# Global Behavior of the Difference Equationxn+1=xn-1g(xn)

Authors: Hongjian Xi; Taixiang Sun; Bin Qin; Hui Wu
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

We consider the following difference equationxn+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.


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## Body

## 1. Introduction

Recently there have been published quite a lot of works concerning global behavior of the difference equations [1–8]. These results are not only valuable in their own right, but they can provide insight into their differential counterparts.In [9], 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. [10] 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 Theorem1(1).
Let{xn}n=-1∞ be a positive solution of (2). Then x2n and x2n+1 are decreasing sequences since g(x)≤1. Let limn→∞x2n=p and limn→∞x2n-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).WriteD=[0,+∞)×[0,+∞) and define f:D→D 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 n≥0. 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:(i)
fis a homeomorphism;(ii)
f(L1)=L0;(iii)
f(R0)=R1 and f(R1)=R0.Proof.
(i) Sincef(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, x≥0, and
(8)u=y=a,v=xg(y)=xg(a)≥0,
which implies f(L1)⊂L0.
Let(u,v)∈L0. Then u=a and v≥0. 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 Lemma2.In order to show Theorem1(2), we will construct two families of strictly increasing functions y=h2n(x) and x=g2n+1(y) (n≥1) as follows. Set
(9)x=g2(y)=ag(y)(y≥0).
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)(y≥a).
Then x=g3(y) is a strictly increasing function which maps [a,+∞) onto [0,+∞).Assume that, for some positive integern, 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)(y≥0).
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) (n≥1).SetP0=[a,+∞)×[0,+∞) and Q0=[0,+∞)×[a,+∞). For any n≥1, 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=u≥0.
Thus x=g2(y)=a/g(y) and L2={(x,y):y=h2(x),x≥a}.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=u≥a.
Thus x=g3(y)=h2(y)/g(y)(y≥a) and L3={(x,y):x=g3(y),y≥a}. Using induction, one can easily show that, for any n≥1,
(15)L2n={(x,y):y=h2n(x),x≥a},L2n+1={(x,y):x=g2n+1(y),y≥a}.
Since f is a homeomorphism and Pn=f-2(Pn-1) with L2n∪R0 is the boundary of Pn, we have that, for any n≥1,
(16)Pn={(x,y):0≤y≤h2n(x),x≥a}.
In a similar fashion, we may show that
(17)Qn={(x,y):0≤x≤g2n+1(y),y≥a}.
Since L2⊂P0, L3⊂Q0, and f is a homeomorphism, we have that P1⊂P0 and Q1⊂Q0, which implies that, for any n≥1,
(18)L2n⊂Pn-1,L2n+1⊂Qn-1,Pn⊂Pn-1,Qn⊂Qn-1.
It follows from (12) and (18) that, for x≥a,
(19)0≤⋯≤h4(x)≤h2(x)
and for y≥a,
(20)0≤⋯≤g5(y)≤g3(y).
Noting (19) and (20), we may assume that, for every x≥a,
(21)H(x)=limn→∞h2n(x)
and for every y≥a,
(22)G(y)=limn→∞g2n+1(y).Set(23)L={(x,y):y=H(x),x≥a},M={(x,y):x=G(y),y≥a}.Lemma 3.
The following statements are true:(i)
f(L)=M and f(M)=L;(ii)
bothy=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=limn→∞h2n(x0), which follows that
(24)f(x0,y0)=f(x0,limn→∞h2n(x0))=limn→∞f(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)limn→∞yn=x0g(y0),limn→∞g2n-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=limn→∞g2n+1(y0), which follows that
(29)f-1(x0,y0)=f-1(limn→∞g2n+1(y0),y0)=limn→∞f-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)limn→∞zn=y0g(x0),limn→∞h2n+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) Sincey=h2n(x)(n≥1) are strictly increasing functions, we have that y=H(x) is an increasing function. For any x0>a, let
(34)limx→x0+H(x)=y0+,limx→x0-H(x)=y0-;
then y0+≥H(x0)≥y0-.
Now we claim thaty0+=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 thatlimx→a+H(x)=H(a)=0. Thus y=H(x)(x≥a) is an increasing continuous function. In a similar fashion, we may show that x=G(y)(y≥a) 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 a≤b. 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 Theorem1(2).
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)(x≥a) 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 n≥0,
(47)y2n-1≥rx2n-1,y2n≤x2n.
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 x0≠0 and {xn}n=-1∞ be a solution of (2) with initial value (x-1,x0).
Ifx-1<a, then it follows from Theorem 1(1) and (2) that limn→∞x2n-1<a which implies limn→∞(x2n-1,x2n)≠(a,0).
Ifx-1≥a and x0>H(x-1), then there exists n≥0 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).
Ifx-1≥a 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 n≥1,
(51)y2n≥rx2n,x2n-1≥y2n-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).



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