1 Introduction

$\begin{array}{c}{u}_{tt}+{\Delta }^{2}u-\underset{0}{\overset{t}{\int }}g\left(t-s\right){\Delta }^{2}u\left(s\right)ds={\left|u\right|}^{p}u,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}x\in \Omega ,\tau >0\\ u\left(x,t\right)={\partial }_{\nu }u\left(x,t\right)=0,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}x\in \partial \Omega ,t\ge 0\\ u\left(x,0\right)={u}_{0}\left(x\right),\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}{u}_{t}\left(x,0\right)={u}_{1}\left(x\right),\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}x\in \Omega \end{array}$
(1.1)

where ΩRnis a bounded domain with smooth boundary ∂Ω in order that the divergence theorem can be applied. ν is the unit normal vector pointing toward the exterior of Ω and p > 0. Here, g represents the kernel of the memory term satisfying some conditions to be specified later.

In the absence of the viscoelastic term, i.e., (g = 0), we motivate our article by presenting some results related to initial-boundary value Petrovsky problem

$\begin{array}{c}{u}_{tt}+{{\Delta }^{2}}_{u}=f\left(u,{u}_{t}\right),\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}x\in \Omega ,\phantom{\rule{1em}{0ex}}t>0\\ u\left(x,t\right)={\partial }_{\nu }u\left(x,t\right)=0,\phantom{\rule{1em}{0ex}}x\in \partial \Omega ,t\ge 0\\ u\left(x,0\right)={u}_{0}\left(x\right),\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}{u}_{t}\left(x,0\right)={u}_{1}\left(x\right),\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}x\in \Omega .\end{array}$
(1.2)

Research of global existence, blow-up and energy decay of solutions for the initial boundary value problem (1.2) has attracted a lot of articles (see [14] and references there in).

Amroun and Benaissa [1] investigated (1.2) with f(u, u t ) = b|u|p-2u-h(u t ) and proved the global existence of solutions by means of the stable set method in ${H}_{0}^{2}\left(\Omega \right)$ combined with the Faedo-Galerkin procedure. In [3], Messaoudi studied problem (1.2) with f(u, u t ) = b|u|p-2u-a|u t |m-2u t . He proved the existence of a local weak solution and showed that this solution blows up in finite time with negative initial energy if p > m.

In the presence of the viscoelastic terms, Rivera et al. [5] considered the plate model:

${u}_{tt}+{\Delta }^{2}u-\underset{0}{\overset{t}{\int }}g\left(t-s\right){\Delta }^{2}u\left(s\right)ds=0$

in a bounded domain ΩRNand showed that the energy of solution decay exponentially provided the kernel function also decay exponentially. For more related results about the existence, finite time blow-up and asymptotic properties, we refer the reader to [516].

In the present article, we devote our study to problem (1.1). We will prove the existence of weak solutions under some appropriate assumptions on the function g and blow-up behavior of solutions. In order to obtain the existence of solutions, we use the Faedo-Galerkin method and to get the blow-up properties of solutions with non-positive and positive initial energy, we modify the method in [17]. Estimates for the blow-up time T* are also given.

2 Preliminaries

We define the energy function related with problem (1.1) is given by

$E\left(t\right)=\frac{1}{2}\left[{∥{u}_{t}∥}^{2}+\left(1-\underset{0}{\overset{t}{\int }}g\left(s\right)ds\right){∥\Delta u∥}^{2}+\left(g\odot \Delta u\right)\left(t\right)\right]-\frac{1}{p+2}{∥u∥}_{p+2}^{p+2},$
(2.1)

where

$\left(g\odot v\right)\left(t\right)=\underset{0}{\overset{t}{\int }}g\left(t-s\right){∥v\left(t\right)-v\left(s\right)∥}_{2}^{2}ds.$

We denote by ∥.∥ k , the Lk-norm over Ω. In particular, the L2-norm is denoted ∥.∥2. We use the familiar function spaces ${H}_{0}^{2},{H}^{4}$ and throughout this article we assume ${u}_{0}\in {H}_{0}^{2}\left(\Omega \right)\cap {H}^{4}\left(\Omega \right)$ and ${u}_{1}\in {H}_{0}^{2}\left(\Omega \right)\cap {L}^{2}\left(\Omega \right)$.

In the sequel, we state some hypotheses and three well-known lemmas that will be needed later.

(A 1) p satisfies

$\begin{array}{c}0

(A 2) g is a positive bounded C1 function satisfying g(0) > 0, and for all t > 0

$1-\underset{0}{\overset{\infty }{\int }}g\left(t\right)ds=l>0,$

also there exists positive constants L0, L1 such that

(A 3)

$-{L}_{0}\le {g}^{\prime }\left(t\right)\le 0,\phantom{\rule{1em}{0ex}}0\le {g}^{″}\left(t\right)\le {L}_{1}.$

Lemma 1 (Sobolev-Poincare's inequality). Let p be a number that satisfies (A 1), then there is a constant C* = C(Ω, p) such that

${∥u∥}_{p}\le {C}_{*}{∥\Delta u∥}_{2},\phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}u\in {H}_{0}^{2}\left(\Omega \right)$
(2.2)

Lemma 2 [4]. Let δ > 0 and B(t) ∈ C2(0, ∞) be a nonnegative function satisfying

${B}^{″}\left(t\right)-4\left(\delta +1\right){B}^{\prime }\left(t\right)+4\left(\delta +1\right)B\left(t\right)\ge 0.$
(2.3)

If

${B}^{\prime }\left(0\right)>{r}_{2}B\left(0\right)+{K}_{0},$
(2.4)

with ${r}_{2}=2\left(\delta +1\right)-2\sqrt{\delta \left(\delta +1\right)}$, then B'(t) > K0 for t > 0, where K0 is a constant.

Lemma 3 [4]. If Y(t) is a non-increasing function on [t0, ∞) and satisfies the differential inequality

${Y}^{\prime }{\left(t\right)}^{2}\ge a+bY{\left(t\right)}^{2+{\delta }^{-1}}\phantom{\rule{1em}{0ex}}for\phantom{\rule{1em}{0ex}}t\ge {t}_{0}\ge 0,$
(2.5)

where a > 0, δ > 0 and bR, then there exists a finite time T* such that

$\underset{t\to {T}^{*-}}{\mathsf{\text{lim}}}Y\left(t\right)=0.$

Upper bounds for T* is estimated as follows:

(i) If b < 0, then

${T}^{*}\le {t}_{0}+\frac{1}{\sqrt{-b}}ln\frac{\sqrt{\frac{-a}{b}}}{\sqrt{\frac{-a}{b}-Y\left({t}_{0}\right)}}.$

(ii) If b = 0, then

${T}^{*}\le {t}_{0}+\frac{Y\left({t}_{0}\right)}{{Y}^{\prime }\left({t}_{0}\right)}.$

(iii) If b > 0, then

${T}^{*}\le \frac{Y\left({t}_{0}\right)}{\sqrt{a}},$

or

${T}^{*}\le {t}_{0}+{2}^{\frac{3\delta +1}{2\delta }}\frac{c\delta }{\sqrt{a}}\left\{1-{\left[1+cY\left({t}_{0}\right)\right]}^{\frac{-1}{2\delta }}\right\},$

where $c={\left(\frac{a}{b}\right)}^{2+\frac{1}{\delta }}$.

3 Existence of solutions

In this section, we are going to obtain the existence of weak solutions to the problem (1.1) using Faedo-Galerkin's approximation.

Theorem 1 Let the assumptions (A 1)-(A 3) hold. Then there exists at least a solution u of (1.1) satisfying

$\begin{array}{c}u\in {L}^{\infty }\left(0,\infty ;{H}_{0}^{2}\left(\Omega \right)\cap {H}^{4}\left(\Omega \right)\right),\phantom{\rule{1em}{0ex}}{u}^{\prime }\in {L}^{\infty }\left(0,\infty ;{H}_{0}^{2}\left(\Omega \right)\cap {L}^{2}\left(\Omega \right)\right),\\ {u}^{″}\in {L}^{\infty }\left(0,\infty ;{L}^{2}\left(\Omega \right)\right)\end{array}$
(3.1)

and

$\begin{array}{c}u\left(x,t\right)\to {u}_{0}\left(x\right)\phantom{\rule{1em}{0ex}}in\phantom{\rule{2.77695pt}{0ex}}{H}_{0}^{2}\left(\Omega \right)\cap {H}^{4}\left(\Omega \right)\\ {u}^{\prime }\left(x,t\right)\to {u}_{1}\left(x\right)\phantom{\rule{1em}{0ex}}in\phantom{\rule{2.77695pt}{0ex}}{H}_{0}^{2}\left(\Omega \right)\cap {L}^{2}\left(\Omega \right)\end{array}$

as t → 0.

Proof We choose a basis {ω k } (k = 1, 2, ...) in ${H}_{0}^{2}\left(\Omega \right)\cap {H}^{4}\left(\Omega \right)$ which is orthonormal in L2(Ω) and ω k being the eigenfunctions of biharmonic operator subject to the homogeneous Dirichlet boundary condition.

Let V m be the subspace of ${H}_{0}^{2}\left(\Omega \right)\cap {H}^{4}\left(\Omega \right)$ generated by the first m vectors. Define

${u}_{m}\left(t\right)=\sum _{k=1}^{m}{d}_{m}^{k}\left(t\right){\omega }_{k},$
(3.2)

where u m (t) is the solution of the following Cauchy problem

$\begin{array}{c}\left({u}_{m}^{″}\left(t\right),{\omega }_{k}\right)+\left(\Delta {u}_{m}\left(t\right),\Delta {\omega }_{k}\right)-\underset{0}{\overset{t}{\int }}\left(t-s\right)\left(\Delta {u}_{m}\left(s\right),\Delta {\omega }_{k}\right)ds\\ -\left({\left|{u}_{m}\left(t\right)\right|}^{p}{u}_{m}\left(t\right),{\omega }_{k}\right)=0\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\forall k=1,m.\end{array}$
(3.3)

with the initial conditions (when m → ∞)

$\left\{\begin{array}{c}\hfill {u}_{m}\left(0\right)={\sum }_{k=1}^{m}\left({u}_{m}\left(0\right),{\omega }_{k}\right){\omega }_{k}\to {u}_{0}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}in\phantom{\rule{2.77695pt}{0ex}}{H}_{0}^{2}\left(\Omega \right)\cap {H}^{4}\left(\Omega \right)\hfill \\ \hfill {u}_{m}^{\prime }\left(0\right)={\sum }_{k=1}^{m}\left({u}_{m}^{\prime }\left(0\right),{\omega }_{k}\right){\omega }_{k}\to {u}_{1}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}in\phantom{\rule{2.77695pt}{0ex}}{H}_{0}^{2}\left(\Omega \right)\cap {L}^{2}\left(\Omega \right)\hfill \end{array}\right\$
(3.4)

The approximate systems (3.3) and (3.4) are the normal one of differential equations which has a solution in [0, T m ) for some T m > 0. The solution can be extended to the [0, T] for any given T > 0 by the first estimate below.

First estimation. Substituting ${u}_{m}^{\prime }\left(t\right)$ instead of ω k in (3.3), we find

$\frac{d}{dt}\left(\frac{1}{2}{∥{u}_{m}^{\prime }∥}^{2}+\frac{1}{2}{∥\Delta {u}_{m}∥}^{2}-\frac{{∥{u}_{m}∥}_{p+2}^{p+2}}{p+2}\right)-\underset{0}{\overset{t}{\int }}g\left(t-s\right)\left(\Delta {u}_{m}\left(s\right),\Delta {u}_{m}^{\prime }\left(t\right)\right)ds=0.$
(3.5)

Simple calculation similar to [11] yield

$\begin{array}{l}\phantom{\rule{1em}{0ex}}-\underset{0}{\overset{t}{\int }}g\left(t-s\right)\left(\Delta {u}_{m}\left(s\right),\Delta {u}_{m}^{\prime }\left(t\right)\right)ds=-\underset{0}{\overset{t}{\int }}g\left(t-s\right)\underset{\Omega }{\int }\Delta {u}_{m}\left(t\right)\Delta {u}_{m}^{\prime }\left(t\right)dxds\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}-\underset{0}{\overset{t}{\int }}g\left(t-s\right)\underset{\Omega }{\int }\left(\Delta {u}_{m}\left(s\right)-\Delta {u}_{m}\left(t\right)\right)\Delta {u}_{m}^{\prime }\left(t\right)dxds\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}=\frac{1}{2}\underset{0}{\overset{t}{\int }}g\left(t-s\right)\frac{d}{dt}{∥\Delta {u}_{m}\left(s\right)-\Delta {u}_{m}\left(t\right)∥}^{2}ds-\frac{1}{2}\underset{0}{\overset{t}{\int }}g\left(t-s\right)\frac{d}{dt}{∥\Delta {u}_{m}\left(t\right)∥}^{2}ds\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}=\frac{1}{2}\frac{d}{dt}\left(g\odot \Delta {u}_{m}\right)\left(t\right)-\frac{1}{2}\left({g}^{\prime }\odot \Delta {u}_{m}\right)\left(t\right)-\frac{1}{2}\frac{d}{dt}\underset{0}{\overset{t}{\int }}g\left(s\right)ds{∥\Delta {u}_{m}\left(t\right)∥}^{2}ds\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+\frac{1}{2}g\left(t\right){∥\Delta {u}_{m}\left(t\right)∥}^{2}.\phantom{\rule{2em}{0ex}}\end{array}$
(3.6)

Combining (3.5) and (3.6), we find

$\begin{array}{l}\frac{d}{dt}\left(\frac{1}{2}{∥{u}_{m}^{\prime }∥}^{2}+\frac{1}{2}\left(1-\underset{0}{\overset{t}{\int }}g\left(s\right)ds\right){∥\Delta {u}_{m}∥}^{2}+\frac{1}{2}\left(g\odot \Delta {u}_{m}\right)\left(t\right)-\frac{{∥{u}_{m}∥}_{p+2}^{p+2}}{p+2}\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}=\frac{1}{2}\left({g}^{\prime }\odot \Delta {u}_{m}\right)\left(t\right)-\frac{1}{2}g\left(t\right){∥\Delta {u}_{m}\left(t\right)∥}^{2},\phantom{\rule{2em}{0ex}}\end{array}$
(3.7)

integrating (3.7) over (0, t) and using assumption (A3) we infer that

${∥{u}_{m}^{\prime }∥}^{2}+{∥\Delta {u}_{m}∥}^{2}+\left(g\odot \Delta {u}_{m}\right)\left(t\right)-{∥{u}_{m}∥}_{p+2}^{p+2}\le {C}_{1},$
(3.8)

where C1 is a positive constant depending only on ∥u0∥, ∥u1∥, p, and l. It follows from (3.8) that

$\left\{\begin{array}{c}\hfill \left\{{u}_{m}\right\}\phantom{\rule{2.77695pt}{0ex}}is\phantom{\rule{2.77695pt}{0ex}}uniformly\phantom{\rule{2.77695pt}{0ex}}bounded\phantom{\rule{2.77695pt}{0ex}}in\phantom{\rule{2.77695pt}{0ex}}{L}^{\infty }\left(0,T;{H}_{0}^{2}\left(\Omega \right)\right)\hfill \\ \hfill \phantom{\rule{2.77695pt}{0ex}}\left\{{u}_{m}^{\prime }\right\}\phantom{\rule{2.77695pt}{0ex}}is\phantom{\rule{2.77695pt}{0ex}}uniformly\phantom{\rule{2.77695pt}{0ex}}bounded\phantom{\rule{2.77695pt}{0ex}}in\phantom{\rule{2.77695pt}{0ex}}{L}^{\infty }\left(0,T;{L}^{2}\left(\Omega \right)\right)\hfill \end{array}\right\$
(3.9)

Second estimation. Differentiating (3.3) with respect to t, we get

$\begin{array}{l}\left({u}_{m}^{‴}\left(t\right),{\omega }_{k}\right)+\left(\Delta {u}_{m}^{\prime }\left(t\right),\Delta {\omega }_{k}\right)-\underset{0}{\overset{t}{\int }}{g}^{\prime }\left(t-s\right)\left(\Delta {u}_{m}\left(s\right),\Delta {\omega }_{k}\right)ds\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}-g\left(0\right)\left(\Delta {u}_{m}\left(t\right),\Delta {\omega }_{k}\right)-\left(p+1\right)\left({\left|{u}_{m}\left(t\right)\right|}^{p}{u}_{m}^{\prime }\left(t\right),{\omega }_{k}\right)=0.\phantom{\rule{2em}{0ex}}\end{array}$
(3.10)

If we substitute ${u}_{m}^{″}\left(t\right)$ instead of ω k in (3.10), it holds that

$\begin{array}{l}\frac{d}{dt}\left(\frac{1}{2}{∥{u}_{m}^{″}∥}^{2}+\frac{1}{2}{∥\Delta {u}_{m}^{\prime }∥}^{2}\right)-\frac{d}{dt}\underset{0}{\overset{t}{\int }}{g}^{\prime }\left(t-s\right)\left(\Delta {u}_{m}\left(s\right),\Delta {u}_{m}^{\prime }\left(t\right)\right)ds\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}+\underset{0}{\overset{t}{\int }}{g}^{″}\left(t-s\right)\left(\Delta {u}_{m}\left(s\right),\Delta {u}_{m}^{\prime }\left(t\right)\right)ds+{g}^{\prime }\left(0\right)\left(\Delta {u}_{m}\left(t\right),\Delta {u}_{m}^{\prime }\left(t\right)\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}-g\left(0\right)\frac{d}{dt}\left(\Delta {u}_{m}\left(t\right),\Delta {u}_{m}^{\prime }\left(t\right)\right)+g\left(0\right)\left(\Delta {u}_{m}^{\prime }\left(t\right),\Delta {u}_{m}^{\prime }\left(t\right)\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}-\left(p+1\right)\left({\left|{u}_{m}\left(t\right)\right|}^{p}{u}_{m}^{\prime }\left(t\right),{u}_{m}^{″}\left(t\right)\right)=0.\phantom{\rule{2em}{0ex}}\end{array}$
(3.11)

Since H2(Ω) ↪ L2p+2(Ω), using Lemma 2, Hölder and Young's inequalities and (3.8)

$\begin{array}{cc}\hfill \left|\left(p+1\right)\left({\left|{u}_{m}\left(t\right)\right|}^{p}{u}_{m}^{\prime }\left(t\right),{u}_{m}^{″}\left(t\right)\right)\right|& \le \left(p+1\right){∥{u}_{m}\left(t\right)∥}_{2p+2}^{p}.{∥{u}_{m}^{\prime }\left(t\right)∥}_{2p+2}.{∥{u}_{m}^{″}\left(t\right)∥}_{2}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\le C\left(\gamma \right){∥\Delta {u}_{m}^{\prime }\left(t\right)∥}^{2}+\gamma {∥{u}_{m}^{″}\left(t\right)∥}^{2}.\hfill \end{array}$
(3.12)

Combining the relations (3.11), (3.12) and integrating over (0, t) for all t ∈ [0, T] with arbitrary fixed T, we obtain

$\begin{array}{c}\frac{1}{2}{∥{u}_{m}^{″}∥}^{2}+\frac{1}{2}{∥\Delta {u}_{m}^{\prime }∥}^{2}\le \frac{1}{2}{∥{u}_{m}^{″}\left(0\right)∥}^{2}+\underset{0}{\overset{t}{\int }}{g}^{\prime }\left(t-s\right)\left(\Delta {u}_{m}\left(s\right),\Delta {u}_{m}^{\prime }\left(t\right)\right)ds\\ +\frac{1}{2}{∥\Delta {u}_{m}^{\prime }\left(0\right)∥}^{2}-\underset{0}{\overset{t}{\int }}\underset{0}{\overset{\tau }{\int }}{g}^{″}\left(\tau -s\right)\left(\Delta {u}_{m}\left(s\right),\Delta {u}_{m}^{\prime }\left(\tau \right)\right)dsd\tau \\ -{g}^{\prime }\left(0\right)\underset{0}{\overset{t}{\int }}\left(\Delta {u}_{m}\left(s\right),\Delta {u}_{m}^{\prime }\left(s\right)\right)+g\left(0\right)\left(\Delta {u}_{m}\left(t\right),\Delta {u}_{m}^{\prime }\left(t\right)\right)\\ -g\left(0\right)\left(\Delta {u}_{m}\left(0\right),\Delta {u}_{m}^{\prime }\left(0\right)\right)-g\left(0\right)\underset{0}{\overset{t}{\int }}{∥\Delta {u}_{m}^{\prime }\left(s\right)∥}^{2}ds\\ +C\left(\gamma \right)\underset{0}{\overset{t}{\int }}{∥\Delta {u}_{m}^{\prime }\left(s\right)∥}^{2}ds+\gamma \underset{0}{\overset{t}{\int }}{∥{u}_{m}^{″}\left(s\right)∥}^{2}ds.\end{array}$
(3.13)

From (3.4) and (3.8), we deduce that

$|\frac{1}{2}{∥\Delta {u}_{m}^{\prime }\left(0\right)∥}^{2}-g\left(0\right)\left(\Delta {u}_{m}\left(0\right),\Delta {u}_{m}^{\prime }\left(0\right)\right)|\le {L}_{2},$
(3.14)

where L2 is a positive constant independent of m. In the following, we find the upper bound for ${∥{u}_{m}^{″}\left(0\right)∥}^{2}$. Again we substitute ${u}_{m}^{″}\left(t\right)$ instead of ω k in (3.3), and choosing t = 0, we arrive at

$\left({u}_{m}^{″}\left(0\right),{u}_{m}^{″}\left(0\right)\right)+\left(\Delta {u}_{m}\left(0\right),\Delta {u}_{m}^{″}\left(0\right)\right)-\left({\left|{u}_{m}\left(0\right)\right|}^{p}{u}_{m}\left(0\right),{u}_{m}^{″}\left(0\right)\right)=0,$

which combined with the Green's formula imply

${∥{u}_{m}^{″}\left(0\right)∥}^{2}+\left({\Delta }^{2}{u}_{m}\left(0\right),{u}_{m}^{″}\left(0\right)\right)-\left({\left|{u}_{m}\left(0\right)\right|}^{p}{u}_{m}\left(0\right),{u}_{m}^{″}\left(0\right)\right)=0.$
(3.15)

By using (A1), (3.4) and Young's inequality, we deduce that

$∥{u}_{m}^{″}\left(0\right)∥\le {L}_{3},$
(3.16)

where L3 > 0 is a constant independent of m.

Owing to (3.8), (3.5) and Young's inequality with (A3), we deduce that

$\begin{array}{c}\left|\underset{0}{\overset{t}{\int }}{g}^{\prime }\left(t-s\right)\left(\Delta {u}_{m}\left(s\right),\Delta {u}_{m}^{\prime }\left(t\right)\right)ds\right)\right|=\left|\left(\Delta {u}_{m}^{\prime }\left(t\right),\underset{0}{\overset{t}{\int }}{g}^{\prime }\left(t-s\right)\Delta {u}_{m}\left(s\right)ds\right)\right|\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\le \gamma {∥\Delta {u}_{m}^{\prime }\left(t\right)∥}^{2}+\frac{1}{4\gamma }\underset{\Omega }{\int }{\left(\underset{0}{\overset{t}{\int }}{g}^{\prime }\left(t-s\right)\Delta {u}_{m}\left(s\right)ds\right)}^{2}dx\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\le \gamma {∥\Delta {u}_{m}^{\prime }\left(t\right)∥}^{2}+\frac{{L}_{0}^{2}}{4\gamma }\underset{0}{\overset{t}{\int }}{∥\Delta {u}_{m}\left(s\right)∥}^{2}ds\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\le \gamma {∥\Delta {u}_{m}^{\prime }\left(t\right)∥}^{2}+{L}_{4}\left(T\right),\end{array}$
(3.17)
$\begin{array}{c}\left|-\underset{0}{\overset{t}{\int }}\underset{0}{\overset{\tau }{\int }}{g}^{″}\left(\tau -s\right)\left(\Delta {u}_{m}\left(s\right),\Delta {u}_{m}^{\prime }\left(\tau \right)\right)dsd\tau \right|\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}=\underset{0}{\overset{t}{\int }}\left(\Delta {u}_{m}^{\prime }\left(\tau \right),\underset{0}{\overset{\tau }{\int }}{g}^{″}\left(\tau -s\right)\Delta {u}_{m}\left(s\right)ds\right)\phantom{\rule{2.77695pt}{0ex}}d\tau \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\le \frac{1}{2}\underset{0}{\overset{t}{\int }}{∥\Delta {u}_{m}^{\prime }\left(s\right)∥}^{2}ds+\frac{1}{2}\underset{0}{\overset{t}{\int }}{\underset{\Omega }{\int }\left(\underset{0}{\overset{\tau }{\int }}{g}^{″}\left(\tau -s\right)\Delta {u}_{m}\left(s\right)ds\right)}^{2}dxd\tau \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\le \frac{1}{2}\underset{0}{\overset{t}{\int }}{∥\Delta {u}_{m}^{\prime }\left(s\right)∥}^{2}ds+\frac{T{L}_{1}^{2}}{2}\underset{0}{\overset{t}{\int }}{∥\Delta {u}_{m}\left(s\right)∥}^{2}ds\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\le \frac{1}{2}\underset{0}{\overset{t}{\int }}{∥\Delta {u}_{m}^{\prime }\left(s\right)∥}^{2}ds+{L}_{5}\left(T\right),\end{array}$
(3.18)
$\left|-{g}^{\prime }\left(0\right)\underset{0}{\overset{t}{\int }}\left(\Delta {u}_{m}\left(s\right),\Delta {u}_{m}^{\prime }\left(s\right)\right)ds\right|\le {L}_{0}\underset{0}{\overset{t}{\int }}{∥\Delta {u}_{m}^{\prime }\left(s\right)∥}^{2}ds+{L}_{6}\left(T\right),$
(3.19)

and

$\left|g\left(0\right)\left(\Delta {u}_{m}\left(t\right),\Delta {u}_{m}^{\prime }\left(t\right)\right)\right|\le \gamma {∥\Delta {u}_{m}^{\prime }\left(t\right)∥}^{2}+{L}_{7}\left(\gamma \right).$
(3.20)

Now we choose γ > 0 small enough and combining (A3), (3.8), (3.13), (3.14), and (3.16)-(3.20), we get

$\frac{1}{2}{∥{u}_{m}^{″}∥}^{2}+\frac{1}{2}{∥\Delta {u}_{m}^{\prime }∥}^{2}\le {L}_{8}\left(\underset{0}{\overset{t}{\int }}{∥{u}_{m}^{″}\left(s\right)∥}^{2}ds+\underset{0}{\overset{t}{\int }}{∥\Delta {u}_{m}^{\prime }\left(s\right)∥}^{2}ds\right)+{L}_{9}.$
(3.21)

By using Gronwall's lemma we arrive at

$\frac{1}{2}{∥{u}_{m}^{″}∥}^{2}+\frac{1}{2}{∥\Delta {u}_{m}^{\prime }∥}^{2}\le {L}_{10},$
(3.22)

for all t ∈ [0, T], and L10 is a positive constant independent of m. Estimate (3.22) implies

$\left\{\begin{array}{c}\hfill \left\{{u}_{m}^{″}\right\}\phantom{\rule{2.77695pt}{0ex}}is\phantom{\rule{2.77695pt}{0ex}}uniformly\phantom{\rule{2.77695pt}{0ex}}bounded\phantom{\rule{2.77695pt}{0ex}}in\phantom{\rule{2.77695pt}{0ex}}{L}^{\infty }\left(0,T;{L}^{2}\left(\Omega \right)\right)\hfill \\ \hfill \left\{{u}_{m}^{\prime }\right\}\phantom{\rule{2.77695pt}{0ex}}is\phantom{\rule{2.77695pt}{0ex}}uniformly\phantom{\rule{2.77695pt}{0ex}}bounded\phantom{\rule{2.77695pt}{0ex}}in\phantom{\rule{2.77695pt}{0ex}}{L}^{\infty }\left(0,T;{H}_{0}^{2}\left(\Omega \right)\right)\hfill \end{array}\right\$
(3.23)

By attention to (3.9) and (3.23), there exists a subsequence {u i } of {u m } and a function u such that

$\left\{\begin{array}{c}\hfill {u}_{i}⇀u\phantom{\rule{2.77695pt}{0ex}}weakly\phantom{\rule{2.77695pt}{0ex}}star\phantom{\rule{2.77695pt}{0ex}}in\phantom{\rule{2.77695pt}{0ex}}{L}^{\infty }\left(0,T;{H}_{0}^{2}\left(\Omega \right)\right)\hfill \\ \hfill {u}_{i}^{\prime }⇀{u}^{\prime }\phantom{\rule{2.77695pt}{0ex}}weakly\phantom{\rule{2.77695pt}{0ex}}star\phantom{\rule{2.77695pt}{0ex}}in\phantom{\rule{2.77695pt}{0ex}}{L}^{\infty }\left(0,T;{H}_{0}^{2}\left(\Omega \right)\right)\hfill \\ \hfill {u}_{i}^{″}⇀{u}^{″}weakly\phantom{\rule{2.77695pt}{0ex}}star\phantom{\rule{2.77695pt}{0ex}}in\phantom{\rule{2.77695pt}{0ex}}{L}^{\infty }\left(0,T;{L}^{2}\left(\Omega \right)\right)\hfill \end{array}\right\$
(3.24)

By Aubin-Lions compactness lemma [18], it follows from (3.24) that

$\left\{\begin{array}{c}\hfill {u}_{i}\to u\phantom{\rule{2.77695pt}{0ex}}strongly\phantom{\rule{2.77695pt}{0ex}}in\phantom{\rule{2.77695pt}{0ex}}C\left(\left[0,T\right];{H}_{0}^{2}\left(\Omega \right)\right)\hfill \\ \hfill {u}_{i}^{\prime }\to u\phantom{\rule{2.77695pt}{0ex}}strongly\phantom{\rule{2.77695pt}{0ex}}in\phantom{\rule{2.77695pt}{0ex}}C\left(\left[0,T\right]\right);{L}^{2}\left(\Omega \right)\right)\hfill \end{array}\right\$
(3.25)

In the sequel we will deal with the nonlinear term. By (3.9) and Sobolev embedding theorem, we obtain

$\left\{{\left|{u}_{m}\right|}^{p}{u}_{m}\right\}\phantom{\rule{2.77695pt}{0ex}}is\phantom{\rule{2.77695pt}{0ex}}uniformly\phantom{\rule{2.77695pt}{0ex}}bounded\phantom{\rule{2.77695pt}{0ex}}in\phantom{\rule{2.77695pt}{0ex}}{L}^{\infty }\left(0,T;{L}^{2}\left(\Omega \right)\right),$
(3.26)

and therefore we can extract a subsequence {u i } of {u m } such that

${\left|{u}_{i}\right|}^{p}{u}_{i}⇀{\left|u\right|}^{p}u\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}weakly\phantom{\rule{2.77695pt}{0ex}}star\phantom{\rule{2.77695pt}{0ex}}in\phantom{\rule{2.77695pt}{0ex}}{L}^{\infty }\left(0,T;{L}^{2}\left(\Omega \right)\right).$
(3.27)

Applying (3.24), (3.27) and letting i → ∞ in (3.3), we see that u satisfies the equation. For the initial conditions by using (3.4), (3.25) and the simple inequality

${∥u-{u}_{0}∥}_{{H}_{0}^{2}\left(\Omega \right)}\le {∥u-{u}_{i}∥}_{{H}_{0}^{2}\left(\Omega \right)}+{∥{u}_{i}-{u}_{i}\left(0\right)∥}_{{H}_{0}^{2}\left(\Omega \right)}+{∥{u}_{i}\left(0\right)-{u}_{0}∥}_{{H}_{0}^{2}\left(\Omega \right)},$

we get the first initial condition immediately. In the similar way, we can show the second initial condition and the proof is complete.

4 Blow-up of solutions

In this section, we study blow-up property of solutions with non-positive initial energy as well as positive initial energy, and estimate the lifespan of solutions. For this purpose, we assume that g is positive and C1 function satisfying

(A 4)

$g\left(0\right)>0,\phantom{\rule{1em}{0ex}}{g}^{\prime }\left(s\right)\le 0,\phantom{\rule{1em}{0ex}}1-\underset{0}{\overset{\infty }{\int }}g\left(s\right)ds=l>0,$

and we make the following extra assumption on g

(A 5)

$\underset{0}{\overset{\infty }{\int }}g\left(s\right)ds<\frac{p}{1+p}.$

From (2.1), (A4) and Lemma 1, we have

$\begin{array}{c}E\left(t\right)\ge \frac{1}{2}\left[\left(1-\underset{0}{\overset{t}{\int }}g\left(s\right)ds\right){∥\Delta u∥}^{2}+\left(g\odot \Delta u\right)\left(t\right)\right]-\frac{1}{p+2}{∥u∥}_{p+2}^{p+2}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\ge \frac{1}{2}\left[l{∥\Delta u∥}^{2}+g\odot \Delta u\right)\left(t\right)\right]-\frac{{C}_{1}^{p+2}{l}^{\frac{p+2}{2}}}{p+2}{∥\Delta u∥}^{p+2}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\ge G\left(\sqrt{l{∥\Delta u∥}^{2}+\left(g\odot \Delta u\right)\left(t\right)}\right),\phantom{\rule{1em}{0ex}}t\ge 0,\end{array}$

where $G\left(\lambda \right)=\frac{1}{2}{\lambda }^{2}-\frac{{C}_{1}^{p+2}}{p+2}{\lambda }^{p+2},\phantom{\rule{1em}{0ex}}{C}_{1}=\frac{{C}_{*}}{\sqrt{l}}$. It is easy to verify that G(λ) has a maximum at ${\lambda }_{1}={C}_{1}^{-\frac{p+2}{p}}$ and the maximum value is ${E}_{1}=\frac{p}{2p+4}{C}_{1}^{-\frac{2p+4}{p}}$.

Lemma 4 Let (A4) hold andu be a local solution of (1.1). Then E(t) is a non-increasing function on [0, T] and

$\frac{d}{dt}E\left(t\right)=\frac{1}{2}\left({g}^{\prime }\odot \Delta u\right)\left(t\right)-\frac{1}{2}g\left(t\right){∥\Delta u∥}^{2}\le 0,$
(4.2)

for almost every t ∈ [0, T].

Proof Multiplying (1.1) by u t , integrating over Ω, and finally integrating by parts, we obtain (4.2) for any regular solution. Then by density arguments, we have the result.

Lemma 5 Let (A4) hold and u be a local solution of (1.1) with initial data satisfying E(0) < E1 and ${l}^{\frac{1}{2}}∥\Delta {u}_{0}∥>{\lambda }_{1}$. Then there exists λ2 > λ1 such that

$l{∥\Delta u∥}^{2}+\left(g\odot \Delta u\right)\left(t\right)\ge {\lambda }_{2}^{2}.$
(4.3)

Proof See Li and Tsai [11].

The choice of the functional is standard (see [19])

$\psi \left(t\right)={∥u∥}^{2}.$
(4.4)

It is clear that

${\psi }^{\prime }\left(t\right)=2\left(u,{u}_{t}\right),$
(4.5)

and from (1.1)

${\psi }^{″}\left(t\right)=2{∥{u}_{t}∥}^{2}-2{∥\Delta u∥}^{2}+2{∥u∥}_{p+2}^{p+2}+2\underset{0}{\overset{t}{\int }}g\left(t-s\right)\left(\Delta u\left(t\right),\Delta u\left(s\right)\right)ds.$
(4.6)

Lemma 6 Let u be a solution of (1.1) and (A4), (A5) hold, then we have

${\psi }^{″}\left(t\right)-\left(4+p\right)\underset{\Omega }{\int }{u}_{t}^{2}dx\ge m\left(l{∥\Delta u∥}^{2}+\left(g\odot \Delta u\right)\left(t\right)\right)-\left(4+2p\right)E\left(0\right),$
(4.7)

where $m=\left(1+p\right)-\frac{1}{l}>0$.

Proof Using the Hölder and Young's inequalities, we arrive at

$\begin{array}{c}\underset{0}{\overset{t}{\int }}g\left(t-s\right)\left(\Delta u\left(t\right),\Delta u\left(s\right)\right)ds\ge -\left[\frac{1}{2}\left(g\odot \Delta u\right)\left(t\right)+\frac{1}{2}\underset{0}{\overset{t}{\int }}g\left(s\right)ds{∥\Delta u∥}^{2}\right]\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}+\underset{0}{\overset{t}{\int }}g\left(s\right)ds{∥\Delta u∥}^{2},\end{array}$

therefore (4.6) becomes

$\begin{array}{c}{\psi }^{″}\left(t\right)-\left(4+p\right){∥{u}_{t}∥}^{2}\ge -\left(2+p\right){∥{u}_{t}∥}^{2}-2{∥\Delta u∥}^{2}-\left(g\odot \Delta u\right)\left(t\right)\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}+\underset{0}{\overset{t}{\int }}g\left(s\right)ds{∥\Delta u∥}^{2}+2{∥u∥}_{p+2}^{p+2}.\end{array}$

Then, using (4.2), we obtain

$\begin{array}{c}{\psi }^{″}\left(t\right)-\left(4+p\right){∥{u}_{t}∥}^{2}\ge -\left(4+2p\right)E\left(0\right)+p{∥\Delta u∥}^{2}+\left(1+p\right)\left(g\odot \Delta u\right)\left(t\right)\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}-\left(1+p\right)\underset{0}{\overset{t}{\int }}g\left(s\right)ds{∥\Delta u∥}^{2}-\left(2+p\right)\underset{0}{\overset{t}{\int }}\left({g}^{\prime }\odot \Delta u\right)\left(s\right)ds,\end{array}$

and so by (2.5) and (A5), we deduce

$\begin{array}{c}{\psi }^{″}\left(t\right)-\left(4+p\right){∥{u}_{t}∥}^{2}\ge -\left(4+2p\right)E\left(0\right)+\left(p-\left(1+p\right)\left(1-l\right)\right){∥\Delta u∥}^{2}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}+\left(1+p\right)\left(g\odot \Delta u\right)\left(t\right),\end{array}$
(4.8)

if we set $m:=\left(1+p\right)-\frac{1}{l}$ then inequality (4.8) yields the desired result.

Consequently, we have the following result.

Lemma 7 Assume that (A4) and (A5) hold. u be a local solution of (1.1) and that either one of the following four conditions is satisfied:

(i) E(0) < 0

(ii) E(0) = 0 and ψ'(0) > 0

(iii) $0 and ${l}^{\frac{1}{2}}∥\Delta {u}_{0}∥>{\lambda }_{1}$

(iv) $\frac{m}{p}{E}_{1}\le E\left(0\right)$ and ${\psi }^{\prime }\left(0\right)>{r}_{2}\left[\psi \left(0\right)+\frac{\left(4+2p\right)E\left(0\right)}{4+p}\right]$.

Then ψ' (t) > 0 for t > t*, where

in case (i)

${t}^{*}=\mathsf{\text{max}}\left\{0,\frac{{\psi }^{\prime }\left(0\right)}{\left(4+2p\right)E\left(0\right)}\right\},$
(4.9)

in cases (ii), (iv)

${t}^{*}=0,$
(4.10)

and in case (iii)

${t}^{*}=\mathsf{\text{max}}\left\{0,\frac{-{\psi }^{\prime }\left(0\right)}{\left(4+2p\right)\left(\frac{m}{p}{E}_{1}-E\left(0\right)\right)}\right\}.$
(4.11)

Proof Suppose that condition (i) is satisfied. Then from (4.5), we have

${\psi }^{\prime }\left(t\right)\ge {\psi }^{\prime }\left(0\right)-\left(4+2p\right)E\left(0\right)t,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}t\ge 0.$

Thus ψ'(t) > 0 for t > t*, and it is easy to see that t* satisfies (4.9).

If E(0) = 0, then by using (4.3) we have ψ" (t) ≥ 0, and since ψ'(0) > 0 we arrive at

${\psi }^{\prime }\left(t\right)>0,\phantom{\rule{1em}{0ex}}for\phantom{\rule{1em}{0ex}}t>0.$

If $0 and ${l}^{\frac{1}{2}}∥\Delta {u}_{0}∥>{\lambda }_{1}$ then by Lemma 4, we see that

$\begin{array}{c}m\left(l{∥\Delta u∥}^{2}+\left(g\odot \Delta u\right)\left(t\right)\right)-\left(4+2p\right)E\left(0\right)\ge m{\lambda }_{2}^{2}-\left(4+2p\right)E\left(0\right)\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}>m\frac{4+2p}{p}{E}_{1}-\left(4+2p\right)E\left(0\right)\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}=\left(4+2p\right)\left[\frac{m}{p}{E}_{1}-E\left(0\right)\right].\end{array}$

Thus from (4.5), we have

$\begin{array}{c}{\psi }^{″}\left(t\right)\ge m\left(l{∥\Delta u∥}^{2}+\left(g\odot \Delta u\right)\left(t\right)\right)-\left(4+2p\right)E\left(0\right)\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}>\left(4+2p\right)\left[\frac{m}{p}{E}_{1}-E\left(0\right)\right]>0,\end{array}$
(4.12)

and integrating (4.12) from 0 to t gives

${\psi }^{\prime }\left(t\right)>0,\phantom{\rule{1em}{0ex}}for\phantom{\rule{1em}{0ex}}t\ge {t}^{*},$

where t* satisfies (4.11).

Let $\frac{m}{p}{E}_{1}\le E\left(0\right)$, this assumption causes that

${\psi }^{″}\left(t\right)-\left(4+p\right){∥{u}_{t}∥}^{2}+\left(4+2p\right)E\left(0\right)\ge 0,$

and by using Hölder and Young's inequalities, we get

${∥{u}_{t}∥}^{2}\ge {\psi }^{\prime }\left(t\right)-\psi \left(t\right),$

thus

${\psi }^{″}\left(t\right)-\left(4+p\right){\psi }^{\prime }\left(t\right)+\left(4+p\right)\psi \left(t\right)+\left(4+2p\right)E\left(0\right)\ge 0.$
(4.13)

We see that the hypotheses of Lemma 2 are fulfilled with

$\delta =\frac{p}{4}\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}B\left(t\right)=\psi \left(t\right)+\frac{\left(4+2p\right)E\left(0\right)}{4+p}$

and the conclusion of Lemma 2.2 gives us

${\psi }^{\prime }\left(t\right)>0,\phantom{\rule{1em}{0ex}}for\phantom{\rule{1em}{0ex}}t>0.$

Therefore the proof is complete.

To estimate the life-span of ψ(t), we define the following functional

$Y\left(t\right)=\psi {\left(t\right)}^{-\frac{p}{4}},\phantom{\rule{1em}{0ex}}for\phantom{\rule{1em}{0ex}}t\ge 0.$
(4.14)

Then we have

${Y}^{\prime }\left(t\right)=\frac{p}{4}Y{\left(t\right)}^{1+\frac{4}{p}}{\psi }^{\prime }\left(t\right),$
(4.15)
${Y}^{″}\left(t\right)=-\frac{p}{4}Y{\left(t\right)}^{1+\frac{8}{p}}\left[{\psi }^{″}\left(t\right)\psi \left(t\right)-\left(1+\frac{p}{4}\right){\left({\psi }^{\prime }\left(t\right)\right)}^{2}\right].$
(4.16)

Using (4.4)-(4.6) and exploiting Holder's inequality on ψ'(t), we get

$\begin{array}{c}{\psi }^{″}\left(t\right)\psi \left(t\right)-\left(1+\frac{p}{4}\right){\left({\psi }^{\prime }\left(t\right)\right)}^{2}\\ \phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\ge \left[\left(l{∥\Delta u∥}^{2}+\left(g\odot \Delta u\right)\left(t\right)\right)-\left(4+2p\right)E\left(0\right)+\left(4+p\right){∥{u}_{t}∥}^{2}\right]\psi \left(t\right)\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}-4\left(1+\frac{p}{4}\right){∥{u}_{t}∥}^{2}\psi \left(t\right)\\ \phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}=\left[\left(l{∥\Delta u∥}^{2}+\left(g\odot \Delta u\right)\left(t\right)\right)-\left(4+2p\right)E\left(0\right)\right]Y{\left(t\right)}^{\frac{-4}{p}}.\end{array}$

Utilizing the last inequality into (4.16) yields

${Y}^{″}\left(t\right)\le -\frac{p}{4}\left[\left(l{∥\Delta u∥}^{2}+\left(g\odot \Delta u\right)\left(t\right)\right)-\left(4+2p\right)E\left(0\right)\right]Y{\left(t\right)}^{1+\frac{4}{p}}.$
(4.17)

Now we should assume different values for initial energy E(0).

1. (1)

At first if E(0) ≤ 0 then from (4.17) we have

${Y}^{″}\left(t\right)\le \frac{p}{4}\left(4+2p\right)E\left(0\right)Y{\left(t\right)}^{1+\frac{4}{p}},$
(4.18)

on the other hand by Lemma 7, Y'(t) < 0 for t > t*. Multiplying (4.18) by Y'(t) and integrating from t* to t, we deduce that

${Y}^{\prime }{\left(t\right)}^{2}\ge \alpha +\beta Y{\left(t\right)}^{2+\frac{4}{p}}\phantom{\rule{1em}{0ex}}for\phantom{\rule{1em}{0ex}}t\ge {t}^{*},$

where

$\alpha =\frac{{p}^{2}}{16}Y{\left({t}^{*}\right)}^{2+\frac{8}{p}}\left[{\psi }^{\prime }{\left({t}^{*}\right)}^{2}-8E\left(0\right)Y{\left({t}^{*}\right)}^{-\frac{4}{p}}\right]>0,$
(4.19)

and

$\beta =\frac{{p}^{2}}{2}E\left(0\right).$
(4.20)

Then the hypotheses of Lemma 3 are fulfilled with $\delta =\frac{p}{4},{t}_{0}={t}^{*}$ and using the conclusion of Lemma 3, there exists a finite time T* such that ${\mathsf{\text{lim}}}_{t\to {T}^{*-}}Y\left(t\right)=0$, i.e., in this case some solutions blow up in finite time T*.

1. (2)

If $0, then from (4.17) and (4.12) we have

${Y}^{″}\left(t\right)\le -\frac{p}{4}\left(4+2p\right)\left[\frac{m}{p}{E}_{1}-E\left(0\right)\right]Y{\left(t\right)}^{1+\frac{4}{p}}.$

Then using the same arguments as in (1), we get

${Y}^{\prime }{\left(t\right)}^{2}\ge {\alpha }_{1}+{\beta }_{1}Y{\left(t\right)}^{2+\frac{4}{p}}\phantom{\rule{1em}{0ex}}for\phantom{\rule{1em}{0ex}}t\ge {t}^{*},$

where

${\alpha }_{1}=\frac{{p}^{2}}{16}Y{\left({t}^{*}\right)}^{2+\frac{8}{p}}\left({\psi }^{\prime }{\left({t}^{*}\right)}^{2}+8\left[\frac{m}{p}{E}_{1}-E\left(0\right)\right]Y{\left({t}^{*}\right)}^{-\frac{4}{p}}\right)>0,$
(4.21)

and

${\beta }_{1}=\frac{{p}^{2}}{2}\left[E\left(0\right)-\frac{m}{p}{E}_{1}\right].$
(4.22)

Thus by Lemma 3, there exists a finite time T* such that

$\underset{t\to T*-}{\mathsf{\text{lim}}}\psi \left(t\right)=\infty .$
1. (3)

$\frac{m}{p}{E}_{1}\le E\left(0\right)$. In this case, it is easy to see that by using (4.19) and (4.20) into discussion in part (1), we obtain

$\alpha >0\phantom{\rule{1em}{0ex}}if\phantom{\rule{2.77695pt}{0ex}}and\phantom{\rule{2.77695pt}{0ex}}only\phantom{\rule{2.77695pt}{0ex}}if\phantom{\rule{1em}{0ex}}E\left(0\right)<\frac{{\psi }^{\prime }{\left({t}^{*}\right)}^{2}}{8\psi \left({t}^{*}\right)}.$

Hence, Lemma 3 yields the blow-up property in this case.

Therefore, we proved the following theorem.

Theorem 2 Assume that (A4) and (A5) hold. u be a local solution of (1.1) and that either one of the following four conditions is satisfied:

(i) E(0) < 0

(ii) E(0) = 0 and ψ'(0) > 0

(iii) $0 and ${l}^{\frac{1}{2}}∥\Delta {u}_{0}∥>{\lambda }_{1}$

(iv) $\frac{m}{p}{E}_{1}\le E\left(0\right)$ and ${\psi }^{\prime }\left(0\right)>{r}_{2}\left[\psi \left(0\right)+\frac{\left(4+2p\right)E\left(0\right)}{4+p}\right]$ holds.

Then the solution u blows up at finite time T*. Moreover, the upper bounds for T* can be estimated according to the sign of E(0):

in case (i)

${T}^{*}\le {t}^{*}-\frac{Y\left({t}^{*}\right)}{{Y}^{\prime }\left({t}^{*}\right)}.$

Furthermore, if $Y\left({t}^{*}\right)<\mathsf{\text{min}}\left\{1,\sqrt{\frac{\alpha }{-\beta }}\right\}$, then

${T}^{*}\le {t}^{*}+\frac{1}{\sqrt{-\beta }}ln\frac{\sqrt{\frac{\alpha }{-\beta }}}{\sqrt{\frac{\alpha }{-\beta }}-Y\left({t}^{*}\right)}$

in cases (ii)

${T}^{*}\le {t}^{*}-\frac{Y\left({t}^{*}\right)}{{Y}^{\prime }\left({t}^{*}\right)}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}or\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}{T}^{*}\le {t}^{*}+\frac{Y\left({t}^{*}\right)}{\sqrt{\alpha }}$

in case (iii)

${T}^{*}\le {t}^{*}-\frac{Y\left({t}^{*}\right)}{{Y}^{\prime }\left({t}^{*}\right)}.$

Furthermore, if $Y\left({t}^{*}\right)<\mathsf{\text{min}}\left\{1,\sqrt{\frac{\alpha }{-\beta }}\right\}$, then

${T}^{*}\le {t}^{*}+\frac{1}{\sqrt{-{\beta }_{1}}}ln\frac{\sqrt{\frac{{\alpha }_{1}}{-{\beta }_{1}}}}{\sqrt{\frac{{\alpha }_{1}}{-{\beta }_{1}}}-Y\left({t}^{*}\right)}$

and in case (iv)

${T}^{*}\le \frac{Y\left({t}^{*}\right)}{\sqrt{\alpha }}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}or\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}{T}^{*}\le {t}^{*}+{2}^{\frac{3p+4}{2p}}\frac{pc}{4\sqrt{\alpha }}\left[1-{\left(1+cY\left({t}^{*}\right)\right)}^{\frac{-2}{p}}\right],$

where $d={\left(\frac{\beta }{\alpha }\right)}^{\frac{p}{p+8}}$. Here α, β, α1, and β1 are given in (4.19)-(4.22), respectively. Note that each t* in the above cases satisfy the same case in Lemma 7.