Initial boundary value problem for a viscoelastic wave equation with Balakrishnan–Taylor damping and a delay term: decay estimates and blow-up result

Abstract

In this paper, we study the initial boundary value problem for the following viscoelastic wave equation with Balakrishnan–Taylor damping and a delay term where the relaxation function satisfies $g'(t)\leq -\xi (t)g^{r}(t)$, $t\geq 0$, $1\leq r< \frac{3}{2}$. The main goal of this work is to study the global existence, general decay, and blow-up result. The global existence has been obtained by potential-well theory, the decay of solutions of energy has been established by introducing suitable energy and Lyapunov functionals, and a blow-up result has been obtained with negative initial energy.

1 Introduction

In this paper, we consider the following initial-boundary value problem with a delay term

$$ \textstyle\begin{cases} v_{tt}- (a+b \Vert \nabla v \Vert _{2}^{2}+\alpha \int _{\Omega}\nabla v \nabla v_{t}\,dx )\Delta v\\ \quad {}+\int _{0}^{t}g(t-s)\Delta v(s)\,ds +\mu _{1}v_{t}+\mu _{2}v_{t}(t-\tau )= \vert v \vert ^{p-2}v,& x\in \Omega , t>0, \\ v(x,t)=0,& x\in \partial \Omega , t>0, \\ v(x,0)=v_{0}(x),\qquad v_{t}(x,0)=v_{1}(x),& x\in \Omega , \\ v_{t}(x,t-\tau )=f_{0}(x,t-\tau),& x\in \Omega , t\in [0, \tau ) , \end{cases} $$

(1.1)

where $\Omega \subset \mathbb{R}^{n}$ is a bounded domain with sufficiently smooth boundary ∂Ω. $p\geq 4, a, b, \alpha , \mu _{1}$ are fixed positive constants, $\mu _{2}$ is a real number, $\tau >0$ represents the time delay, and g is a positive function.

In the absence of the Balakrishnan–Taylor damping $(\alpha =0)$, Problem (1.1) is reduced to the well-known nonlinear wave equation with $b=g=0$ and a Kirchhof-type wave equation with $g=0$, which has been extensively studied, see for instance [5, 8, 13, 24, 30, 31, 35, 38, 41, 42] and the references therein. Balakrishnan–Taylor damping $(\alpha \neq 0)$, $g=0$, and $\mu _{1}=\mu _{2}=0$, was initially proposed by Balakrishnan and Taylor [2], and Bass and Zes [3]. It is related to the panel flutter equation and to the spillover problem. So far, it has been studied by many authors, we refer the interested readers to [12, 15, 32, 39, 43, 44] and the references therein. Zarai and Tatar [44] studied the following problem

$$ v_{tt}- \biggl(a+b \Vert \nabla v \Vert _{2}^{2}+\sigma \int _{\Omega}\nabla v \nabla v_{t}\,dx \biggr)\Delta v+ \int _{0}^{t}h(t-s)\Delta v(s)\,ds=0. $$

(1.2)

They proved the global existence and the polynomial decay of the problem. Exponential decay and blow up of the solution to the problem were established in Tatar and Zarai [39].

It is well known that time-delay effects often appear in many chemical, physical, and economical phenomena because these phenomena depend not only on the present state but also on the past history of the system. Nicaise and Pignotti [33] considered the following wave equation with a delay term

$$ v_{tt}-\Delta v+\mu _{1}v_{t}+ \mu _{2}v_{t}(t-\tau )=0. $$

(1.3)

They obtained some stability results in the case $0<\mu _{2}<\mu _{1}$. Then, they extended the result to the time-dependent delay case in the work of Nicaise and Pignotti [34]. Kirane and Said-Houari [23] considered a viscoelastic wave equation with time delay

$$ v_{tt}-\Delta v+ \int _{0}^{t}g(t-s)\Delta v(s)\,ds+\mu _{1}v_{t}+\mu _{2}v_{t}(t- \tau )=0. $$

(1.4)

They proved the global well posedness of solutions and established the decay rate of energy for $0<\mu _{2}<\mu _{1}$. Kafini et al. [17] investigated the following nonlinear wave equation with delay

$$ v_{tt}-\operatorname{div} \bigl( \vert \nabla v \vert ^{m-2}\nabla v \bigr)+\mu _{1}v_{t}+\mu _{1}v_{t}(t- \tau )=b \vert v \vert ^{p-2}v. $$

(1.5)

They proved the blow-up result of solutions with negative initial energy and $p\geq m$, and we refer the interested readers to [9, 10, 18, 27] and the references therein. For the viscoelastic wave equation with Balakrishnan–Taylor damping and time delay, Lee et al. [25] studied the following equation

$$ \begin{aligned}[b] &v_{tt}- \biggl(a+b \Vert \nabla v \Vert _{2}^{2}+\sigma \int _{\Omega}\nabla v \nabla v_{t}\,dx \biggr)\Delta v \\ &\quad {}+ \int _{0}^{t}g(t-s)\Delta v(s)\,ds+\mu _{0}v_{t}+ \mu _{1}v_{t}(t-\tau )=0 \end{aligned} $$

(1.6)

and established a general energy decay result by suitable Lyapunov functionals. Gheraibia et al. [14] considered the following equation

$$ \begin{aligned}[b] &v_{tt}- \biggl(a+b \Vert \nabla v \Vert _{2}^{2}+\alpha \int _{\Omega}\nabla v \nabla v_{t}\,dx \biggr)\Delta v +\sigma (t) \int _{0}^{t}g(t-s)\Delta v(s)\,ds+ \mu _{1} \vert v_{t} \vert ^{m-2}v_{t} \\ &\quad{} +\mu _{2} \bigl\vert v_{t}(t-\tau ) \bigr\vert ^{m-2}v_{t}(t-\tau )=0 \end{aligned} $$

(1.7)

and proved the general decay result of the solution in the case $|\mu _{2}|< \mu _{1}$. For the related works of PDEs with time delay, see for instance [6, 7, 11, 16, 19–22, 26, 28, 36, 37, 40] and the references therein.

Motivated by the previous work, in this paper, we consider the problem (1.1) and under suitable assumptions on the relaxation functions g, we prove the global existence, general decay and the finite-time blow-up results of the solutions.

The outline of this paper is as follows: In Sect. 2, we give some preliminary results. In Sect. 3, we obtain the global existence of the solution of (1.1). Section 4 and Sect. 5 cover the general decay and blow-up of solutions, respectively.

2 Some preliminaries

In this section, we give some notation for function spaces and preliminary lemmas. Denote by $\|\cdot \|_{p}$ and $\|\cdot \|_{H^{1}}$ to the usual $L^{p}(\Omega )$ norm and $H^{1}(\Omega )$ norm, respectively.

For the relaxation function g, we assume

$(A_{1})$: $g:\mathbb{R}^{+}\rightarrow \mathbb{R}^{+}$ is a nonincreasing differentiable function satisfying

$$ a- \int _{0}^{\infty}g(s)\,ds:= l\geq 0. $$

(2.1)

$(A_{2})$: There exist a nonincreasing differentiable function ξ with $\xi (0)>0$ satisfying

$$ g(t)\geq 0,\qquad g'(t)\leq -\xi (t)g^{r}(t),\quad t\geq 0, 1 \leq r< \frac{3}{2}. $$

(2.2)

$(A_{3})$: The constant p satisfies

$$ p\geq 4,\quad \text{if}\,n=1,2,\qquad 4\leq p\leq \frac{2(n-1)}{n-2},\quad \text{if}\,n\geq 3. $$

(2.3)

$(A_{4})$: The constants $\mu _{1}$ and $\mu _{2}$ satisfy

$$ \vert \mu _{2} \vert < \mu _{1}. $$

Assume further that g satisfies

$$ \int _{0}^{\infty}g(s)\,ds< \frac{a(p-2)}{p-2+(1/2\eta )}. $$

(2.4)

Lemma 2.1

(Sobolev–Poincare inequality [1]). Let q be a number with $2\leq q<\infty \,(n=1,2)$ or $2\leq q<\frac{2n}{n-2}\,(n\geq 3)$, then, there is a constant $c_{*}=c_{*}(\Omega ,q)$ such that

$$ \Vert v \Vert _{q}\leq c_{*} \Vert \nabla v \Vert _{2}\quad \textit{for}\ v\in H_{0}^{1}( \Omega ). $$

By using direct calculations, we have

$$ \begin{aligned}[b] {} \int _{0}^{t}g(t-s) \int _{\Omega}v(s)\,dsv_{t}(t)\,dx ={}&{-} \frac{1}{2} \frac{d}{dt} \biggl[ (g\circ v ) (t)- \bigl\Vert v(t) \bigr\Vert _{2}^{2} \int _{0}^{t}g(s)\,ds \biggr] \\ &{} -\frac{1}{2}g(t) \bigl\Vert v(t) \bigr\Vert _{2}^{2}+\frac{1}{2} \bigl(g' \circ v \bigr) (t), \end{aligned} $$

(2.5)

where

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

To deal with the time-delay term, motivated by Nicaise and Pignotti [33], we introduce a new variable

$$ z(x,\rho ,t)=v_{t}(x,t-\tau \rho ),\quad x\in \Omega , \rho \in (0,1), \,t>0, $$

(2.6)

which gives us

$$ \tau z_{t}(x,\rho ,t)+z_{\rho}(x,\rho ,t)=0,\quad \text{in}\ \Omega \times (0,1)\times (0,\infty ). $$

(2.7)

Then, problem (1.1)is equivalent to

$$ \textstyle\begin{cases} v_{tt}- (a+b \Vert \nabla v \Vert _{2}^{2}+\alpha \int _{\Omega}\nabla v \nabla v_{t}\,dx )\Delta v\\ \quad {}+\int _{0}^{t}g(t-s)\Delta v(s)\,ds +\mu _{1}v_{t}+\mu _{2}z(1,t)= \vert v \vert ^{p-2}v,& \text{$x\in \Omega ,\,t>0$}, \\ \tau z_{t}(\rho ,t)+z_{\rho}(\rho ,t)=0,& \text{$x\in \Omega ,\,\rho \in (0,1),\,t>0$}, \\ z(\rho ,0)=f_{0}(-\tau \rho ),& \text{$x\in \Omega ,\,\rho \in (0,1)$}, \\ v(x,t)=0,& \text{$x\in \partial \Omega ,\,t>0$}, \\ v(x,0)=v_{0}(x),\qquad v_{t}(x,0)=v_{1}(x),& \text{$x\in \Omega $}. \end{cases} $$

(2.8)

Let ζ be a positive constant satisfying

$$ \tau \vert \mu _{2} \vert \leq \zeta \leq \tau \bigl(2\mu _{1}- \vert \mu _{2} \vert \bigr). $$

(2.9)

We first state a local existence theorem that can be established.

Theorem 2.2

Let $(A_{1})$–$(A_{4})$ hold. Then, for every $(v_{0},v_{1})\in H^{1}_{0}(\Omega )\times L^{2}(\Omega )$, $f_{0}\in L^{2}(( \Omega )\times (0,1))$, there exists a unique local solution of the problem (1.1) in the class

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

Now, we define the energy associated with problem (2.8) by

$$ \begin{aligned}[b] E(t)={}&\frac{1}{2} \Vert v_{t} \Vert _{2}^{2}+\frac{1}{2} \biggl(a- \int _{0}^{t}g(s)\,ds \biggr) \Vert \nabla v \Vert _{2}^{2} +\frac{b}{4} \Vert \nabla v \Vert _{2}^{4}+ \frac{1}{2} (g\circ \nabla v ) (t) \\ &{} +\frac{\zeta}{2} \int _{0}^{1} \bigl\Vert z(\rho ,t) \bigr\Vert _{2}^{2}\,d\rho - \frac{1}{p} \Vert v \Vert _{p}^{p}. \end{aligned} $$

(2.10)

Lemma 2.3

Let $(v,z)$ be a solution of problem (2.8). Then,

$$ E'(t)\leq \frac{1}{2} \bigl(g'\circ \nabla v \bigr) (t)-c_{0} \bigl( \Vert v_{t} \Vert _{2}^{2} + \bigl\Vert z(1,t) \bigr\Vert _{2}^{2} \bigr). $$

(2.11)

Proof

Multiplying the first equation in (2.8) by $v_{t}$, integrating over Ω, and using (2.5), we obtain

$$ \begin{aligned}[b] &\frac{d}{dt} \biggl[ \frac{1}{2} \Vert v_{t} \Vert _{2}^{2}+ \frac{1}{2} \biggl(a- \int _{0}^{t}g(s)\,ds \biggr) \Vert \nabla v \Vert _{2}^{2} +\frac{b}{4} \Vert \nabla v \Vert _{2}^{4}+\frac{1}{2} (g\circ \nabla v ) (t)-\frac{1}{p} \Vert v \Vert _{p}^{p} \biggr] \\ &\quad =-\alpha \biggl(\frac{1}{2}\frac{d}{dt} \Vert \nabla v \Vert _{2}^{2} \biggr)^{2}- \frac{1}{2}g(t) \Vert \nabla v \Vert _{2}^{2}- \frac{1}{2} \bigl(g' \circ \nabla v \bigr) (t)\\ &\qquad {} -\mu _{1} \Vert v_{t} \Vert _{2}^{2}- \mu _{2} \int _{ \Omega}z(1,t)v_{t}\,dx. \end{aligned} $$

(2.12)

Multiplying the second equation in (2.8) by ζz and integrating over $\Omega \times (0,1)$, we obtain

$$ \begin{aligned}[b] {}\frac{\zeta}{2} \frac{d}{dt} \int _{\Omega} \int _{0}^{1} \bigl\vert z(\rho ,t) \bigr\vert ^{2}\,d \rho\,dx&= -\frac{\zeta}{2\tau} \int _{\Omega} \int _{0}^{1} \frac{\partial}{\partial \rho} \bigl\vert z(\rho ,t) \bigr\vert ^{2}\,d\rho\,dx \\ &=\frac{\zeta}{2\tau} \bigl( \Vert v_{t} \Vert _{2}^{2}- \bigl\Vert z(1,t) \bigr\Vert _{2}^{2} \bigr). \end{aligned} $$

(2.13)

Using Young’s inequality, we have

$$ -\mu _{2} \int _{\Omega}z(1,t)v_{t}\,dx\leq \frac{ \vert \mu _{2} \vert }{2} \bigl\Vert z(1,t) \bigr\Vert _{2}^{2}+ \frac{ \vert \mu _{2} \vert }{2} \Vert v_{t} \Vert _{2}^{2}. $$

(2.14)

Combining (2.12), (2.13), and (2.14), we obtain

$$\begin{aligned} E'(t)\leq {}&{-}\alpha \biggl(\frac{1}{2} \frac{d}{dt} \Vert \nabla v \Vert _{2}^{2} \biggr)^{2} +\frac{1}{2} \bigl(g'\circ \nabla v \bigr) (t)-\frac{1}{2}g(t) \Vert \nabla v \Vert _{2}^{2} \\ &{}-c_{0} \bigl( \Vert v_{t} \Vert _{2}^{2}+ \bigl\Vert z(1,t) \bigr\Vert _{2}^{2} \bigr), \end{aligned}$$

(2.15)

where $c_{0}=\min \{\mu _{1}-\frac{\zeta}{2\tau}-\frac{|\mu _{2}|}{2}, \frac{\zeta}{2\tau}-\frac{|\mu _{2}|}{2} \}$, which is positive by (2.9). The proof is complete. □

Next, we define the functionals

$$\begin{aligned} I(t)={}& \biggl(a- \int _{0}^{t}g(s)\,ds \biggr) \Vert \nabla v \Vert _{2}^{2} + \frac{b}{2} \Vert \nabla v \Vert _{2}^{4}+ (g\circ \nabla v ) (t) \\ &{} + \zeta \int _{0}^{1} \bigl\Vert z(\rho ,t) \bigr\Vert _{2}^{2}\,d\rho - \Vert v \Vert _{p}^{p} \end{aligned}$$

(2.16)

and

$$\begin{aligned} J(t)={}&\frac{1}{2} \biggl(a- \int _{0}^{t}g(s)\,ds \biggr) \Vert \nabla v \Vert _{2}^{2} +\frac{b}{4} \Vert \nabla v \Vert _{2}^{4}+\frac{1}{2} (g\circ \nabla v ) (t) \\ &{}+\frac{\zeta}{2} \int _{0}^{1} \bigl\Vert z(\rho ,t) \bigr\Vert _{2}^{2}\,d\rho - \frac{1}{p} \Vert v \Vert _{p}^{p}. \end{aligned}$$

(2.17)

Then, it is obvious that

$$ E(t)=\frac{1}{2} \Vert v_{t} \Vert _{2}^{2}+J(t). $$

(2.18)

3 Global existence

In this section, we will prove that the global existence of the solution to (1.1) is in time.

Lemma 3.1

Assume that $(A_{1})$, $(A_{3})$–$(A_{4})$ hold, and for any $(v_{0},v_{1})\in H^{1}_{0}(\Omega )\times L^{2}(\Omega )$, such that

$$ I(0)>0\quad \textit{and}\quad \beta= \frac{c_{*}^{p}}{l} \biggl[\frac{2p}{l(p-2)}E(0) \biggr]^{\frac{p-2}{2}}< 1, $$

(3.1)

then,

$$ I(t)>0,\quad \forall t>0. $$

(3.2)

Proof

Since $I(0) > 0$, then by the continuity of v, there exists a time $T_{m}>0$ such that

$$ I(t)\geq 0,\quad \forall t\in [0,T_{m}]. $$

(3.3)

From (2.16) and (2.17), we have

$$\begin{aligned} J(t) =&\frac{p-2}{2p} \biggl[ \biggl(a- \int _{0}^{t}g(s)\,ds \biggr) \Vert \nabla v \Vert _{2}^{2}+\frac{b}{2} \Vert \nabla v \Vert _{2}^{4}+ (g\circ \nabla v ) (t)+\zeta \int _{0}^{1} \bigl\Vert z(\rho ,t) \bigr\Vert _{2}^{2}\,d\rho \biggr] \\ &{}+\frac{1}{p}I(t) \\ \geq& \frac{p-2}{2p} \biggl[ \biggl(a- \int _{0}^{t}g(s)\,ds \biggr) \Vert \nabla v \Vert _{2}^{2}+\frac{b}{2} \Vert \nabla v \Vert _{2}^{4}+ (g\circ \nabla v ) (t)+\zeta \int _{0}^{1} \bigl\Vert z(\rho ,t) \bigr\Vert _{2}^{2}\,d\rho \biggr] \\ \geq& \frac{p-2}{2p} \biggl[ \biggl(a- \int _{0}^{t}g(s)\,ds \biggr) \Vert \nabla v \Vert _{2}^{2} \biggr]. \end{aligned}$$

(3.4)

Thus, from $(A_{1})$, (2.11), (2.18), and (3.4), we obtain

$$ \begin{aligned}[b] l \Vert \nabla v \Vert _{2}^{2}&\leq \biggl(a- \int _{0}^{t}g(s)\,ds \biggr) \Vert \nabla v \Vert _{2}^{2}\\ &\leq \frac{2p}{p-2}J(t) \leq \frac{2p}{p-2}E(t) \leq \frac{2p}{p-2}E(0),\quad \forall t\in [0,T_{m}]. \end{aligned} $$

(3.5)

Exploiting Lemma 2.1, (3.1), and (3.5), we obtain

$$ \begin{aligned}[b] \Vert v \Vert _{p}^{p}&\leq c_{*}^{p} \Vert \nabla v \Vert _{2}^{p}\leq \frac{c_{*}^{p}}{l} \biggl(\frac{2p}{l(p-2)}E(0) \biggr)^{ \frac{p-2}{2}}l \Vert \nabla v \Vert _{2}^{2} \\ &= \beta l \Vert \nabla v \Vert _{2}^{2}< \biggl(a- \int _{0}^{t}g(s)\,ds \biggr) \Vert \nabla v \Vert _{2}^{2}. \end{aligned} $$

(3.6)

Hence, we can obtain

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

By repeating the procedure, $T_{m}$ is extended to T. The proof is complete. □

Theorem 3.2

Assume that the conditions of Lemma 3.1hold, then the solution (1.1) is global and bounded.

Proof

It suffices to show that $\|v_{t}\|_{2}^{2}+\|\nabla v\|_{2}^{2}$ is bounded independently of t. By using (2.11), (2.18), and (3.5), we obtain

$$ E(0)\geq E(t)=J(t)+\frac{1}{2} \Vert v_{t} \Vert _{2}^{2}\geq \frac{p-2}{2p} \bigl(l \Vert \nabla v \Vert _{2}^{2} \bigr)+\frac{1}{2} \Vert v_{t} \Vert _{2}^{2}. $$

(3.7)

Therefore, we have

$$ \Vert v_{t} \Vert _{2}^{2}+ \Vert \nabla v \Vert _{2}^{2}\leq K_{1}E(0), $$

(3.8)

where $K_{1}$ is a positive constant. □

4 General decay

In this section, we prove the general decay result by constructing a suitable Lyapunov functional.

Theorem 4.1

Let $(v_{0},v_{1})\in H_{0}^{1}(\Omega )\times L^{2}(\Omega )$. Assume that $(A_{1})$–$(A_{4})$ hold. Then, there exist two positive constants K and k such that the solution of problem (1.1) satisfies, for all $\forall t\geq t_{0}$,

$$\begin{aligned} E(t) \leq& Ke^{-k\int _{t_{0}}^{t}\xi (s)\,ds},\quad r=1, \end{aligned}$$

(4.1)

$$\begin{aligned} E(t) \leq& K \biggl[\frac{1}{\int _{t_{0}}^{t}\xi ^{2r-1}(s)\,ds+1} \biggr]^{1/(2r-2)},\quad r>1 . \end{aligned}$$

(4.2)

Moreover, if

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

(4.3)

then

$$ E(t)\leq K \biggl[\frac{1}{\int _{t_{0}}^{t}\xi ^{r}(s)\,ds+1} \biggr]^{1/r-1}, \quad r>1. $$

(4.4)

For this goal, we set

$$ F(t):=E(t)+\varepsilon \Psi (t), $$

(4.5)

where ε is a positive constant to be specified later and

$$ \Psi (t)= \int _{\Omega}vv_{t}\,dx+\frac{\alpha}{4} \Vert \nabla v \Vert _{2}^{4}. $$

(4.6)

In order to show our stability result, we need the following lemmas:

Lemma 4.2

Let $(v,z)$ be a solution of problem (2.8). Then, there exist two positive constants $\alpha _{1}$ and $\alpha _{2}$ such that

$$ \alpha _{1}F(t)\leq E(t)\leq \alpha _{2} F(t), $$

(4.7)

for $\varepsilon >0$ small enough.

Lemma 4.3

Assume that g satisfies $(A_{1})$ and $(A_{2})$, then

$$ \int _{0}^{\infty}\xi (t)g^{1-\theta}(t)\,dt\leq +\infty ,\quad \forall \theta < 2-r . $$

Corollary 4.4

([4]) Assume that g satisfies $(A_{1})$ and $(A_{2})$, and v is the solution of (1.1), then

$$ \xi (t) (g\circ \nabla u ) (t)\leq \bigl[-E'(t) \bigr]^{ \frac{1}{2r-1}}. $$

Lemma 4.5

Let $(v,z)$ be a solution of problem (2.8). Then, the functional $F(t)$ satisfies

$$ F'(t)\leq -k_{1}E(t)+k_{2} (g\circ \nabla v ) (t),\quad \forall t\geq t_{0}, $$

(4.8)

where $k_{1}$ and $k_{2}$ are some positive constants.

Proof

Taking a derivation of (4.5), using (2.8), and Lemma 2.3, we obtain

$$ \begin{aligned}[b] F'(t)={}&E'(t)+ \varepsilon \int _{\Omega}v_{t}^{2}\,dx+\varepsilon \int _{ \Omega}vv_{tt}\,dx+\varepsilon \alpha \Vert \nabla v \Vert _{2}^{2} \int _{\Omega} \nabla v \nabla v_{t}\,dx \\ \leq{}&{-}(c_{0}-\varepsilon ) \Vert v_{t} \Vert _{2}^{2}-c_{0} \bigl\Vert z(1,t) \bigr\Vert _{2}^{2}- \varepsilon a \Vert \nabla v \Vert _{2}^{2} -\epsilon b \Vert \nabla v \Vert _{2}^{4}+ \varepsilon \Vert v \Vert _{p}^{p} \\ &{} +\varepsilon \int _{\Omega}\nabla v \int _{0}^{t}g(t-s)\nabla v(s)\,ds\,dx- \varepsilon \mu _{1} \int _{\Omega}vv_{t}\,dx -\varepsilon \mu _{2} \int _{ \Omega}z(1,t)v\,dx. \end{aligned} $$

(4.9)

By using Hölder’s, Young’s, Sobolev–Poincare inequalities, and $(A_{1})$, we obtain

$$ \int _{\Omega}\nabla v \int _{0}^{t}g(t-s)\nabla v(s)\,ds\,dx\leq \bigl( \eta +(a-l) \bigr) \Vert \nabla v \Vert _{2}^{2}+ \frac{(a-l)}{4\eta} (g \circ \nabla v ) (t) $$

(4.10)

and

$$ \mu _{1} \int _{\Omega}vv_{t}\,dx\leq \eta \mu _{1}^{2}c_{*}^{2} \Vert \nabla v \Vert _{2}^{2}+\frac{1}{4\eta} \Vert v_{t} \Vert _{2}^{2} $$

(4.11)

and

$$ \mu _{2} \int _{\Omega}z(1,t)v\,dx\leq \eta \mu _{2}^{2}c_{*}^{2} \Vert \nabla v \Vert _{2}^{2}+\frac{1}{4\eta} \bigl\Vert z(1,t) \bigr\Vert _{2}^{2}. $$

(4.12)

Combining (4.10)–(4.12) and (4.9), we obtain

$$ \begin{aligned}[b] F'(t)\leq{}&{-} \biggl\{ c_{0}-\varepsilon \biggl(1+\frac{1}{4\eta} \biggr) \biggr\} \Vert v_{t} \Vert _{2}^{2} - \biggl\{ c_{0}-\frac{\varepsilon}{4\eta} \biggr\} \bigl\Vert z(1,t) \bigr\Vert _{2}^{2}-\varepsilon b \Vert \nabla v \Vert _{2}^{4} \\ &{} -\varepsilon \bigl\{ l-\eta \bigl(1+\mu _{1}^{2}c_{*}^{2} \mu _{2}^{2}c_{*}^{2}\bigr) \bigr\} \Vert \nabla v \Vert _{2}^{2} + \frac{(a-l)}{4\eta} (g\circ \nabla v ) (t)+\varepsilon \Vert v \Vert _{p}^{p}. \end{aligned} $$

(4.13)

At this point, we choose η and ε so small that (4.7) remains valid and

$$ l-\eta \bigl(1+\mu _{1}^{2}c_{*}^{2} \mu _{2}^{2}c_{*}^{2}\bigr)>0, \qquad c_{0}- \varepsilon \biggl(1+\frac{1}{4\eta} \biggr)>0, \qquad c_{0}- \frac{\varepsilon}{4\eta}>0 . $$

Consequently, inequality (4.13) becomes

$$ F'(t)\leq -k_{1}E(t)+k_{2} (g\circ \nabla v ) (t),\quad \forall t\geq t_{0}, $$

(4.14)

where $k_{i},\,i=1,2$. are some positive constants. □

Now, we are ready to prove Theorem 4.1.

Proof of Theorem 4.1. Multiplying (4.14) by $\xi (t)$, we obtain

$$ \xi (t) F'(t)\leq -k_{1}\xi (t)E(t)+k_{2}\xi (t) (g\circ \nabla u ) (t),\quad \forall t\geq t_{0}. $$

(4.15)

4.1 Case: $r=1$

Using $(A_{2})$ and (2.11), then inequality (4.14) becomes

$$ \begin{aligned} \xi (t)F'(t)&\leq -k_{1}\xi (t)E(t)+k_{2}\xi (t) (g\circ \nabla v ) (t) \\ &\leq -k_{1}\xi (t)E(t)-k_{2} \bigl(g' \circ \nabla v \bigr) (t) \\ &\leq -k_{1}\xi (t)E(t)-2k_{2}E'(t). \end{aligned} $$

(4.16)

We choose $G(t)=\xi (t)F(t)+2k_{2}E(t)$ that is equivalent to $E(t)$ because of (4.7). Then, from (4.16) we can obtain

$$ G'(t)\leq -k_{0}\xi (t)E(t)\leq -k \xi (t)G(t),\quad \forall t\geq t_{0}. $$

(4.17)

A simple integration of (4.17), leads to

$$ G(t)\leq G(t_{0})e^{-k\int _{t_{0}}^{t}\xi (s)\,ds},\quad \forall t \geq t_{0}, $$

(4.18)

which implies

$$ E(t)\leq Ke^{-k\int _{t_{0}}^{t}\xi (s)\,ds},\quad \forall t\geq t_{0}. $$

(4.19)

4.2 Case: $r>1$

Applying Corollary 4.4, then inequality (4.15) becomes

$$ \xi (t)F'(t)\leq -k_{1}\xi (t)E(t)+k_{2} \bigl[-E'(t) \bigr]^{1/(2r-1)}, \quad \forall t\geq t_{0}. $$

(4.20)

Multiplying (4.20) by $\xi ^{\nu}(t)E^{\nu}(t)$ where $\nu =2r-2$, we have

$$ \begin{aligned}[b] & \xi ^{\nu +1}(t)E^{\nu}(t)F'(t) \\ &\quad\leq -k_{1}\xi ^{\nu +1}(t)E^{\nu +1}(t)+k_{2} \xi ^{\nu}(t)E^{\nu}(t) \bigl[-E'(t) \bigr]^{1/(\nu +1)},\quad \forall t\geq t_{0}. \end{aligned} $$

(4.21)

Using Young’s inequality with $q=\nu +1$ and $q^{*}=\frac{\nu +1}{\nu}$, yields

$$ \begin{aligned}[b] &\xi ^{\nu +1}(t)E^{\nu}(t)F'(t) \\ &\quad \leq -k_{1}\xi ^{\nu +1}(t)E^{\nu +1}(t) +k_{2} \bigl[\eta \xi ^{ \nu +1}(t)E^{\nu +1}(t)-C_{\eta}E'(t) \bigr] \\ &\quad =- (k_{1}-\eta k_{2} )\xi ^{\nu +1}(t)E^{\nu +1}(t)-C_{ \eta}E'(t), \quad \forall t\geq t_{0}. \end{aligned} $$

(4.22)

At this point, we choose $\eta <\frac{k_{1}}{k_{2}}$ and recall that $\xi'(t)\leq0$ and $E'(t)\leq0$, we obtain

$$ \begin{aligned} \bigl(\xi ^{\nu +1}E^{\nu}F \bigr)'(t) &\leq \xi ^{\nu +1}(t)E^{\nu}(t)F'(t) \\ &\leq -k_{3}\xi ^{\nu +1}(t)E^{\nu +1}(t)-k_{4}E'(t), \quad \forall t \geq t_{0}, \end{aligned} $$

which implies

$$ \bigl(\xi ^{\nu +1}E^{\nu}F+k_{4}F \bigr)'(t)\leq -k_{3}\xi ^{\nu +1}(t)E^{ \nu +1}(t), \quad \forall t\geq t_{0}. $$

(4.23)

We choose $G(t)=\xi ^{\nu +1}(t)E^{\nu}(t)F(t)+k_{4}E(t)$ that is equivalent to $E(t)$. Then,

$$ \begin{aligned}[b] G'(t)&\leq -k_{3}\xi ^{\nu +1}(t)G^{\nu +1}(t) \\ &=-k_{3}\xi ^{2r-1}(t)G^{2r-1}(t),\quad \forall t\geq t_{0}. \end{aligned} $$

(4.24)

A simple integration of (4.24) and using the fact that $G(t)\sim E(t)$, leads to

$$ E(t)\leq K \biggl[\frac{1}{\int _{t_{0}}^{t}\xi ^{2r-1}(s)\,ds+1} \biggr]^{1/(2r-2)}, \quad \forall t\geq t_{0}. $$

(4.25)

4.3 Case: $1< r<3/2$

To establish (4.4), we note that from simple calculations show that (4.2) and (4.3) yield

$$ \int _{t_{0}}^{\infty}E(t)< \infty . $$

Next, let

$$ \sigma (t)= \int _{0}^{t} \bigl\Vert \nabla v(t)-\nabla v(t-s) \bigr\Vert _{2}^{2}\,ds, $$

then, we have

$$ \begin{aligned} \sigma (t)&\leq c \int _{0}^{t} \bigl[ \bigl\Vert \nabla v(t) \bigr\Vert _{2}^{2}+ \bigl\Vert \nabla v(t-s) \bigr\Vert _{2}^{2} \bigr]\,ds \leq c \int _{0}^{t} \bigl[E(t)+E(t-s) \bigr]\,ds \leq 2c \int _{0}^{t}E(t-s)\,ds \\ &=2c \int _{0}^{t}E(s)\,ds\leq 2c \int _{0}^{\infty}E(s)\,ds< \infty . \end{aligned} $$

Applying Jensens’s inequality for the second term on the right-hand side of (4.15) and using $(A_{2})$, we obtain

$$ \begin{aligned}[b] \xi(t) F'(t)&\leq -k_{1}\xi (t)E(t)+k_{2}\xi (t) (g\circ \nabla v ) (t) \\ &=-k_{1}\xi (t)E(t)+k_{2}\frac{\sigma (t)}{\sigma (t)} \int _{0}^{t} \bigl[\xi ^{r}(s)g^{r}(s) \bigr]^{\frac{1}{r}} \bigl\Vert \nabla v(t)-\nabla v(t-s) \bigr\Vert _{2}^{2}\,ds \\ &\leq -k_{1}\xi (t)E(t)+k_{2}\sigma (t) \biggl[ \frac{1}{\sigma (t)} \int _{0}^{t}\xi ^{r}(s)g^{r}(s) \bigl\Vert \nabla v(t)-\nabla v(t-s) \bigr\Vert _{2}^{2}\,ds \biggr]^{\frac{1}{r}} \\ &\leq -k_{1}\xi (t)E(t)+k_{2}\sigma ^{\frac{r-1}{r}}(t) \xi ^{r-1}(0) \biggl[ \int _{0}^{t}\xi (s)g^{r}(s) \bigl\Vert \nabla v(t)-\nabla v(t-s) \bigr\Vert _{2}^{2}\,ds \biggr]^{\frac{1}{r}} \\ &\leq -k_{1}\xi (t)E(t)+k_{2} \biggl[ \int _{0}^{t}-g'(s) \bigl\Vert \nabla v(t)- \nabla v(t-s) \bigr\Vert _{2}^{2}\,ds \biggr]^{\frac{1}{r}} \\ &\leq -k_{1}\xi (t)E(t)+k_{2} \bigl[-E'(t) \bigr]^{\frac{1}{r}}. \end{aligned} $$

(4.26)

Multiplying (4.26) by $\xi ^{\nu}(t)E^{\nu}(t)$, where $\nu =r-1$, we have

$$ \xi ^{\nu +1}(t)E^{\nu}(t)F'(t) \leq -k_{1}\xi ^{\nu +1}(t)E^{\nu +1}(t)+k_{2} \xi ^{\nu}(t)E^{\nu}(t) \bigl[-E'(t) \bigr]^{\frac{1}{\nu +1}},\quad \forall t\geq t_{0}. $$

(4.27)

The remainder of the proof is similar to (4.2). The proof is complete.

5 Blow up

In this section, we state and prove the blow up of the solution to problem (1.1) with negative initial energy.

Let

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

(5.1)

where $E(0)<0$. From (5.1) and (2.11) we have

$$ H'(t)=-E'(t)\geq c_{0} \bigl( \Vert v_{t} \Vert _{2}^{2}+ \bigl\Vert z(1,t) \bigr\Vert _{2}^{2} \bigr)\geq 0 $$

(5.2)

and $H(t)$ is an increasing function. Using (2.10) and (5.1), we obtain

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

(5.3)

Moreover, similar to the work of Messaoudi [29], we can obtain the following lemma that is needed later.

Lemma 5.1

Suppose that $(A_{1})$, $(A_{3})$, $(A_{4})$, (2.4), and $E(0)<0$ hold. Then, we have, for any $2\leq s\leq p$,

$$ \Vert v \Vert _{p}^{s}\leq C \biggl(-H(t)- \Vert v_{t} \Vert _{2}^{2}- \Vert \nabla v \Vert ^{4}_{2}-(g \circ \nabla v) (t)- \int _{0}^{1} \bigl\Vert z(\rho ,t) \bigr\Vert _{2}^{2}\,d\rho + \Vert v \Vert _{p}^{p} \biggr), $$

where C is a positive constant.

Theorem 5.2

Let the conditions of Lemma 5.1hold. Then, the solution of problem (1.1) blows up in finite time.

Proof

Set

$$ \Gamma (t)=H^{1-\sigma}(t)+\varepsilon \int _{\Omega}vv_{t}\,dx+ \frac{\alpha}{4} \Vert \nabla v \Vert _{2}^{4}, $$

(5.4)

where $\varepsilon >0$ is a small constant that will be chosen later, and

$$ 0< \sigma \leq \min \biggl\{ \frac{p-2}{2p}, \frac{p-2}{p} \biggr\} . $$

(5.5)

Taking a derivative of (5.4) and using the first equation in (2.8), we have

$$ \begin{aligned}[b] \Gamma '(t)={}&(1-\sigma )H^{-\sigma}(t)H'(t)+\varepsilon \int _{\Omega}v_{t}^{2}\,dx+ \varepsilon \int _{\Omega}vv_{tt}\,dx +\alpha \Vert \nabla u \Vert _{2}^{2} \int _{ \Omega}\nabla u\nabla u_{t}\,dx \\ ={}&(1-\sigma )H^{-\sigma}(t)H'(t)+\varepsilon \Vert v_{t} \Vert _{2}^{2}- \varepsilon a \Vert \nabla v \Vert _{2}^{2} -\epsilon b \Vert \nabla v \Vert _{2}^{4}+ \varepsilon \Vert v \Vert _{p}^{p} \\ &{} +\varepsilon \int _{\Omega}\nabla v \int _{0}^{t}g(t-s)\nabla v(s)\,ds\,dx- \varepsilon \mu _{1} \int _{\Omega}vv_{t}\,dx -\varepsilon \mu _{2} \int _{ \Omega}z(1,t)v\,dx. \end{aligned} $$

(5.6)

Applying Hölder’s and Young’s inequalities, for $\eta ,\delta >0$, we have

$$\begin{aligned}& \int _{\Omega}\nabla v \int _{0}^{t}g(t-s)\nabla v(s)\,ds\,dx \geq \biggl(1- \frac{1}{4\eta} \biggr) \biggl( \int _{0}^{t}g(s)\,ds \biggr) \Vert \nabla v \Vert _{2}^{2}- \eta (g\circ \nabla v) (t), \end{aligned}$$

(5.7)

$$\begin{aligned}& \mu _{1} \int _{\Omega}vv_{t}\,dx\leq \delta \mu _{1}^{2} \Vert v \Vert _{2}^{2} + \frac{1}{4\delta} \Vert v_{t} \Vert _{2}^{2} \leq \delta \mu _{1}^{2} \Vert v \Vert _{2}^{2}+ \frac{1}{4 c_{0}\delta}H'(t) \end{aligned}$$

(5.8)

and

$$ \mu _{2} \int _{\Omega}z(1,t)v\,dx\leq \delta \mu _{2}^{2} \Vert v \Vert _{2}^{2}+ \frac{1}{4\delta} \bigl\Vert z(1,t) \bigr\Vert _{2}^{2}\leq \delta \mu _{2}^{2} \Vert v \Vert _{2}^{2}+ \frac{1}{4 c_{0}\delta}H'(t). $$

(5.9)

Combining these estimates (5.7)–(5.9) and (5.6), we obtain

$$\begin{aligned} \Gamma '(t) \geq& \biggl\{ (1-\sigma )H^{-\sigma}(t)- \frac{\varepsilon}{2c_{0}\delta} \biggr\} H'(t)+\varepsilon \Vert v_{t} \Vert _{2}^{2} -\epsilon b \Vert \nabla v \Vert _{2}^{4}+\varepsilon \Vert v \Vert _{p}^{p} \\ &{} -\varepsilon \biggl\{ a- \biggl(1-\frac{1}{4\eta} \biggr) \biggl( \int _{0}^{t}g(s)\,ds \biggr) \biggr\} \Vert \nabla v \Vert _{2}^{2} - \varepsilon \delta \bigl(\mu _{1}^{2}+\mu _{2}^{2}\bigr) \Vert v \Vert _{2}^{2} \\ &{} -\varepsilon \eta (g\circ \nabla v) (t). \end{aligned}$$

(5.10)

Applying (2.10) to the last term $\|v\|_{p}^{p}$ on the right-hand side of (5.10) and using (5.1), we see that

$$\begin{aligned} \Gamma '(t) \geq &\biggl\{ (1-\sigma )H^{-\sigma}(t)- \frac{\varepsilon}{2c_{0}\delta} \biggr\} H'(t)+\varepsilon \biggl( \frac{p}{2}+1 \biggr) \Vert v_{t} \Vert _{2}^{2} +\varepsilon b \biggl( \frac{p}{4}-1 \biggr) \Vert \nabla v \Vert _{2}^{4} \\ &{} +\varepsilon \biggl\{ a \biggl(\frac{p}{2}-1 \biggr)- \biggl( \frac{p}{2} -1+\frac{1}{4\eta} \biggr) \int _{0}^{t}g(s)\,ds \biggr\} \Vert \nabla v \Vert _{2}^{2}+\varepsilon \biggl(\frac{p}{2}- \eta \biggr) (g \circ \nabla v) (t) \\ &{} -\varepsilon \delta \bigl(\mu _{1}^{2}+\mu _{2}^{2}\bigr) \Vert v \Vert _{2}^{2} + \varepsilon \frac{p\zeta}{2} \int _{0}^{1} \bigl\Vert z(\rho ,t) \bigr\Vert _{2}^{2}\,d\rho + \varepsilon pH(t), \end{aligned}$$

(5.11)

for some number η with $0<\eta <p/2$. By recalling (2.4), the estimate (5.11) reduces to

$$ \begin{aligned}[b] \Gamma '(t)\geq {}&\biggl\{ (1-\sigma )H^{-\sigma}(t)- \frac{\varepsilon}{2c_{0}\delta} \biggr\} H'(t) +\varepsilon \biggl( \frac{p}{2}+1 \biggr) \Vert v_{t} \Vert _{2}^{2} +\varepsilon c_{1} \Vert \nabla v \Vert _{2}^{4}\\ &{}+ \varepsilon c_{2} \Vert \nabla v \Vert _{2}^{2} +\varepsilon c_{3}(g\circ \nabla v) (t) -\varepsilon \delta \bigl( \mu _{1}^{2}+\mu _{2}^{2} \bigr) \Vert v \Vert _{2}^{2}\\ &{}+\varepsilon \frac{p\zeta}{2} \int _{0}^{1} \bigl\Vert z(\rho ,t) \bigr\Vert _{2}^{2}\,d\rho +\varepsilon pH(t), \end{aligned} $$

(5.12)

where

$$ c_{1}=b \biggl(\frac{p}{4}-1 \biggr)>0,\quad c_{2}=a \biggl(\frac{p}{2}-1 \biggr) - \biggl( \frac{p}{2}-1+\frac{1}{4\eta} \biggr) \int _{0}^{t}g(s)\,ds>0, \quad c_{3}= \frac{p}{2}-\eta >0. $$

Therefore, by taking $\delta =H(t)^{\sigma}/2c_{0}k$, where $k>0$ is to be specified later, and exploiting (5.3), we se that

$$ H(t)^{\sigma} \Vert v \Vert _{2}^{2} \leq \frac{1}{p^{\sigma}} \Vert v \Vert _{p}^{\sigma p} \Vert v \Vert _{2}^{2} \leq \frac{c_{p}^{2}}{p^{\sigma}} \Vert v \Vert _{p}^{\sigma p+2}. $$

(5.13)

Substituting (5.13) into (5.12), we obtain

$$ \begin{aligned}[b] \Gamma '(t)\geq{}& \bigl\{ (1-\sigma )-\varepsilon k \bigr\} H^{-\sigma}(t)H'(t) + \varepsilon \biggl(\frac{p}{2}+1 \biggr) \Vert v_{t} \Vert _{2}^{2} + \varepsilon c_{1} \Vert \nabla v \Vert _{2}^{4}\\ &{}+\varepsilon c_{2} \Vert \nabla v \Vert _{2}^{2} +\varepsilon c_{3}(g\circ \nabla v) (t)+\varepsilon c_{4} \Vert v \Vert _{p}^{p} - \varepsilon \frac{c_{5}}{k} \Vert v \Vert _{p}^{\sigma p+2}\\ &{}+ \varepsilon \frac{p\zeta}{2} \int _{0}^{1} \bigl\Vert z(\rho ,t) \bigr\Vert _{2}^{2}\,d\rho +\varepsilon pH(t), \end{aligned} $$

(5.14)

where $c_{5}=(c_{p}^{2}(\mu _{1}^{2}+\mu _{2}^{2}))/2c_{0}p^{\sigma}$. From (5.5) and Lemma 5.1, for $s=\sigma p+2\leq p$, we deduce

$$ \Vert v \Vert _{p}^{\sigma p+2}\leq C \biggl(-H(t)- \Vert v_{t} \Vert _{2}^{2}- \Vert \nabla v \Vert _{2}^{4}-(g\circ \nabla v) (t)- \int _{0}^{1} \bigl\Vert z(\rho ,t) \bigr\Vert _{2}^{2}\,d \rho + \Vert v \Vert _{p}^{p} \biggr). $$

(5.15)

Combining (5.15) with (5.14), we obtain

$$ \begin{aligned}[b] \Gamma '(t)\geq{}& \bigl\{ (1-\sigma )-\varepsilon k \bigr\} H^{-\sigma}(t)H'(t) + \varepsilon \biggl(\frac{p}{2}+1+\frac{c_{5}}{k}C \biggr) \Vert v_{t} \Vert _{2}^{2}+ \varepsilon c_{2} \Vert \nabla v \Vert _{2}^{2} \\ &{} +\varepsilon \biggl(c_{1}+\frac{c_{5}}{k}C \biggr) \Vert \nabla v \Vert _{2}^{4}+ \varepsilon \biggl(c_{3}+\frac{c_{5}}{k}C \biggr) (g\circ \nabla v) (t)- \frac{c_{5}}{k}C \Vert v \Vert _{p}^{p} \\ &{} +\varepsilon \biggl(\frac{p\zeta}{2}+\frac{c_{5}}{k}C \biggr) \int _{0}^{1} \bigl\Vert z(\rho ,t) \bigr\Vert _{2}^{2}\,d\rho +\varepsilon \biggl(p+ \frac{c_{5}}{k}C \biggr)H(t). \end{aligned} $$

(5.16)

Subtracting and adding $\varepsilon \gamma H(t)$ on the right-hand side of (5.16), using (2.10) and (5.1), we deduce

$$\begin{aligned} \Gamma '(t) \geq &\bigl\{ (1-\sigma )-\varepsilon k \bigr\} H^{-\sigma}(t)H'(t) + \varepsilon \biggl(\frac{p}{2}-\frac{\gamma}{2}+1+ \frac{c_{5}}{k}C \biggr) \Vert v_{t} \Vert _{2}^{2} \\ &{}+\varepsilon \biggl(c_{2}- a\frac{\gamma}{2} \biggr) \Vert \nabla v \Vert _{2}^{2} +\varepsilon \biggl(c_{1}-b\frac{\gamma}{4}+ \frac{c_{5}}{k}C \biggr) \Vert \nabla v \Vert _{2}^{4} \\ &{}+ \varepsilon \biggl(c_{3}- \frac{\gamma}{2}+\frac{c_{5}}{k}C \biggr) (g\circ \nabla v) (t)+ \biggl( \frac{\gamma}{p}- \frac{c_{5}}{k}C \biggr) \Vert v \Vert _{p}^{p} \\ &{} +\varepsilon \biggl(\frac{p\zeta}{2}-\frac{\gamma \zeta}{2}+ \frac{c_{5}}{k}C \biggr) \int _{0}^{1} \bigl\Vert z(\rho ,t) \bigr\Vert _{2}^{2}\,d\rho \\ &{}+ \varepsilon \biggl(p-\gamma + \frac{c_{5}}{k}C \biggr)H(t)+ \varepsilon \gamma E_{1}. \end{aligned}$$

(5.17)

First, we fix γ such that

$$ 0< \gamma < \min \biggl\{ p,\frac{2c_{2}}{a},\frac{4c_{1}}{b},2c_{3}, \biggr\} . $$

Secondly, we take k large enough such that

$$ \frac{\gamma}{p}-\frac{c_{5}}{k}C>0. $$

Once k is fixed, we select $\varepsilon >0$ small enough so that

$$ (1-\sigma )-\varepsilon k>0,\quad \text{and}\quad \Gamma (0)=H^{1-\sigma}(0)+ \varepsilon \int _{\Omega}v_{0}v_{1}\,dx+ \frac{\alpha}{4} \Vert \nabla v_{0} \Vert _{2}^{4}>0. $$

Therefore, we obtain from (5.17) that

$$\begin{aligned} \Gamma '(t) \geq& \omega \biggl( \Vert v_{t} \Vert _{2}^{2}+ \Vert \nabla v \Vert _{2}^{2}+ \Vert \nabla v \Vert _{2}^{4}+(g\circ \nabla v) (t) \\ &{}+ \int _{0}^{1} \bigl\Vert z(\rho ,t) \bigr\Vert _{2}^{2}\,d \rho + \Vert v \Vert _{p}^{p}+H(t) \biggr), \end{aligned}$$

(5.18)

where ω is a positive constant.

We now estimate $\Gamma (t)^{\frac{1}{1-\sigma}}$. By Hölder’s inequality, we have

$$ \biggl\vert \int _{\Omega}vv_{t}\,dx \biggr\vert \leq \Vert v \Vert _{2} \Vert v_{t} \Vert _{2}\leq C_{1} \Vert v \Vert _{p} \Vert v_{t} \Vert _{2}, $$

(5.19)

which implies

$$ \biggl\vert \int _{\Omega}vv_{t}\,dx \biggr\vert ^{\frac{1}{1-\sigma}}\leq C_{1} \Vert v \Vert _{p}^{\frac{1}{1-\sigma}} \Vert v_{t} \Vert _{2}^{\frac{1}{1-\sigma}}. $$

(5.20)

Young’s inequality yields

$$ \biggl\vert \int _{\Omega}vv_{t}\,dx \biggr\vert ^{\frac{1}{1-\sigma}}C_{1} \bigl( \Vert v \Vert _{p}^{\frac{\mu}{1-\sigma}}+ \Vert v_{t} \Vert _{2}^{ \frac{\vartheta}{1-\sigma}} \bigr), $$

(5.21)

for $\frac{1}{\mu}+\frac{1}{\vartheta}=1$. To obtain $\frac{\mu}{1-\sigma}= \frac{2}{1-2\sigma}\leq p$, by (5.5), we take $\vartheta= 2(1-\sigma )$. Therefore, (5.21) becomes

$$ \biggl\vert \int _{\Omega}vv_{t}\,dx \biggr\vert ^{\frac{1}{1-\sigma}}C_{1} \bigl( \Vert v \Vert _{p}^{s}+ \Vert v_{t} \Vert _{2}^{2} \bigr), $$

where $s=\frac{2}{1-2\sigma}$. Using Lemma 5.1, we obtain

$$\begin{aligned} \biggl\vert \int _{\Omega}vv_{t}\,dx \biggr\vert ^{\frac{1}{1-\sigma}} \leq& C_{1} \biggl(H(t)+ \Vert v_{t} \Vert _{2}^{2}+ \Vert \nabla v \Vert _{2}^{4}+(g\circ \nabla v) (t) \\ &{}+ \int _{0}^{1} \bigl\Vert z(\rho ,t) \bigr\Vert _{2}^{2}\,d\rho + \Vert v \Vert _{p}^{p} \biggr). \end{aligned}$$

(5.22)

Combining (5.4) and (5.22), we obtain

$$ \begin{aligned}[b] \Gamma ^{\frac{1}{1-\sigma}}(t)={}& \biggl(H^{1-\sigma}(t)+\varepsilon \int _{\Omega}vv_{t}\,dx+\frac{\alpha}{4} \Vert \nabla v \Vert _{2}^{4} \biggr)^{ \frac{1}{1-\sigma}} \\ \leq{}& c_{6} \biggl(H(t)+ \Vert v_{t} \Vert _{2}^{2}+ \Vert \nabla v \Vert _{2}^{4}+(g \circ \nabla v) (t)\\ &{}+ \int _{0}^{1} \bigl\Vert z(\rho ,t) \bigr\Vert _{2}^{2}\,d\rho + \Vert v \Vert _{p}^{p}+ \Vert \nabla v \Vert _{2}^{\frac{4}{1-\sigma}} \biggr). \end{aligned} $$

(5.23)

We note from (3.8) and (5.3) that

$$ \Vert \nabla v \Vert _{2}^{\frac{4}{1-\sigma}}\leq \bigl(K_{1}E(0) \bigr)^{ \frac{2}{1-\sigma}}\leq \bigl(K_{1}E(0) \bigr)^{\frac{2}{1-\sigma}} \frac{H(t)}{H(0)}. $$

(5.24)

It follows from (5.23) and (5.24) that

$$ \Gamma ^{\frac{1}{1-\sigma}}(t)\leq c_{7} \biggl(H(t)+ \Vert v_{t} \Vert _{2}^{2}+ \Vert \nabla v \Vert _{2}^{4}+(g\circ \nabla v) (t)+ \int _{0}^{1} \bigl\Vert z(\rho ,t) \bigr\Vert _{2}^{2}\,d \rho + \Vert v \Vert _{p}^{p} \biggr). $$

(5.25)

Combining (5.25) with (5.18), we find that

$$ \Gamma '(t)\geq \kappa \Gamma ^{\frac{1}{1-\sigma}}(t), \quad t\geq 0. $$

(5.26)

A simple integration of (5.26) over $(0,t)$ yields

$$ \Gamma ^{\frac{\sigma}{1-\sigma}}(t)\geq \frac{1}{\Gamma ^{-\frac{\sigma}{1-\sigma}}(0)-\frac{\kappa \sigma t}{1-\sigma}}. $$

Consequently, the solution of problem (1.1) blows up in finite time $T^{*}$ and $T^{*}\leq \frac{1-\sigma}{\kappa \sigma \Gamma ^{\frac{\sigma}{1-\sigma}}(0)}$. □