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 ,

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 23. 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., 789101112131415. 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 ,

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 ,

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 ,

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 1719 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.

Lemma 2.1: Given z 0 = u 0 ρ 0 ∈ H s × H s - 1 , s > 5/2, then there exists a maximal T = T z 0 H s × H s - 1 > 0 , and a unique solution z = u ρ 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 .

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

is continuous.

The proof is similar with Theorem 4.1 in 21.

Let ρ ¯ = ρ - 1 , 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 .

Consider the following initial value problem,

q t = u t , q , t ∈ 0 , T , q 0 , x = x , x ∈ R .

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 × H^s-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 .

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 .

This completes the proof of this lemma.

Theorem 2.1: Let z 0 = u 0 ρ ¯ 0 ∈ H s × H s - 1 , (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 x₀ ∈ R such that ρ ¯ 0 x 0 = - 1 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 .

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 ¯ 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 .

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 .

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 ¯ .

Note that γ 0 = ρ ¯ 0 x 0 + 1 = 0 , 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 .

Since Λ - 2 1 2 u x 2 + 1 2 γ 2 ≥ 0 , and - Λ - 2 3 m + 1 u m + 1 - 1 2 u 2 ≤ 3 m + 1 u m + 1 + 1 2 u 2 , n ′ t ≤ - 1 2 n 2 + u 2 + 3 m + 1 u m + 1 .

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 . 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 , then we have n ′ t ≤ - 1 2 n 2 t + K 2 .

Since n 0 < - 2 K , we obtain that n t < - 2 t , ∀ t ∈ 0 , T .

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 .

Since 0 < n 0 + 2 K n 0 - 2 K < 1 , there exists 0 < T < 1 2 K ln n 0 + 2 K n 0 - 2 K , such that lim t ↑ T n t = - ∞ , i.e., lim t ↑ T u x t = - ∞ .

This completes the proof of the theorem.

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 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 ,

for all ψ ∈ C c ∞ 0 , T × R × C c ∞ 0 , T × R , where ψ ∈ C c ∞ 0 , T × R × C c ∞ 0 , T × R , 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 ,

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 .

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

and

D 2 = 1 + ε ∂ x 2 - 1 : H s → H s + 2

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 .

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

Theorem 3.1: For each initial data u₀ ∈ H^s (s ≥ 1), ρ₀ ∈ H^s-1(s ≥ 2), there exists a T > 0 depending only on the norm of u 0 H s , ρ 0 H s - 1 , and m such that there exists a unique solution (u, ρ) ∈ C([0,T];H^s) × C([0, T];H^s-1) of system (3.1) and (3.2) in the sense of distribution. If u₀ ∈ H^s (s ≥ 2), ρ₀ ∈ H^s-1(s ≥ 3), the solution (u, ρ) ∈ C([0,∞];H^s) × C([0, ∞];H^s-1) exists for all time, in particular, when u₀ ∈ H^s (s ≥ 4), ρ₀ ∈ H^s-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

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 u₀ ∈ H^s (s ≥ 4); ρ₀ ∈ H^s-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 .

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 .

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 .

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 .

Proof. It is obvious that (3.3) holds. In order to prove (5.4), let Λ = 1 - ∂ x 2 1 / 2 . 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 ,

For any q ∈ (1, s] (s ≥ 5), applying (Λ^q u)Λ^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 ,

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 ,

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 ,

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 .

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 .

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 .

For any q ∈ (1, s-1] (s ≥ 4), applying (Λ^q u_t)Λ^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 ,

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 ,

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 ,

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 .

For any q ∈ (1, s-2] (s ≥ 5), applying (Λ^q-1 ρ_t)Λ^q-1 to 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 .

This complete the proof of the theorem.

Suppose u₀ ∈ H^s (s ≥ 1), ρ₀ ∈ H^s-1(s ≥ 2), and let u_ε0, ρ_ε0 be the convolution u_ε0 = φ_ε*u₀, ρ_ε0 = φ_ε*ρ₀, where φ ε x = ε - 1 4 ⋅ φ ^ ε - 1 4 x such that the Fourier transform φ ^ of φ satisfies φ ^ ∈ C 0 ∞ , φ ^ ξ ≥ 0 , and φ ^ ξ = 1 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 ,

u 0 , x = u ε 0 x , t ≥ 0 , x ∈ R ρ 0 , x = ρ ε 0 x , t ≥ 0 , x ∈ R

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 u_ε0, ρ_ε0 in 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 .

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 .

for any ε with 0 < ε < 1 4 , 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) ∈ H^s-1(R),

s-1 ∈ [1, 3/2] such that u 0 x L ∞ < ∞ , ρ 0 x L ∞ < ∞ . Let u_ε0 = φ_ε*u₀, ρ_ε0 = 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 , ρ ε x L ∞ ≤ c 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 , 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 ,

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 ,

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 .

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 .