# A Note on Periodic Solutions of Second Order Nonautonomous Singular Coupled Systems

Authors: Zhongwei Cao; Chengjun Yuan; Daqing Jiang; Xiaowei Wang
Journal: Mathematical Problems in Engineering
 (2010)

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## Abstract

We establish the existence of periodic solutions
of the second order nonautonomous singular coupled systemsx′′+a1(t)x=f1(t,y(t))+e1(t) for a.e. t∈[0,T], y′′+a2(t)y=f2(t,x(t))+e2(t) for a.e. t∈[0,T]. The proof relies on Schauder's fixed point theorem.


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## Body

## 1. Introduction

Some classical tools have been used in the literature to study the positive solutions for two-point boundary value problems of a coupled system of differential equations. These classical tools include some fixed point theorems in cones for completely continuous operators and Leray-Schauder fixed point theorem; for examples, see [1–3] and literatures therein.Recently, Schauder's fixed point theorem has been used to study the existence of positive solutions of periodic boundary value problems in several papers; see, for example, Torres [4], Chu et al. [5, 6], Cao and Jiang [7], and the references contained therein. However, there are few works on periodic solutions of second-order nonautonomous singular coupled systems. In these papers above, there are the major assumption that their associated Green's functions are positive. Since Green's functions are positive, in the paper, we continue to study the existence of periodic solutions to second-order nonautonomous singular coupled systems in the following form: (1.1)x′′+a1(t)x=f1(t,y(t))+e1(t)fora.e.t∈[0,T],y′′+a2(t)y=f2(t,x(t))+e2(t)fora.e.t∈[0,T],
with a1,a2,e1,e2∈L1[0,T], f1,f2∈Car([0,T]×(0,+∞),(0,+∞)). Here we write f∈Car([0,T]×(0,+∞),(0,+∞)) if f:[0,T]×(0,+∞)→(0,+∞) is an L1-caratheodory function, that is, the map x↦f(t,x) is continuous for a.e. t∈(0,1) and the map t↦f(t,x) is measurable for all x∈(0,+∞), and for every 0<r<s there exists hr,s∈L1(0,T) such that |f(t,x)|≤hr,s(t) for all x∈[r,s] and a.e. t∈[0,T]; here “for a.e." means “for almost every".This paper is mainly motivated by the recent papers [4–6, 8, 9], in which the periodic singular problems have been studied. Some results in [4–6, 9] prove that in some situations weak singularities may help create periodic solutions. In [6], the authors consider the periodic solutions of second-order nonautonomous singular dynamical systems, in which the scalar periodic singular problems have been studied by Leray-Schauder alternative principle, a well-known fixed point theorem in cones, and Schauder's fixed point theorem, respectively.The remaining part of the paper is organized as follows. In Section2, some preliminary results will be given. In Sections 3–5, by employing a basic application of Schauder's fixed point theorem, we state and prove the existence results for (1.1) under the nonnegative of the Green's function associated with (2.1)-(2.2). Our view point sheds some new light on problems with weak force potentials and proves that in some situations weak singularities may stimulate the existence of periodic solutions, just as pointed out in [9] for the scalar case.To illustrate our results, for example, we can select the system(1.2)x′′+a1(t)x=y-α1+e1(t),y′′+a2(t)y=x-α2+e2(t),
with a1,a2,e1,e2∈ℂ[0,T], 0<αi<1, i=1,2. Here we emphasize that in the new results e1,e2 do not need to be positive.Let us fix some notation to be used in the following: givena∈L1(0,1), we write a≻0 if a≥0 for a.e. t∈[0,1] and it is positive in a set of positive measures. For a given function p∈L1[0,T], we denote the essential supremum and infimum by p* and p*, if they exist. The usual Lp-norm is denoted by ∥·∥p. The conjugate exponent of p is denoted by p̃:1/p+1/p̃=1.


## 2. Preliminaries

We consider the scalar equation(2.1)x′′+ai(t)x=ei(t),i=1,2,
with periodic boundary conditions  (2.2)x(0)=x(T),x′(0)=x′(T).
In this paper, we assume that the following standing hypothesis is satisfied.  (H1)
The Green functionGi(t,s), associated with  (2.1)-(2.2), is nonnegative for all (t,s)∈[0,T]×[0,T], i=1,2.In other words, the (strict) antimaximum principle holds for  (2.1)-(2.2). Under the conditions (H1), the solution of  (2.1)-(2.2) is given by(2.3)x(t)=∫0TGi(t,s)ei(s)ds.For a nonconstant functiona(t), there is an Lp-criterion proved in [9], which is given in the following lemma for the sake of completeness. Let K(q) denote the best Sobolev constant in the following inequality:(2.4)C∥u∥q2≤∥u′∥22,∀u∈H01(0,T).
The explicit formula for K(q) is(2.5)K(q)={2πqT1+2/q(22+q)1-2/q(Γ(1/q)Γ(1/2+1/q))2if1≤q<∞,4Tifq=∞,
where Γ is the Gamma function. See [10].Lemma 2.1.
For eachi=1,2, assume that ai(t)≻0 and ai∈Lp[0,T] for some 1≤p≤∞. If
(2.6)∥ai∥p≤K(2p̃),
then the standing hypothesis (H1) holds.We define the functionγi:ℝ→ℝ by(2.7)γi(t)=∫0TGi(t,s)ei(s)ds,i=1,2,
which is the unique T-periodic solution of(2.8)x′′+ai(t)x=ei(t).
Throughout this paper, we use the following notations:(2.9)γi*=mini,tγi(t),γi*=maxi,tγi(t).


## 3. The Caseγ1*≥0,γ2*≥0

Theorem 3.1.
Assume that(H1) is satisfied; furthermore, we assume that there exist bi≻0, b̂i≻0, and 0<αi<1 such that (H2)
(3.1)0≤b̂i(t)xαi≤fi(t,x)≤bi(t)xαi,∀x>0,a.e.t∈(0,T),i=1,2.If γ1*≥0, γ2*≥0, then there exists a positive T-periodic solution of (1.1).Proof.
AT-periodic solution of (1.1) is just a fixed point of the completely continuous map A(x,y)=(Ax,Ay):CT×CT→CT×CT defined as
(3.2)(Ax)(t):=∫0TG1(t,s)[f1(s,y(s))+e1(s)]ds=∫0TG1(t,s)f1(s,y(s))ds+γ1(t),(Ay)(t):=∫0TG2(t,s)[f2(s,x(s))+e2(s)]ds=∫0TG2(t,s)f2(s,x(s))ds+γ2(t).
By a direct application of Schauder's fixed point theorem, the proof is finished if we prove thatA maps the closed convex set defined as
(3.3)K={(x,y)∈CT×CT:r1≤x(t)≤R1,r2≤y(t)≤R2,∀t∈[0,T]},
into itself, where R1>r1>0, R2>r2>0 are positive constants to be fixed properly. For convenience, we introduce the following notations:
(3.4)βi(t)=∫0TGi(t,s)bi(s)ds,β̂i(t)=∫0TGi(t,s)b̂i(s)ds,i=1,2.
Given (x,y)∈K, by the nonnegative sign of Gi and fi, i=1,2, we have
(3.5)(Ax)(t)=∫0TG1(t,s)f1(s,y(s))ds+γ1(t)≥∫0TG1(t,s)b̂1(s)yα1(s)ds≥∫0TG1(t,s)b̂1(s)R2α1ds≥β̂1*·1R2α1,
and note for every (x,y)∈K that
(3.6)(Ax)(t)=∫0TG1(t,s)f1(s,y(s))ds+γ1(t)≤∫0TG1(t,s)b1(s)yα1(s)ds+γ1*≤∫0TG1(t,s)b1(s)r2α1ds+γ1*≤β1*·1r2α1+γ1*.
Also, follow the same strategy,(3.7)(Ay)(t)=∫0TG2(t,s)f2(s,x(s))ds+γ2(t)≥∫0TG2(t,s)b̂2(s)xα2(s)ds≥∫0TG2(t,s)b̂2(s)R1α2ds≥β̂2*·1R1α2,(Ay)(t)=∫0TG2(t,s)f2(s,x(s))ds+γ2(t)≤∫0TG2(t,s)b2(s)xα2(s)ds+γ2*≤∫0TG2(t,s)b2(s)r1α2ds+γ2*≤β2*·1r1α2+γ2*.
Thus (Ax,Ay)∈K if r1,r2,R1, and R2 are chosen so that
(3.8)β̂1*·1R2α1≥r1,β1*·1r2α1+γ1*≤R1,β̂2*·1R1α2≥r2,β2*·1r1α2+γ2*≤R2.
Note that β̂i*,βi*>0 and taking R=R1=R2, r=r1=r2, r=1/R, it is sufficient to find R>1 such that
(3.9)β̂1*·R1-α1≥1,β1*·Rα1+γ1*≤R,β̂2*·R1-α2≥1,β2*·Rα2+γ2*≤R,
and these inequalities hold for R being big enough because αi<1.


## 4. The Caseγ1*<0<γ1*,γ2*<0<γ2*

Theorem 4.1.
Assume(H1) and (H2) are satisfied. If γ1*<0<γ1*, γ2*<0<γ2*, and
(4.1)γ1*≥r10-β̂1*·r10α1α2(β2*+γ2*r10α2)α1,γ2*≥r20-β̂2*·r20α1α2(β1*+γ1*r20α1)α2,
where 0<r10<+∞ is a unique positive solution of the equation
(4.2)r11-α1α2(β2*+γ2*·r1α2)1+α1=α1α2β2*β̂1*,
and 0<r20<+∞ is a unique positive solution of the equation
(4.3)r21-α1α2(β1*+γ1*·r2α1)1+α2=α1α2β1*β̂2*,
then there exists a positive T-periodic solution of (1.1).Proof.
We follow the same strategy and notation as in the proof of ahead theorem. In this case, to prove thatA:K→K, it is sufficient to find r1<R1, r2<R2 such that
(4.4)β̂1*R2α1+γ1*≥r1,β1*r2α1+γ1*≤R1,(4.5)β̂2*R1α2+γ2*≥r2,β2*r1α2+γ2*≤R2.
If we fixR1=β1*/r2α1+γ1*, R2=β2*/r1α2+γ2*, then the first inequality of (4.5) holds if r2 satisfies
(4.6)γ2*≥g(r2)≔r2-β̂2*·r2α1α2(β1*+γ1*·r2α1)α2.
According to
(4.7)g′(r2)=1-β̂2*·1(β1*+γ1*·r2α1)2α2·[α1α2r2α1α2-1(β1*+γ1*·r2α1)α2-r2α1α2α2(β1*+γ1*·r2α1)α2-1α1γ1*r2α1-1]=1-β̂2*α1α2r2α1α2-1(β1*+γ1*·r2α1)α2[1-r2α1γ1*β1*+γ1*·r2α1]=1-α1α2β1*β̂2*r2α1α2-1(β1*+γ1*·r2α1)-1-α2,
we have g′(0)=-∞, g′(+∞)=1; then there exists r20 such that g'(r20)=0, and
(4.8)g′′(r2)=-[α1α2β1*β̂2*(α1α2-1)r2α1α2-2(β1*+γ1*·r2α1)-1-α2+α1α2β1*β̂2*r2α1α2-1(-1-α2)(β1*+γ1*·r2α1)-2-α2γ1*α1r2α1-1]>0.
Then the function g(r2) possesses a minimum at r20, that is, g(r20)=minr2ϵ(0,+∞)g(r2). Note g′(r20)=0; then we have
(4.9)1-α1α2β1*β̂2*r21α1α2-1(β1*+γ1*·r21α1)-1-α2=0,
or equivalently,
(4.10)r201-α1α2(β1*+γ1*·r20α1)1+α2=α1α2β1*β̂2*.
Similarly,
(4.11)γ1*≥g(r1):=r1-β̂1*·r1α1α2(β2*+γ2*·r1α2)α1.g(r10)=minr1ϵ(0,+∞)g(r1), and
(4.12)r101-α1α2(β2*+γ2*·r10α2)1+α1=α1α2β2*β̂1*.
Takingr1=r10 and r2=r20, then the first inequality in (4.4) and (4.5) holds if γ1*≥g(r10), γ2*≥g(r20), which are just condition (4.1). The second inequalities hold directly by the choice of R1 and R2, and it would remain to prove that r10<R1 and r20<R2. This is easily verified through elementary computations
(4.13)R1=β1*r20α1+γ1*=β1*+γ1*·r20α1r20α1=(α1α2β1*β̂2*)1/(1+α2)·r20(α1α2-1)/(1+α2)r20α1=(α1α2β1*β̂2*)1/(1+α2)·r20-(1+α1)/(1+α2).
The proof is the same as that in R1,R2=(α1α2β2*β̂1*)1/(1+α1)·r10-(1+α2)/(1+α1).
Next,we will prover10<R1, r20<R2, or equivalently,
(4.14)r10r20(1+α1)/(1+α2)<(α1α2β1*β̂2*)1/(1+α2),r20r10(1+α2)/(1+α1)<(α1α2β2*β̂1*)1/(1+α1).
Namely,
(4.15)r101+α2r201+α1<α1α2β1*β̂2*,r201+α1r101+α2<α1α2β2*β̂1*.
On the other hand,
(4.16)r201-α1α2(β1*)1+α2≤α1α2β1*β̂2*.
Then
(4.17)r20≤(α1α2(β1*)-α2β̂2*)1/(1-α1α2).
Similarly,
(4.18)r10≤(α1α2(β2*)-α1β̂1*)1/(1-α1α2).
By (4.17) and (4.18),
(4.19)r101+α2r201+α1≤(α1α2(β2*)-α1β̂1*)(1+α2)/(1-α1α2)×(α1α2(β1*)-α2β̂2*)(1+α1)/(1-α1α2).
Now if we can prove
(4.20)(α1α2(β2*)-α1β̂1*)(1+α2)/(1-α1α2)(α1α2(β1*)-α2β̂2*)(1+α1)/(1-α1α2)<α1α2β1*β̂2*,
then
(4.21)r101+α2r201+α1<α1α2β1*β̂2*.
In fact,
(4.22)(α1α2)(2+α2+α1-1)/(1-α1α2)·(β̂1*β1*)(1+α2)/(1-α1α2)·(β̂2*β2*)α1(1+α2)/(1-α1α2)<1,
since β̂i*≤βi*, i=1,2. Similarly, we have r201+α1r101+α2<α1α2β2*β̂1*; we omit the details. Now we can obtain r10<R1, r20<R2. The proof is complete.


## 5. The Caseγ1*≤0,γ2*<0<γ2*  (γ2*≤0,γ1*<0<γ1*)

Theorem 5.1.
Assume(H1) and (H2) are satisfied. If γ1*≤0, γ2*<0<γ2*, and
(5.1)γ2*≥(1-1α1α2)[α1α2β̂2*(β1*)α2]1/(1-α1α2),γ1*≥r11-β̂1*·r11α1α2(β2*+γ2*r11α2)α1,
where 0<r11<+∞ is a unique positive solution of the equation
(5.2)r11-α1α2(β2*+γ2*·r1α2)1+α1=α1α2β2*β̂1*,
then there exists a positive T-periodic solution of (1.1).Proof.
In this case, to prove thatA:K→K, it is sufficient to find r1<R1,r2<R2 such that
(5.3)β̂1*R2α1+γ1*≥r1,β1*r2α1≤R1,β̂2*R1α2+γ2*≥r2,β2*r1α2+γ2*≤R2.
If we fixR1=β1*/r2α1, R2=β2*/r1α2+γ2*, then the first inequality of (6.4) holds if r2 satisfies
(5.4)γ2*≥r2-β̂2*R1α2=r2-β̂2*(β1*)α2·r2α1α2,
or equivalently
(5.5)γ2*≥f(r2):=r2-β̂2*(β1*)α2·r2α1α2.
Then the function f(r2) possesses a minimum at
(5.6)r21=[α1α2·β̂2*(β1*)α2]1/(1-α1α2),
that is, f(r21)=minr2ϵ(0,+∞)f(r2).
On the analogy of (5.4), we obtain
(5.7)γ1*≥r1-β̂1*·r1α1α2(β2*+γ2*r1α2)α1,
or equivalently,
(5.8)γ1*≥h(r1):=r1-β̂1*·r1α1α2(β2*+γ2*r1α2)α1.
According to
(5.9)h′(r1):=1-α1α2β2*β̂1*r1α1α2-1(β2*+γ2*r1α2)-1-α1,
we have h′(0)=-∞, h′(+∞)=1; then there exists r11 such that h'(r11)=0, and
(5.10)h′′(r1)=-[α1α2β2*β̂1*(α1α2-1)r1α1α2-2(β2*+γ2*·r1α2)-1-α1+α1α2β2*β̂1*r1α1α2-1(-1-α1)(β2*+γ2*·r1α2)-2-α1γ2*α2r1α2-1]>0.
Then the function h(r1) possesses a minimum at r11, that is, h(r11)=minr1ϵ(0,+∞)f(r1).
Noteh′(r11)=0; then we have
(5.11)1-α1α2β2*β̂1*r11α1α2-1(β2*+γ2*·r11α2)-1-α1=0.
Namely,
(5.12)r111-α1α2(β2*+γ2*·r11α2)1+α1=α1α2β2*β̂1*.
Taking r2=r21 and r1=r11, then the first inequality in (5.3) hold if γ2*≥h(r21) and γ1*≥h(r11) which are just condition (5.1). The second inequalities hold directly by the choice of R2 and R1, so it would remain to prove that R1=β1*/r21α1>r11, R2=β2*/r11α2+γ2*>r21. Now we turn to prove that R1>r11, R2>r21.
First,(5.13)R1=β1*r21α1=β1*{[α1α2·β̂2*/(β1*)α2]1/(1-α1α2)}α1=β1*[α1α2·β̂2*/(β1*)α2]α1/(1-α1α2)=(β1*)1+(α1α2)/(1-α1α2)(α1α2·β̂2*)α1/(1-α1α2)=(β1*)1/(1-α1α2)[(α1α2·β̂2*)α1]1/(1-α1α2)=[β1*(α1α2·β̂2*)α1]1/(1-α1α2)=[1(α1α2)α1·β1*(β̂2*)α1]1/(1-α1α2)>[α1α2·β̂1*(β2*)α1]1/(1-α1α2)=r11,
since β̂i*≤βi*, i=1,2.
On the other hand,(5.14)R2=β2*r11α2+γ2*=β2*+γ2*·r11α2r11α2.
By (5.2), we have
(5.15)β2*+γ2*·r11α2=(α1α2β2*β̂1*)1/(1+α1)r11(α1α2-1)/(1+α1).
Combing (5.14) and (5.15),
(5.16)R2=(α1α2β2*β̂1*)1/(1+α1)r11-(1+α2)/(1+α1).
In what follows, we will verify that R2>r21. In fact,
(5.17)(α1α2)(2+α2+α1)/(1-α1α2)-1·(β̂2*β2*)(1+α1)/(1-α1α2)·(β̂1*β1*)α2(1+α1)/(1-α1α2)<1,
since β̂i*≤βi*,i=1,2. Thus
(5.18)(α1α2β1*(-α2)β̂2*)(1+α1)/(1-α1α2)·(α1α2β2*(-α1)β̂1*)(1+α2)/(1-α1α2)<α1α2β2*β̂1*.
On the other hand,
(5.19)r211-α1α2β1*(1+α2)≤α1α2β1*β̂2*,r111-α1α2β2*(1+α1)≤α1α2β2*β̂1*.
Thus one can see easily that
(5.20)r21≤(α1α2β1*(-α2)β̂2*)1/(1-α1α2),r11≤(α1α2β2*(-α1)β̂1*)1/(1-α1α2).
From (5.20),
(5.21)r111+α2r211+α1≤(α1α2β2*(-α1)β̂1*)(1+α2)/(1-α1α2)(α1α2β1*(-α2)β̂2*)(1+α1)/(1-α1α2).
Combing (5.18) and (5.21),
(5.22)r111+α2r211+α1<α1α2β2*β̂1*.
Therefore,
(5.23)r21r11(1+α2)/(1+α1)<(α1α2β2*β̂1*)1/(1+α1).
Recall (5.16), we obtain r21<R2 immediately. The proof is complete.Similarly, we have the following theorem.Theorem 5.2.
Assume(H1) and (H2) are satisfied. If γ2*≤0, γ1*<0<γ1*, and
(5.24)γ1*≥(1-1α1α2)·[α1α2β̂1*(β2*)α1]1/(1-α1α2),γ2*≥r21-β̂2*·r21α1α2(β1*+γ1*r21α1)α2,
where 0<r21<+∞ is a unique positive solution of the equation
(5.25)r21-α1α2(β1*+γ1*·r2α1)1+α2=α1α2β1*β̂2*,
then there exists a positive T-periodic solution of (1.1).


## 6. The Caseγ1*≥0,γ2*<0<γ2*(γ2*≥0,γ1*<0<γ1*)

Theorem 6.1.
Assume(H1) and (H2) are satisfied. If γ1*≥0, γ2*<0<γ2*, and
(6.1)γ2*≥r22-β̂2*·r22α1α2(β1*+γ1*r22α1)α2,
where 0<r22<+∞ is a unique positive solution of the equation
(6.2)r21-α1α2(β1*+γ1*·r2α1)1+α2=α1α2β1*β̂2*,
then there exists a positive T-periodic solution of (1.1).Proof.
The following proof is the same as the proof of ahead theorem. In this case, to prove thatA:K→K, it is sufficient to find r1<R1, r2<R2 such that
(6.3)β̂1*R2α1≥r1,β1*r2α1+γ1*≤R1,(6.4)β̂2*R1α2+γ2*≥r2,β2*r1α2+γ2*≤R2.
If we fixR1=β1*/r2α1+γ1*, R2=β2*/r1α2+γ2*, then the first inequality of  (6.4) satisfies
(6.5)β̂2*·(β1*r2α1+γ1*)-α2+γ2*≥r2,
or equivalently
(6.6)γ2*≥l(r2):=r2-β̂2*(β1*+γ1*r2α1)α2·r2α1α2.
Then the function l(r2) possesses a minimum at r22, that is, l(r22)=minr2ϵ(0,+∞)l(r2).
Notel′(r22)=0; then we have
(6.7)1-α1α2β1*β̂2*r22α1α2-1(β1*+γ1*·r22α1)-1-α2=0.
Therefore,
(6.8)r221-α1α2(β1*+γ1*·r22α1)1+α2=α1α2β1*β̂2*.
Note that β̂i*,βi*>0, i=1,2. And taking r2=r22, R1=β1*/r22α1+γ1*, r1=1/R2, it is sufficient to find r1<R1, r2<R2 such that
(6.9)R2α1-1≤β1*,R2α2β2*+γ2*≤R2,
and these inequalities hold for R2 being big enough because αi<1. The proof is completed.Similarly, we have the following theorem.Theorem 6.2.
Assume(H1) and (H2) are satisfied. If γ2*≥0, γ1*<0<γ1*, and
(6.10)γ1*≥r12-β̂1*·r12α1α2(β2*+γ2*r12α2)α1,
where 0<r12<+∞ is a unique positive solution of the equation
(6.11)r11-α1α2(β2*+γ2*·r1α2)1+α1=α1α2β2*β̂1*,
then there exists a positive T-periodic solution of (1.1).



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