Research ArticleSymmetries Associated First Integrals and Double Reduction ofDifference Equations

L Ndlovu M Folly-Gbetoula A H Kara and A Love

School of Mathematics University of the Witwatersrand Private Bag 3 Johannesburg 2050 South Africa

Correspondence should be addressed to A H Kara abdulkarawitsacza

Received 11 April 2014 Accepted 16 July 2014 Published 23 July 2014

Academic Editor Igor Leite Freire

Copyright copy 2014 L Ndlovu et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

We determine the symmetry generators of some ordinary difference equations and proceeded to find the first integral and reducethe order of the difference equations We show that in some cases the symmetry generator and first integral are associated via theldquoinvariance conditionrdquo That is the first integral may be invariant under the symmetry of the original difference equation Whenthis condition is satisfied we may proceed to double reduction of the difference equation

1 Introduction

The theory reasoning and algebraic structures dealing withthe construction of symmetries for differential equations(DEs) are now well established and documented Moreoverthe application of these in the analysis of DEs in particularfor finding exact solutions is widely used in a variety of areasfrom relativity to fluid mechanics (see [1ndash4]) Secondly therelationship between symmetries and conservation laws hasbeen a subject of interest since Noetherrsquos celebrated work [5]for variational DEs The extension of this relationship to DEswhich may not be variational has been done more recently[6 7] The first consequence of this interplay has led to thedouble reduction of DEs [8ndash10]

A vast amount of work has been done to extend the ideasand applications of symmetries to difference equations (ΔEs)in a number of waysmdashsee [11ndash15] and references therein Insome cases the ΔEs are constructed from the DEs in sucha way that the algebra of Lie symmetries remains the same[16] As far as conservation laws of ΔEs go the work is morerecentmdashsee [12 17] Here we construct symmetries and con-servation laws for some ordinary ΔEs utilise the symmetriesto obtain reductions of the equations and show in fact thatthe notion of ldquoassociationrdquo between these concepts can beanalogously extended to ordinaryΔEsThat is an associationbetween a symmetry and first integral exists if and only if thefirst integral is invariant under the symmetryThus a ldquodoublereductionrdquo of the ΔE is possible

2 Preliminaries and Definitions

Consider the following119873th-order OΔE

119906119899+119873

= 120596 (119899 119906119899 119906119899+1

119906119899+119873minus1

) (1)

where 120596 is a smooth function such that (120597120596120597119906119899) = 0 and

integer 119899 is an independent variable The general solution of(1) can be written in the form

119906119899= 119865 (119899 119888

1 119888

119873) (2)

and depends on119873 arbitrary independent constants 119888119894

Definition 1 We define S to be the shift operator acting on 119899as follows

S 119899 997891997888rarr 119899 + 1 (3)

That is if 119906119899= 119865(119899 119888

1 119888

119873) then

S (119906119899) = 119906119899+1

(4)

In the same way

S (119906119899+119896

) = 119906119899+119896+1

119896 = 0 119873 minus 2 (5)

Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 490165 6 pageshttpdxdoiorg1011552014490165

2 Abstract and Applied Analysis

Definition 2 A symmetry generator119883 of (1) is given by

119883 = 119876 (119899 119906119899 119906

119899+119873minus1)

120597

120597119906119899

+ (S119876 (119899 119906119899 119906

119899+119873minus1))

120597

120597119906119899+1

+ sdot sdot sdot + (S119873minus1

119876 (119899 119906119899 119906

119899+119873minus1))

120597

120597119906119899+119873minus1

(6)

and satisfies the symmetry condition

119878119873

119876 (119899 119906119899 119906

119899+119873minus1) minus 119883120596 = 0 (7)

where 119876 = 119876(119899 119906119899 119906

119899+119873minus1) is a function called the

characteristic of the one-parameter group

Definition 3 If 120601 is a first integral then it is constant on thesolutions of the OΔE and hence satisfies

S (120601 (119899 119906119899 119906

119899+119873minus1)) = 120601 (119899 119906

119899 119906

119899+119873minus1)

120601 (119899 + 1 119906119899+1

120596 (119899 119906119899 119906

119899+119873minus1))

= 120601 (119899 119906119899 119906

119899+119873minus1)

(8)

where S is the shift operator defined in (3)

21 First Integral In [11] Hydon presents a methodology toconstruct the first integrals of OΔEs directly For thismethodthe symmetries of the OΔE need not be known Here we willonly consider second-order OΔErsquos

We construct first integrals using (8) and an additionalcondition that is

120601 (119899 + 1 119906119899+1

120596 (119899 119906119899 119906119899+1

)) = 120601 (119899 119906119899 119906119899+1

)

120597120601

120597119906119899+1

= 0

(9)

Now let

1198751(119899 119906119899 119906119899+1

) =120597120601

120597119906119899

(119899 119906119899 119906119899+1

)

1198752(119899 119906119899 119906119899+1

) =120597120601

120597119906119899+1

(10)

Next we differentiate (9) with respect to 119906119899 we obtain

1198751= S1198752

120597120596

120597119906119899

(11)

Differentiating (9) with respect to 119906119899+1

we get

1198752= S1198751+

120597120596

120597119906119899+1

S1198752 (12)

Thus 1198752satisfies the second-order linear functional equation

or first integral condition

S(120597120596

120597119906119899

)S2

1198752+

120597120596

120597119906119899+1

S1198752minus 1198752= 0 (13)

After solving for 1198752and constructing 119875

1 we need to check

that the integrability condition1205971198751

120597119906119899+1

=1205971198752

120597119906119899

(14)

is satisfiedHence if (14) holds the first integral takes the form

120601 = int (1198751119889119906119899+ 1198752119889119906119899+1

) + 119866 (119899) (15)

To solve for 119866(119899) we substitute (15) into (9) and solve for theresulting first-order OΔE

22 Using Symmetries to Obtain the General Solution ofan OΔE We begin this section by providing some usefuldefinitions We consider the theory and example provided byHydon in [11]

Definition 4 The commutator of two symmetry generators119883119873and119883

119872is denoted by [119883

119873 119883119872] and defined by

[119883119873 119883119872] = 119883

119873119883119872minus 119883119872119883119873= minus [119883

119872 119883119873] (16)

Definition 5 Given a symmetry generator for a second-orderOΔE

119883 = 119876 (119899 119906119899 119906119899+1

)120597

120597119906119899

+ 119876 (119899 + 1 119906119899+1

120596 (119899 119906119899 119906119899+1

))

times120597

120597119906119899+1

(17)

there exists an invariantV119899= V (119899 119906

119899 119906119899+1

) (18)

satisfying

119883V119899= 0

120597V119899

120597119906119899+1

= 0 (19)

To determine the invariant we use themethod of charac-teristics Note that the invariant satisfies

[119876120597

120597119906119899

+ S119876120597

120597119906119899+1

] V119899= 0 (20)

We make the assumption that (18) can be inverted to obtain119906119899+1

= 120596 (119899 119906119899 V119899) (21)

for some function120596 Solving (21) requires finding a canonicalcoordinate

119904119899= 119904 (119899 119906

119899) (22)

which satisfies 119883119904119899

= 1 The most obvious choice [11] ofcanonical coordinate is

119904 (119899 119906119899) = int

119889119906119899

119876 (119899 119906119899 120596 (119899 119906

119899 119891 (119899 119888

1)))

(23)

with a general solution of the form

119904119899= 1198882+

119899minus1

sum

119896=1198990

119892 (119896 119891 (119896 1198881)) (24)

where 1198990is any integer

Abstract and Applied Analysis 3

3 Application

The aim of this section is to consider two examples and findtheir symmetries first integrals and general solution Wealso briefly discuss what is meant by double reduction andassociation

31 Example 1 Consider the second-order OΔE [11]

120596 = 119906119899+2

=119899

119899 + 1119906119899+

1

119906119899+1

(25)

311 Symmetry Generator Suppose that we seek characteris-tics of the form119876 = 119876(119899 119906

119899) To do this we use the symmetry

condition and solve for 119876 = 119876(119899 119906119899) Here the symmetry

condition given by (7) becomes

119876 (119899 + 2 120596) + 119876 (119899 + 1 119906119899+1

)1

1199062

119899+1

minus 119876 (119899 119906119899) (

119899

119899 + 1) = 0

(26)

Firstly we differentiate (26) with respect to 119906119899(keeping 120596

fixed) and we consider 119906119899+1

to be a function of 119899 119906119899 and

120596 By the implicit function theorem differentiating 119906119899+1

withrespect to 119906

119899yields

120597119906119899+1

120597119906119899

= minus(120597120596120597119906

119899)

(120597120596120597119906119899+1

)=1198991199062

119899+1

119899 + 1 (27)

Secondly we apply the differential operator given by

119871 =120597

120597119906119899

+120597119906119899+1

120597119906119899

120597

120597119906119899+1

(28)

to (26) to get

minus2119899

(119899 + 1) 119906119899+1

119876 (119899 + 1 119906119899+1

) +119899

119899 + 11198761015840

(119899 + 1 119906119899+1

)

minus119899

119899 + 11198761015840

(119899 119906119899) = 0

(29)

To solve (29) we differentiate it with respect to 119906119899keeping

119906119899+1

fixed As a result we obtain the ODE

119889

119889119906119899

(119899

119899 + 11198761015840

(119899 119906119899)) = 0 (30)

whose solution is given by

119876 (119899 119906119899) = (

119899 + 1

119899)119860 (119899) 119906

119899+ 119861 (119899) (31)

We suppose that 119861(119899) = 0 for ease of computation Next wesubstitute (31) into (29) andwe simplify the resulting equationto obtain

[minus119899 (119899 + 2)

(119899 + 1)2]119860 (119899 + 1) = 119860 (119899) (32)

Thus

119860 (119899) = (119899

119899 + 1) 2119888(minus1)

119899minus1

(33)

where 119888 is a constant Substituting (33) into (31) leads to

119876 (119899 119906119899) = (

119899 + 1

119899) (

119899

119899 + 1) 2119888(minus1)

119899minus1

119906119899= 2119888(minus1)

119899minus1

119906119899

(34)

Therefore the symmetry generator is given by

119883 = 2119888(minus1)119899minus1

119906119899

120597

120597119906119899

(35)

312 First Integral Suppose that 1198752= 1198752(119899 119906119899) then (13) can

be rewritten to give

(119899 + 1

119899 + 2)1198752(119899 + 2 119906

119899+2) minus

1

1199062

119899+1

1198752(119899 + 1 119906

119899+1)

minus 1198752(119899 119906119899) = 0

(36)

We apply the differential operator 119871 given by (28) to (36) toget

119899

119899 + 1

2

119906119899+1

1198752(119899 + 1 119906

119899+1) minus

119899

119899 + 11198751015840

2(119899 + 1 119906

119899+1)

minus 1198751015840

2(119899 119906119899) = 0

(37)

Next we differentiate (37) with respect to 119906119899keeping 119906

119899+1

constant to obtain (119889119889119906119899)(1198751015840

2(119899 119906119899)) = 0 whose solution is

given by

1198752(119899 119906119899) = 119861 (119899) 119906

119899+ 119888 = 119861 (119899) 119906

119899(38)

if we take 119888 = 0 We substitute (38) into (37) to obtain thedifference equation

119861 (119899 + 1) =119899 + 1

119899119861 (119899) (39)

We choose 119861(1) = 1 to get

119861 (119899) = 119899 (40)

The next step consists of substituting (40) into (38) to get

1198752(119899 119906119899) = 119899119906

119899 (41)

From (11) we get

1198751(119899 119906119899 119906119899+1

) = S1198752

120597120596

120597119906119899

= 119899119906119899+1

= 1198751(119899 119906119899+1

) (42)

Since the integrability condition holds we can calculate thefirst integral 120601 From (41) and (42) we have

120601 = int (1198751119889119906119899+ 1198752119889119906119899+1

) + 119866 (119899) = 119899119906119899119906119899+1

+ 119866 (119899) (43)

To find 119866(119899) we substitute (43) into (9) We obtain

119866 (119899 + 1) minus 119866 (119899) + 119899 + 1 = 0 (44)

4 Abstract and Applied Analysis

whose solution is given by

119866 (119899) = minus119899 (119899 + 1)

2 (45)

Finally we substitute (45) into (43) to obtain the first integral

120601 = 119899119906119899119906119899+1

minus119899 (119899 + 1)

2 (46)

Note The symmetry generator given by (35) acts on the firstintegral 120601 to produce the following equation

119883120601 = 119876 (119899 119906119899)120597120601

120597119906119899

+ 119876 (119899 + 1 119906119899+1

)120597120601

120597119906119899+1

= 2119888(minus1)119899minus1

(119899119906119899119899119906119899+1

minus 119899119906119899119899119906119899+1

)

= 0

(47)

We say119883 and 120601 are associated and this property has far reach-ing consequences on ldquofurtherrdquo reduction of the equation

313 Symmetry Reduction Recall that in Section 311 wecalculated the symmetry generator119883 to be

119883 = 2119888(minus1)119899minus1

119906119899

120597

120597119906119899

(48)

given by (35) Suppose V119899= V(119899 119906

119899 119906119899+1

) is an invariant of119883Then

119883V119899= (119876 (119899 119906

119899)

120597

120597119906119899

+S119876 (119899 119906119899)

120597

120597119906119899+1

) V119899= 0 (49)

We can use the characteristics119889119906119899

2119888(minus1)119899minus1

119906119899

=119889119906119899+1

2119888(minus1)119899

119906119899+1

=119889V119899

0(50)

to solve for V119899and construct the equation The independent

and dependent variables are given by120572 = 119906

119899119906119899+1

120574 = V119899 (51)

respectively Therefore by (51) the dependent variable V119899 is

given byV119899= 119906119899119906119899+1

(52)Applying the shift operator on V

119899and solving the resulting

equation we get

V119899=119899 + 1

2+119888

119899 (53)

where 119888 is a constant Then by (52) and (53) and solving for119906119899+1

we obtain

119906119899+1

=119899 + 1

2119906119899

+119888

119899119906119899

(54)

Note Equation (25) has been reduced by one order into (54)Solving (54) for 119888 gives

119888 = 119899119906119899119906119899+1

minus119899 (119899 + 1)

2= 120601 (55)

The first integral 120601 given by (46) and the reduction are thesame This is another indication of a relationship between 120601

and 119883 In fact this is the association that is 120601 is invariantunder119883

32 Example 2 Consider the following linear differenceequation [11]

120596 = 119906119899+2

= 2119906119899+1

minus 119906119899 (56)

321 Symmetry Suppose that 119876 = 119876(119899 119906119899) then the

symmetry condition becomes

119876 (119899 + 2 120596) minus 2119876 (119899 + 1 119906119899+1

) + 119876 (119899 119906119899) = 0 (57)

Similarly we apply the operator 119871 to (57) and we differentiatethe resulting equation

1198761015840

(119899 119906119899) minus 1198761015840

(119899 + 1 119906119899+1

) = 0 (58)

with respect to 119906119899to get 11987610158401015840(119899 119906

119899) = 0 Therefore

119876 (119899 119906119899) = 119860 (119899) 119906

119899+ 119861 (119899) (59)

Next we solve for 119860(119899) by substituting (59) into (58) Thisgives

119860 (119899 + 1) = 119860 (119899) = 119886 (60)

where 119886 is a constant Substituting 119860(119899) = 119886 into (59) yields

119876 (119899 119906119899) = 119886119906

119899+ 119861 (119899) (61)

The substitution of (61) into (57) yields

119861 (119899 + 2) minus 2119861 (119899 + 1) + 119861 (119899) = 0 (62)

Thus

119861 (119899) = 119887119899 + 119888 (63)

where 119887 and 119888 are arbitrary constants Finally we substitute(63) into (61) and obtain the characteristic

119876 (119899 119906119899) = 119886119906

119899+ 119887119899 + 119888 (64)

Therefore the Lie symmetry generators are

1198831= 119906119899

120597

120597119906119899

1198832= 119899

120597

120597119906119899

1198833=

120597

120597119906119899

(65)

322 First Integral Suppose that 1198752= 1198752(119899 119906119899) The first

integral condition is given by

1198752(119899 + 2 120596) minus 2119875

2(119899 + 1 119906

119899+1) + 1198752(119899 119906119899) = 0 (66)

The solution to (66) is given by

1198752(119899 119906119899) = 119896119906

119899+ 119901119899 + 119902 (67)

where 119896 119901 and 119902 are constants Then by (11) we have

1198751(119899 119906119899+1

) = S1198752(119899 119906119899)120597120596

120597119906119899

= minus119896119906119899+1

minus 119901119899 minus 119901 minus 119902

(68)

Abstract and Applied Analysis 5

Substituting (67) and (68) into (15) we obtain the first integral

120601 = 119901119899 (119906119899+1

minus 119906119899) + 119902 (119906

119899+1minus 119906119899) minus 119901119906

119899+ 119866 (119899) (69)

Then we have

S120601 = 119901119899 (119906119899+1

minus 119906119899) + 119902 (119906

119899+1minus 119906119899) minus 119901119906

119899+ 119866 (119899 + 1)

(70)

To satisfy (8) we equate (69) and (70) This gives

119866 (119899 + 1) = 119866 (119899) = 119903 (71)

where 119903 is a constant We thus write 120601 as

120601 = (119901119899 + 119902) 119906119899+1

minus (119901119899 + 119901 + 119902) 119906119899+ 119903 (72)

Next we check if120601 is associatedwith the symmetry generatorsgiven in (65)

(i) Consider that 1198831= 119906119899120597120597119906119899 One can readily verify

that

1198831120601 = minus119906

119899(119901119899 + 119902 + 119901) + 119906

119899+1(119901119899 + 119902) (73)

Thus 120601 is associated with 1198831 that is 119883120601 = 0 if the

following equations are satisfied

119901119899 + 119902 + 119901 = 0 119901119899 + 119902 = 0 (74)

Solving the above equations simultaneously gives 119901 =

119902 = 0 Hence for 120601 to be associated with1198831

120601 = 119903 (75)

(ii) Consider that1198832= 119899120597120597119906

119899 We have

1198832120601 = 119899

120597120601

120597119906119899

+ (119899 + 1)120597120601

120597119906119899+1

= 119902 (76)

Hence 120601 is associated with1198832if 119902 = 0 that is if

120601 = 119901119899119906119899+1

minus (119901119899 + 119901) 119906119899+ 119903 (77)

(iii) Consider that1198833= 119888120597120597119906

119899 Then

1198833120601 = 119888

120597120601

120597119906119899

+ 119888120597120601

120597119906119899+1

= minus119888119901 (78)

Here 120601 is associated with1198833if 119901 = 0 Therefore

120601 = 119902 (119906119899+1

minus 119906119899) + 119903 (79)

323 General Solution We now find the general solutionof (56) We determine the commutators of the symmetriesto indicate the order of the symmetries in the reductionprocedure

(i) Since

[1198831 1198832] = minus119883

2 (80)

(56) will be reduced using 1198832first Suppose that V

119899=

V(119899 119906119899 119906119899+1

) is the invariant of1198832 Then

1198832V119899= [119899

120597V119899

120597119906119899

+ (119899 + 1)120597V119899

120597119906119899+1

] = 0 (81)

Using the method of characteristic we get

V119899= 119899119906119899+1

minus (119899 + 1) 119906119899 (82)

Applying the shift operator on V119899yields

S (V119899) = V119899+1

= V119899 (83)

that is

V119899+1

= V119899= 1198881 (84)

where 1198881is a constant Equating (82) and (84) and

solving for 119906119899+1

we have

119906119899+1

=1198881

119899+ (1 +

1

119899) 119906119899

(85)

whose solution is given by

119906119899= 1198991198882+ 1198881(119899 minus 1) (86)

where 1198882is an arbitrary constant Equation (86) is the

general solution of (56)Note that solving for 119888

1in (85) yields

1198881= 119899119906119899+1

minus (119899 + 1) 119906119899 (87)

Therefore 120601 (given by (77)) and the reduction are thesame if 119901 = 1 and 119903 = 0 That is 120601 = 119888

1 If this

condition holds then 120601 is invariant under1198832

(ii) We can also find a general solution of (56) by using adifferent symmetry generator Here

[1198831 1198833] = minus119883

1 (88)

so that (56) will be reduced using 1198831first Again

suppose that V119899= V(119899 119906

119899 119906119899+1

) is invariant of 1198831

Then

1198831V119899= [119906119899

120597V119899

120597119906119899

+ 119906119899+1

120597V119899

120597119906119899+1

] = 0 (89)

Using the method of characteristics we get

V119899=119906119899+1

119906119899

(90)

Therefore applying the shift operator on V119899gives

V119899+1

= 2 minus1

V119899

(91)

6 Abstract and Applied Analysis

whose solution is given by

V119899=1 + 2119888

1+ 1198991198881

1 + 1198881+ 1198991198881

(92)

where 1198881is a constant Equating (90) and (92) results

in

119906119899+1

= (1 + 2119888

1+ 1198991198881

1 + 1198881+ 1198991198881

)119906119899 (93)

Therefore the general solution of (56) is given by

119906119899=(1 + 1198881+ 1198991198881) 1198882

1 + 1198881

(94)

where 1198882is a constant

(iii) Finally we consider the commutator of1198832and119883

3We

have

[1198832 1198833] = 0 (95)

Since the commutator is 0 we can first reduce theOΔE with either 119883

2or 1198833 However since we have

already reduced (56) with 1198832 we will use 119883

3 As

before suppose that V119899= V(119899 119906

119899 119906119899+1

) is invariant of1198833 Then

1198833V119899= [119888

120597V119899

120597119906119899

+ 119888120597V119899

120597119906119899+1

] = 0 (96)

Applying the method of characteristics we have

V119899= 119906119899+1

minus 119906119899 (97)

Applying the shift factor S on (97) and solving theresulting equation we get

S (V119899) = V119899+1

= V119899= 1198881 (98)

Equating (97) and (98) gives

119906119899+1

= 119906119899+ 1198881 (99)

We solve (99) and find

119906119899= 1198991198881+ 1198882

(100)

which is a general solution of (56) It has to be notedthat (99) is the same as 120601 (given by (75)) if 119902 = 1 and119903 = 0 If this is true then 120601 is invariant under119883

3

4 Conclusion

We have recalled the procedure to calculate the symmetrygenerators of some ordinary difference equations and pro-ceeded to find the first integral and reduce the order of thedifference equations We have shown that in some cases thesymmetry generator 119883 and first integral 120601 are associatedvia the invariance condition 119883120601 = 0 When this condition issatisfied wemay proceed to double reduction of the equation

Conflict of Interests

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

References

[1] G W Bluman and S Kumei Symmetries and DifferentialEquations vol 81 of Applied Mathematical Sciences SpringerNew York NY USA 1989

[2] P J Olver Applications of Lie Groups to Differential Equationsvol 107 of Graduate Texts in Mathematics Springer New YorkNY USA Second edition 1993

[3] H Stephani Differential Equations Their Solutions Using Sym-metries Cambridge University Press Cambridge Mass USA1989

[4] G W Bluman A F Cheviakov and S C Anco Applications ofSymmetry Methods to Partial Differential Equations vol 168 ofApplied Mathematical Sciences Springer New York NY USA2010

[5] E Noether ldquoInvariante variationsproblemerdquo Mathematisch-Physikalische Klasse vol 2 pp 235ndash257 1918

[6] A H Kara and FMMahomed ldquoRelationship between symme-tries and conservation lawsrdquo International Journal ofTheoreticalPhysics vol 39 no 1 pp 23ndash40 2000

[7] A H Kara and F M Mahomed ldquoA basis of conservationlaws for partial differential equationsrdquo Journal of NonlinearMathematical Physics vol 9 supplement 2 pp 60ndash72 2002

[8] W Sarlet and F Cantrijn ldquoGeneralizations of Noetherrsquos theoremin classical mechanicsrdquo SIAM Review vol 23 no 4 pp 467ndash494 1981

[9] A Biswas P Masemola R Morris and A H Kara ldquoOn theinvariances conservation laws and conserved quantities ofthe damped-driven nonlinear Schrodinger equationrdquoCanadianJournal of Physics vol 90 no 2 pp 199ndash206 2012

[10] R Morris A H Kara A Chowdhury and A Biswas ldquoSolitonsolutions conservation laws and reductions of certain classesof nonlinearwave equationsrdquo Zeitschrift fur NaturforschungSection A vol 67 no 10-11 pp 613ndash620 2012

[11] P E Hydon ldquoSymmetries and first integrals of ordinary differ-ence equationsrdquo Proceedings of the Royal Society A vol 456 no2004 pp 2835ndash2855 2000

[12] O G Rasin and P E Hydon ldquoConservation laws of discreteKorteweg-de Vries equationrdquo SIGMA vol 1 pp 1ndash6 2005

[13] D Levi L Vinet and P Winternitz ldquoLie group formalism fordifference equationsrdquo Journal of Physics A Mathematical andGeneral vol 30 no 2 pp 633ndash649 1997

[14] D Levi and P Winternitz ldquoSymmetrie s of discrete dynamicalsystemsrdquo Tech Rep CRM-2312 Centre de recherches mathe-matiques Universite de Montreal 1995

[15] G R W Quispel and R Sahadevan ldquoLie symmetries and theintegration of difference equationsrdquo Physics Letters A vol 184no 1 pp 64ndash70 1993

[16] P Tempesta ldquoIntegrable maps fromGalois differential algebrasBorel transforms and number sequencesrdquo Journal of DifferentialEquations vol 255 no 10 pp 2981ndash2995 2013

[17] P E Hydon and E L Mansfield ldquoA variational complex for dif-ference equationsrdquo Foundations of Computational Mathematicsvol 4 no 2 pp 187ndash217 2004

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Stochastic AnalysisInternational Journal of

2 Abstract and Applied Analysis

Definition 2 A symmetry generator119883 of (1) is given by

119883 = 119876 (119899 119906119899 119906

119899+119873minus1)

120597

120597119906119899

+ (S119876 (119899 119906119899 119906

119899+119873minus1))

120597

120597119906119899+1

+ sdot sdot sdot + (S119873minus1

119876 (119899 119906119899 119906

119899+119873minus1))

120597

120597119906119899+119873minus1

(6)

and satisfies the symmetry condition

119878119873

119876 (119899 119906119899 119906

119899+119873minus1) minus 119883120596 = 0 (7)

where 119876 = 119876(119899 119906119899 119906

119899+119873minus1) is a function called the

characteristic of the one-parameter group

Definition 3 If 120601 is a first integral then it is constant on thesolutions of the OΔE and hence satisfies

S (120601 (119899 119906119899 119906

119899+119873minus1)) = 120601 (119899 119906

119899 119906

119899+119873minus1)

120601 (119899 + 1 119906119899+1

120596 (119899 119906119899 119906

119899+119873minus1))

= 120601 (119899 119906119899 119906

119899+119873minus1)

(8)

where S is the shift operator defined in (3)

21 First Integral In [11] Hydon presents a methodology toconstruct the first integrals of OΔEs directly For thismethodthe symmetries of the OΔE need not be known Here we willonly consider second-order OΔErsquos

We construct first integrals using (8) and an additionalcondition that is

120601 (119899 + 1 119906119899+1

120596 (119899 119906119899 119906119899+1

)) = 120601 (119899 119906119899 119906119899+1

)

120597120601

120597119906119899+1

= 0

(9)

Now let

1198751(119899 119906119899 119906119899+1

) =120597120601

120597119906119899

(119899 119906119899 119906119899+1

)

1198752(119899 119906119899 119906119899+1

) =120597120601

120597119906119899+1

(10)

Next we differentiate (9) with respect to 119906119899 we obtain

1198751= S1198752

120597120596

120597119906119899

(11)

Differentiating (9) with respect to 119906119899+1

we get

1198752= S1198751+

120597120596

120597119906119899+1

S1198752 (12)

Thus 1198752satisfies the second-order linear functional equation

or first integral condition

S(120597120596

120597119906119899

)S2

1198752+

120597120596

120597119906119899+1

S1198752minus 1198752= 0 (13)

After solving for 1198752and constructing 119875

1 we need to check

that the integrability condition1205971198751

120597119906119899+1

=1205971198752

120597119906119899

(14)

is satisfiedHence if (14) holds the first integral takes the form

120601 = int (1198751119889119906119899+ 1198752119889119906119899+1

) + 119866 (119899) (15)

To solve for 119866(119899) we substitute (15) into (9) and solve for theresulting first-order OΔE

22 Using Symmetries to Obtain the General Solution ofan OΔE We begin this section by providing some usefuldefinitions We consider the theory and example provided byHydon in [11]

Definition 4 The commutator of two symmetry generators119883119873and119883

119872is denoted by [119883

119873 119883119872] and defined by

[119883119873 119883119872] = 119883

119873119883119872minus 119883119872119883119873= minus [119883

119872 119883119873] (16)

Definition 5 Given a symmetry generator for a second-orderOΔE

119883 = 119876 (119899 119906119899 119906119899+1

)120597

120597119906119899

+ 119876 (119899 + 1 119906119899+1

120596 (119899 119906119899 119906119899+1

))

times120597

120597119906119899+1

(17)

there exists an invariantV119899= V (119899 119906

119899 119906119899+1

) (18)

satisfying

119883V119899= 0

120597V119899

120597119906119899+1

= 0 (19)

To determine the invariant we use themethod of charac-teristics Note that the invariant satisfies

[119876120597

120597119906119899

+ S119876120597

120597119906119899+1

] V119899= 0 (20)

We make the assumption that (18) can be inverted to obtain119906119899+1

= 120596 (119899 119906119899 V119899) (21)

for some function120596 Solving (21) requires finding a canonicalcoordinate

119904119899= 119904 (119899 119906

119899) (22)

which satisfies 119883119904119899

= 1 The most obvious choice [11] ofcanonical coordinate is

119904 (119899 119906119899) = int

119889119906119899

119876 (119899 119906119899 120596 (119899 119906

119899 119891 (119899 119888

1)))

(23)

with a general solution of the form

119904119899= 1198882+

119899minus1

sum

119896=1198990

119892 (119896 119891 (119896 1198881)) (24)

where 1198990is any integer

Abstract and Applied Analysis 3

3 Application

The aim of this section is to consider two examples and findtheir symmetries first integrals and general solution Wealso briefly discuss what is meant by double reduction andassociation

31 Example 1 Consider the second-order OΔE [11]

120596 = 119906119899+2

=119899

119899 + 1119906119899+

1

119906119899+1

(25)

311 Symmetry Generator Suppose that we seek characteris-tics of the form119876 = 119876(119899 119906

119899) To do this we use the symmetry

condition and solve for 119876 = 119876(119899 119906119899) Here the symmetry

condition given by (7) becomes

119876 (119899 + 2 120596) + 119876 (119899 + 1 119906119899+1

)1

1199062

119899+1

minus 119876 (119899 119906119899) (

119899

119899 + 1) = 0

(26)

Firstly we differentiate (26) with respect to 119906119899(keeping 120596

fixed) and we consider 119906119899+1

to be a function of 119899 119906119899 and

120596 By the implicit function theorem differentiating 119906119899+1

withrespect to 119906

119899yields

120597119906119899+1

120597119906119899

= minus(120597120596120597119906

119899)

(120597120596120597119906119899+1

)=1198991199062

119899+1

119899 + 1 (27)

Secondly we apply the differential operator given by

119871 =120597

120597119906119899

+120597119906119899+1

120597119906119899

120597

120597119906119899+1

(28)

to (26) to get

minus2119899

(119899 + 1) 119906119899+1

119876 (119899 + 1 119906119899+1

) +119899

119899 + 11198761015840

(119899 + 1 119906119899+1

)

minus119899

119899 + 11198761015840

(119899 119906119899) = 0

(29)

To solve (29) we differentiate it with respect to 119906119899keeping

119906119899+1

fixed As a result we obtain the ODE

119889

119889119906119899

(119899

119899 + 11198761015840

(119899 119906119899)) = 0 (30)

whose solution is given by

119876 (119899 119906119899) = (

119899 + 1

119899)119860 (119899) 119906

119899+ 119861 (119899) (31)

We suppose that 119861(119899) = 0 for ease of computation Next wesubstitute (31) into (29) andwe simplify the resulting equationto obtain

[minus119899 (119899 + 2)

(119899 + 1)2]119860 (119899 + 1) = 119860 (119899) (32)

Thus

119860 (119899) = (119899

119899 + 1) 2119888(minus1)

119899minus1

(33)

where 119888 is a constant Substituting (33) into (31) leads to

119876 (119899 119906119899) = (

119899 + 1

119899) (

119899

119899 + 1) 2119888(minus1)

119899minus1

119906119899= 2119888(minus1)

119899minus1

119906119899

(34)

Therefore the symmetry generator is given by

119883 = 2119888(minus1)119899minus1

119906119899

120597

120597119906119899

(35)

312 First Integral Suppose that 1198752= 1198752(119899 119906119899) then (13) can

be rewritten to give

(119899 + 1

119899 + 2)1198752(119899 + 2 119906

119899+2) minus

1

1199062

119899+1

1198752(119899 + 1 119906

119899+1)

minus 1198752(119899 119906119899) = 0

(36)

We apply the differential operator 119871 given by (28) to (36) toget

119899

119899 + 1

2

119906119899+1

1198752(119899 + 1 119906

119899+1) minus

119899

119899 + 11198751015840

2(119899 + 1 119906

119899+1)

minus 1198751015840

2(119899 119906119899) = 0

(37)

Next we differentiate (37) with respect to 119906119899keeping 119906

119899+1

constant to obtain (119889119889119906119899)(1198751015840

2(119899 119906119899)) = 0 whose solution is

given by

1198752(119899 119906119899) = 119861 (119899) 119906

119899+ 119888 = 119861 (119899) 119906

119899(38)

if we take 119888 = 0 We substitute (38) into (37) to obtain thedifference equation

119861 (119899 + 1) =119899 + 1

119899119861 (119899) (39)

We choose 119861(1) = 1 to get

119861 (119899) = 119899 (40)

The next step consists of substituting (40) into (38) to get

1198752(119899 119906119899) = 119899119906

119899 (41)

From (11) we get

1198751(119899 119906119899 119906119899+1

) = S1198752

120597120596

120597119906119899

= 119899119906119899+1

= 1198751(119899 119906119899+1

) (42)

Since the integrability condition holds we can calculate thefirst integral 120601 From (41) and (42) we have

120601 = int (1198751119889119906119899+ 1198752119889119906119899+1

) + 119866 (119899) = 119899119906119899119906119899+1

+ 119866 (119899) (43)

To find 119866(119899) we substitute (43) into (9) We obtain

119866 (119899 + 1) minus 119866 (119899) + 119899 + 1 = 0 (44)

4 Abstract and Applied Analysis

whose solution is given by

119866 (119899) = minus119899 (119899 + 1)

2 (45)

Finally we substitute (45) into (43) to obtain the first integral

120601 = 119899119906119899119906119899+1

minus119899 (119899 + 1)

2 (46)

Note The symmetry generator given by (35) acts on the firstintegral 120601 to produce the following equation

119883120601 = 119876 (119899 119906119899)120597120601

120597119906119899

+ 119876 (119899 + 1 119906119899+1

)120597120601

120597119906119899+1

= 2119888(minus1)119899minus1

(119899119906119899119899119906119899+1

minus 119899119906119899119899119906119899+1

)

= 0

(47)

We say119883 and 120601 are associated and this property has far reach-ing consequences on ldquofurtherrdquo reduction of the equation

313 Symmetry Reduction Recall that in Section 311 wecalculated the symmetry generator119883 to be

119883 = 2119888(minus1)119899minus1

119906119899

120597

120597119906119899

(48)

given by (35) Suppose V119899= V(119899 119906

119899 119906119899+1

) is an invariant of119883Then

119883V119899= (119876 (119899 119906

119899)

120597

120597119906119899

+S119876 (119899 119906119899)

120597

120597119906119899+1

) V119899= 0 (49)

We can use the characteristics119889119906119899

2119888(minus1)119899minus1

119906119899

=119889119906119899+1

2119888(minus1)119899

119906119899+1

=119889V119899

0(50)

to solve for V119899and construct the equation The independent

and dependent variables are given by120572 = 119906

119899119906119899+1

120574 = V119899 (51)

respectively Therefore by (51) the dependent variable V119899 is

given byV119899= 119906119899119906119899+1

(52)Applying the shift operator on V

119899and solving the resulting

equation we get

V119899=119899 + 1

2+119888

119899 (53)

where 119888 is a constant Then by (52) and (53) and solving for119906119899+1

we obtain

119906119899+1

=119899 + 1

2119906119899

+119888

119899119906119899

(54)

Note Equation (25) has been reduced by one order into (54)Solving (54) for 119888 gives

119888 = 119899119906119899119906119899+1

minus119899 (119899 + 1)

2= 120601 (55)

The first integral 120601 given by (46) and the reduction are thesame This is another indication of a relationship between 120601

and 119883 In fact this is the association that is 120601 is invariantunder119883

32 Example 2 Consider the following linear differenceequation [11]

120596 = 119906119899+2

= 2119906119899+1

minus 119906119899 (56)

321 Symmetry Suppose that 119876 = 119876(119899 119906119899) then the

symmetry condition becomes

119876 (119899 + 2 120596) minus 2119876 (119899 + 1 119906119899+1

) + 119876 (119899 119906119899) = 0 (57)

Similarly we apply the operator 119871 to (57) and we differentiatethe resulting equation

1198761015840

(119899 119906119899) minus 1198761015840

(119899 + 1 119906119899+1

) = 0 (58)

with respect to 119906119899to get 11987610158401015840(119899 119906

119899) = 0 Therefore

119876 (119899 119906119899) = 119860 (119899) 119906

119899+ 119861 (119899) (59)

Next we solve for 119860(119899) by substituting (59) into (58) Thisgives

119860 (119899 + 1) = 119860 (119899) = 119886 (60)

where 119886 is a constant Substituting 119860(119899) = 119886 into (59) yields

119876 (119899 119906119899) = 119886119906

119899+ 119861 (119899) (61)

The substitution of (61) into (57) yields

119861 (119899 + 2) minus 2119861 (119899 + 1) + 119861 (119899) = 0 (62)

Thus

119861 (119899) = 119887119899 + 119888 (63)

where 119887 and 119888 are arbitrary constants Finally we substitute(63) into (61) and obtain the characteristic

119876 (119899 119906119899) = 119886119906

119899+ 119887119899 + 119888 (64)

Therefore the Lie symmetry generators are

1198831= 119906119899

120597

120597119906119899

1198832= 119899

120597

120597119906119899

1198833=

120597

120597119906119899

(65)

322 First Integral Suppose that 1198752= 1198752(119899 119906119899) The first

integral condition is given by

1198752(119899 + 2 120596) minus 2119875

2(119899 + 1 119906

119899+1) + 1198752(119899 119906119899) = 0 (66)

The solution to (66) is given by

1198752(119899 119906119899) = 119896119906

119899+ 119901119899 + 119902 (67)

where 119896 119901 and 119902 are constants Then by (11) we have

1198751(119899 119906119899+1

) = S1198752(119899 119906119899)120597120596

120597119906119899

= minus119896119906119899+1

minus 119901119899 minus 119901 minus 119902

(68)

Abstract and Applied Analysis 5

Substituting (67) and (68) into (15) we obtain the first integral

120601 = 119901119899 (119906119899+1

minus 119906119899) + 119902 (119906

119899+1minus 119906119899) minus 119901119906

119899+ 119866 (119899) (69)

Then we have

S120601 = 119901119899 (119906119899+1

minus 119906119899) + 119902 (119906

119899+1minus 119906119899) minus 119901119906

119899+ 119866 (119899 + 1)

(70)

To satisfy (8) we equate (69) and (70) This gives

119866 (119899 + 1) = 119866 (119899) = 119903 (71)

where 119903 is a constant We thus write 120601 as

120601 = (119901119899 + 119902) 119906119899+1

minus (119901119899 + 119901 + 119902) 119906119899+ 119903 (72)

Next we check if120601 is associatedwith the symmetry generatorsgiven in (65)

(i) Consider that 1198831= 119906119899120597120597119906119899 One can readily verify

that

1198831120601 = minus119906

119899(119901119899 + 119902 + 119901) + 119906

119899+1(119901119899 + 119902) (73)

Thus 120601 is associated with 1198831 that is 119883120601 = 0 if the

following equations are satisfied

119901119899 + 119902 + 119901 = 0 119901119899 + 119902 = 0 (74)

Solving the above equations simultaneously gives 119901 =

119902 = 0 Hence for 120601 to be associated with1198831

120601 = 119903 (75)

(ii) Consider that1198832= 119899120597120597119906

119899 We have

1198832120601 = 119899

120597120601

120597119906119899

+ (119899 + 1)120597120601

120597119906119899+1

= 119902 (76)

Hence 120601 is associated with1198832if 119902 = 0 that is if

120601 = 119901119899119906119899+1

minus (119901119899 + 119901) 119906119899+ 119903 (77)

(iii) Consider that1198833= 119888120597120597119906

119899 Then

1198833120601 = 119888

120597120601

120597119906119899

+ 119888120597120601

120597119906119899+1

= minus119888119901 (78)

Here 120601 is associated with1198833if 119901 = 0 Therefore

120601 = 119902 (119906119899+1

minus 119906119899) + 119903 (79)

323 General Solution We now find the general solutionof (56) We determine the commutators of the symmetriesto indicate the order of the symmetries in the reductionprocedure

(i) Since

[1198831 1198832] = minus119883

2 (80)

(56) will be reduced using 1198832first Suppose that V

119899=

V(119899 119906119899 119906119899+1

) is the invariant of1198832 Then

1198832V119899= [119899

120597V119899

120597119906119899

+ (119899 + 1)120597V119899

120597119906119899+1

] = 0 (81)

Using the method of characteristic we get

V119899= 119899119906119899+1

minus (119899 + 1) 119906119899 (82)

Applying the shift operator on V119899yields

S (V119899) = V119899+1

= V119899 (83)

that is

V119899+1

= V119899= 1198881 (84)

where 1198881is a constant Equating (82) and (84) and

solving for 119906119899+1

we have

119906119899+1

=1198881

119899+ (1 +

1

119899) 119906119899

(85)

whose solution is given by

119906119899= 1198991198882+ 1198881(119899 minus 1) (86)

where 1198882is an arbitrary constant Equation (86) is the

general solution of (56)Note that solving for 119888

1in (85) yields

1198881= 119899119906119899+1

minus (119899 + 1) 119906119899 (87)

Therefore 120601 (given by (77)) and the reduction are thesame if 119901 = 1 and 119903 = 0 That is 120601 = 119888

1 If this

condition holds then 120601 is invariant under1198832

(ii) We can also find a general solution of (56) by using adifferent symmetry generator Here

[1198831 1198833] = minus119883

1 (88)

so that (56) will be reduced using 1198831first Again

suppose that V119899= V(119899 119906

119899 119906119899+1

) is invariant of 1198831

Then

1198831V119899= [119906119899

120597V119899

120597119906119899

+ 119906119899+1

120597V119899

120597119906119899+1

] = 0 (89)

Using the method of characteristics we get

V119899=119906119899+1

119906119899

(90)

Therefore applying the shift operator on V119899gives

V119899+1

= 2 minus1

V119899

(91)

6 Abstract and Applied Analysis

whose solution is given by

V119899=1 + 2119888

1+ 1198991198881

1 + 1198881+ 1198991198881

(92)

where 1198881is a constant Equating (90) and (92) results

in

119906119899+1

= (1 + 2119888

1+ 1198991198881

1 + 1198881+ 1198991198881

)119906119899 (93)

Therefore the general solution of (56) is given by

119906119899=(1 + 1198881+ 1198991198881) 1198882

1 + 1198881

(94)

where 1198882is a constant

(iii) Finally we consider the commutator of1198832and119883

3We

have

[1198832 1198833] = 0 (95)

Since the commutator is 0 we can first reduce theOΔE with either 119883

2or 1198833 However since we have

already reduced (56) with 1198832 we will use 119883

3 As

before suppose that V119899= V(119899 119906

119899 119906119899+1

) is invariant of1198833 Then

1198833V119899= [119888

120597V119899

120597119906119899

+ 119888120597V119899

120597119906119899+1

] = 0 (96)

Applying the method of characteristics we have

V119899= 119906119899+1

minus 119906119899 (97)

Applying the shift factor S on (97) and solving theresulting equation we get

S (V119899) = V119899+1

= V119899= 1198881 (98)

Equating (97) and (98) gives

119906119899+1

= 119906119899+ 1198881 (99)

We solve (99) and find

119906119899= 1198991198881+ 1198882

(100)

which is a general solution of (56) It has to be notedthat (99) is the same as 120601 (given by (75)) if 119902 = 1 and119903 = 0 If this is true then 120601 is invariant under119883

3

4 Conclusion

We have recalled the procedure to calculate the symmetrygenerators of some ordinary difference equations and pro-ceeded to find the first integral and reduce the order of thedifference equations We have shown that in some cases thesymmetry generator 119883 and first integral 120601 are associatedvia the invariance condition 119883120601 = 0 When this condition issatisfied wemay proceed to double reduction of the equation

Conflict of Interests

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

References

[1] G W Bluman and S Kumei Symmetries and DifferentialEquations vol 81 of Applied Mathematical Sciences SpringerNew York NY USA 1989

[2] P J Olver Applications of Lie Groups to Differential Equationsvol 107 of Graduate Texts in Mathematics Springer New YorkNY USA Second edition 1993

[3] H Stephani Differential Equations Their Solutions Using Sym-metries Cambridge University Press Cambridge Mass USA1989

[4] G W Bluman A F Cheviakov and S C Anco Applications ofSymmetry Methods to Partial Differential Equations vol 168 ofApplied Mathematical Sciences Springer New York NY USA2010

[5] E Noether ldquoInvariante variationsproblemerdquo Mathematisch-Physikalische Klasse vol 2 pp 235ndash257 1918

[6] A H Kara and FMMahomed ldquoRelationship between symme-tries and conservation lawsrdquo International Journal ofTheoreticalPhysics vol 39 no 1 pp 23ndash40 2000

[7] A H Kara and F M Mahomed ldquoA basis of conservationlaws for partial differential equationsrdquo Journal of NonlinearMathematical Physics vol 9 supplement 2 pp 60ndash72 2002

[8] W Sarlet and F Cantrijn ldquoGeneralizations of Noetherrsquos theoremin classical mechanicsrdquo SIAM Review vol 23 no 4 pp 467ndash494 1981

[9] A Biswas P Masemola R Morris and A H Kara ldquoOn theinvariances conservation laws and conserved quantities ofthe damped-driven nonlinear Schrodinger equationrdquoCanadianJournal of Physics vol 90 no 2 pp 199ndash206 2012

[10] R Morris A H Kara A Chowdhury and A Biswas ldquoSolitonsolutions conservation laws and reductions of certain classesof nonlinearwave equationsrdquo Zeitschrift fur NaturforschungSection A vol 67 no 10-11 pp 613ndash620 2012

[11] P E Hydon ldquoSymmetries and first integrals of ordinary differ-ence equationsrdquo Proceedings of the Royal Society A vol 456 no2004 pp 2835ndash2855 2000

[12] O G Rasin and P E Hydon ldquoConservation laws of discreteKorteweg-de Vries equationrdquo SIGMA vol 1 pp 1ndash6 2005

[13] D Levi L Vinet and P Winternitz ldquoLie group formalism fordifference equationsrdquo Journal of Physics A Mathematical andGeneral vol 30 no 2 pp 633ndash649 1997

[14] D Levi and P Winternitz ldquoSymmetrie s of discrete dynamicalsystemsrdquo Tech Rep CRM-2312 Centre de recherches mathe-matiques Universite de Montreal 1995

[15] G R W Quispel and R Sahadevan ldquoLie symmetries and theintegration of difference equationsrdquo Physics Letters A vol 184no 1 pp 64ndash70 1993

[16] P Tempesta ldquoIntegrable maps fromGalois differential algebrasBorel transforms and number sequencesrdquo Journal of DifferentialEquations vol 255 no 10 pp 2981ndash2995 2013

[17] P E Hydon and E L Mansfield ldquoA variational complex for dif-ference equationsrdquo Foundations of Computational Mathematicsvol 4 no 2 pp 187ndash217 2004

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Mathematical Problems in Engineering

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Differential EquationsInternational 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

Abstract and Applied Analysis 3

3 Application

The aim of this section is to consider two examples and findtheir symmetries first integrals and general solution Wealso briefly discuss what is meant by double reduction andassociation

31 Example 1 Consider the second-order OΔE [11]

120596 = 119906119899+2

=119899

119899 + 1119906119899+

1

119906119899+1

(25)

311 Symmetry Generator Suppose that we seek characteris-tics of the form119876 = 119876(119899 119906

119899) To do this we use the symmetry

condition and solve for 119876 = 119876(119899 119906119899) Here the symmetry

condition given by (7) becomes

119876 (119899 + 2 120596) + 119876 (119899 + 1 119906119899+1

)1

1199062

119899+1

minus 119876 (119899 119906119899) (

119899

119899 + 1) = 0

(26)

Firstly we differentiate (26) with respect to 119906119899(keeping 120596

fixed) and we consider 119906119899+1

to be a function of 119899 119906119899 and

120596 By the implicit function theorem differentiating 119906119899+1

withrespect to 119906

119899yields

120597119906119899+1

120597119906119899

= minus(120597120596120597119906

119899)

(120597120596120597119906119899+1

)=1198991199062

119899+1

119899 + 1 (27)

Secondly we apply the differential operator given by

119871 =120597

120597119906119899

+120597119906119899+1

120597119906119899

120597

120597119906119899+1

(28)

to (26) to get

minus2119899

(119899 + 1) 119906119899+1

119876 (119899 + 1 119906119899+1

) +119899

119899 + 11198761015840

(119899 + 1 119906119899+1

)

minus119899

119899 + 11198761015840

(119899 119906119899) = 0

(29)

To solve (29) we differentiate it with respect to 119906119899keeping

119906119899+1

fixed As a result we obtain the ODE

119889

119889119906119899

(119899

119899 + 11198761015840

(119899 119906119899)) = 0 (30)

whose solution is given by

119876 (119899 119906119899) = (

119899 + 1

119899)119860 (119899) 119906

119899+ 119861 (119899) (31)

We suppose that 119861(119899) = 0 for ease of computation Next wesubstitute (31) into (29) andwe simplify the resulting equationto obtain

[minus119899 (119899 + 2)

(119899 + 1)2]119860 (119899 + 1) = 119860 (119899) (32)

Thus

119860 (119899) = (119899

119899 + 1) 2119888(minus1)

119899minus1

(33)

where 119888 is a constant Substituting (33) into (31) leads to

119876 (119899 119906119899) = (

119899 + 1

119899) (

119899

119899 + 1) 2119888(minus1)

119899minus1

119906119899= 2119888(minus1)

119899minus1

119906119899

(34)

Therefore the symmetry generator is given by

119883 = 2119888(minus1)119899minus1

119906119899

120597

120597119906119899

(35)

312 First Integral Suppose that 1198752= 1198752(119899 119906119899) then (13) can

be rewritten to give

(119899 + 1

119899 + 2)1198752(119899 + 2 119906

119899+2) minus

1

1199062

119899+1

1198752(119899 + 1 119906

119899+1)

minus 1198752(119899 119906119899) = 0

(36)

We apply the differential operator 119871 given by (28) to (36) toget

119899

119899 + 1

2

119906119899+1

1198752(119899 + 1 119906

119899+1) minus

119899

119899 + 11198751015840

2(119899 + 1 119906

119899+1)

minus 1198751015840

2(119899 119906119899) = 0

(37)

Next we differentiate (37) with respect to 119906119899keeping 119906

119899+1

constant to obtain (119889119889119906119899)(1198751015840

2(119899 119906119899)) = 0 whose solution is

given by

1198752(119899 119906119899) = 119861 (119899) 119906

119899+ 119888 = 119861 (119899) 119906

119899(38)

if we take 119888 = 0 We substitute (38) into (37) to obtain thedifference equation

119861 (119899 + 1) =119899 + 1

119899119861 (119899) (39)

We choose 119861(1) = 1 to get

119861 (119899) = 119899 (40)

The next step consists of substituting (40) into (38) to get

1198752(119899 119906119899) = 119899119906

119899 (41)

From (11) we get

1198751(119899 119906119899 119906119899+1

) = S1198752

120597120596

120597119906119899

= 119899119906119899+1

= 1198751(119899 119906119899+1

) (42)

Since the integrability condition holds we can calculate thefirst integral 120601 From (41) and (42) we have

120601 = int (1198751119889119906119899+ 1198752119889119906119899+1

) + 119866 (119899) = 119899119906119899119906119899+1

+ 119866 (119899) (43)

To find 119866(119899) we substitute (43) into (9) We obtain

119866 (119899 + 1) minus 119866 (119899) + 119899 + 1 = 0 (44)

4 Abstract and Applied Analysis

whose solution is given by

119866 (119899) = minus119899 (119899 + 1)

2 (45)

Finally we substitute (45) into (43) to obtain the first integral

120601 = 119899119906119899119906119899+1

minus119899 (119899 + 1)

2 (46)

Note The symmetry generator given by (35) acts on the firstintegral 120601 to produce the following equation

119883120601 = 119876 (119899 119906119899)120597120601

120597119906119899

+ 119876 (119899 + 1 119906119899+1

)120597120601

120597119906119899+1

= 2119888(minus1)119899minus1

(119899119906119899119899119906119899+1

minus 119899119906119899119899119906119899+1

)

= 0

(47)

We say119883 and 120601 are associated and this property has far reach-ing consequences on ldquofurtherrdquo reduction of the equation

313 Symmetry Reduction Recall that in Section 311 wecalculated the symmetry generator119883 to be

119883 = 2119888(minus1)119899minus1

119906119899

120597

120597119906119899

(48)

given by (35) Suppose V119899= V(119899 119906

119899 119906119899+1

) is an invariant of119883Then

119883V119899= (119876 (119899 119906

119899)

120597

120597119906119899

+S119876 (119899 119906119899)

120597

120597119906119899+1

) V119899= 0 (49)

We can use the characteristics119889119906119899

2119888(minus1)119899minus1

119906119899

=119889119906119899+1

2119888(minus1)119899

119906119899+1

=119889V119899

0(50)

to solve for V119899and construct the equation The independent

and dependent variables are given by120572 = 119906

119899119906119899+1

120574 = V119899 (51)

respectively Therefore by (51) the dependent variable V119899 is

given byV119899= 119906119899119906119899+1

(52)Applying the shift operator on V

119899and solving the resulting

equation we get

V119899=119899 + 1

2+119888

119899 (53)

where 119888 is a constant Then by (52) and (53) and solving for119906119899+1

we obtain

119906119899+1

=119899 + 1

2119906119899

+119888

119899119906119899

(54)

Note Equation (25) has been reduced by one order into (54)Solving (54) for 119888 gives

119888 = 119899119906119899119906119899+1

minus119899 (119899 + 1)

2= 120601 (55)

The first integral 120601 given by (46) and the reduction are thesame This is another indication of a relationship between 120601

and 119883 In fact this is the association that is 120601 is invariantunder119883

32 Example 2 Consider the following linear differenceequation [11]

120596 = 119906119899+2

= 2119906119899+1

minus 119906119899 (56)

321 Symmetry Suppose that 119876 = 119876(119899 119906119899) then the

symmetry condition becomes

119876 (119899 + 2 120596) minus 2119876 (119899 + 1 119906119899+1

) + 119876 (119899 119906119899) = 0 (57)

Similarly we apply the operator 119871 to (57) and we differentiatethe resulting equation

1198761015840

(119899 119906119899) minus 1198761015840

(119899 + 1 119906119899+1

) = 0 (58)

with respect to 119906119899to get 11987610158401015840(119899 119906

119899) = 0 Therefore

119876 (119899 119906119899) = 119860 (119899) 119906

119899+ 119861 (119899) (59)

Next we solve for 119860(119899) by substituting (59) into (58) Thisgives

119860 (119899 + 1) = 119860 (119899) = 119886 (60)

where 119886 is a constant Substituting 119860(119899) = 119886 into (59) yields

119876 (119899 119906119899) = 119886119906

119899+ 119861 (119899) (61)

The substitution of (61) into (57) yields

119861 (119899 + 2) minus 2119861 (119899 + 1) + 119861 (119899) = 0 (62)

Thus

119861 (119899) = 119887119899 + 119888 (63)

where 119887 and 119888 are arbitrary constants Finally we substitute(63) into (61) and obtain the characteristic

119876 (119899 119906119899) = 119886119906

119899+ 119887119899 + 119888 (64)

Therefore the Lie symmetry generators are

1198831= 119906119899

120597

120597119906119899

1198832= 119899

120597

120597119906119899

1198833=

120597

120597119906119899

(65)

322 First Integral Suppose that 1198752= 1198752(119899 119906119899) The first

integral condition is given by

1198752(119899 + 2 120596) minus 2119875

2(119899 + 1 119906

119899+1) + 1198752(119899 119906119899) = 0 (66)

The solution to (66) is given by

1198752(119899 119906119899) = 119896119906

119899+ 119901119899 + 119902 (67)

where 119896 119901 and 119902 are constants Then by (11) we have

1198751(119899 119906119899+1

) = S1198752(119899 119906119899)120597120596

120597119906119899

= minus119896119906119899+1

minus 119901119899 minus 119901 minus 119902

(68)

Abstract and Applied Analysis 5

Substituting (67) and (68) into (15) we obtain the first integral

120601 = 119901119899 (119906119899+1

minus 119906119899) + 119902 (119906

119899+1minus 119906119899) minus 119901119906

119899+ 119866 (119899) (69)

Then we have

S120601 = 119901119899 (119906119899+1

minus 119906119899) + 119902 (119906

119899+1minus 119906119899) minus 119901119906

119899+ 119866 (119899 + 1)

(70)

To satisfy (8) we equate (69) and (70) This gives

119866 (119899 + 1) = 119866 (119899) = 119903 (71)

where 119903 is a constant We thus write 120601 as

120601 = (119901119899 + 119902) 119906119899+1

minus (119901119899 + 119901 + 119902) 119906119899+ 119903 (72)

Next we check if120601 is associatedwith the symmetry generatorsgiven in (65)

(i) Consider that 1198831= 119906119899120597120597119906119899 One can readily verify

that

1198831120601 = minus119906

119899(119901119899 + 119902 + 119901) + 119906

119899+1(119901119899 + 119902) (73)

Thus 120601 is associated with 1198831 that is 119883120601 = 0 if the

following equations are satisfied

119901119899 + 119902 + 119901 = 0 119901119899 + 119902 = 0 (74)

Solving the above equations simultaneously gives 119901 =

119902 = 0 Hence for 120601 to be associated with1198831

120601 = 119903 (75)

(ii) Consider that1198832= 119899120597120597119906

119899 We have

1198832120601 = 119899

120597120601

120597119906119899

+ (119899 + 1)120597120601

120597119906119899+1

= 119902 (76)

Hence 120601 is associated with1198832if 119902 = 0 that is if

120601 = 119901119899119906119899+1

minus (119901119899 + 119901) 119906119899+ 119903 (77)

(iii) Consider that1198833= 119888120597120597119906

119899 Then

1198833120601 = 119888

120597120601

120597119906119899

+ 119888120597120601

120597119906119899+1

= minus119888119901 (78)

Here 120601 is associated with1198833if 119901 = 0 Therefore

120601 = 119902 (119906119899+1

minus 119906119899) + 119903 (79)

323 General Solution We now find the general solutionof (56) We determine the commutators of the symmetriesto indicate the order of the symmetries in the reductionprocedure

(i) Since

[1198831 1198832] = minus119883

2 (80)

(56) will be reduced using 1198832first Suppose that V

119899=

V(119899 119906119899 119906119899+1

) is the invariant of1198832 Then

1198832V119899= [119899

120597V119899

120597119906119899

+ (119899 + 1)120597V119899

120597119906119899+1

] = 0 (81)

Using the method of characteristic we get

V119899= 119899119906119899+1

minus (119899 + 1) 119906119899 (82)

Applying the shift operator on V119899yields

S (V119899) = V119899+1

= V119899 (83)

that is

V119899+1

= V119899= 1198881 (84)

where 1198881is a constant Equating (82) and (84) and

solving for 119906119899+1

we have

119906119899+1

=1198881

119899+ (1 +

1

119899) 119906119899

(85)

whose solution is given by

119906119899= 1198991198882+ 1198881(119899 minus 1) (86)

where 1198882is an arbitrary constant Equation (86) is the

general solution of (56)Note that solving for 119888

1in (85) yields

1198881= 119899119906119899+1

minus (119899 + 1) 119906119899 (87)

Therefore 120601 (given by (77)) and the reduction are thesame if 119901 = 1 and 119903 = 0 That is 120601 = 119888

1 If this

condition holds then 120601 is invariant under1198832

(ii) We can also find a general solution of (56) by using adifferent symmetry generator Here

[1198831 1198833] = minus119883

1 (88)

so that (56) will be reduced using 1198831first Again

suppose that V119899= V(119899 119906

119899 119906119899+1

) is invariant of 1198831

Then

1198831V119899= [119906119899

120597V119899

120597119906119899

+ 119906119899+1

120597V119899

120597119906119899+1

] = 0 (89)

Using the method of characteristics we get

V119899=119906119899+1

119906119899

(90)

Therefore applying the shift operator on V119899gives

V119899+1

= 2 minus1

V119899

(91)

6 Abstract and Applied Analysis

whose solution is given by

V119899=1 + 2119888

1+ 1198991198881

1 + 1198881+ 1198991198881

(92)

where 1198881is a constant Equating (90) and (92) results

in

119906119899+1

= (1 + 2119888

1+ 1198991198881

1 + 1198881+ 1198991198881

)119906119899 (93)

Therefore the general solution of (56) is given by

119906119899=(1 + 1198881+ 1198991198881) 1198882

1 + 1198881

(94)

where 1198882is a constant

(iii) Finally we consider the commutator of1198832and119883

3We

have

[1198832 1198833] = 0 (95)

Since the commutator is 0 we can first reduce theOΔE with either 119883

2or 1198833 However since we have

already reduced (56) with 1198832 we will use 119883

3 As

before suppose that V119899= V(119899 119906

119899 119906119899+1

) is invariant of1198833 Then

1198833V119899= [119888

120597V119899

120597119906119899

+ 119888120597V119899

120597119906119899+1

] = 0 (96)

Applying the method of characteristics we have

V119899= 119906119899+1

minus 119906119899 (97)

Applying the shift factor S on (97) and solving theresulting equation we get

S (V119899) = V119899+1

= V119899= 1198881 (98)

Equating (97) and (98) gives

119906119899+1

= 119906119899+ 1198881 (99)

We solve (99) and find

119906119899= 1198991198881+ 1198882

(100)

which is a general solution of (56) It has to be notedthat (99) is the same as 120601 (given by (75)) if 119902 = 1 and119903 = 0 If this is true then 120601 is invariant under119883

3

4 Conclusion

We have recalled the procedure to calculate the symmetrygenerators of some ordinary difference equations and pro-ceeded to find the first integral and reduce the order of thedifference equations We have shown that in some cases thesymmetry generator 119883 and first integral 120601 are associatedvia the invariance condition 119883120601 = 0 When this condition issatisfied wemay proceed to double reduction of the equation

Conflict of Interests

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

References

[1] G W Bluman and S Kumei Symmetries and DifferentialEquations vol 81 of Applied Mathematical Sciences SpringerNew York NY USA 1989

[2] P J Olver Applications of Lie Groups to Differential Equationsvol 107 of Graduate Texts in Mathematics Springer New YorkNY USA Second edition 1993

[3] H Stephani Differential Equations Their Solutions Using Sym-metries Cambridge University Press Cambridge Mass USA1989

[4] G W Bluman A F Cheviakov and S C Anco Applications ofSymmetry Methods to Partial Differential Equations vol 168 ofApplied Mathematical Sciences Springer New York NY USA2010

[5] E Noether ldquoInvariante variationsproblemerdquo Mathematisch-Physikalische Klasse vol 2 pp 235ndash257 1918

[6] A H Kara and FMMahomed ldquoRelationship between symme-tries and conservation lawsrdquo International Journal ofTheoreticalPhysics vol 39 no 1 pp 23ndash40 2000

[7] A H Kara and F M Mahomed ldquoA basis of conservationlaws for partial differential equationsrdquo Journal of NonlinearMathematical Physics vol 9 supplement 2 pp 60ndash72 2002

[8] W Sarlet and F Cantrijn ldquoGeneralizations of Noetherrsquos theoremin classical mechanicsrdquo SIAM Review vol 23 no 4 pp 467ndash494 1981

[9] A Biswas P Masemola R Morris and A H Kara ldquoOn theinvariances conservation laws and conserved quantities ofthe damped-driven nonlinear Schrodinger equationrdquoCanadianJournal of Physics vol 90 no 2 pp 199ndash206 2012

[10] R Morris A H Kara A Chowdhury and A Biswas ldquoSolitonsolutions conservation laws and reductions of certain classesof nonlinearwave equationsrdquo Zeitschrift fur NaturforschungSection A vol 67 no 10-11 pp 613ndash620 2012

[11] P E Hydon ldquoSymmetries and first integrals of ordinary differ-ence equationsrdquo Proceedings of the Royal Society A vol 456 no2004 pp 2835ndash2855 2000

[12] O G Rasin and P E Hydon ldquoConservation laws of discreteKorteweg-de Vries equationrdquo SIGMA vol 1 pp 1ndash6 2005

[13] D Levi L Vinet and P Winternitz ldquoLie group formalism fordifference equationsrdquo Journal of Physics A Mathematical andGeneral vol 30 no 2 pp 633ndash649 1997

[14] D Levi and P Winternitz ldquoSymmetrie s of discrete dynamicalsystemsrdquo Tech Rep CRM-2312 Centre de recherches mathe-matiques Universite de Montreal 1995

[15] G R W Quispel and R Sahadevan ldquoLie symmetries and theintegration of difference equationsrdquo Physics Letters A vol 184no 1 pp 64ndash70 1993

[16] P Tempesta ldquoIntegrable maps fromGalois differential algebrasBorel transforms and number sequencesrdquo Journal of DifferentialEquations vol 255 no 10 pp 2981ndash2995 2013

[17] P E Hydon and E L Mansfield ldquoA variational complex for dif-ference equationsrdquo Foundations of Computational Mathematicsvol 4 no 2 pp 187ndash217 2004

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

4 Abstract and Applied Analysis

whose solution is given by

119866 (119899) = minus119899 (119899 + 1)

2 (45)

Finally we substitute (45) into (43) to obtain the first integral

120601 = 119899119906119899119906119899+1

minus119899 (119899 + 1)

2 (46)

Note The symmetry generator given by (35) acts on the firstintegral 120601 to produce the following equation

119883120601 = 119876 (119899 119906119899)120597120601

120597119906119899

+ 119876 (119899 + 1 119906119899+1

)120597120601

120597119906119899+1

= 2119888(minus1)119899minus1

(119899119906119899119899119906119899+1

minus 119899119906119899119899119906119899+1

)

= 0

(47)

We say119883 and 120601 are associated and this property has far reach-ing consequences on ldquofurtherrdquo reduction of the equation

313 Symmetry Reduction Recall that in Section 311 wecalculated the symmetry generator119883 to be

119883 = 2119888(minus1)119899minus1

119906119899

120597

120597119906119899

(48)

given by (35) Suppose V119899= V(119899 119906

119899 119906119899+1

) is an invariant of119883Then

119883V119899= (119876 (119899 119906

119899)

120597

120597119906119899

+S119876 (119899 119906119899)

120597

120597119906119899+1

) V119899= 0 (49)

We can use the characteristics119889119906119899

2119888(minus1)119899minus1

119906119899

=119889119906119899+1

2119888(minus1)119899

119906119899+1

=119889V119899

0(50)

to solve for V119899and construct the equation The independent

and dependent variables are given by120572 = 119906

119899119906119899+1

120574 = V119899 (51)

respectively Therefore by (51) the dependent variable V119899 is

given byV119899= 119906119899119906119899+1

(52)Applying the shift operator on V

119899and solving the resulting

equation we get

V119899=119899 + 1

2+119888

119899 (53)

where 119888 is a constant Then by (52) and (53) and solving for119906119899+1

we obtain

119906119899+1

=119899 + 1

2119906119899

+119888

119899119906119899

(54)

Note Equation (25) has been reduced by one order into (54)Solving (54) for 119888 gives

119888 = 119899119906119899119906119899+1

minus119899 (119899 + 1)

2= 120601 (55)

The first integral 120601 given by (46) and the reduction are thesame This is another indication of a relationship between 120601

and 119883 In fact this is the association that is 120601 is invariantunder119883

32 Example 2 Consider the following linear differenceequation [11]

120596 = 119906119899+2

= 2119906119899+1

minus 119906119899 (56)

321 Symmetry Suppose that 119876 = 119876(119899 119906119899) then the

symmetry condition becomes

119876 (119899 + 2 120596) minus 2119876 (119899 + 1 119906119899+1

) + 119876 (119899 119906119899) = 0 (57)

Similarly we apply the operator 119871 to (57) and we differentiatethe resulting equation

1198761015840

(119899 119906119899) minus 1198761015840

(119899 + 1 119906119899+1

) = 0 (58)

with respect to 119906119899to get 11987610158401015840(119899 119906

119899) = 0 Therefore

119876 (119899 119906119899) = 119860 (119899) 119906

119899+ 119861 (119899) (59)

Next we solve for 119860(119899) by substituting (59) into (58) Thisgives

119860 (119899 + 1) = 119860 (119899) = 119886 (60)

where 119886 is a constant Substituting 119860(119899) = 119886 into (59) yields

119876 (119899 119906119899) = 119886119906

119899+ 119861 (119899) (61)

The substitution of (61) into (57) yields

119861 (119899 + 2) minus 2119861 (119899 + 1) + 119861 (119899) = 0 (62)

Thus

119861 (119899) = 119887119899 + 119888 (63)

where 119887 and 119888 are arbitrary constants Finally we substitute(63) into (61) and obtain the characteristic

119876 (119899 119906119899) = 119886119906

119899+ 119887119899 + 119888 (64)

Therefore the Lie symmetry generators are

1198831= 119906119899

120597

120597119906119899

1198832= 119899

120597

120597119906119899

1198833=

120597

120597119906119899

(65)

322 First Integral Suppose that 1198752= 1198752(119899 119906119899) The first

integral condition is given by

1198752(119899 + 2 120596) minus 2119875

2(119899 + 1 119906

119899+1) + 1198752(119899 119906119899) = 0 (66)

The solution to (66) is given by

1198752(119899 119906119899) = 119896119906

119899+ 119901119899 + 119902 (67)

where 119896 119901 and 119902 are constants Then by (11) we have

1198751(119899 119906119899+1

) = S1198752(119899 119906119899)120597120596

120597119906119899

= minus119896119906119899+1

minus 119901119899 minus 119901 minus 119902

(68)

Abstract and Applied Analysis 5

Substituting (67) and (68) into (15) we obtain the first integral

120601 = 119901119899 (119906119899+1

minus 119906119899) + 119902 (119906

119899+1minus 119906119899) minus 119901119906

119899+ 119866 (119899) (69)

Then we have

S120601 = 119901119899 (119906119899+1

minus 119906119899) + 119902 (119906

119899+1minus 119906119899) minus 119901119906

119899+ 119866 (119899 + 1)

(70)

To satisfy (8) we equate (69) and (70) This gives

119866 (119899 + 1) = 119866 (119899) = 119903 (71)

where 119903 is a constant We thus write 120601 as

120601 = (119901119899 + 119902) 119906119899+1

minus (119901119899 + 119901 + 119902) 119906119899+ 119903 (72)

Next we check if120601 is associatedwith the symmetry generatorsgiven in (65)

(i) Consider that 1198831= 119906119899120597120597119906119899 One can readily verify

that

1198831120601 = minus119906

119899(119901119899 + 119902 + 119901) + 119906

119899+1(119901119899 + 119902) (73)

Thus 120601 is associated with 1198831 that is 119883120601 = 0 if the

following equations are satisfied

119901119899 + 119902 + 119901 = 0 119901119899 + 119902 = 0 (74)

Solving the above equations simultaneously gives 119901 =

119902 = 0 Hence for 120601 to be associated with1198831

120601 = 119903 (75)

(ii) Consider that1198832= 119899120597120597119906

119899 We have

1198832120601 = 119899

120597120601

120597119906119899

+ (119899 + 1)120597120601

120597119906119899+1

= 119902 (76)

Hence 120601 is associated with1198832if 119902 = 0 that is if

120601 = 119901119899119906119899+1

minus (119901119899 + 119901) 119906119899+ 119903 (77)

(iii) Consider that1198833= 119888120597120597119906

119899 Then

1198833120601 = 119888

120597120601

120597119906119899

+ 119888120597120601

120597119906119899+1

= minus119888119901 (78)

Here 120601 is associated with1198833if 119901 = 0 Therefore

120601 = 119902 (119906119899+1

minus 119906119899) + 119903 (79)

323 General Solution We now find the general solutionof (56) We determine the commutators of the symmetriesto indicate the order of the symmetries in the reductionprocedure

(i) Since

[1198831 1198832] = minus119883

2 (80)

(56) will be reduced using 1198832first Suppose that V

119899=

V(119899 119906119899 119906119899+1

) is the invariant of1198832 Then

1198832V119899= [119899

120597V119899

120597119906119899

+ (119899 + 1)120597V119899

120597119906119899+1

] = 0 (81)

Using the method of characteristic we get

V119899= 119899119906119899+1

minus (119899 + 1) 119906119899 (82)

Applying the shift operator on V119899yields

S (V119899) = V119899+1

= V119899 (83)

that is

V119899+1

= V119899= 1198881 (84)

where 1198881is a constant Equating (82) and (84) and

solving for 119906119899+1

we have

119906119899+1

=1198881

119899+ (1 +

1

119899) 119906119899

(85)

whose solution is given by

119906119899= 1198991198882+ 1198881(119899 minus 1) (86)

where 1198882is an arbitrary constant Equation (86) is the

general solution of (56)Note that solving for 119888

1in (85) yields

1198881= 119899119906119899+1

minus (119899 + 1) 119906119899 (87)

Therefore 120601 (given by (77)) and the reduction are thesame if 119901 = 1 and 119903 = 0 That is 120601 = 119888

1 If this

condition holds then 120601 is invariant under1198832

(ii) We can also find a general solution of (56) by using adifferent symmetry generator Here

[1198831 1198833] = minus119883

1 (88)

so that (56) will be reduced using 1198831first Again

suppose that V119899= V(119899 119906

119899 119906119899+1

) is invariant of 1198831

Then

1198831V119899= [119906119899

120597V119899

120597119906119899

+ 119906119899+1

120597V119899

120597119906119899+1

] = 0 (89)

Using the method of characteristics we get

V119899=119906119899+1

119906119899

(90)

Therefore applying the shift operator on V119899gives

V119899+1

= 2 minus1

V119899

(91)

6 Abstract and Applied Analysis

whose solution is given by

V119899=1 + 2119888

1+ 1198991198881

1 + 1198881+ 1198991198881

(92)

where 1198881is a constant Equating (90) and (92) results

in

119906119899+1

= (1 + 2119888

1+ 1198991198881

1 + 1198881+ 1198991198881

)119906119899 (93)

Therefore the general solution of (56) is given by

119906119899=(1 + 1198881+ 1198991198881) 1198882

1 + 1198881

(94)

where 1198882is a constant

(iii) Finally we consider the commutator of1198832and119883

3We

have

[1198832 1198833] = 0 (95)

Since the commutator is 0 we can first reduce theOΔE with either 119883

2or 1198833 However since we have

already reduced (56) with 1198832 we will use 119883

3 As

before suppose that V119899= V(119899 119906

119899 119906119899+1

) is invariant of1198833 Then

1198833V119899= [119888

120597V119899

120597119906119899

+ 119888120597V119899

120597119906119899+1

] = 0 (96)

Applying the method of characteristics we have

V119899= 119906119899+1

minus 119906119899 (97)

Applying the shift factor S on (97) and solving theresulting equation we get

S (V119899) = V119899+1

= V119899= 1198881 (98)

Equating (97) and (98) gives

119906119899+1

= 119906119899+ 1198881 (99)

We solve (99) and find

119906119899= 1198991198881+ 1198882

(100)

which is a general solution of (56) It has to be notedthat (99) is the same as 120601 (given by (75)) if 119902 = 1 and119903 = 0 If this is true then 120601 is invariant under119883

3

4 Conclusion

We have recalled the procedure to calculate the symmetrygenerators of some ordinary difference equations and pro-ceeded to find the first integral and reduce the order of thedifference equations We have shown that in some cases thesymmetry generator 119883 and first integral 120601 are associatedvia the invariance condition 119883120601 = 0 When this condition issatisfied wemay proceed to double reduction of the equation

Conflict of Interests

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

References

[1] G W Bluman and S Kumei Symmetries and DifferentialEquations vol 81 of Applied Mathematical Sciences SpringerNew York NY USA 1989

[2] P J Olver Applications of Lie Groups to Differential Equationsvol 107 of Graduate Texts in Mathematics Springer New YorkNY USA Second edition 1993

[3] H Stephani Differential Equations Their Solutions Using Sym-metries Cambridge University Press Cambridge Mass USA1989

[4] G W Bluman A F Cheviakov and S C Anco Applications ofSymmetry Methods to Partial Differential Equations vol 168 ofApplied Mathematical Sciences Springer New York NY USA2010

[5] E Noether ldquoInvariante variationsproblemerdquo Mathematisch-Physikalische Klasse vol 2 pp 235ndash257 1918

[6] A H Kara and FMMahomed ldquoRelationship between symme-tries and conservation lawsrdquo International Journal ofTheoreticalPhysics vol 39 no 1 pp 23ndash40 2000

[7] A H Kara and F M Mahomed ldquoA basis of conservationlaws for partial differential equationsrdquo Journal of NonlinearMathematical Physics vol 9 supplement 2 pp 60ndash72 2002

[8] W Sarlet and F Cantrijn ldquoGeneralizations of Noetherrsquos theoremin classical mechanicsrdquo SIAM Review vol 23 no 4 pp 467ndash494 1981

[9] A Biswas P Masemola R Morris and A H Kara ldquoOn theinvariances conservation laws and conserved quantities ofthe damped-driven nonlinear Schrodinger equationrdquoCanadianJournal of Physics vol 90 no 2 pp 199ndash206 2012

[10] R Morris A H Kara A Chowdhury and A Biswas ldquoSolitonsolutions conservation laws and reductions of certain classesof nonlinearwave equationsrdquo Zeitschrift fur NaturforschungSection A vol 67 no 10-11 pp 613ndash620 2012

[11] P E Hydon ldquoSymmetries and first integrals of ordinary differ-ence equationsrdquo Proceedings of the Royal Society A vol 456 no2004 pp 2835ndash2855 2000

[12] O G Rasin and P E Hydon ldquoConservation laws of discreteKorteweg-de Vries equationrdquo SIGMA vol 1 pp 1ndash6 2005

[13] D Levi L Vinet and P Winternitz ldquoLie group formalism fordifference equationsrdquo Journal of Physics A Mathematical andGeneral vol 30 no 2 pp 633ndash649 1997

[14] D Levi and P Winternitz ldquoSymmetrie s of discrete dynamicalsystemsrdquo Tech Rep CRM-2312 Centre de recherches mathe-matiques Universite de Montreal 1995

[15] G R W Quispel and R Sahadevan ldquoLie symmetries and theintegration of difference equationsrdquo Physics Letters A vol 184no 1 pp 64ndash70 1993

[16] P Tempesta ldquoIntegrable maps fromGalois differential algebrasBorel transforms and number sequencesrdquo Journal of DifferentialEquations vol 255 no 10 pp 2981ndash2995 2013

[17] P E Hydon and E L Mansfield ldquoA variational complex for dif-ference equationsrdquo Foundations of Computational Mathematicsvol 4 no 2 pp 187ndash217 2004

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

Abstract and Applied Analysis 5

Substituting (67) and (68) into (15) we obtain the first integral

120601 = 119901119899 (119906119899+1

minus 119906119899) + 119902 (119906

119899+1minus 119906119899) minus 119901119906

119899+ 119866 (119899) (69)

Then we have

S120601 = 119901119899 (119906119899+1

minus 119906119899) + 119902 (119906

119899+1minus 119906119899) minus 119901119906

119899+ 119866 (119899 + 1)

(70)

To satisfy (8) we equate (69) and (70) This gives

119866 (119899 + 1) = 119866 (119899) = 119903 (71)

where 119903 is a constant We thus write 120601 as

120601 = (119901119899 + 119902) 119906119899+1

minus (119901119899 + 119901 + 119902) 119906119899+ 119903 (72)

Next we check if120601 is associatedwith the symmetry generatorsgiven in (65)

(i) Consider that 1198831= 119906119899120597120597119906119899 One can readily verify

that

1198831120601 = minus119906

119899(119901119899 + 119902 + 119901) + 119906

119899+1(119901119899 + 119902) (73)

Thus 120601 is associated with 1198831 that is 119883120601 = 0 if the

following equations are satisfied

119901119899 + 119902 + 119901 = 0 119901119899 + 119902 = 0 (74)

Solving the above equations simultaneously gives 119901 =

119902 = 0 Hence for 120601 to be associated with1198831

120601 = 119903 (75)

(ii) Consider that1198832= 119899120597120597119906

119899 We have

1198832120601 = 119899

120597120601

120597119906119899

+ (119899 + 1)120597120601

120597119906119899+1

= 119902 (76)

Hence 120601 is associated with1198832if 119902 = 0 that is if

120601 = 119901119899119906119899+1

minus (119901119899 + 119901) 119906119899+ 119903 (77)

(iii) Consider that1198833= 119888120597120597119906

119899 Then

1198833120601 = 119888

120597120601

120597119906119899

+ 119888120597120601

120597119906119899+1

= minus119888119901 (78)

Here 120601 is associated with1198833if 119901 = 0 Therefore

120601 = 119902 (119906119899+1

minus 119906119899) + 119903 (79)

323 General Solution We now find the general solutionof (56) We determine the commutators of the symmetriesto indicate the order of the symmetries in the reductionprocedure

(i) Since

[1198831 1198832] = minus119883

2 (80)

(56) will be reduced using 1198832first Suppose that V

119899=

V(119899 119906119899 119906119899+1

) is the invariant of1198832 Then

1198832V119899= [119899

120597V119899

120597119906119899

+ (119899 + 1)120597V119899

120597119906119899+1

] = 0 (81)

Using the method of characteristic we get

V119899= 119899119906119899+1

minus (119899 + 1) 119906119899 (82)

Applying the shift operator on V119899yields

S (V119899) = V119899+1

= V119899 (83)

that is

V119899+1

= V119899= 1198881 (84)

where 1198881is a constant Equating (82) and (84) and

solving for 119906119899+1

we have

119906119899+1

=1198881

119899+ (1 +

1

119899) 119906119899

(85)

whose solution is given by

119906119899= 1198991198882+ 1198881(119899 minus 1) (86)

where 1198882is an arbitrary constant Equation (86) is the

general solution of (56)Note that solving for 119888

1in (85) yields

1198881= 119899119906119899+1

minus (119899 + 1) 119906119899 (87)

Therefore 120601 (given by (77)) and the reduction are thesame if 119901 = 1 and 119903 = 0 That is 120601 = 119888

1 If this

condition holds then 120601 is invariant under1198832

(ii) We can also find a general solution of (56) by using adifferent symmetry generator Here

[1198831 1198833] = minus119883

1 (88)

so that (56) will be reduced using 1198831first Again

suppose that V119899= V(119899 119906

119899 119906119899+1

) is invariant of 1198831

Then

1198831V119899= [119906119899

120597V119899

120597119906119899

+ 119906119899+1

120597V119899

120597119906119899+1

] = 0 (89)

Using the method of characteristics we get

V119899=119906119899+1

119906119899

(90)

Therefore applying the shift operator on V119899gives

V119899+1

= 2 minus1

V119899

(91)

6 Abstract and Applied Analysis

whose solution is given by

V119899=1 + 2119888

1+ 1198991198881

1 + 1198881+ 1198991198881

(92)

where 1198881is a constant Equating (90) and (92) results

in

119906119899+1

= (1 + 2119888

1+ 1198991198881

1 + 1198881+ 1198991198881

)119906119899 (93)

Therefore the general solution of (56) is given by

119906119899=(1 + 1198881+ 1198991198881) 1198882

1 + 1198881

(94)

where 1198882is a constant

(iii) Finally we consider the commutator of1198832and119883

3We

have

[1198832 1198833] = 0 (95)

Since the commutator is 0 we can first reduce theOΔE with either 119883

2or 1198833 However since we have

already reduced (56) with 1198832 we will use 119883

3 As

before suppose that V119899= V(119899 119906

119899 119906119899+1

) is invariant of1198833 Then

1198833V119899= [119888

120597V119899

120597119906119899

+ 119888120597V119899

120597119906119899+1

] = 0 (96)

Applying the method of characteristics we have

V119899= 119906119899+1

minus 119906119899 (97)

Applying the shift factor S on (97) and solving theresulting equation we get

S (V119899) = V119899+1

= V119899= 1198881 (98)

Equating (97) and (98) gives

119906119899+1

= 119906119899+ 1198881 (99)

We solve (99) and find

119906119899= 1198991198881+ 1198882

(100)

which is a general solution of (56) It has to be notedthat (99) is the same as 120601 (given by (75)) if 119902 = 1 and119903 = 0 If this is true then 120601 is invariant under119883

3

4 Conclusion

We have recalled the procedure to calculate the symmetrygenerators of some ordinary difference equations and pro-ceeded to find the first integral and reduce the order of thedifference equations We have shown that in some cases thesymmetry generator 119883 and first integral 120601 are associatedvia the invariance condition 119883120601 = 0 When this condition issatisfied wemay proceed to double reduction of the equation

Conflict of Interests

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

References

[1] G W Bluman and S Kumei Symmetries and DifferentialEquations vol 81 of Applied Mathematical Sciences SpringerNew York NY USA 1989

[2] P J Olver Applications of Lie Groups to Differential Equationsvol 107 of Graduate Texts in Mathematics Springer New YorkNY USA Second edition 1993

[3] H Stephani Differential Equations Their Solutions Using Sym-metries Cambridge University Press Cambridge Mass USA1989

[4] G W Bluman A F Cheviakov and S C Anco Applications ofSymmetry Methods to Partial Differential Equations vol 168 ofApplied Mathematical Sciences Springer New York NY USA2010

[5] E Noether ldquoInvariante variationsproblemerdquo Mathematisch-Physikalische Klasse vol 2 pp 235ndash257 1918

[6] A H Kara and FMMahomed ldquoRelationship between symme-tries and conservation lawsrdquo International Journal ofTheoreticalPhysics vol 39 no 1 pp 23ndash40 2000

[7] A H Kara and F M Mahomed ldquoA basis of conservationlaws for partial differential equationsrdquo Journal of NonlinearMathematical Physics vol 9 supplement 2 pp 60ndash72 2002

[8] W Sarlet and F Cantrijn ldquoGeneralizations of Noetherrsquos theoremin classical mechanicsrdquo SIAM Review vol 23 no 4 pp 467ndash494 1981

[9] A Biswas P Masemola R Morris and A H Kara ldquoOn theinvariances conservation laws and conserved quantities ofthe damped-driven nonlinear Schrodinger equationrdquoCanadianJournal of Physics vol 90 no 2 pp 199ndash206 2012

[10] R Morris A H Kara A Chowdhury and A Biswas ldquoSolitonsolutions conservation laws and reductions of certain classesof nonlinearwave equationsrdquo Zeitschrift fur NaturforschungSection A vol 67 no 10-11 pp 613ndash620 2012

[11] P E Hydon ldquoSymmetries and first integrals of ordinary differ-ence equationsrdquo Proceedings of the Royal Society A vol 456 no2004 pp 2835ndash2855 2000

[12] O G Rasin and P E Hydon ldquoConservation laws of discreteKorteweg-de Vries equationrdquo SIGMA vol 1 pp 1ndash6 2005

[13] D Levi L Vinet and P Winternitz ldquoLie group formalism fordifference equationsrdquo Journal of Physics A Mathematical andGeneral vol 30 no 2 pp 633ndash649 1997

[14] D Levi and P Winternitz ldquoSymmetrie s of discrete dynamicalsystemsrdquo Tech Rep CRM-2312 Centre de recherches mathe-matiques Universite de Montreal 1995

[15] G R W Quispel and R Sahadevan ldquoLie symmetries and theintegration of difference equationsrdquo Physics Letters A vol 184no 1 pp 64ndash70 1993

[16] P Tempesta ldquoIntegrable maps fromGalois differential algebrasBorel transforms and number sequencesrdquo Journal of DifferentialEquations vol 255 no 10 pp 2981ndash2995 2013

[17] P E Hydon and E L Mansfield ldquoA variational complex for dif-ference equationsrdquo Foundations of Computational Mathematicsvol 4 no 2 pp 187ndash217 2004

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

6 Abstract and Applied Analysis

whose solution is given by

V119899=1 + 2119888

1+ 1198991198881

1 + 1198881+ 1198991198881

(92)

where 1198881is a constant Equating (90) and (92) results

in

119906119899+1

= (1 + 2119888

1+ 1198991198881

1 + 1198881+ 1198991198881

)119906119899 (93)

Therefore the general solution of (56) is given by

119906119899=(1 + 1198881+ 1198991198881) 1198882

1 + 1198881

(94)

where 1198882is a constant

(iii) Finally we consider the commutator of1198832and119883

3We

have

[1198832 1198833] = 0 (95)

Since the commutator is 0 we can first reduce theOΔE with either 119883

2or 1198833 However since we have

already reduced (56) with 1198832 we will use 119883

3 As

before suppose that V119899= V(119899 119906

119899 119906119899+1

) is invariant of1198833 Then

1198833V119899= [119888

120597V119899

120597119906119899

+ 119888120597V119899

120597119906119899+1

] = 0 (96)

Applying the method of characteristics we have

V119899= 119906119899+1

minus 119906119899 (97)

Applying the shift factor S on (97) and solving theresulting equation we get

S (V119899) = V119899+1

= V119899= 1198881 (98)

Equating (97) and (98) gives

119906119899+1

= 119906119899+ 1198881 (99)

We solve (99) and find

119906119899= 1198991198881+ 1198882

(100)

which is a general solution of (56) It has to be notedthat (99) is the same as 120601 (given by (75)) if 119902 = 1 and119903 = 0 If this is true then 120601 is invariant under119883

3

4 Conclusion

We have recalled the procedure to calculate the symmetrygenerators of some ordinary difference equations and pro-ceeded to find the first integral and reduce the order of thedifference equations We have shown that in some cases thesymmetry generator 119883 and first integral 120601 are associatedvia the invariance condition 119883120601 = 0 When this condition issatisfied wemay proceed to double reduction of the equation

Conflict of Interests

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

References

[1] G W Bluman and S Kumei Symmetries and DifferentialEquations vol 81 of Applied Mathematical Sciences SpringerNew York NY USA 1989

[2] P J Olver Applications of Lie Groups to Differential Equationsvol 107 of Graduate Texts in Mathematics Springer New YorkNY USA Second edition 1993

[3] H Stephani Differential Equations Their Solutions Using Sym-metries Cambridge University Press Cambridge Mass USA1989

[4] G W Bluman A F Cheviakov and S C Anco Applications ofSymmetry Methods to Partial Differential Equations vol 168 ofApplied Mathematical Sciences Springer New York NY USA2010

[5] E Noether ldquoInvariante variationsproblemerdquo Mathematisch-Physikalische Klasse vol 2 pp 235ndash257 1918

[6] A H Kara and FMMahomed ldquoRelationship between symme-tries and conservation lawsrdquo International Journal ofTheoreticalPhysics vol 39 no 1 pp 23ndash40 2000

[7] A H Kara and F M Mahomed ldquoA basis of conservationlaws for partial differential equationsrdquo Journal of NonlinearMathematical Physics vol 9 supplement 2 pp 60ndash72 2002

[8] W Sarlet and F Cantrijn ldquoGeneralizations of Noetherrsquos theoremin classical mechanicsrdquo SIAM Review vol 23 no 4 pp 467ndash494 1981

[9] A Biswas P Masemola R Morris and A H Kara ldquoOn theinvariances conservation laws and conserved quantities ofthe damped-driven nonlinear Schrodinger equationrdquoCanadianJournal of Physics vol 90 no 2 pp 199ndash206 2012

[10] R Morris A H Kara A Chowdhury and A Biswas ldquoSolitonsolutions conservation laws and reductions of certain classesof nonlinearwave equationsrdquo Zeitschrift fur NaturforschungSection A vol 67 no 10-11 pp 613ndash620 2012

[11] P E Hydon ldquoSymmetries and first integrals of ordinary differ-ence equationsrdquo Proceedings of the Royal Society A vol 456 no2004 pp 2835ndash2855 2000

[12] O G Rasin and P E Hydon ldquoConservation laws of discreteKorteweg-de Vries equationrdquo SIGMA vol 1 pp 1ndash6 2005

[13] D Levi L Vinet and P Winternitz ldquoLie group formalism fordifference equationsrdquo Journal of Physics A Mathematical andGeneral vol 30 no 2 pp 633ndash649 1997

[14] D Levi and P Winternitz ldquoSymmetrie s of discrete dynamicalsystemsrdquo Tech Rep CRM-2312 Centre de recherches mathe-matiques Universite de Montreal 1995

[15] G R W Quispel and R Sahadevan ldquoLie symmetries and theintegration of difference equationsrdquo Physics Letters A vol 184no 1 pp 64ndash70 1993

[16] P Tempesta ldquoIntegrable maps fromGalois differential algebrasBorel transforms and number sequencesrdquo Journal of DifferentialEquations vol 255 no 10 pp 2981ndash2995 2013

[17] P E Hydon and E L Mansfield ldquoA variational complex for dif-ference equationsrdquo Foundations of Computational Mathematicsvol 4 no 2 pp 187ndash217 2004

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