Int. J. Anal. Appl. (2024), 22:50

Dynamical Analysis of Thirtieth-Order Difference Equations

Tshenolo Thomas, Mensah Folly-Gbetoula∗

School of Mathematics, University of the Witwatersrand, Wits 2050, Johannesburg, South Africa

∗Corresponding author: Mensah.Folly-Gbetoula@wits.ac.za

Abstract. The main goal of this paper is to determine exact solutions of a family of thirtieth-order difference equations

with variable coefficients. We use similarity variables obtained via symmetries to lower the order of the equations. We

then reverse the transformations and obtain closed form solutions. We compare our solutions to those found in the

literature for special cases. We investigate the periodic nature of the solutions and present some numerical examples to

confirm the results. Finally, we analyze the stability of the equilibrium points. The method employed in this work can

be applied to equations of higher order provided that they admit non zero characteristics.

1. Introduction

Difference equations are used to investigate the evolution of a state of a particular system over

a period of time. In essence, difference equations are used to determine the future value of a

system in relation to its past value and this is done through a recurrence relationship. Difference

equations are useful in various mathematical disciplines such as control theory, integrable systems

theory, and number theory. However, these equations can also be applied when solving real-

life problems in areas such as economics, biology, physics, engineering, computer science, and

others. Applications in economics include using difference equations to model market trends

and economic growth and determining the price of certain commodities. In biology, for example,

researchers use difference equations to build models to study population dynamics epidemiology

to understand and predict the spread of diseases. In physics, difference equations are particularly

useful for simplifying complex computations that would result from investigating systems that

involve many elements. Difference equations have also been used to solve numerical solutions to

some differential equations and this is done by discretizing the given differential equation.

Various techniques can be used to solve difference equations such as numerical techniques or iter-

ative methods. Analytical methods involve finding a closed form expression satisfying the given

Received: Dec. 29, 2023.

2020 Mathematics Subject Classification. 39A10, 39A13, 39A23, 39A05, 39A99.

Key words and phrases. difference equation; symmetry; reduction; exact solutions; periodicity.

https://doi.org/10.28924/2291-8639-22-2024-50
ISSN: 2291-8639

© 2024 the author(s).

https://doi.org/10.28924/2291-8639-22-2024-50


2 Int. J. Anal. Appl. (2024), 22:50

difference equation. Other methods involve using a combination of analytical and numerical

techniques. For instance, Maeda’s method for solving difference equations uses both analyti-

cal and iterative techniques to solve non-homogeneous difference equations with linear-constant

coefficients [12]. Some other methods include direct substitution, matrix methods, generating

functions, and difference calculus. In this study, we use a symmetry-based method to find the

analytic solution of the following equation

un+1 =
un−29

An + Bnun−4un−9un−14un−19un−24un−29
(1.1)

where An and Bn are sequences with initial conditions u−29, u−28, . . . , u0.

Lama [1] studied the solutions and the dynamical properties, that is, boundedness, local and

global points, and the equilibrium points of the difference equation

ζr+1 =
ζr−29

±1± ζr−29ζr−24ζr−19ζr−14ζr−9ζr−4
, r = 0, 1, 2, · · · . (1.2)

where the initial conditions ζ−29, ζ−28, . . . , ζ0 are real numbers.

Our main objective is to find the solutions of (1.1) using its symmetry. In our investigation, we

expect our solutions to be more general with less restrictions on An and Bn. For definiteness, we

study the equivalent equation

un+30 =
un

An + Bnunun+5un+10un+15un+20un+25
(1.3)

and demonstrate how one uses the solution of (1.3) to recover those of (1.1). For more results on

difference equations, the reader can refer to [2–8, 10]

1.1. Preliminaries. In this section, we define the terms used and the notation according to [9]. Let

un+30 = ω(n, un, un+5, un+10, un+15, un+20, un+25) (1.4)

be a difference equation of order thirty. We assume that the partial derivative of ω with respect to

un is non-zero.

Definition 1.1. The forward shift operator is defined as S : n 7→ n + 1 and so, Si(un) = un+i.

Consider the point transformations

Γε : (n, un) 7→ (ñ, ũn). (1.5)

It is known [9] that Γε is a one-parameter Lie group of transformations if:

• Γ0 is the identity map, so that x̃ = x when ε = 0.

• ΓaΓb = Γa+b for every a and b sufficiently close to 0.

• Each x̃i can be represented as a Taylor series in ε (in a neighborhood of ε = 0 that is

determined by x), and therefore x̃i(x; ε) = xi + εξi(x) + O(ε2), i = 1, . . . , p.



Int. J. Anal. Appl. (2024), 22:50 3

In this paper, we shall assume that the Lie group of point transformations is of the form

ñ = n; ũn = un + εQ(n, un) (1.6)

and that the corresponding (prolonged) infinitesimal generator is given by

X =Q(n, un)
∂
∂un

+ S5Q(n, un)
∂
∂un+5

+ S10Q(n, un)
∂
∂un+10

+ S15Q(n, un)
∂
∂un+15

+ S20Q(n, un)
∂
∂un+20

+ S25Q(n, un)
∂
∂un+25

. (1.7)

The characteristics Q(n, un) can be found by solving the linearized symmetry condition

S
(30)Q(n, un) −Xω = 0 (1.8)

as long as (1.4) holds.

2. Main results

2.1. Symmetries. In the first part of the investigation, we focus on finding the symmetries of (1.3).

Applying the symmetry condition given in (1.8), we obtain the following:

S30Q(n, un) −Q(n, un)
∂ω
∂un
− S5Q(n, un)

∂w
∂un+5

− S10Q(n, un)
∂w
∂un+10

−

S15Q(n, un)
∂w
∂un+15

− S20Q(n, un)
∂w
∂un+20

− S25Q(n, un)
∂w
∂un+25

= 0. (2.1)

We apply the differential operator ∂
∂un
−

ω,un
ω,un+5

∂
∂un+5

to (2.1) and obtain

Q
′

(n, un)ω,un −Q(n, un)ω,unun −S5Q(n, un)ω,unun+5 −S10Q(n, un)ω,unun+10 −S15Q(n, un)ω,unun+15

− S20Q(n, un)ω,unun+20 −S25Q(n, un)ω,unun+25 −
ω,un

ω,un+5

−Q(n, un)ω,unun+5 −S5Q
′

(n, un)ω,un+5

− S5Q(n, un)ω,un+5un+5 −S10Q(n, un)ω,un+5un+10 −S15Q(n, un)ω,un+5un+15 −S20Q(n, un)ω,un+5un+20

− S25Q(n, un)ω,un+5un+25

 = 0. (2.2)

Note that f,x denotes the partial derivative of f with respect to x. Substituting the partial derivatives

into (2.2) with a bit of simplification, we get

2Q(n, un)un+10un+15un+20un+25 − (Q′(n, un) − S5Q′(n, un))unun+10un+15un+20un+25

+ S10Q(n, un)unun+15un+20un+25 + S15Q(n, un)unun+10un+20un+25 + S20Q(n, un)unun+10un+15un+25

+ S25Q(n, un)unun+10un+15un+20 = 0. (2.3)

Now, differentiating (2.3) twice with respect to un and a bit of simplification, we get

Q′′′(n, un) = 0. (2.4)



4 Int. J. Anal. Appl. (2024), 22:50

It follows that the characteristic must be of the form

Q(n, un) = αn + βnun + γnu2
n, (2.5)

where αn, βn and γn are some functions of n. Using (2.5) in (2.1) and employing the method of

separation (see [3, 9] for more details), we obtain the following system of constraints:

αn = 0, (2.6)

βn+5 + βn+10 + βn+15 + βn+20 + βn+25 + βn+30 = 0, (2.7)

βn − βn+30 = 0, (2.8)

γn = 0. (2.9)

Consequently,

βn + βn+5 + βn+10 + βn+15 + βn+20 + βn+25 = 0. (2.10)

This is a linear difference equation with constant coefficients. It follows that

βn =

(
exp

i(2kπ)
30

)n

, (2.11)

k = 1, . . . , 29 and k < {6, 12, 18, 24}. Hence, the 25 symmetries are as follows:

Xk = exp
(

2kπn
30

i
)

un
∂
∂un

, (2.12)

k = 1, . . . , 29 and k < {6, 12, 18, 24}.

2.2. Reduction and exact solutions. We introduce the canonical coordinate

Sn =

∫
dun

Q(n, un)
=

1
βn

ln |un| (2.13)

and let

Ṽn =Snβn + Sn+5βn+5 + Sn+10βn+10 + Sn+15βn+15 + Sn+20βn+20 + Sn+25βn+25

= ln (unun+5un+10un+15un+20un+25). (2.14)

Setting Vn = exp(−Ṽn), we obtain

Vn =
1

unun+5un+10un+15un+20un+25
. (2.15)

Now shifting equation (2.15) five times and substituting for un+30, we get

Vn+5 = AnVn + Bn. (2.16)

By iterating (2.16), we get

V5n+ j = V j

n−1∏
t

A5t+ j

+ n−1∑
i=0

B5i+ j

n−1∏
k2=i+1

A5k2+ j

 , (2.17)



Int. J. Anal. Appl. (2024), 22:50 5

j = 0, 1, 2, 3, 4. From (2.15), we have that

un+30 =
Vn

Vn+5
un. (2.18)

By iteration, we get

u30n+ j =u j

n−1∏
t=0

V30t+ j

V30t+ j+5

=u j

n−1∏
t=0

V30t+5b j
5 c+τ( j)

V30t+5+5b j
5 c+τ( j)

=u j

n−1∏
t=0

V5(6t+b j
5 c)+τ( j)

V5(6t+1+b j
5 c)+τ( j)

. (2.19)

Here, b·c denotes the floor function and τ( j) is the remainder when j is divided by 5. Substituting

(2.17) in equation (2.19), we obtain

u30n+ j =u j

n−1∏
t=0

Vτ( j)

6t+b j
5 c−1∏

s=0
A5s+τ( j)

+ 6t+b j
5 c−1∑

i=0
B5i+τ( j)

6t+b j
5 c−1∏

k2=i+1
A5k2+τ( j)


Vτ( j)

6t+b j
5 c∏

s=0
A5s+τ( j)

+ 6t+b j
5 c∑

i=0
B5i+τ( j)

6t+b j
5 c∏

k2=i+1
A5k2+τ( j)



=u j

n−1∏
t=0


6t+
b

j
5 c−1∏
s=0

A5s+τ( j)

+
6t+
b

j
5 c−1∑
i=0

B5i+τ( j)

Vτ( j)


6t+
b

j
5 c−1∏

k2=i+1
A5k2+τ( j)

6t+b j
5 c∏

s=0
A5s+τ( j)

+ 6t+b j
5 c∑

i=0

B5i+τ( j)

Vτ( j)

6t+b j
5 c∏

k2=i+1
A5k2+τ( j)


, (2.20)

where 1/Vh = uhuh+5uh+10uh+15uh+20uh+25. Equation (2.20) gives the solution of (1.3) in terms of

n and the initial conditions. Consequently, the solution of (1.1) is obtained by shifting back (2.20)

29 times as follows:

u30n+ j−29 =u j−20

n−1∏
t=0


6t+
b

j
5 c−1∏
s=0

a5s+τ( j)

+
6t+
b

j
5 c−1∑
i=0

b5i+τ( j)

Vτ( j)−29


6t+
b

j
5 c−1∏

k2=i+1
a5k2+τ( j)

6t+b j
5 c∏

s=0
a5s+τ( j)

+ 6t+b j
5 c∑

i=0

b5i+τ( j)

Vτ( j)−29

6t+b j
5 c∏

k2=i+1
a5k2+τ( j)


. (2.21)

2.3. The case where the sequences An and Bn are one-periodic. Here, we set An = A and Bn = B.

Equation (2.20) simplifies to



6 Int. J. Anal. Appl. (2024), 22:50

u30n+ j =u j

n−1∏
t=0

A6t+b j
5 c + B

Vτ( j)

6t+b j
5 c−1∑

i=0
Ai

A6t+b j
5 c+1 + B

Vτ( j)

6t+b j
5 c∑

i=0
Ai

. (2.22)

Now, using (2.15) in the equation above, we get the closed form solution of (1.3) for constant

coefficients given by

u30n+ j =

u j

n−1∏
t=0

A6t+b j
5 c + Buτ( j)uτ( j)+5uτ( j)+10uτ( j)+15uτ( j)+20uτ( j)+25

6t+b j
5 c−1∑

i=0
Ai

A6t+b j
5 c+1 + Buτ( j)uτ( j)+5uτ( j)+10uτ( j)+15uτ( j)+20uτ( j)+25

6t+b j
5 c∑

i=0
Ai

(2.23)

for j = 0, 1, . . . , 29. Note that τ( j) ∈ {0, 1, 2, 3}when j ∈ {0, 1, . . . , 29}.

2.3.1. Case where A = 1. When we replace A = 1 in (2.23), we get

u30n+ j =

u j

n−1∏
t=0

1 + (6t + b j
5c)Buτ( j)uτ( j)+5uτ( j)+10uτ( j)+15uτ( j)+20uτ( j)+25

1 + (6t + b j
5c+ 1)Buτ( j)uτ( j)+5uτ( j)+10uτ( j)+15uτ( j)+20uτ( j)+25

. (2.24)

2.3.2. Case where A , 1. In this instance, (2.23) becomes

u30n+ j =

u j

n−1∏
t=0

A6t+b j
5 c + 1−A6t+b

j
5 c

1−A Buτ( j)uτ( j)+5uτ( j)+10uτ( j)+15uτ( j)+20uτ( j)+25

A6t+b j
5 c+1 + 1−A6t+b

j
5 c+1

1−A Buτ( j)uτ( j)+5uτ( j)+10uτ( j)+15uτ( j)+20uτ( j)+25

. (2.25)

Recall that we shifted equation (1.1) forward 29 times to obtain (1.3) whose solution is given by

(2.24) and (2.25). Now we shift backwards 29 times the equations (2.24) and (2.25) to obtain the

solution of the difference equation (1.1) which is given by

u30n+ j−29 = u j−29

n−1∏
t=0

1 + (6t + b j
5c)Buτ( j)−29uτ( j)−24uτ( j)−19uτ( j)−14uτ( j)−9uτ( j)−4

1 + (6t + b j
5c+ 1)Buτ( j)−29uτ( j)−24uτ( j)−19uτ( j)−14uτ( j)−9uτ( j)−4

(2.26)

when a = 1; and

u30n+ j−29 = u j−29

n−1∏
t=0

A6t+b j
5 c +

(
1−A6t+b

j
5 c

1−A

)
Buτ( j)−29uτ( j)−24uτ( j)−19uτ( j)−14uτ( j)−9uτ( j)−4

A6t+b j
5 c+1 +

(
1−A6t+b

j
5 c+1

1−A

)
Buτ( j)−29uτ( j)−24uτ( j)−19uτ( j)−14uτ( j)−9uτ( j)−4

(2.27)

when a , 1.



Int. J. Anal. Appl. (2024), 22:50 7

3. Some special cases leading to some results in the existing literature

Setting k := 29 − j, we have that b j
5c = 5 − b k

5c and τ( j) = 4 − τ(k) when j = 0, 1, . . . , 20,

k = j− 29. Hence, equations (2.26) and (2.27) become

u30n−k = u−k

n−1∏
t=0

1 + (6t + 5− b k
5c)Bu−τ(k)−25u−τ(k)−20u−τ(k)−15u−τ(k)−10u−τ(k)−5u−τ(k)

1 + (6t + 6− b k
5c)Bu−τ(k)−25u−τ(k)−20u−τ(k)−15u−τ(k)−10u−τ(k)−5u−τ(k)

(3.1)

when a = 1; and

u30n−k = u−k

n−1∏
t=0

A6t+5−b k
5 c +

(
1−A6t+5−b k

5 c

1−A

)
Bu−τ(k)−25u−τ(k)−20u−τ(k)−15u−τ(k)−10u−τ(k)−5u−τ(k)

A6t+6−b k
5 c +

(
1−A6t+6−b k

5 c

1−A

)
Bu−τ(k)−25u−τ(k)−20u−τ(k)−15u−τ(k)−10u−τ(k)−5u−τ(k)

(3.2)

when A , 1. If we replace B = 1 in (3.1), we get the results in [1] (see Theorems 1 and 6). In fact,

for B = 1, using (2.26), we have

u30n−k = u−k

n−1∏
t=0

1 + (6t + 5− b k
5c)u−τ(k)−25u−τ(k)−20u−τ(k)−15u−τ(k)−10u−τ(k)−5u−τ(k)

1 + (6t + 6− b k
5c)u−τ(k)−25u−τ(k)−20u−τ(k)−15u−τ(k)−10u−τ(k)−5u−τ(k)


= εk

n−1∏
t=0

1 + (6t + ηk − 1)µk

1 + (6t + ηk)µk

, (3.3)

where µk =
5∏

j=0
εmod (k,5)+5 j, ηk = 6− b k

5c and u−k = εk with k = 0, 1, . . . , 29.

Similarly, for B = −1, using (2.26), we have

u30n−k = u−k

n−1∏
t=0

1− (6t + 5− b k
5c)u−τ(k)−25u−τ(k)−20u−τ(k)−15u−τ(k)−10u−τ(k)−5u−τ(k)

1− (6t + 6− b k
5c)u−τ(k)−25u−τ(k)−20u−τ(k)−15u−τ(k)−10u−τ(k)−5u−τ(k)


= εk

n−1∏
t=0

1− (6t + ηk − 1)µk

1− (6t + ηk)µk

, (3.4)

where µk =
5∏

j=0
εmod (k,5)+5 j , ηk = 6− b k

5c and u−k = εk with k = 0, 1, . . . , 29.

If we replace A = −1 and B = ±1 in (3.2), we get the results in [1] (see Theorems 8 and 12). In fact,

for A = −1 and B = 1, using (3.2), we have

u30n−k = u−k

n−1∏
t=0

(−1)1+b k
5 c +

(
1−(−1)1+b k

5 c

2

)
u−τ(k)−25u−τ(k)−20u−τ(k)−15u−τ(k)−10u−τ(k)−5u−τ(k)

−(−1)1+b k
5 c +

(
1+(−1)1+b k

5 c

2

)
u−τ(k)−25u−τ(k)−20u−τ(k)−15u−τ(k)−10u−τ(k)−5u−τ(k)

=
εk

(−1 + µk)tαk
, (3.5)

where µk =
5∏

j=0
εmod (k,5)+5 j, αk = (−1)b

k
5 c+1 and u−k = εk with k = 0, 1, . . . , 29.



8 Int. J. Anal. Appl. (2024), 22:50

Similarly, for A = −1 and B = −1, using (3.2), we have

u30n−k = u−k

n−1∏
t=0

(−1)1+b k
5 c −

(
1−(−1)1+b k

5 c

2

)
u−τ(k)−25u−τ(k)−20u−τ(k)−15u−τ(k)−10u−τ(k)−5u−τ(k)

−(−1)1+b k
5 c −

(
1+(−1)1+b k

5 c

2

)
u−τ(k)−25u−τ(k)−20u−τ(k)−15u−τ(k)−10u−τ(k)−5u−τ(k)

=
εk

(−1− µk)tαk
, (3.6)

where µ
5∏

j=0
εmod (k,5)+5 j, αk = (−1)b

k
5 c+1 and u−k = εk with k = 0, 1, . . . , 29.

4. Behavior of the solutions

In this section, we give the conditions for periodic solutions with periods 2, 3, 6, 10, 15 and 30.

Additionally, we show that the equilibrium points are non-hyperbolic.

Theorem 4.1. Let un be a solution of

un+30 =
un

A + Bunun+5un+10un+15un+20un+25
(4.1)

with initial conditions ui, i = 0, . . . , 29; and A , 1 and B some non-zero constants. If the initial conditions
satisfy the conditions

uiui+5ui+10ui+15ui+20ui+25 =
1−A

B
(4.2)

and

ui , ui+2, ui , ui+3, ui , ui+6, ui , ui+10, ui , ui+15, (4.3)

then we have a 30-periodic solution.

Proof. Using the assumption uiui+5ui+10ui+15ui+20ui+25 = 1−A
B stated in Theorem 4.1, equation

(2.25) reduces to

u30n+ j =u j

n−1∏
t=0

A6t+b j
5 c +

(
1−A6t+b

j
5 c

1−A

)
B
(1−A

B

)
A6t+b j

5 c+1 +

(
1−A6t+b

j
5 c+1

1−A

)
B
(1−A

B

) (4.4)

=u j. (4.5)

Adding the condition ui , ui+2, ui , ui+3, ui , ui+6, ui , ui+10, ui , ui+15, we conclude that the

solution is periodic with period 30. �



Int. J. Anal. Appl. (2024), 22:50 9

Figure 1. Graph of un+30 = un
−3+2unun+5un+10un+15un+20un+25

.

Figure 1 illustrates the graph of (1.3) with the initial conditions, u0 = 1; u1 = 4; u2 = −3/2; u3 =

−1/2; u4 = 2/3; u5 = 4; u6 = 1/2; u7 = 2; u8 = 1/2; u9 = 3; u10 = 1/2; u11 = 2; u12 = −1/6; u13 =

−8; u14 = 2; u15 = 1/3; u16 = −1/2; u17 = 2; u18 = −1/2; u19 = 1/2; u20 = −3; u21 = 1/5; u22 =

1/2; u23 = 1/2; u24 = 8; u25 = −1; u26 = −5; u27 = 4; u28 = −4; u29 = 1/8, satisfying the conditions

in Theorem 4.1. As expected, the solution is 30-periodic.

Theorem 4.2. Let un be a solution of

un+30 =
un

A + Bunun+5un+10un+15un+20un+25
(4.6)

with initial conditions ui, i = 0, . . . , 29; and A , 1 and B some non-zero constants. If the initial conditions
satisfy the conditions

u2
i u2

i+5u2
i+10 =

1−A
B

(4.7)

and

ui , ui+2, ui , ui+3, ui , ui+6, ui , ui+10, ui = ui+15, (4.8)

then we have a periodic solution with period 15.

Proof. The proof is similar to the proof of Theorem 4.1 and is omitted. �



10 Int. J. Anal. Appl. (2024), 22:50

Figure 2. Graph of (1.3) when a = 2 and b = −1/4.

Figure 2 illustrates the graph of (1.3) with the initial conditions, u0 = 1; u1 = 2; u2 = −4; u3 =

−1/2; u4 = 1/8; u5 = 4; u6 = 1/2; u7 = 1/2; u8 = 1/2; u9 = 8u10 = 1/2; u11 = 2; u12 = −1; u13 =

−8; u14 = 2; u15 = 1; u16 = 2; u17 = −4; u18 = −1/2; u19 = 1/8; u20 = 4; u21 = 1/2; u22 = 1/2; u23 =

1/2; u24 = 8; u25 = 1/2; u26 = 2; u27 = −1; u28 = −8; u29 = 2, satisfying the conditions in Theorem

4.2. As expected, the solution is 15-periodic.

Theorem 4.3. Let un be a solution of

un+30 =
un

A + Bunun+5un+10un+15un+20un+25
(4.9)

with initial conditions ui, i = 0, . . . , 29; and A , 1 and B some non-zero constants. If the initial conditions
satisfy the conditions

u3
i u3

i+5 =
1−A

B
(4.10)

and

ui , ui+2, ui , ui+3, ui , ui+6, ui = ui+10, (4.11)

then we have a periodic solution with period 10.

Proof. The proof is similar to the proof of Theorem 4.1 and is omitted. �



Int. J. Anal. Appl. (2024), 22:50 11

Figure 3. Graph of un+30 = un
2− 1

8 unun+5un+10un+15un+20un+25
.

Figure 3 illustrates the graph of (1.3) with the initial conditions, u0 = 1; u1 = 4; u2 = −1/3; u3 =

−1/4; u4 = 1/5; u5 = 2; u6 = 1/2; u7 = −6; u8 = −8; u9 = 10; u10 = 1; u11 = 4; u12 = −1/3; u13 =

−1/4; u14 = 1/5; u15 = 2; u16 = 1/2; u17 = −6; u18 = −8; u19 = 10; u20 = 1; u21 = 4; u22 =

−1/3; u23 = −1/4; u24 = 1/5; u25 = 2; u26 = 1/2; u27 = −6; u28 = −8; u29 = 10, satisfying the

following conditions:

uτ( j)uτ( j)+5uτ( j)+10uτ( j)+15uτ( j)+20uτ( j)+25 =
1−A

B
(4.12)

and

ui = ui+10. (4.13)

As expected, the solution is 10-periodic.

Theorem 4.4. Let un be a solution of

un+30 =
un

A + Bunun+5un+10un+15un+20un+25
(4.14)

with initial conditions ui, i = 0, . . . , 29; and A , 1 and B some non-zero constants. If the initial conditions
satisfy the conditions

u0u1u2u3u4u5 =
1−A

B
(4.15)



12 Int. J. Anal. Appl. (2024), 22:50

and

ui , ui+2, ui , ui+3, ui = ui+6, (4.16)

then we have a periodic solution with period 6.

Proof. The proof is similar to the proof of Theorem 4.1 and is omitted. �

Figure 4. Graph of (1.3) when a = 2 and b = −1/4.

Figure 4 illustrates the graph of (1.3) with the initial conditions, u0 = 1; u1 = 2; u2 = 3; u3 =

4; u4 = −1/6; u5 = −1; u6 = 1; u7 = 2; u8 = 3; u9 = 4; u10 = −1/6; u11 = −1; u12 = 1; u13 = 2; u14 =

3; u15 = 4; u16 = −1/6; u17 = −1; u18 = 1; u19 = 2; u20 = 3; u21 = 4; u22 = −1/6; u23 = −1; u24 =

1; u25 = 2; u26 = 3; u27 = 4; u28 = −1/6; u29 = −1, satisfying the conditions in Theorem 4.4. As

expected, the solution is 6-periodic.

Theorem 4.5. Let un be a solution of

un+30 =
un

A + Bunun+5un+10un+15un+20un+25
(4.17)

with initial conditions ui, i = 0, . . . , 29; and A , 1 and B some non-zero constants. If the initial conditions
satisfy the conditions

u2
0u2

1u2
2 =

1−A
B

(4.18)



Int. J. Anal. Appl. (2024), 22:50 13

and

ui , ui+2, ui = ui+3, (4.19)

then we have a periodic solution with period 3.

Proof. The proof is similar to the proof of Theorem 4.1 and is omitted. �

Figure 5. Graph of un+30 = un
2− 1

4 unun+5un+10un+15un+20un+25
.

Figure 5 illustrate the graph of (1.3) with the initial conditions, u0 = 1; u1 = 2; u2 = −1; u3 =

1; u4 = 2; u5 = −1; u6 = 1; u7 = 2; u8 = −1; u9 = 1; u10 = 2; u11 = −1; u12 = 1; u13 = 2; u14 =

−1; u15 = 1; u16 = 2; u17 = −1; u18 = 1; u19 = 2; u20 = −1; u21 = 1; u22 = 2; u23 = −1; u24 = 1; u25 =

2; u26 = −1; u27 = 1; u28 = 2; u29 = −1, satisfying the conditions in Theorem 4.5. As expected, the

solution is 3-periodic.

Theorem 4.6. Let un be a solution of

un+30 =
un

A + Bunun+5un+10un+15un+20un+25
(4.20)

with initial conditions ui, i = 0, . . . , 29; and A , 1 and B some non-zero constants. If the initial conditions
satisfy the conditions

u3
0u3

1 =
1−A

B
(4.21)



14 Int. J. Anal. Appl. (2024), 22:50

and

ui = ui+2, (4.22)

then we have a periodic solution with period 2.

Proof. The proof is similar to the proof of Theorem 4.1 and is omitted. �

Figure 6. Graph of (1.3) when a = 2 and b = −1/27.

Figure 6 illustrate the graph of (1.3) with the initial conditions, u0 = 1; u1 = 3; u2 = 1; u3 =

3; u4 = 1; u5 = 3u6 = 1; u7 = 3; u8 = 1; u9 = 3; u10 = 1; u11 = 3; u12 = 1; u13 = 3; u14 = 1; u15 =

3; x[16] = 1; u17 = 3; u18 = 1; u19 = 3; u20 = 1; u21 = 3; u22 = 1; u23 = 3u24 = 1; u25 = 3; u26 =

1; u27 = 3; u28 = 1; u29 = 3, satisfying the conditions in Theorem 4.6. As expected, the solution is

2-periodic.

Theorem 4.7. Given the equation

un+30 =
un

1 + Bunun+5un+10un+15un+20un+25
, (4.23)

the sole equilibrium point ū = 0 is non-hyperbolic.

Proof. The fixed point condition u = u/(1 + Bu6) yields the point u = 0 and the characteristic

equation of (4.23) near 0 reads λ30
− 1 = 0. The roots of λ30

− 1 = 0 are the thirtieth roots of unity,

their moduli are all equal to 1. Thus, u = 0 is non-hyperbolic. �



Int. J. Anal. Appl. (2024), 22:50 15

Theorem 4.8. The fixed point ū = 0 of (4.1) is (locally) asymptotically stable for |A| > 1. In addition, the
non-zero fixed points of (4.1) are non-hyperbolic for all A , 1.

Proof. Imposing the fixed points condition on (4.1) yields ū(A + Bū6
− 1) = 0. For the first part of

the proof, the characteristic equation of (4.1) near ū = 0 given by λ30
−

1
A = 0. Consequently, the

roots λi of the latter are in such a way that |λi| < 1 for |A| > 1. Thus, u = 0 is locally asymptotically
stable when |A| > 1. For the second part, we find the non-zero fixed points by solving the equation
A + Bu6

− 1 = 0 for u. Here, the characteristic equation of (4.1) near a non-zero equilibrium point
is given by

0 =λ30
− (A− 1)λ25

− (A− 1)λ20
− (A− 1)λ15

− (A− 1)λ10
− (A− 1)λ5

−A (4.24)

=(λ25 + λ20 + λ15 + λ10 + λ5 + 1)(λ5
−A) (4.25)

=
1− λ30

1− λ5 (λ5
−A), with λ5 , 1. (4.26)

There exists a root of the characteristic equation with modulus equals 1, for instance, λ = eiπ/15. �

5. Conclusion

We investigated the solutions of the thirtieth-order difference equations (3) by the method of

symmetry analysis. In order to use this method, the shift operator was applied to equation (3)

thereby shifting the equation 29 times to give equation (5). Then the Lie symmetries of the shifted

equation were found and they were used to lower the order of the reduced equation. Our results

were further verified by substituting different values for A and B in (1.1) and then comparing our

solutions to the solutions in the literature. We also studied the periodic nature of the solutions ,

stability of the fixed points and used graphical illustrations to confirm our results.

Conflicts of Interest: The authors declare that there are no conflicts of interest regarding the

publication of this paper.

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Equations, Alexandria Eng. J. 74 (2023), 269–283. https://doi.org/10.1016/j.aej.2023.05.026.

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https://doi.org/10.1093/imamat/38.2.129

	1. Introduction
	1.1. Preliminaries

	2. Main results
	2.1. Symmetries
	2.2. Reduction and exact solutions
	2.3. The case where the sequences An and Bn are one-periodic

	3. Some special cases leading to some results in the existing literature
	4. Behavior of the solutions
	5. Conclusion
	References