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# Local well-posedness and persistence properties for a model containing both Camassa-Holm and Novikov equation

Shouming Zhou1*, Chun Wu1 and Baoshuai Zhang2

Author Affiliations

1 College of Mathematics Science, Chongqing Normal University, Chongqing, 401331, P.R. China

2 School of Economics Management, Chongqing Normal University, Chongqing, 401331, P.R. China

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Boundary Value Problems 2014, 2014:9  doi:10.1186/1687-2770-2014-9

The electronic version of this article is the complete one and can be found online at: http://www.boundaryvalueproblems.com/content/2014/1/9

 Received: 25 June 2013 Accepted: 10 December 2013 Published: 8 January 2014

© 2014 Zhou et al.; licensee Springer.

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.

### Abstract

This paper deals with the Cauchy problem for a generalized Camassa-Holm equation with high-order nonlinearities,

u t u x x t + k u x + a u m u x = ( n + 2 ) u n u x u x x + u n + 1 u x x x ,

where k , a R and m , n Z + . This equation is a generalization of the famous equation of Camassa-Holm and the Novikov equation. The local well-posedness of strong solutions for this equation in Sobolev space H s ( R ) with s > 3 2 is obtained, and persistence properties of the strong solutions are studied. Furthermore, under appropriate hypotheses, the existence of its weak solutions in low order Sobolev space H s ( R ) with 1 < s 3 2 is established.

##### Keywords:
persistence properties; local well-posedness; weak solution

### 1 Introduction

This work is concerned with the following one-dimensional nonlinear dispersive PDE:

{ u t u x x t + k u x + a u m u x = ( n + 2 ) u n u x u x x + u n + 1 u x x x , t > 0 , x R , u ( x , 0 ) = u 0 ( x ) , x R , (1.1)

where k , a R and m , n Z + .

Obviously, if n = 0 , m = 1 , a = 3 , equation (1.1) becomes the Camassa-Holm equation,

u t u x x t + k u x + 3 u u x 2 u x u x x u u x x x = 0 , (1.2)

where the variable u ( t , x ) represents the fluid velocity at time t and in the spatial direction x, and k is a nonnegative parameter related to the critical shallow water speed [1]. The Camassa-Holm equation (1.2) is also a model for the propagation of axially symmetric waves in hyperelastic rods (cf.[2]). It is well known that equation (1.2) has also a bi-Hamiltonian structure [3,4] and is completely integrable (see [5,6] and the in-depth discussion in [7,8]). In [9], Qiao has shown that the Camassa-Holm spectral problem yields two different integrable hierarchies of nonlinear evolution equations, one is of negative order CH hierachy while the other one is of positive order CH hierarchy. Its solitary waves are smooth if k > 0 and peaked in the limiting case c 0 = 0 (cf.[1]). The orbital stability of the peaked solitons is proved in [10], and the stability of the smooth solitons is considered in [11]. It is worth pointing out that solutions of this type are not mere abstractions: the peakons replicate a feature that is characteristic for the waves of great height - waves of largest amplitude that are exact solutions of the governing equations for irrotational water waves (cf.[12-14]). The explicit interaction of the peaked solitons is given in [15] and all possible explicit single soliton solutions are shown in [16]. The Cauchy problem for the Camassa-Holm equation (1.2) has been studied extensively. It has been shown that this problem is locally well-posed for initial data u 0 H s ( R ) with s > 3 2 [17-19]. Moreover, it has global strong solutions and also admits finite time blow-up solutions [17,18,20,21]. On the other hand, it also has global weak solutions in H 1 ( R ) [22-25]. The advantage of the Camassa-Holm equation in comparison with the KdV equation (1.2) lies in the fact that the Camassa-Holm equation has peaked solitons and models the peculiar wave breaking phenomena [1,21].

For n = 0 , m Z + , a R , equation (1.1) becomes a generalized Camassa-Holm equation,

u t u x x t + k u x + a u m u x = 2 u x u x x + u u x x x , (1.3)

Wazwaz [26,27] studied the solitary wave solutions for the generalized Camassa-Holm equation (1.3) with m = 2 , a = 3 , and the peakon wave solutions for this equation were studied in [28-30], and the periodic blow-up solutions and limit forms for (1.3) were obtained in [31]. In [30,32], the authors have given the traveling waves solution, peaked solitary wave solutions for (1.3).

On the other hand, taking m = 1 , a = 4 , k = 0 in (1.1) we found the Novikov equation [33]:

u t u x x t + 4 u 2 u x = 3 u u x u x x + u 2 u x x x , t > 0 , x R , (1.4)

The Novikov equation (1.4) possesses a matrix Lax pair, many conserved densities, a bi-Hamiltonian structure as well as peakon solutions [34]. These apparently exotic waves replicate a feature that is characteristic of the waves of great height-waves of largest amplitude that are exact solutions of the governing equations for water waves, as far as the details are concerned [13,35,36]. The Novikov equation possesses the explicit formulas for multipeakon solutions [37]. It has been shown that the Cauchy problem for the Novikov equation is locally well-posed in the Besov spaces and in Sobolev spaces and possesses the persistence properties [38,39]. In [40,41], the authors showed that the data-to-solution map for equation (1.4) is not uniformly continuous on bounded subsets of H s for s > 3 / 2 . Analogous to the Camassa-Holm equation, the Novikov equation shows the blow-up phenomenon [42] and has global weak solutions [43]. Recently, Zhao and Zhou [44] discussed the symbolic analysis and exact traveling wave solutions of a modified Novikov equation, which is new in that it has a nonlinear term u 4 u x instead of u 2 u x .

Other integrable CH-type equations with cubic nonlinearity have been discovered:

{ m t + ( u 2 u x 2 ) m x + 2 u x m 2 + γ u x = 0 , m = u u x x , t > 0 , x R , u ( 0 , x ) = u 0 ( x ) , x R , (1.5)

where γ is a constant. equation (1.5) was independently proposed by Fokas [45], by Fuchssteiner [46], and Olver and Rosenau [47] as a new generalization of integrable system by using the general method of tri-Hamiltonian duality to the bi-Hamiltonian representation of the modified Korteweg-de Vries equation. Later, it was obtained by Qiao [48,49] from the two-dimensional Euler equations, where the variables u ( t , x ) and m ( t , x ) represent, respectively, the velocity of the fluid and its potential density. Ivanov and Lyons [50] obtain a class of soliton solutions of the integrable hierarchy which has been put forward in a series of woks by Qiao [48,49]. It was shown that equation (1.5) admits the Lax-pair and the Cauchy problem (1.5) may be solved by the inverse scattering transform method. The formation of singularities and the existence of peaked traveling-wave solutions for equation (1.5) was investigated in [51]. The well-posedness, blow-up mechanism, and persistence properties are given in [52]. It was also found that equation (1.5) is related to the short-pulse equation derived by Schäfer and Wayne [53].

Applying the method of pseudoparabolic regularization, Lai and Wu [54] investigated the local well-posedness and existence of weak solutions for the following generalized Camassa-Holm equation with dissipative term:

u t u x x t + 2 k u x + a u m u x = ( n 2 u n 1 u x 2 + u n u x x ) x + β x [ ( u x ) 2 N 1 ] , (1.6)

where m , n , N Z + , and a, k, β are constants. Hakkaev and Kirchev [55] studied the local well-posedness and orbital stability of solitary wave solution for equation (1.6) with a = ( m + 2 ) ( m + 1 ) 2 , n = m and β = 0 .

Motivated by the results mentioned above, the goal of this paper is to establish the well-posedness of strong solutions and weak solutions for problem (1.1). First, we use Kato’s theorem to obtain the existence and uniqueness of strong solutions for equation (1.1).

Theorem 1.1Let u 0 H s ( R ) with s > 3 / 2 . Then there exists a maximal T = T ( u 0 H s ( R ) ) , and a unique solution u ( x , t ) to the problem (1.1) such that

u = u ( , u 0 ) C ( [ 0 , T ) ; H s ( R ) ) C 1 ( [ 0 , T ) ; H s 1 ( R ) ) .

Moreover, the solution depends continuously on the initial data, i.e. the mapping

u 0 u ( , u 0 ) : H s ( R ) C ( [ 0 , T ) ; H s ( R ) ) C 1 ( [ 0 , T ) ; H s 1 ( R ) )

is continuous.

In [38,56,57], the spatial decay rates for the strong solution to the Camassa-Holm Novikov equation were established provided that the corresponding initial datum decays at infinity. This kind of property is so-called the persistence property. Similarly, for equation (1.1), we also have the following persistence properties for the strong solution.

Theorem 1.2Assume that u 0 C ( [ 0 , T ) ; H s ( R ) ) with s > 3 / 2 satisfies

| u 0 ( x ) | , | u 0 x ( x ) | O ( e θ x ) as  x ( respectively , | u 0 ( x ) | , | u 0 x ( x ) | O ( ( 1 + x ) α )  as  x )

for some θ ( 0 , 1 ) (respectively, α max { 1 m + 1 , 1 n } ), then the corresponding strong solution u C ( [ 0 , T ) ; H s ( R ) ) to equation (1.1) satisfies for some T > 0

| u ( x , t ) | , | u x ( x , t ) | O ( e θ x ) as  x ( respectively , | u ( x ) | O ( ( 1 + x ) α )  as  x )

uniformly in the time interval [ 0 , T ] .

Theorem 1.3Assume that k = 0 , m = n , and u 0 C ( [ 0 , T ) ; H s ( R ) ) with s > 3 / 2 satisfies

| u 0 ( x ) | O ( e x ) , | u 0 x ( x ) | O ( e θ x ) as  x ( respectively , | u 0 ( x ) | O ( ( 1 + x ) α ) , | u 0 x ( x ) | O ( ( 1 + x ) β )  as  x )

for some θ ( 1 / ( m + 1 ) , 1 ) (respectively, α 1 m + 1 , β ( α m + 1 , α ) ), then the corresponding strong solution u C ( [ 0 , T ) ; H s ( R ) ) to equation (1.1) satisfies for some T > 0

| u ( x , t ) | O ( e x ) as  x ( respectively , | u ( x ) | O ( ( 1 + x ) α )  as  x )

uniformly in the time interval [ 0 , T ] .

Remark 1.1 The notations mean that

| f ( x ) | O ( e θ x ) as  x  if  lim x f ( x ) e θ x = L .

Finally, we have the following theorem for the existence of a weak solution for equation (1.1).

Theorem 1.4Suppose that u 0 ( x ) H s ( R ) with 1 < s 3 2 and u 0 x L ( R ) < . Then there exists a life span T > 0 such that problem (1.1) has a weak solution u ( x , t ) L 2 ( [ 0 , T ] , H s ( R ) ) in the sense of a distribution and u x L ( [ 0 , T ] × R ) .

The plan of this paper is as follows. In the next section, the local well-posedness and persistence properties of strong solutions for the problem (1.1) are established, and Theorems 1.1-1.3 are proved. The existence of weak solutions for the problem (1.1) is proved in Section 3, and this proves Theorem 1.4.

### 2 Well-posedness and persistence properties of strong solutions

Notation The space of all infinitely differentiable functions f ( x , t ) with compact support in R × [ 0 , + ) is denoted by C 0 . Let p be any constant with 1 p < and denote L p = L p ( R ) to be the space of all measurable functions f such that f L p p = R | f ( x ) | p d x < . The space L = L ( R ) with the standard norm f L = inf m ( e ) = 0 sup x R / e | f ( x ) | . For any real number s, let H s = H s ( R ) denote the Sobolev space with the norm defined by

f H s = ( R ( 1 + | ξ | 2 ) s | f ˆ ( ξ , t ) | 2 d ξ ) 1 2 < ,

where f ˆ ( ξ , t ) = R e i x ξ f ( x , t ) d x . Let C ( [ 0 , T ] ; H s ( R ) ) denote the class of continuous functions from [ 0 , T ] to H s ( R ) and Λ = ( 1 x 2 ) 1 2 .

Proof of Theorem 1.1 To prove well-posedness we apply Kato’s semigroup approach [58]. For this, we rewrite the Cauchy problem of equation (1.1) as follows for the transport equation:

{ u t + u m + 1 u x + F ( u ) = 0 , u ( x , 0 ) = u 0 ( x ) , (2.1)

where F ( u ) : = P E ( u ) . E ( u ) = 2 k u x + a u m u x u n + 1 u x + 2 n + 1 2 x ( u n u x 2 ) + n 2 u n 1 u x 3 and P ( x ) = 1 2 e | x | . Let A ( u ) : = u m + 1 x , Y = H s , X = H s 1 and Q = Λ = ( 1 x 2 ) 1 2 . Following closely the considerations made in [17,54,59], we obtain the statement of Theorem 1.1. □

Proof of Theorem 1.2 We introduce the notation M = sup t [ 0 , T ] u ( t ) H s . The first step we will give estimates on u ( x , t ) L . Integrating the both sides with respect to x variable by multiplying the first equation of (2.1) by u 2 p 1 with p Z + , we get

R u 2 p 1 u t d x + R u 2 p 1 ( u n + 1 u x ) d x + R u 2 p 1 ( P E ( u ) ) d x = 0 . (2.2)

Note that the estimates

R u 2 p 1 u t d x = 1 2 p d d t u ( x , t ) L 2 p 2 p = u ( x , t ) L 2 p 2 p 1 d d t u ( x , t ) L 2 p ,

and

| R u 2 p 1 ( u n + 1 u x ) d x | u x ( x , t ) L u ( x , t ) L 2 p 2 p + n

are true. Moreover, using Hölder’s inequality

| R u 2 p 1 ( P E ( u ) ) d x | u ( x , t ) L 2 p 2 p 1 P E ( u ) L 2 p .

From equation (2.2) we can obtain

d d t u ( x , t ) L 2 p u x ( x , t ) L u ( x , t ) L 2 p n + 1 + P E ( u ) L 2 p .

Since f L p f L as p for any f L L 2 . From the above inequality we deduce that

d d t u ( x , t ) L M n + 1 u ( x , t ) L + P E ( u ) L ,

where we use

u x ( x , t ) L u ( x , t ) L n u x ( x , t ) H 1 2 + u ( x , t ) H 1 2 + n u ( x , t ) H s n + 1 M n + 1 .

Because of Gronwall’s inequality, we get

u ( x , t ) L exp ( M n + 1 t ) ( u 0 ( x ) L + 0 t ( P E ( u ) ) ( x , τ ) L d τ ) .

Next, we will give estimates on u x ( x , t ) L . Differentiating (2.1) with respect to the x-variable produces the equation

u x t + u n + 1 u x x + ( n + 1 ) u n u x 2 + x ( P E ( u ) ) = 0 . (2.3)

Multiplying this equation by ( u x ) 2 p 1 with p Z + , integrating the result in the x-variable, and using integration by parts:

R ( u x ) 2 p 1 u x t d x = 1 2 p d d t u x ( x , t ) L 2 p 2 p = u x ( x , t ) L 2 p 2 p 1 d d t u x ( x , t ) L 2 p , | R ( u x ) 2 p 1 ( u n u x 2 ) d x | u ( x , t ) L n u x ( x , t ) L u x ( x , t ) L 2 p 2 p , | R ( u x ) 2 p 1 ( u n + 1 u x x ) d x | = | n + 1 2 p R u n u x 2 p + 1 d x | n + 1 2 p u ( x , t ) L n u x ( x , t ) L u x ( x , t ) L 2 p 2 p .

From the above inequalities, we also can get the following inequality:

d d t u x ( x , t ) L 2 p ( n + 1 + n + 1 2 p ) M n + 1 u x ( x , t ) L 2 p + x ( P E ( u ) ) L 2 p ,

where we use u x ( x , t ) L u ( t ) L n M n + 1 . Then passing to the limit in this inequality and using Gronwall’s inequality one can obtain

u x ( x , t ) L exp ( ( n + 1 ) M n + 1 t ) ( u 0 x ( x ) L + 0 t x ( P E ( u ) ) ( x , τ ) L d τ ) .

We shall now repeat the arguments using the weight

φ N ( x ) = { 1 , x 0 , e θ x , 0 < x < N , e θ x , x N ,

where N Z . Observe that for all N we have

0 φ N ( x ) φ N ( x ) , for all  x R . (2.4)

Using the notation E ( u ) , from (2.1) we get

t ( u φ N ) + ( u n + 1 φ N ) u x + φ N ( P E ( u ) ) = 0 ,

and from (2.3), we also obtain

t ( φ N x u ) + u n + 1 φ N x 2 u + ( n + 1 ) u n ( φ N x u ) x u + φ N x ( P E ( u ) ) = 0 .

We need to eliminate the second derivatives in the second term in the above equality. Thus, combining integration by parts and equation (2.4) we find

| R u n + 1 φ N x 2 u ( x u φ N ) 2 p 1 | = | R u n + 1 ( x u φ N ) 2 p 1 ( x ( φ N x u ) x u φ N ) d x | = | R 1 2 p u n + 1 x ( ( x u φ N ) 2 p ) u n + 1 ( x u φ N ) 2 p 1 x u φ N d x | ( u L + x u L ) u L n x u φ N L 2 p 2 p .

Hence, as in the weightless case, we have

u φ N L + x u φ N L exp ( ( n + 1 ) M n + 1 t ) ( u 0 ( x ) φ N L + u 0 x ( x ) φ N L ) + exp ( ( n + 1 ) M n + 1 t ) 0 t ( φ N x ( E ( u ) ) L + φ N ( E ( u ) ) L ) d τ .

A simple calculation shows that there exists C > 0 , depending only on θ ( 0 , 1 ) such that for any N Z + ,

φ N R 1 φ N ( y ) d y C = 4 1 θ .

Thus, we have

| φ N ( 1 x 2 ) 1 ( u n 1 x 3 u ) | = 1 2 | φ N R e | x y | ( u n 1 x 3 u ) ( y ) d y | = 1 2 | φ N R e | x y | 1 φ N ( y ) ( φ N x u ) ( u n 1 x 2 u ) ( y ) d y | 1 2 ( φ N R e | x y | 1 φ N ( y ) d y ) φ N x u L u n 1 x 2 u L c φ N x u L u n 1 x 2 u L ,

and

| φ N ( 1 x 2 ) 1 x ( u n 1 x 3 u ) | = 1 2 | φ N R sgn ( x y ) e | x y | ( u n 1 x 3 u ) ( y ) d y | c φ N x u L u n 1 x 2 u L .

Using the same method, we can estimate the other terms:

| φ N ( 1 x 2 ) 1 ( 2 k u x + a u m u x u n + 1 u x ) | c ( 1 + u L m + u L n + 1 ) φ N x u L , | φ N ( 1 x 2 ) 1 x ( 2 k u x + a u m u x u n + 1 u x ) | = | φ N ( 1 x 2 ) 1 x 2 ( 2 k u + a m + 1 u m + 1 1 n + 2 u n + 2 ) | | φ N ( 2 k u + a m + 1 u m + 1 1 n + 2 u n + 2 ) | + | φ N ( 1 x 2 ) 1 ( 2 k u + a m + 1 u m + 1 1 n + 2 u n + 2 ) | c ( 1 + u L m + u L n + 1 ) φ N u L ,

and

| φ N ( 1 x 2 ) 1 x ( u n u x 2 ) | c φ N u L u n 1 x 2 u L , | φ N ( 1 x 2 ) 1 x 2 ( u n u x 2 ) | | φ N u n u x 2 | + | φ N ( 1 x 2 ) 1 ( u n u x 2 ) | c φ N u L u n 1 x 2 u L .

Thus, it follows that there exists a constant C > 0 which depends only on M, m, n, k, a, and T, such that

u φ N L + x u φ N L C ( u 0 φ N L + u 0 x φ N L ) + C 0 t ( ( 1 + u L m + u L n + 1 + u n 1 x 2 u L ) ( φ N x u L + φ N u L ) ) d τ C ( u 0 φ N L + u 0 x φ N L ) + C 0 t ( φ N x u L + φ N u L ) d τ .

Hence, for any n Z and any t [ 0 , T ] we have

u φ N L + x u φ N L C ( u 0 φ N L + u 0 x φ N L ) C ( u 0 max ( 1 , e θ x ) L + u 0 x max ( 1 , e θ x ) L ) .

Finally, taking the limit as N goes to infinity we find that for any t [ 0 , T ] ,

u e θ x L + x u e θ x L C ( u 0 max ( 1 , e θ x ) L + u 0 x max ( 1 , e θ x ) L ) ,

which completes the proof of Theorem 1.2. □

Next, we give a simple proof for Theorem 1.3.

Proof of Theorem 1.3 We should use Theorem 1.3 to prove this theorem.

For any t 1 [ 0 , T ] , integrating equation (2.1) over the time interval [ 0 , t 1 ] we get

u ( x , t 1 ) u ( x , 0 ) + 0 t 1 ( u n + 1 u x ) ( x , τ ) d τ + 0 t 1 ( P E ( u ) ) ( x , τ ) d τ = 0 . (2.5)

From Theorem 1.2, it follows that

0 t 1 ( u n + 1 u x ) ( x , τ ) d τ O ( e ( n + 2 ) α x ) as  x

and so

0 t 1 ( u n + 1 u x ) ( x , τ ) d τ O ( e x ) as  x .

We shall show that the last term in equation (2.5) is O ( e x ) ; thus we have

0 t 1 ( P E ( u ) ) ( x , τ ) d τ = P ( x ) 0 t 1 ( E ( u ) ) ( x , τ ) d τ P ( x ) ρ ( x ) .

From the given condition and Theorem 1.2. we know ρ ( x ) O ( e x ) as x . Since

P ( x ) ρ ( x ) = 1 2 R e | x y | ρ ( y ) d y = 1 2 e x x e y ρ ( y ) d y + 1 2 e x x e y ρ ( y ) d y

we have

e x x e y ρ ( y ) d y = O ( 1 ) e x x e 2 y d y O ( 1 ) e x O ( e x ) as  x , e x x e y | ρ ( y ) d y = O ( 1 ) e x x e 2 y d y O ( 1 ) e x O ( e x ) as  x .

Thus

0 t 1 ( P E ( u ) ) ( x , τ ) d τ O ( e x ) as  x .

From equation (2.5) and | u 0 ( x ) | O ( e x ) as x , we know

| u ( x , t 1 ) | O ( e x ) as  x .

By the arbitrariness of t 1 [ 0 , T ] , we get

| u ( x , t ) | O ( e x ) as  x

uniformly in the time interval [ 0 , T ] . This completes the proof of Theorem 1.3. □

### 3 Existence of solution of the regularized equation

In order to prove Theorem 1.4, we consider the regularized problem for equation (1.1) in the following form:

{ u t u x x t + ϵ u x x x t = x ( 2 k u a m + 1 u m + 1 ) + 1 n + 2 x 3 ( u n + 2 ) u t u x x t + ϵ u x x x t = 2 n + 1 2 x ( u n u x 2 ) n 2 u n 1 u x 3 , u ( x , 0 ) = u 0 ( x ) , (3.1)

where 0 < ϵ < 1 4 , m 1 , n 1 and a, k are constants. One can easily check that when ϵ = 0 , equation (3.1) is equivalent to the IVP (1.1).

Before giving the proof of Theorem 1.4, we give several lemmas.

Lemma 3.1 (See [54])

Letpandqbe real numbers such that p < q p . Then

f g H q c f H p g H q , if  p > 1 2 , f g H p + q 1 2 c f H p g H q , if  p < 1 2 .

Lemma 3.2Let u 0 ( x ) H s ( R ) with s > 3 / 2 . Then the Cauchy problem (3.1) has a unique solution u ( x , t ) C ( [ 0 , T ] ; H s ( R ) ) where T > 0 depends on u 0 H s ( R ) . If s 2 , the solution u ( x , t ) C ( [ 0 , T ] ; H s ( R ) ) exists for all time. In particular, when s 4 , the corresponding solution is a classical globally defined solution of problem (3.1).

Proof First, we note that, for any 0 < ϵ < 1 4 and any s, the integral operator

D = ( 1 x 2 + ϵ x 4 ) 1 : H s H s + 4

defines a bounded linear operator on the indicated Sobolev spaces.

To prove the existence of a solution to the problem (3.1), we apply the operator to both sides of equation (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 [ x ( 2 k u a m + 1 u m + 1 ) + 1 n + 2 x 3 ( u n + 2 ) 2 n + 1 2 x ( u n u x 2 ) n 2 u n 1 u x 3 ] d τ . (3.2)

Suppose that is the operator in the right-hand side of equation (3.2). For fixed t [ 0 , T ] , we get

0 t D [ x ( 2 k u a m + 1 u m + 1 ) + x 2 ( u n + 1 u x ) 2 n + 1 2 x ( u n u x 2 ) n 2 u n 1 u x 3 ] d τ 0 t D [ x ( 2 k v a m + 1 v m + 1 ) + x 3 ( v n + 1 v x ) 2 n + 1 2 x ( v n v x 2 ) n 2 v n 1 v x 3 ] d τ H s C T ( sup 0 t T u v H s + sup 0 t T u m + 1 v m + 1 H s + sup 0 t T u n + 2 v n + 2 H s + sup 0 t T D x [ u n u x 2 v n v x 2 ] H s + sup 0 t T D [ u n 1 u x 3 v n 1 v x 3 ] H s ) .

Since H s 0 is an algebra for s 0 1 2 , we have the inequalities

u m + 1 v m + 1 H s = ( u v ) j = 0 m ( u m j v j ) H s C 1 ( u v H s j = 0 m u H s m j v H s j ) .

Since s > 3 / 2 , by Lemma 3.1, we get

D x [ u n u x 2 v n v x 2 ] H s = 1 n + 1 D x [ x ( u n + 1 ) x u x ( v n + 1 ) x v ] H s C ( D x [ x ( u n + 1 ) x ( u v ) ] H s + D x [ x ( u n + 1 v n + 1 ) x v ] H s ) C ( x ( u n + 1 ) x ( u v ) H s 2 + x ( u n + 1 v n + 1 ) x v H s 2 ) C ( u n + 1 H s u v H s + u n + 1 v n + 1 H s v H s ) C 2 ( u H s n + 1 u v H s + ( u v H s j = 0 n u H s n j v H s j ) v H s ) ,

and

D [ u n 1 u x 3 v n 1 v x 3 ] H s D [ u n 1 ( u x 3 v x 3 ) ] H s + D [ ( u n 1 v n 1 ) v x 3 ] H s C [ u n 1 ( u x 3 v x 3 ) H s 2 + ( u n 1 v n 1 ) v x 3 H s 2 ] C [ u n 1 H s 1 u x 3 v x 3 H s 2 + u n 1 v n 1 H s 1 v x 3 H s 2 ] C 3 [ u H s n 1 u v H s j = 0 2 u H s 2 j v H s j + ( u v H s j = 0 n 2 u H s n 2 j v H s j ) v H s 3 ] ,

where C 2 , C 3 only depend on n. Suppose that both u and v are in the closed ball B R ( 0 ) of radius R about the zero function in C ( [ 0 , T ] ; H s ( R ) ) ; by the above inequalities, we obtain

A u A v C ( [ 0 , T ] ; H s ) θ u v C ( [ 0 , T ] ; H s ) ,

where θ = T C ( R m + R n + 1 ) and C only depend on a, k, m, n. Choosing T sufficiently small such that θ < 1 , we know that is a contraction. Applying the above inequality yields

A u C ( [ 0 , T ] ; H s ) u 0 H s + θ u C ( [ 0 , T ] ; H s ) .

Taking T sufficiently small so that θ R + u 0 H s < R , we deduce that maps B R ( 0 ) to itself. It follows from the contraction-mapping principle that the mapping has a unique fixed point u in B R ( 0 ) .

For s 2 , multiplying the first equation of the system (3.1) by 2u, integrating with respect to x, one derives

d d t R ( u 2 + u x 2 + ϵ u x x 2 ) d x = R 2 u ( 2 k u x a u m u x + ( n + 2 ) u n u x u x x + u n + 1 u x x x ) d x = R ( 2 ( n + 2 ) u n + 1 u x u x x + 2 u n + 2 u x x x ) d x = 0 ,

from which we have the conservation law

R ( u 2 + u x 2 + ϵ u x x 2 ) d x = R ( u 0 2 + u 0 x 2 + ϵ u 0 x x 2 ) d x . (3.3)

The global existence result follows from the integral from equation (3.2) and equation (3.3). □

Now we study the norms of solutions of equation (3.1) using energy estimates. First, recall the following two lemmas.

Lemma 3.3 (See [58])

If r > 0 , then H r L is an algebra, and

f g H r c ( f L g H r + g L h H r ) ,

herecis a constant depending only onr.

Lemma 3.4 (See [58])

If r > 0 , then

[ Λ r , f ] g L 2 c ( x f L Λ r 1 g L 2 + Λ r f L 2 g L ) ,

where [ A , B ] denotes the commutator of the linear operatorsAandB, andcis a constant depending only onr.

Theorem 3.1Suppose that, for some s 4 , the functions u ( t , x ) are a solution of equation (3.1) corresponding to the initial data u 0 H s ( R ) . Then the following inequality holds:

u H 1 2 c R ( u 2 + u x 2 + ϵ u x x 2 ) d x = c R ( u 0 2 + u 0 x 2 + ϵ u 0 x x 2 ) d x . (3.4)

For any real number q ( 0 , s 1 ] , there exists a constantcdepending only onqsuch that

R ( Λ q + 1 u ) 2 d x R [ ( Λ q + 1 u 0 ) 2 + ε ( Λ q + 1 u 0 x x ) 2 ] d x + c 0 t u H q 2 u L n 2 u x L 3 d τ + c 0 t u x L ( u H q 2 ( u L m 1 + u L n ) + 2 u L n u H q + 1 2 ) d τ . (3.5)

For q [ 0 , s 1 ] , there is a constantcindependent ofϵsuch that

( 1 2 ϵ ) u t H q c u H q + 1 ( 1 + ( u L m 1 + u L n ) u H 1 + u L n 1 u x L 2 ) . (3.6)

Proof Using u H 1 2 c R ( u 2 + u x 2 ) d x and (3.3) derives (3.4).

Since x 2 = Λ 2 + 1 and the Parseval equality gives rise to

R ( Λ q u ) Λ q x 2 f d x = R ( Λ q + 1 u ) Λ q + 1 f d x + R ( Λ q u ) Λ q f d x .

For any q ( 0 , s 1 ] , applying ( Λ q u ) Λ q to both sides of the first equation of (3.1), respectively, and integrating with regard to x again,using integration by parts, one obtains

1 2 d d t R ( ( Λ q u ) 2 + ( Λ q u x ) 2 + ϵ ( Λ q u x x ) 2 ) d x = a R ( Λ q u ) Λ q ( u m u x ) d x R ( Λ q + 1 u ) Λ q + 1 ( u n + 1 u x ) d x + 2 n + 1 2 R ( Λ q u x ) Λ q ( u n u x 2 ) d x + R ( Λ q u ) Λ q ( u n + 1 u x ) d x n 2 R ( Λ q u ) Λ q ( u n 1 u x 3 ) d x . (3.7)

We will estimate the terms on the right-hand side of (3.7) separately. For the first term, by using the Cauchy-Schwartz inequality and Lemmas 3.3 and 3.4, we have

R ( Λ q u ) Λ q ( u m u x ) d x = R ( Λ q u ) [ Λ q ( u m u x ) u m Λ q u x ] d x + R ( Λ q u ) u m Λ q u x d x c u H q ( m u L m 1 u x L u H q + u L m 1 u x L u H q ) + m 2 u L m 1 u x L Λ q u L 2 2 c u L m 1 u x L u H q 2 . (3.8)

Using the above estimate to the second term on the right-hand side of equation (3.7) yields

R ( Λ q + 1 u ) Λ q + 1 ( u n + 1 u x ) d x = c u L n u x L u H q + 1 2 . (3.9)

For the fourth term on the right-hand side of equation (3.7), using the Cauchy-Schwartz inequality and Lemma 3.3, we obtain

R ( Λ q u x ) Λ q ( u n u x 2 ) d x Λ q u x L 2 Λ q ( u n u x 2 ) L 2 c u H q + 1 ( u n u x L u x H q + u x L u n u x H q ) c u L n u x L u H q + 1 2 . (3.10)

For the last term on the right-hand side of equation (3.7), using Lemma 3.3 repeatedly results in

R ( Λ q u ) Λ q ( u n 1 u x 3 ) d x u H q u n 1 u x 3 H q u H q 2 u L n 2 u x L 3 . (3.11)

It follows from equations (3.7)-(3.11) that there exists a constant c depending only on a, m, n, s such that

1 2 d d t R ( ( Λ q u ) 2 + ( Λ q u x ) 2 + ϵ ( Λ q u x x ) 2 ) d x c u x L ( u H q 2 ( u L m 1 + u L n ) + 2 u L n u H q + 1 2 ) + u H q 2 u L n 2 u x L 3 .

Integrating both sides of the above inequality with respect to t results in inequality (3.5).

To estimate the norm of u t , we apply the operator ( 1 x 2 ) 1 to both sides of the first equation of the system (3.1) to obtain the equation

( 1 ϵ ) u t ϵ u x x t = ( 1 x 2 ) 1 [ ϵ u t x ( 2 k u + a m + 1 u m + 1 ) + 1 n + 2 x 3 ( u n + 2 ) 2 n + 1 2 x ( u n u x 2 ) n 2 u n 1 u x 3 ] . (3.12)

Applying ( Λ q u t ) Λ q to both sides of equation (3.12) for q ( 0 , s 1 ] gives rise to

( 1 ϵ ) R ( Λ q u t ) 2 d x + ϵ R ( Λ q u x t ) 2 d x = R ( Λ q u t ) Λ q 2 [ ϵ u t + 1 n + 2 x 3 ( u n + 2 ) x ( 2 k u + a m + 1 u m + 1 ) 2 n + 1 2 x ( u n u x 2 ) n 2 u n 1 u x 3 ] d x . (3.13)

For the right-hand of equation (3.13), we have

R ( Λ q u t ) Λ q 2 ( ϵ u t 2 k u x ) d x ϵ u t H q 2 + 2 k u t H q u H q , (3.14)

and

R ( Λ q u t ) ( 1 x 2 ) 1 Λ q x ( a m + 1 u m + 1 2 n + 1 2 u n u x 2 ) d x c u t H q { R ( 1 + ξ 2 ) q 1 [ R ( a m + 1 u m ˆ ( ξ η ) u ˆ ( η ) 2 n + 1 2 u n u x ˆ ( ξ η ) u x ˆ ( η ) ) d η ] 2 } 1 2 c u t H q u H 1 u H q + 1 ( u L m 1 + u L n ) . (3.15)

Since

R ( Λ q u t ) ( 1 x 2 ) 1 Λ q x 2 ( u n + 1 u x 2 ) d x = R ( Λ q u t ) Λ q ( u n + 1 u x ) d x + R ( Λ q u t ) ( 1 x 2 ) 1 Λ q ( u n + 1 u x 2 ) d x . (3.16)

Using Lemma 3.3, u n u x H q c ( u n + 1 ) x H q c u L n u H q + 1 and u L c u H 1 , we have

R ( Λ q u t ) Λ q ( u n + 1 u x ) d x c u t H q u n + 1 u x H q c u t H q u L n u H q + 1 u H 1 , (3.17)

and

R ( Λ q u t ) ( 1 x 2 ) 1 Λ q ( u n + 1 u x 2 ) d x c u t H q u L n u H q + 1 u H 1 . (3.18)

By the Cauchy-Schwartz inequality and Lemma 3.3, we get

R ( Λ q u t ) ( 1 x 2 ) 1 Λ q ( u n 1 u x 3 ) d x c u t H q u x L 2 u L n 1 u H q + 1 . (3.19)

Substituting equations (3.14)-(3.19) into equation (3.13) yields the inequality

( 1 2 ϵ ) u t H q c u H q + 1 ( 1 + ( u L m 1 + u L n ) u H 1 + u L n 1 u x L 2 ) (3.20)

with a constant c > 0 . This completes the proof of Theorem 3.1. □

For a real number s with s > 0 , suppose that the function u 0 ( x ) is in H s ( R ) , and let u ϵ 0 be the convolution u ϵ 0 = ϕ ϵ u 0 of the function ϕ ϵ ( x ) = ϵ 1 4 ϕ ( ϵ 1 4 x ) and u 0 be such that the Fourier transform ϕ ˆ of ϕ satisfies ϕ ˆ C 0 , ϕ ( ξ ) ˆ 0 and ϕ ( ξ ) ˆ = 1 for any ξ ( 1 , 1 ) . Thus we have u ϵ 0 ( x ) C . It follows from Theorem 3.1 that for each ϵ satisfying 0 < ϵ < 1 4 , the Cauchy problem

{ u t u x x t + ϵ u x x x t = x ( 2 k u a m + 1 u m + 1 ) + 1 n + 2 x 3 ( u n + 2 ) u t u x x t + ϵ u x x x t = 2 n + 1 2 x ( u n u x 2 ) n 2 u n 1 u x 3 , u ( x , 0 ) = u ϵ 0 ( x ) (3.21)

has a unique solution u ϵ C ( [ 0 , T ϵ ) , H ( R ) ) , in which T ϵ may depend on ϵ.

For an arbitrary positive Sobolev exponent s > 0 , we give the following lemma.

Lemma 3.5For u 0 H s ( R ) with s > 0 and u ϵ 0 = ϕ ϵ u 0 , the following estimates hold for anyϵwith 0 < ϵ < 1 4 :

u ϵ 0 x L c u 0 x L , if  q s , (3.22)

u ϵ 0 H q c ϵ s q 4 , if  q > s , (3.23)

u ϵ 0 u 0 H q c ϵ s q 4 , if  q s , (3.24)

u ϵ 0 u 0 H s = o ( 1 ) , (3.25)

wherecis a constant independent ofϵ.

Proof This proof is similar to that of Lemma 5 in [60] and Lemma 4.5 in [61], we omit it here. □

Remark 3.1 For s 1 , using u ϵ L c u ϵ H 1 , u ϵ H 1 2 c R ( u ϵ 2 + u ϵ x 2 ) d x , equations (3.4), (3.22), and (3.23), we obtain

u ϵ L 2 c u ϵ H 1 c R ( u ϵ 0 2 + u ϵ 0 x 2 + u ϵ 0 x x 2 ) d x c ( u ϵ 0 H 1 2 + ϵ u ϵ 0 H 2 2 ) c ( c + c ϵ × ϵ s 2 2 ) c 0 , (3.26)

where c 0 is independent of ϵ.

Theorem 3.2If u 0 ( x ) H s ( R ) with s [ 1 , 3 2 ] such that u 0 x L ( R ) < . Let u ϵ 0 be defined as in the system (3.21). Then there exist two constants, cand T > 0 , which are independent ofϵ, such that u ϵ of problem (3.21) satisfies u ϵ x L ( R ) c for any t [ 0 , T ) .

Proof Using the notation u = u ϵ and differentiating equation (3.21) or equation (3.12) with respect to x give rise to

( 1 ϵ ) u x t ϵ u x x x t 2 n + 1 2 u n u x 2 + 1 n + 2 x 2 u n + 2 = 2 k u + a m + 1 u m + 1 1 n + 2 u n + 2 ( 1 x 2 ) 1 [ ϵ u x t + 2 k u + a m + 1 u m + 1 1 n + 2 u n + 2 + 2 n + 1 2 u n u x 2 + n 2 x ( u n 1 u x 3 ) ] .

Letting p > 0 be an integer and multiplying the above equality by ( u x ) 2 p + 1 , then integrating the resulting equation with respect to x, and using

1 n + 2 R x 2 ( u n + 2 ) ( u x ) 2 p + 1 d x = R ( ( n + 1 ) u n u x 2 + u n + 1 u x x ) ( u x ) 2 p + 1 d x = ( n + 1 ) R u n u x 2 p + 3 d x + 1 2 p + 2 R u n + 1 x ( u x ) 2 p + 2 d x = ( n + 1 ) ( 2 p + 1 ) 2 p + 2 R u n u x 2 p + 3 d x ,

we find the equality

d d t 1 ϵ 2 p + 2 R ( u x ) 2 p + 2 d x ϵ R ( u x ) 2 p + 1 u x x x t d x + p n 2 p + 2 R u n u x 2 p + 3 d x = R ( u x ) 2 p + 1 ( 2 k u + a m + 1 u m + 1 1 n + 2 u n + 2 ) d x R ( u x ) 2 p + 1 ( 1 x 2 ) 1 [ ϵ u x t + 2 k u + a m + 1 u m + 1 1 n + 2 u n + 2 + 2 n + 1 2 u n u x 2 + n 2 x ( u n 1 u x 3 ) ] d x . (3.27)

Applying Hölder’s inequality, we get

1 ϵ 2 p + 2 d d t R ( u x ) 2 p + 2 d x { ϵ ( R | u x x x t | 2 p + 2 d x ) 1 2 p + 2 + 2 | k | ( R | u | 2 p + 2 d x ) 1 2 p + 2 + ( R | G | 2 p + 2 d x ) 1 2 p + 2 + a m + 1 ( R | u m + 1 | 2 p + 2 d x ) 1 2 p + 2 + 1 n + 2 ( R | u n + 2 | 2 p + 2 d x ) 1 2 p + 2 } ( R ( u x ) 2 p + 2 d x ) 2 p + 1 2 p + 2 + | p n | 2 p + 2 u x L u L n R | u x | 2 p + 2 d x ,

where G = ( 1 x 2 ) 1 [ ϵ u x t + 2 k u + a m + 1 u m + 1 1 n + 2 u n + 2 + 2 n + 1 2 u n u x 2 + n 2 x ( u n 1 u x 3 ) ] . Furthermore

1 ϵ 2 p + 2 d d t ( R ( u x ) 2 p + 2 d x ) 1 2 p + 2 ϵ ( R | u x x x t | 2 p + 2 d x ) 1 2 p + 2 + 2 | k | ( R | u | 2 p + 2 d x ) 1 2 p + 2 + a m + 1 ( R | u m + 1 | 2 p + 2 d x ) 1 2 p + 2 + 1 n + 2 ( R | u n + 2 | 2 p + 2 d x ) 1 2 p + 2 + ( R | G | 2 p + 2 d x ) 1 2 p + 2 + | p n | 2 p + 2 u x L u L n ( R ( u x ) 2 p + 2 d x ) 1 2 p + 2 .

Since f L p f L as p for any f L L 2 , integrating the above inequality with respect to t and taking the limit as p result in the estimate

( 1 ϵ ) u x L u 0 x L + 0 t [ ϵ u x x x t L + 1 2 u x L u L n u x L 2 + c ( u L + u m + 1 L + u n + 2 L + G L ) ] d τ . (3.28)

Using the algebraic property of H s ( R ) with s > 1 2 and the inequality (3.26) leads to

u n + 2 L c u n + 2 H 1 2 + c u n + 2 H 1 c u H 1 n + 2