# Existence and Uniqueness of Positive Solutions to Nonlinear Second Order Impulsive Differential Equations with Concave or Convex Nonlinearities

Authors: Lingling Zhang; Chengbo Zhai
Journal: Discrete Dynamics in Nature and Society
 (2013)

---

## Abstract

Using a new fixed point theorem of generalized concave operators, we present in this paper criteria which guarantee the existence and uniqueness of positive solutions to nonlinear two-point boundary value problems for second-order impulsive differential equations with concave or convex nonlinearities.


---

## Body

## 1. Introduction

In this paper, we study the existence and uniqueness of positive solutions to the following two-point boundary value problems for second-order impulsive differential equations:(1)-x′′(t)=f(t,x(t)),t∈J,t≠tk,k=1,2,…,m,Δx∣t=tk=Ik(x(tk)),k=1,2,…,m,Δx′∣t=tk=I¯k(x(tk)),k=1,2,…,m,x(0)=x′(0),x(1)=x′(1),
where f∈C[J×R,R], J=[0,1], 0<t1<t2<⋯<tk<⋯<tm<1, Δx∣t=tk=x(tk+)-x(tk-), Δx′∣t=tk=x′(tk+)-x′(tk-), x′(tk+), x′(tk-), x(tk+), x(tk-) denote the right limit (left limit) of x′(t) and x(t) at t=tk, respectively.  Ik,I¯k∈C[R,R], k=1,2,…,m.Impulsive differential equations have been studied extensively in recent years. Such equations arise in many applications such as spacecraft control, impact mechanics, chemical engineering, and inspection process in operations research. It is now recognized that the theory of impulsive differential equations is a natural framework for a mathematical modelling of many natural phenomena. There have appeared numerous papers on impulsive differential equations during the last ten years. Many of them are on boundary value problems, see [1–18], and it is interesting to note that some of them are about comparatively new applications like ecological competition, respiratory dynamics, and vaccination strategies, see [12, 19–25].Second-order impulsive differential equations have been studied by many authors with much of the attention given to positive solutions. For a small sample of such work, we refer the reader to works by Feng and  Xie [6], Hu et al. [8], Jankowski [10, 11], E. K. Lee and Y.-H. Lee [12], Lin and  Jiang [13], Liu et al. [14], Agarwal and O’Regan [26], Wang et al. [27], Zhang [28], and Chu et al. [29]. The results of these papers are based on the Schauder fixed point theorem, Leggett-Williams theorem, fixed point index theorems in cones, Krasnoselski fixed point theorem, the method of upper-lower solutions, fixed point theorems in cones, and so on. But, in most of the existing works, in order to establish the existence of positive solutions, a key condition is the existence of upper-lower solutions. However, as we know, it is difficult to verify the existence of upper-lower solutions for concrete impulsive differential equations. In addition, few papers can be found in the literature on the existence and uniqueness of positive solutions for second-order impulsive differential equations. In this paper, we will study the problem (1) with concave or convex nonlinearities and not suppose the existence of upper-lower solutions and compactness condition. Different from the previously mentioned works, in this paper we will use a new fixed point theorem of generalized concave operators to show the existence and uniqueness of positive solutions for the problem (1).For convenience, we list the following assumptions on the functionsf(t,x),Ik(x), and I¯k(x):(H1)
f(t,0)≤0, f(t,1/2)<0, t∈[0,1], and f(t,x) is decreasing in x∈[0,∞) for each t∈[0,1],(H2)
Ik(0)≤0, I¯k(0)≥0, and Ik(x) is decreasing, and I¯k(x) is increasing in x∈[0,∞), k=1,2,…,m,(H3)
for anyλ∈(0,1), t∈[0,1], and x≥0, there exist α1(λ),α2(λ),α3(λ)∈(λ,1) such that
(2)f(t,λx)≤α1(λ)f(t,x),Ik(λx)≤α2(λ)Ik(x),I¯k(λx)≥α3(λ)I¯k(x),k=1,2,…,m,(H4)
∑k=1m[-2Ik(3/2)+(1+tk)I¯k(3/2)]>0,(H1)′
f(t,3/2)<0, t∈[0,1], and f(t,x) is increasing in x∈[0,∞) for each t∈[0,1] and f(t,x)≤0 for [0,1]×[0,∞),(H2)′
Ik(x)≤0, I¯k(x)≥0 for [0,∞), and Ik(x) is increasing, and I¯k(x) is decreasing in x∈[0,∞), k=1,2,…,m,(H3)′
for anyλ∈(0,1), t∈[0,1], and x≥0, there exist β1(λ),β2(λ),β3(λ)∈(0,1) such that
(3)f(t,λx)≥λ-β1(λ)f(t,x),Ik(λx)≥λ-β2(λ)Ik(x),I¯k(λx)≤λ-β3(λ)I¯k(x),k=1,2,…,m,(H4)′
∑k=1m[-2Ik(1/2)+(1+tk)I¯k(1/2)]>0.


## 2. Preliminaries

In this section, we state some definitions, notations, and known results. For convenience of readers, we suggest that one refers to [30] and references therein for details.Suppose thatE is a real Banach space which is partially ordered by a cone P⊂E. That is, x≤y if and only if y-x∈P. By θ we denote the zero element of E. Recall that a nonempty closed convex set P⊂E is called a cone if it satisfies (i) x∈P,λ≥0⇒λx∈P, (ii) x∈P,-x∈P⇒x=θ.Moreover,P is called normal if there exists a constant N>0 such that, for all x,y∈E,θ≤x≤y implies ∥x∥≤N∥y∥. In this case N is called the normality constant of P. We say that an operator A:E→E is increasing (decreasing) if x≤y implies Ax≤Ay(Ax≥Ay).For allx,y∈E, the notation x~y means that there exist λ>0 and μ>0 such that λx≤y≤μx. Clearly, ~ is an equivalence relation. Given h>θ(i.e.,h≥θ and h≠θ), we denote by Ph the set Ph={x∈E∣x~h}. Clearly, Ph⊂P is convex and λPh=Ph for all λ>0.We now present a fixed point theorem of generalized concave operators which will be used in the latter proof. See [30] for further information.Theorem 1 (from [30, Lemma 2.1, and Theorem 2.1]).
Leth>θ, and let P be a normal cone. Assume that (D1)A:P→P is increasing and Ah∈Ph; (D2) for any x∈P and t∈(0,1), there exists α(t)∈(t,1) with respect to t such that A(tx)≥α(t)Ax. Then (i) there are u0,v0∈Ph and r∈(0,1) such that rv0≤u0<v0,u0≤Au0≤Av0≤v0;  (ii) operator equation x=Ax has a unique solution in Ph.Remark 2.
An operatorA is said to be generalized concave if A satisfies condition (D2).
In what follows, for the sake of convenience, letPC[J,R]={x∣x:J→R, x(t) be continuous at t≠tk and left continuous at t=tk, x(tk+) exists, k=1,2,…,m}, and let PC1[J,R]={x∈PC[J,R]∣x′(t) be continuous at t≠tk and left continuous at t=tk, x′(tk+) exists, k=1,2,…,m}. Evidently, PC[J,R] is a Banach space with the norm ∥x∥PC=sup{|x(t)|:t∈J}, and PC1[J,R] is a Banach space with the norm ∥x∥PC1=sup{∥x∥PC,∥x′∥PC}. Let J′=J∖{t1,t2,…,tm}.Definition 3.
A functionx∈PC1[J,R]⋂‍C2[J′,R] is called a solution of the problem (1) if it satisfies problem (1).Lemma 4.
x
∈
P
C
1
[
J
,
R
]
⋂
‍
C
2
[
J
′
,
R
] is a solution of the problem (1) if and only if x∈PC1[J,R] is the solution of the following integral equation:
(4)x(t)=-∫01‍G(t,s)f(s,x(s))ds-(1+t)∑k=1m‍Ik(x(tk))+∑0<tk<t‍Ik(x(tk))+∑0<tk<t‍(t-tk)I¯k(x(tk))+(1+t)∑k=1m‍tkI¯k(x(tk)),
where
(5)G(t,s)={s(1+t),0≤t≤s≤1,t(1+s),0≤s≤t≤1.Proof.
First suppose thatx∈PC1[J,R]⋂‍C2[J′,R] is a solution of the problem (1). It is easy to see by integration of (1) that
(6)x′(t)=x′(0)-∫0t‍f(s,x(s))ds+∑0<tk<t‍[x′(tk+)-x′(tk)]=x′(0)-∫0t‍f(s,x(s))ds+∑0<tk<t‍I¯k(x(tk)).
Integrate again, we can get
(7)x(t)=x(0)+x′(0)t-∫0t‍(t-s)f(s,x(s))ds+∑0<tk<t‍I¯k(x(tk))(t-tk)+∑0<tk<t‍[x(tk+)-x(tk)]=x(0)+x′(0)t-∫0t‍(t-s)f(s,x(s))ds+∑0<tk<t‍I¯k(x(tk))(t-tk)+∑0<tk<t‍Ik(x(tk)).
Letting t=1 in (6) and (7), we find
(8)x′(1)=x′(0)-∫01‍f(s,x(s))ds+∑k=1m‍I¯k(x(tk)),x(1)=x(0)+x′(0)-∫01‍(1-s)f(s,x(s))ds+∑k=1m‍I¯k(x(tk))(1-tk)+∑k=1m‍Ik(x(tk)).
From the boundary conditions x(0)=x′(0), and x(1)=x′(1), we have
(9)x(1)=x(0)-∫01‍f(s,x(s))ds+∑k=1m‍I¯k(x(tk)),x(1)=2x(0)-∫01‍(1-s)f(s,x(s))ds+∑k=1m‍I¯k(x(tk))(1-tk)+∑k=1m‍Ik(x(tk)).
Then we obtain
(10)x(0)=-∫01‍sf(s,x(s))ds+∑k=1m‍I¯k(x(tk))-∑k=1m‍I¯k(x(tk))(1-tk)-∑k=1m‍Ik(x(tk)).
Substituting (10) into (7), we have
(11)x(t)=-(1+t)∫01‍sf(s,x(s))ds-∫0t‍(t-s)f(s,x(s))ds-(1+t)∑k=1m‍Ik(x(tk))+∑0<tk<t‍Ik(x(tk))+∑0<tk<t‍(t-tk)I¯k(x(tk))+(1+t)∑k=1m‍tkI¯k(x(tk))=-∫01‍G(t,s)f(s,x(s))ds-(1+t)∑k=1m‍Ik(x(tk))+∑0<tk<t‍Ik(x(tk))+∑0<tk<t‍(t-tk)I¯k(x(tk))+(1+t)∑k=1m‍tkI¯k(x(tk)).
Thus, the proof of sufficient is complete.
Conversely, ifx is a solution of (4). Then we can easily get Δx|t=tk=x(tk+)-x(tk-)=Ik(x(tk)). Direct differentiation of (4) implies that, for t≠tk,
(12)x′(t)=-∫0t‍(1+s)f(s,x(s))ds-t(1+t)f(t,x(t))-∫t1‍sf(s,x(s))ds+t(1+t)f(t,x(t))-∑k=1m‍Ik(x(tk))+∑0<tk<t‍I¯k(x(tk))+∑k=1m‍tkI¯k(x(tk)).
Further
(13)x′′(t)=-f(t,x(t)),Δx′|t=tk=x′(tk+)-x′(tk-)=I¯k(x(tk)).
So x∈C2[J′,R] and it is easy to verify that x(0)=x′(0),x(1)=x′(1), and the lemma is proved.Define an operatorA:PC[J,R]→PC[J,R] by
(14)Ax(t)=-∫01‍G(t,s)f(s,x(s))ds-(1+t)∑k=1m‍Ik(x(tk))+∑0<tk<t‍Ik(x(tk))+∑0<tk<t‍(t-tk)I¯k(x(tk))+(1+t)∑k=1m‍tkI¯k(x(tk)).Lemma 5.
x
∈
P
C
1
[
J
,
R
]
⋂
‍
C
2
[
J
′
,
R
] is a solution of problem (1) if and only if x∈PC1[J,R] is a fixed point of the operator A.


## 3. Existence and Uniqueness of Positive Solutions for Problem (1)

In this section, we apply Theorem1 to study the problem (1), and we obtain a new result on the existence and uniqueness of positive solutions. The method used in this paper is new to the literature and so is the existence and uniqueness result to the second-order impulsive differential equations. This is also the main motivation for the study of (1) in the present work.SetP~={u∈PC[J,R]∣u(t)≥0,t∈J}, the standard cone. It is clear that P~ is a normal cone in PC[J,R] and the normality constant is 1. Our main result is summarized in the following theorem.Theorem 6.
Assume that (H1)–(H4)  hold. Then (i)
there existu0,v0∈P~h such that
(15)u0(t)≤-∫01‍G(t,s)f(s,u0(s))ds-(1+t)∑k=1m‍Ik(u0(tk))+∑0<tk<t‍Ik(u0(tk))+∑0<tk<t‍(t-tk)I¯k(u0(tk))+(1+t)∑k=1m‍tkI¯k(u0(tk)),t∈J,v0(t)≥-∫01‍G(t,s)f(s,v0(s))ds-(1+t)∑k=1m‍Ik(v0(tk))+∑0<tk<t‍Ik(v0(tk))+∑0<tk<t‍(t-tk)I¯k(v0(tk))+(1+t)∑k=1m‍tkI¯k(v0(tk)),t∈J,(ii)
the nonlinear impulsive problem (1) has a unique positive solution x* in P~h⋂‍PC1[J,R], where h(t)=(1/2)(t2+t+1),t∈[0,1].Remark 7.
It is easy to see that1/2≤h(t)≤3/2,t∈[0,1].Proof of Theorem6.
Firstly, we show thatA:P~→P~ is increasing, generalized concave. For any x∈P~,
(16)Ax(t)=-∫01‍G(t,s)f(s,x(s))ds-(1+t)∑k=1m‍Ik(x(tk))+∑0<tk<t‍Ik(x(tk))+∑0<tk<t‍(t-tk)I¯k(x(tk))+(1+t)∑k=1m‍tkI¯k(x(tk))=-∫01‍G(t,s)f(s,x(s))ds+∑0<tk<t‍Ik(x(tk))-(1+t)[∑0<tk<t‍Ik(x(tk))+∑t≤tk<1‍Ik(x(tk))]+∑0<tk<t‍(t-tk)I¯k(x(tk))+(1+t)∑k=1m‍tkI¯k(x(tk))=-∫01‍G(t,s)f(s,x(s))ds-[t∑0<tk<t‍Ik(x(tk))+(1+t)∑t≤tk<1‍Ik(x(tk))]+∑0<tk<t‍(t-tk)I¯k(x(tk))+(1+t)∑k=1m‍tkI¯k(x(tk)).
From (H1)and(H2), we know that f(t,x(s))≤0,Ik(x(tk))≤0, and I¯k(x(tk))≥0. So we have Ax(t)≥0 for x∈P~. By Lemma 4, A:P~→P~. It follows from (H1)and(H2) that A:P~→P~ is increasing. Now we prove that A:P~→P~ is generalized concave. Set α(t)=min{α1(t),α2(t),α3(t)},t∈(0,1). Then α(t)∈(t,1). For any x∈P~ and λ∈(0,1), from (H3) we have
(17)A(λx)(t)=-∫01‍G(t,s)f(s,λx(s))ds-[t∑0<tk<t‍Ik(λx(tk))+(1+t)∑t≤tk<1‍Ik(λx(tk))]+∑0<tk<t‍(t-tk)I¯k(λx(tk))+(1+t)∑k=1m‍tkI¯k(λx(tk))≥α1(λ)[-∫01‍G(t,s)f(s,x(s))ds]+α2(λ)×[-t∑0<tk<t‍Ik(x(tk))-(1+t)∑t≤tk<1‍Ik(x(tk))]+α3(λ)[∑0<tk<t‍(t-tk)I¯k(x(tk))++α3(λ)+(1+t)∑k=1m‍tkI¯k(x(tk))]≥α(λ){∑k=1m‍-∫01‍G(t,s)f(s,x(s))ds-[t∑0<tk<t‍Ik(x(tk))--+(1+t)∑t≤tk<1‍Ik(x(tk))]+∑0<tk<t‍(t-tk)I¯k(x(tk))+(1+t)∑k=1m‍tkI¯k(x(tk))}=α(λ)Ax(t).
That is, A(λx)≥α(λ)Ax,x∈P~,λ∈(0,1).
Secondly, we prove thatAh∈P~h. Set
(18)r1=mint∈[0,1][-f(t,12)],r2=maxt∈[0,1][-f(t,32)].
Then from (H1), we have r2≥r1>0. Further, from (H1),(H2), and (H4),
(19)Ah(t)=-∫01‍G(t,s)f(s,h(s))ds-(1+t)∑k=1m‍Ik(h(tk))+∑0<tk<t‍Ik(h(tk))+∑0<tk<t‍(t-tk)I¯k(h(tk))+(1+t)∑k=1m‍tkI¯k(h(tk))≥-∫01‍G(t,s)f(s,12)ds≥r1∫01‍G(t,s)ds=r1h(t),Ah(t)=-∫01‍G(t,s)f(s,h(s))ds-(1+t)∑k=1m‍Ik(h(tk))+∑0<tk<t‍Ik(h(tk))+∑0<tk<t‍(t-tk)I¯k(h(tk))+(1+t)∑k=1m‍tkI¯k(h(tk))≤-∫01‍G(t,s)f(s,32)ds-(1+t)∑k=1m‍Ik(h(tk))+∑k=1m‍(1-tk)I¯k(h(tk))+2∑k=1m‍tkI¯k(h(tk))≤r2h(t)+2(-∑k=1m‍Ik(32))+∑k=1m‍(1+tk)I¯k(32)=r2h(t)+∑k=1m‍[-2Ik(32)+(1+tk)I¯k(32)]≤r2h(t)+2∑k=1m‍[-2Ik(32)+(1+tk)I¯k(32)]·h(t)=(r2+2∑k=1m‍[-2Ik(32)+(1+tk)I¯k(32)])h(t).
Hence,
(20)r1h≤Ah≤(r2+2∑k=1m‍[-2Ik(32)+(1+tk)I¯k(32)])h.
That is, Ah∈P~h. Finally, an application of Theorem 1 implies that (i) there are u0,v0∈P~h such that u0≤Au0,Av0≤v0, (ii) operator equation x=Ax has a unique solution in P~h. That is,
(21)u0(t)≤-∫01‍G(t,s)f(s,u0(s))ds-(1+t)∑k=1m‍Ik(u0(tk))+∑0<tk<t‍Ik(u0(tk))+∑0<tk<t‍(t-tk)I¯k(u0(tk))+(1+t)∑k=1m‍tkI¯k(u0(tk)),t∈J,v0(t)≥-∫01‍G(t,s)f(s,v0(s))ds-(1+t)∑k=1m‍Ik(v0(tk))+∑0<tk<t‍Ik(v0(tk))+∑0<tk<t‍(t-tk)I¯k(v0(tk))+(1+t)∑k=1m‍tkI¯k(v0(tk)),t∈J,
and the problem (1) has a unique positive solution x* in P~h. Moreover, from Lemmas 4 and 5 we know that x*∈PC1[J,R]. Evidently, x* is a positive solution of the problem (1).Theorem 8.
Assume that  (H1)′–(H4)′ hold. Then (i)
there existu0,v0∈P~h such that
(22)u0(t)≤-∫01‍G(t,s)f(s,u¯0(s))ds-(1+t)∑k=1m‍Ik(u¯0(tk))+∑0<tk<t‍Ik(u¯0(tk))+∑0<tk<t‍(t-tk)I¯k(u¯0(tk))+(1+t)∑k=1m‍tkI¯k(u¯0(tk)),t∈J,v0(t)≥-∫01‍G(t,s)f(s,v¯0(s))ds-(1+t)∑k=1m‍Ik(v¯0(tk))+∑0<tk<t‍Ik(v¯0(tk))+∑0<tk<t‍(t-tk)I¯k(v¯0(tk))+(1+t)∑k=1m‍tkI¯k(v¯0(tk)),t∈J,where
(23)u¯0(t)=-∫01‍G(t,s)f(s,u0(s))ds-(1+t)∑k=1m‍Ik(u0(tk))+∑0<tk<t‍Ik(u0(tk))+∑0<tk<t‍(t-tk)I¯k(u0(tk))+(1+t)∑k=1m‍tkI¯k(u0(tk)),t∈J,v¯0(t)=-∫01‍G(t,s)f(s,v0(s))ds-(1+t)∑k=1m‍Ik(v0(tk))+∑0<tk<t‍Ik(v0(tk))+∑0<tk<t‍(t-tk)I¯k(v0(tk))+(1+t)∑k=1m‍tkI¯k(v0(tk)),t∈J,(ii)
the nonlinear impulsive problem (1) has a unique positive solution x* in P~h⋂‍PC1[J,R], where h(t)=(1/2)(t2+t+1), t∈[0,1].Proof.
From the proof of Theorem6, for any x∈P~,
(24)Ax(t)=-∫01‍G(t,s)f(s,x(s))ds-[t∑0<tk<t‍Ik(x(tk))+(1+t)∑t≤tk<1‍Ik(x(tk))]+∑0<tk<t‍(t-tk)I¯k(x(tk))+(1+t)∑k=1m‍tkI¯k(x(tk)).
From (H1)′and(H2)′, we know that Ax(t)≥0, t∈[0,1]. By Lemma 4, A:P~→P~. It follows from (H1)′and(H2)′ that A:P~→P~ is decreasing. Set β(t)=max{β1(t),β2(t),β3(t)}, t∈(0,1). Then β(t)∈(0,1). For any x∈P~ and λ∈(0,1), from (H3)′ we have
(25)A(λx)(t)=-∫01‍G(t,s)f(s,λx(s))ds--[t∑0<tk<t‍Ik(λx(tk))+(1+t)∑t≤tk<1‍Ik(λx(tk))]-+∑0<tk<t‍(t-tk)I¯k(λx(tk))-+(1+t)∑k=1m‍tkI¯k(λx(tk))≤λ-β1(λ)[-∫01‍G(t,s)f(s,x(s))ds]+λ-β2(λ)×[-t∑0<tk<t‍Ik(x(tk))-(1+t)∑t≤tk<1‍Ik(x(tk))]+λ-β3(λ)[∑0<tk<t‍(t-tk)I¯k(x(tk))+λ-β3(λ)++(1+t)∑k=1m‍tkI¯k(x(tk))]≤λ-β(λ){∑k=1m‍-∫01‍G(t,s)f(s,x(s))ds-λ-β(λ)-[t∑0<tk<t‍Ik(x(tk))-λ-β(λ)-++(1+t)∑t≤tk<1‍Ik(x(tk))]-λ-β(λ)+∑0<tk<t‍(t-tk)I¯k(x(tk))-λ-β(λ)+(1+t)∑k=1m‍tkI¯k(x(tk))}=λ-β(λ)Ax(t).
That is, A(λx)≤λ-β(λ)Ax,x∈P~,λ∈(0,1). Further, for λ∈(0,1) and x∈P~,
(26)Ax=A(λ·1λx)≤λ-β(λ)A(1λx).
So we obtain A((1/λ)x)≥λβ(λ)Ax,x∈P~,λ∈(0,1). Consequently, A2:P~→P~ is increasing, and, for x∈P~,λ∈(0,1),
(27)A2(λx)=A(A(λx))≥A(λ-β(λ)Ax)=A(1λβ(λ)Ax)≥(λβ(λ))β(λβ(λ))A2x≥λβ(λ)A2x.
Letα(t)=tβ(t),t∈(0,1). Then α(t)∈(t,1) and A2(λx)≥α(λ)A2x,x∈P~,λ∈(0,1). So the operator A2:P~→P~ is generalized concave. Next we prove that A2h∈P~h. Set
(28)r1=mint∈[0,1][-f(t,32)],r2=maxt∈[0,1][-f(t,12)].
Then from  (H1)′, we have r2≥r1>0. Further, from (H1)′,(H2)′, and (H4)′,
(29)Ah(t)=-∫01‍G(t,s)f(s,h(s))ds-(1+t)∑k=1m‍Ik(h(tk))+∑0<tk<t‍Ik(h(tk))+∑0<tk<t‍(t-tk)I¯k(h(tk))+(1+t)∑k=1m‍tkI¯k(h(tk))≥-∫01‍G(t,s)f(s,32)ds≥r1∫01‍G(t,s)ds=r1h(t),Ah(t)=-∫01‍G(t,s)f(s,h(s))ds-(1+t)∑k=1m‍Ik(h(tk))+∑0<tk<t‍Ik(h(tk))+∑0<tk<t‍(t-tk)I¯k(h(tk))+(1+t)∑k=1m‍tkI¯k(h(tk))≤-∫01‍G(t,s)f(s,12)ds-(1+t)∑k=1m‍Ik(h(tk))+∑k=1m‍(1-tk)I¯k(h(tk))+2∑k=1m‍tkI¯k(h(tk))≤r2h(t)+2(-∑k=1m‍Ik(12))+∑k=1m‍(1+tk)I¯k(12)=r2h(t)+∑k=1m‍[-2Ik(12)+(1+tk)I¯k(12)]≤r2h(t)+2∑k=1m‍[-2Ik(12)+(1+tk)I¯k(12)]·h(t)=(r2+2∑k=1m‍[-2Ik(12)+(1+tk)I¯k(12)])h(t).
Hence,
(30)r1h≤Ah≤(r2+2∑k=1m‍[-2Ik(12)+(1+tk)I¯k(12)])h.
We can choose a sufficiently small number r0∈(0,1) such that
(31)r0≤r1<r2+2∑k=1m‍[-2Ik(12)+(1+tk)I¯k(12)]≤1r0.
Then we get r0h≤Ah≤(1/r0)h. Further,
(32)A2h=A(Ah)≤A(r0h)≤r0-β(r0)Ah≤1r01+β(r0)h,A2h=A(Ah)≥A(1r0h)≥r0β(r0)Ah≥r01+β(r0)h.
That is, A2h∈P~h. Finally, an application of Theorem 1 implies that (i) there are u0,v0∈P~h such that u0≤A2u0,A2v0≤v0, (ii) operator equation x=A2x has a unique solution in P~h. Let u¯0=Au0,v¯0=Av0. Then, u0≤Au¯0,Av¯0≤v0. That is,
(33)u0(t)≤-∫01‍G(t,s)f(s,u¯0(s))ds-(1+t)∑k=1m‍Ik(u¯0(tk))+∑0<tk<t‍Ik(u¯0(tk))+∑0<tk<t‍(t-tk)I¯k(u¯0(tk))+(1+t)∑k=1m‍tkI¯k(u¯0(tk)),t∈J,v0(t)≥-∫01‍G(t,s)f(s,v¯0(s))ds-(1+t)∑k=1m‍Ik(v¯0(tk))+∑0<tk<t‍Ik(v¯0(tk))+∑0<tk<t‍(t-tk)I¯k(v¯0(tk))+(1+t)∑k=1m‍tkI¯k(v¯0(tk)),t∈J.
Next we show that x* is the unique fixed point of A in P~h. In view of A2(Ax*)=A(A2x*)=Ax*, and by the uniqueness of solutions for the operator equation x=A2x, we have Ax*=x*. Suppose that y* is another fixed point of A in P~h. Then A2y*=A(A(y*))=Ay*=y*. Hence, by the uniqueness of solutions for the operator equation x=A2x, we obtain x*=y*. So the problem (1) has a unique positive solution x* in P~h. Moreover, from Lemmas 4 and 5 we know that x*∈PC1[J,R]. Evidently, x* is a positive solution of the problem (1).Remark 9.
Here, we provide an alternative approach to study the same type of problems under different conditions. Our result can guarantee the existence of a unique positive solution without supposing the existence of upper-lower solutions. The method used in this paper is relatively new to the literature and so is the existence and uniqueness result to the impulsive differential equations.
In the following we consider two special cases of the problem (1):
(34)-x′′(t)=f(t,x(t)),t≠tk,k=1,2,…,m,Δx|t=tk=Ik(x(tk)),k=1,2,…,m,x(0)=x′(0),x(1)=x′(1),(35)-x′′(t)=f(t,x(t)),t≠tk,k=1,2,…,m,Δx′|t=tk=I¯k(x(tk)),k=1,2,…,m,x(0)=x′(0),x(1)=x′(1),
where f∈C[J×R,R], Ik, I¯k∈C[R,R], k=1,2,…,m.
From Theorems6 and 8, we have the following conclusions.Corollary 10.
Assume that(H1) holds and(G1)
Ik(0)≤0 and Ik(x) is decreasing in x∈[0,∞),k=1,2,…,m with
(36)∑k=1m‍Ik(32)<0,(G2)
for anyλ∈(0,1),t∈[0,1] and x≥0, there exist α1(λ),α2(λ)∈(λ,1) such that
(37)f(t,λx)≤α1(λ)f(t,x),Ik(λx)≤α2(λ)Ik(x),k=1,2,…,m.Then (i)
there existu0,v0∈P~h such that
(38)u0(t)≤-∫01‍G(t,s)f(s,u0(s))ds-(1+t)∑k=1m‍Ik(u0(tk))+∑0<tk<t‍Ik(u0(tk)),t∈J,v0(t)≥-∫01‍G(t,s)f(s,v0(s))ds-(1+t)∑k=1m‍Ik(v0(tk))+∑0<tk<t‍Ik(v0(tk)),t∈J,(ii)
the nonlinear impulsive problem (34) has a unique positive solution x* in P~h, where h(t)=(1/2)(t2+t+1), t∈[0,1], and G(t,s) is given as in Lemma 4.Corollary 11.
Assume that(H1)′ hold and(G1)′
Ik(x)≤0 for [0,∞) and Ik(x) is increasing in x∈[0,∞),k=1,2,…,m with
(39)∑k=1m‍Ik(12)<0,(G2)′
for anyλ∈(0,1),t∈[0,1] and x≥0, there exist β1(λ),β2(λ)∈(0,1) such that
(40)f(t,λx)≥λ-β1(λ)f(t,x),Ik(λx)≥λ-β2(λ)Ik(x),k=1,2,…,m.
Then(i)
there existu0,v0∈P~h such that
(41)u0(t)≤-∫01‍G(t,s)f(s,u¯0(s))ds-(1+t)∑k=1m‍Ik(u¯0(tk))+∑0<tk<t‍Ik(u¯0(tk)),t∈J,v0(t)≥-∫01‍G(t,s)f(s,v¯0(s))ds-(1+t)∑k=1m‍Ik(v¯0(tk))+∑0<tk<t‍Ik(v¯0(tk)),t∈J,where
(42)u¯0(t)=-∫01‍G(t,s)f(s,u0(s))ds-(1+t)∑k=1m‍Ik(u0(tk))+∑0<tk<t‍Ik(u0(tk)),t∈J,v¯0(t)=-∫01‍G(t,s)f(s,v0(s))ds-(1+t)∑k=1m‍Ik(v0(tk))+∑0<tk<t‍Ik(v0(tk)),t∈J,(ii)
the nonlinear impulsive problem (34) has a unique positive solution x* in P~h, where h(t)=(1/2)(t2+t+1) and  t∈[0,1] and G(t,s) is given as in Lemma 4.Corollary 12.
Assume that(H1) holds and(G3)
I¯k(0)≥0 and I¯k(x) is increasing in x∈[0,∞),k=1,2,…,m with
(43)∑k=1m‍[(1+tk)I¯k(32)]>0,(G4)
for anyλ∈(0,1),t∈[0,1] and x≥0, there exist α1(λ),α2(λ)∈(λ,1) such that
(44)f(t,λx)≤α1(λ)f(t,x),I¯k(λx)≥α2(λ)I¯k(x),k=1,2,…,m.Then (i)
there existu0,v0∈P˘h such that
(45)u0(t)≤-∫01‍G(t,s)f(s,u0(s))ds+∑0<tk<t‍(t-tk)I¯k(u0(tk))+(1+t)∑k=1m‍tkI¯k(u0(tk)),t∈J,v0(t)≥-∫01‍G(t,s)f(s,v0(s))ds+∑0<tk<t‍(t-tk)I¯k(v0(tk))+(1+t)∑k=1m‍tkI¯k(v0(tk)),t∈J,(ii)
the nonlinear impulsive problem (35) has a unique positive solution x* in P˘h⋂‍PC1[J,R], where h(t)=(1/2)(t2+t+1), t∈[0,1], P˘={x∈C[J,R]∣x(t)≥0,t∈J}, and PC1[J,R]={x∈C[J,R]∣x′(t) is continuous at t≠tk and left continuous at t=tk, x′(tk+) exists, k=1,2,…,m}.Corollary 13.
Assume that(H1)′ hold and(G3)′
I¯k(x)≥0 for [0,∞) and I¯k(x) is decreasing in x∈[0,∞), k=1,2,…,m with
(46)∑k=1m‍(1+tk)I¯k(12)>0,(G4)′
for anyλ∈(0,1), t∈[0,1], and x≥0, there exist β1(λ),β2(λ)∈(0,1) such that
(47)f(t,λx)≥λ-β1(λ)f(t,x),I¯k(λx)≤λ-β2(λ)I¯k(x),k=1,2,…,m.
Then(i)
there existu0,v0∈P˘h such that
(48)u0(t)≤-∫01‍G(t,s)f(s,u¯0(s))ds+∑0<tk<t‍(t-tk)I¯k(u¯0(tk))+(1+t)∑k=1m‍tkI¯k(u¯0(tk)),t∈J,v0(t)≥-∫01‍G(t,s)f(s,v¯0(s))ds+∑0<tk<t‍(t-tk)I¯k(v¯0(tk))+(1+t)∑k=1m‍tkI¯k(v¯0(tk)),t∈J,where
(49)u¯0(t)=-∫01‍G(t,s)f(s,u0(s))ds+∑0<tk<t‍(t-tk)I¯k(u0(tk))+(1+t)∑k=1m‍tkI¯k(u0(tk)),t∈J,v¯0(t)=-∫01‍G(t,s)f(s,v0(s))ds+∑0<tk<t‍(t-tk)I¯k(v0(tk))+(1+t)∑k=1m‍tkI¯k(v0(tk)),t∈J,(ii)
the nonlinear impulsive problem (35) has a unique positive solution x* in P˘h⋂‍PC1[J,R], where h(t)=(1/2)(t2+t+1), t∈[0,1], P˘={x∈C[J,R]∣x(t)≥0,t∈J}, and PC1[J,R]={x∈C[J,R]∣x′(t) is continuous at t≠tk and left continuous at t=tk,x′(tk+) exists, k=1,2,…,m}.


## 4. An Example

To illustrate how our main results can be used in practice we present an example.Example 1.
Consider the following boundary value problem:(50)-x′′(t)=-tx+4,t∈J,t≠12,Δx|t=1/2=-x1/3(12),Δx′|t=1/2=x1/4(12),x(0)=x′(0),x(1)=x′(1).Conclusion. BVP (50) has a unique positive solution in P~h, where h(t)=(1/2)(t2+t+1), t∈[0,1].Proof.
BVP (50) can be regarded as a BVP of the form (1), where t1=(1/2), f(t,x)=-tx+4, I1(x)=-x1/3, and I¯1(x)=x1/4. It is not difficult to see that the conditions (H1),(H2), and (H4) hold. In addition, let α1(λ)=λ1/2, α2(λ)=λ1/3, and α3(λ)=λ1/4. Then, the condition (H3) of Theorem 6 holds. Hence, by Theorem 6, the conclusion follows, and the proof is complete.Remark 14.
Example1 implies that there is a large number of functions that satisfy the conditions of Theorem 6. In addition, the conditions of Theorem 6 are also easy to check.



---

*Source: 259730-2013-06-05.xml*