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Convergence Results on a Second-Order Rational Difference Equation with Quadratic Terms
Advances in Difference Equations volume 2009, Article number: 985161 (2009)
Abstract
We investigate the global behavior of the second-order difference equation , where initial conditions and all coefficients are positive. We find conditions on
under which the even and odd subsequences of a positive solution converge, one to zero and the other to a nonnegative number; as well as conditions where one of the subsequences diverges to infinity and the other either converges to a positive number or diverges to infinity. We also find initial conditions where the solution monotonically converges to zero and where it diverges to infinity.
1. Introduction and Preliminaries
There are a number of studies published on second-order rational difference equations (see, e.g., [1–9]). We investigate the global behavior of the second-order difference equation

where the numerator is quadratic and the denominator is linear with . Under various hypotheses on the parameters, we establish the existence of different behaviors of even and odd subsequences of solutions of (1.1). Our results are summarized below.
-
(i)
Let
and
, then we have the following.
-
(a)
There are infinitely many solutions,
, such that for each, one of its subsequences,
,
, converges to zero and the other diverges to infinity.
-
(b)
There exist solutions,
, which
-
(1)
converge to zero if
;
-
(2)
diverge to infinity if
;
-
(3)
are constant if
.
-
(1)
-
(a)
-
(i)
Let
and
. Then for each positive solution
, one of the subsequences,
,
, diverges to infinity and the other to a positive number that can be arbitrarily large depending on initial values. Further there, are positive initial values for which the corresponding solution,
, increases monotonically to infinity.
-
(ii)
Let
and
. Then for each positive solution
, one of the subsequences,
,
, converges to zero and the other to a nonnegative number. Further, there are positive initial values for which the corresponding solution,
, decreases monotonically to zero.
We note that the following results address and solve the first five conjectures posed by Sedaghat in [10].
2. Results
In order to establish this first result, we reduce (1.1) to a first-order equation by means of the substitution This transforms (1.1) to

Theorem 2.1.
Let and
in (1.1). Then one has the following.
-
(i)
There are infinitely many solutions,
, such that for each, one of its subsequences,
,
, converges to zero and the other to infinity.
-
(ii)
There exist solutions,
, which
-
(a)
converge to zero if
;
-
(b)
diverge to infinity if
;
-
(c)
are constant if
.
-
(a)
Proof.
Starting with (2.1), let the function be defined as
. Note that for
,
is a decreasing function since
. Also note that
and
. Hence
has a unique positive fixed point
.
We next compute the expression and simplify, it including canceling the common factor
from the numerator and denominator, thereby obtaining the following:

where

Note that since ,
and
. Thus the numerator of
has one and only one sign change. Therefore, by Descartes' rule of signs, the numerator of
has exactly one positive root,
.
In addition, we see that and so, given that
is the only positive root of the numerator of
, we have
for
. Thus, since
and
is continuous, we must have
for
. Therefore,

We consider two cases depending on the initial value for (2.1).
Case 1 ().
Using induction and the fact that is a decreasing function so that
is an increasing function, we have

Thus, Since
is the only positive fixed point of
, then we must have
and
Case 2 ().
The argument is similar to that in Case 1 in showing and
In both cases, the solution,
, of (2.1) is divided into even and odd subsequences,
and
, where one subsequence converges monotonically to zero and the other to infinity.
We now go back to (1.1) by inferring the behavior of from
. To do this we first consider
. Without loss of generality, we will assume that
and so
and
.
Next, observe that

From this and our assumption with , we have

Hence, for , there exists
such that

for all . We then have

and by induction, for ,

This, in turn, implies that

The argument is similar in showing that since

Hence, result (i) is true.
Now consider . Then
for all
, and so
for all
. Induction then gives us
for all
. We thus have one of the following:
-
(1)
If
(
), then
-
(2)
If
(
), then
-
(3)
If
(
), then
is a constant solution
Thus the result (ii) is true and this completes the proof.
For the next couple of results we rewrite (1.1) in the form

Note that if either and
, or
and
, then
satisfies the following properties:
-
(P1)
, with
undefined when
.
-
(P2)
-
(P3)
if
.
-
If
we consider the addition restriction that
and
, we also obtain
-
(P4)
if
, then
, or
.
Lemma 2.2.
Let be a positive solution of (1.1) with
and
. Then there exist
and
such that the following statements are true:
-
(1)
as
,
-
(2)
as
,
-
(3)
, and
and
are undefined; or if either
or
is not zero, then
is a solution of (1.1).
-
(4)
.
Proof.
Statements 1 and 2 follow from the fact that

by properties (P2) and (P3). Statement 3 follows from the fact that either , and so
and
are undefined by property (P1); or
and

where Statements 1 and 2 and the continuity of (Property (P1) hold. Finally, Statement 4 follows immediately from Statement 3 and Property (P4).
In the first three results, we characterize the convergence of the odd and even subsequences of solutions of (1.1).
Theorem 2.3.
Let and
in (1.1). Then for each positive solution,
, one of the subsequences,
,
, converges to zero and the other to a nonnegative number.
Proof.
Consider (1.1) with ,
, and
. Then it follows from Lemma 2.2 that for each positive solution of (1.1),
, one of the subsequences,
,
, converges to zero and the other to a nonnegative number.
Theorem 2.4.
Let and
in (1.1). Then for each positive solution
, one of the subsequences,
,
, diverges to infinity and the other to a positive number or diverges to infinity.
Proof.
Consider (1.1) with and
. Using the transformation
convert (1.1) to the equation

Then , and so it follows from Lemma 2.2 that for each positive solution of (2.16),
, one of the subsequences,
,
, converges to zero and the other to a nonnegative number. Hence, for each positive solution of (1.1),
, one of the subsequences,
,
, diverges to infinity and the other to a positive number or diverges to infinity.
In the following results, we show the existence of monotonic solutions for (1.1). As with Theorem 2.1 we use the substitution
Theorem 2.5.
Let and
in (1.1). Then there are positive initial values for which the corresponding solutions,
, decrease monotonically to zero.
Proof.
Note that an equilibrium equation for (2.1) satisfies,

Set Given Descartes' rule of signs, we have that there exists a unique positive equilibrium,
, where
and
Recall that
and let
for all
. Then
for all
. It follows from induction that
for all
. Since
,
, with
, decreases monotonically to zero.
Theorem 2.6.
Let and
in (1.1). Then there are positive initial values for which the corresponding solution,
, increases monotonically to infinity.
Proof.
As in the previous proof, an equilibrium equation for (2.1) satisfies (2.17). Setting we obtain from Descartes' rule of signs, a unique positive equilibrium,
, where
and
Recall that
and let
for all
. Then
for all
. It follows from induction that
for all
. Since
,
, with
, increases monotonically to infinity.
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Chan, D.M., Kent, C.M. & Ortiz-Robinson, N.L. Convergence Results on a Second-Order Rational Difference Equation with Quadratic Terms. Adv Differ Equ 2009, 985161 (2009). https://doi.org/10.1155/2009/985161
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DOI: https://doi.org/10.1155/2009/985161