MPEMathematical Problems in Engineering1563-51471024-123XHindawi Publishing Corporation25809010.1155/2009/258090258090Research ArticleMultiPoint BVPs for Second-Order Functional Differential Equations with ImpulsesYangXuxin1, 2HeZhimin3ShenJianhua4PereiraFernando Lobo1Department of MathematicsHunan Normal UniversityChangsha, Hunan 410081Chinahunnu.edu.cn2Department of MathematicsHunan First Normal UniversityChangsha, Hunan 410205Chinahnfnu.edu.cn3College of ScienceZhejiang Forestry UniversityHangzhou, Zhejiang 311300Chinazjfc.edu.cn4Department of MathematicsHangzhou Normal UniversityHangzhou, Zhejiang 310036Chinahztc.edu.cn200919072009200914042009100620092009Copyright © 2009This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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.

1. Introduction

In this paper, we consider the multipoint boundary value problems for the impulsive functional differential equation: -u′′(t)=f(t,u(t),u(θ(t))),tJ=[0,1],ttk,Δu(tk)=Ik(u(tk)),k=1,,m,u(0)-au(0)=cu(η),u(1)+bu(1)=du(ξ), where fC(J×R2,R),    0θ(t)t,tJ,  θC(J),a0,b0,  0c1,  0d1, 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,·,·); IkC(R,R),Δu(tk)=u(tk+)-u(tk-). Denote J-=J{ti,i=1,2,,m}. Let PC(J,R)={u:JR;u(t)|J- is continuous, u(tk+) and u(tk-) exist with u(tk-)=u(tk),k=1,2,,m}; PC1(J,R)={u:JR;  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 uE 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 . 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, 46] for ordinary differential equations,  for functional differential equations, and  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, 1621]. These works motivate that we study the multipoint boundary value problems for the impulsive functional differential equation (1.1).

We also note that when Ik=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, 1618] 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:

a0,  b0,  0c1,0d1,  0<η,ξ<1,  a+c>0,  b+d>0.

For any given function gE, we denote 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 that  uE satisfies -u′′(t)+Mu(t)+Nu(θ(t))0,tJ,ttk,Δu(tk)Lku(tk),k=1,,m,u(0)-au(0)cu(η),u(1)+bu(1)du(ξ), where a0,b0, 0c1,  0d1,0<η,  ξ<1, Lk0 and constants M,  N satisfy M>0,N0,M+N2+k=1mLk1. Then u(t)0 for tJ.

Proof.

Suppose, to the contrary, that u(t)>0 for some tJ.

If u(1)=maxtJu(t)>0, then u(1)0, u(1)u(ξ), and du(ξ)u(1)u(1)+bu(1)du(ξ). So d=1 and u(ξ) is a maximum value.

If u(0)=maxtJu(t)>0, then u(0)0, u(0)u(η), and cu(η)u(0)u(0)-bu(0)cu(η). So c=1 and u(η) is a maximum value.

Therefore, there is a ρ(0,1) such that u(ρ)=maxtJ  u(t)>0,u(ρ+)0.

Suppose that u(t)0 for tJ. From the first inequality of (2.2), we obtain that u(t)0 for tJ. Hence u(0)=maxtJu(t)oru(1)=maxtJu(t).

If u(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],0<MKMu(t)+Nu(θ(t))u′′(t)=0, which is a contradiction.

If u(0)0, then u(0)=maxtJu(t)>0, or u(1)=maxtJu(t)>0. If u(0)=maxtJu(t)>0, then u(t)K (K>0) for t[0,η]. If u(1)=maxtJu(t)>0, then u(t)K for t[ξ,1].

From the first inequality of (2.2), we have that when t[ξ,1],0<MKMu(t)+Nu(θ(t))u′′(t)=0, which is a contradiction.

Suppose that there exist t1,  t2J such that u(t1)>0 and u(t2)<0. We consider two possible cases.Case 1 (<inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M88"><mml:mi>u</mml:mi><mml:mrow><mml:mo stretchy="false">(</mml:mo><mml:mrow><mml:mn>0</mml:mn></mml:mrow><mml:mo stretchy="false">)</mml:mo></mml:mrow><mml:mo>></mml:mo><mml:mn>0</mml:mn></mml:math></inline-formula>).

Since u(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 u(t)(M+N)u(t*),t[0,t*),  ttk,Δu(tk)Lku(tk),k=1,,m. Integrating the above inequality from s(t*st*) to t*, we obtain 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 -u(s)[(t*-s)(M+N)+k=1mLku(t*),t*st*, and then integrate from t* to t* to obtain -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 (<inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M113"><mml:mi>u</mml:mi><mml:mrow><mml:mo stretchy="false">(</mml:mo><mml:mrow><mml:mn>0</mml:mn></mml:mrow><mml:mo stretchy="false">)</mml:mo></mml:mrow><mml:mo>≤</mml:mo><mml:mn>0</mml:mn></mml:math></inline-formula>).

Let t*[0,t*) such that u(t*)=mint[0,t*)u(t)0. From the first inequality of (2.2), we have u(t)(M+N)u(t*),t[0,t*),ttk,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 uE satisfies -u(t)+Mu(t)+Nu(θ(t))+[(M+r)cu(t)+Ncu(θ(t))]0,tJ,ttk,Δu(tk)Lku(tk)+Lkcu(tk),k=1,,m, where constants M,  N satisfy (2.3), and Lk0, then u(t)0 for tJ.

Proof.

Assume that u(0)-au(0)cu(η),u(1)+bu(1)du(ξ), then cu(t)0. By Theorem 2.1, u(t)0.

Assume that u(0)-au(0)cu(η),u(1)+bu(1)>du(ξ), then

cu(t)=sin(πt)bπ+dsin(πξ)(u(1)+bu(1)-du(ξ)).

Put y(t)=u(t)+cu(t),  tJ, then y(t)u(t) for all tJ, and

y(t)=u(t)+πcos(πt)bπ+dsin(πξ)(u(1)+bu(1)-du(ξ)),tJ,y′′(t)=u′′(t)-rcu(t),tJ. Hence 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 tJ, which implies that u(t)0 for tJ.

Assume that u(0)-au(0)>cu(η),u(1)+bu(1)du(ξ), then

cu(t)=sinπtaπ+csin(πη)(u(0)-au(0)-cu(η)). Put y(t)=u(t)+cu(t),  tJ, then y(t)u(t) for all tJ, and y(t)=u(t)+πcos(πt)aπ+csin(πη)(u(0)-au(0)-cu(η)),tJ,y(t)=u(t)-rcu(t),tJ. Hence 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 Theorem 2.1, y(t)0 for all tJ, which implies that u(t)0 for tJ.

Assume that u(0)-au(0)>cu(η),u(1)+bu(1)>du(ξ), then cu(t)=Ausin(πt).

Put y(t)=u(t)+cu(t),  tJ, then y(t)u(t) for all tJ, and y(t)=u(t)+Auπcos(πt),tJ,y′′(t)=u(t)-rcu(t),tJ.   Hence 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 Theorem 2.1, y(t)0 for all tJ, which implies that u(t)0 for tJ. The proof is complete.

3. Linear Problem

In this section, we consider the linear boundary value problem -u(t)+Mu(t)+Nu(θ(t))=σ(t),tJ,ttk,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, σC(J), ekR, and constants M,  N satisfy (2.3) with 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,βE such that

αβ on J,

-α′′(t)+Mα(t)+Nα(θ(t))+[(M+r)cα(t)+Ncα(θ(t))]σ(t),tJ,ttk,Δα(tk)Lkα(tk)+Lkcα(tk)+ek,k=1,,m,

-β′′(t)+Mβ(t)+Nβ(θ(t))-[(M+r)c-β(t)+Nc-β(θ(t))]σ(t),tJ,ttk,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 αuβ for tJ.

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, -v(t)+Mv(t)+Nv(θ(t))=0,tJ,ttk,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 v0 for tJ, that is, u1u2 on J. Similarly, one can obtain that u2u1 on J. Hence u1=u2.

Next, we prove that if u is a solution of (3.1), then αuβ. Let p=α-u. From boundary conditions, we have that cα(t)=cp(t) for all tJ. From (h2) and (3.1), we have

-p′′(t)+Mp(t)+Np(θ(t))+[(M+r)cp(t)+Ncp(θ(t))]0,tJ,ttk,Δp(tk)Lkp(tk)+Lkcp(tk),k=1,,m. By Theorem 2.1, we have that p=α-u0 on J. Analogously, uβ on J.

Finally, we show that the boundary value problem (3.1) has a solution by five steps as follows.

Step 1.

Let α̅(t)=α(t)+cα(t),β̅(t)=β(t)-c-β(t). We claim that

-α̅′′(t)+Mα̅(t)+Nα̅(θ(t))+[(M+r)cα̅(t)+Ncα̅(θ(t))]σ(t)for  tJ,ttk,Δα̅(tk)Lkα̅(tk)+ek,k=1,,m,

-β̅′′(t)+Mβ̅(t)+Nβ̅(θ(t))-[(M+r)c-β̅(t)+Nc-β̅(θ(t))]σ(t)for  tJ,ttk,Δβ̅(tk)Lkβ̅(tk)+ek,k=1,,m,

α(t)α̅(t)β̅(t)β(t) for tJ.

From (h2) and (h3), we have

-α̅′′(t)+Mα̅(t)+Nα̅(θ(t))σ(t),tJ,ttk,Δα̅(tk)Lkα̅(tk)+ek,k=1,,m.-β̅′′(t)+Mβ̅(t)+Nβ̅(θ(t))σ(t),tJ,ttk,Δβ̅(tk)Lkβ̅(tk)+ek,k=1,,m,α̅(0)-aα̅(0)-cα̅(η)=α(0)-aα(0)-cα(η)-(aπ+csin(πη))Bα0,α̅(1)+bα̅(1)-dα̅(ξ)=α(1)+bα(0)-dα(ξ)-(bπ+dsin(πξ))Bα0,-[β̅(0)-aβ̅(0)-cβ̅(η)]=-β(0)+aβ(0)+cβ(η)-(aπ+csin(πη))B-β0,-[β̅(1)+bβ̅(1)-dβ̅(ξ)]=-β(1)-bβ(0)+dβ(ξ)-(bπ+dsin(πξ))B-β0. From (3.9)–(3.14), we obtain that cα̅(t)=c-β̅(t)0,tJ. Combining (3.9) and (3.10), we obtain that ( 1) and ( 2) hold.

It is easy to see that αα̅,  β̅β on J. We show that α̅β̅ 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

-p′′(t)+Mp(t)+Np(θ(t))0,tJ,ttk,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-β0,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-β0,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 p0 on J, that is, α̅β̅ on J. So ( 3) holds.

Step 2.

Consider the boundary value problem -u′′(t)+Mu(t)+Nu(θ(t))=σ(t),tJ,  ttk,Δu(tk)=Lku(tk)+ek,k=1,,m,u(0)-au(0)=λ,u(1)+bu(1)=δ, where λR,  δR. 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:

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 G(t,s)=1a+b+1{(a+t)(1+b-s),0ts1,(a+s)(1+b-t),0st1.

It is easy to see that PC(J,R) with norm u=maxtJ|u(t)| is a Banach space. Define a mapping Φ:PC(J,R)PC(J,R) by

(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,  yPC(J,R), we have |(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=1mLkx-y  01G(t,s)dsx-y(M+N)+(1+(1+b)2a+b+1)k=1mLkx-y. Since maxtJ01G(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 uPC(J) such that Φu=u. The boundary value problem (3.16) has a unique solution.

Step 3.

We show that for any tJ, 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 -u′′(t)+Mu(t)+Nu(θ(t))=σ(t),tJ,ttk,Δu(tk)=Lku(tk)+ek,k=1,,m,u(0)-au(0)=λi,u(1)+bu(1)=δi,i=1,2. Then 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

u(t,λ1,δ1)-u(t,λ2,δ2)|λ1-λ2|+|δ1-δ2|+(M+N)u(t,λ1,δ1)-u(t,λ2,δ2)maxtJ01G(t,s)ds+u(t,λ1,δ1)-u(t,λ2,δ2)(1+(1+b)2a+b+1)k=1mLkx-y    |λ1-λ2|+|δ1-δ2|+μu(t,λ1,δ1)-u(t,λ2,δ2).

Hence

u(t,λ1,δ1)-u(t,λ2,δ2)011-μ(|λ1-λ2|+|δ1-δ2|).

Step 4.

We show that α̅(t)u(t,λ,δ)β̅(t) for any tJ, 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).

Let m(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

-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 m0 on J, that is, α̅(t)u(t,λ,δ) on J. Similarly, u(t,λ,δ)β̅(t) on J.

Step 5.

Let D=[cα̅(η),cβ̅(η)]×[dα̅(ξ),  dβ̅(ξ)]. Define a mapping F:DR2 by 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 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 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

Let MR, NR. We first give the following definition.

Definition 4.1.

A function αE is called a lower solution of the boundary value problem (1.2) if -α(t)+(M+r)cα(t)+Ncα(θ(t))f(t,α(t),α(θ(t))),tJ,ttk,Δα(tk)Ik(α(tk))+Lkcα(tk),k=1,,m.

Definition 4.2.

A function βE is called an upper solution of the boundary value problem (1.2) if -β′′(t)-(M+r)c-β(t)-Nc-β(θ(t))f(t,β(t),β(θ(t)))      tJ,  tJ,ttk,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:

α,  β are lower and upper solutions for boundary value problem (1.2) respectively, and α(t)β(t) on J,

the constants M,  N in definition of upper and lower solutions satisfy (2.3), (3.2), and f(t,x,y)-f(t,x̅,y̅)-M(x-x̅)-N(y-y̅),Ik(x)-Ik(y)Lk(x-y),xy, for α(t)x̅xβ(t),α(θ(t))y̅yβ(θ(t)),  tJ.

Then, there exist monotone sequences {αn},  {βn} with α0=α,  β0=β such that limnαn(t)=ρ(t),limnβn(t)=ϱ(t) uniformly on J, and ρ,  ϱ are the minimal and the maximal solutions of (1.2), respectively, such that

α0α1α2αnρxϱβnβ2β1β0 on J, where x is any solution of (1.1) such that α(t)x(t)β(t) on J.

Proof.

Let [α,β]={uE:α(t)u(t)β(t),  tJ}. For any γ[α,β], we consider the boundary value problem -u′′(t)+Mu(t)+Nu(θ(t))=f(t,γ(t),γ(θ(t)))+Mγ(t)+Nγ(θ(t)),      tJ,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).

Since α is a lower solution of (1.2), from (H2), we have that

-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

-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 Theorem 3.1, 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

αΨα,      Ψββ,

Ψ is nondecreasing in [α,β].

From Ψα[α,β] and Ψβ[α,β], we have that (a) holds. To prove (b), we show that Ψν1Ψν2 if αν1ν2β.

Let ν1*=Ψν1,  ν2*=Ψν2 ,and p=ν1*-ν2*, then by (H2) and boundary conditions, we have that

-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 Ψν1Ψν2.

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

α0α1α2αnβnβ2β1β0 on tJ, and each αn,  βnE satisfies -αn′′(t)+Mαn(t)+Nαn(θ(t))=f(t,αn-1(t),αn-1(θ(t)))+Mαn-1(t)+Nαn-1(θ(t)),tJ,ttk,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)),tJ,ttk,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 limnαn(t)=ρ(t), limnβn(t)=ϱ(t) uniformly on J. Clearly, ρ,  ϱ are solutions of (1.1).

Finally, we prove that if x[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

αn+1(t)x(t)βn+1(t),tJ. Hence, we can conclude that αn(t)x(t)βn(t),n. Passing to the limit as n, we obtain ρ(t)x(t)ϱ(t),tJ. This completes the proof.

Acknowledgments

This work is supported by the NNSF of China (10571050;10871062) and Hunan Provincial Natural Science Foundation of China (NO:09JJ3010), and Science Research Fund of Hunan provincial Education Department (No: 06C052 ).

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