Keywords: differential inclusions, subdifferential operator, Lipschitz func- tions, set-valued map, delay, perturbation, absolutely continuous map

doi:10.7151/dmdico.1159

DELAY PERTURBED EVOLUTION PROBLEMS INVOLVING TIME DEPENDENT SUBDIFFERENTIAL

OPERATORS

Soumia Sa¨ıdi and Mustapha Fateh Yarou Laboratoire LMPA, Department of Mathematics

Jijel University, Algeria e-mail: soumiass@hotmail.fr

mfyarou@yahoo.com

Abstract

We investigate in the present paper, the existence and uniqueness of solutions for functional differential inclusions involving a subdifferential op- erator in the infinite dimensional setting. The perturbation which contains the delay is single-valued, separately measurable, and separately Lipschitz.

We prove, without any compactness condition, that the problem has one and only one solution.

Keywords: differential inclusions, subdifferential operator, Lipschitz func- tions, set-valued map, delay, perturbation, absolutely continuous map.

2010 Mathematics Subject Classification:34A60, 34K09, 49A52.

1. Introduction

Our objective in this paper is to study in an infinite dimensional Hilbert space the existence of solutions for a perturbed evolution problem involving time dependent subdifferential operator whose perturbation is single-valued and contains a delay.

Let T > 0, and for each t ∈ [0, T ] the (set-valued) operator ∂ϕ(t, ·) is the subdif- ferential of a time-dependent proper lower semi-continuous (lsc) convex function ϕ(t, ·) of a Hilbert space H into [0, +∞]. Given a finite delay r ≥ 0, one considers the space C₀ := CH([−r, 0]) endowed with the norm of the uniform convergence k · kC0. With each t ∈ [0, T ], one associates a map τ (t) from C_H([−r, t]) into C₀ defined by,

(τ (t)x(·))(s) := x(t + s) for all s ∈ [−r, 0].

Let f : [0, T ] × C₀ → H be a single-valued map and let ψ be a fixed member of C0 such that ψ(0) ∈ dom ϕ(0, ·) (where for all t ∈ [0, T ] dom ϕ(t, ·) denotes the effective domain of the function ϕ(t, ·)), then the problem is the following:

(Pψ)

( − ˙x(t) ∈ ∂ϕ(t, x(t)) + f (t, τ (t)x(·)) a.e. t ∈ [0, T ], x(s) = ψ(s) ∀s ∈ [−r, 0].

The main purpose of the present work is to show the existence of solutions for (Pψ). Results related to (Pψ) can be found in [2, 3, 5, 6, 11, 7, 8, 14]. Recently, existence (and uniqueness) was obtained for the sweeping process with delay ([6, 10]), i.e., for ϕ(t, ·) taken as the indicator function of a nonempty closed moving set C(t) which is ρ-prox-regular.

Here, we establish, in an infinite-dimensional setting, an existence and unique- ness result without any compactness condition for the perturbed problem with a map f measurable with respect to the first argument and Lipschitz with respect to the second one, and

kf (t, φ(·))k ≤ β(t)(1 + kφ(·)kC0)

for all (t, φ(·)) ∈ [0, T ] × C₀. Note that as in [10], this growth condition in- volves kφ(·)kC0 instead of kφ(0)k. This result is obtained thanks to the one proved recently in [14] concerning perturbed problem governed by the subdif- ferential operator without delay, and through some ideas of Edmond [10]. We proceed as follows: We consider, for each n ∈ N, a partition of [0, T ] given by tⁿ_i := ^iT_n (i = 0, · · ·, n). Then, on each subinterval [tⁿ_i, tⁿ_i+1], we replace f by the map f_iⁿ: [tⁿ_i, tⁿ_i+1] × H → H defined by f_iⁿ(t, x) := f (t, τ (t)hⁿ_i(·, x)), where

hⁿ₀(t, x) =

( ψ(t) if t ∈ [−r, 0]

ψ(0) +_Tⁿt(x − ψ(0)) if t ∈ [0, tⁿ₁],

and hⁿ_i(·, ·) (i ≥ 1) are defined in a quasi similar way. Doing so, we obtain a perturbed problem without delay for which our result in [14] insures the existence of a solution xn(·). This approach is slightly different from the classic idea in that, in our definition of f_iⁿ, we allow the second argument to depend on each t∈ [tⁿ_i, tⁿ_i+1]. In addition to other techniques used to overcome the absence of the compactness, this adaptation enables the proof of the convergence of the sequence x_n(·) to a solution of the original problem.

We need the result of this paper, to prove the existence of a solution for an optimal control problem of the type

ζ(·)∈SinfΓ

L(x^ζ(T )) + Z T

0

J(t, x^ζ(t), ζ(t)) dt,

where x^ζ(·) is the unique solution of the delay perturbed problem ( − ˙x(t) ∈ ∂ϕ(t, x(t)) + g(t, τ (t)x(·), ζ(t)) a.e. t ∈ [0, T ],

x(s) = ψ(s) ∀s ∈ [−r, 0].

we address this problem in a forthcoming paper.

The content of the paper is as follows. In Section 2, we recall some definitions and concepts which are used throughout the paper. Then, we state results of perturbed evolution equation governed by the subdifferential operator ∂ϕ(t, ·) of a proper lsc convex function without delay. Existence and uniqueness for the considered problem (Pψ) are established in Section 3, and we conclude by some properties of the solution.

2. Preliminaries

The purpose of this section is to provide basic notions that will be used in the following sections.

In all the paper I := [0, T ] (T > 0) is an interval of R and H is a real Hilbert space whose scalar product is denoted by h·, ·i and the associated norm by k · k.

We will use the following definitions and notations. B is the closed unit ball of H and for η > 0, B[0, η], the closed ball of radius η centered at 0. On the space CH(I) of continuous maps x : I → H we consider the norm of uniform convergence kxk∞ = sup_t∈Ikx(t)k. By L^p_H(I) for p ∈ [0, +∞[ (resp. p = +∞) we denote the space of measurable maps x : I → H such thatR

Ikf (t)k^pdt <+∞

(resp. which are essentially bounded) endowed with the usual norm kxk_L^p

H(I) = (R

Ikx(t)k^pdt)^1p, 1 ≤ p < +∞ (resp. endowed with the usual essential supremum norm k k). We recall that the topological dual of L¹_H(I) is L^∞_H(I).

Let ϕ be a lsc convex function from H into R ∪ {+∞} which is proper in the sense that its effective domain (dom ϕ) defined by

dom ϕ = {x ∈ H; ϕ(x) < +∞}

is nonempty. As usual, its Fenchel conjugate is defined by ϕ^∗(v) := sup

x∈H

[hv, xi − ϕ(x)].

The subdifferential ∂ϕ(x) of ϕ at x ∈ dom ϕ is

∂ϕ(x) = {v ∈ H : ϕ(y) ≥ hv, y − xi + ϕ(x) ∀y ∈ dom ϕ}

and its effective domain is Dom ∂ϕ = {x ∈ H : ∂ϕ(x) 6= ∅}. It is well known (see, e.g., [4]) that if ϕ is a proper lsc convex function, then its subdifferential

operator ∂ϕ is a maximal monotone operator and satisfies the closure property, that is, for any maximal monotone operator A, if x = lim

n→∞xn strongly in H and y = lim

n→∞y_n weakly in H, where x_n ∈ dom A and yn∈ A(xn), then, x ∈ dom A and y ∈ A(x).

Concerning the properties of maximal monotone operators in Hilbert spaces, we refer to [4] and [1]. We also refer to [9] and [12] for details concerning convex analysis and measurable multifunctions.

Let us recall the following straightforward consequence of Gronwall’s lemma proved in [10].

Lemma 2.1. Let I = [T₀, T] and let (x_n(·)) be a sequence of non-negative contin- uous functions defined on I,(α_n) a sequence of real numbers, and β(·) ∈ L¹_R+(I).

Assume that limnα_n= 0 and, for all n,

(2.1) x_n(t) ≤

Z t T0

β(s)xn(s) ds + αn. Then, for all t∈ [T0, T],

limn x_n(t) = 0.

We will close this section by recalling the two following results concerning per- turbed problems without delay. The first one is an adaptation of Theorem 1 in Peralba [13] (see [14]).

Proposition 2.2. Let H be a Hilbert space and let ϕ: [T₀, T] × H −→ [0, +∞]

be such that

(H₁) for each t ∈ [T₀, T], the function x 7→ ϕ(t, x) is proper, lsc, and convex;

(H₂) there exist a non negative ρ-Lipschitz function k : H −→ R₊ and an abso- lutely continuous function a : [T0, T] → R, with a non-negative derivative

˙a ∈ L²R([T₀, T]), such that

(2.2) ϕ^∗(t, x) ≤ ϕ^∗(s, x) + k(x)|a(t) − a(s)|

for every (t, s, x) ∈ [T₀, T] × [T₀, T] × H, where ϕ^∗(t, ·) is the conjugate function of ϕ(t, ·) (recalled above).

If h∈ L²_H([T₀, T]) and x₀∈ dom ϕ(T0,·), then the differential equation

(Ph)

( − ˙x(t) ∈ ∂ϕ(t, x(t)) + h(t) x(T₀) = x₀∈ dom ϕ(T0,·)

admits an unique absolutely continuous solution x(·) that satisfies k ˙xk_L²

H ≤ 1

2(ρ + 1)k ˙a + |h|k_L²

R + khk_L² (2.3) H

+ [p

T − T0k(0)k ˙a + |h|k_L²

R + (ρ + 1)²

4 k ˙a + |h|k²_L²

R + ϕ(T₀, x₀) − ϕ(T, x(T ))]¹². Now, we address the case of perturbation without delay depending on both time variable and state variable proved in [14]

Theorem 2.3. Assume that (H₁) and (H₂) of Proposition 2.2 hold, and let f : [T₀, T] × H → H be a map such that

(i) f is separately measurable on [T₀, T];

(ii) for every η > 0, there exists a non-negative function γη(·) ∈ L²R([T₀, T]) such that, for all t∈ [T0, T] and for any x, y ∈ B[0, η]

(2.4) kf (t, x) − f (t, y)k ≤ γη(t)kx − yk;

(iii) there exists a non-negative function β(·) ∈ L²R([T₀, T]) such that, for all t∈ [T0, T] and for all x ∈ H, one has

(2.5) kf (t, x)k ≤ β(t)(1 + kxk).

Then, for any x₀∈ dom ϕ(T0,·), the following problem

(P_f(·,·))

( − ˙x(t) ∈ ∂ϕ(t, x(t)) + f (t, x(t)) a.e. t ∈ [T0, T] x(T0) = x0

has one and only one absolutely continuous solution x(·) which satisfies (2.6)

Z T T0

k ˙x(t)k²dt≤ αT0,T + σ Z T

T0

kf (t, x(t))k² dt,

where

αT0,T = (k²(0) + 3(ρ + 1)²) Z T

T0

˙a²(t) dt + 2[T − T₀+ ϕ(T₀, x₀) − ϕ(T, x(T ))], σ = k²(0) + 3(ρ + 1)²+ 4.

3. Perturbation with Delay

This section is devoted to the study of a perturbed problem involving a subdif- ferential operator whose perturbation is single-valued and contains a delay.

We are going to investigate the existence of solutions for the following prob- lem

(P_ψ)

( − ˙x(t) ∈ ∂ϕ(t, x(t)) + f (t, τ (t)x(·)) a.e. t ∈ [0, T ], x(s) = ψ(s) ∀s ∈ [−r, 0].

We mean by a solution of (Pψ) any map x(·) : [−r, T ] → H that satisfies (a) for any s ∈ [−r, 0], one has x(s) = ψ(s);

(b) the restriction x|_{[0,T ]}(·) of x(·) is absolutely continuous and its derivative, denoted by ˙x(·), satisfies the inclusion

− ˙x(t) ∈ ∂ϕ(t, x(t)) + f (t, τ (t)x(·)) a.e. t ∈ [0, T ].

Now, we are going to state and prove our existence result concerning the prob- lem (Pψ).

Theorem 3.1. Assume that ϕ satisfies (H₁) and (H₂) of Proposition 2.2. Let f : I × C₀→ H be a map satisfying:

(i) for any φ ∈ C₀, f(·, φ) is measurable;

(ii) for any η > 0, there exists a non-negative function γ_η(·) ∈ L²R(I) such that, for all φ₁, φ₂ ∈ C0 with kφikC0 ≤ η (i = 1, 2) and for all t ∈ I,

kf (t, φ₁) − f (t, φ₂)k ≤ γη(t)kφ₁− φ₂kC0;

(iii) there exists a non-negative function β(·) ∈ L²R(I) such that, for all t ∈ I and for all φ∈ C0,

kf (t, φ)k ≤ β(t)(1 + kφkC0).

Then, for any ψ∈ C0 with ψ(0) ∈ dom ϕ(0, ·), the problem (P_ψ) has one and only one solution which satisfies

(3.1)

Z T 0

k ˙x(t)k² dt≤ α + σ Z T

0

kf (t, τ (t)x(·))k² dt, with

α = (k²(0) + 3(ρ + 1)²) Z T

0

˙a²(t) dt + 2[T + ϕ(0, ψ(0)) − ϕ(T, x(T ))]

σ = k²(0) + 3(ρ + 1)²+ 4.

Proof. Let us consider

m= 1

4T (k²(0) + 3(ρ + 1)²+ 4) >0.

I) Suppose first, that (3.2)

Z T 0

β²(s) ds < m,

and let construct a sequence of maps (xn(·)) in CH([−r, T ]) which converges uniformly on [−r, T ] to a solution of (P_ψ). For each n ≥ 1, consider the partition of [0, T ] defined by the points tⁿ_i := ^iT_n (i = 0, . . . , n) and define the map f₀ⁿ : [0, tⁿ₁] × H → H by

f₀ⁿ(t, x) := f (t, τ (t)hⁿ₀(·, x)).

where hⁿ₀ : [−r, tⁿ₁] × H → H is given by

hⁿ₀(t, x) =

( ψ(t) if t ∈ [−r, 0]

ψ(0) +_Tⁿt(x − ψ(0)) if t ∈ [0, tⁿ₁].

The map x 7→ τ (t)hⁿ₁(·, x) is 1-Lipschitz, indeed we have for any t ∈ [0, tⁿ₁] and for any x, y ∈ H,

kτ (t)hⁿ₀(·, x) − τ (t)hⁿ₀(·, y)kC0 = sup

s∈[−r,0]

khⁿ₀(t + s, x) − hⁿ₀(t + s, y)k

= sup

s∈[−r+t,t]

khⁿ₀(s, x) − hⁿ₀(s, y)k

≤ sup

s∈[0,t]

khⁿ₀(s, x) − hⁿ₀(s, y)k

= sup

s∈[0,t]

n

Tskx − yk

≤ kx − yk.

On the other hand,

kτ (t)hⁿ₀(·, x)kC0 = sup

s∈[−r+t,t]

khⁿ₀(s, x)k

≤ max{kψkC0, sup

s∈[0,t]

kψ(0) + n

Ts(x − ψ(0))k}

≤ max{kψkC0, sup

s∈[0,t]

((1 − n

Ts)kψ(0)k + n Tskxk)}

≤ max{kψkC0, kψ(0)k + kxk}

≤ kψkC0 + kxk.

Thanks to (ii), there exists a non-negative function denoted by γη(·) ∈ L²R(I) such that for all t ∈ [0, tⁿ₁], and for any x, y ∈ B[0, η],

kf₀ⁿ(t, x) − f₀ⁿ(t, y)k ≤ γη(t)kx − yk.

and by (iii), for all (t, x) ∈ [0, tⁿ₁] × H,

kf₀ⁿ(t, x)k = kf (t, τ (t)hⁿ₀(·, x))k ≤ β(t)(1 + kτ (t)hⁿ₀(·, x)kC0)

≤ β(t)(1 + kψkC0 + kxk)

≤ (1 + kψkC0)β(t)(1 + kxk).

Note also that, due to the fact that hⁿ₀(·, x) is uniformly continuous on [0, tⁿ₁], the map t 7→ τ (t)hⁿ₀(·, x) is continuous from [0, tⁿ₁] into (C₀,k · kC0) and hence f₀ⁿ(·, x) is measurable. Consequently, according to Theorem 2.3, there exists one and only one absolutely continuous map xⁿ₀(·) : [0, tⁿ₁] → H such that xⁿ₀(0) = ψ(0) and, for almost all t ∈ [0, tⁿ₁],

˙xⁿ₀(t) + f₀ⁿ(t, xⁿ₀(t)) ∈ −∂ϕ(t, xⁿ₀(t)) a.e. t ∈ [0, tⁿ₁], with

(3.3)

Z tⁿ₁ 0

k ˙xⁿ₀(t)k²dt≤ α0+ σ Z tⁿ₁

0

kf (t, τ (t)hⁿ₀(·, xⁿ₀(t)))k² dt, where

α₀ = (k²(0) + 3(ρ + 1)²) Z tⁿ₁

0

˙a²(t) dt + 2[tⁿ₁ + ϕ(0, ψ(0)) − ϕ(tⁿ₁, xⁿ₀(tⁿ₁))]

σ = k²(0) + 3(ρ + 1)²+ 4.

Now, define : hⁿ₁ : [−r, tⁿ₂] × H → H by

hⁿ₁(t, x) =











ψ(t) if t ∈ [−r, 0],

xⁿ₀(t) if t ∈ [0, tⁿ₁],

xⁿ₀(tⁿ₁) +_Tⁿ(t − tⁿ₁)(x − xⁿ₀(tⁿ₁)) if t ∈ [tⁿ₁, tⁿ₂].

As previously, we show that, for any t ∈ [0, tⁿ₂], the map x 7→ τ (t)hⁿ₁(·, x) is 1-Lipschitz and

kτ (t)hⁿ₁(·, x)kC0 ≤ max{kψkC0, kxⁿ₀kC_H([0,tⁿ₁])} + kxk.

Therefore, the map f₁ⁿ: [tⁿ₁, tⁿ₂] × H → H defined by f₁ⁿ(t, x) := f (t, τ (t)hⁿ₁(·, x))

satisfies the assumptions of Theorem 2.3, and hence there exists one and only one absolutely continuous map xⁿ₁(·) : [tⁿ₁, tⁿ₂] → H such that xⁿ₁(tⁿ₁) = xⁿ₀(tⁿ₁) and,

˙xⁿ₁(t) + f₁ⁿ(t, xⁿ₁(t)) ∈ −∂ϕ(t, xⁿ₁(t)) a.e. t ∈ [tⁿ₁, tⁿ₂], with

(3.4)

Z tⁿ₂ tⁿ₁

k ˙xⁿ₁(t)k²dt≤ α1+ σ Z tⁿ₂

tⁿ₁

kf (t, τ (t)hⁿ₁(·, xⁿ₁(t)))k² dt,

where

α₁ = (k²(0) + 3(ρ + 1)²) Z tⁿ₂

tⁿ₁

˙a²(t) dt

+ 2[tⁿ₂ − tⁿ₁ + ϕ(tⁿ₁, xⁿ₁(tⁿ₁)) − ϕ(tⁿ₂, xⁿ₁(tⁿ₂))].

Now, suppose that xⁿ₀(·), . . . , xⁿ_i−1(·) (1 ≤ i ≤ n − 1) are defined similarly and define hⁿ_i : [−r, tⁿ_i+1] × H → H by

hⁿ_i(t, x) =











ψ(t) if t ∈ [−r, 0],

xⁿ_j(t) if t ∈ [tⁿ_j, tⁿ_j+1], j ∈ 0, i − 1, xⁿ_i−1(tⁿ_i) +_Tⁿ(t − tⁿ_i)(x − xⁿ_i−1(tⁿ_i)) if t ∈ [tⁿ_i, tⁿ_i+1],

and f_iⁿ: [tⁿ_i, tⁿ_i+1] × H → H by

f_iⁿ(t, x) := f (t, τ (t)hⁿ_i(·, x)).

We have for all t ∈ [tⁿ_i, tⁿ_i+1] and x, y ∈ H,

kτ (t)hⁿ_i(·, x) − τ (t)hⁿ_i(·, y)kC0 ≤ kx − yk and

(3.5) kτ (t)hⁿ_i(·, x)kC0 ≤ Aⁿ_i + kxk, where

Aⁿ_i := maxn

kψkC0, max

0≤j≤i−1kxⁿ_jkC_H([tⁿ_j,tⁿ_j+1])

o .

Due to Theorem 2.3, there exists one and only one absolutely continuous map xⁿ_i(·) : [tⁿ_i, tⁿ_i+1] → H such that xⁿ_i(tⁿ_i) = xⁿ_i−1(tⁿ_i), and

˙xⁿ_i(t) + f_iⁿ(t, xⁿ_i(t)) ∈ −∂ϕ(t, xⁿ_i(t)) a.e. t ∈ [tⁿ_i, tⁿ_i+1].

In this way, we define xⁿ₀(·), . . . , xⁿ_n−1(·) such that, for each i ∈ {0, . . . , n − 1}, xⁿ_i(·) is absolutely continuous on [tⁿ_i, tⁿ_i+1], xⁿ_i(tⁿ_i) = xⁿ_i−1(tⁿ_i) (with the convention xⁿ₋₁(0) := ψ(0)),

˙xⁿ_i(t) + f_iⁿ(t, xⁿ_i(t)) ∈ −∂ϕ(t, xⁿ_i(t)) a.e. t ∈ [tⁿ_i, tⁿ_i+1], with

(3.6)

Z tⁿ_i+1 tⁿ_i

k ˙xⁿ_i(t)k²dt≤ αi+ σ Z tⁿ_i+1

tⁿ_i

kf (t, τ (t)hⁿ_i(·, xⁿ_i(t)))k² dt,

where

α_i = (k²(0) + 3(ρ + 1)²) Z tⁿ_i+1

tⁿ_i

˙a²(t) dt

+ 2[tⁿ_i+1− tⁿ_i + ϕ(tⁿ_i, xⁿ_i(tⁿ_i)) − ϕ(tⁿ_i+1, xⁿ_i(tⁿ_i+1))].

Let us define x_n(·) : [−r, T ] → H by

x_n(t) :=

( ψ(t) if t ∈ [−r, 0],

xⁿ_i(t) if t ∈ [tⁿ_i, tⁿ_i+1], i ∈ {0, . . . , n − 1}.

Then, for each i ∈ {0, . . . , n − 1},

hⁿ_i(t, x) =

( xn(t) if t ∈ [−r, tⁿ_i],

xn(tⁿ_i) +_Tⁿ(t − tⁿ_i)(x − xn(tⁿ_i)) if t ∈ [tⁿ_i, tⁿ_i+1].

Put

θ_n(t) :=

( 0 if t = 0,

tⁿ_i if t ∈ ]tⁿ_i, tⁿ_i+1], i ∈ {0, . . . , n − 1}.

One has, by construction, x_n(0) = ψ(0) and, for almost all t ∈ I, (3.7) ˙xn(t) + f (t, τ (t)hⁿⁿ

Tθn(t)(·, xn(t))) ∈ −∂ϕ(t, xn(t)).

and

x_n(s) = ψ(s) for all s ∈ [−r, 0].

By (3.6), one has (3.8)

Z T 0

k ˙xn(t)k² dt≤ αn+ σ Z T

0

kf (t, τ (t)hⁿⁿ

Tθn(t)(·, xn(t)))k² dt,

where

α_n = (k²(0) + 3(ρ + 1)²) Z T

0

˙a²(t) dt + 2[T + ϕ(0, ψ(0)) − ϕ(T, x_n(T ))].

As −ϕ(T, xn(T )) ≤ 0, setting

η₀ = (k²(0) + 3(ρ + 1)²) Z T

0

˙a²(t) dt + 2[T + ϕ(0, ψ(0))], we may write,

(3.9)

Z T 0

k ˙xn(t)k² dt≤ η0+ σ Z T

0

kf (t, τ (t)hⁿⁿ

Tθn(t)(·, xn(t)))k² dt.

Thanks to (3.5)

(3.10) kτ (t)hⁿⁿ

Tθn(t)(·, xn(t))kC0 ≤ 2kxn(·)kC_H([−r,T ]). This, along with (iii), implies

(3.11) kf (t, τ (t)hⁿⁿ

Tθn(t)(·, xn(t)))kC0 ≤ β(t)(1 + 2kxn(·)kC_H([−r,T ])) a.e. t ∈ I.

Now, let prove that (xn(·)) converges uniformly in CH([−r, T ]). In view of (3.9) and (3.11), we may write

Z T 0

k ˙xn(t)k²dt ≤ η₀+ σ(1 + 2kxn(·)kC_H([−r,T ]))² Z T

0

β²(t) dt,

≤ η0+ 2σ(1 + 4kx_n(·)k²_C

H([−r,T ])) Z T

0

β²(t) dt,

≤ l + 8σkxn(·)k²_C

H([−r,T ])

Z T 0

β²(t) dt, where

l:= η₀+ 2σ Z T

0

β²(t) dt.

Making use of the absolute continuity of x_n(·) on [0, T ], for any s ∈ I, one has kxn(s) − ψ(0)k² ≤ s

Z s 0

k ˙xn(t)k² dt,

≤ T Z T

0

k ˙xn(t)k²dt,

we may also write,

kxn(s)k² ≤ 2kxn(s) − ψ(0)k²+ 2kψ(0)k²,

≤ 2kxn(s) − ψ(0)k²+ 2kψk²_C0. Then, using the preceding inequalities

kxn(s)k² ≤ 2T l + 16T σkxn(·)k²_C

H([−r,T ])

Z T 0

β²(t) dt + 2kψk²_C0. Taking (3.2) into account, that is, 16σTRT

0 β²(t) dt < 1, it follows that (3.12) kxn(·)kC_H([−r,T ])≤ M

2 , where

M :=

v u u t

8(T l + kψk²_C0) 1 − 16σTRT

0 β²(t) dt.

Now, we proceed to prove that (xn(·)) is a Cauchy sequence in CH([0, T ]). Thanks to (3.7), and the monotonicity property of the subdifferential operator, for p, q ≥ 1 and for almost all t ∈ I, we have

h− ˙xp(t) − z_p(t) + ˙x_q(t) + z_q(t), x_p(t) − x_q(t)i ≥ 0 where

z_p(t) = f (t, τ (t)h^pp

Tθp(t)(·, xp(t))).

Hence,

h ˙xp(t) − ˙xq(t), xp(t) − xq(t)i ≤ h−zp(t) + zq(t), xp(t) − xq(t)i

≤ kzp(t) − zq(t)kkxp(t) − xq(t)k and then

(3.13) 1

2 d

dt(kx_p(t) − x_q(t)k²) ≤ B_p,q(t)kx_p(t) − x_q(t)k, where

B_p,q(t) := kf (t, τ (t)h^pp

Tθp(t)(·, xp(t))) − f (t, τ (t)h^qq

Tθq(t)(·, xq(t)))k.

According to (ii), (3.10), and (3.12), we have, for some non-negative function γ_M(·) ∈ L²R(I) and for all t ∈ I

B_p,q(t) ≤ γ_M(t)kτ (t)h^pp

Tθp(t)(·, x_p(t)) − τ (t)h^qq

Tθq(t)(·, x_q(t))kC0.

Then,

B_p,q(t) ≤ γ_M(t)kτ (t)h^pp

Tθp(t)(·, x_p(t)) − τ (t)h^pp

Tθp(t)(·, x_q(t))kC0

+ γM(t)kτ (t)h^pp

Tθp(t)(·, xq(t)) − τ (t)h^qq

Tθq(t)(·, xq(t))kC0. The map x 7→ τ (t)h^pp

Tθp(t)(·, x) being 1-Lipschitz, one has B_p,q(t) ≤ γ_M(t)kx_p(t) − x_q(t)k

+ γM(t)kτ (t)h^pp

Tθp(t)(·, xq(t)) − τ (t)h^qq

Tθq(t)(·, xq(t))kC0. (3.14)

Let i ∈ {0, . . . , p−1} and j ∈ {0, . . . , q −1} such that t ∈]t^p_i, t^p_i+1] and t ∈]t^q_j, t^q_j+1].

Then,

kτ (t)h^pp

Tθp(t)(·, xq(t)) − τ (t)h^qq

Tθq(t)(·, xq(t))kC0

= sup

s∈[−r+t,t]

kh^p_i(s, xq(t)) − h^q_j(s, xq(t))k

≤ sup

s∈[0,t]

kh^p_i(s, x_q(t)) − h^q_j(s, x_q(t))k.

In the case t^p_i ≤ t^q_j one has sup

s∈[0,t]

kh^p_i(s, xq(t)) − h^q_j(s, xq(t))k = max{A¹_p,q(t), A²_p,q(t), A³_p,q(t)},

with

A¹_p,q(t) := sup

s∈[0,t^p_i]

kxp(s) − xq(s)k, A²_p,q(t) := sup

s∈[t^p_i,t^q_j]

kxp(t^p_i) +_T^p(s − t^p_i)(xq(t) − xp(t^p_i)) − xq(s)k,

and

A³_p,q(t) := sup

s∈[t^q_j,t]

kxp(t^p_i) + p

T(s − t^p_i)(xq(t) − xp(t^p_i))

− xq(t^q_j) − q

T(s − t^q_j)(xq(t) − xq(t^q_j))k.

We have

A²_p,q(t) ≤ sup

s∈[t^p_i,t^q_j]

kxp(t^p_i) − xp(s)k + kxp(s) − xq(s)k

+ p

T(s − t^p_i)(kxq(t) − xp(t)k + kxp(t) − xp(t^p_i)k).

Taking (3.11) and (3.12) into account, it follows that, for all n kf (t, τ (t)hⁿⁿ

Tθn(t)(·, x_n(t)))k ≤ β(t)(1 + M ).

Coming back to (3.9), we get

(3.15)

Z T 0

k ˙xn(t)k² dt≤ η0+ σ(1 + M )² Z T

0

β²(t) dt < +∞.

Hence, setting

S = sup

n k ˙xnk_L²

H([0,T ]),

along with the absolute continuity of x_p,one has for all p, and for all s ∈ [t^p_i, t^q_j], (t defined as above)

kxp(t^p_i) − xp(s)k ≤ Z s

t^p_i

k ˙xp(τ )k dτ

≤ Z t

t^p_i

k ˙xp(τ )k dτ

≤ S(t − t^p_i)¹².

Thus,

A²_p,q(t) ≤ sup

s∈[t^p_i,t^q_j]

 Z t t^p_i

k ˙xp(τ )k dτ + kxp(s) − xq(s)k + kxq(t) − xp(t)k

+ Z t

t^p_i

k ˙xp(τ )k dτ



≤ 2S(t − t^p_i)¹² + sup

s∈[t^p_i,t]

kxp(s) − x_q(s)k.

A³_p,q(t) ≤ sup

s∈[t^q_j,t]

kxp(t^p_i) − x_p(t)k + kx_p(t) − x_q(t)k + kx_q(t) − x_q(t^q_j)k

+ p

T(s − t^p_i)(kxq(t) − xp(t)k + kxp(t) − xp(t^p_i)k) + q

T(s − t^q_j)kx_q(t) − x_q(t^q_j)k

≤ Z t

t^p_i

k ˙xp(τ )k dτ + kxp(t) − xq(t)k + Z t

t^q_j

k ˙xq(τ )k dτ

+ kx_q(t) − x_p(t)k + Z _t

t^p_i

k ˙xp(τ )k dτ + Z _t

t^q_j

k ˙xq(τ )k dτ,

then,

A³_p,q(t) ≤ 2S[(t − t^p_i)¹² + (t − t^q_j)¹²] + 2kx_p(t) − x_q(t)k.

Thus, if t^p_i ≤ t^q_j, we have sup

s∈[0,t]

kh^p_i(s, xq(t)) − h^q_j(s, xq(t))k

≤ max

 sup

s∈[0,t^p_i]

kxp(s) − x_q(s)k, sup

s∈[t^p_i,t]

kxp(s) − x_q(s)k, 2kx_p(t) − x_q(t)k



+ 2S[(t − t^p_i)¹² + (t − t^q_j)¹²]

≤ 2kxp(·) − xq(·)kC_H([0,t])+ 2S[(t − t^p_i)¹² + (t − t^q_j)¹²].

Likewise, if t^q_j ≤ t^p_i, interchanging t^q_j and t^p_i, we obtain the same previous inequal- ity. Therefore, for any t ∈ [−r, T ], we get

kτ (t)h^pp

Tθp(t)(·, xq(t)) − τ (t)h^qq

Tθq(t)(·, xq(t))kC0 ≤ 2kxp(·) − xq(·)kC_H([0,t])

+ 2S[(t − θp(t))¹² + (t − θq(t))¹²].

Coming back to (3.14), we obtain

B_p,q(t) ≤ 3γ_M(t)kx_p(·) − x_q(·)k_CH([0,t])+ 2γ_M(t)b_p,q(t), where

b_p,q(t) = S[(t − θp(t))¹² + (t − θq(t))¹²].

Taking (3.13) into account, it follows that, for almost all t ∈ I 1

2 d

dt(kxp(t) − xq(t)k²) ≤ 3γM(t)kxp(·) − xq(·)k²_C

H([0,t])

+ 2kxp(·) − xq(·)kC_H([0,t])γ_M(t)bp,q(t) and, using (3.12), it results that

1 2

d

dt(kx_p(t) − x_q(t)k²) ≤ 3γ_M(t)kx_p(·) − x_q(·)k²_C

H([0,t])

+ 2M γ_M(t)b_p,q(t).

(3.16)

In the following, we use the fact that the map t 7→ kx_p(·) − x_q(·)kC_H([0,t]) is continuous. By integration on [0, t], one has

1

2(kx_p(t) − x_q(t)k²) ≤ 3 Z t

0

γ_M(s)kx_p(·) − x_q(·)k²_C

H([0,s])ds + 2M

Z t 0

γM(s)bp,q(s) ds.

The above inequality being true for any t ∈ [0, T ], it follows that

kxp(·) − x_q(·)k²_C

H([0,t]) ≤ ap,q+ 6 Z t

0

γ_M(s)kx_p(·) − x_q(·)k²_C

H([0,s])ds, where

a_p,q:= 4M Z T

0

γ_M(s)bp,q(s) ds.

Note that lim_p→∞θ_p(t) = t, lim_q→∞θ_q(t) = t, for any t and hence lim_p,q→∞b_p,q(t)

= 0. Therefore, by the dominated convergence theorem, we get limp,q→∞a_p,q = 0 and, according to Lemma 2.1

p,q→∞lim kxp(·) − xq(·)k∞= 0,

which proves that the sequence (xn(·)) converges uniformly in CH([−r, T ]) to some map x(·) ∈ C_H([−r, T ]) with x(s) = ψ(s) for all s ∈ [−r, 0], and then, (3.17) x_n(·) → x(·) strongly in L²_H([0, T ]).

Moreover, thanks to (3.15), we may suppose that ( ˙xn(·)) converges weakly in L²_H([0, T ]) to some map g(·) ∈ L²_H([0, T ]). It results that, for all t ∈ [0, T ], x(t) = ψ(0) +Rt

0g(s) ds and hence x(·) is absolutely continuous on [0, T ] with

˙x(t) = g(t) for almost t ∈ [0, T ]. Consequently,

(3.18) ˙xn(·) → ˙x(·) weakly in L²_H([0, T ]).

Now, we aim at proving that x(·) is a solution of (Pψ), first, let us prove that, for any t ∈]0, T ], one has

limn f(t, τ (t)hⁿⁿ

Tθn(t)(·, xn(t))) = f (t, τ (t)x(·)).

Fix t ∈]0, T ]. For each n ≥ 1, there exists j ∈ {0, . . . , n − 1} such that t ∈]tⁿ_j, tⁿ_j+1] and thus θn(t) = tⁿ_j. Then,

kτ (t)hⁿⁿ

Tθn(t)(·, xn(t)) − τ (t)x(·)kC0 = sup

s∈[−r,0]

khⁿ_j(t + s, xn(t)) − x(t + s)k

= sup

s∈[−r+t,t]

khⁿ_j(s, xn(t)) − x(s)k

≤ maxn sup

s∈[0,tⁿ_j]

kxn(s) − x(s)k, B_n,m¹ (t)o ,

where

B_n,m¹ (t) := sup

s∈[tⁿ_j,t]

kxn(tⁿ_j) + n

T(s − tⁿ_j)(xn(t) − xn(tⁿ_j)) − x(s)k.

We have

B_n,m¹ (t) ≤ sup

s∈[tⁿ_j,t]

(kxn(tⁿ_j) − x(s)k + kxn(t) − xn(tⁿ_j)k)

≤ sup

s∈[tⁿ_j,t]

(kx_n(tⁿ_j) − x_n(s)k + kx_n(s) − x(s)k + kx_n(t) − x_n(tⁿ_j)k).

It follows that

B_n,m¹ (t) ≤ sup

s∈[tⁿ_j,t]

kxn(s) − x(s)k + 2 Z t

θn(t)

k ˙xn(τ )k dτ.

As a result, kτ (t)hⁿⁿ

Tθn(t)(·, xn(t)) − τ (t)x(·)kC0 ≤ kxn(·) − x(·)k∞+ 2S(t − θn(t))¹². Therefore,

kτ (t)hⁿⁿ

Tθn(t)(·, xn(t)) − τ (t)x(·)kC0 → 0.

The Lipschitz behavior of the map f (t, ·) for each fixed t in [0, T ] leads to

n→∞lim kf (t, τ (t)hⁿⁿ

Tθn(t)(·, xn(t))) − f (t, τ (t)x(·))k = 0, along with

n→∞lim Z T

0

kf (t, τ (t)hⁿⁿ

Tθn(t)(·, xn(t))) − f (t, τ (t)x(·))k² dt= 0, by Lebesgue’s convergence theorem. Then,

(3.19) lim

n→∞

Z T 0

kf (t, τ (t)hⁿⁿ

Tθn(t)(·, x_n(t)))k² dt= Z T

0

kf (t, τ (t)x(·))k² dt.

Thus, taking the superior limit on n in (3.8), and using (3.18) and (3.19) yield

(3.20)

Z T 0

k ˙x(t)k² dt≤ lim sup

n

α_n+ σ Z T

0

kf (t, τ (t)x(·))k² dt.

Since xn(t) → x(t), by the lower semi-continuity of ϕ(t, ·), one has lim sup

n −ϕ(T, xn(T )) = − lim inf

n ϕ(T, x_n(T ))

≤ −ϕ(T, x(T )).

Hence, (3.21)

Z T 0

k ˙x(t)k² dt≤ α + σ Z T

0

kf (t, τ (t)x(·))k² dt, where

α = (k²(0) + 3(ρ + 1)²) Z T

0

˙a²(t) dt + 2[T + ϕ(0, ψ(0)) − ϕ(T, x(T ))].

Let introduce now, the operator A : L²_H([0, T ]) → L²_H([0, T ]) defined by Ax = {y ∈ L²_H([0, T ]) : y(t) ∈ ∂ϕ(t, x(t)) a.e. on [0, T ]}.

It’s well known (see [13]) that if ϕ satisfies assumptions (H₁) and (H₂), then A is maximal monotone. Thanks to (3.17), (3.18) and (3.19), then, using the closure property of A, entails that,

− ˙x(t) ∈ ∂ϕ(t, x(t)) + f (t, τ (t)x(·)) a.e. t ∈ [0, T ], that is, the mapping x(·) is a solution of the differential inclusion (P_ψ).

II) Now, assume that RT

0 β²(s) ds ≥ m.

Consider a partition 0 = T₀ < T₁ < · · · < Tn = T of [0, T ] such that, for any i∈ {0, . . . , n − 1},

(3.22)

Z Ti+1

T_i

β²(s) ds < m.

According to the part I), there exists a map x₀(·) : [−r, T₁] → H absolutely continuous on [0, T1] such that

x₀(s) = ψ(s) for all s ∈ [−r, 0]

and

˙x₀(t) + f (t, τ (t)x₀(·)) ∈ −∂ϕ(t, x₀(t)) a.e. t ∈ [0, T₁].

Assume that, for any i ∈ {0, . . . , n − 2}, there exists a map xi(·) : [−r, T_i+1] → H absolutely continuous on [0, T_i+1] such that

(3.23) x_i(s) = ψ(s) for all s ∈ [−r, 0]

and

(3.24) ˙x_i(t) + f (t, τ (t)x_i(·)) ∈ −∂ϕ(t, x_i(t)) a.e. t ∈ [0, T_i+1].

According to (3.21), this inequality holds true Z Ti+1

0

k ˙xi(t)k² dt≤ ξi+ σ Z Ti+1

0

kf (t, τ (t)xi(·))k² dt, where

ξ_i = (k²(0) + 3(ρ + 1)²) Z Ti+1

0

˙a²(t) dt + 2[T_i+1+ ϕ(0, ψ(0)) − ϕ(T_i+1, x_i(T_i+1))].

Let us define ˜f : [0, T_i+2 − Ti+1] × C₀ → H, ˜ϕ: [0, T_i+2− Ti+1] × H → [0, +∞]

and ˜ψ(·) : [−r, 0] → H by

(3.25) f˜(t, φ) := f (t + T_i+1, φ), ˜ϕ(t, x) := ϕ(t + T_i+1, x), and

ψ(s) := x˜ _i(s + T_i+1).

Set ˜a(·) : [0, T_i+2− T_i+1] → R with

(3.26) ˜a(t) := a(t + T_i+1).

Define also ˜β(·) : [0, T_i+2− T_i+1] → R by

β(t) := β(t + T˜ _i+1).

Obviously, for all t ∈ [0, T_i+2− T_i+1] and for all φ ∈ C₀ k ˜f(t, φ)k ≤ ˜β(t)(1 + kφkC0) and, due to (3.22),

Z Ti+2−Ti+1

0

β(s) ds < m.˜

According to the part I) again, there exist a map ˜x(·) : [−r, T_i+2− T_i+1] → H which is absolutely continuous on [0, T_i+2− Ti+1] such that

(3.27) x(s) = ˜˜ ψ(s) for all s ∈ [−r, 0]

and

(3.28) ˙˜x(t) + ˜f(t, τ (t)˜x(·)) ∈ −∂ ˜ϕ(t, ˜x(t)) a.e. t ∈ [0, Ti+2− Ti+1].

Consider the map x_i+1(·) : [−r, T_i+2] → H defined by

x_i+1(t) =

( x_i(t) if t ∈ [−r, T_i+1],

˜

x(t − T_i+1) if t ∈ [T_i+1, T_i+2].

It follows from (3.25) and (3.28) that

(3.29) ˙x_i+1(t) + f (t, τ (t)x_i+1(·)) ∈ −∂ϕ(t, x_i+1(t)) a.e. t ∈ [T_i+1, T_i+2].

Thanks to (3.23) and (3.24), along with (3.29), we obtain x_i+1(s) = ψ(s) for all s ∈ [−r, 0]

and

˙x_i+1(t) + f (t, τ (t)x_i+1(·)) ∈ −∂ϕ(t, x_i+1(t)) a.e. t ∈ [0, T_i+2].

In view of (3.28) and the preceding inequality (3.21), it follows that Z Ti+2

Ti+1

k ˙˜x(t − T_i+1)k² dt =

Z Ti+2−Ti+1

0

k ˙˜x(t)k² dt

≤ ξi+1+ σ

Z Ti+2−Ti+1

0

k ˜f(t, τ (t)˜x(·))k²(t) dt, where

ξ_i+1 = (k²(0) + 3(ρ + 1)²)

Z Ti+2−Ti+1

0

˙˜a²(t) dt

+ 2[T_i+2− Ti+1+ ˜ϕ(0, ˜x(0)) − ˜ϕ(T_i+2− Ti+1,x(T˜ _i+2− Ti+1))].

Taking (3.25) and (3.26) into account, we may write ξ_i+1 = (k²(0) + 3(ρ + 1)²)

Z Ti+2

Ti+1

˙a²(t) dt + 2[Ti+2− Ti+1

+ ϕ(T_i+1,x(0)) − ϕ(T˜ _i+2,x(T˜ _i+2− Ti+1))].

As x_i+1(t + T_i+1) = ˜x(t), then, it results that ξ_i+1 = (k²(0) + 3(ρ + 1)²)

Z Ti+2

Ti+1

˙a²(t) dt

+ 2[T_i+2− Ti+1+ ϕ(T_i+1, x_i+1(T_i+1)) − ϕ(T_i+2, x_i+1(T_i+2))].

Combining this results with the definition of x_i+1,and the fact that xi(T_i+1) = x_i+1(T_i+1), one has

Z Ti+2

0

k ˙xi+1(t)k² dt =

Z Ti+1

0

k ˙xi(t)k² dt+ Z Ti+2

Ti+1

k ˙˜x(t)k² dt

≤ ξi+2+ σ Z Ti+2

0

kf (t, τ (t)xi+1(·))k² dt, (3.30)

where

ξ_i+2 = (k²(0) + 3(ρ + 1)²) Z Ti+2

0

˙a²(t) dt

+ 2[Ti+2+ ϕ(0, ψ(0)) − ϕ(Ti+2, x_i+1(Ti+2))].

By repeating the process we obtain a solution on the whole interval [−r, T ].