Open Access

Energy decay and blow-up of solution for a Kirchhoff equation with dynamic boundary condition

Boundary Value Problems20132013:166

DOI: 10.1186/1687-2770-2013-166

Received: 16 November 2012

Accepted: 27 June 2013

Published: 12 July 2013

Abstract

The energy decay and blow-up of a solution for a Kirchhoff equation with dynamic boundary condition are considered. With the help of Nakao’s inequality and a stable set, the energy decay of the solution is given. By the convexity inequality lemma and an unstable set, the sufficient condition of blow-up of the solution with negative and small positive initial energy are obtained, respectively.

1 Introduction

The aim of this article is to study the energy decay and blow-up of a solution of the following Kirchhoff-type equation with nonlinear dynamic boundary condition:
u t t + u t + u x x x x M ( u x 2 ) u x x = 0 , 0 < x < l , t > 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ1_HTML.gif
(1)
u ( 0 , t ) = u x x ( 0 , t ) , t > 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ2_HTML.gif
(2)
u x x ( l , t ) + u x ( l , t ) = 0 , t > 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ3_HTML.gif
(3)
u t t ( l , t ) + u t ( l , t ) u x x x ( l , t ) + M ( u x 2 ) u x ( l , t ) = f ( u ( l , t ) ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ4_HTML.gif
(4)
u ( x , 0 ) = u 0 ( x ) , u t ( x , 0 ) = u 1 ( x ) , 0 < x < l , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ5_HTML.gif
(5)

here f ( s ) = | s | p 2 s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq1_HTML.gif, M ( s ) = 1 + s m https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq2_HTML.gif, p > 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq3_HTML.gif, m 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq4_HTML.gif are positive constants and u x 2 = 0 l u x 2 d x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq5_HTML.gif.

This problem is based on the equation
u t t + u x x x x ( α + β 0 l u x 2 ( x , t ) d x ) u x x = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ6_HTML.gif
(6)
which was proposed by Woinowsky-Krieger [1, 2] as a model for a vibrating beam with hinged ends, where u ( x , t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq6_HTML.gif is the lateral displacement at the space coordinate x and time t. Equation (6) was studied by many authors such as Dickey [3], Ball Rivera [4], Tucsnak [5], Kouemou Patchen [6], Aassila [7], Oliveira and Lima [8]; Wu and Tsai [9] considered the following beam equation:
u t t + α Δ 2 u M ( u 2 ) Δ u + g ( u t ) = f ( u ) , in  Ω R n . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ7_HTML.gif
(7)
They obtained the existence and uniqueness, as well as decay estimates, of global solutions and blow-up of solutions for the initial boundary value problem of equation (7) through various approaches and assumptive conditions. Feireisl [10] and Fitzgibbon et al. [11] showed the existence of a global attractor and an inertial manifold of equation (6) with damping u t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq7_HTML.gif. Ma [12] studied the existence and decay rates for the solution of equation (6) with nonlinear boundary conditions
u x x x ( l , t ) M ( 0 l u x 2 ( x , t ) d x ) u x ( l , t ) = f ( u ( l , t ) ) + g ( u t ( l , t ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ8_HTML.gif
(8)
Pazoto and Menzala [13] were concerned with equation (6) with rotational inertia term u x x t t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq8_HTML.gif and nonlinear boundary condition (8). Santos et al. [14] established the existence and exponential decay of the Kirchhoff systems with nonlocal boundary condition. Guedda and Labani [15] gave the sufficient condition of the blow-up of the solution to equation (7) with g ( u t ) = u t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq9_HTML.gif and dynamic boundary condition. As the related problem, we mention the following:
u t t M ( u 2 ) Δ u + Q ( t , x , u , u t ) + f ( x , u ) = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equa_HTML.gif

we refer the reader to [1619].

When f = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq10_HTML.gif and M ( s ) = β + k s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq11_HTML.gif, problem (1)-(5) comes from the reference [2022]. In this case, the model describes the weakly damped vibrations of an extensible beam whose ends are a fixed distance apart if one end is hinged while a load is attached to the other end [21]. One can find many references on problem (1)-(5) with M = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq12_HTML.gif and f = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq10_HTML.gif, for example, Littman and Markus [23], Andrews et al. [24], Conrad and Morgul [25], Rao [26].

Dalsen [21, 22] showed the exponential stability of problem (1)-(5) with m = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq13_HTML.gif and f = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq10_HTML.gif. Park et al. [27] discussed the existence of the solution of the Kirchhoff equation with dynamic boundary conditions and boundary differential inclusion. Doronin and Larkin [28] and Gerbi and Said-Houari [29] were concerned with the wave equation with dynamic boundary conditions. Recently, Autuori and Pucci [30] studied the global nonexistence of solutions of the p-Kirchhoff system with dynamic boundary condition.

In this paper, we use the idea of references [31] to get the energy decay and blow-up of the solution for problem (1)-(5). We construct a stable set and an unstable set, which is similar to [32]. By the help of Nakao’s inequality, combining it with the stable set, we get the decay estimate. We find that the set of initial data such that the solution of problem (1)-(5) is decay, is smaller than the potential well in [32]. The blow-up properties of the solution of problem (1)-(5) with small positive initial energy and negative initial energy are obtained by using the convexity lemma [33]. These results are different from the results in [29, 30].

2 Assumptions and preliminaries

In this section, we give some preliminaries which are used throughout this work.

We use the standard space L p [ 0 , l ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq14_HTML.gif and the Sobolev space H 0 1 ( 0 , l ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq15_HTML.gif, H 2 ( 0 , l ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq16_HTML.gif with their usual scalar products and norms. Especially, p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq17_HTML.gif denotes the norm of L p [ 0 , l ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq14_HTML.gif and https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq18_HTML.gif the norms L 2 [ 0 , l ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq19_HTML.gif.

We denote V = { u | u H 2 ( 0 , l ) , u ( 0 ) = u x x ( 0 ) = 0 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq20_HTML.gif.

Lemma 2.1
  1. (1)
    If u H 0 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq21_HTML.gif, then
    | u ( l ) | 2 u x 2 , u 2 C 0 u x 2 ; https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ9_HTML.gif
    (9)
     
  2. (2)
    If u V https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq22_HTML.gif and u x x ( l ) + u x ( l ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq23_HTML.gif, then
    u 2 C 1 ( u x x 2 + u x 2 ( l , t ) ) C 1 ( u x x 2 + u x x 2 ( l , t ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ10_HTML.gif
    (10)
     
Proof Since u ( 0 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq24_HTML.gif, we have
| u ( x ) | 2 = | u ( 0 ) + 0 x u x ( s ) d s | 2 | 0 l u x 2 ( s ) d s | . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equb_HTML.gif
Take x = l https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq25_HTML.gif, we get the first inequality of (9). Integrating the above inequality over [ 0 , l ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq26_HTML.gif, we get the second part of (9). From the following equation
u x ( x ) = u x ( l ) x l u x x ( s ) d s = u x x ( l ) x l u x x ( s ) d s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equc_HTML.gif

and the Cauchy inequality, we can get the result of (10) with the help of (9). □

Lemma 2.2 [34]

Let ϕ ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq27_HTML.gif be a non-increasing and nonnegative function on [ 0 , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq28_HTML.gif such that
sup s [ t , t + 1 ] ϕ ( s ) C ( ϕ ( t ) ϕ ( t + 1 ) ) , t > 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ11_HTML.gif
(11)
then
ϕ ( t ) C e ω t , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equd_HTML.gif

where C, ω are positive constants depending on ϕ ( 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq29_HTML.gif and other known qualities.

Lemma 2.3 [33]

Suppose that a positive, twice-differentiable function H ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq30_HTML.gif satisfies on t 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq31_HTML.gif the inequality
H ( t ) H ( t ) ( β + 1 ) ( H ( t ) ) 2 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ12_HTML.gif
(12)

where β > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq32_HTML.gif, then there is a t 1 < t 2 = H ( 0 ) β H ( 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq33_HTML.gif such that H ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq34_HTML.gif as t t 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq35_HTML.gif.

A solution u of problem (1)-(5) means that there exists T > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq36_HTML.gif such that
u C ( [ 0 , T ] , V ) C 1 ( [ 0 , T ] , H 0 1 ( 0 , l ) H 2 ( 0 , l ) ) , u t t C ( 0 , T ; L 2 [ 0 , l ] ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Eque_HTML.gif
and it satisfies the following identity
0 t { ( u t t , φ ) M ( u x 2 ) ( u x , φ x ) + ( u x x , φ x x ) + ( u t , φ ) + u t t ( l , τ ) φ ( l , τ ) + u t ( l , τ ) φ ( l , τ ) f ( u ( l , τ ) φ ( l , τ ) } d τ = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ13_HTML.gif
(13)

for all φ C ( ( 0 , T ) , V ) C 1 ( 0 , T ; L 2 ( 0 , l ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq37_HTML.gif.

In this paper, we always assume that a local solution exists for problem (1)-(5). In order to study the energy decay or the blow-up phenomenon of the solution of problem (1)-(5), we define the energy of the solution u of problem (1)-(5) by
E ( t ) = E ( u ( t ) ) = 1 2 u t 2 + 1 2 u x x 2 + 1 2 u x 2 + k 2 ( m + 1 ) u x 2 ( m + 1 ) + 1 2 | u t ( l , t ) | 2 + 1 2 | u x x ( l , t ) | 2 1 p | u ( l , t ) | p . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ14_HTML.gif
(14)
The initial energy is defined by
E ( 0 ) = 1 2 u 1 2 + 1 2 u 0 x x 2 + 1 2 u 0 x 2 + k 2 ( m + 1 ) u 0 x 2 ( m + 1 ) + 1 2 | u t ( l , t ) | 2 + 1 2 | u 0 x x ( l ) | 2 1 p | u ( l , t ) | p . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ15_HTML.gif
(15)
Then, after some simple computation, we have
d d t E ( t ) = u t 2 | u t ( l , t ) | 2 < 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ16_HTML.gif
(16)
That is to say, E ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq38_HTML.gif is a non-increasing function on [ 0 , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq28_HTML.gif. Moreover, we have
E ( t ) + 0 t [ u t 2 + | u t ( l , s ) | 2 ] d s = E ( 0 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ17_HTML.gif
(17)
We denote
B 1 1 = inf u V u x x | u ( l ) | , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ18_HTML.gif
(18)
λ 0 = B 1 p p 2 , E 0 = ( 1 2 1 p ) λ 0 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ19_HTML.gif
(19)
We can now define a stable set and an unstable set [31]
Σ 0 = { ( λ , E ) R 2 | 0 < λ < λ 0 , 0 < E E 0 } , Σ e = { ( λ , E ) R 2 | λ > λ 0 , 0 < E < E 0 } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equf_HTML.gif

3 Energy decay of the solution

In order to get the energy decay of the solution, we introduce the following set:
Σ 1 = { ( λ , E ) R 2 | 0 < λ < λ 1 , 0 < E < E 1 } , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ20_HTML.gif
(20)

where λ 1 = ( p B 1 p ) 1 p 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq39_HTML.gif, E 1 = ( 1 2 1 p ) λ 1 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq40_HTML.gif. Obviously, Σ 1 Σ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq41_HTML.gif.

Adapting the idea of Vitillaro [35], we have the following lemma.

Lemma 3.1 Suppose that u is the solution of (1)-(5), u 0 V https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq42_HTML.gif, u 1 L 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq43_HTML.gif and ( u 0 x x , E ( 0 ) ) Σ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq44_HTML.gif, then ( u x x ( t ) , E ( t ) ) Σ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq45_HTML.gif, for t 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq31_HTML.gif.

Lemma 3.2 Under the condition of Lemma  3.1 and p > 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq3_HTML.gif, then, for t 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq31_HTML.gif,
u x x 2 2 | u ( l , t ) | p , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ21_HTML.gif
(21)
E ( t ) p 1 2 p u x x 2 p 1 p | u ( l , t ) | p , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ22_HTML.gif
(22)
E ( t ) C 3 ( u x x 2 + | u x x ( l , t ) | 2 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ23_HTML.gif
(23)
Proof By (14) and (18), we have
E ( t ) 1 2 u x x 2 1 p | u ( l , t ) | p 1 2 u x x 2 | u ( l , t ) | p 1 2 u x x 2 B 1 p u x x p = G ( u x x ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ24_HTML.gif
(24)

where G ( λ ) = 1 2 λ 2 B 1 p λ p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq46_HTML.gif. Note that G ( λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq47_HTML.gif has the maximum at λ 1 = ( p B 1 p ) 1 p 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq39_HTML.gif and the maximum value G ( λ 1 ) = E 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq48_HTML.gif. We see that G ( λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq47_HTML.gif is increasing in ( 0 , λ 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq49_HTML.gif, decreasing in ( λ 1 , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq50_HTML.gif and G ( λ ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq51_HTML.gif as λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq52_HTML.gif. Since u x x < λ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq53_HTML.gif, E ( 0 ) < E 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq54_HTML.gif, then u x x < λ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq53_HTML.gif for any t 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq31_HTML.gif, so G ( u x x ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq55_HTML.gif.

By (24), we have
u x x 2 | u ( l , t ) | p = 1 2 u x x 2 + ( 1 2 u x x 2 | u ( l , t ) | p ) 1 2 u x x 2 + G ( u x x ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equg_HTML.gif
then (21) holds since G ( u x x ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq56_HTML.gif. Furthermore, we have
E ( t ) 1 2 u x x 2 1 p | u ( l , t ) | p p 1 2 p u x x 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ25_HTML.gif
(25)
So (22) holds. Similar to (25), the above equality becomes
E ( t ) 1 2 u x x 2 + 1 2 | u x x ( l , t ) | 2 1 p | u ( l , t ) | p p 1 2 p u x x 2 + 1 2 | u x x ( l , t ) | 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equh_HTML.gif

so (23) holds. □

Theorem 3.3 Let p > 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq3_HTML.gif, ( u 0 x x , E ( 0 ) ) Σ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq57_HTML.gif, and u be the solution of problem (1)-(5), then there exist two positive constants l and θ independent of t such that
E ( t ) l e θ t , t 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equi_HTML.gif
Proof From (16), we have
t t + 1 [ u t ( s ) 2 + | u t ( l , s ) | 2 ] d s = E ( t ) E ( t + 1 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ26_HTML.gif
(26)
Now, for the above estimate and the mean value theorem, we choose t 1 [ t , t + 1 4 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq58_HTML.gif and t 2 [ t + 3 4 , t + 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq59_HTML.gif such that, for i = 1 , 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq60_HTML.gif,
u t ( t i ) ( t t + 1 u t ( s ) 2 d s ) 1 2 C ( E ( t ) E ( t + 1 ) ) 1 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ27_HTML.gif
(27)
| u ( l , t i ) | ( t t + 1 | u ( l , s ) | 2 d s ) 1 2 C ( E ( t ) E ( t + 1 ) ) 1 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ28_HTML.gif
(28)
Multiplying equation (1) by u and integrating over [ 0 , l ] × [ t 1 , t 2 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq61_HTML.gif, by the boundary conditions (2) and (3), we have
t 1 t 2 [ u x x 2 + u x 2 + k u x 2 ( m + 1 ) + | u x x ( l , s ) | 2 | u ( l , s ) | p ] d s = t 1 t 2 [ ( u t t , u ) + u t ( l , s ) u ( l , s ) + ( u t , u ) + u t t ( l , s ) u ( l , s ) ] d s i = 1 2 [ u t ( t i ) u ( t i ) + | u t ( l , t i ) | | u ( l , t i ) | ] + t 1 t 2 [ u t 2 + u t 2 ( l , s ) ] d s + t 1 t 2 [ u t u + | u t ( l , s ) | | u ( l , s ) | ] d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ29_HTML.gif
(29)
Now, we estimate the terms of the right-hand side of (29). By (10), (23), (27) and the Young inequality, we have
I 1 = i = 1 2 u t ( t i ) u ( t i ) C ( E ( t ) E ( t + 1 ) ) 1 2 i = 1 2 u ( t i ) ε E ( t ) + C ( ε ) ( E ( t ) E ( t + 1 ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ30_HTML.gif
(30)
It follows from (9), (10), (23), (28) and the Young inequality that
I 2 = i = 1 2 | u t ( l , t i ) | | u ( l , t i ) | C ( E ( t ) E ( t + 1 ) ) 1 2 i = 1 2 | u ( l , t i ) | ε E ( t ) + C ( ε ) ( E ( t ) E ( t + 1 ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ31_HTML.gif
(31)
From (26), we get
I 3 = t 1 t 2 [ u t 2 + u t 2 ( l , s ) ] d s C ( E ( t ) E ( t + 1 ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ32_HTML.gif
(32)
From the Young inequality, (10), (23), (26) and from the fact that E ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq38_HTML.gif is non-increasing, we arrive at
I 4 = t 1 t 2 u t u d s ε E ( t ) + C ( ε ) ( E ( t ) E ( t + 1 ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ33_HTML.gif
(33)
By the Young inequality, (9), (10), (23), (26) and the fact that E ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq38_HTML.gif is non-increasing, we obtain
I 5 = t 1 t 2 | u ( l , s ) | | u t ( l , s ) | d s ε E ( t ) + C ( ε ) ( E ( t ) E ( t + 1 ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ34_HTML.gif
(34)
Substituting (30)-(34) into (29), we get the estimate
t 1 t 2 [ u x x 2 + u x 2 + k u x 2 ( m + 1 ) + | u x x ( l , s ) | 2 | u ( l , s ) | p ] d s ε E ( t ) + C ( ε ) ( E ( t ) E ( t + 1 ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ35_HTML.gif
(35)
On the other hand, it follows from (22) that
E ( t ) = 1 2 ( u x x 2 | u ( l , t ) | p ) + 1 2 u x 2 + k 2 ( m + 1 ) u x 2 ( m + 1 ) + 1 2 u t 2 + 1 2 | u t ( l , t ) | 2 + 1 2 | u x x ( l , t ) | 2 + p 2 2 p | u ( l , t ) | p 1 2 ( u x x 2 | u ( l , t ) | p ) + 1 2 u x 2 + k 2 ( m + 1 ) u x 2 ( m + 1 ) + 1 2 u t 2 + 1 2 | u t ( l , t ) | 2 + 1 2 | u x x ( l , t ) | 2 + p 2 2 ( p 1 ) E ( t ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ36_HTML.gif
(36)
Then we have
p 2 ( p 1 ) E ( t ) 1 2 [ u x x 2 + u x 2 + k u x 2 ( m + 1 ) + | u x x ( l , t ) | 2 | u ( l , t ) | p ] + 1 2 u t 2 + 1 2 | u t ( l , t ) | 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ37_HTML.gif
(37)
Therefore, by (37), (35) and (26), we arrive at
t 1 t 2 E ( s ) d s C t 1 t 2 [ u x x 2 + u x 2 + k u x 2 ( m + 1 ) + | u x x ( l , s ) | 2 | u ( l , s ) | p ] d s + C t 1 t 2 ( u t 2 + | u t ( l , s ) | 2 ) d s ε 1 E ( t ) + C ( ε 1 ) ( E ( t ) E ( t + 1 ) ) ε 1 sup s [ t , t + 1 ] E ( s ) + C ( ε 1 ) ( E ( t ) E ( t + 1 ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ38_HTML.gif
(38)
Since E ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq38_HTML.gif is non-increasing, we choose t 3 [ t 1 , t 2 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq62_HTML.gif such that
E ( t 3 ) C t 1 t 2 E ( s ) d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ39_HTML.gif
(39)
Then, using (26), t 3 < t + 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq63_HTML.gif, and the fact that E ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq38_HTML.gif is non-increasing, we have
E ( t ) = E ( t + 1 ) + t t + 1 [ u t ( s ) 2 + | u t ( l , s ) | 2 ] d s E ( t 3 ) + t t + 1 [ u t ( s ) 2 + | u t ( l , s ) | 2 ] d s C t 1 t 2 E ( s ) d s + t t + 1 [ u t ( s ) 2 + | u t ( l , s ) | 2 ] d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ40_HTML.gif
(40)
Since E ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq38_HTML.gif is non-increasing, combining this with (40), (38) and (26), we have
sup s [ t , t + 1 ] E ( s ) ε 1 sup s [ t , t + 1 ] E ( s ) + C ( ε 1 ) ( E ( t ) E ( t + 1 ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ41_HTML.gif
(41)
Choosing ε 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq64_HTML.gif sufficiently small, (41) leads to
sup s [ t , t + 1 ] E ( s ) C ( E ( t ) E ( t + 1 ) ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ42_HTML.gif
(42)

then, applying Lemma 2.2, we obtain the energy decay. □

4 Blow-up property

In this section, we show that the solution of problem (1)-(5) blows up in finite time if E ( 0 ) < E 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq65_HTML.gif.

Lemma 4.1 Suppose p > 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq3_HTML.gif, ( u 0 x x , E ( 0 ) ) Σ e https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq66_HTML.gif, then
u x x λ 0 , 0 < E ( t ) < E 0 , t 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ43_HTML.gif
(43)
Proof Since E ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq38_HTML.gif is non-increasing, and 0 < E ( 0 ) < E 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq67_HTML.gif, then 0 < E ( t ) < E 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq68_HTML.gif for t 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq31_HTML.gif. Similar to the proof of Lemma 3.2, we have
E ( t ) 1 2 u x x 2 1 p | u ( l , t ) | p 1 2 u x x 2 1 p B 1 p u x x p = g ( u x x ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ44_HTML.gif
(44)
where g ( λ ) = 1 2 λ 2 1 p B 1 p λ p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq69_HTML.gif. Note that g ( λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq70_HTML.gif has the maximum at λ 0 = B 1 2 p p 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq71_HTML.gif and the maximum value is g ( λ 0 ) = E 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq72_HTML.gif. It is easy to verify that g ( λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq70_HTML.gif is increasing for 0 < λ < λ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq73_HTML.gif, decreasing in λ > λ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq74_HTML.gif and g ( λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq75_HTML.gif as λ + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq76_HTML.gif. Therefore, since E ( 0 ) < E 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq65_HTML.gif, there exists λ 2 > λ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq77_HTML.gif such that g ( λ 2 ) > E ( 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq78_HTML.gif. By (44), we have g ( u 0 x x ) E ( 0 ) = g ( λ 2 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq79_HTML.gif, which implies u 0 x x λ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq80_HTML.gif. We claim that
u x x ( t ) λ 2 , t > 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ45_HTML.gif
(45)

Otherwise, we suppose that u x x ( t 0 ) < λ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq81_HTML.gif for some t 0 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq82_HTML.gif and by the continuity of u x x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq83_HTML.gif, we can choose t 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq84_HTML.gif such that u x x ( t 0 ) < λ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq85_HTML.gif.

Again the use of (44) leads to
E ( t 0 ) g ( u x x ( t 0 ) ) > g ( λ 2 ) = E ( 0 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ46_HTML.gif
(46)

This is impossible since E ( t ) E ( 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq86_HTML.gif for all t 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq31_HTML.gif. Hence, (45) holds. Furthermore, (43) is established since λ 2 > λ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq77_HTML.gif. □

Theorem 4.2 Suppose that u is the local solution of problem (1)-(5), p > 2 ( m + 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq87_HTML.gif, E ( 0 ) < 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq88_HTML.gif, then the solution u blows up at some finite time.

Proof Let
F ( t ) = u 2 + u 2 ( l , t ) + 0 t [ u ( s ) 2 + u 2 ( l , s ) ] d s + ( T 0 t ) ( u 0 2 ( l ) + u 0 x x 2 ( l ) ) + β ( t + t 0 ) 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ47_HTML.gif
(47)
where t 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq84_HTML.gif, T 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq89_HTML.gif, β are positive constants which will be fixed later (see Levine [33]). Then one finds
F ( t ) = 2 ( u , u t ) + 2 u ( l , t ) u t ( l , t ) + 2 0 t [ ( u ( s ) , u s ( s ) ) + u ( l , s ) u s ( l , s ) ] d s + 2 β ( t + t 0 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ48_HTML.gif
(48)
F ( t ) = 2 u t 2 + 2 u t 2 ( l , t ) + 2 ( u , u t ) + 2 u ( l , t ) u t ( l , t ) + 2 β + 2 ( u , u t t ) + 2 u ( l , t ) u t t ( l , t ) F ( t ) = 2 u t 2 + 2 u t 2 ( l , t ) 2 u x x 2 2 u x x 2 ( l , t ) 2 u x 2 F ( t ) = 2 k u x 2 ( m + 1 ) + 2 β + 2 | u ( l , t ) | p . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ49_HTML.gif
(49)
By (17) and (14), we have
F ( t ) = F ( t ) + 2 p ( E ( t ) E ( 0 ) ) + 2 p 0 t [ u t 2 + | u t ( l , s ) | 2 ] d s = ( p + 2 ) [ u t 2 + u t 2 ( l , t ) + 0 t [ u t ( s ) 2 + | u t ( l , s ) | 2 ] d s + β ] + ( p 2 ) [ u x x 2 + u x x 2 ( l , t ) + u x 2 + k u x 2 ( m + 1 ) ] p ( β + 2 E ( 0 ) ) + ( p 2 ) 0 t [ u t 2 + | u t ( l , s ) | 2 ] d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ50_HTML.gif
(50)
Taking 0 < β < 2 E ( 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq90_HTML.gif and noticing p > 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq3_HTML.gif, we get
F ( t ) ( p + 2 ) [ u t 2 + u t 2 ( l , t ) + 0 t [ u t ( s ) 2 + u t 2 ( l , s ) ] d s + β ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ51_HTML.gif
(51)
By the Hölder inequality, we have
[ F ( t ) ] 2 4 [ u 2 u t 2 + u ( l , t ) u t ( l , t ) + ( 0 t u 2 d s ) 1 2 ( 0 t u t 2 d s ) 1 2 + ( 0 t u 2 ( l , s ) d s ) 1 2 ( 0 t u t 2 ( l , s ) d s ) 1 2 + β ( t + t 0 ) 2 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ52_HTML.gif
(52)
Denote
A 1 = u 2 , A 2 = 0 t u 2 ( l , s ) d s , A 3 = 0 t u 2 d s , A 4 = u 2 ( l , t ) , B 1 = u t 2 , B 2 = 0 t u t 2 ( l , s ) d s , B 3 = 0 t u t 2 d s , B 4 = u t 2 ( l , t ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equj_HTML.gif
Then, by (48), (51), (52) and the Cauchy-Schwarz inequality, we arrive at
F ( t ) F ( t ) p + 2 4 [ F ( t ) ] 2 ( p + 2 ) [ A 1 + A 2 + A 3 + A 4 + β ( t + t 0 ) 2 ] [ B 1 + B 2 + B 3 + B 4 + β ] ( p + 2 ) [ A 1 1 2 B 1 1 2 + A 2 1 2 B 2 1 2 + A 3 1 2 B 3 1 2 + A 4 1 2 B 4 1 2 + β ( t + t 0 ) 2 ] 2 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ53_HTML.gif
(53)
Take t 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq84_HTML.gif sufficiently large such that
F ( 0 ) = 2 ( u 0 , u 1 ) + 2 u 0 ( l ) u 1 ( l ) + 2 β t 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ54_HTML.gif
(54)

Noticing F ( 0 ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq91_HTML.gif, by Lemma 2.3, we get the result. □

Theorem 4.3 Suppose that u ( x , t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq6_HTML.gif is the local solution of problem (1)-(5), p > 2 ( m + 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq87_HTML.gif, and that either of the following conditions is satisfied:
  1. (i)

    E ( 0 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq92_HTML.gif and ( u 0 , u 1 ) + u 0 ( l ) u 1 ( l ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq93_HTML.gif;

     
  2. (ii)

    0 < E ( 0 ) < E 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq67_HTML.gif and u 0 x x > λ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq94_HTML.gif (or ( u 0 x x , E ( 0 ) ) Σ e https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq66_HTML.gif);

     

then the solution u blows up at some finite time.

Proof (i) For E ( 0 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq92_HTML.gif, similar to the proof of Theorem 4.2, we take β = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq95_HTML.gif in (51), then (53) holds. Since F ( 0 ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq91_HTML.gif, F ( 0 ) = 2 ( u 0 , u 1 ) + 2 u 0 ( l ) u 1 ( l ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq96_HTML.gif, then the result holds by Lemma 2.3.
  1. (ii)
    For the case of 0 < E ( 0 ) < E 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq67_HTML.gif, from (48), (49), (50) and (14), we get
    F ( t ) = 2 [ u t 2 + u t 2 ( l , t ) u x x 2 u x x 2 ( l , t ) u x 2 k u x 2 ( m + 1 ) + β ] + p [ u t 2 + u t 2 ( l , t ) + u x x 2 + u x x 2 ( l , t ) + u x 2 + k u x 2 ( m + 1 ) 2 E ( t ) ] = ( p + 2 ) [ u t 2 + u t 2 ( l , t ) ] + ( p 2 ) [ u x x 2 + u x x 2 ( l , t ) + u x 2 + ( p m + 1 2 ) k u x 2 ( m + 1 ) ] + 2 β 2 p E ( t ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ55_HTML.gif
    (55)
     
By Lemma 4.1,
( p 2 ) u x x 2 ( p 2 ) λ 0 2 = 2 p E 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ56_HTML.gif
(56)
Combining (55) with (56), E ( 0 ) < E 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq65_HTML.gif and (17), we get
F ( t ) ( p + 2 ) [ u t 2 + u t 2 ( l , t ) ] + ( p 2 ) u x x 2 + 2 β 2 p E ( t ) ( p + 2 ) [ u t 2 + u t 2 ( l , t ) ] + 2 p ( E 0 E ( 0 ) ) + 2 β + 2 p 0 t [ u t 2 + u t 2 ( l , s ) ] d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ57_HTML.gif
(57)
Take β = 2 ( E 0 E ( 0 ) ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq97_HTML.gif, then 2 p ( E 0 E ( 0 ) ) + 2 β = ( p + 2 ) β > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq98_HTML.gif, since p > 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq3_HTML.gif and 2 p > p + 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_IEq99_HTML.gif, then (57) can be rewritten
F ( t ) ( p + 2 ) [ u t 2 + u t 2 ( l , t ) + β + 0 t [ u t 2 + u t 2 ( l , s ) d s ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-166/MediaObjects/13661_2012_Article_421_Equ58_HTML.gif
(58)

The remainder of the proof is the same as the proof of Theorem 4.2. □

Declarations

Acknowledgements

We thank the referees for their valuable suggestions which helped us improve the paper so much. This work was supported by the National Natural Science Foundation of China (11171311) and the Key Science Foundation of Henan University of Technology (09XZD009).

Authors’ Affiliations

(1)
Department of Mathematics, Henan University of Technology

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