# Existence of Some Semilinear Nonlocal Functional Differential Equations of Neutral Type

Authors: Hsiang Liu; Cheng-Wen Liao; Chin-Tzong Pang
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

This paper is concerned with the existence of mild and strong solutions on
the interval[0,T] for some neutral partial differential
equations with nonlocal conditions. The linear part of the
equations is assumed to generate a compact analytic semigroup of
bounded linear operators, whereas the nonlinear part satisfies the
Carathëodory condition and is bounded by some suitable
functions. We first employ the Schauder fixed-point theorem to prove
the existence of solution on the interval [δ,T] for
δ>0 that is small enough, and, then, by letting δ→0 and using a diagonal argument, we have the existence results on
the interval [0,T]. This approach allows one to drop the
compactness assumption on a nonlocal condition, which generalizes
recent conclusions on this topic. The obtained results will be
applied to a class of functional partial differential equations with nonlocal conditions.


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## Body

## 1. Introduction

The purpose of this paper is to study the existence of mild and strong solutions for the following neutral evolution problem with nonlocal initial conditions:(1)ddt[u(t)+F(t,u(t))]=-Au(t)+G(t,u(t)),t∈[0,T],u(0)+g(u)=u0,
in a Banach space(X,∥·∥), whereT>0and-Agenerates an analytic compact semigroupT(·)onX. The functionsF,G, andgwill be specified later. The Cauchy problem with the nonlocal conditionx(0)+g(x)=x0was first considered by Byszewski [1], and since it reflects physical phenomena more precisely than the classical initial conditionx(0)=x0does, this issue has gained enormous attention in the past several years. For more detailed information about the importance of nonlocal initial conditions in applications, we refer to the works of Byszewski [2], Byszewski and Lakshmikantham [3], and to many other authors [4–7] and the references therein.Equation (1) has been studied by many authors under various assumptions on the linear partA, the nonlinear termsF,G, and the nonlocal conditiongsee, for example, [8–14]. A basic approach to this problem is to define the solution operatorΦ:C([0,T],X)→C([0,T],X)by
(2)(Φu)(t)=T(t)[u0+F(0,u(0))-g(u)]-F(t,u(t))+∫0t‍AT(t-s)F(s,u(s))ds+∫0t‍T(t-s)G(s,u(s))ds
and to use various fixed-point theorems, including Schauder fixed-point theorem, Banach contraction principle, Leray-Schauder alternative, and Sadovskii fixed-point theorem, to show thatΦhas a fixed point, which is the mild solution of (1). When using fixed-point theorems, it is necessary that the semigroupT(t)generated by the linear part of (1) be compact; that is,T(t)is a compact operator, for allt>0, so that the norm continuity ofT(t), fort>0, becomes a key point in the study of the existence of mild solutions. Thus, because of the absence of compactness of the solution operator att=0, most of the papers on the relevant topics (e.g., [8–10, 14]) assume complete continuity on the nonlocal termg. However, it is too restrictive in terms of applications.Recently, Liang et al. [15] observed the nonlocal Cauchy problem [1, 3, 4, 6] that the nonlocal conditiongis completely determined on[δ,T], for someδ>0; that is, suchgignores the fact thatt=0; for instance, in [4, 6], the functiong(u)is given by
(3)g(u)=∑i=0p‍ciu(ti),
whereci’s are given constants, and in this case, we have measurements att=0≤t0<t1<⋯<tp≤Trather than just att=0. Thus, by assuming that there is aδ∈(0,T)such that
(4)g(ϕ)=g(ψ),∀ϕ,φ∈Yr≔{φ∈C([0,T],X);∥φ(t)∥mmmmmmmmmmmmm≤r,∀t∈[0,T]},mmwithϕ(s)=ψ(s),s∈[δ,T],
the authors utilize fixed-point theorem twice to deduce the existence results. More recently, Liu and Yuan [16] gave existence results using Schauder fixed-point theorem and a limiting process under the following hypothesis.There is aδ∈(0,T)such thatF(ϕ)=F(ψ)andg(ϕ)=g(ψ), for allϕ,φ∈Yr, withϕ(s)=ψ(s)ands∈[δ,T],with the nonlinear termFbeing bounded by an integrable function.Motivated by the works in [15, 16], we drop the compactness assumption on the nonlocal conditiongand discuss the existence of solutions for (1). The obtained results generalize recent conclusions on this topic.The present work is organized as follows. Section1 is devoted the introduction of the problem we studied. Section 2, we explain some known notations and results we will use. The basic hypotheses on (1) are also given in this section. In Section 3, we study the existence of mild solutions to (1) and in Section 4, we investigate some conditions for (1) to come up with strong solutions. In Section 5, an example is given to illustrate the existence results.


## 2. Preliminaries

Throughout this paper,T>0will be a fixed real number,Xwill be a Banach space with norm∥·∥, and-A:D(A)⊂X→Xis the infinitesimal generator of a compact analytic semigroup of uniformly bounded linear operatorsT(·)such that0∈ρ(A). Then, there exists a constantM≥1such that∥T(t)∥≤M, fort≥0and it is possible to define the fractional powerAα, for0<α≤1, as a closed linear operator on its domainD(Aα)with inverseA-α(see [17]). The followings are the basic properties ofAα.Theorem 1 (see [17], pages 69–75).
The following assertions hold:(i)
D(Aα)is a Banach space with the norm|x|α:=∥Aαx∥, forx∈D(Aα).(ii)
T(t):X→D(Aα), for eacht>0.(iii)
AαT(t)x=T(t)Aαxfor eachx∈D(Aα)andt≥0.(iv)
For everyt>0,AαT(t)is bounded onX, and there existMα>0andδ>0such that
(5)∥AαT(t)∥≤Mαtαe-δt≤Mαtα.(v)
A-αis a bounded linear operator inXwithD(Aα)=Im(A-α).(vi)
If0<α≤β, thenD(Aβ)↪D(Aα).LetXαbe the Banach spaceD(Aα)endowed with the norm∥·∥α. Then, we denote byCαthe operator norm ofA-α, that is,Cα:=∥A-α∥, and letEbe the Banach spaceC([0,T],X)endowed with the supnorm given by
(6)∥u∥E:=sup0≤t≤T|u(t)|,foru∈E,
and, for anyδ∈(0,T), setEδ:=C([δ,T],X). Moreover, letEαbe the Banach spaceC([0,T],Xα)endowed with the supnorm given by
(7)∥u∥Eα:=sup0≤t≤T∥x(t)∥αforu∈Eα.
The following hypotheses are the basic assumptions of this paper.(H1) There existβ∈(0,1)andL1>0such that the functionF:[0,T]×X→Xβsatisfies
(8)∥AβF(t,x)-AβF(s,y)∥≤L1(|t-s|+∥x-y∥),
for allt,s∈[0,T]andx,y∈X.(H2) The functionG:[0,T]×X→Xsatisfies the following conditions.(i)
For eacht∈[0,T], the functionG(t,·):X→Xis continuous, and, for eachx∈X, the functionG(·,x):[0,T]→Xis strongly measurable.(ii)
For eachk∈ℕandt∈[0,T], there exists a positive functiongk∈L1([0,T],ℝ+)such that
(9)sup∥x∥≤k∥G(t,x)∥≤gk(t),
and there is aγ≥0such that
(10)liminfk→∞1k∫0T‍gk(s)ds=γ<∞.(H3) The functiong:L1([0,T],X)→Xis continuous, and there exist constantsL2,L3>0such that
(11)∥g(u)∥≤L2∥u∥E+L3,
foru∈E.


## 3. Mild Solutions

Definition 2.
A continuous functionu:[0,T]→Xis called a mild solution of (1) on[0,T]if, for eacht∈[0,T], the functions↦AT(t-s)F(s,u(s))is integrable on[0,t), and the following equation is satisfied:
(12)u(t)=T(t)[u0+F(0,u(0))-g(u)]-F(t,u(t))+∫0t‍AT(t-s)F(s,u(s))ds+∫0t‍T(t-s)G(s,u(s))ds,
for allt∈[0,T].
To see the existence of mild solution of nonlocal problem (1), we will, in view of (12), locate the fixed point of a mappingΦdefined onEby
(13)(Φu)(t):=T(t)[u0+F(0,u(0))-g(u)]-F(t,u(t))+∫0t‍AT(t-s)F(s,u(s))ds+∫0t‍T(t-s)G(s,u(s))ds,0≤t≤T.
For this, we first observe the following result, where for allk∈ℕ, we letBk={u∈E:∥u(t)∥≤k,t∈[0,T]}.Lemma 3.
Assume that hypotheses (H1)–(H3) are satisfied, and, in addition, there holds the following inequality:
(
H
4
)
(14)
M
(
L
2
+
γ
+
C
β
L
1
)
+
L
1
(
C
β
+
M
1
-
β
T
β
β
)
<
1
.
Then,ΦBk⊂Bk, for somek∈ℕ.Proof.
Suppose, on the contrary, that, for eachk>0, there existuk∈Bkandtk∈[0,T]such that∥(Φuk)(tk)∥>k. Then, we have
(15)k<∥(Φuk)(tk)∥≤∥T(tk)[u0-g(uk)+F(0,uk(0))]∥+∥F(tk,uk(tk))∥+∫0tk‍∥AT(tk-s)F(s,uk(s))∥ds+∫0tk‍∥T(tk-s)G(s,uk(s))∥ds≤∥T(tk)∥×(∥u0∥+∥g(uk)∥+∥A-βAβF(0,uk(0))∥)+∥A-βAβF(tk,uk(tk))∥+∫0tk‍∥A1-βT(tk-s)AβF(s,uk(s))∥ds+∫0tk‍∥T(tk-s)G(s,uk(s))∥ds≤M(∥u0∥+L2∥uk∥E+L3)+(Cβ(1+M)+M1-βTββ)×(L1(∥uk∥E+T)+∥AβF(0,0)∥)+M∫0tk‍gk(s)ds.
Dividing the two sides bykand taking the lower limit ask→+∞, we have
(16)M(L2+γ+CβL1)+L1(Cβ+M1-βTββ)≥1,
which is a contradiction. This completes the proof.By Lemma3, we see that the mappingΦ:E→Edefined by (13) mapsBkinto itself. We will show thatΦhas a fixed point inBk. To see this, note first thatΦis continuous by the continuity ofF,Gandg. We decomposeΦasΦ=Φ1+Φ2, where
(17)(Φ1u)(t)=-F(t,u(t))+∫0t‍AT(t-s)F(s,u(s))ds,(Φ2u)(t)=T(t)[u0+F(0,u(0))-g(u)]+∫0t‍T(t-s)G(s,u(s))ds.
We show thatΦ1is a contraction inBkandΦ2is a compact operator inBk.Lemma 4.
Assume that hypotheses (H1)–(H4) are satisfied. Ifu0∈X, thenΦ1is a contraction inBk.Proof.
Observe that, fort∈[0,T]andu,v∈Bk, we have the assumption (H1) as follows:
(18)∥(Φ1u)(t)-(Φ1v)(t)∥≤∥F(t,u(t))-F(t,v(t))∥+∫0t‍∥AT(t-s)[F(s,u(s))mmmmmmmmmmm-F(s,v(s))]∥ds≤CβL1∥u(t)-v(t)∥+L1(∫0t‍M1-β(t-s)1-βds)sup0≤s≤T∥u(s)-v(s)∥≤L1(Cβ+M1-βTββ)sup0≤s≤T∥u(s)-v(s)∥.
Hence,
(19)∥Φ1u-Φ1v∥E≤L~∥u-v∥E,
whereL~:=L1(Cβ+M1-β(Tβ/β))which is, by (H4), less than 1. Thus,Φ1is a contraction.Lemma 5.
Assume that hypotheses (H1)–(H4) are satisfied, and, in addition, the following is given.
(
H
5
)
There exists aδ∈(0,T)such thatF(·,u(·))=F(·,v(·)),G(·,u(·))=G(·,v(·))andg(u)=g(v), for anyu,v∈Bk, withu(s)=v(s)ands∈[δ,T].
Then the problem (1) has at least one mild solution inBkfor somek∈ℕ.Proof.
Letδbe given by (H5), and let
(20)Bk(δ):={u∈C([δ,T],X):∥u(t)∥≤k,∀t∈[δ,T]}.
For anyu∈Bk(δ), letu-∈Bkbe defined by
(21)u-(t):={u(t),t∈[δ,T],u(δ),t∈[0,δ).
Now, we defineΦδonEδby
(22)(Φδu)(t):=T(t)[u0+F(0,u-(0))-g(u-)]mm-F(t,u-(t))mm+∫0t‍AT(t-s)F(s,u-(s))dsmm+∫0t‍T(t-s)G(s,u-(s))ds,mmmmmmmmmmlδ≤t≤T.
Then, by Lemma 3, we see thatΦδBk(δ)⊂Bk(δ). ConsiderΦδas the sumΦδ=Φ1,δ+Φ2,δ, where
(23)(Φ1,δu)(t)=-F(t,u-(t))m+∫0t‍AT(t-s)F(s,u-(s))ds,mmmm∀u∈Bk(δ),t∈[δ,T],
andΦ2,δis defined onBk(δ)by
(24)(Φ2,δu)(t):=T(t)(u0+F(0,u-(0))-g(u-))mm+∫0t‍T(t-s)G(s,u-(s))ds,mmmmmmmlmmmlt∈[δ,T].
With a similar argument as in the proof of Lemma 4, one sees thatΦ1,δis a contraction onBk(δ).
For the compactness ofΦ2,δ, note first thatΦ2,δis continuous by the continuity ofGandg. Now, to show that the set{Φ2,δu:u∈Bk(δ)}is relatively compact inC([δ,T],X), we will prove that, for eacht∈[δ,T], the two sets
(25){T(t)(u0+F(0,u-(0))-g(u-)):u∈Bk(δ)},V(t):={∫0t‍T(t-s)G(s,u-(s))ds:u∈Bk(δ)}
are relatively compact inXand that
(26){T(·)(u0+F(0,u-(0))-g(u-)):u∈Bk(δ)},{∫0·‍T(·-s)G(s,u-(s))ds:u∈Bk(δ)}
are equicontinuous families of functions on[δ,T]. In fact, it follows from (H3) and the compactness ofT(t), fort∈[δ,T]that, for eacht∈[δ,T],
(27){T(t)(u0+F(0,u-(0))-g(u-)):u∈Bk(δ)}isrelativelycompactinX.
Moreover, since for eacht>0, andϵ∈(0,t), the set
(28){T(ε)∫0t-ε‍T(t-ε-s)G(s,u-(s))ds:u∈Bk(δ)}
is relatively compact, then, in view of
(29)∥∫0t‍T(t-s)G(s,u-(s))ds-T(ε)∫0t-ε‍T(t-ε-s)G(s,u-(s))ds∥≤∫t-ϵt‍∥T(t-s)∥L(X)∥G(s,u-(s))∥ds≤M∫t-ϵt‍gk(s)ds,
we see by (H2) that there are relative compact sets arbitrarily close toV(t), and, hence,V(t)is also relatively compact inX. Now, by the norm continuity ofT(t), fort>0, we see that
(30)∥[T(t+h)-T(t)](u0+F(0,u-(0))-g(u-))∥⟶0mmmmmmmmmmmmmmmmmllllmllllash⟶0
independently ofu∈Bk, and, hence,{T(·)(u0+F(0,u-(0)-g(u-)):u∈Bk(δ)}is an equicontinuous family of functions on[δ,T]. Finally, letϵ>0be arbitrarily small, and we see that
(31)∥∫0t+h‍T(t+h-s)G(s,u-(s))dsmm-∫0t‍T(t-s)G(s,u-(s))ds∫0t+h‍∥mm≤∫0t-ϵ‍∥[T(t+h-s)mmmmmmm-T(t-s)]G(s,u-(s))∥dsmmm+∫t-ϵt‍∥[T(t+h-s)-T(t-s)]mmmmmmmm×G(s,u-(s))∥dsmmm+∫tt+h‍∥T(t+h-s)G(s,u-(s))∥dsmm≤∫0t-ϵ‍∥T(t-s+h)-T(t-s)∥L(X)mmmmm×gk(s)ds+2M∫t-ϵt‍gk(s)dsmmm+M∫tt+h‍gk(s)ds,
which is, by the norm continuity ofT(t), fort>0, arbitrarily small and independent ofu∈Bkash→0. Therefore,{∫0·‍T(·-s)G(s,u(s))ds:u∈Bk(δ)}is an equicontinuous family of functions on[δ,T]and so is{Φ2,δu:u∈Bk(δ)}. It follows from Arzela-Ascoli’s theorem that{Φ2,δu:u∈Bk(δ)}is relatively compact onEδ. Thus, the mappingΦ2,δdefined by (24) is compact.
By the fixed-point theorem of Sadovskiĭ [18], this shows thatΦδhas a fixed point inBk(δ); that is, there is aφ∈Bk(δ)such that
(32)φ(t):=T(t)[u0+F(0,φ-(0))-g(φ-)]mm-F(t,φ-(t))+∫0t‍AT(t-s)F(s,φ-(s))dsmm+∫0t‍T(t-s)G(s,φ-(s))ds,δ≤t≤T.
Now, define a functionψon[0,T]by
(33)ψ(t):=T(t)[u0+F(0,φ-(0))-g(φ-)]mm-F(t,φ-(t))+∫0t‍AT(t-s)F(s,φ-(s))dsmm+∫0t‍T(t-s)G(s,φ-(s))ds,0≤t≤T.
Then,ψ=φon[δ,T], andψ∈Bk. Consequently, (H5) guarantees that
(34)ψ(t):=T(t)[u0+F(0,ψ(0))-g(ψ)]mm-F(t,ψ(t))+∫0t‍AT(t-s)F(s,ψ(s))dsmm+∫0t‍T(t-s)G(s,ψ(s))ds,0≤t≤T.
That is,ψis a mild solution of (1).For the main results in this section, we introduce a family of nonlocal neutral problems as follows. Firstly, we define, for eachδ∈(0,T), an operatorℬδonEby
(35)(ℬδu)(t):={u(δ),0≤t≤δ,u(t),δ<t≤T,
for allu∈E. It is clear thatℬδis bounded onEand∥ℬδ∥L(E)≤1, and, hence,ℬδBk⊂Bk. Now, for eachδ∈[0,T], we definegδ:E→Xby
(36)gδ(u)=g(ℬδu),∀u∈E,Fδ:[0,T]×X→Xby
(37)Fδ(t,u(t))=F(t,ℬδu(t)),∀u∈E,
andGδ:[0,T]×X→Xby
(38)Gδ(t,u(t))=G(t,ℬδu(t)),∀u∈E.
Consider the following nonlocal neutral problem:
(NNPδ)u′(t)+Fδ(t,u(t))=Au(t)+Gδ(t,u(t)),t∈[0,T],u(0)+g(u)=u0∈X.
In view of (35)–(38), the following result is an immediate corollary of Lemmas 4 and 5.Lemma 6.
Suppose that (H1)–(H4) are satisfied. Then, for anyδ∈(0,T], the problem (NNPδ) has at least one mild solution inBk.Theorem 7.
Suppose that, hypotheses (H1)–(H4) are satisfied. Then, problem (1) has at least one mild solution inBkfor somek∈ℕ.Proof.
Choose a decreasing sequence{δn}n∈ℕ⊂(0,T)so thatlimn→∞δn=0, and, then, by Lemma 6, we see that for eachn∈ℕ, there is anunsuch that
(39)un(t):=T(t)[u0+F(0,(ℬδnun)(0))mmmmm-g(ℬδnun)(0,(ℬδnun)(0))]-F(t,(ℬδnun)(t))mm+∫0t‍AT(t-s)F(s,(ℬδnun)(s))dsmm+∫0t‍T(t-s)G(s,(ℬδnun)(s))ds,mmmmmmmmmmmmmlllll0≤t≤T.
Now, for eachn∈ℕ, we defineΦδnonBkby
(40)(Φδnu)(t):=T(t)[u0+F(0,(ℬδnu)(0))mmmmml-gδn(u)(0,(ℬδnu)(0))]mm-F(t,(ℬδnu)(t))mm+∫0t‍AT(t-s)F(s,(ℬδnu)(s))dsmm+∫0t‍T(t-s)G(s,(ℬδnu)(s))ds,mmmmmmmmmmmmllllll0≤t≤T.
Then, (39) implies thatΦδnhas a fixed point inBkwhich is a mild solution for the nonlocal Cauchy problem(NCPδn). DecomposeΦδnasΦδn=Φ1,δn+Φ2,δn, where
(41)(Φ1,δnu)(t)=-F(t,(ℬδnu)(t))m+∫0t‍AT(t-s)F(s,(ℬδnu)(s))ds,mmmmmmmmmmmmmm0≤t≤T,Φ2,δnu(t)=T(t)[u0+F(0,(ℬδnu)(0))mmmml-gδn(u)(0,(ℬδnu)(0))]mm+∫0t‍T(t-s)G(s,(ℬδnu)(s))ds,mmmmmmmmmmmmlmm0≤t≤T.
With the same argument as in the proof of Lemma 4, we have thatΦ1,δnis a contraction. Furthermore, since the sequence{ℬδnun}n∈ℕlies inBk, then a similar argument as in the proof of Lemma 5 (see (27)–(31)) shows that, for eacht∈[0,T], the sets
(42){T(t)(u0+F(0,u(0))-g(ℬδnun))}n∈ℕ,{∫0t‍T(t-s)G(s,(ℬδnun)(s))ds}n∈ℕ
are both relatively compact inXand that the sequence of functions
(43){∫0·‍T(·-s)G(s,(ℬδnun)(s))ds}n∈ℕ
is equicontinuous on[0,T]. Hence, it follows from Ascoli-Arzela theorem that
(44){∫0·‍T(·-s)G(s,(ℬδnun)(s))ds}n∈ℕisrelativelycompactonE.
Now, let{ϵn}n∈ℕ⊂(0,T)be a decreasing sequence such thatlimn→∞ϵn=0, and let{un0}n0∈ℕbe a subsequence of{un}n∈ℕ. Then, a similar argument as in the proof of Lemma 5 insures that{T(·)(u0+F(0,u(0))-g(ℬδn0un0))}n0∈ℕis an equicontinuous sequence of functions on[ϵ1,T]. Thus, Ascoli-Arzela theorem guarantees that the sequence
(45){T(·)(u0+F(0,u(0))-g(ℬδn0un0))}n0∈ℕmllllisrelativelycompactinC([ϵ1,T],X).
Thus, by (44) and (45), we see that{un0}n0∈ℕis relatively compact inC([ϵ1,T],X), and, hence, we can select a subsequence of{un0}n0∈ℕdenoted by{un1}n1∈ℕ, which is a Cauchy sequence inC([ϵ1,T],X). By a similar process, we can select a subsequence of{un1}n1∈ℕdenoted by{un2}n2∈ℕ, which is a Cauchy sequence inC([ϵ2,T],X). Repeat the above argument, and use a diagonal argument to obtain a subsequence of{un0}n0∈ℕdenoted by{vn}n∈ℕ. Then, for everyt∈(0,T],{vn(t)}n∈ℕis a Cauchy sequence inX, and thus, we can define the functionv∞by
(46)v∞(t)={0,t=0,limn→∞vn(t),0<t≤T.
It is clear thatv∞is strongly measurable,v∞∈Bk, and
(47)∫0T‍∥vn(t)-v∞(t)∥dt≤2Tk.
It therefore follows from Lebesgue’s dominated convergence theorem that there is a subsequence{τn}n∈ℕof{δn0}n0∈ℕsuch that
(48)∫0T‍∥(ℬτnvn)(t)-v∞(t)∥dtmm≤∫0T‍∥(ℬτnvn)(t)-vn(t)∥dtmmm+∫0T‍∥vn(t)-v∞(t)∥dt⟶0mmmmmmmmmmmmlasn⟶∞.
This shows that the sequence{ℬδnun}n∈ℕis relatively compact onE, and, hence, by the continuity ofg, it follows that
(49){T(·)(u0+F(0,u(0))-g(ℬδnun))}n∈ℕmmmmmmlisrelativelycompactonE.
By (44) and (49), we see the relative compactness of{un}n∈ℕonE. Thus, there is a subsequence of{un}n∈ℕdenoted by{uν}ν∈ℕand a functionu∞∈Esuch that
(50)limν→∞∥uν-u∞∥E=0.
It is clear thatu∞∈Bk. Since
(51)∥ℬδνuν-u∞∥E=sup0≤t≤T∥ℬδνuν(t)-u∞(t)∥≤sup0≤t≤δν∥uν(δν)-u∞(t)∥≤+supδν≤t≤T∥uν(t)-u∞(t)∥≤∥uν(δν)-u∞(δν)∥≤+sup0≤t≤δν∥u∞(δν)-u∞(t)∥≤+∥uν-u∞∥E≤2∥uν-u∞∥E≤+sup0≤t≤δν∥u∞(δν)-u∞(t)∥,
then (50) and the uniform continuity ofu∞imply thatlimν→∞∥ℬδνuν-u∞∥E=0. By taking limits in (39), we see thatu∞is a mild solution of (1) and this completes the proof.We will consider the case more generally; that is, the nonlocal conditiongis defined onErather thanL1([0,T],X).Theorem 8.
Suppose that, hypotheses (H1) and (H2) are satisfied, and, in addition, there hold the following hypotheses.
(
H
6
)
The functiong:E→Xis continuous, and inequality (11) also holds.
(
H
7
)
limϵ→0supϕ∈Bk∥g(ϕ)-g(ϕϵ)∥E=0, where
(52)ϕϵ(t)={ϕ(ϵ),0≤t≤ϵ,ϕ(t),ϵ<t≤T.
If inequality (H4) holds, then, problem (1) has at least one mild solution inBk, for somek∈ℕ.Proof.
Let{un}n∈ℕand{ϵn}n∈ℕbe the sequences defined as in the proof of Theorem 7. With the same arguments as in the proof of Lemma 4, we see thatΦBk⊂Bk, for somek∈ℕ. Moreover, it follows from the same arguments as in the proof of Theorem 7 that (44), and (45) also hold, and, for every subsequence{un0}n0∈ℕof{un}n∈ℕ, there exist a subsequence{vn}n=1∞and a functionv∞:(0,T]→Xsuch thatv∞is continuous on(0,T]and, for everyϵk,
(53)limn→∞maxt∈[ϵk,T]∥vn(t)-v∞(t)∥=0,orlimn→∞∥ℬϵk[vn-v∞]∥E=0.
Letε>0be given. It follows from (H7) and (53) that there is aδ>0such that
(54)∥g(ϕ)-g(ψ)∥<ε4wheneverϕ=ψon[δ,T]
and that, for everyk∈ℕ, there is anNksuch thatn>Nkimplies that
(55)∥g(ℬϵk[vn-v∞])∥<ε4.
ChooseKthat is large enough so thatϵK<δ, and defineφ:[0,T]→Xby
(56)φ(t)={v∞(ϵK),0≤t≤ϵK,v∞(t),ϵK≤t≤T.
Thus, (H7), (54), and (55) insure that
(57)∥g(vn)-g(vm)∥≤∥g(vn)-g(ℬϵKvn)∥mm+∥g(ℬϵKvn)-g(φ)∥mm+∥g(φ)-g(BϵKvm)∥mm+∥g(ℬϵKvm)-g(vm)∥<ε.
And, hence, by the continuity ofgand the compactness ofT(t), fort>0, (49) is also valid in this case. Therefore, a similar argument as in the last paragraph of the proof of Theorem 7 shows the existence of a mild solution for (1).


## 4. Strong Solutions

Definition 9.
A mild solutionuis called a strong solution ifuis continuously differentiable on(0,T]withu′∈L1([0,T],X)and satisfies (1).
In the following, we establish a result of a strong solution for (1).Theorem 10.
LetXbe a reflexive Banach space. Suppose that there hold the following hypotheses.
(
H
8
)
The functionF:[0,T]×X→X1is a continuous function and there existsL4such that
(58)∥AF(t,x)-AF(s,y)∥≤L4(|t-s|+∥x-y∥),
for allt,s∈[0,T]andx,y∈X.
(
H
9
)
G(·,·)is Lipschitz continuous; that is, there exists a constantL0>0such that
(59)∥G(t,x)-G(s,y)∥≤L0(|t-s|+∥x-y∥),
for all(t,x),(s,y)∈[0,T]×X.
(
H
10
)
The functiong:L1([0,T],X)→Xis continuous,g(u)∈X1, for allu∈E, and
(60)∥g(u)∥1≤L5∥u∥E+L6,
for someL5,L6>0.
(
H
11
)
There holds the following inequality:
(61)L4(C1+MT)+2MTL0<1.
Ifu0∈X1and inequality (H4) also holds withL1,L2;Cβis replaced byL4,L5; andC1:=∥A-1∥, respectively, then (1) has a strong solution on[0,T].Proof.
LetΦbe the operator defined by (13). By (H8), (H9) and (H10), one can use a similar argument as in the proof of Lemma 3 to deduce that there is ak∈ℕsuch thatΦBk⊂Bk. For thisk, consider the set
(62)B={u∈E:∥u∥E≤k,∥u(t)-u(s)∥mmm≤L|t-s|,∀t,s∈[0,T]∥u(t)-u(s)∥},
for somekandLthat are large enough. It is clear thatBis nonempty, convex, and closed. We will prove thatΦhas a fixed point onB. Obviously, from the proofs of Lemmas 4 and 5 and Theorem 7, it is sufficient to show that, for anyx∈B,
(63)∥(Φu)(t2)-(Φu)(t1)∥≤L|t2-t1|,∀t2,t1∈[0,T].
We first fix an elementw∈Band observe that, for anys∈[0,T],
(64)∥AF(s,u(s))∥mm≤∥A[F(s,u(s))-F(0,(0))]∥mmml+∥AF(0,u(0))∥mm≤L4(s+∥u(s)-u(0)∥)+∥AF(0,u(0))∥mm≤L4(T+2k)+k0,∥G(s,u(s))∥mm≤∥G(s,u(s))-G(s,w(s))∥mmml+∥G(s,w(s))-G(0,w(0))∥mmml+∥G(0,w(0))∥≤L0∥u(s)-w(s)∥mmml+L0(1+L)s+∥G(0,w(0))∥mm≤L0[2k+(1+L)T]+k1,
wherek0:=∥AF(0,w(0))∥andk1:=∥G(0,w(0))∥. Now,
(65)∥(Φu)(t2)-(Φu)(t1)∥mm≤∥∫t1t2‍AT(s)[u0+F(0,u(0))-g(u)]ds∥mmml+∥F(t2,u(t2))-F(t1,u(t1))∥mmml+∥∫0t2‍AT(t2-s)F(s,u(s))dsmmmmmm-∫0t1‍AT(t1-s)F(s,u(s))ds∥mmml+∥∫0t2‍T(t2-s)G(s,u(s))dsmmmmmm-∫0t1‍T(t1-s)G(s,u(s))ds∥mm≤∥∫t1t2‍AT(s)u0dsmmmmm+∫t1t2‍T(s)AF(0,u(0))dsmmmmm-∫t1t2‍T(s)Ag(u)ds∥mmml+C1∥A[F(t2,u(t2))-F(t1,u(t1))]∥mmml+∥∫0t1‍T(s)A[F(t2-s,u(t2-s))mmmmmlmmmmmm-F(t1-s,u(t1-s))]dsmmmmm+∫t1t2‍T(s)AF(t2-s,u(t2-s))ds∥mmml+∥∫0t1‍T(s)[G(t2-s,u(t2-s))mmmmmmmmmml-G(t1-s,u(t1-s))]dsmmmmm+∫t1t2‍T(t2-s)G(s,u(s))ds∥.
Thus, from (H8), (H9), and (H10), it follows that
(66)∥(Φu)(t2)-(Φu)(t1)∥≤M∥Au0∥|t2-t1|+M∥AF(0,u(0))∥|t2-t1|+M(kL5+L6)|t2-t1|+C1L4(1+L)|t2-t1|+MTL4(1+L)|t2-t1|+M[L4(T+2k)+k0]|t2-t1|+MTL0(1+L)|t2-t1|+M{L0[2k+(1+L)T]+k1}|t2-t1|≤{K0+L[(C1+MT)L4+2MTL0]}|t2-t1|,
whereK0is a constant independent ofL. Since (H11) implies that
(67)K*:=L4(C1+MT)+2MTL0<1,
then
(68)∥(Φu)(t2)-(Φu)(t1)∥≤L|t2-t1|,∀t1,t2∈[0,T],
whenever
(69)L≥K01-K*.
Therefore,Φhas a fixed pointuwhich is a mild solution of (1). By the above calculation, we see that, for thisu(·), all of the functions
(70)p(t)=F(t,u(t)),q(t)=T(t)[u0+F(0,u(0))-g(u)],x(t)=∫0t‍AT(t-s)F(s,u(s))ds,y(t)=∫0t‍T(t-s)G(s,u(s))ds
are Lipschitz continuous, respectively. Sinceuis Lipschitz continuous on[0,T]and the spaceXis reflexive, then a result of [19] asserts thatu(·)is a.e. differentiable on(0,T]andu′(·)∈L1([0,T],X). A similar argument shows thatp(·),q(·),x(·), andy(·)also have this property. Furthermore, with a standard argument as in [17] (Theorem4.2.4), we have
(71)x′(t)=AF(t,u(t))-A∫0t‍AT(t-s)F(s,u(s))ds,y′(t)=G(t,u(t))-A∫0t‍T(t-s)G(s,u(s))ds.
So the following holds, for almost allt∈[0,T]:
(72)ddt[u(t)+F(t,u(t))]mm=ddt[T(t)(u0+F(0,u(0))-g(u))]mmml+x′(t)+y′(t)mm=-AT(t)(x0+F(0,u(0))-g(u))mmml+AF(t,u(t))-Ax(t)+G(t,u(t))-Ay(t)mm=-A[T(t)(u0+F(0,u(0))-g(u))mmmmmm-F(t,u(t))+x(t)+y(t)(u0+F(0,u(0))-g(u))]+G(t,u(t))mm=-Au(t)+G(t,u(t)).
This shows thatu(·)is also a strong solution to the nonlocal Cauchy problem (1), and the proof is completed.The following result is an immediate corollary of Theorems8 and 10.Corollary 11.
Suppose that the hypotheses (H7)–(H9), and (H11) are satisfied, and in addition, there holds the following hypotheses.
(
H
12
)
The functiong:E→Xis continuous,g(u)∈X1, for allu∈E, and inequality (60) sustains.
Ifu0∈X1and inequality (H4) also holds withL1,L2; andCβis replaced byL4,L5;C1:=∥A-1∥, respectively, then (1) has a strong solution on[0,T].


## 5. An Example

In the last section, our existence results will be applied to solve the following system:(73)∂∂t[u(t,x)+∫01‍b(x,s)u(t,s)ds]mm=∂2∂x2u(t,x)+h(t,x,∫01‍u(t,s)ds),mmmmlllmmmmmmmmmmm0≤t≤T,u(t,0)=u(t,1)=0,0≤t≤T,u(0,x)+∫0T‍k1(t,x)∫01‍k2(r,u(t,r))drdt=u0(x),mmmmmmmmmmmmmmmmmmmmll0≤x≤1,
where0<T≤1andu0∈X:=L2([0,1],ℝ)equipped withL2norm∥·∥.The operatorA:D(A)⊂X→Xdefined by
(74)D(A)={f∈X:f,f′′∈X,f(0)=f(1)=0},Af=-f′′.
Then,-Agenerates a compact, analytic semigroupT(·)of uniformly bounded linear operators. It is well known that0∈ρ(A), and, thus, the fractional powers ofAare well-defined where the eigenvalues ofAaren2π2and the corresponding normalized eigenvectors areen(x)=2sin(nπx),n=1,2,…. Moreover,
(75)Az=∑n=1∞‍n2π2〈z,en〉en,∀z∈D(A),A-1/2z=∑n=1∞‍1n〈z,en〉en,∀z∈X,
with∥A-1/2∥=1, and the operatorA1/2is given by
(76)A1/2z=∑n=1∞‍n〈z,en〉en,
with domainD(A1/2):={f∈X:∑n=1∞‍n〈f,en〉en∈X}.We need the following assumptions to solve (73) with our results.(A1) The functionb:[0,1]×[0,1]→ℝsatisfies the following conditions.(a)
(x,y)↦(∂/∂x)b(x,y)is welldefined and measurable with
(77)C:=∫01‍∫01‍(∂∂xb(x,y))2dydx<∞.(b)
b(0,x)=b(1,x)=0, for eachx∈[0,1].(A2) The functionh:[0,T]×[0,1]×ℝ→ℝsatisfies the following conditions.(a)
For each(x,y)∈[0,1]×ℝ, the functionh(·,x,y)is measurable.(b)
For each(t,y)∈[0,T]×ℝ, the functionh(t,·,y)is continuous.(c)
There is anl∈ℝ+such that
(78)|h(t1,x,y1)-h(t2,x,y2)|≤l(|t1-t2|+|y1-y2|),mmmmmm...∀x∈[0,1].(A3) The functionsk1andk2satisfy the following conditions, respectively:(a)
k1∈L2([0,1]×[0,T]),(b)
k2∈L2([0,T]×ℝ)and there is anl0such that
(79)|k2(t,y)-k2(s,z)|≤l0(|t-s|+|y-z|),mmmmllmmmmlll∀t,s∈[0,T],y,z∈ℝ.LetEbe the Banach spaceC([0,T],X)equipped with supnorm, letF:[0,T]×X→Xbe defined by
(80)(F(t,φ))(x)=∫01‍b(x,s)φ(s)dsmml(t,φ)∈[0,T]×X,x∈[0,1],
and letG:[0,T]×X→Xbe defined by
(81)(G(t,ϕ))(x)=h(t,x,∫01‍ϕ(s)ds)mmlll(t,ϕ)∈[0,T]×X,x∈[0,1].
Moreover, ifu:[0,T]×[0,1]→ℝ,we defineu:[0,T]→Xby
(82)u(t)(·)=u(t,·).Assumptions (A1) and (A2) imply the following conclusions.Theorem 12.
The functionsFandGhave the following inequalities. (a)
Fsatisfies hypothesis (H1) withβ=1/2andL1=Cthat is,
(83)∥A1/2F(t1,φ1)-A1/2F(t2,φ2)∥≤C∥ϕ1-ϕ2∥.(b)
Gsatisfies the hypothesis (H2) withgk=l(T+k)+∥G(0,0)∥andγ=l; that is,
(84)∥G(t1,φ1)-G(t2,φ2)∥≤l(|t1-t2|+∥φ1-φ2∥),sup∥ϕ∥≤k∥G(t,ϕ)∥≤l(T+k)+∥G(0,0)∥.Proof.
(a) By the definition ofFand assumption (A1), we see thatF(·,·)∈D(A)and
(85)∥A1/2F(t1,φ1)-A1/2F(t2,φ2)∥2mm=∫01‍∫01‍(∂∂xb(x,s)(φ1(y)-φ2(y)))2dydxmm≤∫01‍∫01‍(∂∂xb(x,y))2dydx·∥ϕ1-ϕ2∥2mm≤C∥ϕ1-ϕ2∥2,
for all(t1,ϕ1),(t2,ϕ2)∈[0,T]×X. Hence,Fsatisfies hypothesis (H1).
(b) By the part of (c) of assumption (A2) and Hölder’s inequality, we have(86)∥G(t1,φ1)-G(t2,φ2)∥mm=(-h(t2,x,∫01‍φ2(s)ds)|2∫01‍|h(t1,x,∫01‍φ1(s)ds)mmmmmmml-h(t2,x,∫01‍φ2(s)ds)|2dx)1/2mm≤(+|∫01‍φ1(s)ds-∫01‍φ2(s)ds||∫01‍φ1(s)ds-∫01‍φ2(s)ds|)2∫01‍(|∫01‍φ1(s)ds-∫01‍φ2(s)ds|l2|t1-t2|mmmmmml+|∫01‍φ1(s)ds-∫01‍φ2(s)ds||∫01‍φ1(s)ds-∫01‍φ2(s)ds|)2dx)1/2mm≤l(|t1-t2|+(∫01‍|φ1(s)-φ2(s)|ds))mm≤l(|t1-t2|+∥φ1-φ2∥).
SoG(t,·)is a continuous function fromXintoX, for eacht∈[0,T]. Moreover, letk∈ℕbe arbitrary, and it follows that
(87)sup∥ϕ∥≤k∥G(t,ϕ)∥mm≤sup∥ϕ∥≤k(∥G(t,ϕ)-G(0,0)∥+∥G(0,0)∥)mm≤sup∥ϕ∥≤k(l(t+∥ϕ∥)+∥G(0,0)∥)mm≤l(T+k)+∥G(0,0)∥.
SoG(·,·)satisfies (H2).Now, we defineg:E→Xby
(88)(g(u))(x)=∫0T‍∫01‍k1(t,x)k2(r,u(t,r))drdt,mmmmmml∀u∈E,0≤x≤1.Theorem 13.
gsatisfies the following properties. (a)
gis a continuous function fromEintoX.(b)
∥g(u)∥≤l0∥k1∥L2([0,1]×[0,T])∥u∥E+∥g(0)∥.(c)
Ifϕ∈E, thenlimϵ→0supϕ∈Bk∥g(ϕ)-g(ϕϵ)∥=0, where
(89)ϕϵ(t)={ϕ(ϵ),0≤t≤ϵ,ϕ(t),ϵ<t≤T.Proof.
(a) This follows since(90)∥g(u)-g(v)∥mm=(×dt∫0T‍)2dx∫0T‍∫01‍(∫0T‍∫01‍k1(t,x)mmmmmmmmm×[k2(r,u(t,r))mmmmmmmmllmm-k2(r,v(t,r))]drdt∫0T‍)2dx∫0T‍)1/2mm≤(×dt∫0T‍)2dx∫0T‍∫01‍(∫0T‍∫01‍k1(t,x)mmmmmmm×(l0|u(t,r)-v(t,r)|)drdt∫0T‍)2dx)1/2mm≤(×dt∫0T‍)2dx∫0T‍∫01‍(∫0T‍k1(t,x)mmmmmm×(l0supt∈[0,T]∫01‍|u(t,r)-v(t,r)|dr)mmmmmm×dt∫0T‍)2dx∫0T‍)1/2mm≤l0∥u-v∥E(∫01‍(∫0T‍k1(t,x)dt)2dx)1/2mm≤l0∥k1∥L2([0,1]×[0,T])∥u-v∥E.
(b) This is clear from the proof of part (a).
(c) Letk∈ℕandη>0be arbitrary. Sinceϕ∈E, thenϕis uniformly continuous on[0,T], and, hence, there is aδ>0such that∥ϕ(t)-ϕ(ϵ)∥<η, for allt∈[0,ϵ], whenever0<ϵ<δ. Thus,0<ϵ<δimplies that∥ϕ-ϕϵ∥E<η. Sincegis continuous fromEintoXby the part of (a), the assertion follows.Theorem13 show thatgsatisfies the hypotheses (H6) and (H7) withL4=l0∥k1∥L2([0,1]×[0,T])andL5=∥g(0)∥respectively. Consequently, since (73) is transformed into
(91)ddt(u(t)+F(t,u(t)))=Au(t)+G(t,u(t)),mmmmmmt∈[0,T],u(0)+g(u)=z0,
the following result is deduced by Theorem 8.Theorem 14.
If(92)(l0∥k1∥L2([0,1]×[0,T])+l)+2C(1+M1/2)<1,
then (73) has a mild solution.Theorem12 also shows thatGsatisfies (H9) withL0=l. Ifk1also satisfies.(A4)k1(x,y)is twice differentiable with respect tox,(∂2/∂2x)k1(x,y)∈L2([0,1]×[0,T]), and
(93)C¯:=∫01‍∫01‍(∂2∂2xb(x,y))2dydx<∞.
then Corollary 11 indicates the following result.Theorem 15.
Assume that assumptions (A2)–(A4) are satisfied and the function(94)u0(x):=u(0,x)∈W2,2([0,T]).
If inequalities (92) andC¯(1+M)+2l<1hold, then (73) has a strong solution.



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*Source: 503656-2013-11-26.xml*