EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 10, No. 4, 2017, 668-701 ISSN 1307-5543 – www.ejpam.com Published by New York Business Global Existence, nonexistence and decay estimate of global solutions for a viscoelastic wave equation with nonlinear boundary damping and internal source terms Huafei Di1, Yadong Shang1,∗ 1 School of Mathematics and Information Science, Guangzhou University, Guangdong, P R China Abstract. In this paper, we consider the initial boundary value problem for a viscoelastic wave equation with nonlinear boundary damping and internal source terms. We first prove the exis- tence of global weak solutions by the combination of Galerkin approximation, potential well and monotonicity-compactness methods. Then, we give an explicit decay rate estimate of the energy by making use of the perturbed energy method. Finally, the finite time blow up result of the solutions is investigated under certain assumptions on the relaxation function g and initial data. 2010 Mathematics Subject Classifications: 35L35, 35L75, 35R15 Key Words and Phrases: Viscoelastic equation, Nonlinear boundary damping, Internal source, Blow up, Decay rate estimate, Perturbed energy method 1. Introduction We are concerned with the following initial boundary value problem of the viscoelastic wave equation with nonlinear boundary damping and internal source terms |ut|ρ−1utt −4u+ ∫ t 0 g(t− s)4u(s)ds = |u|p−1u, in Ω× (0,∞), u(x, t) = 0, on Γ0 × (0,∞), ∂u ∂ν − ∫ t 0 g(t− s)∂u∂ν (s)ds+ |ut|q−1ut = 0, on Γ1 × (0,∞), u(x, 0) = u0(x), ut(x, 0) = u1(x), x ∈ Ω, (1.1) where ρ, p, q ≥ 1, and Ω is a bounded domain of Rn with a smooth boundary Γ. Let {Γ0,Γ1} be a partition of its boundary Γ such that Γ = Γ0 ∪ Γ1, Γ0 ∩ Γ1 = ∅ and meas(Γ0) > 0. Here, ν is the unit outward normal to Γ, and g represents the kernel of memory term, namely the relaxation function, satisfying certain conditions to be specified later. ∗Corresponding author. Email addresses: dihuafei@yeah.net (H.F. Di), gzydshang@126.com (Y.D. Shang) http://www.ejpam.com 668 c© 2017 EJPAM All rights reserved. H.F. Di, Y.D. Shang / Eur. J. Pure Appl. Math, 10 (4) (2017), 668-701 669 It is well known that viscoelastic materials present nature damping, which is due to some special properties of these materials to keep memory of their past trace. From the mathematical point of view, these damping effects are modeled by integro-differential operators. This type of equations with viscoelastic term describes a variety of important physical processes, such as the analysis of heat conduction in viscoelastic materials, electric signals in nonlinear telegraph line with nonlinear damping, viscous flow in viscoelastic materials [1], vibration of nonlinear elastic rod with viscosity [2], nonlinear bidirectional shallow water waves [3], and the velocity evolution of ion-acoustic waves in a collision less plasma when a ion viscosity is invoked [4] and so on. For the nonlinear viscoelastic wave equations with homogeneous Dirichlet boundary condition, many authors have given attention to them for quite a long time. There are extensive literature on the existence or nonexistence of global solutions, blow up results in finite time, and the asymptotic behavior of the solutions for this type of problems. Berrimi and Messaoudi [5] considered the following nonlinear viscoelastic wave equation utt −4u+ ∫ t 0 g(t− s)4u(s)ds = u|u|p−2, in Ω× (0,∞), (1.2) in a bounded domain and p ≥ 2. They established a local existence result and showed that the local solution is global and decays uniformly if the initial data are small enough. Kim and Han [6] proved that any weak solution with negative initial energy blows up in finite time under suitable conditions on the relaxation function g for the equation (1.2). In [7], Wang et al. studied the following nonlinear viscoelastic wave equation utt −4u+ ∫ t 0 g(t− s)4u(s)ds+ ut = u|u|p−2, in Ω× (0,∞). (1.3) Under some appropriate assumptions on g, by introducing potential wells they obtained the existence of global solution and the explicit exponential energy decay estimates. Later, Wang [8] proved that solution with arbitrary positive initial energy blows up in finite time under some appropriate assumptions on the relaxation function g and the initial data. Messaoudi [9] changed the linear damping term ut into the nonlinear damping term aut|ut|m−2. Under suitable conditions on g, he proved that the solution with negative initial energy blows up in finite time. This blow up result was extended by the same author [10] to certain solution with positive initial energy. Song and Zhong [11] considered the nonlinear viscoelastic wave equation with strong damping term utt −4u+ ∫ t 0 g(t− s)4u(s)ds−4ut = u|u|p−2, in Ω× (0,∞), (1.4) with homogeneous Dirichlet boundary condition. They proved that the solution with positive initial energy blows up in finite time. In [12], Han and Wang studied the general decay of energy for the following nonlinear viscoelastic equation without source term utt −4u+ ∫ t 0 g(t− s)4u(s)ds−4utt + ut|ut|m−2 = 0, in Ω× (0,∞). (1.5) H.F. Di, Y.D. Shang / Eur. J. Pure Appl. Math, 10 (4) (2017), 668-701 670 More recently, Xu, Yang and Liu [13] investigated the following strongly damped vis- coelastic wave equation utt −4u+ ∫ t 0 g(t− s)4u(s)ds−4ut −4utt + ut = u|u|p−1, in Ω× (0,∞). (1.6) They proved the existence and nonexistence of global weak solution with low initial energy by introducing a family of potential wells. Then, they established a blow up result for certain solutions with arbitrary positive initial energy. Messaoudi and Tatar [14] considered the following nonlinear viscoelastic equation |ut|ρutt −4u+ ∫ t 0 g(t− s)4u(s)ds−4utt = bu|u|p−1, in Ω× (0,∞), (1.7) with Dirichlet boundary condition. By using the potential well method, they proved that the viscoelastic term is enough to ensure the global existence and uniform decay of solutions provided that the initial data are in same stable set. Liu [15] proved that for certain class of relaxation function g and certain initial data in the unstable set, there are the solutions with positive initial energy that blow up in finite time. Cavalcanti et al. [16] considered the following nonlinear viscoelastic equation without source and weak damping terms |ut|ρutt −4u+ ∫ t 0 g(t− s)4u(s)ds− γ4ut −4utt = 0, in Ω× (0,∞). (1.8) They obtained the global existence of weak solution and uniform decay rates of the energy by assuming that the relaxation g has a exponential decay. In [17], Wu studied the following viscoelastic equation with nonlinear source and weak damping terms |ut|ρutt −4u−4utt + ∫ t 0 g(t− s)4u(s)ds+ |ut|mut = |u|p−1u, in Ω× (0,∞). (1.9) He discussed the general uniform decay estimate of solution energy under suitable condi- tions on the relaxation function g, the initial data and the parameters ρ,m, p. We also note that the potential well method is a very popular and important way to study the global existence and finite time blow up of solutions for nonlinear evolution equations. This method was first introduced by Sattinger [26] to study the global existence of solutions for nonlinear hyperbolic equations. And it also plays a very vital role in deriving the threshold results between the global existence and nonexistence of solutions. Hence, the potential well method has been widely used and extended by many authors to study different kinds of evolution equations, we refer the reader to see [25,27-29] and the papers cited therein. For the viscoelastic equation with nonlinear boundary condition, there are also some results about the well-posedness for this type of problems. We refer readers to see [18]- [22] and the papers cited therein. In [18]-[20], the initial boundary value problem of the H.F. Di, Y.D. Shang / Eur. J. Pure Appl. Math, 10 (4) (2017), 668-701 671 viscoelastic equation with a nonlinear boundary damping term utt −4u+ ∫ t 0 g(t− s)4u(s)ds = 0, in Ω× (0,∞), u(x, t) = 0, on Γ0 × (0,∞), ∂u ∂ν − ∫ t 0 g(t− s)∂u∂ν (s)ds+ h(ut) = 0, on Γ1 × (0,∞), u(x, 0) = u0(x), ut(x, 0) = u1(x), x ∈ Ω, (1.10) was studied. Cavalcanti et al. [18] obtained a global existence result for strong and weak solutions under the classical assumptions on g. Some uniform decay rate results were established under quite restrictive assumptions on both the damping term h and the relaxation function g. Later, Cavalcanti et al. [19] studied the problem (1.10) under weaker conditions on the relaxation function g and without imposing a growth assumption on the function h. They obtained the decay rate estimates of the energy depending on the behavior of h near zero and on the behavior of the relaxation g at infinity. For a wider class of relaxation function g and without imposing any restrictive growth assumptions on the damping term h, Messaoudi and Mustafa [20] also established an explicit and general decay rate result for the problem (1.10). Lu et al. [21] considered the following initial boundary value problem of the viscoelastic wave equation with nonlinear boundary damping and source terms utt −4u+ ∫ t 0 g(t− s)4u(s)ds = 0, in Ω× (0,∞), u(x, t) = 0, on Γ0 × (0,∞), ∂u ∂ν − ∫ t 0 g(t− s)∂u∂ν (s)ds+ ut|ut|m−2 = u|u|p−2, on Γ1 × (0,∞), u(x, 0) = u0(x), ut(x, 0) = u1(x), x ∈ Ω. (1.11) They obtained the global existence of solution and a general decay of the energy under some appropriate assumptions on the function g and certain initial data. In [22], Liu and Yu first extended the decay result obtained by Lu et al. Then, they established two blow up results: one is for certain solutions with nonpositive initial energy as well as positive initial energy in the case m ≥ 2, the other is for certain solutions with arbitrary positive initial energy in the case m = 2. Motivated by the above researches, in the present work we consider the viscoelastic wave equation with internal nonlinear terms |ut|ρ−1utt, |u|p−1u and boundary nonlinear damping term |ut|q−1ut. First of all, we prove the existence of global weak solutions by the combination of Galerkin approximation, potential well and monotonicity-compactness methods. Then, we give an explicit decay rate estimate of the energy by making use of the perturbed energy method introduced by Cavalcanti et al.[16,18,23], Messaoudi and Tatar [14,24] and Liu [22] coupled with some technical Lemmas. Finally, the finite time blow up result of the solutions is investigated under certain assumptions on the relaxation function g and initial data. The rest of this paper is organized as follows: In Section 2, we give some preliminaries and state our main results. The proof of the existence of global weak solutions and an exponential decay result will be given in Sections 3 and 4. In the last Section, we investigate the finite time blow up result of solutions under certain conditions. H.F. Di, Y.D. Shang / Eur. J. Pure Appl. Math, 10 (4) (2017), 668-701 672 2. Preliminaries In order to state our results precisely, we first give some notations, basic definitions and important Lemmas which will be needed in the course of this paper. Let Ω be a bounded open domain of Rn with a smooth boundary Γ. We consider m(x) = x − x0 (x0 is a fixed point of Rn), and introduce a partition of the boundary Γ such that Γ0 = {x ∈ Γ : m(x) · ν(x) ≤ 0} , Γ1 = {x ∈ Γ : m(x) · ν(x) > 0} . We define some inner products and norms (u, v) = ∫ Ω u(x)v(x)dx, (u, v)Γ1 = ∫ Γ1 u(x)v(x)dΓ, ‖u‖pp = ∫ Ω |u(x)|pdx, ‖u‖pΓ1,p = ∫ Γ1 |u(x)|pdΓ, ‖u‖∞ = ess sup x∈Ω |u(x)| and the Hilbert space H1 Γ0 (Ω) = { u ∈ H1(Ω) ∣∣ u = 0 on Γ0 } . Since Γ0 has positive (n − 1) dimensional Lebesgue measure, by Poincaré inequality, we can endow H1 Γ0 (Ω) with the equivalent norm ‖u‖H1 Γ0 = ‖∇u‖2 (see [25] for details). Now, we state the general hypotheses. (A1) The relaxation function g: [0,∞)→ (0,∞) is a C1 function satisfying g′(t) ≤ 0, b = 1− ∫ ∞ 0 g(s)ds ≤ 1− ∫ t 0 g(s)ds = l(t). (A2) There exists a positive differentiable function ξ(t) such that g′(t) ≤ −ξ(t)g(t), t ≥ 0, and for some positive constant k, ξ(t) satisfies∣∣∣∣ξ′(t)ξ(t) ∣∣∣∣ ≤ k, ξ′(t) ≤ 0, ∀ t > 0. (A3) We also assume that 1 < p <∞ if n ≤ 2, 1 < p ≤ n+ 2 n− 2 if n ≥ 3, 1 < q <∞ if n ≤ 2, 1 < q ≤ n n− 2 if n ≥ 3. H.F. Di, Y.D. Shang / Eur. J. Pure Appl. Math, 10 (4) (2017), 668-701 673 Next, we shall define some functionals and study their some basic properties which are related with potential well. Firstly, let us consider the functionals E(t) = 1 ρ+ 1 ‖ut‖ρ+1 ρ+1 + 1 2 ( 1− ∫ t 0 g(s)ds ) ‖∇u‖22 + 1 2 (g ◦ ∇u)(t)− 1 p+ 1 ‖u‖p+1 p+1, (2.1) J(u) = 1 2 ( 1− ∫ t 0 g(s)ds ) ‖∇u‖22 + 1 2 (g ◦ ∇u)(t)− 1 p+ 1 ‖u‖p+1 p+1, (2.2) I(u) = ( 1− ∫ t 0 g(s)ds ) ‖∇u‖22 + (g ◦ ∇u)(t)− ‖u‖p+1 p+1, (2.3) where (g ◦ ∇u)(t) = ∫ t 0 g(t− s)‖∇u(t)−∇u(s)‖22ds, ∀ u ∈ H1 Γ0 (Ω). Lemma 1. Let the assumptions (A1), (A3) hold, then for any u ∈ H1 Γ0 (Ω), ‖u‖H1 Γ0 6= 0, it follows that (1) limλ→0+ J(λu) = 0, limλ→+∞ J(λu) = −∞; (2) On the interval 0 < λ <∞, there exists a unique λ∗ = λ∗(u) such that d dλ J(λu) ∣∣ λ=λ∗ = 0; (3) J(λu) is increasing on 0 ≤ λ ≤ λ∗, decreasing on λ∗ ≤ λ <∞, and takes the maximum at λ = λ∗; (4) I(λu) > 0 for 0 ≤ λ < λ∗, I(λu) < 0 for λ∗ < λ <∞, and I(λ∗u) = 0. Proof. (1) From the definition of the functional (2.2), we have J(λu) = 1 2 l(t)λ2‖∇u‖22 + 1 2 λ2(g ◦ ∇u)(t)− λp+1 p+ 1 ‖u‖p+1 p+1. Hence, the conclusion holds. (2) The conclusion follows from d dλ J(λu) = λl(t)‖∇u‖22 + λ(g ◦ ∇u)(t)− λp‖u‖p+1 p+1 = 0. (2.4) (3) From the conclusion of (2), we can easily get d dλ J(λu) ≥ 0, for 0 ≤ λ ≤ λ∗, d dλ J(λu) ≤ 0, for λ∗ ≤ λ <∞. (4) The conclusion follows from I(λu) = λ2l(t)‖∇u‖22 + λ2(g ◦ ∇u)(t)− λp‖u‖p+1 p+1 = λ d dλ J(λu). (2.5) H.F. Di, Y.D. Shang / Eur. J. Pure Appl. Math, 10 (4) (2017), 668-701 674 Then, for t ≥ 0, we define d(t) = inf u∈H1 Γ0 (Ω)\{0} { sup λ>0 J(λu) } . (2.6) In fact (see [26,27] for details), d(t) is positive and equal to inf I(u)=0,u6=0 J(u). (2.7) Lemma 2. Let the assumptions (A1), (A3) hold, then for all t ∈ [0,∞), we have 0 < d̃ ≤ d(t) ≤ ˜̃ d(u) = sup λ>0 J(λu), (2.8) where d̃ = p−1 2(p+1)( b B2 p+1 ) p+1 p−1 , and Bp+1 is the optimal constant satisfying the Sobolev in- equality ‖u‖p+1 ≤ Bp+1‖∇u‖2. Proof. From the definition of d(t), we get d(t) ≤ ˜̃ d = supλ>0 J(λu). By the Sobolev inequality, it follows that J(λu) = 1 2 l(t)‖∇λu‖22 + 1 2 (g ◦ ∇λu)(t)− 1 p+ 1 ‖λu‖p+1 p+1 ≥ 1 2 b‖∇λu‖22 − 1 p+ 1 Bp+1 p+1‖∇λu‖ p+1 2 . (2.9) Here, we define the function h(λ) = 1 2bλ 2 − 1 p+1B p+1 p+1λ p+1, λ > 0. By the direct com- putation, we deduce that h is increasing for 0 < λ < λ1, decreasing for λ > λ1 and λ1 = ( b Bp+1 p+1 ) 1 p−1 is the absolute maximum point of h such that d̃ = h(λ1) = p− 1 2(p+ 1) ( b B2 p+1 ) p+1 p−1 . By the combination of (2.9) and the definition of d̃, it follows that d̃ = h(λ1) ≤ J(λu). Moreover, from the definition of d(t), we have d̃ ≤ infu∈H1 Γ0 (Ω)\{0} { supλ>0 J(λu) } = d(t). The proof is completed. To obtain the results of this paper, we will construct the potential wells associated with the functionals J(u) and I(u). Next, let us introduce the stable and unstable sets: W = { u ∈ H1 Γ0 (Ω) ∣∣ I(u) > 0, J(u) < d̃ } ∪ {0}, (2.10) and V = { u ∈ H1 Γ0 (Ω) ∣∣ I(u) < 0, J(u) < d̃ } . (2.11) For simplicity, we define the weak solutions of (1.1) over the interval Ω× [0, T ), but it is to be understood that T is either infinity or the limit of the existence interval. H.F. Di, Y.D. Shang / Eur. J. Pure Appl. Math, 10 (4) (2017), 668-701 675 Definition 1. We say that u(x, t) is called a weak solution of the problem (1.1) on the in- terval Ω×[0, T ). If u ∈ L∞(0, T ;H1 Γ0 (Ω)) with ut ∈ L∞(0, T ;Lρ+1(Ω))∩Lq+1(0, T ;Lq+1(Γ1)) satisfy the following conditions (i) For any v ∈ H1 Γ0 (Ω) ∩ Lq+1(Γ1) ∩ Lρ+1(Ω) and a.e 0 ≤ t ≤ T , such that 1 ρ (|ut|ρ−1ut, v) + ∫ t 0 b1(u, v)ds+ ∫ t 0 b2(u, v)ds − ∫ t 0 (|u|p−1u, v)ds+ ∫ t 0 (|ut|q−1ut, v)Γ1ds = 1 ρ (|u1|ρ−1u1, v), (2.12) where b1(u, v) = (∇u,∇v), b2(u, v) = −( ∫ s 0 g(s− τ)∇u(τ)dτ,∇v); (ii) u(x, 0) = u0(x) in H1 Γ0 (Ω), ut(x, 0) = u1(x) in Lρ+1(Ω) ∩ Lq+1(Γ1). (iii) The following energy inequality holds E(t) ≤ E(0), (2.13) for any 0 ≤ t < T . The following Lemma is similar to the Lemmas of [28,29] with slight modification. Lemma 3. Let the assumptions (A1), (A3) hold and u be a solution of problem (1.1). Further assume that u0(x) ∈ H1 Γ0 (Ω), u1(x) ∈ Lρ+1(Ω) ∩ Lq+1(Γ1), we have (1) If E(0) < d̃, I(u0) > 0 or ‖u0‖H1 Γ0 = 0, then the solution u(t) ∈W for all t ∈ [0, T ); (2) If E(0) < d̃, I(u0) < 0, then the solution u(t) ∈ V for all t ∈ [0, T ). Proof. (1) Let u be any solution of problem (1.1) with E(0) < d̃ and I(u0) > 0 or ‖u0‖H1 Γ0 = 0. If ‖u0‖H1 Γ0 = 0, then u0(x) ∈W . If I(u0) > 0, from the inequality 1 ρ+ 1 ‖u1‖ρ+1 ρ+1 + J(u0) = E(0) < d̃, (2.14) we have u0(x) ∈W . We prove u(t) ∈W for 0 < t < T . Arguing by contradiction and considering the time continuity of I(u), we suppose that there exists a time t0 ∈ (0, T ) such that u(t0) ∈ ∂W , which means that I(u(t0)) = 0, ‖u(t0)‖H1 Γ0 6= 0 or J(u(t0)) = d̃. From (2.13), it follows that 1 ρ+ 1 ‖ut‖ρ+1 ρ+1 + J(u) ≤ E(0) < d̃, 0 < t < T. (2.15) H.F. Di, Y.D. Shang / Eur. J. Pure Appl. Math, 10 (4) (2017), 668-701 676 Thus, we see that J(u(t0)) 6= d̃. If I(u(t0)) = 0, ‖u(t0)‖H1 Γ0 6= 0, then by the definition of d we have J(u(t0)) ≥ d which contradicts (2.15). The proof of (1) is completed. (2) Let u(t) be any solution of problem (1.1) with E(0) < d̃ and I(u0) < 0. From (2.11) we get that u0(x) ∈ V . We prove u(t) ∈ V for 0 < t < T . Arguing by contradiction, we suppose that there exists a time t0 ∈ (0, T ) such that u(t0) ∈ ∂V which means that I(u(t0)) = 0 or J(u(t0)) = d̃. Again (2.15) shows that J(u(t0)) 6= d̃. If I(u(t0)) = 0, then by the definition of d we have J(u(t0)) ≥ d which contradicts (2.15). Lemma 4. Let the assumptions (A1), (A3) hold. For any fixed positive number β < 1, assume that I(u0) < 0, E(0) < βd̃, then we have I(u(t)) < 0 for all t ∈ [0, T ) and d̃ < p− 1 2(p+ 1) [l(t)‖∇u‖22 + (g ◦ ∇u)(t)] < p− 1 2(p+ 1) ‖u‖p+1 p+1. (2.16) Proof. Arguing by contradiction, we can get I(u(t)) < 0 for all t ∈ [0, T ). In fact, suppose this is not true, then there exist some t0 ∈ [0, T ) such that I(u(t0)) = 0 and I(u(t)) < 0 for 0 ≤ t < t0. Hence, l(t)‖∇u‖22 + (g ◦ ∇u)(t) < ‖u‖p+1 p+1, 0 ≤ t < t0. (2.17) From the definition of d̃, it follows that d̃ = p− 1 2(p+ 1) ( b B2 p+1 ) p+1 p−1 ≤ p− 1 2(p+ 1) { l(t)‖∇u‖22 + (g ◦ ∇u)(t) ‖u‖2p+1 } p+1 p−1 < p− 1 2(p+ 1) { l(t)‖∇u‖22 + (g ◦ ∇u)(t) (l(t)‖∇u‖22 + (g ◦ ∇u)(t)) 2 p+1 } p+1 p−1 = p− 1 2(p+ 1) [l(t)‖∇u‖22 + (g ◦ ∇u)(t)], 0 ≤ t < t0. (2.18) We deduce from (2.17) and (2.18) that ‖u‖p+1 p+1 > 2(p+ 1) p− 1 d̃ > 0, 0 ≤ t < t0. (2.19) Since t→ ‖u(t)‖p+1 p+1 is continuous, from (2.19) we have d̃ ≤ p− 1 2(p+ 1) ‖u(t0)‖p+1 p+1 = J(u(t0)). (2.20) This is impossible since J(u(t0)) ≤ E(t0) ≤ E(0) < d̃. Hence, we obtain I(u(t)) < 0 for all t ∈ [0, T ). Furthermore, from (2.18) again, we have d̃ < p− 1 2(p+ 1) [l(t)‖∇u‖22 + (g ◦ ∇u)(t)] H.F. Di, Y.D. Shang / Eur. J. Pure Appl. Math, 10 (4) (2017), 668-701 677 < (p− 1) 2(p+ 1) ‖u‖p+1 p+1, 0 ≤ t < T. (2.21) Thus, the proof is completed. 3. Existence of global weak solutions In this section, we are going to obtain the existence of global weak solutions for the problem (1.1) with the initial conditions E(0) < d̃ and I(u0) > 0 or ‖u0‖H1 Γ0 = 0 by the combination of Galerkin approximation, potential well and monotonicity-compactness methods. Theorem 5. Let the assumptions (A1), (A3) hold, u0(x) ∈ H1 Γ0 (Ω), u1(x) ∈ Lρ+1(Ω) ∩ Lq+1(Γ1). Further assume that E(0) < d̃ and I(u0) > 0 or ‖u0‖H1 Γ0 = 0, then the problem (1.1) admits a global weak solution satisfying u ∈ L∞(0,∞;H1 Γ0 (Ω)), ut ∈ L∞(0,∞;Lρ+1(Ω)) ∩ Lq+1(0,∞;Lq+1(Γ1)), u(t) ∈W for 0 ≤ t <∞, and the energy identity E(t) + ∫ t 0 ‖ut(s)‖q+1 Γ1,q+1ds− 1 2 ∫ t 0 (g′ ◦ ∇u)(s)ds+ 1 2 ∫ t 0 g(s)‖∇u(s)‖22ds = E(0), (3.1) holds for 0 ≤ t <∞. Remark 1. From (3.1), we can easily obtain E′(t) = −‖ut(t)‖q+1 Γ1,q+1 + 1 2 (g′ ◦ ∇u)(t)− 1 2 g(t)‖∇u(t)‖22 ≤ 0. (3.2) Proof. Let {wj(x)} be a complete orthogonal system in H1 Γ0 (Ω)∩Lq+1(Γ1)∩Lρ+1(Ω). We suppose that the approximate weak solution um of the problem (1.1) can be written um(t) = m∑ j=1 dmj(t)wj(x), m = 1, 2, · · · · · · . (3.3) According to Galerkin’s method, these coefficients dmj need to satisfy the following initial value problem of nonlinear ordinary integro-differential equations 1 ρ (|u′m|ρ−1u′m, wj) + ∫ t 0 b1(um, wj)ds+ ∫ t 0 b2(um, wj)ds − ∫ t 0 (|um|p−1um, wj)ds+ ∫ t 0 (|u′m|q−1u′m, wj)Γ1ds = 1 ρ (|u′m(0)|ρ−1u′m(0), wj), j = 1, 2, · · · ,m, (3.4) H.F. Di, Y.D. Shang / Eur. J. Pure Appl. Math, 10 (4) (2017), 668-701 678 um(x, 0) = m∑ j=1 dmj(0)ωj(x)→ u0(x), in H1 Γ0 (Ω), (3.5) u′m(x, 0) = m∑ j=1 d′mj(0)ωj(x)→ u1(x), in Lρ+1(Ω) ∩ Lq+1(Γ1), (3.6) where b1(um, wj) = (∇um,∇wj), b2(um, wj) = −( ∫ s 0 g(s− τ)∇um(τ)dτ,∇wj). We will prove that the initial value problem (3.4)-(3.6) of the nonlinear integro-differential equations have global weak solutions in the interval [0,∞). Furthermore, we show that the solutions of the problem (1.1) can be approximated by the functions um. Now, differentiating (3.4) with respect to t, and multiplying the obtained equation by d′mj(t), summing for j = 1, · · · ,m, then we have (|u′m|ρ−1u′′m, u ′ m) + b1(um, u ′ m) + b2(um, u ′ m) + (|u′m|q−1u′m, u ′ m)Γ1 = (|um|p−1um, u ′ m).(3.7) By a direct calculation, it follows that (|u′m|ρ−1u′′m, u ′ m) = 1 ρ+ 1 d dt ‖u′m‖ ρ+1 ρ+1, (3.8) b1(um, u ′ m) = (∇um,∇u′m) = 1 2 d dt ‖∇um‖22, (3.9) (|um|p−1um, u ′ m) = 1 p+ 1 d dt ‖um‖p+1 p+1, (3.10) and b2(um, u ′ m) = − ∫ Ω ∫ t 0 g(t− s)∇um(s)∇u′m(t)dsdx =− ∫ Ω ∫ t 0 g(t− s)[∇um(s)−∇um(t)]∇u′m(t)dsdx − ∫ Ω ∫ t 0 g(t− s)∇um(t)∇u′m(t)dsdx = 1 2 ∫ Ω ∫ t 0 g(t− s) d dt [∇um(s)−∇um(t)]2dsdx H.F. Di, Y.D. Shang / Eur. J. Pure Appl. Math, 10 (4) (2017), 668-701 679 − 1 2 ∫ Ω ∫ t 0 g(t− s) d dt [∇um(t)]2dsdx = 1 2 d dt ∫ Ω ∫ t 0 g(t− s)[∇um(s)−∇um(t)]2dsdx − 1 2 ∫ Ω ∫ t 0 g′(t− s)[∇um(s)−∇um(t)]2dsdx − 1 2 d dt ∫ t 0 g(s)ds‖∇um(t)‖22 + 1 2 g(t)‖∇um(t)‖22. (3.11) Inserting (3.8)-(3.11) into (3.7), we have 1 ρ+ 1 d dt ‖u′m‖ ρ+1 ρ+1 + 1 2 d dt [(1− ∫ t 0 g(s)ds)‖∇um(t)‖22] + 1 2 d dt (g ◦ ∇um)(t)− 1 p+ 1 d dt ‖um‖p+1 p+1 =− ‖u′m‖ q+1 Γ1,q+1 + 1 2 (g′ ◦ ∇um)(t)− 1 2 g(t)‖∇um(t)‖22 ≤ 0, (3.12) which implies that E′m(t) = −‖u′m‖ q+1 Γ1,q+1 + 1 2 (g′ ◦ ∇um)(t)− 1 2 g(t)‖∇um(t)‖22 ≤ 0, (3.13) where Em(t) = 1 ρ+ 1 ‖u′m‖ ρ+1 ρ+1 + 1 2 (1− ∫ t 0 g(s)ds)‖∇um(t)‖22 + 1 2 (g ◦ ∇um)(t)− 1 p+ 1 ‖um‖p+1 p+1 = 1 ρ+ 1 ‖u′m‖ ρ+1 ρ+1 + J(um), 0 ≤ t <∞. (3.14) From E(0) < d̃ and I(u0) > 0 or ‖u0‖H1 Γ0 = 0, it follows that u0(x) ∈ W . Hence, we obtain from (3.5) and (3.6) that Em(0) < d̃, I(um(0)) > 0 and um(0) ∈W for sufficiently large m. In what follows, from the (3.14) and the arguments in the proof of Lemma 3 (1), we can obtain um(t) ∈W for sufficiently large m and 0 ≤ t <∞ such that J(um) = 1 2 l(t)‖∇um‖22 + 1 2 (g ◦ ∇um)(t)− 1 p+ 1 ‖um‖p+1 p+1 = p− 1 2(p+ 1) [l(t)‖∇um‖22 + (g ◦ ∇um)(t)] + 1 p+ 1 I(um) ≥ p− 1 2(p+ 1) [l(t)‖∇um‖22 + (g ◦ ∇um)(t)]. (3.15) By the combination of (3.13) and (3.15), we get 1 ρ+ 1 ‖u′m‖ ρ+1 ρ+1 + p− 1 2(p+ 1) [l(t)‖∇um‖22 + (g ◦ ∇um)(t)] ≤ Em(t) ≤ Em(0) < d̃, (3.16) H.F. Di, Y.D. Shang / Eur. J. Pure Appl. Math, 10 (4) (2017), 668-701 680 for sufficiently large m and 0 ≤ t <∞. Integrating (3.13) with respect to t, then we have Em(t) + ∫ t 0 ‖u′m‖ q+1 Γ1,q+1ds = 1 2 ∫ t 0 (g′ ◦ ∇um)(s)ds− 1 2 ∫ t 0 g(s)‖∇um(s)‖22ds+ Em(0). (3.17) Combining (3.16) and (3.17), we obtain 1 ρ+ 1 ‖u′m‖ ρ+1 ρ+1 + ∫ t 0 ‖u′m‖ q+1 Γ1,q+1ds+ p− 1 2(p+ 1) [l(t)‖∇um‖22 + (g ◦ ∇um)] ≤ Em(0) < d̃,(3.18) for sufficiently large m and 0 ≤ t <∞. From (3.18), we have l(t)‖∇um‖22 + (g ◦ ∇um)(t) < 2(p+ 1) p− 1 d̃, 0 ≤ t <∞, (3.19) ‖u′m‖ ρ+1 ρ+1 < (ρ+ 1)d̃, 0 ≤ t <∞, (3.20) ∫ t 0 ‖u′m‖ q+1 Γ1,q+1ds < d̃, 0 ≤ t <∞. (3.21) Using the Sobolev inequality and (3.19), it follows that ‖um‖2p+1 ≤ B2 p+1‖∇um‖22 < B2 p+1 2(p+ 1) (p− 1)b d̃, 0 ≤ t <∞. (3.22) Furthermore, by (3.20) and (3.22), we get |(|u′m|ρ−1u′m, u ′ m)| ≤ ‖u′m‖ ρ+1 ρ+1 < (ρ+ 1)d̃, 0 ≤ t <∞, (3.23) |(|um|p−1um, um)| ≤ ‖um‖p+1 p+1 < Bp+1 p+1 ( 2(p+ 1) (p− 1)b d̃ ) p+1 2 , 0 ≤ t <∞. (3.24) The estimates (3.19)-(3.24) permit us to obtain a subsequences of {um} which from now on will be also denoted by {um} and functions u, χ1, χ2, χ3 such that um → u in L∞(0,∞;H1 Γ0 (Ω)) weakly star, m −→∞, (3.25) u′m → u′ in L∞(0,∞;Lρ+1(Ω)) weakly star, m −→∞, (3.26) |u′m|q−1u′m → χ1 in L q+1 q (0,∞;L q+1 q (Γ1)) weakly, m −→∞, (3.27) H.F. Di, Y.D. Shang / Eur. J. Pure Appl. Math, 10 (4) (2017), 668-701 681 |um|p−1um → χ2 in L ∞(0,∞;L p+1 p (Ω)) weakly star, m −→∞, (3.28) |u′m|ρ−1u′m → χ3 in L ∞(0,∞;L ρ+1 ρ (Ω)) weakly star, m −→∞. (3.29) Since H1 Γ0 (Ω) ↪→ L2(Ω) are compact(see [25] for details), we have, thanks to Aubin-Lions theorem, that um → u in L2(0,∞;L2(Ω)) strongly, m −→∞, (3.30) and consequently, making use of the Lemma 1.3 in [30], we deduce |um|p−1um → χ2 = |u|p−1u in L∞(0,∞;L p+1 p (Ω)) weakly star, m −→∞. (3.31) From the trace Theorem and (3.25), we deduce that ∂um ∂ν ∈ L∞(0,∞;H − 1 2 Γ0 (Ω)), which implies that |u′m|q−1u′m = −∂um ∂ν + ∫ t 0 g(t− s)∂um ∂ν (s)ds→ −∂u ∂ν + ∫ t 0 g(t− s)∂u ∂ν (s)ds = |u′|q−1u′ in L∞(0,∞;H − 1 2 Γ0 (Ω)) weakly star, m −→∞. (3.32) Combining (3.27) and the above convergence, we have |u′m|q−1u′m → χ1 = |u′|q−1u′ in L q+1 q (0,∞;L q+1 q (Γ1)) weakly, m −→∞, (3.33) Passing to the limit in (3.4) and making use of (3.25)-(3.27), (3.29), (3.31) and (3.33), we obtain 1 ρ (χ3, wj) + ∫ t 0 b1(u,wj)ds+ ∫ t 0 b2(u,wj)ds + ∫ t 0 (|u′|q−1u′, wj)Γ1ds− ∫ t 0 (|u|p−1u,wj)ds = 1 ρ (|u1|ρ−1u1, wj). (3.34) Since {wj(x)} is a basic of H1 Γ0 (Ω) ∩ Lq+1(Γ1) ∩ Lρ+1(Ω), then for all t > 0, multiplying (3.34) by d′mj(t), and summing for j = 1, · · · · · · , then we have 1 ρ (χ3, u ′) + ∫ t 0 b1(u, u′)ds+ ∫ t 0 b2(u, u′)ds + ∫ t 0 (|u′|q−1u′, u′)Γ1ds− ∫ t 0 (|u|p−1u, u′)ds = 1 ρ (|u1|ρ−1u1, u ′). (3.35) In what follows, multiplying (3.4) by d′mj(t), and summing for j = 1, · · · ,m, then we obtain 1 ρ (|u′m|ρ−1u′m, u ′ m) + ∫ t 0 b1(um, u ′ m)ds+ ∫ t 0 (|u′m|q−1u′m, u ′ m)Γ1ds H.F. Di, Y.D. Shang / Eur. J. Pure Appl. Math, 10 (4) (2017), 668-701 682 + ∫ t 0 b2(um, u ′ m)ds− ∫ t 0 (|um|p−1um, u ′ m)ds = 1 ρ (|u′m(0)|ρ−1u′m(0), u′m). (3.36) Taking m→∞ in (3.36), it follows that 1 ρ lim m→∞ (|u′m|ρ−1u′m, u ′ m) + ∫ t 0 b1(u, u′)ds+ ∫ t 0 (|u′|q−1u′, u′)Γ1ds + ∫ t 0 b2(u, u′)ds− ∫ t 0 (|u|p−1u, u′)ds = 1 ρ (|u1|ρ−1u1, u ′). (3.37) Combining (3.35) and (3.37), we deduce that (χ3, u ′) = lim m→∞ (|u′m|ρ−1u′m, u ′ m). (3.38) On the other hand, utilizing the non-decreasing monotonicity of the function |s|ρ−1s, s ∈ R, we have (|u′m|ρ−1u′m − |ψ|ρ−1ψ, u′m − ψ) ≥ 0, (3.39) for all ψ ∈ Lρ+1(Ω). Thus, we get from the inequality (3.39) that (|u′m|ρ−1u′m, ψ) + (|ψ|ρ−1ψ, u′m − ψ) ≤ (|u′m|ρ−1u′m, u ′ m). (3.40) Passing to the limit in (3.40) as m→∞, it follows that (χ3 − |ψ|ρ−1ψ, u′ − ψ) ≥ 0. (3.41) In order to prove χ3 = |u′|ρ−1u′ from (3.41), we use the semi-continuity of the function |s|ρ−1s, s ∈ R ([30], Chapter 2). Let ψ = u′ − µφ, µ ≥ 0 and ∀ φ ∈ Lρ+1(Ω), then (χ3 − |u′ − µφ|ρ−1(u′ − µφ), φ) ≥ 0. (3.42) Passing to the limit in (3.42) as µ→ 0, we have (χ3 − |u′|ρ−1u′, φ) ≥ 0, ∀ φ ∈ Lρ+1(Ω). (3.43) In a similar way, let ψ = u′ − µφ, µ ≤ 0 and ∀ φ ∈ Lρ+1(Ω), then we obtain (χ3 − |u′|ρ−1u′, φ) ≤ 0, ∀ φ ∈ Lρ+1(Ω). (3.44) From the combination of (3.43) and (3.44), we see that χ3 = |u′|ρ−1u′. (3.45) Next, we shall prove that u satisfies (2.13). From the discussion above, we obtain for each fixed t > 0 that |(g ◦ ∇u)(t)− (g ◦ ∇um)(t)| H.F. Di, Y.D. Shang / Eur. J. Pure Appl. Math, 10 (4) (2017), 668-701 683 = ∣∣ ∫ t 0 g(t− s) ∫ Ω |∇u(s)−∇u(t)|2dxds− ∫ t 0 g(t− s) ∫ Ω |∇um(s)−∇um(t)|2dxds ∣∣ ≤ ∫ t 0 g(t− s)‖∇u(s)−∇um(s)‖2‖∇u(s) +∇um(s)‖2ds + ∫ t 0 g(t− s)‖∇u(s)−∇um(s)‖2ds‖∇u(t) +∇um(t)‖2 + ∫ t 0 g(t− s)‖∇u(s) +∇um(s)‖2ds‖∇u(t)−∇um(t)‖2 + ∫ t 0 g(s)ds‖∇u(t) +∇um(t)‖2‖∇u(t)−∇um(t)‖2 ≤ C ∫ t 0 g(t− s)‖∇u(s)−∇um(s)‖2ds+ C ∫ t 0 g(s)ds‖∇u(t)−∇um(t)‖2 → 0, (3.46) as m→∞. Taking into account the nonlinear term of the functional J(u), we deduce ‖um‖p+1 p+1 − ‖u‖ p+1 p+1 ≤ (p+ 1)| ∫ Ω |u+ θmum|p−1(u+ θmum)(um − u)dx| ≤ (p+ 1)‖u+ θmum‖pp+1‖um − u‖p+1 ≤ C‖um − u‖p+1 → 0, (3.47) as m→∞, where 0 < θm < 1. Hence, we have lim m→∞ (g ◦ ∇um)(t) = (g ◦ ∇u)(t), lim m→∞ ‖um‖p+1 p+1 = ‖u‖p+1 p+1. (3.48) From (3.5),(3.6), it follows that Em(0) → E(0) as m → ∞. Therefore, making use of Fatou’s Lemma and (3.14), we deduce 1 ρ+ 1 ‖u′‖ρ+1 ρ+1 + 1 2 l‖∇u‖22 ≤ lim inf m→∞ [ 1 ρ+ 1 ‖um‖ρ+1 ρ+1 + 1 2 l(t)‖∇um‖22] = lim inf m→∞ [Em(t) + 1 p+ 1 ‖um‖p+1 p+1 − 1 2 (g ◦ ∇um)(t)] ≤ lim m→∞ [Em(0) + 1 p+ 1 ‖um‖p+1 p+1 − 1 2 (g ◦ ∇um)(t)] = E(0) + 1 p+ 1 ‖u‖p+1 p+1 − 1 2 (g ◦ ∇u)(t). (3.49) which yields (2.13). Thus, we obtain that u is a global weak solution of problem (1.1). Then, making use of Lemma 3 (1) again, we get u(t) ∈ W for 0 ≤ t <∞. Finally, taking m→∞ in (3.17), we deduce that the energy identity (3.1) also holds for 0 ≤ t <∞. H.F. Di, Y.D. Shang / Eur. J. Pure Appl. Math, 10 (4) (2017), 668-701 684 4. Decay estimate In this section, we shall prove the energy decay estimate of the global solutions obtained in the previous section by making use of the perturbed energy method introduced by Cavalcanti et al.[16,18,23], Messaoudi and Tatar [14,24] and Liu [22] coupled with some new technical Lemmas. Theorem 6. Let the assumptions (A1)− (A3) hold, u0(x) ∈ H1 Γ0 (Ω), u1(x) ∈ Lρ+1(Ω) ∩ Lq+1(Γ1). Further assume that 1 < ρ <∞ if n ≤ 2, 1 < ρ ≤ n+2 n−2 if n ≥ 3, E(0) < d̃ and I(u0) > 0, then for each t0 > 0, there exist two positive constants L and η such that the solutions of the problem (1.1) satisfies E(t) ≤ Le−η ∫ t t0 ξ(s)ds , t ≥ t0. For this purpose, we introduce the functional F (t) = ME(t) + εΨ(t) + Φ(t), (4.1) where ε, M are positive constants which shall be determined later, and Ψ(t) = ξ(t) ρ ∫ Ω |ut|ρ−1utudx, (4.2) Φ(t) = −ξ(t) ρ ∫ Ω |ut|ρ−1ut ∫ t 0 g(t− s)[u(t)− u(s)]dsdx. (4.3) Remark 2. This functional was first introduced in [14] but choose ξ(t) ≡ 1 and in [22,24] for ξ(t) 6≡ 1. Here, we can choose ε sufficiently small and M sufficiently large (if needed) in (4.1) so that F (t) ∼ E(t). Firstly, we state several Lemmas to prove the decay rate estimate of the energy. Lemma 7. Let u ∈ L∞(0,∞;H1 Γ0 (Ω)) be the solution of (1.1) and E(0) < d̃, I(u0) > 0, then we have∫ Ω (∫ t 0 g(t− s)[u(t)− u(s)]ds )ρ+1 dx ≤ Bρ+1 ρ+1(1− b)ρ ( 4(p+ 1)E(0) (p− 1)b ) ρ−1 2 (g ◦ ∇u)(t), (4.4) where Bρ+1 is the optimal constant satisfying the Sobolev inequality ‖u‖ρ+1 ≤ Bρ+1‖∇u‖2. Proof. From E(0) < d̃, I(u0) > 0 and Lemma 3 (1), we can obtain u(t) ∈ W for 0 ≤ t <∞. Thus we have 1 ρ+ 1 ‖u′‖ρ+1 ρ+1 + p− 1 2(p+ 1) [l(t)‖∇u‖22 + (g ◦ ∇u)(t)] ≤ 1 ρ+ 1 ‖u′‖ρ+1 ρ+1 + p− 1 2(p+ 1) [l(t)‖∇u‖22 + (g ◦ ∇u)(t)] + 1 p+ 1 I(u) H.F. Di, Y.D. Shang / Eur. J. Pure Appl. Math, 10 (4) (2017), 668-701 685 = 1 ρ+ 1 ‖u′‖ρ+1 ρ+1 + J(u) = E(t) ≤ E(0) < d̃. (4.5) Taking the Hölder inequality and (4.5) into account, we have∫ Ω (∫ t 0 g(t− s)[u(t)− u(s)]ds )ρ+1 dx = ∫ Ω (∫ t 0 [g(t− s)] ρ ρ+1 [g(t− s)] 1 ρ+1 [u(t)− u(s)]ds )ρ+1 dx ≤ ( ∫ t 0 g(s)ds)ρ ∫ t 0 g(t− s) ∫ Ω |u(t)− u(s)|ρ+1dxds ≤ (1− l(t))ρBρ+1 ρ+1 ∫ t 0 g(t− s)‖∇u(t)−∇u(s)‖ρ+1 2 ds ≤ Bρ+1 ρ+1(1− b)ρ ( 4(p+ 1)E(0) (p− 1)b ) ρ−1 2 (g ◦ ∇u)(t). (4.6) Lemma 8. For ε > 0 is small enough and M > 0 is large enough, the inequality C1F (t) ≤ E(t) ≤ C2F (t) (4.7) holds for two positive constants C1 and C2. Proof. By using Young inequality, Sobolev embedding theorem and (4.5), we deduce that |1 ρ ∫ Ω |ut|ρ−1utudx| ≤ 1 ρ+ 1 ‖ut‖ρ+1 ρ+1 + 1 ρ(ρ+ 1) ‖u‖ρ+1 ρ+1 ≤ 1 ρ+ 1 ‖ut‖ρ+1 ρ+1 + Bρ+1 ρ+1 ρ(ρ+ 1) ‖∇u‖ρ+1 2 ≤ 1 ρ+ 1 ‖ut‖ρ+1 ρ+1 + Bρ+1 ρ+1 ρ(ρ+ 1) ( 2(p+ 1)E(0) (p− 1)b ) ρ−1 2 ‖∇u‖22. (4.8) From the Young inequality, (4.5) and lemma 5, we get that |1 ρ ∫ Ω |ut|ρ−1ut ∫ t 0 g(t− s)[u(t)− u(s)]dsdx| ≤ 1 ρ+ 1 ‖ut‖ρ+1 ρ+1 + 1 ρ(ρ+ 1) ∫ Ω (∫ t 0 g(t− s)[u(t)− u(s)]ds )ρ+1 dx H.F. Di, Y.D. Shang / Eur. J. Pure Appl. Math, 10 (4) (2017), 668-701 686 ≤ 1 ρ+ 1 ‖ut‖ρ+1 ρ+1 + Bρ+1 ρ+1 ρ(ρ+ 1) (1− b)ρ ( 4(p+ 1)E(0) (p− 1)b ) ρ−1 2 (g ◦ ∇u)(t). (4.9) Considering the expressions of F (t), E(t), Ψ(t), Φ(t) and the conditions (A2), it follows that F (t) ≤ME(t) + ( 1 ρ+ 1 + ε ρ+ 1 ) ξ(t)‖ut‖ρ+1 ρ+1 + ε Bρ+1 ρ+1 ρ(ρ+ 1) ( 2(p+ 1)E(0) (p− 1)b ) ρ−1 2 ξ(t)‖∇u‖22 + Bρ+1 ρ+1 ρ(ρ+ 1) (1− l(t))ρ ( 4(p+ 1)E(0) (p− 1)b ) ρ−1 2 ξ(t)(g ◦ ∇u)(t) ≤ME(t) + ( 1 ρ+ 1 + ε ρ+ 1 ) ξ(0)‖ut‖ρ+1 ρ+1 + ε Bρ+1 ρ+1 ρ(ρ+ 1) ( 2(p+ 1)E(0) (p− 1)b ) ρ−1 2 ξ(0)‖∇u‖22 + Bρ+1 ρ+1 ρ(ρ+ 1) (1− b)ρ ( 4(p+ 1)E(0) (p− 1)b ) ρ−1 2 ξ(0)(g ◦ ∇u)(t) ≤ 1 C1 E(t), (4.10) and F (t) ≥ME(t)− ( 1 ρ+ 1 + ε ρ+ 1 ) ξ(0)‖ut‖ρ+1 ρ+1 − ε Bρ+1 ρ+1 ρ(ρ+ 1) ( 2(p+ 1)E(0) (p− 1)b ) ρ−1 2 ξ(0)‖∇u‖22 − Bρ+1 ρ+1 ρ(ρ+ 1) (1− l(t))ρ ( 4(p+ 1)E(0) (p− 1)b ) ρ−1 2 ξ(0)(g ◦ ∇u)(t) ≥ [ M ρ+ 1 − ( 1 ρ+ 1 + ε ρ+ 1 ) ξ(0) ] ‖ut‖ρ+1 ρ+1 + [ M(p− 1)b 2(p+ 1) − ε Bρ+1 ρ+1 ρ(ρ+ 1) ( 2(p+ 1)E(0) (p− 1)b ) ρ−1 2 ξ(0) ] ‖∇u‖22 + [ M(p− 1) 2(p+ 1) − Bρ+1 ρ+1 ρ(ρ+ 1) (1− b)ρ ( 4(p+ 1)E(0) (p− 1)b ) ρ−1 2 ξ(0) ] (g ◦ ∇u)(t) ≥ 1 C2 E(t), (4.11) where ε > 0 is small enough and M > 0 is large enough. Lemma 9. Let the assumptions (A1)-(A3) hold and 1 < ρ < ∞ if n ≤ 2, 1 < ρ ≤ n+2 n−2 if n ≥ 3. Furthermore assume that E(0) < d̃ and I(u0) > 0, then the functional Ψ(t) = ξ(t) ρ ∫ Ω |ut| ρ−1utudx satisfies the following inequality Ψ′(t) ≤ [ 1 ρ + 1 (ρ+ 1)ρα k ] ξ(t)‖ut‖ρ+1 ρ+1 − [ b 2 − α ρ+ 1 kBρ+1 ρ+1 ( 2(p+ 1)E(0) (p− 1)b ) ρ−1 2 H.F. Di, Y.D. Shang / Eur. J. Pure Appl. Math, 10 (4) (2017), 668-701 687 − qα q + 1 Bq+1 q+1 ( 2(p+ 1)E(0) (p− 1)b ) q−1 2 ] ξ(t)‖∇u‖22 + 1− b 2b ξ(t)(g ◦ ∇u)(t) + ξ(t)‖u‖p+1 p+1 + 1 (q + 1)α ξ(t)‖ut‖q+1 Γ1,q+1. (4.12) Proof. By using the equation of (1.1), we deduce that Ψ′(t) = ξ(t) ρ ‖ut‖ρ+1 ρ+1 + ξ(t) ∫ Ω |ut|ρ−1uttudx+ ξ′(t) ρ ∫ Ω |ut|ρ−1utudx = ξ(t) ρ ‖ut‖ρ+1 ρ+1 − ξ(t)‖∇u‖ 2 2 + ξ(t) ∫ Ω ∇u(t) · ∫ t 0 g(t− s)∇u(s)dsdx + ξ(t)‖u‖p+1 p+1 + ξ′(t) ρ ∫ Ω |ut|ρ−1utudx− ξ(t) ∫ Γ1 u|ut|q−1utdΓ. (4.13) From the Young inequality and the fact that ∫ t 0 g(s)ds ≤ ∫∞ 0 g(s)ds = 1− b, we have∫ Ω ∇u(t) · ∫ t 0 g(t− s)∇u(s)dsdx ≤ 1 2 ‖∇u‖22 + 1 2 ∫ Ω (∫ t 0 g(t− s)(|∇u(s)−∇u(t)|+ |∇u(t)|)ds )2 dx ≤ 1 2 ‖∇u‖22 + 1 2 (1 + η) ∫ Ω (∫ t 0 g(t− s)|∇u(t)|ds )2 dx + 1 2 (1 + 1 η ) ∫ Ω (∫ t 0 g(t− s)|∇u(s)−∇u(t)|ds )2 dx ≤ 1 2 ‖∇u‖22 + 1 2 (1 + η)(1− b)2‖∇u‖22 + 1 2 (1 + 1 η )(1− b)(g ◦ ∇u)(t) (4.14) for any η > 0. We choose η = b 1−b , then (4.14) yields∫ Ω ∇u(t) · ∫ t 0 g(t− s)∇u(s)dsdx ≤ 2− b 2 ‖∇u‖22 + 1− b 2b (g ◦ ∇u)(t). (4.15) Applying the Young inequality, Hölder inequality and (4.5), it is easy to see that∫ Ω |ut|ρ−1utudx ≤ ‖ut‖ρρ+1‖u‖ρ+1 ≤ ρα ρ+ 1 ‖u‖ρ+1 ρ+1 + 1 (ρ+ 1)α ‖ut‖ρ+1 ρ+1 ≤ ρα ρ+ 1 Bρ+1 ρ+1 ( 2(p+ 1)E(0) (p− 1)b ) ρ−1 2 ‖∇u‖22 + 1 (ρ+ 1)α ‖ut‖ρ+1 ρ+1, (4.16) for any α > 0. By the Young inequality, trace theorem and (4.5), it follows that∫ Γ1 |ut|q−1utudΓ H.F. Di, Y.D. Shang / Eur. J. Pure Appl. Math, 10 (4) (2017), 668-701 688 ≤ qα q + 1 ‖u‖q+1 Γ1,q+1 + 1 (q + 1)α ‖ut‖q+1 Γ1,q+1 ≤ qα q + 1 Bq+1 q+1 ( 2(p+ 1)E(0) (p− 1)b ) q−1 2 ‖∇u‖22 + 1 (q + 1)α ‖ut‖q+1 Γ1,q+1. (4.17) where Bq+1 is the optimal constant satisfying the inequality ‖u‖Γ1,q+1 ≤ Bq+1‖∇u‖2. Inserting (4.15)-(4.17) into (4.13) and applying the conditions (A2), we deduce that Ψ′(t) ≤ ξ(t) ρ ‖ut‖ρ+1 ρ+1 − ξ(t)‖∇u‖ 2 2 + 2− b 2 ξ(t)‖∇u‖22 + 1− b 2b ξ(t)(g ◦ ∇u)(t) + ξ(t)‖u‖p+1 p+1 + α ρ+ 1 kξ(t)Bρ+1 ρ+1 ( 2(p+ 1)E(0) (p− 1)b ) ρ−1 2 ‖∇u‖22 + 1 (q + 1)α ξ(t)‖ut‖q+1 Γ1,q+1 + 1 (ρ+ 1)ρα kξ(t)‖ut‖ρ+1 ρ+1 + qα q + 1 Bq+1 q+1 ( 2(p+ 1)E(0) (p− 1)b ) q−1 2 ξ(t)‖∇u‖22 = [ 1 ρ + 1 (ρ+ 1)ρα k ] ξ(t)‖ut‖ρ+1 ρ+1 − [ b 2 − α ρ+ 1 kBρ+1 ρ+1 ( 2(p+ 1)E(0) (p− 1)b ) ρ−1 2 − qα q + 1 Bq+1 q+1 ( 2(p+ 1)E(0) (p− 1)b ) q−1 2 ] ξ(t)‖∇u‖22 + 1− b 2b ξ(t)(g ◦ ∇u)(t) + ξ(t)‖u‖p+1 p+1 + 1 (q + 1)α ξ(t)‖ut‖q+1 Γ1,q+1. Lemma 10. Let the assumptions (A1)-(A3) hold and 1 < ρ < ∞ if n ≤ 2, 1 < ρ ≤ n+2 n−2 if n ≥ 3. Furthermore assume that E(0) < d̃ and I(u0) > 0, then the functional Φ(t) = − ξ(t) ρ ∫ Ω |ut| ρ−1ut ∫ t 0 g(t− s)[u(t)− u(s)]dsdx satisfies the following inequality Φ′(t) ≤ δ [ 1 + 2(1− b)2 + p p+ 1 Bp+1 p+1 ( 2(p+ 1)E(0) (p− 1)b ) p−1 2 ] ξ(t)‖∇u‖22 + [ δ ρ+ 1 + kδ ρ+ 1 − ∫ t 0 g(s)ds ρ ] ξ(t)‖ut‖ρ+1 ρ+1 + [ (2δ + 1 2δ )(1− b) + Bp+1 p+1(1− b)p (p+ 1)δ ( 4(p+ 1)E(0) (p− 1)b ) p−1 2 + Bq+1 q+1(1− b)q (q + 1)δ ( 4(p+ 1)E(0) (p− 1)b ) q−1 2 + kBρ+1 ρ+1(1− b)ρ ρ(ρ+ 1)δ ( 4(p+ 1)E(0) (p− 1)b ) ρ−1 2 ] ξ(t)(g ◦ ∇u)(t) + qδ q + 1 ξ(t)‖ut‖q+1 Γ1,q+1 − g(0)ρBρ+1 ρ+1 ρ(ρ+ 1)δ ( 4(p+ 1)E(0) (p− 1)b ) ρ−1 2 ξ(t)(g′ ◦ ∇u)(t). (4.18) Proof. Applying the equation of (1.1) and integrating by parts, we deduce that Φ′(t) = −ξ(t) ∫ Ω |ut|ρ−1utt ∫ t 0 g(t− s)[u(t)− u(s)]dsdx H.F. Di, Y.D. Shang / Eur. J. Pure Appl. Math, 10 (4) (2017), 668-701 689 − ξ′(t) ρ ∫ Ω |ut|ρ−1ut ∫ t 0 g(t− s)[u(t)− u(s)]dsdx − ξ(t) ρ ∫ Ω |ut|ρ−1ut ∫ t 0 g′(t− s)[u(t)− u(s)]dsdx − ξ(t) ρ ∫ t 0 g(s)ds ∫ Ω |ut|ρ+1dx = ξ(t) ∫ Ω ∇u(t) ∫ t 0 g(t− s)[∇u(t)−∇u(s)]dsdx − ξ(t) ∫ Ω ∫ t 0 g(t− s)∇u(s)ds ∫ t 0 g(t− s)[∇u(t)−∇u(s)]dsdx − ξ(t) ∫ Ω |u|p−1u ∫ t 0 g(t− s)[u(t)− u(s)]dsdx + ξ(t) ∫ Γ1 |ut|q−1ut ∫ t 0 g(t− s)[u(t)− u(s)]dsdΓ − ξ′(t) ρ ∫ Ω |ut|ρ−1ut ∫ t 0 g(t− s)[u(t)− u(s)]dsdx − ξ(t) ρ ∫ Ω |ut|ρ−1ut ∫ t 0 g′(t− s)[u(t)− u(s)]dsdx − ξ(t) ρ ∫ t 0 g(s)ds ∫ Ω |ut|ρ+1dx. (4.19) From the Young inequality and Hölder inequality, for any δ > 0, we have∫ Ω ∇u(t) ∫ t 0 g(t− s)[∇u(t)−∇u(s)]dsdx ≤ δ‖∇u‖22 + 1− b 4δ (g ◦ ∇u)(t). (4.20) By the calculation similar to (4.14), we get − ∫ Ω ∫ t 0 g(t− s)∇u(s)ds ∫ t 0 g(t− s)[∇u(t)−∇u(s)]dsdx ≤ δ ∫ Ω (∫ t 0 g(t− s)[|∇u(s)−∇u(t)|+ |∇u(t)|]ds )2 dx + 1 4δ ∫ Ω (∫ t 0 g(t− s)|∇u(t)−∇u(s)|ds )2 dx ≤ (2δ + 1 4δ ) ∫ Ω (∫ t 0 g(t− s)|∇u(s)−∇u(t)|ds )2 dx+ 2δ(1− b)2‖∇u‖22 ≤ (2δ + 1 4δ )(1− b)(g ◦ ∇u)(t) + 2δ(1− b)2‖∇u‖22. (4.21) Using the Young inequality and Sobolev inequality, it follows that − ∫ Ω |u|p−1u ∫ t 0 g(t− s)[u(t)− u(s)]dsdx H.F. Di, Y.D. Shang / Eur. J. Pure Appl. Math, 10 (4) (2017), 668-701 690 ≤ pδ p+ 1 ‖u‖p+1 p+1 + 1 (p+ 1)δ ∫ Ω (∫ t 0 g(t− s)[u(t)− u(s)]ds )p+1 dx ≤ pδ p+ 1 ‖u‖p+1 p+1 + Bp+1 p+1(1− b)p (p+ 1)δ ( 4(p+ 1)E(0) (p− 1)b ) p−1 2 (g ◦ ∇u)(t). (4.22) Taking the Young inequality and trace theorem into account, we deduce∫ Γ1 |ut|q−1ut ∫ t 0 g(t− s)[u(t)− u(s)]dsdΓ ≤ qδ q + 1 ‖ut‖q+1 Γ1,q+1 + 1 (q + 1)δ ∫ Γ1 (∫ t 0 g(t− s)[u(t)− u(s)]ds )q+1 dΓ ≤ qδ q + 1 ‖ut‖q+1 Γ1,q+1 + Bq+1 q+1(1− b)q (q + 1)δ ( 4(p+ 1)E(0) (p− 1)b ) q−1 2 (g ◦ ∇u)(t). (4.23) Making use of the Young inequality and lemma 5, we have −1 ρ ∫ Ω |ut|ρ−1ut ∫ t 0 g(t− s)[u(t)− u(s)]dsdx ≤ δ ρ+ 1 ‖ut‖ρ+1 ρ+1 + 1 ρ(ρ+ 1)δ ∫ Ω (∫ t 0 g(t− s)[u(t)− u(s)]ds )ρ+1 dx ≤ δ ρ+ 1 ‖ut‖ρ+1 ρ+1 + Bρ+1 ρ+1(1− b)ρ ρ(ρ+ 1)δ ( 4(p+ 1)E(0) (p− 1)b ) ρ−1 2 (g ◦ ∇u)(t). (4.24) Furthermore, similar calculation give that −1 ρ ∫ Ω |ut|ρ−1ut ∫ t 0 g′(t− s)[u(t)− u(s)]dsdx ≤ δ ρ+ 1 ‖ut‖ρ+1 ρ+1 + 1 ρ(ρ+ 1)δ ∫ Ω (∫ t 0 g′(t− s)[u(t)− u(s)]ds )ρ+1 dx ≤ δ ρ+ 1 ‖ut‖ρ+1 ρ+1 + g(0)ρBρ+1 ρ+1 ρ(ρ+ 1)δ ( 4(p+ 1)E(0) (p− 1)b ) ρ−1 2 (−g′ ◦ ∇u)(t). (4.25) Finally, Inserting (4.20)-(4.25) into (4.19) and applying the conditions (A2), we can obtain that the conclusion of Lemma holds. Now, we are ready to give the proof of the Theorem 2. Proof. Since the function g is positive, continuous and g(0) > 0, for any t0 > 0 we have ∫ t 0 g(s)ds ≥ ∫ t0 0 g(s)ds = g0 > 0, ∀ t ≥ t0. (4.26) Combining (4.1),(4.12),(4.18) and lemma 4, by a series of computations, we have that F ′(t) ≤ME′(t) + ε [ 1 ρ + k (ρ+ 1)ρα ] ξ(t)‖ut‖ρ+1 ρ+1 − ε [ b 2 − αkBρ+1 ρ+1 ρ+ 1 H.F. Di, Y.D. Shang / Eur. J. Pure Appl. Math, 10 (4) (2017), 668-701 691 × ( 2(p+ 1)E(0) (p− 1)b ) ρ−1 2 − qαBq+1 q+1 q + 1 ( 2(p+ 1)E(0) (p− 1)b ) q−1 2 ] ξ(t)‖∇u‖22 + ε 1− b 2b ξ(t)(g ◦ ∇u)(t) + εξ(t)‖u‖p+1 p+1 + ε 1 (q + 1)α ξ(t)‖ut‖q+1 Γ1,q+1 + δ [ 1 + 2(1− b)2 + p p+ 1 Bp+1 p+1 ( 2(p+ 1)E(0) (p− 1)b ) p−1 2 ] ξ(t)‖∇u‖22 + [ δ ρ+ 1 + kδ ρ+ 1 − ∫ t 0 g(s)ds ρ ] ξ(t)‖ut‖ρ+1 ρ+1 + [ (2δ + 1 2δ )(1− b) + Bp+1 p+1(1− b)p (p+ 1)δ ( 4(p+ 1)E(0) (p− 1)b ) p−1 2 + Bq+1 q+1(1− b)q (q + 1)δ ( 4(p+ 1)E(0) (p− 1)b ) q−1 2 + k ρ(ρ+ 1)δ Bρ+1 ρ+1(1− b)ρ ( 4(p+ 1)E(0) (p− 1)b ) ρ−1 2 ] ξ(t)(g ◦ ∇u)(t) − g(0)ρBρ+1 ρ+1 ρ(ρ+ 1)δ ( 4(p+ 1)E(0) (p− 1)b ) ρ−1 2 ξ(t)(g′ ◦ ∇u)(t) + qδ q + 1 ξ(t)‖ut‖q+1 Γ1,q+1 ≤ − { ε [ b 2 − αkBρ+1 ρ+1 ρ+ 1 ( 2(p+ 1)E(0) (p− 1)b ) ρ−1 2 − qαBq+1 q+1 q + 1 ( 2(p+ 1)E(0) (p− 1)b ) q−1 2 ] − δ [ 1 + 2(1− b)2 + p p+ 1 Bp+1 p+1 ( 2(p+ 1)E(0) (p− 1)b ) p−1 2 ]} ξ(t)‖∇u‖22 − { M 2 − 1 ρ(ρ+ 1)δ g(0)ρBρ+1 2 ( 4(p+ 1)E(0) (p− 1)b ) ρ−1 2 ξ(0) } (−g′ ◦ ∇u)(t) + { ε 1− b 2b + [ (2δ + 1 2δ )(1− b) + Bp+1 p+1(1− b)p (p+ 1)δ ( 4(p+ 1)E(0) (p− 1)b ) p−1 2 + Bq+1 q+1(1− b)q (q + 1)δ ( 4(p+ 1)E(0) (p− 1)b ) q−1 2 + kBρ+1 2 (1− b)ρ ρ(ρ+ 1)δ × ( 4(p+ 1)E(0) (p− 1)b ) ρ−1 2 ]} ξ(t)(g ◦ ∇u)(t) + εξ(t)‖u‖p+1 p+1 − { M − ε 1 (q + 1)α ξ(0)− qδ q + 1 ξ(0) } ‖ut‖q+1 Γ1,q+1 − { g0 ρ − ε [ 1 ρ + k (ρ+ 1)ρα ] − [ δ ρ+ 1 + kδ ρ+ 1 ]} ξ(t)‖ut‖ρ+1 ρ+1, (4.27) for all t ≥ t0. At this point, we choose α > 0 so small that b 2 − αkBρ+1 ρ+1 ρ+ 1 ( 2(p+ 1)E(0) (p− 1)b ) ρ−1 2 − αqBq+1 q+1 q + 1 ( 2(p+ 1)E(0) (p− 1)b ) q−1 2 > 0. (4.28) H.F. Di, Y.D. Shang / Eur. J. Pure Appl. Math, 10 (4) (2017), 668-701 692 When α is fixed, we choose ε > 0 small enough so that lemma 9 holds and that ε < g0(ρ+ 1)α (ρ+ 1)α+ k . (4.29) Once α and ε are fixed, we choose a positive constant δ small enough such that ε [ b 2 − αkBρ+1 ρ+1 ρ+ 1 ( 2(p+ 1)E(0) (p− 1)b ) ρ−1 2 − αqBq+1 q+1 q + 1 ( 2(p+ 1)E(0) (p− 1)b ) q−1 2 ] − δ [ 1 + 2(1− b)2 + pBp+1 p+1 p+ 1 ( 2(p+ 1)E(0) (p− 1)b ) p−1 2 ] > 0, (4.30) and g0 ρ − ε [ 1 ρ + k (ρ+ 1)ρα ] − [ δ ρ+ 1 + kδ ρ+ 1 ] > 0. (4.31) Then, we pick M sufficiently large such that lemma 9 holds and that{ M 2 − g(0)ρBρ+1 ρ+1 ρ(ρ+ 1)δ ( 4(p+ 1)E(0) (p− 1)b ) ρ−1 2 ξ(0) } − { ε 1− b 2b + [ (2δ + 1 2δ )(1− b) + Bp+1 p+1(1− b)p (p+ 1)δ ( 4(p+ 1)E(0) (p− 1)b ) p−1 2 + Bq+1 q+1(1− b)q (q + 1)δ ( 4(p+ 1)E(0) (p− 1)b ) q−1 2 + kBρ+1 ρ+1(1− b)ρ ρ(ρ+ 1)δ ( 4(p+ 1)E(0) (p− 1)b ) ρ−1 2 ]} > 0, (4.32) and M − ε (q + 1)α ξ(0)− qδ q + 1 ξ(0) > 0. (4.33) Therefore, from the conditions (A2), we obtain that there exists a positive constant β1 > 0 such that{ M 2 − g(0)ρBρ+1 ρ+1 ρ(ρ+ 1)δ ( 4(p+ 1)E(0) (p− 1)b ) ρ−1 2 ξ(0) } (−g′ ◦ ∇u)(t)− { ε 1− b 2b + [ (2δ + 1 2δ )(1− b) + Bp+1 p+1(1− b)p (p+ 1)δ ( 4(p+ 1)E(0) (p− 1)b ) p−1 2 + Bq+1 q+1(1− b)q (q + 1)δ ( 4(p+ 1)E(0) (p− 1)b ) q−1 2 + kBρ+1 ρ+1(1− b)ρ ρ(ρ+ 1)δ ( 4(p+ 1)E(0) (p− 1)b ) ρ−1 2 ]} ξ(t)(g ◦ ∇u)(t) > β1ξ(t)(g ◦ ∇u)(t). (4.34) Combining (4.27)-(4.34), the definition of E(t) and lemma 8, we deduce that there exists a positive constant β2 > 0 such that F ′(t) ≤ −β2ξ(t)E(t) ≤ −C1β2ξ(t)F (t), ∀ t ≥ t0. (4.35) H.F. Di, Y.D. Shang / Eur. J. Pure Appl. Math, 10 (4) (2017), 668-701 693 A simple integration of (4.35) over (t0, t), it follows that F (t) ≤ F (t0)e −C1β2 ∫ t t0 ξ(s)ds , ∀ t ≥ t0. (4.36) Furthermore, by lemma 6 and (4.36), we obtain E(t) ≤ C2F (t0)e −C1β2 ∫ t t0 ξ(s)ds = Le −η ∫ t t0 ξ(s)ds , ∀ t ≥ t0. (4.37) where L = C2F (t0) and η = C1β2. This completes the proof. 5. Finite time blow up of the solutions To prove the blow up result for certain solutions with nonpositive initial energy as well as positive initial energy, we modified and improved the methods of [9,21]. Theorem 11. Let the assumptions (A1), (A3) hold. For any fixed positive number β < 1, assume that u0(x) ∈ H1 Γ0 (Ω), u1(x) ∈ Lρ+1(Ω) ∩ Lq+1(Γ1), and satisfy I(u0) < 0, E(0) < βd̃. (5.1) Further assume that ρ < p and the relaxation function g satisfies∫ ∞ 0 g(s)ds < [(p− 1)(1− β)− γ]2 + 2[(p− 1)(1− β)− γ] [(p− 1)(1− β)− γ + 2]2 + 1 , (5.2) where 0 < γ < (p−1)(1−β). Then, the solutions of problem (1.1) blows up in finite time, that is, the maximum existence time Tmax of u(t) is finite and lim t→Tmax ( ‖ut‖ρ+1 ρ+1 + ‖∇u‖22 + ‖u‖p+1 p+1 ) = +∞. (5.3) First of all, we introduce the following Lemma which will be needed in the course of this section. Lemma 12. Let the assumptions (A3) hold. Then there exists a positive constant C > 1 depending on Ω only such that ‖u‖sp+1 ≤ C(‖∇u‖22 + ‖u‖p+1 p+1) (5.4) for any u ∈ H1 Γ0 (Ω) and 2 ≤ s ≤ p+ 1. Now, we are ready to prove blow up result of the solutions for the problem (1.1). Proof. In Lemma 3 (2), we have proved that if I(u0) < 0 then I(u) < 0 for any t ∈ [0, Tmax) in the case of E(0) < βd̃. By contradiction, we assume that the solution of problem (1.1) is global. Then, for any T > 0 we may consider functional θ : [0, T ]→ R+ defined by θ(t) = ‖ut‖ρ+1 ρ+1 + ‖∇u‖22 + ‖u‖p+1 p+1. (5.5) H.F. Di, Y.D. Shang / Eur. J. Pure Appl. Math, 10 (4) (2017), 668-701 694 As θ(t) is continuous on [0, T ], there exist δ1, δ2 > 0 such that δ1 ≤ θ(t) ≤ δ2. First, we set N(t) = βd̃− E(t) (5.6) for all t ∈ [0, T ]. Differentiating the identity (5.6) with respect to t, we have N ′(t) = −E′(t) = ‖ut‖q+1 Γ1,q+1 + 1 2 g(t)‖∇u(t)‖22 − 1 2 ∫ Ω ∫ t 0 g′(t− s)[∇u(s)−∇u(t)]2dsdx ≥ 0. (5.7) Hence N(t) ≥ N(0) = βd̃− E(0) > 0. (5.8) From the Lemma 4 and (2.1), it follows that N(t) ≤ βd̃+ 1 p+ 1 ‖u‖p+1 p+1 ≤ (β(p− 1) 2(p+ 1) + 1 p+ 1 ) ‖u‖p+1 p+1, (5.9) for all t ∈ [0, T ]. Next, we define G(t) = N1−σ(t) + ε ρ ∫ Ω u|ut|ρ−1utdx, ∀ t ∈ [0, T ], (5.10) where 0 < ε� 1 to be chosen later and 0 < σ < min { 1 ρ+ 1 , 1 q } , (5.11) which will be used later. Differentiating the identity (5.10) with respect to t and using equation (1.1), we obtain G′(t) = (1− σ)N−σ(t)N ′(t) + ε ρ ‖ut‖ρ+1 ρ+1 + ε(|ut|ρ−1utt, u) = (1− σ)N−σ(t)N ′(t) + ε ρ ‖ut‖ρ+1 ρ+1 − ε‖∇u‖ 2 2 + ε‖u‖p+1 p+1 + ε ∫ Ω ∇u(t) ∫ t 0 g(t− s)∇u(s)dsdx− ε ∫ Γ1 |ut|q−1utudΓ. (5.12) Considering the relation (p+ 1− γ)N(t) = (p+ 1− γ)βd̃− p+ 1− γ ρ+ 1 ‖ut‖ρ+1 ρ+1 − p+ 1− γ 2 l(t)‖∇u‖22 − p+ 1− γ 2 (g ◦ ∇u)(t) + (p+ 1− γ) p+ 1 ‖u‖p+1 p+1 (5.13) H.F. Di, Y.D. Shang / Eur. J. Pure Appl. Math, 10 (4) (2017), 668-701 695 and Young inequality∫ Ω ∇u(t) ∫ t 0 g(t− s)[∇u(s)−∇u(t)]dsdx ≤ 1 4ξ ∫ t 0 g(s)ds‖∇u(t)‖22 + ξ ∫ t 0 g(t− s) ∫ Ω |∇u(s)−∇u(t)|2dsdx, (5.14) ∫ Γ1 |ut|q−1utudΓ ≤ µq+1 q + 1 ‖u‖q+1 Γ1,q+1 + q q + 1 µ − q+1 q ‖ut‖q+1 Γ1,q+1, (5.15) where γ, ξ, µ > 0 to be determined later, we get from (5.12) that G′(t) = (1− σ)N−σ(t)N ′(t) + ε(p+ 1− γ)N(t)− ε(p+ 1− γ)βd̃+ ε ρ ‖ut‖ρ+1 ρ+1 + ε p+ 1− γ ρ+ 1 ‖ut‖ρ+1 ρ+1 + ε p+ 1− γ 2 l(t)‖∇u‖22 + ε p+ 1− γ 2 (g ◦ ∇u)(t) − ε(p+ 1− γ) p+ 1 ‖u‖p+1 p+1 − ε‖∇u‖ 2 2 + ε ∫ Ω ∇u(t) ∫ t 0 g(t− s)∇u(s)dsdx+ ε‖u‖p+1 p+1 − ε µ q+1 q + 1 ‖u‖q+1 Γ1,q+1 − ε q q + 1 µ − q+1 q ‖ut‖q+1 Γ1,q+1 ≥ [ (1− σ)N−σ(t)− ε q q + 1 µ − q+1 q ] ‖ut‖q+1 Γ1,q+1 + ε(p+ 1− γ)N(t) + ε [ ( p+ 1− γ 2 − 1)− ( p+ 1− γ 2 + 1 4ξ ) ∫ t 0 g(s)ds ] ‖∇u‖22 + ε [ 1 ρ + p+ 1− γ ρ+ 1 ] ‖ut‖ρ+1 ρ+1 + ε [ p+ 1− γ 2 − ξ ] (g ◦ ∇u)(t)− ε(p+ 1− γ)βd̃ + ε γ p+ 1 ‖u‖p+1 p+1 − ε µq+1 q + 1 ‖u‖q+1 Γ1,q+1. (5.16) As in [9], we take µ − q+1 q = DN−σ(t) for some large D to be specified later and if this is substituted in (5.16), we have G′(t) ≥ [ (1− σ)− ε q q + 1 D ] N−σ(t)‖ut‖q+1 Γ1,q+1 + ε [ 1 ρ + p+ 1− γ ρ+ 1 ] ‖ut‖ρ+1 ρ+1 + ε [ (p+ 1− γ)N(t)− D−q q + 1 Nσq(t)‖u‖q+1 Γ1,q+1 ] − ε(p+ 1− γ)βd̃ + ε [ ( p+ 1− γ 2 − 1)− ( p+ 1− γ 2 + 1 4ξ ) ∫ t 0 g(s)ds ] ‖∇u‖22 + ε [ p+ 1− γ 2 − ξ ] (g ◦ ∇u)(t) + ε γ p+ 1 ‖u‖p+1 p+1. (5.17) Taking into account the Lemma 4, we deduce −ε(p+ 1− γ)βd̃ ≥ −ε(p+ 1)β( 1 2 − 1 p+ 1 )[l(t)‖∇u‖22 + (g ◦ ∇u)(t)] H.F. Di, Y.D. Shang / Eur. J. Pure Appl. Math, 10 (4) (2017), 668-701 696 = −εβ(p− 1) 2 [l(t)‖∇u‖22 + (g ◦ ∇u)(t)]. (5.18) From (5.5),(5.9) and trace Theorem, we get Nσq(t)‖u‖q+1 Γ1,q+1 ≤ ( β(p− 1) + 2 2(p+ 1) )σq ‖u‖(p+1)σq p+1 ‖u‖q+1 Γ1,q+1 ≤ ( β(p− 1) + 2 2(p+ 1) )σq ‖u‖(p+1)σq p+1 Bq+1 q+1‖∇u‖ q+1 2 ≤ ( β(p− 1) + 2 2(p+ 1) )σq ‖u‖(p+1)σq p+1 Bq+1 q+1δ q+1 2 2 . (5.19) Using the following inequality zθ ≤ z + 1 ≤ (1 + 1 ς )(z + ς), 0 < θ ≤ 1, ∀ z > 0, ς > 0, and taking ς = N(0), then we have Nσq(t)‖u‖q+1 Γ1,q+1 ≤ ( β(p− 1) + 2 2(p+ 1) )σq Bq+1 q+1δ q+1 2 2 k ( ‖u‖(p+1) p+1 +N(t) ) , (5.20) where k = 1 + 1 N(0) . Inserting (5.18) and (5.20) into (5.17), we obtain G′(t) ≥ [ (1− σ)− ε q q + 1 D ] N−σ(t)‖ut‖q+1 Γ1,q+1 + ε [ 1 ρ + p+ 1− γ ρ+ 1 ] ‖ut‖ρ+1 ρ+1 + ε [ (p+ 1− γ)− D−q q + 1 ( β(p− 1) + 2 2(p+ 1) )σq Bq+1 q+1δ q+1 2 2 k ] N(t) + ε [ p− 1− γ 2 − β(p− 1) 2 − ( p+ 1− γ 2 − β(p− 1) 2 + 1 4ξ ) ∫ t 0 g(s)ds ] ‖∇u‖22 + ε [ p+ 1− γ 2 − β(p− 1) 2 − ξ ] (g ◦ ∇u)(t) + ε [ γ p+ 1 − D−q q + 1 ( β(p− 1) + 2 2(p+ 1) )σq Bq+1 q+1δ q+1 2 2 k ] ‖u‖p+1 p+1 = K1N −σ(t)‖ut‖q+1 Γ1,q+1 + ε [ 1 ρ + p+ 1− γ ρ+ 1 ] ‖ut‖ρ+1 ρ+1 +K2N(t) +K3‖u‖p+1 p+1 +K4‖∇u‖22 +K5(g ◦ ∇u)(t). (5.21) Utilizing (5.2) and taking 0 < ξ ≤ (1−β)(p−1) 2 + 2−γ 2 , we have K4 = (p− 1)(1− β)− γ 2 − ( (1− β)(p− 1) 2 + 2− γ 2 + 1 4ξ ) ∫ t 0 g(s)ds > 0, (5.22) K5 = (1− β)(p− 1) 2 + 2− γ 2 − ξ ≥ 0, (5.23) H.F. Di, Y.D. Shang / Eur. J. Pure Appl. Math, 10 (4) (2017), 668-701 697 where the positive constant γ satisfies 0 < γ < (p− 1)(1− β). At this point, we choose D large enough such that K2 = (p+ 1− γ)− D−q q + 1 ( β(p− 1) + 2 2(p+ 1) )σq Bq+1 q+1δ q+1 2 2 k > 0, (5.24) K3 = γ p+ 1 − D−q q + 1 ( β(p− 1) + 2 2(p+ 1) )σq Bq+1 q+1δ q+1 2 2 k > 0. (5.25) Once D is fixed, we pick ε small enough so that K1 = (1− σ)− ε q q + 1 D > 0, (5.26) and G(0) = N1−σ(0) + ε ρ ∫ Ω u0|u1|ρ−1u1dx > 0. (5.27) Thus, we have G′(t) ≥ εη [ N(t) + ‖ut‖ρ+1 ρ+1 + ‖∇u‖22 + (g ◦ ∇u)(t) + ‖u‖p+1 p+1 ] ≥ 0, (5.28) for some small number η > 0. Consequently, from (5.27) and (5.28) we obtain G(t) ≥ G(0) > 0, t ≥ 0. (5.29) Now, by the Hölder inequality and Sobolev inequality, we estimate∣∣ ∫ Ω u|ut|ρ−1ut ∣∣ 1 1−σ ≤ ‖ut‖ ρ 1−σ ρ+1 ‖u‖ 1 1−σ ρ+1 ≤ C‖ut‖ ρ 1−σ ρ+1 ‖u‖ 1 1−σ p+1 ≤ C(‖ut‖ ρ% 1−σ ρ+1 + ‖u‖ %′ 1−σ p+1 ), (5.30) where 1 % + 1 %′ = 1. We choose % = (ρ+1)(1−σ) ρ (> 1), then %′ 1− σ = ρ+ 1 (ρ+ 1)(1− σ)− ρ . (5.31) By using lemma 9 and (5.31), then (5.30) becomes∣∣ ∫ Ω u|ut|ρ−1ut ∣∣ 1 1−σ ≤ C ( ‖ut‖ρ+1 ρ+1 + ‖∇u‖22 + ‖u‖p+1 p+1 ) . (5.32) Hence, combining (5.10) and (5.32) , we deduce that G 1 1−σ (t) = ( N1−σ(t) + ε ρ ∫ Ω |ut|ρ−1utudx ) 1 1−σ H.F. Di, Y.D. Shang / Eur. J. Pure Appl. Math, 10 (4) (2017), 668-701 698 ≤ C ( N(t) + ‖ut‖ρ+1 ρ+1 + ‖∇u‖22 + ‖u‖p+1 p+1 ) . (5.33) By the combination of (5.28) and (5.33), we obtain G′(t) ≥ QG 1 1−σ (t), ∀ t ∈ [0, T ], (5.34) where Q is a positive constant depending only on C and εη. A simple integration of (5.34) over (0,t), it follows that G σ 1−σ (t) ≥ 1 G−σ/(1−σ)(0)−Qσt/(1− σ) , ∀ t ∈ [0, T ]. (5.35) This shows that G(t) blows up in finite time T ∗ ≤ 1− σ Gσ/1−σ(0)Qσ . (5.36) Furthermore, we have from (5.33) that there exists a finite time T ∗ ∈ (0, T ) such that lim t→T ∗− ( ‖ut‖ρ+1 ρ+1 + ‖∇u‖22 + ‖u‖p+1 p+1 ) = +∞, which contradicts Tmax =∞. Hence, the solutions of the problem (1.1) blows up in finite time. Remark 3. Noting that from 1 ρ+ 1 ‖u1‖ρ+1 ρ+1 + p− 1 2(p+ 1) ‖∇u0‖22 + 1 p+ 1 I(u0) ≤ 1 ρ+ 1 ‖u1‖ρ+1 ρ+1 + J(u0) = E(0), (5.37) we see that if E(0) < 0, then I(u0) ≥ 0 is impossible. If E(0) = 0, then either I(u0) > 0 or I(u0) = 0 with ‖∇u0‖22 6= 0 is impossible. If 0 < E(0) < βd̃ (βd̃ < d̃), it follows from the definition of d that I(u0) = 0 with ‖∇u0‖22 6= 0 is impossible. Otherwise, we have J(u0) ≥ d which contradicts (5.37). Thus, all possible cases already have been considered in Theorems 1, 3. From the discussion above in Sections 3, 5, a threshold result of global existence and nonexistence of solutions for problem (1.1) has been obtained as follows. 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