# Global Attractivity of a Family of Max-Type Difference Equations

Authors: Xiaofan Yang; Wanping Liu; Jiming Liu
Journal: Discrete Dynamics in Nature and Society
 (2011)

---

## Abstract

We propose to study a generalized family of max-type difference
equations and then prove the global attractivity of a particular case of it under some
parameter conditions. Through some numerical results of other cases, we finally pose
a generic conjecture.


---

## Body

## 1. Introduction

The study of max-type difference equations is a hotspot in the area of discrete dynamics because such equations are often closely related to automatic control theory and competitive dynamics. For recent advances in this direction see [1–8] and the references therein.Motivated by [9], Liu et al. [10] studied the following nonautonomous max-type difference equation:(1.1)yn=p+ryn-sq+ϕn(yn-1,…,yn-m)+yn-s,n∈N0,
where p≥0, r,q>0, s,m∈ℕ, and ϕn:(ℝ+)m→ℝ+, n∈ℕ0 are mappings satisfying the condition βmin{x1,…,xm}≤ϕn(x1,x2,…,xm)≤βmax{x1,…,xm}, for some fixed β∈(0,+∞). When p=0, β∈(0,1), they proved that every positive solution to (1.1) converges to zero if r≤q, while (r-q)/(1+β) if r>q. If p>0 and rq≥p, then each positive solution to (1.1) converges to ((q-r)2+4p(1+β)-(q-r))/(2(1+β)), for some β∈(0,+∞), except for the case q<r, β∈(β0,+∞), where β0=4p/(q-r)2+1. Note that the behavior of positive solutions to (1.1) for the case q<r, β∈(β0,+∞), is still an unsolved open problem as was mentioned in [10].Here, we propose to investigate the asymptotic behavior of positive solutions to the generalized family of max-type difference equations(1.2)xn=max⁡1≤i≤k{pi+rixn-sqi+xn-s+fi(xn-1,…,xn-m)},n∈N0,
where pi≥0, ri,qi>0, s,m,k∈ℕ, k≥2 and the functions fi:[0,+∞)m→[0,+∞), i=1,2,…,k satisfy the condition(1.3)βmin⁡{u1,…,um}≤fi(u1,u2,…,um)≤βmax⁡{u1,…,um},
for some fixed β∈(0,1).In this paper, we mainly consider the particular case that allpi are zero, and then obviously (1.2) reduces to the following form:(1.4)xn=xn-s×max⁡1≤i≤k{riqi+xn-s+fi(xn-1,…,xn-m)},n∈N0.
Let x* be a nonnegative equilibrium point of (1.4), then we have(1.5)x*=x*×max⁡1≤i≤k{riqi+(1+β)x*}.
It follows directly from (1.5) that if 0<ri≤qi for all i=1,2,…,k, then (1.4) has the unique nonnegative equilibrium x*=0, while if there exists at least one j∈{1,2,…,k} such that rj>qj, then (1.4) has a zero equilibrium x*=0 and a unique positive equilibrium x*=max1≤i≤k{ri-qi}/(1+β).Finally, the following two beautiful theorems are derived.Theorem 1.1.
Consider (1.4) with condition (1.3). If 0<ri≤qi for all i=1,2,…,k, then every positive solution to (1.4) converges to the unique nonnegative equilibrium zero.Theorem 1.2.
Consider (1.4) with positive initial values and positive ri and qi. Let fi:[0,+∞)m→[0,+∞) be functions such that for some fixed β∈(0,1), there hold
(1.6)βmin⁡{u1,…,um}≤fi(u1,…,um)≤βmax⁡{u1,…,um},i=1,2,…,k.
If there exists at least one j∈{1,2,…,k} such that rj>qj, then the unique positive equilibrium of (1.4) is a global attractor.


## 2. Preliminary Lemmas

For the purpose of establishing the main results, some auxiliary lemmas are essential.Lemma 2.1.
Consider the first-order difference equation(2.1)xn=xn-1×max⁡1≤i≤k{riqi+xn-1},n∈N0,
with positive initial value x-1 and positive ri and qi. If there exists at least one j∈{1,2,…,k} such that rj>qj, then
(2.2)lim⁡n→∞xn=max⁡{ri-qi:i=1,2,…,k}.Proof.
Suppose thatmax{ri-qi:i=1,2,…,k}=rτ-qτ, which is positive, for some τ∈{1,2,…,k}. By making the variable change xn=(rτ-qτ)yn into (2.1) and then canceling the positive term rτ-qτ from the resulting equation, we can derive
(2.3)yn=yn-1×max⁡1≤i≤k{riqi+(rτ-qτ)yn-1},n∈N0.
Note that min{a1/b1,a2/b2}≤(a1+a2)/(b1+b2)≤max{a1/b1,a2/b2} for ai,bi>0, i=1,2. Then it follows from (2.3) that
(2.4)yn+1=max⁡1≤i≤k{qiyn+(ri-qi)ynqi+(rτ-qτ)yn}≤max⁡1≤i≤k{qiyn+(rτ-qτ)ynqi+(rτ-qτ)yn}≤max⁡{yn,1}.
In addition, the following two inequalities hold:
(2.5)yn+1-1=max⁡1≤i≤k{riynqi+(rτ-qτ)yn-1}≥rτynqτ+(rτ-qτ)yn-1=qτ(yn-1)qτ+(rτ-qτ)yn,(2.6)yn+1-yn=max⁡1≤i≤k{riynqi+(rτ-qτ)yn-yn}≥rτynqτ+(rτ-qτ)yn-yn=(rτ-qτ)yn(1-yn)qτ+(rτ-qτ)yn.
In the following, we are confronted with three possibilities.Case 1.
If there existsn0≥-1 such that yn0=1, then it follows from (2.4) and (2.5) that yn=1 holds for all n≥n0.Case 2.
If there existsn0≥-1 such that yn0>1, then it follows from (2.5) and (2.6) that
(2.7)yn0≥yn0+1≥yn0+2≥⋯>1.
Thus there is a finite limit γ=limn→∞yn≥1. By taking the limits on both sides of (2.3) and canceling the positive factor γ from the resulting equation, we obtain
(2.8)1=max⁡1≤i≤k{riqi+(rτ-qτ)γ},
which implies γ=1. Because if γ>1, then
(2.9)1=max⁡1≤i≤k{riqi+(rτ-qτ)γ}<max⁡1≤i≤k{riqi+(rτ-qτ)}=1,
leading to a contradiction.Case 3.
Ifyn<1 for all n≥-1, then it follows from (2.5) and (2.6) that
(2.10)y-1<y0<y1<⋯<yn<⋯<1.
Therefore, the limit of yn exists, denoted by 0<γ=limn→∞yn≤1. By taking the limits on both sides of (2.3) and canceling the nonzero factor γ from the resulting equation, there hold
(2.11)1=max⁡1≤i≤k{riqi+(rτ-qτ)γ},
which implies γ=1. Because if 0<γ<1, then
(2.12)1=max⁡1≤i≤k{riqi+(rτ-qτ)γ}>max⁡1≤i≤k{riqi+(rτ-qτ)}=1,
which is a contradiction.In either of the above three cases, we get limn→∞yn=1, implying limn→∞xn=rτ-qτ.From Lemma2.1, we have the following result.Lemma 2.2.
Consider thes-order difference equation
(2.13)xn=xn-s×max⁡1≤i≤k{riqi+xn-s},n∈N0,
with positive initial values and ri,qi>0. If there exists at least one j∈{1,2,…,k} such that rj>qj, then
(2.14)lim⁡n→∞xn=max⁡{ri-qi:i=1,2,…,k}.Proof.
Let{xn}n≥-s be an arbitrary positive solution to (2.13). Apparently we know that the sequence {xn}n≥-s can be divided into s subsequences {xj+sk}k≥0, j=-s,-s+1,…,-1, which are, respectively, positive solutions to the first-order equation (2.1) with positive initial values x-s,x-s+1,…,x-1. According to Lemma 2.1, we derive limk→∞xj+sk=max{ri-qi:i=1,2,…,k} for all j=-s,-s+1,…,-1, which directly lead to limn→∞xn=max{ri-qi:i=1,2,…,k}.Lemma 2.3.
Leta>b>0, 0<β<1, and 0<ϵ<((1-β)/(1+β))(a-b). Define two sequences {mk} and {Mk} in the following way:
(2.15)M1=a-b,mk=M1-β(Mk+ϵk),k=1,2,…,Mk=M1-β(mk-1-ϵ(k-1)),k=2,3,….
Then limk→∞mk=limk→∞Mk.Proof.
Observe that(2.16)M2-M1=-β((1-β)(a-b)-(β+1)ϵ)<0,mk+1-mk=β[Mk-Mk+1+ϵk(k+1)],k=1,2,…,Mk+1-Mk=-β[mk-mk-1+ϵk(k-1)],k=2,3,….
It follows by induction that {mk} is increasing and {Mk} is decreasing. Again by induction we derive mk<a-b and Mk>0, k=1,2,…. Hence there are two finite limits ξ=limk→∞mk and η=limk→∞Mk. By taking limits on both sides of (2.15), we derive
(2.17)ξ=a-b-βη,η=a-b-βξ,
which imply (1-β)(ξ-η)=0. Therefore ξ=η=(a-b)/(1+β).


## 3. Proofs of Main Theorems

In this section, we are in a position to prove the main theorems presented in Section1.Proof of Theorem1.1.
Note that for the caseri<qi, i=1,2,…,k, the behavior of positive solutions to (1.4) is quite simple. In this case, we have that
(3.1)xn≤xn-s×max⁡1≤i≤k{riqi}=μxn-s,
where μ=max1≤i≤k{ri/qi}<1. Easily the subsequences {xls+j}l∈ℕ0, j∈{0,1,…,s-1} converge to zero, hence the sequence {xn} also converges to zero.
For the caseri≤qi, i=1,2,…,k with at least one j∈{1,2,…,k} such that rj=qj, we can obtain that(3.2)xn≤xn-s×max⁡1≤i≤k{riqi}=xn-s.
In this case, the subsequences {xls+j}l∈ℕ0, j=0,1,…,s-1 are all positive and nonincreasing, thus they converge, respectively, to some nonnegative limits ψj:=liml→∞xls+j, j=0,1,…,s-1.
If we replacen in (1.4) by sl+j, l∈ℕ0 for an arbitrary fixed j∈{0,1,…,s-1} and let l→∞, we can get(3.3)ψj=ψj×max⁡1≤i≤k{riqi+ψj+fi(ψv1,…,ψvm)},
where vi∈{0,1,…,s-1}, i=1,…,m. Without loss of generality, assume that ψj≠0, then we obtain that
(3.4)1=rτqτ+ψj+fτ(ψv1,…,ψvm),
with some fixed number τ∈{1,2,…,k}. Because rτ≤qτ, then it follows from (3.4) that
(3.5)qτ+ψj+fτ(ψv1,…,ψvm)=rτ≤qτ.
Therefore we have
(3.6)ψj+fτ(ψv1,…,ψvm)=0,
leading to ψj=0, which is a contradiction. Hence we have that ψj=0, j=0,1,…,s-1, and every positive solution to (1.4) converges to zero, if ri≤qi for all i=1,2,…,k.Proof of Theorem1.2.
Suppose thatmax{ri-qi:i=1,2,…,k}=rτ-qτ>0 for some τ∈{1,2,…,k}. Let ϵ be an arbitrary fixed real number with 0<ϵ<((1-β)/(1+β))(rτ-qτ). Define two sequences {Mk} and {mk} in the way shown  in (2.15) with a=rτ,b=qτ.Let{xn} be an arbitrary positive solution to (1.4). Next, we proceed by proving two claims.Claim 1.
There existsN1∈ℕ such that m1-ϵ≤xn≤M1+ϵ for all n≥N1.Proof.
Note that(3.7)xn≤xn-s×max⁡1≤i≤k{riqi+xn-s},n=0,1,2,….
Consider the following difference equation:
(3.8)zn(1)=zn-s(1)×max⁡1≤i≤k{riqi+zn-s(1)},n=0,1,2,….
Let {zn(1)} be a positive solution to (3.7) with the initial values z-1(1)=x-1, z-2(1)=x-2,…,z-s(1)=x-s.
Note that the mappingh(x)=rx/(q+x) is strictly increasing on the interval (0,+∞). It follows by induction that xn≤zn(1) for all n≥-s. By Lemma 2.2, we have limn→∞zn(1)=rτ-qτ=M1. Hence there is an integer N1′∈ℕ such that xn≤M1+ϵ for n≥N1′.
Lett=max{s,m}. Note that(3.9)xn≥xn-s×max⁡1≤i≤k{riqi+xn-s+β(M1+ϵ)},n≥N1′+t.
Consider the difference equation
(3.10)yn(1)=yn-s(1)×max⁡1≤i≤k{riqi+yn-s(1)+β(M1+ϵ)},n≥N1′+t,
with yN1′+t-1(1)=xN1′+t-1, yN1′+t-2(1)=xN1′+t-2,…,yN1′(1)=xN1′. Note the monotonicity of h(x), it follows by induction that xn≥yn(1) for all n≥N1′. By Lemma 2.2, we get that limn→∞yn(1)=m1. Thus there exists an integer N1≥N1′ such that xn≥m1-ϵ for all n≥N1.Working inductively, we will reach the following claim.Claim 2.
For everyk∈ℕ, there exists Nk∈ℕ such that
(3.11)mk-ϵk≤xn≤Mk+ϵk,
for all n≥Nk.Proof.
Obviously, the casek=1 follows directly from Claim 1. In the following, we proceed by induction. Assume that the assertion is true for k=ω(ω≥1). Then it suffices to prove the assertion is also true for k=ω+1.
Note that(3.12)xn≤xn-s×max⁡1≤i≤k{riqi+xn-s+β(mω-ϵ/ω)},n≥Nω+t.
Consider the difference equation
(3.13)zn(ω+1)=zn-s(ω+1)×max⁡1≤i≤k{riqi+zn-s(ω+1)+β(mω-ϵ/ω)},n≥Nω+t,
with zNω+t-1(ω+1)=xNω+t-1, zNω+t-2(ω+1)=xNω+t-2,…,zNω(ω+1)=xNω. Note the monotonicity of h(x), it follows by induction that xn≤zn(ω+1) for all n≥Nω. By Lemma 2.2, we have that limn→∞zn(ω+1)=Mω+1. So there is an integer Nω+1′∈ℕ such that xn≤Mω+1+ϵ/(ω+1) for all n≥Nω+1′. Then note that
(3.14)xn≥xn-s×max⁡1≤i≤k{riqi+xn-s+β(Mω+1+ϵ/(ω+1))},n≥Nω+1′+t.
Consider the following difference equation
(3.15)yn(ω+1)=yn-s(ω+1)×max⁡1≤i≤k{riqi+yn-s(ω+1)+β(Mω+1+ϵ/(ω+1))},n≥Nω+1′+t,
with yNω+1′+t-1(ω+1)=xNω+1′+t-1, zNω+1′+t-2(ω+1)=xNω+1′+t-2,…,zNω+1′(ω+1)=xNω+1′. By the monotonicity of h(x), it follows by induction that xn≥yn(ω+1) for all n≥Nω+1′. By Lemma 2.2, we have that limn→∞yn(ω+1)=mω+1. So there is an integer Nω+1≥Nω+1′ such that xn≥mω+1-ϵ/(ω+1) for all n≥Nω+1.From Claim2, we derive(3.16)lim⁡k→∞mk=lim⁡k→∞(mk-ϵk)≤lim⁡̲n→∞xn≤lim⁡¯n→∞xn≤lim⁡k→∞(Mk+ϵk)=lim⁡k→∞Mk.
This plus Lemma 2.3 leads to that(3.17)lim⁡n→∞xn=lim⁡k→∞mk=lim⁡k→∞Mk=rτ-qτ1+β.


## 4. Simulations and Future Work

In the previous section, we proved the global attractivity of (1.2) when all pi are zero. In this section, we investigate the dynamic behavior of (1.2) provided that all pi are not zero. First, it is trivial to confirm that when all pi are not zero, (1.2) has the following unique positive equilibrium point x*=max1≤i≤k{(qi-ri)2+4pi(1+β)+ri-qi}/(2(1+β)). In the following, some numerical results are presented.Experiment 1.
Consider the first-order difference equation(4.1)xn=max⁡{0.2+0.6xn-10.6+xn-1+0.3xn-1,rxn-1q+xn-1+0.3xn-1},n∈N,
where r,q>0 and the initial value x0>0. (See Figures 1 and 2).r=1, q=2; x*=26/13≈0.3922.
(a)
Initial valuex0=0.2(b)
Initial valuex0=0.7r=2, q=1; x*=10/13≈0.7692.
(a)
Initial valuex0=0.5(b)
Initial valuex0=0.9Experiment 2.
Consider the second-order difference equation(4.2)xn=max⁡{0.5+xn-21+xn-2+0.5xn-1,0.8+rxn-2q+xn-2+0.5xn-1},n≥2,
where r,q>0 and the initial values x0,x1>0. (See Figures 3 and 4).r=1, q=2; x*=3/3≈0.5774.
(a)
Initial valuesx0=0.2, x1=0.4(b)
Initial valuesx0=0.8, x1=0.3r=2, q=1; x*=(5.8+1)/3≈1.1361.
(a)
Initial valuesx0=1.3, x1=1.0(b)
Initial valuesx0=0.5, x1=0.9Experiment 3.
Consider the third-order difference equation(4.3)xn=max⁡{0.5+xn-31+xn-3+0.9(xn-12+xn-22)/2,3xn-32+xn-3+0.9(xn-12+xn-22)/2},n≥3,
where the initial values x0,x1,x2>0. (See Figure 5).x*=10/19≈0.5263.
(a)
Initial valuesx0=0.1, x1=0.3, x2=0.5(b)
Initial valuesx0=0.3, x1=0.7, x2=1.0(c)
Initial valuesx0=0.8, x1=0.3, x2=0.1(d)
Initial valuesx0=1.1, x1=0.8, x2=0.6Inspired by this work and the results of [10], here we pose the following conjecture.Conjecture 4.1.
Consider (1.2) with nonnegative pi and positive ri and qi. Let fi:[0,+∞)m→[0,+∞), i=1,2,…,k be k functions such that for some fixed β∈(0,1), there hold
(4.4)βmin⁡{u1,…,uk}≤fi(u1,…,uk)≤βmax⁡{u1,…,uk}.
If riqi≥pi for all i=1,2,…,k, then every positive solution to (1.2) converges to the equilibrium point
(4.5)x*=12(1+β)max⁡1≤i≤k{(qi-ri)2+4pi(1+β)+ri-qi}.



---

*Source: 506373-2011-02-06.xml*