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In this paper, the dispersive long wave equation is studied by Lie symmetry group theory. Firstly, the Lie symmetries of this system are calculated. Secondly, one dimensional optimal systems of Lie algebra and all the symmetry reductions are obtained. Fina l ly, based on the power series method and the extended Tanh function method, some new explicit solutions of this system are constructed.

In mathematical physics, many significant phenomena and dynamic processes can be represented by nonlinear partial differential equations (NLPDEs) [

At present, there is no general method for solving NLPDEs. Although the symmetry method has a wide range of applications in solving methods, it still faces many difficulties and challenges to promote its development. However, the symmetry method and other methods (e.g. generalized simple equation method [

In the present paper, based on the Lie group method, we will investigate the dispersive long wave equations

{ u t + v x + 1 2 ( u 2 ) x = 0 , v t + ( u v + u + u x x ) x = 0 , (1)

where u represents the amplitude of a surface wave, propagating along the x-axis with a horizontal velocity. It plays an important role in nonlinear physics [

The outline of this paper is as follows: in Section 2, the Lie symmetry analysis is performed for the dispersive long wave equations; in Section 3, the optimal systems and the similarity reductions of Equation (1) are researched employing Lie group analysis in the last section; in Section 4, the exact solutions for the reduced equation are obtained by using the power series method and the extended Tanh method; and in Section 5, a brief summary is done to the full text.

We first do some preparatory work on the concept of classical Lie symmetry of general NLPDEs. Consider the kth-order scalar NLPDEs of the form

f α ( x , u , u ( 1 ) , ⋯ , u ( k ) ) = 0 , α = 1 , 2 , ⋯ , m , (2)

where x = ( x 1 , x 2 , ⋯ , x n ) denotes n independent variables, u = ( u 1 , u 2 , ⋯ , u m ) denotes m independent variables, and u ( j ) = u i 1 i 2 ⋯ i j α ( j = 1 , 2 , ⋯ , k , i s = 1 , 2 , ⋯ , j ) denote the partial derivatives of u α with respect to x i = ( i = 1 , 2 , ⋯ , n ) up to jth-order, i.e.

u ( j ) α = ∂ j u α ∂ x i 1 ∂ x i 2 ⋯ ∂ x i j .

Suppose that the one-parameter Lie group of point transformations

x i ∗ = X ( x , u ; ε ) = x + ε ξ i ( x , u ) + O ( ε 2 ) , u q ∗ = U ( x , u ; ε ) = u + ε η α ( x , u ) + O ( ε 2 ) , (3)

where i = 1 , 2 , ⋯ , n , q = 1 , 2 , ⋯ , m . ε is an infinitesimal parameter, ξ i , η α are some smooth function with variables x , u .

Theorem 1. [

X = ξ i ∂ ∂ x i + η α ∂ ∂ u α (4)

is the infinitesimal generator of the one-parameter Lie group of transformations for (3), and the k-th prolongation of the infinitesimal generator is

X ( k ) = X + η i ( 1 ) α ∂ ∂ u i α + ⋯ + η i 1 , i 2 ⋯ i k ( k ) α ∂ ∂ u i 1 , i 2 ⋯ i k α (5)

where the prolongation of the infinitesimals satisfy the following recurrence relation

η i ( 1 ) α = D i η α − ( D i ξ j ) u j α , i = 1 , 2 , ⋯ , n ,

η i 1 , i 2 ⋯ i k ( k ) α = D i k η i 1 , i 2 ⋯ i k − 1 ( k − 1 ) α − ( D i k ξ j ) u i 1 , i 2 ⋯ i k − 1 j α , i l = 1 , 2 , ⋯ , k ( k ≥ 2 )

where D i denotes the total derivative operator defined as

D i = ∂ ∂ x i + u i α ∂ ∂ u + u i j α ∂ ∂ u j + ⋯ + u i 1 i 2 ⋯ i j α ∂ ∂ u i 1 i 2 ⋯ i j + ⋯ , i = 1 , 2 , ⋯ , n .

That one-parameter Lie group of transformations (3) is the Lie symmetry of Equation (2), if and only if

X ( k ) f α ( x , u , u ( 1 ) , ⋯ , u ( k ) ) | f α ( x , u , u ( 1 ) , ⋯ , u ( k ) ) = 0 = 0.

Next, we calculate the Lie symmetry of Equation (1). With regard to the infinitesimal generator of Equation (1), it can be expressed from (4) as the following form

X = ξ ∂ ∂ x + τ ∂ ∂ t + η ∂ ∂ u + ϕ ∂ ∂ v . (6)

Applying the Theorem 1 to Equation (1), we have

{ X ( 3 ) [ u t + v x + 1 2 ( u 2 ) x ] | u t + v x + 1 2 ( u 2 ) x = 0 , v t + ( u v + u + u x x ) x = 0 = 0 , X ( 3 ) [ v t + ( u v + u + u x x ) x ] | u t + v x + 1 2 ( u 2 ) x = 0 , v t + ( u v + u + u x x ) x = 0 = 0. (7)

By simplifying (7), we can get the following overdetermined equations about ξ , τ , η , ϕ

{ ξ u = ξ v = 0 , τ x = τ u = τ v = 0 , η x = η t = η v = 0 , ϕ x = ϕ t = ϕ u = 0 , 2 η u + 2 v η u − ϕ = 0 , 2 ξ x + 2 v ξ x + ϕ = 0 , τ t + v τ t + ϕ = 0 , 2 η + 2 v η − 2 ξ t − 2 v ξ t − u ϕ = 0 , ϕ − ϕ v − v ϕ v = 0. (8)

From (8) it is easy to caculate that the only solution of this system is

ξ = k 4 x + k 3 t + k 1 , τ = 2 k 4 t + k 2 , η = − k 4 u + k 3 , ϕ = − 2 k 4 v − 2 k 4 , (9)

where k 1 , k 2 and k 3 are arbitrary constants. Accordingly, the symmetry groups of Equation (1) can be written as

X 1 = ∂ ∂ x , X 2 = ∂ ∂ t , X 3 = t ∂ ∂ x + ∂ ∂ u , X 4 = x ∂ ∂ x + 2 t ∂ ∂ t − u ∂ ∂ u − 2 ( v + 1 ) ∂ ∂ v . (10)

The infinitesimal generators (10) correspond to a four-parameter Lie group of nontrivial point transformations acting on ( x , t , u , v ) -space.

In this section, we study how to construct the one-dimensional optimal system of Equation (1) in order to obtain more abundant group invariant solutions. The basic method of constructing it is to simplify the expression of Lie algebra by using a variety of adjoint transformations on the most general expression of Lie algebra. The adjoint transformation is expressed as the following series form

A d ( exp ( ε X i ) ) X j = X j − ε [ X i , X j ] + ε 2 2 [ X i , [ X i , X j ] ] − ε 3 3 ! [ X i , [ X i , [ X i , X j ] ] ] + ⋯ ,

where ε is a parameter, and [ X i , X j ] is the usual commutator, given by

[ X i , X j ] = X i X j − X j X i .

Hence we can get the following commutator

According to the method of constructing one dimensional optimal system in [

X = a 1 X 1 + a 2 X 2 + a 3 X 3 + a 4 X 4 ,

and simplify the coefficients of the vector as much as possible. Without loss of generality, suppose first that a 4 ≠ 0 and set up a 4 = 1 , then the vector X becomes X = a 1 X 1 + a 2 X 2 + a 3 X 3 + X 4 To eliminate the coefficient of X 1 , we use X 1 to act on X by means of the adjoint operation, i.e.

X ′ = A d ( exp ( ε 1 X 1 ) ) X = a 2 X 2 + a 3 X 3 + X 4 ,

where the group parameter ε 1 = a 1 . Then continue to eliminate X 2 , X 3 by using one after another X 2 , X 3 to act on X ′ , the vector becomes

X ″ = A d ( exp ( ε 3 X 3 ) ) A d ( exp ( ε 2 X 2 ) ) X ′ = X 4 ,

where the group parameters ε 2 = a 2 / 2 , ε 3 = − a 3 . It can be seen easily that the vector form can not be simplified much more. Secondly, suppose that a 4 = 0 , a 3 ≠ 0 and set up a 3 = 1 , the vector X becomes X = a 1 X 1 + a 2 X 2 + X 3 . To eliminate the coefficient of the vector X 1 we use X 2 to act on X by means of the adjoint operation, i.e.

X ‴ = A d ( exp ( ε 4 X 2 ) ) X = a 2 X 2 + X 3 ,

[ X i , X j ] | X 1 | X 2 | X 3 | X 4 |
---|---|---|---|---|

X 1 | 0 | 0 | 0 | X 1 |

X 2 | 0 | 0 | X 1 | 2 X 2 |

X 3 | 0 | − X 1 | 0 | − X 3 |

X 4 | − X 1 | − 2 X 2 | X 3 | 0 |

A d | X 1 | X 2 | X 3 | X 4 |
---|---|---|---|---|

X 1 | X 1 | X 2 | X 3 | X 4 − ε X 1 |

X 2 | X 1 | X 2 | X 3 − ε X 1 | X 4 − 2 ε X 2 |

X 3 | X 1 | X 2 + ε X 1 | X 3 | X 4 + ε X 3 |

X 4 | X 1 e ε | X 2 e 2 ε | X 3 e − ε | X 4 |

where the group parameter ε 4 = a 1 . Obviously, it can not continue to simplify by using adjoint operators. Thirdly, suppose that a 4 = 0 , a 3 = 0 , a 2 ≠ 0 and set up a 2 = 1 , the vector is already the simplest form as X = a 1 X 1 + X 2 . Last suppose that a 4 = 0 , a 3 = 0 , a 2 ≠ 0 , a 1 ≠ 0 and set up a 1 = 1 , that can only be X = X 1 .

To summarize, we state the result that the one-dimensional optimal system of symmetry groups (10) is

{ X 1 , X 2 , X 3 , X 4 , a X 1 + X 2 , a X 2 + X 3 }

where a is arbitrary constant.

In the present section, we present all possible similarity reduction forms of Equation (1), which is an indispensable step to solve the NLPDEs by the symmetry method.

For the symmetry X 4 , the corresponding characteristic equation is

d x x = d t 2 t = d u − u = d v − 2 ( v + 1 ) , (11)

hence we can get a similarity independent variable from (11) defined as ς = x t − 1 / 2 and group invariant solutions defined as

u ( x , t ) = t − 1 / 2 F ( ς ) , v ( x , t ) = − 1 + t − 1 H ( ς ) ,

which satisfy the following reduced equation

{ F + ς F ′ − 2 F F ′ − 2 H ′ = 0 , 2 H − 2 H F ′ + ς H ′ − 2 F H ′ − 2 F ( 3 ) = 0 ,

where F ′ = d F d ς , H ′ = d H d ς .

For other symmetries in the optimal system, the reduction method is the same as X 4 . The results are shown in

In the third section, we obtain the one-dimensional optimal system of Equation (1), and give the reduction equation corresponding to each symmetry in the optimal system in

The power series method is a useful approach to solve higher order ordinary differential equations. A large number of solutions for ordinary differential equations can be constructed by utilizing the method.

Suppose that the power series solution is the following form

F ( ς ) = ∑ n = 0 ∞ c n ς n , H ( ς ) = ∑ n = 0 ∞ s n ς n , (12)

where c n , s n is undetermined coefficient. Substituting (12) into (A), we get

{ ∑ n = 1 ∞ c n ς n + c 0 + ∑ n = 1 ∞ n c n ς n − 2 ∑ n = 1 ∞ ∑ k = 0 n ( n + 1 − k ) c k c n + 1 − k ς n − 2 c 0 c 1 − 2 ∑ n = 1 ∞ ( n + 1 ) s n + 1 ς n − 2 s 1 = 0 , 2 ∑ n = 1 ∞ s n ς n + 2 s 0 − 2 ∑ n = 1 ∞ ∑ k = 0 n ( n + 1 − k ) s k c n + 1 − k ς n − 2 s 0 c 1 + ∑ n = 1 ∞ n s n ς n − 2 ∑ n = 1 ∞ ∑ k = 0 n ( n + 1 − k ) c k s n + 1 − k ς n − 2 c 0 s 1 − 2 ∑ n = 1 ∞ ( n + 1 ) ( n + 2 ) ( n + 3 ) c n + 3 ς n − 12 c 3 = 0.

Through comparing the coefficients of ς, we can easily get the following results

when n = 0 ,

s 1 = c 0 − 2 c 0 c 1 2 , c 3 = s 0 − s 0 c 1 − c 0 s 1 6 , (13)

when n ≥ 1 ,

s n + 1 = 1 2 ( n + 1 ) [ ( n + 1 ) c n − 2 ∑ k = 0 n ( n + 1 − k ) c k c n + 1 − k ] , c n + 3 = 1 2 ( n + 1 ) ( n + 2 ) ( n + 3 ) [ ( 2 + n ) s n − 2 ∑ k = 0 n ( n + 1 − k ) s k c n + 1 − k − 2 ∑ k = 0 n ( n + 1 − k ) c k s n + 1 − k ] . (14)

The sequence { s n } 1 ∞ , { c n } 3 ∞ can be uniquely determined by (13) and (14) and depend on the other undetermined coefficients s 0 , c i ( i = 0 , 1 , 2 ) . It is easy to

Infinitesimal generator | Similarity variables | Reduction equation |
---|---|---|

X 1 | ς = t , u ( x , t ) = F ( ς ) , v ( x , t ) = H ( ς ) . | F ″ = 0 , H ″ = 0. |

X 2 | ς = x , u ( x , t ) = F ( ς ) , v ( x , t ) = H ( ς ) . | F F ′ + H ′ = 0 , F ′ + H F ′ + F H ′ + F ( 3 ) = 0. (D) |

X 3 | ς = t , u ( x , t ) = x / t + F ( ς ) , v ( x , t ) = H ( ς ) . | F + ς F ′ = 0 , H + ς H ′ + 1 = 0. |

X 4 | ς = x t − 1 / 2 , u ( x , t ) = t − 1 / 2 F ( ς ) , v ( x , t ) = − 1 + t − 1 H ( ς ) . | F + ς F ′ − 2 F F ′ − 2 H ′ = 0 , 2 H − 2 H F ′ + ς H ′ − 2 F H ′ − 2 F ( 3 ) = 0. (A) |

X 1 + X 2 | ς = x − t , u ( x , t ) = F ( ς ) , v ( x , t ) = H ( ς ) . | − F ′ + F F ′ + H ′ = 0 , F ′ + H F ′ − H ′ + F H ′ + F ( 3 ) = 0. (B) |

X 2 + X 3 | ς = − t 2 + 2 x / 2 , u ( x , t ) = t + F ( ς ) , v ( x , t ) = H ( ς ) . | 1 + F F ′ + H ′ = 0 , F ′ + H F ′ + F H ′ + F ( 3 ) = 0. (C) |

prove that the power series solution is convergent by references [

F ( ς ) = c 0 + c 1 ς + c 2 ς 2 + c 3 ς 3 + ∑ n = 1 ∞ c n + 3 ς n + 3 = c 0 + c 1 ς + c 2 ς 2 + s 0 − s 0 c 1 − c 0 s 1 6 ς 3 + 1 2 ( n + 1 ) ( n + 2 ) ( n + 3 ) ∑ n = 1 ∞ ( 2 s n + n s n − 2 ∑ k = 0 n ( n + 1 − k ) s k c n + 1 − k − 2 ∑ k = 0 n ( n + 1 − k ) c k s n + 1 − k ) ς n + 3 ,

H ( ς ) = s 0 + s 1 ς + ∑ n = 1 ∞ s n + 1 ς n + 1 = s 0 + c 0 − 2 c 0 c 1 2 ς + 1 2 ( n + 1 ) ∑ n = 1 ∞ ( c n + n c n − 2 ∑ k = 0 n ( n + 1 − k ) c k c n + 1 − k ) ς n + 1 .

And then we get the following power series solution of Equation (1)

u ( x , t ) = t − 1 / 2 [ c 0 + c 1 ( x t − 1 / 2 ) + c 2 ( x t − 1 / 2 ) 2 + s 0 − s 0 c 1 − c 0 s 1 6 ( x t − 1 / 2 ) 3 + 1 2 ( n + 1 ) ( n + 2 ) ( n + 3 ) ∑ n = 1 ∞ ( 2 s n + n s n − 2 ∑ k = 0 n ( n + 1 − k ) s k c n + 1 − k − 2 ∑ k = 0 n ( n + 1 − k ) c k s n + 1 − k ) ( x t − 1 / 2 ) n + 3 ] ,

v ( x , t ) = − 1 + t − 1 [ s 0 + c 0 − 2 c 0 c 1 2 ( x t − 1 / 2 )

+ 1 2 ( n + 1 ) ∑ n = 1 ∞ ( ( n + 1 ) c n − 2 ∑ k = 0 n ( n + 1 − k ) c k c n + 1 − k ) ( x t − 1 / 2 ) n + 1 ] ,

where s 0 , c n ( n = 0 , 1 , 2 ) are arbitrary constant.

The extended Tanh function method is a very effective method for solving some nonlinear evolution equations proposed in recent years [

Suppose that the solution of the reduction Equation (B) can be expressed as the form

{ F ( ς ) = a 0 + ∑ i = 1 m ( a i ϕ i ( ς ) + A i ϕ − i ( ς ) ) , H ( ς ) = b 0 + ∑ j = 1 n ( b j ϕ j ( ς ) + B j ϕ − j ( ς ) ) , (15)

where a 0 , b 0 , a i , A i ( i = 1 , ⋯ , m ) , b j , B j ( j = 0 , 1 , ⋯ , n ) are undetermined constants, and function ϕ = ϕ ( ς ) satisfies

ϕ ′ = λ + ρ ϕ + ω ϕ 2 , (16)

where λ , ρ , ω are arbitrary constant. By solving Equation (16), we can know that the solution of function ϕ can be divided into 4 categories, and amount to 27 solutions [

1) when ρ 2 − 4 λ ω > 0 and ρ ω ≠ 0 (or λ ω ≠ 0 ),

ϕ 1 = − 1 2 ω ( ρ + θ tanh [ θ 2 ς ] ) ,

ϕ 2 = − 1 2 ω ( ρ + θ ( tanh [ θ ς ] ± i sech [ θ ς ] ) ) ,

ϕ 3 = − 1 2 ω ( ρ + θ coth [ θ 2 ς ] ) ,

ϕ 4 = − 1 2 ω ( ρ + θ ( coth [ θ ς ] ± i csch [ θ ς ] ) ) ,

ϕ 5 = − 1 4 ω [ 2 ρ + θ ( tanh [ θ 4 ς ] + coth [ θ 4 ς ] ) ] ,

ϕ 6 = 1 2 ω ( − ρ + ( B 2 − A 2 ) ( θ ) − A θ cosh [ θ ς ] A sinh [ θ ς ] + B ) ,

ϕ 7 = 1 2 ω ( − ρ + ( B 2 − A 2 ) ( θ ) + A θ sinh [ θ ς ] A cosh [ θ ς ] + B ) ,

where A , B are two nonzero constants, and satisfy B 2 − A 2 > 0 .

ϕ 8 = 2 λ cosh [ θ ς / 2 ] θ sinh [ θ ς / 2 ] − ρ cosh [ θ ς / 2 ] ,

ϕ 9 = − 2 λ sinh [ θ ς / 2 ] ρ sinh [ θ ς / 2 ] − θ cosh [ θ ς / 2 ] ,

ϕ 10 = 2 λ cosh [ θ ς ] θ sinh [ θ ς ] − ρ cosh [ θ ς ] ± i θ ,

ϕ 11 = 2 λ sinh [ θ ς ] − ρ sinh [ θ ς ] + θ cosh [ θ ς ] ± θ ,

ϕ 12 = 4 λ sinh [ θ ς / 4 ] cosh [ θ ς / 4 ] − 2 ρ sinh [ θ ς / 4 ] cosh [ θ ς / 4 ] + 2 θ cosh 2 [ θ ς / 4 ] − θ .

2) when ρ 2 − 4 λ ω < 0 and ρ ω ≠ 0 (or λ ω ≠ 0 ),

ϕ 13 = 1 2 ω ( − ρ + − θ tan [ − θ 2 ς ] ) ,

ϕ 14 = 1 2 ω [ − ρ + − θ ( tan [ − θ ς ] ± sec [ − θ ς ] ) ] ,

ϕ 15 = − 1 2 ω ( ρ + − θ cot [ − θ 2 ς ] ) ,

ϕ 16 = − 1 2 ω ( ρ + − θ ( cot [ − θ ς ] ± csc [ − θ ς ] ) ) ,

ϕ 17 = 1 4 ω [ − 2 ρ + − θ ( tan [ − θ 4 ς ] − cot [ − θ 4 ς ] ) ] ,

ϕ 18 = 1 2 ω ( − ρ + ± ( A 2 − B 2 ) ( − θ ) − A − θ cos [ − θ ς ] A sinh [ − θ ς ] + B ) ,

ϕ 19 = 1 2 ω ( − ρ + ( A 2 − B 2 ) ( − θ ) + A − θ sinh [ − θ ς ] A cosh [ − θ ς ] + B ) ,

where A , B are two nonzero constants, and satisfy A 2 − B 2 > 0 .

ϕ 20 = 2 λ cos [ − θ ς / 2 ] − θ sin [ − θ ς / 2 ] + ρ cos [ − θ ς / 2 ] ,

ϕ 21 = − 2 λ sin [ − θ ς / 2 ] − ρ sin [ − θ ς / 2 ] − − θ cos [ − θ ς / 2 ] ,

ϕ 22 = 2 λ cos [ − θ ς ] − θ sin [ − θ ς ] + ρ cos [ − θ ς ] ± − θ ,

ϕ 23 = 2 λ sin [ − θ ς ] − ρ sin [ − θ ς ] + − θ cosh [ − θ ς ] ± − θ ,

ϕ 24 = 4 λ sin [ − θ ς / 4 ] cos [ − θ ς / 4 ] − 2 ρ sin [ − θ ς / 4 ] cos [ − θ ς / 4 ] + 2 − θ cos 2 [ − θ ς / 4 ] − − θ .

Above formula ϕ 1 ∼ ϕ 24 , the symbol θ is expressed as θ = ρ 2 − 4 λ ω .

3) when λ = 0 and ρ ω ≠ 0 ,

ϕ 25 = − ρ b ω ( b + cosh [ ρ ς ] − sinh [ ρ ς ] ) ,

ϕ 26 = − ρ ( cosh [ ρ ς ] + sinh [ ρ ς ] ) ω ( b + cosh [ ρ ς ] − sinh [ ρ ς ] ) ,

where b is a arbitrary constant.

4) when ω ≠ 0 and λ = ρ = 0 ,

ϕ 27 = 1 ω ς + c ,

where c is a arbitrary constant, and ς = x − t .

Considering the homogeneous equilibrium between the highest order linear term and the nonlinear term in the reduction Equation (B), we can obtain m = 1 , n = 2 . As a result, the trial Equations (16) reduces to

{ F ( ς ) = a 0 + a 1 ϕ ( ς ) + A 1 ϕ ( ς ) , H ( ς ) = b 0 + b 1 ϕ ( ς ) + b 2 ϕ 2 ( ς ) + B 1 ϕ ( ς ) + B 2 ϕ 2 ( ς ) . (17)

In order to determine the values of undetermined coefficients a 0 , a 1 , b 0 , b 1 , b 2 , substituting (16) and (17) into the reduction Equation (B) and merging the polynomial of the same power of ϕ , and setting up each polynomial coefficient to zero, we can get the following nonlinear algebraic equations

{ ϕ 0 : − λ A 1 2 − 2 λ B 2 = 0 , − 6 λ 3 A 1 − 3 λ A 1 B 2 = 0 ϕ 1 : − λ A 1 − λ a 0 A 1 − ρ A 1 2 − λ B 1 − 2 ρ B 2 = 0 , − 12 λ 2 ρ A 1 − 2 λ A 1 B 1 + 2 λ B 2 − 2 λ a 0 B 2 − 3 ρ A 1 B 2 = 0 ϕ 2 : ρ A 1 − ρ a 0 A 1 − ω A 1 2 − ρ B 1 − 2 ω B 2 = 0 , − λ A 1 − 7 λ ρ 2 A 1 − 8 λ 2 ω A 1 − λ A 1 b 0 + λ B 1 − λ a 0 B 1 − 2 ρ A 1 B 1 + 2 ρ B 2 − 2 ρ a 0 B 2 − λ a 1 B 2 − 3 ω A 1 B 2 = 0 ⋮ ϕ 6 : ω a 1 2 + 2 ω b 2 = 0 , ω a 1 + 7 ρ 2 ω a 1 + 8 λ ω 2 a 1 + ω a 1 b 0 − ω b 1 + ω a 0 b 1 + 2 ρ a 1 b 1 − 2 ρ b 2 + 2 ρ a 0 b 2 + 3 λ a 1 b 2 + ω A 1 b 2 = 0 ϕ 7 : 0 = 0 , 12 ρ ω 2 a 1 + 2 ω a 1 b 1 − 2 ω b 2 + 2 ω a 0 b 2 + 3 ρ a 1 b 2 = 0 ϕ 8 : 0 = 0 , 6 ω 3 a 1 + 3 ω a 1 b 2 = 0

By solving the above system with the help of Mathematic, we can get the following results

a 0 = 1 ± ρ , a 1 = ± 2 ω , A 1 = ± 2 λ , b 0 = − 1 , b 1 = − 2 ρ ω , b 2 = − 2 ω 2 , B 1 = − 2 λ ρ , B 2 = − 2 λ 2 . (18)

Now, substituting (18) into (19), we obtain explicit solutions of Equation (1) as follow.

{ u k ( x , t ) = 1 ± ρ ± 2 ω ϕ k ( ς ) ± 2 λ ϕ k ( ς ) , v k ( x , t ) = − 1 − 2 ρ ω ϕ k ( ς ) − 2 ω 2 ϕ k 2 ( ς ) − 2 λ ρ ϕ k ( ς ) − 2 λ 2 ϕ k 2 ( ς ) , (19)

where k = 1 , 2 , ⋯ , 27 , ς = x − t , and selecting any hyperbolic function in ϕ 1 ∼ ϕ 27 , for example,

ϕ ( ς ) = 1 2 ω ( − ρ + − θ tan [ − θ 2 ς ] ) .

The explicit solutions (19) become as

{ u ( x , t ) = 1 ± − θ tan [ − θ ς 2 ] ∓ 4 λ ω ρ − − θ tan [ − θ ς 2 ] , v ( x , t ) = − 1 + ρ 2 2 + 1 2 θ tan 2 [ − θ ς 2 ] − 8 λ 2 ω 2 ( − ρ + − θ tan [ − θ ς 2 ] ) 2 − 4 λ ρ ω − ρ + − θ tan [ − θ ς 2 ] ,

where ς = x − t , θ = ρ 2 − 4 λ ω < 0 (see

In this section, we study the power series solution of the reduction Equation (C)

in the form of (12). Substituting (12) into the reduction Equation (C), we get

{ 1 + ∑ n = 1 ∞ ∑ k = 0 n ( n + 1 − k ) c k c n + 1 − k ς n + c 0 c 1 + ∑ n = 1 ∞ ( n + 1 ) s n + 1 ς n + s 1 = 0 , ∑ n = 1 ∞ ( n + 1 ) c n + 1 ς n + c 1 + ∑ n = 1 ∞ ∑ k = 0 n ( n + 1 − k ) s k c n + 1 − k ς n + s 0 c 1 + ∑ n = 1 ∞ ∑ k = 0 n ( n + 1 − k ) c k s n + 1 − k ς n + c 0 s 1 + ∑ n = 1 ∞ ( n + 1 ) ( n + 2 ) ( n + 3 ) c n + 3 ς n + 6 c 3 = 0.

Through comparing the coefficients of ς , we can easily get the following results.

where n = 0 ,

s 1 = − 1 − c 0 c 1 , c 3 = − c 1 + s 0 c 1 + c 0 s 1 6 ,

where n ≥ 1 ,

s n + 1 = − 1 ( n + 1 ) ∑ k = 0 n ( n + 1 − k ) c k c n + 1 − k , c n + 3 = − 1 ( n + 1 ) ( n + 2 ) ( n + 3 ) [ ( n + 1 ) c n + 1 + ∑ k = 0 n ( n + 1 − k ) s k c n + 1 − k + ∑ k = 0 n ( n + 1 − k ) c k s n + 1 − k ] .

Accordingly, the power series solution of the reduction Equation (C) is as follows

F ( ς ) = c 0 + c 1 ς + c 2 ς 2 − c 1 + s 0 c 1 + c 0 s 1 6 ς 3 − 1 ( n + 1 ) ( n + 2 ) ( n + 3 ) ∑ n = 1 ∞ [ ( n + 1 ) c n + 1 + ∑ k = 0 n ( n + 1 − k ) s k c n + 1 − k + ∑ k = 0 n ( n + 1 − k ) c k s n + 1 − k ] ς n + 3 , H ( ς ) = s 0 − ( 1 + c 0 c 1 ) ς − 1 ( n + 1 ) ∑ n = 1 ∞ ∑ k = 0 n ( n + 1 − k ) c k c n + 1 − k ς n + 1 .

And then we get the following power series solution of Equation (1)

u ( x , t ) = t + c 0 + c 1 ( − t 2 + 2 x 2 ) + c 2 ( − t 2 + 2 x 2 ) 2 − c 1 + s 0 c 1 + c 0 s 1 6 ( − t 2 + 2 x 2 ) 3 − 1 ( n + 1 ) ( n + 2 ) ( n + 3 ) ∑ n = 1 ∞ [ ( n + 1 ) c n + 1 + ∑ k = 0 n ( n + 1 − k ) s k c n + 1 − k + ∑ k = 0 n ( n + 1 − k ) c k s n + 1 − k ] ( − t 2 + 2 x 2 ) n + 3 ,

v ( x , t ) = s 0 − ( 1 + c 0 c 1 ) ( − t 2 + 2 x 2 ) − 1 ( n + 1 ) ∑ n = 1 ∞ ∑ k = 0 n ( n + 1 − k ) c k c n + 1 − k ( − t 2 + 2 x 2 ) n + 1 .

Using extended tanh function method, similar to the solving of the reduction Equation (B), we obtain the following results:

a 0 = ± ρ , a 1 = ± 2 ω , A 1 = ± 2 λ , b 0 = − 1 , b 1 = − 2 ρ ω , b 2 = − 2 ω 2 , B 1 = − 2 λ ρ , B 2 = − 2 λ 2 .

and selecting the following hyperbolic function

ϕ = − 1 2 ω ( ρ + θ coth [ θ 2 ς ] )

We obtain explicit solutions of the Equation (1.1) as follow

{ u ( x , t ) = ∓ θ coth [ θ ς 2 ] ∓ 4 λ ω ρ − θ coth [ θ ς 2 ] , v ( x , t ) = − 1 + ρ 2 2 − 1 2 θ coth 2 [ θ ς 2 ] − 8 λ 2 ω 2 ( − ρ + θ coth [ θ ς 2 ] ) 2 + 4 λ ρ ω ρ + θ coth [ θ ς 2 ] ,

where ς = x − t , θ = ρ 2 − 4 λ ω < 0 (see

In the field of physics and engineering mechanics, it is particularly important to solve nonlinear differential equations. In the work, the Lie group analysis method has been employed to investigate the dispersive long wave equations. Based on this method, the vector fields and symmetry reductions have been obtained for the system. Since it is difficult to solve the reduction equations directly, the power series method and the extended Tanh function method have been used to construct more explicit solutions, which can enrich the exact solutions of the dispersive long wave equations. The basic idea is efficient and powerful in solving wide classes of nonlinear differential equations.

The authors would like to express their thanks to the unknown referees for their careful reading and helpful comments. Project supported by the Natural Science Foundation of Inner Mongolia, China (Grant No. 2016MS0116).

The authors declare no conflicts of interest regarding the publication of this paper.

Xue, X.M. and Bai, Y.S. (2018) Lie Symmetry Analysis, Optimal Systems and Explicit Solutions of the Dispersive Long Wave Equations. Journal of Applied Mathematics and Physics, 6, 2681-2696. https://doi.org/10.4236/jamp.2018.612222