J. COMPUTATIONAL ANALYSIS AND APPLICATIONS, VOL. 24, NO.8, 2018, COPYRIGHT 2018 EUDOXUS PRESS, LLC
Formulas and properties of some class of nonlinear di erence equations E. M. Elsayed1 2, Faris Alzahrani1 , and H. S. Alayachi1 1 Mathematics Department, Faculty of Science, King Abdulaziz University, P. O. Box 80203, Jeddah 21589, Saudi Arabia. 2 Department of Mathematics, Faculty of Science, Mansoura University, Mansoura 35516, Egypt. E-mails:
[email protected],
[email protected],
[email protected]. ABSTRACT We obtain the formulas of the solutions of the recursive sequences +1
5
=
4 (±1 ±
5)
=0 1
where the initial conditions are arbitrary non zero real numbers. Also, we discuss and illustrate the stability of the solutions in the neighborhood of the critical points and the periodicity of the considered equations. Keywords: equilibrium point, recursive sequences, periodicity. Mathematics Subject Classification: 39A10.
– – – – – – – – – – – – – – – – – – – – – –
1. INTRODUCTION In recent years, the qualitative study of di erence equations has become an active research area among a considerable number of mathematicians. Some economical and biological examples can be seen in [9,36,40,47,48,54]. It is commonly known that nonlinear di erence equations are able to produce and present sophisticated behaviors regardless their orders. Some articles show that a great e ort has been done to demonstrate and explore the dynamics of nonlinear di erence equations (see [40]-[61]). In fact, investigating these equations is a challenge and still new in the mathematical world. It is strongly believed that the rational di erence equations are signiÞcant in their own right. Abo-Zeid and Cinar [1] illustrated the global stability, cyclical behavior, oscillation of all acceptable solutions of the equation 1 +1 = 2 In [7], [8] Cinar considered the solutions of the equations +1
=
1
1+
+1
1
=
1
1+
1
A. El-Moneam, and Alamoudy [16] examined the positive solutions of the equation in terms of its periodicity, boundedness and the global stability. The considered di erence equation is given by +1
=
+
1+
2+
3+
4
1+
2+
3+
4
Khatibzadeh and Ibrahim [42] studied the boundedness, asymptotic stability, oscillatory behavior and discovered the closed form of solutions of the equation
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J. COMPUTATIONAL ANALYSIS AND APPLICATIONS, VOL. 24, NO.8, 2018, COPYRIGHT 2018 EUDOXUS PRESS, LLC
+1
=
1
+
+
1
Simsek et al. [49] has found and explored solutions for the recursive formula 3
=
+1
1+
1
For other related papers, see [25— 46]. We analyze and explore the solutions of the following nonlinear recursive equation +1
5
=
4(±1
±
=0 1
5)
(1)
with conditions posed on the initial values are arbitrary non zero real numbers. Also, we will survey some dynamic behaviors of its solutions. The linearized equation of equation +1
about the equilibrium
= (
)
1
=0 1
(2)
is the linear di erence equation +1
=
P
(
)
=0
Theorem A [43]: Assume that =1 2 and {0 1 2 condition for the asymptotic stability of the di erence equation +
+
1
+
1
+
+
=0
2. THE FIRST EQUATION
+1
P
}. Then
=1 |
|
1 is a su cient
=0 1 5
=
4 (1+
5)
This section is devoted to give a speciÞc solution of the Þrst di erence equation which is +1
Theorem 2.1. Let { 10
5
=
10
3
=
10
1
=
10 +1
=
}
5
=
4 (1
=
¶ Y1 µ 1 + (10 ) 1 + (10 + 5) =0 µ 1 Y 1 + (10 + 2) ¶ 1 + (10 + 7) =0 µ 1 Y 1 + (10 + 4) ¶ 1 + (10 + 9) =0 Y1 µ 1 + (10 + 6) (1 + ) 1 + (10 + 11)
where we put
(1 + 3 5
=
(3)
5)
be a solution of Eq.(3). Then
= 5
Y1 µ 1 + (10 + 1) ¶ 1 + (10 + 6) =0 µ 1 Y 1 + (10 + 3) ¶ 10 2 = 1 + (10 + 8) =0 µ 1 Y 1 + (10 + 5) ¶ = 10 1 + (10 + 10) =0 ¶ Y1 µ 1 + (10 + 7) = 10 +2 (1 + 2 ) 1 + (10 + 12) 10
=0
10 +3
+
¶
=0
=
3
=
=
=0
Y1 µ 1 + (10 + 8) ) 1 + (10 + 13) 4
4
2
=
1518
10 +4
1
=
=
(1 + 4 0
Y1 µ 1 + (10 + 9) ) 1 + (10 + 14) =0
¶ ¶
=
E. M. Elsayed et al 1517-1531
J. COMPUTATIONAL ANALYSIS AND APPLICATIONS, VOL. 24, NO.8, 2018, COPYRIGHT 2018 EUDOXUS PRESS, LLC
Proof: The result holds for
10
10
10
15
=
13
=
11
=
10
9
=
10
7
=
= 0. Assume that
Y2 µ 1 + (10 ) 1 + (10 + 5) =0 µ 2 Y 1 + (10 + 2) 1 + (10 + 7) =0 µ 2 Y 1 + (10 + 4) 1 + (10 + 9) =0
0 and our assumption true for
¶
1. Then;
Y2 µ 1 + (10 + 1) ¶ 1 + (10 + 6) =0 µ 2 Y 1 + (10 + 3) ¶ = 10 12 1 + (10 + 8) =0 ¶ 2 Y µ 1 + (10 + 5) = 10 10 1 + (10 + 10) =0
10
¶
¶
Y2 µ 1 + (10 + 6) ¶ (1 + ) =0 1 + (10 + 11) Y2 µ 1 + (10 + 8) ¶ (1 + 3 ) =0 1 + (10 + 13)
10
10
14
=
Y2 µ 1 + (10 + 7) ¶ 8 = (1 + 2 ) =0 1 + (10 + 12) Y2 µ 1 + (10 + 9) ¶ 6 = (1 + 4 ) =0 1 + (10 + 14)
Now, it follows from Eq.(3) that 10
5
10
=
10(1
10
+
6 10
11
10
6 10
(1+4
=
=
=
=
=
"
Y2 ³
)
11) Y2 ³
# "=0 ´ 1+
1+(10 +5) 1+(10 +10)
=0
´ Y2 ³ 1+(10 +4)
1+(10 +9) 1+(10 +14)
1+(10 +9)
=0
(1+4
)
Y2 ³
1+(10 +9) 1+(10 +14)
=0
´
´ Y2 ³ ´ 1+(10 +4) 1+(10 +9)
=0
¶ Y2 µ 1 + (10 + 4) (1 + 4 ) =0 1 + (10 + 14) " 2µ #" Y 1 + (10 + 5) ¶ Y2 µ 1 + (10 + 4) 1+ 1 + (10 + 10) (1 + 4 ) =0 1 + (10 + 14) =0 µ ¶ " "
1 + (10 6) # Y2 µ 1 + (10 + 5) ¶ 1+ 1 + (10 + 10) 1 + (10
6)
=0
Y2 µ 1 + (10 + 5) 1 + (10 + 10) =0
[1 + (10
¶#
[1 + (10
6)
Y2 µ 1 + (10 + 10) ] 1 + (10 + 5)
5)
=0
+
¶
=
#
¶#
¸ ]
¶ Y1 µ 1 + (10 ) 1 + (10 + 5) =0
Also, we have 10
4
10
= 10
9 (1
1519
+
5 10
10
10
5 10
10)
E. M. Elsayed et al 1517-1531
J. COMPUTATIONAL ANALYSIS AND APPLICATIONS, VOL. 24, NO.8, 2018, COPYRIGHT 2018 EUDOXUS PRESS, LLC
= "
Y1 ³
1+(10 ) 1+(10 +5)
=0
(1+
Y2 ³
)
1+(10 +6) 1+(10 +11)
=0
"
=
(1+
)
Y2 ³ =0
#" ´
´ Y2 ³
1+(10 +5) 1+(10 +10)
=0
1+
µ
Y1 ³
´ Y2 ³
1+(10 ) 1+(10 +5)
=0
1 + (10 5) # ´ 1+(10 +6) 1+ 1+(10 +11)
1+(10 +5) 1+(10 +10)
=0
¶
1
´
1 + (10
´
#
¸
5)
"
# Y2 µ 1 + (10 + 6) ¶ [1 + (10 5) + ] (1 + ) =0 1 + (10 + 11) Y2 µ 1 + (10 + 11) ¶ Y1 µ 1 + (10 + 1) (1 + ) = [1 + (10 4) ] =0 1 + (10 + 6) 1 + (10 + 6) =0
=
=
¶
Similarly 10
3
10
= 10
=
=
"
"
=
=
=
(1+2
)
8(1
"
1 + (10 + 6) #" 2 ³ Y 1+(10 +7) ´ Y1 µ 1 + 1+(10 +12)
"
(1+2
)
=0
Y2 ³ =0
1 (1+2
)
Y2 ³
µ
9)
1+(10 +7) 1+(10 +12)
# ´
1+(10 +7) 1+(10 +12)
# ´
)
Y2 ³ =0
[1 + (10
(1 + (10 + 6)
=0
1 + (10
4)
3)
]
(1 + (10 + 6)
=0
)
#
¶¸
4)
+
#
)
¶ Y2
1 + (10 + 6)
= [1 + (10
´
=0
¶
1 + (10 4) # µ ´ 1+(10 +7) 1+ 1+(10 +12)
=0
1 (1+2
4 10
¶ Y2
=0
"
9
10
=0
Y1 µ =0
)
4 10
Y1 µ 1 + (10 + 1) ¶ Y2 µ 1 + (10 + 6) ¶ (1+ ) 1 + (10 + 6) 1 + (10 + 11) =0 =0 # " 2 1 ³ ´ ³ Y 1+(10 +7) Y 1+(10 +1) ´ Y2 ³ 1+(10 +6) 1 + 1+(10 +12) 1+(10 +6) (1+ ) 1+(10 +11) =0
(1+2
+
] Y1 µ 1 + (10 + 2) 1 + (10 + 7) =0
¶
Similarly, one can simply find the other relations. Thus, the proof is done. Theorem 2.2. The unique equilibrium point of Eq.(3) is the number zero which is not locally asymptotically stable.
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E. M. Elsayed et al 1517-1531
J. COMPUTATIONAL ANALYSIS AND APPLICATIONS, VOL. 24, NO.8, 2018, COPYRIGHT 2018 EUDOXUS PRESS, LLC
Proof: The equilibrium points of Eq.(3) obtained by 2
¡
= 4
Arranging the previous equation gives Let
: (0
)3
(0
= 0 Thus
3
1+
=0
¢
) be a function takes the form (
)=
(1 +
)
Therefore (
)=
(
2
(1 +
)
)=
2 (1
+
(
)
)=
2
(1 +
)
So (
)=1
(
)=1
(
)=1
Then by using Theorem A the proof follows. Example 1. We assume
5
=6
4
= 11
3
=3
2
=2
1
=18
0
=
7. See Fig. 1.
plot of x(n+1)=x(n)x(n 5)/(x(n 4)(1+x(n)x(n 5)) 12 10 8 6 4 2 0 2 4 6 8
0
10
20
30
40
50 n
60
70
80
90
100
Figure 1. Example 2. See Fig. 2, since
5
=16
4
=12
1521
3
=
3
2
= 7
1
=18
0
= 3.
E. M. Elsayed et al 1517-1531
J. COMPUTATIONAL ANALYSIS AND APPLICATIONS, VOL. 24, NO.8, 2018, COPYRIGHT 2018 EUDOXUS PRESS, LLC
plot of x(n+1)=x(n)x(n 5)/(x(n 4)(1+x(n)x(n 5)) 3
2
1
0
1
2
3
0
10
20
30
40
50 n
60
70
80
90
100
Figure 2.
3. THE SECOND EQUATION
+1
=
5 4(
1+
5)
This section is devoted to obtain the solution of the di erence equation which is +1
where
0
5
6= 1
Theorem 3.1. Let {
}
= 5
4(
1+
(4)
5)
be a solution of Eq.(4). Then for
10
5
=
10
3
=
10 1
=
10 +1
=
10 +3
=
Proof: The result holds for
5
=
10
4
( 1+
)
( 1+
)
10
( 1+
)
10
( 1+
)
+1
( 1+
)
+1
= 0. Assume that
= ( 1+ 2
= ( 1+
= ( 1+
10 +2
)
=
10 +4
=
) )
( 1+
)
( 1+
)
0 and that our assumption true for
1522
1. Then;
E. M. Elsayed et al 1517-1531
J. COMPUTATIONAL ANALYSIS AND APPLICATIONS, VOL. 24, NO.8, 2018, COPYRIGHT 2018 EUDOXUS PRESS, LLC
10
15
=
10
13
=
10 11
=
10
9
=
10
7
=
10
14
= ( 1+
1
)
( 1+
)
1
)
1
10
12
= ( 1+
)
1
( 1+
)
1
10
10
= ( 1+
)
1
( 1+
( 1+
1
( 1+
10
)
( 1+
10
)
=
8
6
)
( 1+
=
1
)
It follows from (4) that 10
5
10
=
10(
10
10
4
6 10
1+
11
10
10
= 10
9(
= ( 1+
5 10
1+
=
( 1+ ( 1+
)
1+
1[
)
1
1+
]
=
( 1+
)
10)
5 10
) ( 1+ )
)
( 1+
10
10
( 1+ ) ¸ 1+
= ( 1+
11)
6 10
( 1+
)
)
( 1+
)
1
¸
( 1+ ) = ( 1+ [1 + ]
¸=
)
Similarly one can simply prove the other relations. Theorem 3.2. Eq.(4) has a period ten solution i ½
= 2 and will be in the following form ¾
Proof: Firstly, assume that there exists a period ten solution ½
¾
of Eq.(4). Then, we can notice from the solution of Eq.(4) that
= = =
( 1+
)
( 1+ ) ( 1+ )
= ( 1+
)
= ( 1+ =
=
( 1+
)
( 1+
=
)
+1
= ( 1+
)
( 1+
) =
)
+1
( 1+
)
or, ( 1+
1523
) =1
E. M. Elsayed et al 1517-1531
J. COMPUTATIONAL ANALYSIS AND APPLICATIONS, VOL. 24, NO.8, 2018, COPYRIGHT 2018 EUDOXUS PRESS, LLC
Then =2 Secondly, suppose that
10
5 10
= 2 Then, it is easily seen from the solution of Eq.(4) that
=
10
4
=
=
10 +1
=
10
3
=
10
10 +2
2
=
=
10 1
10 +3
=
=
10 +4
=
Thus, the periodic solution of period ten is obtained and this proves the theorem. 3
Theorem 3.3. Eq.(4) has two equilibrium points which are 0 asymptotically stable.
2 and these equilibrium points are not locally
Proof: The equilibrium points of Eq.(4) can be written in the following form 2
¡
= Arranging this gives 2
¡
1+
2
¢
2
1+
2
=
2
Therefore, the fixed points are 0 ± 2 Let
: (0
)3
(0
¢
¡
¢ 2 =0
2
) be a function defined by (
)=
( 1+
)
Then it follows that (
) =
(
) =
(
) =
( 1+ 2(
2
)
1+
( 1+
) )2
It can be seen that (
)=
1
(
) = ±1
(
)=
1
Then by using Theorem A the proof follows. Example 3. We consider
5
= 8
4
=17
3
= 3
1524
2
=2
1
=18
0
= 7 See Fig. 3.
E. M. Elsayed et al 1517-1531
J. COMPUTATIONAL ANALYSIS AND APPLICATIONS, VOL. 24, NO.8, 2018, COPYRIGHT 2018 EUDOXUS PRESS, LLC
plot of x(n+1)=x(n)x(n 5)/(x(n 4)( 1+x(n)x(n 5)) 50
40
30
20
10
0
10
20
30
0
5
10
15
20
25 n
30
35
=2
1
40
45
50
Figure 3. Example 4. See Fig. 4, since
5
=8
4
=17
3
= 3
2
=18
0
= 1 4.
plot of x(n+1)=x(n)x(n 5)/(x(n 4)( 1+x(n)x(n 5)) 8
7
6
5
4
3
2
1
0
0
5
10
15
20
25 n
30
35
40
45
50
Figure 4.
4. THE THIRD EQUATION
+1
=
5 4 (1
5)
In this section we will obtain and present the solution of the third di erence equation which is +1
=
5 4 (1
1525
(5)
5)
E. M. Elsayed et al 1517-1531
J. COMPUTATIONAL ANALYSIS AND APPLICATIONS, VOL. 24, NO.8, 2018, COPYRIGHT 2018 EUDOXUS PRESS, LLC
Theorem 4.1. Let {
}
10
5
=
10
3
=
10
1
=
10 +1
=
10 +2
=
10 +3
=
10 +4
=
= 5
be a solution of Eq.(5). Then for
=0 1
¶ Y1 µ 1 (10 ) Y1 µ 1 (10 + 1) ¶ = 10 4 1 (10 + 5) 1 (10 + 6) =0 =0 µ ¶ µ Y1 1 (10 + 2) Y1 1 (10 + 3) ¶ = 10 2 1 (10 + 7) 1 (10 + 8) =0 =0 µ ¶ µ ¶ Y1 1 (10 + 4) Y1 1 (10 + 5) 10 = 1 (10 + 9) 1 (10 + 10) =0 =0 Y1 µ 1 (10 + 6) ¶ (1 ) =0 1 (10 + 11) Y1 µ 1 (10 + 7) ¶ (1 2 ) =0 1 (10 + 12) Y1 µ 1 (10 + 8) ¶ (1 3 ) =0 1 (10 + 13) Y1 µ 1 (10 + 9) ¶ (1 4 ) =0 1 (10 + 14)
Theorem 4.2. The unique critical point of Eq.(5) is the number zero which is not locally asymptotically stable. Example 5. Suppose that
5
=8
4
=17
3
= 3
2
=2
1
=18
0
= 1 4 see Fig. 5.
plot of x(n+1)=x(n)x(n 5)/(x(n 4)(1 x(n)x(n 5)) 8
6
4
2
0
2
4
0
5
10
15
20
25 n
30
35
40
45
50
Figure 5. Example 6. See Fig. 6 since
5
=
7
4
=15
3
1526
=
3
2
=2
1
= 12
0
= 4.
E. M. Elsayed et al 1517-1531
J. COMPUTATIONAL ANALYSIS AND APPLICATIONS, VOL. 24, NO.8, 2018, COPYRIGHT 2018 EUDOXUS PRESS, LLC
plot of x(n+1)=x(n)x(n 5)/(x(n 4)(1 x(n)x(n 5)) 12 10 8 6 4 2 0 2 4 6 8
0
5
10
15
20
25 n
30
35
40
45
50
Figure 6.
5. THE FOURTH EQUATION
+1
5
=
4(
1
5)
Now, we will explore and discover the solution of the following di erence equation +1
where
5 0
6=
=
5 4(
1
=0 1
5)
(6)
1
Theorem 5.1. Let { } = 5 be a solution of Eq.(6). Then Eq.(6) has unboundedness solution (except in the case if = 2) and for = 0 1 10
5
=
10
3
=
10 1
=
10 +1
=
10 +3
=
10
4
( 1
)
( 1
)
10
( 1
)
10
( 1
)
+1
( 1
)
+1
= ( 1 2
= ( 1
= ( 1
10 +2
)
=
10 +4
=
Theorem 5.2. Eq.(6) has a periodic ½ ¾ solution of period ten i
=
Example 7. Consider
1
) )
( 1
)
( 1
)
2 and written in the following form
Theorem 5.3. The unique equilibrium of Eq.(6) is the number zero which is not locally asymptotically stable. 5
=
7
4
=15
3
=
3
1527
2
=2
= 12
0
= 4 see Fig. 7.
E. M. Elsayed et al 1517-1531
J. COMPUTATIONAL ANALYSIS AND APPLICATIONS, VOL. 24, NO.8, 2018, COPYRIGHT 2018 EUDOXUS PRESS, LLC
6
5
plot of x(n+1)=x(n)x(n 5)/(x(n 4)( 1 x(n)x(n 5))
x 10
4
3
2
1
0
1
2
0
5
10
15
20
25 n
30
35
40
45
50
Figure 7. Example 8. Fig. 8 illustrates the solutions when 2 7.
5
=
7
4
=15
3
=
3
2
=2
1
= 12
plot of x(n+1)=x(n)x(n 5)/(x(n 4)( 1 x(n)x(n 5)) 12 10 8 6 4 2 0 2 4 6 8
0
5
10
15
20
25 n
30
35
40
45
50
Figure 8.
1528
E. M. Elsayed et al 1517-1531
0
=
J. COMPUTATIONAL ANALYSIS AND APPLICATIONS, VOL. 24, NO.8, 2018, COPYRIGHT 2018 EUDOXUS PRESS, LLC
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+1
1
=
Paran. Mat. (3s.), 31 (1) (2013), 43— 49. 2. A. M. Ahmed, and N. A. Eshtewy, Basin of attraction of the recursive sequence 3. 4. 5. 6.
Bol. Soc. 2
+1
=
+ +
+ +
1 1
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+1
=
1
1+
Appl. Math. Comp., 1
158(3) (2004), 809-812. 8. C. Cinar, On the positive solutions of the di erence equation
+1
=
1
1+
Appl. Math. Comp., 1
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+1
=
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