# MultiPoint BVPs for Second-Order Functional Differential Equations with Impulses

Authors: Xuxin Yang; Zhimin He; Jianhua Shen
Journal: Mathematical Problems in Engineering
 (2009)

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## Abstract

This paper is concerned about the existence of extreme solutions of multipoint boundary value problem for a class of second-order impulsive functional differential
equations. We introduce a new concept of lower and upper solutions. Then, by using the
method of upper and lower solutions introduced and monotone iterative technique, we obtain
the existence results of extreme solutions.


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## Body

## 1. Introduction

In this paper, we consider the multipoint boundary value problems for the impulsive functional differential equation:(1.1)-u′′(t)=f(t,u(t),u(θ(t))),t∈J=[0,1],t≠tk,Δu′(tk)=Ik(u(tk)),k=1,…,m,u(0)-au′(0)=cu(η),u(1)+bu′(1)=du(ξ),
where f∈C(J×R2,R),    0≤θ(t)≤t,t∈J,θ∈C(J),a≥0,b≥0,0≤c≤1,0≤d≤1, 0<η,ξ<1. 0<t1<t2<⋯<tm<1, f is continuous everywhere except at {tk}×R2; f(tk+,·,·), and f(tk-,·,·) exist with f(tk-,·,·)=f(tk,·,·); Ik∈C(R,R),Δu′(tk)=u′(tk+)-u′(tk-). Denote J-=J∖{ti,i=1,2,…,m}. Let PC(J,R)={u:J→R;u(t)|J- is continuous, u(tk+) and u(tk-) exist with u(tk-)=u(tk),k=1,2,…,m}; PC1(J,R)={u:J→R;u(t)|J- is continuous differentiable, u′(tk+) and u′(tk-) exist with u′(tk-)=u′(tk),k=1,2,…,m}. Let E=PC1(J,R)∩C2(J,R). A function u∈E is called a solution of BVP(1.1) if it satisfies (1.1).The method of upper and lower solutions combining monotone iterative technique offers an approach for obtaining approximate solutions of nonlinear differential equations [1–3]. There exist much literature devoted to the applications of this technique to general boundary value problems and periodic boundary value problems, for example, see [1, 4–6] for ordinary differential equations, [7–11] for functional differential equations, and [12] for differential equations with piecewise constant arguments. For the studies about some special boundary value problems, for example, Lidston boundary value problems and antiperiodic boundary value problems, one may see [13, 14] and the references cited therein.Here, we hope to mention some papers where existence results of solutions of certain boundary value problems of impulsive differential equations were studied [11, 15] and certain multipoint boundary value problems also were studied [6, 16–21]. These works motivate that we study the multipoint boundary value problems for the impulsive functional differential equation (1.1).We also note that whenIk=0 and θ(t)=t, the boundary value problem (1.1) reduces to multi-point boundary value problems for ordinary differential equations which have been studied in many papers, see, for example, [6, 16–18] and the references cited therein. To our knowledge, only a few papers paid attention to multi-point boundary value problems for impulsive functional differential equations.In this paper, we are concerned with the existence of extreme solutions for the boundary value problem (1.1). The paper is organized as follows. In Section 2, we establish two comparison principles. In Section 3, we consider a linear problem associated to (1.1) and then give a proof for the existence theorem. In Section 4, we first introduce a new concept of lower and upper solutions. By using the method of upper and lower solutions with a monotone iterative technique, we obtain the existence of extreme solutions for the boundary value problem (1.1).


## 2. Comparison Principles

In the following, we always assume that the following condition(H) is satisfied:(H)
a≥0,b≥0,0≤c≤1,0≤d≤1,0<η,ξ<1,a+c>0,b+d>0.For any given functiong∈E, we denote
(2.1)Ag=max{g(0)-ag′(0)-cg(η)aπ+csinπη,g(1)+bg′(1)-dg(ξ)bπ+dsinπξ},Bg=max{Ag,0},cg(t)=Bgsin(πt),r=π2.
We now present main results of this section.Theorem 2.1.
Assume thatu∈E satisfies
(2.2)-u′′(t)+Mu(t)+Nu(θ(t))≤0,t∈J,t≠tk,Δu′(tk)≥Lku(tk),k=1,…,m,u(0)-au′(0)≤cu(η),u(1)+bu′(1)≤du(ξ),
where a≥0,b≥0,   0≤c≤1,0≤d≤1,0<η,   ξ<1, Lk≥0 and constants M,N satisfy
(2.3)M>0,N≥0,M+N2+∑k=1mLk≤1.
Then u(t)≤0 for t∈J.Proof.
Suppose, to the contrary, thatu(t)>0 for some t∈J.
Ifu(1)=maxt∈Ju(t)>0, then u′(1)≥0, u(1)≥u(ξ), and
(2.4)du(ξ)≤u(1)≤u(1)+bu′(1)≤du(ξ).
So d=1 and u(ξ) is a maximum value.
Ifu(0)=maxt∈Ju(t)>0, then u′(0)≤0, u(0)≥u(η), and
(2.5)cu(η)≤u(0)≤u(0)-bu′(0)≤cu(η).
So c=1 and u(η) is a maximum value.
Therefore, there is aρ∈(0,1) such that
(2.6)u(ρ)=maxt∈Ju(t)>0,u′(ρ+)≤0.
Suppose thatu(t)≥0 for t∈J. From the first inequality of (2.2), we obtain that u′′(t)≥0 for t∈J. Hence
(2.7)u(0)=maxt∈Ju(t)oru(1)=maxt∈Ju(t).
Ifu′(0)≥0, then u′′(t)≥0, t∈(ti,ti+1], it is easy to obtain that u(t) is nondecreasing. Since u(1)≤du(ξ)≤u(1), it follows that u(t)≡K (K>0) for t∈[ξ,1]. From the first inequality of (2.2), we have that when t∈[ξ,1],(2.8)0<MK≤Mu(t)+Nu(θ(t))≤u′′(t)=0,
which is a contradiction.
Ifu′(0)≤0, then u(0)=maxt∈Ju(t)>0, or u(1)=maxt∈Ju(t)>0. If u(0)=maxt∈Ju(t)>0, then u(t)≡K (K>0) for t∈[0,η]. If u(1)=maxt∈Ju(t)>0, then u(t)≡K for t∈[ξ,1].
From the first inequality of (2.2), we have that when t∈[ξ,1],(2.9)0<MK≤Mu(t)+Nu(θ(t))≤u′′(t)=0,
which is a contradiction.
Suppose that there existt1,t2∈J such that u(t1)>0 and u(t2)<0. We consider two possible cases.Case 1 (u(0)>0).
Sinceu(t2)<0, there is κ>0,ε>0 such that u(κ)=0, u(t)≥0 for t∈[0,κ) and u(t)<0 for all t∈(κ,κ+ε]. It is easy to obtain that u′′(t)≥0 for t∈[0,κ]. If t*<κ, then 0<Mu(t*)≤u′′(t*)≤0, a contradiction. Hence t*>κ+ε. Let t*∈[0,t*) such that u(t*)=mint∈[0,t*)u(t), then u(t*)<0. From the first inequality of (2.2), we have
(2.10)u′′(t)≥(M+N)u(t*),t∈[0,t*),t≠tk,Δu′(tk)≥Lku(tk),k=1,…,m.
Integrating the above inequality from s(t*≤s≤t*) to t*, we obtain
(2.11)u′(t*)-u′(s)≥(t*-s)(M+N)u(t*)+∑s<tk<t*Lku(tk)≥(t*-s)(M+N)u(t*)+∑k=1mLku(t*).
Hence
(2.12)-u′(s)≥[(t*-s)(M+N)+∑k=1mLku(t*),t*≤s≤t*,
and then integrate from t* to t* to obtain
(2.13)-u(t*)<u(t*)-u(t*)≤∫t*t*(s-t*)(M+N)u(t*)ds-∑k=1mLku(t*)≤-(M+N2(t*-t*)2+∑k=1mLk)u(t*)≤-(M+N2+∑k=1mLk)u(t*).
From (2.3), we have that u(t*)>0. This is a contradiction.Case 2 (u(0)≤0).
Lett*∈[0,t*) such that u(t*)=mint∈[0,t*)u(t)≤0. From the first inequality of (2.2), we have
(2.14)u′′(t)≥(M+N)u(t*),t∈[0,t*),t≠tk,Δu′(tk)≥Lku(tk),k=1,…,m.
The rest proof is similar to that of Case 1. The proof is complete.Theorem 2.2.
Assume that(H) holds and  u∈E satisfies
(2.15)-u′′(t)+Mu(t)+Nu(θ(t))+[(M+r)cu(t)+Ncu(θ(t))]≤0,t∈J,t≠tk,Δu′(tk)≥Lku(tk)+Lkcu(tk),k=1,…,m,
where constants M,N satisfy (2.3), and Lk≥0, then u(t)≤0 for t∈J.Proof.
Assume thatu(0)-au′(0)≤cu(η),u(1)+bu′(1)≤du(ξ), then cu(t)≡0. By Theorem 2.1, u(t)≤0.
Assume thatu(0)-au′(0)≤cu(η),u(1)+bu′(1)>du(ξ), then
(2.16)cu(t)=sin(πt)bπ+dsin(πξ)(u(1)+bu′(1)-du(ξ)).
Puty(t)=u(t)+cu(t),t∈J, then y(t)≥u(t) for all t∈J, and
(2.17)y′(t)=u′(t)+πcos(πt)bπ+dsin(πξ)(u(1)+bu′(1)-du(ξ)),t∈J,y′′(t)=u′′(t)-rcu(t),t∈J.
Hence
(2.18)y(0)=u(0),y(1)=u(1),y(ξ)=u(ξ)+sin(πξ)bπ+dsin(πξ)(u(1)+bu′(1)-du(ξ)),y′(0)=u′(0)+πbπ+dsin(πξ)(u(1)+bu′(1)-du(ξ)),y′(1)=u′(1)-πbπ+dsin(πξ)(u(1)+bu′(1)-du(ξ)),-y′′(t)+My(t)+Ny(θ(t))=-u′′(t)+Mu(t)+Nu(θ(t))+[(M+r)cu(t)+Ncu(θ(t))]≤0,y(0)-ay′(0)=u(0)-au′(0)-aπbπ+dsin(πξ)(u(1)+bu′(1)-du(ξ))≤cu(η)≤cy(η),y(1)+by′(1)-dy(ξ)=u(1)+bu′(1)-du(ξ)-bπbπ+dsin(πξ)(u(1)+bu′(1)-du(ξ))-dsin(πξ)bπ+dsinπξ(u(1)+bu′(1)-du(ξ))≤0,Δy′(tk)=Δu′(tk)Δcu′(tk)≥Lku(tk)+Lkcu(tk)=Lky(tk).
By Theorem 2.1, y(t)≤0 for all t∈J, which implies that u(t)≤0 for t∈J.
Assume thatu(0)-au′(0)>cu(η),u(1)+bu′(1)≤du(ξ), then
(2.19)cu(t)=sinπtaπ+csin(πη)(u(0)-au′(0)-cu(η)).
Put y(t)=u(t)+cu(t),t∈J, then y(t)≥u(t) for all t∈J, and
(2.20)y′(t)=u′(t)+πcos(πt)aπ+csin(πη)(u(0)-au′(0)-cu(η)),t∈J,y′′(t)=u′′(t)-rcu(t),t∈J.
Hence
(2.21)y(0)=u(0),y(1)=u(1),y(η)=u(η)+sin(πη)aπ+csin(πη)(u(0)-au′(0)-cu(η)),y′(0)=u′(0)+πaπ+csin(πη)(u(0)-au′(0)-cu(η)),y′(1)=u′(1)-πaπ+csin(πη)(u(0)-au′(0)-cu(η)),-y′′(t)+My(t)+Ny(θ(t))=-u′′(t)+Mu(t)+Nu(θ(t))+[(M+r)cu(t)+Ncu(θ(t))]≤0,y(0)-ay′(0)-cy(η)=u(0)-au′(0)-cu(η)-aπaπ+csin(πη)(u(0)-au′(0)-cu(η))-csin(πη)aπ+csin(πη)(u(0)-au′(0)-cu(η))≤0,y(1)+by′(1)=u(1)+bu′(1)-bπaπ+csin(πη)(u(0)-au′(0)-cu(η))≤du(ξ)≤dy(ξ),Δy′(tk)=Δu′(tk)+Δcu′(tk)≥Lku(tk)+Lkcu(tk)=Lky(tk).
By Theorem2.1, y(t)≤0 for all t∈J, which implies that u(t)≤0 for t∈J.
Assume thatu(0)-au′(0)>cu(η),u(1)+bu′(1)>du(ξ), then cu(t)=Ausin(πt).
Puty(t)=u(t)+cu(t),t∈J, then y(t)≥u(t) for all t∈J, and
(2.22)y′(t)=u′(t)+Auπcos(πt),t∈J,y′′(t)=u′′(t)-rcu(t),t∈J.
Hence
(2.23)y(0)=u(0),y(1)=u(1),y(η)=u(η)+Ausin(πη),y(ξ)=u(ξ)+Ausin(πξ),y′(0)=u′(0)+Auπ,y′(1)=u′(1)-Auπ,-y′′(t)+My(t)+Ny(θ(t))=-u′′(t)+Mu(t)+Nu(θ(t))+[(M+r)cu(t)+Ncu(θ(t))]≤0,y(0)-ay′(0)-cy(η)=u(0)-au′(0)-cu(η)-aAuπ-cAusin(πη)≤0,y(1)+by′(1)-dy(ξ)=u(1)+bu′(1)-du(ξ)-bAuπ-dAusin(πξ)≤0,Δy′(tk)=Δu′(tk)+Δcu′(tk)≥Lku(tk)+Lkcu(tk)=Lky(tk).
By Theorem2.1, y(t)≤0 for all t∈J, which implies that u(t)≤0 for t∈J. The proof is complete.


## 3. Linear Problem

In this section, we consider the linear boundary value problem(3.1)-u′′(t)+Mu(t)+Nu(θ(t))=σ(t),t∈J,t≠tk,Δu′(tk)=Lku(tk)+ek,k=1,…,m,u(0)-au′(0)=cu(η),u(1)+bu′(1)=du(ξ).Theorem 3.1.
Assume that(H) holds,  σ∈C(J),  ek∈R,  and constants M,N  satisfy (2.3) with
(3.2)μ=(a(1+2b)2(a+b+1)+18(1+2ba+b+1)2)(M+N)+(1+(1+b)2a+b+1)∑k=1mLk<1.
Further suppose that there exist α,β∈E such that                (h1)
α≤β on J,(h2)
(3.3)-α′′(t)+Mα(t)+Nα(θ(t))+[(M+r)cα(t)+Ncα(θ(t))]≤σ(t),t∈J,t≠tk,Δα′(tk)≥Lkα(tk)+Lkcα(tk)+ek,k=1,…,m,(h3)
(3.4)-β′′(t)+Mβ(t)+Nβ(θ(t))-[(M+r)c-β(t)+Nc-β(θ(t))]≥σ(t),t∈J,t≠tk,Δβ′(tk)≤Lkβ(tk)-Lkc-β(tk)+ek,k=1,…,m.Then the boundary value problem (3.1) has one unique solution u(t) and α≤u≤β for t∈J.Proof.
We first show that the solution of (3.1) is unique. Let u1,u2 be the solution of (3.1) and set v=u1-u2. Thus,
(3.5)-v′′(t)+Mv(t)+Nv(θ(t))=0,t∈J,t≠tk,Δv′(tk)=Lkv(tk),k=1,…,m,v(0)-av′(0)=cv(η),v(1)+bv′(1)=dv(ξ).
By Theorem 2.1, we have that v≤0 for t∈J, that is, u1≤u2 on J. Similarly, one can obtain that u2≤u1 on J. Hence u1=u2.
Next, we prove that ifu is a solution of (3.1), then α≤u≤β. Let p=α-u. From boundary conditions, we have that cα(t)=cp(t) for all t∈J. From (h2) and (3.1), we have
(3.6)-p′′(t)+Mp(t)+Np(θ(t))+[(M+r)cp(t)+Ncp(θ(t))]≤0,t∈J,t≠tk,Δp′(tk)≥Lkp(tk)+Lkcp(tk),k=1,…,m.
By Theorem 2.1, we have that p=α-u≤0 on J. Analogously, u≤β on J.
Finally, we show that the boundary value problem (3.1) has a solution by five steps as follows.
Step 1.
Letα̅(t)=α(t)+cα(t),β̅(t)=β(t)-c-β(t). We claim that(1)
(3.7)-α̅′′(t)+Mα̅(t)+Nα̅(θ(t))+[(M+r)cα̅(t)+Ncα̅(θ(t))]≤σ(t)fort∈J,t≠tk,Δα̅′(tk)≥Lkα̅(tk)+ek,k=1,…,m,(2)
(3.8)-β̅′′(t)+Mβ̅(t)+Nβ̅(θ(t))-[(M+r)c-β̅(t)+Nc-β̅(θ(t))]≥σ(t)fort∈J,t≠tk,Δβ̅′(tk)≤Lkβ̅(tk)+ek,k=1,…,m,(3)
α(t)≤α̅(t)≤β̅(t)≤β(t) for t∈J.
From(h2) and (h3), we have
(3.9)-α̅′′(t)+Mα̅(t)+Nα̅(θ(t))≤σ(t),t∈J,t≠tk,Δα̅′(tk)≥Lkα̅(tk)+ek,k=1,…,m.(3.10)-β̅′′(t)+Mβ̅(t)+Nβ̅(θ(t))≥σ(t),t∈J,t≠tk,Δβ̅′(tk)≤Lkβ̅(tk)+ek,k=1,…,m,(3.11)α̅(0)-aα̅′(0)-cα̅(η)=α(0)-aα′(0)-cα(η)-(aπ+csin(πη))Bα≤0,(3.12)α̅(1)+bα̅′(1)-dα̅(ξ)=α(1)+bα′(0)-dα(ξ)-(bπ+dsin(πξ))Bα≤0,(3.13)-[β̅(0)-aβ̅′(0)-cβ̅(η)]=-β(0)+aβ′(0)+cβ(η)-(aπ+csin(πη))B-β≤0,(3.14)-[β̅(1)+bβ̅′(1)-dβ̅(ξ)]=-β(1)-bβ′(0)+dβ(ξ)-(bπ+dsin(πξ))B-β≤0.
From (3.9)–(3.14), we obtain that cα̅(t)=c-β̅(t)≡0,t∈J. Combining (3.9) and (3.10), we obtain that ( 1) and ( 2) hold.
It is easy to see thatα≤α̅,β̅≤β on J. We show that α̅≤β̅ on J. Let p=α̅-β̅, then p(t)=α(t)-β(t)+cα(t)+c-β(t). From (3.9)–(3.14), we have
(3.15)-p′′(t)+Mp(t)+Np(θ(t))≤0,t∈J,t≠tk,Δp′(tk)≥Lkp(tk),k=1,…,m,p(0)-ap′(0)-cp(η)=α(0)-aα′(0)-cα(η)-(aπ+csinI(πη))Bα-β(0)+aβ′(0)+cβ(η)-(aπ+csin(πη))B-β≤0,p(1)+bp′(1)-dp(ξ)=α(1)+bα′(1)-dα(ξ)-(bπ+dsin(πξ))Bα-β(1)-bβ′(1)+dβ(η)-(bπ+dsin(πξ))B-β≤0,Δp′(tk)=Δα′(tk)-Δβ′(tk)+Δca′(tk)+Δc-β′(tk)≥Lk(α(tk)-β(tk))+Lk(cα(tk)+c-β(tk))=Lkp(tk).
By Theorem 2.1, we have that p≤0 on J, that is, α̅≤β̅ on J. So ( 3) holds.Step 2.
Consider the boundary value problem(3.16)-u′′(t)+Mu(t)+Nu(θ(t))=σ(t),t∈J,t≠tk,Δu′(tk)=Lku(tk)+ek,k=1,…,m,u(0)-au′(0)=λ,u(1)+bu′(1)=δ,
where λ∈R,δ∈R. We show that the boundary value problem (3.16) has one unique solution u(t,λ,δ).
It is easy to check that the boundary value problem (3.16) is equivalent to the integral equation:
(3.17)u(t)=δt+(1-t)λ+bλ+aδa+b+1+∫01G(t,s)[σ(s)-Mu(s)-Nu(θ(s))]ds+∑0<tk<t(t-tk)[Lku(tk)+ek]-1a+b+1(t+b)∑k=1m[(1-tk)+b][Lku(tk)+ek],
where
(3.18)G(t,s)=1a+b+1{(a+t)(1+b-s),0≤t≤s≤1,(a+s)(1+b-t),0≤s≤t≤1.
It is easy to see thatPC(J,R) with norm ∥u∥=maxt∈J|u(t)| is a Banach space. Define a mapping Φ:PC(J,R)→PC(J,R) by
(3.19)(Φu)(t)=δt+(1-t)λ+bλ+aδa+b+1+∫01G(t,s)[σ(s)-Mu(s)-Nu(θ(s))]ds+∑0<tk<t(t-tk)[Lku(tk)+ek]-1a+b+1(t+b)∑k=1m[(1-tk)+b][Lku(tk)+ek].
For any x,y∈PC(J,R), we have
(3.20)|(Φx)(t)-(Φy)(t)|≤∫01G(t,s)[M(y(s)-x(s))+N(y(θ(s))-x(θ(s)))]ds+(1+(1+b)2a+b+1)∑k=1mLk∥x-y∥≤∫01G(t,s)ds∥x-y∥(M+N)+(1+(1+b)2a+b+1)∑k=1mLk∥x-y∥.
Since
(3.21)maxt∈J∫01G(t,s)ds=a(1+2b)2(a+b+1)+18(1+2ba+b+1)2,
the inequality (3.2) implies that Φ:PC(J)→PC(J) is a contraction mapping. Thus there exists a unique u∈PC(J) such that Φu=u. The boundary value problem (3.16) has a unique solution.Step 3.
We show that for anyt∈J, the unique solution u(t,λ,δ) of the boundary value problem (3.16) is continuous in λ and δ. Let u(t,λi,δi),i=1,2, be the solution of
(3.22)-u′′(t)+Mu(t)+Nu(θ(t))=σ(t),t∈J,t≠tk,Δu′(tk)=Lku(tk)+ek,k=1,…,m,u(0)-au′(0)=λi,u(1)+bu′(1)=δi,i=1,2.
Then
(3.23)u(t,λi,δi)=δit+(1-t)λi+bλi+aδia+b+1+∫01G(t,s)[σ(s)-Mu(s,λi,δi)-Nu(θ(s),λi,δi)]ds+∑0<tk<t(t-tk)[Lku(tk)+ek]-1a+b+1(t+b)×∑k=1m[(1-tk)+b][Lku(tk)+ek],i=1,2.
From (3.23), we have that
(3.24)∥u(t,λ1,δ1)-u(t,λ2,δ2)∥≤|λ1-λ2|+|δ1-δ2|+(M+N)∥u(t,λ1,δ1)-u(t,λ2,δ2)∥maxt∈J∫01G(t,s)ds+∥u(t,λ1,δ1)-u(t,λ2,δ2)∥(1+(1+b)2a+b+1)∑k=1mLk∥x-y∥≤|λ1-λ2|+|δ1-δ2|+μ∥u(t,λ1,δ1)-u(t,λ2,δ2)∥.
Hence
(3.25)∥u(t,λ1,δ1)-u(t,λ2,δ2)∥0≤11-μ(|λ1-λ2|+|δ1-δ2|).Step 4.
We show that(3.26)α̅(t)≤u(t,λ,δ)≤β̅(t)
for any t∈J, λ∈[cα̅(η),cβ̅(η)], and δ∈[dα̅(ξ),dβ̅(ξ)], where u(t,λ,δ) is unique solution of the boundary value problem (3.16).
Letm(t)=α̅(t)-u(t,λ,δ). From (3.9), (3.11), (3.12), and (3.16), we have that m(0)-am′(0)≤cm(η),m(1)+bm′(1)≤dm(ξ), and
(3.27)-m′′(t)+Mm(t)+Nm(θ(t))=-α̅′′(t)+Mα̅(t)+Nα̅(θ(t))+u′′(t,λ)-Mu(t,λ,δ)-Nu(θ(t),λ,δ)≤σ(t)-σ(t)≤0,Δm′(tk)≥Lku(tk).
By Theorem 2.1, we obtain that m≤0 on J, that is, α̅(t)≤u(t,λ,δ) on J. Similarly, u(t,λ,δ)≤β̅(t) on J.Step 5.
LetD=[cα̅(η),cβ̅(η)]×[dα̅(ξ),dβ̅(ξ)]. Define a mapping F:D→R2 by
(3.28)F(λ,δ)=(u(η,λ,δ),u(ξ,λ,δ)),
where u(t,λ,δ) is unique solution of the boundary value problem (3.16). From Step 4, we have
(3.29)F(D)⊂D.
Since D is a compact convex set and F is continuous, it follows by Schauder’s fixed point theorem that F has a fixed point (λ0,δ0)∈D such that
(3.30)u(η,λ0,δ0)=λ0,u(ξ,λ0,δ0)=δ0.
Obviously, u(t,λ0,δ0) is unique solution of the boundary value problem (3.1). This completes the proof.


## 4. Main Results

LetM∈R, N∈R. We first give the following definition.Definition 4.1.
A functionα∈E is called a lower solution of the boundary value problem (1.2) if
(4.1)-α′′(t)+(M+r)cα(t)+Ncα(θ(t))≤f(t,α(t),α(θ(t))),t∈J,t≠tk,Δα′(tk)≥Ik(α(tk))+Lkcα(tk),k=1,…,m.Definition 4.2.
A functionβ∈E is called an upper solution of the boundary value problem (1.2) if
(4.2)-β′′(t)-(M+r)c-β(t)-Nc-β(θ(t))≥f(t,β(t),β(θ(t)))t∈J,t∈J,t≠tk,Δβ′(tk)≤Ikβ(tk)-Lkc-β(tk),k=1,…,m.
Our main result is the following theorem.Theorem 4.3.
Assume that(H)  holds. If the following conditions are satisfied:                                                   (H1)
α,β are lower and upper solutions for boundary value problem (1.2) respectively, and α(t)≤β(t) on J,(H2)
the constantsM,N in definition of upper and lower solutions satisfy (2.3), (3.2), and
(4.3)f(t,x,y)-f(t,x̅,y̅)≥-M(x-x̅)-N(y-y̅),Ik(x)-Ik(y)≥Lk(x-y),x≤y,
for α(t)≤x̅≤x≤β(t),α(θ(t))≤y̅≤y≤β(θ(t)),t∈J.
Then, there exist monotone sequences{αn},{βn} with α0=α,β0=β such that limn→∞αn(t)=ρ(t),limn→∞βn(t)=ϱ(t) uniformly on J, and ρ,ϱ are the minimal and the maximal solutions of (1.2), respectively, such that
(4.4)α0≤α1≤α2≤⋯αn≤ρ≤x≤ϱ≤βn≤⋯≤β2≤β1≤β0
on J, where x is any solution of (1.1) such that α(t)≤x(t)≤β(t) on J.Proof.
Let[α,β]={u∈E:α(t)≤u(t)≤β(t),t∈J}. For any γ∈[α,β], we consider the boundary value problem
(4.5)-u′′(t)+Mu(t)+Nu(θ(t))=f(t,γ(t),γ(θ(t)))+Mγ(t)+Nγ(θ(t)),t∈J,Δu′(tk)=Ik(γ(tk))-Lk(u(tk)-γ(tk)),k=1,…,m.u(0)-ax′(0)=cu(η),u(1)+bu′(1)=du(ξ).
Sinceα is a lower solution of (1.2), from (H2), we have that
(4.6)-α′′(t)+Mα(t)+Nα(θ(t))≤f(t,α(t),α(θ(t)))+Mα(t)+Nα(θ(t))-(M+r)cα(t)-Ncα(θ(t))≤f(t,γ(t),γ(θ(t)))+Mγ(t)+Nγ(θ(t))-(M+r)cα(t)-Ncα(θ(t)),Δα′(tk)≥Ik(α(tk))+Lkcα(tk)≥Ik(γ(tk))+Lkα(tk)-Lkγ(tk)+Lkcα(tk).
Similarly, we have that
(4.7)-β′′(t)+Mβ(t)+Nβ(θ(t))≥f(t,γ(t),γ(θ(t)))+Mγ(t)+Nγ(θ(t))+(M+r)c-β(t)+Nc-β(θ(t)),Δβ′(tk)≤Ik(β(tk))-Lkc-β(tk)≤Ik(γ(tk))+Lkβ(tk)-Lkγ(tk)-Lkc-β(tk).
By Theorem3.1, the boundary value problem (4.5) has a unique solution u∈[α,β]. We define an operator Ψ by u=Ψγ, then Ψ is an operator from [α,β] to [α,β].
We will show that
(a)
α≤Ψα,Ψβ≤β,(b)
Ψ is nondecreasing in [α,β].
FromΨα∈[α,β] and Ψβ∈[α,β], we have that (a) holds. To prove (b), we show that Ψν1≤Ψν2 if α≤ν1≤ν2≤β.
Letν1*=Ψν1,ν2*=Ψν2 ,and p=ν1*-ν2*, then by (H2) and boundary conditions, we have that
(4.8)-p′′(t)+Mp(t)+Np(θ(t))=f(t,ν1(t),ν1(θ(t)))+Mν1(t)+Nν1(θ(t))-f(t,ν2(t),ν2(θ(t)))-Mν2(t)-Nν2(θ(t))≤0,Δp′(tk)≥Lkp(tk),p(0)-ap′(0)=cp(η),p(1)+pu′(1)=dp(ξ).
By Theorem 2.1, p(t)≤0 on J, which implies that Ψν1≤Ψν2.
Define the sequences{αn},{βn} with α0=α,β0=β such that αn+1=Ψαn,βn+1=Ψβn for n=0,1,2,… From (a) and (b), we have
(4.9)α0≤α1≤α2≤⋯≤αn≤βn≤⋯≤β2≤β1≤β0
on t∈J, and each αn,βn∈E satisfies
(4.10)-αn′′(t)+Mαn(t)+Nαn(θ(t))=f(t,αn-1(t),αn-1(θ(t)))+Mαn-1(t)+Nαn-1(θ(t)),t∈J,t≠tk,Δαn′(tk)=Ik(αn-1(tk))+Lk(αn(tk)-αn-1(tk)),k=1,2,…,m,αn(0)-aαn′(0)=cαn(η),αn(1)+bαn′(1)=dαn(ξ),-βn′′(t)+Mβn(t)+Nβn(θ(t))=f(t,βn-1(t),βn-1(θ(t)))+Mβn-1(t)+Nβn-1(θ(t)),t∈J,t≠tk,Δβn(tk)=Ik(βn-1(tk))+Lk(βn(tk)-βn-1(tk)),k=1,2,…,m,βn(0)-aβn′(0)=cβn(η),βn(1)+bβn′(1)=dβn(ξ).
Therefore, there exist ρ,ϱ such that such that limn→∞αn(t)=ρ(t), limn→∞βn(t)=ϱ(t) uniformly on J. Clearly, ρ,ϱ are solutions of (1.1).
Finally, we prove that ifx∈[α0,β0] is any solution of (1.1), then ρ(t)≤x(t)≤ϱ(t) on J. To this end, we assume, without loss of generality, that αn(t)≤x(t)≤βn(t) for some n. Since α0(t)≤x(t)≤β0(t), from property (b), we can obtain
(4.11)αn+1(t)≤x(t)≤βn+1(t),t∈J.
Hence, we can conclude that
(4.12)αn(t)≤x(t)≤βn(t),∀n.
Passing to the limit as n→∞, we obtain
(4.13)ρ(t)≤x(t)≤ϱ(t),t∈J.
This completes the proof.



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*Source: 258090-2009-08-12.xml*