# Existence of Positive Periodic Solutions for a Class of Higher-Dimension Functional Differential Equations with Impulses

Authors: Zhang Suping; Jiang Wei
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

By employing the Krasnoselskii fixed point theorem, we establish some criteria for the existence of positive periodic solutions of a class ofn-dimension periodic functional differential equations with impulses, which improve the results of the literature.


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## Body

## 1. Introduction

Some evolution processes are distinguished by the circumstance that the evolutions change very rapidly at certain instants. In mathematical simulations, impulsive delay differential equations may express several simulation processes in real world which depend on their prehistory and are subject to short time disturbances. Such processes occur in the theory of optional control, population dynamics, biotechnologies, economics, and so forth. In recent years, the existence theory of positive periodic solutions of delay differential equations with impulsive effects or without impulsive effects has been an object of active research; we refer the reader to [1–4]. For other related works on studying for impulsive delay differential equations, we refer the reader to [5–7].In [8], Zeng et al. studied the following functional differential equations without impulses:
(1)x˙(t)=A(t,x(t))x(t)+λf(t,xt),
and obtained sufficient conditions for the existence of positive periodic solutions of (1).Zhang et al. [9] investigated the following form:
(2)x˙(t)=A(t)x(t)+f(t,xt),t≠τk,k∈Z+,Δx|t=τk=Ik(x(τk)).In this paper, we will consider then-dimension differential equation with impulses as follows:
(3)x˙(t)=A(t,x(t))x(t)+λf(t,xt),t≠τk,k∈Z+,Δx|t=τk=Ik(x(τk)),
where λ>0 is a parameter, A(t,x(t))=diag[a1(t,x(t)),a2(t,x(t)),…,an(t,x(t))], ai∈C(R×R,R) is ω-periodic, and f(t,xt) is an operator defined on R×BC(R,Rn) (here BC(R,Rn) denotes the Banach space of bounded continuous operator ϕ:R→Rn with the norm ∥ϕ∥=∑i=1supθ∈R|ϕi(θ)|, where ϕ=(ϕ1,ϕ2,…,ϕn)T). For x∈BC and t∈R, xt∈BC is defined by xt(θ)=x(t+θ) for θ∈R (see [10], Zheng). Consider that f(t+ω,xt)=f(t,xt) and Δx|t=τk=x(τk+)-x(τk) (here x(τk+) represents the right limit of x at the point τk), Ik=(Ik1,Ik2,…,Ikn)∈C(R+n,R-n), that is, x changes decreasingly suddenly at τk, ω>0 is a constant, R+ and R- are the sets of all nonnegative and nonpositive real numbers, respectively. We assume that there exists an integer p>0 such that τk+p=τk+ω,Ik+p=Ik, where 0<τ1<τ2<⋯<τp<ω.


## 2. Some Preliminaries

P
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n, ϕ is continuous everywhere except at a finite number of points τk at which ϕ(τk+) and ϕ(τk-) exist and ϕ(τk-)=ϕ(τk)}, J⊂R. For each x=(x1,x2,…,xn)T∈Rn, the norm of x is defined as |x|=∑i=1|xi|.Throughout the paper, we make the following assumptions:(H1)f(t,φt)≤0 for all (t,φ)∈R×BC(R,R+n);(H2)fi(t,φt) is a continuous function of t for each φ∈BC(R,R+n), i=1,2,…,n;(H3) for any L>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
(4)|fi(t,ϕt)-fi(t,ψt)|<ϵ,∀t∈[0,ω],i=1,2,…,n.To conclude this section, we summarize in the following a few concepts and results that will be needed in our arguments.Definition 1.
LetX be a Banach space, and let P be a closed, nonempty subset of X; P is a cone if(i)
αx+βx∈P for all x,y∈P and all α,β≥0;(ii)
x,-x∈P imply x=0.
LetX={x=(x1(t),x2(t),…,xn(t))T∈PC(R,Rn)∣x(t+ω)=x(t)} with the norm ∥x∥=∑i=1|xi|0, where |xi|0=supt∈[0,ω]|xi(t)|; then X is a Banach space.
Ifx(t)=(x1(t),x2(t),…,xn(t))T∈X is a solution of (3), then
(5)xi(t)=λ∫tt+ωGi(t,u)fi(u,xu)du+∑j=1j=pGi(t,τmj+nω)Iji(x(τmj)),
where
(6)Gi(t,u)=exp(-∫tu‍ai(s,x(s))ds)exp(-∫0ω‍ai(s,x(s))ds)-1,i=1,2,…,n.
See [9], Zhang et al.
It is clear thatGi(t+ω,u+ω)=Gi(t,u), for all (t,u)∈R2, and by (H1),
(7)Gi(t,u)fi(u,φu)≥0
for (t,u)∈R2 and (u,φ)∈R×BC(R,R+n).
Define fori=1,2,…,n,
(8)Ai:=min0≤t≤u≤ω|Gi(t,u)|=exp(-∫0ω‍ai(s,x(s))ds)1-exp(-∫0ω‍ai(s,x(s))ds),Bi:=max0≤t≤u≤ω|Gi(t,u)|=11-exp(-∫0ω‍ai(s,x(s))ds),A:=min1≤i≤nAi,B:=max1≤i≤nBi.
Let
(9)K={x(t)=(x1(t),x2(t),…,xn(t))T∈X:xi(t)≥σ|xi|0,00000i=1,2,…,n(x1(t),x2(t),…,xn(t))T},
where σ=A/B∈(0,1). It is not difficult to verify that K is a cone in X. We define an operator Φ:X→X as follows:
(10)(Φx)(t)=((Φ1x)(t),(Φ2x)(t),…,(Φnx)(t))T,
where
(11)(Φix)(t)=λ∫tt+ωGi(t,u)fi(u,xu)du+∑j=1j=pGi(t,τmj+nω)Iji(x(τmj)).
Then, it can be immediately obtained from the assumptions (H2) and (H3) that the operator Φ is completely continuous. On the other hand, it is not difficult to check that x*(t)=(x1*(t),x2*(t),…,xn*(t))T is a positive ω-periodic solution of (3) if and only if x*(t) is a fixed point of the operator Φ.Before stating the main results, we shall give some important lemmas.Lemma 2.
The mappingΦ maps K into K, that is, ΦK⊂K.Proof.
For anyx∈K, it is easy to see that Φx∈X. From (11), we have
(12)|(Φix)|0≤λBi∫0ω|fi(u,xu)|du+Bi∑j=1j=p|Iji(x(τmj))|.
Noting that Gi(t,u)fi(u,xu)≥0, we can also obtain
(13)(Φix)(t)≥λAi∫0ω|fi(u,xu)|du+Ai∑j=1j=p|Iji(x(τmj))|≥AiBi|(Φix)|0≥σ|(Φix)|0.
Hence, ΦK⊂K. The proof is complete.Lemma 3.
LetX be a Banach space, and let K be a cone in X. Suppose that Ω1 and Ω2 are open subsets of X such that 0∈Ω1⊂Ω1¯⊂Ω2. Suppose that
(14)Φ:K∩(Ω2¯∖Ω1)⟶K
is a completely continuous operator and satisfies either (1)
(15)∥Φu∥≤∥u∥,∀u∈K∩∂Ω1;∥Φu∥≥∥u∥,∀u∈K∩∂Ω2;or(2)
(16)∥Φu∥≤∥u∥,∀u∈K∩∂Ω2;∥Φu∥≥∥u∥,∀u∈K∩∂Ω1.Then, Φ has a fixed point in K∩(Ω2¯∖Ω1).
The proof of Lemma3 can be found in [11], Guo et al.Lemma 4.
Assume that(H1)–(H3) hold and there exists η>0, such that
(17)∫0ω|f(s,ϕs)|ds≥η∥ϕ∥,forϕ∈K.
Then,
(18)∥Φx∥≥λAη∥x∥.Proof.
Ifx∈K, then
(19)(Φix)(t)≥λAi∫tt+ω‍|fi(u,xu)|du+Ai∑j=1j=p‍|Iji(x(τmj))|≥λAi∫tt+ω‍|fi(u,xu)|du.
Thus, we have
(20)∥Φx∥=supt∈R∑i=1|(Φix)(t)|≥∑i=1‍λAi∫0ω|fi(u,xu)|du≥λA∫0ω|f(u,xu)|du≥λAη∥x∥.Lemma 5.
Assume that(H1)–(H3) hold and let r>0, if there exists a sufficiently small ϵ>0 such
(21)∫0ω|f(s,ϕs)|ds≤ϵr,∑j=1j=p|Ij(ϕ)|≤ϵr,0000000000000000000forϕ∈K∩∂Ωr.
Then,
(22)∥Φx∥≤(λ+1)Bϵ∥x∥,forx∈K∩∂Ωr.Proof.
For anyx∈K∩∂Ωr,
(23)∥Φx∥=supt∈R∑i=1|(Φix)(t)|≤∑i=1λBi∫0ω|fi(s,xs)|ds+∑i=1Bi∑j=1j=p|Iji(x(τmj))|≤λB∫0ω|f(s,xs)|ds+B∑j=1j=p|Ij(x(τmj))|≤(λ+1)Bϵ∥x∥.


## 3. Main Results

For the sake of convenience, we introduce the following notations:(24)fα=limx∈Ksup∥x∥→α∫0ω‍|f(s,xs)|ds∥x∥,fα=limx∈Kinf∥x∥→α∫0ω‍|f(s,xs)|ds∥x∥,Iα=limx∈Ksup∥x∥→α∑j=1j=p|Ij(x)|∥x∥,Iα=limx∈Kinf∥x∥→α∑j=1j=p|Ij(x)|∥x∥,
where α denotes either 0 or ∞.Theorem 6.
Assume that(H1)–(H3) hold and
(25)(P1)f∞=∞,(P2)f0=I0=0;
then (3) has at least one positive ω-periodic solution.Proof.
By(P2), for any ϵ1,ϵ1>0 there exists r2>0, such that
(26)∫0ω|f(s,ϕs)|ds≤ϵ1∥ϕ∥≤ϵ1r2,∑j=1j=p|Ij(ϕ)|≤ϵ2∥ϕ∥≤ϵ2r2.
Choose ϵ=max{ϵ1,ϵ1}, satisfying 0<ϵ<(1/(λ+1)B), by Lemma 5, we have
(27)∥Φx∥≤(λ+1)Bϵ∥x∥≤∥x∥,0000000000forx∈K∩∂Ω2.
Next, by (P2), there exists r3>r2>0, such that
(28)∫0ω‍|f(s,ϕs)|ds≥η∥ϕ∥,000forϕ∈K,∥ϕ∥≥r3,
where η>0 is chosen, so that λAη>1. It follows from Lemma 4 that
(29)∥Φx∥≥λAη∥x∥>∥x∥,000000forx∈K∩∂Ω3.
It follows from Lemma 3 that (3) has a positive ω-periodic solution satisfying r2≤∥x∥≤r3.Theorem 7.
Assume that(H1)–(H3) hold and
(30)(P3)f0=∞,(P4)f∞=I∞=0;
then (3) has at least one positive ω-periodic solution.Proof.
Sincef0=∞, one can find an r0>0, such that
(31).∫0ω|f(s,ϕs)|ds≥η∥ϕ∥,forϕ∈K,0<∥ϕ∥≤r0,
where η>0 is chosen so that λAη>1. It follows from Lemma 4 that
(32)∥Φx∥≥λAη∥x∥>∥x∥,00000forx∈K∩∂Ωr0.
By (P4), we know that there exists N1>r0 and ϵ1,ϵ2>0 such that
(33)∫0ω|f(s,ϕs)|ds≤ϵ1∥ϕ∥,∑j=1j=p|Ij(ϕ)|≤ϵ2∥ϕ∥,0000000000000000000000forϕ∈K,∥ϕ∥≥N1.
Choose ϵ=max{ϵ1,ϵ1}, satisfying 0<ϵ<1/(2(λ+1)B); then
(34)∫0ω|f(s,ϕs)|ds≤ϵ∥ϕ∥,∑j=1j=p|Ij(ϕ)|≤ϵ∥ϕ∥.
Take
(35)r1>N1+1m+2Bsupϕ∈K,∥ϕ∥<N1[λ∫0ω‍|f(s,ϕs)|ds+∑j=1j=p‍|Ij(ϕ)|],(36)∥Φx∥≤B[λ∫0ω‍|f(s,ϕs)|ds+∑j=1j=p‍|Ij(ϕ)|]=B[ρ(I1)+ρ(I2)]≤r12+∥x∥2=∥x∥,for  anyx∈K∩∂Ωr1,
where ρ(I1)=[λ∫0ω‍|f(s,ϕs)|ds+∑j=1j=p|Ij(ϕ)|]|x∈Ii, i=1,2, and I1={x∈K,∥x∥<N1}, I2={x∈K,∥x∥≥N1}.
This implies that∥Φx∥≤∥x∥, for any x∈K∩∂Ωr1.
Therefore, (3) has at least one positive ω-periodic solution.Theorem 8.
Suppose that(P5) there exists d2>0, such that ∫0ω‍|f(s,ϕs)|ds<d1/λA, for σd1≤∥ϕ∥≤d1,(P6) there  exists d2>0, such that∫0ω‍|f(s,ϕs)|ds<d2/2λB, ∑j=1j=p|Ij(ϕ)|≤d2/2B, for ∥ϕ∥<d2hold; then (3) has at least one positive ω-periodic solution.Proof.
Without loss of generality, we may assume thatd2<d1. If x∈K∩∂Ωd2, then by (P6), one can get
(37)∥Φx∥≤λBd22λB+Bd22B=d2=∥x∥,
in particular, ∥Φx∥<∥x∥, for allx∈K∩∂Ωd2.
On the other hand, by(P5), one has
(38)∥Φx∥≥λA∫0ω|f(s,xs)|ds>λAd1λA=d1=∥x∥,00000000000000000000000.forx∈K∩∂Ωd1.
Therefore, (3) has at least one positive ω-periodic solution.Theorem 9.
If(P7)f0=α1∈[0,1/2λB), I0=α2∈[0,1/2B);(P8)f∞=β1∈(1/λAσ,∞)hold, then (3) has at least one positive ω-periodic solution.Proof.
By assumption(P7), for ϵ=min{(1/2λB)-α1,1/2B-α2}>0, there exists a sufficiently small d2>0 such that
(39)∫0ω‍|f(s,xs)|ds∥x∥<α1+ϵ<12λB,∑j=1j=p|Ij(x)|∥x∥<α2+ϵ<12B,for∥x∥≤d2;
that is
(40)∫0ω‍|f(s,xs)|ds<d22λB,∑j=1j=p‍|Ij(x)|<d22B,for∥x∥≤d2.
So, (P6) is satisfied.
By assumption(P8), for ϵ=β1-1/λAσ, there exists a sufficiently large d1>0 such that
(41)∫0ω‍|f(s,xs)|ds∥x∥>β1-ϵ=1λAσ,forσd1≤∥x∥≤d1,
that is
(42)∫0ω|f(s,xs)|ds>1λAσ∥x∥≥1λAσσd1=d1λA,
therefore, (P5) holds. By Theorem 8, we complete the proof.Theorem 10.
If(P9)f0=α3∈(1/λAσ,∞);(P10)f∞=β2∈[0,1/2λB), I∞=β3∈[0,1/2B)hold, then (3) has at least one positive ω-periodic solution.Proof.
By(P9), for ϵ=α3-(1/λAσ)>0, there exists a sufficiently small d1>0, such that
(43)∫0ω‍|f(s,xs)|ds∥x∥>α3-ϵ=1λAσ,for0<∥x∥≤d1,
that is
(44)∫0ω|f(s,xs)|ds>σd1λA=d1λA,forσd1≤∥x∥≤d1.
Again, By (P10), for ϵ=min{(1/2λB)-β2,(1/2B)-β3}>0, there exists a sufficiently small d>0 such that
(45)∫0ω|f(s,xs)|ds∥x∥<β2+ϵ,∑j=1j=p|Ij(x)|∥x∥<β3+ϵ,
that is
(46)∫0ω|f(s,xs)|ds∥x∥<12λB,∑j=1j=p|Ij(x)|∥x∥<12B,00000000000000000000000for∥x∥>d.
In the following, we consider two cases to prove (P6) to be satisfied:(i)
∫0ω|f(s,xs)|ds+∑j=1j=p|Ij(x)|<∞;(ii)
∫0ω|f(s,xs)|ds=∞, ∑j=1j=p|Ij(x)|=∞.The bounded case is clear. If case (ii) is valid, then there exists y∈BC(R,R+n), ∥y∥=d2>d such that
(47)∫0ω|f(s,xs)|ds≤∫0ω|f(s,ys)|ds,0000∑j=1j=p|Ij(x)|<∑j=1j=p|Ij(y)|,00.for0<∥x∥≤∥y∥=d2.
Since ∥y∥=d2>d, then we have
(48)∫0ω|f(s,xs)|ds<∥y∥2λB=d22λB,00.∑j=1j=p|Ij(x)|<∥y∥2B=d22B,000000.for0<∥x∥≤d2,
which implies that condition (P6) holds. By Theorem 8, we complete the proof.Corollary 11.
If one of the following pairs(P1) and (P2); (P3) and (P4); (P5) and (P6); (P7) and (P8); (P9) and (P10);(P1) and (P7); (P2) and (P8); (P3) and (P10); (P4) and (P9)is valid, then system (3) has at least one positive ω-periodic solution.



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*Source: 396509-2013-11-11.xml*