On a class of functional differential equations in

The aim of this paper is to establish the existence of solutions and some properties of set solutions for a Cauchy problem with causal operator in a separable Banach space.


Introduction
The study of functional equations with causal operators has a rapid development in the last years and some results are assembled in a recent monograph [4].The term of causal operators is adopted from engineering literature and the theory of these operators has the powerful quality of unifying ordinary differential equations, integrodifferential equations, differential equations with finite or infinite delay, Volterra integral equations, and neutral functional equations, to name a few (see, [1], [3], [10], [11], [17], [19], [21], [22]).
The purpose of this article is to study the topological properties of the initial value problem (1.3) in a Banach space.For this we will use ideas from papers [8], [9].Also, we give an existence result for this problem, assuming only the continuity of the operator Q.In the paper [12] is also obtained an existence result assuming that the operator Q is a locally Lipschitz operator.
If dim(A) = sup{||x − y||; x, y ∈ A} is the diameter of the bounded set A, then we have that α(A) ≤ dim(A) and α(A) ≤ 2d if sup x∈A ||x|| ≤ d.We recall the some properties for α (see [14]).
If A, B are bounded subsets of E and A denotes the closure of A, then Two significant examples of causal operators are: the Niemytzki operator and the Volterra-Hammerstein integral operator For i = 0, 1, ..., p, we consider the functions F i : R × E → E, (t, u) → F (t, u), that are measurable in t and continuous in u.Set σ := max i=1,p σ i , where σ i ≥ 0, and let Then, the operator Q, so defined, is a causal operator (for details, see [12]).For other concrete examples which serve to illustrate that the class of causal operators is very large, we refer to the monograph [4].
We consider the initial-valued problem with causal operator under the following assumptions: (h 1 ) Q is continuous; (h 2 ) for each r > 0 and each τ ∈ (0, b), there exists M > 0 such that, for all The existence of solutions for this kind of Cauchy problem has been studied in [12] for the case when Q ) is a locally Lipschitz operator.This problem has been studied in [6] for a Lipschitz causal operator , where E is a real Banach space.
The aim of this paper is to establish the existence of solutions and some properties of set solutions for Cauchy problem (2.1).To prove the properties of set solutions, we use the same method as in [9] and [8], accordingly adapted.

Existence of solutions
In the first half of this section, we present an existence result of the solutions for Cauchy problem (2.1), under conditions (h 1 )-(h 5 ).
Proof.Let δ > 0 be any number and let r : We choose T ∈ (0, τ ] such that T 0 µ(t)dt < δ and we consider the set B defined as follows Further on, we consider the integral operator and we prove that this is a continuous operator from B into B.
First, we observe that u(•) ∈ B, then sup Further on, let u m → u in B. We have By (h 1 ) and (h 3 ) it follows that sup 0≤t≤T Since u m | [−σ,0] = ϕ for every m ∈ N, we deduce that P : B → B is a continuous operator.Moreover, it follows that P (B) is uniformly bounded.Next, we show that P (B) is uniformly equicontinuous on [−σ, T ].Let ε > 0. On the closed set [0, T ], the function t → t 0 µ(s)ds is uniformly continuous, and so there exists On the other hand, since ϕ ∈ C σ is a continuous function on [−σ, 0], then there exists η ′′ > 0 such that Finally, if 0 ≤ s ≤ t ≤ T then, for each u(•) ∈ B, we have Therefore, we conclude that P (B) is uniformly equicontinuous on [−σ, T ].Further on, for each m ≥ 1, we consider the following classical approximations Then, for all m ≥ 1 we have u m (•) ∈ B.Moreover, for 0 ≤ t ≤ T /m, we have and for T /m ≤ t ≤ T , we have Therefore, it follows that Note that, given ε > 0, we can find m(ε) > 0 such that t t−T /m µ(s)ds < ε/2 for t ∈ [0, T ] and m ≥ m(ε).Hence we have that Using the last inequality, we obtain that Since for every t ∈ [0, T ], A(t) is bounded then, by Lemma 2.1, (h 4 ) and (h 5 ), we have that is a solution of Cauchy problem (2.1) because, for a.e.t ∈ [β, β + τ ], we have that Therefore, the solution u(•).can be continued beyond β, contradicting the assumption that β cannot be increased.It follows that β = b.

Some properties of solution sets
In the following, for a fixed ϕ ∈ C σ and a bounded set K ⊂ E, by S T (ϕ, K) we denote the set of all solutions u(•) of Cauchy problem (2.1) on [−σ, T ] with T ∈ (0, b] and such that u(t) ∈ K for all t ∈ [−σ, T ].By A T (ϕ, K) we denote the attainable set; that is,

EJQTDE, 2010 No. 64, p. 10
Proof .We consider a sequence {u m (•)} m≥1 in S T (ϕ, K) and we shall show that this sequence contains a subsequence which converges, uniformly on [−σ, T ], to a solution u(•) ∈ S T (ϕ, K).Since K is a bounded set, then there exists r > 0 such that K ⊂ B r (0).By (h 2 ), there exists Since ||F (t, u m (t), (Qu m )(t))|| ≤ µ(t) for almost all t ∈ [0, T ] and all m ≥ 1, by the Lebesgue dominated convergence theorem, we have ) is a causal operator such that the condition (h 1 ) − (h 5 ) hold.Then the multifunction S T : As in proof of Theorem 3.1 we can show that {u m (•)} n≥1 converges uniformly on This proves that G is closed and so ϕ → S T (ϕ, K) is upper semicontinuous.
) is a causal operator such that the conditions (h 1 ) − (h 5 ) hold.Then, for any ϕ ∈ C σ and any t ∈ [0, T ] the attainable set A t (ϕ, K) is compact in C([−σ, t], E) and the multifunction (t, ϕ) → A t (ϕ, K) is upper semicontinuous.
In the following, we consider a control problem: where g : E → R is a given function.

Monotone iterative technique
In this section, we suppose, in addition, that E is an ordered Banach space with partial order ≤, whose positive cone P = {x ∈ E; x ≥ 0} is normal with normal constant N .Evidently, C([0, b], E) is also an ordered Banach space with the partial order ≤ defined by the positive function cone In the following, consider the initial-valued problem under the following assumptions: where A(t) = {u(t); u ∈ A} and (QA)(t) = {(Qu)(t); u ∈ A}.
(5.3) Also, u(•) is said to be an upper solution of (5.1), if the inequalities of (5.3) are reversed.
for every bounded sets B 1 , B 2 ⊂ E.

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Then the initial value problem (5.1) has minimal and maximal solutions between v 0 and w 0 .
Proof.First, for any h(•) ∈ [v 0 , w 0 ], consider the differential equation where σ(t) = F (s, u(t), (Qu)(t)) + M u(t), M > 0 and ¿From (5.4) we have that u is a solution of (5.1) if and only if Au = u.Obviously, A is a continuous operator.By ( h 4 ), the operator A is increasing in [v 0 , w 0 ], and maps any bounded set in [v 0 , w 0 ] into a bounded set.We shall show that v 0 ≤ Av 0 and Aw 0 ≤ w 0 .If we put σ and so, v 0 ≤ Av 0 .Similarly, we can show that Aw 0 ≤ w 0 .Therefore, since A is an increasing operator in [v 0 , w 0 ], we obtain that A maps [v 0 , w 0 ] into itself.Further, we define the sequences {v m (•)} m≥0 and {w m (•)} m≥0 by iterative scheme Then from monotonicity property of A, it follows that and so, by Gronwall's lemma, we have that α(V (t)) = 0 for every t ∈ [0, T ].Moreover, since (see [14], Theorem 1.4.2) α(V ) = sup (5.7), we infer that v 0 ≤ v ≤ w ≤ w 0 and v = Av, w = Aw.By the monotonicity of A, it is easy to see that v and w are the minimal and maximal fixed points of A in [v 0 , w 0 ], respectively.This completes the proof of our theorem.

. 1 )
Let A = {u m (•); m ≥ 1}.Denote by I the identity mapping on B. From 3.1 it follows that (I − P )(A) is a uniformly equicontinuous subset of B. Since A ⊂ (I − P )(A) + P (A) and the set P (A) is uniformly equicontinuous, then we infer that the set A is also uniformly equicontinuous on [−σ, T ].Set A(t) = EJQTDE, 2010 No. 64, p. 7 {u m (t); m ≥ 1} for t ∈ [0, T ].Then, by (2.5) and property (v ) of the measure of non-compactness we have