# On the Global Asymptotic Stability of a Second-Order System of Difference Equations

Authors: Ibrahim Yalcinkaya
Journal: Discrete Dynamics in Nature and Society
 (2008)

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## Abstract

A sufficient condition is obtained for the global asymptotic
stability of the following system of difference equationszn+1=(tnzn−1+a)/(tn+zn−1),tn+1=(zntn−1+a)/(zn+tn−1),n=0,1,2,…, where the parameter aϵ(0,∞) and the initial values (zk,tk)ϵ(0,∞) (for k=−1,0).


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## Body

## 1. Introduction

Recently,  there has been an increasing interest in the
study of qualitative analyses of rational difference
equations and systems of difference equations. Difference equations appear
naturally as discrete analogues and as numerical solutions of differential and
delay differential equations having applications in biology, ecology, economy, physics, and so forth. Although difference
equations are very simple in form, it is extremely difficult to understand
thoroughly the global behaviors of their solutions (see [1–15] and the
references cited therein).In [9, 10] Papaschinopoluos and Schinas studied
the behavior of the positive solutions of the system of two Lyness difference
equations(1.1)xn+1=byn+cxn−1,yn+1=dxn+eyn−1,n=0,1,2,…,where b,c,d,e are positive constants and initial values x−1,x0,y−1,y0  are positive.Iričanin and Stević  [6] studied, among others, the following
system:(1.2)xn+1(1)=1+xn(k)xn−1(k−1),xn+1(2)=1+xn(1)xn−1(k),…,xn+1(k)=1+xn(k−1)xn−1(k−2),where k∈N.In [2], Amleh et al. proved that all positive
solutions of the difference equations(1.3)xn+1=xn+xn−1xn−2xnxn−1+xn−2,xn+1=xn−1+xnxn−2xnxn−1+xn−2,xn+1=xn+xn−1xn−2xnxn−2+xn−1,where initial values x−2,x−1,x0 are positive, converge to 1 as n→∞.Moreover, Xianyi and Deming  [8] proved that the unique
positive equilibrium  of the difference equation(1.4)xn+1=xnxn−1+axn+xn−1,n=0,1,2,…,where a∈[0,∞) and x−1,x0 are positive, is globally asymptotically
stable.In [1], we extended the
results obtained in [8] to the following difference
equation:(1.5)xn+1=xnxn−k+axn+xn−k,n=0,1,2,…,where k is nonnegative integer, a∈[0,∞) and x−k,…,x0 are positive and
are globally asymptotically stable.Also in [14], we extended
the results obtained in [8] to the following system of difference
equations:(1.6)zn+1=zntn−1+azn+tn−1,tn+1=tnzn−1+atn+zn−1,n=0,1,2,…,where a∈(0,∞) and the initial values (zk,tk)∈(0,∞) (for k=−1,0)
are globally asymptotically stable.In this paper, we consider the following system of
difference equations:(1.7)zn+1=tnzn−1+atn+zn−1,tn+1=zntn−1+azn+tn−1,n=0,1,2,…,where a∈(0,∞) and the initial values (zk,tk)∈(0,∞)  (for k=−1,0). Our main aim is to investigate the global
asymptotic behavior of its solutions.It is clear that the change of
variables(1.8)(zn,tn)=(axn,ayn)reduces the system (1.7) to the
system(1.9)xn+1=ynxn−1+1yn+xn−1,yn+1=xnyn−1+1xn+yn−1,n=0,1,2,…,where the initial values (xk,yk)∈(0,∞) for (k=−1,0).We need the following definitions and theorem.LetI be some interval of real numbers and
let(1.10)f,g:I×I→Ibe continuously differentiable
functions. Then, for all initial values (xk,yk)ϵI,k=−1,0,
the system of difference equations(1.11)xn+1=f(xn,yn−1),yn+1=g(yn,xn−1),n=0,1,2,…,has a unique solution {(xn,yn)}n=−1∞.Definition 1.1.
A point(x¯,y¯) is called an equilibrium point of the system
(1.11) if(1.12)x¯=f(x¯,y¯),y¯=g(x¯,y¯).It is easy to see that the system (1.9) has the
unique positive equilibrium (x¯,y¯)=(1,1) [7].Definition 1.2.
Let(x¯,y¯) be an equilibrium point of the system (1.11).
(a) An equilibrium point(x¯,y¯) is said to be stable if for any ε>0 there is δ>0 such that for every initial points (x−1,y−1) and (x0,y0) for which ∥(x−1,y−1)−(x¯,y¯)∥+∥(x0,y0)−(x¯,y¯)∥<δ, the iterates (xn,yn) of (x−1,y−1) and (x0,y0) satisfy ∥(xn,yn)−(x¯,y¯)∥<ε  for all n>0. An equilibrium point (x¯,y¯) is said to be unstable if it is not stable
(the Euclidean norm in ℝ2 given by ∥(x,y)∥=x2+y2) is denoted by
∥⋅∥.
(b) An equilibrium point(x¯,y¯) is said to be asymptotically stable if there
exists r>0 such that (xn,yn)→(x¯,y¯) as n→∞ for all (x−1,y−1) and (x0,y0) that satisfy ∥(x−1,y−1)−(x¯,y¯)∥+∥(x0,y0)−(x¯,y¯)∥<r  [7].Definition 1.3.
Let(x¯,y¯) be an equilibrium point of a map F=(f,g), where f and g are continuously differentiable functions at (x¯,y¯). The Jacobian matrix of F at (x¯,y¯) is the matrix(1.13)JF(x¯,y¯)=[∂f∂x(x¯,y¯)∂f∂y(x¯,y¯)∂g∂x(x¯,y¯)∂g∂y(x¯,y¯)].
The linear mapJF(x¯,y¯):ℝ2→ℝ2 given by(1.14)JF(p,q)(x¯,y¯)=[∂f∂x(x¯,y¯)x+∂f∂y(x¯,y¯)y∂g∂x(x¯,y¯)x+∂g∂y(x¯,y¯)y]is called the linearization of the mapF at (x¯,y¯) [7].Theorem 1.4 (linearized stability theorem [7]).
LetF=(f,g) be a continuously differentiable function
defined on an open set I in ℝ2,
and let (x¯,y¯) in I be an equilibrium point of the map F=(f,g).
(a)
If all the eigenvalues of the Jacobian matrixJF(x¯,y¯) have modulus less than one, then the
equilibrium point (x¯,y¯) is asymptotically stable.(b)
If at least one of the eigenvalues of the Jacobian
matrixJF(x¯,y¯) has modulus greater than one, then the
equilibrium point (x¯,y¯) is unstable.(c)
An equilibrium point(x¯,y¯) of the map F=(f,g) is locally asymptotically stable if and only
if every solution of the characteristic equation(1.15)λ2−trJF(x¯,y¯)λ+detJF(x¯,y¯)=0
lies inside the unit circle, that is, if and only
if(1.16)|trJF(x¯,y¯)|<1+detJF(x¯,y¯)<2.Definition 1.5.
Let(x¯,y¯) be positive equilibrium point of the system (1.11).
A  “string” of consecutive terms{xs,…,xm} (resp., {ys,…,ym}),s≥−1,m≤∞  is said to be a positive semicycle if xi⩾x¯ (resp., yi⩾y¯), iϵ{s,…,m},xs−1<x¯  (resp., ys−1<y¯), and xm+1<x¯ (resp., ym+1<y¯).
A  “string” of consecutive terms{xs,…,xm} (resp., {ys,…,ym}),s≥−1,m≤∞ is said to be a negative semicycle if xi<x¯ (resp., yi<y¯), iϵ{s,…,m},xs−1⩾x¯  (resp.,  ys−1⩾y¯), and xm+1⩾x¯  (resp., ym+1⩾y¯).
A “string”  of consecutive terms{(xs,ys),…,(xm,ym)} is said to be a positive (resp., negative)
semicycle if {xs,…,xm},{ys,…,ym} are positive (resp., negative) semicycles.
Finally, a “string” of consecutive terms {(xs,ys),…,(xm,ym)} is said to be a semicycle positive (resp.,
negative) with respect to xn and negative (resp., positive) with respect to yn if {xs,…,xm} is a positive (resp., negative) semicycle and {ys,…,ym}  is a negative (resp., positive) semicycle [9].We now make new definitions. These definitions can be
used for different subsequences of{xn} (resp., {yn}).Definition 1.6.
Let(x¯,y¯) be positive equilibrium point of the system (1.11).
A “string” of consecutive terms{x2s,x2s+2,…,x2m} ( resp., {y2s,…,y2m}),s≥1,m≤∞ is said to be a positive sub-semicycle
associated with {x2n} (resp., {y2n})  if xi⩾x¯ (resp., yi⩾y¯), iϵ{2s,2s+2,…,2m},x2s−2<x¯  (resp.,  y2s−2<y¯),  and x2m+2<x¯ (resp., y2m+2<y¯).
A “string” of consecutive terms{x2s,x2s+2,…,x2m} ( resp., {y2s,…,y2m}),s≥1,m≤∞ is said to be a negative sub-semicycle
associated with {x2n} (resp., {y2n}) if xi<x¯ (resp., yi<y¯), iϵ{2s,2s+2,…,2m},x2s−2⩾x¯  (resp., y2s−2⩾y¯), and x2m+2⩾x¯ (resp., y2m+2⩾y¯).
A “string” of consecutive terms{(x2s,y2s),(x2s+2,y2s+2),…,(x2m,y2m)} is said to be a positive (resp., negative)
sub-semicycle if {x2s,x2s+2,…,x2m},{y2s,…,y2m} are positive (resp., negative) sub-semicycles.
Finally, a “string” of consecutive terms {(x2s,y2s),(x2s+2,y2s+2),…,(x2m,y2m)} is said to be a sub-semicycle positive (resp.,
negative) with respect to x2n and negative (resp., positive) with respect to y2n if {x2s,x2s+2,…,x2m} is a positive (resp., negative) sub-semicycle
and {y2s,…,y2m} is a negative (resp., positive) sub-semicycle.Definition 1.7.
Let(x¯,y¯) be positive equilibrium point of the system (1.11).
A “string” of consecutive terms{x2s−1,x2s+1,…,x2m+1} (resp., {y2s−1,…,y2m+1}),s≥1,m≤∞,
is said to be a positive sub-semicycle associated with {x2n−1} (resp., {y2n−1}) if xi⩾x¯ (resp., yi⩾y¯), iϵ{2s+1,2s+1,…,2m+1},x2s−3<x¯ (resp., y2s−3<y¯), and x2m+3<x¯ (resp., y2m+3<y¯).
A “string” of consecutive terms{x2s−1,x2s+1,…,x2m+1} (resp., {y2s−1,…,y2m+1}),s≥1,m≤∞,
is said to be a negative sub-semicycle associated with {x2n−1} (resp., {y2n−1}) if xi<x¯ (resp., yi<y¯), iϵ{2s−1,2s+1,…,2m+1},x2s−3⩾x¯ (resp., y2s−3⩾y¯), and x2m+3⩾x¯ (resp., y2m+3⩾y¯).
A “string” of consecutive terms{(x2s−1,y2s−1),(x2s+1,y2s+1),…,(x2m+1,y2m+1)} is said to be a positive (resp., negative)
sub-semicycle if {x2s−1,x2s+1,…,x2m+1},{y2s−1,…,y2m+1} are positive (resp., negative) sub-semicycles.
Finally, a “string” of consecutive terms {(x2s−1,y2s−1),(x2s+1,y2s+1),…,(x2m+1,y2m+1)} is said to be a sub-semicycle positive (resp.,
negative) with respect to x2n−1 and negative (resp., positive) with respect to y2n−1,
if {x2s−1,x2s+1,…,x2m+1} is a positive (resp., negative) sub-semicycle
and {y2s−1,y2s+1,…,y2m+1} is a negative (resp., positive) sub-semicycle.


## 2. Some Auxiliary Results

In this section, we give the following lemmas which
show us the behavior of semicycles of positive solutions of system (1.9). The
proof of Lemma 2.1 is clear from (1.9). So, it will be omitted.Lemma 2.1.
Assume
that{(xn,yn)}n=−1∞ is a solution of the system (1.9) and consider
the following cases:
(Case a)x0=x−1=1;
(Case b)y0=y−1=1;
(Case c)x0=y0=1;
(Case d)x−1=y−1=1.
If one of the above cases occurs, then every positive
solution of system (1.9) is equal to (1,1).Lemma 2.2.
Assume that{(xn,yn)}n=−1∞ is a positive solution of the system (1.9) which
is not eventually equal to (1,1).
Then the following statements are true:
(i)
(xn+1−xn−1)(xn−1−1)<0and(yn+1−yn−1)(yn−1−1)<0foralln≥0;(ii)
(xn+1−1)(xn−1−1)(yn−1)>0and(yn+1−1)(yn−1−1)(xn−1)>0foralln≥0.Proof.
In view
of system (1.9), we obtain(2.1)xn+1−xn−1=(1−xn−1)(1+xn−1)yn+xn−1,yn+1−yn−1=(1−yn−1)(1+yn−1)xn+yn−1,xn+1−1=(xn−1−1)(yn−1)yn+xn−1,yn+1−1=(yn−1−1)(xn−1)xn+yn−1 for n=0,1,2,…, from which the inequalities in (i) and (ii)
follow.Lemma 2.3.
Assume that{(xn,yn)}n=−1∞ is a solution of system (1.9) and suppose that the case,
(Case 1)xk,yk>1 ( for k=−1,0),  holds.
Then,(x2n−1,y2n−1) and (x2n,y2n) are positive sub-semicycles of system (1.9) with
an infinite number of terms and they monotonically tend to the positive
equilibrium (x¯,y¯)=(1,1).Proof.
Ifxk,yk>1  (for k=−1,0), then by Lemma  2.2(ii), it follows that(2.2)x2n−1,x2n,y2n−1,y2n>1∀n≥0,that is, these positive
sub-semicycles have an infinite number of terms. Furthermore, according to
Lemma 2.2(i), we know that (x2n−1,y2n−1) and (x2n,y2n) are strictly decreasing for all n≥0. So, the limits(2.3)limn→∞x2n−1=l1,limn→∞x2n=l2,limn→∞y2n−1=l3,limn→∞y2n=l4. exist and are finite. From (1.9), we can
write(2.4)x2n+1=y2nx2n−1+1y2n+x2n−1,y2n+1=x2ny2n−1+1x2n+y2n−1,n=0,1,2,…, taking limits on both sides of (2.4), we
have(2.5)l1=l4l1+1l4+l1,l3=l2l3+1l2+l3, and thus l1=l3=1. Similarly, one can see that l2=l4=1. Therefore, the proof is complete.Lemma 2.4.
Assume that{(xn,yn)}n=−1∞ is a solution of system (1.9), and consider the following cases:
(Case 2)x−1,y−1>1 and  x0,y0<1;
(Case 3)x−1,y−1<1  and x0,y0>1;
(Case 4)x−1,x0<1 and y−1,y0>1;
(Case 5)x−1,x0>1 and y−1,y0<1;
(Case 6)x−1,y−1,y0<1 and x0>1;
(Case 7)x0,y−1,y0<1 and x−1>1;
(Case 8)x−1,x0,y−1<1 and y0>1;
(Case 9)x−1,x0,y0<1 and y−1>1;
(Case 10)x−1,x0,y−1,y0<1.
If one of the above cases occurs, then the following
hold.
(i)
Every positive sub-semicycle associated with{x2n−1} and {x2n} (resp., {y2n−1} and {y2n}) of system (1.9) consists of one term.(ii)
Every negative sub-semicycle associated with{x2n−1} and {x2n} (resp., {y2n−1} and {y2n}) of system (1.9) consists of two terms.(iii)
Every positive sub-semicycle of length one is
followed by a negative sub-semicycle of length two.(iv)
Every negative sub-semicycle of length two is
followed by a positive sub-semicycle of length one.Proof.
If Case 2 occurs, then in view of
inequality (ii) of Lemma2.2 we have: x1,y1<1;x3,y3<1 and(2.6)x6n+5,y6n+5>1;x6n+7,y6n+7<1,x6n+9,y6n+9<1∀n≥0,which imply that every positive
sub-semicycle associated with {x2n−1} and {y2n−1} of system (1.9) of length one is followed by a negative
sub-semicycle of length two, which in turn is followed by a positive
sub-semicycle of length one.
Similarly, if Case 2 occurs, then in view of
inequality (ii) of Lemma2.2 we have(2.7)x6n+2,y6n+2>1;x6n+4,y6n+4<1,x6n+6,y6n+6<1∀n≥0,which imply that every positive
sub-semicycle associated with {x2n} and {y2n} of system (1.9) of length two is followed by a negative
sub-semicycle of length four, which in turn is followed by a positive
sub-semicycle of length two.
Proofs of the other cases are similar, so they will be
omitted. Therefore, the proof is complete.We omit the proofs of the following two results since
they can easily be obtained in a way similar to the proof of Lemma2.4.Lemma 2.5.
Assume that{(xn,yn)}n=−1∞ is a solution of system (1.9) and consider the
following cases:
(Case 11)x−1,y−1,y0>1 and   x0<1;
(Case 12)x−1,x0,y0>1 and y−1<1;
(Case 13)x−1,y0>1 and x0,y−1<1.
If one of the above cases occurs, then the following
hold.
(i)
(x2n−1,y2n) is a positive sub-semicycle of system (1.9) with
an infinite number of terms  (monotonically tend to the positive equilibrium (x¯,y¯)=(1,1)).(ii)
Every positive sub-semicycle associated with{x2n} and {y2n−1} of system (1.9) consists of one term.(iii)
Every negative
sub-semicycle associated with{x2n} and {y2n−1} of system (1.9) consists of two terms.(iv)
Every positive sub-semicycle of length one is
followed by a negative sub-semicycle of length two.(v)
Every negative sub-semicycle of length two is
followed by a positive sub-semicycle of length one.Lemma 2.6.
Assume
that{(xn,yn)}n=−1∞ is a solution of system (1.9) and consider the
following cases:
(Case 14)x−1,x0,y−1>1 and y0<1;
(Case 15)x0,y−1,y0>1 and x−1<1;
(Case 16)x0,y−1>1 and x−1,y0<1.
If one of the above cases occurs, then the following
hold.
(i)
(x2n,y2n−1) is a positive sub-semicycle of system (1.9) with
an infinite number of terms  (monotonically tend to the positive equilibrium (x¯,y¯)=(1,1)).(ii)
Every positive sub-semicycle associated with{x2n−1} and {y2n} of system (1.9) consists of one term.(iii)
Every negative
sub-semicycle associated with{x2n−1} and {y2n} of system (1.9) consists of two terms.(iv)
Every positive sub-semicycle of length one is
followed by a negative sub-semicycle of length two.(v)
Every negative sub-semicycle of length two is
followed by a positive sub-semicycle of length one.


## 3. Main Result

Theorem 3.1.
The positive equilibrium point(x¯,y¯)=(1,1) of the system (1.9) is globally asymptotically
stable.Proof.
We must show that the positive
equilibrium point(x¯,y¯)=(1,1) of the system (1.9) is both locally
asymptotically stable and (xn,yn)→(x¯,y¯) as n→∞   (or equivalently (x2n−1,y2n−1)→(x¯,y¯) and (x2n,y2n)→(x¯,y¯) as n→∞). The characteristic equation of the system (1.9)
about the positive equilibrium point (x¯,y¯)=(1,1) is(3.1)λ2−0.λ+0=0,and so it is clear from Theorem
1.4 that positive equilibrium point (x¯,y¯)=(1,1) of the system (1.9) is locally asymptotically
stable. It remains to verify that every positive solution {(xn,yn)}n=−1∞ of the system (1.9) converges to (x¯,y¯)=(1,1) as n→∞. Namely, we want to prove(3.2)limn→∞x2n=limn→∞x2n−1=x¯=1,limn→∞y2n=limn→∞y2n−1=y¯=1.
If the solution{(xn,yn)}n=−1∞ of (1.9) is nonoscillatory about the positive
equilibrium point (x¯,y¯)=(1,1) of the system (1.9), then according to Lemmas 2.1
and 2.3, respectively, we know that the solution is either eventually equal to (1,1) or an eventually positive one which has an
infinite number of terms and monotonically tends the positive equilibrium point (x¯,y¯)=(1,1) of the
system (1.9) and so (3.2) holds. Therefore, it suffices to prove that (3.2) holds for
strictly oscillatory solutions. Now, let {(xn,yn)}n=−1∞ be strictly oscillatory about the positive
equilibrium point (x¯,y¯)=(1,1) of the system (1.9). By virtue of Lemmas 2.2(ii)
and 2.4, one can see that every positive sub-semicycle associated with {x2n−1} (resp. {x2n},{y2n−1},{y2n}) of this solution has one term, and every
negative sub-semicycle associated with {x2n−1} (resp., {x2n},{y2n−1},{y2n}) except perhaps for the first has exactly two
terms. Every positive sub-semicycle of length one is followed by a negative
sub-semicycle of length two.
We consider the sub-semicycles associated with{x2n} and {y2n}.
For the convenience of statement, without loss of
generality, we use the following notation. We denote byx2p  and y2p the terms of a positive sub-semicycle of
length one, followed by x2p+2,x2p+4  and y2p+2,y2p+4 which are the terms of a negative
sub-semicycle of length two. Afterwards, there are the positive sub-semicycles x2p+6 and y2p+6 in turn followed by the negative
sub-semicycles, and so on.
Therefore, we have the following sequences consisting
of positive and negative sub-semicycles (forn=0,1,…):(3.3){x2p+6n}n=0∞,{x2p+6n+2,xp+6n+4}n=0∞,{y2p+6n}n=0∞,{y2p+6n+2,y2p+6n+4}n=0∞.
We have the following assertions:
(i)
x2p+6n+2<x2p+6n+4 and y2p+6n+2<y2p+6n+4;(ii)
x2p+6nx2p+6n+2>1 and y2p+6ny2p+6n+2>1;(iii)
x2p+6n+4x2p+6n+6<1 and y2p+6n+4y2p+6n+6<1.
In fact, inequality (i) immediately follows from Lemma2.2(i). From the observations that(3.4)x2p+6n+2=y2p+6n+1x2p+6n+1y2p+6n+1+x2p+6n>y2p+6n+1x2p+6n+1y2p+6n+1x2p+6n2+x2p+6n=1x2p+6n,y2p+6n+2=x2p+6n+1y2p+6n+1x2p+6n+1+y2p+6n>x2p+6n+1y2p+6n+1x2p+6n+1y2p+6n2+y2p+6n=1y2p+6n, one can see that (ii) is valid.
As for (iii), it is obtained from(3.5)x2p+6n+6=y2p+6n+5x2p+6n+4+1y2p+6n+5+x2p+6n+4<y2p+6n+5x2p+6n+4+1y2p+6n+5x2p+6n+42+x2p+6n+4=1x2p+6n+4,y2p+6n+6=x2p+6n+5y2p+6n+4+1x2p+6n+5+y2p+6n+4<x2p+6n+5y2p+6n+4+1x2p+6n+5y2p+6n+42+y2p+6n+4=1y2p+6n+4, for n=0,1,2,….
Combining the above inequalities, we
derive(3.6)1x2p+6n<x2p+6n+2<x2p+6n+4<1x2p+6n+6,1y2p+6n<y2p+6n+2<y2p+6n+4<1y2p+6n+6.
From (3.6), one can see that {x2p+6n+2}n=0∞ and {y2p+6n+2}n=0∞ are increasing with upper bound 1. So the
limits(3.7)limn→∞x2p+6n+2=L1,limn→∞y2p+6n+2=L2 exist and are finite. Accordingly, in view of (3.6), we
obtain(3.8)limn→∞x2p+6n+4=L1,limn→∞x2p+6n+6=1L1,limn→∞y2p+6n+4=L2,limn→∞y2p+6n+6=1L2.
Now,  we consider the sub-semicycles associated with{x2n−1}  and  {y2n−1}.
Similarly, for the convenience of statement, without
loss of generality, we use the following notation. We denote byx2p+1 and y2p+1 the terms of a positive sub-semicycle of length
one, followed by x2p+3,x2p+5 and y2p+3,y2p+5 which are the terms of a negative
sub-semicycle of length two. Afterwards, there are the positive sub-semicycles x2p+7 and y2p+7 in turn followed by the negative
sub-semicycles, and so on.
Therefore, we have the following sequences consisting
of positive and negative sub-semicycles (forn=0,1,…):(3.9){x2p+6n+1}n=0∞,{x2p+6n+3,xp+6n+5}n=0∞,{y2p+6n+1}n=0∞,{y2p+6n+3,y2p+6n+5}n=0∞.
We have the following assertions:
(i)
x2p+6n+3<x2p+6n+5 and  y2p+6n+3<y2p+6n+5;(ii)
x2p+6n+1x2p+6n+3>1 and  y2p+6n+1y2p+6n+3>1;(iii)
x2p+6n+5x2p+6n+7<1 and  y2p+6n+5y2p+6n+7<1.
Combining the above inequalities, we
derive(3.10)1x2p+6n+1<x2p+6n+3<x2p+6n+5<1x2p+6n+7,1y2p+6n+1<y2p+6n+3<y2p+6n+5<1y2p+6n+7.
From (3.10), one can see that {x2p+6n+3}n=0∞ and {y2p+6n+3}n=0∞  are increasing with upper bound 1. So the
limits(3.11)limn→∞x2p+6n+3=L3,limn→∞y2p+6n+3=L4 exist and are finite. Accordingly, in view of (3.10), we
obtain(3.12)limn→∞x2p+6n+5=L3,limn→∞x2p+6n+7=1L3,limn→∞y2p+6n+5=L4,limn→∞y2p+6n+7=1L4.
It suffices to verify that(3.13)L1=L2=L3=L4=1.
To this end, note that(3.14)x2p+6n+6=y2p+6n+5x2p+6n+4+1y2p+6n+5+x2p+6n+4,y2p+6n+6=x2p+6n+5y2p+6n+4+1x2p+6n+5+y2p+6n+4.
Take the limits on both sides of the above equality
and obtain(3.15)1L1=L4⋅L1+1L4+L1,1L2=L3⋅L2+1L3+L2, which imply that L1=L2=1. Similarly, one can see that L3=L4=1.
Moreover, by virtue of Lemmas2.2(ii) and 2.5 (resp., 2.6),
one can see that (3.2) holds. Therefore, the proof is complete.



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*Source: 860152-2008-11-10.xml*