General Existence Results for Third-Order Nonconvex State-Dependent Sweeping Process with Unbounded Perturbations

We prove the existence of solutions for third-order nonconvex state-dependent sweeping process with unbounded perturbations of the form: −A x 3 t ∈ N K t, ẋ t ; A ẍ t F t, x t , ẋ t , ẍ t G x t , ẋ t , ẍ t a.e. 0, T , A ẍ t ∈ K t, ẋ t , a.e. t ∈ 0, T , x 0 x0, ẋ 0 u0, ẍ 0 υ0, where T > 0, K is a nonconvex Lipschitz set-valued mapping, F is an unbounded scalarly upper semicontinuous convex set-valued mapping, and G is an unbounded uniformly continuous nonconvex set-valued mapping in a separable Hilbert space H.


Introduction
In the seventies, Moreau introduced and studied in 1 the following differential inclusion: a.e. on I, x 0 x 0 ∈ K 0 , SP where I 0, T T > 0 , K : I → H is a set-valued mapping defined from I to a Hilbert space H and takes closed convex values, N K t x t denotes the normal cone to the set K t at the position x t .The differential inclusion SP is called the Moreau's sweeping process problem.In 2-7 , the authors studied the existence of solutions for various extensions and variants of SP .In 6 , the author studied for the first time the existence of solutions for the following type of second-order differential inclusions ẍ t ∈ −N K x t ; ẋ t , x 0 x 0 , ẋ t ∈ K x t , SSP with convex set-valued mapping K.The problem SSP has been extended in several ways.For instance, in 8 , the authors studied a variant of SSP with a perturbation, −Ax t ∈ N K t ; ẋ t ẍ t F t, ẋ t a.e. on 0, T , 1.1 when the set-valued mapping K is not necessarily convex and A is a linear and bounded operator on a separable Hilbert space.In 9 , the author studied existence results for the following general problem: x 0 x 0 , ẋ 0 u 0 , ẍ 0 υ 0 , TSPMP where K is a nonconvex Lipschitz set-valued mapping, A is surjective bounded linear operator, F is an unbounded scalarly upper semicontinuous convex set-valued mapping, and G is an unbounded uniformly continuous nonconvex set-valued mapping in a separable Hilbert space H.We will call it third-order nonconvex state-dependent sweeping process with mixed perturbations in short TSPMP .Problem TSPMP includes as a special case the following differential variational inequality DVI : given a convex compact set D in H, three points Find T > 0 and a Lipschitz mapping x : 0, T −→ H such that where α > 0, Λ : H → R is inf-compact and lower-C 2 function, a •, • is a real bilinear, symmetric, bounded, and elliptic form on H × H, and f : I × H → H is a Lipschitz function.We use our main theorem to prove that DVI has at least one Lipschitz solution.
This paper is organized as follows.Section 2 contains some definitions, notations, and important results needed in the paper.In Section 3, we prove an existence result for TSPMP when the set-valued mapping K is not necessarily convex, by using ideas and techniques from Nonsmooth Analysis.The result is proved by showing that a sequence of approximate solutions converges to a solution of TSPMP .Then, we deduce from our main theorem an existence result for a second-order nonconvex differential inclusion ẍ t ∈ K t, ẋ t for almost all t ∈ I, where the right-hand side is not convex.In Section 4, we study the closedness and the compactness of the solution sets of TSPMP .In Section 5, we state an application to differential variational inequalities DVI .

Preliminaries
Throughout the paper, H will denote a Hilbert space.We need to recall some notation and definitions that will be used in all the paper.Let S be a nonempty closed subset of H.We denote by d S • the usual distance function to the subset S, that is, d S x : inf z∈S x − z .We recall see 10 that the proximal normal cone of S at x is given by Equivalently N P S; x can be defined by 2 as the set of all ξ ∈ H for which there exist σ > 0 such that Now, let f : H → R ∪ { ∞} be a function and x any point in H where f is finite.We recall that the proximal subdifferential ∂ P f x is the set of all ξ ∈ H for which there exist δ, σ > 0 such that for all x ∈ x δB Here B denotes the closed unit ball centered at the origin of H. Recall that for a given r ∈ 0, ∞ , a subset S is uniformly prox-regular with respect to r we will say uniformly r-proxregular see 10, 11 if and only if for all x ∈ S and all 0 / ξ ∈ N P S; x one has for all x ∈ S. We make the convention 1/r 0 for r ∞.Recall that for r ∞, the uniform r-prox-regularity of S is equivalent to the convexity of S, which makes this class of great importance.
In 12 , the authors established the following characterization of the uniform proxregularity in terms of the subdifferential of the distance function.We recall here a consequence of their result needed in the sequel.

2.6
The following proposition summarizes some important consequences of the uniform prox-regularity needed in the sequel of the paper see 11, 13 .Proposition 2.2.Let S be a nonempty closed subset of H and x ∈ S. The following assertions hold.Now, we give the following proposition.It proves the result of closedness of the proximal subdifferential of the distance function of prox-regular set.In 12 , the authors proved the result when the set-valued mapping depends only on t ∈ I.We adapt the proof to the case of set-valued mapping depending on two variables t and x.For the completeness of the paper, we give the proof.
Proposition 2.3.Let r > 0; K : I × Ω → H be a Hausdorff-continuous set-valued mapping with uniformly r-prox regular values.For a given 0 < δ < r, the following holds: "for any t ∈ I and z ∈ Ω, and hence the Hausdorff-continuity of K yields for n large enough

2.11
For n large enough and by Proposition 2.1, we have

2.13
Consequently, by the continuity of the distance function with respect to t, z, x because of 2.7 , the inequality 2.12 gives, by letting n → ∞,

2.14
This ensures that ξ ∈ ∂ P d K t,z x , and so the proof is complete.

Existence Results for Third-Order Nonconvex State-Dependent Sweeping Process with Perturbation.
Throughout this section, H will denote a separable Hilbert space.Let x 0 , u 0 ∈ H, A υ 0 ∈ K 0, u 0 , ς, > 0, T > 0, U 0 , V 0 be open neighborhoods of u 0 , x 0 resp. in H such that x 0 ςB ⊂ U 0 , u 0 ςB ⊂ V 0 , and K : 0, ς/ × cl U 0 → H be a Lipschitz set-valued mapping with ratio λ 1 , λ 2 taking nonempty closed uniformly r-prox regular values in H. Assume that A is a surjective bounded linear operator.Our aim is to prove the local existence of solution of TSPMP , that is, there exist T > 0 and Lipschitz mappings x : 0, T → cl V 0 , u : 0, T → cl U 0 , and υ : 0, T → H such that ii for any n ≥ 1 and any x, x ∈ X, one has We prove our main theorem in this section.
Proof.We give the proof in four steps.
Step 1. Construction of the approximants.
Let T ∈ 0, ς/ and put I 0, T and κ I × βB × αB × bB.Then by the linear growth condition of F and G we have Let n 1/2 n , n 1, 2, 3, . . . .Then by the uniform continuity of G and Lemma 3.1, there is a strictly decreasing sequence of positive numbers e n converging to 0 such that e n ≤ 1, 1/e n−1 and e n−1 /e n are integers ≥ 2, and the following implication holds: for every t, x, u, υ , t , x , u , υ ∈ κ where η For all n ≥ n 0 , we consider the following partition of I. Without loss of generality we assume that T is integer: where υ n,0 υ 0 and for all i 0, . . ., μ n − 1.Since A is surjective, we can choose where This algorithm is well defined.Indeed, as therefore, by 1 and 3.11 , we get b υ n,i ≤ A υ n,i ≤ b, 3.12 so that υ n,i ≤ .

3.13
Then by the Lipschitz property of K and the relations 3.4 , 3.6 , 3.8 , 3.10 , and 3.13 , we get

3.14
Therefore, as K has uniformly r-prox-regular values, by Proposition 2.2 one can choose a point ω n,i 1 Proj A υ n,i e n f n,i g n,i , K t n,i 1 , u n t n,i 1 .

3.15
Define θ n : I → I by θ n 0 0, and θ n t t n,i , ∀t ∈ I n,i .

Then 3.11 becomes
A υ n t ∈ K θ n t , u n θ n t .

3.17
So that, all the mappings u n are Lipschitz with ratio , and they are also equibounded, with u n t ≤ u 0 T. Observe that for all n ≥ n 0 and all t ∈ I, one has

3.19
We note that x n θ n t x n t n,i x 0 t n,i 0 u n s ds Now we define the affine approximants

3.21
Observe that A z n θ n t A υ n,i ∈ K θ n t , u n θ n t ⊂ bB.

3.25
So for any t, s ∈ I n,i

3.26
By addition on all the interval I, we obtain the Lipschitz property of z n on all I. Clearly, by the definition of z n • and 3.25 , we have 3.27 and so − ω n,i e n f n,i g n,i .

3.31
Then, by properties of proximal normal cone, we have for a.e.t ∈ I, −N K ρ n t , u n ρ n t ; A υ n ρ n t .

3.32
On the other hand, by 3.25 and 3.30 , we have Therefore, the relations 3.32 and 3.33 and Proposition 2.2 entail for a.e.t ∈ I A żn t − f n t g n t ∈ −δ∂d K ρ n t ,u n ρ n t A υ n ρ n t .

3.34
Step 2. We will prove the uniform convergence of u n and z n .Since e −1 n t − t n,i ≤ 1 for all t ∈ I n,i and υ n,i 1 , υ n,i ∈ κ 1 and A −1 κ 1 is a convex set in H, one gets z n t ∈ κ 1 for all t ∈ I so that, for every t ∈ I, the set {z n t : n ≥ n 0 } is relatively compact.By using Arzela-Ascoli theorem, there exists a Lipschitz mapping υ : I → H with

3.35
By 3.28 , we have and so

3.37
This ensures that

3.38
Thus, u n , x n , and v n uniformly converge to u, x, and v, respectively.
Step 3 relative strong compactness of g n .The points g n,i defining the step function g n were chosen arbitrarily in our construction.Nevertheless, by using the uniform continuity of the set-valued mapping G over κ and the techniques of 14 see also 15, 16 , the sequence g n can be constructed relatively strongly compact for the uniform convergence in the space of bounded functions.Therefore, there exists a bounded mapping g : I → H such that g n − g → 0.
Step 4 existence of solutions .Since |ρ n t − t| ≤ e n on 0, T , then ρ n t → t, as well as θ n t → t.Also x n •θ n , u n •θ n , and υ n •θ n converge uniformly to x, u, and υ, respectively.
Then by continuity of G on κ, the closedness of the set G t, x t , u t , A υ t , and by 3.23 , we obtain g t ∈ G t, x t , u t , A υ t a.e. on I.
Since A υ n θ n t ∈ K θ n t , u n θ n t and so by the closedness and the continuity of K and the continuity of A, we have A υ t ∈ K t, u t a.e. on I.By 3.23 , the sequence

3.40
Using the same technique, we get

3.41
Indeed, for every measurable set Ω in I and every ξ ∈ H, we have

3.42
Thus, as the set-valued mapping t → −δ∂d K t,u t A v t is measurable with convex weakly compact values see 17 , it follows that and since A υ t ∈ K t, u t a.e. on I, we get for a.e.t ∈ I,

3.44
Thus the proof of the theorem is complete.
We deduce from our main theorem an existence result for a second-order nonconvex differential inclusion.
) in H such that x 0 ςB ⊂ U 0 , u 0 ςB ⊂ V 0 , and K : 0, ς/ × cl U 0 → H is a Lipschitz set-valued mapping with ratio (λ 1 , λ 2 taking nonempty closed uniformly r-prox regular values in H. Assume that K t, u ⊂ κ ⊂ B, for all t, x ∈ 0, ς/ ×cl U 0 , for some convex compact set κ ⊂ H. Then for any t ∈ 0, ς/ , there exists a Lipschitz solution of the second-order differential inclusion ẍ t ∈ K t, ẋ t a.e. on I.

3.45
Proof.Take F G 0, A Id in Theorem 3.2.Then there is a Lipschitz solution x : 0, T → H to the cauchy problem for the third-order differential inclusion −x 3 t ∈ N K t, ẋ t ẍ t a.e.0, T ; x 0 x 0 , ẋ 0 u 0 , ẍ t ∈ K t, ẋ t a.e. on 0, T , 3.46 which proves that ẍ t ∈ K t, ẋ t has a solution almost everywhere on 0, T .
Remark 3.4.The existence result proved in Theorem 3.2 cannot be covered by the recent existence result for third-order differential inclusions established in 18 in finite-dimensional case and extended in 19 in Hilbert spaces.Indeed, The right-hand side in Theorem 3.2 contains the normal cone, which cannot be bounded nor u.s.c. as a set-valued mapping.These two assumptions are essential in the proof of the results in 18, 19 .

Solution Set
Throughout this section, let T > 0 and I 0, T , and let F : subsets in H, and K : I × cl Ω → H a Lipschitz set-valued mapping with ratio λ 1 , λ 2 taking nonempty closed uniformly r-prox regular values in H. Let x 0 ∈ Ω 1 , u 0 ∈ Ω 2 , A υ 0 ∈ K 0, u 0 with u 0 T ∈ Ω 2 .We denote S x 0 , u 0 , υ 0 the set of all triple x, u, υ of Lipschitz mappings x, u, υ : I → H such that 0 , u n 0 , υ n 0 such that x n 0 , u n 0 , υ n 0 n uniformly converges to some x 0 , u 0 , υ 0 ∈ Ω 1 × Ω 2 × Im K and x n , u n ,υ n n uniformly converges to some x, u, υ ∈ C I; H × H × H .We have to show that x, u, υ ∈ S x 0 , u 0 , υ 0 .First, observe that for n sufficiently large, u n 0 ∈ u 0 ςB ⊂ U 0 and x n 0 ∈ x 0 α ς/ B ⊂ V 0 , where α is given as in the proof of Theorem 3.2.Now, it is not difficult to check that the continuity of K, the continuity of A, the uniform convergence of both sequences x n 0 , u n 0 , υ n 0 n and x n , u n , υ n n , and A υ n t ∈ K t, u n t for almost all t ∈ I imply that A υ t ∈ K t, u t for almost all t ∈ I. and so the proof is complete.

Applications to Differential Variational Inequalities
In this section, we are interested with an application of the main result proved in Theorem 3.2 to differential variational inequalities DVI : given a convex compact set D in H, three points x 0 , u 0 ∈ H, v 0 ∈ D f 0, u 0 with Λ v 0 − f 0, u 0 ≤ 0: Find T > 0 and a Lipschitz mapping x : 0, T → H such that 1 x 0 x 0 , ẋ 0 u 0 , ẍ 0 v 0 ; Proof.Let S : {x ∈ D : Λ x ≤ 0} and define K t, y S f t, y , for all t, y ∈ R × H. Since Λ is inf-compact, the set S is compact in H and so K has compact values.Also, the lower C 2 property of the function Λ and the assumption inf{ ξ : ξ ∈ ∂Λ x with x ∈ D} / 0 ensure by Theorem 3.3 in 20 the uniform prox regularity for some r > 0, and so K has uniform r-prox-regular values.The Lipschitz behavior of the set-valued mapping K is inherited from the function f.Now, we use the definition of proximal normal cones for uniform prox-regular sets to rewrite DVI in the form of TSPMP as follows: x 3 t − A x t ẋ t − ẍ t − αx t ∈ N K t, ẋ t ẍ t , a.e. on 0, T , x 0 x 0 , ẋ 0 u 0 , ẍ 0 v 0 , ẍ t ∈ K t, ẋ t a.e. on 0, T .

5.3
Let, now, G t, x, y, z {A x y − z − αx}, for all t, x, y, z ∈ R × H × H × H. Clearly G is uniformly continuous since A is a bounded linear operator with compact values.Also G satisfies the linear growth condition with ρ 2 α A .Indeed, G t, x, y, z ≤ α x A x y − z ≤ α A 1 x y z .Consequently, all the assumptions of Theorem 3.2 are satisfied, and so we have the existence of a Lipschitz solution of DVI .

ratio 1 /b λ 1 λ 2 2 ζ 1 ζ
2 such that, z n converges uniformly to υ on I; żn weakly star converges to υ in L ∞ I, H . Now, we define the Lipschitz mappings u : I → H and x : I → H as follows: , ∀t ∈ I.

Proposition 2.1. Let
S be nonempty closed subset in H, and let r > 0. Assume that S is r-proxregular.Then the following holds: , and let K : I × H → H be a set-valued mapping.We will say that K is Hausdorff-continuous resp., Lipschitz with ratio λ 1 , λ 2 if for any t ∈ I and x ∈ X one has lim B a , sup b∈B d A b .2.9 For every n ≥ n 0 , we define the following approximating mappings on each interval I n,i as i , t n,i 1 for all i 0, . . ., μ n − 1 and I n,μ n {T }.
H is the dual of the separable Banach space L 1 I, H . Therefore, by integrating, we have for every measurable set Ω in I and every x ∈ H.By assumption, the set-valued mapping t → F t, x t , u t , A υ t is measurable with convex weakly compact values, which ensures the last inequality F t, x t , u t , A υ t , a.e. on I.
Assume that the hypothesis on F and K in Theorem 3.2 are satisfied and let G 0. Then the graph of the set-valued mapping S is closed in On the other hand, we have Now takingδ A /b λ 1 λ 2 2 ρ 1 ρ 2 ζ 2 ρ 1 ρ 2 ζand using Proposition 2.2 yieldA υn t − f n t ∈ −δ∂d K t,u n t A υ n t .
A υ t ∈ −N K t,u t A υ t F t,x t , u t , A υ t , a.e. on I. Therefore, we get the σ L ∞ I; H , L 1 I; H -convergence of subsequences of both υn and f n to υ and f, respectively, in L ∞ I; H . Using the same techniques in the proof of Theorem 3.2, we can prove that f t ∈ F t, x t , u t , A υ t a.e.t ∈ I.
− ẍ t 2 , a.e. on 0, T ,5.1where α > 0, Λ : H → R is inf-compact and lower-C 2 function, a •, • is a real bilinear, symmetric, bounded, and elliptic form on H × H, and f : I × H → H is a Lipschitz function.Let A be a linear and bounded operator on H associated with a •, • , that is, a u, v Au, v , for all u, v ∈ H.We use Theorem 3.2 to prove that DVI has at least one Lipschitz solution.Assume that H is a separable Hilbert space and inf{ ξ : ξ ∈ ∂Λ x with x ∈ D} / 0.5.2Then(DVI) has at least one Lipschitz solution.