ON THE RATIONAL RECURSIVE SEQUENCE . P

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133 (2008)
MATHEMATICA BOHEMICA
No. 3, 225–239
ON THE RATIONAL RECURSIVE SEQUENCE
. P
k
k
P
βi xn−i
αi xn−i
xn+1 = A +
i=0
i=0
E. M. E. Zayed, El-Taif, M. A. El-Moneam, Zagazig
(Received December 17, 2006)
Abstract. The main objective of this paper is to study the boundedness character, the
periodic character, the convergence and the global stability of positive solutions of the
difference equation
xn+1
. X
k
k
X
= A+
αi xn−i
βi xn−i , n = 0, 1, 2, . . .
i=0
i=0
where the coefficients A, αi , βi and the initial conditions x−k , x−k+1 , . . . , x−1 , x0 are positive real numbers, while k is a positive integer number.
Keywords: difference equations, boundedness character, period two solution, convergence, global stability
MSC 2000 : 39A10, 39A11, 39A99, 34C99
1. Introduction
Our goal in this paper is to investigate the boundedness character, the periodic
character, the convergence and the global stability of positive solutions of the difference equation
(1)
xn+1 =
A+
k
X
i=0
αi xn−i
. X
k
βi xn−i ,
n = 0, 1, 2, . . .
i=0
where the coefficients A, αi , βi and the initial conditions x−k , x−k+1 , . . . , x−1 , x0 are
positive real numbers, while k is a positive integer number. The case when any of
A, αi , βi is allowed to be zero gives different special cases of the equation (1) which
225
have been studied by many authors (see for example [1]–[14]). For related work see
[15]–[27]. The study of these equations is challenging and rewarding and is still in
its infancy. We believe that nonlinear rational difference equations are of paramount
importance in their own right. Furthermore, the results about such equations offer
prototypes for the development of the basic theory of the global behavior of nonlinear
difference equations.
Definition 1. A difference equation of order (k + 1) is of the form
xn+1 = F (xn , xn−1 , . . . , xn−k ),
n = 0, 1, 2, . . .
where F is a continuous function which maps some set J k+1 into J and J is a set
of real numbers. An equilibrium point x̃ of this equation is a point that satisfies the
condition x̃ = F (x̃, x̃, . . . , x̃). That is, the constant sequence {xn }∞
n=−k with xn = x̃
for all n > −k is a solution of that equation.
Definition 2. Let x̃ ∈ (0, ∞) be an equilibrium point of the difference equation (1). Then
(i) An equilibrium point x̃ of the difference equation (1) is called locally stable
if for every ε > 0 there exists δ > 0 such that, if x−k , . . . , x−1 , x0 ∈ (0, ∞) with
|x−k − x̃| + . . . + |x−1 − x̃| + |x0 − x̃| < δ, then |xn − x̃| < ε for all n > −k.
(ii) An equilibrium point x̃ of the difference equation (1) is called locally asymptotically stable if it is locally stable and there exists γ > 0 such that, if x−k , . . . , x−1 , x0 ∈
(0, ∞) with |x−k − x̃| + . . . + |x−1 − x̃| + |x0 − x̃| < γ, then
lim xn = x̃.
n→∞
(iii) An equilibrium point x̃ of the difference equation (1) is called a global attractor
if for every x−k , . . . , x−1 , x0 ∈ (0, ∞) we have
lim xn = x̃.
n→∞
(iv) An equilibrium point x̃ of the equation (1) is called globally asymptotically
stable if it is locally stable and a global attractor.
(v) An equilibrium point x̃ of the difference equation (1) is called unstable if it is
not locally stable.
Definition 3. We say that a sequence {xn }∞
n=−k is bounded and persists if there
exist positive constants m and M such that
m 6 xn 6 M
for all n > −k.
Definition 4. A sequence {xn }∞
n=−k is said to be periodic with period p if
xn+p = xn for all n > −k. A sequence {xn }∞
n=−k is said to be periodic with prime
period p if p is the smallest positive integer having this property.
226
Assume that ã =
k
P
i=0
αi , ā =
k
P
(−1)i αi , b̃ =
k
P
βi and b̄ =
(−1)i βi . Since
i=0
i=0
i=0
k
P
the coefficients A, αi , βi are positive, a positive equilibrium point x̃ of Eq. (1) is a
solution of the equation
(2)
x̃ =
A + ãx̃
b̃x̃
.
Consequently, the positive equilibrium point x̃ of the difference equation (1) is given
by
p
ã ± ã2 + 4Ab̃
x̃ = x̃1,2 =
.
2b̃
Let F : (0, ∞)k+1 −→ (0, ∞) be a continuous function defined by
. X
k
k
X
βi u i .
αi ui
(3)
F (u0 , u1 , . . . , uk ) = A +
i=0
i=0
We have
yn+1 =
k
X
∂F (x̃, . . . , x̃)
j=0
∂uj
yn−j ,
and then the linearized equation is
(4)
yn+1 =
k
X
bj yn−j ,
j=0
where
(5)
bj = (αj − βj x̃)/b̃ x̃.
The characteristic equation of the linearized equation (4) is given by
(6)
λn+1 =
k
X
bj λn−j ,
j=0
which can be rewritten in the form
(7)
k
X
bj λ−j−1 = 1.
j=0
2. Main results
In this section we establish some results which show that the positive equilibrium
point x̃ of the difference equation (1) is globally asymptotically stable and every
positive solution of the difference equation (1) is bounded and has prime period two.
227
Theorem 1 ([13] The linearized stability theorem).
Suppose F is a continuously differentiable function defined on an open neighbourhood of the equilibrium x̃. Then the following statements are true.
(i) If all roots of the characteristic equation (6) of the linearized equation (4) have
absolute value less than one, then the equilibrium point x̃ is locally asymptotically stable.
(ii) If at least one root of Eq. (6) has absolute value greater than one, then the
equilibrium point x̃ is unstable.
(iii) If all roots of Eq. (6) have absolute value greater than one, then the equilibrium
point x̃ is a source.
Theorem 2 (See [4], [10], [13], [17]). Assume that a, b ∈ R and k ∈ {0, 1, 2, . . .}.
Then
(8)
|a| + |b| < 1
is a sufficient condition for the asymptotic stability of the difference equation
(9)
xn+1 + axn + bxn−k = 0, n = 0, 1, . . .
R e m a r k 1 (See [13]). Theorem 1 can be easily extended to a general linear
difference equation of the form
(10)
xn+k + p1 xn+k−1 + . . . + pk xn = 0,
n = 0, 1, 2, . . .
where p1 , p2 , . . . , pk ∈ R and k ∈ {1, 2, . . .}. We can see that the equation (10) is
asymptotically stable provided that
(11)
k
X
i=1
|pi | < 1.
Theorem 3 (See [13]). Consider the difference equation
xn+1 = F (xn , xn−1 , . . . , xn−k )
where F ∈ C(I k+1 , R) where I is an open interval of real numbers and R is the set
of real numbers. Let x̃ ∈ I be an equilibrium of this equation. Suppose also that
(i) F is a nondecreasing function in each of its arguments.
228
(ii) F satisfies the negative feedback property
(x − x̃)[F (x, x, . . . , x) − x] < 0
for all x ∈ I − {x̃}.
Then the equilibrium point x̃ is a global attractor.
The following lemma is an extension of that obtained in [16], [22] which is needed
here.
Lemma 4. Suppose that bj (j = 0, 1, . . . , k) are real numbers such that
k
P
j=0
0 and σj (j = 0, 1, . . . , k) are positive integers. Then the equation
k
P
j=0
has a unique solution in x ∈ (0, ∞).
|bj | =
6
|bj |x−σj = 1
Theorem 5. If all roots of the polynomial equation (6) lie in the open unit disk
|λ| < 1, then
k
X
(12)
j=0
|bj | < 1.
P r o o f. Assume that µ is a nonzero root of the equation (6) satisfying |µ| < 1.
√
Let us write µ = r exp(iθ), i = −1 and then write (7) in the form
k
X
(13)
bj r−j−1 cos(j + 1)θ = 1
j=0
and
k
X
(14)
bj r−j−1 sin(j + 1)θ = 0.
j=0
Let us now discuss the following cases:
C a s e 1. If bj > 0 (j = 0, 1, . . . , k), then by virtue of Lemma 4 we see that the
k
P
bj ̺1−j−1 = 1 has a unique solution ̺1 ∈ (0, ∞). Thus, (r, θ) = (̺1 , nπ)
equation
j=0
where n = 0, 2, 4, . . . is a solution of the equations (13), (14). This implies that
̺1 = r = |µ| < 1. But then we get
1=
k
X
j=0
bj ̺1−j−1 >
k
X
j=0
|bj |.
229
C a s e 2. If bj < 0 (j = 0, 2, 4, . . .) and bj > 0 (j = 1, 3, 5, . . .), then by virtue
k
P
|bj |̺2−j−1 = 1 has a unique solution ̺2 ∈
of Lemma 4 we see that the equation
j=0
(0, ∞). Thus, (r, θ) = (̺2 , nπ) where n = 1, 3, 5, . . . is a solution of the equations
(13), (14). This implies that ̺2 = r = |µ| < 1. But then we get
1=
k
X
j=0
|bj |̺2−j−1 >
k
X
j=0
|bj |.
Thus, the proof of Theorem 5 is completed.
Theorem 6. Let {xn }∞
n=−k be a positive solution of the difference equation (1)
such that for some n0 > 0,
(15)
either xn > x̃1
(16)
or xn 6 x̃1
for all n > n0
for all n > n0 .
Then {xn } converges to the equilibrium point x̃1 as n → ∞.
P r o o f. Assume that (15) holds. The case when (16) holds is similar and will
be omitted. Then for n > n0 + k we deduce that
xn+1 =
. X
k
k
X
βi xn−i
αi xn−i
A+
i=0
i=0
=
X
k
1+
αi xn−i
X
k
i=0
6
A
k
P
αi xn−i
i=0
αi xn−i
i=0
[1 + (A/ãx̃1 )]
b̃x̃1
=
. X
k
βi xn−i
i=0
X
k
αi xn−i
i=0
(A + ãx̃1 )
ãb̃x̃21
With the aid of (2) the last inequality becomes
xn+1 6
X
k
i=0
αi xn−i
(A + ãx̃1 ) 1 X
αi xn−i /ã,
6
ãx̃1
b̃x̃1
i=0
k
and so
(17)
230
xn+1 6 max {xn−i }
06i6k
for n > n0 + k.
.
Set
(18)
yn = max {xn−i }
06i6k
for n > n0 + k.
Then clearly
(19)
yn > xn+1 > x̃1
for n > n0 + k.
Next we claim that
(20)
yn+1 6 yn
for n > n0 + k.
We have
yn+1 = max {xn+1−i } = max{xn+1 , max {xn−i }} 6 max{xn+1 , yn } = yn .
06i6k
06i6k−1
From (19) and (20) it follows that the sequence {yn } is convergent and that
(21)
y = lim yn > x̃1 .
n→∞
To complete the proof, it suffices to prove that y 6 x̃1 . To this end, we note that
k
X
αi xn−i /b̃x̃1 6 (A + ã yn )/b̃x̃1 .
xn+1 6 A +
i=0
From this and by using (20) we obtain
xn+i 6 (A + ãyn+i−1 )/b̃x̃1 6 (A + ãyn )/b̃x̃1
for i = 1, . . . , k + 1.
Then
(22)
yn+k+1 =
max {xn+i } 6 (A + ãyn )/b̃x̃1 ,
16i6k+1
and letting n −→ ∞, we have
y6
A + ãy
.
b̃x̃1
Consequently, we obtain
(23)
ã A
y 1−
6
.
b̃x̃1
b̃x̃1
From (2) and (23) we deduce that
y b̃x̃1 − ã b̃x̃1 − ã 6
.
x̃1
b̃
b̃
Since x̃1 > ã/b̃, the term in the two brackets is positive. Thus, we have y 6 x̃1 .
Therefore, we have lim yn = x̃1 and with help of (19) we obtain lim xn = x̃1 . The
n→∞
n→∞
proof of Theorem 6 is completed.
231
Theorem 7. If {xn }∞
n=−k is a positive solution of Eq. (1) which is monotonic
increasing, then it is bounded and persists.
P r o o f. Let {xn }∞
n=−k be a positive solution of the difference equation (1). It
follows from Eq. (1) that
xn+1 = (A + α0 xn + α1 xn−1 + . . . + αk xn−k )/(β0 xn + β1 xn−1 + . . . + βk xn−k ).
Since β0 xn < β0 xn + β1 xn−1 + . . . + βk xn−k , we have
A/(β0 xn + β1 xn−1 + . . . + βk xn−k ) < A/(β0 xn ),
and also we note that
(α0 xn )/(β0 xn + β1 xn−1 + . . . + βk xn−k ) <
α0
.
β0
Similarly, we can show that
(α1 xn−1 )/(β0 xn + β1 xn−1 + . . . + βk xn−k ) <
α1
,
β1
and so on. Now, we deduce that
k
(24)
xn+1 6
X αi
A
+
,
β0 xn i=0 βi
n > 0.
Since the sequence {xn }∞
n=−k is positive and monotonic increasing, we have xn+1 >
xn and hence (24) can be rewritten in the form
x2n − xn
k
X
αi
i=0
βi
6
A
.
β0
Consequently, we have
k
xn −
1 X αi
2 i=0 βi
2
6
1
4
X
k
i=0
αi
βi
2
2
#
4A
= M,
+
β0
+
A
.
β0
From this we deduce that
(25)
232
1
xn 6
2
"
k
X
αi
i=0
βi
+
Ã
X
k
i=0
αi
βi
where M is a positive constant. On the other hand, the change of variables xn = 1/zn
transforms the equation (1) to
1
(26)
zn+1
=
k
k
X
αi . X βi
A+
.
z
z
i=0 n−i
i=0 n−i
Consequently, we get
zn+1 = β0 zn−1 . . . zn−k + β1 zn zn−2 . . . zn−k + . . . + βk zn zn−1 . . . zn−k+1
× (Azn zn−1 . . . zn−k + α0 zn−1 . . . zn−k
+ α1 zn zn−2 . . . zn−k + . . . + αk zn zn−1 . . . zn−k+1 )−1 ,
from which we deduce that
α0 zn−1 . . . zn−k < Azn zn−1 . . . zn−k + α0 zn−1 . . . zn−k + . . . + αk zn zn−1 . . . zn−k+1 ,
and hence
β0 zn−1 . . . zn−k
× (Azn zn−1 . . . zn−k + α0 zn−1 . . . zn−k + . . . + αk zn zn−1 . . . zn−k+1 )−1 <
β0
.
α0
Similarly, we see that
β1 zn zn−2 . . . zn−k
× (Azn zn−1 . . . zn−k + α0 zn−1 . . . zn−k + . . . + αk zn zn−1 . . . zn−k+1 )−1 <
β1
,
α1
and so on. Now, we deduce that
zn+1 6
k
X
βi
= H,
α
i=0 i
for all n > 0.
Thus, we obtain
(27)
xn =
1
1
>
= m,
zn
H
where H and m are positive constants. From (25) and (27) we get
m 6 xn 6 M.
Therefore, the solution of the difference equation (1) is bounded and persists. The
proof of Theorem 7 is completed.
233
Theorem 8. The positive equilibrium points x̃1,2 of the difference equation (1)
are globally asymptotically stable.
P r o o f. The linearized equation (4) with the equation (5) can be written in the
form
k
X
βj x̃i − αj
yn+1 +
yn−j = 0 (i = 1, 2),
b̃x̃i
j=0
and its characteristic equation is
λn+1 +
k
X
βj x̃i − αj
j=0
b̃x̃i
λn−j = 0
(i = 1, 2).
Now, we discuss the following cases:
C a s e 1. Since x̃1 > ã/b̃, we have
p
∞ ∞
X
b̃x̃1 − ã
ã2 + 4Ab̃ − ã
βj x̃1 − αj X βj x̃1 − αj
=
= p
< 1.
=
b̃x̃1
b̃x̃1
b̃x̃1
ã2 + 4Ab̃ + ã
j=0
j=0
C a s e 2. Since x̃2 < ã/b̃, we have
p
∞
∞ X
ã − b̃x̃2
ã − ã2 + 4Ab̃
βj x̃2 − αj X αj − βj x̃2
p
=
=
< 1.
=
b̃x̃2
b̃x̃2
b̃x̃2
ã + ã2 + 4Ab̃
j=0
j=0
Applying Theorem 1 we deduce that the equilibrium points x̃1,2 are locally asymptotically stable. It remains to prove that x̃1,2 are global attractors. To this end, we
apply Theorem 3 to the function F (u0 , u1 , . . . , uk ) given by the formula (3) as follows:
The function F : (0, ∞)k+1 −→ (0, ∞) given by (3) is continuous and nondecreasing
in each of its arguments. In addition, we deduce for x ∈ (0, ∞) that
h A + ãx
i
[F (x, x, . . . , x) − x](x − x̃1 ) =
− x (x − x̃1 )
b̃x
b̃x2 − ãx − A (x − x̃1 )2 (x − x̃2 )
=−
(x − x̃1 ) = −
<0
x
b̃x
for all x > x̃2 . Thus, the conditions of Theorem 3 are satisfied. This proves that the
equilibrium point x̃1 is a global attractor. Similarly, we can show that
[F (x, x, . . . , x) − x](x − x̃2 ) = −
(x − x̃1 )(x − x̃2 )2
<0
x
for all x > x̃1 . This proves that the equilibrium point x̃2 is a global attractor. Now,
we have shown that the equilibrium points x̃ = x
e1,2 are global attractors. The proof
of Theorem 8 is completed.
234
Theorem 9. A necessary and sufficient condition for the difference equation (1)
to have a positive prime period two solution is that the inequality
A(b̃ − b̄)2 − ā(ã + ā)(b̃ − b̄) < b̄ā2
(28)
is valid, provided ā < 0 and b̄ > 0.
P r o o f. First, suppose that there exists a positive prime period two solution
. . . , P, Q, P, Q, . . .
of the difference equation (1). We shall prove that the condition (28) holds. It follows
from the difference equation (1) that if k is even, then xn = xn−k and we have
P =
A + α0 Q + α1 P + α2 Q + α3 P + . . . + αk Q
β 0 Q + β1 P + β 2 Q + β 3 P + . . . + β k Q
Q=
A + α0 P + α1 Q + α2 P + α3 Q + . . . + αk P
,
β 0 P + β1 Q + β2 P + β 3 Q + . . . + β k P
and
while if k is odd, then xn+1 = xn−k and we have
P =
A + α0 Q + α1 P + α2 Q + α3 P + . . . + αk P
β0 Q + β1 P + β2 Q + β3 P + . . . + β k P
Q=
A + α0 P + α1 Q + α2 P + α3 Q + . . . + αk Q
.
β 0 P + β 1 Q + β2 P + β3 Q + . . . + β k Q
and
Now, we discuss the case when k is even (and in a similar way we can discuss the
case when k is odd which is omitted here). Consequently, we obtain
(29) A + α0 Q + α1 P + α2 Q + . . . + αk Q = β0 P Q + β1 P 2 + β2 P Q + . . . + βk P Q
and
(30) A + α0 P + α1 Q + α2 P + . . . + αk P = β0 P Q + β1 Q2 + β2 P Q + . . . + βk P Q.
By subtracting, we deduce after some reduction that
(31)
P +Q=
−ā
,
β1 + β3 + . . . + βk−1
while by adding we obtain
(32)
PQ =
A(β1 + β3 + . . . + βk−1 ) − ā(α0 + α2 + . . . + αk )
,
b̄(β1 + β3 + . . . + βk−1 )
235
where βi > 0, ā < 0 and b̄ > 0. Assume that P and Q are two positive distinct real
roots of the quadratic equation
t2 − (P + Q)t + P Q = 0.
(33)
We deduce that
(34)
2
−ā
β1 + β3 + . . . + βk−1
A(β1 + β3 + . . . + βk−1 ) − ā(α0 + α2 + . . . + αk )
.
>4
b̄(β1 + β3 + . . . + βk−1 )
From (34), we obtain
A(b̃ − b̄)2 − (ã + ā)(b̃ − b̄)ā < b̄ā2 ,
and hence the condition (28) is valid. Conversely, suppose that the condition (28)
is valid. Then we deduce immediately from (28) that the inequality (34) holds.
Consequently, there exist two positive distinct real numbers P and Q such that
(35)
P =
−ā
1p
T1
−
2(β1 + β3 + . . . + βk−1 ) 2
Q=
1p
−ā
+
T1 ,
2(β1 + β3 + . . . + βk−1 ) 2
and
(36)
where T1 > 0 is given by the formula
(37)
T1 =
2
−ā
β1 + β3 + . . . + βk−1
A(β1 + β3 + . . . + βk−1 ) − ā(α0 + α2 + . . . + αk )
−4
.
b̄(β1 + β3 + . . . + βk−1 )
Thus, P and Q represent two positive distinct real roots of the quadratic equation
(33). Now, we are going to prove that P and Q are positive solutions of prime period
two for the difference equation (1). To this end, we assume that
x−k = P, x−k+1 = Q, . . . , x−1 = Q,
We wish to show that
x1 = Q
236
and
x2 = P.
and
x0 = P.
To this end, we deduce from the difference equation (1) that
A + α0 x0 + α1 x−1 + . . . + αk x−k
β0 x0 + β1 x−1 + . . . + βk x−k
A + P (α0 + α2 + . . . + αk ) + Q(α1 + α3 + . . . + αk−1 )
.
=
P (β0 + β2 + . . . + βk ) + Q(β1 + β3 + . . . + βk−1 )
x1 =
(38)
Multiplying the denominator and numerator of (38) by γ = −(β1 + β3 + . . .+ βk−1 )/ā
and using (35)–(37) we obtain
√
2Aγ + [1 + K1 ](α0 + α2 + . . . + αk )
√
√
(39) x1 =
[1 + K1 ](β0 + β2 + . . . + βk ) + [1 − K1 ](β1 + β3 + . . . + βk−1 )
√
[1 − K1 ](α1 + α3 + . . . + αk−1 )
√
√
+
[1 + K1 ](β0 + β2 + . . . + βk ) + [1 − K1 ](β1 + β3 + . . . + βk−1 )
[(α0 + α2 + . . . + αk ) + (α1 + α3 + . . . + αk−1 ) + 2Aγ
√
=
[(β0 + β2 + . . . + βk ) + (β1 + β3 + . . . + βk−1 )] + b̄ K1
√
[(α0 + α2 + . . . + αk ) − (α1 + α3 + . . . + αk−1 )] K1
√
+
[(β0 + β2 + . . . + βk ) + (β1 + β3 + . . . + βk−1 )] + b̄ K1
√
[ã + 2Aγ] + ā K1
,
=
√
b̃ + b̄ K1
where
(40)
K1 = 1 −
h A(b̃ − b̄)2 − ā(ã + ā)(b̃ − b̄) i
,
b̄ā2
and from the condition (28) we deduce that K1 > 0. Multiplying the denominator
and numerator of (39) by
p
b̃ − b̄ K1 ,
we have
x1 =
b̃[ã + 2Aγ] − b̄āK1
+
√
[b̃ā − ãb̄ + −2Ab̄γ] K1
b̃2 − b̄2 K1
b̃2 − b̄2 K1
After some reduction, we deduce that
√
√
(1 + K1 )T2
−ā(1 + K1 )
x1 =
=
2γT2
2(β1 + β3 + . . . + βk−1 )
1p
−ā
T1 = Q,
+
=
2(β1 + β3 + . . . + βk−1 ) 2
.
where
T2 = 2(α1 + α3 + . . . + αk−1 )(β0 + β2 + . . . + βk )
−2(α0 + α2 + . . . + αk )(β1 + β3 + . . . + βk−1 ) −
2Ab̄(β1 + β3 + . . . + βk−1 )
.
B + ā
237
Similarly, we can show that
x2 =
=
A + α0 x1 + α1 x0 + . . . + αk x−(k−1)
β0 x1 + β1 x0 + . . . + βk x−(k−1)
A + Q(α0 + α2 + . . . + αk ) + P (α1 + α3 + . . . + αk−1 )
= P.
Q(β0 + β2 + . . . + βk ) + P (β1 + β3 + . . . + βk−1 )
By using the mathematical induction, we have
xn = P
and xn+1 = Q
for all n > −k.
Thus the difference eqution (1) has a positive prime period two solution
. . . , P, Q, P, Q, . . .
The proof of Theorem 9 is completed.
A c k n o w l e d g e m e n t. The authors would like to express their deep thanks to
the referee for his interesting suggestions and comments on this paper.
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Authors’ addresses: E. M. E. Zayed, Mathematics Department, Faculty of Science, Taif
University, El-Taif, Hawai, P.O. Box 888, Kingdom of Saudi Arabia, e-mail: emezayed
@hotmail.com; M. A. El-Moneam, Mathematics Department, Faculty of Science, Zagazig
University, Zagazig, Egypt, e-mail: mabdelmeneam2004@yahoo.com.
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