# Blow up of Coupled Nonlinear Klein-Gordon System with Distributed Delay, Strong Damping, and Source Terms

Authors: Djamel Ouchenane; Salah Mahmoud Boulaaras; Asma Alharbi; Bahri Cherif
Journal: Journal of Function Spaces
 (2020)

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## Abstract

This present work deals with the blow up of the coupled Klein-Gordon system with strong damping, distributed delay, and source terms, under suitable conditions.


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## Body

## 1. Introduction

In the present paper, we consider the following system:(1)utt+m1u2−Δu−ω1Δut+∫0tgt−sΔusds+μ1ut+∫τ2τ1μ2ϱutx,t−ϱdϱ=f1u,v,x,t∈Ω×ℝ+,vttm2v2−Δv−ω2Δvt+∫0tht−sΔvsds+μ3vt+∫τ1τ2μ4ϱvtx,t−ϱdϱ=f2u,v,x,t∈Ω×ℝ+,ux,t=0,vx,t=0,x∈∂Ω,utx,−t=f0x,t,vtx,−t=k0x,tx,t∈Ω×0,τ2,ux,0=u0x,utx,0=u1x,x∈Ω,vx,0=v0x,vtx,0=v1x,x∈Ω,where(2)f1u,v=a1u+v2p+1u+v+b1up⋅u⋅vp+2,f2u,v=a1u+v2p+1u+v+b1vp⋅v⋅up+2,and m1,m2,ω1,ω2,μ1,μ3,a1,b1>0, and τ1,τ2 are the time delay with 0≤τ1<τ2, and μ2,μ4 are a L∞ functions, and g,h are differentiable functions.Viscous materials are the opposite of flexible materials that have dissipate mechanical energy and the ability to store. The mechanical properties of viscous materials are so important that we find them in many applications of natural sciences. Many authors have been concerned with this problem in recent decades.If there is only one equation and ifw1=0, that is, for absence of Δut, and μ1=μ2=0. Our problem (1) has been studied by Berrimi and Messaoudi [1]. Using Galerkin’s method they proved the result of local existence. They also made it clear that the local solution is global in time under suitable conditions and at the same rate of decaying (exponential or polynomial) of the kernel g. In addition, the authors themselves demonstrated that the dissipation can be deduced by the term viscous integral and that it is strong enough to stabilize the solution oscillations. Their results were also obtained under weaker conditions than those used by Cavalcanti et al. [2].In [3], the authors considered the following problem:(3)utt−Δu+∫0tgt−sΔusds+axut+uγ,u=0,where the authors proved the exponential decay result. This subsequent result was improved by Berrimi et al. in [1], as they showed that the viscosity elastic dissipation alone is strong enough to stabilize the problem even with the exponential rate with respect to the kernel g assumptions. In the case μ1≠0, in problem (1), Kafini and Messaoudi in [4] proved a blow up result for the following problem:(4)utt−Δu+∫0∞gt−sΔusds+ut=up−2⋅u,x,t∈ℝn×0,∞,ux,0=u0x,utx,0=u1x,where g satisfies(5)∫0∞gsds<2p−4/2p−3.The initial data was backed by negative energy as(6)∫u0u1dx>0.In [5], Song and Xue considered the following problem:(7)utt−Δu+∫0∞gt−sΔusds−Δutup−2⋅u,x,t∈Ω×0,∞,ux,0=u0x,utx,0=u1x,where the authors showed that there were solutions of (7) with initial energy according to suitable assumptions on g. Moreover, they showed the blow up in a finite time. Then, the same authors in [6] continued to prove that there were solutions of (7) with positive initial energy that blow up in finite time. In [7], the author studied the following problem:(8)utt−Δu−ωΔut+∫0tgt−sΔusds+autm−2⋅ut=up−2⋅u,x,t∈Ω0,∞,ux,0=u0x,utx,0=u1x,x∈Ω,ux,t=0,x∈∂Ω,where they proved the exponential growth result under suitable assumptions. The authors in [8] studied the following problem:(9)utt−Δu+∫0∞gsΔut−sds−ε1Δut+ε2ututm−2=ε3uup−2,ux,t=0,x∈∂Ω,t>0,ux,0=u0x,utx,0=u1x,x∈Ω,where they showed a blow up result if p>m and established the global existence. In the coupled equation case, the authors in [9] studied the following system:(10)utt−Δu+ututm−2=f1u,v,vtt−Δv+vtvtr−2=f2u,v,with f1 and f2 nonlinear functions satisfying appropriate conditions. According to certain restrictions imposed on the initial data and parameters, they obtained numerous results on the existence of weak solutions. They obtained many results on the presence of weak solutions. In addition, by using the same techniques similar to that in [10] with negative initial, energy blows up for a finite period of time.In [11], the authors have proved the solution of the problem:(11)utt−Δu+auk+bvlututm−2=f1u,v,vtt−Δv+auθ+bvϑvtvtr−2=f2u,v,where under some restrictions on positive initial energy for certain conditions on the functions f1 and f2, the authors proved the blows up in finite time of solution.The result of [11] has been extended by the authors in [12], where they studied the following system:(12)utt−Δu+∫0∞gsΔut−sds+auk+bvlututm−2=f1u,v,vtt−Δv+∫0∞hsΔvt−sds+auθ+bvϱvtvtr−2=f2u,v,they proved that the solutions of a system of wave equations with degenerate damping, viscoelastic term and strong nonlinear sources acting in both equations at the same time are globally nonexisting provided that the initial data are sufficiently large in a bounded domain of Ω.As complement to these works, we are working to prove the blow up result with distributed delay of problem (1), under appropriate assumptions, and we prove these results using the energy method. In the following, let c,ci>0,i=1,⋯,12.The present paper is organized as follows. In Section2, we give some necessarily assumptions for the main result. In Section 3, we prove the blow up result.


## 2. Assumptions

We consider the following suitable assumptions.(A1)
g,h:ℝ+⟶ℝ+ are differentiable and decreasing functions such that(13)gt≥0,1−∫0∞gsds=l1>0,(14)ht≥0,1−∫0∞hsds=l2>0.(A2)
There exists a constantsξ1,ξ2>0 such that(15)g′t≤−ξ1gt,t≥0,(16)h′t≤−ξ2ht,t≥0.(A3)
μ2,μ4:τ1,τ2⟶ℝ are a L∞ functions so that(17)2δ−12∫τ1τ2μ2ϱdϱ<μ1,δ>12,(18)2δ−12∫τ1τ2μ4ϱdϱ<μ3,δ>12.


## 3. Blow up

In this section, we obtain the proof of the blow up result of the solution of problem (1). First, of all in [13], we introduce the new variables(19)yx,ρ,ϱ,t=utx,t−ϱρ,zx,ρ,ϱ,t=vtx,t−ϱρ,then, we obtain(20)ϱytx,ρ,ϱ,t+yρx,ρ,ϱ,t=0,yx,0,ϱ,t=utx,t,ϱztx,ρ,ϱ,t+zρx,ρ,ϱ,t=0,zx,0,ϱ,t=vtx,t.Let us denote by(21)gou=∫Ω∫0tgt−sut−us2dsdx.Therefore, problem (1) get the following form:(22)utt+m1u2−Δu−ω1Δut+∫0tgt−sΔusds+μ1ut+∫τ1τ2μ2ϱyx,1,ϱ,tdϱ=f1u,v,x∈Ω,t≥0,vtt+m2v2−Δv−ω2Δvt+∫0tht−sΔvsds+μ3vt+∫τ1τ2μ4ϱzx,1,ϱ,tdϱ=f2u,v,x∈Ω,t≥0,ϱytx,ρ,ϱ,t+ypx,ρ,ϱ,t=0,ϱztx,ρ,ϱ,t+zpx,ρ,ϱ,t=0,with initial and boundary conditions(23)ux,t=0,vx,t=0,x∈∂Ω,yx,ρ,ϱ,0=f0x,ϱρ,zx,ρ,ϱ,0=k0x,ϱρ,ux,0=u0x,utx,0=u1x,vx,0=v0x,vtx,0=v1x,where(24)x,ρ,ϱ,t∈Ω×0,1×τ1,τ2×0,∞.Theorem 1.
Assume (14), (16), and (17) hold. Let(25)−1<p<4−nn−2,n≥3,p≥−1,n=1,2.For any initial data,(26)uo,u1,v0,v1,f0,k0∈H,where(27)H=H01Ω×L2Ω×H01Ω×L2Ω×L2Ω×0,1×τ1,τ2×L2Ω×0,1×τ1,τ2,then, problem (22) has a unique solution(28)u∈C0,T;H,for some T>0.Lemma 2.
There exists a functionFu,v such that(29)Fu,v=12ρ+2uf1u,v+vf2u,v,=12ρ+2a1u+v2p+2+2b1uvp+2≥0,where(30)∂F∂u=f1u,v,∂F∂v=f2u,v,we take a1=b1=1 for convenience.Lemma 3. (see [12]).
There exist two positive constantsc0 and c1 such that(31)c02ρ+2u2p+2+v2p+2≤Fu,v≤c12ρ+2u2p+2+v2p+2.We define the energy functional (see, e.g., [14–16] and reference therein).Lemma 4.
Assume (14), (16), (17), and (25) hold, let u,v,y,z be a solution of (22), then Et is nonincreasing, that is,(32)Et=12ut22+12vt22+m12u22+m22v22+12l1∇u22+12l2∇v22+12go∇u+12ho∇u+12Ky,z−∫ΩFu,vdx,satisfies(33)E′t≤−c3ut22+vt22+u22+v22+∫Ω∫τ1τ2μ2ϱy2x,1,ϱ,tdϱdx+∫Ω∫τ1τ2μ4ϱz2x,1,ϱ,tdϱdx≤0,where(34)Ky,z=∫Ω∫01∫τ1τ2ϱμ2ϱy2x,ρ,,ϱ,t+μ4ϱz2x,ρ,,ϱ,tdϱdρdx.Proof.
By multiplying (3.4)1, (3.4)2 by ut, vt and integrating over Ω, we get(35)ddt12ut22+12vt22+m12u22+m22v22+12l1∇u22+12l2∇v22+12go∇u+12ho∇u−∫ΩFu,vdx=−μ1ut22−m1u22−∫Ωut∫τ1τ2μ2ϱyx,1,ϱ,tdϱdx−μ3vt22−m2v22−∫Ωvt∫τ1τ2μ4ϱzx,1,ϱ,tdϱdx+12g′o∇u−12gt∇u22−ω1∇ut22+12h′o∇u−12ht∇v22−ω2∇vt22,and, from (3.4)3, (3.4)4, we have(36)ddt12∫Ω∫01∫τ1τ2ϱμ2ϱy2x,ρ,ϱ,tdϱdρdx=−12∫Ω∫01∫τ1τ22μ2ϱyyρdϱdρdx,+12∫Ω∫τ1τ2μ2ϱy2x,0,ϱ,tdϱdx−12∫Ω∫τ1τ2μ2ϱy2x,1,ϱ,tdϱdx,=12∫τ1τ2μ2ϱdϱut22−12∫Ω∫τ1τ2μ2ϱy2x,1,ϱ,tdϱdx,(37)ddt12∫Ω∫01∫τ1τ2ϱμ4ϱz2x,ρ,ϱ,tdϱdρdx=−12∫Ω∫01∫τ1τ22μ4ϱzzρdϱdρdx,+12∫Ω∫τ1τ2μ4ϱz2x,0,ϱ,tdϱdx−12∫Ω∫τ1τ2μ4ϱz2x,1,ϱ,tdϱdx,=12∫τ1τ2μ4ϱdϱvt22−12∫Ω∫τ1τ2μ4ϱz2x,1,ϱ,tdϱdx,then, we get(38)ddtEt=−μ1ut22−m1u22−∫Ω∫τ1τ2μ2ϱutyx,1,ϱ,tdϱdx+12g′o∇u−12gt∇u22−ω1∇ut22+12∫τ1τ2μ2ϱdϱut22−12∫Ω∫τ1τ2μ2ϱy2x,1,ϱ,tdϱdx−μ3vt22−m2v22−∫Ω∫τ1τ2μ4ϱvtzx,1,ϱ,tdϱdx+12h′o∇v−12ht∇v22−ω2∇vt22+12∫τ1τ2μ4ϱdϱvt22−12∫Ω∫τ1τ2μ4ϱz2x,1,ϱ,tdϱdx.By (35)–(37), we get (32).And by using Young’s inequality, (14), (16), and (17) in (38), we obtain (33).Now, we define the functional(39)ℍt=−Et=−12ut22−12vt22−m12u22−m22v22−12l1∇u22−12l2∇v22−12go∇u−12ho∇u−12Ky,z+12p+2u+v2p+22p+2+2uvp+2p+2.Theorem 5.
Assume (14)–(17) and (25) hold. Assume further that E0<0, then the solution of problem (22) blow up in finite time.Proof.
From (32), we have(40)Et≤E0≤0.
Therefore,(41)ℍ′t=−E′t≥c3ut22+u22+∫Ω∫τ1τ2μ2ϱy2x,1,ϱ,tdϱdx+vt22+v22+∫Ω∫τ1τ2μ4ϱz2x,1,ϱ,tdϱdx,hence,(42)ℍ′t≥c3∫Ω∫τ1τ2μ2ϱy2x,1,ϱ,tdϱdx≥0,(43)ℍ′t≥c3∫Ω∫τ1τ2μ2ϱz2x,1,ϱ,tdϱdx≥0,(44)0≤ℍ0ℍt≤12p+2u+v2p+22p+2+2uvp+2p+2,≤c12p+2u2p+22p+2+v2p+22p+2.
We set(45)Kt=ℍ1−α+ε∫Ωuut+vvtdx+32∫Ωμ1u2+μ3v2dx+ε2∫Ωω1∇u2+ω2∇u2dx,where ε>0 to be assigned later and(46)0<α<2p+24p+2<1.By multiplying (3.4)1, (3.4)2 by u,v and with a derivative of (45), we get(47)K′t=1−αℍ−αℍ′t+εut22+vt22+u22+v22−ε∇u22+∇v22+ε∫Ω∇u∫01gt−s∇usdsdx+ε∫Ω∇v∫01ht−s∇vsdsdx−ε∫Ω∫τ1τ2μ2ϱuyx,1,ϱ,tdϱdx−ε∫Ω∫τ1τ2μ4ϱvzx,1,ϱ,tdϱdx+εu+v2p+22p+2+2uvp+2p+2.Using Young’s inequality, we get(48)ε∫Ω∫τ1τ2μ2ϱuyx,1,ϱ,tdϱdx≤εδ1∫τ1τ2μ2ϱdϱu22+14δ1∫Ω∫τ1τ2μ2ϱy2x,1,ϱ,tdϱdx,ε∫Ω∫τ1τ2μ4ϱvzx,1,ϱ,tdϱdx≤εδ2∫τ1τ2μ4ϱdϱu22+14δ2∫Ω∫τ1τ2μ4ϱz2x,1,ϱ,tdϱdx,and we have(49)ε∫0tgt−sds∫Ω∇u⋅∇usdxds=ε∫0tgt−sds∫Ω∇u⋅∇us−∇utdxds+ε∫0tgsds∇u22≥ε2∫0tgsds∇u22−ε2go∇u,ε∫0tht−sds∫Ω∇v⋅∇vsdxds=ε∫0tht−sds∫Ω∇v⋅∇vs−∇vtdxds+ε∫0thsds∇v22≥ε2∫0thsds∇v22−ε2ho∇v.We obtain, from (47),(50)K′t≥1−αℍ−αℍ′t+εut22+ut22+u22+v22−ε1−12∫0tgsds∇u22+1−12∫0thsds∇v22−εδ1∫τ1τ2μ2ϱdϱu22−εδ2∫τ1τ2μ4ϱdϱv22−ε2go∇u−ε4δ1∫Ω∫τ1τ2μ2ϱy2x,1,ϱ,tdϱdx−ε2ho∇v−ε4δ2∫Ω∫τ1τ2μ4ϱz2x,1,ϱ,tdϱdx+εu+v2p+22p+2+2uvp+2p+2.Therefore, using (43) and by setting δ1, δ1 so that, 1/4δ1c3=κℍ−αt/2 and 1/4δ2c3=κℍ−αt/2, substituting in (50), we get(51)K′t≥1−α−εκℍ−αℍ′t+εut22+vt22+u22+v22−ε1−12∫0tgsds∇u22−ε1−12∫0thsds∇v22−εℍαt2c3κ∫τ1τ2μ2ϱdϱu22−ε2go∇u−εℍαt2c3κ∫τ1τ2μ4ϱdϱv22−ε2ho∇u+εu+v2p+22p+2+2uvp+2p+2.For0<a<1, from (39),(52)εu+v2p+22p+2+2uvp+2p+2=εau+v2p+22p+2+2uvp+2p+2+ε2p+21−aℍt+εp+21−aut22+vt22+εp+21−a1−∫0tgsds∇u22+εp+21−a1−∫0thsds∇v22−εp+21−ago∇u−εp+21−aho∇v+εp+21−aKy,z,substituting in (51), we get(53)K′t≥1−α−εκℍ−αℍ′t+εp+21−α+1ut22+vt22+u22+v22+εp+21−α1−∫0tgsds−1−12∫0tgsds∇u22+εp+21−α1−∫0thsds−1−12∫0thsds∇v22−εℍαt2c3κ∫τ1τ2μ2ϱdϱu22−εℍαt2c3κ∫τ1τ2μ4ϱdϱv22+εp+21−αKy,z+εp+21−α−12go∇u+ho∇v+εau+v2p+22p+2+2uvp+2p+2+ε2p+21−αℍt.Since (25) hold, we obtain by using (44) and (46)(54)ℍαtu22≤c4u2p+22αp+2+2+v2p+22αp+2+2u22,(55)ℍαtv22≤c5v2p+22αp+2+2+u2p+22αp+2+2v22,for some positive constants c4, c5. By using (46) and the algebraic inequality,(56)Bθ≤B+1≤1+1bB+b,∀B>0,0<θ<1,b>0,we have, ∀t>0(57)u2p+22αp+2+2≤du2p+22p+2+ℍ0≤du2p+22p+2+ℍt,(58)v2p+22αp+2+2≤dv2p+22p+2+ℍt≤dv2p+22p+2+ℍt,where d=1+1/ℍ0. Also, since(59)x+yγ≤Cxγ+yγ,∀x,y>0,γ>0,we conclude(60)v2p+22αp+2u22≤c6v2p+22p+2+u22p+2≤c7v2p+22p+2+u2p+22p+2,(61)u2p+22αp+2v22≤c8u2p+22p+2+v22p+2≤c9u2p+22p+2+v2p+22p+2,substituting (58) and (61) in (55), we get(62)ℍαtu22≤c10v2p+22p+2+u2p+22p+2+c10ℍt,(63)ℍαtv22≤c11u2p+22p+2+v2p+22p+2+c11ℍt,Combining (53) and (63), using (31), we get(64)K′t≥1−α−εκℍ−αℍ′t+εp+21−α+1ut22+vt22+u22+v22+εp+21−α−1−∫0tgsdsp+21−α−12∇u22+εp+21−α−1−∫0thsdsp+21−α−12∇v22+εp+21−αKy,z+εp+21−α−12go∇u+ho∇v+εc0a−λ1+λ22c3κu2p+22p+2+v2p+22p+2+ε2p+21−α−λ1+λ22c3κℍt,where λ1=c10∫τ1τ2μ2ϱdϱ, λ2=c11∫τ1τ2μ4ϱdϱ.In this case, we takea>0 small enough, then(65)α1=p+21−a−1>0,assuming(66)max∫0∞gsds,∫0∞hsds<p+21−a−1p+21−a−1/2=2α12α1+1,we have(67)α2=p+21−a−1−∫0tgsdsp+21−a−12>0,α3=p+21−a−1−∫0thsdsp+21−a−12>0,choose κ so large that(68)α4=ac0−λ1+λ22c3κ>0,α5=2p+21−a−λ1+λ22c3κ>0,fix κ and a, we appoint ε small enough so that(69)α6=1−α−εκ>0.Then, forβ>0, we estimate (64) and it becomes(70)K′t≥βℍt+ut22+vt22+u22+v22+∇u22+∇v22+go∇u+ho∇v+Ky,z+u2p+22p+2+u2p+22p+2.By (31), for β1>0, we get(71)K′t≥β1ℍt+ut22+vt22+u22+v22+∇u22+∇v22+go∇u+ho∇v+Ky,z+u+v2p+22p+2+2uvp+2p+2,Kt≥K0>0,t>0.Using Holder’s and Young’s inequalities, we have(72)∫Ωuut+vvtdx1/1−α≥Cu2p+2θ/1−α+ut2μ/1−α+v2p+2θ/1−α+vt2μ/1−α,where 1/μ+1/θ=1 put θ=21−α, to get(73)μ1−α=21−2α≤2p+2.Subsequently, fors=2/1−2α and by using (39), we get(74)u2p+22/1−2α≤du2p+22p+2+ℍt,v2p+22/1−2α≤dv2p+22p+2+ℍt,∀t≥0.Therefore,(75)∫Ωuut+vvtdx1/1−α≥c12u2p+22p+2+v2p+22p+2+ut22+vt22+ℍt.Subsequently,(76)K1/1−αt=ℍ1−α+ε∫Ωuut+vvtdx+ε2∫Ωμ1u2+μ3v2dx+ε2∫Ωω1∇u2+ω2∇v2dx1/1−α≤cℍt∫Ωuut+vvtdx1/1−α+u22/1−α+∇u22/1−α+v22/1−α+∇v22/1−α≤cℍt+ut22+vt22+u22+v22+∇u22+∇v22+go∇u+ho∇v+u2p+22p+2+v2p+22p+2.From (70) and (76), gives(77)K′t≥λK11−αt,with λ>0, this quantity depends on β and c. By simple integration of (77), we obtain(78)Kα/1−αt≥1K−α/1−α0−λα/1−αt,Hence,Kt in a situation of blow up in time, when(79)T≤T∗=1−αλαKα/1−α0,Then, this completes the proof of the theorem.


## 4. Conclusion

In this work, we have studied the blow up of the coupled Klein-Gordon system with strong damping, distributed delay, and source terms, under suitable conditions which are so important that we find them in many applications of natural sciences. Many authors have been concerned with this problem in recent decades (see, for example, [17–19]). In the next work, we will try to apply the same technique with a new class of Boussinesq equations which are nonlinear partial differential equation that arises in hydrodynamics and some physical applications. It was subsequently applied to problems in the percolation of water in porous subsurface strata (see, for example, [20, 21]).



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