# Stability of Nonhyperbolic Equilibrium Solution of Second Order Nonlinear Rational Difference Equation

Authors: S. Atawna; R. Abu-Saris; E. S. Ismail; I. Hashim
Journal: Journal of Difference Equations
 (2015)

---

## Abstract

This is a continuation part of our investigation in which the second order nonlinear rational difference equationxn+1=(α+βxn+γxn-1)/(A+Bxn+Cxn-1), n=0,1,2,…, where the parameters A≥0 and B, C, α, β, γ are positive real numbers and the initial conditions x-1, x0 are nonnegative real numbers such that A+Bx0+Cx-1>0, is considered. The first part handled the global asymptotic stability of the hyperbolic equilibrium solution of the equation. Our concentration in this part is on the global asymptotic stability of the nonhyperbolic equilibrium solution of the equation.


---

## Body

## 1. Introduction

In part 1 of this investigation [1], we have established the global stability of the hyperbolic equilibrium solution of the second order rational difference equation:(1)xn+1=α+βxn+γxn-1A+Bxn+Cxn-1,n=0,1,2,…,where the parameters A≥0 and B, C, α, β, γ are positive real numbers and the initial conditions x-1, x0 are nonnegative real numbers such that A+Bx0+Cx-1>0. Our aim in this part is on the global attractivity of the nonhyperbolic equilibrium solution of (1).The periodic character of positive solutions of (1) has been investigated by the authors in [2]. They showed that the period-two solution is locally asymptotically stable if it exists.Many rational difference equations were studied extensively in [3]. A systematic study of the second order rational difference equation of form (1), where the parameters α, β, γ, A, B, C and the initial conditions x-1, x0 are nonnegative real numbers, was considered in the monograph of Kulenovic and Ladas [4]. They presented the known results up to 2002. Next, Kulenovic and Ladas [4] derived several ones on the boundedness, the global stability, and the periodicity of solutions of all rational difference equations of form (1). Furthermore, they posed several open problems and conjectures related to this equation and its functional generalization.Even after a sustained effort by many researchers such as [5–9], there were some difference equations of form (1) that have not been investigated till 2007.Amleh et al. in [10, 11] give an up-to-date account on recent developments related to (1) up to 2007. Furthermore, they reposed several open problems and conjectures related to this equation.Camouzis and Ladas in [12] summarize the progress up to 2008. Recently, the work done by many researchers such as [13–21] have solved many open problems and conjectures proposed in [4, 10–12] related to (1) and have led to the development of some general theory about difference equation. However, as confirmed by Professor Kulenovic (personal communication, August, 24, 2014), the case A=0 remains open.Our approach handles the aforementioned case as well as other cases. Furthermore, the results in this paper, together with the established results in [1, 2, 4], give a complete picture of the nature of solutions of the second order rational difference equation of form (1).It is worth mentioning that there are very few results in the literature regarding the stability of nonhyperbolic equilibrium solution of a general difference equation of the form(2)xn+1=fxn,xn-1,n=0,1,….We believe that our result is an important stepping stone in understanding the behavior of solutions of rational difference equation of form (1) which provides prototypes towards the development of the basic theory of the global behavior of solutions of nonlinear difference equations of form (2).The transformation(3)xn=γCynreduces (1) to the following equation: (4)yn+1=r+pyn+yn-1z+qyn+yn-1,n=0,1,2,…,where (5)r=αCγ2,p=βγ,z=Aγ,q=BCare positive real numbers and the initial conditions y-1, y0 are nonnegative real numbers.That being said, the remainder of this paper is organized as follows. In the next section, a brief description of some definitions and results from the literature that are needed to prove the main results in this paper is given. Section3 gives necessary and sufficient conditions for (4) to have nonhyperbolic solution. Next, Section 4 examines the existence of intervals which attract all solutions of (4) and shows that the nonhyperbolic equilibrium solution of (4) is globally asymptotically stable. In Section 5 we consider several numerical examples generated by MATLAB to illustrate the results of the previous sections and to support our theoretical discussion. Finally, we conclude in Section 6 with suggestions for future research.


## 2. Preliminaries

For the sake of self-containment and convenience, we recall the following definitions and results from [4].LetI be a nondegenerate interval of real numbers and let f:I×I→I be a continuously differentiable function. Then for every set of initial conditions x0,x-1∈I, the difference equation of form (2) has a unique solution {xn}n=-1∞.A constant sequence,xn=x¯ for all n where x¯∈I, is called an equilibrium solution of (2) if (6)x¯=fx¯,x¯.Definition 1.
Letx¯ be an equilibrium solution of (2).(i)
x¯ is called locally stable if for every ϵ>0, there exists δ>0 such that, for all x0,x-1∈I, with |x0-x¯|+|x-1-x¯|<δ, we have (7)xn-x¯<ϵ∀n≥-1.(ii)
x¯ is called locally asymptotically stable if it is locally stable, and if there exists γ>0, such that, for all x0,x-1∈I, with |x0-x¯|+|x-1-x¯|<γ, we have (8)limn→∞⁡xn=x¯.(iii)
x¯ is called a global attractor if for every x0,x-1∈I, we have (9)limn→∞⁡xn=x¯.(iv)
x¯ is called globally asymptotically stable if it is locally stable and a global attractor.(v)
x¯ is calledunstable if it is not stable.(vi)
x¯ is called asource, or arepeller, if there exists r>0 such that, for all x0,x-1∈I, with 0<|x0-x¯|+|x-1-x¯|<r, there exists N≥1 such that(10)xN-x¯≥r.Clearly a source is an unstable equilibrium.Definition 2.
Let(11)a=∂f∂ux¯,x¯,b=∂f∂vx¯,x¯denote the partial derivatives of f(u,v) evaluated at the equilibrium x¯ of (2). Then the equation(12)yn+1=ayn+byn-1,n=0,1,…is called the linearized equation associated with (2) about the equilibrium solution x¯.Theorem 3 (linearized stability).
(a) If both roots of the quadratic equation(13)λ2-aλ-b=0lie in the open unit disk |λ|<1, then the equilibrium x¯ of (2) is locally asymptotically stable.
(b) If at least one of the roots of (13) has absolute value greater than one, then the equilibrium x¯ of (2) is unstable.
(c) A necessary and sufficient condition for both roots of (13) to lie in the open unit disk |λ|<1 is(14)a<1-b<2.In this case the locally asymptotically stable equilibrium x¯ is also called a sink.
(d) A necessary and sufficient condition for both roots of (13) to have absolute value greater than one is(15)b>1,a<1-b.In this case x¯ is a repeller.
(e) A necessary and sufficient condition for one root of (13) to have absolute value greater than one and for the other to have absolute value less than one is(16)a2+4b>0,a>1-b.In this case the unstable equilibrium x¯ is called a saddle point.
(f) A necessary and sufficient condition for a root of (13) to have absolute value equal to one is(17)a=1-bor(18)b=-1,a≤2.In this case the equilibrium x¯ is called a nonhyperbolic point.Theorem 4.
Consider the difference equation (2). Let I=[a,b] be some interval of real numbers and assume that(19)f:a,b×a,b⟶a,bis a continuous function satisfying the following properties: (a)
f(x,y) is nonincreasing in x∈[a,b] for each y∈[a,b], and f(x,y) is nondecreasing in y∈[a,b] for each x∈[a,b];(b)
the difference equation (2) has no solutions of prime period two in [a,b];then (2) has a unique equilibrium x¯∈[a,b] and every solution of (2) converges to x¯.The following result from [12] will become handy in the sequel.Theorem 5.
LetI be a set and let(20)f:I×I⟶Ibe a function f(u,v) which decreases in u and increases in v. Then for every solution {xn}n=-1∞ of the equation(21)xn+1=fxn,xn-1,n=0,1,…the subsequences {x2n}n=0∞ and {x2n+1}n=-1∞ of even and odd terms of the solution do exactly one of the following:(i)
they are both monotonically increasing;(ii)
they are both monotonically decreasing;(iii)
eventually, one of them is monotonically increasing and the other is monotonically decreasing.The following result was established in [2] and will prove to be useful in our investigation.Theorem 6.
(a) When(22)p+z≥1,(4) has no nonnegative prime period-two solution.
(b) When(23)p+z<1,(4) has prime period-two solution, (24)…,ϕ,ψ,ϕ,ψ,…;if and only if condition(25)r<1-p-zq1-p-z-1+3p-z4,where ϕ and ψ are the positive and distinct solutions of the quadratic equation (26)t2-1-z-pt+p1-z-p+rq-1=0,q>1.The following two results were established in part 1 of this investigation [1] and will prove to be useful in our investigation.Theorem 7.
Assume thatp<q and r≤z; then one has two cases to be considered.(1)
Ifpz-qr≤0, then [p/q,1] is invariant.(2)
Ifpz-qr>0, then we have two subcases to be considered:
(a)
ifpz-qr/q-p≤p/q, then every positive solution of (4) eventually enters and remains in the interval p/q,1;(b)
ifpz-qr/q-p>p/q, then every positive solution of (4) eventually enters and remains in the interval 0,p/q.Theorem 8.
Assume thatp<q and r>z; then one has two cases to be considered. (1)
Ifr-z/q-p≤1, then every positive solution of (4) eventually enters and remains in the interval r-z/q-p,1.(2)
Ifr-z/q-p>1, then every positive solution of (4) eventually enters and remains in the interval 1,r-z/q-p.


## 3. Existence of Nonhyperbolic Equilibrium Solution

In this section, we give explicit conditions on the parameter values of (4) for the equilibrium y¯ to be nonhyperbolic.Equation (4) has a unique positive equilibrium given by(27)y¯=1+p-z+1+p-z2+4rq+12q+1.The linearized equation associated with (4) about the equilibrium solution is given by(28)zn+1=p-qy¯z+q+1y¯zn+1-y¯z+q+1y¯zn-1.Therefore, its characteristic equation is(29)λ2-p-qy¯z+q+1y¯λ-1-y¯z+q+1y¯=0.By applying Theorem3(f) we have the following result.Theorem 9.
Assume that(30)p+z<1,then the positive equilibrium of (4) is nonhyperbolic if and only if(31)r=1-p-zq1-p-z-1+3p-z4.Proof.
By employing Theorem3(f), conditions (17) and (18) are equivalent to the following two inequalities:(32)p-qy¯z+q+1y¯=1-1-y¯z+q+1y¯,(33)1-y¯z+q+1y¯︸1=-1,p-qy¯z+q+1y¯≤2,respectively. Notice that Part (1) of (33) implies -(1+z)=qy¯, which is impossible to be satisfied since z,q,y¯>0, while (32) is equivalent to the following two inequalities:(34)p-qy¯z+q+1y¯=1-1-y¯z+q+1y¯or(35)p-qy¯z+q+1y¯=1-y¯z+q+1y¯-1.Equation (34) implies 1+p-z=2q+1y¯, which contradicts (27), while (35) is equivalent to (36)p-qy¯z+q+1y¯-1-y¯z+q+1y¯=-1.From which we have(37)1-p-z=2y¯.Clearly the equilibrium y¯ is the positive solution of the quadratic equation(38)q+1y¯2+z-p-1y¯-r=0.Now set(39)Fu=q+1u2+z-p-1u-r,and (37) holds if and only if(40)p+z<1,F1-p-z2=0.That is, (41)q+11-p-z22+z-p-11-p-z2-r=0,⟺q+11-p-z2+2z-p-11-p-z-4r=0,⟺1-p-zq+11-p-z+2z-p-1-4r=0,⟺1-p-zq1-p-z-1+3p-z-4r=0,from which (31) follows.
The proof is complete.


## 4. Global Stability Analysis

In this section, we give necessary and sufficient conditions for the nonhyperbolic solution of (4) to be globally attractive.The characteristic polynomial associated with (4) about the positive equilibrium is given by(42)fλ=λ2-p-qy¯z+q+1y¯λ-1-y¯z+q+1y¯=0⟹λ+1λ-1-y¯z+q+1y¯=0⟹λ=-1orλ=1-y¯z+q+1y¯<1.By the Stable Manifold Theorem, there is a manifold of solutions that converge to the equilibrium solution.Now, sincer>0, condition (31) implies(43)q>1+3p-z1-p-z>1.Indeed, since p+z<1 we have(44)p<q.With that in mind we examine the existence of intervals which attract all solutions of (1) in the next section.


### 4.1. Invariant Intervals

Table1 gives the signs of ∂f(yn,yn-1)/∂yn and ∂f(yn,yn-1)/∂yn-1 of (4) in all possible nondegenerate cases when p<q.Table 1
Signs of∂f/∂yn and ∂f/∂yn-1 of (4) when p<q.
Case
p
z
-
q
r
r
-
z
∂
f
y
n
,
y
n
-
1
∂
y
n
=
q
-
p
z
+
q
y
n
+
y
n
-
1
2
p
z
-
q
r
q
-
p
-
y
n
-
1
∂
f
y
n
,
y
n
-
1
∂
y
n
-
1
=
q
-
p
z
+
q
y
n
+
y
n
-
1
2
y
n
-
r
-
z
q
-
p
1
+
−
− ifyn-1>pz-qr/q-p and + if yn-1<pz-qr/q-p
+
2
0
−
−
+
3
−
−
−
+
4
−
0
−
+
5
−
+
−
+ ifyn>r-z/q-p and − if yn<r-z/q-pThe following two lemmas will be useful in investigating the attracting intervals of solutions of (4).Lemma 10.
Assume that condition (31) holds. (a)
Ifr≤z and pz-qr>0, then pz-qr/q-p<p/q.(b)
Ifr>z, then r-z/q-p<1.Proof.
(a) Considerr≤z and pz-qr>0.
Assume for the sake of contradiction that(45)pz-qrq-p≥pq.Then(46)r≤pzq-pq+p2q2.By condition (31), inequality (46) is equivalent to(47)q31-p-z2-q21-p-z1+3p-z≤4pzq-pq+p2.Since p+z<1 and q>1, the right-hand side of inequality (47) is equivalent to(48)4pqz-1+p2≤4pq-p+p2=-4p2q-1<0.Thus, inequality (47) implies(49)⟹q31-p-z2-q21-p-z1+3p-z≤0,⟹q31-p-z2≤q21-p-z1+3p-z,⟹q≤1+3p-z1-p-z,which contradicts condition (43).
(b) Considerr>z.
Assume for the sake of contradiction that(50)r-zq-p≥1.Then(51)r≥q-p+z.By condition (31), inequality (51) is equivalent to(52)1-p-zq1-p-z-1+3p-z≥4q+4z-p.Since p+z<1, inequality (52) implies (53)q1-p-z2-1-p-z1+3p-z≥4q+4z-1-z>4q-4,⟹q4-1-p-z2<4-1-p-z1+3p-z,⟹q<4-1-p-z1+3p-z4-1-p-z2,⟹q<-1-p-z1+3p-z-1-p-z2,⟹q<1+3p-z1-p-z,which contradicts condition (43).
The proof is complete.Lemma 11.
Assume that condition (31) holds. (a)
Ifr≤z, then p/q≤y¯<1.(b)
Ifr>z, then r-z/q-p≤y¯<1.Proof.
Condition (37) implies(54)y¯=1-p-z2.Since p+z<1, it is clear that y¯<1. To complete the proof we have two cases to be considered.(a)
Considerr≤z.To show thaty¯=1-p-z/2>p/q, it is enough to show q>2p/1-p-z which is clearly satisfied since condition (43) holds.(b)
Considerr>z.Our interest is to show that(55)y¯≥r-zq-p.Assume for the sake of contradiction that(56)y¯<r-zq-p⟹1-p-z2<r-zq-p⟹q-p1-p-z<2r-z⟹2r>2z+q-p1-p-z.By condition (31), the last inequality is equivalent to (57)1-p-zq1-p-z-1+3p-z2>2z+q-p1-p-z⟹1-p-zq1-p-z-1+3p-z>4z+2q-p1-p-z⟹1-p-z·q1-p-z-1+3p-z-2q-p>4z⟹1-p-z-q-qp+z-1-p+z>4z⟹1-p-z-q1+p+z-1+p-z︸-ve>4z,which is impossible.
The proof is complete.By Theorem7, Lemmas 10 and 11, we obtain the following key result.Theorem 12.
Assume that condition (31) holds. (a)
Ifr≤z, then every positive solution of (4) eventually enters and remains in the interval [p/q,1].(b)
Ifr>z, then every positive solution of (4) eventually enters and remains in the interval [r-z/q-p,1].Proof.
We have two cases to be considered.(a)
Considerr≤z.Sincer≤z and condition (44) holds, then we have two subcases to be considered:
(1)
ifpz-qr≤0, Theorem 7 part 1 implies that the interval [p/q,1] is invariant;(2)
Ifpz-qr>0, Lemma 10 implies that pz-qr/q-p<p/q, and by Theorem 7 part 2(a) every positive solution of (4) eventually enters and remains in the interval p/q,1.(b)
Considerr>z.Sincer>z and condition (44) holds, Lemma 10 implies that r-z/q-p<1. By Theorem 8 part 1. every positive solution of (4) eventually enters and remains in the interval r-z/q-p,1.
The proof is complete.


### 4.2. Global Stability of the Nonhyperbolic Equilibrium Solution

The following lemma will be useful in investigating the global stability of the nonhyperbolic equilibrium solution of (4).Lemma 13.
Assumer>z. Then yn+1≤yn-1.Proof.
By (4), we have (58)yn+1-yn-1=r+pyn+yn-1z+qyn+yn-1-yn-1=r+pyn+yn-1-zyn-1-qynyn-1-yn-12z+qyn+yn-1=r+pyn+1-zyn-1-qynyn-1-yn-12z+qyn+yn-1.Let (59)hyn,yn-1=r+pyn+1-zyn-1-qynyn-1-yn-12.Our interest now is to show that h(yn,yn-1)≤0 for all yn,yn-1∈r-z/q-p,1.
Observe thath(yn,yn-1) is continuously differentiable map defined on a compact interval. As such, either the maximum is attained at an interior stationary point or on the boundary. Figure 1 depicts the region where h(yn,yn-1) is defined.
Furthermore,(60)∂hyn,yn-1∂yn=p-qyn-1,∂hyn,yn-1∂yn-1=1-z-qyn-2yn-1.As such, there is one interior stationary point, namely, 1-z-2p/q/q,p/q.
Recall that we only want stationary point in the regionR, so 1-z-2p/q/q,p/q will be ignored, because p/q<r-z/q-p.
Now, we need to find the absolute maximum of the functionh(yn,yn-1) along the boundary of the rectangle R.
The boundary of this rectangle is given by the following.(1)
Upper side:yn=1, r-z/q-p≤yn-1≤1.Define(61)gyn-1=h1,yn-1=r+p+1-z-qyn-1-yn-12.Now, finding the absolute maximum of the functionh(yn,yn-1) a long the right side of the rectangle R will be equivalent to finding the absolute maximum of the function g(yn-1) in the range r-z/q-p≤yn-1≤1.Hence,(62)g′=1-z-q-2yn-1⟹yn-1=1-z-q2<0.This is not in the ranger-z/q-p≤yn-1≤1, so we will ignore it.The value of this function at the end points is(63)gr-zq-p=h1,r-zq-p=r+p+1-z-qr-zq-p-r-zq-p2,g1=h1,1=r+p+1-z-q-1=r-z-q-p.Sincer-z/q-p<1 by Lemma 10(b), then (64)h1,r-zq-p<r+p+1-z-q1-1=r-z-q-p<0,h1,1=r-z-q-p<0.(2)
Lower side:yn=r-z/q-p, r-z/q-p≤yn-1≤1.Define(65)gyn-1=hr-zq-p,yn-1=r+pr-zq-p+1-z-qr-zq-pyn-1-yn-12,g′=1-z-qr-zq-p-2yn-1⟹yn-1=1-z-qr-z/q-p2<0.This is not in the ranger-z/q-p≤yn-1≤1, so we will ignore it.The value of this function at the end points is(66)gr-zq-p=hr-zq-p,r-zq-p=r+p+1-zr-zq-p-q+1r-zq-p2,g1=hr-zq-p,1=r-z-q-pr-zq-p.Sincer-z/q-p<1 by Lemma 10(b), then (67)hr-zq-p,r-zq-p<r+p+1-z-q-1=r-z-q-p<0.Furthermore,(68)hr-zq-p,1=0.(3)
Right side:yn-1=1, r-z/q-p≤yn≤1.Define(69)fyn=hyn,1=r-z+p-qyn,f′=p-q<0.Thenf decreases and its maximum occurs at yn=r-z/q-p. Furthermore, (70)fr-zq-p=hr-zq-p,1=0.(4)
Left side:yn-1=r-z/q-p, r-z/q-p≤yn≤1.Define(71)fyn=hyn,r-zq-p=r+pyn+1-zr-zq-p-qr-zq-pyn-r-zq-p2,f′=p-qr-zq-p<0.Thenf decreases and the maximum occurs at yn=r-z/q-p. Furthermore, (72)fr-zq-p=hr-zq-p,r-zq-p<0Now, collect up all the function values for h(yn,yn-1): (73)hr-zq-p,r-zq-p<0,h1,r-zq-p<0,hr-zq-p,1=0,h1,1<0.Clearly the maximum value of h is 0 and it occurs at r-z/q-p,1.
The proof is complete.Figure 1
The region whereh(yn,yn-1) is defined.The result about the global stability of the positive nonhyperbolic equilibrium solution of (4) is given in the following theorem.Theorem 14.
The positive nonhyperbolic equilibrium solution of (4) is globally asymptotically stable.Proof.
Let(74)fyn,yn-1=r+pyn+yn-1z+qyn+yn-1.We have two cases to be considered. (a)
Considerr≤z.Here we distinguish between two subcases.
(1)
Considerpz-qr≤0.Let{yn}n=0∞ be a positive solution of (4). It follows from Theorem 12(a) that every positive solution of (4) eventually enters and remains in the interval p/q,1. Furthermore, p/q≤y¯<1 by Lemma 11. Indeed, function (74) is decreasing in yn and increasing in yn-1, in the interval [p/q,1]. By applying Theorem 5 the subsequences {y2n} and {y2n+1} of the solution converge monotonically to finite limits l1 and l2. Set(75)l1=limn→∞y2n,l2=limn→∞y2n+1.By (4) we have (76)y2n+2=r+py2n+1+y2nz+qy2n+1+y2n,y2n+1=r+py2n+y2n-1z+qy2n+y2n-1.Hence,(77)l1=r+pl2+l1z+ql2+l1,l2=r+pl1+l2z+ql1+l2.Thus,(78)zl1+ql1l2+l12=r+pl2+l1,zl2+ql1l2+l22=r+pl1+l2and by subtracting we have(79)l1-l2z+l1+l2+p-1=0.This is true if and only if(80)l1=l2orl1+l2=1-p-z.However, ifl1=l2=l then by (77) we have (81)l=r+pl+lz+ql+l⟹l=r/l+p+1z/l+q+1⟹z+q+1l=rl+p+1⟹q+1l-rl=1+p-z⟹q+1l2-r=1+p-zl⟹q+1l2-1+p-zl-r=0⟹l=1+p-z+1+p-z2+4rq+12q+1⟹l=y¯,whereasl1+l2=1-p-z=2y¯ is impossible in this case since by Theorem 6, (4) does not possess a period-two solution.(2)
Considerpz-qr>0.Let{yn}n=0∞ be a positive solution of (4). It follows from Lemma 10(a) that pz-qr/q-p<p/q. Furthermore, Theorem 12(a) implies that every positive solution of (4) eventually enters and remains in the interval p/q,1. Indeed, p/q≤y¯<1 by Lemma 11. Figure 2 depicts the region where f(yn,yn-1) is defined.With the understanding that function (74) is decreasing in yn and increasing in yn-1 in the interval [p/q,1] by Table 1 case 1, and since condition (31) holds, (4) does not possess a period-two solution by Theorem 6. Thus all conditions of Theorem 4 are satisfied and we conclude that y¯ is globally asymptotically stable.(b)
Considerr>z.Let{yn}n=0∞ be a positive solution of (4). It follows from  Theorem 12(b) that every positive solution of (4) eventually enters and remains in the interval r-z/p-q,1. Furthermore, r-z/q-p≤y¯<1 by Lemma 11.Lemma13 shows that yn+1≤yn-1. Replacing n by n-1 in the previous inequality then yn≤yn-2. As such, the odd and even terms of any solution yn of (4) form two monotonic nondecreasing subsequences. Furthermore, both of these subsequences are bounded because yn is bounded. Hence by Monotone Convergence Theorem the subsequences {y2n} and {y2n+1} of the solution converge to finite limits ȷ1 and ȷ2. Set(82)ȷ1=limn→∞⁡y2n,ȷ2=limn→∞⁡y2n+1.Using technique similar to the one used in proofing Case (a)(1), we have(83)y2n+2=r+py2n+1+y2nz+qy2n+1+y2n,y2n+1=r+py2n+y2n-1z+qy2n+y2n-1.Hence,(84)ȷ1=r+pȷ2+ȷ1z+qȷ2+ȷ1,ȷ2=r+pȷ1+ȷ2z+qȷ1+ȷ2.Thus,(85)zȷ1+qȷ1ȷ2+ȷ12=r+pȷ2+ȷ1,zȷ2+qȷ1ȷ2+ȷ22=r+pȷ1+ȷ2and by subtracting we have(86)ȷ1-ȷ2z+ȷ1+ȷ2+p-1=0.This is true if and only if(87)ȷ1=ȷ2orȷ1+ȷ2=1-p-z.However, ifȷ1=ȷ2=ȷ then by (84) we have (88)ȷ=r+pȷ+ȷz+qȷ+ȷ⟹ȷ=r/ȷ+p+1z/ȷ+q+1⟹z+q+1ȷ=rȷ+p+1⟹q+1ȷ-rȷ=1+p-z⟹q+1ȷ2-r=1+p-zȷ⟹q+1ȷ2-1+p-zȷ-r=0⟹ȷ=1+p-z+1+p-z2+4rq+12q+1⟹ȷ=y¯,whereas,ȷ1+ȷ2=1-p-z=2y¯ is impossible in this case since by Theorem 6, (4) does not possess a period-two solution.The proof is complete.Figure 2
The region wheref(yn,yn-1) is defined.Remark 15.
The papers [22–24] give the proof of the existence of both stable and unstable manifolds for second order difference equations decreasing in first and increasing in second argument in nonhyperbolic case of stable type, that is, of the type when second characteristic value is in (-1,1). In such a way one can avoid the use of center manifold for such equations.


## 4.1. Invariant Intervals

Table1 gives the signs of ∂f(yn,yn-1)/∂yn and ∂f(yn,yn-1)/∂yn-1 of (4) in all possible nondegenerate cases when p<q.Table 1
Signs of∂f/∂yn and ∂f/∂yn-1 of (4) when p<q.
Case
p
z
-
q
r
r
-
z
∂
f
y
n
,
y
n
-
1
∂
y
n
=
q
-
p
z
+
q
y
n
+
y
n
-
1
2
p
z
-
q
r
q
-
p
-
y
n
-
1
∂
f
y
n
,
y
n
-
1
∂
y
n
-
1
=
q
-
p
z
+
q
y
n
+
y
n
-
1
2
y
n
-
r
-
z
q
-
p
1
+
−
− ifyn-1>pz-qr/q-p and + if yn-1<pz-qr/q-p
+
2
0
−
−
+
3
−
−
−
+
4
−
0
−
+
5
−
+
−
+ ifyn>r-z/q-p and − if yn<r-z/q-pThe following two lemmas will be useful in investigating the attracting intervals of solutions of (4).Lemma 10.
Assume that condition (31) holds. (a)
Ifr≤z and pz-qr>0, then pz-qr/q-p<p/q.(b)
Ifr>z, then r-z/q-p<1.Proof.
(a) Considerr≤z and pz-qr>0.
Assume for the sake of contradiction that(45)pz-qrq-p≥pq.Then(46)r≤pzq-pq+p2q2.By condition (31), inequality (46) is equivalent to(47)q31-p-z2-q21-p-z1+3p-z≤4pzq-pq+p2.Since p+z<1 and q>1, the right-hand side of inequality (47) is equivalent to(48)4pqz-1+p2≤4pq-p+p2=-4p2q-1<0.Thus, inequality (47) implies(49)⟹q31-p-z2-q21-p-z1+3p-z≤0,⟹q31-p-z2≤q21-p-z1+3p-z,⟹q≤1+3p-z1-p-z,which contradicts condition (43).
(b) Considerr>z.
Assume for the sake of contradiction that(50)r-zq-p≥1.Then(51)r≥q-p+z.By condition (31), inequality (51) is equivalent to(52)1-p-zq1-p-z-1+3p-z≥4q+4z-p.Since p+z<1, inequality (52) implies (53)q1-p-z2-1-p-z1+3p-z≥4q+4z-1-z>4q-4,⟹q4-1-p-z2<4-1-p-z1+3p-z,⟹q<4-1-p-z1+3p-z4-1-p-z2,⟹q<-1-p-z1+3p-z-1-p-z2,⟹q<1+3p-z1-p-z,which contradicts condition (43).
The proof is complete.Lemma 11.
Assume that condition (31) holds. (a)
Ifr≤z, then p/q≤y¯<1.(b)
Ifr>z, then r-z/q-p≤y¯<1.Proof.
Condition (37) implies(54)y¯=1-p-z2.Since p+z<1, it is clear that y¯<1. To complete the proof we have two cases to be considered.(a)
Considerr≤z.To show thaty¯=1-p-z/2>p/q, it is enough to show q>2p/1-p-z which is clearly satisfied since condition (43) holds.(b)
Considerr>z.Our interest is to show that(55)y¯≥r-zq-p.Assume for the sake of contradiction that(56)y¯<r-zq-p⟹1-p-z2<r-zq-p⟹q-p1-p-z<2r-z⟹2r>2z+q-p1-p-z.By condition (31), the last inequality is equivalent to (57)1-p-zq1-p-z-1+3p-z2>2z+q-p1-p-z⟹1-p-zq1-p-z-1+3p-z>4z+2q-p1-p-z⟹1-p-z·q1-p-z-1+3p-z-2q-p>4z⟹1-p-z-q-qp+z-1-p+z>4z⟹1-p-z-q1+p+z-1+p-z︸-ve>4z,which is impossible.
The proof is complete.By Theorem7, Lemmas 10 and 11, we obtain the following key result.Theorem 12.
Assume that condition (31) holds. (a)
Ifr≤z, then every positive solution of (4) eventually enters and remains in the interval [p/q,1].(b)
Ifr>z, then every positive solution of (4) eventually enters and remains in the interval [r-z/q-p,1].Proof.
We have two cases to be considered.(a)
Considerr≤z.Sincer≤z and condition (44) holds, then we have two subcases to be considered:
(1)
ifpz-qr≤0, Theorem 7 part 1 implies that the interval [p/q,1] is invariant;(2)
Ifpz-qr>0, Lemma 10 implies that pz-qr/q-p<p/q, and by Theorem 7 part 2(a) every positive solution of (4) eventually enters and remains in the interval p/q,1.(b)
Considerr>z.Sincer>z and condition (44) holds, Lemma 10 implies that r-z/q-p<1. By Theorem 8 part 1. every positive solution of (4) eventually enters and remains in the interval r-z/q-p,1.
The proof is complete.


## 4.2. Global Stability of the Nonhyperbolic Equilibrium Solution

The following lemma will be useful in investigating the global stability of the nonhyperbolic equilibrium solution of (4).Lemma 13.
Assumer>z. Then yn+1≤yn-1.Proof.
By (4), we have (58)yn+1-yn-1=r+pyn+yn-1z+qyn+yn-1-yn-1=r+pyn+yn-1-zyn-1-qynyn-1-yn-12z+qyn+yn-1=r+pyn+1-zyn-1-qynyn-1-yn-12z+qyn+yn-1.Let (59)hyn,yn-1=r+pyn+1-zyn-1-qynyn-1-yn-12.Our interest now is to show that h(yn,yn-1)≤0 for all yn,yn-1∈r-z/q-p,1.
Observe thath(yn,yn-1) is continuously differentiable map defined on a compact interval. As such, either the maximum is attained at an interior stationary point or on the boundary. Figure 1 depicts the region where h(yn,yn-1) is defined.
Furthermore,(60)∂hyn,yn-1∂yn=p-qyn-1,∂hyn,yn-1∂yn-1=1-z-qyn-2yn-1.As such, there is one interior stationary point, namely, 1-z-2p/q/q,p/q.
Recall that we only want stationary point in the regionR, so 1-z-2p/q/q,p/q will be ignored, because p/q<r-z/q-p.
Now, we need to find the absolute maximum of the functionh(yn,yn-1) along the boundary of the rectangle R.
The boundary of this rectangle is given by the following.(1)
Upper side:yn=1, r-z/q-p≤yn-1≤1.Define(61)gyn-1=h1,yn-1=r+p+1-z-qyn-1-yn-12.Now, finding the absolute maximum of the functionh(yn,yn-1) a long the right side of the rectangle R will be equivalent to finding the absolute maximum of the function g(yn-1) in the range r-z/q-p≤yn-1≤1.Hence,(62)g′=1-z-q-2yn-1⟹yn-1=1-z-q2<0.This is not in the ranger-z/q-p≤yn-1≤1, so we will ignore it.The value of this function at the end points is(63)gr-zq-p=h1,r-zq-p=r+p+1-z-qr-zq-p-r-zq-p2,g1=h1,1=r+p+1-z-q-1=r-z-q-p.Sincer-z/q-p<1 by Lemma 10(b), then (64)h1,r-zq-p<r+p+1-z-q1-1=r-z-q-p<0,h1,1=r-z-q-p<0.(2)
Lower side:yn=r-z/q-p, r-z/q-p≤yn-1≤1.Define(65)gyn-1=hr-zq-p,yn-1=r+pr-zq-p+1-z-qr-zq-pyn-1-yn-12,g′=1-z-qr-zq-p-2yn-1⟹yn-1=1-z-qr-z/q-p2<0.This is not in the ranger-z/q-p≤yn-1≤1, so we will ignore it.The value of this function at the end points is(66)gr-zq-p=hr-zq-p,r-zq-p=r+p+1-zr-zq-p-q+1r-zq-p2,g1=hr-zq-p,1=r-z-q-pr-zq-p.Sincer-z/q-p<1 by Lemma 10(b), then (67)hr-zq-p,r-zq-p<r+p+1-z-q-1=r-z-q-p<0.Furthermore,(68)hr-zq-p,1=0.(3)
Right side:yn-1=1, r-z/q-p≤yn≤1.Define(69)fyn=hyn,1=r-z+p-qyn,f′=p-q<0.Thenf decreases and its maximum occurs at yn=r-z/q-p. Furthermore, (70)fr-zq-p=hr-zq-p,1=0.(4)
Left side:yn-1=r-z/q-p, r-z/q-p≤yn≤1.Define(71)fyn=hyn,r-zq-p=r+pyn+1-zr-zq-p-qr-zq-pyn-r-zq-p2,f′=p-qr-zq-p<0.Thenf decreases and the maximum occurs at yn=r-z/q-p. Furthermore, (72)fr-zq-p=hr-zq-p,r-zq-p<0Now, collect up all the function values for h(yn,yn-1): (73)hr-zq-p,r-zq-p<0,h1,r-zq-p<0,hr-zq-p,1=0,h1,1<0.Clearly the maximum value of h is 0 and it occurs at r-z/q-p,1.
The proof is complete.Figure 1
The region whereh(yn,yn-1) is defined.The result about the global stability of the positive nonhyperbolic equilibrium solution of (4) is given in the following theorem.Theorem 14.
The positive nonhyperbolic equilibrium solution of (4) is globally asymptotically stable.Proof.
Let(74)fyn,yn-1=r+pyn+yn-1z+qyn+yn-1.We have two cases to be considered. (a)
Considerr≤z.Here we distinguish between two subcases.
(1)
Considerpz-qr≤0.Let{yn}n=0∞ be a positive solution of (4). It follows from Theorem 12(a) that every positive solution of (4) eventually enters and remains in the interval p/q,1. Furthermore, p/q≤y¯<1 by Lemma 11. Indeed, function (74) is decreasing in yn and increasing in yn-1, in the interval [p/q,1]. By applying Theorem 5 the subsequences {y2n} and {y2n+1} of the solution converge monotonically to finite limits l1 and l2. Set(75)l1=limn→∞y2n,l2=limn→∞y2n+1.By (4) we have (76)y2n+2=r+py2n+1+y2nz+qy2n+1+y2n,y2n+1=r+py2n+y2n-1z+qy2n+y2n-1.Hence,(77)l1=r+pl2+l1z+ql2+l1,l2=r+pl1+l2z+ql1+l2.Thus,(78)zl1+ql1l2+l12=r+pl2+l1,zl2+ql1l2+l22=r+pl1+l2and by subtracting we have(79)l1-l2z+l1+l2+p-1=0.This is true if and only if(80)l1=l2orl1+l2=1-p-z.However, ifl1=l2=l then by (77) we have (81)l=r+pl+lz+ql+l⟹l=r/l+p+1z/l+q+1⟹z+q+1l=rl+p+1⟹q+1l-rl=1+p-z⟹q+1l2-r=1+p-zl⟹q+1l2-1+p-zl-r=0⟹l=1+p-z+1+p-z2+4rq+12q+1⟹l=y¯,whereasl1+l2=1-p-z=2y¯ is impossible in this case since by Theorem 6, (4) does not possess a period-two solution.(2)
Considerpz-qr>0.Let{yn}n=0∞ be a positive solution of (4). It follows from Lemma 10(a) that pz-qr/q-p<p/q. Furthermore, Theorem 12(a) implies that every positive solution of (4) eventually enters and remains in the interval p/q,1. Indeed, p/q≤y¯<1 by Lemma 11. Figure 2 depicts the region where f(yn,yn-1) is defined.With the understanding that function (74) is decreasing in yn and increasing in yn-1 in the interval [p/q,1] by Table 1 case 1, and since condition (31) holds, (4) does not possess a period-two solution by Theorem 6. Thus all conditions of Theorem 4 are satisfied and we conclude that y¯ is globally asymptotically stable.(b)
Considerr>z.Let{yn}n=0∞ be a positive solution of (4). It follows from  Theorem 12(b) that every positive solution of (4) eventually enters and remains in the interval r-z/p-q,1. Furthermore, r-z/q-p≤y¯<1 by Lemma 11.Lemma13 shows that yn+1≤yn-1. Replacing n by n-1 in the previous inequality then yn≤yn-2. As such, the odd and even terms of any solution yn of (4) form two monotonic nondecreasing subsequences. Furthermore, both of these subsequences are bounded because yn is bounded. Hence by Monotone Convergence Theorem the subsequences {y2n} and {y2n+1} of the solution converge to finite limits ȷ1 and ȷ2. Set(82)ȷ1=limn→∞⁡y2n,ȷ2=limn→∞⁡y2n+1.Using technique similar to the one used in proofing Case (a)(1), we have(83)y2n+2=r+py2n+1+y2nz+qy2n+1+y2n,y2n+1=r+py2n+y2n-1z+qy2n+y2n-1.Hence,(84)ȷ1=r+pȷ2+ȷ1z+qȷ2+ȷ1,ȷ2=r+pȷ1+ȷ2z+qȷ1+ȷ2.Thus,(85)zȷ1+qȷ1ȷ2+ȷ12=r+pȷ2+ȷ1,zȷ2+qȷ1ȷ2+ȷ22=r+pȷ1+ȷ2and by subtracting we have(86)ȷ1-ȷ2z+ȷ1+ȷ2+p-1=0.This is true if and only if(87)ȷ1=ȷ2orȷ1+ȷ2=1-p-z.However, ifȷ1=ȷ2=ȷ then by (84) we have (88)ȷ=r+pȷ+ȷz+qȷ+ȷ⟹ȷ=r/ȷ+p+1z/ȷ+q+1⟹z+q+1ȷ=rȷ+p+1⟹q+1ȷ-rȷ=1+p-z⟹q+1ȷ2-r=1+p-zȷ⟹q+1ȷ2-1+p-zȷ-r=0⟹ȷ=1+p-z+1+p-z2+4rq+12q+1⟹ȷ=y¯,whereas,ȷ1+ȷ2=1-p-z=2y¯ is impossible in this case since by Theorem 6, (4) does not possess a period-two solution.The proof is complete.Figure 2
The region wheref(yn,yn-1) is defined.Remark 15.
The papers [22–24] give the proof of the existence of both stable and unstable manifolds for second order difference equations decreasing in first and increasing in second argument in nonhyperbolic case of stable type, that is, of the type when second characteristic value is in (-1,1). In such a way one can avoid the use of center manifold for such equations.


## 5. Numerical Examples

In order to illustrate the results of the previous sections and to support our theoretical discussion, we consider several numerical examples generated by MATLAB.Example 1.
Consider the following equation:(89)yn+1=0.01+0.6yn+yn-10.2+14yn+yn-1.Since r=0.01 satisfies condition (31), by Theorem 9, the equilibrium is nonhyperbolic. Indeed, r≤z and pz-qr≤0, and Theorem 12 implies that every positive solution of (89) eventually enters and remains in the interval [p/q,1]. Furthermore, the equilibrium y¯ is globally asymptotically stable by Theorem 14. The dynamics of (89) is shown in Figure 3.Figure 3
Dynamics ofyn+1=0.01+0.6yn+yn-1/0.2+14yn+yn-1.Example 2.
Consider the following equation:(90)yn+1=0.0025+0.3yn+yn-10.6+14yn+yn-1.Since r=0.0025 satisfies condition (31), by Theorem 9, the equilibrium is nonhyperbolic. Indeed, r<z and pz-qr>0, and Theorem 12 implies that every positive solution of (90) eventually enters and remains in the interval [p/q,1]. Furthermore, the equilibrium y¯ is globally asymptotically stable by Theorem 14. The dynamics of (90) is shown in Figure 4.Figure 4
Dynamics ofyn+1=0.0025+0.3yn+yn-1/0.6+14yn+yn-1.Example 3.
Consider the following example:(91)yn+1=0.031+0.3yn+yn-10.01+3yn+yn-1.Since r=0.0025 satisfies condition (31), by Theorem 9, the equilibrium is nonhyperbolic. Indeed, r>z, and Theorem 12 implies that every positive solution of (91) eventually enters and remains in the interval [r-z/q-p,1]. Furthermore, the equilibrium y¯ is globally asymptotically stable by Theorem 14. The dynamics of (91) is shown in Figure 5.Figure 5
Dynamics ofyn+1=0.031+0.3yn+yn-1/0.01+3yn+yn-1.


## 6. Conclusion

In this paper, we have established the global stability of the nonhyperbolic equilibrium solutions of the second order rational difference equation:(92)xn+1=α+βxn+γxn-1A+Bxn+Cxn-1,n=0,1,2,…,where the parameters α, β, γ, A, B, C are positive real numbers and the initial conditions x-1, x0 are nonnegative real numbers.We believe that our result is an important stepping stone in understanding the behavior of solutions of rational difference equations which provides prototypes towards the development of the basic theory of the global behavior of solutions of nonlinear difference equations of higher order.



---

*Source: 486985-2015-05-06.xml*