SUBHARMONIC AND MULTIPLE PERIODIC SOLUTIONS FOR HAMILTONIAN SYSTEMS WITH LOCAL PARTIAL SUBLINEAR NONLINEARITY

— The existence of subharmonic and multiple periodic solutions as well as the minimality of periods are obtained for the nonautonomous Hamiltonian systems ẋ = JH ′(t, x) with locally and partially sublinear Hamiltonian H; that is, there exist a decomposition IR = A ⊕ B of IR , an α ∈]0, 1[ and two periodic functions a ∈ L 2 1−α ([0, T ], IR) and b ∈ L([0, T ], IR) such that |H ′(t, u+ v)| ≤ a(t) |v| + b(t) for all (t, u, v) ∈ [0, T ] × A × B and H(t,u+v) |v|2α −→ +∞ or − ∞ as |v| −→ ∞ in B, uniformly in u ∈ A for a.e. t in some non empty open subset C of [0, T ]. For the resolution we use an analogy of Egorov’s theorem and a generalized saddle point theorem.


Introduction
Let H : R × R 2N → R, (t, x) → H(t, x) be a continuous function Tperiodic (T > 0) in the first variable, differentiable in the second variable, and H ′ (t, x) = ∂H ∂x (t, x) is continuous. Consider the Hamiltonian system of ordinary differential equations: where x ∈ C 1 (R, R 2N ), J is the standard symplectic matrix: I N being the identity matrix of order N . It has been proved that the system (H) has subharmonic solutions by using many different techniques, for example Morse theory, minimax theory. Many solvability conditions are given, such as the coercivity condition (see [3,6,7,10]), the convexity condition (see [2,9,11]), the boundedness condition (see [8]), the sublinear condition (see [1,8]). Specially, under the conditions (1) There exist α ∈]0, 1[ and two constants a, b > 0 such that H ′ (t, x) ≤ a |x| α + b, ∀x ∈ R 2N , a.e. t ∈ R.
(2) lim |x|→∞ H(t, x) |x| 2α → +∞ or − ∞, uniformly in t ∈ R, Silva [8] proved the existence of subharmonics for problem (H) (see theorem 1.2 in [8]). Recently, Daouas and Timoumi [1] generalized the result mentioned above. In this paper, we suppose that the nonlinearity H ′ (t, x) is partially sublinear; that is, there exists a decomposition R 2N = A ⊕ B of R 2N such that H is periodic in A and there exist 0 < α < 1 and two periodic functions a ∈ L Under these conditions, we obtain some existence of subharmonics and multiple periodic solutions for the system (H). Furthermore, we study the minimality of periods. For the resolution, we use an analogy of Egorov's theorem, the Least Action Principle and a Generalized Saddle Point Theorem [4].

Preliminaries
We will recall a minimax theorem: "Generalized Saddle Point Theorem [4]", which will be useful in the proof of our results.
Given a Banach space E and a complete connected Finsler manifold V of class C 2 , we consider the space X = E × V . Let E = W ⊕ Z (topological direct sum) and E n ⊕ Z n be a sequence of closed subspaces with Z n ⊂ Z, For f ∈ C 1 (X, R), we denote by f n = f |X n . Then we have f n ∈ C 1 (X n , R), for all n ≥ 1.
. The function f satisfies the Palais-Smale condition with respect to (X n ) at a level c ∈ R if every sequence (x n ) satisfying x n ∈ X n , f n (x n ) → c, f ′ n (x n ) → 0 has a subsequence which converges in X to a critical point of f . The above property will be referred as the (P S) * c condition with respect to (X n ).
Periodic solutions for Hamiltonian systems 355 Definition 2.2. Let Y be a closed subset of a space X. The cuplength of X relative to Y is the largest integer n such that there exist α 0 ∈ H * (X, Y ), * ≥ 0 and α 1 , . . . , α n ∈ H * (X), * ≥ 1 with We write then cuplenth(X,Y)=n. We set cuplength(X, Y ) = −∞ if no such integer exists. Here H * denotes the singular cohomology over the real field R.
Theorem 2.1 (Generalized saddle point theorem). Assume that there exist r > 0 and α < β ≤ γ such that Then f −1 ([β, γ]) contains at least cuplength (V ) + 1 critical points of f . Now, for giving a variational formulation of (H), some preliminary materials on functional spaces and norms are needed.
Consider the Hilbert space where ., . inside the sign integral is the inner product in R 2N . If x ∈ E, then x has a Fourier expansion wherex m ∈ R 2N and m∈Z (1 + |m|)|x m | 2 < ∞. By an easy calculation, we obtain Therefore Q is a continuous quadratic form on E.
Consider the subspaces of E: T mtJ x m a.e. ,

M. Timoumi
Then E = E 0 ⊕ E + ⊕ E − . It is not difficult to verify that E 0 , E − , E + are respectively the subspaces of E on which Q is null, negative definite and positive definite, and these subspaces are orthogonal with respect to the bilinear form: B(x, y) = 1 2 T 0 < Jẋ, y > dt, x, y ∈ E associated with Q. If x ∈ E + and y ∈ E − , then B(x, y) = 0 and Q(x + y) = Q(x) + Q(y). It is also easy to check that E 0 , E − and E + are mutually orthogonal in is an equivalent norm in E.
In particular there is a constant α > 0 such that for all x ∈ E.

Consider a decomposition
where 0 ≤ p ≤ 2N − 1 and (e i ) 1≤i≤2N is the standard basis of R 2N . Here, P A (resp. P B ) denotes the projection of R 2N into A (resp.B).
x) be a continuous function Tperiodic (T > 0) in the first variable, differentiable with respect to the second variable, and H ′ (t, x) = ∂H ∂x (t, x) is continuous. Consider the following assumptions: and Our first main result conserns the multiplicity of periodic solutions and is: hold. Then the Hamiltonian system possesses at least (p + 1) T -periodic solutions geometrically distinct.
If 0 ≤ p ≤ 2N − 2, we consider the assumption: Our second main result conserns the subharmonics and is: , then for all integer k ≥ 1, the system (H) possesses at least (p + 1) kT -periodic solutions Remark 3.1. The theorem 3.2 generalizes the theorem 1.2 in [8].
Now, consider the assumptions: Our third main result conserns the minimality of periods and is: Then for all integer k ≥ 1, the system (H) possesses at least (p + 1) kT -periodic solutions x 1 k , . . . , x p+1 k geometrically distinct such that for all i = 1, . . . , p + 1, lim k→∞ x i k ∞ = +∞. Moreover, for all i = 1, . . . , p + 1 and for any sufficiently large prime number k, kT is the minimal period of x i k .
be a continuous periodic function in (t, r), differentiable in r and ∂A ∂r (t, r) is continuous. Let 0 < α < 1, 0 < ǫ < 1 − α and consider the Hamiltonian: where a : R → R is a continuous periodic function not identically null and having a constant sign. Then the Hamiltonian H satisfies all the above assumptions.
Proof of Theorem 3.1. To begin let us observe the following The following two lemmas, which are analogeous to Egorov's lemma, will be needed in the proof of our results. The first lemma treats the sequence case and the second does the continuous function case. They deal with tending to +∞.
Lemma 3.1. Suppose that C is a non empty open subset of R with meas(C) < ∞, F is a given set and f n (t, u) is a sequence of functions defined in C ×F , continuous in t such that f n (t, u) → +∞ as n → +∞, uniformly in u ∈ F , for a.e. t ∈ C. Then, for any ρ > 0, there exists a subset C ρ of C with Proof. Without loss of generality, we may assume that f n (t, u) → +∞ as n → +∞, uniformly in u ∈ F , for all t ∈ C. For every r > 0 and every positive integer n, define Then C(n, r) is measurable and C(m, r) ⊂ C(n, r) if m < n. Hence we have because that f n (t, u) → +∞ as n → +∞, uniformly in u ∈ F , for all t ∈ C. By the properties of Lebesgue's measure one has meas(C) = lim n→∞ meas(C(n, r)) which implies that lim n→∞ meas(C − C(n, r)) = 0.
Hence for any ρ > 0 and for every integer i there exists n i ∈ N such that Then one has for all integer n ≥ 1, t ∈ [0, T ] and u ∈ A. By the continuity of f (t, u, v) in v for almost every t ∈ C and all u ∈ A one has for all n ≥ 1, almost every t ∈ C and all u ∈ A, which implies that f n (., u) is mesurable for all positive integer n and u ∈ A. Now the fact f n (t, u) → +∞ as n → ∞ uniformly in u ∈ A, for almost every t ∈ C follows from the same property of f (t, u, v). By lemma 3.1 there exists, for every ρ > 0, a subset C ρ with meas(C − C ρ ) < ρ such that f n (t, u) → +∞ as n → +∞ uniformly for all (t, u) ∈ C ρ × A, which implies the desired property of f (t, u, v).
Assume that 0 ≤ p ≤ 2N − 1. We are interested in the existence and multiplicity of periodic solutions to the system (H). By making the change of variables t → t k , the system (H) transforms to (H k )u = kJH ′ (kt, u).
Hence, to find kT -periodic solutions of (H), it suffises to find T -periodic solutions of (H k ).
Associate to the systems (H k ) the family of functionals (φ k ) defined on It is well known that every functional φ k is continuously differentiable in E and critical points of φ k on E correspond to the T -periodic solutions of the system (H k ), moreover one has which is nothing but the torus T p . Here, B(inZ) consists of constant Bvalued functions. We regard the functional φ k as defined on the space X = (W ⊕ Z) × V as follows To find critical points of φ k we will apply Theorem 2.1 to this functional with respect to the sequence of subspaces X n = E n × V , where 2π T mtJ û m a.e. , n ≥ 0.
Let us check the Palais-Smale condition. Proof. Let (u n , v n ) n∈N be a sequence such that for all n ∈ N, (u n , v n ) ∈ X n and We have the relation Since φ ′ k,n (u n + v n ) → 0 as n → ∞, there exists a constant c 1 > 0 such that (3.4) ∀n ∈ N, φ ′ k,n (u n + v n ).u + n ≤ c 1 u + n . By assumption (H 1 ) and Hölder's inequality, with p = 1 α , q = 1 1−α , we have  u − n ≤ c 2 P B (u n ) α + c 3 . We conclude from (3.6) and (3.7) that the sequence (u n ) is bounded if and only if the sequence (P B (u n )) is bounded. Assume that (P B (u n )) is not bounded, we can assume, by going to a subsequence if necessary, that P B (u n ) → ∞ as n → ∞. Since 0 < α < 1, we conclude by (3.6) and (3.7) that Therefore, we have (3.9) y n = u n P B (u n ) → y ∈ B, |y| = 1 as n → ∞.
It follows that Now, we apply the fact that (φ k (u n )) is bounded to get On the other hand, by mean value theorem, assumption (H 1 ) and Hölder's inequality, we have By considering (3.14) and Sobolev's embedding E ֒→ L 2 (0, T ; R 2N ), we can find a constant c 7 > 0 such that . After combining (3.11), (3.13) and (3.15), we get n | 2α dt ≤ c 8 for some positive constant c 8 . However, the condition (3.16) contradicts (H 2 )(i) because u 0 n → ∞ as n → ∞. Consequently, (u n ) is bounded in X. Going if necessary to a subsequence, we can assume that u n ⇀ u, u 0 n → u 0 and v n → v. Notice that condition for all c ∈ R. The lemma 3.3 is proved. Now, let us prove that for all k ≥ 1, the functional φ k satisfies the conditions a), b) and c) of Theorem 2.1. a) Let (u, v) ∈ W × V , with u = u − ∈ E − , we have by using mean value Theorem, assumption (H 1 ) and Proposition 2.2 where c 9 , c 10 , c 11 are three positive constants. So we have by using mean value theorem By assumption (H 1 ) and Proposition 2.2, we can find a constant c 12 > 0 such that Therefore, by using (3.20) and (3.21) we obtain Now let d > c 2 12 2 , then by assumption (H 2 )(i), there exists a constant e > 0 such that So by (3.22) and (3.23), we have Hence by lemma 3.3 and properties (3.19), (3.25), we deduce that, for all k ≥ 1, the functional φ k satisfies all the assumptions of Theorem 2.1. Therefore, for all integer k ≥ 1, the Hamiltonian system (H k ) possesses at least (p + 1) T -periodic solutions u 1 k , . . . , u p+1 k geometrically distincts. The proof of theorem 3.1 and part 1 of theorem 3.2 are proved.
Proof of Theorem 3.2. In the following, we will suppose that 0 ≤ p ≤ 2N − 2.
By Theorem 2.1 and Remark 2.1, the sequences (u i k ) obtained in the proof of theorem 3.1 satisfy for all i = 1, . . . , p + 1 We will prove, that for all i = 1, . . . , p + 1, the sequence (u i k ) has the following property This will be obtained by using the estimates (3.26) on the critical levels of φ k , and implies that for all i = 1, . . . , p + 1, the sequence ( u i k ∞ ) k∈N goes to infinity as k goes to infinity. For this, we will need the following two lemmas.
Lemma 3.4. Let i 0 ∈ {1, . . . , N } be such that e i 0 , e i 0 +N ∈ B and given Proof. Arguing by contradiction and assume that P B (u(t)) = 0 for a.e. But P B (u(t)) = 0 implies P e i 0 (u(t)) = 0 and P e i 0 +N (u(t)) = 0, which gives us where ϕ(t) = 1 √ π exp( 2π T tJ)e i 0 , with i 0 ∈ {1, . . . , N } is such that e i 0 , e i 0 +N ∈ B. Proof. Arguing by contradiction and assume that there exist sequences k j → ∞, (u j ) ⊂ Z × V , and a constant c 1 ∈ R such that Taking we obtain, by an easy calculation By assumption (H 2 )(i), the Hamiltonian H is bounded from below, so we deduce from (3.31) that there exists a constant c 2 > 0 such that . Combining this with (3.30), we conclude that (u + j ) is bounded in X. Taking a subsequence if necessary we can find u + ∈ E + such that We claim that (u 0 j ) is also bounded in X. Indeed, if we suppose otherwise, (3.33) implies that . Let ρ > 0 and C ρ be a mesurable subset of C defined as in lemma 3.2, we have In the other hand, we have Therefore, we have by (3.35) and (3.36) and we deduce from equality (3.31) that which contradicts (3.30) and proves our claim. Taking a subsequence if necessary, we can assume that there exists u 0 ∈ B and v ∈ V such that for almost every t ∈ [0, T ] (3.39). u + j (t) + u 0 j + v j → u(t) = u + (t) + u 0 + v, as j → ∞. By lemma 3.4, we know that P B (ϕ(t) + u(t)) = 0 for almost every t ∈ [0, T ]. Therefore (3.40) k j |P B (ϕ(t)+u + j (t)+u 0 j +v j )| → +∞ as j → ∞, for a.e. t ∈ [0, T ]. As above, by using (3.40), (H 2 )(i) and lemma 3.2, we obtain (3.37), which contradicts (3.30). That concludes the proof of lemma 3.5.
We claim that x i k ∞ = u i k ∞ → ∞ as k → ∞. Indeed, if we suppose otherwise, (u i k ) possesses a bounded subsequence (u i k n ). Since Proof of theorem 3.3. As in remark 3.2, we can assume without loss of generality that the Hamiltonian H satisfies (H ′ 2 )(i). The following estimate will be needed later.
Lemma 3.6. If assumptions (H 0 ), (H ′ 1 ) and (H ′ 2 ) hold, then for all x ∈ R 2N such that |P B (x)| ≥ 1 and almost every t ∈ C, we have Proof. For x ∈ R 2N such that |P B (x)| ≥ 1 and for almost every t ∈ C, we have By (H ′ 1 ), we have On the other hand by (H ′ 2 )(i), we have Then property (3.43) follows from properties (3.44)-(3.46), which completes the proof of lemma 3.6.
not bounded, we can assume, by going to a subsequence if necessary, that P B (x k ) → ∞ as k → ∞. We deduce as in the proof of lemma 3.3 that → y ∈ B, y = 1, as k → ∞.
Since the embedding X ֒→ L 2 , x → x is compact, we may assume without loss of generality that On the other hand, by (H ′ 2 )(i) we have By lemma 3.7, assumption (H ′ 1 ) and Hölder's inequality, we can find two positive constants c 3 , c 4 such that 93.57) On the other hand, as in (3.11), there exists a constant c 5 > 0 such that Therefore, by Proposition 2.2, there exist two positive constants c 6 , c 7 such that (3.59) x k L 2 ≤ c 6 P B (x k ) α L β + c 7 . By Hölder's inequality, (3.57) and (3.59), we have L β + c 9 where c 8 , c 9 are two positive constants.
Combining (3.56), (3.57) and (3.60), we can find two constants c 10 , c 11 > 0 such that (3.61) P B (x k ) β L β ≤ c 10 P B (x k ) 2α L β + c 11 . However (3.61) contradicts (3.55) because β > 2α. Hence S T is bounded and as a consequence ψ 1 (S T ) is bounded. Since for any x ∈ S T one has ψ k (x) = kψ 1 (x), there exists a positive constant M such that (3.62) ∀x ∈ S T , ∀k ≥ 1, Consequently, (3.48) and (3.62) show that for all i = 1, . . . , p+1 and for all k sufficiently large, we have x i k / ∈ S T . So if k is chosen to be prime number, the minimal period of x i k has to be kT and the proof of theorem 3.3 is complete.