# Positive Periodic Solutions of Second-Order Differential Equations with Delays

Authors: Yongxiang Li
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

The existence results of positiveω-periodic solutions
are obtained for the second-order differential equation with delays −u″+a(t)=f(t,u(t−τ1),...,u(t−τn)), where a∈C(ℝ,(0,∞)) is a ω-periodic function, f:ℝ×[0,∞)n→[0,∞) is a continuous function, which is ω-periodic in t, and τ1,τ2,...,τn are positive constants. Our discussion is based on the fixed point index theory in cones.


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## Body

## 1. Introduction and Main Results

In this paper, we discuss the existence of positiveω-periodic solutions of the second-order differential equation with delays
(1.1)-u′′(t)+a(t)u(t)=f(t,u(t-τ1),…,u(t-τn)),
where a∈C(ℝ,(0,∞)) is a ω-periodic function, f:ℝ×[0,∞)×ℝn→ℝ is a continuous function, which is ω-periodic in t, and τ1,τ2,…,τn are positive constants.In recent years, the existence of periodic solutions for second-order delay differential equations has been researched by many authors, see [1–8] and references therein. In some practice models, only positive periodic solutions are significant. In [4, 5, 7], the authors obtained the existence of positive periodic solutions for some delay second-order differential equations by using Krasnoselskii's fixed-point theorem of cone mapping. For the second-order differential equations without delay, the existence of positive periodic solutions has been discussed by more authors, see [9–14].Motivated by the papers mentioned above, we research the existence of positive periodic solutions of (1.1) with multiple delays. We aim to obtain the essential conditions on the existence of positive periodic solutions of (1.1) via the theory of the fixed-point index in cones. The conditions concern with the relation of the coefficient function a(t) and nonlinearity f(t,x1,…,xn). Let
(1.2)m=min⁡0≤t≤ωa(t),M=max⁡0≤t≤ωa(t).
Obviously, 0<m≤M. Our main results are as follows.Theorem 1.1.
Leta∈C(ℝ,(0,∞)) be a ω-periodic function, f∈C(ℝ×[0,∞)n,[0,∞)), and f(t,x1,…,xn)ω-periodic in t. If f satisfies the following conditions:(F1)
there exist positive constantsc1,…,cn satisfying c1+⋯+cn<m and δ>0 such that
(1.3)f(t,x1,…,xn)≤c1x1+⋯+cnxn,
for t∈ℝ and x1,…,xn∈[0,δ];(F2)
there exist positive constantsd1,…,dn satisfying d1+⋯+dn>M and H>0 such that
(1.4)f(t,x1,…,xn)≥d1x1+⋯+dnxn,
for t∈ℝ and x1,…,xn≥H,
then (1.1) has at least one positive ω-periodic solution.Theorem 1.2.
Leta∈C(ℝ,(0,∞)) be a ω-periodic function, f∈C(ℝ×[0,∞)n,[0,∞)), and f(t,x1,…,xn)ω-periodic in t. If f satisfies the following conditions:(F3)
there exist positive constantsd1,…,dn satisfying d1+⋯+dn>M and δ>0 such that
(1.5)f(t,x1,…,xn)≥d1x1+⋯+dnxn,
for t∈ℝ and x1,…,xn∈[0,δ];(F4)
there exist positive constantsc1,…,cn satisfying c1+⋯+cn<m and H>0 such that
(1.6)f(t,x1,…,xn)≤c1x1+⋯+cnxn,
for t∈ℝ and x1,…,xn≥H,
then (1.1) has at least one positive ω-periodic solution.In Theorem1.1, the conditions (F1) and (F2) allow f(t,x1,…,xn) to be superlinear growth on x1,…,xn. For example,
(1.7)f(t,x1,…,xn)=a1(t)x12+⋯+an(t)xn2
satisfies (F1) and (F2), where a1(t),…,an(t) are positive and continuous ω-periodic functions.In Theorem1.2, the conditions (F3) and (F4) allow f(t,x1,…,xn) to be sublinear growth on x1,…,xn. For example,
(1.8)f(t,x1,…,xn)=b1(t)|x1|+⋯+bn(t)|xn|
satisfies (F3) and (F4), where b1(t),…,bn(t) are positive and continuous ω-periodic functions.Our results are different from those in the references mentioned above. The conditions (F1) and (F2) in Theorem1.1 and the conditions (F3) and (F4) in Theorem 1.2 are optimal for the existence of positive periodic solutions of (1.1). This fact can been shown from the differential equation with linear delays
(1.9)-u′′(t)+a0u(t)=a1u(t-τ1)+⋯+anu(t-τn)+h(t),
where a0,a1,…an are positive constants and h∈C(ℝ) is a positive ω-periodic function. If a1,…an satisfy
(1.10)a1+a2+⋯+an=a0.
Equation (1.9) has no positive ω-periodic solutions. In fact, if (1.9) has a positive ω-periodic solution, integrating the equation on [0,ω] and using the periodicity of u(t), we can obtain that ∫0ω‍h(t)dt=0, which contradicts to the positivity of h(t). Hence, (1.9) has no positive ω-periodic solution. For a(t)≡a0 and f(t,x1,…,xn)=a1x1+⋯+anxn+h(t), if Condition (1.10) holds, the conditions (F1) and (F2) in Theorem 1.1 and the conditions (F3) and (F4) in Theorem 1.2 have just not been satisfied. From this, we see that the conditions in Theorems 1.1–1.2 are optimal.The proofs of Theorems1.1–1.2 are based on the fixed point index theory in cones, which will be given in Section 3. Some preliminaries to discuss (1.1) are presented in Section 2.


## 2. Preliminaries

LetCω(ℝ) denote the Banach space of all continuous ω-periodic function u(t) with norm ∥u∥C=max0≤t≤ω|u(t)|. Let Cω+(ℝ) be the cone of all nonnegative functions in Cω(ℝ). Generally, Cωm(ℝ) denotes the mth-order continuous differentiable ω-periodic function space for m∈ℕ.LetM be the positive constant defined by (1.2). For h∈Cω(ℝ), we consider the linear second-order differential equation
(2.1)-u′′(t)+Mu(t)=h(t),t∈R.
The ω-periodic solutions of (2.1) are can been expressed by the solution of the linear second-order boundary value problem
(2.2)-u′′(t)+Mu(t)=0,0≤t≤ω,u(0)-u(ω)=0,u˙(0)-u˙(ω)=-1,
see [11]. Problem (2.2) has a unique solution, which is explicitly given by
(2.3)Φ(t)=cosh⁡β(t-ω/2)2βsinh⁡(βω/2),0≤t≤ω,
where β=M.By a direct calculation, we easily prove the following lemma.Lemma 2.1.
LetM>0. For every h∈Cω(ℝ), the linear equation (2.1) has a unique ω-periodic solution u(t), which is given by
(2.4)u(t)=∫t-ωt‍Φ(t-s)h(s)ds:=Th(t),t∈R.
Moreover, T:Cω(ℝ)→Cω(ℝ) is a completely continuous linear operator.SinceΦ(t)>0 for every t∈[0,ω], if h∈Cω+(ℝ) and h(t)≢0, by (2.4) the ω-periodic solution of (2.1) u=Th(t) is positive. Moreover, we can show that the ω-periodic solution has the following strong positivity:
(2.5)Th(t)≥σ‖Th‖C,t∈R,h∈Cω+(R),
where σ=Φ_/Φ¯=1/cosh(βω/2), in which
(2.6)Φ_=min⁡0≤t≤ωΦ(t)=12βsinh⁡(βω/2),Φ¯=max⁡0≤t≤ωΦ(t)=cosh⁡(βω/2)2βsinh⁡(βω/2).
In fact, for h∈Cω+(ℝ) and t∈ℝ, from (2.4) it follows that
(2.7)Th(t)=∫t-ωtΦ(t-s)h(s)ds‍≤Φ¯∫t-ωth(s)ds‍=Φ¯∫0ωh(s)ds,‍
and therefore,
(2.8)‖Th‖C≤Φ-∫0ω‍h(s)ds.
Using (2.4) and this inequality, we have that
(2.9)Th(t)=∫t-ωt‍Φ(t-s)h(s)ds≥Φ_∫t-ωt‍h(s)ds=Φ_∫0ω‍h(s)ds=(Φ_/Φ-)⋅Φ-∫0ω‍h(s)ds≥σ‖Th‖C.
Hence, (2.5) holds.Now we consider the periodic solution problem of the linear differential equation with variable coefficient(2.10)-u′′(t)+a(t)u(t)=h(t),t∈R.Lemma 2.2.
Leta∈Cω(ℝ) be a positive ω-periodic function. For every h∈Cω(ℝ), the linear equation (2.10) has a unique ω-periodic solution u∶=Sh. Moreover, S:Cω(ℝ)→Cω(ℝ) is a completely continuous linear operator and with strong positivity
(2.11)Sh(t)≥mσM‖Sh‖C,t∈R,h∈Cω+(R).Proof.
LetM and m be the positive constants defined by (1.2). Then 0<m≤a(t)≤M, t∈ℝ. Let T:Cω(ℝ)→Cω(ℝ) be the ω-periodic solution operator of (2.1) given by (2.4). We rewrite (2.10) to the form of
(2.12)-u′′(t)+Mu(t)=(M-a(t))u(t)+h(t),t∈R.
Then it is easy to see that the ω-periodic solution problem of (2.10) is equivalent to the operator equation in Banach space Cω(ℝ)(2.13)(I-T∘B)u=Th,
where I is the identity operator in Cω(ℝ) and B:Cω(ℝ)→Cω(ℝ) is the product operator defined by
(2.14)Bu(t)=(M-a(t))u(t),u∈Cω(R),
which is a positive linear bounded operator. We prove that the norm of T∘B in ℒ(Cω(ℝ),Cω(ℝ)) satisfies ∥T∘B∥<1.
For everyu∈Cω(ℝ) and t∈ℝ, by the definition (2.4) of T and the positivity of Φ, we have
(2.15)|(T∘B)u(t)|=|T(Bu)(t)|=|∫t-ωtΦ(t-s)(M-a(s))u(s)ds|≤∫t-ωt‍Φ(t-s)|(M-a(s))u(s)|ds≤(M-m)∥u∥C∫t-ωt‍Φ(t-s)ds=(M-m)∥u∥C∫0ω‍Φ(s)ds=(1-mM)∥u∥C.
Therefore, ∥(T∘B)u∥C≤(1-m/M)∥u∥C. By the arbitrariness of u∈Cω(ℝ), we have ∥T∘B∥≤1-m/M<1.
Thus,I-T∘B has a bounded inverse operator given by the series
(2.16)(I-T∘B)-1=∑n=0∞‍(T∘B)n
with the norm estimate
(2.17)‖(I-T∘B)-1‖≤11-‖T∘B‖≤Mm.
Consequently, (2.13), equivalently (2.10), has a unique ω-periodic solution
(2.18)u=(I-T∘B)-1(Th):=Sh,
where
(2.19)S=(I-T∘B)-1∘T=∑n=0∞‍(T∘B)nT.
By the complete continuity of T, S:Cω(ℝ)→Cω(ℝ) is a completely continuous linear operator.
For everyh∈Cω(ℝ), by the expression (2.19) of S, we have
(2.20)‖Sh‖C≤‖(I-T∘B)-1‖⋅‖Th‖C≤Mm‖Th‖C.
If h∈Cω+(ℝ), by the series expression of S and the positivity of T and B, we have
(2.21)Sh=(∑n=0∞‍(T∘B)nT)h=∑n=0∞‍(T∘B)n(Th)≥Th.
Hence, form (2.5) and (2.20), it follows that
(2.22)Sh(t)≥Th(t)≥σ‖Th‖≥mσM‖Sh‖C,t∈R.
Namely, (2.11) holds.Letf∈C(ℝ×[0,∞)n,[0,∞)). For every u∈Cω+(ℝ), set
(2.23)F(u)(t)∶=f(t,u(t-τ1),…,u(t-τn)),t∈R.
Then F:Cω+(ℝ)→Cω+(ℝ) is continuous. Define a mapping A:Cω+(ℝ)→Cω+(ℝ) by
(2.24)A=S∘F.
By the definition of operator S, the ω-periodic solution of (1.1) is equivalent to the fixed point of A. Choose a subcone of Cω+(ℝ) by
(2.25)K={u∈Cω+(R)∣u(t)≥mσM‖u‖C,t∈R}.
From the strong positivity of S in Lemma 2.2 and the definition of A, we easily obtain the following lemma.Lemma 2.3.
A
(
C
ω
+
(
ℝ
)
)
⊂
K, and A:K→K is completely continuous.Hence, the positiveω-periodic solution of (1.1) is equivalent to the nontrivial fixed point of A. We will find the nonzero fixed point of A by using the fixed point index theory in cones.We recall some concepts and conclusions on the fixed point index in [15, 16]. Let E be a Banach space and K⊂E be a closed convex cone in E. Assume Ω is a bounded open subset of E with boundary ∂Ω, and K∩Ω≠∅. Let A:K∩Ω¯→K be a completely continuous mapping. If Au≠u for any u∈K∩∂Ω, then the fixed point index i(A,K∩Ω,K) has definition. One important fact is that if i(A,K∩Ω,K)≠0, then A has a fixed point in K∩Ω. The following two lemmas are needed in our argument.Lemma 2.4 (see [16]).
LetΩ be a bounded open subset of E with θ∈Ω and A:K∩Ω¯→K a completely continuous mapping. If λAu≠u for every u∈K∩∂Ω and 0<λ≤1, then i(A,K∩Ω,K)=1.Lemma 2.5 (see [16]).
LetΩ be a bounded open subset of E and A:K∩Ω¯→K a completely continuous mapping. If there exists an e∈K∖{θ} such that u-Au≠μe for every u∈K∩∂Ω and μ≥0, then i(A,K∩Ω,K)=0.In next section, we will use Lemmas2.4 and 2.5 to prove Theorems 1.1 and 1.2.


## 3. Proofs of Main Results

Proof of Theorem1.1.
Choose the working spaceE=Cω(ℝ). Let K⊂Cω+(ℝ) be the closed convex cone in Cω(ℝ) defined by (2.25) and A:K→K the operator defined by (2.24). Then the positive ω-periodic solution of (1.1) is equivalent to the nontrivial fixed point of A. Let 0<r<R<+∞ and set
(3.1)Ω1={u∈Cω(R)∣‖u‖C<r},Ω2={u∈Cω(R)∣‖u‖C<R}.
We show that the operator A has a fixed point in K∩(Ω2∖Ω¯1) when r is small enough and R large enough.
Letr∈(0,δ), where δ is the positive constant in Condition (F1). We prove that A satisfies the condition of Lemma 2.4 in K∩∂Ω1, namely, λAu≠u for every u∈K∩∂Ω1 and 0<λ≤1. In fact, if there exist u0∈K∩∂Ω1 and 0<λ0≤1 such that λ0Au0=u0, then by the definition of A and Lemma 2.2, u0∈Cω2(ℝ) satisfies the delay differential equation
(3.2)-u0′′(t)+a(t)u0(t)=λ0f(t,u0(t-τ1),…,u0(t-τn)),t∈R.
Since u0∈K∩∂Ω1, by the definitions of K and Ω1, we have
(3.3)0≤u0(t-τk)≤‖u0‖C=r<δ,k=1,…,n,t∈R.
Hence from condition (F1), it follows that
(3.4)f(t,u0(t-τ1),…,u0(t-τn))≤c1u0(t-τ1)+⋯+cnu0(t-τn),t∈R.
By this and (3.2), we get that
(3.5)-u0′′(t)+a(t)u0(t)≤c1u0(t-τ1)+⋯+cnu0(t-τn),t∈R.
Integrating both sides of this inequality from 0 to ω and using the periodicity of u0, we have
(3.6)∫0ωa(t)u0(t)dt‍≤c1∫0ωu0(t-τ1)dt‍+⋯+cn∫0ωu0(t-τn)dt‍=(c1+⋯+cn)∫0ωu0(t)ds.‍
Hence, we obtain that
(3.7)m∫0ωu0(t)dt‍≤∫0ωa(t)u0(t)dt‍≤(c1+⋯+cn)∫0ωu0(t)ds.
By the definition of cone K, ∫0ω‍u0(t)dt≥(mσ/M)∥u0∥C·ω>0. From (3.7), it follows that m≤c1+⋯+cn, which contradicts to the assumption in Condition (F1). Hence A satisfies the condition of Lemma 2.4 in K∩∂Ω1. By Lemma 2.4, we have
(3.8)i(A,K∩Ω1,K)=1.
On the other hand, choose R>max{(M/mσ)H,δ}, where H is the positive constant in condition (F2), and let e(t)≡1. Clearly, e∈K∖{θ}. We show that A satisfies the condition of Lemma 2.5 in K∩∂Ω2, namely, u-Au≠μv for every u∈K∩∂Ω2 and μ≥0. In fact, if there exist u1∈K∩∂Ω2 and μ1≥0 such that u1-Au1=μ1e, since u1-μ1e=Au1, by definition of A and Lemma 2.2, u1∈Cω2(ℝ) satisfies the differential equation
(3.9)-u1′′(t)+a(t)(u1(t)-μ1)=f(t,u1(t-τ1),…,u1(t-τn)),t∈R.
Since u1∈K∩∂Ω2, by the definition of K, we have
(3.10)u1(t-τk)≥mσM‖u1‖C=mσMR>H,t∈I,k=1,…,n.
From this and Condition (F2), it follows that
(3.11)f(t,u1(t-τ1),…,u1(t-τn))≥d1u1(t-τ1)+⋯+dnun(t-τn),t∈I.
By this inequality and (3.9), we have
(3.12)-u1′′(t)+a(t)(u1(t)-μ1)≥d1u1(t-τ1)+⋯+dnu1(t-τn),t∈I.
Integrating this inequality on [0,ω] and using the periodicity of u1, we obtain that
(3.13)∫0ω‍a(t)(u1(t)-μ1)dt≥d1∫0ω‍u1(t-τ1)dt+⋯+dn∫0ω‍u1(t-τn)dt=(d1+⋯+dn)∫0ω‍u1(t)ds.
Consequently, we have that
(3.14)M∫0ωu1(t)dt‍≥∫0ωa(t)u1(t)dt‍≥∫0ωa(t)(u1(t)-μ1)dt‍≥(d1+⋯+dn)∫0ω‍u1(t)ds.
Since ∫0ω‍u1(t)dt≥(mσ/M)∥u1∥C·ω>0, form this inequality it follows that M≥d1+⋯+dn, which contradicts to the assumption in Condition (F2). This means that A satisfies the condition of Lemma 2.5 in K∩∂Ω2. By Lemma 2.5,
(3.15)i(A,K∩Ω2,K)=0.
Now by the additivity of fixed point index, (3.8), and (3.15) we have
(3.16)i(A,K∩(Ω2∖Ω¯1),K)=i(A,K∩Ω2,K)-i(A,K∩Ω1,K)=-1.
Hence A has a fixed point in K∩(Ω2∖Ω¯1), which is a positive ω-periodic solution of (1.1).Proof of Theorem1.2.
LetΩ1,Ω2⊂Cω(ℝ) be defined by (3.1). We prove that the operator A defined by (2.24) has a fixed point in K∩(Ω2∖Ω¯1) if r is small enough and R large enough.
Letr∈(0,δ), where δ is the positive constant in Condition (F2), and choose e(t)≡1. We prove that A satisfies the condition of Lemma 2.5 in K∩∂Ω1, namely, u-Au≠μe for every u∈K∩∂Ω1 and μ≥0. In fact, if there exist u0∈K∩∂Ω1 and μ0≥0 such that u0-Au0=μ0e, since u0-μ0e=Au0, by definition of A and Lemma 2.2, u0∈Cω2(ℝ) satisfies the differential equation
(3.17)-u0′′(t)+a(t)(u0(t)-μ0)=f(t,u0(t-τ1),…,u0(t-τn)),t∈R.
Since u0∈K∩∂Ω1, by the definitions of K and Ω1, u0 satisfies (3.3). From (3.3) and Condition (F3), it follows that
(3.18)f(t,u0(t-τ1),…,u0(t-τn))≥d1u0(t-τ1)+⋯+dnu0(t-τn),t∈R.
From this and (3.17), we see that
(3.19)-u0′′(t)+a(t)(u0(t)-μ0)≥d1u0(t-τ1)+⋯+dnu0(t-τn),t∈R.
Integrating this inequality on [0,ω] and using the periodicity of u0(t), we have
(3.20)∫0ω‍a(t)(u0(t)-μ0)dt≥d1∫0ω‍u0(t-τ1)dt+⋯+dn∫0ωu0(t-τn)dt‍=(d1+⋯+dn)∫0ωu0(t)ds‍.
From this we obtain that
(3.21)M∫0ω‍u0(t)dt≥∫0ω‍a(t)u0(t)dt≥∫0ω‍a(t)(u0(t)-μ0)dt≥(d1+⋯+dn)∫0ω‍u0(t)ds.
Since ∫0ω‍u0(t)dt≥(mσ/M)∥u0∥C·ω>0, from the inequality above, it follows that M≥d1+⋯+dn, which contradicts to the assumption in (F3). Hence A satisfies the condition of Lemma 2.5 in K∩∂Ω1. By Lemma 2.5, we have
(3.22)i(A,K∩Ω1,K)=0.
Then, choosing R>max{(M/mσ)H,δ}, we show that A satisfies the condition of Lemma 2.4 in K∩∂Ω2, namely, λAu≠u for every u∈K∩∂Ω2 and 0<λ≤1. In fact, if there exist u1∈K∩∂Ω2 and 0<λ1≤1 such that λ1Au1=u1, then by the definition of A and Lemma 2.2, u1∈Cω2(ℝ) satisfies the differential equation
(3.23)-u1′′(t)+a(t)u1(t)=λ1f(t,u1(t-τ1),…,u1(t-τn)),t∈R.
Since u1∈K∩∂Ω2, by the definition of K, u1 satisfies (3.10). From (3.10) and condition (F4), it follows that
(3.24)f(t,u1(t-τ1),…,u1(t-τn))≤c1u1(t-τ1)+⋯+cnu1(t-τn),t∈R.
By this and (3.23), we have
(3.25)-u1′′(t)+a(t)u1(t)≤c1u1(t-τ1)+⋯+cnu1(t-τn),t∈R.
Integrating this inequality on [0,ω] and using the periodicity of u1(t), we have
(3.26)∫0ωa(t)u1(t)dt‍≤c1∫0ωu1(t-τ1)dt‍+⋯+cn∫0ωu1(t-τn)dt‍=(c1+⋯+cn)∫0ωu1(t)ds.‍
From this we obtain that
(3.27)m∫0ωu1(t)dt‍≤∫0ωa(t)u1(t)dt‍≤(c1+⋯+cn)∫0ωu1(t)ds.‍
Since ∫0ω‍u1(t)dt≥(mσ/M)∥u0∥C·ω>0, from the inequality (3.27), it follows that m≤c1+⋯+cn, which contradicts to the assumption in Condition (F4). Hence A satisfies the condition of Lemma 2.4 in K∩∂Ω1. By Lemma 2.4, we have
(3.28)i(A,K∩Ω2,K)=1.
Now, from (3.22) and (3.28), it follows that
(3.29)i(A,K∩(Ω2∖Ω¯1),K)=i(A,K∩Ω2,K)-i(A,K∩Ω1,K)=1.
Hence A has a fixed point in K∩(Ω2∖Ω¯1), which is a positive ω-periodic solution of  (1.1).


## 4. Remarks

In Theorems1.1 and 1.2, the conditions (F1) and (F4) can be replaced by the following condition: (F5)
there exist positive constantsc1,…,cn satisfying c1+⋯+cn<m and H>0 such that
(4.1)f(t,x1,…,xn)≤c1x1+⋯+cnxn,
for t∈ℝ and x1,…,xn∈[(mσ/M)H,H];and (F2) and (F3) can be replaced by the(F6)
there exist positive constantsd1,…,dn satisfying d1+⋯+dn>M and H>0 such that
(4.2)f(t,x1,…,xn)≥d1x1+⋯+dnxn,
for t∈ℝ and x1,…,xn∈[(mσ/M)H,H].In fact, if condition (F5) holds, setting(4.3)Ω3={u∈Cω(R)∣‖u‖C<H},
similar to the proof of (3.28), we can prove that
(4.4)i(A,K∩Ω3,K)=1,
and if condition (F6) holds, similar to the proof of (3.15), we can prove that
(4.5)i(A,K∩Ω3,K)=0.
Therefore, by the proofs of Theorems 1.1 and 1.2, we have the following theorem.Theorem 4.1.
Leta∈C(ℝ,(0,∞)) be a ω-periodic function, f∈C(ℝ×[0,∞)n,[0,∞)) and f(t,x1,…,xn)ω-periodic in t. Then in each case of the following:    (1)
(F1) and (F6) hold,(2)
(F2) and (F5) hold,(3)
(F3) and (F5) hold,(4)
(F4) and (F6) hold. Equation (1.1) has at least one positive ω-periodic solution.Now we consider the existence of two positive periodic solutions of (1.1). If the conditions (F2), (F3), and (F5) hold, by the proof of Theorem 1.1, condition (F2) implies that (3.15) holds when R is large enough and R>H, and by the proof of Theorem 1.2, condition (F3) implies that (3.22) holds when r is small enough and r<H. Since Ω¯1⊂Ω3 and Ω¯3⊂Ω2, by (3.15), (3.22), and (4.4), we have
(4.6)i(A,K∩(Ω3∖Ω¯1),K)=i(A,K∩Ω3,K)-i(A,K∩Ω1,K)=1,i(A,K∩(Ω2∖Ω¯3),K)=i(A,K∩Ω2,K)-i(A,K∩Ω3,K)=-1.
This means that A has fixed-points u1∈K∩(Ω3∖Ω¯1) and u2∈K∩(Ω2∖Ω¯3), and u1 and u2 are two positive ω-periodic solution of (1.1). Consequently, we have the following theorem.Theorem 4.2.
Leta∈C(ℝ,(0,∞)) be a ω-periodic function and f∈C(ℝ×[0,∞)n,[0,∞)) and f(t,x1,…,xn) be ω-periodic in t. If (F2), (F3), and (F5) hold, then (1.1) has two positive ω-periodic solutions.Similar to Theorem4.2, we have the following theorem.Theorem 4.3.
Leta∈C(ℝ,(0,∞)) be a ω-periodic function, f∈C(ℝ×[0,∞)n,[0,∞)), and f(t,x1,…,xn)ω-periodic in t. If (F1), (F4), and (F6) hold, then (1.1) has two positive ω-periodic solutions.



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*Source: 829783-2012-07-03.xml*