# On a Couple of Nonlocal Singular Viscoelastic Equations with Damping and General Source Terms: Blow-Up of Solutions

Authors: Erhan Piskin; Salah Mahmoud Boulaaras; Hasan Kandemir; Bahri Belkacem Cherif; Mohamed Biomy
Journal: Journal of Function Spaces
 (2021)

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## Abstract

Under some given conditions, we prove the explosion result of the solution of the system of nonlocal singular viscoelastic with damping and source terms on general case. This current study is a general case of the previous work of Boulaaras.


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## Body

## 1. Introduction

During the last decades, many nonlocal problems of deterministic and parabolic partial differential equations have been studied. These equations and their systems represent the modeling of many physical phenomena related to time. These constraints can be data measured directly at the boundary or give integral boundary conditions (for instance, see [1–25]).In this work, we investigate the blow-up of the following system of nonlinear damping term:(1)utt−1xxuxx+∫0tg1t−s1xxuxx,sxds+utm−1ut=f1u,v,Q,vtt−1xxvxx+∫0tg2t−s1xxvxx,sxds+vtm−1vt=f2u,v,Q,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,wheref1u,v,f2u,v:R2⟶Rgiven by
(2)f1u,v=au+v2r+1u+v+buruvr+2,f2u,v=au+v2r+1u+v+bvrvur+2,with a,b∈R, r≥−1 (we get a=b=1), Q=0,α×0,T, α<∞, T<∞, and
(3)g1.,g2.:R+⟶R+are given functions which will be specified later. The motivation of our work is because of some results regarding the following research paper: in [12], under some conditions suitable for the relaxation function, the author explained that solutions with initial negative energy explode in a finite time if p>m and continue to find if m≥p, for the following studied problem:
(4)utt−Δu+∫0tgt−sΔu+ututm−2=up−2u,inΩ×0,∞,u=0x∈∂Ω,t∈0,∞,ux,0=u0x,utx,0=u1x,x∈Ω.In [4], the author studied a model describing the movement of a flexible two-dimensional viscous body on the unit disk (i.e., radial solutions) and by using some density arguments and some prior estimates, the authors demonstrated the existence and uniqueness of a generalized solution to the following problem:
(5)utt−1xxuxx+∫0tgt−s1xxuxx,sxds=fx,t,u,ux,inQ,ux1,t=0,∫01xux,tdx=0,t∈0,T,ux,0=φx,utx,0=ψx,x∈0,1,where
(6)Q=0,1×0,Tand f is the right-hand side that satisfied the Lipschitzian condition. Recently, in [3], the authors demonstrated the decay result of energy for a small enough initial data together with the explosion result of large initial data of the following singular problem:
(7)utt−1xxuxx+∫0tgt−s1xxuxx,sxds=up−2u,ua,t=0,∫0axux,tdx=0,ux,0=φx,utx,0=ψx.That is, they obtained the blow-up properties of local solution by Georgiev-Todorova method with nonpositive initial energy. More work followed up on similar nonlocal singular viscoelastic equations and systems in [8, 9].In this work, we continue the study on system (1). According to some given conditions, we prove the explosion result of the solution of the system of nonlocal singular viscoelastic with damping and source terms on general case, where we begin by giving basic definitions and theories about the function spaces we need, and then, we mention the theorem of local existence. Finally, we announce and prove the main result of our studied problem in (1).


### 1.1. Preliminaries

In this section, we introduce some functional spaces and give some lemma’s need for the remaining of this paper. LetLxp=Lxp0,α be the weighed Banach space equipped with the norm
(8)uLxp=∫0αxupdx1/p.H=Lx20,αis, in particular, the Hilbert space of square integral functions having the finite norm
(9)uH=∫0αxu2dx1/2.V=Vx10,αis the Hilbert space equipped with the norm
(10)uV=uH2+uxH21/2,V0=u∈V:uα=0.Lemma 1 (Poincare-type inequality).
For anyu∈V0,
(11)∫0αxu2dx≤CP∫0αxux2dx,where CP is some positive constant.Remark 2.
It is clear thatuV0=uxHdefines an equivalent norm onV0.Lemma 3.
For anyu∈V0and2<p<4, we have(12)uLxp0,αp≤C∗uxH=Lx20,αp,where C∗ is a constant depending on α and p only. For the g1 and g2 functions, assumptions are as follows:(G1): g1.,g2.:R+⟶R+ are two differentiable and nonincreasing functions with
(13)g1t≥0,1−∫0∞g1sds=I1≥0,g2t≥0,1−∫0∞g2sds=I2≥0,
(G2): For allt≥0,(14)g1t≥0,g1′t≤0,g2t≥0,g2′t≤0.
(G3):r≥−1.Theorem 4.
Suppose that (G1), (G2), and (G3) hold. Then, for allu0,v0∈V02and allu1,v1∈H2, problem ((1)) admits a unique local solution (u, v):(15)u,v∈C0,T;V0∩C10,T;H,for T>0 small enough.Lemma 5.
Assume that (G1), (G2), and (G3) hold and (u, v) is a solution of problem((1)); then, the energy functional(16)Et=12∫0αxut2dx+12∫0αxvt2dx+121−∫0tg1sds∫0αxux2dx+121−∫0tg2sds∫0αxvx2dx+12g1∘uxt+12g2∘vxt−∫0αxFu,vdx,where
(17)g1∘uxt=∫0α∫0txg1t−suxx,s−uxx,t2dsdx,g2∘vxt=∫0α∫0txg2t−svxx,s−vxx,t2dsdx,Fu,v=12r+2u+v∣2r+2+2uvr+2.Remark 6.
Multiplying the first equation in((1))byxutand the second equation in((1))byxvtintegrating over0,α, we obtain the following equation:(18)ddtEt=−∫0αxutm+1dx−∫0αxvtm+1dx,=−utLxm+1m+1+vtLxm+1m+1.The definition of the norm is as follows:(19)utLxm+1m+1+vtLxm+1m+1≥0.From here,(20)−utLxm+1m+1+vtLxm+1m+1≤0.Thus,(21)ddtEt=−utLxm+1m+1+vtLxm+1m+1≤0.Lemma 7.
There existc0andc1positive constants such that(22)c02r+2u2r+2+v2r+2≤Fu,v≤c12r+2u2r+2+v2r+2.Lemma 8.
If2≤s≤p,
(23)uLxps≤CuxH2+uLxpp.


## 1.1. Preliminaries

In this section, we introduce some functional spaces and give some lemma’s need for the remaining of this paper. LetLxp=Lxp0,α be the weighed Banach space equipped with the norm
(8)uLxp=∫0αxupdx1/p.H=Lx20,αis, in particular, the Hilbert space of square integral functions having the finite norm
(9)uH=∫0αxu2dx1/2.V=Vx10,αis the Hilbert space equipped with the norm
(10)uV=uH2+uxH21/2,V0=u∈V:uα=0.Lemma 1 (Poincare-type inequality).
For anyu∈V0,
(11)∫0αxu2dx≤CP∫0αxux2dx,where CP is some positive constant.Remark 2.
It is clear thatuV0=uxHdefines an equivalent norm onV0.Lemma 3.
For anyu∈V0and2<p<4, we have(12)uLxp0,αp≤C∗uxH=Lx20,αp,where C∗ is a constant depending on α and p only. For the g1 and g2 functions, assumptions are as follows:(G1): g1.,g2.:R+⟶R+ are two differentiable and nonincreasing functions with
(13)g1t≥0,1−∫0∞g1sds=I1≥0,g2t≥0,1−∫0∞g2sds=I2≥0,
(G2): For allt≥0,(14)g1t≥0,g1′t≤0,g2t≥0,g2′t≤0.
(G3):r≥−1.Theorem 4.
Suppose that (G1), (G2), and (G3) hold. Then, for allu0,v0∈V02and allu1,v1∈H2, problem ((1)) admits a unique local solution (u, v):(15)u,v∈C0,T;V0∩C10,T;H,for T>0 small enough.Lemma 5.
Assume that (G1), (G2), and (G3) hold and (u, v) is a solution of problem((1)); then, the energy functional(16)Et=12∫0αxut2dx+12∫0αxvt2dx+121−∫0tg1sds∫0αxux2dx+121−∫0tg2sds∫0αxvx2dx+12g1∘uxt+12g2∘vxt−∫0αxFu,vdx,where
(17)g1∘uxt=∫0α∫0txg1t−suxx,s−uxx,t2dsdx,g2∘vxt=∫0α∫0txg2t−svxx,s−vxx,t2dsdx,Fu,v=12r+2u+v∣2r+2+2uvr+2.Remark 6.
Multiplying the first equation in((1))byxutand the second equation in((1))byxvtintegrating over0,α, we obtain the following equation:(18)ddtEt=−∫0αxutm+1dx−∫0αxvtm+1dx,=−utLxm+1m+1+vtLxm+1m+1.The definition of the norm is as follows:(19)utLxm+1m+1+vtLxm+1m+1≥0.From here,(20)−utLxm+1m+1+vtLxm+1m+1≤0.Thus,(21)ddtEt=−utLxm+1m+1+vtLxm+1m+1≤0.Lemma 7.
There existc0andc1positive constants such that(22)c02r+2u2r+2+v2r+2≤Fu,v≤c12r+2u2r+2+v2r+2.Lemma 8.
If2≤s≤p,
(23)uLxps≤CuxH2+uLxpp.


## 2. Blow-Up of Solution

In this section, we shall deal with the blow-up behavior of solutions for problem (1). We derive the blow-up properties of solutions of problem (1) with nonpositive initial energy by the method given in [1].Theorem 9.
Assume that (G1), (G2), and (G3) hold.E0<0and(24)∫0∞gisds<r+1r+1+1/4r+2,i=1,2.Then, the solution of problem (1) blows up in finite time.Proof.
Sinced/dtEt=E′t≤0,
(25)Et≤E0<0,∀t≥0.
We defineHt=−Et; then,
(26)0<H0≤Ht=−Et,∀t≥0.☐We obviously substituteEt in (26); then,
(27)0<H0≤Ht=−12∫0αxut2dx−12∫0αxvt2dx−121−∫0tg1sds∫0αxux2dx−121−∫0tg2sds∫0αxvx2dx−12g1∘uxt−12g2∘vxt+∫0αxFu,vdx.From (22) and (27),
(28)0<H0≤Ht≤∫0αxFu,vdx≤c12r+2∫0αxu2r+2dx+∫0αxv2r+2dx=c12r+2uLx2r+22r+2+vLx2r+22r+2.Thus,(29)Ht≤c12r+2uLx2r+22r+2+vLx2r+22r+2.Equation (29) will then be used as an important data for proof of the theorem. Now, we define
(30)Lt=H1−σt+ε∫0αxuutdx+∫0αxvvtdxfor ε small enough and
(31)0<σ≤min2r+2−m2mr+2,2r+2−m2mr+2,2r+24r+2.By differentiating (30), using (1) and H′t=utLxm+1m+1+vtLxm+1m+1, we obtain
(32)L′t=1−σH−σtutLxm+1m+1+vtLxm+1m+1+ε∫0αxut2dx+ε∫0αxvt2dx−ε1−∫0tg1sds∫0αxux2dx−ε1−∫0tg2sds∫0αxvx2dx+ε∫0α∫0txg1t−suxx,tuxx,s−uxx,tdsdx+ε∫0α∫0txg2t−svxx,tvxx,s−vxx,tdsdx−ε∫0αxuututm−1dx−ε∫0αxvvtvtm−1dx+ε2r+2∫0αxFu,vdx.By using Young inequality and fromHt=−Et,
(33)∫0αxFu,vdx=Ht+12∫0αxut2dx+12∫0αxvt2dx+121−∫0tg1sds∫0αxux2dx+121−∫0tg2sds∫0αxvx2dx+12g1∘uxt+12g2∘vxt,we obtain
(34)L′t≥1−σH−σtutLxm+1m+1+vtLxm+1m+1+εr+3∫0αxut2dx+εr+3∫0αxvt2dx+εr+1−r+1+14θ∫0tg1sds∫0αxux2dx+εr+1−r+1+14θ∫0tg2sds∫0αxvx2dx+εr−θ+2g1∘uxt+εr−θ+2g2∘vxt+ε2r+2Ht−ε∫0αxuututm−1dx−ε∫0αxvvtvtm−1dx.where
(35)α3=r−θ+2>0⇒r+2>θ>0,α4=r−θ+2>0⇒r+2>θ>0,α1=r+1−r+1+14θ∫0tg1sds>0,α2=r+1−r+1+14θ∫0tg2sds>0.From (34),
(36)L′t≥1−σH−σtutLxm+1m+1+vtLxm+1m+1+εr+3∫0αxut2dx+εr+3∫0αxvt2dx+εα1∫0αxux2dx+εα2∫0αxvx2dx+εα3g1∘uxt+εα4g2∘vxt+ε2r+2Ht−ε∫0αxuutut∣m−1dx−ε∫0αxvvtvtm−1dx,(37)Ht=−Et=−12∫0αxut2dx−12∫0αxvt2dx−121−∫0tg1sds∫0αxux2dx−121−∫0tg2sds∫0αxvx2dx−12g1∘uxt−12g2∘vxt+∫0αxFu,vdx,and for a5<minα1,α2,α3,α4,2r+2,
(38)ε2r+2Ht=εa5+2r+2−a5Ht=εa5Ht+ε2r+2−a5Ht=−ε2a5∫0αxut2dx−ε2a5∫0αxvt2dx−ε2a51−∫0tg1sds∫0αxux2dx−ε2a51−∫0tg2sds∫0αxvx2dx−ε2a5g1∘uxt−ε2a5g2∘vxt+εa5∫0αxFu,vdx+ε2r+2−a5Ht.To estimate the last term in (36), we apply the three-parameter Young inequality: a,b≥0,1/r+1/q=1,ab≤δr/rar+δ−qbq/q,∀δ>0. We take
(39)r=m+1veq=m+1m,in this case:
(40)−ε∫0αxuututm−1dx≥−εδ1m+1m+1uLxm+1m+1−εmm+1δ1−m+1/mutLxm+1m+1.Similarly(41)−ε∫0αxvvtvtm−1dx≥−εδ2m+1m+1vLxm+1m+1−εmm+1δ2−m+1/mvtLxm+1m+1.Substituting (38), (40), and (41) into (36), by organizing, we obtain
(42)L′t≥1−σH−σt−mm+1εδ1−m+1/mutLxm+1m+1+1−σH−σt−mm+1εδ2−m+1/mvtLxm+1m+1+εr+3−a52∫0αxut2dx+εr+3−a52∫0αxvt2dx+εα1−a521−∫0tg1sds∫0αxux2dx+εα2−a521−∫0tg2sds∫0αxvx2dx+εα3−a52g1∘uxt+εα4−a52g2∘vxt+εa5∫0αxFu,vdx+ε2r+2−a5Ht−εδ1m+1m+1uLxm+1m+1−εδ2m+1m+1vLxm+1m+1.Since integration in estimate (40) and (41) is performed over the space, the parameter δ1 and δ2 can be a function of time; we get them as follows:
(43)δ1−m+1/m=k1H−σt⇒δ1m+1=k1−mHσmt,(44)δ2−m+1/m=k2H−σt⇒δ2m+1=k2−mHσmt,where k1>0 and k2>0 are sufficiently large constants to be specified further. By using (43) and (44) in (42), we have
(45)L′t≥1−σ−mm+1εk1H−σtutLxm+1m+1+1−σ−mm+1εk2H−σtvtLxm+1m+1+εr+3−a52∫0αxut2dx+εr+3−a52∫0αxvt2dx+εα1−a521−∫0tg1sds∫0αxux2dx+εα2−a521−∫0tg2sds∫0αxvx2dx+εα3−a52g1∘uxt+εα4−a52g2∘vxt+εa5∫0αxFu,vdx+ε2r+2−a5Ht−εk1−mm+1HσmtuLxm+1m+1−εk2−mm+1HσmtvLxm+1m+1.To estimate the last two terms in (45), we use (29); then,
(46)εk1−mm+1HtσmuLxm+1m+1≤εk1−mc2σmm+1uLx2r+22r+2+vLx2r+22r+2σmuLxm+1m+1.On the other hand, sincer>maxm,m from Lx2r+2°Lxm+1,
(47)uLxm+1m+1≤CuLx2r+2m+1≤CuLx2r+2+vLx2r+2m+1.Substituting (47) into (46),
(48)εk1−mm+1HtσmuLxm+1m+1≤εk1−mc2σmCm+1uLx2r+22r+2+vLx2r+22r+2σmuLx2r+2+vLx2r+2m+1.By using(49)a,b≥0,1≤p<∞,ap+bp≤a+bp,we can estimate the following:
(50)uLx2r+22r+2+vLx2r+22r+2≤uLx2r+2+vLx2r+22r+2.Consequently, we have(51)εk1−mm+1HtσmuLxm+1m+1≤εk1−mc2σmCm+1uLx2r+2+vLx2r+22r+2σm+m+1.Similarly(52)εk2−mm+1HtσmvLxm+1m+1≤εk2−mc2σmCm+1uLx2r+2+vLx2r+22r+2σm+m+1.By using (51) and (52),
(53)a,b≥0,1≤p<∞,a+bp≤cap+bp,c=2p−1,forc·C=C′; we have
(54)εk1−mm+1HtσmuLxm+1m+1≤εk1−mc2σmC′m+1uLx2r+22r+2σm+m+1+vLx2r+22r+2σm+m+1,(55)εk2−mm+1HtσmvLxm+1m+1≤εk2−mc2σmC′m+1uLx2r+22r+2σm+m+1+vLx2r+22r+2σm+m+1.From (31),
(56)2r+2σm+m+1≤2r+22r+2−m2mr+2m+m+1=2r+2+1,r≥−1,2r+2≥2,2r+2σm+m+1≥2.From here,(57)2≤2r+2σm+m+1≤2r+2+1.Thus, by applying (23), we obtain
(58)uLx2r+22r+2σm+m+1≤uxH2+uLx2r+22r+2,vLx2r+22r+2σm+m+1≤vxH2+vLx2r+22r+2.Substituting these inequalities in (54) and (55), in this case,
(59)εk1−mm+1HtσmuLxm+1m+1≤εk1−mc2σmC′m+1uxH2+uLx2r+22r+2+vxH2+vLx2r+22r+2,(60)εk2−mm+1HtσmvLxm+1m+1≤εk2−mc2σmC′m+1uxH2+uLx2r+22r+2+vxH2+vLx2r+22r+2.With the combination of (59) and (60), we obtain
(61)−εk1−mm+1HσmtuLxm+1m+1−εk2−mm+1HσmtvLxm+1m+1≥−εk1−mc2σmC′m+1−εk2−mc2σmC′m+1uLx2r+22r+2+vLx2r+22r+2+−εk1−mc2σmC′m+1−εk2−mc2σmC′m+1uxH2+vxH2.Finally,(62)uxH2=∫0αxux2dx,vxH2=∫0αxvx2dx,uLx2r+22r+2=∫0αxu2r+2dx,vLx2r+22r+2=∫0αxv2r+2dx,c′·∫0αxu2r+2dx+∫0αxv2r+2dx≤∫0αxFu,vdx,c02r+2=c′,and by considering (61), thus by organizing (45), we have
(63)L′t≥1−σ−mm+1εk1H−σtutLxm+1m+1+1−σ−mm+1εk2H−σtvtLxm+1m+1+εr+3−a52∫0αxut2dx+εr+3−a52∫0αxvt2dx+εα1−a521−∫0tg1sds−k1−mc2σmC′m+1−k2−mc2σmC′m+1∫0αxux2dx+εα2−a521−∫0tg2sds−k1−mc2σmC′m+1−k2−mc2σmC′m+1∫0αxvx2dx+εc′a5−k1−mc2σmC′m+1−k2−mc2σmC′m+1∫0αxu2r+2dx+εc′a5−k1−mc2σmC′m+1−k2−mc2σmC′m+1∫0αxv2r+2dx+εα3−a52g1∘uxt+εα4−a52g2∘uxt+ε2r+2−a5Ht,which introduce the constant
(64)γ=ε·minr+3−a52,α1−a521−∫0tg1sds−k1−mc2σmC′m+1−k2−mc2σmC′m+1,α2−a521−∫0tg2sds−k1−mc2σmC′m+1−k2−mc2σmC′m+1,c′a5−k1−mc2σmC′m+1−k2−mc2σmC′m+1,α3−a52,α4−a52,2r+2−a5.Taking sufficiently largek1>0 and k2>0 for the positive constant γ, we simplify (63)
(65)L′t≥1−σ−mm+1εk1H−σtutLxm+1m+1+1−σ−mm+1εk2H−σtvtLxm+1m+1+εγ∫0αxut2dx+∫0αxvt2dx+∫0αxux2dx+∫0αxvx2dx+∫0αxu2r+2dx+∫0αxv2r+2dx+g1∘uxt+g2∘vxt+Ht.For fixedk1>0, k2>0, and γ>0, we choose ε>0 so small that the following inequality holds:
(66)1−σ−mm+1εk1≥0,1−σ−mm+1εk2≥0.Moreover, we assume that the initial data satisfy the estimate(67)L0=H1−σ0+ε∫0αxu0u1dx+∫0αxv0v1dx>0.Then, from (65), we obtain the following inequality:
(68)L′t≥εγ∫0αxut2dx+∫0αxvt2dx+∫0αxux2dx+∫0αxvx2dx+∫0αxu2r+2dx+∫0αxv2r+2dx+g1∘uxt+g2∘vxt+Ht.On the other hand, in Equation (30), we take the1/1−σ·powerof each side
(69)Lt1/1−σ=H1−σt+ε∫0αxuutdx+∫0αxvvtdx1/1−σ.Twice by applying the following inequality to (69)
(70)a,b≥0,1≤p<∞,a+bp≤2p−1ap+bp,we have
(71)Lt1/1−σ≤2σ/1−σHt+ε1/1−σ∫0αxuutdx+∫0αxvvtdx1/1−σ≤2σ/1−σHt+ε1/1−σ2σ/1−σ∫0αxuutdx1/1−σ+∫0αxvvtdx1/1−σ≤CHt+∫0αxuutdx1/1−σ+∫0αxvvtdx1/1−σ,where C>0. Now, to estimate the last two terms in (71), we, respectively, apply Holder inequality, Lx2r+2°LxH, and Young inequality; thus,
(72)∫0αxuutdx1/1−σ≤uH1/1−σutH1/1−σ≤CuLx2r+21/1−σutH1/1−σ≤CuLx2r+2θ/1−σ+utHμ/1−σ.Similarly,(73)∫0αxvvtdx1/1−σ≤CvLx2r+2θ/1−σ+vtHμ/1−σ,where 1/θ+1/μ=1. In these inequalities by collecting side by side, we obtain
(74)∫0αxuutdx1/1−σ+∫0αxvvtdx1/1−σ≤CuLx2r+2θ/1−σ+utHμ/1−σ+vLx2r+2θ/1−σ+vtHμ/1−σ.We chooseμ=21−σ, to get
(75)θ=21−σ1−2σ≤2r+2,then
(76)∫0αxuutdx1/1−σ+∫0αxvvtdx1/1−σ≤CuLx2r+22/1−2σ+utH2+vLx2r+22/1−2σ+vtH2.By applying (23), we can write
(77)uLx2r+22/1−2σ≤CuxH2+uLx2r+22r+2,vLx2r+22/1−2σ≤CvxH2+vLx2r+22r+2.From here, we obtain(78)∫0αxuutdx1/1−σ+∫0αxvvtdx1/1−σ≤CutH2+vtH2+uxH2+vxH2+uLx2r+22r+2+vLx2r+22r+2.Thus, by considering (78) and the following in (71),
(79)utH2=∫0αxut2dx,vtH2=∫0αxvt2dx,uxH2=∫0αxux2dx,vxH2=∫0αxvx2dx,uLx2r+22r+2=∫0αxu2r+2dx,vLx2r+22r+2=∫0αxv2r+2dx,g1∘uxt≥0,g2∘vxt≥0,we obtain
(80)Lt1/1−σ≤CHt+∫0αxut2dx+∫0αxvt2dx+∫0αxux2dx+∫0αxvx2dx+∫0αxu2r+2dx+∫0αxv2r+2dx+g1∘uxt+g2∘vxt.Finally, by combining (68) and (80), we obtain the following ordinary differential inequality:
(81)L′t≥λL1/1−σt∀t≥0,obviously, where λ>0 is a constant depending only C, ε, and γ. This differential inequality integration over 0,t gives
(82)L1/1−σt≥1L−σ/1−σ0−λσ/1−σt,where we choose
(83)t≤T∗=1−σλσLσ/1−σ0.Hence,(84)limt⟶T∗−Lt⟶∞.


## 3. Conclusions

The purpose of this paper is to study the explosion result of the solution of the system of nonlocal singular viscoelastic with damping and source terms on general case. This current study is a general case of the previous work of Boulaaras in ([5]). In the next work, we will try to obtain the same result for the two-dimensional problem that allows a reasonable description of the phenomenon occurring in a three-dimensional domain. Then, we will try to prove uniqueness results of the weak solution.



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*Source: 9914386-2021-08-23.xml*