Arbitrary decays for a viscoelastic equation

Abstract

In this paper, we consider the nonlinear viscoelastic equation $\mid u_t{\mid }^{\rho }{u}_{tt}-\Delta u-\Delta u_{tt}+{\int }_0^{t}g\left(t-s\right)\Delta u\left(s\right)\mathsf{\text{d}}s+\mid u{\mid }^{p}u=0$, in a bounded domain with initial conditions and Dirichlet boundary conditions. We prove an arbitrary decay result for a class of kernel function g without setting the function g itself to be of exponential (polynomial) type, which is a necessary condition for the exponential (polynomial) decay of the solution energy for the viscoelastic problem. The key ingredient in the proof is based on the idea of Pata (Q Appl Math 64:499-513, 2006) and the work of Tatar (J Math Phys 52:013502, 2010), with necessary modification imposed by our problem.

Mathematical Subject Classification (2010): 35B35, 35B40, 35B60

1 Introduction

It is well known that viscoelastic materials have memory effects. These properties are due to the mechanical response influenced by the history of the materials themselves. As these materials have a wide application in the natural sciences, their dynamics are of great importance and interest. From the mathematical point of view, their memory effects are modeled by an integro-differential equations. Hence, questions related to the behavior of the solutions for the PDE system have attracted considerable attention in recent years. Many authors have focused on this problem for the last two decades and several results concerning existence, decay and blow-up have been obtained, see [1–28] and the reference therein.

In [3], Cavalcanti et al. studied the following problem

$$\begin{array}{c}\mid u_t{\mid }^{\rho }{u}_{tt}-\Delta u-\Delta u_{tt}+\underset{0}{\overset{t}{\int }}g\left(t-s\right)\Delta u\left(s\right)\mathsf{\text{d}}s-\gamma \Delta u_t=0,\mathsf{\text{in }}\Omega \times \left(0,\infty \right),\\ u\left(x,0\right)=u_0\left(x\right),u_t\left(x,0\right)=u_1\left(x\right),x\in \Omega ,\\ u\left(x,t\right)=0,x\in \partial \Omega ,t\ge 0,\end{array}$$

(1.1)

where Ω ⊂ R^N, N ≥ 1, is a bounded domain with a smooth boundary ∂Ω, γ ≥ 0, $0<\rho \le \frac{2}{N-2}$ if N ≥ 3 or ρ > 0 if N = 1, 2, and the function g: R⁺ → R⁺ is a nonincreasing function. This type of equations usually arise in the theory of viscoelasticity when the material density varies according to the velocity. In that paper, they proved a global existence result of weak solutions for γ ≥ 0 and a uniform decay result for γ > 0. Precisely, they showed that the solutions goes to zero in an exponential rate for γ > 0 and g is a positive bounded C¹-function satisfying

$$1-\underset{0}{\overset{\infty }{\int }}g\left(s\right)\mathsf{\text{d}}s=1-l>0,$$

(1.2)

and

$$-{\xi }_{1}g\left(t\right)\le g^{\prime }\left(t\right)\le -{\xi }_{2}g\left(t\right),$$

(1.3)

for all t ≥ 0 and some positive constants ξ₁ and ξ₂. Later, this result was extended by Messaoudi and Tatar [15] to a situation where a nonlinear source term is competing with the dissipation terms induced by both the viscoelasticity and the viscosity. Recently Messaoudi and Tatar [14] studied problem (1.1) for the case of γ = 0, they improved the result in [3] by showing that the solution goes to zero with an exponential or polynomial rate, depending on the decay rate of the relaxation function g.

The assumptions (1.2) and (1.3), on g, are frequently encountered in the linear case (ρ = 0), see [1, 2, 4–6, 13, 22, 23, 29–31]. Lately, these conditions have been weakened by some researchers. For instance, instead of (1.3) Furati and Tatar [8] required the functions e^αt g(t) and e^αtg'(t) to have sufficiently small L¹-norm on (0, ∞) for some α > 0 and they can also have an exponential decay of solutions. In particular, they do not impose a rate of decreasingness for g. Later on Messaoudi and Tatar [21] improved this result further by removing the condition on g'. They established an exponential decay under the conditions g'(t) ≤ 0 and e^αt g(t) ∈ L¹(0, ∞) for some large α > 0. This last condition was shown to be necessary condition for exponential decay [7]. More recently Tatar [25] investigated the asymptotic behavior to problem (1.1) with ρ = γ = 0 when h(t)g(t) ∈ L¹(0, ∞) for some nonnegative function h(t). He generalized earlier works to an arbitrary decay not necessary of exponential or polynomial rate.

Motivated by previous works [21, 25], in this paper, we consider the initial boundary value problem for the following nonlinear viscoelastic equation:

$$\mid u_t{\mid }^{\rho }{u}_{tt}-\Delta u-\Delta u_{tt}+\underset{0}{\overset{t}{\int }}g\left(t-s\right)\Delta u\left(s\right)\mathsf{\text{d}}s+\mid u{\mid }^{p}u=0,\mathsf{\text{in }}\Omega \times \left(0,\infty \right),$$

(1.4)

with initial conditions

$$u\left(x,0\right)=u_0\left(x\right),u_t\left(x,0\right)=u_1\left(x\right),x\in \Omega ,$$

(1.5)

and boundary condition

$$u\left(x,t\right)=0,x\in \partial \Omega ,t\ge 0,$$

(1.6)

where Ω ⊂ R^N, N ≥ 1, is a bounded domain with a smooth boundary ∂Ω. Here ρ, p > 0 and g represents the kernel of the memory term, with conditions to be stated later [see assumption (A1)-(A3)].

We intend to study the arbitrary decay result for problem (1.4)-(1.6) under the weaker assumption on g, which is not necessarily decaying in an exponential or polynomial fashion. Indeed, our result will be established under the conditions g'(t) ≤ 0 and ${\int }_0^{\infty }\xi \left(s\right)g\left(s\right)\mathsf{\text{d}}s<\infty$ for some nonnegative function ξ(t). Therefore, our result allows a larger class of relaxation functions and improves some earlier results concerning the exponential decay or polynomial decay.

The content of this paper is organized as follows. In Section 2, we give some lemmas and assumptions which will be used later, and we mention the local existence result in Theorem 2.2. In Section 3, we establish the statement and proof of our result related to the arbitrary decay.

2 Preliminary results

In this section, we give some assumptions and lemmas which will be used throughout this work. We use the standard Lebesgue space L^p(Ω) and Sobolev space $H_0^1\left(\Omega \right)$ with their usual inner products and norms.

Lemma 2.1. (Sobolev-Poincaré inequality) Let $2\le p\le \frac{2N}{N-2}$, the inequality

$$\parallel u{\parallel }_p\le c_s\parallel \nabla u{\parallel }_2\mathsf{\text{for }}u\in H_0^1\left(\Omega \right),$$

holds with the optimal positive constant c _s , where || · || _p denotes the norm of L^p(Ω).

Assume that ρ satisfies

$$0<\rho \le \frac{2}{N-2}\mathsf{\text{if }}N\ge 3\mathsf{\text{or}}\rho >0\mathsf{\text{if}}N=1,2.$$

(2.1)

With regards to the relaxation function g(t), we assume that it verifies

(A1) g(t) ≥ 0, for all t ≥ 0, is a continuous function satisfying

$$0<\underset{0}{\overset{\infty }{\int }}g\left(s\right)\mathsf{\text{d}}s=l<1.$$

(2.2)

(A2) g'(t) ≤ 0 for almost all t > 0.

(A3) There exists a positive nondecreasing function ξ(t): [0, ∞) → (0, ∞) such that $\frac{{\xi }^{\prime }\left(t\right)}{\xi \left(t\right)}=\eta \left(t\right)$ is a decreasing function and

$$\underset{0}{\overset{\infty }{\int }}\xi \left(s\right)g\left(s\right)\mathsf{\text{d}}s<\infty .$$

(2.3)

Now, we state, without a proof, the existence result of the problem (1.4)-(1.6) which can be established by Faedo-Galerkin methods, we refer the reader to [3, 5].

Theorem 2.2. Suppose that (2.1) and (A1) hold, and that $\left(u_0,u_1\right)\in H_0^1\left(\Omega \right)\times H_0^1\left(\Omega \right)$. Assume $0<\rho \le \frac{2}{N-2}$, if N ≥ 3, p > 0, if N = 1, 2. Then there exists at least one global solution u of (1.4)-(1.6) satisfying

$$u\in L^{\infty }\left(\left[0,\infty \right);H_0^1\left(\Omega \right)\right),u_t\in L^{\infty }\left(\left[0,\infty \right);H_0^1\left(\Omega \right)\right),u_{tt}\in L^2\left(\left[0,\infty \right);L^2\left(\Omega \right)\right).$$

Next, we introduce the modified energy functional for problem (1.4)-(1.6)

$$\begin{array}{cc}\hfill E\left(t\right)& =\frac{1}{\rho +2}\parallel u_t{\parallel }_{\rho +2}^{\rho +2}+\frac{1}{2}\left(1-\underset{0}{\overset{t}{\int }}g\left(s\right)\mathsf{\text{d}}s\right)\parallel \nabla u\left(t\right){\parallel }_2^2+\frac{1}{2}\parallel \nabla u_t\left(t\right){\parallel }_2^2\hfill \\ +\frac{1}{2}\left(g\circ \nabla u\right)\left(t\right)+\frac{1}{p+2}\parallel u\left(t\right){\parallel }_{p+2}^{p+2},\hfill \end{array}$$

(2.4)

where

$$\left(g\circ \nabla u\right)\left(t\right)=\underset{0}{\overset{t}{\int }}\underset{\Omega }{\int }g\left(t-s\right)\mid \nabla u\left(t\right)-\nabla u\left(s\right){\mid }^2\mathsf{\text{d}}x\mathsf{\text{d}}s.$$

(2.5)

Lemma 2.3. Let u be the solution of (1.4)-(1.6), then the modified energy E(t) satisfies

$$E^{\prime }\left(t\right)=\frac{1}{2}\left(g^{\prime }\circ \nabla u\right)\left(t\right)-\frac{1}{2}g\left(t\right)\parallel \nabla u\left(t\right){\parallel }_2^2\le \frac{1}{2}\left(g^{\prime }\circ \nabla u\right)\left(t\right)\le 0.$$

(2.6)

Proof. Multiplying Eq. (1.4) by u _t and integrating it over Ω, then using integration by parts and the assumption (A1)-(A2), we obtain (2.6).

Remark. It follows from Lemma 2.3 that the energy is uniformly bounded by E(0) and decreasing in t. Besides, from the definition of E(t) and (2, 2), we note that

$$\left(1-l\right)\parallel \nabla u{\parallel }_2^2+\parallel \nabla u_t\left(t\right){\parallel }_2^2+\left(g\circ \nabla u\right)\left(t\right)\le 2E\left(0\right),\forall t\ge 0.$$

(2.7)

3 Decay of the solution energy

In this section, we shall state and prove our main result. For this purpose, we first define the functional

$$L\left(t\right)=E\left(t\right)+\sum _{i=1}^3{\lambda }_i{\Phi }_i\left(t\right),$$

(3.1)

where λ _i are positive constants, i = 1, 2, 3 to be specified later and

$${\Phi }_1\left(t\right)=\frac{1}{\rho +1}\underset{\Omega }{\int }\mid u_t{\mid }^{\rho }{u}_{t}u\mathsf{\text{d}}x+\underset{\Omega }{\int }\nabla u_t\left(t\right)\nabla u\left(t\right)\mathsf{\text{d}}x,$$

(3.2)

$${\Phi }_2\left(t\right)=\underset{\Omega }{\int }\left(\Delta u_t-\frac{1}{\rho +1}{\mid u_t\mid }^{\rho }{u}_t\right)\underset{0}{\overset{t}{\int }}g\left(t-s\right)\left(u\left(t\right)-u\left(s\right)\right)\mathsf{\text{d}}s\mathsf{\text{d}}x,$$

(3.3)

$${\Phi }_3\left(t\right)=\underset{\Omega }{\int }\underset{0}{\overset{t}{\int }}H\left(t-s\right)\mid \nabla u\left(s\right){\mid }^2\mathsf{\text{d}}s\mathsf{\text{d}}x,$$

(3.4)

here

$$H\left(t\right)=\xi {\left(t\right)}^{-1}\underset{t}{\overset{\infty }{\int }}g\left(s\right)\xi \left(s\right)\mathsf{\text{d}}s.$$

Remark. This functional was first introduced by Tatar [25] for the case of ρ = 0 and without imposing the dispersion term and forcing term as far as (1.4) is concerned.

The following Lemma tells us that L(t) and E(t) + Φ₃(t) are equivalent.

Lemma 3.1. There exists two positive constants β₁ and β₂ such that the relation

$${\beta }_1\left(E\left(t\right)+{\Phi }_3\left(t\right)\right)\le L\left(t\right)\le {\beta }_2\left(E\left(t\right)+{\Phi }_3\left(t\right)\right),$$

(3.5)

holds for all t ≥ 0 and λ _i small, i = 1, 2.

Proof. By Hölder inequality Young's inequality Lemma 2.1, (2.7) and (2.2), we deduce that

$$\begin{array}{cc}\hfill \left|\underset{\Omega }{\int }\mid u_t{\mid }^{\rho }{u}_{t}u\mathsf{\text{d}}x\right|& \le \frac{1}{2}\parallel u_t{\parallel }_{2\left(\rho +1\right)}^{2\left(\rho +1\right)}+\frac{1}{2}\parallel u{\parallel }_2^2\hfill \\ \le \frac{c_s^{2\left(\rho +1\right)}}{2}\parallel \nabla u_t{\parallel }_2^{2\left(\rho +1\right)}+\frac{c_s^2}{2}\parallel \nabla u{\parallel }_2^2\hfill \\ \le \frac{{\alpha }_1}{2}\parallel \nabla u_t{\parallel }_2^2+\frac{c_s^2}{2}\parallel \nabla u{\parallel }_2^2,\hfill \\ \hfill \left|\underset{\Omega }{\int }\nabla u_t\left(t\right)\nabla u\left(t\right)\mathsf{\text{d}}x\right|& \le \frac{1}{2}\left(\parallel \nabla u_t{\parallel }_2^2+\parallel \nabla u{\parallel }_2^2\right),\hfill \end{array}$$

$$\begin{array}{c}\left|\underset{\Omega }{\int }\nabla u_t\underset{0}{\overset{t}{\int }}g\left(t-s\right)\left(\nabla u\left(t\right)-\nabla u\left(s\right)\right)\mathsf{\text{d}}s\mathsf{\text{d}}x\right|\\ \le \frac{1}{2}\parallel \nabla u_t{\parallel }_2^2+\frac{1}{2}\underset{\Omega }{\int }{\left(\underset{0}{\overset{t}{\int }}g\left(t-s\right)\left(\nabla u\left(t\right)-\nabla u\left(s\right)\right)\right)}^2\mathsf{\text{d}}s\mathsf{\text{d}}x\\ \le \frac{1}{2}\parallel \nabla u_t{\parallel }_2^2+\frac{l}{2}\left(g\circ \nabla u\right)\left(t\right),\end{array}$$

and

$$\begin{array}{c}\left|\underset{\Omega }{\int }\mid u_t{\mid }^{\rho }{u}_t\underset{0}{\overset{t}{\int }}g\left(t-s\right)\left(u\left(t\right)-u\left(s\right)\right)\mathsf{\text{d}}s\mathsf{\text{d}}x\right|\\ \le \frac{1}{2}\parallel u_t{\parallel }_{2\left(\rho +1\right)}^{2\left(\rho +1\right)}+\frac{1}{2}\underset{\Omega }{\int }{\left(\underset{0}{\overset{t}{\int }}g\left(t-s\right)\left(u\left(t\right)-u\left(s\right)\right)\right)}^2\mathsf{\text{d}}s\mathsf{\text{d}}x\\ \le \frac{{\alpha }_1}{2}\parallel \nabla u_t{\parallel }_2^2+\frac{l{c}_s^2}{2}\left(g\circ \nabla u\right)\left(t\right),\end{array}$$

where ${\alpha }_1=c_s^{2\left(\rho +1\right)}{\left(2E\left(0\right)\right)}^{\rho }$. Therefore, from above estimates, the definition of E(t) by (2.4) and (2.2), we have

$$\begin{array}{cc}\hfill L\left(t\right)& =E\left(t\right)+\sum _{i=1}^3{\lambda }_i{\Phi }_i\left(t\right)\hfill \\ \le E\left(t\right)+c_1\left|\right|\nabla u|{|}_2^2+c_2||\nabla u_t|{|}_2^2+c_3\left(g\circ \nabla u\right)\left(t\right)+{\lambda }_3{\Phi }_3\left(t\right)\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill L\left(t\right)& \ge E\left(t\right)-c_1\parallel \nabla u{\parallel }_2^2-c_2\parallel \nabla u_t{\parallel }_2^2-c_3\left(g\circ \nabla u\right)\left(t\right)+{\lambda }_3{\Phi }_3\left(t\right)\hfill \\ \ge \frac{1}{\rho +2}\parallel u_t{\parallel }_{\rho +2}^{\rho +2}+\left(\frac{1}{2}\left(1-l\right)-c_1\right)\parallel \nabla u\left(t\right){\parallel }_2^2\hfill \\ +\left(\frac{1}{2}-c_2\right)\parallel \nabla u_t\left(t\right){\parallel }_2^2+\left(\frac{1}{2}-c_3\right)\left(g\circ \nabla u\right)\left(t\right)\hfill \\ +\frac{1}{p+2}\parallel u\left(t\right){\parallel }_{p+2}^{p+2}+{\lambda }_3{\Phi }_3\left(t\right),\hfill \end{array}$$

where $c_1=\frac{{\lambda }_1\left(c_s^2+\rho +1\right)}{2\left(\rho +1\right)}$, $c_2=\frac{\left({\lambda }_1+{\lambda }_2\right)\left({\alpha }_1+\rho +1\right)}{2\left(\rho +1\right)}$, and $c_3=\frac{l\left(\rho +1+c_s^2\right){\lambda }_2}{2\left(\rho +1\right)}$. Hence, selecting λ _i , i = 1, 2 such that

$$\begin{array}{c}{\lambda }_1<min\left\{\frac{\left(\rho +1\right)\left(1-l\right)}{\underset{s}{\overset{2}{c}}+\rho +1},\frac{\rho +1}{{\alpha }_1+\rho +1}\right\},\\ {\lambda }_2<min\left\{\frac{\rho +1}{l\left(\underset{s}{\overset{2}{c}}+\rho +1\right)},\frac{\rho +1}{{\alpha }_1+\rho +1}-{\lambda }_1\right\},\end{array}$$

and again from the definition of E(t), there exist two positive constants β₁ and β₂ such that

$${\beta }_1\left(E\left(t\right)+{\Phi }_3\left(t\right)\right)\le L\left(t\right)\le {\beta }_2\left(E\left(t\right)+{\Phi }_3\left(t\right)\right),t\ge 0.$$

To obtain a better estimate for ${\int }_{\Omega }\nabla u{\int }_0^{t}g\left(t-s\right)\nabla u\left(s\right)\mathsf{\text{d}}s\mathsf{\text{d}}x$, we need the following Lemma which repeats Lemma 2 in [25].

Lemma 3.2. For t ≥ 0, we have

$$\begin{array}{cc}\hfill \underset{\Omega }{\int }\nabla u\underset{0}{\overset{t}{\int }}g\left(t-s\right)\nabla u\left(s\right)\mathsf{\text{d}}s\mathsf{\text{d}}x& =\frac{1}{2}\left(\underset{0}{\overset{t}{\int }}g\left(s\right)\mathsf{\text{d}}s\right)\parallel \nabla u{\parallel }_2^2+\frac{1}{2}\underset{0}{\overset{t}{\int }}g\left(t-s\right)\parallel \nabla u\left(s\right){\parallel }_2^2\mathsf{\text{d}}s\hfill \\ -\frac{1}{2}\left(g\circ \nabla u\right)\left(t\right).\hfill \end{array}$$

(3.6)

Proof. Straightforward computations yield this identity.

Now, we are ready to state and prove our result. First, we introduce the following notations as in [24, 25]. For every measurable set A ⊂ R⁺, we define the probability measure $\hat{g}$ by

$$\hat{g}\left(A\right)=\frac{1}{l}\underset{A}{\int }g\left(s\right)\mathsf{\text{d}}s.$$

The flatness set and the flatness rate of g are defined by

$$F_g=\left\{s\in R^{+}\mid g\left(s\right)>0\mathsf{\text{and }}{g}^{\prime }\left(s\right)=0\right\}$$

(3.7)

and

$$R_g=\hat{g}\left(F_g\right)=\frac{1}{l}\underset{F_g}{\int }g\left(s\right)\mathsf{\text{d}}s.$$

(3.8)

Before proceeding, we note that there exists t₀ > 0 such that

$$\underset{0}{\overset{t}{\int }}g\left(s\right)\mathsf{\text{d}}s\ge \underset{0}{\overset{t_0}{\int }}g\left(s\right)\mathsf{\text{d}}s=g_{*}>0,\forall t\ge t_0,$$

(3.9)

since g is nonnegative and continuous.

Theorem 3.3. Let $\left(u_0,u_1\right)\in H_0^1\left(\Omega \right)\times H_0^1\left(\Omega \right)$ be given. Suppose that (A1)-(A3), (2, 1) and the hypothesis on p hold. Assume further that $R_g<\frac{g_{*}\alpha }{l\left(2\alpha +2c_s^2\right)}$, $H\left(0\right)<\frac{g_{*}\left(4+l\right)-3l}{4}$ and $g_{*}>\frac{3l}{4+l}$ with

$$\alpha =\frac{\left(g_{*}\left(4+l\right)-3l\right){\left(1-l\right)}^p}{8c_s^{2\left(p+1\right)}{\left(2E\left(0\right)\right)}^p}.$$

Then the solution energy of (1.4)-(1.6) satisfies

$$E\left(t\right)\le K\xi {\left(t\right)}^{-\mu },t\ge 0,$$

where μ and K are positive constants.

Proof. In order to obtain the decay result of E(t), it suffices to prove that of L(t). To this end, we need to estimate the derivative of L(t). It follows from (3.2) and Eq. (1.4) that

$$\begin{array}{cc}\hfill {\Phi }_1^{\prime }\left(t\right)& =\frac{1}{\rho +1}\parallel u_t{\parallel }_{\rho +2}^{\rho +2}+\parallel \nabla u_t{\parallel }_2^2-\parallel \nabla u{\parallel }_2^2\hfill \\ +\underset{\Omega }{\int }\nabla u\underset{0}{\overset{t}{\int }}g\left(t-s\right)\nabla u\left(s\right)\mathsf{\text{d}}s\mathsf{\text{d}}x-\parallel u{\parallel }_{p+2}^{p+2},\hfill \end{array}$$

which together with the identity (3.6) and (2.2) gives

$$\begin{array}{cc}\hfill {\Phi }_1^{\prime }\left(t\right)& \le \frac{1}{\rho +1}\parallel u_t{\parallel }_{\rho +2}^{\rho +2}+\parallel \nabla u_t{\parallel }_2^2-\left(1-\frac{l}{2}\right)\parallel \nabla u{\parallel }_2^2\hfill \\ +\frac{1}{2}\underset{0}{\overset{t}{\int }}g\left(t-s\right)\parallel \nabla u\left(s\right){\parallel }_2^2\mathsf{\text{d}}s-\frac{1}{2}\left(g\circ \nabla u\right)\left(t\right)-\parallel u{\parallel }_{p+2}^{p+2}.\hfill \end{array}$$

(3.10)

Next, we would like to estimate ${\Phi }_2^{\prime }\left(t\right)$. Taking a derivative of Φ₂ in (3.3) and using Eq. (1.4) to get

$$\begin{array}{cc}\hfill {\Phi }_2^{\prime }\left(t\right)& =\left(1-\underset{0}{\overset{t}{\int }}g\left(s\right)\mathsf{\text{d}}s\right)\underset{\Omega }{\int }\nabla u\left(t\right)\underset{0}{\overset{t}{\int }}g\left(t-s\right)\left(\nabla u\left(t\right)-\nabla u\left(s\right)\right)\mathsf{\text{d}}s\mathsf{\text{d}}x\hfill \\ +\underset{\Omega }{\int }{\left(\underset{0}{\overset{t}{\int }}g\left(t-s\right)\left(\nabla u\left(t\right)-\nabla u\left(s\right)\right)\mathsf{\text{d}}s\right)}^2\mathsf{\text{d}}x-\frac{1}{\rho +1}\left(\underset{0}{\overset{t}{\int }}g\left(s\right)\mathsf{\text{d}}s\right)\parallel u_t{\parallel }_{\rho +2}^{\rho +2}\hfill \\ -\left(\underset{0}{\overset{t}{\int }}g\left(s\right)\mathsf{\text{d}}s\right)\parallel \nabla u_t{\parallel }_2^2-\underset{\Omega }{\int }\nabla u_t\left(t\right)\underset{0}{\overset{t}{\int }}{g}^{\prime }\left(t-s\right)\left(\nabla u\left(t\right)-\nabla u\left(s\right)\right)\mathsf{\text{d}}s\mathsf{\text{d}}x\hfill \\ -\frac{1}{\rho +1}\underset{\Omega }{\int }\mid u_t{\mid }^{\rho }{u}_t\underset{0}{\overset{t}{\int }}{g}^{\prime }\left(t-s\right)\left(u\left(t\right)-u\left(s\right)\right)\mathsf{\text{d}}s\mathsf{\text{d}}x\hfill \\ +\underset{\Omega }{\int }\mid u{\mid }^{p}u\underset{0}{\overset{t}{\int }}g\left(t-s\right)\left(u\left(t\right)-u\left(s\right)\right)\mathsf{\text{d}}s\mathsf{\text{d}}x.\hfill \end{array}$$

(3.11)

We now estimate the first two terms on the right-hand side of (3.11) as in [25].

Indeed, for all measure set A and F such that A = R⁺ - F, we have

$$\begin{array}{c}\underset{\Omega }{\int }\nabla u\left(t\right)\underset{0}{\overset{t}{\int }}g\left(t-s\right)\left(\nabla u\left(t\right)-\nabla u\left(s\right)\right)\mathsf{\text{d}}s\mathsf{\text{d}}x\\ =\underset{\Omega }{\int }\nabla u\left(t\right)\underset{A\cap \left[0,t\right]}{\int }g\left(t-s\right)\left(\nabla u\left(t\right)-\nabla u\left(s\right)\right)\mathsf{\text{d}}s\mathsf{\text{d}}x\\ +\underset{\Omega }{\int }\nabla u\left(t\right)\underset{F\cap \left[0,t\right]}{\int }g\left(t-s\right)\left(\nabla u\left(t\right)-\nabla u\left(s\right)\right)\mathsf{\text{d}}s\mathsf{\text{d}}x\\ \le \underset{\Omega }{\int }\nabla u\left(t\right)\underset{A\cap \left[0,t\right]}{\int }g\left(t-s\right)\left(\nabla u\left(t\right)-\nabla u\left(s\right)\right)\mathsf{\text{d}}s\mathsf{\text{d}}x\\ +\left(\underset{F\cap \left[0,t\right]}{\int }g\left(s\right)\mathsf{\text{d}}s\right)\parallel \nabla u{\parallel }_2^2-\underset{\Omega }{\int }\nabla u\left(t\right)\underset{F\cap \left[0,t\right]}{\int }g\left(t-s\right)\nabla u\left(s\right)\mathsf{\text{d}}s\mathsf{\text{d}}x.\end{array}$$

(3.12)

To simplify notations, we denote

$$A_t=A\cap \left[0,t\right]\mathsf{\text{and }}{F}_t=F\cap \left[0,t\right].$$

Using Hölder inequality Young's inequality and (2.2), we see that, for δ₁ > 0,

$$\begin{array}{c}\underset{\Omega }{\int }\nabla u\left(t\right)\underset{A_t}{\int }g\left(t-s\right)\left(\nabla u\left(t\right)-\nabla u\left(s\right)\right)\mathsf{\text{d}}s\mathsf{\text{d}}x\\ \le {\delta }_1\parallel \nabla u{\parallel }_2^2+\frac{l}{4{\delta }_1}\underset{\Omega }{\int }\underset{A_t}{\int }g\left(t-s\right)\mid \nabla u\left(t\right)-\nabla u\left(s\right){\mid }^2\mathsf{\text{d}}s\mathsf{\text{d}}x\end{array}$$

and

$$\begin{array}{c}\underset{\Omega }{\int }\nabla u\left(t\right)\underset{F_t}{\int }g\left(t-s\right)\nabla u\left(s\right)\mathsf{\text{d}}s\mathsf{\text{d}}x\\ \le \frac{1}{2}\left(\underset{F_t}{\int }g\left(s\right)\mathsf{\text{d}}s\right)\parallel \nabla u{\parallel }_2^2+\frac{1}{2}\underset{F_t}{\int }g\left(t-s\right)\parallel \nabla u\left(s\right){\parallel }_2^2\mathsf{\text{d}}s.\end{array}$$

Thus, from the definition of $\hat{g}\left(F\right)$ by (3.8), (3.12) becomes

$$\begin{array}{c}\underset{\Omega }{\int }\nabla u\left(t\right)\underset{0}{\overset{t}{\int }}g\left(t-s\right)\left(\nabla u\left(t\right)-\nabla u\left(s\right)\right)\mathsf{\text{d}}s\mathsf{\text{d}}x\\ \le \left({\delta }_1+\frac{3}{2}l\hat{g}\left(F\right)\right)\parallel \nabla u{\parallel }_2^2+\frac{l}{4{\delta }_1}\underset{\Omega }{\int }\underset{A_t}{\int }g\left(t-s\right)\mid \nabla u\left(t\right)-\nabla u\left(s\right){\mid }^2\mathsf{\text{d}}s\mathsf{\text{d}}x\\ +\frac{1}{2}\underset{F_t}{\int }g\left(t-s\right)\parallel \nabla u\left(s\right){\parallel }_2^2\mathsf{\text{d}}s.\end{array}$$

(3.13)

The second term on the right-hand side of (3.11) can be estimated as follows (see [25]), for δ₂ > 0,

$$\begin{array}{cc}\hfill \underset{\Omega }{\int }& {\left(\underset{0}{\overset{t}{\int }}g\left(t-s\right)\left(\nabla u\left(t\right)-\nabla u\left(s\right)\right)\mathsf{\text{d}}s\right)}^2\mathsf{\text{d}}x\hfill \\ =\underset{\Omega }{\int }{\left(\underset{A_t}{\int }g\left(t-s\right)\left(\nabla u\left(t\right)-\nabla u\left(s\right)\right)\mathsf{\text{d}}s+\underset{F_t}{\int }g\left(t-s\right)\left(\nabla u\left(t\right)-\nabla u\left(s\right)\right)\mathsf{\text{d}}s\right)}^2\mathsf{\text{d}}x\hfill \\ \le \left(1+\frac{1}{{\delta }_2}\right)l\underset{\Omega }{\int }\underset{A_t}{\int }g\left(t-s\right)|\nabla u\left(t\right)-\nabla u\left(s\right){|}^2\mathsf{\text{d}}s\mathsf{\text{d}}x\hfill \\ +\left(1+{\delta }_2\right)l\hat{g}\left(F\right)\underset{\Omega }{\int }\underset{F_t}{\int }g\left(t-s\right)|\nabla u\left(t\right)-\nabla u\left(s\right){|}^2\mathsf{\text{d}}s\mathsf{\text{d}}x.\hfill \end{array}$$

(3.14)

Using Hölder inequality Young's inequality and (A2) to deal with the fifth term, for δ₃ > 0,

$$\begin{array}{c}\left|\underset{\Omega }{\int }\nabla u_t\left(t\right)\underset{0}{\overset{t}{\int }}{g}^{\prime }\left(t-s\right)\left(\nabla u\left(t\right)-\nabla u\left(s\right)\right)\mathsf{\text{d}}s\mathsf{\text{d}}x\right|\\ \le {\delta }_3\parallel \nabla u_t{\parallel }_2^2+\frac{1}{4{\delta }_3}\underset{\Omega }{\int }{\left(\underset{0}{\overset{t}{\int }}{g}^{\prime }\left(t-s\right)\left(\nabla u\left(t\right)-\nabla u\left(s\right)\right)\mathsf{\text{d}}s\right)}^2\mathsf{\text{d}}x\\ \le {\delta }_3\parallel \nabla u_t{\parallel }_2^2-\frac{g\left(0\right)}{4{\delta }_3}\left(g^{\prime }\circ \nabla u\right)\left(t\right).\end{array}$$

(3.15)

Exploiting Hölder inequality Young's inequality Lemma 2.1 and (A2) to estimate the sixth term, for δ₄ > 0,

$$\begin{array}{c}\left|\frac{1}{\rho +1}\underset{\Omega }{\int }\mid u_t{\mid }^{\rho }{u}_t\underset{0}{\overset{t}{\int }}{g}^{\prime }\left(t-s\right)\left(u\left(t\right)-u\left(s\right)\right)\mathsf{\text{d}}s\mathsf{\text{d}}x\right|\\ \le \frac{1}{\rho +1}\left({\delta }_4\parallel u_t{\parallel }_{2\left(\rho +1\right)}^{2\left(\rho +1\right)}+\frac{1}{4{\delta }_4}\underset{\Omega }{\int }{\left(\underset{0}{\overset{t}{\int }}{g}^{\prime }\left(t-s\right)\left(u\left(t\right)-u\left(s\right)\right)\mathsf{\text{d}}s\right)}^2\mathsf{\text{d}}x\right)\\ \le \frac{1}{\rho +1}\left({\delta }_4{c}_s^{2\left(\rho +1\right)}\parallel \nabla u_t{\parallel }_2^{2\left(\rho +1\right)}-\frac{g\left(0\right)c_s^2}{4{\delta }_4}\left(g^{\prime }\circ \nabla u\right)\left(t\right)\right)\\ \le \frac{1}{\rho +1}\left({\alpha }_1{\delta }_4\parallel \nabla u_t{\parallel }_2^2-\frac{g\left(0\right)c_s^2}{4{\delta }_4}\left(g^{\prime }\circ \nabla u\right)\left(t\right)\right).\end{array}$$

(3.16)

For the last term, thanks to Hölder inequality Young's inequality Lemma 2.1, (2.7), (2.2) and (3.8), we have, for δ₅ > 0,

$$\begin{array}{c}\left|\underset{\Omega }{\int }\mid u{\mid }^{p}u\underset{0}{\overset{t}{\int }}g\left(t-s\right)\left(u\left(t\right)-u\left(s\right)\right)\mathsf{\text{d}}s\mathsf{\text{d}}x\right|\\ \le {\delta }_5\underset{\Omega }{\int }\mid u{\mid }^{2\left(p+1\right)}\mathsf{\text{d}}x+\frac{1}{4{\delta }_5}\underset{\Omega }{\int }{\left(\underset{0}{\overset{t}{\int }}g\left(t-s\right)\left(u\left(t\right)-u\left(s\right)\right)\mathsf{\text{d}}s\right)}^2\mathsf{\text{d}}x\\ \le {\delta }_5{\alpha }_2\parallel \nabla u{\parallel }_2^2+\frac{l{c}_s^2}{2{\delta }_5}\underset{\Omega }{\int }\underset{A_t}{\int }g\left(t-s\right)\mid \nabla u\left(t\right)-\nabla u\left(s\right){\mid }^2\mathsf{\text{d}}s\mathsf{\text{d}}x\\ +\frac{l{c}_s^2}{2{\delta }_5}\hat{g}\left(F\right)\underset{\Omega }{\int }\underset{F_t}{\int }g\left(t-s\right)\mid \nabla u\left(t\right)-\nabla u\left(s\right){\mid }^2\mathsf{\text{d}}s\mathsf{\text{d}}x,\end{array}$$

(3.17)

where ${\alpha }_2=c_s^{2\left(p+1\right)}{\left(\frac{2E\left(0\right)}{1-l}\right)}^p$. Thus, gathering these estimates (3.13)-(3.17) and using (3.9), we obtain, for t ≥ t₀,

$$\begin{array}{cc}\hfill {\Phi }_2^{\prime }\left(t\right)& \le \left(\left(1-g_{*}\right)\left({\delta }_1+\frac{3}{2}l\hat{g}\left(F\right)\right)+{\delta }_5{\alpha }_2\right)\parallel \nabla u{\parallel }_2^2\hfill \\ -\left(\frac{g\left(0\right)}{4{\delta }_3}+\frac{g\left(0\right)c_s}{4\left(\rho +1\right){\delta }_4}\right)\left(g^{\prime }\circ \nabla u\right)\left(t\right)+\frac{1-g_{*}}{2}\underset{F_t}{\int }g\left(t-s\right)\parallel \nabla u\left(s\right){\parallel }_2^2\mathsf{\text{d}}s\hfill \\ +\left({\delta }_3+\frac{{\alpha }_1{\delta }_4}{\rho +1}-g_{*}\right)\parallel \nabla u_t{\parallel }_2^2-\frac{g_{*}}{\left(\rho +1\right)}\parallel u_t{\parallel }_{\rho +2}^{\rho +2}\hfill \\ +l\left(1+\frac{1}{{\delta }_2}+\frac{c_s^2}{2{\delta }_5}+\frac{1-g_{*}}{4{\delta }_1}\right)\underset{\Omega }{\int }\underset{A_t}{\int }g\left(t-s\right)\mid \nabla u\left(t\right)-\nabla u\left(s\right){\mid }^2\mathsf{\text{d}}s\mathsf{\text{d}}x\hfill \\ +\left(1+{\delta }_2+\frac{c_s^2}{2{\delta }_5}\right)l\hat{g}\left(F\right)\underset{\Omega }{\int }\underset{F_t}{\int }g\left(t-s\right)\mid \nabla u\left(t\right)-\nabla u\left(s\right){\mid }^2\mathsf{\text{d}}s\mathsf{\text{d}}x.\hfill \end{array}$$

(3.18)

Further, taking a derivative of Φ₃(t), using the fact that $\frac{{\xi }^{\prime }\left(t\right)}{\xi \left(t\right)}=\eta \left(t\right)$ is a decreasing function and the definition of Φ₃(t) by (3.4), we derive that (see [25])

$$\begin{array}{cc}\hfill {{\Phi }^{\prime }}_3\left(t\right)& =H\left(0\right)\parallel \nabla u{\parallel }_2^2-\underset{0}{\overset{t}{\int }}\frac{{\xi }^{\prime }\left(t-s\right)}{\xi \left(t-s\right)}H\left(t-s\right)\parallel \nabla u\left(s\right){\parallel }_2^2\mathsf{\text{d}}s-\underset{0}{\overset{t}{\int }}g\left(t-s\right)\parallel \nabla u\left(s\right){\parallel }_2^2\mathsf{\text{d}}s\hfill \\ \le H\left(0\right)\parallel \nabla u{\parallel }_2^2-\eta \left(t\right)\underset{0}{\overset{t}{\int }}H\left(t-s\right)\parallel \nabla u\left(s\right){\parallel }_2^2\mathsf{\text{d}}s-\underset{0}{\overset{t}{\int }}g\left(t-s\right)\parallel \nabla u\left(s\right){\parallel }_2^2\mathsf{\text{d}}s\hfill \\ =H\left(0\right)\parallel \nabla u{\parallel }_2^2-\eta \left(t\right){\Phi }_3\left(t\right)-\underset{0}{\overset{t}{\int }}g\left(t-s\right)\parallel \nabla u\left(s\right){\parallel }_2^2\mathsf{\text{d}}s.\hfill \end{array}$$

(3.19)

Hence, we conclude from (2.6), (3.10), (3.18) and (3.19) that for any t ≥ t₀ > 0,

$$\begin{array}{cc}\hfill L^{\prime }\left(t\right)& =E^{\prime }\left(t\right)+{\lambda }_1{{\Phi }^{\prime }}_1\left(t\right)+{\lambda }_2{{\Phi }^{\prime }}_2\left(t\right)+{\lambda }_3{{\Phi }^{\prime }}_3\left(t\right)\hfill \\ \le \left(\frac{1}{2}-{\lambda }_2\left(\frac{g\left(0\right)}{4{\delta }_3}+\frac{g\left(0\right)c_s^2}{4\left(\rho +1\right){\delta }_4}\right)\right)\left(g^{\prime }\circ \nabla u\right)\left(t\right)\hfill \\ +\frac{1}{\rho +1}\left({\lambda }_1-{\lambda }_2{g}_{*}\right)\parallel u_t{\parallel }_{\rho +2}^{\rho +2}-{\lambda }_1\parallel u{\parallel }_{p+2}^{p+2}-\frac{{\lambda }_1}{2}\left(g\circ \nabla u\right)\left(t\right)\hfill \\ +\left({\lambda }_1+{\lambda }_2\left({\delta }_3+\frac{{\alpha }_1{\delta }_4}{\rho +1}-g_{*}\right)\right)\parallel \nabla u_t{\parallel }_2^2+\left(\frac{{\lambda }_1}{2}-{\lambda }_3\right)\underset{0}{\overset{t}{\int }}g\left(t-s\right)\parallel \nabla u\left(s\right){\parallel }_2^2\mathsf{\text{d}}s\hfill \\ +\left({\lambda }_2\left(1-g_{*}\right)\left({\delta }_1+\frac{3}{2}l\hat{g}\left(F\right)\right)+{\lambda }_2{\delta }_5{\alpha }_2-{\lambda }_1\left(1-\frac{l}{2}\right)+{\lambda }_{3}H\left(0\right)\right)\parallel \nabla u{\parallel }_2^2\hfill \\ +{\lambda }_{2}l\left(1+\frac{1}{{\delta }_2}+\frac{c_s^2}{2{\delta }_5}+\frac{1-g_{*}}{4{\delta }_1}\right)\underset{\Omega }{\int }\underset{A_t}{\int }g\left(t-s\right)\mid \nabla u\left(t\right)-\nabla u\left(s\right){\mid }^2\mathsf{\text{d}}s\mathsf{\text{d}}x\hfill \\ +{\lambda }_2\left(1+{\delta }_2+\frac{c_s^2}{2{\delta }_5}\right)l\hat{g}\left(F\right)\underset{\Omega }{\int }\underset{F_t}{\int }g\left(t-s\right)\mid \nabla u\left(t\right)-\nabla u\left(s\right){\mid }^2\mathsf{\text{d}}s\mathsf{\text{d}}x\hfill \\ +{\lambda }_2\left(\frac{1-g_{*}}{2}\right)\underset{F_t}{\int }g\left(t-s\right)\parallel \nabla u\left(s\right){\parallel }_2^2\mathsf{\text{d}}s-{\lambda }_3\eta \left(t\right){\Phi }_3\left(t\right).\hfill \end{array}$$

(3.20)

For $n\in \mathcal{N}$, we consider the sets (see [24, 25])

$$A_n=\left\{s\in R^{+}\mid n{g}^{\prime }\left(s\right)+g\left(s\right)\le 0\right\}$$

and observe that

$$\underset{n}{\cup }{A}_n=R^{+}-\left\{F_g\cup N_g\right\},$$

where F _g is given in (3.7) and N _g is the null set where g' is not defined. In addition, denoting F _n = R⁺ - A _n , then

$$\underset{n\to \infty }{lim}\hat{g}\left(F_n\right)=\hat{g}\left(F_g\right),$$

because A _n are increasingly nested. Thus, choosing A = A _n , F = F _n and λ₁ = (g _* - ε) λ₂ for some ε > 0 in (3.20), we obtain

$$\begin{array}{cc}\hfill L^{\prime }\left(t\right)& \le \left(\frac{1}{2}-{\lambda }_2\left(\frac{g\left(0\right)}{4{\delta }_3}+\frac{g\left(0\right)c_s^2}{4\left(\rho +1\right){\delta }_4}\right)\right)\left(g^{\prime }\circ \nabla u\right)\left(t\right)\hfill \\ -\frac{{\lambda }_2\epsilon }{\rho +1}\parallel u_t{\parallel }_{\rho +2}^{\rho +2}+{\lambda }_2\left({\delta }_3+\frac{{\alpha }_1{\delta }_4}{\rho +1}-\epsilon \right)\parallel \nabla u_t{\parallel }_2^2-\left(g_{*}-\epsilon \right){\lambda }_2\parallel u{\parallel }_{p+2}^{p+2}\hfill \\ -\frac{\left(g_{*}-\epsilon \right){\lambda }_2}{2}\left(g\circ \nabla u\right)\left(t\right)+\left(\frac{\left(g_{*}-\epsilon \right){\lambda }_2}{2}-{\lambda }_3\right)\underset{0}{\overset{t}{\int }}g\left(t-s\right)\parallel \nabla u\left(s\right){\parallel }_2^2\mathsf{\text{d}}s\hfill \\ +\left({\lambda }_2\left(1-g_{*}\right)\left({\delta }_1+\frac{3}{2}l\hat{g}\left(F_n\right)\right)+{\lambda }_2{\delta }_5{\alpha }_2-{\lambda }_2\left(g_{*}-\epsilon \right)\left(1-\frac{l}{2}\right)+{\lambda }_{3}H\left(0\right)\right)\parallel \nabla u{\parallel }_2^2\hfill \\ +{\lambda }_{2}l\left(1+\frac{1}{{\delta }_2}+\frac{c_s^2}{2{\delta }_5}+\frac{1-g_{*}}{4{\delta }_1}\right)\underset{\Omega }{\int }\underset{A_{nt}}{\int }g\left(t-s\right)\mid \nabla u\left(t\right)-\nabla u\left(s\right){\mid }^2\mathsf{\text{d}}s\mathsf{\text{d}}x\hfill \\ +{\lambda }_2\left(1+{\delta }_2+\frac{c_s^2}{2{\delta }_5}\right)l\hat{g}\left(F_n\right)\underset{\Omega }{\int }\underset{F_{nt}}{\int }g\left(t-s\right)\mid \nabla u\left(t\right)-\nabla u\left(s\right){\mid }^2\mathsf{\text{d}}s\mathsf{\text{d}}x\hfill \\ +{\lambda }_2\left(\frac{1-g_{*}}{2}\right)\underset{F_{nt}}{\int }g\left(t-s\right)\parallel \nabla u\left(s\right){\parallel }_2^2\mathsf{\text{d}}s-{\lambda }_3\eta \left(t\right){\Phi }_3\left(t\right).\hfill \end{array}$$

(3.21)

At this point, we take ${\delta }_3={\delta }_4<\frac{\left(\rho +1\right)\epsilon }{2\left(\rho +1+{\alpha }_1\right)}$ and select λ₂ so that

$$\frac{1}{2}-{\lambda }_2\left(\frac{g\left(0\right)}{4{\delta }_3}+\frac{g\left(0\right)c_s^2}{4\left(\rho +1\right){\delta }_4}\right)\ge \frac{1}{4},$$

then (3.21) becomes

$$\begin{array}{cc}\hfill L^{\prime }\left(t\right)& \le -\frac{{\lambda }_2\epsilon }{\rho +1}\parallel u_t{\parallel }_{\rho +2}^{\rho +2}-\frac{{\lambda }_2\epsilon }{2}\parallel \nabla u_t{\parallel }_2^2-\left(g_{*}-\epsilon \right){\lambda }_2\parallel u{\parallel }_{p+2}^{p+2}\hfill \\ -\frac{\left(g_{*}-\epsilon \right){\lambda }_2}{2}\left(g\circ \nabla u\right)\left(t\right)+\left(\frac{\left(g_{*}-\epsilon \right){\lambda }_2}{2}-{\lambda }_3\right)\underset{0}{\overset{t}{\int }}g\left(t-s\right)\parallel \nabla u\left(s\right){\parallel }_2^2\mathsf{\text{d}}s\hfill \\ +\left({\lambda }_2\left(1-g_{*}\right)\left({\delta }_1+\frac{3}{2}l\hat{g}\left(F_n\right)\right)+{\lambda }_2{\delta }_5{\alpha }_2-{\lambda }_2\left(g_{*}-\epsilon \right)\left(1-\frac{l}{2}\right)+{\lambda }_{3}H\left(0\right)\right)\parallel \nabla u{\parallel }_2^2\hfill \\ +\left({\lambda }_{2}l\left(1+\frac{1}{{\delta }_2}+\frac{c_s^2}{2{\delta }_5}+\frac{1-g_{*}}{4{\delta }_1}\right)-\frac{1}{4n}\right)\underset{\Omega }{\int }\underset{A_{nt}}{\int }g\left(t-s\right)\mid \nabla u\left(t\right)-\nabla u\left(s\right){\mid }^2\mathsf{\text{d}}s\mathsf{\text{d}}x\hfill \\ +{\lambda }_2\left(1+{\delta }_2+\frac{c_s^2}{2{\delta }_5}\right)l\hat{g}\left(F_n\right)\underset{\Omega }{\int }\underset{F_{nt}}{\int }g\left(t-s\right)\mid \nabla u\left(t\right)-\nabla u\left(s\right){\mid }^2\mathsf{\text{d}}s\mathsf{\text{d}}x\hfill \\ +{\lambda }_2\left(\frac{1-g_{*}}{2}\right)\underset{F_{nt}}{\int }g\left(t-s\right)\parallel \nabla u\left(s\right){\parallel }_2^2\mathsf{\text{d}}s-{\lambda }_3\eta \left(t\right){\Phi }_3\left(t\right).\hfill \end{array}$$

For ε, δ₂ small enough and large value of n and t₀, we see that if

$${\delta }_5=\alpha \mathsf{\text{and }}\hat{g}\left(F_n\right)<\frac{g_{*}\alpha }{l\left(2\alpha +2c_s^2\right)},$$

then

$$\left(\left(1+{\delta }_2\right)+\frac{c_s^2}{2{\delta }_5}\right)l\hat{g}\left(F_n\right)-\frac{g_{*}-\epsilon }{2}\le 0$$

and

$$\frac{3}{2}\left(1-g_{*}\right)l\hat{g}\left(F_n\right)<\delta \left(g_{*}-\epsilon \right)\left(1-\frac{l}{2}\right),$$

(3.22)

where

$$\alpha =\frac{g_{*}\left(4+l\right)-3l}{8{\alpha }_2}=\frac{\left(g_{*}\left(4+l\right)-3l\right){\left(1-l\right)}^p}{8c_s^{2\left(p+1\right)}{\left(2E\left(0\right)\right)}^p}$$

and

$$\delta =\frac{3\left(1-g_{*}\right)l}{g_{*}\left(4-2l\right)}.$$

Note that α > 0 and 0 < δ < 1 due to $g_{*}>\frac{3l}{4+l}$. Furthermore, we require λ₂ and λ₃ satisfying

$${\lambda }_{2}l\left(1+\frac{1}{{\delta }_2}+\frac{c_s^2}{2{\delta }_5}+\frac{1-g_{*}}{4{\delta }_1}\right)-\frac{1}{4n}<0$$

and

$$\frac{{\lambda }_2}{2}<{\lambda }_3<\frac{{\lambda }_2\left(g_{*}\left(4+l\right)-3l\right)}{8H\left(0\right)},$$

this is possible because of $H\left(0\right)<\frac{g_{*}\left(4+l\right)-3l}{4}$. Then, letting δ₁ be small enough and using (3.22), we see that

$${\lambda }_2\left(1-g_{*}\right)\left({\delta }_1+\frac{3}{2}l\hat{g}\left(F_n\right)\right)+{\lambda }_2{\delta }_5{\alpha }_2-{\lambda }_2\left(g_{*}-\epsilon \right)\left(1-\frac{l}{2}\right)+{\lambda }_{3}H\left(0\right)\le 0.$$

Hence, from the definition of E(t) by (2.4), we have, for all t ≥ t₀,

$$L^{\prime }\left(t\right)\le -c_{4}E\left(t\right)-{\lambda }_3\eta \left(t\right){\Phi }_3\left(t\right),$$

for some positive constant c₄. As η(t) is decreasing, we have η(t) ≤ c₄ after some t_* ≥ t₀. Hence, with the help of the right hand side inequality in (3.5), we find

$$L^{\prime }\left(t\right)\le -c_5\eta \left(t\right)L\left(t\right),\forall t\ge t_{*},$$

(3.23)

for some positive constant c₅ > 0. An integration of (3.23) over (t_*, t) gives

$$L\left(t\right)\le L\left(t_{*}\right){\mathsf{\text{e}}}^{-c_5{\int }_{t_{*}}^t\eta \left(s\right)\mathsf{\text{d}}s},\forall t\ge t_{*}.$$

Then using the left hand side inequality in (3.5) leads to

$${\beta }_1\left(E\left(t\right)+{\Phi }_3\left(t\right)\right)\le L\left(t_{*}\right){\mathsf{\text{e}}}^{-c_5{\int }_{t_{*}}^t\eta \left(s\right)\mathsf{\text{d}}s},\forall t\ge t_{*}.$$

Therefore, by virtue of the continuity and boundedness of E(t) and ξ(t) on the interval [0, t_*], we infer that

$$E\left(t\right)\le K\xi {\left(t\right)}^{-\mu },t\ge 0,$$

for some positive constants K and μ.

Similar to those remarks as in [25], we have the following remark.

Remark. Note that there is a wide class of relaxation functions satisfying (A3). More precisely, if ξ(t) = e^αt, α > 0, then η(t) = α, this gives the exponential decay estimate $E\left(t\right)\le c_1{\mathsf{\text{e}}}^{-c_2\alpha t}$, for some positive constants c₁ and c₂. Similarly, if ξ(t) = (1 + t)^α , α > 0, then we obtain the polynomial decay estimate E (t) ≤ c₃ (1 + t)^-μ, for some positive constants c₃ and μ.