PERIODIC SOLUTIONS OF A CLASS OF NON-AUTONOMOUS SECOND-ORDER DIFFERENTIAL INCLUSIONS SYSTEMS

Using an abstract framework due to Clarke (1999), we prove the 
existence of periodic solutions for second-order differential inclusions systems.

Wu and Tang in [4] proved the existence of solutions for problem (1.1) when F = F 1 + F 2 and F 1 , F 2 satisfy some assumptions.Now we will consider problem (1.1) in a more general sense.More precisely, our results represent the extensions to systems with discontinuity (we consider the generalized gradients unlike continuously gradient in classical results).
152 Periodic solutions

Main results
Consider the second-order differential inclusions systems ü(t) ∈ ∂F t, u(t) a.e.t ∈ [0, T ], where T > 0, F : [0, T ] × R n → R and ∂ denotes the Clarke subdifferential.We suppose that F = F 1 + F 2 and F 1 , F 2 satisfy the following assumption: (A ) F 1 , F 2 are measurable in t for each x ∈ R n , at least F 1 or F 2 are strictly differentiable in x and there exist where F 1 , F 2 satisfy assumption (A ) and the following conditions: for all x ∈ R n and a.e.t ∈ [0, T ]; Then problem (2.1) Daniel Paşca 153 for all x ∈ R n and all t ∈ [0, T ], and (2.9) (2.10) (2.11) Then problem (2.1) has at least one solution which minimizes ϕ on H 1 T .

Preliminary results
We introduce some functional spaces.Let [0, T ] be a fixed real interval (0 < T < ∞) and 1 < p < ∞.We denote by W We denote by H 1 T the Hilbert space W 1,2 T .We recall that For our aims, it is necessary to recall some very well-known results (for proof and details see [2]): Let X be a Banach space.Now, following [1], for each x, v ∈ X, we define the generalized directional derivative at x in the direction v of a given f ∈ Lip loc (X, R) as and denote x by the generalized gradient of f at x (the Clarke subdifferential).We recall the Lebourg's mean value theorem (see [1,Theorem 2.3.7]).Let x and y be points in X, and suppose that f is Lipschitz on an open set containing the line segment [x, y].Then there exists a point u in (x, y) such that Clarke considered in [1] the following abstract framework: • let (T ,-, µ) be a positive complete measure space with µ(T ) < ∞, and let Y be a separable Banach space; where f t : Y → R, (t ∈ T ) is a given family of functions; • we suppose that for each y in Y the function t → f t (y) is measurable, and that x is a point at which f (x) is defined (finitely).

Hypothesis 3.3.
There is a function k in L q (T , R), (1/p + 1/q = 1) such that, for all t ∈ T , Then, the functional f : Z ∈ R, where (3.17) For f 1 we can apply Theorem 3.5 under Hypothesis 3.3, with the following cast of characters: • (T , -, µ) = [0, T ] with Lebesgue measure, Y = R n × R n is the Hilbert product space (hence is separable); • p = 2 and ; in our assumptions it results that the integrand L 1 (t, x, y) is measurable in t for a given element (x, y) of Y and there exists k ∈ L 2 (0, T ; R) such that Moreover, Corollary 1 of Proposition 2.3.3 in [1] implies that, if at least one of the functions Remark 3.8.The interpretation of expression (3.25) is that if (u 0 , v 0 ) is an element of Z (so that v 0 = u0 ) and if ζ ∈ ∂f (u 0 , v 0 ), we deduce the existence of a measurable function (q(t), p(t)) such that and for any (u, v) in Z, one has ]dt), it then follows easily that q(t) = ṗ(t) a.e., or taking into account (3.26) so that u 0 satisfies the inclusions system (2.1).

Proofs of the theorems
Proof of Theorem 2.1.From assumption (A ) it follows immediately that there exist a for all x ∈ R n and all t ∈ [0, T ].Like, in [4], we obtain for all x ∈ R n and all t ∈ [0, T ], where β < 2 and a 0 = max 0≤s≤1 a(s).
for all u ∈ H 1 T and some positive constants C 1 , C 2 , and C 3 .Hence we have for all u ∈ H 1 T , which implies that ϕ(u) → ∞ as u → ∞ by (2.4) because α < 1, β < 2, and the norm u = ( ū 2 + u 2 L 2 ) 1/2 is an equivalent norm on H 1  T .Now we write ϕ(u) = ϕ 1 (u) + ϕ 2 (u) where The function ϕ 1 is weakly lower semi-continuous (w.l.s.c.) on H 1 T .From (i), (ii), and Theorem 3.5, taking into account Remark 3.6 and Proposition 3.2, it follows that ϕ 2 is w.l.s.c. on H 1 T .By [2, Theorem 1.1], it follows that ϕ has a minimum u 0 on H 1 T .Evidently, Z H 1 T and ϕ(u) = f (u, v) for all (u, v) ∈ Z. From Theorem 3.7, it results that f is uniformly Lipschitz on bounded subsets of Z, and therefore ϕ possesses the same properties relative to H 1 T .Proposition 2.3.2 in [1] implies that 0 ∈ ∂ϕ(u 0 ) (so that u 0 is a critical point for ϕ).Now from Theorem 3.7 and Remark 3.8 it follows that problem (2.1) has at least one solution u ∈ H 1 T .
Proof of Theorem 2.3.Let (u k ) be a minimizing sequence of ϕ.It follows from (iv), (v), Lebourg's mean value theorem, and Sobolev inequality, that for all u ∈ H 1 T and some positive constants C 1 , C 3 , and C 4 .Now it follows like in the proof of Theorem 2.1 that ϕ is coercive by (ix), which completes the proof.
.10) Hypothesis 3.4.Each function f t is Lipschitz (of some rank) near each point of Y , and for some constant c, for all t ∈ T , y ∈ Y , one has ζ ∈ ∂f t (y) ⇒ ζ Y * ≤ c 1 + y Theorem 3.5.Under either of Hypotheses 3.3 or 3.4, f is uniformly Lipschitz on bounded subsets of Z, and there is ∂f (x) ⊂ T ∂f t x(t) µ(dt).(3.12)Further, if each f t is regular at x(t) then f is regular at x and equality holds.Remark 3.6.The function f is globally Lipschitz on Z when Hypothesis 3.3 holds.Now we can prove the following result.Theorem 3.7.Let F : [0, T ] × R n → R such that F = F 1 + F 2 where F 1 , F 2 are measurable in t for each x ∈ R n , and there exist k 1 [1]follows that f = f 1 + f 2 is uniformly Lipschitz on the bounded subsets of Z.Propositions 2.3.3 and 2.3.15in[1]imply that .6) for all k and some constants c 2 , c 3 , which implies that ( ũk ) is bounded.On the other hand, in a way similar to the proof of Theorem 2.1, one hasL 2 + C 1 u L 2 , (4.7)for all k and some positive constant C 1 , which implies that and some positive constant C 1 .It follows from (vi) and the boundedness of ( ũk ) that ( ūk ) is bounded.Hence ϕ has a bounded minimizing sequence (u k ).This completes the proof.Proof of Theorem 2.4.From (vii), (3.26), and Sobolev's inequality it follows that T 0 F 2 t, u(t) − F 2 (t, ū) dt ≤ 1 4 u 2 T 0 h(t), u(t) dt + T 0 γ (t)dt + T 0 F 2 (t, ū)dt + T 0 F 2 t, u(t) − F 2 (t, ū) dt