Electronic Journal of Qualitative Theory of Differential Equations 2007, No. 7, 1-14; http://www.math.u-szeged.hu/ejqtde/

VIABILITY PROBLEM WITH PERTURBATION IN HILBERT SPACE

MYELKEBIR AITALIOUBRAHIM AND SAID SAJID

University Hassan II-Mohammedia, U.F.R Mathematics and applications, F.S.T, BP 146, Mohammedia, Morocco. E-mail:saidsajid@hotmail.com and aitalifr@yahoo.fr

Abstract. This paper deals with the existence result of viable solutions of the differential inclusion

ẋ(t) ∈ f(t, x(t)) + F (x(t))
x(t) ∈ K on [0, T],

where K is a locally compact subset in separable Hilbert space H, (f(s, ·))_{s} is an equicontinuous family of measurable functions with respect to s and F is an upper semi-continuous set-valued mapping with compact values contained in the Clarke subdifferential ∂_{c}V (x) of an uniformly regular function V.

Key words: Regularity, upper semi-continuous, equicontinuous perturbation, Clarke subdifferential. Mathematics subject classification: 34A60, 49J52.

1. Introduction

Existence result of local solution for differential inclusion with upper semi-conti-nuous and cyclically monotone right hand-side whose values in finite-dimensional space, was first established by Bressan, Cellina and Colombo (see [6]). The authors exploited rich properties of subdifferential of convex lower semi-continuous function; in order to overcome the weakly convergence of derivatives of approximate solutions, they used the basic relation (see [7])

\frac{d}{dt}V (x(t)) = ∥ẋ(t)∥^{2}.

Later, Ancona, Cellina and Colombo (see [1]), under the same hypotheses as the above paper, extend this result to the perturbed problem

ẋ(t) ∈ f(t, x(t)) + F (x(t))

where f(·, ·) is a Carathéodory function.

This program of research was pursued by a series of works. In the first one (see [9]), Truong proved a viability result for similar problem, where the perturbation f is replaced by a globally continuous set-valued mapping G with values in finitedimensional space. This result was extended by Bounkhel (see [4]) for a similar

EJQTDE, 2007 No. 7, p. 1 problem, where F is not cyclically monotone but contained in the Clarke subdifferential of locally Lipschitz uniformly regular function. However under very strong assumptions namely, the space of states is finite-dimensional and the following tangential condition

(G(t, x) + F (x)) ⊂ T_{K}(x)

where T_{K}(x) is the contingent cone at x to K.

Recently, Morchadi and Sajid (see [8]) proved an exact viability version of the work of Ancona and Colombo assuming the same hypotheses and the following tangential condition

∀(t, x) ∈ ℝ × K, ∃v ∈ F (x) such that

lim_{h↦0^{+}} inf \frac{1}{h}d_{K} (x + hv + ∫_{t}^{t+h} f(s, x)ds) = 0.  (1.1)

Remark that in all the above works, the convexity assumption of V and/or the finite-dimensional hypothesis of the space of states were widely used in the proof.

This paper is devoted to establish a local solution of the problem

ẋ(t) ∈ f(t, x(t)) + F (x(t)), F (x(t)) ⊂ ∂_{c}V (x(t))

x(t) ∈ K ⊂ H,

where K is a locally compact subset of a separable Hilbert space H, F is an upper semi-continuous multifunction, ∂_{c}V denotes the Clarke subdifferential of a locally lipschitz function V and the set {f(s, .) : s ∈ ℝ} is equicontinuous, where for each x ∈ K, s ↦ f(s, x) is measurable and the same tangential condition (1.1). One case deserves mentioning: when f is globally continuous, the condition (1.1) is weaker than the following

(f(t, x) + F (x)) ∩ T_{K}(x) ≠ ∅.

To remove the convexity assumption of V and the finite-dimensional hypothesis of H, we rely on some properties of Clarke subdifferential of uniformly regular function and the local compactness of K.

2. Preliminaries and statement of the main result

Let H be a real separable Hilbert space with the norm ∥·∥ and the scalar product < ·, · > . For x ∈ H and r > 0 let B(x, r) be the open ball centered at x with radius r and B̄(x, r) be its closure and put B = B(0, 1).

Let us recall the definition of the Clarke subdifferential and the concept of regularity that will be used in the sequel.

Definition 2.1. Let V : H → ℝ ∪ {+∞} be a lower semi-continuous function and x be any point where V is finite. The Clarke subdifferential of V at x is defined by

∂_{c}V (x) := {y ∈ H : < y, h > ≤ V^{↑}(x, h), for all h ∈ H},

where V^{↑}(x, h) is the generalized Rockafellar directional derivative given by

V^{↑}(x, h) := lim sup_{x′→x,V (x′)→V (x),t→0} inf_{h′→h} \frac{V (x′ + th′) − V (x′)}{t}.

Definition 2.2. Let V : H → ℝ ∪ {+∞} be a lower semi-continuous function and let U ⊂ DomV be a nonempty open subset. We will say that V is uniformly regular over U if there exists a positive number β such that for all x ∈ U and for all ξ ∈ ∂_{p}V (x) one has

< ξ, x′ − x > ≤ V (x′) − V (x) + β∥x′ − x∥^{2} for all x′ ∈ U.

∂_{p}V (x) denotes the proximal subdifferential of V at x which is the set of all y ∈ H for which there exist δ, σ > 0 such that for all x′ ∈ x + δB̄

< y, x′ − x > ≤ V (x′) − V (x) + σ ∥ x′ − x ∥^{2} .

We say that V is uniformly regular over closed set S if there exists an open set U containing S such that V is uniformly regular over U. For more details on the concept of regularity, we refer the reader to [4].

Proposition 2.3. [3, 4] Let V : H → ℝ be a locally Lipschitz function and S a nonempty closed set. If V is uniformly regular over S, then the following conditions holds:

(a) The proximal subdifferential of V is closed over S, that is, for every x_{n} → x ∈ S with x_{n} ∈ S and every ξ_{n} → ξ with ξ_{n} ∈ ∂_{p}V (x_{n}) one has ξ ∈ ∂_{p}V (x).

(b) The proximal subdifferential of V coincides with the Clarke subdifferential of V for any point x.

(c) The proximal subdifferential of V is upper hemicontinuous over S, that is, the support function x ↦ σ(v, ∂_{p}V (x)) is u.s.c. over S for every v ∈ H.

Now let us state the main result.

Let V : H → ℝ be a locally Lipschitz function and β-uniformly regular over K ⊂ H. Assume that

(H1) K is a nonempty locally compact subset in H;

(H2) F : K → 2^{H} is an upper semi-continuous set valued map with compact

values satisfying

F (x) ⊂ ∂_{c}V (x) for all x ∈ K;

(H3) f : ℝ × H → H is a function with the following properties:

(1) For all x ∈ H, t ↦ f(t, x) is measurable,

(2) The family {f(s, .) : s ∈ ℝ} is equicontinuous,

(3) For all bounded subset S of H, there exists M > 0 such that

∥ f(t, x) ∥≤ M, ∀(t, x) ∈ ℝ × S;

(H4) (Tangential condition) ∀(t, x) ∈ ℝ × K, ∃v ∈ F (x) such that

lim_{h↦0^{+}} inf \frac{1}{h}d_{K} (x + hv + ∫_{t}^{t+h} f(s, x)ds) = 0.

For any x_{0} ∈ K, consider the problem:

\{ \begin{matrix}ẋ(t) ∈ f(t, x(t)) + F (x(t)) & a.e; \\ x(0) = x_{0}; \\ x(t) ∈ K. \end{matrix}  (2.1)

Theorem 2.4. If assumptions (H1)-(H4) are satisfied, then there exists T > 0 such that the problem (2.1) admits a solution on [0, T].

3. Proof of the main result

Choose r > 0 such that K_{0} = K ∩ (x_{0} + rB̄) is compact and V is Lipschitz continuous on x_{0} + rB̄ with Lipschitz constant λ > 0. Then ∂_{c}V (x) ⊂ λB̄ for every x ∈ K_{0}. Let M > 0 such that

∥ f(t, x) ∥≤ M, ∀(t, x) ∈ ℝ × (x_{0} + rB̄).  (3.1)

Set

T = \frac{r}{2(λ + 1 + M)}.  (3.2)

In the sequel, we will use the following important Lemma. It will play a crucial role in the proof of the main result.

Lemma 3.1. If assumptions (H1)-(H4) are satisfied, then for all 0 < ε < inf(T, 1), there exists η > 0 (η < ε) such that ∀(t, x) ∈ [0, T]×K_{0}, ∃h_{t,x} ∈ [η, ε], u ∈ F (x)+ \frac{1}{h_{t,x}} ∫_{t}^{t+h_{t,x}} f(s, x)ds+ \frac{ε}{T} B, y_{t,x} ∈ K_{0} and v ∈ F (y_{t,x}) such that

(x + h_{t,x}u) ∈ K ∩ B(x + h_{t,x}v + ∫_{t}^{t+h_{t,x}} f(s, x)ds, λ + M + 1).

Proof. Let (t, x) ∈ [0, T] × K_{0}, be fixed, let 0 < ε < inf(T, 1). Since F is u.s.c on x, then there exists δ_{x} > 0 such that

F (y) ⊂ F (x) + \frac{ε}{2T} B, for all y ∈ B(x, δ_{x}).

Let (s, y) ∈ [0, T] × K_{0}. By the tangential condition, there exists v ∈ F (y) and h_{s,y} ∈]0, ε] such that

d_{K} (y + h_{s,y}v + ∫_{s}^{s+h_{s,y}} f(τ, y)dτ) < h_{s,y} \frac{ε}{4T} .

Consider the subset

N(s, y) = \{(t, z) ∈ ℝ × H/d_{K}(z + h_{s,y}v + ∫_{t}^{t+h_{s,y}} f(τ, z)dτ) < h_{s,y} \frac{ε}{4T} \} .

Since

∥ f(τ, z) ∥≤ M, ∀(τ, z) ∈ ℝ × B̄(x_{0}, r),

EJQTDE, 2007 No. 7, p. 4 then the dominated convergence theorem applied to the sequence (χ_{[t,t+h_{s,y}]}f(·, ·))_{t} of functions shows that the function

(l, z) ↦ z + h_{s,y}v + ∫_{l}^{l+h_{s,y}} f(τ, z)dτ

is continuous. So that, the function

(l, z) ↦ d_{K} (z + h_{s,y}v + ∫_{l}^{l+h_{s,y}} f(τ, z)dτ)

is continuous and consequently N(s, y) is open. Moreover, since (s, y) belongs to N(s, y), there exists a ball B((s, y), η_{s,y}) of radius η_{s,y} < δ_{x} contained in N(s, y), therefore, the compact subset [0, T] × K_{0} can be covered by q such balls B((s_{i}, y_{i}), η_{s_{i},y_{i}}). For simplicity, we set

h_{s_{i},y_{i}} := h_{i} and η_{i} := η_{s_{i},y_{i}}, i = 1, …, q.

Put η = min{h_{i}/1 ≤ i ≤ q} and let i ∈ {1, …, q} such that (t, x) ∈ B((s_{i}, y_{i}), η_{i}), hence (t, x) ∈ N(s_{i}, y_{i}). Then there exists v_{i} ∈ F (y_{i}) such that

d_{K} (x + h_{i}v_{i} + ∫_{t}^{t+h_{i}} f(τ, x)dτ) < h_{i} \frac{ε}{4T} .

Let x_{i} ∈ K such that

\frac{1}{h_{i}} ∥x_{i} − (x + h_{i}v_{i} + ∫_{t}^{t+h_{i}} f(τ, x)dτ)∥ ≤ \frac{1}{h_{i}} d_{K} (x + h_{i}v_{i} + ∫_{t}^{t+h_{i}} f(τ, x)dτ)+\frac{ε}{4T} .

Hence

∥\frac{x_{i} − x}{h_{i}} − (v_{i} + \frac{1}{h_{i}} ∫_{t}^{t+h_{i}} f(τ, x)dτ)∥ < \frac{ε}{2T} .

Set

u = \frac{x_{i} − x}{h_{i}},

then x_{i} = x + h_{i}u ∈ K and

u ∈ (\frac{1}{h_{i}} ∫_{t}^{t+h_{i}} f(τ, x)dτ + F (y_{i}) + \frac{ε}{2T} B) .

Since ∥ x − y_{i} ∥< η_{i} < δ_{x} we have

F (y_{i}) ⊂ F (x) + \frac{ε}{2T} B,

then

u ∈ (\frac{1}{h_{i}} ∫_{t}^{t+h_{i}} f(τ, x)dτ + F (x) + \frac{ε}{T} B) .

On the other hand, since x ∈ K, we have

∥x_{i} − (x + h_{i}v_{i} + ∫_{t}^{t+h_{i}} f(τ, x)dτ)∥ ≤ d_{K} (x + h_{i}v_{i} + ∫_{t}^{t+h_{i}} f(τ, x)dτ) + \frac{ε}{4T}

≤ ∥h_{i}v_{i} + ∫_{t}^{t+h_{i}} f(τ, x)dτ∥ + \frac{ε}{4T}
≤ h_{i}(λ + M) + 1 < λ + M + 1.

Thus x_{i} ∈ B (x + h_{i}v_{i} + ∫_{t}^{t+h_{i}} f(τ, x)dτ, λ + M + 1) . □

Now, we are able to prove the main result. Our approach consists of constructing, in a first step, a sequence of approximate solutions and deduce, in a second step, from available estimates that a subsequence converges to a solution of (2.1).

Step 1. Approximate solutions. Let x_{0} ∈ K_{0} and 0 < ε < inf(T, 1). By Lemma 3.1, there exist η > 0, h_{0} ∈ [η, ε], u_{0} ∈ (\frac{1}{h_{0}} ∫_{0}^{h_{0}} f(s, x_{0})ds + F (x_{0}) + \frac{ε}{T} B), y_{0} ∈ K_{0} and v_{0} ∈ F (y_{0}) such that

x_{1} = x_{0} + h_{0}u_{0} ∈ K ∩ B (x_{0} + h_{0}v_{0} + ∫_{0}^{h_{0}} f(s, x_{0})ds, λ + M + 1) .

Then by (H2), (3.1) and (3.2), we have

∥ x_{1} − x_{0} ∥ = ∥ h_{0}u_{0} ∥ ≤ (λ + 1 + M)T < r

and thus x_{1} ∈ K_{0}. Set h_{−1} = 0. By induction, for q ≥ 2 and for every p = 1, …, q − 1, we construct the sequences (h_{p})_{p} ⊂ [η, ε], ((x_{p})_{p}, (y_{p})_{p}) ⊂ K_{0} × K_{0} and ((u_{p})_{p}, (v_{p})_{p}) ⊂ H × H such that ∑_{p=1}^{q−1} h_{p} ≤ T and

\{ \begin{matrix}x_{p} = x_{p−1} + h_{p−1}u_{p−1}; \\ x_{p} ∈ K ∩ B (x_{p−1} + h_{p−1}v_{p−1} + ∑_{i=0}^{p−1} h_{i} ; ∫ ; ∑_{i=0}^{p−2} h_{i}f(s, x_{p−1})ds, λ + M + 1); \\ u_{p} ∈ (\frac{1}{h_{p}} ∑_{i=0}^{p} h_{i} ; ∫ ; ∑_{i=0}^{p−1} h_{i} f(s, x_{p})ds + F (x_{p}) + \frac{ε}{T} B); \\ v_{p} ∈ F (y_{p}). \end{matrix}

Since h_{i} ≥ η > 0 there exists an integer s such that

∑_{i=0}^{s−1} h_{i} < T ≤ ∑_{i=0}^{s} h_{i}.

Then we have constructed the sequences (h_{p})_{p} ⊂ [η, ε], ((x_{p})_{p}, (y_{p})_{p}) ⊂ K_{0} × K_{0} and ((u_{p})_{p}, (v_{p})_{p}) ⊂ H × H such that for every p = 1, …, s, we have

(i) x_{p} = x_{p−1} + h_{p−1}u_{p−1};

(ii) x_{p} ∈ K ∩ B (x_{p−1} + h_{p−1}v_{p−1} + ∑_{i=0}^{p−1} h_{i} ; ∫ ; ∑_{i=0}^{p−2} h_{i}f(s, x_{p−1})ds, λ + M + 1);

(iii) u_{p} ∈ F (x_{p}) + \frac{1}{h_{p}} ∑_{i=0}^{p} h_{i} ; ∫ ; ∑_{i=0}^{p−1} h_{i} f(s, x_{p})ds + \frac{ε}{T} B;

(iv) v_{p} ∈ F (y_{p}).

By induction, for all p = 1, …, s we have

x_{p} = x_{0} + ∑_{i=0}^{p−1} h_{i}u_{i}.

Moreover by (iii), (H2), (3.1), (3.2) and because ∑_{i=0}^{p−1} h_{i} < T, we have

∥x_{p} − x_{0}∥ = ∥∑_{i=0}^{p−1} h_{i}u_{i}∥ ≤ ∑_{i=0}^{p−1} h_{i}∥u_{i}∥ ≤ ∑_{i=0}^{p−1} h_{i}(λ + 1 + M) < r,  (3.3)

hence x_{p} ∈ K_{0}.

For any nonzero integ k and for every integer q = 0, …, s − 1 denote by h_{q}^{k} a real associated to ε = \frac{1}{k} and x = x_{q} given by Lemma 3.1. Consider the sequence (τ_{k}^{q})_{k} defined as the following

\{ \begin{matrix}τ_{k}^{0} = 0, τ_{k}^{s+1} = T; \\ τ_{k}^{q} = h_{0}^{k} + … + h_{q−1}^{k} & if 1 ≤ q ≤ s, \end{matrix}

and define on [0, T] the sequence of functions (x_{k}(.))_{k} by

\{ \begin{matrix}x_{k}(t) = x_{q−1} + (t − τ_{k}^{q−1}) u_{q−1}, & ∀t ∈ [τ_{k}^{q−1}, τ_{k}^{q}]; \\ x_{k}(0) = x_{0}. \end{matrix}

Step 2. Convergence of approximate solutions. By definition of x_{k}(.), for all t ∈ [τ_{k}^{q−1}, τ_{k}^{q}] we have ẋ_{k}(t) = u_{q−1}. By (iii), (H2), (3.1), for a. e. t ∈ [0, T], we have

∥ ẋ_{k}(t) ∥ ≤ λ + 1 + M.

On the other hand, by (ii), (iv), (H2), (3.1) and (3.3) we have

∥x_{q}∥ ≤ ∥x_{q} − (x_{q−1} + h_{q−1}^{k}v_{q−1} + ∫_{τ_{k}^{q−1}}^{τ_{k}^{q}} f(s, x_{q−1})ds)∥

+ ∥x_{q−1} + h_{q−1}^{k}v_{q−1} + ∫_{τ_{k}^{q−1}}^{τ_{k}^{q}} f(s, x_{q−1})ds∥

≤ λ + M + 1 + ∥x_{0} − (x_{0} − x_{q−1}) + h_{q−1}^{k}v_{q−1} + ∫_{τ_{k}^{q−1}}^{τ_{k}^{q}} f(s, x_{q−1})ds∥
≤ λ + M + 1 + ∥x_{0}∥ + ∥x_{0} − x_{q−1}∥ + h_{q−1}^{k}∥v_{q−1}∥ + h_{q−1}^{k}M
≤ λ + M + 1 + ∥x_{0}∥ + r + λ + M
< 2(λ + M + 1) + ∥x_{0}∥ + r = R.

Then x_{q} ∈ K_{0} ∩ B̄(0, R) = K_{1}. By construction, for all t ∈ [τ_{k}^{q−1}, τ_{k}^{q}] we have

x_{k}(t) = x_{q−1} + (t − τ_{k}^{q−1})u_{q−1} = x_{q−1} + \frac{(t − τ_{k}^{q−1})}{h_{q−1}^{k}}(x_{q} − x_{q−1}).

Also since 0 ≤ t − τ_{k}^{q−1} ≤ τ_{k}^{q} − τ_{k}^{q−1} = h_{q−1}^{k}, we have

0 ≤ \frac{(t − τ_{k}^{q−1})}{h_{q−1}^{k}} ≤ 1.

Then

\frac{(t − τ_{k}^{q−1})}{h_{q−1}^{k}}(x_{q} − x_{q−1}) ∈ c̄o\{\{0\} ∪ (K_{1} − K_{0})\},

hence x_{k}(t) ∈ K_{0} + c̄o{{0} ∪ (K_{1} − K0)} which is compact. Therefore, we can select a subsequence, again denoted by (x_{k}(.))_{k} which converges uniformly to an absolutely continuous function x(.) on [0, T], moreover ẋ_{k}(.) converges weakly to ẋ(.) in L^{2}([0, T], H). The family of approximate solution x_{k}(.) satisfies the following

property.

Proposition 3.2. For every t ∈ [0, T], there exists q ∈ {1, …, s + 1} such that

lim_{k→+∞} d_{grF} ((x_{k}(t), ẋ_{k}(t) − \frac{1}{h_{q−1}^{k}} ∫_{τ_{k}^{q−1}}^{τ_{k}^{q}} f(s, x_{q−1})ds)) = 0.

Proof. Let t ∈ [0, T], there exists q ∈ {1, …, s + 1} such that t ∈ [τ_{k}^{q−1}, τ_{k}^{q}] and lim_{k→+∞} τ_{k}^{q−1} = t. Since

ẋ_{k}(t) − \frac{1}{h_{q−1}^{k}} ∫_{τ_{k}^{q−1}}^{τ_{k}^{q}} f(s, x_{q−1})ds ∈ F (x_{q−1}) + \frac{1}{kT} B,  (3.4)

we have

d_{grF} ((x_{k}(t), ẋ_{k}(t) − \frac{1}{h_{q−1}^{k}} ∫_{τ_{k}^{q−1}}^{τ_{k}^{q}} f(s, x_{q−1})ds)) ≤ ∥x_{k}(t) − x_{k}(τ_{k}^{q−1})∥ + \frac{1}{kT} ,

hence

lim_{k→+∞} d_{grF} ((x_{k}(t), ẋ_{k}(t) − \frac{1}{h_{q−1}^{k}} ∫_{τ_{k}^{q−1}}^{τ_{k}^{q}} f(s, x_{q−1})ds)) = 0.

□

Claim 3.3.

lim_{k→+∞} \frac{1}{h_{q−1}^{k}} ∫_{τ_{k}^{q−1}}^{τ_{k}^{q}} f(s, x_{q−1})ds = f(t, x(t)).

Proof. Fix any t ∈ [0, T], there exists q ∈ {1, …, s + 1} such that t ∈ [τ_{k}^{q−1}, τ_{k}^{q}], lim_{k→+∞} τ_{k}^{q−1} = lim_{k→+∞} τ_{k}^{q} = t and lim_{k→+∞} x_{k}(τ_{k}^{q−1}) = x(t). Put

G(t, y) = ∫_{0}^{t} f(s, y)ds.

Note that the function G is differentiable on t and

\frac{dG}{dt}(t, y) = f(t, y).

EJQTDE, 2007 No. 7, p. 8 We have

∥\frac{G(τ_{k}^{q}, x(t)) − G(τ_{k}^{q−1}, x(t))}{τ_{k}^{q} − τ_{k}^{q−1}} − f(t, x(t))∥

≤ ∥\frac{G(τ_{k}^{q}, x(t)) − G(τ_{k}^{q−1}, x(t))}{τ_{k}^{q} − τ_{k}^{q−1}} − \frac{G(t, x(t)) − G(τ_{k}^{q−1}, x(t))}{t − τ_{k}^{q−1}} ∥

+ ∥\frac{G(t, x(t)) − G(τ_{k}^{q−1}, x(t))}{t − τ_{k}^{q−1}} − f(t, x(t))∥ .

On the other hand

∥\frac{G(τ_{k}^{q}, x(t)) − G(τ_{k}^{q−1}, x(t))}{τ_{k}^{q} − τ_{k}^{q−1}} − \frac{G(t, x(t)) − G(τ_{k}^{q−1}, x(t))}{t − τ_{k}^{q−1}} ∥

= ∥\frac{τ_{k}^{q} − t}{τ_{k}^{q} − τ_{k}^{q−1}} (\frac{G(τ_{k}^{q}, x(t)) − G(t, x(t))}{τ_{k}^{q} − t} − \frac{G(τ_{k}^{q−1}, x(t)) − G(t, x(t))}{τ_{k}^{q−1} − t})∥

≤ ∥\frac{G(τ_{k}^{q}, x(t)) − G(t, x(t))}{τ_{k}^{q} − t} − f(t, x)∥

+ ∥\frac{G(τ_{k}^{q−1}, x(t)) − G(t, x(t))}{τ_{k}^{q−1} − t} − f(t, x(t))∥ .

Hence

∥\frac{G(τ_{k}^{q}, x(t)) − G(τ_{k}^{q−1}, x(t))}{τ_{k}^{q} − τ_{k}^{q−1}} − f(t, x(t))∥

≤ ∥\frac{G(τ_{k}^{q}, x(t)) − G(t, x(t))}{τ_{k}^{q} − t} − f(t, x(t))∥

+2 ∥\frac{G(t, x(t)) − G(τ_{k}^{q−1}, x(t))}{t − τ_{k}^{q−1}} − f(t, x(t))∥ .

As

lim_{k→+∞} \frac{G(τ_{k}^{q}, x(t)) − G(t, x(t))}{τ_{k}^{q} − t} = \frac{dG}{dt}(t, x(t)) = f(t, x(t))

and

lim_{k→+∞} \frac{G(τ_{k}^{q−1}, x(t)) − G(t, x(t))}{τ_{k}^{q−1} − t} = \frac{dG}{dt}(t, x(t)) = f(t, x(t)),

we have

lim_{k→+∞} \frac{G(τ_{k}^{q}, x(t)) − G(τ_{k}^{q−1}, x(t))}{τ_{k}^{q} − τ_{k}^{q−1}} = f(t, x(t)).  (3.5)

Put

ρ_{k} = ∥\frac{G(τ_{k}^{q}, x(t)) − G(τ_{k}^{q−1}, x(t))}{τ_{k}^{q} − τ_{k}^{q−1}} − f(t, x(t))∥ .

On the other hand we have

∥\frac{1}{h_{q−1}^{k}} ∫_{τ_{k}^{q−1}}^{τ_{k}^{q}} f(s, x_{k}(τ_{k}^{q−1}))ds − f(t, x(t))∥

= ∥\frac{G(τ_{k}^{q}, x_{k}(τ_{k}^{q−1})) − G(τ_{k}^{q−1}, x_{k}(τ_{k}^{q−1}))}{τ_{k}^{q} − τ_{k}^{q−1}} − f(t, x(t))∥
≤ ∥\frac{G(τ_{k}^{q}, x_{k}(τ_{k}^{q−1})) − G(τ_{k}^{q−1}, x_{k}(τ_{k}^{q−1}))}{τ_{k}^{q} − τ_{k}^{q−1}} − \frac{G(τ_{k}^{q}, x(t)) − G(τ_{k}^{q−1}, x(t))}{τ_{k}^{q} − τ_{k}^{q−1}} ∥ + ρ_{k}
= ∥\frac{1}{τ_{k}^{q} − τ_{k}^{q−1}} ∫_{τ_{k}^{q−1}}^{τ_{k}^{q}}(f(s, x_{k}(τ_{k}^{q−1})) − f(s, x(t)))ds∥ + ρ_{k}.

Since the family {f(s, ·) : s ∈ ℝ} is equicontinuous, then there exists k_{0} such that

∥f(s, x_{k}(τ_{k}^{q−1})) − f(s, x(t))∥ ≤ \frac{1}{k} for all k ≥ k_{0} and for all s ∈ ℝ,

consequently we have for k ≥ k_{0}

∥\frac{G(τ_{k}^{q}, x_{k}(τ_{k}^{q−1})) − G(τ_{k}^{q−1}, x_{k}(τ_{k}^{q−1}))}{τ_{k}^{q} − τ_{k}^{q−1}} − f(t, x(t))∥ ≤ \frac{1}{k} + ρ_{k}.

By (3.5), the last term converges to 0. This completes the proof of the Claim.

□

The function x(.) has the following property Proposition 3.4. For all t ∈ [0, T], we have ẋ(t) − f(t, x(t)) ∈ ∂_{c}V (x(t)). Proof. The weak convergence of ẋ_{k}(.) to ẋ(.) in L^{2}([0, T], H) and the Mazur’s Lemma entail

ẋ(t) ∈ ⋂_{k} c̄o{ẋ_{m}(t) : m ≥ k}, for a.e. on [0, T].

Fix any t ∈ [0, T], there exists q ∈ {1, …, s + 1} such that t ∈ [τ_{k}^{q−1}, τ_{k}^{q}] and lim_{k→+∞} τ_{k}^{q−1} = t. Then for all y ∈ H

< y, ẋ(t) >≤ inf_{m} sup_{k≥m} < y, ẋ_{k}(t) > .

Since F (x) ⊂ ∂_{c}V (x), then by (3.4), one has

ẋ_{k}(t) ∈ ∂_{c}V (x_{q−1}) + \frac{1}{h_{q−1}^{k}} ∫_{τ_{k}^{q−1}}^{τ_{k}^{q}} f(s, x_{q−1})ds + \frac{1}{kT} B.

Thus for all m

< y, ẋ(t) >≤ sup_{k≥m} σ (y, ∂_{c}V (x_{q−1}) + \frac{1}{h_{q−1}^{k}} ∫_{τ_{k}^{q−1}}^{τ_{k}^{q}} f(s, x_{q−1})ds + \frac{1}{kT} B) ,

from which we deduce that

< y, ẋ(t) >≤ lim sup_{k→+∞} σ (y, ∂_{c}V (x_{q−1}) + \frac{1}{h_{q−1}^{k}} ∫_{τ_{k}^{q−1}}^{τ_{k}^{q}} f(s, x_{q−1})ds + \frac{1}{kT} B) .

EJQTDE, 2007 No. 7, p. 10 By Proposition 2.3, the function x ↦ σ(y, ∂_{c}V (x)) is u.s.c and hence we get

< y, ẋ(t) >≤ σ(y, ∂_{c}V (x(t)) + f(t, x(t))).

So, the convexity and the closedness of the set ∂_{c}V (x(t)) ensure

ẋ(t) − f(t, x(t)) ∈ ∂_{c}V (x(t)).
□

Proposition 3.5. The application x(.) is a solution of the problem (2.1).

Proof. As x(.) is an absolutely continuous function and V is uniformly regular locally Lipschitz function over K (hence directionally regular over K (see [5])), by Theorem 2 in Valadier [10, 11] and by Proposition 3.4, we obtain

\frac{d}{dt}V (x(t)) = < ẋ(t), ẋ(t) − f(t, x(t)) > a. e. on [0, T],

therefore,

V (x(T)) − V (x_{0}) = ∫_{0}^{T} ∥ ẋ(s) ∥^{2} ds − ∫_{0}^{T} < ẋ(s), f(s, x(s)) > ds.  (3.6)

On the other hand, by construction, for all q = 1, …, s + 1, we have

ẋ_{k}(t) − \frac{1}{h_{q−1}^{k}} ∫_{τ_{k}^{q−1}}^{τ_{k}^{q}} f(s, x_{q−1})ds ∈ ∂_{c}V (x_{q−1}) + \frac{1}{kT} B.

Let b_{q} such that

ẋ_{k}(t) − \frac{1}{h_{q−1}^{k}} ∫_{τ_{k}^{q−1}}^{τ_{k}^{q}} f(s, x_{q−1})ds + \frac{1}{kT} b_{q} ∈ ∂_{c}V (x_{q−1}).

Since V is β−uniformly regular over K, we have

V (x_{k}(τ_{k}^{q})) − V (x_{k}(τ_{k}^{q−1})) ≥ < x_{k}(τ_{k}^{q}) − x_{k}(τ_{k}^{q−1}), ẋ_{k}(t)

−\frac{1}{h_{q−1}^{k}} ∫_{τ_{k}^{q−1}}^{τ_{k}^{q}} f(s, x_{q−1})ds + \frac{1}{kT} b_{q} >

−β ∥x_{k}(τ_{k}^{q}) − x_{k}(τ_{k}^{q−1})∥^{2}

= < ∫_{τ_{k}^{q−1}}^{τ_{k}^{q}} ẋ_{k}(s)ds, ẋ_{k}(t)

−\frac{1}{h_{q−1}^{k}} ∫_{τ_{k}^{q−1}}^{τ_{k}^{q}} f(s, x_{q−1})ds + \frac{1}{kT} b_{q} >

−β ∥x_{k}(τ_{k}^{q}) − x_{k}(τ_{k}^{q−1})∥^{2}

= ∫_{τ_{k}^{q−1}}^{τ_{k}^{q}} < ẋ_{k}(s), ẋ_{k}(s) > ds

− ∫_{τ_{k}^{q−1}}^{τ_{k}^{q}} < ẋ_{k}(s), \frac{1}{h_{q−1}^{k}} ∫_{τ_{k}^{q−1}}^{τ_{k}^{q}} f(τ, x_{q−1})dτ > ds

+\frac{1}{kT} ∫_{τ_{k}^{q−1}}^{τ_{k}^{q}} < ẋ_{k}(s), b_{q} > ds

−β ∥x_{k}(τ_{k}^{q}) − x_{k}(τ_{k}^{q−1})∥^{2} .

By adding, we obtain

\begin{aligned}V (x_{k}(T)) − V (x_{0}) & ≥ ∫_{0}^{T} ∥ ẋ_{k}(s) ∥^{2} ds \\ & − ∑_{q=1}^{s+1} ∫_{τ_{k}^{q−1}}^{τ_{k}^{q}} < ẋ_{k}(s), \frac{1}{h_{q−1}^{k}} ∫_{τ_{k}^{q−1}}^{τ_{k}^{q}} f(τ, x_{q−1})dτ > ds \\ & +\frac{1}{kT} ∑_{q=1}^{s+1} ∫_{τ_{k}^{q−1}}^{τ_{k}^{q}} < ẋ_{k}(s), b_{q} > ds \\ & − ∑_{q=1}^{s+1} β ∥ x_{k}(τ_{k}^{q}) − x_{k}(τ_{k}^{q−1}) ∥^{2} . \end{aligned} (3.7)

Claim 3.6.

lim_{k→+∞} ∑_{q=1}^{s+1} ∫_{τ_{k}^{q−1}}^{τ_{k}^{q}} < ẋ_{k}(s), \frac{1}{h_{q−1}^{k}} ∫_{τ_{k}^{q−1}}^{τ_{k}^{q}} f(τ, x_{q−1})dτ > ds = ∫_{0}^{T} < ẋ(s), f(s, x(s)) > ds.

Proof. We have

∥∑_{q=1}^{s+1} ∫_{τ_{k}^{q−1}}^{τ_{k}^{q}} < ẋ_{k}(s), \frac{1}{h_{q−1}^{k}} ∫_{τ_{k}^{q−1}}^{τ_{k}^{q}} f(τ, x_{q−1})dτ > ds − ∫_{0}^{T} < ẋ(s), f(s, x(s)) > ds∥

= ∥∑_{q=1}^{s+1} ∫_{τ_{k}^{q−1}}^{τ_{k}^{q}}(< ẋ_{k}(s), \frac{1}{h_{q−1}^{k}} ∫_{τ_{k}^{q−1}}^{τ_{k}^{q}} f(τ, x_{q−1})dτ > − < ẋ(s), f(s, x(s)) >)ds∥
≤ ∥∑_{q=1}^{s+1} ∫_{τ_{k}^{q−1}}^{τ_{k}^{q}}(< ẋ_{k}(s), \frac{1}{h_{q−1}^{k}} ∫_{τ_{k}^{q−1}}^{τ_{k}^{q}} f(τ, x_{q−1})dτ > − < ẋ_{k}(s), f(s, x(s)) >)ds∥

+ ∥∑_{q=1}^{s+1} ∫_{τ_{k}^{q−1}}^{τ_{k}^{q}}(< ẋ_{k}(s), f(s, x(s)) > − < ẋ(s), f(s, x(s)) >)ds∥

≤ ∑_{q=1}^{s+1} ∫_{τ_{k}^{q−1}}^{τ_{k}^{q}} ∥< ẋ_{k}(s), \frac{1}{h_{q−1}^{k}} ∫_{τ_{k}^{q−1}}^{τ_{k}^{q}} f(τ, x_{q−1})dτ > − < ẋ_{k}(s), f(s, x(s)) >∥ ds

+ ∥∫_{0}^{T} (< ẋ_{k}(s), f(s, x(s)) > − < ẋ(s), f(s, x(s)) >)ds∥ . Since

∥ẋ_{k}(t)∥ ≤ λ + M + 1, lim_{k→+∞} \frac{1}{h_{q−1}^{k}} ∫_{τ_{k}^{q−1}}^{τ_{k}^{q}} f(s, x_{q−1})ds = f(t, x(t))

and ẋ_{k}(.) converges weakly to ẋ(.), the last term converges to 0. This completes the proof of the Claim. □ Claim 3.7.

lim_{k→+∞} ∑_{q=1}^{s+1} β ∥ x_{k}(τ_{k}^{q}) − x_{k}(τ_{k}^{q−1}) ∥^{2}= 0.

EJQTDE, 2007 No. 7, p. 12 Proof. By construction we have

∥ x_{k}(τ_{k}^{q}) − x_{k}(τ_{k}^{q−1}) ∥ = ∥ (τ_{k}^{q} − τ_{k}^{q−1})u_{q−1} ∥
≤ (τ_{k}^{q} − τ_{k}^{q−1}) ∥ u_{q−1} ∥
≤ (τ_{k}^{q} − τ_{k}^{q−1})(λ + 1 + M).

Hence

∥ x_{k}(τ_{k}^{q}) − x_{k}(τ_{k}^{q−1}) ∥^{2} ≤ (τ_{k}^{q} − τ_{k}^{q−1})^{2}(λ + 1 + M)^{2}
≤ (τ_{k}^{q} − τ_{k}^{q−1})h_{q−1}^{k}(λ + 1 + M)^{2}
≤ (τ_{k}^{q} − τ_{k}^{q−1})\frac{1}{k} (λ + 1 + M)^{2}.

Then

∑_{q=1}^{s+1} β ∥ x_{k}(τ_{k}^{q}) − x_{k}(τ_{k}^{q−1}) ∥^{2}≤ \frac{βT (λ + 1 + M)^{2}}{k} ,

hence

lim_{k→+∞} ∑_{q=1}^{s+1} β ∥ x_{k}(τ_{k}^{q}) − x_{k}(τ_{k}^{q−1}) ∥^{2}= 0.

□

Note that

lim_{k→+∞} \frac{1}{kT} ∑_{q=1}^{s+1} ∫_{τ_{k}^{q−1}}^{τ_{k}^{q}} < ẋ_{k}(s), b_{q} > ds = 0.

By passing to the limit for k → ∞ in (3.7) and using the continuity of the function V on the ball B(x_{0}, r), we obtain

V (x(T)) − V (x_{0}) ≥ lim sup_{k→+∞} ∫_{0}^{T} ∥ ẋ_{k}(s) ∥^{2} ds − ∫_{0}^{T} < ẋ(s), f(s, x(s)) > ds.

Moreover, by (3.6), we have

∥ ẋ ∥_{2}^{2}≥ lim sup_{k→+∞} ∥ ẋ_{k} ∥_{2}^{2}

and by the weak l.s.c of the norm ensures

∥ ẋ ∥_{2}^{2}≤ lim inf_{k→+∞} ∥ ẋ_{k} ∥_{2}^{2} .

Hence we get

∥ ẋ ∥_{2}^{2}= lim_{k→+∞} ∥ ẋ_{k} ∥_{2}^{2} .

Finally, there exists a subsequence of (ẋ_{k}(.))_{k} (still denoted (ẋ_{k}(.))_{k}) converges pointwisely to ẋ(.). In view of Proposition (3.2), we conclude that

d_{grF} ((x(t), ẋ(t) − f(t, x(t)))) = 0

and as F has a closed graph, we obtain

ẋ(t) ∈ f(t, x(t)) + F (x(t)) a.e on [0, T].

EJQTDE, 2007 No. 7, p. 13 Now, let t ∈ [0, T], there exists q ∈ {1, …, s + 1} such that t ∈ [τ_{k}^{q−1}, τ_{k}^{q}] and

lim_{k→+∞} τ_{k}^{q−1} = t. Since

lim_{k→+∞} ∥ x(t) − x_{k}(τ_{k}^{q−1}) ∥= 0,

x_{k}(τ_{k}^{q−1}) ∈ K_{0} and K_{0} is closed we obtain x(t) ∈ K_{0} ⊂ K. The proof is complete.

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(Received November 27, 2006)