# Infinite Boundary Value Problems for Second-Order Nonlinear Impulsive Differential Equations with Supremum

Authors: Piao-Piao Shi; Wen-Xia Wang
Journal: Discrete Dynamics in Nature and Society
 (2010)

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## Abstract

We investigate the infinite boundary value problems for second-order
impulsive differential equations with supremum by establishing a new comparison result
and using the lower and upper solution method, and obtain the existence results for their
maximal and minimal solutions.


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## Body

## 1. Introduction

Differential equations with supremum are used modelling different real processes, and have been receiving much attention in recent years (see [1, 2]). In the theory of automatic regulation, for example, they are used in describing the system for regulation of the voltage of generator with constant current: T0u′(t)+u(t)+qmaxs∈[t-h,t]u(s)=f(t) (see [1]). If the equation is impulsive, periodic boundary value problem for first-order differential equation with supremum on finite domain was studied in [2], and on infinite domain, infinite boundary value problem for the same equation was investigated in [3]. Such equations with supremum are about first-order in the previous literature [1–3], but little is about second-order. Motivated by [2–5], we discuss in this paper the existence of maximal and minimal solutions of the system (IBVP): (1.1)x′′(t)=f(t,x(t),x′(t),sups∈[t-h,t]x′(s)),t≠tk,t∈J,k=1,2,…,Δx|t=tk=akx'(tk),k=1,2,…,Δx'|t=tk=Ĩk(x(tk),x′(tk)),k=1,2,…,x(0)=x0,x'(0)=x'(∞),x'(t)=x'(0),t∈[-h,0],
where J=[0,+∞);f∈C[J×R×R×R,R]; ak,h∈R+, Ĩk∈C[R×R,R], 0<t1<t2<⋯<tk<⋯<+∞, and tk→+∞ as k→+∞,k=1,2,…;x0∈R and ∑k=1∞ak is convergent, x'(∞)=limt→+∞x'(t);Δx|t=tk=x(tk+)-x(tk-) denotes the jump of x(t) at t=tk, where x(tk+) and x(tk-) resent the right-hand and left-hand limit of x(t) at t=tk, respectively. Δx'|t=tk has similar meaning for x'(t). Denote J0=[0,t1], J1=(t1,t2],…,Jk=(tk,tk+1]…, J′=J∖{t1,t2,…,tk…}, Jh=[-h,+∞).LetPC[Jh,R]={x:Jh→R∣x(t)=x(0), for t∈[-h,0]; x(t) is continuous at t∈J', left continuous at t=tk, and each x(tk+) exists, for k=1,2,…}, PC1[Jh,R]={x:Jh→R∣x'(t)=x'(0), for t∈[-h,0]; x'(t) is continuous at t≠tk, x(tk-), x(tk+), x'(tk+) and x'(tk-) exist, and x(tk-)=x(tk), k=1,2,…}, BPC[Jh,R]={x∈PC[Jh,R]|supt∈J∥x(t)∥<+∞}, TPC[Jh,R]={x∈BPC[Jh,R]∣limt→+∞x(t)=x(∞)exists}, BPC1[Jh,R]={x∈PC1[Jh,R]|supt∈J∥x'(t)∥<+∞}, and TPC1[Jh,R]={x∈BPC1[Jh,R]∣limt→+∞x′(t)=x′(∞)exists}.We get from [4] that x'-(tk)=x'(tk-). In the following, x'(tk) is understood as x'-(tk). Evidently, BPC[Jh,R] equipped with the norm ∥x∥B=supt∈J∥x(t)∥ is a Banach space and TPC[Jh,R]⊂BPC[Jh,R].We sayx∈TPC1[Jh,R]∩C2[J',R] is a solution of IBVP(1.1), if it is satisfies (1.1).In Section2, we prove the existence result of minimal and maximal solutions for first-order impulsive differential equations which nonlinearly involve the operator B, that is, Theorem 2.5. In special case of IBVP(2.1) where f=f(t,u(t),sups∈[t-h,t]u(s)) and Ĩk=Ĩk(u(tk)), the infinite boundary value problems for first-order impulsive differential equations were studied in [3]. In Section 3, by applying Theorem 2.5, the main result (Theorem 3.1) of this paper is obtained, that is the existence theorem of minimal and maximal solutions of IBVP(1.1).


## 2. Result for First-Order Impulsive Differential Equation with Nonlinear Operator Terms

Consider the existence of solutions for the following first-order impulsive differential equations:(2.1)y'(t)=f(t,(By)(t),y(t),sups∈[t-h,t]y(s)),t∈J,t≠tk,Δy|t=tk=Ĩk((By)(tk),y(tk)),k=1,2,…,y(0)=y(∞),y(t)=y(0),t∈[-h,0],
where f,Ĩk(k=1,2,…) are the same as IBVP (1.1), and (By)(t)=x0+∫0ty(s)ds+∑0<tk<taky(tk).Lemma 2.1 (Comparison Result).
Letx∈TPC[Jh,R]∩C1[J',R]. Assume that there exist a,b,c∈C[J,R+]∩L1(J), ta∈L1(J), b≢0, constants Lk≥0,k=1,2,…, and ∑k=1∞Lk<∞ such that
(2.2)x'(t)≥-a(t)(Dx)(t)-b(t)x(t)-c(t)sups∈[t-h,t]x(s),t∈J,t≠tk,Δx|t=tk≥-Lkx(tk),k=1,2,…,x(0)≥x(∞),x(t)=x(0),t∈[-h,0].
Then x(t)≥0 for t∈Jh provided that
(2.3)e∫0∞b(τ)dτ{∫0∞e∫0tb(τ)dτ[c(t)+a(t)(t+∑k=1∞ak)]dt+∑k=1∞Lk}≤1,
where (Dx)(t)=∫0tx(s)ds+∑0<tk<takx(tk).Proof.
Setm(t)=x(t)e∫0tb(τ)dτ, then we have from (2.2) that
(2.4)m'(t)≥-a(t)e∫0tb(τ)dτ[∫0tm(s)e-∫0sb(τ)dτds+∑0<tk<takm(tk)e-∫0tkb(τ)dτ]-c(t)e∫0tb(τ)dτsups∈[t-h,t]m(s)e-∫0sb(τ)dτ,t∈J,t≠tk,Δm|t=tk≥-Lkm(tk),k=1,2,…,m(0)≥m(∞)e-∫0∞b(τ)dτ,m(t)=m(0)e∫0tb(τ)dτ,t∈[-h,0].
We claim thatm(t)≥0 for t∈J, moreover m(t)≥0 for t∈Jh. Otherwise, we will consider two cases.
Case 1.
m(t)≤0 for t∈J, and there exists t1*∈J such that m(t1*)<0.Case 2.
there existt1*,t2*∈J such that m(t1*)<0, m(t2*)>0.
In Case1, we see from (2.4) that m'(t)≥0 for t∈J, t≠tk. On the other hand m(tk+)=m(tk)+Δm|t=tk≥m(tk), thus m(t) is increasing on J, and m(0)≤m(t1*)<0,m(0)≤m(∞)≤m(0)e∫0∞b(τ)dτ. Hence e∫0∞b(τ)dτ≤1, which is a contradiction.
In Case2, denote supt∈Jm(t)=λ, then λ>0, and it is clear that supt∈Jhm(t)=supt∈Jm(t)=λ. Then we have either that (a): there exists some Ji such that m(t0*)=λ for some t0*∈Ji, or m(ti+)=λ, or that (b):m(∞)=λ.
In subcase(a), we only discuss the case of m(t0*)=λ for t0*∈Ji, since the discussion of the case of m(ti+)=λ is similar.
If there exists someJi such that m(t0*)=λ for t0*∈Ji, then we have from (2.4) that for t∈J, t≠tk, (2.5)m'(t)≥λe∫0tb(τ)dτ[-c(t)-a(t)(t+∑k=1∞ak)].
For any integer l≥i, the calculus fundamental principle implies that
(2.6)m(tl+1)-m(t0*)=∫t0*tl+1m′(t)dt+∑k=i+1lΔm|t=tk≥-λ∫t0*tl+1e∫0tb(τ)dτ[c(t)+a(t)(t+∑k=1∞ak)]dt-λ∑k=i+1lLk.
Let l→+∞, we have
(2.7)m(∞)-λ≥-λ∫t0*∞e∫0tb(τ)dτ[c(t)+a(t)(t+∑k=1∞ak)]dt-λ∑k=i+1∞Lk.
This means that
(2.8)m(∞)≥λ{1-∫t0*∞e∫0tb(τ)dτ[c(t)+a(t)(t+∑k=1∞ak)]dt-∑k=i+1∞Lk}.
From (2.3), we have m(∞)≥0 and m(0)≥m(∞)e-∫0∞b(τ)dτ≥0. Therefore 0<t1*<+∞. Without loss of generality, we assume that t1*∈Jj.
Ift0*<t1*, then i≤j. Hence using the same method as is used above, we have(2.9)-λ>m(t1*)-m(t0*)≥-λ{∫t0*t1*e∫0tb(τ)dτ[c(t)+a(t)(t+∑k=1∞ak)]dt+∑k=i+1jLk}≥-λ{∫0∞e∫0tb(τ)dτ[c(t)+a(t)(t+∑k=1∞ak)]dt+∑k=1∞Lk},
hence,
(2.10)∫0∞e∫0tb(τ)dτ[c(t)+a(t)(t+∑k=1∞ak)]dt+∑k=1∞Lk>1,
which is a contradiction to (2.3).
Ift0*>t1*, then i≥j. Similar argument shows that(2.11)-m(0)>m(t1*)-m(0)≥-λ{∫0t1*e∫0tb(τ)dτ[c(t)+a(t)(t+∑k=1∞ak)]dt+∑k=1jLk},
which, noticing m(0)≥m(∞)e-∫0∞b(τ)dτ, implies that
(2.12)-m(∞)>-λe∫0∞b(τ)dτ{∫0t1*e∫0tb(τ)dτ[c(t)+a(t)(t+∑k=1∞ak)]dt+∑k=1jLk}.
Adding (2.8) and (2.12), we show that
(2.13)1<∫t0*∞e∫0tb(τ)dτ[c(t)+a(t)(t+∑k=1∞ak)]dt+∑k=i+1∞Lk+e∫0∞b(τ)dτ{∫0t1*e∫0tb(τ)dτ[c(t)+a(t)(t+∑k=1∞ak)]dt+∑k=1jLk}<e∫0∞b(τ)dτ{∫0∞e∫0tb(τ)dτ[c(t)+a(t)(t+∑k=1∞ak)]dt+∑k=1∞Lk},
which also contradicts (2.3).
In subcase(b), m(∞)=λ, then it follows from (2.12) that(2.14)-λ>-λe∫0∞b(τ)dτ{∫0∞e∫0tb(τ)dτ[c(t)+a(t)(t+∑k=1∞ak)]dt+∑k=1∞Lk}.
This also leads to a contradiction with (2.3).
Therefore, the Case2 is also impossible. Then, we conclude that m(t)≥0 on Jh, and hence x(t)≥0 on Jh. The proof is complete.We first consider the following linear impulsive differential equations:(2.15)y'(t)=f(t,Bη(t),η(t),sups∈[t-h,t]η(s))-a(t)((By)(t)-(Bη)(t))-b(t)(y(t)-η(t))-c(t)(sups∈[t-h,t]y(s)-sups∈[t-h,t]η(s)),t∈J,t≠tk,Δy|t=tk=Ĩk((Bη)(tk),η(tk))-Lk(y(tk)-η(tk)),k=1,2,…,y(0)=y(∞),y(t)=y(0),t∈[-h,0].Let us list some conditions for convenience.(H1)
There existp,q,l∈C[J,R+]∩L1(J),tp∈L1(J), such that
(2.16)|f(t,u,v,w)|≤p(t)|u|+q(t)|v|+l(t)|w|,u,v,w∈R,t∈J.(H2)
There existL̃k≥0,k=1,2,…, such that ∑k=1∞L̃k is convergent and
(2.17)|Ik(u,v)|≤L̃k|v|,u,v∈R,∀t∈J,k=1,2,….Lemma 2.2.
Leta,b,c,ta∈C[J,R+]∩L1(J), b≢0, Lk≥0, k=1,2,…, with ∑k=1∞Lk<∞, and assume also that conditions (H1) and (H2) hold. Then for any η∈BPC[Jh,R], y∈TPC[Jh,R]∩C1[J',R] is a solution of the linear impulsive differential equations (2.15) if and only if y∈BPC[Jh,R] is a solution of the following impulsive integral equation:
(2.18)y(t)=e-∫0tb(τ)dτ×{(e∫0∞b(τ)dτ-1)-1[∫0∞e∫0sb(τ)dτ[f(s,Bη(s),η(s),supr∈[s-h,s]η(r))-a(s)((By)(s)-(Bη)(s))+b(s)η(s)-c(s)(supr∈[s-h,s]y(r)-supr∈[s-h,s]η(r))]ds+∑k=1∞e∫0tkb(τ)dτ[Ĩk((Bη)(tk),η(tk))-Lk(y(tk)-η(tk))]]+∫0te∫0sb(τ)dτ[f(s,Bη(s),η(s),supr∈[s-h,s]η(r))-a(s)((By)(r)-(Bη)(r))+b(s)η(s)-c(s)(supr∈[s-h,s]y(r)-supr∈[s-h,s]η(r))]ds}+∑0<tk<te∫ttkb(τ)dτ[Ĩk((Bη)(tk),η(tk))-Lk(y(tk)-η(tk))],∀t∈J,
with the initial condition y(t)=y(0), for t∈[-h,0].Proof.
By the definition ofB, we have |Bη(s)|≤|x0|+(s+∑k=1∞ak)∥η∥B. Together with (H1), (H2), we have
(2.19)|f(s,Bη(s),η(s),supr∈[s-h,s]η(r))-a(s)((By)(s)-(Bη)(s))+b(s)η(s)-c(s)(supr∈[s-h,s]y(r)-supr∈[s-h,s]η(r))|≤[p(s)(s+∑k=1∞ak)+q(s)+l(s)+a(s)(s+∑k=1∞ak)+b(s)+c(s)]∥η∥B+[c(s)+a(s)(s+∑k=1∞ak)]∥y∥B+(p(s)+2a(s))|x0|=:M(s),|Ĩk((Bη)(tk),η(tk))-Lk(y(tk)-η(tk))|≤(L̃k+Lk)∥η∥B+Lk∥y∥B=:Nk,
which, noticing M(s)∈L1(J) and ∑k=1∞Nk<∞, implies that the right hand of (2.18) is well defined. Moreover, we show by direct computation that y∈TPC[Jh,R]∩C1[J',R] is a solution of (2.15).
We next prove the uniqueness of solution. Lety1,y2 be any two solutions of (2.15), and y=y1-y2, then we have (2.20)y′(t)=y1′(t)-y2′(t)=-a(t)(Dy)(t)-b(t)y(t)-c(t)sups∈[t-h,t]y(s),t∈J,t≠tk,Δy|t=tk=-Lky(tk),k=1,2,…,y(0)=y(∞),y(t)=y(0),t∈[-h,0].
Hence Lemma 2.1 implies that y≥0, that is, y1≥y2. Similar argument shows that y1≤y2. Therefore y1=y2. We complete the proof.Lemma 2.3.
Let(H1) and (H2) be satisfied. Assume further that
(2.21)g=1e∫0∞b(τ)dτ-1[(2e∫0∞b(τ)dτ-1)∫0∞[c(s)+a(s)(s+∑k=1∞ak)]ds+∑k=1∞Lk(e∫0∞b(τ)dτ+e∫0tkb(τ)dτ-1)]<1,
then the integral equation (2.18) possesses a unique solution y∈BPC[Jh,R].Proof.
For anyη∈BPC[Jh,R], we define the operator T by (Ty)(t) being the right hand of (2.18) and (Ty)(0)=(Ty)(t), t∈[-h,0]. By virtue of (H1), (H2), it is obvious that T:BPC[Jh,R]→BPC[Jh,R]. Then for any y1,y2∈BPC[Jh,R], we have
(2.22)|(By2)(s)-(By1)(s)|≤(s+∑k=1∞ak)∥y1-y2∥B.
Moreover,
(2.23)|(Ty1)(t)-(Ty2)(t)|≤1e∫0∞b(τ)dτ-1[|∫0∞e∫0sb(τ)dτ[a(s)(By2(s)-By1(s))+c(s)supr∈[s-h,s](y2(r)-y1(r))]ds|+∑k=1∞e∫0tkb(τ)dτLk|y2(tk)-y1(tk)|]+|∫0te-∫stb(τ)dτ[a(s)((By2)(s)-(By1)(s))+c(s)supr∈[s-h,s](y2(r)-y1(r))]ds|+∑0<tk<te-∫tktb(τ)dτLk|y2(tk)-y1(tk)|≤e∫0∞b(τ)dτe∫0∞b(τ)dτ-1∫0∞[a(s)|(By2)(s)-(By1)(s)|+c(s)supr∈[s-h,s]|y2(r)-y1(r)|]ds+1e∫0∞b(τ)dτ-1∑k=1∞e∫0tkb(τ)dτLk|y2(tk)-y1(tk)|+∫0∞[a(s)|(By2)(s)-(By1)(s)|+c(s)supr∈[s-h,s]|y2(r)-y1(r)|]ds+∑k=1∞Lk|y2(tk)-y1(tk)|≤1e∫0∞b(τ)dτ-1[(2e∫0∞b(τ)dτ-1)∫0∞[c(s)+a(s)(s+∑k=1∞ak)]ds+∑k=1∞Lk(e∫0∞b(τ)dτ+e∫0tkb(τ)dτ-1)]∥y1-y2∥B.
Thus, ∥Ty1-Ty2∥B≤g∥y1-y2∥B. Hence, Banach's fixed point theorem implies that T has a unique fixed point, that is, a unique solution of (2.18).
For anyη∈BPC[Jh,R], define an operator A by A:(Aη)(t)= the right hand of (2.18) on J, and (Aη)(t)=(Aη)(0) for t∈[-h,0].Lemmas2.2 and 2.3 immediately yield the following result.Lemma 2.4.
y∈TPC[Jh,R]∩C1[J',R] is a solution of (2.1) if and only if y∈BPC[Jh,R] is a fixed point of A.Let us list some conditions for convenience.(H3) There exist the upper and lower solutions of (2.1), that is, u0,v0∈TPC[Jh,R]∩C1[J',R], satisfying u0(t)≤v0(t),
(2.24)u0′(t)≤f(t,(Bu0)(t),u0(t),sups∈[t-h,t]u0(s)),t∈J,t≠tk,Δu0|t=tk≤Ĩk((Bu0)(tk),u0(tk)),k=1,2,…,u0(0)≤u0(∞),u0(t)=u0(0),t∈[-h,0],
and v0(t) satisfies inverse inequalities above.Define the sets[u0,v0]={u∈PC[Jh,R]:u0(t)≤u(t)≤v0(t),t∈Jh}, Ω={(t,x,y,z):t∈J,(Bu0)(t)≤x(t)≤(Bv0)(t), u0(t)≤y(t)≤v0(t), sups∈[t-h,t]u0(s)≤z(t)≤sups∈[t-h,t]v0(s)}.(H4) There exist a,b,c,ta∈C[J,R+]∩L1(J) with b≢0, such that
(2.25)f(t,x,y,z)-f(t,x¯,y¯,z¯)≥-a(t)(x-x¯)-b(t)(y-y¯)-c(t)(z-z¯),∀t∈J,Ĩk(x,y)-Ĩk(x¯,y¯)≥-Lk(y-y¯),k=1,2,…,
where (t,x,y,z),(t,x¯,y¯,z¯)∈Ω,x¯≤x,y¯≤y,z¯≤z.Theorem 2.5.
Assume that conditions(H1)–(H4), (2.3), and (2.21) hold. Then (2.1) has minimal and maximal solutions u*,v*∈[u0,v0]; moreover, the iterative sequences {vn(t)} and {un(t)} converge uniformly on each Jk to v*(t) and u*(t), where
(2.26)un(t)=Aun-1(t),vn(t)=Avn-1(t),∀t∈J,un(t)=un(0),vn(t)=vn(0),t∈[-h,0],n=1,2,….Proof.
Firstly, the proof of Lemma2.2 implies that the operator A is well defined.Next, we will show thatu0≤Au0,Av0≤v0 and A is nondecreasing in [u0,v0].Indeed, for anyη∈[u0,v0], we have by Lemmas 2.2 and 2.3 that Aη∈TPC[Jh,R]∩C1[J',R] is a unique solution of (2.15), together with (2.26), we deduce that (2.27)un(t)=e-∫0tb(s)ds{(e∫0∞b(τ)dτ-1)-1×[∫0∞e∫0sb(τ)dτ[f(s,(Bun-1)(s),un-1(s),supr∈[s-h,s]un-1(r))-a(s)((Bun)(s)-(Bun-1)(s))+b(s)un-1(s)-c(s)(supr∈[s-h,s]un(r)-supr∈[s-h,s]un-1(r))]ds+∑k=1∞e∫0tkb(τ)dτ[Ĩk((Bun-1)(tk),un-1(tk))-Lk(un(tk)-un-1(tk))]]+∫0te∫0sb(τ)dτ[f(s,(Bun-1)(s),un-1(s),supr∈[s-h,s]un-1(r))-a(s)((Bun)(s)-(Bun-1)(s))+b(s)un-1(s)-c(s)(supr∈[s-h,s]un(r)-supr∈[s-h,s]un-1(r))]ds}+∑0<tk<te∫ttkb(τ)dτ[Ĩk((Bun-1)(tk),un-1(tk))-Lk(un(tk)-un-1(tk))].
Let u1-u0=u, then by (H3), (2.15), and the definition of A, we have (2.28)u'(t)=u1′(t)-u0′(t)≥-a(t)(Du)(t)-b(t)u(t)-c(t)sups∈[t-h,t]u(s),t∈J,t≠tk,Δu|t=tk≥-Lku(tk),k=1,2,…,u(0)≥u(∞),u(t)=u(0),t∈[-h,0].
This implies by Lemma 2.1 that u(t)≥0, that is, Au0=u1≥u0. Analogously, we get Av0≤v0. Similar argument by the facts that Aη is a solution of (2.15) and (H4), shows that A is nondecreasing. Moreover, together with (2.26), we have (2.29)u0(t)≤u1(t)≤⋯≤un(t)≤⋯≤vn(t)≤⋯≤v1(t)≤v0(t),t∈Jh.
Therefore it follows from (2.29) that (2.30)limn→∞un(t)=u*(t),t∈Jh,
and then there exists a constant L*>0, such that ∥un∥B≤L*,n=1,2,…. Hence, for s∈J, by (H1), (H2), we have (2.31)|f(s,(Bun-1)(s),un-1(s),supr∈[s-h,s]un-1(r))-a(s)((Bun)(s)-(Bun-1)(s))+b(s)un-1(s)-c(s)(supr∈[s-h,s]un(r)-supr∈[s-h,s]un-1(r))|≤[(p(s)+2a(s))(s+∑k=1∞ak)+q(s)+l(s)+b(s)+2c(s)]L*+(2a(s)+p(s))|x0|,|Ĩk((Bun-1)(tk),un-1(tk))-Lk(un(tk)-un-1(tk))|≤L*(L̃k+2Lk),n,k=1,2,….
Hence it follows from (2.26), (2.31) that {un(t)} is equicontinuous on each Jk. So in view of (2.30), an application of Arzela-Ascoli's theorem and diagonal method implies that there exists a subsequence {uni(t)}⊂{un(t)} such that {uni(t)} converges uniformly on each Jk to u*(t). Then the whole sequence {un(t)} converges uniformly on each Jk to u*(t). Thus u*(t)∈PC[Jh,R], and the fact that ∥un∥B≤L* implies ∥u*∥B≤L*. Hence u*∈BPC[Jh,R]. In view of (2.30), the continuity of f and Ĩk gives that (2.32)f(s,(Bun-1)(s),un-1(s),supr∈[s-h,s]un-1(r))-a(s)(un(s)-un-1(s))+b(s)un-1(s)-c(s)(supr∈[s-h,s]un(r)-supr∈[s-h,s]un-1(r))→f(s,(Bu*)(s),u*(s),supr∈[s-h,s]u*(r))+b(s)u*(s),s∈J,n→∞,Ĩk((Bun-1)(tk),un-1(tk))-Lk(un(tk)-un-1(tk))→Ĩk((Bu*)(tk),u*(tk)),n→∞,k=1,2,….
By the facts that a(s),b(s),c(s),sa(s),p(s),q(s),l(s),sp(s)∈L1(J), and ∑k=1∞L*(L̃k+2Lk) is convergent, observing (2.27) and taking limits as n→∞, the dominated convergence theorem yields that (2.33)u*(t)=(e∫0∞b(τ)dτ-1)-1×{∫0∞e∫tsb(τ)dτ[f(s,(Bu*)(s),u*(s),supr∈[t-h,t]u*(r))+b(s)u*(s)]ds+∑k=1∞e∫ttkb(τ)dτĨk((Bu*)(tk),u*(tk))}+∫0te∫tsb(τ)dτ[f(s,(Bu*)(s),u*(s),supr∈[t-h,t]u*(r))+b(s)u*(s)]ds+∑0<tk<te∫ttkb(τ)dτĨk((Bu*)(tk),u*(tk)),t∈J,
that is, u*(t)=Au*(t),u*(t) is a fixed point of A. It is easy to check that u*(t)∈TPC[Jh,R]∩C1[J',R]. Therefore we conclude by Lemma 2.4 that u*(t) is a solution of (2.1).Similarly, we can show that{vn(t)} converges uniformly on each Jk to v*(t), and v*∈TPC[Jh,R]∩C1[J',R] is also a solution of (2.1).Clearly,u*,v*∈[u0,v0]. Using a standard method, we can show that u*,v* is the minimal and maximal solutions of (2.1) in [u0,v0].Remark 2.6.
Theorem of [3] is a special case of Theorem 2.5 in this paper, where f and Ĩk did not involve the operator B. Hence Theorem 2.5 in this paper extends and improves the result of [3].Remark 2.7.
In system 2.1, if the interval is finite[0,m], then the conditions of (H1), (H2) can be deleted. Thus Theorem 2.5 in this paper extends and improves the result of [2].


## 3. Main Result for Second-Order Impulsive Differential Equation

Let us list other conditions for convenience.(H3′) There exist y0,z0∈TPC1[Jh,R]∩C2[J',R], and y0(t)≤z0(t),y0′(t)≤z0′(t) such that
(3.1)z0′′(t)≥f(t,z0(t),z0′(t),supr∈[t-h,t]z0′(r)),t∈J,t≠tk,Δz0|t=tk=akz0′(tk),k=1,2,…,Δz0′|t=tk≥Ĩk(z0(tk),z0′(tk)),k=1,2,…,z0(0)=x0,z0′(0)≥z0′(∞),z0′(0)=z0′(t),t∈[-h,0],
and y0(t) satisfies inverse inequalities above.(H4′) There exist a,b,c,ta∈C[J,R+]∩L1(J) with b≢0, such that
(3.2)f(t,x,y,z)-f(t,x¯,y¯,z¯)≥-a(t)(x-x¯)-b(t)(y-y¯)-c(t)(z-z¯),∀t∈J,Ĩk(x,y)≥Ĩk(x¯,y¯),k=1,2,…,
where (t,x,y,z),(t,x¯,y¯,z¯)∈Ω', x¯≤x,y¯≤y,z¯≤z, Ω'={(t,x,y,z):t∈J, y0(t)≤x(t)≤z0(t), y0′(s)≤y(t)≤z0′(s), sups∈[t-h,t]y0′(s)≤z(t)≤sups∈[t-h,t]z0′(s)}.Theorem 3.1.
Assume that conditions(H1), (H2), (H3′), (H4′) and (2.3), (2.21) hold. Then IBVP(1.1) has minimal and maximal solutions y*,z*∈TPC1[Jh,R]∩C2[J',R].Proof.
Letx'(t)=y(t). Then IBVP(1.1) is equivalent to the following system:
(3.3)x'(t)=y(t),y'(t)=f(t,x(t),y(t),sups∈[t-h,t]y(s)),t∈J,t≠tk,Δx|t=tk=aky(tk),Δy|t=tk=Ĩk(x(tk),y(tk))k=1,2,…,x(0)=x0,y(0)=y(∞),y(t)=y(0),t∈[-h,0].
Clearly, the system
(3.4)x'(t)=y(t),t∈J,Δx|t=tk=aky(tk),k=1,2,…,x(0)=x0,
has a unique solution x∈PC[Jh,R]∩C1[J',R] and x(t)=x0+∫0ty(s)ds+∑0<tk<taky(tk). Let
(3.5)(By)(t)=x0+∫0ty(s)ds+∑0<tk<taky(tk),
we have x(t)=(By)(t), and then IBVP (1.1) is transformed into first-order impulsive equations (2.1).
Lety0′(t)=u0(t),z0′(t)=v0(t), we have u0≤v0. By the condition (H3′) and the definition of B, we get that y0(t)=(Bu0)(t), z0(t)=(Bv0)(t), and u0,v0 satisfy (H3). By the condition (H4′), it is easy to see that (H4) holds. Hence, it follows from Theorem 2.5 that (2.1) has minimal and maximal solutions u*,v*∈TPC[Jh,R]∩C1[J',R].
Lety*(t)=(Bu*)(t),z*(t)=(Bv*)(t), then y*,z*∈TPC1[Jh,R]∩C2[J',R]. It follows by simple calculation that (3.6)y*′(t)=u*(t),t∈J,t≠tk,Δy*|t=tk=aku*(tk),k=1,2,…,y*(0)=x0,(3.7)(z*)'(t)=v*(t),t∈J,t≠tk,Δz*|t=tk=akv*(tk),k=1,2,…,z*(0)=x0.
The facts that u*,v* satisfies (2.1) and y*,z* satisfies (3.7) imply that y*,z*∈TPC1[Jh,R]∩C2[J',R] are solutions of IBVP(1.1).
Finally, it is easy to show thaty*,z* are the minimal and maximal solutions of IBVP(1.1), respectively. We complete the proof.



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*Source: 393757-2010-06-28.xml*