Minimization properties of Hill's orbits and applications to ... - CiteSeerX

Ann. Inst. Henri Poincaré, Analyse non linéaire 17, 5 (2000) 617–650  2000 Éditions scientifiques et médicales Elsevier SAS. All rights reserved

Minimization properties of Hill’s orbits and applications to some N -body problems * by

Gianni ARIOLIa,1 , Filippo GAZZOLAa,2 , Susanna TERRACINIb,3 a Dipartimento di Scienze e T.A. C.so Borsalino 54, 15100 Alessandria, Italy b Dipartimento di Matematica, Via Bonardi 9, 20133 Milano, Italy

Manuscript received 11 November 1999, revised 22 February 2000

A BSTRACT. – We consider the periodic problem for a class of planar N-body systems in Celestial Mechanics. Our goal is to give a variational characterization of the Hill’s (retrograde) orbits as minima of the action functional under some geometrical and topological constraints. The method developed here also turns out to be useful in the study of the full problem with N primaries each having at most two satellites.  2000 Éditions scientifiques et médicales Elsevier SAS Key words: N-body problem, Non-collision orbits AMS classification: 70F10, 47J30

R ÉSUMÉ. – On considère le problème périodique pour une certaine classe de systèmes de N-corps en Mécanique Céleste. Notre but est de donner une caractérisation variationnelle des orbites (rétrogrades) de Hill comme minima de la fonctionnelle d’action sous certaines contraintes géométriques et topologiques. La méthode ici développée est également * This research was supported by MURST project “Metodi variazionali ed Equazioni Differenziali Non Lineari”. 1 E-mail: [email protected]. 2 E-mail: [email protected]. 3 E-mail: [email protected].

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utile pour l’étude du problème complet avec N corps primaires ayant chacun au plus deux satellites.  2000 Éditions scientifiques et médicales Elsevier SAS

1. INTRODUCTION This paper concerns the periodic problem for a class of planar N body systems in Celestial Mechanics. We mainly deal with a class of 3-body problems, though our method may be applied in some particular cases of more general N -body systems. We consider the case of two major bodies and a satellite and we seek orbits when the whole system revolves with a frequence θ , while the third mass rotates around one of the other bodies with a T -periodic motion. In the restricted case, when the motion of the satellite takes place close to one of the primaries, this problem is known as Hill’s problem and a simple argument based on the inverse function theorem shows the existence of periodic orbits (in the rotating frame) for small values of the quantity θT , see [12]. Our goal is to give a variational characterization of the Hill’s (retrograde) orbits as minima of the action functional under some geometrical and topological constraints. The method developed here also turns out to be useful in the study of the (full) problem with N primaries each having at most two satellites. The periodic problem for both restricted and full N -body systems has such a long story that it is impossible to give an extensive bibliography here; we refer the reader to the classical texts [12–14]. In the last two decades, a new method for finding periodic motions has been provided by the use of variational techniques, see, e.g., the book [3] and the references therein. The first variational characterization of the periodic solutions of the 2-body problem goes back to a paper by Gordon [11], where it is shown that the periodic orbits are local minima of the action under the topological constraint of non triviality of the rotation index. This constraint is used to overcome the lack of coercivity of the action integral in the space of periodic functions. However, from the functional point of view, the minimization problem (even in a local sense) is degenerate, that is it possesses a continuum of solutions. This is due to the fact that every solution to the associated differential equation is periodic, provided its energy is negative, and the period (and the associated action

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value) depends only on the energy. In particular all the periodic orbits, including the degenerate ellipses, where the two bodies collide, share the same variational characterization; in other words, it is impossible to distinguish them by looking at their functional levels. On the other hand, even though they can be extended as global solutions to the differential equations, the motions of collision type are periodic only in a mathematical sense. Starting from the subsequent paper by Gordon [10], different kinds of assumptions have been considered in order to rule out the collision solutions, in the case of 2-body and N -body problems. In the case of Keplerian interaction potentials, a fundamental remark is that the minimization problem may become non degenerate by imposing further symmetry constraints in the space of periodic functions. This fact has been first pointed out in the 2-body case in [9], and then exploited in order to obtain noncollision periodic orbits in various situations [4,6–8,16,17]. In the planar N -body problem this idea has led to associate the boundary condition xi (t + τ ) = Rϕ xi (t) with the equations system, where xi (t) is the position of the i-mass, and Rϕ a rotation of the plane of angle ϕ and τ is the period of the mutual distances between the bodies see [4]. In this setting, it can be shown (see [5]) that the simple minimization argument in the space of symmetric functions leads to the relative equilibrium motions that are well known periodic solutions to the system [1]. In order to avoid such a triviality, Bessi and Coti Zelati [4] imposed a further topological constraint, that is one of the body couples has a non trivial rotation index in the rotating frame. However, though they ruled out the simultaneous collisions of the whole system, they were unable to avoid periodic solutions having partial collisions. The aim of this paper is to go further in the analysis in [4], and prove, by level estimates, that the minimum of the action functional under both symmetry and topological constraints is free of any collision. We first deal with the 3-body problem in the restricted and full cases: to this end, we exploit the variational structure of the problem, looking for minimizers of the action integral among the functions which satisfy the above mentioned constraints. More precisely, we require the system coordinates X(t) = (x1 (t), x2 (t), x3 (t)) to satisfy xi (t + T ) = RθT xi (t), where RθT is a rotation of angle θT in the plane; in addition we require the motion of R−θt x3 (t) to have a negative rotation index with respect to the first body, that is, to be retrograde. Finally, we use similar arguments to study more general problems with more major bodies each one having at most two retrograde satellites.

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2. STATEMENT OF THE RESULTS Throughout this paper we denote by G the universal gravitation constant, by B1 , . . . , Bn (n = 3 or n = 4) the bodies of the problem we consider and by m1 , . . . , mn their respective masses; their positions in the plane R2 are described by the n functions xi = xi (t) (i = 1, . . . , n). D EFINITION 1. – We say that (x 1 , . . . , x n ) is a noncollision orbit on the interval [0, T ] if x i (t) 6= x j (t) for all 1 6 i < j 6 n and all t ∈ [0, T ]. We first deal with a restricted 3-body problem {B1 , B2 , B3 }, which we briefly describe. Consider for a moment only the system {B1 , B2 }: if we assume the center of mass to be fixed in the origin and we set the period to be 2π/ϑ , then a solution of the following equations of motion 

xi − xj −mi x¨i = Gm1 m2 , |xi − xj |3

xi (0) = xi



2π , ϑ

i, j = 1, 2, i 6= j,

is given by x1 (t) = −R1 (cos ϑt, sin ϑt),

x2 (t) = R2 (cos ϑt, sin ϑt),

(2.1)

where R1 = m2 G1/3 (m1 + m2 )−2/3 ϑ −2/3 , R2 = m1 G1/3 (m1 + m2 )−2/3 ϑ −2/3 . The restricted problem we consider consists in assigning x1 (t), x2 (t) as in (2.1), while the motion of B3 satisfies the equation −m3 x¨3 = V 0 (x3 ),

(2.2)

where V (x3 ) = −

Gm1 m3 Gm2 m3 − . |x3 − x1 | |x3 − x2 |

Fix T , ϑ > 0 so that ϑT < 2π . Let Rα be the rotation in R2 by an angle α; we denote by Rα the corresponding matrix as well 

cos α Rα = sin α



− sin α . cos α

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Note that x1 (t) = Rϑt x1 (0) and x2 (t) = Rϑt x2 (0). We look for solutions of (2.2) satisfying x3 (T ) = RϑT x3 (0); to this end we introduce the space 






H = x ∈ H 1 [0, T ], R2 : x(T ) = RϑT x(0) . Moreover, we require B3 to orbit around B1 without colliding with neither B1 nor B2 ; more precisely we consider the noncollision set 



Λ0 = x ∈ H : x(t) 6= xi (t) (i = 1, 2) for all t ∈ [0, T ] and 



Λ1 = x ∈ Λ0 : ind(x − x1 ) 6= 0 , where ind(y) denotes the winding number of y in the interval [0, T ]. Consider the following Lagrangian functional L(x) =

ZT 0

m3 2 |x| ˙ − V (x), 2

(2.3)

whose critical points correspond to solutions of (2.2). Then, we have THEOREM 1. – There exists a continuum of periodic and quasiperiodic noncollision solutions of (2.2). More precisely, there exists ν ∈ (0, 2π ) (depending only on the ratio m2 /m1 ) such that if ϑT 6 ν then problem (2.2) admits a solution x3 ∈ Λ1 ; moreover, x3 minimizes L over Λ1 .

We prove Theorem 1 in Section 3. Of course, a major problem concerning the statement of Theorem 1 is to estimate ν: in order to show that our results are not perturbative we take the masses of B1 , B2 and B3 to be respectively the masses of the Earth, the Sun and the Moon. In this case m2 /m1 ≈ (3.3)105 : moreover, the real (direct) motion of this 3-body system has the corresponding ϑT ≈ 0.46; hence, it is of some interest to show that our method avoids collisions (for the retrograde motion) when ν = 0.46. In fact, our next result states that even larger values are allowed: 2. – Assume that m2 /m1 = (3.3)105 ; then, if ϑT 6 0.8 problem (2.2) admits a solution x3 ∈ Λ1 ; moreover, x3 minimizes L over Λ1 . THEOREM

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This result is proved in Section 6 where we also give the pictures of some numerical experiments: such results lead us to conjecture that the upper bound for ν could be even larger. Next we consider a full planar 3-body problem {B1 , B2 , B3 }; their motion is described by the equations −mi x¨i = Vi (x),

i = 1, 2, 3,

(2.4)

where the potential is given by V (x) = −

X

Gmi mj |x − xi | 16i 0 depending only on m1 , m2 such that if ϑT 6 ν and m3 6 M, then problem (2.4) admits a solution x ∈ Λ1 ; moreover, x minimizes Φ over Λ1 .

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This result enables us to study the planar 4-body problem {B1 , B2 , B3 , B4 } where B3 and B4 are two satellites of B1 . Their motion is described by the equations −mi x¨i = Vi (x),

i = 1, 2, 3, 4,

(2.6)

where the potential is given by V (x) = −

X

Gmi mj . |x − xi | 16i 0 depending only on m1 , m2 such that if ϑT 6 ν and m3 , m4 6 M, then problem (2.6) admits a solution x ∈ Λ1 ; moreover, x minimizes Φ over Λ1 .

Remark 1. – The proofs of our results may be naturally modified in order to obtain similar statements for N -body problems {B1 , . . . , BN } with k major bodies (k < N ), each one having at most two satellites. Remark 2. – Although we seek solutions in Λ1 (a set having nontrivial index for some couples of bodies), in fact by construction all the solutions

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we find have the index equal to −1, therefore they represent a clockwise motion, while the two major bodies are assumed to rotate anticlockwise.

3. THE RESTRICTED PROBLEM In this section we prove Theorem 1. First assume that ϑT < 2π . We switch to a rotating coordinate frame (0, e1 , e2 ) so that the bodies B1 and B2 , whose motions in the original coordinate frame are described by (2.1), are at rest. More precisely, the position of B1 is −R1 e1 and the position of B2 is R2 e1 . If we set q(t) = R−ϑt x3 (t) and if 

J=

0 −1 1 0



denotes the standard symplectic matrix, then we get x˙3 (t) = R˙ ϑt q(t) + Rϑt q(t) ˙ and |x˙3 (t)|2 = |q(t)+ϑJ ˙ q(t)|2 ; the condition x3 (T ) = RϑT x3 (0) becomes q(T ) = q(0). Next we rescale the period T and we translate the system in order to have B1 at the origin by setting y(t/T ) = q(t) + R1 e1 , so that we get the standard Lagrangian functional of the restricted 3-body problem: 1 I (y) = T

Z1 0

2 m3 y˙ + T ϑJ (y − R1 e1 ) − T 2 V (y), 2

where V (y) = −

Gm1 m3 Gm2 m3 − . |y| |y − (R1 + R2 )e1 |

Up to the addition of a constant, we may redefine the Lagrangian I as follows: 1 I (y) = T

Z1 0

m3 |y˙ + T ϑJy|2 − m3 ϑ 2 T 2 R1 (y, e1 ) − T 2 V (y); 2

we remark that the corresponding Euler–Lagrange equations are unchanged. We set x(t) = (T 2 Gm1 )−1/3 y(t), we introduce the adimen-

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sional parameters ρ = m2 /m1 and ν = ϑT and we define −1/3 Φ(x) = m−1 (Gm1 )−2/3 I (x) − 3 T

ρν 2/3 , (ρ + 1)1/3

so that we get Φ(x) =

Z1

1 1 |x˙ + T ϑJ x|2 + 2 |x|

0

−2/3

T 2/3 G1/3 m2 m1 + |(T 2 Gm1 )1/3x − (m1 + m2 )1/3 G1/3 ϑ −2/3 e1 | − =

Z1

1 1 m2 /m1 |x˙ + T ϑJ x|2 + + 2 |x| |x − (m2 /m1 + 1)1/3 (ϑT )−2/3 e1 |

0

− =

ϑ 2 T 4/3 m2 G1/3 ρν 2/3 (x, e ) − 1 (Gm1 )1/3 (m1 + m2 )2/3 ϑ 2/3 (ρ + 1)1/3

Z1

ϑ 4/3 T 4/3m2 1/3

m1 (m1 + m2 )2/3

(x, e1 ) −

ρν 2/3 (ρ + 1)1/3

1 1 ρ |x˙ + νJ x|2 + + 2 |x| |x − (ρ + 1)1/3 ν −2/3 e1 |

0

−

ρν 4/3 ρν 2/3 (x, e ) − . 1 (ρ + 1)2/3 (ρ + 1)1/3

For all ξ ∈ R2 define the functions g(ξ ) :=

ρ |ξ − (ρ + 1)1/3 ν −2/3 e1 |

and f (ξ ) := g(ξ ) −

ρν 2/3 ρν 4/3 − (ξ, e1 ), (ρ + 1)1/3 (ρ + 1)2/3

so that Φ(x) =

Z1 0

Note that

1 1 |x˙ + νJ x|2 + + f (x). 2 |x|

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ξ − (ρ + 1)1/3 ν −2/3 e1 , |ξ − (ρ + 1)1/3 ν −2/3 e1 |3 ρ|ζ |2 ∇ 2 g(ξ )[ζ, ζ ] = − |ξ − (ρ + 1)1/3 ν −2/3 e1 |3 3ρ(ξ − (ρ + 1)1/3 ν −2/3 e1 , ζ )2 + , |ξ − (ρ + 1)1/3 ν −2/3 e1 |5 and, by expanding in McLaurin polynomial, there exists s ∈ (0, 1) (s = s(ξ )) such that ∇g(ξ ) = −ρ

ρν 2/3 ρν 4/3 ρ|ξ |2 + (ξ, e ) − 1 (ρ + 1)1/3 (ρ + 1)2/3 2|sξ − (ρ + 1)1/3 ν −2/3 e1 |3 1/3 −2/3 2 e1 , ξ ) 3ρ(sξ − (ρ + 1) ν + ; 1/3 −2/3 2|sξ − (ρ + 1) ν e1 |5 therefore, we also have for some s = s(ξ ) ∈ (0, 1) g(ξ ) =

ρ|ξ |2 2|sξ − (ρ + 1)1/3 ν −2/3 e1 |3 3ρ(sξ − (ρ + 1)1/3 ν −2/3 e1 , ξ )2 + . (3.1) 2|sξ − (ρ + 1)1/3 ν −2/3 e1 |5 The Lagrangian functional L defined in (2.3) is therefore transformed into Φ while the original space H is transformed in the Hilbert space H11 of 1-periodic functions in H 1 . Consider the noncollision open subset f (ξ ) = −





Λ0 = x ∈ H11 : x(t) 6= 0 and x(t) 6= (ρ + 1)1/3 ν −2/3 e1 for all t ∈ [0, 1] and





Λ1 = x ∈ Λ0 : ind(x) = −1 .

(3.2)

Theorem 1 is proved if we show that the problem inf Φ(x)

x∈Λ1

admits a solution x ∈ Λ1 and no solutions in ∂Λ1 . L EMMA 1. – Let {x n } ⊂ Λ1 be a minimizing sequence for Φ; then there exists x ∈ Λ1 such that x n * x, up to a subsequence, and Φ(x) = infx∈Λ1 Φ(x). Proof. – In the following c represents a generic positive constant which may vary within the same formula. Let x0 = R(sin 2π t, cos 2π t) with

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R = (2π )−2/3 ; then x0 achieves the minimum of the functional Ψ (x) =

Z1 0

|x| ˙2 1 + 2 |x|

under the constraint ind(x) = −1; x0 is not a minimum of Φ, therefore for large values of n we have Φ(x n ) 6 Φ(x0 ) and Z1 0

2
1 n ρν 2/3 x˙ + νJ x n − c x n , e1 6 Φ(x0 ) + = c. 2 (ρ + 1)1/3

Note that if ν 6 ν then Z1

inf

kxk ˙ 2 =1



|x˙ + νJ x|2 = 1 −

0

ν 2π

2



> 1−

ν 2π

2

;

(3.3)

so the last inequality yields

2



c x˙ n 2 − c x˙ n 2 6 c and finally kx˙ n k2 6 c. By the topological requirements and Poincaré inequality we infer kx n k∞ 6 ckx˙ n k2 6 c, therefore {x n } is bounded in H and, up to a subsequence, it admits a weak limit x ∈ H , which is a minimum of Φ by the weak lower semicontinuity of the functional Φ. Finally, x ∈ Λ1 by uniform convergence of {x n }. 2 To complete the proof of Theorem 1 we exclude the case x ∈ ∂Λ1 in the statement of the previous lemma; first of all it is possible to obtain a contradiction to a collision between B3 and B2 , because they are not linked by a topological constraint. Indeed, if there exists τ ∈ [0, 1] such that x(τ ) = (ρ + 1)1/3 ν −2/3 e1 , we can modify the trajectory of x = x(t) in a neighborhood of τ in order to lower both the kinetic part and the potential part of Φ and to avoid a collision. Since the proof is standard, we omit it. L EMMA 2. – Let x be as in Lemma 1; then x(t) 6= (ρ + 1)1/3 ν −2/3 e1 for all t ∈ [0, 1]. The proof that B3 does not collide with B1 is more subtle and requires some estimates; we exclude collisions for sufficiently small ν:

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L EMMA 3. – Let x denote an orbit obtained in Lemma 1. There exists ν ∈ (0, 2π ) such that if ν 6 ν then x(t) 6= 0 for all t ∈ [0, 1] and inf Φ < inf Φ; Λ1

∂Λ1

in particular, x ∈ Λ1 . Proof. – In this proof we denote by Ki (i = 1, . . . , 4) positive constants depending (eventually) on ρ. By contradiction, assume that x ∈ ∂Λ1 , where x is determined by the statement of Lemma 1; without loss of generality, we may assume that x(0) = 0 and hence kxk2 6

1 kxk ˙ 2, π

1 kxk∞ 6 kxk ˙ 2. 2

(3.4)

Consider again the function x0 = R(sin 2π t, cos 2π t) with R = (2π )−2/3 ; if ν is sufficiently small, then we have |sx0 (t) − (ρ + 1)1/3 ν −2/3 e1 | > 1 (ρ + 1)1/3 ν −2/3 for all t ∈ [0, 1] and s ∈ (0, 1), hence, (3.1) yields 2 f (x0 ) 6

ρR 2 6 K1 ν 2 , 2|sx0 − (ρ + 1)1/3 ν −2/3 e1 |3

∀t ∈ [0, 1],

and Φ(x0 ) 6

(2π − ν)2 3 2/3 2 + (2π ) + K ν < (2π )2/3 − K2 ν + K3 ν 2 . (3.5) 1 2(2π )4/3 2

Since x minimizes Φ, by (3.3) (3.4) and (3.5) we get 

2

ρν 4/3 ρν 2/3 k xk ˙ − 2 π(ρ + 1)2/3 (ρ + 1)1/3 3 6 Φ(x) 6 Φ(x0 ) < (2π )2/3 − K2 ν + K3 ν 2 ; 2

1 ν 1− 2 2π

kxk ˙ 22 −

therefore, we obtain (independently of ν 6 ν 6 π ) kxk ˙ 2 6 K4 : in turn, if ν is sufficiently small, by (3.4) this yields kxk∞ 6

1 (ρ + 1)1/3 . 2 ν 2/3

(3.6)

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On the other hand, by (3.1) Φ(x) >

Z1 0

1 1 ρ|x|2 |x˙ + νJ x|2 + − 2 |x| 2|sx − (ρ + 1)1/3 ν −2/3 e1 |3

which, together with (3.6), gives Φ(x) >

Z1 0

1 1 |x˙ + νJ x|2 + − 4ν 2 |x|2 ; 2 |x|

(3.7)

finally, note that Z1

inf

x∈H01

0

1 1 |x˙ + νJ x|2 + − 4ν 2 |x|2 2 |x| Z1

= inf

x∈H01

0

1 2 1 |x| ˙ + − 4ν 2 |x|2 . 2 |x|

(3.8)

Therefore, since we assume that x ∈ H01 minimizes Φ, by (3.4), (3.5), (3.7) and (3.8) 3 (2π )2/3 − K2 ν + K3 ν 2 2 > Φ(x0 ) > Φ(x) > inf

x∈H01

> inf

y∈H01



Z1 0



Z1 0

1 2 1 |x| ˙ + − 4ν 2 |x|2 2 |x|

2

1 8ν 1 1 − 2 |y| ˙ 2+ 2 π |y|

8ν 2 = 1− 2 π

1/3

Z1

inf

z∈H01

0



1 2 1 8ν 2 |˙z| + = 1− 2 2 |z| π

1/3

3 (2π )2/3 2

which is impossible if ν > 0 is small enough: hence, for such ν, we get a contradiction and the lemma is proved. 2 Theorem 1 is proved: if θT 6 ν we obtain a solution of (2.2); such solution is periodic if θT /π ∈ Q, while it is quasi-periodic if θT /π ∈ R \ Q.

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4. THE FULL 3-BODY PROBLEM In this section we prove Theorem 3. To this end, we show that the Lagrangian functional Φ defined in (2.5) admits a global minimum in the (open) set 



Λ1 = x ∈ Λ0 : ind(x3 − x1 ) = −1 , where Λ0 is the noncollision open set 



Λ0 = x ∈ H : xi (t) 6= xj (t) for all t ∈ [0, T ] and i 6= j . We first prove L EMMA 4. – Let {x n } ⊂ Λ1 be a minimizing sequence for Φ; then {x n } is bounded in H . P R

Proof. – Since Φ(x) > i 0T m2i |x˙i |2 for all x ∈ Λ0 , then kx˙ n k2 is bounded. Since xi (T ) = RϑT xi (0), then |xi (t + T ) R− xi (t)|2 = |(RϑT − R0 )xi (t)|2 = c|xi (t)|2 . But xi (t + T ) − xi (t) = tt +T x˙i (s) ds, hence |xi (t)|2 6 c−1 kx˙in k21 6 T c−1 kx˙in k22 for all t, so kx n k∞ is bounded and finally kx n k is bounded. 2 We use again a rotating coordinate system by setting Qi (t) = R−ϑt xi (t) for all x = (x1 , x2 , x3 ) ∈ H so that x˙i (t) = R˙ ϑt Qi (t) + Rϑt Q˙ i (t) and ˙ i (t) + ϑJ Qi (t)|2 . Next we rescale the period by setting |x˙i (t)|2 = |Q y(t/T ) = Q(t). Up to a multiplication by T , the Lagrangian becomes I (y) =

Z1 X 3 0

i=1

mi |y˙i + T ϑJyi |2 − T 2 V (y); 2

P

P

clearly, the constraint i mi xi (t) ≡ 0 transforms into i mi yi (t) ≡ 0. We introduce a new Hilbert space of periodic functions (which we still denote by H ), defined by (




H = y = (y1 , y2 , y3 ) ∈ H [0, 1], R : y(1) = y(0), 1

6

3 X

)

mi yi (t) ≡ 0 ,

i=1

the corresponding noncollision open set 

Λ0 = y ∈ H : yi (t) 6= yj (t) for all t ∈ [0, 1] and i 6= j



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and its subset 



Λ1 = y ∈ Λ0 : ind(y3 − y1 ) = −1 . Therefore, the original problem of minimizing Φ is reduced to the following minimization problem: inf I (y).

(4.1)

y∈Λ1

We prove that the infimum in (4.1) is achieved: L EMMA 5. – There exists y ∈ Λ1 such that I (y) = infy∈Λ1 I (y). Proof. – Let {y n } ⊂ Λ1 be a minimizing sequence for I ; then, the corresponding sequence {x n } defined by setting xin (t) = Rϑt yin (t/T ) is minimizing for Φ: by Lemma 4, it is bounded and so is {y n }. Therefore, up to a subsequence, {y n } admits a weak limit y ∈ H , which is a minimum of I by the lower semicontinuity of the functional I ; finally, y ∈ Λ1 by uniform convergence of the sequence {y n }. 2 As in Section 3, one can easily exclude collisions between B2 and B3 : we have to exclude the other possible collisions. Let I0 be the Lagrangian I corresponding to m3 = 0: as I0 does not depend on the third component y3 of y, we may identify it with its restriction to the subspace H0 corresponding to y3 = 0, namely (


4

H0 = y = (y1 , y2 ) ∈ H 1 [0, 1], R : y(1) = y(0),

2 X

)

mi yi (t) ≡ 0

i=1

and we study the minimization (2-body) problem inf I0 (x),

(4.2)

x∈Ω0

where Ω0 is the corresponding noncollision open set. Consider the function f : R4 → R defined by f (ξ1 , ξ2 ) =

2 X mi i=1

2

ϑ 2 |ξi |2 +

m 1 m2 |ξ1 − ξ2 |

(ξi ∈ R2 );

one can easily check that f has a unique strict global minimum, up to a SO(2)-symmetry: more precisely, there exists Σ ⊂ R4 (which is a SO(2)

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orbit), such that f attains its global minima on Σ . One such minimum is 

ϑ −2/3 −



m2 m1 , 0, ,0 2/3 (m1 + m2 ) (m1 + m2 )2/3

and the corresponding Hessian matrix of f has rank 3 with 3 strictly positive eigenvalues (we denote by C > 0 the smallest of these eigenvalues); the 0 eigenvalue corresponds to the direction tangent to Σ . Therefore, if we denote by ξ Σ = (ξ1Σ , ξ2Σ ) the projection on Σ of any ξ = (ξ1 , ξ2 ) ∈ R4 sufficiently close to Σ we have 2 X mi i=1

>

2

ϑ 2 |ξi |2 +

2 X mi

2

i=1



m 1 m2 |ξ1 − ξ2 | 2

ϑ 2 ξiΣ +

2 m 1 m2 CX ξi − ξ Σ 2 ; + i Σ Σ |ξ1 − ξ2 | 2 i=1

(4.3)

in particular, any (stationary) point in Σ is also a minimum for the functional I0 . Fix ϑ > 0: we prove that if m3 and T are small enough and y = (y1 , y2 , y3 ) is the minimum obtained in Lemma 5, then (y1 , y2 ) is close to Σ in the H norm topology. In particular, this shows that B1 and B2 do not collide. L EMMA 6. – There exist two constants T , M > 0 depending only on θ, m1 , m2 and a constant c > 0 such that for all T 6 T , all m3 6 M and all y = (y1 , y2 , y3 ) ∈ H achieving the minimum in (4.1) (as given by Lemma 5) we have 2 X



2


ky˙i k22 + T 2 yi − yiΣ 2 6 cT 4/3 m3 ,

i=1

where yiΣ = yiΣ (t) is the (pointwise) projection of yi (t) onto Σ . Proof. – Take any (yE , yS ) ∈ Σ and consider the function Y ∈ H defined by   Y1 (t) ≡ yE , 

Y2 (t) ≡ yS , Y3 (t) = yE + T 2/3 (sin 2π t, cos 2π t),

so that V23 (yE , Y3 ) = −

m 2 m3 T 2/3

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and Y˙3 (t) = 2π T 2/3(cos 2π t, − sin 2π t). Then the kinetic part of Y3 is given by

|Y˙3 + T ϑJ Y3 |2 = T 4/3 2π cos 2π t, − sin 2π t





2 + T 1/3 ϑJ (yE + T 2/3 sin 2π t, cos 2π t) .

If T is small enough, then Y ∈ Λ1 , and by taking into account that infΛ1 I 6 I (Y ) we obtain inf I0 (y) 6 inf I 6 inf I0 + c1 T 4/3 m3 + c2 T 2 m3 . Λ1

Λ1

Λ1

(4.4)

Let {mn3 } be a vanishing sequence, let In be the functional correspondn ing to mn3 and let y n = y n (mn3 , T ) ∈ Λ1 be the minimum of In obtained in Lemma 5. If yin (i = 1, 2, 3) denote the components of y n , then by (4.4) we infer that (y1n , y2n ) is a minimizing sequence for I0 , hence it converges weakly in H and uniformly to some (yE , yS ) ∈ Σ , up to a subsequence. In particular, this proves that for any given ε > 0 there exists mε3 > 0 such that if m3 6 mε3 , then y1 (t) − yE 6 ε

and

y2 (t) − yS 6 ε

for all t ∈ [0, 1].

(4.5)

R

For all y ∈ H 1 (S 1 , R2 ) satisfying 01 y = 0 we have kyk∞ 6 kyk ˙ 16 γ kyk ˙ 2 (the first is Wirtinger inequality and the second is Hölder’s). Then Z1 Z1 Z1 ! ˙ y) = J y, ˙ y − y 6 kyk ˙ 22 . (J y, 0

0

(4.6)

0

1 1 4 R 1 Now let y = (yE , yS ), fix y ∈ H (S , R ) and let y˜ = y − y; since 0

Z1

(J y, ˙ y) = 0, then |y˙i +T ϑJyi | =

Z1

2

0




|y˙i |2 +T 2 ϑ 2 |y i + y˜i |2 −2T ϑ(J y˙i , y˜i ) ,

i = 1, 2.

0

Recall that f (y) = f (y Σ (t)) for all t ∈ [0, 1], using (4.3), (4.5) and (4.6) we get: I0 (y) − I0 (y) =

Z1 X 2 0

i=1


mi |y˙i |2 + T 2 ϑ 2 |yi |2 − 2T ϑ(J y˙i , y˜i ) 2

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T 2 m1 m2 + − |y1 − y2 | >

Z1 X 2 0

>

i=1

2 X mi i=1

4

Z1 X 2 0

i=1

mi 2 2 Σ 2 T 2 m1 m2 T ϑ yi + Σ 2 |y1 − y2Σ |

2
C mi |y˙i |2 − 2T ϑ(J y˙i , y˜i ) + T 2 yi − yiΣ 2 2

ky˙i k22 +

2 C 2 T yi − yiΣ 2 ; 2

the last inequality follows by Wirtinger inequality and (4.6) since, for T small enough, we have Z1 X 2 0

mi T ϑ(y˙i , J y˜i ) 6

2 X mi

i=1

i=1

4

ky˙i k22 .

Finally, from (4.4) we get infΛ1 I − infΛ1 I0 6 c1 T 4/3m3 + c2 T 2 m3 and therefore 2 X mi i=1

4

ky˙i k22 +

2 C 2 T yi − yiΣ 2 6 c1 T 4/3 m3 + c2 T 2 m3 , 2

which proves the estimate.

2

Finally, to prove that the minimum obtained in Lemma 5 is a noncollision minimum, we show that B1 and B3 do not collide: L EMMA 7. – If T and m3 are sufficiently small, then inf I (y) < inf I (y);

y∈Λ1

y∈∂Λ1

in particular, there exists y ∈ Λ1 such that I (y) = infy∈Λ1 I (y). Proof. – Fix T > 0, let IT be the corresponding functional and let y T ∈ Λ1 be the minimum of IT over Λ1 obtained in Lemma 5: we claim that y T ∈ / ∂Λ1 for T and m3 sufficiently small. By contradiction, let y ∈ ∂Λ1 be a collision minimum, that is, I (y) = infΛ1 I , where we have set I = IT . By Lemma 6 we know that (y1 , y2 ) is close in the norm topology of H 1 ([0, 1], R4 ) to some point in Σ ; in particular, y1 (t) 6= y2 (t) for all t ∈ [0, 1]. Moreover, if we set V (q) = −

m1 m2 − , |q − y1 | |q − y2 |

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then y3 minimizes the restricted functional Ψ (q) =

Z1 0

1 |q˙ + T ϑJ q|2 − T 2 V (q) 2

on the set Ω of functions q ∈ H 1 ([0, 1], R2 ) satisfying q(0) = q(1), q(t) 6= yi (t) for all t ∈ [0, 1] and i = 1, 2 and ind(q − y1 ) = −1. We set e(t) = T −2/3 y1 (t), q(x) = T 2/3 (x + e) and Φ(x) = T −4/3 Ψ (q(x)); then, we infer that y0 (t) = T −2/3 (y3 (t) − y1 (t)) minimizes the functional Φ(x) =

Z1  0

+

1 2 m1 |x| ˙ + + (x, ˙ e˙) + T ϑ(x, ˙ J x) + 2T ϑ(e, ˙ J x) 2 |x| 

T 2ϑ 2 2 m2 + c(e), |x| + T 2 ϑ 2 (e, x) + T 2/3 2/3 2 |T (x + e) − y2 |

where c(e) =

Z1 0

|e| ˙2 T 2ϑ 2 2 + T ϑ(e, ˙ J e) + |e| . 2 2

Next we define Φ(x) = Φ(x) − c(e) −

Z1 0

m2 T 2/3 , |y2 |

so that y0 also minimizes the functional Φ(x) =

Z1  0

1 2 m1 |x| ˙ + + (x, ˙ e˙) + T ϑ(x, ˙ J x) + 2T ϑ(e, ˙ J x) 2 |x| 

T 2ϑ 2 2 + |x| + T 2 ϑ 2 (e, x) + T 4/3 m2 G(x) , 2 where G(x) is the smooth function given by G(x) = T

−2/3





1 1 ; − 2/3 |T (x + e) − y2 | |y2 |

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moreover, since y2 (t) 6= 0 for all t, by the asymptotic expansion 

1 (u, v) 1 1+ε = + o(ε) |εu − v| |v| |v|



(which holds for all u, v ∈ R2 \ {0} as ε → 0) we have as T → 0 

G(x) = x + e,

y2 |y2 |3



+ o(1).

Let Ξ = {t ∈ [0, 1]: y0 (t) = 0}. It is well known [2] that Ξ has measure zero and that y0 (t) satisfies the Euler equation corresponding to Φ for all t∈ / Ξ , namely y0 y¨0 = −2T ϑJ y˙0 − T 2 ϑ 2 y0 + T 2 ϑ 2 e − 2T ϑJ e˙ − e¨ − m1 |y0 |3


+ T 4/3 m2 ∇G(y0 ) and if we compute the scalar product of this equation by Jy0 we obtain
d ¨ Jy0 (y˙0 , Jy0 ) = (y¨0 , Jy0 ) = −2T ϑJ y˙0 + T 2 ϑ 2 e − 2T ϑJ e˙ − e, dt
+ T 4/3 m2 ∇G(y0 ), Jy0 ; since y0 ∈ ∂Λ1 we may assume that 0 is a collision time, i.e. (y˙0 , Jy0 )(0) = 0: hence, by integrating the previous equation on the interval [0, t] with 0 < t 6 1, we get (y˙0 , Jy0 )(t) = −2T ϑ

Zt

(y˙0 , y0 ) + T ϑ 2

Zt 2

0

+

Zt 0

(e, Jy0 ) − 2T ϑ

0

(e, ˙ J y˙0 ) + T

Zt 4/3

m2

Zt

(e, ˙ y0 )

0




∇G(y0 ), Jy0 .

(4.7)

0

Now we estimate the integrals in (4.7): since, y0 = y0 (T ) is bounded in H 1 as T → 0, we have Zt 2T ϑ (y˙0 , y0 ) 6 2T ϑky˙0 k2 ky0 k2 6 2λT

(4.8)

0

for a.e. t ∈ [0, 1]. Choose a constant µ > 0 and let m3 6 µT 8/3 ;

(4.9)

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since e(t) = T −2/3 y1 (t), by Lemma 6 we have kek ˙ 2 6 µT 4/3 and therefore Zt 2

T ϑ

2

(e, Jy0 ) − 2T ϑ

0

Zt

(e, ˙ y0 ) +

0

Zt

(e, ˙ J y˙0 ) = o(T )

(4.10)

0

for a.e. t ∈ [0, 1]. Hence, by (4.7), (4.8) and (4.10) we obtain for small T

(y˙0 , Jy0 )

∞

6 λT .

Consider the function X(t) = R(sin 2π t, cos 2π t),

where R =

q 3

m1 /(4π 2 ),

which minimizes the functional Iˆ(x) =

Z1 0

|x| ˙ 2 m1 + 2 |x|

˙ J X) ≡ −2π R 2 and under the constraint ind(x) = −1; we have (X, therefore Φ(X) − Φ(y0 ) = Iˆ(X) − Iˆ(y0 ) + T ϑ

Z1





˙ J X) − (y˙0 , Jy0 ) + o(T ) < 0 (X,

0

for small T . This contradicts the assumption that y0 minimizes Φ and proves the lemma. 2 To complete the proof of Theorem 3, note that a noncollision critical point x = (x1 , x2 , x3 ) of Φ satisfies (2.4); then, it also satisfies








x˙i (T ), p(T ) = x˙i (0), p(0) = RϑT x˙i (0), p(T ) ,

i = 1, 2, 3,

for all p ∈ H 1 ([0, T ], R2 ); hence, x˙i (T ) = RϑT x˙i (0). This proves that the motion x is periodic if θT /π ∈ Q while it is quasi-periodic if θT /π ∈ R \ Q.

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5. THE 4-BODY PROBLEM In this section we prove Theorem 4. We may assume that m4 6 m3 . As in the previous section, by rotating and rescaling, the Lagrangian becomes I (y) =

Z1 X 4 0

i=1

mi |y˙i + T ϑJyi |2 − T 2 V (y), 2

where V (y) = −

X

Gmi mj ; |y − yi | 16i 0 depending only on θ, m1 , m2 and a constant c > 0 such that for all T 6 T , all m4 6 m3 6 M and all y = (y1 , y2 , y3 , y4 ) ∈ H achieving the minimum in (4.1) (as given by Lemma 8) the following holds 2 X



2


ky˙i k22 + T 2 yi − yiΣ 2 6 cT 4/3 m3 ,

i=1

where yiΣ = yiΣ (t) is the (pointwise) projection of yi (t) onto Σ .

G. ARIOLI ET AL. / Ann. Inst. Henri Poincaré 17 (2000) 617–650

639

Proof. – Take any (yE , yS ) ∈ Σ and consider the function Y ∈ H defined by  Y1 (t) ≡ yE ,   

Y2 (t) ≡ yS ,  Y (t) = yE + T 2/3 (sin 2π t, cos 2π t),   3 Y4 (t) = yE − T 2/3 (sin 2π t, cos 2π t), so that (4.4) holds (recall that m4 6 m3 ). The proof may now be completed as in Lemma 6. 2 In particular, Lemma 9 excludes collisions between B1 and B2 . Now we return to the original problem: in order to prove Theorem 4 it suffices to show that the functional Φ(x) =

ZT X 4 0

i=1

X Gmi mj mi |x˙i |2 + 2 |x − xi | 16i 0 sufficiently small and m3 6 µT 8/3 as in (4.9); then, by Lemma 7, there exists CT > 0 (independent of m3 and m4 ) such that inf F (x) − inf F (x) > CT ,

x∈∂Ω1

x∈Ω1

the two infima being in fact two minima: let X be one minimum of F over Ω1 . By definition of Ψ we also have inf Ψ > inf Ψ + CT m4

∂Ω2

Ω2

(5.1)

and the minimum of Ψ over Ω2 is achieved by (X(t), X(t + s)) for any s ∈ [0, T ]. Now take X3 (t) = X(t) and take X4 (t) = X(t + T /2) so that X4 is also a noncollision minimum of F . Since X minimizes F , then a

G. ARIOLI ET AL. / Ann. Inst. Henri Poincaré 17 (2000) 617–650

641

simpleR argument shows that X3 (t) 6= X4 (t) for all t and we can define G C0 = 0T |X3 −X . Assume by contradiction that (x 3 , x 4 ) ∈ ∂Ω2 ; then, by 4| (5.1), we have Φ(x 3 , x 4 ) = Ψ (x 3 , x 4 ) +

ZT 0

> Φ(X3 , X4 ) −

Gm3 m4 > inf Ψ + |x 4 − x 3 | ∂Ω2

ZT 0

Gm3 m4 + |X4 − X3 |

ZT 0

ZT 0

Gm3 m4 |x 4 − x 3 |

Gm3 m4 + C T m4 |x 4 − x 3 |

> Φ(X3 , X4 ) − C0 m3 m4 + CT m4 and therefore Φ(x 3 , x 4 ) > Φ(X3 , X4 ) for sufficiently small m3 (and m4 ): this contradicts the assumption that (x 3 , x 4 ) minimizes Φ over Ω 2 and proves the lemma. 2 Finally, we exclude the case where the two satellites collide with each other: L EMMA 11. – x 3 (t) 6= x 4 (t) for all t ∈ [0, T ]. Proof. – We make the following change of variables: let  

m3 x3 + m4 x4 , m3 + m 4  r = x4 − x3 , X=

and we denote by (X, r) the couple corresponding to (x 3 , x 4 ). We focus our attention on x3 and x4 and we consider the restricted Lagrangian Ψ (X, r) =

ZT 0

m3 + m 4 ˙ 2 m 3 m4 Gm3 m4 |X| + |˙r |2 + − VR (X, r), 2 2(m3 + m4 ) |r|

where VR (X, r) = −

Gm1 m3 Gm1 m4 Gm2 m3 Gm2 m4 − − − . |x3 − x 1 | |x4 − x 1 | |x3 − x 2 | |x4 − x 2 |

Obviously, Φ(x 1 , x 2 , x3 , x4 ) = Ψ (X, r) +

ZT 0

m1 m2 Gm1 m2 |x˙ 1 |2 + |x˙ 2 |2 + . 2 2 |x 2 − x 1 |

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By contradiction, assume that in the motion described by x = x(t) the satellites B3 and B4 collide with each other, but not with B1 , at t = 0, that is r(0) = 0 and X(0) 6= x 1 (0). Let ε > 0 satisfy ε
Ψ (X, R). We may assume that r(tε ) = (ε, 0) and r(t ε ) = εeiα for some α ∈ (−π, π ]; let p = ε 3/2 eiα/2 . Let `=

|r(tε ) − p| |r(t ε ) − p| = ; ε ε

since |α|/2 6 π/2, we have |r(tε )|2 + |p|2 − 2(r(tε ), p) ε 2 + ε 3 − 2(r(tε ), p) = 6 1 + ε. ε2 ε2 (5.2) Let S1 and S2 be respectively the segments connecting r(tε ) and r(t ε ) with p and let rˆ (t) be such that `2 =

rˆ (t) ∈ S1

and

rˆ (t) ∈ S2

and

rˆ (t) − p = ` r(t) rˆ (t) − p = ` r(t)

∀t ∈ [tε , 0], ∀t ∈ [0, t ε ].

Since the motion of rˆ is straight, for all t ∈ [tε , 0] we get  
rˆ (t) − p · d rˆ (t) = rˆ (t) − p, d rˆ (t) − p dt dt 2 1 d rˆ (t) − p 2 = ` d r(t) 2 = 2 dt 2 dt d 6 `2 r(t) · r(t) , dt

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and by (5.2) 2 2 2 d rˆ (t) 6 `2 d r(t) 6 (1 + ε) d r(t) ; dt dt dt

(5.3)

similarly, we obtain the same inequality when t ∈ [0, t ε ]. Next, note that





VR X(t), rˆ (t) − VR X(t), r(t) > −cε

for all t ∈ [tε , t ε ];

this, together with (5.3) yields



Zt ε





dr 2 dˆr 2 m3 m4 − dt 2(m3 + m4 ) dt

tε







+ VR X(t), rˆ (t) − VR X(t), r(t) > −cε.

(5.4)

1 To estimate |r| − |ˆ1r | , we argue as in [17]: by conservation of the total energy E we obtain for all t ∈ [tε , t ε ] 2 m1 x˙ (t) 2 + m2 x˙ (t) 2 + m3 x˙ (t) 2 + m4 x˙ (t) 2 c r˙ (t) 6 1 2 3 4 2 2 2 2 Gm1 m2 Gm1 m3 Gm1 m4 =E + + + |x 2 (t) − x 1 (t)| |x 3 (t) − x 1 (t)| |x 4 (t) − x 1 (t)| Gm2 m3 Gm2 m4 Gm3 m4 c + + + 6 , |x 3 (t) − x 2 (t)| |x 4 (t) − x 2 (t)| |r(t)| |r(t)|

the latter inequality being consequence of the fact that there exists K > 0 such that for all ε ∈ (0, ε) we have 





inf ε min x j (t) − x i (t) , 1 6 i < j 6 4, (i, j ) 6= (3, 4) > K.

t ∈(tε ,t )

Therefore, if we let ρ(t) = |r(t)|, we have ρ ρ˙ =


1 d 2 c ρ = r(t), r˙ (t) 6 r(t) · r˙ (t) 6 ρ √ ; 2 dt ρ

this proves that ρ(t) 6 ct 2/3

∀t ∈ [tε , t ε ].

(5.5)

Let tε < s1 < 0 < s2 < t ε satisfy |r(si )| = ε 3/2 /2 and |r(t)| < ε 3/2/2 for all t ∈ (s1 , s2 ); by (5.5) we obtain |si | > cε 9/4 . Moreover, since

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|ˆr (t)| > |r(t)| for all t ∈ (tε , t ε ) we infer that Zs1 tε

1 > |r|

Zs1 tε

Zt ε

1 |ˆr |

and s2

1 > |r|

Zt ε s2

1 ; |ˆr |

on the other hand, |ˆr (t)| > ε 3/2 > 2|r(t)| for all t ∈ (s1 , s2 ) so that, by (5.5), we obtain Zt ε

Gm3 m4 tε

1 1 − > Gm3 m4 |r| |ˆr | >c

cε Z 9/4

−cε 9/4

Zs2 s1

1 t 2/3

1 1 − >c |r| |ˆr |

cε Z 9/4

−cε 9/4

1 2|r|

> cε 3/4 .

This, together with (5.4), implies that Ψ (X, r) − Ψ (X, R) > cε 3/4 − cε > 0 if ε is small. 2 The proof of Theorem 4 may now be completed as in the previous section. 6. NUMERICAL RESULTS In this section we prove Theorem 2 and we give some numerical results illustrating the orbits determined in Theorem 1. Since all the solutions we obtained are minima, it is possible to numerically compute them by a rather simple procedure. The technique is standard: one chooses a finite dimensional vector space Hf which approximates the Hilbert space H , introduces a functional Φf : Hf → R approximating Φ and looks for minima of Φf by choosing an arbitrary starting point x0 ∈ Hf and defining a sequence {xn } by setting xn+1 = xn − hn ∇Φf (xn ), where ∇Φf : Hf → Hf is the gradient of Φ and represents the maximum slope direction of the functional Φf , while hn is computed at each step in order to minimize the function h 7→ Φ(xn+1 (h)). If the approximated functional maintains the properties of Φ, then the sequence {xn } converges to a minimum point of the functional Φf . We only treat the restricted problem (although there are no obstructions to the treatment of the complete problem). As an approximate space Hf we chose the set of closed m-gonals and we let m vary between 100 and 300, depending on the values of the parameters. A function x ∈ Hf is

G. ARIOLI ET AL. / Ann. Inst. Henri Poincaré 17 (2000) 617–650

645

uniquely characterized by the coordinates of the vertices, therefore it can be represented by a point in R2m . Given x ∈ Hf by xi ∈ R2 we denote the coordinates of the ith vertex. The (full) functional we consider is Φ(x) =

Z1 0

−

1 1 ρ |x˙ + νJ x|2 + + 2 |x| |x − (ρ + 1)1/3 ν −2/3 e1 | ρν 4/3 (x, e1 ), (ρ + 1)2/3

and its representation in Hf is given by 2 m X 1 xi+1 − xi 1 ρ Φf (x) = + νJ xi + + 2 h |xi | |xi − (ρ + 1)1/3 ν −2/3 e1 | i=1

−

ρν 4/3 (xi , e1 ), (ρ + 1)2/3

where h = 1/m. We also have ∂Φf (x) xi+1 − 2xi + xi−1 xi xi+1 − xi =− − − 2νJ + ν 2 xi 2 3 ∂xi h |xi | h 1/3 −2/3 4/3 e1 xi − (ρ + 1) ν ρν −ρ − e1 . |xi − (ρ + 1)1/3 ν −2/3 e1 |3 (ρ + 1)2/3 Clearly, also in the numerical approximation we have to cope with the presence of the singularity in the potential; furthermore is it not very clear how to implement the topological constraint. In order to overcome these problems we introduced a naive method, i.e. we checked that at every step no vertices of xn were too close to the singularities. More precisely we checked that the minimum of the distances of the vertices from the singularity was larger than the maximum of the length of the sides of the m-gonal. In fact this condition was satisfied at all times during all our computations. The following pictures represent the results we obtained for various values of the parameters ν and ρ. As a starting point x0 we chose the orbit corresponding to the solution for the case ν = 0, that is the circular orbit of radius R = (2π )−2/3 . Although analytically we could exclude collisions in the case ρ = (3.3)105 only for values of ν smaller than 0.8, numerically it clearly appears that the minima of the functional is very close to a circular orbit even for values of ν up to 3. For larger values the orbit becomes more similar to an ellipse, but still it does not come

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Fig. 1.

close to a collision orbit. In order to show some different behavior, we also show a picture (Fig. 1) in the case ρ = 1 and ν = 5. Proof of Theorem 2. – Let ρ = (3.3)105 , ν 6 0.8 and let x denote a corresponding orbit found in Lemma 1. We argue as in the proof of Lemma 3 making finer estimates. By contradiction, assume that x ∈ ∂Λ1

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and consider again the function x0 = (2π )−2/3 (sin 2π t, cos 2π t). By arguing corresponding to (3.5) we find that (if ν 6 0.8) 3 3 Φ(x0 ) < (2π )2/3 − (0.541)ν + (0.175)ν 2 < (2π )2/3. 2 2

(6.1)

Since x minimizes Φ, by (3.3) and (6.1) we get 

1 ν 1− 2 2π

2

kxk ˙ 22 −

ρν 4/3 ρν 2/3 3 k xk ˙ − 6 (2π )2/3 ; (6.2) 2 2/3 1/3 π(ρ + 1) (ρ + 1) 2

by (3.4) this proves that kxk ˙ 2 < 130,

kxk∞ < 65 < (0.811)

(ρ + 1)1/3 . ν 2/3

(6.3)

Now we claim that for all t ∈ [0, 1] we have
ρ ρν 4/3 x(t), e − 1 |x(t) − (ρ + 1)1/3 ν −2/3 e1 | (ρ + 1)2/3 ρν 2/3 > −(0.035) . (6.4) (ρ + 1)1/3

h(x(t)) :=

If (x(t), e1 ) 6 0, (6.4) follows readily. If (x(t), e1 ) > 0, then by (6.3) we get 2/3 (ρ + 1)2/3 x(t) − (ρ + 1)1/3 ν −2/3 e1 2 < kxk2 + (ρ + 1) < (1.658) ; ∞ 4/3 4/3

ν

ν

then (6.4) follows by estimating (x(t), e1 ) with kxk∞ and by (6.3). Since ν 6 0.8, by (6.4) we may replace (6.2) with (0.3807)kxk ˙ 22 − (1.035)

ρν 2/3 3 6 (2π )2/3 , (ρ + 1)1/3 2

(6.5)

which, by (3.4), proves that kxk ˙ 2 < 106,

kxk∞ < 53 < (0.661)

(ρ + 1)1/3 ; ν 2/3

(6.6)

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G. ARIOLI ET AL. / Ann. Inst. Henri Poincaré 17 (2000) 617–650

this allows to prove that


h x(t) > (0.173)

ρν 2/3 (ρ + 1)1/3

∀t ∈ [0, 1].

(6.7)

Indeed, if (x(t), e1 ) 6 0, we have


h x(t) >

ρ ρν 2/3 > (0.6) . kxk∞ + (ρ + 1)1/3 ν −2/3 (ρ + 1)1/3

If (x(t), e1 ) > 0, then by arguing as for (6.4) and by taking into account (6.6) we get (6.7). By (6.7) the inequality (6.5) becomes (0.3807)kxk ˙ 22 − (0.827)

ρν 2/3 3 6 (2π )2/3 , 1/3 (ρ + 1) 2

(6.8)

which, by (3.4), proves that kxk ˙ 2 < 94.8,

kxk∞ < 47.4 < (0.592)

(ρ + 1)1/3 . ν 2/3

Finally, repeating once more the whole procedure, we get h(x(t)) > (0.267)ρν 2/3 (ρ + 1)−1/3 which yields kxk ˙ 2 < 89.1,

kxk∞ < 44.55 < (0.556)

(ρ + 1)1/3 . ν 2/3

Hence, |sx(t) − (ρ + 1)1/3 ν −2/3 e1 | > (0.444)(ρ + 1)1/3 ν −2/3 and Z1 0

ρ|x|2 ˙ 22 ν 2 kxk < 6 (0.3705)kxk ˙ 22 . 2|sx − (ρ + 1)1/3 ν −2/3 e1 |3 2(0.444)3 π 2

Therefore, by (3.1) and (6.1) we obtain 3 ˙ 22 − (2π )2/3 > Φ(x) > (0.3807)kxk 2 > (0.0102)kxk ˙ 22 ;

Z1 0

ρ|x|2 2|sx − (ρ + 1)1/3 ν −2/3 e1 |3

G. ARIOLI ET AL. / Ann. Inst. Henri Poincaré 17 (2000) 617–650

649

this proves that kxk ˙ 2 < 22.4,

kxk∞ < 11.2 < (0.14)

(ρ + 1)1/3 . ν 2/3

Repeating this procedure we get Z1 0

ρ|x|2 ˙ 22 ν 2 kxk < 6 (0.051)kxk ˙ 22 . 2|sx − (ρ + 1)1/3 ν −2/3 e1 |3 2(0.86)3 π 2

Therefore, by (3.1), (6.1) and Jensen’s inequality we obtain 3 π ˙ 22 + , (2π )2/3 > (0.3297)kxk 2 kxk ˙ 2 which proves that kxk ˙ 2 < 3.6,

kxk∞ < 1.8 < (0.023)

(ρ + 1)1/3 ν 2/3

and Z1 0

ρ|x|2 ν 2 kxk22 < 6 (0.537)ν 2 kxk22 . 2|sx − (ρ + 1)1/3 ν −2/3 e1 |3 2(0.977)3

Therefore, by arguing as in the proof of Lemma 3 we get 3 (2π )2/3 − (0.541)ν + (0.175)ν 2 2 > Φ(x0 ) > Φ(x) >

Z1 0



(1.074)ν > 1− π2 

= 1−

(1.074)ν π2

1 1 |x˙ + νJ x|2 + − (0.537)ν 2 |x|2 2 |x|

2 1/3

Z1

inf

z∈H01 2 1/3

0

1 2 1 |˙z| + 2 |z|

3 (2π )2/3 . 2 2

Since ν 6 0.8 we have (1 − (1.074)ν )1/3 > 1 − (0.038)ν 2 and therefore the π2 last sequence of inequalities yields (0.37)ν > 0.541 which contradicts ν 6 0.8 and proves the statement. 2

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G. ARIOLI ET AL. / Ann. Inst. Henri Poincaré 17 (2000) 617–650

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