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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;
[email protected]
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
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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.
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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.
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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)
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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)
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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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