# Positive Periodic Solution of Second-Order Coupled Systems with Singularities

Authors: Tiantian Ma
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

This paper establishes the existence of periodic solution for a kind of second-order singular nonautonomous coupled systems. Our approach is based on fixed point theorem in cones. Examples are given to illustrate the main result.


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## Body

## 1. Introduction

We are concerned with the existence of positiveT-periodic solution for the second-order nonautonomous singular coupled systems:
(1)u′′(t)=f1(t,v(t))+c1(t),t∈(0,T),v′′(t)=f2(t,u(t))+c2(t),t∈(0,T),u(0)=u(T),u′(0)=u′(T),v(0)=v(T),v′(0)=v′(T),
where fi∈C((0,T)×[0,+∞),[0,+∞)), ci:(0,T)→[0,+∞), i=1,2, are Lebesgue integrable, fi may be singular at t=0, T and ci can have finitely many singularities.Singular differential equations or systems arise from many branches of applied mathematics and physics such as gas dynamics, Newtonian fluid mechanics, and nuclear physics, which have been widely studied by many authors (see [1–7] and references therein). Some classical-tools have been used to study the positive solutions for two point nonperiodic boundary value problems of coupled systems [8, 9]. However, there are few works on periodic solutions of second order nonautonomous singular coupled systems of type (1).In the recent papers [10, 11], the periodic solutions of singular coupled systems
(2)x′′+a1(t)x=f1(t,y)+c1(t),y′′+a2(t)y=f2(t,x)+c2(t),
were proved by using some fixed point theorems in cones for completely continuous operators, where a1,a2,c1,c2∈L1(0,T), f1,f2∈Car([0,T]×(0,+∞),(0,+∞)). When the Green's function Gi(t,s) (i=1,2), associated with the periodic boundary problem
(3)x′′+ai(t)x=ci(t),x(0)=x(T),x′(0)=x′(T),
is nonnegative for all (t,s)∈[0,T]×[0,T] and fi (i=1,2) satisfies weak singularities
(4)0≤b^i(t)xαi≤fi(t,x)≤bi(t)xαi,∀x>0,a.e.t∈(0,T),
where 0<αi<1, b^i,bi∈L1(0,T), b^i≥0, bi≥0, and b^i, bi are strictly positive on some positive measure subsets of (0,T), some sufficient conditions for the existence of periodic solutions of (2) were obtained in [10, 11].Motivated by the papers [9–11], we consider the existence of positive T-periodic solution of (1). Owing to the disappearing of the terms ai(t) (i=1,2) in (2), the methods in [9, 10] are no longer valid. In present paper, we will deal with the periodic solutions of (1) under new conditions. Let k be a constant satisfying 0<k<π/T. Denote by G(t,s) the Green function of
(5)x′′(t)+k2x(t)=0,x(0)=x(T),x′(0)=x′(T),
which can be expressed by
(6)G(t,s)=12k(1-coskT){sink(t-s)+sink(T-t+s),mmmmmmmm0≤s≤t≤T,sink(s-t)+sink(T-s+t),mmmmmmmm0≤t≤s≤T.
By a direct computation, we can get
(7)m∶=min0≤t,s≤TG(t,s)=12kcotkT2,M∶=max0≤t,s≤TG(t,s)=12ksin(kT/2),σ∶=mM=coskT2∈(0,1).
Assume fi (i=1,2) satisfies the following conditions.(H1)
Fort∈(0,T),fi(t,1)>0 and there exist constants μ1≥μ2>1, λ1≥λ2>1 such that, for any constants 0≤ρi≤1, i=1,2,
(8)ρ1μ1f1(t,u)≤f1(t,ρ1u)≤ρ1μ2f1(t,u),(t,u)∈(0,T)×[0,+∞),(9)ρ2λ1f2(t,u)≤f2(t,ρ2u)≤ρ2λ2f2(t,u),(t,u)∈(0,T)×[0,+∞).(H2)
∫0T‍c1-(t)dt=r1>0,∫0T‍c2-(t)dt=r2>0 and
(10)∫0T‍(f1(t,1)+c1+(t))dt<k2Tmr1(2M2r2/m+1)μ1,∫0T‍(f2(t,1)+c2+(t))dt<k2Tmr2(2M2r1/m+1)λ1,where ci+(t)=max{ci(t),0}, ci-(t)=max{-ci(t),0}, i=1,2.Remark 1.
For anyρ1≥1, we get from (8) that
(11)ρ1μ2f1(t,u)≤f1(t,ρ1u)≤ρ1μ1f1(t,u),(t,u)∈(0,T)×[0,+∞).
For anyρ2≥1, we get from (9) that
(12)ρ2λ2f2(t,u)≤f2(t,ρ2u)≤ρ2λ1f2(t,u),(t,u)∈(0,T)×[0,+∞).Typical functions that satisfy (8) or (9) are those taking the form
(13)fi(t,u)=∑k=1n‍aik(t)ulik,t∈(0,T),
where aik∈C(0,T), aik(t)>0, lik>1, i=1,2; k=1,2,…,n.Definition 2.
Supposing that(u,v)∈C1[0,T]∩C2(0,T)×C1[0,T]∩C2(0,T) satisfies (1) and u(t)>0, v(t)>0 for any t∈[0,T], then one says that (u,v) is a C1[0,T]×C1[0,T] positive solution of system (1).
By using fixed point theorem in cones, we are able to prove the following result.Theorem 3.
Assume that (H1), (H2) hold. Then (1) has at least one positive T-periodic solution.The proof of Theorem3 will be given in Section 3 of this paper.


## 2. Preliminaries

Lemma 4 (see [12]).
LetX be a Banach space, K a cone in X, Ω1, Ω2 two nonempty bounded open sets in K, θ∈Ω1⊂Ω1¯⊂Ω2. T:Ω2¯/Ω1→K is a completely continuous operator. If(i)
T(x)≠λx, x∈∂Ω1, λ>1,(ii)
T(x)≠λx, x∈∂Ω2, 0<λ<1, infx∈∂Ω2∥Tx∥>0,then T has a fixed point in Ω2¯/Ω1.Lemma 5.
Iffi(t,u) satisfy (H1), then, for t∈(0,T), fi(t,u) are increasing on u and, for any [α,β]⊂(0,T),
(14)limu→+∞fi(t,u)u=+∞,(i=1,2)
uniformly with respect to t∈[α,β].Proof.
We only deal withf1. Without loss of generality, let 0≤x≤y. If y=0, we get f1(t,x)≤f1(t,y). If y≠0, let c0=x/y, then 0≤c0≤1. From (8), we get
(15)f1(t,x)=f1(t,c0y)≤c0μ2f1(t,y)≤f1(t,y),
which means f1(t,u) is increasing on u.
Assumeu>1. It follows from (11) that f1(t,u)≥uμ2f1(t,1). Thus
(16)f1(t,u)u≥uμ2-1f1(t,1),t∈(0,T).
From (H1), for any [α,β]⊂(0,T), we obtain
(17)f1(t,u)u≥uμ2-1f1(t,1)≥maxt∈[α,β]f1(t,1)uμ2-1>0.
Therefore
(18)limu→+∞f1(t,u)u=+∞
uniformly with respect to t∈[α,β].LetX=C[0,T]. We know X is a Banach space with the norm ∥u∥=maxt∈[0,T]|u(t)|. Define the sets
(19)P={u∈X:u(t)≥0},Q={u∈P:min0≤t≤Tu(t)≥σ∥u∥}.
It is easy to check that P, Q are cones in X and Q⊂P. Throughout this paper, we consider the space X×X. It is easy to see X×X is a Banach space with the norm,
(20)∥(u,v)∥0=∥u∥+∥v∥,(u,v)∈X×X.We can get the conclusion thatP×P, Q×Q are cones in X×X and Q×Q⊂P×P.For anyu∈X, define the function
(21)[u(t)]*={u(t),u(t)≥0,0,u(t)<0.
Then the solution of periodic boundary value problem
(22)x′′(t)+k2x(t)=-c1-(t),x(0)=x(T),x′(0)=x′(T)
can be expressed by x~1(t)=-∫0T‍G(t,s)c1-(s)ds, and the solution of periodic boundary value problem
(23)x′′(t)+k2x(t)=-c2-(t),x(0)=x(T),x′(0)=x′(T)
can be expressed by x~2(t)=-∫0T‍G(t,s)c2-(s)ds. Obviously, x~i(t)<0, i=1,2. Then
(24)[u(t)+x~1(t)]*≤u(t)≤∥u∥,[v(t)+x~2(t)]*≤v(t)≤∥v∥.
From (11), (12), and Lemma 5, we have
(25)f1(t,[v(t)+x~2(t)]*)≤f1(t,∥v∥)≤f1(t,∥v∥+1)≤(∥v∥+1)μ1f1(t,1),(26)f2(t,[u(t)+x~1(t)]*)≤f2(t,∥u∥)≤f2(t,∥u∥+1)≤(∥u∥+1)λ1f2(t,1).
Then, for any fixed (u,v)∈P×P, it follows from (H2) that
(27)0≤∫0T‍G(t,s)×{k2[u(s)+x~1(s)]*mmmm+f1(s,[v(s)+x~2(s)]*)+c1+(s)}ds≤M∫0T‍{k2∥u∥+(∥v∥+1)μ1f1(s,1)+c1+(s)}ds≤k2MT∥u∥+M(∥v∥+1)μ1∫0T‍{f1(s,1)+c1+(s)}ds<+∞,0≤∫0T‍G(t,s)×{k2[v(s)+x~2(s)]*mmmm+f2(s,[u(s)+x~1(s)]*)+c2+(s)}ds≤M∫0T‍{k2∥v∥+(∥u∥+1)λ1f2(s,1)+c2+(s)}ds≤k2MT∥v∥+M(∥u∥+1)λ1∫0T‍{f2(s,1)+c2+(s)}ds<+∞.Thus, we can define the operatorT:P×P→P×P, T(u,v)=(T1u,T2v) by
(28)T1u(t)=∫0T‍G(t,s)×{k2[u(s)+x~1(s)]*,mmmm+f1(s,[v(s)+x~2(s)]*)+c1+(s)}ds,T2v(t)=∫0T‍G(t,s)×{k2[v(s)+x~2(s)]*mmmm+f2(s,[u(s)+x~1(s)]*)+c2+(s)}ds,
for (u,v)∈P×P. Then, we have the following lemma.Lemma 6.
Assuming that (H1), (H2) hold, then T has a fixed point if and only if
(29)u′′(t)+k2u(t)=k2[u(t)+x~1(t)]*+f1(t,[v(t)+x~2(t)]*)+c1+(t),t∈(0,T),v′′(t)+k2v(t)=k2[v(t)+x~2(t)]*+f2(t,[u(t)+x~1(t)]*)+c2+(t),t∈(0,T),u(0)=u(T),u′(0)=u′(T),v(0)=v(T),v′(0)=v′(T)
has one positive T-periodic solution.Lemma 7.
Assuming that (H1), (H2) hold, then T(Q×Q)⊂Q×Q and T:Q×Q→Q×Q is completely continuous.Proof.
For(u,v)∈Q×Q, we have
(30)(T1u)(t)≥m∫0T{k2[u(s)+x~1(s)]*+f1‍mmmm×(s,[v(s)+x~2(s)]*)+c1+(s)[u(s)+x~1(s)]*[u(s)+x~1(s)]*}ds=σ∫0Tmax0≤t,s≤TG(t,s){k2[u(s)+x~1(s)]*‍mmmmmmmmmmm+f1(s,[v(s)+x~2(s)]*)mmmmmmmmmmm+c1+(s)[u(s)+x~1(s)]*}ds=σ∥T1u∥,(T2v)(t)≥m∫0T{k2[v(s)+x~2(s)]*‍mmmm,,+f2(s,[u(s)+x~1(s)]*)mmmm+c2+(s)[v(s)+x~2(s)]*}ds=σ∫0T‍max0≤t,s≤TG(t,s)mmm×{k2[v(s)+x~2(s)]*mmmmm+f2(s,[u(s)+x~1(s)]*)+c2+(s)}ds=σ∥T2v∥.
Then, we can get
(31)∥T(u,v)∥0=∥(T1u,T2v)∥0=∥T1u∥+∥T2v∥≥σ(∥T1u∥+∥T2v∥)=σ∥T(u,v)∥0,
which means T(Q×Q)⊂Q×Q.
LetB⊂Q×Q be any bounded set. Then there exists a constant N such that, for any (u,v)∈B,
(32)∥(u,v)∥0=∥u∥+∥v∥≤N.
From (27), we have
(33)(T1u)(t)≤k2MT∥u∥+M(∥v∥+1)μ1×∫0T‍{f1(s,1)+c1+(s)}ds≤k2MTN+M(N+1)μ1×∫0T‍{f1(s,1)+c1+(s)}ds,(T2v)(t)≤k2MT∥v∥+M(∥u∥+1)λ1×∫0T‍{f2(s,1)+c2+(s)}ds≤k2MTN+M(N+1)λ1×∫0T‍{f2(s,1)+c2+(s)}ds.
Let
(34)N~=k2MTN+M(N+1)λ1+μ1×(∫0T‍{f1(s,1)+c1+(s)}ds+∫0T‍{f2(s,1)+c2+(s)}ds)<+∞.
Thus
(35)∥T(u,v)∥0=∥T1u∥+∥T2v∥≤N~,
which implies that T(B) is bounded.
Next we prove thatT(B) is equicontinuous. For any (u,v)∈B, t∈[0,T], we know
(36)(T1u)(t)=∫0T‍G(t,s)mmm×{k2[u(s)+x~1(s)]*mmmmm+f1(s,[v(s)+x~2(s)]*)+c1+(s)}ds=12k(1-coskT)mm×∫0t‍(sink(t-s)mmmmm+sink(T-t+s))mmmm×{k2[u(s)+x~1(s)]*mmmmmm+f1(s,[v(s)+x~2(s)]*)mmmmm1+c1+(s)[u(s)+x~1(s)]*[u(s)+x~1(s)]*}dsmm+12k(1-coskT)×∫tT‍(sink(s-t)+sink(T-s+t))mmmmm×{k2[u(s)+x~1(s)]*mmmmmmm,+f1(s,[v(s)+x~2(s)]*)mmmmmm11,+c1+(s)[u(s)+x~1(s)]*}ds,(T1u)′(t)=-12sin(kT/2)×∫0t‍sin2k(t-s)-kT2mmmmm×{k2[u(s)+x~1(s)]*mmmmmm,,+f1(s,[v(s)+x~2(s)]*)mmmmm11,+c1+(s)[u(s)+x~1(s)]*}ds+12sin(kT/2)×∫tT‍sin2k(s-t)-kT2mmmmm×{k2[u(s)+x~1(s)]*mmmmmm,+f1(s,[v(s)+x~2(s)]*)mmmmm11,+c1+(s)[u(s)+x~1(s)]*}ds.
From (25) and (H2), we have
(37)|(T1u)′(t)|≤12sin(kT/2)×∫0t‍{k2[u(s)+x~1(s)]*mmmmm+f1(s,[v(s)+x~2(s)]*)+c1+(s)[u(s)+x~1(s)]*}ds+12sin(kT/2)×∫tT‍{k2[u(s)+x~1(s)]*mmmmm+f1(s,[v(s)+x~2(s)]*)+c1+(s)[u(s)+x~1(s)]*}ds=12sin(kT/2)×∫0T‍{k2[u(s)+x~1(s)]*mmmmm+f1(s,[v(s)+x~2(s)]*)+c1+(s)[u(s)+x~1(s)]*}ds≤12sin(kT/2)×∫0T‍{k2∥u∥+(∥v∥+1)μ1mmmmm×f1(s,1)+c1+(s)k2∥u∥+(∥v∥+1)μ1}ds≤k2T2sin(kT/2)∥u∥+12sin(kT/2)(∥v∥+1)μ1×∫0T‍{f1(s,1)+c1+(s)}ds≤k2T2sin(kT/2)N+12sin(kT/2)(N+1)μ1×∫0T‍{f1(s,1)+c1+(s)}ds<+∞.
Using the same method, we can obtain|(T2v)′(t)|<+∞. Therefore, T(B) is equicontinuous. According to Ascoli-Arzela theorem, T(B) is a relatively compact set.
Next, we prove thatT:Q×Q→Q×Q is continuous. Suppose (un,vn),(u0,v0)∈Q×Q, (un,vn)→(u0,v0), n→+∞, that is, un→u0, vn→v0, n→+∞. We know that there exists a constant L>0 such that
(38)∥u0∥≤L,∥un∥≤L,∥v0∥≤L,∥vn∥≤L,n=1,2….
We shall proveT(un,vn)→T(u0,v0), n→+∞, that is,
(39)T1un⟶T1u0,T2vn⟶T2v0,n⟶+∞.
We first deal withT1un→T1u0, n→+∞. Otherwise, there exist ε0>0, {tn}∈[0,T] such that |T1un(tn)-T1u0(tn)|≥ε0. Without loss of generality, we can assume tn→t0∈[0,T]. We know
(40)T1un(tn)-T1un(t0)⟶0,T1u0(tn)-T1u0(t0)⟶0,n⟶∞.
Next, we showT1un(t0)-T1u0(t0)→0, n→∞. In fact,
(41)T1un(t0)-T1u0(t0)=∫0T‍G(t0,s)×{k2[un(s)+x~1(s)]*mmmm-k2[u0(s)+x~1(s)]*mmmm+f1(s,[vn(s)+x~2(s)]*)mmmm-f1(s,[v0(s)+x~2(s)]*)}ds.
Let
(42)rn(s)=G(t0,s)×{k2[un(s)+x~1(s)]*-k2[u0(s)+x~1(s)]*+f1(s,[vn(s)+x~2(s)]*)-f1(s,[v0(s)+x~2(s)]*)}.
Since
(43)|[vn(s)+x~2(s)]*-[v0(s)+x~2(s)]*|=||vn(s)+x~2(s)|+vn(s)+x~2(s)2-|v0(s)+x~2(s)|+v0(s)+x~2(s)2|=||vn(s)+x~2(s)|-|v0(s)+x~2(s)|2+vn(s)-v0(s)2|vn(s)+x~2(s)|-|v0(s)+x~2(s)|2|≤|vn(s)-v0(s)|⟶0,n⟶+∞,
and f1 is continuous, we know rn(s)→0, n→∞. From (24), we know
(44)[un(s)+x~1(s)]*≤L,[u0(s)+x~1(s)]*≤L,[vn(s)+x~2(s)]*≤L<L+1,[v0(s)+x~2(s)]*≤L<L+1.
It can be inferred from (8) that
(45)|rn(s)|≤M{|k2[un(s)+x~1(s)]*-k2[u0(s)+x~1(s)]*|mmmm+|f1(s,[vn(s)+x~2(s)]*)-f1(s,[v0(s)+x~2(s)]*)|}≤2Mk2L+2M(L+1)μ1f1(s,1).
SetF(s)=2Mk2L+2M(L+1)μ1(f1(s,1)+c1+(s)). Thus, we get
(46)|rn(s)|≤F(s),s∈(0,T),n=1,2,….
From (H2), we know ∫0T‍F(s)ds<+∞. Using Lebesgue-dominated convergence theorem, we get
(47)T1un(t0)-T1u0(t0)=∫0T‍rn(s)ds⟶0,n⟶+∞.
From (40) and (47), we obtain
(48)ε0≤|T1un(tn)-T1u0(tn)|=|T1un(tn)-T1un(t0)+T1un(t0)-T1u0(t0)+T1u0(t0)-T1u0(tn)|≤|T1un(tn)-T1un(t0)|+|T1un(t0)-T1u0(t0)|+|T1u0(t0)-T1u0(tn)|⟶0,n⟶∞,
which is a contradiction. Thus, we know T1un→T1u0, n→+∞. Using the same method, we can obtain T2vn→T2v0, n→+∞. Then
(49)T(un,vn)=(T1un,T2vn)⟶(T1u0,T2v0)=T(u0,v0),n⟶+∞,
which means T:Q×Q→Q×Q is continuous. Therefore T:Q×Q→Q×Q is a completely continuous operator.


## 3. Proof of Theorem3

We proceed to prove Theorem3 in two steps.(1) LetΩ1={(u,v)∈Q×Q:∥u∥<2M2r1/m,∥v∥<2M2r2/m}. We can get
(50)T(u,v)≠λ(u,v),(u,v)∈∂Ω1,λ>1.
For (u,v)∈∂Ω1, we have the following two cases.Case I.
One has{(u,v)∈Q×Q:∥u∥=2M2r1/m,∥v∥≤2M2r2/m}. Under this condition, we can get
(51)T1u≠λu,(u,v)∈∂Ω1,λ>1.
Otherwise, there exist (u0,v0)∈∂Ω1, λ0>1 such that T1u0=λ0u0. As
(52)u0(t)+x~1(t)=u0(t)-∫0T‍G(t,s)c1-(s)ds≥σ∥u0∥-Mr1>0,
then [u0(t)+x~1(t)]*=u0(t)+x~1(t). By a direct computation, we know u0=(1/λ0)T1u0 satisfies
(53)u0′′(t)+k2u0(t)=1λ0{k2(u0(t)+x~1(t))+f1(t,[v0(t)+x~2(t)]*)1λ01λ01λ0+c1+(t)k2(u0(t)+x~1(t))+f1(t,[v0(t)+x~2(t)]*)},u0(0)=u0(T),u0′(0)=u0′(T).
Then, we get
(54)u0′′(t)=1λ0{k2(u0(t)+x~1(t))+f1(t,[v0(t)+x~2(t)]*)mmmm.+c1+(t)k2(u0(t)+x~1(t))+f1(t,[v0(t)+x~2(t)]*)}-k2u0(t)≤k2u0(t)+k2x~1(t)+f1(t,[v0(t)+x~2(t)]*)+c1+(t)-k2u0(t)=k2x~1(t)+f1(t,[v0(t)+x~2(t)]*)+c1+(t).
Since[v0(t)+x~2(t)]*≤v0(t)<2M2r2/m+1, integrating both sides of (54) on [0,T], we get
(55)0≤k2∫0T‍x~1(t)dt+∫0T‍{f1(t,[v0(t)+x~2(t)]*)+c1+(t)}dt≤k2∫0T‍x~1(t)dt+(2M2r2m+1)μ1×∫0T‍{f1(t,1)+c1+(t)}dt.
Then
(56)(2M2r2m+1)μ1∫0T‍[f1(t,1)+c1+(t)]dt≥-k2∫0T‍x~1(t)dt=k2∫0T‍∫0T‍G(t,s)c1-(s)dsdt≥k2Tmr1.
That is,
(57)∫0T‍[f1(t,1)+c1+(t)]dt≥k2Tmr1(2M2r2/m+1)μ1,
which contradicts with (H2).Case II.
One has{(u,v)∈Q×Q:∥u∥≤2M2r1/m,∥v∥=2M2r2/m}. Under this condition, we can get
(58)T2v≠λv,(u,v)∈∂Ω1,λ>1.
Otherwise, there exist (u0,v0)∈∂Ω1, λ0>1 such that T2v0=λ0v0. As
(59)v0(t)+x~2(t)=v0(t)-∫0T‍G(t,s)c2-(s)ds≥σ∥v0∥-Mr2>0,
then [v0(t)+x~2(t)]*=v0(t)+x~2(t). By a direct computation, we know v0=(1/λ0)T2v0 satisfies
(60)v0′′(t)+k2v0(t)=1λ0{k2(v0(t)+x~2(t))mm,,,m+f2(t,[u0(t)+x~1(t)]*)mm,,,m+c2+(t)k2(v0(t)+x~2(t))},v0(0)=v0(T),v0′(0)=v0′(T).
For[u0(t)+x~1(t)]*≤u0(t)<2M2r1/m+1, using the same method as condition I, we obtain
(61)∫0T‍[f2(t,1)+c2+(t)]dt≥k2Tmr2(2M2r1/m+1)λ1,
which is also a contradiction.(2) Choose an interval[α,β]⊂(0,T) satisfying β-α=T/2. Set M~>4M/Tm2. From Lemma 5, there exists R1>2M2r/m, r=max{r1,r2} such that
(62)fi(t,u)≥M~u,u≥R1,t∈[α,β],i=1,2.LetR≥(2M/m)R1>R1>2M2r/m. Define Ω2={(u,v)∈Q×Q:∥u∥<R,∥v∥<R}. We can get
(63)T(u,v)≠λ(u,v),(u,v)∈∂Ω2,0<λ<1.For(u,v)∈∂Ω2, we have the following two cases.Case I.
One has{(u,v)∈Q×Q:∥u∥=R,∥v∥≤R}. Under this condition, we know
(64)u(t)+x~1(t)≥σ∥u∥-Mr1≥σR-Mr>m2MR(>R1>0).
Thus [u(t)+x~1(t)]*=u(t)+x~1(t). Furthermore, we obtain from (62) that
(65)T2v(t)=∫0T‍G(t,s){k2[v(s)+x~2(s)]*mm11+f2(s,u(s)+x~1(s))mmmmmm+c2+(s)k2[v(s)+x~2(s)]*}ds≥∫αβ‍G(t,s)f2(s,u(s)+x~1(s))ds≥M~TRm24M.
On the other hand, for 0<λ<1, we know λv(t)<v(t)≤R. From the choice of M~, we get T2v≠λv.Case II.
One has{(u,v)∈Q×Q:∥u∥≤R,∥v∥=R}. Under this condition, we get
(66)v(t)+x~2(t)≥σ∥v∥-Mr2≥σR-Mr>m2MR(>R1>0).
Thus [v(t)+x~2(t)]*=v(t)+x~2(t). From (62), we get
(67)T1u(t)=∫0T‍G(t,s){k2[u(s)+x~1(s)]*mmm111+f1(s,v(s)+x~2(s))mmmmmm,,+c1+(s)k2[u(s)+x~1(s)]*}ds≥∫αβ‍G(t,s)f1(s,u(s)+x~1(s))ds≥M~TRm24M.
For 0<λ<1, λu(t)<u(t)≤R, from the choice of M~, we know T1u≠λu.
Furthermore, we can obtain(68)∥T(u,v)∥0=∥T1u∥+∥T2v∥≥M~TRm24M,(u,v)∈∂Ω2.
It implies inf(u,v)∈∂Ω2∥T(u,v)∥0>0.
From Lemma4, we know T has a fixed point (u~,v~) in Ω2¯/Ω1. For (u~,v~)∈Ω2¯/Ω1, we have the following three cases.Case 1.
One has(69)(u~,v~)∈{Q×Q:2M2rm≤∥u~∥≤R,∥v~∥<2M2rm}(u~,v~)∶=S1.Case 2.
One has(70)(u~,v~)∈{Q×Q:∥u~∥<2M2rm,2M2rm≤∥v~∥≤R}(u~,v~)∶=S2.Case 3.
One has(71)(u~,v~)∈{Q×Q:2M2rm≤∥u~∥≤R,2M2rm≤∥v~∥≤R}(u~,v~)∶=S3.
Next, we show Cases1 and 2 are impossible. In Case 1, we have
(72)u~(t)+x~1(t)=u~(t)-∫0T‍G(t,s)c1-(s)ds≥σ∥u~∥-Mr1>0.
It follows that [u~(t)+x~1(t)]*=u~(t)+x~1(t). By a direct computation, we know u~=T1u~ satisfies
(73)u~′′(t)+k2u~(t)=k2(u~(t)+x~1(t))+f1(t,[v~(t)+x~2(t)]*)+c1+(t),u~(0)=u~(T),u~′(0)=u~′(T).
Then, we get
(74)u~′′(t)=k2x~1(t)+f1(t,[v~(t)+x~2(t)]*)+c1+(t).
Since [v~(t)+x~2(t)]*≤v~(t)<2M2r2/m+1, integrating both sides of (74) on [0,T], we get
(75)0=k2∫0T‍x~1(t)dt+∫0T‍{f1(t,[v~(t)+x~2(t)]*)+c1+(t)}dt≤k2∫0T‍x~1(t)dt+(2M2r2m+1)μ1×∫0T‍{f1(t,1)+c1+(t)}dt.
Then
(76)(2M2r2m+1)μ1∫0T‍[f1(t,1)+c1+(t)]dt≥-k2∫0T‍x~1(t)dt=k2∫0T‍∫0T‍G(t,s)c1-(s)dsdt≥k2Tmr1.
That is,
(77)∫0T‍[f1(t,1)+c1+(t)]dt≥k2Tmr1(2M2r2/m+1)μ1,
which contradicts with (H2). Using the same method, we can prove that Case 2 is also impossible. Therefore, Case 3 is satisfied and T has a fixed point (u~,v~) in Ω2¯/Ω1 satisfying
(78)2M2rm≤∥u~∥≤R,2M2rm≤∥v~∥≤R.
Since
(79)u~(t)+x~1(t)≥σ∥u~∥-Mr1>0,v~(t)+x~2(t)≥σ∥v~∥-Mr2>0,
from Lemma 6, we know u~(t),v~(t) satisfy
(80)u~′′(t)+k2u~(t)=k2(u~(t)+x~1(t))+f1(t,v~(t)+x~2(t))+c1+(t),v~′′(t)+k2v~(t)=k2(v~(t)+x~2(t))+f2(t,u~(t)+x~1(t))+c2+(t),u~(0)=u~(T),u~′(0)=u~′(T),v~(0)=v~(T),v~′(0)=v~′(T).
Let(u*(t),v*(t))=(u~(t)+x~1(t),v~(t)+x~2(t)). For
(81)c1(t)=c1+(t)-c1-(t),c2(t)=c2+(t)-c2-(t),
we obtain
(82)u*′′(t)=f(t,v*(t))+c1(t),t∈(0,T),v*′′(t)=f(t,u*(t))+c2(t),t∈(0,T),u*(0)=u*(T),u*′(0)=u*′(T),v*(0)=v*(T),v*′(0)=v*′(T).
This means(u*,v*) is one positive T-periodic solution of (1).


## 4. Applications of Theorem3

Finally, we give some examples as the applications of Theorem3:
(83)(e1){u′′(t)=k2Tmv3/2(2M2T/m+1)2t-2t,v′′(t)=k2Tmu3/2(2M2T/m+1)2t-2t,u(0)=u(T),u′(0)=u′(T),v(0)=v(T),v′(0)=v′(T).(e2){u′′(t)=k2Tmv3/2(2M2T/m+1)2T-t-2t,v′′(t)=k2Tmu3/2(2M2T/m+1)2T-t-2t,u(0)=u(T),u′(0)=u′(T),v(0)=v(T),v′(0)=v′(T).Choosingf1(t,v)=k2Tmv3/2/(2M2T/m+1)2t, f2(t,u)=k2Tmu3/2/(2M2T/m+1)2T-t, ci-(t)=2/t, ci+(t)≡0, ri=∫0T‍ci-(t)dt4T, i=1,2, and μ1=2>μ2=5/4>1, λ1=2>λ2=5/4>1, then (H1) is satisfied. Notice (H2) also holds, since
(84)∫0T‍(fi(t,1)+ci+(t))dt=2k2TTm(2M2T/m+1)2,i=1,2.Existence of the positiveT-periodic solutions is guaranteed from Theorem 3. We can also consider the following examples and the same result can be obtained:
(85)(e3){u′′(t)=k2Tmv3/2(2M2T/m+1)2T-t-2t,v′′(t)=k2Tmu3/2(2M2T/m+1)2t-2T-t,u(0)=u(T),u′(0)=u′(T),v(0)=v(T),v′(0)=v′(T),(86)(e4){u′′(t)=k2Tmv3/2(2M2T/m+1)2T-t-2T-t,v′′(t)=k2Tmu3/2(2M2T/m+1)2T-t-2T-t,u(0)=u(T),u′(0)=u′(T),v(0)=v(T),v′(0)=v′(T).



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*Source: 504573-2013-07-16.xml*