# Decay Estimates for a Type of Fuzzy Viscoelastic Integro-Differential Model

Authors: Fengyun Zhang; Funing Lin; Guangwang Su; Guangming Xue
Journal: Complexity
 (2021)

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## Abstract

We consider a type of fuzzy viscoelastic integro-differential model in this paper. With the aid of some appropriate hypotheses, a unified method and the multiplier technique are implemented to get priori estimates precisely without constructing any auxiliary function. By establishing the estimation of energy function, we derive the stability result of the global solution, and we calculate the estimations of energy attenuation in exponential and polynomial forms, respectively.


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## Body

## 1. Introduction

In this work, the following fuzzy viscoelastic integro-differential model is considered in a real Hilbert spaceX=L2Ω:(1)utt−Δu+∫0tgt−ζΔuζdζ−uγu+η2u=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 Ω is an open bounded neighbourhood in RN with N≥3, 0<γ≤2/N−2. Meanwhile, Γ:=∂Ω is smooth enough. The memory kernel gt and the fuzzy number η are both positive, and gt is locally and absolutely continuous.As far as the viscoelastic equation is concerned, profound research works have been made in many literature studies [1–7]. For example, the authors in [3] proved a local existence theorem for the next equation:(2)utt−Δu+∫0tgt−ζΔuζdζ=uγu,which is subject to some proper initial data and conditions. In [4], an appropriate Lyapunov-type function was introduced by Nasser-eddine Tatar to prove the decay of solutions for the wave equation:(3)utt−Δutt−Δu+∫0tht−ζΔuζdζ=0.The key contribution of ref. [6] is that the authors demonstrated the decay of the energy function for the next wave equation:(4)utt−k0Δu+∫0tgt−ζdivax∇uζdζ+bxhut=0,and some Lyapunov functions were exploited felicitously to deduct more general energy decay results. In [8], a nonlinear hereditary memory evolution equation was considered, and several stability results were given just by means of a simple auxiliary function. The authors in [9] attained analytical and approximate solutions for the cubic Boussinesq equations and modified ones with the aid of the He–Laplace method. Besides, fuzzy synchronization problems have captured the intensive interests of scholars (see, e.g., [10, 11]), where an adaptive fuzzy backstepping control method was developed in ref. [11] for a sort of uncertain fractional-order nonlinear system.Generally speaking, in most of the existing works, the presence of auxiliary functions is inevitable, which is exploited to seek the attenuation result of the solution. Accordingly, in the discussion of energy attenuation of solutions for the fuzzy viscoelastic integro-differential model, how to reduce the construction of auxiliary functions has become a problem worth discussing. Taking the integro-differential abstract equation into account led to fruitful excellent results (see, e.g., [12–14]). In [12], Boussouira et al. proposed a unified method creatively. They derived the decay results for second-order integro-differential equations in the following abstract form:(5)u″+Au−∫0tβt−ζAuζdζ=∇Fu.Also, they put forward an exquisite unified method. With the help of the multiplier method, they accurately described the energy attenuation of the solution of the abstract equation mentioned above.Inspired by these works, system (1) involved in this paper is an extension of the equation appeared in [12], in which a term with fuzzy coefficient is creatively added. The decay rates in exponential and polynomial forms, respectively, are straightly derived through the unified method. The specific arrangement is made as follows: firstly, in Section 2, several preliminary materials and essential assumptions are listed, and secondly, Section 3 mainly concentrates on the global solution and the estimation of energy attenuation, which are derived by letting t⟶∞, and the priori estimates are deduced without constructing any auxiliary function. Such outcomes reflect the reliability and effectiveness of the unified method in practice.


## 2. Preliminaries

Throughout this work, the inner product⋅,⋅ of X will be utilized in its usual sense, and the norm is defined as follows:(6)u=∫Ωux2dx,∀u∈X.Note that(7)ux,t=0,x,t∈Γ×0,∞,Taking the operator(8)−Δ:D−Δ⟶X,into consideration, we can verify that(9)D−Δ=H2Ω∩H01Ω⊂X,D∇=H01Ω,where D−Δ is a dense domain. It is evident that, for some positive constant M, the linear operator −Δ is self-adjoint on the real Hilbert space L2Ω and satisfies an inequality similar to the Poincaré inequality [15]:(10)−Δu,u=∇u,∇u=∇u2≥Mu2.What is more, we find that−Δ is accretive due to −Δu,u≥0.Now, we give the following assumptions and preliminary materials about the memory kernelgt.H1: as far as g:0,+∞⟶0,+∞ is concerned, for some 2<p≤∞ and k>0, the function g fulfills the following conditions:(11)g0>0,∫0∞gζdζ<1,g′≤−kgp+1/p.Remark 1.
Ifp=∞, then g′≤−kg yields gt≤Ce−kt with t≥0, which indicates that gt will decay exponentially.
If2<p<∞, then g′≤−kgp+1/p yields(12)g′g−1−1/p≤−k,i.e.,(13)−pg−1−1/p+1′=−pg−1/p′≤−k.
Next, based on the aforementioned results, the following expression can be obtained by integrating from 0 tot:(14)−pg−1/pt+pg−1/p0≤−kt,namely,(15)gt≤1kt/p+g−1/p0p.
This means thatgt will decay polynomially. Simultaneously, by generalized integral property, if pϑ>1, it may imply that gϑ∈L10,∞.Lemma 1.
Suppose that(16)Fu=1γ+2∫Ωuγ+2dx−η22∫Ωu2dx.The function is Gateaux differentiable for everyu∈D∇,∇Fu=uγu−η2u, and(17)Fu≤C∇uγ/2+2.Indeed, it is straightforward to see thatF0=0,∇F0=0. For any pair u,v∈D∇, there exists cu>0 such that(18)DFuv=∇Fu,v=∫Ω∇Fuvdξ=∫Ωuγ−η2uvdξ≤C∇uγ+1v+η2∫Ωuvdξ≤C∇uγ+1v+η2uv≤C∇uγ+1v+η2M∇uv≤cuv,in whichcu=C∇uγ+1+η2/M∇u.For allu,v∈D∇ with ∇u,∇v≤R, where R>0, it is easily seen that u−v∈D∇. Combining the mean value formula applied in [12] and (10), one can verify that(19)∇Fu−∇Fv2=∫Ωuγu−η2u−vγv+η2v2dξ=∫Ωuγu−vγv−η2u−v2dξ=∫Ωuγu−vγv2dξ+η2∫Ωu−v2dξ≤C∫Ω∇u2+∇v2dξγ∫Ω∇u−∇v2dξ+η2M∫Ω∇u−∇v2dξ≤CR2∇u−∇v2.That is, some positive constantCR=C∫Ω∇u2+∇v2dξγ+η2/M1/2 can be found to satisfy(20)∇Fu−∇Fv≤CR∇u−∇v.By lettingψu=uγ/2, it is easy to see that ψ:0,∞⟶0,∞ is continuous and strictly increasing. Now, we suppose that ∇Fu,u≤C∇uγ/2+2, that is,(21)∇Fu,u=uγ−η2u,u=∫Ωuξγ−η2uξ2dξ≤C∇uγ/2∇u2=C∇uγ/2+2,∀u∈D∇.For everyu∈D∇,(22)Fu≤∫01∇Ftu,udt≤∇ut2∫01t∇utγ/2tdt≤C∇ut2+γ/2,which yields(23)1γ+2∫Ωuγ+2dx−η22∫Ωu2dx≤∇ut2+γ/2.Remark 2.
For any0<T≤∞, by taking the measurable function u:0,T⟶X into consideration, it is known that both u1=∫0Tutdt and u∞=esssupt∈0,Tut are finite.
For anyf∈L10,T and u∈L10,T;X, we denote the convolution as follows:(24)f∗ut=∫0tft−susds,0≤t≤T.
Aiming to facilitate the subsequent narration, we proceed to present the next useful lemmas.Lemma 2.
Consider a nonnegative nonincreasing functionEt with 0≤t<∞. If there exists a negative constant T such that(25)∫t∞Esds≤TEt,then(26)Et≤E0e1−t/T,∀t≥T.Proof.
Let(27)fx=ex/T∫x∞Esds,∀x≥0.
Then its derivative is calculated as(28)f′x=ex/TT∫x∞Esds−TEx.
Considering that∫t∞Esds≤TEt, we have(29)fx≤f0=∫0∞Esds≤TE0.
This implies that(30)∫x∞Esds≤TE0e−x/T.
On the other hand, sinceE is nonnegative and nonincreasing,(31)∫x∞Esds≥∫xx+T1⋅Esds≥TEx+T.
Combining (30) with (31), we get(32)Ex+T≤E0e−x/T.
Takingt=x+T, by x≥0, formula (26) is obtained naturally.Lemma 3.
LetEt be a nonnegative and nonincreasing function on 0,∞. If(33)∫T∞E1+mtdt≤CEm0ET,∀T≥T0,where m,C, and T0 are all positive constants. Then, for arbitrary t∈0,∞, it holds that(34)Et≤E0C+T01+mmt+C+T01/m.The proof of Lemma3 is analogous to that of Lemma 2, and hence, it is omitted here.Letui∈Xi=0,1. Now, let us discuss the problem as follows:(35)utt−Δu+∫0tgt−ζΔuζdζ−uγu+η2u=0,0<t<∞,u0=u0,u|t0=u1.For any0≤t≤T,T>0, with the aid of the description in [12], a mild solution of (35) can be described as follows:(36)ut=u0St+∫0tSζu1dζ+∫0t1∗St−ζuζγ−η2uζdζ,where(37)1∗St−ζ=∫0t−ζSτdτ,ζ≤t≤T+ζ,and St is the resolvent for the corresponding linear problem of (35).As far as the weak solution is concerned,u is a function in C10,T;X∩C0,T;D∇ and satisfies(38)ddtutt,v+∇u,∇v−∫0tgt−ζ∇uζdζ,∇v=uγ−η2u,v,∀v∈D∇,utt,v∈C10,T and 0≤t≤T.Local existence, uniqueness, and regularity for (1) are naturally guaranteed by the result in [12].Considering a mild solutionu of (1) t∈0,T, and using ut as a multiplier, the multiplier method can be used to get the energy of u as follows:(39)Eut=12ut2+1−∫0tgζdζ∇u2+∫0tgt−ζ∇uζ−∇ut2dζ−1γ+2∫Ωuγ+2dx+η22∫Ωu2dx.Next, it is necessary to discuss the decay ofEut.Consider thatu is a strong solution of problem (1) on an interval 0,T. By taking derivative of (39), we obtain(40)ddtEut=utt,ut+∇u,∇ut−gt2∇u2−12∫0tgζdζddt∇u2+12∫0tg′t−ζ∇ut−∇uζ2dζ−∫0tgt−ζ∇uζ−∇ut,∇uttdζ−uγ−η2u,ut=utt,ut+∇u,∇ut−gt2∇u2−12∫0tgζdζddt∇u2+12∫0tg′t−ζ∇ut−∇uζ2dζ−∫0tgt−ζ∇uζdζ,∇ut+12∫0tgζdζddt∇u2−uγ−η2u,ut=utt,ut+−Δu,ut+∫0tgt−ζΔuζdζ,ut−uγ−η2u,ut−gt2∇u2+12∫0tg′t−ζ∇ut−∇uζ2dζ=−gt2∇u2+12∫0tg′t−ζ∇ut−∇uζ2dζ.In view of the facts thatg≤0 and g′≥0, it follows from these assumptions that(41)ddtEut=−gt2∇u2+12∫0tg′t−ζ∇ut−∇uζ2dζ≤0,that is, Eut is decreasing. One can draw a similar conclusion for mild solutions. In a word, if the initial conditions are small sufficiently, the solution of model (1) exists globally.Theorem 1.
Assume thatH1 holds. For any u0∈D∇ and u1∈X, if there is a positive scalar ρ0 such that(42)∇u0+u1<ρ0,then there is a unique mild solution u for problem (1). Besides, for arbitrary t∈0,∞,(43)Eut>0,(44)Eut≤Eu0≤ρ02,(45)Eut≥12ut2+1−∫0∞gζdζ4∇ut2,(46)∇ut2+γ/2≤1−∫0∞gζdζ4.Furthermore,u is a strong solution of (1), provided that u0∈D−Δ and u1∈D∇.Proof.
Assume that a maximal definition interval for the mild solution of problem (1) is 0,T, and ℓ=1−∫0∞gζdζ/2. According to Lemma 1, one gets 1/γ+2∫Ωu0γ+2dx≤∇u02+γ/2. Besides, equation (39) implies(47)Eu0=12u12+∇u02−1γ+2∫Ωu0γ+2dx+η22∫Ωu02dx.
If∇u0γ/2<ℓ/2, we get(48)1γ+2∫Ωu0γ+2dx−η22∫Ωu02dx≤1γ+2∫Ωu0γ+2dx+η22∫Ωu02dx≤∇u02+γ/2+η22M∇u02<ℓ2∇u02+η22M∇u02≤ℓ˜2∇u02,where ℓ˜≥ℓ+η2/M.
Thus, it is naturally acquired that(49)−1γ+2∫Ωu0γ+2dx+η22∫Ωu02dx≥−ℓ˜2∇u02.
Consequently,Eu0≥1/2u12+1−ℓ˜∇u02≥0.
Putρ0=ℓ˜/21/2+2/γ. Suppose that u0∈D∇ and u1∈X satisfy(50)∇u0+u1<ρ0.
Utilizing the amplification method, i.e.,(51)Eu0≤12u12+∇u02≤u12+∇u02≤u1+∇u02≤ρ02=ℓ˜21+4/γ,we derive that 2Eu0/ℓ˜γ/4<ℓ˜/2 and(52)Eut≥12ut2+ℓ˜2∇ut2≥12ut2+1−∫0∞gζdζ4+η22M∇ut2≥12ut2+1−∫0∞gζdζ4∇ut2,∀t∈0,T.
So, the energyEu is well bounded and the solution u exists globally. The proof of the aforementioned formula is based on the idea of reductio ad absurdum, and the detailed process is omitted here.


## 3. Main Results

In this sequel, without invoking any auxiliary function, we put forward the main result as follows.Theorem 2.
Assume thatH1 holds. Given S≥S0>0. For each pair u0,u1∈D∇×X, if u1+∇u0<ρ0 with ρ0 being a positive constant, then there is some positive constant C ensuring that the mild solution of model (1) satisfies the next property:(53)∫S∞Eu1+1/ptdt≤CEu1/p0EuS.Specifically,(54)Eut≤Eu0e1−Ct,p=∞,Eut≤Eu0p+1p+Ctp,2<p<∞.Proof.
By Theorem1, we know that the solution of (1) is global. Moreover, it is easy to check that the solution is strong if u0∈D−Δ and u1∈D∇. Aiming to show (53), we begin to focus on the formula as follows:(55)∫STEu1+1/ptdt=12∫STEu1/ptut2dt+12∫STEu1/pt1−∫0tgζdζ∇u2dt+12∫STEu1/pt∫0tgt−ζ∇uζ−∇ut2dζdt−1γ+2∫STEu1/pt∫Ωuγ+2dxdt+η22∫STEu1/pt∫Ωu2dxdt.
Next, our task is introducing an approach for controlling every term of the right hand of equation (55) via multiplier methods. At the beginning, we propose the following lemma.Lemma 4.
Suppose thatφt:R+⟶R+ is a multiplier, fulfilling that φ′t<0. Then for any positive constant T with T≥S≥S0, there exists C>0 such that(56)12∫STφtut2dt+12∫STφt1−∫0tgζdζ∇ut2dt−1γ+2∫STφt∫Ωuγ+2dxdt+η22∫STφt∫Ωu2dxdt≤Cφ0EuS.Proof.
Firstly, an inner product of model (1) with the multiplication of u and φt should be taken. Next, integrating it on the closed interval S,T, the following description is now obtained:(57)∫STφtutt−Δu+∫0tgt−ζΔuζdζ,utdt=∫STφtuγ−η2ut,utdt.
Integrating by parts, we get(58)∫STφtutt,utdt=∫STφtddtut,utdt−∫STφtddtut,utdt=φtut,utST−∫STφ′tddtut,utdt−∫STφtddtut2dt,(59)∫STφt−Δu,utdt=∫STφt∇u,∇udt=∫STφt∇u2dt,(60)∫STφt∫0tgt−ζΔuζdζ,utdt=−∫STφt∫0tgt−ζ∇uζdζ,∇utdt=−∫STφt∫0tgt−ζ∇uζ−∇utdζ,∇utdt,(61)−∫STφt∫0tgt−ζ∇ut,∇utdζdt=−∫STφt∫0tgt−ζ∇uζ−∇utdζ,∇utdt−∫STφt∫0tgζdζ∇u2dt,(62)∫STφt1−∫0tgζdζ∇ut2dt=∫STφtut2dt+∫STφt∫0tgt−ζ∇uζ−∇utdζ,∇utdt+∫STφ′tut,utdt+∫STφtuγ−η2ut,utdt−φtutt,utST.
Applying Schwartz inequality∀ε1>0, we have(63)∫STφt∇ut∫0tgt−ζ∇uζ−∇utdζdt=∫STφtφt∇ut∫0tgt−ζ∇uζ−∇utdζdt≤ε12∫STφt∇ut2dt+12ε1∫STφt∫0tgt−ζ∇uζ−∇utdζ2dt.
Taking the integrability ofgϑ and the assumption that g′≤−kgp+1/p into consideration, with the help of Hölder inequality and the description of Eu′t, we have(64)∫STφt∫0tgt−ζ∇uζ−∇utdζ2dt=∫STφt∫0tgp−1/2pt−ζgp+1/2pt−ζ∇uζ−∇utdζ2dt≤∫STφt∫0tgp−1/pζdζ∫0tgp+1/pt−ζ∇uζ−∇ut2dζdt≤∫0∞gp−1/pζdζ∫STφt∫0tgp+1/pt−ζ∇uζ−∇ut2dζdt≤−1k∫0∞gp−1/pζdζ∫STφt∫0tg′t−ζ∇uζ−∇ut2dζdt≤−2k∫0∞gp−1/pζdζ∫STφ0Eu′tdt=−2k∫0∞gp−1/pζdζφ0EuT−EuS≤2kφ0EuS∫0∞gp−1/pζdζ.
Combining (62) and (63), we get(65)∫STφt∫0tgt−ζ∇uζ−∇utdζ,∇utdt=∫STφt∇ut∫0tgt−ζ∇uζ−∇utdζdt≤ε12∫STφt∇ut2dt+1kε1φ0EuS∫0∞gp−1/pζdζ.
In view of (45), we have(66)12ut2≤Eut,12∇ut2≤2Eut1−∫0∞gζdζ.
Thus, we obtain(67)12ut2≤1M12∇ut2≤2/MEut1−∫0∞gζdζ.
Therefore, the following result is arrived:(68)utt,ut=∫0Tuttutdt≤12∫0Tutt2dt+12∫0Tut2dt≤12ut2+2/MEut1−∫0∞gζdζ≤1+2/M1−∫0∞gζdζEut.
Byφtt<0 and Eut≤EuS, we get(69)−∫STφttdt=φS−φT≤φS≤φ0.
Therefore,(70)∫STφttutt,utdt≤−∫STφttutt,utdt≤−1+2/M1−∫0∞gζdζ∫STφttEutdt≤EuS1+2/M1−∫0∞gζdζ−∫STφttdt≤EuSφ01+2/M1−∫0∞gζdζ.
By the condition imposed on the proof of Theorem1,(71)∇utγ/2≤1−∫0∞gζdζ4,one has(72)∫STφtuγ−η2ut,utdt≤∫STφt∇ut2+γ/2dt≤14∫STφt1−∫0tgζdζ∇ut2dt.
Considering that bothφt and Et are decreasing, from (67), we deduce(73)−φtutt,utST≤2φ01+2/M1−∫0∞gζdζEuS.
Based on equation (61), it is trivially shown that(74)∫STφt1−∫0tgζdζ∇ut2dt=∫STφtut2dt+∫STφt∫0tgt−ζ∇uζ−∇utdζ,∇utdt+∫STφ′tut,utdt+∫STφtuγ−η2ut,utdt−φtutt,utST≤∫STφtut2dt+ε12∫STφt∇ut2dt+1kε1φ0EuS∫0∞gp−1/pζdζ+EuSφ01+2/M1−∫0∞gζdζ+14∫STφt1−∫0tgζdζ∇ut2dt+2φ01+2/M1−∫0∞gζdζEuS,which means that(75)34∫STφt1−∫0tgζdζ∇ut2dt≤∫STφtut2dt+ε12∫STφt∇ut2dt+1kε1φ0EuS∫0∞gp−1/pζdζ+3EuSφ01+2/M1−∫0∞gζdζ.
For simplicity, selectingε1=1−∫0∞gζdζ, we get(76)∫STφt1−∫0tgζdζ∇ut2dt≤4∫STφtut2dt+4∫0∞gp−1/pζdζk1−∫0tgζdζ+31+2/M1−∫0∞gζdζEuSφ0≤4∫STφtut2dt+4M∫0∞gp−1/pζdζ+12k2+M1−∫0∞gζdζMk1−∫0∞gζdζEuSφ0=ΔC1∫STφtut2dt+C2EuSφ0,where C1=4 and(77)C2=4M∫0∞gp−1/pζdζ+12k2+M1−∫0∞gζdζMk1−∫0∞gζdζare both positive.
Next, multiplying both sides of the original equation (1) by φt at the same time, taking ∫0tgt−ζuζ−utdζ as a multiplier, and integrating on the closed interval S,T, the following equation can be obtained:(78)∫STφtutt−Δu+∫0tgt−ζΔuζdζ−uγ−η2ut,∫0tgt−ζuζ−utdζdt=0.
Taking integration by parts, one obtains(79)∫STφtutt,∫0tgt−ζuζ−utdζdt=∫STuttφt∫0tgt−ζuζ−utdζdt=utφt∫0tgt−ζuζ−utdζ|ST−∫STutφt∫0tgt−ζuζ−utdζ′dt=φtut,∫0tgt−ζuζ−utdζ|ST−∫STφ′tutt,∫0tgt−ζuζ−utdζdt−∫STφtutt,∫0tg′t−ζuζ−utdζdt+∫STφtut2∫0tgζdζdt,(80)∫STφt−Δu+∫0tgt−ζΔuζζ,∫0tgt−ζuζ−utdζdt=∫STφt∇u,∫0tgt−ζuζ−utdζdt−∫STφt∫0tgt−ζ∇uζdζ,∫0tgt−ζuζ−utdζdt=∫STφt∇ut∫0tgt−ζuζ−utdζdt−∫STφt∫0tgt−ζ∇uζ−∇ut+∇utdζ∫0tgt−ζ∇uζ−∇utdζdt=∫STφt∇ut∫0tgt−ζuζ−utdζdt−∫STφt∫0tgt−ζ∇uζ−∇utdζ2dt−∫STφt∫0tgt−ζ∇utdζ∫0tgt−ζ∇uζ−∇utdζdt=∫STφt1−∫0tgζdζ∇ut,∫0tgt−ζ∇uζ−∇utdζdt−∫STφt∫0tgt−ζ∇uζ−∇utdζ2dt.
Substituting (79) and (80) into (78) leads to the following:(81)∫STφt∫0tgζdζut2dt=−φtut,∫0tgt−ζuζ−utdζST+∫STφ′tutt,∫0tgt−ζuζ−utdζdt+∫STφtutt,∫0tg′t−ζuζ−utdζdt+∫STφt∫0tgt−ζ∇uζ−∇utdζ2dt+∫STφt−1+∫0tgζdζ∇ut,∫0tgt−ζ∇uζ−∇utdζdt+∫STφtuγ−η2ut,∫0tgt−ζuζ−utdζdt.
Now, let us consider the term∫STφtut2dt and evaluate it. First of all, according to (10), we have(82)ut,∫0tgt−ζuζ−utdζ≤12ut2+∫0tgt−ζuζ−utdζ2≤Eut+12M∫0tgt−ζ∇uζ−∇utdζ2≤Eut+12M∫0tgt−ζ∇uζ−∇utdζ∫0tgζdζ≤1+MMEut.
Consequently,(83)−φtut,∫0tgt−ζuζ−utdζST≤2φ01+MMEuS.
As a result,(84)∫STφ′tutt,∫0tgt−ζuζ−utdζdt≤1+MM∫ST−φ′tEutdt≤1+MMEuS∫ST−φ′tdt≤1+MMEuSφ0.
Using the Cauchy inequality, we have(85)∫STφtutt,∫0tg′t−ζuζ−utdζdt≤δ12∫STφtut2dt+12δ1∫STφt∫0tg′t−ζ⋅uζ−utdζ2dt.
Recall thatg′ζ=−g′ζ, which is deduced from g′t≤0. Hence,(86)∫0tg′t−ζ⋅uζ−utdζ2=∫0tg′t−ζ⋅g′t−ζuζ−utdζ2≤−∫0tg′ζdζ∫0tg′t−ζ⋅uζ−ut2dζ=g0−gt∫0t−g′t−ζ⋅uζ−ut2dζ≤g0∫0t−g′t−ζ⋅uζ−ut2dζ≤g0M∫0t−g′t−ζ⋅uζ−ut2dζ≤−2Eu′tg0M.
Then, (85) is transformed into the following form:(87)∫STφtutt,∫0tg′t−ζuζ−utdζdt≤δ12∫STφtut2dt+12δ1∫STφt∫0tg′t−ζ⋅uζ−utdζ2dt≤δ12∫STφtut2dt+12δ1∫ST−2g0φtEu′tMdt≤δ12∫STφtut2dt+12δ12g0φ0M−Eut|ST≤δ12∫STφtut2dt+g0φ0Mδ1EuS.
In view of the assumption ofgt and ∫0∞gζdζ<1, the estimates of (64) can be arrived as follows:(88)∫STφt−1+∫0tgζdζ∫0tgt−ζ∇uζ−∇utdζ,∇utdt≤∫STφt∫0tgt−ζ∇uζ−∇utdζ,∇utdt≤ε12∫STφt∇ut2dt+1kε1φ0EuS∫0∞gp−1/pζdζ.
Observing thatFu≤C∇ut2+γ/2, by combining (44) and (45), we can attain that ∀t≥0,(89)∇ut≤2Eut1−∫0∞gζdζ≤2Eu01−∫0∞gζdζ≤2ρ01−∫0∞gζdζ=2ℓρ0.
Therefore,(90)utγ−η2ut=utγ−η2ut−u0γ−η2u0≤C∇ut−∇0=C∇ut.
So, combining (62) with (63) and taking a part of (78) into consideration, we obtain(91)∫STφtuγ−η2ut,∫0tgt−ζuζ−utdζdt≤C∫STφt∇ut,∫0tgt−ζuζ−utdζdt=C∫STφt∇ut∫0tgt−ζuζ−utdζdt≤∫STφtφt∇ut∫0tgt−ζuζ−utdζdt=∫STφt∇utCMφt∫0tgt−ζ∇uζ−∇utdζdt≤ε22∫STφt∇ut2dt+C2Mkε2φ0EuS∫0∞gp−1/pζdζ.
Since∫0∞gζdζ<1, it is natural to see that ∫0tgζdζ can be regarded as a small number tending to 0. Now, we consider the existence of such a t0∈0,t, which guarantees the positiveness of ∫0tgζdζ. Further, by a combination of equations (63), (83), (84), (87), (88), and (91), the variant of (81) can be obtained, which satisfies the following estimation:(92)∫STφt∫0tgζdζ−δ2∇utt2dt≤ε3∫STφt∇utt2dt+g0Mδ2+32M+12M+1k2+1ε3+C2Mε3∫0∞gp−1/pζdζφ0EuS.
For anyS0∈0,S, with the help of the fact g0>0 and the continuity of g, it can be acquired that(93)∫0Sgζdζ≥∫0S0gζdζ>0.
Choosing a positive constantδ3 which is small enough so that δ3<∫0S0gζdζ, and considering(94)∫0tgζdζ<∫0Sgζdζ,we can check that(95)∫0tgζdζ−δ32<δ3−δ32=δ32.
Now, for anyS∈S0,T, we have(96)δ32∫STφtutt2dt<12∫0S0gζdζ∫STφtutt2dt≤ε3∫STφt∇utt2dt+g0Mδ3+32M+12M+1k2+1ε3+C2Mε3∫0∞gp−1/pζdζφ0EuS=Δε3∫STφt∇utt2dt+C01φ0EuS.
Thus, we can conclude that(97)∫STφtutt2dt≤ε4∫STφt∇utt2dt+C1φ0EuS,where ε4=2ε3/δ3, and(98)C1=2δ3g0Mδ3+32M+12M+1k2+1ε3+C2Mε3∫0∞gp−1/pζdζ.
Consequently,(99)∫STφt1−∫0tgζdζ∇ut2dt≤ε5C1∫STφt∇ut2dt+C2φ0EuS.
Ifε5 is small enough in the above formula, then(100)∫STφt1−∫0tgζdζ∇ut2dt≤C2φ0EuS.
Alternatively, ifC3 is taken properly, the estimation can be arrived as(101)∫STφtutt2dt≤C3φ0EuS.
Considering the third term and the fourth one of (56), we get(102)−∫STφt1γ+2∫Ωuγ+2dx−η22∫Ωu2dxdt≤C∫STφt∇ut2+γ/2dt≤C4∫STφt1−∫0∞gζdζ∇ut2dt≤C4∫STφt1−∫0tgζdζ∇ut2dt≤C4φ0EuS.
By combining equations (100)–(102), it is shown that (56) is true, which concludes the proof.
Next, it remains to complete the Proof of Theorem 2.Proof of Theorem 2.
Let us consider the case wherep equals infinity firstly. For any t, if c represents any positive constant, then by taking φt=c in (56), we can get the following result:(103)12∫STut2dt+12∫ST1−∫0tgζdζ∇ut2dt−1γ+2∫ST∫Ωuγ+2dxdt+η22∫ST∫Ωu2dxdt≤C5EuS.
Again,gt≤−g′t/k follows from g′t≤−kgt. Invoking Lemma 3, one may deduce that(104)12∫ST∫0tgt−ζ∇uζ−∇ut2dζdt≤−1k12∫ST∫0tg′t−ζ∇uζ−∇ut2dζdt≤−1k∫STEu′tdt≤1kEuS.
Then,∫STEup+1/ptdt≤CEuS can be derived from (55). This fact further explains the attenuation of Eut according to a polynomial form.
Secondly, it is valuable to consider the case of2<p<∞. Aiming to evaluate the last term of Eut, we will put forward the following lemmas.Lemma 5.
For any0≤S≤T and t>0, the following inequality holds:(105)∫STεu1/pt∫0tgt−ζ∇uζ−∇ut2dζdt≤C6εup/p+1S∫STεup+1/pt∫0t∇uζ−∇ut2dζdt1/p+1.Proof.
Let(106)Φ1t≔∫0t∇uζ−∇ut2dζ.
In view of the assumptionH1 and the Hölder inequality, we get(107)∫0Tgt−ζ∇uζ−∇ut2dζ=∫0T∇uζ−∇ut21/p+1gp+1/pt−ζ∇uζ−∇ut2p/p+1dζ≤∫0T∇uζ−∇ut21/p+1p+1dζ1/p+1×∫0Tgp+1/pt−ζ∇uζ−∇ut2p/p+1p+1/pdζp/p+1=Φ1p/p+1t∫0Tgp+1/pt−ζ∇uζ−∇ut2dζp/p+1.
Hence,(108)∫STεu1/pt∫0Tgt−ζ∇uζ−∇ut2dζdt≤∫STεu1/ptΦ11/p+1t∫0Tg1+1/pt−ζ∇uζ−∇ut2dζp/p+1dt=∫STεup+1/ptΦ1t1/p+1∫0Tg1+1/pt−ζ∇uζ−∇ut2dζp/p+1dt≤∫STεup+1/ptΦ1tdt1/p+1∫ST∫0Tg1+1/pt−ζ∇uζ−∇ut2dζdtp/p+1≤∫STεup+1/ptΦ1tdt1/p+1∫ST∫0T−g′t−ζk∇uζ−∇ut2dζdtp/p+1=k−p/p+1∫STεup+1/ptΦ1tdt1/p+1∫ST∫0T−g′t−ζ∇uζ−∇ut2dζdtp/p+1≤k−p/p+1∫STεup+1/ptΦ1tdt1/p+1−2∫STεu′tdtp/p+1=k2−p/p+1∫STεup+1/ptΦ1tdt1/p+1εuS−εuTp/p+1≤k2−p/p+1∫STεup+1/ptΦ1tdt1/p+1εup/p+1S=C6εup/p+1S∫STεup+1/ptΦ1tdt1/p+1,where C6=k/2−p/p+1. This completes the proof.Lemma 6.
Let(109)Φ2t≔∫0tgt−ζ∇uζ−∇ut2dζ,∀t≤0.Then, for0≤S≤T, it holds that(110)∫STεu2/pt∫0tgt−ζ∇uζ−∇ut2dζdt≤C7εup/p+2S∫STεu1+2/ptΦ2tdt2/p+2.The proof of Lemma6 can be deduced similar to the one of Lemma 5, and this process will not be stated here. Besides, it is easy to see that Φ2t is bounded.Indeed,gζ∈L10,∞ can be ensured from p>2. With the assistance of (45), we find(111)∇ut2≤4Eut1−∫0∞gζdζ,and(112)Φ2t≤C2a∫0tgt−ζ∇uζ2+∇ut2dζ≤C2b∫0tgt−ζEuζ+Eutdζ≤C2c∫0tgt−ζEu0dζ≤C2c∫0∞gt−ζEu0dζ≤C2d∫0∞gt−ζEu0dζ≤C2eEu0,where C2e≥C2d∫0∞gt−ζdζ. Now,(113)Φ2t∞2/p≤C2e2/pEu02/p,which follows from (112).Lemma 7.
For anyS0∈0,∞, if S≥S0, then there is a constant C8∈0,∞ such that(114)∫S∞εup+2/ptdt≤C8εuSεu2/p0+Φ2t∞2/p.Proof.
GivenS∈S0,T. By means of (39), we get(115)∫STEup+2/ptdt=∫STEu2/ptEutdt=∫STEu2/pt12ut2+1−∫0tgζdζ∇u2+∫0tgt−ζ∇uζ−∇ut2dζ−1γ+2∫Ωuγ+2dx+η22∫Ωu2dxdt=12∫STEu2/ptut2dt+12∫STEu2/pt1−∫0tgζdζ∇u2dt−1γ+2∫STEu2/pt∫Ωuγ+2dxdt+η22∫STEu2/pt∫Ωu2dxdt+12∫STEu2/pt∫0tgt−ζ∇uζ−∇ut2dζdt=ΔI+II+III+IV+V.
Takingεu2/pt as a multiplier and replacing the position of φt in (56), we have(116)I+II+III+IV≤Cεu2/p0εuS.
Applying (110) and Young inequality, it is inferred that(117)V≤C7εup/p+2S∫STεu1+2/ptΦ2tdt2/p+2≤C7εup/p+2S∫STεu1+2/ptdt2/p+2∫STΦ2tdt2/p+2≤C7εup/p+2S∫STεu1+2/ptdt2/p+2Φ2t∞2/p+2≤ε6∫STεu1+2/ptdt2/p+2.p+2/2+2ε6/p+2−p/2p/p+2εup/p+2Sp+2/pΦ2t∞2/p+2.p+2/p=ε6∫STεu1+2/ptdt+2ε6/p+2−p/2p/p+2εuSΦ2t∞2/p.
Combining (116) and (117), one gets(118)∫STεup+2/ptdt≤ε6∫STεu1+2/ptdt+εuS2ε6/p+2−p/2p/p+2Φ2t∞2/p+Cεu2/p0.
Let the positive numberε6 be infinitely close to zero, and C8≥maxC,2ε6/p+2−p/2/p/p+2. Then(119)∫STεup+2/ptdt≤C8εuSΦ2t∞2/p+εu2/p0.
AsT tends to infinity, the limit result of long-time memory is easily seen, and thus, (114) is true.Remark 3.
With the aid of paper [12], it is trivial to show that(120)∫S∞εup+1/ptdt≤C8′εuSεu1/p0+Φ1t∞1/p.
The proof of (120) is entirely similar to that of (114) and so it is omitted here. Now, let us turn back to complete the verification of Theorem 2.
Continued Proof of Theorem 2. By equation (113), we have(121)∫S∞εup+2/ptdt≤C8εuSεu2/p0+C2e2/pεu02/p≤C9εuSεu2/p0,where C9≥C81+C2e2/p.
Taking2/p<1, T=S and T0=S0, it is inferred from (121) that(122)Eut≤Eu0S0+C1+2/p2t/p+S0+Cp/2,∀t∈0,+∞.
From the representation of (45), it is not difficult to examine that for any t∈0,∞,(123)Φ1t=∫0t∇uζ−∇ut2dζ≤C10∫0tEuζdζ+∫0tEutdt≤C10∫0∞Euζdζ+tEut≤C10∫0tEuζdζ+C11tEu0≤C12Eu0,where C12≥t1+C11C10. That is, Φ1t∞≤C12Eu0.
The application of (120) and S≥S0 yields(124)∫S∞εup+1/ptdt≤C13εuSεu1/p0.
Thus, employingp>2, we have(125)Eut≤Eu0S0+C1+1/pt/p+S0+Cp=Eu0p+1p+Ctp.
Besides, under the condition thatp=∞, equation (124) turns into(126)∫S∞Eutdt≤C13EuS,which yields(127)Eut≤Eu0e1−t/C13′=Eu0e1−Ct.
This completes the proof.


## 4. Conclusion

Based on the proposed appropriate assumptions of the convolution kernels along with the discussion about the fuzzy numberη, the exponential and polynomial aspects of the energy decay rates for system (1) are estimated only through the application of the multiplier method and the unified technique. In this process, the most valuable point is that our research has avoided the construction of auxiliary functions perfectly. The appearance of the term with fuzzy coefficient makes the expression form of Eu richer and it leads to some difficulties in calculation. At the same time, more efforts have been spent on discussing the integro-differential inequalities and the discussion is quite interesting. Considering the case of η=0, we can see that the results coincide with that of reference [12]. In summary, the result in this paper reveals the wide applicability of the unified method, and further discussion for the blow-up problems may be considered in the future.



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