# Blow-Up of Solutions for a Coupled Nonlinear Viscoelastic Equation with Degenerate Damping Terms: Without Kirchhoff Term

Authors: Salah Mahmoud Boulaaras; Abdelbaki Choucha; Mohamed Abdalla; Karthikeyan Rajagopal; Sahar Ahmed Idris
Journal: Complexity
 (2021)

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## Abstract

In this work, we consider a quasilinear system of viscoelastic equations with degenerate damping and source terms without the Kirchhoff term. Under suitable hypothesis, we study the blow-up of solutions.


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## Body

## 1. Introduction

In this paper, we consider the following problem:(1)utηutt−Δu+∫0th1t−sΔusds+uk+vlutj−1ut=f1u,v,x,t∈Ω×0,T,vtηvtt−Δv+∫0th2t−sΔvsds+vθ+uvts−1vt=f2u,v,x,t∈Ω×0,T,ux,t=vx,t=0,x,t∈∂Ω×0,T,ux,0=u0x,utx,0=u1x,x∈Ω,vx,0=v0x,vtx,0=v1x,x∈Ω,where k,l,θ,ϱ≥0; j,s≥1 for N=1,2, and 0≤j,s≤N+2/N−2 for N≥3; and η≥0 for N=1,2, and 0<η≤2/N−2 for N≥3; and hi.:R+⟶R+i=1,2 are positive relaxation functions which will be specified later. .a+.b.tτ−1.t is the degenerate damping term, and(2)f1u,v=a1u+v2p+1u+v+b1up.u.vp+2,f2u,v=a1u+v2p+1u+v+b1vp.v.up+2.The motivation of our problem firstly is by the initial boundary value problem for the quasilinear equation of the form(3)utηutt−Δu+∫0tht−sΔusds+gu,ut=fu.This type of problem is frequently found in some mathematical models in applied sciences, especially in the theory of viscoelasticity. Problem (3) has been studied by various authors, and several results concerning asymptotic behavior and blow-up have been studied (case η≥0). For example, in the case (gu,ut=0), problem (3) has been investigated in [1] and the author proved the blow-up result. In the case (gu,ut=0) of boundary value problem and in the presence of the dispersion term (−Δutt), Liu [2] studied a general decay of solutions. And, in [3], the authors applied the potential well method to indicate the global existence and uniform decay of solutions (gu;ut) = 0 instead of Δut). Furthermore, the authors obtained a blow-up result. In the case (gu,ut=utmut), in [4], Wu studied a general decay of solution. Later, the same author in [5] considered the same problem but (gu,ut=ut) and discussed the decay rate of solution. Recently, in [6], the authors proved the existence of global solution and a general stability result.There are several works in case (η=0), where the authors have studied the blow-up of solutions of problem (3) (for example, see [3, 7–12]).For a coupled system, He [13] considered the following problem:(4)utηutt−Δu+∫0th1t−sΔusds−Δutt+g1u,ut=f1u,v,vtηvtt−Δv+∫0th2t−sΔvsds−Δvtt+g2v,vt=f2u,v,where η>0;j,s≥2; and g1u,ut=utj−2utandg2v,vt=vts−2vt. The author proved general and optimal decay of solutions. Then, in [14], the author investigated the same problem without damping term and established a general decay of solutions. Furthermore, the author obtained a blow-up of solutions. In addition, in problem (1) with η=0, in [15], Wu proved a general decay of solutions. Later, in [16], Piskin and Ekinci established a general decay and blow-up of solutions with nonpositive initial energy for problem (1) case (Kirchhoff type). In recent years, some other authors investigate the hyperbolic type system with degenerate damping terms (see [17–20]). Very recently, in the presence of the dispersion term −Δutt, our problem (1) has been studied in [21]. Under some restrictions on the initial datum and standard conditions on relaxation functions, the authors have established the global existence and proved the general decay of solutions.Based on all of the abovementioned discussion, we believe that the combination of these terms of damping (memory term, degenerate damping, and source terms) constitutes a new problem worthy of study and research, different from the above that we will try to shed light on, especially the blow-up of solutions.Our paper is divided into several sections: In Section2, we lay down the hypotheses, concepts, and lemmas we need. In Section 3, we prove our main result. Finally, we give some concluding remarks in the last section.


## 2. Preliminaries

We prove the blow-up result under the following suitable assumptions:(A1)hi:ℝ+⟶ℝ+ are differentiable and decreasing functions such that(5)hit≥0,1−∫0∞hisds=li>0,i=1,2.(A2) There exist a constantsξ1,ξ2>0 such that(6)hi′t≤−ξihit,t≥0,i=1,2.Lemma 1.
There exists a functionFu,v such that(7)Fu,v=12p+2uf1u,v+vf2u,v,12p+2a1u+v2p+2+2b1uvp+2≥0,where(8)∂F∂u=f1u,v,∂F∂v=f2u,v.We takea1=b1=1 for convenience.Lemma 2 (see [18]).
There exist two positive constantsc0 and c1 such that(9)c02p+2u2p+2+v2p+2≤Fu,v≤c12p+2u2p+2+v2p+2.Now, we state the local existence theorem that can be established by combining arguments of [13, 16].Theorem 1.
Assume (5) and (6) hold. Let(10)−1<p<4−nn−2,n≥3,p≥−1,n=1,2.
Then, for any initial datum,(11)u0,u1,v0,v1∈ℋ.
Problem (1) has a unique solution, for some T>0:(12)u,v∈C0,T;H2Ω∩H01Ω,ut∈C0,T;H01Ω∩Lj+1Ω,vt∈C0,T;H01Ω∩Ls+1Ω,where(13)ℋ=H01Ω×L2Ω×H01Ω×L2Ω.Now, we define the energy functional.Lemma 3.
Assume (5), (6), and (10) hold; let u,v be a solution of (1); then, Et is nonincreasing, that is,(14)Et=1η+2utη+2η+2+vtη+2η+2,+121−∫0th1sds∇u22+1−∫0th2sds∇u22,+12h1o∇ut+h2o∇ut−∫ΩFu,vdx,which satisfies(15)E′t≤12h1′o∇ut+h2′o∇ut−12h1t∇u22+h2t∇u22,−∫Ωuk+vlutj+1dx−∫Ωvθ+uϱvts+1dx,≤0.Proof.
By multiplying the first and second equations in (1) by ut,vt and integrating over Ω, we get(16)ddt1η+2utη+2η+2+1η+2vtη+2η+2+12−1∫0th1sds∇u22+12−1∫0th2sds∇u22+12h1o∇ut+12h2o∇ut−∫ΩFu,vdx,=−∫Ωuk+vlutj+1dx−∫Ωvθ+uϱvts+1dx+12h1′o∇u−12h1t∇u22+12h2′o∇u−12h2t∇u22.
We obtain (14) and (15).


## 3. Blow-Up

In this section, we prove the blow-up result of solution of problem (1).First, we define the functional as(17)ℍt=−Et=−1η+2utη+2η+2+vtη+2η+2−121−∫0th1sds∇u22+121−∫0th2sds∇v22−12h1o∇ut+h2o∇ut+12p+2u+v2p+22p+2+uv2p+22p+2.Theorem 4.
Assume that (5), (6), and (10) hold, and suppose that E0<0 and(18)2p+2>maxk+j+1;l+j+1;θ+s+1;ϱ+s+1.
Then, the solution of problem (1) blows up in finite time.Proof.
From (14), we have(19)Et≤E0≤0.
Therefore,(20)ℍ′t=−E′t≥∫Ωuk+vlutj+1dx+∫Ωvθ+uϱvts+1dx.Hence,(21)ℍ′t≥∫Ωuk+vlutj+1dx≥0,ℍ′t≥∫Ωvθ+uϱvts+1dx≥0.
By (9) and (17), we have(22)0≤ℍ0≤ℍt≤12p+2u+v2p+22p+2+2uv2p+22p+2,≤c12p+2utη+2η+2+vtη+2η+2.
We set(23)Kt=ℍ1−α+εη+1∫Ωuutηut+vvtηvtdx,where ε>0 is to be assigned later and(24)0<α<min1−12p+2−1η+2,2p+3−k+j2jp+2,2p+3−l+j2jp+2,2p+3−θ+s2sp+2,2p+3−ϱ+s2sp+2<1.
By multiplying the first and second equations in (1) by u,v and with a derivative of (23), we get(25)K′t=1−αℍ−αℍ′t+εη+1utη+2η+2+vtη+2η+2+ε∫Ω∇u∫0tgt−s∇usdsdx︸J1+ε∫Ω∇v∫0tht−s∇usdsdx︸J2−ε∫Ωuk+vlutj−1ut.udx︸J3−ε∫Ωvθ+uϱvts−1vt.vdx︸J4−ε∇u22+∇v22+εu+v2p+22p+2+2uv2p+22p+2︸J5,where we have(26)J1=ε∫0th1t−sds∫Ω∇u.∇us−∇utdxds+ε∫0th1sds∇u22,≥ε2∫0th1sds∇u22−ε2h1o∇u,(27)J2=ε∫0th2t−sds∫Ω∇u.∇us−∇utdxds+ε∫0th2sds∇u22,≥ε2∫0th2sds∇u22−ε2h2o∇u.
From (25), we find that(28)K′t≥1−αℍ−αℍ′t+εη+1utη+2η+2+vtη+2η+2−ε1−12∫0th1sds∇u22+1−12∫0th2sds∇v22−ε2h1o∇u−ε2h2o∇u−J3−J4+J5.
At this point, we use Young’s inequality; forδ>0,(29)XY≤δαXαα+δ−βXββ,α,β>0,1α+1β=1,
We get that forδ1,δ2>0,(30)uutj−1ut≤δ1j+1j+1utj+1+jj+1δ1−j+1/jutj+1,vvts−1vt≤δ2s+1s+1vts+1+ss+1δ2−s+1/svts+1.
Hence, we have(31)J3≤εδ1j+1j+1∫Ωuk+vlutj+1dx+εjδ1−j+1/jj+1∫Ωuk+vlutj+1dxdx,J4≤εδ2s+1s+1∫Ωvθ+uϱvts+1dx+εsδ2−s+1/ss+1∫Ωvθ+uϱvts+1dx.
Therefore, using (21) and by setting δ1,δ2 so that(32)jδ1−j+1/jj+1=κℍ−αt2,sδ2−s+1/ss+1=κℍ−αt2,and substituting in (28), we get(33)K′t≥1−α−εκℍ−αℍ′t+εη+1utη+2η+2+vtη+2η+2−ε1−12∫0th1sds∇u22+1−12∫0th2sds∇v22−ε2h1o∇u−ε2h2o∇u−εC1κℍαjt∫Ωuk+vlutj+1dx−εC2κℍαjt∫Ωvθ+uϱvts+1+J5,where(34)C1κ:=2jκj+1j+11j+1,C2κ:=2sκs+1s+11s+1.We have(35)∫Ωuk+vlutj+1dx=uk+j+1k+j+1+∫Ωvluj+1dx,∫Ωvθ+uϱvts+1dx=vθ+s+1θ+s+1+∫Ωuϱvs+1dx.
By Young’s inequality, we find that forδ3,δ4>0,(36)∫Ωvluj+1dx≤ll+j+1δ3l+j+1/lvl+j+1l+j+1+j+1l+j+1δ3−l+j+1/lul+j+1l+j+1,∫Ωuϱvs+1dx≤ϱϱ+s+1δ4ϱ+s+1/ϱuϱ+s+1ϱ+s+1+s+1ϱ+s+1δ4−ϱ+s+1/ϱvϱ+s+1ϱ+s+1.
Hence,(37)ℍαjt∫Ωuk+vlutj+1dx≤ℍαjtuk+j+1k+j+1+lℍαjtl+j+1δ3l+j+1/lvl+j+1l+j+1+j+1ℍαjtl+j+1δ3−l+j+1/lul+j+1l+j+1,ℍαjt∫Ωvθ+uϱvts+1dx≤ℍαstvθ+s+1θ+s+1+ϱℍαstϱ+s+1δ4ϱ+s+1/ϱuϱ+s+1ϱ+s+1+s+1ℍαstϱ+s+1δ4−ϱ+s+1/ϱvϱ+s+1ϱ+s+1.
Since (10) holds, we obtain the following by using (22) and (24):(38)ℍαjtuk+j+1k+j+1≤c1u2p+22αjp+2+k+j+1+v2p+22αjp+2uk+j+1k+j+1,ℍαjtvk+j+1k+j+1≤c2v2p+22αjp+2+k+j+1+u2p+22αjp+2vk+j+1k+j+1,ℍαstvθ+s+1θ+s+1≤c3v2p+22αsp+2+θ+s+1+u2p+22αsp+2vθ+s+1θ+s+1,ℍαstuθ+s+1θ+s+1≤c4u2p+22αsp+2+θ+s+1+v2p+22αsp+2uθ+s+1θ+s+1,for some positive constants ci,i=1,..4. By using (24) and the algebraic inequality(39)Bς≤B+1≤1+1bB+b,∀B>0,0<ς<1,b>0,we have, ∀t>0,(40)u2p+22αjp+2+k+j+1≤du2p+22p+2+ℍ0≤du2p+22p+2+ℍt,v2p+22αjp+2+k+j+1≤dv2p+22p+2+ℍ0≤dv2p+22p+2+ℍt,v2p+22αsp+2+θ+s+1≤dv2p+22p+2+ℍ0≤dv2p+22p+2+ℍt,u2p+22αsp+2+θ+s+1≤du2p+22p+2+ℍ0≤du2p+22p+2+ℍt,where d=1+1/ℍ0. Also, since(41)X+Yγ≤CXγ+Yγ,X,Y>0,γ>0,we conclude that(42)v2p+22αjp+2uk+j+1k+j+1≤c5v2p+22p+2+u2p+22p+2,u2p+22αjp+2vk+j+1k+j+1≤c6u2p+22p+2+v2p+22p+2,u2p+22αsp+2vθ+s+1θ+s+1≤c7v2p+22p+2+u2p+22p+2,v2p+22αsp+2uθ+s+1θ+s+1≤c8v2p+22p+2+u2p+22p+2.
Substituting (40) and (42) in (38), we get(43)ℍαjtuk+j+1k+j+1≤c9v2p+22p+2+u2p+22p+2+ℍt,ℍαjtvk+j+1k+j+1≤c10v2p+22p+2+u2p+22p+2+ℍt,ℍαjtvk+j+1k+j+1≤c11v2p+22p+2+u2p+22p+2+ℍt,ℍαjtuk+j+1k+j+1≤c12v2p+22p+2+u2p+22p+2+ℍt.
Hence, by fixingδ3,δ4>0, we get(44)ℍαjt∫Ωuk+vlutj+1dx≤M11+lδ3l+j+1/ll+j+1+j+1δ3−l+j+1/ll+j+1v2p+22p+2+u2p+22p+2+ℍt,ℍαst∫Ωvθ+uϱvts+1dx≤M21+ϱδ4ϱ+s+1/ϱϱ+s+1+s+1δ4−ϱ+s+1/ϱϱ+s+1v2p+22p+2+u2p+22p+2+ℍt,for some constants M1,M2>0.
Now, for0<a<1, from (17),(45)J5=εu+v2p+22p+2+2uvp+2p+2=εau+v2p+22p+2+2uvp+2p+2+2εp+21−aη+2utη+2η+2+vtη+2η+2+εp+21−a1−∫0tgsds∇u22+εp+21−a1−∫0thsds∇v22−εp+21−ah1o∇u+h2o∇u+ε2p+21−aℍt.
Substituting in (33) and by using (9), we get(46)K′t≥1−a−εκℍ−αℍ′t+ε2εp+21−aη+2+1η+1utη+2η+2+vtη+2η+2,+εp+21−a1−∫0th1sds−1−12∫0th1sds∇u22+εp+21−a1−∫0th2sds−1−12∫0th2sds∇v22+εp+21−a−12h1o∇u+h2o∇v+εc0a−M3C1κ+M4C2κu2p+22p+2+v2p+22p+2+ε2p+21−a−M3C1κ+M4C2κℍt,where(47)M3:=M11+lδ3l+j+1/ll+j+1+j+1δ3−l+j+1/ll+j+1>0,M4:=M21+ϱδ4ϱ+s+1/ϱϱ+s+1+s+1δ4−ϱ+s+1/ϱϱ+s+1>0.
In this stage, we takea>0 small enough so that(48)λ1=p+21−a−1>0,and we assume that(49)max∫0∞h1sds,∫0∞h2sds<p+21−a−1p+21−a−1/2=2λ12λ1+1gives(50)λ2=p+21−a−1−∫0th1sdsp+21−a−12>0,λ3=p+21−a−1−∫0th2sdsp+21−a−12>0.Then, we choose κ so large such that(51)λ4=ac0−M3C1κ+M4C2κ>0,λ5=2p+21−a−M3C1κ+M4C2κ>0.
Finally, we fixκanda and we appoint ε small enough so that(52)λ6=1−α−εκ>0.
Thus, for someβ>0, estimate (46) becomes(53)K′t≥βℍt+utη+2η+2+vtη+2η+2+∇u22+∇v22+h1o∇u+h2o∇v+u2p+22p+2+u2p+22p+2.
By (9), for some β1>0, we obtain(54)K′t≥β1ℍt+utη+2η+2+vtη+2η+2+∇u22+∇v22+h1o∇u+h2o∇v+u+v2p+22p+2+2uvp+2p+2and(55)Kt≥K0>0,t>0.
Next, using Hölder’s and Young’s inequalities, we have(56)∫Ωuutηut+vvtηvtdx1/1−α≤Cu2p+2θ/1−α+utη+2μ/1−α+v2p+2θ/1−α+vtη+2μ/1−α,where 1/μ+1/θ=1.
We takeμ=η+21−α, to get(57)θ1−α=η+21−αη+2−1≤2p+2.
Subsequently, by using (24), (22), and (39), we obtain(58)u2p+2η+2/1−αη+2−1≤du2p+22p+2+ℍt,v2p+2η+2/1−αη+2−1≤dv2p+22p+2+ℍt,∀t≥0.
Therefore,(59)∫Ωuutηut+vvtηvtdx1/1−α≤c13u2p+22p+2+v2p+22p+2+utη+2η+2+vtη+2η+2+Ht.
Hence, by substituting (59) into (23), we get(60)K1/1−αt=ℍ1−α+εη+1∫Ωuutηut+vvtηvtdx1/1−α,≤cℍt+∫Ωuutηut+vvtηvtdx1/1−α,≤cℍt+utη+2η+2+vtη+2η+2+u2p+22p+2+v2p+22p+2,≤cℍt+utη+2η+2+vtη+2η+2+∇u22+∇v22+h1o∇u+h2o∇v+u2p+22p+2+v2p+22p+2.
From (53) and (60), we get(61)K′t≥λK1/1−αt,where λ>0, and this depends only on β and c.
By integration of (61), we obtain(62)Kα/1−αt≥1K−α/1−α0−λα/1−αt.
Hence,Kt blows up in time(63)T≤T∗=1−αλαKα/1−α0.
Then, the proof is completed.


## 4. Conclusion

The objective of this work is the study of the blow-up of solutions for a quasilinear viscoelastic system with degenerate damping. This type of problem is frequently found in some mathematical models in applied sciences, especially in the theory of viscoelasticity. What interests us in this current work is the combination of these terms of damping (memory term, degenerate damping, and source terms), which dictates the emergence of these terms in the system.In the next work, we will try using the same method with the same problem in addition to other damping terms (dispersion term, Balakrishnan–Taylor damping, and delay term).



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