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

Authors: L. Ndlovu; M. Folly-Gbetoula; A. H. Kara; A. Love
Journal: Abstract and Applied Analysis
 (2014)

---

## 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 areassociated 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.


---

## Body

## 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 [1–4]). 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 [8–10].A vast amount of work has been done to extend the ideas and applications of symmetries to difference equations (ΔEs) in a number of ways—see [11–15] and references therein. In some cases, the ΔEs 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 ΔEs go, the work is more recent—see [12, 17]. Here, we construct symmetries and conservation laws for some ordinary ΔEs, 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 ΔEs. 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 ΔE is possible.


## 2. Preliminaries and Definitions

Consider the followingNth-order OΔE:
(1)
u
n
+
N
=
ω
(
n
,
u
n
,
u
n
+
1
,
…
,
u
n
+
N
-
1
)
,
where ω is a smooth function such that (
∂
ω
/
∂
u
n
)
≠
0 and integer n is an independent variable. The general solution of (1) can be written in the form
(2)
u
n
=
F
(
n
,
c
1
,
…
,
c
N
)
and depends on N arbitrary independent constants c
i.Definition 1.
We defineS to be the shift operator acting on n as follows:
(3)
S
:
n
⟼
n
+
1
.That is, ifu
n
=
F
(
n
,
c
1
,
…
,
c
N
) then(4)
S
(
u
n
)
=
u
n
+
1
.
In the same way,
(5)
S
(
u
n
+
k
)
=
u
n
+
k
+
1
,
k
=
0
,
…
,
N
-
2
.Definition 2.
A symmetry generator,X, of (1) is given by
(6)
X
=
Q
(
n
,
u
n
,
…
,
u
n
+
N
-
1
)
∂
∂
u
n
+
(
S
Q
(
n
,
u
n
,
…
,
u
n
+
N
-
1
)
)
∂
∂
u
n
+
1
+
⋯
+
(
S
N
-
1
Q
(
n
,
u
n
,
…
,
u
n
+
N
-
1
)
)
∂
∂
u
n
+
N
-
1
and satisfies the symmetry condition
(7)
S
N
Q
(
n
,
u
n
,
…
,
u
n
+
N
-
1
)
-
X
ω
=
0
,
where Q
=
Q
(
n
,
u
n
,
…
,
u
n
+
N
-
1
) is a function called the characteristic of the one-parameter group.Definition 3.
Ifϕ is a first integral, then it is constant on the solutions of the OΔE and hence satisfies
(8)
S
(
ϕ
(
n
,
u
n
,
…
,
u
n
+
N
-
1
)
)
=
ϕ
(
n
,
u
n
,
…
,
u
n
+
N
-
1
)
,
ϕ
(
n
+
1
,
u
n
+
1
,
…
,
ω
(
n
,
u
n
,
…
,
u
n
+
N
-
1
)
)
=
ϕ
(
n
,
u
n
,
…
,
u
n
+
N
-
1
)
,
where S is the shift operator defined in (3).


### 2.1. First Integral

In [11], Hydon presents a methodology to construct the first integrals of OΔEs directly. For this method, the symmetries of the OΔE need not be known. Here, we will only consider second-order OΔE’s.We construct first integrals using (8) and an additional condition; that is,
(9)
ϕ
(
n
+
1
,
u
n
+
1
,
ω
(
n
,
u
n
,
u
n
+
1
)
)
=
ϕ
(
n
,
u
n
,
u
n
+
1
)
,
∂
ϕ
∂
u
n
+
1
≠
0
.
Now let
(10)
P
1
(
n
,
u
n
,
u
n
+
1
)
=
∂
ϕ
∂
u
n
(
n
,
u
n
,
u
n
+
1
)
,
P
2
(
n
,
u
n
,
u
n
+
1
)
=
∂
ϕ
∂
u
n
+
1
.
Next we differentiate (9) with respect to u
n; we obtain
(11)
P
1
=
S
P
2
∂
ω
∂
u
n
.
Differentiating (9) with respect to u
n
+
1 we get
(12)
P
2
=
S
P
1
+
∂
ω
∂
u
n
+
1
S
P
2
.
Thus, P
2 satisfies the second-order linear functional equation orfirst integral condition,
(13)
S
(
∂
ω
∂
u
n
)
S
2
P
2
+
∂
ω
∂
u
n
+
1
S
P
2
-
P
2
=
0
.
After solving for P
2 and constructing P
1, we need to check that theintegrability condition
(14)
∂
P
1
∂
u
n
+
1
=
∂
P
2
∂
u
n
is satisfied. Hence if (14) holds, the first integral takes the form
(15)
ϕ
=
∫
‍
(
P
1
d
u
n
+
P
2
d
u
n
+
1
)
+
G
(
n
)
.
To solve for G
(
n
), we substitute (15) into (9) and solve for the resulting first-order OΔE.


### 2.2. Using Symmetries to Obtain the General Solution of an OΔE

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 generatorsX
N and X
M is denoted by [
X
N
,
X
M
] and defined by
(16)
[
X
N
,
X
M
]
=
X
N
X
M
-
X
M
X
N
=
-
[
X
M
,
X
N
]
.Definition 5.
Given a symmetry generator for a second-order OΔE,
(17)
X
=
Q
(
n
,
u
n
,
u
n
+
1
)
∂
∂
u
n
+
Q
(
n
+
1
,
u
n
+
1
,
ω
(
n
,
u
n
,
u
n
+
1
)
)
×
∂
∂
u
n
+
1
;
there exists an invariant,
(18)
v
n
=
v
(
n
,
u
n
,
u
n
+
1
)
,
satisfying
(19)
X
v
n
=
0
,
∂
v
n
∂
u
n
+
1
≠
0
.To determine the invariant, we use themethod of characteristics. Note that the invariant satisfies
(20)
[
Q
∂
∂
u
n
+
S
Q
∂
∂
u
n
+
1
]
v
n
=
0
.
We make the assumption that (18) can be inverted to obtain
(21)
u
n
+
1
=
ω
(
n
,
u
n
,
v
n
)
for some function ω. Solving (21) requires finding a canonical coordinate
(22)
s
n
=
s
(
n
,
u
n
)
which satisfies X
s
n
=
1. The most obvious choice [11] of canonical coordinate is
(23)
s
(
n
,
u
n
)
=
∫
‍
d
u
n
Q
(
n
,
u
n
,
ω
(
n
,
u
n
,
f
(
n
;
c
1
)
)
)
with a general solution of the form
(24)
s
n
=
c
2
+
∑
k
=
n
0
n
-
1
‍
g
(
k
,
f
(
k
;
c
1
)
)
,
where n
0 is any integer.


## 2.1. First Integral

In [11], Hydon presents a methodology to construct the first integrals of OΔEs directly. For this method, the symmetries of the OΔE need not be known. Here, we will only consider second-order OΔE’s.We construct first integrals using (8) and an additional condition; that is,
(9)
ϕ
(
n
+
1
,
u
n
+
1
,
ω
(
n
,
u
n
,
u
n
+
1
)
)
=
ϕ
(
n
,
u
n
,
u
n
+
1
)
,
∂
ϕ
∂
u
n
+
1
≠
0
.
Now let
(10)
P
1
(
n
,
u
n
,
u
n
+
1
)
=
∂
ϕ
∂
u
n
(
n
,
u
n
,
u
n
+
1
)
,
P
2
(
n
,
u
n
,
u
n
+
1
)
=
∂
ϕ
∂
u
n
+
1
.
Next we differentiate (9) with respect to u
n; we obtain
(11)
P
1
=
S
P
2
∂
ω
∂
u
n
.
Differentiating (9) with respect to u
n
+
1 we get
(12)
P
2
=
S
P
1
+
∂
ω
∂
u
n
+
1
S
P
2
.
Thus, P
2 satisfies the second-order linear functional equation orfirst integral condition,
(13)
S
(
∂
ω
∂
u
n
)
S
2
P
2
+
∂
ω
∂
u
n
+
1
S
P
2
-
P
2
=
0
.
After solving for P
2 and constructing P
1, we need to check that theintegrability condition
(14)
∂
P
1
∂
u
n
+
1
=
∂
P
2
∂
u
n
is satisfied. Hence if (14) holds, the first integral takes the form
(15)
ϕ
=
∫
‍
(
P
1
d
u
n
+
P
2
d
u
n
+
1
)
+
G
(
n
)
.
To solve for G
(
n
), we substitute (15) into (9) and solve for the resulting first-order OΔE.


## 2.2. Using Symmetries to Obtain the General Solution of an OΔE

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 generatorsX
N and X
M is denoted by [
X
N
,
X
M
] and defined by
(16)
[
X
N
,
X
M
]
=
X
N
X
M
-
X
M
X
N
=
-
[
X
M
,
X
N
]
.Definition 5.
Given a symmetry generator for a second-order OΔE,
(17)
X
=
Q
(
n
,
u
n
,
u
n
+
1
)
∂
∂
u
n
+
Q
(
n
+
1
,
u
n
+
1
,
ω
(
n
,
u
n
,
u
n
+
1
)
)
×
∂
∂
u
n
+
1
;
there exists an invariant,
(18)
v
n
=
v
(
n
,
u
n
,
u
n
+
1
)
,
satisfying
(19)
X
v
n
=
0
,
∂
v
n
∂
u
n
+
1
≠
0
.To determine the invariant, we use themethod of characteristics. Note that the invariant satisfies
(20)
[
Q
∂
∂
u
n
+
S
Q
∂
∂
u
n
+
1
]
v
n
=
0
.
We make the assumption that (18) can be inverted to obtain
(21)
u
n
+
1
=
ω
(
n
,
u
n
,
v
n
)
for some function ω. Solving (21) requires finding a canonical coordinate
(22)
s
n
=
s
(
n
,
u
n
)
which satisfies X
s
n
=
1. The most obvious choice [11] of canonical coordinate is
(23)
s
(
n
,
u
n
)
=
∫
‍
d
u
n
Q
(
n
,
u
n
,
ω
(
n
,
u
n
,
f
(
n
;
c
1
)
)
)
with a general solution of the form
(24)
s
n
=
c
2
+
∑
k
=
n
0
n
-
1
‍
g
(
k
,
f
(
k
;
c
1
)
)
,
where n
0 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 OΔE [11]:
(25)
ω
=
u
n
+
2
=
n
n
+
1
u
n
+
1
u
n
+
1
.


#### 3.1.1. Symmetry Generator

Suppose that we seek characteristics of the formQ
=
Q
(
n
,
u
n
). To do this, we use the symmetry condition and solve for Q
=
Q
(
n
,
u
n
). Here, the symmetry condition, given by (7), becomes
(26)
Q
(
n
+
2
,
ω
)
+
Q
(
n
+
1
,
u
n
+
1
)
1
u
n
+
1
2
-
Q
(
n
,
u
n
)
(
n
n
+
1
)
=
0
.
Firstly, we differentiate (26) with respect to u
n (keeping ω fixed) and we consider u
n
+
1 to be a function of n,u
n, and ω. By the implicit function theorem differentiating u
n
+
1 with respect to u
n yields
(27)
∂
u
n
+
1
∂
u
n
=
-
(
∂
ω
/
∂
u
n
)
(
∂
ω
/
∂
u
n
+
1
)
=
n
u
n
+
1
2
n
+
1
.
Secondly, we apply thedifferential operator, given by
(28)
L
=
∂
∂
u
n
+
∂
u
n
+
1
∂
u
n
∂
∂
u
n
+
1
,
to (26) to get
(29)
-
2
n
(
n
+
1
)
u
n
+
1
Q
(
n
+
1
,
u
n
+
1
)
+
n
n
+
1
Q
′
(
n
+
1
,
u
n
+
1
)
-
n
n
+
1
Q
′
(
n
,
u
n
)
=
0
.
To solve (29), we differentiate it with respect to u
n keeping u
n
+
1 fixed. As a result we obtain the ODE:
(30)
d
d
u
n
(
n
n
+
1
Q
′
(
n
,
u
n
)
)
=
0
whose solution is given by
(31)
Q
(
n
,
u
n
)
=
(
n
+
1
n
)
A
(
n
)
u
n
+
B
(
n
)
.
We suppose that B
(
n
)
=
0 for ease of computation. Next we substitute (31) into (29) and we simplify the resulting equation to obtain
(32)
[
-
n
(
n
+
2
)
(
n
+
1
)
2
]
A
(
n
+
1
)
=
A
(
n
)
.
Thus,
(33)
A
(
n
)
=
(
n
n
+
1
)
2
c
(
-
1
)
n
-
1
,
where c is a constant. Substituting (33) into (31) leads to
(34)
Q
(
n
,
u
n
)
=
(
n
+
1
n
)
(
n
n
+
1
)
2
c
(
-
1
)
n
-
1
u
n
=
2
c
(
-
1
)
n
-
1
u
n
.
Therefore, the symmetry generator is given by
(35)
X
=
2
c
(
-
1
)
n
-
1
u
n
∂
∂
u
n
.


#### 3.1.2. First Integral

Suppose thatP
2
=
P
2
(
n
,
u
n
); then (13) can be rewritten to give
(36)
(
n
+
1
n
+
2
)
P
2
(
n
+
2
,
u
n
+
2
)
-
1
u
n
+
1
2
P
2
(
n
+
1
,
u
n
+
1
)
-
P
2
(
n
,
u
n
)
=
0
.
We apply the differential operator L, given by (28), to (36) to get(37)
n
n
+
1
2
u
n
+
1
P
2
(
n
+
1
,
u
n
+
1
)
-
n
n
+
1
P
2
′
(
n
+
1
,
u
n
+
1
)
-
P
2
′
(
n
,
u
n
)
=
0
.
Next we differentiate (37) with respect to u
n keeping u
n
+
1 constant to obtain (
d
/
d
u
n
)
(
P
2
′
(
n
,
u
n
)
)
=
0 whose solution is given by
(38)
P
2
(
n
,
u
n
)
=
B
(
n
)
u
n
+
c
=
B
(
n
)
u
n
if we take c
=
0. We substitute (38) into (37) to obtain the difference equation
(39)
B
(
n
+
1
)
=
n
+
1
n
B
(
n
)
.
We choose B
(
1
)
=
1 to get
(40)
B
(
n
)
=
n
.
The next step consists of substituting (40) into (38) to get
(41)
P
2
(
n
,
u
n
)
=
n
u
n
.
From (11) we get
(42)
P
1
(
n
,
u
n
,
u
n
+
1
)
=
S
P
2
∂
ω
∂
u
n
=
n
u
n
+
1
=
P
1
(
n
,
u
n
+
1
)
.
Since the integrability condition holds, we can calculate the first integral ϕ. From (41) and (42) we have
(43)
ϕ
=
∫
‍
(
P
1
d
u
n
+
P
2
d
u
n
+
1
)
+
G
(
n
)
=
n
u
n
u
n
+
1
+
G
(
n
)
.
To find G
(
n
) we substitute (43) into (9). We obtain
(44)
G
(
n
+
1
)
-
G
(
n
)
+
n
+
1
=
0
whose solution is given by
(45)
G
(
n
)
=
-
n
(
n
+
1
)
2
.
Finally we substitute (45) into (43) to obtain the first integral(46)
ϕ
=
n
u
n
u
n
+
1
-
n
(
n
+
1
)
2
.Note. The symmetry generator given by (35) acts on the first integral, ϕ, to produce the following equation:
(47)
X
ϕ
=
Q
(
n
,
u
n
)
∂
ϕ
∂
u
n
+
Q
(
n
+
1
,
u
n
+
1
)
∂
ϕ
∂
u
n
+
1
=
2
c
(
-
1
)
n
-
1
(
n
u
n
n
u
n
+
1
-
n
u
n
n
u
n
+
1
)
=
0
.
We say X and ϕ areassociated and this property has far reaching consequences on “further” reduction of the equation.


#### 3.1.3. Symmetry Reduction

Recall that, in Section3.1.1, we calculated the symmetry generator, X, to be
(48)
X
=
2
c
(
-
1
)
n
-
1
u
n
∂
∂
u
n
given by (35). Suppose v
n
=
v
(
n
,
u
n
,
u
n
+
1
) is an invariant of X. Then
(49)
X
v
n
=
(
Q
(
n
,
u
n
)
∂
∂
u
n
+
S
Q
(
n
,
u
n
)
∂
∂
u
n
+
1
)
v
n
=
0
.
We can use the characteristics
(50)
d
u
n
2
c
(
-
1
)
n
-
1
u
n
=
d
u
n
+
1
2
c
(
-
1
)
n
u
n
+
1
=
d
v
n
0
to solve for v
n and construct the equation. The independent and dependent variables are given by
(51)
α
=
u
n
u
n
+
1
,
γ
=
v
n
,
respectively. Therefore by (51), the dependent variable, v
n, is given by
(52)
v
n
=
u
n
u
n
+
1
.
Applying the shift operator on v
n and solving the resulting equation we get
(53)
v
n
=
n
+
1
2
+
c
n
,
where c is a constant. Then by (52) and (53) and solving for u
n
+
1 we obtain(54)
u
n
+
1
=
n
+
1
2
u
n
+
c
n
u
n
.Note. Equation (25) has been reduced by one order into (54). Solving (54) for c gives
(55)
c
=
n
u
n
u
n
+
1
-
n
(
n
+
1
)
2
=
ϕ
.
The first integral ϕ, given by (46), and the reduction are the same. This is another indication of a relationship between ϕ and X. In fact, this is the association; that is, ϕ isinvariant under X.


### 3.2. Example 2

Consider the following linear difference equation [11]:
(56)
ω
=
u
n
+
2
=
2
u
n
+
1
-
u
n
.


#### 3.2.1. Symmetry

Suppose thatQ
=
Q
(
n
,
u
n
); then the symmetry condition becomes
(57)
Q
(
n
+
2
,
ω
)
-
2
Q
(
n
+
1
,
u
n
+
1
)
+
Q
(
n
,
u
n
)
=
0
.
Similarly, we apply the operator L to (57) and we differentiate the resulting equation:
(58)
Q
′
(
n
,
u
n
)
-
Q
′
(
n
+
1
,
u
n
+
1
)
=
0
,
with respect to u
n to get Q
′′
(
n
,
u
n
)
=
0. Therefore,
(59)
Q
(
n
,
u
n
)
=
A
(
n
)
u
n
+
B
(
n
)
.
Next we solve for A
(
n
) by substituting (59) into (58). This gives
(60)
A
(
n
+
1
)
=
A
(
n
)
=
a
,
where a is a constant. Substituting A
(
n
)
=
a into (59) yields
(61)
Q
(
n
,
u
n
)
=
a
u
n
+
B
(
n
)
.
The substitution of (61) into (57) yields
(62)
B
(
n
+
2
)
-
2
B
(
n
+
1
)
+
B
(
n
)
=
0
.
Thus,
(63)
B
(
n
)
=
b
n
+
c
,
where b and c are arbitrary constants. Finally we substitute (63) into (61) and obtain the characteristic
(64)
Q
(
n
,
u
n
)
=
a
u
n
+
b
n
+
c
.
Therefore, the Lie symmetry generators are
(65)
X
1
=
u
n
∂
∂
u
n
,
X
2
=
n
∂
∂
u
n
,
X
3
=
∂
∂
u
n
.


#### 3.2.2. First Integral

Suppose thatP
2
=
P
2
(
n
,
u
n
). Thefirst integral condition is given by
(66)
P
2
(
n
+
2
,
ω
)
-
2
P
2
(
n
+
1
,
u
n
+
1
)
+
P
2
(
n
,
u
n
)
=
0
.
The solution to (66) is given by
(67)
P
2
(
n
,
u
n
)
=
k
u
n
+
p
n
+
q
,
where k,p, and q are constants. Then by (11), we have
(68)
P
1
(
n
,
u
n
+
1
)
=
S
P
2
(
n
,
u
n
)
∂
ω
∂
u
n
=
-
k
u
n
+
1
-
p
n
-
p
-
q
.
Substituting (67) and (68) into (15) we obtain the first integral
(69)
ϕ
=
p
n
(
u
n
+
1
-
u
n
)
+
q
(
u
n
+
1
-
u
n
)
-
p
u
n
+
G
(
n
)
.
Then we have
(70)
S
ϕ
=
p
n
(
u
n
+
1
-
u
n
)
+
q
(
u
n
+
1
-
u
n
)
-
p
u
n
+
G
(
n
+
1
)
.
To satisfy (8), we equate (69) and (70). This gives
(71)
G
(
n
+
1
)
=
G
(
n
)
=
r
,
where r is a constant. We thus write ϕ as
(72)
ϕ
=
(
p
n
+
q
)
u
n
+
1
-
(
p
n
+
p
+
q
)
u
n
+
r
.
Next we check if ϕ is associated with the symmetry generators given in (65). (i)
Consider thatX
1
=
u
n
∂
/
∂
u
n. One can readily verify that
(73)
X
1
ϕ
=
-
u
n
(
p
n
+
q
+
p
)
+
u
n
+
1
(
p
n
+
q
)
.
Thus ϕ is associated with X
1; that is, X
ϕ
=
0, if the following equations are satisfied:
(74)
p
n
+
q
+
p
=
0
,
p
n
+
q
=
0
.
Solving the above equations simultaneously gives p
=
q
=
0. Hence, for ϕ to be associated with X
1,
(75)
ϕ
=
r
.
(ii)
Consider thatX
2
=
n
∂
/
∂
u
n. We have
(76)
X
2
ϕ
=
n
∂
ϕ
∂
u
n
+
(
n
+
1
)
∂
ϕ
∂
u
n
+
1
=
q
.
Hence ϕ is associated with X
2 if q
=
0, that is, if
(77)
ϕ
=
p
n
u
n
+
1
-
(
p
n
+
p
)
u
n
+
r
.
(iii)
Consider thatX
3
=
c
∂
/
∂
u
n. Then,
(78)
X
3
ϕ
=
c
∂
ϕ
∂
u
n
+
c
∂
ϕ
∂
u
n
+
1
=
-
c
p
.
Here ϕ is associated with X
3 if p
=
0. Therefore
(79)
ϕ
=
q
(
u
n
+
1
-
u
n
)
+
r
.


#### 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(80)
[
X
1
,
X
2
]
=
-
X
2
,
(56) will be reduced using X
2 first. Suppose that v
n
=
v
(
n
,
u
n
,
u
n
+
1
) is the invariant of X
2. Then
(81)
X
2
v
n
=
[
n
∂
v
n
∂
u
n
+
(
n
+
1
)
∂
v
n
∂
u
n
+
1
]
=
0
.
Using the method of characteristic we get(82)
v
n
=
n
u
n
+
1
-
(
n
+
1
)
u
n
.
Applying the shift operator onv
n yields
(83)
S
(
v
n
)
=
v
n
+
1
=
v
n
;
that is,(84)
v
n
+
1
=
v
n
=
c
1
,
wherec
1 is a constant. Equating (82) and (84) and solving for u
n
+
1, we have
(85)
u
n
+
1
=
c
1
n
+
(
1
+
1
n
)
u
n
whose solution is given by(86)
u
n
=
n
c
2
+
c
1
(
n
-
1
)
,
wherec
2 is an arbitrary constant. Equation (86) is the general solution of (56).
Note that solving forc
1 in (85) yields
(87)
c
1
=
n
u
n
+
1
-
(
n
+
1
)
u
n
.
Therefore,ϕ (given by (77)) and the reduction are the same if p
=
1 and r
=
0. That is, ϕ
=
c
1. If this condition holds then ϕ is invariant under X
2.
(ii)
We can also find a general solution of (56) by using a different symmetry generator. Here,
(88)
[
X
1
,
X
3
]
=
-
X
1
,
so that (56) will be reduced using X
1 first. Again suppose that v
n
=
v
(
n
,
u
n
,
u
n
+
1
) is invariant of X
1. Then
(89)
X
1
v
n
=
[
u
n
∂
v
n
∂
u
n
+
u
n
+
1
∂
v
n
∂
u
n
+
1
]
=
0
.
Using the method of characteristics we get(90)
v
n
=
u
n
+
1
u
n
.
Therefore applying the shift operator onv
n gives
(91)
v
n
+
1
=
2
-
1
v
n
whose solution is given by(92)
v
n
=
1
+
2
c
1
+
n
c
1
1
+
c
1
+
n
c
1
,
wherec
1 is a constant. Equating (90) and (92) results in
(93)
u
n
+
1
=
(
1
+
2
c
1
+
n
c
1
1
+
c
1
+
n
c
1
)
u
n
.
Therefore the general solution of (56) is given by
(94)
u
n
=
(
1
+
c
1
+
n
c
1
)
c
2
1
+
c
1
,
wherec
2 is a constant.
(iii)
Finally we consider the commutator ofX
2 and X
3. We have
(95)
[
X
2
,
X
3
]
=
0
.
Since the commutator is0, we can first reduce the OΔE with either X
2 or X
3. However, since we have already reduced (56) with X
2, we will use X
3. As before, suppose that v
n
=
v
(
n
,
u
n
,
u
n
+
1
) is invariant of X
3. Then
(96)
X
3
v
n
=
[
c
∂
v
n
∂
u
n
+
c
∂
v
n
∂
u
n
+
1
]
=
0
.
Applying the method of characteristics, we have(97)
v
n
=
u
n
+
1
-
u
n
.
Applying the shift factor,S, on (97) and solving the resulting equation we get
(98)
S
(
v
n
)
=
v
n
+
1
=
v
n
=
c
1
.
Equating (97) and (98) gives
(99)
u
n
+
1
=
u
n
+
c
1
.
We solve (99) and find
(100)
u
n
=
n
c
1
+
c
2
which is a general solution of (56). It has to be noted that (99) is the same as ϕ (given by (75)) if q
=
1 and r
=
0. If this is true then ϕ is invariant under X
3.


## 3.1. Example 1

Consider the second-order OΔE [11]:
(25)
ω
=
u
n
+
2
=
n
n
+
1
u
n
+
1
u
n
+
1
.


### 3.1.1. Symmetry Generator

Suppose that we seek characteristics of the formQ
=
Q
(
n
,
u
n
). To do this, we use the symmetry condition and solve for Q
=
Q
(
n
,
u
n
). Here, the symmetry condition, given by (7), becomes
(26)
Q
(
n
+
2
,
ω
)
+
Q
(
n
+
1
,
u
n
+
1
)
1
u
n
+
1
2
-
Q
(
n
,
u
n
)
(
n
n
+
1
)
=
0
.
Firstly, we differentiate (26) with respect to u
n (keeping ω fixed) and we consider u
n
+
1 to be a function of n,u
n, and ω. By the implicit function theorem differentiating u
n
+
1 with respect to u
n yields
(27)
∂
u
n
+
1
∂
u
n
=
-
(
∂
ω
/
∂
u
n
)
(
∂
ω
/
∂
u
n
+
1
)
=
n
u
n
+
1
2
n
+
1
.
Secondly, we apply thedifferential operator, given by
(28)
L
=
∂
∂
u
n
+
∂
u
n
+
1
∂
u
n
∂
∂
u
n
+
1
,
to (26) to get
(29)
-
2
n
(
n
+
1
)
u
n
+
1
Q
(
n
+
1
,
u
n
+
1
)
+
n
n
+
1
Q
′
(
n
+
1
,
u
n
+
1
)
-
n
n
+
1
Q
′
(
n
,
u
n
)
=
0
.
To solve (29), we differentiate it with respect to u
n keeping u
n
+
1 fixed. As a result we obtain the ODE:
(30)
d
d
u
n
(
n
n
+
1
Q
′
(
n
,
u
n
)
)
=
0
whose solution is given by
(31)
Q
(
n
,
u
n
)
=
(
n
+
1
n
)
A
(
n
)
u
n
+
B
(
n
)
.
We suppose that B
(
n
)
=
0 for ease of computation. Next we substitute (31) into (29) and we simplify the resulting equation to obtain
(32)
[
-
n
(
n
+
2
)
(
n
+
1
)
2
]
A
(
n
+
1
)
=
A
(
n
)
.
Thus,
(33)
A
(
n
)
=
(
n
n
+
1
)
2
c
(
-
1
)
n
-
1
,
where c is a constant. Substituting (33) into (31) leads to
(34)
Q
(
n
,
u
n
)
=
(
n
+
1
n
)
(
n
n
+
1
)
2
c
(
-
1
)
n
-
1
u
n
=
2
c
(
-
1
)
n
-
1
u
n
.
Therefore, the symmetry generator is given by
(35)
X
=
2
c
(
-
1
)
n
-
1
u
n
∂
∂
u
n
.


### 3.1.2. First Integral

Suppose thatP
2
=
P
2
(
n
,
u
n
); then (13) can be rewritten to give
(36)
(
n
+
1
n
+
2
)
P
2
(
n
+
2
,
u
n
+
2
)
-
1
u
n
+
1
2
P
2
(
n
+
1
,
u
n
+
1
)
-
P
2
(
n
,
u
n
)
=
0
.
We apply the differential operator L, given by (28), to (36) to get(37)
n
n
+
1
2
u
n
+
1
P
2
(
n
+
1
,
u
n
+
1
)
-
n
n
+
1
P
2
′
(
n
+
1
,
u
n
+
1
)
-
P
2
′
(
n
,
u
n
)
=
0
.
Next we differentiate (37) with respect to u
n keeping u
n
+
1 constant to obtain (
d
/
d
u
n
)
(
P
2
′
(
n
,
u
n
)
)
=
0 whose solution is given by
(38)
P
2
(
n
,
u
n
)
=
B
(
n
)
u
n
+
c
=
B
(
n
)
u
n
if we take c
=
0. We substitute (38) into (37) to obtain the difference equation
(39)
B
(
n
+
1
)
=
n
+
1
n
B
(
n
)
.
We choose B
(
1
)
=
1 to get
(40)
B
(
n
)
=
n
.
The next step consists of substituting (40) into (38) to get
(41)
P
2
(
n
,
u
n
)
=
n
u
n
.
From (11) we get
(42)
P
1
(
n
,
u
n
,
u
n
+
1
)
=
S
P
2
∂
ω
∂
u
n
=
n
u
n
+
1
=
P
1
(
n
,
u
n
+
1
)
.
Since the integrability condition holds, we can calculate the first integral ϕ. From (41) and (42) we have
(43)
ϕ
=
∫
‍
(
P
1
d
u
n
+
P
2
d
u
n
+
1
)
+
G
(
n
)
=
n
u
n
u
n
+
1
+
G
(
n
)
.
To find G
(
n
) we substitute (43) into (9). We obtain
(44)
G
(
n
+
1
)
-
G
(
n
)
+
n
+
1
=
0
whose solution is given by
(45)
G
(
n
)
=
-
n
(
n
+
1
)
2
.
Finally we substitute (45) into (43) to obtain the first integral(46)
ϕ
=
n
u
n
u
n
+
1
-
n
(
n
+
1
)
2
.Note. The symmetry generator given by (35) acts on the first integral, ϕ, to produce the following equation:
(47)
X
ϕ
=
Q
(
n
,
u
n
)
∂
ϕ
∂
u
n
+
Q
(
n
+
1
,
u
n
+
1
)
∂
ϕ
∂
u
n
+
1
=
2
c
(
-
1
)
n
-
1
(
n
u
n
n
u
n
+
1
-
n
u
n
n
u
n
+
1
)
=
0
.
We say X and ϕ areassociated and this property has far reaching consequences on “further” reduction of the equation.


### 3.1.3. Symmetry Reduction

Recall that, in Section3.1.1, we calculated the symmetry generator, X, to be
(48)
X
=
2
c
(
-
1
)
n
-
1
u
n
∂
∂
u
n
given by (35). Suppose v
n
=
v
(
n
,
u
n
,
u
n
+
1
) is an invariant of X. Then
(49)
X
v
n
=
(
Q
(
n
,
u
n
)
∂
∂
u
n
+
S
Q
(
n
,
u
n
)
∂
∂
u
n
+
1
)
v
n
=
0
.
We can use the characteristics
(50)
d
u
n
2
c
(
-
1
)
n
-
1
u
n
=
d
u
n
+
1
2
c
(
-
1
)
n
u
n
+
1
=
d
v
n
0
to solve for v
n and construct the equation. The independent and dependent variables are given by
(51)
α
=
u
n
u
n
+
1
,
γ
=
v
n
,
respectively. Therefore by (51), the dependent variable, v
n, is given by
(52)
v
n
=
u
n
u
n
+
1
.
Applying the shift operator on v
n and solving the resulting equation we get
(53)
v
n
=
n
+
1
2
+
c
n
,
where c is a constant. Then by (52) and (53) and solving for u
n
+
1 we obtain(54)
u
n
+
1
=
n
+
1
2
u
n
+
c
n
u
n
.Note. Equation (25) has been reduced by one order into (54). Solving (54) for c gives
(55)
c
=
n
u
n
u
n
+
1
-
n
(
n
+
1
)
2
=
ϕ
.
The first integral ϕ, given by (46), and the reduction are the same. This is another indication of a relationship between ϕ and X. In fact, this is the association; that is, ϕ isinvariant under X.


## 3.1.1. Symmetry Generator

Suppose that we seek characteristics of the formQ
=
Q
(
n
,
u
n
). To do this, we use the symmetry condition and solve for Q
=
Q
(
n
,
u
n
). Here, the symmetry condition, given by (7), becomes
(26)
Q
(
n
+
2
,
ω
)
+
Q
(
n
+
1
,
u
n
+
1
)
1
u
n
+
1
2
-
Q
(
n
,
u
n
)
(
n
n
+
1
)
=
0
.
Firstly, we differentiate (26) with respect to u
n (keeping ω fixed) and we consider u
n
+
1 to be a function of n,u
n, and ω. By the implicit function theorem differentiating u
n
+
1 with respect to u
n yields
(27)
∂
u
n
+
1
∂
u
n
=
-
(
∂
ω
/
∂
u
n
)
(
∂
ω
/
∂
u
n
+
1
)
=
n
u
n
+
1
2
n
+
1
.
Secondly, we apply thedifferential operator, given by
(28)
L
=
∂
∂
u
n
+
∂
u
n
+
1
∂
u
n
∂
∂
u
n
+
1
,
to (26) to get
(29)
-
2
n
(
n
+
1
)
u
n
+
1
Q
(
n
+
1
,
u
n
+
1
)
+
n
n
+
1
Q
′
(
n
+
1
,
u
n
+
1
)
-
n
n
+
1
Q
′
(
n
,
u
n
)
=
0
.
To solve (29), we differentiate it with respect to u
n keeping u
n
+
1 fixed. As a result we obtain the ODE:
(30)
d
d
u
n
(
n
n
+
1
Q
′
(
n
,
u
n
)
)
=
0
whose solution is given by
(31)
Q
(
n
,
u
n
)
=
(
n
+
1
n
)
A
(
n
)
u
n
+
B
(
n
)
.
We suppose that B
(
n
)
=
0 for ease of computation. Next we substitute (31) into (29) and we simplify the resulting equation to obtain
(32)
[
-
n
(
n
+
2
)
(
n
+
1
)
2
]
A
(
n
+
1
)
=
A
(
n
)
.
Thus,
(33)
A
(
n
)
=
(
n
n
+
1
)
2
c
(
-
1
)
n
-
1
,
where c is a constant. Substituting (33) into (31) leads to
(34)
Q
(
n
,
u
n
)
=
(
n
+
1
n
)
(
n
n
+
1
)
2
c
(
-
1
)
n
-
1
u
n
=
2
c
(
-
1
)
n
-
1
u
n
.
Therefore, the symmetry generator is given by
(35)
X
=
2
c
(
-
1
)
n
-
1
u
n
∂
∂
u
n
.


## 3.1.2. First Integral

Suppose thatP
2
=
P
2
(
n
,
u
n
); then (13) can be rewritten to give
(36)
(
n
+
1
n
+
2
)
P
2
(
n
+
2
,
u
n
+
2
)
-
1
u
n
+
1
2
P
2
(
n
+
1
,
u
n
+
1
)
-
P
2
(
n
,
u
n
)
=
0
.
We apply the differential operator L, given by (28), to (36) to get(37)
n
n
+
1
2
u
n
+
1
P
2
(
n
+
1
,
u
n
+
1
)
-
n
n
+
1
P
2
′
(
n
+
1
,
u
n
+
1
)
-
P
2
′
(
n
,
u
n
)
=
0
.
Next we differentiate (37) with respect to u
n keeping u
n
+
1 constant to obtain (
d
/
d
u
n
)
(
P
2
′
(
n
,
u
n
)
)
=
0 whose solution is given by
(38)
P
2
(
n
,
u
n
)
=
B
(
n
)
u
n
+
c
=
B
(
n
)
u
n
if we take c
=
0. We substitute (38) into (37) to obtain the difference equation
(39)
B
(
n
+
1
)
=
n
+
1
n
B
(
n
)
.
We choose B
(
1
)
=
1 to get
(40)
B
(
n
)
=
n
.
The next step consists of substituting (40) into (38) to get
(41)
P
2
(
n
,
u
n
)
=
n
u
n
.
From (11) we get
(42)
P
1
(
n
,
u
n
,
u
n
+
1
)
=
S
P
2
∂
ω
∂
u
n
=
n
u
n
+
1
=
P
1
(
n
,
u
n
+
1
)
.
Since the integrability condition holds, we can calculate the first integral ϕ. From (41) and (42) we have
(43)
ϕ
=
∫
‍
(
P
1
d
u
n
+
P
2
d
u
n
+
1
)
+
G
(
n
)
=
n
u
n
u
n
+
1
+
G
(
n
)
.
To find G
(
n
) we substitute (43) into (9). We obtain
(44)
G
(
n
+
1
)
-
G
(
n
)
+
n
+
1
=
0
whose solution is given by
(45)
G
(
n
)
=
-
n
(
n
+
1
)
2
.
Finally we substitute (45) into (43) to obtain the first integral(46)
ϕ
=
n
u
n
u
n
+
1
-
n
(
n
+
1
)
2
.Note. The symmetry generator given by (35) acts on the first integral, ϕ, to produce the following equation:
(47)
X
ϕ
=
Q
(
n
,
u
n
)
∂
ϕ
∂
u
n
+
Q
(
n
+
1
,
u
n
+
1
)
∂
ϕ
∂
u
n
+
1
=
2
c
(
-
1
)
n
-
1
(
n
u
n
n
u
n
+
1
-
n
u
n
n
u
n
+
1
)
=
0
.
We say X and ϕ areassociated and this property has far reaching consequences on “further” reduction of the equation.


## 3.1.3. Symmetry Reduction

Recall that, in Section3.1.1, we calculated the symmetry generator, X, to be
(48)
X
=
2
c
(
-
1
)
n
-
1
u
n
∂
∂
u
n
given by (35). Suppose v
n
=
v
(
n
,
u
n
,
u
n
+
1
) is an invariant of X. Then
(49)
X
v
n
=
(
Q
(
n
,
u
n
)
∂
∂
u
n
+
S
Q
(
n
,
u
n
)
∂
∂
u
n
+
1
)
v
n
=
0
.
We can use the characteristics
(50)
d
u
n
2
c
(
-
1
)
n
-
1
u
n
=
d
u
n
+
1
2
c
(
-
1
)
n
u
n
+
1
=
d
v
n
0
to solve for v
n and construct the equation. The independent and dependent variables are given by
(51)
α
=
u
n
u
n
+
1
,
γ
=
v
n
,
respectively. Therefore by (51), the dependent variable, v
n, is given by
(52)
v
n
=
u
n
u
n
+
1
.
Applying the shift operator on v
n and solving the resulting equation we get
(53)
v
n
=
n
+
1
2
+
c
n
,
where c is a constant. Then by (52) and (53) and solving for u
n
+
1 we obtain(54)
u
n
+
1
=
n
+
1
2
u
n
+
c
n
u
n
.Note. Equation (25) has been reduced by one order into (54). Solving (54) for c gives
(55)
c
=
n
u
n
u
n
+
1
-
n
(
n
+
1
)
2
=
ϕ
.
The first integral ϕ, given by (46), and the reduction are the same. This is another indication of a relationship between ϕ and X. In fact, this is the association; that is, ϕ isinvariant under X.


## 3.2. Example 2

Consider the following linear difference equation [11]:
(56)
ω
=
u
n
+
2
=
2
u
n
+
1
-
u
n
.


### 3.2.1. Symmetry

Suppose thatQ
=
Q
(
n
,
u
n
); then the symmetry condition becomes
(57)
Q
(
n
+
2
,
ω
)
-
2
Q
(
n
+
1
,
u
n
+
1
)
+
Q
(
n
,
u
n
)
=
0
.
Similarly, we apply the operator L to (57) and we differentiate the resulting equation:
(58)
Q
′
(
n
,
u
n
)
-
Q
′
(
n
+
1
,
u
n
+
1
)
=
0
,
with respect to u
n to get Q
′′
(
n
,
u
n
)
=
0. Therefore,
(59)
Q
(
n
,
u
n
)
=
A
(
n
)
u
n
+
B
(
n
)
.
Next we solve for A
(
n
) by substituting (59) into (58). This gives
(60)
A
(
n
+
1
)
=
A
(
n
)
=
a
,
where a is a constant. Substituting A
(
n
)
=
a into (59) yields
(61)
Q
(
n
,
u
n
)
=
a
u
n
+
B
(
n
)
.
The substitution of (61) into (57) yields
(62)
B
(
n
+
2
)
-
2
B
(
n
+
1
)
+
B
(
n
)
=
0
.
Thus,
(63)
B
(
n
)
=
b
n
+
c
,
where b and c are arbitrary constants. Finally we substitute (63) into (61) and obtain the characteristic
(64)
Q
(
n
,
u
n
)
=
a
u
n
+
b
n
+
c
.
Therefore, the Lie symmetry generators are
(65)
X
1
=
u
n
∂
∂
u
n
,
X
2
=
n
∂
∂
u
n
,
X
3
=
∂
∂
u
n
.


### 3.2.2. First Integral

Suppose thatP
2
=
P
2
(
n
,
u
n
). Thefirst integral condition is given by
(66)
P
2
(
n
+
2
,
ω
)
-
2
P
2
(
n
+
1
,
u
n
+
1
)
+
P
2
(
n
,
u
n
)
=
0
.
The solution to (66) is given by
(67)
P
2
(
n
,
u
n
)
=
k
u
n
+
p
n
+
q
,
where k,p, and q are constants. Then by (11), we have
(68)
P
1
(
n
,
u
n
+
1
)
=
S
P
2
(
n
,
u
n
)
∂
ω
∂
u
n
=
-
k
u
n
+
1
-
p
n
-
p
-
q
.
Substituting (67) and (68) into (15) we obtain the first integral
(69)
ϕ
=
p
n
(
u
n
+
1
-
u
n
)
+
q
(
u
n
+
1
-
u
n
)
-
p
u
n
+
G
(
n
)
.
Then we have
(70)
S
ϕ
=
p
n
(
u
n
+
1
-
u
n
)
+
q
(
u
n
+
1
-
u
n
)
-
p
u
n
+
G
(
n
+
1
)
.
To satisfy (8), we equate (69) and (70). This gives
(71)
G
(
n
+
1
)
=
G
(
n
)
=
r
,
where r is a constant. We thus write ϕ as
(72)
ϕ
=
(
p
n
+
q
)
u
n
+
1
-
(
p
n
+
p
+
q
)
u
n
+
r
.
Next we check if ϕ is associated with the symmetry generators given in (65). (i)
Consider thatX
1
=
u
n
∂
/
∂
u
n. One can readily verify that
(73)
X
1
ϕ
=
-
u
n
(
p
n
+
q
+
p
)
+
u
n
+
1
(
p
n
+
q
)
.
Thus ϕ is associated with X
1; that is, X
ϕ
=
0, if the following equations are satisfied:
(74)
p
n
+
q
+
p
=
0
,
p
n
+
q
=
0
.
Solving the above equations simultaneously gives p
=
q
=
0. Hence, for ϕ to be associated with X
1,
(75)
ϕ
=
r
.
(ii)
Consider thatX
2
=
n
∂
/
∂
u
n. We have
(76)
X
2
ϕ
=
n
∂
ϕ
∂
u
n
+
(
n
+
1
)
∂
ϕ
∂
u
n
+
1
=
q
.
Hence ϕ is associated with X
2 if q
=
0, that is, if
(77)
ϕ
=
p
n
u
n
+
1
-
(
p
n
+
p
)
u
n
+
r
.
(iii)
Consider thatX
3
=
c
∂
/
∂
u
n. Then,
(78)
X
3
ϕ
=
c
∂
ϕ
∂
u
n
+
c
∂
ϕ
∂
u
n
+
1
=
-
c
p
.
Here ϕ is associated with X
3 if p
=
0. Therefore
(79)
ϕ
=
q
(
u
n
+
1
-
u
n
)
+
r
.


### 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(80)
[
X
1
,
X
2
]
=
-
X
2
,
(56) will be reduced using X
2 first. Suppose that v
n
=
v
(
n
,
u
n
,
u
n
+
1
) is the invariant of X
2. Then
(81)
X
2
v
n
=
[
n
∂
v
n
∂
u
n
+
(
n
+
1
)
∂
v
n
∂
u
n
+
1
]
=
0
.
Using the method of characteristic we get(82)
v
n
=
n
u
n
+
1
-
(
n
+
1
)
u
n
.
Applying the shift operator onv
n yields
(83)
S
(
v
n
)
=
v
n
+
1
=
v
n
;
that is,(84)
v
n
+
1
=
v
n
=
c
1
,
wherec
1 is a constant. Equating (82) and (84) and solving for u
n
+
1, we have
(85)
u
n
+
1
=
c
1
n
+
(
1
+
1
n
)
u
n
whose solution is given by(86)
u
n
=
n
c
2
+
c
1
(
n
-
1
)
,
wherec
2 is an arbitrary constant. Equation (86) is the general solution of (56).
Note that solving forc
1 in (85) yields
(87)
c
1
=
n
u
n
+
1
-
(
n
+
1
)
u
n
.
Therefore,ϕ (given by (77)) and the reduction are the same if p
=
1 and r
=
0. That is, ϕ
=
c
1. If this condition holds then ϕ is invariant under X
2.
(ii)
We can also find a general solution of (56) by using a different symmetry generator. Here,
(88)
[
X
1
,
X
3
]
=
-
X
1
,
so that (56) will be reduced using X
1 first. Again suppose that v
n
=
v
(
n
,
u
n
,
u
n
+
1
) is invariant of X
1. Then
(89)
X
1
v
n
=
[
u
n
∂
v
n
∂
u
n
+
u
n
+
1
∂
v
n
∂
u
n
+
1
]
=
0
.
Using the method of characteristics we get(90)
v
n
=
u
n
+
1
u
n
.
Therefore applying the shift operator onv
n gives
(91)
v
n
+
1
=
2
-
1
v
n
whose solution is given by(92)
v
n
=
1
+
2
c
1
+
n
c
1
1
+
c
1
+
n
c
1
,
wherec
1 is a constant. Equating (90) and (92) results in
(93)
u
n
+
1
=
(
1
+
2
c
1
+
n
c
1
1
+
c
1
+
n
c
1
)
u
n
.
Therefore the general solution of (56) is given by
(94)
u
n
=
(
1
+
c
1
+
n
c
1
)
c
2
1
+
c
1
,
wherec
2 is a constant.
(iii)
Finally we consider the commutator ofX
2 and X
3. We have
(95)
[
X
2
,
X
3
]
=
0
.
Since the commutator is0, we can first reduce the OΔE with either X
2 or X
3. However, since we have already reduced (56) with X
2, we will use X
3. As before, suppose that v
n
=
v
(
n
,
u
n
,
u
n
+
1
) is invariant of X
3. Then
(96)
X
3
v
n
=
[
c
∂
v
n
∂
u
n
+
c
∂
v
n
∂
u
n
+
1
]
=
0
.
Applying the method of characteristics, we have(97)
v
n
=
u
n
+
1
-
u
n
.
Applying the shift factor,S, on (97) and solving the resulting equation we get
(98)
S
(
v
n
)
=
v
n
+
1
=
v
n
=
c
1
.
Equating (97) and (98) gives
(99)
u
n
+
1
=
u
n
+
c
1
.
We solve (99) and find
(100)
u
n
=
n
c
1
+
c
2
which is a general solution of (56). It has to be noted that (99) is the same as ϕ (given by (75)) if q
=
1 and r
=
0. If this is true then ϕ is invariant under X
3.


## 3.2.1. Symmetry

Suppose thatQ
=
Q
(
n
,
u
n
); then the symmetry condition becomes
(57)
Q
(
n
+
2
,
ω
)
-
2
Q
(
n
+
1
,
u
n
+
1
)
+
Q
(
n
,
u
n
)
=
0
.
Similarly, we apply the operator L to (57) and we differentiate the resulting equation:
(58)
Q
′
(
n
,
u
n
)
-
Q
′
(
n
+
1
,
u
n
+
1
)
=
0
,
with respect to u
n to get Q
′′
(
n
,
u
n
)
=
0. Therefore,
(59)
Q
(
n
,
u
n
)
=
A
(
n
)
u
n
+
B
(
n
)
.
Next we solve for A
(
n
) by substituting (59) into (58). This gives
(60)
A
(
n
+
1
)
=
A
(
n
)
=
a
,
where a is a constant. Substituting A
(
n
)
=
a into (59) yields
(61)
Q
(
n
,
u
n
)
=
a
u
n
+
B
(
n
)
.
The substitution of (61) into (57) yields
(62)
B
(
n
+
2
)
-
2
B
(
n
+
1
)
+
B
(
n
)
=
0
.
Thus,
(63)
B
(
n
)
=
b
n
+
c
,
where b and c are arbitrary constants. Finally we substitute (63) into (61) and obtain the characteristic
(64)
Q
(
n
,
u
n
)
=
a
u
n
+
b
n
+
c
.
Therefore, the Lie symmetry generators are
(65)
X
1
=
u
n
∂
∂
u
n
,
X
2
=
n
∂
∂
u
n
,
X
3
=
∂
∂
u
n
.


## 3.2.2. First Integral

Suppose thatP
2
=
P
2
(
n
,
u
n
). Thefirst integral condition is given by
(66)
P
2
(
n
+
2
,
ω
)
-
2
P
2
(
n
+
1
,
u
n
+
1
)
+
P
2
(
n
,
u
n
)
=
0
.
The solution to (66) is given by
(67)
P
2
(
n
,
u
n
)
=
k
u
n
+
p
n
+
q
,
where k,p, and q are constants. Then by (11), we have
(68)
P
1
(
n
,
u
n
+
1
)
=
S
P
2
(
n
,
u
n
)
∂
ω
∂
u
n
=
-
k
u
n
+
1
-
p
n
-
p
-
q
.
Substituting (67) and (68) into (15) we obtain the first integral
(69)
ϕ
=
p
n
(
u
n
+
1
-
u
n
)
+
q
(
u
n
+
1
-
u
n
)
-
p
u
n
+
G
(
n
)
.
Then we have
(70)
S
ϕ
=
p
n
(
u
n
+
1
-
u
n
)
+
q
(
u
n
+
1
-
u
n
)
-
p
u
n
+
G
(
n
+
1
)
.
To satisfy (8), we equate (69) and (70). This gives
(71)
G
(
n
+
1
)
=
G
(
n
)
=
r
,
where r is a constant. We thus write ϕ as
(72)
ϕ
=
(
p
n
+
q
)
u
n
+
1
-
(
p
n
+
p
+
q
)
u
n
+
r
.
Next we check if ϕ is associated with the symmetry generators given in (65). (i)
Consider thatX
1
=
u
n
∂
/
∂
u
n. One can readily verify that
(73)
X
1
ϕ
=
-
u
n
(
p
n
+
q
+
p
)
+
u
n
+
1
(
p
n
+
q
)
.
Thus ϕ is associated with X
1; that is, X
ϕ
=
0, if the following equations are satisfied:
(74)
p
n
+
q
+
p
=
0
,
p
n
+
q
=
0
.
Solving the above equations simultaneously gives p
=
q
=
0. Hence, for ϕ to be associated with X
1,
(75)
ϕ
=
r
.
(ii)
Consider thatX
2
=
n
∂
/
∂
u
n. We have
(76)
X
2
ϕ
=
n
∂
ϕ
∂
u
n
+
(
n
+
1
)
∂
ϕ
∂
u
n
+
1
=
q
.
Hence ϕ is associated with X
2 if q
=
0, that is, if
(77)
ϕ
=
p
n
u
n
+
1
-
(
p
n
+
p
)
u
n
+
r
.
(iii)
Consider thatX
3
=
c
∂
/
∂
u
n. Then,
(78)
X
3
ϕ
=
c
∂
ϕ
∂
u
n
+
c
∂
ϕ
∂
u
n
+
1
=
-
c
p
.
Here ϕ is associated with X
3 if p
=
0. Therefore
(79)
ϕ
=
q
(
u
n
+
1
-
u
n
)
+
r
.


## 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(80)
[
X
1
,
X
2
]
=
-
X
2
,
(56) will be reduced using X
2 first. Suppose that v
n
=
v
(
n
,
u
n
,
u
n
+
1
) is the invariant of X
2. Then
(81)
X
2
v
n
=
[
n
∂
v
n
∂
u
n
+
(
n
+
1
)
∂
v
n
∂
u
n
+
1
]
=
0
.
Using the method of characteristic we get(82)
v
n
=
n
u
n
+
1
-
(
n
+
1
)
u
n
.
Applying the shift operator onv
n yields
(83)
S
(
v
n
)
=
v
n
+
1
=
v
n
;
that is,(84)
v
n
+
1
=
v
n
=
c
1
,
wherec
1 is a constant. Equating (82) and (84) and solving for u
n
+
1, we have
(85)
u
n
+
1
=
c
1
n
+
(
1
+
1
n
)
u
n
whose solution is given by(86)
u
n
=
n
c
2
+
c
1
(
n
-
1
)
,
wherec
2 is an arbitrary constant. Equation (86) is the general solution of (56).
Note that solving forc
1 in (85) yields
(87)
c
1
=
n
u
n
+
1
-
(
n
+
1
)
u
n
.
Therefore,ϕ (given by (77)) and the reduction are the same if p
=
1 and r
=
0. That is, ϕ
=
c
1. If this condition holds then ϕ is invariant under X
2.
(ii)
We can also find a general solution of (56) by using a different symmetry generator. Here,
(88)
[
X
1
,
X
3
]
=
-
X
1
,
so that (56) will be reduced using X
1 first. Again suppose that v
n
=
v
(
n
,
u
n
,
u
n
+
1
) is invariant of X
1. Then
(89)
X
1
v
n
=
[
u
n
∂
v
n
∂
u
n
+
u
n
+
1
∂
v
n
∂
u
n
+
1
]
=
0
.
Using the method of characteristics we get(90)
v
n
=
u
n
+
1
u
n
.
Therefore applying the shift operator onv
n gives
(91)
v
n
+
1
=
2
-
1
v
n
whose solution is given by(92)
v
n
=
1
+
2
c
1
+
n
c
1
1
+
c
1
+
n
c
1
,
wherec
1 is a constant. Equating (90) and (92) results in
(93)
u
n
+
1
=
(
1
+
2
c
1
+
n
c
1
1
+
c
1
+
n
c
1
)
u
n
.
Therefore the general solution of (56) is given by
(94)
u
n
=
(
1
+
c
1
+
n
c
1
)
c
2
1
+
c
1
,
wherec
2 is a constant.
(iii)
Finally we consider the commutator ofX
2 and X
3. We have
(95)
[
X
2
,
X
3
]
=
0
.
Since the commutator is0, we can first reduce the OΔE with either X
2 or X
3. However, since we have already reduced (56) with X
2, we will use X
3. As before, suppose that v
n
=
v
(
n
,
u
n
,
u
n
+
1
) is invariant of X
3. Then
(96)
X
3
v
n
=
[
c
∂
v
n
∂
u
n
+
c
∂
v
n
∂
u
n
+
1
]
=
0
.
Applying the method of characteristics, we have(97)
v
n
=
u
n
+
1
-
u
n
.
Applying the shift factor,S, on (97) and solving the resulting equation we get
(98)
S
(
v
n
)
=
v
n
+
1
=
v
n
=
c
1
.
Equating (97) and (98) gives
(99)
u
n
+
1
=
u
n
+
c
1
.
We solve (99) and find
(100)
u
n
=
n
c
1
+
c
2
which is a general solution of (56). It has to be noted that (99) is the same as ϕ (given by (75)) if q
=
1 and r
=
0. If this is true then ϕ is invariant under X
3.


## 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,X, and first integral, ϕ, are associated via the invariance condition X
ϕ
=
0. When this condition is satisfied, we may proceed to double reduction of the equation.



---

*Source: 490165-2014-07-23.xml*