# Global Existence for Two Singular One-Dimensional Nonlinear Viscoelastic Equations with respect to Distributed Delay Term

Authors: Abdelbaki Choucha; Salah Mahmoud Boulaaras; Djamel Ouchenane; Ali Allahem
Journal: Journal of Function Spaces
 (2021)

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## Abstract

In this current work, we are interested in a system of two singular one-dimensional nonlinear equations with a viscoelastic, general source and distributed delay terms. The existence of a global solution is established by the theory of potential well, and by using the energy method with the function of Lyapunov, we prove the general decay result of our system.


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## Body

## 1. Introduction

We are interested in the following system:(1)utt−1xxuxx+∫0tg1t−s1xxuxx,sxds+μ1ut+∫τ1τ2μ2ρutx,t−ρdρ=f1u,v,inQ,vtt−1xxvxx+∫0tg2t−s1xxvxx,sxds+μ3vt+∫τ1τ2μ4ρvtx,t−ρdρ=f2u,v,inQ,with
(2)ux,0=u0x,utx,0=u1x,x∈0,L,vx,0=v0x,vtx,0=v1x,x∈0,L,utx,−t=f0x,t,vtx,−t=g0x,t,t∈0,τ2,uL,t=vL,t=0,∫0Lxux,tdx=∫0Lxvx,tdx=0,where Q=0,L×0,T, L<∞, T<∞, g1., g2.:ℝ+⟶ℝ+, μ1,μ3>0, the second integral represents the distributed delay and μ2,μ4:τ1,τ2⟶ℝ are bounded functions, where τ1,τ2 are two real numbers satisfying 0≤τ1<τ2, and f1.,., f2.,.:ℝ2⟶ℝ are defined functions later.Three decades ago, these problems that arise in one-dimensional elasticity have been studied and developed with regard to viscosity with long-term memory. And it has been studied in many fields of science, engineering, medical sciences, and chemistry, as well as population and other matters; see, for example, [1–24]. Recently, in the absence of delay (μi=0,i=1..4), problem (1) was studied in [25], and also later in [26], the authors considered problem (1) with localized frictional damping term. We also know that delay, especially distributed delay, is a phenomenon in our life and is almost found in various fields, and its inclusion in any problem makes it more important. The distributed delay in many works has been studied and many authors have taken care of it, for example, [5, 9, 27, 28]. Based on all this and the results of the research papers [14, 15, 17, 28–30, 31], the introduction of the term distributed delay as a damping mechanism in problem (1) makes it a new problem from what has been previously studied.And we have divided this paper into the following. We present in the second section the definitions, basics, and theories of function spaces that are required throughout the rest of the paper. In Section3, we present the energy function while proving to be decreasing. And in the final section, the general decay is obtained by applying the energy method and the function of Lyapunov.


## 2. Preliminaries

LetLxp=Lxp0,L be the weighted Banach space equipped with the norm
(3)uLxp=∫0Lxupdx1/p,H=Lx20,L be the Hilbert space of square integral functions having the finite norm
(4)uH=∫0Lxu2dx1/2,and K=Lx20,L×0,1×τ1,τ2 be the Hilbert space equipped with the norm
(5)zK,μ2=∫01∫τ1τ2∣μ2ρ∣zHdρdρ.V=Vx1 is the Hilbert space equipped with the norm
(6)uV=uH2+uxH21/2,(7)V0=u∈VsuchthatuL=0.Theorem 1 [27].
For2<p<4 and ∀v in V0, we have
(8)∫0Lxvpdx≤C∗vxH=Lx20,Lp,where C∗ is a constant depending on L and p only.As in [18], introducing the new variables
(9)zx,ρ,ϱ,t=utx,t−ϱρ,yx,ρ,ϱ,t=vtx,t−ϱρ,yields
(10)ϱztx,ρ,ϱ,t+zρx,ρ,ϱ,t=0,zx,0,ϱ,t=utx,t,ϱytx,ρ,ϱ,t+yρx,ρ,ϱ,t=0,yx,0,ϱ,t=vtx,t.Problem (1) arrives at
(11)utt−1xxuxx+∫0tg1t−s1xxuxx,sxds+μ1ut+∫τ1τ2∣μ2ϱ∣utx,t−ϱdϱ=f1u,v,vtt−1xxvxx+∫0tg2t−s1xxvxx,sxds+μ3vt+∫τ1τ2∣μ4ϱ∣vtx,t−ϱdϱ=f2u,v,ϱztx,ρ,ϱ,t+zρx,ρ,ϱ,t=0,ϱytx,ρ,ϱ,t+yρx,ρ,ϱ,t=0,where
(12)x,ρ,s,t∈0,L×0,1×τ1,τ2×0,∞.With the initial data and boundary conditions(13)ux,0,vx,0=u0x,v0x,in0,L,utx,0,vtx,0=u1x,v1x,in0,L,utx,−t,vtx,−t=f0x,t,g0x,t,in0,L×0,τ2,u0,t=uL,t=v0,t=vL,t=0,zx,ρ,ϱ,0=f0x,ρϱ,in0,L×0,1×0,τ2,yx,ρ,ϱ,0=g0x,ρϱ,uL,t=vL,t=0,∫0Lxux,tdx=∫0Lxvx,tdx=0.We have the following assumptions:(G1)git:ℝ+⟶ℝ+ are C1, nonincreasing functions satisfying
(14)gis≥0,gi′s≤0,gi0>0,1−∫0∞gisds=li>0,i=1,2,(G2)∃ξt>0 a differentiable function, such that
(15)gi′t≤−ξtgiσt,i=1,2,t≥0,1≤σ<32,and ξt satisfies for some l<1(16)ξ′t≤0,ξ′tξt≤l,∫0∞ξsds=+∞,∀t>0.And also, where1<σ<3/2, ∀t0>0 fixed, ∃Cσσ>0, such that
(17)t1+∫t0tξsds1/2σ−1≤Cσ,∀t≥t0.(G3) we take(18)f1u,v=au+v2r+1u+v+buruvr+2,f2u,v=au+v2r+1u+v+bvrvur+2,where a,b>0 and r>−1.We have(19)uf1u,v+vf2u,v=2r+2Fu,v,∀u,v∈ℝ2,where
(20)Fu,v=12r+2au+v2r+2+2buvr+2.(G4)μ2,μ4:τ1,τ2⟶ℝ satisfying
(21)∫τ1τ2μ2ϱdϱ<μ1,(22)∫τ1τ2∣μ4ϱ∣dϱ<μ3.Theorem 2.
Assume (14) and p<3. Then, ∀u0,v0∈V02, v1,v2∈H2 and f0,g0∈K2 problem (1) has a unique local solution
(23)u,v,z,y∈C0,t∗;V02×K2∩C10,t∗;H2×K2,for t∗>0 small enough.Lemma 3.
Forr>−1, ∃η>0 such that ∀u,v∈V∩V00,L, we have
(24)u+vLx2r+22r+2+2uvLxr+2r+2≤ηl1uxH2+l2vxH2r+2.Proof.
It is clear that by using the Minkowski inequality we get(25)u+vLx2r+22≤2uLx2r+22+vLx2r+22.
Also, Hȯlder’s and Young’s inequalities give us
(26)uvLxr+2r+2≤uLx2r+2vLx2r+2(27)≤cl1uxH2+l2vxH2.
By applying the embeddingV∩V00,L↪Lx2r+20,L and (25), (27) gives (15).Lemma 4.
∃Λ1,Λ2>0 such that
(28)∫0Lxfiu,v2dx≤Λil1∫0Lxux2dx+l2∫0Lxvx2dx2r+3,∀x∈0,L,i=1,2.Proof.
We prove inequality forf1 and the same result also holds for f2.
It is clear that(29)f1u,v≤Cu+v2r+3+ur+1vr+2≤Cu2r+3+v2r+3+ur+1vr+2.
By Young’s inequality, with(30)q=2r+3r+1,q′=2r+3r+2,we get
(31)ur+1vr+2≤c1u2r+3+c2v2r+3.Therefore,
(32)f1u,v≤Cu2r+3+v2r+3.
Hence, by Poincaré’s inequality and (11), we obtain(33)∫0Lxfiu,v2dx≤CuxH22r+3+vxH22r+3≤Λ1l1uxH2+l2vxH22r+3.
The proof of lemma is complete.The energy function (see, e.g., [8, 19] and reference therein) is defined by
(34)Et=12∫0Lxut2dx+12∫0Lxvt2dx+121−∫0tg1sds∫0Lxux2dx+121−∫0tg2sds∫0Lxvx2x,tdx+12Kz,y+12g1∘uxt+12g2∘vxt−∫0LFu,vdx,where
(35)Kz,y=∫0L∫01∫τ1τ2xϱ∣μ2ϱ∣z2x,ρ,ϱ,t+∣μ4ϱ∣y2x,ρ,ϱ,tdϱdρdx,g∘uxt=∫0L∫0txgt−suxx,t−uxx,s2dsdx.Lemma 5.
Letu,v,z,y be the solution of system (11); then, Et is a nonincreasing function, that is, ∀t≥0(36)E′t≤−d1∫0Lxut2dx−d2∫0Lxvt2dx+12g1′∘uxt+12g2′∘vxt−12g1t∫0Lxux2dx−12g2t∫0Lxvx2dx≤0,where
(37)d1=μ1−∫τ1τ2μ2ϱdϱ>0,d2=μ3−∫τ1τ2∣μ4ϱ∣dϱ>0.Proof.
Multiplying equation (11)1,2 by xut,xvt, and integrating over 0,L, we find
(38)∫0Lxuttutdx−∫0Lxuxxutdx+μ1∫0Lxut2dx+∫0Lxut∫τ1τ2∣μ2ϱ∣zx,1,ϱ,tdρdx+∫0L∫0tg1t−sxuxx,sxdsutdx+∫0Lxvttvtdx−∫0Lxvxxvtdx+μ3∫0Lxvt2dx+∫0Lxvt∫τ1τ2∣μ4ϱ∣yx,1,ϱ,tdϱdx+∫0L∫0tg2t−sxvxx,sxdsvtdx=∫0Lau+v2r+1u+v+buruvr+2xutdx+∫0Lau+v2r+1u+v+bvrvur+2xvtdx.Using integration by parts, we get
(39)∫0Lxuttutdx=12ddt∫0Lxut2dx,(40)∫0Lxvttvtdx=12ddt∫0Lxvt2dx,(41)−∫0Lxuxxutdx=12ddt∫0Lxux2dx,(42)−∫0Lxvxxvtdx=12ddt∫0Lxvx2dx,(43)12r+2∫0Lxf1u,vuutdx+12r+2∫0Lxf2u,vvvtdx=12r+2ddt∫0Lau+v2r+2+2buvr+2xdx,(44)∫0L∫0tg1t−sxuxsxdsuttdx=12ddtg1∘uxt−∫0tg1sds∫0Lxux2dx(45)−12g1′∘uxt+12g1t∫0Lxux2dx,(46)∫0L∫0tg2t−sxvxsxdsvttdx=12ddtg2∘vxt−∫0tg2tds∫0Lxvx2dx−12g2′∘vxt+12g2t∫0Lxvx2dx.
Now, multiplying equation (11)3 by xz∣μ2ϱ∣ and integrating over 0,L×0,1×τ1,τ2, we get
(47)ddt12∫0L∫01∫τ1τ2ϱ∣μ2ϱ∣xz2dϱdρdx=−∫0L∫01∫τ1τ2∣μ2ϱ∣xzzρdϱdρdx=−12∫0L∫01∫τ1τ2x∣μ2ϱ∣ddρz2dϱdρdx=12∫0L∫τ1τ2x∣μ2ϱ∣zx,0,ϱ,t2−zx,1,ϱ,t2dϱdx=12∫τ1τ2∣μ2ϱ∣dϱ∫0Lxut2dx−12∫0L∫τ1τ2x∣μ2ϱ∣zx,1,ϱ,t2dϱdx.
Similarly, by multiplying equation (11)4 by xy∣μ4ϱ∣ and integrating over 0,L×0,1×τ1,τ2, we get
(48)ddt12∫0L∫01∫τ1τ2ϱμ4ϱxy2dϱdρdx=12∫τ1τ2μ4ϱdϱ∫0Lxvt2dx−12∫0L∫τ1τ2xμ4ϱy2x,1,ϱ,tdϱdx.Using Young’s and Cauchy-Schwartz inequalities, we have
(49)−∫0Lxut∫τ1τ2∣μ2ϱ∣zx,1,ϱ,tdϱdx≤12∫τ1τ2∣μ2ϱ∣dϱ∫0Lxut2dx+12∫0L∫τ1τ2∣μ2ϱ∣xz2x,1,ϱ,tdϱdx.
Similarly, we get(50)−∫0Lxvt∫τ1τ2μ4ϱyx,1,ϱ,tdϱdx≤12∫τ1τ2μ4ϱdϱ∫0Lxvt2dx+12∫0L∫τ1τ2μ4ϱxy2x,1,ϱ,tdϱdx.By combining (39), (40), (41), (42), (43), (45), (46), (47), (48), (49), and (50) in (38), we get (34) and (36).


## 3. Global Existence

In this section, we showed the global existence of the solutions of the system (11).First, introducing the following notation(51)It≔Iut,vt=1−∫0tg1sds∫0Lxux2dx+g1∘uxt+1−∫0tg2sds∫0Lxvx2dx+g2∘vxt+Kz,y−2r+2∫0Lxau+v2r+2+2buvr+2dx,(52)Jt≔Jut,vt=121−∫0tg1sds∫0Lxux2dx+12g1∘uxt+121−∫0tg2sds∫0Lxvx2dx+12g2∘vxt+12Kz,y−∫0Lxau+v2r+2+2buvr+2dx,note that
(53)Et=Jt+12∫0Lxut2dx+12∫0Lxvt2dx.Lemma 6.
Assume that (24), (14), (15), (16), (17), and (22) hold, and ∀u0,v0∈V02, u1,v1∈H2 and f0,g0∈Lx20,L,0,1,τ1,τ2 satisfying
(54)I0>0,β≔η2r+2r+1E0r+1<1.Then,∃t∗>0 such that
(55)It>0,∀t∈0,t∗,where
(56)E0=J0+12∫0Lxu12dx+12∫0Lxv12dx.Proof.
AsI0>0, then by continuity of It, ∃Tm≤t∗ such that It≥0, ∀t∈0,Tm; this implies that we have a maximum time value noting Tm such that
(57)ITm=0andIt>0,forall0≤t<Tm.
This, with (51), (52), and (14), we have
(58)Jt=r+12r+21−∫0tg1sds∫0Lxux2dx+1−∫0tg2sds∫0Lxvx2dx+r+12r+2g1∘uxt+g2∘vxt+Kz,y+12r+2It≥r+12r+2l1∫0Lxux2dx+l2∫0Lxvx2dx+g1∘uxt+g2∘vxt+Kz,y.Hence,
(59)l1∫0Lxux2dx+l2∫0Lxvx2dx≤2r+2r+1Jt≤2r+2r+1Et≤2r+2r+1E0,∀t∈0,Tm.
By (24) and (54), we get
(60)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+Kz,y.Hence,
(61)1−∫0tg1sds∫0Lxux2dx+1−∫0tg2sds∫0Lxvx2dx+g1∘uxt+g2∘vxt+Kz,y−2r+2∫0LxFu,vdx>0.This proves that It>0,∀t∈0,Tm. By repeating the procedure, Tm is extended to t∗.Theorem 7.
Let (14), (15), (16), (17), (22), and (24) hold. Then, ∀u0,v0∈V02, u1,v1∈H2, and f0,g0∈Lx20,L,0,1,τ1,τ2 satisfying (54) the solution of system (11) is bounded and global.Proof.
To prove thatuxH2+vxH2+utH2+vtH2+zK,μ22+yK,μ42 is bounded independently of t, using (36) yields
(62)E0≥Et.Using (52), we find
(63)−2r+2∫0Lxau+v2r+2+2buvr+2dx=It−1−∫0tg1sds∫0Lxux2dx−1−∫0tg2sds∫0Lxvx2dx−g1∘uxt−g2∘vxt−Kz,y.By using (62) in (63), we get
(64)E0≥Et=12∫0Lxut2dx+12∫0Lxvt2dx+121−∫0tg1sds∫0Lxux2dx+121−∫0tg2sds∫0Lxvx2x,tdx+12g1∘uxt+12g2∘vxt+12Kz,y+It,and using (14), (15), and (54) in (64), we get
(65)E0≥Et≥12∫0Lxut2dx+12∫0Lxvt2dx+r+12r+2l1∫0Lxux2dx+l2∫0Lxvx2dx+Kz,y≥μ0∫0Lxut2dx+∫0Lxvt2dx+∫0Lxux2dx+∫0Lxvx2dx+Kz,y.
So(66)uxH2+vxH2+utH2+vtH2+zK,μ22+yK,μ42≤μE0/μ≔1μ0,where
(67)μ0≔min12,r+12r+2l1,r+12r+2l2,r+12r+2.
Hence, the solution of system (11) is bounded and global.


## 4. Decay of Solutions

In this section, the decay result is showed by using several lemmas.As, we let(68)Ft≔Et+ε1Φt+ε2χt+ε3Ψt,where ε1,ε2,ε3>0, and
(69)Φt≔ξt∫0Lxutudx+ξt∫0Lxvtvdx,(70)χt≔−ξt∫0Lxut∫0tg1t−sut−usdsdx−ξt∫0Lxvt∫0tg2t−svt−vsdsdx,(71)Ψz,y≔ξt∫0L∫01∫τ1τ2xϱe−ρϱμ2ϱz2+μ4ϱy2dϱdρdx.Lemma 8.
There existα1,α2>0, such that
(72)α1Ft≤Et≤α2Ft,
forε1, ε2, and ε3 small enough.Proof.
Using the inequality of Young and the Poincaré-type inequality and0<ξt≤ξ0, we find
(73)ε1ξt∫0Lxutudx≤ε12ξ0∫0Lxut2dx+ε12Cpξ0∫0Lxux2dx,(74)ε1ξt∫0Lxvtvdx≤ε12ξ0∫0Lxvt2dx+ε12Cpξ0∫0Lxvx2dx,(75)−ε2ξt∫0Lxut∫0tg1t−sut−usdsdx≤ε22ξ0∫0Lxut2dx+ε22Cpξ01−l1g1∘uxt,(76)−ε2ξt∫0Lxvt∫0tg2t−svt−vsdsdx≤ε22ξ0∫0Lxvt2dx+ε22Cpξ01−l2g2∘vxt,(77)Ψz,y≤ξ0∫0L∫01∫τ1τ2xϱ∣μ2ϱ∣z2+∣μ4ϱ∣y2dϱdρdx,where Cp>0.
A combination of (73), (74), (75), (76), and (77) in (68) gives
(78)Ft≤Et+ε1+ε22ξ0∫0Lxut2dx+ε1+ε22ξ0∫0Lxvt2dx+ε12Cpξ0∫0Lxux2dx+ε12Cpξ0∫0Lxvx2dx+ε22Cpξ01−l1g1∘uxt+ε22Cpξ01−l2g2∘vxt+ε3ξ0∫0L∫01∫τ1τ2xρ∣μ2ϱ∣z2+∣μ4ϱ∣y2dϱdρdx.
Then,∃α1>0, for ε1, ε2, and ε3 small enough, such that
(79)Ft≤1α1Et.
Similarly, thanks to the inequalities of Young and Poincaré-type and using0<ξt≤ξ0 gives
(80)ε1ξt∫0Lxutudx≥−ε12ξ0∫0Lxut2dx−ε12Cpξ0∫0Lxux2dx,(81)ε1ξt∫0Lxvtvdx≥−ε12ξ0∫0Lxvt2dx−ε12Cpξ0∫0Lxvx2dx,(82)−ε2ξt∫0Lxut∫0tg1t−sut−usdsdx≥−ε22ξ0∫0Lxut2dx−ε22Cpξ01−l1g1∘uxt,(83)−ε2ξt∫0Lxvt∫0tg2t−svt−vsdsdx≥−ε22ξ0∫0Lxvt2dx−ε22Cpξ01−l2g2∘vxt,and
(84)−ε3Ψz,y≥−ε3ξ0∫0L∫01∫τ1τ2xϱ∣μ2ϱ∣z2+∣μ4ρϱ∣y2dϱdρdx.
By combining (80), (81), (82), (83), and (84) in (68), we find
(85)Ft≥Et−ε1+ε22ξ0∫0Lxut2dx−ε1+ε22ξ0∫0Lxvt2dx−ε12Cpξ0∫0Lxux2dx−ε12Cpξ0∫0Lxvx2dx−ε22Cpξ01−l1g1∘uxt−ε22Cpξ01−l2g2∘vxt−ε3ξ0∫0L∫01∫τ1τ2xρ∣μ2ϱ∣z2+∣μ4ϱ∣y2dϱdρdx.
Then,∃α2>0, for ε1, ε2, and ε3 small enough, such that
(86)Ft≥1α2Et.
This completes the proof.Lemma 9.
Forσ>1 and 0<θ<1, we have
(87)∫0tgt−sws2ds≤∫0tg1−θt−sws2ds1/σ×∫0tgσ−1+θ/σ−1t−sws2dsσ−1/σ,
∀w∈H.Proof.
It suffices to note that(88)∫0tgt−sws2ds=∫0tg1−θ/rt−sws2/rgσ−1+θ/σt−sws2σ−1/σds,using Hȯlder’s inequality for
(89)p=σ,q=σσ−1,r>1.
This completes the proof.Lemma 10.
Letv∈L∞0,T;H be such that vx∈L∞0,t;H and g be a continuous function on 0,T and suppose that 0<θ<1 and ρ>1. Then, ∃C>0 so that
(90)∫0tgt−svx.,t−vx.,sH2ds≤Csup0<s<Tv.,sH2∫0tg1−θsdsρ−1/ρ−1+θ×∫0tgρt−svx.,t−vx.,sH2dsθ/ρ−1+θ.Proof.
By applying Lemma8 with σ=ρ−1+θ/ρ−1 gives
(91)∫0tgt−svx.,t−vx.,sH2ds≤∫0tg1−θt−svx.,t−vx.,sH2dsρ−1/ρ−1+θ×∫0tgρt−svx.,t−vx.,sH2θ/ρ−1+θ.
We also have(92)∫0tg1−θt−svx.,t−vx.,sH2ds≤Csup0<s<Tvx.,sH2∫0tg1−θsds,
by combining (82) and (83). This completes the proof.Lemma 11.
Suppose thatv∈L∞0,T;H be such that vx∈L∞0,T;H and g be a continuous function on 0,T and assume ρ>1. Then, ∃C>0 so that
(93)∫0tgt−svx.,t−vx.,sH2ds≤ctvx.,tH2+∫0tvx.,sH2dsρ−1/ρ×∫0tgρt−svx.,t−vx.,sH2ds1/ρ.Proof.
By using (82) for θ=1 gives
(94)∫0tgt−svx.,t−vx.,sH2ds≤∫0tvx.,t−vx.,sH2dsρ−1/ρ×∫0tgρt−svx.,t−vx.,sH2ds1/ρ,where
(95)∫0tvx.,t−vx.,sH2ds≤2tvx.,tH2+2∫0tvx.,sH2ds,to obtain (93). Hence, this ends the proof.Lemma 12.
Suppose thatr satisfies (15) and (52) hold. Then, the functional Φt, given by (69), satisfies
(96)Φ′t≤1+l2δ+μ12δ1ξt∫0Lxut2dx+1+l2δ+μ32δ2ξt∫0Lxvt2dx−ξtl1−Cpδl−2δ1μ12∫0Lxux2dx−ξtl2−Cpδl−2δ2μ32∫0Lxvx2dx+ξt2l1∫0tg12−σsdsg1σ∘uxt+ξt2l2∫0tg22−σsdsg2σ∘vxt+ξt2δ1∫0L∫τ1τ2x∣μ2ϱ∣z2x,1,ϱ,tdϱdx+ξt2δ2∫0L∫τ1τ2x∣μ4ϱ∣y2x,1,ϱ,tdϱdx+ξt2r+2au+v2r+2+2buvr+2dx.For anyδ,δ1,δ2>0.Proof.
The derivation of (11) gives
(97)Φ′t=ξ′t∫0Lxutudx+ξt∫0Lxut2dx+ξt∫0Lxuttudx+ξ′t∫0Lxvtvdx+ξt∫0Lxvt2dx+ξt∫0Lxvttvdx=ξ′t∫0Lxutudx+ξt∫0Lxut2dx−ξt∫0Lxux2dx−ξtμ1∫0Lxuutdx−ξt∫0Lxu∫τ1τ2∣μ2ϱ∣z2x,1,ϱ,tdϱdx+ξt∫0Lxux∫0tg1t−suxsdsdx+ξ′t∫0Lxvtvdx+ξt∫0Lxvt2dx−ξt∫0Lxvx2dx−ξtμ3∫0Lxvvtdx−ξt∫0Lxv∫τ1τ2∣μ4ϱ∣y2x,1,ϱ,tdϱdx+ξt∫0Lxvx∫0tg2t−svxsdsdx+ξt2r+2au+v2r+2+2buvr+2dx.
By Young’s and Poincaré inequalities and (14) and (15), we find
(98)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
(99)∫0Lxvxt∫0tg1t−svxsdsdx≤ξt1+1+η21−l222∫0Lxvx2dx+ξt21+1η2∫0tg22−σsdsg2σ∘vxt.
∀η1,η2>0. As we have
(100)ξ′t∫0Lxutudx≤ξt2ξ′tξtCpδ∫0Lxux2dx+1δ∫0Lxut2dx≤ξt2Cplδ∫0Lxux2dx+lδ∫0Lxut2dx,∀δ>0,and similarly, we find
(101)ξ′t∫0Lxvtvdx≤ξt2Cplδ∫0Lxvx2dx+lδ∫0Lxvt2dx.By using Young’s and Poincaré’s inequalities and (22) gives
(102)−ξt∫0Lxu∫τ1τ2∣μ2ϱ∣z2x,1,ϱ,tdϱdx≤ξt2Cpδ1μ1∫0Lxux2dx+1δ1∫0L∫τ1τ2∣μ2ϱ∣z2x,1,ϱ,tdϱdx,(103)−ξt∫0Lxv∫τ1τ2∣μ4ϱ∣y2x,1,ϱ,tdϱdx≤ξt2Cpδ2μ3∫0Lxvx2dx+1δ2∫0L∫τ1τ2∣μ4ϱ∣y2x,1,ϱ,tdϱdx.
Similarly, we get(104)ξt∫0Lxutudx≤ξt2Cpδ1μ1∫0Lxux2dx+μ1δ1∫0Lxut2dxξt∫0Lxvtvdx≤ξt2Cpδ2μ3∫0Lxvx2dx+μ3δ2∫0Lxvt2dx.
In a combination of (98), (99), (100), (101), (102), (103), and (104) in (97), we obtain
(105)Φ′t≤1+l2δ+μ12δ1ξt∫0Lxut2dx+1+l2δ+μ32δ2ξt∫0Lxvt2dx−ξt21−1+η11−l12−δCpl−2δ1Cpμ1∫0Lxux2dx−ξt21−1+η21−l22−δCpl−2δ2Cpμ3∫0Lxvx2dx+ξt21+1η1∫0tg12−σsdsg1σ∘uxt+ξt21+1η2∫0tg22−σsdsg2σ∘vxt+ξt2∫0L∫τ1τ21δ1∣μ2ϱ∣z2x,1,ϱ,t+1δ2∣μ4ϱ∣y2x,1,ϱ,tdϱdx+ξt2r+2∫0Lau+v2r+2+2buvr+2dx,by choosing η1,η1, so that η1=l1/1−l1; hence, 1/2−1+1+η11−l12=−l1/2 and 1+1/η1=1/l1,and η2=l2/1−l2; therefore, 1/2−1+1+η21−l22=−l2/2 and 1+1/η2=1/l2.
Then, (96) is proved.Lemma 13.
Assuming thatr satisfies (15), (14), and (15) and (22) and (52) hold. Then, the functional χt given by (70) satisfies along the solution of (11)
(106)χ′t≤ξtθ1+c1+c1′+21−l12∫0Lxux2dx+ξtθ1+c2+c2′+21−l22∫0Lxvx2dx+ξtθ−∫0tg1sds+θl+θ1μ1∫0Lxut2dx+ξtθ−∫0tg2sds+θl+θ2μ3∫0Lxvt2dx+12θ+2θ+μ1Cp2θ1+Cp1+l4θ×ξt∫0tg12−σsdsg1σ∘uxt+12θ+2θ+μ3Cp2θ2+Cp1+l4θ×ξt∫0tg22−σsdsg2σ∘vxt−Cp4θξtg10g1′∘uxt−Cp4θξtg20g2′∘vxt+ξt∫0L∫τ1τ2xθ1∣μ2ϱ∣z2x,1,ϱ,t+θ2∣μ4ϱ∣y2x,1,ϱ,tdϱdx,for any θ,θ1,θ2>0.Proof.
Direct calculation gives(107)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
(108)ddt∫αtβtft,sds=∫αtβt∂ft,s∂tds+∂βt∂tft,βt−∂αt∂tft,αt.As we have u,v,z,y the solution of (11), we find
(109)χ′t=−ξ′t∫0Lxut∫0tg1t−sut−usdsdx+ξt∫0Lxux∫0tg1t−suxt−uxsdsdx−ξt∫0Lx∫0tg1t−suxsds∫0tg1t−suxt−uxsdsdx−ξtμ1∫0Lxut∫0tg1t−sut−usdsdx−ξt∫0Lx∫τ1τ2∣μ2ρ∣z2x,1,ρ,tdρ∫0tg1t−sut−usdsdx−ξt∫0Lxf1u,v∫0tg1t−sut−usdsdx−ξt∫0Lxut∫0tg1′t−sut−usdsdx−ξt∫0tg1sds∫0Lxut2dx−ξt∫0tg2sds∫0Lxvt2dx−ξ′t∫0Lxvt∫0tg2t−svt−vsdsdx+ξt∫0Lxvx∫0tg2t−svxt−vxsdsdx−ξt∫0Lx∫0tg2t−svxsds∫0tg2t−svxt−vxsdsdx−ξtμ3∫0Lxvt∫0tg2t−svt−vsdsdx−ξt∫0Lx∫τ1τ2∣μ4ρ∣y2x,1,ρ,tdρ∫0tg2t−svt−vsdsdx−ξt∫0Lxf2u,v∫0tg2t−svt−vsdsdx−ξt∫0Lxvt∫0tg2′t−svt−vsdsdx.
By Young’s inequality and (14) and (15), we arrive to
(110)−ξ′t∫0Lxut∫0tg1t−sut−usdsdx≤ξtξ′tξtθ∫0Lxut2dx+Cp4θ∫0tg12−σsdsg1σ∘uxt≤θlξt∫0Lxut2dx+Cpl4θξt∫0tg12−σsdsg1σ∘uxt,(111)ξt∫0Lxux∫0tg1t−suxt−uxsdsdx≤θξt∫0Lxux2dx+14θξt∫0tg12−σsdsg1σ∘uxt.
Similarly, we get(112)ξtμ1∫0Lxut∫0tg1t−sut−usdsdx≤θ1μ1ξt∫0Lxut2dx+14θ1Cpξt∫0tg12−σsdsg1σ∘uxt,(113)ξtμ3∫0Lxvt∫0tg2t−svt−vsdsdx≤θ2μ3ξt∫0Lxvt2dx+14θ2Cpξt∫0tg22−σsdsg2σ∘vxt,with
(114)−ξt∫0Lx∫0tg1t−suxsds∫0tg1t−suxt−uxsdsdx≤2θ1−l12ξt∫0Lxux2dx+2θ+14θξt∫0tg12−σsdsg1σ∘uxt.
So(115)−ξt∫0Lxf1u,v∫0tg1t−sut−usdsdx≤Cp4θξt∫0tg12−σsdsg1σ∘uxt+c1θξt∫0Lxux2dx+c2θξt∫0Lxvx2dx,where
(116)c1≔Λ12r+2r+1E02r+1,c2≔Λ22r+2r+1E02r+1,(117)−ξt∫0Lxut∫0tg1′t−sut−usdsdx≤θξt∫0Lxut2dx−g104θCpξtg1′∘uxt.Then,
(118)−ξ′t∫0Lxvt∫0tg2t−svt−vsdsdx≤θlξt∫0Lxvt2dx+Cpl4θξt∫0tg22−σsdsg2σ∘vxt,(119)ξt∫0Lxvx∫0tg2t−svxt−vxsdsdx≤θξt∫0Lxvx2dx+14θξt∫0tg22−σsdsg2σ∘vxt.Thus,
(120)−ξt∫0Lx∫0tg2t−svxsds∫0tg2t−svxt−vxsdsdx≤2θ1−l22ξt∫0Lxvx2dx+2θ+14θξt∫0tg22−σsdsg2σ∘vxt,(121)−ξt2r+2∫0Lxf2u,v∫0tg2t−svt−vsdsdx≤Cp4θξt∫0tg22−σsdsg2σ∘vxt+c1′θξt∫0Lxux2dx+c2′θξt∫0Lxvx2dx,where
(122)c1′≔Λ1′2r+2r+1E02r+1,c2′≔Λ2′2r+2r+1E02r+1,(123)−ξt∫0Lxvt∫0tg2′t−svt−vsdsdx≤θξt∫0Lxvt2dx−g204θCpξtg2′∘vxt.
Similarly, we have(124)−ξt∫0Lx∫τ1τ2∣μ2ρ∣z2x,1,ρ,tdρ∫0tg1t−sut−usdsdxθ1ξt∫0L∫τ1τ2x∣μ2ρ∣z2x,1,ρ,tdρdx+14θ1μ1Cp∫0tg12−σsdsg1σ∘uxt,(125)−ξt∫0Lx∫τ1τ2∣μ4ϱ∣y2x,1,ϱ,tdϱ∫0tg2t−svt−vsdsdxθ2ξt∫0L∫τ1τ2x∣μ4ϱ∣y2x,1,ϱ,tdϱdx+14θ2μ3Cp∫0tg22−σsdsg2σ∘vxt.
A combination of (110), (111), (112), (113), (114), (115), (117), (118), (119), (120), (121), (123), (124), and (125) into (109) gives (106).Lemma 14.
Letu,v,z,y be the solution of (11). Then, for η3>0, the functional Ψt satisfies
(126)Ψ′t≤−ξtη4∫0L∫01∫τ1τ2xϱ∣μ2ϱ∣z2+∣μ4ϱ∣y2dϱdρdx+ξtμ1∫0Lxut2dx+ξtμ3∫0Lxvt2dx−ξtη3∫0L∫τ1τ2x∣μ2ϱ∣z2x,1,ϱ,t+∣μ4ϱ∣y2x,1,ϱ,tdϱdx,where η3>0 and η4=η31−l>0>0.Proof.
By differentiatingΨt and using equations (11)3 and (11)4, we get
(127)Ψ′t=ξ′t∫0L∫01∫τ1τ2xρe−ϱρ∣μ2ϱ∣z2+∣μ4ϱ∣z2dϱdρdx−2ξt∫0L∫01∫τ1τ2e−ϱρ∣μ2ϱ∣zzρx,ρ,ϱ,tdϱdρdx−2ξt∫0L∫01∫τ1τ2e−ϱρ∣μ4ϱ∣yyρx,ρ,ϱ,tdϱdρdx=ξ′t∫0L∫01∫τ1τ2xϱe−ϱρ∣μ2ϱ∣z2+∣μ4ϱ∣z2dϱdρdx−ξt∫0L∫01∫τ1τ2xϱe−ϱρ∣μ2ϱ∣z2dϱdρdx−ξt∫0L∫τ1τ2x∣μ2ϱ∣e−ϱz2x,1,ϱ,t−z2x,0,ϱ,tdϱdx−ξt∫0L∫01∫τ1τ2xϱe−ϱρ∣μ4p∣y2dρϱρdx−ξt∫0L∫τ1τ2x∣μ4ϱ∣e−ϱy2x,1,ϱ,t−y2x,0,ϱ,tdϱdx.Using the equalityzx,0,ϱ,t=utx,t, yx,0,ϱ,t=vtx,t, and e−ϱ≤e−ρϱ≤1, for any 0<ρ<1, we find
(128)Ψ′t≤ξtl∫0L∫01∫τ1τ2xϱμ2ϱz2+μ4ϱz2dϱdρdx−ξt∫0L∫01∫τ1τ2xϱe−ϱρμ2ϱz2+μ4ϱy2dϱdρdx−ξt∫0L∫τ1τ2xe−ϱμ2ϱz2x,1,ϱ,t+μ4ϱy2x,1,ϱ,tdϱdx+∫τ1τ2μ2ϱdϱξt∫0Lxut2dx+∫τ1τ2μ4ϱdϱξt∫0Lxvt2dx.As−e−ϱ is an increasing function, we have −e−ϱ≤−e−τ2, for any ρ∈τ1,τ2.Then, settingη3=e−τ2 and (22), we obtain (126).Theorem 15.
Letu0,v0∈V02, u1,v1∈H2, and f0,g0∈Lx20,L×0,1×τ1,τ2 be defined and satisfy (163). Assume that r satisfies (24), (14), (15), (16), (17), and (22) hold. Then, for each t0>0, ∃K and k such that the solution of (11) satisfies ∀t≥t0, we have the following inequality for the energy function
(129)Et≤Ke−k∫t0tξsds,σ=1,K1+∫t0tξsds−1/σ−1,1<σ<32.Proof.
Asg1,g2 is continuous and g10,g20>0, hence ∀t0>0; we have
(130)∫0tg1sds≥∫t0tg1sds=g1,0>0,∀t≥t0,∫0tg2sds≥∫t0tg2sds=g2,0>0,∀t≥t0.
By using (36), (96), (106), (126), and (130) and 0<ξt≤ξ0henceξt/ξ0<1, we get
(131)F′t=E′t+ε1Φ′t+ε2χ′t+ε3Ψ′t≤−d1−ε11+12δ+μ12δ1+ε2g1,0−θ−θl−μ1θ1−ε3μ1ξt∫0Lxut2dx−d2−ε11+12δ+μ32δ2+ε2g2,0−θ−θl−μ1θ1−ε3μ3ξt∫0Lxvt2dx+2ε1ξt∫0Lxau+v2r+2+2buvr+2dx+12−ε2ξ04θCpg10g1′∘uxt+12−ε2ξ04θCpg20g2′∘vxt−ε12l1−δCpl−2δ1μ1Cp−ε2θ1+c1+c1′+21−l12ξt∫0Lxux2dx−ε12l2−δCpl−2δ2μ3Cp−ε2θ1+c2+c2′+21−l22ξt∫0Lxvx2dx+ε12l1+ε212θ+2θ+μ1Cp2θ1+Cp+lCp4θξt∫0tg12−σsdsg1σ∘uxt+ε12l2+ε212θ+2θ+μ1Cp2θ1+Cp+lCp4θξt∫0tg22−σsdsg2σ∘vxt−ε3η3−ε112δ1−ε2θ1ξt∫0L∫τ1τ2x∣μ2ϱ∣z2x,1,ϱ,tdϱdx−ε3η3−ε112δ2−ε2θ2ξt∫0L∫τ1τ2x∣μ4ϱ∣y2x,1,ϱ,tdϱdx−ε3η4ξt∫0L∫01∫τ1τ2xρ∣μ2ϱ∣z2+∣μ4ϱ∣y2dϱdρdx.
By choosingδ, δ1, and δ2 so small that
(132)l1−δCpl−2μ1δ1Cp>l12,l2−δCpl−2μ3δ2Cp>l22.Then,
(133)δ<14Cplminl1,l2,δ1<18μ1Cpminl1,l2,δ2<18μ3Cpminl1,l2.
At this point, we chooseθ small enough, such that
(134)k3≔ε1l14−ε2θ1+c1+c1′+21−l12>0,k4≔ε1l24−ε2θ1+c2+c2′+21−l22>0.Then,
(135)θ<minε1l14ε21+c1+c1′+21−l12,ε1l24ε21+c2+c2′+21−l22.
Now,δ,δ1,δ2, and θ are fixed. Then, we select ε1,ε2,ε3,θ1, and θ2 so small that (72) and (162) remain correct and
(136)k1≔d1−ε11+12δ+μ12δ1+ε2g1,0−μ1θ1−1+lθ−ε3μ1>0,k2≔d2−ε11+12δ+μ32δ2+ε2g2,0−μ3θ2−1+lθ−ε3μ3>0,k5≔12−ε2ξ04θCpg10−ε12l1+ε212θ+2θ+μ1Cp2θ1+Cp+lCp4θ∫0tg12−σsds>0,k6≔12−ε2ξ04θCpg20−ε12l2+ε212θ+2θ+μ3Cp2θ2+Cp+lCp4θ∫0tg22−σsds>0,k7≔ε3η3−ε112δ1−ε2θ1>0,k8≔ε3η3−ε112δ2−ε2θ2>0.
Hence, by using (15) gives, for some σ>0,
(137)F′t≤−σξt∫0Lxut2dx+∫0Lxvt2dx−∫0Lxau+v2r+2+2buvr+2dx+∫0Lxux2dx+∫0Lxvx2dx+g1σ∘uxt+g2σ∘vxt+Kz,y.
We chooseθ, θ1, and θ2 so small that
(138)g1,0−μ1θ1−1+lθ>12g1,0,g2,0−μ3θ2−1+lθ>12g2,0.By (134), we get
(139)θ<minε1l14ε21+c1+c1′+21−l12,ε1l24ε21+c2+c2′+21−l22,141+lg1,0,141+lg2,0,θ1<14μ11+lg1,0,θ2<14μ31+lg2,0,4θ1+c1+c1′+21−l12l1<g1,02+l/δ+μ1/δ1,4θ1+c2+c2′+21−l22l2<g2,02+l/δ+μ3/δ2.
Withθ,θ1,θ2, and α′ fixed, we pick ε1,ε2, and ε3 such that
(140)max4θ1+c1+c1′+21−l12l1,4θ1+c2+c2′+21−l22l2ε2<ε1<12+l/δ+minμ1/δ1,μ3/δ2mind1,d2+ε2ming1,0,2,0+ε3minμ1+μ3.We will make
(141)k1≔d1−ε11+12δ+μ12δ1+ε2g1,0−μ1θ1−1+lθ−ε3μ1>0,k2≔−ε11+12δ+μ32δ2+ε2g2,0−μ3θ2−θ−θl−ε3μ3>0,k3≔ε12l1−2μ1δ1Cp−δCpl−ε2θ1+c1+c1′+21−l12>0,k4≔ε12l2−2μ3δ2Cp−δCpl−ε2θ1+c2+c2′+21−l22>0,k7≔ε3η3−ε112δ1−ε2θ1>0,k8≔ε3η3−ε112δ2−ε2θ2>0.
Then, we selectε1, ε2, and ε3 so small that (72) and (137) remain correct and
(142)k5≔12−ε2ξ04θCpg10−ε12l1+ε212θ+2θ+μ1Cp2θ1+Cp+lCp4θ∫0tg12−σsds>0,k6≔12−ε2ξ04θCpg20−ε12l2+ε212θ+2θ+μ3Cp2θ2+Cp+lCp4θ∫0tg22−σsds>0.Next, as (137) is showed, according to the different ranges of r, we give the following two cases.Case 1.
σ=1.
By choosingε1,ε2,ε3,θ1,θ2, and θ, (137) gives, for γ>0 is constant so that,
(143)F′t≤−γξtEt,∀t≥t0.
Therefore, with the help of the LHS of (72) and (143), we obtain
(144)F′t≤−γα1ξtFt,∀t≥t0.
By integration of (144) over t0,t gives
(145)F′t≤Ft0e−γα1∫t0tξsds,∀t≥t0.
Therefore, (129)1 is proved by (72) as well.Case 2.
1<σ<3/2.
We use (11), which gives
(146)g1t1−σ≥σ−1∫t0tξsds+g1t01−σ,g2t1−σ≥σ−1∫t0tξsds+g2t01−σ.We have, for 0<τ<1,
(147)∫0∞g11−τsds≤∫0∞1σ−1∫t0tξsds+g1t01−σ1−τ/σ−1ds,∫0∞g21−τsds≤∫0∞1σ−1∫t0tξsds+g2t01−σ1−τ/σ−1ds.For 0<τ<2−σ<1, we have 1−τ/σ−1>1 and (15), we find
(148)∫0∞g11−τsds<∞,∀0<τ<2−σ,∫0∞g21−τsds<∞,∀0<τ<2−σ.
From (72) forθ=τandρ=σ and (55) gives
(149)g1∘uxt≤C1E0∫0∞g11−τsdsσ−1/σ−1+τg1σ∘uxtτ/σ−1+τ≤C1′g1σ∘vxtτ/σ−1+τ.Similarly, we have
(150)g2∘vxt≤C2′g2σ∘vxtτ/σ−1+τ,for some C1′,C2′>0. Hence, ∀σ1>1, we find
(151)Eσ1t≤C′′Eσ1−10∫0Lxut2dx+∫0Lxvt2dx+∫0Lxux2dx+∫0Lxvx2dx−∫0Lxau+v2r+2+2buvr+2dx+Kz,y+C1′′g1∘uxtσ1+C2′′g2∘vxtσ1≤C′′Eσ1−10∫0Lxut2dx+∫0Lxvt2dx+∫0Lxux2dx+∫0Lxvx2dx−∫0Lxau+v2r+2+2buvr+2dx+Kz,y+C1′′′g1σ∘uxtτσ1/σ−1+τ+C2′′′g2σ∘vxtτσ1/σ−1+τ.
We chooseτ=1/2 and σ1=2σ−1therefore,τσ1/σ−1+τ=1 and (144); we get, for some Γ>0,
(152)Eσ1t≤Γ∫0Lxut2dx+∫0Lxvt2dx+∫0Lxux2dx+∫0Lxvx2dx+Kz,y−∫0Lxau+v2r+2+2buvr+2dx+g1σ∘uxt+g2σ∘vxt.
By combining (72), (137), and (151), we find
(153)F′t≤−σΓξtEσ1t≤−σΓα1σ1Fσ1tξt,∀t≥t0.
By integrating (153) gives
(154)Ft≤C1∗1+∫t0tξsds−1/σ1−1,∀t≥t0.
Hence,(155)∫t0∞Ftdt≤C1∗∫t0∞11+∫t0tξsds1/σ1−1dt.
From1/σ1−1>0 and 1+∫t0tξsds⟶+∞ as t⟶+∞, we find
(156)∫t0∞Ftdt<∞.
Also, we use (24), and we get
(157)tFt≤C1∗t1+∫t0tξsds1/σ1−1≤Cσ.
Hence, we find(158)supt≥t0tFt<∞.
FromEt which is bounded, using (72), (156), and (158) to get
(159)∫t0∞Ftdt+supt≥0tFt<∞.
Therefore, using (55) and Lemma 10forρ=σ gives
(160)g1∘uxt≤C2∗tuxx,tH2+∫0tuxx,sH2dsσ−1/σ×∫0tgσt−suxx,t−uxx,sH2ds≤C2∗tFt+∫t0tFsdsσ−1/σg1σ∘uxt1/σ≤C3∗g1σ∘uxt1/σ.This means
(161)g1σ∘uxt≥C4g1∘uxtσ,(162)g2σ∘vxt≥C5g2∘vxtσ,for some C4,C5>0.
Then, combining (137), (161), and (162) yields
(163)F′t≤−C6ξt∫0Lxut2dx+∫0Lxvt2dx+∫0Lxux2dx+∫0Lxvx2dx−∫0Lxau+v2r+2+2buvr+2dx+Kz,y+g1∘uxtσ+g2∘vxtσ,for some C6>0.
As in [1], we obtain
(164)Eσt≤C7ξt∫0Lxut2dx+∫0Lxvt2dx+∫0Lxux2dx+∫0Lxvx2dx−∫0Lxau+v2r+2+2buvr+2dx+Kz,y+g1∘uxttσ+g2∘vxtσ,
∀t≥0 and some C7>0.
Combining (163), (164), and (72), we find
(165)F′t≤−C8ξtFσt,∀t≥t0,for some C8>0.
By integrating (163) over t0,t, we get
(166)Ft≤C91+∫t0tξsds−1/σ−1,∀t≥t0.
Hence, (129)2 is showed by (72) as well.



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*Source: 6683465-2021-02-16.xml*