AAA Abstract and Applied Analysis 1687-0409 1085-3375 Hindawi Publishing Corporation 705893 10.1155/2014/705893 705893 Research Article Global Behavior of the Difference Equation xn+1=xn-1g(xn) Xi Hongjian 1, 2 http://orcid.org/0000-0001-9819-1615 Sun Taixiang 1 Qin Bin 2 Wu Hui 1 Li Tongxing 1 College of Mathematics and Information Science Guangxi University Nanning, Guangxi 530004 China gxu.edu.cn 2 College of Information and Statistics Guangxi University of Finance and Economics Nanning, Guangxi 530003 China gxufe.cn 2014 322014 2014 17 11 2013 23 12 2013 3 2 2014 2014 Copyright © 2014 Hongjian Xi 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 the following difference equation xn+1=xn-1g(xn), n=0,1,, where initial values x-1,x0[0,+) and g:[0,+)(0,1] is a strictly decreasing continuous surjective function. We show the following. (1) Every positive solution of this equation converges to a,0,a,0,, or 0,a,0,a, for some a[0,+). (2) Assume a(0,+). Then the set of initial conditions (x-1,x0)(0,+)×(0,+) such that the positive solutions of this equation converge to a,0,a,0,, or 0,a,0,a, is a unique strictly increasing continuous function or an empty set.

1. Introduction

Recently there have been published quite a lot of works concerning global behavior of the difference equations . These results are not only valuable in their own right, but they can provide insight into their differential counterparts.

In , Kulenović and Ladas considered the positive solutions for difference equation (1)xn+1=xn-11+Axn with A>0. They gave some partial results on the convergence of this equation.

Kalikow et al.  studied the following difference equation: (E1)xn+1=xn-11+f(xn),n=0,1,2,, where initial values x-1,x0[0,+) and f is in a certain class of increasing continuous functions. They showed that the set of initial conditions (x-1,x0) of (E1) in the first quadrant that converge to any given boundary point of the first quadrant forms a unique strictly increasing continuous function.

Motivated by the above studies, in this paper, we consider the following difference equation: (2)xn+1=xn-1g(xn),n=0,1,, where initial values x-1,x0[0,+) and g:[0,+)(0,1] is a strictly decreasing continuous surjective function. Our main result is the following theorem.

Theorem 1.

( 1 ) Every positive solution of (2) converges to (3)a,0,a,0,,or0,a,0,a, for some a[0,+).

( 2 ) Assume a(0,+). Then the set of initial conditions (x-1,x0)(0,+)×(0,+) such that the positive solutions of (2) converge to (4)a,0,a,0,,or0,a,0,a, is a unique strictly increasing continuous function or an empty set.

2. The Main Result Proof of Theorem <xref ref-type="statement" rid="thm1">1</xref><inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M30"><mml:mo stretchy="false">(</mml:mo><mml:mn>1</mml:mn><mml:mo stretchy="false">)</mml:mo></mml:math></inline-formula>.

Let {xn}n=-1 be a positive solution of (2). Then x2n and x2n+1 are decreasing sequences since g(x)1. Let limnx2n=p and limnx2n-1=q. Then we have (5)p=pg(q), which implies p=0 or g(q)=1. If g(q)=1, then q=0 since g:[0,+)(0,1] is a strictly decreasing continuous surjective function with g(0)=1. This completes proof of Theorem 1(1).

Write D=[0,+)×[0,+) and define f:DD by (6)f(x,y)=(y,xg(y)), for all (x,y)D. It is easy to see that if {xn}n=-1 is a solution of (2), then fn(x-1,x0)=(xn-1,xn) for any n0. In the following, let (7)L0={a}×[0,+),L1=[0,+)×{a},R0=[a,+)×{0},R1={0}×[a,+) for some a(0,+).

Lemma 2.

The following statements are true:

f  is a homeomorphism;

f(L1)=L0;

f(R0)=R1 and f(R1)=R0.

Proof.

(i) Since f(x1,y1)f(x2,y2) for any (x1,y1),(x2,y2)D with (x1,y1)(x2,y2) and f-1(u,v)=(v/g(u),u) is continuous for any (u,v)D, f is a homeomorphism.

(ii) Let (x,y)L1 and (u,v)=f(x,y)=(y,xg(y)). Then y=a, x0, and (8)u=y=a,v=xg(y)=xg(a)0, which implies f(L1)L0.

Let (u,v)L0. Then u=a and v0. Choose (x,y)=(v/g(a),a)L1. Then f(x,y)=(u,v). Thus f(L1)=L0.

The proof of (iii) is similar to that of (ii). This completes the proof of Lemma 2.

In order to show Theorem 1(2), we will construct two families of strictly increasing functions y=h2n(x) and x=g2n+1(y) (n1) as follows. Set (9)x=g2(y)=ag(y)(y0). Then y=h2(x)=g2-1(x)=g-1(a/x) is a strictly increasing function which maps [a,+) onto [0,+). Set (10)x=g3(y)=h2(y)g(y)(ya). Then x=g3(y) is a strictly increasing function which maps [a,+) onto [0,+).

Assume that, for some positive integer n, we already define strictly increasing functions y=h2n(x) and x=g2n+1(y) such that both h2n and g2n+1 map [a,+) onto [0,+). Set (11)x=g2n+2(y)=g2n+1-1(y)g(y)(y0). Then both y=h2n+2(x)=g2n+2-1(x) and x=g2n+3(y)=h2n+2(y)/g(y) are strictly increasing functions which map [a,+) onto [0,+). In such a way, we construct two families of strictly increasing functions y=h2n(x) and x=g2n+1(y) (n1).

Set P0=[a,+)×[0,+) and Q0=[0,+)×[a,+). For any n1, write (12)Pn=f-2(Pn-1),Qn=f-2(Qn-1),Ln=f-1(Ln-1).

Let (x,y)L2. Since f(L2)=L1 and (u,v)=f(x,y)=(y,xg(y))L1, it follows that (13)xg(y)=v=a,y=u0. Thus x=g2(y)=a/g(y) and L2={(x,y):y=h2(x),xa}.

Let (x,y)L3. Since f(L3)=L2 and (u,v)=f(x,y)=(y,xg(y))L2, it follows that (14)xg(y)=v=h2(u)=h2(y),y=ua. Thus x=g3(y)=h2(y)/g(y)  (ya) and L3={(x,y):x=g3(y),ya}. Using induction, one can easily show that, for any n1, (15)L2n={(x,y):y=h2n(x),xa},L2n+1={(x,y):x=g2n+1(y),ya}. Since f is a homeomorphism and Pn=f-2(Pn-1) with L2nR0 is the boundary of Pn, we have that, for any n1, (16)Pn={(x,y):0yh2n(x),xa}. In a similar fashion, we may show that (17)Qn={(x,y):0xg2n+1(y),ya}. Since L2P0, L3Q0, and f is a homeomorphism, we have that P1P0   and Q1Q0, which implies that, for any n1, (18)L2nPn-1,  L2n+1Qn-1,PnPn-1,  QnQn-1. It follows from (12) and (18) that, for xa, (19)0h4(x)h2(x) and for ya, (20)0g5(y)g3(y). Noting (19) and (20), we may assume that, for every xa, (21)H(x)=limnh2n(x) and for every ya, (22)G(y)=limng2n+1(y).

Set (23)L={(x,y):y=H(x),xa},M={(x,y):x=G(y),ya}.

Lemma 3.

The following statements are true:

f(L)=M and f(M)=L;

both y=H(x) and x=G(y) are increasing continuous functions which map [a,+) onto [0,+).

Proof.

(i)  Let (x0,y0)L. Then we have y0=limnh2n(x0), which follows that (24)f(x0,y0)=f(x0,limnh2n(x0))=limnf(x0,h2n(x0)). Since f(L2n)=L2n-1, we have (25)f(x0,h2n(x0))=(h2n(x0),x0g(h2n(x0)))=(g2n-1(x0g(h2n(x0))),x0g(h2n(x0))). Let yn=x0g(h2n(x0)). It follows from (24) and (25) that (26)f(x0,y0)=limn(g2n-1(yn),yn)=(y0,x0g(y0)), so we have (27)limnyn=x0g(y0),limng2n-1(yn)=G(x0g(y0)). It follows from (25) and (27) that (28)f(x0,y0)=(G(x0g(y0)),x0g(y0))M. Thus we have f(L)M.

Let (x0,y0)M. Then we have x0=limng2n+1(y0), which follows that (29)f-1(x0,y0)=f-1(limng2n+1(y0),y0)=limnf-1(g2n+1(y0),y0). Since f-1(L2n+1)=L2n+2, we have (30)f-1(g2n+1(y0),y0)=(y0g(g2n+1(y0)),g2n+1(y0))=(y0g(g2n+1(y0)),h2n+2(y0g(g2n+1(y0)))). Let zn=y0/g(g2n+1(y0)). It follows from (29) and (30) that (31)f-1(x0,y0)=limn(zn,h2n+2(zn))=(y0g(x0),x0), so we have (32)limnzn=y0g(x0),limnh2n+2(zn)=H(y0g(x0)). It follows from (31) and (32) that (33)f-1(x0,y0)=(y0g(x0),H(y0g(x0)))L. Thus we have f(L)=M. In a similar fashion, we can show that f(M)=L.

(ii) Since y=h2n(x)  (n1) are strictly increasing functions, we have that y=H(x) is an increasing function. For any x0>a, let (34)limxx0+H(x)=y0+,limxx0-H(x)=y0-; then y0+H(x0)y0-.

Now we claim that y0+=y0-. Indeed, if y0+>y0-, then it follows from (6) that (35)f2(x0,y0+)=(x0g(y0+),y0+g[x0g(y0+)]),f2(x0,y0-)=(x0g(y0-),y0-g[x0g(y0-)]). So we have that (36)x0g(y0+)<x0g(y0-),y0+g[x0g(y0+)]>y0-g[x0g(y0-)]. It follows from (34) and (36) that there exist (x1,y1),(x2,y2)L such that (37)f2(x1,y1)=(x1g(y1),y1g[x1g(y1)]),f2(x2,y2)=(x2g(y2),y2g[x2g(y2)]),(38)x1g(y1)<x2g(y2),y1g[x1g(y1)]>y2g[x2g(y2)]. It follows from Lemma 3(i) and (37) that (39)(x1g(y1),y1g[x1g(y1)]),(x2g(y2),y2g[x2g(y2)])L, and this is a contradiction. The claim is proven.

In a similar fashion, we may show that limxa+H(x)=H(a)=0. Thus y=H(x)  (xa) is an increasing continuous function. In a similar fashion, we may show that x=G(y)  (ya) is an increasing continuous function. Lemma 3 is proven.

Let (40)L1={(x,y):y=H(x)=0,x[a,b]},    f(L1)=M1,L2={(x,y):y=H(x)>0,x(b,+)},f(L2)=M2, where a(0,+) and ab. It follows from Lemma 2(iii) and Lemma 3(ii) that (41)M1={(x,y):x=G(y)=0,y[a,b]},f(M1)=L1,M2={(x,y):x=G(y)>0,y(b,+)},f(M2)=L2.

Proof of Theorem <xref ref-type="statement" rid="thm1">1</xref><inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M190"><mml:mo stretchy="false">(</mml:mo><mml:mn>2</mml:mn><mml:mo stretchy="false">)</mml:mo></mml:math></inline-formula>.

Noting (40), we consider the following two cases.

Case  1 ( a = b ) . It follows from (40) that (42)L=L2{(a,0)}. Let (x-1,x0)L2 and {xn}n=-1 be a solution of (2) with initial value (x-1,x0); it follows from Lemma 3(i) that (43)(x2n-1,x2n)=f2n(x-1,x0)L, which implies that limn(x2n-1,x2n)L. It follows from (42) and Theorem 1(1) that (44)limn(x2n-1,x2n)=(a,0). Next we claim that y=H(x)  (xa) is a strictly increasing function. Indeed, if there exists (x-1,x0), (y-1,y0)L such that y-1>x-1 and x0=y0, then there exist r(1,+) such that y-1=rx-1. Set (45)fn(x-1,x0)=(xn-1,xn),fn(y-1,y0)=(yn-1,yn),gfgfgfgfgfgfgfgfgfgfggfgfgiiiiiin=1,2,. Then we have (46)y1=y-1g(y0)rx-1g(x0)=rx1,y2=y0g(y1)x0g(x1)=x2. Using induction, one can show that, for any n0, (47)y2n-1rx2n-1,y2nx2n. It follows from (44) and (47) that (48)(a,0)=limn(y2n-1,y2n)limn(x2n-1,x2n)=(a,0). This is a contradiction. The claim is proven.

Now let (x-1,x0)D-L with x00 and {xn}n=-1 be a solution of (2) with initial value (x-1,x0).

If x-1<a, then it follows from Theorem 1(1) and (2) that limnx2n-1<a which implies limn(x2n-1,x2n)(a,0).

If x-1a and x0>H(x-1), then there exists n0 such that (49)(x-1,x0)Pn-Pn+1, from which it follows that (50)f2n(x-1,x0)=(x2n-1,x2n)P0-P1. Then we have x2n+1<a, which implies limn(x2n-1,x2n)(a,0).

If x-1a and x0<H(x-1), then let y-1=x-1 and y0=H(x-1), and there exists r(1,+) such that y0=rx0. We can show that, for any n1, (51)y2nrx2n,x2n-1y2n-1,x2n+1y2n+1=x2n-1g(x2n)y2n-1g(y2n)>x2n-1y2n-1>>x1y1=x-1g(x0)y-1g(y0)=g(x0)g(y0)>1, which implies (52)limn(x2n-1,x2n)limn(y2n-1,y2n)=(a,0). From all abovementioned, the set of initial conditions (x-1,x0) such that the positive solutions of (2) converge to (53)a,0,a,0, is y=H(x)  (x>a).

In a similar fashion, we also may show that the set of initial conditions (x-1,x0) such that the positive solutions of (2) converge to (54)0,a,0,a, is x=G(y)  (y>a).

Case  2 (a<b). It follows from (41) and Case 1 that the set of initial conditions such that the positive solutions of (2) converge to (55)a,0,a,0,,or0,a,0,a, is an empty set. This completes the proof of Theorem 1(2).

Conflict of Interests

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

Acknowledgments

This project was supported by NNSF of China (11261005), NSF of Guangxi (2012GXNSFDA276040), and SF of ED of Guangxi (2013ZD061).

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