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ON GENERAL PROPERTIES OF RETARDED FUNCTIONAL DIFFERENTIAL

. We investigate general properties, such as existence and uniqueness, continuous dependence on data and continuation, of solutions to retarded functional diﬀerential equations with inﬁnite delay on a diﬀerentiable manifold.


1.
Introduction. The purpose of this paper is to analyze the general properties of retarded functional differential equations (RFDEs for short) with infinite delay on a smooth boundaryless manifold. We will discuss existence and uniqueness of solutions, continuous dependence on data and continuation of solutions to such equations.
Let N ⊆ R k be a boundaryless smooth manifold. Denote by BU ((−∞, 0], N ) the set of bounded and uniformly continuous maps from (−∞, 0] into N with the topology of the uniform convergence and let Ω be an open subset of R × BU ((−∞, 0], N ). A continuous map g : Ω → R k will be called a functional field over N provided that g(t, ϕ) ∈ T ϕ(0) N for all (t, ϕ) ∈ Ω, where, given p ∈ N , by T p N ⊆ R k we denote the tangent space of N at p.
In this paper we will consider RFDEs of the type where g : Ω → R k is a functional field over N . A solution of the above equation is a function x : J → N , defined on an open real interval J with inf J = −∞, p n → p and α n (p n − p) → v.
The set C p A of the inward vectors to A at p is called the tangent cone of A at p. The tangent space T p A of A at p is the vector subspace of R k spanned by C p A. A vector v of R k is said to be tangent to A at p if v ∈ T p A. One can easily check that the tangent cone is always closed in R k .
To simplify some statements and definitions we put C p A = T p A = ∅ whenever p ∈ R k does not belong to A (this can be regarded as a consequence of Definition 2.3 if one replaces the assumption p ∈ A with p ∈ R k ). Observe that T p A is the trivial subspace {0} of R k if and only if p is an isolated point of A. In fact, if p is a limit point, then, given any {p n } in A\{p} such that p n → p, the sequence α n (p n − p) , with α n = 1/|p n − p|, admits a convergent subsequence whose limit is a unit vector.
One can show that, in the special and important case when A is a smooth manifold with (possibly empty) boundary ∂A (a ∂-manifold for short), this definition of tangent space is equivalent to the classical one (see for instance [13], [7]). Moreover, if p ∈ ∂A, C p A is a closed half-space in T p A (delimited by T p ∂A), while C p A = T p A if p ∈ A\∂A. Finally, a vector v is said to be strictly inward at p ∈ ∂A if v ∈ C p A\T p ∂A.
Let g : Ω → R k be a continuous map on an open subset Ω of R × A. We say that g is a (retarded) functional (tangent vector) field (over A) if g(t, ϕ) ∈ T ϕ(0) A for all (t, ϕ) ∈ Ω. In particular, g will be said inward (to A) if g(t, ϕ) ∈ C ϕ(0) A for all (t, ϕ).
Let us consider a retarded functional differential equation (RFDE for short) of the type where g : Ω ⊆ R × A → R k is a functional field over A. Here, as usual and whenever it makes sense, given t ∈ R, by x t ∈ A we mean the function θ → x(t + θ). By a solution of 3 we mean a function x : J → A, defined on an open real interval J with inf J = −∞, bounded and uniformly continuous on any closed half-line (−∞, b] ⊂ J, and which verifies eventually the equality x (t) = g(t, x t ). That is, x is a solution of 3 if there exists τ , with −∞ ≤ τ < sup J, such that (t, x t ) ∈ Ω for all t ∈ (τ, sup J), x is C 1 on the subinterval (τ, sup J) of J, and verifies x (t) = g(t, x t ) for all t ∈ (τ, sup J). Observe that the derivative of a solution x may not exist at t = τ . However, the right derivative D + x(τ ) of x at τ always exists and is equal to g(τ, x τ ). Also, notice that, since x is uniformly continuous on any closed half-line (−∞, b] of J, then t → x t is a continuous curve in A.
A solution of 3 is said to be maximal if it is not a proper restriction of another solution to the same equation. As in the case of ODEs, Zorn's lemma implies that any solution is the restriction of a maximal solution.
Given (τ, η) ∈ Ω, we will also be interested in the following initial value problem: A solution of problem 4 is a solution x : J → A of 3 such that sup J > τ , x (t) = g(t, x t ) for t > τ , and x τ = η.
Clearly, x : J → A is a solution of 4 if and only if Below, we will be concerned with the uniqueness of the solution of equation 3 and of problem 4.
In proving our uniqueness results, we will make use of the following folk result, whose proof is given here for the sake of completeness.
We have Being δc < 1, this inequality is verified if and only if α(t 0 ) = 0. Thus α(t) = 0 for any t ∈ [τ, τ + δ], and the assertion follows in a finite number of steps.
Let g : Ω ⊆ R × A → R k be a functional field and let U be an open subset of Ω. We will say that g is compactly Lipschitz in U or, for short, c-Lipschitz in U if, given any compact subset Q of U , there exists L ≥ 0 such that Moreover, we will say that g is locally c-Lipschitz in Ω if for any (τ, η) ∈ Ω there exists an open neighborhood of (τ, η) in Ω in which g is c-Lipschitz. In spite of the fact that a locally Lipschitz map is not necessarily (globally) Lipschitz, one could actually show that if g is locally c-Lipschitz in Ω, then it is also (globally) c-Lipschitz in Ω. As a consequence, if g is locally Lipschitz in the second variable, then it is additionally c-Lipschitz.
Theorem 2.5 below shows that, if g is c-Lipschitz in Ω, then one gets the uniqueness of the solution of the initial value problem.
Theorem 2.5 (uniqueness). Let A be a subset of R k , Ω an open subset of R × A and g : Ω → R k a c-Lipschitz functional field. Let Then, each one of the sets Since g is c-Lipschitz in Ω, there exists L ≥ 0, corresponding to the compact set Q = Q 1 ∪ Q 2 , such that for any t ∈ [τ, τ + h] we have we get α(τ ) = 0 and, by the above inequality, This shows that x 1 and x 2 coincide in any right neighborhood of τ contained in J 1 ∩ J 2 , proving the uniqueness of the maximal solution of problem 4.
3. Existence of solutions. In this paper we will be mainly interested in RFDEs on manifolds. From now on, we will assume that A is a boundaryless smooth differentiable manifold in R k which will be denoted by N . Let g : Ω → R k be a functional field over N defined on an open subset Ω of R× N . Given (τ, η) ∈ Ω, we will be concerned with the following initial value problem: Moreover, given a relatively closed subset X of N , we will investigate under what assumptions at least one solution of 5, possibly the only one, remains confined into X. For instance, X could be a ∂-manifold of the type {p ∈ N : φ(p) ≤ 0}, where the "cutting function" φ : N → R is smooth, having 0 ∈ R as a regular value. In particular, X could be an open subset of R k (in this case N = X), or a closed subset of R k (in this case N could be an open neighborhood of X, possibly coinciding with R k ), or a compact ∂-manifold. In the last case, one may consider as N the double of X (see e.g. [11], [14]).
To be more precise, in what follows, given a relatively closed subset X of N , the problem will be called the confined problem and an X-valued solution of 6 a confined solution.
In the first part of this section we are interested in obtaining existence results for the problems 5 and 6. More precisely, in Theorems 3.1 and 3.4 we will be concerned with the existence of a local solution. Then, suitable assumptions on g will ensure the existence of a global solution (see Theorems 3.7 and 3.9). In Lemma 3.2 below, we will consider the equation x (t) = γ(t, x t−r ), where γ : R × W → R k is continuous, W is an open subset of R k and r is a given positive constant. We point out that this equation can be still regarded as an RFDE, since it can be written in the form x (t) = γ r (t, x t ), where γ r is the continuous map γ r (t, ϕ) = γ(t, ϕ −r ), which is clearly defined on R × W . Therefore, by a solution of we mean a solution (in the RFDE meaning introduced previously) of where γ r denotes the restriction of γ r to R × W .
Proof. Let (τ, η) ∈ R × W . Notice that, because of the delay r, the R k -valued function . This clearly happens if t * = τ + h, since in this case one has |x(t) − η(0)| ≤ µ|t − τ | < µh and, thus, x(t) cannot reach the boundary of W for all t < t * . Thus, the function x : (−∞, τ + h) → R k takes values in W and is actually a solution of problem 7.
The following result is known (see e.g. [9,Chapter 12] and [10, Chapter 2]). Here we provide an independent proof. Proof. Let V be an open neighborhood of (τ, η) in R× W such that V ⊂ Ω and g(V ) is bounded. By the Tietze Extension Theorem, there exists a continuous extension g : R × W → R k of the restriction g| V of g to V with bounded image. Let {ε n } be a sequence of positive numbers converging to 0 and consider the following auxiliary problem depending on n ∈ N: By applying Lemma 3.2 to problem 8 with γ = g and r = ε n , we get, for any n ∈ N , a solution x n : As already observed, any x n solves the integral problem Since x n (t) = η(t − τ ) for t ≤ τ and for all n ∈ N, and since g has bounded image, because of Ascoli's Theorem we may assume, without loss of generality, that the sequence {x n (t)} converges uniformly on any half-line Thus, in any such half-line, the functions x n are equi-uniformly continuous. This implies that, for any t > τ , x n s−εn −x n s → 0 uniformly with respect to s ∈ [τ, t]. Moreover, observe that x n s − x s → 0. Consequently, x n s−εn − x s → 0 uniformly for s ∈ [τ, t] and, thus, because of Remark 2.2, the sequence { g(s, x n s−εn )} converges uniformly to g(s, x s ) on [τ, t]. Therefore, passing to the limit in the integral of 9, we obtain Proof of Theorem 3.1. Let W ⊆ R k be a tubular neighborhood of N with associated retraction ρ : W → N (if N is an open subset of R k , then W = N and ρ is the identity). Hence, N is relatively closed in W and N is a (relatively) closed subset of W . Therefore, there exists an open subset Ω of R× W such that Ω = Ω∩(R× N ) and Ω is closed in Ω. By the Tietze Extension Theorem, g admits a continuous extension g : Ω → R k . Moreover, we may assume that g has the following additional property: In fact, if this is not the case, it is sufficient to consider the orthogonal projection of g(t, ϕ) onto the space T ρ(ϕ(0)) N .
Therefore, by applying Lemma 3.3 to W, Ω and g, we obtain the existence of δ > 0 such that the problem ∈ N for some τ < t < τ + δ. This means that α(t) > 0 for some τ < t < τ + δ and, consequently, its derivative must be positive at some τ < θ < τ + δ. That is, where ·, · denotes the inner product in R k , and w(θ) is the derivative at t = θ of the curve t → ρ( x(t)). This is a contradiction, since both the vectors g(θ, x θ ) and w(θ) are tangent to N at ρ( x(θ)) and, consequently, orthogonal to x(θ) − ρ( x(θ)). This proves that x(t) ∈ N for all t < τ + δ.
In the next result we will be concerned with the local existence of the solutions in the confined case. Let X be a relatively closed subset of N and let p ∈ X be given. We will say that the functional field g : Ω → R k is away from N at p ∈ X if either g(t, ϕ) ∈ C p (N \X) for all (t, ϕ) ∈ Ω with ϕ(0) = p or g(t, ϕ) = 0, for all (t, ϕ) ∈ Ω such that ϕ(0) = p. We point out that this condition is obviously satisfied whenever p, which is a point of X, is not in the topological boundary of X relative to N since, in that case, C p (N \X) = ∅. Notice that this condition is also satisfied when X = N , since C p (∅) = ∅. Theorem 3.4 (confined local existence). Let N ⊆ R k be a boundaryless smooth manifold, Ω an open subset of R × N and g : Ω → R k a functional field. Let X be a relatively closed subset of N . Then, there exists δ > 0 such that the confined problem 6 admits at least one (X-valued) solution on (−∞, τ + δ), provided that g is away from N at η(0).
Proof. Let (τ, η) ∈ R× X be the initial condition of the confined problem 6. Consider first the case when g(t, ϕ) = 0, for all (t, ϕ) such that ϕ(0) = η(0). We claim that the X valued function , t ≥ τ satisfies 6. In fact, for any t > τ one has x(t) = x t (0) = η(0), which clearly implies x (t) = 0 and, by assumption, also g(t, x t ) = 0. Moreover, the initial condition x τ = η is obviously verified. This shows that the above function x, which is actually defined up to +∞, is the required solution.
Remark 3.5. In the non-uniqueness case, the only assumption g away from N at η(0) ∈ X does not guarantee that any solution of 5 with η ∈ X is necessarily X-valued for any t > τ . To see this, take, for example, N = R, X = (−∞, 0] and g(t, ϕ) = 3 3 ϕ(0) 2 . Then, the function In what follows, the continuous map H : Ω → N that associates to any (t, ϕ) ∈ Ω the "head" H(t, ϕ) = ϕ(0) ∈ N of ϕ will be called, for brevity, the head map.
The set will be called the pre-support of g and its closure in N , i.e. S g ∩ N , the support of g. In other words, p ∈ N belongs to the pre-support of g if and only if H −1 (p) is nonempty and g is not identically zero in this set. One can easily check that, because of the continuity of g (and recalling that Ω is open), the pre-support S g is a relatively open subset of N . This fact will be used below in proving the continuous dependence of the solutions on the initial data.
The following proposition shows how the pre-support S g of g is related to the solutions of problem 5.
Proposition 3.6. Let (τ, η) be the initial condition of problem 5. If η(0) / ∈ S g , then 5 has a unique maximal solution x : R → N that is constantly equal to η(0), for any t > τ . If η(0) ∈ S g , then any solution x : J → N of 5 satisfies x(t) ∈ S g for all τ < t < sup J.
Proof. If η(0) / ∈ S g , by an argument similar to that used in the first part of the proof of Theorem 3.4, we get that the function which is clearly defined for any t ∈ R, is a solution of 5. Notice that this solution is unique, since N \S g is open in N . Otherwise, if η(0) ∈ S g , the fact that g(t, ϕ) = 0 whenever ϕ(0) belongs to N \S g implies that x(t) cannot enter the relatively open set N \S g .
The following result shows that any maximal solution of 5, whose existence is ensured by Theorem 3.1 and Zorn's Lemma, is globally defined provided that g has complete support and bounded image. Theorem 3.7 (global existence). Let N ⊆ R k be a boundaryless smooth manifold and g : R × N → R k a functional field. Assume that g has complete support and bounded image. Then, any maximal solution of problem 5 is defined on the whole real line.
Proof. Let (τ, η) be the initial condition of problem 5. If η(0) / ∈ S g , by Proposition 3.6 the N valued function is the unique global solution of 5. Otherwise, if η(0) ∈ S g , we will prove that any maximal solutionx of 5 is defined on the whole real line. By contradiction, supposex defined up to b > τ , with b < +∞. Since g has bounded image, then the derivativex is bounded in the interval (τ, b). Moreover, again by Proposition 3.6, we getx(t) ∈ S g for τ < t < b and, since S g is complete, lim t→b −x(t) exists and belongs to S g ⊆ N . Consequently, the functionη : (−∞, 0] → N given bȳ is uniformly continuous. Thus, by applying Theorem 3.1 to the initial value problem we get the existence of a solution of 10 defined up to b + δ, for some δ > 0. This contradicts the maximality ofx.
Remark 3.8. The assumptions of Theorem 3.7 above are clearly satisfied if g has compact support and bounded image. Moreover, when N = R k , the support of g is, by definition, a closed set. Thus, as well known, any maximal solution is globally defined, provided that g has bounded image.
Our purpose now is to obtain a global existence result for the confined problem 6. To this end, similarly to Theorem 3.4, we need to impose an extra assumption on the functional field g. Namely, we will assume that g is away from N in X, meaning that g is away from N at any p ∈ X. Notice that to obtain this condition, it is enough to verify that p ∈ S g ∩ X implies g(t, ϕ) ∈ C p (N \X), for all (t, ϕ) with ϕ(0) = p (see page 34). Observe that, if g is away from N in X and σ : N → [0, +∞) is continuous, then the product (t, ϕ) → σ(ϕ(0))g(t, ϕ) is still away from N in X and S σg ⊆ S g .
As an example, let X be a compact ∂-manifold in R k and N be the double of X. In this case, a functional field g : R × N → R k is said to be strictly inward at a point p ∈ ∂X if g(t, ϕ) is strictly inward at p whenever ϕ(0) = p. Clearly, a functional field g is away from N at p ∈ ∂X if and only if either p ∈ S g , or g is strictly inward at p. In addition, an inward functional field g is said to be strictly inward if it is strictly inward at any p ∈ ∂X.
Theorem 3.9 below is a global existence result for the confined case. Its proof is analogous to the one of Theorem 3.7, provided that, in the contradiction argument, one replaces the local existence in N (Theorem 3.1) with the local existence in X (Theorem 3.4). Theorem 3.9 (confined global existence). Let X be a relatively closed subset of a boundaryless smooth manifold N ⊆ R k , and g : R × N → R k a functional field away from N in X. Assume that g has complete support in X (i.e. S g ∩ X is closed in R k ) and that g(R × X) is bounded. Then, any maximal solution of the confined problem 6 is defined on the whole real line.

Continuous dependence on data and continuation of solutions.
In what follows we will denote by BU (R, R k ) the Fréchet space of those functions which are bounded and uniformly continuous on any left half-line, with the topology generated by the family of seminorms {P b : b ∈ R}, where P b (x) = sup t≤b |x(t)|. Clearly, for any b ∈ R, one has P b (x) = x b , where, as previously, · denotes the norm in the Banach space R k . Moreover, given any subset A ⊆ R k , we will denote by BU (R, A) the subset of BU (R, R k ) of the A-valued functions.
In Lemmas 4.1, 4.2 and 4.3 below we will be concerned with upper semicontinuous multivalued maps. We recall that a multivalued map F between two metric spaces X and Y is said to be upper semicontinuous if it is compact valued and for any open subset U of Y the upper inverse image of U , i.e. the set F −1 (U ) = {x ∈ X : F (x) ⊆ U }, is an open subset of X . Equivalently, F is upper semicontinuous if and only if for any x 0 ∈ X and sequences {x n } in X , x n → x 0 and {y n } in Y, y n ∈ F (x n ) for all n ∈ N, there exists a subsequence of {y n } converging to some element y ∈ F (x 0 ). Recall that the composition of upper semicontinuous maps is upper semicontinuous.
The next lemma, in which the uniqueness of the solution of the initial value problem is not assumed, regards the continuous dependence on the initial data of the set of solutions of problem 5. Its statement concerning multivalued maps is justified by its application in the proof of Theorem 4.4 below. In fact, although in Theorem 4.4 the uniqueness of the solution of 5 is assumed for the sake of simplicity, the use in its proof of the Tietze Extension Theorem does not guarantee the same property for the extended problem to which Lemma 4.1 is applied. Lemma 4.1. Let N ⊆ R k be a boundaryless smooth manifold and let g : R× N → R k be a functional field. Assume that g has complete support and bounded image. Then, the multivalued map Σ : R × N BU (R, N ) that associates to any (τ, η) the set Σ(τ, η) of the (global) solutions x of 5 is upper semicontinuous.
Proof. Take sequences {(τ n , η n )} in R× N converging to (τ, η) and {x n } in BU (R, N ) such that x n ∈ Σ(τ n , η n ) for any n ∈ N. We have to prove that {x n } admits a subsequence converging to some x ∈ Σ(τ, η).
Suppose that {η n } has a subsequence, again denoted by {η n }, such that η n (0) / ∈ S g for any n. Since, as already observed, the pre-support S g is an open subset of N , then η(0) / ∈ S g . By Proposition 3.6, the solution x n is unique and given by Clearly, {x n } converges in BU (R, N ) to the function Consequently, x is a solution of problem 5 (recall that η(0) / ∈ S g ). That is, x ∈ Σ(τ, η), proving the assertion in the case η n (0) / ∈ S g . Otherwise, without loss of generality, we may assume that η n (0) ∈ S g for any n ∈ N. By the definition of Σ, any x n is a (globally defined) solution of the problem Therefore, as already observed, one has Now, set y n (s) = x n (τ n + s). By 11, for s ≥ 0, we obtain y n (s) = x n (τ n + s) = η n (0) + τ n +s τ n g(ρ, x n ρ ) dρ = η n (0) + τ n +s τ n g(ρ, y n ρ−τ n ) dρ .
Since g has bounded image, say |g(t, ϕ)| ≤ c, we get |(y n ) (s)| ≤ |g(τ n + s, y n s )| ≤ c, s > 0. Hence, by Ascoli's Theorem, {y n } has a subsequence converging to a function y ∈ BU (R, R k ) that clearly satisfies y(s) = η(s) for s ≤ 0. Without loss of generality we may assume that {y n } itself converges to y.
Define x(t) = y(t − τ ). Let us show that x n → x in BU (R, R k ). This follows from the inequality |x n (t)−x(t)| = |y n (t−τ n )−y(t−τ )| ≤ |y n (t−τ n )−y(t−τ n )|+|y(t−τ n )−y(t−τ )|, taking into account that y belongs to BU (R, R k ) and y n → y (in BU (R, R k )). Observe now that, since we are assuming η n (0) ∈ S g , by Proposition 3.6 we get x n (t) ∈ S g for any t. Thus, the limit x(t) belongs to S g as well. Now, from the completeness of S g , we get S g ⊆ N , so that x(t) ∈ N for any t. Moreover, the same fact obviously holds for any y n and for y. To complete our proof, it remains to show that x is a solution of problem 5.
Since, for 0 ≤ σ ≤ s, {y n σ } converges to y σ as n → ∞, and again using the fact that g has bounded image, by the Lebesgue Theorem, we get in 12 On the other hand, the convergence of {y n } to y in BU (R, N ) implies, in particular, that {y n (s)} converges to y(s) for any s. Thus, we obtain Therefore, for t ≥ τ and again by a change of variable, we get

ON GENERAL PROPERTIES OF RFDE 39
Consequently, 13 becomes This proves that x is a solution of problem 5, that is x ∈ Σ(τ, η).
Our purpose now is to remove the assumptions ensuring that the solutions of 5 are globally defined. To this end we need two preliminary results: Lemmas 4.2 and 4.3 below. The first one is a folk result and its simple proof will be omitted.
Lemma 4.2. Let C(X , Y) be the metric space of the continuous maps from a compact metric space X to a metric space Y. Then, the multivalued map that associates to any f ∈ C(X , Y) its image Img(f ) ⊆ Y is upper semicontinuous.

Lemma 4.3.
Let N ⊆ R k be a boundaryless smooth manifold and let g : R× N → R k be a functional field. Assume that g has complete support and bounded image. Then, given a compact real interval I, the multivalued map is upper semicontinuous.
Proof. Let A I : BU (R, R k ) → C(I, R× R k ) be the single valued map that associates to any y ∈ BU (R, R k ) the (continuous) curve t ∈ I → (t, y t ) ∈ R × R k . Let us show that A I is continuous. In fact, one has where b = max I. This proves the continuity of A I . Consequently, by applying Lemma 4.1 and Lemma 4.2 with X = I and Y = R × N , we get that the map Img •A I • Σ : R × N R × N , being composition of upper semicontinuous maps, is upper semicontinuous. Now, it is enough to observe that This completes our proof.
We are now in a position to state our result on the continuous dependence of the solutions of problem 5 on the initial data, in the general case when g is a functional field over N defined on an open subset Ω of R × N . Proof. Let (τ 0 , η 0 ) ∈ D b and fix a < τ 0 . Denote by x 0 the restriction to (−∞, b] of the unique maximal solution of problem 5 with initial conditions (τ 0 , η 0 ). Set be an open neighborhood of K 0 in Ω such that V ⊂ Ω and g(V ) is bounded. Because of the Tietze Extension Theorem, there exists a continuous extension g : R × N → R k of the restriction g| V of g to V with bounded image. We may also assume that g is a functional field over N , by taking if necessary the orthogonal projection onto the tangent space. As previously, let H denote the head map (t, ϕ) → ϕ(0) and consider the set Since N is locally compact, we may also assume that U 2 is a relatively compact subset of N . Let σ : N → [0, 1] be a continuous function such that σ(p) = 1 if p ∈ U 1 , σ(p) = 0 if p / ∈ U 2 . Define g : R × N → R k by g(t, ϕ) = σ(ϕ(0)) g(t, ϕ). The map g is clearly continuous being g(t, ϕ) = σ(H(t, ϕ)) g(t, ϕ), and has bounded image, since g itself has bounded image. Moreover, its support, that is the closure (in R k ) of the set S g = p ∈ N : ∃ (t, ϕ) ∈ R × N such that ϕ(0) = p, g(t, ϕ) = 0 , is clearly contained in U 2 and, thus, is compact. Therefore, by applying Lemma 4.3 to the N -functional field g with I = [a, b], we obtain that the multivalued map is upper semicontinuous. Let us show now that Clearly, K 0 ⊆ K I (τ 0 , η 0 ). On the other hand, since g = g on the neighborhood x t ) with initial condition (τ 0 , η 0 ) coincides with x 0 for t ∈ [a, b]. Therefore, K I (τ 0 , η 0 ) ⊆ K 0 , proving 14. Consequently, from the equality 14, V ∩ H −1 (U 1 ) is actually an open neighborhood of K I (τ 0 , η 0 ) and, thus, by the upper semicontinuity of K I , K −1 is an open subset of R × N containing (τ 0 , η 0 ). Let us prove that the set is clearly an open neighborhood of (τ 0 , η 0 ) in R × N , is in fact contained in D b . To this end, take (τ, η) ∈ W . Then, for t ∈ [a, b], we have (t, x t ) ∈ V and x t (0) ∈ U 1 , where x(·) satisfies x (t) = g(t, x t ), x τ = η. Recalling again that g = g on V ∩ H −1 (U 1 ), we get g(t, x t ) = g(t, x t ) for t ∈ [a, b]. Hence, in particular, g(t, x t ) = g(t, x t ) for t ∈ [τ, b]. On the other hand, because of the uniqueness assumption, x must coincide with the solution of problem 5 that, therefore, is defined up to b. Thus, (τ, η) ∈ D b . This proves that D b is an open subset of Ω, as claimed.
It remains to show that the map s b is continuous in D b . To this end, given any compact interval I ⊂ (−∞, b] and (τ, η) ∈ D b define K I (τ, η) = {(t, x t ) : t ∈ I, x the maximal solution of 5}.
The proof of the continuous dependence in the confined case (Corollary 4.5 below) is a straightforward consequence of Theorem 4.4. Indeed, the assumption that g is away from N in X together with the uniqueness of the solutions of problem 5 imply that any solution of 5 with η ∈ X is actually a confined solution (recall Remark 3.5).
Corollary 4.5 (confined continuous dependence). Let N, Ω, g be as in Theorem 4.4 and assume that problem 5 has a unique maximal solution for any (τ, η) ∈ Ω. Let X be a relatively closed subset of N and assume that g is away from N in X. Then, given b ∈ R, the restriction of the continuous map s b to the relatively open Let us now discuss the continuation property of the solutions. Analogous results in Euclidean spaces can be found e.g. in [9,Chapter 12] and [10,Chapter 2].
Our first result states that, given a solution x : J → N of equation 3, if the curve t → (t, x t ) lies eventually in a bounded and complete subset C of Ω such that g(C) is bounded, then x is not a maximal solution. Proof. Since (t, x t ) belongs eventually to C, g(t, x t ) is eventually bounded, and so is x (t). Thus, lim t→b − x(t) exists and is finite. Therefore, the continuous function t → x(t) can be extended to the uniformly continuous function ). Since C is complete, (b,x b ) belongs to C and, thus, to Ω. Consequently, by applying Theorem 3.1 with (τ, η) = (b,x b ), we get that x is continuable on the right hand side of b.
The following two corollaries are straightforward consequences of Theorem 4.6. Therefore, their proofs will be omitted.    Proof. The proof is analogous to the one of Theorem 4.6, except that here one applies Theorem 3.4 (confined local existence) instead of Theorem 3.1. Moreover, since X is a relatively closed subset of N , it should be observed that now lim t→b − x(t) belongs to X.
The following consequence of Theorem 4.6 can be regarded as a Kamke-type result for RFDEs. Corollary 4.10. Let N be a boundaryless smooth manifold, and g : R × N → R k a functional field. Assume that g sends bounded subsets of R × N into bounded subsets of R k . If x : J → N is a solution of equation 3 whose graph-curve t → (t, x(t)) belongs eventually to a compact subset of R × N , then x is continuable.
Proof. Suppose that there exists τ ∈ J and a compact subset K of R × N such that (t, x(t)) ∈ K when τ ≤ t < sup J. Clearly b = sup J < +∞. Consider the following subset of R × R k : Notice that C is closed in R × R k and contained in R × N . Thus, it is complete. Moreover g(C) is bounded, since so is C. Observe, finally, that (t, x t ) ∈ C for t ≥ τ . Therefore, all the assumptions in Theorem 4.6 are satisfied, so that x is continuable.
The following example shows that, if the assumption that g sends bounded sets into bounded sets is removed, then the continuation property of the solutions may fail.
Example 4.11. Let N = R and let x : (−∞, 1) → R given by Consider the continuous injective curve γ : [0, 1) → R that to any t ∈ [0, 1) associates x t . We claim that the image Γ = {x t : t ∈ [0, 1)} of γ is a closed subset of R, and that γ is a homeomorphism onto Γ. We need to show that, if C ⊆ [0, 1) is relatively closed in [0, 1), then γ(C) is closed in R. If sup C < 1, then C is compact and, thus, γ(C) is closed. Therefore, it is sufficient to show that, if {t n } is a sequence in [0, 1) converging to 1, then {γ(t n )} is divergent (i.e. it does not admit a convergent subsequence). Given such a {t n }, for any θ < 0 we get Thus, {γ(t n )} is divergent, since the function θ → sin 1 θ is not uniformly continuous on (−∞, 0), and this proves our claim.
Define the continuous function g : Γ → R by g(ϕ) = x (γ −1 (ϕ)). Since Γ is closed, by the Tietze Extension Theorem g can be extended to a continuous map g on R.
We point out that this fact does not contradicts the assertion of Corollary 4.10 since g(Γ) is not bounded and, thus, g does not send bounded subsets of R into bounded subsets of R.

5.
Reduction to a confined problem: an example. We close the paper with an example in which we illustrate how the results about the confined problem 6 can be applied to retarded functional motion equations. In particular we would like to highlight how the relatively closed subset X, introduced in the confined problem, will be interpreted in Example 5.1. In fact, as we will show below, the difficulty arising from the noncompactness of the tangent bundle will be removed by restricting the search of T -periodic solutions to a convenient compact manifold with boundary that plays exactly the role that the set X has in the general context.
be the tangent bundle of M . Given q ∈ M , let (T q M ) ⊥ ⊆ R s denote the normal space of M at q. Since R s = T q M ⊕ (T q M ) ⊥ , any vector u ∈ R s can be uniquely decomposed into the sum of the parallel (or tangential) component u π ∈ T q M of u at q and the normal component u ν ∈ (T q M ) ⊥ of u at q.
Consider the retarded functional motion equation on the constraint M where x π (t) stands for the parallel component of the acceleration x (t) ∈ R s at the point x(t), the parameter ε > 0 is the frictional coefficient, and the map F : R× M → R s is a (continuous) functional field, T -periodic in the first variable, with bounded image, and which verifies conditions ensuring the uniqueness of the associated initial value problems. Any T -periodic solution of 15 is called a forced oscillation.
In [2] we proved that the equation 15 admits at least one forced oscillation, provided that the constraint M is compact with nonzero Euler-Poincaré characteristic and assuming the stronger hypothesis of the continuity of the functional field F on R × C((−∞, 0], M ) instead on R × M . We believe that the same result is true with the sole assumption that F is continuous on R × M -recall that M has a finer topology than C((−∞, 0], M ). Unfortunately the proof in [2] does not fit in this context. However, in the attempt of proving the existence of a forced oscillation of 15, one could still define a Poincaré-type operator, acting on a suitable topological space, and with the property that its fixed points correspond to the T -periodic solutions of the equation, and then prove the existence of a fixed point of such an operator by means of topological methods. Let us sketch now this argument.
A crucial step to define a Poincaré-type operator is to write 15 as a confined first order RFDE, and then use the confined global existence theorem (Theorem 3.9) together with the uniqueness theorem (Theorem 2.5) proved above. The possibility of reducing 15 to a first order RFDE is motivated by the fact that any second order differential equation on M is equivalent to a first order system on the tangent bundle T M of M . To be more precise, observe that the equation 15 can be equivalently written as x (t) = R(x(t), x (t)) + F (t, x t ) − εx (t), (16) where R : T M → R s is a smooth map -the so-called reactive force (or inertial reaction) -with the following properties: (a) R(q, v) ∈ (T q M ) ⊥ for any (q, v) ∈ T M ; (b) R is quadratic in the second variable; (c) given (q, v) ∈ T M , R(q, v) is the unique vector such that (v, R(q, v)) belongs to T (q,v) (T M ); (d) any C 2 curve γ : (a, b) → M verifies the condition γ ν (t) = R(γ(t), γ (t)) for any t ∈ (a, b), i.e. for each t ∈ (a, b), the normal component γ ν (t) of γ (t) at γ(t) equals R(γ(t), γ (t)). Now, the second order equation 16 can be transformed in the first order system x (t) = y(t), y (t) = R(x(t), y(t)) + F (t, x t ) − εy(t) (17) and is equivalent to 17 in the following sense: a function x : J → M is a solution of 16 if and only if the pair (x, x ) is a solution of 17. Observe that system 17 is actually a first order RFDE on the noncompact manifold T M , since it can be written as (x (t), y (t)) = G(t, (x t , y t )), where the map G : R × T M → R s × R s is the functional field over T M given by G(t, (ϕ, ψ)) = (ψ(0), R(ϕ(0), ψ(0)) + F (t, ϕ) − εψ(0)).
Given c > 0, consider the closed subset As in [2], one can show that if c is sufficiently large, then G is strictly inward to X c and, thus, away from T M in X c (observe that the tangent cone of X c at (q, v) ∈ ∂X c is the half subspace of T (q,v) X c given by where ·, · denotes the inner product in R s ). Consequently, we are reduced to the context of the confined problem 6 with R k = R s × R s , N = T M , g = G and the confining set X given by the ∂-manifold X c . Now, given (ϕ, ψ) ∈ X c , consider the following initial value problem depending on a real parameter λ: (x (t), y (t)) = λG(t, (x t , y t )), (x 0 , y 0 ) = (ϕ, ψ).
Let λ ≥ 0 be given. Theorems 2.5 and 3.9 yield a unique global X c -valued solution (x, y) of 18. Define the Poincaré-type operator P λ : X c → X c by P λ (ϕ, ψ)(s) = x(s + T ), y(s + T ) , s ∈ (−∞, 0]. Our conjecture is that the assumption ε > 0 ensures the existence of an unbounded connected subset of (λ, (ϕ, ψ)) ∈ [0, +∞) × X c : P λ (ϕ, ψ) = (ϕ, ψ) emanating from the slice {0} × X c . This would imply the existence of a fixed point of P 1 , due to the boundedness of X c . As a consequence one would get the existence of a forced oscillation of the motion equation 15 above.