# Integral BVPs for a Class of First-Order Impulsive Functional Differential Equations

Authors: Xiaofei He; Jingli Xie; Guoping Chen; Jianhua Shen
Journal: International Journal of Differential Equations
 (2010)

---

## Abstract

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.


---

## Body

## 1. Introduction and Preliminaries

In this paper, we study the following integral boundary value problems (BVPs for short) of the impulsive functional differential equation(1.1)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}.LetPC1(J)={u∈PC(J):u be continuously differentiable for t∈J,t≠tk}. PC(J) and PC1(J) are Banach spaces with the norms(1.2)∥u∥PC(J)=sup{|u(t)|:t∈J},∥u∥PC1(J)=max{∥u∥PC(J),∥u′∥PC(J)}.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(1.3)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.Whenμ=0, b(t)≡0, (1.1) reduces to(1.4)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.Whenμ=0, b(t)≡0, [Kx](t)=x(t), the equation (1.1) reduces to the periodic boundary value problem of the impulsive differential equation(1.5)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 functionsα,β∈PC1(J) are called lower solution and upper solution of (1.1), respectively, if
(1.6)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(1.7)[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.(H1)
α(t),β(t) are lower and upper solutions of (1.1) such that α(t)≤β(t).(H2)
There existsM≥0 such that
(1.8)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.(H3)
There exist0≤Lk<1,k=1,2,…,m such that
(1.9)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 [9]).
Suppose that the following conditions are satisfied.(A0)
Sequence{tk} satisfies 0≤t0<t1<t2⋯, and limn→∞tn=∞.(A1)
m∈PC1[J,R] and m(t) is left continuous at tk,k=1,2,….(A2)
Fork=1,2,…,t≥t0,
(2.1)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
(2.2)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 [12]).
Ifm∈PC1(J) and
(2.3)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.
Ifx∈PC(J),M>0,0<Lk≤1,k=1,2,…,m, and (1/(1-e-MT))∑k=0mLk<1, then the equation
(2.4)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
(2.5)x(t)=-e-Mt1-e-MTd+∫0TG(t,s)σ(s)ds+∑k=0mG(t,tk)(-Lkx(tk)+dk),
where
(2.6)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.
Ifx(t)∈PC1(J) is solution of (2.4), then, by directly integrating we obtain
(2.7)x(t)=-e-Mt1-e-MTd+∫0TG(t,s)σ(s)ds+∑k=0mG(t,tk)(-Lkx(tk)+dk).
Ifx(t)∈PC1(J) is solution of the above-mentioned integral equation, then
(2.8)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),t≠tk,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 yieldsx(0)+d=x(T). So (2.4) is equivalent to the integral equation
(2.9)x(t)=-e-Mt1-e-MTd+∫0TG(t,s)σ(s)ds+∑k=0mG(t,tk)(-Lkx(tk)+dk).
Now, we define operatorA:PC(J)→PC(J) as
(2.10)(Ax)(t)=-e-Mt1-e-MTd+∫0TG(t,s)σ(s)ds+∑k=0mG(t,tk)(-Lkx(tk)+dk).
For eachx,y∈PC(J),
(2.11)|(Ax)(t)-(Ay)(t)|≤11-e-MT∑k=0mLk|x-y|≤11-e-MT∑k=0mLk∥x(tk)-y(tk)∥,
and so
(2.12)∥(Ax)(t)-(Ay)(t)∥≤11-e-MT∑k=0mLk∥x(tk)-y(tk)∥.
This indicates thatA: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 eachψ∈[α,β], we consider the equation
(2.13)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.
(a)
α0≤Aα0,Aβ0≤β0.(b)
A is monotone nondecreasing on [α0,β0].
Proofs of properties(a),(b) are divided into three steps to proceed.
Step 1.
Suppose thatp=α0-α1, then
(2.14)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 thatp=β1-β0, then
(2.15)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 thatα1(t)≤β1(t), hence α0(t)≤α1(t)≤β1(t)≤β0(t).
Step 3.
Ifn=m,αm-1≤αm≤βm≤βm-1, then when n=m+1, let p=αm-αm+1. Then
(2.16)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),(2.17)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 thatp=βm+1-βm. When t≠tk,
(2.18)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,
(2.19)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 thatαm+1(t)≤βm+1(t).
Thus by mathematical induction we can know that
(2.20)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 thatαn,βn,n=0,1,2,…, are lower and upper solutions of (1.1). Similarly, we can use mathematical induction to prove this.
Whenn=0, α0,β0 are already lower and upper solutions of (1.1).
Whenn=1,
(2.21)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),t≠tk,t∈J,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)+μ∫0Tα1(s)ds≤α1(0)+μ∫0Tα(s)ds=α1(T),μ≤0.
Thusα1 is lower solution of (1.1).
Suppose thatαn is lower solution of (1.1) when n=m.
Then whenn=m+1,
(2.22)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),t≠tk,t∈J,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)+μ∫0Tαm+1(s)ds≤αm+1(0)+μ∫0Tαm(s)ds=αm+1(T).
Thus by mathematical induction we can know thatαn is lower solution of (1.1). In the same way we can prove that βn is upper solution of (1.1).
Byαn-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 toαn,βn satisfying (2.13), that is,
(2.23)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
(2.24)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 thatρ(t),r(t) are minimal and maximal solutions of the equation (1.1) in [α,β].
Suppose thatx(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.
Ifp(t)=αn+1-x, then
(2.25)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),t≠tk.
And sinceb(t)≤0, αm-x≤0, p'≤-Mp(t),
(2.26)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(2.27)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)].



---

*Source: 908960-2010-03-21.xml*