# Existence Results for Second-Order Impulsive Neutral Functional Differential Equations with Nonlocal Conditions

Authors: Meili Li; Chunhai Kou
Journal: Discrete Dynamics in Nature and Society
 (2009)

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## Abstract

The existence of mild solutions for second-order impulsive semilinear neutral functional differential equations with nonlocal conditions in Banach spaces is investigated. The results are obtained by using fractional power of operators and Sadovskii's fixed point theorem.


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## Body

## 1. Introduction

The study of impulsive functional differential equations is linked to their utility in simulating processes and phenomena subject to short-time perturbations during their evolution. The perturbations are performed discretely and their duration is negligible in comparison with the total duration of the processes. That is why the perturbations are considered to take place “instantaneously” in the form of impulses. The theory of impulsive differential and functional differential equations has been extensively developed; see the monographs of Bainov and Simeonov [1], Lakshmikantham et al. [2], and Samoilenko and Perestyuk [3], where numerous properties of their solutions are studied, and detailed bibliographies are given.This paper is devoted to extending existing results to second-order differential equations. To be precise, in [4], the authors used Sadovsii's fixed point theorem for a condensing map to establish existence results for first-order impulsive semilinear neutral functional differential inclusions with nonlocal conditions. Here, we obtain existence results for second-order semilinear impulsive differential equations with nonlocal conditions of the form(1.1)ddt[x′(t)-F(t,x(h1(t)))]=Ax(t)+G(t,x(h2(t))),t∈J=[0,b],t≠tk,Δx|t=tk=Ik(x(tk-)),k=1,…,m,Δx′|t=tk=I̅k(x(tk-)),k=1,…,m,x(0)+g(x)=x0,x′(0)=η,
where A is the infinitesimal generator of a strongly continuous cosine family C(t),t∈R, of bounded linear operators in X. Also,0=t0<t1<⋯<tm<tm+1=b,Δx|t=tk=x(tk+)-x(tk-),Δx′|t=tk=x′(tk+)-x′(tk-).Finally,F,G,g,Ik,I̅k(k=1,…,m) and h1,h2 are given functions to be specified later.Other results on second order functional differential equations with and without impulsive effect can be founded in the monographs [5–8].This paper is organized as follows. In Section2, we recall briefly some basic definitions and lemmas. The existence theorem for (1.1) and its proof are arranged in Section 3. Our approaches are based on Sadovskii's fixed point theorem, and the theory of strongly continuous cosine families.


## 2. Preliminaries

Definition 2.1 (see [9]).
A one-parameter familyC(t),t∈R, of bounded linear operators in the Banach space X is called a strongly continuous cosine family if and only if(i)
C(s+t)+C(s-t)=2C(s)C(t) for all s,t∈R;(ii)
C(0)=I;(iii)
C(t)x is strongly continuous in t on R for each fixed x∈X.We define the associated sine family S(t),t∈R, by
(2.1)S(t)x=∫0tC(s)xds,x∈X,t∈R.We make the following assumption onA:(H1)
A is the infinitesimal generator of a strongly continuous cosine family C(t),t∈R, of bounded linear operators from X into itself.The infinitesimal generator of a strongly continuous cosine familyC(t),t∈R is the operator A:X→X defined by(2.2)Ax=d2dt2C(t)x|t=0,x∈D(A),
where(2.3)D(A)={x∈X:C(t)xistwicecontinuouslydifferentiableint}.We define(2.4)E={x∈X:C(t)xisoncecontinuouslydifferentiableint}.Lemma 2.2 (see [9]).
IfC(t),t∈R, be a strongly continuous cosine family in X, then(i)
there exist constantsK≥1 and ω≥0 so that ∥C(t)∥≤Keω|t|, for all t∈R, and
(2.5)∥S(t1)-S(t2)∥≤K|∫t1t2eω|s|ds|,∀t1,t2∈R;(ii)
ifx∈E, then S(t)x∈D(A) and (d/dt)C(t)x=AS(t)x.It is proved in [10] that for 0≤α≤1, the fractional powers (-A)α exist as close linear operator in X, D((-A)α)⊂D((-A)β), for 0≤β≤α≤1, and (-A)α(-A)β=(-A)α+β for 0≤α+β≤1.We assume in addition the following assumption:(H2)
for0≤α≤1,(-A)α maps onto X and is 1-1, so that D((-A)α) is a Banach space when endowed with the form ∥x∥α=∥(-A)αx∥,x∈D((-A)α). We denote this Banach space by Xα.DenoteJ0=[0,t1],Jk=(tk,tk+1],k=1,2,…,m. We define the following classes of functions:PC(J,Xα)={x:J→Xα:xk∈C(Jk,Xα),k=0,1,…,m and there exist x(tk+),x(tk-),k=1,…,m with x(tk)=x(tk-)}:PC1(J,Xα)={x∈PC(J,Xα):xk′∈C(Jk,Xα),k=0,1,…,m and there exist x′(tk+),x′(tk-),k=1,…,m with x′(tk)=x′(tk-)}, where xk and xk′ represent the restriction of x and x′ to Jk, respectively, (k=0,…,m), and ∥xk∥Jk=sups∈Jk∥xk(s)∥α.Obviously,PC(J,Xα) is a Banach space with the norm ∥x∥PC=max{∥xk∥Jk,k=0,…,m}, and PC1(J,Xα) is also a Banach space with the norm ∥x∥PC1=max{∥x∥PC,∥x′∥PC}.Definition 2.3.
A functionx(·)∈PC1(J,Xα) is said to be a mild solution of (1.1) if(i)
x(0)+g(x)=x0,x′(0)=η;(ii)
Δx|t=tk=Ik(x(tk-)),k=1,…,m;(iii)
Δx′|t=tk=I̅k(x(tk-)),k=1,…,m;(iv)
the restriction ofx(·) to the interval Jk(k=0,…,m) is continuous and the following integral equation is verified:(2.6)x(t)=C(t)[x0-g(x)]+S(t)[η-F(0,x(h1(0)))]+∫0tC(t-s)F(s,x(h1(s)))ds+∫0tS(t-s)G(s,x(h2(s)))ds+∑0<tk<tC(t-tk)Ik(x(tk-))+∑0<tk<tS(t-tk)I̅k(x(tk-)),t∈J.For (1.1), we assume that the following hypotheses are satisfied: for some α∈(0,1),(H3)
there exists a constantβ∈(0,1) such that F:J×Xα→Xβ is a continuous function, and (-A)βF:J×Xα→Xα satisfies the Lipschitz condition, that is, there exists a constant L>0 such that(2.7)∥(-A)βF(t1,x1)-(-A)βF(t2,x2)∥α≤L(|t1-t2|+∥x1-x2∥α),
for any 0≤t1,t2≤b,x1,x2∈Xα. Moreover, there exists a constant L1>0 such that the inequality(2.8)∥(-A)βF(t,x)∥α≤L1(∥x∥α+1)
holds for any x∈Xα;(H4)
the functionG:J×Xα→X satisfies the following conditions:
(i)
for eacht∈J, the function G(t,·):Xα→X is continuous, and for each x∈Xα, the function G(·,x):J→X is strongly measurable,(ii)
for each positive numberl∈N, there is a positive function wl∈L1(J) such that(2.9)sup∥x∥α≤l∥G(t,x)∥≤wl(t)a.e.onJ,lim infl→∞1l∫0bwl(s)ds=γ<∞,
where
(2.10)∥x∥α=sup0≤s≤b∥x(s)∥α;(H5)
hi∈C(J,J),i=1,2.g:PC1(J,Xα)→Xα is continuous and satisfies that
(i)
there exist positive constantsL2 and L2′ such that
(2.11)∥g(u)∥α≤L2∥u∥PC1+L2′∀u∈PC1(J,Xα),(ii)
g is a completely continuous map;(H6)
Ik,I̅k∈C(Xα,Xα),k=1,…,m are all bounded, that is, there exist constants dk,d̅k,k=1,…,m, such that ∥Ik(x)∥α≤dk,∥I̅k(x)∥α≤d̅k,  for x∈Xα;(H7)
C(t),t∈J, is completely continuous.


## 3. Main Result

Theorem 3.1.
Letx0∈Xα. If the hypotheses (H1)–(H7) are satisfied, then (1.1) has a mild solution provided that
(3.1)L0:=2M0LMb<1,(3.2)M(L2+2M0bL1+bγ)<1,
where
(3.3)M=sup{∥C(t)∥:t∈J},M′=sup{∥C′(t)∥:t∈J},M0=∥(-A)-β∥.Proof.
Consider the spaceB=PC1(J,Xα) with morm ∥x∥PC1=max{∥x∥PC,∥x′∥PC}. We should now show that the operator P defined by
(3.4)(Px)(t)=C(t)[x0-g(x)]+S(t)[η-F(0,x(h1(0)))]+∫0tC(t-s)F(s,x(h1(s)))ds+∫0tS(t-s)G(s,x(h2(s)))ds+∑0<tk<tC(t-tk)Ik(x(tk-))+∑0<tk<tS(t-tk)I̅k(x(tk-))
has a fixed point. This fixed point is then a solution of (2.6).
For each positive numberl, let Bl={x∈B:∥x(t)∥α≤l,t∈J}. Then for each l,Bl is clearly a bounded close convex set in B. We claim that there exists a positive integer l such that PBl⊆Bl. If it is not true, then for each positive integer l, there is a function xl(·)∈Bl, but (Pxl)(·)∉Bl, that is, ∥(Pxl)(t)∥α>l for some t(l)∈J, where t(l) denotes t is dependent on l. However, on the other hand, we have
(3.5)l<∥(Pxl)(t)∥α=∥∑0<tk<tC(t)[x0-g(xl)]+S(t)[η-F(0,xl(h1(0)))]+∫0tC(t-s)F(s,xl(h1(s)))ds+∫0tS(t-s)G(s,xl(h2(s)))ds+∑0<tk<tC(t-tk)Ik(xl(tk-))+∑0<tk<tS(t-tk)I¯k(xl(tk-))∥α≤∥C(t)[x0-g(xl)]∥α+∥S(t)[η-(-A)-β(-A)βF(0,xl(h1(0)))]∥α+∥∫0tC(t-s)(-A)-β(-A)βF(s,xl(h1(s)))ds∥α+∥∫0tS(t-s)G(s,xl(h2(s)))ds∥α+∑0<tk<t∥C(t-tk)Ik(xl(tk-))∥α+∑0<tk<t∥S(t-tk)I¯k(xl(tk-))∥α≤M[∥x0∥α+L2l+L2′]+Mb[∥η∥α+M0L1(l+1)]+MM0bL1(l+1)+Mb∫0bwl(s)ds+M∑k=1mdk+M∑k=1m(b-tk)d¯k.
Dividing on both sides byl and taking the lower limits as l→+∞, we get M(L2+2M0bL1+bγ)≥1. This is a contradiction with the formula (3.2). Hence for some positive integer l,PBl⊆Bl.
Next we will show that the operatorP has a fixed point on Bl, which implies that (1.1) has a mild solution. For this purpose, we decompose P as P=P1+P2, where the operators P1,P2 are defined on Bl, respectively, by
(3.6)(P1x)(t)=∫0tC(t-s)F(s,x(h1(s)))ds-S(t)F(0,x(h1(0))),(P2x)(t)=C(t)[x0-g(x)]+S(t)η+∫0tS(t-s)G(s,x(h2(s)))ds+∑0<tk<tC(t-tk)Ik(x(tk-))+∑0<tk<tS(t-tk)I¯k(x(tk-)),
for t∈J, and we will verify that P1 is a contraction while P2 is a completely continuous operator.
To prove thatP1 is a contraction, we take x1,x2∈Bl arbitrarily. Then for each t∈J and by condition (H3), we have that
(3.7)∥(P1x1)(t)-(P1x2)(t)∥α≤∥∫0tC(t-s)[F(s,x1(h1(s)))-F(s,x2(h1(s)))]ds∥α+∥S(t)[F(0,x1(h1(0)))-F(0,x2(h1(0)))]∥α=∥∫0tC(t-s)(-A)-β(-A)β[F(s,x1(h1(s)))-F(s,x2(h1(s)))]ds∥α+∥S(t)(-A)-β{(-A)β[F(0,x1(h1(0)))-F(0,x2(h1(0)))]}∥α≤2M0LMbsup0≤s≤b∥x1(s)-x2(s)∥α=L0sup0≤s≤b∥x1(s)-x2(s)∥α.
Thus∥P1x1-P1x2∥α≤L0∥x1-x2∥α. Therefore, by assumption 0<L0<1 (see (3.1)), we see that P1 is a contraction.
To prove thatP2 is completely continuous, firstly we prove that P2 is continuous on Bl. Let xn→x*,xn∈Bl, then by (H4)(i), we have G(s,xn(h2(s)))→G(s,x*(h2(s))),n→∞. Since ∥G(s,xn(h2(s)))-G(s,x*(h2(s)))∥≤2wl(s), by the dominated convergence theorem, we have
(3.8)∥P2xn-P2x*∥PC=supt∈J∥∑0<tk<tC(t)[g(x*)-g(xn)]+∫0tS(t-s)[G(s,xn(h2(s)))-G(s,x*(h2(s)))]ds+∑0<tk<tC(t-tk)[Ik(xn(tk-))-Ik(x*(tk-))]+∑0<tk<tS(t-tk)[I¯k(xn(tk-))-I¯k(x*(tk-))]∥α≤M∥g(x*)-g(xn)∥α+∫0b∥S(t-s)[G(s,xn(h2(s)))-G(s,x*(h2(s)))]∥αds+∑k=1mM∥Ik(xn(tk-))-Ik(x*(tk-))∥α+∑k=1mM(b-tk)∥I¯k(xn(tk-))-I¯k(x*(tk-))∥α→0asn→∞,∥(P2xn)′-(P2x*)′∥PC=supt∈J∥∑0<tk<tC′(t)[g(x*)-g(xn)]+∫0tC(t-s)[G(s,xn(h2(s)))-G(s,x*(h2(s)))]ds+∑0<tk<tC′(t-tk)[Ik(xn(tk-))-Ik(x*(tk-))]+∑0<tk<tC(t-tk)[I¯k(xn(tk-))-I¯k(x*(tk-))]∥α≤M′∥g(x*)-g(xn)∥α+∫0b∥C(t-s)[G(s,xn(h2(s)))-G(s,x*(h2(s)))]∥αds+∑k=1mM′∥Ik(xn(tk-))-Ik(x*(tk-))∥α+∑k=1mM∥I¯k(xn(tk-))-I¯k(x*(tk-))∥α→0asn→∞.
Thus, P2 is continuous.
Next, we prove that{P2x:x∈Bl} is a family of equicontinuous functions. Let τ1,τ2∈J,τ1<τ2. Then for each t∈J, we have
(3.9)∥(P2x)(τ2)-(P2x)(τ1)∥α≤∥[C(τ2)-C(τ1)](x0-g(x))∥α+∥S(τ2)η-S(τ1)η∥α+∥∫0τ1[S(τ2-s)-S(τ1-s)]G(s,x(h2(s)))ds∥α+∥∫τ1τ2S(τ2-s)G(s,x(h2(s)))ds∥α+∥∑0<tk<τ1[C(τ2-tk)-C(τ1-tk)]Ik(x(tk-))∥α+∥∑τ1≤tk<τ2C(τ2-tk)Ik(x(tk-))∥α+∥∑0<tk<τ1[S(τ2-tk)-S(τ1-tk)]I¯k(x(tk-))∥α+∥∑τ1≤tk<τ2S(τ2-tk)I¯k(x(tk-))∥α≤∥[C(τ2)-C(τ1)](x0-g(x))∥α+∥S(τ2)η-S(τ1)η∥α+∫0τ1∥[S(τ2-s)-S(τ1-s)]G(s,x(h2(s)))∥αds+∥∫τ1τ2S(τ2-s)G(s,x(h2(s)))ds∥α+∑0<tk<τ1∥C(τ2-tk)-C(τ1-tk)∥dk+∑τ1≤tk<τ2∥C(τ2-tk)∥dk+∑0<tk<τ1∥S(τ2-tk)-S(τ1-tk)∥d¯k+∑τ1≤tk<τ2∥S(τ2-tk)∥d¯k,
and similarly
(3.10)∥(P2x)′(τ2)-(P2x)′(τ1)∥α≤∥[C′(τ2)-C′(τ1)](x0-g(x))∥α+∥[S′(τ2)-S′(τ1)]η∥α+∥∫0τ1[C(τ2-s)-C(τ1-s)]G(s,x(h2(s)))ds∥α+∥∫τ1τ2C(τ2-s)G(s,x(h2(s)))ds∥α+∥∑0<tk<τ1[C′(τ2-tk)-C′(τ1-tk)]Ik(x(tk-))∥α+∥∑τ1≤tk<τ2C′(τ2-tk)Ik(x(tk-))∥α+∥∑0<tk<τ1[S′(τ2-tk)-S′(τ1-tk)]I¯k(x(tk-))∥α+∥∑τ1≤tk<τ2S′(τ2-tk)I¯k(x(tk-))∥α≤∥[C′(τ2)-C′(τ1)](x0-g(x))∥α+∥[S′(τ2)-S′(τ1)]η∥α+∫0τ1∥[C(τ2-s)-C(τ1-s)]G(s,x(h2(s)))∥αds+∥∫τ1τ2C(τ2-s)G(s,x(h2(s)))ds∥α+∑0<tk<τ1∥C′(τ2-tk)-C′(τ1-tk)∥dk+∑τ1≤tk<τ2∥C′(τ2-tk)∥dk+∑0<tk<τ1∥S′(τ2-tk)-S′(τ1-tk)∥d¯k+∑τ1≤tk<τ2∥S′(τ2-tk)∥d¯k.
The right-hand sides are independent of x∈Bl and tend to zero as τ2-τ1→0, since C(t),S(t),C′(t),S′(t) are uniformly continuous for t∈J and the compactness of C(t),S(t) for t>0 implies the continuity in the uniform operator topology.
The compactness ofS(t) follows from that of C(t) and Lemma 2.2.
This shows thatP2 maps Bl into a family of equicontinuous functions.
It remains to prove thatV(t)={(P2x)(t):x∈Bl} is relatively compact in X.
Obviously, by condition(H5)(ii), V(0) is relatively compact in B. Let 0<t≤b be fixed and 0<ϵ<t. For x∈Bl, we define
(3.11)(P2,ϵx)(t)=C(t)[x0-g(x)]+S(t)η+∫0t-ϵS(t-s)G(s,x(h2(s)))ds+∑0<tk<tC(t-tk)Ik(x(tk-))+∑0<tk<tS(t-tk)I¯k(x(tk-)).
Since C(t),S(t) are compact operators, the set Vϵ(t)={(P2,ϵx)(t):x∈Bl} is relatively compact in B for every ϵ,0<ϵ<t. Moreover, for every x∈Bl, we have
(3.12)∥(P2x)(t)-(P2,ϵx)(t)∥α=∥∫t-ϵtS(t-s)G(s,x(h2(s)))ds∥α≤∫t-ϵt∥S(t-s)∥wl(s)ds,∥(P2x)′(t)-(P2,ϵx)′(t)∥α=∥∫t-ϵtC(t-s)G(s,x(h2(s)))ds∥α≤∫t-ϵt∥C(t-s)∥wl(s)ds.
Therefore, there are relatively compact sets arbitrarily close to the set V(t). Hence, the set V(t) is relatively compact in B.Thus, by Arzela-Ascoli theorem,P2 is a completely continuous operator. Those arguments enable us to conclude that P=P1+P2 is a condensing map on Bl, and by the fixed point theorem of Sadovskii, there exists a fixed point x(·) for P on Bl. Therefore, the nonlocal Cauchy problem with impulsive effect (1.1) has a mild solution. The proof is completed.



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*Source: 641368-2009-11-17.xml*