Existence and blow up of solutions to a Petrovsky equation with memory and nonlinear source term

  • Faramarz Tahamtani1Email author and

    Affiliated with

    • Mohammad Shahrouzi1

      Affiliated with

      Boundary Value Problems20122012:50

      DOI: 10.1186/1687-2770-2012-50

      Received: 3 December 2011

      Accepted: 26 April 2012

      Published: 26 April 2012

      Abstract

      We consider the semilinear Petrovsky equation

      u t t + Δ 2 u - 0 t g ( t - s ) Δ 2 u ( s ) d s = u p u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equa_HTML.gif

      in a bounded domain and prove the existence of weak solutions. Furthermore, we show that there are solutions under some conditions on initial data which blow up in finite time with non-positive initial energy as well as positive initial energy. Estimates of the lifespan of solutions are also given.

      Mathematics Subject Classification (2000): 35L35; 35L75; 37B25.

      Keywords

      viscoelasticity existence blow-up life-span negative initial energy positive initial energy

      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 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ1_HTML.gif
      (1.1)

      where ΩR n is 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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ2_HTML.gif
      (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 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_IEq1_HTML.gif 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 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equb_HTML.gif

      in a bounded domain ΩR N and 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 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ3_HTML.gif
      (2.1)
      where
      ( g v ) ( t ) = 0 t g ( t - s ) v ( t ) - v ( s ) 2 2 d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equc_HTML.gif

      We denote by ∥.∥ k , the L k -norm over Ω. In particular, the L2-norm is denoted ∥.∥2. We use the familiar function spaces H 0 2 , H 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_IEq2_HTML.gif and throughout this article we assume u 0 H 0 2 ( Ω ) H 4 ( Ω ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_IEq3_HTML.gif and u 1 H 0 2 ( Ω ) L 2 ( Ω ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_IEq4_HTML.gif.

      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 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equd_HTML.gif
      (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 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Eque_HTML.gif

      also there exists positive constants L0, L1 such that

      (A 3)
      - L 0 g ( t ) 0 , 0 g ( t ) L 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equf_HTML.gif
      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 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ4_HTML.gif
      (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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ5_HTML.gif
      (2.3)
      If
      B ( 0 ) > r 2 B ( 0 ) + K 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ6_HTML.gif
      (2.4)

      with r 2 = 2 ( δ + 1 ) - 2 δ ( δ + 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_IEq5_HTML.gif, 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 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ7_HTML.gif
      (2.5)
      where a > 0, δ > 0 and bR, then there exists a finite time T* such that
      lim t T * - Y ( t ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equg_HTML.gif

      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 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equh_HTML.gif
      (ii) If b = 0, then
      T * t 0 + Y ( t 0 ) Y ( t 0 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equi_HTML.gif
      (iii) If b > 0, then
      T * Y ( t 0 ) a , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equj_HTML.gif
      or
      T * t 0 + 2 3 δ + 1 2 δ c δ a 1 - [ 1 + c Y ( t 0 ) ] - 1 2 δ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equk_HTML.gif

      where c = a b 2 + 1 δ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_IEq6_HTML.gif.

      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 ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ8_HTML.gif
      (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 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equl_HTML.gif

      as t → 0.

      Proof We choose a basis {ω k } (k = 1, 2, ...) in H 0 2 ( Ω ) H 4 ( Ω ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_IEq7_HTML.gif 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 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_IEq7_HTML.gif generated by the first m vectors. Define
      u m ( t ) = k = 1 m d m k ( t ) ω k , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ9_HTML.gif
      (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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ10_HTML.gif
      (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 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ11_HTML.gif
      (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 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_IEq8_HTML.gif 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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ12_HTML.gif
      (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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ13_HTML.gif
      (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 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ14_HTML.gif
      (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 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ15_HTML.gif
      (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 ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ16_HTML.gif
      (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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ17_HTML.gif
      (3.10)
      If we substitute u m ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_IEq9_HTML.gif 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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ18_HTML.gif
      (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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ19_HTML.gif
      (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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ20_HTML.gif
      (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 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ21_HTML.gif
      (3.14)
      where L2 is a positive constant independent of m. In the following, we find the upper bound for u m ( 0 ) 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_IEq10_HTML.gif. Again we substitute u m ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_IEq9_HTML.gif 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 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equm_HTML.gif
      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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ22_HTML.gif
      (3.15)
      By using (A1), (3.4) and Young's inequality, we deduce that
      u m ( 0 ) L 3 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ23_HTML.gif
      (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 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ24_HTML.gif
      (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 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ25_HTML.gif
      (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 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ26_HTML.gif
      (3.19)
      and
      g ( 0 ) ( Δ u m ( t ) , Δ u m ( t ) ) γ Δ u m ( t ) 2 + L 7 ( γ ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ27_HTML.gif
      (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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ28_HTML.gif
      (3.21)
      By using Gronwall's lemma we arrive at
      1 2 u m 2 + 1 2 Δ u m 2 L 10 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ29_HTML.gif
      (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 ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ30_HTML.gif
      (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 ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ31_HTML.gif
      (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 ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ32_HTML.gif
      (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 ) ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ33_HTML.gif
      (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 ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ34_HTML.gif
      (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 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equn_HTML.gif

      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 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equo_HTML.gif

      and we make the following extra assumption on g

      (A 5)
      0 g ( s ) d s < p 1 + p . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equp_HTML.gif
      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 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equq_HTML.gif

      where G ( λ ) = 1 2 λ 2 - C 1 p + 2 p + 2 λ p + 2 , C 1 = C * l http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_IEq11_HTML.gif. It is easy to verify that G(λ) has a maximum at λ 1 = C 1 - p + 2 p http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_IEq12_HTML.gif and the maximum value is E 1 = p 2 p + 4 C 1 - 2 p + 4 p http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_IEq13_HTML.gif.

      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 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ35_HTML.gif
      (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 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_IEq14_HTML.gif. Then there exists λ2 > λ1 such that
      l Δ u 2 + ( g Δ u ) ( t ) λ 2 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ36_HTML.gif
      (4.3)

      Proof See Li and Tsai [11].

      The choice of the functional is standard (see [19])
      ψ ( t ) = u 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ37_HTML.gif
      (4.4)
      It is clear that
      ψ ( t ) = 2 ( u , u t ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ38_HTML.gif
      (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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ39_HTML.gif
      (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 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ40_HTML.gif
      (4.7)

      where m = ( 1 + p ) - 1 l > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_IEq15_HTML.gif.

      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 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equr_HTML.gif
      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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equs_HTML.gif
      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 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equt_HTML.gif
      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 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ41_HTML.gif
      (4.8)

      if we set m : = ( 1 + p ) - 1 l http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_IEq16_HTML.gif 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 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_IEq17_HTML.gif and l 1 2 Δ u 0 > λ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_IEq18_HTML.gif

      (iv) m p E 1 E ( 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_IEq19_HTML.gif and ψ ( 0 ) > r 2 ψ ( 0 ) + ( 4 + 2 p ) E ( 0 ) 4 + p http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_IEq20_HTML.gif.

      Then ψ' (t) > 0 for t > t*, where

      in case (i)
      t * = max 0 , ψ ( 0 ) ( 4 + 2 p ) E ( 0 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ42_HTML.gif
      (4.9)
      in cases (ii), (iv)
      t * = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ43_HTML.gif
      (4.10)
      and in case (iii)
      t * = max 0 , - ψ ( 0 ) ( 4 + 2 p ) m p E 1 - E ( 0 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ44_HTML.gif
      (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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equu_HTML.gif

      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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equv_HTML.gif
      If 0 < E ( 0 ) < m p E 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_IEq17_HTML.gif and l 1 2 Δ u 0 > λ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_IEq18_HTML.gif 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 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equw_HTML.gif
      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 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ45_HTML.gif
      (4.12)
      and integrating (4.12) from 0 to t gives
      ψ ( t ) > 0 , f o r t t * , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equx_HTML.gif

      where t* satisfies (4.11).

      Let m p E 1 E ( 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_IEq19_HTML.gif, this assumption causes that
      ψ ( t ) - ( 4 + p ) u t 2 + ( 4 + 2 p ) E ( 0 ) 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equy_HTML.gif
      and by using Hölder and Young's inequalities, we get
      u t 2 ψ ( t ) - ψ ( t ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equz_HTML.gif
      thus
      ψ ( t ) - ( 4 + p ) ψ ( t ) + ( 4 + p ) ψ ( t ) + ( 4 + 2 p ) E ( 0 ) 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ46_HTML.gif
      (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 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equaa_HTML.gif
      and the conclusion of Lemma 2.2 gives us
      ψ ( t ) > 0 , f o r t > 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equab_HTML.gif

      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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ47_HTML.gif
      (4.14)
      Then we have
      Y ( t ) = p 4 Y ( t ) 1 + 4 p ψ ( t ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ48_HTML.gif
      (4.15)
      Y ( t ) = - p 4 Y ( t ) 1 + 8 p ψ ( t ) ψ ( t ) - 1 + p 4 ( ψ ( t ) ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ49_HTML.gif
      (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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equac_HTML.gif
      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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ50_HTML.gif
      (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 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ51_HTML.gif
        (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 * , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equad_HTML.gif
      where
      α = p 2 16 Y ( t * ) 2 + 8 p ψ ( t * ) 2 - 8 E ( 0 ) Y ( t * ) - 4 p > 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ52_HTML.gif
      (4.19)
      and
      β = p 2 2 E ( 0 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ53_HTML.gif
      (4.20)
      Then the hypotheses of Lemma 3 are fulfilled with δ = p 4 , t 0 = t * http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_IEq21_HTML.gif and using the conclusion of Lemma 3, there exists a finite time T* such that lim t T * - Y ( t ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_IEq22_HTML.gif, i.e., in this case some solutions blow up in finite time T*.
      1. (2)
        If 0 < E ( 0 ) < m p E 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_IEq17_HTML.gif, 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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equae_HTML.gif
         
      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 * , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equaf_HTML.gif
      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 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ54_HTML.gif
      (4.21)
      and
      β 1 = p 2 2 E ( 0 ) - m p E 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equ55_HTML.gif
      (4.22)
      Thus by Lemma 3, there exists a finite time T* such that
      lim t T * - ψ ( t ) = . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equag_HTML.gif
      1. (3)
        m p E 1 E ( 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_IEq19_HTML.gif. 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 * ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equah_HTML.gif
         

      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 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_IEq17_HTML.gif and l 1 2 Δ u 0 > λ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_IEq18_HTML.gif

      (iv) m p E 1 E ( 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_IEq19_HTML.gif and ψ ( 0 ) > r 2 ψ ( 0 ) + ( 4 + 2 p ) E ( 0 ) 4 + p http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_IEq20_HTML.gif 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 * ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equai_HTML.gif
      Furthermore, if Y ( t * ) < min { 1 , α - β } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_IEq23_HTML.gif, then
      T * t * + 1 - β l n α - β α - β - Y ( t * ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equaj_HTML.gif
      in cases (ii)
      T * t * - Y ( t * ) Y ( t * ) o r T * t * + Y ( t * ) α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equak_HTML.gif
      in case (iii)
      T * t * - Y ( t * ) Y ( t * ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equal_HTML.gif
      Furthermore, if Y ( t * ) < min { 1 , α - β } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_IEq23_HTML.gif, then
      T * t * + 1 - β 1 l n α 1 - β 1 α 1 - β 1 - Y ( t * ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equam_HTML.gif
      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 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_Equan_HTML.gif

      where d = β α p p + 8 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-50/MediaObjects/13661_2011_Article_141_IEq24_HTML.gif. 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.

      Declarations

      Acknowledgements

      The authors would like to thank the referees for the careful reading of this article and for the valuable suggestions to improve the presentation and style of the article. This study was supported by Shiraz University.

      Authors’ Affiliations

      (1)
      Department of Mathematics, College of Sciences, Shiraz University

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