Discrete Dynamics in Nature and Society
Volume 2015 (2015), Article ID 195247, 6 pages
http://dx.doi.org/10.1155/2015/195247

Research Article

Global Attractor of Solutions of a Rational System in the Plane

Miron B. Bekker,¹ Martin J. Bohner,² and Hristo D. Voulov³

¹Department of Mathematics, University of Pittsburgh at Johnstown, Johnstown, PA, USA
²Department of Mathematics and Statistics, Missouri University of Science and Technology, Rolla, MO, USA
³Department of Mathematics and Statistics, University of Missouri-Kansas City, Kansas City, MO, USA

Received 22 March 2015; Revised 14 July 2015; Accepted 15 July 2015

Academic Editor: Antonia Vecchio

Copyright © 2015 Miron B. Bekker et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract.

We consider a two-dimensional autonomous system of rational difference equations with three positive parameters. It was conjectured that every positive solution of this system converges to a finite limit. Here we confirm this conjecture, subject to an additional assumption.

1. Introduction

Rational systems of first-order difference equations in the plane have been receiving increasing attention in the last decade. Recently, in [1–15] (see also the references therein), efforts have been made for a more systematic approach. In particular, the rational system

with nonnegative coefficients and initial conditions was studied in [4]. Along with the results published in [4], several conjectures about some nontrivial cases were also posed. In the case when

we have confirmed in [2] that [4, Conjecture 2.4 on page 1223] is true. Our goal here is to confirm another conjecture, in the case when

Conjecture 1 (see [4, Conjecture 2.1 on page 1217]). Let and . Then every positive solution of the system

converges to a finite limit.

By utilizing the relation

it is clear that the -component of any positive solution of (4) must satisfy the inequality as well as the difference equation

where the function is defined by

We consider the open subset of the first quadrant defined by

and the map defined on by

It is easy to see that is invariant for , that is, , and that has a unique fixed point in . We will prove that, in the case

every solution of (6), with positive initial values and such that , is positive, satisfies the inequality for all , and converges to the equilibrium . Thus, the fixed point is a global attractor for all trajectories with initial point . Then, in view of (5), the point is a global attractor for all positive solutions of (4), which confirms the conjecture in case (10) holds.

2. Preliminaries

We recall the following three useful theorems.

Theorem 2 (see [5, page 11]). Let and suppose that is nonincreasing in the first variable and nondecreasing in the second variable. Then, for every solution of

both subsequences and are eventually monotone.

Theorem 3 (see [3]). Every positive solution of (4) is bounded.

Theorem 4 (see [11, Theorem on page 13]). Let be a bounded interval and suppose that is continuous and nonincreasing in each of its arguments. Assume that if with and , then . Then there exists such that every solution of converges to .

For the proof of our main result in the next section, we also use the following auxiliary result.

Lemma 5. Let and . Let be defined by (7) for , , and define

Then the following statements are true:

(i) is decreasing in provided , and is increasing in provided .(ii) is increasing in provided , and is decreasing in provided .(iii)If , then .(iv)If is a positive solution of (6) such that , then for all , and the sequence is a positive and bounded solution of (4).(v)If and , then

Proof. By direct computation of the partial derivatives

the proofs of (i) and (ii) follow. From (7), we obtain

which implies (iii). The proof of (iv) follows from (iii), (6), and Theorem 3. It remains to prove (v). Since , (i) implies that is decreasing in , and (13) follows from . Finally, for , it follows from (ii) that is increasing in and . Similarly, for , it follows from (ii) that is decreasing in and . Thus, (14) follows and the proof is complete.

3. Main Results

Lemma 6. Let and . Let be defined by (7) and let be a positive solution of (6) such that . Assume that the subsequence converges to . Then the following statements are true:

(i)The subsequence converges to some .(ii)The sequence converges to some , and is the unique root of the characteristic equation

 such that .

Proof. We first show (i). It follows from Lemma 5(iv) that the sequence is bounded. In view of (6) and (7), we have

If we suppose that , then from (18) and the boundedness of , by letting , we obtain , which is a contradiction. Hence, , and, from (18), it follows that

Since , by Lemma 5(iii), it follows that for all and, by letting , we obtain . If we suppose that , then (20) implies , which is a contradiction. Therefore, . The proof of (i) is complete.
Now we show (ii). Since the sequence is also a positive solution of (6) such that , it follows from (i) that besides (20) we also have

By subtracting (21) from (20), we obtain

and, in view of , this yields . Therefore,

and (21) implies that is a zero of the characteristic polynomial

Observe that has a unique root in the interval , and the inequality follows from . The proof is complete.

Lemma 7. Let and . Let be defined by (7) and let be a positive solution of (6) such that . Then there exists such that

Proof. Define and by (12). For , inequality (25) is trivial. Now assume that that is, . Since , it follows from Lemma 5(ii) that is increasing in , and implies

Hence, (25) will follow if each of the subsequences and enters the interval . For the sake of contradiction, suppose that, for some , we have

Then, by Lemma 5(i), is increasing in , and, for all , the inequality implies that

Since , it follows that

In view of (29) and (30), (28) implies that the subsequence is increasing and must converge to some that is,

On the other hand, Lemma 6 implies that , which is a contradiction. Hence, (28) cannot be true, and the proof of (25) is complete. Then, for every , we have

If for some , then it follows from (32) that

It remains to show that both subsequences and eventually enter the interval . For the sake of contradiction, suppose that, for some , we have

Then, for every such that , it follows from (32) that . Therefore, the subsequence is nondecreasing and must converge to some . However, Lemma 6 implies that , which is a contradiction. Hence, (34) cannot be true and the proof of (26) follows. The proof is complete.

Now we are ready to prove our main result.

Theorem 8. Let and and assume that (10) holds. Assume that is a positive solution of (6) such that . Then, converges to a finite limit , which is the unique root of the characteristic equation (17) such that .

Proof. From Lemma 5(iv), it follows that there exists such that for all . Then, (6) implies that

Thus,

First, assume that . If , then we have that for all and, by Lemma 5, is decreasing in and increasing in . Therefore, in this case, the proof follows from (36), Theorem 2 with , and Lemma 6. Now assume that that is, . Then , and, by Lemma 7, we have eventually. It is easy to see that implies . From the inequalities and , we obtain by Lemma 5 that is decreasing in and increasing in on . Hence, in this case, the proof follows from (36), Theorem 2 with , and Lemma 6.
Now we assume that . Define and as in (12). Since , from Lemma 5(v), it follows that, for all ,

If and , then (37) and (38) imply

and, in view of (39), we obtain

In the same fashion, we obtain

Now, assume that for some . Then, by (41), we have

From (41), Lemma 5(i)(ii), and Theorem 2 with , it follows that the subsequence is monotone. But it is also bounded, in view of (41), and must converge to some . Therefore, by Lemma 6, we obtain and

Next, assume that for some . Then, by (42), we have

In this case, , the function defined by (7) is continuous, and, by Lemma 5(i)(ii), it is nonincreasing in each of its arguments. In order to verify the second condition of Theorem 4, we suppose for the sake of contradiction that there exist with , , and , which implies

Then, in view of , we obtain

Thus, and are different positive roots of the quadratic polynomial

Therefore, and , which contradicts (10). Since is the unique positive root of the characteristic equation (17), it follows from (42) and Theorem 4 with that (44) holds. It remains to prove (44) in the last case, when

where

In view of Lemma 7, (36), and (49), there exists some such that

where

Since the sequence is also a positive solution of (6), in view of (37) and (38), we may assume without loss of generality that

Then, by Lemma 5(i)(ii), is decreasing in , while is increasing in and is decreasing in . Hence, for every , we obtain that

The equation has a unique solution for :

Observe that the inequality is equivalent to , provided . Since (52) and (53) imply that

it follows from (54) that

Set , , , , and, for ,

Then, by induction, we obtain that

Since the sequences , , , and are monotone and bounded, there exist such that

By letting in (58) and (59), we obtain

Therefore,

But (63) implies and and, in view of (64), we obtain

Finally, (65), (7), and (55) imply that

Thus,

Hence and, in view of (62), . Then, from (59), it follows that the subsequence converges to , and the proof is complete by Lemma 6.

Example 9. Let , , and ; that is, we consider the example of system (4):

Then (10) is satisfied. The root of the characteristic equation (17), that is,

satisfying is . By Theorem 8, if , then any positive solution of (68) satisfies

(a)Let and . Then and . Hence . In fact, and for all that is, (70) holds true.(b)Let and . Then . Hence . From Figure 1, it can be seen that (70) holds true.(c)Let and . Then . Hence . From Figure 2, it can be seen that (70) holds true.

Figure 1: Plot of for Example 9(b).

Figure 2: Plot of for Example 9(c).

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

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