GLOBAL BEHAVIOR OF SOLUTIONS OF
xn+1 =
max{xn , A}
xn xn−1
J. FEUER1, E. J. JANOWSKI2, G. LADAS2,3
and
C. TEIXEIRA2
Suggested running head:
GLOBAL BEHAVIOR OF SOLUTIONS OF xn+1 =
max{xn ,A}
xn xn−1
Contact author:
G. LADAS
1
Department of Mathematics and Computer Science, Merrimack College,
North Andover, MA. 01845, USA.
2
Department of Mathematics, University of Rhode Island, Kingston, R.I.
02881-0816, USA.
3
E Mail: [email protected]
Telephone: (401)874-5592
Fax: (401)874-4617
ABSTRACT
We investigate the asymptotic behavior, the oscillatory character
and the periodic nature of solutions of the difference equation
xn+1 =
max{xn , A}
,
xn xn−1
n = 0, 1, . . .
where A is a real parameter and the initial conditions are arbitrary
nonzero real numbers.
Key Words: asymptotic behavior, invariant, invariant region, periodic solutions, oscillatory solutions.
AMS 1990 Mathematics Subject Classification: 39A10
GLOBAL BEHAVIOR OF SOLUTIONS OF xn+1 =
max{xn ,A}
xn xn−1
3
1. INTRODUCTION
Our aim in this paper is to investigate the difference equation
xn+1 =
max{xn , A}
,
xn xn−1
n = 0, 1, . . .
(1.1.1)
where A is a real parameter and the initial conditions are nonzero
real numbers. In particular, we study the asymptotic behavior, the
oscillatory character, and the periodic nature of its solutions.
Observe that when A = 0, every nontrivial solution of Eq(1.1.1)
is periodic with prime period four.
It is interesting to note that when A > 0, the change of variables
xn =

1+yn

 A
eyn

 A−1+yn
if A > 1
if A=1
if A < 1
reduces Eq(1.1.1) to the piecewise linear difference equation
yn+1 =
1
1
| yn | − yn − yn−1 + δ,
2
2
where
δ=
n = 0, 1, . . .
(1.1.2)


 −2 if A > 1
0

 2
if A=1
if A < 1
which is a special case of the Lozi Map (see [7]).
We will show that when A=1, every positive solution of Eq(1.1.1)
is periodic with period 7.
When A is a positive real number, we will show that Eq(1.1.1)
possesses an invariant (see [3],[5]) and then use it to establish that
every positive solution is bounded and persists. No proof has yet
been found to show this result without using the invariant [6]. We will
exhibit invariant regions in the first quadrant [4] inside of which every
solution is periodic of period 3 or 4. This, in particular, implies that
the equilibrium of Eq(1.1.1) is stable but not asymptotically stable.
By using semicycle [1] analysis, we will show that every positive
GLOBAL BEHAVIOR OF SOLUTIONS OF xn+1 =
max{xn ,A}
xn xn−1
4
solution is strictly oscillatory about the positive equilibrium and that
a solution which lies outside of the invariant region achieves its upper
bound in every positive semicycle (except possibly the first). We
also give a detailed description of the solutions of Eq(1.1.1) which
are outside of the first quadrant.
When A is a negative real number, we will use a change of variables to show that the solutions of Eq(1.1.1) in the third quadrant
have the same behavior as the solutions of Eq(1.1.1) in the first quadrant when A > 0. Finally, we describe the behavior of the solutions
of Eq(1.1.1) in the remaining quadrants.
2. THE CASE OF POSITIVE A
In the following sections we will assume A is a positive real number.
2.1. THE CASE A = 1
Theorem 2.1.1 Assume A = 1. If x−1 , x0 ∈ (0, ∞), then every
nontrivial solution of Eq(1.1.1) is periodic with prime period 7.
Proof. Let x−1 = α and x0 = β. The proof is a consequence of the
following analysis:
Case 1: α ≥ 1, β ≥ 1 and α ≤ β. The solution is {α, β, α1 , αβ , β, αβ , β1 }.
Case 2:
Case 3:
Case 4:
Case 5:
Case 6:
Case 7:
2
α ≥ 1, β
α ≥ 1, β
α > 1, β
α < 1, β
α < 1, β
α < 1, β
≥ 1 and α > β.
< 1 and αβ < 1.
< 1 and αβ ≥ 1.
≥ 1 and αβ ≥ 1.
≥ 1 and αβ < 1.
< 1 and αβ < 1.
The solution is {α, β, α1 , αβ , α, αβ , β1 }.
1 1
The solution is {α, β, αβ
, β , αβ, α1 , β1 }.
1
The solution is {α, β, αβ
, α, αβ, α1 , β1 }.
1
The solution is {α, β, α1 , β1 , αβ, β, αβ
}.
1 1
1 1
The solution is {α, β, α , β , αβ, α , αβ }.
1 1
1
The solution is {α, β, αβ
, β , αβ, α1 , αβ
}.
GLOBAL BEHAVIOR OF SOLUTIONS OF xn+1 =
max{xn ,A}
xn xn−1
5
2.2. BOUNDEDNESS AND PERSISTENCE OF
SOLUTIONS
We will first show that Eq(1.1.1) posseses an invariant.
Theorem 2.2.1 Let A > 0. Then Eq(1.1.1) possesses the invariant
In = max{A, xn−1 , xn } max{1,
1
} = constant,
xn xn−1
(2.2.1)
for all n ≥ 0.
Proof. Indeed,
In+1 = max{A, xn , xn+1 } max{1,
1
}
xn+1 xn


max{xn , A}
= max A, xn ,
max 1,

xn xn−1
max{xn , A}
xn xn−1
= max max{xn , A},
= max{1,
1
max{xn ,A}
xn−1



max{xn−1 , xn , A}
max{xn , A}
1
}max{xn−1 , xn , A}
xn xn−1
= In .
2
Corollary 2.2.1 Let A > 0. Then every positive solution of Eq(1.1.1)
is bounded and persists. Furthermore, if {xn } is a positive solution
of Eq(1.1.1), then
1
xn ∈
, I0
I0
for n = 1, 2, . . . .
GLOBAL BEHAVIOR OF SOLUTIONS OF xn+1 =
max{xn ,A}
xn xn−1
6
2.3. INVARIANT REGIONS
Define the sets,
1
R1 = {(x, y) ∈ <+ x <+ : x, y ∈ [ , A] and xy ≥ 1}
A
and
R2 = {(x, y) ∈ <+ x <+ : x, y ∈ [A,
1
]}.
A
Theorem 2.3.1
(a) Suppose A > 1. Then R1 is an invariant region. Furthermore
if {xn } is a nontrivial solution of Eq(1.1.1) with (x−1 , x0 ) ∈ R1 ,
then {xn } is periodic with prime period three.
(b) Suppose 0 < A < 1. Then R2 is an invariant region.
Furthermore, if {xn } is a nontrivial solution of Eq(1.1.1) with
(x−1 , x0 ) ∈ R2 , then {xn } is periodic with prime period four.
Proof. We will prove (a). The proof of (b) follows similarly and will
be omitted. Suppose (x−1 , x0 ) ∈ R1 . Then
x1 =
max{x0 , A}
A
=
x0 x−1
x0 x−1
and clearly x1 ∈ [ A1 , A]. Also note that
x0 x1 =
A
A
≥
= 1.
x−1
A
Hence, (x0 , x1 ) ∈ R1 . To complete the proof, observe that
x2 =
max{x1 , A}
A
=
= x−1
x0 x1
x0 x1
x3 =
A
max{x2 , A}
=
= x0 .
x1 x2
x1 x2
and
Therefore, the solution is periodic with period three and
(x1 , x2 ) ∈ R1 since x2 x1 ≥ 1.
2
GLOBAL BEHAVIOR OF SOLUTIONS OF xn+1 =
max{xn ,A}
xn xn−1
7
Lemma 2.3.1 Eq(1.1.1) has a unique positive equilibrium x.
Furthermore,
(
1√
if 1 ≥ A > 0
x=
3
A if A > 1 .
It is interesting to note that if A > 1, then (x, x) ∈ R1 and if
0 < A < 1, then (x, x) ∈ R2 . Thus the following result is clear.
Corollary 2.3.1 Assume that A > 0. Then the equilibrium x of
Eq(1.1.1) is stable.
2.4. BEHAVIOR OF SOLUTIONS IN THE FIRST
QUADRANT
Throughout this section we shall assume that the initial conditions
x−1 and x0 of Eq(1.1.1) are positive and outside the invariant regions.
The following two lemmas characterize these solutions for A > 1 by
giving partial explicit form solutions.
Define the set,
R = {(x, y) ∈ <+ x <+ : x < A and y ≤ A−1 }.
Lemma 2.4.1 Let {xn } be a solution of Eq(1.1.1) such that (xN −1 , xN ) ∈ R
for some nonnegative integer N. Let l be a positive integer such that
A−1−2l < xN ≤ A1−2l .
Then for k = 0, 1, . . . , l − 1,
x7k+N −1 =
x7k+N
xN −1
A2k
< A1−2k
= A2k xN ≤
x7k+N +1 =
A
xN −1 xN
1
A
>A
GLOBAL BEHAVIOR OF SOLUTIONS OF xn+1 =
x7k+N +2 =
1
≥A
A2k xN
x7k+N +3 =
xN −1 xN
1
<
A
A
x7k+N +4 =
A2k+2
>A
xN −1
x7k+N +5 =
A
>A.
xN −1 xN
max{xn ,A}
xn xn−1
8
Furthermore,
x7l+N −1 =
xN −1
1
<
2l
A
A
and
x7l+N
1
= A2l xN ∈ ( , A].
A
Proof. Proof is by induction and will be omitted.
2
Note that x7l+N −1 x7l+N < 1. Hence a solution satisfying the hypothesis of the previous lemma will now satisfy the hypothesis of the
following lemma. Its proof is also by induction and will be omitted.
Define the set,
C = {(x, y) ∈ <+ x <+ : x < A and A−1 < y ≤ A and xy < 1}.
Lemma 2.4.2 Let {xn } be a solution of Eq(1.1.1) such that (xN −1 , xN ) ∈ C
for some nonnegative integer N. Let m be a positive integer such that
1
< xN (xN −1 xN )m−1 ≤ A
A
and
xN (xN −1 xN )m ≤
Then for k = 0, 1, . . . , m − 1,
x3k+N
1
= (xN xN −1 )k xN ∈ ( , A]
A
1
.
A
GLOBAL BEHAVIOR OF SOLUTIONS OF xn+1 =
x3k+N +1 =
x3k+N +2 =
A
xN −1 xN
max{xn ,A}
xn xn−1
9
>A
1
1
∈ [ , A) .
xN (xN xN −1 )k
A
Furthermore,
x3m+N −1 =
xN −1
1
∈ [ , A)
m
(xN xN −1 )
A
and
x3m+N
= (xN xN −1 )m xN ≤
1
.
A
Note that the solution now satifies the hypothesis of lemma 2.4.1 .
Define the sets,
S = {(x, y) ∈ <+ x <+ : x ≤ A−1 and xy > 1} ,
T
= {(x, y) ∈ <+ x <+ : A ≤ y < x} ,
U
= {(x, y) ∈ <+ x <+ : x ≥ A and xy < 1} ,
V
= {(x, y) ∈ <+ x <+ : y > A and xy ≤ 1} ,
W
= {(x, y) ∈ <+ x <+ : A < x ≤ y} ,
Z = {(x, y) ∈ <+ x <+ : y < A−1 and xy ≥ 1} ,
D = {(x, y) ∈ <+ x <+ : A−1 < x ≤ A and y > A},
E = {(x, y) ∈ <+ x <+ : x > A and A−1 ≤ y < A} .
The next result follows directly from lemma 2.4.1 and lemma 2.4.2 .
Lemma 2.4.3 Let {xn } be a solution of Eq(1.1.1) such that (xN −1 , xN ) ∈ R
for some nonnegative integer N. Then there exists positive integers l
and m such that
A−1−2l < xN ≤ A1−2l
and
1
< xN (xN −1 xN )m−1 ≤ A
A
and
xN (xN −1 xN )m ≤
1
.
A
GLOBAL BEHAVIOR OF SOLUTIONS OF xn+1 =
max{xn ,A}
xn xn−1
10
Furthermore for k = 0, 1, . . . , l − 1,
(xN +7k−1 , xN +7k ) ∈ R,
(xN +7k , xN +7k+1 ) ∈ S,
(xN +7k+1 , xN +7k+2 ) ∈ T,
(xN +7k+2 , xN +7k+3 ) ∈ U,
(xN +7k+3 , xN +7k+4 ) ∈ V,
(xN +7k+4 , xN +7k+5 ) ∈ W,
(xN +7k+5 , xN +7k+6 ) ∈ Z,
and for k = 0, 1, . . . , m − 1,
(xN +7l+3k−1 , xN +7l+3k ) ∈ C,
(xN +7l+3k , xN +7l+3k+1 ) ∈ D,
(xN +7l+3k+1 , xN +7l+3k+2 ) ∈ E,
(xN +7l+3m−1 , xN +7l+3m ) ∈ R .
The following two lemmas characterize solutions outside the invariant region in the first quadrant for A < 1 by giving partial explicit
form solutions. Their proofs are by induction and will be omitted.
Define the set,
F = {(x, y) ∈ <+ x <+ : xy < 1 and y ≥ A−1 }.
Lemma 2.4.4 Let {xn } be a solution of Eq(1.1.1) such that (xN −1 , xN ) ∈ F
for some nonnegative integer N. Let l be a positive integer such that
A1−2l ≤ xN < A−1−2l .
Then for k = 0, 1, . . . , l − 1,
x7k+N −1 = xN −1 < A
x7k+N
= A2k xN ≥
1
>A
A
GLOBAL BEHAVIOR OF SOLUTIONS OF xn+1 =
x7k+N +1 =
x7k+N +2 =
1
xN −1
>
max{xn ,A}
xn xn−1
11
1
>A
A
1
≤A
A2k xN
x7k+N +3 = A2k+1 xN −1 xN < A2k+1 ≤ A
x7k+N +4 =
x7k+N +5 =
1
1
>
>A
xN −1
A
1
A2k+1 xN −1 xN
> A−2k−1 ≥
1
.
A
Furthermore,
x7l+N −1 = xN −1 < A
and
x7l+N
= A2l xN ∈ [A,
1
).
A
Note that the solution now satisfies the hypothesis of the next lemma.
Define the set,
M = {(x, y) ∈ <+ x <+ : x < A and A ≤ y < A−1 }.
Lemma 2.4.5 Let {xn } be a solution of Eq(1.1.1) such that (xN −1 , xN ) ∈ M
for some nonnegative integer N. Then since
A
> 1,
xN −1
there exists a positive integer m such that
A≤(
1
A m−1
)
xN <
xN −1
A
and
(
A m
1
) xN ≥
.
xN −1
A
Then for k = 0, 1, . . . , m − 1,
x4k+N
=
Ak
1
xN ∈ [A, )
A
xkN −1
GLOBAL BEHAVIOR OF SOLUTIONS OF xn+1 =
x4k+N +1 =
1
1
>
>A
xN −1
A
x4k+N +2 =
xkN −1
1
∈ (A, ]
k
A xN
A
max{xn ,A}
xn xn−1
12
x4k+N +3 = xN −1 < A .
Furthermore,
x4m+N −1 = xN −1 < A
and
x4m+N
=
1
Am
xN ≥ .
m
xN −1
A
The solution now satifies the hypothesis of lemma 2.4.4 .
Define the sets,
G = {(x, y) ∈ <+ x <+ : x ≥ A−1 and y > x} ,
H = {(x, y) ∈ <+ x <+ : xy > 1 and y ≤ A} ,
I = {(x, y) ∈ <+ x <+ : x ≤ A and y < A} ,
J
= {(x, y) ∈ <+ x <+ : x < A and xy ≥ 1} ,
K = {(x, y) ∈ <+ x <+ : A−1 < y ≤ x} ,
L = {(x, y) ∈ <+ x <+ : x > A−1 and xy ≤ 1} ,
O = {(x, y) ∈ <+ x <+ : A ≤ x < A−1 and y > A−1 },
P
= {(x, y) ∈ <+ x <+ : x > A−1 and A < y ≤ A−1 },
Q = {(x, y) ∈ <+ x <+ : A < x ≤ A−1 and y < A}.
The next result follows directly from lemma 2.4.4 and lemma 2.4.5 .
Lemma 2.4.6 Let {xn } be a solution of Eq(1.1.1) such that (xN −1 , xN ) ∈ F
for some nonnegative integer N. Then there exists positive integers l
and m such that
A1−2l ≤ xN < A−1−2l
and
A ≤ xN (
A
xN −1
)m−1 <
1
A
and
xN (
A
xN −1
)m ≥
1
.
A
GLOBAL BEHAVIOR OF SOLUTIONS OF xn+1 =
max{xn ,A}
xn xn−1
13
Furthermore for k = 0, 1, . . . , l − 1,
(xN +7k−1 , xN +7k ) ∈ F,
(xN +7k , xN +7k+1 ) ∈ G,
(xN +7k+1 , xN +7k+2 ) ∈ H,
(xN +7k+2 , xN +7k+3 ) ∈ I,
(xN +7k+3 , xN +7k+4 ) ∈ J,
(xN +7k+4 , xN +7k+5 ) ∈ K,
(xN +7k+5 , xN +7k+6 ) ∈ L,
and for k = 0, 1, . . . , m − 1,
(xN +7l+4k−1 , xN +7l+4k ) ∈ M,
(xN +7l+4k , xN +7l+4k+1 ) ∈ O,
(xN +7l+4k+1 , xN +7l+4k+2 ) ∈ P,
(xN +7l+4k+1 , xN +7l+4k+2 ) ∈ Q,
(xN +7l+4m−1 , xN +7l+4m ) ∈ F.
From an analysis of the semicycles of Eq(1.1.1) we obtain the following lemmas whose proofs are straight forward and will be omitted.
Lemma 2.4.7 No nontrivial solutions of Eq(1.1.1) may eventually
become identically equal to the equilibrium x.
Lemma 2.4.8 Assume that A > 0. Let {xn } be a nontrivial solution
of Eq.(1.1.1). Then the following are true:
(a) If A ≤ 1, every positive semicycle (except for possibly the
first) has 2 or 3 terms. Furthermore, if it has 3 terms,
then the first and the third are equal to the equilibrium x.
(b) If A > 1, every positive semicycle has at most 2 terms.
(c) Every negative semicycle has at most 2 terms.
As an immediate consequence of the above lemmas, we have the
following result.
GLOBAL BEHAVIOR OF SOLUTIONS OF xn+1 =
max{xn ,A}
xn xn−1
14
Theorem 2.4.1 Let A > 0. Then every nontrivial solution of Eq(1.1.1)
is strictly oscillatory about the equilibrium x.
Our next result shows that not only are the solutions of Eq(1.1.1)
outside the invariant regions bounded from above but that they actually achieve this upper bound.
Theorem 2.4.2 Assume that the initial conditions x−1 and x0 of
Eq(1.1.1) are such that
(x−1 , x0 ) 6∈ R1
when
A>1
(x−1 , x0 ) 6∈ R2
when
0 < A < 1.
and
Then the solution of Eq(1.1.1) achieves its maximum value I0 in
every positive semicycle (except possibly the first).
Proof. Let xm be the largest term in a positive semicycle.
√
We shall show the case when A > 1 and xm+1 > 3 A. The other
cases are similar and will be omitted.
Since xm > A, then xm = max{A, xm , xm+1 }.
Observe that xm xm+1 > 1. Hence 1 = max{1, xm x1m+1 }.
Therefore,
I0 = Im+1 = max{A, xm , xm+1 }max{1, xm x1m+1 } = xm .
2
The following theorem shows that there are periodic solutions of
Eq(1.1.1) outside the invariant regions.
Theorem 2.4.3 Let A > 0. Assume that for k = 2, 3, . . .
x−1 = A
k
1
and x0 ∈
,A
A
or
h
x−1 ∈ Ak−1 , Ak
i
and x0 = Ak .
(a) If A > 1, then the solution of Eq(1.1.1) is periodic with period
(
7k − 1
7k−1
2
if k is even
if k is odd.
GLOBAL BEHAVIOR OF SOLUTIONS OF xn+1 =
max{xn ,A}
xn xn−1
15
(b) If A < 1, then the solution of Eq(1.1.1) is periodic with period
(
7k + 1
7k+1
2
if k is even
if k is odd.
Proof. One can see that with the above initial conditions the solution of Eq(1.1.1) can be exhibited in a closed form from which the
result will follow.
2
2.5. SOLUTIONS OUTSIDE THE FIRST QUADRANT
In this section we will see that the period 7 phenomenon is unique to
the first quadrant. The proof of the next lemma is straight forward
and will be omitted.
Lemma 2.5.1 Let {xn } be a solution of Eq(1.1.1). Then the following statements are true:
(a)If x−1 < 0 and x0 < 0, then x1 > 0, x2 < 0 and x3 < 0.
(b)If x−1 < 0 and x0 > 0, then x1 < 0, x2 < 0 and x3 > 0.
(c)If x−1 > 0 and x0 < 0, then x1 < 0, x2 > 0 and x3 < 0.
From lemma 2.5.1 we see that if we wish to characterize solutions with one or both of the initial conditions negative, it suffices to
consider only the case where both initial conditions are negative.
The following theorem implies that some solutions which begin in
the third quadrant have subsequences diverging to negative infinity
and other subsequences converging to zero.
Theorem 2.5.1 Let A > 0 and let {xn } be a solution of Eq(1.1.1)
with x−1 = α < 0 and x0 = β < 0. Assume that αβ < 1. Then
for k = 0,1,. . .
x3k = αk β k+1
A
x3k+1 =
αβ
1
x3k+2 =
.
k
α β k+1
GLOBAL BEHAVIOR OF SOLUTIONS OF xn+1 =
max{xn ,A}
xn xn−1
16
Proof. For k = 0 the result holds. Now suppose that k > 0 and
that our assumption holds for k − 1. We shall show that the result
holds for k.
x3k =
x3k+1 =
x3k+2 =
1
max{ αk−1
, A}
βk
A
αk β k+1
max{αk β k+1 , A}
αβ
A
max{ αβ
, A}
k β k+1
A α αβ
=
= αk β k+1
=
A
αβ
1
αk β k+1
.
2
The next result shows that the remaining solutions which begin
in the third quadrant are periodic of period 3.
Theorem 2.5.2 Let A > 0 and let {xn } be a solution of Eq(1.1.1)
with x−1 = α < 0 and x0 = β < 0. Assume that αβ ≥ 1. Then {xn }
is periodic with period three.
Proof. Let x−1 = α and x0 = β. Then
x1 =
x2 =
max{β, A}
A
=
αβ
αβ
A
max{ αβ
, A}
=α
A
α
x3 =
max{α, A}
A
β
=β
and the proof is complete.
2
3. THE CASE OF NEGATIVE A
In the remaining sections we will assume that A < 0. One can see
that when A < 0 and x−1 , x0 ∈ (−∞, 0) the change of variables
xn = −
1
1
and B = −
yn
A
GLOBAL BEHAVIOR OF SOLUTIONS OF xn+1 =
max{xn ,A}
xn xn−1
17
reduces Eq(1.1.1) to
yn+1 =
max{yn , B}
,
yn yn−1
n = 0, 1, . . .
(3.0.1)
with y−1 , y0 , B ∈ (0, ∞). Hence the behavior of the solutions of
Eq(1.1.1) which begin in the third quadrant when A < 0 can be easily deduced from our previous investigation.
3.1. SOLUTIONS OUTSIDE THE THIRD QUADRANT
Observe that Eq(1.1.1) has the unique positive equilibrium x = 1.
The proofs of the following theorems are straight forward and will
be omitted.
Theorem 3.1.1 Assume that A < 0 and that x−1 , x0 ∈ (0, ∞).
Then a nontrivial solution of Eq(1.1.1) is periodic with prime period 4.
When x−1 x0 < 0 and A < 0 the orbit of a solution of Eq(1.1.1)
alternates between the second and fourth quadrants. Hence without
loss of generality, we will assume that
x−1 < 0
and x0 > 0 .
Theorem 3.1.2 Let A < 0 and let {xn } be a solution of Eq(1.1.1)
with x−1 < 0 and x0 > 0. Then the following are true:
(a) Suppose that x−1 = −1. If A ≤ −1 and x0 = 1, then {xn } is
periodic with prime period 2. Otherwise, {xn } is periodic with
prime period 4.
1
(b) Suppose that x−1 6= −1. If min{x−1 , x−1
} ≥ A, then {xn } is
periodic with prime period 4. Otherwise {xn } is unbounded
and does not persist.
GLOBAL BEHAVIOR OF SOLUTIONS OF xn+1 =
max{xn ,A}
xn xn−1
18
4. REFERENCES
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tional Recursive Sequence xn+1 = (γxαx
, Communications on
n +δ)xn−1
Applied Nonlinear Analysis 1(1994), 61-72.
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Publishers, Dordrecht, 1993.
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sequences, J. Math. Anal. Appl. 173(1993), 127-157.
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, Proceedings of the First Inxn−1
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(1995), 413-419.
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