1 Introduction

In this article, we concerned with the problem

u t t + Δ 2 u - 0 t g ( t - s ) Δ 2 u ( s ) d s = u p u , x Ω , τ > 0 u ( x , t ) = ν u ( x , t ) = 0 , x Ω , t 0 u ( x , 0 ) = u 0 ( x ) , u t ( x , 0 ) = u 1 ( x ) , x Ω
(1.1)

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

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

u t t + Δ 2 u = f ( u , u t ) , x Ω , t > 0 u ( x , t ) = ν u ( x , t ) = 0 , x Ω , t 0 u ( x , 0 ) = u 0 ( x ) , u t ( x , 0 ) = u 1 ( x ) , x Ω .
(1.2)

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

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

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

u t t + Δ 2 u - 0 t g ( t - s ) Δ 2 u ( s ) d s = 0

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

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

2 Preliminaries

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

E ( t ) = 1 2 u t 2 + 1 - 0 t g ( s ) d s Δ u 2 + ( g Δ u ) ( t ) - 1 p + 2 u p + 2 p + 2 ,
(2.1)

where

( g v ) ( t ) = 0 t g ( t - s ) v ( t ) - v ( s ) 2 2 d s .

We denote by ∥.∥ k , the Lk-norm over Ω. In particular, the L2-norm is denoted ∥.∥2. We use the familiar function spaces H 0 2 , H 4 and throughout this article we assume u 0 H 0 2 ( Ω ) H 4 ( Ω ) and u 1 H 0 2 ( Ω ) L 2 ( Ω ) .

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

(A 1) p satisfies

0 < p ( N 4 ) , 0 < p 2 ( N - 2 ) N - 4 ( N 5 ) .

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

1 - 0 g ( t ) d s = l > 0 ,

also there exists positive constants L0, L1 such that

(A 3)

- L 0 g ( t ) 0 , 0 g ( t ) L 1 .

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

u p C * Δ u 2 , u H 0 2 ( Ω )
(2.2)

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

B ( t ) - 4 ( δ + 1 ) B ( t ) + 4 ( δ + 1 ) B ( t ) 0 .
(2.3)

If

B ( 0 ) > r 2 B ( 0 ) + K 0 ,
(2.4)

with r 2 =2 ( δ + 1 ) -2 δ ( δ + 1 ) , then B'(t) > K0 for t > 0, where K0 is a constant.

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

Y ( t ) 2 a + b Y ( t ) 2 + δ - 1 f o r t t 0 0 ,
(2.5)

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

lim t T * - Y ( t ) = 0 .

Upper bounds for T* is estimated as follows:

(i) If b < 0, then

T * t 0 + 1 - b l n - a b - a b - Y ( t 0 ) .

(ii) If b = 0, then

T * t 0 + Y ( t 0 ) Y ( t 0 ) .

(iii) If b > 0, then

T * Y ( t 0 ) a ,

or

T * t 0 + 2 3 δ + 1 2 δ c δ a 1 - [ 1 + c Y ( t 0 ) ] - 1 2 δ ,

where c= a b 2 + 1 δ .

3 Existence of solutions

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

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

u L ( 0 , ; H 0 2 ( Ω ) H 4 ( Ω ) ) , u L ( 0 , ; H 0 2 ( Ω ) L 2 ( Ω ) ) , u L ( 0 , ; L 2 ( Ω ) )
(3.1)

and

u ( x , t ) u 0 ( x ) i n H 0 2 ( Ω ) H 4 ( Ω ) u ( x , t ) u 1 ( x ) i n H 0 2 ( Ω ) L 2 ( Ω )

as t → 0.

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

Let V m be the subspace of H 0 2 ( Ω ) H 4 ( Ω ) generated by the first m vectors. Define

u m ( t ) = k = 1 m d m k ( t ) ω k ,
(3.2)

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

u m ( t ) , ω k + ( Δ u m ( t ) , Δ ω k ) - 0 t ( t - s ) ( Δ u m ( s ) , Δ ω k ) d s - u m ( t ) p u m ( t ) , ω k = 0 k = 1 , m .
(3.3)

with the initial conditions (when m → ∞)

u m ( 0 ) = k = 1 m ( u m ( 0 ) , ω k ) ω k u 0 i n H 0 2 ( Ω ) H 4 ( Ω ) u m ( 0 ) = k = 1 m u m ( 0 ) , ω k ω k u 1 i n H 0 2 ( Ω ) L 2 ( Ω )
(3.4)

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

First estimation. Substituting u m ( t ) instead of ω k in (3.3), we find

d d t 1 2 u m 2 + 1 2 Δ u m 2 - u m p + 2 p + 2 p + 2 - 0 t g ( t - s ) ( Δ u m ( s ) , Δ u m ( t ) ) d s = 0 .
(3.5)

Simple calculation similar to [11] yield

- 0 t g ( t - s ) ( Δ u m ( s ) , Δ u m ( t ) ) d s = - 0 t g ( t - s ) Ω Δ u m ( t ) Δ u m ( t ) d x d s - 0 t g ( t - s ) Ω ( Δ u m ( s ) - Δ u m ( t ) ) Δ u m ( t ) d x d s = 1 2 0 t g ( t - s ) d d t Δ u m ( s ) - Δ u m ( t ) 2 d s - 1 2 0 t g ( t - s ) d d t Δ u m ( t ) 2 d s = 1 2 d d t ( g Δ u m ) ( t ) - 1 2 ( g Δ u m ) ( t ) - 1 2 d d t 0 t g ( s ) d s Δ u m ( t ) 2 d s + 1 2 g ( t ) Δ u m ( t ) 2 .
(3.6)

Combining (3.5) and (3.6), we find

d d t 1 2 u m 2 + 1 2 1 - 0 t g ( s ) d s Δ u m 2 + 1 2 ( g Δ u m ) ( t ) - u m p + 2 p + 2 p + 2 = 1 2 ( g Δ u m ) ( t ) - 1 2 g ( t ) Δ u m ( t ) 2 ,
(3.7)

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

u m 2 + Δ u m 2 + ( g Δ u m ) ( t ) - u m p + 2 p + 2 C 1 ,
(3.8)

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

{ u m } i s u n i f o r m l y b o u n d e d i n L ( 0 , T ; H 0 2 ( Ω ) ) { u m } i s u n i f o r m l y b o u n d e d i n L ( 0 , T ; L 2 ( Ω ) )
(3.9)

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

u m ( t ) , ω k + ( Δ u m ( t ) , Δ ω k ) - 0 t g ( t - s ) ( Δ u m ( s ) , Δ ω k ) d s - g ( 0 ) ( Δ u m ( t ) , Δ ω k ) - ( p + 1 ) u m ( t ) p u m ( t ) , ω k = 0 .
(3.10)

If we substitute u m ( t ) instead of ω k in (3.10), it holds that

d d t 1 2 u m 2 + 1 2 Δ u m 2 - d d t 0 t g ( t - s ) Δ u m ( s ) , Δ u m ( t ) d s + 0 t g ( t - s ) Δ u m ( s ) , Δ u m ( t ) d s + g ( 0 ) Δ u m ( t ) , Δ u m ( t ) - g ( 0 ) d d t Δ u m ( t ) , Δ u m ( t ) + g ( 0 ) Δ u m ( t ) , Δ u m ( t ) - ( p + 1 ) u m ( t ) p u m ( t ) , u m ( t ) = 0 .
(3.11)

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

( p + 1 ) u m ( t ) p u m ( t ) , u m ( t ) ( p + 1 ) u m ( t ) 2 p + 2 p . u m ( t ) 2 p + 2 . u m ( t ) 2 C ( γ ) Δ u m ( t ) 2 + γ u m ( t ) 2 .
(3.12)

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

1 2 u m 2 + 1 2 Δ u m 2 1 2 u m ( 0 ) 2 + 0 t g ( t - s ) ( Δ u m ( s ) , Δ u m ( t ) ) d s + 1 2 Δ u m ( 0 ) 2 - 0 t 0 τ g ( τ - s ) ( Δ u m ( s ) , Δ u m ( τ ) ) d s d τ - g ( 0 ) 0 t Δ u m ( s ) , Δ u m ( s ) + g ( 0 ) Δ u m ( t ) , Δ u m ( t ) - g ( 0 ) Δ u m ( 0 ) , Δ u m ( 0 ) - g ( 0 ) 0 t Δ u m ( s ) 2 d s + C ( γ ) 0 t Δ u m ( s ) 2 d s + γ 0 t u m ( s ) 2 d s .
(3.13)

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

| 1 2 Δ u m ( 0 ) 2 - g ( 0 ) ( Δ u m ( 0 ) , Δ u m ( 0 ) ) | L 2 ,
(3.14)

where L2 is a positive constant independent of m. In the following, we find the upper bound for u m ( 0 ) 2 . Again we substitute u m ( t ) instead of ω k in (3.3), and choosing t = 0, we arrive at

u m ( 0 ) , u m ( 0 ) + Δ u m ( 0 ) , Δ u m ( 0 ) - u m ( 0 ) p u m ( 0 ) , u m ( 0 ) = 0 ,

which combined with the Green's formula imply

u m ( 0 ) 2 + Δ 2 u m ( 0 ) , u m ( 0 ) - u m ( 0 ) p u m ( 0 ) , u m ( 0 ) = 0 .
(3.15)

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

u m ( 0 ) L 3 ,
(3.16)

where L3 > 0 is a constant independent of m.

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

0 t g ( t - s ) ( Δ u m ( s ) , Δ u m ( t ) ) d s ) = Δ u m ( t ) , 0 t g ( t - s ) Δ u m ( s ) d s γ Δ u m ( t ) 2 + 1 4 γ Ω 0 t g ( t - s ) Δ u m ( s ) d s 2 d x γ Δ u m ( t ) 2 + L 0 2 4 γ 0 t Δ u m ( s ) 2 d s γ Δ u m ( t ) 2 + L 4 ( T ) ,
(3.17)
- 0 t 0 τ g ( τ - s ) ( Δ u m ( s ) , Δ u m ( τ ) ) d s d τ = 0 t Δ u m ( τ ) , 0 τ g ( τ - s ) Δ u m ( s ) d s d τ 1 2 0 t Δ u m ( s ) 2 d s + 1 2 0 t Ω 0 τ g ( τ - s ) Δ u m ( s ) d s 2 d x d τ 1 2 0 t Δ u m ( s ) 2 d s + T L 1 2 2 0 t Δ u m ( s ) 2 d s 1 2 0 t Δ u m ( s ) 2 d s + L 5 ( T ) ,
(3.18)
- g ( 0 ) 0 t ( Δ u m ( s ) , Δ u m ( s ) ) d s L 0 0 t Δ u m ( s ) 2 d s + L 6 ( T ) ,
(3.19)

and

g ( 0 ) ( Δ u m ( t ) , Δ u m ( t ) ) γ Δ u m ( t ) 2 + L 7 ( γ ) .
(3.20)

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

1 2 u m 2 + 1 2 Δ u m 2 L 8 0 t u m ( s ) 2 d s + 0 t Δ u m ( s ) 2 d s + L 9 .
(3.21)

By using Gronwall's lemma we arrive at

1 2 u m 2 + 1 2 Δ u m 2 L 10 ,
(3.22)

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

{ u m } i s u n i f o r m l y b o u n d e d i n L ( 0 , T ; L 2 ( Ω ) ) { u m } i s u n i f o r m l y b o u n d e d i n L ( 0 , T ; H 0 2 ( Ω ) )
(3.23)

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

u i u w e a k l y s t a r i n L ( 0 , T ; H 0 2 ( Ω ) ) u i u w e a k l y s t a r i n L ( 0 , T ; H 0 2 ( Ω ) ) u i u w e a k l y s t a r i n L ( 0 , T ; L 2 ( Ω ) )
(3.24)

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

u i u s t r o n g l y i n C ( [ 0 , T ] ; H 0 2 ( Ω ) ) u i u s t r o n g l y i n C ( [ 0 , T ] ) ; L 2 ( Ω ) )
(3.25)

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

u m p u m i s u n i f o r m l y b o u n d e d i n L ( 0 , T ; L 2 ( Ω ) ) ,
(3.26)

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

u i p u i u p u w e a k l y s t a r i n L ( 0 , T ; L 2 ( Ω ) ) .
(3.27)

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

u - u 0 H 0 2 ( Ω ) u - u i H 0 2 ( Ω ) + u i - u i ( 0 ) H 0 2 ( Ω ) + u i ( 0 ) - u 0 H 0 2 ( Ω ) ,

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

4 Blow-up of solutions

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

(A 4)

g ( 0 ) > 0 , g ( s ) 0 , 1 - 0 g ( s ) d s = l > 0 ,

and we make the following extra assumption on g

(A 5)

0 g ( s ) d s < p 1 + p .

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

E ( t ) 1 2 1 - 0 t g ( s ) d s Δ u 2 + ( g Δ u ) ( t ) - 1 p + 2 u p + 2 p + 2 1 2 l Δ u 2 + g Δ u ) ( t ) - C 1 p + 2 l p + 2 2 p + 2 Δ u p + 2 G l Δ u 2 + ( g Δ u ) ( t ) , t 0 ,

where G ( λ ) = 1 2 λ 2 - C 1 p + 2 p + 2 λ p + 2 , C 1 = C * l . It is easy to verify that G(λ) has a maximum at λ 1 = C 1 - p + 2 p and the maximum value is E 1 = p 2 p + 4 C 1 - 2 p + 4 p .

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

d d t E ( t ) = 1 2 ( g Δ u ) ( t ) - 1 2 g ( t ) Δ u 2 0 ,
(4.2)

for almost every t ∈ [0, T].

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

Lemma 5 Let (A4) hold and u be a local solution of (1.1) with initial data satisfying E(0) < E1 and l 1 2 Δ u 0 > λ 1 . Then there exists λ2 > λ1 such that

l Δ u 2 + ( g Δ u ) ( t ) λ 2 2 .
(4.3)

Proof See Li and Tsai [11].

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

ψ ( t ) = u 2 .
(4.4)

It is clear that

ψ ( t ) = 2 ( u , u t ) ,
(4.5)

and from (1.1)

ψ ( t ) = 2 u t 2 - 2 Δ u 2 + 2 u p + 2 p + 2 + 2 0 t g ( t - s ) ( Δ u ( t ) , Δ u ( s ) ) d s .
(4.6)

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

ψ ( t ) - ( 4 + p ) Ω u t 2 d x m ( l Δ u 2 + ( g Δ u ) ( t ) ) - ( 4 + 2 p ) E ( 0 ) ,
(4.7)

where m= ( 1 + p ) - 1 l >0.

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

0 t g ( t - s ) ( Δ u ( t ) , Δ u ( s ) ) d s - 1 2 ( g Δ u ) ( t ) + 1 2 0 t g ( s ) d s Δ u 2 + 0 t g ( s ) d s Δ u 2 ,

therefore (4.6) becomes

ψ ( t ) - ( 4 + p ) u t 2 - ( 2 + p ) u t 2 - 2 Δ u 2 - ( g Δ u ) ( t ) + 0 t g ( s ) d s Δ u 2 + 2 u p + 2 p + 2 .

Then, using (4.2), we obtain

ψ ( t ) - ( 4 + p ) u t 2 - ( 4 + 2 p ) E ( 0 ) + p Δ u 2 + ( 1 + p ) ( g Δ u ) ( t ) - ( 1 + p ) 0 t g ( s ) d s Δ u 2 - ( 2 + p ) 0 t ( g Δ u ) ( s ) d s ,

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

ψ ( t ) - ( 4 + p ) u t 2 - ( 4 + 2 p ) E ( 0 ) + ( p - ( 1 + p ) ( 1 - l ) ) Δ u 2 + ( 1 + p ) ( g Δ u ) ( t ) ,
(4.8)

if we set m:= ( 1 + p ) - 1 l then inequality (4.8) yields the desired result.

Consequently, we have the following result.

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

(i) E(0) < 0

(ii) E(0) = 0 and ψ'(0) > 0

(iii) 0<E ( 0 ) < m p E 1 and l 1 2 Δ u 0 > λ 1

(iv) m p E 1 E ( 0 ) and ψ ( 0 ) > r 2 ψ ( 0 ) + ( 4 + 2 p ) E ( 0 ) 4 + p .

Then ψ' (t) > 0 for t > t*, where

in case (i)

t * = max 0 , ψ ( 0 ) ( 4 + 2 p ) E ( 0 ) ,
(4.9)

in cases (ii), (iv)

t * = 0 ,
(4.10)

and in case (iii)

t * = max 0 , - ψ ( 0 ) ( 4 + 2 p ) m p E 1 - E ( 0 ) .
(4.11)

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

ψ ( t ) ψ ( 0 ) - ( 4 + 2 p ) E ( 0 ) t , t 0 .

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

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

ψ ( t ) > 0 , f o r t > 0 .

If 0<E ( 0 ) < m p E 1 and l 1 2 Δ u 0 > λ 1 then by Lemma 4, we see that

m ( l Δ u 2 + ( g Δ u ) ( t ) ) - ( 4 + 2 p ) E ( 0 ) m λ 2 2 - ( 4 + 2 p ) E ( 0 ) > m 4 + 2 p p E 1 - ( 4 + 2 p ) E ( 0 ) = ( 4 + 2 p ) m p E 1 - E ( 0 ) .

Thus from (4.5), we have

ψ ( t ) m ( l Δ u 2 + ( g Δ u ) ( t ) ) - ( 4 + 2 p ) E ( 0 ) > ( 4 + 2 p ) m p E 1 - E ( 0 ) > 0 ,
(4.12)

and integrating (4.12) from 0 to t gives

ψ ( t ) > 0 , f o r t t * ,

where t* satisfies (4.11).

Let m p E 1 E ( 0 ) , this assumption causes that

ψ ( t ) - ( 4 + p ) u t 2 + ( 4 + 2 p ) E ( 0 ) 0 ,

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

u t 2 ψ ( t ) - ψ ( t ) ,

thus

ψ ( t ) - ( 4 + p ) ψ ( t ) + ( 4 + p ) ψ ( t ) + ( 4 + 2 p ) E ( 0 ) 0 .
(4.13)

We see that the hypotheses of Lemma 2 are fulfilled with

δ = p 4 a n d B ( t ) = ψ ( t ) + ( 4 + 2 p ) E ( 0 ) 4 + p

and the conclusion of Lemma 2.2 gives us

ψ ( t ) > 0 , f o r t > 0 .

Therefore the proof is complete.

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

Y ( t ) = ψ ( t ) - p 4 , f o r t 0 .
(4.14)

Then we have

Y ( t ) = p 4 Y ( t ) 1 + 4 p ψ ( t ) ,
(4.15)
Y ( t ) = - p 4 Y ( t ) 1 + 8 p ψ ( t ) ψ ( t ) - 1 + p 4 ( ψ ( t ) ) 2 .
(4.16)

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

ψ ( t ) ψ ( t ) - 1 + p 4 ( ψ ( t ) ) 2 [ ( l Δ u 2 + ( g Δ u ) ( t ) ) - ( 4 + 2 p ) E ( 0 ) + ( 4 + p ) u t 2 ] ψ ( t ) - 4 1 + p 4 u t 2 ψ ( t ) = [ ( l Δ u 2 + ( g Δ u ) ( t ) ) - ( 4 + 2 p ) E ( 0 ) ] Y ( t ) - 4 p .

Utilizing the last inequality into (4.16) yields

Y ( t ) - p 4 [ ( l Δ u 2 + ( g Δ u ) ( t ) ) - ( 4 + 2 p ) E ( 0 ) ] Y ( t ) 1 + 4 p .
(4.17)

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

  1. (1)

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

    Y ( t ) p 4 ( 4 + 2 p ) E ( 0 ) Y ( t ) 1 + 4 p ,
    (4.18)

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

Y ( t ) 2 α + β Y ( t ) 2 + 4 p f o r t t * ,

where

α = p 2 16 Y ( t * ) 2 + 8 p ψ ( t * ) 2 - 8 E ( 0 ) Y ( t * ) - 4 p > 0 ,
(4.19)

and

β = p 2 2 E ( 0 ) .
(4.20)

Then the hypotheses of Lemma 3 are fulfilled with δ= p 4 , t 0 = t * and using the conclusion of Lemma 3, there exists a finite time T* such that lim t T * - Y ( t ) = 0 , i.e., in this case some solutions blow up in finite time T*.

  1. (2)

    If 0<E ( 0 ) < m p E 1 , then from (4.17) and (4.12) we have

    Y ( t ) - p 4 ( 4 + 2 p ) m p E 1 - E ( 0 ) Y ( t ) 1 + 4 p .

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

Y ( t ) 2 α 1 + β 1 Y ( t ) 2 + 4 p f o r t t * ,

where

α 1 = p 2 16 Y ( t * ) 2 + 8 p ( ψ ( t * ) 2 + 8 m p E 1 - E ( 0 ) Y ( t * ) - 4 p ) > 0 ,
(4.21)

and

β 1 = p 2 2 E ( 0 ) - m p E 1 .
(4.22)

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

lim t T * - ψ ( t ) = .
  1. (3)

    m p E 1 E ( 0 ) . In this case, it is easy to see that by using (4.19) and (4.20) into discussion in part (1), we obtain

    α > 0 i f a n d o n l y i f E ( 0 ) < ψ ( t * ) 2 8 ψ ( t * ) .

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

Therefore, we proved the following theorem.

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

(i) E(0) < 0

(ii) E(0) = 0 and ψ'(0) > 0

(iii) 0<E ( 0 ) < m p E 1 and l 1 2 Δ u 0 > λ 1

(iv) m p E 1 E ( 0 ) and ψ ( 0 ) > r 2 ψ ( 0 ) + ( 4 + 2 p ) E ( 0 ) 4 + p holds.

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

in case (i)

T * t * - Y ( t * ) Y ( t * ) .

Furthermore, if Y ( t * ) < min { 1 , α - β } , then

T * t * + 1 - β l n α - β α - β - Y ( t * )

in cases (ii)

T * t * - Y ( t * ) Y ( t * ) o r T * t * + Y ( t * ) α

in case (iii)

T * t * - Y ( t * ) Y ( t * ) .

Furthermore, if Y ( t * ) < min { 1 , α - β } , then

T * t * + 1 - β 1 l n α 1 - β 1 α 1 - β 1 - Y ( t * )

and in case (iv)

T * Y ( t * ) α o r T * t * + 2 3 p + 4 2 p p c 4 α 1 - ( 1 + c Y ( t * ) ) - 2 p ,

where d= β α p p + 8 . Here α, β, α1, and β1 are given in (4.19)-(4.22), respectively. Note that each t* in the above cases satisfy the same case in Lemma 7.