Global Attractor of Solutions of a Rational System in the Plane
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.
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 systemwith nonnegative coefficients and initial conditions was studied in . Along with the results published in , several conjectures about some nontrivial cases were also posed. In the case whenwe have confirmed in  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 systemconverges to a finite limit.
By utilizing the relationit is clear that the -component of any positive solution of (4) must satisfy the inequality as well as the difference equationwhere the function is defined byWe consider the open subset of the first quadrant defined byand the map defined on byIt is easy to see that is invariant for , that is, , and that has a unique fixed point in . We will prove that, in the caseevery 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.
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 ofboth subsequences and are eventually monotone.
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 defineThen 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 derivativesthe 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 haveIf we suppose that , then from (18) and the boundedness of , by letting , we obtain , which is a contradiction. Hence, , and, from (18), it follows thatSince , 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 haveBy subtracting (21) from (20), we obtainand, in view of , this yields . Therefore,and (21) implies that is a zero of the characteristic polynomialObserve that has a unique root in the interval , and the inequality follows from . The proof is complete.
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 impliesHence, (25) will follow if each of the subsequences and enters the interval . For the sake of contradiction, suppose that, for some , we haveThen, by Lemma 5(i), is increasing in , and, for all , the inequality implies thatSince , it follows thatIn 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 haveIf for some , then it follows from (32) thatIt remains to show that both subsequences and eventually enter the interval . For the sake of contradiction, suppose that, for some , we haveThen, 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 thatThus,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) implyand, in view of (39), we obtainIn the same fashion, we obtainNow, assume that for some . Then, by (41), we haveFrom (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 andNext, assume that for some . Then, by (42), we haveIn 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 impliesThen, in view of , we obtainThus, and are different positive roots of the quadratic polynomialTherefore, 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, whenwhereIn view of Lemma 7, (36), and (49), there exists some such thatwhereSince the sequence is also a positive solution of (6), in view of (37) and (38), we may assume without loss of generality thatThen, by Lemma 5(i)(ii), is decreasing in , while is increasing in and is decreasing in . Hence, for every , we obtain thatThe equation has a unique solution for :Observe that the inequality is equivalent to , provided . Since (52) and (53) imply thatit follows from (54) thatSet , , , , and, for ,Then, by induction, we obtain thatSince the sequences , , , and are monotone and bounded, there exist such thatBy letting in (58) and (59), we obtainTherefore,But (63) implies and and, in view of (64), we obtainFinally, (65), (7), and (55) imply thatThus,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.
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
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