MULTIPLE PERIODIC SOLUTIONS OF HAMILTONIAN SYSTEMS CONFINED IN A BOX

. We consider a nonautonomous Hamiltonian system, T -periodic in time, possibly deﬁned on a bounded space region, the boundary of which consists of singularity points which can never be attained. Assuming that the system has an interior equilibrium point, we prove the existence of inﬁnitely many T -periodic solutions, by the use of a generalized version of the Poincar´e– Birkhoﬀ theorem.

The Hamiltonian function H(t, x, y) is assumed to be continuous, T -periodic in t, and continuously differentiable in x and y. It is defined for (x, y) belonging to a "generalized rectangle" Assumption A3. The function U : R × R → R is continuous and has a bounded continuous gradient with respect to (x, y) belonging to R.
We are now ready to state our main result for the planar case.
Theorem 1.1. Let Assumptions A1, A2, and A3 be satisfied. Then, there exists an integer K 0 such that, for every integer K > K 0 , the Hamiltonian system (1) has at least two T -periodic solutions performing exactly K clockwise rotations around (x 0 , y 0 ) in the time interval [0, T ].
The above theorem thus provides the existence of infinitely many T -periodic solutions of system (1). Its proof will be carried out in Section 2, by the use of a generalized version of the Poincaré -Birkhoff theorem [7,Theorem 1.2], recently obtained by the first author and A. J. Ureña., after a suitable modification of the Hamiltonian function, so to guarantee the global existence of the solutions to the associated initial value problems. Notice that we are not assuming the uniqueness of such solutions, so that the Poincaré map could be multivalued. The meaning of the conclusion concerning the K clockwise rotations of the T -periodic solutions around (x 0 , y 0 ) will be clarified in Section 2, after introducing the so called rotation number.
Let us mention that conditions like (2) have been introduced in [4] in order to treat scalar second order equations with a nonlinearity having either one or two repulsive singularities, or a superlinear growth at infinity. It has been shown, indeed, that the singularities of this type provide a behavior of the solutions which resembles the situation encountered while dealing with superlinear systems (see, e.g. [1,2,3,6,7,8,9,10,11,12,13]). In the same spirit, the existence of subharmonic solutions can also be easily obtained, but we will avoid such a discussion, for briefness.
The above situation will be generalized in Section 3 to a Hamiltonian system in R 2N . Again, the proof will be based on the above mentioned higher dimensional generalized version of the Poincaré -Birkhoff theorem.
2. The proof of Theorem 1.1. There is no loss of generality in assuming that the equilibrium (x 0 , y 0 ) coincides with the origin (0, 0). Moreover, by Assumption A3, we can assume that U is the restriction of a continuous function defined on the whole space R × R 2 , still having a bounded continuous gradient. Indeed, in the case when R is a strict subset of R 2 , we can choose a larger open set S, containing the closure of R, and extend U so that it vanishes outside R × S, thus keeping the gradient bounded. Hence, let C > 0 be a constant for which 2.1. The modified problem. In order to ensure the existence of the solutions on the whole interval [0, T ], we are going to modify our system, and, whenever R does not already coincide with R 2 , extend it on the whole plane. Define, for every i, j ∈ {1, 2} and every p ∈ ]0, 1[ , the numbers and the rectangle whose closure is contained in R. We will be interested in taking p near to 1, so that the rectangle R(p) "approaches" R, so to speak. By Assumption A1, we can consider two auxiliary continuous functions f 1 : R× ]a 1,1 , a 1,2 [ → R and f 2 : R× ]a 2,1 , a 2,2 [ → R, so that We thus obtain a new Hamiltonian function H p : x, y) , which coincides with H on R × R(p). We then consider the auxiliary Hamiltonian system It will be useful to write the solutions z(t) = x(t), y(t) of (7), whenever they do not attain the origin, using the polar coordinates Then, the radial velocity is given by and the angular velocity by If z(t) = 0 for every t in an interval [τ 0 , τ 1 ], the corresponding clockwise rotation number will be denoted by

2.2.
The small amplitude solutions. We first estimate the number of rotations of the solutions starting near the origin. So, we fix a positive radius ε such that the ball B ε , centered at the origin, is fully contained in R( 1 2 ) ∩ V, where V is the neighborhood of the origin introduced in Assumption A1. The desired estimate is given by the following proposition. Proposition 1. There exist a positive integer K 0 and some positive constants η, δ, with η < δ < ε, having the following property: if z(t) = x(t), y(t) is a solution to system (7), with p ∈ 1 2 , 1 , and |z(t 0 )| = δ at a certain time t 0 , then η < |z(t)| < ε , Proof. Consider a solution z(t) = x(t), y(t) to (7), with p > 1 2 , such that 0 < |z(t)| < ε in a certain time interval I, and parametrize it in polar coordinates, as in (8). Recalling Assumption A1, since , it is easy to see that, for every t ∈ I, the radial velocity (9) satisfies the inequality Choose δ < εe −αT and η < δe −αT , and assume that |z(t 0 )| = δ at a certain time t 0 ∈ I. We can now take I as the maximal interval of time in which 0 < |z(t)| < ε. By Gronwall's Lemma we thus have At this point, the last part of the statement can be deduced by just noticing that the angular velocity (10) is a continuous function in the compact set B ε \ B η .
2.3. The large amplitude solutions. We now estimate the number of rotations of the solutions starting sufficiently far from the origin.
In order to prove the above proposition, we first need two lemmas.
Proof. It is an immediate consequence of Assumption A1, and the compactness of R(p), as a subset of R.
Proof. We just prove the first inequality, the other one being analogous. Given β > 0, by (2) in Assumption A2 it is possible to find p ∈ 1 2 , 1 such that x) > βx 2 + C|x| , for every t ∈ [0, T ] and x / ∈ ]a p 1,1 , a p 1,2 [ , where C > 0 is the constant introduced in (4). Equivalently, recalling (6), we can write f 1 (t, x) > β + C/|x|. Consequently, for every q > p, one has for every t ∈ [0, T ], x / ∈ ]a p 1,1 , a p 1,2 [ and y ∈ R, thus concluding the proof. We can now proceed with the proof of Proposition 2. Given the integer K, we fix a constant b, so to have Use Lemma 2.2, with β = b, to find the value p 1 = p(b). Then use Lemma 2.1, with p = p 1 , and find A = A(p 1 ). Now, set where C is the constant introduced in (4). Notice that b > b. Use again Lemma 2.2, with β = b , in order to find p 2 = p(b ) > p 1 . Let z(t) = x(t), y(t) be a solution of (7), with p > p 2 , which remains outside R(p 2 ) for some time. As long as t varies, z(t) may cross different regions in the plane, and we distinguish three situations, as shown in Figure 1: ∈ a p1 1,1 , a p1 1,2 and y(t) / ∈ a p1 2,1 , a p1 2,2 , it is easy to check that −ϑ (t) > b .
(b) If x(t) ∈ a p1 1,1 , a p1 1,2 and y(t) / ∈ a p2 2,1 , a p2 2,2 , then one has |x(t)| < d |y(t)| and |x(t)| < y 2 (t) d/∆, so that (c) If x(t) / ∈ a p2 1,1 , a p2 1,2 and y(t) ∈ a p1 2,1 , a p1 2,2 , one has similarly |y(t)| < d |x(t)| and |y(t)| < x 2 (t) d/∆, so that As a consequence, every solution z(t) to (7), with p > p 2 , which remains outside R(p 2 ), has to rotate clockwise, with an angular velocity bounded below by b. In particular, by (11), if z(t) / ∈ R(p 2 ) for every t ∈ [0, T ], then rot z, [0, T ] > K. Let us now focus our attention on the solutions which enter the region R(p 2 ), coming from the outside, and remain inside this region in a time interval [τ 0 , τ 1 ], with τ 1 − τ 0 ≤ T . Applying Proposition 1, such solutions cannot reach the origin: more precisely, they remain outside the ball B η , for every t ∈ [τ 0 , τ 1 ]. Following one of these solutions, we see that, after entering the set R(p 2 ), it could perform a certain number of counter-clockwise rotations while remaining inside it. However, since R(p 2 ) \ B η is compact, the angular velocity will remain bounded. So, there exists a constant D ≥ 0 such that, for every such solution z(t) which remains inside R(p 2 ) in a time interval [τ 0 , τ 1 ], it has to be rot z, Now, we will provide the existence of a guiding curve γ which controls the solutions of system (1). As a consequence, we will see that this curve guides also the solutions of system (7), when p is chosen large enough. By Assumption A2, we can find four functions g i, such that for every (t, x, y) ∈ [0, T ] × R, and it is not restrictive to assume these functions to be strictly increasing. Define, then, their primitives It is possible to choose all the functions g i, in order to ensure that lim ξ→ai,j Being the functions g i, strictly increasing, there are x 1 < 0 < x 2 and y 1 < 0 < y 2 such that g 1,1 (x 2 ) = g 1,2 (x 1 ) = g 2,1 (y 2 ) = g 2,2 (y 1 ) = 0 , and these points are unique. Define the four regions in R, depicted in Figure 2,

and the energy functions
where µ, ν ∈ {1, 2}. Then, for every solution z(t) to (1), Now, starting from a point not belonging to R(p 2 ), we construct a guiding curve γ, glueing together different branches of the level sets of the energy functions E µ,ν , in the corresponding regions R µ,ν . Precisely, such a curve starts from a point P 0 , belonging to the segment R 2,1 ∩ R 2,2 , and proceeds counter-clockwise, as shown in Figure 2, remaining in the rectangle R 2,2 until it reaches the point P 1 , on the segment R 2,2 ∩ R 1,2 . Then, it enters and remains in R 1,2 till it reaches the point P 2 on R 1,2 ∩ R 1,1 . And so on, crossing the four regions, and thus completing a first lap at the point P 4 , which again belongs to the segment R 2,1 ∩ R 2,2 . Since we have that P 4 necessarily lies above P 0 .
Iterating such a construction, we can obtain a curve having the shape of a spiral, which rotates counter-clockwise around the origin while becoming larger and larger, having the following property: if a solution z(t) to system (1) intersects the guiding curve γ at a certain time, then it must cross γ from the inner part to the outer part. As a consequence we have that, roughly speaking, a solution coming from some point near the boundary of R, and approaching R(p 2 ), will have to rotate clockwise several times, guided by the curve γ.
It is possible to verify that the curve γ, so to speak, gets nearer and nearer the boundary of R while it rotates counter-clockwise in the plane. This is certainly true when R is a bounded rectangle. If not, in the case when R is unbounded in some directions, the curve will become larger and larger in those directions. Moreover, we can choose the starting point of the curve in order that the first lap, and hence the whole curve, lays outside R(p 2 ). Hence, we can draw K + D + 1 laps of the curve γ, starting outside the region R(p 2 ), and then find some p K ∈ ]p 2 , 1[ such that all the K + D + 1 laps of γ are contained in the set R(p K ). Notice that, being H p = H in R(p), all the previous considerations on the guiding curve γ still hold for every solution to system (7), when p is chosen greater than p K .
Let us show that this constant p K verifies the property that, if z(t) is a solution to (7), with p > p K , and z(t 0 ) / ∈ R(p K ) at a certain time t 0 , then rot z, [t 0 , t 0 + T ] > K. By the above computation on the angular velocity, we only need to examine the case when there exists a time t 1 ∈ [t 0 , t 0 + T ] at which the solution enters the rectangle R(p 2 ), since otherwise the solution would perform more than K clockwise rotations around the origin in a period time T . Therefore, in some interval [t 0 , t 1 ], the solution goes from outside R(p K ) to inside R(p 2 ), guided by the curve γ, thus performing at least K +D+1 clockwise rotations around the origin. Recalling that, as long as z(t) remains inside R(p 2 ) in a time interval [τ 0 , τ 1 ], with τ 1 − τ 0 ≤ T , it never attains the origin and rot z, The proof of Proposition 2 is thus completed.
2.4. The Poincaré -Birkhoff setting. The proof of Theorem 1.1 will now be concluded by the use of a generalized version of the Poincaré -Birkhoff theorem provided in [7], which does not require uniqueness for the Cauchy problems associated with our equations.
Let K 0 and δ be given by Proposition 1, and fix an integer K > K 0 . Once K has been fixed, let p K ∈ 1 2 , 1 be given by Proposition 2, and fix a p ∈ ]p K , 1[ . Notice that, necessarily, the closed ball B δ is contained in R(p K ). We want to apply [7, Theorem 1.2] to system (7), with respect to the annulus Let us check that the twist condition holds. With this aim, let z(t) be a solution of (7). If |z(0)| = δ, then Proposition 1 guarantees that rot z, [0, T ] ≤ K 0 < K. On the other hand, if z(0) ∈ ∂R(p K ), then Proposition 2 tells us that rot z, [0, T ] > K. The twist condition is thus verified.
Hence, by [7, Theorem 1.2], system (7) has at least two T -periodic solutions z(t),z(t), with rot z, [0, T ] = rot z, [0, T ] = K. By Proposition 2 again, it has to be that z(t),z(t) ∈ R(p K ), for every t ∈ [0, T ], so that these are indeed solutions of the original system (1). The proof of Theorem 1.1 is thus completed. Remark 1. In the case when the rectangle R reduces to a strip, like e.g. ]a, b[ ×R, with a and b real numbers, the superlinear growth assumption in the y variable can be weakended, in the spirit of [4]. Indeed, being b − a finite, the time needed for a large amplitude solution to go from one side to the other will be small, even if the angular speed does not necessarily approach infinity.
3. Higher dimensional Hamiltonian systems. Let us consider a higher dimensional system of the type x = ∇ y H(t, x, y) , y = −∇ x H(t, x, y) .
We are dealing with a Hamiltonian function H : R × R → R which is T -periodic in its first variable and continuously differentiable in x and y. Here, R is a subset of R 2N of the type R = R 1 × · · · × R N , where the sets R m = ]a m 1,1 , a m 1,2 [ × ]a m 2,1 , a m 2,2 [ are some "generalized rectangles", in the above sense. A solution z(t) = x(t), y(t) of system (14) is such that x(t) = x 1 (t), . . . , x N (t) , y(t) = y 1 (t), . . . , y N (t) , with the component z m (t) = x m (t), y m (t) varying in R m , for every m = 1, . . . , N . We will assume that the Hamiltonian function can be decomposed as H(t, x, y) = N m=1 H 1,m (t, x m ) + H 2,m (t, y m ) + U(t, x, y) .
Let us introduce the analogues of Assumptions A1, A2, and A3.