Download Global Behavior of a Higher Order Rational Difference Equation

Filomat 30:12 (2016), 3265–3276
DOI 10.2298/FIL1612265A
Published by Faculty of Sciences and Mathematics,
University of Niš, Serbia
Available at: http://www.pmf.ni.ac.rs/filomat
Global Behavior of a Higher Order Rational Difference Equation
R. Abo-Zeida
a Department
of Basic Science, The Egyptian Academy for Engineering and Advanced Technology, Cairo, Egypt
Abstract.
In this paper, we derive the forbidden set and discuss the global behavior of all solutions of the difference
equation
Axn−k
xn+1 =
,
n = 0, 1, . . .
Q
B − C ki=0 xn−i
where A, B, C are positive real numbers and the initial conditions x−k , ..., x−1 , x0 are real numbers.
1. Introduction
No one can deny that, Difference equations have played an important role in analysis of mathematical
models of biology, physics and engineering. Recently, there has been a great interest in studying properties
of nonlinear and rational difference equations. One can see [1]-[25] and the references therein.
In [4], M. Aloqeili discussed the stability properties and semicycle behavior of the solutions of the difference equation
xn+1 =
xn−1
,
a − xn xn−1
n = 0, 1, . . .
(1)
with real initial conditions and positive real number a.
In [23], D. Simsek et al. introduced the solution of the difference equation
xn+1 =
xn−5
,
1 + xn−1 xn−3
n = 0, 1, ...
with positive initial conditions.
In [11], E.M. Elsayed discussed the solutions of the difference equation
xn+1 =
xn−5
,
−1 + xn−2 xn−5
n = 0, 1, ...
2010 Mathematics Subject Classification. Primary 39A10
Keywords. (Difference Equation, Periodic Solution, Unbounded Solutions)
Received: 25 September 2014; Accepted: 15 May 2015
Communicated by Svetlana Janković
Email address: abuzead73@yahoo.com (R. Abo-Zeid)
R. Abo-Zeid / Filomat 30:12 (2016), 3265–3276
3266
where the initial conditions are nonzero real numbers with x−5 x−2 , 1, x−4 x−1 , 1 and x−3 x0 , 1.
He also in [9], determined the solutions to some difference equations. He obtained the solution to the
difference equation
xn−3
xn+1 =
, n = 0, 1, ...
1 + xn−1 xn−3
where the initial conditions are nonzero positive real numbers.
R. Karatas et al. [15] discussed the positive solutions and the attractivity of the difference equation
xn+1 =
xn−5
,
1 + xn−2 xn−5
n = 0, 1, ...
where the initial conditions are nonnegative real numbers.
The authors in [14], discussed the solutions and attractivity of the difference equation
axn−(2k+2)
,
Q
−a + 2k+2
i=0 xn−i
xn+1 =
n = 0, 1, ...
where a, x−(2k−2) , ..., x0 are real numbers such that x−(2k−2) x−(2k−1) ...x0 , a and k is a nonnegative integer.
Elabbasy et al. [8] determined and discussed the solutions of the difference equation
xn+1 =
αxn−k
,
Q
β + γ ki=0 xn−i
n = 0, 1, ...
with nonnegative real numbers α,β, γ, positive real initial conditions and positive integer k.
In [16], we investigated the behavior and periodic nature of the two difference equations
xn+1 =
xn−2
,
±1 + xn xn−1 xn−2
n = 0, 1, . . .
In [2], we have also discussed the oscillation, periodicity, boundedness and the global behavior of all
admissible solutions of the difference equation
xn+1 =
Axn−2r−1
,
Q
B − C ki=l xn−2i
n = 0, 1, . . .
where A, B, C are positive real numbers and l, r, k are nonnegative integers, such that l ≤ k.
Also in [1], we discussed the global stability of all solutions of the difference equation
xn+1 =
Axn−2
,
B + Cxn xn−1 xn−2
n = 0, 1, . . .
where A, B, C are positive real numbers and the initial conditions x−2 , x−1 , x0 are real numbers.
In this paper, we discuss the global behavior of all solutions of the difference equation
xn+1 =
Axn−k
,
Q
B − C ki=0 xn−i
n = 0, 1, . . .
(2)
where A, B, C are positive real numbers and the initial conditions x−k , ..., x−1 , x0 are real numbers. The
difference equation (2) is a more general case of the difference equation (1).
2. Linearized Stability and Solutions of Equation (2)
In this section we introduce an explicit formula for the solutions of the difference equation (2) and study
its linearized stability.
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It is convenient to reduce the parameters on which equation (2) depends on.
q
The change of variables k+1 AC xn = yn reduces equation (2) to the equation
yn+1 =
p−
yn−k
Qk
i=0
yn−i
,
n = 0, 1, . . .
(3)
where p = AB .
We will deal with equation (3) rather than equation (2).
To start navigating the global behavior of the difference equation (3), we classify the nontrivial solutions
of equation (3) into two types of solutions:
• Solutions with initial points (y−k , y−k+1 , ..., y0 ) such that y−i = 0, for some but not all i ∈ {0, 1, ..., k}.
• Solutions with initial points (y−k , y−k+1 , ..., y0 ) such that y−i , 0, for all i ∈ {0, 1, ..., k}.
These two types of solutions exhibit a global behavior different from each other.
Theorem 2.1. Let y−k , y−k+1 , ..., y−1 and y0 be real numbers such that y−i = 0 for some but not all i ∈ {0, 1, ..., k}.
Then the solution {yn }∞
of equation (3) is
n=−k


















yn = 
















( p1 ) k+1 +1 y−k
n−1
( p1 )
n−2
k+1 +1
, n = 1, k + 2, 2k + 3, ...
, n = 2, k + 3, 2k + 4, ...
y−k+1
.....
.....
.....
(4)
( p1 ) k+1 +1 y−1
n−k
n−k−1
( 1p ) k+1 +1 y0
, n = k, 2k + 1, 3k + 2, ...
, n = k + 1, 2k + 2, 3k + 3, ...
Proof. Let {yn }∞
be a solution of equation (3) such that y−i = 0 for some but not all i ∈ {0, 1, ..., k}. Using equation
n=−k
(3), we can write
Qk
k
Y
l=0 yn−l
yn+1−l =
, n = 0, 1, ....
Qk
p − l=0 yn−l
l=0
Q
Q
But as kl=0 y−l = 0, we get kl=0 yn−l = 0 for all n ≥ 1.
It follows that
yn−k
yn−k
yn+1 =
=
Qk
p
p−
yn−l
l=0
for all n ≥ 0, from which the result follows.
Now suppose that y−i , 0, for all i ∈ {0, 1, ..., k}. From equation (3) and using the substitution tn =
1
yn yn−1 ...yn−k , we can obtain the linear nonhomogeneous difference equation
tn+1 = ptn − 1,
t0 =
1
.
y0 y−1 ...y−k
It is clear that the mapping h(x) = px − 1 is invertible and its inverse is h−1 (x) = 1p x + p1 .
We try to deduce the forbidden set of equation (3).
(5)
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For,
suppose that we start from an initial point (y−k , y−k+1 , ..., y0 ) such that y0 y−1 ...y−k = p.
The backward orbits, vn = yn yn−11...yn−k satisfy the equation
vn = h−1 (vn−1 ) =
then we obtain vn =
That is
1
yn yn−1 ...yn−k
p
yn yn−1 ...yn−k = Pn ( 1 )l .
l=0 p
1
1
vn−1 +
p
p
= h−n (v0 ) =
1
p
v0 =
with
1
1
= ,
y0 y−1 ...y−k
p
Pn
1 l
l=0 ( p ) .
On the other hand, we can observe that if we start from an initial point (y−k , ..., y−1 , y0 ) such that y0 y−1 ...y−k =
p
Pn0 1 l for some n0 ∈ N, then according to equation (5) we obtain
( )
l=0 p
tn0 =
1
yn0 yn0 −1 ...yn0 −k
=
1
.
p
This implies that p − yn0 yn0 −1 ...yn0 −k = 0.
Therefore, yn0 +1 is undefined.
These observations lead us to conclude the following result.
Proposition 2.2. The forbidden set F of equation (3) is
F=
∞
[
{(u0 , u1 , ..., uk ) :
n=0
k
Y
ui = Pn
p
1 l
l=0 ( p )
i=0
}.
Theorem 2.3. Let y−k , y−k+1 , ..., y−1 and y0 be real numbers such that α = y0 y−1 ...y−k ,
p
Pn
1 l
l=0 ( p )
for any n ∈ N.
Then the solution of equation (3) is





















yn = 



















Q n−1
k+1
p(k+1) j −α
P(k+1) j−1
pl
l=0
j=0 p(k+1) j+1 −α P(k+1)j pl
l=0
P(k+1)
j
Q n−2
p(k+1)j+1 −α l=0 pl
k+1
y−k+1 j=0 (k+1) j+2 P(k+1) j+1 l
p
p
−α
y−k
, n = 1, k + 2, 2k + 3, ...
, n = 2, k + 3, 2k + 4, ...
l=0
y−1
.....
.....
.....
P(k+1) j+k−2 l
Q n−k
p(k+1) j+k−1 −α l=0
p
k+1
j=0 p(k+1)j+k −α P(k+1)j+k−1 pl
Pl=0
(k+1) j+k−1 l
Q n−k−1
p(k+1) j+k −α l=0
p
k+1
y0 j=0
P(k+1) j+k l
(k+1)
j+k+1
p
−α
p
(6)
, n = k, 2k + 1, 3k + 2, ...
, n = k + 1, 2k + 2, 3k + 3, ...
l=0
Proof. Let y−k , y−k+1 , ..., y−1 and y0 be real numbers such that α = y0 y−1 ...y−k ,
The solution of the linear nonhomogeneous difference equation (5) is
tn+1 = pn+1 t0 −
n
X
pr ,
r=0
t0 =
1
.
y0 y−1 ...y−k
If we set α = y0 y−1 ...y−k , then we can write
k
Y
l=0
yn+1−l =
pn+1
α
.
P
− α nr=0 pr
p
Pn
1 l
l=0 ( p )
for any n ∈ N.
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It follows that
Qk
l=0 yn+1−l
Qk
l=0 yn−l
P
r
pn − α n−1
r=0 p
= n+1
Pn r .
p − α r=0 p
This implies that
yn+1
P
r
pn − α n−1
r=0 p
= yn−k n+1
Pn r ,
p − α r=0 p
from which we can write the form (6).
1
Corollary 2.4. Assume that p = 1 and α = y0 y−1 ...y−k , n+1
for any n ∈ N. Then the solution of equation (3) is

Q n−1
1−(k+1)jα

k+1

, n = 1, k + 2, 2k + 3, ...
y−k j=0


1−((k+1)j+1)α


n−2
Q

1−((k+1)j+1)α

k+1

y−k+1 j=0
, n = 2, k + 3, 2k + 4, ...

1−((k+1)j+2)α




.....



.....
(7)
yn = 



.....




Q n−k

k+1 1−((k+1) j+k−1)α

y−1 j=0
, n = k, 2k + 1, 3k + 2, ...


1−((k+1) j+k)α


n−k−1
Q

1−((k+1)
j+k)α

k+1

, n = k + 1, 2k + 2, 3k + 3, ...
y0 j=0
1−((k+1) j+k+1)α
P
P
Proof. It is sufficient to note that, nr=0 pr = nr=0 ( p1 )r = n + 1 when p = 1.
Using this fact, the solution form (6) reduced to the form (7) and the result follows.
Corollary 2.5. Assume that p < 1 and let {yn }∞
be a nontrivial solution of equation (3). If α = y0 y−1 ...y−k = 0,
n=−k
then the solution {yn }∞
is
unbounded.
n=−k
Example (1) Figure 1. shows that if {yn }∞
is a solution of the equation
n=−2
yn+1 =
yn−2
,
0.5 − yn yn−1 yn−2
n = 0, 1, . . .
with initial conditions y−2 = 2, y−1 = 0, y0 = 1 (α = 0) where k = 2 and p = 0.5, then the solution {yn }∞
is
n=−2
unbounded.
4 ´ 109
3 ´ 109
2 ´ 109
1 ´ 109
0
0
20
40
60
80
Figure 1: The difference equation yn+1 =
yn−2
0.5−yn yn−1 yn−2
We end this section with the discussion of the local stability of the equilibrium points of equation (3).
It is clear that the equilibrium point ȳ = 0 is always an equilibrium point of equation (3) and the nonzero
equilibrium points depend on whether k is even or odd.
R. Abo-Zeid / Filomat 30:12 (2016), 3265–3276
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p
When k is odd, we have the nonzero equilibrium points ȳ = ± k+1 p − 1 if p > 1.
p
When k is even, we have the nonzero equilibrium point ȳ = k+1 p − 1, p , 1.
Lemma 2.6. Assume that P(x) is the polynomial
xk + xk−1 + ... + x + 1.
Then the zeros of P(x) are of modulus one.
The following theorem describes the local behavior of the equilibrium points.
Theorem 2.7. The following statements are true.
1. The equilibrium point ȳ = 0 is locally asymptotically stable if p > 1 and unstable if p < 1.
p
2. If k is even, then ȳ = k+1 p − 1 is unstable if p > 1 and nonhyperbolic if p < 1.
p
3. If k is odd, then the equilibrium points ȳ = ± k+1 p − 1 are unstable equilibrium points.
Proof. The linearized equation associated with equation (3) about an equilibrium point ȳ is
zn+1 −
k−1
X
p
ȳk+1
zn−k = 0
zn−i −
k+1
2
(p − ȳ ) i=0
(p − ȳk+1 )2
, n = 0, 1, 2, . . .
(8)
Its characteristic equation associated with this equation is
λk+1 −
k−1
X
ȳk+1
p
λk−i −
= 0.
k+1
2
(p − ȳ ) i=0
(p − ȳk+1 )2
(9)
Therefore, (1) follows directly.
Equation (8) about a nonzero equilibrium point ȳ is
zn+1 − (p − 1)
k−1
X
zn−i − pzn−k = 0
, n = 0, 1, 2, . . .
(10)
i=0
Also equation (9) becomes
λk+1 − (p − 1)
k−1
X
λk−i − p = 0.
(11)
i=0
Let
f (λ) = λk+1 − (p − 1)
k−1
X
λk−i − p.
i=0
We can see that
f (λ) = (λ − p)
k
X
λl = (λ − p)P(λ).
l=0
Then the roots of equation (11) are the zeros of f (λ). Using lemma (2.6), we see that, the roots of equation (11) are p
and k other roots with modulus 1.
Therefore, (2) and (3) follow directly.
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3. Global Behavior of Equation (3)
The solution of equation (3) can be written as
y(k+1)m+i
But as
P(k+1) j+i−2 l
m
Y
p(k+1) j+i−1 − α l=0
p
= y−(k+1)+i
,
P
(k+1)
j+i−1
(k+1) j+i − α
l
p
j=0 p
l=0
i = 1, 2, ..., k + 1
P(k+1)j+i−2 l
p(k+1) j+i−1 − α l=0
p
p(k+1) j+i−1 µ − α
=
,
P
(k+1)j+i−1 l
p(k+1) j+i µ − α
p(k+1) j+i − α l=0
p
We can write
y(k+1)m+i = y−(k+1)+i
m
Y
βi (j),
and m = 0, 1, ...
(12)
where µ = 1 − p + α.
i = 1, 2, ..., k + 1
and m = 0, 1, ...
j=0
where
βi ( j) =
p(k+1) j+i−1 µ − α
p(k+1) j+i µ − α
,
i = 1, 2, ..., k + 1.
Theorem 3.1. Assume that {yn }∞
is a solution of equation (3) such that α ,
n=−k
then
{yn }∞
n=−k
is a periodic solution with period k + 1.
p
Pn
1 l
l=0 ( p )
for any n ∈ N. If α = p − 1,
Proof. It is sufficient to see that if α = p − 1, then µ = 0. Therefore,
y(k+1)m+i = y−(k+1)+i
m
Y
p(k+1) j+i−1 µ − α
j=0
p(k+1)j+i µ − α
Proposition 3.2. Assume that p < 1 and let α ,
p
Pn
1 l
l=0 ( p )
= y−(k+1)+i ,
i = 1, 2, ..., k + 1.
for any n ∈ N. Then there exists j0 ∈ N such that βi ( j) > 0
for all j ≥ j0 .
Proof. We have three situations:
1. If α < p − 1 < 0, then 0 < µ − α < −α. Hence for each j ∈ N,
p(k+1)j+i−1 µ − α > µ − α > 0, i = 1, 2, ..., k + 1. Then
βi ( j) =
p(k+1) j+i−1 µ − α
p(k+1) j+i µ − α
2. If p − 1 < α < 0, then 0 < −α < µ − α.
But
lim βi (j) = lim
j→∞
j→∞
>0
for all
p(k+1) j+i−1 µ − α
p(k+1) j+i µ − α
j ≥ 0.
= 1.
Then there exists j0 ∈ N such that βi (j) > 0 for all j ≥ j0 .
3. When p − 1 < 0 < α, the situation is similar to that in (2).
In all cases there exists j0 ∈ N such that βi ( j) > 0 for all j ≥ j0 .
Theorem 3.3. Assume that {yn }∞
is a solution of equation (3) such that α , p − 1 and α ,
n=−k
Then the following statements are true.
p
Pn
1 l
l=0 ( p )
for any n ∈ N.
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1. If p > 1, then {yn }∞
converges to ȳ = 0.
n=−k
2. If p < 1 and α , 0, then {yn }∞
is bounded.
n=−k
Proof. Let {yn }∞
be a solution of equation (3) such that α ,
n=−k
The condition α , p − 1 ensures that the solution
{yn }∞
n=−k
p
Pn
1 l
l=0 ( p )
for any n ∈ N.
is not a (k + 1)-periodic solution.
1. Suppose that p > 1. It is clear that, as the equilibrium point
1
p−1
of equation (5) is repelling, every non-constant
solution of equation (5) approaches ∞ or −∞ according to the value of t0 = α1 .
We shall consider the following situations:
(a) If α = t10 < 0, then according to equation (5), we have
Qk
1
i=0 yn−i = tn < 0, for each n ∈ N. Therefore,
| yn+1 |=
| yn−k |
| yn−k |
<
,
Qk
p
| p − i=0 yn−i |
n = 0, 1, . . .
Q
< p − 1, then according to equation (5), ki=0 yn−i = t1n → 0 as n → ∞. Then there exists
Q
n0 ∈ N such that 0 < ki=0 yn−i < p − 1 for each n > n0 . Therefore,
(b) If 0 < α =
1
t0
| yn+1 |=
| yn−k |
<| yn−k |,
Q
| p − ki=0 yn−i |
n ≥ n0 .
(c) If p − 1 < α = t10 < p, then according to equation (5), there exists n0 ∈ N such that
for each n ≥ n0 . Therefore,
| yn+1 |=
(d) If α =
1
t0
| yn−k |
| yn−k |
<
,
Qk
p
| p − i=0 yn−i |
> p > 0, then according to equation (5),
| yn+1 |=
Qk
i=0
yn−i =
1
tn
| yn−k |
| yn−k |
<
,
Qk
p
| p − i=0 yn−i |
Qk
i=0
yn−i =
1
tn
<0
n ≥ n0 .
< 0 for each n > 0. Therefore,
n = 0, 1, . . .
In all cases, yn → 0 as n → ∞.
2. Suppose that p < 1. Using proposition (3.2), there exists j0 ∈ N such that βi ( j) > 0 for all j ≥ j0 . Hence for
each i ∈ {1, 2, ..., k + 1}, we have for large m
y(k+1)m+i = y−(k+1)+i
m
Y
βi ( j) = y−(k+1)+i
j=0
= y−(k+1)+i
βi ( j) exp(ln
j=0
= y−(k+1)+i
j=0
βi ( j)
j=0
j0 −1
Y
j0 −1
Y
j0 −1
Y
m
Y
βi ( j))
j=j0
βi ( j) exp (
m
X
ln βi (j)).
j= j0
P
It is sufficient to test the convergence of the series ∞
j= j0 | ln βi ( j) |.
But
ln βi ( j + 1) 0
lim
= .
j→∞
ln βi ( j)
0
m
Y
j= j0
βi ( j)
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Then
ln βi ( j + 1)
= lim
j→∞
j→∞
ln βi ( j)
lim
= lim
j→∞
d
d j (ln βi ( j
+ 1))
d
d j (ln βi ( j))
(p−1)(ln p)µ(k+1)p(k+1)(j+1)+i−1
(p(k+1)(j+1)+i−1 µ−α)(p(k+1)(j+1)+i µ−α)
(p−1)(ln p)µ(k+1)p(k+1) j+i−1
(p(k+1)j+i−1 µ−α)(p(k+1)j+i µ−α)
= pk+1 < 1.
P
It follows from D’ Alemberts’ test that the series ∞
j= j0 | ln βi ( j) | is convergent.
This ensures that the solution {yn }∞
is
bounded.
n=−k
Example (2) Figure 2. shows that if {yn }∞
is the solution of the equation
n=−2
yn+1 =
yn−2
,
2 − yn yn−1 yn−2
n = 0, 1, . . .
with initial conditions y−2 = 2, y−1 = 1, y0 = 2 (α , p − 1 and α ,
p
Pn
p = 2, then the solution {yn }∞
converges to zero.
n=−2
1 l
l=0 ( p )
for any n ∈ N) where k = 2 and
0.8
0.6
0.4
0.2
0.0
-0.2
-0.4
0
10
20
30
40
50
60
Figure 2: The difference equation yn+1 =
yn−2
2−yn yn−1 yn−2
We can observe in case p < 1 that, the behavior of the solution {yn }∞
is totally different according to
n=−k
whether α = 0 or α , 0. This is obvious in corollary (2.5) and theorem (3.3).
Theorem 3.4. Assume that p < 1 and let {yn }∞
be a solution of equation (3) such that α ,
n=−k
Then
{yn }∞
n=−k
p
Pn
1 l
l=0 ( p )
for any n ∈ N.
converges to a (k + 1)-periodic solution {ρ0 , ρ1 , ..., ρk } of equation (3) with ρ0 ρ1 ...ρk = p − 1.
Proof. By theorem (3.3), there exist k + 1 real numbers ρi ∈ R such that
lim y(k+1)m+i = ρi ,
j→∞
i ∈ {0, 1, ..., k}.
If we set n = (k + 1)m + i − 1, i = 0, 1, ..., k in equation (3), we get
y(k+1)m+i =
y(k+1)(m−1)+i
,
Qk
p − l=0 y(k+1)(m−1)+i−l+k
i = 0, 1, ..., k
and
m = 0, 1, ...
R. Abo-Zeid / Filomat 30:12 (2016), 3265–3276
3274
By taking the limit as m → ∞, we obtain
ρi =
p−
ρi
Qk
l=0
ρi−l+k
,
i = 0, 1, ..., k.
Q
But from equation (5) we have kl=0 yn−l = yn yn−1 ...yn−k = t1n → p − 1 as n → ∞.
Qk
This implies that i=0 y(k+1)m+i → ρ0 ρ1 ...ρk = p − 1 as m → ∞.
Therefore, {yn }∞
converges to the (k + 1)-periodic solution
n=−k
{..., ρ0 , ρ1 , ..., ρk−1 ,
p−1
p−1
, ρ0 , ρ1 , ..., ρk−1 ,
, ...}
ρ0 ρ1 ...ρk−1
ρ0 ρ1 ...ρk−1
Example (3) Figure 3. shows that if {yn }∞
is the solution of the equation
n=−3
yn+1 =
yn−3
,
0.8 − yn yn−1 yn−2 yn−3
n = 0, 1, . . .
with initial conditions y−3 = 2, y−2 = 2.6, y−1 = 0.2, y0 = 2.2 (α , 0 and α ,
k = 3 and p = 0.8, then the solution
is bounded.
Moreover, the solution {yn }∞
converges
to
4-periodic solution.
n=−2
{yn }∞
n=−2
p
Pn
1 l
l=0 ( p )
for any n ∈ N) where
2
1
0
-1
0
20
40
60
80
Figure 3: The difference equation yn+1 =
yn−3
0.8−yn yn−1 yn−2 yn−3
4. Case p = 1
We end this work with the discussion of the case p = 1.
If we set p = 1 in equation (12), we get
y(k+1)m+i = y−(k+1)+i
m
Y
ζi ( j),
i = 1, 2, ..., k + 1
and m = 0, 1, ...
j=0
where
ζi ( j) =
1 − α((k + 1)j + i − 1)
,
1 − α((k + 1)j + i)
i = 1, 2, ..., k + 1.
(13)
R. Abo-Zeid / Filomat 30:12 (2016), 3265–3276
Proposition 4.1. Assume that p = 1 and let α ,
for all j ≥ j0 .
3275
for any n ∈ N. Then there exists j0 ∈ N such that ζi ( j) > 0
1
n+1
Proof. When α < 0, the result is obvious where ζi ( j) > 0 for each j ∈ N.
When α > 0, It is sufficient to see that,
lim ζi (j) = lim
j→∞
j→∞
1 − α((k + 1)j + i − 1)
= 1.
1 − α((k + 1)j + i)
This implies that, there exists j0 ∈ N such that ζi (j) > 0 for all j ≥ j0 .
Theorem 4.2. Assume that p = 1. Then any solution {yn }∞
of equation (3) with α , 0 and α ,
n=−k
converges to zero.
1
n+1
for any n ∈ N
1
Proof. Let {yn }∞
be a solution of equation (3) such that α , n+1
for any n ∈ N.
n=−k
∞
The condition α , 0 ensures that the solution {yn }n=−k is not a (k + 1)-periodic solution.
Using proposition (4.1), there exists j0 ∈ N such that ζi ( j) > 0 for all j ≥ j0 . Hence for each i ∈ {1, 2, ..., k + 1}, we
have for large m
j0 −1
m
m
Y
Y
Y
y(k+1)m+i = y−(k+1)+i
ζi (j) = y−(k+1)+i
ζi ( j)
ζi ( j)
j=0
= y−(k+1)+i
j=0
j0 −1
Y
ζi (j) exp(ln
j=0
= y−(k+1)+i
j0 −1
Y
j0 −1
Y
ζi ( j) exp (
m
X
We shall show that
j= j0
ln ζi1( j) =
ln ζi ( j))
j= j0
ζi ( j) exp (−
j=0
P∞
ζi ( j))
j= j0
j=0
= y−(k+1)+i
m
Y
j= j0
m
X
ln
j= j0
1
).
ζi ( j)
1−α((k+1) j+i)
P∞
j= j0
ln 1−α((k+1) j+i−1) = ∞, by considering the series
α
j=j0 −1+α((k+1) j+i) .
P∞
But as
ln (1 − α((k + 1)j + i))/(1 − α((k + 1)j + i − 1))
= 1,
j→∞
α/(−1 + α((k + 1)j + i))
lim
using the limit comparison test, we get
Therefore,
P∞
j= j0
ln ζi1( j) = ∞.
y(k+1)m+i = y−(k+1)+i
j0 −1
Y
j=0
ζi ( j) exp (−
m
X
j= j0
ln
1
)
ζi ( j)
converges to zero as m → ∞.
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