YU Hao-yang CHONG Ge-zi

(1- School of Mathematics, Northwest University, Xi’an 710127;2- Center for Nonlinear Studies, Northwest University, Xi’an 710069)

Abstract: The long time behavior of solutions is one of important problems in the study of partial differential equations. To a great degree, the properties of solutions will depend on those of initial values. The persistence properties imply that the solutions of the equations decay at infinity when the initial data satisfies the condition of decaying at infinity. In this paper,we consider the persistence properties of solutions to the initial value problem for a new generalized two-component Camassa-Holmtype system by using weight functions. Furthermore, we give an optimal decaying estimate.

Keywords: generalized Camassa-Holm-type system; optimal decaying estimate; persistence properties

1 Introduction

In this paper, we consider with the persistence properties of the initial value problem associated with a new generalized two-component Camassa-Holm-type equations[1]

wherem=u?uxx, n=v ?vxxandp, q ∈N, a, b ∈R. As shown in[1], this system in(1)has waltzing peakons and wave braking. It is easy to see that ifa=b, p=q, u=v,the system in (1) contains a lot of famous shallow water wave equations, namely, the Camassa-Holm equation (the system in (1) withu=v, p=q= 0, a=b= 2), the Degasperis-Procesi equation (the system in (1) withu=v, p=q= 0, a=b= 3),and the Novikov equation (the system in (1) withu=v, p=q= 1, a=b= 3).On the other hand, one can see that the system in (1) includes the two-component cross-coupled Camassa-Holm equations and Novikov system.

Takingp=q= 0, a=b= 2 in the system in (1), we find the cross-coupled Camassa-Holm equations

Forp=q= 1, a=b= 3, the system in (1) becomes the two-component Novikov equations

System (2) admits waltzing peakon solutions, but it is not in the two-component Camassa-Holm equations integrable hierarchy[2]. The global existence, blow up phenomenon and persistence properties of the system were discussed in [3,4]. The local well-posedness for the system (3) was established in a range of Besov spaces[5]and in the critical Besov spaces. The results of the persistence properties for the Camassa-Holm equation,the Degasperis-Procesi equation and the Novikov equation can be found in [6–8].

In this paper,we consider the persistence properties of solutions to the initial value problem (1). The solutions of this problem (1) decay at infinity in the spatial variable provided that the initial data satisfies the condition of decaying at infinity. Because all spaces of functions are over R, for simplicity, we drop R in our notation of function space if there is no ambiguity. Before presenting our main results, let us first present the following well-posedness theorem given in [1].

Firstly, we introduce the following spaces[1]

withT>0, s ∈R and 1≤l, r ≤∞.

Lemma 1[1]Assume that 1≤ l,r ≤+∞ands> max{2+1/l,5/2}, let(u0,v0)×. There exists a timeT> 0, and a unique solution (u,v)(T)×(T) to the problem (1). Moreover, the mapping (u0,v0)(u,v) is continuous from a neighborhood of (u0,v0) in×into

for everys′

Remark 1[1]Thanks toBs2,2=Hs, by takingl=r=2 in Lemma 1 one can get the local well-posedness of the problem (1) in

C([0,T];Hs)∩C1([0,T];Hs?1)×C([0,T];Hs)∩C1([0,T];Hs?1) withs>5/2.

We may now utilize the well-posedness result to study the persistence properties of the problem(1). We have the following results based on the work for the Camassa-Holm equation[7].

Theorem 1Assume thats> 5/2, T> 0, andz= (u,v)∈C([0,T];Hs)×C([0,T];Hs) is a solution of the initial value problem (1). If the initial datez0(x) =(u0,v0) = (u(x,0),v(x,0)) decays at infinity, more precisely, if there exists someθ ∈(0,1) such that as|x|→∞,

Then as|x|→∞, we have

Theorem 1 tells us that the solutions of the problem (1) decay at infinity in the spatial variable when the initial data satisfies the condition of decaying at infinity. It is natural to ask the question of whether the decay is optimal. The next result tells us some information.

Theorem 2Assume that for someT>0 ands>5/2,

Then as|x|→∞,

2 The proof of Theorem 1

Note thatG(x)=e?|x|is the kernel of the operator(1??x2)?1,then(1??x2)?1f=G?f,for allf ∈L2(R).G?m=uandG?n=v. The problem(1)can be reformulated as follows

where

Assume thatz ∈C([0,T];Hs)×C([0,T];Hs) is a solution of the initial value problem(1) withs>5/2. Let

hence by Sobolev imbedding theorem, we have

Multiplying the first equation of the problem (4) byu2k?1withk ∈N and integrating the result in thex-variable, we obtain

(5), (6) and H¨older’s inequality lead us to achieve the following estimates

From (7) and the above estimates, we get

and therefore, by Gronwall’s inequality

For any functionf ∈L1(R)∩L∞(R),

SinceF1,F2∈L1∩L∞andG ∈W1,1, we know that?xG?F1, G?F2∈L1∩L∞. Thus,by taking the limit of (8) ask →∞, we get

Next, differentiating the first equation of the problem (4) in thex-variable produces the equation

Multiplying (10) by (ux)2k?1withk ∈N and integrating the result in thex-variable,we obtain

This leads us to obtain the following estimates

and using integration by parts

we achieve the following differential inequality

by Gronwall’s inequality, (11) implies the following estimate

Since=G ?δ, we can use (9) and pass to the limit in (12) to obtain

We shall now repeat the above argument using the weight

whereN ∈N andθ ∈(0,1). Observe that for allNwe have

Multiplying the first equation of the problem (4) by (uφN)2k?1φNwithk ∈N and integrating the result in thex-variable, we obtain

Multiplying (10) by (uxφN)2k?1φNwithk ∈N and integrating the result in thexvariable, we obtain

Using integration by parts

we get

By adding (13) and (14), we have

A simple calculation shows that forθ ∈(0,1),

For any functionf, g, h ∈L∞, we have

and

Since?2xG ?f=G ?f ?f,

Thus, inserting (16) and (17) into (15) we obtain

whereCis a constant depending onC0,M, andT.

Multiplying the second equation of the problem (4) by (vφN)2k?1φNwithk ∈N and integrating the result in thex-variable, then differentiating the second equation of the problem (4) with respect to the spacial variablex, multiplying by (vxφN)2k?1φNand integrating the result in thex-variable, using the similar steps above, we get

Adding (18) and (19) up, we have

Hence, for anyN ∈N and anyt ∈[0,T], by Gronwall’s inequality, we get

which completes the proof of Theorem 1.

3 The proof of Theorem 2

Integrating the first equation of the problem (4) over the time interval [0,t1], we get

In view of the assumption, we get as|x|→∞,

So we obtain

and so

For

we get

Moreover

Thus

Form (23), it follows that

Hence

Thus, we have

Form (21), we get fort1∈[0,T],

Then

Using the similar steps above, we also get

Then as|x|→∞,|v(x,t)|～O(e?x), which completes the proof of Theorem 2.