A two-point boundary value problem associated with Hamiltonian systems on a cylinder

We prove the existence of multiple solutions for a two-point boundary value problem associated with Hamiltonian systems on a cylinder. Unlike the periodic problem, where the Poincaré–Birkhoff Theorem plays a central role, no twist condition is needed here.


Introduction and main result
We consider the problem x = ∂ y H (t, x, y), y = −∂ x H (t, x, y), y(a) = 0 = y(b), (1) where H : [a, b] × R 2N → R is a continuous function, with continuous partial derivatives ∂ x H (t, x, y) and ∂ y H (t, x, y).We use the notation z = (x, y), with x = (x 1 , . . ., x N ) and y = (y 1 , . . ., y N ).Here are our assumptions.A1.For every i ∈ {1, . . ., N }, the function H (t, x, y) is τ i -periodic in x i , for some τ i > 0. A2.The solutions of the differential system in (1), with initial value are defined on the whole time interval [a, b].
By assumption A1, once a solution z(t) = (x(t), y(t)) of (1) has been found, infinitely many others appear by just adding an integer multiple of τ i to one of the components x i (t).We will call geometrically distinct two solutions which cannot be obtained from each other in this way.
Here is our main result.
Different explicit conditions on ∇ H can be imposed in order to have assumption A2 satisfied.In the "Appendix" at the end of the paper we will present one of them.
In the literature on Hamiltonian systems it is rather unusual to find multiplicity results for two-point boundary value problems.Starting with Rabinowitz [9] in 1978, a lot of effort has been devoted to the study of the periodic problem.Since then, we can identify two main lines of research: variational methods, and the symplectic approach.
Variational methods have to face the great difficulty that the action functional associated with the problem is strongly indefinite.In [9], it was shown that the natural setting for the Tperiodic problem is the space H 1/2 T .However, the elements of this space are not necessarily continuous, and probably this is the main issue when one wants to deal with two-point boundary conditions.
The symplectic approach emerged from the famous Poincaré-Birkhoff Theorem [8] for the periodic problem associated with a Hamiltonian system.It is very well suited in one degree of freedom (i.e., when N = 1), providing the existence of two T -periodic solutions as fixed points of the Poincaré map, under some twist assumptions.The possibility of a higher dimensional version of the Poincaré-Birkhoff Theorem has been investigated, starting with Birkhoff himself [1], by many authors (see [4] and the references therein).
There is a striking issue in our Theorem 1.1: no twist condition is assumed!We will try to explain the geometrical idea behind this theorem in Sect.3, in the one-degree of freedom case.
When the Hamiltonian function has the special form H (t, x, y) = 1 2 |y| 2 + G(t, x) and N = 1, problem (1) becomes a Neumann boundary value problem for a scalar second order differential equation.We can find a multiplicity result for the Neumann problem by Castro [2] in 1980 and a similar one by Rabinowitz [10] in 1988, where the result is mentioned as a remark at the end of the paper, after a detailed study of the periodic problem associated with a Lagrangian system.Both papers use variational methods.
The proof of Theorem 1.1 is provided in Sect. 2. The main novelty lies in the fact that, while for the periodic problem x and y are usually both taken in the same space H 1/2 T , here we assume x and y to belong to some complementary spaces X α and Y β , which are closely related to fractional Sobolev spaces.We will explain in detail how these spaces are defined in Sect.2.1.
The variational setting will then be developed in Sects.2.2 and 2.3.Then, as in [4], we will be able to apply a multiplicity result by Szulkin [11], based on an infinite-dimensional extension of Lusternik-Schnirelmann theory.
Notation.In the following, we will use the notation e 1 , . . ., e N for the vectors of the canonical basis of R N .The Euclidean scalar product will be denoted by •, • , and the corresponding norm by | • |.Moreover, we will denote the average of a function f ∈ L p (0, T ) by

Proof of Theorem 1.1
Without loss of generality, we can assume [a, b] = [0, π].Indeed, we can reduce to this case with a simple change of variable t → π b−a (t − a) .By assumption A2 and a standard compactness argument, there is a constant R > 0 such that every solution satisfying (2), with a = 0, is such that and consider the modified Hamiltonian function We claim that our goal will be attained assuming also, without loss of generality, that Indeed, replacing in (1) the function H by H , the solutions (x, y) of and y(0) = 0, as long as they satisfy |y(t)| ≤ R, they are also solutions of (1), hence they satisfy the same estimate (3).By A1, the assumption x(0) ∈ N i=1 [0, τ i ] is not restrictive.The Claim is thus proved.As a consequence, by A1, we can assume that there exists a constant c > 0 such that The rest of the proof is divided in three parts.In the first one we introduce the function spaces where to build our variational setting, which will be carried out in the further two parts.As in [4], the conclusion will follow by applying the following special case of a multiplicity result by Szulkin [11], where we denote by T N the N -dimensional torus.In the following, we will treat T N as being lifted to R N .Theorem 2.1 Let E be a real Hilbert space, and L : E → E be an invertible bounded selfadjoint operator.Denote by M the set E × T N , as being lifted to E × R N , and let ψ : M → R be a continuously differentiable function such that dψ(M ), the image of its differential, is relatively compact in the dual space L (M , R).Then, the function ϕ : M → R defined as ϕ(e, x) = 1  2 Le, e + ψ(e, x) has at least N + 1 critical points.

The function spaces
In this section, divided into three subsections, we introduce our function spaces.

Some remarks on L p spaces
It is well known that any function f ∈ L p (−π, π), with p ∈ ]1, +∞[ , has a real Fourier expansion and a complex expansion with the corresponding coefficients Given p > 1, q > 1, with p ≤ 2 ≤ q and (1/ p) + (1/q) = 1, we know from Hausdorff-Young inequality (see [12, The above inequality becomes an identity if p = q = 2. Let us now work in the subinterval ]0, π[ .A function f ∈ L p (0, π) could be expressed as a traditional Fourier series, but if we enlarge the set of admissible frequencies it can also be expanded in a series involving only cosines, or in a series involving only sines, Indeed, it is sufficient to extend the function f to the interval ] − π, π[ , either as an even function, or as an odd function, so to obtain the above expansions.Let us focus on the first case.We will need the following.
Proposition 2.2 Let p > 1 and q > 1 be such that p ≤ 2 ≤ q and (1/ p) Proof In this case, after having extended f to an even function on ] − π, π[ , we have Hence, by he Hausdorff-Young inequality (5), , so that (7) holds true.
Concerning the second case, when the function f is expanded in sines, one can similarly prove that Both ( 7) and ( 8) become identities when p = q = 2.

The space X
For any α ∈ ]0, 1[, we define X α as the set of those real valued functions x ∈ L 2 (0, π) such that The space X α is endowed with the inner product and corresponding norm Notice that the functions in X α have a zero mean on [0, π].Moreover, Proof We can define a linear mapping L : X α → 2 by setting It is a bijective function, and since we have indeed defined an isometric isomorphism from X α to 2 .Consequently, X α is a separable Hilbert space.It is readily verified that (e m ) m≥1 is an orthonormal family, and for each x ∈ X α we have the expansion thus proving the second part of the statement.
the statement clearly holds true.
Proof Since every function x in C 1 ([0, π]) can be extended to an even function on [−π, π] with finite left and right derivatives at 0, it admits a uniformly convergent expansion on [0, π], Hence, the set and the sequence (s n ) n converges to x in X α .

The space Y
For any β ∈ ]0, 1[, we define Y β as the set of those real valued functions y ∈ L 2 (0, π) such that The space Y β is endowed with the inner product and corresponding norm Notice that   1  2 , we can define the real number This shows that the space Y β is continuously embedded in C([0, π]).Since the series of sines defining y is convergent everywhere, we also deduce that y(0) = 0 = y(π).
Proof The argument in the proof of Proposition 2.5 applies the same also here, since every function y in C 1 0 ([0, π]) admits a uniformly convergent expansion The statement then holds true.
In the following, we will consider functions having values in R N .They can be seen as elements of the Cartesian products With the natural scalar products and associated norms they become separable Hilbert spaces.
For simplicity, we will still denote them by X α and Y β , respectively, and the same will be done for the other spaces involved, like, e.g., L p (0, π), W k, p (0, π),

The variational setting: I
We choose two positive numbers α < 1 2 < β such that α + β = 1, and consider the space E = X α × Y β .It is a separable Hilbert space, being endowed with the scalar product and the corresponding norm Let us introduce the torus The solutions of problem (1) will be written as dt, and we will assume that We will thus search the solutions in the space E × T N .Set M = E × T N and let ψ : M → R be the functional defined as It is well defined, since the function H : [0, π] × R 2N → R is continuous and globally bounded, in view of the comments made at the beginning of this section.In the following, we will treat T N as being lifted to R N , so M will often be identified with E × R N .
Proposition 2.9 The functional ψ : M → R is continuously differentiable.
Having identified M with E ×R N , we can consider the gradient function ∇ψ : M → M , so that and Let We can write for some F 0 and F m ∈ R N .From (12) we see that F 0 = w ∈ T N , showing us that ∇ x ψ has a relatively compact image.Let us also write for some w m ∈ R N .Selecting ũ(t) = e (α) m (t)e i in (10), with i = 1, . . ., N , we obtain Take P > 1 and Q > 1 such that (1/P) + (1/Q) = 1 and 2αP > 1, and set We can write, using the Hölder Inequality, .
By the Hausdorff-Young Inequality (7) with p = 2Q/(2Q − 1) and q = 2Q, (we have used the triangle inequality in L 2Q 2Q −1 (0, π)).Hence, by (4), We have thus proved that the image of ∇ x ψ is bounded in X α , precisely In the same way we see that, for every M ≥ 1, Hence, by Proposition 3.2 applied with the complete orthonormal system In a similar way, using (11), one shows that the image of ∇ y ψ is relatively compact, as well, thus ending the proof.

The variational setting: II
We want to define a continuous symmetric bilinear form B : E × E → R. We first define it on D × D, where ).For each ( x, y) and ( ũ, v) in D, we set Let us see that Let us now prove that, if we consider on D × D the topology of E × E, then B : D × D → R is continuous.We write Then, Similarly one proves that Then, since by Propositions 2.5 and 2.8 we know that D is a dense subspace of E, the function B : D × D → R can be extended by continuity to E × E. We will still denote by B : E × E → R this continuous (and therefore continuously differentiable) symmetric bilinear form.

Remark 2.11
It should be noticed that the equality in (13) still holds if x, y are in H 1 (0, π) and ũ, v are in L 2 (0, π).In view of the identities in ( 14), if x, y are in L 2 (0, π) and ũ, v are in H 1 (0, π) we can also write Remark 2.12 When writing this paper, in a first attempt we defined the functional spaces making use of the classical Fourier expansions in both sines and cosines, but we had to face rather technical proofs.We then realized that the choice of the systems {cos(mt)} and {sin(mt)}, associated with X α and Y β , respectively, enjoy an important property: they diagonalize the bilinear form B. For this reason, the computations are now very natural.
On the other hand, taking ū = 0, ũ = 0 and v = φ, after noticing that x(t) = x + x(t) differs from x(t) by a constant, we see that the equality holds in the sense of distributions.Again by (4), this yields that x belongs to W 1,∞ (0, π).
Since both x(t) and y(t) are continuous, from the differential equations they satisfy we deduce that they are continuously differentiable.

The continuous symmetric bilinear form
Proposition 2.14 The operator L : E → E is invertible, with a continuous inverse.

Proof Let us first prove that
Let Lz = ω, with z = ( x, y) ∈ E and ω = ( p, q) ∈ E. We know that Taking v = 0 and writing the expansions Then, using the fact that α + β = 1, we conclude that On the other hand, taking ũ = 0 in (16) and proceeding in the same way as above, we find that By the definition of the norm in E, the equalities (17) and ( 18) imply (15).
As a consequence of (15), we have that ker L = {0}.Let us prove that the image of L is closed.Let (ω n ) n be a sequence in the image of L having a limit ω ∈ E. Let (z n ) n be a sequence in E such that Lz n = ω n .By (15), for every m, n we have that Hence, (z n ) n is a Cauchy sequence, thus converging to some z ∈ E. Passing to the limit in Lz n = ω n we obtain Lz = ω, proving that w belongs to the image of L. We have thus proved that Im(L), the image of L, is closed.Since L is selfadjoint, with ker L = {0}, and its image is closed, we conclude that We can finally apply Theorem 2.1 to obtain N + 1 critical points of the functional ϕ.By Proposition 2.13, these critical points determine the N + 1 solutions (x, y) of (1) we are looking for.
The proof is thus completed.
The Poincaré map induces a diffeomorphism It is well known that this map is exact symplectic, meaning that the differential form y 1 dx 1 − y 0 dx 0 is exact in the cylinder T × R (see, e.g.[5,Appendix 1]).As a consequence we have the identity for two-forms dy 1 ∧ dx 1 = dy 0 ∧ dx 0 , and P preserves Haar measure on the cylinder.
Indeed, P satisfies a more subtle property.Given a Jordan curve ⊆ T × R which is C 1 , regular and non-contractible, the image 1 = P( ) enjoys the same properties.Let us fix y 0 ∈ R such that and 1 both lie in {y > y 0 }, and let A and A 1 be the bounded components of {y > y 0 } \ and {y > y 0 } \ 1 , respectively.Then μ(A) = μ(A 1 ), where μ denotes the Haar measure.To prove this, it is sufficient to apply Stokes-Cartan Theorem to the differential form η = y dx, so to obtain and From this property we deduce a well known principle: given any Jordan C 1 -curve with the previous properties, the intersection of with 1 has at least two points.The proof of Theorem 1.1 for N = 1 and H smooth then follows from this principle, with = {( x, 0) : x ∈ T}.
It is interesting to observe that the same ideas can be applied to very general nonlinear boundary value problems.Assume that σ : T × R → R is a C 1 -function such that 0 is a regular value and = {( x, y) ∈ T × R : σ ( x, y) = 0} defines a non-contractible Jordan curve.We are interested in the problem Let us keep assumption A1, with τ 1 = 2π, and replace A2 with [a, b] = [0, 1] by A2 .
The solutions of the differential system in (19), with initial conditions lying in , are defined on the whole time interval [0, 1].

Theorem 3.1 Under assumptions A1 and A2 , problem (19) has at least two geometrically distinct solutions.
As a particular case, we can deal with a problem like where : R → R is any T -periodic C 1 -function.Indeed, taking σ (x, y) = y − (x), we can apply Theorem 3.1 and get the existence of at least two geometrically distinct solutions.
Funding Open access funding provided by Universitá degli Studi di Trieste within the CRUI-CARE Agreement.

Declarations
Conflict of interest The authors declare that there are no conflict of interest.
Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made.The images or other third party material in this article are included in the article's Creative Commons licence, unless indicated otherwise in a credit line to the material.If material is not included in the article's Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder.

A sufficient condition for assumption A2
Here we provide an explicit condition guaranteeing the validity of assumption A2.
(Here, T N is the N -dimensional torus defined in (9).)Under this condition, for every solution of the Hamiltonian system in (1) we have the linear differential inequality As a consequence, using the Gronwall inequality, assumption A1 and the compactness of T N , the solutions cannot blow up in the future, and A2 is satisfied.

A compactness criterion
For the reader's convenience, we report here a characterization of relatively compact sets in a general separable Hilbert space H with a complete orthonormal system (e m ) m .For any h ∈ H , let us write h = ∞ m=0 h m e m .One can find the following characterization in [3, page 338, Ex. 3] and [7, Proof Assume A to be relatively compact.Given ε > 0 and M ∈ N, define It is an open set in H .Then, H = M≥1 H M,ε , and A must be covered by a finite number of these sets, say Setting M = max{M 1 , . . ., M r }, one has that A ⊆ H M,ε , proving in particular that m≥M |h m | 2 < ε, for every h ∈ A.
On the other way around, given a sequence (h (n) ) n in A, we extract a subsequence (maintaining the same notation as for the original sequence) which weakly converges to some h ∈ H .In particular, h (n) m → h m , for every m.We want to prove that h (n) → h strongly.Given ε > 0, select M such that Then, Since M is fixed and M m=0 |h

Proposition 2 . 6 Y
β is a separable Hilbert space.A Hilbert basis in Y β is provided by the functions ( (β) m ) m≥1 defined as

Proposition 2 . 10
with the natural scalar product •, • M .123 The gradient function ∇ψ has a relatively compact image.Proof If ∇ψ(( x, y), x) = (( w, ζ ), w), then, for every page 80, Th. 2].Let H be a separable Hilbert space.A bounded set A ⊆ H is relatively compact if and only if