthy of attention.Also, there are many works that have studied this type of problems, of which [11, 15–24].The basic evolution equations for one-dimensional theories of swelling porous materials with temperature and microtemperature [25–28] are given by
(6)ρuutt=Tx,(7)ρφϕtt=Hx+G,(8)ρηt=qx,(9)ρEt=Px∗+q−Q.Here,T,H,G,q,η,P∗,Q,andE represent the stress, the equilibrated stress, the equilibrated body force, the heat flux vector, the entropy, the first heat flux moment, the mean heat flux, and the first moment of energy. The constitutive equations are
(10)T=P1+G1+H1P∗=−k2wx,(11)H=P2+P3ρη=γux+c0θ+mϕ,(12)G=G2+H2Q=−k3w−k1θx,(13)q=κθxρE=−αw−dϕx,where w is the microtemperature vector, ρu,ρϕ,k1,k2,k3,a1,a2,a3,α,κ,c0,μ1>0. As coupling is considered, a2≠0 and satisfies
(14)a=a3−a22a1>0.And the coefficientsγ,m,d>0.The goal of this work is the thermal effects, so we suppose that the heat capacityc0>0, and for more excitement in posing the problem, we suppose that the thermal conductivity is nonexistent κ≠0.And by introducing the distributed delay term, form a new problem different from previous studies. Under appropriate suppositions, the well posedness of the system is established, and we prove the exponential stability result by the energy method.We consider in this work:(15)G1=G2=0,P3=−dw,(16)H1=−γθ,(17)H2=mθ−μ1ϕt−∫τ1τ2μ2σϕtx,t−σdσ.Now, by substituting (13)–(17) into (9), we arrive at the following problem:
(18)ρuutt−a1uxx−a2ϕxx−γθx=0,ρϕϕtt−a3ϕxx−a2uxx−dwx+mθ+μ1ϕt+∫τ1τ2∣μ2σ∣ϕtx,t−σdσ=0,c0θt=−γutx−mϕt−k1wx,αwt=k2wxx−k3w−k1θx−dϕtx,where
(19)x,σ,t∈H=0,1×τ1,τ2×0,∞,under the initial and boundary conditions
(20)ux,0=u0x,utx,0=u1x,θx,0=θ0xϕx,0=ϕ0x,ϕtx,0=ϕ1x,wx,0=w0x,x∈0,1,ϕtx,−t=f0x,t,x,t∈0,1×0,τ2,u0,t=u1,t=ϕ0,t=ϕ1,t=0,θ0,t=θ1,t=wx0,t=wx1,t=0,t≥0.First, as in [24], we introduce the new variable
(21)Yx,ρ,σ,t=ϕtx,t−σρ,Then, we get(22)σYtx,ρ,σ,t+Yρx,ρ,σ,t=0,Yx,0,σ,t=ϕtx,t.Consequently, our problem is written in the form(23)ρuutt−a1uxx−a2ϕxx−γθx=0,ρϕϕtt−a3ϕxx−a2uxx−dwx+mθ+μ1ϕt+∫τ1τ2∣μ2σ∣Yx,1,σ,tdσ=0,c0θt=−γutx−mϕt−k1wx,αwt=k2wxx−k3w−k1θx−dϕtx,σYtx,ρ,σ,t+Yρx,ρ,σ,t=0.where
(24)x,ρ,σ,t∈0,1×H,with the initial data
(25)ux,0=u0x,utx,0=u1x,θx,0=θ0x,ϕx,0=ϕ0x,ϕtx,0=ϕ1x,wx,0=w0xx∈0,1,Yx,ρ,σ,0=f0x,ρσ,x,ρ,σ∈0,1×0,1×0,τ2,and the boundary conditions
(26)u0,t=u1,t=ϕ0,t=ϕ1,t=0,θ0,t=θ1,t=wx0,t=wx1,t=0,t≥0.Here, the integral represent the distributed delay terms withτ1,τ2>0 are a time delay; μ2 is an L∞ function satisfying:(H1) μ2:τ1,τ2⟶ℝ is a bounded function satisfying
(27)∫τ1τ2∣μ2σ∣dσ<μ1.Meanwhile, from (23)4 and (26), it follows that
(28)ddt∫01ωx,tdx+k3α∫01ωx,tdx=0.So, by solving (28) and using the initial data of u, we get
(29)∫01ωx,tdx=∫01ω0xdxe−t/αk3.Consequently, if we let(30)ω−x,t=ωx,t−∫01ω0xdxe−t/αk3,we get
(31)∫01ω−x,tdx=0,∀t≥0.Therefore, the use of Poincare’s inequality forω− is justified. In addition, simple substitution shows that u,ϕ,θ,ω−,Y satisfies system (23). Henceforth, we work with ω− instead of ω but write ω for simplicity of notation.In this paper, we consideru,ϕ,θ,w,Y to be a solution of system (23)–(26) with the regularity needed to justify the calculations. In Section 2, the well posedness is established, and in Section 3, the exponential stability is proved. In all of the following, we mention that c>0.Remark 1.
The coupling that we have proposed in this work with the presence of microtemperatures and distributed delay in problems of swelling in porous elasticity we believe constitutes a new contribution and differs from the previous studies.


## 2. Well Posedness

In this section, we established the well posedness of the system (23)–(26).First, introducing the vector function(32)U=u,ut,ϕ,ϕt,θ,w,YT,and the variables v=ut,φ=ϕt, then the system (23) writes as follows:
(33)Ut=AU,U0=U0=u0,u1,ϕ0,ϕ1,θ0,w0,f0T,where A:DA⊂H:⟶H is the linear operator given by
(34)AU=v−1ρua1uxx+a2ϕxx−γθxφ1ρϕa3ϕxx+a2uxx−dwx+mθ−μ1φ−∫τ1τ2∣μ2σ∣Yx,1,σ,tdσ−1c0γvx+mφ+k1wx1αk2wxx−k3w−k1θx−dφx−1σYρ,and H is the energy space given by
(35)H=H010,1×L20,1×H010,1×L20,1×L20,1×L∗20,1×L20,1×0,1×τ1,τ2,where
(36)L∗20,1=ψ∈L20,1∫01ψxdx=0,H∗10,1=H10,1∩L∗20,1,H∗20,1=ψ∈H20,1ψx1=ψx0=0.For any,(37)U=u,v,ϕ,φ,θ,w,YT∈H,U^=u∧,v∧,ϕ∧,φ∧,θ∧,w∧,Y∧T∈H,we equip H with the inner product defined by
(38)<U,U^>H=ρu∫01vv^dx+a1∫01uxu^xdx+ρϕ∫01φφ^dx+a3∫01ϕxϕ^xdx+c0∫01θθ^dx+α∫01ww^dx+a2∫01uxϕ^+u^xϕdx+∫01∫01∫τ1τ2σ∣μ2σ∣YY^dσdρdx.The domain ofA is given by
(39)DA=U∈H/u,ϕ∈H20,1∩H010,1,v,φ,θ∈H010,1,w∈H∗20,1∩H∗10,1,Y,Yρ∈L20,1×0,1×τ1,τ2,Yx,0,σ,t=φ.Clearly,DA is dense in H.Theorem 2.
LetU0∈H and assume that (27) holds. Then, there exists a unique solution U∈Cℝ+,H of problem (33).
Moreover, ifU0∈DA, then
(40)U∈Cℝ+,DA∩C1ℝ+,H.Proof.
First, we prove that the operatorA is dissipative. For any U0∈DA and by using (38), we have
(41)<AU,U>H=−μ1∫01φ2dx−∫01∫τ1τ2μ2σφYx,1,σ,tdσdx−∫01∫01∫τ1τ2∣μ2σ∣YρYdσdρdx−k2∫01wx2dx−k3∫01w2dx.
For the third term of the RHS of (41), we have
(42)−∫01∫01∫τ1τ2∣μ2σ∣YρYdσdρdx=−12∫01∫τ1τ2∫01∣μ2σ∣ddρY2dρdσdx=−12∫01∫τ1τ2∣μ2σ∣Y2x,1,σ,tdσdx+12∫01∫τ1τ2∣μ2σ∣Y2x,0,σ,tdσdx.
By using Young’s inequality, we get(43)−∫01∫τ1τ2∣μ2σ∣φYx,1,σ,tdσdx≤12∫τ1τ2∣μ2σ∣dσ∫01φ2dx+12∫01∫τ1τ2∣μ2σ∣Y2x,1,σ,tdσdx.
Substituting (42), (43) into (41), using Yx,0,σ,t=φx,t and (27), we find
(44)<AU,U>H≤−η0∫01φ2dx−k2∫01wx2dx−k3∫01w2dx≤0,where η0=μ1−∫τ1τ2∣μ2σ∣dσ>0. Hence, A is dissipative operator.
Next, we proveA is maximal operator. It is sufficient to show that λI−A is surjective operator.
Indeed, for anyF=f1,f2,f3,f4,f5,f6,f7T∈H, we prove that there exists a unique U=u,v,ϕ,φ,θ,w,z∈DA such that
(45)λI−AU=F.
That is,(46)λu−v=f1∈H010,1,ρuλv−a1uxx−a2ϕxx+γθx=ρuf2∈L20,1,λϕ−φ=f3∈H010,1,ρϕλφ−a3ϕxx−a2uxx+dwx−mθ+μ1φ+∫τ1τ2μ2σYx,1,σ,tdσ=ρϕf4∈L2,c0λθ+γvx+mφ+k1wx=c0f5∈L20,1,αλw−k2wxx+k3w+k1θx+dφx=αf6∈H∗10,1,σλYtx,ρ,σ,t+Yρx,ρ,σ,t=σf7∈L20,1×0,1×τ1,τ2.
We note that Equation (46)7 with Yx,0,σ,t=φx,t has a unique solution defined by
(47)Yx,ρ,σ,t=e−λρσφ+σeσρλ∫0ρeλσρf7x,ρ,σ,tdρ,then
(48)Yx,1,σ,t=e−λσφ+σeλσ∫01eλσρf7x,ρ,σ,tdρ,and we have
(49)v=λu−f1,φ=λϕ−f3.
Inserting (48) and (49) in (46)2, (46)4, (46)5, and (46)6, we get
(50)ρuλ2u−a1uxx−a2ϕxx+γθx=h1,μ3ϕ−a3ϕxx−a2uxx+dwx−mθ=h2,c0θ+γux+mϕ+k1λwx=h3,αλ+k3λw−k2λwxx+k1λθx+dϕx=h4,where
(51)h1=ρuλf1+f2,h2=ρϕλ+μ1+∫τ1τ2∣μ2σ∣e−σλdσf3−∫τ1τ2σμ2σeσλ∫01eλσρf7x,ρ,σ,tdρdσ,h3=1λγf1x+mf3+c0f5,h4=1λαf6+dλf3,μ3=ρϕλ2+μ1λ+λ∫τ1τ2∣μ2σ∣e−λσdσ.
We multiply (50) by u^,ϕ^,θ^,w^, respectively, and integrate their sum over 0,1 to find the following variational formulation:
(52)Bu,ϕ,θ,w,u^,ϕ^,θ^,w^=Γu^,ϕ^,θ^,w^,where
(53)B:H010,1×H010,1×L20,1×H∗10,12⟶ℝ,is the bilinear form given by
(54)Bu,ϕ,θ,w,u^,ϕ^,θ^,w^=ρuλ2∫01uu^dx+a1∫01uxu^xdx+a2∫01ϕxu^xdx+γ∫01θxu^dx+μ3∫01ϕϕ^dx+a3∫01ϕxϕ^xdx+a2∫01uxϕ^xdx+d∫01wxϕ^dx−m∫01θϕ^dx+c0∫01θθ^dx+γ∫01uxθ^dx+m∫01ϕθ^dx+k1λ∫01wxθ^dx+αλ+k3λ∫01ww^dx+k2λ∫01wxw^xdx+k1λ∫01θxw^dx+d∫01ϕxw^dx,and
(55)Γ:H010,1×H010,1×L20,1×H∗10,1⟶ℝ,is the linear functional defined by
(56)Γu^,ϕ^,θ^,w^=∫01h1u^dx+∫01h2ϕ^dx+∫01h3θ^dx+∫01h4w^dx.
Now, forV=H010,1×H010,1×L20,1×H∗10,1, equipped with the norm,
(57)u,ϕ,θ,wV2=u22+ux22+ϕ22+ϕx22+θ22+wx22+w22,then we have
(58)Bu,ϕ,θ,w,u,ϕ,θ,w=ρuλ2∫01u2dx+a1∫01ux2dx+μ3∫01ϕ2dx+a3∫01ϕx2dx+2a2∫01uxϕxdx+c0∫01θ2dx+αλ+k3λ∫01w2dx+k2λ∫01wx2dx.
On the other hand, we can write(59)a1ux2+2a2uxϕx+a3ϕx2=12a1ux+a2a3ϕx2+a3ϕx+a2a1ux2+ux2a1+a22a3+ϕx2a3+a22a1.
Since (14), we deduce
(60)a1ux2+2a2uxϕx+a3ϕx2>12ux2a1+a22a3+ϕx2a3+a22a1,
then, for someM0>0(61)Bu,ϕ,θ,w,u,ϕ,θ,w≥M0∥u,ϕ,θ,w∥V2.
Thus,B is coercive. Hence, we use the Lax-Milgram theorem to conclude that (52) has a unique solution:
(62)u,ϕ∈H010,1,w∈H∗10,1,θ∈L20,1.
Substitutingu,ϕ,θ, and w into (46)1,3, we have
(63)v,φ∈H010,1.
Similarly, the compensation ofv in (47) with (46)7, gives
(64)Y,Yρ∈L20,1×0,1×τ1,τ2.
Moreover, if we takeu^=θ^=w^=0 in (54), we get
(65)a3∫01ϕxϕ^xdx+μ3∫01ϕϕ^dx+a2∫01uxϕ^xdx+d∫01wxϕ^dx−m∫01θϕ^dx=∫01h2ϕ^dx,∀ϕ^∈H010,1,which implies
(66)a3∫01ϕxϕ^xdx=∫01h2−μ3ϕ+a2uxx−dwx+mθϕ^dx,∀ϕ^∈H010,1,that is
(67)a3ϕxx=μ3ϕ−a2uxx+dwx−mθ−h2∈L20,1.
Consequently(68)ϕ∈H20,1∩H010,1.
Similarly, we get(69)u∈H20,1∩H010,1,θ∈H010,1,and, if we let u^=θ^=ϕ^=0 in (54), we get
(70)αλ+k3λ∫01ww^dx+k2λ∫01wxw^xdx+k1λ∫01θxw^dx+d∫01ϕxw^dx−m∫01θϕ^dx=∫01h4w^dx,∀w^∈H∗10,1,which implies
(71)k2λ∫01wxψxdx=∫01−h4+αλ+k3λw+k1λθx+dϕxψdx,∀ψ∈C10,1⊂H∗10,1,
Thus, using integration by parts, we get(72)wx1ψ1−wx0ψ0=0,∀ψ∈C10,1.
Therefore,(73)wx1=wx0=0.
Consequently,(74)w∈H∗20,1∩H∗10,1.
Finally, the application of regularity theory for the linear elliptic equations guarantees the existence of uniqueU∈DA such that (45) is satisfied.
Consequently, we conclude thatA is a maximal dissipative operator. Hence, by Lumer-Philips theorem (see [29]), we have the well-posedness result. This completes the proof.☐


## 3. Exponential Decay

In this section, we prove our stability result of the system (23)–(26).For this, we have the following lemmas.Lemma 3.
The energy functionalE, defined by
(75)Et=12∫01ρuut2+a1ux2+ρϕϕt2+a3ϕx2+2a2uxϕx+c0θ2+αw2dx+12∫01∫01∫τ1τ2σ∣μ2σ∣Y2x,ρ,σ,tdσdρdx,satisfies
(76)E′t≤−k2∫01wx2dx−k3∫01w2dx−η0∫01ϕt2dx≤0,where η0=μ1−∫τ1τ2∣μ2σ∣dσ>0.Proof.
Multiplying Equation (23)1,2,3,4 by ut,ϕt,θ, and w, integrating by parts over 0,1, and using (26), we get
(77)12ddt∫01ρuut2+a1ux2+ρϕϕt2+a3ϕx2+2a2uxϕx+c0θ2+αw2dx+μ1∫01ϕt2dx+∫01ϕt∫τ1τ2∣μ2σ∣Yx,1,σ,tdσdx+k3∫01w2dx+k2∫01wx2dx=0.
Now, multiplying Equation (23)5 by Y∣μ2σ∣, and integrating the result over 0,1×0,1×τ1,τ2(78)ddt12∫01∫01∫τ1τ2σ∣μ2σ∣Y2x,ρ,σ,tdσdρdx=−∫01∫01∫τ1τ2∣μ2σ∣YYρx,ρ,σ,tdσdρdx=−12∫01∫01∫τ1τ2∣μ2σ∣ddρY2x,ρ,σ,tdσdρdx=12∫01∫τ1τ2∣μ2σ∣Y2x,0,σ,t−Y2x,1,σ,tdσdx=12∫τ1τ2∣μ2σ∣dσ∫01ϕt2dx−12∫01∫τ1τ2∣μ2σ∣Y2x,1,σ,tdσdx.
Now, by substituting (78) into (77), and using Young’s inequality, we have
(79)E′t≤−k3∫01w2dx−k2∫01wx2dx−μ1−∫τ1τ2∣μ2σ∣dσ∫01ϕt2dx,then, by (27), ∃η0>0 so that
(80)E′t≤−k3∫01w2dx−k2∫01wx2dx−η0∫01ϕt2dx,then we obtain (76) (E is a nonincreasing function).☐Remark 4.
Using (14), we conclude that Et satisfies
(81)Et>12∫01ρuut2+a4ux2+ρϕϕt2+a5ϕx2+c0θ2+αw2dx+12∫01∫01∫τ1τ2σ∣μ2σ∣Y2x,ρ,σ,tdσdρdx,where
(82)a4=12a1−a22a3>0,a5=12a3−a22a1>0.
Then, the functionEt is nonnegative.Lemma 5.
The functional(83)D1t≔ρϕ∫01ϕtϕdx−a2a1ρu∫01ϕutdx+μ12∫01ϕ2dx,satisfies, for any ε1>0(84)D1′t≤−a2∫01ϕx2dx+ε1∫01ut2dx+c1+1ε1∫01ϕt2dx+c∫01w2dx+c∫01θ2dx+c∫01∫τ1τ2∣μ2σ∣Y2x,1,σ,tdσdx.Proof.
Direct computation using integration by parts and Young’s inequality yields(85)D1′t=−a3∫01ϕx2dx+ρϕ∫01ϕt2dx+a22a1∫01ϕx2dx+a2a1ρu∫01ϕtutdx−d∫01wxϕdx+m∫01θϕdx+a2γa1∫01θxϕdx−∫01ϕ∫τ1τ2μ2σYx,1,σ,tdσdx≤−a3−a22a1∫01ϕx2dx+ρϕ∫01ϕt2dx+a2a1ρu∫01ϕtutdx−d∫01wxϕdx+m∫01θϕdx+a2γa1∫01θxϕdx−∫01ϕ∫τ1τ2μ2σYx,1,σ,tdσdx.
We use Cauchy-Schwartz, Young’s, and Poincare’s inequalities; forδ1,ε1>0, we obtain
(86)D1′t≤−a3−a22a1−μ1cδ1∫01ϕx2dx+ε1∫01ut2dx+c1+1ε1∫01ϕt2dx+d∫01wϕxdx+m∫01θϕdx−a2γa1∫01θϕxdx+14δ1∫0t∫τ1τ2∣μ2σ∣Y2x,1,σ,tdσdx.
Bearing in mind (14), and letting δ1=a/2c, we obtain the estimate (84).☐Lemma 6.
The functional(87)D2t≔a2∫01ϕtudx−∫01ϕutdx,satisfies,
(88)D2′t≤−a222ρϕ∫01ux2dx+c∫01ϕx2dx+c∫01ϕt2dx+c∫01w2dx+c∫01θ2dx+c∫01∫τ1τ2∣μ2σ∣Y2x,1,σ,tdσdx.Proof.
By differentiatingD2, then using (23), integration by parts, and (26), we obtain
(89)D2′t=−a22ρϕ∫01ux2dx+a22ρu∫01ϕx2dx−a2a3ρϕ−a1a2ρu∫01ϕxuxdx−a2μ1ρϕ∫01uϕtdx+a2dρϕ∫01wuxdx+a2mρϕ∫01θudx−a2γρu∫01θϕxdx−a2ρϕ∫01u∫τ1τ2∣μ2σ∣Yx,1,σ,tdσdx.
Now, we estimate the last six terms in the RHS of (89), using Young’s, Cauchy-Schwartz, and Poincare’s inequalities. For δ2,δ3,δ4,δ5,δ6>0, we have
(90)−a2a3ρϕ−a1a2ρu∫01ϕxuxdx≤δ2∫01ux2dx+a2a3ρϕ−a1a2ρu214δ2∫01ϕ2dx.a2dρϕ∫01wuxdx≤δ3∫01ux2dx+c4δ3∫01w2dx,−a2μ1ρϕ∫01uϕtdx≤cδ4∫01ux2dx+c4δ4∫01ϕt2dx,and
(91)a2mρϕ∫01θudx≤δ5c∫01ux2dx+c4δ5∫01θ2dx,−a2ρϕ∫01u∫τ1τ2μ2σYx,1,σ,tdσdx≤cδ6∫01ux2dx,−c4δ6∫01∫τ1τ2∣μ2σ∣Y2x,1,σ,tdσdx.
By lettingδ2=δ3=a2/10ρϕ,δ4=δ5=δ6=a2/10cρϕ, and substituting into (89), we get (88).☐Lemma 7.
The functional(92)D3t≔−ρu∫01utudx,satisfies
(93)D3′t≤−ρu∫01ut2dx+3a1∫01ux2dx+a34∫01ϕx2dx+γ24a1∫01θ2dx.Proof.
Direct computations give(94)D3′t=−ρu∫01ut2dx+a1∫01ux2dx+a2∫01uxϕxdx−γ∫01uxθdx.
Estimating (93) easily follows by using Young’s inequality and (14).☐Lemma 8.
The functional(95)D4t≔−c0α∫01θ∫0xwydydx,satisfies
(96)D4′t≤−c0k12∫01θ2dx+ε2∫01ut2dx+c1+1ε2∫01w2dx+c∫01ϕt2dx+c∫01wx2dx.Proof.
Direct computations give(97)D4′t=−c0k1∫01θ2dx+αk1∫01w2dx+αγ∫01utwdx+c0d∫01θϕtdx−αm∫01ϕt∫0xwydydx+c0k2∫01wxθdx−c0k3∫01θ∫0xwydydx.
Estimate (96) easily follows by using Young’s and Cauchy-Schwartz inequalities.☐Now, let us introduce the following functional used by.Lemma 9.
The functional(98)D5t≔∫01∫01∫τ1τ2σe−σρ∣μ2σ∣Y2x,ρ,σ,tdσdρdx,satisfies
(99)D5′t≤−η1∫01∫01∫τ1τ2σ∣μ2σ∣Y2x,ρ,σ,tdσdρdx+μ1∫01ϕt2dx−η1∫01∫τ1τ2∣μ2σ∣Y2x,1,σ,tdσdx,where η1>0.Proof.
By differentiatingD5, with respect to t and using the last equation in (23), we have
(100)D5′t=−2∫01∫01∫τ1τ2e−σρ∣μ2σ∣YYρx,ρ,σ,tdσdρdx=−∫01∫01∫τ1τ2σe−σρ∣μ2σ∣Y2x,ρ,σ,tdσdρdx−∫01∫τ1τ2∣μ2σ∣e−σY2x,1,σ,t−Y2x,0,σ,tdσdx.
Using the fact thatYx,0,σ,t=ϕtx,t, and e−σ≤e−σρ≤1, for all 0<ρ<1, we obtain
(101)D5′t=−η1∫01∫01∫τ1τ2σ∣μ2σ∣Y2x,ρ,σ,tdσdρdx−∫01∫τ1τ2e−σ∣μ2σ∣Y2x,1,σ,tdσdx+∫τ1τ2∣μ2σ∣dσ∫01ϕt2dx.☐Because−e−σ is a increasing function, we have −e−σ≤−e−τ2, for all σ∈τ1,τ2.Finally, settingη1=e−τ2 and recalling (27), we find (99). We are now ready to prove the main result.Theorem 10.
Assume (27) holds. Then, ∀t0>0, there exist β1,β2>0 such that the energy functional given by (75) satisfies
(102)Et≤β1e−β2t,∀t≥0.Proof.
We define the functional of Lyapunov(103)Lt≔NEt+N1D1t+N2D2t+D3t+N4D4t+N5D5t,where N,N1,N2,N4,N5>0 we will assign them later.
By differentiating (103) and using (75), (84), (88), (93), (96), and (99), we have
(104)L′t≤−aN12−cN2−a34∫01ϕx2dx−ρu−N1ε1−N4ε2∫01ut2dx−a22N22ρϕ−3a1∫01ux2dx−k2N−cN4∫01wx2dx−η0N−cN11+1ε1−N2c−N4c−μ1N5∫01ϕt2dx−k3N−cN1−cN2−cN41+1ε2∫01w2dx−c0k1N42−cN1−cN2−γ24a1∫01θ2dx−N5η1−cN1−cN2∫01∫τ1τ2∣μ2σ∣Y2x,1,σ,tdσdx−N5η1∫01∫01∫τ1τ2σ∣μ2σ∣Y2x,ρ,σ,tdσdρdx.
By setting(105)ε1=ρu4N1,ε2=ρu4N4,we obtain
(106)L′t≤−aN12−cN2−a34∫01ϕx2dx−ρu2∫01ut2dx−a22N22ρϕ−3a1∫01ux2dx−k2N−cN4∫01wx2dx−η0N−cN11+N1−N2c−N4c−μ1N5∫01ϕt2dx−k3N−cN1−cN2−cN41+N4∫01w2dx−c0k1N42−cN1−cN2−γ24a1∫01θ2dx−N5η1−cN1−cN2∫01∫τ1τ2∣μ2σ∣Y2x,1,σ,tdσdx−N5η1∫01∫01∫τ1τ2σ∣μ2σ∣Y2x,ρ,σ,tdσdρdx.
At this point, we choose our constants.
We chooseN2 large enough so that
(107)α1=a22N22ρϕ−3a1>0,then we pick N1 large enough such that
(108)α2=aN12−cN2−a34>0,then we select N4 and N5 large enough such that
(109)α3=c0k1N42−cN1−cN2−γ24a1>0,α4=N5η1−cN1−cN2>0.
Thus, we arrive at(110)L′t≤−α2∫01ϕx2dx−ρu2∫01ut2dx−α1∫01ux2dx−η0N−c∫01ϕt2dx−k3N−c∫01w2dx−k2N−c∫01wx2dx−α3∫01θ2dx−α4∫01∫τ1τ2∣μ2σ∣Y2x,1,σ,tdσdx−α5∫01∫01∫τ1τ2σ∣μ2σ∣Y2x,ρ,σ,tdσdρdx.where α5=η1N5.
On the other hand, if we let(111)Lt=N1D1t+N2D2t+D3t+N4D4t+N5D5t,then
(112)Lt≤N1ρϕ∫01ϕϕtdx+N1a2a1ρu∫01ϕutdx+N1μ12∫01ϕ2dx+N2a2∫01ϕut−uϕtdx+ρu∫01utudx+N4c0α∫01∣θ∫0xwydy∣dx+N5∫01∫01∫τ1τ2σe−σρ∣μ2σ∣Y2x,ρ,σ,tdσdρdx.
According Young’s, Cauchy-Schwartz, and Poincaré inequalities, we find(113)Lt≤c∫01ut2+ϕt2+ϕx2+ux2+θ2+w2dx+c∫01∫01∫τ1τ2σ∣μ2σ∣Y2x,ρ,σ,tdσdρ.
On the other hand, we can write(114)a1ux2+2a2ϕxux+a4ϕx2=12a1ux+a2a1ϕx2+a4ϕx+a2a4ux2+a1−a22a4ux2+a4−a22a1ϕx2.
Sincea1a3>a22 and (27), we deduce that
(115)a1ux2+2a2ϕxux+a4ϕx2>12a1−a22a4ux2+a4−a22a1ϕx2.
Hence, we get(116)Lt=Lt−NEt≤cEt,that is,
(117)N−cEt≤Lt≤N+cEt.
At this point, we chooseN large enough such that
(118)N−c>0,Nη0−c>0,Nk3−c>0,Nk2−c>0,and exploiting (75), the estimates (110) and (117), respectively, gives
(119)c2Et≤Lt≤c3Et,∀t≥0,(120)L′t≤−d1Et,∀t≥0,for some d1,c2,c3>0.
Consequently, for someβ2>0, we find
(121)L′t≤−β2Lt,∀t≥0.
Integration of (120) over 0,t gives
(122)Lt≤L0e−β2t,∀t≥0.
Consequently, (102) is established by virtue of (117) and (121).☐



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