# Global Existence and Decay for a System of Two Singular Nonlinear Viscoelastic Equations with General Source and Localized Frictional Damping Terms

Authors: Salah Mahmoud Boulaaras; Rafik Guefaifia; Nadia Mezouar; Ahmad Mohammed Alghamdi
Journal: Journal of Function Spaces
 (2020)

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## Abstract

The current paper deals with the proof of a global solution of a viscoelasticity singular one-dimensional system with localized frictional damping and general source terms, taking into consideration nonlocal boundary condition. Moreover, similar to that in Boulaaras’ recent studies by constructing a Lyapunov functional and use it together with the perturbed energy method in order to prove a general decay result.


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## Body

## 1. Introduction

The evolution problem with integral conditions is related with many branches of sciences ([1–6]). Cause of this, interest in it occurs naturally in inflation cosmology, nuclear physics, supersymmetric field theories, and quantum mechanics (see for example [2, 7]). Later, by the motivation of this work, some authors gave necessary and sufficient conditions for the hyperbolic equation with source term (see, e.g., [8–10]).This manuscript is devoted to the study of the global existence and decay for a system of two singular one-dimensional nonlinear viscoelastic equations with general source terms(1)utt−1xxuxx+∫0tg1t−s1xxuxx,sxds+μxut=f1u,v,in0,L×0,T,vtt−1xxvxx+∫0tg2t−s1xxvxx,sxds+μxvt=f2u,v,in0,L×0,T,ux,0=u0x,utx,0=u1x,x∈0,L,vx,0=v0x,vtx,0=v1x,x∈0,L,uL,t=vL,t=0,∫0Lxux,tdx=∫0Lxvx,tdx=0,where L<∞,T<∞,μ∈C10,α,g1.,g2.:ℝ+⟶ℝ+ and f1.,.,f2.,.:ℝ2⟶ℝ are given functions, which will be specified later.The problems related with localized frictional damping have extensively studied by many teams as [11], where the authors obtained an exponential rate of decay for the solution of the viscoelastic nonlinear wave equation:
(2)utt−Δu+fx,t,u+∫0tg1t−sΔusds+axut=0,in0,L×0,T,for damping term axut may be null for some part of the domain.We used the techniques in [11]; we have proven in [8] the existence of a global solution using the potential well theory for the following viscoelastic system with nonlocal boundary condition and localized frictional damping
(3)utt−1xxuxx+∫0tg1t−s1xxuxx,sxds+axut=vq+1up−1u,in0,L×0,T,vtt−1xxvxx+∫0tg2t−s1xxvxx,sxds+axvt=vq+1up−1v,in0,L×0,T,ux,0=u0x,utx,0=u1x,x∈0,α,vx,0=v0x,vtx,0=v1x,x∈0,α,uα,t=vα,t=0,∫0αxux,tdx=∫0αxvx,tdx=0.Very recently, in ([9]), we study the following singular one-dimensional nonlinear equations that arise in generalized viscoelasticity with long-term memory:
(4)utt−1xxuxx+∫0tg1t−s1xxuxx,sxds=f1u,v,in0,L×0,T,vtt−1xxvxx+∫0tg2t−s1xxvxx,sxds=f2u,v,in0,L×0,T,ux,0=u0x,utx,0=u1x,x∈0,L,vx,0=v0x,vtx,0=v1x,x∈0,L,uL,t=vL,t=0,∫0Lxux,tdx=∫0Lxvx,tdx=0.In view of the articles mentioned above in ([8, 9, 11]), much less effort has been devoted to the existence of a global solution to the system of two singular nonlinear equations which arise in generalized viscoelasticity with localized frictional damping terms using the potential-well theory. Moreover, we prove a general decay result by constructing a Lyapunov functional and use it together with the perturbed energy method.The structure of the work is as follows: To facilitate the description, firstly in Section2, we give the fundamental definitions and theorems on function spaces that will be needed in the body of the paper and state the local existence theorem. In Section 3, the energy function Et is defined and proved to be a nonincreasing function of time. Finally, the main result is obtained, which gives the general decay conditions:
(5)gi′t≤−ξtgirt,i=1,2.


## 2. Preliminaries

LetLxp=Lxp0,L be the weighted Banach space equipped with the norm
(6)uLxp=∫0Lxupdx1/p,when p=2, we get a Hilbert space, and we denote by H=Lx2, it provided with the finite norm
(7)uH=∫0Lxu2dx1/2.V=Vx10,L be the Hilbert space equipped with the norm
(8)uV=uH2+uxH21/2.We get the following lemma by combining the Poincare inequality and (see [8]).Lemma 1.
LetV0 space defined as follows
(9)V0=u∈VsuchthatuL=0.
Then, for2≤p4, we have
(10)∫0Lxvpdx≤C∗vxH=Lx20,Lp,∀u∈V0,where C∗ is a constant depending on L and p only and for p=2, C∗=Cp is the Poincare constant.Remark 2.
It is clear thatuV0=uxH defines an equivalent norm on V0.
The next theorem confirms that our problem has a local solution under some condition onp and the relaxation functions gi, the proof can be established by following the argument of [12].Theorem 3.
We takeu0,v0∈V02 and v1,v2∈H2. If p<3 and
(11)gi0>0,1−∫0∞gisds=l>0,fori=1,2,then, there exists t∗>0 small enough such that the problem (1) has a unique local solution
(12)u∈C0,t∗;V0∩C10,t∗;H.Remark 4.
The condition onp is needed so that the embedding of V0 in Lx2 is Lipchitz.
We need the following assumptions to get our results.
(G1) For i=1,2,git:ℝ+⟶ℝ+ is a nonincreasing C2 function such that
(13)gis≥0,gi′s≤0and,gi0≥0,1−∫0∞gisds=l1>0,and
(G2)
(14)gi′t≤−ξtgiσt,t≥0,1≤σ<32,where ξt is a positive differentiable function. It satisfies for some positive constant l(15)ξ′tξt≤l,ξ′t≤0,∫0∞ξsds=+∞,∀t>0.
Furthermore, for anyt0>0 and 1<σ<3/2, there exists a positive constant Cσ such that
(16)t1+∫t0tξsds1/2σ−1≤Cσ,∀t≥t0.
(G3) We take
(17)f1u,v=au+v2r+1u+v+buruvr+2,f2u,v=au+v2r+1u+v+bvrvur+2,where a,b>0 are constants and r>−1.
One can easily verify that(18)uf1u,v+vf2u,v=2r+2Fu,v,∀u,v∈ℝ2,where
(19)Fu,v=12r+2au+v2r+2+2buvr+2.
(G4) μ≥0,μ>0 in L0,L, where 0≤L0≤L.Lemma 5.
Forr>−1, there exist η>0 such that for any u,v∈V∩V00,L, we have
(20)u+vLx2r+22r+2+2uvLxr+2r+2≤ηl1uxH2+l2vxH2r+2.Proof.
From Minkowski inequality, we have(21)u+vLx2r+22≤2uLx2r+22+vLx2r+22.
We apply successively Holder’s and Young’s inequalities we obtain(22)uvLxr+2r+2≤uLx2r+2vLx2r+2≤cl1uxH2+l2vxH2.
We combine the two previous inequalities and the embeddingV∩V00,L,↪Lx2r+20,L, we get (20).Lemma 6.
There exist two positive constantsΛ1 and Λ2 such that
(23)xfiu,v2dx≤Λil1∫0Lxux2dx+l2∫0Lxvx2dx2r+3,∀x∈0,L,i=1,2.Proof.
It is clear that(24)f1u,v≤Cu+v2r+3+ur+1vr+2≤Cu2r+3v2r+3+ur+1vr+2,
Applying Young’s inequality with exponentsq=2r+3/r+1,q′=2r+3/r+2, in the last term in the above inequality, we get
(25)∣f1u,v∣≤Cu2r+3+v2r+3.
Consequently, by using Poincaré’s inequality and (20), we obtain
(26)∫0Lxfiu,v2dx≤CuxH22r+3+vxH22r+3≤Λ1l1uxH2+l2uxH22r+3.
Similarly, we get the inequality forf2. The proof is completed.We define the energy function as(27)Et=12∫0Lxut2dx+12∫0Lxvt2dx+121−∫0tg1sds∫0Lxux2dx+121−∫0tg1sds∫0Lxvx2x,tdx−∫0LFu,vdx+12g1∘uxt+12g2∘uxt,where
(28)g2∘uxt=∫0L∫0txgt−suxx,t−uxx,s2dsdx.Lemma 7.
Letu,v be the solution of system (1) then for all t≥0(29)ddtEt=−∫0Lxμxut2dx−∫0Lxμxvt2dx+12g′∘uxt−12g1t∫0Lxux2+12g′∘vxt−12g2∫0Lxvx2dx.
Hence,Et is a nonincreasing function.Proof.
Multiplying the first and the second equations in (1) by xut and xvt, respectively, integrating over 0,L, summing up, we obtain (30)
(30)∫0Lxuttutdx−∫0Lxuxxutdx+∫0L∫0tg1t−sxuxx,sxdsutdx+∫0Lxvttvtdx−∫0Lxvxvt+∫oL∫0tg2t−sxvxx,sxdsvtdx=−∫0Lxμxut2dx−∫0Lxμxvt2xμxvt2dx+∫0Lau+v2r+1u+v+buruvr+2xutdx+∫0Lau+v2r+1u+v+bvrvur+2xvtdx.
By integration by parts, we obtain(31)∫0Lxuttutdx=12ddt∫0Lxut2dx,(32)∫0Lxvttvtdx=12ddt∫0Lxvt2dx,(33)−∫0Lxuxxutdx=12ddt∫0Lxux2dx,(34)−∫0Lxvxxvtdx=12ddt∫0Lxvx2dx,(35)12r+2∫0Lxf1u,vuutdx+12r+t∫0Lxf2u,vvvtdx=12r+2ddt∫0Lau+v2r+2+2buvr+2xdx,(36)∫0L∫0tg1t−sxuxsxdsuttdx=12ddtg1∘uxt−∫0tg1sds∫0Lxux2dx−12g′1∘uxt+12g1t∫0Lxux2dx,(37)∫0L∫0tg2t−sxvxsxdsvttdx=12ddtg2∘uxt−∫0tg2tds∫0Lxvx2dx−12g′2∘uxt+12g2t∫0Lxvx2dx,
Combining (32)−(2.22) in (31), we get (30).


## 3. Global Existence

In order to state and prove the global existence, we set the following notation(38)It≔Iut,vt=1−∫0tg1sds∫0Lxux2dx+1−∫0tg2sds∫0Lxvx2dx+g1∘uxt+g2∘vxt−2r+2∫0Lxau+v2r+2+2buvr+2dx,(39)Jt≔Jut,vt=121−∫0tg1sds∫0Lxux2dx+121−∫0tg2sds∫0Lxvx2dx+12g1∘uxt+12g2∘vxt−∫0Lxau+v2r+2+2buvr+2dx,we remark that
(40)Et=Jt12∫0Lxut2dx+∫0Lxvt2dx.Lemma 8.
Assume that (G1), (G2), and (20) hold also for any u0,v0∈V02 and u1,v1∈H2 satisfying
(41)I0>0,β≔η2r+2r+1E0r+1<1,where
(42)E0=J0+12∫0Lxu12dx+12∫0Lxv12dx.
Then, there existst∗>0 such that
(43)It>0,∀t∈0,t∗.Proof.
SinceI0>0, then from the continuity of I(t), there exist tm≤t∗ such that It≥0 for all t∈0,tm; this implies that we have a maximum time value noting Tm such that
(44)ITm=0andIt>0,forall0≤t<Tm.
From formulas ofJt and It together with (G1), we have
(45)Jt=r+12r+21−∫0tg1sds∫0Lxux2ds+1−∫0tg2sds∫0Lxvx2dx+r+12r+2g1∘uxt+12g2∘vxt+12r+2It≥r+12r+2l1∫0Lxux2dx+l2∫0Lxvx2dx+g1∘uxt+12g2∘vxt,hence,
(46)l1∫0Lxux2dx+l2∫0Lxvx2dx≤2r+2r+1Jt≤2r+2r+1Et≤2r+2r+1E0,∀t∈0,Tm,
Recalling Lemma5 and (41), we get
(47)2r+2∫0LFuTm,vTmdx≤ηl1∫0Lxux2dx+l2∫0Lxvx2dxr+2≤η2r+2r+1E0r+1l1∫0Lxux2dx+l2∫0Lxvx2dx=βl1∫0Lxux2dx+l2∫0Lxvx2dx<1−∫0tg1sds∫0Lxux2dx+1−∫0tg2sds∫0Lxvx2dx+g1∘uxt+g2∘vxt,consequently
(48)1−∫0tg1sds∫0Lxux2dx+1−∫0tg2sds∫0Lxvx2dx+g1∘uxt+g2∘vxt−2r+2∫0LxFu,vdx>0,we deduce that It>0,∀t∈0,Tm. By repeating the procedure, Tm is extended to t∗.Theorem 9.
Suppose that (G1), (G2), and (20) hold. Then, for any u0,v0∈V02 and u1,v1∈H2 satisfying (41), the solution of system (1) is a bounded and globally in time.Proof.
To achieve the proof of this theorem, it suffices to show thatuxH2+vxH2+utH2+vtH2 is bounded independently of t. As Et is a nonincreasing function, we have
(49)E0≥Et,in the other hand and for the definition of It, we have
(50)xau+v2r+2+2buvr+2dx=It−1−∫0tg1sds∫0Lxux2dx−1−∫0tg2sds∫0Lxvx2dx−g1∘uxt−g2∘vxt,we introduce (49) into (50), we get
(51)E0≥Et=12∫0Lxut2dx+12∫0Lxvt2dx+121−∫0tg1sds∫0Lxux2dx+121−∫0tg2sds∫0Lxvx2x,tdx+12g1∘uxt+12g2∘vxt+It,by using (14), (15), and (41), (51) yields
(52)E0≥Et≥12∫0Lxut2dx+12∫0Lxvt2dx+r+12r+2l1∫0Lxux2dx+r+12r+2l2∫0Lxvx2dx≥μ0∫0Lxut2dx+∫0Lxvt2dx+∫Lxux2dx+∫0Lxvx2dx.
So(53)uxH2+vxH2+utH2+vtH2≤τE0,where
(54)τ≔max2,2r+2r+1l1,2r+2r+1l2.
The proof is completed.


## 4. Decay of Solutions

Throughout this section, we will study the asymptotic behavior of solutions’ decay by constructing a suitable Lyapunov function; to do so, forN, ε1, and ε2 are positive constants, we define the following function as
(55)Ft≔Et+ε1Φt+ε2χt+ψt,where
(56)Φt≔ξt∫0Lxutudx+ξt∫0Lxvtvdx+ξt2∫0Lxμxu2+v2dx,(57)xt≔−ξt∫0Lxut∫0tg1t−sut−usdsdx−ξt∫0Lxvt∫0tg2t−svt−vsdsdx,and
(58)ψt≔ξt∫0Lxuthxuxdx+ξt∫0Lxvthxvxdx,with h∈C10,L,h0=hL=0,xhx′≤x,In the first step, we prove the equivalence betweenFt and Et given in the following lemma.Lemma 10.
For a choice ofε1 and ε2 small enough, we find two positive constants α1 and α2 such that
(59)α1Ft≤Et≤α2Ft.Proof.
By using Young inequality, follow by recalling Lemma1 and the fact that 0<ξt≤ξ0, we get
(60)ε1ξt∫0Lxutudx≤ε12ξ0∫0Lxut2dx+Cp∫0Lxux2dx,(61)ε1ξt∫0Lxvtvdx≤ε12ξ0∫0Lxvt2dx+Cp∫0Lxvx2dx,(62)−ε2ξt∫0Lxut∫0tg1t−sut−usdsdx≤ε22ξ0∫0Lxut2dx+Cp1−l1g1∘uxt,(63)−ε2ξt∫0Lxvt∫0tg2t−svt−vsdsdx≤ε22ξ0∫0Lxvx2dx+Cp1−l2g2∘vxt,(64)ξt∫0Lxuthxuxdx≤ξ02h∞∫0Lxux2dx+∫0Lxut2dx,(65)ξt∫0Lxvthxvxdx≤ξ02h∞∫0Lxvx2dx+∫0Lxvt2dx,(66)ξt2∫0Lxμxu2+v2dx≤ξ02μ∞Cp∫0Lxux2dx+∫0Lxvx2dx,combining (60)−(66) in (55), we get
(67)Ft−NEt≤ε1+ε22ξ0+ξ02h∞∫0Lxut2dx+ε1+ε22ξ0+ξ02h∞∫0Lxvt2dx+1+μ∞2ε1Cpξ0+ξ02h∞p+1∫0Lxux2dx+1+μ∞2ε1Cpξ0+ξ02h∞q+1∫0Lxvx2dx+ε22Cpξ01−l1g1∘uxt+1−l2g2∘vxt.
If we chooseε1, ε2 small enough, and N large enough we find α1, α2>0 such that (59) holds true.Now, we state a Lemma corresponding to the boundness ofgovxt. It will be used in the calculus.Lemma 11.
Letw∈L∞0,T;H be such that wx∈L∞0,t;H and g be a continuous function on 0,T and suppose that. Then, there exists a constant C>0 such that
(68)gowxt≤Csup0<s<Tw.,sH2∫0tg1−θsdsρ−1/ρ−1+θ×∫0tgρt−swx.,sH2dsθ/ρ−1+θ,∀0<θ<1andρ>1.(69)gowxt≤ctwx.,tH2+∫0twx.,sH2dsρ−1/ρ×∫0tgρt−swx.,t−wx.,sH2ds1/2forallρ>1.Proof.
(1)
For anyσ>1, we have(70)gowxt=∫0tgt−s1−θ/σwx.,t−wx.,s2/σgt−sσ−1+θ/σwx.,t−wx.,s2σ−1/σds.
Applying Holder’s inequality with exponentsσ and σ/σ−1, we get
(71)gowxt≤∫0tg1−θt−swx.,t−wx.,s2ds1/σ×∫0tgσ−1+θ/σ−1t−swx.,t−wx.,s2dsσ−1/σ,
We setσ=ρ−1+θ/ρ−1, (71) yields
(72)gowxt≤∫0tg1−θt−swx.,t−wx.,sH2dsρ−1/ρ−1+θ×∫0tgρt−swx.,t−wx.,sH2θ/ρ−1+θ.
It is easy to see that(73)∫0tg1−θt−swx.,t−wx.,sH2ds≤Csup0<s<Twx.,sH2∫0tg1−θsds.
We obtain (68) by combining (72) and (73).
(2)
We setθ=1 in (72) and it suffices to note that
(74)∫0twx.,t−wx.,sH2ds≤2twx.,tH2+2∫0twx.,sH2ds,to arrive at (69).In the next, we present three lemmas in which we give an upper bound of each derivative’s functions inFt.Lemma 12.
Suppose thatr>−1 and (39) hold. Then
(75)Φ′t≤1+l2δξt∫0Lxut2dx+1+l2δξt∫0Lxvt2dx−ξtl1−δCpl2−Cpμ∞2∫0Lxux2dx−ξtl2−δCpl2−Cpμ∞2∫0Lxvx2dx−ξt+ξt2l1∫0tg12−σsdsg1σ∘uxt+ξt2l2∫0tg22−σsdsg2σ∘vxt+ξt2r+2au+v2r+2+2buvr+2dx.
For anyδ>0.Proof.
After derivation of (56), we recall the differential equations in (1), we get
(76)Φ′t=ξ′t∫0Lxutudx+ξt∫0Lxut2dx+t∫0Lxuttudx+ξ′t∫0Lxvtvdx+ξ′t∫0Lxvt2dx+ξt∫0Lxvttvdx+ξ′t2∫0Lxμxu2+v2dx+ξt∫0Lxμxutu+vtvdx=ξ′t∫0Lxutudx+ξt∫0Lxut2dx−ξt∫0Lxux2dx+ξt∫0Lxux∫0tg1t−suxsdsdx+ξ′t∫0Lxvtvdx+ξt∫0Lxvt2dx−ξt∫0Lxvx2dx+ξt∫0Lxvx∫0tg2t−svxsdsdx+ξt2r+2au+v2r+2+2buvr+2dx.
By Young’s inequality and from, (G1), (G2), and Lemma 1, we arrive at
(77)ξt∫0Lxuxt∫0tg1t−suxsdsdx≤ξt2∫0Lxux2dx+ξt2∫0Lx∫0tg1t−suxs−uxt+uxtds2dx≤ξt2∫0Lxux2dx+ξt21+η11+l12∫0Lxux2tdx+ξt21+1η1∫0tg12−σsdsg1σ∘uxt=ξt1+1+η11−l122∫0Lxux2dx+ξt21+1η1∫0tg12−σsdsg1σ∘uxt+ξtr+2∫0Lau+v2r+2+2buvr+2dx,similarly, we get
(78)∫0Lxvxt∫0tg1t−svxsdsdx≤ξt1+1+η21−l222∫0Lxvx2dx+ξt21+1η2∫0tg22−σsdsg2σ∘vxt.
For anyη1 and η2>0. We also have
(79)ξ′t∫0Lxutudx≤ξt2ξ′tξtCpδ∫0Lxux2dx+1δ∫0Lxut2dx≤ξt2Cplδ∫0Lxux2dx+lδ∫0Lxut2dx,∀δ>0,and similarly, we get
(80)ξ′t∫0Lxvtvdx≤ξt2Cplδ∫0Lxvx2dx+lδ∫0Lxvt2dx.
Also, by Lemma1, we have
(81)ξ′t2∫0Lxμxu2+v2dx≤μ∞ξt2Cp∫0Lxux2+vx2dx.
Combining (77)−(81) in (76) leads to
(82)Φ′t≤1+l2δξt∫0Lxut2dx+1+l2δξt∫0Lxvt2dx−ξt21−1+η11−l12−δCpl−Cpu∞∫0Lxut2dx−ξt21−1+η21−l22−δCpl−Cpu∞∫0Lxut2dx+ξt21+1η1∫0tg12−σsdsg1σ∘uxt+ξt21+1η2∫0tg22−σsdsg2σ∘vxt+ξt2r+2∫0Lau+v2r+2+2buvr+2dx,we choose η1=l1/1−l1 and η2=l2/1−l2, hence (75) is established.Lemma 13.
Suppose thatr>−1, (G1), (G2), and (41) hold. Then
(83)χ′t≤ξtθ2+c1+c1′+21−l12∫0Lxux2dx+ξtθ2+c2+c2′+21−l22∫0Lxvx2dx+ξtθ−∫0tg1sds+θl∫0Lxut2dx+ξtθ−∫0tg2sds+θl∫0Lxvt2dx+12θ+2θ+Cp2+l4θ×ξt∫0tg12−σsdsg1σ∘uxt+12θ+2θ+Cp2+l4θ×ξt∫0tg22−σsdsg2σ∘vxt−Cp4θξtg10g1′∘uxt−Cp4θξtg20g2′∘vxt,for any θ>0.Proof.
A derivation of (57) gives
(84)χ′t=−ξ′t∫0Lxut∫0tg1t−sut−usdsdx−ξt∫0Lxutt∫0tg1t−sut−usdsdx−ξt∫0Lxutddt∫0tg1t−sut−usdsdx−ξ′t∫0Lxvt∫0tg2t−svt−vsdsdx−ξt∫0Lxvtt∫0tg2t−svt−vsdsdx−ξt∫0Lxvtddt∫0tg2t−svt−vsdsdx,by using Liebniz’s formula, we get
(85)χ′t=−ξ′t∫0Lxut∫0tg1t−sut−usdsdx−ξt∫0Lxutt∫0tg1t−sut−usdsdx−ξt∫0Lxut∫0tg1′t−sut−usdsdx−ξt∫0tg1sds∫0Lxut2dx−ξ′t∫0Lxvt∫0tg2t−svt−vsdsdx−ξt∫0Lxvtt∫0tg2t−svt−vsdsdx−ξt∫0Lxvt∫0tg2′t−svt−vsdsdx−ξt∫0tg2sds∫0Lxvt2dx.
Recalling the differentials equation in (1), we get
(86)χ′t=−ξ′t∫0Lxut∫0tg1t−sut−usdsdx+ξt∫0Lxμxut∫0tg1t−sut−usdsdx+ξt∫0Lxux∫0tg1t−suxt−uxsdsdx−ξt∫0Lx∫0tg1t−suxsds∫0tg1t−suxt−uxsdsdx−ξt∫0Lxf1u,v∫0tg1t−sut−usdsdx−ξt∫0Lxut∫0tg1′t−sut−usdsdx−ξt∫0tg1sds∫0Lxut2dx−ξ′t∫0Lxvt∫0tg2t−svt−vsdsdx+ξt∫0Lxμxvt∫0tg2t−svt−vsdsdx+ξt∫0Lxvx∫0tg2t−svxt−vxsdsdx−ξt∫0Lx∫0tg2t−svxsds∫0tg2t−svxt−vxsdsdx−ξt∫0Lxf2u,v∫0tg2t−svt−vsdsdx−ξt∫0Lxvt∫0tg2′t−svt−vsdsdx−ξt∫0tg2sds∫0Lxvt2dx.
We will estimate all term in (86) by Young’s inequality, Lemma 1, (G1), and (G2).
(87)−ξ′t∫0Lxut∫0tg1t−sut−usdsdx≤ξtξ′tξtθ∫0Lxut2dx+Cp4θ∫0tg12−σsdsg1σ∘uxt≤θlξt∫0Lxut2dx+Cp4θξt∫0tg12−σsdsg1σ∘uxt,(88)−ξt∫0Lxμxut∫0tg1t−sut−usdsdx≤ξtμ∞θ∫0Lxut2dx+Cp4θ∫0tg12−rsdsg1r∘uxt,and
(89)−ξt∫0Lxux∫0tg1t−suxt−uxsdsdx≤θξt∫0Lxux2dx+14θξt∫0tg12−σsdsg1σ∘uxt,−ξt∫0Lx∫0tg1t−suxsds∫0tg1t−suxt−uxsdsdx≤2θ1−l12ξt∫0Lxux2dx+2θ+14θξt∫0tg12−σsdsg1σ∘uxt,−ξt∫0Lxut∫0tg1′t−sut−usdsdx≤θξt∫0Lxut2dx−g104θCpξtg1′∘uxt,and
(90)−ξt∫0Lxf1u,v∫0tg1t−sut−usdsdx≤Cp4θξt∫0tg12−σsdsg1σ∘uxt+c1θξt∫0Lxux2dx+c2θξt∫0Lxvx2dx,where
(91)c1≔Λ12r+2r+1E02r+1,c2≔Λ22r+2r+1E02r+1.
By the same technique, we obtain estimations on integrals corresponding tov, g2, and f2(92)−ξ′t∫0Lxvt∫0tg2t−svt−vsdsdx≤θlξt∫0Lxvt2dx+Cpl4θξt∫0tg22−σsdsg2σ∘vxt,−ξt∫0Lxμxvt∫0tg2t−svt−vsdsdx≤ξtμ∞θ∫0Lxvt2dx+Cp4θ∫0tg22−σsdsg2σ∘vxt,ξt∫0Lxvx∫0tg2t−svxt−vxsdsdx≤θξt∫0Lxvx2dx+14θξt∫0tg22−σsdsg2σ∘vxt,−ξt∫0Lx∫0tg2t−svxsds∫0tg2t−svxt−vxsdsdx≤2θ1−l22ξt∫0Lxvx2dx+2θ+14θξt∫0tg22−σsdsg2σ∘uxt,−ξt∫0Lxvt∫0tg2′t−svt−vsdsdx≤θξt∫0Lxvt2dx−g204θCpξtg2′∘vxt,and
(93)−ξt2r+2∫0Lxf2u,v∫0tg2t−svt−vsdsdx≤Cp4θξt∫0tg22−σsdsg2σ∘vxt+c1′θξt∫0Lxux2dx+c2′θξt∫0Lxvx2dx,where
(94)c1′≔Λ1′2r+2r+1E02r+1,c2′≔Λ2′2r+2r+1E02r+1.
A combination of (87)−(93) into (86) yields (83).Lemma 14.
Suppose thatr>−1, (G1), (G2), (G4), and (41) hold. Then
(95)ψt′≤ξt1+h∞θ2+μ∞∫0Lxux2dx+ξt1+h∞θ2+μ∞∫0Lxvx2dx+ξt∫0L1+h∞θ12θ+14θμxxut2dx+ξt∫0L1+h∞θ12θ+14θμxxvt2dx+ξth∞Cp4θ∫0tg12−σsdsg1σ∘uxt+ξth∞Cp4θ∫0tg22−σsdsg2σ∘vxt+ξt2r+2∫0Lxau+v2r+2+2buvr+2dx,
For anyθ>0.Proof.
We derivate (58) and integrate by part and we finish by using the differential equations in (1), we get
(96)ψ′t=ξ′t∫0Lutxhxuxdx+ξ′t∫0Lvtxhxvxdx+ξt∫0Lutxhxutxdx+q+1ξ′t∫0Lvtxhxvtxdx+ξt∫0Luttxhxuxdx+ξt∫0Lvttxhxvtxdx=ξ′t∫0Lutxhxuxdx+ξ′t∫0Lvtxhxvxdx−12ξt∫0Lxhx′ut2dx−12ξt∫0Lxhx′vt2dx−12ξt∫0Lxhx′ux2dx−12ξt∫0Lxhx′vx2dx−ξt∫0Lxμxhxutuxdx−ξt∫0Lxμxhxvtvxdx−ξt∫0Lxhxux∫0tg1t−sut−usdsdx−ξt∫0Lxhxvx∫0tg2t−svt−vsdsdx+ξt2r+2∫0Lxhx′au+v2r+2+2buvr+2dx,
Applying Young’s and Poincare’s inequalities, we get(97)ξ′t∫0Lutxhxuxdx≤ξth∞θ∫0Lxux2dx+14θ∫0Lxut2dx,(98)∫0Lxμxhxutuxdx≤h∞θ∫0Lxμxux2dx+14θ∫0Lxμxut2dx,finally
(99)∫0Lxhxux∫0tg1t−sut−usdsdx≤h∞θ∫0Lxux2dx+Cp4θ∫0tg12−rsdsg1r∘uxt,
similarly(100)ξ′t∫0Lvtxhxvxdx≤ξth∞θ∫0Lxvx2dx+14θ∫0Lxvt2dx,∫0Lxμxhxvtvxdx≤h∞θ∫0Lxμxvx2dx+14θ∫0Lxμxvt2dx,and
(101)∫0Lxhxvx∫0tg1t−svt−vsdsdx≤h∞θ∫0Lxvx2dx+Cp4θ∫0tg12−rsdsg1r∘vxt,
Using (97)–(101) and xhx′≤x, we obtain (95).Next Theorem show that solutions decreases exponentially with respect toξt and σ.Theorem 15.
Suppose thatr>−1, (G1), and (G2) hold and taking u0,v0∈V02, and u1,v1∈H2 such that (41) hold true. Then, for each t0>0, there exist positive constants K and k such that
(102)Et≤Ke−k∫t0tξsds,σ=1,K1+∫t0tξsds−1/σ−1,1<σ<32,∀t≥t0.Proof.
Sinceg1 and g2 is continuous and g10>0, g20>0 then for any t0>0, we have
(103)∫0tgisds≥∫t0tgisds=g1,0>0,∀t≥t0,i=1,2.
Asξ is a positive decreasing function hence 1<ξt/ξ0 and by recalling Lemmas 7, 12, 13, 14, and (101), we get
(104)F′t=E′t+ε1Φ′t+ε2χ′+ψ′t≤−∫0LN−h∞4θμx−1+h∞4θ−ε11+12δ+ε2g1,0−θ−θlξtxut2dx−∫0LN−h∞4θμx−1+h∞4θ−ε11+12δ+ε2g2,0−θ−θlξtxvt2dx+2ε1+1ξt∫0Lxau+v2r+2+2buvr+2dx+12−ε2ξ04θCpg10g1′∘uxt+12−ε2ξ04θCpg20g2′∘vxt−ε12l2−Cpδl+μ∞−ε2θ1+c2+c2′+21−l22−1+h∞θ2+μ∞×ξt∫0Lxvx2dx+ε12l1+ε212θ+2θ+2+l+Cph∞+4θ4θ×ξt∫0Lg12−σsdsg1σ∘uxt+ε12l2+ε212θ+2θ+2+l+Cph∞+4θ4θ×ξt∫0Lg22−σsdsg2σ∘uxt.
We takeδ<1/2Cplminl1,l2, hence
(105)l1−δCpl>l12,l2−δCpl>l22,
Also, we takeθ>maxg1,0/21+l,g2,0/21+l, hence
(106)g1,0−1+lθ<12g1,0,g2,0−1+lθ<12g2,0.
Now, we chooseε2 small enough such
(107)k1≔12−ε2ξ04θCpg10>0,k1≔12−ε2ξ04θCpg20>0,
As far asδ, θ, and ε2 are fixed, we then pick ε1 so small that
(108)k3≔ε14l1−2Cpμ∞−ε2θ1+c1+c1′+21−l12−1+h∞θ2+μ∞>0,k4≔ε14l2−2Cpμ∞−ε2θ1+c2+c2′+21−l22−1+h∞θ2+μ∞>0,k5≔−ε11+12δ+ε2g1,0−1+lθ>0,k6≔−ε11+12δ+ε2g2,0−1+lθ>0,−ε12l1+ε212θ+2θ+2+l+Cph∞+4θ4θ∫0tg12−σsds>0,and
(109)−ε12l3+ε212θ+2θ+2+l+Cph∞+4θ4θ∫0tg22−σsds>0,
Finally we chooseN>h∞/4θ large enough such that
(110)N−h∞4θμx−1+h∞4θ>0.
After fixed all this choices then from (14) and (27), we obtain for some σ>0,
(111)F′t≤−σξt∫0Lxut2dx+∫0Lxvt2dx−∫0Lxau+v2r+2+2buvr+2dx+∫0Lxux2dx+∫0Lxvx2dx+g1σ∘uxt+g2σ∘vxt≤−γξtEt,∀t≥t0.We distingue two cases accordingσCase 16.
σ=1
We recall lemma10, and the estimate (111) gives
(112)F′t≤−γα1ξtFt,∀t≥t0.
A simple integration of the above inequality over (t0,t) leads to
(113)F′t≤Ft0e−γα1∫t0tξsds,∀t≥t0.
Hence, (102) is established.Case 17.
1<σ<3/2
By using (16), we get
(114)git1−σ≥σ−1∫t0tξsds+git01−σ,i=1,2.
So, for∀0<τ<2−σ<1, (hence 1−τ/σ−1>1), we have
(115)∫0∞gi1−τsds≤∫0∞1σ−1∫t0tξsds+git01−σ1−τ/σ−1ds.In the other hand, by using the fact that∫0∞ξsds=+∞, we obtain
(116)∫0∞gi1−τsds<∞,∀0<τ<2−σ,fori=1,2.So form (i) of Lemma11 with θ=τ and ρ=σ and (41) yield
(117)gi∘wxt≤CiE0∫0∞gi1−τsdsσ−1/σ−1+τgiσ∘wxtτ/σ−1+τ≤Ci′giσ∘wxtτ/σ−1+τ,for i=1,2 and w=u,v, respectively, with Ci′ are positive constants.Therefore, for anyσ1>1, we arrive at
(118)Eσ1t≤C″Eσ1−10∫0Lxut2dx+∫0Lxvt2dx+∫0Lxux2dx+∫0Lxvx2dx−∫0Lxau+v2r+2+2buvr+2dx+C1″g1∘uxtσ1+C2″g2∘vxtσ1≤C″Eσ1−10∫0Lxut2dx+∫0Lxvt2dx+∫0Lxux2dx+∫0Lxvx2dx−∫0Lxau+v2r+2+2buvr+2dx+C1‴g1σ∘uxtτσ1/σ−1+τ+C2‴g2σ∘vxtτσ1/σ−1+τ.By settingτ=1/2 and σ1=2σ−1 which give τσ1/σ−1+τ=1, the estimate (119) becomes,
(119)Eσ1t≤Γ∫0Lxut2dx+∫0Lxvt2dx+∫0Lxux2dx+∫0Lxvx2dx−∫0Lxau+v2r+2+2buvr+2dx+g1σ∘uxt+g2σ∘vxt,for some Γ>0. Form the equivalence between F and E and by combining the above inequality with (111), we obtain
(120)F′t≤−σΓξtEσ1t≤−σΓα1σ1Fσ1t,∀t≥t0.By twice integration overt0,t and t0,∞ of (119) successively leads to
(121)Ft≤C1∗1+∫t0tξsds−1/σ1−1,∀t≥t0,(122)∫t0∞Ftdt≤C1∗∫t0∞11+∫t0tξsds1/σ1−1dt.Since1/σ1−1>0 and 1+∫t0tξsds⟶+∞ as t⟶+∞, we get
(123)∫t0∞Ftdt≤∞.Again from (121), we have
(124)tFt≤C1∗t1+∫t0tξsds1/σ1−1≤Cσ,∀t≥t0,which implies that
(125)supt≥t0tFt<∞.Summing (123) and (125), we get
(126)∫t0∞Ftdt+supt≥t0tFt<∞.By recalling (ii) of Lemma11 with ρ=σ, Lemma 10, and by using (27) and the above bounded, we have
(127)gi∘wxt≤Ci2∗twxx,tH2+∫0twxx,sH2dsσ−1/σ×∫0tgiσt−swxx,t−wxx,sH2ds1/σ≤Ci2∗tFt+∫t0tFsdsσ−1/σgiσ∘wxt1/σ≤Ci3∗giσ∘wxt1/σ,Hence,(128)giσ∘wxt≥Ci3∗−σgi∘wxtσ,for i=1,2, w=u,v, respectively, and some positive constants Ci3∗.Consequently, introduce (128) in (111) and in, (118), we find
(129)F′t≤−C4ξt∫0Lxut2dx+∫0Lxvt2dx+∫0Lxux2dx+∫0Lxvx2dx−∫0Lxau+v2r+2+2buvr+2dx+g1∘uxtσ+g2∘vxtσ,and
(130)Eσt≤C5ξt∫0Lxut2dx+∫0Lxvt2dx+∫0Lxux2dx+∫0Lxvx2dx−∫0Lxau+v2r+2+2buvr+2dx+g1∘uxtσ+g2∘vxtσ,for all t≥0 and some positive constant C4,C5.By combining the last two inequalities and along the equivalence betweenF and E, we obtain
(131)F′t≤−C6ξtFσt,∀t≥t0,for some constant C6>0.A simple integration of (131) over t0,t gives
(132)Ft≤C91+∫t0tξsds−1/σ−1,∀t≥t0.The proof is completed by using (59).



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