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)),t≠tk,t∈J=[0,T],Δx(tk)=Ik(x(tk)),k=1,2,…,m,x(0)+μ∫0Tx(s)ds=x(T),μ≤0,
where f∈C(J×R2,R),Ik∈C(R,R),(1≤k≤m),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:J→R,u is continuous for t∈J,t≠tk,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 A⊆PC(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)={u∈PC(J):u be continuously differentiable for t∈J,t≠tk}. PC(J) and PC1(J) are Banach spaces with the norms

By a solution of (1.1) we mean a u∈PC1(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, Ik≡0, (1.1) reduces to

x'(t)+b(t)x(t)=f(t,x(t),[Kx](t)),t∈J=[0,T],x(0)=x(T).
In [1], 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)),t≠tk,t∈J=[0,T],Δx(tk)=Ik(x(tk)),k=1,2,…,m,x(0)=x(T).
In [2], 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)),t≠tk,t∈J=[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 [3–8]). 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 [4] 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, [1–3, 5–14]. 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 α≤β, 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,β∈PC1(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)),t≠tk,t∈J,Δα(tk)≤Ik(α(tk)),k=1,2,…,m,α(0)+μ∫0Tα(s)ds≤α(T),μ≤0.β'(t)+b(t)β(t)≥f(t,β(t),[Kβ](t)),t≠tk,t∈J,Δβ(tk)≥Ik(β(tk)),k=1,2,…,m,β(0)+μ∫0Tβ(s)ds≥β(T),μ≤0.

In what follows we define the set

[α,β]={w∈PC(J,R):α(t)≤w(t)≤β(t),t∈J}
for α,β∈PC(J,R) and α≤β.

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 M≥0 such that
f(t,x,y)-f(t,x¯,y¯)≥-M(x-x¯),
for α(t)≤x¯≤x≤β(t),[Tα](t)≤y¯≤y≤[Tβ](t),t∈J.

There exist 0≤Lk<1,k=1,2,…,m such that
Ik(x)-Ik(y)≥-Lk(x-y),k=1,2,…,m,
for α(t)≤y≤x≤β(t),t∈J.

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 0≤t0<t1<t2⋯, and limn→∞tn=∞.

m∈PC1[J,R] and m(t) is left continuous at tk,k=1,2,….

For k=1,2,…,t≥t0,
m'(t)≤p(t)m(t)+q(t),t≠tk,t∈J,m(tk+)≤dkm(tk)+bk,k=1,2,…,m,
where q,p∈C[R+,R],bk,dk≥0 are constants, then
m(t)≤m(t0)∏t0<tk≤tdkexp(∫t0tp(s)ds)+∑t0<tk≤t(∏tk<tj≤tdjexp(∫tktp(s)ds))bk+∫t0t∏s<tk<tdkexp(∫stp(σ)dσ)q(s)ds.

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

If m∈PC1(J) and
m'(t)≤-Mm(t),t≠tk,t∈J=[0,T],Δm(tk)≤-Lkm(tk),k=1,2,…,m,m(0)≤m(T),
where M>0,0<Lk≤1, then m(t)≤0,t∈J.

Lemma 2.3.

If x∈PC(J),M>0,0<Lk≤1,k=1,2,…,m, and (1/(1-e-MT))∑k=0mLk<1, then the equation
x'(t)+Mx(t)=σ(t),t≠tk,t∈J=[0,T],Δx(tk)=-Lkx(tk)+dk,k=1,2,…,m,x(0)+d=x(T),d∈R
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,0≤s<t≤T,e-M(T+t-s)1-e-MT,0≤t≤s≤T.

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

|(Ax)(t)-(Ay)(t)|≤11-e-MT∑k=0mLk|x-y|≤11-e-MT∑k=0mLk∥x(tk)-y(tk)∥,
and so
∥(Ax)(t)-(Ay)(t)∥≤11-e-MT∑k=0mLk∥x(tk)-y(tk)∥.

This indicates that A:PC(J)→PC(J) is a contraction mapping. Then there is one unique x∈PC(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<Lk≤1,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)),t≠tk,t∈J,Δx(tk)=Ik(ψ(tk))-Lk(x(tk)-ψ(tk)),k=1,2,…,m,x(0)+μ∫0Tψ(s)ds=x(T),μ≤0.
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.

α0≤Aα0,Aβ0≤β0.

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),t≠tk,Δp(tk)=Δα0-Δα1≤Ik(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,t∈J, so α0(t)≤α1(t).

Step 2.

Suppose that p=β1-β0, then
p'=β1′-β0′≤b(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,t∈J, 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-1≤αm≤βm≤βm-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),t≠tk.

Furthermore, b(t)≤0,αm-1-αm≤0, 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)=μ∫0T(αm(s)-αm-1(s))ds+p(T)≤p(T).
By Lemma 2.2, we obtain p(t)≤0,t∈J, so αm(t)≤αm+1(t).

Similarly, we can assume that p=βm+1-βm. When t≠tk,

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-1≤0, 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)=μ∫0T(βm-1(s)-βm(s))ds+p(T)≤p(T),
hence by Lemma 2.2, we obtain p(t)≤0,t∈J, 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-1≤αn≤βn≤βn-1,n=0,1,2,…,t∈J.
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).

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-1≤αn≤βn≤βn-1,n=0,1,2,…,t∈J, 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 αn≤ρ(t)≤r(t)≤βn≤βn-1,n=0,1,2,…,t∈J.

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)),t≠tk,t∈J,Δαn(tk)=Ik(αn-1(tk))-Lk(αn(tk)-αn-1(tk)),k=1,2,…,m,αn(0)+μ∫0Tαn-1(s)ds=αn(T),μ≤0,βn′(t)=f(t,βn-1(t),[Kβn-1](t))-b(t)βn-1(t)-M(βn(t)-βn-1(t)),t≠tk,t∈J,Δβn(tk)=Ik(βn-1(tk))-Lk(βn(tk)-βn-1(tk)),k=1,2,…,m,βn(0)+μ∫0Tβn-1(s)ds=βn(T),μ≤0,
when n→+∞, we have
ρ′(t)=f(t,ρ(t),[Kρ](t))-b(t)ρ(t),t≠tk,t∈J,Δρ(tk)=Ik(ρ(tk)),k=1,2,…,m,ρ(0)+μ∫0Tρ(s)ds=ρ(T),μ≤0.r′(t)=f(t,r(t),[Kr](t))-b(t)r(t),t≠tk,t∈J,Δr(tk)=Ik(r(tk)),k=1,2,…,m,r(0)+μ∫0Tr(s)ds=r(T),μ≤0.
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],t∈J, obviously, we can assume that there is an n such that αn≤x≤βn.

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)=μ∫0T(x(s)-αn(s))ds+p(T)≤p(T).
Hence by Lemma 2.2, we can obtain p(t)≤0,t∈J, so αn+1(t)≤x(t). Similarly, we can obtain: x(t)≤βn+1(t),t∈J. This indicates that αn(t)≤x(t)≤βn+1(t),t∈J,n=0,1,2,…. Hence when n→+∞, we can obtain that ρ(t)≤x(t)≤r(t),t∈J. 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,t≠t1,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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