Citation: Hao Dong, Yue Liu, Jiaqi Chang. The heterogeneous linkage of economic policy uncertainty and oil return risks[J]. Green Finance, 2019, 1(1): 46-66. doi: 10.3934/GF.2019.1.46
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The Moore-Gibson-Thompson (MGT) equation is one of the equations of nonlinear acoustics describing acoustic wave propagation in gases and liquids [13,15,30] and arising from modeling high frequency ultrasound waves [9,18] accounting for viscosity and heat conductivity as well as effect of the radiation of heat on the propagation of sound. This research field is highly active due to a wide range of applications such as the medical and industrial use of high intensity ultrasound in lithotripsy, thermotherapy, ultraound cleaning, etc. The classical nonlinear acoustics models include the Kuznetson's equation, the Westervelt equation and the Kokhlov-Zabolotskaya-Kuznetsov equation.
In order to gain a better understanding of the nonlinear MGT equation, we shall begin with the linearized model. In [15], Kaltenbacher, Lasiecka and Marchand investigated the following linearized MGT equation
τuttt+αutt+c2Au+bAut=0. | (1.1) |
For equation (1.1), they disclosed a critical parameter
τuttt+αutt+c2Au+bAut=f(u,ut,utt). | (1.2) |
They proved that the underlying PDE generates a well-posed dynamical system which admits a global and finite dimensional attractor. They also overcomed the difficulty of lacking the Lyapunov function and the lack of compactness of the trajectory.
Now, we concentrate on the stabilization of MGT equation with memory which has received a considerable attention recently. For instance, Lasiecka and Wang [17] studied the following equation:
τuttt+αutt+bAut+c2Au−∫t0g(t−s)Aw(s)ds=0, | (1.3) |
where
g′(t)≤−c0g(t), | (1.4) |
they discussed the effect of memory described by three types on decay rates of the energy when
τuttt+αutt+bAut+c2Au−∫∞0g(s)Aw(t−s)ds=0. | (1.5) |
Alves et al. [1] investigated the uniform stability of equation (1.5) encompassing three different types of memory in a history space set by the linear semigroup theory. Moreover, we refer the reader to [3,6,7,12,24,25,26,28] for other works of the equation(s) with memory.
More recently, Filippo and Vittorino [11] considered the fourth-order MGT equation
utttt+αuttt+βutt+γAutt+δAut+ϱAu=0. | (1.6) |
They investigated the stability properties of the related solution semigroup. And, according to the values of certain stability numbers depending on the strictly positive parameters
Motivated by the above results, we intend to study the following abstract version of the fourth-order Moore-Gibson-Thompson (MGT) equation with a memory term
utttt+αuttt+βutt+γAutt+δAut+ϱAu−∫t0g(t−s)Au(s)ds=0, | (1.7) |
where
u(0)=u0,ut(0)=u1,utt(0)=u2,uttt(0)=u3. | (1.8) |
A natural question that arised in dealing with the general decay of fourth-order MGT equation with memory:
● Can we get a general decay result for a class of relaxation functions satisfying
Mustafa answered this question for viscoelastic wave equations in [31,32]. Messaoudi and Hassan [29] considered the similar question for memory-type Timoshenko system in the cases of equal and non-equal speeds of wave propagation. Moreover, they extended the range of polynomial decay rate optimality from
The aim of this paper is to establish the well-posedness and answer the above mention question for fourth-order MGT equation with memory (1.7). We first use the Faedo-Galerkin method to prove the well-posedness result. We then use the idea developed by Mustafa in [31,32], taking into consideration the nature of fourth-order MGT equation, to prove new general decay results for the case
The rest of our paper is organized as follows. In Section
In this section, we consider the following assumptions and state our main results. We use
First, we consider the following assumptions as in [11] for
0<g(0)<2αϱδ(αγ−δ), ϱ−∫+∞0g(s)ds=l>0. |
g′(t)≤−ξ(t)M(g(t)),∀ t≥0. | (2.1) |
Remark 1. ([31,Remark 2.8])
(1) From assumption
g(t)→0ast→+∞andg(t)≤ϱ−lt,∀ t>0. |
Furthermore, from the assumption
g(t0)=randg(t)≤r,∀ t≥t0. |
The non-increasing property of
0<g(t0)≤g(t)≤g(0)and0<ξ(t0)≤ξ(t)≤ξ(0),∀ t∈[0,t0]. |
A combination of these with the continuity of
a≤ξ(t)M(g(t))≤d,∀ t∈[0,t0]. |
Consequently, for any
g′(t)≤−ξ(t)M(g(t))≤−a=−ag(0)g(0)≤−ag(0)g(t) |
and, hence,
g(t)≤−g(0)ag′(t),∀ t∈[0,t0]. | (2.2) |
(2) If
¯M=C2t2+(B−Cr)t+(A+C2r2−Br). |
Then, inspired by the notations in [11], we define the Hilbert spaces
Hr:=D(Ar2),r∈R. |
In order to simplify the notation, we denote the usual space
H=D(A12)×D(A12)×D(A12)×H. |
Moreover, we will denote the inner product of
After that, we introduce the following energy functional
E(t)=12[‖uttt+αutt+αϱδut‖2+δα(ϱ−G(t)ϱ)‖A12utt+αA12ut+αϱδA12u‖2+δαϱG(t)‖A12utt+αA12ut‖2+(γ−δα)‖A12utt‖2+(γ−δα)αϱδ‖A12ut‖2+2∫t0g(t−s)(A12u(t)−A12u(s),A12utt+αA12ut)ds+αϱδ(g∘A12u)(t)−α(g′∘A12u)(t)+αg(t)‖A12u‖2+(β−αϱδ)‖utt‖2+αϱδ(β−αϱδ)‖ut‖2], |
where
(g∘v)(t):=∫Ω∫t0g(t−s)(v(t)−v(s))2dsdx. |
As in [31], we set, for any
Cν=∫∞0g2(s)νg(s)−g′(s)dsandh(t)=νg(t)−g′(t). |
The following lemmas play an important role in the proof of our main results.
Lemma 2.1. ([31] )
Assume that condition
∫Ω(∫t0g(t−s)(A12u(s)−A12u(t))ds)2dx≤Cν(h∘A12u)(t),∀ t≥0. |
Lemma 2.2. (Jensen's inequality) Let
P(1k∫Ωf(x)h(x)dx)≤1k∫ΩP(f(x))h(x)dx. |
Lemma 2.3. ([2])(The generalized Young inequality) If
AB≤f∗(A)+f(B), | (2.3) |
where
f∗(s)=s(f′)−1(s)−f[(f′)−1(s)]. | (2.4) |
We are now in a position to state the well-posedness and the general decay result for problem (1.7)-(1.8).
Theorem 2.4. (Well-posedness)
Assume that
u∈C([0,T];D(A12))∩C1([0,T];H). |
Theorem 2.5. (General decay)
Let
E(t)≤k2M−11(k1∫tg−1(r)ξ(s)ds),∀ t≥g−1(t), | (2.5) |
where
Remark 2. Assume that
E(t)≤{Cexp(−˜k∫t0ξ(s)ds),ifp=1,¯k(1+∫t0ξ(s)ds)−1p−1,if1<p<2. | (2.6) |
In this section, we will prove the global existence and uniqueness of the solution of problem (1.7)-(1.8). Firstly, we give the following lemmas.
Lemma 3.1.
If
Proof. Since
2αϱδ(α−σ)(γ−δα)→2αϱδ(αγ−δ)asσ→0, |
which is trivially true.
Lemma 3.2.
Assume that
12[‖uttt+αutt+αϱδut‖2+δα(ϱ−G(t)ϱ)‖A12utt+αA12ut+αϱδA12u‖2+(γ−δα)‖A12utt‖2+(γ−δα)αϱδ‖A12ut‖2−α(g′∘A12u)(t)+αg(t)‖A12u‖2+(β−αϱδ)‖utt‖2+αϱδ(β−αϱδ)‖ut‖2]≤E(t)≤12[‖uttt+αutt+αϱδut‖2+δα(ϱ−G(t)ϱ)‖A12utt+αA12ut+αϱδA12u‖2+(γ−δα)‖A12utt‖2+(γ−δα)αϱδ‖A12ut‖2−α(g′∘A12u)(t)+2αϱδ(g∘A12u)(t)+αg(t)‖A12u‖2+(β−αϱδ)‖utt‖2+αϱδ(β−αϱδ)‖ut‖2+2δαϱG(t)‖A12utt+αA12ut‖2]. |
Proof. From the definition of
E(t)=12[‖uttt+αutt+αϱδut‖2+δα(ϱ−G(t)ϱ)‖A12utt+αA12ut+αϱδA12u‖2+δαϱG(t)‖A12utt+αA12ut‖2+(γ−δα)‖A12utt‖2+(γ−δα)αϱδ‖A12ut‖2+2∫t0g(t−s)(A12u(t)−A12u(s),A12utt+αA12ut)ds+αϱδ(g∘A12u)(t)−α(g′∘A12u)(t)+αg(t)‖A12u‖2+(β−αϱδ)‖utt‖2+αϱδ(β−αϱδ)‖ut‖2]. |
Then, we estimate the sixth term of the above equality
2|∫t0g(t−s)(A12u(t)−A12u(s),A12utt+αA12ut)ds|≤∫t0g(t−s)[αϱδ‖A12u(t)−A12u(s)‖2+δαϱ‖A12utt+αA12ut‖2]ds=αϱδ(g∘A12u)(t)+δαϱG(t)‖A12utt+αA12ut‖2. |
A combination of the above results, we complete the proof of lemma.
Now, we prove the well-posedness result of problem (1.7)-(1.8).
Proof of Theorem 2.1. The proof is given by Faedo-Galerkin method and combines arguments from [16,39,38]. We present only the main steps.
Step
We construct approximations of the solution
um(t)=m∑j=1amj(t)wj(x), | (3.1) |
where
∫Ωumtttt(t)wjdx+α∫Ωumttt(t)wjdx+β∫Ωumtt(t)wjdx+γ∫ΩA12umtt(t)A12wjdx+δ∫ΩA12umt(t)A12wjdx+ϱ∫ΩA12um(t)A12wjdx−∫t0g(t−s)∫ΩA12um(t)A12wjdxds=0 | (3.2) |
with initial conditions
(um(0),umt(0),umtt(0),umttt(0))=(um0,um1,um2,um3). | (3.3) |
According to the standard theory of ordinary differential equation, the finite dimensional problem (3.2)-(3.3) has a local solution
Step
Multiplying equation (3.2) by
ddtEm(t)+α(β−αϱδ)‖umtt‖2−αg′(t)2‖A12um‖2+δ2αϱg(t)‖A12umtt+αϱδA12um‖2+α2(g″∘A12um)(t)=−[α(γ−δα)−δg(t)2αϱ]‖A12umtt‖2+∫t0g′(t−s)(A12um(t)−A12um(s),A12umtt)ds+αϱ2δ(g′∘A12um)(t), |
where
Em(t)=12[‖umttt+αumtt+αϱδumt‖2+δα(ϱ−G(t)ϱ)‖A12umtt+αA12umt+αϱδA12um‖2+δαϱG(t)‖A12umtt+αA12umt‖2+(γ−δα)‖A12umtt‖2+(γ−δα)αϱδ‖A12umt‖2+2∫t0g(t−s)(A12um(t)−A12um(s),A12umtt+αA12umt)ds+αϱδ(g∘A12um)(t)−α(g′∘A12um)(t)+αg(t)‖A12um‖2+(β−αϱδ)‖umtt‖2+αϱδ(β−αϱδ)‖umt‖2]. | (3.4) |
From assumptions
∫t0g′(t−s)(A12um(t)−A12um(s),A12umtt)ds≤−(α−ε)ϱ2δ(g′∘A12um)(t)−δ2(α−ε)ϱ‖A12umtt‖2∫t0g′(t−s)ds=−(α−ε)ϱ2δ(g′∘A12um)(t)+δ(g(0)−g(t))2(α−ε)ϱ‖A12umtt‖2 |
and so
−[α(γ−δα)−δg(t)2αϱ]‖A12umtt‖2+∫t0g′(t−s)(A12um(t)−A12um(s),A12umtt)ds+αϱ2δ(g′∘A12um)(t)≤−[α(γ−δα)−δg(t)2αϱ]‖A12umtt‖2−(α−ε)ϱ2δ(g′∘A12um)(t)+δ(g(0)−g(t))2(α−ε)ϱ‖A12umtt‖2+αϱ2δ(g′∘A12um)(t)=−[α(γ−δα)−δg(0)2(α−ε)ϱ]‖A12umtt‖2+εϱ2δ(g′∘A12um)(t)−[δg(t)2(α−ε)ϱ−δg(t)2αϱ]‖A12umtt‖2≤0. | (3.5) |
Therefore, we have
ddtEm(t)+α(β−αϱδ)‖umtt‖2−αg′(t)2‖A12um‖2+δ2αϱg(t)‖A12umtt+αϱδA12um‖2+α2(g″∘A12um)(t)≤0. | (3.6) |
Integrating (3.6) from
Em(t)+∫t0[α(β−αϱδ)‖umtt‖2−αg′(τ)2‖A12um‖2+δ2αϱg(τ)‖A12umtt+αϱδA12um‖2+α2(g″∘A12um)(τ)]dτ≤Em(0). | (3.7) |
Now, since the sequences
Em(t)≤C. | (3.8) |
Therefore, using the fact
(um)m∈NisboundedinL∞(0,T;D(A12))(umt)m∈NisboundedinL∞(0,T;D(A12))(umtt)m∈NisboundedinL∞(0,T;D(A12))(umttt)m∈NisboundedinL∞(0,T;H). | (3.9) |
Consequently, we may conclude that
um⇀uweak∗inL∞(0,T;D(A12))umt⇀utweak∗inL∞(0,T;D(A12))umtt⇀uttweak∗inL∞(0,T;D(A12))umttt⇀utttweak∗inL∞(0,T;H). |
From (3.9), we get that
Since the embedding
un→ustronglyinL2(0,T;H(Ω)). |
Therefore,
un→ustronglyanda.e.on(0,T)×Ω. |
The proof now can be completed arguing as in [21].
Step
It is sufficient to show that the only weak solution of (1.7)-(1.8) with
u≡0. | (3.10) |
According to the energy estimate (3.8) and noting that
E(u(t))=0,∀t∈[0,T]. |
So, we have
‖uttt+αutt+αϱδut‖2=‖A12utt+αA12ut+αϱδA12u‖2=‖A12utt‖2=‖A12ut‖2=‖A12u‖2=‖utt‖2=0,∀t∈[0,T]. |
And this implies (3.10). Thus, we conclude that problem (1.7)-(1.8) has at most one solution.
In this section, we state and prove some lemmas needed to establish our general decay result.
Lemma 4.1.
Let
ddtE(t)≤−α(β−αϱδ)‖utt‖2−[α(γ−δα)−δg(0)2(α−ε)ϱ]‖A12utt‖2+αg′(t)2‖A12u‖2−[δg(t)2(α−ε)ϱ−δg(t)2αϱ]‖A12utt‖2−δ2αϱg(t)‖A12utt+αϱδA12u‖2−α2(g″∘A12u)(t)+εϱ2δ(g′∘A12u)≤0. |
Proof. Multiplying (1.7) by
ddtE(t)=−α(β−αϱδ)‖utt‖2−[α(γ−δα)−δg(t)2αϱ]‖A12utt‖2+αg′(t)2‖A12u‖2−δ2αϱg(t)‖A12utt+αϱδA12u‖2+∫t0g′(t−s)(A12u(t)−A12u(s),A12utt)ds−α2(g″∘A12u)(t)+αϱ2δ(g′∘A12u)(t). | (4.1) |
We proceed to show that, for a constant
|∫t0g′(t−s)(A12u(t)−A12u(s),A12utt)ds|≤−(α−ε)ϱ2δ(g′∘A12u)(t)+δ(g(0)−g(t))2(α−ε)ϱ‖A12utt‖2. | (4.2) |
Then, combining (4.1) and (4.2), we can obtain
ddtE(t)≤−α(β−αϱδ)‖utt‖2−[α(γ−δα)−δg(0)2(α−ε)ϱ]‖A12utt‖2+αg′(t)2‖A12u‖2−[δg(t)2(α−ε)ϱ−δg(t)2αϱ]‖A12utt‖2−δ2αϱg(t)‖A12utt+αϱδA12u‖2−α2(g″∘A12u)(t)+εϱ2δ(g′∘A12u). |
According to
Lemma 4.2.
Assume that
F1(t)=∫Ω(utt+αut+αϱδu)(uttt+αutt+αϱδut)dx |
satisfies the estimate
F′1(t)≤−δ2α‖A12utt+αA12ut+αϱδA12u‖2+2αλ0δ(β−αϱδ)2‖utt‖2+2αδ(γ−δα)2‖A12utt‖2+‖uttt+αutt+αϱδut‖2+2α(ϱ−l)2δ‖A12u‖2+2αδCν(h∘A12u)(t). | (4.3) |
Proof. Taking the derivative of
F′1(t)=∫Ω[−(β−αϱδ)utt](utt+αut+αϱδu)dx−δα‖A12utt+αA12ut+αϱδA12u‖2+‖uttt+αutt+αϱδut‖2−∫Ω(γ−δα)A12utt(A12utt+αA12ut+αϱδA12u)dx+∫Ω(∫t0g(t−s)A12u(s)ds)(A12utt+αA12ut+αϱδA12u)dx. |
Using Young's inequality, Lemma 2.1,
∫Ω[−(β−αϱδ)utt](utt+αut+αϱδu)dx≤2αλ0δ(β−αϱδ)2‖utt‖2+δ8αλ0‖utt+αut+αϱδu‖2≤2αλ0δ(β−αϱδ)2‖utt‖2+δ8α‖A12utt+αA12ut+αϱδA12u‖2 |
and
∫Ω(∫t0g(t−s)A12u(s)ds)(A12utt+αA12ut+αϱδA12u)dx=∫Ω(∫t0g(t−s)(A12u(s)−A12u(t))ds)(A12utt+αA12ut+αϱδA12u)dx+∫Ω(∫t0g(t−s)A12u(t)ds)(A12utt+αA12ut+αϱδA12u)dx≤2αδCν(h∘A12u)(t)+δ4α‖A12utt+αA12ut+αϱδA12u‖2+2α(ϱ−l)2δ‖A12u‖2. |
Also, we have
−∫Ω(γ−δα)A12utt(A12utt+αA12ut+αϱδA12u)dx≤2αδ(γ−δα)2‖A12utt‖2+δ8α‖A12utt+αA12ut+αϱδA12u‖2. |
Then, combining the above inequalities, we complete the proof of (4.3).
Lemma 4.3.
Assume that
F2(t)=−∫Ω(uttt+αutt+αϱδut)∫t0g(t−s)[(utt+αut+αϱδu)(t)−αϱδu(s)]dsdx |
satisfies the estimate
F′2(t)≤−G(t)4‖uttt+αutt+αϱδut‖2+[(ϱ−l)2α22ε1+4λ0g2(0)α2G(t)]‖A12ut‖2+[λ0(ϱ−l)22+(δ2α2+ϱ2)+3(ϱ−l)2]ε1‖A12utt+αA12ut+αϱδA12u‖2+[α2ϱ2(ϱ−l)2λ0ε12δ2+(ϱ−l)22ε1+3(ϱ−l)2(αϱδ)2ε1+(ϱ−l)22(αϱδ)2]‖A12u‖2+[(ϱ−l)22ε1+12ε1(γ−δα)2+12(αϱδ)2(γ−δα)2]‖A12utt‖2+[(ε1α2ϱ2λ04δ2+34ε1+1+αϱδ+2α4ϱ2λ0G(t)δ2)Cν+2α2ϱ2λ0G(t)δ2](h∘A12u)(t)+[2(β−αϱδ)2ε1+4g2(0)G(t)]‖utt‖2, | (4.4) |
where
Proof. By differentiating
F′2(t)=∫Ω[βutt+γAutt+δAut+ϱAu−∫t0g(t−s)Au(s)ds−αϱδutt]×∫t0g(t−s)[(utt+αut+αϱδu)(t)−αϱδu(s)]dsdx−g(0)∫Ω(uttt+αutt+αϱδut)(utt+αut)dx−∫Ω(uttt+αutt+αϱδut)∫t0g′(t−s)[(utt+αut+αϱδu)(t)−αϱδu(s)]dsdx−∫t0g(s)ds‖uttt+αutt+αϱδut‖2=∫Ω[(β−αϱδ)utt+δα(Autt+αAut+αϱδAu)+(γ−δα)Autt−∫t0g(s)dsAu(t)]∫t0g(t−s)[(utt+αut+αϱδu)(t)−αϱδu(s)]dsdx−∫t0g(s)ds‖uttt+αutt+αϱδut‖2−∫Ω(uttt+αutt+αϱδut)∫t0g′(t−s)[(utt+αut+αϱδu)(t)−αϱδu(s)]dsdx+(∫t0g(s)ds)∫Ω∫t0g(t−s)(Au(t)−Au(s))ds(utt+αut)dx+αϱδ∫Ω(∫t0g(t−s)(A12u(t)−A12u(s))ds)2dx−g(0)∫Ω(uttt+αutt+αϱδut)(utt+αut)dx. |
Now, we estimate the terms in the right-hand side of the above identity.
Using Young's inequality, we obtain, for
∫Ω[(β−αϱδ)utt+δα(Autt+αAut+αϱδAu)+(γ−δα)Autt−∫t0g(s)dsAu(t)]∫t0g(t−s)[(utt+αut+αϱδu)−αϱδu(s)]dsdx≤[λ0(ϱ−l)22+(δ2α2+ϱ2)+2(ϱ−l)2]ε1‖A12utt+αA12ut+αϱδA12u‖2+2(β−αϱδ)2ε1‖utt‖2+[α2ϱ2(ϱ−l)2λ0ε12δ2+(ϱ−l)22ε1+2(ϱ−l)2(αϱδ)2ε1+(ϱ−l)22(αϱδ)2]‖A12u‖2+[(ϱ−l)22ε1+12ε1(γ−δα)2+12(αϱδ)2(γ−δα)2]‖A12utt‖2+(ϱ−l)2α22ε1‖A12ut‖2+(ε1α2ϱ2λ04δ2+14ε1+1)Cν(h∘A12u)(t) |
and
(∫t0g(s)ds)∫Ω∫t0g(t−s)(Au(t)−Au(s))ds(utt+αut)dx=(∫t0g(s)ds)∫Ω∫t0g(t−s)(A12u(t)−A12u(s))ds(A12utt+αA12ut)dx≤12ε1Cν(h∘A12u)(t)+(ϱ−l)22ε1‖A12utt+αA12ut‖2≤12ε1Cν(h∘A12u)(t)+(ϱ−l)2ε1‖A12utt+αA12ut+αϱδA12u‖2+(ϱ−l)2(αϱδ)2ε1‖A12u‖2. |
Also, we have
αϱδ∫Ω(∫t0g(t−s)(A12u(t)−A12u(s))ds)2dx≤αϱδCν(h∘A12u)(t) |
and
−g(0)∫Ω(uttt+αutt+αϱδut)(utt+αut)dx≤G(t)4‖uttt+αutt+αϱδut‖2+g2(0)G(t)‖utt+αut‖2≤G(t)4‖uttt+αutt+αϱδut‖2+2g2(0)G(t)‖utt‖2+2λ0g2(0)α2G(t)‖A12ut‖2. |
Exploiting Young's inequality and
−∫Ω(uttt+αutt+αϱδut)∫t0g′(t−s)[(utt+αut+αϱδu)(t)−αϱδu(s)]dsdx=−∫Ω(uttt+αutt+αϱδut)∫t0g′(t−s)(utt+αut)(t)dsdx−αϱδ∫Ω(uttt+αutt+αϱδut)∫t0g′(t−s)(u(t)−u(s))dsdx≤G(t)2‖uttt+αutt+αϱδut‖2+2g2(0)G(t)‖utt‖2+2λ0α2g2(0)G(t)‖A12ut‖2+2α2ϱ2λ0G(t)δ2(α2Cν+1)(h∘A12u)(t). |
A combination of all the above estimates gives the desired result.
As in [11], we introduce the following auxiliary functional
F3(t)=∫Ω(uttt+αutt)utdx+ϱ2‖A12u‖2. |
Lemma 4.4.
Assume that
F′3(t)≤−(3δ8−ε2δ4)‖A12ut‖2+ε2δ38α2ϱ2λ0‖uttt+αutt+αϱδut‖2+2γ2δ‖A12utt‖2+(ε2δ34ϱ2λ0+4α2ϱ2λ0ε2δ3+2β2λ0δ)‖utt‖2+1δCν(h∘A12u)(t)+2(ϱ−l)2δ‖A12u‖2, | (4.5) |
where
Proof. Using the equation (1.7), a direct computation leads to the following identity
F′3(t)=∫Ω(uttt+αutt)uttdx+∫Ω(utttt+αuttt)utdx+ϱ(A12u,A12ut)=(uttt,utt)+α‖utt‖2−β(utt,ut)−γ(A12utt,A12ut)−δ‖A12ut‖2+(∫t0g(t−s)A12u(s)ds,A12ut). | (4.6) |
Now, the first and third terms in the right-hand side of (4.6) can be estimated as follows:
(uttt,utt)≤ε2δ316α2ϱ2λ0‖uttt‖2+4α2ϱ2λ0ε2δ3‖utt‖2≤ε2δ38α2ϱ2λ0‖uttt+αutt+αϱδut‖2+ε2δ38α2ϱ2λ0‖αutt+αϱδut‖2+4α2ϱ2λ0ε2δ3‖utt‖2≤ε2δ38α2ϱ2λ0‖uttt+αutt+αϱδut‖2+α2(ε2δ34α2ϱ2λ0+4ϱ2λ0ε2δ3)‖utt‖2+ε2δ4‖A12ut‖2 |
and
−β(utt,ut)≤2β2λ0δ‖utt‖2+δ8λ0‖ut‖2≤2β2λ0δ‖utt‖2+δ8‖A12ut‖2, |
where
Using Young's inequality and Lemma 2.1, we get
−γ(A12utt,A12ut)≤2γ2δ‖A12utt‖2+δ8‖A12ut‖2 |
and
(∫t0g(t−s)A12u(s)ds,A12ut)=(∫t0g(t−s)(A12u(s)−A12u(t)+A12u(t))ds,A12ut)≤1δ∫Ω(∫t0g(t−s)(A12u(t)−A12u(s))ds)2dx+δ4‖A12ut‖2+2(ϱ−l)2δ‖A12u‖2+δ8‖A12ut‖2≤1δCν(h∘A12u)(t)+3δ8‖A12ut‖2+2(ϱ−l)2δ‖A12u‖2. |
Then, combining the above inequalities, we obtain the desired result.
Lemma 4.5.
Assume that
F4(t)=∫Ω∫t0f(t−s)|A12u(s)|2dsdx |
satisfies the estimate
F′4(t)≤−12(g∘A12u)(t)+3(ϱ−l)‖A12u‖2, | (4.7) |
where
Proof. Noting that
F′4(t)=f(0)‖A12u‖2−∫Ω∫t0g(t−s)|A12u(s)|2dsdx=f(0)‖A12u‖2−∫Ω∫t0g(t−s)|A12u(s)−A12u(t)|2dsdx−2∫ΩA12u∫t0g(t−s)(A12u(s)−A12u(t))dsdx−(∫t0g(s)ds)‖A12u‖2=−(g∘A12u)(t)−2∫ΩA12u∫t0g(t−s)(A12u(s)−A12u(t))dsdx+f(t)‖A12u‖2. |
Exploiting Young's inequality and the fact
−2∫ΩA12u∫t0g(t−s)(A12u(s)−A12u(t))dsdx≤2(ϱ−l)‖A12u‖2+12(ϱ−l)(∫t0g(t−s)ds)∫Ω∫t0g(t−s)(A12u(s)−A12u(t))2dsdx≤2(ϱ−l)‖A12u‖2+12(g∘A12u)(t). |
Moreover, taking account of
f(t)‖A12u‖2≤(ϱ−l)‖A12u‖2. |
Combining the above estimates, we arrive at the desired result.
Lemma 4.6.
Assume that
L(t)=NE(t)+F1(t)+N2F2(t)+N3F3(t) |
satisfies, for a suitable choice of
L(t)∼E(t) |
and the estimate, for all
L′(t)≤−c[‖utt‖2+‖A12utt‖2+‖uttt+αutt+αϱδut‖2+‖A12utt+αA12ut+αϱδu‖2]−4(ϱ−l)‖A12u‖2+18(g∘A12u)(t), | (4.8) |
where
Proof. Combining Lemmas 4.1-4.4 and recalling that
L′(t)≤−[α(β−αϱδ)N−2αλ0δ(β−αϱδ)2−(2(β−αϱδ)2ε1+4g2(0)G(t))N2−(ε2δ34ϱ2λ0+4α2ϱ2λ0ε2δ3+2β2λ0δ)N3]‖utt‖2−[(α(γ−δα)−δg(0)2(α−ε)ϱ)N+(δg(t)2(α−ε)ϱ−δg(t)2αϱ)N−2αδ(γ−δα)2−((ϱ−l)22ε1+12ε1(γ−δα)2+12(αϱδ)2(γ−δα)2)N2−2γ2δN3]‖A12utt‖2−α2N(g″∘A12u)(t)−[(3δ8−ε2δ4)N3−((ϱ−l)2α22ε1+4λ0g2(0)α2G(t))N2]‖A12ut‖2−(G(t)4N2−1−ε2δ3N38α2ϱ2λ0)‖uttt+αutt+αϱδut‖2+εϱν2δN(g∘A12u)(t)−[δ2α−(λ0(ϱ−l)22+(δ2α2+ϱ2)+3(ϱ−l)2)ε1N2]×‖A12utt+αA12ut+αϱδA12u‖2−[−αg′(t)2N−2α(ϱ−l)2δ−(α2ϱ2(ϱ−l)2λ0ε12δ2+(ϱ−l)22ε1+3(ϱ−l)2(αϱδ)2ε1+(ϱ−l)22(αϱδ)2)N2−2(ϱ−l)2δN3]‖A12u‖2−[εϱ2δN−2αδCν−((ε1α2ϱ2λ04δ2+34ε1+1+αϱδ+2α4ϱ2λ0G(t)δ2)Cν+2α2ϱ2λ0G(t)δ2)N2−N3δCν](h∘A12u)(t). |
At this point, we need to choose our constants very carefully. First, we choose
\begin{align*} \varepsilon_{1} = \frac{\alpha\delta}{2N_{2}\left[\lambda_{0}\alpha^{2}(\varrho-l)^{2}+2(\delta^{2}+\alpha^{2}\varrho^{2})+6\alpha^{2}(\varrho-l)^{2}\right]}\quad\quad and \quad\quad \varepsilon_{2} = \frac{1}{N_{3}}. \end{align*} |
The above choice yields
\begin{align*} &\mathcal{L'}(t)\nonumber\\ &\leq-\left[\alpha\left(\beta-\frac{\alpha\varrho}{\delta}\right)N-\frac{2\alpha\lambda_{0}}{\delta}\left(\beta-\frac{\alpha\varrho}{\delta}\right)^{2}-\left(\frac{2\left(\beta-\frac{\alpha\varrho}{\delta}\right)^{2}}{\varepsilon_{1}}+\frac{4g^{2}(0)}{G(t)}\right)N_{2}\right. \nonumber\\ &\left.-\frac{\delta^{3}}{4\varrho^{2}\lambda_{0}}-\frac{4\alpha^{2}\varrho^{2}\lambda_{0}}{\delta^{3}}N_{3}^{2}-\frac{2\beta^{2}\lambda_{0}}{\delta}N_{3}\right]\|u_{tt}\|^{2}-\left[\left(\alpha\left(\gamma-\frac{\delta}{\alpha}\right)-\frac{\delta g(0)}{2(\alpha-\varepsilon)\varrho}\right)N\right. \nonumber\\ &\left.+\left(\frac{\delta g(t)}{2(\alpha-\varepsilon)\varrho}-\frac{\delta g(t)}{2\alpha\varrho}\right)N-\frac{2\alpha}{\delta}\left(\gamma-\frac{\delta}{\alpha}\right)^{2}-\left(\frac{(\varrho-l)^{2}}{2\varepsilon_{1}}+\frac{1}{2\varepsilon_{1}}\left(\gamma-\frac{\delta}{\alpha}\right)^{2}\right.\right. \nonumber\\ &\left.\left.+\frac{1}{2}\left(\frac{\alpha\varrho}{\delta}\right)^{2}\left(\gamma-\frac{\delta}{\alpha}\right)^{2}\right)N_{2}-\frac{2\gamma^{2}}{\delta}N_{3}\right]\left\|\mathcal{A}^{\frac{1}{2}}u_{tt}\right\|^{2}-\frac{\alpha}{2}N\left(g''\circ\mathcal{A}^{\frac{1}{2}}u\right)(t) \nonumber\\ &-\left[\frac{3\delta}{8}N_{3}-\frac{\delta}{4}-\left(\frac{(\varrho-l)^{2}\alpha^{2}}{2\varepsilon_{1}}+\frac{4\lambda_{0}g^{2}(0) \alpha^{2}}{G(t)}\right)N_{2}\right]\left\|\mathcal{A}^{\frac{1}{2}}u_{t}\right\|^{2} \nonumber\\ &+\frac{\varepsilon\varrho\nu}{2\delta}N\left(g\circ\mathcal{A}^{\frac{1}{2}}u\right)(t)-\left(\frac{G(t)}{4}N_{2}-1-\frac{\delta^{3}}{8\alpha^{2}\varrho^{2}\lambda_{0}}\right)\left\|u_{ttt}+\alpha u_{tt}+\frac{\alpha\varrho}{\delta}u_{t}\right\|^{2}\nonumber\\ & -\frac{\delta}{4\alpha}\left\|\mathcal{A}^{\frac{1}{2}}u_{tt}+\alpha\mathcal{A}^{\frac{1}{2}}u_{t}+\frac{\alpha\varrho}{\delta}\mathcal{A}^{\frac{1}{2}}u\right\|^{2} \nonumber\\ &-\left[-\frac{\alpha g'(t)}{2}N-\frac{2\alpha(\varrho-l)^{2}}{\delta}-\left(\frac{\alpha^{2}\varrho^{2}(\varrho-l)^{2} \lambda_{0}\varepsilon_{1}}{2\delta^{2}}+\frac{(\varrho-l)^{2}}{2\varepsilon_{1}}+3(\varrho-l)^{2} \left(\frac{\alpha\varrho}{\delta}\right)^{2}\varepsilon_{1}\right.\right. \nonumber\\ &\left.\left.+\frac{(\varrho-l)^{2}}{2}\left(\frac{\alpha\varrho}{\delta}\right)^{2}\right)N_{2} -\frac{2(\varrho-l)^{2}}{\delta}N_{3}\right]\left\|\mathcal{A}^{\frac{1}{2}}u\right\|^{2} -\left[\frac{\varepsilon\varrho}{2\delta}N-\frac{2\alpha}{\delta}C_{\nu} \right. \nonumber\\ &-\left(\left(\frac{\varepsilon_{1}\alpha^{2}\varrho^{2}\lambda_{0}}{4\delta^{2}}\right.\right.\left. \left.\left.+\frac{3}{4\varepsilon_{1}}+1+\frac{\alpha\varrho}{\delta}+\frac{2\alpha^{4}\varrho^{2}\lambda_{0}}{G(t)\delta^{2}}\right) C_{\nu}+\frac{2\alpha^{2}\varrho^{2}\lambda_{0}}{G(t)\delta^{2}}\right)N_{2}\right. \nonumber\\ &\left.-\frac{N_{3}}{\delta}C_{\nu}\right]\left(h\circ\mathcal{A}^{\frac{1}{2}}u\right)(t). \end{align*} |
Then, we choose
\begin{align*} \frac{G(t)}{4}N_{2}-1-\frac{\delta^{3}}{8\alpha^{2}\varrho^{2}\lambda_{0}} > 0. \end{align*} |
Next, we choose
\begin{align*} \frac{3\delta}{8}N_{3}-\frac{\delta}{4}-\left(\frac{(\varrho-l)^{2}\alpha^{2}}{2\varepsilon_{1}}+\frac{4\lambda_{0}g^{2}(0)\alpha^{2}}{G(t)}\right)N_{2} > 0. \end{align*} |
Now, as
\begin{align*} \nu C_{\nu} = \int_{0}^{\infty}\frac{\nu^{2}g(s)}{\nu g(s)-g'(s)}ds\rightarrow 0, \quad\quad as \quad\quad \nu\rightarrow 0. \end{align*} |
Hence, there is
\begin{align*} \nu C_{\nu} < \frac{1}{16\left(\frac{2\alpha}{\delta}+\left(\frac{\varepsilon_{1}\alpha^{2}\varrho^{2}\lambda_{0}}{4\delta^{2}}+\frac{3}{4\varepsilon_{1}}+1+\frac{\alpha\varrho}{\delta}+\frac{2\alpha^{4}\varrho^{2}\lambda_{0}}{G(t)\delta^{2}}\right)N_{2}+\frac{N_{3}}{\delta}\right)}. \end{align*} |
Now, let us choose
\begin{align*} \frac{\varepsilon\varrho}{4\delta}N-\frac{2\alpha^{2}\varrho^{2}\lambda_{0}}{G(t)\delta^{2}}N_{2} > 0 \quad\quad and \quad\quad \nu = \frac{\delta}{4\varepsilon\varrho N} < \nu_{0}, \end{align*} |
which means
\begin{align*} &\frac{\varepsilon\varrho}{2\delta}N-\frac{2\alpha^{2}\varrho^{2}\lambda_{0}}{G(t)\delta^{2}}N_{2}\nonumber\\ &-C_{\nu}\left(\frac{2\alpha}{\delta}+\left(\frac{\varepsilon_{1}\alpha^{2}\varrho^{2}\lambda_{0}}{4\delta^{2}}+\frac{3}{4\varepsilon_{1}} +1+\frac{\alpha\varrho}{\delta}+\frac{2\alpha^{4}\varrho^{2}\lambda_{0}}{G(t)\delta^{2}}\right)N_{2}+\frac{N_{3}}{\delta}\right) \nonumber\\ > &\frac{\varepsilon\varrho}{2\delta}N-\frac{2\alpha^{2}\varrho^{2}\lambda_{0}}{G(t)\delta^{2}}N_{2}-\frac{1}{16\nu} = \frac{\varepsilon\varrho}{4\delta}N-\frac{2\alpha^{2}\varrho^{2}\lambda_{0}}{G(t)\delta^{2}}N_{2} > 0 \end{align*} |
and
\begin{align*} &\alpha\left(\beta-\frac{\alpha\varrho}{\delta}\right)N-\frac{2\alpha\lambda_{0}}{\delta}\left(\beta-\frac{\alpha\varrho}{\delta}\right)^{2}-\left(\frac{2\left(\beta-\frac{\alpha\varrho}{\delta}\right)^{2}}{\varepsilon_{1}}+\frac{4g^{2}(0)}{G(t)}\right)N_{2} \nonumber\\ &-\frac{\delta^{3}}{4\varrho^{2}\lambda_{0}}-\frac{4\alpha^{2}\varrho^{2}\lambda_{0}}{\delta^{3}}N_{3}^{2}-\frac{2\beta^{2}\lambda_{0}}{\delta}N_{3} > 0, \nonumber\\ &\left(\alpha\left(\gamma-\frac{\delta}{\alpha}\right)-\frac{\delta g(0)}{2(\alpha-\varepsilon)\varrho}\right)N +\left(\frac{\delta g(t)}{2(\alpha-\varepsilon)\varrho}-\frac{\delta g(t)}{2\alpha\varrho}\right)N-\frac{2\alpha}{\delta}\left(\gamma-\frac{\delta}{\alpha}\right)^{2} \nonumber\\ &-\left(\frac{(\varrho-l)^{2}}{2\varepsilon_{1}}+\frac{1}{2\varepsilon_{1}}\left(\gamma-\frac{\delta}{\alpha}\right)^{2} +\frac{1}{2}\left(\frac{\alpha\varrho}{\delta}\right)^{2}\left(\gamma-\frac{\delta}{\alpha}\right)^{2}\right)N_{2}-\frac{2\gamma^{2}}{\delta}N_{3} > 0, \nonumber\\ &-\frac{\alpha g'(t)}{2}N-\frac{2\alpha(\varrho-l)^{2}}{\delta}-\left(\frac{\alpha^{2}\varrho^{2}(\varrho-l)^{2}\lambda_{0}\varepsilon_{1}}{2\delta^{2}}+\frac{(\varrho-l)^{2}}{2\varepsilon_{1}}+3(\varrho-l)^{2}\left(\frac{\alpha\varrho}{\delta}\right)^{2}\varepsilon_{1}\right. \nonumber\\ &\left.+\frac{(\varrho-l)^{2}}{2}\left(\frac{\alpha\varrho}{\delta}\right)^{2}\right)N_{2}-\frac{2(\varrho-l)^{2}}{\delta}N_{3} > 4(\varrho-l). \end{align*} |
So we arrive at, for positive constant
\begin{align*} \mathcal{L'}(t)\leq&-c\left[\|u_{tt}\|^{2}+\left\|\mathcal{A}^{\frac{1}{2}}u_{tt}\right\|^{2}+\left\|u_{ttt}+\alpha u_{tt}+\frac{\alpha\varrho}{\delta}u_{t}\right\|^{2} \right. \nonumber\\ &\left.+\left\|\mathcal{A}^{\frac{1}{2}}u_{tt}+\alpha\mathcal{A}^{\frac{1}{2}}u_{t}+\frac{\alpha\varrho}{\delta}u\right\|^{2}\right]-4(\varrho-l)\left\|\mathcal{A}^{\frac{1}{2}}u\right\|^{2}+\frac{1}{8}\left(g\circ\mathcal{A}^{\frac{1}{2}}u\right)(t). \end{align*} |
On the other hand, from Lemma 3.2, we find that
\begin{align*} &\left|\mathcal{L}(t)-NE(t)\right| \nonumber\\ \leq&\int_{\Omega}\left|u_{tt}+\alpha u_{t}+\frac{\alpha\varrho}{\delta}u\right|\left|u_{ttt}+\alpha u_{tt}+\frac{\alpha\varrho}{\delta}u_{t}\right|dx \nonumber\\ &+N_{2}\int_{\Omega}\left|u_{ttt}+\alpha u_{tt}+\frac{\alpha\varrho}{\delta}u_{t}\right|\int_{0}^{t}g(t-s) \nonumber\\ & \times\left|\left(u_{tt}+\alpha u_{t}+\frac{\alpha\varrho}{\delta}u\right)(t)-\frac{\alpha\varrho}{\delta} u(s)\right|dsdx \nonumber\\ &+N_{3}\int_{\Omega}\left|u_{ttt}+\alpha u_{tt}\right||u_{t}|dx+N_{3}\frac{\varrho}{2}\left\|\mathcal{A}^{\frac{1}{2}}u\right\|^{2} \nonumber\\ \leq&c E(t). \end{align*} |
Therefore, we can choose
In this section, we will give an estimate to the decay rate for the problem (1.7)-(1.8).
Proof of Theorem 2.2. Our proof starts with the observation that, for any
\begin{align*} &\int_{0}^{t_{0}}g(s)\int_{\Omega}\left|\mathcal{A}^{\frac{1}{2}}u(t)-\mathcal{A}^{\frac{1}{2}}u(t-s)\right|^{2}dxds\nonumber\\ \leq&-\frac{g(0)}{a}\int_{0}^{t_{0}}g'(s)\int_{\Omega}\left|\mathcal{A}^{\frac{1}{2}}u(t)-\mathcal{A}^{\frac{1}{2}}u(t-s)\right|^{2}dxds \nonumber\\ \leq&-\frac{g(0)}{a}\int_{0}^{t}g'(s)\int_{\Omega}\left|\mathcal{A}^{\frac{1}{2}}u(t)-\mathcal{A}^{\frac{1}{2}}u(t-s)\right|^{2}dxds \nonumber\\ \leq&-c E'(t), \end{align*} |
which are derived from (2.2) and Lemma 4.1 and can be used in (4.8).
Taking
\begin{align*} &\mathcal{L'}(t)\nonumber\\ \leq&-c\left[\|u_{tt}\|^{2}+\left\|\mathcal{A}^{\frac{1}{2}}u_{tt}\right\|^{2}+\left\|u_{ttt}+\alpha u_{tt}+\frac{\alpha\varrho}{\delta}u_{t}\right\|^{2}+\left\|\mathcal{A}^{\frac{1}{2}}u_{tt}+\alpha\mathcal{A}^{\frac{1}{2}}u_{t}+\frac{\alpha\varrho}{\delta}u\right\|^{2}\right] \nonumber\\ &-4(\varrho-l)\left\|\mathcal{A}^{\frac{1}{2}}u\right\|^{2}+\frac{1}{8}\left(g\circ\mathcal{A}^{\frac{1}{2}}u\right)(t) \nonumber\\ \leq&-mE(t)+c\left(g\circ\mathcal{A}^{\frac{1}{2}}u\right)(t) \nonumber\\ \leq&-mE(t)-cE'(t)+c\int_{t_{0}}^{t}g(s)\int_{\Omega}\left|\mathcal{A}^{\frac{1}{2}}u(t)-\mathcal{A}^{\frac{1}{2}}u(t-s)\right|^{2}dxds, \end{align*} |
where
\begin{align} \mathcal{F'}(t)& = \mathcal{L'}(t)+cE'(t)\\ & \leq-mE(t)+c\int_{t_{0}}^{t}g(s)\int_{\Omega}\left|\mathcal{A}^{\frac{1}{2}}u(t)-\mathcal{A}^{\frac{1}{2}}u(t-s)\right|^{2}dxds. \end{align} | (5.1) |
We consider the following two cases relying on the ideas presented in [31].
(ⅰ)
We multiply (5.1) by
\begin{align*} \xi(t)\mathcal{F'}(t)\leq&-m\xi(t)E(t)+c\xi(t)\int_{t_{0}}^{t}g(s)\int_{\Omega}\left|\mathcal{A}^{\frac{1}{2}}u(t)-\mathcal{A}^{\frac{1}{2}}u(t-s)\right|^{2}dxds \nonumber\\ \leq&-m\xi(t)E(t)+c\int_{t_{0}}^{t}\xi(s)g(s)\int_{\Omega}\left|\mathcal{A}^{\frac{1}{2}}u(t)-\mathcal{A}^{\frac{1}{2}}u(t-s)\right|^{2}dxds \nonumber\\ \leq&-m\xi(t)E(t)-c\int_{t_{0}}^{t}g'(s)\int_{\Omega}\left|\mathcal{A}^{\frac{1}{2}}u(t)-\mathcal{A}^{\frac{1}{2}}u(t-s)\right|^{2}dxds \nonumber\\ \leq&-m\xi(t)E(t)-c\int_{0}^{t}g'(s)\int_{\Omega}\left|\mathcal{A}^{\frac{1}{2}}u(t)-\mathcal{A}^{\frac{1}{2}}u(t-s)\right|^{2}dxds \nonumber\\ \leq&-m\xi(t)E(t)-cE'(t). \end{align*} |
Therefore,
\begin{align*} \xi(t)\mathcal{F'}(t)+cE'(t)\leq-m\xi(t)E(t). \end{align*} |
As
\begin{align*} \xi(t)\mathcal{F}(t)+cE(t)\sim E(t) \end{align*} |
and
\begin{align*} (\xi\mathcal{F}+cE)'(t)\leq -m\xi(t)E(t), \quad\quad \forall\ t\geq t_{0}. \end{align*} |
It follows immediately that
\begin{align*} E'(t)\leq -m\xi(t)E(t), \quad\quad \forall\ t\geq t_{0}. \end{align*} |
We may now integrate over
\begin{align*} E(t)\leq k_{2}\exp\left(-k_{1}\int_{t_{0}}^{t}\xi(s)ds\right), \quad\quad \forall\ t\geq t_{0}. \end{align*} |
By the continuity of
\begin{align*} E(t)\leq k_{2}\exp\left(-k_{1}\int_{0}^{t}\xi(s)ds\right), \quad\quad \forall\ t > 0. \end{align*} |
(ⅱ)
First, we define the functional
\begin{align*} L(t) = \mathcal{L}(t)+F_{4}(t). \end{align*} |
Obviously,
\begin{align*} L'(t) = &\mathcal{L}'(t)+F_{4}'(t) \nonumber\\ \leq&-c\left[\|u_{tt}\|^{2}+\left\|\mathcal{A}^{\frac{1}{2}}u_{tt}\right\|^{2}+\left\|u_{ttt}+\alpha u_{tt}+\frac{\alpha\varrho}{\delta}u_{t}\right\|^{2} \right. \nonumber\\ &\left.+\left\|\mathcal{A}^{\frac{1}{2}}u_{tt}+\alpha\mathcal{A}^{\frac{1}{2}}u_{t}+\frac{\alpha\varrho}{\delta}u\right\|^{2}\right]-(\varrho-l)\left\|\mathcal{A}^{\frac{1}{2}}u\right\|^{2}-\frac{3}{8}\left(g\circ\mathcal{A}^{\frac{1}{2}}u\right)(t) \nonumber\\ \leq& -b E(t). \end{align*} |
Therefore, integrating the above inequality over
\begin{align*} -L(t_{0})\leq L(t)-L(t_{0})\leq-b\int_{t_{0}}^{t}E(s){\rm d}s. \end{align*} |
It is sufficient to show that
\begin{align} \int_{0}^{\infty}E(s){\rm d}s < \infty \end{align} | (5.2) |
and
\begin{align*} E(t)\leq \frac{c}{t-t_{0}}, \quad\quad\quad \forall t > t_{0}. \end{align*} |
Now, we define a functional
\begin{align*} \lambda(t): = -\int_{t_{0}}^{t}g'(s)\left\|\mathcal{A}^{\frac{1}{2}}u(t)-\mathcal{A}^{\frac{1}{2}}u(t-s)\right\|^{2}{\rm d}s. \end{align*} |
Clearly, we have
\begin{align} \lambda(t)\leq&-\int_{0}^{t}g'(s)\left\| \mathcal{A}^{\frac{1}{2}}u(t)-\mathcal{A}^{\frac{1}{2}}u(t-s)\right\|^{2}ds \\ \leq&-cE'(t), \quad\quad\quad \forall\ t\geq t_{0}. \end{align} | (5.3) |
After that, we define another functional
\begin{align*} I(t): = q\int_{t_{0}}^{t}\left\|\mathcal{A}^{\frac{1}{2}}u(t)-\mathcal{A}^{\frac{1}{2}}u(t-s)\right\|^{2}{\rm d}s. \end{align*} |
Now, the following inequality holds under Lemma 4.1 and (5.2) that
\begin{align} \int_{t_{0}}^{t}\left\|\mathcal{A}^{\frac{1}{2}}u(t)-\mathcal{A}^{\frac{1}{2}}u(t-s)\right\|^{2}{\rm d}s \leq & 2\int_{t_{0}}^{t}\left(\left\|\mathcal{A}^{\frac{1}{2}}u(t)\right\|^{2}+\left\|\mathcal{A}^{\frac{1}{2}}u(t-s)\right\|^{2}\right)ds \\ \leq& 4\int_{t_{0}}^{t}\left(E(t)+E(t-s)\right)ds \\ \leq& 8\int_{t_{0}}^{t}E(0)ds \\ < & \infty. \end{align} | (5.4) |
Then (5.4) allows for a constant
\begin{align} 0 < I(t) < 1; \end{align} | (5.5) |
otherwise we get an exponential decay from (5.1).
Moreover, recalling that
\begin{align*} M(\theta x)\leq \theta M(x), \quad\quad\quad for \quad\quad 0\leq \theta\leq 1 \quad\quad and \quad\quad x\in(0,r]. \end{align*} |
From assumptions
\begin{align*} \lambda(t) = &-\int_{t_{0}}^{t}g'(s)\left\|\mathcal{A}^{\frac{1}{2}}u(t)-\mathcal{A}^{\frac{1}{2}}u(t-s)\right\|^{2}{\rm d}s \nonumber\\ = &\frac{1}{qI(t)}\int_{t_{0}}^{t}I(t)(-g'(s))q\left\|\mathcal{A}^{\frac{1}{2}}u(t)-\mathcal{A}^{\frac{1}{2}}u(t-s)\right\|^{2}{\rm d}s \nonumber\\ \geq& \frac{1}{qI(t)}\int_{t_{0}}^{t}I(t)\xi(s)M(g(s))q\left\|\mathcal{A}^{\frac{1}{2}}u(t)-\mathcal{A}^{\frac{1}{2}}u(t-s)\right\|^{2}{\rm d}s \nonumber\\ \geq&\frac{\xi(t)}{qI(t)}\int_{t_{0}}^{t}M(I(t)g(s))q\left\|\mathcal{A}^{\frac{1}{2}}u(t)-\mathcal{A}^{\frac{1}{2}}u(t-s)\right\|^{2}{\rm d}s \nonumber\\ \geq&\frac{\xi(t)}{q}M\left(\frac{1}{I(t)}\int_{t_{0}}^{t}I(t)g(s)q\left\|\mathcal{A}^{\frac{1}{2}}u(t)-\mathcal{A}^{\frac{1}{2}}u(t-s)\right\|^{2}{\rm d}s\right) \nonumber\\ = &\frac{\xi(t)}{q}M\left(q\int_{t_{0}}^{t}g(s)\left\|\mathcal{A}^{\frac{1}{2}}u(t)-\mathcal{A}^{\frac{1}{2}}u(t-s)\right\|^{2}{\rm d}s\right). \end{align*} |
According to
\begin{align*} \lambda(t)\geq&\frac{\xi(t)}{q}\overline{M}\left(q\int_{t_{0}}^{t}g(s)\left\|\mathcal{A}^{\frac{1}{2}}u(t)-\mathcal{A}^{\frac{1}{2}}u(t-s)\right\|^{2}{\rm d}s\right). \end{align*} |
In this way,
\begin{align*} \int^{t}_{t_{0}}g(s)\left\|\mathcal{A}^{\frac{1}{2}}u(t)-\mathcal{A}^{\frac{1}{2}}u(t-s)\right\|^{2}{\rm d}s\leq\frac{1}{q}\overline{M}^{-1}\left(\frac{q\lambda(t)}{\xi(t)}\right) \end{align*} |
and (5.1) becomes
\begin{align} \mathcal{F'}(t) \leq&-mE(t)+c\int_{t_{0}}^{t}g(s)\int_{\Omega}\left|\mathcal{A}^{\frac{1}{2}}u(t)-\mathcal{A}^{\frac{1}{2}}u(t-s)\right|^{2}dxds \\ \leq&-mE(t)+c\overline{M}^{-1}\left(\frac{q\lambda(t)}{\xi(t)}\right), \quad\quad\quad \forall\ t\geq t_{0}. \end{align} | (5.6) |
Let
\begin{align*} \mathcal{F}_{1}(t): = \overline{M}'\left(\varepsilon_{0}\frac{E(t)}{E(0)}\right)\mathcal{F}(t)+E(t), \quad\quad\quad \forall\ t\geq 0. \end{align*} |
Then, recalling that
\begin{align} \mathcal{F}'_{1}(t) \leq& -mE(t)\overline{M}'\left(\varepsilon_{0}\frac{E(t)}{E(0)}\right)+c\overline{M}'\left(\varepsilon_{0}\frac{E(t)}{E(0)}\right)\overline{M}^{-1}\left(\frac{q\lambda(t)}{\xi(t)}\right)+E'(t). \end{align} | (5.7) |
Taking account of Lemma 2.3, we obtain
\begin{align} &\overline{M}'\left(\varepsilon_{0}\frac{E(t)}{E(0)}\right)\overline{M}^{-1}\left(\frac{q\lambda(t)}{\xi(t)}\right) \\ \leq&\overline{M}^{*}\left(\overline{H}'\left(\varepsilon_{0}\frac{E(t)}{E(0)}\right)\right)+\overline{M}\left(\overline{M}^{-1}\left(\frac{q\lambda(t)}{\xi(t)}\right)\right) \\ = &\overline{M}^{*}\left(\overline{M}'\left(\varepsilon_{0}\frac{E(t)}{E(0)}\right)\right)+\frac{q\lambda(t)}{\xi(t)} \end{align} | (5.8) |
where
\begin{align} &\overline{M}^{*}\left(\overline{M}'\left(\varepsilon_{0}\frac{E(t)}{E(0)}\right)\right) \\ = &\overline{M}'\left(\varepsilon_{0}\frac{E(t)}{E(0)}\right)\left(\overline{M}'\right)^{-1}\left(\overline{M}'\left(\varepsilon_{0}\frac{E(t)}{E(0)}\right)\right)-\overline{M}\left[\left(\overline{M}'\right)^{-1}\left(\overline{M}'\left(\varepsilon_{0}\frac{E(t)}{E(0)}\right)\right)\right] \\ = &\varepsilon_{0}\frac{E(t)}{E(0)}\overline{M}'\left(\varepsilon_{0}\frac{E(t)}{E(0)}\right)-\overline{M}\left(\varepsilon_{0}\frac{E(t)}{E(0)}\right) \\ \leq&\varepsilon_{0}\frac{E(t)}{E(0)}\overline{M}'\left(\varepsilon_{0}\frac{E(t)}{E(0)}\right). \end{align} | (5.9) |
So, combining (5.7), (5.8) and (5.9), we obtain
\begin{align*} \mathcal{F}'_{1}(t)\leq-(mE(0)-c\varepsilon_{0})\frac{E(t)}{E(0)}\overline{M}'\left(\varepsilon_{0}\frac{E(t)}{E(0)}\right)+c\frac{q\lambda(t)}{\xi(t)}+E'(t). \end{align*} |
From this, we multiply the above inequality by
\begin{align*} \xi(t)\mathcal{F}'_{1}(t)\leq-(mE(0)-c\varepsilon_{0})\xi(t)\frac{E(t)}{E(0)}\overline{M}'\left(\varepsilon_{0}\frac{E(t)}{E(0)}\right)+cq\lambda(t)+\xi(t)E'(t). \end{align*} |
Then, using the fact that, as
\begin{align*} \xi(t)\mathcal{F}'_{1}(t) \leq&-(mE(0)-c\varepsilon_{0})\xi(t)\frac{E(t)}{E(0)}M'\left(\varepsilon_{0}\frac{E(t)}{E(0)}\right)-cE'(t). \end{align*} |
Consequently, defining
\begin{align} \mathcal{F}_{2}(t)\sim E(t), \end{align} | (5.10) |
and with a suitable choice of
\begin{align} \mathcal{F}'_{2}(t)\leq-k\xi(t)\left(\frac{E(t)}{E(0)}\right)M'\left(\varepsilon_{0}\frac{E(t)}{E(0)}\right). \end{align} | (5.11) |
Define
\begin{align*} R(t) = \frac{\lambda_{1}\mathcal{F}_{2}(t)}{E(0)}, \quad\quad\lambda_{1} > 0\quad\quad\quad and \quad\quad\quad M_{2}(t) = tM'(\varepsilon_{0} t). \end{align*} |
Moreover, it suffices to show that
\begin{align} \mathcal{F}'_{2}(t) \leq -k\xi(t)M_{2}\left(\frac{E(t)}{E(0)}\right). \end{align} | (5.12) |
According to (5.10) and (5.12), there exist
\begin{align} \lambda_{2}R(t)\leq E(t)\leq\lambda_{3}R(t). \end{align} | (5.13) |
Then, it follows that there exists
\begin{align} k_{1}\xi(t)\leq-\frac{R'(t)}{M_{2}(R(t))}, \quad\quad\quad \forall\ t\geq t_{0}. \end{align} | (5.14) |
Next, we define
\begin{align*} M_{1}(t): = \int_{t}^{r}\frac{1}{sM'(s)}{\rm d}s. \end{align*} |
And based on the properties of
Now, we integrate (5.14) over
\begin{align*} -\int_{t_{0}}^{t}\frac{R'(s)}{M_{2}(R(s))}{\rm d}s\geq k_{1}\int_{t_{0}}^{t}\xi(s){\rm d}s \end{align*} |
so
\begin{align*} k_{1}\int_{t_{0}}^{t}\xi(s){\rm d}s\leq M_{1}\left(\varepsilon_{0}R(t)\right)-M_{1}\left(\varepsilon_{0}R(t_{0})\right), \end{align*} |
which implies that
\begin{align*} M_{1}\left(\varepsilon_{0}R(t)\right)\geq k_{1}\int_{t_{0}}^{t}\xi(s){\rm d}s. \end{align*} |
It is easy to obtain that
\begin{align} R(t)\leq\frac{1}{\varepsilon_{0}}M^{-1}_{1}\left(k_{1}\int_{t_{0}}^{t}\xi(s){\rm d}s\right), \quad\quad\quad \forall\ t\geq t_{0}. \end{align} | (5.15) |
A combining of (5.13) and (5.15) gives the proof.
The authors are grateful to the anonymous referees and the editor for their useful remarks and comments.
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