Viability and Filippov-type Lemma for Stieltjes Differential Inclusions

We prove a viability result for differential inclusions involving the Stieltjes derivative with respect to a left-continuous non-decreasing function with time dependent state constraints. A tangential condition using a generalized notion of the contingent derivative is imposed. Classical viability results (for usual differential inclusions) are thus generalized and, at the same time, the gate to new viability results for difference inclusions, impulsive differential inclusions or dynamic inclusions on time scales is open. As a consequence, in the particular case where the state constraint is a tube, a Filippov-type lemma is obtained for this very general setting, of differential problems driven by Stieltjes derivatives.


Introduction
Originating with the paper of Nagumo ([23]), the viability matter for various kinds of differential problems has been investigated (see [1] or [13] for a comprehensive regard on the related literature).
In this work, we focus on the viability of differential inclusions involving the Stieltjes derivative ( [25]) instead of the usual derivative, aware of the significant recent growing of this theory (we refer the reader to [10] and the references therein).
By μ g one denotes the Lebesgue-Stieltjes measure induced by the map g (e.g.[28]).We get a viability result (Theorem 4) when K : [0, 1] → P cc (R d ) is g-absolutely continuous from the left w.r.t. the topology generated by g ( [14]) and at every discontinuity point of g, K has limit at the right for the usual topology in the sense of Pompeiu-Hausdorff distance.
The map F : [0, 1] × R d → P cc (R d ) is assumed to be upper semicontinuous for the product of the g-topology of [0, 1] with the usual topology of R d .As in the classical case, a tangential condition is imposed, namely that for every t ∈ [0, 1] in a full-measure set and every (t, x) ∈ G(K ) (G(K ) is the graph of K ), where D g K (t, x) is the contingent g-derivative (introduced in [27]).
A viability result (for the case where g is right-continuous and all assumptions are in the mirror) can be found in [3, Proposition 5.5] but under rather technical hypotheses.We note that the appearance of the Stieltjes derivative places us in a very wide framework.
Other particular instances lead to impulsive problems (when g is the sum of the identical function with a step function, see [14,22,25]) and difference inclusions (in the case where g is a step function).
Besides, it allows one to study systems whose dynamic is stationary on some intervals of time or rapidly changing at some moments (like those in [14] or [26]).What is more, since the theory of Stieltjes differential equations offers a tool to study complex hybrid systems which is (under natural assumptions) equivalent with the theory of dynamic equations on time scales, to that of generalized differential equations ( [11], [15], [19], [33]) and also to the theory of differential equations driven by Borel measures ( [5], [7], [9], [10], [21], [25], [30], [31], [32]), our result gives an opportunity to get new (as far as the authors know) viability results for all the above mentioned problems.

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One thus complements the Filippov-type result recently given in [17] under the hypothesis that F is Lipschitz only with respect to the state; it is proved that imposing stronger hypothesis allows us to get a better estimation r (t) (Proposition 6).

Notations and Auxiliary Results
Let g : [0, 1] → R be a left-continuous nondecreasing function.Without restricting the generality, we may suppose that g is continuous at 0 and that g(0) = 0.The measurability with respect to (bref, w.r.t.) the σ -algebra defined by g will be called g-measurability, μ g stands for the Stieltjes measure generated by g (see [28,Example 6.14]) and the Lebesgue-Stieltjes (shortly, LS-) integrability w.r.t.g means the abstract Lebesgue integrability w.r.t. the Stieltjes measure μ g .Let L 1 g ([0, 1]) be the space of LS-integrable functions w.r.t.g.The g− topology on [0, 1] is the topology with basis the class of all sets In [25] a notion of differentiability with respect to a left-continuous non-decreasing function was introduced (following an idea in [34], see also [35]).The Stieltjes or g-derivative of a function f : [0, 1] → R d (d ∈ N, d ≥ 1) with respect to g at a point t ∈ [0, 1] is (in the case of existence of limits) if g is continuous at t, otherwise.
It does not make sense on the set Note that if t is a discontinuity point of g, it suffices that f have right limit f (t+) in order to get .
The particular cases of the theory of Stieltjes differential equations are consequences of particular non-decreasing functions g.To be more specific: when g(t) = t we get ordinary differential equations, in the case where g is a sum of step functions we retrieve difference equations, while for the situation where g is a sum of the identical function with step functions we get impulsive problems.So, the advantage of using Stieltjes derivatives is that it allows one to study various problems through a unified theory.
This generalized derivative has intensively been used in solving various problems where abrupt modifications (corresponding to discontinuity points of g) and dead times intervals (corresponding to intervals where g is constant) can occur, such as [14], [16] or [26].
(This follows automatically from the definition of the g− derivative.)A useful result on the g-derivability of a product was given in [18,Proposition 2.5]: A function u : [0, 1] → R d is g-absolutely continuous (shortly g-AC, see [25] or [35]) if for every ε > 0 there is δ ε > 0 such that for any set It is known ( [14]) that such a function is continuous at the continuity points of g.
The Fundamental Theorem of Calculus for LS-integrals is given below.
e. on [a, c) and • Assume that g is continuous at a.
• Assume that g is discontinuous at a. Then a / ∈ E so, ϕ is g− differentiable at a.For each t ∈ (a, b) with g(t) = g(a) we have lim which implies that u is g− differentiable at a and u g (a) = ϕ g (a).Hence, 123 So, in both cases, we have u g = ϕ g , μ g − a.e. on [a, b).
Arguing in a similar way we may also prove that u is g− differentiable μ g − a.e. on [b, c) and u g = ψ g , μ g − a.e. on [b, c).
Since ϕ g , ψ g are μ g − integrable on [a, b), [b, c) respectively, it turns out that u g is μ g − integrable on [a, c).
(ii) By virtue of the Fundamental Theorem of Calculus (Theorem 1), it suffices to prove that e. on [a, t) and hence, and hence, Consequently, , which implies that ũ is g− differentiable at a and .
The above arguments combined with the Fundamental Theorem of Calculus (Theorem 1) ensure that ũ is g− AC.
The next result concerning the g-derivative of a limit of a sequence of g-absolutely continuous functions will be used in what follows.
Lemma 4 ([27, Lemma 3.8]) Let (u k ) k be a sequence of g-absolutely continuous R d -valued functions on [0, 1], pointwisely convergent to some u : then u is also g-absolutely continuous and We remind that a function u : [0, 1] → R d is g-continuous ( [14]) if it is continuous w.r.t. the g-topology, i.e. if for any t ∈ [0, 1] and any ε > 0 there exists δ t,ε > 0 such that Obviously, g-absolute continuity involves g-continuity.
Let us recall the following compactness result in the space BC g ([0, 1], R d ) of bounded and g-continuous functions endowed with the sup-norm topology ([14, Proposition 5.6]).
In [14] it was proved that a g-continuous function might not have right limit u(t+) at the discontinuity points of g and it might be not even necessarily bounded.Related to this remark, we introduce the following notion.

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Obviously, any g-absolutely continuous map is g-uniformly continuous; also, any guniformly continuous function is g-continuous but the converse is not true ([14, Example 3.3] shows a g-continuous function for which the right limit does not exist at a discontinuity point of g, thus, using Corollary 1 below, it is not g-uniformly continuous).
The next result is not needed in the sequel in its whole generality, but we think it could be of a larger interest.
Lemma 5 Let u : [0, 1] \ E → R d be g-uniformly continuous, where E ⊂ [0, 1] is a g-null set.Then at any right-accumulation point of [0, 1) \ E, t 0 , there exists the right limit Proof For every ε > 0 there is δ ε > 0 such that for any t, t k is a Cauchy sequence, therefore it has a limit.We can see that this limit does not depend on the choice of (t k ) k : if we take another sequence and the limit of (u(t k )) k is, indeed, independent on the choice of (t k ) k .
Then lim We claim that (u(t n )) n is Cauchy.Indeed, fix ρ > 0 and choose s ∈ (0, δ) as postulated in the hypothesis.We may find n 0 ≥ 1 such that Indeed, fix again ρ > 0 and choose s ∈ (0, δ) as postulated in the hypothesis.Then It turns out that l does not depend on the choice of the sequence

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We finally remind the reader of some basic elements of set-valued analysis ([2], [6]).The open ball centered at 0 ∈ R d of radius r > 0 is denoted by B r (0).
Let us denote by P cc (R d ) the family of nonempty compact convex subsets of R d .If A, B ∈ P cc (R d ), the Hausdorff-Pompeiu distance is defined as follows: where the (Pompeiu-) excess of the set A over the set B is defined by Let (X , τ ) be a topological space.A compact convex -valued mapping : (X , τ ) Below is the notion of contingent g-derivative, recently introduced in [27] (and applied e.g. in [20]). . Equivalently: Similarly to the g-derivative in the single-valued case, the notion of contingent g-derivative does not make sense at the points t 0 at the right of which g is constant on some interval; this is why in the hypothesis 4) below a condition involving the contingent g-derivative is imposed only g-almost everywhere.
Let us recall the existence and uniqueness result for linear Stieltjes differential equations studied in [14, Proposition 6.8] and [18].

Viability Result
We aim to prove a viability result for the Stieltjes differential inclusion (1) ) is g-absolutely continuous from the left w.r.t. the g-topology in the following sense (see [13,Definition 4.1] in a particular case): for every ε > 0 there exists δ ε > 0 satisfying, for any set Moreover, we assume that at every discontinuity point t 1 ∈ [0, 1) of g, K has limit at the right for the usual topology in the sense of Pompeiu-Hausdorff distance: there exists K (t 1 +) ∈ P cc (R d ) such that for every ε > 0 there exists δ t 1 ,ε > 0 satisfying, for any t ∈ [0, 1] with t 1 < t < t 1 + δ t 1 ,ε : The first part of hypothesis 1) implies that for every ε > 0 there exists δ ε > 0 satisfying, for any set In particular, for every ε > 0, there exists δ ε > 0 such that for each t 0 ∈ (0, 1]: ) is upper semicontinuous for the product of the g-topology of [0, 1] with the usual topology of R d .

Remark 3 This can be equivalently expressed as: for each
3).There is M > 0 such that for every 4).There exists E ⊂ [0, 1] with μ g (E) = 0 satisfying the following condition: for every t ∈ [0, 1] \ E and every (t, x) ∈ G(K ), Without restricting the generality we may suppose that C g ⊂ E.
A key auxiliary result is the following.
(2) Proof Case I. Suppose t 0 is a continuity point of g.Then we apply the hypothesis 4) for There exists y ∈ F(t 0 , x 0 ) ∩ D g K (t 0 , x 0 ), so one can find δ ∈ (0, ε) such that for some z ∈ K (t 0 + δ), As g is continuous at t 0 , one can choose δ such that g(t which is g-absolutely continuous on [t 0 , t 0 + δ] and clearly satisfies the first assertion of (2) .The function g being non-decreasing, for all t ∈ [t 0 , t 0 + δ], g(t) ≤ g(t 0 + δ), so for every t in this interval, This implies that for all t ∈ [t 0 , and thus, the second assertion of (2) holds.Besides, by the definition of the g-derivative, on [t 0 , t 0 + δ), and so, (see also (3).)Hence, the fourth assertion of ( 2) is valid too.Finally, which is the third assertion of (2).
Case II.Suppose t 0 is a discontinuity point of g.Then by the assumption 4) applied for for some h k ↓ 0 and x k ∈ K (t 0 + h k ) for all k ∈ N. It follows that there exists the limit lim k→∞ x k and Moreover, lim k→∞ x k belongs to K (t 0 +).To check this, fix ρ > 0. Hypothesis 1) gives rise to some θ ∈ (0, 1 − t 0 ) such that Since t 0 + h k ∈ (t 0 , t 0 + θ) eventually, we get that If g is constant on some interval (t 0 , t 0 + δ], then let us remark that, by hypothesis 1), In addition, To this end, fix t ∈ (t 0 , t 0 + δ], ρ > 0 and choose θ ∈ (0, 1 − t 0 ) such that Then choose s ∈ (t 0 , min{t 0 + θ, t}) to obtain that Since ρ > 0 is arbitrary, we get the desired property.We can define v as v(t 0 ) = x 0 and on (t 0 , t 0 + δ] as a constant function: The condition v(t 0 +δ) ∈ K (t 0 +δ) follows again from the fact that v(t To finish, while on (t 0 , t 0 + δ) it does not matter since this is a g-null set.II.2.If at the right of t 0 the function g is not constant, then for each k, let us choose is, by hypothesis, nonempty) and let k ε be such that We shall prove that v satisfies (2) for δ = h k ε .
The function v is g-absolutely continuous (see Lemma 3), so the first condition in (2) is checked.
To prove the second one, fix t ∈ (t 0 , Besides, and choose s ∈ (t 0 , min{t 0 + θ, t}) to obtain that For the third condition in (2), 123 It remains to check the last one.
At the point t 0 , by the definition of the g-derivative, Since y ∈ F(t 0 , x 0 ) = F(t 0 , v(t 0 )), the assertion is proved.
Proof We follow the ideas of proof used in [13, Theorem 4.3] (see also, in a particular case, [8, Lemma 5.1]).
By assumption 1), there exists 0 < r ε < ε satisfying, for any set In particular, for every t 0 ∈ (0, 1] and t ∈ [0, t 0 ] with g(t 0 ) − g(t) < r ε : Since μ g (E) = 0 and since μ g is regular (see for example [12, Th.1.18,p.36]), one can find an open set E ε such that ( 123 Let us prove that there exists v : [0, 1] → R d a g− absolutely continuous function satisfying the following conditions: (unless I is a singleton), where I is finite or countable such that for each i ∈ I , To this aim, we consider the set M of all triplets with the following properties: (unless I is a singleton), where I is finite or countable such that for each i ∈ I , Claim 1: The set M defined above is nonempty.

Proof of Claim 1:
If 0 / ∈ E, this follows from Lemma 7 applied for t 0 = 0 and x 0 = x 0 ∈ K (0).Indeed, in this case, if v, δ have the properties obtained by Lemma 7, it is clear that If 0 ∈ E, we have two cases: a) there is an interval [0, β] ⊂ [0, 1] (β ∈ (0, 1]) such that g is constant on this interval.b) g is not constant on any interval at the right of 0 .
It turns out that In case b), since E ε is open, the continuity of g at 0 gives rise to some e ε ∈ (0, Choose x ε ∈ K (e ε ) such that Let us check that Obviously, v is g− AC and v(0 Hence, the first and the third assertions of (7) hold.
. Thus, the second assertion of (7) also holds. Finally, which verifies the fourth assertion of ( 7) and finishes the proof of Claim 1.
Next, we consider on M a partial ordering ( ) in the following way: Claim 2 : Every totally ordered subset of ( M, ) has an upper bound.

Proof of the Claim 2:
be a totally ordered subset of ( M, ), where is a nonempty index set.Set δ = sup ϕ∈ δ ϕ .Clearly, Consider the family Since C is a chain, the third assertion of the definition of ( ) (( 8)) yields that any two distinct intervals of J belong to a partition of some interval [0, δ ϕ 0 ), ϕ 0 ∈ and thus, they are disjoint.It turns out that J is at most countable.(To see this, for each J ∈ J choose exactly one rational number in J .)Now the family J may be written in the form where is a nonempty at most countable index set.

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For each t ∈ [0, δ), (Here X A denotes the characteristic function of a nonempty set A.) and so Lebesgue Dominated Convergence Theorem assures that lim Due to Theorem 1, ṽ is g− AC.
Let's check the third assertion.
Finally, the above arguments also imply that the triplet ṽ, δ, The proofs of Subclaim 2.3 and also of Claim 2 are now complete.
The above Claim combined with Zorn's lemma ensure that the set M admits a maximal element where I is at most countable.

Proof of Claim 3:
Arguing by contradiction, assume on the contrary that δ < 1.
• if δ is a discontinuity point of g, we have that μ g ({δ}) > 0, so δ / ∈ E. Then we apply Lemma 7 for x 0 = v(δ), t 0 = δ in order to get that there exist δ > 0 and v : [δ, δ + δ] → R d a g-absolutely continuous function with The concatenated function is by Lemma 2 g-absolutely continuous and

Since δ /
∈ E (see also the fourth assertion of ( 11)), it is easily checked that the triplet lies in M and it is strictly ( )− bigger than the maximal element Thus, we arrive at a contradiction.
• if δ is a continuity point of g, we can proceed exactly as we did at the origin for x 0 = δ (see the beginning of the proof of Claim 1 above) to obtain δ > 0 and v : [δ, δ + δ] → R d a g-absolutely continuous satisfying the first three assertions in (11), accompanied by the following: - -if δ / ∈ E, we have Concatening v, v as before, we arrive again to a contradiction and we finish the proof of Claim 3. Therefore, one can infer that δ = 1 so, there exists at least one function v : [0, 1] → R d satisfying (6).
Step II.Let (ε n ) n be a sequence of positive numbers convergent to 0. Hypothesis 1) (see also Remark 2) gives rise to a sequence r ε n > 0, n ∈ N satisfying, for any set {(t γ , t γ )} γ ∈ of non-overlapping subintervals of [0, 1] with at most countable, 123 For each n ∈ N, choose an open set E n such that: We may suppose that (E n ) n is a decreasing sequence.Let (v n ) n be the corresponding functions obtained in Step I satisfying the set of conditions (6) for ε n instead of ε, i.e. for each n ∈ N, , where I n finite or countable and : • otherwise, Claim 4 : The sequence )-bounded and μ g − uniformly integrable.To this end, for each n ∈ N set Clearly, for each n ∈ N, we have μ g (T n ) = 0 and also ( see (13) ).

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Then for every n ≥ n η , by virtue of the fourth assertion of ( 14), we have (12), ( 15) and hypothesis 3)) Applying the above inequality for A = [0, 1), we obtain that Moreover, applying the above inequality for any g− measurable set A ⊆ [0, 1) with , we deduce that for every n ≥ n η , which implies that ((v n ) g ) n is μ g − uniformly integrable and finishes the proof of Claim 4. Now Claim 4 assures that the sequence ((v n ) g ) n is weakly L 1 g ([0, 1])-relatively compact.On a subsequence (not relabelled) it weakly converges in L 1 g ([0, 1]) to some function w ∈ L 1 g ([0, 1]).By the fundamental theorem of calculus, (v n ) n pointwisely converges to the g-absolutely continuous function v : We shall prove that v(t) ∈ K (t), for every t ∈ [0, 1].For this purpose, let t ∈ [0, 1] be fixed and let δ > 0 be arbitrary.Using hypothesis 1), there is 0 < η < δ 2 + M (depending on t and δ) such that and the assertion v(t) ∈ K (t) is proved, since δ > 0 is arbitrary.
Claim 5: There exists a positive constant C such that for each N ∈ N, To check this, first let (Note that for i ∈ n , we have that (combine the fourth assertion of ( 14) satisfied by Then S is also g− null and the above arguments imply that (v n ) g (t) ≤ C, for all t ∈ ([0, 1) \ E N ) \ S and for all n ≥ N which finishes the proof of Claim 5.
It remains to prove that v g (t) ∈ F(t, v(t)), μ g − a.e. in [0, 1).To this end, fix N ∈ N. By virtue of Claim 5, on [0, 1) \ E N , the sequence (v n ) n≥N satisfies the hypotheses of Theorem 2, so a subsequence (again relabelled by In order to prove that v g (t) ∈ F(t, v(t)) μ g − a.e. on [0, 1) \ E N , let us fix t for which the preceding relation holds (i.e.μ g -a.e.).Take an arbitrary δ > 0. By assumption 2), there exist 0 < η < δ 2+M such that and n η ∈ N satisfying 123 One can see that whenever p ≥ n η , and as δ > 0 was arbitrary, it follows (the values of F being compact and convex) that Finally, as this holds for every N ∈ N and is over.

Remark 4
In the classical case where g(t) = t we fall onto known viability results (for instance [13], [4]) since the notions related to the g-topology are to be expressed in terms of the usual topology of the real line, while the conditions imposed at the discontinuity points of g are superfluous.
We end this section with the following observation.
Remark 5 Unlike in the classical case, we cannot replace the boundedness assumption 3) on the multifunction F by the condition | F(t, x) |≤ M(t) where M is a LS-integrable map by using the upper semicontinuity of F on a compact since even if we truncate it to a domain [0, 1] × R where R ⊂ R d is bounded, this set is not compact in the product of the g-topology of [0, 1] with the usual topology of R d .What is more, in the classical case ( [13], [4], [8]) F is a Carathéodory set-valued mapping; the reduction to the case of a jointly upper semicontinuous multifunction is made through a Scorza-Dragoni result ( [29]); using this idea in the setting of Stieltjes differential problems requires a special attention and will be followed in a further work.
Proof Let us first note that, by [18,Theorem 3.5], r (t) ≥ 0 for every t ∈ [0, 1].Let E be the g-null set such that on [0, 1) \ E the maps y and r are g-differentiable.
The idea is to check that the multifunction K : [0, 1] → P cc (R d ), satisfies the assumptions of the viability Theorem 4.
As y and r are g-absolutely continuous, K is upper semicontinuous from the left for the g-topology and by Corollary 1 it has limit at the right for the usual topology in the sense of Hausdorff distance at any discontinuity point of g, so 1) is valid.The hypothesis 2) is obviously checked.As for 3), we note that r and y are g-absolutely continuous, therefore bounded; denote by m = max(sup{ y(t) ; t ∈ [0, 1]}, sup{r (t); t ∈ [0, 1]}).Then for any (t, x) ∈ G(K ): Now we see that only hypothesis 4) remains to be verified.We have to prove first that for μ g -almost every t ∈ [0, 1] and every (t, x) ∈ G(K ), Case I. if u < 1, we can find ourselves in one of the two following situations: I.1.t is a continuity point of g; then for any z ∈ R d , one can find a sequence due to the continuity of g at t.It follows that by choosing 123 we can easily check that In other words, D g K (t, x) = R d and obviously in this case since once z ∈ D g K (t, x), there are h k ↓ 0 and for some u k ∈ B(0, 1) and so, indeed, • (1 + u ).
One can find w ∈ F(t, y(t)) with and also one can find v ∈ F(t, x) such that We may see that From ( 18) and ( 19) we infer that (17)) so this intersection is nonempty.
Case II.if u = 1 then by definition Indeed, as before, for any z ∈ D g K (t, x), for some u k ∈ B(0, 1) and so, and also v ∈ F(t, x) such that (20)) so again the intersection is nonempty.
We in fact get, by Theorem 3, the following Filippov type result for Stieltjes differential inclusions.

Corollary 2 If F satisfies the hypotheses of the preceding Theorem 5, then given a g-absolutely continuous function y
for every t ∈ [0, 1], where γ (t) = d(y g (t), F(t, y(t))) and L is defined as in Theorem 3.

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Another Filippov type result for measure differential inclusions was proved in [17].Namely, [17,Theorem 2.2] states that if F is measurable w.r.t.t and L(t)-Lipschitz w.r.t.x and L, γ are LS-integrable w.r.t.g, then given a g-absolutely continuous function y : [0, 1] → R d with d(y g (t), F(t, y(t))) ≤ γ (t) for all t, there exists a solution x of (1) such that for every t ∈ [0, 1]: Denote by q(t) the term on the right-hand side of the estimation given in [17, Theorem 2.2] , i.e.
We shall prove that the estimation of the distance x(t) − y(t) obtained in our Theorem 5 under stronger assumptions, is better than the above one.
For sake of generality, in the following Proposition we consider the case where L = L(t) and L(•) is non negative LS-integrable w.r.t.g.

Proposition 6 For every
L is defined as in Theorem 3 and q is given by (21).
Let us remind the reader that r g (t) = L(t)r (t) + γ (t), μ g − a.e. in [0, 1), in other words Choose a g− null set S ⊂ [0, 1) such that for each t ∈ [0, 1) \ S, the LS -primitive of L is g− differentiable at t and also (22) holds.
Summarizing the above arguments, we deduce that μ g − a.e.r g (t) ≤ q g (t).