DDNS Discrete Dynamics in Nature and Society 1607-887X 1026-0226 Hindawi Publishing Corporation 10.1155/2015/195247 195247 Research Article Global Attractor of Solutions of a Rational System in the Plane Bekker Miron B. 1 Bohner Martin J. 2 Voulov Hristo D. 3 Vecchio Antonia 1 Department of Mathematics, University of Pittsburgh at Johnstown, Johnstown, PA USA pitt.edu 2 Department of Mathematics and Statistics, Missouri University of Science and Technology, Rolla, MO USA mst.edu 3 Department of Mathematics and Statistics, University of Missouri-Kansas City, Kansas City, MO USA umkc.edu 2015 11102015 2015 22 03 2015 14 07 2015 15 07 2015 11102015 2015 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.

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  (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 . Along with the results published in , several conjectures about some nontrivial cases were also posed. In the case when(2)α1=α2=β2=0,we 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 (3)α10,β2=C2=0.

Conjecture 1 (see [<xref ref-type="bibr" rid="B4">4</xref>, Conjecture 2.1 on page 1217]).

Let a0 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,nN0,it is clear that the x-component of any positive solution {(xn,yn)}nN0 of (4) must satisfy the inequality xnxn+1>a as well as the difference equation(6)xn+2=fxn+1,xn,nN0,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,vG.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)p2a+q+11+a+1,every solution {xn}nN0 of (6), with positive initial values x0 and x1 such that x0x1>a, is positive, satisfies the inequality xnxn+1>a for all nN0, and converges to the equilibrium x. Thus, the fixed point (x,x) is a global attractor for all trajectories {Tn(u,v)}nN0 with initial point (u,v)G. Then, in view of (5), the point (x,x2-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 [<xref ref-type="bibr" rid="B5">5</xref>, page 11]).

Let IR and suppose that F:I×II is nonincreasing in the first variable and nondecreasing in the second variable. Then, for every solution {xn}nN0 of(11)xn+2=Fxn+1,xn,nN0,both subsequences {x2n}nN0 and {x2n+1}nN0 are eventually monotone.

Theorem 3 (see [<xref ref-type="bibr" rid="B3">3</xref>]).

Every positive solution of (4) is bounded.

Theorem 4 (see [<xref ref-type="bibr" rid="B11">11</xref>, Theorem <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M43"><mml:mn>1.4</mml:mn><mml:mo>.</mml:mo><mml:mn>7</mml:mn></mml:math></inline-formula> on page 13]).

Let IR be a bounded interval and suppose that f:I×II is continuous and nonincreasing in each of its arguments. Assume that if m,MI with f(m,m)=M and f(M,M)=m, then m=M. Then there exists xI 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.

Let a0 and p,q>0. Let f(u,v) be defined by (7) for u>0, v0, and define(12)A=p-a+aq,B=p-a.Then the following statements are true:

f(u,v) is decreasing in u provided av+A>0, and f(u,v) is increasing in u provided av+A<0.

f(u,v) is increasing in v provided qu>B, and f(u,v) is decreasing in v provided qu<B.

If uv>a, then f(u,v)u>a.

If {xn}nN0 is a positive solution of (6) such that x0x1>a, then yn=xn+1xn-a>0 for all nN0, and the sequence {(xn,yn)}nN0 is a positive and bounded solution of (4).

If B,u,v>0 and uqB, 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)fuu,v=a-p-aq-avq+vu2,fvu,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.

Let a0 and p,q>0. Let f be defined by (7) and let {xn}nN0 be a positive solution of (6) such that x0x1>a. Assume that the subsequence {x2n}nN0 converges to xe>0. Then the following statements are true:

The subsequence {x2n+1}nN0 converges to some xo>axe-1.

The sequence {xn}nN0 converges to some x, and x is the unique root of the characteristic equation(17)λ3+λ2q-1-aλ+a-aq-p=0

such that x>max{a,1-q}.

Proof.

We first show (i). It follows from Lemma 5(iv) that the sequence {xn}nN0 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}nN0, by letting n, we obtain 0=axe-a(1-q)+p=p, which is a contradiction. Hence, xe1-q, and, from (18), it follows that(19)0limnx2n+1=axe-a+aq+pxeq+xe-1xo,(20)xoxeq+xe-1=axe+q-1+p.Since x0x1>a, by Lemma 5(iii), it follows that x2n+1x2n>a for all nN0 and, by letting n, we obtain xoxea. If we suppose that xoxe=a, then (20) implies p=0, which is a contradiction. Therefore, xoxe>a0. The proof of (i) is complete.

Now we show (ii). Since the sequence {xn+1}nN0 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=xox. Therefore,(23)limnxn=x>a,and (21) implies that x is a zero of the characteristic polynomial(24)gλλ3+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.

Let a0 and p,q>0. Let f be defined by (7) and let {xn}nN0 be a positive solution of (6) such that x0x1>a. Then there exists n0N0 such that(25)axna-aq-p,n>n0,(26)xn>1-q,n>n0.

Proof.

Define A and B by (12). For A0, 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,xnfxn+1,-Aa-1=-Aa-p>-Aa.Hence, (25) will follow if each of the subsequences {x2n}nN0 and {x2n+1}nN0 enters the interval [-Aa-1,). For the sake of contradiction, suppose that, for some m{0,1}, we have(28)xm+2n0,-Aa-1,nN0.Then, by Lemma 5(i), f(u,xm+2n) is increasing in u, and, for all nN0, 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}nN0 is increasing and must converge to some x(0,-Aa-1] that is, (31)x1-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+qxnq+xn.If xn>1-q for some n>n0, then it follows from (32) that(33)xn+2xnq+xn=1-qq+xn>1-q.It remains to show that both subsequences {x2n}nN0 and {x2n+1}nN0 eventually enter the interval (1-q,). For the sake of contradiction, suppose that, for some m{0,1}, we have(34)xm+2n1-q,nN0.Then, for every nN0 such that m+2n>n0, it follows from (32) that xm+2n+2xm+2n. Therefore, the subsequence {xm+2n}nN0 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.

Let a0 and p,q>0 and assume that (10) holds. Assume that {xn}nN0 is a positive solution of (6) such that x0x1>a. Then, {xn}nN0 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 Lemma 5(iv), it follows that there exists L>0 such that xn(0,L) for all nN0. Then, (6) implies that(35)xn+2>pxn+1q+xn>pLq+L=b>0.Thus,(36)0<b<xn<L,n3.First, assume that ap. 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-p0 that is, 0<pa(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)>0p-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 that a<p. Define A and B as in (12). Since AB>0, from Lemma 5(v), it follows that, for all u,v>0,(37)u-Bq-1fu,v-AB-1<0provideduBq-1,(38)u-Bq-1fu,v-Aqu-1>0provideduBq-1,(39)fBq-1,v=AB-1.If Bq-1<uAB-1 and v>0, then (37) and (38) imply(40)AB-1>fu,v>Aqu-1Bq-1,and, in view of (39), we obtain(41)fu,vBq-1,AB-1,uBq-1,AB-1,v>0.In the same fashion, we obtain(42)fu,vAB-1,Bq-1,uAB-1,Bq-1,v>0.Now, assume that xm[Bq-1,AB-1] for some m>1. Then, by (41), we have(43)xnBq-1,AB-1,nm.From (41), Lemma 5(i)(ii), and Theorem 2 with I=[Bq-1,AB-1], it follows that the subsequence {x2n}nN0 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)limnxn=x.Next, assume that xm[AB-1,Bq-1] for some m>1. Then, by (42), we have(45)xnAB-1,Bq-1,nm.In this case, I=[AB-1,Bq-1], the function f:I×II 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,MI with f(m,m)=M, f(M,M)=m, and mM, 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)xnc,d,n2,where(50)c=minBq-1,AB-1,d=maxBq-1,AB-1.In view of Lemma 7, (36), and (49), there exists some n0 such that(51)xnm,cd,L,nn0,where(52)m=maxb,1-q.Since the sequence {xn0+n}nN0 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<cd<x2n+1<L,nN0.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 nN0, 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-vprovidedzvq+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>11m+q,d+q>1,it follows from (54) that(57)x2n+1>gm,c,x2n+2<gd,dnN0.Set m1=m, c1=c, d1=d, L1=L, and, for sN,(58)ms+1=maxms,fLs,ms,cs+1=mincs,gds,ds,ds+1=maxds,gms,cs,Ls+1=minLs,fms,ds.Then, by induction, we obtain that(59)0<ms<x2k<csds<x2k+1<Ls,ks.Since the sequences {ms}, {cs}, {ds}, and {Ls} are monotone and bounded, there exist m,c,d,L>0 such that(60)limsms=m,limscs=c,limsds=d,limsLs=L.By letting s in (58) and (59), we obtain(61)m=maxm,fL,m,c=minc,gd,d,d=maxd,gm,c,L=minL,fm,d,(62)0<mcdL.Therefore,(63)mfL,m,dgm,c,(64)cgd,d,Lfm,d.But (63) implies Lg(m,m) and f(d,m)c and, in view of (64), we obtain(65)fd,mgd,d,fm,dm,m.Finally, (65), (7), and (55) imply that(66)q-1md-a+ppm-p-ad-md2,q-1dm-a+ppd-p-am-dm2.Thus,(67)pd-p-am-dm2pm-p-ad-md2,2p-a+dmd-m0.Hence dm and, in view of (62), m=c=d. Then, from (59), it follows that the subsequence {x2n}nN0 converges to d>0, and the proof is complete by Lemma 6.

Example 9.

Let a=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)limnxn=2,limnyn=1.

Let x0=2 and y0=1. Then x1=2 and y1=1. Hence x0x1=4>3. In fact, xn=2 and yn=1 for all nN0 that is, (70) holds true.

Let x0=1 and y0=4. Then x1=7. Hence x0x1=7>3. From Figure 1, it can be seen that (70) holds true.

Let x0=4 and y0=1. Then x1=1. Hence x0x1=4>3. From Figure 2, it can be seen that (70) holds true.

Plot of (xn,yn) for Example 9(b).

Plot of (xn,yn) 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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