# Global Attractor of Solutions of a Rational System in the Plane

Authors: Miron B. Bekker; Martin J. Bohner; Hristo D. Voulov
Journal: Discrete Dynamics in Nature and Society
 (2015)

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## 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.


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## Body

## 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(1)xn+1=α1+ynxn,yn+1=α2+β2xn+γ2ynA2+B2xn+C2ynwith 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(2)α1=α2=β2=0,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 (3)α1≥0,β2=C2=0.Conjecture 1 (see [4, Conjecture 2.1 on page 1217]).
Leta≥0 and p,q>0. Then every positive solution of the system(4)xn+1=a+ynxn,yn+1=p+ynq+xnconverges to a finite limit.By utilizing the relation(5)yn=xnxn+1-a,n∈N0,it is clear that the x-component of any positive solution {(xn,yn)}n∈N0 of (4) must satisfy the inequality xnxn+1>a as well as the difference equation(6)xn+2=fxn+1,xn,n∈N0,where the function f is defined by(7)fu,v=uv+av+p+aq-auq+v.We consider the open subset G of the first quadrant defined by(8)G=u,v:u,v>0,uv>aand the map T defined on G by(9)Tu,v=fu,v,u,u,v∈G.It is easy to see that G is invariant for T, that is, T(G)⊂G, and that T has a unique fixed point (x∗,x∗) in G. We will prove that, in the case(10)p2≤a+q+11+a+1,every solution {xn}n∈N0 of (6), with positive initial values x0 and x1 such that x0x1>a, is positive, satisfies the inequality xnxn+1>a for all n∈N0, and converges to the equilibrium x∗. Thus, the fixed point (x∗,x∗) is a global attractor for all trajectories {Tn(u,v)}n∈N0 with initial point (u,v)∈G. Then, in view of (5), the point (x∗,x∗2-a) 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]).
LetI⊂R and suppose that F:I×I→I is nonincreasing in the first variable and nondecreasing in the second variable. Then, for every solution {xn}n∈N0 of(11)xn+2=Fxn+1,xn,n∈N0,both subsequences {x2n}n∈N0 and {x2n+1}n∈N0 are eventually monotone.Theorem 3 (see [3]).
Every positive solution of (4) is bounded.Theorem 4 (see [11, Theorem 1.4.7 on page 13]).
LetI⊂R be a bounded interval and suppose that f:I×I→I is continuous and nonincreasing in each of its arguments. Assume that if m,M∈I with f(m,m)=M and f(M,M)=m, then m=M. Then there exists x∗∈I such that every solution of xn+2=f(xn+1,xn) converges to x∗.For the proof of our main result in the next section, we also use the following auxiliary result.Lemma 5.
Leta≥0 and p,q>0. Let f(u,v) be defined by (7) for u>0, v≥0, and define(12)A=p-a+aq,B=p-a.Then the following statements are true:(i)
f(u,v) is decreasing in u provided av+A>0, and f(u,v) is increasing in u provided av+A<0.(ii)
f(u,v) is increasing in v provided qu>B, and f(u,v) is decreasing in v provided qu<B.(iii)
Ifuv>a, then f(u,v)u>a.(iv)
If{xn}n∈N0 is a positive solution of (6) such that x0x1>a, then yn=xn+1xn-a>0 for all n∈N0, and the sequence {(xn,yn)}n∈N0 is a positive and bounded solution of (4).(v)
IfB,u,v>0 and uq≠B, then(13)u-Bq-1fu,v-AB-1<0,(14)u-Bq-1fu,v-Aqu-1>0.Proof.
By direct computation of the partial derivatives(15)∂f∂uu,v=a-p-aq-avq+vu2,∂f∂vu,v=a-p+quuq+v2,the proofs of (i) and (ii) follow. From (7), we obtain (16)ufu,v-a=uv+av+p+aq-aq+v-a=uv-a+pq+vwhich implies (iii). The proof of (iv) follows from (iii), (6), and Theorem 3. It remains to prove (v). Since A=B+aq>0, (i) implies that f(u,v) is decreasing in u, and (13) follows from f(Bq-1,v)=AB-1. Finally, for u>Bq-1, it follows from (ii) that f(u,v) is increasing in v and f(u,v)>f(u,0)=A(qu)-1. Similarly, for u<Bq-1, it follows from (ii) that f(u,v) is decreasing in v and f(u,v)<f(u,0)=A(qu)-1. Thus, (14) follows and the proof is complete.


## 3. Main Results

Lemma 6.
Leta≥0 and p,q>0. Let f be defined by (7) and let {xn}n∈N0 be a positive solution of (6) such that x0x1>a. Assume that the subsequence {x2n}n∈N0 converges to xe>0. Then the following statements are true:(i)
The subsequence{x2n+1}n∈N0 converges to some xo>axe-1.(ii)
The sequence{xn}n∈N0 converges to some x∗, and x∗ is the unique root of the characteristic equation(17)λ3+λ2q-1-aλ+a-aq-p=0such thatx∗>max{a,1-q}.Proof.
We first show (i). It follows from Lemma5(iv) that the sequence {xn}n∈N0 is bounded. In view of (6) and (7), we have(18)x2n+1x2n+2q+x2n-x2n=ax2n-a+aq+p.If we suppose that xe=1-q, then from (18) and the boundedness of {x2n+1}n∈N0, by letting n→∞, we obtain 0=axe-a(1-q)+p=p, which is a contradiction. Hence, xe≠1-q, and, from (18), it follows that(19)0≤limn→∞⁡x2n+1=axe-a+aq+pxeq+xe-1≕xo,(20)xoxeq+xe-1=axe+q-1+p.Since x0x1>a, by Lemma 5(iii), it follows that x2n+1x2n>a for all n∈N0 and, by letting n→∞, we obtain xoxe≥a. If we suppose that xoxe=a, then (20) implies p=0, which is a contradiction. Therefore, xoxe>a≥0. The proof of (i) is complete.
Now we show (ii). Since the sequence{xn+1}n∈N0 is also a positive solution of (6) such that x0x1>a, it follows from (i) that besides (20) we also have(21)xexoq+xo-1=axo+q-1+p.By subtracting (21) from (20), we obtain(22)xexoxe-xo=axe-xoand, in view of xexo>a, this yields xe=xo≕x∗. Therefore,(23)limn→∞⁡xn=x∗>a,and (21) implies that x∗ is a zero of the characteristic polynomial(24)gλ≔λ3+q-1λ2-aλ+a1-q-p.Observe that g has a unique root x∗ in the interval (a,∞), and the inequality x∗>max{a,1-q} follows from g(a)=g(1-q)=-p<0. The proof is complete.Lemma 7.
Leta≥0 and p,q>0. Let f be defined by (7) and let {xn}n∈N0 be a positive solution of (6) such that x0x1>a. Then there exists n0∈N0 such that(25)axn≥a-aq-p,n>n0,(26)xn>1-q,n>n0.Proof.
DefineA and B by (12). For A≥0, inequality (25) is trivial. Now assume that A<0 that is, 0<p<a(1-q)<a. Since B=p-a<0, it follows from Lemma 5(ii) that f(u,v) is increasing in v, and xn≥-Aa-1 implies(27)xn+2=fxn+1,xn≥fxn+1,-Aa-1=-Aa-p>-Aa.Hence, (25) will follow if each of the subsequences {x2n}n∈N0 and {x2n+1}n∈N0 enters the interval [-Aa-1,∞). For the sake of contradiction, suppose that, for some m∈{0,1}, we have(28)xm+2n∈0,-Aa-1,n∈N0.Then, by Lemma 5(i), f(u,xm+2n) is increasing in u, and, for all n∈N0, the inequality xm+2n+1xm+2n>a implies that(29)xm+2n+2=fxm+2n+1,xm+2n>faxm+2n-1,xm+2n.Since p>0, it follows that(30)fax-1,x=xaq+ax+paq+x>x,x>0.In view of (29) and (30), (28) implies that the subsequence {xm+2n}n∈N0 is increasing and must converge to some x∗∈(0,-Aa-1] that is, (31)x∗≤1-q-pa-1<1-q.On the other hand, Lemma 6 implies that x∗>1-q, which is a contradiction. Hence, (28) cannot be true, and the proof of (25) is complete. Then, for every n>n0, we have(32)xn+2=fxn+1,xn=xn+1xn+axn+Axn+1xn+q≥xnq+xn.If xn>1-q for some n>n0, then it follows from (32) that(33)xn+2≥xnq+xn=1-qq+xn>1-q.It remains to show that both subsequences {x2n}n∈N0 and {x2n+1}n∈N0 eventually enter the interval (1-q,∞). For the sake of contradiction, suppose that, for some m∈{0,1}, we have(34)xm+2n≤1-q,n∈N0.Then, for every n∈N0 such that m+2n>n0, it follows from (32) that xm+2n+2≥xm+2n. Therefore, the subsequence {xm+2n}n∈N0 is nondecreasing and must converge to some x∗∈(0,1-q]. However, Lemma 6 implies that x∗>1-q, 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.
Leta≥0 and p,q>0 and assume that (10) holds. Assume that {xn}n∈N0 is a positive solution of (6) such that x0x1>a. Then, {xn}n∈N0 converges to a finite limit x∗, which is the unique root of the characteristic equation (17) such that x∗>max{a,1-q}.Proof.
From Lemma5(iv), it follows that there exists L>0 such that xn∈(0,L) for all n∈N0. Then, (6) implies that(35)xn+2>pxn+1q+xn>pLq+L=b>0.Thus,(36)0<b<xn<L,n≥3.First, assume that a≥p. If a-aq-p<0, then we have that f(u,v)>0 for all u,v>0 and, by Lemma 5, f(u,v) is decreasing in u and increasing in v. Therefore, in this case, the proof follows from (36), Theorem 2 with I=(0,∞), and Lemma 6. Now assume that a-aq-p≥0 that is, 0<p≤a(1-q)<a. Then q<1, and, by Lemma 7, we have xn>1-q eventually. It is easy to see that u,v∈(1-q,∞) implies f(u,v)∈(1-q,∞). From the inequalities av>a(1-q)-p and qu>q(1-q)>0≥p-a, we obtain by Lemma 5 that f(u,v) is decreasing in u and increasing in v on (1-q,∞). Hence, in this case, the proof follows from (36), Theorem 2 with I=(1-q,∞), and Lemma 6.
Now we assume thata<p. Define A and B as in (12). Since A≥B>0, from Lemma 5(v), it follows that, for all u,v>0,(37)u-Bq-1fu,v-AB-1<0providedu≠Bq-1,(38)u-Bq-1fu,v-Aqu-1>0providedu≠Bq-1,(39)fBq-1,v=AB-1.If Bq-1<u≤AB-1 and v>0, then (37) and (38) imply(40)AB-1>fu,v>Aqu-1≥Bq-1,and, in view of (39), we obtain(41)fu,v∈Bq-1,AB-1,u∈Bq-1,AB-1,v>0.In the same fashion, we obtain(42)fu,v∈AB-1,Bq-1,u∈AB-1,Bq-1,v>0.Now, assume that xm∈[Bq-1,AB-1] for some m>1. Then, by (41), we have(43)xn∈Bq-1,AB-1,n≥m.From (41), Lemma 5(i)(ii), and Theorem 2 with I=[Bq-1,AB-1], it follows that the subsequence {x2n}n∈N0 is monotone. But it is also bounded, in view of (41), and must converge to some xe∈[b,L]. Therefore, by Lemma 6, we obtain xe=x∗ and(44)limn→∞⁡xn=x∗.Next, assume that xm∈[AB-1,Bq-1] for some m>1. Then, by (42), we have(45)xn∈AB-1,Bq-1,n≥m.In this case, I=[AB-1,Bq-1]≠∅, the function f:I×I→I 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 m,M∈I with f(m,m)=M, f(M,M)=m, and m≠M, which implies(46)M-mM+m+a-mM=0,mM=M+m+a.Then, in view of f(m,m)=M, we obtain(47)m+M=B-aq+1,mM=Aq+1.Thus, m and M are different positive roots of the quadratic polynomial(48)z2-B-aq+1z+Aq+1.Therefore, B-a>0 and (B-a)2-4A(q+1)>0, which contradicts (10). Since x∗ is the unique positive root of the characteristic equation (17), it follows from (42) and Theorem 4 with I=[AB-1,Bq-1] that (44) holds. It remains to prove (44) in the last case, when(49)xn∉c,d,n≥2,where(50)c=min⁡Bq-1,AB-1,d=max⁡Bq-1,AB-1.In view of Lemma 7, (36), and (49), there exists some n0 such that(51)xn∈m,c∪d,L,n≥n0,where(52)m=max⁡b,1-q.Since the sequence {xn0+n}n∈N0 is also a positive solution of (6), in view of (37) and (38), we may assume without loss of generality that(53)0<m<x2n<c≤d<x2n+1<L,n∈N0.Then, by Lemma 5(i)(ii), f(x2k,v) is decreasing in v, while f(x2k+1,v) is increasing in v and f(u,v) is decreasing in u. Hence, for every n∈N0, we obtain that(54)c>x2n+2=fx2n+1,x2n>fx2n+1,m>fL,m,d<x2n+3=fx2n+2,x2n+1<fx2n+2,d<fm,d.The equation f(u,v)=z has a unique solution for u:(55)u=gv,z=av+aq+p-azq+v-vprovidedz≠vq+v.Observe that the inequality f(u,v)<z is equivalent to u>g(v,z), provided z(q+v)-v>0. Since (52) and (53) imply that(56)cm>1≥1m+q,d+q>1,it follows from (54) that(57)x2n+1>gm,c,x2n+2<gd,dn∈N0.Set m1=m, c1=c, d1=d, L1=L, and, for s∈N,(58)ms+1=max⁡ms,fLs,ms,cs+1=min⁡cs,gds,ds,ds+1=max⁡ds,gms,cs,Ls+1=min⁡Ls,fms,ds.Then, by induction, we obtain that(59)0<ms<x2k<cs≤ds<x2k+1<Ls,k≥s.Since the sequences {ms}, {cs}, {ds}, and {Ls} are monotone and bounded, there exist m∗,c∗,d∗,L∗>0 such that(60)lims→∞⁡ms=m∗,lims→∞⁡cs=c∗,lims→∞⁡ds=d∗,lims→∞⁡Ls=L∗.By letting s→∞ in (58) and (59), we obtain(61)m∗=max⁡m∗,fL∗,m∗,c∗=min⁡c∗,gd∗,d∗,d∗=max⁡d∗,gm∗,c∗,L∗=min⁡L∗,fm∗,d∗,(62)0<m∗≤c∗≤d∗≤L∗.Therefore,(63)m∗≥fL∗,m∗,d∗≥gm∗,c∗,(64)c∗≤gd∗,d∗,L∗≤fm∗,d∗.But (63) implies L∗≥g(m∗,m∗) and f(d∗,m∗)≤c∗ and, in view of (64), we obtain(65)fd∗,m∗≤gd∗,d∗,fm∗,d∗≥m∗,m∗.Finally, (65), (7), and (55) imply that(66)q-1m∗d∗-a+p≤pm∗-p-ad∗-m∗d∗2,q-1d∗m∗-a+p≥pd∗-p-am∗-d∗m∗2.Thus,(67)pd∗-p-am∗-d∗m∗2≤pm∗-p-ad∗-m∗d∗2,2p-a+d∗m∗d∗-m∗≤0.Hence d∗≤m∗ and, in view of (62), m∗=c∗=d∗. Then, from (59), it follows that the subsequence {x2n}n∈N0 converges to d∗>0, and the proof is complete by Lemma 6.Example 9.
Leta=3, q=2, and p=3; that is, we consider the example of system (4):(68)xn+1=3+ynxn,yn+1=3+yn2+xn.Then (10) is satisfied. The root x∗ of the characteristic equation (17), that is, (69)λ3+λ2-3λ-6=0,satisfying x∗>max{a,1-q} is x∗=2. By Theorem 8, if x0x1>3, then any positive solution of (68) satisfies(70)limn→∞⁡xn=2,limn→∞⁡yn=1.(a)
Letx0=2 and y0=1. Then x1=2 and y1=1. Hence x0x1=4>3. In fact, xn=2 and yn=1 for all n∈N0 that is, (70) holds true.(b)
Letx0=1 and y0=4. Then x1=7. Hence x0x1=7>3. From Figure 1, it can be seen that (70) holds true.(c)
Letx0=4 and y0=1. Then x1=1. Hence x0x1=4>3. From Figure 2, it can be seen that (70) holds true.Figure 1
Plot of(xn,yn) for Example 9(b).Figure 2
Plot of(xn,yn) for Example 9(c).



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