Research Article Traveling Wave Solutions of a Generalized...
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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 zsdzln126com
Received 1 August 2015 Accepted 14 September 2015
Academic Editor Maria Gandarias
Copyright copy 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 119862(3 2 2) 119906119905+ 119896119906119909+ 1205741119906119909119909119905
+ 1205742(1199063)119909+ 1205743119906119909(1199062)119909119909
+ 1205743119906(1199062)119909119909119909
= 0 Weshow that the 119862(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 cusponsmooth soliton solutions of the generalized Camassa-Holm equation under inhomogeneous boundary condition are obtainedTheparametric conditions of existence of the single peak soliton solutions are given by using the phase portrait analytical techniqueAsymptotic analysis and numerical simulations are provided for single peak soliton kink wave and kink compacton solutions ofthe 119862(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 interestnamely 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 Backlund 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 Howevernot 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[6ndash9]
To study the traveling wave solutions of a nonlinear PDE
Φ(119906 119906119905 119906119909 119906119909119909 119906119909119905 119906119905119905 ) = 0 (1)
let 120585 = 119909 minus 119888119905 and 119906(119909 119905) = 120593(120585) where 119888 is the wave speedSubstituting them into (1) leads the PDE to the followingordinary differential equation
Φ1(120593 1205931015840 12059310158401015840 ) = 0 (2)
Here we consider the case of (2) which can be reduced to thefollowing planar dynamical system
119889120593
119889120585= 1205931015840= 119910
119889119910
119889120585= 119865 (120593 119910)
(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 pageshttpdxdoiorg1011552015610979
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]
119906119905minus 119906119905119909119909
+ 3119906119906119909= 2119906119909119906119909119909
+ 119906119906119909119909119909
(4)
arising as a model for nonlinear waves in cylindrical axiallysymmetric hyperelastic rods with 119906(119909 119905) representing theradial stretch relative to a prestressed state where Camassaand Holm showed that (4) has a peakon of the form 119906(119909 119905) =
119888119890minus|119909minus119888119905| 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
119906119905+ (119906119898)119909+ (119906119899)119909119909119909
= 0 (5)
This equation denoted by 119870(119898 119899) owns the property thatfor certain119898 and 119899 its solitary wave solutions have compactsupport [12] That is they identically vanish outside a finitecore region For instance the 119870(2 2) equation admits thefollowing compacton solution
119906 (119909 119905) =
4119888
3cos2 (119909 minus 119888119905
4) |119909 minus 119888119905| le 2120587
0 otherwise(6)
The Camassa-Holm equation the 119870(2 2) equation andalmost all integrable dispersive equations have the sameclass of traveling wave systems which can be written in thefollowing form [13]
119889120593
119889120585= 119910 =
1
1198632 (120593)
120597119867
120597119910
119889119910
119889120585= minus
1
1198632 (120593)
120597119867
120597120593= minus
1198631015840(120593) 1199102+ 119892 (120593)
1198632 (120593)
(7)
where 119867 = 119867(120593 119910) = (12)11991021198632(120593) + int119863(120593)119892(120593)119889120593 is the
first integral It is easy to see that (4) is actually a special case of(3) with 119865(120593 119910) = minus(1119863
2(120593))(120597119867120597120593) If there is a function
120593 = 120593119904such that 119863(120593
119904) = 0 then 120593 = 120593
119904is a vertical straight
line solution of the system
119889120593
119889120577= 119910119863 (120593)
119889119910
119889120577= minus1198631015840(120593) 1199102minus 119892 (120593)
(8)
where 119889120585 = 119863(120593)119889120577 for 120593 = 120593119904 The two systems have
the same topological phase portraits except for the verticalstraight line 120593 = 120593
119904and 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 120593 = 120593
119904 However the orbits
which do intersect with the vertical straight line 120593 = 120593119904or
are unbounded but can approach the vertical straight linecorrespond 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 [6ndash9] But tillnow there have been few works on the integrable nonlinearequations with two singular straight lines or other types ofsingular curves [13ndash15]
In 2004 Tian and Yin [16] introduced the following fullynonlinear generalized Camassa-Holm equation 119862(119898 119899 119901)
119906119905+ 119896119906119909+ 1205741119906119909119909119905
+ 1205742(119906119898)119909+ 1205743119906119909(119906119899)119909119909
+ 1205744119906 (119906119901)119909119909119909
= 0
(9)
where 119896 1205741 1205742 1205743 and 120574
4are arbitrary real constants and 119898
119899 and 119901 are positive integers By using four direct ansatzsthey obtained kink compacton solutions nonsymmetry com-pacton solutions and solitary wave solutions for the119862(2 1 1)and 119862(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 119898 = 3119899 = 119901 = 2 and 120574
3= 1205744 Then (9) reduces to the 119862(3 2 2)
equation
119906119905+ 119896119906119909+ 1205741119906119909119909119905
+ 1205742(1199063)119909+ 1205743119906119909(1199062)119909119909
+ 1205743119906 (1199062)119909119909119909
= 0
(10)
Actually we have already considered a special 119862(3 2 2)equation in [17] namely 120574
1= minus1 120574
2= minus3 and 120574
3= minus1
where the bifurcation of peakons are obtained by applyingthe qualitative theory of dynamical systems In this worka more general 119862(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 instancepeakons cuspons compactons kinks and kink-compactons
This paper is organized as follows In Section 2we 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 solutionscuspon 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 119906(119909 119905) = 120593(120585)
with 120585 = 119909 minus 119888119905 be the solution of (10) then it follows that
(119896 minus 119888) 1205931015840minus 1205741119888120593101584010158401015840
+ 1205742(1205933)1015840
+ 12057431205931015840(1205932)10158401015840
+ 1205743120593 (1205932)101584010158401015840
= 0
(11)
where 1205931015840 = 120593120585 12059310158401015840 = 120593
120585120585 and 120593101584010158401015840 = 120593
120585120585120585 Integrating (11) once
and setting the integration constant as 119892 we have
(119896 minus 119888) 120593 minus 120574111988812059310158401015840+ 12057421205933+ 1205743120593 (1205932)10158401015840
= minus119892 (12)
Clearly (12) is equivalent to the planar system
119889120593
119889120585= 119910
119889119910
119889120585=1205731205933+ 1205931199102+ 120598120593 + 120590
120588 minus 1205932
(13)
where 120588 = 119888120574121205743 120573 = 120574
221205743 120590 = 1198922120574
3 and 120598 = (119896minus119888)2120574
3
(1205743
= 0) System (13) has the first integral
119867(120593 119910) =120573
41205934+1
21205981205932+ 120590120593 +
1
212059321199102minus1
21205881199102= ℎ (14)
Obviously for 120588 gt 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 120588 gt 0 in the rest of thispaper (120588 = 984858
2 984858 gt 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 120593 = 984858 and 120593 = minus984858 on the right-hand side of the secondequation of (13) To avoid the singular lines temporarily wedefine a new independent variable 120577 by setting (119889120585119889120577) =
9848582minus1205932 then system (13) is changed to a Hamiltonian system
written as119889120593
119889120577= (9848582minus 1205932) 119910
119889119910
119889120577= 1205931199102+ 1205731205933+ 120598120593 + 120590
(15)
System (15) has the same topological phase portraits as system(13) except for the singular lines 120593 = 984858 and 120593 = minus984858
We now investigate the bifurcation of phase portraits ofthe system (15) Denote that
119891 (120593) = 1205731205933+ 120598120593 + 120590 (16)
Let 119872(120593119890 119910119890) be the coefficient matrix of the linearized
system of (15) at the equilibrium point (120593119890 119910119890) then
119872(120593119890 119910119890) = (
minus21205931198901199101198901199102
120598+ 1198911015840(120593119890)
9848582minus 1205932
1198902120593119890119910119890
) (17)
and at this equilibrium point we have
119869 (120593119890 119910119890) = det119872(120593
119890 119910119890)
= (1205932
119890minus 9848582) 1198911015840(120593119890) minus (3120593
2
119890+ 9848582) 1199102
119890
(18)
By the theory of planar dynamical systems for an equilibriumpoint of a Hamiltonian system if 119869 lt 0 then it is a saddlepoint a center point if 119869 gt 0 and a degenerate equilibriumpoint if 119869 = 0
From the above analysis we can obtain the bifurcationcurves and phase portraits under different parameter condi-tions
Let
1205731(120590) = minus
41205983
271205902
1205732(120590) =
120590 minus 120598984858
9848583
1205733(120590) =
minus120590 minus 120598984858
9848583
(19)
Clearly for 120573 gt 1205731(120590) the function 119891(120593) = 0 has three
real roots 1205931 1205932 and 120593
3(1205931
gt 1205932
gt 1205933) that is system
(15) has three equilibrium points 119864119894(120593119894 0) 119894 = 1 2 3 on
the 120593-axis When 120573 lt 1205733(120590) (15) has two equilibrium
points 11987812(984858 plusmn119884
1) on the straight line 120593 = radic119886 where 119884
1=
radic(minus120590 minus 984858(120598 + 1205739848582))984858 When 120573 lt 1205732(120590) system (15) has two
equilibrium points 11987834(minus984858 plusmn119884
2) on the straight line 120593 = minus984858
where 1198842= radic(120590 minus 984858(120598 + 1205739848582))984858 Notice that on making the
transformation 120590 rarr minus120590 120593 rarr minus120593 120577 rarr minus120577 system (15)is invariant This means that for 120590 lt 0 the phase portraitsof (15) are just the reflections of the corresponding phaseportraits of (15) in the case 120590 gt 0 with respect to the 119910-axisThus we only need to consider the case 120590 ge 0 To knowthe dynamical behavior of the orbits of system (15) we willdiscuss two cases 120598 gt 0 and 120598 lt 0
21 Case I 120598 gt 0
Lemma 1 Suppose that 120598 gt 0 Denote ℎ+984858= 119867(984858 plusmn119884
1) ℎminus984858=
119867(minus984858 plusmn1198842) and ℎ
119894= 119867(120593
119894 0) 119894 = 1 2 3
(1) For 120590 gt 0 and 1205731(120590) lt 120573 lt 120573
3(120590) there exists one
and only one curve 120573 = 1205734(120590) on the (120590 120573)-plane on which
ℎ+
984858= ℎ2 for 120590 gt 0 and 120573
1(120590) lt 120573 lt 120573
2(120590) there exists one
and only one curve 120573 = 1205735(120590) on the (120590 120573)-plane on which
ℎminus
984858= ℎ2or ℎminus984858= ℎ3Moreover the curves120573 = 120573
1(120590)120573 = 120573
2(120590)
120573 = 1205734(120590) and 120573 = 120573
5(120590) are tangent at the point (120590
1119904 1205731119904)
The curve 120573 = 1205733(120590) intersects with the curves 120573 = 120573
1(120590)
120573 = 1205734(120590) and 120573 = 120573
5(120590) at the points (120590
2119904 1205732119904) (1205903119904 1205733119904)
and (1205904119904 1205734119904) (0 lt 120590
4119904lt 1205903119904lt 1205902119904lt 1205901119904) respectively
(2) For 120590 = 0 there exists a bifurcation point 119860(0 minus21205989848582)on (120590 120573)-plane on which ℎ
+
984858= ℎminus
984858= ℎ2when 120573 = minus2120598984858
2ℎ+
984858= ℎminus
984858lt ℎ2when 120573 lt minus2120598984858
2 and ℎ+
984858= ℎminus
984858gt ℎ2when
minus21205989848582lt 120573 lt minus120598984858
2
4 Mathematical Problems in Engineering
1205732(120590)
(1205901s 1205731s)
1205733(120590)
1205734(120590)1205735(120590)
120573
minus2
minus2
minus4
minus4
minus6
minus8
minus10
minus12
minus14
1205731(120590)
1205902 4
Figure 1 The bifurcation curves in the (120590 120573)-parameter plane for120598 gt 0
According to the above analysis and Lemma 1 we obtainthe following proposition on the bifurcation curves of thephase portraits of system (15) for 120598 gt 0
Proposition 2 When 120598 gt 0 for system (15) in (120590 120573)-parameter plane there exist five bifurcation curves (see Fig-ure 1)
120573 = 1205731(120590) = minus
41205983
271205902
120573 = 1205732(120590) =
120590 minus 120598984858
9848583
120573 = 1205733(120590) =
minus120590 minus 120598984858
9848583
120573 = 1205734(120590) (|120590| le 120590
3119904)
120573 = 1205735(120590)
(20)
These five curves divide the right-half (120590 120573)-parameter planeinto thirty-one regions as follows
1198601 (120590 120573) | 1205901s lt 120590 lt 120598984858 120573 = 1205732(120590)
1198602 (120590 120573) | 120590 gt 1205901s 1205735(120590) lt 120573 lt 1205732(120590)
1198603 (120590 b) | 120590 gt 1205901s 120573 = 1205735(120590)
1198604 (120590 120573) | 120590 gt 1205901s 1205731(120590) lt 120573 lt 1205735(120590)
1198605 (120590 120573) | 120590 gt 1205901s 120573 = 1205731(120590)
1198606 (120590 120573) | 120590 gt 1205902s 1205733(120590) lt 120573 lt 1205731(120590)
1198607 (120590 120573) | 120590 lt 1205902s 120573 = 1205733(120590)
1198608 (120590 120573) | 120590 gt 0 120573 lt max1205731(120590) 1205733(120590)
1198609 (120590 120573) = (1205902s 1205732s)
11986010 (120590 120573) | 0 lt 120590 lt 1205902s 120573 = 1205731(120590)
11986011 (120590 120573) | 120590 lt 1205902s 1205731(120590) lt 120573 lt min1205733(120590)
1205734(120590)
11986012 (120590 120573) | 120590 lt 1205903s 120573 = 1205734(120590)
11986013 (120590 120573) | 120590 gt 0 1205734(120590) lt 120573 lt 1205735(120590)
11986014 (120590 120573) | 120590 lt 1205904s 120573 = 1205735(120590)
11986015 (120590 120573) | 120590 gt 0 1205735(120590) lt 120573 lt 1205733(120590)
11986016 (120590 120573) | 120590 = 0 120573 lt minus21205989848582
11986017 (120590 120573) = (0 minus21205989848582)
11986018 (120590 120573) | 120590 = 0 minus21205989848582 lt 120573 lt minus120598984858
2
11986019 (120590 120573) = (0 minus1205989848582)
11986020 (120590 120573) | 0 lt 120590 lt 1205904s 120573 = 1205733(120590)
11986021 (120590 120573) = (1205904s 1205734s)
11986022 (120590 120573) | 1205904s lt 120590 lt 1205902s 120573 = 1205733(120590)
11986023 (120590 120573) | 1205904s lt 120590 lt 1205901s max1205731(120590) 1205733(120590) lt
120573 lt 1205735(120590)11986024 (120590 120573) | 120590
s4 lt 120590 lt 1205901s 120573 = 1205735(120590)
11986025 (120590 120573) | 1205904s lt 120590 lt 1205901s max1205733(120590) 1205735(120590) lt
120573 lt 1205732(120590)11986026 (120590 120573) | 0 lt 120590 lt 1205901s 120573 = 1205732(d)
11986027 (120590 120573) = (1205901s 1205731s)
11986028 (120590 120573) | 0 le 120590 lt 120598984858 120573 gt 1205732(120590)
11986029 (120590 120573) | 120590 isin R 120573 gt max0 1205732(120590)
11986030 (120590 120573) | 120590 gt 120598984858 120573 = 1205732(120590)
11986031 (120590 120573) | 120590 gt 120598984858 0 lt 120573 lt 1205732(120590)
In this case the phase portraits of system (15) can beshown in Figure 2
22 Case II 120598 lt 0 In this case we have the following
Proposition 3 When 120598 lt 0 for system (15) in (120590 120573)-parameter plane there exist four parametric bifurcation curves(see Figure 3)
120573 = 1205731(120590) = minus
41205983
271205902
120573 = 1205732(120590) =
120590 minus 120598984858
9848583
120573 = 1205733(120590) =
minus120590 minus 120598984858
9848583 120590 = 0
(21)
These four curves divide the right-half (120590 120573)-parameter planeinto twenty-two regions
1198611 (120590 120573) | 1205901s lt 120590 lt 120598984858 120573 = 1205733(120590)
1198612 (120590 120573) | 120590 gt 1205901s max0 1205733(120590) lt 120573 lt 1205731(120590)
1198613 (120590 120573) | 120590 gt 1205901s 120573 = 1205731(120590)
1198614 (120590 120573) | 120590 gt 1205902s 1205731(120590) lt 120573 lt 1205732(120590)
1198615 (120590 120573) = (1205902s 1205732s)
1198616 (120590 120573) | 120590 gt 1205902s 120573 = 1205732(120590)
1198617 (120590 120573) | 120590 gt 0 120573 gt max1205731(120590) 1205732(120590)
1198618 (120590 120573) | 0 lt 120590 lt 1205902s 120573 = 1205731(120590)
1198619 (120590 120573) | 0 lt 120590 lt 1205902s 1205732(120590) lt 120573 lt 1205731(120590)
11986110 (120590 120573) | 120590 = 0 120573 gt minus120598984858
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
00 05 10
0
2
4
00 05 10
0
2
4
00 05 10
0
1
2
3
00 05 10
0
2
4
6
00 05 10
0
2
4
00 05 10
0
2
4
minus3
minus3
minus2 minus2 minus2minus1minus4
minus4
minus2 minus2
minus2
minus1
minus1
minus4minus2minus4
minus4 minus4
minus2 minus2 minus2
minus4
minus2
minus4
minus2
minus4
minus2
minus15 minus10 minus05
minus10 minus05 minus10 minus05 minus10 minus05
minus15 minus10 minus05 minus15 minus10 minus05
minus6
minus2
minus4
minus2
minus4
minus2
minus4
minus2
minus4
minus2
minus4
120593120593120593
120593 120593 120593
120593120593120593
120593 120593 120593
yy
y
y yy
yyy
y
y y
(1) (120590 120573) isin A1
(4) (120590 120573) isin A4
(7) (120590 120573) isin A7 (8) (120590 120573) isin A8 (9) (120590 120573) isin A9
(5) (120590 120573) isin A5
(2) (120590 120573) isin A2 (3) (120590 120573) isin A3
(6) (120590 120573) isin A6
(10) (120590 120573) isin A10 (11) (120590 120573) isin A11 (12) (120590 120573) isin A12
Figure 2 Continued
6 Mathematical Problems in Engineering
00 05 10
0
1
2
3
00 05 10
0
2
4
00 05 10
0
1
2
3
00 05 10
0
1
2
3
00 05 10
0
1
2
00 05 10
0
1
2
00 05 10
0
1
2
00 05 10
0
1
2
00 05 10
0
1
2
3
00 05 10
0
1
2
3
00 05 10
0
1
2
3
00 05 10
0
1
2
3
minus3
minus2
minus1
minus3
minus2
minus1
minus2
minus1
minus2
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus1
minus2
minus1
minus3
minus2
minus1
minus10 minus05
minus10 minus05
minus10 minus05
minus10 minus05
minus10 minus05
minus10minus15 minus05 minus10minus15 minus05
minus10 minus05
minus10 minus05 minus10 minus05
minus10 minus05 minus10 minus05
minus2
minus4
y
120593120593120593
120593 120593 120593
120593120593120593
120593 120593120593
yy
y yy
yy
y
yy
y
(13) (120590 120573) isin A13
(16) (120590 120573) isin A16
(19) (120590 120573) isin A19 (20) (120590 120573) isin A20
(23) (120590 120573) isin A23(24) (120590 120573) isin A24
(21) (120590 120573) isin A21
(22) (120590 120573) isin A22
(17) (120590 120573) isin A17 (18) (120590 120573) isin A18
(14) (120590 120573) isin A14 (15) (120590 120573) isin A15
Figure 2 Continued
Mathematical Problems in Engineering 7
00 05 10 15
0
1
2
3
00 05 10 15
0
1
2
3
0 1 2
0
1
2
3
0 1 2
0
1
2
3
00 05 10 15
0
1
2
3
0 1 2
0
2
4
0 1 2
0
2
4
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus2
minus4
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2 minus1
minus2minus3 minus1
minus2
minus2
minus3
minus4
minus1
minus2 minus1
minus10minus15 minus05 minus10minus15 minus05
minus10minus15 minus05
y
yyy
y y y
120593
120593120593120593
120593 120593 120593
(25) (120590 120573) isin A25
(28) (120590 120573) isin A28 (29) (120590 120573) isin A29
(31) (120590 120573) isin A31
(30) (120590 120573) isin A30
(26) (120590 120573) isin A26 (27) (120590 120573) isin A27
Figure 2 The bifurcation of phase portraits of system (15) when 120598 gt 0 and 120590 ge 0
11986111 (120590 120573) = (0 minus1205989848582)
11986112 (120590 120573) | 0 lt 120590 lt 1205902s 120573 = 1205732(120590)
11986113 (120590 120573) | 0 lt 120590 lt 1205901s 1205733(120590) lt 120573 lt min1205731(120590)
1205732(120590)
11986114 (120590 120573) | 1205904s lt 120590 lt 1205901s 120573 = 1205731(120590)
11986115 (120590 120573) = (1205901s 1205731s)
11986116 (120590 120573) | 0 lt 120590 lt 1205901s 120573 = 1205733(120590)
11986117 (120590 120573) | 0 lt 120590 lt 120598984858 120573 lt 1205733(120590)
11986118 (120590 120573) | 120590 = 0 120573 = minus120598984858
2
11986119 (120590 120573) | 120590 = 0 120573 lt 0
11986120 (120590 120573) | 0 lt 120590 lt 120598984858 120573 lt min0 1205733(120590)
11986121 (120590 120573) | 120590 gt 120598984858 120573 = 1205733(120590)
11986122 (120590 120573) | 120590 gt 120598984858 1205733(120590) lt 120573 lt 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
1205732(120590)
(1205901s 1205731s)
(1205902s 1205732s)
1205733(120590)
120573
minus2minus4minus6
1205731(120590)
120590
Figure 3 The bifurcation curves in the (120590 120573)-parameter plane for120598 lt 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 Let119862119896(Ω) denote the set of all 119896 times continuouslydifferentiable functions on the open setΩ 119871119901loc(R) refer to theset of all functions whose restriction on any compact subsetis 119871119901 integrable1198671loc(R) stands for1198671loc(R) = 120593 isin 119871
2
loc(R) |
1205931015840isin 1198712
loc(R)To study single peak soliton solutions we impose the
boundary condition
lim120585rarrplusmninfin
120593 = 119860 (22)
where 119860 is a constant In fact the constant 119860 is equal to thehorizontal coordinate of saddle point 119864(120593
119890 0) Substituting
the boundary condition (22) into (14) generates the followingconstant
ℎ = 119860(120590 +119860120598
2+1198603120573
4) (23)
So the ODE (14) becomes
(1205931015840)2
=
(120593 minus 119860)2
(1205731205932+ 2119860120573120593 + 3119860
2120573 + 2120598)
2 (9848582 minus 1205932) (24)
If 120573(1198602120573 + 120598) le 0 then (24) reduces to
(1205931015840)2
=120573 (120593 minus 119860)
2
(120593 minus 1198611) (120593 minus 119861
2)
2 (9848582 minus 1205932) (25)
where
1198611= minus119860 minus
radicminus2120573 (1198602120573 + 120598)
120573
1198612= minus119860 +
radicminus2120573 (1198602120573 + 120598)
120573
(26)
From (26) we know that 1198611gt 1198612if 120573 lt 0 and 119861
1lt 1198612if
120573 gt 0
Definition 4 A function120593(120585) is said to be a single peak solitonsolution for the 119862(3 2 2) equation (10) if 120593(120585) satisfies thefollowing conditions
(C1) 120593(120585) is continuous on R and has a unique peakpoint 120585
0 where 120593(120585) attains its global maximum or
minimum value
(C2) 120593(120585) isin 119862
3(R minus 120585
0) satisfies (24) on R minus 120585
0
(C3) 120593(120585) satisfies the boundary condition (22)
Definition 5 A wave function 120593 is called smooth solitonsolution if 120593 is smooth locally on either side of 120585
0and
lim120585uarr1205850
1205931015840(120585) = lim
120585darr12058501205931015840(120585) = 0
Definition 6 A wave function 120593 is called peakon if 120593 issmooth locally on either side of 120585
0and lim
120585uarr12058501205931015840(120585) =
minuslim120585darr1205850
1205931015840(120585) = 119886 119886 = 0 119886 = plusmninfin
Definition 7 Awave function120593 is called cuspon if120593 is smoothlocally on either side of 120585
0and lim
120585uarr12058501205931015840(120585) = minuslim
120585darr12058501205931015840(120585) =
plusmninfin
Without any loss of generality we choose the peak point1205850as vanishing 120585
0= 0
Theorem 8 Assume that 119906(119909 119905) = 120593(120585) = 120593(119909 minus 119888119905) is a singlepeak soliton solution of the 119862(3 2 2) equation (10) at the peakpoint 120585
0= 0 Then we have the following
(i) If 120573(1198602120573 + 120598) gt 0 then 120593(0) = 984858 or 120593(0) = minus984858
(ii) If 120573(1198602120573 + 120598) le 0 then 120593(0) = 984858 or 120593(0) = minus984858 or120593(0) = 119861
1or 120593(0) = 119861
2
Proof If 120593(0) = plusmn984858 then 120593(120585) = plusmn984858 for any 120585 isin R since120593(120585) isin 119862
3(R minus 0) Differentiating both sides of (24) yields
120593 isin 119862infin(R)
(i) When 120573(1198602120573 + 120598) gt 0 if 120593(0) = 984858 and 120593(0) = minus984858 then
120593 isin 119862infin(R) By the definition of single peak soliton we have
1205931015840(0) = 0 However by (24) we must have 120593(0) = 119860 which
contradicts the fact that 0 is the unique peak point(ii) When 120573(119860
2120573 + 120598) le 0 if 120593(0) = 984858 and 120593(0) = minus984858
by (24) we know 1205931015840(0) exists and 120593
1015840(0) = 0 since 0 is a peak
point Thus we obtain 120593(0) = 1198611or 120593(0) = 119861
2from (25)
since 120593(0) = 119860 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 119906(119909 119905) = 120593(119909 minus 119888119905) is a single peaksoliton solution of the 119862(3 2 2) equation (10) at the peak point1205850= 0 Then we have the following solution classification(i) If |120593(0)| = 984858 then 120593(120585) isin 119862
infin(R) and 120593 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
00 05 10 15
0
2
4
0 1
0
2
4
6
00 05 10 15
0
2
4
00 05 10
0
1
2
3
00 05 10 15
0
1
2
3
00 05 10
0
1
2
00 05 10 15
0
1
2
00 05 10 15
0
1
2
3
minus3 minus1minus4
minus4
minus2
minus4
minus2
minus4
minus2
minus4
minus2minus4
minus6
minus2
minus4
minus2
minus4
minus2
minus2 minus4 minus4minus2 minus2
minus3
minus1
minus2
minus1
minus2
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus1minus2 minus1minus2
120593
120593
120593
120593 120593 120593
120593 120593
120593 120593
120593 120593
y
y
y
yy
y
y y
y y
y y
minus15 minus10 minus05
minus15minus20 minus10 minus05
minus15 minus10 minus05 minus15 minus10 minus05minus10 minus05
minus15 minus10 minus05minus10 minus05
(10) (120590 120573) isin B10 (11) (120590 120573) isin B11 (12) (120590 120573) isin B12
(1) (120590 120573) isin B1
(4) (120590 120573) isin B4
(7) (120590 120573) isin B7 (8) (120590 120573) isin B8 (9) (120590 120573) isin B9
(5) (120590 120573) isin B5 (6) (120590 120573) isin B6
(2) (120590 120573) isin B2 (3) (120590 120573) isin B3
Figure 4 Continued
10 Mathematical Problems in Engineering
00 05 10 15
0
1
2
3
0 1
0
1
2
3
0 1 2
0
2
4
00 05 10 15
0
1
2
3
0 1 2
0
1
2
3
00 05 10 15
0
1
2
00 05 10 15
0
1
2
3
0 1 2
0
1
2
3
00 05 10 15
0
2
4
00 05 10 15
0
2
4
6
minus4
minus2
minus4
minus2
minus4
minus6
minus2
minus3 minus1minus2
minus3 minus1minus2
minus1minus2
minus1minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus1
minus2
minus3
minus1
minus2
120593 120593 120593
120593120593120593
120593 120593 120593
120593
y y y
yy
y
y yy
y
minus15 minus10 minus05
minus15minus20 minus10 minus05
minus15 minus10 minus05
minus15 minus10 minus05
minus15 minus10 minus05
minus15 minus10 minus05
(13) (120590 120573) isin B13 (14) (120590 120573) isin B14 (15) (120590 120573) isin B15
(16) (120590 120573) isin B16 (17) (120590 120573) isin B17 (18) (120590 120573) isin B18
(22) (120590 120573) isin B22
(20) (120590 120573) isin B20 (21) (120590 120573) isin B21(19) (120590 120573) isin B19
Figure 4 The bifurcation of phase portraits of system (15) when 120598 lt 0 and 120590 ge 0
Mathematical Problems in Engineering 11
(ii) If 120593(0) = 984858 = 1198611 then 120593 is a cuspon solution and 120593 has
the following asymptotic behavior
120593 (120585) minus 984858 = 1205821
10038161003816100381610038161205851003816100381610038161003816
23
+ 119874(10038161003816100381610038161205851003816100381610038161003816
43
) 120585 997888rarr 0
1205931015840(120585) =
2
31205821
10038161003816100381610038161205851003816100381610038161003816
minus13 sgn (120585) + 119874 (10038161003816100381610038161205851003816100381610038161003816
13
)
120585 997888rarr 0
(27)
where 1205821= plusmn(9|120573984858
2+ 2119860120573984858 + 3119860
2120573 + 2120598|(984858 minus 119860)
216984858)13
Thus 120593(120585) notin 1198671
119897119900119888(R)
(iii) If 120593(0) = minus984858 = 1198612 then 120593 is a cuspon solution and 120593
has the following asymptotic behavior
120593 (120585) + 984858 = 1205822
10038161003816100381610038161205851003816100381610038161003816
23
+ 119874(10038161003816100381610038161205851003816100381610038161003816
43
) 120585 997888rarr 0
1205931015840(120585) =
2
31205822
10038161003816100381610038161205851003816100381610038161003816
minus13 sgn (120585) + 119874 (10038161003816100381610038161205851003816100381610038161003816
13
)
120585 997888rarr 0
(28)
where 1205822= plusmn(9|120573984858
2minus 2119860120573984858 + 3119860
2120573 + 2120598|(984858 + 119860)
216984858)13
Thus 120593(120585) notin 1198671
119897119900119888(R)
(iv) If 120593(0) = 984858 = 1198611and 119860 = 0 then 120593 is a peakon-like
solution and
120593 (120585) minus 984858 = 1205823
10038161003816100381610038161205851003816100381610038161003816 + 119874 (
10038161003816100381610038161205851003816100381610038161003816
2
) 120585 997888rarr 0
1205931015840(120585) = 120582
3sgn (120585) + 119874 (
10038161003816100381610038161205851003816100381610038161003816) 120585 997888rarr 0
(29)
where 1205823= plusmn|119860 minus 984858|radic120573(119861
2minus 984858)4984858
(v) If 120593(0) = minus984858 = 1198612and 119860 = 0 then 120593 is a peakon-like
solution and
120593 (120585) + 984858 = 1205824
10038161003816100381610038161205851003816100381610038161003816 + 119874 (
10038161003816100381610038161205851003816100381610038161003816
2
) 120585 997888rarr 0
1205931015840(120585) = 120582
4sgn (120585) + 119874 (
10038161003816100381610038161205851003816100381610038161003816) 120585 997888rarr 0
(30)
where 1205824= plusmn|119860 + 984858|radicminus120573(119861
1+ 984858)4984858
(vi) If 120593(0) = 984858 = 1198611and 119860 = 0 then 120593 gives the peakon
solution 984858 exp(minusradicminus1205732|119909 minus 119888119905|)(vii) If 120593(0) = minus984858 = 119861
2and 119860 = 0 then 120593 gives the peakon
solution minus984858 exp(minusradicminus1205732|119909 minus 119888119905|)
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 |120593(0)| = 984858 then 120593 isin 119862
infin(R) and 120593 is a smooth soliton
solution(ii) If 120593(0) = 984858 = 119861
1 then by the definition of single peak
soliton we have 119860 = 984858 thus 1205731205932 + 2119860120573120593 + 31198602120573 + 2120598 does
not contain the factor 120593 minus 984858 From (24) we obtain
1205931015840= sgn (119860 minus 984858)
1003816100381610038161003816120593 minus 1198601003816100381610038161003816
radic210038161003816100381610038161205932 minus 9848582
1003816100381610038161003816
sdot radic10038161003816100381610038161205731205932 + 2119860120573120593 + 31198602120573 + 2120598
1003816100381610038161003816 sgn (120585)
(31)
Let 1198971(120593) = radic2(120593 + 984858)|120593 minus 119860|radic|1205731205932 + 2119860120573120593 + 31198602120573 + 2120598|
then 1198971(984858) = 2radic984858|984858 minus 119860|radic|1205739848582 + 2119860120573984858 + 31198602120573 + 2120598| and
int 1198971(120593)radic
1003816100381610038161003816120593 minus 9848581003816100381610038161003816119889120593 = int sgn (119860 minus 984858) sgn (120585) 119889120585 (32)
Inserting 1198971(120593) = 119897
1(984858) + 119874(|120593 minus 984858|) into (32) and using the
initial condition 120593(0) = 984858 we obtain
21198971(984858)
3
1003816100381610038161003816120593 minus 9848581003816100381610038161003816
32
(1 + 119874 (1003816100381610038161003816120593 minus 984858
1003816100381610038161003816)) =10038161003816100381610038161205851003816100381610038161003816
(33)
thus
120593 minus 984858 = plusmn(3
21198971(984858)
)
23
10038161003816100381610038161205851003816100381610038161003816
23
(1 + 119874 (1003816100381610038161003816120593 minus 984858
1003816100381610038161003816))minus23
= plusmn(3
21198971(984858)
)
23
10038161003816100381610038161205851003816100381610038161003816
23
(1 + 119874 (1003816100381610038161003816120593 minus 984858
1003816100381610038161003816))
(34)
which implies 120593 minus 984858 = 119874(|120585|23
) Therefore we have
120593 (120585) = 984858 plusmn (3
21198971(984858)
)
23
10038161003816100381610038161205851003816100381610038161003816
23
+ 119874(10038161003816100381610038161205851003816100381610038161003816
43
)
= 984858 + 1205821
10038161003816100381610038161205851003816100381610038161003816
23
+ 119874(10038161003816100381610038161205851003816100381610038161003816
43
) 120585 997888rarr 0
1205821= plusmn(
3
21198971(984858)
)
23
= plusmn(
9100381610038161003816100381610038161205739848582+ 2119860120573984858 + 3119860
2120573 + 2120598
10038161003816100381610038161003816(984858 minus 119860)
2
16984858)
13
1205931015840(120585) =
2
31205821
10038161003816100381610038161205851003816100381610038161003816
minus13 sgn (120585) + 119874 (10038161003816100381610038161205851003816100381610038161003816
13
) 120585 997888rarr 0
(35)
So 120593(120585) notin 1198671
loc(R)(iii) Similar to the proof of (ii) we ignore it in this paper(iv) If 120593(0) = 984858 = 119861
1and 119860 = 0 then from (25) we obtain
1205931015840= radicminus
120573
2sgn (119860 minus 984858)
1003816100381610038161003816120593 minus 1198601003816100381610038161003816radic120593 minus 1198612
120593 + 984858sgn (120585) (36)
Let 1198972(120593) = (1|120593minus119860|)radic(120593 + 984858)(120593 minus 119861
2) then 119897
2(984858) = (1|984858minus
119860|)radic2984858(984858 minus 1198612) and
int 1198972(120593) 119889120593 = radicminus
120573
2int sgn (119860 minus 984858) sgn (120585) 119889120585 (37)
Inserting 1198972(120593) = 119897
2(984858) + 119874(|120593 minus 984858|) into (37) and using the
initial condition 120593(0) = 984858 we obtain
1198972(984858) (120593 minus 984858) (1 + 119874 (
1003816100381610038161003816120593 minus 9848581003816100381610038161003816))minus1
= sgn (119860 minus 984858)10038161003816100381610038161205851003816100381610038161003816
(38)
Sincesgn (120593 minus 984858) sgn (119860 minus 984858) ge 0
1
1 + 119874 (120593 minus 984858)= 1 + 119874 (120593 minus 984858)
(39)
12 Mathematical Problems in Engineering
we get
1003816100381610038161003816120593 minus 9848581003816100381610038161003816 =
radicminus120573
2
1
1198972(984858)
10038161003816100381610038161205851003816100381610038161003816 (1 + 119874 (120593 minus 984858)) (40)
which implies |120593 minus 984858| = 119874(|120585|) Therefore we have
120593 (120585) = 984858 + 1205823
10038161003816100381610038161205851003816100381610038161003816 + 119874 (
10038161003816100381610038161205851003816100381610038161003816
2
) 120585 997888rarr 0
1205931015840(120585) = 120582
3sgn (120585) + 119874 (
10038161003816100381610038161205851003816100381610038161003816) 120585 997888rarr 0
(41)
where 1205823= plusmn|119860 minus 984858|radic120573(119861
2minus 984858)4984858
(v) Similar to the proof of (iv) we ignore it in this paper
By virtue of Theorem 9 any single peak soliton for the119862(3 2 2) equation (10) must satisfy the following initial andboundary values problem
(1205931015840)2
=120573 (120593 minus 119860)
2
(120593 minus 1198611) (120593 minus 119861
2)
2 (9848582 minus 1205932)fl119871 (120593)
120593 (0) isin 984858 minus984858 1198611 1198612
lim|120585|rarrinfin
120593 (120585) = 119860
(42)
119871(120593) ge 0 and the boundary condition (24) imply thefollowing
(a) If 120573(9848582 minus 1205932) ge 0 then 120593 ge max119861
1 1198612 or 120593 le
min1198611 1198612
(b) If120573(9848582minus1205932) le 0 thenmin1198611 1198612 le 120593 le max119861
1 1198612
Below we will present some implicit formulas for thesingle peak soliton solutions in the case of specific 120590 and 120573
Case 1 ((120590 120573) isin 11986012) In this case we have minus984858 lt 119861
2lt 119860 lt
1198611= 984858 From the standard phase analysis and Theorem 9
we know that if 120593 is a single peak soliton of the 119862(3 2 2)
equation then
1205931015840= minusradicminus
120573
2(120593 minus 119860)radic
120593 minus 1198612
120593 + 984858sgn (120585) (43)
From the separation of variables we get
intℎ (120593) 119889120593 = radicminus120573
2sgn (120585) 119889120585 (44)
where ℎ(120593) = (1(119860 minus 120593))radic(120593 + 984858)(120593 minus 1198612) After a lengthy
calculation of integral we obtain the implicit solution 120593
defined by
119867(120593)flradic119860 + 984858
119860 minus 1198612
1198681(120593) minus 119868
2(120593) = radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870 (45)
where
1198681(120593) = ln
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
(119860 + 984858) (120593 minus 1198612) + (119860 minus 119861
2) (120593 + 984858) + 2radic(119860 + 984858) (119860 minus 119861
2) (120593 minus 119861
2) (120593 + 984858)
radic(119860 + 984858) (119860 minus 1198612) (119860 + 984858) (119860 minus 120593)
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
1198682(120593) = ln
1003816100381610038161003816100381610038161003816(120593 + 984858) + (120593 minus 119861
2) + 2radic(120593 + 984858) (120593 minus 119861
2)
1003816100381610038161003816100381610038161003816
(46)
and 119870 is an arbitrary integration constant For 120593(0) = 984858 theconstant119870 = 119867(120593(0)) is defined by
119870 = radic119860 + 984858
119860 minus 1198612
1198681(984858) minus 119868
2(984858) (47)
and for 120593(0) = 1198612
119870 = radic119860 + 984858
119860 minus 1198612
1198681(1198612) minus 1198682(1198612) (48)
(i) If 120593(0) = 984858 then 119860 lt 120593 le 984858 Since 1198671015840(120593) = ℎ(120593) weknow that119867(120593) strictly decreases on the interval (119860 984858] thus1198671(120593) = 119867|
(119860984858](120593) gives a single peak solitonwith119867
1(984858) = 119870
and1198671(119860+) = infin Therefore 120593
1(120585) = 119867
minus1
1(radicminus1205732|120585| + 119870) is
the solution satisfying
1205931(0) = 984858
lim|120585|rarrinfin
1205931(120585) = 119860
1205931015840
1(0plusmn) = plusmn (119860 minus 984858)radic
120573 (1198612minus 984858)
4984858
(49)
So 1205931(120585) is a peakon-like solution (see Figure 5)
(ii) If 120593(0) = 1198612 then 119861
2le 120593 lt 119860 By 119867
1015840(120593) = 119891(120593)
we know that 119867(120593) strictly increases on the interval [1198612 119860)
Thus
1198672(120593) = 119867|
[1198612 119860)(120593) (50)
Mathematical Problems in Engineering 13
minus3 minus2 minus1
120593
minus02
1 2 3
02
04
06
08
10
x
(a)
02
02
minus2 minus2 x
10
05
00
t
120593
22
(b)
Figure 5 Two- and three-dimensional graphs of the peakon-like solution
minus03
minus04
minus05
1 2 3minus3 minus2 minus1
120593
minus02
x
(a)
0
2
0
2
minus2minus2
minus02
minus03
minus04
minus05
minus06
t x
120593
0
2
0
2
minus2minus2
t x
(b)
Figure 6 Two- and three-dimensional graphs of the smooth soliton solution
has the inverse denoted by 1205932(120585) = 119867
minus1
2(radicminus1205732|120585| + 119870)
1205932(120585) gives a kind of smooth soliton solution (see Figure 6)
satisfying
1205932(0) = 119861
2
lim|120585|rarrinfin
1205932(120585) = 119860
1205931015840
2(0) = 0
(51)
Case 2 ((120590 120573) isin 11986014) In this case we have minus984858 = 119861
2lt 119860 lt
984858 lt 1198611and (25) is equivalent to
1205931015840= minusradicminus
120573
2(120593 minus 119860)radic
120593 minus 1198611
120593 minus 984858sgn (120585) (52)
Let
119892 (120593) =1
119860 minus 120593radic120593 minus 1198611
120593 minus 984858 (53)
14 Mathematical Problems in Engineering
minus1 1 2 3
02
04
06
08
10
minus3 minus2
120593
x
(a)
02
02minus2
minus2
120593
x t
00
05
10
(b)
Figure 7 Two- and three-dimensional graphs of the cuspon solution
then (52) is converted to
119892 (120593) 119889120593 =1
119860 minus 120593radic120593 minus 1198611
120593 minus 984858119889120593 = radicminus
120573
2sgn (120585) 119889120585 (54)
Integrating (54) on the interval [0 120585] (or [120585 0]) leads to thefollowing implicit solutions
119866 (120593)flradic119860 minus 984858
119860 minus 1198611
1198683(120593) + 119868
4(120593) = radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870 (55)
where
1198683(120593) = ln
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
(119860 minus 984858) (120593 minus 1198611) + (119860 minus 119861
1) (120593 minus 984858) + 2radic(119860 minus 984858) (119860 minus 119861
1) (120593 minus 984858) (120593 minus 119861
1)
radic(119860 minus 984858) (119860 minus 1198611) (119860 minus 984858) (120593 minus 119860)
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
1198684(120593) = ln
1003816100381610038161003816100381610038161003816(120593 minus 984858) + (120593 minus 119861
1) + 2radic(120593 minus 984858) (120593 minus 119861
1)
1003816100381610038161003816100381610038161003816
(56)
And119870 is an arbitrary integration constant It is obvious thatfor 120593(0) = 984858 the constant119870 = 119867(120593(0)) is defined by
119870 = radic119860 minus 984858
119860 minus 1198611
1198683(984858) + 119868
4(984858) (57)
and for 120593(0) = minus984858
119870 = radic119860 minus 984858
119860 minus 1198611
1198683(minus984858) + 119868
4(minus984858) (58)
(i) If 120593(0) = 984858 then 119860 lt 120593 le 984858 From 119892(120593) lt 0 weknow that 119866(120593) strictly decreases on the interval (119860 984858] with119866(984858) = 119870 and 119866(119860+) = infin Define
1198661(120593) = 119866|
(119860984858](120593) (59)
Since 1198661(120593) is a strictly decreasing function from (119860 984858] onto
[119870infin) we can solve for 120593 uniquely from (59) and obtain
1205931(120585) = 119866
minus1
1(radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870) (60)
It is easy to check that 120593 satisfies
1205931(0) = 984858
lim|120585|rarrinfin
1205931(120585) = 119860
1205931015840
1(0plusmn) = ∓infin
(61)
Therefore the solution1205931defined by (60) is a cuspon solution
for the 119862(3 2 2) equation (see Figure 7)
Mathematical Problems in Engineering 15
(ii) If 120593(0) = minus984858 then minus984858 le 120593 lt 119860 Through a similaranalysis we get a strictly increasing function 119866(120593) on theinterval [minus984858 119860) satisfying
119866 (120593) = radicminus120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870 (62)
where 119866(120593) is defined by (55) Let
1198662(120593) = 119866|
[minus984858119860)(120593) (63)
then 1198662(120593) is a strictly increasing function from [minus984858 119860) onto
[119870infin) so that we can solve for 120593 and obtain
1205932(120585) = 119866
minus1
2(radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870) (64)
It is easy to check that 1205932satisfies
1205932(0) = minus984858
lim|120585|rarrinfin
1205932(120585) = 119860
1205931015840
2(0plusmn) = plusmn (119860 + 984858)radicminus
120573 (1198611+ 984858)
4984858
(65)
Therefore the solution 1205932defined by (64) is a peakon-like
solution whose graph is similar to those in Figure 5
Case 3 ((120590 120573) isin 11986016) In this case we have minus984858 lt 119861
2lt 119860 =
0 lt 1198611lt 984858 119861
2= minus1198611 and
1205931015840= minusradicminus
120573
2120593radic
120593 minus 1198612
1
1205932 minus 9848582sgn (120585) (66)
Hence from the separation of variables we have
radic9848582 minus 1205932
120593radic1198612
1minus 1205932
119889120593 = minusradicminus120573
2sgn (120585) 119889120585 (67)
Integrating (67) on the interval [0 120585] (or [120585 0]) leads tothe following implicit formula for the two smooth solitonsolutions
radic9848582 minus 1205932 + radic1198612
1minus 1205932
radic9848582 minus 1198612
1
(
radic(9848582 minus 1198612
1) 1205932
984858radic1198612
1minus 1205932 + 119861
1radic9848582 minus 1205932
)
984858
= exp(minusradicminus120573
2|119909 minus 119888119905|)
(68)
where 120593 isin (119860 1198611] Consider
radic9848582 minus 1198612
2
radic9848582 minus 1205932 + radic1198612
2minus 1205932
(
984858radic1198612
2minus 1205932 minus 119861
2radic9848582 minus 1205932
radic(9848582 minus 1198612
2) 1205932
)
984858
= exp(radicminus120573
2|119909 minus 119888119905|)
(69)
where 120593 isin [1198612 119860)
Case 4 ((120590 120573) isin 11986017) In this case we have minus984858 = 119861
2lt 119860 =
0 lt 1198611= 984858 and
1205931015840= minusradicminus
120573
2120593 sgn (120585) (70)
Choosing 120593(0) = 984858 (or minus984858) as initial value we get
int
120593
984858
1
120593119889120593 = minusint
120585
0
radicminus120573
2sgn (120585) 119889120585 for 119860 lt 120593 le 984858
int
120593
minus984858
1
120593119889120593 = minusint
120585
0
radicminus120573
2sgn (120585) 119889120585 for minus 984858 le 120593 lt 119860
(71)
which immediately yield the peakon solutions
120593 (119909 minus 119888119905) = plusmn984858 exp(minusradicminus120573
2|119909 minus 119888119905|) (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 the119862(3 2 2) equation (10) In order to study kinkwave solutionswe assume that
lim120585rarrinfin
120593 (120585) = 1198601
lim120585rarrminusinfin
120593 (120585) = 1198602
(73)
16 Mathematical Problems in Engineering
2 4
02
04
06
08
10
minus2minus4
120593
x
(a)
0 2
02
00
05
10
minus2
120593
x
t
0
2
minus2
(b)
Figure 8 Two- and three-dimensional graphs of the peakon solution
where 1198601gt 1198602 Substituting the boundary condition (73)
into (14) generates
(1205931015840)2
=
120573 (120593 minus 1198601) (120593 minus 119860
2) [1205932+ (1198601+ 1198602) 120593 + 119860
2
1+ 11986011198602+ 1198602
2+ 2120598120573]
2 (9848582 minus 1205932)fl119865 (120593) (74)
The nonlinear differential equation (74) may sustaindifferent kinds of nonlinear excitations In what follows weconfine our attention to the cases 119860
2= minus119860
1and 120598 =
minus1198602
1120573 which describe kinks and kink compactons which
play an important role in the dynamics systems Under theseconsiderations (74) reduces to
(1205931015840)2
=120573 (120593 minus 119860
1)2
(120593 minus 1198602)2
2 (9848582 minus 1205932) (75)
If 1198601lt 984858 then from the phase analysis in Section 2 (see
Figure 4(10)) we know that (1198601 0) and (119860
2 0) are two saddle
points of (13) and the kink solutions can be obtained fromthe two heteroclinic orbits connecting (120601 119910) = (119860
1 0) and
(1198602 0) When 119860
1increases upon reaching 984858 that is 119860
1= 984858
(75) becomes
(1205931015840)2
= minus120573 (120593 minus 119860
1) (120593 minus 119860
2)
2 (76)
and the ellipse 21199102+120573(120593minus1198601)(120593minus119860
2) = 0 (see Figure 4(11))
which is tangent to the singular lines 120593 = 984858 and 120593 = minus984858 atpoints (119860
1 0) and (119860
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 120593 is a kink wave solutions of (75)
or (76) we have 1205931015840
rarr 0 as 120593 rarr 1198601and as 120593 rarr 119860
2
Moreover we have 119865(120593) ge 0 for 1198602le 120593 le 119860
1and 120593 is
strictly monotonic in any interval where 119865(120593) gt 0 Thus if1205931015840gt 0 at some points 120593will be strictly increasing until it gets
close to the next zero of 119865 Denoting this zero 1198601 we have
120593 uarr 1198601 What will happen to the solution when it approaches
1198601 Depending on whether the zero is double or simple 120593
has a different behavior We explore the two cases in turn
Theorem 11 (i) If 120593 has a simple zero at 120593 = 1198601 so that
119865(1198601) = 0 and 1198651015840(119860
1) lt 0 then the solution 120593 of (75) satisfies
120593 (120585) minus 1198601=1198651015840(1198601)
4(120585 minus 120578)
2
+ 119874 ((120585 minus 120578)4
)
119886119904 120585 uarr 120578
(77)
where 120593(120578) = 1198601
(ii) When 120593 approaches the double zero 1198601of 119865(120593) so that
1198651015840(1198601) = 0 and11986510158401015840(119860
1) gt 0 then the solution120593 of (75) satisfies
120593 (120585) minus 1198601sim 120572 exp(minus120585radic11986510158401015840 (119860
1)) 119886119904 120585 997888rarr infin (78)
for some constant 120572 Thus 120593 uarr 1198601exponentially as 120585 rarr infin
Mathematical Problems in Engineering 17
Proof (i) When 1198601= minus119860
2= 984858 and 120598 = minus119860
2
1120573 from (76)
119865(120593) has a simple zero at 120593 = 1198601 Then
(1205931015840)2
= (120593 minus 1198601) 1198651015840(1198601) + 119874 ((120593 minus 119860
1)2
)
as 120593 uarr 1198601
(79)
Using the fact that (1205931015840)2 ge 0 we know that 1198651015840(1198601) lt 0
Moreover
119889120585
119889120593=
1
radic(120593 minus 1198601) 1198651015840 (119860
1) + 119874 ((120593 minus 119860
1)2
)
(80)
Because
radic(120593 minus 1198601) 1198651015840 (119860
1) + 119874 ((120593 minus 119860
1)2
)
= radic1198601minus 120593(radicminus1198651015840 (119860
1) + 119874 (119860
1minus 120593))
1
radicminus1198651015840 (1198601) + 119874 (119860
1minus 120593)
=1
radicminus1198651015840 (1198601)
+ 119874 (1198601minus 120593)
(81)
we obtain119889120585
119889120593=
1
radicminus (1198601minus 120593) 1198651015840 (119860
1)
+ 119874 ((1198601minus 120593)12
) (82)
Integrating (82) yields
120578 minus 120585 =2
radicminus1198651015840 (1198601)
(1198601minus 120593)12
+ 119874((1198601minus 120593)32
) (83)
where 120578 satisfies 120593(120578) = 1198601 Thus
(120578 minus 120585)2
=4
minus1198651015840 (1198601)(1198601minus 120593) + 119874 ((119860
1minus 120593)2
) (84)
which implies 119874((1198601minus 120593)2) = 119874((120578 minus 120585)
4) Therefore we get
120593 (120585) = 1198601+1198651015840(1198601)
4(120585 minus 120578)
2
+ 119874 ((120585 minus 120578)4
)
as 120585 uarr 120578
(85)
where 120593(120578) = 1198601
(ii) When 1198601= minus119860
2lt 984858 and 120598 = minus119860
2
1120573 from (75) 119865(120593)
has a double zero at 120593 = 1198601 Then
(1205931015840)2
= (120593 minus 1198601)2
11986510158401015840(1198601) + 119874 ((120593 minus 119860
1)3
)
as 120593 uarr 1198601
(86)
Furthermore we get
119889120585
119889120593=
1
radic(120593 minus 1198601)2
11986510158401015840 (1198601) + 119874 ((120593 minus 119860
1)3
)
(87)
Observing that
radic(120593 minus 1198601)2
11986510158401015840 (1198601) + 119874 ((120593 minus 119860
1)3
)
= (1198601minus 120593) (radic11986510158401015840 (119860
1) + 119874 (119860
1minus 120593))
1
radic11986510158401015840 (1198601) + 119874 (119860
1minus 120593)
=1
radic11986510158401015840 (1198601)
+ 119874 (1198601minus 120593)
(88)
we obtain119889120585
119889120593=
1
(1198601minus 120593) 11986510158401015840 (119860
1)+ 119874 (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 119860
1= minus119860
2lt 984858 and 120598 = minus119860
2
1120573 Then (75)
becomes
radic9848582 minus 1205932
1205932 minus 1198602
1
119889120593 = ∓radic120573
2119889120585 (90)
Integrating both sides of (90) gives the following implicitexpressions of kink and antikink wave solutions
arctan120593
radic9848582 minus 1205932
+
radic9848582 minus 1198602
1
1198601
arctanhradic9848582 minus 119860
2
1
1198601
120593
radic9848582 minus 1205932
= plusmnradic120573
2120585
(91)
By letting 1198601rarr 984858 in (91) we get two kink compactons
which are given by
plusmn120593 =
984858 sinradic120573
2120585
10038161003816100381610038161205851003816100381610038161003816 le
120587
radic2120573
984858 120585 gt120587
radic2120573
minus984858 120585 lt minus120587
radic2120573
(92)
The graphs for the kink wave solutions (91) and kinkcompacton solutions (92) are shown in Figures 9 and 10respectively
Remark 12 The two kink compacton solutions (92) aredifferent from the well-known smooth kink wave solutionsIn comparison with kink wave solutions (91) the kink com-pacton solutions (92) have no exponential decay propertiesbut have compact supportThat is they minus a constant thedifferences identically vanish outside a finite core region
18 Mathematical Problems in Engineering
120593
120585minus1 1
05
minus05
(a)
minus1minus1
minus05
05
00120593
x t
0
1
01
11
(b)
Figure 9 Two- and three-dimensional graphs of the kink wave solution
1
1120593
120585
minus1
minus1
(a)
0
1
0
1
000510
120593
minus1 minus1
minus05
minus10
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 the119862(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 Newtonrsquos equation which we write in the form 119910
2=
119865(120593) where 119910 = 1205931015840
120585(120585) we solve Newtonrsquos equation 119910
2=
119865(120593) 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 the119862(3 2 2)equation
Actually for 119898 = 2119896 + 1 119896 isin N+ in 119862(119898 2 2) equation(9) the dynamical behavior of traveling wave solutions of(9) is similar to the case 119898 = 3 for 119898 = 2119896 119896 isin N+ in119862(119898 2 2) equation (9) the dynamical behavior of travelingwave solutions of (9) is similar to the case 119898 = 2 We areapplying the approach mentioned in this work to 119862(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
References
[1] C Rogers and W R Shadwick Backlund Transformation andTheir Applications Academic Press New York NY USA 1982
[2] W Malfliet and W Hereman ldquoThe tanh method I Exactsolutions of nonlinear evolution and wave equationsrdquo PhysicaScripta vol 54 no 6 pp 563ndash568 1996
[3] R Hirota Direct Method in Soliton Theory Springer BerlinGermany 1980
[4] J S Cohen Computer Algebra and Symbolic ComputationMathematical Methods A K Peters 2003
[5] S Lie ldquoUber die integration durch bestimmte integrale voneiner klasse linearer partieller differential eichungenrdquoArchiv derMathematik vol 6 pp 328ndash368 1881
[6] H-H Dai and X-H Zhao ldquoTraveling waves in a rod composedof amodifiedMooney-Rivlinmaterial I bifurcation of equilibraand non-singular caserdquo Proceedings of the Royal Society AMathematical Physical and Engineering Sciences vol 455 no1990 pp 3845ndash3874 1999
[7] H-H Dai and F-F Wang ldquoAsymptotic bifurcation solutionsfor compressions of a clamped nonlinearly elastic rectangletransition region and barrelling to a corner-like profilerdquo SIAMJournal on Applied Mathematics vol 70 no 7 pp 2673ndash26922010
[8] J Li and Z Liu ldquoSmooth and non-smooth traveling waves ina nonlinearly dispersive equationrdquo Applied Mathematical Mod-elling vol 25 no 1 pp 41ndash56 2000
[9] A Chen and J Li ldquoSingle peak solitary wave solutions forthe osmosis K(22) equation under inhomogeneous boundaryconditionrdquo Journal of Mathematical Analysis and Applicationsvol 369 no 2 pp 758ndash766 2010
[10] R Camassa and D D Holm ldquoAn integrable shallow waterequation with peaked solitonsrdquo Physical Review Letters vol 71no 11 pp 1661ndash1664 1993
[11] P Rosenau ldquoOn nonanalytic solitary waves formed by a nonlin-ear dispersionrdquo Physics Letters A vol 230 no 5-6 pp 305ndash3181997
[12] P Rosenau and J M Hyman ldquoCompactons solitons with finitewavelengthrdquo Physical Review Letters vol 70 no 5 pp 564ndash5671993
[13] L Zhang L-Q Chen and X Huo ldquoThe effects of horizontalsingular straight line in a generalized nonlinear Klein-Gordonmodel equationrdquo Nonlinear Dynamics vol 72 no 4 pp 789ndash801 2013
[14] J Li and H Dai On the Study of Singular Nonlinear TravelingWave Equations Dynamical System Approach Science PressBeijing China 2007
[15] A Chen S Wen S Tang W Huang and Z Qiao ldquoEffects ofquadratic singular curves in integrable equationsrdquo Studies inApplied Mathematics vol 134 no 1 pp 24ndash61 2015
[16] L Tian and J Yin ldquoNew compacton solutions and solitary wavesolutions of fully nonlinear generalized Camassa-Holm equa-tionsrdquo Chaos Solitons amp Fractals vol 20 no 2 pp 289ndash2992004
[17] L Zhang ldquoThe bifurcation and peakons for the specialC(3 2 2)equationrdquo Pramana vol 83 no 3 pp 331ndash340 2014
[18] J Lenells ldquoTraveling wave solutions of the CamassandashHolmequationrdquo Journal of Differential Equations vol 217 no 2 pp393ndash430 2005
[19] Z J Qiao and G P Zhang ldquoOn peaked and smooth solitons forthe Camassa-Holm equationrdquo Europhysics Letters vol 73 no 5pp 657ndash663 2006
Submit your manuscripts athttpwwwhindawicom
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Mathematical Problems in Engineering
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Stochastic AnalysisInternational Journal of
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]
119906119905minus 119906119905119909119909
+ 3119906119906119909= 2119906119909119906119909119909
+ 119906119906119909119909119909
(4)
arising as a model for nonlinear waves in cylindrical axiallysymmetric hyperelastic rods with 119906(119909 119905) representing theradial stretch relative to a prestressed state where Camassaand Holm showed that (4) has a peakon of the form 119906(119909 119905) =
119888119890minus|119909minus119888119905| 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
119906119905+ (119906119898)119909+ (119906119899)119909119909119909
= 0 (5)
This equation denoted by 119870(119898 119899) owns the property thatfor certain119898 and 119899 its solitary wave solutions have compactsupport [12] That is they identically vanish outside a finitecore region For instance the 119870(2 2) equation admits thefollowing compacton solution
119906 (119909 119905) =
4119888
3cos2 (119909 minus 119888119905
4) |119909 minus 119888119905| le 2120587
0 otherwise(6)
The Camassa-Holm equation the 119870(2 2) equation andalmost all integrable dispersive equations have the sameclass of traveling wave systems which can be written in thefollowing form [13]
119889120593
119889120585= 119910 =
1
1198632 (120593)
120597119867
120597119910
119889119910
119889120585= minus
1
1198632 (120593)
120597119867
120597120593= minus
1198631015840(120593) 1199102+ 119892 (120593)
1198632 (120593)
(7)
where 119867 = 119867(120593 119910) = (12)11991021198632(120593) + int119863(120593)119892(120593)119889120593 is the
first integral It is easy to see that (4) is actually a special case of(3) with 119865(120593 119910) = minus(1119863
2(120593))(120597119867120597120593) If there is a function
120593 = 120593119904such that 119863(120593
119904) = 0 then 120593 = 120593
119904is a vertical straight
line solution of the system
119889120593
119889120577= 119910119863 (120593)
119889119910
119889120577= minus1198631015840(120593) 1199102minus 119892 (120593)
(8)
where 119889120585 = 119863(120593)119889120577 for 120593 = 120593119904 The two systems have
the same topological phase portraits except for the verticalstraight line 120593 = 120593
119904and 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 120593 = 120593
119904 However the orbits
which do intersect with the vertical straight line 120593 = 120593119904or
are unbounded but can approach the vertical straight linecorrespond 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 [6ndash9] But tillnow there have been few works on the integrable nonlinearequations with two singular straight lines or other types ofsingular curves [13ndash15]
In 2004 Tian and Yin [16] introduced the following fullynonlinear generalized Camassa-Holm equation 119862(119898 119899 119901)
119906119905+ 119896119906119909+ 1205741119906119909119909119905
+ 1205742(119906119898)119909+ 1205743119906119909(119906119899)119909119909
+ 1205744119906 (119906119901)119909119909119909
= 0
(9)
where 119896 1205741 1205742 1205743 and 120574
4are arbitrary real constants and 119898
119899 and 119901 are positive integers By using four direct ansatzsthey obtained kink compacton solutions nonsymmetry com-pacton solutions and solitary wave solutions for the119862(2 1 1)and 119862(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 119898 = 3119899 = 119901 = 2 and 120574
3= 1205744 Then (9) reduces to the 119862(3 2 2)
equation
119906119905+ 119896119906119909+ 1205741119906119909119909119905
+ 1205742(1199063)119909+ 1205743119906119909(1199062)119909119909
+ 1205743119906 (1199062)119909119909119909
= 0
(10)
Actually we have already considered a special 119862(3 2 2)equation in [17] namely 120574
1= minus1 120574
2= minus3 and 120574
3= minus1
where the bifurcation of peakons are obtained by applyingthe qualitative theory of dynamical systems In this worka more general 119862(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 instancepeakons cuspons compactons kinks and kink-compactons
This paper is organized as follows In Section 2we 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 solutionscuspon 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 119906(119909 119905) = 120593(120585)
with 120585 = 119909 minus 119888119905 be the solution of (10) then it follows that
(119896 minus 119888) 1205931015840minus 1205741119888120593101584010158401015840
+ 1205742(1205933)1015840
+ 12057431205931015840(1205932)10158401015840
+ 1205743120593 (1205932)101584010158401015840
= 0
(11)
where 1205931015840 = 120593120585 12059310158401015840 = 120593
120585120585 and 120593101584010158401015840 = 120593
120585120585120585 Integrating (11) once
and setting the integration constant as 119892 we have
(119896 minus 119888) 120593 minus 120574111988812059310158401015840+ 12057421205933+ 1205743120593 (1205932)10158401015840
= minus119892 (12)
Clearly (12) is equivalent to the planar system
119889120593
119889120585= 119910
119889119910
119889120585=1205731205933+ 1205931199102+ 120598120593 + 120590
120588 minus 1205932
(13)
where 120588 = 119888120574121205743 120573 = 120574
221205743 120590 = 1198922120574
3 and 120598 = (119896minus119888)2120574
3
(1205743
= 0) System (13) has the first integral
119867(120593 119910) =120573
41205934+1
21205981205932+ 120590120593 +
1
212059321199102minus1
21205881199102= ℎ (14)
Obviously for 120588 gt 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 120588 gt 0 in the rest of thispaper (120588 = 984858
2 984858 gt 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 120593 = 984858 and 120593 = minus984858 on the right-hand side of the secondequation of (13) To avoid the singular lines temporarily wedefine a new independent variable 120577 by setting (119889120585119889120577) =
9848582minus1205932 then system (13) is changed to a Hamiltonian system
written as119889120593
119889120577= (9848582minus 1205932) 119910
119889119910
119889120577= 1205931199102+ 1205731205933+ 120598120593 + 120590
(15)
System (15) has the same topological phase portraits as system(13) except for the singular lines 120593 = 984858 and 120593 = minus984858
We now investigate the bifurcation of phase portraits ofthe system (15) Denote that
119891 (120593) = 1205731205933+ 120598120593 + 120590 (16)
Let 119872(120593119890 119910119890) be the coefficient matrix of the linearized
system of (15) at the equilibrium point (120593119890 119910119890) then
119872(120593119890 119910119890) = (
minus21205931198901199101198901199102
120598+ 1198911015840(120593119890)
9848582minus 1205932
1198902120593119890119910119890
) (17)
and at this equilibrium point we have
119869 (120593119890 119910119890) = det119872(120593
119890 119910119890)
= (1205932
119890minus 9848582) 1198911015840(120593119890) minus (3120593
2
119890+ 9848582) 1199102
119890
(18)
By the theory of planar dynamical systems for an equilibriumpoint of a Hamiltonian system if 119869 lt 0 then it is a saddlepoint a center point if 119869 gt 0 and a degenerate equilibriumpoint if 119869 = 0
From the above analysis we can obtain the bifurcationcurves and phase portraits under different parameter condi-tions
Let
1205731(120590) = minus
41205983
271205902
1205732(120590) =
120590 minus 120598984858
9848583
1205733(120590) =
minus120590 minus 120598984858
9848583
(19)
Clearly for 120573 gt 1205731(120590) the function 119891(120593) = 0 has three
real roots 1205931 1205932 and 120593
3(1205931
gt 1205932
gt 1205933) that is system
(15) has three equilibrium points 119864119894(120593119894 0) 119894 = 1 2 3 on
the 120593-axis When 120573 lt 1205733(120590) (15) has two equilibrium
points 11987812(984858 plusmn119884
1) on the straight line 120593 = radic119886 where 119884
1=
radic(minus120590 minus 984858(120598 + 1205739848582))984858 When 120573 lt 1205732(120590) system (15) has two
equilibrium points 11987834(minus984858 plusmn119884
2) on the straight line 120593 = minus984858
where 1198842= radic(120590 minus 984858(120598 + 1205739848582))984858 Notice that on making the
transformation 120590 rarr minus120590 120593 rarr minus120593 120577 rarr minus120577 system (15)is invariant This means that for 120590 lt 0 the phase portraitsof (15) are just the reflections of the corresponding phaseportraits of (15) in the case 120590 gt 0 with respect to the 119910-axisThus we only need to consider the case 120590 ge 0 To knowthe dynamical behavior of the orbits of system (15) we willdiscuss two cases 120598 gt 0 and 120598 lt 0
21 Case I 120598 gt 0
Lemma 1 Suppose that 120598 gt 0 Denote ℎ+984858= 119867(984858 plusmn119884
1) ℎminus984858=
119867(minus984858 plusmn1198842) and ℎ
119894= 119867(120593
119894 0) 119894 = 1 2 3
(1) For 120590 gt 0 and 1205731(120590) lt 120573 lt 120573
3(120590) there exists one
and only one curve 120573 = 1205734(120590) on the (120590 120573)-plane on which
ℎ+
984858= ℎ2 for 120590 gt 0 and 120573
1(120590) lt 120573 lt 120573
2(120590) there exists one
and only one curve 120573 = 1205735(120590) on the (120590 120573)-plane on which
ℎminus
984858= ℎ2or ℎminus984858= ℎ3Moreover the curves120573 = 120573
1(120590)120573 = 120573
2(120590)
120573 = 1205734(120590) and 120573 = 120573
5(120590) are tangent at the point (120590
1119904 1205731119904)
The curve 120573 = 1205733(120590) intersects with the curves 120573 = 120573
1(120590)
120573 = 1205734(120590) and 120573 = 120573
5(120590) at the points (120590
2119904 1205732119904) (1205903119904 1205733119904)
and (1205904119904 1205734119904) (0 lt 120590
4119904lt 1205903119904lt 1205902119904lt 1205901119904) respectively
(2) For 120590 = 0 there exists a bifurcation point 119860(0 minus21205989848582)on (120590 120573)-plane on which ℎ
+
984858= ℎminus
984858= ℎ2when 120573 = minus2120598984858
2ℎ+
984858= ℎminus
984858lt ℎ2when 120573 lt minus2120598984858
2 and ℎ+
984858= ℎminus
984858gt ℎ2when
minus21205989848582lt 120573 lt minus120598984858
2
4 Mathematical Problems in Engineering
1205732(120590)
(1205901s 1205731s)
1205733(120590)
1205734(120590)1205735(120590)
120573
minus2
minus2
minus4
minus4
minus6
minus8
minus10
minus12
minus14
1205731(120590)
1205902 4
Figure 1 The bifurcation curves in the (120590 120573)-parameter plane for120598 gt 0
According to the above analysis and Lemma 1 we obtainthe following proposition on the bifurcation curves of thephase portraits of system (15) for 120598 gt 0
Proposition 2 When 120598 gt 0 for system (15) in (120590 120573)-parameter plane there exist five bifurcation curves (see Fig-ure 1)
120573 = 1205731(120590) = minus
41205983
271205902
120573 = 1205732(120590) =
120590 minus 120598984858
9848583
120573 = 1205733(120590) =
minus120590 minus 120598984858
9848583
120573 = 1205734(120590) (|120590| le 120590
3119904)
120573 = 1205735(120590)
(20)
These five curves divide the right-half (120590 120573)-parameter planeinto thirty-one regions as follows
1198601 (120590 120573) | 1205901s lt 120590 lt 120598984858 120573 = 1205732(120590)
1198602 (120590 120573) | 120590 gt 1205901s 1205735(120590) lt 120573 lt 1205732(120590)
1198603 (120590 b) | 120590 gt 1205901s 120573 = 1205735(120590)
1198604 (120590 120573) | 120590 gt 1205901s 1205731(120590) lt 120573 lt 1205735(120590)
1198605 (120590 120573) | 120590 gt 1205901s 120573 = 1205731(120590)
1198606 (120590 120573) | 120590 gt 1205902s 1205733(120590) lt 120573 lt 1205731(120590)
1198607 (120590 120573) | 120590 lt 1205902s 120573 = 1205733(120590)
1198608 (120590 120573) | 120590 gt 0 120573 lt max1205731(120590) 1205733(120590)
1198609 (120590 120573) = (1205902s 1205732s)
11986010 (120590 120573) | 0 lt 120590 lt 1205902s 120573 = 1205731(120590)
11986011 (120590 120573) | 120590 lt 1205902s 1205731(120590) lt 120573 lt min1205733(120590)
1205734(120590)
11986012 (120590 120573) | 120590 lt 1205903s 120573 = 1205734(120590)
11986013 (120590 120573) | 120590 gt 0 1205734(120590) lt 120573 lt 1205735(120590)
11986014 (120590 120573) | 120590 lt 1205904s 120573 = 1205735(120590)
11986015 (120590 120573) | 120590 gt 0 1205735(120590) lt 120573 lt 1205733(120590)
11986016 (120590 120573) | 120590 = 0 120573 lt minus21205989848582
11986017 (120590 120573) = (0 minus21205989848582)
11986018 (120590 120573) | 120590 = 0 minus21205989848582 lt 120573 lt minus120598984858
2
11986019 (120590 120573) = (0 minus1205989848582)
11986020 (120590 120573) | 0 lt 120590 lt 1205904s 120573 = 1205733(120590)
11986021 (120590 120573) = (1205904s 1205734s)
11986022 (120590 120573) | 1205904s lt 120590 lt 1205902s 120573 = 1205733(120590)
11986023 (120590 120573) | 1205904s lt 120590 lt 1205901s max1205731(120590) 1205733(120590) lt
120573 lt 1205735(120590)11986024 (120590 120573) | 120590
s4 lt 120590 lt 1205901s 120573 = 1205735(120590)
11986025 (120590 120573) | 1205904s lt 120590 lt 1205901s max1205733(120590) 1205735(120590) lt
120573 lt 1205732(120590)11986026 (120590 120573) | 0 lt 120590 lt 1205901s 120573 = 1205732(d)
11986027 (120590 120573) = (1205901s 1205731s)
11986028 (120590 120573) | 0 le 120590 lt 120598984858 120573 gt 1205732(120590)
11986029 (120590 120573) | 120590 isin R 120573 gt max0 1205732(120590)
11986030 (120590 120573) | 120590 gt 120598984858 120573 = 1205732(120590)
11986031 (120590 120573) | 120590 gt 120598984858 0 lt 120573 lt 1205732(120590)
In this case the phase portraits of system (15) can beshown in Figure 2
22 Case II 120598 lt 0 In this case we have the following
Proposition 3 When 120598 lt 0 for system (15) in (120590 120573)-parameter plane there exist four parametric bifurcation curves(see Figure 3)
120573 = 1205731(120590) = minus
41205983
271205902
120573 = 1205732(120590) =
120590 minus 120598984858
9848583
120573 = 1205733(120590) =
minus120590 minus 120598984858
9848583 120590 = 0
(21)
These four curves divide the right-half (120590 120573)-parameter planeinto twenty-two regions
1198611 (120590 120573) | 1205901s lt 120590 lt 120598984858 120573 = 1205733(120590)
1198612 (120590 120573) | 120590 gt 1205901s max0 1205733(120590) lt 120573 lt 1205731(120590)
1198613 (120590 120573) | 120590 gt 1205901s 120573 = 1205731(120590)
1198614 (120590 120573) | 120590 gt 1205902s 1205731(120590) lt 120573 lt 1205732(120590)
1198615 (120590 120573) = (1205902s 1205732s)
1198616 (120590 120573) | 120590 gt 1205902s 120573 = 1205732(120590)
1198617 (120590 120573) | 120590 gt 0 120573 gt max1205731(120590) 1205732(120590)
1198618 (120590 120573) | 0 lt 120590 lt 1205902s 120573 = 1205731(120590)
1198619 (120590 120573) | 0 lt 120590 lt 1205902s 1205732(120590) lt 120573 lt 1205731(120590)
11986110 (120590 120573) | 120590 = 0 120573 gt minus120598984858
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
00 05 10
0
2
4
00 05 10
0
2
4
00 05 10
0
1
2
3
00 05 10
0
2
4
6
00 05 10
0
2
4
00 05 10
0
2
4
minus3
minus3
minus2 minus2 minus2minus1minus4
minus4
minus2 minus2
minus2
minus1
minus1
minus4minus2minus4
minus4 minus4
minus2 minus2 minus2
minus4
minus2
minus4
minus2
minus4
minus2
minus15 minus10 minus05
minus10 minus05 minus10 minus05 minus10 minus05
minus15 minus10 minus05 minus15 minus10 minus05
minus6
minus2
minus4
minus2
minus4
minus2
minus4
minus2
minus4
minus2
minus4
120593120593120593
120593 120593 120593
120593120593120593
120593 120593 120593
yy
y
y yy
yyy
y
y y
(1) (120590 120573) isin A1
(4) (120590 120573) isin A4
(7) (120590 120573) isin A7 (8) (120590 120573) isin A8 (9) (120590 120573) isin A9
(5) (120590 120573) isin A5
(2) (120590 120573) isin A2 (3) (120590 120573) isin A3
(6) (120590 120573) isin A6
(10) (120590 120573) isin A10 (11) (120590 120573) isin A11 (12) (120590 120573) isin A12
Figure 2 Continued
6 Mathematical Problems in Engineering
00 05 10
0
1
2
3
00 05 10
0
2
4
00 05 10
0
1
2
3
00 05 10
0
1
2
3
00 05 10
0
1
2
00 05 10
0
1
2
00 05 10
0
1
2
00 05 10
0
1
2
00 05 10
0
1
2
3
00 05 10
0
1
2
3
00 05 10
0
1
2
3
00 05 10
0
1
2
3
minus3
minus2
minus1
minus3
minus2
minus1
minus2
minus1
minus2
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus1
minus2
minus1
minus3
minus2
minus1
minus10 minus05
minus10 minus05
minus10 minus05
minus10 minus05
minus10 minus05
minus10minus15 minus05 minus10minus15 minus05
minus10 minus05
minus10 minus05 minus10 minus05
minus10 minus05 minus10 minus05
minus2
minus4
y
120593120593120593
120593 120593 120593
120593120593120593
120593 120593120593
yy
y yy
yy
y
yy
y
(13) (120590 120573) isin A13
(16) (120590 120573) isin A16
(19) (120590 120573) isin A19 (20) (120590 120573) isin A20
(23) (120590 120573) isin A23(24) (120590 120573) isin A24
(21) (120590 120573) isin A21
(22) (120590 120573) isin A22
(17) (120590 120573) isin A17 (18) (120590 120573) isin A18
(14) (120590 120573) isin A14 (15) (120590 120573) isin A15
Figure 2 Continued
Mathematical Problems in Engineering 7
00 05 10 15
0
1
2
3
00 05 10 15
0
1
2
3
0 1 2
0
1
2
3
0 1 2
0
1
2
3
00 05 10 15
0
1
2
3
0 1 2
0
2
4
0 1 2
0
2
4
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus2
minus4
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2 minus1
minus2minus3 minus1
minus2
minus2
minus3
minus4
minus1
minus2 minus1
minus10minus15 minus05 minus10minus15 minus05
minus10minus15 minus05
y
yyy
y y y
120593
120593120593120593
120593 120593 120593
(25) (120590 120573) isin A25
(28) (120590 120573) isin A28 (29) (120590 120573) isin A29
(31) (120590 120573) isin A31
(30) (120590 120573) isin A30
(26) (120590 120573) isin A26 (27) (120590 120573) isin A27
Figure 2 The bifurcation of phase portraits of system (15) when 120598 gt 0 and 120590 ge 0
11986111 (120590 120573) = (0 minus1205989848582)
11986112 (120590 120573) | 0 lt 120590 lt 1205902s 120573 = 1205732(120590)
11986113 (120590 120573) | 0 lt 120590 lt 1205901s 1205733(120590) lt 120573 lt min1205731(120590)
1205732(120590)
11986114 (120590 120573) | 1205904s lt 120590 lt 1205901s 120573 = 1205731(120590)
11986115 (120590 120573) = (1205901s 1205731s)
11986116 (120590 120573) | 0 lt 120590 lt 1205901s 120573 = 1205733(120590)
11986117 (120590 120573) | 0 lt 120590 lt 120598984858 120573 lt 1205733(120590)
11986118 (120590 120573) | 120590 = 0 120573 = minus120598984858
2
11986119 (120590 120573) | 120590 = 0 120573 lt 0
11986120 (120590 120573) | 0 lt 120590 lt 120598984858 120573 lt min0 1205733(120590)
11986121 (120590 120573) | 120590 gt 120598984858 120573 = 1205733(120590)
11986122 (120590 120573) | 120590 gt 120598984858 1205733(120590) lt 120573 lt 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
1205732(120590)
(1205901s 1205731s)
(1205902s 1205732s)
1205733(120590)
120573
minus2minus4minus6
1205731(120590)
120590
Figure 3 The bifurcation curves in the (120590 120573)-parameter plane for120598 lt 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 Let119862119896(Ω) denote the set of all 119896 times continuouslydifferentiable functions on the open setΩ 119871119901loc(R) refer to theset of all functions whose restriction on any compact subsetis 119871119901 integrable1198671loc(R) stands for1198671loc(R) = 120593 isin 119871
2
loc(R) |
1205931015840isin 1198712
loc(R)To study single peak soliton solutions we impose the
boundary condition
lim120585rarrplusmninfin
120593 = 119860 (22)
where 119860 is a constant In fact the constant 119860 is equal to thehorizontal coordinate of saddle point 119864(120593
119890 0) Substituting
the boundary condition (22) into (14) generates the followingconstant
ℎ = 119860(120590 +119860120598
2+1198603120573
4) (23)
So the ODE (14) becomes
(1205931015840)2
=
(120593 minus 119860)2
(1205731205932+ 2119860120573120593 + 3119860
2120573 + 2120598)
2 (9848582 minus 1205932) (24)
If 120573(1198602120573 + 120598) le 0 then (24) reduces to
(1205931015840)2
=120573 (120593 minus 119860)
2
(120593 minus 1198611) (120593 minus 119861
2)
2 (9848582 minus 1205932) (25)
where
1198611= minus119860 minus
radicminus2120573 (1198602120573 + 120598)
120573
1198612= minus119860 +
radicminus2120573 (1198602120573 + 120598)
120573
(26)
From (26) we know that 1198611gt 1198612if 120573 lt 0 and 119861
1lt 1198612if
120573 gt 0
Definition 4 A function120593(120585) is said to be a single peak solitonsolution for the 119862(3 2 2) equation (10) if 120593(120585) satisfies thefollowing conditions
(C1) 120593(120585) is continuous on R and has a unique peakpoint 120585
0 where 120593(120585) attains its global maximum or
minimum value
(C2) 120593(120585) isin 119862
3(R minus 120585
0) satisfies (24) on R minus 120585
0
(C3) 120593(120585) satisfies the boundary condition (22)
Definition 5 A wave function 120593 is called smooth solitonsolution if 120593 is smooth locally on either side of 120585
0and
lim120585uarr1205850
1205931015840(120585) = lim
120585darr12058501205931015840(120585) = 0
Definition 6 A wave function 120593 is called peakon if 120593 issmooth locally on either side of 120585
0and lim
120585uarr12058501205931015840(120585) =
minuslim120585darr1205850
1205931015840(120585) = 119886 119886 = 0 119886 = plusmninfin
Definition 7 Awave function120593 is called cuspon if120593 is smoothlocally on either side of 120585
0and lim
120585uarr12058501205931015840(120585) = minuslim
120585darr12058501205931015840(120585) =
plusmninfin
Without any loss of generality we choose the peak point1205850as vanishing 120585
0= 0
Theorem 8 Assume that 119906(119909 119905) = 120593(120585) = 120593(119909 minus 119888119905) is a singlepeak soliton solution of the 119862(3 2 2) equation (10) at the peakpoint 120585
0= 0 Then we have the following
(i) If 120573(1198602120573 + 120598) gt 0 then 120593(0) = 984858 or 120593(0) = minus984858
(ii) If 120573(1198602120573 + 120598) le 0 then 120593(0) = 984858 or 120593(0) = minus984858 or120593(0) = 119861
1or 120593(0) = 119861
2
Proof If 120593(0) = plusmn984858 then 120593(120585) = plusmn984858 for any 120585 isin R since120593(120585) isin 119862
3(R minus 0) Differentiating both sides of (24) yields
120593 isin 119862infin(R)
(i) When 120573(1198602120573 + 120598) gt 0 if 120593(0) = 984858 and 120593(0) = minus984858 then
120593 isin 119862infin(R) By the definition of single peak soliton we have
1205931015840(0) = 0 However by (24) we must have 120593(0) = 119860 which
contradicts the fact that 0 is the unique peak point(ii) When 120573(119860
2120573 + 120598) le 0 if 120593(0) = 984858 and 120593(0) = minus984858
by (24) we know 1205931015840(0) exists and 120593
1015840(0) = 0 since 0 is a peak
point Thus we obtain 120593(0) = 1198611or 120593(0) = 119861
2from (25)
since 120593(0) = 119860 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 119906(119909 119905) = 120593(119909 minus 119888119905) is a single peaksoliton solution of the 119862(3 2 2) equation (10) at the peak point1205850= 0 Then we have the following solution classification(i) If |120593(0)| = 984858 then 120593(120585) isin 119862
infin(R) and 120593 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
00 05 10 15
0
2
4
0 1
0
2
4
6
00 05 10 15
0
2
4
00 05 10
0
1
2
3
00 05 10 15
0
1
2
3
00 05 10
0
1
2
00 05 10 15
0
1
2
00 05 10 15
0
1
2
3
minus3 minus1minus4
minus4
minus2
minus4
minus2
minus4
minus2
minus4
minus2minus4
minus6
minus2
minus4
minus2
minus4
minus2
minus2 minus4 minus4minus2 minus2
minus3
minus1
minus2
minus1
minus2
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus1minus2 minus1minus2
120593
120593
120593
120593 120593 120593
120593 120593
120593 120593
120593 120593
y
y
y
yy
y
y y
y y
y y
minus15 minus10 minus05
minus15minus20 minus10 minus05
minus15 minus10 minus05 minus15 minus10 minus05minus10 minus05
minus15 minus10 minus05minus10 minus05
(10) (120590 120573) isin B10 (11) (120590 120573) isin B11 (12) (120590 120573) isin B12
(1) (120590 120573) isin B1
(4) (120590 120573) isin B4
(7) (120590 120573) isin B7 (8) (120590 120573) isin B8 (9) (120590 120573) isin B9
(5) (120590 120573) isin B5 (6) (120590 120573) isin B6
(2) (120590 120573) isin B2 (3) (120590 120573) isin B3
Figure 4 Continued
10 Mathematical Problems in Engineering
00 05 10 15
0
1
2
3
0 1
0
1
2
3
0 1 2
0
2
4
00 05 10 15
0
1
2
3
0 1 2
0
1
2
3
00 05 10 15
0
1
2
00 05 10 15
0
1
2
3
0 1 2
0
1
2
3
00 05 10 15
0
2
4
00 05 10 15
0
2
4
6
minus4
minus2
minus4
minus2
minus4
minus6
minus2
minus3 minus1minus2
minus3 minus1minus2
minus1minus2
minus1minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus1
minus2
minus3
minus1
minus2
120593 120593 120593
120593120593120593
120593 120593 120593
120593
y y y
yy
y
y yy
y
minus15 minus10 minus05
minus15minus20 minus10 minus05
minus15 minus10 minus05
minus15 minus10 minus05
minus15 minus10 minus05
minus15 minus10 minus05
(13) (120590 120573) isin B13 (14) (120590 120573) isin B14 (15) (120590 120573) isin B15
(16) (120590 120573) isin B16 (17) (120590 120573) isin B17 (18) (120590 120573) isin B18
(22) (120590 120573) isin B22
(20) (120590 120573) isin B20 (21) (120590 120573) isin B21(19) (120590 120573) isin B19
Figure 4 The bifurcation of phase portraits of system (15) when 120598 lt 0 and 120590 ge 0
Mathematical Problems in Engineering 11
(ii) If 120593(0) = 984858 = 1198611 then 120593 is a cuspon solution and 120593 has
the following asymptotic behavior
120593 (120585) minus 984858 = 1205821
10038161003816100381610038161205851003816100381610038161003816
23
+ 119874(10038161003816100381610038161205851003816100381610038161003816
43
) 120585 997888rarr 0
1205931015840(120585) =
2
31205821
10038161003816100381610038161205851003816100381610038161003816
minus13 sgn (120585) + 119874 (10038161003816100381610038161205851003816100381610038161003816
13
)
120585 997888rarr 0
(27)
where 1205821= plusmn(9|120573984858
2+ 2119860120573984858 + 3119860
2120573 + 2120598|(984858 minus 119860)
216984858)13
Thus 120593(120585) notin 1198671
119897119900119888(R)
(iii) If 120593(0) = minus984858 = 1198612 then 120593 is a cuspon solution and 120593
has the following asymptotic behavior
120593 (120585) + 984858 = 1205822
10038161003816100381610038161205851003816100381610038161003816
23
+ 119874(10038161003816100381610038161205851003816100381610038161003816
43
) 120585 997888rarr 0
1205931015840(120585) =
2
31205822
10038161003816100381610038161205851003816100381610038161003816
minus13 sgn (120585) + 119874 (10038161003816100381610038161205851003816100381610038161003816
13
)
120585 997888rarr 0
(28)
where 1205822= plusmn(9|120573984858
2minus 2119860120573984858 + 3119860
2120573 + 2120598|(984858 + 119860)
216984858)13
Thus 120593(120585) notin 1198671
119897119900119888(R)
(iv) If 120593(0) = 984858 = 1198611and 119860 = 0 then 120593 is a peakon-like
solution and
120593 (120585) minus 984858 = 1205823
10038161003816100381610038161205851003816100381610038161003816 + 119874 (
10038161003816100381610038161205851003816100381610038161003816
2
) 120585 997888rarr 0
1205931015840(120585) = 120582
3sgn (120585) + 119874 (
10038161003816100381610038161205851003816100381610038161003816) 120585 997888rarr 0
(29)
where 1205823= plusmn|119860 minus 984858|radic120573(119861
2minus 984858)4984858
(v) If 120593(0) = minus984858 = 1198612and 119860 = 0 then 120593 is a peakon-like
solution and
120593 (120585) + 984858 = 1205824
10038161003816100381610038161205851003816100381610038161003816 + 119874 (
10038161003816100381610038161205851003816100381610038161003816
2
) 120585 997888rarr 0
1205931015840(120585) = 120582
4sgn (120585) + 119874 (
10038161003816100381610038161205851003816100381610038161003816) 120585 997888rarr 0
(30)
where 1205824= plusmn|119860 + 984858|radicminus120573(119861
1+ 984858)4984858
(vi) If 120593(0) = 984858 = 1198611and 119860 = 0 then 120593 gives the peakon
solution 984858 exp(minusradicminus1205732|119909 minus 119888119905|)(vii) If 120593(0) = minus984858 = 119861
2and 119860 = 0 then 120593 gives the peakon
solution minus984858 exp(minusradicminus1205732|119909 minus 119888119905|)
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 |120593(0)| = 984858 then 120593 isin 119862
infin(R) and 120593 is a smooth soliton
solution(ii) If 120593(0) = 984858 = 119861
1 then by the definition of single peak
soliton we have 119860 = 984858 thus 1205731205932 + 2119860120573120593 + 31198602120573 + 2120598 does
not contain the factor 120593 minus 984858 From (24) we obtain
1205931015840= sgn (119860 minus 984858)
1003816100381610038161003816120593 minus 1198601003816100381610038161003816
radic210038161003816100381610038161205932 minus 9848582
1003816100381610038161003816
sdot radic10038161003816100381610038161205731205932 + 2119860120573120593 + 31198602120573 + 2120598
1003816100381610038161003816 sgn (120585)
(31)
Let 1198971(120593) = radic2(120593 + 984858)|120593 minus 119860|radic|1205731205932 + 2119860120573120593 + 31198602120573 + 2120598|
then 1198971(984858) = 2radic984858|984858 minus 119860|radic|1205739848582 + 2119860120573984858 + 31198602120573 + 2120598| and
int 1198971(120593)radic
1003816100381610038161003816120593 minus 9848581003816100381610038161003816119889120593 = int sgn (119860 minus 984858) sgn (120585) 119889120585 (32)
Inserting 1198971(120593) = 119897
1(984858) + 119874(|120593 minus 984858|) into (32) and using the
initial condition 120593(0) = 984858 we obtain
21198971(984858)
3
1003816100381610038161003816120593 minus 9848581003816100381610038161003816
32
(1 + 119874 (1003816100381610038161003816120593 minus 984858
1003816100381610038161003816)) =10038161003816100381610038161205851003816100381610038161003816
(33)
thus
120593 minus 984858 = plusmn(3
21198971(984858)
)
23
10038161003816100381610038161205851003816100381610038161003816
23
(1 + 119874 (1003816100381610038161003816120593 minus 984858
1003816100381610038161003816))minus23
= plusmn(3
21198971(984858)
)
23
10038161003816100381610038161205851003816100381610038161003816
23
(1 + 119874 (1003816100381610038161003816120593 minus 984858
1003816100381610038161003816))
(34)
which implies 120593 minus 984858 = 119874(|120585|23
) Therefore we have
120593 (120585) = 984858 plusmn (3
21198971(984858)
)
23
10038161003816100381610038161205851003816100381610038161003816
23
+ 119874(10038161003816100381610038161205851003816100381610038161003816
43
)
= 984858 + 1205821
10038161003816100381610038161205851003816100381610038161003816
23
+ 119874(10038161003816100381610038161205851003816100381610038161003816
43
) 120585 997888rarr 0
1205821= plusmn(
3
21198971(984858)
)
23
= plusmn(
9100381610038161003816100381610038161205739848582+ 2119860120573984858 + 3119860
2120573 + 2120598
10038161003816100381610038161003816(984858 minus 119860)
2
16984858)
13
1205931015840(120585) =
2
31205821
10038161003816100381610038161205851003816100381610038161003816
minus13 sgn (120585) + 119874 (10038161003816100381610038161205851003816100381610038161003816
13
) 120585 997888rarr 0
(35)
So 120593(120585) notin 1198671
loc(R)(iii) Similar to the proof of (ii) we ignore it in this paper(iv) If 120593(0) = 984858 = 119861
1and 119860 = 0 then from (25) we obtain
1205931015840= radicminus
120573
2sgn (119860 minus 984858)
1003816100381610038161003816120593 minus 1198601003816100381610038161003816radic120593 minus 1198612
120593 + 984858sgn (120585) (36)
Let 1198972(120593) = (1|120593minus119860|)radic(120593 + 984858)(120593 minus 119861
2) then 119897
2(984858) = (1|984858minus
119860|)radic2984858(984858 minus 1198612) and
int 1198972(120593) 119889120593 = radicminus
120573
2int sgn (119860 minus 984858) sgn (120585) 119889120585 (37)
Inserting 1198972(120593) = 119897
2(984858) + 119874(|120593 minus 984858|) into (37) and using the
initial condition 120593(0) = 984858 we obtain
1198972(984858) (120593 minus 984858) (1 + 119874 (
1003816100381610038161003816120593 minus 9848581003816100381610038161003816))minus1
= sgn (119860 minus 984858)10038161003816100381610038161205851003816100381610038161003816
(38)
Sincesgn (120593 minus 984858) sgn (119860 minus 984858) ge 0
1
1 + 119874 (120593 minus 984858)= 1 + 119874 (120593 minus 984858)
(39)
12 Mathematical Problems in Engineering
we get
1003816100381610038161003816120593 minus 9848581003816100381610038161003816 =
radicminus120573
2
1
1198972(984858)
10038161003816100381610038161205851003816100381610038161003816 (1 + 119874 (120593 minus 984858)) (40)
which implies |120593 minus 984858| = 119874(|120585|) Therefore we have
120593 (120585) = 984858 + 1205823
10038161003816100381610038161205851003816100381610038161003816 + 119874 (
10038161003816100381610038161205851003816100381610038161003816
2
) 120585 997888rarr 0
1205931015840(120585) = 120582
3sgn (120585) + 119874 (
10038161003816100381610038161205851003816100381610038161003816) 120585 997888rarr 0
(41)
where 1205823= plusmn|119860 minus 984858|radic120573(119861
2minus 984858)4984858
(v) Similar to the proof of (iv) we ignore it in this paper
By virtue of Theorem 9 any single peak soliton for the119862(3 2 2) equation (10) must satisfy the following initial andboundary values problem
(1205931015840)2
=120573 (120593 minus 119860)
2
(120593 minus 1198611) (120593 minus 119861
2)
2 (9848582 minus 1205932)fl119871 (120593)
120593 (0) isin 984858 minus984858 1198611 1198612
lim|120585|rarrinfin
120593 (120585) = 119860
(42)
119871(120593) ge 0 and the boundary condition (24) imply thefollowing
(a) If 120573(9848582 minus 1205932) ge 0 then 120593 ge max119861
1 1198612 or 120593 le
min1198611 1198612
(b) If120573(9848582minus1205932) le 0 thenmin1198611 1198612 le 120593 le max119861
1 1198612
Below we will present some implicit formulas for thesingle peak soliton solutions in the case of specific 120590 and 120573
Case 1 ((120590 120573) isin 11986012) In this case we have minus984858 lt 119861
2lt 119860 lt
1198611= 984858 From the standard phase analysis and Theorem 9
we know that if 120593 is a single peak soliton of the 119862(3 2 2)
equation then
1205931015840= minusradicminus
120573
2(120593 minus 119860)radic
120593 minus 1198612
120593 + 984858sgn (120585) (43)
From the separation of variables we get
intℎ (120593) 119889120593 = radicminus120573
2sgn (120585) 119889120585 (44)
where ℎ(120593) = (1(119860 minus 120593))radic(120593 + 984858)(120593 minus 1198612) After a lengthy
calculation of integral we obtain the implicit solution 120593
defined by
119867(120593)flradic119860 + 984858
119860 minus 1198612
1198681(120593) minus 119868
2(120593) = radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870 (45)
where
1198681(120593) = ln
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
(119860 + 984858) (120593 minus 1198612) + (119860 minus 119861
2) (120593 + 984858) + 2radic(119860 + 984858) (119860 minus 119861
2) (120593 minus 119861
2) (120593 + 984858)
radic(119860 + 984858) (119860 minus 1198612) (119860 + 984858) (119860 minus 120593)
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
1198682(120593) = ln
1003816100381610038161003816100381610038161003816(120593 + 984858) + (120593 minus 119861
2) + 2radic(120593 + 984858) (120593 minus 119861
2)
1003816100381610038161003816100381610038161003816
(46)
and 119870 is an arbitrary integration constant For 120593(0) = 984858 theconstant119870 = 119867(120593(0)) is defined by
119870 = radic119860 + 984858
119860 minus 1198612
1198681(984858) minus 119868
2(984858) (47)
and for 120593(0) = 1198612
119870 = radic119860 + 984858
119860 minus 1198612
1198681(1198612) minus 1198682(1198612) (48)
(i) If 120593(0) = 984858 then 119860 lt 120593 le 984858 Since 1198671015840(120593) = ℎ(120593) weknow that119867(120593) strictly decreases on the interval (119860 984858] thus1198671(120593) = 119867|
(119860984858](120593) gives a single peak solitonwith119867
1(984858) = 119870
and1198671(119860+) = infin Therefore 120593
1(120585) = 119867
minus1
1(radicminus1205732|120585| + 119870) is
the solution satisfying
1205931(0) = 984858
lim|120585|rarrinfin
1205931(120585) = 119860
1205931015840
1(0plusmn) = plusmn (119860 minus 984858)radic
120573 (1198612minus 984858)
4984858
(49)
So 1205931(120585) is a peakon-like solution (see Figure 5)
(ii) If 120593(0) = 1198612 then 119861
2le 120593 lt 119860 By 119867
1015840(120593) = 119891(120593)
we know that 119867(120593) strictly increases on the interval [1198612 119860)
Thus
1198672(120593) = 119867|
[1198612 119860)(120593) (50)
Mathematical Problems in Engineering 13
minus3 minus2 minus1
120593
minus02
1 2 3
02
04
06
08
10
x
(a)
02
02
minus2 minus2 x
10
05
00
t
120593
22
(b)
Figure 5 Two- and three-dimensional graphs of the peakon-like solution
minus03
minus04
minus05
1 2 3minus3 minus2 minus1
120593
minus02
x
(a)
0
2
0
2
minus2minus2
minus02
minus03
minus04
minus05
minus06
t x
120593
0
2
0
2
minus2minus2
t x
(b)
Figure 6 Two- and three-dimensional graphs of the smooth soliton solution
has the inverse denoted by 1205932(120585) = 119867
minus1
2(radicminus1205732|120585| + 119870)
1205932(120585) gives a kind of smooth soliton solution (see Figure 6)
satisfying
1205932(0) = 119861
2
lim|120585|rarrinfin
1205932(120585) = 119860
1205931015840
2(0) = 0
(51)
Case 2 ((120590 120573) isin 11986014) In this case we have minus984858 = 119861
2lt 119860 lt
984858 lt 1198611and (25) is equivalent to
1205931015840= minusradicminus
120573
2(120593 minus 119860)radic
120593 minus 1198611
120593 minus 984858sgn (120585) (52)
Let
119892 (120593) =1
119860 minus 120593radic120593 minus 1198611
120593 minus 984858 (53)
14 Mathematical Problems in Engineering
minus1 1 2 3
02
04
06
08
10
minus3 minus2
120593
x
(a)
02
02minus2
minus2
120593
x t
00
05
10
(b)
Figure 7 Two- and three-dimensional graphs of the cuspon solution
then (52) is converted to
119892 (120593) 119889120593 =1
119860 minus 120593radic120593 minus 1198611
120593 minus 984858119889120593 = radicminus
120573
2sgn (120585) 119889120585 (54)
Integrating (54) on the interval [0 120585] (or [120585 0]) leads to thefollowing implicit solutions
119866 (120593)flradic119860 minus 984858
119860 minus 1198611
1198683(120593) + 119868
4(120593) = radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870 (55)
where
1198683(120593) = ln
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
(119860 minus 984858) (120593 minus 1198611) + (119860 minus 119861
1) (120593 minus 984858) + 2radic(119860 minus 984858) (119860 minus 119861
1) (120593 minus 984858) (120593 minus 119861
1)
radic(119860 minus 984858) (119860 minus 1198611) (119860 minus 984858) (120593 minus 119860)
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
1198684(120593) = ln
1003816100381610038161003816100381610038161003816(120593 minus 984858) + (120593 minus 119861
1) + 2radic(120593 minus 984858) (120593 minus 119861
1)
1003816100381610038161003816100381610038161003816
(56)
And119870 is an arbitrary integration constant It is obvious thatfor 120593(0) = 984858 the constant119870 = 119867(120593(0)) is defined by
119870 = radic119860 minus 984858
119860 minus 1198611
1198683(984858) + 119868
4(984858) (57)
and for 120593(0) = minus984858
119870 = radic119860 minus 984858
119860 minus 1198611
1198683(minus984858) + 119868
4(minus984858) (58)
(i) If 120593(0) = 984858 then 119860 lt 120593 le 984858 From 119892(120593) lt 0 weknow that 119866(120593) strictly decreases on the interval (119860 984858] with119866(984858) = 119870 and 119866(119860+) = infin Define
1198661(120593) = 119866|
(119860984858](120593) (59)
Since 1198661(120593) is a strictly decreasing function from (119860 984858] onto
[119870infin) we can solve for 120593 uniquely from (59) and obtain
1205931(120585) = 119866
minus1
1(radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870) (60)
It is easy to check that 120593 satisfies
1205931(0) = 984858
lim|120585|rarrinfin
1205931(120585) = 119860
1205931015840
1(0plusmn) = ∓infin
(61)
Therefore the solution1205931defined by (60) is a cuspon solution
for the 119862(3 2 2) equation (see Figure 7)
Mathematical Problems in Engineering 15
(ii) If 120593(0) = minus984858 then minus984858 le 120593 lt 119860 Through a similaranalysis we get a strictly increasing function 119866(120593) on theinterval [minus984858 119860) satisfying
119866 (120593) = radicminus120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870 (62)
where 119866(120593) is defined by (55) Let
1198662(120593) = 119866|
[minus984858119860)(120593) (63)
then 1198662(120593) is a strictly increasing function from [minus984858 119860) onto
[119870infin) so that we can solve for 120593 and obtain
1205932(120585) = 119866
minus1
2(radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870) (64)
It is easy to check that 1205932satisfies
1205932(0) = minus984858
lim|120585|rarrinfin
1205932(120585) = 119860
1205931015840
2(0plusmn) = plusmn (119860 + 984858)radicminus
120573 (1198611+ 984858)
4984858
(65)
Therefore the solution 1205932defined by (64) is a peakon-like
solution whose graph is similar to those in Figure 5
Case 3 ((120590 120573) isin 11986016) In this case we have minus984858 lt 119861
2lt 119860 =
0 lt 1198611lt 984858 119861
2= minus1198611 and
1205931015840= minusradicminus
120573
2120593radic
120593 minus 1198612
1
1205932 minus 9848582sgn (120585) (66)
Hence from the separation of variables we have
radic9848582 minus 1205932
120593radic1198612
1minus 1205932
119889120593 = minusradicminus120573
2sgn (120585) 119889120585 (67)
Integrating (67) on the interval [0 120585] (or [120585 0]) leads tothe following implicit formula for the two smooth solitonsolutions
radic9848582 minus 1205932 + radic1198612
1minus 1205932
radic9848582 minus 1198612
1
(
radic(9848582 minus 1198612
1) 1205932
984858radic1198612
1minus 1205932 + 119861
1radic9848582 minus 1205932
)
984858
= exp(minusradicminus120573
2|119909 minus 119888119905|)
(68)
where 120593 isin (119860 1198611] Consider
radic9848582 minus 1198612
2
radic9848582 minus 1205932 + radic1198612
2minus 1205932
(
984858radic1198612
2minus 1205932 minus 119861
2radic9848582 minus 1205932
radic(9848582 minus 1198612
2) 1205932
)
984858
= exp(radicminus120573
2|119909 minus 119888119905|)
(69)
where 120593 isin [1198612 119860)
Case 4 ((120590 120573) isin 11986017) In this case we have minus984858 = 119861
2lt 119860 =
0 lt 1198611= 984858 and
1205931015840= minusradicminus
120573
2120593 sgn (120585) (70)
Choosing 120593(0) = 984858 (or minus984858) as initial value we get
int
120593
984858
1
120593119889120593 = minusint
120585
0
radicminus120573
2sgn (120585) 119889120585 for 119860 lt 120593 le 984858
int
120593
minus984858
1
120593119889120593 = minusint
120585
0
radicminus120573
2sgn (120585) 119889120585 for minus 984858 le 120593 lt 119860
(71)
which immediately yield the peakon solutions
120593 (119909 minus 119888119905) = plusmn984858 exp(minusradicminus120573
2|119909 minus 119888119905|) (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 the119862(3 2 2) equation (10) In order to study kinkwave solutionswe assume that
lim120585rarrinfin
120593 (120585) = 1198601
lim120585rarrminusinfin
120593 (120585) = 1198602
(73)
16 Mathematical Problems in Engineering
2 4
02
04
06
08
10
minus2minus4
120593
x
(a)
0 2
02
00
05
10
minus2
120593
x
t
0
2
minus2
(b)
Figure 8 Two- and three-dimensional graphs of the peakon solution
where 1198601gt 1198602 Substituting the boundary condition (73)
into (14) generates
(1205931015840)2
=
120573 (120593 minus 1198601) (120593 minus 119860
2) [1205932+ (1198601+ 1198602) 120593 + 119860
2
1+ 11986011198602+ 1198602
2+ 2120598120573]
2 (9848582 minus 1205932)fl119865 (120593) (74)
The nonlinear differential equation (74) may sustaindifferent kinds of nonlinear excitations In what follows weconfine our attention to the cases 119860
2= minus119860
1and 120598 =
minus1198602
1120573 which describe kinks and kink compactons which
play an important role in the dynamics systems Under theseconsiderations (74) reduces to
(1205931015840)2
=120573 (120593 minus 119860
1)2
(120593 minus 1198602)2
2 (9848582 minus 1205932) (75)
If 1198601lt 984858 then from the phase analysis in Section 2 (see
Figure 4(10)) we know that (1198601 0) and (119860
2 0) are two saddle
points of (13) and the kink solutions can be obtained fromthe two heteroclinic orbits connecting (120601 119910) = (119860
1 0) and
(1198602 0) When 119860
1increases upon reaching 984858 that is 119860
1= 984858
(75) becomes
(1205931015840)2
= minus120573 (120593 minus 119860
1) (120593 minus 119860
2)
2 (76)
and the ellipse 21199102+120573(120593minus1198601)(120593minus119860
2) = 0 (see Figure 4(11))
which is tangent to the singular lines 120593 = 984858 and 120593 = minus984858 atpoints (119860
1 0) and (119860
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 120593 is a kink wave solutions of (75)
or (76) we have 1205931015840
rarr 0 as 120593 rarr 1198601and as 120593 rarr 119860
2
Moreover we have 119865(120593) ge 0 for 1198602le 120593 le 119860
1and 120593 is
strictly monotonic in any interval where 119865(120593) gt 0 Thus if1205931015840gt 0 at some points 120593will be strictly increasing until it gets
close to the next zero of 119865 Denoting this zero 1198601 we have
120593 uarr 1198601 What will happen to the solution when it approaches
1198601 Depending on whether the zero is double or simple 120593
has a different behavior We explore the two cases in turn
Theorem 11 (i) If 120593 has a simple zero at 120593 = 1198601 so that
119865(1198601) = 0 and 1198651015840(119860
1) lt 0 then the solution 120593 of (75) satisfies
120593 (120585) minus 1198601=1198651015840(1198601)
4(120585 minus 120578)
2
+ 119874 ((120585 minus 120578)4
)
119886119904 120585 uarr 120578
(77)
where 120593(120578) = 1198601
(ii) When 120593 approaches the double zero 1198601of 119865(120593) so that
1198651015840(1198601) = 0 and11986510158401015840(119860
1) gt 0 then the solution120593 of (75) satisfies
120593 (120585) minus 1198601sim 120572 exp(minus120585radic11986510158401015840 (119860
1)) 119886119904 120585 997888rarr infin (78)
for some constant 120572 Thus 120593 uarr 1198601exponentially as 120585 rarr infin
Mathematical Problems in Engineering 17
Proof (i) When 1198601= minus119860
2= 984858 and 120598 = minus119860
2
1120573 from (76)
119865(120593) has a simple zero at 120593 = 1198601 Then
(1205931015840)2
= (120593 minus 1198601) 1198651015840(1198601) + 119874 ((120593 minus 119860
1)2
)
as 120593 uarr 1198601
(79)
Using the fact that (1205931015840)2 ge 0 we know that 1198651015840(1198601) lt 0
Moreover
119889120585
119889120593=
1
radic(120593 minus 1198601) 1198651015840 (119860
1) + 119874 ((120593 minus 119860
1)2
)
(80)
Because
radic(120593 minus 1198601) 1198651015840 (119860
1) + 119874 ((120593 minus 119860
1)2
)
= radic1198601minus 120593(radicminus1198651015840 (119860
1) + 119874 (119860
1minus 120593))
1
radicminus1198651015840 (1198601) + 119874 (119860
1minus 120593)
=1
radicminus1198651015840 (1198601)
+ 119874 (1198601minus 120593)
(81)
we obtain119889120585
119889120593=
1
radicminus (1198601minus 120593) 1198651015840 (119860
1)
+ 119874 ((1198601minus 120593)12
) (82)
Integrating (82) yields
120578 minus 120585 =2
radicminus1198651015840 (1198601)
(1198601minus 120593)12
+ 119874((1198601minus 120593)32
) (83)
where 120578 satisfies 120593(120578) = 1198601 Thus
(120578 minus 120585)2
=4
minus1198651015840 (1198601)(1198601minus 120593) + 119874 ((119860
1minus 120593)2
) (84)
which implies 119874((1198601minus 120593)2) = 119874((120578 minus 120585)
4) Therefore we get
120593 (120585) = 1198601+1198651015840(1198601)
4(120585 minus 120578)
2
+ 119874 ((120585 minus 120578)4
)
as 120585 uarr 120578
(85)
where 120593(120578) = 1198601
(ii) When 1198601= minus119860
2lt 984858 and 120598 = minus119860
2
1120573 from (75) 119865(120593)
has a double zero at 120593 = 1198601 Then
(1205931015840)2
= (120593 minus 1198601)2
11986510158401015840(1198601) + 119874 ((120593 minus 119860
1)3
)
as 120593 uarr 1198601
(86)
Furthermore we get
119889120585
119889120593=
1
radic(120593 minus 1198601)2
11986510158401015840 (1198601) + 119874 ((120593 minus 119860
1)3
)
(87)
Observing that
radic(120593 minus 1198601)2
11986510158401015840 (1198601) + 119874 ((120593 minus 119860
1)3
)
= (1198601minus 120593) (radic11986510158401015840 (119860
1) + 119874 (119860
1minus 120593))
1
radic11986510158401015840 (1198601) + 119874 (119860
1minus 120593)
=1
radic11986510158401015840 (1198601)
+ 119874 (1198601minus 120593)
(88)
we obtain119889120585
119889120593=
1
(1198601minus 120593) 11986510158401015840 (119860
1)+ 119874 (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 119860
1= minus119860
2lt 984858 and 120598 = minus119860
2
1120573 Then (75)
becomes
radic9848582 minus 1205932
1205932 minus 1198602
1
119889120593 = ∓radic120573
2119889120585 (90)
Integrating both sides of (90) gives the following implicitexpressions of kink and antikink wave solutions
arctan120593
radic9848582 minus 1205932
+
radic9848582 minus 1198602
1
1198601
arctanhradic9848582 minus 119860
2
1
1198601
120593
radic9848582 minus 1205932
= plusmnradic120573
2120585
(91)
By letting 1198601rarr 984858 in (91) we get two kink compactons
which are given by
plusmn120593 =
984858 sinradic120573
2120585
10038161003816100381610038161205851003816100381610038161003816 le
120587
radic2120573
984858 120585 gt120587
radic2120573
minus984858 120585 lt minus120587
radic2120573
(92)
The graphs for the kink wave solutions (91) and kinkcompacton solutions (92) are shown in Figures 9 and 10respectively
Remark 12 The two kink compacton solutions (92) aredifferent from the well-known smooth kink wave solutionsIn comparison with kink wave solutions (91) the kink com-pacton solutions (92) have no exponential decay propertiesbut have compact supportThat is they minus a constant thedifferences identically vanish outside a finite core region
18 Mathematical Problems in Engineering
120593
120585minus1 1
05
minus05
(a)
minus1minus1
minus05
05
00120593
x t
0
1
01
11
(b)
Figure 9 Two- and three-dimensional graphs of the kink wave solution
1
1120593
120585
minus1
minus1
(a)
0
1
0
1
000510
120593
minus1 minus1
minus05
minus10
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 the119862(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 Newtonrsquos equation which we write in the form 119910
2=
119865(120593) where 119910 = 1205931015840
120585(120585) we solve Newtonrsquos equation 119910
2=
119865(120593) 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 the119862(3 2 2)equation
Actually for 119898 = 2119896 + 1 119896 isin N+ in 119862(119898 2 2) equation(9) the dynamical behavior of traveling wave solutions of(9) is similar to the case 119898 = 3 for 119898 = 2119896 119896 isin N+ in119862(119898 2 2) equation (9) the dynamical behavior of travelingwave solutions of (9) is similar to the case 119898 = 2 We areapplying the approach mentioned in this work to 119862(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
References
[1] C Rogers and W R Shadwick Backlund Transformation andTheir Applications Academic Press New York NY USA 1982
[2] W Malfliet and W Hereman ldquoThe tanh method I Exactsolutions of nonlinear evolution and wave equationsrdquo PhysicaScripta vol 54 no 6 pp 563ndash568 1996
[3] R Hirota Direct Method in Soliton Theory Springer BerlinGermany 1980
[4] J S Cohen Computer Algebra and Symbolic ComputationMathematical Methods A K Peters 2003
[5] S Lie ldquoUber die integration durch bestimmte integrale voneiner klasse linearer partieller differential eichungenrdquoArchiv derMathematik vol 6 pp 328ndash368 1881
[6] H-H Dai and X-H Zhao ldquoTraveling waves in a rod composedof amodifiedMooney-Rivlinmaterial I bifurcation of equilibraand non-singular caserdquo Proceedings of the Royal Society AMathematical Physical and Engineering Sciences vol 455 no1990 pp 3845ndash3874 1999
[7] H-H Dai and F-F Wang ldquoAsymptotic bifurcation solutionsfor compressions of a clamped nonlinearly elastic rectangletransition region and barrelling to a corner-like profilerdquo SIAMJournal on Applied Mathematics vol 70 no 7 pp 2673ndash26922010
[8] J Li and Z Liu ldquoSmooth and non-smooth traveling waves ina nonlinearly dispersive equationrdquo Applied Mathematical Mod-elling vol 25 no 1 pp 41ndash56 2000
[9] A Chen and J Li ldquoSingle peak solitary wave solutions forthe osmosis K(22) equation under inhomogeneous boundaryconditionrdquo Journal of Mathematical Analysis and Applicationsvol 369 no 2 pp 758ndash766 2010
[10] R Camassa and D D Holm ldquoAn integrable shallow waterequation with peaked solitonsrdquo Physical Review Letters vol 71no 11 pp 1661ndash1664 1993
[11] P Rosenau ldquoOn nonanalytic solitary waves formed by a nonlin-ear dispersionrdquo Physics Letters A vol 230 no 5-6 pp 305ndash3181997
[12] P Rosenau and J M Hyman ldquoCompactons solitons with finitewavelengthrdquo Physical Review Letters vol 70 no 5 pp 564ndash5671993
[13] L Zhang L-Q Chen and X Huo ldquoThe effects of horizontalsingular straight line in a generalized nonlinear Klein-Gordonmodel equationrdquo Nonlinear Dynamics vol 72 no 4 pp 789ndash801 2013
[14] J Li and H Dai On the Study of Singular Nonlinear TravelingWave Equations Dynamical System Approach Science PressBeijing China 2007
[15] A Chen S Wen S Tang W Huang and Z Qiao ldquoEffects ofquadratic singular curves in integrable equationsrdquo Studies inApplied Mathematics vol 134 no 1 pp 24ndash61 2015
[16] L Tian and J Yin ldquoNew compacton solutions and solitary wavesolutions of fully nonlinear generalized Camassa-Holm equa-tionsrdquo Chaos Solitons amp Fractals vol 20 no 2 pp 289ndash2992004
[17] L Zhang ldquoThe bifurcation and peakons for the specialC(3 2 2)equationrdquo Pramana vol 83 no 3 pp 331ndash340 2014
[18] J Lenells ldquoTraveling wave solutions of the CamassandashHolmequationrdquo Journal of Differential Equations vol 217 no 2 pp393ndash430 2005
[19] Z J Qiao and G P Zhang ldquoOn peaked and smooth solitons forthe Camassa-Holm equationrdquo Europhysics Letters vol 73 no 5pp 657ndash663 2006
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Mathematical Problems in Engineering
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Stochastic AnalysisInternational Journal of
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 119906(119909 119905) = 120593(120585)
with 120585 = 119909 minus 119888119905 be the solution of (10) then it follows that
(119896 minus 119888) 1205931015840minus 1205741119888120593101584010158401015840
+ 1205742(1205933)1015840
+ 12057431205931015840(1205932)10158401015840
+ 1205743120593 (1205932)101584010158401015840
= 0
(11)
where 1205931015840 = 120593120585 12059310158401015840 = 120593
120585120585 and 120593101584010158401015840 = 120593
120585120585120585 Integrating (11) once
and setting the integration constant as 119892 we have
(119896 minus 119888) 120593 minus 120574111988812059310158401015840+ 12057421205933+ 1205743120593 (1205932)10158401015840
= minus119892 (12)
Clearly (12) is equivalent to the planar system
119889120593
119889120585= 119910
119889119910
119889120585=1205731205933+ 1205931199102+ 120598120593 + 120590
120588 minus 1205932
(13)
where 120588 = 119888120574121205743 120573 = 120574
221205743 120590 = 1198922120574
3 and 120598 = (119896minus119888)2120574
3
(1205743
= 0) System (13) has the first integral
119867(120593 119910) =120573
41205934+1
21205981205932+ 120590120593 +
1
212059321199102minus1
21205881199102= ℎ (14)
Obviously for 120588 gt 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 120588 gt 0 in the rest of thispaper (120588 = 984858
2 984858 gt 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 120593 = 984858 and 120593 = minus984858 on the right-hand side of the secondequation of (13) To avoid the singular lines temporarily wedefine a new independent variable 120577 by setting (119889120585119889120577) =
9848582minus1205932 then system (13) is changed to a Hamiltonian system
written as119889120593
119889120577= (9848582minus 1205932) 119910
119889119910
119889120577= 1205931199102+ 1205731205933+ 120598120593 + 120590
(15)
System (15) has the same topological phase portraits as system(13) except for the singular lines 120593 = 984858 and 120593 = minus984858
We now investigate the bifurcation of phase portraits ofthe system (15) Denote that
119891 (120593) = 1205731205933+ 120598120593 + 120590 (16)
Let 119872(120593119890 119910119890) be the coefficient matrix of the linearized
system of (15) at the equilibrium point (120593119890 119910119890) then
119872(120593119890 119910119890) = (
minus21205931198901199101198901199102
120598+ 1198911015840(120593119890)
9848582minus 1205932
1198902120593119890119910119890
) (17)
and at this equilibrium point we have
119869 (120593119890 119910119890) = det119872(120593
119890 119910119890)
= (1205932
119890minus 9848582) 1198911015840(120593119890) minus (3120593
2
119890+ 9848582) 1199102
119890
(18)
By the theory of planar dynamical systems for an equilibriumpoint of a Hamiltonian system if 119869 lt 0 then it is a saddlepoint a center point if 119869 gt 0 and a degenerate equilibriumpoint if 119869 = 0
From the above analysis we can obtain the bifurcationcurves and phase portraits under different parameter condi-tions
Let
1205731(120590) = minus
41205983
271205902
1205732(120590) =
120590 minus 120598984858
9848583
1205733(120590) =
minus120590 minus 120598984858
9848583
(19)
Clearly for 120573 gt 1205731(120590) the function 119891(120593) = 0 has three
real roots 1205931 1205932 and 120593
3(1205931
gt 1205932
gt 1205933) that is system
(15) has three equilibrium points 119864119894(120593119894 0) 119894 = 1 2 3 on
the 120593-axis When 120573 lt 1205733(120590) (15) has two equilibrium
points 11987812(984858 plusmn119884
1) on the straight line 120593 = radic119886 where 119884
1=
radic(minus120590 minus 984858(120598 + 1205739848582))984858 When 120573 lt 1205732(120590) system (15) has two
equilibrium points 11987834(minus984858 plusmn119884
2) on the straight line 120593 = minus984858
where 1198842= radic(120590 minus 984858(120598 + 1205739848582))984858 Notice that on making the
transformation 120590 rarr minus120590 120593 rarr minus120593 120577 rarr minus120577 system (15)is invariant This means that for 120590 lt 0 the phase portraitsof (15) are just the reflections of the corresponding phaseportraits of (15) in the case 120590 gt 0 with respect to the 119910-axisThus we only need to consider the case 120590 ge 0 To knowthe dynamical behavior of the orbits of system (15) we willdiscuss two cases 120598 gt 0 and 120598 lt 0
21 Case I 120598 gt 0
Lemma 1 Suppose that 120598 gt 0 Denote ℎ+984858= 119867(984858 plusmn119884
1) ℎminus984858=
119867(minus984858 plusmn1198842) and ℎ
119894= 119867(120593
119894 0) 119894 = 1 2 3
(1) For 120590 gt 0 and 1205731(120590) lt 120573 lt 120573
3(120590) there exists one
and only one curve 120573 = 1205734(120590) on the (120590 120573)-plane on which
ℎ+
984858= ℎ2 for 120590 gt 0 and 120573
1(120590) lt 120573 lt 120573
2(120590) there exists one
and only one curve 120573 = 1205735(120590) on the (120590 120573)-plane on which
ℎminus
984858= ℎ2or ℎminus984858= ℎ3Moreover the curves120573 = 120573
1(120590)120573 = 120573
2(120590)
120573 = 1205734(120590) and 120573 = 120573
5(120590) are tangent at the point (120590
1119904 1205731119904)
The curve 120573 = 1205733(120590) intersects with the curves 120573 = 120573
1(120590)
120573 = 1205734(120590) and 120573 = 120573
5(120590) at the points (120590
2119904 1205732119904) (1205903119904 1205733119904)
and (1205904119904 1205734119904) (0 lt 120590
4119904lt 1205903119904lt 1205902119904lt 1205901119904) respectively
(2) For 120590 = 0 there exists a bifurcation point 119860(0 minus21205989848582)on (120590 120573)-plane on which ℎ
+
984858= ℎminus
984858= ℎ2when 120573 = minus2120598984858
2ℎ+
984858= ℎminus
984858lt ℎ2when 120573 lt minus2120598984858
2 and ℎ+
984858= ℎminus
984858gt ℎ2when
minus21205989848582lt 120573 lt minus120598984858
2
4 Mathematical Problems in Engineering
1205732(120590)
(1205901s 1205731s)
1205733(120590)
1205734(120590)1205735(120590)
120573
minus2
minus2
minus4
minus4
minus6
minus8
minus10
minus12
minus14
1205731(120590)
1205902 4
Figure 1 The bifurcation curves in the (120590 120573)-parameter plane for120598 gt 0
According to the above analysis and Lemma 1 we obtainthe following proposition on the bifurcation curves of thephase portraits of system (15) for 120598 gt 0
Proposition 2 When 120598 gt 0 for system (15) in (120590 120573)-parameter plane there exist five bifurcation curves (see Fig-ure 1)
120573 = 1205731(120590) = minus
41205983
271205902
120573 = 1205732(120590) =
120590 minus 120598984858
9848583
120573 = 1205733(120590) =
minus120590 minus 120598984858
9848583
120573 = 1205734(120590) (|120590| le 120590
3119904)
120573 = 1205735(120590)
(20)
These five curves divide the right-half (120590 120573)-parameter planeinto thirty-one regions as follows
1198601 (120590 120573) | 1205901s lt 120590 lt 120598984858 120573 = 1205732(120590)
1198602 (120590 120573) | 120590 gt 1205901s 1205735(120590) lt 120573 lt 1205732(120590)
1198603 (120590 b) | 120590 gt 1205901s 120573 = 1205735(120590)
1198604 (120590 120573) | 120590 gt 1205901s 1205731(120590) lt 120573 lt 1205735(120590)
1198605 (120590 120573) | 120590 gt 1205901s 120573 = 1205731(120590)
1198606 (120590 120573) | 120590 gt 1205902s 1205733(120590) lt 120573 lt 1205731(120590)
1198607 (120590 120573) | 120590 lt 1205902s 120573 = 1205733(120590)
1198608 (120590 120573) | 120590 gt 0 120573 lt max1205731(120590) 1205733(120590)
1198609 (120590 120573) = (1205902s 1205732s)
11986010 (120590 120573) | 0 lt 120590 lt 1205902s 120573 = 1205731(120590)
11986011 (120590 120573) | 120590 lt 1205902s 1205731(120590) lt 120573 lt min1205733(120590)
1205734(120590)
11986012 (120590 120573) | 120590 lt 1205903s 120573 = 1205734(120590)
11986013 (120590 120573) | 120590 gt 0 1205734(120590) lt 120573 lt 1205735(120590)
11986014 (120590 120573) | 120590 lt 1205904s 120573 = 1205735(120590)
11986015 (120590 120573) | 120590 gt 0 1205735(120590) lt 120573 lt 1205733(120590)
11986016 (120590 120573) | 120590 = 0 120573 lt minus21205989848582
11986017 (120590 120573) = (0 minus21205989848582)
11986018 (120590 120573) | 120590 = 0 minus21205989848582 lt 120573 lt minus120598984858
2
11986019 (120590 120573) = (0 minus1205989848582)
11986020 (120590 120573) | 0 lt 120590 lt 1205904s 120573 = 1205733(120590)
11986021 (120590 120573) = (1205904s 1205734s)
11986022 (120590 120573) | 1205904s lt 120590 lt 1205902s 120573 = 1205733(120590)
11986023 (120590 120573) | 1205904s lt 120590 lt 1205901s max1205731(120590) 1205733(120590) lt
120573 lt 1205735(120590)11986024 (120590 120573) | 120590
s4 lt 120590 lt 1205901s 120573 = 1205735(120590)
11986025 (120590 120573) | 1205904s lt 120590 lt 1205901s max1205733(120590) 1205735(120590) lt
120573 lt 1205732(120590)11986026 (120590 120573) | 0 lt 120590 lt 1205901s 120573 = 1205732(d)
11986027 (120590 120573) = (1205901s 1205731s)
11986028 (120590 120573) | 0 le 120590 lt 120598984858 120573 gt 1205732(120590)
11986029 (120590 120573) | 120590 isin R 120573 gt max0 1205732(120590)
11986030 (120590 120573) | 120590 gt 120598984858 120573 = 1205732(120590)
11986031 (120590 120573) | 120590 gt 120598984858 0 lt 120573 lt 1205732(120590)
In this case the phase portraits of system (15) can beshown in Figure 2
22 Case II 120598 lt 0 In this case we have the following
Proposition 3 When 120598 lt 0 for system (15) in (120590 120573)-parameter plane there exist four parametric bifurcation curves(see Figure 3)
120573 = 1205731(120590) = minus
41205983
271205902
120573 = 1205732(120590) =
120590 minus 120598984858
9848583
120573 = 1205733(120590) =
minus120590 minus 120598984858
9848583 120590 = 0
(21)
These four curves divide the right-half (120590 120573)-parameter planeinto twenty-two regions
1198611 (120590 120573) | 1205901s lt 120590 lt 120598984858 120573 = 1205733(120590)
1198612 (120590 120573) | 120590 gt 1205901s max0 1205733(120590) lt 120573 lt 1205731(120590)
1198613 (120590 120573) | 120590 gt 1205901s 120573 = 1205731(120590)
1198614 (120590 120573) | 120590 gt 1205902s 1205731(120590) lt 120573 lt 1205732(120590)
1198615 (120590 120573) = (1205902s 1205732s)
1198616 (120590 120573) | 120590 gt 1205902s 120573 = 1205732(120590)
1198617 (120590 120573) | 120590 gt 0 120573 gt max1205731(120590) 1205732(120590)
1198618 (120590 120573) | 0 lt 120590 lt 1205902s 120573 = 1205731(120590)
1198619 (120590 120573) | 0 lt 120590 lt 1205902s 1205732(120590) lt 120573 lt 1205731(120590)
11986110 (120590 120573) | 120590 = 0 120573 gt minus120598984858
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
00 05 10
0
2
4
00 05 10
0
2
4
00 05 10
0
1
2
3
00 05 10
0
2
4
6
00 05 10
0
2
4
00 05 10
0
2
4
minus3
minus3
minus2 minus2 minus2minus1minus4
minus4
minus2 minus2
minus2
minus1
minus1
minus4minus2minus4
minus4 minus4
minus2 minus2 minus2
minus4
minus2
minus4
minus2
minus4
minus2
minus15 minus10 minus05
minus10 minus05 minus10 minus05 minus10 minus05
minus15 minus10 minus05 minus15 minus10 minus05
minus6
minus2
minus4
minus2
minus4
minus2
minus4
minus2
minus4
minus2
minus4
120593120593120593
120593 120593 120593
120593120593120593
120593 120593 120593
yy
y
y yy
yyy
y
y y
(1) (120590 120573) isin A1
(4) (120590 120573) isin A4
(7) (120590 120573) isin A7 (8) (120590 120573) isin A8 (9) (120590 120573) isin A9
(5) (120590 120573) isin A5
(2) (120590 120573) isin A2 (3) (120590 120573) isin A3
(6) (120590 120573) isin A6
(10) (120590 120573) isin A10 (11) (120590 120573) isin A11 (12) (120590 120573) isin A12
Figure 2 Continued
6 Mathematical Problems in Engineering
00 05 10
0
1
2
3
00 05 10
0
2
4
00 05 10
0
1
2
3
00 05 10
0
1
2
3
00 05 10
0
1
2
00 05 10
0
1
2
00 05 10
0
1
2
00 05 10
0
1
2
00 05 10
0
1
2
3
00 05 10
0
1
2
3
00 05 10
0
1
2
3
00 05 10
0
1
2
3
minus3
minus2
minus1
minus3
minus2
minus1
minus2
minus1
minus2
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus1
minus2
minus1
minus3
minus2
minus1
minus10 minus05
minus10 minus05
minus10 minus05
minus10 minus05
minus10 minus05
minus10minus15 minus05 minus10minus15 minus05
minus10 minus05
minus10 minus05 minus10 minus05
minus10 minus05 minus10 minus05
minus2
minus4
y
120593120593120593
120593 120593 120593
120593120593120593
120593 120593120593
yy
y yy
yy
y
yy
y
(13) (120590 120573) isin A13
(16) (120590 120573) isin A16
(19) (120590 120573) isin A19 (20) (120590 120573) isin A20
(23) (120590 120573) isin A23(24) (120590 120573) isin A24
(21) (120590 120573) isin A21
(22) (120590 120573) isin A22
(17) (120590 120573) isin A17 (18) (120590 120573) isin A18
(14) (120590 120573) isin A14 (15) (120590 120573) isin A15
Figure 2 Continued
Mathematical Problems in Engineering 7
00 05 10 15
0
1
2
3
00 05 10 15
0
1
2
3
0 1 2
0
1
2
3
0 1 2
0
1
2
3
00 05 10 15
0
1
2
3
0 1 2
0
2
4
0 1 2
0
2
4
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus2
minus4
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2 minus1
minus2minus3 minus1
minus2
minus2
minus3
minus4
minus1
minus2 minus1
minus10minus15 minus05 minus10minus15 minus05
minus10minus15 minus05
y
yyy
y y y
120593
120593120593120593
120593 120593 120593
(25) (120590 120573) isin A25
(28) (120590 120573) isin A28 (29) (120590 120573) isin A29
(31) (120590 120573) isin A31
(30) (120590 120573) isin A30
(26) (120590 120573) isin A26 (27) (120590 120573) isin A27
Figure 2 The bifurcation of phase portraits of system (15) when 120598 gt 0 and 120590 ge 0
11986111 (120590 120573) = (0 minus1205989848582)
11986112 (120590 120573) | 0 lt 120590 lt 1205902s 120573 = 1205732(120590)
11986113 (120590 120573) | 0 lt 120590 lt 1205901s 1205733(120590) lt 120573 lt min1205731(120590)
1205732(120590)
11986114 (120590 120573) | 1205904s lt 120590 lt 1205901s 120573 = 1205731(120590)
11986115 (120590 120573) = (1205901s 1205731s)
11986116 (120590 120573) | 0 lt 120590 lt 1205901s 120573 = 1205733(120590)
11986117 (120590 120573) | 0 lt 120590 lt 120598984858 120573 lt 1205733(120590)
11986118 (120590 120573) | 120590 = 0 120573 = minus120598984858
2
11986119 (120590 120573) | 120590 = 0 120573 lt 0
11986120 (120590 120573) | 0 lt 120590 lt 120598984858 120573 lt min0 1205733(120590)
11986121 (120590 120573) | 120590 gt 120598984858 120573 = 1205733(120590)
11986122 (120590 120573) | 120590 gt 120598984858 1205733(120590) lt 120573 lt 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
1205732(120590)
(1205901s 1205731s)
(1205902s 1205732s)
1205733(120590)
120573
minus2minus4minus6
1205731(120590)
120590
Figure 3 The bifurcation curves in the (120590 120573)-parameter plane for120598 lt 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 Let119862119896(Ω) denote the set of all 119896 times continuouslydifferentiable functions on the open setΩ 119871119901loc(R) refer to theset of all functions whose restriction on any compact subsetis 119871119901 integrable1198671loc(R) stands for1198671loc(R) = 120593 isin 119871
2
loc(R) |
1205931015840isin 1198712
loc(R)To study single peak soliton solutions we impose the
boundary condition
lim120585rarrplusmninfin
120593 = 119860 (22)
where 119860 is a constant In fact the constant 119860 is equal to thehorizontal coordinate of saddle point 119864(120593
119890 0) Substituting
the boundary condition (22) into (14) generates the followingconstant
ℎ = 119860(120590 +119860120598
2+1198603120573
4) (23)
So the ODE (14) becomes
(1205931015840)2
=
(120593 minus 119860)2
(1205731205932+ 2119860120573120593 + 3119860
2120573 + 2120598)
2 (9848582 minus 1205932) (24)
If 120573(1198602120573 + 120598) le 0 then (24) reduces to
(1205931015840)2
=120573 (120593 minus 119860)
2
(120593 minus 1198611) (120593 minus 119861
2)
2 (9848582 minus 1205932) (25)
where
1198611= minus119860 minus
radicminus2120573 (1198602120573 + 120598)
120573
1198612= minus119860 +
radicminus2120573 (1198602120573 + 120598)
120573
(26)
From (26) we know that 1198611gt 1198612if 120573 lt 0 and 119861
1lt 1198612if
120573 gt 0
Definition 4 A function120593(120585) is said to be a single peak solitonsolution for the 119862(3 2 2) equation (10) if 120593(120585) satisfies thefollowing conditions
(C1) 120593(120585) is continuous on R and has a unique peakpoint 120585
0 where 120593(120585) attains its global maximum or
minimum value
(C2) 120593(120585) isin 119862
3(R minus 120585
0) satisfies (24) on R minus 120585
0
(C3) 120593(120585) satisfies the boundary condition (22)
Definition 5 A wave function 120593 is called smooth solitonsolution if 120593 is smooth locally on either side of 120585
0and
lim120585uarr1205850
1205931015840(120585) = lim
120585darr12058501205931015840(120585) = 0
Definition 6 A wave function 120593 is called peakon if 120593 issmooth locally on either side of 120585
0and lim
120585uarr12058501205931015840(120585) =
minuslim120585darr1205850
1205931015840(120585) = 119886 119886 = 0 119886 = plusmninfin
Definition 7 Awave function120593 is called cuspon if120593 is smoothlocally on either side of 120585
0and lim
120585uarr12058501205931015840(120585) = minuslim
120585darr12058501205931015840(120585) =
plusmninfin
Without any loss of generality we choose the peak point1205850as vanishing 120585
0= 0
Theorem 8 Assume that 119906(119909 119905) = 120593(120585) = 120593(119909 minus 119888119905) is a singlepeak soliton solution of the 119862(3 2 2) equation (10) at the peakpoint 120585
0= 0 Then we have the following
(i) If 120573(1198602120573 + 120598) gt 0 then 120593(0) = 984858 or 120593(0) = minus984858
(ii) If 120573(1198602120573 + 120598) le 0 then 120593(0) = 984858 or 120593(0) = minus984858 or120593(0) = 119861
1or 120593(0) = 119861
2
Proof If 120593(0) = plusmn984858 then 120593(120585) = plusmn984858 for any 120585 isin R since120593(120585) isin 119862
3(R minus 0) Differentiating both sides of (24) yields
120593 isin 119862infin(R)
(i) When 120573(1198602120573 + 120598) gt 0 if 120593(0) = 984858 and 120593(0) = minus984858 then
120593 isin 119862infin(R) By the definition of single peak soliton we have
1205931015840(0) = 0 However by (24) we must have 120593(0) = 119860 which
contradicts the fact that 0 is the unique peak point(ii) When 120573(119860
2120573 + 120598) le 0 if 120593(0) = 984858 and 120593(0) = minus984858
by (24) we know 1205931015840(0) exists and 120593
1015840(0) = 0 since 0 is a peak
point Thus we obtain 120593(0) = 1198611or 120593(0) = 119861
2from (25)
since 120593(0) = 119860 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 119906(119909 119905) = 120593(119909 minus 119888119905) is a single peaksoliton solution of the 119862(3 2 2) equation (10) at the peak point1205850= 0 Then we have the following solution classification(i) If |120593(0)| = 984858 then 120593(120585) isin 119862
infin(R) and 120593 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
00 05 10 15
0
2
4
0 1
0
2
4
6
00 05 10 15
0
2
4
00 05 10
0
1
2
3
00 05 10 15
0
1
2
3
00 05 10
0
1
2
00 05 10 15
0
1
2
00 05 10 15
0
1
2
3
minus3 minus1minus4
minus4
minus2
minus4
minus2
minus4
minus2
minus4
minus2minus4
minus6
minus2
minus4
minus2
minus4
minus2
minus2 minus4 minus4minus2 minus2
minus3
minus1
minus2
minus1
minus2
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus1minus2 minus1minus2
120593
120593
120593
120593 120593 120593
120593 120593
120593 120593
120593 120593
y
y
y
yy
y
y y
y y
y y
minus15 minus10 minus05
minus15minus20 minus10 minus05
minus15 minus10 minus05 minus15 minus10 minus05minus10 minus05
minus15 minus10 minus05minus10 minus05
(10) (120590 120573) isin B10 (11) (120590 120573) isin B11 (12) (120590 120573) isin B12
(1) (120590 120573) isin B1
(4) (120590 120573) isin B4
(7) (120590 120573) isin B7 (8) (120590 120573) isin B8 (9) (120590 120573) isin B9
(5) (120590 120573) isin B5 (6) (120590 120573) isin B6
(2) (120590 120573) isin B2 (3) (120590 120573) isin B3
Figure 4 Continued
10 Mathematical Problems in Engineering
00 05 10 15
0
1
2
3
0 1
0
1
2
3
0 1 2
0
2
4
00 05 10 15
0
1
2
3
0 1 2
0
1
2
3
00 05 10 15
0
1
2
00 05 10 15
0
1
2
3
0 1 2
0
1
2
3
00 05 10 15
0
2
4
00 05 10 15
0
2
4
6
minus4
minus2
minus4
minus2
minus4
minus6
minus2
minus3 minus1minus2
minus3 minus1minus2
minus1minus2
minus1minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus1
minus2
minus3
minus1
minus2
120593 120593 120593
120593120593120593
120593 120593 120593
120593
y y y
yy
y
y yy
y
minus15 minus10 minus05
minus15minus20 minus10 minus05
minus15 minus10 minus05
minus15 minus10 minus05
minus15 minus10 minus05
minus15 minus10 minus05
(13) (120590 120573) isin B13 (14) (120590 120573) isin B14 (15) (120590 120573) isin B15
(16) (120590 120573) isin B16 (17) (120590 120573) isin B17 (18) (120590 120573) isin B18
(22) (120590 120573) isin B22
(20) (120590 120573) isin B20 (21) (120590 120573) isin B21(19) (120590 120573) isin B19
Figure 4 The bifurcation of phase portraits of system (15) when 120598 lt 0 and 120590 ge 0
Mathematical Problems in Engineering 11
(ii) If 120593(0) = 984858 = 1198611 then 120593 is a cuspon solution and 120593 has
the following asymptotic behavior
120593 (120585) minus 984858 = 1205821
10038161003816100381610038161205851003816100381610038161003816
23
+ 119874(10038161003816100381610038161205851003816100381610038161003816
43
) 120585 997888rarr 0
1205931015840(120585) =
2
31205821
10038161003816100381610038161205851003816100381610038161003816
minus13 sgn (120585) + 119874 (10038161003816100381610038161205851003816100381610038161003816
13
)
120585 997888rarr 0
(27)
where 1205821= plusmn(9|120573984858
2+ 2119860120573984858 + 3119860
2120573 + 2120598|(984858 minus 119860)
216984858)13
Thus 120593(120585) notin 1198671
119897119900119888(R)
(iii) If 120593(0) = minus984858 = 1198612 then 120593 is a cuspon solution and 120593
has the following asymptotic behavior
120593 (120585) + 984858 = 1205822
10038161003816100381610038161205851003816100381610038161003816
23
+ 119874(10038161003816100381610038161205851003816100381610038161003816
43
) 120585 997888rarr 0
1205931015840(120585) =
2
31205822
10038161003816100381610038161205851003816100381610038161003816
minus13 sgn (120585) + 119874 (10038161003816100381610038161205851003816100381610038161003816
13
)
120585 997888rarr 0
(28)
where 1205822= plusmn(9|120573984858
2minus 2119860120573984858 + 3119860
2120573 + 2120598|(984858 + 119860)
216984858)13
Thus 120593(120585) notin 1198671
119897119900119888(R)
(iv) If 120593(0) = 984858 = 1198611and 119860 = 0 then 120593 is a peakon-like
solution and
120593 (120585) minus 984858 = 1205823
10038161003816100381610038161205851003816100381610038161003816 + 119874 (
10038161003816100381610038161205851003816100381610038161003816
2
) 120585 997888rarr 0
1205931015840(120585) = 120582
3sgn (120585) + 119874 (
10038161003816100381610038161205851003816100381610038161003816) 120585 997888rarr 0
(29)
where 1205823= plusmn|119860 minus 984858|radic120573(119861
2minus 984858)4984858
(v) If 120593(0) = minus984858 = 1198612and 119860 = 0 then 120593 is a peakon-like
solution and
120593 (120585) + 984858 = 1205824
10038161003816100381610038161205851003816100381610038161003816 + 119874 (
10038161003816100381610038161205851003816100381610038161003816
2
) 120585 997888rarr 0
1205931015840(120585) = 120582
4sgn (120585) + 119874 (
10038161003816100381610038161205851003816100381610038161003816) 120585 997888rarr 0
(30)
where 1205824= plusmn|119860 + 984858|radicminus120573(119861
1+ 984858)4984858
(vi) If 120593(0) = 984858 = 1198611and 119860 = 0 then 120593 gives the peakon
solution 984858 exp(minusradicminus1205732|119909 minus 119888119905|)(vii) If 120593(0) = minus984858 = 119861
2and 119860 = 0 then 120593 gives the peakon
solution minus984858 exp(minusradicminus1205732|119909 minus 119888119905|)
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 |120593(0)| = 984858 then 120593 isin 119862
infin(R) and 120593 is a smooth soliton
solution(ii) If 120593(0) = 984858 = 119861
1 then by the definition of single peak
soliton we have 119860 = 984858 thus 1205731205932 + 2119860120573120593 + 31198602120573 + 2120598 does
not contain the factor 120593 minus 984858 From (24) we obtain
1205931015840= sgn (119860 minus 984858)
1003816100381610038161003816120593 minus 1198601003816100381610038161003816
radic210038161003816100381610038161205932 minus 9848582
1003816100381610038161003816
sdot radic10038161003816100381610038161205731205932 + 2119860120573120593 + 31198602120573 + 2120598
1003816100381610038161003816 sgn (120585)
(31)
Let 1198971(120593) = radic2(120593 + 984858)|120593 minus 119860|radic|1205731205932 + 2119860120573120593 + 31198602120573 + 2120598|
then 1198971(984858) = 2radic984858|984858 minus 119860|radic|1205739848582 + 2119860120573984858 + 31198602120573 + 2120598| and
int 1198971(120593)radic
1003816100381610038161003816120593 minus 9848581003816100381610038161003816119889120593 = int sgn (119860 minus 984858) sgn (120585) 119889120585 (32)
Inserting 1198971(120593) = 119897
1(984858) + 119874(|120593 minus 984858|) into (32) and using the
initial condition 120593(0) = 984858 we obtain
21198971(984858)
3
1003816100381610038161003816120593 minus 9848581003816100381610038161003816
32
(1 + 119874 (1003816100381610038161003816120593 minus 984858
1003816100381610038161003816)) =10038161003816100381610038161205851003816100381610038161003816
(33)
thus
120593 minus 984858 = plusmn(3
21198971(984858)
)
23
10038161003816100381610038161205851003816100381610038161003816
23
(1 + 119874 (1003816100381610038161003816120593 minus 984858
1003816100381610038161003816))minus23
= plusmn(3
21198971(984858)
)
23
10038161003816100381610038161205851003816100381610038161003816
23
(1 + 119874 (1003816100381610038161003816120593 minus 984858
1003816100381610038161003816))
(34)
which implies 120593 minus 984858 = 119874(|120585|23
) Therefore we have
120593 (120585) = 984858 plusmn (3
21198971(984858)
)
23
10038161003816100381610038161205851003816100381610038161003816
23
+ 119874(10038161003816100381610038161205851003816100381610038161003816
43
)
= 984858 + 1205821
10038161003816100381610038161205851003816100381610038161003816
23
+ 119874(10038161003816100381610038161205851003816100381610038161003816
43
) 120585 997888rarr 0
1205821= plusmn(
3
21198971(984858)
)
23
= plusmn(
9100381610038161003816100381610038161205739848582+ 2119860120573984858 + 3119860
2120573 + 2120598
10038161003816100381610038161003816(984858 minus 119860)
2
16984858)
13
1205931015840(120585) =
2
31205821
10038161003816100381610038161205851003816100381610038161003816
minus13 sgn (120585) + 119874 (10038161003816100381610038161205851003816100381610038161003816
13
) 120585 997888rarr 0
(35)
So 120593(120585) notin 1198671
loc(R)(iii) Similar to the proof of (ii) we ignore it in this paper(iv) If 120593(0) = 984858 = 119861
1and 119860 = 0 then from (25) we obtain
1205931015840= radicminus
120573
2sgn (119860 minus 984858)
1003816100381610038161003816120593 minus 1198601003816100381610038161003816radic120593 minus 1198612
120593 + 984858sgn (120585) (36)
Let 1198972(120593) = (1|120593minus119860|)radic(120593 + 984858)(120593 minus 119861
2) then 119897
2(984858) = (1|984858minus
119860|)radic2984858(984858 minus 1198612) and
int 1198972(120593) 119889120593 = radicminus
120573
2int sgn (119860 minus 984858) sgn (120585) 119889120585 (37)
Inserting 1198972(120593) = 119897
2(984858) + 119874(|120593 minus 984858|) into (37) and using the
initial condition 120593(0) = 984858 we obtain
1198972(984858) (120593 minus 984858) (1 + 119874 (
1003816100381610038161003816120593 minus 9848581003816100381610038161003816))minus1
= sgn (119860 minus 984858)10038161003816100381610038161205851003816100381610038161003816
(38)
Sincesgn (120593 minus 984858) sgn (119860 minus 984858) ge 0
1
1 + 119874 (120593 minus 984858)= 1 + 119874 (120593 minus 984858)
(39)
12 Mathematical Problems in Engineering
we get
1003816100381610038161003816120593 minus 9848581003816100381610038161003816 =
radicminus120573
2
1
1198972(984858)
10038161003816100381610038161205851003816100381610038161003816 (1 + 119874 (120593 minus 984858)) (40)
which implies |120593 minus 984858| = 119874(|120585|) Therefore we have
120593 (120585) = 984858 + 1205823
10038161003816100381610038161205851003816100381610038161003816 + 119874 (
10038161003816100381610038161205851003816100381610038161003816
2
) 120585 997888rarr 0
1205931015840(120585) = 120582
3sgn (120585) + 119874 (
10038161003816100381610038161205851003816100381610038161003816) 120585 997888rarr 0
(41)
where 1205823= plusmn|119860 minus 984858|radic120573(119861
2minus 984858)4984858
(v) Similar to the proof of (iv) we ignore it in this paper
By virtue of Theorem 9 any single peak soliton for the119862(3 2 2) equation (10) must satisfy the following initial andboundary values problem
(1205931015840)2
=120573 (120593 minus 119860)
2
(120593 minus 1198611) (120593 minus 119861
2)
2 (9848582 minus 1205932)fl119871 (120593)
120593 (0) isin 984858 minus984858 1198611 1198612
lim|120585|rarrinfin
120593 (120585) = 119860
(42)
119871(120593) ge 0 and the boundary condition (24) imply thefollowing
(a) If 120573(9848582 minus 1205932) ge 0 then 120593 ge max119861
1 1198612 or 120593 le
min1198611 1198612
(b) If120573(9848582minus1205932) le 0 thenmin1198611 1198612 le 120593 le max119861
1 1198612
Below we will present some implicit formulas for thesingle peak soliton solutions in the case of specific 120590 and 120573
Case 1 ((120590 120573) isin 11986012) In this case we have minus984858 lt 119861
2lt 119860 lt
1198611= 984858 From the standard phase analysis and Theorem 9
we know that if 120593 is a single peak soliton of the 119862(3 2 2)
equation then
1205931015840= minusradicminus
120573
2(120593 minus 119860)radic
120593 minus 1198612
120593 + 984858sgn (120585) (43)
From the separation of variables we get
intℎ (120593) 119889120593 = radicminus120573
2sgn (120585) 119889120585 (44)
where ℎ(120593) = (1(119860 minus 120593))radic(120593 + 984858)(120593 minus 1198612) After a lengthy
calculation of integral we obtain the implicit solution 120593
defined by
119867(120593)flradic119860 + 984858
119860 minus 1198612
1198681(120593) minus 119868
2(120593) = radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870 (45)
where
1198681(120593) = ln
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
(119860 + 984858) (120593 minus 1198612) + (119860 minus 119861
2) (120593 + 984858) + 2radic(119860 + 984858) (119860 minus 119861
2) (120593 minus 119861
2) (120593 + 984858)
radic(119860 + 984858) (119860 minus 1198612) (119860 + 984858) (119860 minus 120593)
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
1198682(120593) = ln
1003816100381610038161003816100381610038161003816(120593 + 984858) + (120593 minus 119861
2) + 2radic(120593 + 984858) (120593 minus 119861
2)
1003816100381610038161003816100381610038161003816
(46)
and 119870 is an arbitrary integration constant For 120593(0) = 984858 theconstant119870 = 119867(120593(0)) is defined by
119870 = radic119860 + 984858
119860 minus 1198612
1198681(984858) minus 119868
2(984858) (47)
and for 120593(0) = 1198612
119870 = radic119860 + 984858
119860 minus 1198612
1198681(1198612) minus 1198682(1198612) (48)
(i) If 120593(0) = 984858 then 119860 lt 120593 le 984858 Since 1198671015840(120593) = ℎ(120593) weknow that119867(120593) strictly decreases on the interval (119860 984858] thus1198671(120593) = 119867|
(119860984858](120593) gives a single peak solitonwith119867
1(984858) = 119870
and1198671(119860+) = infin Therefore 120593
1(120585) = 119867
minus1
1(radicminus1205732|120585| + 119870) is
the solution satisfying
1205931(0) = 984858
lim|120585|rarrinfin
1205931(120585) = 119860
1205931015840
1(0plusmn) = plusmn (119860 minus 984858)radic
120573 (1198612minus 984858)
4984858
(49)
So 1205931(120585) is a peakon-like solution (see Figure 5)
(ii) If 120593(0) = 1198612 then 119861
2le 120593 lt 119860 By 119867
1015840(120593) = 119891(120593)
we know that 119867(120593) strictly increases on the interval [1198612 119860)
Thus
1198672(120593) = 119867|
[1198612 119860)(120593) (50)
Mathematical Problems in Engineering 13
minus3 minus2 minus1
120593
minus02
1 2 3
02
04
06
08
10
x
(a)
02
02
minus2 minus2 x
10
05
00
t
120593
22
(b)
Figure 5 Two- and three-dimensional graphs of the peakon-like solution
minus03
minus04
minus05
1 2 3minus3 minus2 minus1
120593
minus02
x
(a)
0
2
0
2
minus2minus2
minus02
minus03
minus04
minus05
minus06
t x
120593
0
2
0
2
minus2minus2
t x
(b)
Figure 6 Two- and three-dimensional graphs of the smooth soliton solution
has the inverse denoted by 1205932(120585) = 119867
minus1
2(radicminus1205732|120585| + 119870)
1205932(120585) gives a kind of smooth soliton solution (see Figure 6)
satisfying
1205932(0) = 119861
2
lim|120585|rarrinfin
1205932(120585) = 119860
1205931015840
2(0) = 0
(51)
Case 2 ((120590 120573) isin 11986014) In this case we have minus984858 = 119861
2lt 119860 lt
984858 lt 1198611and (25) is equivalent to
1205931015840= minusradicminus
120573
2(120593 minus 119860)radic
120593 minus 1198611
120593 minus 984858sgn (120585) (52)
Let
119892 (120593) =1
119860 minus 120593radic120593 minus 1198611
120593 minus 984858 (53)
14 Mathematical Problems in Engineering
minus1 1 2 3
02
04
06
08
10
minus3 minus2
120593
x
(a)
02
02minus2
minus2
120593
x t
00
05
10
(b)
Figure 7 Two- and three-dimensional graphs of the cuspon solution
then (52) is converted to
119892 (120593) 119889120593 =1
119860 minus 120593radic120593 minus 1198611
120593 minus 984858119889120593 = radicminus
120573
2sgn (120585) 119889120585 (54)
Integrating (54) on the interval [0 120585] (or [120585 0]) leads to thefollowing implicit solutions
119866 (120593)flradic119860 minus 984858
119860 minus 1198611
1198683(120593) + 119868
4(120593) = radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870 (55)
where
1198683(120593) = ln
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
(119860 minus 984858) (120593 minus 1198611) + (119860 minus 119861
1) (120593 minus 984858) + 2radic(119860 minus 984858) (119860 minus 119861
1) (120593 minus 984858) (120593 minus 119861
1)
radic(119860 minus 984858) (119860 minus 1198611) (119860 minus 984858) (120593 minus 119860)
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
1198684(120593) = ln
1003816100381610038161003816100381610038161003816(120593 minus 984858) + (120593 minus 119861
1) + 2radic(120593 minus 984858) (120593 minus 119861
1)
1003816100381610038161003816100381610038161003816
(56)
And119870 is an arbitrary integration constant It is obvious thatfor 120593(0) = 984858 the constant119870 = 119867(120593(0)) is defined by
119870 = radic119860 minus 984858
119860 minus 1198611
1198683(984858) + 119868
4(984858) (57)
and for 120593(0) = minus984858
119870 = radic119860 minus 984858
119860 minus 1198611
1198683(minus984858) + 119868
4(minus984858) (58)
(i) If 120593(0) = 984858 then 119860 lt 120593 le 984858 From 119892(120593) lt 0 weknow that 119866(120593) strictly decreases on the interval (119860 984858] with119866(984858) = 119870 and 119866(119860+) = infin Define
1198661(120593) = 119866|
(119860984858](120593) (59)
Since 1198661(120593) is a strictly decreasing function from (119860 984858] onto
[119870infin) we can solve for 120593 uniquely from (59) and obtain
1205931(120585) = 119866
minus1
1(radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870) (60)
It is easy to check that 120593 satisfies
1205931(0) = 984858
lim|120585|rarrinfin
1205931(120585) = 119860
1205931015840
1(0plusmn) = ∓infin
(61)
Therefore the solution1205931defined by (60) is a cuspon solution
for the 119862(3 2 2) equation (see Figure 7)
Mathematical Problems in Engineering 15
(ii) If 120593(0) = minus984858 then minus984858 le 120593 lt 119860 Through a similaranalysis we get a strictly increasing function 119866(120593) on theinterval [minus984858 119860) satisfying
119866 (120593) = radicminus120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870 (62)
where 119866(120593) is defined by (55) Let
1198662(120593) = 119866|
[minus984858119860)(120593) (63)
then 1198662(120593) is a strictly increasing function from [minus984858 119860) onto
[119870infin) so that we can solve for 120593 and obtain
1205932(120585) = 119866
minus1
2(radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870) (64)
It is easy to check that 1205932satisfies
1205932(0) = minus984858
lim|120585|rarrinfin
1205932(120585) = 119860
1205931015840
2(0plusmn) = plusmn (119860 + 984858)radicminus
120573 (1198611+ 984858)
4984858
(65)
Therefore the solution 1205932defined by (64) is a peakon-like
solution whose graph is similar to those in Figure 5
Case 3 ((120590 120573) isin 11986016) In this case we have minus984858 lt 119861
2lt 119860 =
0 lt 1198611lt 984858 119861
2= minus1198611 and
1205931015840= minusradicminus
120573
2120593radic
120593 minus 1198612
1
1205932 minus 9848582sgn (120585) (66)
Hence from the separation of variables we have
radic9848582 minus 1205932
120593radic1198612
1minus 1205932
119889120593 = minusradicminus120573
2sgn (120585) 119889120585 (67)
Integrating (67) on the interval [0 120585] (or [120585 0]) leads tothe following implicit formula for the two smooth solitonsolutions
radic9848582 minus 1205932 + radic1198612
1minus 1205932
radic9848582 minus 1198612
1
(
radic(9848582 minus 1198612
1) 1205932
984858radic1198612
1minus 1205932 + 119861
1radic9848582 minus 1205932
)
984858
= exp(minusradicminus120573
2|119909 minus 119888119905|)
(68)
where 120593 isin (119860 1198611] Consider
radic9848582 minus 1198612
2
radic9848582 minus 1205932 + radic1198612
2minus 1205932
(
984858radic1198612
2minus 1205932 minus 119861
2radic9848582 minus 1205932
radic(9848582 minus 1198612
2) 1205932
)
984858
= exp(radicminus120573
2|119909 minus 119888119905|)
(69)
where 120593 isin [1198612 119860)
Case 4 ((120590 120573) isin 11986017) In this case we have minus984858 = 119861
2lt 119860 =
0 lt 1198611= 984858 and
1205931015840= minusradicminus
120573
2120593 sgn (120585) (70)
Choosing 120593(0) = 984858 (or minus984858) as initial value we get
int
120593
984858
1
120593119889120593 = minusint
120585
0
radicminus120573
2sgn (120585) 119889120585 for 119860 lt 120593 le 984858
int
120593
minus984858
1
120593119889120593 = minusint
120585
0
radicminus120573
2sgn (120585) 119889120585 for minus 984858 le 120593 lt 119860
(71)
which immediately yield the peakon solutions
120593 (119909 minus 119888119905) = plusmn984858 exp(minusradicminus120573
2|119909 minus 119888119905|) (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 the119862(3 2 2) equation (10) In order to study kinkwave solutionswe assume that
lim120585rarrinfin
120593 (120585) = 1198601
lim120585rarrminusinfin
120593 (120585) = 1198602
(73)
16 Mathematical Problems in Engineering
2 4
02
04
06
08
10
minus2minus4
120593
x
(a)
0 2
02
00
05
10
minus2
120593
x
t
0
2
minus2
(b)
Figure 8 Two- and three-dimensional graphs of the peakon solution
where 1198601gt 1198602 Substituting the boundary condition (73)
into (14) generates
(1205931015840)2
=
120573 (120593 minus 1198601) (120593 minus 119860
2) [1205932+ (1198601+ 1198602) 120593 + 119860
2
1+ 11986011198602+ 1198602
2+ 2120598120573]
2 (9848582 minus 1205932)fl119865 (120593) (74)
The nonlinear differential equation (74) may sustaindifferent kinds of nonlinear excitations In what follows weconfine our attention to the cases 119860
2= minus119860
1and 120598 =
minus1198602
1120573 which describe kinks and kink compactons which
play an important role in the dynamics systems Under theseconsiderations (74) reduces to
(1205931015840)2
=120573 (120593 minus 119860
1)2
(120593 minus 1198602)2
2 (9848582 minus 1205932) (75)
If 1198601lt 984858 then from the phase analysis in Section 2 (see
Figure 4(10)) we know that (1198601 0) and (119860
2 0) are two saddle
points of (13) and the kink solutions can be obtained fromthe two heteroclinic orbits connecting (120601 119910) = (119860
1 0) and
(1198602 0) When 119860
1increases upon reaching 984858 that is 119860
1= 984858
(75) becomes
(1205931015840)2
= minus120573 (120593 minus 119860
1) (120593 minus 119860
2)
2 (76)
and the ellipse 21199102+120573(120593minus1198601)(120593minus119860
2) = 0 (see Figure 4(11))
which is tangent to the singular lines 120593 = 984858 and 120593 = minus984858 atpoints (119860
1 0) and (119860
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 120593 is a kink wave solutions of (75)
or (76) we have 1205931015840
rarr 0 as 120593 rarr 1198601and as 120593 rarr 119860
2
Moreover we have 119865(120593) ge 0 for 1198602le 120593 le 119860
1and 120593 is
strictly monotonic in any interval where 119865(120593) gt 0 Thus if1205931015840gt 0 at some points 120593will be strictly increasing until it gets
close to the next zero of 119865 Denoting this zero 1198601 we have
120593 uarr 1198601 What will happen to the solution when it approaches
1198601 Depending on whether the zero is double or simple 120593
has a different behavior We explore the two cases in turn
Theorem 11 (i) If 120593 has a simple zero at 120593 = 1198601 so that
119865(1198601) = 0 and 1198651015840(119860
1) lt 0 then the solution 120593 of (75) satisfies
120593 (120585) minus 1198601=1198651015840(1198601)
4(120585 minus 120578)
2
+ 119874 ((120585 minus 120578)4
)
119886119904 120585 uarr 120578
(77)
where 120593(120578) = 1198601
(ii) When 120593 approaches the double zero 1198601of 119865(120593) so that
1198651015840(1198601) = 0 and11986510158401015840(119860
1) gt 0 then the solution120593 of (75) satisfies
120593 (120585) minus 1198601sim 120572 exp(minus120585radic11986510158401015840 (119860
1)) 119886119904 120585 997888rarr infin (78)
for some constant 120572 Thus 120593 uarr 1198601exponentially as 120585 rarr infin
Mathematical Problems in Engineering 17
Proof (i) When 1198601= minus119860
2= 984858 and 120598 = minus119860
2
1120573 from (76)
119865(120593) has a simple zero at 120593 = 1198601 Then
(1205931015840)2
= (120593 minus 1198601) 1198651015840(1198601) + 119874 ((120593 minus 119860
1)2
)
as 120593 uarr 1198601
(79)
Using the fact that (1205931015840)2 ge 0 we know that 1198651015840(1198601) lt 0
Moreover
119889120585
119889120593=
1
radic(120593 minus 1198601) 1198651015840 (119860
1) + 119874 ((120593 minus 119860
1)2
)
(80)
Because
radic(120593 minus 1198601) 1198651015840 (119860
1) + 119874 ((120593 minus 119860
1)2
)
= radic1198601minus 120593(radicminus1198651015840 (119860
1) + 119874 (119860
1minus 120593))
1
radicminus1198651015840 (1198601) + 119874 (119860
1minus 120593)
=1
radicminus1198651015840 (1198601)
+ 119874 (1198601minus 120593)
(81)
we obtain119889120585
119889120593=
1
radicminus (1198601minus 120593) 1198651015840 (119860
1)
+ 119874 ((1198601minus 120593)12
) (82)
Integrating (82) yields
120578 minus 120585 =2
radicminus1198651015840 (1198601)
(1198601minus 120593)12
+ 119874((1198601minus 120593)32
) (83)
where 120578 satisfies 120593(120578) = 1198601 Thus
(120578 minus 120585)2
=4
minus1198651015840 (1198601)(1198601minus 120593) + 119874 ((119860
1minus 120593)2
) (84)
which implies 119874((1198601minus 120593)2) = 119874((120578 minus 120585)
4) Therefore we get
120593 (120585) = 1198601+1198651015840(1198601)
4(120585 minus 120578)
2
+ 119874 ((120585 minus 120578)4
)
as 120585 uarr 120578
(85)
where 120593(120578) = 1198601
(ii) When 1198601= minus119860
2lt 984858 and 120598 = minus119860
2
1120573 from (75) 119865(120593)
has a double zero at 120593 = 1198601 Then
(1205931015840)2
= (120593 minus 1198601)2
11986510158401015840(1198601) + 119874 ((120593 minus 119860
1)3
)
as 120593 uarr 1198601
(86)
Furthermore we get
119889120585
119889120593=
1
radic(120593 minus 1198601)2
11986510158401015840 (1198601) + 119874 ((120593 minus 119860
1)3
)
(87)
Observing that
radic(120593 minus 1198601)2
11986510158401015840 (1198601) + 119874 ((120593 minus 119860
1)3
)
= (1198601minus 120593) (radic11986510158401015840 (119860
1) + 119874 (119860
1minus 120593))
1
radic11986510158401015840 (1198601) + 119874 (119860
1minus 120593)
=1
radic11986510158401015840 (1198601)
+ 119874 (1198601minus 120593)
(88)
we obtain119889120585
119889120593=
1
(1198601minus 120593) 11986510158401015840 (119860
1)+ 119874 (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 119860
1= minus119860
2lt 984858 and 120598 = minus119860
2
1120573 Then (75)
becomes
radic9848582 minus 1205932
1205932 minus 1198602
1
119889120593 = ∓radic120573
2119889120585 (90)
Integrating both sides of (90) gives the following implicitexpressions of kink and antikink wave solutions
arctan120593
radic9848582 minus 1205932
+
radic9848582 minus 1198602
1
1198601
arctanhradic9848582 minus 119860
2
1
1198601
120593
radic9848582 minus 1205932
= plusmnradic120573
2120585
(91)
By letting 1198601rarr 984858 in (91) we get two kink compactons
which are given by
plusmn120593 =
984858 sinradic120573
2120585
10038161003816100381610038161205851003816100381610038161003816 le
120587
radic2120573
984858 120585 gt120587
radic2120573
minus984858 120585 lt minus120587
radic2120573
(92)
The graphs for the kink wave solutions (91) and kinkcompacton solutions (92) are shown in Figures 9 and 10respectively
Remark 12 The two kink compacton solutions (92) aredifferent from the well-known smooth kink wave solutionsIn comparison with kink wave solutions (91) the kink com-pacton solutions (92) have no exponential decay propertiesbut have compact supportThat is they minus a constant thedifferences identically vanish outside a finite core region
18 Mathematical Problems in Engineering
120593
120585minus1 1
05
minus05
(a)
minus1minus1
minus05
05
00120593
x t
0
1
01
11
(b)
Figure 9 Two- and three-dimensional graphs of the kink wave solution
1
1120593
120585
minus1
minus1
(a)
0
1
0
1
000510
120593
minus1 minus1
minus05
minus10
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 the119862(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 Newtonrsquos equation which we write in the form 119910
2=
119865(120593) where 119910 = 1205931015840
120585(120585) we solve Newtonrsquos equation 119910
2=
119865(120593) 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 the119862(3 2 2)equation
Actually for 119898 = 2119896 + 1 119896 isin N+ in 119862(119898 2 2) equation(9) the dynamical behavior of traveling wave solutions of(9) is similar to the case 119898 = 3 for 119898 = 2119896 119896 isin N+ in119862(119898 2 2) equation (9) the dynamical behavior of travelingwave solutions of (9) is similar to the case 119898 = 2 We areapplying the approach mentioned in this work to 119862(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
References
[1] C Rogers and W R Shadwick Backlund Transformation andTheir Applications Academic Press New York NY USA 1982
[2] W Malfliet and W Hereman ldquoThe tanh method I Exactsolutions of nonlinear evolution and wave equationsrdquo PhysicaScripta vol 54 no 6 pp 563ndash568 1996
[3] R Hirota Direct Method in Soliton Theory Springer BerlinGermany 1980
[4] J S Cohen Computer Algebra and Symbolic ComputationMathematical Methods A K Peters 2003
[5] S Lie ldquoUber die integration durch bestimmte integrale voneiner klasse linearer partieller differential eichungenrdquoArchiv derMathematik vol 6 pp 328ndash368 1881
[6] H-H Dai and X-H Zhao ldquoTraveling waves in a rod composedof amodifiedMooney-Rivlinmaterial I bifurcation of equilibraand non-singular caserdquo Proceedings of the Royal Society AMathematical Physical and Engineering Sciences vol 455 no1990 pp 3845ndash3874 1999
[7] H-H Dai and F-F Wang ldquoAsymptotic bifurcation solutionsfor compressions of a clamped nonlinearly elastic rectangletransition region and barrelling to a corner-like profilerdquo SIAMJournal on Applied Mathematics vol 70 no 7 pp 2673ndash26922010
[8] J Li and Z Liu ldquoSmooth and non-smooth traveling waves ina nonlinearly dispersive equationrdquo Applied Mathematical Mod-elling vol 25 no 1 pp 41ndash56 2000
[9] A Chen and J Li ldquoSingle peak solitary wave solutions forthe osmosis K(22) equation under inhomogeneous boundaryconditionrdquo Journal of Mathematical Analysis and Applicationsvol 369 no 2 pp 758ndash766 2010
[10] R Camassa and D D Holm ldquoAn integrable shallow waterequation with peaked solitonsrdquo Physical Review Letters vol 71no 11 pp 1661ndash1664 1993
[11] P Rosenau ldquoOn nonanalytic solitary waves formed by a nonlin-ear dispersionrdquo Physics Letters A vol 230 no 5-6 pp 305ndash3181997
[12] P Rosenau and J M Hyman ldquoCompactons solitons with finitewavelengthrdquo Physical Review Letters vol 70 no 5 pp 564ndash5671993
[13] L Zhang L-Q Chen and X Huo ldquoThe effects of horizontalsingular straight line in a generalized nonlinear Klein-Gordonmodel equationrdquo Nonlinear Dynamics vol 72 no 4 pp 789ndash801 2013
[14] J Li and H Dai On the Study of Singular Nonlinear TravelingWave Equations Dynamical System Approach Science PressBeijing China 2007
[15] A Chen S Wen S Tang W Huang and Z Qiao ldquoEffects ofquadratic singular curves in integrable equationsrdquo Studies inApplied Mathematics vol 134 no 1 pp 24ndash61 2015
[16] L Tian and J Yin ldquoNew compacton solutions and solitary wavesolutions of fully nonlinear generalized Camassa-Holm equa-tionsrdquo Chaos Solitons amp Fractals vol 20 no 2 pp 289ndash2992004
[17] L Zhang ldquoThe bifurcation and peakons for the specialC(3 2 2)equationrdquo Pramana vol 83 no 3 pp 331ndash340 2014
[18] J Lenells ldquoTraveling wave solutions of the CamassandashHolmequationrdquo Journal of Differential Equations vol 217 no 2 pp393ndash430 2005
[19] Z J Qiao and G P Zhang ldquoOn peaked and smooth solitons forthe Camassa-Holm equationrdquo Europhysics Letters vol 73 no 5pp 657ndash663 2006
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Stochastic AnalysisInternational Journal of
4 Mathematical Problems in Engineering
1205732(120590)
(1205901s 1205731s)
1205733(120590)
1205734(120590)1205735(120590)
120573
minus2
minus2
minus4
minus4
minus6
minus8
minus10
minus12
minus14
1205731(120590)
1205902 4
Figure 1 The bifurcation curves in the (120590 120573)-parameter plane for120598 gt 0
According to the above analysis and Lemma 1 we obtainthe following proposition on the bifurcation curves of thephase portraits of system (15) for 120598 gt 0
Proposition 2 When 120598 gt 0 for system (15) in (120590 120573)-parameter plane there exist five bifurcation curves (see Fig-ure 1)
120573 = 1205731(120590) = minus
41205983
271205902
120573 = 1205732(120590) =
120590 minus 120598984858
9848583
120573 = 1205733(120590) =
minus120590 minus 120598984858
9848583
120573 = 1205734(120590) (|120590| le 120590
3119904)
120573 = 1205735(120590)
(20)
These five curves divide the right-half (120590 120573)-parameter planeinto thirty-one regions as follows
1198601 (120590 120573) | 1205901s lt 120590 lt 120598984858 120573 = 1205732(120590)
1198602 (120590 120573) | 120590 gt 1205901s 1205735(120590) lt 120573 lt 1205732(120590)
1198603 (120590 b) | 120590 gt 1205901s 120573 = 1205735(120590)
1198604 (120590 120573) | 120590 gt 1205901s 1205731(120590) lt 120573 lt 1205735(120590)
1198605 (120590 120573) | 120590 gt 1205901s 120573 = 1205731(120590)
1198606 (120590 120573) | 120590 gt 1205902s 1205733(120590) lt 120573 lt 1205731(120590)
1198607 (120590 120573) | 120590 lt 1205902s 120573 = 1205733(120590)
1198608 (120590 120573) | 120590 gt 0 120573 lt max1205731(120590) 1205733(120590)
1198609 (120590 120573) = (1205902s 1205732s)
11986010 (120590 120573) | 0 lt 120590 lt 1205902s 120573 = 1205731(120590)
11986011 (120590 120573) | 120590 lt 1205902s 1205731(120590) lt 120573 lt min1205733(120590)
1205734(120590)
11986012 (120590 120573) | 120590 lt 1205903s 120573 = 1205734(120590)
11986013 (120590 120573) | 120590 gt 0 1205734(120590) lt 120573 lt 1205735(120590)
11986014 (120590 120573) | 120590 lt 1205904s 120573 = 1205735(120590)
11986015 (120590 120573) | 120590 gt 0 1205735(120590) lt 120573 lt 1205733(120590)
11986016 (120590 120573) | 120590 = 0 120573 lt minus21205989848582
11986017 (120590 120573) = (0 minus21205989848582)
11986018 (120590 120573) | 120590 = 0 minus21205989848582 lt 120573 lt minus120598984858
2
11986019 (120590 120573) = (0 minus1205989848582)
11986020 (120590 120573) | 0 lt 120590 lt 1205904s 120573 = 1205733(120590)
11986021 (120590 120573) = (1205904s 1205734s)
11986022 (120590 120573) | 1205904s lt 120590 lt 1205902s 120573 = 1205733(120590)
11986023 (120590 120573) | 1205904s lt 120590 lt 1205901s max1205731(120590) 1205733(120590) lt
120573 lt 1205735(120590)11986024 (120590 120573) | 120590
s4 lt 120590 lt 1205901s 120573 = 1205735(120590)
11986025 (120590 120573) | 1205904s lt 120590 lt 1205901s max1205733(120590) 1205735(120590) lt
120573 lt 1205732(120590)11986026 (120590 120573) | 0 lt 120590 lt 1205901s 120573 = 1205732(d)
11986027 (120590 120573) = (1205901s 1205731s)
11986028 (120590 120573) | 0 le 120590 lt 120598984858 120573 gt 1205732(120590)
11986029 (120590 120573) | 120590 isin R 120573 gt max0 1205732(120590)
11986030 (120590 120573) | 120590 gt 120598984858 120573 = 1205732(120590)
11986031 (120590 120573) | 120590 gt 120598984858 0 lt 120573 lt 1205732(120590)
In this case the phase portraits of system (15) can beshown in Figure 2
22 Case II 120598 lt 0 In this case we have the following
Proposition 3 When 120598 lt 0 for system (15) in (120590 120573)-parameter plane there exist four parametric bifurcation curves(see Figure 3)
120573 = 1205731(120590) = minus
41205983
271205902
120573 = 1205732(120590) =
120590 minus 120598984858
9848583
120573 = 1205733(120590) =
minus120590 minus 120598984858
9848583 120590 = 0
(21)
These four curves divide the right-half (120590 120573)-parameter planeinto twenty-two regions
1198611 (120590 120573) | 1205901s lt 120590 lt 120598984858 120573 = 1205733(120590)
1198612 (120590 120573) | 120590 gt 1205901s max0 1205733(120590) lt 120573 lt 1205731(120590)
1198613 (120590 120573) | 120590 gt 1205901s 120573 = 1205731(120590)
1198614 (120590 120573) | 120590 gt 1205902s 1205731(120590) lt 120573 lt 1205732(120590)
1198615 (120590 120573) = (1205902s 1205732s)
1198616 (120590 120573) | 120590 gt 1205902s 120573 = 1205732(120590)
1198617 (120590 120573) | 120590 gt 0 120573 gt max1205731(120590) 1205732(120590)
1198618 (120590 120573) | 0 lt 120590 lt 1205902s 120573 = 1205731(120590)
1198619 (120590 120573) | 0 lt 120590 lt 1205902s 1205732(120590) lt 120573 lt 1205731(120590)
11986110 (120590 120573) | 120590 = 0 120573 gt minus120598984858
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
00 05 10
0
2
4
00 05 10
0
2
4
00 05 10
0
1
2
3
00 05 10
0
2
4
6
00 05 10
0
2
4
00 05 10
0
2
4
minus3
minus3
minus2 minus2 minus2minus1minus4
minus4
minus2 minus2
minus2
minus1
minus1
minus4minus2minus4
minus4 minus4
minus2 minus2 minus2
minus4
minus2
minus4
minus2
minus4
minus2
minus15 minus10 minus05
minus10 minus05 minus10 minus05 minus10 minus05
minus15 minus10 minus05 minus15 minus10 minus05
minus6
minus2
minus4
minus2
minus4
minus2
minus4
minus2
minus4
minus2
minus4
120593120593120593
120593 120593 120593
120593120593120593
120593 120593 120593
yy
y
y yy
yyy
y
y y
(1) (120590 120573) isin A1
(4) (120590 120573) isin A4
(7) (120590 120573) isin A7 (8) (120590 120573) isin A8 (9) (120590 120573) isin A9
(5) (120590 120573) isin A5
(2) (120590 120573) isin A2 (3) (120590 120573) isin A3
(6) (120590 120573) isin A6
(10) (120590 120573) isin A10 (11) (120590 120573) isin A11 (12) (120590 120573) isin A12
Figure 2 Continued
6 Mathematical Problems in Engineering
00 05 10
0
1
2
3
00 05 10
0
2
4
00 05 10
0
1
2
3
00 05 10
0
1
2
3
00 05 10
0
1
2
00 05 10
0
1
2
00 05 10
0
1
2
00 05 10
0
1
2
00 05 10
0
1
2
3
00 05 10
0
1
2
3
00 05 10
0
1
2
3
00 05 10
0
1
2
3
minus3
minus2
minus1
minus3
minus2
minus1
minus2
minus1
minus2
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus1
minus2
minus1
minus3
minus2
minus1
minus10 minus05
minus10 minus05
minus10 minus05
minus10 minus05
minus10 minus05
minus10minus15 minus05 minus10minus15 minus05
minus10 minus05
minus10 minus05 minus10 minus05
minus10 minus05 minus10 minus05
minus2
minus4
y
120593120593120593
120593 120593 120593
120593120593120593
120593 120593120593
yy
y yy
yy
y
yy
y
(13) (120590 120573) isin A13
(16) (120590 120573) isin A16
(19) (120590 120573) isin A19 (20) (120590 120573) isin A20
(23) (120590 120573) isin A23(24) (120590 120573) isin A24
(21) (120590 120573) isin A21
(22) (120590 120573) isin A22
(17) (120590 120573) isin A17 (18) (120590 120573) isin A18
(14) (120590 120573) isin A14 (15) (120590 120573) isin A15
Figure 2 Continued
Mathematical Problems in Engineering 7
00 05 10 15
0
1
2
3
00 05 10 15
0
1
2
3
0 1 2
0
1
2
3
0 1 2
0
1
2
3
00 05 10 15
0
1
2
3
0 1 2
0
2
4
0 1 2
0
2
4
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus2
minus4
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2 minus1
minus2minus3 minus1
minus2
minus2
minus3
minus4
minus1
minus2 minus1
minus10minus15 minus05 minus10minus15 minus05
minus10minus15 minus05
y
yyy
y y y
120593
120593120593120593
120593 120593 120593
(25) (120590 120573) isin A25
(28) (120590 120573) isin A28 (29) (120590 120573) isin A29
(31) (120590 120573) isin A31
(30) (120590 120573) isin A30
(26) (120590 120573) isin A26 (27) (120590 120573) isin A27
Figure 2 The bifurcation of phase portraits of system (15) when 120598 gt 0 and 120590 ge 0
11986111 (120590 120573) = (0 minus1205989848582)
11986112 (120590 120573) | 0 lt 120590 lt 1205902s 120573 = 1205732(120590)
11986113 (120590 120573) | 0 lt 120590 lt 1205901s 1205733(120590) lt 120573 lt min1205731(120590)
1205732(120590)
11986114 (120590 120573) | 1205904s lt 120590 lt 1205901s 120573 = 1205731(120590)
11986115 (120590 120573) = (1205901s 1205731s)
11986116 (120590 120573) | 0 lt 120590 lt 1205901s 120573 = 1205733(120590)
11986117 (120590 120573) | 0 lt 120590 lt 120598984858 120573 lt 1205733(120590)
11986118 (120590 120573) | 120590 = 0 120573 = minus120598984858
2
11986119 (120590 120573) | 120590 = 0 120573 lt 0
11986120 (120590 120573) | 0 lt 120590 lt 120598984858 120573 lt min0 1205733(120590)
11986121 (120590 120573) | 120590 gt 120598984858 120573 = 1205733(120590)
11986122 (120590 120573) | 120590 gt 120598984858 1205733(120590) lt 120573 lt 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
1205732(120590)
(1205901s 1205731s)
(1205902s 1205732s)
1205733(120590)
120573
minus2minus4minus6
1205731(120590)
120590
Figure 3 The bifurcation curves in the (120590 120573)-parameter plane for120598 lt 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 Let119862119896(Ω) denote the set of all 119896 times continuouslydifferentiable functions on the open setΩ 119871119901loc(R) refer to theset of all functions whose restriction on any compact subsetis 119871119901 integrable1198671loc(R) stands for1198671loc(R) = 120593 isin 119871
2
loc(R) |
1205931015840isin 1198712
loc(R)To study single peak soliton solutions we impose the
boundary condition
lim120585rarrplusmninfin
120593 = 119860 (22)
where 119860 is a constant In fact the constant 119860 is equal to thehorizontal coordinate of saddle point 119864(120593
119890 0) Substituting
the boundary condition (22) into (14) generates the followingconstant
ℎ = 119860(120590 +119860120598
2+1198603120573
4) (23)
So the ODE (14) becomes
(1205931015840)2
=
(120593 minus 119860)2
(1205731205932+ 2119860120573120593 + 3119860
2120573 + 2120598)
2 (9848582 minus 1205932) (24)
If 120573(1198602120573 + 120598) le 0 then (24) reduces to
(1205931015840)2
=120573 (120593 minus 119860)
2
(120593 minus 1198611) (120593 minus 119861
2)
2 (9848582 minus 1205932) (25)
where
1198611= minus119860 minus
radicminus2120573 (1198602120573 + 120598)
120573
1198612= minus119860 +
radicminus2120573 (1198602120573 + 120598)
120573
(26)
From (26) we know that 1198611gt 1198612if 120573 lt 0 and 119861
1lt 1198612if
120573 gt 0
Definition 4 A function120593(120585) is said to be a single peak solitonsolution for the 119862(3 2 2) equation (10) if 120593(120585) satisfies thefollowing conditions
(C1) 120593(120585) is continuous on R and has a unique peakpoint 120585
0 where 120593(120585) attains its global maximum or
minimum value
(C2) 120593(120585) isin 119862
3(R minus 120585
0) satisfies (24) on R minus 120585
0
(C3) 120593(120585) satisfies the boundary condition (22)
Definition 5 A wave function 120593 is called smooth solitonsolution if 120593 is smooth locally on either side of 120585
0and
lim120585uarr1205850
1205931015840(120585) = lim
120585darr12058501205931015840(120585) = 0
Definition 6 A wave function 120593 is called peakon if 120593 issmooth locally on either side of 120585
0and lim
120585uarr12058501205931015840(120585) =
minuslim120585darr1205850
1205931015840(120585) = 119886 119886 = 0 119886 = plusmninfin
Definition 7 Awave function120593 is called cuspon if120593 is smoothlocally on either side of 120585
0and lim
120585uarr12058501205931015840(120585) = minuslim
120585darr12058501205931015840(120585) =
plusmninfin
Without any loss of generality we choose the peak point1205850as vanishing 120585
0= 0
Theorem 8 Assume that 119906(119909 119905) = 120593(120585) = 120593(119909 minus 119888119905) is a singlepeak soliton solution of the 119862(3 2 2) equation (10) at the peakpoint 120585
0= 0 Then we have the following
(i) If 120573(1198602120573 + 120598) gt 0 then 120593(0) = 984858 or 120593(0) = minus984858
(ii) If 120573(1198602120573 + 120598) le 0 then 120593(0) = 984858 or 120593(0) = minus984858 or120593(0) = 119861
1or 120593(0) = 119861
2
Proof If 120593(0) = plusmn984858 then 120593(120585) = plusmn984858 for any 120585 isin R since120593(120585) isin 119862
3(R minus 0) Differentiating both sides of (24) yields
120593 isin 119862infin(R)
(i) When 120573(1198602120573 + 120598) gt 0 if 120593(0) = 984858 and 120593(0) = minus984858 then
120593 isin 119862infin(R) By the definition of single peak soliton we have
1205931015840(0) = 0 However by (24) we must have 120593(0) = 119860 which
contradicts the fact that 0 is the unique peak point(ii) When 120573(119860
2120573 + 120598) le 0 if 120593(0) = 984858 and 120593(0) = minus984858
by (24) we know 1205931015840(0) exists and 120593
1015840(0) = 0 since 0 is a peak
point Thus we obtain 120593(0) = 1198611or 120593(0) = 119861
2from (25)
since 120593(0) = 119860 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 119906(119909 119905) = 120593(119909 minus 119888119905) is a single peaksoliton solution of the 119862(3 2 2) equation (10) at the peak point1205850= 0 Then we have the following solution classification(i) If |120593(0)| = 984858 then 120593(120585) isin 119862
infin(R) and 120593 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
00 05 10 15
0
2
4
0 1
0
2
4
6
00 05 10 15
0
2
4
00 05 10
0
1
2
3
00 05 10 15
0
1
2
3
00 05 10
0
1
2
00 05 10 15
0
1
2
00 05 10 15
0
1
2
3
minus3 minus1minus4
minus4
minus2
minus4
minus2
minus4
minus2
minus4
minus2minus4
minus6
minus2
minus4
minus2
minus4
minus2
minus2 minus4 minus4minus2 minus2
minus3
minus1
minus2
minus1
minus2
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus1minus2 minus1minus2
120593
120593
120593
120593 120593 120593
120593 120593
120593 120593
120593 120593
y
y
y
yy
y
y y
y y
y y
minus15 minus10 minus05
minus15minus20 minus10 minus05
minus15 minus10 minus05 minus15 minus10 minus05minus10 minus05
minus15 minus10 minus05minus10 minus05
(10) (120590 120573) isin B10 (11) (120590 120573) isin B11 (12) (120590 120573) isin B12
(1) (120590 120573) isin B1
(4) (120590 120573) isin B4
(7) (120590 120573) isin B7 (8) (120590 120573) isin B8 (9) (120590 120573) isin B9
(5) (120590 120573) isin B5 (6) (120590 120573) isin B6
(2) (120590 120573) isin B2 (3) (120590 120573) isin B3
Figure 4 Continued
10 Mathematical Problems in Engineering
00 05 10 15
0
1
2
3
0 1
0
1
2
3
0 1 2
0
2
4
00 05 10 15
0
1
2
3
0 1 2
0
1
2
3
00 05 10 15
0
1
2
00 05 10 15
0
1
2
3
0 1 2
0
1
2
3
00 05 10 15
0
2
4
00 05 10 15
0
2
4
6
minus4
minus2
minus4
minus2
minus4
minus6
minus2
minus3 minus1minus2
minus3 minus1minus2
minus1minus2
minus1minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus1
minus2
minus3
minus1
minus2
120593 120593 120593
120593120593120593
120593 120593 120593
120593
y y y
yy
y
y yy
y
minus15 minus10 minus05
minus15minus20 minus10 minus05
minus15 minus10 minus05
minus15 minus10 minus05
minus15 minus10 minus05
minus15 minus10 minus05
(13) (120590 120573) isin B13 (14) (120590 120573) isin B14 (15) (120590 120573) isin B15
(16) (120590 120573) isin B16 (17) (120590 120573) isin B17 (18) (120590 120573) isin B18
(22) (120590 120573) isin B22
(20) (120590 120573) isin B20 (21) (120590 120573) isin B21(19) (120590 120573) isin B19
Figure 4 The bifurcation of phase portraits of system (15) when 120598 lt 0 and 120590 ge 0
Mathematical Problems in Engineering 11
(ii) If 120593(0) = 984858 = 1198611 then 120593 is a cuspon solution and 120593 has
the following asymptotic behavior
120593 (120585) minus 984858 = 1205821
10038161003816100381610038161205851003816100381610038161003816
23
+ 119874(10038161003816100381610038161205851003816100381610038161003816
43
) 120585 997888rarr 0
1205931015840(120585) =
2
31205821
10038161003816100381610038161205851003816100381610038161003816
minus13 sgn (120585) + 119874 (10038161003816100381610038161205851003816100381610038161003816
13
)
120585 997888rarr 0
(27)
where 1205821= plusmn(9|120573984858
2+ 2119860120573984858 + 3119860
2120573 + 2120598|(984858 minus 119860)
216984858)13
Thus 120593(120585) notin 1198671
119897119900119888(R)
(iii) If 120593(0) = minus984858 = 1198612 then 120593 is a cuspon solution and 120593
has the following asymptotic behavior
120593 (120585) + 984858 = 1205822
10038161003816100381610038161205851003816100381610038161003816
23
+ 119874(10038161003816100381610038161205851003816100381610038161003816
43
) 120585 997888rarr 0
1205931015840(120585) =
2
31205822
10038161003816100381610038161205851003816100381610038161003816
minus13 sgn (120585) + 119874 (10038161003816100381610038161205851003816100381610038161003816
13
)
120585 997888rarr 0
(28)
where 1205822= plusmn(9|120573984858
2minus 2119860120573984858 + 3119860
2120573 + 2120598|(984858 + 119860)
216984858)13
Thus 120593(120585) notin 1198671
119897119900119888(R)
(iv) If 120593(0) = 984858 = 1198611and 119860 = 0 then 120593 is a peakon-like
solution and
120593 (120585) minus 984858 = 1205823
10038161003816100381610038161205851003816100381610038161003816 + 119874 (
10038161003816100381610038161205851003816100381610038161003816
2
) 120585 997888rarr 0
1205931015840(120585) = 120582
3sgn (120585) + 119874 (
10038161003816100381610038161205851003816100381610038161003816) 120585 997888rarr 0
(29)
where 1205823= plusmn|119860 minus 984858|radic120573(119861
2minus 984858)4984858
(v) If 120593(0) = minus984858 = 1198612and 119860 = 0 then 120593 is a peakon-like
solution and
120593 (120585) + 984858 = 1205824
10038161003816100381610038161205851003816100381610038161003816 + 119874 (
10038161003816100381610038161205851003816100381610038161003816
2
) 120585 997888rarr 0
1205931015840(120585) = 120582
4sgn (120585) + 119874 (
10038161003816100381610038161205851003816100381610038161003816) 120585 997888rarr 0
(30)
where 1205824= plusmn|119860 + 984858|radicminus120573(119861
1+ 984858)4984858
(vi) If 120593(0) = 984858 = 1198611and 119860 = 0 then 120593 gives the peakon
solution 984858 exp(minusradicminus1205732|119909 minus 119888119905|)(vii) If 120593(0) = minus984858 = 119861
2and 119860 = 0 then 120593 gives the peakon
solution minus984858 exp(minusradicminus1205732|119909 minus 119888119905|)
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 |120593(0)| = 984858 then 120593 isin 119862
infin(R) and 120593 is a smooth soliton
solution(ii) If 120593(0) = 984858 = 119861
1 then by the definition of single peak
soliton we have 119860 = 984858 thus 1205731205932 + 2119860120573120593 + 31198602120573 + 2120598 does
not contain the factor 120593 minus 984858 From (24) we obtain
1205931015840= sgn (119860 minus 984858)
1003816100381610038161003816120593 minus 1198601003816100381610038161003816
radic210038161003816100381610038161205932 minus 9848582
1003816100381610038161003816
sdot radic10038161003816100381610038161205731205932 + 2119860120573120593 + 31198602120573 + 2120598
1003816100381610038161003816 sgn (120585)
(31)
Let 1198971(120593) = radic2(120593 + 984858)|120593 minus 119860|radic|1205731205932 + 2119860120573120593 + 31198602120573 + 2120598|
then 1198971(984858) = 2radic984858|984858 minus 119860|radic|1205739848582 + 2119860120573984858 + 31198602120573 + 2120598| and
int 1198971(120593)radic
1003816100381610038161003816120593 minus 9848581003816100381610038161003816119889120593 = int sgn (119860 minus 984858) sgn (120585) 119889120585 (32)
Inserting 1198971(120593) = 119897
1(984858) + 119874(|120593 minus 984858|) into (32) and using the
initial condition 120593(0) = 984858 we obtain
21198971(984858)
3
1003816100381610038161003816120593 minus 9848581003816100381610038161003816
32
(1 + 119874 (1003816100381610038161003816120593 minus 984858
1003816100381610038161003816)) =10038161003816100381610038161205851003816100381610038161003816
(33)
thus
120593 minus 984858 = plusmn(3
21198971(984858)
)
23
10038161003816100381610038161205851003816100381610038161003816
23
(1 + 119874 (1003816100381610038161003816120593 minus 984858
1003816100381610038161003816))minus23
= plusmn(3
21198971(984858)
)
23
10038161003816100381610038161205851003816100381610038161003816
23
(1 + 119874 (1003816100381610038161003816120593 minus 984858
1003816100381610038161003816))
(34)
which implies 120593 minus 984858 = 119874(|120585|23
) Therefore we have
120593 (120585) = 984858 plusmn (3
21198971(984858)
)
23
10038161003816100381610038161205851003816100381610038161003816
23
+ 119874(10038161003816100381610038161205851003816100381610038161003816
43
)
= 984858 + 1205821
10038161003816100381610038161205851003816100381610038161003816
23
+ 119874(10038161003816100381610038161205851003816100381610038161003816
43
) 120585 997888rarr 0
1205821= plusmn(
3
21198971(984858)
)
23
= plusmn(
9100381610038161003816100381610038161205739848582+ 2119860120573984858 + 3119860
2120573 + 2120598
10038161003816100381610038161003816(984858 minus 119860)
2
16984858)
13
1205931015840(120585) =
2
31205821
10038161003816100381610038161205851003816100381610038161003816
minus13 sgn (120585) + 119874 (10038161003816100381610038161205851003816100381610038161003816
13
) 120585 997888rarr 0
(35)
So 120593(120585) notin 1198671
loc(R)(iii) Similar to the proof of (ii) we ignore it in this paper(iv) If 120593(0) = 984858 = 119861
1and 119860 = 0 then from (25) we obtain
1205931015840= radicminus
120573
2sgn (119860 minus 984858)
1003816100381610038161003816120593 minus 1198601003816100381610038161003816radic120593 minus 1198612
120593 + 984858sgn (120585) (36)
Let 1198972(120593) = (1|120593minus119860|)radic(120593 + 984858)(120593 minus 119861
2) then 119897
2(984858) = (1|984858minus
119860|)radic2984858(984858 minus 1198612) and
int 1198972(120593) 119889120593 = radicminus
120573
2int sgn (119860 minus 984858) sgn (120585) 119889120585 (37)
Inserting 1198972(120593) = 119897
2(984858) + 119874(|120593 minus 984858|) into (37) and using the
initial condition 120593(0) = 984858 we obtain
1198972(984858) (120593 minus 984858) (1 + 119874 (
1003816100381610038161003816120593 minus 9848581003816100381610038161003816))minus1
= sgn (119860 minus 984858)10038161003816100381610038161205851003816100381610038161003816
(38)
Sincesgn (120593 minus 984858) sgn (119860 minus 984858) ge 0
1
1 + 119874 (120593 minus 984858)= 1 + 119874 (120593 minus 984858)
(39)
12 Mathematical Problems in Engineering
we get
1003816100381610038161003816120593 minus 9848581003816100381610038161003816 =
radicminus120573
2
1
1198972(984858)
10038161003816100381610038161205851003816100381610038161003816 (1 + 119874 (120593 minus 984858)) (40)
which implies |120593 minus 984858| = 119874(|120585|) Therefore we have
120593 (120585) = 984858 + 1205823
10038161003816100381610038161205851003816100381610038161003816 + 119874 (
10038161003816100381610038161205851003816100381610038161003816
2
) 120585 997888rarr 0
1205931015840(120585) = 120582
3sgn (120585) + 119874 (
10038161003816100381610038161205851003816100381610038161003816) 120585 997888rarr 0
(41)
where 1205823= plusmn|119860 minus 984858|radic120573(119861
2minus 984858)4984858
(v) Similar to the proof of (iv) we ignore it in this paper
By virtue of Theorem 9 any single peak soliton for the119862(3 2 2) equation (10) must satisfy the following initial andboundary values problem
(1205931015840)2
=120573 (120593 minus 119860)
2
(120593 minus 1198611) (120593 minus 119861
2)
2 (9848582 minus 1205932)fl119871 (120593)
120593 (0) isin 984858 minus984858 1198611 1198612
lim|120585|rarrinfin
120593 (120585) = 119860
(42)
119871(120593) ge 0 and the boundary condition (24) imply thefollowing
(a) If 120573(9848582 minus 1205932) ge 0 then 120593 ge max119861
1 1198612 or 120593 le
min1198611 1198612
(b) If120573(9848582minus1205932) le 0 thenmin1198611 1198612 le 120593 le max119861
1 1198612
Below we will present some implicit formulas for thesingle peak soliton solutions in the case of specific 120590 and 120573
Case 1 ((120590 120573) isin 11986012) In this case we have minus984858 lt 119861
2lt 119860 lt
1198611= 984858 From the standard phase analysis and Theorem 9
we know that if 120593 is a single peak soliton of the 119862(3 2 2)
equation then
1205931015840= minusradicminus
120573
2(120593 minus 119860)radic
120593 minus 1198612
120593 + 984858sgn (120585) (43)
From the separation of variables we get
intℎ (120593) 119889120593 = radicminus120573
2sgn (120585) 119889120585 (44)
where ℎ(120593) = (1(119860 minus 120593))radic(120593 + 984858)(120593 minus 1198612) After a lengthy
calculation of integral we obtain the implicit solution 120593
defined by
119867(120593)flradic119860 + 984858
119860 minus 1198612
1198681(120593) minus 119868
2(120593) = radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870 (45)
where
1198681(120593) = ln
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
(119860 + 984858) (120593 minus 1198612) + (119860 minus 119861
2) (120593 + 984858) + 2radic(119860 + 984858) (119860 minus 119861
2) (120593 minus 119861
2) (120593 + 984858)
radic(119860 + 984858) (119860 minus 1198612) (119860 + 984858) (119860 minus 120593)
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
1198682(120593) = ln
1003816100381610038161003816100381610038161003816(120593 + 984858) + (120593 minus 119861
2) + 2radic(120593 + 984858) (120593 minus 119861
2)
1003816100381610038161003816100381610038161003816
(46)
and 119870 is an arbitrary integration constant For 120593(0) = 984858 theconstant119870 = 119867(120593(0)) is defined by
119870 = radic119860 + 984858
119860 minus 1198612
1198681(984858) minus 119868
2(984858) (47)
and for 120593(0) = 1198612
119870 = radic119860 + 984858
119860 minus 1198612
1198681(1198612) minus 1198682(1198612) (48)
(i) If 120593(0) = 984858 then 119860 lt 120593 le 984858 Since 1198671015840(120593) = ℎ(120593) weknow that119867(120593) strictly decreases on the interval (119860 984858] thus1198671(120593) = 119867|
(119860984858](120593) gives a single peak solitonwith119867
1(984858) = 119870
and1198671(119860+) = infin Therefore 120593
1(120585) = 119867
minus1
1(radicminus1205732|120585| + 119870) is
the solution satisfying
1205931(0) = 984858
lim|120585|rarrinfin
1205931(120585) = 119860
1205931015840
1(0plusmn) = plusmn (119860 minus 984858)radic
120573 (1198612minus 984858)
4984858
(49)
So 1205931(120585) is a peakon-like solution (see Figure 5)
(ii) If 120593(0) = 1198612 then 119861
2le 120593 lt 119860 By 119867
1015840(120593) = 119891(120593)
we know that 119867(120593) strictly increases on the interval [1198612 119860)
Thus
1198672(120593) = 119867|
[1198612 119860)(120593) (50)
Mathematical Problems in Engineering 13
minus3 minus2 minus1
120593
minus02
1 2 3
02
04
06
08
10
x
(a)
02
02
minus2 minus2 x
10
05
00
t
120593
22
(b)
Figure 5 Two- and three-dimensional graphs of the peakon-like solution
minus03
minus04
minus05
1 2 3minus3 minus2 minus1
120593
minus02
x
(a)
0
2
0
2
minus2minus2
minus02
minus03
minus04
minus05
minus06
t x
120593
0
2
0
2
minus2minus2
t x
(b)
Figure 6 Two- and three-dimensional graphs of the smooth soliton solution
has the inverse denoted by 1205932(120585) = 119867
minus1
2(radicminus1205732|120585| + 119870)
1205932(120585) gives a kind of smooth soliton solution (see Figure 6)
satisfying
1205932(0) = 119861
2
lim|120585|rarrinfin
1205932(120585) = 119860
1205931015840
2(0) = 0
(51)
Case 2 ((120590 120573) isin 11986014) In this case we have minus984858 = 119861
2lt 119860 lt
984858 lt 1198611and (25) is equivalent to
1205931015840= minusradicminus
120573
2(120593 minus 119860)radic
120593 minus 1198611
120593 minus 984858sgn (120585) (52)
Let
119892 (120593) =1
119860 minus 120593radic120593 minus 1198611
120593 minus 984858 (53)
14 Mathematical Problems in Engineering
minus1 1 2 3
02
04
06
08
10
minus3 minus2
120593
x
(a)
02
02minus2
minus2
120593
x t
00
05
10
(b)
Figure 7 Two- and three-dimensional graphs of the cuspon solution
then (52) is converted to
119892 (120593) 119889120593 =1
119860 minus 120593radic120593 minus 1198611
120593 minus 984858119889120593 = radicminus
120573
2sgn (120585) 119889120585 (54)
Integrating (54) on the interval [0 120585] (or [120585 0]) leads to thefollowing implicit solutions
119866 (120593)flradic119860 minus 984858
119860 minus 1198611
1198683(120593) + 119868
4(120593) = radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870 (55)
where
1198683(120593) = ln
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
(119860 minus 984858) (120593 minus 1198611) + (119860 minus 119861
1) (120593 minus 984858) + 2radic(119860 minus 984858) (119860 minus 119861
1) (120593 minus 984858) (120593 minus 119861
1)
radic(119860 minus 984858) (119860 minus 1198611) (119860 minus 984858) (120593 minus 119860)
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
1198684(120593) = ln
1003816100381610038161003816100381610038161003816(120593 minus 984858) + (120593 minus 119861
1) + 2radic(120593 minus 984858) (120593 minus 119861
1)
1003816100381610038161003816100381610038161003816
(56)
And119870 is an arbitrary integration constant It is obvious thatfor 120593(0) = 984858 the constant119870 = 119867(120593(0)) is defined by
119870 = radic119860 minus 984858
119860 minus 1198611
1198683(984858) + 119868
4(984858) (57)
and for 120593(0) = minus984858
119870 = radic119860 minus 984858
119860 minus 1198611
1198683(minus984858) + 119868
4(minus984858) (58)
(i) If 120593(0) = 984858 then 119860 lt 120593 le 984858 From 119892(120593) lt 0 weknow that 119866(120593) strictly decreases on the interval (119860 984858] with119866(984858) = 119870 and 119866(119860+) = infin Define
1198661(120593) = 119866|
(119860984858](120593) (59)
Since 1198661(120593) is a strictly decreasing function from (119860 984858] onto
[119870infin) we can solve for 120593 uniquely from (59) and obtain
1205931(120585) = 119866
minus1
1(radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870) (60)
It is easy to check that 120593 satisfies
1205931(0) = 984858
lim|120585|rarrinfin
1205931(120585) = 119860
1205931015840
1(0plusmn) = ∓infin
(61)
Therefore the solution1205931defined by (60) is a cuspon solution
for the 119862(3 2 2) equation (see Figure 7)
Mathematical Problems in Engineering 15
(ii) If 120593(0) = minus984858 then minus984858 le 120593 lt 119860 Through a similaranalysis we get a strictly increasing function 119866(120593) on theinterval [minus984858 119860) satisfying
119866 (120593) = radicminus120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870 (62)
where 119866(120593) is defined by (55) Let
1198662(120593) = 119866|
[minus984858119860)(120593) (63)
then 1198662(120593) is a strictly increasing function from [minus984858 119860) onto
[119870infin) so that we can solve for 120593 and obtain
1205932(120585) = 119866
minus1
2(radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870) (64)
It is easy to check that 1205932satisfies
1205932(0) = minus984858
lim|120585|rarrinfin
1205932(120585) = 119860
1205931015840
2(0plusmn) = plusmn (119860 + 984858)radicminus
120573 (1198611+ 984858)
4984858
(65)
Therefore the solution 1205932defined by (64) is a peakon-like
solution whose graph is similar to those in Figure 5
Case 3 ((120590 120573) isin 11986016) In this case we have minus984858 lt 119861
2lt 119860 =
0 lt 1198611lt 984858 119861
2= minus1198611 and
1205931015840= minusradicminus
120573
2120593radic
120593 minus 1198612
1
1205932 minus 9848582sgn (120585) (66)
Hence from the separation of variables we have
radic9848582 minus 1205932
120593radic1198612
1minus 1205932
119889120593 = minusradicminus120573
2sgn (120585) 119889120585 (67)
Integrating (67) on the interval [0 120585] (or [120585 0]) leads tothe following implicit formula for the two smooth solitonsolutions
radic9848582 minus 1205932 + radic1198612
1minus 1205932
radic9848582 minus 1198612
1
(
radic(9848582 minus 1198612
1) 1205932
984858radic1198612
1minus 1205932 + 119861
1radic9848582 minus 1205932
)
984858
= exp(minusradicminus120573
2|119909 minus 119888119905|)
(68)
where 120593 isin (119860 1198611] Consider
radic9848582 minus 1198612
2
radic9848582 minus 1205932 + radic1198612
2minus 1205932
(
984858radic1198612
2minus 1205932 minus 119861
2radic9848582 minus 1205932
radic(9848582 minus 1198612
2) 1205932
)
984858
= exp(radicminus120573
2|119909 minus 119888119905|)
(69)
where 120593 isin [1198612 119860)
Case 4 ((120590 120573) isin 11986017) In this case we have minus984858 = 119861
2lt 119860 =
0 lt 1198611= 984858 and
1205931015840= minusradicminus
120573
2120593 sgn (120585) (70)
Choosing 120593(0) = 984858 (or minus984858) as initial value we get
int
120593
984858
1
120593119889120593 = minusint
120585
0
radicminus120573
2sgn (120585) 119889120585 for 119860 lt 120593 le 984858
int
120593
minus984858
1
120593119889120593 = minusint
120585
0
radicminus120573
2sgn (120585) 119889120585 for minus 984858 le 120593 lt 119860
(71)
which immediately yield the peakon solutions
120593 (119909 minus 119888119905) = plusmn984858 exp(minusradicminus120573
2|119909 minus 119888119905|) (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 the119862(3 2 2) equation (10) In order to study kinkwave solutionswe assume that
lim120585rarrinfin
120593 (120585) = 1198601
lim120585rarrminusinfin
120593 (120585) = 1198602
(73)
16 Mathematical Problems in Engineering
2 4
02
04
06
08
10
minus2minus4
120593
x
(a)
0 2
02
00
05
10
minus2
120593
x
t
0
2
minus2
(b)
Figure 8 Two- and three-dimensional graphs of the peakon solution
where 1198601gt 1198602 Substituting the boundary condition (73)
into (14) generates
(1205931015840)2
=
120573 (120593 minus 1198601) (120593 minus 119860
2) [1205932+ (1198601+ 1198602) 120593 + 119860
2
1+ 11986011198602+ 1198602
2+ 2120598120573]
2 (9848582 minus 1205932)fl119865 (120593) (74)
The nonlinear differential equation (74) may sustaindifferent kinds of nonlinear excitations In what follows weconfine our attention to the cases 119860
2= minus119860
1and 120598 =
minus1198602
1120573 which describe kinks and kink compactons which
play an important role in the dynamics systems Under theseconsiderations (74) reduces to
(1205931015840)2
=120573 (120593 minus 119860
1)2
(120593 minus 1198602)2
2 (9848582 minus 1205932) (75)
If 1198601lt 984858 then from the phase analysis in Section 2 (see
Figure 4(10)) we know that (1198601 0) and (119860
2 0) are two saddle
points of (13) and the kink solutions can be obtained fromthe two heteroclinic orbits connecting (120601 119910) = (119860
1 0) and
(1198602 0) When 119860
1increases upon reaching 984858 that is 119860
1= 984858
(75) becomes
(1205931015840)2
= minus120573 (120593 minus 119860
1) (120593 minus 119860
2)
2 (76)
and the ellipse 21199102+120573(120593minus1198601)(120593minus119860
2) = 0 (see Figure 4(11))
which is tangent to the singular lines 120593 = 984858 and 120593 = minus984858 atpoints (119860
1 0) and (119860
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 120593 is a kink wave solutions of (75)
or (76) we have 1205931015840
rarr 0 as 120593 rarr 1198601and as 120593 rarr 119860
2
Moreover we have 119865(120593) ge 0 for 1198602le 120593 le 119860
1and 120593 is
strictly monotonic in any interval where 119865(120593) gt 0 Thus if1205931015840gt 0 at some points 120593will be strictly increasing until it gets
close to the next zero of 119865 Denoting this zero 1198601 we have
120593 uarr 1198601 What will happen to the solution when it approaches
1198601 Depending on whether the zero is double or simple 120593
has a different behavior We explore the two cases in turn
Theorem 11 (i) If 120593 has a simple zero at 120593 = 1198601 so that
119865(1198601) = 0 and 1198651015840(119860
1) lt 0 then the solution 120593 of (75) satisfies
120593 (120585) minus 1198601=1198651015840(1198601)
4(120585 minus 120578)
2
+ 119874 ((120585 minus 120578)4
)
119886119904 120585 uarr 120578
(77)
where 120593(120578) = 1198601
(ii) When 120593 approaches the double zero 1198601of 119865(120593) so that
1198651015840(1198601) = 0 and11986510158401015840(119860
1) gt 0 then the solution120593 of (75) satisfies
120593 (120585) minus 1198601sim 120572 exp(minus120585radic11986510158401015840 (119860
1)) 119886119904 120585 997888rarr infin (78)
for some constant 120572 Thus 120593 uarr 1198601exponentially as 120585 rarr infin
Mathematical Problems in Engineering 17
Proof (i) When 1198601= minus119860
2= 984858 and 120598 = minus119860
2
1120573 from (76)
119865(120593) has a simple zero at 120593 = 1198601 Then
(1205931015840)2
= (120593 minus 1198601) 1198651015840(1198601) + 119874 ((120593 minus 119860
1)2
)
as 120593 uarr 1198601
(79)
Using the fact that (1205931015840)2 ge 0 we know that 1198651015840(1198601) lt 0
Moreover
119889120585
119889120593=
1
radic(120593 minus 1198601) 1198651015840 (119860
1) + 119874 ((120593 minus 119860
1)2
)
(80)
Because
radic(120593 minus 1198601) 1198651015840 (119860
1) + 119874 ((120593 minus 119860
1)2
)
= radic1198601minus 120593(radicminus1198651015840 (119860
1) + 119874 (119860
1minus 120593))
1
radicminus1198651015840 (1198601) + 119874 (119860
1minus 120593)
=1
radicminus1198651015840 (1198601)
+ 119874 (1198601minus 120593)
(81)
we obtain119889120585
119889120593=
1
radicminus (1198601minus 120593) 1198651015840 (119860
1)
+ 119874 ((1198601minus 120593)12
) (82)
Integrating (82) yields
120578 minus 120585 =2
radicminus1198651015840 (1198601)
(1198601minus 120593)12
+ 119874((1198601minus 120593)32
) (83)
where 120578 satisfies 120593(120578) = 1198601 Thus
(120578 minus 120585)2
=4
minus1198651015840 (1198601)(1198601minus 120593) + 119874 ((119860
1minus 120593)2
) (84)
which implies 119874((1198601minus 120593)2) = 119874((120578 minus 120585)
4) Therefore we get
120593 (120585) = 1198601+1198651015840(1198601)
4(120585 minus 120578)
2
+ 119874 ((120585 minus 120578)4
)
as 120585 uarr 120578
(85)
where 120593(120578) = 1198601
(ii) When 1198601= minus119860
2lt 984858 and 120598 = minus119860
2
1120573 from (75) 119865(120593)
has a double zero at 120593 = 1198601 Then
(1205931015840)2
= (120593 minus 1198601)2
11986510158401015840(1198601) + 119874 ((120593 minus 119860
1)3
)
as 120593 uarr 1198601
(86)
Furthermore we get
119889120585
119889120593=
1
radic(120593 minus 1198601)2
11986510158401015840 (1198601) + 119874 ((120593 minus 119860
1)3
)
(87)
Observing that
radic(120593 minus 1198601)2
11986510158401015840 (1198601) + 119874 ((120593 minus 119860
1)3
)
= (1198601minus 120593) (radic11986510158401015840 (119860
1) + 119874 (119860
1minus 120593))
1
radic11986510158401015840 (1198601) + 119874 (119860
1minus 120593)
=1
radic11986510158401015840 (1198601)
+ 119874 (1198601minus 120593)
(88)
we obtain119889120585
119889120593=
1
(1198601minus 120593) 11986510158401015840 (119860
1)+ 119874 (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 119860
1= minus119860
2lt 984858 and 120598 = minus119860
2
1120573 Then (75)
becomes
radic9848582 minus 1205932
1205932 minus 1198602
1
119889120593 = ∓radic120573
2119889120585 (90)
Integrating both sides of (90) gives the following implicitexpressions of kink and antikink wave solutions
arctan120593
radic9848582 minus 1205932
+
radic9848582 minus 1198602
1
1198601
arctanhradic9848582 minus 119860
2
1
1198601
120593
radic9848582 minus 1205932
= plusmnradic120573
2120585
(91)
By letting 1198601rarr 984858 in (91) we get two kink compactons
which are given by
plusmn120593 =
984858 sinradic120573
2120585
10038161003816100381610038161205851003816100381610038161003816 le
120587
radic2120573
984858 120585 gt120587
radic2120573
minus984858 120585 lt minus120587
radic2120573
(92)
The graphs for the kink wave solutions (91) and kinkcompacton solutions (92) are shown in Figures 9 and 10respectively
Remark 12 The two kink compacton solutions (92) aredifferent from the well-known smooth kink wave solutionsIn comparison with kink wave solutions (91) the kink com-pacton solutions (92) have no exponential decay propertiesbut have compact supportThat is they minus a constant thedifferences identically vanish outside a finite core region
18 Mathematical Problems in Engineering
120593
120585minus1 1
05
minus05
(a)
minus1minus1
minus05
05
00120593
x t
0
1
01
11
(b)
Figure 9 Two- and three-dimensional graphs of the kink wave solution
1
1120593
120585
minus1
minus1
(a)
0
1
0
1
000510
120593
minus1 minus1
minus05
minus10
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 the119862(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 Newtonrsquos equation which we write in the form 119910
2=
119865(120593) where 119910 = 1205931015840
120585(120585) we solve Newtonrsquos equation 119910
2=
119865(120593) 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 the119862(3 2 2)equation
Actually for 119898 = 2119896 + 1 119896 isin N+ in 119862(119898 2 2) equation(9) the dynamical behavior of traveling wave solutions of(9) is similar to the case 119898 = 3 for 119898 = 2119896 119896 isin N+ in119862(119898 2 2) equation (9) the dynamical behavior of travelingwave solutions of (9) is similar to the case 119898 = 2 We areapplying the approach mentioned in this work to 119862(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
References
[1] C Rogers and W R Shadwick Backlund Transformation andTheir Applications Academic Press New York NY USA 1982
[2] W Malfliet and W Hereman ldquoThe tanh method I Exactsolutions of nonlinear evolution and wave equationsrdquo PhysicaScripta vol 54 no 6 pp 563ndash568 1996
[3] R Hirota Direct Method in Soliton Theory Springer BerlinGermany 1980
[4] J S Cohen Computer Algebra and Symbolic ComputationMathematical Methods A K Peters 2003
[5] S Lie ldquoUber die integration durch bestimmte integrale voneiner klasse linearer partieller differential eichungenrdquoArchiv derMathematik vol 6 pp 328ndash368 1881
[6] H-H Dai and X-H Zhao ldquoTraveling waves in a rod composedof amodifiedMooney-Rivlinmaterial I bifurcation of equilibraand non-singular caserdquo Proceedings of the Royal Society AMathematical Physical and Engineering Sciences vol 455 no1990 pp 3845ndash3874 1999
[7] H-H Dai and F-F Wang ldquoAsymptotic bifurcation solutionsfor compressions of a clamped nonlinearly elastic rectangletransition region and barrelling to a corner-like profilerdquo SIAMJournal on Applied Mathematics vol 70 no 7 pp 2673ndash26922010
[8] J Li and Z Liu ldquoSmooth and non-smooth traveling waves ina nonlinearly dispersive equationrdquo Applied Mathematical Mod-elling vol 25 no 1 pp 41ndash56 2000
[9] A Chen and J Li ldquoSingle peak solitary wave solutions forthe osmosis K(22) equation under inhomogeneous boundaryconditionrdquo Journal of Mathematical Analysis and Applicationsvol 369 no 2 pp 758ndash766 2010
[10] R Camassa and D D Holm ldquoAn integrable shallow waterequation with peaked solitonsrdquo Physical Review Letters vol 71no 11 pp 1661ndash1664 1993
[11] P Rosenau ldquoOn nonanalytic solitary waves formed by a nonlin-ear dispersionrdquo Physics Letters A vol 230 no 5-6 pp 305ndash3181997
[12] P Rosenau and J M Hyman ldquoCompactons solitons with finitewavelengthrdquo Physical Review Letters vol 70 no 5 pp 564ndash5671993
[13] L Zhang L-Q Chen and X Huo ldquoThe effects of horizontalsingular straight line in a generalized nonlinear Klein-Gordonmodel equationrdquo Nonlinear Dynamics vol 72 no 4 pp 789ndash801 2013
[14] J Li and H Dai On the Study of Singular Nonlinear TravelingWave Equations Dynamical System Approach Science PressBeijing China 2007
[15] A Chen S Wen S Tang W Huang and Z Qiao ldquoEffects ofquadratic singular curves in integrable equationsrdquo Studies inApplied Mathematics vol 134 no 1 pp 24ndash61 2015
[16] L Tian and J Yin ldquoNew compacton solutions and solitary wavesolutions of fully nonlinear generalized Camassa-Holm equa-tionsrdquo Chaos Solitons amp Fractals vol 20 no 2 pp 289ndash2992004
[17] L Zhang ldquoThe bifurcation and peakons for the specialC(3 2 2)equationrdquo Pramana vol 83 no 3 pp 331ndash340 2014
[18] J Lenells ldquoTraveling wave solutions of the CamassandashHolmequationrdquo Journal of Differential Equations vol 217 no 2 pp393ndash430 2005
[19] Z J Qiao and G P Zhang ldquoOn peaked and smooth solitons forthe Camassa-Holm equationrdquo Europhysics Letters vol 73 no 5pp 657ndash663 2006
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
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Differential EquationsInternational Journal of
Volume 2014
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OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
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Discrete Dynamics in Nature and Society
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Decision SciencesAdvances in
Discrete MathematicsJournal of
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
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
00 05 10
0
2
4
00 05 10
0
2
4
00 05 10
0
1
2
3
00 05 10
0
2
4
6
00 05 10
0
2
4
00 05 10
0
2
4
minus3
minus3
minus2 minus2 minus2minus1minus4
minus4
minus2 minus2
minus2
minus1
minus1
minus4minus2minus4
minus4 minus4
minus2 minus2 minus2
minus4
minus2
minus4
minus2
minus4
minus2
minus15 minus10 minus05
minus10 minus05 minus10 minus05 minus10 minus05
minus15 minus10 minus05 minus15 minus10 minus05
minus6
minus2
minus4
minus2
minus4
minus2
minus4
minus2
minus4
minus2
minus4
120593120593120593
120593 120593 120593
120593120593120593
120593 120593 120593
yy
y
y yy
yyy
y
y y
(1) (120590 120573) isin A1
(4) (120590 120573) isin A4
(7) (120590 120573) isin A7 (8) (120590 120573) isin A8 (9) (120590 120573) isin A9
(5) (120590 120573) isin A5
(2) (120590 120573) isin A2 (3) (120590 120573) isin A3
(6) (120590 120573) isin A6
(10) (120590 120573) isin A10 (11) (120590 120573) isin A11 (12) (120590 120573) isin A12
Figure 2 Continued
6 Mathematical Problems in Engineering
00 05 10
0
1
2
3
00 05 10
0
2
4
00 05 10
0
1
2
3
00 05 10
0
1
2
3
00 05 10
0
1
2
00 05 10
0
1
2
00 05 10
0
1
2
00 05 10
0
1
2
00 05 10
0
1
2
3
00 05 10
0
1
2
3
00 05 10
0
1
2
3
00 05 10
0
1
2
3
minus3
minus2
minus1
minus3
minus2
minus1
minus2
minus1
minus2
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus1
minus2
minus1
minus3
minus2
minus1
minus10 minus05
minus10 minus05
minus10 minus05
minus10 minus05
minus10 minus05
minus10minus15 minus05 minus10minus15 minus05
minus10 minus05
minus10 minus05 minus10 minus05
minus10 minus05 minus10 minus05
minus2
minus4
y
120593120593120593
120593 120593 120593
120593120593120593
120593 120593120593
yy
y yy
yy
y
yy
y
(13) (120590 120573) isin A13
(16) (120590 120573) isin A16
(19) (120590 120573) isin A19 (20) (120590 120573) isin A20
(23) (120590 120573) isin A23(24) (120590 120573) isin A24
(21) (120590 120573) isin A21
(22) (120590 120573) isin A22
(17) (120590 120573) isin A17 (18) (120590 120573) isin A18
(14) (120590 120573) isin A14 (15) (120590 120573) isin A15
Figure 2 Continued
Mathematical Problems in Engineering 7
00 05 10 15
0
1
2
3
00 05 10 15
0
1
2
3
0 1 2
0
1
2
3
0 1 2
0
1
2
3
00 05 10 15
0
1
2
3
0 1 2
0
2
4
0 1 2
0
2
4
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus2
minus4
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2 minus1
minus2minus3 minus1
minus2
minus2
minus3
minus4
minus1
minus2 minus1
minus10minus15 minus05 minus10minus15 minus05
minus10minus15 minus05
y
yyy
y y y
120593
120593120593120593
120593 120593 120593
(25) (120590 120573) isin A25
(28) (120590 120573) isin A28 (29) (120590 120573) isin A29
(31) (120590 120573) isin A31
(30) (120590 120573) isin A30
(26) (120590 120573) isin A26 (27) (120590 120573) isin A27
Figure 2 The bifurcation of phase portraits of system (15) when 120598 gt 0 and 120590 ge 0
11986111 (120590 120573) = (0 minus1205989848582)
11986112 (120590 120573) | 0 lt 120590 lt 1205902s 120573 = 1205732(120590)
11986113 (120590 120573) | 0 lt 120590 lt 1205901s 1205733(120590) lt 120573 lt min1205731(120590)
1205732(120590)
11986114 (120590 120573) | 1205904s lt 120590 lt 1205901s 120573 = 1205731(120590)
11986115 (120590 120573) = (1205901s 1205731s)
11986116 (120590 120573) | 0 lt 120590 lt 1205901s 120573 = 1205733(120590)
11986117 (120590 120573) | 0 lt 120590 lt 120598984858 120573 lt 1205733(120590)
11986118 (120590 120573) | 120590 = 0 120573 = minus120598984858
2
11986119 (120590 120573) | 120590 = 0 120573 lt 0
11986120 (120590 120573) | 0 lt 120590 lt 120598984858 120573 lt min0 1205733(120590)
11986121 (120590 120573) | 120590 gt 120598984858 120573 = 1205733(120590)
11986122 (120590 120573) | 120590 gt 120598984858 1205733(120590) lt 120573 lt 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
1205732(120590)
(1205901s 1205731s)
(1205902s 1205732s)
1205733(120590)
120573
minus2minus4minus6
1205731(120590)
120590
Figure 3 The bifurcation curves in the (120590 120573)-parameter plane for120598 lt 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 Let119862119896(Ω) denote the set of all 119896 times continuouslydifferentiable functions on the open setΩ 119871119901loc(R) refer to theset of all functions whose restriction on any compact subsetis 119871119901 integrable1198671loc(R) stands for1198671loc(R) = 120593 isin 119871
2
loc(R) |
1205931015840isin 1198712
loc(R)To study single peak soliton solutions we impose the
boundary condition
lim120585rarrplusmninfin
120593 = 119860 (22)
where 119860 is a constant In fact the constant 119860 is equal to thehorizontal coordinate of saddle point 119864(120593
119890 0) Substituting
the boundary condition (22) into (14) generates the followingconstant
ℎ = 119860(120590 +119860120598
2+1198603120573
4) (23)
So the ODE (14) becomes
(1205931015840)2
=
(120593 minus 119860)2
(1205731205932+ 2119860120573120593 + 3119860
2120573 + 2120598)
2 (9848582 minus 1205932) (24)
If 120573(1198602120573 + 120598) le 0 then (24) reduces to
(1205931015840)2
=120573 (120593 minus 119860)
2
(120593 minus 1198611) (120593 minus 119861
2)
2 (9848582 minus 1205932) (25)
where
1198611= minus119860 minus
radicminus2120573 (1198602120573 + 120598)
120573
1198612= minus119860 +
radicminus2120573 (1198602120573 + 120598)
120573
(26)
From (26) we know that 1198611gt 1198612if 120573 lt 0 and 119861
1lt 1198612if
120573 gt 0
Definition 4 A function120593(120585) is said to be a single peak solitonsolution for the 119862(3 2 2) equation (10) if 120593(120585) satisfies thefollowing conditions
(C1) 120593(120585) is continuous on R and has a unique peakpoint 120585
0 where 120593(120585) attains its global maximum or
minimum value
(C2) 120593(120585) isin 119862
3(R minus 120585
0) satisfies (24) on R minus 120585
0
(C3) 120593(120585) satisfies the boundary condition (22)
Definition 5 A wave function 120593 is called smooth solitonsolution if 120593 is smooth locally on either side of 120585
0and
lim120585uarr1205850
1205931015840(120585) = lim
120585darr12058501205931015840(120585) = 0
Definition 6 A wave function 120593 is called peakon if 120593 issmooth locally on either side of 120585
0and lim
120585uarr12058501205931015840(120585) =
minuslim120585darr1205850
1205931015840(120585) = 119886 119886 = 0 119886 = plusmninfin
Definition 7 Awave function120593 is called cuspon if120593 is smoothlocally on either side of 120585
0and lim
120585uarr12058501205931015840(120585) = minuslim
120585darr12058501205931015840(120585) =
plusmninfin
Without any loss of generality we choose the peak point1205850as vanishing 120585
0= 0
Theorem 8 Assume that 119906(119909 119905) = 120593(120585) = 120593(119909 minus 119888119905) is a singlepeak soliton solution of the 119862(3 2 2) equation (10) at the peakpoint 120585
0= 0 Then we have the following
(i) If 120573(1198602120573 + 120598) gt 0 then 120593(0) = 984858 or 120593(0) = minus984858
(ii) If 120573(1198602120573 + 120598) le 0 then 120593(0) = 984858 or 120593(0) = minus984858 or120593(0) = 119861
1or 120593(0) = 119861
2
Proof If 120593(0) = plusmn984858 then 120593(120585) = plusmn984858 for any 120585 isin R since120593(120585) isin 119862
3(R minus 0) Differentiating both sides of (24) yields
120593 isin 119862infin(R)
(i) When 120573(1198602120573 + 120598) gt 0 if 120593(0) = 984858 and 120593(0) = minus984858 then
120593 isin 119862infin(R) By the definition of single peak soliton we have
1205931015840(0) = 0 However by (24) we must have 120593(0) = 119860 which
contradicts the fact that 0 is the unique peak point(ii) When 120573(119860
2120573 + 120598) le 0 if 120593(0) = 984858 and 120593(0) = minus984858
by (24) we know 1205931015840(0) exists and 120593
1015840(0) = 0 since 0 is a peak
point Thus we obtain 120593(0) = 1198611or 120593(0) = 119861
2from (25)
since 120593(0) = 119860 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 119906(119909 119905) = 120593(119909 minus 119888119905) is a single peaksoliton solution of the 119862(3 2 2) equation (10) at the peak point1205850= 0 Then we have the following solution classification(i) If |120593(0)| = 984858 then 120593(120585) isin 119862
infin(R) and 120593 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
00 05 10 15
0
2
4
0 1
0
2
4
6
00 05 10 15
0
2
4
00 05 10
0
1
2
3
00 05 10 15
0
1
2
3
00 05 10
0
1
2
00 05 10 15
0
1
2
00 05 10 15
0
1
2
3
minus3 minus1minus4
minus4
minus2
minus4
minus2
minus4
minus2
minus4
minus2minus4
minus6
minus2
minus4
minus2
minus4
minus2
minus2 minus4 minus4minus2 minus2
minus3
minus1
minus2
minus1
minus2
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus1minus2 minus1minus2
120593
120593
120593
120593 120593 120593
120593 120593
120593 120593
120593 120593
y
y
y
yy
y
y y
y y
y y
minus15 minus10 minus05
minus15minus20 minus10 minus05
minus15 minus10 minus05 minus15 minus10 minus05minus10 minus05
minus15 minus10 minus05minus10 minus05
(10) (120590 120573) isin B10 (11) (120590 120573) isin B11 (12) (120590 120573) isin B12
(1) (120590 120573) isin B1
(4) (120590 120573) isin B4
(7) (120590 120573) isin B7 (8) (120590 120573) isin B8 (9) (120590 120573) isin B9
(5) (120590 120573) isin B5 (6) (120590 120573) isin B6
(2) (120590 120573) isin B2 (3) (120590 120573) isin B3
Figure 4 Continued
10 Mathematical Problems in Engineering
00 05 10 15
0
1
2
3
0 1
0
1
2
3
0 1 2
0
2
4
00 05 10 15
0
1
2
3
0 1 2
0
1
2
3
00 05 10 15
0
1
2
00 05 10 15
0
1
2
3
0 1 2
0
1
2
3
00 05 10 15
0
2
4
00 05 10 15
0
2
4
6
minus4
minus2
minus4
minus2
minus4
minus6
minus2
minus3 minus1minus2
minus3 minus1minus2
minus1minus2
minus1minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus1
minus2
minus3
minus1
minus2
120593 120593 120593
120593120593120593
120593 120593 120593
120593
y y y
yy
y
y yy
y
minus15 minus10 minus05
minus15minus20 minus10 minus05
minus15 minus10 minus05
minus15 minus10 minus05
minus15 minus10 minus05
minus15 minus10 minus05
(13) (120590 120573) isin B13 (14) (120590 120573) isin B14 (15) (120590 120573) isin B15
(16) (120590 120573) isin B16 (17) (120590 120573) isin B17 (18) (120590 120573) isin B18
(22) (120590 120573) isin B22
(20) (120590 120573) isin B20 (21) (120590 120573) isin B21(19) (120590 120573) isin B19
Figure 4 The bifurcation of phase portraits of system (15) when 120598 lt 0 and 120590 ge 0
Mathematical Problems in Engineering 11
(ii) If 120593(0) = 984858 = 1198611 then 120593 is a cuspon solution and 120593 has
the following asymptotic behavior
120593 (120585) minus 984858 = 1205821
10038161003816100381610038161205851003816100381610038161003816
23
+ 119874(10038161003816100381610038161205851003816100381610038161003816
43
) 120585 997888rarr 0
1205931015840(120585) =
2
31205821
10038161003816100381610038161205851003816100381610038161003816
minus13 sgn (120585) + 119874 (10038161003816100381610038161205851003816100381610038161003816
13
)
120585 997888rarr 0
(27)
where 1205821= plusmn(9|120573984858
2+ 2119860120573984858 + 3119860
2120573 + 2120598|(984858 minus 119860)
216984858)13
Thus 120593(120585) notin 1198671
119897119900119888(R)
(iii) If 120593(0) = minus984858 = 1198612 then 120593 is a cuspon solution and 120593
has the following asymptotic behavior
120593 (120585) + 984858 = 1205822
10038161003816100381610038161205851003816100381610038161003816
23
+ 119874(10038161003816100381610038161205851003816100381610038161003816
43
) 120585 997888rarr 0
1205931015840(120585) =
2
31205822
10038161003816100381610038161205851003816100381610038161003816
minus13 sgn (120585) + 119874 (10038161003816100381610038161205851003816100381610038161003816
13
)
120585 997888rarr 0
(28)
where 1205822= plusmn(9|120573984858
2minus 2119860120573984858 + 3119860
2120573 + 2120598|(984858 + 119860)
216984858)13
Thus 120593(120585) notin 1198671
119897119900119888(R)
(iv) If 120593(0) = 984858 = 1198611and 119860 = 0 then 120593 is a peakon-like
solution and
120593 (120585) minus 984858 = 1205823
10038161003816100381610038161205851003816100381610038161003816 + 119874 (
10038161003816100381610038161205851003816100381610038161003816
2
) 120585 997888rarr 0
1205931015840(120585) = 120582
3sgn (120585) + 119874 (
10038161003816100381610038161205851003816100381610038161003816) 120585 997888rarr 0
(29)
where 1205823= plusmn|119860 minus 984858|radic120573(119861
2minus 984858)4984858
(v) If 120593(0) = minus984858 = 1198612and 119860 = 0 then 120593 is a peakon-like
solution and
120593 (120585) + 984858 = 1205824
10038161003816100381610038161205851003816100381610038161003816 + 119874 (
10038161003816100381610038161205851003816100381610038161003816
2
) 120585 997888rarr 0
1205931015840(120585) = 120582
4sgn (120585) + 119874 (
10038161003816100381610038161205851003816100381610038161003816) 120585 997888rarr 0
(30)
where 1205824= plusmn|119860 + 984858|radicminus120573(119861
1+ 984858)4984858
(vi) If 120593(0) = 984858 = 1198611and 119860 = 0 then 120593 gives the peakon
solution 984858 exp(minusradicminus1205732|119909 minus 119888119905|)(vii) If 120593(0) = minus984858 = 119861
2and 119860 = 0 then 120593 gives the peakon
solution minus984858 exp(minusradicminus1205732|119909 minus 119888119905|)
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 |120593(0)| = 984858 then 120593 isin 119862
infin(R) and 120593 is a smooth soliton
solution(ii) If 120593(0) = 984858 = 119861
1 then by the definition of single peak
soliton we have 119860 = 984858 thus 1205731205932 + 2119860120573120593 + 31198602120573 + 2120598 does
not contain the factor 120593 minus 984858 From (24) we obtain
1205931015840= sgn (119860 minus 984858)
1003816100381610038161003816120593 minus 1198601003816100381610038161003816
radic210038161003816100381610038161205932 minus 9848582
1003816100381610038161003816
sdot radic10038161003816100381610038161205731205932 + 2119860120573120593 + 31198602120573 + 2120598
1003816100381610038161003816 sgn (120585)
(31)
Let 1198971(120593) = radic2(120593 + 984858)|120593 minus 119860|radic|1205731205932 + 2119860120573120593 + 31198602120573 + 2120598|
then 1198971(984858) = 2radic984858|984858 minus 119860|radic|1205739848582 + 2119860120573984858 + 31198602120573 + 2120598| and
int 1198971(120593)radic
1003816100381610038161003816120593 minus 9848581003816100381610038161003816119889120593 = int sgn (119860 minus 984858) sgn (120585) 119889120585 (32)
Inserting 1198971(120593) = 119897
1(984858) + 119874(|120593 minus 984858|) into (32) and using the
initial condition 120593(0) = 984858 we obtain
21198971(984858)
3
1003816100381610038161003816120593 minus 9848581003816100381610038161003816
32
(1 + 119874 (1003816100381610038161003816120593 minus 984858
1003816100381610038161003816)) =10038161003816100381610038161205851003816100381610038161003816
(33)
thus
120593 minus 984858 = plusmn(3
21198971(984858)
)
23
10038161003816100381610038161205851003816100381610038161003816
23
(1 + 119874 (1003816100381610038161003816120593 minus 984858
1003816100381610038161003816))minus23
= plusmn(3
21198971(984858)
)
23
10038161003816100381610038161205851003816100381610038161003816
23
(1 + 119874 (1003816100381610038161003816120593 minus 984858
1003816100381610038161003816))
(34)
which implies 120593 minus 984858 = 119874(|120585|23
) Therefore we have
120593 (120585) = 984858 plusmn (3
21198971(984858)
)
23
10038161003816100381610038161205851003816100381610038161003816
23
+ 119874(10038161003816100381610038161205851003816100381610038161003816
43
)
= 984858 + 1205821
10038161003816100381610038161205851003816100381610038161003816
23
+ 119874(10038161003816100381610038161205851003816100381610038161003816
43
) 120585 997888rarr 0
1205821= plusmn(
3
21198971(984858)
)
23
= plusmn(
9100381610038161003816100381610038161205739848582+ 2119860120573984858 + 3119860
2120573 + 2120598
10038161003816100381610038161003816(984858 minus 119860)
2
16984858)
13
1205931015840(120585) =
2
31205821
10038161003816100381610038161205851003816100381610038161003816
minus13 sgn (120585) + 119874 (10038161003816100381610038161205851003816100381610038161003816
13
) 120585 997888rarr 0
(35)
So 120593(120585) notin 1198671
loc(R)(iii) Similar to the proof of (ii) we ignore it in this paper(iv) If 120593(0) = 984858 = 119861
1and 119860 = 0 then from (25) we obtain
1205931015840= radicminus
120573
2sgn (119860 minus 984858)
1003816100381610038161003816120593 minus 1198601003816100381610038161003816radic120593 minus 1198612
120593 + 984858sgn (120585) (36)
Let 1198972(120593) = (1|120593minus119860|)radic(120593 + 984858)(120593 minus 119861
2) then 119897
2(984858) = (1|984858minus
119860|)radic2984858(984858 minus 1198612) and
int 1198972(120593) 119889120593 = radicminus
120573
2int sgn (119860 minus 984858) sgn (120585) 119889120585 (37)
Inserting 1198972(120593) = 119897
2(984858) + 119874(|120593 minus 984858|) into (37) and using the
initial condition 120593(0) = 984858 we obtain
1198972(984858) (120593 minus 984858) (1 + 119874 (
1003816100381610038161003816120593 minus 9848581003816100381610038161003816))minus1
= sgn (119860 minus 984858)10038161003816100381610038161205851003816100381610038161003816
(38)
Sincesgn (120593 minus 984858) sgn (119860 minus 984858) ge 0
1
1 + 119874 (120593 minus 984858)= 1 + 119874 (120593 minus 984858)
(39)
12 Mathematical Problems in Engineering
we get
1003816100381610038161003816120593 minus 9848581003816100381610038161003816 =
radicminus120573
2
1
1198972(984858)
10038161003816100381610038161205851003816100381610038161003816 (1 + 119874 (120593 minus 984858)) (40)
which implies |120593 minus 984858| = 119874(|120585|) Therefore we have
120593 (120585) = 984858 + 1205823
10038161003816100381610038161205851003816100381610038161003816 + 119874 (
10038161003816100381610038161205851003816100381610038161003816
2
) 120585 997888rarr 0
1205931015840(120585) = 120582
3sgn (120585) + 119874 (
10038161003816100381610038161205851003816100381610038161003816) 120585 997888rarr 0
(41)
where 1205823= plusmn|119860 minus 984858|radic120573(119861
2minus 984858)4984858
(v) Similar to the proof of (iv) we ignore it in this paper
By virtue of Theorem 9 any single peak soliton for the119862(3 2 2) equation (10) must satisfy the following initial andboundary values problem
(1205931015840)2
=120573 (120593 minus 119860)
2
(120593 minus 1198611) (120593 minus 119861
2)
2 (9848582 minus 1205932)fl119871 (120593)
120593 (0) isin 984858 minus984858 1198611 1198612
lim|120585|rarrinfin
120593 (120585) = 119860
(42)
119871(120593) ge 0 and the boundary condition (24) imply thefollowing
(a) If 120573(9848582 minus 1205932) ge 0 then 120593 ge max119861
1 1198612 or 120593 le
min1198611 1198612
(b) If120573(9848582minus1205932) le 0 thenmin1198611 1198612 le 120593 le max119861
1 1198612
Below we will present some implicit formulas for thesingle peak soliton solutions in the case of specific 120590 and 120573
Case 1 ((120590 120573) isin 11986012) In this case we have minus984858 lt 119861
2lt 119860 lt
1198611= 984858 From the standard phase analysis and Theorem 9
we know that if 120593 is a single peak soliton of the 119862(3 2 2)
equation then
1205931015840= minusradicminus
120573
2(120593 minus 119860)radic
120593 minus 1198612
120593 + 984858sgn (120585) (43)
From the separation of variables we get
intℎ (120593) 119889120593 = radicminus120573
2sgn (120585) 119889120585 (44)
where ℎ(120593) = (1(119860 minus 120593))radic(120593 + 984858)(120593 minus 1198612) After a lengthy
calculation of integral we obtain the implicit solution 120593
defined by
119867(120593)flradic119860 + 984858
119860 minus 1198612
1198681(120593) minus 119868
2(120593) = radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870 (45)
where
1198681(120593) = ln
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
(119860 + 984858) (120593 minus 1198612) + (119860 minus 119861
2) (120593 + 984858) + 2radic(119860 + 984858) (119860 minus 119861
2) (120593 minus 119861
2) (120593 + 984858)
radic(119860 + 984858) (119860 minus 1198612) (119860 + 984858) (119860 minus 120593)
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
1198682(120593) = ln
1003816100381610038161003816100381610038161003816(120593 + 984858) + (120593 minus 119861
2) + 2radic(120593 + 984858) (120593 minus 119861
2)
1003816100381610038161003816100381610038161003816
(46)
and 119870 is an arbitrary integration constant For 120593(0) = 984858 theconstant119870 = 119867(120593(0)) is defined by
119870 = radic119860 + 984858
119860 minus 1198612
1198681(984858) minus 119868
2(984858) (47)
and for 120593(0) = 1198612
119870 = radic119860 + 984858
119860 minus 1198612
1198681(1198612) minus 1198682(1198612) (48)
(i) If 120593(0) = 984858 then 119860 lt 120593 le 984858 Since 1198671015840(120593) = ℎ(120593) weknow that119867(120593) strictly decreases on the interval (119860 984858] thus1198671(120593) = 119867|
(119860984858](120593) gives a single peak solitonwith119867
1(984858) = 119870
and1198671(119860+) = infin Therefore 120593
1(120585) = 119867
minus1
1(radicminus1205732|120585| + 119870) is
the solution satisfying
1205931(0) = 984858
lim|120585|rarrinfin
1205931(120585) = 119860
1205931015840
1(0plusmn) = plusmn (119860 minus 984858)radic
120573 (1198612minus 984858)
4984858
(49)
So 1205931(120585) is a peakon-like solution (see Figure 5)
(ii) If 120593(0) = 1198612 then 119861
2le 120593 lt 119860 By 119867
1015840(120593) = 119891(120593)
we know that 119867(120593) strictly increases on the interval [1198612 119860)
Thus
1198672(120593) = 119867|
[1198612 119860)(120593) (50)
Mathematical Problems in Engineering 13
minus3 minus2 minus1
120593
minus02
1 2 3
02
04
06
08
10
x
(a)
02
02
minus2 minus2 x
10
05
00
t
120593
22
(b)
Figure 5 Two- and three-dimensional graphs of the peakon-like solution
minus03
minus04
minus05
1 2 3minus3 minus2 minus1
120593
minus02
x
(a)
0
2
0
2
minus2minus2
minus02
minus03
minus04
minus05
minus06
t x
120593
0
2
0
2
minus2minus2
t x
(b)
Figure 6 Two- and three-dimensional graphs of the smooth soliton solution
has the inverse denoted by 1205932(120585) = 119867
minus1
2(radicminus1205732|120585| + 119870)
1205932(120585) gives a kind of smooth soliton solution (see Figure 6)
satisfying
1205932(0) = 119861
2
lim|120585|rarrinfin
1205932(120585) = 119860
1205931015840
2(0) = 0
(51)
Case 2 ((120590 120573) isin 11986014) In this case we have minus984858 = 119861
2lt 119860 lt
984858 lt 1198611and (25) is equivalent to
1205931015840= minusradicminus
120573
2(120593 minus 119860)radic
120593 minus 1198611
120593 minus 984858sgn (120585) (52)
Let
119892 (120593) =1
119860 minus 120593radic120593 minus 1198611
120593 minus 984858 (53)
14 Mathematical Problems in Engineering
minus1 1 2 3
02
04
06
08
10
minus3 minus2
120593
x
(a)
02
02minus2
minus2
120593
x t
00
05
10
(b)
Figure 7 Two- and three-dimensional graphs of the cuspon solution
then (52) is converted to
119892 (120593) 119889120593 =1
119860 minus 120593radic120593 minus 1198611
120593 minus 984858119889120593 = radicminus
120573
2sgn (120585) 119889120585 (54)
Integrating (54) on the interval [0 120585] (or [120585 0]) leads to thefollowing implicit solutions
119866 (120593)flradic119860 minus 984858
119860 minus 1198611
1198683(120593) + 119868
4(120593) = radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870 (55)
where
1198683(120593) = ln
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
(119860 minus 984858) (120593 minus 1198611) + (119860 minus 119861
1) (120593 minus 984858) + 2radic(119860 minus 984858) (119860 minus 119861
1) (120593 minus 984858) (120593 minus 119861
1)
radic(119860 minus 984858) (119860 minus 1198611) (119860 minus 984858) (120593 minus 119860)
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
1198684(120593) = ln
1003816100381610038161003816100381610038161003816(120593 minus 984858) + (120593 minus 119861
1) + 2radic(120593 minus 984858) (120593 minus 119861
1)
1003816100381610038161003816100381610038161003816
(56)
And119870 is an arbitrary integration constant It is obvious thatfor 120593(0) = 984858 the constant119870 = 119867(120593(0)) is defined by
119870 = radic119860 minus 984858
119860 minus 1198611
1198683(984858) + 119868
4(984858) (57)
and for 120593(0) = minus984858
119870 = radic119860 minus 984858
119860 minus 1198611
1198683(minus984858) + 119868
4(minus984858) (58)
(i) If 120593(0) = 984858 then 119860 lt 120593 le 984858 From 119892(120593) lt 0 weknow that 119866(120593) strictly decreases on the interval (119860 984858] with119866(984858) = 119870 and 119866(119860+) = infin Define
1198661(120593) = 119866|
(119860984858](120593) (59)
Since 1198661(120593) is a strictly decreasing function from (119860 984858] onto
[119870infin) we can solve for 120593 uniquely from (59) and obtain
1205931(120585) = 119866
minus1
1(radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870) (60)
It is easy to check that 120593 satisfies
1205931(0) = 984858
lim|120585|rarrinfin
1205931(120585) = 119860
1205931015840
1(0plusmn) = ∓infin
(61)
Therefore the solution1205931defined by (60) is a cuspon solution
for the 119862(3 2 2) equation (see Figure 7)
Mathematical Problems in Engineering 15
(ii) If 120593(0) = minus984858 then minus984858 le 120593 lt 119860 Through a similaranalysis we get a strictly increasing function 119866(120593) on theinterval [minus984858 119860) satisfying
119866 (120593) = radicminus120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870 (62)
where 119866(120593) is defined by (55) Let
1198662(120593) = 119866|
[minus984858119860)(120593) (63)
then 1198662(120593) is a strictly increasing function from [minus984858 119860) onto
[119870infin) so that we can solve for 120593 and obtain
1205932(120585) = 119866
minus1
2(radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870) (64)
It is easy to check that 1205932satisfies
1205932(0) = minus984858
lim|120585|rarrinfin
1205932(120585) = 119860
1205931015840
2(0plusmn) = plusmn (119860 + 984858)radicminus
120573 (1198611+ 984858)
4984858
(65)
Therefore the solution 1205932defined by (64) is a peakon-like
solution whose graph is similar to those in Figure 5
Case 3 ((120590 120573) isin 11986016) In this case we have minus984858 lt 119861
2lt 119860 =
0 lt 1198611lt 984858 119861
2= minus1198611 and
1205931015840= minusradicminus
120573
2120593radic
120593 minus 1198612
1
1205932 minus 9848582sgn (120585) (66)
Hence from the separation of variables we have
radic9848582 minus 1205932
120593radic1198612
1minus 1205932
119889120593 = minusradicminus120573
2sgn (120585) 119889120585 (67)
Integrating (67) on the interval [0 120585] (or [120585 0]) leads tothe following implicit formula for the two smooth solitonsolutions
radic9848582 minus 1205932 + radic1198612
1minus 1205932
radic9848582 minus 1198612
1
(
radic(9848582 minus 1198612
1) 1205932
984858radic1198612
1minus 1205932 + 119861
1radic9848582 minus 1205932
)
984858
= exp(minusradicminus120573
2|119909 minus 119888119905|)
(68)
where 120593 isin (119860 1198611] Consider
radic9848582 minus 1198612
2
radic9848582 minus 1205932 + radic1198612
2minus 1205932
(
984858radic1198612
2minus 1205932 minus 119861
2radic9848582 minus 1205932
radic(9848582 minus 1198612
2) 1205932
)
984858
= exp(radicminus120573
2|119909 minus 119888119905|)
(69)
where 120593 isin [1198612 119860)
Case 4 ((120590 120573) isin 11986017) In this case we have minus984858 = 119861
2lt 119860 =
0 lt 1198611= 984858 and
1205931015840= minusradicminus
120573
2120593 sgn (120585) (70)
Choosing 120593(0) = 984858 (or minus984858) as initial value we get
int
120593
984858
1
120593119889120593 = minusint
120585
0
radicminus120573
2sgn (120585) 119889120585 for 119860 lt 120593 le 984858
int
120593
minus984858
1
120593119889120593 = minusint
120585
0
radicminus120573
2sgn (120585) 119889120585 for minus 984858 le 120593 lt 119860
(71)
which immediately yield the peakon solutions
120593 (119909 minus 119888119905) = plusmn984858 exp(minusradicminus120573
2|119909 minus 119888119905|) (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 the119862(3 2 2) equation (10) In order to study kinkwave solutionswe assume that
lim120585rarrinfin
120593 (120585) = 1198601
lim120585rarrminusinfin
120593 (120585) = 1198602
(73)
16 Mathematical Problems in Engineering
2 4
02
04
06
08
10
minus2minus4
120593
x
(a)
0 2
02
00
05
10
minus2
120593
x
t
0
2
minus2
(b)
Figure 8 Two- and three-dimensional graphs of the peakon solution
where 1198601gt 1198602 Substituting the boundary condition (73)
into (14) generates
(1205931015840)2
=
120573 (120593 minus 1198601) (120593 minus 119860
2) [1205932+ (1198601+ 1198602) 120593 + 119860
2
1+ 11986011198602+ 1198602
2+ 2120598120573]
2 (9848582 minus 1205932)fl119865 (120593) (74)
The nonlinear differential equation (74) may sustaindifferent kinds of nonlinear excitations In what follows weconfine our attention to the cases 119860
2= minus119860
1and 120598 =
minus1198602
1120573 which describe kinks and kink compactons which
play an important role in the dynamics systems Under theseconsiderations (74) reduces to
(1205931015840)2
=120573 (120593 minus 119860
1)2
(120593 minus 1198602)2
2 (9848582 minus 1205932) (75)
If 1198601lt 984858 then from the phase analysis in Section 2 (see
Figure 4(10)) we know that (1198601 0) and (119860
2 0) are two saddle
points of (13) and the kink solutions can be obtained fromthe two heteroclinic orbits connecting (120601 119910) = (119860
1 0) and
(1198602 0) When 119860
1increases upon reaching 984858 that is 119860
1= 984858
(75) becomes
(1205931015840)2
= minus120573 (120593 minus 119860
1) (120593 minus 119860
2)
2 (76)
and the ellipse 21199102+120573(120593minus1198601)(120593minus119860
2) = 0 (see Figure 4(11))
which is tangent to the singular lines 120593 = 984858 and 120593 = minus984858 atpoints (119860
1 0) and (119860
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 120593 is a kink wave solutions of (75)
or (76) we have 1205931015840
rarr 0 as 120593 rarr 1198601and as 120593 rarr 119860
2
Moreover we have 119865(120593) ge 0 for 1198602le 120593 le 119860
1and 120593 is
strictly monotonic in any interval where 119865(120593) gt 0 Thus if1205931015840gt 0 at some points 120593will be strictly increasing until it gets
close to the next zero of 119865 Denoting this zero 1198601 we have
120593 uarr 1198601 What will happen to the solution when it approaches
1198601 Depending on whether the zero is double or simple 120593
has a different behavior We explore the two cases in turn
Theorem 11 (i) If 120593 has a simple zero at 120593 = 1198601 so that
119865(1198601) = 0 and 1198651015840(119860
1) lt 0 then the solution 120593 of (75) satisfies
120593 (120585) minus 1198601=1198651015840(1198601)
4(120585 minus 120578)
2
+ 119874 ((120585 minus 120578)4
)
119886119904 120585 uarr 120578
(77)
where 120593(120578) = 1198601
(ii) When 120593 approaches the double zero 1198601of 119865(120593) so that
1198651015840(1198601) = 0 and11986510158401015840(119860
1) gt 0 then the solution120593 of (75) satisfies
120593 (120585) minus 1198601sim 120572 exp(minus120585radic11986510158401015840 (119860
1)) 119886119904 120585 997888rarr infin (78)
for some constant 120572 Thus 120593 uarr 1198601exponentially as 120585 rarr infin
Mathematical Problems in Engineering 17
Proof (i) When 1198601= minus119860
2= 984858 and 120598 = minus119860
2
1120573 from (76)
119865(120593) has a simple zero at 120593 = 1198601 Then
(1205931015840)2
= (120593 minus 1198601) 1198651015840(1198601) + 119874 ((120593 minus 119860
1)2
)
as 120593 uarr 1198601
(79)
Using the fact that (1205931015840)2 ge 0 we know that 1198651015840(1198601) lt 0
Moreover
119889120585
119889120593=
1
radic(120593 minus 1198601) 1198651015840 (119860
1) + 119874 ((120593 minus 119860
1)2
)
(80)
Because
radic(120593 minus 1198601) 1198651015840 (119860
1) + 119874 ((120593 minus 119860
1)2
)
= radic1198601minus 120593(radicminus1198651015840 (119860
1) + 119874 (119860
1minus 120593))
1
radicminus1198651015840 (1198601) + 119874 (119860
1minus 120593)
=1
radicminus1198651015840 (1198601)
+ 119874 (1198601minus 120593)
(81)
we obtain119889120585
119889120593=
1
radicminus (1198601minus 120593) 1198651015840 (119860
1)
+ 119874 ((1198601minus 120593)12
) (82)
Integrating (82) yields
120578 minus 120585 =2
radicminus1198651015840 (1198601)
(1198601minus 120593)12
+ 119874((1198601minus 120593)32
) (83)
where 120578 satisfies 120593(120578) = 1198601 Thus
(120578 minus 120585)2
=4
minus1198651015840 (1198601)(1198601minus 120593) + 119874 ((119860
1minus 120593)2
) (84)
which implies 119874((1198601minus 120593)2) = 119874((120578 minus 120585)
4) Therefore we get
120593 (120585) = 1198601+1198651015840(1198601)
4(120585 minus 120578)
2
+ 119874 ((120585 minus 120578)4
)
as 120585 uarr 120578
(85)
where 120593(120578) = 1198601
(ii) When 1198601= minus119860
2lt 984858 and 120598 = minus119860
2
1120573 from (75) 119865(120593)
has a double zero at 120593 = 1198601 Then
(1205931015840)2
= (120593 minus 1198601)2
11986510158401015840(1198601) + 119874 ((120593 minus 119860
1)3
)
as 120593 uarr 1198601
(86)
Furthermore we get
119889120585
119889120593=
1
radic(120593 minus 1198601)2
11986510158401015840 (1198601) + 119874 ((120593 minus 119860
1)3
)
(87)
Observing that
radic(120593 minus 1198601)2
11986510158401015840 (1198601) + 119874 ((120593 minus 119860
1)3
)
= (1198601minus 120593) (radic11986510158401015840 (119860
1) + 119874 (119860
1minus 120593))
1
radic11986510158401015840 (1198601) + 119874 (119860
1minus 120593)
=1
radic11986510158401015840 (1198601)
+ 119874 (1198601minus 120593)
(88)
we obtain119889120585
119889120593=
1
(1198601minus 120593) 11986510158401015840 (119860
1)+ 119874 (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 119860
1= minus119860
2lt 984858 and 120598 = minus119860
2
1120573 Then (75)
becomes
radic9848582 minus 1205932
1205932 minus 1198602
1
119889120593 = ∓radic120573
2119889120585 (90)
Integrating both sides of (90) gives the following implicitexpressions of kink and antikink wave solutions
arctan120593
radic9848582 minus 1205932
+
radic9848582 minus 1198602
1
1198601
arctanhradic9848582 minus 119860
2
1
1198601
120593
radic9848582 minus 1205932
= plusmnradic120573
2120585
(91)
By letting 1198601rarr 984858 in (91) we get two kink compactons
which are given by
plusmn120593 =
984858 sinradic120573
2120585
10038161003816100381610038161205851003816100381610038161003816 le
120587
radic2120573
984858 120585 gt120587
radic2120573
minus984858 120585 lt minus120587
radic2120573
(92)
The graphs for the kink wave solutions (91) and kinkcompacton solutions (92) are shown in Figures 9 and 10respectively
Remark 12 The two kink compacton solutions (92) aredifferent from the well-known smooth kink wave solutionsIn comparison with kink wave solutions (91) the kink com-pacton solutions (92) have no exponential decay propertiesbut have compact supportThat is they minus a constant thedifferences identically vanish outside a finite core region
18 Mathematical Problems in Engineering
120593
120585minus1 1
05
minus05
(a)
minus1minus1
minus05
05
00120593
x t
0
1
01
11
(b)
Figure 9 Two- and three-dimensional graphs of the kink wave solution
1
1120593
120585
minus1
minus1
(a)
0
1
0
1
000510
120593
minus1 minus1
minus05
minus10
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 the119862(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 Newtonrsquos equation which we write in the form 119910
2=
119865(120593) where 119910 = 1205931015840
120585(120585) we solve Newtonrsquos equation 119910
2=
119865(120593) 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 the119862(3 2 2)equation
Actually for 119898 = 2119896 + 1 119896 isin N+ in 119862(119898 2 2) equation(9) the dynamical behavior of traveling wave solutions of(9) is similar to the case 119898 = 3 for 119898 = 2119896 119896 isin N+ in119862(119898 2 2) equation (9) the dynamical behavior of travelingwave solutions of (9) is similar to the case 119898 = 2 We areapplying the approach mentioned in this work to 119862(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
References
[1] C Rogers and W R Shadwick Backlund Transformation andTheir Applications Academic Press New York NY USA 1982
[2] W Malfliet and W Hereman ldquoThe tanh method I Exactsolutions of nonlinear evolution and wave equationsrdquo PhysicaScripta vol 54 no 6 pp 563ndash568 1996
[3] R Hirota Direct Method in Soliton Theory Springer BerlinGermany 1980
[4] J S Cohen Computer Algebra and Symbolic ComputationMathematical Methods A K Peters 2003
[5] S Lie ldquoUber die integration durch bestimmte integrale voneiner klasse linearer partieller differential eichungenrdquoArchiv derMathematik vol 6 pp 328ndash368 1881
[6] H-H Dai and X-H Zhao ldquoTraveling waves in a rod composedof amodifiedMooney-Rivlinmaterial I bifurcation of equilibraand non-singular caserdquo Proceedings of the Royal Society AMathematical Physical and Engineering Sciences vol 455 no1990 pp 3845ndash3874 1999
[7] H-H Dai and F-F Wang ldquoAsymptotic bifurcation solutionsfor compressions of a clamped nonlinearly elastic rectangletransition region and barrelling to a corner-like profilerdquo SIAMJournal on Applied Mathematics vol 70 no 7 pp 2673ndash26922010
[8] J Li and Z Liu ldquoSmooth and non-smooth traveling waves ina nonlinearly dispersive equationrdquo Applied Mathematical Mod-elling vol 25 no 1 pp 41ndash56 2000
[9] A Chen and J Li ldquoSingle peak solitary wave solutions forthe osmosis K(22) equation under inhomogeneous boundaryconditionrdquo Journal of Mathematical Analysis and Applicationsvol 369 no 2 pp 758ndash766 2010
[10] R Camassa and D D Holm ldquoAn integrable shallow waterequation with peaked solitonsrdquo Physical Review Letters vol 71no 11 pp 1661ndash1664 1993
[11] P Rosenau ldquoOn nonanalytic solitary waves formed by a nonlin-ear dispersionrdquo Physics Letters A vol 230 no 5-6 pp 305ndash3181997
[12] P Rosenau and J M Hyman ldquoCompactons solitons with finitewavelengthrdquo Physical Review Letters vol 70 no 5 pp 564ndash5671993
[13] L Zhang L-Q Chen and X Huo ldquoThe effects of horizontalsingular straight line in a generalized nonlinear Klein-Gordonmodel equationrdquo Nonlinear Dynamics vol 72 no 4 pp 789ndash801 2013
[14] J Li and H Dai On the Study of Singular Nonlinear TravelingWave Equations Dynamical System Approach Science PressBeijing China 2007
[15] A Chen S Wen S Tang W Huang and Z Qiao ldquoEffects ofquadratic singular curves in integrable equationsrdquo Studies inApplied Mathematics vol 134 no 1 pp 24ndash61 2015
[16] L Tian and J Yin ldquoNew compacton solutions and solitary wavesolutions of fully nonlinear generalized Camassa-Holm equa-tionsrdquo Chaos Solitons amp Fractals vol 20 no 2 pp 289ndash2992004
[17] L Zhang ldquoThe bifurcation and peakons for the specialC(3 2 2)equationrdquo Pramana vol 83 no 3 pp 331ndash340 2014
[18] J Lenells ldquoTraveling wave solutions of the CamassandashHolmequationrdquo Journal of Differential Equations vol 217 no 2 pp393ndash430 2005
[19] Z J Qiao and G P Zhang ldquoOn peaked and smooth solitons forthe Camassa-Holm equationrdquo Europhysics Letters vol 73 no 5pp 657ndash663 2006
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Decision SciencesAdvances in
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
6 Mathematical Problems in Engineering
00 05 10
0
1
2
3
00 05 10
0
2
4
00 05 10
0
1
2
3
00 05 10
0
1
2
3
00 05 10
0
1
2
00 05 10
0
1
2
00 05 10
0
1
2
00 05 10
0
1
2
00 05 10
0
1
2
3
00 05 10
0
1
2
3
00 05 10
0
1
2
3
00 05 10
0
1
2
3
minus3
minus2
minus1
minus3
minus2
minus1
minus2
minus1
minus2
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus1
minus2
minus1
minus3
minus2
minus1
minus10 minus05
minus10 minus05
minus10 minus05
minus10 minus05
minus10 minus05
minus10minus15 minus05 minus10minus15 minus05
minus10 minus05
minus10 minus05 minus10 minus05
minus10 minus05 minus10 minus05
minus2
minus4
y
120593120593120593
120593 120593 120593
120593120593120593
120593 120593120593
yy
y yy
yy
y
yy
y
(13) (120590 120573) isin A13
(16) (120590 120573) isin A16
(19) (120590 120573) isin A19 (20) (120590 120573) isin A20
(23) (120590 120573) isin A23(24) (120590 120573) isin A24
(21) (120590 120573) isin A21
(22) (120590 120573) isin A22
(17) (120590 120573) isin A17 (18) (120590 120573) isin A18
(14) (120590 120573) isin A14 (15) (120590 120573) isin A15
Figure 2 Continued
Mathematical Problems in Engineering 7
00 05 10 15
0
1
2
3
00 05 10 15
0
1
2
3
0 1 2
0
1
2
3
0 1 2
0
1
2
3
00 05 10 15
0
1
2
3
0 1 2
0
2
4
0 1 2
0
2
4
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus2
minus4
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2 minus1
minus2minus3 minus1
minus2
minus2
minus3
minus4
minus1
minus2 minus1
minus10minus15 minus05 minus10minus15 minus05
minus10minus15 minus05
y
yyy
y y y
120593
120593120593120593
120593 120593 120593
(25) (120590 120573) isin A25
(28) (120590 120573) isin A28 (29) (120590 120573) isin A29
(31) (120590 120573) isin A31
(30) (120590 120573) isin A30
(26) (120590 120573) isin A26 (27) (120590 120573) isin A27
Figure 2 The bifurcation of phase portraits of system (15) when 120598 gt 0 and 120590 ge 0
11986111 (120590 120573) = (0 minus1205989848582)
11986112 (120590 120573) | 0 lt 120590 lt 1205902s 120573 = 1205732(120590)
11986113 (120590 120573) | 0 lt 120590 lt 1205901s 1205733(120590) lt 120573 lt min1205731(120590)
1205732(120590)
11986114 (120590 120573) | 1205904s lt 120590 lt 1205901s 120573 = 1205731(120590)
11986115 (120590 120573) = (1205901s 1205731s)
11986116 (120590 120573) | 0 lt 120590 lt 1205901s 120573 = 1205733(120590)
11986117 (120590 120573) | 0 lt 120590 lt 120598984858 120573 lt 1205733(120590)
11986118 (120590 120573) | 120590 = 0 120573 = minus120598984858
2
11986119 (120590 120573) | 120590 = 0 120573 lt 0
11986120 (120590 120573) | 0 lt 120590 lt 120598984858 120573 lt min0 1205733(120590)
11986121 (120590 120573) | 120590 gt 120598984858 120573 = 1205733(120590)
11986122 (120590 120573) | 120590 gt 120598984858 1205733(120590) lt 120573 lt 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
1205732(120590)
(1205901s 1205731s)
(1205902s 1205732s)
1205733(120590)
120573
minus2minus4minus6
1205731(120590)
120590
Figure 3 The bifurcation curves in the (120590 120573)-parameter plane for120598 lt 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 Let119862119896(Ω) denote the set of all 119896 times continuouslydifferentiable functions on the open setΩ 119871119901loc(R) refer to theset of all functions whose restriction on any compact subsetis 119871119901 integrable1198671loc(R) stands for1198671loc(R) = 120593 isin 119871
2
loc(R) |
1205931015840isin 1198712
loc(R)To study single peak soliton solutions we impose the
boundary condition
lim120585rarrplusmninfin
120593 = 119860 (22)
where 119860 is a constant In fact the constant 119860 is equal to thehorizontal coordinate of saddle point 119864(120593
119890 0) Substituting
the boundary condition (22) into (14) generates the followingconstant
ℎ = 119860(120590 +119860120598
2+1198603120573
4) (23)
So the ODE (14) becomes
(1205931015840)2
=
(120593 minus 119860)2
(1205731205932+ 2119860120573120593 + 3119860
2120573 + 2120598)
2 (9848582 minus 1205932) (24)
If 120573(1198602120573 + 120598) le 0 then (24) reduces to
(1205931015840)2
=120573 (120593 minus 119860)
2
(120593 minus 1198611) (120593 minus 119861
2)
2 (9848582 minus 1205932) (25)
where
1198611= minus119860 minus
radicminus2120573 (1198602120573 + 120598)
120573
1198612= minus119860 +
radicminus2120573 (1198602120573 + 120598)
120573
(26)
From (26) we know that 1198611gt 1198612if 120573 lt 0 and 119861
1lt 1198612if
120573 gt 0
Definition 4 A function120593(120585) is said to be a single peak solitonsolution for the 119862(3 2 2) equation (10) if 120593(120585) satisfies thefollowing conditions
(C1) 120593(120585) is continuous on R and has a unique peakpoint 120585
0 where 120593(120585) attains its global maximum or
minimum value
(C2) 120593(120585) isin 119862
3(R minus 120585
0) satisfies (24) on R minus 120585
0
(C3) 120593(120585) satisfies the boundary condition (22)
Definition 5 A wave function 120593 is called smooth solitonsolution if 120593 is smooth locally on either side of 120585
0and
lim120585uarr1205850
1205931015840(120585) = lim
120585darr12058501205931015840(120585) = 0
Definition 6 A wave function 120593 is called peakon if 120593 issmooth locally on either side of 120585
0and lim
120585uarr12058501205931015840(120585) =
minuslim120585darr1205850
1205931015840(120585) = 119886 119886 = 0 119886 = plusmninfin
Definition 7 Awave function120593 is called cuspon if120593 is smoothlocally on either side of 120585
0and lim
120585uarr12058501205931015840(120585) = minuslim
120585darr12058501205931015840(120585) =
plusmninfin
Without any loss of generality we choose the peak point1205850as vanishing 120585
0= 0
Theorem 8 Assume that 119906(119909 119905) = 120593(120585) = 120593(119909 minus 119888119905) is a singlepeak soliton solution of the 119862(3 2 2) equation (10) at the peakpoint 120585
0= 0 Then we have the following
(i) If 120573(1198602120573 + 120598) gt 0 then 120593(0) = 984858 or 120593(0) = minus984858
(ii) If 120573(1198602120573 + 120598) le 0 then 120593(0) = 984858 or 120593(0) = minus984858 or120593(0) = 119861
1or 120593(0) = 119861
2
Proof If 120593(0) = plusmn984858 then 120593(120585) = plusmn984858 for any 120585 isin R since120593(120585) isin 119862
3(R minus 0) Differentiating both sides of (24) yields
120593 isin 119862infin(R)
(i) When 120573(1198602120573 + 120598) gt 0 if 120593(0) = 984858 and 120593(0) = minus984858 then
120593 isin 119862infin(R) By the definition of single peak soliton we have
1205931015840(0) = 0 However by (24) we must have 120593(0) = 119860 which
contradicts the fact that 0 is the unique peak point(ii) When 120573(119860
2120573 + 120598) le 0 if 120593(0) = 984858 and 120593(0) = minus984858
by (24) we know 1205931015840(0) exists and 120593
1015840(0) = 0 since 0 is a peak
point Thus we obtain 120593(0) = 1198611or 120593(0) = 119861
2from (25)
since 120593(0) = 119860 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 119906(119909 119905) = 120593(119909 minus 119888119905) is a single peaksoliton solution of the 119862(3 2 2) equation (10) at the peak point1205850= 0 Then we have the following solution classification(i) If |120593(0)| = 984858 then 120593(120585) isin 119862
infin(R) and 120593 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
00 05 10 15
0
2
4
0 1
0
2
4
6
00 05 10 15
0
2
4
00 05 10
0
1
2
3
00 05 10 15
0
1
2
3
00 05 10
0
1
2
00 05 10 15
0
1
2
00 05 10 15
0
1
2
3
minus3 minus1minus4
minus4
minus2
minus4
minus2
minus4
minus2
minus4
minus2minus4
minus6
minus2
minus4
minus2
minus4
minus2
minus2 minus4 minus4minus2 minus2
minus3
minus1
minus2
minus1
minus2
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus1minus2 minus1minus2
120593
120593
120593
120593 120593 120593
120593 120593
120593 120593
120593 120593
y
y
y
yy
y
y y
y y
y y
minus15 minus10 minus05
minus15minus20 minus10 minus05
minus15 minus10 minus05 minus15 minus10 minus05minus10 minus05
minus15 minus10 minus05minus10 minus05
(10) (120590 120573) isin B10 (11) (120590 120573) isin B11 (12) (120590 120573) isin B12
(1) (120590 120573) isin B1
(4) (120590 120573) isin B4
(7) (120590 120573) isin B7 (8) (120590 120573) isin B8 (9) (120590 120573) isin B9
(5) (120590 120573) isin B5 (6) (120590 120573) isin B6
(2) (120590 120573) isin B2 (3) (120590 120573) isin B3
Figure 4 Continued
10 Mathematical Problems in Engineering
00 05 10 15
0
1
2
3
0 1
0
1
2
3
0 1 2
0
2
4
00 05 10 15
0
1
2
3
0 1 2
0
1
2
3
00 05 10 15
0
1
2
00 05 10 15
0
1
2
3
0 1 2
0
1
2
3
00 05 10 15
0
2
4
00 05 10 15
0
2
4
6
minus4
minus2
minus4
minus2
minus4
minus6
minus2
minus3 minus1minus2
minus3 minus1minus2
minus1minus2
minus1minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus1
minus2
minus3
minus1
minus2
120593 120593 120593
120593120593120593
120593 120593 120593
120593
y y y
yy
y
y yy
y
minus15 minus10 minus05
minus15minus20 minus10 minus05
minus15 minus10 minus05
minus15 minus10 minus05
minus15 minus10 minus05
minus15 minus10 minus05
(13) (120590 120573) isin B13 (14) (120590 120573) isin B14 (15) (120590 120573) isin B15
(16) (120590 120573) isin B16 (17) (120590 120573) isin B17 (18) (120590 120573) isin B18
(22) (120590 120573) isin B22
(20) (120590 120573) isin B20 (21) (120590 120573) isin B21(19) (120590 120573) isin B19
Figure 4 The bifurcation of phase portraits of system (15) when 120598 lt 0 and 120590 ge 0
Mathematical Problems in Engineering 11
(ii) If 120593(0) = 984858 = 1198611 then 120593 is a cuspon solution and 120593 has
the following asymptotic behavior
120593 (120585) minus 984858 = 1205821
10038161003816100381610038161205851003816100381610038161003816
23
+ 119874(10038161003816100381610038161205851003816100381610038161003816
43
) 120585 997888rarr 0
1205931015840(120585) =
2
31205821
10038161003816100381610038161205851003816100381610038161003816
minus13 sgn (120585) + 119874 (10038161003816100381610038161205851003816100381610038161003816
13
)
120585 997888rarr 0
(27)
where 1205821= plusmn(9|120573984858
2+ 2119860120573984858 + 3119860
2120573 + 2120598|(984858 minus 119860)
216984858)13
Thus 120593(120585) notin 1198671
119897119900119888(R)
(iii) If 120593(0) = minus984858 = 1198612 then 120593 is a cuspon solution and 120593
has the following asymptotic behavior
120593 (120585) + 984858 = 1205822
10038161003816100381610038161205851003816100381610038161003816
23
+ 119874(10038161003816100381610038161205851003816100381610038161003816
43
) 120585 997888rarr 0
1205931015840(120585) =
2
31205822
10038161003816100381610038161205851003816100381610038161003816
minus13 sgn (120585) + 119874 (10038161003816100381610038161205851003816100381610038161003816
13
)
120585 997888rarr 0
(28)
where 1205822= plusmn(9|120573984858
2minus 2119860120573984858 + 3119860
2120573 + 2120598|(984858 + 119860)
216984858)13
Thus 120593(120585) notin 1198671
119897119900119888(R)
(iv) If 120593(0) = 984858 = 1198611and 119860 = 0 then 120593 is a peakon-like
solution and
120593 (120585) minus 984858 = 1205823
10038161003816100381610038161205851003816100381610038161003816 + 119874 (
10038161003816100381610038161205851003816100381610038161003816
2
) 120585 997888rarr 0
1205931015840(120585) = 120582
3sgn (120585) + 119874 (
10038161003816100381610038161205851003816100381610038161003816) 120585 997888rarr 0
(29)
where 1205823= plusmn|119860 minus 984858|radic120573(119861
2minus 984858)4984858
(v) If 120593(0) = minus984858 = 1198612and 119860 = 0 then 120593 is a peakon-like
solution and
120593 (120585) + 984858 = 1205824
10038161003816100381610038161205851003816100381610038161003816 + 119874 (
10038161003816100381610038161205851003816100381610038161003816
2
) 120585 997888rarr 0
1205931015840(120585) = 120582
4sgn (120585) + 119874 (
10038161003816100381610038161205851003816100381610038161003816) 120585 997888rarr 0
(30)
where 1205824= plusmn|119860 + 984858|radicminus120573(119861
1+ 984858)4984858
(vi) If 120593(0) = 984858 = 1198611and 119860 = 0 then 120593 gives the peakon
solution 984858 exp(minusradicminus1205732|119909 minus 119888119905|)(vii) If 120593(0) = minus984858 = 119861
2and 119860 = 0 then 120593 gives the peakon
solution minus984858 exp(minusradicminus1205732|119909 minus 119888119905|)
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 |120593(0)| = 984858 then 120593 isin 119862
infin(R) and 120593 is a smooth soliton
solution(ii) If 120593(0) = 984858 = 119861
1 then by the definition of single peak
soliton we have 119860 = 984858 thus 1205731205932 + 2119860120573120593 + 31198602120573 + 2120598 does
not contain the factor 120593 minus 984858 From (24) we obtain
1205931015840= sgn (119860 minus 984858)
1003816100381610038161003816120593 minus 1198601003816100381610038161003816
radic210038161003816100381610038161205932 minus 9848582
1003816100381610038161003816
sdot radic10038161003816100381610038161205731205932 + 2119860120573120593 + 31198602120573 + 2120598
1003816100381610038161003816 sgn (120585)
(31)
Let 1198971(120593) = radic2(120593 + 984858)|120593 minus 119860|radic|1205731205932 + 2119860120573120593 + 31198602120573 + 2120598|
then 1198971(984858) = 2radic984858|984858 minus 119860|radic|1205739848582 + 2119860120573984858 + 31198602120573 + 2120598| and
int 1198971(120593)radic
1003816100381610038161003816120593 minus 9848581003816100381610038161003816119889120593 = int sgn (119860 minus 984858) sgn (120585) 119889120585 (32)
Inserting 1198971(120593) = 119897
1(984858) + 119874(|120593 minus 984858|) into (32) and using the
initial condition 120593(0) = 984858 we obtain
21198971(984858)
3
1003816100381610038161003816120593 minus 9848581003816100381610038161003816
32
(1 + 119874 (1003816100381610038161003816120593 minus 984858
1003816100381610038161003816)) =10038161003816100381610038161205851003816100381610038161003816
(33)
thus
120593 minus 984858 = plusmn(3
21198971(984858)
)
23
10038161003816100381610038161205851003816100381610038161003816
23
(1 + 119874 (1003816100381610038161003816120593 minus 984858
1003816100381610038161003816))minus23
= plusmn(3
21198971(984858)
)
23
10038161003816100381610038161205851003816100381610038161003816
23
(1 + 119874 (1003816100381610038161003816120593 minus 984858
1003816100381610038161003816))
(34)
which implies 120593 minus 984858 = 119874(|120585|23
) Therefore we have
120593 (120585) = 984858 plusmn (3
21198971(984858)
)
23
10038161003816100381610038161205851003816100381610038161003816
23
+ 119874(10038161003816100381610038161205851003816100381610038161003816
43
)
= 984858 + 1205821
10038161003816100381610038161205851003816100381610038161003816
23
+ 119874(10038161003816100381610038161205851003816100381610038161003816
43
) 120585 997888rarr 0
1205821= plusmn(
3
21198971(984858)
)
23
= plusmn(
9100381610038161003816100381610038161205739848582+ 2119860120573984858 + 3119860
2120573 + 2120598
10038161003816100381610038161003816(984858 minus 119860)
2
16984858)
13
1205931015840(120585) =
2
31205821
10038161003816100381610038161205851003816100381610038161003816
minus13 sgn (120585) + 119874 (10038161003816100381610038161205851003816100381610038161003816
13
) 120585 997888rarr 0
(35)
So 120593(120585) notin 1198671
loc(R)(iii) Similar to the proof of (ii) we ignore it in this paper(iv) If 120593(0) = 984858 = 119861
1and 119860 = 0 then from (25) we obtain
1205931015840= radicminus
120573
2sgn (119860 minus 984858)
1003816100381610038161003816120593 minus 1198601003816100381610038161003816radic120593 minus 1198612
120593 + 984858sgn (120585) (36)
Let 1198972(120593) = (1|120593minus119860|)radic(120593 + 984858)(120593 minus 119861
2) then 119897
2(984858) = (1|984858minus
119860|)radic2984858(984858 minus 1198612) and
int 1198972(120593) 119889120593 = radicminus
120573
2int sgn (119860 minus 984858) sgn (120585) 119889120585 (37)
Inserting 1198972(120593) = 119897
2(984858) + 119874(|120593 minus 984858|) into (37) and using the
initial condition 120593(0) = 984858 we obtain
1198972(984858) (120593 minus 984858) (1 + 119874 (
1003816100381610038161003816120593 minus 9848581003816100381610038161003816))minus1
= sgn (119860 minus 984858)10038161003816100381610038161205851003816100381610038161003816
(38)
Sincesgn (120593 minus 984858) sgn (119860 minus 984858) ge 0
1
1 + 119874 (120593 minus 984858)= 1 + 119874 (120593 minus 984858)
(39)
12 Mathematical Problems in Engineering
we get
1003816100381610038161003816120593 minus 9848581003816100381610038161003816 =
radicminus120573
2
1
1198972(984858)
10038161003816100381610038161205851003816100381610038161003816 (1 + 119874 (120593 minus 984858)) (40)
which implies |120593 minus 984858| = 119874(|120585|) Therefore we have
120593 (120585) = 984858 + 1205823
10038161003816100381610038161205851003816100381610038161003816 + 119874 (
10038161003816100381610038161205851003816100381610038161003816
2
) 120585 997888rarr 0
1205931015840(120585) = 120582
3sgn (120585) + 119874 (
10038161003816100381610038161205851003816100381610038161003816) 120585 997888rarr 0
(41)
where 1205823= plusmn|119860 minus 984858|radic120573(119861
2minus 984858)4984858
(v) Similar to the proof of (iv) we ignore it in this paper
By virtue of Theorem 9 any single peak soliton for the119862(3 2 2) equation (10) must satisfy the following initial andboundary values problem
(1205931015840)2
=120573 (120593 minus 119860)
2
(120593 minus 1198611) (120593 minus 119861
2)
2 (9848582 minus 1205932)fl119871 (120593)
120593 (0) isin 984858 minus984858 1198611 1198612
lim|120585|rarrinfin
120593 (120585) = 119860
(42)
119871(120593) ge 0 and the boundary condition (24) imply thefollowing
(a) If 120573(9848582 minus 1205932) ge 0 then 120593 ge max119861
1 1198612 or 120593 le
min1198611 1198612
(b) If120573(9848582minus1205932) le 0 thenmin1198611 1198612 le 120593 le max119861
1 1198612
Below we will present some implicit formulas for thesingle peak soliton solutions in the case of specific 120590 and 120573
Case 1 ((120590 120573) isin 11986012) In this case we have minus984858 lt 119861
2lt 119860 lt
1198611= 984858 From the standard phase analysis and Theorem 9
we know that if 120593 is a single peak soliton of the 119862(3 2 2)
equation then
1205931015840= minusradicminus
120573
2(120593 minus 119860)radic
120593 minus 1198612
120593 + 984858sgn (120585) (43)
From the separation of variables we get
intℎ (120593) 119889120593 = radicminus120573
2sgn (120585) 119889120585 (44)
where ℎ(120593) = (1(119860 minus 120593))radic(120593 + 984858)(120593 minus 1198612) After a lengthy
calculation of integral we obtain the implicit solution 120593
defined by
119867(120593)flradic119860 + 984858
119860 minus 1198612
1198681(120593) minus 119868
2(120593) = radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870 (45)
where
1198681(120593) = ln
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
(119860 + 984858) (120593 minus 1198612) + (119860 minus 119861
2) (120593 + 984858) + 2radic(119860 + 984858) (119860 minus 119861
2) (120593 minus 119861
2) (120593 + 984858)
radic(119860 + 984858) (119860 minus 1198612) (119860 + 984858) (119860 minus 120593)
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
1198682(120593) = ln
1003816100381610038161003816100381610038161003816(120593 + 984858) + (120593 minus 119861
2) + 2radic(120593 + 984858) (120593 minus 119861
2)
1003816100381610038161003816100381610038161003816
(46)
and 119870 is an arbitrary integration constant For 120593(0) = 984858 theconstant119870 = 119867(120593(0)) is defined by
119870 = radic119860 + 984858
119860 minus 1198612
1198681(984858) minus 119868
2(984858) (47)
and for 120593(0) = 1198612
119870 = radic119860 + 984858
119860 minus 1198612
1198681(1198612) minus 1198682(1198612) (48)
(i) If 120593(0) = 984858 then 119860 lt 120593 le 984858 Since 1198671015840(120593) = ℎ(120593) weknow that119867(120593) strictly decreases on the interval (119860 984858] thus1198671(120593) = 119867|
(119860984858](120593) gives a single peak solitonwith119867
1(984858) = 119870
and1198671(119860+) = infin Therefore 120593
1(120585) = 119867
minus1
1(radicminus1205732|120585| + 119870) is
the solution satisfying
1205931(0) = 984858
lim|120585|rarrinfin
1205931(120585) = 119860
1205931015840
1(0plusmn) = plusmn (119860 minus 984858)radic
120573 (1198612minus 984858)
4984858
(49)
So 1205931(120585) is a peakon-like solution (see Figure 5)
(ii) If 120593(0) = 1198612 then 119861
2le 120593 lt 119860 By 119867
1015840(120593) = 119891(120593)
we know that 119867(120593) strictly increases on the interval [1198612 119860)
Thus
1198672(120593) = 119867|
[1198612 119860)(120593) (50)
Mathematical Problems in Engineering 13
minus3 minus2 minus1
120593
minus02
1 2 3
02
04
06
08
10
x
(a)
02
02
minus2 minus2 x
10
05
00
t
120593
22
(b)
Figure 5 Two- and three-dimensional graphs of the peakon-like solution
minus03
minus04
minus05
1 2 3minus3 minus2 minus1
120593
minus02
x
(a)
0
2
0
2
minus2minus2
minus02
minus03
minus04
minus05
minus06
t x
120593
0
2
0
2
minus2minus2
t x
(b)
Figure 6 Two- and three-dimensional graphs of the smooth soliton solution
has the inverse denoted by 1205932(120585) = 119867
minus1
2(radicminus1205732|120585| + 119870)
1205932(120585) gives a kind of smooth soliton solution (see Figure 6)
satisfying
1205932(0) = 119861
2
lim|120585|rarrinfin
1205932(120585) = 119860
1205931015840
2(0) = 0
(51)
Case 2 ((120590 120573) isin 11986014) In this case we have minus984858 = 119861
2lt 119860 lt
984858 lt 1198611and (25) is equivalent to
1205931015840= minusradicminus
120573
2(120593 minus 119860)radic
120593 minus 1198611
120593 minus 984858sgn (120585) (52)
Let
119892 (120593) =1
119860 minus 120593radic120593 minus 1198611
120593 minus 984858 (53)
14 Mathematical Problems in Engineering
minus1 1 2 3
02
04
06
08
10
minus3 minus2
120593
x
(a)
02
02minus2
minus2
120593
x t
00
05
10
(b)
Figure 7 Two- and three-dimensional graphs of the cuspon solution
then (52) is converted to
119892 (120593) 119889120593 =1
119860 minus 120593radic120593 minus 1198611
120593 minus 984858119889120593 = radicminus
120573
2sgn (120585) 119889120585 (54)
Integrating (54) on the interval [0 120585] (or [120585 0]) leads to thefollowing implicit solutions
119866 (120593)flradic119860 minus 984858
119860 minus 1198611
1198683(120593) + 119868
4(120593) = radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870 (55)
where
1198683(120593) = ln
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
(119860 minus 984858) (120593 minus 1198611) + (119860 minus 119861
1) (120593 minus 984858) + 2radic(119860 minus 984858) (119860 minus 119861
1) (120593 minus 984858) (120593 minus 119861
1)
radic(119860 minus 984858) (119860 minus 1198611) (119860 minus 984858) (120593 minus 119860)
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
1198684(120593) = ln
1003816100381610038161003816100381610038161003816(120593 minus 984858) + (120593 minus 119861
1) + 2radic(120593 minus 984858) (120593 minus 119861
1)
1003816100381610038161003816100381610038161003816
(56)
And119870 is an arbitrary integration constant It is obvious thatfor 120593(0) = 984858 the constant119870 = 119867(120593(0)) is defined by
119870 = radic119860 minus 984858
119860 minus 1198611
1198683(984858) + 119868
4(984858) (57)
and for 120593(0) = minus984858
119870 = radic119860 minus 984858
119860 minus 1198611
1198683(minus984858) + 119868
4(minus984858) (58)
(i) If 120593(0) = 984858 then 119860 lt 120593 le 984858 From 119892(120593) lt 0 weknow that 119866(120593) strictly decreases on the interval (119860 984858] with119866(984858) = 119870 and 119866(119860+) = infin Define
1198661(120593) = 119866|
(119860984858](120593) (59)
Since 1198661(120593) is a strictly decreasing function from (119860 984858] onto
[119870infin) we can solve for 120593 uniquely from (59) and obtain
1205931(120585) = 119866
minus1
1(radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870) (60)
It is easy to check that 120593 satisfies
1205931(0) = 984858
lim|120585|rarrinfin
1205931(120585) = 119860
1205931015840
1(0plusmn) = ∓infin
(61)
Therefore the solution1205931defined by (60) is a cuspon solution
for the 119862(3 2 2) equation (see Figure 7)
Mathematical Problems in Engineering 15
(ii) If 120593(0) = minus984858 then minus984858 le 120593 lt 119860 Through a similaranalysis we get a strictly increasing function 119866(120593) on theinterval [minus984858 119860) satisfying
119866 (120593) = radicminus120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870 (62)
where 119866(120593) is defined by (55) Let
1198662(120593) = 119866|
[minus984858119860)(120593) (63)
then 1198662(120593) is a strictly increasing function from [minus984858 119860) onto
[119870infin) so that we can solve for 120593 and obtain
1205932(120585) = 119866
minus1
2(radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870) (64)
It is easy to check that 1205932satisfies
1205932(0) = minus984858
lim|120585|rarrinfin
1205932(120585) = 119860
1205931015840
2(0plusmn) = plusmn (119860 + 984858)radicminus
120573 (1198611+ 984858)
4984858
(65)
Therefore the solution 1205932defined by (64) is a peakon-like
solution whose graph is similar to those in Figure 5
Case 3 ((120590 120573) isin 11986016) In this case we have minus984858 lt 119861
2lt 119860 =
0 lt 1198611lt 984858 119861
2= minus1198611 and
1205931015840= minusradicminus
120573
2120593radic
120593 minus 1198612
1
1205932 minus 9848582sgn (120585) (66)
Hence from the separation of variables we have
radic9848582 minus 1205932
120593radic1198612
1minus 1205932
119889120593 = minusradicminus120573
2sgn (120585) 119889120585 (67)
Integrating (67) on the interval [0 120585] (or [120585 0]) leads tothe following implicit formula for the two smooth solitonsolutions
radic9848582 minus 1205932 + radic1198612
1minus 1205932
radic9848582 minus 1198612
1
(
radic(9848582 minus 1198612
1) 1205932
984858radic1198612
1minus 1205932 + 119861
1radic9848582 minus 1205932
)
984858
= exp(minusradicminus120573
2|119909 minus 119888119905|)
(68)
where 120593 isin (119860 1198611] Consider
radic9848582 minus 1198612
2
radic9848582 minus 1205932 + radic1198612
2minus 1205932
(
984858radic1198612
2minus 1205932 minus 119861
2radic9848582 minus 1205932
radic(9848582 minus 1198612
2) 1205932
)
984858
= exp(radicminus120573
2|119909 minus 119888119905|)
(69)
where 120593 isin [1198612 119860)
Case 4 ((120590 120573) isin 11986017) In this case we have minus984858 = 119861
2lt 119860 =
0 lt 1198611= 984858 and
1205931015840= minusradicminus
120573
2120593 sgn (120585) (70)
Choosing 120593(0) = 984858 (or minus984858) as initial value we get
int
120593
984858
1
120593119889120593 = minusint
120585
0
radicminus120573
2sgn (120585) 119889120585 for 119860 lt 120593 le 984858
int
120593
minus984858
1
120593119889120593 = minusint
120585
0
radicminus120573
2sgn (120585) 119889120585 for minus 984858 le 120593 lt 119860
(71)
which immediately yield the peakon solutions
120593 (119909 minus 119888119905) = plusmn984858 exp(minusradicminus120573
2|119909 minus 119888119905|) (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 the119862(3 2 2) equation (10) In order to study kinkwave solutionswe assume that
lim120585rarrinfin
120593 (120585) = 1198601
lim120585rarrminusinfin
120593 (120585) = 1198602
(73)
16 Mathematical Problems in Engineering
2 4
02
04
06
08
10
minus2minus4
120593
x
(a)
0 2
02
00
05
10
minus2
120593
x
t
0
2
minus2
(b)
Figure 8 Two- and three-dimensional graphs of the peakon solution
where 1198601gt 1198602 Substituting the boundary condition (73)
into (14) generates
(1205931015840)2
=
120573 (120593 minus 1198601) (120593 minus 119860
2) [1205932+ (1198601+ 1198602) 120593 + 119860
2
1+ 11986011198602+ 1198602
2+ 2120598120573]
2 (9848582 minus 1205932)fl119865 (120593) (74)
The nonlinear differential equation (74) may sustaindifferent kinds of nonlinear excitations In what follows weconfine our attention to the cases 119860
2= minus119860
1and 120598 =
minus1198602
1120573 which describe kinks and kink compactons which
play an important role in the dynamics systems Under theseconsiderations (74) reduces to
(1205931015840)2
=120573 (120593 minus 119860
1)2
(120593 minus 1198602)2
2 (9848582 minus 1205932) (75)
If 1198601lt 984858 then from the phase analysis in Section 2 (see
Figure 4(10)) we know that (1198601 0) and (119860
2 0) are two saddle
points of (13) and the kink solutions can be obtained fromthe two heteroclinic orbits connecting (120601 119910) = (119860
1 0) and
(1198602 0) When 119860
1increases upon reaching 984858 that is 119860
1= 984858
(75) becomes
(1205931015840)2
= minus120573 (120593 minus 119860
1) (120593 minus 119860
2)
2 (76)
and the ellipse 21199102+120573(120593minus1198601)(120593minus119860
2) = 0 (see Figure 4(11))
which is tangent to the singular lines 120593 = 984858 and 120593 = minus984858 atpoints (119860
1 0) and (119860
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 120593 is a kink wave solutions of (75)
or (76) we have 1205931015840
rarr 0 as 120593 rarr 1198601and as 120593 rarr 119860
2
Moreover we have 119865(120593) ge 0 for 1198602le 120593 le 119860
1and 120593 is
strictly monotonic in any interval where 119865(120593) gt 0 Thus if1205931015840gt 0 at some points 120593will be strictly increasing until it gets
close to the next zero of 119865 Denoting this zero 1198601 we have
120593 uarr 1198601 What will happen to the solution when it approaches
1198601 Depending on whether the zero is double or simple 120593
has a different behavior We explore the two cases in turn
Theorem 11 (i) If 120593 has a simple zero at 120593 = 1198601 so that
119865(1198601) = 0 and 1198651015840(119860
1) lt 0 then the solution 120593 of (75) satisfies
120593 (120585) minus 1198601=1198651015840(1198601)
4(120585 minus 120578)
2
+ 119874 ((120585 minus 120578)4
)
119886119904 120585 uarr 120578
(77)
where 120593(120578) = 1198601
(ii) When 120593 approaches the double zero 1198601of 119865(120593) so that
1198651015840(1198601) = 0 and11986510158401015840(119860
1) gt 0 then the solution120593 of (75) satisfies
120593 (120585) minus 1198601sim 120572 exp(minus120585radic11986510158401015840 (119860
1)) 119886119904 120585 997888rarr infin (78)
for some constant 120572 Thus 120593 uarr 1198601exponentially as 120585 rarr infin
Mathematical Problems in Engineering 17
Proof (i) When 1198601= minus119860
2= 984858 and 120598 = minus119860
2
1120573 from (76)
119865(120593) has a simple zero at 120593 = 1198601 Then
(1205931015840)2
= (120593 minus 1198601) 1198651015840(1198601) + 119874 ((120593 minus 119860
1)2
)
as 120593 uarr 1198601
(79)
Using the fact that (1205931015840)2 ge 0 we know that 1198651015840(1198601) lt 0
Moreover
119889120585
119889120593=
1
radic(120593 minus 1198601) 1198651015840 (119860
1) + 119874 ((120593 minus 119860
1)2
)
(80)
Because
radic(120593 minus 1198601) 1198651015840 (119860
1) + 119874 ((120593 minus 119860
1)2
)
= radic1198601minus 120593(radicminus1198651015840 (119860
1) + 119874 (119860
1minus 120593))
1
radicminus1198651015840 (1198601) + 119874 (119860
1minus 120593)
=1
radicminus1198651015840 (1198601)
+ 119874 (1198601minus 120593)
(81)
we obtain119889120585
119889120593=
1
radicminus (1198601minus 120593) 1198651015840 (119860
1)
+ 119874 ((1198601minus 120593)12
) (82)
Integrating (82) yields
120578 minus 120585 =2
radicminus1198651015840 (1198601)
(1198601minus 120593)12
+ 119874((1198601minus 120593)32
) (83)
where 120578 satisfies 120593(120578) = 1198601 Thus
(120578 minus 120585)2
=4
minus1198651015840 (1198601)(1198601minus 120593) + 119874 ((119860
1minus 120593)2
) (84)
which implies 119874((1198601minus 120593)2) = 119874((120578 minus 120585)
4) Therefore we get
120593 (120585) = 1198601+1198651015840(1198601)
4(120585 minus 120578)
2
+ 119874 ((120585 minus 120578)4
)
as 120585 uarr 120578
(85)
where 120593(120578) = 1198601
(ii) When 1198601= minus119860
2lt 984858 and 120598 = minus119860
2
1120573 from (75) 119865(120593)
has a double zero at 120593 = 1198601 Then
(1205931015840)2
= (120593 minus 1198601)2
11986510158401015840(1198601) + 119874 ((120593 minus 119860
1)3
)
as 120593 uarr 1198601
(86)
Furthermore we get
119889120585
119889120593=
1
radic(120593 minus 1198601)2
11986510158401015840 (1198601) + 119874 ((120593 minus 119860
1)3
)
(87)
Observing that
radic(120593 minus 1198601)2
11986510158401015840 (1198601) + 119874 ((120593 minus 119860
1)3
)
= (1198601minus 120593) (radic11986510158401015840 (119860
1) + 119874 (119860
1minus 120593))
1
radic11986510158401015840 (1198601) + 119874 (119860
1minus 120593)
=1
radic11986510158401015840 (1198601)
+ 119874 (1198601minus 120593)
(88)
we obtain119889120585
119889120593=
1
(1198601minus 120593) 11986510158401015840 (119860
1)+ 119874 (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 119860
1= minus119860
2lt 984858 and 120598 = minus119860
2
1120573 Then (75)
becomes
radic9848582 minus 1205932
1205932 minus 1198602
1
119889120593 = ∓radic120573
2119889120585 (90)
Integrating both sides of (90) gives the following implicitexpressions of kink and antikink wave solutions
arctan120593
radic9848582 minus 1205932
+
radic9848582 minus 1198602
1
1198601
arctanhradic9848582 minus 119860
2
1
1198601
120593
radic9848582 minus 1205932
= plusmnradic120573
2120585
(91)
By letting 1198601rarr 984858 in (91) we get two kink compactons
which are given by
plusmn120593 =
984858 sinradic120573
2120585
10038161003816100381610038161205851003816100381610038161003816 le
120587
radic2120573
984858 120585 gt120587
radic2120573
minus984858 120585 lt minus120587
radic2120573
(92)
The graphs for the kink wave solutions (91) and kinkcompacton solutions (92) are shown in Figures 9 and 10respectively
Remark 12 The two kink compacton solutions (92) aredifferent from the well-known smooth kink wave solutionsIn comparison with kink wave solutions (91) the kink com-pacton solutions (92) have no exponential decay propertiesbut have compact supportThat is they minus a constant thedifferences identically vanish outside a finite core region
18 Mathematical Problems in Engineering
120593
120585minus1 1
05
minus05
(a)
minus1minus1
minus05
05
00120593
x t
0
1
01
11
(b)
Figure 9 Two- and three-dimensional graphs of the kink wave solution
1
1120593
120585
minus1
minus1
(a)
0
1
0
1
000510
120593
minus1 minus1
minus05
minus10
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 the119862(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 Newtonrsquos equation which we write in the form 119910
2=
119865(120593) where 119910 = 1205931015840
120585(120585) we solve Newtonrsquos equation 119910
2=
119865(120593) 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 the119862(3 2 2)equation
Actually for 119898 = 2119896 + 1 119896 isin N+ in 119862(119898 2 2) equation(9) the dynamical behavior of traveling wave solutions of(9) is similar to the case 119898 = 3 for 119898 = 2119896 119896 isin N+ in119862(119898 2 2) equation (9) the dynamical behavior of travelingwave solutions of (9) is similar to the case 119898 = 2 We areapplying the approach mentioned in this work to 119862(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
References
[1] C Rogers and W R Shadwick Backlund Transformation andTheir Applications Academic Press New York NY USA 1982
[2] W Malfliet and W Hereman ldquoThe tanh method I Exactsolutions of nonlinear evolution and wave equationsrdquo PhysicaScripta vol 54 no 6 pp 563ndash568 1996
[3] R Hirota Direct Method in Soliton Theory Springer BerlinGermany 1980
[4] J S Cohen Computer Algebra and Symbolic ComputationMathematical Methods A K Peters 2003
[5] S Lie ldquoUber die integration durch bestimmte integrale voneiner klasse linearer partieller differential eichungenrdquoArchiv derMathematik vol 6 pp 328ndash368 1881
[6] H-H Dai and X-H Zhao ldquoTraveling waves in a rod composedof amodifiedMooney-Rivlinmaterial I bifurcation of equilibraand non-singular caserdquo Proceedings of the Royal Society AMathematical Physical and Engineering Sciences vol 455 no1990 pp 3845ndash3874 1999
[7] H-H Dai and F-F Wang ldquoAsymptotic bifurcation solutionsfor compressions of a clamped nonlinearly elastic rectangletransition region and barrelling to a corner-like profilerdquo SIAMJournal on Applied Mathematics vol 70 no 7 pp 2673ndash26922010
[8] J Li and Z Liu ldquoSmooth and non-smooth traveling waves ina nonlinearly dispersive equationrdquo Applied Mathematical Mod-elling vol 25 no 1 pp 41ndash56 2000
[9] A Chen and J Li ldquoSingle peak solitary wave solutions forthe osmosis K(22) equation under inhomogeneous boundaryconditionrdquo Journal of Mathematical Analysis and Applicationsvol 369 no 2 pp 758ndash766 2010
[10] R Camassa and D D Holm ldquoAn integrable shallow waterequation with peaked solitonsrdquo Physical Review Letters vol 71no 11 pp 1661ndash1664 1993
[11] P Rosenau ldquoOn nonanalytic solitary waves formed by a nonlin-ear dispersionrdquo Physics Letters A vol 230 no 5-6 pp 305ndash3181997
[12] P Rosenau and J M Hyman ldquoCompactons solitons with finitewavelengthrdquo Physical Review Letters vol 70 no 5 pp 564ndash5671993
[13] L Zhang L-Q Chen and X Huo ldquoThe effects of horizontalsingular straight line in a generalized nonlinear Klein-Gordonmodel equationrdquo Nonlinear Dynamics vol 72 no 4 pp 789ndash801 2013
[14] J Li and H Dai On the Study of Singular Nonlinear TravelingWave Equations Dynamical System Approach Science PressBeijing China 2007
[15] A Chen S Wen S Tang W Huang and Z Qiao ldquoEffects ofquadratic singular curves in integrable equationsrdquo Studies inApplied Mathematics vol 134 no 1 pp 24ndash61 2015
[16] L Tian and J Yin ldquoNew compacton solutions and solitary wavesolutions of fully nonlinear generalized Camassa-Holm equa-tionsrdquo Chaos Solitons amp Fractals vol 20 no 2 pp 289ndash2992004
[17] L Zhang ldquoThe bifurcation and peakons for the specialC(3 2 2)equationrdquo Pramana vol 83 no 3 pp 331ndash340 2014
[18] J Lenells ldquoTraveling wave solutions of the CamassandashHolmequationrdquo Journal of Differential Equations vol 217 no 2 pp393ndash430 2005
[19] Z J Qiao and G P Zhang ldquoOn peaked and smooth solitons forthe Camassa-Holm equationrdquo Europhysics Letters vol 73 no 5pp 657ndash663 2006
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
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Differential EquationsInternational Journal of
Volume 2014
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Mathematical PhysicsAdvances in
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OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Decision SciencesAdvances in
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Mathematical Problems in Engineering 7
00 05 10 15
0
1
2
3
00 05 10 15
0
1
2
3
0 1 2
0
1
2
3
0 1 2
0
1
2
3
00 05 10 15
0
1
2
3
0 1 2
0
2
4
0 1 2
0
2
4
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus2
minus4
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2 minus1
minus2minus3 minus1
minus2
minus2
minus3
minus4
minus1
minus2 minus1
minus10minus15 minus05 minus10minus15 minus05
minus10minus15 minus05
y
yyy
y y y
120593
120593120593120593
120593 120593 120593
(25) (120590 120573) isin A25
(28) (120590 120573) isin A28 (29) (120590 120573) isin A29
(31) (120590 120573) isin A31
(30) (120590 120573) isin A30
(26) (120590 120573) isin A26 (27) (120590 120573) isin A27
Figure 2 The bifurcation of phase portraits of system (15) when 120598 gt 0 and 120590 ge 0
11986111 (120590 120573) = (0 minus1205989848582)
11986112 (120590 120573) | 0 lt 120590 lt 1205902s 120573 = 1205732(120590)
11986113 (120590 120573) | 0 lt 120590 lt 1205901s 1205733(120590) lt 120573 lt min1205731(120590)
1205732(120590)
11986114 (120590 120573) | 1205904s lt 120590 lt 1205901s 120573 = 1205731(120590)
11986115 (120590 120573) = (1205901s 1205731s)
11986116 (120590 120573) | 0 lt 120590 lt 1205901s 120573 = 1205733(120590)
11986117 (120590 120573) | 0 lt 120590 lt 120598984858 120573 lt 1205733(120590)
11986118 (120590 120573) | 120590 = 0 120573 = minus120598984858
2
11986119 (120590 120573) | 120590 = 0 120573 lt 0
11986120 (120590 120573) | 0 lt 120590 lt 120598984858 120573 lt min0 1205733(120590)
11986121 (120590 120573) | 120590 gt 120598984858 120573 = 1205733(120590)
11986122 (120590 120573) | 120590 gt 120598984858 1205733(120590) lt 120573 lt 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
1205732(120590)
(1205901s 1205731s)
(1205902s 1205732s)
1205733(120590)
120573
minus2minus4minus6
1205731(120590)
120590
Figure 3 The bifurcation curves in the (120590 120573)-parameter plane for120598 lt 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 Let119862119896(Ω) denote the set of all 119896 times continuouslydifferentiable functions on the open setΩ 119871119901loc(R) refer to theset of all functions whose restriction on any compact subsetis 119871119901 integrable1198671loc(R) stands for1198671loc(R) = 120593 isin 119871
2
loc(R) |
1205931015840isin 1198712
loc(R)To study single peak soliton solutions we impose the
boundary condition
lim120585rarrplusmninfin
120593 = 119860 (22)
where 119860 is a constant In fact the constant 119860 is equal to thehorizontal coordinate of saddle point 119864(120593
119890 0) Substituting
the boundary condition (22) into (14) generates the followingconstant
ℎ = 119860(120590 +119860120598
2+1198603120573
4) (23)
So the ODE (14) becomes
(1205931015840)2
=
(120593 minus 119860)2
(1205731205932+ 2119860120573120593 + 3119860
2120573 + 2120598)
2 (9848582 minus 1205932) (24)
If 120573(1198602120573 + 120598) le 0 then (24) reduces to
(1205931015840)2
=120573 (120593 minus 119860)
2
(120593 minus 1198611) (120593 minus 119861
2)
2 (9848582 minus 1205932) (25)
where
1198611= minus119860 minus
radicminus2120573 (1198602120573 + 120598)
120573
1198612= minus119860 +
radicminus2120573 (1198602120573 + 120598)
120573
(26)
From (26) we know that 1198611gt 1198612if 120573 lt 0 and 119861
1lt 1198612if
120573 gt 0
Definition 4 A function120593(120585) is said to be a single peak solitonsolution for the 119862(3 2 2) equation (10) if 120593(120585) satisfies thefollowing conditions
(C1) 120593(120585) is continuous on R and has a unique peakpoint 120585
0 where 120593(120585) attains its global maximum or
minimum value
(C2) 120593(120585) isin 119862
3(R minus 120585
0) satisfies (24) on R minus 120585
0
(C3) 120593(120585) satisfies the boundary condition (22)
Definition 5 A wave function 120593 is called smooth solitonsolution if 120593 is smooth locally on either side of 120585
0and
lim120585uarr1205850
1205931015840(120585) = lim
120585darr12058501205931015840(120585) = 0
Definition 6 A wave function 120593 is called peakon if 120593 issmooth locally on either side of 120585
0and lim
120585uarr12058501205931015840(120585) =
minuslim120585darr1205850
1205931015840(120585) = 119886 119886 = 0 119886 = plusmninfin
Definition 7 Awave function120593 is called cuspon if120593 is smoothlocally on either side of 120585
0and lim
120585uarr12058501205931015840(120585) = minuslim
120585darr12058501205931015840(120585) =
plusmninfin
Without any loss of generality we choose the peak point1205850as vanishing 120585
0= 0
Theorem 8 Assume that 119906(119909 119905) = 120593(120585) = 120593(119909 minus 119888119905) is a singlepeak soliton solution of the 119862(3 2 2) equation (10) at the peakpoint 120585
0= 0 Then we have the following
(i) If 120573(1198602120573 + 120598) gt 0 then 120593(0) = 984858 or 120593(0) = minus984858
(ii) If 120573(1198602120573 + 120598) le 0 then 120593(0) = 984858 or 120593(0) = minus984858 or120593(0) = 119861
1or 120593(0) = 119861
2
Proof If 120593(0) = plusmn984858 then 120593(120585) = plusmn984858 for any 120585 isin R since120593(120585) isin 119862
3(R minus 0) Differentiating both sides of (24) yields
120593 isin 119862infin(R)
(i) When 120573(1198602120573 + 120598) gt 0 if 120593(0) = 984858 and 120593(0) = minus984858 then
120593 isin 119862infin(R) By the definition of single peak soliton we have
1205931015840(0) = 0 However by (24) we must have 120593(0) = 119860 which
contradicts the fact that 0 is the unique peak point(ii) When 120573(119860
2120573 + 120598) le 0 if 120593(0) = 984858 and 120593(0) = minus984858
by (24) we know 1205931015840(0) exists and 120593
1015840(0) = 0 since 0 is a peak
point Thus we obtain 120593(0) = 1198611or 120593(0) = 119861
2from (25)
since 120593(0) = 119860 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 119906(119909 119905) = 120593(119909 minus 119888119905) is a single peaksoliton solution of the 119862(3 2 2) equation (10) at the peak point1205850= 0 Then we have the following solution classification(i) If |120593(0)| = 984858 then 120593(120585) isin 119862
infin(R) and 120593 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
00 05 10 15
0
2
4
0 1
0
2
4
6
00 05 10 15
0
2
4
00 05 10
0
1
2
3
00 05 10 15
0
1
2
3
00 05 10
0
1
2
00 05 10 15
0
1
2
00 05 10 15
0
1
2
3
minus3 minus1minus4
minus4
minus2
minus4
minus2
minus4
minus2
minus4
minus2minus4
minus6
minus2
minus4
minus2
minus4
minus2
minus2 minus4 minus4minus2 minus2
minus3
minus1
minus2
minus1
minus2
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus1minus2 minus1minus2
120593
120593
120593
120593 120593 120593
120593 120593
120593 120593
120593 120593
y
y
y
yy
y
y y
y y
y y
minus15 minus10 minus05
minus15minus20 minus10 minus05
minus15 minus10 minus05 minus15 minus10 minus05minus10 minus05
minus15 minus10 minus05minus10 minus05
(10) (120590 120573) isin B10 (11) (120590 120573) isin B11 (12) (120590 120573) isin B12
(1) (120590 120573) isin B1
(4) (120590 120573) isin B4
(7) (120590 120573) isin B7 (8) (120590 120573) isin B8 (9) (120590 120573) isin B9
(5) (120590 120573) isin B5 (6) (120590 120573) isin B6
(2) (120590 120573) isin B2 (3) (120590 120573) isin B3
Figure 4 Continued
10 Mathematical Problems in Engineering
00 05 10 15
0
1
2
3
0 1
0
1
2
3
0 1 2
0
2
4
00 05 10 15
0
1
2
3
0 1 2
0
1
2
3
00 05 10 15
0
1
2
00 05 10 15
0
1
2
3
0 1 2
0
1
2
3
00 05 10 15
0
2
4
00 05 10 15
0
2
4
6
minus4
minus2
minus4
minus2
minus4
minus6
minus2
minus3 minus1minus2
minus3 minus1minus2
minus1minus2
minus1minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus1
minus2
minus3
minus1
minus2
120593 120593 120593
120593120593120593
120593 120593 120593
120593
y y y
yy
y
y yy
y
minus15 minus10 minus05
minus15minus20 minus10 minus05
minus15 minus10 minus05
minus15 minus10 minus05
minus15 minus10 minus05
minus15 minus10 minus05
(13) (120590 120573) isin B13 (14) (120590 120573) isin B14 (15) (120590 120573) isin B15
(16) (120590 120573) isin B16 (17) (120590 120573) isin B17 (18) (120590 120573) isin B18
(22) (120590 120573) isin B22
(20) (120590 120573) isin B20 (21) (120590 120573) isin B21(19) (120590 120573) isin B19
Figure 4 The bifurcation of phase portraits of system (15) when 120598 lt 0 and 120590 ge 0
Mathematical Problems in Engineering 11
(ii) If 120593(0) = 984858 = 1198611 then 120593 is a cuspon solution and 120593 has
the following asymptotic behavior
120593 (120585) minus 984858 = 1205821
10038161003816100381610038161205851003816100381610038161003816
23
+ 119874(10038161003816100381610038161205851003816100381610038161003816
43
) 120585 997888rarr 0
1205931015840(120585) =
2
31205821
10038161003816100381610038161205851003816100381610038161003816
minus13 sgn (120585) + 119874 (10038161003816100381610038161205851003816100381610038161003816
13
)
120585 997888rarr 0
(27)
where 1205821= plusmn(9|120573984858
2+ 2119860120573984858 + 3119860
2120573 + 2120598|(984858 minus 119860)
216984858)13
Thus 120593(120585) notin 1198671
119897119900119888(R)
(iii) If 120593(0) = minus984858 = 1198612 then 120593 is a cuspon solution and 120593
has the following asymptotic behavior
120593 (120585) + 984858 = 1205822
10038161003816100381610038161205851003816100381610038161003816
23
+ 119874(10038161003816100381610038161205851003816100381610038161003816
43
) 120585 997888rarr 0
1205931015840(120585) =
2
31205822
10038161003816100381610038161205851003816100381610038161003816
minus13 sgn (120585) + 119874 (10038161003816100381610038161205851003816100381610038161003816
13
)
120585 997888rarr 0
(28)
where 1205822= plusmn(9|120573984858
2minus 2119860120573984858 + 3119860
2120573 + 2120598|(984858 + 119860)
216984858)13
Thus 120593(120585) notin 1198671
119897119900119888(R)
(iv) If 120593(0) = 984858 = 1198611and 119860 = 0 then 120593 is a peakon-like
solution and
120593 (120585) minus 984858 = 1205823
10038161003816100381610038161205851003816100381610038161003816 + 119874 (
10038161003816100381610038161205851003816100381610038161003816
2
) 120585 997888rarr 0
1205931015840(120585) = 120582
3sgn (120585) + 119874 (
10038161003816100381610038161205851003816100381610038161003816) 120585 997888rarr 0
(29)
where 1205823= plusmn|119860 minus 984858|radic120573(119861
2minus 984858)4984858
(v) If 120593(0) = minus984858 = 1198612and 119860 = 0 then 120593 is a peakon-like
solution and
120593 (120585) + 984858 = 1205824
10038161003816100381610038161205851003816100381610038161003816 + 119874 (
10038161003816100381610038161205851003816100381610038161003816
2
) 120585 997888rarr 0
1205931015840(120585) = 120582
4sgn (120585) + 119874 (
10038161003816100381610038161205851003816100381610038161003816) 120585 997888rarr 0
(30)
where 1205824= plusmn|119860 + 984858|radicminus120573(119861
1+ 984858)4984858
(vi) If 120593(0) = 984858 = 1198611and 119860 = 0 then 120593 gives the peakon
solution 984858 exp(minusradicminus1205732|119909 minus 119888119905|)(vii) If 120593(0) = minus984858 = 119861
2and 119860 = 0 then 120593 gives the peakon
solution minus984858 exp(minusradicminus1205732|119909 minus 119888119905|)
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 |120593(0)| = 984858 then 120593 isin 119862
infin(R) and 120593 is a smooth soliton
solution(ii) If 120593(0) = 984858 = 119861
1 then by the definition of single peak
soliton we have 119860 = 984858 thus 1205731205932 + 2119860120573120593 + 31198602120573 + 2120598 does
not contain the factor 120593 minus 984858 From (24) we obtain
1205931015840= sgn (119860 minus 984858)
1003816100381610038161003816120593 minus 1198601003816100381610038161003816
radic210038161003816100381610038161205932 minus 9848582
1003816100381610038161003816
sdot radic10038161003816100381610038161205731205932 + 2119860120573120593 + 31198602120573 + 2120598
1003816100381610038161003816 sgn (120585)
(31)
Let 1198971(120593) = radic2(120593 + 984858)|120593 minus 119860|radic|1205731205932 + 2119860120573120593 + 31198602120573 + 2120598|
then 1198971(984858) = 2radic984858|984858 minus 119860|radic|1205739848582 + 2119860120573984858 + 31198602120573 + 2120598| and
int 1198971(120593)radic
1003816100381610038161003816120593 minus 9848581003816100381610038161003816119889120593 = int sgn (119860 minus 984858) sgn (120585) 119889120585 (32)
Inserting 1198971(120593) = 119897
1(984858) + 119874(|120593 minus 984858|) into (32) and using the
initial condition 120593(0) = 984858 we obtain
21198971(984858)
3
1003816100381610038161003816120593 minus 9848581003816100381610038161003816
32
(1 + 119874 (1003816100381610038161003816120593 minus 984858
1003816100381610038161003816)) =10038161003816100381610038161205851003816100381610038161003816
(33)
thus
120593 minus 984858 = plusmn(3
21198971(984858)
)
23
10038161003816100381610038161205851003816100381610038161003816
23
(1 + 119874 (1003816100381610038161003816120593 minus 984858
1003816100381610038161003816))minus23
= plusmn(3
21198971(984858)
)
23
10038161003816100381610038161205851003816100381610038161003816
23
(1 + 119874 (1003816100381610038161003816120593 minus 984858
1003816100381610038161003816))
(34)
which implies 120593 minus 984858 = 119874(|120585|23
) Therefore we have
120593 (120585) = 984858 plusmn (3
21198971(984858)
)
23
10038161003816100381610038161205851003816100381610038161003816
23
+ 119874(10038161003816100381610038161205851003816100381610038161003816
43
)
= 984858 + 1205821
10038161003816100381610038161205851003816100381610038161003816
23
+ 119874(10038161003816100381610038161205851003816100381610038161003816
43
) 120585 997888rarr 0
1205821= plusmn(
3
21198971(984858)
)
23
= plusmn(
9100381610038161003816100381610038161205739848582+ 2119860120573984858 + 3119860
2120573 + 2120598
10038161003816100381610038161003816(984858 minus 119860)
2
16984858)
13
1205931015840(120585) =
2
31205821
10038161003816100381610038161205851003816100381610038161003816
minus13 sgn (120585) + 119874 (10038161003816100381610038161205851003816100381610038161003816
13
) 120585 997888rarr 0
(35)
So 120593(120585) notin 1198671
loc(R)(iii) Similar to the proof of (ii) we ignore it in this paper(iv) If 120593(0) = 984858 = 119861
1and 119860 = 0 then from (25) we obtain
1205931015840= radicminus
120573
2sgn (119860 minus 984858)
1003816100381610038161003816120593 minus 1198601003816100381610038161003816radic120593 minus 1198612
120593 + 984858sgn (120585) (36)
Let 1198972(120593) = (1|120593minus119860|)radic(120593 + 984858)(120593 minus 119861
2) then 119897
2(984858) = (1|984858minus
119860|)radic2984858(984858 minus 1198612) and
int 1198972(120593) 119889120593 = radicminus
120573
2int sgn (119860 minus 984858) sgn (120585) 119889120585 (37)
Inserting 1198972(120593) = 119897
2(984858) + 119874(|120593 minus 984858|) into (37) and using the
initial condition 120593(0) = 984858 we obtain
1198972(984858) (120593 minus 984858) (1 + 119874 (
1003816100381610038161003816120593 minus 9848581003816100381610038161003816))minus1
= sgn (119860 minus 984858)10038161003816100381610038161205851003816100381610038161003816
(38)
Sincesgn (120593 minus 984858) sgn (119860 minus 984858) ge 0
1
1 + 119874 (120593 minus 984858)= 1 + 119874 (120593 minus 984858)
(39)
12 Mathematical Problems in Engineering
we get
1003816100381610038161003816120593 minus 9848581003816100381610038161003816 =
radicminus120573
2
1
1198972(984858)
10038161003816100381610038161205851003816100381610038161003816 (1 + 119874 (120593 minus 984858)) (40)
which implies |120593 minus 984858| = 119874(|120585|) Therefore we have
120593 (120585) = 984858 + 1205823
10038161003816100381610038161205851003816100381610038161003816 + 119874 (
10038161003816100381610038161205851003816100381610038161003816
2
) 120585 997888rarr 0
1205931015840(120585) = 120582
3sgn (120585) + 119874 (
10038161003816100381610038161205851003816100381610038161003816) 120585 997888rarr 0
(41)
where 1205823= plusmn|119860 minus 984858|radic120573(119861
2minus 984858)4984858
(v) Similar to the proof of (iv) we ignore it in this paper
By virtue of Theorem 9 any single peak soliton for the119862(3 2 2) equation (10) must satisfy the following initial andboundary values problem
(1205931015840)2
=120573 (120593 minus 119860)
2
(120593 minus 1198611) (120593 minus 119861
2)
2 (9848582 minus 1205932)fl119871 (120593)
120593 (0) isin 984858 minus984858 1198611 1198612
lim|120585|rarrinfin
120593 (120585) = 119860
(42)
119871(120593) ge 0 and the boundary condition (24) imply thefollowing
(a) If 120573(9848582 minus 1205932) ge 0 then 120593 ge max119861
1 1198612 or 120593 le
min1198611 1198612
(b) If120573(9848582minus1205932) le 0 thenmin1198611 1198612 le 120593 le max119861
1 1198612
Below we will present some implicit formulas for thesingle peak soliton solutions in the case of specific 120590 and 120573
Case 1 ((120590 120573) isin 11986012) In this case we have minus984858 lt 119861
2lt 119860 lt
1198611= 984858 From the standard phase analysis and Theorem 9
we know that if 120593 is a single peak soliton of the 119862(3 2 2)
equation then
1205931015840= minusradicminus
120573
2(120593 minus 119860)radic
120593 minus 1198612
120593 + 984858sgn (120585) (43)
From the separation of variables we get
intℎ (120593) 119889120593 = radicminus120573
2sgn (120585) 119889120585 (44)
where ℎ(120593) = (1(119860 minus 120593))radic(120593 + 984858)(120593 minus 1198612) After a lengthy
calculation of integral we obtain the implicit solution 120593
defined by
119867(120593)flradic119860 + 984858
119860 minus 1198612
1198681(120593) minus 119868
2(120593) = radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870 (45)
where
1198681(120593) = ln
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
(119860 + 984858) (120593 minus 1198612) + (119860 minus 119861
2) (120593 + 984858) + 2radic(119860 + 984858) (119860 minus 119861
2) (120593 minus 119861
2) (120593 + 984858)
radic(119860 + 984858) (119860 minus 1198612) (119860 + 984858) (119860 minus 120593)
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
1198682(120593) = ln
1003816100381610038161003816100381610038161003816(120593 + 984858) + (120593 minus 119861
2) + 2radic(120593 + 984858) (120593 minus 119861
2)
1003816100381610038161003816100381610038161003816
(46)
and 119870 is an arbitrary integration constant For 120593(0) = 984858 theconstant119870 = 119867(120593(0)) is defined by
119870 = radic119860 + 984858
119860 minus 1198612
1198681(984858) minus 119868
2(984858) (47)
and for 120593(0) = 1198612
119870 = radic119860 + 984858
119860 minus 1198612
1198681(1198612) minus 1198682(1198612) (48)
(i) If 120593(0) = 984858 then 119860 lt 120593 le 984858 Since 1198671015840(120593) = ℎ(120593) weknow that119867(120593) strictly decreases on the interval (119860 984858] thus1198671(120593) = 119867|
(119860984858](120593) gives a single peak solitonwith119867
1(984858) = 119870
and1198671(119860+) = infin Therefore 120593
1(120585) = 119867
minus1
1(radicminus1205732|120585| + 119870) is
the solution satisfying
1205931(0) = 984858
lim|120585|rarrinfin
1205931(120585) = 119860
1205931015840
1(0plusmn) = plusmn (119860 minus 984858)radic
120573 (1198612minus 984858)
4984858
(49)
So 1205931(120585) is a peakon-like solution (see Figure 5)
(ii) If 120593(0) = 1198612 then 119861
2le 120593 lt 119860 By 119867
1015840(120593) = 119891(120593)
we know that 119867(120593) strictly increases on the interval [1198612 119860)
Thus
1198672(120593) = 119867|
[1198612 119860)(120593) (50)
Mathematical Problems in Engineering 13
minus3 minus2 minus1
120593
minus02
1 2 3
02
04
06
08
10
x
(a)
02
02
minus2 minus2 x
10
05
00
t
120593
22
(b)
Figure 5 Two- and three-dimensional graphs of the peakon-like solution
minus03
minus04
minus05
1 2 3minus3 minus2 minus1
120593
minus02
x
(a)
0
2
0
2
minus2minus2
minus02
minus03
minus04
minus05
minus06
t x
120593
0
2
0
2
minus2minus2
t x
(b)
Figure 6 Two- and three-dimensional graphs of the smooth soliton solution
has the inverse denoted by 1205932(120585) = 119867
minus1
2(radicminus1205732|120585| + 119870)
1205932(120585) gives a kind of smooth soliton solution (see Figure 6)
satisfying
1205932(0) = 119861
2
lim|120585|rarrinfin
1205932(120585) = 119860
1205931015840
2(0) = 0
(51)
Case 2 ((120590 120573) isin 11986014) In this case we have minus984858 = 119861
2lt 119860 lt
984858 lt 1198611and (25) is equivalent to
1205931015840= minusradicminus
120573
2(120593 minus 119860)radic
120593 minus 1198611
120593 minus 984858sgn (120585) (52)
Let
119892 (120593) =1
119860 minus 120593radic120593 minus 1198611
120593 minus 984858 (53)
14 Mathematical Problems in Engineering
minus1 1 2 3
02
04
06
08
10
minus3 minus2
120593
x
(a)
02
02minus2
minus2
120593
x t
00
05
10
(b)
Figure 7 Two- and three-dimensional graphs of the cuspon solution
then (52) is converted to
119892 (120593) 119889120593 =1
119860 minus 120593radic120593 minus 1198611
120593 minus 984858119889120593 = radicminus
120573
2sgn (120585) 119889120585 (54)
Integrating (54) on the interval [0 120585] (or [120585 0]) leads to thefollowing implicit solutions
119866 (120593)flradic119860 minus 984858
119860 minus 1198611
1198683(120593) + 119868
4(120593) = radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870 (55)
where
1198683(120593) = ln
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
(119860 minus 984858) (120593 minus 1198611) + (119860 minus 119861
1) (120593 minus 984858) + 2radic(119860 minus 984858) (119860 minus 119861
1) (120593 minus 984858) (120593 minus 119861
1)
radic(119860 minus 984858) (119860 minus 1198611) (119860 minus 984858) (120593 minus 119860)
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
1198684(120593) = ln
1003816100381610038161003816100381610038161003816(120593 minus 984858) + (120593 minus 119861
1) + 2radic(120593 minus 984858) (120593 minus 119861
1)
1003816100381610038161003816100381610038161003816
(56)
And119870 is an arbitrary integration constant It is obvious thatfor 120593(0) = 984858 the constant119870 = 119867(120593(0)) is defined by
119870 = radic119860 minus 984858
119860 minus 1198611
1198683(984858) + 119868
4(984858) (57)
and for 120593(0) = minus984858
119870 = radic119860 minus 984858
119860 minus 1198611
1198683(minus984858) + 119868
4(minus984858) (58)
(i) If 120593(0) = 984858 then 119860 lt 120593 le 984858 From 119892(120593) lt 0 weknow that 119866(120593) strictly decreases on the interval (119860 984858] with119866(984858) = 119870 and 119866(119860+) = infin Define
1198661(120593) = 119866|
(119860984858](120593) (59)
Since 1198661(120593) is a strictly decreasing function from (119860 984858] onto
[119870infin) we can solve for 120593 uniquely from (59) and obtain
1205931(120585) = 119866
minus1
1(radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870) (60)
It is easy to check that 120593 satisfies
1205931(0) = 984858
lim|120585|rarrinfin
1205931(120585) = 119860
1205931015840
1(0plusmn) = ∓infin
(61)
Therefore the solution1205931defined by (60) is a cuspon solution
for the 119862(3 2 2) equation (see Figure 7)
Mathematical Problems in Engineering 15
(ii) If 120593(0) = minus984858 then minus984858 le 120593 lt 119860 Through a similaranalysis we get a strictly increasing function 119866(120593) on theinterval [minus984858 119860) satisfying
119866 (120593) = radicminus120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870 (62)
where 119866(120593) is defined by (55) Let
1198662(120593) = 119866|
[minus984858119860)(120593) (63)
then 1198662(120593) is a strictly increasing function from [minus984858 119860) onto
[119870infin) so that we can solve for 120593 and obtain
1205932(120585) = 119866
minus1
2(radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870) (64)
It is easy to check that 1205932satisfies
1205932(0) = minus984858
lim|120585|rarrinfin
1205932(120585) = 119860
1205931015840
2(0plusmn) = plusmn (119860 + 984858)radicminus
120573 (1198611+ 984858)
4984858
(65)
Therefore the solution 1205932defined by (64) is a peakon-like
solution whose graph is similar to those in Figure 5
Case 3 ((120590 120573) isin 11986016) In this case we have minus984858 lt 119861
2lt 119860 =
0 lt 1198611lt 984858 119861
2= minus1198611 and
1205931015840= minusradicminus
120573
2120593radic
120593 minus 1198612
1
1205932 minus 9848582sgn (120585) (66)
Hence from the separation of variables we have
radic9848582 minus 1205932
120593radic1198612
1minus 1205932
119889120593 = minusradicminus120573
2sgn (120585) 119889120585 (67)
Integrating (67) on the interval [0 120585] (or [120585 0]) leads tothe following implicit formula for the two smooth solitonsolutions
radic9848582 minus 1205932 + radic1198612
1minus 1205932
radic9848582 minus 1198612
1
(
radic(9848582 minus 1198612
1) 1205932
984858radic1198612
1minus 1205932 + 119861
1radic9848582 minus 1205932
)
984858
= exp(minusradicminus120573
2|119909 minus 119888119905|)
(68)
where 120593 isin (119860 1198611] Consider
radic9848582 minus 1198612
2
radic9848582 minus 1205932 + radic1198612
2minus 1205932
(
984858radic1198612
2minus 1205932 minus 119861
2radic9848582 minus 1205932
radic(9848582 minus 1198612
2) 1205932
)
984858
= exp(radicminus120573
2|119909 minus 119888119905|)
(69)
where 120593 isin [1198612 119860)
Case 4 ((120590 120573) isin 11986017) In this case we have minus984858 = 119861
2lt 119860 =
0 lt 1198611= 984858 and
1205931015840= minusradicminus
120573
2120593 sgn (120585) (70)
Choosing 120593(0) = 984858 (or minus984858) as initial value we get
int
120593
984858
1
120593119889120593 = minusint
120585
0
radicminus120573
2sgn (120585) 119889120585 for 119860 lt 120593 le 984858
int
120593
minus984858
1
120593119889120593 = minusint
120585
0
radicminus120573
2sgn (120585) 119889120585 for minus 984858 le 120593 lt 119860
(71)
which immediately yield the peakon solutions
120593 (119909 minus 119888119905) = plusmn984858 exp(minusradicminus120573
2|119909 minus 119888119905|) (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 the119862(3 2 2) equation (10) In order to study kinkwave solutionswe assume that
lim120585rarrinfin
120593 (120585) = 1198601
lim120585rarrminusinfin
120593 (120585) = 1198602
(73)
16 Mathematical Problems in Engineering
2 4
02
04
06
08
10
minus2minus4
120593
x
(a)
0 2
02
00
05
10
minus2
120593
x
t
0
2
minus2
(b)
Figure 8 Two- and three-dimensional graphs of the peakon solution
where 1198601gt 1198602 Substituting the boundary condition (73)
into (14) generates
(1205931015840)2
=
120573 (120593 minus 1198601) (120593 minus 119860
2) [1205932+ (1198601+ 1198602) 120593 + 119860
2
1+ 11986011198602+ 1198602
2+ 2120598120573]
2 (9848582 minus 1205932)fl119865 (120593) (74)
The nonlinear differential equation (74) may sustaindifferent kinds of nonlinear excitations In what follows weconfine our attention to the cases 119860
2= minus119860
1and 120598 =
minus1198602
1120573 which describe kinks and kink compactons which
play an important role in the dynamics systems Under theseconsiderations (74) reduces to
(1205931015840)2
=120573 (120593 minus 119860
1)2
(120593 minus 1198602)2
2 (9848582 minus 1205932) (75)
If 1198601lt 984858 then from the phase analysis in Section 2 (see
Figure 4(10)) we know that (1198601 0) and (119860
2 0) are two saddle
points of (13) and the kink solutions can be obtained fromthe two heteroclinic orbits connecting (120601 119910) = (119860
1 0) and
(1198602 0) When 119860
1increases upon reaching 984858 that is 119860
1= 984858
(75) becomes
(1205931015840)2
= minus120573 (120593 minus 119860
1) (120593 minus 119860
2)
2 (76)
and the ellipse 21199102+120573(120593minus1198601)(120593minus119860
2) = 0 (see Figure 4(11))
which is tangent to the singular lines 120593 = 984858 and 120593 = minus984858 atpoints (119860
1 0) and (119860
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 120593 is a kink wave solutions of (75)
or (76) we have 1205931015840
rarr 0 as 120593 rarr 1198601and as 120593 rarr 119860
2
Moreover we have 119865(120593) ge 0 for 1198602le 120593 le 119860
1and 120593 is
strictly monotonic in any interval where 119865(120593) gt 0 Thus if1205931015840gt 0 at some points 120593will be strictly increasing until it gets
close to the next zero of 119865 Denoting this zero 1198601 we have
120593 uarr 1198601 What will happen to the solution when it approaches
1198601 Depending on whether the zero is double or simple 120593
has a different behavior We explore the two cases in turn
Theorem 11 (i) If 120593 has a simple zero at 120593 = 1198601 so that
119865(1198601) = 0 and 1198651015840(119860
1) lt 0 then the solution 120593 of (75) satisfies
120593 (120585) minus 1198601=1198651015840(1198601)
4(120585 minus 120578)
2
+ 119874 ((120585 minus 120578)4
)
119886119904 120585 uarr 120578
(77)
where 120593(120578) = 1198601
(ii) When 120593 approaches the double zero 1198601of 119865(120593) so that
1198651015840(1198601) = 0 and11986510158401015840(119860
1) gt 0 then the solution120593 of (75) satisfies
120593 (120585) minus 1198601sim 120572 exp(minus120585radic11986510158401015840 (119860
1)) 119886119904 120585 997888rarr infin (78)
for some constant 120572 Thus 120593 uarr 1198601exponentially as 120585 rarr infin
Mathematical Problems in Engineering 17
Proof (i) When 1198601= minus119860
2= 984858 and 120598 = minus119860
2
1120573 from (76)
119865(120593) has a simple zero at 120593 = 1198601 Then
(1205931015840)2
= (120593 minus 1198601) 1198651015840(1198601) + 119874 ((120593 minus 119860
1)2
)
as 120593 uarr 1198601
(79)
Using the fact that (1205931015840)2 ge 0 we know that 1198651015840(1198601) lt 0
Moreover
119889120585
119889120593=
1
radic(120593 minus 1198601) 1198651015840 (119860
1) + 119874 ((120593 minus 119860
1)2
)
(80)
Because
radic(120593 minus 1198601) 1198651015840 (119860
1) + 119874 ((120593 minus 119860
1)2
)
= radic1198601minus 120593(radicminus1198651015840 (119860
1) + 119874 (119860
1minus 120593))
1
radicminus1198651015840 (1198601) + 119874 (119860
1minus 120593)
=1
radicminus1198651015840 (1198601)
+ 119874 (1198601minus 120593)
(81)
we obtain119889120585
119889120593=
1
radicminus (1198601minus 120593) 1198651015840 (119860
1)
+ 119874 ((1198601minus 120593)12
) (82)
Integrating (82) yields
120578 minus 120585 =2
radicminus1198651015840 (1198601)
(1198601minus 120593)12
+ 119874((1198601minus 120593)32
) (83)
where 120578 satisfies 120593(120578) = 1198601 Thus
(120578 minus 120585)2
=4
minus1198651015840 (1198601)(1198601minus 120593) + 119874 ((119860
1minus 120593)2
) (84)
which implies 119874((1198601minus 120593)2) = 119874((120578 minus 120585)
4) Therefore we get
120593 (120585) = 1198601+1198651015840(1198601)
4(120585 minus 120578)
2
+ 119874 ((120585 minus 120578)4
)
as 120585 uarr 120578
(85)
where 120593(120578) = 1198601
(ii) When 1198601= minus119860
2lt 984858 and 120598 = minus119860
2
1120573 from (75) 119865(120593)
has a double zero at 120593 = 1198601 Then
(1205931015840)2
= (120593 minus 1198601)2
11986510158401015840(1198601) + 119874 ((120593 minus 119860
1)3
)
as 120593 uarr 1198601
(86)
Furthermore we get
119889120585
119889120593=
1
radic(120593 minus 1198601)2
11986510158401015840 (1198601) + 119874 ((120593 minus 119860
1)3
)
(87)
Observing that
radic(120593 minus 1198601)2
11986510158401015840 (1198601) + 119874 ((120593 minus 119860
1)3
)
= (1198601minus 120593) (radic11986510158401015840 (119860
1) + 119874 (119860
1minus 120593))
1
radic11986510158401015840 (1198601) + 119874 (119860
1minus 120593)
=1
radic11986510158401015840 (1198601)
+ 119874 (1198601minus 120593)
(88)
we obtain119889120585
119889120593=
1
(1198601minus 120593) 11986510158401015840 (119860
1)+ 119874 (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 119860
1= minus119860
2lt 984858 and 120598 = minus119860
2
1120573 Then (75)
becomes
radic9848582 minus 1205932
1205932 minus 1198602
1
119889120593 = ∓radic120573
2119889120585 (90)
Integrating both sides of (90) gives the following implicitexpressions of kink and antikink wave solutions
arctan120593
radic9848582 minus 1205932
+
radic9848582 minus 1198602
1
1198601
arctanhradic9848582 minus 119860
2
1
1198601
120593
radic9848582 minus 1205932
= plusmnradic120573
2120585
(91)
By letting 1198601rarr 984858 in (91) we get two kink compactons
which are given by
plusmn120593 =
984858 sinradic120573
2120585
10038161003816100381610038161205851003816100381610038161003816 le
120587
radic2120573
984858 120585 gt120587
radic2120573
minus984858 120585 lt minus120587
radic2120573
(92)
The graphs for the kink wave solutions (91) and kinkcompacton solutions (92) are shown in Figures 9 and 10respectively
Remark 12 The two kink compacton solutions (92) aredifferent from the well-known smooth kink wave solutionsIn comparison with kink wave solutions (91) the kink com-pacton solutions (92) have no exponential decay propertiesbut have compact supportThat is they minus a constant thedifferences identically vanish outside a finite core region
18 Mathematical Problems in Engineering
120593
120585minus1 1
05
minus05
(a)
minus1minus1
minus05
05
00120593
x t
0
1
01
11
(b)
Figure 9 Two- and three-dimensional graphs of the kink wave solution
1
1120593
120585
minus1
minus1
(a)
0
1
0
1
000510
120593
minus1 minus1
minus05
minus10
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 the119862(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 Newtonrsquos equation which we write in the form 119910
2=
119865(120593) where 119910 = 1205931015840
120585(120585) we solve Newtonrsquos equation 119910
2=
119865(120593) 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 the119862(3 2 2)equation
Actually for 119898 = 2119896 + 1 119896 isin N+ in 119862(119898 2 2) equation(9) the dynamical behavior of traveling wave solutions of(9) is similar to the case 119898 = 3 for 119898 = 2119896 119896 isin N+ in119862(119898 2 2) equation (9) the dynamical behavior of travelingwave solutions of (9) is similar to the case 119898 = 2 We areapplying the approach mentioned in this work to 119862(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
References
[1] C Rogers and W R Shadwick Backlund Transformation andTheir Applications Academic Press New York NY USA 1982
[2] W Malfliet and W Hereman ldquoThe tanh method I Exactsolutions of nonlinear evolution and wave equationsrdquo PhysicaScripta vol 54 no 6 pp 563ndash568 1996
[3] R Hirota Direct Method in Soliton Theory Springer BerlinGermany 1980
[4] J S Cohen Computer Algebra and Symbolic ComputationMathematical Methods A K Peters 2003
[5] S Lie ldquoUber die integration durch bestimmte integrale voneiner klasse linearer partieller differential eichungenrdquoArchiv derMathematik vol 6 pp 328ndash368 1881
[6] H-H Dai and X-H Zhao ldquoTraveling waves in a rod composedof amodifiedMooney-Rivlinmaterial I bifurcation of equilibraand non-singular caserdquo Proceedings of the Royal Society AMathematical Physical and Engineering Sciences vol 455 no1990 pp 3845ndash3874 1999
[7] H-H Dai and F-F Wang ldquoAsymptotic bifurcation solutionsfor compressions of a clamped nonlinearly elastic rectangletransition region and barrelling to a corner-like profilerdquo SIAMJournal on Applied Mathematics vol 70 no 7 pp 2673ndash26922010
[8] J Li and Z Liu ldquoSmooth and non-smooth traveling waves ina nonlinearly dispersive equationrdquo Applied Mathematical Mod-elling vol 25 no 1 pp 41ndash56 2000
[9] A Chen and J Li ldquoSingle peak solitary wave solutions forthe osmosis K(22) equation under inhomogeneous boundaryconditionrdquo Journal of Mathematical Analysis and Applicationsvol 369 no 2 pp 758ndash766 2010
[10] R Camassa and D D Holm ldquoAn integrable shallow waterequation with peaked solitonsrdquo Physical Review Letters vol 71no 11 pp 1661ndash1664 1993
[11] P Rosenau ldquoOn nonanalytic solitary waves formed by a nonlin-ear dispersionrdquo Physics Letters A vol 230 no 5-6 pp 305ndash3181997
[12] P Rosenau and J M Hyman ldquoCompactons solitons with finitewavelengthrdquo Physical Review Letters vol 70 no 5 pp 564ndash5671993
[13] L Zhang L-Q Chen and X Huo ldquoThe effects of horizontalsingular straight line in a generalized nonlinear Klein-Gordonmodel equationrdquo Nonlinear Dynamics vol 72 no 4 pp 789ndash801 2013
[14] J Li and H Dai On the Study of Singular Nonlinear TravelingWave Equations Dynamical System Approach Science PressBeijing China 2007
[15] A Chen S Wen S Tang W Huang and Z Qiao ldquoEffects ofquadratic singular curves in integrable equationsrdquo Studies inApplied Mathematics vol 134 no 1 pp 24ndash61 2015
[16] L Tian and J Yin ldquoNew compacton solutions and solitary wavesolutions of fully nonlinear generalized Camassa-Holm equa-tionsrdquo Chaos Solitons amp Fractals vol 20 no 2 pp 289ndash2992004
[17] L Zhang ldquoThe bifurcation and peakons for the specialC(3 2 2)equationrdquo Pramana vol 83 no 3 pp 331ndash340 2014
[18] J Lenells ldquoTraveling wave solutions of the CamassandashHolmequationrdquo Journal of Differential Equations vol 217 no 2 pp393ndash430 2005
[19] Z J Qiao and G P Zhang ldquoOn peaked and smooth solitons forthe Camassa-Holm equationrdquo Europhysics Letters vol 73 no 5pp 657ndash663 2006
Submit your manuscripts athttpwwwhindawicom
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Mathematical Problems in Engineering
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Stochastic AnalysisInternational Journal of
8 Mathematical Problems in Engineering
2 4 6
5
10
15
1205732(120590)
(1205901s 1205731s)
(1205902s 1205732s)
1205733(120590)
120573
minus2minus4minus6
1205731(120590)
120590
Figure 3 The bifurcation curves in the (120590 120573)-parameter plane for120598 lt 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 Let119862119896(Ω) denote the set of all 119896 times continuouslydifferentiable functions on the open setΩ 119871119901loc(R) refer to theset of all functions whose restriction on any compact subsetis 119871119901 integrable1198671loc(R) stands for1198671loc(R) = 120593 isin 119871
2
loc(R) |
1205931015840isin 1198712
loc(R)To study single peak soliton solutions we impose the
boundary condition
lim120585rarrplusmninfin
120593 = 119860 (22)
where 119860 is a constant In fact the constant 119860 is equal to thehorizontal coordinate of saddle point 119864(120593
119890 0) Substituting
the boundary condition (22) into (14) generates the followingconstant
ℎ = 119860(120590 +119860120598
2+1198603120573
4) (23)
So the ODE (14) becomes
(1205931015840)2
=
(120593 minus 119860)2
(1205731205932+ 2119860120573120593 + 3119860
2120573 + 2120598)
2 (9848582 minus 1205932) (24)
If 120573(1198602120573 + 120598) le 0 then (24) reduces to
(1205931015840)2
=120573 (120593 minus 119860)
2
(120593 minus 1198611) (120593 minus 119861
2)
2 (9848582 minus 1205932) (25)
where
1198611= minus119860 minus
radicminus2120573 (1198602120573 + 120598)
120573
1198612= minus119860 +
radicminus2120573 (1198602120573 + 120598)
120573
(26)
From (26) we know that 1198611gt 1198612if 120573 lt 0 and 119861
1lt 1198612if
120573 gt 0
Definition 4 A function120593(120585) is said to be a single peak solitonsolution for the 119862(3 2 2) equation (10) if 120593(120585) satisfies thefollowing conditions
(C1) 120593(120585) is continuous on R and has a unique peakpoint 120585
0 where 120593(120585) attains its global maximum or
minimum value
(C2) 120593(120585) isin 119862
3(R minus 120585
0) satisfies (24) on R minus 120585
0
(C3) 120593(120585) satisfies the boundary condition (22)
Definition 5 A wave function 120593 is called smooth solitonsolution if 120593 is smooth locally on either side of 120585
0and
lim120585uarr1205850
1205931015840(120585) = lim
120585darr12058501205931015840(120585) = 0
Definition 6 A wave function 120593 is called peakon if 120593 issmooth locally on either side of 120585
0and lim
120585uarr12058501205931015840(120585) =
minuslim120585darr1205850
1205931015840(120585) = 119886 119886 = 0 119886 = plusmninfin
Definition 7 Awave function120593 is called cuspon if120593 is smoothlocally on either side of 120585
0and lim
120585uarr12058501205931015840(120585) = minuslim
120585darr12058501205931015840(120585) =
plusmninfin
Without any loss of generality we choose the peak point1205850as vanishing 120585
0= 0
Theorem 8 Assume that 119906(119909 119905) = 120593(120585) = 120593(119909 minus 119888119905) is a singlepeak soliton solution of the 119862(3 2 2) equation (10) at the peakpoint 120585
0= 0 Then we have the following
(i) If 120573(1198602120573 + 120598) gt 0 then 120593(0) = 984858 or 120593(0) = minus984858
(ii) If 120573(1198602120573 + 120598) le 0 then 120593(0) = 984858 or 120593(0) = minus984858 or120593(0) = 119861
1or 120593(0) = 119861
2
Proof If 120593(0) = plusmn984858 then 120593(120585) = plusmn984858 for any 120585 isin R since120593(120585) isin 119862
3(R minus 0) Differentiating both sides of (24) yields
120593 isin 119862infin(R)
(i) When 120573(1198602120573 + 120598) gt 0 if 120593(0) = 984858 and 120593(0) = minus984858 then
120593 isin 119862infin(R) By the definition of single peak soliton we have
1205931015840(0) = 0 However by (24) we must have 120593(0) = 119860 which
contradicts the fact that 0 is the unique peak point(ii) When 120573(119860
2120573 + 120598) le 0 if 120593(0) = 984858 and 120593(0) = minus984858
by (24) we know 1205931015840(0) exists and 120593
1015840(0) = 0 since 0 is a peak
point Thus we obtain 120593(0) = 1198611or 120593(0) = 119861
2from (25)
since 120593(0) = 119860 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 119906(119909 119905) = 120593(119909 minus 119888119905) is a single peaksoliton solution of the 119862(3 2 2) equation (10) at the peak point1205850= 0 Then we have the following solution classification(i) If |120593(0)| = 984858 then 120593(120585) isin 119862
infin(R) and 120593 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
00 05 10 15
0
2
4
0 1
0
2
4
6
00 05 10 15
0
2
4
00 05 10
0
1
2
3
00 05 10 15
0
1
2
3
00 05 10
0
1
2
00 05 10 15
0
1
2
00 05 10 15
0
1
2
3
minus3 minus1minus4
minus4
minus2
minus4
minus2
minus4
minus2
minus4
minus2minus4
minus6
minus2
minus4
minus2
minus4
minus2
minus2 minus4 minus4minus2 minus2
minus3
minus1
minus2
minus1
minus2
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus1minus2 minus1minus2
120593
120593
120593
120593 120593 120593
120593 120593
120593 120593
120593 120593
y
y
y
yy
y
y y
y y
y y
minus15 minus10 minus05
minus15minus20 minus10 minus05
minus15 minus10 minus05 minus15 minus10 minus05minus10 minus05
minus15 minus10 minus05minus10 minus05
(10) (120590 120573) isin B10 (11) (120590 120573) isin B11 (12) (120590 120573) isin B12
(1) (120590 120573) isin B1
(4) (120590 120573) isin B4
(7) (120590 120573) isin B7 (8) (120590 120573) isin B8 (9) (120590 120573) isin B9
(5) (120590 120573) isin B5 (6) (120590 120573) isin B6
(2) (120590 120573) isin B2 (3) (120590 120573) isin B3
Figure 4 Continued
10 Mathematical Problems in Engineering
00 05 10 15
0
1
2
3
0 1
0
1
2
3
0 1 2
0
2
4
00 05 10 15
0
1
2
3
0 1 2
0
1
2
3
00 05 10 15
0
1
2
00 05 10 15
0
1
2
3
0 1 2
0
1
2
3
00 05 10 15
0
2
4
00 05 10 15
0
2
4
6
minus4
minus2
minus4
minus2
minus4
minus6
minus2
minus3 minus1minus2
minus3 minus1minus2
minus1minus2
minus1minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus1
minus2
minus3
minus1
minus2
120593 120593 120593
120593120593120593
120593 120593 120593
120593
y y y
yy
y
y yy
y
minus15 minus10 minus05
minus15minus20 minus10 minus05
minus15 minus10 minus05
minus15 minus10 minus05
minus15 minus10 minus05
minus15 minus10 minus05
(13) (120590 120573) isin B13 (14) (120590 120573) isin B14 (15) (120590 120573) isin B15
(16) (120590 120573) isin B16 (17) (120590 120573) isin B17 (18) (120590 120573) isin B18
(22) (120590 120573) isin B22
(20) (120590 120573) isin B20 (21) (120590 120573) isin B21(19) (120590 120573) isin B19
Figure 4 The bifurcation of phase portraits of system (15) when 120598 lt 0 and 120590 ge 0
Mathematical Problems in Engineering 11
(ii) If 120593(0) = 984858 = 1198611 then 120593 is a cuspon solution and 120593 has
the following asymptotic behavior
120593 (120585) minus 984858 = 1205821
10038161003816100381610038161205851003816100381610038161003816
23
+ 119874(10038161003816100381610038161205851003816100381610038161003816
43
) 120585 997888rarr 0
1205931015840(120585) =
2
31205821
10038161003816100381610038161205851003816100381610038161003816
minus13 sgn (120585) + 119874 (10038161003816100381610038161205851003816100381610038161003816
13
)
120585 997888rarr 0
(27)
where 1205821= plusmn(9|120573984858
2+ 2119860120573984858 + 3119860
2120573 + 2120598|(984858 minus 119860)
216984858)13
Thus 120593(120585) notin 1198671
119897119900119888(R)
(iii) If 120593(0) = minus984858 = 1198612 then 120593 is a cuspon solution and 120593
has the following asymptotic behavior
120593 (120585) + 984858 = 1205822
10038161003816100381610038161205851003816100381610038161003816
23
+ 119874(10038161003816100381610038161205851003816100381610038161003816
43
) 120585 997888rarr 0
1205931015840(120585) =
2
31205822
10038161003816100381610038161205851003816100381610038161003816
minus13 sgn (120585) + 119874 (10038161003816100381610038161205851003816100381610038161003816
13
)
120585 997888rarr 0
(28)
where 1205822= plusmn(9|120573984858
2minus 2119860120573984858 + 3119860
2120573 + 2120598|(984858 + 119860)
216984858)13
Thus 120593(120585) notin 1198671
119897119900119888(R)
(iv) If 120593(0) = 984858 = 1198611and 119860 = 0 then 120593 is a peakon-like
solution and
120593 (120585) minus 984858 = 1205823
10038161003816100381610038161205851003816100381610038161003816 + 119874 (
10038161003816100381610038161205851003816100381610038161003816
2
) 120585 997888rarr 0
1205931015840(120585) = 120582
3sgn (120585) + 119874 (
10038161003816100381610038161205851003816100381610038161003816) 120585 997888rarr 0
(29)
where 1205823= plusmn|119860 minus 984858|radic120573(119861
2minus 984858)4984858
(v) If 120593(0) = minus984858 = 1198612and 119860 = 0 then 120593 is a peakon-like
solution and
120593 (120585) + 984858 = 1205824
10038161003816100381610038161205851003816100381610038161003816 + 119874 (
10038161003816100381610038161205851003816100381610038161003816
2
) 120585 997888rarr 0
1205931015840(120585) = 120582
4sgn (120585) + 119874 (
10038161003816100381610038161205851003816100381610038161003816) 120585 997888rarr 0
(30)
where 1205824= plusmn|119860 + 984858|radicminus120573(119861
1+ 984858)4984858
(vi) If 120593(0) = 984858 = 1198611and 119860 = 0 then 120593 gives the peakon
solution 984858 exp(minusradicminus1205732|119909 minus 119888119905|)(vii) If 120593(0) = minus984858 = 119861
2and 119860 = 0 then 120593 gives the peakon
solution minus984858 exp(minusradicminus1205732|119909 minus 119888119905|)
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 |120593(0)| = 984858 then 120593 isin 119862
infin(R) and 120593 is a smooth soliton
solution(ii) If 120593(0) = 984858 = 119861
1 then by the definition of single peak
soliton we have 119860 = 984858 thus 1205731205932 + 2119860120573120593 + 31198602120573 + 2120598 does
not contain the factor 120593 minus 984858 From (24) we obtain
1205931015840= sgn (119860 minus 984858)
1003816100381610038161003816120593 minus 1198601003816100381610038161003816
radic210038161003816100381610038161205932 minus 9848582
1003816100381610038161003816
sdot radic10038161003816100381610038161205731205932 + 2119860120573120593 + 31198602120573 + 2120598
1003816100381610038161003816 sgn (120585)
(31)
Let 1198971(120593) = radic2(120593 + 984858)|120593 minus 119860|radic|1205731205932 + 2119860120573120593 + 31198602120573 + 2120598|
then 1198971(984858) = 2radic984858|984858 minus 119860|radic|1205739848582 + 2119860120573984858 + 31198602120573 + 2120598| and
int 1198971(120593)radic
1003816100381610038161003816120593 minus 9848581003816100381610038161003816119889120593 = int sgn (119860 minus 984858) sgn (120585) 119889120585 (32)
Inserting 1198971(120593) = 119897
1(984858) + 119874(|120593 minus 984858|) into (32) and using the
initial condition 120593(0) = 984858 we obtain
21198971(984858)
3
1003816100381610038161003816120593 minus 9848581003816100381610038161003816
32
(1 + 119874 (1003816100381610038161003816120593 minus 984858
1003816100381610038161003816)) =10038161003816100381610038161205851003816100381610038161003816
(33)
thus
120593 minus 984858 = plusmn(3
21198971(984858)
)
23
10038161003816100381610038161205851003816100381610038161003816
23
(1 + 119874 (1003816100381610038161003816120593 minus 984858
1003816100381610038161003816))minus23
= plusmn(3
21198971(984858)
)
23
10038161003816100381610038161205851003816100381610038161003816
23
(1 + 119874 (1003816100381610038161003816120593 minus 984858
1003816100381610038161003816))
(34)
which implies 120593 minus 984858 = 119874(|120585|23
) Therefore we have
120593 (120585) = 984858 plusmn (3
21198971(984858)
)
23
10038161003816100381610038161205851003816100381610038161003816
23
+ 119874(10038161003816100381610038161205851003816100381610038161003816
43
)
= 984858 + 1205821
10038161003816100381610038161205851003816100381610038161003816
23
+ 119874(10038161003816100381610038161205851003816100381610038161003816
43
) 120585 997888rarr 0
1205821= plusmn(
3
21198971(984858)
)
23
= plusmn(
9100381610038161003816100381610038161205739848582+ 2119860120573984858 + 3119860
2120573 + 2120598
10038161003816100381610038161003816(984858 minus 119860)
2
16984858)
13
1205931015840(120585) =
2
31205821
10038161003816100381610038161205851003816100381610038161003816
minus13 sgn (120585) + 119874 (10038161003816100381610038161205851003816100381610038161003816
13
) 120585 997888rarr 0
(35)
So 120593(120585) notin 1198671
loc(R)(iii) Similar to the proof of (ii) we ignore it in this paper(iv) If 120593(0) = 984858 = 119861
1and 119860 = 0 then from (25) we obtain
1205931015840= radicminus
120573
2sgn (119860 minus 984858)
1003816100381610038161003816120593 minus 1198601003816100381610038161003816radic120593 minus 1198612
120593 + 984858sgn (120585) (36)
Let 1198972(120593) = (1|120593minus119860|)radic(120593 + 984858)(120593 minus 119861
2) then 119897
2(984858) = (1|984858minus
119860|)radic2984858(984858 minus 1198612) and
int 1198972(120593) 119889120593 = radicminus
120573
2int sgn (119860 minus 984858) sgn (120585) 119889120585 (37)
Inserting 1198972(120593) = 119897
2(984858) + 119874(|120593 minus 984858|) into (37) and using the
initial condition 120593(0) = 984858 we obtain
1198972(984858) (120593 minus 984858) (1 + 119874 (
1003816100381610038161003816120593 minus 9848581003816100381610038161003816))minus1
= sgn (119860 minus 984858)10038161003816100381610038161205851003816100381610038161003816
(38)
Sincesgn (120593 minus 984858) sgn (119860 minus 984858) ge 0
1
1 + 119874 (120593 minus 984858)= 1 + 119874 (120593 minus 984858)
(39)
12 Mathematical Problems in Engineering
we get
1003816100381610038161003816120593 minus 9848581003816100381610038161003816 =
radicminus120573
2
1
1198972(984858)
10038161003816100381610038161205851003816100381610038161003816 (1 + 119874 (120593 minus 984858)) (40)
which implies |120593 minus 984858| = 119874(|120585|) Therefore we have
120593 (120585) = 984858 + 1205823
10038161003816100381610038161205851003816100381610038161003816 + 119874 (
10038161003816100381610038161205851003816100381610038161003816
2
) 120585 997888rarr 0
1205931015840(120585) = 120582
3sgn (120585) + 119874 (
10038161003816100381610038161205851003816100381610038161003816) 120585 997888rarr 0
(41)
where 1205823= plusmn|119860 minus 984858|radic120573(119861
2minus 984858)4984858
(v) Similar to the proof of (iv) we ignore it in this paper
By virtue of Theorem 9 any single peak soliton for the119862(3 2 2) equation (10) must satisfy the following initial andboundary values problem
(1205931015840)2
=120573 (120593 minus 119860)
2
(120593 minus 1198611) (120593 minus 119861
2)
2 (9848582 minus 1205932)fl119871 (120593)
120593 (0) isin 984858 minus984858 1198611 1198612
lim|120585|rarrinfin
120593 (120585) = 119860
(42)
119871(120593) ge 0 and the boundary condition (24) imply thefollowing
(a) If 120573(9848582 minus 1205932) ge 0 then 120593 ge max119861
1 1198612 or 120593 le
min1198611 1198612
(b) If120573(9848582minus1205932) le 0 thenmin1198611 1198612 le 120593 le max119861
1 1198612
Below we will present some implicit formulas for thesingle peak soliton solutions in the case of specific 120590 and 120573
Case 1 ((120590 120573) isin 11986012) In this case we have minus984858 lt 119861
2lt 119860 lt
1198611= 984858 From the standard phase analysis and Theorem 9
we know that if 120593 is a single peak soliton of the 119862(3 2 2)
equation then
1205931015840= minusradicminus
120573
2(120593 minus 119860)radic
120593 minus 1198612
120593 + 984858sgn (120585) (43)
From the separation of variables we get
intℎ (120593) 119889120593 = radicminus120573
2sgn (120585) 119889120585 (44)
where ℎ(120593) = (1(119860 minus 120593))radic(120593 + 984858)(120593 minus 1198612) After a lengthy
calculation of integral we obtain the implicit solution 120593
defined by
119867(120593)flradic119860 + 984858
119860 minus 1198612
1198681(120593) minus 119868
2(120593) = radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870 (45)
where
1198681(120593) = ln
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
(119860 + 984858) (120593 minus 1198612) + (119860 minus 119861
2) (120593 + 984858) + 2radic(119860 + 984858) (119860 minus 119861
2) (120593 minus 119861
2) (120593 + 984858)
radic(119860 + 984858) (119860 minus 1198612) (119860 + 984858) (119860 minus 120593)
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
1198682(120593) = ln
1003816100381610038161003816100381610038161003816(120593 + 984858) + (120593 minus 119861
2) + 2radic(120593 + 984858) (120593 minus 119861
2)
1003816100381610038161003816100381610038161003816
(46)
and 119870 is an arbitrary integration constant For 120593(0) = 984858 theconstant119870 = 119867(120593(0)) is defined by
119870 = radic119860 + 984858
119860 minus 1198612
1198681(984858) minus 119868
2(984858) (47)
and for 120593(0) = 1198612
119870 = radic119860 + 984858
119860 minus 1198612
1198681(1198612) minus 1198682(1198612) (48)
(i) If 120593(0) = 984858 then 119860 lt 120593 le 984858 Since 1198671015840(120593) = ℎ(120593) weknow that119867(120593) strictly decreases on the interval (119860 984858] thus1198671(120593) = 119867|
(119860984858](120593) gives a single peak solitonwith119867
1(984858) = 119870
and1198671(119860+) = infin Therefore 120593
1(120585) = 119867
minus1
1(radicminus1205732|120585| + 119870) is
the solution satisfying
1205931(0) = 984858
lim|120585|rarrinfin
1205931(120585) = 119860
1205931015840
1(0plusmn) = plusmn (119860 minus 984858)radic
120573 (1198612minus 984858)
4984858
(49)
So 1205931(120585) is a peakon-like solution (see Figure 5)
(ii) If 120593(0) = 1198612 then 119861
2le 120593 lt 119860 By 119867
1015840(120593) = 119891(120593)
we know that 119867(120593) strictly increases on the interval [1198612 119860)
Thus
1198672(120593) = 119867|
[1198612 119860)(120593) (50)
Mathematical Problems in Engineering 13
minus3 minus2 minus1
120593
minus02
1 2 3
02
04
06
08
10
x
(a)
02
02
minus2 minus2 x
10
05
00
t
120593
22
(b)
Figure 5 Two- and three-dimensional graphs of the peakon-like solution
minus03
minus04
minus05
1 2 3minus3 minus2 minus1
120593
minus02
x
(a)
0
2
0
2
minus2minus2
minus02
minus03
minus04
minus05
minus06
t x
120593
0
2
0
2
minus2minus2
t x
(b)
Figure 6 Two- and three-dimensional graphs of the smooth soliton solution
has the inverse denoted by 1205932(120585) = 119867
minus1
2(radicminus1205732|120585| + 119870)
1205932(120585) gives a kind of smooth soliton solution (see Figure 6)
satisfying
1205932(0) = 119861
2
lim|120585|rarrinfin
1205932(120585) = 119860
1205931015840
2(0) = 0
(51)
Case 2 ((120590 120573) isin 11986014) In this case we have minus984858 = 119861
2lt 119860 lt
984858 lt 1198611and (25) is equivalent to
1205931015840= minusradicminus
120573
2(120593 minus 119860)radic
120593 minus 1198611
120593 minus 984858sgn (120585) (52)
Let
119892 (120593) =1
119860 minus 120593radic120593 minus 1198611
120593 minus 984858 (53)
14 Mathematical Problems in Engineering
minus1 1 2 3
02
04
06
08
10
minus3 minus2
120593
x
(a)
02
02minus2
minus2
120593
x t
00
05
10
(b)
Figure 7 Two- and three-dimensional graphs of the cuspon solution
then (52) is converted to
119892 (120593) 119889120593 =1
119860 minus 120593radic120593 minus 1198611
120593 minus 984858119889120593 = radicminus
120573
2sgn (120585) 119889120585 (54)
Integrating (54) on the interval [0 120585] (or [120585 0]) leads to thefollowing implicit solutions
119866 (120593)flradic119860 minus 984858
119860 minus 1198611
1198683(120593) + 119868
4(120593) = radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870 (55)
where
1198683(120593) = ln
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
(119860 minus 984858) (120593 minus 1198611) + (119860 minus 119861
1) (120593 minus 984858) + 2radic(119860 minus 984858) (119860 minus 119861
1) (120593 minus 984858) (120593 minus 119861
1)
radic(119860 minus 984858) (119860 minus 1198611) (119860 minus 984858) (120593 minus 119860)
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
1198684(120593) = ln
1003816100381610038161003816100381610038161003816(120593 minus 984858) + (120593 minus 119861
1) + 2radic(120593 minus 984858) (120593 minus 119861
1)
1003816100381610038161003816100381610038161003816
(56)
And119870 is an arbitrary integration constant It is obvious thatfor 120593(0) = 984858 the constant119870 = 119867(120593(0)) is defined by
119870 = radic119860 minus 984858
119860 minus 1198611
1198683(984858) + 119868
4(984858) (57)
and for 120593(0) = minus984858
119870 = radic119860 minus 984858
119860 minus 1198611
1198683(minus984858) + 119868
4(minus984858) (58)
(i) If 120593(0) = 984858 then 119860 lt 120593 le 984858 From 119892(120593) lt 0 weknow that 119866(120593) strictly decreases on the interval (119860 984858] with119866(984858) = 119870 and 119866(119860+) = infin Define
1198661(120593) = 119866|
(119860984858](120593) (59)
Since 1198661(120593) is a strictly decreasing function from (119860 984858] onto
[119870infin) we can solve for 120593 uniquely from (59) and obtain
1205931(120585) = 119866
minus1
1(radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870) (60)
It is easy to check that 120593 satisfies
1205931(0) = 984858
lim|120585|rarrinfin
1205931(120585) = 119860
1205931015840
1(0plusmn) = ∓infin
(61)
Therefore the solution1205931defined by (60) is a cuspon solution
for the 119862(3 2 2) equation (see Figure 7)
Mathematical Problems in Engineering 15
(ii) If 120593(0) = minus984858 then minus984858 le 120593 lt 119860 Through a similaranalysis we get a strictly increasing function 119866(120593) on theinterval [minus984858 119860) satisfying
119866 (120593) = radicminus120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870 (62)
where 119866(120593) is defined by (55) Let
1198662(120593) = 119866|
[minus984858119860)(120593) (63)
then 1198662(120593) is a strictly increasing function from [minus984858 119860) onto
[119870infin) so that we can solve for 120593 and obtain
1205932(120585) = 119866
minus1
2(radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870) (64)
It is easy to check that 1205932satisfies
1205932(0) = minus984858
lim|120585|rarrinfin
1205932(120585) = 119860
1205931015840
2(0plusmn) = plusmn (119860 + 984858)radicminus
120573 (1198611+ 984858)
4984858
(65)
Therefore the solution 1205932defined by (64) is a peakon-like
solution whose graph is similar to those in Figure 5
Case 3 ((120590 120573) isin 11986016) In this case we have minus984858 lt 119861
2lt 119860 =
0 lt 1198611lt 984858 119861
2= minus1198611 and
1205931015840= minusradicminus
120573
2120593radic
120593 minus 1198612
1
1205932 minus 9848582sgn (120585) (66)
Hence from the separation of variables we have
radic9848582 minus 1205932
120593radic1198612
1minus 1205932
119889120593 = minusradicminus120573
2sgn (120585) 119889120585 (67)
Integrating (67) on the interval [0 120585] (or [120585 0]) leads tothe following implicit formula for the two smooth solitonsolutions
radic9848582 minus 1205932 + radic1198612
1minus 1205932
radic9848582 minus 1198612
1
(
radic(9848582 minus 1198612
1) 1205932
984858radic1198612
1minus 1205932 + 119861
1radic9848582 minus 1205932
)
984858
= exp(minusradicminus120573
2|119909 minus 119888119905|)
(68)
where 120593 isin (119860 1198611] Consider
radic9848582 minus 1198612
2
radic9848582 minus 1205932 + radic1198612
2minus 1205932
(
984858radic1198612
2minus 1205932 minus 119861
2radic9848582 minus 1205932
radic(9848582 minus 1198612
2) 1205932
)
984858
= exp(radicminus120573
2|119909 minus 119888119905|)
(69)
where 120593 isin [1198612 119860)
Case 4 ((120590 120573) isin 11986017) In this case we have minus984858 = 119861
2lt 119860 =
0 lt 1198611= 984858 and
1205931015840= minusradicminus
120573
2120593 sgn (120585) (70)
Choosing 120593(0) = 984858 (or minus984858) as initial value we get
int
120593
984858
1
120593119889120593 = minusint
120585
0
radicminus120573
2sgn (120585) 119889120585 for 119860 lt 120593 le 984858
int
120593
minus984858
1
120593119889120593 = minusint
120585
0
radicminus120573
2sgn (120585) 119889120585 for minus 984858 le 120593 lt 119860
(71)
which immediately yield the peakon solutions
120593 (119909 minus 119888119905) = plusmn984858 exp(minusradicminus120573
2|119909 minus 119888119905|) (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 the119862(3 2 2) equation (10) In order to study kinkwave solutionswe assume that
lim120585rarrinfin
120593 (120585) = 1198601
lim120585rarrminusinfin
120593 (120585) = 1198602
(73)
16 Mathematical Problems in Engineering
2 4
02
04
06
08
10
minus2minus4
120593
x
(a)
0 2
02
00
05
10
minus2
120593
x
t
0
2
minus2
(b)
Figure 8 Two- and three-dimensional graphs of the peakon solution
where 1198601gt 1198602 Substituting the boundary condition (73)
into (14) generates
(1205931015840)2
=
120573 (120593 minus 1198601) (120593 minus 119860
2) [1205932+ (1198601+ 1198602) 120593 + 119860
2
1+ 11986011198602+ 1198602
2+ 2120598120573]
2 (9848582 minus 1205932)fl119865 (120593) (74)
The nonlinear differential equation (74) may sustaindifferent kinds of nonlinear excitations In what follows weconfine our attention to the cases 119860
2= minus119860
1and 120598 =
minus1198602
1120573 which describe kinks and kink compactons which
play an important role in the dynamics systems Under theseconsiderations (74) reduces to
(1205931015840)2
=120573 (120593 minus 119860
1)2
(120593 minus 1198602)2
2 (9848582 minus 1205932) (75)
If 1198601lt 984858 then from the phase analysis in Section 2 (see
Figure 4(10)) we know that (1198601 0) and (119860
2 0) are two saddle
points of (13) and the kink solutions can be obtained fromthe two heteroclinic orbits connecting (120601 119910) = (119860
1 0) and
(1198602 0) When 119860
1increases upon reaching 984858 that is 119860
1= 984858
(75) becomes
(1205931015840)2
= minus120573 (120593 minus 119860
1) (120593 minus 119860
2)
2 (76)
and the ellipse 21199102+120573(120593minus1198601)(120593minus119860
2) = 0 (see Figure 4(11))
which is tangent to the singular lines 120593 = 984858 and 120593 = minus984858 atpoints (119860
1 0) and (119860
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 120593 is a kink wave solutions of (75)
or (76) we have 1205931015840
rarr 0 as 120593 rarr 1198601and as 120593 rarr 119860
2
Moreover we have 119865(120593) ge 0 for 1198602le 120593 le 119860
1and 120593 is
strictly monotonic in any interval where 119865(120593) gt 0 Thus if1205931015840gt 0 at some points 120593will be strictly increasing until it gets
close to the next zero of 119865 Denoting this zero 1198601 we have
120593 uarr 1198601 What will happen to the solution when it approaches
1198601 Depending on whether the zero is double or simple 120593
has a different behavior We explore the two cases in turn
Theorem 11 (i) If 120593 has a simple zero at 120593 = 1198601 so that
119865(1198601) = 0 and 1198651015840(119860
1) lt 0 then the solution 120593 of (75) satisfies
120593 (120585) minus 1198601=1198651015840(1198601)
4(120585 minus 120578)
2
+ 119874 ((120585 minus 120578)4
)
119886119904 120585 uarr 120578
(77)
where 120593(120578) = 1198601
(ii) When 120593 approaches the double zero 1198601of 119865(120593) so that
1198651015840(1198601) = 0 and11986510158401015840(119860
1) gt 0 then the solution120593 of (75) satisfies
120593 (120585) minus 1198601sim 120572 exp(minus120585radic11986510158401015840 (119860
1)) 119886119904 120585 997888rarr infin (78)
for some constant 120572 Thus 120593 uarr 1198601exponentially as 120585 rarr infin
Mathematical Problems in Engineering 17
Proof (i) When 1198601= minus119860
2= 984858 and 120598 = minus119860
2
1120573 from (76)
119865(120593) has a simple zero at 120593 = 1198601 Then
(1205931015840)2
= (120593 minus 1198601) 1198651015840(1198601) + 119874 ((120593 minus 119860
1)2
)
as 120593 uarr 1198601
(79)
Using the fact that (1205931015840)2 ge 0 we know that 1198651015840(1198601) lt 0
Moreover
119889120585
119889120593=
1
radic(120593 minus 1198601) 1198651015840 (119860
1) + 119874 ((120593 minus 119860
1)2
)
(80)
Because
radic(120593 minus 1198601) 1198651015840 (119860
1) + 119874 ((120593 minus 119860
1)2
)
= radic1198601minus 120593(radicminus1198651015840 (119860
1) + 119874 (119860
1minus 120593))
1
radicminus1198651015840 (1198601) + 119874 (119860
1minus 120593)
=1
radicminus1198651015840 (1198601)
+ 119874 (1198601minus 120593)
(81)
we obtain119889120585
119889120593=
1
radicminus (1198601minus 120593) 1198651015840 (119860
1)
+ 119874 ((1198601minus 120593)12
) (82)
Integrating (82) yields
120578 minus 120585 =2
radicminus1198651015840 (1198601)
(1198601minus 120593)12
+ 119874((1198601minus 120593)32
) (83)
where 120578 satisfies 120593(120578) = 1198601 Thus
(120578 minus 120585)2
=4
minus1198651015840 (1198601)(1198601minus 120593) + 119874 ((119860
1minus 120593)2
) (84)
which implies 119874((1198601minus 120593)2) = 119874((120578 minus 120585)
4) Therefore we get
120593 (120585) = 1198601+1198651015840(1198601)
4(120585 minus 120578)
2
+ 119874 ((120585 minus 120578)4
)
as 120585 uarr 120578
(85)
where 120593(120578) = 1198601
(ii) When 1198601= minus119860
2lt 984858 and 120598 = minus119860
2
1120573 from (75) 119865(120593)
has a double zero at 120593 = 1198601 Then
(1205931015840)2
= (120593 minus 1198601)2
11986510158401015840(1198601) + 119874 ((120593 minus 119860
1)3
)
as 120593 uarr 1198601
(86)
Furthermore we get
119889120585
119889120593=
1
radic(120593 minus 1198601)2
11986510158401015840 (1198601) + 119874 ((120593 minus 119860
1)3
)
(87)
Observing that
radic(120593 minus 1198601)2
11986510158401015840 (1198601) + 119874 ((120593 minus 119860
1)3
)
= (1198601minus 120593) (radic11986510158401015840 (119860
1) + 119874 (119860
1minus 120593))
1
radic11986510158401015840 (1198601) + 119874 (119860
1minus 120593)
=1
radic11986510158401015840 (1198601)
+ 119874 (1198601minus 120593)
(88)
we obtain119889120585
119889120593=
1
(1198601minus 120593) 11986510158401015840 (119860
1)+ 119874 (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 119860
1= minus119860
2lt 984858 and 120598 = minus119860
2
1120573 Then (75)
becomes
radic9848582 minus 1205932
1205932 minus 1198602
1
119889120593 = ∓radic120573
2119889120585 (90)
Integrating both sides of (90) gives the following implicitexpressions of kink and antikink wave solutions
arctan120593
radic9848582 minus 1205932
+
radic9848582 minus 1198602
1
1198601
arctanhradic9848582 minus 119860
2
1
1198601
120593
radic9848582 minus 1205932
= plusmnradic120573
2120585
(91)
By letting 1198601rarr 984858 in (91) we get two kink compactons
which are given by
plusmn120593 =
984858 sinradic120573
2120585
10038161003816100381610038161205851003816100381610038161003816 le
120587
radic2120573
984858 120585 gt120587
radic2120573
minus984858 120585 lt minus120587
radic2120573
(92)
The graphs for the kink wave solutions (91) and kinkcompacton solutions (92) are shown in Figures 9 and 10respectively
Remark 12 The two kink compacton solutions (92) aredifferent from the well-known smooth kink wave solutionsIn comparison with kink wave solutions (91) the kink com-pacton solutions (92) have no exponential decay propertiesbut have compact supportThat is they minus a constant thedifferences identically vanish outside a finite core region
18 Mathematical Problems in Engineering
120593
120585minus1 1
05
minus05
(a)
minus1minus1
minus05
05
00120593
x t
0
1
01
11
(b)
Figure 9 Two- and three-dimensional graphs of the kink wave solution
1
1120593
120585
minus1
minus1
(a)
0
1
0
1
000510
120593
minus1 minus1
minus05
minus10
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 the119862(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 Newtonrsquos equation which we write in the form 119910
2=
119865(120593) where 119910 = 1205931015840
120585(120585) we solve Newtonrsquos equation 119910
2=
119865(120593) 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 the119862(3 2 2)equation
Actually for 119898 = 2119896 + 1 119896 isin N+ in 119862(119898 2 2) equation(9) the dynamical behavior of traveling wave solutions of(9) is similar to the case 119898 = 3 for 119898 = 2119896 119896 isin N+ in119862(119898 2 2) equation (9) the dynamical behavior of travelingwave solutions of (9) is similar to the case 119898 = 2 We areapplying the approach mentioned in this work to 119862(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
References
[1] C Rogers and W R Shadwick Backlund Transformation andTheir Applications Academic Press New York NY USA 1982
[2] W Malfliet and W Hereman ldquoThe tanh method I Exactsolutions of nonlinear evolution and wave equationsrdquo PhysicaScripta vol 54 no 6 pp 563ndash568 1996
[3] R Hirota Direct Method in Soliton Theory Springer BerlinGermany 1980
[4] J S Cohen Computer Algebra and Symbolic ComputationMathematical Methods A K Peters 2003
[5] S Lie ldquoUber die integration durch bestimmte integrale voneiner klasse linearer partieller differential eichungenrdquoArchiv derMathematik vol 6 pp 328ndash368 1881
[6] H-H Dai and X-H Zhao ldquoTraveling waves in a rod composedof amodifiedMooney-Rivlinmaterial I bifurcation of equilibraand non-singular caserdquo Proceedings of the Royal Society AMathematical Physical and Engineering Sciences vol 455 no1990 pp 3845ndash3874 1999
[7] H-H Dai and F-F Wang ldquoAsymptotic bifurcation solutionsfor compressions of a clamped nonlinearly elastic rectangletransition region and barrelling to a corner-like profilerdquo SIAMJournal on Applied Mathematics vol 70 no 7 pp 2673ndash26922010
[8] J Li and Z Liu ldquoSmooth and non-smooth traveling waves ina nonlinearly dispersive equationrdquo Applied Mathematical Mod-elling vol 25 no 1 pp 41ndash56 2000
[9] A Chen and J Li ldquoSingle peak solitary wave solutions forthe osmosis K(22) equation under inhomogeneous boundaryconditionrdquo Journal of Mathematical Analysis and Applicationsvol 369 no 2 pp 758ndash766 2010
[10] R Camassa and D D Holm ldquoAn integrable shallow waterequation with peaked solitonsrdquo Physical Review Letters vol 71no 11 pp 1661ndash1664 1993
[11] P Rosenau ldquoOn nonanalytic solitary waves formed by a nonlin-ear dispersionrdquo Physics Letters A vol 230 no 5-6 pp 305ndash3181997
[12] P Rosenau and J M Hyman ldquoCompactons solitons with finitewavelengthrdquo Physical Review Letters vol 70 no 5 pp 564ndash5671993
[13] L Zhang L-Q Chen and X Huo ldquoThe effects of horizontalsingular straight line in a generalized nonlinear Klein-Gordonmodel equationrdquo Nonlinear Dynamics vol 72 no 4 pp 789ndash801 2013
[14] J Li and H Dai On the Study of Singular Nonlinear TravelingWave Equations Dynamical System Approach Science PressBeijing China 2007
[15] A Chen S Wen S Tang W Huang and Z Qiao ldquoEffects ofquadratic singular curves in integrable equationsrdquo Studies inApplied Mathematics vol 134 no 1 pp 24ndash61 2015
[16] L Tian and J Yin ldquoNew compacton solutions and solitary wavesolutions of fully nonlinear generalized Camassa-Holm equa-tionsrdquo Chaos Solitons amp Fractals vol 20 no 2 pp 289ndash2992004
[17] L Zhang ldquoThe bifurcation and peakons for the specialC(3 2 2)equationrdquo Pramana vol 83 no 3 pp 331ndash340 2014
[18] J Lenells ldquoTraveling wave solutions of the CamassandashHolmequationrdquo Journal of Differential Equations vol 217 no 2 pp393ndash430 2005
[19] Z J Qiao and G P Zhang ldquoOn peaked and smooth solitons forthe Camassa-Holm equationrdquo Europhysics Letters vol 73 no 5pp 657ndash663 2006
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Mathematical Problems in Engineering
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
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
00 05 10 15
0
2
4
0 1
0
2
4
6
00 05 10 15
0
2
4
00 05 10
0
1
2
3
00 05 10 15
0
1
2
3
00 05 10
0
1
2
00 05 10 15
0
1
2
00 05 10 15
0
1
2
3
minus3 minus1minus4
minus4
minus2
minus4
minus2
minus4
minus2
minus4
minus2minus4
minus6
minus2
minus4
minus2
minus4
minus2
minus2 minus4 minus4minus2 minus2
minus3
minus1
minus2
minus1
minus2
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus1minus2 minus1minus2
120593
120593
120593
120593 120593 120593
120593 120593
120593 120593
120593 120593
y
y
y
yy
y
y y
y y
y y
minus15 minus10 minus05
minus15minus20 minus10 minus05
minus15 minus10 minus05 minus15 minus10 minus05minus10 minus05
minus15 minus10 minus05minus10 minus05
(10) (120590 120573) isin B10 (11) (120590 120573) isin B11 (12) (120590 120573) isin B12
(1) (120590 120573) isin B1
(4) (120590 120573) isin B4
(7) (120590 120573) isin B7 (8) (120590 120573) isin B8 (9) (120590 120573) isin B9
(5) (120590 120573) isin B5 (6) (120590 120573) isin B6
(2) (120590 120573) isin B2 (3) (120590 120573) isin B3
Figure 4 Continued
10 Mathematical Problems in Engineering
00 05 10 15
0
1
2
3
0 1
0
1
2
3
0 1 2
0
2
4
00 05 10 15
0
1
2
3
0 1 2
0
1
2
3
00 05 10 15
0
1
2
00 05 10 15
0
1
2
3
0 1 2
0
1
2
3
00 05 10 15
0
2
4
00 05 10 15
0
2
4
6
minus4
minus2
minus4
minus2
minus4
minus6
minus2
minus3 minus1minus2
minus3 minus1minus2
minus1minus2
minus1minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus1
minus2
minus3
minus1
minus2
120593 120593 120593
120593120593120593
120593 120593 120593
120593
y y y
yy
y
y yy
y
minus15 minus10 minus05
minus15minus20 minus10 minus05
minus15 minus10 minus05
minus15 minus10 minus05
minus15 minus10 minus05
minus15 minus10 minus05
(13) (120590 120573) isin B13 (14) (120590 120573) isin B14 (15) (120590 120573) isin B15
(16) (120590 120573) isin B16 (17) (120590 120573) isin B17 (18) (120590 120573) isin B18
(22) (120590 120573) isin B22
(20) (120590 120573) isin B20 (21) (120590 120573) isin B21(19) (120590 120573) isin B19
Figure 4 The bifurcation of phase portraits of system (15) when 120598 lt 0 and 120590 ge 0
Mathematical Problems in Engineering 11
(ii) If 120593(0) = 984858 = 1198611 then 120593 is a cuspon solution and 120593 has
the following asymptotic behavior
120593 (120585) minus 984858 = 1205821
10038161003816100381610038161205851003816100381610038161003816
23
+ 119874(10038161003816100381610038161205851003816100381610038161003816
43
) 120585 997888rarr 0
1205931015840(120585) =
2
31205821
10038161003816100381610038161205851003816100381610038161003816
minus13 sgn (120585) + 119874 (10038161003816100381610038161205851003816100381610038161003816
13
)
120585 997888rarr 0
(27)
where 1205821= plusmn(9|120573984858
2+ 2119860120573984858 + 3119860
2120573 + 2120598|(984858 minus 119860)
216984858)13
Thus 120593(120585) notin 1198671
119897119900119888(R)
(iii) If 120593(0) = minus984858 = 1198612 then 120593 is a cuspon solution and 120593
has the following asymptotic behavior
120593 (120585) + 984858 = 1205822
10038161003816100381610038161205851003816100381610038161003816
23
+ 119874(10038161003816100381610038161205851003816100381610038161003816
43
) 120585 997888rarr 0
1205931015840(120585) =
2
31205822
10038161003816100381610038161205851003816100381610038161003816
minus13 sgn (120585) + 119874 (10038161003816100381610038161205851003816100381610038161003816
13
)
120585 997888rarr 0
(28)
where 1205822= plusmn(9|120573984858
2minus 2119860120573984858 + 3119860
2120573 + 2120598|(984858 + 119860)
216984858)13
Thus 120593(120585) notin 1198671
119897119900119888(R)
(iv) If 120593(0) = 984858 = 1198611and 119860 = 0 then 120593 is a peakon-like
solution and
120593 (120585) minus 984858 = 1205823
10038161003816100381610038161205851003816100381610038161003816 + 119874 (
10038161003816100381610038161205851003816100381610038161003816
2
) 120585 997888rarr 0
1205931015840(120585) = 120582
3sgn (120585) + 119874 (
10038161003816100381610038161205851003816100381610038161003816) 120585 997888rarr 0
(29)
where 1205823= plusmn|119860 minus 984858|radic120573(119861
2minus 984858)4984858
(v) If 120593(0) = minus984858 = 1198612and 119860 = 0 then 120593 is a peakon-like
solution and
120593 (120585) + 984858 = 1205824
10038161003816100381610038161205851003816100381610038161003816 + 119874 (
10038161003816100381610038161205851003816100381610038161003816
2
) 120585 997888rarr 0
1205931015840(120585) = 120582
4sgn (120585) + 119874 (
10038161003816100381610038161205851003816100381610038161003816) 120585 997888rarr 0
(30)
where 1205824= plusmn|119860 + 984858|radicminus120573(119861
1+ 984858)4984858
(vi) If 120593(0) = 984858 = 1198611and 119860 = 0 then 120593 gives the peakon
solution 984858 exp(minusradicminus1205732|119909 minus 119888119905|)(vii) If 120593(0) = minus984858 = 119861
2and 119860 = 0 then 120593 gives the peakon
solution minus984858 exp(minusradicminus1205732|119909 minus 119888119905|)
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 |120593(0)| = 984858 then 120593 isin 119862
infin(R) and 120593 is a smooth soliton
solution(ii) If 120593(0) = 984858 = 119861
1 then by the definition of single peak
soliton we have 119860 = 984858 thus 1205731205932 + 2119860120573120593 + 31198602120573 + 2120598 does
not contain the factor 120593 minus 984858 From (24) we obtain
1205931015840= sgn (119860 minus 984858)
1003816100381610038161003816120593 minus 1198601003816100381610038161003816
radic210038161003816100381610038161205932 minus 9848582
1003816100381610038161003816
sdot radic10038161003816100381610038161205731205932 + 2119860120573120593 + 31198602120573 + 2120598
1003816100381610038161003816 sgn (120585)
(31)
Let 1198971(120593) = radic2(120593 + 984858)|120593 minus 119860|radic|1205731205932 + 2119860120573120593 + 31198602120573 + 2120598|
then 1198971(984858) = 2radic984858|984858 minus 119860|radic|1205739848582 + 2119860120573984858 + 31198602120573 + 2120598| and
int 1198971(120593)radic
1003816100381610038161003816120593 minus 9848581003816100381610038161003816119889120593 = int sgn (119860 minus 984858) sgn (120585) 119889120585 (32)
Inserting 1198971(120593) = 119897
1(984858) + 119874(|120593 minus 984858|) into (32) and using the
initial condition 120593(0) = 984858 we obtain
21198971(984858)
3
1003816100381610038161003816120593 minus 9848581003816100381610038161003816
32
(1 + 119874 (1003816100381610038161003816120593 minus 984858
1003816100381610038161003816)) =10038161003816100381610038161205851003816100381610038161003816
(33)
thus
120593 minus 984858 = plusmn(3
21198971(984858)
)
23
10038161003816100381610038161205851003816100381610038161003816
23
(1 + 119874 (1003816100381610038161003816120593 minus 984858
1003816100381610038161003816))minus23
= plusmn(3
21198971(984858)
)
23
10038161003816100381610038161205851003816100381610038161003816
23
(1 + 119874 (1003816100381610038161003816120593 minus 984858
1003816100381610038161003816))
(34)
which implies 120593 minus 984858 = 119874(|120585|23
) Therefore we have
120593 (120585) = 984858 plusmn (3
21198971(984858)
)
23
10038161003816100381610038161205851003816100381610038161003816
23
+ 119874(10038161003816100381610038161205851003816100381610038161003816
43
)
= 984858 + 1205821
10038161003816100381610038161205851003816100381610038161003816
23
+ 119874(10038161003816100381610038161205851003816100381610038161003816
43
) 120585 997888rarr 0
1205821= plusmn(
3
21198971(984858)
)
23
= plusmn(
9100381610038161003816100381610038161205739848582+ 2119860120573984858 + 3119860
2120573 + 2120598
10038161003816100381610038161003816(984858 minus 119860)
2
16984858)
13
1205931015840(120585) =
2
31205821
10038161003816100381610038161205851003816100381610038161003816
minus13 sgn (120585) + 119874 (10038161003816100381610038161205851003816100381610038161003816
13
) 120585 997888rarr 0
(35)
So 120593(120585) notin 1198671
loc(R)(iii) Similar to the proof of (ii) we ignore it in this paper(iv) If 120593(0) = 984858 = 119861
1and 119860 = 0 then from (25) we obtain
1205931015840= radicminus
120573
2sgn (119860 minus 984858)
1003816100381610038161003816120593 minus 1198601003816100381610038161003816radic120593 minus 1198612
120593 + 984858sgn (120585) (36)
Let 1198972(120593) = (1|120593minus119860|)radic(120593 + 984858)(120593 minus 119861
2) then 119897
2(984858) = (1|984858minus
119860|)radic2984858(984858 minus 1198612) and
int 1198972(120593) 119889120593 = radicminus
120573
2int sgn (119860 minus 984858) sgn (120585) 119889120585 (37)
Inserting 1198972(120593) = 119897
2(984858) + 119874(|120593 minus 984858|) into (37) and using the
initial condition 120593(0) = 984858 we obtain
1198972(984858) (120593 minus 984858) (1 + 119874 (
1003816100381610038161003816120593 minus 9848581003816100381610038161003816))minus1
= sgn (119860 minus 984858)10038161003816100381610038161205851003816100381610038161003816
(38)
Sincesgn (120593 minus 984858) sgn (119860 minus 984858) ge 0
1
1 + 119874 (120593 minus 984858)= 1 + 119874 (120593 minus 984858)
(39)
12 Mathematical Problems in Engineering
we get
1003816100381610038161003816120593 minus 9848581003816100381610038161003816 =
radicminus120573
2
1
1198972(984858)
10038161003816100381610038161205851003816100381610038161003816 (1 + 119874 (120593 minus 984858)) (40)
which implies |120593 minus 984858| = 119874(|120585|) Therefore we have
120593 (120585) = 984858 + 1205823
10038161003816100381610038161205851003816100381610038161003816 + 119874 (
10038161003816100381610038161205851003816100381610038161003816
2
) 120585 997888rarr 0
1205931015840(120585) = 120582
3sgn (120585) + 119874 (
10038161003816100381610038161205851003816100381610038161003816) 120585 997888rarr 0
(41)
where 1205823= plusmn|119860 minus 984858|radic120573(119861
2minus 984858)4984858
(v) Similar to the proof of (iv) we ignore it in this paper
By virtue of Theorem 9 any single peak soliton for the119862(3 2 2) equation (10) must satisfy the following initial andboundary values problem
(1205931015840)2
=120573 (120593 minus 119860)
2
(120593 minus 1198611) (120593 minus 119861
2)
2 (9848582 minus 1205932)fl119871 (120593)
120593 (0) isin 984858 minus984858 1198611 1198612
lim|120585|rarrinfin
120593 (120585) = 119860
(42)
119871(120593) ge 0 and the boundary condition (24) imply thefollowing
(a) If 120573(9848582 minus 1205932) ge 0 then 120593 ge max119861
1 1198612 or 120593 le
min1198611 1198612
(b) If120573(9848582minus1205932) le 0 thenmin1198611 1198612 le 120593 le max119861
1 1198612
Below we will present some implicit formulas for thesingle peak soliton solutions in the case of specific 120590 and 120573
Case 1 ((120590 120573) isin 11986012) In this case we have minus984858 lt 119861
2lt 119860 lt
1198611= 984858 From the standard phase analysis and Theorem 9
we know that if 120593 is a single peak soliton of the 119862(3 2 2)
equation then
1205931015840= minusradicminus
120573
2(120593 minus 119860)radic
120593 minus 1198612
120593 + 984858sgn (120585) (43)
From the separation of variables we get
intℎ (120593) 119889120593 = radicminus120573
2sgn (120585) 119889120585 (44)
where ℎ(120593) = (1(119860 minus 120593))radic(120593 + 984858)(120593 minus 1198612) After a lengthy
calculation of integral we obtain the implicit solution 120593
defined by
119867(120593)flradic119860 + 984858
119860 minus 1198612
1198681(120593) minus 119868
2(120593) = radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870 (45)
where
1198681(120593) = ln
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
(119860 + 984858) (120593 minus 1198612) + (119860 minus 119861
2) (120593 + 984858) + 2radic(119860 + 984858) (119860 minus 119861
2) (120593 minus 119861
2) (120593 + 984858)
radic(119860 + 984858) (119860 minus 1198612) (119860 + 984858) (119860 minus 120593)
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
1198682(120593) = ln
1003816100381610038161003816100381610038161003816(120593 + 984858) + (120593 minus 119861
2) + 2radic(120593 + 984858) (120593 minus 119861
2)
1003816100381610038161003816100381610038161003816
(46)
and 119870 is an arbitrary integration constant For 120593(0) = 984858 theconstant119870 = 119867(120593(0)) is defined by
119870 = radic119860 + 984858
119860 minus 1198612
1198681(984858) minus 119868
2(984858) (47)
and for 120593(0) = 1198612
119870 = radic119860 + 984858
119860 minus 1198612
1198681(1198612) minus 1198682(1198612) (48)
(i) If 120593(0) = 984858 then 119860 lt 120593 le 984858 Since 1198671015840(120593) = ℎ(120593) weknow that119867(120593) strictly decreases on the interval (119860 984858] thus1198671(120593) = 119867|
(119860984858](120593) gives a single peak solitonwith119867
1(984858) = 119870
and1198671(119860+) = infin Therefore 120593
1(120585) = 119867
minus1
1(radicminus1205732|120585| + 119870) is
the solution satisfying
1205931(0) = 984858
lim|120585|rarrinfin
1205931(120585) = 119860
1205931015840
1(0plusmn) = plusmn (119860 minus 984858)radic
120573 (1198612minus 984858)
4984858
(49)
So 1205931(120585) is a peakon-like solution (see Figure 5)
(ii) If 120593(0) = 1198612 then 119861
2le 120593 lt 119860 By 119867
1015840(120593) = 119891(120593)
we know that 119867(120593) strictly increases on the interval [1198612 119860)
Thus
1198672(120593) = 119867|
[1198612 119860)(120593) (50)
Mathematical Problems in Engineering 13
minus3 minus2 minus1
120593
minus02
1 2 3
02
04
06
08
10
x
(a)
02
02
minus2 minus2 x
10
05
00
t
120593
22
(b)
Figure 5 Two- and three-dimensional graphs of the peakon-like solution
minus03
minus04
minus05
1 2 3minus3 minus2 minus1
120593
minus02
x
(a)
0
2
0
2
minus2minus2
minus02
minus03
minus04
minus05
minus06
t x
120593
0
2
0
2
minus2minus2
t x
(b)
Figure 6 Two- and three-dimensional graphs of the smooth soliton solution
has the inverse denoted by 1205932(120585) = 119867
minus1
2(radicminus1205732|120585| + 119870)
1205932(120585) gives a kind of smooth soliton solution (see Figure 6)
satisfying
1205932(0) = 119861
2
lim|120585|rarrinfin
1205932(120585) = 119860
1205931015840
2(0) = 0
(51)
Case 2 ((120590 120573) isin 11986014) In this case we have minus984858 = 119861
2lt 119860 lt
984858 lt 1198611and (25) is equivalent to
1205931015840= minusradicminus
120573
2(120593 minus 119860)radic
120593 minus 1198611
120593 minus 984858sgn (120585) (52)
Let
119892 (120593) =1
119860 minus 120593radic120593 minus 1198611
120593 minus 984858 (53)
14 Mathematical Problems in Engineering
minus1 1 2 3
02
04
06
08
10
minus3 minus2
120593
x
(a)
02
02minus2
minus2
120593
x t
00
05
10
(b)
Figure 7 Two- and three-dimensional graphs of the cuspon solution
then (52) is converted to
119892 (120593) 119889120593 =1
119860 minus 120593radic120593 minus 1198611
120593 minus 984858119889120593 = radicminus
120573
2sgn (120585) 119889120585 (54)
Integrating (54) on the interval [0 120585] (or [120585 0]) leads to thefollowing implicit solutions
119866 (120593)flradic119860 minus 984858
119860 minus 1198611
1198683(120593) + 119868
4(120593) = radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870 (55)
where
1198683(120593) = ln
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
(119860 minus 984858) (120593 minus 1198611) + (119860 minus 119861
1) (120593 minus 984858) + 2radic(119860 minus 984858) (119860 minus 119861
1) (120593 minus 984858) (120593 minus 119861
1)
radic(119860 minus 984858) (119860 minus 1198611) (119860 minus 984858) (120593 minus 119860)
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
1198684(120593) = ln
1003816100381610038161003816100381610038161003816(120593 minus 984858) + (120593 minus 119861
1) + 2radic(120593 minus 984858) (120593 minus 119861
1)
1003816100381610038161003816100381610038161003816
(56)
And119870 is an arbitrary integration constant It is obvious thatfor 120593(0) = 984858 the constant119870 = 119867(120593(0)) is defined by
119870 = radic119860 minus 984858
119860 minus 1198611
1198683(984858) + 119868
4(984858) (57)
and for 120593(0) = minus984858
119870 = radic119860 minus 984858
119860 minus 1198611
1198683(minus984858) + 119868
4(minus984858) (58)
(i) If 120593(0) = 984858 then 119860 lt 120593 le 984858 From 119892(120593) lt 0 weknow that 119866(120593) strictly decreases on the interval (119860 984858] with119866(984858) = 119870 and 119866(119860+) = infin Define
1198661(120593) = 119866|
(119860984858](120593) (59)
Since 1198661(120593) is a strictly decreasing function from (119860 984858] onto
[119870infin) we can solve for 120593 uniquely from (59) and obtain
1205931(120585) = 119866
minus1
1(radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870) (60)
It is easy to check that 120593 satisfies
1205931(0) = 984858
lim|120585|rarrinfin
1205931(120585) = 119860
1205931015840
1(0plusmn) = ∓infin
(61)
Therefore the solution1205931defined by (60) is a cuspon solution
for the 119862(3 2 2) equation (see Figure 7)
Mathematical Problems in Engineering 15
(ii) If 120593(0) = minus984858 then minus984858 le 120593 lt 119860 Through a similaranalysis we get a strictly increasing function 119866(120593) on theinterval [minus984858 119860) satisfying
119866 (120593) = radicminus120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870 (62)
where 119866(120593) is defined by (55) Let
1198662(120593) = 119866|
[minus984858119860)(120593) (63)
then 1198662(120593) is a strictly increasing function from [minus984858 119860) onto
[119870infin) so that we can solve for 120593 and obtain
1205932(120585) = 119866
minus1
2(radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870) (64)
It is easy to check that 1205932satisfies
1205932(0) = minus984858
lim|120585|rarrinfin
1205932(120585) = 119860
1205931015840
2(0plusmn) = plusmn (119860 + 984858)radicminus
120573 (1198611+ 984858)
4984858
(65)
Therefore the solution 1205932defined by (64) is a peakon-like
solution whose graph is similar to those in Figure 5
Case 3 ((120590 120573) isin 11986016) In this case we have minus984858 lt 119861
2lt 119860 =
0 lt 1198611lt 984858 119861
2= minus1198611 and
1205931015840= minusradicminus
120573
2120593radic
120593 minus 1198612
1
1205932 minus 9848582sgn (120585) (66)
Hence from the separation of variables we have
radic9848582 minus 1205932
120593radic1198612
1minus 1205932
119889120593 = minusradicminus120573
2sgn (120585) 119889120585 (67)
Integrating (67) on the interval [0 120585] (or [120585 0]) leads tothe following implicit formula for the two smooth solitonsolutions
radic9848582 minus 1205932 + radic1198612
1minus 1205932
radic9848582 minus 1198612
1
(
radic(9848582 minus 1198612
1) 1205932
984858radic1198612
1minus 1205932 + 119861
1radic9848582 minus 1205932
)
984858
= exp(minusradicminus120573
2|119909 minus 119888119905|)
(68)
where 120593 isin (119860 1198611] Consider
radic9848582 minus 1198612
2
radic9848582 minus 1205932 + radic1198612
2minus 1205932
(
984858radic1198612
2minus 1205932 minus 119861
2radic9848582 minus 1205932
radic(9848582 minus 1198612
2) 1205932
)
984858
= exp(radicminus120573
2|119909 minus 119888119905|)
(69)
where 120593 isin [1198612 119860)
Case 4 ((120590 120573) isin 11986017) In this case we have minus984858 = 119861
2lt 119860 =
0 lt 1198611= 984858 and
1205931015840= minusradicminus
120573
2120593 sgn (120585) (70)
Choosing 120593(0) = 984858 (or minus984858) as initial value we get
int
120593
984858
1
120593119889120593 = minusint
120585
0
radicminus120573
2sgn (120585) 119889120585 for 119860 lt 120593 le 984858
int
120593
minus984858
1
120593119889120593 = minusint
120585
0
radicminus120573
2sgn (120585) 119889120585 for minus 984858 le 120593 lt 119860
(71)
which immediately yield the peakon solutions
120593 (119909 minus 119888119905) = plusmn984858 exp(minusradicminus120573
2|119909 minus 119888119905|) (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 the119862(3 2 2) equation (10) In order to study kinkwave solutionswe assume that
lim120585rarrinfin
120593 (120585) = 1198601
lim120585rarrminusinfin
120593 (120585) = 1198602
(73)
16 Mathematical Problems in Engineering
2 4
02
04
06
08
10
minus2minus4
120593
x
(a)
0 2
02
00
05
10
minus2
120593
x
t
0
2
minus2
(b)
Figure 8 Two- and three-dimensional graphs of the peakon solution
where 1198601gt 1198602 Substituting the boundary condition (73)
into (14) generates
(1205931015840)2
=
120573 (120593 minus 1198601) (120593 minus 119860
2) [1205932+ (1198601+ 1198602) 120593 + 119860
2
1+ 11986011198602+ 1198602
2+ 2120598120573]
2 (9848582 minus 1205932)fl119865 (120593) (74)
The nonlinear differential equation (74) may sustaindifferent kinds of nonlinear excitations In what follows weconfine our attention to the cases 119860
2= minus119860
1and 120598 =
minus1198602
1120573 which describe kinks and kink compactons which
play an important role in the dynamics systems Under theseconsiderations (74) reduces to
(1205931015840)2
=120573 (120593 minus 119860
1)2
(120593 minus 1198602)2
2 (9848582 minus 1205932) (75)
If 1198601lt 984858 then from the phase analysis in Section 2 (see
Figure 4(10)) we know that (1198601 0) and (119860
2 0) are two saddle
points of (13) and the kink solutions can be obtained fromthe two heteroclinic orbits connecting (120601 119910) = (119860
1 0) and
(1198602 0) When 119860
1increases upon reaching 984858 that is 119860
1= 984858
(75) becomes
(1205931015840)2
= minus120573 (120593 minus 119860
1) (120593 minus 119860
2)
2 (76)
and the ellipse 21199102+120573(120593minus1198601)(120593minus119860
2) = 0 (see Figure 4(11))
which is tangent to the singular lines 120593 = 984858 and 120593 = minus984858 atpoints (119860
1 0) and (119860
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 120593 is a kink wave solutions of (75)
or (76) we have 1205931015840
rarr 0 as 120593 rarr 1198601and as 120593 rarr 119860
2
Moreover we have 119865(120593) ge 0 for 1198602le 120593 le 119860
1and 120593 is
strictly monotonic in any interval where 119865(120593) gt 0 Thus if1205931015840gt 0 at some points 120593will be strictly increasing until it gets
close to the next zero of 119865 Denoting this zero 1198601 we have
120593 uarr 1198601 What will happen to the solution when it approaches
1198601 Depending on whether the zero is double or simple 120593
has a different behavior We explore the two cases in turn
Theorem 11 (i) If 120593 has a simple zero at 120593 = 1198601 so that
119865(1198601) = 0 and 1198651015840(119860
1) lt 0 then the solution 120593 of (75) satisfies
120593 (120585) minus 1198601=1198651015840(1198601)
4(120585 minus 120578)
2
+ 119874 ((120585 minus 120578)4
)
119886119904 120585 uarr 120578
(77)
where 120593(120578) = 1198601
(ii) When 120593 approaches the double zero 1198601of 119865(120593) so that
1198651015840(1198601) = 0 and11986510158401015840(119860
1) gt 0 then the solution120593 of (75) satisfies
120593 (120585) minus 1198601sim 120572 exp(minus120585radic11986510158401015840 (119860
1)) 119886119904 120585 997888rarr infin (78)
for some constant 120572 Thus 120593 uarr 1198601exponentially as 120585 rarr infin
Mathematical Problems in Engineering 17
Proof (i) When 1198601= minus119860
2= 984858 and 120598 = minus119860
2
1120573 from (76)
119865(120593) has a simple zero at 120593 = 1198601 Then
(1205931015840)2
= (120593 minus 1198601) 1198651015840(1198601) + 119874 ((120593 minus 119860
1)2
)
as 120593 uarr 1198601
(79)
Using the fact that (1205931015840)2 ge 0 we know that 1198651015840(1198601) lt 0
Moreover
119889120585
119889120593=
1
radic(120593 minus 1198601) 1198651015840 (119860
1) + 119874 ((120593 minus 119860
1)2
)
(80)
Because
radic(120593 minus 1198601) 1198651015840 (119860
1) + 119874 ((120593 minus 119860
1)2
)
= radic1198601minus 120593(radicminus1198651015840 (119860
1) + 119874 (119860
1minus 120593))
1
radicminus1198651015840 (1198601) + 119874 (119860
1minus 120593)
=1
radicminus1198651015840 (1198601)
+ 119874 (1198601minus 120593)
(81)
we obtain119889120585
119889120593=
1
radicminus (1198601minus 120593) 1198651015840 (119860
1)
+ 119874 ((1198601minus 120593)12
) (82)
Integrating (82) yields
120578 minus 120585 =2
radicminus1198651015840 (1198601)
(1198601minus 120593)12
+ 119874((1198601minus 120593)32
) (83)
where 120578 satisfies 120593(120578) = 1198601 Thus
(120578 minus 120585)2
=4
minus1198651015840 (1198601)(1198601minus 120593) + 119874 ((119860
1minus 120593)2
) (84)
which implies 119874((1198601minus 120593)2) = 119874((120578 minus 120585)
4) Therefore we get
120593 (120585) = 1198601+1198651015840(1198601)
4(120585 minus 120578)
2
+ 119874 ((120585 minus 120578)4
)
as 120585 uarr 120578
(85)
where 120593(120578) = 1198601
(ii) When 1198601= minus119860
2lt 984858 and 120598 = minus119860
2
1120573 from (75) 119865(120593)
has a double zero at 120593 = 1198601 Then
(1205931015840)2
= (120593 minus 1198601)2
11986510158401015840(1198601) + 119874 ((120593 minus 119860
1)3
)
as 120593 uarr 1198601
(86)
Furthermore we get
119889120585
119889120593=
1
radic(120593 minus 1198601)2
11986510158401015840 (1198601) + 119874 ((120593 minus 119860
1)3
)
(87)
Observing that
radic(120593 minus 1198601)2
11986510158401015840 (1198601) + 119874 ((120593 minus 119860
1)3
)
= (1198601minus 120593) (radic11986510158401015840 (119860
1) + 119874 (119860
1minus 120593))
1
radic11986510158401015840 (1198601) + 119874 (119860
1minus 120593)
=1
radic11986510158401015840 (1198601)
+ 119874 (1198601minus 120593)
(88)
we obtain119889120585
119889120593=
1
(1198601minus 120593) 11986510158401015840 (119860
1)+ 119874 (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 119860
1= minus119860
2lt 984858 and 120598 = minus119860
2
1120573 Then (75)
becomes
radic9848582 minus 1205932
1205932 minus 1198602
1
119889120593 = ∓radic120573
2119889120585 (90)
Integrating both sides of (90) gives the following implicitexpressions of kink and antikink wave solutions
arctan120593
radic9848582 minus 1205932
+
radic9848582 minus 1198602
1
1198601
arctanhradic9848582 minus 119860
2
1
1198601
120593
radic9848582 minus 1205932
= plusmnradic120573
2120585
(91)
By letting 1198601rarr 984858 in (91) we get two kink compactons
which are given by
plusmn120593 =
984858 sinradic120573
2120585
10038161003816100381610038161205851003816100381610038161003816 le
120587
radic2120573
984858 120585 gt120587
radic2120573
minus984858 120585 lt minus120587
radic2120573
(92)
The graphs for the kink wave solutions (91) and kinkcompacton solutions (92) are shown in Figures 9 and 10respectively
Remark 12 The two kink compacton solutions (92) aredifferent from the well-known smooth kink wave solutionsIn comparison with kink wave solutions (91) the kink com-pacton solutions (92) have no exponential decay propertiesbut have compact supportThat is they minus a constant thedifferences identically vanish outside a finite core region
18 Mathematical Problems in Engineering
120593
120585minus1 1
05
minus05
(a)
minus1minus1
minus05
05
00120593
x t
0
1
01
11
(b)
Figure 9 Two- and three-dimensional graphs of the kink wave solution
1
1120593
120585
minus1
minus1
(a)
0
1
0
1
000510
120593
minus1 minus1
minus05
minus10
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 the119862(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 Newtonrsquos equation which we write in the form 119910
2=
119865(120593) where 119910 = 1205931015840
120585(120585) we solve Newtonrsquos equation 119910
2=
119865(120593) 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 the119862(3 2 2)equation
Actually for 119898 = 2119896 + 1 119896 isin N+ in 119862(119898 2 2) equation(9) the dynamical behavior of traveling wave solutions of(9) is similar to the case 119898 = 3 for 119898 = 2119896 119896 isin N+ in119862(119898 2 2) equation (9) the dynamical behavior of travelingwave solutions of (9) is similar to the case 119898 = 2 We areapplying the approach mentioned in this work to 119862(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
References
[1] C Rogers and W R Shadwick Backlund Transformation andTheir Applications Academic Press New York NY USA 1982
[2] W Malfliet and W Hereman ldquoThe tanh method I Exactsolutions of nonlinear evolution and wave equationsrdquo PhysicaScripta vol 54 no 6 pp 563ndash568 1996
[3] R Hirota Direct Method in Soliton Theory Springer BerlinGermany 1980
[4] J S Cohen Computer Algebra and Symbolic ComputationMathematical Methods A K Peters 2003
[5] S Lie ldquoUber die integration durch bestimmte integrale voneiner klasse linearer partieller differential eichungenrdquoArchiv derMathematik vol 6 pp 328ndash368 1881
[6] H-H Dai and X-H Zhao ldquoTraveling waves in a rod composedof amodifiedMooney-Rivlinmaterial I bifurcation of equilibraand non-singular caserdquo Proceedings of the Royal Society AMathematical Physical and Engineering Sciences vol 455 no1990 pp 3845ndash3874 1999
[7] H-H Dai and F-F Wang ldquoAsymptotic bifurcation solutionsfor compressions of a clamped nonlinearly elastic rectangletransition region and barrelling to a corner-like profilerdquo SIAMJournal on Applied Mathematics vol 70 no 7 pp 2673ndash26922010
[8] J Li and Z Liu ldquoSmooth and non-smooth traveling waves ina nonlinearly dispersive equationrdquo Applied Mathematical Mod-elling vol 25 no 1 pp 41ndash56 2000
[9] A Chen and J Li ldquoSingle peak solitary wave solutions forthe osmosis K(22) equation under inhomogeneous boundaryconditionrdquo Journal of Mathematical Analysis and Applicationsvol 369 no 2 pp 758ndash766 2010
[10] R Camassa and D D Holm ldquoAn integrable shallow waterequation with peaked solitonsrdquo Physical Review Letters vol 71no 11 pp 1661ndash1664 1993
[11] P Rosenau ldquoOn nonanalytic solitary waves formed by a nonlin-ear dispersionrdquo Physics Letters A vol 230 no 5-6 pp 305ndash3181997
[12] P Rosenau and J M Hyman ldquoCompactons solitons with finitewavelengthrdquo Physical Review Letters vol 70 no 5 pp 564ndash5671993
[13] L Zhang L-Q Chen and X Huo ldquoThe effects of horizontalsingular straight line in a generalized nonlinear Klein-Gordonmodel equationrdquo Nonlinear Dynamics vol 72 no 4 pp 789ndash801 2013
[14] J Li and H Dai On the Study of Singular Nonlinear TravelingWave Equations Dynamical System Approach Science PressBeijing China 2007
[15] A Chen S Wen S Tang W Huang and Z Qiao ldquoEffects ofquadratic singular curves in integrable equationsrdquo Studies inApplied Mathematics vol 134 no 1 pp 24ndash61 2015
[16] L Tian and J Yin ldquoNew compacton solutions and solitary wavesolutions of fully nonlinear generalized Camassa-Holm equa-tionsrdquo Chaos Solitons amp Fractals vol 20 no 2 pp 289ndash2992004
[17] L Zhang ldquoThe bifurcation and peakons for the specialC(3 2 2)equationrdquo Pramana vol 83 no 3 pp 331ndash340 2014
[18] J Lenells ldquoTraveling wave solutions of the CamassandashHolmequationrdquo Journal of Differential Equations vol 217 no 2 pp393ndash430 2005
[19] Z J Qiao and G P Zhang ldquoOn peaked and smooth solitons forthe Camassa-Holm equationrdquo Europhysics Letters vol 73 no 5pp 657ndash663 2006
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Mathematical Problems in Engineering
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
10 Mathematical Problems in Engineering
00 05 10 15
0
1
2
3
0 1
0
1
2
3
0 1 2
0
2
4
00 05 10 15
0
1
2
3
0 1 2
0
1
2
3
00 05 10 15
0
1
2
00 05 10 15
0
1
2
3
0 1 2
0
1
2
3
00 05 10 15
0
2
4
00 05 10 15
0
2
4
6
minus4
minus2
minus4
minus2
minus4
minus6
minus2
minus3 minus1minus2
minus3 minus1minus2
minus1minus2
minus1minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus3
minus1
minus2
minus1
minus2
minus3
minus1
minus2
120593 120593 120593
120593120593120593
120593 120593 120593
120593
y y y
yy
y
y yy
y
minus15 minus10 minus05
minus15minus20 minus10 minus05
minus15 minus10 minus05
minus15 minus10 minus05
minus15 minus10 minus05
minus15 minus10 minus05
(13) (120590 120573) isin B13 (14) (120590 120573) isin B14 (15) (120590 120573) isin B15
(16) (120590 120573) isin B16 (17) (120590 120573) isin B17 (18) (120590 120573) isin B18
(22) (120590 120573) isin B22
(20) (120590 120573) isin B20 (21) (120590 120573) isin B21(19) (120590 120573) isin B19
Figure 4 The bifurcation of phase portraits of system (15) when 120598 lt 0 and 120590 ge 0
Mathematical Problems in Engineering 11
(ii) If 120593(0) = 984858 = 1198611 then 120593 is a cuspon solution and 120593 has
the following asymptotic behavior
120593 (120585) minus 984858 = 1205821
10038161003816100381610038161205851003816100381610038161003816
23
+ 119874(10038161003816100381610038161205851003816100381610038161003816
43
) 120585 997888rarr 0
1205931015840(120585) =
2
31205821
10038161003816100381610038161205851003816100381610038161003816
minus13 sgn (120585) + 119874 (10038161003816100381610038161205851003816100381610038161003816
13
)
120585 997888rarr 0
(27)
where 1205821= plusmn(9|120573984858
2+ 2119860120573984858 + 3119860
2120573 + 2120598|(984858 minus 119860)
216984858)13
Thus 120593(120585) notin 1198671
119897119900119888(R)
(iii) If 120593(0) = minus984858 = 1198612 then 120593 is a cuspon solution and 120593
has the following asymptotic behavior
120593 (120585) + 984858 = 1205822
10038161003816100381610038161205851003816100381610038161003816
23
+ 119874(10038161003816100381610038161205851003816100381610038161003816
43
) 120585 997888rarr 0
1205931015840(120585) =
2
31205822
10038161003816100381610038161205851003816100381610038161003816
minus13 sgn (120585) + 119874 (10038161003816100381610038161205851003816100381610038161003816
13
)
120585 997888rarr 0
(28)
where 1205822= plusmn(9|120573984858
2minus 2119860120573984858 + 3119860
2120573 + 2120598|(984858 + 119860)
216984858)13
Thus 120593(120585) notin 1198671
119897119900119888(R)
(iv) If 120593(0) = 984858 = 1198611and 119860 = 0 then 120593 is a peakon-like
solution and
120593 (120585) minus 984858 = 1205823
10038161003816100381610038161205851003816100381610038161003816 + 119874 (
10038161003816100381610038161205851003816100381610038161003816
2
) 120585 997888rarr 0
1205931015840(120585) = 120582
3sgn (120585) + 119874 (
10038161003816100381610038161205851003816100381610038161003816) 120585 997888rarr 0
(29)
where 1205823= plusmn|119860 minus 984858|radic120573(119861
2minus 984858)4984858
(v) If 120593(0) = minus984858 = 1198612and 119860 = 0 then 120593 is a peakon-like
solution and
120593 (120585) + 984858 = 1205824
10038161003816100381610038161205851003816100381610038161003816 + 119874 (
10038161003816100381610038161205851003816100381610038161003816
2
) 120585 997888rarr 0
1205931015840(120585) = 120582
4sgn (120585) + 119874 (
10038161003816100381610038161205851003816100381610038161003816) 120585 997888rarr 0
(30)
where 1205824= plusmn|119860 + 984858|radicminus120573(119861
1+ 984858)4984858
(vi) If 120593(0) = 984858 = 1198611and 119860 = 0 then 120593 gives the peakon
solution 984858 exp(minusradicminus1205732|119909 minus 119888119905|)(vii) If 120593(0) = minus984858 = 119861
2and 119860 = 0 then 120593 gives the peakon
solution minus984858 exp(minusradicminus1205732|119909 minus 119888119905|)
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 |120593(0)| = 984858 then 120593 isin 119862
infin(R) and 120593 is a smooth soliton
solution(ii) If 120593(0) = 984858 = 119861
1 then by the definition of single peak
soliton we have 119860 = 984858 thus 1205731205932 + 2119860120573120593 + 31198602120573 + 2120598 does
not contain the factor 120593 minus 984858 From (24) we obtain
1205931015840= sgn (119860 minus 984858)
1003816100381610038161003816120593 minus 1198601003816100381610038161003816
radic210038161003816100381610038161205932 minus 9848582
1003816100381610038161003816
sdot radic10038161003816100381610038161205731205932 + 2119860120573120593 + 31198602120573 + 2120598
1003816100381610038161003816 sgn (120585)
(31)
Let 1198971(120593) = radic2(120593 + 984858)|120593 minus 119860|radic|1205731205932 + 2119860120573120593 + 31198602120573 + 2120598|
then 1198971(984858) = 2radic984858|984858 minus 119860|radic|1205739848582 + 2119860120573984858 + 31198602120573 + 2120598| and
int 1198971(120593)radic
1003816100381610038161003816120593 minus 9848581003816100381610038161003816119889120593 = int sgn (119860 minus 984858) sgn (120585) 119889120585 (32)
Inserting 1198971(120593) = 119897
1(984858) + 119874(|120593 minus 984858|) into (32) and using the
initial condition 120593(0) = 984858 we obtain
21198971(984858)
3
1003816100381610038161003816120593 minus 9848581003816100381610038161003816
32
(1 + 119874 (1003816100381610038161003816120593 minus 984858
1003816100381610038161003816)) =10038161003816100381610038161205851003816100381610038161003816
(33)
thus
120593 minus 984858 = plusmn(3
21198971(984858)
)
23
10038161003816100381610038161205851003816100381610038161003816
23
(1 + 119874 (1003816100381610038161003816120593 minus 984858
1003816100381610038161003816))minus23
= plusmn(3
21198971(984858)
)
23
10038161003816100381610038161205851003816100381610038161003816
23
(1 + 119874 (1003816100381610038161003816120593 minus 984858
1003816100381610038161003816))
(34)
which implies 120593 minus 984858 = 119874(|120585|23
) Therefore we have
120593 (120585) = 984858 plusmn (3
21198971(984858)
)
23
10038161003816100381610038161205851003816100381610038161003816
23
+ 119874(10038161003816100381610038161205851003816100381610038161003816
43
)
= 984858 + 1205821
10038161003816100381610038161205851003816100381610038161003816
23
+ 119874(10038161003816100381610038161205851003816100381610038161003816
43
) 120585 997888rarr 0
1205821= plusmn(
3
21198971(984858)
)
23
= plusmn(
9100381610038161003816100381610038161205739848582+ 2119860120573984858 + 3119860
2120573 + 2120598
10038161003816100381610038161003816(984858 minus 119860)
2
16984858)
13
1205931015840(120585) =
2
31205821
10038161003816100381610038161205851003816100381610038161003816
minus13 sgn (120585) + 119874 (10038161003816100381610038161205851003816100381610038161003816
13
) 120585 997888rarr 0
(35)
So 120593(120585) notin 1198671
loc(R)(iii) Similar to the proof of (ii) we ignore it in this paper(iv) If 120593(0) = 984858 = 119861
1and 119860 = 0 then from (25) we obtain
1205931015840= radicminus
120573
2sgn (119860 minus 984858)
1003816100381610038161003816120593 minus 1198601003816100381610038161003816radic120593 minus 1198612
120593 + 984858sgn (120585) (36)
Let 1198972(120593) = (1|120593minus119860|)radic(120593 + 984858)(120593 minus 119861
2) then 119897
2(984858) = (1|984858minus
119860|)radic2984858(984858 minus 1198612) and
int 1198972(120593) 119889120593 = radicminus
120573
2int sgn (119860 minus 984858) sgn (120585) 119889120585 (37)
Inserting 1198972(120593) = 119897
2(984858) + 119874(|120593 minus 984858|) into (37) and using the
initial condition 120593(0) = 984858 we obtain
1198972(984858) (120593 minus 984858) (1 + 119874 (
1003816100381610038161003816120593 minus 9848581003816100381610038161003816))minus1
= sgn (119860 minus 984858)10038161003816100381610038161205851003816100381610038161003816
(38)
Sincesgn (120593 minus 984858) sgn (119860 minus 984858) ge 0
1
1 + 119874 (120593 minus 984858)= 1 + 119874 (120593 minus 984858)
(39)
12 Mathematical Problems in Engineering
we get
1003816100381610038161003816120593 minus 9848581003816100381610038161003816 =
radicminus120573
2
1
1198972(984858)
10038161003816100381610038161205851003816100381610038161003816 (1 + 119874 (120593 minus 984858)) (40)
which implies |120593 minus 984858| = 119874(|120585|) Therefore we have
120593 (120585) = 984858 + 1205823
10038161003816100381610038161205851003816100381610038161003816 + 119874 (
10038161003816100381610038161205851003816100381610038161003816
2
) 120585 997888rarr 0
1205931015840(120585) = 120582
3sgn (120585) + 119874 (
10038161003816100381610038161205851003816100381610038161003816) 120585 997888rarr 0
(41)
where 1205823= plusmn|119860 minus 984858|radic120573(119861
2minus 984858)4984858
(v) Similar to the proof of (iv) we ignore it in this paper
By virtue of Theorem 9 any single peak soliton for the119862(3 2 2) equation (10) must satisfy the following initial andboundary values problem
(1205931015840)2
=120573 (120593 minus 119860)
2
(120593 minus 1198611) (120593 minus 119861
2)
2 (9848582 minus 1205932)fl119871 (120593)
120593 (0) isin 984858 minus984858 1198611 1198612
lim|120585|rarrinfin
120593 (120585) = 119860
(42)
119871(120593) ge 0 and the boundary condition (24) imply thefollowing
(a) If 120573(9848582 minus 1205932) ge 0 then 120593 ge max119861
1 1198612 or 120593 le
min1198611 1198612
(b) If120573(9848582minus1205932) le 0 thenmin1198611 1198612 le 120593 le max119861
1 1198612
Below we will present some implicit formulas for thesingle peak soliton solutions in the case of specific 120590 and 120573
Case 1 ((120590 120573) isin 11986012) In this case we have minus984858 lt 119861
2lt 119860 lt
1198611= 984858 From the standard phase analysis and Theorem 9
we know that if 120593 is a single peak soliton of the 119862(3 2 2)
equation then
1205931015840= minusradicminus
120573
2(120593 minus 119860)radic
120593 minus 1198612
120593 + 984858sgn (120585) (43)
From the separation of variables we get
intℎ (120593) 119889120593 = radicminus120573
2sgn (120585) 119889120585 (44)
where ℎ(120593) = (1(119860 minus 120593))radic(120593 + 984858)(120593 minus 1198612) After a lengthy
calculation of integral we obtain the implicit solution 120593
defined by
119867(120593)flradic119860 + 984858
119860 minus 1198612
1198681(120593) minus 119868
2(120593) = radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870 (45)
where
1198681(120593) = ln
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
(119860 + 984858) (120593 minus 1198612) + (119860 minus 119861
2) (120593 + 984858) + 2radic(119860 + 984858) (119860 minus 119861
2) (120593 minus 119861
2) (120593 + 984858)
radic(119860 + 984858) (119860 minus 1198612) (119860 + 984858) (119860 minus 120593)
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
1198682(120593) = ln
1003816100381610038161003816100381610038161003816(120593 + 984858) + (120593 minus 119861
2) + 2radic(120593 + 984858) (120593 minus 119861
2)
1003816100381610038161003816100381610038161003816
(46)
and 119870 is an arbitrary integration constant For 120593(0) = 984858 theconstant119870 = 119867(120593(0)) is defined by
119870 = radic119860 + 984858
119860 minus 1198612
1198681(984858) minus 119868
2(984858) (47)
and for 120593(0) = 1198612
119870 = radic119860 + 984858
119860 minus 1198612
1198681(1198612) minus 1198682(1198612) (48)
(i) If 120593(0) = 984858 then 119860 lt 120593 le 984858 Since 1198671015840(120593) = ℎ(120593) weknow that119867(120593) strictly decreases on the interval (119860 984858] thus1198671(120593) = 119867|
(119860984858](120593) gives a single peak solitonwith119867
1(984858) = 119870
and1198671(119860+) = infin Therefore 120593
1(120585) = 119867
minus1
1(radicminus1205732|120585| + 119870) is
the solution satisfying
1205931(0) = 984858
lim|120585|rarrinfin
1205931(120585) = 119860
1205931015840
1(0plusmn) = plusmn (119860 minus 984858)radic
120573 (1198612minus 984858)
4984858
(49)
So 1205931(120585) is a peakon-like solution (see Figure 5)
(ii) If 120593(0) = 1198612 then 119861
2le 120593 lt 119860 By 119867
1015840(120593) = 119891(120593)
we know that 119867(120593) strictly increases on the interval [1198612 119860)
Thus
1198672(120593) = 119867|
[1198612 119860)(120593) (50)
Mathematical Problems in Engineering 13
minus3 minus2 minus1
120593
minus02
1 2 3
02
04
06
08
10
x
(a)
02
02
minus2 minus2 x
10
05
00
t
120593
22
(b)
Figure 5 Two- and three-dimensional graphs of the peakon-like solution
minus03
minus04
minus05
1 2 3minus3 minus2 minus1
120593
minus02
x
(a)
0
2
0
2
minus2minus2
minus02
minus03
minus04
minus05
minus06
t x
120593
0
2
0
2
minus2minus2
t x
(b)
Figure 6 Two- and three-dimensional graphs of the smooth soliton solution
has the inverse denoted by 1205932(120585) = 119867
minus1
2(radicminus1205732|120585| + 119870)
1205932(120585) gives a kind of smooth soliton solution (see Figure 6)
satisfying
1205932(0) = 119861
2
lim|120585|rarrinfin
1205932(120585) = 119860
1205931015840
2(0) = 0
(51)
Case 2 ((120590 120573) isin 11986014) In this case we have minus984858 = 119861
2lt 119860 lt
984858 lt 1198611and (25) is equivalent to
1205931015840= minusradicminus
120573
2(120593 minus 119860)radic
120593 minus 1198611
120593 minus 984858sgn (120585) (52)
Let
119892 (120593) =1
119860 minus 120593radic120593 minus 1198611
120593 minus 984858 (53)
14 Mathematical Problems in Engineering
minus1 1 2 3
02
04
06
08
10
minus3 minus2
120593
x
(a)
02
02minus2
minus2
120593
x t
00
05
10
(b)
Figure 7 Two- and three-dimensional graphs of the cuspon solution
then (52) is converted to
119892 (120593) 119889120593 =1
119860 minus 120593radic120593 minus 1198611
120593 minus 984858119889120593 = radicminus
120573
2sgn (120585) 119889120585 (54)
Integrating (54) on the interval [0 120585] (or [120585 0]) leads to thefollowing implicit solutions
119866 (120593)flradic119860 minus 984858
119860 minus 1198611
1198683(120593) + 119868
4(120593) = radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870 (55)
where
1198683(120593) = ln
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
(119860 minus 984858) (120593 minus 1198611) + (119860 minus 119861
1) (120593 minus 984858) + 2radic(119860 minus 984858) (119860 minus 119861
1) (120593 minus 984858) (120593 minus 119861
1)
radic(119860 minus 984858) (119860 minus 1198611) (119860 minus 984858) (120593 minus 119860)
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
1198684(120593) = ln
1003816100381610038161003816100381610038161003816(120593 minus 984858) + (120593 minus 119861
1) + 2radic(120593 minus 984858) (120593 minus 119861
1)
1003816100381610038161003816100381610038161003816
(56)
And119870 is an arbitrary integration constant It is obvious thatfor 120593(0) = 984858 the constant119870 = 119867(120593(0)) is defined by
119870 = radic119860 minus 984858
119860 minus 1198611
1198683(984858) + 119868
4(984858) (57)
and for 120593(0) = minus984858
119870 = radic119860 minus 984858
119860 minus 1198611
1198683(minus984858) + 119868
4(minus984858) (58)
(i) If 120593(0) = 984858 then 119860 lt 120593 le 984858 From 119892(120593) lt 0 weknow that 119866(120593) strictly decreases on the interval (119860 984858] with119866(984858) = 119870 and 119866(119860+) = infin Define
1198661(120593) = 119866|
(119860984858](120593) (59)
Since 1198661(120593) is a strictly decreasing function from (119860 984858] onto
[119870infin) we can solve for 120593 uniquely from (59) and obtain
1205931(120585) = 119866
minus1
1(radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870) (60)
It is easy to check that 120593 satisfies
1205931(0) = 984858
lim|120585|rarrinfin
1205931(120585) = 119860
1205931015840
1(0plusmn) = ∓infin
(61)
Therefore the solution1205931defined by (60) is a cuspon solution
for the 119862(3 2 2) equation (see Figure 7)
Mathematical Problems in Engineering 15
(ii) If 120593(0) = minus984858 then minus984858 le 120593 lt 119860 Through a similaranalysis we get a strictly increasing function 119866(120593) on theinterval [minus984858 119860) satisfying
119866 (120593) = radicminus120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870 (62)
where 119866(120593) is defined by (55) Let
1198662(120593) = 119866|
[minus984858119860)(120593) (63)
then 1198662(120593) is a strictly increasing function from [minus984858 119860) onto
[119870infin) so that we can solve for 120593 and obtain
1205932(120585) = 119866
minus1
2(radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870) (64)
It is easy to check that 1205932satisfies
1205932(0) = minus984858
lim|120585|rarrinfin
1205932(120585) = 119860
1205931015840
2(0plusmn) = plusmn (119860 + 984858)radicminus
120573 (1198611+ 984858)
4984858
(65)
Therefore the solution 1205932defined by (64) is a peakon-like
solution whose graph is similar to those in Figure 5
Case 3 ((120590 120573) isin 11986016) In this case we have minus984858 lt 119861
2lt 119860 =
0 lt 1198611lt 984858 119861
2= minus1198611 and
1205931015840= minusradicminus
120573
2120593radic
120593 minus 1198612
1
1205932 minus 9848582sgn (120585) (66)
Hence from the separation of variables we have
radic9848582 minus 1205932
120593radic1198612
1minus 1205932
119889120593 = minusradicminus120573
2sgn (120585) 119889120585 (67)
Integrating (67) on the interval [0 120585] (or [120585 0]) leads tothe following implicit formula for the two smooth solitonsolutions
radic9848582 minus 1205932 + radic1198612
1minus 1205932
radic9848582 minus 1198612
1
(
radic(9848582 minus 1198612
1) 1205932
984858radic1198612
1minus 1205932 + 119861
1radic9848582 minus 1205932
)
984858
= exp(minusradicminus120573
2|119909 minus 119888119905|)
(68)
where 120593 isin (119860 1198611] Consider
radic9848582 minus 1198612
2
radic9848582 minus 1205932 + radic1198612
2minus 1205932
(
984858radic1198612
2minus 1205932 minus 119861
2radic9848582 minus 1205932
radic(9848582 minus 1198612
2) 1205932
)
984858
= exp(radicminus120573
2|119909 minus 119888119905|)
(69)
where 120593 isin [1198612 119860)
Case 4 ((120590 120573) isin 11986017) In this case we have minus984858 = 119861
2lt 119860 =
0 lt 1198611= 984858 and
1205931015840= minusradicminus
120573
2120593 sgn (120585) (70)
Choosing 120593(0) = 984858 (or minus984858) as initial value we get
int
120593
984858
1
120593119889120593 = minusint
120585
0
radicminus120573
2sgn (120585) 119889120585 for 119860 lt 120593 le 984858
int
120593
minus984858
1
120593119889120593 = minusint
120585
0
radicminus120573
2sgn (120585) 119889120585 for minus 984858 le 120593 lt 119860
(71)
which immediately yield the peakon solutions
120593 (119909 minus 119888119905) = plusmn984858 exp(minusradicminus120573
2|119909 minus 119888119905|) (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 the119862(3 2 2) equation (10) In order to study kinkwave solutionswe assume that
lim120585rarrinfin
120593 (120585) = 1198601
lim120585rarrminusinfin
120593 (120585) = 1198602
(73)
16 Mathematical Problems in Engineering
2 4
02
04
06
08
10
minus2minus4
120593
x
(a)
0 2
02
00
05
10
minus2
120593
x
t
0
2
minus2
(b)
Figure 8 Two- and three-dimensional graphs of the peakon solution
where 1198601gt 1198602 Substituting the boundary condition (73)
into (14) generates
(1205931015840)2
=
120573 (120593 minus 1198601) (120593 minus 119860
2) [1205932+ (1198601+ 1198602) 120593 + 119860
2
1+ 11986011198602+ 1198602
2+ 2120598120573]
2 (9848582 minus 1205932)fl119865 (120593) (74)
The nonlinear differential equation (74) may sustaindifferent kinds of nonlinear excitations In what follows weconfine our attention to the cases 119860
2= minus119860
1and 120598 =
minus1198602
1120573 which describe kinks and kink compactons which
play an important role in the dynamics systems Under theseconsiderations (74) reduces to
(1205931015840)2
=120573 (120593 minus 119860
1)2
(120593 minus 1198602)2
2 (9848582 minus 1205932) (75)
If 1198601lt 984858 then from the phase analysis in Section 2 (see
Figure 4(10)) we know that (1198601 0) and (119860
2 0) are two saddle
points of (13) and the kink solutions can be obtained fromthe two heteroclinic orbits connecting (120601 119910) = (119860
1 0) and
(1198602 0) When 119860
1increases upon reaching 984858 that is 119860
1= 984858
(75) becomes
(1205931015840)2
= minus120573 (120593 minus 119860
1) (120593 minus 119860
2)
2 (76)
and the ellipse 21199102+120573(120593minus1198601)(120593minus119860
2) = 0 (see Figure 4(11))
which is tangent to the singular lines 120593 = 984858 and 120593 = minus984858 atpoints (119860
1 0) and (119860
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 120593 is a kink wave solutions of (75)
or (76) we have 1205931015840
rarr 0 as 120593 rarr 1198601and as 120593 rarr 119860
2
Moreover we have 119865(120593) ge 0 for 1198602le 120593 le 119860
1and 120593 is
strictly monotonic in any interval where 119865(120593) gt 0 Thus if1205931015840gt 0 at some points 120593will be strictly increasing until it gets
close to the next zero of 119865 Denoting this zero 1198601 we have
120593 uarr 1198601 What will happen to the solution when it approaches
1198601 Depending on whether the zero is double or simple 120593
has a different behavior We explore the two cases in turn
Theorem 11 (i) If 120593 has a simple zero at 120593 = 1198601 so that
119865(1198601) = 0 and 1198651015840(119860
1) lt 0 then the solution 120593 of (75) satisfies
120593 (120585) minus 1198601=1198651015840(1198601)
4(120585 minus 120578)
2
+ 119874 ((120585 minus 120578)4
)
119886119904 120585 uarr 120578
(77)
where 120593(120578) = 1198601
(ii) When 120593 approaches the double zero 1198601of 119865(120593) so that
1198651015840(1198601) = 0 and11986510158401015840(119860
1) gt 0 then the solution120593 of (75) satisfies
120593 (120585) minus 1198601sim 120572 exp(minus120585radic11986510158401015840 (119860
1)) 119886119904 120585 997888rarr infin (78)
for some constant 120572 Thus 120593 uarr 1198601exponentially as 120585 rarr infin
Mathematical Problems in Engineering 17
Proof (i) When 1198601= minus119860
2= 984858 and 120598 = minus119860
2
1120573 from (76)
119865(120593) has a simple zero at 120593 = 1198601 Then
(1205931015840)2
= (120593 minus 1198601) 1198651015840(1198601) + 119874 ((120593 minus 119860
1)2
)
as 120593 uarr 1198601
(79)
Using the fact that (1205931015840)2 ge 0 we know that 1198651015840(1198601) lt 0
Moreover
119889120585
119889120593=
1
radic(120593 minus 1198601) 1198651015840 (119860
1) + 119874 ((120593 minus 119860
1)2
)
(80)
Because
radic(120593 minus 1198601) 1198651015840 (119860
1) + 119874 ((120593 minus 119860
1)2
)
= radic1198601minus 120593(radicminus1198651015840 (119860
1) + 119874 (119860
1minus 120593))
1
radicminus1198651015840 (1198601) + 119874 (119860
1minus 120593)
=1
radicminus1198651015840 (1198601)
+ 119874 (1198601minus 120593)
(81)
we obtain119889120585
119889120593=
1
radicminus (1198601minus 120593) 1198651015840 (119860
1)
+ 119874 ((1198601minus 120593)12
) (82)
Integrating (82) yields
120578 minus 120585 =2
radicminus1198651015840 (1198601)
(1198601minus 120593)12
+ 119874((1198601minus 120593)32
) (83)
where 120578 satisfies 120593(120578) = 1198601 Thus
(120578 minus 120585)2
=4
minus1198651015840 (1198601)(1198601minus 120593) + 119874 ((119860
1minus 120593)2
) (84)
which implies 119874((1198601minus 120593)2) = 119874((120578 minus 120585)
4) Therefore we get
120593 (120585) = 1198601+1198651015840(1198601)
4(120585 minus 120578)
2
+ 119874 ((120585 minus 120578)4
)
as 120585 uarr 120578
(85)
where 120593(120578) = 1198601
(ii) When 1198601= minus119860
2lt 984858 and 120598 = minus119860
2
1120573 from (75) 119865(120593)
has a double zero at 120593 = 1198601 Then
(1205931015840)2
= (120593 minus 1198601)2
11986510158401015840(1198601) + 119874 ((120593 minus 119860
1)3
)
as 120593 uarr 1198601
(86)
Furthermore we get
119889120585
119889120593=
1
radic(120593 minus 1198601)2
11986510158401015840 (1198601) + 119874 ((120593 minus 119860
1)3
)
(87)
Observing that
radic(120593 minus 1198601)2
11986510158401015840 (1198601) + 119874 ((120593 minus 119860
1)3
)
= (1198601minus 120593) (radic11986510158401015840 (119860
1) + 119874 (119860
1minus 120593))
1
radic11986510158401015840 (1198601) + 119874 (119860
1minus 120593)
=1
radic11986510158401015840 (1198601)
+ 119874 (1198601minus 120593)
(88)
we obtain119889120585
119889120593=
1
(1198601minus 120593) 11986510158401015840 (119860
1)+ 119874 (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 119860
1= minus119860
2lt 984858 and 120598 = minus119860
2
1120573 Then (75)
becomes
radic9848582 minus 1205932
1205932 minus 1198602
1
119889120593 = ∓radic120573
2119889120585 (90)
Integrating both sides of (90) gives the following implicitexpressions of kink and antikink wave solutions
arctan120593
radic9848582 minus 1205932
+
radic9848582 minus 1198602
1
1198601
arctanhradic9848582 minus 119860
2
1
1198601
120593
radic9848582 minus 1205932
= plusmnradic120573
2120585
(91)
By letting 1198601rarr 984858 in (91) we get two kink compactons
which are given by
plusmn120593 =
984858 sinradic120573
2120585
10038161003816100381610038161205851003816100381610038161003816 le
120587
radic2120573
984858 120585 gt120587
radic2120573
minus984858 120585 lt minus120587
radic2120573
(92)
The graphs for the kink wave solutions (91) and kinkcompacton solutions (92) are shown in Figures 9 and 10respectively
Remark 12 The two kink compacton solutions (92) aredifferent from the well-known smooth kink wave solutionsIn comparison with kink wave solutions (91) the kink com-pacton solutions (92) have no exponential decay propertiesbut have compact supportThat is they minus a constant thedifferences identically vanish outside a finite core region
18 Mathematical Problems in Engineering
120593
120585minus1 1
05
minus05
(a)
minus1minus1
minus05
05
00120593
x t
0
1
01
11
(b)
Figure 9 Two- and three-dimensional graphs of the kink wave solution
1
1120593
120585
minus1
minus1
(a)
0
1
0
1
000510
120593
minus1 minus1
minus05
minus10
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 the119862(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 Newtonrsquos equation which we write in the form 119910
2=
119865(120593) where 119910 = 1205931015840
120585(120585) we solve Newtonrsquos equation 119910
2=
119865(120593) 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 the119862(3 2 2)equation
Actually for 119898 = 2119896 + 1 119896 isin N+ in 119862(119898 2 2) equation(9) the dynamical behavior of traveling wave solutions of(9) is similar to the case 119898 = 3 for 119898 = 2119896 119896 isin N+ in119862(119898 2 2) equation (9) the dynamical behavior of travelingwave solutions of (9) is similar to the case 119898 = 2 We areapplying the approach mentioned in this work to 119862(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
References
[1] C Rogers and W R Shadwick Backlund Transformation andTheir Applications Academic Press New York NY USA 1982
[2] W Malfliet and W Hereman ldquoThe tanh method I Exactsolutions of nonlinear evolution and wave equationsrdquo PhysicaScripta vol 54 no 6 pp 563ndash568 1996
[3] R Hirota Direct Method in Soliton Theory Springer BerlinGermany 1980
[4] J S Cohen Computer Algebra and Symbolic ComputationMathematical Methods A K Peters 2003
[5] S Lie ldquoUber die integration durch bestimmte integrale voneiner klasse linearer partieller differential eichungenrdquoArchiv derMathematik vol 6 pp 328ndash368 1881
[6] H-H Dai and X-H Zhao ldquoTraveling waves in a rod composedof amodifiedMooney-Rivlinmaterial I bifurcation of equilibraand non-singular caserdquo Proceedings of the Royal Society AMathematical Physical and Engineering Sciences vol 455 no1990 pp 3845ndash3874 1999
[7] H-H Dai and F-F Wang ldquoAsymptotic bifurcation solutionsfor compressions of a clamped nonlinearly elastic rectangletransition region and barrelling to a corner-like profilerdquo SIAMJournal on Applied Mathematics vol 70 no 7 pp 2673ndash26922010
[8] J Li and Z Liu ldquoSmooth and non-smooth traveling waves ina nonlinearly dispersive equationrdquo Applied Mathematical Mod-elling vol 25 no 1 pp 41ndash56 2000
[9] A Chen and J Li ldquoSingle peak solitary wave solutions forthe osmosis K(22) equation under inhomogeneous boundaryconditionrdquo Journal of Mathematical Analysis and Applicationsvol 369 no 2 pp 758ndash766 2010
[10] R Camassa and D D Holm ldquoAn integrable shallow waterequation with peaked solitonsrdquo Physical Review Letters vol 71no 11 pp 1661ndash1664 1993
[11] P Rosenau ldquoOn nonanalytic solitary waves formed by a nonlin-ear dispersionrdquo Physics Letters A vol 230 no 5-6 pp 305ndash3181997
[12] P Rosenau and J M Hyman ldquoCompactons solitons with finitewavelengthrdquo Physical Review Letters vol 70 no 5 pp 564ndash5671993
[13] L Zhang L-Q Chen and X Huo ldquoThe effects of horizontalsingular straight line in a generalized nonlinear Klein-Gordonmodel equationrdquo Nonlinear Dynamics vol 72 no 4 pp 789ndash801 2013
[14] J Li and H Dai On the Study of Singular Nonlinear TravelingWave Equations Dynamical System Approach Science PressBeijing China 2007
[15] A Chen S Wen S Tang W Huang and Z Qiao ldquoEffects ofquadratic singular curves in integrable equationsrdquo Studies inApplied Mathematics vol 134 no 1 pp 24ndash61 2015
[16] L Tian and J Yin ldquoNew compacton solutions and solitary wavesolutions of fully nonlinear generalized Camassa-Holm equa-tionsrdquo Chaos Solitons amp Fractals vol 20 no 2 pp 289ndash2992004
[17] L Zhang ldquoThe bifurcation and peakons for the specialC(3 2 2)equationrdquo Pramana vol 83 no 3 pp 331ndash340 2014
[18] J Lenells ldquoTraveling wave solutions of the CamassandashHolmequationrdquo Journal of Differential Equations vol 217 no 2 pp393ndash430 2005
[19] Z J Qiao and G P Zhang ldquoOn peaked and smooth solitons forthe Camassa-Holm equationrdquo Europhysics Letters vol 73 no 5pp 657ndash663 2006
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Mathematical Problems in Engineering
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Stochastic AnalysisInternational Journal of
Mathematical Problems in Engineering 11
(ii) If 120593(0) = 984858 = 1198611 then 120593 is a cuspon solution and 120593 has
the following asymptotic behavior
120593 (120585) minus 984858 = 1205821
10038161003816100381610038161205851003816100381610038161003816
23
+ 119874(10038161003816100381610038161205851003816100381610038161003816
43
) 120585 997888rarr 0
1205931015840(120585) =
2
31205821
10038161003816100381610038161205851003816100381610038161003816
minus13 sgn (120585) + 119874 (10038161003816100381610038161205851003816100381610038161003816
13
)
120585 997888rarr 0
(27)
where 1205821= plusmn(9|120573984858
2+ 2119860120573984858 + 3119860
2120573 + 2120598|(984858 minus 119860)
216984858)13
Thus 120593(120585) notin 1198671
119897119900119888(R)
(iii) If 120593(0) = minus984858 = 1198612 then 120593 is a cuspon solution and 120593
has the following asymptotic behavior
120593 (120585) + 984858 = 1205822
10038161003816100381610038161205851003816100381610038161003816
23
+ 119874(10038161003816100381610038161205851003816100381610038161003816
43
) 120585 997888rarr 0
1205931015840(120585) =
2
31205822
10038161003816100381610038161205851003816100381610038161003816
minus13 sgn (120585) + 119874 (10038161003816100381610038161205851003816100381610038161003816
13
)
120585 997888rarr 0
(28)
where 1205822= plusmn(9|120573984858
2minus 2119860120573984858 + 3119860
2120573 + 2120598|(984858 + 119860)
216984858)13
Thus 120593(120585) notin 1198671
119897119900119888(R)
(iv) If 120593(0) = 984858 = 1198611and 119860 = 0 then 120593 is a peakon-like
solution and
120593 (120585) minus 984858 = 1205823
10038161003816100381610038161205851003816100381610038161003816 + 119874 (
10038161003816100381610038161205851003816100381610038161003816
2
) 120585 997888rarr 0
1205931015840(120585) = 120582
3sgn (120585) + 119874 (
10038161003816100381610038161205851003816100381610038161003816) 120585 997888rarr 0
(29)
where 1205823= plusmn|119860 minus 984858|radic120573(119861
2minus 984858)4984858
(v) If 120593(0) = minus984858 = 1198612and 119860 = 0 then 120593 is a peakon-like
solution and
120593 (120585) + 984858 = 1205824
10038161003816100381610038161205851003816100381610038161003816 + 119874 (
10038161003816100381610038161205851003816100381610038161003816
2
) 120585 997888rarr 0
1205931015840(120585) = 120582
4sgn (120585) + 119874 (
10038161003816100381610038161205851003816100381610038161003816) 120585 997888rarr 0
(30)
where 1205824= plusmn|119860 + 984858|radicminus120573(119861
1+ 984858)4984858
(vi) If 120593(0) = 984858 = 1198611and 119860 = 0 then 120593 gives the peakon
solution 984858 exp(minusradicminus1205732|119909 minus 119888119905|)(vii) If 120593(0) = minus984858 = 119861
2and 119860 = 0 then 120593 gives the peakon
solution minus984858 exp(minusradicminus1205732|119909 minus 119888119905|)
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 |120593(0)| = 984858 then 120593 isin 119862
infin(R) and 120593 is a smooth soliton
solution(ii) If 120593(0) = 984858 = 119861
1 then by the definition of single peak
soliton we have 119860 = 984858 thus 1205731205932 + 2119860120573120593 + 31198602120573 + 2120598 does
not contain the factor 120593 minus 984858 From (24) we obtain
1205931015840= sgn (119860 minus 984858)
1003816100381610038161003816120593 minus 1198601003816100381610038161003816
radic210038161003816100381610038161205932 minus 9848582
1003816100381610038161003816
sdot radic10038161003816100381610038161205731205932 + 2119860120573120593 + 31198602120573 + 2120598
1003816100381610038161003816 sgn (120585)
(31)
Let 1198971(120593) = radic2(120593 + 984858)|120593 minus 119860|radic|1205731205932 + 2119860120573120593 + 31198602120573 + 2120598|
then 1198971(984858) = 2radic984858|984858 minus 119860|radic|1205739848582 + 2119860120573984858 + 31198602120573 + 2120598| and
int 1198971(120593)radic
1003816100381610038161003816120593 minus 9848581003816100381610038161003816119889120593 = int sgn (119860 minus 984858) sgn (120585) 119889120585 (32)
Inserting 1198971(120593) = 119897
1(984858) + 119874(|120593 minus 984858|) into (32) and using the
initial condition 120593(0) = 984858 we obtain
21198971(984858)
3
1003816100381610038161003816120593 minus 9848581003816100381610038161003816
32
(1 + 119874 (1003816100381610038161003816120593 minus 984858
1003816100381610038161003816)) =10038161003816100381610038161205851003816100381610038161003816
(33)
thus
120593 minus 984858 = plusmn(3
21198971(984858)
)
23
10038161003816100381610038161205851003816100381610038161003816
23
(1 + 119874 (1003816100381610038161003816120593 minus 984858
1003816100381610038161003816))minus23
= plusmn(3
21198971(984858)
)
23
10038161003816100381610038161205851003816100381610038161003816
23
(1 + 119874 (1003816100381610038161003816120593 minus 984858
1003816100381610038161003816))
(34)
which implies 120593 minus 984858 = 119874(|120585|23
) Therefore we have
120593 (120585) = 984858 plusmn (3
21198971(984858)
)
23
10038161003816100381610038161205851003816100381610038161003816
23
+ 119874(10038161003816100381610038161205851003816100381610038161003816
43
)
= 984858 + 1205821
10038161003816100381610038161205851003816100381610038161003816
23
+ 119874(10038161003816100381610038161205851003816100381610038161003816
43
) 120585 997888rarr 0
1205821= plusmn(
3
21198971(984858)
)
23
= plusmn(
9100381610038161003816100381610038161205739848582+ 2119860120573984858 + 3119860
2120573 + 2120598
10038161003816100381610038161003816(984858 minus 119860)
2
16984858)
13
1205931015840(120585) =
2
31205821
10038161003816100381610038161205851003816100381610038161003816
minus13 sgn (120585) + 119874 (10038161003816100381610038161205851003816100381610038161003816
13
) 120585 997888rarr 0
(35)
So 120593(120585) notin 1198671
loc(R)(iii) Similar to the proof of (ii) we ignore it in this paper(iv) If 120593(0) = 984858 = 119861
1and 119860 = 0 then from (25) we obtain
1205931015840= radicminus
120573
2sgn (119860 minus 984858)
1003816100381610038161003816120593 minus 1198601003816100381610038161003816radic120593 minus 1198612
120593 + 984858sgn (120585) (36)
Let 1198972(120593) = (1|120593minus119860|)radic(120593 + 984858)(120593 minus 119861
2) then 119897
2(984858) = (1|984858minus
119860|)radic2984858(984858 minus 1198612) and
int 1198972(120593) 119889120593 = radicminus
120573
2int sgn (119860 minus 984858) sgn (120585) 119889120585 (37)
Inserting 1198972(120593) = 119897
2(984858) + 119874(|120593 minus 984858|) into (37) and using the
initial condition 120593(0) = 984858 we obtain
1198972(984858) (120593 minus 984858) (1 + 119874 (
1003816100381610038161003816120593 minus 9848581003816100381610038161003816))minus1
= sgn (119860 minus 984858)10038161003816100381610038161205851003816100381610038161003816
(38)
Sincesgn (120593 minus 984858) sgn (119860 minus 984858) ge 0
1
1 + 119874 (120593 minus 984858)= 1 + 119874 (120593 minus 984858)
(39)
12 Mathematical Problems in Engineering
we get
1003816100381610038161003816120593 minus 9848581003816100381610038161003816 =
radicminus120573
2
1
1198972(984858)
10038161003816100381610038161205851003816100381610038161003816 (1 + 119874 (120593 minus 984858)) (40)
which implies |120593 minus 984858| = 119874(|120585|) Therefore we have
120593 (120585) = 984858 + 1205823
10038161003816100381610038161205851003816100381610038161003816 + 119874 (
10038161003816100381610038161205851003816100381610038161003816
2
) 120585 997888rarr 0
1205931015840(120585) = 120582
3sgn (120585) + 119874 (
10038161003816100381610038161205851003816100381610038161003816) 120585 997888rarr 0
(41)
where 1205823= plusmn|119860 minus 984858|radic120573(119861
2minus 984858)4984858
(v) Similar to the proof of (iv) we ignore it in this paper
By virtue of Theorem 9 any single peak soliton for the119862(3 2 2) equation (10) must satisfy the following initial andboundary values problem
(1205931015840)2
=120573 (120593 minus 119860)
2
(120593 minus 1198611) (120593 minus 119861
2)
2 (9848582 minus 1205932)fl119871 (120593)
120593 (0) isin 984858 minus984858 1198611 1198612
lim|120585|rarrinfin
120593 (120585) = 119860
(42)
119871(120593) ge 0 and the boundary condition (24) imply thefollowing
(a) If 120573(9848582 minus 1205932) ge 0 then 120593 ge max119861
1 1198612 or 120593 le
min1198611 1198612
(b) If120573(9848582minus1205932) le 0 thenmin1198611 1198612 le 120593 le max119861
1 1198612
Below we will present some implicit formulas for thesingle peak soliton solutions in the case of specific 120590 and 120573
Case 1 ((120590 120573) isin 11986012) In this case we have minus984858 lt 119861
2lt 119860 lt
1198611= 984858 From the standard phase analysis and Theorem 9
we know that if 120593 is a single peak soliton of the 119862(3 2 2)
equation then
1205931015840= minusradicminus
120573
2(120593 minus 119860)radic
120593 minus 1198612
120593 + 984858sgn (120585) (43)
From the separation of variables we get
intℎ (120593) 119889120593 = radicminus120573
2sgn (120585) 119889120585 (44)
where ℎ(120593) = (1(119860 minus 120593))radic(120593 + 984858)(120593 minus 1198612) After a lengthy
calculation of integral we obtain the implicit solution 120593
defined by
119867(120593)flradic119860 + 984858
119860 minus 1198612
1198681(120593) minus 119868
2(120593) = radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870 (45)
where
1198681(120593) = ln
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
(119860 + 984858) (120593 minus 1198612) + (119860 minus 119861
2) (120593 + 984858) + 2radic(119860 + 984858) (119860 minus 119861
2) (120593 minus 119861
2) (120593 + 984858)
radic(119860 + 984858) (119860 minus 1198612) (119860 + 984858) (119860 minus 120593)
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
1198682(120593) = ln
1003816100381610038161003816100381610038161003816(120593 + 984858) + (120593 minus 119861
2) + 2radic(120593 + 984858) (120593 minus 119861
2)
1003816100381610038161003816100381610038161003816
(46)
and 119870 is an arbitrary integration constant For 120593(0) = 984858 theconstant119870 = 119867(120593(0)) is defined by
119870 = radic119860 + 984858
119860 minus 1198612
1198681(984858) minus 119868
2(984858) (47)
and for 120593(0) = 1198612
119870 = radic119860 + 984858
119860 minus 1198612
1198681(1198612) minus 1198682(1198612) (48)
(i) If 120593(0) = 984858 then 119860 lt 120593 le 984858 Since 1198671015840(120593) = ℎ(120593) weknow that119867(120593) strictly decreases on the interval (119860 984858] thus1198671(120593) = 119867|
(119860984858](120593) gives a single peak solitonwith119867
1(984858) = 119870
and1198671(119860+) = infin Therefore 120593
1(120585) = 119867
minus1
1(radicminus1205732|120585| + 119870) is
the solution satisfying
1205931(0) = 984858
lim|120585|rarrinfin
1205931(120585) = 119860
1205931015840
1(0plusmn) = plusmn (119860 minus 984858)radic
120573 (1198612minus 984858)
4984858
(49)
So 1205931(120585) is a peakon-like solution (see Figure 5)
(ii) If 120593(0) = 1198612 then 119861
2le 120593 lt 119860 By 119867
1015840(120593) = 119891(120593)
we know that 119867(120593) strictly increases on the interval [1198612 119860)
Thus
1198672(120593) = 119867|
[1198612 119860)(120593) (50)
Mathematical Problems in Engineering 13
minus3 minus2 minus1
120593
minus02
1 2 3
02
04
06
08
10
x
(a)
02
02
minus2 minus2 x
10
05
00
t
120593
22
(b)
Figure 5 Two- and three-dimensional graphs of the peakon-like solution
minus03
minus04
minus05
1 2 3minus3 minus2 minus1
120593
minus02
x
(a)
0
2
0
2
minus2minus2
minus02
minus03
minus04
minus05
minus06
t x
120593
0
2
0
2
minus2minus2
t x
(b)
Figure 6 Two- and three-dimensional graphs of the smooth soliton solution
has the inverse denoted by 1205932(120585) = 119867
minus1
2(radicminus1205732|120585| + 119870)
1205932(120585) gives a kind of smooth soliton solution (see Figure 6)
satisfying
1205932(0) = 119861
2
lim|120585|rarrinfin
1205932(120585) = 119860
1205931015840
2(0) = 0
(51)
Case 2 ((120590 120573) isin 11986014) In this case we have minus984858 = 119861
2lt 119860 lt
984858 lt 1198611and (25) is equivalent to
1205931015840= minusradicminus
120573
2(120593 minus 119860)radic
120593 minus 1198611
120593 minus 984858sgn (120585) (52)
Let
119892 (120593) =1
119860 minus 120593radic120593 minus 1198611
120593 minus 984858 (53)
14 Mathematical Problems in Engineering
minus1 1 2 3
02
04
06
08
10
minus3 minus2
120593
x
(a)
02
02minus2
minus2
120593
x t
00
05
10
(b)
Figure 7 Two- and three-dimensional graphs of the cuspon solution
then (52) is converted to
119892 (120593) 119889120593 =1
119860 minus 120593radic120593 minus 1198611
120593 minus 984858119889120593 = radicminus
120573
2sgn (120585) 119889120585 (54)
Integrating (54) on the interval [0 120585] (or [120585 0]) leads to thefollowing implicit solutions
119866 (120593)flradic119860 minus 984858
119860 minus 1198611
1198683(120593) + 119868
4(120593) = radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870 (55)
where
1198683(120593) = ln
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
(119860 minus 984858) (120593 minus 1198611) + (119860 minus 119861
1) (120593 minus 984858) + 2radic(119860 minus 984858) (119860 minus 119861
1) (120593 minus 984858) (120593 minus 119861
1)
radic(119860 minus 984858) (119860 minus 1198611) (119860 minus 984858) (120593 minus 119860)
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
1198684(120593) = ln
1003816100381610038161003816100381610038161003816(120593 minus 984858) + (120593 minus 119861
1) + 2radic(120593 minus 984858) (120593 minus 119861
1)
1003816100381610038161003816100381610038161003816
(56)
And119870 is an arbitrary integration constant It is obvious thatfor 120593(0) = 984858 the constant119870 = 119867(120593(0)) is defined by
119870 = radic119860 minus 984858
119860 minus 1198611
1198683(984858) + 119868
4(984858) (57)
and for 120593(0) = minus984858
119870 = radic119860 minus 984858
119860 minus 1198611
1198683(minus984858) + 119868
4(minus984858) (58)
(i) If 120593(0) = 984858 then 119860 lt 120593 le 984858 From 119892(120593) lt 0 weknow that 119866(120593) strictly decreases on the interval (119860 984858] with119866(984858) = 119870 and 119866(119860+) = infin Define
1198661(120593) = 119866|
(119860984858](120593) (59)
Since 1198661(120593) is a strictly decreasing function from (119860 984858] onto
[119870infin) we can solve for 120593 uniquely from (59) and obtain
1205931(120585) = 119866
minus1
1(radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870) (60)
It is easy to check that 120593 satisfies
1205931(0) = 984858
lim|120585|rarrinfin
1205931(120585) = 119860
1205931015840
1(0plusmn) = ∓infin
(61)
Therefore the solution1205931defined by (60) is a cuspon solution
for the 119862(3 2 2) equation (see Figure 7)
Mathematical Problems in Engineering 15
(ii) If 120593(0) = minus984858 then minus984858 le 120593 lt 119860 Through a similaranalysis we get a strictly increasing function 119866(120593) on theinterval [minus984858 119860) satisfying
119866 (120593) = radicminus120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870 (62)
where 119866(120593) is defined by (55) Let
1198662(120593) = 119866|
[minus984858119860)(120593) (63)
then 1198662(120593) is a strictly increasing function from [minus984858 119860) onto
[119870infin) so that we can solve for 120593 and obtain
1205932(120585) = 119866
minus1
2(radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870) (64)
It is easy to check that 1205932satisfies
1205932(0) = minus984858
lim|120585|rarrinfin
1205932(120585) = 119860
1205931015840
2(0plusmn) = plusmn (119860 + 984858)radicminus
120573 (1198611+ 984858)
4984858
(65)
Therefore the solution 1205932defined by (64) is a peakon-like
solution whose graph is similar to those in Figure 5
Case 3 ((120590 120573) isin 11986016) In this case we have minus984858 lt 119861
2lt 119860 =
0 lt 1198611lt 984858 119861
2= minus1198611 and
1205931015840= minusradicminus
120573
2120593radic
120593 minus 1198612
1
1205932 minus 9848582sgn (120585) (66)
Hence from the separation of variables we have
radic9848582 minus 1205932
120593radic1198612
1minus 1205932
119889120593 = minusradicminus120573
2sgn (120585) 119889120585 (67)
Integrating (67) on the interval [0 120585] (or [120585 0]) leads tothe following implicit formula for the two smooth solitonsolutions
radic9848582 minus 1205932 + radic1198612
1minus 1205932
radic9848582 minus 1198612
1
(
radic(9848582 minus 1198612
1) 1205932
984858radic1198612
1minus 1205932 + 119861
1radic9848582 minus 1205932
)
984858
= exp(minusradicminus120573
2|119909 minus 119888119905|)
(68)
where 120593 isin (119860 1198611] Consider
radic9848582 minus 1198612
2
radic9848582 minus 1205932 + radic1198612
2minus 1205932
(
984858radic1198612
2minus 1205932 minus 119861
2radic9848582 minus 1205932
radic(9848582 minus 1198612
2) 1205932
)
984858
= exp(radicminus120573
2|119909 minus 119888119905|)
(69)
where 120593 isin [1198612 119860)
Case 4 ((120590 120573) isin 11986017) In this case we have minus984858 = 119861
2lt 119860 =
0 lt 1198611= 984858 and
1205931015840= minusradicminus
120573
2120593 sgn (120585) (70)
Choosing 120593(0) = 984858 (or minus984858) as initial value we get
int
120593
984858
1
120593119889120593 = minusint
120585
0
radicminus120573
2sgn (120585) 119889120585 for 119860 lt 120593 le 984858
int
120593
minus984858
1
120593119889120593 = minusint
120585
0
radicminus120573
2sgn (120585) 119889120585 for minus 984858 le 120593 lt 119860
(71)
which immediately yield the peakon solutions
120593 (119909 minus 119888119905) = plusmn984858 exp(minusradicminus120573
2|119909 minus 119888119905|) (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 the119862(3 2 2) equation (10) In order to study kinkwave solutionswe assume that
lim120585rarrinfin
120593 (120585) = 1198601
lim120585rarrminusinfin
120593 (120585) = 1198602
(73)
16 Mathematical Problems in Engineering
2 4
02
04
06
08
10
minus2minus4
120593
x
(a)
0 2
02
00
05
10
minus2
120593
x
t
0
2
minus2
(b)
Figure 8 Two- and three-dimensional graphs of the peakon solution
where 1198601gt 1198602 Substituting the boundary condition (73)
into (14) generates
(1205931015840)2
=
120573 (120593 minus 1198601) (120593 minus 119860
2) [1205932+ (1198601+ 1198602) 120593 + 119860
2
1+ 11986011198602+ 1198602
2+ 2120598120573]
2 (9848582 minus 1205932)fl119865 (120593) (74)
The nonlinear differential equation (74) may sustaindifferent kinds of nonlinear excitations In what follows weconfine our attention to the cases 119860
2= minus119860
1and 120598 =
minus1198602
1120573 which describe kinks and kink compactons which
play an important role in the dynamics systems Under theseconsiderations (74) reduces to
(1205931015840)2
=120573 (120593 minus 119860
1)2
(120593 minus 1198602)2
2 (9848582 minus 1205932) (75)
If 1198601lt 984858 then from the phase analysis in Section 2 (see
Figure 4(10)) we know that (1198601 0) and (119860
2 0) are two saddle
points of (13) and the kink solutions can be obtained fromthe two heteroclinic orbits connecting (120601 119910) = (119860
1 0) and
(1198602 0) When 119860
1increases upon reaching 984858 that is 119860
1= 984858
(75) becomes
(1205931015840)2
= minus120573 (120593 minus 119860
1) (120593 minus 119860
2)
2 (76)
and the ellipse 21199102+120573(120593minus1198601)(120593minus119860
2) = 0 (see Figure 4(11))
which is tangent to the singular lines 120593 = 984858 and 120593 = minus984858 atpoints (119860
1 0) and (119860
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 120593 is a kink wave solutions of (75)
or (76) we have 1205931015840
rarr 0 as 120593 rarr 1198601and as 120593 rarr 119860
2
Moreover we have 119865(120593) ge 0 for 1198602le 120593 le 119860
1and 120593 is
strictly monotonic in any interval where 119865(120593) gt 0 Thus if1205931015840gt 0 at some points 120593will be strictly increasing until it gets
close to the next zero of 119865 Denoting this zero 1198601 we have
120593 uarr 1198601 What will happen to the solution when it approaches
1198601 Depending on whether the zero is double or simple 120593
has a different behavior We explore the two cases in turn
Theorem 11 (i) If 120593 has a simple zero at 120593 = 1198601 so that
119865(1198601) = 0 and 1198651015840(119860
1) lt 0 then the solution 120593 of (75) satisfies
120593 (120585) minus 1198601=1198651015840(1198601)
4(120585 minus 120578)
2
+ 119874 ((120585 minus 120578)4
)
119886119904 120585 uarr 120578
(77)
where 120593(120578) = 1198601
(ii) When 120593 approaches the double zero 1198601of 119865(120593) so that
1198651015840(1198601) = 0 and11986510158401015840(119860
1) gt 0 then the solution120593 of (75) satisfies
120593 (120585) minus 1198601sim 120572 exp(minus120585radic11986510158401015840 (119860
1)) 119886119904 120585 997888rarr infin (78)
for some constant 120572 Thus 120593 uarr 1198601exponentially as 120585 rarr infin
Mathematical Problems in Engineering 17
Proof (i) When 1198601= minus119860
2= 984858 and 120598 = minus119860
2
1120573 from (76)
119865(120593) has a simple zero at 120593 = 1198601 Then
(1205931015840)2
= (120593 minus 1198601) 1198651015840(1198601) + 119874 ((120593 minus 119860
1)2
)
as 120593 uarr 1198601
(79)
Using the fact that (1205931015840)2 ge 0 we know that 1198651015840(1198601) lt 0
Moreover
119889120585
119889120593=
1
radic(120593 minus 1198601) 1198651015840 (119860
1) + 119874 ((120593 minus 119860
1)2
)
(80)
Because
radic(120593 minus 1198601) 1198651015840 (119860
1) + 119874 ((120593 minus 119860
1)2
)
= radic1198601minus 120593(radicminus1198651015840 (119860
1) + 119874 (119860
1minus 120593))
1
radicminus1198651015840 (1198601) + 119874 (119860
1minus 120593)
=1
radicminus1198651015840 (1198601)
+ 119874 (1198601minus 120593)
(81)
we obtain119889120585
119889120593=
1
radicminus (1198601minus 120593) 1198651015840 (119860
1)
+ 119874 ((1198601minus 120593)12
) (82)
Integrating (82) yields
120578 minus 120585 =2
radicminus1198651015840 (1198601)
(1198601minus 120593)12
+ 119874((1198601minus 120593)32
) (83)
where 120578 satisfies 120593(120578) = 1198601 Thus
(120578 minus 120585)2
=4
minus1198651015840 (1198601)(1198601minus 120593) + 119874 ((119860
1minus 120593)2
) (84)
which implies 119874((1198601minus 120593)2) = 119874((120578 minus 120585)
4) Therefore we get
120593 (120585) = 1198601+1198651015840(1198601)
4(120585 minus 120578)
2
+ 119874 ((120585 minus 120578)4
)
as 120585 uarr 120578
(85)
where 120593(120578) = 1198601
(ii) When 1198601= minus119860
2lt 984858 and 120598 = minus119860
2
1120573 from (75) 119865(120593)
has a double zero at 120593 = 1198601 Then
(1205931015840)2
= (120593 minus 1198601)2
11986510158401015840(1198601) + 119874 ((120593 minus 119860
1)3
)
as 120593 uarr 1198601
(86)
Furthermore we get
119889120585
119889120593=
1
radic(120593 minus 1198601)2
11986510158401015840 (1198601) + 119874 ((120593 minus 119860
1)3
)
(87)
Observing that
radic(120593 minus 1198601)2
11986510158401015840 (1198601) + 119874 ((120593 minus 119860
1)3
)
= (1198601minus 120593) (radic11986510158401015840 (119860
1) + 119874 (119860
1minus 120593))
1
radic11986510158401015840 (1198601) + 119874 (119860
1minus 120593)
=1
radic11986510158401015840 (1198601)
+ 119874 (1198601minus 120593)
(88)
we obtain119889120585
119889120593=
1
(1198601minus 120593) 11986510158401015840 (119860
1)+ 119874 (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 119860
1= minus119860
2lt 984858 and 120598 = minus119860
2
1120573 Then (75)
becomes
radic9848582 minus 1205932
1205932 minus 1198602
1
119889120593 = ∓radic120573
2119889120585 (90)
Integrating both sides of (90) gives the following implicitexpressions of kink and antikink wave solutions
arctan120593
radic9848582 minus 1205932
+
radic9848582 minus 1198602
1
1198601
arctanhradic9848582 minus 119860
2
1
1198601
120593
radic9848582 minus 1205932
= plusmnradic120573
2120585
(91)
By letting 1198601rarr 984858 in (91) we get two kink compactons
which are given by
plusmn120593 =
984858 sinradic120573
2120585
10038161003816100381610038161205851003816100381610038161003816 le
120587
radic2120573
984858 120585 gt120587
radic2120573
minus984858 120585 lt minus120587
radic2120573
(92)
The graphs for the kink wave solutions (91) and kinkcompacton solutions (92) are shown in Figures 9 and 10respectively
Remark 12 The two kink compacton solutions (92) aredifferent from the well-known smooth kink wave solutionsIn comparison with kink wave solutions (91) the kink com-pacton solutions (92) have no exponential decay propertiesbut have compact supportThat is they minus a constant thedifferences identically vanish outside a finite core region
18 Mathematical Problems in Engineering
120593
120585minus1 1
05
minus05
(a)
minus1minus1
minus05
05
00120593
x t
0
1
01
11
(b)
Figure 9 Two- and three-dimensional graphs of the kink wave solution
1
1120593
120585
minus1
minus1
(a)
0
1
0
1
000510
120593
minus1 minus1
minus05
minus10
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 the119862(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 Newtonrsquos equation which we write in the form 119910
2=
119865(120593) where 119910 = 1205931015840
120585(120585) we solve Newtonrsquos equation 119910
2=
119865(120593) 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 the119862(3 2 2)equation
Actually for 119898 = 2119896 + 1 119896 isin N+ in 119862(119898 2 2) equation(9) the dynamical behavior of traveling wave solutions of(9) is similar to the case 119898 = 3 for 119898 = 2119896 119896 isin N+ in119862(119898 2 2) equation (9) the dynamical behavior of travelingwave solutions of (9) is similar to the case 119898 = 2 We areapplying the approach mentioned in this work to 119862(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
References
[1] C Rogers and W R Shadwick Backlund Transformation andTheir Applications Academic Press New York NY USA 1982
[2] W Malfliet and W Hereman ldquoThe tanh method I Exactsolutions of nonlinear evolution and wave equationsrdquo PhysicaScripta vol 54 no 6 pp 563ndash568 1996
[3] R Hirota Direct Method in Soliton Theory Springer BerlinGermany 1980
[4] J S Cohen Computer Algebra and Symbolic ComputationMathematical Methods A K Peters 2003
[5] S Lie ldquoUber die integration durch bestimmte integrale voneiner klasse linearer partieller differential eichungenrdquoArchiv derMathematik vol 6 pp 328ndash368 1881
[6] H-H Dai and X-H Zhao ldquoTraveling waves in a rod composedof amodifiedMooney-Rivlinmaterial I bifurcation of equilibraand non-singular caserdquo Proceedings of the Royal Society AMathematical Physical and Engineering Sciences vol 455 no1990 pp 3845ndash3874 1999
[7] H-H Dai and F-F Wang ldquoAsymptotic bifurcation solutionsfor compressions of a clamped nonlinearly elastic rectangletransition region and barrelling to a corner-like profilerdquo SIAMJournal on Applied Mathematics vol 70 no 7 pp 2673ndash26922010
[8] J Li and Z Liu ldquoSmooth and non-smooth traveling waves ina nonlinearly dispersive equationrdquo Applied Mathematical Mod-elling vol 25 no 1 pp 41ndash56 2000
[9] A Chen and J Li ldquoSingle peak solitary wave solutions forthe osmosis K(22) equation under inhomogeneous boundaryconditionrdquo Journal of Mathematical Analysis and Applicationsvol 369 no 2 pp 758ndash766 2010
[10] R Camassa and D D Holm ldquoAn integrable shallow waterequation with peaked solitonsrdquo Physical Review Letters vol 71no 11 pp 1661ndash1664 1993
[11] P Rosenau ldquoOn nonanalytic solitary waves formed by a nonlin-ear dispersionrdquo Physics Letters A vol 230 no 5-6 pp 305ndash3181997
[12] P Rosenau and J M Hyman ldquoCompactons solitons with finitewavelengthrdquo Physical Review Letters vol 70 no 5 pp 564ndash5671993
[13] L Zhang L-Q Chen and X Huo ldquoThe effects of horizontalsingular straight line in a generalized nonlinear Klein-Gordonmodel equationrdquo Nonlinear Dynamics vol 72 no 4 pp 789ndash801 2013
[14] J Li and H Dai On the Study of Singular Nonlinear TravelingWave Equations Dynamical System Approach Science PressBeijing China 2007
[15] A Chen S Wen S Tang W Huang and Z Qiao ldquoEffects ofquadratic singular curves in integrable equationsrdquo Studies inApplied Mathematics vol 134 no 1 pp 24ndash61 2015
[16] L Tian and J Yin ldquoNew compacton solutions and solitary wavesolutions of fully nonlinear generalized Camassa-Holm equa-tionsrdquo Chaos Solitons amp Fractals vol 20 no 2 pp 289ndash2992004
[17] L Zhang ldquoThe bifurcation and peakons for the specialC(3 2 2)equationrdquo Pramana vol 83 no 3 pp 331ndash340 2014
[18] J Lenells ldquoTraveling wave solutions of the CamassandashHolmequationrdquo Journal of Differential Equations vol 217 no 2 pp393ndash430 2005
[19] Z J Qiao and G P Zhang ldquoOn peaked and smooth solitons forthe Camassa-Holm equationrdquo Europhysics Letters vol 73 no 5pp 657ndash663 2006
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Mathematical Problems in Engineering
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Stochastic AnalysisInternational Journal of
12 Mathematical Problems in Engineering
we get
1003816100381610038161003816120593 minus 9848581003816100381610038161003816 =
radicminus120573
2
1
1198972(984858)
10038161003816100381610038161205851003816100381610038161003816 (1 + 119874 (120593 minus 984858)) (40)
which implies |120593 minus 984858| = 119874(|120585|) Therefore we have
120593 (120585) = 984858 + 1205823
10038161003816100381610038161205851003816100381610038161003816 + 119874 (
10038161003816100381610038161205851003816100381610038161003816
2
) 120585 997888rarr 0
1205931015840(120585) = 120582
3sgn (120585) + 119874 (
10038161003816100381610038161205851003816100381610038161003816) 120585 997888rarr 0
(41)
where 1205823= plusmn|119860 minus 984858|radic120573(119861
2minus 984858)4984858
(v) Similar to the proof of (iv) we ignore it in this paper
By virtue of Theorem 9 any single peak soliton for the119862(3 2 2) equation (10) must satisfy the following initial andboundary values problem
(1205931015840)2
=120573 (120593 minus 119860)
2
(120593 minus 1198611) (120593 minus 119861
2)
2 (9848582 minus 1205932)fl119871 (120593)
120593 (0) isin 984858 minus984858 1198611 1198612
lim|120585|rarrinfin
120593 (120585) = 119860
(42)
119871(120593) ge 0 and the boundary condition (24) imply thefollowing
(a) If 120573(9848582 minus 1205932) ge 0 then 120593 ge max119861
1 1198612 or 120593 le
min1198611 1198612
(b) If120573(9848582minus1205932) le 0 thenmin1198611 1198612 le 120593 le max119861
1 1198612
Below we will present some implicit formulas for thesingle peak soliton solutions in the case of specific 120590 and 120573
Case 1 ((120590 120573) isin 11986012) In this case we have minus984858 lt 119861
2lt 119860 lt
1198611= 984858 From the standard phase analysis and Theorem 9
we know that if 120593 is a single peak soliton of the 119862(3 2 2)
equation then
1205931015840= minusradicminus
120573
2(120593 minus 119860)radic
120593 minus 1198612
120593 + 984858sgn (120585) (43)
From the separation of variables we get
intℎ (120593) 119889120593 = radicminus120573
2sgn (120585) 119889120585 (44)
where ℎ(120593) = (1(119860 minus 120593))radic(120593 + 984858)(120593 minus 1198612) After a lengthy
calculation of integral we obtain the implicit solution 120593
defined by
119867(120593)flradic119860 + 984858
119860 minus 1198612
1198681(120593) minus 119868
2(120593) = radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870 (45)
where
1198681(120593) = ln
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
(119860 + 984858) (120593 minus 1198612) + (119860 minus 119861
2) (120593 + 984858) + 2radic(119860 + 984858) (119860 minus 119861
2) (120593 minus 119861
2) (120593 + 984858)
radic(119860 + 984858) (119860 minus 1198612) (119860 + 984858) (119860 minus 120593)
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
1198682(120593) = ln
1003816100381610038161003816100381610038161003816(120593 + 984858) + (120593 minus 119861
2) + 2radic(120593 + 984858) (120593 minus 119861
2)
1003816100381610038161003816100381610038161003816
(46)
and 119870 is an arbitrary integration constant For 120593(0) = 984858 theconstant119870 = 119867(120593(0)) is defined by
119870 = radic119860 + 984858
119860 minus 1198612
1198681(984858) minus 119868
2(984858) (47)
and for 120593(0) = 1198612
119870 = radic119860 + 984858
119860 minus 1198612
1198681(1198612) minus 1198682(1198612) (48)
(i) If 120593(0) = 984858 then 119860 lt 120593 le 984858 Since 1198671015840(120593) = ℎ(120593) weknow that119867(120593) strictly decreases on the interval (119860 984858] thus1198671(120593) = 119867|
(119860984858](120593) gives a single peak solitonwith119867
1(984858) = 119870
and1198671(119860+) = infin Therefore 120593
1(120585) = 119867
minus1
1(radicminus1205732|120585| + 119870) is
the solution satisfying
1205931(0) = 984858
lim|120585|rarrinfin
1205931(120585) = 119860
1205931015840
1(0plusmn) = plusmn (119860 minus 984858)radic
120573 (1198612minus 984858)
4984858
(49)
So 1205931(120585) is a peakon-like solution (see Figure 5)
(ii) If 120593(0) = 1198612 then 119861
2le 120593 lt 119860 By 119867
1015840(120593) = 119891(120593)
we know that 119867(120593) strictly increases on the interval [1198612 119860)
Thus
1198672(120593) = 119867|
[1198612 119860)(120593) (50)
Mathematical Problems in Engineering 13
minus3 minus2 minus1
120593
minus02
1 2 3
02
04
06
08
10
x
(a)
02
02
minus2 minus2 x
10
05
00
t
120593
22
(b)
Figure 5 Two- and three-dimensional graphs of the peakon-like solution
minus03
minus04
minus05
1 2 3minus3 minus2 minus1
120593
minus02
x
(a)
0
2
0
2
minus2minus2
minus02
minus03
minus04
minus05
minus06
t x
120593
0
2
0
2
minus2minus2
t x
(b)
Figure 6 Two- and three-dimensional graphs of the smooth soliton solution
has the inverse denoted by 1205932(120585) = 119867
minus1
2(radicminus1205732|120585| + 119870)
1205932(120585) gives a kind of smooth soliton solution (see Figure 6)
satisfying
1205932(0) = 119861
2
lim|120585|rarrinfin
1205932(120585) = 119860
1205931015840
2(0) = 0
(51)
Case 2 ((120590 120573) isin 11986014) In this case we have minus984858 = 119861
2lt 119860 lt
984858 lt 1198611and (25) is equivalent to
1205931015840= minusradicminus
120573
2(120593 minus 119860)radic
120593 minus 1198611
120593 minus 984858sgn (120585) (52)
Let
119892 (120593) =1
119860 minus 120593radic120593 minus 1198611
120593 minus 984858 (53)
14 Mathematical Problems in Engineering
minus1 1 2 3
02
04
06
08
10
minus3 minus2
120593
x
(a)
02
02minus2
minus2
120593
x t
00
05
10
(b)
Figure 7 Two- and three-dimensional graphs of the cuspon solution
then (52) is converted to
119892 (120593) 119889120593 =1
119860 minus 120593radic120593 minus 1198611
120593 minus 984858119889120593 = radicminus
120573
2sgn (120585) 119889120585 (54)
Integrating (54) on the interval [0 120585] (or [120585 0]) leads to thefollowing implicit solutions
119866 (120593)flradic119860 minus 984858
119860 minus 1198611
1198683(120593) + 119868
4(120593) = radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870 (55)
where
1198683(120593) = ln
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
(119860 minus 984858) (120593 minus 1198611) + (119860 minus 119861
1) (120593 minus 984858) + 2radic(119860 minus 984858) (119860 minus 119861
1) (120593 minus 984858) (120593 minus 119861
1)
radic(119860 minus 984858) (119860 minus 1198611) (119860 minus 984858) (120593 minus 119860)
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
1198684(120593) = ln
1003816100381610038161003816100381610038161003816(120593 minus 984858) + (120593 minus 119861
1) + 2radic(120593 minus 984858) (120593 minus 119861
1)
1003816100381610038161003816100381610038161003816
(56)
And119870 is an arbitrary integration constant It is obvious thatfor 120593(0) = 984858 the constant119870 = 119867(120593(0)) is defined by
119870 = radic119860 minus 984858
119860 minus 1198611
1198683(984858) + 119868
4(984858) (57)
and for 120593(0) = minus984858
119870 = radic119860 minus 984858
119860 minus 1198611
1198683(minus984858) + 119868
4(minus984858) (58)
(i) If 120593(0) = 984858 then 119860 lt 120593 le 984858 From 119892(120593) lt 0 weknow that 119866(120593) strictly decreases on the interval (119860 984858] with119866(984858) = 119870 and 119866(119860+) = infin Define
1198661(120593) = 119866|
(119860984858](120593) (59)
Since 1198661(120593) is a strictly decreasing function from (119860 984858] onto
[119870infin) we can solve for 120593 uniquely from (59) and obtain
1205931(120585) = 119866
minus1
1(radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870) (60)
It is easy to check that 120593 satisfies
1205931(0) = 984858
lim|120585|rarrinfin
1205931(120585) = 119860
1205931015840
1(0plusmn) = ∓infin
(61)
Therefore the solution1205931defined by (60) is a cuspon solution
for the 119862(3 2 2) equation (see Figure 7)
Mathematical Problems in Engineering 15
(ii) If 120593(0) = minus984858 then minus984858 le 120593 lt 119860 Through a similaranalysis we get a strictly increasing function 119866(120593) on theinterval [minus984858 119860) satisfying
119866 (120593) = radicminus120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870 (62)
where 119866(120593) is defined by (55) Let
1198662(120593) = 119866|
[minus984858119860)(120593) (63)
then 1198662(120593) is a strictly increasing function from [minus984858 119860) onto
[119870infin) so that we can solve for 120593 and obtain
1205932(120585) = 119866
minus1
2(radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870) (64)
It is easy to check that 1205932satisfies
1205932(0) = minus984858
lim|120585|rarrinfin
1205932(120585) = 119860
1205931015840
2(0plusmn) = plusmn (119860 + 984858)radicminus
120573 (1198611+ 984858)
4984858
(65)
Therefore the solution 1205932defined by (64) is a peakon-like
solution whose graph is similar to those in Figure 5
Case 3 ((120590 120573) isin 11986016) In this case we have minus984858 lt 119861
2lt 119860 =
0 lt 1198611lt 984858 119861
2= minus1198611 and
1205931015840= minusradicminus
120573
2120593radic
120593 minus 1198612
1
1205932 minus 9848582sgn (120585) (66)
Hence from the separation of variables we have
radic9848582 minus 1205932
120593radic1198612
1minus 1205932
119889120593 = minusradicminus120573
2sgn (120585) 119889120585 (67)
Integrating (67) on the interval [0 120585] (or [120585 0]) leads tothe following implicit formula for the two smooth solitonsolutions
radic9848582 minus 1205932 + radic1198612
1minus 1205932
radic9848582 minus 1198612
1
(
radic(9848582 minus 1198612
1) 1205932
984858radic1198612
1minus 1205932 + 119861
1radic9848582 minus 1205932
)
984858
= exp(minusradicminus120573
2|119909 minus 119888119905|)
(68)
where 120593 isin (119860 1198611] Consider
radic9848582 minus 1198612
2
radic9848582 minus 1205932 + radic1198612
2minus 1205932
(
984858radic1198612
2minus 1205932 minus 119861
2radic9848582 minus 1205932
radic(9848582 minus 1198612
2) 1205932
)
984858
= exp(radicminus120573
2|119909 minus 119888119905|)
(69)
where 120593 isin [1198612 119860)
Case 4 ((120590 120573) isin 11986017) In this case we have minus984858 = 119861
2lt 119860 =
0 lt 1198611= 984858 and
1205931015840= minusradicminus
120573
2120593 sgn (120585) (70)
Choosing 120593(0) = 984858 (or minus984858) as initial value we get
int
120593
984858
1
120593119889120593 = minusint
120585
0
radicminus120573
2sgn (120585) 119889120585 for 119860 lt 120593 le 984858
int
120593
minus984858
1
120593119889120593 = minusint
120585
0
radicminus120573
2sgn (120585) 119889120585 for minus 984858 le 120593 lt 119860
(71)
which immediately yield the peakon solutions
120593 (119909 minus 119888119905) = plusmn984858 exp(minusradicminus120573
2|119909 minus 119888119905|) (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 the119862(3 2 2) equation (10) In order to study kinkwave solutionswe assume that
lim120585rarrinfin
120593 (120585) = 1198601
lim120585rarrminusinfin
120593 (120585) = 1198602
(73)
16 Mathematical Problems in Engineering
2 4
02
04
06
08
10
minus2minus4
120593
x
(a)
0 2
02
00
05
10
minus2
120593
x
t
0
2
minus2
(b)
Figure 8 Two- and three-dimensional graphs of the peakon solution
where 1198601gt 1198602 Substituting the boundary condition (73)
into (14) generates
(1205931015840)2
=
120573 (120593 minus 1198601) (120593 minus 119860
2) [1205932+ (1198601+ 1198602) 120593 + 119860
2
1+ 11986011198602+ 1198602
2+ 2120598120573]
2 (9848582 minus 1205932)fl119865 (120593) (74)
The nonlinear differential equation (74) may sustaindifferent kinds of nonlinear excitations In what follows weconfine our attention to the cases 119860
2= minus119860
1and 120598 =
minus1198602
1120573 which describe kinks and kink compactons which
play an important role in the dynamics systems Under theseconsiderations (74) reduces to
(1205931015840)2
=120573 (120593 minus 119860
1)2
(120593 minus 1198602)2
2 (9848582 minus 1205932) (75)
If 1198601lt 984858 then from the phase analysis in Section 2 (see
Figure 4(10)) we know that (1198601 0) and (119860
2 0) are two saddle
points of (13) and the kink solutions can be obtained fromthe two heteroclinic orbits connecting (120601 119910) = (119860
1 0) and
(1198602 0) When 119860
1increases upon reaching 984858 that is 119860
1= 984858
(75) becomes
(1205931015840)2
= minus120573 (120593 minus 119860
1) (120593 minus 119860
2)
2 (76)
and the ellipse 21199102+120573(120593minus1198601)(120593minus119860
2) = 0 (see Figure 4(11))
which is tangent to the singular lines 120593 = 984858 and 120593 = minus984858 atpoints (119860
1 0) and (119860
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 120593 is a kink wave solutions of (75)
or (76) we have 1205931015840
rarr 0 as 120593 rarr 1198601and as 120593 rarr 119860
2
Moreover we have 119865(120593) ge 0 for 1198602le 120593 le 119860
1and 120593 is
strictly monotonic in any interval where 119865(120593) gt 0 Thus if1205931015840gt 0 at some points 120593will be strictly increasing until it gets
close to the next zero of 119865 Denoting this zero 1198601 we have
120593 uarr 1198601 What will happen to the solution when it approaches
1198601 Depending on whether the zero is double or simple 120593
has a different behavior We explore the two cases in turn
Theorem 11 (i) If 120593 has a simple zero at 120593 = 1198601 so that
119865(1198601) = 0 and 1198651015840(119860
1) lt 0 then the solution 120593 of (75) satisfies
120593 (120585) minus 1198601=1198651015840(1198601)
4(120585 minus 120578)
2
+ 119874 ((120585 minus 120578)4
)
119886119904 120585 uarr 120578
(77)
where 120593(120578) = 1198601
(ii) When 120593 approaches the double zero 1198601of 119865(120593) so that
1198651015840(1198601) = 0 and11986510158401015840(119860
1) gt 0 then the solution120593 of (75) satisfies
120593 (120585) minus 1198601sim 120572 exp(minus120585radic11986510158401015840 (119860
1)) 119886119904 120585 997888rarr infin (78)
for some constant 120572 Thus 120593 uarr 1198601exponentially as 120585 rarr infin
Mathematical Problems in Engineering 17
Proof (i) When 1198601= minus119860
2= 984858 and 120598 = minus119860
2
1120573 from (76)
119865(120593) has a simple zero at 120593 = 1198601 Then
(1205931015840)2
= (120593 minus 1198601) 1198651015840(1198601) + 119874 ((120593 minus 119860
1)2
)
as 120593 uarr 1198601
(79)
Using the fact that (1205931015840)2 ge 0 we know that 1198651015840(1198601) lt 0
Moreover
119889120585
119889120593=
1
radic(120593 minus 1198601) 1198651015840 (119860
1) + 119874 ((120593 minus 119860
1)2
)
(80)
Because
radic(120593 minus 1198601) 1198651015840 (119860
1) + 119874 ((120593 minus 119860
1)2
)
= radic1198601minus 120593(radicminus1198651015840 (119860
1) + 119874 (119860
1minus 120593))
1
radicminus1198651015840 (1198601) + 119874 (119860
1minus 120593)
=1
radicminus1198651015840 (1198601)
+ 119874 (1198601minus 120593)
(81)
we obtain119889120585
119889120593=
1
radicminus (1198601minus 120593) 1198651015840 (119860
1)
+ 119874 ((1198601minus 120593)12
) (82)
Integrating (82) yields
120578 minus 120585 =2
radicminus1198651015840 (1198601)
(1198601minus 120593)12
+ 119874((1198601minus 120593)32
) (83)
where 120578 satisfies 120593(120578) = 1198601 Thus
(120578 minus 120585)2
=4
minus1198651015840 (1198601)(1198601minus 120593) + 119874 ((119860
1minus 120593)2
) (84)
which implies 119874((1198601minus 120593)2) = 119874((120578 minus 120585)
4) Therefore we get
120593 (120585) = 1198601+1198651015840(1198601)
4(120585 minus 120578)
2
+ 119874 ((120585 minus 120578)4
)
as 120585 uarr 120578
(85)
where 120593(120578) = 1198601
(ii) When 1198601= minus119860
2lt 984858 and 120598 = minus119860
2
1120573 from (75) 119865(120593)
has a double zero at 120593 = 1198601 Then
(1205931015840)2
= (120593 minus 1198601)2
11986510158401015840(1198601) + 119874 ((120593 minus 119860
1)3
)
as 120593 uarr 1198601
(86)
Furthermore we get
119889120585
119889120593=
1
radic(120593 minus 1198601)2
11986510158401015840 (1198601) + 119874 ((120593 minus 119860
1)3
)
(87)
Observing that
radic(120593 minus 1198601)2
11986510158401015840 (1198601) + 119874 ((120593 minus 119860
1)3
)
= (1198601minus 120593) (radic11986510158401015840 (119860
1) + 119874 (119860
1minus 120593))
1
radic11986510158401015840 (1198601) + 119874 (119860
1minus 120593)
=1
radic11986510158401015840 (1198601)
+ 119874 (1198601minus 120593)
(88)
we obtain119889120585
119889120593=
1
(1198601minus 120593) 11986510158401015840 (119860
1)+ 119874 (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 119860
1= minus119860
2lt 984858 and 120598 = minus119860
2
1120573 Then (75)
becomes
radic9848582 minus 1205932
1205932 minus 1198602
1
119889120593 = ∓radic120573
2119889120585 (90)
Integrating both sides of (90) gives the following implicitexpressions of kink and antikink wave solutions
arctan120593
radic9848582 minus 1205932
+
radic9848582 minus 1198602
1
1198601
arctanhradic9848582 minus 119860
2
1
1198601
120593
radic9848582 minus 1205932
= plusmnradic120573
2120585
(91)
By letting 1198601rarr 984858 in (91) we get two kink compactons
which are given by
plusmn120593 =
984858 sinradic120573
2120585
10038161003816100381610038161205851003816100381610038161003816 le
120587
radic2120573
984858 120585 gt120587
radic2120573
minus984858 120585 lt minus120587
radic2120573
(92)
The graphs for the kink wave solutions (91) and kinkcompacton solutions (92) are shown in Figures 9 and 10respectively
Remark 12 The two kink compacton solutions (92) aredifferent from the well-known smooth kink wave solutionsIn comparison with kink wave solutions (91) the kink com-pacton solutions (92) have no exponential decay propertiesbut have compact supportThat is they minus a constant thedifferences identically vanish outside a finite core region
18 Mathematical Problems in Engineering
120593
120585minus1 1
05
minus05
(a)
minus1minus1
minus05
05
00120593
x t
0
1
01
11
(b)
Figure 9 Two- and three-dimensional graphs of the kink wave solution
1
1120593
120585
minus1
minus1
(a)
0
1
0
1
000510
120593
minus1 minus1
minus05
minus10
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 the119862(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 Newtonrsquos equation which we write in the form 119910
2=
119865(120593) where 119910 = 1205931015840
120585(120585) we solve Newtonrsquos equation 119910
2=
119865(120593) 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 the119862(3 2 2)equation
Actually for 119898 = 2119896 + 1 119896 isin N+ in 119862(119898 2 2) equation(9) the dynamical behavior of traveling wave solutions of(9) is similar to the case 119898 = 3 for 119898 = 2119896 119896 isin N+ in119862(119898 2 2) equation (9) the dynamical behavior of travelingwave solutions of (9) is similar to the case 119898 = 2 We areapplying the approach mentioned in this work to 119862(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
References
[1] C Rogers and W R Shadwick Backlund Transformation andTheir Applications Academic Press New York NY USA 1982
[2] W Malfliet and W Hereman ldquoThe tanh method I Exactsolutions of nonlinear evolution and wave equationsrdquo PhysicaScripta vol 54 no 6 pp 563ndash568 1996
[3] R Hirota Direct Method in Soliton Theory Springer BerlinGermany 1980
[4] J S Cohen Computer Algebra and Symbolic ComputationMathematical Methods A K Peters 2003
[5] S Lie ldquoUber die integration durch bestimmte integrale voneiner klasse linearer partieller differential eichungenrdquoArchiv derMathematik vol 6 pp 328ndash368 1881
[6] H-H Dai and X-H Zhao ldquoTraveling waves in a rod composedof amodifiedMooney-Rivlinmaterial I bifurcation of equilibraand non-singular caserdquo Proceedings of the Royal Society AMathematical Physical and Engineering Sciences vol 455 no1990 pp 3845ndash3874 1999
[7] H-H Dai and F-F Wang ldquoAsymptotic bifurcation solutionsfor compressions of a clamped nonlinearly elastic rectangletransition region and barrelling to a corner-like profilerdquo SIAMJournal on Applied Mathematics vol 70 no 7 pp 2673ndash26922010
[8] J Li and Z Liu ldquoSmooth and non-smooth traveling waves ina nonlinearly dispersive equationrdquo Applied Mathematical Mod-elling vol 25 no 1 pp 41ndash56 2000
[9] A Chen and J Li ldquoSingle peak solitary wave solutions forthe osmosis K(22) equation under inhomogeneous boundaryconditionrdquo Journal of Mathematical Analysis and Applicationsvol 369 no 2 pp 758ndash766 2010
[10] R Camassa and D D Holm ldquoAn integrable shallow waterequation with peaked solitonsrdquo Physical Review Letters vol 71no 11 pp 1661ndash1664 1993
[11] P Rosenau ldquoOn nonanalytic solitary waves formed by a nonlin-ear dispersionrdquo Physics Letters A vol 230 no 5-6 pp 305ndash3181997
[12] P Rosenau and J M Hyman ldquoCompactons solitons with finitewavelengthrdquo Physical Review Letters vol 70 no 5 pp 564ndash5671993
[13] L Zhang L-Q Chen and X Huo ldquoThe effects of horizontalsingular straight line in a generalized nonlinear Klein-Gordonmodel equationrdquo Nonlinear Dynamics vol 72 no 4 pp 789ndash801 2013
[14] J Li and H Dai On the Study of Singular Nonlinear TravelingWave Equations Dynamical System Approach Science PressBeijing China 2007
[15] A Chen S Wen S Tang W Huang and Z Qiao ldquoEffects ofquadratic singular curves in integrable equationsrdquo Studies inApplied Mathematics vol 134 no 1 pp 24ndash61 2015
[16] L Tian and J Yin ldquoNew compacton solutions and solitary wavesolutions of fully nonlinear generalized Camassa-Holm equa-tionsrdquo Chaos Solitons amp Fractals vol 20 no 2 pp 289ndash2992004
[17] L Zhang ldquoThe bifurcation and peakons for the specialC(3 2 2)equationrdquo Pramana vol 83 no 3 pp 331ndash340 2014
[18] J Lenells ldquoTraveling wave solutions of the CamassandashHolmequationrdquo Journal of Differential Equations vol 217 no 2 pp393ndash430 2005
[19] Z J Qiao and G P Zhang ldquoOn peaked and smooth solitons forthe Camassa-Holm equationrdquo Europhysics Letters vol 73 no 5pp 657ndash663 2006
Submit your manuscripts athttpwwwhindawicom
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Mathematical Problems in Engineering
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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Stochastic AnalysisInternational Journal of
Mathematical Problems in Engineering 13
minus3 minus2 minus1
120593
minus02
1 2 3
02
04
06
08
10
x
(a)
02
02
minus2 minus2 x
10
05
00
t
120593
22
(b)
Figure 5 Two- and three-dimensional graphs of the peakon-like solution
minus03
minus04
minus05
1 2 3minus3 minus2 minus1
120593
minus02
x
(a)
0
2
0
2
minus2minus2
minus02
minus03
minus04
minus05
minus06
t x
120593
0
2
0
2
minus2minus2
t x
(b)
Figure 6 Two- and three-dimensional graphs of the smooth soliton solution
has the inverse denoted by 1205932(120585) = 119867
minus1
2(radicminus1205732|120585| + 119870)
1205932(120585) gives a kind of smooth soliton solution (see Figure 6)
satisfying
1205932(0) = 119861
2
lim|120585|rarrinfin
1205932(120585) = 119860
1205931015840
2(0) = 0
(51)
Case 2 ((120590 120573) isin 11986014) In this case we have minus984858 = 119861
2lt 119860 lt
984858 lt 1198611and (25) is equivalent to
1205931015840= minusradicminus
120573
2(120593 minus 119860)radic
120593 minus 1198611
120593 minus 984858sgn (120585) (52)
Let
119892 (120593) =1
119860 minus 120593radic120593 minus 1198611
120593 minus 984858 (53)
14 Mathematical Problems in Engineering
minus1 1 2 3
02
04
06
08
10
minus3 minus2
120593
x
(a)
02
02minus2
minus2
120593
x t
00
05
10
(b)
Figure 7 Two- and three-dimensional graphs of the cuspon solution
then (52) is converted to
119892 (120593) 119889120593 =1
119860 minus 120593radic120593 minus 1198611
120593 minus 984858119889120593 = radicminus
120573
2sgn (120585) 119889120585 (54)
Integrating (54) on the interval [0 120585] (or [120585 0]) leads to thefollowing implicit solutions
119866 (120593)flradic119860 minus 984858
119860 minus 1198611
1198683(120593) + 119868
4(120593) = radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870 (55)
where
1198683(120593) = ln
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
(119860 minus 984858) (120593 minus 1198611) + (119860 minus 119861
1) (120593 minus 984858) + 2radic(119860 minus 984858) (119860 minus 119861
1) (120593 minus 984858) (120593 minus 119861
1)
radic(119860 minus 984858) (119860 minus 1198611) (119860 minus 984858) (120593 minus 119860)
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
1198684(120593) = ln
1003816100381610038161003816100381610038161003816(120593 minus 984858) + (120593 minus 119861
1) + 2radic(120593 minus 984858) (120593 minus 119861
1)
1003816100381610038161003816100381610038161003816
(56)
And119870 is an arbitrary integration constant It is obvious thatfor 120593(0) = 984858 the constant119870 = 119867(120593(0)) is defined by
119870 = radic119860 minus 984858
119860 minus 1198611
1198683(984858) + 119868
4(984858) (57)
and for 120593(0) = minus984858
119870 = radic119860 minus 984858
119860 minus 1198611
1198683(minus984858) + 119868
4(minus984858) (58)
(i) If 120593(0) = 984858 then 119860 lt 120593 le 984858 From 119892(120593) lt 0 weknow that 119866(120593) strictly decreases on the interval (119860 984858] with119866(984858) = 119870 and 119866(119860+) = infin Define
1198661(120593) = 119866|
(119860984858](120593) (59)
Since 1198661(120593) is a strictly decreasing function from (119860 984858] onto
[119870infin) we can solve for 120593 uniquely from (59) and obtain
1205931(120585) = 119866
minus1
1(radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870) (60)
It is easy to check that 120593 satisfies
1205931(0) = 984858
lim|120585|rarrinfin
1205931(120585) = 119860
1205931015840
1(0plusmn) = ∓infin
(61)
Therefore the solution1205931defined by (60) is a cuspon solution
for the 119862(3 2 2) equation (see Figure 7)
Mathematical Problems in Engineering 15
(ii) If 120593(0) = minus984858 then minus984858 le 120593 lt 119860 Through a similaranalysis we get a strictly increasing function 119866(120593) on theinterval [minus984858 119860) satisfying
119866 (120593) = radicminus120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870 (62)
where 119866(120593) is defined by (55) Let
1198662(120593) = 119866|
[minus984858119860)(120593) (63)
then 1198662(120593) is a strictly increasing function from [minus984858 119860) onto
[119870infin) so that we can solve for 120593 and obtain
1205932(120585) = 119866
minus1
2(radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870) (64)
It is easy to check that 1205932satisfies
1205932(0) = minus984858
lim|120585|rarrinfin
1205932(120585) = 119860
1205931015840
2(0plusmn) = plusmn (119860 + 984858)radicminus
120573 (1198611+ 984858)
4984858
(65)
Therefore the solution 1205932defined by (64) is a peakon-like
solution whose graph is similar to those in Figure 5
Case 3 ((120590 120573) isin 11986016) In this case we have minus984858 lt 119861
2lt 119860 =
0 lt 1198611lt 984858 119861
2= minus1198611 and
1205931015840= minusradicminus
120573
2120593radic
120593 minus 1198612
1
1205932 minus 9848582sgn (120585) (66)
Hence from the separation of variables we have
radic9848582 minus 1205932
120593radic1198612
1minus 1205932
119889120593 = minusradicminus120573
2sgn (120585) 119889120585 (67)
Integrating (67) on the interval [0 120585] (or [120585 0]) leads tothe following implicit formula for the two smooth solitonsolutions
radic9848582 minus 1205932 + radic1198612
1minus 1205932
radic9848582 minus 1198612
1
(
radic(9848582 minus 1198612
1) 1205932
984858radic1198612
1minus 1205932 + 119861
1radic9848582 minus 1205932
)
984858
= exp(minusradicminus120573
2|119909 minus 119888119905|)
(68)
where 120593 isin (119860 1198611] Consider
radic9848582 minus 1198612
2
radic9848582 minus 1205932 + radic1198612
2minus 1205932
(
984858radic1198612
2minus 1205932 minus 119861
2radic9848582 minus 1205932
radic(9848582 minus 1198612
2) 1205932
)
984858
= exp(radicminus120573
2|119909 minus 119888119905|)
(69)
where 120593 isin [1198612 119860)
Case 4 ((120590 120573) isin 11986017) In this case we have minus984858 = 119861
2lt 119860 =
0 lt 1198611= 984858 and
1205931015840= minusradicminus
120573
2120593 sgn (120585) (70)
Choosing 120593(0) = 984858 (or minus984858) as initial value we get
int
120593
984858
1
120593119889120593 = minusint
120585
0
radicminus120573
2sgn (120585) 119889120585 for 119860 lt 120593 le 984858
int
120593
minus984858
1
120593119889120593 = minusint
120585
0
radicminus120573
2sgn (120585) 119889120585 for minus 984858 le 120593 lt 119860
(71)
which immediately yield the peakon solutions
120593 (119909 minus 119888119905) = plusmn984858 exp(minusradicminus120573
2|119909 minus 119888119905|) (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 the119862(3 2 2) equation (10) In order to study kinkwave solutionswe assume that
lim120585rarrinfin
120593 (120585) = 1198601
lim120585rarrminusinfin
120593 (120585) = 1198602
(73)
16 Mathematical Problems in Engineering
2 4
02
04
06
08
10
minus2minus4
120593
x
(a)
0 2
02
00
05
10
minus2
120593
x
t
0
2
minus2
(b)
Figure 8 Two- and three-dimensional graphs of the peakon solution
where 1198601gt 1198602 Substituting the boundary condition (73)
into (14) generates
(1205931015840)2
=
120573 (120593 minus 1198601) (120593 minus 119860
2) [1205932+ (1198601+ 1198602) 120593 + 119860
2
1+ 11986011198602+ 1198602
2+ 2120598120573]
2 (9848582 minus 1205932)fl119865 (120593) (74)
The nonlinear differential equation (74) may sustaindifferent kinds of nonlinear excitations In what follows weconfine our attention to the cases 119860
2= minus119860
1and 120598 =
minus1198602
1120573 which describe kinks and kink compactons which
play an important role in the dynamics systems Under theseconsiderations (74) reduces to
(1205931015840)2
=120573 (120593 minus 119860
1)2
(120593 minus 1198602)2
2 (9848582 minus 1205932) (75)
If 1198601lt 984858 then from the phase analysis in Section 2 (see
Figure 4(10)) we know that (1198601 0) and (119860
2 0) are two saddle
points of (13) and the kink solutions can be obtained fromthe two heteroclinic orbits connecting (120601 119910) = (119860
1 0) and
(1198602 0) When 119860
1increases upon reaching 984858 that is 119860
1= 984858
(75) becomes
(1205931015840)2
= minus120573 (120593 minus 119860
1) (120593 minus 119860
2)
2 (76)
and the ellipse 21199102+120573(120593minus1198601)(120593minus119860
2) = 0 (see Figure 4(11))
which is tangent to the singular lines 120593 = 984858 and 120593 = minus984858 atpoints (119860
1 0) and (119860
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 120593 is a kink wave solutions of (75)
or (76) we have 1205931015840
rarr 0 as 120593 rarr 1198601and as 120593 rarr 119860
2
Moreover we have 119865(120593) ge 0 for 1198602le 120593 le 119860
1and 120593 is
strictly monotonic in any interval where 119865(120593) gt 0 Thus if1205931015840gt 0 at some points 120593will be strictly increasing until it gets
close to the next zero of 119865 Denoting this zero 1198601 we have
120593 uarr 1198601 What will happen to the solution when it approaches
1198601 Depending on whether the zero is double or simple 120593
has a different behavior We explore the two cases in turn
Theorem 11 (i) If 120593 has a simple zero at 120593 = 1198601 so that
119865(1198601) = 0 and 1198651015840(119860
1) lt 0 then the solution 120593 of (75) satisfies
120593 (120585) minus 1198601=1198651015840(1198601)
4(120585 minus 120578)
2
+ 119874 ((120585 minus 120578)4
)
119886119904 120585 uarr 120578
(77)
where 120593(120578) = 1198601
(ii) When 120593 approaches the double zero 1198601of 119865(120593) so that
1198651015840(1198601) = 0 and11986510158401015840(119860
1) gt 0 then the solution120593 of (75) satisfies
120593 (120585) minus 1198601sim 120572 exp(minus120585radic11986510158401015840 (119860
1)) 119886119904 120585 997888rarr infin (78)
for some constant 120572 Thus 120593 uarr 1198601exponentially as 120585 rarr infin
Mathematical Problems in Engineering 17
Proof (i) When 1198601= minus119860
2= 984858 and 120598 = minus119860
2
1120573 from (76)
119865(120593) has a simple zero at 120593 = 1198601 Then
(1205931015840)2
= (120593 minus 1198601) 1198651015840(1198601) + 119874 ((120593 minus 119860
1)2
)
as 120593 uarr 1198601
(79)
Using the fact that (1205931015840)2 ge 0 we know that 1198651015840(1198601) lt 0
Moreover
119889120585
119889120593=
1
radic(120593 minus 1198601) 1198651015840 (119860
1) + 119874 ((120593 minus 119860
1)2
)
(80)
Because
radic(120593 minus 1198601) 1198651015840 (119860
1) + 119874 ((120593 minus 119860
1)2
)
= radic1198601minus 120593(radicminus1198651015840 (119860
1) + 119874 (119860
1minus 120593))
1
radicminus1198651015840 (1198601) + 119874 (119860
1minus 120593)
=1
radicminus1198651015840 (1198601)
+ 119874 (1198601minus 120593)
(81)
we obtain119889120585
119889120593=
1
radicminus (1198601minus 120593) 1198651015840 (119860
1)
+ 119874 ((1198601minus 120593)12
) (82)
Integrating (82) yields
120578 minus 120585 =2
radicminus1198651015840 (1198601)
(1198601minus 120593)12
+ 119874((1198601minus 120593)32
) (83)
where 120578 satisfies 120593(120578) = 1198601 Thus
(120578 minus 120585)2
=4
minus1198651015840 (1198601)(1198601minus 120593) + 119874 ((119860
1minus 120593)2
) (84)
which implies 119874((1198601minus 120593)2) = 119874((120578 minus 120585)
4) Therefore we get
120593 (120585) = 1198601+1198651015840(1198601)
4(120585 minus 120578)
2
+ 119874 ((120585 minus 120578)4
)
as 120585 uarr 120578
(85)
where 120593(120578) = 1198601
(ii) When 1198601= minus119860
2lt 984858 and 120598 = minus119860
2
1120573 from (75) 119865(120593)
has a double zero at 120593 = 1198601 Then
(1205931015840)2
= (120593 minus 1198601)2
11986510158401015840(1198601) + 119874 ((120593 minus 119860
1)3
)
as 120593 uarr 1198601
(86)
Furthermore we get
119889120585
119889120593=
1
radic(120593 minus 1198601)2
11986510158401015840 (1198601) + 119874 ((120593 minus 119860
1)3
)
(87)
Observing that
radic(120593 minus 1198601)2
11986510158401015840 (1198601) + 119874 ((120593 minus 119860
1)3
)
= (1198601minus 120593) (radic11986510158401015840 (119860
1) + 119874 (119860
1minus 120593))
1
radic11986510158401015840 (1198601) + 119874 (119860
1minus 120593)
=1
radic11986510158401015840 (1198601)
+ 119874 (1198601minus 120593)
(88)
we obtain119889120585
119889120593=
1
(1198601minus 120593) 11986510158401015840 (119860
1)+ 119874 (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 119860
1= minus119860
2lt 984858 and 120598 = minus119860
2
1120573 Then (75)
becomes
radic9848582 minus 1205932
1205932 minus 1198602
1
119889120593 = ∓radic120573
2119889120585 (90)
Integrating both sides of (90) gives the following implicitexpressions of kink and antikink wave solutions
arctan120593
radic9848582 minus 1205932
+
radic9848582 minus 1198602
1
1198601
arctanhradic9848582 minus 119860
2
1
1198601
120593
radic9848582 minus 1205932
= plusmnradic120573
2120585
(91)
By letting 1198601rarr 984858 in (91) we get two kink compactons
which are given by
plusmn120593 =
984858 sinradic120573
2120585
10038161003816100381610038161205851003816100381610038161003816 le
120587
radic2120573
984858 120585 gt120587
radic2120573
minus984858 120585 lt minus120587
radic2120573
(92)
The graphs for the kink wave solutions (91) and kinkcompacton solutions (92) are shown in Figures 9 and 10respectively
Remark 12 The two kink compacton solutions (92) aredifferent from the well-known smooth kink wave solutionsIn comparison with kink wave solutions (91) the kink com-pacton solutions (92) have no exponential decay propertiesbut have compact supportThat is they minus a constant thedifferences identically vanish outside a finite core region
18 Mathematical Problems in Engineering
120593
120585minus1 1
05
minus05
(a)
minus1minus1
minus05
05
00120593
x t
0
1
01
11
(b)
Figure 9 Two- and three-dimensional graphs of the kink wave solution
1
1120593
120585
minus1
minus1
(a)
0
1
0
1
000510
120593
minus1 minus1
minus05
minus10
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 the119862(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 Newtonrsquos equation which we write in the form 119910
2=
119865(120593) where 119910 = 1205931015840
120585(120585) we solve Newtonrsquos equation 119910
2=
119865(120593) 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 the119862(3 2 2)equation
Actually for 119898 = 2119896 + 1 119896 isin N+ in 119862(119898 2 2) equation(9) the dynamical behavior of traveling wave solutions of(9) is similar to the case 119898 = 3 for 119898 = 2119896 119896 isin N+ in119862(119898 2 2) equation (9) the dynamical behavior of travelingwave solutions of (9) is similar to the case 119898 = 2 We areapplying the approach mentioned in this work to 119862(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
References
[1] C Rogers and W R Shadwick Backlund Transformation andTheir Applications Academic Press New York NY USA 1982
[2] W Malfliet and W Hereman ldquoThe tanh method I Exactsolutions of nonlinear evolution and wave equationsrdquo PhysicaScripta vol 54 no 6 pp 563ndash568 1996
[3] R Hirota Direct Method in Soliton Theory Springer BerlinGermany 1980
[4] J S Cohen Computer Algebra and Symbolic ComputationMathematical Methods A K Peters 2003
[5] S Lie ldquoUber die integration durch bestimmte integrale voneiner klasse linearer partieller differential eichungenrdquoArchiv derMathematik vol 6 pp 328ndash368 1881
[6] H-H Dai and X-H Zhao ldquoTraveling waves in a rod composedof amodifiedMooney-Rivlinmaterial I bifurcation of equilibraand non-singular caserdquo Proceedings of the Royal Society AMathematical Physical and Engineering Sciences vol 455 no1990 pp 3845ndash3874 1999
[7] H-H Dai and F-F Wang ldquoAsymptotic bifurcation solutionsfor compressions of a clamped nonlinearly elastic rectangletransition region and barrelling to a corner-like profilerdquo SIAMJournal on Applied Mathematics vol 70 no 7 pp 2673ndash26922010
[8] J Li and Z Liu ldquoSmooth and non-smooth traveling waves ina nonlinearly dispersive equationrdquo Applied Mathematical Mod-elling vol 25 no 1 pp 41ndash56 2000
[9] A Chen and J Li ldquoSingle peak solitary wave solutions forthe osmosis K(22) equation under inhomogeneous boundaryconditionrdquo Journal of Mathematical Analysis and Applicationsvol 369 no 2 pp 758ndash766 2010
[10] R Camassa and D D Holm ldquoAn integrable shallow waterequation with peaked solitonsrdquo Physical Review Letters vol 71no 11 pp 1661ndash1664 1993
[11] P Rosenau ldquoOn nonanalytic solitary waves formed by a nonlin-ear dispersionrdquo Physics Letters A vol 230 no 5-6 pp 305ndash3181997
[12] P Rosenau and J M Hyman ldquoCompactons solitons with finitewavelengthrdquo Physical Review Letters vol 70 no 5 pp 564ndash5671993
[13] L Zhang L-Q Chen and X Huo ldquoThe effects of horizontalsingular straight line in a generalized nonlinear Klein-Gordonmodel equationrdquo Nonlinear Dynamics vol 72 no 4 pp 789ndash801 2013
[14] J Li and H Dai On the Study of Singular Nonlinear TravelingWave Equations Dynamical System Approach Science PressBeijing China 2007
[15] A Chen S Wen S Tang W Huang and Z Qiao ldquoEffects ofquadratic singular curves in integrable equationsrdquo Studies inApplied Mathematics vol 134 no 1 pp 24ndash61 2015
[16] L Tian and J Yin ldquoNew compacton solutions and solitary wavesolutions of fully nonlinear generalized Camassa-Holm equa-tionsrdquo Chaos Solitons amp Fractals vol 20 no 2 pp 289ndash2992004
[17] L Zhang ldquoThe bifurcation and peakons for the specialC(3 2 2)equationrdquo Pramana vol 83 no 3 pp 331ndash340 2014
[18] J Lenells ldquoTraveling wave solutions of the CamassandashHolmequationrdquo Journal of Differential Equations vol 217 no 2 pp393ndash430 2005
[19] Z J Qiao and G P Zhang ldquoOn peaked and smooth solitons forthe Camassa-Holm equationrdquo Europhysics Letters vol 73 no 5pp 657ndash663 2006
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14 Mathematical Problems in Engineering
minus1 1 2 3
02
04
06
08
10
minus3 minus2
120593
x
(a)
02
02minus2
minus2
120593
x t
00
05
10
(b)
Figure 7 Two- and three-dimensional graphs of the cuspon solution
then (52) is converted to
119892 (120593) 119889120593 =1
119860 minus 120593radic120593 minus 1198611
120593 minus 984858119889120593 = radicminus
120573
2sgn (120585) 119889120585 (54)
Integrating (54) on the interval [0 120585] (or [120585 0]) leads to thefollowing implicit solutions
119866 (120593)flradic119860 minus 984858
119860 minus 1198611
1198683(120593) + 119868
4(120593) = radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870 (55)
where
1198683(120593) = ln
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
(119860 minus 984858) (120593 minus 1198611) + (119860 minus 119861
1) (120593 minus 984858) + 2radic(119860 minus 984858) (119860 minus 119861
1) (120593 minus 984858) (120593 minus 119861
1)
radic(119860 minus 984858) (119860 minus 1198611) (119860 minus 984858) (120593 minus 119860)
1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816
1198684(120593) = ln
1003816100381610038161003816100381610038161003816(120593 minus 984858) + (120593 minus 119861
1) + 2radic(120593 minus 984858) (120593 minus 119861
1)
1003816100381610038161003816100381610038161003816
(56)
And119870 is an arbitrary integration constant It is obvious thatfor 120593(0) = 984858 the constant119870 = 119867(120593(0)) is defined by
119870 = radic119860 minus 984858
119860 minus 1198611
1198683(984858) + 119868
4(984858) (57)
and for 120593(0) = minus984858
119870 = radic119860 minus 984858
119860 minus 1198611
1198683(minus984858) + 119868
4(minus984858) (58)
(i) If 120593(0) = 984858 then 119860 lt 120593 le 984858 From 119892(120593) lt 0 weknow that 119866(120593) strictly decreases on the interval (119860 984858] with119866(984858) = 119870 and 119866(119860+) = infin Define
1198661(120593) = 119866|
(119860984858](120593) (59)
Since 1198661(120593) is a strictly decreasing function from (119860 984858] onto
[119870infin) we can solve for 120593 uniquely from (59) and obtain
1205931(120585) = 119866
minus1
1(radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870) (60)
It is easy to check that 120593 satisfies
1205931(0) = 984858
lim|120585|rarrinfin
1205931(120585) = 119860
1205931015840
1(0plusmn) = ∓infin
(61)
Therefore the solution1205931defined by (60) is a cuspon solution
for the 119862(3 2 2) equation (see Figure 7)
Mathematical Problems in Engineering 15
(ii) If 120593(0) = minus984858 then minus984858 le 120593 lt 119860 Through a similaranalysis we get a strictly increasing function 119866(120593) on theinterval [minus984858 119860) satisfying
119866 (120593) = radicminus120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870 (62)
where 119866(120593) is defined by (55) Let
1198662(120593) = 119866|
[minus984858119860)(120593) (63)
then 1198662(120593) is a strictly increasing function from [minus984858 119860) onto
[119870infin) so that we can solve for 120593 and obtain
1205932(120585) = 119866
minus1
2(radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870) (64)
It is easy to check that 1205932satisfies
1205932(0) = minus984858
lim|120585|rarrinfin
1205932(120585) = 119860
1205931015840
2(0plusmn) = plusmn (119860 + 984858)radicminus
120573 (1198611+ 984858)
4984858
(65)
Therefore the solution 1205932defined by (64) is a peakon-like
solution whose graph is similar to those in Figure 5
Case 3 ((120590 120573) isin 11986016) In this case we have minus984858 lt 119861
2lt 119860 =
0 lt 1198611lt 984858 119861
2= minus1198611 and
1205931015840= minusradicminus
120573
2120593radic
120593 minus 1198612
1
1205932 minus 9848582sgn (120585) (66)
Hence from the separation of variables we have
radic9848582 minus 1205932
120593radic1198612
1minus 1205932
119889120593 = minusradicminus120573
2sgn (120585) 119889120585 (67)
Integrating (67) on the interval [0 120585] (or [120585 0]) leads tothe following implicit formula for the two smooth solitonsolutions
radic9848582 minus 1205932 + radic1198612
1minus 1205932
radic9848582 minus 1198612
1
(
radic(9848582 minus 1198612
1) 1205932
984858radic1198612
1minus 1205932 + 119861
1radic9848582 minus 1205932
)
984858
= exp(minusradicminus120573
2|119909 minus 119888119905|)
(68)
where 120593 isin (119860 1198611] Consider
radic9848582 minus 1198612
2
radic9848582 minus 1205932 + radic1198612
2minus 1205932
(
984858radic1198612
2minus 1205932 minus 119861
2radic9848582 minus 1205932
radic(9848582 minus 1198612
2) 1205932
)
984858
= exp(radicminus120573
2|119909 minus 119888119905|)
(69)
where 120593 isin [1198612 119860)
Case 4 ((120590 120573) isin 11986017) In this case we have minus984858 = 119861
2lt 119860 =
0 lt 1198611= 984858 and
1205931015840= minusradicminus
120573
2120593 sgn (120585) (70)
Choosing 120593(0) = 984858 (or minus984858) as initial value we get
int
120593
984858
1
120593119889120593 = minusint
120585
0
radicminus120573
2sgn (120585) 119889120585 for 119860 lt 120593 le 984858
int
120593
minus984858
1
120593119889120593 = minusint
120585
0
radicminus120573
2sgn (120585) 119889120585 for minus 984858 le 120593 lt 119860
(71)
which immediately yield the peakon solutions
120593 (119909 minus 119888119905) = plusmn984858 exp(minusradicminus120573
2|119909 minus 119888119905|) (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 the119862(3 2 2) equation (10) In order to study kinkwave solutionswe assume that
lim120585rarrinfin
120593 (120585) = 1198601
lim120585rarrminusinfin
120593 (120585) = 1198602
(73)
16 Mathematical Problems in Engineering
2 4
02
04
06
08
10
minus2minus4
120593
x
(a)
0 2
02
00
05
10
minus2
120593
x
t
0
2
minus2
(b)
Figure 8 Two- and three-dimensional graphs of the peakon solution
where 1198601gt 1198602 Substituting the boundary condition (73)
into (14) generates
(1205931015840)2
=
120573 (120593 minus 1198601) (120593 minus 119860
2) [1205932+ (1198601+ 1198602) 120593 + 119860
2
1+ 11986011198602+ 1198602
2+ 2120598120573]
2 (9848582 minus 1205932)fl119865 (120593) (74)
The nonlinear differential equation (74) may sustaindifferent kinds of nonlinear excitations In what follows weconfine our attention to the cases 119860
2= minus119860
1and 120598 =
minus1198602
1120573 which describe kinks and kink compactons which
play an important role in the dynamics systems Under theseconsiderations (74) reduces to
(1205931015840)2
=120573 (120593 minus 119860
1)2
(120593 minus 1198602)2
2 (9848582 minus 1205932) (75)
If 1198601lt 984858 then from the phase analysis in Section 2 (see
Figure 4(10)) we know that (1198601 0) and (119860
2 0) are two saddle
points of (13) and the kink solutions can be obtained fromthe two heteroclinic orbits connecting (120601 119910) = (119860
1 0) and
(1198602 0) When 119860
1increases upon reaching 984858 that is 119860
1= 984858
(75) becomes
(1205931015840)2
= minus120573 (120593 minus 119860
1) (120593 minus 119860
2)
2 (76)
and the ellipse 21199102+120573(120593minus1198601)(120593minus119860
2) = 0 (see Figure 4(11))
which is tangent to the singular lines 120593 = 984858 and 120593 = minus984858 atpoints (119860
1 0) and (119860
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 120593 is a kink wave solutions of (75)
or (76) we have 1205931015840
rarr 0 as 120593 rarr 1198601and as 120593 rarr 119860
2
Moreover we have 119865(120593) ge 0 for 1198602le 120593 le 119860
1and 120593 is
strictly monotonic in any interval where 119865(120593) gt 0 Thus if1205931015840gt 0 at some points 120593will be strictly increasing until it gets
close to the next zero of 119865 Denoting this zero 1198601 we have
120593 uarr 1198601 What will happen to the solution when it approaches
1198601 Depending on whether the zero is double or simple 120593
has a different behavior We explore the two cases in turn
Theorem 11 (i) If 120593 has a simple zero at 120593 = 1198601 so that
119865(1198601) = 0 and 1198651015840(119860
1) lt 0 then the solution 120593 of (75) satisfies
120593 (120585) minus 1198601=1198651015840(1198601)
4(120585 minus 120578)
2
+ 119874 ((120585 minus 120578)4
)
119886119904 120585 uarr 120578
(77)
where 120593(120578) = 1198601
(ii) When 120593 approaches the double zero 1198601of 119865(120593) so that
1198651015840(1198601) = 0 and11986510158401015840(119860
1) gt 0 then the solution120593 of (75) satisfies
120593 (120585) minus 1198601sim 120572 exp(minus120585radic11986510158401015840 (119860
1)) 119886119904 120585 997888rarr infin (78)
for some constant 120572 Thus 120593 uarr 1198601exponentially as 120585 rarr infin
Mathematical Problems in Engineering 17
Proof (i) When 1198601= minus119860
2= 984858 and 120598 = minus119860
2
1120573 from (76)
119865(120593) has a simple zero at 120593 = 1198601 Then
(1205931015840)2
= (120593 minus 1198601) 1198651015840(1198601) + 119874 ((120593 minus 119860
1)2
)
as 120593 uarr 1198601
(79)
Using the fact that (1205931015840)2 ge 0 we know that 1198651015840(1198601) lt 0
Moreover
119889120585
119889120593=
1
radic(120593 minus 1198601) 1198651015840 (119860
1) + 119874 ((120593 minus 119860
1)2
)
(80)
Because
radic(120593 minus 1198601) 1198651015840 (119860
1) + 119874 ((120593 minus 119860
1)2
)
= radic1198601minus 120593(radicminus1198651015840 (119860
1) + 119874 (119860
1minus 120593))
1
radicminus1198651015840 (1198601) + 119874 (119860
1minus 120593)
=1
radicminus1198651015840 (1198601)
+ 119874 (1198601minus 120593)
(81)
we obtain119889120585
119889120593=
1
radicminus (1198601minus 120593) 1198651015840 (119860
1)
+ 119874 ((1198601minus 120593)12
) (82)
Integrating (82) yields
120578 minus 120585 =2
radicminus1198651015840 (1198601)
(1198601minus 120593)12
+ 119874((1198601minus 120593)32
) (83)
where 120578 satisfies 120593(120578) = 1198601 Thus
(120578 minus 120585)2
=4
minus1198651015840 (1198601)(1198601minus 120593) + 119874 ((119860
1minus 120593)2
) (84)
which implies 119874((1198601minus 120593)2) = 119874((120578 minus 120585)
4) Therefore we get
120593 (120585) = 1198601+1198651015840(1198601)
4(120585 minus 120578)
2
+ 119874 ((120585 minus 120578)4
)
as 120585 uarr 120578
(85)
where 120593(120578) = 1198601
(ii) When 1198601= minus119860
2lt 984858 and 120598 = minus119860
2
1120573 from (75) 119865(120593)
has a double zero at 120593 = 1198601 Then
(1205931015840)2
= (120593 minus 1198601)2
11986510158401015840(1198601) + 119874 ((120593 minus 119860
1)3
)
as 120593 uarr 1198601
(86)
Furthermore we get
119889120585
119889120593=
1
radic(120593 minus 1198601)2
11986510158401015840 (1198601) + 119874 ((120593 minus 119860
1)3
)
(87)
Observing that
radic(120593 minus 1198601)2
11986510158401015840 (1198601) + 119874 ((120593 minus 119860
1)3
)
= (1198601minus 120593) (radic11986510158401015840 (119860
1) + 119874 (119860
1minus 120593))
1
radic11986510158401015840 (1198601) + 119874 (119860
1minus 120593)
=1
radic11986510158401015840 (1198601)
+ 119874 (1198601minus 120593)
(88)
we obtain119889120585
119889120593=
1
(1198601minus 120593) 11986510158401015840 (119860
1)+ 119874 (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 119860
1= minus119860
2lt 984858 and 120598 = minus119860
2
1120573 Then (75)
becomes
radic9848582 minus 1205932
1205932 minus 1198602
1
119889120593 = ∓radic120573
2119889120585 (90)
Integrating both sides of (90) gives the following implicitexpressions of kink and antikink wave solutions
arctan120593
radic9848582 minus 1205932
+
radic9848582 minus 1198602
1
1198601
arctanhradic9848582 minus 119860
2
1
1198601
120593
radic9848582 minus 1205932
= plusmnradic120573
2120585
(91)
By letting 1198601rarr 984858 in (91) we get two kink compactons
which are given by
plusmn120593 =
984858 sinradic120573
2120585
10038161003816100381610038161205851003816100381610038161003816 le
120587
radic2120573
984858 120585 gt120587
radic2120573
minus984858 120585 lt minus120587
radic2120573
(92)
The graphs for the kink wave solutions (91) and kinkcompacton solutions (92) are shown in Figures 9 and 10respectively
Remark 12 The two kink compacton solutions (92) aredifferent from the well-known smooth kink wave solutionsIn comparison with kink wave solutions (91) the kink com-pacton solutions (92) have no exponential decay propertiesbut have compact supportThat is they minus a constant thedifferences identically vanish outside a finite core region
18 Mathematical Problems in Engineering
120593
120585minus1 1
05
minus05
(a)
minus1minus1
minus05
05
00120593
x t
0
1
01
11
(b)
Figure 9 Two- and three-dimensional graphs of the kink wave solution
1
1120593
120585
minus1
minus1
(a)
0
1
0
1
000510
120593
minus1 minus1
minus05
minus10
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 the119862(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 Newtonrsquos equation which we write in the form 119910
2=
119865(120593) where 119910 = 1205931015840
120585(120585) we solve Newtonrsquos equation 119910
2=
119865(120593) 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 the119862(3 2 2)equation
Actually for 119898 = 2119896 + 1 119896 isin N+ in 119862(119898 2 2) equation(9) the dynamical behavior of traveling wave solutions of(9) is similar to the case 119898 = 3 for 119898 = 2119896 119896 isin N+ in119862(119898 2 2) equation (9) the dynamical behavior of travelingwave solutions of (9) is similar to the case 119898 = 2 We areapplying the approach mentioned in this work to 119862(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
References
[1] C Rogers and W R Shadwick Backlund Transformation andTheir Applications Academic Press New York NY USA 1982
[2] W Malfliet and W Hereman ldquoThe tanh method I Exactsolutions of nonlinear evolution and wave equationsrdquo PhysicaScripta vol 54 no 6 pp 563ndash568 1996
[3] R Hirota Direct Method in Soliton Theory Springer BerlinGermany 1980
[4] J S Cohen Computer Algebra and Symbolic ComputationMathematical Methods A K Peters 2003
[5] S Lie ldquoUber die integration durch bestimmte integrale voneiner klasse linearer partieller differential eichungenrdquoArchiv derMathematik vol 6 pp 328ndash368 1881
[6] H-H Dai and X-H Zhao ldquoTraveling waves in a rod composedof amodifiedMooney-Rivlinmaterial I bifurcation of equilibraand non-singular caserdquo Proceedings of the Royal Society AMathematical Physical and Engineering Sciences vol 455 no1990 pp 3845ndash3874 1999
[7] H-H Dai and F-F Wang ldquoAsymptotic bifurcation solutionsfor compressions of a clamped nonlinearly elastic rectangletransition region and barrelling to a corner-like profilerdquo SIAMJournal on Applied Mathematics vol 70 no 7 pp 2673ndash26922010
[8] J Li and Z Liu ldquoSmooth and non-smooth traveling waves ina nonlinearly dispersive equationrdquo Applied Mathematical Mod-elling vol 25 no 1 pp 41ndash56 2000
[9] A Chen and J Li ldquoSingle peak solitary wave solutions forthe osmosis K(22) equation under inhomogeneous boundaryconditionrdquo Journal of Mathematical Analysis and Applicationsvol 369 no 2 pp 758ndash766 2010
[10] R Camassa and D D Holm ldquoAn integrable shallow waterequation with peaked solitonsrdquo Physical Review Letters vol 71no 11 pp 1661ndash1664 1993
[11] P Rosenau ldquoOn nonanalytic solitary waves formed by a nonlin-ear dispersionrdquo Physics Letters A vol 230 no 5-6 pp 305ndash3181997
[12] P Rosenau and J M Hyman ldquoCompactons solitons with finitewavelengthrdquo Physical Review Letters vol 70 no 5 pp 564ndash5671993
[13] L Zhang L-Q Chen and X Huo ldquoThe effects of horizontalsingular straight line in a generalized nonlinear Klein-Gordonmodel equationrdquo Nonlinear Dynamics vol 72 no 4 pp 789ndash801 2013
[14] J Li and H Dai On the Study of Singular Nonlinear TravelingWave Equations Dynamical System Approach Science PressBeijing China 2007
[15] A Chen S Wen S Tang W Huang and Z Qiao ldquoEffects ofquadratic singular curves in integrable equationsrdquo Studies inApplied Mathematics vol 134 no 1 pp 24ndash61 2015
[16] L Tian and J Yin ldquoNew compacton solutions and solitary wavesolutions of fully nonlinear generalized Camassa-Holm equa-tionsrdquo Chaos Solitons amp Fractals vol 20 no 2 pp 289ndash2992004
[17] L Zhang ldquoThe bifurcation and peakons for the specialC(3 2 2)equationrdquo Pramana vol 83 no 3 pp 331ndash340 2014
[18] J Lenells ldquoTraveling wave solutions of the CamassandashHolmequationrdquo Journal of Differential Equations vol 217 no 2 pp393ndash430 2005
[19] Z J Qiao and G P Zhang ldquoOn peaked and smooth solitons forthe Camassa-Holm equationrdquo Europhysics Letters vol 73 no 5pp 657ndash663 2006
Submit your manuscripts athttpwwwhindawicom
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Mathematical Problems in Engineering
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Stochastic AnalysisInternational Journal of
Mathematical Problems in Engineering 15
(ii) If 120593(0) = minus984858 then minus984858 le 120593 lt 119860 Through a similaranalysis we get a strictly increasing function 119866(120593) on theinterval [minus984858 119860) satisfying
119866 (120593) = radicminus120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870 (62)
where 119866(120593) is defined by (55) Let
1198662(120593) = 119866|
[minus984858119860)(120593) (63)
then 1198662(120593) is a strictly increasing function from [minus984858 119860) onto
[119870infin) so that we can solve for 120593 and obtain
1205932(120585) = 119866
minus1
2(radicminus
120573
2
10038161003816100381610038161205851003816100381610038161003816 + 119870) (64)
It is easy to check that 1205932satisfies
1205932(0) = minus984858
lim|120585|rarrinfin
1205932(120585) = 119860
1205931015840
2(0plusmn) = plusmn (119860 + 984858)radicminus
120573 (1198611+ 984858)
4984858
(65)
Therefore the solution 1205932defined by (64) is a peakon-like
solution whose graph is similar to those in Figure 5
Case 3 ((120590 120573) isin 11986016) In this case we have minus984858 lt 119861
2lt 119860 =
0 lt 1198611lt 984858 119861
2= minus1198611 and
1205931015840= minusradicminus
120573
2120593radic
120593 minus 1198612
1
1205932 minus 9848582sgn (120585) (66)
Hence from the separation of variables we have
radic9848582 minus 1205932
120593radic1198612
1minus 1205932
119889120593 = minusradicminus120573
2sgn (120585) 119889120585 (67)
Integrating (67) on the interval [0 120585] (or [120585 0]) leads tothe following implicit formula for the two smooth solitonsolutions
radic9848582 minus 1205932 + radic1198612
1minus 1205932
radic9848582 minus 1198612
1
(
radic(9848582 minus 1198612
1) 1205932
984858radic1198612
1minus 1205932 + 119861
1radic9848582 minus 1205932
)
984858
= exp(minusradicminus120573
2|119909 minus 119888119905|)
(68)
where 120593 isin (119860 1198611] Consider
radic9848582 minus 1198612
2
radic9848582 minus 1205932 + radic1198612
2minus 1205932
(
984858radic1198612
2minus 1205932 minus 119861
2radic9848582 minus 1205932
radic(9848582 minus 1198612
2) 1205932
)
984858
= exp(radicminus120573
2|119909 minus 119888119905|)
(69)
where 120593 isin [1198612 119860)
Case 4 ((120590 120573) isin 11986017) In this case we have minus984858 = 119861
2lt 119860 =
0 lt 1198611= 984858 and
1205931015840= minusradicminus
120573
2120593 sgn (120585) (70)
Choosing 120593(0) = 984858 (or minus984858) as initial value we get
int
120593
984858
1
120593119889120593 = minusint
120585
0
radicminus120573
2sgn (120585) 119889120585 for 119860 lt 120593 le 984858
int
120593
minus984858
1
120593119889120593 = minusint
120585
0
radicminus120573
2sgn (120585) 119889120585 for minus 984858 le 120593 lt 119860
(71)
which immediately yield the peakon solutions
120593 (119909 minus 119888119905) = plusmn984858 exp(minusradicminus120573
2|119909 minus 119888119905|) (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 the119862(3 2 2) equation (10) In order to study kinkwave solutionswe assume that
lim120585rarrinfin
120593 (120585) = 1198601
lim120585rarrminusinfin
120593 (120585) = 1198602
(73)
16 Mathematical Problems in Engineering
2 4
02
04
06
08
10
minus2minus4
120593
x
(a)
0 2
02
00
05
10
minus2
120593
x
t
0
2
minus2
(b)
Figure 8 Two- and three-dimensional graphs of the peakon solution
where 1198601gt 1198602 Substituting the boundary condition (73)
into (14) generates
(1205931015840)2
=
120573 (120593 minus 1198601) (120593 minus 119860
2) [1205932+ (1198601+ 1198602) 120593 + 119860
2
1+ 11986011198602+ 1198602
2+ 2120598120573]
2 (9848582 minus 1205932)fl119865 (120593) (74)
The nonlinear differential equation (74) may sustaindifferent kinds of nonlinear excitations In what follows weconfine our attention to the cases 119860
2= minus119860
1and 120598 =
minus1198602
1120573 which describe kinks and kink compactons which
play an important role in the dynamics systems Under theseconsiderations (74) reduces to
(1205931015840)2
=120573 (120593 minus 119860
1)2
(120593 minus 1198602)2
2 (9848582 minus 1205932) (75)
If 1198601lt 984858 then from the phase analysis in Section 2 (see
Figure 4(10)) we know that (1198601 0) and (119860
2 0) are two saddle
points of (13) and the kink solutions can be obtained fromthe two heteroclinic orbits connecting (120601 119910) = (119860
1 0) and
(1198602 0) When 119860
1increases upon reaching 984858 that is 119860
1= 984858
(75) becomes
(1205931015840)2
= minus120573 (120593 minus 119860
1) (120593 minus 119860
2)
2 (76)
and the ellipse 21199102+120573(120593minus1198601)(120593minus119860
2) = 0 (see Figure 4(11))
which is tangent to the singular lines 120593 = 984858 and 120593 = minus984858 atpoints (119860
1 0) and (119860
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 120593 is a kink wave solutions of (75)
or (76) we have 1205931015840
rarr 0 as 120593 rarr 1198601and as 120593 rarr 119860
2
Moreover we have 119865(120593) ge 0 for 1198602le 120593 le 119860
1and 120593 is
strictly monotonic in any interval where 119865(120593) gt 0 Thus if1205931015840gt 0 at some points 120593will be strictly increasing until it gets
close to the next zero of 119865 Denoting this zero 1198601 we have
120593 uarr 1198601 What will happen to the solution when it approaches
1198601 Depending on whether the zero is double or simple 120593
has a different behavior We explore the two cases in turn
Theorem 11 (i) If 120593 has a simple zero at 120593 = 1198601 so that
119865(1198601) = 0 and 1198651015840(119860
1) lt 0 then the solution 120593 of (75) satisfies
120593 (120585) minus 1198601=1198651015840(1198601)
4(120585 minus 120578)
2
+ 119874 ((120585 minus 120578)4
)
119886119904 120585 uarr 120578
(77)
where 120593(120578) = 1198601
(ii) When 120593 approaches the double zero 1198601of 119865(120593) so that
1198651015840(1198601) = 0 and11986510158401015840(119860
1) gt 0 then the solution120593 of (75) satisfies
120593 (120585) minus 1198601sim 120572 exp(minus120585radic11986510158401015840 (119860
1)) 119886119904 120585 997888rarr infin (78)
for some constant 120572 Thus 120593 uarr 1198601exponentially as 120585 rarr infin
Mathematical Problems in Engineering 17
Proof (i) When 1198601= minus119860
2= 984858 and 120598 = minus119860
2
1120573 from (76)
119865(120593) has a simple zero at 120593 = 1198601 Then
(1205931015840)2
= (120593 minus 1198601) 1198651015840(1198601) + 119874 ((120593 minus 119860
1)2
)
as 120593 uarr 1198601
(79)
Using the fact that (1205931015840)2 ge 0 we know that 1198651015840(1198601) lt 0
Moreover
119889120585
119889120593=
1
radic(120593 minus 1198601) 1198651015840 (119860
1) + 119874 ((120593 minus 119860
1)2
)
(80)
Because
radic(120593 minus 1198601) 1198651015840 (119860
1) + 119874 ((120593 minus 119860
1)2
)
= radic1198601minus 120593(radicminus1198651015840 (119860
1) + 119874 (119860
1minus 120593))
1
radicminus1198651015840 (1198601) + 119874 (119860
1minus 120593)
=1
radicminus1198651015840 (1198601)
+ 119874 (1198601minus 120593)
(81)
we obtain119889120585
119889120593=
1
radicminus (1198601minus 120593) 1198651015840 (119860
1)
+ 119874 ((1198601minus 120593)12
) (82)
Integrating (82) yields
120578 minus 120585 =2
radicminus1198651015840 (1198601)
(1198601minus 120593)12
+ 119874((1198601minus 120593)32
) (83)
where 120578 satisfies 120593(120578) = 1198601 Thus
(120578 minus 120585)2
=4
minus1198651015840 (1198601)(1198601minus 120593) + 119874 ((119860
1minus 120593)2
) (84)
which implies 119874((1198601minus 120593)2) = 119874((120578 minus 120585)
4) Therefore we get
120593 (120585) = 1198601+1198651015840(1198601)
4(120585 minus 120578)
2
+ 119874 ((120585 minus 120578)4
)
as 120585 uarr 120578
(85)
where 120593(120578) = 1198601
(ii) When 1198601= minus119860
2lt 984858 and 120598 = minus119860
2
1120573 from (75) 119865(120593)
has a double zero at 120593 = 1198601 Then
(1205931015840)2
= (120593 minus 1198601)2
11986510158401015840(1198601) + 119874 ((120593 minus 119860
1)3
)
as 120593 uarr 1198601
(86)
Furthermore we get
119889120585
119889120593=
1
radic(120593 minus 1198601)2
11986510158401015840 (1198601) + 119874 ((120593 minus 119860
1)3
)
(87)
Observing that
radic(120593 minus 1198601)2
11986510158401015840 (1198601) + 119874 ((120593 minus 119860
1)3
)
= (1198601minus 120593) (radic11986510158401015840 (119860
1) + 119874 (119860
1minus 120593))
1
radic11986510158401015840 (1198601) + 119874 (119860
1minus 120593)
=1
radic11986510158401015840 (1198601)
+ 119874 (1198601minus 120593)
(88)
we obtain119889120585
119889120593=
1
(1198601minus 120593) 11986510158401015840 (119860
1)+ 119874 (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 119860
1= minus119860
2lt 984858 and 120598 = minus119860
2
1120573 Then (75)
becomes
radic9848582 minus 1205932
1205932 minus 1198602
1
119889120593 = ∓radic120573
2119889120585 (90)
Integrating both sides of (90) gives the following implicitexpressions of kink and antikink wave solutions
arctan120593
radic9848582 minus 1205932
+
radic9848582 minus 1198602
1
1198601
arctanhradic9848582 minus 119860
2
1
1198601
120593
radic9848582 minus 1205932
= plusmnradic120573
2120585
(91)
By letting 1198601rarr 984858 in (91) we get two kink compactons
which are given by
plusmn120593 =
984858 sinradic120573
2120585
10038161003816100381610038161205851003816100381610038161003816 le
120587
radic2120573
984858 120585 gt120587
radic2120573
minus984858 120585 lt minus120587
radic2120573
(92)
The graphs for the kink wave solutions (91) and kinkcompacton solutions (92) are shown in Figures 9 and 10respectively
Remark 12 The two kink compacton solutions (92) aredifferent from the well-known smooth kink wave solutionsIn comparison with kink wave solutions (91) the kink com-pacton solutions (92) have no exponential decay propertiesbut have compact supportThat is they minus a constant thedifferences identically vanish outside a finite core region
18 Mathematical Problems in Engineering
120593
120585minus1 1
05
minus05
(a)
minus1minus1
minus05
05
00120593
x t
0
1
01
11
(b)
Figure 9 Two- and three-dimensional graphs of the kink wave solution
1
1120593
120585
minus1
minus1
(a)
0
1
0
1
000510
120593
minus1 minus1
minus05
minus10
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 the119862(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 Newtonrsquos equation which we write in the form 119910
2=
119865(120593) where 119910 = 1205931015840
120585(120585) we solve Newtonrsquos equation 119910
2=
119865(120593) 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 the119862(3 2 2)equation
Actually for 119898 = 2119896 + 1 119896 isin N+ in 119862(119898 2 2) equation(9) the dynamical behavior of traveling wave solutions of(9) is similar to the case 119898 = 3 for 119898 = 2119896 119896 isin N+ in119862(119898 2 2) equation (9) the dynamical behavior of travelingwave solutions of (9) is similar to the case 119898 = 2 We areapplying the approach mentioned in this work to 119862(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
References
[1] C Rogers and W R Shadwick Backlund Transformation andTheir Applications Academic Press New York NY USA 1982
[2] W Malfliet and W Hereman ldquoThe tanh method I Exactsolutions of nonlinear evolution and wave equationsrdquo PhysicaScripta vol 54 no 6 pp 563ndash568 1996
[3] R Hirota Direct Method in Soliton Theory Springer BerlinGermany 1980
[4] J S Cohen Computer Algebra and Symbolic ComputationMathematical Methods A K Peters 2003
[5] S Lie ldquoUber die integration durch bestimmte integrale voneiner klasse linearer partieller differential eichungenrdquoArchiv derMathematik vol 6 pp 328ndash368 1881
[6] H-H Dai and X-H Zhao ldquoTraveling waves in a rod composedof amodifiedMooney-Rivlinmaterial I bifurcation of equilibraand non-singular caserdquo Proceedings of the Royal Society AMathematical Physical and Engineering Sciences vol 455 no1990 pp 3845ndash3874 1999
[7] H-H Dai and F-F Wang ldquoAsymptotic bifurcation solutionsfor compressions of a clamped nonlinearly elastic rectangletransition region and barrelling to a corner-like profilerdquo SIAMJournal on Applied Mathematics vol 70 no 7 pp 2673ndash26922010
[8] J Li and Z Liu ldquoSmooth and non-smooth traveling waves ina nonlinearly dispersive equationrdquo Applied Mathematical Mod-elling vol 25 no 1 pp 41ndash56 2000
[9] A Chen and J Li ldquoSingle peak solitary wave solutions forthe osmosis K(22) equation under inhomogeneous boundaryconditionrdquo Journal of Mathematical Analysis and Applicationsvol 369 no 2 pp 758ndash766 2010
[10] R Camassa and D D Holm ldquoAn integrable shallow waterequation with peaked solitonsrdquo Physical Review Letters vol 71no 11 pp 1661ndash1664 1993
[11] P Rosenau ldquoOn nonanalytic solitary waves formed by a nonlin-ear dispersionrdquo Physics Letters A vol 230 no 5-6 pp 305ndash3181997
[12] P Rosenau and J M Hyman ldquoCompactons solitons with finitewavelengthrdquo Physical Review Letters vol 70 no 5 pp 564ndash5671993
[13] L Zhang L-Q Chen and X Huo ldquoThe effects of horizontalsingular straight line in a generalized nonlinear Klein-Gordonmodel equationrdquo Nonlinear Dynamics vol 72 no 4 pp 789ndash801 2013
[14] J Li and H Dai On the Study of Singular Nonlinear TravelingWave Equations Dynamical System Approach Science PressBeijing China 2007
[15] A Chen S Wen S Tang W Huang and Z Qiao ldquoEffects ofquadratic singular curves in integrable equationsrdquo Studies inApplied Mathematics vol 134 no 1 pp 24ndash61 2015
[16] L Tian and J Yin ldquoNew compacton solutions and solitary wavesolutions of fully nonlinear generalized Camassa-Holm equa-tionsrdquo Chaos Solitons amp Fractals vol 20 no 2 pp 289ndash2992004
[17] L Zhang ldquoThe bifurcation and peakons for the specialC(3 2 2)equationrdquo Pramana vol 83 no 3 pp 331ndash340 2014
[18] J Lenells ldquoTraveling wave solutions of the CamassandashHolmequationrdquo Journal of Differential Equations vol 217 no 2 pp393ndash430 2005
[19] Z J Qiao and G P Zhang ldquoOn peaked and smooth solitons forthe Camassa-Holm equationrdquo Europhysics Letters vol 73 no 5pp 657ndash663 2006
Submit your manuscripts athttpwwwhindawicom
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Mathematical Problems in Engineering
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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
16 Mathematical Problems in Engineering
2 4
02
04
06
08
10
minus2minus4
120593
x
(a)
0 2
02
00
05
10
minus2
120593
x
t
0
2
minus2
(b)
Figure 8 Two- and three-dimensional graphs of the peakon solution
where 1198601gt 1198602 Substituting the boundary condition (73)
into (14) generates
(1205931015840)2
=
120573 (120593 minus 1198601) (120593 minus 119860
2) [1205932+ (1198601+ 1198602) 120593 + 119860
2
1+ 11986011198602+ 1198602
2+ 2120598120573]
2 (9848582 minus 1205932)fl119865 (120593) (74)
The nonlinear differential equation (74) may sustaindifferent kinds of nonlinear excitations In what follows weconfine our attention to the cases 119860
2= minus119860
1and 120598 =
minus1198602
1120573 which describe kinks and kink compactons which
play an important role in the dynamics systems Under theseconsiderations (74) reduces to
(1205931015840)2
=120573 (120593 minus 119860
1)2
(120593 minus 1198602)2
2 (9848582 minus 1205932) (75)
If 1198601lt 984858 then from the phase analysis in Section 2 (see
Figure 4(10)) we know that (1198601 0) and (119860
2 0) are two saddle
points of (13) and the kink solutions can be obtained fromthe two heteroclinic orbits connecting (120601 119910) = (119860
1 0) and
(1198602 0) When 119860
1increases upon reaching 984858 that is 119860
1= 984858
(75) becomes
(1205931015840)2
= minus120573 (120593 minus 119860
1) (120593 minus 119860
2)
2 (76)
and the ellipse 21199102+120573(120593minus1198601)(120593minus119860
2) = 0 (see Figure 4(11))
which is tangent to the singular lines 120593 = 984858 and 120593 = minus984858 atpoints (119860
1 0) and (119860
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 120593 is a kink wave solutions of (75)
or (76) we have 1205931015840
rarr 0 as 120593 rarr 1198601and as 120593 rarr 119860
2
Moreover we have 119865(120593) ge 0 for 1198602le 120593 le 119860
1and 120593 is
strictly monotonic in any interval where 119865(120593) gt 0 Thus if1205931015840gt 0 at some points 120593will be strictly increasing until it gets
close to the next zero of 119865 Denoting this zero 1198601 we have
120593 uarr 1198601 What will happen to the solution when it approaches
1198601 Depending on whether the zero is double or simple 120593
has a different behavior We explore the two cases in turn
Theorem 11 (i) If 120593 has a simple zero at 120593 = 1198601 so that
119865(1198601) = 0 and 1198651015840(119860
1) lt 0 then the solution 120593 of (75) satisfies
120593 (120585) minus 1198601=1198651015840(1198601)
4(120585 minus 120578)
2
+ 119874 ((120585 minus 120578)4
)
119886119904 120585 uarr 120578
(77)
where 120593(120578) = 1198601
(ii) When 120593 approaches the double zero 1198601of 119865(120593) so that
1198651015840(1198601) = 0 and11986510158401015840(119860
1) gt 0 then the solution120593 of (75) satisfies
120593 (120585) minus 1198601sim 120572 exp(minus120585radic11986510158401015840 (119860
1)) 119886119904 120585 997888rarr infin (78)
for some constant 120572 Thus 120593 uarr 1198601exponentially as 120585 rarr infin
Mathematical Problems in Engineering 17
Proof (i) When 1198601= minus119860
2= 984858 and 120598 = minus119860
2
1120573 from (76)
119865(120593) has a simple zero at 120593 = 1198601 Then
(1205931015840)2
= (120593 minus 1198601) 1198651015840(1198601) + 119874 ((120593 minus 119860
1)2
)
as 120593 uarr 1198601
(79)
Using the fact that (1205931015840)2 ge 0 we know that 1198651015840(1198601) lt 0
Moreover
119889120585
119889120593=
1
radic(120593 minus 1198601) 1198651015840 (119860
1) + 119874 ((120593 minus 119860
1)2
)
(80)
Because
radic(120593 minus 1198601) 1198651015840 (119860
1) + 119874 ((120593 minus 119860
1)2
)
= radic1198601minus 120593(radicminus1198651015840 (119860
1) + 119874 (119860
1minus 120593))
1
radicminus1198651015840 (1198601) + 119874 (119860
1minus 120593)
=1
radicminus1198651015840 (1198601)
+ 119874 (1198601minus 120593)
(81)
we obtain119889120585
119889120593=
1
radicminus (1198601minus 120593) 1198651015840 (119860
1)
+ 119874 ((1198601minus 120593)12
) (82)
Integrating (82) yields
120578 minus 120585 =2
radicminus1198651015840 (1198601)
(1198601minus 120593)12
+ 119874((1198601minus 120593)32
) (83)
where 120578 satisfies 120593(120578) = 1198601 Thus
(120578 minus 120585)2
=4
minus1198651015840 (1198601)(1198601minus 120593) + 119874 ((119860
1minus 120593)2
) (84)
which implies 119874((1198601minus 120593)2) = 119874((120578 minus 120585)
4) Therefore we get
120593 (120585) = 1198601+1198651015840(1198601)
4(120585 minus 120578)
2
+ 119874 ((120585 minus 120578)4
)
as 120585 uarr 120578
(85)
where 120593(120578) = 1198601
(ii) When 1198601= minus119860
2lt 984858 and 120598 = minus119860
2
1120573 from (75) 119865(120593)
has a double zero at 120593 = 1198601 Then
(1205931015840)2
= (120593 minus 1198601)2
11986510158401015840(1198601) + 119874 ((120593 minus 119860
1)3
)
as 120593 uarr 1198601
(86)
Furthermore we get
119889120585
119889120593=
1
radic(120593 minus 1198601)2
11986510158401015840 (1198601) + 119874 ((120593 minus 119860
1)3
)
(87)
Observing that
radic(120593 minus 1198601)2
11986510158401015840 (1198601) + 119874 ((120593 minus 119860
1)3
)
= (1198601minus 120593) (radic11986510158401015840 (119860
1) + 119874 (119860
1minus 120593))
1
radic11986510158401015840 (1198601) + 119874 (119860
1minus 120593)
=1
radic11986510158401015840 (1198601)
+ 119874 (1198601minus 120593)
(88)
we obtain119889120585
119889120593=
1
(1198601minus 120593) 11986510158401015840 (119860
1)+ 119874 (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 119860
1= minus119860
2lt 984858 and 120598 = minus119860
2
1120573 Then (75)
becomes
radic9848582 minus 1205932
1205932 minus 1198602
1
119889120593 = ∓radic120573
2119889120585 (90)
Integrating both sides of (90) gives the following implicitexpressions of kink and antikink wave solutions
arctan120593
radic9848582 minus 1205932
+
radic9848582 minus 1198602
1
1198601
arctanhradic9848582 minus 119860
2
1
1198601
120593
radic9848582 minus 1205932
= plusmnradic120573
2120585
(91)
By letting 1198601rarr 984858 in (91) we get two kink compactons
which are given by
plusmn120593 =
984858 sinradic120573
2120585
10038161003816100381610038161205851003816100381610038161003816 le
120587
radic2120573
984858 120585 gt120587
radic2120573
minus984858 120585 lt minus120587
radic2120573
(92)
The graphs for the kink wave solutions (91) and kinkcompacton solutions (92) are shown in Figures 9 and 10respectively
Remark 12 The two kink compacton solutions (92) aredifferent from the well-known smooth kink wave solutionsIn comparison with kink wave solutions (91) the kink com-pacton solutions (92) have no exponential decay propertiesbut have compact supportThat is they minus a constant thedifferences identically vanish outside a finite core region
18 Mathematical Problems in Engineering
120593
120585minus1 1
05
minus05
(a)
minus1minus1
minus05
05
00120593
x t
0
1
01
11
(b)
Figure 9 Two- and three-dimensional graphs of the kink wave solution
1
1120593
120585
minus1
minus1
(a)
0
1
0
1
000510
120593
minus1 minus1
minus05
minus10
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 the119862(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 Newtonrsquos equation which we write in the form 119910
2=
119865(120593) where 119910 = 1205931015840
120585(120585) we solve Newtonrsquos equation 119910
2=
119865(120593) 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 the119862(3 2 2)equation
Actually for 119898 = 2119896 + 1 119896 isin N+ in 119862(119898 2 2) equation(9) the dynamical behavior of traveling wave solutions of(9) is similar to the case 119898 = 3 for 119898 = 2119896 119896 isin N+ in119862(119898 2 2) equation (9) the dynamical behavior of travelingwave solutions of (9) is similar to the case 119898 = 2 We areapplying the approach mentioned in this work to 119862(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
References
[1] C Rogers and W R Shadwick Backlund Transformation andTheir Applications Academic Press New York NY USA 1982
[2] W Malfliet and W Hereman ldquoThe tanh method I Exactsolutions of nonlinear evolution and wave equationsrdquo PhysicaScripta vol 54 no 6 pp 563ndash568 1996
[3] R Hirota Direct Method in Soliton Theory Springer BerlinGermany 1980
[4] J S Cohen Computer Algebra and Symbolic ComputationMathematical Methods A K Peters 2003
[5] S Lie ldquoUber die integration durch bestimmte integrale voneiner klasse linearer partieller differential eichungenrdquoArchiv derMathematik vol 6 pp 328ndash368 1881
[6] H-H Dai and X-H Zhao ldquoTraveling waves in a rod composedof amodifiedMooney-Rivlinmaterial I bifurcation of equilibraand non-singular caserdquo Proceedings of the Royal Society AMathematical Physical and Engineering Sciences vol 455 no1990 pp 3845ndash3874 1999
[7] H-H Dai and F-F Wang ldquoAsymptotic bifurcation solutionsfor compressions of a clamped nonlinearly elastic rectangletransition region and barrelling to a corner-like profilerdquo SIAMJournal on Applied Mathematics vol 70 no 7 pp 2673ndash26922010
[8] J Li and Z Liu ldquoSmooth and non-smooth traveling waves ina nonlinearly dispersive equationrdquo Applied Mathematical Mod-elling vol 25 no 1 pp 41ndash56 2000
[9] A Chen and J Li ldquoSingle peak solitary wave solutions forthe osmosis K(22) equation under inhomogeneous boundaryconditionrdquo Journal of Mathematical Analysis and Applicationsvol 369 no 2 pp 758ndash766 2010
[10] R Camassa and D D Holm ldquoAn integrable shallow waterequation with peaked solitonsrdquo Physical Review Letters vol 71no 11 pp 1661ndash1664 1993
[11] P Rosenau ldquoOn nonanalytic solitary waves formed by a nonlin-ear dispersionrdquo Physics Letters A vol 230 no 5-6 pp 305ndash3181997
[12] P Rosenau and J M Hyman ldquoCompactons solitons with finitewavelengthrdquo Physical Review Letters vol 70 no 5 pp 564ndash5671993
[13] L Zhang L-Q Chen and X Huo ldquoThe effects of horizontalsingular straight line in a generalized nonlinear Klein-Gordonmodel equationrdquo Nonlinear Dynamics vol 72 no 4 pp 789ndash801 2013
[14] J Li and H Dai On the Study of Singular Nonlinear TravelingWave Equations Dynamical System Approach Science PressBeijing China 2007
[15] A Chen S Wen S Tang W Huang and Z Qiao ldquoEffects ofquadratic singular curves in integrable equationsrdquo Studies inApplied Mathematics vol 134 no 1 pp 24ndash61 2015
[16] L Tian and J Yin ldquoNew compacton solutions and solitary wavesolutions of fully nonlinear generalized Camassa-Holm equa-tionsrdquo Chaos Solitons amp Fractals vol 20 no 2 pp 289ndash2992004
[17] L Zhang ldquoThe bifurcation and peakons for the specialC(3 2 2)equationrdquo Pramana vol 83 no 3 pp 331ndash340 2014
[18] J Lenells ldquoTraveling wave solutions of the CamassandashHolmequationrdquo Journal of Differential Equations vol 217 no 2 pp393ndash430 2005
[19] Z J Qiao and G P Zhang ldquoOn peaked and smooth solitons forthe Camassa-Holm equationrdquo Europhysics Letters vol 73 no 5pp 657ndash663 2006
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Mathematical Problems in Engineering 17
Proof (i) When 1198601= minus119860
2= 984858 and 120598 = minus119860
2
1120573 from (76)
119865(120593) has a simple zero at 120593 = 1198601 Then
(1205931015840)2
= (120593 minus 1198601) 1198651015840(1198601) + 119874 ((120593 minus 119860
1)2
)
as 120593 uarr 1198601
(79)
Using the fact that (1205931015840)2 ge 0 we know that 1198651015840(1198601) lt 0
Moreover
119889120585
119889120593=
1
radic(120593 minus 1198601) 1198651015840 (119860
1) + 119874 ((120593 minus 119860
1)2
)
(80)
Because
radic(120593 minus 1198601) 1198651015840 (119860
1) + 119874 ((120593 minus 119860
1)2
)
= radic1198601minus 120593(radicminus1198651015840 (119860
1) + 119874 (119860
1minus 120593))
1
radicminus1198651015840 (1198601) + 119874 (119860
1minus 120593)
=1
radicminus1198651015840 (1198601)
+ 119874 (1198601minus 120593)
(81)
we obtain119889120585
119889120593=
1
radicminus (1198601minus 120593) 1198651015840 (119860
1)
+ 119874 ((1198601minus 120593)12
) (82)
Integrating (82) yields
120578 minus 120585 =2
radicminus1198651015840 (1198601)
(1198601minus 120593)12
+ 119874((1198601minus 120593)32
) (83)
where 120578 satisfies 120593(120578) = 1198601 Thus
(120578 minus 120585)2
=4
minus1198651015840 (1198601)(1198601minus 120593) + 119874 ((119860
1minus 120593)2
) (84)
which implies 119874((1198601minus 120593)2) = 119874((120578 minus 120585)
4) Therefore we get
120593 (120585) = 1198601+1198651015840(1198601)
4(120585 minus 120578)
2
+ 119874 ((120585 minus 120578)4
)
as 120585 uarr 120578
(85)
where 120593(120578) = 1198601
(ii) When 1198601= minus119860
2lt 984858 and 120598 = minus119860
2
1120573 from (75) 119865(120593)
has a double zero at 120593 = 1198601 Then
(1205931015840)2
= (120593 minus 1198601)2
11986510158401015840(1198601) + 119874 ((120593 minus 119860
1)3
)
as 120593 uarr 1198601
(86)
Furthermore we get
119889120585
119889120593=
1
radic(120593 minus 1198601)2
11986510158401015840 (1198601) + 119874 ((120593 minus 119860
1)3
)
(87)
Observing that
radic(120593 minus 1198601)2
11986510158401015840 (1198601) + 119874 ((120593 minus 119860
1)3
)
= (1198601minus 120593) (radic11986510158401015840 (119860
1) + 119874 (119860
1minus 120593))
1
radic11986510158401015840 (1198601) + 119874 (119860
1minus 120593)
=1
radic11986510158401015840 (1198601)
+ 119874 (1198601minus 120593)
(88)
we obtain119889120585
119889120593=
1
(1198601minus 120593) 11986510158401015840 (119860
1)+ 119874 (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 119860
1= minus119860
2lt 984858 and 120598 = minus119860
2
1120573 Then (75)
becomes
radic9848582 minus 1205932
1205932 minus 1198602
1
119889120593 = ∓radic120573
2119889120585 (90)
Integrating both sides of (90) gives the following implicitexpressions of kink and antikink wave solutions
arctan120593
radic9848582 minus 1205932
+
radic9848582 minus 1198602
1
1198601
arctanhradic9848582 minus 119860
2
1
1198601
120593
radic9848582 minus 1205932
= plusmnradic120573
2120585
(91)
By letting 1198601rarr 984858 in (91) we get two kink compactons
which are given by
plusmn120593 =
984858 sinradic120573
2120585
10038161003816100381610038161205851003816100381610038161003816 le
120587
radic2120573
984858 120585 gt120587
radic2120573
minus984858 120585 lt minus120587
radic2120573
(92)
The graphs for the kink wave solutions (91) and kinkcompacton solutions (92) are shown in Figures 9 and 10respectively
Remark 12 The two kink compacton solutions (92) aredifferent from the well-known smooth kink wave solutionsIn comparison with kink wave solutions (91) the kink com-pacton solutions (92) have no exponential decay propertiesbut have compact supportThat is they minus a constant thedifferences identically vanish outside a finite core region
18 Mathematical Problems in Engineering
120593
120585minus1 1
05
minus05
(a)
minus1minus1
minus05
05
00120593
x t
0
1
01
11
(b)
Figure 9 Two- and three-dimensional graphs of the kink wave solution
1
1120593
120585
minus1
minus1
(a)
0
1
0
1
000510
120593
minus1 minus1
minus05
minus10
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 the119862(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 Newtonrsquos equation which we write in the form 119910
2=
119865(120593) where 119910 = 1205931015840
120585(120585) we solve Newtonrsquos equation 119910
2=
119865(120593) 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 the119862(3 2 2)equation
Actually for 119898 = 2119896 + 1 119896 isin N+ in 119862(119898 2 2) equation(9) the dynamical behavior of traveling wave solutions of(9) is similar to the case 119898 = 3 for 119898 = 2119896 119896 isin N+ in119862(119898 2 2) equation (9) the dynamical behavior of travelingwave solutions of (9) is similar to the case 119898 = 2 We areapplying the approach mentioned in this work to 119862(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
References
[1] C Rogers and W R Shadwick Backlund Transformation andTheir Applications Academic Press New York NY USA 1982
[2] W Malfliet and W Hereman ldquoThe tanh method I Exactsolutions of nonlinear evolution and wave equationsrdquo PhysicaScripta vol 54 no 6 pp 563ndash568 1996
[3] R Hirota Direct Method in Soliton Theory Springer BerlinGermany 1980
[4] J S Cohen Computer Algebra and Symbolic ComputationMathematical Methods A K Peters 2003
[5] S Lie ldquoUber die integration durch bestimmte integrale voneiner klasse linearer partieller differential eichungenrdquoArchiv derMathematik vol 6 pp 328ndash368 1881
[6] H-H Dai and X-H Zhao ldquoTraveling waves in a rod composedof amodifiedMooney-Rivlinmaterial I bifurcation of equilibraand non-singular caserdquo Proceedings of the Royal Society AMathematical Physical and Engineering Sciences vol 455 no1990 pp 3845ndash3874 1999
[7] H-H Dai and F-F Wang ldquoAsymptotic bifurcation solutionsfor compressions of a clamped nonlinearly elastic rectangletransition region and barrelling to a corner-like profilerdquo SIAMJournal on Applied Mathematics vol 70 no 7 pp 2673ndash26922010
[8] J Li and Z Liu ldquoSmooth and non-smooth traveling waves ina nonlinearly dispersive equationrdquo Applied Mathematical Mod-elling vol 25 no 1 pp 41ndash56 2000
[9] A Chen and J Li ldquoSingle peak solitary wave solutions forthe osmosis K(22) equation under inhomogeneous boundaryconditionrdquo Journal of Mathematical Analysis and Applicationsvol 369 no 2 pp 758ndash766 2010
[10] R Camassa and D D Holm ldquoAn integrable shallow waterequation with peaked solitonsrdquo Physical Review Letters vol 71no 11 pp 1661ndash1664 1993
[11] P Rosenau ldquoOn nonanalytic solitary waves formed by a nonlin-ear dispersionrdquo Physics Letters A vol 230 no 5-6 pp 305ndash3181997
[12] P Rosenau and J M Hyman ldquoCompactons solitons with finitewavelengthrdquo Physical Review Letters vol 70 no 5 pp 564ndash5671993
[13] L Zhang L-Q Chen and X Huo ldquoThe effects of horizontalsingular straight line in a generalized nonlinear Klein-Gordonmodel equationrdquo Nonlinear Dynamics vol 72 no 4 pp 789ndash801 2013
[14] J Li and H Dai On the Study of Singular Nonlinear TravelingWave Equations Dynamical System Approach Science PressBeijing China 2007
[15] A Chen S Wen S Tang W Huang and Z Qiao ldquoEffects ofquadratic singular curves in integrable equationsrdquo Studies inApplied Mathematics vol 134 no 1 pp 24ndash61 2015
[16] L Tian and J Yin ldquoNew compacton solutions and solitary wavesolutions of fully nonlinear generalized Camassa-Holm equa-tionsrdquo Chaos Solitons amp Fractals vol 20 no 2 pp 289ndash2992004
[17] L Zhang ldquoThe bifurcation and peakons for the specialC(3 2 2)equationrdquo Pramana vol 83 no 3 pp 331ndash340 2014
[18] J Lenells ldquoTraveling wave solutions of the CamassandashHolmequationrdquo Journal of Differential Equations vol 217 no 2 pp393ndash430 2005
[19] Z J Qiao and G P Zhang ldquoOn peaked and smooth solitons forthe Camassa-Holm equationrdquo Europhysics Letters vol 73 no 5pp 657ndash663 2006
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
18 Mathematical Problems in Engineering
120593
120585minus1 1
05
minus05
(a)
minus1minus1
minus05
05
00120593
x t
0
1
01
11
(b)
Figure 9 Two- and three-dimensional graphs of the kink wave solution
1
1120593
120585
minus1
minus1
(a)
0
1
0
1
000510
120593
minus1 minus1
minus05
minus10
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 the119862(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 Newtonrsquos equation which we write in the form 119910
2=
119865(120593) where 119910 = 1205931015840
120585(120585) we solve Newtonrsquos equation 119910
2=
119865(120593) 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 the119862(3 2 2)equation
Actually for 119898 = 2119896 + 1 119896 isin N+ in 119862(119898 2 2) equation(9) the dynamical behavior of traveling wave solutions of(9) is similar to the case 119898 = 3 for 119898 = 2119896 119896 isin N+ in119862(119898 2 2) equation (9) the dynamical behavior of travelingwave solutions of (9) is similar to the case 119898 = 2 We areapplying the approach mentioned in this work to 119862(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
References
[1] C Rogers and W R Shadwick Backlund Transformation andTheir Applications Academic Press New York NY USA 1982
[2] W Malfliet and W Hereman ldquoThe tanh method I Exactsolutions of nonlinear evolution and wave equationsrdquo PhysicaScripta vol 54 no 6 pp 563ndash568 1996
[3] R Hirota Direct Method in Soliton Theory Springer BerlinGermany 1980
[4] J S Cohen Computer Algebra and Symbolic ComputationMathematical Methods A K Peters 2003
[5] S Lie ldquoUber die integration durch bestimmte integrale voneiner klasse linearer partieller differential eichungenrdquoArchiv derMathematik vol 6 pp 328ndash368 1881
[6] H-H Dai and X-H Zhao ldquoTraveling waves in a rod composedof amodifiedMooney-Rivlinmaterial I bifurcation of equilibraand non-singular caserdquo Proceedings of the Royal Society AMathematical Physical and Engineering Sciences vol 455 no1990 pp 3845ndash3874 1999
[7] H-H Dai and F-F Wang ldquoAsymptotic bifurcation solutionsfor compressions of a clamped nonlinearly elastic rectangletransition region and barrelling to a corner-like profilerdquo SIAMJournal on Applied Mathematics vol 70 no 7 pp 2673ndash26922010
[8] J Li and Z Liu ldquoSmooth and non-smooth traveling waves ina nonlinearly dispersive equationrdquo Applied Mathematical Mod-elling vol 25 no 1 pp 41ndash56 2000
[9] A Chen and J Li ldquoSingle peak solitary wave solutions forthe osmosis K(22) equation under inhomogeneous boundaryconditionrdquo Journal of Mathematical Analysis and Applicationsvol 369 no 2 pp 758ndash766 2010
[10] R Camassa and D D Holm ldquoAn integrable shallow waterequation with peaked solitonsrdquo Physical Review Letters vol 71no 11 pp 1661ndash1664 1993
[11] P Rosenau ldquoOn nonanalytic solitary waves formed by a nonlin-ear dispersionrdquo Physics Letters A vol 230 no 5-6 pp 305ndash3181997
[12] P Rosenau and J M Hyman ldquoCompactons solitons with finitewavelengthrdquo Physical Review Letters vol 70 no 5 pp 564ndash5671993
[13] L Zhang L-Q Chen and X Huo ldquoThe effects of horizontalsingular straight line in a generalized nonlinear Klein-Gordonmodel equationrdquo Nonlinear Dynamics vol 72 no 4 pp 789ndash801 2013
[14] J Li and H Dai On the Study of Singular Nonlinear TravelingWave Equations Dynamical System Approach Science PressBeijing China 2007
[15] A Chen S Wen S Tang W Huang and Z Qiao ldquoEffects ofquadratic singular curves in integrable equationsrdquo Studies inApplied Mathematics vol 134 no 1 pp 24ndash61 2015
[16] L Tian and J Yin ldquoNew compacton solutions and solitary wavesolutions of fully nonlinear generalized Camassa-Holm equa-tionsrdquo Chaos Solitons amp Fractals vol 20 no 2 pp 289ndash2992004
[17] L Zhang ldquoThe bifurcation and peakons for the specialC(3 2 2)equationrdquo Pramana vol 83 no 3 pp 331ndash340 2014
[18] J Lenells ldquoTraveling wave solutions of the CamassandashHolmequationrdquo Journal of Differential Equations vol 217 no 2 pp393ndash430 2005
[19] Z J Qiao and G P Zhang ldquoOn peaked and smooth solitons forthe Camassa-Holm equationrdquo Europhysics Letters vol 73 no 5pp 657ndash663 2006
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
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
References
[1] C Rogers and W R Shadwick Backlund Transformation andTheir Applications Academic Press New York NY USA 1982
[2] W Malfliet and W Hereman ldquoThe tanh method I Exactsolutions of nonlinear evolution and wave equationsrdquo PhysicaScripta vol 54 no 6 pp 563ndash568 1996
[3] R Hirota Direct Method in Soliton Theory Springer BerlinGermany 1980
[4] J S Cohen Computer Algebra and Symbolic ComputationMathematical Methods A K Peters 2003
[5] S Lie ldquoUber die integration durch bestimmte integrale voneiner klasse linearer partieller differential eichungenrdquoArchiv derMathematik vol 6 pp 328ndash368 1881
[6] H-H Dai and X-H Zhao ldquoTraveling waves in a rod composedof amodifiedMooney-Rivlinmaterial I bifurcation of equilibraand non-singular caserdquo Proceedings of the Royal Society AMathematical Physical and Engineering Sciences vol 455 no1990 pp 3845ndash3874 1999
[7] H-H Dai and F-F Wang ldquoAsymptotic bifurcation solutionsfor compressions of a clamped nonlinearly elastic rectangletransition region and barrelling to a corner-like profilerdquo SIAMJournal on Applied Mathematics vol 70 no 7 pp 2673ndash26922010
[8] J Li and Z Liu ldquoSmooth and non-smooth traveling waves ina nonlinearly dispersive equationrdquo Applied Mathematical Mod-elling vol 25 no 1 pp 41ndash56 2000
[9] A Chen and J Li ldquoSingle peak solitary wave solutions forthe osmosis K(22) equation under inhomogeneous boundaryconditionrdquo Journal of Mathematical Analysis and Applicationsvol 369 no 2 pp 758ndash766 2010
[10] R Camassa and D D Holm ldquoAn integrable shallow waterequation with peaked solitonsrdquo Physical Review Letters vol 71no 11 pp 1661ndash1664 1993
[11] P Rosenau ldquoOn nonanalytic solitary waves formed by a nonlin-ear dispersionrdquo Physics Letters A vol 230 no 5-6 pp 305ndash3181997
[12] P Rosenau and J M Hyman ldquoCompactons solitons with finitewavelengthrdquo Physical Review Letters vol 70 no 5 pp 564ndash5671993
[13] L Zhang L-Q Chen and X Huo ldquoThe effects of horizontalsingular straight line in a generalized nonlinear Klein-Gordonmodel equationrdquo Nonlinear Dynamics vol 72 no 4 pp 789ndash801 2013
[14] J Li and H Dai On the Study of Singular Nonlinear TravelingWave Equations Dynamical System Approach Science PressBeijing China 2007
[15] A Chen S Wen S Tang W Huang and Z Qiao ldquoEffects ofquadratic singular curves in integrable equationsrdquo Studies inApplied Mathematics vol 134 no 1 pp 24ndash61 2015
[16] L Tian and J Yin ldquoNew compacton solutions and solitary wavesolutions of fully nonlinear generalized Camassa-Holm equa-tionsrdquo Chaos Solitons amp Fractals vol 20 no 2 pp 289ndash2992004
[17] L Zhang ldquoThe bifurcation and peakons for the specialC(3 2 2)equationrdquo Pramana vol 83 no 3 pp 331ndash340 2014
[18] J Lenells ldquoTraveling wave solutions of the CamassandashHolmequationrdquo Journal of Differential Equations vol 217 no 2 pp393ndash430 2005
[19] Z J Qiao and G P Zhang ldquoOn peaked and smooth solitons forthe Camassa-Holm equationrdquo Europhysics Letters vol 73 no 5pp 657ndash663 2006
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of