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Chaos, Solitons and Fractals xxx (2016) xxx–xxx

Contents lists available at ScienceDirect

Chaos, Solitons and Fractals
Nonlinear Science, and Nonequilibrium and Complex Phenomena
journal homepage: www.elsevier.com/locate/chaos

Group analysis, exact solutions and conservation laws of a
generalized fifth order KdV equation
Gangwei Wang a,∗, Abdul H. Kara b, Kamran Fakhar c, Jose Vega-Guzman d,
Anjan Biswas e,f

Q1

Q2

a

School of Mathematics and Statistics, Beijing Institute of Technology, Beijing 100081, PR China
School of Mathematics, Centre for Differential Equations Continuum Mechanics and Applications, University of the Witwatersrand,
Johannesburg Wits 2050, South Africa
c
Department of Mathematics, University of British Columbia, Vancouver V6T 1Z2, Canada
d
Department of Mathematics, Howard University, Washington, DC 20059, USA
e
Department of Mathematical Sciences, Delaware State University, Dover, DE 19901-2277, USA
f
Department of Mathematics, Faculty of Science, King Abdulaziz University, Jeddah 21589, Saudi Arabia
b

a r t i c l e

i n f o

Article history:
Received 22 January 2016
Accepted 11 February 2016
Available online xxx
PACS:
02.30.Jr
05.45.Yv
02.30.Ik

a b s t r a c t
We study the generalized fifth order KdV equation using group methods and conservation
laws. All of the geometric vector fields of the special fifth order KdV equation are presented. By using the nonclassical Lie group method, it is show that this equation does not
admit nonclassical type symmetries. Then, on the basis of the optimal system, the symmetry reductions and exact solutions to this equation are constructed. For some special cases,
we obtain additional nontrivial conservation laws and scaling symmetries.
© 2016 Published by Elsevier Ltd.

Keywords:
Fifth order KdV equation
Symmetry analysis
Exact solutions
Conservation laws

1
2
3
4
5
6
7
8
9
10

1. Introduction
Nonlinear evolution equations (NLEEs) are of importance in nonlinear science, in particular in applied mathematics and theoretical physics. Their solutions are important in the understanding of nonlinear interaction and
behaviors of complex system. There are various techniques
[1–17] used to deal with NLEEs, some of the commonly
used ways involve the generalized symmetries, nonlocal
symmetries, nonclassical Lie group and classical Lie group
method.

It is well known that differential equations (DEs) have
a number fundamental structures, that is, symmetries and
conservation laws (CLs). CLs play a key roles in DEs analysis, particularly in studies of existence, uniqueness and stability of solutions. Various approaches have been used to
handle symmetries and conservation laws of PDE systems
(see [1–6] and the references therein).
In the present paper, we use the group method and
the multiplier approach to study the fifth order KdV equation

ut + α un ux + β ux uxx + γ uuxxx + uxxxxx = 0.

Corresponding author. Tel.: +86 13683181850.
E-mail address: wanggangwei@bit.edu.cn (G. Wang).

ut + α u2 ux + β ux uxx + γ uuxxx + uxxxxx = 0,

13
14
15
16
17
18
19
20

(1)

In particular, for n = 2, one can get
∗

11
12

21

(2)

http://dx.doi.org/10.1016/j.chaos.2016.02.013
0960-0779/© 2016 Published by Elsevier Ltd.

Please cite this article as: G. Wang et al., Group analysis, exact solutions and conservation laws of a generalized fifth order
KdV equation, Chaos, Solitons and Fractals (2016), http://dx.doi.org/10.1016/j.chaos.2016.02.013

JID: CHAOS
2

22
23
24
25
26

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here α , β and γ are nonzero constants. Clearly, this equation has two dispersive terms uxxx and uxxxxx . By choosing
the real values of the parameters n, α , β and γ , one can
get a variety of fKdV equations [7,8] such as, when n = 2,
the Sawada–Kotera (SK) equation

ut + 5u2 ux + 5ux uxx + 5uuxxx + uxxxxx = 0,
27

29

ut + 2u ux + 6ux uxx + 3uuxxx + uxxxxx = 0.
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46

47
48
49
50

(4)

(6)

(7)

Many authors have been studied these equations using different approaches. There is still, however, a lot of room for
extensions and improvements. in particular exact solutions,
symmetries and conservation law.
The main purpose of this paper is to investigate symmetry and conservation law classification of the generalized KdV equation. We will show that the particular
case n = 2 is special as it is the only case that admit a
scaling symmetry. The paper is organized as follows. In
Section 2, all of the vector fields of Eq. (1) are constructed.
In Section 3, we consider the special case of n, that is
n = 2. In this case, the vector fields and some exact solutions are obtained. In Section 4, we study the soliton solutions of the equation. In Section 5, we find the conservation laws. Finally, conclusions and some remarks are given
in Section 5.
2. Group analysis of the generalized fifth-order KdV
equation
Consider a one-parameter Lie group of infinitesimal
transformation:

x∗ = x + ξ (x, t, u ) + O( 2 ),
u∗ = u + η (x, t, u ) + O( 2 ),
51
52

with a small parameter   1, and the above group of
transformations infinitesimal generator can read

∂
∂
∂
V = τ (x, t, u )
+ ξ (x, t, u )
+ η (x, t, u )
,
∂t
∂x
∂u
53
54
55

(9)

and we need to solve the coefficient functions τ (x, t, u),
ξ (x, t, u), η(x, t, u).
Meanwhile, V must satisfy Lie’s symmetry condition

pr
56

(8)

(5 )

V ( )| =0 = 0,

(10)

where

= ut + α u2 ux + β ux uxx + γ uuxxx + uxxxxx .

(11)

59
60

(12)
61

η = Dt (η ) − ux Dt (ξ ) − ut Dt (τ ),
ηx = Dx (η ) − ux Dx (ξ ) − ut Dx (τ ),
(13)

Here, Di are the total derivative operators defined by

Di =

∂
∂
∂
+ ui
+ ui j
+ · · · i = 1, 2,
∂u
∂uj
∂ xi

and (x1 , x2 ) = (t, x ).
On the basis of the Lie symmetry analysis method, one
can get

τ = c1 , ξ = c2 , η = 0,

∂
∂
,V =
.
∂ x 2 ∂t

63
64
65

(15)

where c1 and c2 are arbitrary constants. So one can have
the geometric vector fields

V1 =

62

(14)

66
67

(16)

3. Classical, nonclassical, potential symmetry and exact
solutions for n = 2

68

In this section, we use the classical and nonclassical
symmetry method to handle the fifth-order equation for
n = 2.

70

3.1. Classical symmetry analysis

73

On the basis of the Lie symmetry analysis method, one
can get

75

τ = 5c1t + c2 , ξ = c1 x + c3 , η = −2c1 u,

V1 =

∂
∂
∂
∂
∂
,V =
, V =x
+ 5t
− 2u
.
∂ x 2 ∂t 3
∂x
∂t
∂u

69

71
72

74

(17)

where c1 , c2 and c3 are arbitrary constants. So one can
have the geometric vector fields

t ∗ = t + τ (x, t, u ) + O( 2 ),

58

t

···
(5)

the Ito equation
2

31

ηt + nαηux + α un ηx + βηx uxx + βηxx ux
+γ ηuxxx + γ ηxxx u + ηxxxxx = 0,

the Kaup–Kupershmidt (KK) equation

ut + 20u2 ux + 25ux uxx + 10uuxxx + uxxxxx = 0,
30

By using the fifth prolongation Pr(5) V to Eq. (1), one can
see that the coefficient functions satisfy the following
equation:

where

the Lax equation

ut + 30u2 ux + 20ux uxx + 10uuxxx + uxxxxx = 0,
Q3

(3)

the Caudrey–Dodd–Gibbon (CDG) equation

ut + 180u2 ux + 30ux uxx + 30uuxxx + uxxxxx = 0,
28

[m3Gdc;February 24, 2016;10:55]

76
77

(18)

3.2. Nonclassical symmetry analysis

78

In the previous subsection, we used the classical symmetry method to deal with the fifth order KdV equation.
Next, we employ the nonclassical symmetry method [9,10]
to study the fifth order KdV equation. The aim is that nonclassical symmetries are much more numerous than classical ones and maybe get more solutions of PDEs. In terms
of the classical symmetry, the invariance surface condition
should been added:

79

1 =

η − ξ ux − τ ut .

80
81
82
83
84
85
86

(19)

57

87

Please cite this article as: G. Wang et al., Group analysis, exact solutions and conservation laws of a generalized fifth order
KdV equation, Chaos, Solitons and Fractals (2016), http://dx.doi.org/10.1016/j.chaos.2016.02.013

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88
89

If the vector field (9) is a nonclassical symmetry of (1),
which should satisfy

pr (5)V ( )| =0, 1 =0 = 0.
90
91
92
93
94
95
96
97

Consider the nature of the invariant surface condition
(19), without loss of generality, there are two cases to
arise: (i)τ = 1; (ii)τ = 0, ξ = 1. More details see [9,10] and
the references therein. In following, we will consider them
respectively.
1. τ = 1.
Then from the invariance surface condition (19), one
can get

ut = η − ξ ux .
98
99

(20)

(21)

After differentiating (21) with respect to x, and then replacing ut by using η − ξ ux , one can get

ηt − ξt ux + ηu (η − ξ ux ) + 2αηuux + α u2 (ηx + (ηu − ξx )ux )
+β uxx (ηx + (ηu − ξx )ux ) + β ux (ηxx + (2ηxu − ξxx )ux
+(ηuu −2ξxu u2x ) + (ηu −2ξx uxx )) + γ ηuxxx
+γ u[ηxxx + 3ηxxu ux + 3ηxuu u2x + 3ηxu uxx + ηuuu u3x + 3ηuu ux uxx
+ηu uxxx − ξxxx ux −3ξxx uxx −3ξx uxxx ] + 5ηxxxxu ux + 10ηxxxu uxx
+10ηxxu uxxx + 3ηxu uxxxx + ηu uxxxxx − ξxxxxx ux −5ξxxxx uxx
−10ξxxx uxxx +10ηuu uxx uxxx + ηxxxxx − 10ξxx uxxxx −5ξx uxxxxx
+10ηxxuu u2x +10ηxxuuu u3x + 5ηxuuuu u4x + 15ηxuuu u2xx + ηuuuuu u5x
+30ηxuuu u2x uxx + 20ηxuu ux uxxx + 30ηxxuu ux uxx + 5ηuu ux uxxxx
+15ηuuu ux u2xx +10ηuuuu u3x uxx +10ηuuu u2x uxxx = 0.
(22)
100

102

104
105
106
107
108

−2u
x + C2
,ξ =
,
5t + C1
5t + C1

x + C2 ∂
∂
−2u ∂
+
+
.
5t + C1 ∂ x
∂ t 5t + C1 ∂ u

(24)

It is cleat that, V4 = V3 + C1V2 + C2V1 , it is the classical symmetry. Also, we can find the equation does not have nonclassical symmetries.
2. τ = 0, ξ = 1.
Now, using the same approach as before, one can
have

η = ux .
109

(23)

where C1 and C2 are arbitrary constants. Consequently, one
can get the corresponding “nonclassical symmetry”is

V4 =
103

(26)

113

That is to say, in this case, we could not get supplementary
symmetries, of non-classical type. This also means no new
explicit solutions can be constructed in the case of τ = 0,
ξ = 1.

114

3.3. Potential symmetry analysis

111
112

115

vx = u,
1

1
vt = − α u3 + (β − γ )u2x + γ uuxx + uxxxx .
3

2

(28)

After repeating previous steps, one can get the coefficient functions τ (x, t, u, v ), ξ (x, t, u, v ), η (x, t, u, v ) and
ψ (x, t, u, v ) are:

τ = c1 t + c2 , ξ =

118
119
120

2 c1
c1
x + c4 , η = −
u, ψ = −c1 v + c3 .
3
3
(29)

One can find out that this equation does not have potential
symmetry.

121
122

3.4. Symmetry reductions and group-invariant solutions for
n=2

123

In the previous section, we use the classical and nonclassical group method to deal with the fifth order KdV
equation In this section, by using the optimal system, we
give some group-invariant solutions.

125

3.4.1. One-dimensional optimal system of subalgebras
In order to get the optimal system, we applying the adjoint transformations formula [1] given by

129

Ad (exp(Vi ))V j = V j −  [Vi , V j ] +

124

126
127
128

130
131

1 2
 [Vi , [Vi , V j ]] − · · ·
2
(30)

where  is a nonzero constant. Here [Vi , Vj ] is the commutator for the Lie algebra given by

[Vi , V j ] = ViV j − V jVi .

V1 , V2 + λV1 , V3 .

132
133

(31)

We can get an optimal system of one-dimensional subalgebras:

134
135

(32)

3.4.2. Symmetry reductions
In the present subsection, we employ the optimal system of one-dimensional subalgebras to deal with (1). and
in the next subsection we will give some exact solutions of
(1).
(1) V1 .
For the generator V1 , one can get the group-invariant
solution is u = f (ξ ), and ξ = t is the group-invariant,
in this case, one can get trivial solution u(x, t ) = C, and
C is a constant quantity.
(2) V2 + λV1 .
For the case of V2 + λV1 , we get the group-invariant solutions

u = f ( ξ ),

136
137
138
139
140

(27)

−λ f  + α f 2 f  + β f  f  + γ f f  + f (5) = 0.

141
142
143
144
145
146
147
148

(33)

where ξ = x − λt. Plugging (33) into (2), one can get
the following ODE:

Suppose (1) can be written an a conservation law,

Dt T (x, t, u ) + Dx X (x, t, u ) = 0.

117

(25)

Then, solving the overdetermined equations, one can get

τ = 0, ξ = 1, η = 0.
110

The PDE system S(x, t, u, v ) = 0 given by

Solving the overdetermined system of equations, leads to

η=
101

3

149
150

(34)

116

151

Please cite this article as: G. Wang et al., Group analysis, exact solutions and conservation laws of a generalized fifth order
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In particular, if β = 2γ , and each term is multiplied by
f, and integral once, one can get

152
153

−

λ 2
2

1 4
α f + γ f 2 f  + f f (4) − f  f 
4

f +

(35)

159
160

1

(36)

Substitution of (36) into (2), one can lead to

−

2
1
f − ξ f  + α f 2 f  + β f  f  + γ f f  + f (5) = 0.
5
5

cn ξ n .

∞


+α

n=1



n


(38)

n


α

∞ 
n 
k


1
2
− ncn − cn
5
5

(n + 1 − k )c j ck− j cn+1−kξ

u(x, t ) =

n

∞ 
n


ξ n+5 .

(42)

165

(k + 1 )

n=1

1

=

2

1

1

1

1

n=0



(39)

×

α

n 
k


(n + 1 − k )c j ck− j cn+1−k

k=0 j=0

(40)

+β

n


(k + 1 )(n + 1 − k )(n + 2 − k )ck+1 cn+2−k

k=0

+γ

1
(n + 1 )(n + 2 )(n + 3 )(n + 4 )(n + 5 )

α

1

cn+5 (xt − 5 )n+5 (t − 5 )

+c4 (xt − 5 )4
∞

1
−
(n + 1 )(n + 2 )(n + 3 )(n + 4 )(n + 5 )

For general case, if n ≥ 1, one can get

n 
k


∞


c0 + c1 (xt − 5 ) + c2 (xt − 5 )2 + c3 (xt − 5 )3

Comparing coefficients for n = 0 in (39), one yields



1

n=0

n=1

c0 − α c02 c1 − 2β c1 c2 − 6γ c0 c3
.
120

1

c0 + c1 (xt − 5 ) + c2 (xt − 5 )2 + c3 (xt − 5 )3
+c4 (xt − 5 )4 +

n=1 k=0

×



1

× (n + 3 − k )ck cn+3−k ξ n
∞
∞
2

1
2
−
ncn ξ n −
c0 +
cn ξ n = 0.
5
5
5

cn+5 = −

(n + 1 − k )(n + 2 − k )(n + 3 − k )ck cn+3−k

Therefore, one can get

× (n + 1 − k )(n + 2 − k )ck+1 cn+2−k ξ n
∞ 
n

+6γ c0 c3 + γ
(n + 1 − k )(n + 2 − k )

2

(k + 1 )(n + 1 − k )(n + 2 − k )ck+1 cn+2−k

k=0

n=1 k=0

c5 = 5

(n + 1 − k )c j ck− j cn+1−k

(n + 1 )(n + 2 )(n + 3 )(n + 4 )(n + 5 )cn+5 ξ n

+2β c1 c2 + β

163

n 
k

k=0 j=0

n=1 k=0 j=0

162

α

+γ

n=1

c02 c1 +

4

k=0

Substituting (38) into (36), one can have

120c5 +

3

c0 − α c02 c1 − 2β c1 c2 − 6γ c0 c3 5
ξ
120
∞

1
−
(n + 1 )(n + 2 )(n + 3 )(n + 4 )(n + 5 )

+β

n=0

161

= c0 + c1 ξ + c2 ξ + c3 ξ + c4 ξ

(37)

Supposing that (34) has the following solutions
∞


n=1
2

×

3.5. Exact group-invariant solutions using power serious
method

f (ξ ) =

∞

cn+5 ξ n+5

+5

u = t − 5 f (ξ ), ξ = xt − 5 .
157

164

2

where k is an integration constant.
155 (3) V3 .
156
In the case of the generator V3 , we get

154

2

In this way, the power series solution of can be rewritten

f ( ξ ) = c0 + c1 ξ + c2 ξ 2 + c3 ξ 3 + c4 ξ 4 + c5 ξ 5 +

1
+ ( f  )2 + k = 0,
2

158

[m3Gdc;February 24, 2016;10:55]

n

k=0

(n + 1 − k )(n + 2 − k )(n + 3 − k )ck cn+3−k



1
2
1
2
− ncn − cn (xt − 5 )n+5 (t − 5 ),
5
5

(n + 1 − k )c j ck− j cn+1−k

(43)

k=0 j=0

+β

n


here ci (i = 0, 1, 2, 3, 4 ) are arbitrary constants.

(k + 1 )(n + 1 − k )(n + 2 − k )ck+1 cn+2−k

166

k=0

+γ

n

k=0

(n + 1 − k )(n + 2 − k )(n + 3 − k )ck cn+3−k



1
2
− ncn − cn .
5
5

(41)

Remark. The exact solution of (43) also can be fixed in
similarly way. The details are omitted here.
In addition, by using the Maple soft, one can get following Jacobi function and Weierstrass elliptic function solutions:

Please cite this article as: G. Wang et al., Group analysis, exact solutions and conservation laws of a generalized fifth order
KdV equation, Chaos, Solitons and Fractals (2016), http://dx.doi.org/10.1016/j.chaos.2016.02.013

167
168
169
170
171

Q4

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−6C32 SN −6

u(x, t ) =

2βγ +β 2 −


2
β +4β 3 γ +4β 2 γ 2 −40αβ 2 −12α C3 5 t
+
C
x
+
C
,
i
3
2
α

√ 4



β 12 γ 2 + 12 β γ + 6 β 2 + 6 β 4 + 4 β 3 γ + 4 β 2 γ 2 − 40 α β 2 − 120 α


−6γ 2 2βγ +β 2 −

×

×



√ 4

β +4β 3 γ +4 β 2 γ 2 −40αβ 2 −12α
α

−60β 2 − 60





−6C32 SN

−6

2βγ +β 2 +

β 4 + 4β 3 γ + 4β 2 γ 2 − 40αβ 2


2
β +4β 3 γ +4β 2 γ 2 −40αβ 2 −12α C3 5 t
+ C3 x + C2 , i
α

×

×



√ 4

β +4β 3 γ +4 β 2 γ 2 −40αβ 2 −12α
α

−60β 2 + 60

1

u(x, t ) = −3℘



2





β 2 + 4βγ + 4γ 2 − 40α C4 α .
−1

2

(46)
173

1

u(x, t ) = −3℘



2

C4 5C2 (−3N + 36 )t + C4 x + C3 , C2 , C1

× 2γ +β +

β

2 +4

βγ +4γ

2 −40





−1

181
182
183
184

There are four types of solutions that are going to be extracted for (48) with the aid of the method of undetermined coefficients. They are in the next few subsections:

p

(49)

with

τ = B(x − vt )

(50)

189
190

−(1 + p)(2 + p)(3 + p)(4 + p)δ B sech
4

τ
τ =0

p+3

p+5

(51)

p=

4
.
n

191
192

(52)

Notice also that the same principle allows to equate 2 p + 3
with p + 5, and 2 p + 1 with p + 3, both situations leading
to

p=2

193
194
195

(53)

Thus, from (52) and (53) we have

196

(54)

Consequently, the system (51) collapses into

In order to obtain solitary waves, the starting hypothesis is:

188

τ

n = 2.

4.1. Solitary waves

u(x, t ) = Asech τ
185

(48)

187

τ

The balancing principle allows to equate the exponents
(n + 1 ) p + 1 with p + 5 from which

This section will obtain solitary wave solutions to the
model equation given by (1). The method of undetermined
coefficients will be adopted to retrieve these solitons. In
this case we rewrite Eq. (1) as follows:

179

(n+1 ) p+1

+2(1 + p)(2 + p)[2 + p(2 + p)]δ B sech

175

186

(v − p4 δ B4 )sech p+1 τ − p2 (β + γ )AB2 sech2 p+1 τ

4

4. Solitons solutions

180

(45)

2 p+3

174

ut + α un ux + β ux uxx + γ uuxxx + δ uxxxxx = 0.

.

+(1 + p)(2γ + p(β + γ ))AB2 sech

α C4 α .
2



where A is the amplitude of the soliton, p > 0 is a parameter that will be obtained with the aid of the balancing principle, B represents the inverse width of the soliton while v is the speed. By substituting (49) into (48) one
obtain

−α An sech

(47)

177

− 72γ 2 + 120βγ

β 4 + 4β 3 γ + 4β 2 γ 2 − 40αβ 2

C4 5C2 (−3M + 36 )t + C4 x + C3 , C2 , C1

× 2γ + β −

(44)



β 12 γ 2 + 12βγ + 6β 2 − 6 β 4 + 4 β 3 γ + 4 β 2 γ 2 − 40 α β 2 − 120 α

172

.



β 12 γ 2 + 12 β γ + 6 β 2 + 6 β 4 + 4 β 3 γ + 4 β 2 γ 2 − 40 α β 2 − 120 α




√ 4

β 12 γ 2 + 12 β γ + 6 β 2 + 6 β 4 + 4 β 3 γ + 4 β 2 γ 2 − 40 α β 2 − 120 α


178

− 72γ 2 + 120βγ

β 12 γ 2 + 12 β γ + 6 β 2 + 6 β 4 + 4 β 3 γ + 4 β 2 γ 2 − 40 α β 2 − 120 α

−6γ 2 2βγ +β 2 +

176



β 12 γ 2 + 12 β γ + 6 β 2 + 6 β 4 + 4 β 3 γ + 4 β 2 γ 2 − 40 α β 2 − 120 α



u(x, t ) =

5

197

(v − 16δ B )sech τ − 4[(β + γ )A − 60δ B ]B sech τ
4

3

2

2

−[360δ B4 + α A2 − 6(2γ + β )AB2 ]sech τ = 0.
7

5

(55)

After equating the coefficients of the linearly independent
functions sechj τ for j = 3, 5, 7 to zero one get

v = 16δ B ,
4

(56)

Please cite this article as: G. Wang et al., Group analysis, exact solutions and conservation laws of a generalized fifth order
KdV equation, Chaos, Solitons and Fractals (2016), http://dx.doi.org/10.1016/j.chaos.2016.02.013

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200

201

G. Wang et al. / Chaos, Solitons and Fractals xxx (2016) xxx–xxx

A=

60δ B2
β +γ

205

4.3. Singular solitary waves (Type-I)

229

For type-I singular solitary wave solution to Eq. (48) the
starting hypothesis is

230

(59)

Therefore, the solitary wave solution to (48) is given by
2

206

(58)

and the identity

u(x, t ) = Asech [B(x − vt )]
204

226

constrained by

(2γ + β )(γ + β ) − (β + γ )2 − 10αδ = 0.
203

where the amplitude is given in (65) while the speed turns
to be (64) along with the corresponding conditions. In addition, the condition (66) has to be satisfied.

(57)

β + γ = 0,
202

(60)

where the speed v is given in (56), while the amplitude A
is provided in (57). For the solution to exist the identity
(59) has to be satisfied along with the condition (58).

u(x, t ) = Acsch τ
p

207

4.2. Shock waves

208

In order to solve the generalized fifth order KdV Eq.
(48) for shock wave the starting hypothesis is taken to be

210

211
212
213

u(x, t ) = A tanh τ
p

(61)

with τ as defined in (50), A and B are free parameters, and
p > 0 is a key parameter to be determined. Substituting
Eq. (49) into Eq. (48) gives in a simplified form

τ − α An tanh(n+1) p−1 τ
p−3
+4( p − 1 )( p − 2 )[2 + p( p − 2 )]δ B4 tanh
τ
p−5
4
−( p − 1 )( p − 2 )( p − 3 )( p − 4 )δ B tanh
τ
2 p−1
2
2
+2 p (β + γ )AB tanh
τ + ( p − 1)
2 p−3
×[2γ − p(β + γ )]AB2 tanh
τ
p+1
+4( p + 1 )( p + 2 )[2 + p(2 + p)]δ B4 tanh
τ
p+3
4
−( p + 1 )( p + 2 )( p + 3 )( p + 4 )δ B tanh
τ
2 p+1
−( p + 1 )[2γ + p(β + γ )]AB2 tanh
τ = 0.
(62)

[v − 2 p2 (5 + 3 p2 )δ B4 ] tanh

214
215
216
217
218

p−1

By the balancing principle, equating the exponents (n +
1 ) p − 1 and p + 3 leads to (52), but also it is possible
to equate 2 p + 1 with p + 3 resulting in (53) and consequently one get (54). In view of this values of p and n, the
Eq. (62) can be rewritten as

[v − 136δ B4 − 2β AB2 ] tanh τ + 8[(β + γ )A
+ 60δ B2 ]B2 tanh τ − [6(β + 2γ )AB2
3

+ 360δ B + α A ] tanh τ = 0.
4

219
220

2

5

(63)

Equating to zero the coefficients of the linearly independent functions tanh j τ for j = 1, 3, 5 lead us to

v = 2(β A + 68δ B2 )B2 ,
221

and the amplitude becomes

A=−
222

60δ B2
.
β +γ

(65)

along with the condition (58). In addition, the identity

224
225

(66)

has to be satisfied in order for the shock waves to exist.
Therefore, the shock wave to the generalized fifth order
KdV Eq. (48) is given by

u(x, t ) = A tanh [B(x − vt )]
2

(67)

231

232
233
234
235

(v − p δ B )csch τ − p (β + γ )AB csch
τ
(n+1 ) p+1
n
−α A csch
τ − 2(1 + p)(2 + p)
p+3
×(2 + p(2 + p))δ B4 csch τ
2 p+3
−(1 + p)[2γ + p(β + γ )]AB2 csch
τ
p+5
−(1 + p)(2 + p)(3 + p)(4 + p)δ B4 csch τ = 0. (69)
p+1

4

2

2 p+1

2

As in Eq. (51), it is possible to equate (n + 1 ) p + 1 with
p + 5, 2 p + 3 with p + 5, and also 2 p + 1 with p + 3, thus
leading to (53), and consequently (54). As a consequence,
the Eq. (69) becomes

236
237
238
239

(v − 16δ B )csch τ − 4[(β + γ )A + 60δ B ]B csch τ
7
−[360δ B4 + α A2 + 6(2γ + β )AB2 ]csch τ = 0.
(70)
3

4

2

2

5

After equating the coefficients of the linearly independent
functions cschj τ for j = 3, 5, 7 to zero one get the wave
speed as in (56), the amplitude as in (65), and the identity
turns to be the same as in (66). Thus, the type-I singular
solitary wave solution to the generalized KdV Eq. (48) is
given by

u(x, t ) = Acsch [B(x − vt )]
2

240
241
242
243
244
245

(71)

where the soliton speed and amplitude are given in (56)
and (65) respectively, while the identity (66) has to be
satisfied in order to preserve the existence of the solitary
wave.

246
247
248
249

4.4. Singular solitary waves (Type-II)

250

To obtain type-II singular solitary wave solutions to Eq.
(48) the starting Ansatz is

252

u(x, t ) = A coth τ
p

[v − 2 p (5 + 3 p )δ B ] coth

(n+1 ) p−1

τ − α A coth
p−3
+4( p − 1 )( p − 2 )[2 + p( p − 2 )]δ B4 coth
τ
p−5
−( p − 1 )( p − 2 )( p − 3 )( p − 4 )δ B4 coth
τ
2 p−1
+2 p2 (β + γ )AB2 coth
τ + ( p − 1)
2 p−3
2
×[2γ − p(β + γ )]AB coth
τ
p+1
+4( p + 1 )( p + 2 )[2 + p(2 + p)]δ B4 coth
τ
p+3
4
−( p + 1 )( p + 2 )( p + 3 )( p + 4 )δ B coth
τ
2 p+1
−( p + 1 )[2γ + p(β + γ )]AB2 coth
τ = 0.
2

251

(72)

where the meaning of A, p and τ are as in the previous
subsections. By inserting (72) into the fifth order KdV Eq.
(48) one get
2

(β + γ )2 − (2γ + β )(γ + β ) + 10αδ = 0.
223

(64)

227
228

(68)

where the parameter p > 0 will be determined with the
help of the balancing principle, A and B are free parameters, while τ retain the same meaning as in (50). The substitution of (68) into (48) leads to
4

209

[m3Gdc;February 24, 2016;10:55]

4

p−1

n

τ

(73)

Please cite this article as: G. Wang et al., Group analysis, exact solutions and conservation laws of a generalized fifth order
KdV equation, Chaos, Solitons and Fractals (2016), http://dx.doi.org/10.1016/j.chaos.2016.02.013

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257

The balancing principle leads to (53) and (54). Both values
reduce the Eq. (73) into

[v − 136δ B4 − 2β AB2 ] coth τ + 8[(β + γ )A
+ 60δ B2 ]B2 coth τ − [6(β + 2γ )AB2

follows:

 ∞

5

Q5

258
259
260

and consequently, the results (64)–(66) reappear. Finally,
the type-II singular solitary wave solution to the generalized fifth order KdV Eq. (48) is given by

u(x, t ) = A coth [B(x − vt )]
2

261
262

 ∞

(74)

(75)

where the amplitude is provided in (65) while the speed
turns to be (64) along with the corresponding conditions.

263

5. Conservation laws

264

In this section, we employ the multipliers method [3–
5] to deal with the conservation law. Firstly, we introduces
some basic definitions and concepts.
From what has been described above, suppose the conservation law is given by Dx T x + Dt T t = on the solutions of
(1).
In Section 2, we got the Lie point symmetry generators
X1 = ∂t and X2 = ∂x . For the case n = 2, we obtain the additional scaling symmetry X3 = x∂x + 5t ∂t − 2u∂u .
In general, we have the only conserved vector based on
the multiplier Q = 1 given by

265
266
267
268
269
270
271
272
273
274

1
1
α un+1 + (β − γ )ux 2 + γ uuxx + uxxxx ,
n+1
2
T t = u.
Tx =

275
276

(76)

In particular, if β = 2γ , one can get the multiplier Q = u,
and get the conserved vector

1
1
α un+2 + γ u2 uxx + uuxxxx − ux uxxx + u2xx ,
n+2
2
1
T t = u2 .
(77)
2
Tx =

277

For n = 2, subject to the condition

10α + 2β 2 − 7βγ + 3γ 2 = 0,
278
279
280

1
(α (8β − 4γ )u5 + 20(2β − γ )γ u3 uxx
200
−20u(5uxt − 2((β + 2γ )uxx 2 + (−2β + γ )ux uxxx ))
−5u2 ((10α + 2β 2 − 7βγ + 3γ 2 )ux 2
+4(−2β + γ )uxxxx ) + 20(5ut ux + 2(2β − γ )ux 2 uxx

−5(uxxx 2 − 2uxx uxxxx ))),
1
Tt =
((2β − γ )u3 + 15uuxx ).
30
281
282
283
284
285

1
I2 =
T t dx =
2
−∞
 ∞

287

 ∞

1
I3 =
T t dx =
30
−∞
=

(80)

2A2
u2 d x =
3B
−∞
 ∞
−∞

(81)
288



(2β − γ )u3 + 15uuxx dx

8A2
{(2β − γ )A + 15B2 }
225B

(82)

Remark. It does not exist other nth order multiplier, in
other words, it only exist, for the general case, second order multiplier.

289

6. Conclusion

292

In the present paper, using the group methods, and the
multiplier approach, the generalized fifth-order KdV equation is studied. Furthermore, we derive the corresponding
Lie algebra and the similarity reductions of special case of
generalized fifth-order KdV equation. Also, we found that
the analyzed model does not admit nonclassical type symmetries for n = 2. In addition, on the basis of symmetries,
the optimal system is constructed, based on the optimal
system, some exact solutions are presented. Meanwhile,
some soliton solutions are presented. Finally, conservations
laws are derived. These results are important for the understanding of nonlinear interaction and behaviors of complex system in some piratical physical problems.
Acknowledgments
This work is supported by the Research Project International Graduate Exchange Program of Beijing Institute of
Technology, National Natural Science Foundation of China
(NNSFC) (Grant no. 11171022), Graduate Student Science
and Technology Innovation Activities of Beijing Institute of
Technology (No. 2014cx10037).

290
291

293
294
295
296
297
298
299
300
301
302
303
304
305
306
307
308
309
310
311
312

(78)

we obtain an additional, second order, multiplier Q =
1
2
10 ((2β − γ )u + 10uxx ) leading to the nontrivial conserved
flow

Tx =

286

 ∞

2A
I1 =
T t dx =
udx =
B
−∞
−∞

3

+ 360δ B4 + α A2 ] coth τ = 0.

7

(79)

We note that the action of X3 on Q satisfies X3 Q = −4Q.
For any solution u(x, t), if u and its derivatives converge to
∞
x −→ ±∞, −∞ T t dx provide conserved quantities.
From the solitary wave solution derived in the previous section, the conserved quantities, with δ = 1, are as

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