# Uniform Energy Decay Rates for the Fuzzy Viscoelastic Model with a Nonlinear Source

Authors: Fengyun Zhang
Journal: Journal of Mathematics
 (2020)

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## Abstract

This paper considers the fuzzy viscoelastic model with a nonlinear sourceutt+Lu+∫0tgt−ζΔuζdζ−uγu−ηΔut=0 in a bounded field Ω. Under weak assumptions of the function gt, with the aid of Mathematica software, the computational technique is used to construct the auxiliary functionals and precise priori estimates. As time goes to infinity, we prove that the solution is global and energy decays to zero in two different ways: the exponential form and the polynomial form.


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## Body

## 1. Introduction

In this paper, we take the following fuzzy viscoelastic model into account:(1)utt+Lu+∫0tgt−ζΔuζdζ−uγu−ηΔut=0,x∈Ω,t∈0,∞,ux,t=0,x,t∈Γ×0,∞,ux,t|t=0=u0x,utx,t|t=0=u1x,x∈Ω,where the fuzzy number η∈1,1000,γ>0:(2)Lut=−Δut+c2ut.InRNN≥1, Ω is a domain which is well bounded. Besides, the boundary of Ω is smooth perfectly and expressed as Γ:=∂Ω. Meanwhile, the memory kernel g (t) is positive and some assumptions will be given in detail.This type of problems has been observed in many areas of scientific and engineering fields. For example, time analyticity for the viscoelastic equation is studied as follows [1]:(3)utt−Δu+∫0tgt−ζΔuζdζ=uγu.S. Berrimi and S. A. Messaoudi applied weak conditions on the memory kernelg. Meanwhile, considering the condition that the energy is positive and relatively small, they obtained the existence of global solutions.Taking(4)utt−Δu+∫0tgt−ζΔuζdζ+axut+uγu=0,into account, they also obtained a decay rate exponentially in [2]. M.M. Cavalcanti and H.P. Oquendo improved this latter result in [3]. In their work, two situations, the internal dissipation and the viscoelastic dissipation, are considered to act on respective part separately. The authors in [3] expanded the internal dissipation to nonlinear cases as much as possible. Simultaneously, the system is well stabilized through the dissipation which is induced by the integral term.As we know, the viscoelastic terms attract many mathematicians; for instance, the authors in [4] studied the energy decay rate for the global solution of a quasilinear viscoelastic model, and the authors in [5–9] considered time analysis of solutions for some viscoelastic models. Moreover, F. Y. Zhang et al. considered a nonlinear viscoelastic equation in [10]:(5)utt+Lu+∫0tgt−ζΔuζdζ=0.The solution is perfectly stabilized through the dissipation, which is induced by the viscoelastic term. The modified energy functional in [10] has been used to prove the energy decay through two different ways: the exponential form and the polynomial form. Additionally, the construction of auxiliary functions is organized by the computational technique of undetermined coefficients. Besides, the authors in [11, 12] discussed the adaptive fuzzy control of nonlinear systems, and the discussion of fuzzy coefficients involved in these papers is quite interesting.Inspired by these works, we consider (1) in this paper, the two optimal decay rates, exponential decay and polynomial decay, are easily and directly established through the application of Mathematica software. The specific arrangement of this work is as follows: in Section 2, we present some notations and necessary materials; in Section 3, in view of the fuzzy number ƞ, we give the whole decay result, and our choice of the “Lyapunov” functional shows the extensive applicability and practical significance of the computational technique.


## 2. Preliminaries

In this section,LpΩ and H01Ω are understood and applied in their usual senses. We impose the following hypotheses and preliminaries on the memory kernel g (t). In addition, the definition of energy function plays a significant role to our main result:H1: as a bounded C1−function, gt:R+⟶R+ satisfies(6)1−∫0∞gζdζ=ℓ,where bothg (0) and ℓ are positive.H2: the existence of a positive constant ξ makes the following formula hold:(7)g′t≤−ξ⋅gpt,t≥0,p∈1,32.Remark 1.
From the assumption above, ifp=1, we have(8)gt≤ce−ξt.
If1<p<3/2, we have(9)gt≤1g1−p0+p−1ξt1/p−1=Δ1d1t+d21/p−1,where d1=p−1ξ>0andd2=g1−p0.
Indeed, the conditionp<3/2 plays an important role to ensure that(10)∫0∞g2−pζdζ<∞.Theorem 1.
Assuming thatu0,u1∈H2Ω∩H01Ω×H01Ω, then u (t) can be found as a unique solution to model (1) with(11)u∈Lloc∞0,∞;H01Ω∩H2Ω,u′∈Lloc∞0,∞;H01Ω,u″∈Lloc∞0,∞;L2Ω.Simultaneously, we get(12)u∈C0,∞;H01Ω,u′∈C10,∞;L2Ω.Taking the initial data, less regularity, and the priori estimate into consideration, the theorem above guarantees the existence of solutions for model (1) as in [7, 13]. In addition, the Galerkin approximation method can be applied to accomplish the proof of the theorem above.Our primary assignment is to find out the energy functionεt. Combining the multiplier method, the integral subsection integration, and (H1, H2), the calculation is provided:(13)0=∫Ωutt−Δu+c2u+∫0tgt−ζΔuζdζ−uγu−ηΔututdx=12ddt∫Ωut2dx+∫Ω∇u2dx+c2∫Ωu2dx+∫0tgt−ζ∫ΩΔuζutdxdζ−∫Ωuγuutdx+η∫Ω∇ut2dx=12ddt∫Ωut2dx+∫Ω∇u2dx+c2∫Ωu2dx−11+γ/2∫Ωuγ+2dx−∫0tgt−ζ∫Ω∇uζ∇utdxdζ+η∫Ω∇ut2dx,(14)−∫0tgt−ζ∫Ω∇uζ∇utdxdζ=−∫Ω∇ut∫0tgt−ζ∇uζdζdx=−∫Ω∇ut∫0tgt−ζ∇uζ−∇utdζdx−∫Ω∇ut∇ut∫0tgt−ζdζdx=12ddt∫Ω∫0tgt−ζ∇uζ−∇ut2dζdx−12∫Ω∫0tg′t−ζ∇uζ−∇ut2dζdx−∫Ω∇ut∇ut∫0tgt−ζdζdx=12ddtg∘∇u−∫0tgζdζ∫Ω∇ut2dx−12g′∘∇u+gt2∫Ω∇ut2dx.Now, (13) yields(15)12ddt∫Ωut2dx+1−∫0tgζdζ∫Ω∇ut2dx+c2∫Ωu2dx+g∘∇u−∫Ωutγ+2dx1+γ/2−12g′∘∇u+gt2∫Ω∇u2dx+η∫Ω∇ut2dx=0.Now, it is obvious that the energy of model (1) can be given directly:(16)εt=12∫Ωut2dx+121−∫0tgζdζ∫Ω∇ut2dx+c22∫Ωu2dx+12g∘∇u−∫Ωutγ+2dxγ+2,where(17)g∘f=∫Ω∫0tgt−ζft−fζ2dζdx.Thereby, we can find(18)ε′t=ddtεt=12g′∘∇u−∫Ω∇u2dx−η∫Ω∇ut2dx≤12g′∘∇u−∫Ω∇u2dx≤12−ξgp∘∇u−gt∫Ω∇u2dx≤0,which means εt is decreasing.


## 3. Main Results

In this chapter, the main result of this work is put forward directly and clearly. We introduce the following auxiliary functionals firstly:(19)φt=∫Ωutudx+∫Ω∇u2dx,χt=∫Ωu+αΔu−βut∫0tgt−ζut−uζdζdx,where the coefficients αandβ will be determined later.Lemma 1.
The existence of a positive constantC makes the following conclusion:(20)φt≤Cεt,∀t≥0.Proof.
Recalling the inequalities of Cauchy and Poincaré, the following estimates can be arrived:(21)φt≤∫Ωut2dx+∫Ωu2dx2+∫Ω∇u2dx≤Cp+12∫Ω∇u2dx+12∫Ωut2dx,where the symbol Cp>0 satisfies u2≤Cp∇u2 for all u∈H01Ω.
Here, it is necessary to indicate that the positive constantsCandCp later in this article represent different meanings in different places.
From the representation ofεt, it is naturally for us to get(22)φt≤Cεt,∀t≥0.
Thus, (20) follows.Lemma 2.
For anyt≥0, the following estimate holds:(23)φ′t≤∫Ωut2dx+14ε1+ε11+12ε21−ℓ2+Cp2γ+222γ+4E0γℓγ−η∫Ω∇u2dx+12−c2∫Ωu2dx+ε11+2ε2∫0tg2−pζdζgp∘∇u+2−η2∫Ω∇ut2dx,where ε1andε2 are selected appropriately.Proof.
Deriving functionφt with respect to time, the following calculation will be given clearly:(24)ddtφt=∫Ωut2dx+∫Ωuttudx+2∫Ω∇u∇utdx=∫Ωut2dx−∫Ω∇u2dx−c2∫Ωu2dx+∫Ω∫0tgt−ζ∇ut∇uζdζdx+∫Ωuγu2dx+2−η∫Ω∇u∇utdx.
Considering the fourth item of (23), we adopt the following estimate from [10] without proof:(25)∫Ω∫0tgt−ζ∇ut∇uζdζdx≤14ε1+ε11+12ε21−ℓ2∫Ω∇u2dx+ε11+2ε2∫0tg2−pζdζgp∘∇u.
Taking the sixth item of (23) into account, we adopt the following estimate from [1] as follows:(26)∫Ωuγu2dx=∫Ωuγ+1udx≤12∫Ωu2γ+1dx+12∫Ωu2dx≤Cp2γ+222γ+4E0γℓγ∫Ω∇u2dx+12∫Ωu2dx.
For the last item of (23), we have(27)2−η∫Ω∇u∇utdx≤1−η2∫Ω∇u2dx+∫Ω∇ut2dx.
Thus, (21) follows.Lemma 3.
Fort≥0, the following estimate holds:(28)χ′t≤βc2ε6+ε9+1−ℓε12∫Ωu2dx+ε3+βε11+14ε12−β1−ℓ∫Ωut2dx+Cp4ε3+βη−α4ε4+c2β4ε6+βε7+14ε7+β4ε8∫0tg2−pζdζgp∘∇u+g0Cp4ε9+α4ε10+βCp4ε11−g′∘∇u+βε5+βε71−ℓ2+βε8Cp2γ+22γ+4E0γℓγ+αε10+αε131−ℓ∫Ω∇u2dx+ε4βη−α+α4ε13∫Ω∇ut2dx,where εi>0,5≤i≤13, are selected appropriately.Proof.
Deriving functionχt with respect to time, the expression will be given distinctly:(29)ddtχt=∫Ωut+αΔut−βutt∫0tgt−ζut−uζdζdx+∫Ωu+αΔu−βut∫0tg′t−ζut−uζdζdx+∫Ωu+αΔu−βut∫0tgt−ζutdζdx=∫Ωut+αΔut−β−Δu+c2u+∫0tgt−ζΔuζdζ−uγu−ηΔut∫0tgt−ζut−uζdζdx+∫Ωu+αΔu−βut∫0tg′t−ζut−uζdζdx+∫0tgζdζ∫Ωu+αΔu−βututdx=∫Ωut+α−βηΔut−βΔu+βc2u+β∫0tgt−ζΔuζdζ−βuγu∫0tgt−ζut−uζdζdx+∫Ωu+αΔu−βut∫0tg′t−ζut−uζdζdx+∫0tgζdζ∫Ωu+αΔu−βututdx=∫Ωut∫0tgt−ζut−uζdζdx+∫Ω∇uttβη−α∫0tgt−ζ∇ut−∇uζdζdx+β∫Ω∇u∫0tgt−ζ∇ut−∇uζdζdx+βc2∫Ωu∫0tgt−ζut−uζdζdx+β∫0tgt−ζΔuζdζ∫0tgt−ζut−uζdζdx−β∫Ωu1+γ∫0tgt−ζut−uζdζdx+∫Ωu+αΔu−βut∫0tg′t−ζut−uζdζdx+∫0tgζdζ∫Ωuut+αutΔu−βut2dx.
For anyα,β≥0, we will estimate every item of the right hand of (29):(30)∫Ωut∫0tgt−ζut−uζdζdx≤ε3∫Ωut2dx+14ε3∫Ω∫0tgt−ζut−uζdζ2dx=ε3∫Ωut2dx+14ε3∫Ω∫0tg1−p/2t−ζgp/2t−ζut−uζdζ2dx≤ε3∫Ωut2dx+Cp4ε3∫0tg2−pζdζgp∘∇u,(31)βη−α∫Ω∇utt∫0tgt−ζ∇ut−∇uζdζdx≤βη−αε4∫Ω∇ut2dx+14ε4∫Ω∫0tgt−ζ∇ut−∇uζdζ2dx≤βη−αε4∫Ω∇ut2dx+βη−α4ε4∫0tg2−pζdζgp∘∇u,(32)β∫Ω∇u∫0tgt−ζ∇ut−∇uζdζdx≤βε5∫Ω∇u2dx+β4ε5∫0tg2−pζdζgp∘∇u.
Similar to the first term, considering the third item of (29), similar estimate will be given:(33)βc2∫Ωu∫0tgt−ζ∇ut−∇uζdζdx≤βc2ε6∫Ωu2dx+βc24ε6∫0tg2−pζdζgp∘∇u.
Applying Young’s inequality properly, the following estimates can be obtained from [10]:(34)β∫0tgt−ζΔuζdζ∫0tgt−ζut−uζdζdx=β∫0tgt−ζ∇uζdζ∫0tgt−ζ∇uζ−∇utdζdx≤β1−ℓ2ε7∫Ω∇u2dx+βε7+14ε7∫0tg2−pζdζgp∘∇u,(35)−β∫Ωuγu∫0tgt−ζut−uζdζdx≤βε8∫Ωu2γ+2dx+β4ε8∫Ω∫0tgt−ζut−uζdζ2dx≤βε8Cp2γ+2∫Ω∇u2γ+2dx+β4ε8∫0tg2−pζdζgp∘∇u≤βε8Cp2γ+22γ+4E0γℓγ∫Ω∇u2dx+β4ε8∫0tg2−pζdζgp∘∇u,(36)∫Ωu∫0tg′t−ζut−uζdζdx≤ε9∫Ωu2dx+g0Cp4ε9∫Ω∫0t−g′t−ζ∇ut−∇uζ2dζdx=ε9∫Ωu2dx+g0Cp4ε9−g′∘∇u,(37)α∫ΩΔu∫0tg′t−ζut−uζdζdx≤αε10∫Ω∇u2dx+αg04ε10∫Ω∫0t−g′t−ζ∇ut−∇uζ2dζdx=αε10∫Ω∇u2dx+αg04ε10−g′∘∇u.
Similarly, we have(38)−β∫Ωut∫0tg′t−ζut−uζdζdx≤βε11∫Ωut2dx+βg0Cp4ε11−g′∘∇u.
Furthermore,(39)∫0tgζdζ∫Ωuut+αutΔu−βut2dx≤∫0tgζdζε12∫Ωu2dx+14ε12−β∫Ωut2dx+αε13∫Ω∇u2dx+α4ε13∫Ω∇ut2dx=1−ℓε12∫Ωu2dx+14ε12−β∫Ωut2dx+αε13∫Ω∇u2dx+α4ε13∫Ω∇ut2dx.
Taking (29)–(39) and H2 into account, for all t≥0, we get (28).Theorem 2.
Let every pairu0,u1∈H01Ω×H01Ω, H1 and H2 hold, and t0∈0,∞ For every t∈t0,∞, there must be some positive constants K1, K2, and k, which would enable the solution of the model (1) to satisfy the following:(40)εt≤K1ε0ekt,p=1,εt≤K2t+11/2p−1,p>1.Proof.
We start the proof by selecting an appropriate auxiliary functional(41)Lt=Mεt+aφt+bχt,where the positive constants M, a, and b will be chosen in the sequel.
Applying Lemma1 and the hypothesis g′t≤−ξ⋅gpt, the equivalence between εt and Lt is achieved. To this point, firstly, a simple calculation shows that(42)Lt−Mεt=aφt+bχt=a∫Ωutudx+a∫Ω∇u2dx+b∫Ωu+αΔu−βut∫0tgt−ζut−uζdζdx=a∫Ωutudx+a∫Ω∇u2dx+b∫Ωu−βut∫0tgt−ζut−uζdζdx−bα∫Ω∇u∫0tgt−ζut−uζdζdx≤a2∫Ωut2dx+aCp+2a2∫Ω∇u2dx+b2∫Ωu−βut2dx+bα2∫Ω∇u2dx+b2∫Ω∫0tgt−ζut−uζdζ2dx+bα2∫Ω∫0tgt−ζ∇ut−∇uζdζ2dx≤a2∫Ωut2dx+aCp+2a+bα2∫Ω∇u2dx+b2∫Ωu2dx+β2b2∫Ωut2dx−βb∫Ωutudx+b212∫Ω∫0tgt−ζdζdx+12∫Ω∫0tgt−ζut−uζ2dζdx+bα212∫Ω∫0tgt−ζdζdx+12∫Ω∫0tgt−ζ∇ut−∇uζ2dζdx≤a+bβ22∫Ωut2dx+aCp+2a+bα2∫Ω∇u2dx+b2∫Ωu2dx+bα+bCp4g∘∇u+b1+α1−ℓ4∫Ωdx≤C1εt.
By (25), we get(43)Lt−Mεt+1γ+2∫Ωuγ+2dx≤a+bβ22∫Ωut2dx+12aCp+2a+bα+Cp2γ+22γ+4E0γℓγ∫Ω∇u2dx+b+12∫Ωu2dx+bα+bCp4g∘∇u+b1+α1−ℓ4∫Ωdx≤C1εt+1γ+2∫Ωuγ+2dx.
Thus, the equivalence betweenLt and εt is complete.
Secondly, differentiating (41) with respect to t, considering the assumption H2 and Lemmas 1 and 2, the estimate of L′t is given naturally:(44)L′t=Mε′t+aφ′t+bχ′t≤M−ξ2gp∘∇u−gt2∫Ω∇u2dx+aφ′t+bχ′t≤c1∫Ωut2dx+c2∫Ω∇u2dx+c3∫Ωu2dx+c4gp∘∇u+c5∫Ω∇ut2dx,where the coefficients c1,c2,c3,andc4 are given as follows:(45)c1=a+bε3+bβε11+b1−ℓ14ε12−β,c2=a14ε1+ε11+12ε21−ℓ2+Cp2γ+222γ+4E0γℓγ−η2+bβε5+βε71−ℓ2+βε8Cp2γ+22γ+4E0γℓγ+αε10+αε131−ℓ−Mgt2,c3=a12−c2+bβc2ε6+ε9+ε121−ℓ,c4=aε11+2ε2+bCp4ε3+βη−α4ε4+c2β4ε6+βε7+14ε7+β4ε8∫0tg2−pζdζ+bξg0Cp4ε9+g0α4ε10+βg0Cp4ε11−Mξ2,c5=a1−η2+bε4βη−α+αε13.
Next, it is time to carry out a discussion on the different values ofp.Case 1.
p=1
In such a case, choosingc1˜=c1,c2˜=c2+h/γ+2Cp2γ+2/22γ+4E0/γℓγ,c3˜=c3+h/2γ+2,c4˜=c4,c5˜=c5=0 and taking the positive constant h as(46)h≤min−2c1˜,−2c2˜1−∫0tgζdζ,−2c3˜c2,−2c4˜,we find(47)L′t−hγ+2∫Ωuγ+2dx≤L′t+hγ+2∫Ωuγ+2dx≤L′t+hγ+2Cp2γ+222γ+4E0γℓγ∫Ω∇u2dx+h2γ+2∫Ωu2dx≤c1˜∫Ωut2dx+c2˜∫Ω∇u2dx+c3˜∫Ωu2dx+c4˜gp∘∇u+c5˜∫Ω∇ut2dx≤−h12∫Ωut2dx+121−∫0tgζdζ∫Ω∇u2dx+c22∫Ωu2dx+12g∘∇u.
Therefore, (45) yields(48)L′t≤−hεt.Remark 2.
Using the computational technique proposed in [10] and selecting some appropriate numerical values by means of the calculation software Mathematica, it is obvious that ci˜<01≤i≤4.
Considering the interval of the fuzzy numberη, taking(49)M=10,η=100,εi=0.01,1≤i≤13,α=1,β=2,a=1,b=−1,γ=2,
for example, with the aid of Mathematica, we easily verify that(50)c1˜=−0.580816−231−ℓ,c2˜=13.4295+5gt+0.611833ℓ−0.300916ℓ2−4.72653−2hE02Cp6ℓ2,c3˜=−0.295408+0.530816c2+h8+0.01ℓ,c4˜=−5100.03+50c21−ℓ−5ξ−100ξg0.
Taking another example into consideration,(51)M=10,η=10,εi=i1000,1≤i≤13,α=1,β=2,a=1,b=−1,γ=2,we find that(52)c1˜=−4.85169−18.83331−ℓ,c2˜=−1183.8+5g0+2.464ℓ−1.2255ℓ2−38.8695−2hE02Cp6ℓ2,c3˜=−2.43435+4.81469c2+h8+0.012ℓ,c4˜=−1321.45+83.3333c2+Cp1−ℓ−5ξ−25ξg0−73.2323Cpg0ξ.
Additionally, we may retrieve more effective data ofci˜, and the data provide an intuitive understanding in the discussion of the auxiliary functionals.
Considering the equivalence betweenLt and εt, it is easy for us to see that(53)M1εt≤Lt≤M2εt,where the coefficients M1 and M2 in (53) have many possibilities to be chosen.
The following expression can be presented through the combination of (48) and (53):(54)L′t≤−hM1εt.
After a simple integral on the intervalt0,t, (54) leads to(55)Lt≤e−ht−t0/M1Lt0.
As a consequence, takingk=−h/M1, (55) leads to(56)εt≤Lt0M1eh/M1t−t0≤K1ε0ekt,where K1 is some appropriate positive constant.Case 2.
p>1.
Using H1, H2, and (9), we easily verify that(57)∫0∞g1−θζdζ≤∫0+∞1d1t+d21−θ/p−1dζ<∞,0≤θ<2−p.
Moreover, considering the inequality in [6], for some constant C,(58)g∘∇u≤CE0∫0∞g1−θζdζp−1/p−1+θgp∘∇uθ/p−1+θ.
Takingθ=1/2,γ=2p−1, we find γθ/p−1+θ=1. Therefore, for some m>1, we get(59)εmt≤C∫Ωu2dx+∫Ω∇u2dx+∫Ωut2dx+gp∘∇u,that is,(60)−∫Ωu2dx+∫Ω∇u2dx+∫Ωut2dx+gp∘∇u≤−1Cεmt.
Taking (48) into account, we have(61)−εmt≤−1M2mLmt.
For eacht≥t0, it is straight for us to get(62)L′t≤hγ+2∫Ωuγ+2dx+L′t≤−h2∫Ωut2dx+ℓ∫Ω∇u2dx+c2∫Ωu2dx+g∘∇u≤−hCεmt≤−hCM2mLmt≤−CLmt.
As we mentioned before, the symbolC denotes different constants in different places. Executing a simple integral on the interval t0,t, (62) yields(63)Lt≤Cat+Cb−1/m−1=Cat+Cb−1/2p−1.
Considering the equivalence betweenLt and εt, it is apparently for us to get(64)εt≤K2t+11/2p−1.
This completes the proof.


## 4. Conclusion

Under the assumptions on the relaxation function and the interval of the fuzzy numberη, applying the computational technique, a lot of auxiliary functionals can be constructed numerically. Two decay results, the exponential one and the polynomial one, are derived for the model (1) eventually. The result shows a new way for the decay rates, which is quite different from other literatures.



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