General decay and blow-up of solutions for a nonlinear viscoelastic wave equation with strong damping

Abstract

This article is concerned with the decay and blow-up properties of a nonlinear viscoelastic wave equation with strong damping. We first show a local existence theorem. Then, we prove the global existence of solutions and establish a general decay rate estimate. Finally, we show the finite time blow-up result for some solutions with negative initial energy and positive initial energy.

1 Introduction

In this work we investigate the decay and blow-up properties of the nonlinear viscoelastic wave equation of the form:

$$\begin{aligned} \textstyle\begin{cases} u_{tt}-\Delta u+\int_{0}^{t}g(t-\tau)\Delta u(\tau)\,d\tau-\Delta u_{t}+u_{t}=u \vert u \vert ^{p-2},&\mbox{in }\Omega\times(0,\infty),\\ u(x,t)=0, &\mbox{on } \partial\Omega\times(0,\infty),\\ u(x,0)=u_{0}(x),\qquad u_{t}(x,0)=u_{1}(x),&\mbox{in } \Omega, \end{cases}\displaystyle \end{aligned}$$

(1.1)

where $\Omega\subset\mathbb{R}^{n}$ is bounded domains with smooth boundary ∂Ω. Problems of this type have been investigated by many authors, and some results in connection with existence and nonexistence have been established. For example, Berrimi and Messaoudi [1] studied the following viscoelastic equation:

$$\begin{aligned} \textstyle\begin{cases} u_{tt}-\Delta u+\int^{t}_{0}g(t-\tau)\Delta u(\tau)\,d\tau=u \vert u \vert ^{\gamma },&\mbox{in }\Omega\times(0,\infty),\\ u(x,t)=0, &\mbox{on } \partial\Omega\times(0,\infty),\\ u(x,0)=u_{0}(x),\qquad u_{t}(x,0)=u_{1}(x),&\mbox{in } \Omega, \end{cases}\displaystyle \end{aligned}$$

where $\gamma>0$. The authors established the local existence and global existence theorems and showed that the solution energy exponentially or polynomially decays. Later, Messaoudi [17] improved the results of [1], he established a general decay result. Inspired by the ideas of Messaoudi [17] and [18], Han and Wang [10] investigated a nonlinear viscoelastic equation with the dispersive term $\Delta u_{tt}$ by modifying the perturbed energy functional; they also obtained that the solution energy is general decay. Recently, Guesmia et al. [8] combined the techniques given in [17] with the character of Kirchhoff equation and obtained the optimal decay rate estimate of solution energy. For the case of wave equation with nonlinear boundary damping and source terms, Vitillaro [21] established the local and global existence of solutions under reasonable conditions on the initial data. In [3], Cavalcanti et al. obtained both well-posedness and the optimal decay rate estimate for solutions.

In article [6], Gazzola and Squassina discussed the following viscoelastic equation with strong damping term $\Delta u_{t}$:

$$\begin{aligned} \textstyle\begin{cases} u_{tt}-\Delta u-\omega\Delta u_{t}+\mu u_{t}=u \vert u \vert ^{p-2},&\mbox{in } \Omega\times(0,\infty),\\ u(x,t)=0, &\mbox{on } \partial\Omega\times(0,\infty),\\ u(x,0)=u_{0}(x),\qquad u_{t}(x,0)=u_{1}(x),&\mbox{in } \Omega, \end{cases}\displaystyle \end{aligned}$$

(1.2)

where $p>2$, $\omega,\mu>0$. The authors established the global existence theorem and proved that the global solution is uniformly bounded. They also constructed the finite time blow-up of solutions for low initial energy or arbitrarily high initial energy. When the linear damping term is replaced by nonlinear damping term in equation (1.2), Chen and Liu [5] obtained a global existence theorem, uniform decay rate estimate, and exponential growth for the solutions.

In paper [2], Cavalcanti et al. dealt with the following problem:

$$\begin{aligned} \textstyle\begin{cases} \vert u_{t} \vert ^{\rho}u_{tt}-\Delta u-\Delta u_{tt}+\int^{t}_{0}g(t-\tau)\Delta u(\tau)\,d\tau-\gamma\Delta u_{t}=0,&\mbox{in } \Omega\times(0,\infty),\\ u(x,t)=0, &\mbox{on }\partial\Omega\times(0,\infty),\\ u(x,0)=u_{0}(x),\qquad u_{t}(x,0)=u_{1}(x),&\mbox{in }\Omega, \end{cases}\displaystyle \end{aligned}$$

(1.3)

under reasonable conditions on g and γ, the authors established the global existence result for $\gamma\geq0$ and the exponential decay result for $\gamma>0$. Cavalcanti et al. [4] discussed equation (1.3) for $\rho\geq0$, $\gamma\geq0$ and obtained that the energy decays to zero with the decay rate which is dominated by the solutions of the ODE quantifying the conduct of $g(t)$.

For the finite time blow-up, Messaoudi [15] studied the following problem:

$$\begin{aligned} \textstyle\begin{cases} u_{tt}-\Delta u+\int_{0}^{t}g(t-\tau)\Delta u(\tau)\,d\tau +au_{t} \vert u_{t} \vert ^{m-2}=bu \vert u \vert ^{p-2},&\mbox{in } \Omega\times(0,\infty),\\ u(x,t)=0, &\mbox{on } \partial\Omega\times(0,\infty),\\ u(x,0)=u_{0}(x),\qquad u_{t}(x,0)=u_{1}(x),&\mbox{in } \Omega. \end{cases}\displaystyle \end{aligned}$$

(1.4)

He showed that the solution blows up in finite time when the initial energy is negative and $p>m$ and the solution exists globally for $m\geq p$. In [16], Messaoudi extended the blow-up result to certain situations in which the initial energy is positive. Later, Song [19] proved the finite time blow-up of solutions whose initial data have arbitrarily high initial energy. It is worth mentioning some other literatures concerning existence and nonexistence of wave equation, namely [7, 9, 11, 13, 14, 20] and the references therein.

At the present time, less results are investigated for the wave equation with strong damping term and many problems are unsolved (see [6]). So, in this paper, we study a nonlinear viscoelastic wave equation with strong damping. We first show a local existence theorem. Then, we prove the global existence of solutions and establish a general decay rate estimate. Finally, we show the finite time blow-up result for some solutions with negative initial energy and positive initial energy.

This article is organized as follows. In Sect. 2, we give some preliminaries. In Sect. 3, we prove the local existence and uniqueness of solutions for problem (1.1). Then, the general decay of the solutions is considered in Sect. 4. In the last section, we discuss the blow-up phenomenon for the equation.

2 Preliminaries

We begin with some materials needed in the proof of the main results. We first recall the following assumptions as in [17]:

(H1):

$g:R_{+}\rightarrow R_{+}$ is a nonincreasing and bounded $C^{1}$ function satisfying

$$g(0)>0, \quad 1- \int^{\infty}_{0}g(\tau)\,d\tau=l>0. $$

(H2):

There exists a positive differentiable function $\xi(t)$ such that

$$g'(t)\leq-\xi(t)g(t),\quad t\geq0, $$

and

$$\biggl\vert \frac{\xi'(t)}{\xi(t)} \biggr\vert \leq k,\quad\xi(t)>0, \xi'(t)\leq 0,\quad\forall t>0,\quad \int_{0}^{+\infty}\xi(t)\,dt=+\infty. $$

(H3):

For the nonlinear term, we let

$$2< p\leq\frac{2n}{n-2},\quad\mbox{if } n\geq3; \qquad 2< p< \infty ,\quad \mbox{if } n=1,2. $$

Remark 2.1

Since ξ is a nonincreasing function, then $\xi(t)\leq\xi(0)=M$.

We will use the embedding $H^{1}_{0}(\Omega)\hookrightarrow L^{s}(\Omega )$ for $2\leq s\leq2n/(n-2)$ if $n\geq3$ or $s\geq2$ if $n=1,2$; and $L^{r}(\Omega)\hookrightarrow L^{s}(\Omega)$ for $s < r$, and we will use the same embedding constant denoted by $C_{*}$ such that

$$\begin{aligned} \Vert u \Vert _{s}\leq C_{*} \Vert \nabla u \Vert _{2},\qquad \Vert u \Vert _{s}\leq C_{*} \Vert u \Vert _{r}. \end{aligned}$$

(2.1)

For our aim, we use the following functionals:

$$\begin{aligned} &I(t)=I\bigl(u(t)\bigr)=\biggl(1- \int^{t}_{0}g(\tau)\,d\tau\biggr) \bigl\Vert \nabla u(t) \bigr\Vert ^{2}_{2}+(g\circ\nabla u) (t)- \bigl\Vert u(t) \bigr\Vert ^{p}_{p}, \\ &J(t)=J\bigl(u(t)\bigr)=\frac{1}{2}\biggl(1- \int^{t}_{0}g(\tau)\,d\tau\biggr) \bigl\Vert \nabla u(t) \bigr\Vert ^{2}_{2}+\frac{1}{2}(g\circ \nabla u) (t)-\frac{1}{p} \bigl\Vert u(t) \bigr\Vert ^{p}_{p}, \\ &E(t)=E\bigl(u(t),u_{t}(t)\bigr)=J(t)+\frac{1}{2} \Vert u_{t} \Vert ^{2}_{2}, \end{aligned}$$

(2.2)

where

$$(g\circ v) (t)= \int^{t}_{0}g(t-\tau) \bigl\Vert v(t)-v(\tau) \bigr\Vert ^{2}_{2}\,d\tau. $$

Lemma 2.2

If (H1), (H2), (H3) hold and $(u_{0},u_{1})\in H^{1}_{0}(\Omega )\times L^{2}(\Omega)$, u is the solution of (1.1), then the energy functional $E(t)$ satisfies

$$\begin{aligned} E'(t)=\frac{1}{2}\bigl(g'\circ\nabla u\bigr) (t)-\frac{1}{2}g(t) \bigl\Vert \nabla u(t) \bigr\Vert ^{2}_{2}- \Vert \nabla u_{t} \Vert ^{2}_{2}- \Vert u_{t} \Vert ^{2}_{2}\leq0 \end{aligned}$$

(2.3)

for $\forall t\in[0,T]$.

Proof

Multiplying (1.1) by $u_{t}$ and integrating over Ω, we obtain

$$\begin{aligned} &\frac{d}{dt} \biggl\{ \frac{1}{2} \int_{\Omega} \vert u_{t} \vert ^{2}\,dx+ \frac {1}{2} \int_{\Omega} \vert \nabla u \vert ^{2}\,dx- \frac{1}{p} \int_{\Omega } \vert u \vert ^{p}\,dx \biggr\} \\ &\quad {}- \int^{t}_{0}g(t-\tau) \int_{\Omega}\nabla u_{t}(t)\cdot\nabla u(\tau )\,dx\,d\tau=- \int_{\Omega} \vert \nabla u_{t} \vert ^{2}\,dx- \int_{\Omega } \vert u_{t} \vert ^{2}\,dx. \end{aligned}$$

(2.4)

For the last term on the left-hand side of (2.4), we get

$$\begin{aligned} & \int^{t}_{0}g(t-\tau) \int_{\Omega}\nabla u_{t}(t)\cdot\nabla u(\tau )\,dx\,d\tau \\ &\quad = \int^{t}_{0}g(t-\tau) \int_{\Omega}\nabla u_{t}(t)\cdot \bigl[\nabla u(\tau)- \nabla u(t) \bigr]\,dx\,d\tau \\ &\qquad {}+ \int^{t}_{0}g(t-\tau) \int_{\Omega }\nabla u_{t}(t)\cdot\nabla u(t)\,dx\,d\tau \\ &\quad = -\frac{1}{2} \int^{t}_{0}g(t-\tau) \biggl(\frac{d}{dt} \int_{\Omega } \bigl\vert \nabla u(\tau)-\nabla u(t) \bigr\vert ^{2}\,dx \biggr)\,d\tau \\ &\qquad {}+ \int^{t}_{0}g(\tau) \biggl(\frac{d}{dt} \frac{1}{2} \int_{\Omega} \bigl\vert \nabla u(t) \bigr\vert ^{2}\,dx \biggr)\,d\tau \\ &\quad =-\frac{1}{2}\,\frac{d}{dt} \biggl[ \int^{t}_{0}g(t-\tau) \int_{\Omega } \bigl\vert \nabla u(\tau)-\nabla u(t) \bigr\vert ^{2}\,dx\,d\tau \biggr] \\ &\qquad {}+\frac{1}{2}\frac {d}{dt} \biggl[ \int^{t}_{0}g(\tau) \int_{\Omega} \bigl\vert \nabla u(t) \bigr\vert ^{2}\,dx\,d\tau \biggr] \\ &\qquad{}+\frac{1}{2} \int_{0}^{t}g'(t-\tau) \int_{\Omega} \bigl\vert \nabla u(\tau )-\nabla u(t) \bigr\vert ^{2}\,dx\,d\tau \\ &\qquad {}-\frac{1}{2}g(t) \int_{\Omega} \bigl\vert \nabla u(t) \bigr\vert ^{2}\,dx. \end{aligned}$$

(2.5)

Inserting (2.5) into (2.4), we obtain

$$\begin{array}{rll}& \frac{d}{dt}\{\frac{1}{2}{\int }_{\mathrm{\Omega }}|u_t{|}^{2}dx+\frac{1}{2}{\int }_{\mathrm{\Omega }}|\mathrm{\nabla }u{|}^{2}dx-\frac{1}{p}{\int }_{\mathrm{\Omega }}|u{|}^{p}dx\}& \\ & -\frac{1}{2}\frac{d}{dt}[{\int }_0^{t}g(\tau )d\tau {\parallel \mathrm{\nabla }u(t)\parallel }_2^2]+\frac{1}{2}\frac{d}{dt}[{\int }_0^{t}g(t-\tau ){\int }_{\mathrm{\Omega }}|\mathrm{\nabla }u(\tau )-\mathrm{\nabla }u(t){|}^{2}dxd\tau ]& \\ & =-{\int }_{\mathrm{\Omega }}|\mathrm{\nabla }{u}_t{|}^{2}dx-{\int }_{\mathrm{\Omega }}|u_t{|}^{2}dx+\frac{1}{2}(g^{\prime }\circ \mathrm{\nabla }u)(t)-\frac{1}{2}g(t){\parallel \mathrm{\nabla }u(t)\parallel }_2^2\le 0.& \end{array}$$

(2.6)

 □

Lemma 2.3

If (H3) holds, then there exists a positive constant $C>1$ such that

$$\begin{aligned} \Vert u \Vert ^{s}_{p}\leq C\bigl( \Vert \nabla u \Vert ^{2}_{2}+ \Vert u \Vert ^{p}_{p} \bigr), \quad2\leq s\leq p, \end{aligned}$$

(2.7)

for any u being a solution of (1.1) on $[0,T]$.

Proof

If $\|u\|_{p}\leq1$, then $\|u\|^{s}_{p}\leq\|u\| ^{2}_{p}\leq C\|\nabla u\|^{2}_{2}$ by using Sobolev embedding theorems. If $\|u\|_{p}>1$, then $\|u\|^{s}_{p}\leq\|u\|^{p}_{p}$. Therefore (2.7) follows. □

We set

$$H(t):=-E(t), $$

and use C to denote a general positive constant depending on Ω only. As a result of (2.2) and (2.7), we have the following.

Corollary 2.4

Let the assumption of the above lemma hold. Then we have the following:

$$\begin{aligned} \Vert u \Vert ^{s}_{p}\leq C \bigl(H(t)+ \Vert u_{t} \Vert ^{2}_{2}+ \Vert \nabla u \Vert ^{2}_{2}+(g\circ\nabla u) (t) \bigr),\quad\forall t \in[0,T], \end{aligned}$$

(2.8)

for any $u\in H^{1}_{0}(\Omega)$ and $2\leq s\leq p$.

Proof

Using (H1) and (2.2) leads to

$$\begin{aligned} \frac{1}{p} \bigl\Vert u(t) \bigr\Vert ^{p}_{p} &\leq-E(t)+\frac{1}{2} \Vert u_{t} \Vert ^{2}_{2}+ \frac{1}{2}\biggl(1- \int_{0}^{\infty}g(\tau)\,d\tau\biggr) \Vert \nabla u \Vert ^{2}_{2}+\frac{1}{2}(g\circ\nabla u) (t) \\ & \leq H(t)+\frac{1}{2} \Vert u_{t} \Vert ^{2}_{2}+\frac{1}{2}\biggl(1- \int_{0}^{\infty }g(\tau)\,d\tau\biggr) \Vert \nabla u \Vert ^{2}_{2}+\frac{1}{2}(g\circ\nabla u) (t). \end{aligned}$$

(2.9)

Finally, a combination of (2.7) and (2.9) gives the needed result. □

3 Local existence

In this section, the aim is to establish the local existence result for (1.1). For this goal, we first discuss a related linear problem. Then, by using the contraction mapping theorem, we obtain the existence of solutions to the nonlinear problem. For v given, the related linear problem is of the form:

$$ \textstyle\begin{cases} u_{tt}-\Delta u+\int_{0}^{t}g(t-\tau)\Delta u(\tau)\,d\tau-\Delta u_{t}+u_{t}=v \vert v \vert ^{p-2},&\mbox{in }\Omega\times(0,\infty),\\ u(x,t)=0, &\mbox{on }\partial\Omega\times(0,\infty),\\ u(x,0)=u_{0}(x),\qquad u_{t}(x,0)=u_{1}(x),&\mbox{in }\Omega. \end{cases} $$

(3.1)

Similar to the proof in [12], we can get the following lemma.

Lemma 3.1

Suppose $v\in C ([0,T]; C^{\infty}_{0}(\Omega) )$ and $u_{0}, u_{1}\in C^{\infty}_{0}(\Omega)$, then problem (3.1) has a unique solution u satisfying

$$\begin{aligned} \begin{aligned} &u\in L^{\infty} \bigl((0,T); H^{1}_{0}(\Omega)\cap H^{2}(\Omega) \bigr),\\ & u_{t}\in L^{\infty} \bigl((0,T); H^{1}_{0}(\Omega) \bigr),\\ &u_{tt}\in L^{\infty} \bigl((0,T); L^{2}(\Omega) \bigr). \end{aligned} \end{aligned}$$

(3.2)

Next, we prove the existence of solutions to equation (3.1) when the initial data is less regular.

Lemma 3.2

If (H3) holds, then given any $(u_{0},u_{1})\in H^{1}_{0}(\Omega )\times L^{2}(\Omega)$ and $v\in C ([0,T]; H_{0}^{1}(\Omega) )$, problem (3.1) has a unique weak solution

$$ u\in C \bigl([0,T]; H^{1}_{0}(\Omega) \bigr),\qquad u_{t}\in C \bigl([0,T]; L^{2}(\Omega) \bigr). $$

(3.3)

Proof

We approximate $u_{0}$, $u_{1}$ by sequences $(u_{0n})$, $(u_{1n})$ in $C^{\infty}_{0}(\Omega)$, and v by a sequence $(v_{n})$ in $C ([0,T]; C^{\infty}_{0}(\Omega) )$. Then from Lemma 3.1 we can obtain a solution $(u_{n})$ satisfying:

$$\begin{aligned} \textstyle\begin{cases} u_{ntt}-\Delta u_{n}+\int_{0}^{t}g(t-\tau)\Delta u_{n}(\tau)\,d\tau -\Delta u_{nt}+u_{nt}=v_{n} \vert v_{n} \vert ^{p-2},&\mbox{in }\Omega\times (0,\infty),\\ u_{n}(x,t)=0,&\mbox{on }\partial\Omega\times(0,\infty), \\ u_{n}(x,0)=u_{0n}(x),\qquad u_{nt}(x,0)=u_{1n}(x),&\mbox{in }\Omega, \end{cases}\displaystyle \end{aligned}$$

(3.4)

and satisfying (3.2). Now we prove that the sequence $(u_{n})$ is Cauchy in

$$ W:= \bigl\{ u: u\in C \bigl([0,T]; H^{1}_{0}(\Omega) \bigr) \cap C^{1} \bigl([0,T]; L^{2}(\Omega) \bigr), u_{t}\in C \bigl([0,T]; L^{2}(\Omega) \bigr) \bigr\} , $$

with the defined norm

$$\begin{aligned} \Vert u \Vert ^{2}_{W}:=\max_{0\leq t\leq T} \frac{1}{2}\bigl( \Vert u_{t} \Vert ^{2}_{2}+l \Vert \nabla u \Vert ^{2}_{2}\bigr)+ \int_{0}^{T} \Vert \nabla u_{t} \Vert ^{2}_{2}\,d\tau+ \int ^{T}_{0} \Vert u_{t} \Vert ^{2}_{2}\leq M^{2}. \end{aligned}$$

(3.5)

For this purpose, we let $U=u_{n}-u_{n'}$, $V=v_{n}-v_{n'}$, then U satisfies

$$ \textstyle\begin{cases} U_{tt}-\Delta U+\int_{0}^{t}g(t-\tau)\Delta U(\tau)\,d\tau-\Delta U_{t}+U_{t}\\ \quad =v_{n} \vert v_{n} \vert ^{p-2}-v_{n'} \vert v_{n'} \vert ^{p-2},\quad x\in\Omega, t>0,\\ U(x,t)=0, \quad x\in\partial\Omega, t\geq0,\\ U(x,0)=U_{0}(x)=u_{0n}-u_{0n'},\qquad U_{t}(x,0)=U_{1}(x)=u_{1n}-u_{1n'},\quad x\in\Omega. \end{cases} $$

(3.6)

Multiplying (3.6) by $U_{t}$ and integrating over $(0,t)\times \Omega$, we obtain

$$\begin{aligned} &\frac{1}{2} \biggl[ \Vert U_{t} \Vert ^{2}_{2}+\biggl(1- \int_{0}^{t}g(\tau)\,d\tau\biggr) \Vert \nabla U \Vert ^{2}_{2} \biggr]-\frac{1}{2} \int_{0}^{t}\bigl(g' \circ\nabla U \bigr) (\tau )\,d\tau+ \int^{t}_{0} \Vert \nabla U_{t} \Vert ^{2}_{2}\,d\tau \\ &\qquad{}+\frac{1}{2}(g\circ\nabla U) (t)+\frac{1}{2} \int_{0}^{t} \int _{\Omega}g(\tau) \bigl\vert \nabla U(\tau) \bigr\vert ^{2}\,dx\,d\tau+ \int^{t}_{0} \Vert U_{t} \Vert ^{2}_{2}\,d\tau \\ &\quad =\frac{1}{2} \bigl( \Vert U_{1} \Vert ^{2}_{2}+ \Vert \nabla U_{0} \Vert ^{2}_{2} \bigr)+ \int ^{t}_{0} \int_{\Omega }\bigl(v_{n} \vert v_{n} \vert ^{p-2}-v_{n'} \vert v_{n'} \vert ^{p-2} \bigr)U_{t}\,dx\,d\tau. \end{aligned}$$

(3.7)

We estimate the second term on the right-hand side of (3.7) as follows:

$$\begin{aligned} & \int_{\Omega}\bigl(v_{n} \vert v_{n} \vert ^{p-2}-v_{n'} \vert v_{n'} \vert ^{p-2} \bigr)U_{t}\,dx \\ &\quad \leq C \Vert U_{t} \Vert _{2} \Vert V \Vert _{\frac{2n}{n-2}} \bigl( \Vert v_{n} \Vert ^{p-2}_{n(p-2)}+ \Vert v_{n'} \Vert ^{p-2}_{n(p-2)} \bigr) \\ &\quad \leq C \Vert U_{t} \Vert _{2} \Vert \nabla V \Vert _{2} \bigl( \Vert \nabla v_{n} \Vert ^{p-2}_{2}+ \Vert \nabla v_{n'} \Vert ^{p-2}_{2} \bigr), \end{aligned}$$

(3.8)

where C is a constant. Using (H1), (H2) and the following fact:

$$-\frac{1}{2} \int_{0}^{t}\bigl(g' \circ\nabla U \bigr) (\tau)\,d\tau+\frac {1}{2}(g\circ\nabla U) (t)+\frac{1}{2} \int_{0}^{t} \int_{\Omega}g(\tau ) \bigl\vert \nabla U(\tau) \bigr\vert ^{2}\,dx\,d\tau\geq0, $$

by estimating (3.7), we obtain

$$\begin{aligned} &\frac{1}{2} \bigl[ \Vert U_{t} \Vert ^{2}_{2}+l \Vert \nabla U \Vert ^{2}_{2} \bigr]+ \int ^{t}_{0} \Vert \nabla U_{t} \Vert ^{2}_{2}\,d\tau+ \int^{t}_{0} \Vert U_{t} \Vert ^{2}_{2}\,d\tau \\ &\quad \leq\frac{1}{2} \bigl( \Vert U_{1} \Vert ^{2}_{2}+ \Vert \nabla U_{0} \Vert ^{2}_{2} \bigr)+C \int^{t}_{0} \Vert U_{t} \Vert _{2} \Vert \nabla V \Vert _{2} \bigl( \Vert \nabla v_{n} \Vert ^{p-2}_{2}+ \Vert \nabla v_{n'} \Vert ^{p-2}_{2} \bigr)\,d\tau \\ &\quad \leq\frac{1}{2} \bigl( \Vert U_{1} \Vert ^{2}_{2}+ \Vert \nabla U_{0} \Vert ^{2}_{2} \bigr)+\Gamma \int^{t}_{0} \Vert U_{t} \Vert _{2} \Vert \nabla V \Vert _{2}\,d\tau, \end{aligned}$$

(3.9)

where $\Gamma>0$ is a constant depending on $\Omega,l,\gamma$ and the radius of the ball in $C ([0,T];H^{1}_{0}(\Omega) )$ containing $(v_{n})$ and $(v_{n'})$. By employing Gronwall’s and Young’s inequalities to the second term of (3.9), we can get

$$\begin{aligned} \Vert U \Vert ^{2}_{W}\leq\Gamma \bigl( \Vert U_{1} \Vert ^{2}_{2}+ \Vert \nabla U_{0} \Vert ^{2}_{2} \bigr)+\Gamma T \Vert V \Vert ^{2}_{W} . \end{aligned}$$

(3.10)

Since $(u_{0n})$ is Cauchy in $H^{1}_{0}(\Omega)$, $(u_{1n})$ is Cauchy in $L^{2}(\Omega)$, and $(v_{n})$ is Cauchy in $C ([0,T]; H^{1}_{0}(\Omega) )$, we obtain that $(u_{n})$ is Cauchy in W, then $u_{n}$ converges to a limit u in W. Now, we prove that the limit u is a weak solution of (3.1). Multiplying equation (3.4) by $\theta\in H^{1}_{0}(\Omega)\cap L^{2}(\Omega)$, we obtain

$$\begin{aligned} &\frac{d}{dt}(u_{nt},\theta)+ \int_{\Omega}\nabla u_{n}\nabla\theta \,dx- \int^{t}_{0} \int_{\Omega}g(t-\tau)\nabla u_{n}\cdot\nabla\theta \,d \tau \,dx+ \int_{\Omega}\nabla u_{nt}\nabla\theta \,dx \\ &\quad {}+ \int_{\Omega}u_{nt}\theta \,dx= \int_{\Omega } \vert v_{n} \vert ^{p-2}v_{n} \theta \,dx. \end{aligned}$$

(3.11)

When $n\rightarrow\infty$, we know that $(\nabla u_{n},\nabla\theta )\rightarrow(\nabla u,\nabla\theta)$, $\int_{\Omega }|v_{n}|^{p-2}v_{n}\theta \,dx\rightarrow\int_{\Omega}|v|^{p-2}v\theta \,dx$ in $C[0, T]$, and $\int^{t}_{0}\int_{\Omega}g(t-\tau)\nabla u_{n}\cdot\nabla\theta \,d\tau \,dx\rightarrow\int^{t}_{0}\int_{\Omega }g(t-\tau)\nabla u \cdot\nabla\theta \,d\tau \,dx$, $\int_{\Omega}\nabla u_{nt}\nabla\theta \,dx\rightarrow\int_{\Omega}\nabla u_{t}\nabla \theta \,dx$, $\int_{\Omega}u_{nt}\theta \,dx\rightarrow\int_{\Omega }u_{t}\theta \,dx$ in $L^{1}(0, T)$, then (3.11) proves that $\lim\limits_{n\rightarrow\infty}(u_{nt}, \theta )=(u_{t},\theta)$ is an absolutely continuous function, so u is a weak solution. To show the uniqueness property, we let $v^{1}$, $v^{2}$ and $u^{1}$, $u^{2}$ be the corresponding solution of (3.1). Take $U= u^{1}-u^{2}$, we have

$$\begin{aligned} &\frac{1}{2} \bigl( \Vert U_{t} \Vert ^{2}_{2}+l \Vert \nabla U \Vert ^{2}_{2} \bigr)-\frac {1}{2} \int_{0}^{t}\bigl(g' \circ\nabla U \bigr) (\tau)\,d\tau+ \int^{t}_{0} \Vert \nabla U_{t} \Vert ^{2}_{2}\,d\tau \\ &\qquad{}+\frac{1}{2}(g\circ\nabla U) (t)+\frac{1}{2} \int_{0}^{t} \int _{\Omega}g(\tau) \bigl\vert \nabla U(\tau) \bigr\vert ^{2}\,dx\,d\tau+ \int^{t}_{0} \Vert U_{t} \Vert ^{2}_{2}\,d\tau \\ &\quad = \int^{t}_{0} \int_{\Omega} \bigl(v^{1} \bigl\vert v^{1} \bigr\vert ^{p-2}-v^{2} \bigl\vert v^{2} \bigr\vert ^{p-2} \bigr)U_{t}\,dx\,d\tau. \end{aligned}$$

(3.12)

Assume $v^{1}=v^{2}$, then (3.12) proves that $U=0$ and the solution is unique. □

Next, we state and prove the local existence result theorem.

Theorem 3.3

If $u_{0}\in H^{1}_{0}(\Omega)$, $u_{1}\in L^{2}(\Omega)$ and $H(3)$ holds, then equation (1.1) has a unique weak solution $u\in W$ for T small enough.

Proof

For $M>0$ large and $T>0$, we define a class of functions $Z(M, T)$ consisting of all functions w in W satisfying the initial data of (1.1) and $\|w\|_{W}\leq M^{2}$. We also define the map f from $Z(M, T)$ into W by $u:=f(w)$.

We will show that f is a contraction from $Z(M, T)$ into itself. Multiplying (3.1) by $u_{t}$ and integrating over $(0, t)\times \Omega$, we obtain

$$\begin{aligned} \Vert u \Vert _{W}^{2}\leq C \bigl( \Vert u_{1} \Vert ^{2}_{2}+ \Vert \nabla u_{0} \Vert ^{2}_{2} \bigr)+CM^{p-1}T \Vert u \Vert _{W}, \end{aligned}$$

(3.13)

where C is independent of M. Choosing M large enough and T small enough, we get u satisfying $\|u\|_{W}\leq M^{2}$, i.e., $u\in Z(M,T)$. This proves that f maps $Z(M,T)$ into itself.

Next, we prove that f is a contraction. For this aim, we let $U=u-\bar {u}$ and $V=v-\bar{v}$, where $u=f(v)$ and $\bar{u}=f(\bar{v})$, then U satisfies

$$ \textstyle\begin{cases} U_{tt}-\Delta U+\int_{0}^{t}g(t-\tau)\Delta U(\tau)\,d\tau-\Delta U_{t}+U_{t}\\ =v \vert v \vert ^{p-2}-\bar{v} \vert \bar{v} \vert ^{p-2}, \quad x\in\Omega, t>0,\\ U(x,t)=0, \quad x\in\partial\Omega, t\geq0,\\ U(x,0)=U_{t}(x,0)=0,\quad x\in\Omega. \end{cases} $$

(3.14)

Similar to the proof of (3.9), we get

$$\begin{aligned} &\frac{1}{2} \bigl( \Vert U_{t} \Vert ^{2}_{2}+l \Vert \nabla u \Vert ^{2}_{2} \bigr)+ \int ^{t}_{0} \Vert \nabla U_{t} \Vert ^{2}_{2}\,d\tau+ \int^{t}_{0} \Vert U_{t} \Vert ^{2}_{2}\,d\tau \\ &\quad \leq C \int^{t}_{0} \Vert U_{t} \Vert _{2} \Vert \nabla V \Vert _{2} \bigl( \Vert \nabla v \Vert ^{p-2}_{2}+ \Vert \nabla\bar{v} \Vert ^{p-2}_{2} \bigr)\,d\tau. \end{aligned}$$

(3.15)

Thus we get

$$\begin{aligned} \Vert U \Vert _{W}\leq CTM^{p-2} \Vert V \Vert _{W}. \end{aligned}$$

(3.16)

We let T small enough such that $CTM^{p-2}\leq\frac{1}{2}$. Then from (3.16) we can get that f is a contraction in $Z(M,T)$. By using the contraction mapping principle, we can obtain that there exists a unique u satisfying $u=f(u)$. Then it is the solution of (1.1). □

4 Decay of global solution

In this section we state and prove the general decay result for global solutions. Firstly, we establish the global existence theorem.

Lemma 4.1

If (H1), (H2), (H3) hold and $(u_{0},u_{1})\in H^{1}_{0}(\Omega )\times L^{2}(\Omega)$ such that

$$\begin{aligned} \beta=\frac{C^{p}_{*}}{l} \biggl(\frac{2p}{l(p-2)}E(0) \biggr)^{(p-2)/2}< 1, \quad I(u_{0})>0, \end{aligned}$$

(4.1)

then $I (u(t) )>0$ for $\forall t>0$. Here $C_{*}$ is given in (2.1).

Proof

For $I(u_{0})>0$, then there is $T_{m}< T$ such that $I (u(t) )\geq0$ for $\forall t \in[0,T_{m})$. So, we get

$$\begin{aligned} J(t) =&\frac{1}{2}\biggl(1- \int^{t}_{0}g(\tau)\,d\tau\biggr) \bigl\Vert \nabla u(t) \bigr\Vert ^{2}_{2}+\frac{1}{2}(g\circ \nabla u) (t)-\frac{1}{p} \bigl\Vert u(t) \bigr\Vert ^{p}_{p} \\ =&\frac{p-2}{2p}\biggl(1- \int^{t}_{0}g(\tau)\,d\tau\biggr) \bigl\Vert \nabla u(t) \bigr\Vert ^{2}_{2}+\frac{p-2}{2p}(g\circ \nabla u) (t)+\frac{1}{p}I\bigl(u(t)\bigr) \\ \geq&\frac{p-2}{2p} \biggl\{ \biggl(1- \int^{t}_{0}g(\tau)\,d\tau\biggr) \bigl\Vert \nabla u(t) \bigr\Vert ^{2}_{2}+(g\circ\nabla u) (t) \biggr\} , \quad \forall t\in[0,T_{m}). \end{aligned}$$

(4.2)

By using (H1), (2.2), (2.3), and (4.2), we arrive at

$$\begin{aligned} l \bigl\Vert \nabla u(t) \bigr\Vert ^{2}_{2} \leq& \biggl(1- \int^{t}_{0}g(\tau)\,d\tau\biggr) \bigl\Vert \nabla u(t) \bigr\Vert ^{2}_{2}\leq\frac{2p}{p-2}J(t) \\ \leq&\frac{2p}{p-2}E(t) \\ \leq&\frac{2p}{p-2}E(0), \quad\forall t\in[0,T_{m}). \end{aligned}$$

(4.3)

By combining (H1), (2.1), (4.1) with (4.3), we get

$$\begin{aligned} \bigl\Vert u(t) \bigr\Vert ^{p}_{p} \leq& C^{p}_{*} \bigl\Vert \nabla u(t) \bigr\Vert ^{p}_{2} \\ \leq&\frac{C^{p}_{*}}{l} \bigl\Vert \nabla u(t) \bigr\Vert ^{p-2}_{2} l \bigl\Vert \nabla u(t) \bigr\Vert ^{2}_{2} \\ \leq&\frac{C^{p}_{*}}{l} \biggl[\frac{2p}{l(p-2)}E(0) \biggr]^{(p-2)/2} l \bigl\Vert \nabla u(t) \bigr\Vert ^{2}_{2} \\ \leq&\beta l \bigl\Vert \nabla u(t) \bigr\Vert ^{2}_{2} \\ < &\biggl(1- \int^{t}_{0}g(\tau)\,d\tau\biggr) \bigl\Vert \nabla u(t) \bigr\Vert ^{2}_{2}. \end{aligned}$$

(4.4)

Therefore,

$$\begin{aligned} I(t)=\biggl(1- \int^{t}_{0}g(s)\,ds\biggr) \bigl\Vert \nabla u(t) \bigr\Vert ^{2}_{2}+(g\circ\nabla u) (t)- \bigl\Vert u(t) \bigr\Vert ^{p}_{p}>0 \end{aligned}$$

for $\forall t\in[0,T_{m})$. Repeating the process and using the fact that

$$\lim_{t\to T_{m}}\frac{C^{p}_{*}}{l} \biggl(\frac{2p}{l(p-2)}E(0) \biggr)^{(p-2)/2}\leq\beta< 1, $$

$T_{m}$ is extended to T. □

Theorem 4.2

If $(u_{0},u_{1})\in H^{1}_{0}(\Omega)\times L^{2}(\Omega)$ and satisfies (4.1), and suppose (H1), (H2), (H3) hold, then the solution is global and bounded.

Proof

The aim is to show that

$$\begin{aligned} \bigl\Vert \nabla u(t) \bigr\Vert ^{2}_{2}+ \bigl\Vert u_{t}(t) \bigr\Vert ^{2}_{2} \end{aligned}$$

is bounded independently of t. For this goal, we use Lemma 4.1 and (2.2) to obtain

$$\begin{aligned} E(0) \geq&E(t)=J(t)+\frac{1}{2} \bigl\Vert u_{t}(t) \bigr\Vert ^{2}_{2} \\ \geq&\frac{p-2}{2p} \biggl\{ \biggl(1- \int^{t}_{0}g(s)\,ds\biggr) \bigl\Vert \nabla u(t) \bigr\Vert ^{2}_{2}+(g\circ\nabla u) (t) \biggr\} + \frac{1}{p}I\bigl(u(t)\bigr)+\frac{1}{2} \bigl\Vert u_{t}(t) \bigr\Vert ^{2}_{2} \\ \geq&\frac{p-2}{2p} \biggl\{ \biggl(1- \int^{t}_{0}g(s)\,ds\biggr) \bigl\Vert \nabla u(t) \bigr\Vert ^{2}_{2}+(g\circ\nabla u) (t) \biggr\} + \frac{1}{2} \bigl\Vert u_{t}(t) \bigr\Vert ^{2}_{2}, \end{aligned}$$

(4.5)

since $I(u(t))\geq0$ and $(g\circ\nabla u)(t)$ are positive. Therefore

$$\begin{aligned} \bigl\Vert \nabla u(t) \bigr\Vert ^{2}_{2}+ \bigl\Vert u_{t}(t) \bigr\Vert ^{2}_{2}\leq CE(0), \end{aligned}$$

where C is a positive constant. □

For establishing the general decay rate estimate, we use the following functional:

$$\begin{aligned} F(t):=E(t)+\epsilon_{1}\Psi(t)+\epsilon_{2} \Phi(t), \end{aligned}$$

(4.6)

where $\epsilon_{1}$ and $\epsilon_{2}$ are positive constants and

$$\begin{aligned} &\Psi(t):=\xi(t) \int_{\Omega}uu_{t}\,dx, \\ &\Phi(t):=\xi(t) \int_{\Omega}(\Delta u-u_{t}) \int^{t}_{0}g(t-\tau) \bigl(u(t)-u(\tau) \bigr)\,d \tau \,dx. \end{aligned}$$

Lemma 4.3

For $\epsilon_{1}$ and $\epsilon_{2}$ small enough, we have

$$\begin{aligned} \alpha_{1}F(t)\leq E(t)\leq\alpha_{2} F(t) \end{aligned}$$

(4.7)

holds, where $\alpha_{1}$ and $\alpha_{2}$ are positive constants.

Proof

By straightforward computations, we obtain

$$\begin{aligned} F(t) \leq&E(t)+\frac{\epsilon_{1}}{2}\xi(t) \int_{\Omega} \vert u \vert ^{2}\,dx+ \frac {\epsilon_{1}}{2}\xi(t) \int_{\Omega} \vert u_{t} \vert ^{2}\,dx+ \frac{\epsilon _{2}}{2}\xi(t) \int_{\Omega} \vert u_{t} \vert ^{2}\,dx \\ &{}+\frac{\epsilon_{2}}{2}C^{2}_{*}(1-l)\xi(t) (g\circ \nabla u)+\frac{\epsilon_{2}}{2}\xi(t) \int_{\Omega} \vert \nabla u \vert ^{2}\,dx+ \frac {\epsilon_{2}}{2}(1-l)\xi(t) (g\circ\nabla u) \\ \leq& E(t)+\frac{1}{2}(\epsilon_{1}+\epsilon_{2}) \xi(t) \int_{\Omega } \vert u_{t} \vert ^{2}\,dx+ \frac{1}{2}\bigl(\epsilon_{1}C_{*}^{2}+ \epsilon_{2}\bigr)\xi(t) \int _{\Omega} \vert \nabla u \vert ^{2}\,dx \\ &{}+\frac{\epsilon_{2}}{2}\bigl(C_{*}^{2}+1\bigr) (1-l) \xi(t) (g\circ\nabla u) (t) \\ \leq& \frac{1}{2} \bigl[1+(\epsilon_{1}+\epsilon_{2})M \bigr] \Vert u_{t} \Vert ^{2}_{2}+ \frac{1}{2} \bigl[l+\bigl(\epsilon_{1}C_{*}^{2}+ \epsilon_{2}\bigr)M \bigr] \Vert \nabla u \Vert ^{2}_{2} \\ &{}+ \biggl[\frac{1}{2}+\frac{\epsilon_{2}}{2}\bigl(C_{*}^{2}+1 \bigr)M(1-l) \biggr](g\circ\nabla u) (t)-\frac{1}{p} \bigl\Vert u(t) \bigr\Vert ^{p}_{p} \\ \leq&\frac{1}{\alpha_{1}}E(t). \end{aligned}$$

Similarly, we have

$$\begin{aligned} F(t) \geq& E(t)-\frac{1}{2}(\epsilon_{1}+\epsilon_{2}) \xi(t) \int_{\Omega } \vert u_{t} \vert ^{2}\,dx- \frac{1}{2}\bigl(\epsilon_{1}C^{2}_{*}+ \epsilon_{2}\bigr)\xi(t) \int _{\Omega} \vert \nabla u \vert ^{2}\,dx \\ &{}-\frac{\epsilon_{2}}{2}\bigl(C_{*}^{2}+1\bigr) (1-l) \xi(t) (g\circ\nabla u) (t) \\ \geq& \biggl[\frac{l}{2}-\frac{1}{2}M\bigl(\epsilon_{1}C^{2}_{*}+ \epsilon _{2}\bigr) \biggr] \Vert \nabla u \Vert ^{2}_{2}+\biggl(\frac{1}{2}-\frac{\epsilon_{1}+\epsilon _{2}}{2}M \biggr) \Vert u_{t} \Vert ^{2}_{2} \\ &{}+ \biggl[\frac{1}{2}-\frac{\epsilon_{2}}{2}M\bigl( C_{*}^{2}+1\bigr) (1-l) \biggr](g\circ\nabla u) (t)- \frac{1}{p} \bigl\Vert u(t) \bigr\Vert ^{p}_{p} \\ \geq&\frac{1}{\alpha_{2}}E(t) \end{aligned}$$

for $\epsilon_{1}$ and $\epsilon_{2}$ small enough. □

Lemma 4.4

If (H1), (H2) hold and u is the solution of (1.1), let $(u_{0},u_{1})\in H^{1}_{0}(\Omega)\times L^{2}(\Omega)$ be given, then the functional

$$ \Psi(t):=\xi(t) \int_{\Omega}uu_{t}\,dx $$

(4.8)

satisfies

$$\begin{aligned} \Psi'(t) \leq& \biggl(1+\frac{1+k}{4\alpha} \biggr)\xi(t) \Vert u_{t} \Vert ^{2}_{2}+ \biggl\{ - \frac{l}{2}+\frac{1}{4\alpha}+(1+k)\alpha C^{2}_{*} \biggr\} \xi(t) \Vert \nabla u \Vert ^{2}_{2} \\ &{}+\frac{1-l}{2l}\xi(t) (g\circ\nabla u) (t)+\alpha\xi(t) \Vert \nabla u_{t} \Vert ^{2}_{2}+\xi(t) \Vert u \Vert ^{p}_{p}. \end{aligned}$$

(4.9)

Proof

Taking a time derivative of (4.8) and using equation (1.1), we have

$$\begin{aligned} \Psi'(t) =&\xi(t) \int_{\Omega} \vert u_{t} \vert ^{2}\,dx+ \xi(t) \int_{\Omega }uu_{tt}\,dx+\xi'(t) \int_{\Omega}uu_{t}\,dx \\ =&\xi(t) \biggl( \Vert u_{t} \Vert ^{2}_{2}- \Vert \nabla u \Vert ^{2}_{2}+ \int_{\Omega }\nabla u(t) \int^{t}_{0}g(t-\tau)\nabla u(\tau)\,d\tau \,dx \\ &{}- \int_{\Omega}\nabla u_{t}\nabla u\,dx- \int_{\Omega}u_{t}u\,dx+ \Vert u \Vert ^{p}_{p} \biggr)+\xi'(t) \int_{\Omega}uu_{t}\,dx. \end{aligned}$$

(4.10)

Now, we estimate the third term on the right-hand side of (4.10) as follows:

$$\begin{aligned} & \int_{\Omega}\nabla u(t) \int^{t}_{0}g(t-\tau)\nabla u(\tau)\,d\tau \,dx \\ &\quad \leq\frac{1}{2} \int_{\Omega} \vert \nabla u \vert ^{2}\,dx+ \frac{1}{2} \int_{\Omega}\biggl( \int^{t}_{0}g(t-\tau) \bigl\vert \nabla u(\tau) \bigr\vert \,d\tau \biggr)^{2}\,dx \\ &\quad \leq \frac{1}{2} \int_{\Omega} \vert \nabla u \vert ^{2}\,dx+ \frac{1}{2} \int_{\Omega}\biggl( \int^{t}_{0}g(t-\tau) \bigl( \bigl\vert \nabla u( \tau)-\nabla u(t) \bigr\vert + \bigl\vert \nabla u(t) \bigr\vert \bigr)\,d \tau \biggr)^{2}\,dx. \end{aligned}$$

(4.11)

We then use Young’s inequality and (H1) to obtain, for $\forall \eta>0$,

$$\begin{aligned} & \int_{\Omega}\biggl( \int^{t}_{0}g(t-\tau) \bigl( \bigl\vert \nabla u( \tau)-\nabla u(t) \bigr\vert + \bigl\vert \nabla u(t) \bigr\vert \bigr)\,d \tau \biggr)^{2}\,dx \\ &\quad \leq \int_{\Omega}\biggl( \int^{t}_{0}g(t-\tau) \bigl\vert \nabla u(\tau)- \nabla u(t) \bigr\vert \,d\tau \biggr)^{2}\,dx+ \int_{\Omega}\biggl( \int^{t}_{0}g(t-\tau) \bigl\vert \nabla u(t) \bigr\vert \,d\tau \biggr)^{2}\,dx \\ &\qquad {}+2 \int_{\Omega}\biggl( \int^{t}_{0}g(t-\tau) \bigl\vert \nabla u(\tau)- \nabla u(t) \bigr\vert \,d\tau \biggr) \biggl( \int^{t}_{0}g(t-\tau) \bigl\vert \nabla u(t) \bigr\vert \,d\tau \biggr)\,dx \\ &\quad \leq (1+\eta) \int_{\Omega}\biggl( \int^{t}_{0}g(t-\tau) \bigl\vert \nabla u(t) \bigr\vert \,d\tau \biggr)^{2}\,dx \\ &\qquad {}+\biggl(1+\frac{1}{\eta}\biggr) \int_{\Omega}\biggl( \int^{t}_{0}g(t-\tau) \bigl\vert \nabla u(\tau )- \nabla u(t) \bigr\vert \,d\tau \biggr)^{2}\,dx \\ &\quad \leq \biggl(1+\frac{1}{\eta}\biggr) (1-l) (g\circ\nabla u) (t)+(1+\eta) (1-l)^{2} \int _{\Omega}\bigl\vert \nabla u(t) \bigr\vert ^{2}\,dx. \end{aligned}$$

(4.12)

By using Young’s and Poincaré’s inequalities and for $\forall\alpha >0$, we have

$$\begin{aligned} &\int_{\Omega}\nabla u_{t}\nabla u\,dx\leq\alpha \Vert \nabla u_{t} \Vert ^{2}_{2}+\frac{1}{4\alpha} \Vert \nabla u \Vert ^{2}_{2}, \end{aligned}$$

(4.13)

$$\begin{aligned} &\int_{\Omega}u_{t}u\,dx\leq\alpha C^{2}_{*} \Vert \nabla u \Vert ^{2}_{2}+\frac {1}{4\alpha} \Vert u_{t} \Vert ^{2}_{2}. \end{aligned}$$

(4.14)

Combining (4.10)–(4.14) yields

$$\begin{aligned} \Psi'(t) \leq& \biggl(1+\frac{1}{4\alpha}+ \biggl\vert \frac{\xi'(t)}{\xi(t)} \biggr\vert \frac{1}{4\alpha} \biggr)\xi(t) \Vert u_{t} \Vert ^{2}_{2} \\ &{}+ \biggl\{ -\frac{1}{2}+\frac{1}{2}(1+\eta) (1-l)^{2}+\frac{1}{4\alpha }+ \biggl(1+ \biggl\vert \frac{\xi'(t)}{\xi(t)} \biggr\vert \biggr)\alpha C^{2}_{*} \biggr\} \xi (t) \Vert \nabla u \Vert ^{2}_{2} \\ &{}+\frac{1}{2}\biggl(1+\frac{1}{\eta}\biggr) (1-l)\xi(t) (g\circ \nabla u) (t)+\alpha\xi(t) \Vert \nabla u_{t} \Vert ^{2}_{2}+\xi(t) \Vert u \Vert ^{p}_{p}. \end{aligned}$$

(4.15)

By choosing $\eta=l/(1-l)$ and using (H2), we can get (4.9). □

Lemma 4.5

Assume that (H1), (H2) hold and $(u_{0},u_{1})\in H^{1}_{0}(\Omega )\times L^{2}(\Omega)$ is given. If u is the solution of (1.1), then the functional

$$\begin{aligned} \Phi(t):=\xi(t) \int_{\Omega}(\Delta u-u_{t}) \int^{t}_{0}g(t-\tau) \bigl(u(t)-u(\tau) \bigr)\,d \tau \,dx \end{aligned}$$

(4.16)

satisfies

$$\begin{aligned} \Phi'(t) \leq&\delta \biggl\{ k+2(1-l)^{2}+3+C^{2(p-1)}_{*} \biggl(\frac {2p}{l(p-2)}E(0) \biggr)^{p-2} \biggr\} \xi(t) \Vert \nabla u \Vert ^{2}_{2} \\ &{}+ \biggl\{ \delta(2+k)- \int^{t}_{0}g(\tau)\,d\tau \biggr\} \xi(t) \Vert u_{t} \Vert ^{2}_{2}+\frac{1}{4\delta}(1-l)^{2} \xi(t) \Vert \nabla u_{t} \Vert ^{2}_{2} \\ &{}+(1-l) \biggl(\frac{k}{4\delta}+\frac{k C^{2}_{*}}{4\delta}+\frac {1}{2\delta}+2 \delta+\frac{C^{2}_{*}}{2\delta} \biggr)\xi(t) (g\circ\nabla u) (t) \\ &{}-\frac{g(0)}{4\delta} \bigl(1+C^{2}_{*} \bigr)\xi(t) \bigl(g'\circ\nabla u\bigr) (t). \end{aligned}$$

(4.17)

Proof

By taking a time derivative of (4.16) and using equation (1.1), we arrive at

$$\begin{aligned} \Phi'(t) =&-\xi'(t) \int_{\Omega}\nabla u \int^{t}_{0}g(t-\tau) \bigl(\nabla u(t)-\nabla u( \tau) \bigr)\,d\tau \,dx \\ &{}-\xi'(t) \int_{\Omega}u_{t} \int^{t}_{0}g(t-\tau) \bigl(u(t)-u(\tau ) \bigr)\,d \tau \,dx \\ &{}+\xi(t) \int_{\Omega}\nabla u \int^{t}_{0}g(t-\tau) \bigl(\nabla u(t)-\nabla u( \tau) \bigr)\,d\tau \,dx \\ &{}-\xi(t) \int_{\Omega} \int^{t}_{0}g(t-\tau)\nabla u(\tau)\,d\tau \int ^{t}_{0}g(t-\tau) \bigl(\nabla u(t)-\nabla u( \tau) \bigr)\,d\tau \,dx \\ &{}+\xi(t) \int_{\Omega}u_{t} \int^{t}_{0}g(t-\tau) \bigl(u(t)-u(\tau ) \bigr)\,d \tau \,dx \\ &{}-\xi(t) \int_{\Omega}u \vert u \vert ^{p-2} \int^{t}_{0}g(t-\tau) \bigl(u(t)-u(\tau) \bigr)\,d \tau \,dx \\ &{}-\xi(t) \int_{\Omega}\nabla u \int^{t}_{0}g'(t-\tau) \bigl(\nabla u(t)-\nabla u(\tau) \bigr)\,d\tau \,dx \\ &{}-\xi(t) \int_{\Omega}u_{t} \int^{t}_{0}g'(t-\tau) \bigl(u(t)-u( \tau ) \bigr)\,d\tau \,dx \\ &{}-\xi(t) \int_{\Omega}\nabla u \int^{t}_{0}g(t-\tau)\nabla u_{t}\,d\tau \,dx -\xi(t) \int^{t}_{0}g(\tau)\,d\tau \Vert u_{t} \Vert ^{2}_{2} \\ :=&\xi(t) \biggl( \biggl\vert \frac{\xi'(t)}{\xi(t)} \biggr\vert I_{1}+ \biggl\vert \frac{\xi '(t)}{\xi(t)} \biggr\vert I_{2}+I_{3}+I_{4}+I_{5}+I_{6} \\ &{}+I_{7}+I_{8}+I_{9}- \int^{t}_{0}g(\tau)\,d\tau \Vert u_{t} \Vert ^{2}_{2} \biggr). \end{aligned}$$

(4.18)

We will estimate $I_{j}$, $j=1,\ldots,9$, on the right-hand side of (4.18). Using Young’s inequality, Cauchy–Schwarz’s inequality, Poincaré’s inequality, (H1) and (H2), for $\forall\delta>0$, we have

$$\begin{aligned} I_{1} \leq&\delta \int_{\Omega} \vert \nabla u \vert ^{2}\,dx+ \frac{1}{4\delta} \int _{\Omega}\biggl\vert \int^{t}_{0}g(t-\tau) \bigl(\nabla u(t)-\nabla u( \tau) \bigr)\,d\tau \biggr\vert ^{2} \,dx \\ \leq&\delta \int_{\Omega} \vert \nabla u \vert ^{2}\,dx+ \frac{1}{4\delta} \int_{\Omega}g(t-\tau) \int^{t}_{0}g(t-\tau) \bigl(\nabla u(t)-\nabla u( \tau) \bigr)^{2}\,d\tau \,dx \\ \leq& \delta \Vert \nabla u \Vert ^{2}_{2}+ \frac{1-l}{4\delta}(g\circ\nabla u) (t). \end{aligned}$$

(4.19)

$$\begin{aligned} I_{2} \leq&\delta \int_{\Omega} \vert u_{t} \vert ^{2}\,dx+ \frac{1}{4\delta} \int_{\Omega}\biggl\vert \int^{t}_{0}g(t-\tau) \bigl(u(t)-u(\tau) \bigr)\,d \tau \biggr\vert ^{2} \,dx \\ \leq&\delta \Vert u_{t} \Vert ^{2}_{2}+ \frac{1-l}{4\delta}C^{2}_{*}(g\circ \nabla u) (t). \end{aligned}$$

(4.20)

$$\begin{aligned} I_{3} \leq&\delta \int_{\Omega} \vert \nabla u \vert ^{2}\,dx+ \frac{1}{4\delta} \int _{\Omega}\biggl\vert \int^{t}_{0}g(t-\tau) \bigl(\nabla u(t)-\nabla u( \tau) \bigr)\,d\tau \biggr\vert ^{2} \,dx \\ \leq&\delta \Vert \nabla u \Vert ^{2}_{2}+ \frac{1-l}{4\delta}(g\circ\nabla u) (t). \end{aligned}$$

(4.21)

By taking $\eta=1$ in (4.12), we can get $I_{4}$ as follows:

$$\begin{aligned} I_{4} =& \int_{\Omega} \int^{t}_{0}g(t-\tau)\nabla u(\tau)\,d\tau \int ^{t}_{0}g(t-\tau) \bigl(\nabla u(t)-\nabla u( \tau) \bigr)\,d\tau \,dx \\ \leq&\delta \int_{\Omega} \biggl\vert \int^{t}_{0}g(t-\tau)\nabla u(\tau)\,d\tau \biggr\vert ^{2}\,dx+\frac{1}{4\delta} \int_{\Omega}\biggl\vert \int^{t}_{0}g(t-\tau) \bigl(\nabla u(t)-\nabla u( \tau) \bigr)\,d\tau \biggr\vert ^{2} \,dx \\ \leq&\frac{1}{4\delta} \int_{\Omega} \biggl\vert \int^{t}_{0}g(t-\tau) \bigl\vert \nabla u(t)- \nabla u(\tau) \bigr\vert \,d\tau \biggr\vert ^{2}\,dx+2 \delta(1-l)^{2} \int_{\Omega} \vert \nabla u \vert ^{2}\,dx \\ &{}+2\delta(1-l) (g\circ\nabla u) (t) \\ \leq&2\delta(1-l)^{2} \Vert \nabla u \Vert ^{2}_{2}+ \biggl(2\delta+\frac{1}{4\delta }\biggr) (1-l) (g\circ\nabla u) (t). \end{aligned}$$

(4.22)

$$\begin{aligned} I_{5} \leq&\delta \int_{\Omega} \vert u_{t} \vert ^{2}\,dx+ \frac{1}{4\delta} \int_{\Omega}\biggl\vert \int^{t}_{0}g(t-\tau) \bigl(u(t)-u(\tau) \bigr)\,d \tau \biggr\vert ^{2} \,dx \\ \leq&\delta \Vert u_{t} \Vert ^{2}_{2}+ \frac{1-l}{4\delta}C^{2}_{*}(g\circ \nabla u) (t). \end{aligned}$$

(4.23)

With the help of the inequalities mentioned above and the Sobolev embedding theorem, we infer that

$$\begin{aligned} I_{6} \leq&\delta \int_{\Omega} \vert u \vert ^{2(p-1)}\,dx+ \frac{1}{4\delta} \int _{\Omega}\biggl\vert \int^{t}_{0}g(t-\tau) \bigl(u(t)-u(\tau) \bigr)\,d \tau \biggr\vert ^{2}\,dx \\ \leq&\delta \Vert u \Vert ^{2(p-1)}_{2(p-1)}+ \frac{1-l}{4\delta }C^{2}_{*}(g\circ\nabla u) (t) \\ \leq&\delta C^{2(p-1)}_{*} \biggl(\frac{2p}{l(p-2)}E(0) \biggr)^{p-2} \Vert \nabla u \Vert ^{2}_{2}+ \frac{1-l}{4\delta}C^{2}_{*}(g\circ\nabla u) (t). \end{aligned}$$

(4.24)

Similarly, using (H2), we obtain $I_{7}$, $I_{8}$, $I_{9}$ as follows:

$$\begin{aligned} I_{7} \leq&\delta \int_{\Omega} \vert \nabla u \vert ^{2}\,dx+ \frac{1}{4\delta} \int _{\Omega}\biggl\vert \int^{t}_{0}g'(t-\tau) \bigl(\nabla u(t)-\nabla u(\tau) \bigr)\,d\tau \biggr\vert ^{2} \,dx \\ \leq&\delta \int_{\Omega} \vert \nabla u \vert ^{2}\,dx+ \frac{1}{4\delta} \int_{\Omega}g'(t-\tau) \int^{t}_{0}g'(t-\tau) \bigl\vert \nabla u(t)-\nabla u(\tau) \bigr\vert ^{2}\,d\tau \,dx \\ \leq&\delta \Vert \nabla u \Vert ^{2}_{2}- \frac{g(0)}{4\delta}\bigl(g'\circ\nabla u\bigr) (t); \end{aligned}$$

(4.25)

$$\begin{aligned} I_{8} \leq&\delta \int_{\Omega} \vert u_{t} \vert ^{2}\,dx+ \frac{1}{4\delta} \int_{\Omega}\biggl\vert \int^{t}_{0}g'(t-\tau) \bigl(u(t)-u( \tau) \bigr)\,d\tau \biggr\vert ^{2}\,dx \\ \leq&\delta \Vert u_{t} \Vert ^{2}_{2}- \frac{g(0)}{4\delta}C^{2}_{*}\bigl(g'\circ \nabla u\bigr) (t); \end{aligned}$$

(4.26)

$$\begin{aligned} I_{9} \leq&\delta \int_{\Omega} \vert \nabla u \vert ^{2}\,dx+ \frac{1}{4\delta} \int _{\Omega}\biggl\vert \int^{t}_{0}g(t-\tau)\nabla u_{t}\,d\tau \biggr\vert ^{2} \,dx \\ \leq&\delta \Vert \nabla u \Vert ^{2}_{2}+ \frac{1}{4\delta}(1-l)^{2} \Vert \nabla u_{t} \Vert ^{2}_{2}. \end{aligned}$$

(4.27)

Combining (4.18)–(4.27), we get estimate (4.17). □

Theorem 4.6

Let $(u_{0},u_{1})\in H^{1}_{0}(\Omega)\times L^{2}(\Omega)$ be given, satisfying (4.1). If (H1), (H2), and (H3) hold, then, for each $t_{0}>0$, there exist strictly positive constants K and λ such that the solution of (1.1) satisfies

$$\begin{aligned} E(t)\leq K e^{-\lambda\int^{t}_{t_{0}}\xi(t)\,ds},\quad\forall t\geq t_{0}. \end{aligned}$$

(4.28)

Proof

For g is positive and continuous, $g(0)>0$, then for $\forall t_{0}>0$ we get

$$\begin{aligned} \int^{t}_{0}g(\tau)\,d\tau\geq \int^{t_{0}}_{0}g(\tau)\,d\tau=g_{0}>0,\quad \forall t\geq t_{0}. \end{aligned}$$

(4.29)

Taking a time derivative of (4.6), using (2.3), (4.6), (4.9), (4.17), (4.29) and Remark 2.1, we get, for $\forall t\geq t_{0}$,

$$\begin{aligned} F'(t) \leq&- \biggl\{ \epsilon_{2} \bigl[g_{0}- \delta(2+k) \bigr]-\epsilon _{1} \biggl(1+\frac{1+k}{4\alpha} \biggr) \biggr\} \xi(t) \Vert u_{t} \Vert ^{2}_{2} \\ &{}- \biggl\{ \epsilon_{1} \biggl[\frac{l}{2}- \frac{1}{4\alpha }-(1+k)\alpha C^{2}_{*} \biggr]- \epsilon_{2}\delta \biggl[k+2(1-l)^{2}+3 \\ &{}+C^{2(p-1)}_{*} \biggl(\frac{2p}{l(p-2)}E(0) \biggr)^{p-2} \biggr] \biggr\} \xi(t) \Vert \nabla u \Vert ^{2}_{2} \\ &{}+ \biggl\{ \frac{\epsilon_{1}(1-l)}{2l}+\epsilon_{2}(1-l) \biggl( \frac {k}{4\delta}+\frac{k C^{2}_{*}}{4\delta}+\frac{1}{2\delta}+2\delta+ \frac {C^{2}_{*}}{2\delta} \biggr) \biggr\} \xi(t) (g\circ\nabla u) (t) \\ &{}+ \biggl(\frac{1}{2}-\frac{g(0)}{4\delta}\bigl(1+C^{2}_{*} \bigr)\epsilon _{2}M \biggr) \bigl(g'\circ\nabla u\bigr) (t) \\ &{}- \biggl(1-\epsilon_{1}\alpha M-\frac{\epsilon_{2}}{4\delta }(1-l)^{2}M \biggr) \Vert \nabla u_{t} \Vert ^{2}_{2}+ \epsilon_{1}\xi(t) \Vert u \Vert ^{p}_{p}. \end{aligned}$$

(4.30)

At this point we choose δ small enough such that

$$\begin{aligned} &\frac{g_{0}-\delta(2+k)}{1+\frac{1+k}{4\alpha}}>\frac{1}{2}g_{0}, \\ &\frac{\delta[k+2(1-l)^{2}+3+C^{2(p-1)}_{*}(\frac {2p}{l(p-2)}E(0))^{p-2}]}{\frac{l}{2}-\frac{1}{4\alpha}-(1+k)\alpha C^{2}_{*}}< \frac{1}{4}g_{0}. \end{aligned}$$

When δ is fixed, we choose any two positive constants $\epsilon _{1}$ and $\epsilon_{2}$ satisfying

$$\begin{aligned} \frac{1}{4}g_{0}\epsilon_{2}< \epsilon_{1}< \frac{1}{2}g_{0}\epsilon _{2} \end{aligned}$$

(4.31)

will make

$$\begin{aligned} &\kappa_{1}:=\epsilon_{2} \bigl[g_{0}- \delta(2+k) \bigr]-\epsilon_{1}\biggl(1+\frac {1+k}{4\alpha}\biggr)>0, \\ &\kappa_{2}:=\epsilon_{1} \biggl[\frac{l}{2}- \frac{1}{4\alpha}-(1+k)\alpha C^{2}_{*} \biggr]\\ &\hphantom{\kappa_{2}:=}{}- \epsilon_{2}\delta \biggl[k+2(1-l)^{2}+3+C^{2(p-1)}_{*} \biggl(\frac{2p}{l(p-2)}E(0)\biggr)^{p-2} \biggr]>0. \end{aligned}$$

We then pick $\epsilon_{1}$ and $\epsilon_{2}$ small enough such that (4.7) and (4.31) remain valid and

$$\begin{aligned} &\kappa_{3}:= \biggl(\frac{1}{2}-\frac{g(0)}{4\delta} \bigl(1+C^{2}_{*}\bigr)\epsilon _{2}M \biggr)\\ &\hphantom{\kappa_{3}:=}{}- \biggl\{ \frac{\epsilon_{1}(1-l)}{2l}+\epsilon_{2}(1-l) \biggl(\frac {k}{4\delta}+ \frac{k C^{2}_{*}}{4\delta}+\frac{1}{2\delta}+2\delta+\frac {C^{2}_{*}}{2\delta}\biggr) \biggr\} >0, \\ &\kappa_{4}:=1-\epsilon_{1}\alpha M-\frac{1}{4\delta}(1-l)^{2} \epsilon_{2}M< 0. \end{aligned}$$

Hence

$$\begin{aligned} &\biggl(\frac{1}{2}-\frac{g(0)}{4\delta}\bigl(1+C^{2}_{*} \bigr)\epsilon_{2}M \biggr) \bigl(g'\circ\nabla u\bigr) (t) \\ &\qquad {}+ \biggl\{ \frac{\epsilon_{1}(1-l)}{2l}+\epsilon _{2}(1-l) \biggl( \frac{k}{4\delta} +\frac{k C^{2}_{*}}{4\delta}+\frac{1}{2\delta}+2\delta+\frac {C^{2}_{*}}{2\delta}\biggr) \biggr\} \xi(t) (g\circ\nabla u) (t) \\ &\quad \leq-\kappa_{3}\xi (t) (g\circ\nabla u) (t). \end{aligned}$$

(4.32)

For $\xi(t)$ is nonincreasing. Therefore, by using (4.7) and (4.30), we arrive at

$$\begin{aligned} F'(t)\leq-\gamma\xi(t)E(t)\leq-\gamma\alpha_{1} \xi(t)F(t),\quad\forall t\geq t_{0}. \end{aligned}$$

(4.33)

By integration of (4.33), we get

$$\begin{aligned} F(t)\leq F(t_{0})e^{-\gamma\alpha_{1}\int^{t}_{t_{0}}\xi(s)\,ds},\quad \forall t\geq t_{0}. \end{aligned}$$

(4.34)

Thus (4.7) and (4.34) yield

$$\begin{aligned} E(t)\leq\alpha_{2}F(t_{0})e^{-\gamma\alpha_{1}\int^{t}_{t_{0}}\xi (s)\,ds}=Ke^{-\lambda\int^{t}_{t_{0}}\xi(s)\,ds}, \quad\forall t\geq t_{0}. \end{aligned}$$

(4.35)

 □

Remark 4.7

We can obtain exponential decay if $\xi(t)=a$ and polynomial decay if $\xi(t)=a(1+t)^{-1}$, where $a>0$ is a constant.

Remark 4.8

For the continuity and boundedness of $E(t)$ and $\xi(t)$, estimates of (4.34) are also true for $t\in[0,t_{0}]$.

5 Blow-up phenomenon

In this section we state and prove the blow-up result.

Theorem 5.1

If (H1), (H2), (H3) hold, $E(0)<0$ and $\int^{\infty}_{0}g(\tau)\,d\tau<\frac{(p/2)-1}{p/2-1+(1/2p)}$, then the solution of (1.1) blows up in finite time.

Proof

For the definition of $H(t)$, we have

$$H'(t)=-\frac{1}{2}\bigl(g'\circ\nabla u\bigr) (t)+\frac{1}{2}g(t) \bigl\Vert \nabla u(t) \bigr\Vert ^{2}_{2}+ \Vert \nabla u_{t} \Vert ^{2}_{2}+ \Vert u_{t} \Vert ^{2}_{2}\geq0 $$

and

$$ 0< H(0)\leq H(t)\leq\frac{1}{p} \Vert u \Vert ^{p}_{p}. $$

(5.1)

Furthermore, we define

$$ L(t)=H^{1-\alpha}(t)+\epsilon \int_{\Omega}uu_{t}\,dx, $$

(5.2)

where ϵ is a small constant and will be chosen later, $0<\alpha <\frac{p-2}{2p}$.

By taking a time derivative of (5.2), we get

$$\begin{aligned} L'(t) =&(1-\alpha)H^{-\alpha}(t)H'(t)+\epsilon \int_{\Omega } \vert u_{t} \vert ^{2}\,dx+ \epsilon \int_{\Omega}uu_{tt}\,dx \\ =&(1-\alpha)H^{-\alpha}(t)H'(t)+\epsilon \int_{\Omega } \vert u_{t} \vert ^{2}\,dx- \epsilon \int_{\Omega} \vert \nabla u \vert ^{2}\,dx-\epsilon \int _{\Omega}u_{t}u\,dx \\ &{}+\epsilon \int_{\Omega}\nabla u \int^{t}_{0}g(t-\tau)\nabla u(\tau )\,d\tau \,dx- \epsilon \int_{\Omega}\nabla u \cdot\nabla u_{t}\,dx+\epsilon \int _{\Omega} \vert u \vert ^{p}\,dx. \end{aligned}$$

(5.3)

Using Young’s and Schwarz’s inequalities, we obtain

$$\begin{aligned} & \int_{\Omega}\nabla u \int^{t}_{0}g(t-\tau)\nabla u(\tau)\,d\tau \,dx \\ &\quad = \int_{\Omega}\nabla u \int^{t}_{0}g(t-\tau) \bigl(\nabla u(\tau)-\nabla u(t) \bigr)\,d\tau \,dx+ \int_{\Omega}\nabla u \int^{t}_{0}g(t-\tau)\nabla u(t)\,d\tau \,dx \\ &\quad \geq -\delta \Vert \nabla u \Vert ^{2}_{2}- \frac{1}{4\delta}\biggl( \int^{t}_{0}g(\tau )\,d\tau\biggr) (g\circ\nabla u) (t)+ \biggl( \int^{t}_{0}g(\tau)\,d\tau \biggr) \Vert \nabla u \Vert ^{2}_{2}, \end{aligned}$$

(5.4)

$$\begin{aligned} &\int_{\Omega}\nabla u\cdot\nabla u_{t}\,dx\geq-\gamma \Vert \nabla u_{t} \Vert ^{2}_{2}- \frac{1}{4\gamma} \Vert \nabla u \Vert ^{2}_{2}, \end{aligned}$$

(5.5)

$$\begin{aligned} &\int_{\Omega}u_{t}u\,dx\geq-\frac{\delta^{2}}{2} \Vert u \Vert ^{2}_{2}-\frac {\delta^{-2}}{2} \Vert u_{t} \Vert ^{2}_{2}, \end{aligned}$$

(5.6)

where δ and γ are positive constants.

Inserting (5.4), (5.5), and (5.6) into (5.3), we deduce

$$\begin{aligned} L'(t) \geq&(1-\alpha)H^{-\alpha}(t)H'(t)+ \epsilon\biggl(1-\frac{\delta ^{-2}}{2}\biggr) \Vert u_{t} \Vert ^{2}_{2}-\gamma\epsilon \Vert \nabla u_{t} \Vert ^{2}_{2} \\ &{}+\epsilon\biggl(-1-\delta+ \int^{t}_{0}g(\tau) \,d\tau-\frac{1}{4\gamma } \biggr) \Vert \nabla u \Vert ^{2}_{2}-\frac{\epsilon}{4\delta} \int^{t}_{0}g(\tau)\,d\tau (g\circ\nabla u) (t) \\ &{}-\frac{\delta^{2}}{2}\epsilon \Vert u \Vert ^{2}_{2}+ \epsilon \Vert u \Vert ^{p}_{p}. \end{aligned}$$

(5.7)

From (2.3), we know that

$$\begin{aligned} - \Vert \nabla u_{t} \Vert ^{2}_{2}\geq E'(t)=-H'(t). \end{aligned}$$

(5.8)

If we set $\delta^{2}=kH^{\alpha}(t)$, $\delta^{-2}=k^{-1}H^{-\alpha }(t)$, where $k>0$, then we have

$$ -\frac{\delta^{2}}{2}\epsilon \Vert u \Vert ^{2}_{2}=- \frac{\epsilon}{2}kH^{\alpha }(t) \Vert u \Vert ^{2}_{2}. $$

(5.9)

Using (2.1) and (5.1), we obtain

$$ H^{\alpha}(t) \Vert u \Vert ^{2}_{2}\leq C \biggl(\frac{1}{p}\biggr)^{\alpha} \Vert u \Vert ^{2+\alpha p}_{p}, $$

(5.10)

where we let $2\leq2+\alpha p\leq p$, then $0<\alpha\leq\frac{p-2}{p}$.

Using Corollary 2.4 and (2.2), inserting (5.10) and (5.8) into (5.7), we obtain

$$\begin{aligned} L'(t) \geq& \bigl[(1-\alpha)H^{-\alpha}(t)-\gamma\epsilon \bigr]H'(t)+\epsilon \biggl(p-\frac{kC}{2}\biggl( \frac{1}{p}\biggr)^{\alpha} \biggr)H(t) \\ &{}+\epsilon \biggl\{ \frac{p-2}{2}\biggl(1- \int^{t}_{0}g(\tau)\,d\tau\biggr)-\delta - \frac{1}{4\gamma}-\frac{kC}{2}\biggl(\frac{1}{p} \biggr)^{\alpha} \biggr\} \Vert \nabla u \Vert ^{2}_{2} \\ &{}+\epsilon \biggl\{ \frac{p}{2}-\frac{1}{4\delta} \int^{t}_{0}g(\tau )\,d\tau-\frac{kC}{2} \biggl(\frac{1}{p}\biggr)^{\alpha} \biggr\} (g\circ\nabla u) (t) \\ &{}+\epsilon \biggl\{ \frac{p+2}{2}-\frac{1}{2k}H^{-\alpha}(t)- \frac {kC}{2}\biggl(\frac{1}{p}\biggr)^{\alpha} \biggr\} \Vert u_{t} \Vert ^{2}_{2}. \end{aligned}$$

(5.11)

By using the hypothesis in Theorem 5.1 and taking $k,\gamma$ and δ suitable such that

$$\begin{aligned} &\frac{p-2}{2}\biggl(1- \int^{t}_{0}g(\tau)\,d\tau\biggr)-\delta- \frac{1}{4\gamma}-\frac {kC}{2}\biggl(\frac{1}{p} \biggr)^{\alpha}>0, \\ &\frac{p}{2}-\frac{1}{4\delta} \int^{t}_{0}g(\tau)\,d\tau-\frac{kC}{2}\biggl( \frac {1}{p}\biggr)^{\alpha}>0, \\ &\frac{p+2}{2}-\frac{1}{2k}H^{-\alpha}(t)-\frac{kC}{2} \biggl(\frac {1}{p}\biggr)^{\alpha}>0. \end{aligned}$$

When $k,\gamma$ is fixed, we choose ϵ small enough such that

$$(1-\alpha)H^{-\alpha}(t)-\gamma\epsilon>0,\qquad L(0)=H^{1-\alpha }(0)+ \epsilon \int_{\Omega}u_{0}u_{1}\,dx>0. $$

Then we can deduce that

$$L'(t)\geq C \bigl[H(t)+ \Vert u_{t} \Vert ^{2}_{2}+ \Vert \nabla u \Vert ^{2}_{2}+(g \circ \nabla u) (t) \bigr]. $$

By using Hölder’s and Young’s inequalities, we get

$$\begin{aligned} \biggl\vert \int_{\Omega}uu_{t}\,dx \biggr\vert ^{1/(1-\alpha)} \leq& \Vert u \Vert ^{1/(1-\alpha )}_{2} \Vert u_{t} \Vert ^{1/(1-\alpha)}_{2} \\ \leq&C \Vert u \Vert ^{1/(1-\alpha)}_{p} \Vert u_{t} \Vert ^{1/(1-\alpha)}_{2} \\ \leq&C\bigl( \Vert u \Vert ^{s}_{p}+ \Vert u_{t} \Vert ^{2}_{2}\bigr) \\ \leq&C \bigl(H(t)+ \Vert u_{t} \Vert ^{2}_{2}+(g \circ\nabla u) (t)+ \Vert \nabla u \Vert ^{2}_{2} \bigr), \end{aligned}$$

(5.12)

where $2\leq s:=\frac{2}{1-2\alpha}\leq p$, then $0<\alpha<\frac {p-2}{2p}$. Hence

$$\begin{aligned} L^{1/(1-\alpha)}(t) =& \biggl(H^{1-\alpha}(t)+\epsilon \int_{\Omega }uu_{t}\,dx \biggr)^{1/(1-\alpha)} \\ \leq&2^{1/(1-\alpha)} \biggl(H(t)+ \biggl\vert \int_{\Omega}uu_{t}\,dx \biggr\vert ^{1/(1-\alpha)} \biggr) \\ \leq&C \bigl(H(t)+ \Vert u_{t} \Vert ^{2}_{2}+ \Vert \nabla u \Vert ^{2}_{2}+(g\circ\nabla u) (t) \bigr). \end{aligned}$$

(5.13)

Then we can get

$$L'(t)\geq\lambda L^{\frac{1}{1-\alpha}}(t),\quad t>0. $$

Therefore

$$L(t)\geq \biggl(L^{\frac{-\alpha}{1-\alpha}}(0)+\frac{-\alpha}{1-\alpha }\lambda t \biggr)^{-\frac{1-\alpha}{\alpha}}. $$

So $L(t)$ tends to infinity when t tends to $(1-\alpha)/ (\alpha \lambda L^{\frac{\alpha}{1-\alpha}}(0) )$. □

To get another blow-up result, we first give the following lemma.

Lemma 5.2

If (H1), (H2) hold, assume further that

$$\Vert u_{0} \Vert _{p}>\lambda_{0}\equiv B_{0}^{\frac{-2}{p-2}},\qquad E(0)< E_{0}=\biggl( \frac{1}{2}-\frac{1}{p}\biggr) B_{0}^{\frac{-2p}{p-2}}. $$

Then

$$\Vert u \Vert _{p}>\lambda_{0},\qquad \Vert \nabla u \Vert _{2}>B_{0}^{\frac {-p}{p-2}},\quad\forall t\geq0, $$

where $B_{0}=\frac{B}{l^{1/2}}$ for $\|u\|_{p}\leq B\|\nabla u\|_{2}$.

Proof

By using (2.2) and the hypothesis, we obtain

$$\begin{aligned} E(t) =&\frac{1}{2} \bigl\Vert u_{t}(t) \bigr\Vert ^{2}_{2}+\frac{1}{2} \biggl(1- \int ^{t}_{0}g(\tau)\,d\tau \biggr) \bigl\Vert \nabla u(t) \bigr\Vert ^{2}_{2}+\frac{1}{2}(g\circ \nabla u) (t)-\frac{1}{p} \bigl\Vert u(t) \bigr\Vert ^{p}_{p} \\ \geq&\frac{1}{2}\biggl(1- \int^{t}_{0}g(\tau)\,d\tau\biggr) \bigl\Vert \nabla u(t) \bigr\Vert ^{2}_{2}-\frac{1}{p} \bigl\Vert u(t) \bigr\Vert ^{p}_{p} \\ \geq&\frac{l}{2} \bigl\Vert \nabla u(t) \bigr\Vert ^{2}_{2}-\frac{1}{p} \bigl\Vert u(t) \bigr\Vert ^{p}_{p} \\ \geq&\frac{1}{2B^{2}_{0}} \bigl\Vert u(t) \bigr\Vert ^{2}_{p}- \frac{1}{p} \bigl\Vert u(t) \bigr\Vert ^{p}_{p}. \end{aligned}$$

(5.14)

We set $h(\xi)=\frac{1}{2B^{2}_{0}}\xi^{2}-\frac{1}{p}\xi^{p}$, $\xi >0$. Then $h(\xi)$ satisfies

• $h(\xi)$ is strictly increasing on $[0,\lambda_{0})$;

• $h(\xi)$ takes its maximum value $(\frac{1}{2}-\frac {1}{p}) B_{0}^{\frac{-2p}{p-2}}$ at $\lambda_{0}$;

• $h(\xi)$ is strictly decreasing on $(\lambda_{0}, \infty)$.

Since $E_{0}>E(0)\geq E(t)\geq h(\|u\|_{p})$ for $\forall t\geq0$, there is no time t such that $\|u\|_{p}=\lambda_{0}$. By the continuity, we obtain

$$\bigl\Vert u(\cdot, t) \bigr\Vert _{p}>\lambda_{0}=B_{0}^{\frac{-2}{p-2}}, \quad\forall t \geq0. $$

Then

$$\bigl\Vert \nabla u(\cdot, t) \bigr\Vert _{2}\geq \frac{1}{l^{1/2}B_{0}} \bigl\Vert u(\cdot, t) \bigr\Vert _{p}> \frac{1}{l^{1/2}}B_{0}^{\frac{-p}{p-2}}>B_{0}^{\frac{-p}{p-2}}. $$

This completes the proof. □

Theorem 5.3

If that (H1), (H2), and (H3) hold, suppose further that

$$\begin{aligned} \int^{\infty}_{0}g(\tau)\,d\tau< \frac{(p/2)-1}{p/2-1+(1/2p)}, \end{aligned}$$

$\|u_{0}\|_{p}>\lambda_{0}$ and $E(0)\leq E_{0}$. Then the solution of (1.1) blows up in finite time.

Proof

Set $G(t)=E_{0}+H(t)$, then

$$G'(t)=-\frac{1}{2}\bigl(g'\circ\nabla u\bigr) (t)+\frac{1}{2}g(t) \bigl\Vert \nabla u(t) \bigr\Vert ^{2}_{2}+ \Vert \nabla u_{t} \Vert ^{2}_{2}+ \Vert u_{t} \Vert ^{2}_{2}\geq0, $$

from which we obtain

$$\begin{aligned} 0< G(t) =&E_{0}+H(t)\\ =&\biggl(\frac{1}{2}-\frac{1}{p} \biggr) B_{0}^{\frac {-2p}{p-2}}+H(t) \\ \leq&\biggl(\frac{1}{2}-\frac{1}{p}\biggr) \Vert \nabla u \Vert ^{2}_{2}+H(t) \\ \leq&C \bigl( \Vert \nabla u \Vert ^{2}_{2}+H(t) \bigr). \end{aligned}$$

By using Lemma 5.2, we have

$$\begin{aligned} 0 < &G(t) \\ =&E_{0}-\frac{1}{2} \Vert u_{t} \Vert ^{2}_{2}-\frac{1}{2}\biggl(1- \int^{t}_{0}g(\tau )\,d\tau\biggr) \bigl\Vert \nabla u(t) \bigr\Vert ^{2}_{2}\\ &{}-\frac{1}{2}(g\circ \nabla u) (t)+\frac {1}{p} \bigl\Vert u(t) \bigr\Vert ^{p}_{p} \\ \leq&E_{0}-\frac{1}{2}\biggl(1- \int^{t}_{0}g(\tau)\,d\tau\biggr) \bigl\Vert \nabla u(t) \bigr\Vert ^{2}_{2}+\frac{1}{p} \bigl\Vert u(t) \bigr\Vert ^{p}_{p} \\ \leq&\biggl(\frac{1}{2}-\frac{1}{p}\biggr)B_{0}^{\frac{-2p}{p-2}}- \frac{l}{2}\biggl(\frac {1}{l^{1/2}}\biggr)^{2}B_{0}^{\frac{-2p}{p-2}}+ \frac{1}{p} \bigl\Vert u(t) \bigr\Vert ^{p}_{p} \\ \leq&\frac{1}{p} \bigl\Vert u(t) \bigr\Vert ^{p}_{p}. \end{aligned}$$

Let

$$Q(t)=G^{1-\alpha}(t)+\epsilon \int_{\Omega}uu_{t}\,dx, $$

with ϵ small to be chosen later and $0<\alpha<\frac{p-2}{2p}$.

By the same computations as in the proof of Theorem 5.1, we can deduce that

$$Q'(t)\geq C \bigl[H(t)+ \Vert u_{t} \Vert ^{2}_{2}+ \Vert \nabla u \Vert ^{2}_{2}+(g \circ \nabla u) (t) \bigr]. $$

Observing (5.13), we see that

$$\begin{aligned} Q^{1/(1-\alpha)}(t) =& \biggl(G^{1-\alpha}(t)+\epsilon \int_{\Omega }uu_{t}\,dx \biggr)^{1/(1-\alpha)} \\ \leq&2^{1/(1-\alpha)} \biggl(G(t)+ \biggl\vert \int_{\Omega}uu_{t}\,dx \biggr\vert ^{1/(1-\alpha)} \biggr) \\ \leq&C \bigl(H(t)+ \Vert u_{t} \Vert ^{2}_{2}+ \Vert \nabla u \Vert ^{2}_{2}+(g\circ\nabla u) (t) \bigr). \end{aligned}$$

(5.15)

Then we can obtain

$$\begin{aligned} Q'(t)\geq\lambda Q^{\frac{1}{1-\alpha}}(t),\quad t>0. \end{aligned}$$

(5.16)

Therefore, we get

$$Q(t)\geq \biggl(Q^{\frac{-\alpha}{1-\alpha}}(0)+\frac{-\alpha}{1-\alpha }\lambda t \biggr)^{-\frac{1-\alpha}{\alpha}}, \quad t>0. $$

So $Q(t)$ tends to infinity when t tends to $(1-\alpha)/ (\alpha \lambda Q^{\frac{\alpha}{1-\alpha}}(0) )$. □