On decay and blow-up of solutions for a system of viscoelastic equations with weak damping and source terms

Abstract

In this article, we investigate a system of two viscoelastic equations with Dirichlet boundary conditions. Under some suitable assumptions on the function $g_{i}(\cdot )$, $f_{i}(\cdot ,\cdot )$ ($i=1,2$) and the initial data, we obtain general and optimal decay results. Moreover, for certain initial data, we establish a finite time blow-up result. This work generalizes and improves earlier results in the literature. The conditions of the relaxation functions $g_{1}(t)$ and $g_{2}(t)$ in our work are weak and seldom appear in previous literature, which is an important breakthrough.

1 Introduction

In this paper, we investigate the following initial-boundary problem:

$$\begin{aligned} \textstyle\begin{cases} u_{tt}-\Delta u+\int _{0}^{t}g_{1}(t-\tau )\Delta u(\tau )\,d\tau +u_{t}=f _{1}(u,v), \quad x\in \varOmega , t>0, \\ v_{tt}-\Delta v+\int _{0}^{t}g_{2}(t-\tau )\Delta v(\tau )\,d\tau +v _{t}=f_{2}(u,v), \quad x\in \varOmega , t>0, \\ u(x,t)=v(x,t)=0, \quad x\in \partial \varOmega , t\geq 0, \\ u(x,0)=u_{0}, \quad\quad u_{t}(x,0)=u_{1}, \quad\quad v(x,0)=v_{0}, \quad\quad v_{t}(x,0)=v_{1}, \quad x \in \varOmega , \end{cases}\displaystyle \end{aligned}$$

(1.1)

where Ω is a bounded domain of $R^{n}$ ($n\geq 1$) with a smooth boundary ∂Ω, u and v represent the transverse displacements of wave. The functions $g_{1}$ and $g_{2}$ denote the kernel of the memory term, and the nonlinear functions $f_{1}$ and $f_{2}$ will be specified later. Problem (1.1) describes the propagation of some material possessing a capacity of storage and dissipation of mechanical energy. Models of this type arise in the theory of viscoelasticity and physics.

Firstly, we present some results related to viscoelastic equations. The single viscoelastic equation of the form

$$ \textstyle\begin{cases} \vert u_{t} \vert ^{\rho }u_{tt}-\Delta u-\Delta u_{tt}+\int _{0}^{t}g(t-\tau ) \Delta u(\tau )\,d\tau +a \vert u_{t} \vert ^{m-2}u_{t}=b \vert u \vert ^{p-2}u, \\ \quad \text{in } \varOmega \times (0,+\infty ), \\ u(x,t)=0, \quad \text{on } \partial \varOmega \times (0,+\infty ), \\ u(x,0)=u_{0}(x), \quad\quad u_{t}(x,0)=u_{1}(x), \quad \text{in } \varOmega , \end{cases} $$

(1.2)

which has been extensively studied by many authors, and related results concerning existence, decay, and blow-up have been recently established (see [6, 11]). Here, we understand $\triangle u_{tt}$, $\vert u _{t} \vert ^{m-2}u_{t}$, $\int _{0}^{t}g(t-\tau )\bigtriangleup u(\tau )d \tau$, and $\vert u \vert ^{p-2}u$ to be the dispersion term, weak damping term, viscoelasticity dissipative term, and source term, respectively. This type of problem usually appears as a model in nonlinear viscoelasticity.

As $a=b=0$ in the presence of the strong damping term, Cavalcanti et al. [1] dealt with the equation

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

(1.3)

where Ω is a bounded domain with smooth boundary ∂Ω in $R^{n}$ ($n\geq 1$) and $\rho >0$. They established a global existence result when the constant $\gamma \geq 0$ and an exponential decay result for the case $\gamma >0$. Later, this result was improved by Messaoudi and Tatar [8] to a situation where a source term is presented. By using perturbation techniques, they established a global existence and an exponential decay result.

When $a=b=1$ in the absence of the dispersion term, the following problem

$$\begin{aligned} \textstyle\begin{cases} \vert u_{t} \vert ^{\rho }u_{tt}-\Delta u+\int _{0}^{t}g(t-\tau )\Delta u(\tau )d \tau + \vert u_{t} \vert ^{m-2}u_{t}= \vert u \vert ^{p-2}u, \quad \text{in }\varOmega \times [0,T], \\ u(x,t)=0,\quad x\in \partial \varOmega , \\ u(x,0)=u_{0}(x), \quad\quad u_{t}(x,0)=u_{1}(x), \quad x\in \varOmega , \end{cases}\displaystyle \end{aligned}$$

(1.4)

in a bounded domain and $m>2$, $p>2$ was studied by Song [10]. The author proved the nonexistence of global solutions of problem (1.4) with bounded positive initial energy.

The case of $\rho =0$ in the absence of the viscoelasticity term problem (1.2) has been discussed by many authors. For example, Chen and Liu [2] studied the following equation:

$$\begin{aligned} u_{tt}-\Delta u-\omega \Delta u_{t}+a \vert u_{t} \vert ^{m-2}u_{t}=b \vert u \vert ^{p-2}u \end{aligned}$$

(1.5)

subject to the same boundary and initial conditions as problem (1.4). Under some suitable conditions on the initial data, they proved the global existence of solutions in both cases which are polynomial and exponential decay in the energy space respectively.

In the case of $a=b=1$, $m=2$, in the presence of the strong damping term and without the dispersion term, the following viscoelastic equation

$$ \textstyle\begin{cases} u_{tt}-\Delta u+\int _{0}^{t}g(t-\tau )\Delta u(\tau )\,d\tau -\Delta u _{t}+u_{t}= \vert u \vert ^{p-2}u, \quad \text{in }\varOmega \times (0,+\infty ), \\ u(x,t)=0, \quad \text{on }\partial \varOmega \times (0,+\infty ), \\ u(x,0)=u_{0}(x), \quad\quad u_{t}(x,0)=u_{1}(x), \quad \text{in }\varOmega , \end{cases} $$

(1.6)

was studied by Li and He [4]. By using some properties of the convex functions, they obtained a general decay rate result. Moreover, they established a finite time blow-up result for solutions with negative initial energy and positive initial energy.

For a coupled system, the following system of viscoelastic equations

$$\begin{aligned} \textstyle\begin{cases} u_{tt}-\Delta u+\int _{0}^{t}g_{1}(t-\tau )\Delta u(\tau )\,d\tau +f_{1}(u,v)=0, \quad x\in \varOmega , t>0, \\ v_{tt}-\Delta v+\int _{0}^{t}g_{2}(t-\tau )\Delta v(\tau )\,d\tau +f _{2}(u,v)=0, \quad x\in \varOmega , t>0, \\ u(x,t)=v(x,t)=0, \quad x\in \partial \varOmega , t\geq 0, \\ u(x,0)=u_{0}(x), \quad\quad u_{t}(x,0)=u_{1}(x), \quad x\in \varOmega , \\ v(x,0)=v_{0}(x), \quad\quad v_{t}(x,0)=v_{1}(x), \quad x\in \varOmega , \end{cases}\displaystyle \end{aligned}$$

(1.7)

was considered by Mustafa [9]. Under some suitable conditions, they proved the well-posedness and obtained a generalized stability result.

In the work [5], Liu considered the following problem:

$$\begin{aligned} \textstyle\begin{cases} \vert u_{t} \vert ^{\rho }u_{tt}-\Delta u-\gamma _{1}\Delta u_{tt}+\int _{0}^{t}g(t- \tau )\Delta u(\tau )\,d\tau +f(u,v)=0, \quad x\in \varOmega , t>0, \\ \vert v_{t} \vert ^{\rho }v_{tt}-\Delta v-\gamma _{2}\Delta v_{tt} +\int _{0}^{t}h(t- \tau )\Delta v(\tau )\,d\tau +k(u,v)=0, \quad x\in \varOmega , t>0, \\ u(x,t)=v(x,t)=0, \quad x\in \partial \varOmega , t\geq 0, \\ u(x,0)=u_{0}(x), \quad\quad u_{t}(x,0)=u_{1}(x), \quad x\in \varOmega , \\ v(x,0)=v_{0}(x), \quad\quad v_{t}(x,0)=v_{1}(x), \quad x\in \varOmega , \end{cases}\displaystyle \end{aligned}$$

(1.8)

where $\gamma _{1},\gamma _{2}>0$ and $0<\rho \leq \frac{2}{n-2}$ if $n\geq 3$ or $\rho >0$ if $n=1,2$. Under the following assumptions on the relaxation functions:

$$ g_{1}'(t)\leq -\zeta _{1}(t)g_{1}^{p}(t), \quad\quad g_{2}'(t)\leq -\zeta _{2}(t)g_{2}^{q}(t), \quad t\geq 0, $$

where $\zeta _{1}$ and $\zeta _{2}$ are positive constants. By exploiting the perturbed energy method, a uniform decay of the energy result was established.

Recently, Houari et al. [3] investigated the following system:

$$\begin{aligned} \textstyle\begin{cases} u_{tt}-\Delta u+\int _{0}^{t}g_{1}(t-\tau )\Delta u(\tau )\,d\tau + \vert u _{t} \vert ^{m-1}u_{t}=f_{1}(u,v), \quad \text{in } \varOmega \times (0,+\infty ), \\ v_{tt}-\Delta v+\int _{0}^{t}g_{2}(t-\tau )\Delta v(\tau )\,d\tau + \vert v _{t} \vert ^{r-1}v_{t} =f_{2}(u,v), \quad \text{in }\varOmega \times (0,+\infty ), \\ u(x,t)=v(x,t)=0,\quad \text{on } \partial \varOmega \times (0,+\infty ), \\ u(x,0)=u_{0}(x), \quad\quad u_{t}(x,0)=u_{1}(x),\quad \text{in }\varOmega , \\ v(x,0)=v_{0}(x), \quad\quad v_{t}(x,0)=v_{1}(x),\quad \text{in }\varOmega , \end{cases}\displaystyle \end{aligned}$$

(1.9)

where Ω is an open bounded domain of $R^{n}$ with a smooth boundary ∂Ω, m, $r\geq 1$. Under some suitable conditions, they obtained a general decay, which depends on the relaxation functions.

As for a single wave equation, in the absence of the source term, the damping term assures global existence. On the other hand, without the damping term, the source term may cause finite time blow-up of solution. Hence, it is valuable to study the viscoelastic equation with damping and source terms.

Our aim in this work is to establish the global existence, decay, and blow-up result of solutions to problem (1.1). By adopting and modifying the methods used in [9], we establish the general and optimal decay and blow-up results, while we should overcome the additional difficulty caused by the changes of the conditions of the relaxation functions $g_{1}(t)$ and $g_{2}(t)$. In [9], the relaxation functions $g_{i}(t)$ ($i=1,2$) satisfy $g_{i}'(t)\leq - \zeta _{i}g_{i}(t)$ for all $t\geq 0$, where $\zeta _{i}(t)$ are positive nonincreasing functions. In this paper, the conditions have been replaced by $g_{i}'(t)\leq -\zeta _{i}(t)g_{i}^{\gamma }(t)$, $\gamma \in [1, 3/2)$. As far as we know, the conditions of the relaxation functions $g_{i}(t)$ in our work seldom appear in previous literature, which is an important breakthrough. Our main novel contribution is an extension and improvement of the previous result from [9].

This paper is organized as follows. In Sect. 2, we give material needed for our work. In Sect. 3, we prove the global existence. In Sect. 4, we present some technical lemmas needed in the proof of our result. Section 5 is devoted to the general decay result. In Sect. 6, we carry out the proof of finite time blow-up result.

2 Preliminaries

In this section, we present some material needed for our work. First, we make the following assumptions.

• (G1) $g_{i}: \mathbb{R_{+}}\rightarrow \mathbb{R_{+}}$ (for $i=1,2$) are $C^{1}$ nonincreasing functions satisfying

  $$ 1- \int _{0}^{\infty }g_{1}(\tau )\,d\tau =l>0,\quad\quad 1- \int _{0}^{\infty }g_{2}(\tau )\,d\tau =k>0, \quad g_{i}(0)>0. $$

  (2.1)

• (G2) There exist two nonincreasing differentiable functions $\zeta _{1}, \zeta _{2}:\mathbb{R_{+}}\rightarrow \mathbb{R_{+}}$, with $\zeta _{1}(0), \zeta _{2}(0)>0$ and satisfying

  $$ g_{i}'(t)\leq -\zeta _{i}(t)g_{i}^{\gamma }(t), \quad t\geq 0, 1\leq \gamma < \frac{3}{2}, \text{for }i=1,2. $$

  (2.2)

• (G3) For the functions $f_{1}$ and $f_{2}$, we note that

  $$\begin{aligned}& f_{1}(u,v)= \vert u+v \vert ^{2(p+1)}(u+v)+ \vert u \vert ^{p}u \vert v \vert ^{p+2}, \end{aligned}$$

  (2.3)
  $$\begin{aligned}& f_{2}(u,v)= \vert u+v \vert ^{2(p+1)}(u+v)+ \vert u \vert ^{p+2}v \vert v \vert ^{p}, \end{aligned}$$

  (2.4)

  where p satisfies

  $$ p>-1 \quad \text{if } n=1,2 \quad \text{and} \quad -1< p\leq \frac{3-n}{n-2} \quad \text{if } n\geq 3. $$

  (2.5)

It is easy to verify that

$$\begin{aligned} uf_{1}+vf_{2}=2(p+2)F(u,v), \end{aligned}$$

(2.6)

where

$$\begin{aligned} F(u,v)=\frac{1}{2(p+2)} \bigl[ \vert u+v \vert ^{2(p+2)}+2 \vert uv \vert ^{p+2} \bigr]. \end{aligned}$$

(2.7)

Remark 2.1

There are many functions $g_{i}(t)$ and $\zeta _{i}(t)$ (for $i=1,2$) satisfying (G1) and (G2). An example of such functions is

$$ \begin{gathered} g_{1}(t)=\frac{1}{(1+t)^{4}}, \quad\quad \zeta _{1}(t)=\frac{4}{1+t}, \\ g_{2}(t)= \frac{1}{16(1+t)^{4}}, \quad\quad \zeta _{2}(t)=\frac{8}{1+t}, \quad\quad \gamma =\frac{5}{4}. \end{gathered} $$

(2.8)

Definition 2.2

A pair of functions $(u,v)$ defined on $[0,T]$ is called a weak solution of system (1.1) if $u,v\in C_{w}([0,T],H_{0}^{1}(\varOmega ))$, $u _{t},v_{t}\in C_{w}([0,T],L^{2}(\varOmega ))$, $(u(x,0),v(x,0))=(u_{0},v _{0})\in H_{0}^{1}(\varOmega )\times H_{0}^{1}(\varOmega )$, $(u_{t}(x,0),v _{t}(x,0))=(u_{1},v_{1})\in H_{0}^{1}(\varOmega )\times H_{0}^{1}(\varOmega )$ and $(u,v)$ satisfies

$$\begin{aligned}& \begin{aligned}[b] & \int _{\varOmega }u_{tt}\phi \,dx- \int _{0}^{t}g_{1}(t-\tau ) \int _{\varOmega } \nabla \phi (\tau )\nabla u(\tau )\,dx\,d\tau \\ &\quad {} + \int _{\varOmega }\nabla \phi \nabla u\,dx+ \int _{\varOmega }\phi u _{t}\,dx= \int _{\varOmega }f_{1}(u,v)\phi \,dx, \end{aligned} \end{aligned}$$

(2.9)

$$\begin{aligned}& \begin{aligned}[b] & \int _{\varOmega }v_{tt}\psi \,dx- \int _{0}^{t}g_{2}(t-\tau ) \int _{\varOmega } \nabla \psi (\tau )\nabla v(\tau )\,dx\,d\tau \\ &\quad {} + \int _{\varOmega }\nabla \psi \nabla v\,dx+ \int _{\varOmega }\psi v _{t}\,dx= \int _{\varOmega }f_{2}(u,v)\psi \,dx \end{aligned} \end{aligned}$$

(2.10)

for a.e. $t\in [0,T]$ and all test functions $\phi ,\psi \in H_{0} ^{1}(\varOmega )$. Here, $C_{w}([0,T],X)$ denotes the space of weakly continuous functions from $[0,T]$ into Banach space X.

Proposition 2.3

Let $(u_{0},v_{0})\in H_{0}^{1}(\varOmega )\times H_{0}^{1}(\varOmega )$ and $(u_{1},v_{1})\in L^{2}(\varOmega )\times L^{2}(\varOmega )$. Assume that (G1)–(G3) hold. Then there exists a unique local weak solution $(u,v)$ of system (1.1) defined in $[0,T_{m}]$ for some $T_{m}>0$ small enough.

3 Global existence

In this section, we prove the global existence of the solution to system (1.1).

First, we define

$$\begin{aligned}& \begin{aligned}[b] I(t) &=\biggl(1- \int _{0}^{t}g_{1}(\tau )\,d\tau \biggr) \Vert \nabla u \Vert _{2}^{2}+\biggl(1- \int _{0}^{t}g_{2}(\tau )\,d\tau \biggr) \Vert \nabla v \Vert _{2}^{2} \\ &\quad {} +(g_{1}\circ \nabla u) (t)+(g_{2}\circ \nabla v) (t)-2(p+2) \int _{\varOmega }F(u,v)\,dx, \end{aligned} \end{aligned}$$

(3.1)

$$\begin{aligned}& \begin{aligned}[b] J(t) &=\frac{1}{2}\biggl(1- \int _{0}^{t}g_{1}(\tau )\,d\tau \biggr) \Vert \nabla u \Vert _{2} ^{2}+ \frac{1}{2}\biggl(1- \int _{0}^{t}g_{2}(\tau )\,d\tau \biggr) \Vert \nabla v \Vert _{2} ^{2} \\ &\quad {} +\frac{1}{2}(g_{1}\circ \nabla u) (t)+ \frac{1}{2}(g_{2}\circ \nabla v) (t)- \int _{\varOmega }F(u,v)\,dx, \end{aligned} \end{aligned}$$

(3.2)

and the energy function associated with system (1.1)

$$\begin{aligned} E(t) &=\frac{1}{2} \Vert u_{t} \Vert _{2}^{2}+\frac{1}{2} \Vert v_{t} \Vert _{2}^{2}+ \frac{1}{2}\biggl(1- \int _{0}^{t}g_{1}(\tau )\,d\tau \biggr) \Vert \nabla u \Vert _{2}^{2}+ \frac{1}{2}(g_{1}\circ \nabla u) (t) \\ &\quad {} +\frac{1}{2}\biggl(1- \int _{0}^{t}g_{2}(\tau )\,d\tau \biggr) \Vert \nabla v \Vert _{2} ^{2}+ \frac{1}{2}(g_{2}\circ \nabla v) (t)- \int _{\varOmega }F(u,v)\,dx, \end{aligned}$$

(3.3)

where

$$\begin{aligned}& (g_{1}\circ \nabla u) (t)= \int _{0}^{t}g_{1}(t-\tau ) \bigl\Vert \nabla u(t)- \nabla u(\tau ) \bigr\Vert _{2}^{2} \,d\tau , \end{aligned}$$

(3.4)

$$\begin{aligned}& (g_{2}\circ \nabla v) (t)= \int _{0}^{t}g_{2}(t-\tau ) \bigl\Vert \nabla v(t)- \nabla v(\tau ) \bigr\Vert _{2}^{2} \,d\tau . \end{aligned}$$

(3.5)

Lemma 3.1

Let (G1)–(G3) hold and $(u,v)$ be the solution of (1.1), then we have

$$\begin{aligned} \frac{dE(t)}{dt} &=\frac{1}{2} \bigl[\bigl(g_{1}' \circ \nabla u\bigr) (t)+\bigl(g_{2}' \circ \nabla v \bigr) (t) \bigr]- \Vert u_{t} \Vert _{2}^{2} \\ &\quad {} - \Vert v_{t} \Vert _{2}^{2}- \frac{1}{2} \bigl[g_{1}(t) \Vert \nabla u \Vert _{2} ^{2}-g_{2}(t) \Vert \nabla v \Vert _{2}^{2} \bigr]\leq 0. \end{aligned}$$

(3.6)

Proof

Multiplying the first equation in system (1.1) by $u_{t}$ and integrating over Ω gives

$$\begin{aligned} & \int _{\varOmega }u_{t}u_{tt}\,dx- \int _{0}^{t}g_{1}(t-\tau ) \int _{\varOmega } \nabla u_{t}(t)\nabla u(\tau )\,dx\,d \tau \\ &\quad {} + \int _{\varOmega }\nabla u_{t}\nabla u\,dx+ \Vert u_{t} \Vert _{2}^{2}=- \int _{\varOmega }u_{t}f_{1}(u,v)\,dx. \end{aligned}$$

(3.7)

Then

$$\begin{aligned} &\frac{d}{dt} \biggl\{ \frac{1}{2} \int _{\varOmega } \vert u_{t} \vert ^{2}\,dx+ \frac{1}{2} \int _{\varOmega } \vert \nabla u \vert ^{2}\,dx \biggr\} + \Vert u_{t} \Vert _{2}^{2} \\ &\quad {} - \int _{0}^{t}g_{1}(t-\tau ) \int _{\varOmega }\nabla u_{t}(t) \nabla u(\tau )\,dx\,d \tau =- \int _{\varOmega }u_{t}f_{1}(u,v)\,dx. \end{aligned}$$

(3.8)

For the last term on the left-hand side of (3.8), we get

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

(3.9)

By combining (3.4), (3.8), and (3.9), we deduce

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

(3.10)

Similarly, multiplying the second equation in (1.1) by $v_{t}$ and integrating over Ω yields

$$\begin{aligned} &\frac{d}{dt} \biggl\{ \frac{1}{2} \int _{\varOmega } \vert v_{t} \vert ^{2}\,dx+ \frac{1}{2} \int _{\varOmega } \vert \nabla v \vert ^{2}\,dx \biggr\} -\frac{1}{2} \frac{d}{dt} \biggl[ \int _{0}^{t}g_{2}(\tau ) \bigl\Vert \nabla v(t) \bigr\Vert _{2}^{2}d \tau \biggr] \\ & \qquad {} +\frac{1}{2}\frac{d}{dt} \biggl[ \int _{0}^{t}g_{2}(t-\tau ) \int _{\varOmega } \bigl\vert \nabla v(\tau )-\nabla v(t) \bigr\vert ^{2}\,dx\,d\tau \biggr] \\ &\quad =- \int _{\varOmega }v_{t}f_{2}(u,v)\,dx+ \frac{1}{2}\bigl(g_{2}'\circ \nabla v\bigr) (t)-\frac{1}{2}g_{2}(t) \bigl\Vert \nabla v(t) \bigr\Vert _{2}^{2}- \Vert v_{t} \Vert _{2} ^{2}. \end{aligned}$$

(3.11)

Finally, by adding (3.10) to (3.11), (3.6) is established. □

The following lemma is important to prove the global existence of solution.

Lemma 3.2

([3])

Assume that (2.5) holds. Then there exists $\eta >0$ such that, for any $(u,v)\in H_{0}^{1}(\varOmega )\times H_{0}^{1}(\varOmega )$, we obtain

$$\begin{aligned} \Vert u+v \Vert _{2(p+2)}^{2(p+2)}+2 \Vert uv \Vert _{p+2}^{p+2}\leq \eta \bigl(l \Vert \nabla u \Vert _{2}^{2}+k \Vert \nabla v \Vert _{2}^{2} \bigr)^{p+2}. \end{aligned}$$

(3.12)

Proof

Exploiting Minkowski’s inequality, we have

$$\begin{aligned} \Vert u+v \Vert _{2(p+2)}^{2}\leq 2 \bigl( \Vert u \Vert _{2(p+2)}^{2}+ \Vert v \Vert _{2(p+2)}^{2} \bigr). \end{aligned}$$

(3.13)

From Hölder’s inequality and Young’s inequality, we derive

$$\begin{aligned} \Vert uv \Vert _{p+2}\leq \Vert u \Vert _{2(p+2)} \Vert v \Vert _{2(p+2)}\leq c \bigl(l \Vert \nabla u \Vert _{2} ^{2}+k \Vert \nabla v \Vert _{2}^{2} \bigr). \end{aligned}$$

(3.14)

Then, combining (3.13), (3.14) and the embedding $H_{0}^{1}( \varOmega )\hookrightarrow L^{2(p+2)}(\varOmega )$ leads to (3.12). □

Lemma 3.3

Let $(u_{0},v_{0})\in H_{0}^{1}(\varOmega )\times H_{0}^{1}(\varOmega )$ and $(u_{1},v_{1})\in L^{2}(\varOmega )\times L^{2}(\varOmega )$. Assume that (G1)–(G3) hold. If

$$ I(0)=I(u_{0},v_{0})>0\quad \textit{and} \quad \beta = \eta \biggl(\frac{2(p+2)}{p+1}E(0) \biggr)^{p+1}< 1, $$

(3.15)

then $I(t)>0$ for $t\in [0,T_{m}]$.

Proof

Since $I(0)>0$, then by the continuity of $I(t)$, there exists $T^{*}< T_{m}$ such that $I(t)>0$, $\forall t\in [0,T^{*}]$. By using (3.1) and (3.2), we have

$$\begin{aligned} J(t) &=\frac{p+1}{2(p+2)} \biggl\{ \biggl(1- \int _{0}^{t}g_{1}(\tau )\,d\tau \biggr) \Vert \nabla u \Vert _{2}^{2}+\biggl(1- \int _{0}^{t}g_{2}(\tau )\,d\tau \biggr) \Vert \nabla v \Vert _{2} ^{2} \\ &\quad {} +(g_{1}\circ \nabla u) (t)+(g_{2}\circ \nabla v) (t) \biggr\} + \frac{1}{2(p+2)}I(t) \\ &\geq \frac{p+1}{2(p+2)} \biggl\{ \biggl(1- \int _{0}^{t}g_{1}(\tau )\,d\tau \biggr) \Vert \nabla u \Vert _{2}^{2}+\biggl(1- \int _{0}^{t}g_{2}(\tau )\,d\tau \biggr) \Vert \nabla v \Vert _{2} ^{2} \\ &\quad {} +(g_{1}\circ \nabla u) (t)+(g_{2}\circ \nabla v) (t) \biggr\} . \end{aligned}$$

(3.16)

From (2.1), (3.1), (3.2), and (3.6), we infer that

$$\begin{aligned} l \Vert \nabla u \Vert _{2}^{2}+k \Vert \nabla v \Vert _{2}^{2} &\leq \biggl(1- \int _{0}^{t}g_{1}( \tau )\,d\tau \biggr) \Vert \nabla u \Vert _{2}^{2} \\ &\quad {} +\biggl(1- \int _{0}^{t}g_{2}(\tau )\,d\tau \biggr) \Vert \nabla v \Vert _{2}^{2} \\ &\leq \frac{2(p+2)}{p+1}J(t) \\ &\leq \frac{2(p+2)}{p+1}E(t) \\ &\leq \frac{2(p+2)}{p+1}E(0). \end{aligned}$$

(3.17)

It follows from (2.7), (3.12), and (3.15) that

$$\begin{aligned} & 2(p+2) \int _{\varOmega }F(u,v)\,dx \\ &\quad \leq \eta \bigl(l \Vert \nabla u \Vert _{2}^{2}+k \Vert \nabla v \Vert _{2}^{2} \bigr)^{p+2} \\ &\quad \leq \eta \bigl(l \Vert \nabla u \Vert _{2}^{2}+k \Vert \nabla v \Vert _{2}^{2} \bigr)^{p+1} \bigl(l \Vert \nabla u \Vert _{2}^{2}+k \Vert \nabla v \Vert _{2}^{2} \bigr) \\ &\quad \leq \eta \biggl[\frac{2(p+2)}{p+1}E(0) \biggr]^{p+1} \bigl(l \Vert \nabla u \Vert _{2} ^{2}+k \Vert \nabla v \Vert _{2}^{2} \bigr),\quad \forall t\in \bigl[0,T^{*}\bigr]. \end{aligned}$$

(3.18)

Combining (3.15) and (3.18), we deduce that

$$\begin{aligned} &2(p+2) \int _{\varOmega }F(u,v)\,dx \\ &\quad \leq \beta \bigl(l \Vert \nabla u \Vert _{2}^{2}+k \Vert \nabla v \Vert _{2}^{2} \bigr) \\ &\quad \leq \beta \biggl(1- \int _{0}^{t}g_{1}(\tau )\,d\tau \biggr) \Vert \nabla u \Vert _{2} ^{2}+\beta \biggl(1- \int _{0}^{t}g_{2}(\tau )\,d\tau \biggr) \Vert \nabla v \Vert _{2}^{2} \\ &\quad \leq \biggl(1- \int _{0}^{t}g_{1}(\tau )\,d\tau \biggr) \Vert \nabla u \Vert _{2}^{2}+\biggl(1- \int _{0}^{t}g_{2}(\tau )\,d\tau \biggr) \Vert \nabla v \Vert _{2}^{2}, \quad \forall t\in \bigl[0,T ^{*}\bigr]. \end{aligned}$$

(3.19)

Therefore

$$\begin{aligned} I(t) &=\biggl(1- \int _{0}^{t}g_{1}(\tau )\,d\tau \biggr) \Vert \nabla u \Vert _{2}^{2}+\biggl(1- \int _{0}^{t}g_{2}(\tau )\,d\tau \biggr) \Vert \nabla v \Vert _{2}^{2}+(g_{1} \circ \nabla u) (t) \\ &\quad {} +(g_{2}\circ \nabla v) (t)-2(p+2) \int _{\varOmega }F(u,v)\,dx>0, \quad \forall t\in \bigl[0,T^{*}\bigr]. \end{aligned}$$

(3.20)

By repeating this procedure, $T^{*}$ is extended to $T_{m}$. □

Lemma 3.4

Assume that (G1)–(G3) hold. If $(u_{0},v_{0})\in H_{0} ^{1}(\varOmega )\times H_{0}^{1}(\varOmega )$ and $(u_{1},v_{1})\in L^{2}( \varOmega )\times L^{2}(\varOmega )$ and satisfy (3.15), then the solution of system (1.1) is bounded and global in time.

Proof

From Lemma 3.3, (3.6), and (3.17), we see that

$$\begin{aligned} E(0) &\geq E(t)=\frac{1}{2} \Vert u_{t} \Vert _{2}^{2}+\frac{1}{2} \Vert v_{t} \Vert _{2} ^{2}+J(t) \\ &\geq \frac{p+1}{2(p+2)} \bigl(l \Vert \nabla u \Vert _{2}^{2}+k \Vert \nabla v \Vert _{2} ^{2} \bigr), \end{aligned}$$

(3.21)

therefore

$$\begin{aligned} l \Vert \nabla u \Vert _{2}^{2}+k \Vert \nabla v \Vert _{2}^{2}\leq \frac{2(p+2)}{p+1}E(0), \end{aligned}$$

(3.22)

which implies that the solution of system (1.1) is global and bounded. □

4 Technical lemmas

In this section, we present some lemmas needed for the proof of our result.

Lemma 4.1

([3])

There exist two constants $a_{0}>0$ and $a_{1}>0$ such that

$$\begin{aligned} \frac{a_{0}}{2(p+2)} \bigl( \vert u \vert ^{2(p+2)}+ \vert v \vert ^{2(p+2)} \bigr)\leq F(u,v) \leq \frac{a_{1}}{2(p+2)} \bigl( \vert u \vert ^{2(p+2)}+ \vert v \vert ^{2(p+2)} \bigr). \end{aligned}$$

(4.1)

Proof

The right-hand side of (4.1) is trivial. If $u=v=0$, for the left-hand side of (4.1), the result is also trivial. If, without loss of generality, $v\neq 0$, then either $\vert u \vert \leq \vert v \vert$ or $\vert u \vert > \vert v \vert$.

If $\vert u \vert \leq \vert v \vert$, we have

$$\begin{aligned} F(u,v)=\frac{1}{2(p+2)} \vert v \vert ^{2(p+2)} \biggl[ \biggl\vert 1+\frac{u}{v} \biggr\vert ^{2(p+2)}+2 \biggl\vert \frac{u}{v} \biggr\vert ^{p+2} \biggr]. \end{aligned}$$

(4.2)

Now we consider the continuous function

$$\begin{aligned} j(\omega )= \vert 1+\omega \vert ^{2(p+2)}+2 \vert \omega \vert ^{p+2} \quad \text{over } [-1,1], \end{aligned}$$

(4.3)

then we obtain that $\min j(\omega )\geq 0$. If $\min j(\omega )=0$, then there exists $\omega _{0}\in [-1,1]$ such that

$$\begin{aligned} j(\omega _{0})= \vert 1+\omega _{0} \vert ^{2(p+2)}+2 \vert \omega _{0} \vert ^{{p+2}}=0. \end{aligned}$$

(4.4)

This infers that $\vert 1+\omega _{0} \vert = \vert \omega _{0} \vert =0$, which is impossible. Hence $\min j(\omega )=2a_{0}>0$. Thus

$$\begin{aligned} F(u,v)\geq \frac{a_{0}}{p+2} \vert v \vert ^{2(p+2)}\geq \frac{a_{0}}{p+2} \vert u \vert ^{2(p+2)}. \end{aligned}$$

(4.5)

It follows from (4.5) that

$$\begin{aligned} 2F(u,v)\geq \frac{a_{0}}{p+2} \bigl\{ { \vert v \vert ^{2(p+2)}+ \vert u \vert ^{2(p+2)}} \bigr\} , \end{aligned}$$

(4.6)

and then

$$\begin{aligned} \frac{a_{0}}{2(p+2)} \bigl\{ { \vert v \vert ^{2(p+2)}+ \vert u \vert ^{2(p+2)}} \bigr\} \leq F(u,v). \end{aligned}$$

(4.7)

If $\vert u \vert > \vert v \vert$, we deduce that

$$\begin{aligned} F(u,v) &=\frac{1}{2(p+2)} \vert u \vert ^{2(p+2)} \biggl[ \biggl\vert 1+\frac{v}{u} \biggr\vert ^{2(p+2)}+2 \biggl\vert \frac{v}{u} \biggr\vert ^{p+2} \biggr] \\ &\geq \frac{a_{0}}{p+2} \vert u \vert ^{2(p+2)}\geq \frac{a_{0}}{p+2} \vert v \vert ^{2(p+2)}. \end{aligned}$$

(4.8)

Hence, this gives the desired result. □

Lemma 4.2

([3])

There exist two positive constants $\lambda _{1}$ and $\lambda _{2}$ such that

$$\begin{aligned} \int _{\varOmega } \bigl\vert f_{i}(u,v) \bigr\vert ^{2}\,dx\leq \lambda _{i} \bigl(l \Vert \nabla u \Vert _{2} ^{2}+k \Vert \nabla v \Vert _{2}^{2} \bigr)^{2p+3},\quad i=1,2. \end{aligned}$$

(4.9)

Proof

From (2.3), we easily get

$$\begin{aligned} \bigl\vert f_{1}(u,v) \bigr\vert &\leq C \bigl( \vert u+v \vert ^{2p+3}+ \vert u \vert ^{p+1} \vert v \vert ^{p+2} \bigr) \\ &\leq C \bigl( \vert u \vert ^{2p+3}+ \vert v \vert ^{2p+3}+ \vert u \vert ^{p+1} \vert v \vert ^{p+2} \bigr). \end{aligned}$$

(4.10)

By (4.10) and Young’s inequality, with $q=\frac{2p+3}{p+1}$ and $q'=\frac{2p+3}{p+2}$, we have

$$\begin{aligned} \vert u \vert ^{p+1} \vert v \vert ^{p+2}\leq c_{1} \vert u \vert ^{2p+3}+c_{2} \vert v \vert ^{2p+3}, \end{aligned}$$

(4.11)

therefore

$$\begin{aligned} \bigl\vert f_{1}(u,v) \bigr\vert &\leq C \bigl( \vert u \vert ^{2p+3}+ \vert v \vert ^{2p+3} \bigr). \end{aligned}$$

(4.12)

Then, by using Poincare’s inequality and (2.5), we observe that

$$\begin{aligned} \int _{\varOmega } \bigl\vert f_{1}(u,v) \bigr\vert ^{2}\,dx &\leq C \bigl( \Vert \nabla u \Vert _{2}^{2(2p+3)}+ \Vert \nabla v \Vert _{2}^{2(2p+3)} \bigr) \\ &\leq \lambda _{1} \bigl(l \Vert \nabla u \Vert _{2}^{2}+k \Vert \nabla v \Vert _{2}^{2} \bigr)^{2p+3}. \end{aligned}$$

(4.13)

In the same way, we have

$$\begin{aligned} \int _{\varOmega } \bigl\vert f_{2}(u,v) \bigr\vert ^{2}\,dx &\leq C \bigl( \Vert \nabla u \Vert _{2}^{2(2p+3)}+ \Vert \nabla v \Vert _{2}^{2(2p+3)} \bigr) \\ &\leq \lambda _{2} \bigl(l \Vert \nabla u \Vert _{2}^{2}+k \Vert \nabla v \Vert _{2}^{2} \bigr)^{2p+3}. \end{aligned}$$

(4.14)

 □

Lemma 4.3

(Jensen inequality)

Suppose that G is a concave function on $[a,b]$, $f:\varOmega \rightarrow [a,b]$ and h are integrable functions on Ω, with $h(x) \geq 0$, and $\int _{\varOmega }h(x)\,dx=r>0$, then

$$\begin{aligned} \frac{1}{r} \int _{\varOmega }G\bigl[f(x)\bigr]h(x)\,dx\leq G \biggl[ \frac{1}{r} \int _{ \varOmega }f(x)h(x)\,dx \biggr]. \end{aligned}$$

(4.15)

For the special case $G(y)=y^{\frac{1}{\gamma }}$, $y\geq 0$, $\gamma >1$, we obtain

$$\begin{aligned} \frac{1}{r} \int _{\varOmega }\bigl[f(x)\bigr]^{\frac{1}{\gamma }}h(x)\,dx\leq \biggl[ \frac{1}{r} \int _{\varOmega }f(x)h(x)\,dx \biggr]^{\frac{1}{\gamma }}. \end{aligned}$$

(4.16)

Lemma 4.4

Assume that $g_{i}$ satisfies (G1) and (G2) for $i=1,2$, then

$$ \int _{0}^{+\infty }\zeta _{i}(t)g_{i}^{1-\theta }(t) \,dt< +\infty ,\quad \forall 0\leq \theta < 2-\gamma . $$

(4.17)

Proof

From (G1) and (G2), we see that

$$\begin{aligned} \zeta _{i}(t)g_{i}^{1-\theta }(t)=\zeta _{i}(t)g_{i}^{\gamma }(t)g_{i} ^{1-\theta -\gamma }(t)\leq -g_{i}'(t)g_{i}^{1-\theta -\gamma }(t). \end{aligned}$$

(4.18)

Integrating (4.18) over $(0,+\infty )$ and using the fact that $0\leq \theta <2-\gamma$, we get

$$\begin{aligned} \int _{0}^{+\infty }\zeta _{i}(t)g_{i}^{1-\theta }(t) \,dt\leq - \int _{0} ^{+\infty }g_{i}'(t)g_{i}^{1-\theta -\gamma }(t) \,dt=\frac{-g_{i}^{2- \theta -\gamma }(t)}{2-\theta -\gamma }\bigg\vert _{0}^{+\infty }< +\infty . \end{aligned}$$

(4.19)

 □

Lemma 4.5

([7])

If (G1)–(G3) hold, $u\in L^{\infty }(0,T,H_{0}^{1}(\varOmega ))$, for $0<\theta <1$, we obtain

$$ (g_{1}\circ \nabla u) (t)\leq C \biggl\{ \biggl( \int _{0}^{+\infty }g_{1} ^{1-\theta }(t)\,dt \biggr)E(0) \biggr\} ^{\frac{\gamma -1}{\gamma -1+\theta }} \bigl\{ \bigl(g_{1}^{\gamma }\circ \nabla u\bigr) \bigr\} ^{\frac{\theta }{ \gamma -1+\theta }}(t). $$

(4.20)

By taking $\theta =\frac{1}{2}$, we have

$$ (g_{1}\circ \nabla u) (t)\leq C \biggl\{ \int _{0}^{t}g_{1}^{\frac{1}{2}}( \tau )\,d\tau \biggr\} ^{\frac{2\gamma -2}{2\gamma -1}} \bigl\{ \bigl(g_{1}^{ \gamma } \circ \nabla u\bigr) \bigr\} ^{\frac{1}{2\gamma -1}}(t). $$

(4.21)

Proof

For $q>1$, we derive

$$ \begin{aligned}[b] (g_{1}\circ \nabla u) (t)&= \int _{0}^{t}g_{1}^{\frac{1-\theta }{q}}(t- \tau ) \bigl\Vert \nabla u(t)-\nabla u(\tau ) \bigr\Vert _{2}^{\frac{2}{q}} \\&\quad{}\times g_{1}^{\frac{q-1+ \theta }{q}}(t- \tau ) \bigl\Vert \nabla u(t)-\nabla u(\tau ) \bigr\Vert _{2}^{ \frac{2q-2}{q}} \,d\tau . \end{aligned} $$

(4.22)

Applying Hölder’s inequality, we deduce

$$\begin{aligned} (g_{1}\circ \nabla u) (t) &\leq \biggl( \int _{0}^{t}g_{1}^{1-\theta }(t- \tau ) \bigl\Vert \nabla u(t)-\nabla u(\tau ) \bigr\Vert _{2}^{2} \,d\tau \biggr)^{\frac{1}{q}} \\ &\quad {}\times \biggl( \int _{0}^{t}g_{1}^{\frac{q-1+\theta }{q-1}}(t- \tau ) \bigl\Vert \nabla u(t)-\nabla u(\tau ) \bigr\Vert _{2}^{2} \,d\tau \biggr)^{\frac{q-1}{q}}. \end{aligned}$$

(4.23)

By taking $q=\frac{\gamma -1+\theta }{\gamma -1}$, we arrive at

$$\begin{aligned} (g_{1}\circ \nabla u) (t) &\leq C \biggl\{ \biggl( \int _{0}^{t}g_{1}^{1- \theta }( \tau )\,d\tau \biggr) \Vert \nabla u \Vert _{L^{\infty }(0,T,H_{0}^{1}( \varOmega ))}^{2} \biggr\} ^{\frac{\gamma -1}{\gamma -1+\theta }} \bigl\{ \bigl(g _{1}^{\gamma }\circ \nabla u\bigr) \bigr\} ^{\frac{\theta }{\gamma -1+\theta }}(t) \\ &\leq C \biggl\{ \biggl( \int _{0}^{t}g_{1}^{1-\theta }( \tau )\,d\tau \biggr) \sup E(t) \biggr\} ^{\frac{\gamma -1}{\gamma -1+\theta }} \bigl\{ \bigl(g_{1} ^{\gamma }\circ \nabla u\bigr) \bigr\} ^{\frac{\theta }{\gamma -1+\theta }}(t) \\ &\leq C \biggl\{ \biggl( \int _{0}^{t}g_{1}^{1-\theta }( \tau )\,d\tau \biggr) \sup E(0) \biggr\} ^{\frac{\gamma -1}{\gamma -1+\theta }} \bigl\{ \bigl(g_{1} ^{\gamma }\circ \nabla u\bigr) \bigr\} ^{\frac{\theta }{\gamma -1+\theta }}(t) \\ &\leq C \biggl\{ \biggl( \int _{0}^{t}g_{1}^{1-\theta }( \tau )\,d\tau \biggr) \biggr\} ^{\frac{\gamma -1}{\gamma -1+\theta }} \bigl\{ \bigl(g_{1}^{\gamma } \circ \nabla u\bigr) \bigr\} ^{\frac{\theta }{\gamma -1+\theta }}(t). \end{aligned}$$

(4.24)

Then, by taking $\theta =\frac{1}{2}$ in (4.24), (4.21) is established. □

Similarly,

$$ (g_{2}\circ \nabla v) (t)\leq C \biggl\{ \biggl( \int _{0}^{t}g_{2}^{1-\theta }( \tau )\,d\tau \biggr)E(0) \biggr\} ^{\frac{\gamma -1}{\gamma -1+\theta }} \bigl\{ \bigl(g_{2}^{\gamma } \circ \nabla v\bigr) \bigr\} ^{\frac{\theta }{\gamma -1+\theta }}(t) $$

(4.25)

and

$$ (g_{2}\circ \nabla v) (t)\leq C \biggl\{ \int _{0}^{t}g_{2}^{\frac{1}{2}}( \tau )\,d\tau \biggr\} ^{\frac{2\gamma -2}{2\gamma -1}} \bigl\{ \bigl(g_{2}^{ \gamma } \circ \nabla v\bigr) \bigr\} ^{\frac{1}{2\gamma -1}}(t). $$

(4.26)

Lemma 4.6

([7])

Suppose that $g_{1}$ satisfies (G1) and (G2), then we have

$$ \zeta _{1}(t) (g_{1}\circ \nabla u) (t)\leq C \bigl[-E'(t)\bigr]^{\frac{1}{2\gamma -1}}. $$

(4.27)

Proof

Multiplying both sides of (4.21) by $\zeta _{1}(t)$ and by using (G2), (3.6), and (4.17) gives

$$\begin{aligned} \zeta _{1}(t) (g_{1}\circ \nabla u) (t) &\leq C\zeta _{1}^{\frac{2\gamma -2}{2 \gamma -1}}(t) \biggl[ \int _{0}^{t}g_{1}^{\frac{1}{2}}( \tau )\,d\tau \biggr]^{\frac{2 \gamma -2}{2\gamma -1}}\zeta _{1}^{\frac{1}{2\gamma -1}}(t) \bigl(g_{1} ^{\gamma }\circ \nabla u \bigr)^{\frac{1}{2\gamma -1}}(t) \\ &\leq C \biggl[ \int _{0}^{t}\zeta _{1}(\tau )g_{1}^{\frac{1}{2}}(\tau )d \tau \biggr]^{\frac{2\gamma -2}{2\gamma -1}} \bigl( \zeta _{1}g_{1}^{\gamma }\circ \nabla u \bigr)^{\frac{1}{2\gamma -1}}(t) \\ &\leq C \biggl[ \int _{0}^{t}\zeta _{1}(\tau )g_{1}^{\frac{1}{2}}(\tau )d \tau \biggr]^{\frac{2\gamma -2}{2\gamma -1}} \bigl(-g_{1}'\circ \nabla u \bigr)^{\frac{1}{2\gamma -1}}(t) \\ &\leq C\bigl[-E'(t)\bigr]^{\frac{1}{2\gamma -1}}. \end{aligned}$$

(4.28)

 □

Likewise,

$$ \zeta _{2}(t) (g_{2}\circ \nabla v) (t)\leq C \bigl[-E'(t)\bigr]^{\frac{1}{2\gamma -1}}. $$

(4.29)

5 The decay result

In this section, we establish three related lemmas before proving our result.

Lemma 5.1

If (G1)–(G3) and (3.15) hold, the functional $\phi (t)$ defined by

$$ \phi (t):= \int _{\varOmega }u_{t}u\,dx+ \int _{\varOmega }v_{t}v\,dx $$

satisfies, along solutions of (1.1),

$$\begin{aligned} \phi '(t) &\leq -\frac{l}{4} \Vert \nabla u \Vert _{2}^{2}+\frac{1-l}{2l}(g_{1} \circ \nabla u) (t)+\biggl(1+\frac{1}{4\beta _{1}}\biggr) \Vert u_{t} \Vert _{2}^{2} \\ &\quad {} -\frac{k}{4} \Vert \nabla v \Vert _{2}^{2}+ \frac{1-k}{2k}(g_{2}\circ \nabla v) (t)+\biggl(1+ \frac{1}{4\beta _{2}}\biggr) \Vert v_{t} \Vert _{2}^{2} \\ &\quad {} +2(p+2) \int _{\varOmega }F(u,v)\,dx. \end{aligned}$$

(5.1)

Proof

A differentiation of $\phi (t)$ with respect to t, it follows from system (1.1) that

$$\begin{aligned} \phi '(t) &= \int _{\varOmega }u_{tt}u\,dx+ \Vert u_{t} \Vert _{2}^{2}+ \int _{\varOmega }v _{tt}v\,dx+ \Vert v_{t} \Vert _{2}^{2} \\ &= \int _{\varOmega }\nabla u(t) \int _{0}^{t}g_{1}(t-\tau )\nabla u(\tau )d \tau \,dx+ \Vert u_{t} \Vert _{2}^{2}- \Vert \nabla u \Vert _{2}^{2}- \int _{\varOmega }uu_{t}\,dx \\ &\quad {} + \int _{\varOmega }\nabla v(t) \int _{0}^{t}g_{2}(t-\tau )\nabla v( \tau )\,d\tau \,dx+ \Vert v_{t} \Vert _{2}^{2}- \Vert \nabla v \Vert _{2}^{2}- \int _{\varOmega }vv _{t}\,dx \\ &\quad {} + \int _{\varOmega }uf_{1}(u,v)\,dx+ \int _{\varOmega }vf_{2}(u,v)\,dx. \end{aligned}$$

(5.2)

For the first term on the right-hand side of (5.2), by using Young’s inequality and the fact that $\int _{0}^{t}g_{1}(s)\,ds\leq \int _{0}^{\infty }g_{1}(s)\,ds =1-l$, for $\eta =\frac{l}{1-l}>0$, we get

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

(5.3)

Similar calculations also yield, for $\eta _{1}=\frac{k}{1-k}>0$,

$$\begin{aligned} \int _{\varOmega }\nabla v(t) \int _{0}^{t}g_{2}(t-\tau )\nabla v(\tau )d \tau \,dx\leq \frac{2-k}{2} \Vert \nabla v \Vert _{2}^{2}+\frac{1-k}{2k}(g_{2} \circ \nabla v) (t). \end{aligned}$$

(5.4)

Applying Young’s inequality and Poincare’s inequality, for some $\beta _{1}>0$, we obtain

$$\begin{aligned} \int _{\varOmega }uu_{t}\,dx &\leq \beta _{1} \Vert u \Vert _{2}^{2}+ \frac{1}{4\beta _{1}} \Vert u_{t} \Vert _{2}^{2} \\ &\leq \beta _{1} C_{*}^{2} \Vert \nabla u \Vert _{2}^{2}+\frac{1}{4\beta _{1}} \Vert u _{t} \Vert _{2}^{2}. \end{aligned}$$

(5.5)

Likewise, for some $\beta _{2}>0$, we have

$$\begin{aligned} \int _{\varOmega }vv_{t}\,dx &\leq \beta _{2} \Vert v \Vert _{2}^{2}+ \frac{1}{4\beta _{2}} \Vert v_{t} \Vert _{2}^{2} \\ &\leq \beta _{2} C_{*}^{2} \Vert \nabla v \Vert _{2}^{2}+\frac{1}{4\beta _{2}} \Vert v _{t} \Vert _{2}^{2}. \end{aligned}$$

(5.6)

Inserting (5.3)–(5.6) into (5.2) yields

$$\begin{aligned} \phi '(t) &\leq - \biggl(\frac{l}{2}-\beta _{1}C_{*}^{2} \biggr) \Vert \nabla u \Vert _{2} ^{2}+\frac{1-l}{2l}(g_{1} \circ \nabla u) (t) \\ &\quad {} +\biggl(1+\frac{1}{4\beta _{1}}\biggr) \Vert u_{t} \Vert _{2}^{2}- \biggl(\frac{k}{2}- \beta _{2}C_{*}^{2} \biggr) \Vert \nabla v \Vert _{2}^{2} \\ &\quad {} +\frac{1-k}{2k}(g_{2}\circ \nabla v) (t)+\biggl(1+ \frac{1}{4\beta _{2}}\biggr) \Vert v_{t} \Vert _{2}^{2}+2(p+2) \int _{\varOmega }F(u,v)\,dx. \end{aligned}$$

(5.7)

Now, we pick $\beta _{1}, \beta _{2}>0$ small enough such that

$$\begin{aligned} \frac{l}{2}-\beta _{1} C_{*}^{2} \geq \frac{l}{4}, \quad\quad \frac{k}{2}-\beta _{2} C_{*}^{2}\geq \frac{k}{4}. \end{aligned}$$

(5.8)

Finally, a combination of (5.7) and (5.8) gives (5.1). □

Lemma 5.2

Suppose that (G1)–(G3) and (3.15) hold. The functional $\psi _{1}(t)$ defined by

$$ \psi _{1}(t):=- \int _{\varOmega }u_{t} \int _{0}^{t}g_{1}(t-\tau ) \bigl(u(t)-u( \tau )\bigr)\,d\tau \,dx $$

satisfies, along solutions of (1.1),

$$\begin{aligned} \psi _{1}'(t) &\leq \bigl[\delta +2(1-l)^{2}\delta +\alpha _{2}\delta l \bigr] \Vert \nabla u \Vert _{2}^{2}-\frac{g_{1}(0)C_{*}^{2}}{4\delta } \bigl(g_{1}' \circ \nabla u\bigr) (t) \\ &\quad {} + \biggl(2\delta +\frac{1}{2\delta }+\frac{C_{*}^{2}}{2\delta } \biggr) (1-l) (g_{1}\circ \nabla u) (t) \\ &\quad {} +\alpha _{2}\delta k \Vert \nabla v \Vert _{2}^{2}+ \biggl(2\delta - \int _{0}^{t}g_{1}(\tau )\,d\tau \biggr) \Vert u_{t} \Vert _{2}^{2}, \end{aligned}$$

(5.9)

where $\alpha _{2}=\lambda _{1} (\frac{2(p+2)}{p+1}E(0) )^{2p+2}$, $\lambda _{1}$ is the constant in Lemma 4.2.

Proof

Taking the derivative of $\psi _{1}(t)$ with respect to t and using system (1.1) gives

$$\begin{aligned} \psi _{1}'(t) &=- \int _{\varOmega }u_{tt} \int _{0}^{t}g_{1}(t-\tau ) \bigl(u(t)-u( \tau )\bigr)\,d\tau \,dx \\ &\quad {} - \int _{\varOmega }u_{t} \int _{0}^{t}g_{1}'(t- \tau ) \bigl(u(t)-u(\tau )\bigr)d \tau \,dx- \int _{0}^{t}g_{1}(\tau )\,d\tau \Vert u_{t} \Vert _{2}^{2} \\ &= \int _{\varOmega }\nabla u(t) \int _{0}^{t}g_{1}(t-\tau ) \bigl( \nabla u(t)- \nabla u(\tau )\bigr)\,d\tau \,dx \\ &\quad {} - \int _{\varOmega } \biggl( \int _{0}^{t}g_{1}(t-\tau )\nabla u(\tau )d \tau \biggr) \biggl( \int _{0}^{t}g_{1}(t-\tau ) \bigl( \nabla u(t)-\nabla u(\tau )\bigr)d \tau \biggr)\,dx \\ &\quad {} + \int _{\varOmega }u_{t} \int _{0}^{t}g_{1}(t-\tau ) \bigl(u(t)-u(\tau )\bigr)d \tau \,dx \\ &\quad {} - \int _{\varOmega }f_{1}(u,v) \int _{0}^{t}g_{1}(t-\tau ) \bigl(u(t)-u( \tau )\bigr)\,d\tau \,dx \\ &\quad {} - \int _{\varOmega }u_{t} \int _{0}^{t}g'_{1}(t- \tau ) \bigl(\nabla u(t)- \nabla u(\tau )\bigr)\,d\tau \,dx- \int _{0}^{t}g_{1}(\tau )\,d\tau \Vert u_{t} \Vert _{2} ^{2}. \end{aligned}$$

(5.10)

For the first term on the right-hand side of (5.10), by exploiting (G1), Young’s inequality, and Cauchy–Schwarz inequality, for any $\delta >0$, we get

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

(5.11)

As for the second term in (5.10), recall that $(a+b)^{2}\leq 2(a ^{2}+b^{2})$, for $\eta _{2}=1$, we obtain

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

(5.12)

The third term can be handled by

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

(5.13)

For the forth term, it follows from (4.9) that

$$\begin{aligned}& \int _{\varOmega }f_{1}(u,v) \int _{0}^{t}g_{1}(t-\tau ) \bigl(u(t)-u(\tau )\bigr)d \tau \,dx \\& \quad \leq \lambda _{1}\delta \bigl(l \Vert \nabla u \Vert _{2}^{2}+k \Vert \nabla v \Vert _{2} ^{2} \bigr)^{2p+3}+\frac{1}{4\delta } \int _{\varOmega } \biggl( \int _{0}^{t}g _{1}(t-\tau ) \bigl\vert u(\tau )-u(t) \bigr\vert \,d\tau \biggr)^{2}\,dx \\& \quad \leq \lambda _{1}\delta \biggl(\frac{2(p+2)}{p+1}E(0) \biggr)^{2p+2} \bigl(l \Vert \nabla u \Vert _{2}^{2}+k \Vert \nabla v \Vert _{2}^{2} \bigr)+ \frac{C_{*}^{2}(1-l)}{4 \delta }(g_{1}\circ \nabla u) (t) \\& \quad =\alpha _{2}\delta \bigl(l \Vert \nabla u \Vert _{2}^{2}+k \Vert \nabla v \Vert _{2} ^{2} \bigr)+\frac{C_{*}^{2}(1-l)}{4\delta }(g_{1}\circ \nabla u) (t), \end{aligned}$$

(5.14)

where $\alpha _{2}=\lambda _{1} (\frac{2(p+2)}{p+1}E(0) )^{2p+2}$.

The fifth term on the right-hand side of (5.10) can be estimated as

$$\begin{aligned}& \int _{\varOmega }u_{t} \int _{0}^{t}g_{1}'(t- \tau ) \bigl(u(t)-u(\tau )\bigr)d \tau \,dx \\& \quad \leq \delta \Vert u_{t} \Vert _{2}^{2}+ \frac{1}{4\delta } \int _{\varOmega } \biggl( \int _{0}^{t}g_{1}'(t- \tau ) \bigl\vert u(t)-u(\tau ) \bigr\vert \,d\tau \biggr)^{2} \,dx \\& \quad \leq \delta \Vert u_{t} \Vert _{2}^{2}- \frac{g_{1}(0)C_{*}^{2}}{4 \delta }\bigl(g_{1}'\circ \nabla u\bigr) (t). \end{aligned}$$

(5.15)

Taking into account estimates (5.11)–(5.13), estimate (5.9) is established. □

Similar computations also yield the following.

Lemma 5.3

Suppose that (G1)–(G3) and (3.15) hold. The functional $\psi _{2}(t)$ defined by

$$ \psi _{2}(t):=-v_{t} \int _{0}^{t}g_{2}(t-\tau ) \bigl(v(t)-v(\tau )\bigr)\,d\tau \,dx $$

satisfies, along solutions of system (1.1),

$$\begin{aligned} \psi _{2}'(t) &\leq \bigl[\delta +2(1-k)^{2}\delta +\alpha _{3}\delta k \bigr] \Vert \nabla v \Vert _{2}^{2}-\frac{g_{2}(0)C_{*}^{2}}{4\delta } \bigl(g_{2}' \circ \nabla v\bigr) (t) \\ &\quad {} + \biggl(2\delta +\frac{1}{2\delta }+\frac{C_{*}^{2}}{2\delta } \biggr) (1-k) (g_{2}\circ \nabla v) (t) \\ &\quad {} +\alpha _{3}\delta l \Vert \nabla u \Vert _{2}^{2}+ \biggl(2\delta - \int _{0}^{t}g_{2}(\tau )\,d\tau \biggr) \Vert v_{t} \Vert _{2}^{2}, \end{aligned}$$

(5.16)

where $\alpha _{3}=\lambda _{2} (\frac{2(p+2)}{p+1}E(0) )^{2p+2}$, $\lambda _{2}$ is the constant in Lemma 4.2.

Now, we define the functional

$$\begin{aligned} W(t)=E(t)+\epsilon _{1}\phi (t)+\epsilon _{2}\psi (t), \end{aligned}$$

(5.17)

where $\epsilon _{1}$ and $\epsilon _{2}$ are positive constants, $\phi (t)$ is given in Lemma 5.1 and $\psi (t):=\psi _{1}(t)+\psi _{2}(t)$.

Lemma 5.4

([5])

Let $(u,v)$ be the solution of system (1.1) and assume that (3.15) holds. Then there exist constant $\epsilon >0$ small enough and $M>0$ large enough such that the following relation

$$\begin{aligned} \beta _{1}W(t)\leq E(t)\leq \beta _{2}W(t), \quad \forall t\geq 0, \end{aligned}$$

(5.18)

holds for two positive constants $\beta _{1}$ and $\beta _{2}$.

Theorem 5.5

Assume (G1)–(G3) and (3.15) hold. Let $(u_{0},v_{0}) \in H_{0}^{1}(\varOmega )\times H_{0}^{1}(\varOmega )$ and $(u_{1},v_{1}) \in L^{2}(\varOmega )\times L^{2}(\varOmega )$. Then, for each $t_{0}>0$, there exist positive constants K, k, $k_{1}$, $k_{2}$ such that the solution of system (1.1) satisfies, for all $t\geq t_{0}$,

$$\begin{aligned}& E(t)\leq Ke^{-k\int _{t_{0}}^{t}\zeta (\tau )\,d\tau }, \quad \gamma =1, \end{aligned}$$

(5.19)

$$\begin{aligned}& E(t)\leq k_{1} \biggl[\frac{1}{1+\int _{t_{0}}^{t}\zeta ^{2\gamma -1}( \tau )\,d\tau } \biggr]^{\frac{1}{2\gamma -2}}, \quad 1< \gamma < \frac{3}{2}. \end{aligned}$$

(5.20)

Furthermore, if

$$ \int _{0}^{+\infty } \biggl[\frac{1}{1+t\zeta ^{2\gamma -1}(t)} \biggr]^{\frac{1}{2 \gamma -2}}\,dt< +\infty , \quad 1< \gamma < \frac{3}{2}, $$

(5.21)

then

$$ E(t)\leq k_{2} \biggl[\frac{1}{1+\int _{t_{0}}^{t}\zeta ^{\gamma }(\tau )d \tau } \biggr]^{\frac{1}{\gamma -1}}, \quad 1< \gamma < \frac{3}{2}, $$

(5.22)

where $\zeta (t)=\min \{\zeta _{1}(t),\zeta _{2}(t) \}$.

Proof

From (G1), we know $g_{1}$ and $g_{2}$ are positive, then for any $t\geq t_{0}>0$, we have

$$\begin{aligned}& \int _{0}^{t}g_{1}(\tau )\,d\tau \geq \int _{0}^{t_{0}}g_{1}(\tau )d \tau =g_{0}, \\& \int _{0}^{t}g_{2}(\tau )\,d\tau \geq \int _{0}^{t_{0}}g_{2}(\tau )d \tau =h_{0}. \end{aligned}$$

Taking derivative of (5.17) with respect to t and using (3.6), (5.1), and (5.16) yields

$$\begin{aligned} W'(t) &=E'(t)+\epsilon _{1}\phi '(t)+\epsilon _{2}\psi '(t) \\ &\leq \biggl(\frac{1}{2}-\frac{g_{1}(0)C_{*}^{2}}{4\delta }\epsilon _{2} \biggr) \bigl(g_{1}'\circ \nabla u \bigr) (t)+ \biggl(\frac{1}{2}-\frac{g_{2}(0)C _{*}^{2}}{4\delta }\epsilon _{2} \biggr) \bigl(g_{2}'\circ \nabla v \bigr) (t) \\ &\quad {} - \biggl[\epsilon _{2}(g_{0}-2\delta )- \epsilon _{1}\biggl(1+\frac{1}{4 \beta _{1}}\biggr) \biggr] \Vert u_{t} \Vert _{2}^{2} \\ &\quad {} - \biggl[\epsilon _{2}(h_{0}-2\delta )- \epsilon _{1}\biggl(1+\frac{1}{4 \beta _{2}}\biggr) \biggr] \Vert v_{t} \Vert _{2}^{2}+2(p+2)\epsilon _{1} \int _{\varOmega }F(u,v)\,dx \\ &\quad {} - \biggl[\frac{l\epsilon _{1}}{4}-\epsilon _{2} \bigl(1+2(1-l)^{2}+ \alpha _{2}l+\alpha _{3}l \bigr)\delta \biggr] \Vert \nabla u \Vert _{2}^{2} \\ &\quad {} - \biggl[\frac{k\epsilon _{2}}{4}-\epsilon _{2} \bigl(1+2(1-k)^{2}+ \alpha _{2}k+\alpha _{3}k \bigr)\delta \biggr] \Vert \nabla v \Vert _{2}^{2} \\ &\quad {} + \biggl[\frac{\epsilon _{1}}{2l}+ \biggl(2\delta +\frac{1}{2\delta }+ \frac{C_{*}^{2}}{2\delta } \biggr)\epsilon _{2} \biggr](1-l) (g_{1}\circ \nabla u) (t) \\ &\quad {} + \biggl[\frac{\epsilon _{2}}{2k}+ \biggl(2\delta +\frac{1}{2\delta }+ \frac{C_{*}^{2}}{2\delta } \biggr)\epsilon _{2} \biggr](1-k) (g_{2}\circ \nabla v) (t). \end{aligned}$$

(5.23)

At this point, we pick $\delta >0$ small enough such that

$$\begin{aligned} \delta \leq \min \biggl\{ \frac{g_{0}}{2}, \frac{h_{0}}{2} \biggr\} \end{aligned}$$

and

$$\begin{aligned}& \frac{4}{l}\delta \bigl(1+2(1-l)^{2}+\alpha _{2}l+\alpha _{3}l \bigr)< \frac{g _{0}}{4(1+\frac{1}{4\beta _{1}})}, \\& \frac{4}{k}\delta \bigl(1+2(1-k)^{2}+\alpha _{2}k+\alpha _{3}k \bigr)< \frac{g _{0}}{4(1+\frac{1}{4\beta _{2}})}. \end{aligned}$$

As long as δ is fixed, the choice of any two positive constants $\epsilon _{1}$ and $\epsilon _{2}$ satisfying

$$\begin{aligned} \frac{g_{0}}{4(1+\frac{1}{4\beta _{1}})}\epsilon _{2}< \epsilon _{1}< \frac{g _{0}}{2(1+\frac{1}{4\beta _{1}})}\epsilon _{2} \end{aligned}$$

and

$$\begin{aligned} \frac{h_{0}}{4(1+\frac{1}{4\beta _{2}})}\epsilon _{2}< \epsilon _{1}< \frac{h _{0}}{2(1+\frac{1}{4\beta _{2}})}\epsilon _{2} \end{aligned}$$

will make

$$\begin{aligned}& k_{1}=\epsilon _{2}(g_{0}-2\delta )- \epsilon _{1}\biggl(1+ \frac{1}{4\beta _{1}}\biggr)>0, \\& k_{2}=\epsilon _{2}(h_{0}-2\delta )- \epsilon _{1}\biggl(1+ \frac{1}{4\beta _{2}}\biggr)>0, \\& k_{3}=\frac{l\epsilon _{1}}{4}-\epsilon _{2} \bigl(1+2(1-l)^{2}+\alpha _{2}l+\alpha _{3}l \bigr)\delta >0, \\& k_{4}=\frac{k\epsilon _{2}}{4}-\epsilon _{2} \bigl(1+2(1-k)^{2}+\alpha _{2}k+\alpha _{3}k \bigr)\delta >0. \end{aligned}$$

Hence, there exist two positive constants m and C such that

$$\begin{aligned} W'(t)\leq -mE(t)+C \bigl[(g_{1}\circ \nabla u) (t)+(g_{2}\circ \nabla v) (t) \bigr],\quad t\geq t_{0}. \end{aligned}$$

(5.24)

Case 1. $\gamma =1$:

Let $\zeta (t)=\min \{\zeta _{1}(t),\zeta _{2}(t) \}$, since $\zeta _{1}(t)$ and $\zeta _{2}(t)$ are nonincreasing differentiable functions, then we get that $\zeta (t)$ is nonincreasing.

When $\gamma =1$, from (2.2), we easily get

$$\begin{aligned} g_{i}'(t)\leq -\zeta _{i}(t)g_{i}^{\gamma }(t)=- \zeta _{i}(t)g_{i}(t) \quad \text{for }i=1,2. \end{aligned}$$

(5.25)

Multiplying both sides of (5.24) by $\zeta (t)$ and applying (3.6), (4.27), (4.29), and (5.25) yields

$$\begin{aligned} \zeta (t)W'(t) &\leq -m\zeta (t)E(t)+C\zeta (t) (g_{1}\circ \nabla u) (t)+C \zeta (t) (g_{2}\circ \nabla v) (t) \\ &\leq -m\zeta (t)E(t)+C\zeta _{1}(t) (g_{1}\circ \nabla u) (t)+C\zeta _{2}(t) (g_{2}\circ \nabla v) (t) \\ &\leq -m\zeta (t)E(t)+C(\zeta _{1}g_{1}\circ \nabla u) (t)+C(\zeta _{2}g _{2}\circ \nabla v) (t) \\ &\leq -m\zeta (t)E(t)-C\bigl(g_{1}'\circ \nabla u \bigr) (t)-C\bigl(g_{2}'\circ \nabla v\bigr) (t) \\ &\leq -m\zeta (t)E(t)-CE'(t). \end{aligned}$$

(5.26)

Setting $F(t)=\zeta (t)W(t)+CE(t)$, then clearly $F\sim E$. Recalling that $W\sim E\geq 0$ by (3.21) and (5.18), $\zeta '(t) \leq 0$ by (G2), we get that $\zeta '(t)W(t)\leq 0$, then together with (5.26), we have, for some $k>0$,

$$\begin{aligned} F'(t) &=\zeta '(t)W(t)+\zeta (t)W'(t)+CE'(t) \\ &\leq \zeta (t)W'(t)+CE'(t) \\ &\leq -m\zeta (t)E(t)-CE'(t)+CE'(t) \\ &\leq -k\zeta (t)F(t). \end{aligned}$$

(5.27)

Integrating (5.27) over $[t_{0},t]$ gives

$$\begin{aligned} F(t)\leq F(t_{0})e^{-k\int _{t_{0}}^{t}\zeta (\tau )\,d\tau }, \quad \forall t \geq t_{0}. \end{aligned}$$

(5.28)

Therefore, by using the fact that $F(t)\sim E(t)$, we derive

$$ E(t)\leq Ke^{-k\int _{t_{0}}^{t}\zeta (\tau )\,d\tau }, \quad \forall t\geq t _{0}. $$

(5.29)

Case 2. $1<\gamma <\frac{3}{2}$:

Multiplying both sides of (5.24) by $\zeta (t)$, using (4.27) and (4.29) leads to

$$\begin{aligned} \zeta (t)W'(t) &\leq -m\zeta (t)E(t)+C\zeta (t) (g_{1}\circ \nabla u) (t)+C \zeta (t) (g_{2}\circ \nabla v) (t) \\ &\leq -m\zeta (t)E(t)+C\zeta _{1}(t) (g_{1}\circ \nabla u) (t)+C\zeta _{2}(t) (g_{2}\circ \nabla v) (t) \\ &\leq -m\zeta (t)E(t)+C\bigl[-E'(t)\bigr]^{\frac{1}{2\gamma -1}}. \end{aligned}$$

(5.30)

Multiplying (5.30) by $\zeta ^{\alpha }(t)E^{\alpha }(t)$, with $\alpha =2\gamma -2$, we get

$$\begin{aligned} \zeta ^{\alpha +1}(t)E^{\alpha }(t)W'(t)\leq -m\zeta ^{\alpha +1}(t)E ^{\alpha +1}(t)+C(\zeta E)^{\alpha }(t) \bigl[-E'(t)\bigr]^{\frac{1}{\alpha +1}}. \end{aligned}$$

(5.31)

Exploiting Young’s inequality with $q=1+\alpha$ and $q'=\frac{1+ \alpha }{\alpha }$ gives

$$\begin{aligned} \zeta ^{\alpha +1}(t)E^{\alpha }(t)W'(t) &\leq -m \zeta ^{\alpha +1}(t)E ^{\alpha +1}(t)+C \bigl[\epsilon \zeta ^{\alpha +1}(t)E^{\alpha +1}(t)-C _{\epsilon }E'(t) \bigr] \\ &\leq -(m-\epsilon C)\zeta ^{\alpha +1}(t)E^{\alpha +1}(t)-CE'(t). \end{aligned}$$

(5.32)

We pick $\epsilon <\frac{m}{C}$ and recall that $\zeta '(t)\leq 0$, $\zeta (t)>0$ by (G2), $E'(t)\leq 0$ by (3.6), and $W\thicksim E\geq 0$ by (3.21) and (5.18), then together with (5.32) we have, for some $c_{1}>0$,

$$\begin{aligned} \bigl(\zeta ^{\alpha +1}E^{\alpha }W \bigr)'(t) &=( \alpha +1)\zeta ^{\alpha }(t)\zeta '(t)E^{\alpha }(t)W(t) \\ &\quad {} +\alpha \zeta ^{\alpha +1}(t)E^{\alpha -1}(t)E'(t)W(t) \\ &\quad {} +\zeta ^{\alpha +1}(t)E^{\alpha }(t)W'(t) \\ &\leq \zeta ^{\alpha +1}(t)E^{\alpha }(t)W'(t) \\ &\leq -c_{1}\zeta ^{\alpha +1}(t)E^{\alpha +1}(t)-CE'(t). \end{aligned}$$

(5.33)

Next, we take $F(t)=\zeta ^{\alpha +1}WE^{\alpha }+CE$, which is clearly equivalent to $E(t)$, then there exists $a_{0}>0$ such that

$$\begin{aligned} F'(t)\leq -c_{1}\zeta ^{\alpha +1}(t)E^{\alpha +1}(t) \leq -a_{0} \zeta ^{\alpha +1}(t)F^{\alpha +1}(t). \end{aligned}$$

(5.34)

Integrating (5.34) over $(t_{0},t)$ and recalling that $F(t)\thicksim E(t)$ and $\alpha =2\gamma -2$, we obtain

$$\begin{aligned} E(t)\leq k_{1} \biggl[\frac{1}{1+\int _{t_{0}}^{t}\zeta ^{2\gamma -1}( \tau )\,d\tau } \biggr]^{\frac{1}{2\gamma -2}}, \quad \forall t\geq t_{0}. \end{aligned}$$

(5.35)

From (5.21) and (5.35), we infer that

$$\begin{aligned} \int _{t_{0}}^{+\infty }E(t)\,dt< +\infty . \end{aligned}$$

(5.36)

Setting $\lambda _{1}(t)=\int _{0}^{t} \Vert \nabla u(t)-\nabla u(t-\tau ) \Vert _{2} ^{2}\,d\tau$, by using (3.3), we deduce

$$\begin{aligned} \lambda _{1}(t)&= \int _{0}^{t} \bigl\Vert \nabla u(t)-\nabla u(t-\tau ) \bigr\Vert _{2}^{2}d \tau \leq C \int _{0}^{t} \bigl( \bigl\Vert \nabla u(t) \bigr\Vert _{2}^{2}+ \bigl\Vert \nabla u(t- \tau ) \bigr\Vert _{2}^{2} \bigr)\,d\tau \\ &\leq C \int _{0}^{t} \bigl[E(t)+E(t-\tau ) \bigr]\,d \tau \\ &\leq 2C \int _{0}^{t}E(t-\tau )\,d\tau \\ &=2C \int _{0}^{t}E(\tau )\,d\tau < 2C \int _{0}^{+\infty }E(\tau )\,d\tau < + \infty . \end{aligned}$$

(5.37)

Similarly, let $\lambda _{2}(t)=\int _{0}^{t} \Vert \nabla v(t)-\nabla v(t- \tau ) \Vert _{2}^{2}\,d\tau$, we have $\lambda _{2}(t)<+\infty$.

From (5.24) and recalling that $\zeta (t)=\min \{\zeta _{1}(t), \zeta _{2}(t) \}$, $\zeta _{1}(t)$ and $\zeta _{2}(t)$ are nonincreasing differentiable functions, we arrive at

$$\begin{aligned} \zeta (t)W'(t) &\leq -m\zeta (t)E(t)+C\zeta (t) \bigl[(g_{1}\circ \nabla u) (t)+(g_{2}\circ \nabla v) (t) \bigr] \\ &\leq -m\zeta (t)E(t)+C\zeta _{1}(t) (g_{1}\circ \nabla u) (t)+C\zeta _{2}(t) (g_{2}\circ \nabla v) (t) \\ &\leq -m\zeta (t)E(t)+C\frac{\lambda _{1}(t)}{\lambda _{1}(t)} \int _{0} ^{t} \bigl(\zeta _{1}^{\gamma }( \tau )g_{1}^{\gamma }(\tau ) \bigr)^{\frac{1}{ \gamma }} \bigl\Vert \nabla u(t)-\nabla u(t-\tau ) \bigr\Vert _{2}^{2} \,d\tau \\ &\quad {} +C\frac{\lambda _{2}(t)}{\lambda _{2}(t)} \int _{0}^{t} \bigl(\zeta _{2}^{\gamma }( \tau )g_{2}^{\gamma }(\tau ) \bigr)^{\frac{1}{\gamma }} \bigl\Vert \nabla v(t)-\nabla v(t-\tau ) \bigr\Vert _{2}^{2} \,d\tau . \end{aligned}$$

(5.38)

Exploiting Jensen’s inequality, for the second term on the right-hand side of (5.38), with $G(y)=y^{\frac{1}{\gamma }}$, $y>0$, $f(\tau )= \zeta ^{\gamma }(\tau )g^{\gamma }(\tau )$ and $h(\tau )= \Vert \nabla u(t)- \nabla u(t-\tau ) \Vert _{2}^{2}$, where we assume that $\lambda _{1}(t), \lambda _{2}(t)>0$, otherwise we get $\Vert \nabla u(t)-\nabla u(t-\tau ) \Vert = \Vert \nabla v(t)-\nabla v(t-\tau ) \Vert =0$, then by using (5.24) we deduce

$$\begin{aligned} E(t)\leq Ce^{-mt}. \end{aligned}$$

Since $\zeta _{1}(t)$ and $\zeta _{2}(t)$ are nonincreasing, then for some $C_{1}>0$, estimate (5.38) becomes

$$\begin{aligned} \zeta (t)W'(t) &\leq -m\zeta (t)E(t)+C\lambda _{1}(t) \biggl[\frac{1}{ \lambda _{1}(t)} \int _{0}^{t}\zeta _{1}^{\gamma }( \tau )g_{1}^{\gamma }( \tau ) \bigl\Vert \nabla u(t)-\nabla u(t-\tau ) \bigr\Vert _{2}^{2}\,d\tau \biggr]^{\frac{1}{ \gamma }} \\ &\quad {} +C\lambda _{2}(t) \biggl[\frac{1}{\lambda _{2}(t)} \int _{0}^{t}\zeta _{2}^{\gamma }( \tau )g_{2}^{\gamma }(\tau ) \bigl\Vert \nabla v(t)-\nabla v(t- \tau ) \bigr\Vert _{2}^{2}\,d\tau \biggr]^{\frac{1}{\gamma }} \\ &\leq -m\zeta (t)E(t)+C\lambda _{1}^{\frac{\gamma -1}{\gamma }}(t) \biggl[ \zeta _{1}^{\gamma -1}(0) \int _{0}^{t}\zeta _{1}(\tau )g_{1}^{ \gamma }(\tau ) \bigl\Vert \nabla u(t)-\nabla u(t- \tau ) \bigr\Vert _{2}^{2}\,d\tau \biggr]^{\frac{1}{ \gamma }} \\ &\quad {} +C\lambda _{2}^{\frac{\gamma -1}{\gamma }}(t) \biggl[\zeta _{2} ^{\gamma -1}(0) \int _{0}^{t}\zeta _{2}(\tau )g_{2}^{\gamma }(\tau ) \bigl\Vert \nabla v(t)-\nabla v(t- \tau ) \bigr\Vert _{2}^{2}\,d\tau \biggr]^{\frac{1}{\gamma }} \\ &=-m\zeta (t)E(t)+C \bigl(\lambda _{1}\zeta _{1}(0) \bigr)^{\frac{\gamma -1}{ \gamma }}\bigl(-g_{1}'\circ \nabla u \bigr)^{\frac{1}{\gamma }}+C \bigl(\lambda _{2}\zeta _{2}(0) \bigr)^{\frac{\gamma -1}{\gamma }}\bigl(-g_{2}' \circ \nabla v\bigr)^{\frac{1}{\gamma }} \\ &\leq -m\zeta (t)E(t)+C_{1}\bigl[-E'(t) \bigr]^{\frac{1}{\gamma }}. \end{aligned}$$

(5.39)

Multiplying both sides of (5.39) by $\zeta ^{\alpha }(t)E^{\alpha }(t)$, with $\alpha =\gamma -1$, we deduce

$$\begin{aligned} \zeta ^{\alpha +1}(t)E^{\alpha }(t)W'(t)\leq -m\zeta ^{\alpha +1}(t)E ^{\alpha +1}(t)+C_{1}\zeta ^{\alpha }(t)E^{\alpha }(t)\bigl[-E'(t) \bigr]^{\frac{1}{ \alpha +1}}. \end{aligned}$$

(5.40)

Applying Young’s inequality, with $q=1+\alpha$, and $q'=\frac{1+ \alpha }{\alpha }$ leads to

$$\begin{aligned} \zeta ^{\alpha +1}(t)E^{\alpha }(t)W'(t) &\leq -m \zeta ^{\alpha +1}(t)E ^{\alpha +1}(t)+C_{1} \bigl(\sigma \zeta ^{\alpha +1}(t)E^{\alpha +1}(t)-C _{\sigma }E'(t) \bigr) \\ &=-(m-C_{1}\sigma )\zeta ^{\alpha +1}(t)E^{\alpha +1}(t)-CE'(t). \end{aligned}$$

(5.41)

Then, by taking $\sigma <\frac{m}{C_{1}}$ and recalling that $\zeta '(t)\leq 0$, $\zeta (t)>0$ by (G2), $E'(t)\leq 0$ by (3.6), and $W(t)\thicksim E(t)\geq 0$ by (3.21) and (5.18), together with (5.41), we get, for some $C_{2}>0$,

$$\begin{aligned} \bigl(\zeta ^{\alpha +1}E^{\alpha }W \bigr)'(t) &=( \alpha +1)\zeta ^{\alpha }(t)\zeta '(t)E^{\alpha }(t)W(t) \\ &\quad {} +\alpha \zeta ^{\alpha +1}(t)E^{\alpha -1}(t)E'(t)W(t) \\ &\quad {} +\zeta ^{\alpha +1}(t)E^{\alpha }(t)W'(t) \\ &\leq \zeta ^{\alpha +1}(t)E^{\alpha }(t)W'(t) \\ &\leq -C_{2}\zeta ^{\alpha +1}(t)E^{\alpha +1}(t)-CE'(t), \end{aligned}$$

(5.42)

which implies

$$\begin{aligned} \bigl(\zeta ^{\alpha +1}E^{\alpha }W+CE \bigr)'(t) \leq -C_{2}\zeta ^{ \alpha +1}(t)E^{\alpha +1}(t). \end{aligned}$$

(5.43)

Let $L=\zeta ^{\alpha +1}E^{\alpha }W+CE$, then clearly $L\thicksim E$, we obtain, for some $C_{3}>0$,

$$\begin{aligned} L'(t)\leq -C_{3}\zeta ^{\alpha +1}(t)L^{\alpha +1}(t), \quad \forall t \geq t_{0}. \end{aligned}$$

(5.44)

Integrating (5.44) over $(t_{0},t)$ and recalling that $L\thicksim E$ and $\alpha =\gamma -1$ yields

$$\begin{aligned} E(t)\leq k_{2} \biggl[\frac{1}{1+\int _{t_{0}}^{t}\zeta ^{\gamma }(\tau )d \tau } \biggr]^{\frac{1}{\gamma -1}}, \quad \forall t\geq t_{0}. \end{aligned}$$

(5.45)

This completes the proof. □

6 The blow-up result

In this section, we carry out the proof of the finite time blow-up result.

Theorem 6.1

If (G1) and (G3) hold and the initial energy $E(0)<0$. Assume that $g_{i}$ satisfies

$$\begin{aligned} \int _{0}^{\infty }g_{i}(\tau )\,d\tau < \frac{p+1}{p+2},\quad i=1,2. \end{aligned}$$

(6.1)

Then the solution of system (1.1) blows up in finite time.

Proof

First, we define

$$\begin{aligned} H(t)=-E(t), \quad\quad G(t)= \int _{\varOmega }F(u,v)\,dx, \end{aligned}$$

(6.2)

where $E(t)$ is defined in (3.3). By (G1) and (3.6), we find that

$$\begin{aligned} H'(t)=-E'(t) &=-\frac{1}{2} \bigl[ \bigl(g_{1}'\circ \nabla u\bigr) (t)+ \bigl(g_{2}' \circ \nabla v\bigr) (t) \bigr]+ \Vert u_{t} \Vert _{2}^{2} \\ &\quad {} + \Vert v_{t} \Vert _{2}^{2}+ \frac{g_{1}(t)}{2} \Vert \nabla u \Vert _{2}^{2}+ \frac{g _{2}(t)}{2} \Vert \nabla v \Vert _{2}^{2} \\ &\geq \Vert u_{t} \Vert _{2}^{2}+ \Vert v_{t} \Vert _{2}^{2} \\ &\geq 0. \end{aligned}$$

(6.3)

Noting the assumption $E(0)<0$, from (4.1), (6.2), and (6.3), we get

$$\begin{aligned} 0< H(0)\leq H(t)\leq G(t)\leq c_{1} \bigl( \Vert u \Vert _{2(p+2)}^{2(p+2)}+ \Vert v \Vert _{2(p+2)} ^{2(p+2)} \bigr). \end{aligned}$$

(6.4)

From (4.1), we see that

$$\begin{aligned} G(t)\geq c_{0} \bigl( \Vert u \Vert _{2(p+2)}^{2(p+2)}+ \Vert v \Vert _{2(p+2)}^{2(p+2)} \bigr). \end{aligned}$$

(6.5)

Let

$$\begin{aligned} L(t)=H^{1-\alpha }(t)+\epsilon \int _{\varOmega }uu_{t}\,dx+\epsilon \int _{\varOmega }vv_{t}\,dx, \end{aligned}$$

(6.6)

where ϵ is a positive constant to be chosen later and

$$\begin{aligned} 0< \alpha < \min \biggl\{ \frac{1}{2}-\frac{1}{2(p+2)}, \frac{p}{2p+2} \biggr\} . \end{aligned}$$

(6.7)

Taking the derivative of $L(t)$ and using system (1.1) gives

$$\begin{aligned} L'(t) &=(1-\alpha )H^{-\alpha }(t)H'(t)+ \epsilon \int _{\varOmega }uu_{tt}\,dx+ \epsilon \int _{\varOmega }vv_{tt}\,dx+\epsilon \Vert u_{t} \Vert _{2}^{2}+\epsilon \Vert v_{t} \Vert _{2}^{2} \\ &=(1-\alpha )H^{-\alpha }(t)H'(t)+\epsilon \int _{0}^{t}g_{1}(t- \tau ) \int _{\varOmega }\nabla u(\tau )\nabla u(t)\,dx\,d\tau \\ &\quad {} +\epsilon \int _{0}^{t}g_{2}(t-\tau ) \int _{\varOmega }\nabla u( \tau )\nabla u(t)\,dx\,d\tau -\epsilon \Vert \nabla u \Vert _{2}^{2}-\epsilon \Vert \nabla v \Vert _{2}^{2} \\ &\quad {} -\epsilon \int _{\varOmega }uu_{t}\,dx-\epsilon \int _{\varOmega }vv _{t}\,dx+2(p+2)\epsilon G(t). \end{aligned}$$

(6.8)

It follows from Young’s inequality that

$$\begin{aligned} & \int _{0}^{t}g_{1}(t-\tau ) \int _{\varOmega }\nabla u(\tau )\nabla u(t)\,dx\,d \tau \\ &\quad = \int _{0}^{t}g_{1}(t-\tau ) \int _{\varOmega }\nabla u(t) \bigl(\nabla u( \tau )-\nabla u(t) \bigr)\,dx\,d\tau \\ & \qquad {} + \int _{0}^{t}g_{1}(\tau )\,d\tau \bigl\Vert \nabla u(t) \bigr\Vert _{2}^{2} \\ &\quad \geq - \int _{0}^{t}g_{1}(\tau )\,d\tau \bigl\Vert \nabla u(t) \bigr\Vert _{2}^{2}- \frac{1}{4}(g_{1}\circ \nabla u) (t) \\ & \qquad {} + \int _{0}^{t}g_{1}(\tau )\,d\tau \bigl\Vert \nabla u(t) \bigr\Vert _{2}^{2} \\ &\quad =-\frac{1}{4}(g_{1}\circ \nabla u) (t). \end{aligned}$$

(6.9)

Similarly,

$$\begin{aligned} \int _{0}^{t}g_{2}(t-\tau ) \int _{\varOmega }\nabla v(\tau )\nabla v(t)\,dx\,d \tau \geq - \frac{1}{4}(g_{2}\circ \nabla v) (t). \end{aligned}$$

(6.10)

A combination of (6.8)–(6.10) leads to

$$\begin{aligned} L'(t) &\geq (1-\alpha )H^{-\alpha }(t)H'(t)- \frac{\epsilon }{4}(g_{1} \circ \nabla u) (t)-\frac{\epsilon }{4}(g_{2} \circ \nabla v) (t)-\epsilon \Vert \nabla u \Vert _{2}^{2} \\ &\quad {} -\epsilon \Vert \nabla v \Vert _{2}^{2}- \epsilon \int _{\varOmega }uu_{t}\,dx- \epsilon \int _{\varOmega }vv_{t}\,dx+\epsilon \Vert u_{t} \Vert _{2}^{2}+\epsilon \Vert v _{t} \Vert _{2}^{2} \\ &\quad{} +2(p+2)\epsilon G(t). \end{aligned}$$

(6.11)

Let $0< r\leq \min \{l, k \}$, then by the expression of $E(t)$ and $H(t)$, we obtain

$$\begin{aligned} - \bigl( \bigl\Vert \nabla u(t) \bigr\Vert _{2}^{2}+ \bigl\Vert \nabla v(t) \bigr\Vert _{2}^{2} \bigr) & \geq \frac{2}{r}H(t)+\frac{1}{r} \bigl( \Vert u_{t} \Vert _{2}^{2}+ \Vert v_{t} \Vert _{2}^{2} \bigr)- \frac{2}{r}G(t) \\ &\quad {} +\frac{1}{r} \bigl[(g_{1}\circ \nabla u) (t)+(g_{2}\circ \nabla v) (t) \bigr]. \end{aligned}$$

(6.12)

Inserting (6.12) into (6.11), we arrive at

$$\begin{aligned} L'(t) &\geq (1-\alpha )H^{-\alpha }(t)H'(t)+ \epsilon \frac{2}{r}H(t)+ \epsilon \biggl(1+\frac{1}{r}\biggr) \bigl( \Vert u_{t} \Vert _{2}^{2}+ \Vert v_{t} \Vert _{2}^{2} \bigr) \\ &\quad {} +\epsilon \biggl(\frac{1}{r}-\frac{1}{4}\biggr) \bigl[(g_{1}\circ \nabla u) (t)+(g _{2}\circ \nabla v) (t) \bigr]+\epsilon \biggl[2(p+2)-\frac{2}{r} \biggr]G(t) \\ &\quad {} -\epsilon \int _{\varOmega }uu_{t}\,dx-\epsilon \int _{\varOmega }vv _{t}\,dx. \end{aligned}$$

(6.13)

By using (G1), (6.1), and $r=\min \{l, k \}$, we infer that

$$\begin{aligned} 2(p+2)-\frac{2}{r}>0, \quad\quad \frac{1}{r}- \frac{1}{4}>0. \end{aligned}$$

(6.14)

For the last two terms of (6.13), by using Hölder’s inequality and (6.3)–(6.5) yields

$$\begin{aligned} \int _{\varOmega }uu_{t}\,dx &\leq \Vert u \Vert _{2} \Vert u_{t} \Vert _{2} \\ &\leq \vert \varOmega \vert ^{\frac{p+1}{2(p+2)}} \Vert u \Vert _{2(p+2)} \Vert u_{t} \Vert _{2} \\ &\leq c_{0}^{\frac{-1}{2(p+2)}} \vert \varOmega \vert ^{\frac{p+1}{2(p+2)}}G^{ \frac{1}{2(p+2)}}(t) \Vert u_{t} \Vert _{2} \\ &\leq K_{1} \Vert u_{t} \Vert _{2}G^{\frac{1}{2}}(t)H^{\frac{1}{2(p+2)}- \frac{1}{2}}(t), \end{aligned}$$

(6.15)

where $K_{1}=c_{0}^{\frac{-1}{2(p+2)}} \vert \varOmega \vert ^{\frac{p+1}{2(p+2)}}$.

By (6.7), we know $\frac{1}{2(p+2)}-\frac{1}{2}+\alpha <0$, it follows from Young’s inequality and (6.3) that

$$\begin{aligned} \int _{\varOmega }uu_{t}\,dx &\leq K_{1} \Vert u_{t} \Vert _{2}G^{\frac{1}{2}}(t)H^{ \frac{1}{2(p+2)}-\frac{1}{2}}(t) \\ &\leq H^{\frac{1}{2(p+2)}-\frac{1}{2}}(t) \bigl(\delta _{1}G(t)+K_{1} ^{2}\delta _{1}^{-\frac{1}{2}} \Vert u_{t} \Vert _{2}^{2} \bigr) \\ &\leq \delta _{1}H^{\frac{1}{2(p+2)}-\frac{1}{2}}(0)G(t)+K_{1}^{2} \delta _{1}^{-\frac{1}{2}}H^{\frac{1}{2(p+2)}-\frac{1}{2}+\alpha }(t)H ^{-\alpha }(t)H'(t) \\ &\leq \delta _{1}H^{\frac{1}{2(p+2)}-\frac{1}{2}}(0)G(t)+K_{1}^{2} \delta _{1}^{-\frac{1}{2}}H^{\frac{1}{2(p+2)}-\frac{1}{2}+\alpha }(0)H ^{-\alpha }(t)H'(t), \quad \forall \delta _{1}>0. \end{aligned}$$

(6.16)

Likewise,

$$\begin{aligned} \int _{\varOmega }vv_{t}\,dx\leq \delta _{2}H^{\frac{1}{2(p+2)}-\frac{1}{2}}(0)G(t)+K _{1}^{2} \delta _{1}^{-\frac{1}{2}}H^{\frac{1}{2(p+2)}-\frac{1}{2}+ \alpha }(0)H^{-\alpha }(t)H'(t), \quad \forall \delta _{2}>0. \end{aligned}$$

(6.17)

We pick

$$\begin{aligned} \delta _{1}=\delta _{2}=\frac{1}{4} \biggl[2(p+2)-\frac{2}{r} \biggr]H^{ \frac{1}{2(p+2)}-\frac{1}{2}}(0), \end{aligned}$$

(6.18)

then it follows from (6.13)–(6.18) that

$$\begin{aligned} L'(t) &\geq (1-\alpha -\epsilon K_{3})H^{-\alpha }(t)H'(t)+ \epsilon \frac{2}{r}H(t)+\epsilon \biggl(1+\frac{1}{r}\biggr) \bigl( \Vert u_{t} \Vert _{2}^{2}+ \Vert v _{t} \Vert _{2}^{2} \bigr) \\ &\quad {} +\epsilon \biggl(\frac{1}{r}-\frac{1}{4}\biggr) \bigl[(g_{1}\circ \nabla u) (t)+(g _{2}\circ \nabla v) (t) \bigr]+\frac{\epsilon }{2} \biggl[2(p+2)- \frac{2}{r} \biggr]G(t), \end{aligned}$$

(6.19)

where

$$\begin{aligned} K_{3} &=K_{1}^{2}\delta _{1}^{-\frac{1}{2}}H^{\frac{1}{2(p+2)}- \frac{1}{2}+\alpha }(0)+K_{1}^{2} \delta _{1}^{-\frac{1}{2}}H^{ \frac{1}{2(p+2)}-\frac{1}{2}+\alpha }(0) \\ &=4 \biggl[2(p+2)-\frac{2}{r} \biggr]^{-\frac{1}{2}}K_{1}^{2}H^{ \frac{1}{4(p+2)}-\frac{1}{4}+\alpha }(0). \end{aligned}$$

(6.20)

Now, we choose $0<\epsilon <1$ small enough such that

$$\begin{aligned} 1-\alpha -\epsilon K_{3}\geq 0. \end{aligned}$$

(6.21)

A combination of (6.14), (6.19), and (6.21) yields

$$\begin{aligned} L'(t) &\geq (1-\alpha -\epsilon K_{3})H^{-\alpha }(t)H'(t)+ \epsilon \frac{2}{r}H(t)+\epsilon \biggl(1+\frac{1}{r}\biggr) \bigl( \Vert u_{t} \Vert _{2}^{2}+ \Vert v _{t} \Vert _{2}^{2} \bigr) \\ &\quad {} +\epsilon \biggl(\frac{1}{r}-\frac{1}{4}\biggr) \bigl[(g_{1}\circ \nabla u) (t)+(g _{2}\circ \nabla v) (t) \bigr]+\frac{\epsilon }{2} \biggl[2(p+2)- \frac{2}{r} \biggr]G(t) \\ &\geq \epsilon C \bigl[H(t)+ \Vert u_{t} \Vert _{2}^{2}+ \Vert v_{t} \Vert _{2}^{2}+(g_{1} \circ \nabla u) (t)+(g_{2}\circ \nabla v) (t)+G(t) \bigr] \\ &>0, \end{aligned}$$

(6.22)

where C is a positive constant. It is clear that $L(t)$ is increasing on $[0,T)$ and

$$\begin{aligned} L(t) &=H^{1-\alpha }(t)+\epsilon \int _{\varOmega }uu_{t}\,dx+\epsilon \int _{\varOmega }vv_{t}\,dx \\ &\geq H^{1-\alpha }(0)+\epsilon \int _{\varOmega }u_{0}u_{1}\,dx+\epsilon \int _{\varOmega }v_{0}v_{1}\,dx. \end{aligned}$$

(6.23)

In the case of $\int _{\varOmega }u_{0}u_{1}\,dx+\epsilon \int _{\varOmega }v _{0}v_{1}\,dx\geq 0$, no further restriction on ϵ is needed. For the case $\int _{\varOmega }u_{0}u_{1}\,dx+\epsilon \int _{\varOmega }v_{0}v _{1}\,dx<0$, we assume that

$$\begin{aligned} 0< \epsilon < -\frac{H^{1-\alpha }(0)}{2\int _{\varOmega }u_{0}u_{1}\,dx+2 \int _{\varOmega }v_{0}v_{1}\,dx}. \end{aligned}$$

(6.24)

Therefore, in either case, we have

$$\begin{aligned} L(t)\geq \frac{1}{2}H^{1-\alpha }(0)>0, \quad \forall t\in [0,T). \end{aligned}$$

(6.25)

Now we prove that $L(t)$ satisfies the following inequality:

$$\begin{aligned} L'(t)\geq C\epsilon L^{\frac{1}{1-\alpha }}(t), \quad t\in [0,T), \end{aligned}$$

(6.26)

where C is a positive constant, $1<\frac{1}{1-\alpha }<2$ assumed by (6.7). At this point, we distinguish two cases.

Case 1. $\int _{\varOmega }uu_{t}\,dx+\int _{\varOmega }vv_{t}\,dx\leq 0$ for some $t\in [0,T)$. For such t, we obtain

$$\begin{aligned} L^{\frac{1}{1-\alpha }}(t)= \biggl(H^{1-\alpha }(t)+\epsilon \int _{ \varOmega }uu_{t}\,dx+\epsilon \int _{\varOmega }vv_{t}\,dx \biggr)^{\frac{1}{1- \alpha }} \leq H(t), \end{aligned}$$

(6.27)

which together with (6.22), then (6.26) follows for all such t.

Case 2. $\int _{\varOmega }uu_{t}\,dx+\int _{\varOmega }vv_{t}\,dx\geq 0$ for all $t\in [0,T)$. Since $0<\alpha <\frac{1}{2}$, $1<\frac{1}{1-\alpha }<2$, and $0<\epsilon <1$, then we deduce

$$\begin{aligned} L^{\frac{1}{1-\alpha }}(t)\leq 2^{\frac{1}{1-\alpha }-1} \biggl(H(t)+ \biggl\vert \int _{\varOmega }uu_{t}\,dx+\epsilon \int _{\varOmega }vv_{t}\,dx \biggr\vert ^{\frac{1}{1- \alpha }} \biggr). \end{aligned}$$

(6.28)

Exploiting Hölder’s inequality and Young’s inequality, we see that

$$\begin{aligned} \biggl\vert \int _{\varOmega }uu_{t}\,dx+\epsilon \int _{\varOmega }vv_{t}\,dx \biggr\vert ^{\frac{1}{1- \alpha }} &\leq C \bigl( \Vert u \Vert _{2}^{\frac{1}{1-\alpha }} \Vert u_{t} \Vert _{2}^{\frac{1}{1- \alpha }}+ \Vert v \Vert _{2}^{\frac{1}{1-\alpha }} \Vert v_{t} \Vert _{2}^{\frac{1}{1- \alpha }} \bigr) \\ &\leq C \bigl( \Vert u \Vert _{2p+2}^{\frac{1}{1-\alpha }} \Vert u_{t} \Vert _{2}^{\frac{1}{1- \alpha }}+ \Vert v \Vert _{2p+2}^{\frac{1}{1-\alpha }} \Vert v_{t} \Vert _{2}^{\frac{1}{1- \alpha }} \bigr) \\ &\leq C \bigl( \Vert u \Vert _{2p+2}^{\frac{2}{1-2\alpha }}+ \Vert u_{t} \Vert _{2}^{2}+ \Vert v \Vert _{2p+2} ^{\frac{2}{1-2\alpha }}+ \Vert v_{t} \Vert _{2}^{2} \bigr). \end{aligned}$$

(6.29)

From (6.7), we get that $\frac{1}{(1-2\alpha )(p+1)}-1<0$. Then, by (6.3) and (6.5), we have

$$\begin{aligned} \Vert u \Vert _{2p+2}^{\frac{2}{1-2\alpha }} &= \bigl( \Vert u \Vert _{2p+2}^{2p+2} \bigr)^{\frac{1}{(1-2 \alpha )(p+1)}} \\ &\leq CG^{\frac{1}{(1-2\alpha )(p+1)}}(t) \\ &\leq CG^{\frac{1}{(1-2\alpha )(p+1)}-1}(t)G(t) \\ &\leq CH^{\frac{1}{(1-2\alpha )(p+1)}-1}(0)G(t) \\ &\leq CG(t). \end{aligned}$$

(6.30)

Similarly,

$$\begin{aligned} \Vert u \Vert _{2p+2}^{\frac{2}{1-2\alpha }}\leq CG(t). \end{aligned}$$

(6.31)

By combining (6.28)–(6.31), we deduce that

$$\begin{aligned} L^{\frac{1}{1-\alpha }}(t)\leq C \bigl(H(t)+ \Vert u_{t} \Vert _{2}^{2}+ \Vert v_{t} \Vert _{2} ^{2}+G(t) \bigr). \end{aligned}$$

(6.32)

It follows from (6.22) and (6.32) that

$$\begin{aligned} L'(t)\geq C\epsilon \bigl(H(t)+ \Vert u_{t} \Vert _{2}^{2}+ \Vert v_{t} \Vert _{2}^{2}+G(t) \bigr)\geq C\epsilon L^{\frac{1}{1-\alpha }}(t), \end{aligned}$$

(6.33)

which shows (6.26) follows for Case 2.

Integrating (6.26) over $(0,t)$ yields

$$\begin{aligned} L^{\frac{\alpha }{1-\alpha }}(t)\geq \frac{1}{L^{\frac{-\alpha }{1- \alpha }}(0)-\frac{\alpha }{1-\alpha }C\epsilon t}. \end{aligned}$$

(6.34)

This shows that $G(t)$ blows up in finite time

$$\begin{aligned} T^{*}\leq \frac{1-\alpha }{\alpha C\epsilon L^{\frac{\sigma }{1- \sigma }}(0)}, \quad \forall t\geq 0. \end{aligned}$$

(6.35)

Therefore, the solution of system (1.1) blows up in finite time. □