# Global Asymptotic Stability of a Nonautonomous Difference Equation

Authors: Mehmet Gümüş; Özkan Öcalan
Journal: Journal of Applied Mathematics
 (2014)

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## Abstract

We study the following nonautonomous difference equation:xn+1=(xnxn-1+pn)/(xn+xn-1), n=0,1,…, where pn>0 is a period-2 sequence and the initial values x-1,x0∈(0,∞). We show that the unique prime period-2 solution of the equation above is globally asymptotically stable.


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## Body

## 1. Introduction

In [1], Xianyi and Deming studied the global asymptotic stability of positive solutions of the difference equation
(1)xn+1=xnxn-1+pxn+xn-1,n=0,1,…,
where p∈[0,∞) and the initial conditions 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). Moreover, a lot of difference equations with periodic coefficients have been applied in mathematical models in biology. In addition, between others in [2–10], we can see some more difference equations with periodic coefficient that have been studied. For constant coefficient, see [1, 11–15] and the references cited therein.In the present paper, we consider the following nonautonomous difference equation:(2)xn+1=xnxn-1+pnxn+xn-1,n=0,1,…,
where {pn} is a two periodic sequence of nonnegative real numbers and the initial conditions x-1,x0 are arbitrary positive numbers.The autonomous case of (2) is (1).It is natural problem to investigate the behavior of the solutions of (1) where we replace the constant p by a nonnegative periodic sequence pn. That is, we will consider the recursive sequence (2) where {pn} is a two-periodic sequence of nonnegative real numbers. What do the solutions of (2) with positive initial conditions x-1 and x0 do?Our aim in this paper is to investigate the question above and extend some results obtained in [1]. More precisely, we study the existence of a unique positive periodic solution of (2) and we investigate the boundedness character and the convergence of the positive solutions to the unique two periodic solution of (2). Finally, we study the global asyptotic stability of the positive periodic solution (2).As far as we examine, there is no paper dealing with (2). Therefore, in this paper, we focus on (2) in order to fill the gap.Now, assume that(3)pn={αifnis evenβifnis odd
in (2). Then,
(4)x2n+1=x2nx2n-1+αx2n+x2n-1,n=0,1,…,x2n+2=x2n+1x2n+βx2n+1+x2n,n=0,1,…,
where x-1,x0∈(0,∞).Here, we review some results which will be useful in our investigation of the behavior of positive solutions of (2).LetI be some interval of real numbers and let
(5)f:I×I⟶I
be a continuously differentiable function. Then for every pair of initial values x-1,x0∈I, the difference equation
(6)xn+1=f(xn,xn-1),n=0,1,…,
has a unique solution {xn}.Definition 1.
Let(x¯,y¯) be an equilibrium point of the map
(7)B=(f,un,un-1,…,un-k,g,zn,…,zn-k),
where f and g are continuously differentiable functions at (x¯,y¯). The linearized system of
(8)un+1=f(un,un-1,…,un-k,zn,…,zn-k),zn+1=g(un,un-1,…,un-k,zn,…,zn-k)
about the equilibrium point (x¯,y¯) is vn+1=B(vn)=Bjvn, where
(9)vn=(un⋮un-kvn⋮vn-k)
and Bj is the Jacobian matrix of system (8) about the equilibrium point (x¯,y¯).Theorem 2 (linearized stability).
Consider the following.(1)
If both roots of the quadratic equation(10)λ2-pλ-q=0
lie in the open unit disk |λ|<1, then the equilibrium x¯ of (6) is locally asymptotically stable.(2)
If at least one of the roots of (10) has absolute value greater than one, then the equilibrium x¯ of (6) is unstable.(3)
A necessary and sufficent condition for both roots of (10) to lie in the open unit disk |λ|<1 is
(11)|p|<1-q<2.(4)
A necessary and sufficent condition for both roots of (10) to have absolute value greater than one is
(12)|q|>1,|p|<|1-q|.(5)
A necessary and sufficent condition for one root of (10) to have absolute value greater than one and for the other to have absolute value less than one is
(13)p2+4q>0,|p|>|1-q|.(6)
A necessary and sufficent condition for a root of (10) to have absolute value equal to one is
(14)|p|=|1-q|
or
(15)q=-1,|p|≤2.


## 2. Boundedness of Solutions to (2)

In this section, we mainly investigate the boundedness character of (2), assuming (3) is satisfied.Theorem 3.
Letα,β>0. Then, every positive solution of (2) is bounded.Proof.
Suppose for the sake of contadiction that{xn} is an unbounded solution of (2). Then, there exist the subsequences {x2n+1} and {x2n+2} of the solution {xn} such that one of the following statements is true.(i)
x2n+1→∞ and x2n+2→a (a is arbitrary positive real number) for all n≥0.(ii)
x2n+1→b (b is arbitrary positive real number) and x2n+2→∞ for all n≥0.(iii)
x2n+1→∞ and x2n+2→∞ for all n≥1.
Now, we will show that the above statements are never satisfied.
From (2), we have the following inequalities:
(16)x2n+1<x2n-1+αx2n,forn≥0,x2n+1<x2n+αx2n-1,forn≥0,x2n+2<x2n+βx2n+1,forn≥0,x2n+2<x2n+1+βx2n,forn≥0.
If we get limits on both sides of the above inequalities, they contradict cases (i) and (ii). From (16) and using the induction, we obtain that
(17)x2n+1<x-1+α(1x0+⋯+1x2n),x2n+2<x0+β(1x1+⋯+1x2n+1).
Also, if we consider the limiting case of the above inequalities, they contradict case (iii) and complete the proof.


## 3. Periodicity and Stability of Solutions to (2)

In this section, we show that (2) has the unique two-periodic solution and the unique prime period-2 solution of this equation is locally asymptotically stable.Theorem 4.
Suppose thatpn>0 is a two periodic sequence such that
(18)pn+2=pn,n=0,1,….
Then (2) has the unique two-periodic solution.Proof.
Letxn be a solution of (2). Using (18), xn is two-periodic solution if and only if x-1 and x0 satisfy
(19)x-1=x1=x0x-1+αx0+x-1,x0=x2=x0x1+βx0+x1.
We setx-1=x and x0=y; then from (19) we obtain the system of equations
(20)x=xy+αx+y,y=xy+βx+y.
We prove that (2) has a solution (x¯,y¯),x¯,y¯>0. From (20) we get
(21)x2+xy=xy+α,y2+xy=xy+β.
It follows that (2) has a prime period-2 solution if and only if α,β∈(0,∞), α≠β, in which case (2) has a unique two-periodic solution
(22)α,β,…,α,β,….Theorem 5.
Assume thatpn>0 is a two periodic sequence and
(23)α,β∈(0,∞).
Then (2) has the unique prime period-2 solution which is locally asymptotically stable.Proof.
Letxn be a positive solution of (2) and let
(24)…,α,β,α,β,…,α,β,…
be the unique prime period-2 solution of (2). For any n≥0, set
(25)un=x2n-1,vn=x2n.
Then, (2) is equivalent to the system
(26)un+1=unvn+αun+vn,vn+1=un(vn2+β)+(α+β)vnvn(2un+vn)+α.
So, in order to study (2) we investigate system (26).
LetH be the function on (0,∞)2 defined by
(27)H(uv)=(uv+αu+vu(v2+β)+(α+β)vv(2u+v)+α)=(g(u,v)h(u,v)).
Then,
(28)(αβ)
is a fixed point of H. Clearly the unique prime period-2 solution is locally asymptotically stable when the eigenvalues of the jakobian matrix JH, evaluated at (αβ), lie inside the unit disk. We have
(29)JH(αβ)=(∂g∂u(α,β)∂g∂v(α,β)∂h∂u(α,β)∂h∂v(α,β)).
The linearized system of (26) about (α,β) is the system
(30)zn+1=Bzn,
where
(31)B=(β-αα+β+2αβ00α-βα+β+2αβ),zn=(unvn).
The characteristic equation of B is
(32)λ2-(β-αα+β+2αβ+α-βα+β+2αβ)λ-(α-β)2(α+β+2αβ)2=0
and thus
(33)λ2-(α-β)2(α+β+2αβ)2=0.
From Theorem 2 and (23) all the roots of (33) are of modulus less than 1. So the system (26) is locally asymptotically stable. Therefore, (2) has the unique prime period-2 solution which is locally asymptotically stable.


## 4. Global Asymptotic Stability for (2)

In this section, we give our main result. Before doing so, in the next theorem, we will provide an alternative proof for the global stability of (1). In the proof of the main theorem, we will use a similar technique.Theorem 6.
The positive equilibrium pointx¯ of (1) is globally asymptotically stable.Proof.
We know from Theorem3 that every positive solution of (1) is bounded for p∈(0,∞) and we know from [1] that the positive equilibrium point x¯ of (1) is locally asymptotically stable for p∈(0,∞).
Let{xn} be a positive solution of (1) and set
(34)limsup⁡⁡xn=K,liminf⁡⁡xn=k.
Then by the boundedness of {xn}; k, K are positive real numbers.
Thus, it is easy to see from (1) that
(35)k≥k2+p2K,K≤K2+p2k,k2+p≤2Kk≤K2+p.
Thus, we have
(36)k2≤K2
and so
(37)k≤K.
From (37), if k=K, the proof is obvious. If k<K, this case contradicts the unique equilibrium solution of (1) which is locally asymptotically stable. Consequently, it must be K=k. This completes the proof.Now, we are ready to give the proof of the main result.Theorem 7.
Assume thatpn>0 is a two-periodic sequence and α,β∈(0,∞); then (2) has the unique prime period-2 solution which is globally asymptotically stable.Proof.
We must prove that the unique prime period-2 solution of (2) is both locally asymptotically stable and globally attractive. We know from Theorem 5 that the unique prime period-2 solution of (2) is locally asymptotically stable for α,β∈(0,∞). It remains to verify that every positive solution of (2) converges the unique prime period-2 solution. Since α,β∈(0,∞), by Theorem 3 every positive solution of (2) is bounded. Let {xn} be a positive solution of (2) and set
(38)limsup⁡⁡x2n+1=M,limsup⁡⁡x2n=S,liminf⁡⁡x2n+1=m,liminf⁡⁡x2n=s.
Then by the boundedness of {xn}, m, M, s, and S are positive real numbers. Moreover, it is obvious that since α≠β, then from (2)
(39)lim⁡n→∞x2n≠lim⁡x2n+1,
and it is clear that if lim⁡⁡x2n+1 exists, then so does lim⁡n→∞⁡x2n.
Thus, it is easy to see from (2) that
(40)m≥ms+αM+S,M≤MS+αm+s,(41)s≥ms+βM+S,S≤MS+βm+s.
From (40), we have
(42)mM+mS-ms≥α≥mM+sM-MS,mM+mS-ms≥mM+sM-MS,S≥s.
From (41), we have
(43)sM+sS-ms≥β≥mS+sS-MS,sM+sS-ms≥mS+sS-MS,M≥m.
With cases (42) and (43), if s=S and m=M, the proof is obvious. If s<S and m<M, this case contradicts the unique prime period-2 solution of (2) which is locally asymptotically stable. Consequently, it must be s=S and M=m. The proof is complete.



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