Citation: Wenyuan Xie, Jason Wei Jun Low, Arunmozhiarasi Armugam, Kandiah Jeyaseelan, Yen Wah Tong. Regulation of Aquaporin Z osmotic permeability in ABA tri-block copolymer[J]. AIMS Biophysics, 2015, 2(3): 381-397. doi: 10.3934/biophy.2015.3.381
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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) < \frac{2\alpha\varrho}{\delta}(\alpha\gamma-\delta),\ \ \ \ \varrho-\int_{0}^{+\infty}g(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
$ ε1=αδ2N2[λ0α2(ϱ−l)2+2(δ2+α2ϱ2)+6α2(ϱ−l)2]andε2=1N3. $ |
The above choice yields
$ L′(t)≤−[α(β−αϱδ)N−2αλ0δ(β−αϱδ)2−(2(β−αϱδ)2ε1+4g2(0)G(t))N2−δ34ϱ2λ0−4α2ϱ2λ0δ3N23−2β2λ0δN3]‖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)−[3δ8N3−δ4−((ϱ−l)2α22ε1+4λ0g2(0)α2G(t))N2]‖A12ut‖2+εϱν2δN(g∘A12u)(t)−(G(t)4N2−1−δ38α2ϱ2λ0)‖uttt+αutt+αϱδut‖2−δ4α‖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). $ |
Then, we choose
$ G(t)4N2−1−δ38α2ϱ2λ0>0. $ |
Next, we choose
$ 3δ8N3−δ4−((ϱ−l)2α22ε1+4λ0g2(0)α2G(t))N2>0. $ |
Now, as
$ νCν=∫∞0ν2g(s)νg(s)−g′(s)ds→0,asν→0. $ |
Hence, there is
$ νCν<116(2αδ+(ε1α2ϱ2λ04δ2+34ε1+1+αϱδ+2α4ϱ2λ0G(t)δ2)N2+N3δ). $ |
Now, let us choose
$ εϱ4δN−2α2ϱ2λ0G(t)δ2N2>0andν=δ4εϱN<ν0, $ |
which means
$ εϱ2δN−2α2ϱ2λ0G(t)δ2N2−Cν(2αδ+(ε1α2ϱ2λ04δ2+34ε1+1+αϱδ+2α4ϱ2λ0G(t)δ2)N2+N3δ)>εϱ2δN−2α2ϱ2λ0G(t)δ2N2−116ν=εϱ4δN−2α2ϱ2λ0G(t)δ2N2>0 $ |
and
$ α(β−αϱδ)N−2αλ0δ(β−αϱδ)2−(2(β−αϱδ)2ε1+4g2(0)G(t))N2−δ34ϱ2λ0−4α2ϱ2λ0δ3N23−2β2λ0δN3>0,(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>0,−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>4(ϱ−l). $ |
So we arrive at, for positive constant
$ L′(t)≤−c[‖utt‖2+‖A12utt‖2+‖uttt+αutt+αϱδut‖2+‖A12utt+αA12ut+αϱδu‖2]−4(ϱ−l)‖A12u‖2+18(g∘A12u)(t). $ |
On the other hand, from Lemma 3.2, we find that
$ |L(t)−NE(t)|≤∫Ω|utt+αut+αϱδu||uttt+αutt+αϱδut|dx+N2∫Ω|uttt+αutt+αϱδut|∫t0g(t−s)×|(utt+αut+αϱδu)(t)−αϱδu(s)|dsdx+N3∫Ω|uttt+αutt||ut|dx+N3ϱ2‖A12u‖2≤cE(t). $ |
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
$ ∫t00g(s)∫Ω|A12u(t)−A12u(t−s)|2dxds≤−g(0)a∫t00g′(s)∫Ω|A12u(t)−A12u(t−s)|2dxds≤−g(0)a∫t0g′(s)∫Ω|A12u(t)−A12u(t−s)|2dxds≤−cE′(t), $ |
which are derived from (2.2) and Lemma 4.1 and can be used in (4.8).
Taking
$ L′(t)≤−c[‖utt‖2+‖A12utt‖2+‖uttt+αutt+αϱδut‖2+‖A12utt+αA12ut+αϱδu‖2]−4(ϱ−l)‖A12u‖2+18(g∘A12u)(t)≤−mE(t)+c(g∘A12u)(t)≤−mE(t)−cE′(t)+c∫tt0g(s)∫Ω|A12u(t)−A12u(t−s)|2dxds, $ |
where
$ F′(t)=L′(t)+cE′(t)≤−mE(t)+c∫tt0g(s)∫Ω|A12u(t)−A12u(t−s)|2dxds. $ | (5.1) |
We consider the following two cases relying on the ideas presented in [31].
(ⅰ)
We multiply (5.1) by
$ ξ(t)F′(t)≤−mξ(t)E(t)+cξ(t)∫tt0g(s)∫Ω|A12u(t)−A12u(t−s)|2dxds≤−mξ(t)E(t)+c∫tt0ξ(s)g(s)∫Ω|A12u(t)−A12u(t−s)|2dxds≤−mξ(t)E(t)−c∫tt0g′(s)∫Ω|A12u(t)−A12u(t−s)|2dxds≤−mξ(t)E(t)−c∫t0g′(s)∫Ω|A12u(t)−A12u(t−s)|2dxds≤−mξ(t)E(t)−cE′(t). $ |
Therefore,
$ ξ(t)F′(t)+cE′(t)≤−mξ(t)E(t). $ |
As
$ ξ(t)F(t)+cE(t)∼E(t) $ |
and
$ (ξF+cE)′(t)≤−mξ(t)E(t),∀ t≥t0. $ |
It follows immediately that
$ E′(t)≤−mξ(t)E(t),∀ t≥t0. $ |
We may now integrate over
$ E(t)≤k2exp(−k1∫tt0ξ(s)ds),∀ t≥t0. $ |
By the continuity of
$ E(t)≤k2exp(−k1∫t0ξ(s)ds),∀ t>0. $ |
(ⅱ)
First, we define the functional
$ L(t)=L(t)+F4(t). $ |
Obviously,
$ L′(t)=L′(t)+F′4(t)≤−c[‖utt‖2+‖A12utt‖2+‖uttt+αutt+αϱδut‖2+‖A12utt+αA12ut+αϱδu‖2]−(ϱ−l)‖A12u‖2−38(g∘A12u)(t)≤−bE(t). $ |
Therefore, integrating the above inequality over
$ −L(t0)≤L(t)−L(t0)≤−b∫tt0E(s)ds. $ |
It is sufficient to show that
$ ∫∞0E(s)ds<∞ $ | (5.2) |
and
$ E(t)≤ct−t0,∀t>t0. $ |
Now, we define a functional
$ λ(t):=−∫tt0g′(s)‖A12u(t)−A12u(t−s)‖2ds. $ |
Clearly, we have
$ λ(t)≤−∫t0g′(s)‖A12u(t)−A12u(t−s)‖2ds≤−cE′(t),∀ t≥t0. $ | (5.3) |
After that, we define another functional
$ I(t):=q∫tt0‖A12u(t)−A12u(t−s)‖2ds. $ |
Now, the following inequality holds under Lemma 4.1 and (5.2) that
$ ∫tt0‖A12u(t)−A12u(t−s)‖2ds≤2∫tt0(‖A12u(t)‖2+‖A12u(t−s)‖2)ds≤4∫tt0(E(t)+E(t−s))ds≤8∫tt0E(0)ds<∞. $ | (5.4) |
Then (5.4) allows for a constant
$ 0<I(t)<1; $ | (5.5) |
otherwise we get an exponential decay from (5.1).
Moreover, recalling that
$ M(θx)≤θM(x),for0≤θ≤1andx∈(0,r]. $ |
From assumptions
$ λ(t)=−∫tt0g′(s)‖A12u(t)−A12u(t−s)‖2ds=1qI(t)∫tt0I(t)(−g′(s))q‖A12u(t)−A12u(t−s)‖2ds≥1qI(t)∫tt0I(t)ξ(s)M(g(s))q‖A12u(t)−A12u(t−s)‖2ds≥ξ(t)qI(t)∫tt0M(I(t)g(s))q‖A12u(t)−A12u(t−s)‖2ds≥ξ(t)qM(1I(t)∫tt0I(t)g(s)q‖A12u(t)−A12u(t−s)‖2ds)=ξ(t)qM(q∫tt0g(s)‖A12u(t)−A12u(t−s)‖2ds). $ |
According to
$ λ(t)≥ξ(t)q¯M(q∫tt0g(s)‖A12u(t)−A12u(t−s)‖2ds). $ |
In this way,
$ ∫tt0g(s)‖A12u(t)−A12u(t−s)‖2ds≤1q¯M−1(qλ(t)ξ(t)) $ |
and (5.1) becomes
$ F′(t)≤−mE(t)+c∫tt0g(s)∫Ω|A12u(t)−A12u(t−s)|2dxds≤−mE(t)+c¯M−1(qλ(t)ξ(t)),∀ t≥t0. $ | (5.6) |
Let
$ F1(t):=¯M′(ε0E(t)E(0))F(t)+E(t),∀ t≥0. $ |
Then, recalling that
$ F′1(t)≤−mE(t)¯M′(ε0E(t)E(0))+c¯M′(ε0E(t)E(0))¯M−1(qλ(t)ξ(t))+E′(t). $ | (5.7) |
Taking account of Lemma 2.3, we obtain
$ ¯M′(ε0E(t)E(0))¯M−1(qλ(t)ξ(t))≤¯M∗(¯H′(ε0E(t)E(0)))+¯M(¯M−1(qλ(t)ξ(t)))=¯M∗(¯M′(ε0E(t)E(0)))+qλ(t)ξ(t) $ | (5.8) |
where
$ ¯M∗(¯M′(ε0E(t)E(0)))=¯M′(ε0E(t)E(0))(¯M′)−1(¯M′(ε0E(t)E(0)))−¯M[(¯M′)−1(¯M′(ε0E(t)E(0)))]=ε0E(t)E(0)¯M′(ε0E(t)E(0))−¯M(ε0E(t)E(0))≤ε0E(t)E(0)¯M′(ε0E(t)E(0)). $ | (5.9) |
So, combining (5.7), (5.8) and (5.9), we obtain
$ F′1(t)≤−(mE(0)−cε0)E(t)E(0)¯M′(ε0E(t)E(0))+cqλ(t)ξ(t)+E′(t). $ |
From this, we multiply the above inequality by
$ ξ(t)F′1(t)≤−(mE(0)−cε0)ξ(t)E(t)E(0)¯M′(ε0E(t)E(0))+cqλ(t)+ξ(t)E′(t). $ |
Then, using the fact that, as
$ ξ(t)F′1(t)≤−(mE(0)−cε0)ξ(t)E(t)E(0)M′(ε0E(t)E(0))−cE′(t). $ |
Consequently, defining
$ F2(t)∼E(t), $ | (5.10) |
and with a suitable choice of
$ F′2(t)≤−kξ(t)(E(t)E(0))M′(ε0E(t)E(0)). $ | (5.11) |
Define
$ R(t)=λ1F2(t)E(0),λ1>0andM2(t)=tM′(ε0t). $ |
Moreover, it suffices to show that
$ F′2(t)≤−kξ(t)M2(E(t)E(0)). $ | (5.12) |
According to (5.10) and (5.12), there exist
$ λ2R(t)≤E(t)≤λ3R(t). $ | (5.13) |
Then, it follows that there exists
$ k1ξ(t)≤−R′(t)M2(R(t)),∀ t≥t0. $ | (5.14) |
Next, we define
$ M1(t):=∫rt1sM′(s)ds. $ |
And based on the properties of
Now, we integrate (5.14) over
$ −∫tt0R′(s)M2(R(s))ds≥k1∫tt0ξ(s)ds $ |
so
$ k1∫tt0ξ(s)ds≤M1(ε0R(t))−M1(ε0R(t0)), $ |
which implies that
$ M1(ε0R(t))≥k1∫tt0ξ(s)ds. $ |
It is easy to obtain that
$ R(t)≤1ε0M−11(k1∫tt0ξ(s)ds),∀ t≥t0. $ | (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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