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Adel M. Al-Mahdi Mohammad M. Al-Gharabli Mohammad Kafini Department of Mathematics and Statistics, King Fahd University of Petroleum and Minerals, Dhahran 31261, Saudi Arabia. Shadi Al-Omari

Stability result for a viscoelastic wave equation in the presence of finite and infinite memories

Adel M. Al-Mahdi Mohammad M. Al-Gharabli Mohammad Kafini Department of Mathematics and Statistics, King Fahd University of Petroleum and Minerals, Dhahran 31261, Saudi Arabia. Shadi Al-Omari
Abstract

In this paper, we are concerned with the following viscoelastic wave equation

uttu+0tg1(ts)𝑑iv(a1(x)u(s))𝑑s+0+g2(s)𝑑iv(a2(x)u(ts))𝑑s=0,u_{tt}-\nabla u+\int_{0}^{t}g_{1}(t-s)~{}div(a_{1}(x)\nabla u(s))~{}ds+\int_{0}^{+\infty}g_{2}(s)~{}div(a_{2}(x)\nabla u(t-s))~{}ds=0,

in a bounded domain Ω\Omega. Under suitable conditions on a1a_{1} and a2a_{2} and for a wide class of relaxation functions g1g_{1} and g2g_{2}. We establish a general decay result. The proof is based on the multiplier method and makes use of convex functions and some inequalities. More specifically, we remove the constraint imposed on the boundedness condition on the initial data u0\nabla u_{0}. This study generalizes and improves previous literature outcomes.

Keywords:  Stability, Finite and Infinite Memories, Relaxation Functions, Multiplier Method, Convexity.

AMS Classification.35B40,74D99,75D05,93D15,93D2035\text{B}40,74\text{D}99,75\text{D}05,93\text{D}15,93\text{D}20.

1 Introduction

Let us consider an nn-dimensional body occupies with a bounded open set Ωn(n1)\Omega\subseteq\mathbb{R}^{n}(n\geq 1) with smooth boundary Ω\partial\Omega. Let xu(x,t)x\rightarrow u(x,t) be the position of the material particle xx at time tt. So that, xΩ,t>0\forall x\in\Omega,\forall t>0, the corresponding motion equation is

uttu+0tg1(ts)𝑑iv(a1(x)u(s))𝑑s+0+g2(s)𝑑iv(a2(x)u(ts))𝑑s=0,u(x,t)=0,u(x,t)=u0(x,t),ut(x,0)=u1(x).\begin{array}[]{ll}u_{tt}-\nabla u+\int_{0}^{t}g_{1}(t-s)div(a_{1}(x)\nabla u(s))ds+\int_{0}^{+\infty}g_{2}(s)div(a_{2}(x)\nabla u(t-s))ds=0,\\ u(x,t)=0,\\ u(x,-t)=u_{0}(x,t),u_{t}(x,0)=u_{1}(x).\end{array} (1.1)

The functions g1g_{1} and g2g_{2} are called the relaxation (kernels) functions and they are positive non-increasing and defined on +\mathbb{R}^{+}. The functions a1a_{1} and a2a_{2} are essentially bounded non-negative defined on Ω\Omega. Here, u0u_{0} and u1u_{1} are the given initial data. This model of materials consisting of an elastic part (without memory) and a viscoelastic, part, where the dissipation given by the memory is effective.
In this paper, we are concerned with the above viscoelastic wave problem (1.1) and mainly interested in the asymptotic behavior of the solution uu when tt tends to infinity. In fact, We prove that the solutions of the corresponding viscoelastic model decay to zero and no matter how small is the viscoelastic part of the material. Note that the above model is dissipative, and the dissipation is given by the memory term only and the memory is effective only in a part of the body. For materials with memory the stress depends not only on the present values but also on the entire temporal history of the motion. Therefore, we have also to prescribe the history of uu before 0. Here, we assume that u(x,t)=u0(x,t),xΩ,t>0.u(x,-t)=u_{0}(x,t),\forall x\in\Omega,\forall t>0. Let us mention some other papers related to the problems we address. We start our literature review with the pioneer work of Dafermos [1], in 1970, where the author discussed a certain one-dimensional viscoelastic problem, established some existence results, and then proved that, for smooth monotone decreasing relaxation functions, the solutions go to zero as tt goes to infinity. However, no rate of decay has been specified. In Dafermos [2], a similar result, under a ity condition on the kernel, has been established. After that a great deal of attention has been devoted to the study of viscoelastic problems and many existence and long-time behavior results have been established. Hrusa [3] considered a one-dimensional nonlinear viscoelastic equation of the form

uttcuxx+0tm(ts)(ψ(ux(x,s)))x𝑑s=f(x,t)u_{tt}-cu_{xx}+\int_{0}^{t}m(t-s)\left(\psi(u_{x}(x,s))\right)_{x}~{}ds=f(x,t) (1.2)

and proved several global existence results for large data. He also proved an exponential decay for strong solutions when m(s)=esm(s)=e^{-s} and ψ\psi satisfies certain conditions. Dassios and Zafiropoulos [4] studied a viscoelastic problem in 3\mathbb{R}^{3} and proved a polynomial decay results for exponentially decaying kernels. After that, a very important contribution by Rivera was introduced. In 1994, Rivera [5] considered equations for linear isotropic homogeneous viscoelastic solids of integral type which occupy a bounded domain or the whole space n\mathbb{R}^{n}. in the bounded domain case, and for exponentially decaying memory kernel and regular solutions, he showed that the sum of the first and the second energy decays exponentially. For the whole-space case and for exponentially decaying memory kernels, he showed that the rate of decay of energy is of algebraic type and depends on the regularity of the solution. this result was later generalized to a situation, where the kernel is decaying algebraically but not exponentially by Cabanillas and Rivera [6]. In the paper, the authors considered the case of bounded domains as well as the case when the material is occupying the entire space and showed that the decay of solutions is also algebraic, at a rate which can be determined by the rate of decay of the relaxation function. This latter result was later improved by Baretto et al. [7], where equations related to linear viscoelastic plates were treated. Precisely, they showed that the solution energy decays at the same rate of the relaxation function. For partially viscoelastic material, Rivera and Salvatierra [8] showed that the energy decays exponentially, provide the relaxation function decays in a similar fashion and the dissipation is acting on a part of the domain near to the boundary. See also, in this direction, the work of Rivera and Oquendo [9]. The uniform decay of solutions for the viscoelastic wave equation

uttk0u+0t𝑑iv[a(x)g(tτ)u(τ)]𝑑τ+b(x)h(ut)+f(u)=0,u_{tt}-k_{0}\nabla u+\int_{0}^{t}div\left[a(x)g(t-\tau)\nabla u(\tau)\right]d\tau+b(x)h(u_{t})+f(u)=0, (1.3)

was investigated by Cavalcanti and Oquendo [10] where they considered the condition a(x)+b(x)δ>0a(x)+b(x)\geq\delta>0. They established exponential and polynomial stability results based on some conditions on gg and the linearity of the function hh. After that, Guesmiaa and Messaoudib [11] extended the work of [10] and they establish a general decay result for (1.1) under the same conditions on a1a_{1} and a2a_{2} used in [10] and for some other conditions for the relaxation functions g1g_{1} and g2g_{2}, from which the usual exponential and polynomial decay rates are only special cases. More precisely, they used the conditions g1(t)ξ(t)g1(t),t0,g_{1}^{\prime}(t)\leq-\xi(t)g_{1}(t),\hskip 10.84006pt\forall t\geq 0, and

0+g2(s)G1(g2(s))𝑑s+sups+g2(s)G1(g2(s))<+,\int_{0}^{+\infty}\frac{g_{2}(s)}{G^{-1}(-g_{2}^{\prime}(s))}ds+\sup_{s\in\mathbb{R}_{+}}\frac{g_{2}(s)}{G^{-1}(-g_{2}^{\prime}(s))}<+\infty, (1.4)

such that

G(0)=G(0)=0andlimt+G(t)=+,G(0)=G^{\prime}(0)=0\hskip 3.61371pt\text{and}\hskip 3.61371pt\lim_{t\rightarrow+\infty}{G^{\prime}(t)}=+\infty, (1.5)

where G:++G:\mathbb{R}_{+}\to\mathbb{R}_{+} is an increasing strictly convex function.
In the present work, we extend the works of [10] and [11]. In fact, we consider (1.1) and under the same conditions on a1a_{1} and a2a_{2} used in [10] and for a large class of the relaxation functions, we prove a general decay result. More precisely, we assume that the relaxations functions g1g_{1} and g2g_{2} are satisfying

gi(t)ξ(t)Hi(gi(t)),t0.g_{i}^{\prime}(t)\leq-\xi(t)H_{i}(g_{i}(t)),\hskip 10.84006pt\forall t\geq 0.

In fact, our result allows a large class of relaxation functions and improves the decay rates in some earlier papers. The paper is organized as follows. In Section 2, we present some material needed for our work. Some essential lemmas are presented and established in Section 3. Section 4 contains the statement and the proof of our main result. We end our paper by giving some illustrating examples in Sections 5.

2 Preliminaries

In this section, we present some material needed in the proof of our main result. Through this paper, we use cc to denote a positive generic constant. Now, we start with the following assumptions:
(A1) gi:++g_{i}:\mathbb{R}^{+}\longrightarrow\mathbb{R}^{+} are differentiable non-increasing functions such that

gi(0)>0,i=1,2,1a10+g1(s)𝑑sa20+g2(s)𝑑s=l>0.g_{i}(0)>0,~{}~{}i=1,2,~{}~{}~{}~{}~{}1-||a_{1}||_{\infty}\int_{0}^{+\infty}g_{1}(s)ds-||a_{2}||_{\infty}\int_{0}^{+\infty}g_{2}(s)ds=l>0.

and there exists C1C^{1} functions Hi:++H_{i}:\mathbb{R}_{+}\to\mathbb{R}_{+} which are linear or it is strictly increasing and strictly convex C2C^{2} function on (0,r](0,r] for some r>0r>0 with Hi(0)=Hi(0)=0H_{i}(0)=H_{i}^{\prime}(0)=0, such that

gi(t)ξ(t)Hi(gi(t)),t0,g_{i}^{\prime}(t)\leq-\xi(t)H_{i}(g_{i}(t)),\hskip 10.84006pt\forall t\geq 0,

where ξi:++\xi_{i}:\mathbb{R}^{+}\longrightarrow\mathbb{R}^{+} are positive nonincreasing differentiable functions.
(A2) ai:Ω¯+a_{i}:\bar{\Omega}\longrightarrow\mathbb{R}^{+} are in C1(Ω¯)C^{1}(\bar{\Omega}) such that, for positive constant δ\delta and a0a_{0} and for Γ1,Γ2Ω\Gamma_{1},\Gamma_{2}\subset\partial\Omega with means (Γi)>0,i=1,2,infxΩ¯(a1(x)+a2(x))δ(\Gamma_{i})>0,~{}i=1,2,~{}\inf_{x\in\bar{\Omega}}(a_{1}(x)+a_{2}(x))\geq\delta and

ai=0orinfΓiai(x)2a0,i=1,2.a_{i}=0~{}~{}~{}\text{or}~{}~{}~{}\inf_{\Gamma_{i}}~{}a_{i}(x)\geq 2a_{0},~{}~{}~{}~{}~{}~{}i=1,2.
Remark 2.1.

If ai0,i=1,2,a_{i}\neq 0,~{}i=1,2, there exist neighborhoods wiw_{i} of Γi,i=1,2,\Gamma_{i},~{}i=1,2, such that

infΩwi¯ai(x)a0>0,i=1,2.\inf_{\bar{\Omega\bigcap w_{i}}}a_{i}(x)\geq a_{0}>0,~{}i=1,2.

As in [10, 11], let d=min{a0,δ}d=\min\{a_{0},\delta\} and let αiC1(Ω¯),i=1,2,\alpha_{i}\in C^{1}(\bar{\Omega}),~{}i=1,2, be such that

{0αi(x)ai(x)αi(x)=0,ifai(x)d4αi(x)=ai(x),ifai(x)d2.\begin{cases}\begin{array}[]{ll}0\leq\alpha_{i}(x)\leq a_{i}(x)&\\ \alpha_{i}(x)=0,&\text{if}~{}~{}a_{i}(x)\leq\frac{d}{4}\\ \alpha_{i}(x)=a_{i}(x),&\text{if}~{}~{}a_{i}(x)\geq\frac{d}{2}.\\ \end{array}\end{cases} (2.1)
Lemma 2.2.

The functions αi,i=1,2,\alpha_{i},~{}i=1,2, are not identically zero and satisfy α1(x)+α2(x)d2\alpha_{1}(x)+\alpha_{2}(x)\geq\frac{d}{2}.

Proof.

(1) For xΩwix\in\Omega\cap w_{i}, we have ai(x)a0da_{i}(x)\geq a_{0}\geq d, which implies, by (2.1), that αi(x)=ai(x)d.\alpha_{i}(x)=a_{i}(x)\geq d. Thus αi\alpha_{i} is not identically zero.
(2) If a1(x)d2a_{1}(x)\geq\frac{d}{2}, then α1(x)=a1(x)\alpha_{1}(x)=a_{1}(x). Consequently α1(x)+α2(x)a1(x)d2\alpha_{1}(x)+\alpha_{2}(x)\geq a_{1}(x)\geq\frac{d}{2}. If a1(x)<d2a_{1}(x)<\frac{d}{2}, then a2(x)>d2a_{2}(x)>\frac{d}{2} which implies, by (2.1), α2(x)=a2(x)>d2\alpha_{2}(x)=a_{2}(x)>\frac{d}{2}. Consequently α1(x)+α2(x)>d2\alpha_{1}(x)+\alpha_{2}(x)>\frac{d}{2}. This completes the proof. ∎

The existence and uniqueness of the solution of problem (1.1) can be established by using the Galerkin method. We define the ”modified” energy functional of the weak solution of problem (1.1), by

E(t)=12Ωut2x+12Ω[1a1(x)0tg1(s)𝑑sa2(x)0+g2(s)𝑑s]|u|2𝑑x+12g1ou+12g2ou,\begin{array}[]{c}E(t)=\frac{1}{2}\int_{\Omega}u_{t}^{2}\partial x+\frac{1}{2}\int_{\Omega}\left[1-a_{1}(x)\int_{0}^{t}g_{1}(s)ds-a_{2}(x)\int_{0}^{+\infty}g_{2}(s)ds\right]|\nabla u|^{2}dx\\ +\frac{1}{2}~{}g_{1}~{}o~{}\nabla u+\frac{1}{2}~{}g_{2}~{}o~{}\nabla u,\end{array} (2.2)

where

g1ou=Ωa1(x)0tg1(ts)|u(t)u(s)|2𝑑s𝑑xg2ou=Ωa2(x)0+g2(s)|u(t)u(ts)|2𝑑s𝑑x.\begin{array}[]{l}g_{1}~{}o~{}\nabla u=\int_{\Omega}a_{1}(x)\int_{0}^{t}g_{1}(t-s)|\nabla u(t)-\nabla u(s)|^{2}ds~{}dx\\ g_{2}~{}o~{}\nabla u=\int_{\Omega}a_{2}(x)\int_{0}^{+\infty}g_{2}(s)|\nabla u(t)-\nabla u(t-s)|^{2}ds~{}dx\\ .\end{array} (2.3)
Lemma 2.3.

The ”modified” energy functional satisfies, along the solution of problem (1.1), the following

E(t)=12g1(t)Ω|u(t)|2𝑑x+12g1ou+12g2ou0,E^{\prime}(t)=-\frac{1}{2}~{}g_{1}(t)\int_{\Omega}|\nabla u(t)|^{2}dx+\frac{1}{2}~{}g^{\prime}_{1}~{}o~{}\nabla u+\frac{1}{2}~{}g^{\prime}_{2}~{}o~{}\nabla u\leq 0, (2.4)

where

g1ou=Ωa1(x)0tg1(ts)|u(t)u(s)|2𝑑s𝑑xg2ou=Ωa2(x)0+g2(s)|u(t)u(ts)|2𝑑s𝑑x.\begin{array}[]{l}g^{\prime}_{1}~{}o~{}\nabla u=\int_{\Omega}a_{1}(x)\int_{0}^{t}g^{\prime}_{1}(t-s)|\nabla u(t)-\nabla u(s)|^{2}~{}ds~{}dx\\ g^{\prime}_{2}~{}o~{}\nabla u=\int_{\Omega}a_{2}(x)\int_{0}^{+\infty}g^{\prime}_{2}(s)|\nabla u(t)-\nabla u(t-s)|^{2}~{}ds~{}dx\\ .\end{array} (2.5)
Proof.

By multiplying the first equation in problem (1.1) by utu_{t} and integrating over Ω\Omega, using integration by parts, hypotheses (A1)(A1) and (A2)(A2) and some manipulations as in [6, 12] and others, we obtain (2.4) for regular solutions. This inequality remains valid for weak solutions by a simple density argument. ∎

3 Technical lemmas

In this section, we introduce some fundamental lemmas. These lemmas will help us for proving our results. We use our equations and assumptions to prove those lemmas.

Lemma 3.1.

[13]For i=1,2,i=1,2, we have

01(0tgi(ts)(u(s)u(ts))𝑑s)2𝑑xCα(hiu)(t),\int_{0}^{1}\left(\int_{0}^{t}g_{i}(t-s)\left(\nabla u(s)-\nabla u(t-s)\right)ds\right)^{2}dx\leq C_{\alpha}(h_{i}\circ\nabla u)(t), (3.1)

where, for any 0<α<10<\alpha<1,

Cα=0tgi2(s)αgi(s)gi(s)𝑑s and hi(t)=αgi(t)gi(t).C_{\alpha}=\int_{0}^{t}\frac{g_{i}^{2}(s)}{\alpha g_{i}(s)-g_{i}^{\prime}(s)}ds\hskip 10.84006pt\text{ and }\hskip 10.84006pth_{i}(t)=\alpha g_{i}(t)-g_{i}^{\prime}(t). (3.2)

Furthermore, using the fact that

αgi2(s)αgi(s)gi(s)<gi(s)\frac{\alpha g_{i}^{2}(s)}{\alpha g_{i}(s)-g_{i}^{\prime}(s)}<g_{i}(s)

and recalling the Lebesgue dominated convergence theorem, one can deduce that

αCα=0αgi2(s)αgi(s)gi(s)𝑑s0 as α0.\alpha C_{\alpha}=\int_{0}^{\infty}\frac{\alpha g_{i}^{2}(s)}{\alpha g_{i}(s)-g_{i}^{\prime}(s)}ds\to 0\text{ as }\alpha\to 0. (3.3)
Lemma 3.2.

There exists a positive constant M1M_{1} such that

Ωa2(x)t+g2(s)|u(t)u(ts)|22𝑑sM1h0(t),\displaystyle\int_{\Omega}a_{2}(x)\int_{t}^{+\infty}g_{2}(s)|\nabla u(t)-\nabla u(t-s)|_{2}^{2}ds\leq M_{1}h_{0}(t), (3.4)

where h0(t)=0+g2(t+s)(1+u0(s)2)𝑑sh_{0}(t)=\int_{0}^{+\infty}g_{2}(t+s)\left(1+||\nabla u_{0}(s)||^{2}\right)ds.

Proof.

The proof is identical to the one in [14]. Indeed, we have

Ωa2(x)t+g2(s)|u(t)u(ts)|22𝑑s\displaystyle\int_{\Omega}a_{2}(x)\int_{t}^{+\infty}g_{2}(s)|\nabla u(t)-\nabla u(t-s)|_{2}^{2}ds\leq (3.5)
2a2u(t)2t+g2(s)𝑑s+2a2t+g2(s)u(ts)2𝑑s\displaystyle 2||a_{2}||_{\infty}||\nabla u(t)||^{2}\int_{t}^{+\infty}g_{2}(s)ds+2||a_{2}||_{\infty}\int_{t}^{+\infty}g_{2}(s)||\nabla u(t-s)||^{2}ds
2a2sups0u(s)20+g2(t+s)𝑑s+2a20+g2(t+s)u(s)2𝑑s\displaystyle\leq 2||a_{2}||_{\infty}\sup_{s\geq 0}||\nabla u(s)||^{2}\int_{0}^{+\infty}g_{2}(t+s)ds+2||a_{2}||_{\infty}\int_{0}^{+\infty}g_{2}(t+s)||\nabla u(-s)||^{2}ds
4a2E(s)0+g2(t+s)𝑑s+2a20+g2(t+s)u0(s)2𝑑s\displaystyle\leq\frac{4||a_{2}||_{\infty}E(s)}{\ell}\int_{0}^{+\infty}g_{2}(t+s)ds+2||a_{2}||_{\infty}\int_{0}^{+\infty}g_{2}(t+s)||\nabla u_{0}(s)||^{2}ds
4a2E(0)0+g2(t+s)𝑑s+2a20+g2(t+s)u0(s)2𝑑s\displaystyle\leq\frac{4||a_{2}||_{\infty}E(0)}{\ell}\int_{0}^{+\infty}g_{2}(t+s)ds+2||a_{2}||_{\infty}\int_{0}^{+\infty}g_{2}(t+s)||\nabla u_{0}(s)||^{2}ds
M10+g2(t+s)(1+u0(s)2)𝑑s.\displaystyle\leq M_{1}\int_{0}^{+\infty}g_{2}(t+s)\left(1+||\nabla u_{0}(s)||^{2}\right)ds.

where M1=max{2a2,4a2E(0)}.M_{1}=\max\big{\{}2||a_{2}||_{\infty},\frac{4||a_{2}||_{\infty}E(0)}{\ell}\big{\}}.

Lemma 3.3.

[11] Assume that (A1)(A1) and (A2)(A2) are hold, the functionals

ψ1(t)=Ωuut𝑑x,\psi_{1}(t)=\int_{\Omega}uu_{t}dx,
ψ2(t)=Ωα1(x)ut0tg1(ts)(u(t)u(s))𝑑s𝑑x,\psi_{2}(t)=-\int_{\Omega}\alpha_{1}(x)u_{t}\int_{0}^{t}g_{1}(t-s)\left(u(t)-u(s)\right)~{}ds~{}dx,
ψ3(t)=Ωα2(x)ut0+g2(s)(u(t)u(ts))𝑑s𝑑x,\psi_{3}(t)=-\int_{\Omega}\alpha_{2}(x)u_{t}\int_{0}^{+\infty}g_{2}(s)\left(u(t)-u(t-s)\right)~{}ds~{}dx,

satisfies, along the solution of (1.1) and for any ϵ1,ϵ2,ϵ3>0\epsilon_{1},\epsilon_{2},\epsilon_{3}>0 and t0\forall t\geq 0 the following estimates

ψ1(t)Ωut2𝑑x[1ϵ1a10+g1(s)𝑑sa20+g2(s)𝑑s]Ω|u|2𝑑x+cCαϵ1(h1ou+h2ou),\begin{array}[]{l}\psi_{1}^{\prime}(t)\leq\int_{\Omega}u_{t}^{2}dx-\left[1-\epsilon_{1}-||a_{1}||_{\infty}\int_{0}^{+\infty}g_{1}(s)ds-||a_{2}||_{\infty}\int_{0}^{+\infty}g_{2}(s)ds\right]\int_{\Omega}|\nabla u|^{2}dx\\ ~{}~{}~{}~{}~{}~{}~{}~{}~{}+\frac{cC_{\alpha}}{\epsilon_{1}}\left(h_{1}~{}o~{}\nabla u+h_{2}~{}o~{}\nabla u\right),\end{array} (3.6)
ψ2(t)[0tg1(s)𝑑sϵ2]Ωα1(x)ut2𝑑x+ϵ32Ω|u|2𝑑xcϵ2g1ou+cCαϵ3(h1ou+h2ou),\begin{array}[]{l}\psi^{\prime}_{2}(t)\leq-\left[\int_{0}^{t}g_{1}(s)ds-\epsilon_{2}\right]\int_{\Omega}\alpha_{1}(x)u_{t}^{2}dx+\frac{\epsilon_{3}}{2}\int_{\Omega}|\nabla u|^{2}dx-\frac{c}{\epsilon_{2}}g^{\prime}_{1}~{}o~{}\nabla u\\ ~{}~{}~{}~{}~{}~{}~{}+\frac{cC_{\alpha}}{\epsilon_{3}}\left(h_{1}~{}o~{}\nabla u+h_{2}~{}o~{}\nabla u\right),\end{array} (3.7)
ψ3(t)[0+g2(s)𝑑sϵ2]Ωα2(x)ut2𝑑x+ϵ32Ω|u|2𝑑xcϵ2g2ou+cCαϵ3(h1ou+h2ou).\begin{array}[]{l}\psi^{\prime}_{3}(t)\leq-\left[\int_{0}^{+\infty}g_{2}(s)ds-\epsilon_{2}\right]\int_{\Omega}\alpha_{2}(x)u_{t}^{2}dx+\frac{\epsilon_{3}}{2}\int_{\Omega}|\nabla u|^{2}dx-\frac{c}{\epsilon_{2}}g^{\prime}_{2}~{}o~{}\nabla u\\ ~{}~{}~{}~{}~{}~{}~{}+\frac{cC_{\alpha}}{\epsilon_{3}}\left(h_{1}~{}o~{}\nabla u+h_{2}~{}o~{}\nabla u\right).\end{array} (3.8)
Lemma 3.4.

Assume that (A1)(A1) and (A2)(A2) are hold, then for a suitable choice of N,M,m>0N,M,m>0 and for all t0t\geq 0, the functional

(t):=NE(t)+Mψ1(t)+ψ2(t)+ψ3(t)\mathcal{L}(t):=NE(t)+M\psi_{1}(t)+{\psi_{2}}(t)+{\psi_{3}}(t)

satisfies for any t>0t>0

(t)mE(t)+14(g1u+g2u))(t).\mathcal{L}^{\prime}(t)\leq-mE(t)+\frac{1}{4}\bigg{(}g_{1}\circ\nabla u+g_{2}\circ\nabla u)\bigg{)}(t). (3.9)
Proof.

Let, for t0>0t_{0}>0 fixed, g0:=min{0t0g1(s)𝑑s,0+g2(s)𝑑s}g_{0}:=\min\{\int_{0}^{t_{0}}g_{1}(s)ds,\int_{0}^{+\infty}g_{2}(s)ds\}. Then direct differentiation of \mathcal{L} and using (2.4), (4.21), (3.7) and (3.8) leads to

(t)(N2cε2)((g1u)(t)+(g2u)(t))\displaystyle\mathcal{L}^{\prime}(t)\leq\bigg{(}\frac{N}{2}-\frac{c}{\varepsilon_{2}}\bigg{)}\bigg{(}(g_{1}^{\prime}\circ\nabla u)(t)+(g_{2}^{\prime}\circ\nabla u)(t)\bigg{)}
Ω[(g0ε2)(α1+α2)M]ut2𝑑x\displaystyle-\int_{\Omega}\bigg{[}(g_{0}-\varepsilon_{2})(\alpha_{1}+\alpha_{2})-M\bigg{]}u_{t}^{2}dx
×(cCαε3+McCαε1)((h1u)(t)+(h2u)(t))\displaystyle\times\bigg{(}\frac{cC_{\alpha}}{\varepsilon_{3}}+\frac{McC_{\alpha}}{\varepsilon_{1}}\bigg{)}\bigg{(}(h_{1}\circ\nabla u)(t)+(h_{2}\circ\nabla u)(t)\bigg{)}
[(ε1)Mε3]Ω|u|2𝑑x,tt0.\displaystyle-\bigg{[}(\ell-\varepsilon_{1})M-\varepsilon_{3}\bigg{]}\int_{\Omega}|\nabla u|^{2}dx,~{}t\geq t_{0}.

Recalling that gi=αgihig_{i}^{\prime}=\alpha g_{i}-h_{i} where i=1,2.i=1,2. We obtain

(t)\displaystyle\mathcal{L}^{\prime}(t) α(N2cε2)(g1u+g2u)(t)\displaystyle\leq\alpha\bigg{(}\frac{N}{2}-\frac{c}{\varepsilon_{2}}\bigg{)}\bigg{(}g_{1}\circ\nabla u+g_{2}\circ\nabla u\bigg{)}(t)
(N2cε2)(h1u+h2u)(t)\displaystyle-\bigg{(}\frac{N}{2}-\frac{c}{\varepsilon_{2}}\bigg{)}\bigg{(}h_{1}\circ\nabla u+h_{2}\circ\nabla u\bigg{)}(t)
Ω[(g0ε2)(α1+α2)M]ut2𝑑x\displaystyle-\int_{\Omega}\bigg{[}(g_{0}-\varepsilon_{2})(\alpha_{1}+\alpha_{2})-M\bigg{]}u_{t}^{2}dx
×(cCαε3+McCαε1)(h1u)(t)+(h2u)(t))\displaystyle\times\bigg{(}\frac{cC_{\alpha}}{\varepsilon_{3}}+\frac{McC_{\alpha}}{\varepsilon_{1}}\bigg{)}\bigg{(}h_{1}\circ\nabla u)(t)+(h_{2}\circ\nabla u)(t)\bigg{)}
[(ε1)Mε3]Ω|u|2𝑑x,tt0.\displaystyle-\bigg{[}(\ell-\varepsilon_{1})M-\varepsilon_{3}\bigg{]}\int_{\Omega}|\nabla u|^{2}dx,~{}t\geq t_{0}.

Simplifying the above estimate, we arrive at

(t)\displaystyle\mathcal{L}^{\prime}(t) α(N2cε2)(g1u+g2u)(t)\displaystyle\leq\alpha\bigg{(}\frac{N}{2}-\frac{c}{\varepsilon_{2}}\bigg{)}\bigg{(}g_{1}\circ\nabla u+g_{2}\circ\nabla u\bigg{)}(t)
[N2cε2Cα(cε3+Mcε1)](h1u+h2u)(t)\displaystyle-\bigg{[}\frac{N}{2}-\frac{c}{\varepsilon_{2}}-C_{\alpha}\big{(}\frac{c}{\varepsilon_{3}}+\frac{Mc}{\varepsilon_{1}}\big{)}\bigg{]}\bigg{(}h_{1}\circ\nabla u+h_{2}\circ\nabla u\bigg{)}(t)
Ω[(g0ε2)(α1+α2)M]ut2𝑑x\displaystyle-\int_{\Omega}\bigg{[}(g_{0}-\varepsilon_{2})(\alpha_{1}+\alpha_{2})-M\bigg{]}u_{t}^{2}dx
[(ε1)Mε3]Ω|u|2𝑑x,tt0.\displaystyle-\bigg{[}(\ell-\varepsilon_{1})M-\varepsilon_{3}\bigg{]}\int_{\Omega}|\nabla u|^{2}dx,~{}t\geq t_{0}.

By using the fact that (α1+α2)d2(\alpha_{1}+\alpha_{2})\geq\frac{d}{2} and choosing

ε1=2,ε2=g02M=dg08ε3<d2dg016\varepsilon_{1}=\frac{\ell}{2},~{}~{}\varepsilon_{2}=\frac{g_{0}}{2}~{}~{}M=\frac{dg_{0}}{8}^{\prime}~{}~{}\varepsilon_{3}<\frac{d\ell^{2}dg_{0}}{16}

we get

(ε1)Mε34(1)(\ell-\varepsilon_{1})M-\varepsilon_{3}\geq 4(1-\ell)

As a consequence of (3.3), there exists 0<α0<10<\alpha_{0}<1 such that if α<α0\alpha<\alpha_{0}, then

αCα<18[cε3+Mcε1].\alpha C_{\alpha}<\frac{1}{8\bigg{[}\frac{c}{\varepsilon_{3}}+\frac{Mc}{\varepsilon_{1}}\bigg{]}}. (3.10)

It is clear that (3.10) gives

Cα[N2cε2Cα(cε3+Mcε1)]<N2.C_{\alpha}\bigg{[}\frac{N}{2}-\frac{c}{\varepsilon_{2}}-C_{\alpha}\big{(}\frac{c}{\varepsilon_{3}}+\frac{Mc}{\varepsilon_{1}}\big{)}\bigg{]}<\frac{N}{2}. (3.11)

Choosing α=12N<α0,\alpha=\frac{1}{2N}<\alpha_{0},, then for some m>0m>0, we have

(t)mE(t)+14(g1u+g2u))(t).\mathcal{L}^{\prime}(t)\leq-mE(t)+\frac{1}{4}\bigg{(}g_{1}\circ\nabla u+g_{2}\circ\nabla u)\bigg{)}(t).

Then (3.9) is established. Moreover, one can choose NN large enough so that E\mathcal{L}\sim E. ∎

Lemma 3.5.

For all t+t\in\mathbb{R}^{+} and fixed positive constants m0m_{0} , we have the following estimates

0tE(s)𝑑s<m0(1+0th0(s)𝑑s).\int_{0}^{t}E(s)ds<m_{0}\left(1+\int_{0}^{t}h_{0}(s)ds\right). (3.12)

where h0h_{0} is defined in (3.4).

Proof.

The proof of this lemma can be established by following the same arguments in [15, 14]. ∎

Lemma 3.6.

If (A1)(A2)(A1)-(A2) are satisfied, then we have, for all t>0t>0 and for i=1,2i=1,2, the following estimates

Ωai(x)0tgi(s)|u(t)u(ts)|2𝑑s𝑑x1Λi(t)Hi1(Λi(t)μi(t)ξi(t))\int_{\Omega}a_{i}(x)\int_{0}^{t}g_{i}(s)|\nabla u(t)-\nabla u(t-s)|^{2}dsdx\leq\frac{1}{\Lambda_{i}(t)}{H_{i}}^{-1}\left(\frac{\Lambda_{i}(t)\mu_{i}(t)}{\xi_{i}(t)}\right) (3.13)

where Λi(t)=λi1+0th0(s)𝑑s\Lambda_{i}(t)=\frac{\lambda_{i}}{1+\int_{0}^{t}h_{0}(s)ds}; λi(0,1)\lambda_{i}\in(0,1), HiH_{i} and ξi\xi_{i} are introduced in (A1)(A1), and

μi(t):=Ωai(x)0tgi(s)|(t)(ts)|2𝑑s𝑑xcE(t),\mu_{i}(t):=-\int_{\Omega}a_{i}(x)\int_{0}^{t}g_{i}^{\prime}(s)|\nabla(t)-\nabla(t-s)|^{2}dsdx\leq-cE^{\prime}(t), (3.14)

.

Proof.

To establish (3.13), we introduce the following functional

χi(t):=Λi(t)Ωai(x)0t|u(t)u(ts)|2𝑑s𝑑x.\chi_{i}(t):=\Lambda_{i}(t)\int_{\Omega}a_{i}(x)\int_{0}^{t}|\nabla u(t)-\nabla u(t-s)|^{2}dsdx. (3.15)

Then, using the fact that EE is nonincreasing and (2.2) to get

χi(t)2Λi(t)(Ωai(x)0t|u(t)|2+Ωai(x)0t|u(ts)|2𝑑s𝑑x).\displaystyle\chi_{i}(t)\leq 2\Lambda_{i}(t)\bigg{(}\int_{\Omega}a_{i}(x)\int_{0}^{t}|\nabla u(t)|^{2}+\int_{\Omega}a_{i}(x)\int_{0}^{t}|\nabla u(t-s)|^{2}dsdx\bigg{)}. (3.16)
4Λi(t)ai(0t(E(t)+E(ts))𝑑s)\displaystyle\hskip 28.90755pt\leq\frac{4\Lambda_{i}(t)||a_{i}||_{\infty}}{\ell}\bigg{(}\int_{0}^{t}\big{(}E(t)+E(t-s)\big{)}ds\bigg{)}
8Λi(t)ai0tE(s)𝑑s\displaystyle\hskip 28.90755pt\leq\frac{8\Lambda_{i}(t)||a_{i}||_{\infty}}{\ell}\int_{0}^{t}E(s)ds
<+.\displaystyle\hskip 28.90755pt<+\infty.

Thus, λi\lambda_{i} can be chosen so small so that, for all t>0t>0,

χi(t)<1.\chi_{i}(t)<1. (3.17)

Without loss of the generality, for all t>0t>0, we assume that ξi(t)>0\xi_{i}(t)>0, otherwise we get an exponential decay from (3.9). The use of Jensen’s inequality and using (3.14) and (3.17) gives

μ1(t)=1Λi(t)0tΛi(t)(gi(s))Ωai(x)|(t)(ts)|2𝑑x𝑑s\displaystyle\mu_{1}(t)=\frac{1}{\Lambda_{i}(t)}\int_{0}^{t}\Lambda_{i}(t)(-g_{i}^{\prime}(s))\int_{\Omega}{a_{i}(x)|\nabla(t)-\nabla(t-s)|}^{2}dxds (3.18)
1Λi(t)0tΛi(t)ξ1(s)Hi(gi(s))Ωai(x)|(t)(ts)|2𝑑x𝑑s\displaystyle\hskip 21.68121pt\geq\frac{1}{\Lambda_{i}(t)}\int_{0}^{t}\Lambda_{i}(t)\xi_{1}(s)H_{i}(g_{i}(s))\int_{\Omega}{a_{i}(x)|\nabla(t)-\nabla(t-s)|}^{2}dxds
ξi(t)Λi(t)0tHi(Λi(t)gi(s))Ωai(x)|(t)(ts)|2𝑑x𝑑s\displaystyle\hskip 21.68121pt\geq\frac{\xi_{i}(t)}{\Lambda_{i}(t)}\int_{0}^{t}H_{i}(\Lambda_{i}(t)g_{i}(s))\int_{\Omega}{a_{i}(x)|\nabla(t)-\nabla(t-s)|}^{2}dxds
ξi(t)Λi(t)Hi(Λi(t)0tgi(s)Ωai(x)|(t)(ts)|2𝑑x𝑑s)\displaystyle\hskip 21.68121pt\geq\frac{\xi_{i}(t)}{\Lambda_{i}(t)}H_{i}\biggl{(}\Lambda_{i}(t)\int_{0}^{t}g_{i}(s)\int_{\Omega}a_{i}(x){|\nabla(t)-\nabla(t-s)|}^{2}dxds\biggr{)}
=ξ1(t)ΛiHi¯(Λi(t)0tgi(s)Ωa1(x)|u(t)u(ts)|2𝑑x𝑑s),\displaystyle\hskip 21.68121pt=\frac{\xi_{1}(t)}{\Lambda_{i}}\overline{H_{i}}\biggl{(}\Lambda_{i}(t)\int_{0}^{t}g_{i}(s)\int_{\Omega}a_{1}(x){|\nabla u(t)-\nabla u(t-s)|}^{2}dxds\biggr{)},

Hence, (3.13) is established. ∎

4 Decay result

In this section, we state, prove our main result and provide some example to illustrate our decay results. Let us start introducing some functions and then establishing several lemmas needed for the proof of our main result. Now, for i=1,2,i=1,2, let us take

ξ=min{ξi},μ=max{μi},λ=max{λi}G=(H11+H21)1.\xi=\min\{\xi_{i}\},~{}~{}\mu=\max\{\mu_{i}\},~{}~{}\lambda=\max\{\lambda_{i}\}~{}~{}G=\left(H_{1}^{-1}+H_{2}^{-1}\right)^{-1}. (4.1)

As in [14], we introduce the following functions:

G1(t):=tr1sG(s)𝑑s,G5(t)=G11(c10tξ(s)𝑑s),G_{1}(t):=\int_{t}^{r}\frac{1}{sG^{\prime}(s)}ds,~{}~{}G_{5}(t)=G_{1}^{-1}\bigg{(}c_{1}\int_{0}^{t}\xi(s)ds\bigg{)}, (4.2)
G2(t)=tG(t),G3(t)=t(G)1(t),G4(t)=G3(t).G_{2}(t)=tG^{\prime}(t),\quad G_{3}(t)=t(G^{\prime})^{-1}(t),\quad G_{4}(t)=G_{3}^{*}(t). (4.3)

One can easily verify that G1G_{1} is decreasing function over (0,r](0,r], and G2,G3G_{2},G_{3} and G4G_{4} be convex and increasing functions on (0,r](0,r]. Further, we introduce the class SS of functions ϑ:+(0,r]\vartheta:\mathbb{R}_{+}\rightarrow(0,r] satisfying for fixed c1,c2>0c_{1},c_{2}>0,

ϑC1(+),ϑ0,\vartheta\in C^{1}(\mathbb{R}_{+}),~{}~{}~{}\vartheta^{\prime}\leq 0, (4.4)

and

c2G4[cdΛ(t)h0(t)]c1(G2(G5(s)ϑ(s))G2(G5(t))ϑ(t)),c_{2}G_{4}\left[\frac{c}{d}\Lambda(t)h_{0}(t)\right]\leq c_{1}\left(G_{2}\bigg{(}\frac{G_{5}(s)}{\vartheta(s)}\bigg{)}-\frac{G_{2}\left(G_{5}(t)\right)}{\vartheta(t)}\right), (4.5)

where h2h_{2} and qq will be defined later in the proof of our main result.

Theorem 4.1.

Assume that (A1)(A2)(A1)-(A2) hold, then there exists a strictly positive constant CC such that the solution of (1.1) satisfies, for all t0t\geq 0,

E(t)CG5(t)ϑ(t)Λ(t),E(t)\leq\frac{CG_{5}(t)}{\vartheta(t)\Lambda(t)}, (4.6)

where G5G_{5} and ϑ\vartheta are defined in (4.2) and (4.4) respectively.

Proof.

Using (3.4), (3.9) and (3.13), then for some positive constant mm, and any t0t\geq 0, we get

L(t)mE(t)+cΛ(t)G1(Λ(t)μ(t)ξ(t))+ch0(t).\displaystyle L^{\prime}(t)\leq-mE(t)+\frac{c}{\Lambda(t)}G^{-1}\left(\frac{\Lambda(t)\mu(t)}{\xi(t)}\right)+ch_{0}(t). (4.7)

Without lose of generality, one can assume that E(0)>0E(0)>0. For ε0<r\varepsilon_{0}<r, let the functional \mathcal{F} defined by

(t):=G(ε0E(t)Λ(t)E(0))L(t),\mathcal{F}(t):=G^{\prime}\left(\varepsilon_{0}\frac{E(t)\Lambda(t)}{E(0)}\right)L(t),

which satisfies E\mathcal{F}\sim E. By noting that G′′0,G^{\prime\prime}\geq 0, Λ0\Lambda^{\prime}\leq 0 and E0E^{\prime}\leq 0, we get

(t)=ε0(qE)(t)E(0)G′′(ε0E(t)Λ(t)E(0))L(t)+G(ε0E(t)Λ(t)E(0))L(t)\displaystyle\mathcal{F}^{\prime}(t)=\varepsilon_{0}\frac{(qE)^{\prime}(t)}{E(0)}G^{\prime\prime}\left(\varepsilon_{0}\frac{E(t)\Lambda(t)}{E(0)}\right)L(t)+G^{\prime}\left(\varepsilon_{0}\frac{E(t)\Lambda(t)}{E(0)}\right)L^{\prime}(t) (4.8)
mE(t)G(ε0E(t)Λ(t)E(0))+cΛ(t)G(ε0E(t)Λ(t)E(0))G1(Λ(t)μ(t)ξ(t))\displaystyle\hskip 21.68121pt\leq-mE(t)G^{\prime}\left(\varepsilon_{0}\frac{E(t)\Lambda(t)}{E(0)}\right)+\frac{c}{\Lambda(t)}G^{\prime}\left(\varepsilon_{0}\frac{E(t)\Lambda(t)}{E(0)}\right)G^{-1}\left(\frac{\Lambda(t)\mu(t)}{\xi(t)}\right)
+ch0(t)G(ε0E(t)Λ(t)E(0)).\displaystyle\hskip 21.68121pt+ch_{0}(t)G^{\prime}\left(\varepsilon_{0}\frac{E(t)\Lambda(t)}{E(0)}\right).

Let GG^{*} be the convex conjugate of GG in the sense of Young (see [16]), then

G(s)=s(G)1(s)G[(G)1(s)],ifs(0,G(r)]G^{*}(s)=s(G^{\prime})^{-1}(s)-G\left[(G^{\prime})^{-1}(s)\right],\hskip 7.22743pt\text{if}\hskip 3.61371pts\in(0,G^{\prime}(r)] (4.9)

and GG^{*} satisfies the following generalized Young inequality

ABG(A)+G(B),ifA(0,G(r)],B(0,r].AB\leq G^{*}(A)+G(B),\hskip 10.84006pt\text{if}\hskip 3.61371ptA\in(0,G^{\prime}(r)],\hskip 3.61371ptB\in(0,r]. (4.10)

So, with A=G(ε0E(t)Λ(t)E(0))A=G^{\prime}\left(\varepsilon_{0}\frac{E(t)\Lambda(t)}{E(0)}\right) and B=G1(Λ(t)μ(t)ξ(t))B=G^{-1}\left(\frac{\Lambda(t)\mu(t)}{\xi(t)}\right) and using (4.8)-(4.10), we arrive at

(t)mE(t)G(ε0E(t)Λ(t)E(0))+cΛ(t)G(G(ε0E(t)Λ(t)E(0)))+c(μ(t)ξ(t))\displaystyle\mathcal{F}^{\prime}(t)\leq-mE(t)G^{\prime}\left(\varepsilon_{0}\frac{E(t)\Lambda(t)}{E(0)}\right)+\frac{c}{\Lambda(t)}G^{*}\left(G^{\prime}\left(\varepsilon_{0}\frac{E(t)\Lambda(t)}{E(0)}\right)\right)+c\left(\frac{\mu(t)}{\xi(t)}\right) (4.11)
+ch0(t)G(ε0E(t)Λ(t)E(0))\displaystyle\hskip 21.68121pt+ch_{0}(t)G^{\prime}\left(\varepsilon_{0}\frac{E(t)\Lambda(t)}{E(0)}\right)
mE(t)G(ε0E(t)Λ(t)E(0))+cε0E(t)E(0)G(ε0E(t)Λ(t)E(0))+c(μ(t)ξ(t))\displaystyle\hskip 21.68121pt\leq-mE(t)G^{\prime}\left(\varepsilon_{0}\frac{E(t)\Lambda(t)}{E(0)}\right)+c\varepsilon_{0}\frac{E(t)}{E(0)}G^{\prime}\left(\varepsilon_{0}\frac{E(t)\Lambda(t)}{E(0)}\right)+c\left(\frac{\mu(t)}{\xi(t)}\right)
+ch0(t)G(ε0E(t)Λ(t)E(0)).\displaystyle\hskip 21.68121pt+ch_{0}(t)G^{\prime}\left(\varepsilon_{0}\frac{E(t)\Lambda(t)}{E(0)}\right).

So, multiplying (4.11) by ξ(t)\xi(t) and using (3.14) and the fact that ε0E(t)Λ(t)E(0)<r\varepsilon_{0}\frac{E(t)\Lambda(t)}{E(0)}<r, gives

ξ(t)(t)mξ(t)E(t)G(ε0E(t)Λ(t)E(0))+cξ(t)ε0E(t)E(0)G(ε0E(t)Λ(t)E(0))\displaystyle\xi(t)\mathcal{F}^{\prime}(t)\leq-m\xi(t)E(t)G^{\prime}\left(\varepsilon_{0}\frac{E(t)\Lambda(t)}{E(0)}\right)+c\xi(t)\varepsilon_{0}\frac{E(t)}{E(0)}G^{\prime}\left(\varepsilon_{0}\frac{E(t)\Lambda(t)}{E(0)}\right)
+cμ(t)Λ(t)+cξ(t)h0(t)G(ε0E(t)Λ(t)E(0))\displaystyle\hskip 21.68121pt+c\mu(t)\Lambda(t)+c\xi(t)h_{0}(t)G^{\prime}\left(\varepsilon_{0}\frac{E(t)\Lambda(t)}{E(0)}\right)
ε0(mE(0)ε0c)ξ(t)E(t)E(0)G(ε0E(t)Λ(t)E(0))cE(t)+cξ(t)h0(t)G(ε0E(t)Λ(t)E(0)).\displaystyle\hskip 21.68121pt\leq-\varepsilon_{0}(\frac{mE(0)}{\varepsilon_{0}}-c)\xi(t)\frac{E(t)}{E(0)}G^{\prime}\left(\varepsilon_{0}\frac{E(t)\Lambda(t)}{E(0)}\right)-cE^{\prime}(t)+c\xi(t)h_{0}(t)G^{\prime}\left(\varepsilon_{0}\frac{E(t)\Lambda(t)}{E(0)}\right).

Consequently, recalling the definition of G2G_{2} and choosing ε0\varepsilon_{0} so that k=(mE(0)ε0c)>0k=(\frac{mE(0)}{\varepsilon_{0}}-c)>0, we obtain, for all t+t\in\mathbb{R}_{+},

1(t)\displaystyle\mathcal{F}_{1}^{\prime}(t) kξ(t)(E(t)E(0))G(ε0E(t)Λ(t)E(0))+cξ(t)h0(t)G(ε0E(t)Λ(t)E(0))\displaystyle\leq-k\xi(t)\left(\frac{E(t)}{E(0)}\right)G^{\prime}\left(\varepsilon_{0}\frac{E(t)\Lambda(t)}{E(0)}\right)+c\xi(t)h_{0}(t)G^{\prime}\left(\varepsilon_{0}\frac{E(t)\Lambda(t)}{E(0)}\right) (4.12)
=kξ(t)Λ(t)G2(E(t)Λ(t)E(0))+cξ(t)h0(t)G(ε0E(t)Λ(t)E(0)),\displaystyle=-k\frac{\xi(t)}{\Lambda(t)}G_{2}\left(\frac{E(t)\Lambda(t)}{E(0)}\right)+c\xi(t)h_{0}(t)G^{\prime}\left(\varepsilon_{0}\frac{E(t)\Lambda(t)}{E(0)}\right),

where 1=ξ+cEE\mathcal{F}_{1}=\xi\mathcal{F}+cE\sim E and satisfies for some α1,α2>0.\alpha_{1},\alpha_{2}>0.

α11(t)E(t)α21(t).\alpha_{1}\mathcal{F}_{1}(t)\leq E(t)\leq\alpha_{2}\mathcal{F}_{1}(t). (4.13)

Since G2(t)=G(t)+tG′′(t)G^{\prime}_{2}(t)=G^{\prime}(t)+tG^{\prime\prime}(t) and GG is strictly increasing and strictly convex on (0,r],(0,r], we find that G2(t),G2(t)>0G_{2}^{\prime}(t),G_{2}(t)>0 on (0,r].(0,r]. Using the general Young inequality (4.10) on the last term in (4.12) with A=G(ε0E(t)Λ(t)E(0))A=G^{\prime}\left(\varepsilon_{0}\frac{E(t)\Lambda(t)}{E(0)}\right) and B=[cdh0(t)]B=[\frac{c}{d}h_{0}(t)], we have for d>0d>0

ch0(t)G(ε0E(t)Λ(t)E(0))\displaystyle ch_{0}(t)G^{\prime}\left(\varepsilon_{0}\frac{E(t)\Lambda(t)}{E(0)}\right) =dΛ(t)[cdΛ(t)h0(t)](G(ε0E(t)Λ(t)E(0)))\displaystyle=\frac{d}{\Lambda(t)}\left[\frac{c}{d}\Lambda(t)h_{0}(t)\right]\bigg{(}G^{\prime}\left(\varepsilon_{0}\frac{E(t)\Lambda(t)}{E(0)}\right)\bigg{)} (4.14)
dΛ(t)G3(G(ε0E(t)Λ(t)E(0)))+dΛ(t)G3[cdΛ(t)h0(t)]\displaystyle\leq\frac{d}{\Lambda(t)}G_{3}\bigg{(}G^{\prime}\left(\varepsilon_{0}\frac{E(t)\Lambda(t)}{E(0)}\right)\bigg{)}+\frac{d}{\Lambda(t)}G_{3}^{*}\left[\frac{c}{d}\Lambda(t)h_{0}(t)\right]
dΛ(t)(ε0E(t)Λ(t)E(0))(G(ε0E(t)Λ(t)E(0)))+dΛ(t)G4[cdΛ(t)h0(t)]\displaystyle\leq\frac{d}{\Lambda(t)}\left(\varepsilon_{0}\frac{E(t)\Lambda(t)}{E(0)}\right)\left(G^{\prime}\left(\varepsilon_{0}\frac{E(t)\Lambda(t)}{E(0)}\right)\right)+\frac{d}{\Lambda(t)}G_{4}\left[\frac{c}{d}\Lambda(t)h_{0}(t)\right]
dΛ(t)G2(ε0E(t)Λ(t)E(0))+dΛ(t)G4[cdΛ(t)h0(t)].\displaystyle\leq\frac{d}{\Lambda(t)}G_{2}\left(\varepsilon_{0}\frac{E(t)\Lambda(t)}{E(0)}\right)+\frac{d}{\Lambda(t)}G_{4}\left[\frac{c}{d}\Lambda(t)h_{0}(t)\right].

Now, combining (4.12) and (4.14) and choosing dd small enough so that k0=(kd)>0k_{0}=(k-d)>0, we arrive at

1(t)kξ(t)Λ(t)G2(ε0E(t)Λ(t)E(0))+dξ(t)Λ(t)G2(ε0E(t)Λ(t)E(0))+dξ(t)Λ(t)G4[cdΛ(t)h0(t)]\displaystyle\mathcal{F}_{1}^{\prime}(t)\leq-k\frac{\xi(t)}{\Lambda(t)}G_{2}\left(\varepsilon_{0}\frac{E(t)\Lambda(t)}{E(0)}\right)+\frac{d\xi(t)}{\Lambda(t)}G_{2}\left(\varepsilon_{0}\frac{E(t)\Lambda(t)}{E(0)}\right)+\frac{d\xi(t)}{\Lambda(t)}G_{4}\left[\frac{c}{d}\Lambda(t)h_{0}(t)\right] (4.15)
k1ξ(t)Λ(t)G2(ε0E(t)Λ(t)E(0))+dξ(t)Λ(t)G4[cdΛ(t)h0(t)].\displaystyle\hskip 7.22743pt\leq-k_{1}\frac{\xi(t)}{\Lambda(t)}G_{2}\left(\varepsilon_{0}\frac{E(t)\Lambda(t)}{E(0)}\right)+\frac{d\xi(t)}{\Lambda(t)}G_{4}\left[\frac{c}{d}\Lambda(t)h_{0}(t)\right].

Using the equivalent property in (4.13) and the increasing of G2G_{2}, we have

G2(ε0E(t)Λ(t)E(0))G2(d01(t)Λ(t)).G_{2}\left(\varepsilon_{0}\frac{E(t)\Lambda(t)}{E(0)}\right)\geq G_{2}\bigg{(}d_{0}\mathcal{F}_{1}(t)\Lambda(t)\bigg{)}.

Letting 2(t):=d01(t)Λ(t)\mathcal{F}_{2}(t):=d_{0}\mathcal{F}_{1}(t)\Lambda(t) and recalling Λ0\Lambda^{\prime}\leq 0, then we arrive at, for some c1,c2>0c_{1},c_{2}>0,

2(t)c1ξ(t)G2(2(t))+c2ξ(t)G4[cdΛ(t)h0(t)].\mathcal{F}_{2}^{\prime}(t)\leq-c_{1}\xi(t)G_{2}(\mathcal{F}_{2}(t))+c_{2}\xi(t)G_{4}\left[\frac{c}{d}\Lambda(t)h_{0}(t)\right]. (4.16)

Since d0Λ(t)d_{0}\Lambda(t) is nonincreasing. Using the equivalent property 1E\mathcal{F}_{1}\sim E implies that there exists b0>0b_{0}>0 such that 2(t)b0E(t)Λ(t)\mathcal{F}_{2}(t)\geq b_{0}E(t)\Lambda(t). Let t+t\in\mathbb{R}_{+} and ϑ(t)\vartheta(t) satisfying (4.4) and (4.5).
If

b0Λ(t)E(t)2G5(t)ϑ(t),b_{0}\Lambda(t)E(t)\leq 2\frac{G_{5}(t)}{\vartheta(t)}, (4.17)

then, we have

E(t)2b0G5(t)ϑ(t)Λ(t).E(t)\leq\frac{2}{b_{0}}\frac{G_{5}(t)}{\vartheta(t)\Lambda(t)}. (4.18)

If

b0Λ(t)E(t)>2G5(t)ϑ(t).b_{0}\Lambda(t)E(t)>2\frac{G_{5}(t)}{\vartheta(t)}. (4.19)

Then, for any 0st0\leq s\leq t, we get

b0q(s)E(s)>2G5(t)ϑ(t),b_{0}q(s)E(s)>2\frac{G_{5}(t)}{\vartheta(t)}, (4.20)

since, Λ(t)E(t)\Lambda(t)E(t) is nonincreasing function. Therefore, for any 0st0\leq s\leq t, we have

2(s)>2G5(t)ϑ(t).\mathcal{F}_{2}(s)>2\frac{G_{5}(t)}{\vartheta(t)}. (4.21)

Using (4.21), recalling the definition of G2G_{2}, the fact that G2G_{2} is convex and 0<ϑ10<\vartheta\leq 1, we have, for any 0st0\leq s\leq t and 0<ϵ110<\epsilon_{1}\leq 1,

G2(ϵ1ϑ(s)2(s)ϵ1G5(s))ϵ1ϑ(s)G2(2(s)G5(s)ϑ(s))\displaystyle G_{2}\bigg{(}\epsilon_{1}\vartheta(s)\mathcal{F}_{2}(s)-\epsilon_{1}G_{5}(s)\bigg{)}\leq\epsilon_{1}\vartheta(s)G_{2}\bigg{(}\mathcal{F}_{2}(s)-\frac{G_{5}(s)}{\vartheta(s)}\bigg{)} (4.22)
ϵ1ϑ(s)2(s)G(2(s)G5(s)ϑ(s))ϵ1ϑ(s)G5(s)ϑ(s)G(2(s)G5(s)ϑ(s))\displaystyle\leq\epsilon_{1}\vartheta(s)\mathcal{F}_{2}(s)G^{\prime}\bigg{(}\mathcal{F}_{2}(s)-\frac{G_{5}(s)}{\vartheta(s)}\bigg{)}-\epsilon_{1}\vartheta(s)\frac{G_{5}(s)}{\vartheta(s)}G^{\prime}\bigg{(}\mathcal{F}_{2}(s)-\frac{G_{5}(s)}{\vartheta(s)}\bigg{)}
ϵ1ϑ(s)2(s)G(2(s))ϵ1ϑ(s)G5(s)ϑ(s)G(G5(s)ϑ(s)).\displaystyle\leq\epsilon_{1}\vartheta(s)\mathcal{F}_{2}(s)G^{\prime}\bigg{(}\mathcal{F}_{2}(s)\bigg{)}-\epsilon_{1}\vartheta(s)\frac{G_{5}(s)}{\vartheta(s)}G^{\prime}\bigg{(}\frac{G_{5}(s)}{\vartheta(s)}\bigg{)}.

Now, we let

3(s)=ϵ1ϑ(s)2(s)ϵ1G5(s),\mathcal{F}_{3}(s)=\epsilon_{1}\vartheta(s)\mathcal{F}_{2}(s)-\epsilon_{1}G_{5}(s), (4.23)

where ϵ1\epsilon_{1} small enough so that 3(0)1\mathcal{F}_{3}(0)\leq 1. Then (4.22) becomes, for any 0st0\leq s\leq t,

G2(3(s))ϵ1ϑ(s)G2(2(s))ϵ1ϑ(s)G2(G5(s)ϑ(s)).\displaystyle G_{2}\bigg{(}\mathcal{F}_{3}(s)\bigg{)}\leq\epsilon_{1}\vartheta(s)G_{2}\bigg{(}\mathcal{F}_{2}(s)\bigg{)}-\epsilon_{1}\vartheta(s)G_{2}\bigg{(}\frac{G_{5}(s)}{\vartheta(s)}\bigg{)}. (4.24)

Further, we have

3(t)=ϵ1ϑ(t)2(t)+ϵ1ϑ(s)2(t)ϵ1G5(t).\displaystyle\mathcal{F}^{\prime}_{3}(t)=\epsilon_{1}\vartheta^{\prime}(t)\mathcal{F}_{2}(t)+\epsilon_{1}\vartheta(s)\mathcal{F}^{\prime}_{2}(t)-\epsilon_{1}G_{5}^{\prime}(t). (4.25)

Since ϑ0\vartheta^{\prime}\leq 0 and using (4.16), then for any 0st0\leq s\leq t, 0<ϵ110<\epsilon_{1}\leq 1, we obtain

3(t)ϵ1ϑ(t)2(t)ϵ1G5(t)\displaystyle\mathcal{F}^{\prime}_{3}(t)\leq\epsilon_{1}\vartheta(t)\mathcal{F}^{\prime}_{2}(t)-\epsilon_{1}G_{5}^{\prime}(t) (4.26)
c1ϵ1ξ(t)ϑ(t)G2(2(t))+c2ϵ1ξ(t)ϑ(s)G4[cdΛ(t)h0(t)]ϵ1G5(t).\displaystyle\leq-c_{1}\epsilon_{1}\xi(t)\vartheta(t)G_{2}(\mathcal{F}_{2}(t))+c_{2}\epsilon_{1}\xi(t)\vartheta(s)G_{4}\left[\frac{c}{d}\Lambda(t)h_{0}(t)\right]-\epsilon_{1}G_{5}^{\prime}(t).

Then, using (4.5) and (4.24), we get

3(t)c1ξ(t)G2(3(t))+c2ϵ1ξ(t)ϑ(t)G4[cdΛ(t)h0(t)]\displaystyle\mathcal{F}^{\prime}_{3}(t)\leq-c_{1}\xi(t)G_{2}(\mathcal{F}_{3}(t))+c_{2}\epsilon_{1}\xi(t)\vartheta(t)G_{4}\left[\frac{c}{d}\Lambda(t)h_{0}(t)\right] (4.27)
c1ϵ1ξ(t)ϑ(t)G2(G5(t)ϑ(t))ϵ1G5(t).\displaystyle-c_{1}\epsilon_{1}\xi(t)\vartheta(t)G_{2}\bigg{(}\frac{G_{5}(t)}{\vartheta(t)}\bigg{)}-\epsilon_{1}G_{5}^{\prime}(t).

From the definition of G1G_{1} and G5G_{5}, we have

G1(G5(s))=c10sξ(τ)𝑑τ,G_{1}\left(G_{5}(s)\right)=c_{1}\int_{0}^{s}\xi(\tau)d\tau,

hence,

G5(s)=c1ξ(s)G2(G5(s)).G_{5}^{\prime}(s)=-c_{1}\xi(s)G_{2}\left(G_{5}(s)\right). (4.28)

Now, we have

c2ϵ1ξ(t)ϑ(t)G4[cdΛ(t)h0(t)]c1ϵ1ξ(t)ϑ(t)G2(G5(t)ϑ(t))ϵ1G5(t)\displaystyle c_{2}\epsilon_{1}\xi(t)\vartheta(t)G_{4}\left[\frac{c}{d}\Lambda(t)h_{0}(t)\right]-c_{1}\epsilon_{1}\xi(t)\vartheta(t)G_{2}\bigg{(}\frac{G_{5}(t)}{\vartheta(t)}\bigg{)}-\epsilon_{1}G_{5}^{\prime}(t) (4.29)
=c2ϵ1ξ(t)ϑ(t)G4[cdΛ(t)h0(t)]ϵ1ξ(t)ϑ(t)G2(G5(t)ϑ(t))+cϵ1ξ(t)G2(G5(t))\displaystyle=c_{2}\epsilon_{1}\xi(t)\vartheta(t)G_{4}\left[\frac{c}{d}\Lambda(t)h_{0}(t)\right]-\epsilon_{1}\xi(t)\vartheta(t)G_{2}\bigg{(}\frac{G_{5}(t)}{\vartheta(t)}\bigg{)}+c\epsilon_{1}\xi(t)G_{2}\left(G_{5}(t)\right)
=ϵ1ξ(t)ϑ(t)(c2G4[cdΛ(t)h0(t)]c1G2(G5(t)ϑ(t))+c1G2(G5(t))ϑ(t)).\displaystyle=\epsilon_{1}\xi(t)\vartheta(t)\bigg{(}c_{2}G_{4}\left[\frac{c}{d}\Lambda(t)h_{0}(t)\right]-c_{1}G_{2}\bigg{(}\frac{G_{5}(t)}{\vartheta(t)}\bigg{)}+c_{1}\frac{G_{2}\left(G_{5}(t)\right)}{\vartheta(t)}\bigg{)}.

Then, according to (4.5), we get

ϵ1ξ(t)ϑ(t)(c2G4[cdΛ(t)h0(t)]c1G2(G5(t)ϑ(t))c1G2(G5(t))ϑ(t))0\epsilon_{1}\xi(t)\vartheta(t)\bigg{(}c_{2}G_{4}\left[\frac{c}{d}\Lambda(t)h_{0}(t)\right]-c_{1}G_{2}\bigg{(}\frac{G_{5}(t)}{\vartheta(t)}\bigg{)}-c_{1}\frac{G_{2}\left(G_{5}(t)\right)}{\vartheta(t)}\bigg{)}\leq 0

Then (4.27) gives

3(t)c1ξ(t)G2(3(t)).\displaystyle\mathcal{F}^{\prime}_{3}(t)\leq-c_{1}\xi(t)G_{2}(\mathcal{F}_{3}(t)). (4.30)

Thus from (4.30) and the definition of G1G_{1} and G2G_{2} in (4.2) and (4.3), we obtain

(G1(3(t)))c1ξ(t).\bigg{(}G_{1}\left(\mathcal{F}_{3}(t)\right)\bigg{)}^{\prime}\geq c_{1}\xi(t). (4.31)

Integrating (4.31) over [0,t][0,t], we get

G1(3(t))c10tξ(s)𝑑s+G1(3(0)).G_{1}\left(\mathcal{F}_{3}(t)\right)\geq c_{1}\int_{0}^{t}\xi(s)ds+G_{1}\left(\mathcal{F}_{3}(0)\right). (4.32)

Since G1G_{1} is decreasing, 3(0)1\mathcal{F}_{3}(0)\leq 1 and G1(1)=0G_{1}(1)=0, then

3(t)G11(c10tξ(s)𝑑s)=G5(t).\mathcal{F}_{3}(t)\leq G_{1}^{-1}\bigg{(}c_{1}\int_{0}^{t}\xi(s)ds\bigg{)}=G_{5}(t). (4.33)

Recalling that 3(t)=ϵ1ϑ(t)2(t)ϵ1G5(t)\mathcal{F}_{3}(t)=\epsilon_{1}\vartheta(t)\mathcal{F}_{2}(t)-\epsilon_{1}G_{5}(t), we have

2(t)(1+ϵ1)ϵ1G5(t)ϑ(t),\mathcal{F}_{2}(t)\leq\frac{(1+\epsilon_{1})}{\epsilon_{1}}\frac{G_{5}(t)}{\vartheta(t)}, (4.34)

Similarly, recall that 2(t):=d01(t)Λ(t)\mathcal{F}_{2}(t):=d_{0}\mathcal{F}_{1}(t)\Lambda(t), then

1(t)(1+ϵ1)d0ϵ1G5(t)ϑ(t)Λ(t),\mathcal{F}_{1}(t)\leq\frac{(1+\epsilon_{1})}{d_{0}\epsilon_{1}}\frac{G_{5}(t)}{\vartheta(t)\Lambda(t)}, (4.35)

Since 1E\mathcal{F}_{1}\sim E, then for some b>0b>0, we have E(t)b1E(t)\leq b\mathcal{F}_{1}; which gives

E(t)b(1+ϵ1)d0ϵ1G5(t)ϑ(t)Λ(t),E(t)\leq\frac{b(1+\epsilon_{1})}{d_{0}\epsilon_{1}}\frac{G_{5}(t)}{\vartheta(t)\Lambda(t)}, (4.36)

From (4.18) and (4.36), we obtain the following estimate

E(t)c3(G5(t)ϑ(t)Λ(t)),E(t)\leq c_{3}\bigg{(}\frac{G_{5}(t)}{\vartheta(t)\Lambda(t)}\bigg{)}, (4.37)

where c3=max{2b0,b(1+ϵ1)d0ϵ1}.c_{3}=\max\{\frac{2}{b_{0}},\frac{b(1+\epsilon_{1})}{d_{0}\epsilon_{1}}\}.

5 Examples

Let g(t)=a(1+t)βg(t)=\frac{a}{(1+t)^{\beta}}, where β>1\beta>1 and 0<a<β10<a<\beta-1 so that (A1)(A1) is satisfied. In this case ξ(t)=βa1β\xi(t)=\beta a^{\frac{-1}{\beta}} and G(t)=tβ+1βG(t)=t^{\frac{\beta+1}{\beta}}. Then, there exist positive constants ai(i=0,,3)a_{i}(i=0,...,3) depending only on a,βa,\beta such that

G4(t)=a0tβ+1β,G2(t)=a1tβ+1β,G1(t)=a2(t1β1),G5(t)=(a3t+1)β.\displaystyle G_{4}(t)=a_{0}t^{\frac{\beta+1}{\beta}},~{}~{}~{}G_{2}(t)=a_{1}t^{\frac{\beta+1}{\beta}},~{}~{}~{}~{}G_{1}(t)=a_{2}(t^{\frac{-1}{\beta}}-1),~{}~{}G_{5}(t)=(a_{3}t+1)^{-\beta}. (5.1)

We will discuss two cases:
Case 1: if

m0(1+t)r1+u02m1(1+t)rm_{0}(1+t)^{r}\leq 1+||\nabla u_{0}||^{2}\leq m_{1}(1+t)^{r} (5.2)

where 0<r<β10<r<\beta-1 and m0,m1>0m_{0},m_{1}>0, then we have, for some positive constants ai(i=4,,7)a_{i}(i=4,...,7) depending only on a,β,m0,m1,ra,\beta,m_{0},m_{1},r, the following:

a4(1+t)β+1+rh0(t)a5(1+t)β+1+r,\displaystyle a_{4}(1+t)^{-\beta+1+r}\leq h_{0}(t)\leq a_{5}(1+t)^{-\beta+1+r}, (5.3)
λΛ(t)a6{1+ln(1+t),βr=2;2,βr>2;(1+t)β+r+2,1<βr<2 .\displaystyle\frac{\lambda}{\Lambda(t)}\geq a_{6}\left\{\begin{array}[]{ll}1+\ln(1+t),&\hbox{$\beta-r=2$;}\\ 2,&\hbox{$\beta-r>2$;}\\ (1+t)^{-\beta+r+2},&\hbox{$1<\beta-r<2$ .}\\ \end{array}\right. (5.4)
λΛ(t)a7{1+ln(1+t),βr=2;2,βr>2;(1+t)β+r+2,1<βr<2 .\displaystyle\frac{\lambda}{\Lambda(t)}\leq a_{7}\left\{\begin{array}[]{ll}1+\ln(1+t),&\hbox{$\beta-r=2$;}\\ 2,&\hbox{$\beta-r>2$;}\\ (1+t)^{-\beta+r+2},&\hbox{$1<\beta-r<2$ .}\\ \end{array}\right. (5.5)

We notice that condition (4.5) is satisfied if

(t+1)βΛ(t)h0(t)ϑ(t)a8(1(ϑ)1β)ββ+1.(t+1)^{\beta}\Lambda(t)h_{0}(t)\vartheta(t)\leq a_{8}\bigg{(}1-(\vartheta)^{\frac{1}{\beta}}\bigg{)}^{\frac{\beta}{\beta+1}}. (5.6)

where a8>0a_{8}>0 depending on a,β,c1a,\beta,c_{1} and c2c_{2}. Choosing ϑ(t)\vartheta(t) as the following

ϑ(t)=λ{(1+t)p,p=r+1βr2;(1+t)p,p=β1,1<βr<2 .\displaystyle\vartheta(t)=\lambda\left\{\begin{array}[]{ll}(1+t)^{-p},~{}~{}~{}~{}~{}p=r+1&\hbox{$\beta-r\geq 2$;}\\ (1+t)^{-p},~{}~{}~{}~{}~{}p=\beta-1,&\hbox{$1<\beta-r<2$ .}\\ \end{array}\right. (5.7)

so that (4.4) is valid. Moreover, using (5.3) and (5.4), we see that (5.6) is satisfied if 0<λ10<\lambda\leq 1 is small enough, and then (4.5) is satisfied. Hence (4.6) and (5.5) imply that, for any t+t\in\mathbb{R}_{+}

E(t)a9{(1+ln(1+t))(1+t)(βr1),βr=2;(1+t)(βr1),βr>2;(1+t)(βr1), 1<βr<2 .\displaystyle E(t)\leq a_{9}\left\{\begin{array}[]{ll}\bigg{(}1+\ln(1+t)\bigg{)}(1+t)^{-(\beta-r-1)},&\hbox{$\beta-r=2$;}\\ (1+t)^{-(\beta-r-1)},&\hbox{$\beta-r>2$;}\\ (1+t)^{-(\beta-r-1)},&\hbox{ $1<\beta-r<2$ .}\\ \end{array}\right. (5.8)

Thus, the estimate (5.8) gives limt+E(t)=0\lim_{t\rightarrow+\infty}E(t)=0.
Case 2: if m01+u02m1m_{0}\leq 1+||\nabla u_{0}||^{2}\leq m_{1}. That is r=0r=0 in (5.2) as it was assumed in many papers in the literature. So, it clear the estimate (5.8) gives limt+E(t)=0\lim_{t\rightarrow+\infty}E(t)=0 when r=0r=0.

Acknowledgment

The authors thank King Fahd University of Petroleum and Minerals (KFUPM) for its continuous supports. This work is supported by KFUPM under project # SB201026.

Conflict of Interest

The authors declare that there is no conflict of interest regarding the publication of this paper.

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