IJDEInternational Journal of Differential Equations1687-96511687-9643Hindawi Publishing Corporation90896010.1155/2010/908960908960Research ArticleIntegral BVPs for a Class of First-Order Impulsive Functional Differential EquationsHeXiaofei1XieJingli1ChenGuoping1ShenJianhua2ChenFengde1Department of MathematicsCollege of Zhangjiajie, Jishou UniversityJishou, Hunan 416000Chinajsu.edu.cn2Department of MathematicsHangzhou Normal UniversityHangzhou, Zhejiang 310036Chinahztc.edu.cn201008032010201011112009250120102010Copyright © 2010This 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.

The methods of lower and upper solutions and monotone iterative technique are employed to the study of integral boundary value problems for a class of first-order impulsive functional differential equations. Sufficient conditions are obtained for the existence of extreme solutions.

1. Introduction and Preliminaries

In this paper, we study the following integral boundary value problems (BVPs for short) of the impulsive functional differential equation

x'(t)+b(t)x(t)=f(t,x(t),[Kx](t)),ttk,tJ=[0,T],Δx(tk)=Ik(x(tk)),k=1,2,,m,x(0)+μ0Tx(s)ds=x(T),μ0, where fC(J×R2,R),IkC(R,R),  (1km),  b(t)C(R),  b(t)0,  J=[0,T],  0=t0<t1<t2<<tm<tm+1=T. K:PC(J)PC(J), where PC(J)={u:JR,u is continuous for tJ,  ttk,  u(ti+),u(ti-) exist, and u(ti-)=u(ti),  i=1,2,,m}. Furthermore, we will assume that K is continuous and monotone nondecreasing, and for any bounded set APC(J),KA is bounded. Δx(tk)=x(tk+)-x(tk-) denotes the jump of x(t) at t=tk; x(tk+) and x(tk-) represent the right and left limits of x(t) at t=tk, respectively. Denote J'=J{t1,t2,,tm}.

Let PC1(J)={uPC(J):u be continuously differentiable for tJ,  ttk}. PC(J) and PC1(J) are Banach spaces with the norms

uPC(J)=sup{|u(t)|:tJ},uPC1(J)=max{uPC(J),uPC(J)}.

By a solution of (1.1) we mean a uPC1(J) for which problem (1.1) is satisfied.

Note that (1.1) has a very general form, as special instances resulting from (1.1), one can have impulsive differential equations with deviating arguments and impulsive differential equations with the Volterra or Fredholm operators. When μ=0, Ik0, (1.1) reduces to

x'(t)+b(t)x(t)=f(t,x(t),[Kx](t)),tJ=[0,T],x(0)=x(T). In , Cao and Li. studied and understood existence and stability of solution of this equation by using fixed theorem and monotone iteration techniques.

When μ=0, b(t)0, (1.1) reduces to

x(t)=f(t,x(t),[Kx](t)),ttk,tJ=[0,T],Δx(tk)=Ik(x(tk)),k=1,2,,m,x(0)=x(T). In , Li discussed and built the existence theorem of solutions of this equation by using fixed theorem, upper and lower solutions methods and monotone iterative techniques.

When μ=0, b(t)0, [Kx](t)=x(t), the equation (1.1) reduces to the periodic boundary value problem of the impulsive differential equation

x'(t)=f(t,x(t)),ttk,tJ=[0,T],Δx(tk)=Ik(x(tk)),  k=1,2,,m,x(0)=x(T). There are plenty of results on studying the periodic boundary value problem of impulsive differential equations (see ). According to author’s know, there are no dependent references for studying the (1.1) yet. To fill in this void, we try to find the conditions on f and Ik, so that make sure that the (1.1) exists extremal solution.

It is well known that the monotone iterative technique offers an approach for obtaining approximate solutions of nonlinear differential equations, for details, see  and the references therein. There also exist several works devoted to the applications of this technique to boundary value problems of impulsive differential equations, see, for example, [13, 514]. In this paper, we consider (1.1) by using the method of upper and lower solutions combined with monotone iterative technique. This technique plays an important role in constructing monotone sequences which converge to the solutions of our problems. In presence of a lower solution α and an upper solution β with αβ, we show under suitable conditions the sequences converge to the solutions of (1.1) by using the method of upper and lower solutions and monotone iterative technique.

Definition 1.1.

The functions α,βPC1(J) are called lower solution and upper solution of (1.1), respectively, if α(t)+b(t)α(t)f(t,α(t),[Kα](t)),ttk,tJ,Δα(tk)Ik(α(tk)),k=1,2,,m,α(0)+μ0Tα(s)dsα(T),μ0.β'(t)+b(t)β(t)f(t,β(t),[Kβ](t)),ttk,tJ,Δβ(tk)Ik(β(tk)),k=1,2,,m,β(0)+μ0Tβ(s)dsβ(T),μ0.

In what follows we define the set

[α,β]={wPC(J,R):α(t)w(t)β(t),  tJ} for α,βPC(J,R) and αβ.

We list the following conditions.

α(t),β(t) are lower and upper solutions of (1.1) such that α(t)β(t).

There exists M0 such that f(t,x,y)-f(t,x¯,y¯)-M(x-x¯), for α(t)x¯xβ(t),  [Tα](t)y¯y[Tβ](t),  tJ.

There exist 0Lk<1,  k=1,2,,m such that Ik(x)-Ik(y)-Lk(x-y),k=1,2,,m, for α(t)yxβ(t),  tJ.

2. Main Results

To obtain our main results, we need the following lemmas.

Lemma 2.1 (see [<xref ref-type="bibr" rid="B1">9</xref>]).

Suppose that the following conditions are satisfied.

Sequence {tk} satisfies 0t0<t1<t2, and limntn=.

mPC1[J,R] and m(t) is left continuous at tk,  k=1,2,.

For k=1,2,,  tt0, m'(t)p(t)m(t)+q(t),ttk,tJ,m(tk+)dkm(tk)+bk,k=1,2,,m, where q,pC[R+,R],  bk,dk0 are constants, then     m(t)m(t0)t0<tktdkexp(t0tp(s)ds)+t0<tkt(tk<tjtdjexp(tktp(s)ds))bk+t0ts<tk<tdkexp(stp(σ)dσ)q(s)ds.

Lemma 2.2 (see [<xref ref-type="bibr" rid="B12">12</xref>]).

If mPC1(J) and m'(t)-Mm(t),ttk,tJ=[0,T],Δm(tk)-Lkm(tk),k=1,2,,m,m(0)m(T), where M>0,  0<Lk1, then m(t)0,  tJ.

Lemma 2.3.

If xPC(J),  M>0,  0<Lk1,  k=1,2,,m, and (1/(1-e-MT))k=0mLk<1, then the equation x'(t)+Mx(t)=σ(t),ttk,tJ=[0,T],Δx(tk)=-Lkx(tk)+dk,k=1,2,,m,x(0)+d=x(T),dR has one unique solution.

Proof.

Firstly, we prove that (2.4) is equivalent to the integral equation x(t)=-e-Mt1-e-MTd+0TG(t,s)σ(s)ds+k=0mG(t,tk)(-Lkx(tk)+dk), where G(t,s)={e-M(t-s)1-e-MT,      0s<tT,e-M(T+t-s)1-e-MT,      0tsT.

If x(t)PC1(J) is solution of (2.4), then, by directly integrating we obtain

x(t)=-e-Mt1-e-MTd+0TG(t,s)σ(s)ds+k=0mG(t,tk)(-Lkx(tk)+dk).

If x(t)PC1(J) is solution of the above-mentioned integral equation, then

x'(t)=-M[-e-Mt1-e-MTd+0TG(t,s)σ(s)ds+k=0mG(t,tk)(-Lkx(tk)+dk)]+σ(t)=-Mx(t)+σ(t),ttk,Δx(tk)=-Lk(x(tk))+dk,k=1,2,,m,x(0)=-11-e-MTd+0Te-M(T-s)1-e-MTσ(s)ds+k=0me-M(T-tk)1-e-MT(-Lkx(tk)+dk),x(T)=-e-MT1-e-MTd+0Te-M(T-s)1-e-MTσ(s)ds+k=0me-M(T-tk)1-e-MT(-Lkx(tk)+dk).

This yields x(0)+d=x(T). So (2.4) is equivalent to the integral equation

x(t)=-e-Mt1-e-MTd+0TG(t,s)σ(s)ds+k=0mG(t,tk)(-Lkx(tk)+dk).

Now, we define operator A:PC(J)PC(J) as

(Ax)(t)=-e-Mt1-e-MTd+0TG(t,s)σ(s)ds+k=0mG(t,tk)(-Lkx(tk)+dk).

For each x,yPC(J),

|(Ax)(t)-(Ay)(t)|11-e-MTk=0mLk|x-y|11-e-MTk=0mLkx(tk)-y(tk), and so (Ax)(t)-(Ay)(t)11-e-MTk=0mLkx(tk)-y(tk).

This indicates that A:PC(J)PC(J) is a contraction mapping. Then there is one unique xPC(J) such that Ax=x, that is, (2.4) has an unique solution x(t). The proof is complete.

Theorem 2.4.

If the conditions (H1),  (H2),  (H3) are all satisfied, and, in addition, if there exist M>0,  0<Lk1,  k=1,2,,m, such that (1/(1-e-MT))k=0mLk<1, then the impulsive equation (1.1) has minimal and maximal solutions ρ(t),r(t)PC1(J) in [α,β], and there are monotone sequences {αn},{βn} convergeing uniformly to ρ(t),r(t) in J, respectively, where α0=α,  β0=β, and αn(t),βn(t) are lower and upper solutions of (1.1), respectively.

Proof.

For each ψ[α,β], we consider the equation x(t)=f(t,ψ(t),[Kψ](t))-b(t)ψ(t)-M(x(t)-ψ(t)),ttk,tJ,Δx(tk)=Ik(ψ(tk))-Lk(x(tk)-ψ(tk)),k=1,2,,m,x(0)+μ0Tψ(s)ds=x(T),μ0. By Lemma 2.3, we know that (2.13) has a unique solution x(t)PC1(J). Now, we define operator A:PC1(J)PC1(J) as Aψ=x.

We will prove that {αn},  {βn} have the following properties.

α0Aα0,Aβ0β0.

A is monotone nondecreasing on [α0,β0].

Proofs of properties (a),(b) are divided into three steps to proceed.

Step 1.

Suppose that p=α0-α1, then p'=α0-α1-b(t)α(t)+f(t,α(t),[Kα](t))-f(t,α(t),[Kα(t)](t))+b(t)α(t)+M(α1(t)-α(t))=-Mp(t),ttk,Δp(tk)=Δα0-Δα1Ik(α(tk))-Ik(α(tk))+Lk(α1(tk)-α(tk))=-Lkp(tk),t=tk,p(0)=α0(0)-α1(0)p(T). By Lemma 2.2, we obtain p(t)0,tJ, so α0(t)α1(t).

Step 2.

Suppose that p=β1-β0, then p'=β1-β0b(t)β(t)-f(t,β(t),[Kβ](t))+f(t,β(t),[Kβ(t)](t))-b(t)β(t)-M(β1(t)-β(t))=-Mp(t),t=tk,Δp(tk)=Δβ1-Δβ0-Ik(β(tk))+Ik(β(tk))-Lk(β1(tk)-β(tk))=-Lkp(tk),t=tk,p(0)=β1(0)-β0(0)p(T). By Lemma 2.2, we obtain p(t)0,tJ, so β1(t)β0(t).

Similary we can show that α1(t)β1(t), hence α0(t)α1(t)β1(t)β0(t).

Step 3.

If n=m,  αm-1αmβmβm-1, then when n=m+1, let p=αm-αm+1. Then p'=αm-αm+1-b(t)αm-1(t)+f(t,αm-1(t),[Kαm-1](t))-M(αm(t)-αm-1(t))-f(t,αm(t),[Kαm(t)](t))+b(t)αm(t)+M(αm+1(t)-αm(t))=-b(t)(αm-1-αm)-Mp(t),ttk.

Furthermore, b(t)0,  αm-1-αm0, thus p'=α'm-α'm+1-Mp(t),Δp(tk)=Δαm-Δαm+1=Ik(αm-1(tk))-Lk(αm(tk)-αm-1(tk))-Ik(αm(tk))+Lk(αm+1(tk)-αm(tk))-Lkp(tk),t=tk,p(0)=αm(0)-αm+1(0)=μ0T(αm(s)-αm-1(s))ds+p(T)p(T). By Lemma 2.2, we obtain p(t)0,tJ, so αm(t)αm+1(t).

Similarly, we can assume that p=βm+1-βm. When ttk,

p'=βm+1-βm-b(t)βm(t)+f(t,βm(t),[Kβm](t))-M(βm+1(t)-βm(t))-f(t,βm-1(t),[Kβm-1(t)](t))+b(t)βm-1(t)+M(βm(t)-βm-1(t))=-b(t)(βm-βm-1)-Mp(t). Furthermore, b(t)0, βm-βm-10, thus p'=βm+1-βm-Mp(t), when t=tk, Δp(tk)=Δβm+1-Δβm=Ik(βm(tk))-Lk(βm+1(tk)-βm(tk))-Ik(βm-1(tk))+Lk(βm(tk)-βm-1(tk))-Lkp(tk),p(0)=βm+1(0)-βm(0)=μ0T(βm-1(s)-βm(s))ds+p(T)p(T), hence by Lemma 2.2, we obtain p(t)0,  tJ, so βm+1(t)βm(t).

In the same way we can prove that αm+1(t)βm+1(t).

Thus by mathematical induction we can know that

αn-1αnβnβn-1,n=0,1,2,,tJ. So far, we finish the proof of the properties (a),(b).

Now we prove that αn,βn,  n=0,1,2,, are lower and upper solutions of (1.1). Similarly, we can use mathematical induction to prove this.

When n=0, α0,β0 are already lower and upper solutions of (1.1).

When n=1,

α1(t)=f(t,α(t),[Kα](t))-b(t)α(t)-M(α1(t)-α(t))-f(t,α1(t),[Kα1](t))+f(t,α1(t),[Kα1](t))f(t,α1(t),[Kα1](t))-b(t)α1(t),ttk,tJ,Δα1(tk)=Ik(α(tk))-Lk(α1(tk)-α(tk))+Ik(α1(tk))-Ik(α1(tk))Ik(α1(tk)),k=1,2,,m,α1(0)+μ0Tα1(s)dsα1(0)+μ0Tα(s)ds=α1(T),μ0.

Thus α1 is lower solution of (1.1).

Suppose that αn is lower solution of (1.1) when n=m.

Then when n=m+1,

αm+1(t)=f(t,αm(t),[Kαm](t))-b(t)αm(t)-M(αm+1(t)-αm(t))-f(t,αm+1(t),[Kαm+1](t))+f(t,αm+1(t),[Kαm+1](t))f(t,αm+1(t),[Kαm+1](t))-b(t)αm+1(t),ttk,tJ,Δαm+1(tk)=Ik(αm(tk))-Lk(αm+1(tk)-αm(tk))+Ik(αm+1(tk))-Ik(αm+1(tk))Ik(αm+1(tk)),k=1,2,,m,αm+1(0)+μ0Tαm+1(s)dsαm+1(0)+μ0Tαm(s)ds=αm+1(T).

Thus by mathematical induction we can know that αn is lower solution of (1.1). In the same way we can prove that βn is upper solution of (1.1).

By αn-1αnβnβn-1,  n=0,1,2,,  tJ, we can know that when n+,{αn},{βn} have limits ρ(t),r(t), respectively. Since they are independent of t when n+{αn}, {βn} converge uniformly to ρ(t),r(t) and αnρ(t)r(t)βnβn-1,  n=0,1,2,,  tJ.

According to αn,βn satisfying (2.13), that is,

αn(t)=f(t,αn-1(t),[Kαn-1](t))-b(t)αn-1(t)-M(αn(t)-αn-1(t)),ttk,tJ,Δαn(tk)=Ik(αn-1(tk))-Lk(αn(tk)-αn-1(tk)),k=1,2,,m,αn(0)+μ0Tαn-1(s)ds=αn(T),μ0,βn(t)=f(t,βn-1(t),[Kβn-1](t))-b(t)βn-1(t)-M(βn(t)-βn-1(t)),ttk,tJ,Δβn(tk)=Ik(βn-1(tk))-Lk(βn(tk)-βn-1(tk)),k=1,2,,m,βn(0)+μ0Tβn-1(s)ds=βn(T),μ0, when n+, we have ρ(t)=f(t,ρ(t),[Kρ](t))-b(t)ρ(t),ttk,tJ,Δρ(tk)=Ik(ρ(tk)),k=1,2,,m,ρ(0)+μ0Tρ(s)ds=ρ(T),μ0.r(t)=f(t,r(t),[Kr](t))-b(t)r(t),ttk,tJ,Δr(tk)=Ik(r(tk)),k=1,2,,m,r(0)+μ0Tr(s)ds=r(T),μ0. Equation (2.24) indicates that ρ(t),r(t) are solutions of (1.1).

Lastly, we prove that ρ(t),r(t) are minimal and maximal solutions of the equation (1.1) in [α,β].

Suppose that x(t) is a solution of the equation and satisfies x(t)[α,β],tJ, obviously, we can assume that there is an n such that αnxβn.

If p(t)=αn+1-x, then

p'=αn+1-x-b(t)αn(t)+f(t,αn(t),[Kαn](t))-M(αn+1(t)-αn(t))-f(t,x(t),[Kx(t)](t))+b(t)x(t)  -b(t)(αn-x)-Mp(t),ttk.

And since b(t)0, αm-x0, p'-Mp(t),

Δp(tk)=Δαn+1-Δx=Ik(αn(tk))-Lk(αn+1(tk)-αn(tk))-Ik(x(tk))-Lkp(tk),t=tk,p(0)=αn+1(0)-x(0)=μ0T(x(s)-αn(s))ds+p(T)p(T). Hence by Lemma 2.2, we can obtain p(t)0,  tJ, so αn+1(t)x(t). Similarly, we can obtain: x(t)βn+1(t),  tJ. This indicates that αn(t)x(t)βn+1(t),  tJ,  n=0,1,2,  . Hence when n+, we can obtain that ρ(t)x(t)r(t),  tJ. This ends the proof.

Finally, we give an example to illustrate the efficiency of our results.

Example 2.5.

Consider the problem of x'(t)-x(t)sint=-x(t)+t+0tx(s)ds,0<t<1,tt1,Δx(t1)=-18x(t1),t1=12,x(0)-01x(s)ds=x(1), where b(t)=sint,f(t,x(t),Kx(t))=-x(t)+t+0tx(s)ds,I1(x)=-x. Obviously, α(t)=0,  β(t)=1-t are the lower solution and upper solution for (2.27) with α(t)β(t), respectively. f(t,x,Kx)-f(t,y,Ky)=-(x-y)-0t(x(s)-y(s))ds, I1(x)-I1(y)=-(x-y). Let T=1,Lk=1/8, the conditions of Theorem 2.4 are all satisfied, so problem (2.27) has the maximal and minimal solutions in the segement [α(t),β(t)].

Acknowledgments

This work was supported by the NNSF of China (10571050; 10871062), the Hunan Provincial Natural Science Foundation of China (09JJ6010), and the Zhejiang Provincial Natural Science Foundation of China (Y6090057).

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