# On Semilinear Integro-Differential Equations with Nonlocal Conditions in Banach Spaces

Authors: Tran Dinh Ke; Valeri Obukhovskii; Ngai-Ching Wong; Jen-Chih Yao
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We study the abstract Cauchy problem for a class of integrodifferential
equations in a Banach space with nonlinear perturbations and nonlocal conditions.
By using MNC estimates, the existence and continuous dependence results are proved.
Under some additional assumptions, we study the topological structure of the solution
set.


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## Body

## 1. Introduction

In this paper, we investigate the following problem:(1.1)x′(t)=A[x(t)+∫0tF(t-s)x(s)ds]+g(t,x(t)),t∈J∶=[0,T],(1.2)x(0)+h(x)=x0.
Here x(t) takes values in a Banach space X; F(t), for each t∈J, is a linear operator on X; maps g:J×X→X and h:C(J;X)→X are given. In this model, A is the generator of a strongly continuous semigroup S(·) on X.It is known that (1.1) with g=g(t) arises from some real applications. For example, the classical heat equation for medium with memory can be written as(1.3)xt(t,y)=∂2∂y2[x(t,y)+∫0tb(t-s)x(s,y)ds]+g(t,y),x(0,y)=x0,
where t∈ℝ+ and y∈[0,a]⊂ℝ+ (for more details, see [1, 2]). In addition, if we replace the initial condition x(0,y)=x0 by the nonlocal condition (1.2), it allows to describe the model more effectively. As an example of h, the following function can be considered:(1.4)h(x)=∑i=1pcix(ti),
where ci(i=1,…,p) are given constants and 0≤t1<⋯<tp≤T. As another example, one can take(1.5)h(x)=∑i=1pKix(ti),
where Ki:X→X are given linear operators. Regarding to (1.3), in the case X=L2(0,a), the operators Ki can be given by(1.6)Kix(ti,y)=∫0aki(ξ,y)x(ti,ξ)dξ,
where ki(i=1,…,p) are continuous kernel functions.Semilinear problem (1.1)-(1.2) with F=0 was studied extensively. In [3–5], the existence and uniqueness results were obtained by using the contraction mapping principle, under the Lipschitz conditions imposed on g and h. Supposing Carathéodory-type conditions on g, the authors in [6] proved the global existence result with the assumption that the semigroup S(t) is compact. However, as it was indicated in [7], if the Lipschitz condition is relaxed, one may get difficulties in proving the compactness of the solution map since the map t↦S(t), in general, is not uniformly continuous in [0,T], even in case when S(t) is compact. Recently, Fan and Li [8] gave an asymptotical method to solve this problem for the case when S(t) is a compact strongly continuous semigroup and the nonlocal function h is supposed to be continuous only.It is known that, in the caseF=0, the mild solution of (1.1)-(1.2) on J is defined via the integral equation(1.7)x(t)=S(t)[x0-h(x)]+∫0tS(t-s)g(s,x(s))ds,t∈J.
Problem (1.1)-(1.2) involving integro-differential equations was introduced in [2]. The complete references to integro-differential equations can be found in [1, 9, 10]. For some additional problems on solvability and controllability of integro-differential equations, we refer the reader to [11–13]. In order to represent the mild solutions via the variation of constants formula for this case, the notion of so-called resolvent for the corresponding linear equation(1.8)x′(t)=A[x(t)+∫0tF(t-s)x(s)ds],t∈J
can be applied. More precisely, an operator-valued function R(·):J↦L(X) is called the resolvent of (7) if it satisfies the following:(1)
R(0)=I, the identity operator on X,(2)
for eachv∈X, the map t↦R(t)v is continuous on J,(3)
ifY is the Banach space formed from D(A), the domain of A, endowed with the graph norm, then R(t)∈L(Y),R(·)y∈C1(J;X)∩C(J;Y) for y∈Y and(1.9)ddtR(t)y=A[R(t)y+∫0tF(t-s)R(s)yds]=R(t)Ay+∫0tR(t-s)AF(s)ds,t∈J.
For the existence of resolvent operators, we refer the reader to [14].It is worth noting that, from definition of resolvent operator and the uniform boundedness principle, there existsCR<+∞ such that(1.10)sup⁡t∈J‖R(t)‖L(X)≤CR.
Then the mild solution on J can be represented as(1.11)x(t)=R(t)[x0-h(x)]+∫0tR(t-s)g(s,x(s))ds,t∈J.
By a similar approach as in [3], the authors in [2] obtained the existence and uniqueness of solutions for (1.11) with the assumptions of the Lipschitz conditions on g and h.In this work, instead of the Lipschitz conditions posed ong and h, we assume the regularity of g and h expressed in terms of the measure of noncompactness. The mentioned regularity can be considered as a generalization of the Lipschitz condition. We first prove the existence of solutions for (1.1)-(1.2) in Section 2. Our method is to find fixed points of a corresponding condensing map, which yields the existence but does not provide the uniqueness of solutions. The arguments in this work are mainly based on the estimates with measure of noncompactness (MNC estimates). It should be noted that this technique was developed in [15], and it has been employed widely for differential inclusions. In Section 3, we prove that the solution set of our problem is continuously dependent on initial data. Section 4 is devoted to a special case when h is a Lipschitz function and R(t) is compact for t>0. We show that, in this case, the solution set to (1.1)-(1.2) has the so-called Rδ-set structure. We end this paper with an example in Section 5.


## 2. Existence Results

We start with the recalling of some notions and facts (see, e.g. [15, 16]).Definition 2.1.
Letℰ be a Banach space with power set 𝒫(ℰ), and (𝒜,≥) a partially ordered set. A function β:𝒫(ℰ)→𝒜 is called a measure of noncompactness (MNC) in ℰ if
(2.1)β(co⁡¯Ω)=β(Ω)foreveryΩ∈P(E),
where co¯Ω is the closure of convex hull of Ω. An MNC β is called (i)
monotone, if Ω0,Ω1∈𝒫(ℰ) such that Ω0⊂Ω1, then β(Ω0)≤β(Ω1);(ii)
nonsingular, if β({a}∪Ω)=β(Ω) for any a∈ℰ,Ω∈𝒫(ℰ);(iii)
invariant with respect to union with compact sets, if β(K∪Ω)=β(Ω) for every relatively compact set K⊂ℰ and Ω∈𝒫(ℰ). If, in addition, 𝒜 is a cone in a normed space, we say that β is (iv)
algebraically semiadditive, if β(Ω0+Ω1)≤β(Ω0)+β(Ω1) for any Ω0,Ω1∈𝒫(ℰ);(v)
regular, if β(Ω)=0 is equivalent to the relative compactness of Ω.
An important example of MNC isthe Hausdorff MNC, which satisfies all properties given in the previous definition:(2.2)χ(Ω)=inf⁡{ε:Ωhasafiniteε-net}.
Other examples of MNC defined on the space C(J;X) of continuous functions on an interval J=[0,T] with values in a Banach space X are the following: (i)
the modulus of fiber noncompactness:(2.3)γ(Ω)=sup⁡t∈Jχ(Ω(t)),
where χ is the Hausdorff MNC on X and Ω(t)={y(t):y∈Ω};(ii)
the modulus of equicontinuity:(2.4)modC(Ω)=lim⁡δ→0sup⁡y∈Ωmax⁡|t1-t2|<δ‖y(t1)-y(t2)‖.
As indicated in [15], these MNCs satisfy all properties mentioned in Definition 2.1 except the regularity.
Let𝒯∈L(ℰ), that is, 𝒯 is a bounded linear operator from ℰ into ℰ. We recall the notion of χ-norm (see e.g., [16]) as follows:
(2.5)‖T‖χ∶=inf⁡{M:χ(TΩ)≤Mχ(Ω),Ω⊂Eisaboundedset}.
The χ-norm of 𝒯 can be evaluated as
(2.6)‖T‖χ=χ(TS1)=χ(TB1),
where S1 and B1 are the unit sphere and the unit ball in ℰ, respectively. It is easy to see that
(2.7)‖T‖χ≤‖T‖L(X).Definition 2.2.
A continuous mapℱ:Z⊆ℰ→ℰ is said to be condensing with respect to a MNC β (β-condensing) if for every bounded set Ω⊂Z that is not relatively compact, we have
(2.8)β(F(Ω))≧β(Ω).
Letβ be a monotone nonsingular MNC in ℰ. The application of the topological degree theory for condensing maps (see, e.g., [15, 16]) yields the following fixed point principles.Theorem 2.3 (cf. [15, Corollary 3.3.1]).
Letℳ be a bounded convex closed subset of ℰ and  ℱ:ℳ→ℳ a  β- condensing map. Then Fixℱ={x=ℱ(x)} is a nonempty compact set.Theorem 2.4 (cf. [15, Corollary 3.3.3]).
Let𝒱⊂ℰ be a bounded open neighborhood of zero, and ℱ:𝒱¯→ℰ a  β- condensing map satisfying the following boundary condition:
(2.9)x≠λF(x)
for all x∈∂𝒱 and 0<λ≤1. Then the fixed point set Fix(ℱ)={x=ℱ(x)}⊂𝒱 is nonempty and compact.Now, returning to problem (1.1)-(1.2), we impose the following assumptions for g and h: (G1)
the mapg:J×X→X is continuous;(G2)
there exist functionμ∈L1(J) and nondecreasing function Υ:ℝ+→ℝ+ such that(2.10)‖g(t,η)‖X≤μ(t)Υ(‖η‖X)for a.e.t∈J and for all η∈X;(G3)
there exists a functionk∈L1(J) such that for each nonempty, bounded set Ω⊂X we have(2.11)χ(g(t,Ω))≤k(t)χ(Ω)for a.e.t∈J, where χ is the Hausdorff MNC in X;(H1)
h:C(J;X)→X is a continuous function and there is a nondecreasing function Θ:ℝ+→ℝ+ such that(2.12)‖h(x)‖X≤Θ(‖x‖C),for allx∈C(J;X), where ∥x∥C=∥x∥C(J;X);(H2)
there is a constantCh such that(2.13)χ(h(Ω))≤Chγ(Ω)for any bounded subsetΩ⊂C(J;X), where γ is defined in (2.3).(H3)
ifΩ⊂C(J;X) is a bounded set, then(2.14)modC(R(⋅)h(Ω))=0.Remark 2.5.
(1) IfX is a finite dimensional space, one can exclude the hypothesis (G3) since it can be deduced from (G2).
(2) It is known (see, e.g, [15, 16]) that condition (G3) is fulfilled if
(2.15)g(t,η)=g1(t,η)+g2(t,η),
where g1 is Lipschitz with respect to the second argument:
(2.16)‖g1(t,ξ)-g1(t,η)‖X≤k(t)‖ξ-η‖X
for a.e. t∈J and ξ,η∈X with k∈L1(J) and g2 is compact in second argument; that is, for each t∈J and bounded Ω⊂X, the set g2(t,Ω) is relatively compact in X.
(3) If we assume thath is completely continuous, that is, it is continuous and compact on bounded sets, then (H2)-(H3) will be satisfied. It is obvious that if the function h in (1.4) obeys (H1)-(H2) and function t↦R(t) is uniformly continuous, (H3) is also satisfied. It is worth noting that the function h given by (1.5)-(1.6) obeys (H1)–(H3).
As in [2], we assume in the sequel that(F1)
F(t)∈L(X) for t∈J and for x(·) continuous with values in Y=D(A),AF(·)x(·)∈L1(J;X);(F2)
for eachx∈X, the function t↦F(t)x is continuously differentiable on J.It is known that under conditions (F1)-(F2), the resolvent operator for (1.8) exists. We assume, in addition, that(HA)
t↦R(t) is uniformly norm continuous for t>0. We define the following operator:
(2.17)Φ:L1(J;X)⟶C(J;X),Φ(f)(t)=∫0tR(t-s)f(s)ds.
Before collecting some properties of Φ, we recall the following definitions.Definition 2.6.
A subset𝒬 of L1(J;X) is said to be integrably bounded if there exists a function μ∈L1(J) such that
(2.18)‖f(t)‖X≤μ(t)fora.e.t∈J,
for all f∈𝒬.Definition 2.7.
The sequence{ξn}⊂L1(J;X) is called semicompact if it is integrably bounded and the set {ξn(t)} is relatively compact in X for a.e. t∈J.
By using hypothesis (HA) and the same arguments as those in [15, Lemma 4.2.1, Theorem 4.2.2, Proposition 4.2.1, and Theorem 5.1.1], one can verify the following properties for Φ:
(Φ1)
the operatorΦ sends any integrably bounded set in L1(J;X) to equicontinuous set in C(J;X);(Φ2)
the following inequality holds:(2.19)‖Φ(ξ)(t)-Φ(η)(t)‖X≤CR∫0t‖ξ(s)-η(s)‖Xds
for every ξ,η∈L1(J;X),t∈J;(Φ3)
for any compactK⊂X and sequence {ξn}⊂L1(J;X) such that {ξn(t)}⊂K for a.e. t∈J, the weak convergence ξn⇀ξ implies the uniform convergence Φ(ξn)→Φ(ξ);(Φ4)
if{ξn}⊂L1(J;X) is an integrably bounded sequence and q∈L1(J) is a nonnegative function such that χ({ξn(t)})≤q(t),fora.e.t∈J, then(2.20)χ({Φ(ξn)(t)})≤2CR∫0tq(s)ds,t∈J;(Φ5)
if{ξn}⊂L1(J;X) is a semicompact sequence, then {ξn} is weakly compact in L1(J;X) and {Φ(ξn)} is relatively compact in C(J;X). Moreover, if ξn⇀ξ0, then Φ(ξn)→Φ(ξ0). Denote
(2.21)Φ*(x)(t)=R(t)[x0-h(x)]
for t∈J and x∈C(J;X). By Ng we denote the Nemytskii operator corresponding to the nonlinearity g, that is,
(2.22)Ng(x)(t)=g(t,x(t))fort∈J,x∈C(J;X).
We see that x is a solution of (1.1)-(1.2) if and only if
(2.23)x=Φ*(x)+ΦNg(x).
Let
(2.24)Ψ(x)=Φ*(x)+ΦNg(x).
Then the solutions of (1.1)-(1.2) can be considered as the fixed points of Ψ, the operator defined on C(J;X).
It follows from (G1) and (H1) thatΨ is continuous on C(J;X). Consider the function
(2.25)ν:P(C(J;X))⟶R+2,ν(Ω)=max⁡D∈Δ(Ω)(γ(D),modC(D)),
where γ and modC are defined in (2.3) and (2.4), respectively, Δ(Ω) denotes the collection of all countable subsets of Ω, and the maximum is taken in the sense of the ordering in the cone ℝ+2. By applying the same arguments as in [15], we have that ν is well defined. That is, the maximum is archiving in Δ(Ω) and so ν is an MNC in the space C(J;X), which satisfies all properties in Definition 2.1 (see [15, Example 2.1.3] for details).Theorem 2.8.
LetF satisfy (F1)-(F2). Assume that conditions (G1)–(G3) and (H1)–(H3) are fulfilled. If
(2.26)l:=CR(Ch+2∫0Tk(s)ds)<1,
then Ψ is ν-condensing.Proof.
LetΩ⊂C(J;X) be such that
(2.27)ν(Ψ(Ω))≥ν(Ω).
We will show that Ω is relatively compact in C(J;X). By the definition of ν, there exists a sequence {zn}⊂Ψ(Ω) such that
(2.28)ν(Ψ(Ω))=(γ({zn}),modC({zn})).
Following the construction of Ψ, one can take a sequence {xn}⊂Ω such that
(2.29)zn=Φ*(xn)+Φ(gn),
where
(2.30)gn(t)=g(t,xn(t)),t∈J,Φ*(xn)(t)=R(t)[x0-h(xn)],Φ(gn)(t)=∫0tR(t-s)gn(s)ds.
Using assumption (G3), we have
(2.31)χ({gn(s)})=χ({g(s,xn(s))})≤k(s)χ({xn(s)})≤k(s)γ({xn}),
for all s∈J. Then by (Φ4), we obtain
(2.32)χ({Φ(gn)(t)})≤2CR(∫0tk(s)ds)γ({xn}).
Noting that
(2.33)Φ*(xn)(t)=R(t)x0-R(t)h(xn),
we have
(2.34)χ({Φ*(xn)(t)})=χ({R(t)h(xn)})≤CRChγ({xn})
due to (2.5)-(2.7) and (H2). Combining (2.29), (2.31), and (2.32), we get
(2.35)γ({zn})≤lγ({xn}).
Combining the last inequality with (2.27), we have
(2.36)γ({xn})≤lγ({xn}),
and therefore
(2.37)γ({xn})=0.
But then (2.35) implies
(2.38)γ({zn})=0.
Putting (2.37) together with (2.31), we obtain that {gn} is semicompact. Hence, by (Φ5) one that has {Φ(gn)} is relatively compact. This yields
(2.39)modC({Φ(gn)})=0.
By (H3), we have
(2.40)modC({Φ*(xn)})=0.
Taking (2.29) into account again, we obtain
(2.41)modC({zn})=0.
Now it follows from (2.38)-(2.41) that
(2.42)ν(Ω)=(0,0).
By regularity of ν, we come to the conclusion that Ω is relatively compact.Remark 2.9.
IfR(t) is compact for t>0, we can drop assumption (G3) in the foregoing theorem. Indeed, for any bounded sequence {xn}⊂C(J;X), by setting ξn(t,s)=R(t-s)g(s,xn(s)), one sees that under hypothesis (G2), {ξn(t,·)} is an integrably bounded sequence in L1(0,t;X). In addition, since R(t),t>0, is compact, we have
(2.43)χ({ξn(t,s)})=0,fora.e.s∈[0,t].
Then by [15, Corollary 4.2.5], we obtain
(2.44)χ({∫0tξn(t,s)ds})=0,
for each t∈J. By this reason, inequality (2.32) becomes
(2.45)χ({Φ(gn)(t)})=0,
without the reference to (G3).We now can formulate the existence result in the following way.Theorem 2.10.
Under assumptions of Theorem2.8, if one has
(2.46)liminf⁡r→∞CRr(Θ(r)+Υ(r)∫0Tμ(s)ds)<1,
then the solution set to problem (1.1)-(1.2) is nonempty and compact.Proof.
We will use Theorem2.3. Applying the results of Theorem 2.8, we only need to verify the existence of a number r>0 such that
(2.47)Ψ(Br)⊆Br,
where Br is the closed ball in C(J;X) centered at origin with radius r. Indeed, assume to the contrary that for each n∈ℕ∖{0}, there is xn∈C(J;X) such that
(2.48)‖xn‖C≤n,but‖Ψ(xn)‖c>n.
Recalling that
(2.49)Ψ(xn)(t)=R(t)[x0-h(xn)]+∫0tR(t-s)g(s,xn(s))ds,
we have
(2.50)‖Ψ(xn)(t)‖X≤CR(‖x0‖X+Θ(‖xn‖C))+CR∫0tμ(s)Υ(‖xn(s)‖X)ds,
due to (H1) and (G2). Then
(2.51)n<‖Ψ(xn)‖C≤CR(‖x0‖X+Θ(n))+CRΥ(n)∫0Tμ(s)ds.
Equivalently,
(2.52)1<‖Ψ(xn)‖Cn≤1n(CR(‖x0‖X+Θ(n))+CRΥ(n)∫0Tμ(s)ds).
Passing in the last inequality to the limit as n→+∞, one gets a contradiction due to assumption (2.46). Thus the proof is completed.We have some special cases related to the growth ofΥ and Θ.Corollary 2.11.
Assume hypotheses of Theorem2.8, in which (G2) and (H1) are replaced by(G2′)
∥g(t,η)∥X≤μ(t)(1+∥η∥p),μ∈L1(J),0≤p<1, for all (t,η)∈J×X;(H1′)
h:C(J;X)→X is continuous and
(2.53)‖h(x)‖X≤h0+h1‖x‖cq,h0,h1>0,0≤q<1,
for all x∈C(J;X), respectively.Then the solution set to problem (1.1)-(1.2) is nonempty and compact.Proof.
Sincep<1 and q<1, condition (2.46) in Theorem 2.10 is testified obviously. Then we get the conclusion.Corollary 2.12.
Assume hypotheses of Theorem2.8, in which (G2) and (H1) are replaced by(G2′′)
∥g(t,η)∥X≤μ(t)(1+∥η∥),μ∈L1(J), for all (t,η)∈J×X;(H1′′)
h:C(J;X)→X is continuous and
(2.54)‖h(x)‖X≤h0+h1‖x‖C,forsomeh0,h1>0,
for all x∈C(J;X), respectively. If one has
(2.55)CR(h0+∫0Tμ(s)ds)<1,
then the solution set to problem (1)-(2) is nonempty and compact.Proof.
Under (G2′′) and (H1′′), condition (2.55) is equivalent to (2.46) and the conclusion of Theorem 2.10 holds.It should be mentioned that ifq=0 in (H1′), that is, the nonlocal function h is uniformly bounded, then one can relax the growth of Υ, by the arguments similar to those in [17].Theorem 2.13.
Assume the hypotheses of Theorem2.8, in which (H1) is replaced by(H1b)
h is a continuous function and ||h(x)||X≤Mh for all x∈C(J;X),  where Mh is a positive constant.If one has(2.56)CR∫0Tμ(s)ds<∫M̃∞dzΥ(z),
where M̃=CR(||x0||X+Mh), then the solution set to problem (1.1)-(1.2) is nonempty and compact.Proof.
In this case we employ Theorem2.4. It suffices to verify the boundary condition in Theorem 2.4. We show that if x=λΨ(x) for λ∈(0,1], then x must belong to a bounded set. Indeed, suppose
(2.57)x(t)=λR(t)[x0-h(x)]+λ∫0tR(t-s)g(s,x(s))ds.
It follows that
(2.58)‖x(t)‖X≤CR(‖x0‖X+Mh)+CR∫0tμ(s)Υ(‖x(s)‖X)ds.
Putting
(2.59)v(t)=CR(‖x0‖X+Mh)+CR∫0tμ(s)Υ(‖x(s)‖X)ds,
we have ∥x(t)∥X≤v(t), for all t∈J, and
(2.60)v′(t)=CRμ(t)Υ(‖x(t)‖X)≤CRμ(t)Υ(v(t)),
due to the fact that Υ is nondecreasing. Then, by using (2.56), we have
(2.61)∫M̃v(t)dzΥ(z)≤CR∫0Tμ(s)ds<∫M̃∞dzΥ(z),
for all t∈J. The last inequalities imply that supt∈Jv(t) is bounded, so is ∥x∥C.


## 3. Continuous Dependence Result

We start with some notions from the theory of multivalued maps (see, e.g. [15] for details).Let(Y,ϱY) and (Z,ϱZ) be metric spaces; 𝒦(Z) denotes the collection of all nonempty compact subsets of Z. A multivalued map (multimap) G:Y→𝒦(Z) is said to be (i)  upper semicontinuous (u.s.c.) if for each y∈Y and ϵ>0 there exists δ=δ(y,ϵ)>0 such that condition ϱY(y,y′)<δ implies G(y′)⊂Uϵ(G(y)), where Uϵ(G(y)) denotes the ϵ-neighborhood of the set G(y) induced by the metric ϱZ;(ii)  closed if its graph {(y,z)∈Y×Z:z∈G(y)} is a closed subset of Y×Z;(iii)  compact if G(Y) is relatively compact in Z;(iv)  quasicompact if its restriction to any compact set is compact.The following assertion gives a sufficient condition for upper semicontinuity.Lemma 3.1 (see[15]).
LetG:Y→𝒦(Z) be a closed quasicompact multimap. Then  G is u.s.c.Consider the solution multimap(3.1)W:X⊸C(J;X),W(v)={x:xisasolutionof(1.1)-(1.2)withinitialvaluex0=v}.
Notice that, as we demonstrated previously, under conditions of our existence theorems, the solution multimap W has compact values. We will study the continuity properties of W.Theorem 3.2.
Under the assumptions of Theorem2.10, the solution map W defined in (3.1) is u.s.c.Proof.
We first prove thatW is a quasicompact multimap. Let 𝒬⊂X be a compact set. We will show that W(𝒬)  is relatively compact in C(J;X). Suppose that {xn}⊂W(𝒬). Then there exists a sequence {vn}⊂𝒬 such that
(3.2)xn(t)=R(t)vn-R(t)h(xn)+Φgn(t),
where gn(t)=g(t,xn(t)).
Notice that the sequence{xn} is bounded. In fact, from (3.2) we have the estimate
(3.3)‖xn‖c≤CR(‖vn‖x+Θ(‖xn‖c))+CRΥ(‖xn‖c)∫0Tμ(s)ds.
Supposing to the contrary that the sequence ∥xn∥C is unbounded, by dividing the last inequality over ∥xn∥C and using condition (2.46) and the boundedness of the sequence {vn}, we get a contradiction.
Since{vn} is relatively compact, we obtain from (3.2) that
(3.4)χ({xn(t)})≤χ({R(t)h(xn)})+χ({Φgn(t)}).
Using (G3) we have
(3.5)χ({gn(s)})≤k(s)χ({xn(s)})≤k(s)γ({xn})
for all s∈J. Referring to (Φ4), one gets
(3.6)χ({Φ(gn)(s)})≤2CR(∫0tk(s)ds)γ({xn}),
and then
(3.7)γ({Φ(gn)})≤2CR(∫0Tk(s)ds)γ({xn}).
On the other hand, by (H2) one has
(3.8)χ({R(t)h(xn)})≤CRChγ({xn}).
Combining the last inequality with (3.4)-(3.7), we have
(3.9)γ({xn})≤lγ({xn}).
This leads to the conclusion that γ({xn})=0.
Now, condition (G2) implies that{gn} is integrably bounded in L1(J;X). Thus (Φ1) ensures that {Φ(gn)} is equicontinuous. Then applying condition (H3), we obtain
(3.10)modC({xn})≤modC({Φ*(xn)})+modC({Φ(gn)})=0.
So we have ν({xn})=(0,0) and therefore {xn} is relatively compact in C(J;X).
In order to prove thatW is u.s.c., it remains, according to Lemma 3.1, to show that W is closed. Let vn→v in X and xn∈W(vn),xn→x in C(J;X). We claim that x∈W(v). Indeed, one has
(3.11)xn(t)=Φ*(xn)(t)+∫0tR(t-s)g(s,xn(s))ds.
We first observe that
(3.12)Φ*(xn)=R(⋅)[vn-h(xn)]⟶R(⋅)[v-h(x)]=Φ*(x)
in C(J;X) in accordance with (H1). In addition, since g is a continuous function, we have g(s,xn(s))→g(s,x(s)) a.e. s∈J. The Lebesgue dominated convergence theorem implies that
(3.13)g(⋅,xn(⋅))-g(⋅,x(⋅))⟶0inL1(J;X)
due to the fact that {g(·,xn(·))} is integrably bounded. Therefore, taking (3.11) into account, we arrive at
(3.14)x(t)=Φ*(x)(t)+∫0tR(t-s)g(s,x(s))ds,t∈J.
The proof is completed.


## 4. Lipschitz Assumption for the Function from Nonlocal Condition

### 4.1. Existence Result

In this section, we assume thath is a Lipschitz function.(H2’)
There exists a constanth0>0 such that(4.1)‖h(x)-h(y)‖X≤h0‖x-y‖C.This implies the growth ofh:(4.2)‖h(x)‖X≤h0‖x‖C+‖h(0)‖X,
and the last inequality covers (H1).LetχC be the Hausdorff MNC in C(J;X). We have(4.3)‖Φ*(x)(t)-Φ*(y)(t)‖X=‖R(t)[h(x)-h(y)]‖X≤CRh0‖x-y‖C,
for all t∈J, where Φ* is given in (2.21). Thus(4.4)‖Φ*(x)-Φ*(y)‖C≤CRh0‖x-y‖C.
Then we know that (see [15]) condition (H2’) implies(4.5)χC(Φ*(Ω))≤CRh0χC(Ω),
for any bounded set Ω⊂C(J;X). We recall the following facts, which will be used in the sequel: for each bounded set Ω⊂C(J;X), one has the following:(i)
χ(Ω(t))≤χC(Ω), for all t∈J;(ii)
ifΩ is an equicontinuous set (modC(Ω)=0), then(4.6)χC(Ω)=sup⁡t∈Jχ(Ω(t))(=γ(Ω)).Theorem 4.1.
Assume thatg satisfies (G1)–(G3) and h obeys (H2’). If the following relations
(4.7)l:=CR(h0+2∫0Tk(s)ds)<1,(4.8)CRh0+CR∫0Tμ(s)dslim⁡inf⁡r→∞Ψ(r)r<1,
hold true, then problem (1.1)-(1.2) has at least one solution.Proof.
As we know from the proof of Theorem2.10, condition (4.8) implies that there exists a ball Br⊂C(J,X), r>0, such that
(4.9)Ψ(Br)⊂Br.
To apply Theorem 2.3, we verify that Ψ is χC-condensing. Let Ω⊂C(J;X) be a bounded set satisfying the inequality
(4.10)χC(Ψ(Ω))≥χC(Ω).
We will show that Ω is relatively compact. Notice that
(4.11)Ψ(Ω)⊂Φ*(Ω)+Φ∘Ng(Ω),
where
(4.12)Φ*(Ω)(t)=R(t)[x0-h(Ω)],Φ∘Ng(Ω)(t)={∫0tR(t-s)g(s,y(s))ds:y∈Ω}.
Then we have
(4.13)χC(Ψ(Ω))≤χC(Φ*(Ω))+χC(Φ∘Ng(Ω)).
The boundedness ofΩ in C(J;X) implies that Ng(Ω) is a bounded set in L1(J;X). By (Φ1), the set Φ∘Ng(Ω) is equicontinuous and therefore we have(4.14)χC(Φ∘Ng(Ω))=sup⁡t∈Jχ(Φ∘Ng(Ω)(t))≤sup⁡t∈J2CR∫0tk(s)χ(Ω(s))ds,
due to (G3) and (Φ4). Thus
(4.15)χC(Φ∘Ng(Ω))≤2CRχC(Ω)∫0Tk(s)ds.
Combining (4.5), (4.13), and (4.15), we obtain
(4.16)χC(Ψ(Ω))≤CR(h0+2∫0Tk(s)ds)χC(Ω).
Relations (4.7) and (4.10) yield
(4.17)χC(Ω)≤lχC(Ω).
Since ℓ<1, we have χC(Ω)=0. The regularity of χC ensures that Ω is relatively compact.Remark 4.2.
(1) Assumption (H2’) allows us to drop (H1)–(H3).(2)
As indicated in Remark2.9, in the case when R(t) is compact for t>0, condition (G3) can be dropped and condition (4.7) is reduced to
(4.18)CRh0<1,
which is covered by (4.8). Recall that in this case we have
(4.19)χ(Φ∘Ng({xn})(t))=0
for any bounded sequence {xn}⊂C(J;X) and for all t∈J.


### 4.2. The Structure of the Solution Set

We are in a position to study the structure of the solution set to (1.1)-(1.2) under the hypotheses of Theorem 4.1 and the assumption that R(t),t>0, is compact. At first, let us recall some notions.Definition 4.3.
A subsetB of a metric space Y is said to be contractible in Y if the inclusion map iB:B→Y is null-homotopic; that is, there exist y0∈Y and a continuous map h:B×[0,1]→Y such that h(y,0)=y and h(y,1)=y0 for every y∈B.The following notion [18] is important for our purposes.Definition 4.4.
LetY be a metric space; a subset B⊂Y is called an Rδ-set if B can be represented as the intersection of a decreasing sequence of compact contractible sets.The next lemma gives us a tool for the recognition ofRδ-set.Lemma 4.5 (see [19]).
LetX be a metric space,  E a Banach space, and  V:X→E a proper map; that is,  V is continuous and  V-1(K) is compact for each compact set  K⊂E. Suppose that there exists a sequence  {Vn} of mappings from  X into  E such that (1)
Vn is proper and {Vn} converges to V uniformly on X;(2)
for a given pointy0∈E and for all y in a neighborhood 𝒩(y0) of y0 in E, there exists exactly one solution xn of the equation Vn(x)=y. Then V-1(y0) is an Rδ-set.In order to use this lemma, we need the following result, which is called Lasota-Yorke Approximation Theorem (see, e.g. [20]).Lemma 4.6.
LetE be a normed space,  X a metric space, and  f:X→E a continuous map. Then for each  ϵ>0, there is a locally Lipschitz map  fϵ:X→E such that
(4.20)‖fϵ(x)-f(x)‖E<ϵ,foreveryx∈X.We now can formulate the main result of this section.Theorem 4.7.
Assume thatg satisfies (G1)-(G2) and h obeys (H2′). If R(t) is compact for t>0 and
(4.21)CRh0+CR∫0Tμ(s)dslim⁡inf⁡r→∞Υ(r)r<1,
then the solution set of problem (1.1)-(1.2) is an Rδ-set.Proof.
By Theorem4.1 and Remark 4.2, the hypotheses of Theorem 4.7 provide the existence result, that is,
(4.22)Fix⁡(Ψ)≠∅.
We will show that Fix(Ψ) is an Rδ-set.
Consider the nonlinearityg. By Lemma 4.6, there exists a sequence of functions {gn} such that(i)
gn:J×X→X is continuous and locally Lipschitz map;(ii)
∥gn(t,η)-g(t,η)∥X<ϵn for all (t,η)∈J×X, where ϵn→0 as n→∞. One can assume, without loss of generality, that
(4.23)‖gn(t,η)‖X≤μ(t)Υ(‖η‖X)+1
for all (t,η)∈J×X and n≥1. Let us consider the following equation:
(4.24)x(t)=y(t)+R(t)[x0-h(x)]+∫0tR(t-s)gn(s,x(s))ds,t∈J,
where y∈C(J;X) is a given function. Define Ψn:C(J;X)→C(J;X) by
(4.25)Ψn(x)(t)=y(t)+R(t)[x0-h(x)]+∫0tR(t-s)gn(s,x(s))ds,t∈J.
By applying the same arguments as in the proof of Theorem4.1 and in Remark 4.2, we can see that Ψn is χC-condensing. In addition, using the similar estimates as in Theorem 2.10, one can find a ball Br, r>0, such that
(4.26)Ψn(Br)⊂Br,
due to (4.23). Therefore, Ψn has a fixed point due to Theorem 2.3 and then (4.24) has at least one solution. Moreover, since h(·) is Lipschitzian and gn(t,·) is a locally Lipschitz function, the solution to (4.24) is unique.
Now by setting(4.27)Vn(x)(t)=x(t)-[R(t)(x0-h(x))+∫0tR(t-s)gn(s,x(s))ds],V(x)(t)=x(t)-[R(t)(x0-h(x))+∫0tR(t-s)g(s,x(s))ds],t∈J,
we see that {Vn} converges to V uniformly on C(J;X). In addition, for a given y∈C(J;X) the equation
(4.28)Vn(x)=y
has a unique solution, which is the fixed point of Ψn mentioned previously.
We now show thatV and Vn are proper. We proceed with Vn; the proof for V is similar. Obviously, Vn is continuous. Let K⊂C(J;X) be a compact set and Vn(Ω)=K. We claim that Ω is a compact set in C(J;X). Since Vn is continuous and K is closed, we deduce that Ω is closed. Assume that {xj} is a sequence in Ω, then one can take a sequence {yj}⊂K such that
(4.29)Vn(xj)=yj.
Equivalently,
(4.30)xj(t)=yj(t)+R(t)[x0-h(xj)]+∫0tR(t-s)gn(s,xj(s))ds,t∈J.
We first show that the sequence {xj} is bounded. We have
(4.31)‖xj(t)‖X≤‖yj(t)‖X+CR(‖x0‖x+h0‖xj‖C+‖h(0)‖X)+CR∫0t(μ(s)ϒ(‖xj(s)‖X)+1)ds,
due to (H2’) and (4.23). Thus
(4.32)‖xj‖C≤‖yj‖C+CR(‖x0‖X+‖h(0)‖X)+TCR+CRh0‖xj‖C+CRϒ(‖xj‖C)∫0Tμ(s)ds.
If {xj} is unbounded, then there is a subsequence (still denoted by {xj}) such that ||xj||C→+∞ as j→+∞. Now from the last inequality, it follows that
(4.33)1≤1‖xj‖C(‖yj‖C+CR(‖x0‖X+‖h(0)‖X)+TCR)+CRh0+CR∫0Tμ(s)dsΥ(‖xj‖C)‖xj‖C.
Passing in the last inequality to limits as j→+∞, one gets a contradiction due to the hypotheses of the Theorem. Now from (4.30), we have
(4.34)χC({xj})≤χC({yj})+χC(R(⋅)h({xj}))+χC(Φ∘Ngn({xj})).
Using the same arguments as in the proof of Theorem 4.1 and Remark 4.2, we obtain that
(4.35)χC(Φ∘Ngn({xj}))=sup⁡t∈Jχ(Φ∘Ngn({xj})(t))=0,χC(R(⋅)h({xj}))=sup⁡t∈Jχ(R(t)h({xj}))≤CRh0χC({xj}).
Substituting the last inequalities into (4.34) and using the fact that {yj} is a compact sequence, we obtain
(4.36)χC({xj})≤CRh0χC({xj}).
Noting that CRh0<1, we deduce χC({xj})=0. Therefore, {xj} is a relatively compact sequence in C(J;X) and we arrive at the conclusion that Ω is compact and then Vn is proper.
Finally, by the observation thatFixΨ=V-1(0), from Lemma 4.5, we obtain that FixΨ is an Rδ-set.Remark 4.8.
The topological structure of the solution set of problem (1.1)-(1.2) for the case of a non-compact resolvent R(t) is an open problem.Further Remarks
Some additional remarks can be given in the case whenR(t),t>0, is compact. Following the technique presented in [8], we can consider the following problem:
(4.37)x′(t)=A[x(t)+∫0tF(t-s)x(s)ds]+g(t,x(t)),t∈J∶=[0,T],(4.38)x(0)+hn(x)=x0,
where hn(x)=R(1/n)h(x),n∈ℕ∖{0}. Since h is continuous, hn is a completely continuous function. Then it satisfies (H2)-(H3). Therefore, under the assumptions (G1)–(G3), and (H1) and (2.46), problem (4.37)-(4.38) has at least one mild solution xn∈C(J;X). Furthermore,
(4.39)χ(ΦNg(Ω)(t))=χ(∫0tR(t-s)g(s,Ω(s))ds)≤2∫0tχ(R(t-s)g(s,Ω(s)))ds=0
for all bounded subset Ω⊂C(J;X). Thus, one can drop assumption (G3), and then the solution operator for (4.37)-(4.38)
(4.40)Ψn(x)(t)=R(t)[x0-hn(x)]+∫0tR(t-s)g(s,x(s))ds
is ν-condensing without assuming the condition
(4.41)CR(Ch+2∫0Tk(s)ds)<1.
In fact, we have the following assertion.Theorem 4.9.
LetR(t) be compact for t>0. Under assumptions (G1)-(G2) and (H1) and (2.46), the solution set of problem (4.37)-(4.38) is a nonempty compact set.Using the same arguments as in [8], one can prove that the sequence {xn} of solutions to (4.37)-(4.38) is relatively compact. Finally, passing to the limit as n→+∞ in the equation(4.42)xn(t)=R(t)x0-R(t)R(1n)h(xn)+∫0tR(t-s)g(s,xn(s))ds,
we obtain the solution of problem (1.1)-(1.2).


## 4.1. Existence Result

In this section, we assume thath is a Lipschitz function.(H2’)
There exists a constanth0>0 such that(4.1)‖h(x)-h(y)‖X≤h0‖x-y‖C.This implies the growth ofh:(4.2)‖h(x)‖X≤h0‖x‖C+‖h(0)‖X,
and the last inequality covers (H1).LetχC be the Hausdorff MNC in C(J;X). We have(4.3)‖Φ*(x)(t)-Φ*(y)(t)‖X=‖R(t)[h(x)-h(y)]‖X≤CRh0‖x-y‖C,
for all t∈J, where Φ* is given in (2.21). Thus(4.4)‖Φ*(x)-Φ*(y)‖C≤CRh0‖x-y‖C.
Then we know that (see [15]) condition (H2’) implies(4.5)χC(Φ*(Ω))≤CRh0χC(Ω),
for any bounded set Ω⊂C(J;X). We recall the following facts, which will be used in the sequel: for each bounded set Ω⊂C(J;X), one has the following:(i)
χ(Ω(t))≤χC(Ω), for all t∈J;(ii)
ifΩ is an equicontinuous set (modC(Ω)=0), then(4.6)χC(Ω)=sup⁡t∈Jχ(Ω(t))(=γ(Ω)).Theorem 4.1.
Assume thatg satisfies (G1)–(G3) and h obeys (H2’). If the following relations
(4.7)l:=CR(h0+2∫0Tk(s)ds)<1,(4.8)CRh0+CR∫0Tμ(s)dslim⁡inf⁡r→∞Ψ(r)r<1,
hold true, then problem (1.1)-(1.2) has at least one solution.Proof.
As we know from the proof of Theorem2.10, condition (4.8) implies that there exists a ball Br⊂C(J,X), r>0, such that
(4.9)Ψ(Br)⊂Br.
To apply Theorem 2.3, we verify that Ψ is χC-condensing. Let Ω⊂C(J;X) be a bounded set satisfying the inequality
(4.10)χC(Ψ(Ω))≥χC(Ω).
We will show that Ω is relatively compact. Notice that
(4.11)Ψ(Ω)⊂Φ*(Ω)+Φ∘Ng(Ω),
where
(4.12)Φ*(Ω)(t)=R(t)[x0-h(Ω)],Φ∘Ng(Ω)(t)={∫0tR(t-s)g(s,y(s))ds:y∈Ω}.
Then we have
(4.13)χC(Ψ(Ω))≤χC(Φ*(Ω))+χC(Φ∘Ng(Ω)).
The boundedness ofΩ in C(J;X) implies that Ng(Ω) is a bounded set in L1(J;X). By (Φ1), the set Φ∘Ng(Ω) is equicontinuous and therefore we have(4.14)χC(Φ∘Ng(Ω))=sup⁡t∈Jχ(Φ∘Ng(Ω)(t))≤sup⁡t∈J2CR∫0tk(s)χ(Ω(s))ds,
due to (G3) and (Φ4). Thus
(4.15)χC(Φ∘Ng(Ω))≤2CRχC(Ω)∫0Tk(s)ds.
Combining (4.5), (4.13), and (4.15), we obtain
(4.16)χC(Ψ(Ω))≤CR(h0+2∫0Tk(s)ds)χC(Ω).
Relations (4.7) and (4.10) yield
(4.17)χC(Ω)≤lχC(Ω).
Since ℓ<1, we have χC(Ω)=0. The regularity of χC ensures that Ω is relatively compact.Remark 4.2.
(1) Assumption (H2’) allows us to drop (H1)–(H3).(2)
As indicated in Remark2.9, in the case when R(t) is compact for t>0, condition (G3) can be dropped and condition (4.7) is reduced to
(4.18)CRh0<1,
which is covered by (4.8). Recall that in this case we have
(4.19)χ(Φ∘Ng({xn})(t))=0
for any bounded sequence {xn}⊂C(J;X) and for all t∈J.


## 4.2. The Structure of the Solution Set

We are in a position to study the structure of the solution set to (1.1)-(1.2) under the hypotheses of Theorem 4.1 and the assumption that R(t),t>0, is compact. At first, let us recall some notions.Definition 4.3.
A subsetB of a metric space Y is said to be contractible in Y if the inclusion map iB:B→Y is null-homotopic; that is, there exist y0∈Y and a continuous map h:B×[0,1]→Y such that h(y,0)=y and h(y,1)=y0 for every y∈B.The following notion [18] is important for our purposes.Definition 4.4.
LetY be a metric space; a subset B⊂Y is called an Rδ-set if B can be represented as the intersection of a decreasing sequence of compact contractible sets.The next lemma gives us a tool for the recognition ofRδ-set.Lemma 4.5 (see [19]).
LetX be a metric space,  E a Banach space, and  V:X→E a proper map; that is,  V is continuous and  V-1(K) is compact for each compact set  K⊂E. Suppose that there exists a sequence  {Vn} of mappings from  X into  E such that (1)
Vn is proper and {Vn} converges to V uniformly on X;(2)
for a given pointy0∈E and for all y in a neighborhood 𝒩(y0) of y0 in E, there exists exactly one solution xn of the equation Vn(x)=y. Then V-1(y0) is an Rδ-set.In order to use this lemma, we need the following result, which is called Lasota-Yorke Approximation Theorem (see, e.g. [20]).Lemma 4.6.
LetE be a normed space,  X a metric space, and  f:X→E a continuous map. Then for each  ϵ>0, there is a locally Lipschitz map  fϵ:X→E such that
(4.20)‖fϵ(x)-f(x)‖E<ϵ,foreveryx∈X.We now can formulate the main result of this section.Theorem 4.7.
Assume thatg satisfies (G1)-(G2) and h obeys (H2′). If R(t) is compact for t>0 and
(4.21)CRh0+CR∫0Tμ(s)dslim⁡inf⁡r→∞Υ(r)r<1,
then the solution set of problem (1.1)-(1.2) is an Rδ-set.Proof.
By Theorem4.1 and Remark 4.2, the hypotheses of Theorem 4.7 provide the existence result, that is,
(4.22)Fix⁡(Ψ)≠∅.
We will show that Fix(Ψ) is an Rδ-set.
Consider the nonlinearityg. By Lemma 4.6, there exists a sequence of functions {gn} such that(i)
gn:J×X→X is continuous and locally Lipschitz map;(ii)
∥gn(t,η)-g(t,η)∥X<ϵn for all (t,η)∈J×X, where ϵn→0 as n→∞. One can assume, without loss of generality, that
(4.23)‖gn(t,η)‖X≤μ(t)Υ(‖η‖X)+1
for all (t,η)∈J×X and n≥1. Let us consider the following equation:
(4.24)x(t)=y(t)+R(t)[x0-h(x)]+∫0tR(t-s)gn(s,x(s))ds,t∈J,
where y∈C(J;X) is a given function. Define Ψn:C(J;X)→C(J;X) by
(4.25)Ψn(x)(t)=y(t)+R(t)[x0-h(x)]+∫0tR(t-s)gn(s,x(s))ds,t∈J.
By applying the same arguments as in the proof of Theorem4.1 and in Remark 4.2, we can see that Ψn is χC-condensing. In addition, using the similar estimates as in Theorem 2.10, one can find a ball Br, r>0, such that
(4.26)Ψn(Br)⊂Br,
due to (4.23). Therefore, Ψn has a fixed point due to Theorem 2.3 and then (4.24) has at least one solution. Moreover, since h(·) is Lipschitzian and gn(t,·) is a locally Lipschitz function, the solution to (4.24) is unique.
Now by setting(4.27)Vn(x)(t)=x(t)-[R(t)(x0-h(x))+∫0tR(t-s)gn(s,x(s))ds],V(x)(t)=x(t)-[R(t)(x0-h(x))+∫0tR(t-s)g(s,x(s))ds],t∈J,
we see that {Vn} converges to V uniformly on C(J;X). In addition, for a given y∈C(J;X) the equation
(4.28)Vn(x)=y
has a unique solution, which is the fixed point of Ψn mentioned previously.
We now show thatV and Vn are proper. We proceed with Vn; the proof for V is similar. Obviously, Vn is continuous. Let K⊂C(J;X) be a compact set and Vn(Ω)=K. We claim that Ω is a compact set in C(J;X). Since Vn is continuous and K is closed, we deduce that Ω is closed. Assume that {xj} is a sequence in Ω, then one can take a sequence {yj}⊂K such that
(4.29)Vn(xj)=yj.
Equivalently,
(4.30)xj(t)=yj(t)+R(t)[x0-h(xj)]+∫0tR(t-s)gn(s,xj(s))ds,t∈J.
We first show that the sequence {xj} is bounded. We have
(4.31)‖xj(t)‖X≤‖yj(t)‖X+CR(‖x0‖x+h0‖xj‖C+‖h(0)‖X)+CR∫0t(μ(s)ϒ(‖xj(s)‖X)+1)ds,
due to (H2’) and (4.23). Thus
(4.32)‖xj‖C≤‖yj‖C+CR(‖x0‖X+‖h(0)‖X)+TCR+CRh0‖xj‖C+CRϒ(‖xj‖C)∫0Tμ(s)ds.
If {xj} is unbounded, then there is a subsequence (still denoted by {xj}) such that ||xj||C→+∞ as j→+∞. Now from the last inequality, it follows that
(4.33)1≤1‖xj‖C(‖yj‖C+CR(‖x0‖X+‖h(0)‖X)+TCR)+CRh0+CR∫0Tμ(s)dsΥ(‖xj‖C)‖xj‖C.
Passing in the last inequality to limits as j→+∞, one gets a contradiction due to the hypotheses of the Theorem. Now from (4.30), we have
(4.34)χC({xj})≤χC({yj})+χC(R(⋅)h({xj}))+χC(Φ∘Ngn({xj})).
Using the same arguments as in the proof of Theorem 4.1 and Remark 4.2, we obtain that
(4.35)χC(Φ∘Ngn({xj}))=sup⁡t∈Jχ(Φ∘Ngn({xj})(t))=0,χC(R(⋅)h({xj}))=sup⁡t∈Jχ(R(t)h({xj}))≤CRh0χC({xj}).
Substituting the last inequalities into (4.34) and using the fact that {yj} is a compact sequence, we obtain
(4.36)χC({xj})≤CRh0χC({xj}).
Noting that CRh0<1, we deduce χC({xj})=0. Therefore, {xj} is a relatively compact sequence in C(J;X) and we arrive at the conclusion that Ω is compact and then Vn is proper.
Finally, by the observation thatFixΨ=V-1(0), from Lemma 4.5, we obtain that FixΨ is an Rδ-set.Remark 4.8.
The topological structure of the solution set of problem (1.1)-(1.2) for the case of a non-compact resolvent R(t) is an open problem.Further Remarks
Some additional remarks can be given in the case whenR(t),t>0, is compact. Following the technique presented in [8], we can consider the following problem:
(4.37)x′(t)=A[x(t)+∫0tF(t-s)x(s)ds]+g(t,x(t)),t∈J∶=[0,T],(4.38)x(0)+hn(x)=x0,
where hn(x)=R(1/n)h(x),n∈ℕ∖{0}. Since h is continuous, hn is a completely continuous function. Then it satisfies (H2)-(H3). Therefore, under the assumptions (G1)–(G3), and (H1) and (2.46), problem (4.37)-(4.38) has at least one mild solution xn∈C(J;X). Furthermore,
(4.39)χ(ΦNg(Ω)(t))=χ(∫0tR(t-s)g(s,Ω(s))ds)≤2∫0tχ(R(t-s)g(s,Ω(s)))ds=0
for all bounded subset Ω⊂C(J;X). Thus, one can drop assumption (G3), and then the solution operator for (4.37)-(4.38)
(4.40)Ψn(x)(t)=R(t)[x0-hn(x)]+∫0tR(t-s)g(s,x(s))ds
is ν-condensing without assuming the condition
(4.41)CR(Ch+2∫0Tk(s)ds)<1.
In fact, we have the following assertion.Theorem 4.9.
LetR(t) be compact for t>0. Under assumptions (G1)-(G2) and (H1) and (2.46), the solution set of problem (4.37)-(4.38) is a nonempty compact set.Using the same arguments as in [8], one can prove that the sequence {xn} of solutions to (4.37)-(4.38) is relatively compact. Finally, passing to the limit as n→+∞ in the equation(4.42)xn(t)=R(t)x0-R(t)R(1n)h(xn)+∫0tR(t-s)g(s,xn(s))ds,
we obtain the solution of problem (1.1)-(1.2).


## 5. Example

We conclude this note with an example, in which we find a representation for the resolvent operator generated by the linear part and impose suitable conditions to demonstrate the existence result and the structure of the solution set. Precisely, consider the following system:(5.1)∂u∂t=∂2∂x2[u(x,t)+∫0ta(t-s)u(x,s)ds]+∫0xf(t,y,u(y,t))dy,x∈(0,π),t∈(0,b],(5.2)u(x,0)+∑i=1mαiu(x,ti)=u0,ti∈(0,b),(5.3)u(0,t)=u(π,t)=0.
Let X=L2(0;π),(5.4)D(A)={z∈X:zandz′areabsolutelycontinuous,z(0)=z(π)=0},
and A:D(A)→X, Az=z′′. Then it is known that A generates a strongly continuous semigroup {S(t)}t≥0 on X. Recall that the functions(5.5){ϕn(x)=2πsin⁡⁡nx:n=1,2,…}
form an orthonormal basis in X and they are the eigenfunctions corresponding to the eigenvalues {λn=n2:n=1,2,…} of -A.We are in a position to consider the linear part of (5.1)–(5.3):(5.6)∂u∂t=∂2∂x2[u(x,t)+∫0ta(t-s)u(x,s)ds],x∈[0,π],t∈[0,b],(5.7)u(x,0)=u0,(5.8)u(0,t)=u(π,t)=0.
Assume that a∈C1(0;b) and u0=∑n=1∞γnϕn. We are searching for the resolvent operator generated by (5.6)–(5.8) in the following form:(5.9)(R(t)u0)(x)=u(x,t)=∑n=1∞γnTn(t)ϕn(x),
where Tn is the solution of the equation(5.10)Tn′(t)=-n2[Tn(t)+∫0ta(t-s)Tn(s)ds]
subject to Tn(0)=1. We know that(5.11)Tn̂(λ)=1λ+n2(1+â(λ)),
where Tn̂ and â are the Laplace transforms of Tn and a, respectively. For the simple case, when a is constant, a<0, we have(5.12)Tn̂(λ)=λλ2+n2λ+n2a.
By some computations, one gets(5.13)Tn(t)=p(n)2p(n)+n2ep(n)t+q(n)2q(n)+n2eq(n)t,
where p(n)=(1/2)(-n2+nn2-4a) and q(n)=(1/2)(-n2-nn2-4a).Taking assumption (HA) into account, we conclude thatTn needs to satisfy the following condition:(5.14)|Tn(t)-Tn(s)|≤cn|t-s|,fort,s∈(0,b],∑n=1∞cn2<+∞.
This condition is obviously fulfilled for Tn given by (5.13).We now verify the compactness ofR(t) for t>0. Since the embedding H01(0;π)⊂X is compact, it is sufficient to find a condition providing that the set(5.15)∂∂xR(t)BX(0,r)={∂∂xR(t)v:‖v‖X≤r}
is bounded in X. It is easy to verify that this condition follows from(5.16)∑n=1∞|nTn(t)|2<+∞,fort>0.
The last inequality also holds for Tn given by (5.13).As far as nonlinear problem (5.1)–(5.3) is concerned, we see that the nonlocal function h(u)(x)=∑i=1mαiu(x,ti) is a Lipschitz function with the constant h0=∑i=1mαi. Let g(t,v)(x)=∫0xf(t,y,v(y))dy. Then the nonlinearity g satisfies (G1)-(G2) if we assume that f:[0,b]×[0,π]×ℝ→ℝ is continuous and there exists a function μ∈L1(0;b) such that(5.17)|f(t,y,η)|≤μ(t)(1+|η|),∀t∈[0,b],y∈[0,π],η∈R.
As indicated in [16], the Hausdorff MNC of a bounded set Ω⊂L2(0;π) can be expressed by(5.18)χ(Ω)=12lim⁡δ→0sup⁡v∈Ωmax⁡0≤ω≤δ‖v-vω‖L2(0;π),
where vω denote the ω-translation of v:(5.19)vω(x)={v(ω+x),0≤x≤π-ω,v(π),π-ω≤x≤π,
or, alternatively, the Steklov function:(5.20)vω(x)=12ω∫x-ωx+ωv(z)dz
(v is extended outside of [0,π] by zero). Therefore, condition (G3) is provided by the following inequality:(5.21)|∫xx+ωf(t,y,v(y))dy|≤k(t)|v(x)-vω(x)|,t∈[0,b];x,ω∈[0,π],
for a nonnegative function k∈L1(0;b).Applying Theorem4.1, we obtain that problem (5.1)-(5.2) has at least one solution if (5.14), (5.17), and (5.21) take place together with the following estimates:(5.22)CR(h0+2∫0bk(s)ds)<1,CR(h0+∫0bμ(s)ds)<1.
If we assume that (5.16) holds and hence R(t) is compact for t>0, then the solution set of (5.1)–(5.3) is an Rδ-set if (5.17) holds and(5.23)CR(h0+∫0bμ(s)ds)<1,
due to Theorem 4.7.



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*Source: 137576-2012-04-01.xml*