# Positive Periodic Solutions of Nonlinear First-Order Functional Difference Equations

Authors: Ruyun Ma; Tianlan Chen; Yanqiong Lu
Journal: Discrete Dynamics in Nature and Society
 (2010)

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## Abstract

We consider the existence, multiplicity, and nonexistence of positiveT-periodic solutions for the difference equations Δx(n)=a(n)g(x(n))x(n)-λb(n)f(x(n-τ(n))), and Δx(n)+a(n)g(x(n))x(n)=λb(n)f(x(n-τ(n))), where a,b:ℤ→[0,∞) are T-periodic, τ:ℤ→ℤ is T-periodic.


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## Body

## 1. Introduction

In the recent years, there has been considerable interest in the existence of periodic solutions of the following equation:(1.1)x′(t)=ã(t)g̃(x(t))x(t)-λb̃(t)f̃(x(t-τ(t))),
where ã,b̃∈C(ℝ,[0,∞)) are ω-periodic functions, ∫0ωã(t)dt>0, ∫0ωb̃(t)dt>0, τ is a continuous ω-periodic function. Equation (1.1) has been proposed as a model for a variety of physiological processes and conditions including production of blood cells, respiration, and cardiac arrhythmias. See, for example, [1–13] and the references therein.Letℤ be the set of all integers. In the present paper, we study the existence of positive T-periodic solutions of discrete analogues to (1.1) of the form (1.2)Δx(n)=a(n)g(x(n))x(n)-λb(n)f(x(n-τ(n))),n∈Z,
and the difference equation (1.3)Δx(n)+a(n)g(x(n))x(n)=λb(n)f(x(n-τ(n))),n∈Z,
where ℤ is the set of integer numbers, T∈ℕ is a fixed integer, a,b:ℤ→[0,+∞) are T-periodic, a(n)≢0, b(n)≢0 on [0,T-1]=:{0,1,…,T-1}, τ:ℤ→ℤ is T-periodic, and f,g∈C([0,+∞),[0,+∞)), λ>0 is a parameter.So far, relatively little is known about the existence of positive periodic solutions of (1.2) and (1.3). To our best knowledge, only Raffoul [14] dealt with the special equations of (1.2) and (1.3) of the form (1.4)Δx(n)=α(n)x(n)-λb(n)f(x(n-τ(n))),
with α(n)=a(n)-1>0, and (1.5)Δx(n)+α(n)x(n)=λb(n)f(x(n-τ(n))),
with α(n)=1-a(n)>0, and determining values of λ, for which there exist positive T-periodic solutions of (1.4) and (1.5), respectively.It is the purpose of this paper to study more general equations (1.2) and (1.3) and generalize the main results of Raffoul [14]. We establish some existence, multiplicity, and nonexistence results of positive periodic solutions for (1.2) and (1.3), respectively. The main tools we will use are fixed point theorem in cones and fixed point index theory [15, 16]. Throughout this paper, we denote the product of y(n) from n=a to n=b by ∏n=aby(n) with the understanding that ∏n=aby(n)=1 for all a>b.The rest of the paper is arranged as follows: in Section2, we give some preliminary results. Section 3 is devoted to generalize the main results of Raffoul [14]. Finally, in Section 4, we state and prove some existence, multiplicity, and nonexistence results of positive periodic solutions for (1.2) and (1.3). For related results on the associated differential equations, see Wang [11].


## 2. Preliminaries

In this paper, we make the following assumptions.(H1)
a,b:ℤ→[0,+∞) are T-periodic, and a(n)≢0, b(n)≢0 on n∈[0,T-1], τ:ℤ→ℤ is T-periodic.(H2)
f,g:[0,+∞)→[0,+∞) are continuous, 0<l≤g(x)≤L<∞ with l,L are positive constants, and f(x)>0, for x>0. Also,
(2.1)σl=∏s=0T-1(1+a(s)l),σL=∏s=0T-1(1+a(s)L),
it is clear that 1<σl≤σL. Let
(2.2)M(r)=max⁡{f(x)∣0≤x≤r},m(r)=min⁡{f(x)∣σl-1(σL-1)σL⋅r≤x≤r}.The following well-known result of fixed point index is crucial to our arguments.Lemma 2.1 (see [15, 16]).
LetE be a Banach space, and let  K be a cone in  E. For  r>0, define  Kr={u∈K:∥u∥<r}. Assume that  T:K¯r→K is completely continuous such that  Tu≠u for  u∈∂Kr={u∈K:∥u∥=r}.(i)
If∥Tu∥≥∥u∥ for u∈∂Kr, then
(2.3)i(T,Kr,K)=0.(ii)
If∥Tu∥≤∥u∥ for u∈∂Kr, then
(2.4)i(T,Kr,K)=1.LetX be the set of all real T-periodic sequences. This set, endowed with the maximum norm ∥x∥=maxn∈[0,T-1]|x(n)|, is a Banach space. The next lemma is essential in obtaining our results.Lemma 2.2.
Assume that (H1)-(H2) hold. Then,x∈X is a solution of (1.2) if and only if
(2.5)x(n)=λ∑u=nn+T-1Gx(n,u)b(u)f(x(u-τ(u))),n∈Z,
where
(2.6)Gx(n,u)=∏s=u+1n+T-1(1+a(s)g(x(s)))∏s=0T-1(1+a(s)g(x(s)))-1,u∈[n,n+T-1].Proof.
Ifx∈X and satisfies (2.5), then
(2.8)x(n+1)=λ∑u=n+1n+TGx(n+1,u)b(u)f(x(u-τ(u)))=λ∑u=n+1n+T-1Gx(n+1,u)b(u)f(x(u-τ(u)))+λGx(n+1,n+T)b(n+T)f(x(n+T-τ(n+T)))=λ∑u=n+1n+T-1Gx(n+1,u)b(u)f(x(u-τ(u)))+λGx(n+1,n+T)b(n)f(x(n-τ(n))),[1+a(n)g(x(n))]x(n)=λ∑u=n+1n+T-1[1+a(n)g(x(n))]Gx(n,u)b(u)f(x(u-τ(u)))+λ[1+a(n)g(x(n))]Gx(n,n)b(n)f(x(n-τ(n))).
From (2.2), we know that
(2.9)Gx(n+1,n+T)=∏s=n+T+1n+T(1+a(s)g(x(s)))∏s=0T-1(1+a(s)g(x(s)))-1=1∏s=0T-1(1+a(s)g(x(s)))-1,[1+a(n)g(x(n))]Gx(n,n)=∏s=nn+T-1(1+a(s)g(x(s)))∏s=0T-1(1+a(s)g(x(s)))-1=∏s=0T-1(1+a(s)g(x(s)))∏s=0T-1(1+a(s)g(x(s)))-1,Gx(n+1,u)=∏s=u+1n+T(1+a(s)g(x(s)))∏s=0T-1(1+a(s)g(x(s)))-1=∏s=u+1n+T-1(1+a(s)g(x(s)))(1+a(n+T)g(x(n+T)))∏s=0T-1(1+a(s)g(x(s)))-1=[1+a(n)g(x(n))]Gx(n,u).
So, by (2.8)-(2.9), we can conclude that
(2.10)x(n+1)-[1+a(n)g(x(n))]x(n)=-λb(n)f(x(n-τ(n))),
thus, x is a T-periodic solutions of (1.2). On the other hand, if x∈X and satisfies (1.2), then (1.2) is equivalent to
(2.11)Δ[∏s=-∞n-1(1+a(s)g(x(s)))-1x(n)]=-λb(n)f(x(n-τ(n)))∏s=-∞n[1+a(s)g(x(s))]-1,
by summing the above equation from u=n to u=n+T-1, we obtain (2.5).Note that sincea(n)≢0 for n∈[0,T-1] and 0<l≤g(x)≤L<∞, we have (2.12)1σL-1≤Gx(n,u)≤σLσl-1,n≤u≤n+T-1,
and 0<(σl-1)/(σL-1)σL<1.DefineK as a cone in X by (2.13)K={x∈X:x(n)≥0,n∈Z,x(n)≥σl-1(σL-1)σL‖x‖}.
Also, define, for r a positive number, Ωr by (2.14)Ωr={x∈K:‖x‖<r},
note that ∂Ωr={x∈K:∥x∥=r}.DefineTλ:X→X by (2.15)Tλx(n)=λ∑u=nn+T-1Gx(n,u)b(u)f(x(u-τ(u))),n∈Z,
where Gx(n,u) is given by (2.6).Lemma 2.3.
Assume that (H1)-(H2) hold. Then,Tλ(K)⊂K and  Tλ:K→K is compact and continuous.Proof.
In the view of the definition ofK, for x∈K, we have
(2.16)(Tλx)(n+T)=λ∑u=n+Tn+2T-1Gx(n+T,u)b(u)f(x(u-τ(u)))=λ∑s=nn+T-1Gx(n+T,s+T)b(s+T)f(x(s+T-τ(s+T)))=λ∑u=nn+T-1Gx(n+T,u+T)b(u)f(x(u-τ(u)))=(Tλx)(n).
In fact,
(2.17)Gx(n+T,u+T)=∏s=u+T+1n+2T-1(1+a(s)g(x(s)))∏s=0T-1(1+a(s)g(x(s)))-1=∏s=u+1n+T-1(1+a(s)g(x(s)))∏s=0T-1(1+a(s)g(x(s)))-1=Gx(n,u).
So, Tλx∈X. One can show that for x∈K and n∈[0,T-1],
(2.18)(Tλx)(n)≥λσL-1∑u=nn+T-1b(u)f(x(u-τ(u)))=σl-1(σL-1)σL⋅σLσl-1⋅λ∑u=nn+T-1b(u)f(x(u-τ(u)))≥σl-1(σL-1)σL‖Tλx‖.
Thus, Tλ(K)⊂K, and it is easy to show that Tλ:K→K is compact and continuous.Consequently,x∈K is a solution of (1.2) if and only if x is a fixed point of Tλ in K.Lemma 2.4.
Assume that (H1)-(H2) hold. Letη>0, if  x∈K and  f(x(n))≥x(n)η for  n∈[0,T-1], then
(2.19)‖Tλx‖≥λη(σl-1)(σL-1)2σL⋅∑u=0T-1b(u)⋅‖x‖.Proof.
Sincex∈K and f(x(n))≥x(n)η for n∈[0,T-1], we have
(2.20)(Tλx)(n)≥λσL-1∑u=nn+T-1b(u)f(x(u-τ(u)))=λησL-1⋅σl-1(σL-1)σL∑u=0T-1b(u)⋅‖x‖=λη(σl-1)(σL-1)2σL⋅∑u=0T-1b(u)⋅‖x‖.
Thus,
(2.21)‖Tλx‖≥λη(σl-1)(σL-1)2σL⋅∑u=0T-1b(u)⋅‖x‖.Lemma 2.5.
Assume that (H1)-(H2) hold. Letr>0, if  x∈∂Ωr, and there exists an  ε>0 such that  f(x(n))≤x(n)ε for  n∈[0,T-1], then
(2.22)‖Tλx‖≤λεσLσl-1⋅∑u=0T-1b(u)⋅‖x‖.Proof.
From the definition ofTλ, for x∈∂Ωr, we have
(2.23)‖Tλx‖≤λσLσl-1⋅∑u=0T-1b(u)f(x(u-τ(u)))≤λεσLσl-1⋅∑u=0T-1b(u)⋅‖x‖.The following two lemmas are weak forms of Lemmas2.4 and 2.5.Lemma 2.6.
Assume that (H1)-(H2) hold. Ifx∈∂Ωr, r>0, then
(2.24)‖Tλx‖≥λ∑u=0T-1b(u)σL-1⋅m(r).Proof.
Sincef(x(n))≥m(r) for n∈[0,T-1], it is easy to see that this lemma can be achieved in a similar manner as in Lemma 2.4.Lemma 2.7.
Assume that (H1)-(H2) hold. Ifx∈∂Ωr, r>0, then
(2.25)‖Tλx‖≤λσLM(r)σl-1⋅∑u=0T-1b(u).Proof.
Sincef(x(n))≤M(r) for n∈[0,T-1], it is easy to see that this lemma can be obtained in a similar manner as in Lemma 2.5.


## 3. Generalization of the Main Results in [14]

Let(3.1)f0=lim⁡x→0+f(x)x,f∞=lim⁡x→∞f(x)x.
In this Section, we make the following assumptions.(L1)
f0=∞.(L2)
f∞=∞.(L3)
f0=0.(L4)
f∞=0.(L5)
f0=m with 0<m<∞.(L6)
f0=M with 0<M<∞. And let
(3.2)A=max⁡0≤n≤T-1∑u=0T-1Gx(n,u)b(u),B=min⁡0≤n≤T-1∑u=0T-1Gx(n,u)b(u).Theorem 3.1.
Assume that (H1), (H2), (L5), and (L6) hold. Then, for eachλ satisfying
(3.3)σL(σL-1)σl-1⋅1BM<λ<1Am,
or
(3.4)σL(σL-1)σl-1⋅1Bm<λ<1AM,
equation (1.2) has at least one positive T-periodic solutions.Proof.
Using the same method to prove[14, Theorem 2.3] with obvious changes, we can prove Theorem 3.1. The process of the proof is omitted.Theorem 3.2.
Assume that (H1) and (H2) hold. If (L1) and (L4) hold, or (L2) and (L3) hold, then (1.2) has at least one positive  T-periodic solution for any  λ>0.Proof.
The proof is similar to arguments to prove [14, Theorem 2.4].The next two corollaries are consequences of the previous two theorems.Corollary 3.3.
Assume that (H1) and (H2) hold. If (L1) and (L6) hold, or (L2) and (L5) hold, then (1.2) has at least one positive  T-periodic solution if  λ satisfies either  0<λ<1/AM or  0<λ<1/Am.Corollary 3.4.
Assume that (H1) and (H2) hold. Also, if (L3) and (L6) hold, or (L4) and (L5) hold, then (1.2) has at least one positive  T-periodic solution if  λ satisfies either  σL(σL-1)/(σl-1)·(1/BM)<λ<∞ or  σL(σL-1)/(σl-1)·(1/Bm)<λ<∞.


## 4. Existence, Multiplicity, and Nonexistence Results of PositivePeriodic Solution(s) for (1.2) and (1.3)

In this section, we use the notations(4.1)i0=numberofzerosintheset{f0,f∞},i∞=numberofinfinitiesintheset{f0,f∞}.
It is clear that i0,i∞=0,1 or 2. Then, we will show that (1.2) has i0 or i∞ positive T-periodic solution(s) for sufficiently large or small λ.Theorem 4.1.
Assume that (H1)-(H2) hold.(a)
Ifi0=1 or 2, then (1.2) has i0 positive T-periodic solutions for λ>(σL-1)/m(1)∑u=0T-1b(u)>0.(b)
Ifi∞=1 or 2, then (1.2) has i∞ positive T-periodic solutions for 0<λ<(σl-1)/σLM(1)∑u=0T-1b(u).(c)
Ifi0=0 or i∞=0, then (1.2) has no positive T-periodic solution for sufficiently large or small λ>0, respectively.Proof.
(a) Chooser1=1. By Lemma 2.6, we infer that there exists a λ0=(σL-1)/m(r1)∑u=0T-1b(u)>0, such that
(4.2)‖Tλx‖>‖x‖forx∈∂Ωr1,λ>λ0.
Iff0=0, we can choose 0<r2<r1 so that f(x)≤εx for 0≤x≤r2, where the constant ε>0 satisfies
(4.3)λεσL∑u=0T-1b(u)σl-1<1.
Thus, f(x(n))≤εx(n) for x∈∂Ωr2 and n∈[0,T-1]. We have by Lemma 2.5 that
(4.4)‖Tλx‖≤λεσL∑u=0T-1b(u)σl-1‖x‖<‖x‖forx∈∂Ωr2.
It follows from Lemma 2.1 that
(4.5)i(Tλ,Ωr1,K)=0,i(Tλ,Ωr2,K)=1.
Thus, i(Tλ,Ωr1∖Ω¯r2,K)=-1 and Tλ has a fixed point in Ωr1∖Ω¯r2, which is a positive T-periodic solution of (1.2) for λ>λ0.
Iff∞=0, there is an Ĥ>0 such that f(x)≤εx for x≥Ĥ, where ε>0 satisfies (4.3). Let r3=max{2r1,Ĥ/((σl-1)/σL(σL-1))}, and it follows that x(n)≥((σl-1)/σL(σL-1))∥x∥≥Ĥ for x∈∂Ωr3 and n∈[0,T-1]. Thus, f(x(n))≤εx(n) for x∈∂Ωr3 and  n∈[0,T-1]. In view of Lemma 2.5, we have
(4.6)‖Tλx‖≤λεσL∑u=0T-1b(u)σl-1‖x‖<‖x‖forx∈∂Ωr3.
Again, it follows from Lemma 2.1 that
(4.7)i(Tλ,Ωr1,K)=0,i(Tλ,Ωr3,K)=1.
Thus, i(Tλ,Ωr3∖Ω¯r1,K)=1 and (1.2) has a positive T-periodic solution for λ>λ0.
Iff0=f∞=0, it is easy to see from the above proof that Tλ has a fixed point x1 in Ωr1∖Ω¯r2 and a fixed point x2 in Ωr3∖Ω¯r1 such that
(4.8)r2<‖x1‖<r1<‖x2‖<r3.
Consequently, (1.2) has two positive T-periodic solutions for λ>λ0 if f0=f∞=0.
(b) Chooser1=1. By Lemma 2.7, we infer that there exists
(4.9)λ0=σl-1σLM(r1)∑u=0T-1b(u)>0,
such that
(4.10)‖Tλx‖<‖x‖forx∈∂Ωr1,0<λ<λ0.
If f0=∞, there is a positive number r2<r1 such that f(x)≥ηx for 0≤x≤r2, where η>0 is chosen so that
(4.11)λη(σl-1)(σL-1)2σL∑u=0T-1b(u)>1.
Then,
(4.12)f(x(n))≥ηx(n)forx∈∂Ωr2,n∈[0,T-1].
By Lemma 2.4, we have that
(4.13)‖Tλx‖≥λη(σl-1)(σL-1)2σL∑u=0T-1b(u)⋅‖x‖>‖x‖forx∈∂Ωr2.
It follows from Lemma 2.1 that
(4.14)i(Tλ,Ωr1,K)=1,i(Tλ,Ωr2,K)=0.
Thus, i(Tλ,Ωr1∖Ω¯r2,K)=1 and Tλ has a fixed point in Ωr1∖Ω¯r2 for 0<λ<λ0, which is a positive T-periodic solution of (1.2).
Iff∞=∞, there is an Ĥ>0 such that f(x)≥ηx for x≥Ĥ, where η>0 satisfies (4.11). Let r3=max{2r1,Ĥ/((σl-1)/σL(σL-1))}. If x∈∂Ωr3, then
(4.15)x(n)≥σl-1σL(σL-1)‖x‖≥Ĥ,
and hence,
(4.16)f(x(n))≥ηx(n)forn∈[0,T-1].
Again, it follows from Lemma 2.4 that
(4.17)‖Tλx‖≥λη(σl-1)(σL-1)2σL∑u=0T-1b(u)‖x‖>‖x‖forx∈∂Ωr3.
It follows from Lemma 2.1 that
(4.18)i(Tλ,Ωr1,K)=1,i(Tλ,Ωr3,K)=0,
and hence, i(Tλ,Ωr3∖Ω¯r1,K)=-1. Thus, Tλ has a fixed point in Ωr3∖Ω¯r1 for 0<λ<λ0, which is a positive T-periodic solution of (1.2).
Iff0=f∞=∞, it is easy to see from the above proof that Tλ has a fixed point x1 in Ωr1∖Ω¯r2 and a fixed point x2 in Ωr3∖Ω¯r1 such that
(4.19)r2<‖x1‖<r1<‖x2‖<r3.
Consequently, (1.2) has two positive T-periodic solutions for 0<λ<λ0 if f0=f∞=∞.
(c) If i0=0, then f0>0 and f∞>0. It follows that there exist positive numbers η1, η2, r1 and r2, such that r1<r2 and
(4.20)f(x)≥η1xforx∈[0,r1],f(x)≥η2xforx∈[r2,∞).
Let c1=min{η1,η2,minr1≤x≤r2{f(x)/x}}>0. Thus, we have
(4.21)f(x)≥c1xforx∈[0,∞).
Assume v is a positive T-periodic solution of (1.2). We will show that this leads to a contradiction for λ>λ0, where λ0=(σL-1)2σL/c1(σl-1)∑u=0T-1b(u). Since Tλv(n)=v(n) for n∈[0,T-1], it follows from Lemma 2.4 that for λ>λ0,
(4.22)‖v‖=‖Tλv‖≥λ(σl-1)∑u=0T-1b(u)(σL-1)2σLc1‖v‖>‖v‖,
which is a contradiction.
Ifi∞=0, then f0<∞ and f∞<∞. It follows that there exist positive numbers ε1, ε2, r1 and r2 such that r1<r2,
(4.23)f(x)≤ε1xforx∈[0,r1],f(x)≤ε2xforx∈[r2,∞).
Let c2=max{ε1,ε2,maxr1≤x≤r2{f(x)/x}}>0. Thus, we have
(4.24)f(x)≤c2xforx∈[0,∞).
Assume v is a positive T-periodic solution of (1.2). We will show that this leads to a contradiction for 0<λ<λ0, where λ0=(σl-1)/c2σL∑u=0T-1b(u). Since Tλv(n)=v(n) for n∈[0,T-1], it follows from Lemma 2.5 that for 0<λ<λ0,
(4.25)‖v‖=‖Tλv‖≤λσL∑u=0T-1b(u)σl-1c2‖v‖<‖v‖,
which is a contradiction.The following result is a direct consequence of Theorem4.1(c).Corollary 4.2.
Assume that (H1)-(H2) hold.(a)
If there exists ac1>0 such that f(x)≥c1x for x∈[0,∞), then there is a λ0=(σL-1)2σL/c1(σl-1)∑u=0T-1b(u) such that for all λ>λ0, (1.2) has no positive T-periodic solutions.(b)
If there exists ac2>0 such that f(x)≤c2x for x∈[0,∞), then there is a λ0=(σl-1)/c2σL∑u=0T-1b(u) such that for all 0<λ<λ0, (1.2) has no positive T-periodic solutions.Theorem 4.3.
Assume that (H1)-(H2) hold andi0=i∞=0. If
(4.26)(σL-1)2σL(σl-1)∑u=0T-1b(u)⋅1max⁡{f0,f∞}<λ<σl-1σL∑u=0T-1b(u)⋅1min⁡{f0,f∞},
then (1.2) has a positive T-periodic solutions.Proof.
Iff∞>f0, then (σL-1)2σL/(σl-1)∑u=0T-1b(u)f∞<λ<(σl-1)/σL∑u=0T-1b(u)f0. It is easy to see that there exists ε:0<ε<f∞ such that
(4.27)(σL-1)2σL(σl-1)∑u=0T-1b(u)(f∞-ε)<λ<σl-1σL∑u=0T-1b(u)(f0+ε).
Now, turning to f0 and f∞, there is an r1>0, such that f(x)≤(f0+ε)x for 0≤x≤r1. Thus, f(x(n))≤(f0+ε)x(n) for x∈∂Ωr1 and n∈[0,T-1]. We have by Lemma 2.5 that
(4.28)‖Tλu‖≤λ(f0+ε)σL∑u=0T-1b(u)σl-1‖x‖<‖x‖forx∈∂Ωr1.
On the other hand, there is an Ĥ>r1 such that f(x)≥(f∞-ε)x for x≥Ĥ. Let r2=max{2r1,Ĥ/((σl-1)/σL(σL-1))}, and it follows that x(n)≥((σl-1)/σL(σL-1))∥x∥≥Ĥ for x∈∂Ωr2 and n∈[0,T-1]. In view of Lemma 2.4, we have
(4.29)‖Tλu‖≥λ(f∞-ε)(σl-1)∑u=0T-1b(u)(σL-1)2σL‖x‖>‖x‖forx∈∂Ωr2.
It follows from Lemma 2.1 that
(4.30)i(Tλ,Ωr1,K)=1,i(Tλ,Ωr2,K)=0.
Thus, i(Tλ,Ωr2∖Ω¯r1,K)=-1. Hence, Tλ has a fixed point in Ωr2∖Ω¯r1. Consequently, (1.2) has a positive T-periodic solution.
Iff∞<f0, then (σL-1)2σL/(σl-1)∑u=0T-1b(u)f0<λ<(σl-1)/σL∑u=0T-1b(u)f∞. It is easy to see that there exists an ε:0<ε<f0 such that
(4.31)(σL-1)2σL(σl-1)∑u=0T-1b(u)(f0-ε)<λ<σl-1σL∑u=0T-1b(u)(f∞+ε).
Now, turning to f0 and f∞, there is an r1>0 such that f(x)≥(f0-ε)x for 0≤x≤r1. Thus, f(x(n))≥(f0-ε)x(n) for x∈∂Ωr1 and n∈[0,T-1]. We have by Lemma 2.4 that
(4.32)‖Tλu‖≥λ(f0-ε)(σl-1)∑u=0T-1b(u)(σL-1)2σL‖x‖>‖x‖forx∈∂Ωr1.
On the other hand, there is an Ĥ>r1 such that f(x)≤(f∞+ε)x for x≥Ĥ. Let r2=max{2r1,Ĥ/((σl-1)/σL(σL-1))}, and it follows that x(n)≥((σl-1)/σL(σL-1))∥x∥≥Ĥ for x∈∂Ωr2 and n∈[0,T-1]. Thus, f(x(n))≤(f∞+ε)x(n) for x∈∂Ωr2 and n∈[0,T-1]. In view of Lemma 2.5, we have
(4.33)‖Tλu‖≤λ(f∞+ε)σL∑u=0T-1b(u)σl-1‖x‖<‖x‖forx∈∂Ωr2.
It follows from Lemma 2.1 that
(4.34)i(Tλ,Ωr1,K)=0,i(Tλ,Ωr2,K)=1.
Thus, i(Tλ,Ωr2∖Ω¯r1,K)=1. Hence, Tλ has a fixed point in Ωr2∖Ω¯r1. Consequently, (1.2) has a positive T-periodic solution.Next, we pay our attention to (1.3), that is (4.35)x(n+1)=[1-a(n)g(x(n))]x(n)+λb(n)f(x(n-τ(n))),n∈Z,
where λ, a, b, f(x), and g(x) satisfy the same assumptions stated for (1.2) except that 0≤la(n)≤La(n)<1 for all n∈[0,T-1]. In view of (1.3), we have that (4.36)x(n)=λ∑u=nn+T-1Kx(n,u)b(u)f(x(u-τ(u))),
where (4.37)Kx(n,u)=∏s=u+1n+T-1(1-a(s)g(x(s)))1-∏s=0T-1(1-a(s)g(x(s))),u∈[n,n+T-1].Note that since0≤la(n)≤La(n)<1 for all n∈[0,T-1], we have (4.38)ρL1-ρL≤Kx(n,u)≤11-ρl,n≤u≤n+T-1,
here (4.39)ρl=∏s=0T-1(1-a(s)l),ρL=∏s=0T-1(1-a(s)L),
and 0<ρL(1-ρl)/(1-ρL)<1.Similarly, we can get the following theorems.Theorem 4.4.
Assume that (H1)-(H2) hold and0≤la(n)≤La(n)<1 for  n∈[0,T-1]. (a)
Ifi0=1 or 2, then (1.3) has i0 positive T-periodic solutions for λ>(1-ρL)/ρLm(1)∑u=0T-1b(u)>0.(b)
Ifi∞=1 or 2, then (1.3) has i∞ positive T-periodic solutions for 0<λ<(1-ρl)/M(1)∑u=0T-1b(u).(c)
Ifi0=0 or i∞=0, then (1.3) has no positive T-periodic solution for sufficiently large or small λ>0, respectively.The following result is a direct consequence of Theorem4.4(c).Corollary 4.5.
Assume that (H1)-(H2) hold and0≤la(n)≤La(n)<1 for  n∈[0,T-1]. (a)
If there exists ac1>0 such that f(x)≥c1x for x∈[0,∞), then there is a λ0=((1-ρL)/ρL)2·(1/c1(1-ρl)∑u=0T-1b(u)) such that for all λ>λ0 (1.3) has no positive T-periodic solutions.(b)
If there exists ac2>0 such that f(x)≤c2x for x∈[0,∞), then there is a λ0=(1-ρl)/c2∑u=0T-1b(u) such that for all 0<λ<λ0 (1.3) has no positive T-periodic solutions.Theorem 4.6.
Assume that (H1)-(H2) hold andi0=i∞=0. Let  0≤la(n)≤La(n)<1 for  n∈[0,T-1]. If
(4.40)(1-ρLρL)2⋅1(1-ρl)∑u=0T-1b(u)⋅1max⁡{f0,f∞}<λ<1-ρl∑u=0T-1b(u)⋅1min⁡{f0,f∞},
then (1.3) has a positive T-periodic solution.



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*Source: 419536-2011-01-12.xml*