Mixed Semicontinuous Perturbation of a Second Order Nonconvex Sweeping Process

We prove a theorem on the existence of solutions of a second order differential inclusion governed by a class of nonconvex sweeping process with a mixed semicontinuous perturbation .


Introduction
The existence of solutions for the second order differential inclusions governed by the sweeping process where N K(u(t)) (•) denotes the normal cone to K(u(t)), has been thoroughly studied (when the sets K(x) are convex or nonconvex) by Castaing for the first time when F ≡ {0} see [5], and later by many other authors see for example [2], [3], [6], [10] and [12].Note that in this literature some existence results are established for the problem (P F ) with lower and upper semicontinuous perturbations.EJQTDE, 2008 No. 37, p. 1 Our aim in this paper is to prove existence results for (P F ) when F is a mixed semicontinuous set-valued map and K(x) are nonconvex sets.For the first order sweeping process with a mixed semicontinuous perturbation we refer the reader to [9], and to [13] for the sweeping process with non regular sets and to [1] for second order differential inclusions with mixed semicontinuous perturbations.
After some preliminaries, we present our main result in the finite dimensional space H whenever the sets K(x) are uniformly ρ-prox-regular (ρ > 0) and the set-valued mapping F is mixed semicontinuous, that is, F (•, •, •) is measurable and for every t ∈ [0, T ], at each (x, y) ∈ H × H where F (t, x, y) is convex the set-valued map F (t, •, •) is upper semicontinuous, and whenever F (t, x, y) is not convex F (t, •, •) is lower semicontinuous on some neighborhood of (x, y).

Definition and preliminary results
Let H be a real Hilbert space and let S be a nonempty closed subset of H.We denote by d(•, S) the usual distance function associated with S, i.e., d(u, S) := inf y∈S u − y .For any x ∈ H and r ≥ 0 the closed ball centered at x with radius r will be denoted by B H (x, r).For x = 0 and r = 1 we will put B H in place of B H (0, 1).L([0, T ]) is the σ-algebra of Lebesgue-measurable sets of [0, T ] and B(H) is the σ-algebra of Borel subsets of H.By L 1 H ([0, T ]) we denote the space of all Lebesgue-Bochner integrable H-valued mappings defined on [0, T ] and by C H ([0, T ]) the Banach space of all continuous mappings u : [0, 1] → H, endowed with the sup norm We need first to recall some notation and definitions that will be used in all the paper.Let x be a point in S. We recall ( see [8]) that the proximal normal cone to S at x is defined by N P S (x) := ∂ P ψ S (x), where ψ S denotes the indicator function of S, i.e., ψ S (x) = 0 if x ∈ S and +∞ otherwise.Note that the proximal normal cone is also given by Recall now that for a given ρ ∈]0, +∞] the subset S is uniformly ρ-prox-regular (see [11]) or equivalently ρ -proximally smooth ( see [8]) if and only if every nonzero proximal normal to S can be realized by ρ-ball, this means that for all x ∈ S and all 0 = ξ ∈ N for all x ∈ S. We make the convention 1 ρ = 0 for ρ = +∞.Recall that for ρ = +∞ the uniform ρ-prox-regularity of S is equivalent to the convexity of S. The following proposition summarizes some important consequences of the uniform prox-regularity needed in the sequel.For the proof of these results we refer the reader to [11].(2.2) the proximal subdifferential of d(•, S) coincides with its Clarke subdifferential at all points x ∈ H satisfying d(x, S) < ρ.So, in such a case, the subdiferential ∂d(x, S) := As a consequence of (2.3) we get that for uniformly ρ-prox-regular sets, the proximal normal cone to S coincides with all the normal cones contained in the Clarke normal cone at all points x ∈ S, i.e., N P S (x) = N C S (x).In such a case, we put N S (x) := N P S (x) = N C S (x).Here and above ∂ C d(x, S) and N C S (x) denote respectively the Clarke subdifferential of d(•, S) and the Clarke normal cone to S (see [8]).Now, we recall some preliminaries concerning set-valued mappings.Let T > 0. Let C : [0, T ] ⇒ H and K : H ⇒ H be two set-valued mappings.We say that C is absolutely continuous provided that there exists an absolutely continuous nonnegative function a : for all x, y ∈ H and all s, t ∈ [0, T ].We will say that K is Hausdorff-continuous (resp.Lipschitz with ratio λ > 0) if for any x ∈ H one has lim We close this section with the following theorem in [4], which is an important closedness property of the subdifferential of the distance function associated with a set-valued mapping.
Theorem 2.1 Let ρ ∈]0, +∞], Ω be an open subset in H, and K : Ω ⇒ H be a Hausdorffcontinuous set-valued mapping.Assume that K(z) is uniformly ρ-prox-regular for all z ∈ Ω.Then for a given 0 < δ < ρ, the following holds: Here → w means the weak convergence in H. Remark 2.1 As a direct consequence of this theorem, we have for every ρ ∈]0, +∞], for a given 0 < δ < ρ, and for every set-valued mapping K : Ω ⇒ H with uniformly ρprox regular values, the set-valued mapping } to H endowed with the weak topology, which is equivalent to the upper semicontinuity of the function Here σ(S, p) denotes the support function to S defined by σ(S, p) = sup s∈S s, p .
3 Existence results under mixed semicontinuous perturbation.
Our existence result is stated in a finite dimensional space H under the following assumptions.
(H 1 ) For each x ∈ H, K(x) is a nonempty closed subset in H and uniformly ρ-prox-regular for some fixed ρ ∈]0, +∞]; The proof of our main theorem uses existence results for the first order sweeping process, the selection theorem proved in Tolstonogov [14] and the Kakutani fixed point theorem for set-valued mappings.We begin by recalling them.Proposition 3.1 (Proposition 1.1 in [7]) Let H be a finite dimensional space, T > 0 and let C : I := [0, T ] ⇒ H be a nonempty closed valued set-valued mapping satisfying the following assumptions.(A 1 ) For each t ∈ I, C(t) is ρ-prox-regular for some fixed ρ ∈]0, +∞]; (A 2 ) C(t) varies in an absolutely continuous way, that is, there exists a nonnegative absolutely continuous function v : for all x, y ∈ H and s, t ∈ I.

Then for any mapping
admits one and only one absolutely continuous solution u(•) and Further, let m be a nonnegative Lebesgue-integrable function defined on [0, T ] and let Then the solutions set {u h : h ∈ K}, where u h is the unique absolutely continuous solution of the above inclusion, is compact in C H ([0, T ]), and the mapping h → u h is continuous on K when K is endowed with the weak topology For the proof of our theorem we will also need the following theorem which is a direct consequence of Theorem 2.1 in [14].(ii) for every t ∈ [0, 1], at each (x, y) ∈ H × H such that M (t, x, y) is convex, M (t, •, •) is upper semicontinuous, and whenever M (t, x, y) is not convex, M (t, •, •) is lower semicontinuous on some neighborhood of (x, y); (iii) there exists a Caratheodory function ζ : [0, 1] × H × H → R + which is integrably bounded and such that M (t, x, y) B H (0, ζ(t, x, y)) = ∅ for all (t, x, y) ∈ [0, 1] × H × H. Then for any ε > 0 and any compact set K ⊂ C H ([0, T ]) there is a nonempty closed convex valued multifunction Φ : K ⇒ L 1 H ([0, T ]) which has a strongly-weakly sequentially closed graph such that for any u ∈ K and φ ∈ Φ(u) one has Now we are able to prove our main result.Theorem 3.2 Let H be a finite dimensional space, K : H ⇒ H be a set-valued mapping satisfying assumptions (H 1 ), (H 2 ) and (H 3 ).Let T > 0 and let F : [0, T ] × H × H ⇒ H be a set-valued mapping satisfying hypotheses (i) and (ii) of Theorem 3.1 and the following one (iv) there exist nonnegative Lebesgue-integrable functions m, p and q defined on [0, T ] such that Then for all u 0 ∈ H and v 0 ∈ K(u 0 ), there exist two Lipschitz mappings u, v : [0, T ] → H such that In other words, there is a Lipschitz solution u : [0, T ] → H to the Cauchy problem (P F ).
Proof.Step 1.Put I := [0, T ], M (t) = m(t) + p(t)( u 0 + lT ) + q(t)l, and let us consider the sets EJQTDE, 2008 No. 37, p. 5 , and by Ascoli-Arzelà theorem X and U are convex compact sets in C H (I). Observe now, that for all f ∈ X the set valued mapping K • f is Lipschitz with ratio λl.Indeed, for all t, t ∈ I By Proposition 3.1, for all (f, h) ∈ X × K, there exists a unique solution u f,h to the problem and for almost all t ∈ I, uf,h (t) ≤ λl + 2M (t), i.e., u f,h ∈ U.

Let us consider the mapping
, where u f,h is the unique solution of (P ).We wish to show that ) to h ∈ K, and since (u fn,hn ) n ⊂ U we may suppose that it converges uniformly to some mapping v ∈ U.For each n ∈ N we have − ufn,hn (t) ∈ N K(fn(t)) (u fn,hn (t)) + h n (t), a.e. on I; u fn,hn (t) ∈ K(f n (t)), ∀t ∈ I; u fn,hn (0) = v 0 .
Since u fn,hn (t) ∈ K(f n (t)) for all t ∈ I, it follows from the Lipschitz property of K ).This can be rephrased as In other words, v is of the form u f,h with We conclude that A is continuous.Hence, the mapping P : X × K → C H (I) defined by P (f, h), where for all t ∈ I is also continuous when X is endowed with the topology of uniform convergence and K is endowed with the weak topology.Observe that for all t ∈ I, u f,h (t) ∈ K(f (t)) and then by (H 3 ), we have u f,h (t) ≤ l, we conclude that P (f, h) ∈ X .
Step 2. By Theorem 3.1, there is a nonempty closed convex valued set-valued mapping Φ : X ⇒ L 1 H (I) such that for any u ∈ X and φ ∈ Φ(u) for almost all t ∈ I. Since u ∈ X , we have u(t) ≤ l and The relation (3.1) shows that Φ has w(L 1 H (I), L ∞ H (I))-compact values in L 1 H (I). Now, let us consider the set-valued mapping Ψ : X ⇒ X defined by It is clear that Ψ has nonempty convex values since Φ has nonempty convex values.Furthermore, for all f ∈ X , Ψ(f ) is compact in X .Indeed, Let (v n ) n be a sequence in Ψ(f ), then, for each n, there is h n ∈ Φ(f ) such that v n = P (f, h n ).Since (h n ) n ⊂ Φ(f ), by extracting a subsequence (that we do not relabel) we may suppose that (h n ) n w(L 1 H (I), L ∞ H (I))-converges to some mapping h ∈ Φ(f ), and by the continuity of P we get This shows the compactness of Ψ(f ).We will prove that Ψ is upper semicontinuous, or equivalently the graph of Ψ gph(Ψ) = {(x, y) ∈ X × X : y ∈ Ψ(x)} is closed.Let (x n , y n ) n be a sequence in gph(Ψ) converging to (x, y) ∈ X × X .For all n ∈ N, y n ∈ Ψ(x n ), so, there is h n ∈ Φ(x n ) such that y n = P (x n , h n ).since (h n ) n ⊂ K, by extracting a subsequence (that we do not relabel) we may suppose that (h n ) n w(L 1 H (I), L ∞ H (I))-converges to some mapping h ∈ K.As the sequence (x n ) n converges uniformly to x ∈ X and since gph(Φ) is strongly-weakly sequentially closed we conclude that h ∈ Φ(x).On the other hand, by the continuity of the mapping P we get y = lim n→+∞ y n = lim n→+∞ P (x n , h n ) = P (x, h).