Blow-up and local weak solution for a modified two-component Camassa-Holm equations

  • Lixin Tian1Email author and

    Affiliated with

    • Minyi Zhu1

      Affiliated with

      Boundary Value Problems20122012:52

      DOI: 10.1186/1687-2770-2012-52

      Received: 14 November 2011

      Accepted: 2 May 2012

      Published: 2 May 2012

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      Abstract

      In this article, we establish some blow-up results for a modified two-component Camassa-Holm system in Sobolev spaces. We also obtain the existence of the weak solutions of this system in H s × Hs-1, s > 5/2.

      Keywords

      the modified two-component Camassa-Holm equations blow up weak solutions.

      1. Introduction

      The well-known two-component Camassa-Holm equations [1]
      m t + 2 u x m + u m x + σ ρ ρ x = 0 , ρ t + u ρ x = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equ1_HTML.gif
      (1.1)

      where m = u - u xx and σ = ± 1. Constantin and Ivanov [2] derived this system in the context of shallow water theory. u can be interpreted as the horizontal fluid velocity and ρ is related to the water elevation in the first approximation [2, 3]. They showed that while small initial data develop into global solutions, for some initial data wave breaking occurs. They also discussed the solitary wave solutions. In Vlasov plasma models, system (1.1) describes the closure of the kinetic moments of the single-particle probability distribution for geodesic motion on the simplectomorphisms. While in the large-deformation diffeomorphic approach to image matching, system (1.1) is summoned in a type of matching procedure called metamorphosis (see [4] and the references therein). This system appeared originally in [5]. Based on the deformation of bi-Hamiltonian structure of the hydrodynamic type, Chen et al. [6] obtained system (1.1) when σ = -1. They show that it has the peakon and multilink solitons, and is integrable in the sense that it has Lax-pair. The mathematical properties of system (1.1) have been studied further in many articles, see, e.g., [715]. In [4], Holm and Ivanov generalized the Lax-pair formulation of system (1.1) to produce an integrable multi-component family, CH(n, k), of equations with n components and 1 ≤ |k| ≤ n velocities. They determined their Lie-Poisson Hamiltonian structures and gave numerical examples of their soliton solution behavior. Recently, a new global existence result and several new blow-up results of strong solutions for the Cauchy problem of Equation (1.1) with σ = 1 were obtained in [8]. Gui and Liu [14] established the local well posedness for the two-component Camassa-Holm system in a range of the Besov spaces. Chen and Liu [16] discussed the wave-breaking phenomenon of a generalized two-component Camassa-Holm system, and determined the exact blow-up rate of such solutions. The existence and uniqueness of global weak solutions to Equation (1.1) have also been discussed by Guan and Yin [17].

      In this article, we consider a two-component generalization of Equation (1.1), that is
      u t - u x x t + 3 u m u x - 2 u x u x x - u u x x x + ρ ρ x = 0 , ρ t + u ρ x = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equ2_HTML.gif
      (1.2)
      with initial data
      u 0 , x = u 0 x H s , s 1 , x R , ρ 0 , x = ρ 0 x H s - 1 , s - 1 1 , x R , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equ3_HTML.gif
      (1.3)

      where m ≥ 1. It can be reduced to (1.1) as m = 1.

      The purpose of this article is to study the well posedness, local weak solution, and blow-up for Cauchy problem (1.2) and (1.3). System (1.2) also conserves conservation laws. Our starting point is to obtain the local well posedness by using Kato's theory, Next, we derive some blow-up results of the solutions by the following transport equation,
      q t = u t , q , t 0 , T , q 0 , x = x , x R , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equa_HTML.gif

      which is a crucial ingredient to obtain the blow-up phenomenon. Last, by using the conserves from laws and the contraction mapping theorem, we obtain the existence of weak solutions of Cauchy problem (1.2) and (1.3). These methods are similar to that was used in [18]. However, because of the asymmetry and the high strength of the nonlinearity of Equation (1.3), it is more difficult to estimate the norm of u, ρ, u x , ρ x in Sobolev space. In addition, also we get Equation (5.10) which is different with that in [18]. As for the blow-up phenomenon, we get some new results of (1.2) and (1.3).

      Guan and Yin [17, 19] got the global weak solutions for two-component Camassa-Holm shallow water system; they first obtained approximate solutions for the system, then they prove the compactness of these solutions, and at last they got the global weak solutions. Using the same way, Liu and Yin [20] also got global weak solutions for a periodic two-component μ-Hunter-Saxton system. However, in this article, we add high-order perturbation terms in this system, and by using the conserves laws and the contraction mapping theorem, we obtain the existence of weak solutions.

      The remainder of this article is organized as follows. Section 2 is the preliminary. In Section 3, the local well posedness for strong solution of Cauchy problem (1.2) and (1.3) is established by Kato's theory. In Section 4, by transport equation, some blow-up results of the solutions of Cauchy problem (1.2) and (1.3) are obtained. The proof of existence of local weak solution is carried out in Section 5.

      2. Blow-up

      Lemma 2.1: Given z 0 = u 0 ρ 0 H s × H s - 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq1_HTML.gif, s > 5/2, then there exists a maximal T = T z 0 H s × H s - 1 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq2_HTML.gif, and a unique solution z = u ρ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq3_HTML.gif to Cauchy problem (1.2) and (1.3) such that
      z , z 0 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%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equb_HTML.gif
      Moreover, the solution depends continuously on the initial data, i.e., the mapping
      z 0 z , z 0 : H s × H s - 1 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%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equc_HTML.gif

      is continuous.

      The proof is similar with Theorem 4.1 in [21].

      Let ρ ¯ = ρ - 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq4_HTML.gif, then (1.2) is equivalent to
      u t - u t x x + 3 u m u x = 2 u x u x x + u u x x x - ρ ¯ ρ ¯ x - ρ ¯ x . ρ ¯ t + u ρ ¯ x = - u x ρ ¯ x - u . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equ4_HTML.gif
      (2.1)
      Consider 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%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equ5_HTML.gif
      (2.2)

      where u is the first component of the solution z to Equation (1.2).

      To prove the blow-up result, we need the following lemma.

      Lemma 2.2: Let z0∈ H s × Hs-1, (s > 5/2), and let T > 0 be the maximal existence time of the corresponding solution z to Equation (2.1), then we have
      ρ ¯ t , q t , x + 1 q x t , x = ρ ¯ 0 x + 1 , t , x 0 , T × R . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equ6_HTML.gif
      (2.3)
      Proof. Differentiating the left-hand side of Equation (2.3) with respect t. It follows from (2.1) and (2.2), that
      d d t ρ ¯ t , q t , x + 1 q x t , x = ρ ¯ t t , q t , x + ρ ¯ x t , q t , x q t t , x q x t , x + ρ ¯ t , q t , x + 1 q x t t , x = ρ ¯ t t , q t , x + ρ ¯ x t , q t , x u t , q + ρ ¯ t , q u x t , q + u x t , q q x t , x = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equd_HTML.gif

      This completes the proof of this lemma.

      Theorem 2.1: Let z 0 = u 0 ρ ¯ 0 H s × H s - 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq5_HTML.gif, (s > 5/2), and T be the maximal time of the solution z to Equation (1.2) with the initial data z0. Assume that there exists x0R such that ρ ¯ 0 x 0 = - 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq6_HTML.gif and
      u 0 x 0 < - 2 u 0 H 1 2 + ρ 0 L 2 2 + 3 m + 1 u 0 H 1 2 + ρ 0 L 2 2 m + 1 2 1 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Eque_HTML.gif

      Then, T is finite and the slope of u tends to negative infinity as t goes to T while u is uniformly bounded on [0, T).

      Proof. Let z = u ρ ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq7_HTML.gif be the solution of Equation (2.1) with the initial data z0, and T be the maximal time of z, and let
      n t = u x t , q t , x 0 , γ t = ρ ¯ t , q t , x 0 + 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equf_HTML.gif
      From (2.1) and (2.2), we have
      d n d t = u t x + u u x x t , q t , x 0 , d γ d t = - γ n . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equg_HTML.gif
      Differentiating the first equation in (2.1) with respect x, we have
      u t x + u u x x = 3 m + 1 u m + 1 - 1 2 u 2 - 1 2 u x 2 + 1 2 ρ ¯ 2 + ρ ¯ - Λ - 2 3 m + 1 u m + 1 - 1 2 u 2 + 1 2 u x 2 + 1 2 ρ ¯ 2 + ρ ¯ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equh_HTML.gif

      Note that γ 0 = ρ ¯ 0 x 0 + 1 = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq8_HTML.gif, Then, by Lemma4.1, we have γ(t) = 0, ∀t ∈[0, T).

      Thus
      n t = - 1 2 n 2 + 1 2 γ 2 + 3 m + 1 u m + 1 - 1 2 u 2 - Λ - 2 3 m + 1 u m + 1 - 1 2 u 2 + 1 2 u x 2 + 1 2 γ 2 t , q . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equi_HTML.gif

      Since Λ - 2 1 2 u x 2 + 1 2 γ 2 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq9_HTML.gif, and - Λ - 2 3 m + 1 u m + 1 - 1 2 u 2 3 m + 1 u m + 1 + 1 2 u 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq10_HTML.gif, n t - 1 2 n 2 + u 2 + 3 m + 1 u m + 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq11_HTML.gif

      Note that u L u H 1 2 + ρ L 2 2 1 2 = u 0 H 1 2 + ρ 0 L 2 2 1 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq12_HTML.gif. If let K = u 0 H 1 2 + ρ 0 L 2 2 + 3 m + 1 u 0 H 1 2 + ρ 0 L 2 2 m + 1 2 1 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq13_HTML.gif then we have n t - 1 2 n 2 t + K 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq14_HTML.gif.

      Since n 0 < - 2 K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq15_HTML.gif, we obtain that n t < - 2 t , t 0 , T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq16_HTML.gif.

      With the inequality above, we get
      n 0 + 2 K n 0 - 2 K exp 2 K t - 1 2 2 K n t - 2 K 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equj_HTML.gif

      Since 0 < n 0 + 2 K n 0 - 2 K < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq17_HTML.gif, there exists 0 < T < 1 2 K ln n 0 + 2 K n 0 - 2 K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq18_HTML.gif, such that lim t T n t = - http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq19_HTML.gif, i.e., lim t T u x t = - http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq20_HTML.gif.

      This completes the proof of the theorem.

      3. Local weak solution

      Definition 3.1: ([22]) Let (u0, ρ0) ∈ H1(R) × H1(R). If (u, ρ) belongs to L loc 0 , T ; H 1 R × L l o c 0 , T ; H 1 R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq21_HTML.gif and satisfies the identity
      0 T R z ψ t + F u ψ x d x d t + R z 0 x ψ 0 , x d x = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equk_HTML.gif

      for all ψ C c 0 , T × R × C c 0 , T × R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq22_HTML.gif, where ψ C c 0 , T × R × C c 0 , T × R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq23_HTML.gif, the set of all the restrictions to ([0,T) × R) × ([0,T) × R) of smooth functions on R2 × R2 with compact support contained in ((-T, T) × R) × ((-T, T) × R). Then, z is called a weak solution to Equation (1.6). If z is a weak solution on [0, T) × [0, T) for every T > 0, then it is called global weak solution to Equation (1.6).

      In this section, we discuss the existence of weak solution of Cauchy problem (1.2) and (1.3). To this purpose, we consider the following Cauchy problem:
      u t - u x x t + ε u x x x x t + 3 u m u x - 2 u x u x x - u u x x x + ρ ρ x = 0 , ρ t + ε ρ x x t + u ρ x = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equ7_HTML.gif
      (3.1)
      u 0 , x = u 0 x H s , s 1 , x R . ρ 0 , x = ρ 0 x H s - 1 , s - 1 1 , x R . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equ8_HTML.gif
      (3.2)

      where ε is a constant satisfying 0 < ε < 1/4. Note that when ε = 0, system (3.1) and (3.2) is just the system (1.2) and (1.3).

      For any 0 < ε < 1/4 and s ≥ 1, the integral operators
      D 1 = 1 - x 2 + ε x 4 - 1 : H s H s + 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equl_HTML.gif
      and
      D 2 = 1 + ε x 2 - 1 : H s H s + 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equm_HTML.gif

      define two bounded linear operator in the indicated Sobolev spaces.

      To prove the existence of solutions to the problem (3.1) and (3.2), we apply the two operators above to both sides of (3.1) and then integrate the resulting equations with regard to t. This leads to the following equations.
      u ( x , t ) = u 0 ( x ) + 0 t D 1 - 3 m + 1 u m + 1 x + u x 2 x + u u x x - 1 2 u x 2 x + 1 2 ρ 2 x d τ . ρ x , t = ρ 0 x + 0 t D 2 - ( u ρ ) x d τ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equn_HTML.gif

      A standard application of the contraction mapping theorem leads to the following existence result.

      Theorem 3.1: For each initial data u0H s (s ≥ 1), ρ0Hs-1(s ≥ 2), there exists a T > 0 depending only on the norm of u 0 H s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq24_HTML.gif, ρ 0 H s - 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq25_HTML.gif, and m such that there exists a unique solution (u, ρ) ∈ C([0,T];H s ) × C([0, T];Hs-1) of system (3.1) and (3.2) in the sense of distribution. If u0H s (s ≥ 2), ρ0Hs-1(s ≥ 3), the solution (u, ρ) ∈ C([0,∞];H s ) × C([0, ∞];Hs-1) exists for all time, in particular, when u0H s (s ≥ 4), ρ0Hs-1(s ≥ 5), the corresponding solution is a classical globally defined solution of (3.1) and (3.2).

      The global existence result follows from the conservation law
      R u 2 + ρ 2 + u x 2 + ε ρ x 2 + ε u x x 2 d x = R u 0 2 + ρ 0 2 + u 0 x 2 + ε ρ 0 x 2 + ε u 0 x x 2 d x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equo_HTML.gif

      admitted by (3.1) in its integral form.

      Theorem 3.2: Suppose that for some s ≥ 4, the function pair u(x, t) and ρ(x, t) in the solution of Equation (3.1) corresponding to the initial data u0H s (s ≥ 4); ρ0Hs-1(s ≥ 5), then the following inequalities hold:
      u H 1 2 , ρ L 2 2 R u 0 2 + ρ 0 2 + u 0 x 2 + ε ρ 0 x 2 + ε u 0 x x 2 d x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equ9_HTML.gif
      (3.3)
      For any real number q ∈ (1, s] (s ≥ 5), there exists a constant c depending only on q, m, such that
      R 1 - ε Λ q u 2 + Λ q - 1 ρ 2 + 1 + ε Λ q - 1 ρ x 2 + ε Λ q u x 2 + ε Λ q - 1 u 2 d x R 1 - ε Λ q u 0 2 + Λ q - 1 ρ 0 2 + 1 + ε Λ q - 1 ρ 0 x 2 + ε Λ q u 0 x 2 + ε Λ q - 1 u 0 2 d x + c 0 t u L m + ρ x L + u x L + ρ L 2 u H q 2 + ρ H q - 1 2 d τ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equ10_HTML.gif
      (3.4)
      For any q ∈ (1, s-1] (s ≥ 4), there exists a constant c such that
      1 - ε u t H q c 1 + u x L + ρ x L u H q + u H q m + ρ H q - 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equ11_HTML.gif
      (3.5)
      And for any q ∈ (1, s-2] (s ≥ 5), there exists a constant c such that
      ρ t H q - 1 c 1 + u x L + ρ x L u H q + u H q m + ρ H q - 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equ12_HTML.gif
      (3.6)
      Proof. It is obvious that (3.3) holds. In order to prove (5.4), let Λ = 1 - x 2 1 / 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq26_HTML.gif. We rewrite Equation (3.1) in the following equivalent form.
      1 - ε u t - ε u x x t + ε Λ - 2 u t = - Λ - 2 ρ ρ x + 3 u m u x - 2 u x u x x - u u x x x , ρ t + 1 + ε Λ - 2 ρ x x t = - Λ - 2 u ρ x , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equ13_HTML.gif
      (3.7)
      For any q ∈ (1, s] (s ≥ 5), applying (Λ q u q to the both sides of the first equation of Equation (3.7), respectively, and integrating with regard to x, we obtain
      1 - ε Λ q u , Λ q u t - ε Λ q u , Λ q u x x t + ε Λ q u , Λ q - 2 u t = 1 2 d d t R 1 - ε Λ q u 2 + ε Λ q u x 2 + ε Λ q - 1 u 2 d x , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equp_HTML.gif
      By using Sobolev embedding theorems, we have
      Λ q u , Λ q - 2 ρ ρ x 0 Λ q - 2 , ρ ρ x , Λ q u 0 + ρ Λ q - 2 ρ x , Λ q u 0 c u H q ρ x L Λ q - 3 ρ x L 2 + Λ q - 2 ρ L 2 ρ x L + u H q ρ L 2 Λ q - 2 ρ x L 2 c u H q ρ x L ρ H q - 2 + ρ H q - 2 ρ x L + ρ L 2 ρ H q - 1 c ρ x L + ρ L 2 u H q 2 + ρ H q - 1 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equq_HTML.gif
      where we have used lemma in [23] with r = q -2 > 0. Also
      Λ q u , Λ q - 2 3 u m u x - 2 u x u x x - u u x x x 0 = x Λ - 2 3 m + 1 u m + 1 - 1 2 u x 2 - u u x x , u q 3 m + 1 u m + 1 - 1 2 u x 2 - u u x x H q - 1 u H q c u m L + u x L u H q 2 + ρ H q - 1 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equr_HTML.gif

      where we have used Lemma in [24] with r = q -1 > 0.

      Then, we get
      1 2 d d t R 1 - ε Λ q u 2 + ε Λ q u x 2 + ε Λ q - 1 u 2 d x c u L m + u x L + ρ x L + ρ L 2 u H q 2 + ρ H q - 1 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equ14_HTML.gif
      (3.8)
      For any q ∈ (1, s-1] (s ≥ 5), applying (Λ q-1 ρ) Λ q-1 to the both sides of the second equation of Equation (3.7), respectively, then we obtain
      1 2 d d t R Λ q - 1 ρ 2 + 1 + ε Λ q - 2 ρ x 2 d x c u L + ρ L u H q 2 + ρ H q - 1 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equ15_HTML.gif
      (3.9)
      Summing up (3.8) and (3.9), we get
      R 1 - ε Λ q u 2 + Λ q - 1 ρ 2 + 1 + ε Λ q - 2 ρ x 2 + ε Λ q u x 2 + ε Λ q - 1 u 2 d x R 1 - ε Λ q u 0 2 + Λ q - 1 ρ 0 2 + 1 + ε Λ q - 2 ρ 0 x 2 + ε Λ q u 0 x 2 + ε Λ q - 1 u 0 2 d x + c 0 t u L m + ρ x L + u x L + ρ L 2 u H q 2 + ρ H q - 1 2 d τ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equs_HTML.gif
      For any q ∈ (1, s-1] (s ≥ 4), applying (Λ q u t q to the both sides of the first equation of Equation (3.7), respectively, and integrating with regard to x, we obtain that
      1 - ε Λ q u t , Λ q u t - ε Λ q u t , Λ q u x x t + ε Λ q u t , Λ q - 2 u t = 1 - ε u t H q 2 + ε u x t H q 2 + ε u t H q - 1 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equt_HTML.gif
      and
      x Λ - 2 ρ 2 , u t q = u t , Λ - 2 ρ ρ x q Λ q - 2 , ρ ρ x , Λ q u t 0 + ρ Λ q - 2 ρ x , Λ q u t 0 c u t H q ρ x L Λ q - 3 ρ x L 2 + Λ q - 2 ρ L 2 ρ x L + ρ L 2 Λ q - 2 ρ x L 2 c u t H q ρ H q - 2 ρ x L + ρ L 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equu_HTML.gif
      u t , Λ - 2 3 u m u x - 2 u x u x x - u u x x x q = u t , x Λ - 2 3 m + 1 u m + 1 - 1 2 u x 2 - u u x x q u t H q 3 m + 1 u m + 1 - 1 2 u x 2 - u u x x H q - 1 c u t H q u L m + u L + u x L u H q + u H q m , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equv_HTML.gif
      where we have used lemma in [24] with r = q -1 > 0. Then, we get
      1 - ε u t H q c 1 + u x L + ρ x L u H q + u H q m + ρ H q - 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equw_HTML.gif
      For any q ∈ (1, s-2] (s ≥ 5), applying (Λq-1ρ t q-1to the both sides of the second equation of Equation (5.7), respectively, then we obtain
      ρ t H q - 1 c 1 + ρ x L + u x L u H q + ρ H q - 1 c 1 + ρ x L + u x L u H q + u H q m + ρ H q - 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equx_HTML.gif

      This complete the proof of the theorem.

      Suppose u0H s (s ≥ 1), ρ0Hs-1(s ≥ 2), and let uε 0, ρε 0be the convolution uε 0= φ ε *u0, ρε 0= φ ε *ρ0, where φ ε x = ε - 1 4 φ ^ ε - 1 4 x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq27_HTML.gif such that the Fourier transform φ ^ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq28_HTML.gif of φ satisfies φ ^ C 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq29_HTML.gif, φ ^ ξ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq30_HTML.gif, and φ ^ ξ = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq31_HTML.gif for any ξ ∈ (-1,1). Then, it follows from Theorem 3.1 that for each ε with 0 < ε <1/4, the Cauchy problem
      u t - u x x t + ε u x x x x t + 3 u m u x - 2 u x u x x - u u x x x + ρ ρ x = 0 , ρ t + ε ρ x x t + u p x = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equ16_HTML.gif
      (3.10)
      u 0 , x = u ε 0 x , t 0 , x R ρ 0 , x = ρ ε 0 x , t 0 , x R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equy_HTML.gif

      has a unique solution u ε (t, x) ∈C([0,∞);H and ρ ε (t, x) ∈C([0,∞);H. We first demonstrate the properties of the initial data uε 0, ρε 0in the following lemma. The proof is similar to Lemma 5 in [25].

      Lemma 3.1: Under the above assumptions, there hold
      u ε 0 H q c , if q s ; ρ ε 0 H q - 1 c , if q - 1 s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equz_HTML.gif
      u ε 0 H q c ε ( s - q ) 4 , if q > s ; ρ ε 0 H q - 1 c ε ( s - q + 1 ) 4 , if q - 1 > s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equaa_HTML.gif

      for any ε with 0 < ε < 1 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq32_HTML.gif, where c is a constant independent of ε. The proof is similar to Lemma 5 in [25].

      Theorem 3.3: Suppose that u0(x) ∈ H s (R), s ∈ [1, 3/2]; ρ0(x) ∈ Hs-1(R),

      s-1 ∈ [1, 3/2] such that u 0 x L < http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq33_HTML.gif, ρ 0 x L < http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq34_HTML.gif. Let uε 0= φ ε *u0, ρε 0= φ ε *ρ0, be defined the same as above. Then, there exist constants T > 0 and c > 0 independent of ε such that the corresponding solution u ε , ρ ε of (3.10) satisfy the inequalities u ε x L c http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq35_HTML.gif, ρ ε x L c http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq36_HTML.gif for any t ∈ [0,T).

      Proof. Use Equation (3.7) with u = u ε , ρ = ρ ε . Differentiating with respect to x on both sides of the first equation in Equation (3.7). Note that x 2 Λ - 2 = Λ - 2 - I http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq37_HTML.gif, we obtain
      1 - ε u x t - ε u x x x t = - 1 2 Λ - 2 ρ 2 + 1 2 ρ 2 - x 2 Λ - 2 3 m + 1 u m + 1 - 1 2 u x 2 - u u x x - ε Λ - 2 u x t = 1 2 ρ 2 + 3 m + 1 u m + 1 - 1 2 u x 2 - u u x x - Λ - 2 1 2 ρ 2 + 3 m + 1 u m + 1 - 1 2 u x 2 - u u x x - ε Λ - 2 u x t , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equab_HTML.gif
      Let n > 0 be an integer. Then, multiplying the above equation by (u x )2n+1 to integrate with respect to x, we get
      R 1 - ε u x t u x 2 n + 1 - ε u x x x t u x 2 n + 1 d x = R 1 2 ρ 2 u x 2 n + 1 d x + R 3 m + 1 u m + 1 u x 2 n + 1 d x - R 1 2 u x 2 n + 3 d x + 1 2 n + 2 R u x 2 n + 3 d x - R g u x 2 n + 1 d x , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equac_HTML.gif

      where g = Λ - 2 1 2 ρ 2 + 3 m + 1 u m + 1 - 1 2 u x 2 - u u x x + ε x Λ - 2 u t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq38_HTML.gif.

      It follows from Hölder inequality that
      1 - ε d d t R u x 2 n + 2 d x 1 2 n + 2 ε R u x x x t 2 n + 2 d x 1 2 n + 2 + 1 2 R ρ 4 n + 4 d x 1 2 n + 2 + 3 m + 1 R u m + 1 2 n + 2 d x 1 2 n + 2 + 1 2 R u x 4 n + 4 d x 1 2 n + 2 + 1 2 n + 2 R u x 4 n + 4 d x 1 2 n + 2 + R g 2 n + 2 d x 1 2 n + 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equad_HTML.gif
      Note that f L p f L http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq39_HTML.gif as p→∞ for any fLL2. Integrating the above inequality over R with respect to t, and taking the limitation as n→∞, we have
      1 - ε u x L 1 - ε u 0 x L + 0 t ε u x x x t L + c ρ 2 L + u m + 1 L + u x 2 L + g L d τ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equae_HTML.gif
      It follows from (3.3) that
      g L c ̃ u t L 2 + ρ L 2 2 + u L 2 m + 1 + u x L 2 2 + u u x x L 2 c ̃ 1 1 + u t L 2 + u u x x L 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equaf_HTML.gif
      For any given r ∈ (1/2,1), we have
      u x x x t L c r u x x x t H r c r u t H r + 3 c r 1 + u x L + ρ x L u H r + 4 + u H r + 4 m + ρ H r + 3 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equag_HTML.gif
      Then from (3.4), we have
      u H r + 4 + ρ H r + 3 2 u H r + 4 2 + ρ H r + 3 2 2 2 1 - ε 1 - ε u 0 H r + 4 2 + ε u 0 x H r + 4 2 + ε u 0 H r + 3 2 + ρ 0 H r + 3 2 + 1 + ε ρ 0 x H r + 3 2 1 2 exp c u L m + ρ L 2 + u x L + ρ x L d τ c ε s - r - 4 4 exp c 0 t 1 + u x L + ρ x L d τ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equah_HTML.gif
      and
      u H r + 4 + u H r + 4 m + ρ H r + 3 u H r + 4 + ρ H r + 3 + u H r + 4 + ρ H r + 3 m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equai_HTML.gif
      Thus,
      u x x x t L c r ε s - r - 4 4 1 + u x L + ρ x L exp c 0 t 1 + u x L + ρ x L d τ + c m ε s - r - 4 m 4 1 + u x L + ρ x L m exp c m 0 t 1 + u x L + ρ x L d τ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equaj_HTML.gif
      and
      ρ 2 L ρ L 2 ρ H 1 2 = R | ρ | 2 d x + R | ρ x | 2 d x = ρ L 2 2 + ρ x L 2 2 c ( 1 + ρ x L 2 2 ) c ( 1 + ρ x L 2 ) u m + 1 L u L m + 1 u H 1 m + 1 c , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equak_HTML.gif
      u t L 2 u t H 1 u t H r + 3 1 + u x L + ρ x L u H r + 4 + u H r + 4 m + ρ H r + 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equal_HTML.gif
      u u x x L 2 u u x x H 1 c u L u x x H 1 + u H 1 u x x L c u H 1 u H r + 4 c u H r + 4 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equam_HTML.gif
      Then, we get
      g L ̃ c ̃ 1 + 1 + u x L + ρ x L u H r + 4 + u H r + 4 m + ρ H r + 3 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equan_HTML.gif
      It follows that
      u x L u 0 x L + 1 1 - ε 0 t c 1 + u x L + ρ x L + u x L + ρ x L 2 + 1 + u x L + ρ x L u H r + 4 + u H r + 4 m + ρ H r + 3 + ε c r 1 + u x L + ρ x L u H r + 4 + u H r + 4 m + ρ H r + 3 d τ u 0 x L + 4 c 3 0 t 1 + u x L + ρ x L + u x L + ρ x L 2 + 1 + u x L + ρ x L http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equao_HTML.gif

      exp c 0 τ 1 + u x L + ρ x L d η + 1 + u x L + ρ x L m exp c m 0 τ 1 + u x L + ρ x L d η d τ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq40_HTML.gif where c is a constant depends on Λ-2 and m.

      Also, we can obtain
      ρ x L ρ 0 x L + c 0 t 1 + u x L + ρ x L + 1 + ε 1 + u x L + ρ x L u H r + 4 + u H r + 4 m + ρ H r + 3 d τ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equap_HTML.gif
      where r 1 2 , 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq41_HTML.gif. From (3.9), we derive
      u x L + ρ x L u 0 x L + ρ 0 x L + 4 c 3 0 t 1 + u x L + ρ x L + u x L + ρ x L 2 + 1 + u x L + ρ x L exp c 0 τ 1 + u x L + ρ x L d η + 1 + u x L + ρ x L m exp c m 0 τ 1 + u x L + ρ x L d η d τ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equaq_HTML.gif
      It follows from the contraction mapping theorem that there exists a constant T > 0 such that the equation
      f t = u 0 x L + ρ 0 x L + 4 c m 3 0 t 1 + f τ + f 2 τ + 1 + f τ exp c 0 τ 1 + f η d η + 1 + f τ m exp c m 0 τ 1 + f η d η d τ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equ17_HTML.gif
      (3.11)

      has a unique solution f(t) ∈ C [0,T]. Theorem II in Section I.1 in [26] shows that u x L + ρ x L f t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq42_HTML.gif for any t ∈ [0,T] which leads to the conclusion of this theorem.

      Let u = u ε , ρ = ρ ε , with (3.4) used
      u ε H q + ρ ε H q - 1 c ε s - r - 4 4 exp c 0 t 1 + u x L + ρ x L d τ c exp c 0 t 1 + f τ d τ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equar_HTML.gif
      where s 1 , 3 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq43_HTML.gif, r 1 2 , 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq44_HTML.gif.
      u ε t H r + ρ ε t H r - 1 = u t H r + ρ t H r - 1 c 1 + f t exp c 0 t 1 + f τ d τ + c 1 + f t m exp c m 0 t 1 + f τ d τ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equas_HTML.gif

      where q ∈ (0, s], r ∈ (0, s-1], t ∈ [0,T].

      Then, it follows from Aubin's compactness theorem [27] that there exist subsequences of {u ε }, {ρ ε } denoted by u ε n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq45_HTML.gif, ρ ε n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq46_HTML.gif such that u ε n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq45_HTML.gif, ρ ε n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq46_HTML.gif are weakly convergent to u(t, x)∈ L2([0,T]; H s ), ρ(t, x)∈ L2([0,T]; Hs-1), respectively, and u ε n t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq47_HTML.gif, ρ ε n t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq48_HTML.gif are weakly convergent to u t (t, x)∈ L2([0,T]; Hs-1), ρ t (t, x)∈ L2([0,T]; Hs-2), respectively. Because u ε n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq45_HTML.gif are weakly convergent to u(t, x)∈ L2([0,T]; H s ), f u ε n - u 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq49_HTML.gif for any f∈ (L2([0,T]; H s ))* = L2([0,T]; H s ) when n → ∞. Applying Riesz lemma, we conclude that there exists f u ε n - u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq50_HTML.gif such that
      f u ε n - u u ε n - u = u ε n - u , u ε n - u = u ε n - u 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equat_HTML.gif
      Since f u ε n - u u ε n - u 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq51_HTML.gif as n → ∞, we have u ε n - u 2 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq52_HTML.gif. Then for any real R > 0, u ε n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq45_HTML.gif converges strongly to uL2([0,T]; H q (-R, R)) for any q ∈ [0, s-1); and u ε n t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq47_HTML.gif converges to u t strongly in L2([0,T]; H r (-R, R)) for any r ∈ [0, s-1]. Similarly, g ρ ε n - ρ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq53_HTML.gif for any g∈ (L2([0,T]; H s ))* = L2([0,T]; H s ) as n → ∞. By Riesz lemma, we conclude that there exists g u ε n - u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq54_HTML.gif such that
      g u ε n - u u ε n - u = u ε n - u , u ε n - u = u ε n - u 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equau_HTML.gif

      Since g u ε n - u u ε n - u 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq55_HTML.gif as n → ∞, we have u ε n - u 2 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq56_HTML.gif. Then for any real R > 0, u ε n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq45_HTML.gif converges strongly to ρL2 ([0,T]; Hq-1(-R, R)) for any q ∈ [0, s-1), and ρ ε n t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq48_HTML.gif converge to u t , ρ t strongly in L2([0,T]; Hr-1(-R, R)) for any r ∈ [0, s-1]. Hence, the existence of a weak solution to the Cauchy problem (1.2) and (1.3) is established.

      Theorem 3.4: Let u0(x) ∈H s (R) s 1 , 3 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq57_HTML.gif and ρ0(x) ∈Hs-1(R) s - 1 1 , 3 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq58_HTML.gif, which satisfy u 0 x L < http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq59_HTML.gif, ρ 0 x L < http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq60_HTML.gif. Then there exists a constant T > 0 such that the Cauchy problem (1.2) and (1.3) with the initial data has a solution
      u t , x , ρ t , x L 2 0 , T ; H s × L 2 0 , T ; H s - 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equav_HTML.gif

      in the sense of distribution. And u x , ρ x L([0,T] × R).

      Proof . It follows from Theorem 3.3 that u ε n x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq61_HTML.gif, ρ ε n x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq62_HTML.gif are bounded in the space L. Hence, the sequences u ε n 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq63_HTML.gif, ρ ε n 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq64_HTML.gif, u ε n x 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq65_HTML.gif, ρ ε n x 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq66_HTML.gif, u ε n ρ ε n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq67_HTML.gif, u ε n x ρ ε n x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq68_HTML.gif, u ε n x ρ ε n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq69_HTML.gif, u ε n ρ ε n x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq70_HTML.gif are also weakly convergent to u2, ρ2, u x 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq71_HTML.gif, ρ x 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq72_HTML.gif, uρ, u x ρ x , u x ρ, uρ x L2([0,T]; H r (-R, R)) for any r ∈ [0, s-1] and R > 0, respectively. Therefore, u, ρ satisfy
      0 T R u f t - f x x t d x d t = - 0 T R 3 m + 1 u m + 1 + 1 2 u x 2 + 1 2 ρ 2 f x + u u x f x x d x d t , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equaw_HTML.gif
      and
      0 T R ρ f t d x d t = - 0 T R u ρ f d x d t , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_Equax_HTML.gif

      with u(0,x) = u0(x), ρ(0,x) = ρ0(x), and any f C 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq73_HTML.gif. Moreover, since X = L1([0,T] × R) is a separable Banach space and u ε n x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq61_HTML.gif, ρ ε n x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq62_HTML.gif are bounded sequences in the dual space X* = L([0,T] × R) of X, there are two subsequences of u ε n x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq61_HTML.gif, ρ ε n x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq62_HTML.gif (still denoted by u ε n x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq61_HTML.gif, ρ ε n x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq62_HTML.gif) weak star convergent to two functions U, PL([0,T] × R), respectively. Because u ε n x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq61_HTML.gif, ρ ε n x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-52/MediaObjects/13661_2011_Article_136_IEq62_HTML.gif are also weakly convergent to u x , ρ x L([0,T] × R), respectively. It follows that u x = U, ρ x = P hold almost everywhere. Hence, u x , ρ x L([0,T] × R).

      Declarations

      Acknowledgements

      The study was supported by the National Nature Science Foundation of China (No. 11171135, 71073072), the Nature Science Foundation of Jiangsu (No. BK 2010329), the Project of Excellent Discipline Construction of Jiangsu Province of China, and the Natural Science Foundation of the Jiangsu Higher Education Institutions of China (No. 09KJB110003).

      Authors’ Affiliations

      (1)
      Nonlinear Scientific Research Center, Faculty of Science, Jiangsu University

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      Copyright

      © Tian and Zhu; licensee Springer. 2012

      This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://​creativecommons.​org/​licenses/​by/​2.​0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.