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Abstract and Applied Analysis
Volume 2014, Article ID 490165, 6 pages
http://dx.doi.org/10.1155/2014/490165
Research Article

Symmetries, Associated First Integrals, and Double Reduction of Difference Equations

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

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

Academic Editor: Igor Leite Freire

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

Abstract

We determine the symmetry generators of some ordinary difference equations and proceeded to find the first integral and reduce the order of the difference equations. We show that, in some cases, the symmetry generator and first integral are associated via the “invariance condition.” That is, the first integral may be invariant under the symmetry of the original difference equation. When this condition is satisfied, we may proceed to double reduction of the difference equation.

1. Introduction

The theory, reasoning, and algebraic structures dealing with the construction of symmetries for differential equations (DEs) are now well established and documented. Moreover, the application of these in the analysis of DEs, in particular, for finding exact solutions, is widely used in a variety of areas from relativity to fluid mechanics (see [14]). Secondly, the relationship between symmetries and conservation laws has been a subject of interest since Noether’s celebrated work [5] for variational DEs. The extension of this relationship to DEs which may not be variational has been done more recently [6, 7]. The first consequence of this interplay has led to the double reduction of DEs [810].

A vast amount of work has been done to extend the ideas and applications of symmetries to difference equations (Es) in a number of ways—see [1115] and references therein. In some cases, the Es are constructed from the DEs in such a way that the algebra of Lie symmetries remains the same [16]. As far as conservation laws of Es go, the work is more recent—see [12, 17]. Here, we construct symmetries and conservation laws for some ordinary Es, utilise the symmetries to obtain reductions of the equations, and show, in fact, that the notion of “association” between these concepts can be analogously extended to ordinary Es. That is, an association between a symmetry and first integral exists if and only if the first integral is invariant under the symmetry. Thus, a “double reduction” of the E is possible.

2. Preliminaries and Definitions

Consider the following th-order OE: where is a smooth function such that and integer is an independent variable. The general solution of (1) can be written in the form and depends on arbitrary independent constants .

Definition 1. We define to be the shift operator acting on as follows:

That is, if then In the same way,

Definition 2. A symmetry generator, , of (1) is given by and satisfies the symmetry condition where is a function called the characteristic of the one-parameter group.

Definition 3. If is a first integral, then it is constant on the solutions of the OE and hence satisfies where is the shift operator defined in (3).

2.1. First Integral

In [11], Hydon presents a methodology to construct the first integrals of OEs directly. For this method, the symmetries of the OE need not be known. Here, we will only consider second-order OE’s.

We construct first integrals using (8) and an additional condition; that is, Now let Next we differentiate (9) with respect to ; we obtain Differentiating (9) with respect to we get Thus, satisfies the second-order linear functional equation or first integral condition, After solving for and constructing , we need to check that the integrability condition is satisfied. Hence if (14) holds, the first integral takes the form To solve for , we substitute (15) into (9) and solve for the resulting first-order OE.

2.2. Using Symmetries to Obtain the General Solution of an OE

We begin this section by providing some useful definitions. We consider the theory and example provided by Hydon in [11].

Definition 4. The commutator of two symmetry generators and is denoted by and defined by

Definition 5. Given a symmetry generator for a second-order OE, there exists an invariant, satisfying

To determine the invariant, we use the method of characteristics. Note that the invariant satisfies We make the assumption that (18) can be inverted to obtain for some function . Solving (21) requires finding a canonical coordinate which satisfies . The most obvious choice [11] of canonical coordinate is with a general solution of the form where is any integer.

3. Application

The aim of this section is to consider two examples and find their symmetries, first integrals, and general solution. We also briefly discuss what is meant by double reduction and association.

3.1. Example 1

Consider the second-order OE [11]:

3.1.1. Symmetry Generator

Suppose that we seek characteristics of the form . To do this, we use the symmetry condition and solve for . Here, the symmetry condition, given by (7), becomes Firstly, we differentiate (26) with respect to (keeping fixed) and we consider to be a function of ,, and . By the implicit function theorem differentiating with respect to yields Secondly, we apply the differential operator, given by to (26) to get To solve (29), we differentiate it with respect to keeping fixed. As a result we obtain the ODE: whose solution is given by We suppose that for ease of computation. Next we substitute (31) into (29) and we simplify the resulting equation to obtain Thus, where is a constant. Substituting (33) into (31) leads to Therefore, the symmetry generator is given by

3.1.2. First Integral

Suppose that ; then (13) can be rewritten to give We apply the differential operator , given by (28), to (36) to get Next we differentiate (37) with respect to keeping constant to obtain whose solution is given by if we take . We substitute (38) into (37) to obtain the difference equation We choose to get The next step consists of substituting (40) into (38) to get From (11) we get Since the integrability condition holds, we can calculate the first integral . From (41) and (42) we have To find we substitute (43) into (9). We obtain whose solution is given by Finally we substitute (45) into (43) to obtain the first integral

Note. The symmetry generator given by (35) acts on the first integral, , to produce the following equation: We say and are associated and this property has far reaching consequences on “further” reduction of the equation.

3.1.3. Symmetry Reduction

Recall that, in Section 3.1.1, we calculated the symmetry generator, , to be given by (35). Suppose is an invariant of . Then We can use the characteristics to solve for and construct the equation. The independent and dependent variables are given by respectively. Therefore by (51), the dependent variable, , is given by Applying the shift operator on and solving the resulting equation we get where is a constant. Then by (52) and (53) and solving for we obtain

Note. Equation (25) has been reduced by one order into (54). Solving (54) for gives The first integral , given by (46), and the reduction are the same. This is another indication of a relationship between and . In fact, this is the association; that is, is invariant under .

3.2. Example 2

Consider the following linear difference equation [11]:

3.2.1. Symmetry

Suppose that ; then the symmetry condition becomes Similarly, we apply the operator to (57) and we differentiate the resulting equation: with respect to to get . Therefore, Next we solve for by substituting (59) into (58). This gives where is a constant. Substituting into (59) yields The substitution of (61) into (57) yields Thus, where and are arbitrary constants. Finally we substitute (63) into (61) and obtain the characteristic Therefore, the Lie symmetry generators are

3.2.2. First Integral

Suppose that . The first integral condition is given by The solution to (66) is given by where ,, and are constants. Then by (11), we have Substituting (67) and (68) into (15) we obtain the first integral Then we have To satisfy (8), we equate (69) and (70). This gives where is a constant. We thus write as Next we check if is associated with the symmetry generators given in (65). (i)Consider that . One can readily verify that Thus is associated with ; that is, , if the following equations are satisfied: Solving the above equations simultaneously gives . Hence, for to be associated with , (ii)Consider that . We have Hence is associated with if , that is, if (iii)Consider that . Then, Here is associated with if . Therefore

3.2.3. General Solution

We now find the general solution of (56). We determine the commutators of the symmetries to indicate the order of the symmetries in the reduction procedure.(i)Since (56) will be reduced using first. Suppose that is the invariant of . Then Using the method of characteristic we get Applying the shift operator on yields that is, where is a constant. Equating (82) and (84) and solving for , we have whose solution is given by where is an arbitrary constant. Equation (86) is the general solution of (56).Note that solving for in (85) yields Therefore, (given by (77)) and the reduction are the same if and . That is, . If this condition holds then is invariant under .(ii)We can also find a general solution of (56) by using a different symmetry generator. Here, so that (56) will be reduced using first. Again suppose that is invariant of . Then Using the method of characteristics we get Therefore applying the shift operator on gives whose solution is given by where is a constant. Equating (90) and (92) results in Therefore the general solution of (56) is given by where is a constant.(iii)Finally we consider the commutator of and . We have Since the commutator is , we can first reduce the OE with either or . However, since we have already reduced (56) with , we will use . As before, suppose that is invariant of . Then Applying the method of characteristics, we have Applying the shift factor, , on (97) and solving the resulting equation we get Equating (97) and (98) givesWe solve (99) and find which is a general solution of (56). It has to be noted that (99) is the same as (given by (75)) if and . If this is true then is invariant under .

4. Conclusion

We have recalled the procedure to calculate the symmetry generators of some ordinary difference equations and proceeded to find the first integral and reduce the order of the difference equations. We have shown that, in some cases, the symmetry generator, , and first integral, , are associated via the invariance condition . When this condition is satisfied, we may proceed to double reduction of the equation.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

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