Research ArticleTraveling Wave Solutions of a Generalized Camassa-HolmEquation: A Dynamical System Approach

Lina Zhang and Tao Song

Department of Mathematics, Huzhou University, Huzhou, Zhejiang 313000, China

Correspondence should be addressed to Lina Zhang; zsdzln@126.com

Received 1 August 2015; Accepted 14 September 2015

Academic Editor: Maria Gandarias

Copyright © 2015 L. Zhang and T. Song. This is an open access article distributed under the Creative Commons AttributionLicense, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properlycited.

We investigate a generalized Camassa-Holm equation 𝐶(3, 2, 2): 𝑢𝑡+ 𝑘𝑢𝑥+ 𝛾1𝑢𝑥𝑥𝑡

+ 𝛾2(𝑢3)𝑥+ 𝛾3𝑢𝑥(𝑢2)𝑥𝑥

+ 𝛾3𝑢(𝑢2)𝑥𝑥𝑥

= 0. Weshow that the 𝐶(3, 2, 2) equation can be reduced to a planar polynomial differential system by transformation of variables. We treatthe planar polynomial differential system by the dynamical systems theory and present a phase space analysis of their singularpoints. Two singular straight lines are found in the associated topological vector field. Moreover, the peakon, peakon-like, cuspon,smooth soliton solutions of the generalized Camassa-Holm equation under inhomogeneous boundary condition are obtained.Theparametric conditions of existence of the single peak soliton solutions are given by using the phase portrait analytical technique.Asymptotic analysis and numerical simulations are provided for single peak soliton, kink wave, and kink compacton solutions ofthe 𝐶(3, 2, 2) equation.

1. Introduction

Mathematical modeling of dynamical systems processing in agreat variety of natural phenomena usually leads to nonlinearpartial differential equations (PDEs).There is a special class ofsolutions for nonlinear PDEs that are of considerable interest,namely, the traveling wave solutions. Such a wave may belocalized or periodic, which propagates at constant speedwithout changing its shape.

Many powerful methods have been presented for findingthe traveling wave solutions, such as the Bäcklund trans-formation [1], tanh-coth method [2], bilinear method [3],symbolic computation method [4], and Lie group analysismethod [5]. Furthermore, a great amount of works focusedon various extensions and applications of the methods inorder to simplify the calculation procedure. The basic ideaof those methods is that, by introducing different types ofAnsatz, the original PDEs can be transformed into a set ofalgebraic equations. Balancing the same order of the Ansatzthen yields explicit expressions for the PDE waves. However,not all of the special forms for the PDE waves can be derivedby those methods. In order to obtain all possible forms ofthe PDEwaves and analyze qualitative behaviors of solutions,

the bifurcation theory plays a very important role in studyingthe evolution of wave patterns with variation of parameters[6–9].

To study the traveling wave solutions of a nonlinear PDE

Φ(𝑢, 𝑢𝑡, 𝑢𝑥, 𝑢𝑥𝑥, 𝑢𝑥𝑡, 𝑢𝑡𝑡, . . .) = 0, (1)

let 𝜉 = 𝑥 − 𝑐𝑡 and 𝑢(𝑥, 𝑡) = 𝜑(𝜉), where 𝑐 is the wave speed.Substituting them into (1) leads the PDE to the followingordinary differential equation:

Φ1(𝜑, 𝜑, 𝜑, . . .) = 0. (2)

Here, we consider the case of (2) which can be reduced to thefollowing planar dynamical system:

𝑑𝜑

𝑑𝜉= 𝜑= 𝑦,

𝑑𝑦

𝑑𝜉= 𝐹 (𝜑, 𝑦) ,

(3)

through integrals. Equation (3) is called the traveling wavesystem of the nonlinear PDE (1). So, we just study the

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2015, Article ID 610979, 19 pageshttp://dx.doi.org/10.1155/2015/610979

2 Mathematical Problems in Engineering

traveling wave system (3) to get the traveling wave solutionsof the nonlinear PDE (1).

Let us begin with some well-known nonlinear waveequations. The first one is the Camassa-Holm (CH) equation[10]

𝑢𝑡− 𝑢𝑡𝑥𝑥

+ 3𝑢𝑢𝑥= 2𝑢𝑥𝑢𝑥𝑥

+ 𝑢𝑢𝑥𝑥𝑥

(4)

arising as a model for nonlinear waves in cylindrical axiallysymmetric hyperelastic rods, with 𝑢(𝑥, 𝑡) representing theradial stretch relative to a prestressed state where Camassaand Holm showed that (4) has a peakon of the form 𝑢(𝑥, 𝑡) =𝑐𝑒−|𝑥−𝑐𝑡|. Among the nonanalytic entities, the peakon, a

soliton with a finite discontinuity in gradient at its crest, isperhaps the weakest nonanalyticity observable by the eye [11].

To understand the role of nonlinear dispersion in theformation of patters in liquid drop, Rosenau and Hyman [12]introduced and studied a family of fully nonlinear dispersionKorteweg-de Vries equations

𝑢𝑡+ (𝑢𝑚)𝑥+ (𝑢𝑛)𝑥𝑥𝑥

= 0. (5)

This equation, denoted by 𝐾(𝑚, 𝑛), owns the property that,for certain𝑚 and 𝑛, its solitary wave solutions have compactsupport [12]. That is, they identically vanish outside a finitecore region. For instance, the 𝐾(2, 2) equation admits thefollowing compacton solution:

𝑢 (𝑥, 𝑡) =

{{

{{

{

4𝑐

3cos2 (𝑥 − 𝑐𝑡

4) , |𝑥 − 𝑐𝑡| ≤ 2𝜋,

0, otherwise.(6)

The Camassa-Holm equation, the 𝐾(2, 2) equation, andalmost all integrable dispersive equations have the sameclass of traveling wave systems which can be written in thefollowing form [13]:

𝑑𝜑

𝑑𝜉= 𝑦 =

1

𝐷2 (𝜑)

𝜕𝐻

𝜕𝑦,

𝑑𝑦

𝑑𝜉= −

1

𝐷2 (𝜑)

𝜕𝐻

𝜕𝜑= −

𝐷(𝜑) 𝑦2+ 𝑔 (𝜑)

𝐷2 (𝜑),

(7)

where 𝐻 = 𝐻(𝜑, 𝑦) = (1/2)𝑦2𝐷2(𝜑) + ∫𝐷(𝜑)𝑔(𝜑)𝑑𝜑 is thefirst integral. It is easy to see that (4) is actually a special case of(3) with 𝐹(𝜑, 𝑦) = −(1/𝐷2(𝜑))(𝜕𝐻/𝜕𝜑). If there is a function𝜑 = 𝜑

𝑠such that 𝐷(𝜑

𝑠) = 0, then 𝜑 = 𝜑

𝑠is a vertical straight

line solution of the system

𝑑𝜑

𝑑𝜁= 𝑦𝐷 (𝜑) ,

𝑑𝑦

𝑑𝜁= −𝐷(𝜑) 𝑦2− 𝑔 (𝜑) ,

(8)

where 𝑑𝜉 = 𝐷(𝜑)𝑑𝜁 for 𝜑 ̸= 𝜑𝑠. The two systems have

the same topological phase portraits except for the verticalstraight line 𝜑 = 𝜑

𝑠and the directions in time. Consequently,

we can obtain bifurcation and smooth solutions of thenonlinear PDE (1) through studying the system (8), if the

corresponding orbits are bounded and do not intersect withthe vertical straight line 𝜑 = 𝜑

𝑠. However, the orbits,

which do intersect with the vertical straight line 𝜑 = 𝜑𝑠or

are unbounded but can approach the vertical straight line,correspond to the non-smooth singular traveling waves. Itis worth of pointing out that traveling waves sometimes losetheir smoothness during the propagation due to the existenceof singular curves within the solution surfaces of the waveequation.

Most of these works are concentrated on the nonlinearwave equationswith only a singular straight line [6–9]. But tillnow there have been few works on the integrable nonlinearequations with two singular straight lines or other types ofsingular curves [13–15].

In 2004, Tian and Yin [16] introduced the following fullynonlinear generalized Camassa-Holm equation 𝐶(𝑚, 𝑛, 𝑝):

𝑢𝑡+ 𝑘𝑢𝑥+ 𝛾1𝑢𝑥𝑥𝑡

+ 𝛾2(𝑢𝑚)𝑥+ 𝛾3𝑢𝑥(𝑢𝑛)𝑥𝑥

+ 𝛾4𝑢 (𝑢𝑝)𝑥𝑥𝑥

= 0,

(9)

where 𝑘, 𝛾1, 𝛾2, 𝛾3, and 𝛾

4are arbitrary real constants and 𝑚,

𝑛, and 𝑝 are positive integers. By using four direct ansatzs,they obtained kink compacton solutions, nonsymmetry com-pacton solutions, and solitary wave solutions for the𝐶(2, 1, 1)and 𝐶(3, 2, 2) equations.

Generally, it is not an easy task to obtain a uniformanalytic first integral of the corresponding traveling wavesystem of (9). In this paper, we consider the cases 𝑚 = 3,𝑛 = 𝑝 = 2, and 𝛾

3= 𝛾4. Then, (9) reduces to the 𝐶(3, 2, 2)

equation

𝑢𝑡+ 𝑘𝑢𝑥+ 𝛾1𝑢𝑥𝑥𝑡

+ 𝛾2(𝑢3)𝑥+ 𝛾3𝑢𝑥(𝑢2)𝑥𝑥

+ 𝛾3𝑢 (𝑢2)𝑥𝑥𝑥

= 0.

(10)

Actually, we have already considered a special 𝐶(3, 2, 2)equation in [17], namely, 𝛾

1= −1, 𝛾

2= −3, and 𝛾

3= −1,

where the bifurcation of peakons are obtained by applyingthe qualitative theory of dynamical systems. In this work,a more general 𝐶(3, 2, 2) equation (10) is studied. Differentbifurcation curves are derived to divide the parameter spaceinto different regions associated with different types of phasetrajectories. Meanwhile, it is interesting to point out that thecorresponding traveling wave system of (10) has two singularstraight lines compared with (4), which therefore gives rise toa variety of nonanalytic traveling wave solutions, for instance,peakons, cuspons, compactons, kinks, and kink-compactons.

This paper is organized as follows. In Section 2,we analyzethe bifurcation sets and phase portraits of correspondingtraveling wave system. In Section 3, we classify single peaksoliton solutions of (10) and give the parametric representa-tions of the smooth soliton solutions, peakon-like solutions,cuspon solutions, and peakon solutions. In Section 4, weobtain the kink wave and kink compacton solutions of (10).A short conclusion is given in Section 5.

Mathematical Problems in Engineering 3

2. Bifurcation Sets and Phase Portraits

In this section, we shall study all possible bifurcations andphase portraits of the vector fields defined by (10) in theparameter space. To achieve such a goal, let 𝑢(𝑥, 𝑡) = 𝜑(𝜉)with 𝜉 = 𝑥 − 𝑐𝑡 be the solution of (10), then it follows that

(𝑘 − 𝑐) 𝜑− 𝛾1𝑐𝜑

+ 𝛾2(𝜑3)

+ 𝛾3𝜑(𝜑2)

+ 𝛾3𝜑 (𝜑2)

= 0,

(11)

where 𝜑 = 𝜑𝜉, 𝜑 = 𝜑

𝜉𝜉, and 𝜑 = 𝜑

𝜉𝜉𝜉. Integrating (11) once

and setting the integration constant as 𝑔, we have

(𝑘 − 𝑐) 𝜑 − 𝛾1𝑐𝜑+ 𝛾2𝜑3+ 𝛾3𝜑 (𝜑2)

= −𝑔. (12)

Clearly, (12) is equivalent to the planar system

𝑑𝜑

𝑑𝜉= 𝑦,

𝑑𝑦

𝑑𝜉=𝛽𝜑3+ 𝜑𝑦2+ 𝜖𝜑 + 𝜎

𝜌 − 𝜑2,

(13)

where 𝜌 = 𝑐𝛾1/2𝛾3, 𝛽 = 𝛾

2/2𝛾3, 𝜎 = 𝑔/2𝛾

3, and 𝜖 = (𝑘−𝑐)/2𝛾

3

(𝛾3

̸= 0). System (13) has the first integral

𝐻(𝜑, 𝑦) =𝛽

4𝜑4+1

2𝜖𝜑2+ 𝜎𝜑 +

1

2𝜑2𝑦2−1

2𝜌𝑦2= ℎ. (14)

Obviously, for 𝜌 > 0, system (13) is a singular travelingwave system [14]. Such a system may possess complicateddynamical behavior and thus generate many new travelingwave solutions. Hence, we assume 𝜌 > 0 in the rest of thispaper (𝜌 = 2, > 0). The phase portraits defined by thevector fields of system (13) determine all possible travelingwave solutions of (10). However, it is not convenient todirectly investigate (13) since there exist two singular straightlines 𝜑 = and 𝜑 = − on the right-hand side of the secondequation of (13). To avoid the singular lines temporarily, wedefine a new independent variable 𝜁 by setting (𝑑𝜉/𝑑𝜁) =2−𝜑2; then, system (13) is changed to a Hamiltonian system,

written as𝑑𝜑

𝑑𝜁= (2− 𝜑2) 𝑦,

𝑑𝑦

𝑑𝜁= 𝜑𝑦2+ 𝛽𝜑3+ 𝜖𝜑 + 𝜎.

(15)

System (15) has the same topological phase portraits as system(13) except for the singular lines 𝜑 = and 𝜑 = −.

We now investigate the bifurcation of phase portraits ofthe system (15). Denote that

𝑓 (𝜑) = 𝛽𝜑3+ 𝜖𝜑 + 𝜎. (16)

Let 𝑀(𝜑𝑒, 𝑦𝑒) be the coefficient matrix of the linearized

system of (15) at the equilibrium point (𝜑𝑒, 𝑦𝑒); then,

𝑀(𝜑𝑒, 𝑦𝑒) = (

−2𝜑𝑒𝑦𝑒𝑦2

𝜖+ 𝑓(𝜑𝑒)

2− 𝜑2

𝑒2𝜑𝑒𝑦𝑒

) , (17)

and at this equilibrium point, we have

𝐽 (𝜑𝑒, 𝑦𝑒) = det𝑀(𝜑

𝑒, 𝑦𝑒)

= (𝜑2

𝑒− 2) 𝑓(𝜑𝑒) − (3𝜑

2

𝑒+ 2) 𝑦2

𝑒.

(18)

By the theory of planar dynamical systems, for an equilibriumpoint of a Hamiltonian system, if 𝐽 < 0, then it is a saddlepoint, a center point if 𝐽 > 0, and a degenerate equilibriumpoint if 𝐽 = 0.

From the above analysis, we can obtain the bifurcationcurves and phase portraits under different parameter condi-tions.

Let

𝛽1(𝜎) = −

4𝜖3

27𝜎2,

𝛽2(𝜎) =

𝜎 − 𝜖

3,

𝛽3(𝜎) =

−𝜎 − 𝜖

3.

(19)

Clearly, for 𝛽 > 𝛽1(𝜎), the function 𝑓(𝜑) = 0 has three

real roots 𝜑1, 𝜑2, and 𝜑

3(𝜑1

> 𝜑2

> 𝜑3); that is, system

(15) has three equilibrium points 𝐸𝑖(𝜑𝑖, 0), 𝑖 = 1, 2, 3 on

the 𝜑-axis. When 𝛽 < 𝛽3(𝜎), (15) has two equilibrium

points 𝑆1,2(, ±𝑌

1) on the straight line 𝜑 = √𝑎, where 𝑌

1=

√(−𝜎 − (𝜖 + 𝛽2))/. When 𝛽 < 𝛽2(𝜎), system (15) has two

equilibrium points 𝑆3,4(−, ±𝑌

2) on the straight line 𝜑 = −,

where 𝑌2= √(𝜎 − (𝜖 + 𝛽2))/. Notice that on making the

transformation 𝜎 → −𝜎, 𝜑 → −𝜑, 𝜁 → −𝜁, system (15)is invariant. This means that, for 𝜎 < 0, the phase portraitsof (15) are just the reflections of the corresponding phaseportraits of (15) in the case 𝜎 > 0 with respect to the 𝑦-axis.Thus, we only need to consider the case 𝜎 ≥ 0. To knowthe dynamical behavior of the orbits of system (15), we willdiscuss two cases: 𝜖 > 0 and 𝜖 < 0.

2.1. Case I: 𝜖 > 0

Lemma 1. Suppose that 𝜖 > 0. Denote ℎ+= 𝐻(, ±𝑌

1), ℎ−=

𝐻(−, ±𝑌2), and ℎ

𝑖= 𝐻(𝜑

𝑖, 0), 𝑖 = 1, 2, 3.

(1) For 𝜎 > 0 and 𝛽1(𝜎) < 𝛽 < 𝛽

3(𝜎), there exists one

and only one curve 𝛽 = 𝛽4(𝜎) on the (𝜎, 𝛽)-plane on which

ℎ+

= ℎ2; for 𝜎 > 0 and 𝛽

1(𝜎) < 𝛽 < 𝛽

2(𝜎), there exists one

and only one curve 𝛽 = 𝛽5(𝜎) on the (𝜎, 𝛽)-plane on which

ℎ−

= ℎ2or ℎ−= ℎ3.Moreover, the curves𝛽 = 𝛽

1(𝜎),𝛽 = 𝛽

2(𝜎),

𝛽 = 𝛽4(𝜎), and 𝛽 = 𝛽

5(𝜎) are tangent at the point (𝜎

1𝑠, 𝛽1𝑠).

The curve 𝛽 = 𝛽3(𝜎) intersects with the curves 𝛽 = 𝛽

1(𝜎),

𝛽 = 𝛽4(𝜎), and 𝛽 = 𝛽

5(𝜎) at the points (𝜎

2𝑠, 𝛽2𝑠), (𝜎3𝑠, 𝛽3𝑠),

and (𝜎4𝑠, 𝛽4𝑠) (0 < 𝜎

4𝑠< 𝜎3𝑠< 𝜎2𝑠< 𝜎1𝑠), respectively.

(2) For 𝜎 = 0, there exists a bifurcation point 𝐴(0, −2𝜖/2)on (𝜎, 𝛽)-plane on which ℎ+

= ℎ−

= ℎ2when 𝛽 = −2𝜖/2,

ℎ+

= ℎ−

< ℎ2when 𝛽 < −2𝜖/2, and ℎ+

= ℎ−

> ℎ2when

−2𝜖/2< 𝛽 < −𝜖/

2.

4 Mathematical Problems in Engineering

𝛽2(𝜎)

(𝜎1s, 𝛽1s)

𝛽3(𝜎)

𝛽4(𝜎)𝛽5(𝜎)

𝛽

−2

−2

−4

−4

−6

−8

−10

−12

−14

𝛽1(𝜎)

𝜎2 4

Figure 1: The bifurcation curves in the (𝜎, 𝛽)-parameter plane for𝜖 > 0.

According to the above analysis and Lemma 1, we obtainthe following proposition on the bifurcation curves of thephase portraits of system (15) for 𝜖 > 0.

Proposition 2. When 𝜖 > 0, for system (15), in (𝜎, 𝛽)-parameter plane, there exist five bifurcation curves (see Fig-ure 1):

𝛽 = 𝛽1(𝜎) = −

4𝜖3

27𝜎2,

𝛽 = 𝛽2(𝜎) =

𝜎 − 𝜖

3,

𝛽 = 𝛽3(𝜎) =

−𝜎 − 𝜖

3,

𝛽 = 𝛽4(𝜎) (|𝜎| ≤ 𝜎

3𝑠) ,

𝛽 = 𝛽5(𝜎) .

(20)

These five curves divide the right-half (𝜎, 𝛽)-parameter planeinto thirty-one regions as follows:

𝐴1: {(𝜎, 𝛽) | 𝜎1s < 𝜎 < 𝜖, 𝛽 = 𝛽2(𝜎)},

𝐴2: {(𝜎, 𝛽) | 𝜎 > 𝜎1s, 𝛽5(𝜎) < 𝛽 < 𝛽2(𝜎)},

𝐴3: {(𝜎, b) | 𝜎 > 𝜎1s, 𝛽 = 𝛽5(𝜎)},

𝐴4: {(𝜎, 𝛽) | 𝜎 > 𝜎1s, 𝛽1(𝜎) < 𝛽 < 𝛽5(𝜎)},

𝐴5: {(𝜎, 𝛽) | 𝜎 > 𝜎1s, 𝛽 = 𝛽1(𝜎)},

𝐴6: {(𝜎, 𝛽) | 𝜎 > 𝜎2s, 𝛽3(𝜎) < 𝛽 < 𝛽1(𝜎)},

𝐴7: {(𝜎, 𝛽) | 𝜎 < 𝜎2s, 𝛽 = 𝛽3(𝜎)},

𝐴8: {(𝜎, 𝛽) | 𝜎 > 0, 𝛽 < max{𝛽1(𝜎), 𝛽3(𝜎)}},

𝐴9: (𝜎, 𝛽) = (𝜎2s, 𝛽2s),

𝐴10: {(𝜎, 𝛽) | 0 < 𝜎 < 𝜎2s, 𝛽 = 𝛽1(𝜎)},

𝐴11: {(𝜎, 𝛽) | 𝜎 < 𝜎2s, 𝛽1(𝜎) < 𝛽 < min{𝛽3(𝜎),

𝛽4(𝜎)}},

𝐴12: {(𝜎, 𝛽) | 𝜎 < 𝜎3s, 𝛽 = 𝛽4(𝜎)},

𝐴13: {(𝜎, 𝛽) | 𝜎 > 0, 𝛽4(𝜎) < 𝛽 < 𝛽5(𝜎)},

𝐴14: {(𝜎, 𝛽) | 𝜎 < 𝜎4s, 𝛽 = 𝛽5(𝜎)},

𝐴15: {(𝜎, 𝛽) | 𝜎 > 0, 𝛽5(𝜎) < 𝛽 < 𝛽3(𝜎)},

𝐴16: {(𝜎, 𝛽) | 𝜎 = 0, 𝛽 < −2𝜖/2},

𝐴17: (𝜎, 𝛽) = (0, −2𝜖/2),

𝐴18: {(𝜎, 𝛽) | 𝜎 = 0, −2𝜖/2 < 𝛽 < −𝜖/2},

𝐴19: (𝜎, 𝛽) = (0, −𝜖/2),

𝐴20: {(𝜎, 𝛽) | 0 < 𝜎 < 𝜎4s, 𝛽 = 𝛽3(𝜎)},

𝐴21: (𝜎, 𝛽) = (𝜎4s, 𝛽4s),

𝐴22: {(𝜎, 𝛽) | 𝜎4s < 𝜎 < 𝜎2s, 𝛽 = 𝛽3(𝜎)},

𝐴23: {(𝜎, 𝛽) | 𝜎4s < 𝜎 < 𝜎1s, max{𝛽1(𝜎), 𝛽3(𝜎)} <

𝛽 < 𝛽5(𝜎)},𝐴24: {(𝜎, 𝛽) | 𝜎

s4 < 𝜎 < 𝜎1s, 𝛽 = 𝛽5(𝜎)},

𝐴25: {(𝜎, 𝛽) | 𝜎4s < 𝜎 < 𝜎1s, max{𝛽3(𝜎), 𝛽5(𝜎)} <

𝛽 < 𝛽2(𝜎)},𝐴26: {(𝜎, 𝛽) | 0 < 𝜎 < 𝜎1s, 𝛽 = 𝛽2(d)},

𝐴27: (𝜎, 𝛽) = (𝜎1s, 𝛽1s),

𝐴28: {(𝜎, 𝛽) | 0 ≤ 𝜎 < 𝜖, 𝛽 > 𝛽2(𝜎)},

𝐴29: {(𝜎, 𝛽) | 𝜎 ∈ R, 𝛽 > max{0, 𝛽2(𝜎)}},

𝐴30: {(𝜎, 𝛽) | 𝜎 > 𝜖, 𝛽 = 𝛽2(𝜎)},

𝐴31: {(𝜎, 𝛽) | 𝜎 > 𝜖, 0 < 𝛽 < 𝛽2(𝜎)}.

In this case, the phase portraits of system (15) can beshown in Figure 2.

2.2. Case II: 𝜖 < 0. In this case, we have the following.

Proposition 3. When 𝜖 < 0, for system (15), in (𝜎, 𝛽)-parameter plane, there exist four parametric bifurcation curves(see Figure 3):

𝛽 = 𝛽1(𝜎) = −

4𝜖3

27𝜎2,

𝛽 = 𝛽2(𝜎) =

𝜎 − 𝜖

3,

𝛽 = 𝛽3(𝜎) =

−𝜎 − 𝜖

3, 𝜎 = 0.

(21)

These four curves divide the right-half (𝜎, 𝛽)-parameter planeinto twenty-two regions:

𝐵1: {(𝜎, 𝛽) | 𝜎1s < 𝜎 < 𝜖, 𝛽 = 𝛽3(𝜎)},

𝐵2: {(𝜎, 𝛽) | 𝜎 > 𝜎1s, max{0, 𝛽3(𝜎)} < 𝛽 < 𝛽1(𝜎)},

𝐵3: {(𝜎, 𝛽) | 𝜎 > 𝜎1s, 𝛽 = 𝛽1(𝜎)},

𝐵4: {(𝜎, 𝛽) | 𝜎 > 𝜎2s, 𝛽1(𝜎) < 𝛽 < 𝛽2(𝜎)},

𝐵5: (𝜎, 𝛽) = (𝜎2s, 𝛽2s),

𝐵6: {(𝜎, 𝛽) | 𝜎 > 𝜎2s, 𝛽 = 𝛽2(𝜎)},

𝐵7: {(𝜎, 𝛽) | 𝜎 > 0, 𝛽 > max{𝛽1(𝜎), 𝛽2(𝜎)}},

𝐵8: {(𝜎, 𝛽) | 0 < 𝜎 < 𝜎2s, 𝛽 = 𝛽1(𝜎)},

𝐵9: {(𝜎, 𝛽) | 0 < 𝜎 < 𝜎2s, 𝛽2(𝜎) < 𝛽 < 𝛽1(𝜎)},

𝐵10: {(𝜎, 𝛽) | 𝜎 = 0, 𝛽 > −𝜖/2},

Mathematical Problems in Engineering 5

0 1 2 3

0

2

4

0 2 4

0

2

4

0 2 4

0

2

4

0 2 4

0

2

4

0 2 4

0

2

4

0 1

0

2

4

0.0 0.5 1.0

0

2

4

0.0 0.5 1.0

0

2

4

0.0 0.5 1.0

0

1

2

3

0.0 0.5 1.0

0

2

4

6

0.0 0.5 1.0

0

2

4

0.0 0.5 1.0

0

2

4

−3

−3

−2 −2 −2−1−4

−4

−2 −2

−2

−1

−1

−4−2−4

−4 −4

−2 −2 −2

−4

−2

−4

−2

−4

−2

−1.5 −1.0 −0.5

−1.0 −0.5 −1.0 −0.5 −1.0 −0.5

−1.5 −1.0 −0.5 −1.5 −1.0 −0.5

−6

−2

−4

−2

−4

−2

−4

−2

−4

−2

−4

𝜑𝜑𝜑

𝜑 𝜑 𝜑

𝜑𝜑𝜑

𝜑 𝜑 𝜑

yy

y

y yy

yyy

y

y y

(1) (𝜎, 𝛽) ∈ A1

(4) (𝜎, 𝛽) ∈ A4

(7) (𝜎, 𝛽) ∈ A7 (8) (𝜎, 𝛽) ∈ A8 (9) (𝜎, 𝛽) ∈ A9

(5) (𝜎, 𝛽) ∈ A5

(2) (𝜎, 𝛽) ∈ A2 (3) (𝜎, 𝛽) ∈ A3

(6) (𝜎, 𝛽) ∈ A6

(10) (𝜎, 𝛽) ∈ A10 (11) (𝜎, 𝛽) ∈ A11 (12) (𝜎, 𝛽) ∈ A12

Figure 2: Continued.

6 Mathematical Problems in Engineering

0.0 0.5 1.0

0

1

2

3

0.0 0.5 1.0

0

2

4

0.0 0.5 1.0

0

1

2

3

0.0 0.5 1.0

0

1

2

3

0.0 0.5 1.0

0

1

2

0.0 0.5 1.0

0

1

2

0.0 0.5 1.0

0

1

2

0.0 0.5 1.0

0

1

2

0.0 0.5 1.0

0

1

2

3

0.0 0.5 1.0

0

1

2

3

0.0 0.5 1.0

0

1

2

3

0.0 0.5 1.0

0

1

2

3

−3

−2

−1

−3

−2

−1

−2

−1

−2

−1

−2

−3

−1

−2

−3

−1

−2

−3

−1

−2

−3

−1

−2

−1

−2

−1

−3

−2

−1

−1.0 −0.5

−1.0 −0.5

−1.0 −0.5

−1.0 −0.5

−1.0 −0.5

−1.0−1.5 −0.5 −1.0−1.5 −0.5

−1.0 −0.5

−1.0 −0.5 −1.0 −0.5

−1.0 −0.5 −1.0 −0.5

−2

−4

y

𝜑𝜑𝜑

𝜑 𝜑 𝜑

𝜑𝜑𝜑

𝜑 𝜑𝜑

yy

y yy

yy

y

yy

y

(13) (𝜎, 𝛽) ∈ A13

(16) (𝜎, 𝛽) ∈ A16

(19) (𝜎, 𝛽) ∈ A19 (20) (𝜎, 𝛽) ∈ A20

(23) (𝜎, 𝛽) ∈ A23 (24) (𝜎, 𝛽) ∈ A24

(21) (𝜎, 𝛽) ∈ A21

(22) (𝜎, 𝛽) ∈ A22

(17) (𝜎, 𝛽) ∈ A17 (18) (𝜎, 𝛽) ∈ A18

(14) (𝜎, 𝛽) ∈ A14 (15) (𝜎, 𝛽) ∈ A15

Figure 2: Continued.

Mathematical Problems in Engineering 7

0.0 0.5 1.0 1.5

0

1

2

3

0.0 0.5 1.0 1.5

0

1

2

3

0 1 2

0

1

2

3

0 1 2

0

1

2

3

0.0 0.5 1.0 1.5

0

1

2

3

0 1 2

0

2

4

0 1 2

0

2

4

−2

−3

−1

−2

−3

−1

−2

−3

−2

−4

−1

−2

−3

−1

−2

−3

−1

−2 −1

−2−3 −1

−2

−2

−3

−4

−1

−2 −1

−1.0−1.5 −0.5 −1.0−1.5 −0.5

−1.0−1.5 −0.5

y

yyy

y y y

𝜑

𝜑𝜑𝜑

𝜑 𝜑 𝜑

(25) (𝜎, 𝛽) ∈ A25

(28) (𝜎, 𝛽) ∈ A28 (29) (𝜎, 𝛽) ∈ A29

(31) (𝜎, 𝛽) ∈ A31

(30) (𝜎, 𝛽) ∈ A30

(26) (𝜎, 𝛽) ∈ A26 (27) (𝜎, 𝛽) ∈ A27

Figure 2: The bifurcation of phase portraits of system (15) when 𝜖 > 0 and 𝜎 ≥ 0.

𝐵11: (𝜎, 𝛽) = (0, −𝜖/2),

𝐵12: {(𝜎, 𝛽) | 0 < 𝜎 < 𝜎2s, 𝛽 = 𝛽2(𝜎)},

𝐵13: {(𝜎, 𝛽) | 0 < 𝜎 < 𝜎1s, 𝛽3(𝜎) < 𝛽 < min{𝛽1(𝜎),

𝛽2(𝜎)}},

𝐵14: {(𝜎, 𝛽) | 𝜎4s < 𝜎 < 𝜎1s, 𝛽 = 𝛽1(𝜎)},

𝐵15: (𝜎, 𝛽) = (𝜎1s, 𝛽1s),

𝐵16: {(𝜎, 𝛽) | 0 < 𝜎 < 𝜎1s, 𝛽 = 𝛽3(𝜎)},

𝐵17: {(𝜎, 𝛽) | 0 < 𝜎 < 𝜖, 𝛽 < 𝛽3(𝜎)},

𝐵18: {(𝜎, 𝛽) | 𝜎 = 0, 𝛽 = −𝜖/2},

𝐵19: {(𝜎, 𝛽) | 𝜎 = 0, 𝛽 < 0},

𝐵20: {(𝜎, 𝛽) | 0 < 𝜎 < 𝜖, 𝛽 < min{0, 𝛽3(𝜎)}},

𝐵21: {(𝜎, 𝛽) | 𝜎 > 𝜖, 𝛽 = 𝛽3(𝜎)},

𝐵22: {(𝜎, 𝛽) | 𝜎 > 𝜖, 𝛽3(𝜎) < 𝛽 < 0}.

Based on Proposition 3, we obtain the phase portraits ofsystem (15) which are shown in Figure 4.

8 Mathematical Problems in Engineering

2 4 6

5

10

15

𝛽2(𝜎)

(𝜎1s, 𝛽1s)

(𝜎2s, 𝛽2s)

𝛽3(𝜎)

𝛽

−2−4−6

𝛽1(𝜎)

𝜎

Figure 3: The bifurcation curves in the (𝜎, 𝛽)-parameter plane for𝜖 < 0.

3. Single Peak Soliton Solutions

In this section, we study classification of single peak solitonsolutions of (10) by using the phase portraits given inSection 2. Let𝐶𝑘(Ω) denote the set of all 𝑘 times continuouslydifferentiable functions on the open setΩ. 𝐿𝑝loc(R) refer to theset of all functions whose restriction on any compact subsetis 𝐿𝑝 integrable.𝐻1loc(R) stands for𝐻

1

loc(R) = {𝜑 ∈ 𝐿2

loc(R) |

𝜑∈ 𝐿2

loc(R)}.To study single peak soliton solutions, we impose the

boundary condition

lim𝜉→±∞

𝜑 = 𝐴, (22)

where 𝐴 is a constant. In fact, the constant 𝐴 is equal to thehorizontal coordinate of saddle point 𝐸(𝜑

𝑒, 0). Substituting

the boundary condition (22) into (14) generates the followingconstant:

ℎ = 𝐴(𝜎 +𝐴𝜖

2+𝐴3𝛽

4) . (23)

So the ODE (14) becomes

(𝜑)2

=

(𝜑 − 𝐴)2

(𝛽𝜑2+ 2𝐴𝛽𝜑 + 3𝐴

2𝛽 + 2𝜖)

2 (2 − 𝜑2). (24)

If 𝛽(𝐴2𝛽 + 𝜖) ≤ 0, then (24) reduces to

(𝜑)2

=𝛽 (𝜑 − 𝐴)

2

(𝜑 − 𝐵1) (𝜑 − 𝐵

2)

2 (2 − 𝜑2), (25)

where

𝐵1= −𝐴 −

√−2𝛽 (𝐴2𝛽 + 𝜖)

𝛽,

𝐵2= −𝐴 +

√−2𝛽 (𝐴2𝛽 + 𝜖)

𝛽.

(26)

From (26) we know that 𝐵1> 𝐵2if 𝛽 < 0 and 𝐵

1< 𝐵2if

𝛽 > 0.

Definition 4. A function𝜑(𝜉) is said to be a single peak solitonsolution for the 𝐶(3, 2, 2) equation (10) if 𝜑(𝜉) satisfies thefollowing conditions:

(C1) 𝜑(𝜉) is continuous on R and has a unique peakpoint 𝜉

0, where 𝜑(𝜉) attains its global maximum or

minimum value.

(C2) 𝜑(𝜉) ∈ 𝐶3(R − {𝜉

0}) satisfies (24) on R − {𝜉

0}.

(C3) 𝜑(𝜉) satisfies the boundary condition (22).

Definition 5. A wave function 𝜑 is called smooth solitonsolution if 𝜑 is smooth locally on either side of 𝜉

0and

lim𝜉↑𝜉0

𝜑(𝜉) = lim

𝜉↓𝜉0𝜑(𝜉) = 0.

Definition 6. A wave function 𝜑 is called peakon if 𝜑 issmooth locally on either side of 𝜉

0and lim

𝜉↑𝜉0𝜑(𝜉) =

−lim𝜉↓𝜉0

𝜑(𝜉) = 𝑎, 𝑎 ̸= 0, 𝑎 ̸= ±∞.

Definition 7. Awave function𝜑 is called cuspon if𝜑 is smoothlocally on either side of 𝜉

0and lim

𝜉↑𝜉0𝜑(𝜉) = −lim

𝜉↓𝜉0𝜑(𝜉) =

±∞.

Without any loss of generality, we choose the peak point𝜉0as vanishing, 𝜉

0= 0.

Theorem 8. Assume that 𝑢(𝑥, 𝑡) = 𝜑(𝜉) = 𝜑(𝑥 − 𝑐𝑡) is a singlepeak soliton solution of the 𝐶(3, 2, 2) equation (10) at the peakpoint 𝜉

0= 0. Then, we have the following:

(i) If 𝛽(𝐴2𝛽 + 𝜖) > 0, then 𝜑(0) = or 𝜑(0) = −.

(ii) If 𝛽(𝐴2𝛽 + 𝜖) ≤ 0, then 𝜑(0) = or 𝜑(0) = − or𝜑(0) = 𝐵

1or 𝜑(0) = 𝐵

2.

Proof. If 𝜑(0) ̸= ±, then 𝜑(𝜉) ̸= ± for any 𝜉 ∈ R since𝜑(𝜉) ∈ 𝐶

3(R − {0}). Differentiating both sides of (24) yields

𝜑 ∈ 𝐶∞(R).

(i) When 𝛽(𝐴2𝛽 + 𝜖) > 0, if 𝜑(0) ̸= and 𝜑(0) ̸= −, then𝜑 ∈ 𝐶

∞(R). By the definition of single peak soliton we have

𝜑(0) = 0. However, by (24) we must have 𝜑(0) = 𝐴, which

contradicts the fact that 0 is the unique peak point.(ii) When 𝛽(𝐴2𝛽 + 𝜖) ≤ 0, if 𝜑(0) ̸= and 𝜑(0) ̸= −,

by (24) we know 𝜑(0) exists and 𝜑(0) = 0 since 0 is a peakpoint. Thus, we obtain 𝜑(0) = 𝐵

1or 𝜑(0) = 𝐵

2from (25),

since 𝜑(0) = 𝐴 contradicts the fact that 0 is the unique peakpoint.

Now we give the following theorem on the classificationof single peak solitons of (10).The idea is inspired by the studyof the traveling waves of Camassa-Holm equation [18, 19].

Theorem 9. Assume that 𝑢(𝑥, 𝑡) = 𝜑(𝑥 − 𝑐𝑡) is a single peaksoliton solution of the 𝐶(3, 2, 2) equation (10) at the peak point𝜉0= 0. Then, we have the following solution classification:(i) If |𝜑(0)| ̸= , then 𝜑(𝜉) ∈ 𝐶∞(R), and 𝜑 is a smooth

soliton solution.

Mathematical Problems in Engineering 9

0 2

0

2

4

0 2

0

2

4

0 1 2

0

2

4

0 1 2

0

2

4

0.0 0.5 1.0 1.5

0

2

4

0 1

0

2

4

6

0.0 0.5 1.0 1.5

0

2

4

0.0 0.5 1.0

0

1

2

3

0.0 0.5 1.0 1.5

0

1

2

3

0.0 0.5 1.0

0

1

2

0.0 0.5 1.0 1.5

0

1

2

0.0 0.5 1.0 1.5

0

1

2

3

−3 −1−4

−4

−2

−4

−2

−4

−2

−4

−2−4

−6

−2

−4

−2

−4

−2

−2 −4 −4−2 −2

−3

−1

−2

−1

−2

−1

−2

−3

−1

−2

−3

−1

−2

−1−2 −1−2

𝜑

𝜑

𝜑

𝜑 𝜑𝜑

𝜑 𝜑

𝜑 𝜑

𝜑 𝜑

y

y

y

yy

y

y y

y y

y y

−1.5 −1.0 −0.5

−1.5−2.0 −1.0 −0.5

−1.5 −1.0 −0.5 −1.5 −1.0 −0.5−1.0 −0.5

−1.5 −1.0 −0.5−1.0 −0.5

(10) (𝜎, 𝛽) ∈ B10 (11) (𝜎, 𝛽) ∈ B11 (12) (𝜎, 𝛽) ∈ B12

(1) (𝜎, 𝛽) ∈ B1

(4) (𝜎, 𝛽) ∈ B4

(7) (𝜎, 𝛽) ∈ B7 (8) (𝜎, 𝛽) ∈ B8 (9) (𝜎, 𝛽) ∈ B9

(5) (𝜎, 𝛽) ∈ B5 (6) (𝜎, 𝛽) ∈ B6

(2) (𝜎, 𝛽) ∈ B2 (3) (𝜎, 𝛽) ∈ B3

Figure 4: Continued.

10 Mathematical Problems in Engineering

0.0 0.5 1.0 1.5

0

1

2

3

0 1

0

1

2

3

0 1 2

0

2

4

0.0 0.5 1.0 1.5

0

1

2

3

0 1 2

0

1

2

3

0.0 0.5 1.0 1.5

0

1

2

0.0 0.5 1.0 1.5

0

1

2

3

0 1 2

0

1

2

3

0.0 0.5 1.0 1.5

0

2

4

0.0 0.5 1.0 1.5

0

2

4

6

−4

−2

−4

−2

−4

−6

−2

−3 −1−2

−3 −1−2

−1−2

−1−2

−3

−1

−2

−3

−1

−2

−3

−1

−2

−3

−1

−2

−3

−1

−2

−1

−2

−3

−1

−2

𝜑 𝜑 𝜑

𝜑𝜑𝜑

𝜑 𝜑𝜑

𝜑

y y y

yy

y

y yy

y

−1.5 −1.0 −0.5

−1.5−2.0 −1.0 −0.5

−1.5 −1.0 −0.5

−1.5 −1.0 −0.5

−1.5 −1.0 −0.5

−1.5 −1.0 −0.5

(13) (𝜎, 𝛽) ∈ B13 (14) (𝜎, 𝛽) ∈ B14 (15) (𝜎, 𝛽) ∈ B15

(16) (𝜎, 𝛽) ∈ B16 (17) (𝜎, 𝛽) ∈ B17 (18) (𝜎, 𝛽) ∈ B18

(22) (𝜎, 𝛽) ∈ B22

(20) (𝜎, 𝛽) ∈ B20 (21) (𝜎, 𝛽) ∈ B21(19) (𝜎, 𝛽) ∈ B19

Figure 4: The bifurcation of phase portraits of system (15) when 𝜖 < 0 and 𝜎 ≥ 0.

Mathematical Problems in Engineering 11

(ii) If 𝜑(0) = ̸= 𝐵1, then 𝜑 is a cuspon solution and 𝜑 has

the following asymptotic behavior:

𝜑 (𝜉) − = 𝜆1

𝜉

2/3

+ 𝑂(𝜉

4/3

) , 𝜉 → 0,

𝜑(𝜉) =

2

3𝜆1

𝜉

−1/3 sgn (𝜉) + 𝑂 (𝜉

1/3

) ,

𝜉 → 0,

(27)

where 𝜆1= ±(9|𝛽

2+ 2𝐴𝛽 + 3𝐴

2𝛽 + 2𝜖|( − 𝐴)

2/16)1/3.

Thus, 𝜑(𝜉) ∉ 𝐻1𝑙𝑜𝑐(R).

(iii) If 𝜑(0) = − ̸= 𝐵2, then 𝜑 is a cuspon solution and 𝜑

has the following asymptotic behavior:

𝜑 (𝜉) + = 𝜆2

𝜉

2/3

+ 𝑂(𝜉

4/3

) , 𝜉 → 0,

𝜑(𝜉) =

2

3𝜆2

𝜉

−1/3 sgn (𝜉) + 𝑂 (𝜉

1/3

) ,

𝜉 → 0,

(28)

where 𝜆2= ±(9|𝛽

2− 2𝐴𝛽 + 3𝐴

2𝛽 + 2𝜖|( + 𝐴)

2/16)1/3.

Thus, 𝜑(𝜉) ∉ 𝐻1𝑙𝑜𝑐(R).

(iv) If 𝜑(0) = = 𝐵1and 𝐴 ̸= 0, then 𝜑 is a peakon-like

solution and

𝜑 (𝜉) − = 𝜆3

𝜉 + 𝑂 (

𝜉

2

) , 𝜉 → 0,

𝜑(𝜉) = 𝜆

3sgn (𝜉) + 𝑂 (𝜉

) , 𝜉 → 0,

(29)

where 𝜆3= ±|𝐴 − |√𝛽(𝐵

2− )/4.

(v) If 𝜑(0) = − = 𝐵2and 𝐴 ̸= 0, then 𝜑 is a peakon-like

solution and

𝜑 (𝜉) + = 𝜆4

𝜉 + 𝑂 (

𝜉

2

) , 𝜉 → 0,

𝜑(𝜉) = 𝜆

4sgn (𝜉) + 𝑂 (𝜉

) , 𝜉 → 0,

(30)

where 𝜆4= ±|𝐴 + |√−𝛽(𝐵

1+ )/4.

(vi) If 𝜑(0) = = 𝐵1and 𝐴 = 0, then 𝜑 gives the peakon

solution exp(−√−𝛽/2|𝑥 − 𝑐𝑡|).(vii) If 𝜑(0) = − = 𝐵

2and 𝐴 = 0, then 𝜑 gives the peakon

solution − exp(−√−𝛽/2|𝑥 − 𝑐𝑡|).

Proof. (vi) and (vii) are obvious. Let us prove (i), (ii), and (iv)in order.

(i) From the process of proofing of Theorem 8, we knowthat if |𝜑(0)| ̸= , then 𝜑 ∈ 𝐶∞(R) and 𝜑 is a smooth solitonsolution.

(ii) If 𝜑(0) = ̸= 𝐵1, then by the definition of single peak

soliton we have 𝐴 ̸= ; thus, 𝛽𝜑2 + 2𝐴𝛽𝜑 + 3𝐴2𝛽 + 2𝜖 doesnot contain the factor 𝜑 − . From (24), we obtain

𝜑= sgn (𝐴 − )

𝜑 − 𝐴

√2𝜑2 − 2

⋅ √𝛽𝜑2 + 2𝐴𝛽𝜑 + 3𝐴2𝛽 + 2𝜖

sgn (𝜉) .

(31)

Let 𝑙1(𝜑) = √2(𝜑 + )/|𝜑 − 𝐴|√|𝛽𝜑2 + 2𝐴𝛽𝜑 + 3𝐴2𝛽 + 2𝜖|;

then, 𝑙1() = 2√/| − 𝐴|√|𝛽

2 + 2𝐴𝛽 + 3𝐴2𝛽 + 2𝜖|, and

∫ 𝑙1(𝜑)√

𝜑 − 𝑑𝜑 = ∫ sgn (𝐴 − ) sgn (𝜉) 𝑑𝜉. (32)

Inserting 𝑙1(𝜑) = 𝑙

1() + 𝑂(|𝜑 − |) into (32) and using the

initial condition 𝜑(0) = , we obtain

2𝑙1()

3

𝜑 −

3/2

(1 + 𝑂 (𝜑 −

)) =𝜉 ;

(33)

thus,

𝜑 − = ±(3

2𝑙1()

)

2/3

𝜉

2/3

(1 + 𝑂 (𝜑 −

))−2/3

= ±(3

2𝑙1()

)

2/3

𝜉

2/3

(1 + 𝑂 (𝜑 −

)) ,

(34)

which implies 𝜑 − = 𝑂(|𝜉|2/3). Therefore, we have

𝜑 (𝜉) = ± (3

2𝑙1()

)

2/3

𝜉

2/3

+ 𝑂(𝜉

4/3

)

= + 𝜆1

𝜉

2/3

+ 𝑂(𝜉

4/3

) , 𝜉 → 0,

𝜆1= ±(

3

2𝑙1()

)

2/3

= ±(

9𝛽2+ 2𝐴𝛽 + 3𝐴

2𝛽 + 2𝜖

( − 𝐴)

2

16)

1/3

,

𝜑(𝜉) =

2

3𝜆1

𝜉

−1/3 sgn (𝜉) + 𝑂 (𝜉

1/3

) , 𝜉 → 0.

(35)

So, 𝜑(𝜉) ∉ 𝐻1loc(R).(iii) Similar to the proof of (ii), we ignore it in this paper.(iv) If 𝜑(0) = = 𝐵

1and 𝐴 ̸= 0, then from (25) we obtain

𝜑= √−

𝛽

2sgn (𝐴 − ) 𝜑 − 𝐴

√𝜑 − 𝐵2

𝜑 + sgn (𝜉) . (36)

Let 𝑙2(𝜑) = (1/|𝜑−𝐴|)√(𝜑 + )/(𝜑 − 𝐵

2); then, 𝑙

2() = (1/|−

𝐴|)√2/( − 𝐵2) and

∫ 𝑙2(𝜑) 𝑑𝜑 = √−

𝛽

2∫ sgn (𝐴 − ) sgn (𝜉) 𝑑𝜉. (37)

Inserting 𝑙2(𝜑) = 𝑙

2() + 𝑂(|𝜑 − |) into (37) and using the

initial condition 𝜑(0) = , we obtain

𝑙2() (𝜑 − ) (1 + 𝑂 (

𝜑 − ))−1

= sgn (𝐴 − ) 𝜉 .

(38)

Sincesgn (𝜑 − ) sgn (𝐴 − ) ≥ 0,

1

1 + 𝑂 (𝜑 − )= 1 + 𝑂 (𝜑 − ) ,

(39)

12 Mathematical Problems in Engineering

we get

𝜑 − =

√−𝛽

2

1

𝑙2()

𝜉 (1 + 𝑂 (𝜑 − )) ,

(40)

which implies |𝜑 − | = 𝑂(|𝜉|). Therefore, we have

𝜑 (𝜉) = + 𝜆3

𝜉 + 𝑂 (

𝜉

2

) , 𝜉 → 0,

𝜑(𝜉) = 𝜆

3sgn (𝜉) + 𝑂 (𝜉

) , 𝜉 → 0,

(41)

where 𝜆3= ±|𝐴 − |√𝛽(𝐵

2− )/4.

(v) Similar to the proof of (iv), we ignore it in this paper.

By virtue of Theorem 9, any single peak soliton for the𝐶(3, 2, 2) equation (10) must satisfy the following initial andboundary values problem:

(𝜑)2

=𝛽 (𝜑 − 𝐴)

2

(𝜑 − 𝐵1) (𝜑 − 𝐵

2)

2 (2 − 𝜑2)fl𝐿 (𝜑) ,

𝜑 (0) ∈ {, −, 𝐵1, 𝐵2} ,

lim|𝜉|→∞

𝜑 (𝜉) = 𝐴.

(42)

𝐿(𝜑) ≥ 0 and the boundary condition (24) imply thefollowing:

(a) If 𝛽(2 − 𝜑2) ≥ 0, then 𝜑 ≥ max{𝐵1, 𝐵2} or 𝜑 ≤

min{𝐵1, 𝐵2}.

(b) If𝛽(2−𝜑2) ≤ 0, thenmin{𝐵1, 𝐵2} ≤ 𝜑 ≤ max{𝐵

1, 𝐵2}.

Below, we will present some implicit formulas for thesingle peak soliton solutions in the case of specific 𝜎 and 𝛽.

Case 1 ((𝜎, 𝛽) ∈ 𝐴12). In this case, we have − < 𝐵

2< 𝐴 <

𝐵1= . From the standard phase analysis and Theorem 9

we know that if 𝜑 is a single peak soliton of the 𝐶(3, 2, 2)equation, then

𝜑= −√−

𝛽

2(𝜑 − 𝐴)√

𝜑 − 𝐵2

𝜑 + sgn (𝜉) . (43)

From the separation of variables we get

∫ℎ (𝜑) 𝑑𝜑 = √−𝛽

2sgn (𝜉) 𝑑𝜉, (44)

where ℎ(𝜑) = (1/(𝐴 − 𝜑))√(𝜑 + )/(𝜑 − 𝐵2). After a lengthy

calculation of integral, we obtain the implicit solution 𝜑defined by

𝐻(𝜑)fl√𝐴 +

𝐴 − 𝐵2

𝐼1(𝜑) − 𝐼

2(𝜑) = √−

𝛽

2

𝜉 + 𝐾,

(45)

where

𝐼1(𝜑) = ln

(𝐴 + ) (𝜑 − 𝐵2) + (𝐴 − 𝐵

2) (𝜑 + ) + 2√(𝐴 + ) (𝐴 − 𝐵

2) (𝜑 − 𝐵

2) (𝜑 + )

√(𝐴 + ) (𝐴 − 𝐵2) (𝐴 + ) (𝐴 − 𝜑)

,

𝐼2(𝜑) = ln

(𝜑 + ) + (𝜑 − 𝐵

2) + 2√(𝜑 + ) (𝜑 − 𝐵

2)

,

(46)

and 𝐾 is an arbitrary integration constant. For 𝜑(0) = , theconstant𝐾 = 𝐻(𝜑(0)) is defined by

𝐾 = √𝐴 +

𝐴 − 𝐵2

𝐼1() − 𝐼

2() , (47)

and for 𝜑(0) = 𝐵2,

𝐾 = √𝐴 +

𝐴 − 𝐵2

𝐼1(𝐵2) − 𝐼2(𝐵2) . (48)

(i) If 𝜑(0) = , then 𝐴 < 𝜑 ≤ . Since 𝐻(𝜑) = ℎ(𝜑), weknow that𝐻(𝜑) strictly decreases on the interval (𝐴, ]; thus,𝐻1(𝜑) = 𝐻|

(𝐴,](𝜑) gives a single peak solitonwith𝐻

1() = 𝐾

and𝐻1(𝐴+) = ∞. Therefore, 𝜑

1(𝜉) = 𝐻

−1

1(√−𝛽/2|𝜉| + 𝐾) is

the solution satisfying

𝜑1(0) = ,

lim|𝜉|→∞

𝜑1(𝜉) = 𝐴,

𝜑

1(0±) = ± (𝐴 − )√

𝛽 (𝐵2− )

4.

(49)

So, 𝜑1(𝜉) is a peakon-like solution (see Figure 5).

(ii) If 𝜑(0) = 𝐵2, then 𝐵

2≤ 𝜑 < 𝐴. By 𝐻(𝜑) = 𝑓(𝜑),

we know that 𝐻(𝜑) strictly increases on the interval [𝐵2, 𝐴).

Thus,

𝐻2(𝜑) = 𝐻|

[𝐵2 ,𝐴)(𝜑) (50)

Mathematical Problems in Engineering 13

−3 −2 −1

𝜑

−0.2

1 2 3

0.2

0.4

0.6

0.8

1.0

x

(a)

02

02

−2 −2 x

1.0

0.5

0.0

t

𝜑

22

(b)

Figure 5: Two- and three-dimensional graphs of the peakon-like solution.

−0.3

−0.4

−0.5

1 2 3−3 −2 −1

𝜑

−0.2

x

(a)

0

2

0

2

−2−2

−0.2

−0.3

−0.4

−0.5

−0.6

t x

𝜑

0

2

0

2

−2−2

t x

(b)

Figure 6: Two- and three-dimensional graphs of the smooth soliton solution.

has the inverse denoted by 𝜑2(𝜉) = 𝐻

−1

2(√−𝛽/2|𝜉| + 𝐾).

𝜑2(𝜉) gives a kind of smooth soliton solution (see Figure 6)

satisfying

𝜑2(0) = 𝐵

2,

lim|𝜉|→∞

𝜑2(𝜉) = 𝐴,

𝜑

2(0) = 0.

(51)

Case 2 ((𝜎, 𝛽) ∈ 𝐴14). In this case, we have − = 𝐵

2< 𝐴 <

< 𝐵1and (25) is equivalent to

𝜑= −√−

𝛽

2(𝜑 − 𝐴)√

𝜑 − 𝐵1

𝜑 − sgn (𝜉) . (52)

Let

𝑔 (𝜑) =1

𝐴 − 𝜑√𝜑 − 𝐵1

𝜑 − ; (53)

14 Mathematical Problems in Engineering

−1 1 2 3

0.2

0.4

0.6

0.8

1.0

−3 −2

𝜑

x

(a)

02

02−2

−2

𝜑

x t

0.0

0.5

1.0

(b)

Figure 7: Two- and three-dimensional graphs of the cuspon solution.

then, (52) is converted to

𝑔 (𝜑) 𝑑𝜑 =1

𝐴 − 𝜑√𝜑 − 𝐵1

𝜑 − 𝑑𝜑 = √−

𝛽

2sgn (𝜉) 𝑑𝜉. (54)

Integrating (54) on the interval [0, 𝜉] (or [𝜉, 0]) leads to thefollowing implicit solutions:

𝐺 (𝜑)fl√𝐴 −

𝐴 − 𝐵1

𝐼3(𝜑) + 𝐼

4(𝜑) = √−

𝛽

2

𝜉 + 𝐾,

(55)

where

𝐼3(𝜑) = ln

(𝐴 − ) (𝜑 − 𝐵1) + (𝐴 − 𝐵

1) (𝜑 − ) + 2√(𝐴 − ) (𝐴 − 𝐵

1) (𝜑 − ) (𝜑 − 𝐵

1)

√(𝐴 − ) (𝐴 − 𝐵1) (𝐴 − ) (𝜑 − 𝐴)

,

𝐼4(𝜑) = ln

(𝜑 − ) + (𝜑 − 𝐵

1) + 2√(𝜑 − ) (𝜑 − 𝐵

1)

.

(56)

And𝐾 is an arbitrary integration constant. It is obvious that,for 𝜑(0) = , the constant𝐾 = 𝐻(𝜑(0)) is defined by

𝐾 = √𝐴 −

𝐴 − 𝐵1

𝐼3() + 𝐼

4() , (57)

and for 𝜑(0) = −,

𝐾 = √𝐴 −

𝐴 − 𝐵1

𝐼3(−) + 𝐼

4(−) . (58)

(i) If 𝜑(0) = , then 𝐴 < 𝜑 ≤ . From 𝑔(𝜑) < 0, weknow that 𝐺(𝜑) strictly decreases on the interval (𝐴, ] with𝐺() = 𝐾 and 𝐺(𝐴+) = ∞. Define

𝐺1(𝜑) = 𝐺|

(𝐴,](𝜑) . (59)

Since 𝐺1(𝜑) is a strictly decreasing function from (𝐴, ] onto

[𝐾,∞), we can solve for 𝜑 uniquely from (59) and obtain

𝜑1(𝜉) = 𝐺

−1

1(√−

𝛽

2

𝜉 + 𝐾) .

(60)

It is easy to check that 𝜑 satisfies

𝜑1(0) = ,

lim|𝜉|→∞

𝜑1(𝜉) = 𝐴,

𝜑

1(0±) = ∓∞.

(61)

Therefore, the solution𝜑1defined by (60) is a cuspon solution

for the 𝐶(3, 2, 2) equation (see Figure 7).

Mathematical Problems in Engineering 15

(ii) If 𝜑(0) = −, then − ≤ 𝜑 < 𝐴. Through a similaranalysis, we get a strictly increasing function 𝐺(𝜑) on theinterval [−, 𝐴) satisfying

𝐺 (𝜑) = √−𝛽

2

𝜉 + 𝐾,

(62)

where 𝐺(𝜑) is defined by (55). Let

𝐺2(𝜑) = 𝐺|

[−,𝐴)(𝜑) , (63)

then 𝐺2(𝜑) is a strictly increasing function from [−, 𝐴) onto

[𝐾,∞) so that we can solve for 𝜑 and obtain

𝜑2(𝜉) = 𝐺

−1

2(√−

𝛽

2

𝜉 + 𝐾) .

(64)

It is easy to check that 𝜑2satisfies

𝜑2(0) = −,

lim|𝜉|→∞

𝜑2(𝜉) = 𝐴,

𝜑

2(0±) = ± (𝐴 + )√−

𝛽 (𝐵1+ )

4.

(65)

Therefore, the solution 𝜑2defined by (64) is a peakon-like

solution, whose graph is similar to those in Figure 5.

Case 3 ((𝜎, 𝛽) ∈ 𝐴16). In this case, we have − < 𝐵

2< 𝐴 =

0 < 𝐵1< , 𝐵

2= −𝐵1, and

𝜑= −√−

𝛽

2𝜑√

𝜑 − 𝐵2

1

𝜑2 − 2sgn (𝜉) . (66)

Hence from the separation of variables we have

√2 − 𝜑2

𝜑√𝐵2

1− 𝜑2

𝑑𝜑 = −√−𝛽

2sgn (𝜉) 𝑑𝜉. (67)

Integrating (67) on the interval [0, 𝜉] (or [𝜉, 0]) leads tothe following implicit formula for the two smooth solitonsolutions:

√2 − 𝜑2 + √𝐵2

1− 𝜑2

√2 − 𝐵2

1

(

√(2 − 𝐵2

1) 𝜑2

√𝐵2

1− 𝜑2 + 𝐵

1√2 − 𝜑2

)

= exp(−√−𝛽

2|𝑥 − 𝑐𝑡|) ,

(68)

where 𝜑 ∈ (𝐴, 𝐵1]. Consider

√2 − 𝐵2

2

√2 − 𝜑2 + √𝐵2

2− 𝜑2

(

√𝐵2

2− 𝜑2 − 𝐵

2√2 − 𝜑2

√(2 − 𝐵2

2) 𝜑2

)

= exp(√−𝛽

2|𝑥 − 𝑐𝑡|) ,

(69)

where 𝜑 ∈ [𝐵2, 𝐴).

Case 4 ((𝜎, 𝛽) ∈ 𝐴17). In this case, we have − = 𝐵

2< 𝐴 =

0 < 𝐵1= , and

𝜑= −√−

𝛽

2𝜑 sgn (𝜉) . (70)

Choosing 𝜑(0) = (or −) as initial value, we get

∫

𝜑

1

𝜑𝑑𝜑 = −∫

𝜉

0

√−𝛽

2sgn (𝜉) 𝑑𝜉, for 𝐴 < 𝜑 ≤ ,

∫

𝜑

−

1

𝜑𝑑𝜑 = −∫

𝜉

0

√−𝛽

2sgn (𝜉) 𝑑𝜉, for − ≤ 𝜑 < 𝐴,

(71)

which immediately yield the peakon solutions

𝜑 (𝑥 − 𝑐𝑡) = ± exp(−√−𝛽

2|𝑥 − 𝑐𝑡|) . (72)

The graphs for the peakon solution (72) are shown in Figure 8.

Remark 10. The classical peakon solution (72) and peakon-like solution (64) admit left-half derivative and right-halfderivative at their crest. But the signs of the left-half derivativeand right-half derivative are opposite, so the peakon andpeakon-like solutions admit the discontinuous first orderderivative at their crest. In comparison with classical peakonsolution (72), the expression of the peakon-like solution (64)is more complex. Moreover, by observing Figures 2(14) and2(17) we find that the phase orbits of the peakon consist ofthree straight lines, but the phase orbits of the peakon-likeconsist of two curves and a straight line. Therefore, we callthe soliton solution (64) the peakon-like solution.

4. Kink Wave and Kink Compacton Solutions

We now turn our attention to the kink wave solutions of the𝐶(3, 2, 2) equation (10). In order to study kinkwave solutions,we assume that

lim𝜉→∞

𝜑 (𝜉) = 𝐴1,

lim𝜉→−∞

𝜑 (𝜉) = 𝐴2,

(73)

16 Mathematical Problems in Engineering

2 4

0.2

0.4

0.6

0.8

1.0

−2−4

𝜑

x

(a)

0 2

02

0.0

0.5

1.0

−2

𝜑

x

t

0

2

−2

(b)

Figure 8: Two- and three-dimensional graphs of the peakon solution.

where 𝐴1> 𝐴2. Substituting the boundary condition (73)

into (14) generates

(𝜑)2

=

𝛽 (𝜑 − 𝐴1) (𝜑 − 𝐴

2) [𝜑2+ (𝐴1+ 𝐴2) 𝜑 + 𝐴

2

1+ 𝐴1𝐴2+ 𝐴2

2+ 2𝜖/𝛽]

2 (2 − 𝜑2)fl𝐹 (𝜑) . (74)

The nonlinear differential equation (74) may sustaindifferent kinds of nonlinear excitations. In what follows, weconfine our attention to the cases 𝐴

2= −𝐴

1and 𝜖 =

−𝐴2

1𝛽 which describe kinks and kink compactons which

play an important role in the dynamics systems. Under theseconsiderations, (74) reduces to

(𝜑)2

=𝛽 (𝜑 − 𝐴

1)2

(𝜑 − 𝐴2)2

2 (2 − 𝜑2). (75)

If 𝐴1< , then from the phase analysis in Section 2 (see

Figure 4(10)), we know that (𝐴1, 0) and (𝐴

2, 0) are two saddle

points of (13) and the kink solutions can be obtained fromthe two heteroclinic orbits connecting (𝜙, 𝑦) = (𝐴

1, 0) and

(𝐴2, 0). When 𝐴

1increases upon reaching , that is 𝐴

1= ,

(75) becomes

(𝜑)2

= −𝛽 (𝜑 − 𝐴

1) (𝜑 − 𝐴

2)

2, (76)

and the ellipse 2𝑦2+𝛽(𝜑−𝐴1)(𝜑−𝐴

2) = 0 (see Figure 4(11)),

which is tangent to the singular lines 𝜑 = and 𝜑 = − atpoints (𝐴

1, 0) and (𝐴

2, 0), respectively, gives rise to two kink

compactons of (10).We next explore the qualitative behavior of kink wave

solutions to (75) and (76). If 𝜑 is a kink wave solutions of (75)

or (76), we have 𝜑 → 0 as 𝜑 → 𝐴1and as 𝜑 → 𝐴

2.

Moreover, we have 𝐹(𝜑) ≥ 0 for 𝐴2≤ 𝜑 ≤ 𝐴

1and 𝜑 is

strictly monotonic in any interval where 𝐹(𝜑) > 0. Thus, if𝜑> 0 at some points, 𝜑will be strictly increasing until it gets

close to the next zero of 𝐹. Denoting this zero 𝐴1, we have

𝜑 ↑ 𝐴1. What will happen to the solution when it approaches

𝐴1? Depending on whether the zero is double or simple, 𝜑

has a different behavior. We explore the two cases in turn.

Theorem 11. (i) If 𝜑 has a simple zero at 𝜑 = 𝐴1, so that

𝐹(𝐴1) = 0 and 𝐹(𝐴

1) < 0, then the solution 𝜑 of (75) satisfies

𝜑 (𝜉) − 𝐴1=𝐹(𝐴1)

4(𝜉 − 𝜂)

2

+ 𝑂 ((𝜉 − 𝜂)4

)

𝑎𝑠 𝜉 ↑ 𝜂,

(77)

where 𝜑(𝜂) = 𝐴1.

(ii) When 𝜑 approaches the double zero 𝐴1of 𝐹(𝜑) so that

𝐹(𝐴1) = 0 and𝐹(𝐴

1) > 0, then the solution𝜑 of (75) satisfies

𝜑 (𝜉) − 𝐴1∼ 𝛼 exp(−𝜉√𝐹 (𝐴

1)) 𝑎𝑠 𝜉 → ∞ (78)

for some constant 𝛼. Thus, 𝜑 ↑ 𝐴1exponentially as 𝜉 → ∞.

Mathematical Problems in Engineering 17

Proof. (i) When 𝐴1= −𝐴

2= and 𝜖 = −𝐴2

1𝛽, from (76),

𝐹(𝜑) has a simple zero at 𝜑 = 𝐴1. Then,

(𝜑)2

= (𝜑 − 𝐴1) 𝐹(𝐴1) + 𝑂 ((𝜑 − 𝐴

1)2

)

as 𝜑 ↑ 𝐴1.

(79)

Using the fact that (𝜑)2 ≥ 0, we know that 𝐹(𝐴1) < 0.

Moreover,

𝑑𝜉

𝑑𝜑=

1

√(𝜑 − 𝐴1) 𝐹 (𝐴

1) + 𝑂 ((𝜑 − 𝐴

1)2

)

. (80)

Because

√(𝜑 − 𝐴1) 𝐹 (𝐴

1) + 𝑂 ((𝜑 − 𝐴

1)2

)

= √𝐴1− 𝜑(√−𝐹 (𝐴

1) + 𝑂 (𝐴

1− 𝜑)) ,

1

√−𝐹 (𝐴1) + 𝑂 (𝐴

1− 𝜑)

=1

√−𝐹 (𝐴1)

+ 𝑂 (𝐴1− 𝜑) ,

(81)

we obtain𝑑𝜉

𝑑𝜑=

1

√− (𝐴1− 𝜑) 𝐹 (𝐴

1)

+ 𝑂 ((𝐴1− 𝜑)1/2

) . (82)

Integrating (82) yields

𝜂 − 𝜉 =2

√−𝐹 (𝐴1)

(𝐴1− 𝜑)1/2

+ 𝑂((𝐴1− 𝜑)3/2

) , (83)

where 𝜂 satisfies 𝜑(𝜂) = 𝐴1. Thus,

(𝜂 − 𝜉)2

=4

−𝐹 (𝐴1)(𝐴1− 𝜑) + 𝑂 ((𝐴

1− 𝜑)2

) , (84)

which implies 𝑂((𝐴1− 𝜑)2) = 𝑂((𝜂 − 𝜉)

4). Therefore, we get

𝜑 (𝜉) = 𝐴1+𝐹(𝐴1)

4(𝜉 − 𝜂)

2

+ 𝑂 ((𝜉 − 𝜂)4

)

as 𝜉 ↑ 𝜂,(85)

where 𝜑(𝜂) = 𝐴1.

(ii) When 𝐴1= −𝐴

2< and 𝜖 = −𝐴2

1𝛽, from (75), 𝐹(𝜑)

has a double zero at 𝜑 = 𝐴1. Then,

(𝜑)2

= (𝜑 − 𝐴1)2

𝐹(𝐴1) + 𝑂 ((𝜑 − 𝐴

1)3

)

as 𝜑 ↑ 𝐴1.

(86)

Furthermore, we get

𝑑𝜉

𝑑𝜑=

1

√(𝜑 − 𝐴1)2

𝐹 (𝐴1) + 𝑂 ((𝜑 − 𝐴

1)3

)

. (87)

Observing that

√(𝜑 − 𝐴1)2

𝐹 (𝐴1) + 𝑂 ((𝜑 − 𝐴

1)3

)

= (𝐴1− 𝜑) (√𝐹 (𝐴

1) + 𝑂 (𝐴

1− 𝜑)) ,

1

√𝐹 (𝐴1) + 𝑂 (𝐴

1− 𝜑)

=1

√𝐹 (𝐴1)

+ 𝑂 (𝐴1− 𝜑) ,

(88)

we obtain𝑑𝜉

𝑑𝜑=

1

(𝐴1− 𝜑) 𝐹 (𝐴

1)+ 𝑂 (1) . (89)

By a similar computation as the one that leads to (85), wearrive at (78). This completes the proof of Theorem 11.

Next we try to find the exact formulas for the kink wavesolutions. Let 𝐴

1= −𝐴

2< and 𝜖 = −𝐴2

1𝛽. Then, (75)

becomes

√2 − 𝜑2

𝜑2 − 𝐴2

1

𝑑𝜑 = ∓√𝛽

2𝑑𝜉. (90)

Integrating both sides of (90) gives the following implicitexpressions of kink and antikink wave solutions:

arctan𝜑

√2 − 𝜑2

+

√2 − 𝐴2

1

𝐴1

arctanh√2 − 𝐴

2

1

𝐴1

𝜑

√2 − 𝜑2

= ±√𝛽

2𝜉.

(91)

By letting 𝐴1→ in (91), we get two kink compactons

which are given by

±𝜑 =

{{{{{{{{{

{{{{{{{{{

{

sin√𝛽

2𝜉,

𝜉 ≤

𝜋

√2𝛽,

, 𝜉 >𝜋

√2𝛽,

−, 𝜉 < −𝜋

√2𝛽.

(92)

The graphs for the kink wave solutions (91) and kinkcompacton solutions (92) are shown in Figures 9 and 10,respectively.

Remark 12. The two kink compacton solutions (92) aredifferent from the well-known smooth kink wave solutions.In comparison with kink wave solutions (91), the kink com-pacton solutions (92) have no exponential decay propertiesbut have compact support.That is, they minus a constant, thedifferences identically vanish outside a finite core region.

18 Mathematical Problems in Engineering

𝜑

𝜉−1 1

0.5

−0.5

(a)

−1−1

−0.5

0.5

0.0𝜑

x t

0

1

01

11

(b)

Figure 9: Two- and three-dimensional graphs of the kink wave solution.

1

1𝜑

𝜉

−1

−1

(a)

0

1

0

1

0.00.51.0

𝜑

−1 −1

−0.5

−1.0

tx

0

1

0

1

(b)

Figure 10: Two- and three-dimensional graphs of the kink compacton solution.

5. Conclusion

In this paper, we investigate the travelingwave solutions of the𝐶(3, 2, 2) equation (10). We show that (10) can be reduced toa planar polynomial differential system by transformation ofvariables. We treat the planar polynomial differential systemby the dynamical systems theory and present a phase spaceanalysis of their singular points. Two singular straight linesare found in the associated topological vector field. Theinfluence of parameters as well as the singular lines onthe smoothness property of the traveling wave solutions isexplored in detail.

Because any traveling wave solution of (10) is determinedfrom Newton’s equation which we write in the form 𝑦2 =𝐹(𝜑), where 𝑦 = 𝜑

𝜉(𝜉), we solve Newton’s equation 𝑦2 =

𝐹(𝜑) for single peak soliton solutions and kink wave andkink compacton solutions. We classify all single peak solitonsolutions of (10). Then peakon, peakon-like, cuspon, smoothsoliton solutions of the generalized Camassa-Holm equation

(10) are obtained. The parametric conditions of existenceof the single peak soliton solutions are given by using thephase portrait analytical technique. Asymptotic analysis andnumerical simulations are provided for single peak solitonand kink wave and kink compacton solutions of the𝐶(3, 2, 2)equation.

Actually, for 𝑚 = 2𝑘 + 1, 𝑘 ∈ N+ in 𝐶(𝑚, 2, 2) equation(9), the dynamical behavior of traveling wave solutions of(9) is similar to the case 𝑚 = 3; for 𝑚 = 2𝑘, 𝑘 ∈ N+ in𝐶(𝑚, 2, 2) equation (9), the dynamical behavior of travelingwave solutions of (9) is similar to the case 𝑚 = 2. We areapplying the approach mentioned in this work to 𝐶(2, 2, 2)equation (9) and already get some new solutions, which wewill report in another paper.

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper.

Mathematical Problems in Engineering 19

Acknowledgments

This work was partially supported by the National NaturalScience Foundation of China (no. 11326131 and no. 61473332)andZhejiangProvincialNatural Science Foundation ofChinaunder Grant nos. LQ14A010009 and LY13A010005.

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