# Multiple Positive Periodic Solutions for Functional Differential Equations with Impulses and a Parameter

Authors: Zhenguo Luo
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

We apply the Krasnoselskii fixed-point theorem to investigate the existence of multiple positive periodic solutions for a class of impulsive functional differential equations with a parameter; some verifiable sufficient results are established easily. In particular, our results extend and improve some previous results.


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## Body

## 1. Introduction

It is well known that impulsive differential equations arise naturally from a wide variety of applications such as aircraft control, the inspection processes in operations research, drug administration, and threshold theory in biology. Therefore, the impulsive differential equations represent a more natural framework for the mathematical model of many real world phenomena than differential equations (see [1–7]). In recent years, many researchers have obtained some properties of impulsive differential equations, such as oscillation, asymptotic behavior, stability and existence of solutions (see [8–16]). However, there are a little work discussing the existence of multiple positive periodic solutions for the high-dimensional functional differential equations with impulse and parameters. Motivated by this, in this paper, we mainly consider the following impulsive functional differential equations with a parameter:
(1)x′(t)=A(t,x(t))x(t)+λB(t,x(t))f(t,xt),hhhht∈R,t≠tk,Δx(tk)=Ik(tk,x(tk)),k∈Z+,
where λ>0 is a parameter, A(t,x(t))=diag⁡[a1(t,x(t)),a2(t,x(t)),…,an(t,x(t))], B(t,x(t))=diag⁡[b1(t,x(t)),b2(t,x(t)),…,bn(t,x(t))], aj,bj∈C(R×R+,R+)(j=1,…,n) are ω-periodic. f=(f1,…,fn)T, f(t,xt) is an operator on R×BC(R,Rn) (here BC(R,Rn) denoting the Banach space of bounded continuous operator φ:R→Rn with the norm ∥φ∥=∑i+1nsup⁡θ∈R|φ(θ)|, where φ=(φ1,…,φn)T)); fi(t+ω,xt)=fi(t,xt). If x∈BC(R,Rn), then xt∈BC(R,Rn) for any t∈R, where xt is defined by xt(θ)=x(t+θ) for θ∈R and Δx(tk)=x(tk+)-x(tk) (here x(tk+) representing the right limit of x(t) at the point tk). Consider that Ik=(Ik1,Ik2,…,Ikn)∈C(R+n,R-n); that is, x changes decreasingly suddenly at times tk. ω>0 is a constant, Z+={1,2,3,…}, R=(-∞,+∞), R+=[0,+∞), and R-=(-∞,0]. We assume that there exists an integer q>0 such that tk+q=tk+ω, Ik+q=Ik, where 0<t1<t2<⋯<tq<ω.It is well known that the functional differential system (1) includes many mathematical ecological models for example:in the case (a),B(t,x(t))≡1, Ik(tk,x(tk))≡0 in [17], Zeng et al. studied the existence of multiple positive periodic solutions of (1) by applying the Krasnoselskii fixed-point theorem.in the case (b),A(t,x(t))=A(t), λB(t,x(t))≡1; in [18], Zhang et al. established the existence of positive periodic solutions of (1) by using the fixed-point theorem in cones.in the case (c),A(t,x(t))=A(t), B(t,x(t))≡1, and Ik(tk,x(tk))≡0; in [19], Jiang et al. investigated the existence, multiplicity, and nonexistence of positive periodic solutions of (1).In this paper, we will study the existence of positive periodic solutions in more cases than the previously mentioned papers and obtain some easily verifiable sufficient criteria.Throughout the paper, we make the following assumptions:(H1)
ai,bi:R×R+→R+ satisfy Caratheodory conditions; that is, ai(t,x), bi(t,x) are locally Lebesgue measurable in t for each fixed x, are continuous in x for each fixed t, and are ω-periodic functions in t. Moreover, there exist ω-periodic functions a1i, a2i, b1i, b2i:R→R+ which are locally bounded Lebesgue measurable such that a1i(t)≤ai(t,x(t))≤a2i(t), b1i(t)≤b1i(t,x(t))≤b2i(t) and ∫0ω‍a1i(t)dt>0, ∫0ω‍b1i(t)dt>0;(H2)
f(t,φt)≤0 for all (t,φ)∈R×BC(R,R+n), and fi(t,φt) is a continuous function of t for each φ∈BC(R,R+n), i=1,2,…,n;(H3)
for anyL>0 and ϵ>0, there exists δ>0 such that for ϕ,ψ∈BC(R,R+n), |ϕ|≤L, |ψ|≤L, and |ϕ-ψ|<δ imply that |fi(s,ϕs)-fi(s,ψs)|<ϵ, s∈[0,ω](i=1,2,…,n);(H4)
{tk}, k∈Z+ satisfies 0<t1<t2<⋯<tk<⋯ and lim⁡k→+∞tk=+∞; Ik:R×R+→R, k∈Z+, satisfy Caratheodory conditions and are ω-periodic functions in t and, moreover, Ik(t,0)=0 for all k∈Z+. There exists a positive constant q such that tk+q=tk+ω, Ik+q(tk+q,x(tk+q))=Ik(tk,x(tk)), k∈Z+. Without loss of generality, we can assume that tk≠0 and [0,ω]∩{tk,k∈Z+}={t1,t2,…,tq}.In addition, the parameters in this paper are assumed to be not identically equal to zero.Furthermore, we will use the following notation. LetJ⊂R denote by PC(J,Rn) the set of operators φ:J→Rn which are continuous for t∈J, t≠tk and have discontinuities of the first kind at the points tk∈J(k∈Z+) but are continuous from the left at these points. For each x=(x1,x2,…,xn)T∈Rn, the norm of x is defined as |x|=∑i=1n|xi|. The matrix A>B(A≤B) means that each pair of corresponding elements of A and B satisfies the inequality “>” (“≤”). In particular, A is called a positive matrix if A>0.The paper is organized as follows. In Section2, we give some definitions and lemmas to prove the main results of this paper. In Section 3, existence theorems for one or two positive periodic solutions of (1) are established by using the Krasnoselskii fixed-point theorem under some conditions.


## 2. Preliminaries

In this section, we make some preparations for the following sections. For(t,s)∈R2, 1≤i≤n, we define
(2)Gi(t,s)=e-∫ts‍ai(ξ,x(ξ))dξe-∫0ω‍ai(ξ,x(ξ))dξ-1,G(t,s)=diag⁡[G1(t,s),G2(t,s),…,Gn(t,s)].
It is clear that Gi(t+ω,s+ω)=Gi(t,s), ∂Gi(t,s)/∂t=ai(t,x(t))Gi(t,s), Gi(t,t+ω)-Gi(t,t)=1. For all (t,s)∈R2 and by (H2), we have
(3)Gi(t,s)fi(s,φs)≥0,forany(t,s)∈R2,a(s,φs)∈R×BC(R,R+n).
In view of (H1), we also define for 1≤i≤n the following:
(4)αi:=min⁡0≤t≤s≤ω|Gi(t,s)|=e-∫0ω‍a2i(ξ,x(ξ))dξ1-e-∫0ω‍a2i(ξ,x(ξ))dξ,βi:=max⁡0≤t≤s≤ω|Gi(t,s)|=e-∫0ω‍a1i(ξ,x(ξ))dξ1-e-∫0ω‍a1i(ξ,x(ξ))dξ,α=min⁡1≤i≤nαi,β=max⁡1≤i≤nβi,σ=αβ∈(0,1),Bi(t)=max⁡{|b1i(t)|,|b2i(t)|},Bi′(t)=min⁡{|b1i(t)|,|b2i(t)|},B(t)=max⁡1≤i≤n{Bi(t)},B′(t)=min⁡1≤i≤n{Bi′(t)}.
Let X={x=(x1(t),x2(t),…,xn(t))T∈PC(R,Rn)∣x(t+ω)=x(t)} with the norm ∥x∥=∑i=1n|xi|0, |xi|0=sup⁡t∈[0,ω]|xi(t)|. It is easy to verify that (X,∥·∥) is a Banach space. Define E as a cone in X by
(5)E={x=(x1(t),x2(t),…,xn(t))T∈X:xi(t)hhh≥σ∥xi∥0,t∈[0,ω](x1(t),x2(t),…,xn(t))T}.
We easily verify that E is a cone in X. We define an operator T:X→X as follows:
(6)(Tx)(t)=((T1x)(t),(T2x)(t),…,(Tnx)(t))T,
where
(7)(Tix)(t)=λ∫tt+ω‍Gi(t,s)bi(s,x(s))fi(s,xs)ds+∑t≤tk<t+ω‍Gi(t,tk)Iki(tk,x(tk)).
The proofs of the main results in this paper are based on an application of the Krasnoselskii fixed-point theorem in cones. To make use of the fixed-point theorem in cones, firstly, we need to introduce some definitions and lemmas.Definition 1 (see [20]).
A functionx:R→(0,+∞) is said to be a positive solution of (1), if the following conditions are satisfied:(a)
x(t) is absolutely continuous on each (tk,tk+1);(b)
for eachk∈Z+,x(tk+) and x(tk-) exist, and x(tk-)=x(tk);(c)
x(t) satisfies the first equation of (1) for almost everywhere in R and x(tk) satisfies the second equation of (1) at impulsive point tk,k∈Z+.Definition 2 (see [21]).
LetX be a real Banach space; E is a cone of X. The semiorder induced by the cone E is denoted by “≤”; that is, x≤y if and only if y-x∈P for any x,y∈E.
Secondly, let us introduce the Krasnoselskii point theorem in cones which will be used in this paper.Lemma 3 (for the Krasnoselskii fixed-point theorem; see [22–24]).
LetE be a cone in a real Banach space X. Assume that Ω1 and Ω2 are open subsets of X with 0∈Ω1⊂Ω1¯⊂Ω2, where Ωi={x∈X:∥x∥<ri}(i=1,2). Let T:E∩(Ω2∖Ω1¯)→E be a completely continuous operator and satisfy either(1)
∥Tx∥≥∥x∥, for any x∈E∩∂Ω1 and ∥Tx∥≤∥x∥, for any x∈E∩∂Ω2, or(2)
∥Tx∥≤∥x∥, for any x∈E∩∂Ω1 and ∥Tx∥≥∥x∥, for any x∈E∩∂Ω2.Then T has a fixed point in E∩(Ω2∖Ω1¯).Lemma 4 (see [25]).
Assume thatf(t) and g(t) are continuous nonnegative functions defined on the interval [α,β]; then there exists ξ∈[α,β] such that
(8)∫αβ‍f(t)g(t)dt=f(ξ)∫αβ‍g(t)dt.Lemma 5.
Assume that(H1)–(H4) hold. The existence of positive ω-periodic solution of (1) is equivalent to that of nonzero fixed point of T in E.Proof.
Assume thatx=(x1(t),x2(t),…,xn(t))T∈X is a periodic solution of (1). Then, we have
(9)[xi(t)e-∫0t‍ai(s,x(s))ds]′=λe-∫0t‍ai(s,x(s))dsbi(t,x(t))fi(t,xt),haaaahhhhhhhhhht≠tk,i=1,2,…,n.
Integrating the above equation over [t,t+ω], we can have
(10)xi(u)e-∫0u‍ai(s,x(s))ds|ttm1+nω+xi(u)e-∫0u‍ai(s,x(s))ds|tm1+nωtm2+nω+⋯+xi(u)e-∫0u‍ai(s,x(s))ds|tmq+nωt+ω=λ∫tt+ω‍e-∫0u‍ai(s,x(s))dsbi(u,x(u))fi(u,xu)du,
where tmk+nω∈(t,t+ω), mk∈{1,2,…,q}, k=1,2,…,q, and n∈Z+. Therefore,
(11)xi(t)e-∫0t‍ai(s,x(s))ds[e-∫tt+ω‍ai(s,x(s))ds-1]-∑t≤tk<t+ω‍Δxi(tmk)e-∫0tmk+nω‍ai(s,x(s))ds=λ∫tt+ω‍e-∫0u‍ai(s,x(s))dsbi(u,x(u))fi(u,xu)du,
which can be transformed into
(12)xi(t)=λ∫tt+ω‍Gi(t,s)bi(s,x(s))fi(s,xs)ds+∑t≤tk<t+ω‍Gi(t,tk)Iki(tk,x(tk))=(Tix)(t).
Thus, xi is a periodic solution for (7).
Ifx=(x1(t),x2(t),…,xn(t))T∈E and Tx=(T1x,T2x,…,Tnx)T=x with x≠0, then, for any t=tk we can get the derivation of (7) about t,
(13)(Tix)′(t)=ddt[λ∫tt+ω‍Gi(t,s)bi(s,x(s))fi(s,xs)ds]=λ[(t,x(t))fiGi(t,t+ω)bi(t+ω,x(t+ω))hhih×fi(t+ω,xt+ω)-Gi(t,t)hhih×bi(t,x(t))fi(t,xt)]+ai(t,x(t))xi(t)=ai(t,x(t))xi(t)+λbi(t,x(t))fi(t,xt)=xi′(t).
For any t=tj, j∈Z+, we have from (7) that
(14)xi(tj+)-xi(tj)=λ∫tjtj+ω‍[Gi(tj+,s)-Gi(tj,s)]aaaaaaaaaaaaa×bi(s,x(s))fi(s,xs)ds+∑tj+≤tk<tj+ω‍Gi(tj+,tk)Iki(tk,x(tk))-∑tj≤tk<tj+ω‍Gi(tj,tk)Iki(tk,x(tk))=Iki(tk,x(tk)).
Hence x(t)=(x1(t),x2(t),…,xn(t))T is a positive ω-periodic solution of (1). Thus we complete the proof of Lemma 5.Lemma 6.
Assume that(H1)–(H4) hold. Then T:E→E is well defined.Proof.
From (7), it is easy to verify that (Tx)(t) is continuous in (tk,tk+1), (Tx)(tk+) and (Tx)(tk-) exist, and (Tx)(tk-)=(Tx)(tk) for each k∈Z+. Moreover, for any x∈E,
(15)(Tx)(t+ω)=λ∫t+ωt+2ω‍G(t+ω,s)b(s,x(s))f(s,xs)ds+∑t+ω≤tk<t+2ω‍G(t+ω,tk)Ik(tk,x(tk))=λ∫tt+ω‍G(t+ω,u+ω)hhhhh×b(u+ω,x(u+ω))hhhhh×f(u+ω,xu+ω)du+∑t≤tk<t+ω‍G(t,tk)Ik(tk,x(tk))=λ∫tt+ω‍G(t,s)b(s,x(s))f(s,xs)ds+∑t≤tk<t+ω‍G(t,tk)Ik(tk,x(tk))=(Tx)(t).
Therefore, (Tx)∈X. From (7), we have
(16)|Tix|0≤βi[∑t≤tk<t+ω‍λ∫0ω‍|bi(s,x(s))fi(s,xs)|ds+∑t≤tk<t+ω‍Iki(tk,x(tk))].
Noticing that Gi(t,s)fi(s,xs)≥0, we obtain
(17)(Tix)(t)≥αi[∑t≤tk<t+ω‍λ∫0ω‍|bi(s,x(s))fi(s,xs)|ds+∑t≤tk<t+ω‍Iki(tk,x(tk))]≥αiβi|Tix|0≥σ|Tix|0.
Therefore, Tx∈E. This completes the proof of Lemma 6.Lemma 7.
Assume that(H1)–(H4) hold. Then T:E→E is completely continuous.Proof.
We first show thatT is continuous. By (H3)-(H4), f and Ik are continuous in x; it follows that, for any ϵ>0, let δ>0 be small enough to satisfy that, if x,y∈E, with |x-y|<δ,
(18)|f(s,xs)-f(s,ys)|<ϵ2B¯λβω,s∈R;|Ik(tk,x(tk))-Ik(tk,y(tk))|<ϵ2βq,k∈Z+.
Therefore,
(19)∥(Tx)(t)-(Ty)(t)∥=∑i=1n|Tix-Tiy|0‍≤β∑i=1n‍λ∫tt+ω‍|bi(s,x(s))fi(s,xs)aaaaaaaaaaaaaaaa-bi(s,y(s))fi(s,ys)|ds+β∑i=1n‍∑t≤tk<t+ω‍|Iki(tk,x(tk))aaaaaaaaaaaaaaaaahh-Iki(tk,y(tk))|<βλB¯ωϵ2B¯λβω+βqϵ2βq=ϵ,
which implies that T is continuous on E.
Next we show thatT maps a bounded set into a bounded set. Indeed, let C⊂E be a bounded set. For any t∈R and x∈C, by (7), we have
(20)∥(Tx)(t)∥=∑i=1n‍|Tix|0≤β[λ∑i=1n‍∫tt+ω‍|bi(s,x(s))fi(s,xs)|dshhhh+∑i=1n‍∑t≤tk<t+ω‍|Iki(tk,x(tk))|]=β[∑t≤tk<t+ω‍∫0ω‍b2i(s)f(s,xs)dshhhh+∑t≤tk<t+ω‍|Ik(tk,x(tk))|].
Since C is bounded, in view of the continuity of T, it follows from (19) that Tx is bounded and {Tx:x∈C} is uniformly bounded. Finally, we show that the family of functions {Tx:x∈C} is equicontinuous on [0,ω]. Let θ1,θ2∈[0,ω] with θ1<θ2. From (7), for any x∈C, we have
(21)∥(Tx)(θ2)-(Tx)(θ1)∥≤λ∑i=1n‍[∫θ1θ2‍(Gi(θ2,s)-Gi(θ1,s))aaaaaaaaaaai×bi(s,x(s))fi(s,xs)dsaaaaaaaaaaai+∫θ2θ1+ω‍(Gi(θ2,s)-Gi(θ1,s))aaaaaaaaaaai×bi(s,x(s))fi(s,xs)dsaaaaaaaaaaai+∑θ1≤tk<θ2‍(Gi(θ2,tk)-Gi(θ1,tk))aaaaaaaaaaaaaaaaaaaaa×|Iki(tk,x(tk))|aaaaaaaaaaai+∑θ2≤tk<θ1+ω‍(Gi(θ2,tk)-Gi(θ1,tk))aaaaaaaaaaaaaaaaaaaaaaa×|Iki(tk,x(tk))|∫θ1θ2‍].
Since for x∈C, t∈[0,ω], 0≤k≤q, bi(t,x(t)), fi(t,x(t-τ1(t,x(t))),…,x(t-τm(t,x(t)))), and Iki(tk,x(tk)) are uniformly bounded in X; in view of (21), it is easy to see that when θ2-θ1 tends to zero, |(Tx)(θ2)-(Tx)(θ1)| tends uniformly to zero in X. Hence, {Tx:x∈C} is a family of uniformly bounded and equicontinuous functions on [0,ω]. By Ascoli-Arzelà theorem, the operator T is completely continuous. The proof of Lemma 7 is complete.For convenience in the following discussion, we introduce the following notations:(22)fa=lim⁡sup⁡x∈P,∥x∥→amax⁡t∈[0,ω]∫0ω‍|f(t,xt)|dt∥x∥,fa=lim⁡inf⁡x∈P,∥x∥→amin⁡t∈[0,ω]∫0ω‍|f(t,xt)|dt∥x∥,f¯r=max⁡0<x≤rmax⁡t∈[0,ω]∫0ω‍|f(t,xt)|dt∥x∥,f_r=min⁡0<x≤rmin⁡t∈[0,ω]∫0ω‍|f(t,xt)|dt∥x∥,Ia=lim⁡sup⁡x∈P,∥x∥→amax⁡t∈[0,ω]∑t≤tk<t+ω|Ik(t,x)|∥x∥,Ia=lim⁡inf⁡x∈P,∥x∥→amin⁡t∈[0,ω]∑t≤tk<t+ω|Ik(t,x)|∥x∥,I¯r=max⁡0<x≤rmax⁡t∈[0,ω],k∈[1,q]∑t≤tk<t+ω|Ik(t,x)|∥x∥,I_r=min⁡0<x≤rmin⁡t∈[0,ω],k∈[1,q]∑t≤tk<t+ω|Ik(t,x)|∥x∥,
where a denotes either 0 or ∞, r denotes a positive number, and ∥x∥=max⁡{|x1|,|x2|,…,|xm|}.


## 3. Main Results

Our main results of this paper are as follows.Theorem 8.
Assume that(H1)–(H4) and the following conditions:(H5)
ασ(λB′(ξ)f_r+I_r)>1, ξ∈[0,ω];(H6)
f0=I0=f∞=I∞=0hold. Then (1) has two positive ω-periodic solutions.Proof.
First, we defineΩr={x∈X:∥x∥<r}; then Ωr is an open subset of X. From (7), (H5), and Lemma 4, for any x∈E∩∂Ωr, we have
(23)∥(Tx)(t)∥=∑i=1n‍|Tix|0=∑i=1n‍[∑t≤tk<t+ω‍λ∫tt+ω‍|Gi(t,s)bi(s,x(s))fi(s,xs)|dsaaaaaa+∑t≤tk<t+ω‍|Gi(t,tk)Iki(tk,x(tk))|]≥λB′(ξ)∑i=1n‍∫tt+ω‍|Gi(t,s)fi(s,xs)|ds+∑i=1n‍∑t≤tk<t+ω‍|Gi(t,tk)Iki(tk,x(tk))|=α[∑t≤tk<t+ω‍λB′(ξ)∫0ω‍|f(s,xs)|dsaaaai+∑t≤tk<t+ω‍|Ik(tk,x(tk))|]≥α(λB′(ξ)f_r+I_r)∥x∥>∥x∥.
This yields
(24)∥(Tx)(t)∥>∥x∥,for anyx∈E∩∂Ωr.
On the other hand, if f0=I0=0 holds, then we can choose 0<r1<r, such that ∫0ω‍|f(t,xt)|dt≤ϵ∥x∥ and ∑t≤tk<t+ω|Ik(tk,x(tk))|≤ϵ∥x∥ for x∈[0,r1], t∈[0,ω], and 1≤k<q, where constant ϵ>0 satisfies ϵβ(λB(ξ)+1)≤1. By (7) and Lemma 4, we can obtain
(25)(Tx)(t)=∑i=1n‍(Tix)=∑i=1n‍[∑t≤tk<t+ω‍λ∫tt+ω‍Gi(t,s)bi(s,x(s))fi(s,xs)dshhhhh+∑t≤tk<t+ω‍Gi(t,tk)Iki(tk,x(tk))]≤λB(ξ)∑i=1n‍∫tt+ω‍|Gi(t,s)fi(s,xs)|ds+∑i=1n‍∑t≤tk<t+ω‍|Gi(t,tk)Iki(tk,x(tk))|=β[∑t≤tk<t+ω‍λB(ξ)∫0ω‍|f(s,xs)|dshhhh+∑t≤tk<t+ω‍|Ik(tk,x(tk))|]≤ϵβ(λB(ξ)+1)∥x∥≤∥x∥.
This yields
(26)∥(Tx)(t)∥≤∥x∥,for anyx∈E∩∂Ωr1.
In view of (24) and (26), by Lemma 3, it follows that T has a fixed point x1∈E∩(Ωr∖Ωr1¯) with r1<∥x1∥<r, which is a positive ω-periodic solution of (1).
Likewise, iff∞=I∞=0 holds, then there is N>0 such that ∫0ω‍|f(t,xt)|dt≤ϵ∥x∥ and ∑t≤tk<t+ω|Ik(tk,x(tk))|≤ϵ∥x∥ for x≥N, t∈[0,ω], and 1≤k<q, where constant ϵ>0 satisfies ϵβ(λB(ξ)+1)≤1. Let r2=max⁡{2r,N/σ} and it follows that x(t)≥σ∥x∥>N for x∈Ωr2, t∈[0,ω], and 0<k<q. Thus
(27)∫0ω‍|f(t,xt)|dt≤ϵ∥x∥;hh∑t≤tk<t+ω‍|Ik(tk,x(tk))|≤ϵ∥x∥,forx∈Ωr2,t∈[0,ω],1≤k<q.
By (7) and Lemma 4, we have
(28)(Tx)(t)≤ϵβ(λB(ξ)+1)∥x∥≤∥x∥;
this yields
(29)∥(Tx)(t)∥≤∥x∥,foranyx∈E∩∂Ωr2.
In view of (24) and (29), by Lemma 3, it follows that T has a fixed point x2∈E∩(Ωr2∖Ωr¯) with r<∥x2∥<r2, which is a positive ω-periodic solution of (1). Therefore (1) has at least two positive periodic solutions; that is, r1<∥x1∥<r<∥x2∥<r2. This proves Theorem 8.Remark 9.
Assume that(H1)–(H4) and the following conditions:(H5)
ασ(λB′(ξ)f_r+I_r)>1;(H7)
f0=I0=0,orf∞=I∞=0hold. Then (1) has a positive ω-periodic solution.Corollary 10.
Assume that(H1)–(H3) and λ>1/ασB′(ξ)f_r hold.(H6)
is satisfied; then (1) has two positive ω-periodic solutions;(H7)
is satisfied; then (1) has a positive ω-periodic solution.Theorem 11.
Assume that(H1)–(H4) and the following conditions:(H8)
β(λB(ξ)f¯r+I¯r)<1;(H9)
f0=I0=f∞=I∞=∞hold. Then (1) has two positive ω-periodic solutions.Proof.
We defineΩr={x∈X:∥x∥<r}, for a positive number r. Then Ωr is an open subset of X and 0∈Ωr. By (7), (H8), and Lemma 4, for any x∈E∩∂Ωr, we have
(30)∥(Tx)(t)∥=∑i=1n‍|Tix|0=∑i=1n‍[∑t≤tk<t+ω‍λ∫tt+ω‍|Gi(t,s)bi(s,x(s))fi(s,xs)|dshhhhh+∑t≤tk<t+ω‍|Gi(t,tk)Iki(tk,x(tk))|]≤λB(ξ)∑i=1n‍∫tt+ω‍|Gi(t,s)fi(s,xs)|ds+∑i=1n‍∑t≤tk<t+ω‍|Gi(t,tk)Iki(tk,x(tk))|=β[∑t≤tk<t+ω‍λB(ξ)∫0ω‍|f(s,xs)|dshhhi+∑t≤tk<t+ω‍|Ik(tk,x(tk))|]≤β(λB(ξ)f¯r+I¯r)∥x∥<∥x∥.
This implies that for any x∈E∩∂Ωr(31)∥(Tx)(t)∥<∥x∥.
On the one hand, since f0=I0=∞, there exists 0<r1<r and small enough 0<ϵ satisfies αδ[λB′(ξ)(f0-ϵ)+(I0-ϵ)]>1 such that, for any x with ∥x∥∈[0,r1],
(32)∫0ω‍|f(t,xt)|dt≥(f0-ε)∥x∥;∑t≤tk<t+ω‍|Ik(tk,x(tk))|≥(I0-ε)∥x∥.
Define Ωr2={x∈X:∥x∥<r2}; then Ωr1 is an open subset of X. For any x∈E∩∂Ωr1, by (7) and Lemma 4, we have
(33)∥(Tx)(t)∥=∑i=1n‍∥(Tix)∥0=∑i=1n‍[∑t≤tk<t+ω‍λ∫tt+ω‍|Gi(t,s)bi(s,x(s))fi(s,xs)|dshhhhh+∑t≤tk<t+ω‍|Gi(t,tk)Iki(tk,x(tk))|]≥λB′(ξ)∑i=1n‍∫tt+ω‍|Gi(t,s)fi(s,xs)|ds+∑i=1n‍∑t≤tk<t+ω‍|Gi(t,tk)Iki(tk,x(tk))|=α[∑t≤tk<t+ω‍λB′(ξ)∫0ω‍|f(s,xs)|dshhhh+∑t≤tk<t+ω‍|Ik(tk,x(tk))|]≥αδ[λB′(ξ)(f0-ϵ)+(I0-ϵ)]∥x∥≥∥x∥.
This yields
(34)∥(Tx)(t)∥≥∥x∥,foranyx∈E∩∂Ωr1.
In view of (31) and (34), by Lemma 3, it follows that T has a fixed point x1∈E∩(Ωr∖Ωr1¯) with r1<∥x1∥<r, which is a positive ω-periodic solution of (1). On the other hand, if f∞=I∞=∞, we can find small enough 0<ϵ that satisfies αδ[λB′(ξ)(f∞-ϵ)+(I∞-ϵ)]>1 and large enough η>r>0, such that ∥x∥≥η,
(35)∫0ω‍|f(t,xt)|dt≥(f∞-ϵ)∥x∥;∑t≤tk<t+ω‍|Ik(tk,x(tk))|≥(I∞-ϵ)∥x∥.
Define r2=η/σ>r and Ωr2={x∈X:∥x∥<r2}; then Ωr2 is an open subset of X. For any x∈E∩∂Ωr2, from (7) and Lemma 3, we have
(36)∥(Tx)(t)∥=∑i=1n‍∥(Tix)∥0=∑i=1n‍[∑t≤tk<t+ω‍λ∫tt+ω‍|Gi(t,s)bi(s,x(s))fi(s,xs)|dshhhhh+∑t≤tk<t+ω‍|Gi(t,tk)Iki(tk,x(tk))|]≥λB′(ξ)∑i=1n‍∫tt+ω‍|Gi(t,s)fi(s,xs)|ds+∑i=1n‍∑t≤tk<t+ω‍|Gi(t,tk)Iki(tk,x(tk))|=α[∑t≤tk<t+ω‍λB′(ξ)∫0ω‍|f(s,xs)|dshhhh+∑t≤tk<t+ω‍|Ik(tk,x(tk))|]≥αδ[λB′(ξ)(f∞-ϵ)+(I∞-ϵ)]∥x∥≥∥x∥.
This yields
(37)∥(Tx)(t)∥≥∥x∥,foranyx∈E∩∂Ωr2.
In view of (31) and (37), by Lemma 3, it follows that T has a fixed point x2∈E∩(Ωr2∖Ωr¯) with r<∥x2∥<r2, which is a positive ω-periodic solution of (1). Therefore, (1) has at least two positive periodic solutions; that is, r1<∥x1∥<r<∥x2∥<r2. This proves Theorem 11.Remark 12.
Assume that(H1)–(H4) and the following conditions:(H8)
β(λB(ξ)f¯r+I¯r)<1;(H10)
f0=I0=∞, or f∞=I∞=∞hold. Then (1) has a positive ω-periodic solution.Corollary 13.
Assume that(H1)–(H3) and λ<1/βB(ξ)f¯r hold.(H9)
is satisfied; then (1) has two positive ω-periodic solutions;(H10)
is satisfied; then (1) has a positive ω-periodic solution.Theorem 14.
Assume that(H1)–(H4) and(H11)
β(λB(ξ)f0+I0)<1;(H12)
ασ(λB′(ξ)f∞+I∞)>1hold. Then (1) has a positive ω-periodic solution, where f0, f∞, I0, and I∞ are positive constants.Proof.
From(H11), we can choose ϵ>0 such that β(λB(xi)(f0+ϵ)+(I0+ϵ))<1. Thus there exists r>0 such that, for x∈[0,r], t∈[0,ω] and 1≤k<q,
(38)∫0ω‍f(t,xt)dt≤(f0+ϵ)∥x∥;∑t≤tk<t+ω‍Ik(tk,x(tk))≤(I0+ϵ)∥x∥,
by (7), (H11), and Lemma 4, we have
(39)∥(Tx)(t)∥=∑i=1n‍|Tix|0=∑i=1n‍[∑t≤tk<t+ω‍λ∫tt+ω‍|Gi(t,s)bi(s,x(s))fi(s,xs)|hhhhh+∑t≤tk<t+ω‍|Gi(t,tk)Iki(tk,x(tk))|]≤λB(ξ)∑i=1n‍∫tt+ω‍|Gi(t,s)fi(s,xs)|ds+∑i=1n‍∑t≤tk<t+ω‍|Gi(t,tk)Iki(tk,x(tk))|=β[∑t≤tk<t+ω‍λB(ξ)∫0ω‍|f(s,xs)|dshhih+∑t≤tk<t+ω‍|Ik(tk,x(tk))|]≤β(λB(ξ)(f0+ϵ)+(I0+ϵ))∥x∥<∥x∥.
This implies that for any x∈E∩∂Ωr(40)∥(Tx)(t)∥<∥x∥.
On the other hand, choose ε>0 such that f∞-ϵ>0 and I∞-ϵ>0, and from (H12), we can obtain
(41)ασ[λB′(ξ)(f∞-ϵ)+(I∞-ϵ)]>1.
It is easy to see that there exists large enough η>r>0, such that ∥x∥≥η,
(42)∫0ω‍|f(t,xt)|dt≥(f∞-ϵ)∥x∥;∑t≤tk<t+ω‍|Ik(tk,x(tk))|≥(I∞-ϵ)∥x∥.
Define R=η/σ>r and ΩR={x∈X:∥x∥<R}; then ΩR is an open subset of X. From (7), (H12), and Lemma 4, for any x∈E∩∂ΩR, we have
(43)∥(Tx)(t)∥=∑i=1n‍|Tix|0=∑i=1n‍[∑t≤tk<t+ω‍λ∫tt+ω‍|Gi(t,s)bi(s,x(s))fi(s,xs)|dshhhhh+∑t≤tk<t+ω‍|Gi(t,tk)Iki(tk,x(tk))|]≥λB′(ξ)∑i=1n‍∫tt+ω‍|Gi(t,s)fi(s,xs)|ds+∑i=1n‍∑t≤tk<t+ω‍|Gi(t,tk)Iki(tk,x(tk))|=α[∑t≤tk<t+ω‍λB′(ξ)∫0ω‍|f(s,xs)|dsaaaaa+∑t≤tk<t+ω‍|Ik(tk,x(tk))|]≥ασ[λB′(ξ)(f∞-ϵ)+(I∞-ϵ)]∥x∥>∥x∥.
This yields
(44)∥Tx∥>∥x∥,for anyx∈E∩∂ΩR.
In view of (40) and (44), by Lemma 3, it follows that T has a fixed point x*∈E∩(ΩR∖Ωr¯) with r<∥x*∥<R, which is a positive ω-periodic solution of (1). This proves Theorem 14.Corollary 15.
Assume that(H1)–(H3) and the following condition:(H13)
1/ασB′(ξ)f∞<λ<1/βB(ξ)f0hold. Then (1) has a positive ω-periodic solution.Similarly, we can prove the following theorem and corollary.Theorem 16.
Assume that(H1)–(H4) and the following conditions:(H14)
β(λB(ξ)f∞+I∞)<1;(H15)
ασ(λB′(ξ)f0+I0)>1hold. Then (1) has a positive ω-periodic solution, where f0, f∞, I0, and I∞ are positive constants.Corollary 17.
Assume that(H1)–(H3) and the following condition:(H16)
1/ασB′(ξ)f0<λ<1/βB(ξ)f∞hold. Then (1) has a positive ω-periodic solution.Theorem 18.
Assume that(H1)–(H4),(H5), and the following condition:(H17)
0<f0<1/2λβB(ξ), 0<I0<1/2β and 0<f∞<1/2λβB(ξ), 0<I∞<1/2β,hold. Then (1) has two positive ω-periodic solutions.Proof.
First, we defineΩr={x∈X:∥x∥<r}; then Ωr is an open subset of X. From (7), (H5), and Lemma 4, for any x∈E∩∂Ωr, we have
(45)∥(Tx)(t)∥=∑i=1n‍|Tix|0=∑i=1n‍[∑t≤tk<t+ω‍λ∫tt+ω‍|Gi(t,s)bi(s,x(s))fi(s,xs)|dshhhhih+∑t≤tk<t+ω‍|Gi(t,tk)Iki(tk,x(tk))|]≥λB′(ξ)∑i=1n‍∫tt+ω‍|Gi(t,s)fi(s,xs)|ds+∑i=1n‍∑t≤tk<t+ω‍|Gi(t,tk)Iki(tk,x(tk))|=α[∑t≤tk<t+ω‍λB′(ξ)∫0ω‍|f(s,xs)|dsaaaaa+∑t≤tk<t+ω‍|Ik(tk,x(tk))|]≥α(λB′(ξ)f_r+I_r)∥x∥>∥x∥.
This yields
(46)∥(Tx)(t)∥>∥x∥,for anyx∈E∩∂Ωr.
On the one hand, since 0<f0<1/2λβB(ξ) and 0<I0<1/2β, there exists 0<r1<r such that for 0<∥x∥<r1(47)∫0ω‍|f(t,xt)|dt≤r12λβB(ξ);∑t≤tk<t+ω‍|Ik(tk,x(tk))|≤r12β.
Set Ωr1={x∈X:∥x∥<r1}; then Ωr1 is an open subset of X. From (7), (H17), and Lemma 4, for any x∈E∩∂Ωr1, t∈[0,ω], and 1≤k<q, we have
(48)(Tx)(t)=∑i=1n‍(Tix)=∑i=1n‍[∑t≤tk<t+ω‍λ∫tt+ω‍Gi(t,s)bi(s,x(s))fi(s,xs)dsaaaaaa+∑t≤tk<t+ω‍Gi(t,tk)Iki(tk,x(tk))]≤λB(ξ)∑i=1n‍∫tt+ω‍|Gi(t,s)fi(s,xs)|ds+∑i=1n‍∑t≤tk<t+ω‍|Gi(t,tk)Iki(tk,x(tk))|=β[∑t≤tk<t+ω‍λB(ξ)∫0ω‍|f(s,xs)|dshhhh+∑t≤tk<t+ω‍|Ik(tk,x(tk))|]≤β[λB(ξ)12λβB(ξ)+r12β]∥x∥=∥x∥.
This yields
(49)∥(Tx)(t)∥≤∥x∥,for anyx∈E∩∂Ωr1.
In view of (46) and (49), by Lemma 3, it follows that T has a fixed point x1∈E∩(Ωr∖Ωr1¯) with r1<∥x1∥<r, which is a positive ω-periodic solution of (1).
On the other hand, if0<f∞<1/2λβB(ξ), and 0<I∞<1/2β hold, then there is N>0 such that
(50)∫0ω‍|f(t,xt)|dt≤N2λβB(ξ);∑t≤tk<t+ω‍|Ik(tk,x(tk))|≤N2β,
for x≥N, t∈[0,ω], 1≤k<q. Let r2=max⁡{2r,N/σ} and it follows that x(t)≥σ∥x∥>N for x∈Ωr2, t∈[0,ω], and 0<k<q. Thus
(51)∫0ω‍|f(t,xt)|dt≤12λβB(ξ)∥x∥;hhi∑t≤tk<t+ω‍|Ik(tk,x(tk))|≤12β∥x∥,forx∈Ωr2,t∈[0,ω],1≤k<q.
By (7), (H17), and Lemma 4, we have
(52)(Tx)(t)=∑i=1n‍(Tix)=∑i=1n‍[∑t≤tk<t+ω‍λ∫tt+ω‍Gi(t,s)bi(s,x(s))fi(s,xs)dshhhihh+∑t≤tk<t+ω‍Gi(t,tk)Iki(tk,x(tk))]≤λB(ξ)∑i=1n‍∫tt+ω‍|Gi(t,s)fi(s,xs)|ds+∑i=1n‍∑t≤tk<t+ω‍|Gi(t,tk)Iki(tk,x(tk))|=β[∑t≤tk<t+ω‍λB(ξ)∫0ω‍|f(s,xs)|dsahhh+∑t≤tk<t+ω‍|Ik(tk,x(tk))|]≤β[λB(ξ)12λβB(ξ)+r12β]∥x∥=∥x∥.
This yields
(53)∥(Tx)(t)∥≤∥x∥,for anyx∈E∩∂Ωr2.
In view of (46) and (53), by Lemma 3, it follows that T has a fixed point x2∈E∩(Ωr2∖Ωr¯) with r<∥x2∥<r2, which is a positive ω-periodic solution of (1). Therefore (1) has at least two positive periodic solutions; that is, r1<∥x1∥<r<∥x2∥<r2. This proves Theorem 18.Remark 19.
Assume that(H1)–(H4),(H5), and the following condition:(H18)
0<f0<1/2λβB(ξ), 0<I0<1/2β or 0<f∞<1/2λβB(ξ), 0<I∞<1/2βhold. Then (1) has a positive ω-periodic solution.Corollary 20.
Assume that(H1)–(H3) and λ>1/ασB′(ξ)f_r hold.(H17)
is satisfied; then (1) has two positive ω-periodic solutions.(H18)
is satisfied; then (1) has a positive ω-periodic solution.From the arguments in the previous proof, we have the following consequences immediately.Theorem 21.
Assume that(H1)–(H4),(H8), and the following condition:(H19)
∞>f0>1/2λασB′(ξ), ∞>I0>1/2ασ and ∞>f∞>1/2λασB′(ξ), ∞>I∞>1/2ασhold. Then (1) has two positive ω-periodic solutions.Remark 22.
Assume that(H1)–(H4),(H8), and the following condition:(H20)
∞>f0>1/2λασB′(ξ), ∞>I0>1/2ασ or ∞>f∞>1/2λασB′(ξ), ∞>I∞>1/2ασhold. Then (1) has a positive ω-periodic solution.Corollary 23.
Assume that(H1)–(H3) and λ<1/βB(ξ)f¯r hold.(H19)
is satisfied; then (1) has two positive ω-periodic solutions.(H20)
is satisfied; then (1) has a positive ω-periodic solution.Theorem 24.
Assume that(H1)–(H4),(H5), and one of the following conditions(H21)
f0=I0=0 and 0<f∞<1/2λβB(ξ), 0<I∞<1/2β,(H22)
0<f0<1/2λβB(ξ), 0<I0<1/2β and f∞=I∞=0hold. Then (1) has two positive ω-periodic solutions.Proof.
We only consider the case(H21). When the case (H22) holds, the conclusion remains true by a similar proof and we will omit it. We define Ωr={x∈X:∥x∥<r}; then Ωr is an open subset of X. From (7), (H5), and Lemma 4, for any x∈E∩∂Ωr, we have
(54)∥(Tx)(t)∥=∑i=1n‍|Tix|0=∑i=1n‍[∑t≤tk<t+ω‍λ∫tt+ω‍|Gi(t,s)bi(s,x(s))fi(s,xs)|dshhhhh+∑t≤tk<t+ω‍|Gi(t,tk)Iki(tk,x(tk))|]≥λB′(ξ)∑i=1n‍∫tt+ω‍|Gi(t,s)fi(s,xs)|ds+∑i=1n‍∑t≤tk<t+ω‍|Gi(t,tk)Iki(tk,x(tk))|=α[∑t≤tk<t+ω‍λB′(ξ)∫0ω‍|f(s,xs)|dshhhh+∑t≤tk<t+ω‍|Ik(tk,x(tk))|]≥α(λB′(ξ)f_r+I_r)∥x∥>∥x∥.
This yields
(55)∥(Tx)(t)∥>∥x∥,for anyx∈E∩∂Ωr.
On the one hand, if f0=I0=0 holds, then we can choose 0<r1<r, such that ∫0ω‍|f(t,xt)|dt≤ϵ∥x∥ and ∑t≤tk<t+ω|Ik(tk,x(tk))|≤ϵ∥x∥ for x∈[0,r1], t∈[0,ω], and 1≤k<q, where constant ϵ>0 satisfies ϵβ(λB(ξ)+1)≤1. By (7), (H21), and Lemma 4, we can obtain
(56)(Tx)(t)=∑i=1n‍(Tix)=∑i=1n‍[∑t≤tk<t+ω‍λ∫tt+ω‍Gi(t,s)bi(s,x(s))fi(s,xs)dshihhhh+∑t≤tk<t+ω‍Gi(t,tk)Iki(tk,x(tk))]≤λB(ξ)∑i=1n‍∫tt+ω‍|Gi(t,s)fi(s,xs)|ds+∑i=1n‍∑t≤tk<t+ω‍|Gi(t,tk)Iki(tk,x(tk))|=β[∑t≤tk<t+ω‍λB(ξ)∫0ω‍|f(s,xs)|dshhhh+∑t≤tk<t+ω‍|Ik(tk,x(tk))|]≤ϵβ(λB(ξ)+1)∥x∥≤∥x∥.
This yields
(57)∥(Tx)(t)∥≤∥x∥,for anyx∈E∩∂Ωr1.
In view of (55) and (57), by Lemma 3, it follows that T has a fixed point x1∈E∩(Ωr∖Ωr1¯) with r1<∥x1∥<r, which is a positive ω-periodic solution of (1).
On the other hand, if0<f∞<1/2λβB(ξ) and 0<I∞<1/2β hold, then there is N>0 such that
(58)∫0ω‍|f(t,xt)|dt≤N2λβB(ξ);∑t≤tk<t+ω‍|Ik(tk,x(tk))|≤N2β,
for x≥N, t∈[0,ω], and 1≤k<q; let r2=max⁡{2r,N/σ} and it follows that x(t)≥σ∥x∥>N for x∈Ωr2, t∈[0,ω], and 0<k<q. Thus
(59)∫0ω‍|f(t,xt)|dt≤12λβB(ξ)∥x∥;aa∑t≤tk<t+ω‍|Ik(tk,x(tk))|≤12β∥x∥,forx∈Ωr2,t∈[0,ω],1≤k<q.
By (7), (H21), and Lemma 4, we have
(60)(Tx)(t)=∑i=1n‍(Tix)=∑i=1n‍[∑t≤tk<t+ω‍λ∫tt+ω‍Gi(t,s)bi(s,x(s))fi(s,xs)dshhhihh+∑t≤tk<t+ω‍Gi(t,tk)Iki(tk,x(tk))]≤λB(ξ)∑i=1n‍∫tt+ω‍|Gi(t,s)fi(s,xs)|ds+∑i=1n‍∑t≤tk<t+ω‍|Gi(t,tk)Iki(tk,x(tk))|=β[∑t≤tk<t+ω‍λB(ξ)∫0ω‍|f(s,xs)|dshhhh+∑t≤tk<t+ω‍|Ik(tk,x(tk))|]≤β[λB(ξ)12λβB(ξ)+r12β]∥x∥=∥x∥.
This yields
(61)∥(Tx)(t)∥≤∥x∥,for anyx∈E∩∂Ωr2.
In view of (55) and (61), by Lemma 3, it follows that T has a fixed point x2∈E∩(Ωr2∖Ωr¯) with r<∥x2∥<r2, which is a positive ω-periodic solution of (1). Therefore (1) has at least two positive periodic solutions; that is, r1<∥x1∥<r<∥x2∥<r2. This proves Theorem 24.Remark 25.
Assume that(H1)–(H4),(H5), and one of the following conditions:(H23)
f0=I0=0 or 0<f∞<1/2λβB(ξ), 0<I∞<1/2β,(H24)
0<f0<1/2λβB(ξ), 0<I0<1/2β or f∞=I∞=0hold. Then (1) has a positive ω-periodic solution.Corollary 26.
Assume that(H1)–(H3) and λ>1/ασB′(ξ)f_r hold.Either(H21) or (H22) is satisfied; then (1) has two positive ω-periodic solutions.Either(H23) or (H24) is satisfied; then (1) has a positive ω-periodic solution.From the arguments in the previous proof, we have the following consequences immediately.Theorem 27.
Assume that(H1)–(H4),(H8), and one of the following conditions:(H25)
f0=I0=0 and ∞>f∞>1/2λασB′(ξ), ∞>I∞>1/2ασ,(H26)
∞>f0>1/2λασB′(ξ), ∞>I0>1/2ασ and f∞=I∞=0hold. Then (1) has two positive ω-periodic solutions.Remark 28.
Assume that(H1)–(H4),(H8), and one of the following conditions(H27)
f0=I0=0 or ∞>f∞>1/2λασB′(ξ), ∞>I∞>1/2ασ,(H28)
∞>f0>1/2λασB′(ξ), ∞>I0>1/2ασ or f∞=I∞=0hold. Then (1) has a positive ω-periodic solution.Corollary 29.
Assume that(H1)–(H3) and λ<1/βB(ξ)f¯r hold.Either(H25) or (H26) is satisfied; then (1) has two positive ω-periodic solutions.Either(H27) or (H28) is satisfied; then (1) has a positive ω-periodic solution.Remark 30.
Suppose thatB(t,x(t))=1 and Ik(tk,x(tk))=0, under some conditions; we can obtain the corresponding results of [17]. Hence, our results generalize and improve the corresponding results of [17].Remark 31.
Assume thatA(t,x(t))=A(t), B(t,x(t))=1, λ=1 under some conditions; we can obtain the corresponding results of [18]. Hence, our results generalize and improve the corresponding results of [18].



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*Source: 812867-2014-03-26.xml*