# Boundedness, Attractivity,  and Stability of a Rational Difference Equation with Two Periodic Coefficients

Authors: G. Papaschinopoulos; G. Stefanidou; C. J. Schinas
Journal: Discrete Dynamics in Nature and Society
 (2009)

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## Abstract

We study the boundedness, the attractivity, and the stability of the
positive solutions of the rational difference equationxn+1=(pnxn−2+xn−3)/(qn+xn−3),  n=0,1,…,
where pn,qn,
n=0,1,… are positive sequences of period 2.


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## Body

## 1. Introduction

In [1],  Camouzis et al. studied the global character of the positive solutions of the difference equation: (1.1)xn+1=δxn−2+xn−3A+xn−3,n=0,1,…, where δ,A are positive
parameters and the initial values x−3,x−2,x−1,x0 are positive
real numbers.The
mathematical modeling of a physical, physiological, or economical problem very
often leads to difference equations (for partial review of the theory of
difference equations and their applications see [2–12]). Moreover, a lot of
difference equations with periodic coefficients have been applied in
mathematical models in biology (see [13–15]). In addition, between others in [16–19], we can see some more
difference equations with periodic coefficients that have been studied.In this paper,
we investigate the difference equation(1.2)xn+1=pnxn−2+xn−3qn+xn−3,n=0,1,…, where pn,qn, n=0,1,…  are positive
sequences of period 2 and the initial
values xi,i=−3,−2,−1,0  are positive
numbers.Our goal in
this paper is to extend some results obtained in [1]. More precisely, we study
the existence of a unique positive periodic solution of (1.2) of prime
period 2. In the sequel, we investigate the boundedness, the
persistence, and the convergence of the positive solutions to the unique
periodic solution of (1.2). Finally, we study the stability of the positive
periodic solution and the zero solution of (1.2).If we setyn=x2n−1, zn=x2n, it is easy to prove that (1.2) is equivalent to
the following system of difference equations:(1.3)yn+1=p0zn−1+yn−1q0+yn−1,zn+1=p1yn+zn−1q1+zn−1,n=0,1,…, where pi,qi, i=0,1 are positive
constants and the initial values yi,zi, i=−1,0 are positive
numbers. So in order to study (1.2) we investigate system (1.3).


## 2. Existence of the Unique Positive
Equilibrium of System (1.3)

In the
following proposition, we study the existence of the unique positive equilibrium
of system (1.3).Proposition 2.1.
Consider system (1.3) where  pi,qi,  i=0,1 are positive
constants and the initial values yi,zi,  i=−1,0  are positive
numbers. Suppose that (2.1)q0−1<p0,q1−1<p1  are satisfied. Then system (1.3) possesses a unique positive equilibrium.Proof.
Let(y,z) be a positive
equilibrium of system (1.3) then(2.2)y=p0z+yq0+y,z=p1y+zq1+z.Equations (2.2) imply that z is a solution
of the equation(2.3)f(x)=x3+2(q1−1)x2+[(q1−1)2+p1(q0−1)]x+(q1−1)(q0−1)p1−p0p12=0.
Suppose
that(2.4)q1≥1. Let λ1,λ2,  and
λ3 be the
solutions of (2.3). Then from (2.1), (2.3), and (2.4) we
take(2.5)λ1+λ2+λ3=2(1−q1)≤0,λ1λ2λ3=−(q1−1)(q0−1)p1+p0p12>0, and so (2.3) has unique positive solution z. Then from (2.2) and (2.4) we have(2.6)z>1−q1,y=z2+(q1−1)zp1>0,and so system (1.3) has a unique positive equilibrium.
Now suppose
that(2.7)q1<1,(q1−1)(q0−1)>p1p0. If λ1,λ2, and
λ3 are the
solutions of (2.3), then from (2.3) and (2.7) we take(2.8)λ1+λ2+λ3=2(1−q1)>0,λ1λ2λ3=−(q1−1)(q0−1)p1+p0p12<0, and so (2.3) has a negative solution, but also (2.3) has a solution in
the interval (0,1−q1), since(2.9)f(0)=(q1−1)(q0−1)p1−p0p12>0,f(1−q1)=−p0p12<0.Moreover, (2.3) has a solution z in the interval (1−q1,∞), since(2.10)limx→∞f(x)=∞, therefore, we get (2.6) and so system (1.3) has a unique positive equilibrium.
Finally,
suppose that(2.11)q1<1,(q1−1)(q0−1)<p1p0. If λ1,λ2, and
λ3 are the
solutions of (2.3), then from (2.3) and (2.11), we take(2.12)λ1+λ2+λ3=2(1−q1)>0,λ1λ2λ3=−(q1−1)(q0−1)p1+p0p12>0. We have limx→∞f(x)=∞, and since f(1−q1)<0, it is obvious that (2.3) has a solution z in the interval (1−q1,∞). From (2.3), we get(2.13)f′(x)=3x2+4x(q1−1)+(q1−1)2+p1(q0−1).
If equationf′(x)=0 has complex
roots, then it is obvious that z is the unique
solution of (2.3). Therefore, we get (2.6), and so system (1.3) has a unique
positive equilibrium.
Now, suppose
that the roots off′(x)=0(2.14)μ1=2(1−q1)−D3,μ2=2(1−q1)+D3,D=(1−q1)2+3p1(1−q0), are real numbers.
Suppose thatq0<1, then it is
obvious that(2.15)μ1<1−q1<μ2,and so we have that (2.3) has a unique solution z∈(1−q1,∞).
Ifq0≥1, then it holds
that(2.16)0<μ1≤μ2≤1−q1, which implies that (2.3) has a unique solution z∈(1−q1,∞).
Therefore, we
can take (2.6) and so system (1.3) has a unique positive equilibrium. This
completes the proof of the proposition.


## 3. Boundedness and Persistence of the
Solutions of System (1.3)

In the
following propositions we study the boundedness and the persistence of the
positive solutions of system (1.3). In the sequel we will use the following
result which has proved in [20].Theorem 3.1.
Assume that all roots of the
polynomial(3.1)P(t)=tN−s1tN−1−⋯−sN, where s1,s2,…,sN≥0 have absolute
value less than 1, and let yn be a
nonnegative solution of the inequality(3.2)yn+N≤s1yn+N−1+⋯+sNyn+zn. Then, the following statements are true.
(i)
Ifzn is a
nonnegative bounded sequence, then yn is also
bounded.(ii)
Iflimn→∞zn=0, then limn→∞yn=0.Proposition 3.2.
One considers the system of difference equations (1.3) where  pi,qi,  i=0,1  are positive
constants and the initial values  yi,zi,    i=−1,0  are positive
numbers. Then the following statements are true.
(i)
If(3.3)q0q1p0p1≥1, then every solution of (1.3) is bounded.(ii)
If(3.4)q0−1<p0≤q0,q1−1<p1≤q1,then every solution of (1.3) is bounded and persists.Proof.
Let(yn,zn) be an arbitrary
solution of (1.3).
(i) From (3.3),
we get that one of the three following conditions holds:(3.5)q0p0>1,(3.6)q1p1>1,(3.7)p0=q0=p,p1=q1=q.
We assume that
(3.5) is satisfied. We prove that there exists a positive integer N such
that(3.8)yn<1,zn<q0p0,n≥N. First, we show that if there exists a positive integer n0 such
that(3.9)zn0<q0p0, then(3.10)zn0+3p<q0p0,p=0,1,…. In contradiction, we assume that(3.11)zn0+3=p1yn0+2+zn0+1q1+zn0+1≥q0p0.Using relations (1.3), (3.5), and (3.11), we get that(3.12)yn0+2=p0zn0+yn0q0+yn0>q0q1p0p1, and so relations (1.3) and (3.3) imply that(3.13)zn0>q02q1p02p1>q0p0, which contradicts (3.9). So zn0+3<q0/p0 and working
inductively, we get (3.10).
Ifz−1<q0/p0, then from the
analogous relations (3.9) and (3.10), we get(3.14)z−1+3p<q0p0,p=0,1,….Now, suppose that(3.15)z−1≥q0p0, we prove that there exists a positive integer q such
that(3.16)z−1+3q<q0p0. From (3.3), there exists a positive integer h such
that(3.17)z−1<(q0q1p0p1)h. If z2<q0/p0, then (3.16) is
true for q=1.
Now, suppose
that(3.18)z2≥q0p0. Then from (1.3), (3.5), and (3.18), we get y1>q0q1/p0p1 and so from
(1.3), (3.3), and (3.5), we have that(3.19)z−1>q1q02p1p02>q1q0p1p0. If z5<q0/p0, then (3.16) is
true for q=2.
Now, suppose
that(3.20)z5≥q0p0. Using (1.3), (3.3), (3.5), (3.20) and arguing as to prove (3.19) we
get(3.21)z−1>(q1q0p1p0)2.Working inductively, we get that(3.22)ifz−1+3w≥q0p0,w=1,2,…,thenz−1>(q1q0p1p0)w. From (3.22) for w=h, we get z−1>(q1q0/p1p0)h which contradicts
(3.17). So z−1+3h<q0/p0 which means
that (3.16) holds for q=h.
Arguing as forz−1, we can prove
that there exist positive integers k,l such
that(3.23)z0+3k<q0p0,z1+3l<q0p0. From (3.16) and (3.23), we get that there exists a positive integer r such
that(3.24)zr<q0p0,n≥r. Finally, from (1.3) and (3.24), we get yr+2<1 and so (3.8) is
true for N=r+2.
Similarly, we
can prove that if (3.6) holds, then there exists a positive integer N such
that(3.25)zn<1,yn<q1p1,n≥N.
Finally,
suppose that (3.7) hold. From (1.3) and (3.7), we have(3.26)yn+1−1=p(zn−1−1)p+yn−1,zn+1−1=q(yn−1)q+zn−1, and so,(3.27)yn+1−1=pp+yn−1qq+zn−3(yn−2−1). From (3.27), we get(3.28)0≤yn+1−1≤yn−2−1,or0≥yn+1−1≥yn−2−1, and so the subsequences y3n,y3n+1,y3n+2 either are
bounded from below by 1 and decreasing
or bounded from above by 1 and increasing.
Hence, yn is bounded and
persists. Similarly, we can prove that zn is bounded and
persists. This completes the proof of part (i) of the proposition.
(ii) In
statement (i), we have already proved that if (3.7) hold, then every solution of
(1.3) is bounded and persists. So, from (3.4), it remains to show that if
either(3.29)q0−1<p0<q0,q1−1<p1≤q1, or(3.30)q0−1<p0≤q0,q1−1<p1<q1,holds, then the solution (yn,zn) persists. From
(3.3), (3.8), (3.25), (3.29), and (3.30), we get that(3.31)yn<q1p1,zn<q0p0,n≥N. We consider the positive number m such
that(3.32)m<min{yN,zN,yN+1,zN+1,p0+1−q0,p1+1−q1}. Moreover, if(3.33)f(y,z)=p0z+yq0+y,g(y,z)=p1y+zq1+z, then it is easy to see that for the functions (3.33), f is increasing
with respect to y for any z, z<q0/p0 and g is increasing
with respect to z for any y,y<q1/p1.
Therefore, from
(1.3), (3.31), and (3.32) we have(3.34)yN+2>(p0+1)mq0+m>m,zN+2>(p1+1)mq1+m>m, and working inductively, we take(3.35)yN+s≥m,zN+s≥m,s=0,1,…. Therefore, (yn,zn) persists and
using statement (i), then (yn,zn) is bounded and
persists. This completes the proof of the proposition.Proposition 3.3.
One  considers the system of difference equations (1.3) where  pi,qi,    i=0,1
are positive
constants, and the initial values  yi,zi,    i=−1,0  are positive numbers.
Then, the following statements are true.
(i)
If(3.36)q0q1p0p1<1, then every solution of (1.3) persists.(ii)
If(3.37)q0≤p0≤q0+1,q1≤p1≤q1+1, then every solution of (1.3) is bounded and persists.Proof.
Let(yn,zn) be an arbitrary
solution of (1.3).
(i)
From (3.36),
we have(3.38)q0p0<1, or(3.39)q1p1<1.
Arguing as in
the proof of statement (i) of Proposition3.2, we can easily prove that if
(3.38) holds, then there exists a positive integer M such
that(3.40)yn>1,zn>q0p0,n≥M, and if (3.39) holds, then there exists a positive integer M such
that(3.41)zn>1,yn>q1p1,n≥M.
(ii)
From
Proposition3.2, we have that if (3.7) holds, then every solution of (1.3) is
bounded and persists. So, from (3.37), it remains to show that if
either(3.42)q0<p0≤q0+1,q1≤p1≤q1+1,or(3.43)q0≤p0≤q0+1,q1<p1≤q1+1, holds, then the solution (yn,zn) is bounded and
persists.
From (3.36),
(3.40), (3.41), (3.42), and (3.43), we get that(3.44)yn>q1p1,zn>q0p0,n≥M. Suppose that(3.45)p0≠q0+1orp1≠q1+1.
From (1.3) and
(3.44), we have(3.46)zM+1>1,yM+3>1.We have for the functions (3.33) that f is decreasing
with respect to y for any z, z>q0/p0 and g is decreasing
with respect to z for any y, y>q1/p1. Therefore, relations (1.3), (3.44), and (3.46) imply
that(3.47)zM+3≤p1yM+2+1q1+1, and so from (1.3) and (3.46),(3.48)yM+5≤p0p1(q0+1)(q1+1)yM+2+p0(q0+1)(q1+1)+1. Working inductively, we can prove that(3.49)yn+5≤p0p1(q0+1)(q1+1)yn+2+p0(q0+1)(q1+1)+1,n≥M.Then from (3.42), (3.43), (3.45), and Theorem 3.1, yn is bounded.
Similarly, we take that zn is bounded.
Therefore, from (3.44), the solution (yn,zn) is bounded and
persists.
Now, suppose
that(3.50)p0=q0+1,p1=q1+1. We claim that yn is bounded. For
the sake of contradiction, we assume that yn is not bounded.
Then, there exists a subsequence ni such
that(3.51)limi→∞yni+1=∞,(3.52)yni+1>max{yj,j<ni}. Moreover, from (1.3) and (3.50), we get(3.53)yni+1<q0+1q0zni−1+1, and so from (3.51),(3.54)limi→∞zni−1=∞. Moreover, from (1.3) and (3.50),(3.55)zni−1<q1+1q1yni−2+1, and so from (3.54),(3.56)limi→∞yni−2=∞. Working inductively, we can prove that(3.57)limi→∞yni+1−3s=∞,limi→∞zni−1−3s=∞,s=0,1,…. We claim that yni−6 is a bounded
sequence. Suppose on the contrary that there exists an unbounded subsequence of yni−6 and without
loss of generality, we may suppose that(3.58)limi→∞yni−6=∞. Arguing as above, we can easily prove that(3.59)limi→∞yni−9=limi→∞yni−12=∞. Also, since from (1.3),(3.60)yni−6=(q0+1)((zni−8)/(yni−8))+1q0/yni−8+1<(q0+1)zni−8yni−8+1, from (3.58), we have that limi→∞(zni−8/yni−8)=∞ and so
eventually,(3.61)zni−8>yni−8. From (1.3), (3.50), and (3.61), we have(3.62)yni+1=(q0+1)zni−1+yni−1q0+yni−1<q0+1q0zni−1+1=q0+1q0((q1+1)yni−2+zni−3q1+zni−3)+1<1+q0+1q0+q0+1q0q1+1q1yni−2<⋯<A+Byni−8<A+Bzni−8,  where(3.63)A=1+q0+1q0+q0+1q0q1+1q1+(q0+1q0)2q1+1q1+(q0+1q0)2(q1+1q1)2+(q0+1q0)3(q1+1q1)2,B=(q0+1q0)3(q1+1q1)3. Therefore, using (1.3) and (3.50), we get(3.64)yni+1<A+B((q1+1)yni−9+zni−10q1+zni−10), and since from (3.57) and (3.59), we have that yni−9→∞,zni−10→∞ as i→∞, we can easily
prove that eventually, (3.65)yni+1<yni−9, which contradicts to (3.52).
Therefore,yni−6 is a bounded
sequence. From (1.3), (3.50), and (3.57), we get(3.66)zni−5=(q1+1)yni−6+zni−7q1+zni−7=(q1+1)(yni−6/zni−7)+1q1/zni−7+1→1,i→∞. Similarly, from (1.3), (3.50) and (3.57) and (3.66) follows,(3.67)yni−3=(q0+1)zni−5+yni−5q0+yni−5=(q0+1)(zni−5/yni−5)+1q0/yni−5+1→1,i→∞. Now, we prove that(3.68)liminfi→∞yni−1>1. Otherwise, and without loss of generality, we may suppose that limi→∞yni−1≤1. So, relations
(1.3), (3.50), and (3.67) imply that(3.69)limi→∞yni−1=(q0+1)limi→∞zni−3+limi→∞yni−3q0+limi→∞yni−3≤1, and so (3.70)limi→∞zni−3≤q0q0+1.Moreover, from (1.3), (3.44), and (3.50), we get eventually(3.71)zni−3=(q1+1)yni−4+zni−5q1+zni−5>(q1+1)(q1/(q1+1))+zni−5q1+zni−5=1, and so from (3.66), limi→∞zni−3≥1 which
contradicts to (3.70).
Hence, (3.68)
is true.
Similarly, we
can prove that(3.72)liminfi→∞zni−3>1. Therefore, from (3.68) and (3.72), we have eventually(3.73)yni−1>1+k,zni−3>1+m, where k,m are positive
real numbers.
Hence, from
(1.3), (3.50), and (3.73) we have(3.74)yni+1=(q0+1)[((q1+1)yni−2+zni−3)/(q1+zni−3)]+yni−1q0+yni−1<(q0+1)(q1+1)(q1+1+m)(q0+1+k)yni−2+q0+1q0+1. Then from (3.57), we can prove that eventually(3.75)yni+1<yni−2, which contradicts to (3.52).
Therefore,yn is a bounded
sequence. Moreover, from (1.3), (3.50), we take that zn is bounded.
Therefore, the solution (yn,zn) is bounded and
persists. This completes the proof of the proposition.


## 4. Attractivity of the Positive
Equilibrium of System (1.3)

In the
following propositions, we study the convergency of the solutions of system
(1.3) to its positive equilibrium.Proposition 4.1.
One considers the system of difference equations (1.3) where  pi,qi,  i=0,1
are positive
constants, and the initial values  yi,zi,  i=−1,0  are positive
numbers. If either (3.29) or (3.30) hold, then every solution of (1.3) tents to
the positive equilibrium of (1.3).Proof.
Let(yn,zn) be an arbitrary
solution of (1.3). From Proposition 3.2, there exist(4.1)L1=limsupn→∞yn,L2=limsupn→∞zn,l1=liminfn→∞yn,l2=liminfn→∞zn,0<L1,L2,l1,l2<∞. From (1.3), (3.31), and the monotony of
functions (3.33), we have (4.2)L1≤p0L2+L1q0+L1,L2≤p1L1+L2q1+L2,l1≥p0l2+l1q0+l1,l2≥p1l1+l2q1+l2, and hence
(4.3)L12+L1(q0−1)−p0L2≤0,L22+L2(q1−1)−p1L1≤0,l12+l1(q0−1)−p0l2≥0,l22+l2(q1−1)−p1l1≥0.The third inequality of (4.3), implies that(4.4)l1≥1−q0+(1−q0)2+4p0l22,and so from the last inequality of (4.3), we have(4.5)2l22+2l2(q1−1)+(q0−1)p1≥p1(1−q0)2+4p0l2. Hence, we get(4.6)(2l22+2l2(q1−1)+(q0−1)p1)2≥(p1(1−q0)2+4p0l2)2, or (4.7)l23+2l22(q1−1)+l2[(q1−1)2+p1(q0−1)]+p1(q1−1)(q0−1)−p0p12≥0. The first inequality of (4.3), implies that(4.8)0<L1≤1−q0+(1−q0)2+4p0L22, and so from second inequality of (4.3), we get(4.9)2L22+2L2(q1−1)+(q0−1)p1≤p1(1−q0)2+4p0L2. Using (4.3), we have(4.10)L1≥l1>1−q0,L2≥l2>1−q1. Therefore, from (4.5) and (4.10), we get(4.11)2L22+2L2(q1−1)+(q0−1)p1=2L2(L2+q1−1)+(q0−1)p1≥2l22+2l2(q1−1)+(q0−1)p1>0.
Using (4.9) and
(4.11), we have(4.12)L23+2L22(q1−1)+L2[(q1−1)2+p1(q0−1)]+p1(q1−1)(q0−1)−p0p12≤0. In Proposition 2.1, we proved that (2.3) has a unique positive solution z,z∈(1−q1,∞). We can write(4.13)f(x)=(x−z)(x2+ax+b),a,b∈ℝ, where f(x) is defined in
(2.3) and x2+ax+b>0 for any x>1−q1. Then from (4.7), (4.12), and (4.13), we have(4.14)(L2−z)(L22+aL2+b)≤0,(l2−z)(l22+al2+b)≥0. Therefore, from (4.10) and (4.14),L2≤z≤l2, which implies that(4.15)L2=l2=z.
In addition,
using (4.15), the first and the third inequalities of (4.3), we
have(4.16)L12+(q0−1)L1≤l12+(q0−1)l1, and so (4.10) implies that(4.17)L1=l1. This completes the proof of the proposition.Proposition 4.2.
One considers the system of difference equations (1.3) where  pi,qi,  i=0,1 are positive
constants, and the initial values yi,zi,  i=−1,0 are positive
numbers. If either (3.42) or (3.43) hold, then every solution of (1.3) tents to
the positive equilibrium of (1.3).Proof.
Let(yn,zn) be an arbitrary
solution of (1.3). From Proposition 3.3, there exist Li,li, i=1,2 such that (4.1)
are satisfied.
From (1.3), the
monotony of functions (3.33) and (3.44), we have(4.18)L1≤p0L2+l1q0+l1,L2≤p1L1+l2q1+l2,l1≥p0l2+L1q0+L1,l2≥p1l1+L2q1+L2, and hence(4.19)L1l1+L1q0≤p0L2+l1,L1l1+l1q0≥p0l2+L1,L2l2+L2q1≤p1L1+l2,L2l2+l2q1≥p1l1+L2, which implies that(4.20)(1+q0)(L1−l1)≤p0(L2−l2),(1+q1)(L2−l2)≤p1(L1−l1). Therefore,(4.21)[(1+q0)(1+q1)−p0p1](L1−l1)≤0. First suppose that (3.45) holds. Then from (3.42) or (3.43), and (3.45), we get L1−l1≤0, which means
that(4.22)L1=l1.Using (4.20), it is obvious that(4.23)L2=l2. So if (3.45) holds, the proof is completed.
Now, suppose
that (3.50) hold. Then from (4.20), we have(4.24)L2−l2=L1−l1. Moreover, from (4.24), it follows that(4.25)(q0+1)l2+L1−l1q0=(q0+1)L2+l1−L1q0. In addition, from (3.50), the first and the second inequalities of (4.19), we
get(4.26)(q0+1)l2+L1−l1q0≤L1l1≤(q0+1)L2+l1−L1q0.Therefore, from (4.25) and (4.26), we have(4.27)L1=(q0+1)L2+l1q0+l1. We may assume that there exists a positive integer ni such
that(4.28)limi→∞yni−j=Aj,limi→∞zni−j=Bj,limi→∞yni+1=L1. Moreover, from (1.3), (3.50), and (4.28), we get(4.29)L1=(q0+1)B1+A1q0+A1. Since f(x,y)=((q0+1)x+y)/(q0+y) is decreasing
with respect to y, for any x>(q0/(q0+1)), if B1<L2 or l1<A1, then from
(3.44), and (3.50), we get(4.30)L1<(q0+1)L2+l1q0+l1, which contradicts to (4.27). So,(4.31)B1=L2,l1=A1. Using the same argument, we can prove that(4.32)A2=L1,B3=l2,B3=l2,A3=L1,B4=L2,A4=l1,A4=l1,B5=L2,B5=L2,A5=l1,A5=L1,B6=l2, and so L1=l1=A. Also, from
(4.24), we have L2=l2=B. Therefore,(4.33)limn→∞yn=A,limn→∞zn=B, where obviously A=B=2. This completes the proof of the proposition.Proposition 4.3.
One  considers the system of difference equations (1.3) where pi,qi,
i=0,1 are positive
constants, and the initial values yi,zi, i=−1,0 are positive
numbers. If relations (3.7) hold, then every solution of (1.3) tents to the
positive equilibrium (1,1) of (1.3).Proof.
Let(yn,zn) be an arbitrary
solution of (1.3). From the proof of Proposition 3.2, the subsequences y3n,y3n+1,y3n+2,z3n,z3n+1, and  z3n+2    are monotone
and yn, zn are bounded and
persist. So, there exist positive numbers L1,L2,L3,M1,M2, and  M3 such
that(4.34)L1=limn→∞y3n,L2=limn→∞y3n+1,L3=limn→∞y3n+2,M1=limn→∞z3n,M2=limn→∞z3n+1,M3=limn→∞z3n+2, and from (1.3) and (3.7), we get(4.35)L1=pM2+L2p+L2,M1=qL3+M2q+M2,L2=pM3+L3p+L3,M2=qL1+M3q+M3,L3=pM1+L1p+L1,M3=qL2+M1q+M1. Then, we have(4.36)L1p+L1L2=pM2+L2,M1q+M1M2=qL3+M2,L2p+L2L3=pM3+L3,M2q+M2M3=qL1+M3,L3p+L1L3=pM1+L1,M3q+M3M1=qL2+M1, and hence,(4.37)(L1−M2)p=L2(1−L1),(M1−L3)q=M2(1−M1),(L2−M3)p=L3(1−L2),(M2−L1)q=M3(1−M2),(L3−M1)p=L1(1−L3),(M3−L2)q=M1(1−M3). Therefore, we take1pL2(1−L1)=1qM3(M2−1),1pL3(1−L2)=1qM1(M3−1),1pL1(1−L3)=1qM2(M1−1). So,(4.38)ifL1≥1(resp.,L1≤1),thenM2≤1(resp.,M2≥1),ifL2≥1(resp.,L2≤1),thenM3≤1(resp.,M3≥1),ifL3≥1(resp.,L3≤1),thenM1≤1(resp.,M1≥1). Therefore, if L1≥1,M2≤1 (resp., L1≤1,M2≥1), we have L1−M2≥0 (resp., L1−M2≤0) and so from
(4.37), L1≤1 (resp., L1≥1). Hence, L1=1 and from (4.37), M2=1. Similarly, we
can prove that L2=1,L3=1,M1=1,M3=1. This completes the proof of the proposition.


## 5. Stability of System (1.3)

In this
section we find conditions so that the positive equilibrium(y,z) and the zero
equilibrium of (1.3) are stable.Proposition 5.1.
Consider system (1.3) where pi,qi, i=0,1 are positive
constants and the initial values yi,zi, i=−1,0 are positive
numbers. Then, the following statements are true.
(i)
If(5.1)q0−1<p0≤q0,q1−1<p1≤q1,q0+q1+p0p1+q0q1<1, then the unique positive equilibrium (y,z) of (1.3) is
globally asymptotically stable.(ii)
If(5.2)q0+q1+p0p1+1<q0q1, then the zero equilibrium of (1.3) is locally asymptotically stable.Proof.
(i) Since(y,z) is the unique
positive positive equilibrium of (1.3), we have(5.3)y=p0z+yq0+y,z=p1y+zq1+z. Then from (5.1) and (5.3), we get(5.4)y≤q0z+yq0+y,z≤q1y+zq1+z. Without loss of generality we assume that z≤y. Then from (5.4), it results that(5.5)y≤q0y+yq0+y, which means that(5.6)y≤1. Moreover, from (5.4) and (5.6), we get(5.7)z≤q1+zq1+z=1. In addition, from (5.3), we have(5.8)y>yq0+y,z>zq1+z, and so(5.9)y>1−q0,z>1−q1. Then the linearized system of (1.3) about the positive equilibrium (y,z) is(5.10)Zn+1=AZn, where(5.11)A=(00100001q0−p0z(q0+y)2p0q0+y000q1−p1y(q1+z)2p1q1+z0),Zn=(wn−1vn−1wnvn). The
characteristic equation of A is(5.12)λ4−(q0−p0z(q0+y)2+q1−p1y(q1+z)2)λ2−p1p0(q0+y)(q1+z)λ+(q0−p0z)(q1−p1y)(q0+y)2(q1+z)2=0. According to Remark 1.3.1 of [7], all the roots of (5.12) are of modulus less
than 1 if and only if(5.13)|q0−p0z(q0+y)2+q1−p1y(q1+z)2|+p1p0(q0+y)(q1+z)+|(q0−p0z)(q1−p1y)(q0+y)2(q1+z)2|<1. From (5.3), we get(5.14)q0−p0z=(1−y)(y+q0),q1−p1y=(1−z)(z+q1). Then from (5.6), (5.7), and (5.14), inequality (5.13) is equivalent
to(5.15)1−yq0+y+1−zq1+z+p1p0(q0+y)(q1+z)+(1−y)(1−z)(q0+y)(q1+z)<1. Using (5.9), inequality (5.15) holds if (5.1)  are  satisfied. Using Propositions
4.1 and 4.3, we have that the unique   positive equilibrium (y,z) of (1.3) is
globally asymptotically stable.
(ii) Arguing as
above, we can prove that the linearized system of (1.3) about the zero
equilibrium is(5.16)Zn+1=AZn, where(5.17)A=(001000011q0p0q00001q1p1q10),Zn=(wn−1vn−1wnvn). The characteristic equation of A is(5.18)λ4−(1q0+1q1)λ2−p1p0q0q1λ+1q0q1=0. Using [7, Remark 1.3.1], all the roots of (5.18) are of modulus less
than 1 if and only if relation (5.2) holds. This completes the proof of the
proposition.


## 6. Conclusion

In this paper,
in order to investigate (1.2), we study the equivalent system (1.3).
Summarizing the results of Sections 2, 3, 4, we get the following statements,
concerning (1.2).(i)
If (2.1)
hold, then (1.2) has a unique positive periodic solution of period 2.(ii)
If
either (3.4) or (3.37) holds, then every positive solution of (1.2) is
bounded and persists and tends to the unique positive periodic solution.(iii)
If
(5.1) hold, then the unique periodic solution of (1.2) is globally
asymptotically stable and if (5.2) holds, then the zero solution of (1.2) is
locally asymptotically stable.Open Problem
Consider the difference equation (1.2) where pn,qn, n=0,1,… are positive
sequences of period 2, and the initial values xi, i=−3,−2,−1,0 are positive
numbers. Prove that
(i)
if(6.1)q0−1<p0≤q0+1,q1−1<p1≤q1+1 are satisfied, then every positive solution of (1.2) is bounded and persists;(ii)
if
relations (6.1) are satisfied, then every positive solution of (1.2) tends to
the unique positive equilibrium (y,z) of (1.2) as n→∞.



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*Source: 973714-2009-03-30.xml*