The ARS Algorithm
and

Invariance Analysis of Ordinary Difference
Equations

JOLLET TRUTH KUBAYI (710005@students.wits.ac.za)

School of Mathematics

Supervisors: Prof S. Jamal and Dr M. Folly-Gbetoula

A dissertation submitted to the faculty of science, University of the
Witwatersrand, in fulfilment of the requirements for the

degree of Master of Science.

April 2021



Declaration

I declare that the contents of this dissertation are original, except where due references
have been made. It is submitted for the degree of Master of Science at the University
of the Witwatersrand, Johannesburg. It was not submitted before for any degree or
examination to any other institution.

JT Kubayi

Signed at Johannesburg on the 1st day of April 2021.

1



Publications

Details of contributions to publications that form part of the research presented in this
dissertation are:

A study on a solvable ordinary difference equations, Journal of Computational Analysis
and Applications, submitted, 2021.

An investigation of the symmetry and singularity properties of some third-order bound-
ary flow problems, has been accepted to appear in the Journal of Mathematical Analysis
and Applications, 2021.

2



Abstract

This dissertation will be divided into two parts. The first part will involve the use of
symmetries to find exact solutions of higher order difference equations. The second
part will deal with important ordinary differential equations, in the search for singu-
larities and integrability testing. We will analyse a Painlevé equation, Ivey’s equation,
the higher-order Lane-Emden type equation and a class of third-order boundary flow
equations.



Acknowledgements

I would like to sincerely thank Professor S Jamal and Dr M Folly-Gbetoula for their
guidance and support throughout the compilation of this work. I would like to thank
the School of Mathematics, University of the Witwatersrand for the opportunity and
support throughout the programme. To Mr Kurt Finhaber and Dr Cindy Firnhaber,
thank you for ensuring I realise my goals, your financial support ensured that my studies
went smoothly. To my family, let us keep aiding each other towards greatness. “Dreams
are a goodness to be achieved”.

1



Contents

1 Introduction 3
1.1 Outline of Chapters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2 Mathematical preliminaries 6
2.1 Lie Point Symmetries . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
2.2 Definitions and Notations for Ordinary Difference Equations . . . . . . 7
2.3 The ARS Algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

3 Invariance Analysis of Ordinary Difference Equations 11
3.1 Symmetry Analysis of Tenth Order O∆Es . . . . . . . . . . . . . . . . 11
3.2 Reduction and Exact solutions . . . . . . . . . . . . . . . . . . . . . . . 14
3.3 The case where an and bn are 6-periodic. . . . . . . . . . . . . . . . . . 23
3.4 The case an and bn are constant coefficients . . . . . . . . . . . . . . . 24

4 Applications of the ARS algorithm - Singularity Analysis 28
4.1 The Painlevé-Ince Equation . . . . . . . . . . . . . . . . . . . . . . . . 28
4.2 Ivey’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
4.3 The Higher-Order Lane-Emden Type Equation . . . . . . . . . . . . . . 33
4.4 Third-Order Boundary Flow Equations . . . . . . . . . . . . . . . . . . 36

5 Conclusion 39
Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

2



Chapter 1

Introduction

During the nineteenth century, the prominent Norwegian mathematician Sophus Lie
(1842 - 1899) established remarkable work that became an important core of the the-
ory of groups of transformations (continuous) that leave differential equations invariant
[1]. Lie wanted to create a theory of integrating ordinary differential equations that
is equivalent to the Abelian theory of computing algebraic equations. He was inspired
by Abel and Galois’ theory, that stated that the procedure in all exceptional cases
of a universal integration on differential equations, is centred on the invariance of the
differential equation under continuous symmetries. Lie’s group analysis classifies ordi-
nary differential equations in terms of the symmetry group associated with them. He
described the set integrable by group-theoretical methods. The concept of invariance
of differential equations under infinitesimal transformations involves various techniques
which makes it possible to best construct solutions of differential equations, that is, the
development permits one to obtain solutions to differential equations systematically by
performing symmetry analysis. In 1918, the German Mathematician Emmy Noether
proved her theorem that uncovered fundamental justification for conservation laws, that
is, conservation laws follow from the symmetry properties of nature.

Back in 1987, Shigeru Maeda showed that Lie’s method can be extended to also solve
ordinary difference equations(O∆E). He showed that the set of functional equations
developed from the linearized symmetry condition of the O∆Es [6]. The philosophy of
difference equations and their applications have cemented a central core in applicable
analysis.
Maeda [6] showed how to use symmetry methods to obtain the solution of the system
of first-order difference equations. In [7], Güven cinar, Ali Gelis and Ozan Özkan
investigated the solution and the properties of the difference equation

xn =
xn−3kxn−4kxn−5k

xn−kxn−2k (±1± xn−3kxn−4kxn−5k)
. (1)

We are going to be using a symmetry based method to solve a generalisation of (1) and
compare the solutions of the corresponding cases to that of [7].

3



In the second part of this dissertation, we investigate the integrability of ordinary
differential equation by an integrability algorithm.

The precise meaning of the solution of a system of differential equations can be
presented in several ways. One way, is the idea of integrability of a differential equation
in terms of a singularity analysis, and this is the focus of our approach. Singularity
analysis can be traced back to Painlevé [8] after its success in application by Sophie
Kowalevskaya [9]. We consider the Painlevé test for integrability. M.J. Ablowitz, A.
Ramani, H. Segur [10, 11, 12] developed an algorithm, called the ARS (Ablowitz-
Ramani-Segur) algorithm that tests whether the solution of an ordinary differential
equation can be expressed in terms of a Laurent expansion. If so, then the ordinary
differential equation is said to pass the Painlevé test and is concluded without proof
to be integrable. A detailed description of the algorithm can be found in the work by
Conte [13]. In essence, one looks at the existence of a Laurent series for each dependent
variable of the equation.

The Laurent series represents an analytic function in a punctured disc with radius
centred on the singular point that is a pole or a specific type of a movable point, that
is the singular point of a function for which it is possible to assign a complex number
in such a way that a function becomes analytic. Once the movable singularity is found,
the arbitrary constants of integration x0 must be determined and the consistency of the
proposed Laurent expansion must be checked.

1.1 Outline of Chapters

In Chapter 2, we will firstly look at definitions and some notations that we will be using
in this dissertation.
In Chapter 3, we will perform a full Lie analysis of a tenth order ordinary difference
equation. We will study the work in [7], where Güven çinar, Ali Geliş and Ozan Özkan
investigated the solution and the properties of the difference equation

xn =
xn−3kxn−4kxn−5k

xn−kxn−2k (±1± xn−3kxn−4kxn−5k)
. (2)

In this chapter, we will derive the solutions of the following difference equation, via the
invariant of their group of transformations:

xn =
xn−6xn−8xn−10

xn−2xn−4 (an + bnxn−6xn−8xn−10)
, (3)

where (an)N0 and (bn)N0 are non-zero real sequences, using Lie group analysis technique.
Our work is inspired by the results in [7] and for the sake of definitions and notations
used in this paper, we shall consider:

un+10 =
unun+2un+4

un+6un+8 (An +Bnunun+2un+4)
, (4)

4



where An and Bn are real arbitrary sequences. We are going to be using a symmetry
based method to solve the general case and compare the solutions of the corresponding
cases to that of [7].

In chapter 4, we investigate the singularity analysis of several important ordinary differ-
ential equations, such as the Painlevé-Ince equation, Ivey’s equation, the higher-order
Lane-Emden type equation and a family third-order boundary flow equations. To this
end, we follow Kowalevskaya, and seek to determine whether or not the above ordinary
differential equations possess movable singularities.

5



Chapter 2

Mathematical preliminaries

We introduce the relevant theory surrounding Lie point symmetries of differential equa-
tions, ordinary difference equations and the ARS algorithm for ordinary differential
equations.

2.1 Lie Point Symmetries

The procedure for determining point symmetries for an arbitrary system of equations
is as follows [14]. Consider q unknown variables ua which depend on p independent
variables xi, i.e. u = (u1, . . . , uq), x = (x1, . . . , xp), with indices α = 1, . . . , q and
i = 1, . . . , p. Let

Gα

(
x, u(k)

)
= 0, (5)

be a system of nonlinear differential equations, where u(k) represents the kth derivative
of u with respect to x.

Definition 1 A one-parameter Lie group of transformations (ε as the group parameter)
that is invariant under (5) is given by

x̄ = Ξ(x, u; ε) ū = Φ(x, u; ε). (6)

Invariance of (5) under the transformation (6) implies that any solution u = Θ(x) of
(5) maps into another solution v = Ψ(x; ε) of (5). Expanding (6) around the identity
ε = 0, generates the following infinitesimal transformations:

x̄i = xi + εξi(x, u) +O(ε2),

ūα = uα + εηα(x, u) +O(ε2).
(7)

The action of the Lie group can be recovered from that of its infinitesimal generators
acting on the space of independent and dependent variables. Hence, we consider the
following vector field

X = ξi∂xi + ηα∂uα . (8)

6



Definition 2 The infinitesimal criterion for invariance is given by

X
[
Gα

(
x, u(k)

)]
= 0, when Gα

(
x, u(k)

)
= 0, (9)

where X is extended to all derivatives appearing in the equation through an appropriate
prolongation.

2.2 Definitions and Notations for Ordinary Differ-

ence Equations

The definitions and notations in this paper are similar to those adopted by Hydon in
[2]. We consider the general form of the ordinary difference equation of order k given
by

un+k = ω (n, un, un+1, un+2, ...., un+k−1) , (10)

for some function ω with k ∈ N .

Definition 3 We define S to be the shift operator acting on n as follows

S : n→ n+ 1. (11)

That is, if un = F (n, c1, ..., cN) then,

S(un) = S(F (n, c1, ..., cN)) (12)

= F (S(n), c1, ..., cN) (13)

= F (n+ 1, c1, ..., cN) (14)

= un+1, (15)

where the ci’s are independent of n, and also S(un+k) = un+k+1, for k = 0, 1, .... There-
fore S is an operator on n and hence on un+k.

Given an equation (or system) whose continuous variables are x = (x1, ...., xN), a point
transformation is a locally defined diffeomorphism

Ψ : x→ x̂(x). (16)

The term ‘point’, is used because x̂ depends only on the point x.

7



Theorem 1 A parametrized set of point transformations,

Ψε : x→ x̂(x; ε), ε ∈ (ε0, ε1),

where ε0 ≤ 0 and ε1 ≥ 0, is a one-parameter local Lie group if the following conditions
are satisfied:

• Ψ0 is the identity map, so that x̂ = x when ε = 0.

• ΨδΨε = Ψδ+ε for every δ,ε sufficiently close to zero.

• Each x̂α can be represented as a Taylor series in ε (in the neighbourhood of ε = 0
that is determined by x), and therefore

x̂α(x, ε) = xα + εξα(x) +O(ε2), α = 1, ..., N.

Now consider the point transformations

Ψε : (n, un)→ (n, un + εξ (n, un)) , (17)

for some continuous characteristic function ξ.

Definition 4 A symmetry generator is denoted by X and is given by

X = ξ
∂

∂un
+ Sξ

∂

∂un+1

+ · · ·+ Sk−1ξ
∂

∂un+k−1
. (18)

To solve for the characteristic, we will need the linearized symmetry condition from [2]

Skξ −Xω = 0, (19)

provided (10) holds.
Given a symmetry generator for the kth order O∆E (10), we have a (k − 1)th order
invariant

vn = v (n, un, . . . , un+k−1) ,

which satisfies

Xvn = 0.

Assume that the characteristic ξ is known, then we can solve for the invariant vn using
the characteristics equation

dun
ξ

=
dun+1

Sξ
= · · · = dun+k−1

Sn+k−1ξ

(
=
dvn
0

)
. (20)

The procedure of finding symmetries is best explained at length in [2], especially for
second order difference equations.

8



Definition 5 For a given characteristic ξ(n, un) 6= 0, we define a canonical coordinate,
sn(locally) by

sn =

∫
du

ξi(n, un)
.

Definition 6 The product operator multiplies the terms of a partial sequence of a se-
quence. It is defined as follows

n∏
k=1

ak = (a1)(a2)(a3)...(an−1)(an). (21)

2.3 The ARS Algorithm

Consider the equation

y(n) = E

(
x; y′; y′′, ..., yn−1

)
. (22)

If there exists a movable singularity, which is a point where the solution of an ordinary
differential equation behave in an unfavorable manner and is movable, that is to say
that its location depends on the initial conditions of the differential equation, then the
solution of (22) will be described by the power- law function y(x) ' (x−x0)p, where p is
a negative number and x0 indicates the singularity’s position. It is the initial conditions,
that provide us with different positions for the singular point. The algorithm can be
described by the following three steps.

Firstly, we substitute y(x) = a(x − x0)
p into (22), to determine its leading-order

behaviour where x0 is the location of the generally considered movable singularity and
look for two or more dominant terms. One way to determine the dominant terms is to
look for balance after the above substitution by considering the powers in the equation.
Another way is to check that Eq. (22) is invariant under the Lie symmetry qx∂x + y∂y,
and the exponent in the leading order term is given by −1

q
in [10] and the constant a

is determined by the equation. If (22) is invariant separately under x∂x and y∂y (it
has two homogeneity symmetries) the value of p can be determined by the equation.
The original ARS algorithm requires that p be a negative integer, or else, the algorithm
terminates. After obtaining a, the leading-order term of the solution, we look at the
behaviour of the “next-to-leading-order” term. In the ARS algorithm we require that
the Laurent series be an increasing series called the right Painlevé series which is of the
form

y(x) =
∞∑
k=0

ak(x− x0)k+p, (23)

9



where the singularity at x0 is a pole of order p. A decreasing series called the left
Painlevé series is of the form (not dealt with in our dissertation)

y(x) =
∞∑
k=0

ak(x− x0)−k+p. (24)

Next, one must locate the powers at which the arbitrary constants needed to make the
solution a general solution, can be introduced. An expression is introduced,

y(x) = a(x− x0)p +m(x− x0)p+s,

where s is called the resonance, and where the coefficient of terms linear in a variable m
is zero. The solution of s = −1 always occurs, since s = −1 is generic in nature and it
is also related to the arbitrariness of the location of the movable singularity. If the rest
of the values of the resonance s are not integral, then the ARS algorithm terminates.
Lastly we substitute a truncated Laurent series into the original equation, to check
that there are no inconsistencies, which involves checking for the correct number of
arbitrary constants which depends on the degree of th equation in discussion. For a
second order equation with two symmetries γ1 = x∂x and γ2 = −qx∂x + y∂y there
can be no inconsistency. We take note of some limitations of the ARS algorithm as
explained in more detail in [15]:

• the exponents of the leading-order term needs to be a negative integer or a non-
integral rational number,

• the resonances have to be rational and real numbers, because of the movable
singularity,

• excluding the resonance s = −1, for a right Painlevé series the resonances must be
nonnegative, while for a left Painlevé series, the resonances must be nonpositive.

• for a full Laurent expansion the resonances have to be mixed, that is, we can have
mixed signs in the nongeneric resonances [16].

10



Chapter 3

Invariance Analysis of Ordinary
Difference Equations

Consider tenth-order difference equations with the form of (4), i.e.,

un+10 =
unun+2nn+4

un+6un+8 (An +Bnunun+2un+4)
= ω(un, un+2, ..., un+8). (25)

where An and Bn are real sequences (arbitrary).

3.1 Symmetry Analysis of Tenth Order O∆Es

By imposing the linearized symmetry condition (19) to (4) with k = 2 and replacing n
by n+ 10 from (3) to obtain (4), we get

ξ(n+ 10, ω)− ξ(n, un)Anun+2un+4

un+6un+8(An +Bnunun+2un+4)2

− ξ(n+ 2, un+2)Anunun+4

un+6un+8(An +Bnunun+2un+4)2
− ξ(n+ 4, un+4)Anun+2

un+6un+8(An +Bnunun+2un+4)2

+
ξ(n+ 6, un+6)unun+2un+4

u2n+6un+8(An +Bnunun+2un+4)
+

ξ(n+ 8, un+8)unun+2un+4

un+6u2n+8(An +Bnunun+2un+4)
= 0.

(26)

We differentiate the functional equation (26) implicitly with respect to un in order to
solve for ξ, while keeping un+6 fixed, it is important to note that we can choose freely
between un+2, un+4, un+6 and un+8. That is, we apply the Linear differential operator,
noting first

∂un+6

∂un
=

∂ω
∂un
∂ω

∂un+6

=
Anun+6

un(An +Bnunun+2un+4)
,

11



then

L =
∂

∂un
+
∂un+6

∂un

∂

∂un+6

=
∂

∂un
+

Anun+6

un(An +Bnunun+2un+4)

∂

∂un+6

on (26). After clearing fractions and then differentiating thrice with respect to un,
keeping un+6 fixed, we obtain the following

−2(An + 2Bnunun+2un+4)un+6ξ
(3)(n, un)

−(An +Bnunun+2un+4)unun+6ξ
(4)(n, un) = 0.

(27)

Now we separate the above, since ξ depends only on un, which is a continuous
variable, then we will have

un+2un+4un+6 :unξ
(4)(n, un) + 4ξ(3)(n, un) = 0 (28)

un+6 :unξ
(4)(n, un) + 2ξ(3)(n, un) = 0 (29)

implying that
ξ(n, un) = βnu2n + γnun + λn (30)

for some functions βn, γn and λn of n.

Now we substitute (30) into (26) and then separate the resulting equation by the coef-
ficients of products of shifts of un and then setting them to zero. We then obtain the
following results

γn + γn+2 + γn+4 = 0 (31)

βn = 0 (32)

λn = 0. (33)

Then the four independent solutions of the fourth order difference equation (31) are
given by

γ1, γ̄1, γ2, γ̄2,

12



where
γ1 = e

inπ
3

and γ̄1 represents its complex conjugate, also

γ2 = e
2inπ
3

and γ̄2 represents its complex conjugate. Then from (30) we have the characteristics
given by

ξ1 = γn1 un,

ξ2 = γ̄1
nun,

ξ3 = γn2 un

and

ξ4 = γ̄2
nun.

Hence we obtain the following prolongation of the spanning vectors of the Lie algebra
of (25) given below

X1 =γn1 un
∂

∂un
+ γn+1

1 un+1
∂

∂un+1

+ γn+2
1 un+2

∂

∂un+2

+ γn+3
1 un+3

∂

∂un+3

+ γn+4
1 un+4

∂

∂un+4

+ γn+5
1 un+5

∂

∂un+5

+ γn+6
1 un+6

∂

∂un+6

+ γn+7
1 un+7

∂

∂un+7

+ γn+8
1 un+8

∂

∂un+8

+ γn+9
1 un+9

∂

∂un+9

.

X2 =γ̄1
nun

∂

∂un
+ γ̄1

n+1un+1
∂

∂un+1

+ γ̄1
n+2un+2

∂

∂un+2

+ γ̄1
n+3un+3

∂

∂un+3

+ γ̄1
n+4un+4

∂

∂un+4

+ γ̄1
n+5un+5

∂

∂un+5

+ γ̄1
n+6un+6

∂

∂un+6

+ γ̄1
n+7un+7

∂

∂un+7

+ γ̄1
n+8un+8

∂

∂un+8

+ γ̄1
n+9un+9

∂

∂un+9

.

X3 =γn2 un
∂

∂un
+ γ2

n+1un+1
∂

∂un+1

+ γ2
n+2un+2

∂

∂un+2

+ γ2
n+3un+3

∂

∂un+3

+ γ2
n+4un+4

∂

∂un+4

+ γ2
n+5un+5

∂

∂un+5

+ γ2
n+6un+6

∂

∂un+6

+ γ2
n+7un+7

∂

∂un+7

+ γ2
n+8un+8

∂

∂un+8

+ γ2
n+9un+9

∂

∂un+9

.

13



X4 =γ̄2
nun

∂

∂un
+ γ̄2

n+1un+1
∂

∂un+1

+ γ̄2
n+2un+2

∂

∂un+2

+ γ̄2
n+3un+3

∂

∂un+3

+ γ̄2
n+4un+4

∂

∂un+4

+ γ̄2
n+5un+5

∂

∂un+5

+ γ̄2
n+6un+6

∂

∂un+6

+ γ̄2
n+7un+7

∂

∂un+7

+ γ̄2
n+8un+8

∂

∂un+8

+ γ̄2
n+9un+9

∂

∂un+9

.

3.2 Reduction and Exact solutions

Given that ξ(n, un) = βnu2n + γnun + λn is a characteristic, with βn = 0 and λn = 0,
then the following condition must be satisfied:

dun
γnun

=
dun+1

γn+1un+1

= ..... =
dun+9

γn+9un+9

(
=
dvn
0

)
. (34)

That is, given ξ1, the canonical coordinate 5 that linearizes (25), meaning we can find
a linear approximation at this coordinate, is given by

Sn =

∫
dun

ξ1(n, un)
=

1

γn
ln|un| (35)

and from the final constraint (31), we then define the invariant function

Ṽn = γnSn + γn+2Sn+2 + γn+4Sn+4

= ln|un|+ ln|un+2|+ ln|un+4|
= ln|unun+2un+4|.

(36)

Thus from the fact that

X1Ṽn = X2Ṽn = X3Ṽn = X4Ṽn = 0.

We must make use of the function Vn defined as

|Vn| = exp(−Ṽn) (37)

that is

Vn = ± 1

unun+2un+4

, (38)

but we will only consider

Vn =
1

unun+2un+4

, (39)

14



for the relevancy of our work. Then we must show that Vn is an invariant, it suffices to
show that X1Ṽn = 0, it follows from

X1Ṽn = γnun
∂

∂un
Ṽn + γn+1un+1

∂

∂un+1

Ṽn

+γn+2un+2
∂

∂un+2

Ṽn + γn+3un+3
∂

∂un+3

Ṽn

+γn+4un+4
∂

∂un+4

Ṽn + γn+5un+5
∂

∂un+5

Ṽn

+γn+6un+6
∂

∂un+6

Ṽn + γn+7un+7
∂

∂un+7

Ṽn

+γn+8un+8
∂

∂un+8

Ṽn + γn+9un+9
∂

∂un+9

Ṽn.

(40)

Therefore we have

X1Ṽn = γnun(
1

un
) + γn+2un+2(

1

un+2

) + γn+4un+4(
1

un+4

), (41)

which gives
X1Ṽn = γn + γn+2 + γn+4. (42)

Then invoking the characteristic, that is we take note that (31) gives γn+γn+2+γn+4 =
0, then we will have that

X1Ṽn = 0.

Now taking (39)

Vn =
1

unun+2un+4

, (43)

and shift it once, yields

Vn+1 =
1

un+1un+3un+5

, (44)

and shifting Vn six times gives

Vn+6 =
1

un+6un+8un+10

. (45)

We know un+10 from (25), which we then substitute into (45) and obtain

Vn+6 =
1

un+6un+8
unun+2un+4

un+6un+8(An+Bnunun+2un+4)

, (46)

which gives

Vn+6 =
An

unun+2un+4

+Bn = AnVn +Bn. (47)

15



From (39)

Vn =
1

unun+2un+4

, (48)

on which we can perform a shift by two to get

Vn+2 =
1

un+2un+4un+6

, (49)

then making un+6 the subject gives

un+6 =
1

un+2un+4Vn+2

. (50)

Making un+2un+4 the subject in (25), gives

un+2un+4 =
un+10un+6un+8(An +Bnunun+2un+4)

un
. (51)

Now substituting (51) into (50) and then simplifying the results, we obtain

un+6 =
Vn
Vn+2

un. (52)

The solution of (47),

Vn+6 = AnVn +Bn, (53)

in closed form is given by

V6n+j = Vj

(
n−1∏
k1=0

A6k1+j

)
+

n−1∑
l=0

(
B6l+j

n−1∏
k2=l+1

A6k2+j

)
. (54)

Starting with

un+6 =
Vn
Vn+2

un,

16



by iterations that for

n = 0 : u6 =
V0
V2
u0.

n = 1 : u7 =
V1
V3
u1

n = 2 : u8 =
V2
V4
u2

n = 3 : u9 =
V3
V5
u3

n = 4 : u10 =
V4
V6
u4

n = 5 : u11 =
V5
V7
u5

n = 6 : u12 =
V6
V8
u6 =

V6V0
V8V2

u0

n = 7 : u13 =
V7
V9
u7 =

V7V1
V9V3

u1

...

From these iterations, we observe that

u6n+j = uj

(
n−1∏
k1=0

V6k1+j
V6k1+j+2

)
. (55)

To solve (53), we note that it depends on the parity of n, meaning it depends on whether
n is odd or even and we have that

V6n+j = A6n+jV6n+j +B6n+j, (56)

for j = 0, 1, 2, 3, 4, 5.
Having that

Vn =
1

unun+2un+4

17



and depending on the parity of n to solve

Vn+6 = AnVn +Bn,

then for j = 0, 1, 2, 3, 4, 5, we can have

V6n+j = A6n+jV6n+j +B6n+j. (57)

Thus the sequence (V6n+j)j=0,1,2,3,4,5 satisfies the tenth-order linear difference equation

ωn+1 = A6n+jωn +B6n+j (58)

for j = 0, 1, 2, 3, 4, 5. After a few iterations and applying mathematical induction, we
have the following solution in closed form as

ωn = V6n+j = Vj

(
n−1∏
k1=0

A6k1+j

)
+

n−1∑
l=0

(
B6l+j

n−1∏
k2=l+1

A6k2+j

)
(59)

for j = 0, 1, 2, 3, 4, 5. From (59) it can be shown that for j = 0,

u6n = u0

n−1∏
k=0

V6k
V6k+2

= u0

n−1∏
k=0

V0

(
k−1∏
k1=0

A6k1

)
+

k−1∑
l=0

(
B6l

k−1∏
k2=l+1

A6k2

)
V2

(
k−1∏
k1=0

A6k1+2

)
+

k−1∑
l=0

(
B6l+2

k−1∏
k2=l+1

A6k2+2

)

= u0

n−1∏
k=0

V0

[(
k−1∏
k1=0

A6k1

)
+ 1

V0

k−1∑
l=0

(
B6l

k−1∏
k2=l+1

A6k2

)]
V2

[(
k−1∏
k1=0

A6k1+2

)
+ 1

V2

k−1∑
l=0

(
B6l+2

k−1∏
k2=l+1

A6k2+2

)]

= u0

n−1∏
k=0

1
u0u2u4

[(
k−1∏
k1=0

A6k1

)
+ u0u2u4

k−1∑
l=0

(
B6l

k−1∏
k2=l+1

A6k2

)]
1

u2u4u6

[(
k−1∏
k1=0

A6k1+2

)
+ u2u4u6

k−1∑
l=0

(
B6l+2

k−1∏
k2=l+1

A6k2+2

)]

= u1−n0 un6

n−1∏
k=0

k−1∏
k1=0

A6k1 + u0u2u4
k−1∑
l=0

(
B6l

k−1∏
k2=l+1

A6k2

)
k−1∏
k1=0

A6k1+2 + u2u4u6
k−1∑
l=0

(
B6l+2

k−1∏
k2=l+1

A6k2+2

) .

18



For j = 1,

u6n+1 = u1

n−1∏
k=0

V6k+1

V6k+3

= u1

n−1∏
k=0

V1

(
k−1∏
k1=0

A6k1+1

)
+

k−1∑
l=0

(
B6l+1

k−1∏
k2=l+1

A6k2+1

)
V3

(
k−1∏
k1=0

A6k1+3

)
+

k−1∑
l=0

(
B6l+3

k−1∏
k2=l+1

A6k2+3

)

= u1

n−1∏
k=0

1
u1u3u5

[(
k−1∏
k1=0

A6k1+1

)
+ u1u3u5

k−1∑
l=0

(
B6l+1

k−1∏
k2=l+1

A6k2+1

)]
1

u3u5u7

[(
k−1∏
k1=0

A6k1+3

)
+ u3u5u7

k−1∑
l=0

(
B6l+3

k−1∏
k2=l+1

A6k2+3

)]

= u1−n1 un7

n−1∏
k=0

k−1∏
k1=0

A6k1+1 + u1u3u5
k−1∑
l=0

(
B6l+1

k−1∏
k2=l+1

A6k2+1

)
k−1∏
k1=0

A6k1+3 + u3u5u7
k−1∑
l=0

(
B6l+3

k−1∏
k2=l+1

A6k2+3

) .

(60)

19



For j = 2,

u6n+2 = u2

n−1∏
k=0

V6k+2

V6k+4

= u2

n−1∏
k=0

V2

(
k−1∏
k1=0

A6k1+2

)
+

k−1∑
l=0

(
B6l+2

k−1∏
k2=l+1

A6k2+12

)
V4

(
k−1∏
k1=0

A6k1+4

)
+

k−1∑
l=0

(
B6l+4

k−1∏
k2=l+1

A6k2+4

)

= u2

n−1∏
k=0

1
u2u4u6

[(
k−1∏
k1=0

A6k1+2

)
+ u2u4u6

k−1∑
l=0

(
B6l+2

k−1∏
k2=l+1

A6k2+2

)]
1

u4u6u8

[(
k−1∏
k1=0

A6k1+4

)
+ u4u6u8

k−1∑
l=0

(
B6l+4

k−1∏
k2=l+1

A6k2+4

)]

= u1−n2 un8

n−1∏
k=0

k−1∏
k1=0

A6k1+2 + u2u4u6
k−1∑
l=0

(
B6l+2

k−1∏
k2=l+1

A6k2+2

)
k−1∏
k1=0

A6k1+4 + u4u6u8
k−1∑
l=0

(
B6l+4

k−1∏
k2=l+1

A6k2+4

) .

(61)

For j = 3,

u6n+3 = u3

n−1∏
k=0

V6k+3

V6k+5

= u3

n−1∏
k=0

V3

(
k−1∏
k1=0

A6k1+3

)
+

k−1∑
l=0

(
B6l+3

k−1∏
k2=l+1

A6k2+3

)
V5

(
k−1∏
k1=0

A6k1+5

)
+

k−1∑
l=0

(
B6l+5

k−1∏
k2=l+1

A6k2+5

)

= u1−n3 un9

n−1∏
k=0

k−1∏
k1=0

A6k1+3 + u3u5u7
k−1∑
l=0

(
B6l+3

k−1∏
k2=l+1

A6k2+3

)
k−1∏
k1=0

A6k1+5 + u5u7u9
k−1∑
l=0

(
B6l+5

k−1∏
k2=l+1

A6k2+5

) .

(62)

20



For j = 4,

u6n+4 = u4

n−1∏
k=0

V6k+4

V6k+6

= u4

n−1∏
k=0

V4

(
k−1∏
k1=0

A6k1+4

)
+

k−1∑
l=0

(
B6l+4

k−1∏
k2=l+1

A6k2+4

)
V0

(
k∏

k1=0

A6k1

)
+

k∑
l=0

(
B6l

k∏
k2=l+1

A6k2

)

=
u4u

n
2u

n
0

un6u
n
8

n−1∏
k=0

k−1∏
k1=0

A6k1+4 + u4u6u8
k−1∑
l=0

(
B6l+4

k−1∏
k2=l+1

A6k2+4

)
k∏

k1=0

A6k1 + u0u2u4
k∑
l=0

(
B6l

k∏
k2=l+1

A6k2

) .

(63)

For j = 5,

u6n+5 = u5

n−1∏
k=0

V6k+5

V6k+7

= u5

n−1∏
k=0

V5

(
k−1∏
k1=0

A6k1+5

)
+

k−1∑
l=0

(
B6l+5

k−1∏
k2=l+1

A6k2+5

)
V1

(
k∏

k1=0

A6k1+1

)
+

k∑
l=0

(
B6l+1

k∏
k2=l+1

A6k2+1

)

=
u5u

n
3u

n
1

un7u
n
9

n−1∏
k=0

k−1∏
k1=0

A6k1+5 + u5u7u9
k−1∑
l=0

(
B6l+5

k−1∏
k2=l+1

A6k2+5

)
k∏

k1=0

A6k1+1 + u1u3u5
k∑
l=0

(
B6l+1

k∏
k2=l+1

A6k2 + 1

) .

(64)

Then our solution in terms of {xn}∞n=1 is given by following

x6n−9 = x1−n−9 x
n
−3

n−1∏
k=0

k−1∏
k1=0

a6k1 + x−9x−7x−5
k−1∑
l=0

(
b6l

k−1∏
k2=l+1

a6k2

)
k−1∏
k1=0

a6k1+2 + x−7x−5x−3
k−1∑
l=0

(
b6l+2

k−1∏
k2=l+1

a6k2+2

) (65a)

21



x6n−8 = x1−n−8 x
n
−2

n−1∏
k=0

k−1∏
k1=0

a6k1+1 + x−8x−6x−4
k−1∑
l=0

(
b6l+1

k−1∏
k2=l+1

a6k2+1

)
k−1∏
k1=0

a6k1+3 + x−6x−4x−2
k−1∑
l=0

(
b6l+3

k−1∏
k2=l+1

a6k2+3

) (65b)

x6n−7 = x1−n−7 x
n
−1

n−1∏
k=0

k−1∏
k1=0

a6k1+2 + x−7x−5x−3
k−1∑
l=0

(
b6l+2

k−1∏
k2=l+1

a6k2+2

)
k−1∏
k1=0

a6k1+4 + x−5x−3x−1
k−1∑
l=0

(
b6l+4

k−1∏
k2=l+1

a6k2+4

) (65c)

x6n−6 = x1−n−6 x
n
0

n−1∏
k=0

k−1∏
k1=0

a6k1+3 + x−6x−4x−2
k−1∑
l=0

(
b6l+3

k−1∏
k2=l+1

a6k2+3

)
k−1∏
k1=0

a6k1+5 + x−4x−2x0
k−1∑
l=0

(
b6l+5

k−1∏
k2=l+1

a6k2+5

) (65d)

x6n−5 =
x−5x

n
−7x

n
−9

xn−3x
n
−1

n−1∏
k=0

k−1∏
k1=0

a6k1+4 + x−5x−3x−1
k−1∑
l=0

(
b6l+4

k−1∏
k2=l+1

a6k2+4

)
k∏

k1=0

a6k1 + x−9x−7x−5
k∑
l=0

(
b6l

k∏
k2=l+1

a6k2

) (65e)

x6n−4 =
x−4x

n
−6x

n
−8

xn−2x
n
0

n−1∏
k=0

k−1∏
k1=0

a6k1+5 + x−4x−2x0
k−1∑
l=0

(
b6l+5

k−1∏
k2=l+1

a6k2+5

)
k∏

k1=0

a6k1+1 + x−8x−6x−4
k∑
l=0

(
b6l+1

k∏
k2=l+1

a6k2+1

) . (65f)

Assuming that the denominators are non zero, we can rearrange the equation (64d)
above as follows

x6n = x−n−6x
n+1
0

n∏
k=0

k−1∏
k1=0

a6k1+3 + x−6x−4x−2
k−1∑
l=0

(
b6l+3

k−1∏
k2=l+1

a6k2+3

)
k−1∏
k1=0

a6k1+5 + x−4x−2x0
k−1∑
l=0

(
b6l+5

k−1∏
k2=l+1

a6k2+5

) , (66)

then the equations in (65) can be rearranged as follows

x6(n+1) = x
−(n+1)
−6 xn+2

0

n+1∏
k=0

k−1∏
k1=0

a6k1+3 + x−6x−4x−2
k−1∑
l=0

(
b6l+3

k−1∏
k2=l+1

a6k2+3

)
k−1∏
k1=0

a6k1+5 + x−4x−2x0
k−1∑
l=0

(
b6l+5

k−1∏
k2=l+1

a6k2+5

) (67a)

22



x6(n+2)−7 = x
−(n+1)
−7 xn+2

−1

n+1∏
k=0

k−1∏
k1=0

a6k1+2 + x−7x−5x−3
k−1∑
l=0

(
b6l+2

k−1∏
k2=l+1

a6k2+2

)
k−1∏
k1=0

a6k1+4 + x−5x−3x−1
k−1∑
l=0

(
b6l+4

k−1∏
k2=l+1

a6k2+4

) (67b)

x6(n+2)−8 = x
−(n+1)
−8 xn+2

−2

n+1∏
k=0

k−1∏
k1=0

a6k1+1 + x−8x−6x−4
k−1∑
l=0

(
b6l+1

k−1∏
k2=l+1

a6k2+1

)
k−1∏
k1=0

a6k1+3 + x−6x−4x−2
k−1∑
l=0

(
b6l+3

k−1∏
k2=l+1

a6k2+3

) (67c)

x6(n+2)−9 = x
−(n+1)
−9 xn+2

−2

n+1∏
k=0

k−1∏
k1=0

a6k1 + x−9x−7x−5
k−1∑
l=0

(
b6l

k−1∏
k2=l+1

a6k2

)
k−1∏
k1=0

a6k1+2 + x−7x−5x−3
k−1∑
l=0

(
b6l+2

k−1∏
k2=l+1

a6k2+2

) (67d)

x6(n+1)−4 =
x−4x

n+1
−6 x

n+1
−8

xn+1
−2 x

n+1
0

n∏
k=0

k−1∏
k1=0

a6k1+5 + x−4x−2x0
k−1∑
l=0

(
b6l+5

k−1∏
k2=l+1

a6k2+5

)
k∏

k1=0

a6k1+1 + x−8x−6x−4
k∑
l=0

(
b6l+1

k∏
k2=l+1

a6k2+1

) . (67e)

x6(n+1)−5 =
x−5x

n+1
−7 x

n+1
−9

xn+1
−3 x

n+1
−1

n∏
k=0

k−1∏
k1=0

a6k1+4 + x−5x−3x−1
k−1∑
l=0

(
b6l+4

k−1∏
k2=l+1

a6k2+4

)
k∏

k1=0

a6k1 + x−9x−7x−5
k∑
l=0

(
b6l

k∏
k2=l+1

a6k2

) (67f)

3.3 The case where an and bn are 6-periodic.

Let x−9 = a, x−8 = b, x−7 = c, x−6 = d, x−5 = e, x−4 = f , x−3 = g, x−2 = h, x−1 = m
and x0 = p. We assume that an = an+6 and bn = bn+6 for n ∈ N0, therefore the solution
will be given by

x6(n+1) = d−(n+1)pn+2

n+1∏
k=0

ak3 + b3dfh
k−1∑
l=0

al3

ak5 + b5fhp
k−1∑
l=0

al5

(68a)

x6(n+2)−7 = c−(n+1)mn+2

n+1∏
k=0

ak2 + b2ceg
k−1∑
l=0

al2

ak4 + b4egm
k−1∑
l=0

al4

(68b)

23



x6(n+2)−8 = b−(n+1)hn+2

n+1∏
k=0

ak1 + b1bdf
k−1∑
l=0

al1

ak3 + b3dfh
k−1∑
l=0

al3

(68c)

x6(n+2)−9 = a−(n+1)hn+2

n+1∏
k=0

ak0 + b0ace
k−1∑
l=0

al0

ak2 + b2ceg
k−1∑
l=0

al2

(68d)

x6(n+1)−4 =
fdn+1bn+1

hn+1mn+1

n∏
k=0

ak5 + b5fhp
k−1∑
l=0

al5

ak1 + b1bdf
k∑
l=0

al1

(68e)

x6(n+1)−5 =
ecn+1an+1

gn+1mn+1

n∏
k=0

ak4 + b4egm
k−1∑
l=0

al4

ak0 + b0ace
k∑
l=0

al0

. (68f)

3.4 The case an and bn are constant coefficients

Let a6k1 = a, a6k1+1 = a, a6k1+2 = a, a6k1+3 = a, a6k1+4 = a, a6k1+5 = a and b6l =
b, b6l+1 = b, b6l+2 = b, b6l+3 = b, b6l+4 = b, b6l+5 = b. Note that this mean an = a and
bn = b for all values of n since both an and bn are constant coefficients. Then we will
have

x6(n+1) = x
−(n+1)
−6 xn+2

0

n+1∏
k=0

ak + x−6x−4x−2
k−1∑
l=0

(
bal
)

ak + x−4x−2x0
k−1∑
l=0

(bal)

(69a)

x6(n+2)−7 = x
−(n+1)
−7 xn+2

−1

n+1∏
k=0

ak + x−7x−5x−3
k−1∑
l=0

(
bal
)

ak + x−5x−3x−1
k−1∑
l=0

(bal)

(69b)

24



x6(n+2)−8 = x
−(n+1)
−8 xn+2

−2

n+1∏
k=0

ak + x−8x−6x−4
k−1∑
l=0

(
bal
)

ak + x−6x−4x−2
k−1∑
l=0

(bal)

(69c)

x6(n+2)−9 = x
−(n+1)
−9 xn+2

−2

n+1∏
k=0

ak + x−9x−7x−5
k−1∑
l=0

(
bal
)

ak + x−7x−5x−3
k−1∑
l=0

(bal)

(69d)

x6(n+1)−4 =
x−4x

n+1
−6 x

n+1
−8

xn+1
−2 x

n+1
0

n∏
k=0

ak + x−4x−2x0
k−1∑
l=0

(
bal
)

ak+1 + x−8x−6x−4
k∑
l=0

(bal)

. (69e)

x6(n+1)−5 =
x−5x

n+1
−7 x

n+1
−9

xn+1
−3 x

n+1
−1

n∏
k=0

ak + x−5x−3x−1
k−1∑
l=0

(
bal
)

ak+1 + x−9x−7x−5
k∑
l=0

(bal)

(69f)

• When an = 1 and bn = 1, then (69) we simplifies to give

x6(n+1) = d−(n+1)pn+2

n+1∏
k=0

1 + (k)dfh

1 + (k)fhp
(70a)

x6(n+2)−7 = c−(n+1)mn+2

n+1∏
k=0

1 + (k)ceg

1 + (k)egm
(70b)

x(6n+2)−8 = b−(n+1)hn+2

n+1∏
k=0

1 + (k)bdf

1 + (k)dfh
(70c)

x6(n+2)−9 = a−(n+1)hn+2

n+1∏
k=0

1 + (k)ace

1 + (k)ceg
(70d)

x6(n+1)−4 =
fdn+1bn+1

hn+1mn+1

n∏
k=0

1 + kfhp

1 + (k + 1)bdf
(70e)

25



x6(n+1)−5 =
ecn+1an+1

gn+1mn+1

n∏
k=0

1 + kegm

1 + (k + 1)ace
. (70f)

Therefore we have obtained the same results that Çinar, Gelisken and Özkan
obtained for corollary 3.1.1 in [7].

• If an = 1 and bn = −1, then

x6(n+1) = d−(n+1)pn+2

n+1∏
k=0

1− (k)dfh

1− (k)fhp
(71a)

x6(n+2)−7 = c−(n+1)mn+2

n+1∏
k=0

1− (k)ceg

1− (k)egm
(71b)

x6(n+2)−8 = b−(n+1)hn+2

n+1∏
k=0

1− (k)bdf

1− (k)dfh
(71c)

x6(n+2)−9 = a−(n+1)hn+2

n+1∏
k=0

1− (k)ace

1− (k)ceg
(71d)

x6n+2 =
fdn+1bn+1

hn+1mn+1

n∏
k=0

1− kfhp
1− (k + 1)bdf

(71e)

x6n+1 =
ecn+1an+1

gn+1mn+1

n∏
k=0

1− kegm
1− (k + 1)ace

. (71f)

Therefore we have obtained the same results that Çinar, Gelisken and Özkan
obtained for corollary 3.2.1 in [7].

• If an = −1, then

x6(2n) = d−(2n)p2n+1

(
−1 + bdfh

−1 + bfhp

)n

(72a)

x6(2n+1) = d−(2n+1)p2n+2

(
−1 + bdfh

−1 + bfhp

)n+1

(72b)

26



x6(2n+2)−7 = c−(2n+1)m2n+2

(
−1 + bceg

−1 + begm

)n+1

(72c)

x6(2n+3)−7 = c−(2n+2)m2n+3

(
−1 + bceg

−1 + begm

)n+1

(72d)

x6(2n+2)−8 = b−(2n+1)h2n+2

(
−1 + bbdf

−1 + bdfh

)n+1

(72e)

x6(2n+3)−8 = b−(2n+2)h2n+3

(
−1 + bbdf

−1 + bdfh

)n+1

(72f)

x6(2n+2)−9 = a−(2n+1)h2n+2

(
−1 + bace

−1 + bceg

)n+1

(72g)

x6(2n+3)−9 = a−(2n+2)h2n+3

(
−1 + bace

−1 + bceg

)n+1

(72h)

x6(2n+1)−4 =
fd2n+1b2n+1

h2n+1m2n+1

(−1 + bfhp)n

(−1 + bbdf)n+1 (72i)

x6(2n+2)−4 =
fd2n+2b2n+2

h2n+2m2n+2

(
−1 + bfhp

−1 + bbdf

)n+1

(72j)

x6(2n+1)−5 =
ec2n+1a2n+1

g2n+1m2n+1

(−1 + begm)n

(−1 + bace)n+1 . (72k)

x6(2n+2)−5 =
ec2n+2a2n+2

g2n+2m2n+2

(
−1 + begm

−1 + bace

)n+1

. (72l)

For an = −1, bn = 1 and bn = −1 we obtain the same results that Çinar, Gelisken
and Özkan obtained for corollary 3.3.1 and corollary 3.4.1 in [7].

27



Chapter 4

Applications of the ARS algorithm
- Singularity Analysis

Many differential equations, do not have solutions in terms of known functions. Over
many centuries, mathematicians have delved into different strategies to solve differential
equations. We take the ARS algorithm, whereby a differential equation is declared
integrable if it possesses the Painlevé property. In other words, if a singularity manifold
is determined by φ(z1, ..., zn) = 0 and u = u(z1, ..., zn) is a solution of the partial
differential equation, then we assume that u = u(z1, ..., zn) = φα

∑∞
j=0 ujφ

j, where φ =
φ(z1, ..., zn) and uj = uj(z1, ..., zn) are analytic functions of (z1, ..., zn) in a neighborhood
of the manifold, and α and defines the recursion relations for uj, j = 0, 1, 2.... The
expansion must be about a movable pole-like singularity in the complex plane of the
independent variable, x in our case, with the required number of arbitrary constants.
The Laurent expansion implies that the solution is analytic except at the singularities.

In this chapter, prime denotes the differentiation of the dependent variable y(x).

4.1 The Painlevé-Ince Equation

We study the Painlevé-Ince equation of second order, which results from the reduction
of the famous Euler-Bernoulli Beam equation and is given by

y′′ + 3yy′ + y3 = 0. (72)

The Painlevé-Ince equation is linearisable by a point transformation and is a member
of the Riccati hierarchy [17].

28



This equation is maximally symmetric and admits the 8 Lie point symmetries

X1 = ∂x,

X2 = x ∂x − y ∂y,

X3 = y ∂x − y3 ∂y,

X4 =
x2

2
∂x + (1− xy) ∂y,

X5 = xy ∂x + (y2 − xy3) ∂y,

X6 =
x2y

2
∂x +

(
−1

2
x2y3 + xy2 − y

)
∂y,

X7 =
1

6
x3y∂x +

(
−1

6
x3y3 +

1

2
x2y2 − xy +

2

3

)
∂y,

X8 =

(
1

24
x4y − 1

12
x3
)
∂x +

(
1

6
x− 1

24
y3x4 +

1

6
x3y2 − 1

4
x2y

)
∂y.

(73)

Below, we show that Eq. (72) satisfies the ARS algorithm and passes the singularity
test [18]. In fact, this equation is among the best examples of the application of the
ARS algorithm. We determine the leading-order terms of the above equation by firstly
substituting y(x) = a(x − x0)p into (72), where x0 is a constant and considered to be
the location of the movable singularity.

We note that

y′(x) = ap(x− x0)p−1 (74)

y′′(x) = ap(p− 1)(x− x0)p−2 (75)

then we have

ap2(x− x0)p−2 − ap(x− x0)p−2 + 3a2p(x− x0)2p−1 + [a(x− x0)p]3 = 0. (76)

Now to find the values of p that balances the equation, we take the powers of (x− x0)
and equate them and then solve for p, i.e

p− 2 = 2p− 1 =⇒ p = −1.

29



Consequently,

a(−1)2(x− x0)−1−2 − a(−1)(x− x0)−1−2 + 3a2(−1)(x− x0)2(−1)−1 + [a(x− x0)]3 = 0

a3 − 3a2 + 2a = 0

a(a− 1)(a− 2) = 0

then we will have that for the leading-order a, we have a = 1 or a = 2. This implies
that the movable singularity is a simple pole and there are two possibilities for the
leading-order behaviour. The arbitrary location of the movable singularity gives one of
the constants of integrations. Since (72) is a second-order equation, the second constant
of integration has to be determined from a series developed about the singularity. Now
for a = 1, we take the truncated Laurent series given by

y(x) = a(x− x0)p +m(x− x0)p+s (77)

and substitute into (72), to find

2
a

(x− x0)3
+
m (x− x0)s−1 (s− 1)2

(x− x0)2
− m (x− x0)s−1 (s− 1)

(x− x0)2

+3

(
a

x− x0
+m (x− x0)s−1

)(
− a

(x− x0)2
+
m (x− x0)s−1 (s− 1)

x− x0

)

+

(
a

x− x0
+m (x− x0)s−1

)3

. (78)

Here we take the coefficients of m and set them to be equal to zero to determine the
resonances, we get

s2 − 1 = 0

and solving for s we obtain s = 1 or s = −1. The latter value for s should always hold
as it is associated with the movable singularity. The value s = 1, gives the term in the
series at which the second constant of integration occurs.
For s = 1 we use the right Painlevé series which is given by

y(χ) = aχp +
∞∑
s=1

asχ
p+s = χ−1 +

∞∑
s=1

asχ
−1+s, (79)

where χ = x − x0, which is substituted into (72). Now we test for consistency by
solving for some a’s in the sum. This involves checking the correct number of arbitrary
constants, of which we will have two arbitrary constants since the degree of the equation
is two. Hence in the series (79), taking the first few terms of y(x), i.e

y(χ) = χ−1 + a1 + a2χ
1 + a3χ

2

30



and substituting them into (72), and then separating according to powers of χ, we have

χ0 : 3a21 + 3a2 = 0 =⇒ a2 = −a21.

χ1 : a31 + 9a1a2 + 8a3 = 0 =⇒ a3 = a31,
then clearly all other constants are a function of a1, because of the consistency test.

Hence we may conclude that a1 is an arbitrary constant, and the consistency test is
passed as we have the correct number of arbitrary constants. Thus, the Painlevé-Ince
equation of second-order possesses the Painlevé property.
For a = 2, one can do a similar analysis to the above.

4.2 Ivey’s Equation

Ivey’s nonlinear equation appears in space-charge theory, and is given by

y′′ − y′2

y
+

2

x
y′ + ky2 = 0, k is a constant. (80)

This equation admits the Lie point symmetry

Y = x∂x − 2y∂y

for k 6= 0 and the set of 8 symmetries

Y1 = x ∂x,

Y2 = x2 ∂x,

Y3 = y ∂y,

Y4 = x2 ln y ∂x,

Y5 =
y

x
∂y,

Y6 = y ln y ∂y,

Y7 = −∂x +
y

x
ln y ∂y,

Y8 = −x ln y ∂x + ln(y)2y ∂y,

(81)

31



for k = 0. This equation has not been previously analysed in terms of singularity
analysis in the literature. We determine the leading-order of the above equation by
firstly substituting

y(x) = a(x− x0)p (82)

into (80), where x0 is a constant and we find

ap2(x− x0)p−2 − ap(x− x0)p−2 −
a2p2(x− x0)2(p−1)

a(x− x0)p

+
2

x
ap(x− x0)p−1 + k[a(x− x0)p]2 = 0. (83)

We need to find the values of p from the dominant terms that balances the equation,
we take the powers of (x− x0), equate them and then solve for p.
Thus

p− 2 = 2p =⇒ p = −2,

where all terms in (80) are dominant excluding the third term. We substitute p = −2
into the leading terms of (83) and then solve for a to find

ka2 + 2a = 0. (84)

Hence we have that a = 0(not relevant) or a = − 2
k
, which are the leading order values.

For the purpose of this work, we must consider a = − 2
k
, and we take the truncated

Laurent series given by

y(x) = aχp +mχp+s, (85)

where χ = (x− x0), that is

y(x) = −2

k
χ−2 +mχ−2+s (86)

and substitute into the dominant terms of (80) to obtain

−12
1

k (x− x0)4
+
m (x− x0)−2+s (−2 + s)2

(x− x0)2
− m (x− x0)−2+s (−2 + s)

(x− x0)2

−

(
4

1

k (x− x0)3
+
m (x− x0)−2+s (−2 + s)

x− x0

)2(
−2

1

k (x− x0)2
+m (x− x0)−2+s

)−1
+k

(
−2

1

k (x− x0)2
+m (x− x0)−2+s

)2

.(87)

We take the coefficient of m to be equal to zero, to get

s2 − s− 2 = 0,

32



where we solve the resulting equation for s. Therefore the resonances are s = −1
and s = 2. For the value of s, which is −1, is generic in nature and as mentioned,
meaning its properties hold for singularity and it is also related to the arbitrariness of
the location of the movable singularity. For s = 2 we use the right Painlevé series to
find y, which is given by

y(χ) = aχp +
∞∑
s=1

asχ
p+s = −2

k
χ−2 +

∞∑
s=1

asχ
−2+s, (88)

which we test for consistency by solving for some a’s in the sum. Unfortunately this
process leads to no arbitrary constants, and we conclude that Ivey’s equation does not
pass the singularity test via the ARS algorithm.

When the singularity test fails for a differential equation, it does not mean that the
equation is not integrable. Instead it means that the solution cannot be expressed by a
Laurent expansion. Also, there exists the possibility that the equation does not admit
any singularities. For example, the linear equation

y′′ − y = 0,

that has solution
y(x) = c1e

x + c2e
−x.

Finally, for k = 0, Ivey’s equation does not possess enough dominant terms (two or
more) to perform the singularity analysis.

4.3 The Higher-Order Lane-Emden Type Equation

A Lane-Emden-Fowler type equation may be defined in the form

y′′ +
k

y
y′ + f(x)g(y) = 0, (89)

where f(x) and g(y) are arbitrary functions, and where x > 0. This equation possesses
many applications in fluids, patterns, spatial ecology and mechanics, etc. The standard
Lane-Emden equation, which arises when f(x) = 1 and g(y) = yn in (89) has been
shown to pass the singularity test [19]. Our interest lies in the higher-order Lane-
Emden equation [20], which is a third-order differential equation that is given by

y′′′ +
8

x
y′′ +

12

x2
y′ + yn = 0. (90)

This equation has not been analysed from a singularity perspective before. For arbitrary
n, this equation possesses the Lie point symmetry

Z1 = x∂x −
3y

n− 1
∂y,

33



but for n = 1, we have the Lie point symmetry Z2 = y∂y and some strange ∂y symme-
tries with Hypergeometric functions for coefficients.
Firstly we have to substitute y(x) = (x− x0)p into (90), noting that

y(x) = a(x− x0)p

y′(x) = ap(x− x0)p−1

y′′(x) = ap(p− 1)(p− 2)(x− x0)p−2

y′′′(x) = ap(p2 − 3p+ 2)(x− x0)p−3

then we will have

ap(p2 − 3p+ 2)x5(x− x0)p−3 + 8ap(p− 1)x4(x− x0)p−2

+12apx3(x− x0)p−1 + anx5(x− x0)np = 0, (91)

from which we will compute the value(s) of p from the dominant terms as follows

ap(p2 − 3p+ 2)x5(x− x0)p−3 + anx5(x− x0)np = 0, (92)

to get

p = − 3

n− 1
,

for all n > 1. We substitute p into (90) and consider the leading terms to get

ap(p2 − 3p+ 2)xp−3 + (axp)n = 0 (93)

which then gives for a,

a =

[
3(2 + n)(1 + 2n)

(n− 1)3

] 1
n−1

. (94)

34



Given a above, and that y = a (x− x0)p +m (x− x0)p+s (77) we have

−27
1

(n− 1)3 (x− x0)3

(
3

(n+ 2) (2n+ 1)

(n− 1)3

)(n−1)−1

(x− x0)−3 (n−1)
−1

−27
1

(n− 1)2 (x− x0)3

(
3

(n+ 2) (2n+ 1)

(n− 1)3

)(n−1)−1

(x− x0)−3 (n−1)
−1

−6
1

(n− 1) (x− x0)3

(
3

(n+ 2) (2n+ 1)

(n− 1)3

)(n−1)−1

(x− x0)−3 (n−1)
−1

+
m

(x− x0)3
(x− x0)−3 (n−1)

−1+s (−3 (n− 1)−1 + s
)3

−3
m

(x− x0)3
(x− x0)−3 (n−1)

−1+s (−3 (n− 1)−1 + s
)2

+2
m

(x− k)3
(x− x0)−3 (n−1)

−1+s (−3 (n− 1)−1 + s
)

+

((
3

(n+ 2) (2n+ 1)

(n− 1)3

)(n−1)−1

(x− x0)−3 (n−1)
−1

+m (x− x0)−3 (n−1)
−1+s

)n

. (95)

Taking the coefficients of m and equating them to zero we obtain

((s− 2)n− s− 1) (sn− s− 3) ((s− 1)n− s− 2)

(n− 1)3
= 0. (96)

Solving, will give the following results

s =
3

n− 1
, s =

n+ 2

n− 1
, s =

2n+ 1

n− 1
.

Since we know that one of the s variables should be equal to -1, then we can find all
the possible values of n. Hence, we get n = 0, −1/2, −2. Clearly all these values of n
give p as a positive integer, where the requisite value of p should be negative. Since we
need p to be negative, we will recalculate its value, also a(94) and s from (77) by using
n = 2 and n = 4
To ensure that p is a negative integer, we may select n = 2, then we will have p = −3,
and n = 4 yields p = −1.
For the first case, p = −3, and n = 4, we find a = 60 and we must solve

s2 − 13s+ 60 = 0, (97)

which unfortunately gives complex conjugate resonances. For n = 4 yields p = −1, we
have that a = 6

1
3

s2 − 7s+ 18 = 0, (98)

which again gives complex conjugate resonances. In such situations the reliability of the
Painlevé test is known to be uncertain. Therefore we state that the integrability analysis
for the higher-order Lane-Emden equation is inconclusive via the ARS algorithm.

35



4.4 Third-Order Boundary Flow Equations

Consider the class of equations related to boundary flow models, given by

hy′′′ + byy′′ + cy′2 + ky′ = 0, h, b, c, k are constants, and h 6= 0, (99)

Some important equations of this general equation are

2y′′′ + yy′′ = 0, (100)

y′′′ + yy′′ − y′2 = 0, (101)

y′′′ + yy′′ − y′2 −M2y′ = 0, M is a magnetic parameter, (102)

y′′′ + yy′′ − βy′2 = 0, β is a constant, (103)

where, for example, (100) is the Blasius equation, (101)-(102) are Blasius type equations
[21] and (103) is the Falkner-Skan equation.

The symmetries of (99) are

∂x, for c = 0, or c = −3

2
b, or h = 0, or b, c, h, k 6= 0, c 6= −3

2
b.

∂x, ∂y, for b = 0, or b, h = 0.

∂x, F (y)∂y, for b, c, h = 0.

∂x, x∂x − y∂y, for k = 0, or k, c = 0, or h, k = 0.

F (x)∂x, G(y, x)∂y, for b, c, h, k = 0.

∂x, ∂y, x∂x − y∂y, for b, k = 0, or b, h, k = 0.

∂x, x
2c∂x − (2cxy + 18h)∂y, x∂x − y∂y, for c = −3

2
b, k = 0.

∂x, x
2∂x − 2xy∂y, x∂x − y∂y, for c = −3

2
b, k, h = 0,

(104)

36



and lastly

∂x, ∂y, y∂y, cos

(√
kx√
h

)
∂y, sin

(√
kx√
h

)
∂y, y

√
k cos

(√
kx√
h

)
∂y

+
√
h sin

(√
kx√
h

)
∂x, y

√
k sin

(√
kx√
h

)
∂y −

√
h cos

(√
kx√
h

)
∂x, for b, c = 0.

(105)

In the specific cases Eq. (100), (101) and Eq. (102) (M = 0), the Lie point
symmetries are ∂x, x∂x − y∂y, while for M 6= 0 in (102) we only have ∂x. Eq. (103)
admits the same two symmetries for β 6= 3

2
but an extra third symmetry (xy − 6)∂y −

1
2
x2∂x for β = 3

2
.

A singularity analysis of (99), reveals that p = −1 where all terms are dominant
excluding the last term in the equation. This p value yields that

a = 6
h

2 b+ c
.

Substitution of the truncated series

y (x) = 6
h

(2 b+ c) (x− x0)
+m (x− x0)−1+s ,

into the dominant terms of (99), gives a polynomial in m. Taking the coefficients of m
as vanishing, we find the equation

2 (s+ 1)
(
(b+ c/2) s2 + (−4 b− 7/2 c) s+ 6 b+ 3 c

)
h = 0,

which one solves to find the three resonances

s = −1, s =
1

4 b+ 2 c

(
8 b+ 7 c+±

√
−32 b2 + 16 bc+ 25 c2

)
.

Recall that the acceptable values for the resonance s, must be real and rational. Note
that there are many possibilities for the second and third resonance to be positive,
negative or complex. To establish the particular values of the resonance and further
progress to the last step the method, we may consider particular values of the free
parameters, which facilitate the consistency test for the constants of integration.

Testing equation (100) - (102) leads to complex conjugate resonances, so we conclude
that the Painlevé test is unreliable in these cases. For equation (103), we find that the
second and third resonances are

s =
1

4− 2 β

(
8− 7 β +±

√
25 β2 − 16 β − 32

)
.

Taking β = −1, since the resonance s must be real and rational, we have that s = 2
and s = 3. Regarding the consistency test, we substitute

y (x) =
2

x− x0
+ a1 + a2(x− x0) + a3(x− x0)2,

37



into (103), and find that a1 = 0 while a2 and a3 are arbitrary constants. Hence, we
have the correct number of constants and we conclude that that Falkner-Skan equation
passes the singularity test via the ARS algorithm.

38



Chapter 5

Conclusion

In our first investigation, we performed a Lie analysis of a tenth-order difference equa-
tion. We then performed group reductions of the equation using the symmetries of the
problem under investigation and the solutions were all given in a unified manner. We
separate the single solutions that we obtained into categories with the aim to realize
some results that were obtained in the existing literature.

In the second part of this dissertation, we listed the Lie point symmetries and stud-
ied integrability analysis of ordinary differential equations via the ARS algorithm. We
investigated the Painlevé-Ince equation, Ivey’s equation, the higher-order Lane-Emden
type equations and some equations related to boundary flow. This algorithm has been
used extensively on a wide range of ordinary differential equations to test for singulari-
ties and also to find general solutions. A catalogue of equations exist, where equations
are deemed Painlevé integrable or not. In the spirit of this catalogue, we have studied
the algorithm and applied the method to several equations whose status was unknown
in terms of singularity analysis. In showcasing the method, we first considered the
famous Painlevé-Ince equation. We then discovered that the singularity test was un-
successful for Ivey’s equation due to a lack of a second arbitrary constant, whilst for
the higher-order Lane-Emden type equation, was inconclusive. The latter, because of
the presence of complex conjugate resonances. Thereafter, the method was successfully
applied to a member of third-order boundary flow equations. We strongly believe that
this work can be of high importance in the study of various differential problems and
their future advancement.

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