Blow-up phenomena and global existence for the periodic two-component Dullin-Gottwald-Holm system

  • Jingjing Liu1Email author and

    Affiliated with

    • Dequan Zhang2

      Affiliated with

      Boundary Value Problems20132013:158

      DOI: 10.1186/1687-2770-2013-158

      Received: 24 April 2013

      Accepted: 16 June 2013

      Published: 1 July 2013

      Abstract

      This paper is concerned with blow-up phenomena and global existence for the periodic two-component Dullin-Gottwald-Holm system. We first obtain several blow-up results and the blow-up rate of strong solutions to the system. We then present a global existence result for strong solutions to the system.

      MSC:35G25, 35L05.

      Keywords

      periodic two-component Dullin-Gottwald-Holm system blow-up blow-up rate global existence

      1 Introduction

      In this paper, we consider the following periodic two-component Dullin-Gottwald-Holm (DGH) system:
      { m t A u x + u m x + 2 u x m + γ u x x x + ρ ρ x = 0 , t > 0 , x R , ρ t + ( u ρ ) x = 0 , t > 0 , x R , u ( 0 , x ) = u 0 ( x ) , x R , ρ ( 0 , x ) = ρ 0 ( x ) , x R , u ( t , x + 1 ) = u ( t , x ) , t 0 , x R , ρ ( t , x + 1 ) = ρ ( t , x ) , t 0 , x R , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equ1_HTML.gif
      (1.1)

      where m = u u x x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq1_HTML.gif, A > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq2_HTML.gif and γ are constants.

      System (1.1) has been recently derived by Zhu et al. in [1] by following Ivanov’s approach [2]. It was shown in [1] that the DGH system is completely integrable and can be written as a compatibility condition of two linear systems
      Ψ x x = ( ξ 2 ρ 2 + ξ ( m A 2 + γ 2 ) + 1 4 ) Ψ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equa_HTML.gif
      and
      Ψ t = ( 1 2 ξ u + γ ) Ψ x + 1 2 u x Ψ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equb_HTML.gif
      where ξ is a spectral parameter. Moreover, this system has the following two Hamiltonians:
      E ( u , ρ ) = 1 2 ( u 2 + u x 2 + ( ρ 1 ) 2 ) d x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equc_HTML.gif
      and
      F ( u , ρ ) = 1 2 ( u 3 + u u x 2 A u 2 γ u x 2 + 2 u ( ρ 1 ) + u ( ρ 1 ) 2 ) d x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equd_HTML.gif
      For ρ = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq3_HTML.gif and m = u α 2 u x x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq4_HTML.gif, (1.1) becomes the DGH equation [3]
      u t α 2 u t x x A u x + 3 u u x + γ u x x x = α 2 ( 2 u x u x x + u u x x x ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Eque_HTML.gif

      where A and α are two positive constants, modeling unidirectional propagation of surface waves on a shallow layer of water which is at rest at infinity, u ( t , x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq5_HTML.gif stands for fluid velocity. It is completely integrable with a bi-Hamiltonian and a Lax pair. Moreover, its traveling wave solutions include both the KdV solitons and the CH peakons as limiting cases [3]. The Cauchy problem of the DGH equation has been extensively studied, cf. [413].

      For ρ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq6_HTML.gif, γ = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq7_HTML.gif, system (1.1) becomes the two-component Camassa-Holm system [2]
      { m t A u x + u m x + 2 u x m + ρ ρ x = 0 , ρ t + ( u ρ ) x = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equ2_HTML.gif
      (1.2)

      where ρ ( t , x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq8_HTML.gif is in connection with the free surface elevation from scalar density (or equilibrium), and the parameter A characterizes a linear underlying shear flow. System (1.2) describes water waves in the shallow water regime with nonzero constant vorticity, where the nonzero vorticity case indicates the presence of an underlying current. A large amount of literature was devoted to the Cauchy problem (1.2); see [1422].

      The Cauchy problem (1.1) has been discussed in [1]. Therein Zhu and Xu established the local well-posedness to system (1.1), derived the precise blow-up scenario and investigated the wave breaking for it. The aim of this paper is to further study the blow-up phenomena for strong solutions to (1.1) and to present a global existence result.

      Our paper is organized as follows. In Section 2, we briefly state some needed results including the local well posedness of system (1.1), the precise blow-up scenario and some useful lemmas to study blow-up phenomena and global existence. In Section 3, we give several new blow-up results and the precise blow-up rate. In Section 4, we present a new global existence result of strong solutions to (1.1).

      Notation Given a Banach space Z, we denote its norm by Z http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq9_HTML.gif. Since all space of functions is over S http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq10_HTML.gif, for simplicity, we drop S http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq10_HTML.gif in our notations if there is no ambiguity.

      2 Preliminaries

      In this section, we will briefly give some needed results in order to pursue our goal.

      With m = u u x x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq1_HTML.gif, we can rewrite system (1.1) as follows:
      { u t u t x x A u x + γ u x x x + 3 u u x 2 u x u x x u u x x x + ρ ρ x = 0 , t > 0 , x R , ρ t + ( u ρ ) x = 0 , t > 0 , x R , u ( 0 , x ) = u 0 ( x ) , x R , ρ ( 0 , x ) = ρ 0 ( x ) , x R , u ( t , x + 1 ) = u ( t , x ) , t 0 , x R , ρ ( t , x + 1 ) = ρ ( t , x ) , t 0 , x R . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equ3_HTML.gif
      (2.1)
      Note that if G ( x ) : = cosh ( x [ x ] 1 / 2 ) 2 sinh ( 1 / 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq11_HTML.gif, x R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq12_HTML.gif is the kernel of ( 1 x 2 ) 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq13_HTML.gif, then ( 1 x 2 ) 1 f = G f http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq14_HTML.gif for all f L 2 ( S ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq15_HTML.gif, G m = u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq16_HTML.gif. Here we denote by ∗ the convolution. Using this identity, we can rewrite system (2.1) as follows:
      { u t + ( u γ ) u x = x G ( u 2 + 1 2 u x 2 + ( γ A ) u + 1 2 ρ 2 ) , t > 0 , x R , ρ t + ( u ρ ) x = 0 , t > 0 , x R , u ( 0 , x ) = u 0 ( x ) , x R , ρ ( 0 , x ) = ρ 0 ( x ) , x R , u ( t , x + 1 ) = u ( t , x ) , t 0 , x R , ρ ( t , x + 1 ) = ρ ( t , x ) , t 0 , x R . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equ4_HTML.gif
      (2.2)

      The local well-posedness of the Cauchy problem (2.1) can be obtained by applying Kato’s theorem. As a result, we have the following well-posedness result.

      Lemma 2.1 [1]

      Given the initial data ( u 0 , ρ 0 ) H s × H s 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq17_HTML.gif, s 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq18_HTML.gif, there exists a maximal T = T ( ( u 0 , ρ 0 ) H s × H s 1 ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq19_HTML.gif and a unique solution
      ( u , ρ ) C ( [ 0 , T ) ; H s × H s 1 ) C 1 ( [ 0 , T ) ; H s 1 × H s 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equf_HTML.gif

      of (2.1). Moreover, the solution ( u , ρ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq20_HTML.gif depends continuously on the initial data ( u 0 , ρ 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq21_HTML.gif and the maximal time of existence T > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq22_HTML.gif is independent of s.

      Consider now the following initial value problem:
      { q t = u ( t , q ) , t [ 0 , T ) , q ( 0 , x ) = x , x R , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equ5_HTML.gif
      (2.3)

      where u denotes the first component of the solution ( u , ρ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq20_HTML.gif to (2.1).

      Lemma 2.2 [1]

      Let ( u , ρ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq23_HTML.gif be the solution of (2.1) with the initial data ( u 0 , ρ 0 ) H s × H s 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq17_HTML.gif, s 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq18_HTML.gif. Then Eq. (2.3) has a unique solution q C 1 ( [ 0 , T ) × R ; R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq24_HTML.gif. Moreover, the map q ( t , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq25_HTML.gif is an increasing diffeomorphism ofwith
      q x ( t , x ) = exp ( 0 t u x ( s , q ( s , x ) ) d s ) > 0 , ( t , x ) [ 0 , T ) × R . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equg_HTML.gif

      Lemma 2.3 [1]

      Let ( u , ρ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq23_HTML.gif be the solution of (2.1) with the initial data ( u 0 , ρ 0 ) H s × H s 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq17_HTML.gif, s 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq18_HTML.gif, and T > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq26_HTML.gif be the maximal time of existence. Then we have
      ρ ( t , q ( t , x ) ) q x ( t , x ) = ρ 0 ( x ) , ( t , x ) [ 0 , T ) × S . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equh_HTML.gif

      Moreover, if there exists an x 0 S http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq27_HTML.gif such that ρ 0 ( x 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq28_HTML.gif, then ρ ( t , q ( t , x 0 ) ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq29_HTML.gif for all t [ 0 , T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq30_HTML.gif.

      Next, we give two useful conservation laws of strong solutions to (2.1).

      Lemma 2.4 [1]

      Let ( u , ρ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq23_HTML.gif be the solution of (2.1) with the initial data ( u 0 , ρ 0 ) H s × H s 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq17_HTML.gif, s 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq18_HTML.gif, and let T > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq26_HTML.gif be the maximal time of existence. Then, for all t [ 0 , T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq31_HTML.gif, we have
      S ( u 2 + u x 2 + ρ 2 ) d x = S ( u 0 2 + u 0 , x 2 + ρ 0 2 ) d x : = E 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equi_HTML.gif
      Lemma 2.5 Let ( u , ρ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq23_HTML.gif be the solution of (2.1) with the initial data ( u 0 , ρ 0 ) H s × H s 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq17_HTML.gif, s 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq18_HTML.gif, and let T > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq26_HTML.gif be the maximal time of existence. Then, for all t [ 0 , T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq31_HTML.gif, we have
      S u ( t , x ) d x = S u 0 ( x ) d x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equj_HTML.gif
      Proof By the first equation in (2.1), we have
      d d t S u ( t , x ) d x = S u t d x = S ( u t x x + A u x γ u x x x 3 u u x + 2 u x u x x + u u x x x ρ ρ x ) d x = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equk_HTML.gif

      This completes the proof of the lemma. □

      Then we state the following precise blow-up mechanism of (2.1).

      Lemma 2.6 [1]

      Let ( u , ρ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq23_HTML.gif be the solution of (2.1) with the initial data ( u 0 , ρ 0 ) H s × H s 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq17_HTML.gif, s 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq18_HTML.gif, and let T > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq26_HTML.gif be the maximal time of existence. Then the solution blows up in finite time if and only if
      lim inf t T { inf x S u x ( t , x ) } = . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equl_HTML.gif

      Lemma 2.7 [23]

      Let t 0 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq32_HTML.gif and v C 1 ( [ 0 , t 0 ) ; H 2 ( R ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq33_HTML.gif. Then, for every t [ 0 , t 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq34_HTML.gif, there exists at least one point ξ ( t ) R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq35_HTML.gif with
      m ( t ) : = inf x R { v x ( t , x ) } = v x ( t , ξ ( t ) ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equm_HTML.gif
      and the function m is almost everywhere differentiable on ( 0 , t 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq36_HTML.gif with
      d d t m ( t ) = v t x ( t , ξ ( t ) ) a.e. on ( 0 , t 0 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equn_HTML.gif
      Lemma 2.8 [24]
      1. (i)
        For every f H 1 ( S ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq37_HTML.gif, we have
        max x [ 0 , 1 ] f 2 ( x ) e + 1 2 ( e 1 ) f H 1 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equo_HTML.gif
         
      where the constant e + 1 2 ( e 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq38_HTML.gif is sharp.
      1. (ii)
        For every f H 3 ( S ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq39_HTML.gif, we have
        max x [ 0 , 1 ] f 2 ( x ) c f H 1 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equp_HTML.gif
         

      with the best possible constant c lying within the range ( 1 , 13 12 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq40_HTML.gif. Moreover, the best constant c is e + 1 2 ( e 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq38_HTML.gif.

      Lemma 2.9 [25]

      If f H 3 ( S ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq41_HTML.gif is such that S f ( x ) d x = a 0 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq42_HTML.gif, then, for every ϵ > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq43_HTML.gif, we have
      max x [ 0 , 1 ] f 2 ( x ) ϵ + 2 24 S f x 2 d x + ϵ + 2 4 ϵ a 0 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equq_HTML.gif
      Moreover,
      max x [ 0 , 1 ] f 2 ( x ) ϵ + 2 24 f H 1 ( S ) 2 + ϵ + 2 4 ϵ a 0 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equr_HTML.gif

      Lemma 2.10 [26]

      Assume that a differentiable function y ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq44_HTML.gif satisfies
      y ( t ) C y 2 ( t ) + K , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equ6_HTML.gif
      (2.4)

      where C, K are positive constants. If the initial datum y ( 0 ) = y 0 < K C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq45_HTML.gif, then the solution to (2.4) goes to −∞ before t tends to 1 C y 0 + K y 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq46_HTML.gif.

      3 Blow-up phenomena

      In this section, we discuss the blow-up phenomena of system (2.1). Firstly, we prove that there exist strong solutions to (2.1) which do not exist globally in time.

      Theorem 3.1 Let ( u 0 , ρ 0 ) H s × H s 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq47_HTML.gif, s 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq18_HTML.gif, and T be the maximal time of the solution ( u , ρ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq20_HTML.gif to (2.1) with the initial data ( u 0 , ρ 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq48_HTML.gif. If there is some x 0 S http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq49_HTML.gif such that ρ 0 ( x 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq28_HTML.gif and
      u 0 ( x 0 ) = inf x S u 0 ( x ) < e + 1 2 ( e 1 ) E 0 + | γ A | 8 ( e + 1 ) e 1 E 0 1 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equs_HTML.gif

      then the corresponding solution to (2.1) blows up in finite time.

      Proof Applying Lemma 2.1 and a simple density argument, we only need to show that the above theorem holds for some s 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq18_HTML.gif. Here we assume s = 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq50_HTML.gif to prove the above theorem.

      Define now
      m ( t ) : = inf x S [ u x ( t , x ) ] , t [ 0 , T ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equt_HTML.gif
      By Lemma 2.7, we let ξ ( t ) S http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq51_HTML.gif be a point where this infimum is attained. It follows that
      m ( t ) = u x ( t , ξ ( t ) ) and u x x ( t , ξ ( t ) ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equu_HTML.gif
      Differentiating the first equation in (2.2) with respect to x and using the identity x 2 G f = G f f http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq52_HTML.gif, we have
      u t x + ( u γ ) u x x = 1 2 u x 2 + 1 2 ρ 2 + u 2 + ( γ A ) u G ( u 2 + 1 2 u x 2 + 1 2 ρ 2 + ( γ A ) u ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equ7_HTML.gif
      (3.1)
      Since the map q ( t , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq25_HTML.gif given by (2.3) is an increasing diffeomorphism of ℝ, there exists a x ( t ) S http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq53_HTML.gif such that q ( t , x ( t ) ) = ξ ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq54_HTML.gif. In particular, x ( 0 ) = ξ ( 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq55_HTML.gif. Note that u 0 ( x 0 ) = inf x S u 0 ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq56_HTML.gif, we can choose x 0 = ξ ( 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq57_HTML.gif. It follows that x ( 0 ) = ξ ( 0 ) = x 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq58_HTML.gif. By Lemma 2.3 and the condition ρ 0 ( x 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq28_HTML.gif, we have
      ρ ( t , ξ ( t ) ) q x ( t , x ) = ρ ( t , q ( t , x ( t ) ) ) q x ( t , x ) = ρ 0 ( x ( 0 ) ) = ρ 0 ( x 0 ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equv_HTML.gif

      Thus ρ ( t , ξ ( t ) ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq59_HTML.gif.

      Valuating (3.1) at ( t , ξ ( t ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq60_HTML.gif and using Lemma 2.7, we obtain
      d m ( t ) d t 1 2 m 2 ( t ) + 1 2 u 2 + ( γ A ) u ( γ A ) G u , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equ8_HTML.gif
      (3.2)
      here we used the relations G ( u 2 + 1 2 u x 2 ) 1 2 u 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq61_HTML.gif and G ρ 2 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq62_HTML.gif. Note that G L 1 = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq63_HTML.gif. By Lemma 2.4 and Lemma 2.8, we get
      u L 2 e + 1 2 ( e 1 ) u H 1 2 e + 1 2 ( e 1 ) E 0 , | ( γ A ) u | | γ A | u L | γ A | e + 1 2 ( e 1 ) E 0 1 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equw_HTML.gif
      and
      | ( γ A ) G u | | γ A | G L 1 u L | γ A | e + 1 2 ( e 1 ) E 0 1 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equx_HTML.gif
      It follows that
      d m ( t ) d t 1 2 m 2 ( t ) + K , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equ9_HTML.gif
      (3.3)
      where K = e + 1 4 ( e 1 ) E 0 + 2 | γ A | e + 1 2 ( e 1 ) E 0 1 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq64_HTML.gif. Since m ( 0 ) < 2 K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq65_HTML.gif, Lemma 2.10 implies
      lim t T m ( t ) = with  T = 2 u 0 ( x 0 ) 2 K ( u 0 ( x 0 ) ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equy_HTML.gif

      Applying Lemma 2.6, the solution blows up in finite time. □

      Theorem 3.2 Let ( u 0 , ρ 0 ) H s × H s 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq47_HTML.gif, s 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq18_HTML.gif, and T be the maximal time of the solution ( u , ρ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq20_HTML.gif to (2.1) with the initial data ( u 0 , ρ 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq48_HTML.gif. Assume that S u 0 ( x ) d x = a 0 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq66_HTML.gif. If there is some x 0 S http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq49_HTML.gif such that ρ 0 ( x 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq28_HTML.gif and for any ϵ > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq43_HTML.gif,
      u 0 ( x 0 ) = inf x S u 0 ( x ) < ϵ + 2 24 E 0 + ϵ + 2 4 ϵ a 0 2 + | γ A | 2 ( ϵ + 2 ) 3 E 0 + 4 ( ϵ + 2 ) ϵ a 0 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equz_HTML.gif

      then the corresponding solution to (2.1) blows up in finite time.

      Proof By Lemma 2.5, we have S u ( t , x ) d x = a 0 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq67_HTML.gif. Using Lemma 2.4 and Lemma 2.9, we obtain
      u L 2 ϵ + 2 24 E 0 + ϵ + 2 4 ϵ a 0 2 , | ( γ A ) u | | γ A | u L | γ A | ϵ + 2 24 E 0 + ϵ + 2 4 ϵ a 0 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equaa_HTML.gif
      and
      | ( γ A ) G u | | γ A | G L 1 u L | γ A | ϵ + 2 24 E 0 + ϵ + 2 4 ϵ a 0 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equab_HTML.gif
      Following a similar proof in Theorem 3.1, we have
      d m ( t ) d t 1 2 m 2 ( t ) + K , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equ10_HTML.gif
      (3.4)

      where K = ϵ + 2 48 E 0 + ϵ + 2 8 ϵ a 0 2 + | γ A | ϵ + 2 6 E 0 + ϵ + 2 ϵ a 0 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq68_HTML.gif. Following the same argument as in Theorem 3.1, we deduce that the solution blows up in finite time. □

      Letting a 0 = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq69_HTML.gif and ϵ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq70_HTML.gif in Theorem 3.2, we have the following result.

      Corollary 3.1 Let ( u 0 , ρ 0 ) H s × H s 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq47_HTML.gif, s 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq18_HTML.gif, and T be the maximal time of the solution ( u , ρ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq20_HTML.gif to (2.1) with the initial data ( u 0 , ρ 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq48_HTML.gif. Assume that S u 0 ( x ) d x = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq71_HTML.gif. If there is some x 0 S http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq72_HTML.gif such that ρ 0 ( x 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq28_HTML.gif and
      u 0 ( x 0 ) = inf x S u 0 ( x ) < E 0 12 + 2 | γ A | E 0 3 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equac_HTML.gif

      then the corresponding solution to (2.1) blows up in finite time.

      Remark 3.1 Note that system (2.1) is variational under the transformation ( u , x ) ( u , x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq73_HTML.gif and ( ρ , x ) ( ρ , x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq74_HTML.gif even γ = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq7_HTML.gif. Thus, we cannot get a blow-up result according to the parity of the initial data ( u 0 , ρ 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq75_HTML.gif as we usually do.

      Next, we will give more insight into the blow-up mechanism for the wave-breaking solution to system (2.1), that is, the blow-up rate for strong solutions to (2.1).

      Theorem 3.3 Let ( u , ρ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq20_HTML.gif be the solution to system (2.1) with the initial data ( u 0 , ρ 0 ) H s × H s 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq47_HTML.gif, s 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq18_HTML.gif, satisfying the assumption of Theorem 3.1, and T be the maximal time of the solution ( u , ρ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq20_HTML.gif. Then we have
      lim t T ( T t ) inf x S u x ( t , x ) = 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equad_HTML.gif

      Proof As mentioned earlier, here we only need to show that the above theorem holds for s = 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq50_HTML.gif.

      Define now
      m ( t ) : = inf x S [ u x ( t , x ) ] , t [ 0 , T ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equae_HTML.gif
      By the proof of Theorem 3.1, there exists a positive constant K = K ( E 0 , γ , A ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq76_HTML.gif such that
      K d d t m + 1 2 m 2 K a.e. on  ( 0 , T ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equ11_HTML.gif
      (3.5)
      Let ε ( 0 , 1 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq77_HTML.gif. Since lim inf t T m ( t ) = http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq78_HTML.gif by Theorem 3.1, there is some t 0 ( 0 , T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq79_HTML.gif with m ( t 0 ) < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq80_HTML.gif and m 2 ( t 0 ) > K ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq81_HTML.gif. Since m is locally Lipschitz, it is then inferred from (3.5) that
      m 2 ( t ) > K ε , t [ t 0 , T ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equ12_HTML.gif
      (3.6)
      A combination of (3.5) and (3.6) enables us to infer
      1 2 + ε d m d t m 2 1 2 ε a.e. on  ( 0 , T ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equ13_HTML.gif
      (3.7)
      Since m is locally Lipschitz on [ 0 , T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq82_HTML.gif and (3.6) holds, it is easy to check that 1 m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq83_HTML.gif is locally Lipschitz on ( t 0 , T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq84_HTML.gif. Differentiating the relation m ( t ) 1 m ( t ) = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq85_HTML.gif, t ( t 0 , T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq86_HTML.gif, we get
      d d t ( 1 m ) = d m d t m 2 a.e. on  ( t 0 , T ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equaf_HTML.gif
      with 1 m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq83_HTML.gif absolutely continuous on ( t 0 , T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq84_HTML.gif. For t ( t 0 , T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq87_HTML.gif. Integrating (3.7) on ( t , T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq88_HTML.gif to obtain
      ( 1 2 + ε ) ( T t ) 1 m ( t ) ( 1 2 ε ) ( T t ) , t ( t 0 , T ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equag_HTML.gif
      that is,
      1 1 2 + ε m ( t ) ( T t ) 1 1 2 ε , t ( t 0 , T ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equah_HTML.gif

      By the arbitrariness of ε ( 0 , 1 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq77_HTML.gif the statement of Theorem 3.3 follows. □

      4 Global existence

      In this section, we will present a global existence result.

      Theorem 4.1 Let ( u 0 , ρ 0 ) H s × H s 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq47_HTML.gif, s 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq18_HTML.gif, and T be the maximal time of the solution ( u , ρ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq20_HTML.gif to (2.1) with the initial data ( u 0 , ρ 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq48_HTML.gif. If ρ 0 ( x ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq89_HTML.gif for all x S http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq90_HTML.gif, then the corresponding solution ( u , ρ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq20_HTML.gif exists globally in time.

      Proof Define
      m ( t ) : = inf x S [ u x ( t , x ) ] , t [ 0 , T ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equai_HTML.gif
      By Lemma 2.7, we let ξ ( t ) S http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq51_HTML.gif be a point where this infimum is attained. It follows that
      m ( t ) = u x ( t , ξ ( t ) ) and u x x ( t , ξ ( t ) ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equaj_HTML.gif

      Since the map q ( t , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq25_HTML.gif given by (2.3) is an increasing diffeomorphism of ℝ, there exists an x ( t ) S http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq53_HTML.gif such that q ( t , x ( t ) ) = ξ ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq54_HTML.gif.

      Set m ( t ) = u x ( t , ξ ( t ) ) = u x ( t , q ( t , x ( t ) ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq91_HTML.gif and α ( t ) = ρ ( t , ξ ( t ) ) = ρ ( t , q ( t , x ( t ) ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq92_HTML.gif. Valuating (3.1) at ( t , ξ ( t ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq60_HTML.gif and using Lemma 2.7, we obtain
      m ( t ) = 1 2 m 2 ( t ) + 1 2 α 2 ( t ) + f and α ( t ) = m ( t ) α ( t ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equ14_HTML.gif
      (4.1)
      where f = u 2 + ( γ A ) u G ( u 2 + 1 2 u x 2 + 1 2 ρ 2 + ( γ A ) u ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq93_HTML.gif. By Lemma 2.4, Lemma 2.8 and 1 2 sinh 1 2 G ( x ) cosh 1 2 2 sinh 1 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq94_HTML.gif, we have
      | f | u L 2 + 2 | γ A | u L + G L u 2 + 1 2 u x 2 + 1 2 ρ 2 L 1 e + 1 2 ( e 1 ) E 0 + 2 | γ A | e + 1 2 ( e 1 ) E 0 1 2 + cosh 1 2 2 sinh 1 2 E 0 : = c 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equak_HTML.gif
      By Lemmas 2.2-2.3, we know that α ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq95_HTML.gif has the same sign with α ( 0 ) = ρ 0 ( x 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq96_HTML.gif for every x R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq97_HTML.gif. Moreover, there is a constant β > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq98_HTML.gif such that | α ( 0 ) | = inf x S | ρ 0 ( x ) | β > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq99_HTML.gif because of ρ 0 ( x ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq100_HTML.gif for all x S http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq90_HTML.gif. Next, we consider the following Lyapunov positive function:
      w ( t ) = α ( 0 ) α ( t ) + α ( 0 ) α ( t ) ( 1 + m 2 ( t ) ) , t [ 0 , T ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equ15_HTML.gif
      (4.2)
      Letting t = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq101_HTML.gif in (4.2), we have
      w ( 0 ) ρ 0 L 2 + 1 + u 0 ( x ) L 2 : = c 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equal_HTML.gif
      Differentiating (4.2) with respect to t and using (4.1), we obtain
      w ( t ) = α ( 0 ) α ( t ) 2 m ( t ) ( f + 1 2 ) α ( 0 ) α ( t ) ( 1 + m 2 ( t ) ) ( | f | + 1 2 ) w ( t ) ( c 1 + 1 2 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equam_HTML.gif
      By Gronwall’s inequality, we have
      w ( t ) w ( 0 ) e ( c 1 + 1 2 ) t c 2 e ( c 1 + 1 2 ) t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equan_HTML.gif
      for all t [ 0 , T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq30_HTML.gif. On the other hand,
      w ( t ) 2 α 2 ( 0 ) ( 1 + m 2 ( t ) ) 2 β | m ( t ) | , t [ 0 , T ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equao_HTML.gif
      Thus,
      | m ( t ) | 1 2 β w ( t ) c 2 2 β e ( c 1 + 1 2 ) t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equap_HTML.gif
      for all t [ 0 , T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_IEq102_HTML.gif. It follows that
      lim inf t T m ( t ) c 2 2 β e ( c 1 + 1 2 ) T . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-158/MediaObjects/13661_2013_Article_412_Equaq_HTML.gif

      This completes the proof by using Lemma 2.6. □

      Declarations

      Acknowledgements

      The authors would like to thank the editors and the referees for their valuable suggestions to improve the quality of this paper. This research is partially supported by the Doctoral Research Foundation of Zhengzhou University of Light Industry.

      Authors’ Affiliations

      (1)
      Department of Mathematics and Information Science, Zhengzhou University of Light Industry
      (2)
      Faculty of Science, Guilin University of Aerospace Industry

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      © Liu and Zhang; licensee Springer. 2013

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