1 adiabatic limits of closed orbits for some newtoniansystems in r^n (malchiodi)

8/14/2019 1 Adiabatic Limits of Closed Orbits for Some NewtonianSystems in R^n (Malchiodi)

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Adiabatic limits of closed orbits for someNewtonian systems in Rn

Andrea Malchiodi

Scuola Internazionale Superiore di Studi Avanzati

via Beirut, 2-4 34014 Trieste, Italy

AbstractWe deal with a Newtonian system like x+V(x) = 0. We suppose thatV : Rn R possessesan (n 1)-dimensional compact manifoldMof critical points, and we prove the existence ofarbitrarity slow periodic orbits. When the period tends to infinity these orbits, rescaled intime, converge to some closed geodesics on M.

Key Words: closed geodesics, slow motion, periodic solutions, limit trajectories

1 Introduction

Let V : Rn R be a smooth function and suppose that V possesses an (n 1)-dimensional compactmanifoldMof critical points which is non degenerate, namely

(1) kerV(x) = TxM x M, or equivalently V[nx, nx] = 0 x M,

wherenx is a normal vector to M atx.We are interested in studying the existence of solutions to the Newtonian system

(T)

x + V(x) = 0;

x() isT periodic,

when T is large and x() is close to M. Equivalently, setting 2 = 1T

, one looks for solutions to theproblem

(P)

x + 1

V(x) = 0;

x() is 1 periodic,

for >0 sufficiently small.

Supported by M.U.R.S.T., Variational Methods and Nonlinear Differential Equations

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AsTvaries, problem (T) can possess some continuous families of solutions parametrized in T, and the factthatV is degenerate (in the sense of Morse) allows these solutions to have a non trivial limit behaviour.The case of solutions approaching a critical manifold ofVhas been considered for example in [6],[8] and[9]. It is known that if some smooth family x(t, ) solves problem (P) and ifx(t, )

M as

0, then

the curve x(t, 0) is a geodesic on M. The curve x(, 0) is called adiabatic limit for the family x(, ).The aim of this paper is to achieve some complemetary result, namely to prove that for some closed

geodesics on Mthere are indeed solutions of (P) which approach these geodesics. Large period orbitswith some limit behaviour have also been studied in [ 4]for planar systems.

Our main results are the following Theorems. The first one treats the case of a non-degenerate closedgeodesic onM, see Definition2.3.

Theorem 1.1 Suppose x0 : S1 M is a non-degenerate closed geodesic, and suppose V is repulsivew.r.t. M, namely that the following condition holds

(2) V(x)[nx, nx]< 0 for allx M.Then there existsT0 > 0 with the following property. For allT

T0 there exists a functionuT such that

(i) uTis solution of problem (T);

(ii) asT +, uT(T) x0() inC1(S1;Rn).

The proof relies on the Local Inversion Theorem, which can be applied by the non-degeneracy of x0.Since in (P) appears the singular term

1

, a quite accurate expansion ofV is needed.If we want to prove the convergenge of a sequence of trajectories, instead of the convergence of a

one-parameter family, then we can remove any non-degeneracy assumption. With abuse of notation wewill again call adiabatic limit the limit trajectory.

Theorem 1.2 If condition (2) holds, then for every sequence Tk + there exists a sequence ofsolutions(uk)k to problem (T) corresponding to T =Tk such that up to subsequenceuk(Tk) converge inC0(S1,Rn). The adiabatic limit ofuk(Tk

) is a non trivial closed geodesicx0 onM.

Remark 1.3 (a) Since the adiabatic limitx0 in Theorem1.2can be degenerate, and so it is possible thatit belongs to a family of degenerate geodesics, it is natural to expect convergence only along sequences oftrajectories.

(b) The limit geodesicx0 can be characterized as follows. If1(M) = 0,x0 realizes the infimum of thesquare lenght in some component of the closed loops inM. From the proof of Theorem1.2it follows thatone can find adiabatic limits belonging to every element of1(M), see also Remark5.6. If1(M) = 0then the energy ofx0 is the infimum in some suitable min-max scheme.

The proof of Theorem1.2is based on a Lyapunov-Schmidt reduction on the Hilbert manifoldH1(S1; M)of the closed loops in Mof class H1. Standard min-max arguments are applied to a suitable functionalon H1(S1; M) which is a perturbation of the square lenght L0, see formula (7). A similar approach hasbeen used for example in[1] to perform reductions on finite-dimensional manifolds. The new feature of

our method is that we perform a reduction of an infinite dimensional manifold.

IfVis of attractive type, namely ifV (x)[nx, nx]> 0 for all x M, then the situation is very different,since some phenomena of resonance may occur. As a consequence our hypotheses become stronger andwe can prove convergence just for some suitable sequence Tk +. Section 6 contains some resultsconcerning this case. As an example we can state the following one.

Theorem 1.4 SupposeV satisfies the following conditions for someb0 > 0

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(i) V (x)[nx, nx] = b0 for allx M;(ii)

3Vn3

x

(x) = 0 for allx M.

Then there exists a sequence Tk + and there exists a sequence of solutions (uk)k to problem (T)corresponding toT =Tk such that up to subsequenceuk(Tk)converge inC0(S1,Rn). The adiabatic limitofuk(Tk) is a non trivial closed geodesicx0 onM.

The paper is organized as follows: Section 2 is devoted to recalling some notations and preliminary facts.In Section 3 we prove Theorem 1.1. In Section 4 we study some linear ordinary differential equations,used to perform the reduction. In Section 5 we reduce the problem on H1(S1; M), we study the reducedfunctional and we prove Theorem1.2. Finally in Section 6 we treat the attractive case.

Acknowledgements

The author whishes to thank Professor A. Ambrosetti for having proposed him the study of thisproblem and for his useful advices.

2 Notations and Preliminaries

In this Section we recall some well known facts in Riemannian Geometry, we refer to [6] or [10] for thedetails. In particular we introduce the Levi-Civita connection, the Gauss equations, the Hilbert manifoldH1(S1; M), and some properties of the square lenght functionalL0 onH

1(S1; M).It is given an orientable manifold M Rn of codimension 1, which inherits naturally a Riemannian

structure from Rn. OnMit is defined the Gauss mapn : M Sn1 which assigns to every pointx Mthe unit versor nx (TxM), whereTxMis the tangent space ofM atx. The differential ofnx is givenby

(3) d nx[v] = H(x)[v]; x M, v TxM,

whereH(x) : TxM TxMis a symmetric operator. Dealing with the operator H(x) we will also identifyit with the corresponding symmetric bilinear form according to the relation

H(x)[v, w] = H(x)[v] w; v, w TxM.

The bilinear form H(x) is called the second fundamental formofM at x.IfX(M) denotes the class of the smooth vector fields on the manifold M, then for every x M the

Levi-Civita connection :TxM X(M) TxM is defined in the following way

(4) XY =DXY (nx DXY) nx.

Here Yis an extension ofYin a neighbourhood ofx in Rn andDX denotes the standard differentiationin Rn along the direction X.

Remark 2.1 The quantityXY defined in (4) is nothing but the projection ofDXY onto the tangentspace TxM. ActuallyXY depends only on Y(x) and on the derivative of Y along the direction X.Hence formula (4) makes sense also whenY is defined just on a curvec onM for whichc(t0) =x andc(t0) = X. In the following this fact will be considered understood.

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TheRiemann tensorR : TxM TxM TxM TxMis defined by

XYZ YX Z [X,Y]Z.

Here X, Y and Z are smooth extensions ofX, Y and Zrespectively, and the symbol [, ] denotes theusual Lie bracket. The above definition does not depend on the extensions ofX, Y andZ.

Leti : M Rn be the inclusion ofM in Rn: there exists a symmetric bilinear form s: TxMTxMR which satisfies

sx(X, Y) = DXY , X, Y TxM.HereYis any smooth extension ofY and X Yis the component ofDXYnormal toTxM. TheGaussequationsare the following

(5) (R(X, Y)Z, W) = (sx(Y, Z), sx(X, W)) (sx(X, Z), sx(Y, W)), X, Y, Z , W TxM.

GivenX, Y TxM, consider a smooth extension Y ofYand an extension ofnxsuch that (p) TpMfor all p Min some neighbourhood ofx. Differentiating the relation (Y, ) = 0 along the direction X,there results 0 = DX(Y, ) = (DXY, ) + (Y, DX).

Hence it follows thatsx(X, Y) is given by

sx(X, Y) = H(x)[X, Y], X, Y TxM.

In particular equation (5) becomes

(6) (R(X, Y)Z, W) = (H(x)(Y, Z), H(x)(X, W)) (H(x)(X, Z), H(x)(Y, W)), X, Y, Z , W TxM.

In order to define the manifold H1(S1; M), we recall first the differentiable structure of a smooth k-dimensional manifold. This is given by a family of local charts (U, ), where : Rk R arediffeomorphisms such that the compositions 1 : 1 (U) 1 (U) Rk Rk are smoothfunctions.

Definition 2.2 A closed curvec: S1 M is said to be of classH1(S1; M) if for some chart(U, ) ofM the map c: S1 Rk is of classH1.

This definition does not depend on the choice of the chart ( U, ), since the composition of an H1 mapin Rk with a smooth diffeomorphism is still of class H1. The class H1(S1; M) constitutes an infinitedimensional Hilbert manifold. We recall briefly its structure. Given a curve cC(S1; M) we denotebycT M the pull-back ofT M trough c, namely the family of vector fields X: S1 T M such that

X(t) Tc(t)M for allt S1.

We define alsoHc to be the sections ofcT M for whichS1

|(t)|2dt +S1

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If M is simply connected, the proof of the existence of a closed geodesic is more involved and in itsmost general form it is due to Lusternik and Fet, see [7], by means of topological methods. The proof ofour Theorem1.2 follows that argument, and we will recall it later. A fundamental tool is the HurewitzTheorem.

Theorem 2.5 (Hurewitz) Suppose M is a finite dimensional compact manifold such that 1(M) = 0.Defineq to be the smallest integer for whichq(M) = 0, and defineq to be the smallest integer such thatHq(M) = 0. Thenqandq are equal.

Remark 2.6 In our caseM is orientable, and there always resultsHn1(M) Z, so it turns out thatq n 1.

We denote byE0 : H1(S1;Rn) R the square length functional for the curves in Rn, namely

E0(u) =1

2

10

|u(t)|2 dt; u H1(S1;Rn),

and more in general, for every >0, we defineE: H

1

(S

1

,Rn

) to be

E(u) =

10

1

2|u(t)|2 1

V(u(t))

dt, u H1(S1;Rn).

The critical points ofE are precisely the solutions of problem (P).Now we introduce some final notations. Given a covariant tensorTand a vector fieldX, we denote by

LXT the Lie derivative. Ifh H1(S1; M), we define the functions Qh, Ph, Bh : S1 R in the followingway

Qh = |H(h())[h()]|2; Ph = h() H(h())[h()]; Bh = b(h()),where we have set, for brevity

b(x) = V(x)[nx, nx], x M.Sinceh

H1(S1; M) it is easy to check that Qh, Ph

L1(S1) while, since h is continuous form S1 toM,

Bh C0(S1). Finally we setH= sup {H(x) :x M} .

3 The case of a non degenerate geodesic

This Section is devoted to prove Theorem1.1. The strategy is the following: since problems (T) and (P)are equivalent, we are reduced to find critical points ofE for small. In order to do this, we first findsome pseudo critical points for E, in Lemma3.1, and we prove the uniform invertibility ofD

2E atthese points in Lemma3.2. Then, in Proposition3.3 we use the Contraction Mapping Theorem to findtrue critical points ofE.

To carry out the first step of our procedure, let us consider the non-degenerate geodesic x0 in thestatement of Theorem1.1. x0induces the smooth map nx0() : S

1 Sn1. Hence every curve y: S1 Rn

can be decomposed into two parts, the first tangential to Tx0M, and the second normal to Tx0M

(10) yT =y (y nx0) nx0 ; yN = (y nx0) nx0 .When y is differentiable we can also decompose the derivatives ofyT and ofyN into a tangential partand a normal part. Setting yn = y nx0 , there results

(11)

d

dtyTT

= x0yT;

d

dtyTN

= (y nx0) nx0 = H(x0)[y, x0] nx0 ;

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and

(12)

d

dtyNN

= yn nx0 ;

d

dtyNT

= (y nx0) nx0 =yn H(x0)x0.

Using equations (10), (11) and (12) one can easily deduce that for some constant C0 > 0 depending onlyonx0 there holds

(13) 1

C0znH1(S1)+ zTTx0H1(S1;M)

zH1(S1;Rn) C0

znH1(S1)+ zTTx0H1(S1;M)

,

for all z H1(S1; M). This means that H1(S1;Rn) Tx0H1(S1; M) H1(S1) and that the two norms H1(S1;Rn) and Tx0H1(S1;M)+ H1(S1) are equivalent.

Now, roughly, we want to solve the equation x =V(x) up to the first order in : expandingV(x) : S

1 Rn asV(x(t)) = (t) + 2 (t) + O(3),

we have to find f andg such that

(14) x0(t) =

(t); f=

(t).

In order to solve equation(14), we first find the explicit expressions of and depending on f and g .SinceMis assumed to be non-degenerate, the function Vcan be written as

V(x) =1

2b(x) d2x; dx = dist(x, M),

where b: Rn R is a smooth negative function, see condition (2). Because of the factor 1

, to expand1

V(x) up to the first order in, we need to take into account the derivatives ofVup to the third order.We fix some pointx0(t), and we consider an orthonormal frame (e1, . . . , en) in x0(t) such thate1, . . . , en1form an orthonormal basis for Tx0(t)M, and en is orthogonal to M. With simple computations one caneasily check that the only non-zero components of the second and the third differential ofV atx0(t) are

(15) D3innV(x0(t)) = Dib(x0(t)); D3ijnV(x0(t)) = b(x0(t)) D2ijd(x0(t)); i, j = 1, . . . n 1;

(16) D2nnV(x0(t)) = b(x0(t)), D3nnnV(x0(t)) = 3Dnb(x0(t)).

The second differential ofdx at x0(t), see[5] Appendix, is given by

D2dx[v, w] =n1i,j=1

Hijvi wj ; v, w Tx0(t)M.

Here the numbers Hij denote the components of H(x0(t)) with respect to the basis (e1, . . . , en1) ofTx0(t)M. In particular from the last formula it follows that

D3ijnV(x0(t)) = b(x0) D2ijd(x0) = b(x0) Hij(x0), i, j = 1, . . . , n 1.Hence by expanding V (x0+ f+

2 g) in powers of we get

(t) z = b(x0(t)) fn zn, z Rn;(17)

(t) z = b(x0) gn zn+ 12 n1i,j=1

b(x0) Hij(fi fjzn+ fi fn zj + fjfn zi)(18)

+

n1i=1

Dib(x0)(2fi fn zn+ f2n zi) + 3 Dnb(x0) f

2n zn

, z Rn.

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Herefi, fn, etc. denote the components of the vectors with respect to the basis (e1, . . . , en). Taking intoaccount (17) and (18), the equations in(14) become

(19) x0 =

b(x0) fn;

f z = b(x0) gn zn 12

b(x0) n1

i,j=1

Hijfifj+ 2n1i=1

Dib(x0)fi fn+ 3Dnb(x0) f2n

zn

+n1i=1

2 n1

j=1

Hijfjfn+ Dib(x0) f2n

zi

, z Rn.(20)

Equation (19) can be solved in fn with

(21) fn(t) = a(t) := 1

b(x0(t))x0(t) H(x0(t))x0(t) = 1

b(x0(t))H(x0(t))[x0(t), x0(t)].

In fact, since x0 is a geodesic, it turns out that (x0)T

=x0x0 = 0; moreover by taking yT = x0 informula (11), we can conclude that

(x0)T

= 0; (x0)n= H(x0)[x0, x0],

hence (21) follows. As far as (20) is concerned, we can write it in variational form, substituting theexpression offn according to (21)

S1fz =

S1

b(x0) gn zn+1

2

S1

b(x0)H(x0)[f

T, fT] + 2a(t)Tb(x0)fT + 3Dnb(x0) a2(t)

zn

+

S1a(t) b(x0(t)) H(x0)[f

T, zT] +1

2 S1a2(t)Tb(x0)zT; z H1(S1;Rn).(22)

HereTbis the tangential derivative ofb on M.The quantity

S1

fz can be expressed in a suitable way by decomposing f and z into their tangentand normal parts. Using equations (11) and (12) corresponding to f andz , and taking into account (21)we have

S1fz =

S1

fTzT +S1

H[fT, x0] H[zT, x0] +

S1

a zn+

S1

a zn H[x0] H[x0]

+

S1

zn H[x0, fT] +S1

a H[x0, zT] S1

H[x0, fT]zn

S1

H[x0, zT]a.

From equations (6) and (9) it follows that

D2

L0(x0)[zT

, fT

] =S1

fT

zT

+S1

H[fT

, x0] H[zT

, x0] S1

H[x0, x0] H[zT

, fT

],

hence there holdsS1

fz = D2L0(x0)[zT, fT] +

S1

H[x0, x0] H[zT, fT] +

S1

a zn+

S1

a zn H2[x0, x0]

+

S1

zn H[x0, fT] +S1

a H[x0, zT] S1

H[x0, fT]zn

S1

H[x0, zT]a.

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Taking into account that by the definition ofa(t), it is

S1 a(t) b(x0(t)) H(x0)[fT, fT] = S1 H(x0)[x0, x0] H(x0)[f

T, fT],

equation (22) assumes the form

D2L0(x0)[zT, fT] +

S1

a zn+

S1

a zn H2[x0, x0] +

S1

a H[x0, zT] S1

H[x0, fT]zn

= 1

2

S1

b(x0)H(x0)[f

T, fT] + 2a(t)Tb(x0)fT + 3Dnb(x0) a2(t) 2H[x0, fT]

zn(23)

+ 1

2

S1

a2(t)Tb(x0)zT +S1

b(x0) gn zn+

S1

H[x0, zT]a; z H1(S1;Rn).

Now we claim that we can find fT satisfying the following conditions

(24) fT, x0Tx0H1(S1;M) = 0;

D2L0(x0)[zT, fT] =

S1

a H[x0, zT] +12

S1

a2(t)Tb(x0)zT +S1

H[x0, zT]a;

for all zT Tx0H1(S1; M).(25)

In fact, sinceD2L0(x0) is non degenerate on (x0), there existsfT satisfying (24) and satisfying (25) for

allzT (x0). But since D2L0(x0)[x0, fT] = 0 and also

S1

a H[x0, x0] +12

S1

a2(t)Tb(x0)x0+S1

H[x0, x0]a= 0,

as one can check with simple computations, indeed fT satisfies equation (25) for allzT

Tx0H

1(S1; M).

We note that from standard regularity theory for ordinary differential equations, it turns out thatfT issmooth. Then, choosing gn such that

b(x0) gn = 12

b(x0)H(x0)[f

T, fT] + 2a(t)Tb(x0)fT + 3Dnb(x0) a2(t) 2H[x0, fT]

,

a + a H2[x0, x0] + ddt

H[x0, f

T]

,(26)

with the above choice offT equation(23) holds true. In conclusion, we have solved (19) and (20), soalso (14) is satisfied.

We can summarize the above discussion in the following Lemma.

Lemma 3.1 Letfn andfT be given by equations (21)and (24)and (25)respectively. Let alsogn satisfy

(26) and letgT

0. Then, setting(27) x = x0+ f+

2 g,

there existsC1 > 0 such that for suffuciently small

DE(x) C1 2.

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Hence, since fn satisfies equation (21), there results

D2E(x)[z, w] = D2L0(x0)[z, w] +

S1

zn wn+ S1

zn wn H2[x0, x0]

n1

i=1

S1

Ci(zi wn+ wi zn)

+

S1

wn H[x0, zT] +S1

zn H[x0, wT] S1

H[x0, zT] wn

S1

H[x0, wT]zn(31)

1

S1

b(x0) zn wn+ O() z w.

Sincex0 is a non-degenerate critical point for L0, there exist subspaces W+, W Tx0H1(S1; M) with

the following properties

(i)W+ W =Tx0H1(S1; M), W+ W = 0;(ii)D2L0(x0) is positive definite (resp. negative definite) on W

+ (resp. W).

Now, taking into account that H1(S1;Rn) can be decomposed as H1(S1;Rn) Tx0H1(S1; M)H1(S1),we equip it with the equivalent scalar product (, )

(v, w) = (vT, wT)Tx0H1(S1;M)+ (vn, wn)H1(S1).

We set alsoX+ =W+ H1(S1); X =W {0},

and we defineP+ :H1(S1;Rn) X+; P :H1(S1;Rn) X

to be the orthogonal projections, with respect to (, ) onto the subspaces X+ andX.From equation (31)there results

(32) D2E(x)|X =D2L0(x0)|W+ o(1).

So, since D2E(x) is self-adjoint it follows that alsoD2E(x)v, w

=

v, D2E(x)w= o(1) v w; v X+, w X.

This implies that

(33) D2E(x)|X+ =P+ D2E(x)|X++ o(1).

Our aim is to prove that there exists C4> 0 and 1< 0 such that the following properties hold

(j)

D2E(x)y, y C4y2; y X, (0, 1);

(jj )

D2E(x)y, y

C4y2; y X+, (0, 1).

There results D2E(x)y, y

= D2E(x)[y, y]; y X,

so condition (j) follows immediately from (32) and (ii). On the other hand, for z W+ H1(S1), onecan write

S1H[x0, z

T] zn = S1

znd

dt

H[x0, z

T]

.

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So, inserting this relation into (31) one can easily see that there exists C >0 such that

(34) D2E(x)[z, z] D2L0(x0)[z, z] + S1 z2n

1

S1 b z2n CS1 z

2n

12

z.

Here we have setb = supxMb(x)< 0. Given an arbitrary >0, by the Newton inequality there resultsS1

z2n

12

z

1

2 z2 +1

2 1

S1

z2n

,

hence, by equation (13) it follows that

(35)

S1

z2n

12

z

C02 zT2Tx0H1(S1;M)+

C02 zn2H1(S1)+

1

2 1

S1

z2n

.

Hence, since D2L0(x0) is positive definite on W+, we can choose to be so small that

C

C0

min1, infD2L0

(x0

)[w, w] : w

W+,

w

= 1 .With this choice of, equations (34) and (35) imply the existence of2 < 1 for which

(36) D2E(x)[z, z] 12

L0(x0)[zT, zT] +

S1

z2n 1

b

S1

z2n

, (0, 2), z X+.

Equation (36) together with (32) implies (jj ) and concludes the Proof of the Lemma.

Proposition 3.3 For small, problem (P) admits an unique solutionx which satisfies

(37) x xC0(S1) C5 2; for someC5 > 0.

Proof. We prove the Proposition by using the Contraction Mapping Theorem. Actually we want to find

y H1

(S1

;Rn

) which satisfies

DE(x+ y) = 0; yH1(S1;Rn) C5 2.We can write

DE(x+ y) = DE(x+ y) DE(x) D2E(x)[y] + DE(x) + D2E(x)[y].Hence it turns out that

DE(x+ y) = 0 y = F(y),whereF: H

1(S1;Rn) H1(S1;Rn) is defined by

F(z) :=

D2E(x)

1

DE(x)

DE(x+ z) DE(x) D2E(x)[z]

.

We show that the map F is a contraction in some ball B =

z H1(S1;Rn) : z . In fact, ifz B, by equation (28) there results(38) F(z) C2

DE(x) + DE(x+ z) DE(x) D2E(x)[z] .With a straightforward calculation one obtains that for all w H1(S1;Rn) there holds

DE(x+ z)[w] DE(x)[w] D2E(x)[z, w] = 1

S1

(V(x)[z, w] V(x+ z)[w] + V(x)[w]) .

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SinceVis a smooth function, there exists C6 > 0 such that

|V(x(t))[z, w] + V(x(t))[w] V(x(t) + z)[w]| C6 z2 w; w, z Rnz 1, t S1,

so it follows that for sufficiently small

(39)DE(x+ z) DE(x) D2E(x)[z] 1

C6 z2H1(S1;Rn); zH1(S1;Rn) .

Hence, by using equations (38), (28) and (39), for sufficienlty small there holds

(40) F(z) C2

C1 2 +

1

C6 2

; z .

Now consider two functions z , z H1(S1;Rn): for all y H1(S1;Rn) there resultsDE(x + z)[y] D2(x)[z, y] DE(x+ z)[y] + D2(x)[z, y]

= 1

S1

(V(x(t))[z, y]

V(x(t) + z)[y]

V(x(t))[z

, y] + V(x(t) + z)[y]) .

So, taking into account that

V(x(t))[z] V(x(t) + z) V(x(t))[z] + V(x(t) + z)

=

10

(V(x(t) + z+ s(z z)) V(x(t)))[z z]ds,

there results

(41)|(F(z) F(z), w)| 1

C2 suptS1,s[0,1]

V(x(t) + z+ s(z z)) V(x(t)) z z w.

Choosing

(42) = C7 2,with C7 sufficiently large, by equations (40) and (41) the map F turns out to be a contraction in B.This concludes the proof.

Remark 3.4 By equation (42), it follows thaty = O(2). On the other hand, the proof of Lemma3.1 determines uniquely the normal component gn of g. In other words, this means that the followingcondition must be satisfied

(43) ynC0(S1)= o(2).Actually we can prove that (43) holds true. In fact, by the proof of Proposition3.3, the fixed point ysolves

(44) y=

D2E(x)1

z,

withz= DE(x) DE(x+ y) DE(x) D2E(x)[y] .

By using equations (32), (33) and (36) one can show thatyn satisfies the inequality

ynL2(S1) C P+zH1(S1;Rn) C zH1(S1;Rn),

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for some fixedC >0. SincezH1(S1;Rn) = O(2), see Proposition3.3,from the Interpolation Inequality(see for example [2]) it follows that

yn

C0(S1)

C

yn

12

L2(S1)

yn

12

H1(S1)= o(2).

Hence (43) is proved.

Proof of Theorem 1.1 We define uT as uT(T) = x(), T , 2 = 1, see Proposition 3.3. Property (i)follows immediately from Lemma3.1and Proposition3.3. As far as property (ii) is concerned, we notethat formulas (27) and (37) imply thatx x0 = O(), hence 1V(x) =O(1), uniformly in .This means thatx= O(1) uniformly in. The conclusion follows from the Ascoli Theorem.

4 About some linear ODEs

The purpose of this section is to perform a preliminary study in order to reduce the problem, in Section 5,

on the manifold H

1

(S

1

; M). The arguments are elementary, and perhaps our estimates are well known,but for the readers convenience we collect here the proofs.We start by studying the equation

(45)

v(t) + 0 v(t) = (t), in [0, 2],v(0) =v(2), v(0) = v(2),

where0 R is a fixed constant and (t) L1([0, 2]). By the Fredholm alternative Theorem, problem(45) admits a unique solution if0 is not an eigenvalue of the associated homogeneous problem. Theeigenvalues are precisely the numbers{k2}, k N. Since the behaviour of the solutions of (45) changesqualitatively when0 is positive or negative, we distinguish the two cases separately. The former (0 < 0)is related to condition (2), namely to the repulsive case. The latter (0 > 0) is instead related to theattractive case.

Case 0 < 0

LetG(t) be the Green function for problem (45), namely the solution v(t) corresponding to(t) = 0(t).One can verify with straightforward computations that G(t) is given by

(46) G(t) = 1

2

|0| sinh(

|0|)cosh

|0|(t )

, t [0, 2].

The solution for a general function is obtained by convolution, namely one has

(47) v(t) =

S1

G(t s)(s)ds, t [0, 2].

In particular the following estimate holds

(48) vL GL L1 = 1

2|0| L1 .

Furthermore, if L one can deduce

(49) vL GL1 L = 1|0| L .

The last two estimates hold true if, more in general, the constant 0 is substituted by a function boundedabove by0. This is the content of the following Lemma.

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Lemma 4.1 Let(t) be a negative continuous and periodic function on [0, 2] such that(t) 0 < 0,and let(t) L1([0, 2]). Then the solutionv () of problem

(50)v(t) + (t)v(t) = (t), in [0, 2],

v(0) =v(2), v(0) = v(2).

satisfies the estimates (48) and (49).

Proof. The existence (and the uniqueness) of a solution is an easy consequence of the Lax-MilgramTheorem. Letv() denote the unique solution of (50) corresponding to 0. We start by supposing thatf 0. In this way, by the maximum principle, it must bev(t), v(t) 0 for allt. Definey(t) = v(t)v(t):it follows immediately by subtraction thaty is a 2-periodic solution of the equation

(51) y(t) = 0v(t) (t)v(t) |0| y(t).We claim that it must bey(t) 0 for allt. Otherwise, there is somet0 for whichy(t0)< 0, andy(t0) = 0,since the function y is periodic. It then follows from (51) that y(t) should be strictly decreasing in t,

contradicting its periodicity. Hence we deduce that

(52) v(t) v(t) 0, for allt [0, 2].For a general , we write = + , where+ and are respectively the positive and the negativepart of. Let alsov, v denote the solutions corresponding to. By linearity it isv(t) = v+(t)v(t)andv(t) = v+(t) v(t), so, since v andv have definite sign, it turns out that

vL max{v+L , vL} max{v+L , vL} vL .This implies immediately the estimates (54) and (55) for v(t).

We want to prove that the estimates in (48) and in (49) are stable under bounded L1 perturbations ofthe function. Precisely we consider the following problem, where L1(S1).

(53)

v(t) + ( + (t)) v(t) = (t), in [0, 2],v(0) =v(2), v(0) = v(2),

for which it is well known the existence and the uniqueness of a solution v(t).

Lemma 4.2 LetA >0 be a fixed constant, 0 < 0 and(t) C0(S1)satisfy(t) 0 for allt. Let also L1(S1): then, given any number > 0, if|0| is sufficiently large the solutionv() of (53) satisfiesthe inequality

(54) v (1 + ) 12|0| L1 .

Proof. The solutionv satisfies the equation

v(t) + v(t) = (t) (t)v(t), in [0, 2],with periodic boundary conditions, then by Lemma4.1 there holds

vL 12|0| (L1+ M vL)

1

2|0| (L1+ A vL) .

This implies immediately (54) and concludes the proof.

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Lemma 4.3 LetA > 0 be a fixed constant, let0 < 0, and let(t) C0(S1) satisfy(t) 0 for allt. Suppose also that L(S1). Then given any number >0, if|0| is sufficiently large the solutionv() of problem (53) satisfies the inequality

(55) vL (1 + ) 1|0| L .

Proof. Letv0(t) be the solution of problem (53) corresponding to 0, so in particular, by Lemma4.1,there holds

(56) v0L 1|0| L .

Lety : [0, 2] R be defined by y (t) =v0(t) v(t). By subtraction one infers that y(t) is a solution ofthe problem

y(t) + by(t) = c(t)v(t), in [0, 2],y(0) =y(2), y(0) = y(2).

Hence, by applying inequality (48) one deduce sthat

(57) yL 12|0| () v()L1

1

2|0| vL L1 .

So, since by Lemma4.2 the function v(t) satisfies inequality(54), it follows that

yL (1 + ) 12 |0| A L

.

Now, taking into account formulas (54) and (56) it follows that

(58) vL v0L+ yL 1|0| L+ (1 + ) 1

2 |0| A L (1 + )|0|

1 + A

2

.

For|0| large this is a better estimate than (55), and inserting it in formula (57) we obtain

yL 12

|0| vL L1 (1 + )

2 A 1|0| 32

1 +A

2

L .

Using this estimate in (58) we finally deduce, if|0| is sufficiently large

vL 1|0| L +

(1 + )

2 A 1|0| 32

1 +A

2

L (1 + 2 ) 1|0| L

.

This concludes the proof.

Case 0 > 0

We recall that the estimates of this case will be applied to the study of the attractive case. We will

always take for simplicity 0 of the form 0 =

k+ 122

. This is in order to assure that 0 is not an

eigenvalue of the problem and that the distance of0 from the spectrum is always of order

0. Let

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G(t) be the Green function for problem (45), namely the solution v(t) corresponding to(t) = 0(t). Oneeasily verifies thatG(t) is given by

(59) G(t) =

1

20 sin0t , t [0, 2].The solution for a general function is obtained again by convolution, namely one has

(60) v(t) =

S1

G(t s)(s)ds, t [0, 2].

In particular the following estimate can be immediately deduced

(61) vL GL L1 = 12

0 L1 .

If moreover L(S1), one can further deduce

(62) vL GL1 L = 20

L .

Remark 4.4 We note that, differently from the case of0 < 0, the constant is changed only by a factor4 from (61) to (62), and is not by a power of 0, as in the preceding case. However, if L, (62)could be a better estimate than (61), sinceL1 2 L .

The following analogue of Lemmas4.2 and 4.3 holds, the proof follows the same arguments.

Lemma 4.5 LetA >0 be a fixed constant, and suppose0 is a constant of the form0 = (m+ 1/2)2.

Suppose that L1(S1). Then if0 is sufficiently large, problem

(63)v(t) + (0+ (t))

v(t) = (t), in [0, 2],

v(0) =v(2), v(0) = v(2).

possesses an unique solution for all(t) L1(S1) withL1 A. Moreover, given any number >0,if0 is sufficiently large then the solutionv() satisfies the inequality

(64) vL (1 + ) 12

0 L1 .

If moreover L(S1), then for0 is sufficiently large there holds

(65) vL (1 + ) 20

L .

5 Proof of Theorem 1.2

The goal of this section is to prove Theorem 1.2. Two are the main ingredients: the first is the reductionon the manifoldH1(S1; M), treated in Subsection 5.1. The second is the study of the reduced functional,carried out Subsection 5.2.

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5.1 The reduction onH1(S1,M)

In this subsection we perform a Lyapunov-Schmidt reduction of problem (T) on the manifoldH1(S1, M)of the closed H1 loops on M. A fundamental tool are the estimates of Section 4.

Solutions of problem (P), and hence of problem (T), can be found as critical points of the functionalE : H

1(S1;Rn) R.Ifu Rn is in a sufficiently small neighbourhood ofM, say ifdist(u, M) 0 for some 0> 0, then

are uniquely defined h(u) M andv(u) R such that(66) u= h(u) + v(u) nh(u); dist(u, M) 0.It is clear that h and vdepend smoothly on u. In particular, ifu() H1(S1;Rn), and ifdist(u(t); M) 0for all t, then h(u()) and v(u()) are of class H1(S1; M) and H1(S1) respectively. In the sequel we willoften omit the dependence on u ofh(u) andv(u). Viceversa, given h M, andv R, |v| 0, then thepointu Rn, u= h + v nh depends smoothly on h and v .

Ifu() H1(S1;Rn), then there holds(67) u= h + v

n + v

n= h + v

n + v

H(h)[h] = (Id + v H(h))[h] + v

n; a. e. in S1.

From the last expression it follows in particular that

(68) |u|2 = |h|2 + |v|2 + v2|H(h)[h]|2 + 2vh H(h)[h].We define E0, E: H1(S1; M) H1(S1) R and V :M (0, 0) R to be

E0(h, v) = E0(u(h, v)), E(h, v) = E(u(h, v)); h H1(S1; M), v H1(S1), vL 0,V(h, v) = V(u(h, v)); h M, v (0, 0).

Hence, by means of formula (68), the functional E0 assumes the expression

E0(h, v) = 1

2S1 |

h

|2 +

|v

|2 + v2

|H(h)[h]

|2 + 2vh

H(h)[h]

= L0(h) +1

2

S1

|v|2 + v2|H(h)[h]|2 + 2vh H(h)[h]

.

If we differentiateE0 with respect to a variation k ThH1(S1; M) ofh, we obtain

DhE0(h, v)[k] = DL0(h)[k] +

S1

v2

H(h)[h] H(k)[k] + H(h)[h] LkH(h)[h]

(69)

+

S1

v

2H(h)[h,k] +h LkH(h)[h]

.

Here LkH, see Notations, denotes the Lie derivative ofHin the direction k. Similarly, if we differentiateE0 with respect to a variation w H1(S1) ofv, we have

(70) DvE0(h, v)[w] =S1

v wdt +S1

v w |H(h)[h]|2dt +S1

wh H(h)[h]dt.

From equations (66) and (67) it follows that if u() H1(S1;Rn), and ifdist(u(t), M) 0 for allt S1, then the condition DE(h, v) = 0 is equivalent to the system

(71)

DhE(h, v)[k] = 0 k ThH1(S1; M),DvE(h, v)[w] = 0 w H1(S1).

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Hence, in order to find critical points ofE (and hence ofE), we first solve the second equation in (71).Then, denoting by v(h) this solution, we solve in h the equation

DhE(u(h, v(h)))[k] = 0,

k

ThH

1(S1; M).

By (70), the equation inv DwE(u(h, v)) = 0 means thatv : S1 Ris solution of the following problem

(72)

v |H(h)[h]|2v= h H(h)[h] 1

Vv

(h, v). in [0, 2],

v(0) =v(2), v(0) = v(2).

Proposition 5.1 Suppose that the potentialVis of repulsive type, namely that (2) holds, and letA bea fixed positive constant. Then there existsA > 0 andCA >0 such that for all(0, A) and for allh LA0 equation (72) admits an unique solutionv(h) which satisfies(73) v(h)C0(S1) CA

.

Moreover the applicationh

v(h)fromLA0 toC

0(S1)is of classC1 and compact, namely if(hm)m

LA0,

then up to a subsequencev(hm) converge inC0(S1).

Proof. Equation (72) can be written it in the form (Qh, Ph andBh are defined in Section 2)

(74) v+

1

Bh Qh

v = Ph+

1

Bhv V

v(h, v)

,

Since V is of repulsive type and since every curve h H1(S1, M) is continuous, the function Bh isa negative continuous function of t and is bounded above by the negative constant b = supxMb(x).From Lemma 4.2 it follows that the resolutive operator h, : L1(S1) C0(S1) for the 2-periodicproblem

v+

1

Bh Qh

v = (t),

is well defined, whenever h H1(S1; M), L1(S1) and > 0 is sufficiently small. Moreover, byLemma4.2 (resp. Lemma4.3) it follows that, given > 0, the solution of problem (72) satisfies thefollowing estimate, provided is small enough

(75) h,L (1 + ) 12

|b| L1(S1)

resp.h,L (1 + ) |b| L(S1)

.

Forv C0(S1), letv = Ph+

1

Bhv V

v(h, v)

,

and let h, : v h,v. Then, as observed in the Notations it is

(76) Ph L1

(S1

); 1

Bhv V

v (h, v) L(S1).

We show that h, is a contraction in some suitable ballB(0) ={v C0(S1) :vL }, where will be chosen appropriately later.

IfvL , from(76), (75) and from the linearity of equation (74) it follows that

h,vL (1 + ) 12

|b|

S1

Ph+ (1 + ) |b|1

Bhv Vv (h, v)

L

.

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21/28

Proof. Setting for brevity v = v(h), equation(72) assumes the form

S1 v w+ S1 v w Qh+ S1 2w Ph 1

S1V

v(h, v) w= 0, w H1(S1).

In particular, by taking w = v in the last formula it turns out that v satisfies the relation

(82)

S1

v2 +

S1

v2 Qh+

S1

2v Ph =1

S1

V

v(h, v) v.

SinceE(h, v) has the expression

E(u(h, v)) =1

2

S1

|h|2 + |v|2 + v2 Qh+ 2v Ph

1

S1

V(h, v)dt,

we can take into account formula (82) to obtain

(83) G(h) =1

S1 1

2v V

v(h, v) V(h, v) .

SinceVis smooth, it turns out that

1

2v V

v(h, v) V(h, v) = (h, v) v3, |v| 0,

for some regular function . Hence, from equations (73) and (83) one infers that

|G(h)| 1 L v(h)3L C3A L

.

This proves the first inequality in (81).To show that G is of classC

1, we start proving that for some CA > 0 and for sufficiently small,

(84) : LA0

C0(S1), h

vh is of class C

1, and

Dh(h)

CA

.

Consider two elements h and l ofH1(S1; M) and the corresponding solutions v(h) and v(l) (which forbrevity we denote with vh andvl) of problem (72). By subtraction of the equations there results

y +

1

Bh Qh

y +

1

B+ Q

vl = P(85)

+ 1

(vh+ vl) (h, vh) y+ 1

[1+ 2] v2l,

where we have sety= vh vl; B = Bh Bl; Q= Qh Ql;

P =Ph Pl; 1= (h, vh) (l, vh); 2= (l, vh) (l, vl).Since is smooth, there holds

2= (l, vh) (l, vl) = (l, vh, vl) y,for another smooth function . Hence we can write equation (85) in the form

y +

1

Bh Qh 1

(vh+ vl) (h, vh) (l, vh, vl)

y(86)

=

1

B+ Q

vl+ P+

1

1a v2l.

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Now fixh H1(S1; M), and letl h; we can suppose that bothh, l LA0. If (0, A) (see Proposition5.1), then vh andvl satisfy inequality (73). Hence by choosing sufficiently small there results

1

Bh

1

(vh+ vl)

(h, vh)

(l, vh, vl)

1

2

b

.

Moreover, with computations similar to (77) one can easily prove that QhL1(S1) 2H2 A, so Lemmas4.2and 4.3 yield, for fixed and sufficiently small

yL (1 + ) 12

2|b|

CA

QL1(S1)

+ (1 + ) 2|b|

CA

1 BC0(S1)+ C2A

1

1C0(S1)

.

The last quantity tends to 0 as l h. This proves the continuity of : h v(h) from LA0 toC0(S1). The differentiability follows from the same reasoning, taking into account that the maps

h Qh and h Ph from H1(S1; M) toL1(S1; M)h Bh from H1(S1; M) toC0(S1; M) and (h, v) from M (0, 0) to R

are differentiable. We remark that the differentials of these functions are uniformly bounded for h LA0.Passing to the incremental ratio in (86), one obtains that the directional derivative D(h)[k] along thedirectionk TvhH1(S1; M) is the unique 2-periodic solution of the equation (linear iny )

y +

1

Bh Qh 2

(h, v) v 1

v2Dv(h, v)

y(87)

= DPh[k] +1

Dh(h, v)[k] v

2 1

DBh[k] v DQh[k] v.

From this formula (using for example a local chart on H1(S1; M)) one deduces immediately that thedifferential D is continuous on L

A0. Now we prove the estimate in (84). From equation(73) and from

the uniform boundedness on the differentials ofPh and Qh one can deduce that, for some constant C

independent onh LA0 there holdsDPh[k]L1(S1) C k;

1 v2 Dh(h, v)[k]C0(S1)

C k;1 v DBh[k]

C0(S1)

C 1 k; v DQh[k]L1(S1) C

k.

Moreover the coefficient ofx in formula (87) for sufficiently small can be estimated as

1

Bh Qh 2

(h, v) v 1

v2Dv(h, v) 1

2 b

,

Hence one can apply Lemmas4.2 and 4.3 to obtain

D(h)[k]

CA

k

,

h

LA0,

k

ThH

1(S1; M),

which is the estimate in (84).To prove the second inequality in (81) one can write G in the form G(h) =

1

S1

(h, vh) v3h, anddifferentiate with respect to h

(88) DG(h)[k] =1

S1

3(h, v) v2 Dv(h)[k] + v3

h[k] + v3

vDv(h)[k]

.

Now it is sufficient to apply the second inequality in (84). This concludes the proof.

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Lemma 5.3 Let(um)m H1(S1;Rn)be a Palais Smale sequence forE which for someR >0 satisfiesthe condition

(89) um(t) R, m N, t S1.Then, passing to a subsequence, um converges strongly inH1(S1;Rn).

Proof. To prove this claim we first note that by condition (89), the sequence of numbersS1

V(um) isbounded. Moreover there holds

1

2

S1

u2m = E(um) +1

S1

V(um),

so from the convergence ofE(um) and condition (89) one deduces that (um)m is bounded inH1(S1;Rn).

Hence, passing to a subsequence,um u0 weakly in H1(S1;Rn), and strongly in C0(S1;Rn).

As a consequence one has

(90) V(um()) V(u0()) and V (um()) V(u0()) uniformly on S1.From the fact DE(um) 0, it follows that

S1um z 1

S1

V(um)[z] = o(1) z, z H1(S1;Rn),

whereo(1) 0 as m +. In particular, taking as test function z = um, one has

(91)

S1

|um|2 1

S1

V(um)[um] = o(1) um =o(1), m +.

On the other hand, taking z = u0

(92)

S1

|u0|2 1

S1

V(u0)[u0] = limm

S1

um u0 1

S1

V(um)[um] = 0, m +.

From equations (91)and (92) it follows that

limmS1 |

um|2 =

S1 |u0

|2,

and henceum u0 strongly in H1(S1;Rn).

Corollary 5.4 The functionalL: LA0 R, (0, A), satisfies the Palais Smale condition.

Proof. Let (hm)m LA0 be a Palais Smale sequence for L, namely a sequence which satisfies(i)L(hm) c R, as n +;(ii)DL(hm) 0, as n +.

We want to prove that hm converges up to a subsequence to some function h0 H1(S1; M). By Lemma5.3it is sufficient to show that the sequenceum = u(hm, vhm) is a Palais Smale sequence for E. In fact,by the continuity ofh(u) the convergence ofum implies that ofhm.

To prove that um is a Palais Smale sequence for E, it is sufficient to take into account that, by the

choice ofvhn, it is DvE(hn, vhn) = 0, so one has

DhL(h)[k] = DhE(h, vh)[k] + DvE(h, vh)[Dv(h)[k]] = DhE(h, vh)[k].

Hence, it turns out that

DE(un)[w] = DhE(hn, vhn)[(Dh(un)[w]] + DvE(hn, vhn)[Dv(un)[w])] =DL(hn)[Dh(un)[w]].

It is immediate to check that the differential Dh(u) is uniformly bounded for |v| 0, so DL(hn) 0implies that alsoDE(un) 0. This conclues the proof.

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Since L is of class C1, then it is possible to prove, see for example [3], that there exists a pseudo

gradiet forL, namely a C1,1 vector field which satisfies the conditions

((h),

DL(h))

DL(h)

2,

2

DL

;

h

LA0.

induces locally a flow t, t 0, for which L is non descreasing, and which is strictly decreasing

whenever DL= 0.

Lemma 5.5 LetA >0 be a fixed constant. Then there existsA such that for all (0,A) and for allh LA0 the flowth is defined for all t 0.

Proof. By Proposition5.1 there exists 2A such that for (0, 2A) the functional L is well defined onL2A0 . Moreover, by Proposition5.2 we can choose 2A 2A such that

(93) |G(h)| C2A

, DG C2A

; (0, 2A), h L2A0 .

We choose A satisfying

A 2A; C2A A 14

A.

So, fixedh L2A0 , by our choice of A and by equation (93), there holds

L0(h) A L(h) A +14

A L(th) A +1

4A L0(th) A +

1

2A 0. By Proposition5.1 there exists20 >0 such that for (0, 20) the functional L is defined on L200 . Hence for sufficiently small itmakes sense to define also

= inf h[]H1(S1;M) L(h).

By using standard arguments, based on Corollary5.4and Lemma5.5, one can prove that the infimum is achieved by some critical point h ofL, and that 0 as 0. Now letk 0: since byProposition5.2 it isDGk(hk) O(

k), thenhk is a Palais Smale sequence for L0. Hence, passing

to a subsequence, hk must converge to a critical x0 point ofL0 at level 0. This concludes the proof inthe case1(M) = 0.

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Now consider the case of1(M) = 0. By Theorem2.5and by Remark2.6,there exists : Sl+1 M

which is not contractible. To the function one can associate the map

F : (Bl, B l)

(H1(S1; M), C0(S

1; M)),

whereC0(S1; M) denotes the class of constant maps from S1 to M. F is defined in the following way.First, identify the closed l-ballB l with the half equator on Sl+1 Rl+2 given by

{y= (y0, . . . , yl+1) Sl+1 :y0 0, y1 = 0}.Denote by cp(t), 0 t 1, the circle which starts out from p Bl orthogonally to the hyper plane{y1 = 0} and enters the half sphere{y1 0}. With this we put

F = {f cp(t)|0 t 1}.The correspondence F is clearly bijective, and hence one can define the value

0 = inf F,[]

suppSl

L0(F(p)).

Since is non contractible, it is possible to prove that 0 > 0 so, as above, for sufficiently small onecan define the quantity

= inf F,[]

suppSl

L(F(p)).

The number turns out to be a critical value for L and, again, ifk 0 we can find critical pointshk ofLk at level k converging to some geodesic x0 with L0(x0) = 0. This concludes the Proof ofTheorem1.2.

Remark 5.6 In [4]is studied the system (T) onR2. It is assumed, roughly, thatV possesses a non-contractible set of maxima, and are proved some existence and multiplicity results for large-period orbits.Dealing with two-dimensional systems, we suppose that V is non-degenerate in the sense of (1). In

particular, this implies thatM is a simple closed curve. WhenM is a manifold of maxima forV, thiscorresponds to the case treated in Theorem 1.1. The non-degeneracy of V allows us to describe in aquite precise way the asymptotic beheviour of the trajectories. Furthermore, according to Remark1.3, wecould also obtain existence of an arbitrarily large number of adiabatic limits, since we can apply the aboveargument to each different element of1(M). Note also that we can address the case in whichM is amanifold of minima, see the next Section.

6 Attractive potentials

In this section we describe how the arguments of Section 5 can be modified to handle the attractive case,namey that in which V (x)[nx, nx] is positive for all x M. Since attractive potentials may cause someresonance phenomena, the reduction procedure could fail. To overcome this difficulty we need to make

stronger assumptions on V . In particular we assume that the following condition holds

(94) 2V

n2x(x) = b0 > 0; for all x M.

We also set

(95) = supxM

3Vn3x (x) .

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Now we show that h, turns out to be a contraction. In fact, given two functions v, v C0(S1),

there holds

v v = 1 ((h, v) (h, v

)) v2 + (h, v) (v+ v) (v v) .Hence from formula (65) it follows that

h,v h,vC0(S1)

1

kDLC2 2 2 (1 + )

kb0

+ 2

b0

k

k

1

1 8 H A

b0

b0

2 A

k+ o(1)

v vC0(S1).

Ifand k are sufficiently small, then the coefficient ofv vC0(S1) in the last formula is strictly lessthan 1, henceF is a contraction in B. This concludes the proof of the existence. The compactness canbe proved in the same way as before.

About the functional L(h) = E(h, v(h)) we have the following analogous of Proposition 5.2.

Proposition 6.3 Suppose condition (94) holds true, and suppose that satisfies inequality (96). Thenfork sufficiently small the functionalG is of classC1 onLA0 and there existsCA > 0 such that

(102) |G(h)| CA

, |DG(h)| CA k; h LA0, (0, A).Proof. The proof is analogous to that of Proposition 5.2. The only difference is the estimate ofDk(h)[k]C0(S1), namely the norm of the 2 periodic solution of equation

y +

1

kBh Qh 2

k(h, v) v 1

kv2Dv(h, v)

y(103)

= DPh[k] + 1

kDh(h, v)[k] v

2 1k

DBh[k] v DQh[k] v,

which under assumption (94)takes the form

y +

1

kb0 Qh 2

k(h, v) v 1

kv2Dv(h, v)

y(104)

= (t) := DPh[k] + 1

kDh(h, v)[k] v

2 DQh[k] v L1(S1).

The study of this equation requires some modifications of the arguments in Section 4. The reason is thatthe coefficient ofy is not uniformly close (inL1(S1)) to the constant function 1

kb0.

Equation (104) can be written in the form

y+ 1

kb0 y = +

Qh+

2

k(h, v) v+

1

kv2Dv(h, v)

y .

So, applying equations (64) and (65) and taking into account of (99) one can deduce

yC0(S1) (1 + ) k2

b0 yC0(S1)

QhL1(S1)+ 4 Ck 2k + 4

1k

C2 k DL

+ (1 + )

k

2

b0 gL1(S1).

If the constant C is given by formula (101), then one can show that for and k sufficiently small thecoefficient ofxC0(S1) on the right hand side is stricly less than 1, so estimate (102) holds true. Thisconcludes the proof.

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Corollary 5.4 holds without changes also for this case, and having Proposition 6.3 one can easilyprove the analogous of Lemma5.5. These facts allow to apply the reduction on H1(S1; M) below thelevel b0

4H. As a consequence we have the following result.

Theorem 6.4 Suppose1(M) = 0 (resp. 1(M) = 0), and let0 (resp. 0) be the value which appearsin the proof of Theorem1.2. Suppose the following condition is satisfied

4 0 H < b0

resp. 4 0 H < b0

.

Then there exists a sequence Tk + and there exists a sequence of solutions (uk)k to problem (T)corresponding toT =Tk such that up to subsequenceuk(Tk)converge inC0(S1,Rn). The adiabatic limitofuk(Tk) is a non trivial closed geodesicx0 onMat level0 (resp. 0).

Proof of Theorem1.4. It is an immediate consequence of Theorem6.4,since condition (ii) implies that = 0.

References

[1] A. Ambrosetti, M. Badiale,Homoclinics: Poincare-Melnikov type results via a variational approach,Ann. Inst. Henri. Poincare Analyse Non Lineaire 15 (1998), 233-252.

[2] Brezis, H.:Analyse Fonctionelle, Masson 1983.

[3] Chang, K.C.: Infinite-dimensional Morse Theory and Multiple Solution Problems, Birkhauser, 1993.

[4] Cingolani, S., Lazzo, M.: Multiple periodic solutions for autonomous conservative systems, to appearon Top. Meth. Nonlin. Anal.

[5] Gilbarg, D., Trudinger, N.:Elliptic Partial Differential equations of the second order, Springer 1983.

[6] Klingenberg, W. : Riemannian Geometry, Walter de Gruyter, 1982.

[7] Lusternik, L.A., Fet, A.I.: Variational problems on closed manifolds, Dokl. Akad. Nauk SSSR (N.S.)81 17-18 (Russian) (1951).

[8] Manton, N.: A remark on the scattering of BPS monopoles, Phys. Lett. B 110 (1982), 54-56.

[9] Manton, N.: Monopole interactions at long range, Phys. Lett. B 154 (1985).

[10] Spivak, M.:A comprehensive introduction to differential geometry, Publish or Perish, 1977.

[11] Uhlenbeck, K.: Moduli Spaces and Adiabatic Limits, Notices A.M.S. 42-1 (1998), 41-42.

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1 adiabatic limits of closed orbits for some newtoniansystems in r^n (malchiodi)

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