On the dissipative spin-orbit problem DOCTOR OF SCIENCES of

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Doctoral Thesis

On the dissipative spin-orbit problem

Author(s): Antognini, Francesco

Publication Date: 2014

Permanent Link: https://doi.org/10.3929/ethz-a-010272525

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DISS. ETH Nr. 21926

On the dissipative spin-orbit problem

A thesis submitted to attain the degree of

DOCTOR OF SCIENCES of ETH ZURICH

(Dr. sc. ETH Zurich)

presented by

FRANCESCO ANTOGNINI

MSc ETH Zurich

born on 27.05.1985

citizen ofChiasso TI, Switzerland

accepted on the recommendation of

Prof. Dr. Daniel Stoffer, examinerProf. Dr. Norbert Hungerbuhler, co-examiner

Prof. Dr. Luigi Chierchia, co-examinerProf. Dr. Luca Biasco, co-examiner

2014

Acknowledgements

First, I would like to thank my supervisor Daniel Stoffer for giving me the opportunityto write my PhD thesis at ETH and for supervising my work. He provided with hiswillingness for a positive research environment and for interesting and very construc-tive discussions.

I want to express my deep gratitude to my long distance supervisors Luca Biasco andLuigi Chierchia for the lively discussions and their constant support throughout mywork on this thesis. With their great mathematical and human skills, they guaranteedme endless high-quality support.

I thank Prof. Norbert Hungerbuhler for acting as co-examiner.

I thank my colleagues and former colleagues from ETH for the enjoyable working en-vironment and pleasant moments spent together; as well as the ETH for the financialsupport.

I thank the IT support group, specially Michele Marcionelli (“San Michele”) for hispatience and very precious help.

A special thank go to Renato and Marco for hosting me, and to all others relativesfor providing an enjoyable atmosphere and food, during my visits in Rom.

I thank Alberto for the Matlab consulting.

Further thanks to my friends Patric and Jacopo, for all their support and all the greattime that we could spend together.

Above all, I would like to thank my parents Luciana and Piero, my sister Caterina,my brother Martino, my aunt Laura and my girlfriend Katharina, whose constantencouragement and love was crucial for the completion of this thesis. Thank you.

i

Contents

1 The physical model 11.1 Two-body problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1.1 Historical introduction (based on [17]) . . . . . . . . . . . . . . 11.1.2 The model (based on [6] and [7]) . . . . . . . . . . . . . . . . . 3

1.2 The spin-orbit problem . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.2.1 The model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.2.2 The Moment of inertia tensor . . . . . . . . . . . . . . . . . . . 121.2.3 Motivation of the rotational assumption about the z−axes . . . 131.2.4 MacCullagh’s Potential (based on [20]) . . . . . . . . . . . . . . 141.2.5 The Euler equations and the equation of motion (based on [20]) 161.2.6 The dissipative case . . . . . . . . . . . . . . . . . . . . . . . . 19

1.3 Definition of spin-orbit resonances . . . . . . . . . . . . . . . . . . . . 20

2 On the Fourier coefficients of the Newtonian Potential 212.1 Two Integral Formulas for αj . . . . . . . . . . . . . . . . . . . . . . . 212.2 Asymptotic behaviour . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

2.2.1 Hansen’s Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . 272.2.2 Alternative Proof of Theorem 2.1 . . . . . . . . . . . . . . . . . 32

3 Generalized Newtonian Potential 453.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

3.1.1 Lyapunov-Schmidt decomposition . . . . . . . . . . . . . . . . 473.1.2 The range equation . . . . . . . . . . . . . . . . . . . . . . . . . 47

3.1.3 The bifurcation equation . . . . . . . . . . . . . . . . . . . . . 483.2 Computing φ(k)(ξ) for k ≥ 0 . . . . . . . . . . . . . . . . . . . . . . . . 513.3 Condition for non-degeneration of φu(ξ) at any degree . . . . . . . . . 54

3.3.1 Condition for non-degeneration of φu(ξ) at zeroth degree . . . . 543.3.2 Condition for the non-degeneration of φu(ξ) at first degree . . . 563.3.3 Condition for the non-degeneration of φu(ξ) at kth degree . . . 58

3.4 Application to the classical spin-orbit problem . . . . . . . . . . . . . 58

4 Quantitative conditions for the existence of of low-order spin-orbit resonances 61

4.1 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 614.2 Proof of Theorem 4.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

5 Numerical analysis of (p, q)-periodic orbits of the DSOP 735.1 Basic notation and functions . . . . . . . . . . . . . . . . . . . . . . . 73

5.1.1 Functions of t in matrix form . . . . . . . . . . . . . . . . . . . 735.1.2 Function of ξ and t in cell array form . . . . . . . . . . . . . . 77

5.2 The DSOP and its representation with Matlab . . . . . . . . . . . . . 815.3 Control and results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

6 The spin-orbit resonances of the Solar System: a simple mathematical theory 93

6.1 Spin–orbit resonances in the Solar System . . . . . . . . . . . . . . . . 93

iii

iv Contents

6.2 Theorem for Solar System spin–orbit resonances . . . . . . . . . . . . 956.3 Proof of Theorem 6.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 956.4 Small bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1056.5 Eccentricity Intervals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106

7 Estimates on the basin of attraction 1097.1 Main results and applications . . . . . . . . . . . . . . . . . . . . . . . 1097.2 Proof of Theorem 7.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

7.2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1117.2.2 The homogeneous equation Lz = 0 . . . . . . . . . . . . . . . . 1137.2.3 Fundamental solutions of the homogeneous equation Lz = 0 . . 1187.2.4 Fixed point argument . . . . . . . . . . . . . . . . . . . . . . . 1197.2.5 Estimates on the fundamental solutions c(t) and s(t) . . . . . . 1237.2.6 Technical estimates . . . . . . . . . . . . . . . . . . . . . . . . . 1257.2.7 Proof of Theorem 7.2 . . . . . . . . . . . . . . . . . . . . . . . 127

7.3 Proof of Theorem 7.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128

8 Numerical estimates on the basin of attraction 1378.1 General setting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1378.2 An explicit integrator of order two . . . . . . . . . . . . . . . . . . . . 1388.3 Phase portrait of Mercury . . . . . . . . . . . . . . . . . . . . . . . . . 139

Abstract

The spin-orbit model studies the movement of a satellite orbiting about a central bodyon a Keplerian elliptical orbit with eccentricity e ∈ [0, 1). The main difference withthe classical two-body problem is, that one takes care of the complex structure of thesatellite. In particular we imagine the central planet as a point of mass and the satel-lite, having an ellipsoidal form, moving about it. Thus rotations of the satellite on itsown axis are considered. We assume that the satellite has fixed spin axis, coincidingwith the shortest physical axis of the ellipsoid, perpendicular to the orbital plane (noobliquity). This (apparently small) distinction makes the study of this problem veryinteresting.One of the most natural question is about possible resonances of the system, i.e., underwhich conditions do (p, q)−periodic orbits exist, p, q ∈ Z. We say that the system has a(p, q)−resonance (or equivalently that the two bodies move on a (p, q)−periodic orbit)if the satellite makes p−revolutions on itself when it completes q−revolutions aboutthe central planet. The most famous example is the Earth-Moon system, which is a(1, 1)−periodic orbit, indeed from the Earth we see always the same half of the Moon.The curious fact is that the (1, 1)−resonance seems to be the most stable one. Indeed,with the important exception of Mercury-Sun system, which is in (3, 2)−resonance,we observe only (1, 1)−resonances in the solar system. But why does this happen? InLow-order resonances in weakly dissipative spin-orbit models L. Biasco and L. Chier-chia give conditions for the existence of periodic orbits for q = 1, 2 and 4 and estimateson the measure of the basins of attraction of stable periodic orbits for q = 1, 2 arediscussed.Since the main motivation of this dissertation is the better understanding of the dis-sipative spin-orbit problem, the principal goals of this thesis are:

• Verifying explicitly that the conditions for the coexistence of periodic orbits forq = 1, 2 and 4, given by L. Biasco and L. Chierchia in Low-order resonancesin weakly dissipative spin-orbit models, are satisfied if the eccentricity is smallenough, providing also concrete applications to all observed systems in the SolarSystem admitting a (p, q)−periodic orbit.

• Finding explicit conditions for the existence of (p, q)-periodic orbits for all q ≥3, q 6= 4, and discussing lower bounds on the basins of attraction for any q ≥ 3;

• Developing a numerical method to compute approximations of (p, q)−periodicorbits.

v

Riassunto

Il problema spin-orbita studia il movimento di un satellite, che orbita su una traietto-ria ellittica con eccentricita e ∈ [0, 1) attorno ad un pianeta principale. La principaledifferenza con il famoso problema dei due corpi, e che in questo problema tiene contodella forma del satellite. Nello specifico si assume che il satellite abbia una formaellissoidale, mentre il pianeta principale si assume essere puntiforme. Di conseguenzavanno considerate le rotazioni del satellite attorno al proprio asse di rotazione. Inoltresi assume che l’asse di rotazione coicida con il semiasse polare dell’ellissoide e cheesso sia perpendicolare al piano orbitale. Questa differenza (apparentemente piccola)rende lo studio del problema spin-orbita molto interessante.Una delle domande piu naturali che ci si pue porre studiando questo problema riguardale condizioni di esistenza di orbite (p, q)−periodiche con p, q ∈ Z. Con orbita o riso-nanza (p, q)−periodica si intende una traiettoria in cui il satellite compie p rotazioni suse stesso nel tempo in cui compie q rivoluzioni attorno al pianeta principale. L’esempiopiu famoso di questo fenomeno e il sistema Terra-Luna. La Luna descrive un’orbita(1, 1)−periodica attorno alla Terra, infatti noi (dalla Terra) vediamo sempre la stessafaccia della Luna. E curioso il fatto che la risonanza (1, 1) sembra essere la piu stabile.Infatti, eccezion fatta per Mercurio che descrive un orbita (3, 2)−periodica con il Sole,nel nostro sistema solare la risonanza (1, 1) e l’unica astronomicamente osservabile.Come si spiega questo comportamento? Nell’articolo Low-order resonances in weaklydissipative spin-orbit models L. Biasco e L. Chierchia danno condizioni per l’esistenzadi orbite periodiche nei casi in cui q = 1, 2 e 4 e stimano i bacini d’attrazione di orbiteperiodiche nei casi in cui q = 1, 2.Siccome la motivazione principale di questa tesi di dottorato e una migliore compren-sione del problema spin-orbita dissipativo ne elenchiamo qui gli obiettivi principali:

• Verificare quantitativamente che le condizioni per la coesistenza di orbite pe-riodiche per q = 1, 2 e 4, date da L. Biasco e L. Chierchia in Low-order res-onances in weakly dissipative spin-orbit models, siano soddisfatte per valoridell’eccentricita sufficientemente piccoli, dando anche applicazioni concrete atutti i sistemi pianeta-satellite nel sistema solare, che ammettono un’orbita(p, q)−periodica;

• Trovare condizioni per l’esistenza di orbite (p, q)−periodiche per q ≥ 3, q 6= 4 estimare inferiormente il bacino d’attrazione di orbite (p, q)−periodiche per q ≥ 3;

• Sviluppare un metodo numerico per calcolare approssimativamente orbite(p, q)−periodiche.

vii

Introduction

The Solar System is fascinating in several different ways. To name only one of which,synchronous moving patterns of moons around their planet has to be mentioned.Earthlings, for example, are facing always the same side of their satellite.In the Solar System we observe 18 moons1 in a so–called 1:1 spin–orbit resonance:the moons show always the same face to their hosting planets, i.e. during one orbitalrevolution around their host, they rotate one time around their proper spin axis. Thisspin axis is approximately perpendicular to the plane of revolution (see Figure 1).

P1

2

3

4

5

6

Figure 1: Ellipsoidal satellite in a 1 : 1 spin-orbit resonance.

The 3:2 resonance of Mecury around the Sun is the only exception of 1:1 spin-orbitresonance observable in the Solar System (see Figure 2).

P1

2

3

4

5

0 ≡ 12

P7

8

9

10

11

6

Figure 2: (3, 2)−periodic orbit.

In general we say that the system has a (p, q)−resonance (or equivalently that the twobodies move on a (p, q)−periodic orbit) if the satellite makes p−revolutions on itself

1The complete list is: Moon (Earth); Io, Europa, Ganymede, Callisto (Jupiter); Mimas, Ence-ladus, Tethys, Dione, Rhea, Titan, Iapetus, (Saturn); Ariel, Umbriel, Titania, Oberon, Miranda(Uranus); Charon (Pluto); minor bodies with mean radius smaller than 100 km are not considered.

ix

x Contents

when it completes q−revolutions about the central planet.

Pfex

S

Figure 3: Spin-orbit problem (equatorial section).

The dissipative spin-orbit problem2 studies the motion of a satellite S orbiting arounda principal planet P (modelled by a point of mass), under the following assumptions(see Figure 3):

• the center of mass of S moves on a Keplerian ellipse with eccentricity e ∈ [0, 1)around P;

• S is a triaxial ellipsoid with axes aS ≥ bS ≥ cS > 0, where aS and bS are the“equatorial radii” and cS the polar radius;

• the polar radius of S coincides with the spin-axis of S and it is perpendicular tothe orbital plane;

• we consider the problem with dissipation.

The equation of motion is

x+ η(x− ν) + εfx(x, t) = 0, (1)

where:

• t is the mean anomaly;

• x is the angle (see Figure 3) formed by the direction of the major equatorial axisof the satellite with the direction of the semi–major axis of the Keplerian ellipseplane; ‘dot’ represents derivative with respect to t and e is the eccentricity ofthe orbital ellipse;

• the dissipation parameters η = ηe and ν = νe are real-analytic functions of theeccentricity e.

• the constant ε measures the oblateness (or “equatorial ellipticity”) of the satel-lite;

• the function f is the (“dimensionless”) Newtonian potential given by

f(x, t) := − 1

2(1− e cos(ue(t)))3cos(2x− 2fe(t)), (2)

where fe(t) is the true anomaly and u = ue(t) is the eccentric anomaly.

2a more detailed description of the problem will be given in Chapter 1.

Contents xi

In the conservative case η = 0 the spin-orbit problem (1) exhibits very interestingdynamics (periodic orbits of any period, KAM tori, AubryMather sets, etc.). Thisproblem is very well studied in [11], [12]. A summary of the results about this problemcan be found in [4].In the dissipative case η > 0 we have to distinguish two different cases. If ε = 0 andη 6= 0 hold, then the solution of (1) to the initial conditions x(0) = x0, x(0) = v0 isgiven by

x(t) = x0 + νt+1− e−ηt

η(v0 − ν),

showing that x ≡ ν is a global attractor. In the second case ε 6= 0 and η > 0the dynamic is much more varied. In [23] probabilities of capturing a planet (orsatellite) into one of the commensurate rotation states as it is being despun by tidalfriction (see [33]) are calculated. In particular in [19] is motivated why Mercury isin a (3, 2)−periodic resonance with the Sun. In [15], [8] it is proved that for thedissipative spin-orbit problem the coexistence of low-order spin-orbit resonances istypical. Furthermore in [16] it is shown that the dissipative spin-orbit problem mayalso admit quasi- periodic solutions (attractors) x(t) = ωt + U(ωt, t), with U(θ, t)analytic on T2 and ω Diophantine, provided ε is small enough and ν satisfies thecompatibility condition ν = ω(1 +

⟨U2θ

⟩). Furthermore it is proved that, if it exits,

the quasi-periodic attractor is unique.In [15] the measure of the basins of attraction associated to periodic and quasi-periodicattractors is numerically investigated.In this thesis we particularly focus our attention on the results of [8]. In that paper theauthors, Luigi Chierchia and Luca Biasco, give explicit conditions for the coexistenceof (p, 1)−, (p, 2)− and (p, 4)−periodic orbits and provide analytical estimates on themeasure of the basins of attraction of stable (p, 1)− and (p, 2)−periodic orbits. InSection 1 of [8] they suggest the following investigations:

1) Provide explicit conditions for the existence of (p, q)−periodic orbits for any q;

2) Discuss lower bounds on the basins of attraction for any q;

3) Prove (or disprove) that for q = 1, 2, the basin of attraction of a (p, q)-periodicorbit is actually “large”;

4) Discuss more general models (nonrestricted, nonplanar, obliquity, more generaldissipations,...);

5) Extending the approach presented here, give lower bounds on the basins ofattraction of the quasi-periodic attractors found in [16]. At this respect, letus remark that, numerically, the basins of attraction of quasi-periodic solutionsappear to be much larger than those of periodic solutions (compare [15]).

The topics treated in this thesis cover the points 1)− 3) and partially 4). 5) is still anopen problem.The manuscript is organised as follows: In Chapter 1 we motivate the equation ofmotion (1) for the dissipative spin-orbit problem using the relation between this prob-lem and the two body problem. Chapter 2 is devoted to the study of the Fouriercoefficients αj = αj(e) of the Newtonian potential f given in equation (2), i.e.

f(x, t; e) =∑

j∈Zαj(e) cos(2x− jt).

xii Contents

We will give integral formulas to compute these coefficients and to understand theirasymptotical behaviour.In Chapter 3 we study a more general problem than the dissipative spin-orbit problem.In particular we let the potential function

f ∈ C∞(R2,R)

be any analytic 2π−periodic function in both variables x, t and having zero average.For this problem we find conditions for the existence of a (p, q)−periodic orbit.In Chapter 4 we prove that the condition, for the existence of (p, q)−periodic orbitsof the dissipative spin-orbit problem for q = 1, 2 and 4 given by L. Biasco and L.Chierchia in [8], is satisfied provided that the eccentricity and, in the case q = 4, alsothe dissipation, is smaller than an explicitly given bound.In Chapter 5 we develop Matlab programs in order to compute numerically the solu-tion of the dissipative spin-orbit problem (in a constructive way).In Chapter 6 we prove the existence of periodic orbits for astronomical parametervalues corresponding to all satellites of the Solar System observed in spin-orbit reso-nance.In Chapter 7 we study the basin of attraction of (p, q)−periodic orbits for the dissi-pative spin-orbit problem giving two improvements of Theorem 1.3 in [8]: in the firstresult we allow q to be any number greater than 2 (Theorem 1.3 in [8] covered onlythe cases q = 1, 2); the second result is a quantitative reformulation of Theorem 7.1,which makes concrete applications possible.Finally, in Chapter 8 we aim to improve the numerical results on the basin of at-traction of (p, q)−periodic orbits for the dissipative spin-orbit problem obtained byCelletti and Chierchia in [15]. Although our purposes are not yet all achieved, wepresent some partial results.

1 The physical model

The spin-orbit problem studies the movement of a satellite S about a principal bodyP. Under the gravitational influence of P the satellite follows a closed elliptical orbit.Furthermore rotations of the satellite are considered. In particular the spin-orbitproblem assumes that the satellite has an ellipsoidal shape. This fact makes the spin-orbit problem a much more difficult problem to solve than the two-body problem.However this two problems are correlated and in order to well understand the physicalorigins of the equations, which govern the spin-orbit problem, one has to know someimportant relations arising from the two-body problem.Chapter 1 is organised as follows: In Subsection 1.1 we analyse the two-body problemin details. In particular we prove the relation between the eccentric anomaly andthe true anomaly (see equation (1.20)) and the Kepler equation (see equation (1.23)).These equations hold also in the spin-orbit problem. In Subsection 1.2 we consider firstthe conservative spin-orbit problem and then the dissipative spin-orbit problem, givingthe equation of motion (see (1.24)), which will be used in the following chapters of thisthesis. Finally in Subsection 1.3 we give a first definition of p : q periodic resonancesof the dissipative spin-orbit problem, which are the main object of interest in thisthesis.

1.1 Two-body problem

1.1.1 Historical introduction (based on [17])

Ever since mankind existed on earth, men had great interest in studying stars, whichwas then in times solely based on observation with the naked eye. People in Mesopotamia,Central America, China, India and Egypt differentiated astronomers from their priests.Besides many other duties, astronomers were responsible to decide on the calendar,based on moon phases, which was fundamentally important for these agricultural peo-ple. The Greeks gave astronomy very important impulses. Following, some especiallyimportant thinkers will be mentioned: Aristarchus of Samos (approximately 310BC - 230 BC) was the first postulating a heliocentric theory of the universe. Ac-cording to his theory the earth was moving on a circular trajectory around the sun.Claudius Ptolemy (approximately 90 AD - 168 AD) crowned the geocentric theoryof the universe to be the absolute truth, in his treatise Almagesto, written in AD 150.This paper served as basis for astronomic knowledge in all of Europe and Asia till theend of the medieval era. The science of astronomy had to wait until the Renaissanceto finally receive new impulses. We will look at them in chronological order: Nico-laus Copernicus (1473-1543) formulated again, in his book De revolutionibus orbiumcoelesticum, published in the year of his death, the hypothesis of a heliocentric uni-verse. This hypothesis was supported by Galileo Galilei (1564-1642), who could, forthe first time, give scientific proof to the heliocentric theory through observing moonphases of Venus with his own invention, the telescope. Johannes Kepler (1571-1630)was able to push the theory forward, after getting access to precise astronomic dataobserved by Tycho Brahe. He formulated the (nowadays very famous) three Kepler’slaws.

1

2 Chapter 1. The physical model

Figure 1.1: A portrait of Johannes Kepler, 1610 (from http://en.wikipedia.org/wiki/

Kepler).

Kepler’s Laws

First Law: The orbit of every planet is an ellipsewith the Sun at one of the two foci.

Second Law: A line joining a planet and the Sunsweeps out equal areas during equal intervalsof time.

Third Law: The square of the orbital period of aplanet is proportional to the cube of the semi-major axis of its orbit.

The credit for the mathematical demonstration of these laws has to be given to thegenius of Isaac Newton (1642-1727), who elaborated the principles of celestial me-chanics and the law of universal gravity in this book Philosophiae Naturalis PrincipiaMathematica, written in 1687. He summarized the principles and postulated threelaws of motion and the law of universal gravitation (see below), through which he wasable to derive Kepler’s laws.

1.1. Two-body problem 3

Newton’s laws of motiona

First law: Every body preserves in its state of rest, or of uniform motion in aright line, unless it is compelled to change that state by forces impressedthereon.

Second law: The alternation of motion is ever proportional to the motiveforce impressed; and is made in the direction of the right line in whichthat force is impressed.

Third law: To every action there is always opposed an equal reaction: or themutual actions of two bodies upon each other are always equal, anddirected to contrary parts.

Newton’s law of universal gravitationb

Every point mass attracts every single other point mass by a force pointingalong the line intersecting both points. The force is proportional to the prod-uct of the two masses and inversely proportional to the square of the distancebetween them.

aQuote of I. Newton from The Mathematical Principles of Natural Philosophy, Book I,1729.

bQuote Proposition 75, Theorem 35: p.956 - I.Bernard Cohen and Anne Whitman, trans-lators: Isaac Newton, The Principia: Mathematical Principles of Natural Philosophy.Preceded by A Guide to Newton’s Principia, by I.Bernard Cohen. University of Cali-fornia Press 1999 ISBN 0-520-08816-6 ISBN 0-520-08817-4.

Figure 1.2: Godfrey Kneller’s 1689 portrait of Isaac Newton, at the age of 46(left, from http://en.wikipedia.org/wiki/Isaac_Newton). Title page of Prin-cipia, first edition 1686/1687 (right, from http://en.wikipedia.org/wiki/

Philosophiæ_Naturalis_Principia_Mathematica).

The laws of Kepler and of Newton will allow us, in the course of this section, to un-derstand the equation for the motion of two punctiform bodies. This is a fundamentalstep forward in the understanding of the spin-orbit problem, which will be tackled inthe following subsection.

1.1.2 The model (based on [6] and [7])

We consider a principal body P (with mass M) and a satellite S (with mass m)interacting with each other under the gravitational force.


x2

x1

r

R

S

P

O

Figure 1.3: Sketch of the two-body problem with origin in O.

Newton’s second law implies

FPS(x1,x2) = M x1, (1.1)

FSP(x1,x2) = mx2, (1.2)

where FPS is the force on P due to S and FSP is the force on S due to P, respectively.By Newton’s third law and by Newton’s law of universal gravitation

FSP = −FPS = −GMm

r3r

holds, where G ≈ 6.67 · 10−11m3kg−1s−2 is the gravitational constant, and where

r = x2 − x1 and r = |r|. (1.3)

The center of mass

Adding up equations (1.1) and (1.2) we get

0 = FPS + FSP =M x1 +mx2 = (M +m)R, (1.4)

where

R =Mx1 +mx2

M +m, (1.5)

gives the position of the center of mass of the two-body system. Notice that the initialtwo-body problem has been simplified to a one-body problem (the motion of centerof mass). Furthermore the condition R = 0 shows that the velocity v = dR

dt of thecenter of mass is constant.

Displacement vector r

Another possibility to reduce the problem to a one-body problem, arises subtractingequation (1.2) to the equation (1.1):

r = x2 − x1 =FSP(x1,x2)

m− FPS(x1,x2)

M= FSP(x1,x2)

(1

M+

1

m

)

.

This is equivalent to

r = − µ

r3r (1.6)

with µ = G(M +m).


Conservation of Mechanical Energy

Scalar multiplication of (1.6) with r implies

r · r+ µ

r3r · r = 0.

Let v = r and recall that w ·w = ww for any vector w, then the previous equation isequivalent to

vv +µ

r2r = 0.

Since ddt

(v2

2

)

= vv and ddt

(−1

r

)= r

r2we get

d

dt

(v2

2− µ

r

)

= 0.

Therefore the total specific1 mechanical energy of a satellite must be constant alongits orbit, i.e.

E = T + U =v2

2− µ

r= const, (1.7)

where T = v2

2 is the specific kinetic energy and U = −µr is the specific potential energy

of the satellite, respectively.

Conservation of Angular Momentum

Recall the specific angular momentum h of a satellite is defined by

h = r× v. (1.8)

In order to show that h is constant, we differentiate equation (1.8) and use (1.6). Sowe have

d

dth = r× r+ r× r

︸︷︷︸

=0

(1.6)= − µ

r3r× r = 0,

which proves the conservation of the angular momentum.

The trajectory of the satellite

Cross multiplication of (1.6) by h leads to

r× h = − µ

r3r× h. (1.9)

For the left hand side of (1.9) the conservation of the angular momentum implies that

r× h = r× h+ r× h︸︷︷︸

=0

=d

dt(r× h)

holds. Recalling the identities w1 × (w2 × w3) = w2(w1 · w3) − w3(w1 · w2) andw · w = ww, the right hand side of (1.9) becomes

− µ

r3r× h =

µ

r3h× r =

µ

r3(v(r · r)− r(r · r))

=µ

r3(r2v − rrr

)=µ

rv − µr

r2r

1per unit mass


= µd

dt

(r

r

)

.

So (1.9) is equivalent to

d

dt(r× h) = µ

d

dt

r

r.

Integrating the last equation leads to

r× h = µr

r+B, (1.10)

where the vector B is a constant. Scalar multiplication of (1.10) by r implies

r · (r× h) = µr · rr

+ r ·B.

Since r · (r× h) = h · (r× r) = h · h = h2 it follows

h2 = µr + rB cos(ω),

where ω := ∠(B, r). In conclusion we find for r the following relation:

r =

h2

µ

1 + Bµ cos(ω)

. (1.11)

Defining σ2 := h2

µ and e := Bµ equation (1.11) can also be rewritten as

r =σ2

1 + e cos(ω), (1.12)

which is the polar equation (with respect to the angle ω) of a conic section (see Figure1.4). Hence, the possible orbits in a two-body problem are: ellipses, if 0 ≤ e < 1;parabolas, if e = 1; hyperbolas, if e > 1.

r

Oω

Figure 1.4: Polar representation of a conic section.

Since we are interested in the case of a satellite orbiting on a closed orbit around aprincipal body, we only consider the case of a closed elliptical curve.


The case M ≫ m and the proof of the 1st and the 2nd Kepler-law.

Consider the case where the mass m of the satellite S is much smaller than the massM of the principal body P. By (1.3) and (1.5), the original trajectories x1 and x2 inFigure 1.3 are given by

x1(t) = R(t)− m

M +mr(t) ≈ R(t), (1.13)

x2(t) = R(t) +M

M +mr(t) ≈ R(t) + r(t). (1.14)

Therefore the center of mass of the system is approximately in P, i.e. the planet liesin one focus of the elliptical orbit. This proves the first Kepler-law.Furthermore by the conservation of angular momentum and by equation (1.8) and thedefinition of σ we have

A(t) =

∫ t

0

1

2|r× v|dt =

∫ t

0

h

2dt =

σ√µ

2t. (1.15)

This proves the second Kepler-law.

A1

A2

A3

Figure 1.5: Second Kepler-law: the sectors A1, A2 and A3 have the same area. Theray from the central body to the satellite sweeps out equal areas in equaltimes.

Elliptic motion

Since we established that in the two body problem the orbit is an ellipse (see equation(1.12)), it is useful to recall some properties of this curve.It is obvious, comparing Figure 1.4 and 1.6 that ω = fe, which is called true anomaly.By (1.12) we have that the apoapsis and the periapsis2 are given by

rapoapsis =σ2

1− eand rperiapsis =

σ2

1 + e.

The semi-major axis a of the elliptical orbit fulfils the relation

2a = rapoapsis + rperiapsis,

hence we getσ2 = a(1− e2). (1.16)

2apogee/aphelion and perigee/perihelion for orbits around the Earth and the Sun, respectively.


a

a

br

PO

S

feue

Figure 1.6: The ellipse.

Inserting (1.16) in (1.12) we get

r =a(1− e2)

1 + e cos(fe). (1.17)

Further interesting facts follows by the elementary geometry of the ellipse. By (1.17)the coordinates of the foci are (±c, 0) = (ae, 0). The semi-minor axis b is then givenby

b2 = a2 − c2 = a2(1− e2). (1.18)

Choosing the directions of a and b as the reference axes of the system with originO, and introducing the eccentric anomaly ue as in Figure 1.6, the satellite S hascoordinates (a cos(ue), b sin(ue)). The theorem of Pythagoras implies

r2 = (a cos(ue)− ae)2 + b2 sin2(ue)

(1.18)= a2

[

(cos(ue)− e)2 + (1− e2) sin2(ue)]

= a2 (1− e cos(ue))2 ,

hence

r = a (1− e cos(ue)) . (1.19)

The Kepler equation

Notice that by (1.19) and (1.17) we get

cos(ue) =1

e− r

ae=

1

e

(

1− (1− e2)

1 + e cos(fe)

)

=e+ cos(fe)

1 + e cos(fe).

The true anomaly fe and the eccentric anomaly ue are related as follows:

tan2(ue2

)

=1− cos(ue)

1 + cos(ue)=

(1− e)(1− cos(fe))

(1 + e)(1 + cos(fe))=

1− e

1 + etan2

(fe2

)

. (1.20)


Therefore fe = 2arctan(√

1+e

1−etan

(ue

2

))

holds. Taking the derivative with respect to

ue of this formula we obtain

dfedue

=

√

1 + e

1− e

1 + tan2(ue

2

)

1 + 1+e

1−etan2

(ue

2

) . (1.21)

We are now able to compute the integral of the second Kepler-law (see (1.15)), recallingthat for the angular velocity v = r dfedt . Hence we have:

σ√µt =

∫ τ

0|r× v|dt =

∫ fe

0r2dfe =

∫ fe

0

a2(1− e2)2

(1 + e cos(fe))2dfe

(1.21)= a2(1− e2)2

√

1 + e

1− e

∫ ue

0

1

(1 + e cos(fe))21 + tan2

(ue

2

)

1 + 1+e

1−etan2

(ue

2

)due.

Using the elementary trigonometric identity

cos(fe) =1− tan2

(fe2

)

1 + tan2(fe2

) =1− 1+e

1−etan2

(ue

2

)

1 + 1+e

1−etan2

(ue

2

)

we have (according to [34] )

σ√µt = a2(1− e2)2

√

1 + e

1− e

∫ ue

0

1(

1 + e1− 1+e

1−etan2(ue

2 )1+ 1+e

1−etan2(ue

2 )

)2

1 + tan2(ue

2

)

1 + 1+e

1−etan2

(ue

2

)due

= a2(1− e2)2√

1 + e

1− e

∫ ue

0

(

1 + 1+e

1−etan2

(ue

2

)) (1 + tan2

(ue

2

))

(

1 + 1+e

1−etan2

(ue

2

)+ e

(

1− 1+e

1−etan2

(ue

2

)))2due

= a2(1− e2)5/2∫ ue

0

(1 + tan2

(ue

2

)− e+ e tan2

(ue

2

)) (1 + tan2

(ue

2

))

(1− e2)2(1 + tan2

(ue

2

)) due

= a2√

1− e2∫ ue

0

(

1− e1− tan2

(ue

2

)

1 + tan2(ue

2

)

)

due

= a2√

1− e2∫ ue

0(1− e cos(ue)) due = a2

√

1− e2 (ue − e sin(ue)) .

So σ√µt = a2

√1− e2 (ue − e sin(ue)) holds, which by (1.16) is equivalent to

nt = ue − e sin(ue), (1.22)

where n :=√

µa3 = 2π

T (where T is the orbital period3) is called mean motion. Defining

τ := nt as the mean anomaly (1.22) is equivalent to

τ = ue − e sin(ue), (1.23)

which we will refer to as the Kepler equation.

3By (1.15) integrating over one complete orbital period T it follows

πab = A(T ) =σ√µ

2T.

Since (1.16) and (1.18) this is equivalent to

4π2 = µT 2

a3= n2T 2.

This proves the third Kepler-law.


Remark 1.1. Consider the Banach space

χ := v(τ) ∈ C(R,R)|v is 2π − periodic and odd

endowed with the sup-norm ‖v(τ)‖∞ := supτ∈R |v(τ)|. For e ∈ (0, 1) the map

Φ : χ→ χ

v 7→ Φ(v(τ)) := sin(τ + ev(τ))

is a contraction with Lipschitz constant e. Hence for u, v ∈ χ

‖Φ(v) −Φ(u)‖∞ = supτ∈R

|sin(τ + ev(τ)) − sin(t+ eu(τ))|

= supτ∈R

∣∣∣∣2 cos

(

τ + ev(τ) + u(τ)

2

)

sin

(

ev(τ)− u(τ)

2

)∣∣∣∣

≤ 2 supτ∈R

∣∣∣∣sin

(

ev(τ) − u(τ)

2

)∣∣∣∣

≤ e supτ∈R

|v(τ)− u(τ)| = e‖v − u‖∞

holds. From the fixed point theorem there exists a unique function Ue(τ) ∈ χ such that

Ue(τ) = sin (τ + eUe(τ)) .

Notice that the function τ + eUe(τ) satisfies the Kepler equation (1.23) and therefore

ue(τ) = ue = τ + eUe(τ)

holds. It follows that ue is continuous and odd 2π−periodic function of the meananomaly τ . Defining fe(π) = π the true anomaly fe in (1.20) is also continuous andodd 2π−periodic function of the mean anomaly τ .

1.2 The spin-orbit problem

1.2.1 The model

As the two-body problem, the spin-orbit problem studies the motion of a satelliterevolving around a principal planet. The difference, which makes the study of thisproblem much more interesting, is that the shape of the satellite is considered, i.e.rotations of the satellite on its own spin axis can occur. In particular we imaginethe central planet as a point of mass and the satellite as triaxial ellipsoid with axesaS ≥ bS ≥ cS > 0, where aS and bS are the “equatorial radii” and cS the polarradius. We assume that the satellite moves about the central planet on a Keplerianelliptical orbit with semi-major axis a. Furthermore the spin-axis is assumed to beperpendicular to the Keplerian orbit plane4.Under the above hypotheses, the differential equation governing the motion of thesatellite is then given by

x+ η(x− ν) + εfx(x, t) = 0, (1.24)

where:

4For the motivation of this assumption see Subsection 1.2.3

1.2. The spin-orbit problem 11

ρe

e

fex

satellite

Figure 1.7: Triaxial satellite revolving on a rescaled Keplerian ellipse (equatorial section).

(a) t is the mean anomaly (also defined mod 2π), i.e. the ellipse area between thesemi–major axis and the orbital radius ρe divided by the total area times 2π.Notice that in (1.23) we used τ for the mean anomaly;

(b) x is the angle (mod 2π) formed by the direction of (say) the major equatorialaxis of the satellite with the direction of the semi–major axis of the Keplerianellipse plane; ‘dot’ represents derivative with respect to t and e is the eccentricityof the ellipse;

(c) the dissipation parameters η = KΩe and ν = νe are real-analytic functionsof the eccentricity e: K ≥ 0 is a physical constant depending on the internal(non-rigid) structure of the satellite and

Ωe :=

(

1 + 3e2 +3

8e4)

1

(1− e2)9/2, (1.25)

Ne :=

(

1 +15

2e2 +

45

8e4 +

5

16e6)

1

(1− e2)6, (1.26)

νe :=Ne

Ωe

; (1.27)

(d) the constant εmeasures the oblateness (or “equatorial ellipticity”) of the satelliteand it is defined as

ε =3

2

B −A

C, (1.28)

where 0 < A ≤ B ≤ C are the principal moments of inertia of the satellite (Cbeing referred to the polar axis);

(e) the function f is the (“dimensionless”) Newtonian potential given by

f(x, t) := − 1

2ρe(t)3cos(2x− 2fe(t)), (1.29)

where ρe(t) and fe(t) are, respectively, the (normalized) orbital radius

ρe(t) := 1− e cos(ue(t)) (1.30)

and the true anomaly

fe(t) := 2 arctan

(√

1 + e

1− etan

(ue(t)

2

))

, (1.31)


while the eccentric anomaly u = ue(t) is defined implicitly by the Kepler equa-tion5

t = u− e sin(u). (1.32)

Notice that the Newtonian potential f(x, t) is a doubly–periodic function of xand t, with periods 2π.

We should remark that the relations (1.31) and (1.32) for the spin-orbit problem comefrom the two-body problem, in particular from the equations (1.20) and (1.23).For further references on this model, see [20], [23], [48], [12] and [13]; for a differentapproach to this problem, see [5].

1.2.2 The Moment of inertia tensor

Let (x, y, z) be a given reference frame. For any 3−dimensional body with volume Vand mass m we define the quantities

I11 =

∫∫∫

V(y2 + z2)dm =: A

I22 =

∫∫∫

V(x2 + z2)dm =: B

I33 =

∫∫∫

V(x2 + y2)dm =: C

as the moments of inertia and

I23 = I32 = −∫∫∫

Vyzdm =: −F

I12 = I21 = −∫∫∫

Vxzdm =: −G

I13 = I31 = −∫∫∫

Vxydm =: −H

as the products of inertia.The matrix

I = I(x,y,z) =

I11 I12 I13I21 I22 I23I31 I32 I33

is called the inertia tensor. Notice that I depends on the choice of a reference frame(x, y, z).We are interested in computing the moment of inertia of a satellite S with an ellipsoidalshape with axes aS ≥ bS ≥ cS > 0, choosing the principal axes of inertia as referenceframe. Hence the ellipsoid is given by the relation

x2

a2S+y2

b2S+z2

c2S= 1.

Assuming the density ρ = m43πaSbScS

of the satellite to be constant, we get

A =

∫∫∫

S(y2 + z2)dm = ρ

∫∫∫

V(y2 + z2)dxdydz.

5It is well known (see [47]) that for every t ∈ R the function C ∋ e → ue(t) is holomorphic for

|e| < r⋆, with r⋆ := maxy∈R

y

cosh(y)=

y⋆cosh(y⋆)

= 0.6627434 · · · and y⋆ = 1.1996786 · · · .


Substituting x′ = xaS

, y′ = ybS, z′ = z

cSwe integrate now on S3 (the sphere of radius

1). The Jacobi determinant implies dxdydz = abc dx′dy′dz′. Thus we have

A = ρaSbScS

∫∫∫

S3

(b2S(y′)2 + c2S(z

′)2)dx′dy′dz′.

Transforming to spherical coordinates x′ = s sin(θ) cos(φ), y′ = s sin(θ) sin(φ), z′ =s cos(θ) with 0 ≤ s ≤ 1, 0 ≤ θ ≤ π and 0 ≤ φ ≤ 2π we obtain

A = ρaSbScS

∫ 2π

0

∫ π

0

∫ 1

0

[b2Ss

2 sin2(θ) sin2(φ) + c2Ss2 cos2(θ)

]s2 sin(θ)dsdθdφ

=4ρaSbScS(b2S + c2S)π

15=m(b2S + c2S)

5.

Analogously we get

B =1

5m(a2S + c2S), C =

1

5m(a2S + b2S).

Furthermore, choosing the principal axes of inertia as reference frame, the products ofinertia are all zero. For example, making the same substitutions as above, let computethe first product of inertia F . By definition we have

F =

∫∫∫

Syzdm = ρ

∫∫∫

Vyzdxdydz

= aSb2Sc

2S

∫ 2π

0

∫ π

0

∫ 1

0s4 sin(θ)2 cos(θ) sin(φ)dsdθdφ = 0.

The same holds for the other two products of inertia: G = H = 0.We conclude that choosing the principal axes of inertia as reference frame correspondsto diagonalise the inertia tensor I, i.e.

I = I(~aS ,~bS ,~cS)

=

A 0 00 B 00 0 C

. (1.33)

1.2.3 Motivation of the rotational assumption about the z−axes

Let Ω be the center of mass of the satellite and let (Ω, x, y, z) be the reference framepointing in the direction of the satellite’s principal axes of inertia. The moment ofinertia tensor is then given by (1.33). In the following lines we explain why, theassumption that “the spin polar axis is perpendicular to the Keplerian orbit plane”,makes sense.By definition the angular momentum is

L =

Lx

Ly

Lz

= I · ω =

A 0 00 B 00 0 C

·

ωx

ωy

ωz

=

Aωx

Bωy

Cωz

. (1.34)

Let L2 := L2x + L2

y + L2z. Furthermore using (1.34) the rotational energy is given by

E =1

2ω · I · ω

=1

2

(Aω2

x +Bω2y + Cω2

z

)


=1

2

((Aωx)

2

A+

(Bωy)2

B+

(Cωz)2

C

)

=1

2

(

L2x

A+L2y

B+L2z

C

)

.

The angular momentum L lies therefore in the intersection of the sphere x2+y2+z2 =

L2 with the ellipsoid E : x2

2A + y2

2B + z2

2C = E. The size of L is conserved, if weneglect external torques. The tidal term dissipates the energy E. So the new system

LL

E⋆ : x2

2A + y2

2B + z2

2C = E⋆ < E

E : x2

2A + y2

2B + z2

2C = E

x2 + y2 + z2 = L2

Ω

Figure 1.8: While the energy is dissipated the angular momentum L tends to align itself to thelongest axis of the ellipsoid.

has rotational energy E⋆ < E. The ellipsoid E shrinks into E⋆ (see Figure 1.8) andtherefore the direction of L tends to the axis of the ellipsoid, with be biggest momentof inertia. Since, in our case A ≤ B ≤ C, this is the z−axis. So the assumption isplausible.

1.2.4 MacCullagh’s Potential (based on [20])

Consider a triaxial ellipsoidal body with center of mass in Ω. Assume the density ofthe body to be constant. In the following V denotes the volume of the body. Wedefine (Ω, x, y, z) to be the reference frame, where the x−, y−, z−axes point in thedirection of the principal axes of inertia of the ellipsoidal body. Let O be a point withcoordinates (λ, µ, ν). The vector r represents ~OΩ. The length of the vector ~OΩ isequal to r =

√

λ2 + µ2 + ν2. Let P(xP, yP, zP) be any point of the triaxial ellipsoidalbody with mass dm. The vector s represents ~ΩP.

r

s

O(λ, µ, ν)

P

Ωθ

Figure 1.9: Gravitational potential due to a triaxial ellipsoidal body on distant unit mass in O.

With these hypotheses the gravitational potential f , the so–called MacCullagh po-tential, on distant unit mass, lying in O, can be given depending on the coordinates


(λ, µ, ν) of the point O (see Figure 1.9).

f(λ, µ, ν) := −G∫∫∫

V

dm

|r+ s| = −Gr

∫∫∫

V

dm√

1− 2 sr cos(θ) +

s2

r2

= −Gr

∫∫∫

Vdm

(

1 +s

rcos(θ)− 1

2

s2

r2+

3

2

s2

r2cos2(θ) +O

(s

r

)3)

= −Gr

∫∫∫

Vdm

(

1 +s

rcos(θ) +

1

2

s2

r2[2− 3 sin2(θ)

]+O

(s

r

)3)

, (1.35)

where the law of cosines has been used. By symmetry the integral∫∫∫

Vsr cos(θ)dm =

0. Furthermore since s << r we can say that

f(λ, µ, ν) = −Gmr

− G

2r3(A+B + C − 3IOΩ), (1.36)

where A,B,C are the principal moments of inertia of the satellite and

IOΩ :=

∫∫∫

Ss2 sin2(θ)dm =

Aλ2 +Bµ2 + Cν2

r2(1.37)

is the moment of inertia of the satellite along the axes OΩ. Equation (1.36) fol-lows directly from (1.35), from the definition of the principal moment of inertia (seeSubsection 1.2.2) and the observation that

A+B + C = 2

∫∫∫

Ss2dm.

Equation (1.37) can be explained with the ellipsoidal of inertia Ax2+By2+Cz2 = 1.This surface has the interesting property (see [24]) that the moment of inertia alongany axis δ through the center of mass Ω of the body is equal to 1/d2, where d is thedistance between D, the intersection point of the axis δ with the ellipsoidal of inertia,and Ω (see Figure 1.10).

x

yAx2 +By2 + Cz2 = 1

Ω

D

δ

Figure 1.10: Two-dimensional section of the ellipsoid of inertia.

With other words the point D lies on the circle with radius

d =1√Iδ, (1.38)

where Iδ is the moment of inertia of the body along the axis δ.


Combining Figure 1.9 and Figure 1.10 one obtains the following figure:

x

y

Ax2 +By2 + Cz2 = 1

Ax2 +By2 + Cz2 = r2

x2 + y2 + z2 = r2

x2 + y2 + z2 = 1IOΩ

Ω

O

O′

Figure 1.11: Ellipsoidal body with its ellipsoid of inertia.

The point O(λ, µ, ν) lies on the circle with radius r, since√

λ2 + µ2 + ν2 = r. LetO′(λ′, µ′, ν ′) be the intersection point between the line ΩO and the ellipsoidal of inertiaAx2 +By2 +Cz2 = 1. By (1.38) O′ lies also on the circle with radius 1/

√IOΩ, where

IOΩ is the moment of inertia along the axis OΩ. Furthermore chose r such that theellipsoid Ax2 + By2 + Cz2 = r2 contains O and is similar to the ellipsoid of inertia(see Figure 1.11). By similarity it follows

r2

1IOΩ

=Aλ2 +Bν2 + Cµ2

A(λ′)2 +B(ν ′)2 + C(µ′)2=Aλ2 +Bν2 + Cµ2

1.

This fact explains why (1.37) holds.

1.2.5 The Euler equations and the equation of motion (based on [20])

Let Ω be the center of mass of the triaxial ellipsoidal satellite and let (Ω, x, y, z) bethe reference frame pointing in the direction of the principal moment of inertia ofsatellite.

r

P

fex

φ

Ω

x

ysatellite

Figure 1.12: Spin-orbit problem with reference frame (Ω, x, y, z) (equatorial section).


By assumption the spin axis of the satellite coincides with is shortest physical axisand is perpendicular to the orbital plane, i.e. (Ω, x, y, z) is a rotating reference framewith angular velocity

ω =

00ωz

.

It is well known that the time derivative in a rotating frame follows the rule (see [18])

d

dtw =

∂w

∂t+ ω ×w

for any vector w. Therefore the derivative of the angular momentum L = I(x,y,z)ω ofthe satellite with respect to t is equal to

d

dtL =

d

dt(I(x,y,z) · ω)

=∂

∂t(I(x,y,z) · ω) + ω × (I(x,y,z) · ω) = ω · I(x,y,z) + ω × (I(x,y,z) · ω)

=

A 0 00 B 00 0 C

·

00ωz

+

00ωz

×

00

Cωz

=

00

Cωz

.(1.39)

Moreover, the angular momentum also satisfies the relation

L(1.8)= mh = r×mv.

Again taking the derivative with respect to t, one obtains

d

dt(r×mv) =

dr

dt×mv + r×m

dv

dt= r×ma = r× FPS = Γ, (1.40)

whereFPS is the gravitational force exerted from the principal planet P on the satelliteS. Γ is called torque and indicates the tendency of a force to rotate an object. Bythe third Newton-law the satellite exerts an equal and opposite force on the planet P,i.e. FPS = −FSP . By definition of the MacCullagh potential f (see equation (1.36))it follows

FSP = −M∇f(λ, µ, ν),where (λ, µ, ν) are the coordinates of the principal planet P in the (Ω, x, y, z) rotatingreference frame, i.e. (see Figure 1.12)

λ = r cos(φ), µ = r sin(φ), ν = 0, (1.41)

holds. So we have

FPS = M∇f(λ, µ, ν) (1.36)=

3MG

2r3∇IΩP

(1.37)=

3MG

r5

AλBµCν

.

The torque Γ in (1.40) can therefore be rewritten as follows:

Γ = r× F =3MG

r5

λµν

×

AλBµCν

=3MG

r5

(C −B)µν(A− C)λν(B −A)λµ

. (1.42)


The Euler equations arise by comparing (1.39), (1.40) and (1.42). The first twocomponents are trivially satisfied since ν = 0 by (1.41). The equation in the thirdcomponent is

Cωz =3MG

r5(B −A)λµ

(1.41)=

3MG

r3(B −A) cos(φ) sin(φ). (1.43)

Let x be the angle (mod 2π) formed by the direction of the major equatorial axis ofthe satellite with the major axis of the Keplerian ellipse plane (as in the Figure 1.12).Since the sum of the angles in a triangle is equal to π, we get the relation

φ+ x+ π − fe = π ⇒ φ = −(x− fe).

The equation (1.43) is taken to

ωz = −MG3(B −A)

2C

sin(2x− 2fe)

r3. (1.44)

Notice that by equation (1.28) we know ε = 32

B−AC , where 0 < A ≤ B ≤ C are the

principal moments of inertia of the satellite. Then (1.44) is equivalent to

ωz = − εMG

a3sin(2x− 2fe)

(ra

)3 , (1.45)

where a is the length of the major semiaxis of the elliptical orbit. Since M >> m thefollowing approximation is possible:

MG

a3≈ (M +m)G

a3=

µ

a3= n2,

where µ := G(M +m) (as in (1.6)) and n is the mean motion defined in (1.22). By(1.23) the mean anomaly satisfies τ = nt and therefore

ωz =dωz

dt=

1

n2dωz

dτ.

So we get (according to [12])

ωz =dx

dτ= − d

dτ(φ− fe)

and from (1.45) one gets the differential equation

d2x

dτ2+ ε

sin(2x− 2fe)(ra

)3 = 0.

Recalling the definition of the Newtonian potential f in (1.29), of the (normalized)orbital radius ρe in (1.30), and of the function ue = ue(τ) in the Kepler equation(1.23), respectively, the differential equation governing the motion of the satellite(also according to [13]) is then given by

d2x

dτ2+ εfx(x, τ) = 0, (1.46)

which is equivalent to (1.24) (where dot means the derivative with respect to τ) forη = 0. The case η > 0 will be treated in the next subsection.


1.2.6 The dissipative case

To the spin-orbit problem (1.46) we include also small dissipative effects due to thepossible internal non–rigid structure of the satellite. According to the “viscous–tidalmodel, with a linear dependence on the tidal frequency” (see [19]), the dissipative termis given by the average over one revolution period (i.e., 2π with our normalization) ofthe so–called MacDonald torque [33]. Hence the tidal torque is

T (x) = −η(x− ν). (1.47)

Compare also [23], [35], and more recently [13].The dissipation parameters

η = ηe = KΩe and ν = νe (1.48)

are real-analytic functions of the eccentricity e, where6 Ωe, Ne and νe are definedin (1.25), (1.26) and (1.27). For the Moon and Mercury an accepted value for η is∼ 10−8. The dissipation constant K ≥ 0 in (1.48) has the form

K = 3nk2Q

M

m

(Req

a

)3

, (1.49)

where:

- M and m are the masses of the principal body and of the satellite, respectively;

- Req is the satellite’s mean equatorial radius;

- a is the semi-major axis of the Keplerian elliptical orbit;

- n =√

GMa3 is the orbital mean motion, where G is the universal gravitational

constant;

- Q is called quality factor. 1/Q is defined as sin ǫ, where ǫ is a phase lag at thisfrequency;

- k2 is called Love number and is defined by

k2 =1.5

1 + 19ς2gReq

, (1.50)

where

• ς is the rigidity of the satellite,

• is the mean density of the satellite,

• g is the surface gravity of the satellite.

There is no universally accepted determination of the tidal ratio k2/Q in (1.49) formost satellites of the Solar System. Significant amount of work on k2/Q of large icysatellites (Europa, Ganymede, Callisto, Titan) has been made in [26]. However, theonly outer Solar System body, for which k2/Q has been measured, is Titan in [28]. Ofcourse, you can make detailed models in individual cases. Some examples are: Io andJupiter see [30], Saturn see [31] and Iapetus see [9].In any case one notice, that k2 in (1.50) is smaller than 1.5. Furthermore, by definitionof Q, it is impossible to get Q < 1. In conclusion we can be sure that k2

Q ≤ 1.5. Wewill use this estimate in further chapters.

Equation (1.24) follows now from (1.46) and (1.47).

6In [19] (see Eq.ns (3) and (4)) Ωe and Ne are denoted, respectively, Ω(e) and N(e), while, in [35],they are denoted, respectively, by f1(e) and f2(e).


1.3 Definition of spin-orbit resonances

As briefly mentioned in the introduction we say (according to [39]) that two bodiesare in a p : q spin-orbit resonance (with p and q co–prime non–vanishing integers) if

TrevTrot

=p

q,

with p, q ∈ Z+, where Trev are the period of revolution of the satellite around thecentral planet, and Trot the period of rotation of the satellite around its spin-axis,respectively (see Figure 1.7).

Definition 1.1. A (p, q)−periodic orbit (with p and q co–prime non–vanishing inte-gers) for the dissipative spin-orbit problem is a solution xpq(t) : R → R of (1.24)satisfying

xpq(t+ 2πq) = xpq(t) + 2πp, ∀t. (1.51)

Remark 1.2. (i) Indeed, for orbits as in (1.51), after q revolutions of the orbital ra-dius, x has made p complete rotations7. This explains why p : q spin-orbit resonancesare (p, q)−periodic orbits and vice versa.(ii)The double periodicity of the Newtonian potential in (1.29) implies that the (ex-tended) phase space for equation (1.24) is

M :=((x, t), x) ∈ T2 × R

,

where T2 = R2/(2πZ2).(iii)Notice that the winding (or rotation) number of a (p, q)−periodic orbit xpq(t) isequal to

ω = lims→∞

1

sxpq(t+ s) =

p

q.

7Of course, in physical space, x and t being angles, are defined modulus 2π, but to keep track of thetopology (windings and rotations) one needs to consider them in the universal cover R of R/(2πZ).

2 On the Fourier coefficients of theNewtonian Potential

In this chapter we analyse the Fourier coefficients αj = αj(e) of the Newtonian po-tential f given in equation (1.29), i.e.

f(x, t; e) =∑

j∈Zαj(e) cos(2x− jt). (2.1)

In particular in Subsection 2.1 we give two equivalent integral formulas to compute αj

for all j ∈ Z. In Subsection 2.2 we study the asymptotic behaviour in the eccentricityof αj(e) for all j ∈ Z. Both results will be very useful in the following chapters of thisthesis.

2.1 Two Integral Formulas for αj

The Fourier coefficients αj coincide with the Fourier coefficients Gj(e) = Gj of

Ge(t) := − e2ife(t)

2ρe(t)3=∑

j∈ZGj exp(ijt), (2.2)

where ρe(t) and fe(t) are defined in (1.30) and (1.31), respectively. The followinglemma states this result and gives a first integral formula for the coefficients αj withj ∈ Z.

Lemma 2.1. The coefficients αj = αj(e) defined in (2.1) satisfy

αj = αj(e) = − 1

4π

∫ 2π

0

cos(2 fe(u)− ju+ ej sin(u))

ρ2edu = Gj , (2.3)

where Gj for j ∈ Z are defined in (2.2) and

fe = fe(u) := 2 arctan

(√

1 + e

1− etan

(u

2

))

,

ρe = ρe(u) := 1− e cos(u).

holds. In particular, α0 = α0(e) = 0 for e ∈ [0, 1) holds.

Remark 2.1. Since ρe(t) in (1.30) is even and fe(t) in (1.31) is odd, by (2.2) we findthat

∑

j∈ZGje

−ijt = Ge(−t) = Ge(t) =∑

j∈ZGje

−ijt ,

holds, proving that the coefficients Gj are real for all j ∈ Z.

21

22 Chapter 2. On the Fourier coefficients of the Newtonian Potential

Proof. (We follow the proof of Lemma 2.1 given in Appendix A of [8])By definition of f in (1.29) it follows

f(x, t; e) = − 1

2ρe(t)3cos(2x− 2 fe(t)) = Re

(

−1

2ei2x · e

−i2 fe(t))

ρe(t)3

)

.

By Remark 2.1 one gets

f(x, t; e) = Re(ei2xGe(−t)

)= Re

ei2x∑

j∈ZGje

−ijt

= Re

∑

j∈ZGje

i(2x−jt)

=∑

j∈ZGj cos(2x− jt),

proving that αj = Gj for all j ∈ Z. Furthermore, by definition of the Fourier coeffi-cients of a 2π−periodic function we have

αj = Gj :=1

2π

∫ 2π

0Ge(t)e

−ijtdt

(2.2)= − 1

4π

∫ 2π

0

ei2 fe(t)−ijt

ρe(t)3dt

= − 1

4π

∫ 2π

0

cos(2 fe(t)− jt)

ρe(t)3+ i

sin(2 fe(t)− jt)

ρe(t)3︸︷︷︸

=:He(t)

dt.

Since He(t) is a 2π−periodic and odd function,∫ 2π0 He(t)dt = 0 holds. Therefore one

gets

αj = − 1

4π

∫ 2π

0

cos(2 fe(t)− jt)

ρe(t)3dt.

From the Kepler equation (1.32) and (1.30), it follows that

ue(t)′ =

1

ρe(t).

Changing the variable of integration from t to u = ue we obtain

αj = αj(e) = − 1

4π

∫ 2π

0

cos(2 fe(u)− ju+ ej sin(u))

ρe(u)2du, (2.4)

for j ∈ Z. In order to finish the proof of Lemma 2.1 we need to check the hypothesisα0 = α0(e) = 0 for e ∈ [0, 1) for the special case j = 0. First we prove the followingclaim.

Claim:

cos(2 fe) =3e2 − 4e cos(u) + (2− e2) cos(2u)

2ρ2e. (2.5)

For every c ∈ R the equality

c2 = tan2 (arctan(c)) =1− cos2(arctan(c))

cos2(arctan(c))

2.1. Two Integral Formulas for αj 23

holds. This implies

cos2 (arctan(c)) =1

1 + c2. (2.6)

In particular choosing c =√

1+e

1−etan

(u2

)we get

cos2

(

arctan

(√

1 + e

1− etan

(u

2

)))

(2.6)=

1

1 + 1+e

1−etan2

(u2

) . (2.7)

Since for every angle β the relation cos(2β) = 2 cos2(β)− 1 holds, it follows

cos(4β) = 2 cos2(2β) − 1 = 2[2 cos2(β)− 1

]2 − 1 = 8 cos4(β)− 8 cos2(β) + 1. (2.8)

From (1.31), (2.7) and (2.8) choosing β = arctan(√

1+e

1−etan

(u2

))

we get

cos(2 fe) = cos

(

4 arctan

(√

1 + e

1− etan

(u

2

)))

= 8

[

1

1 + 1+e

1−etan2

(u2

)

]2

− 8

[

1

1 + 1+e

1−etan2

(u2

)

]

+ 1

=8− 8

[

1 + 1+e

1−etan2

(u2

)]

+[

1 + 1+e

1−etan2

(u2

)]2

[

1 + 1+e

1−etan2

(u2

)]2

=1− 61+e

1−etan2

(u2

)+(1+e

1−e

)2tan4

(u2

)

1(1−e)2

[(1− e) + (1 + e) tan2

(u2

)]2

=(1− e)2 − 6(1− e)(1 + e) tan2

(u2

)+ (1 + e)2 tan4

(u2

)

[(1− e) + (1 + e) tan2

(u2

)]2 .

Let define

ι := tan(u

2

)

. (2.9)

Then we have:

cos(2 fe) =(1− e)2 − 6(1− e2)ι2 + (1 + e)2 ι4

[(1− e) + (1 + e)ι2]2.

Since cos(u) = 1−ι2

1+ι2 and

[

(1− e) + (1 + e) tan2(u

2

)]2=(1 + ι2

)2[1− e cos(u)]2 =

(1 + ι2

)2ρ2e,

hold, we have to prove that

(1− e)2 − 6(1− e2)ι2 + (1 + e)2 ι4

(1 + ι2)2=

3e2 − 4e cos(u) + (2− e2) cos(2u)

2,

which is equivalent to

2(1− e)2 − 12(1 − e2)ι2 + 2 (1 + e)2 ι4 =(1 + ι2

)2 (3e2 − 4e cos(u) + (2− e2) cos(2u)

).

(2.10)


The left hand side of (2.10) can be computed directly

2(1 − e)2 − 12(1− e2)ι2 + 2 (1 + e)2 ι4 =

= 2− 4e+ 2e2 − 12ι2 + 12e2ι2 + 2ι4 + 4eι4 + 2e2ι4. (2.11)

The substitution (2.9) implies cos(u) = 1−ι2

1+ι2, cos(2u) = 2

[1−ι2

1+ι2

]2− 1. Therefore the

right hand side of (2.10) is given by:

(1 + ι2

)2 (3e2 − 4e cos(u) + (2− e2) cos(2u)

)=

=(1 + ι2

)2

(

3e2 − 4e1− ι2

1 + ι2+ (2− e2)

(

2

[1− ι2

1 + ι2

]2

− 1

))

= 3e2(1 + ι2)2 − 4e(1− ι2)(1 + ι2) + 2(2 − e2)(1− ι2)2 − (2− e2)(1 + ι2)2

= 3e2 + 6e2ι2 + 3e2ι4 − 4e+ 4eι4 + 4− 8ι2 + 4ι4 − 2e2 + 4e2ι2 − 2e2ι4 +

+e2 + 2e2ι2 + e2ι4 − 2− 4ι2 − 2ι4

= 2− 4e+ 2e2 − 12ι2 + 12e2ι2 + 2ι4 + 4eι4 + 2e2ι4. (2.12)

(2.11) and (2.12) are equal. This concludes the proof of Claim (2.5).We are now able to prove that:

α0 = α0(e) = 0, for e ∈ [0, 1) . (2.13)

By (2.4) and (2.5) it follows:

α0(e) = − 1

4π

∫ 2π

0

cos(2fe(u))

ρe(u)2du = − 1

4π

∫ π

−π

cos(2fe(u))

ρe(u)2du

= − 1

4π

∫ π

−π

3e2 − 4e cos(u) + (2− e2) cos(2u)

2ρe(u)4du.

The substitution (2.9), cos(u) = 1−ι2

1+ι2, sin(u) = 2ι

1+ι2and du = 2dι

1+ι2implies

α0(e) = − 1

4π

∫ ∞

−∞

3e2 − 4e1−ι2

1+ι2+ (2− e2)

[(1−ι2

1+ι2

)2−(

2ι1+ι2

)2]

2(

1− e1−ι2

1+ι2

)4

2dι

1 + ι2

= − 1

4π

∫ ∞

−∞(1 + ι2)

3e2(1 + ι2)2 − 4e(1 − ι4) + (2− e2)[(1− ι2

)2 − (2ι)2]

2 ((1 + ι2)− e(1− ι2))4dι

= − 1

2π

∫ ∞

−∞(1 + ι2)

1− 6ι2 + ι4 − 2e(1− ι4) + e2(1 + 6ι2 + ι4)

((1 + ι2)− e(1− ι2))4dι

= − 1

2π

∫ ∞

−∞(1 + ι2)

(1− e)2 − 6(1 + e)(1− e)ι2 + (1 + e)2ι4

(1− e+ (1 + e)ι2)4dι.

Let us define 1be

:=√

1−e

1+e∈ (0, 1]. Then:

α0(e) = −(1 + (b−1e )2)2

8π

∫ ∞

−∞(1 + ι2)

(b−1e )4 − 6(b−1

e )2ι2 + ι4

((b−1e )2 + ι2)4

dι.

Define the functions h(z) and h(z) as follow:

h(z) :=(1 + z2)(b−1

e )4 − 6(b−1e )2z2 + z4

((b−1e )2 + z2)4

2.1. Two Integral Formulas for αj 25

=(1 + z2)(b−1

e )4 − 6(b−1e )2z2 + z4

(z − ib−1e )4(z + ib−1

e )4=:

h(z)

(z − ib−1e )4

.

h(z) has a pole of order 4 (3 if b−1e = 1) in z⋆ = ib−1

e . If b−1e 6= 1 we have

Resz⋆h(z) =1

6limz→z⋆

∂3

∂z3[(z − z⋆)

4h(z)]=

1

6limz→z⋆

∂3

∂z3h(z).

Define functions p(z) := (1 + z2)((b−1e )4 − 6(b−1

e )2z2 + z4) and q(z) := (z + i(b−1e ))4.

Then it follows:

∂3

∂z3

(p(z)

q(z)

)

=∂2

∂z2

(p′q − pq′

q2

)

=∂

∂z

(q2(qp′′ − pq′′)− 2qq′(qp′ − pq′)

q4

)

=1

q8[(2qq′(qp′′ − pq′′) + q2(q′p′′ + qp′′′ − p′q′′ − pq′′′)

−2(q′)2(qp′ − pq′)− 2q(q′′(qp′ − pq′) + q′(qp′′ − pq′′)))q4

−4q3q′(q2(qp′′ − pq′′)− 2qq′(qp′ − pq′)

)].

Inserting z = z⋆ we get Resz⋆h(z) = 0 (we performed the computation with Mathe-matica). If b−1

e = 1 we have

Resi h(z) =1

2limz→i

∂2

∂z2[(z − i)3h(z)

]= lim

z→i

∂2

∂z21− 6z2 + z4

(z + i)3.

Define functions p(z) := 1− 6z2 + z4 and q(z) := (z + i)3. Then it follows:

∂2

∂z2

(p(z)

q(z)

)

=∂

∂z

(p′q − pq′

q2

)

=q2(qp′′ − pq′′)− 2qq′(qp′ − pq′)

q4.

Inserting z = i we get Resi h(z) = 0 (we performed the computation with Mathemat-ica).

R

iR

γ1

γ2

−R R

Figure 2.1: Integration’s path γ = γ1 γ2.

Let the path γ := γ1 γ2, with γ1 and γ2 as in Figure 2.1. Then we have

0 = Resi h(z) = Resz⋆h(z)

=

∫

γh(z)dz =

∫

γ1

h(z)dz +

∫

γ2

h(z)dzR→∞

−−−−→∫ ∞

−∞h(z)dz,

since ∣∣∣∣

∫

γ2

h(z)dz

∣∣∣∣=

∣∣∣∣

∫ π

0h(Reiφ)Rieiφdφ

∣∣∣∣≤ O

(1

R

)R→∞

−−−−→ 0.

This proves (2.13) and terminates the proof of Lemma 2.1.


In the next lemma we provide a second equivalent integral formula for the Fouriercoefficients αj of the Newtonian potential f in (1.29).

Lemma 2.2. For j ∈ Z \ 0 the coefficients αj = αj(e) defined in (2.1) satisfy

αj = − 1

4π

∫ 2π

0

1

ρ2e(w2e + 1)2

[

(w4e − 6w2

e + 1)cj(u)− 4we(w2e − 1)sj(u)

]

du , (2.14)

where we = w(u; e) :=√

1+e

1−etan u

2 , ρe := ρe(u) = 1− e cos u and

cj(u) := cos(ju− je sin u) , sj(u) := sin(ju− je sinu) .

Proof. If z = arctanw, then

e2iz =i− w

w + i= −(w − i)2

w2 + 1, (2.15)

so that if we(t) := w(ue(t), e) one has fe(t) = 2 arctan(we(t)) and

Ge(t)(2.2)= − 1

2ρe(t)3

(

e2ife(t)2

)2

(2.15)= − 1

2ρe(t)3(we(t)− i)4

(we(t)2 + 1)2

= − 1

2ρe(t)3we(t)

4 − 6we(t)2 + 1− 4iwe(t)(we(t)

2 − 1)

(we(t)2 + 1)2. (2.16)

By Remark 2.1 we know that Gj ∈ R for every j ∈ Z, i.e. Gj = Re (Gj). Thus, thefollowing holds for j ∈ Z:

αj = Gj = Re

[1

2π

∫ 2π

0Ge(t)e

−ijt dt

]

(2.16)= − 1

4π

∫ 2π

0

(we(t)4 − 6we(t)

2 + 1) cos(jt)− 4we(t)(we(t)2 − 1) sin(jt)

ρe(t)3(we(t)2 + 1)2dt.

Making again the change of variable given by the Kepler equation (1.32), i.e. inte-grating with respect to u instead of t an using ue(t)

′ = 1ρe(t)

we find

αj = − 1

4π

∫ 2π

0

(w4e − 6w2

e + 1) cos(ju − je sinu)− 4we(w2e − 1) sin(ju− je sinu)

ρ2e(w2e + 1)2

du.

Hence equation (2.14) holds. This terminates the proof of Lemma 2.2.

2.2 Asymptotic behaviour

The following theorem studies the expansion of the Fourier coefficients αj = αj(e),0 6= j ∈ Z defined in (2.1) with respect to the eccentricity e.

Theorem 2.1. Let αj(e) be as in (2.1). Then for all 0 6= j ∈ Z we have

αj(e) = αje|j−2| +O(e|j−2|+1), (2.17)

where

αj =

P2−j(j), for j ≤ 2,Qj−2(j), for j > 2.

(2.18)

2.2. Asymptotic behaviour 27

(Pk)k≥0 and (Qk)k≥0 are two families of polynomial functions given by the relations

Pk(x) :=

(−1

2

)k+1 xk

k!, (2.19)

Qk(x) := − 1

2k+1

k∑

l=0

1

l!

(k − l + 3

3

)

xl. (2.20)

Theorem 2.1 can be interpreted as a Corollary of A. Cayley [10], F. Tisserand [43] andW. M. Kaula [29], where the main question was about the two-body problem and notthe spin-orbit problem. Many recent papers use Theorem 2.1 as a well known results,referring to [10], [43] and [29] for the proof.However A. Cayley in [10] presents the problem using an old mathematical approach,which makes the proof difficult to be followed at the current days. On the other handin Chapter 3.3 of [29] (see equations (3.67)-(3.71)) Kaula gives an explicit formula forαj , but there is no proof for it. Anyway one can find those formulas in Remark 2.3.Finally F. Tisserand in [43] gives an elegant proof, due to Hansen, which is postponedto the following subsection. I found equations (2.19) and (2.20) independently and forcompleteness I will give my alternative proof of Theorem 2.1 in Subsection 2.2.2.

2.2.1 Hansen’s Proof

In the proof of Theorem 2.1 one uses the following result, which is due to Hansen (seeChapter XV of [43]).

Lemma 2.3. (Hansen) Let u, t, fe(t) and ρe(t) be as in Chapter 1, m ∈ N, n ∈ Zand

β :=e

1 +√1− e2

=1−

√1− e2

e. (2.21)

Furthermore define Pn,m0 = Qn,m

0 := 1 and for k ≥ 1 define

Pn,mk (ν) :=

νk

k!+

k−1∑

l=0

νl

l!

n(n− 1) . . . (n− (k − l) + 1)

(k − l)!, (2.22)

Qn,mk (ν) := (−1)k

νk

k!+

k−1∑

l=0

νl

l!(−1)l

n(n− 1) . . . (n− (k − l) + 1)

(k − l)!, (2.23)

where

ν :=je

2β, n = n+m+ 1 and n = n−m+ 1. (2.24)

Then the coefficients Xn,ml in the expansion

(ρe(t))neimfe(t) =:

∑

j∈ZXn,m

j eijt, (2.25)

are given, for j 6= 0, by the following formula:

Xn,mj =

(1 + β2)−n−1(−β)j−m∑∞

k=0Qn,mj−m+k(ν)P

n,mk (ν)β2k, if j > m,

(1 + β2)−n−1(−β)m−j∑∞

k=0 Pn,mm−j+k(ν)Q

n,mk (ν)β2k, if j ≤ m.

(2.26)


Remark 2.2. Also the formula for Xn,m0 is known, (see again Chapter XV of [43]),

namely Xn,m0 = 0 if n ≥ −m− 1 and

Xn,m0 =

(−β)m(1− β2)n+1

(n+ 2)(n + 3) . . . (n+m+ 1)

m!F (m− n− 1,−n− 1,m+ 1, β)

(2.27)if n < −m− 1, where F is the hypergeometric function.

Proof. Setx = eife(t), y = eiu and z = eit. (2.28)

From the Kepler equation (1.32) it follows that

z = eit = eiu−ie sin(u) = ye− e

2

(

y− 1y

)

(2.29)

holds. Since tan(α/2) = eiα−1i(eiα+1)

holds for every angle α, we can find the relation

between x and y in (2.28). By (2.21) and (2.28) the equation (1.31) is then equivalentto

y − 1

y + 1=

1− β

1 + β

x− 1

x+ 1,

since1− β

1 + β=

1 +√1− e2 − e

1 +√1− e2 + e

=

√

1− e

1 + e.

Solving this equation with respect to x we get

(1− x)(1 + y − β − βy) = (1 + x)(1− y + β − βy)

x =y − β

1− βy= y

1− βy

1− βy. (2.30)

Now we express the normalized radius ρe(t) in (1.30) as function of y, i.e.

ρe(t) = 1− e

2

(

y +1

y

)

= −ey2 − 2y + e

2y

= − e

2y

(

y − 1 +√1− e2

e

)(

y − 1−√1− e2

e

)

= − e

2y(y − β)

(

y − 1

β

)

=e

2β(1− βy)

(

1− β

y

)

=1

1 + β2(1− βy)

(

1− β

y

)

. (2.31)

Furthermore, from the Kepler equation (1.32) we get

dt = (1− e cos(u))du =r

adu. (2.32)

Now multiplying (2.25) by z−j = e−ijt and integrating from 0 to 2π with respect tot, we get

Xn,mj =

1

2π

∫ 2π

0

( r

a

)nxmz−jdt =

1

2π

∫ 2π

0ρe(t)

nxmz−jdt. (2.33)

Using (2.29), (2.30), (2.31) and (2.32) the equation (2.33) becomes

2πXn,mj =

∫ 2π

0

(1

1 + β2(1− βy)

(

1− β

y

))n+1(

y1− β

y

1− βy

)m(

ye− e

2

(

y− 1y

))−j

du


= (1 + β2)−n−1

∫ 2π

0ym−j(1− βy)n−m+1

(

1− β

y

)n+m+1

ej e

2

(

y− 1y

)

du.

(2.34)

The terms ym−j , (1 − βy)n−m+1,(

1− βy

)n+m+1and e

j e

2

(

y− 1y

)

are functions, which

may be expanded in a power series of y. We define

Φ = (1− βy)n−m+1

(

1− β

y

)n+m+1

ej e

2

(

y− 1y

)

. (2.35)

For q ∈ Z we have

∫ 2π

0yqdu

(2.28)=

∫ 2π

0eiqudu =

2π, if q = 0,0, if q 6= 0.

(2.36)

Let expand Φ in (2.35) in a power series of y and denote the coefficient yj−m by A.By (2.34) and (2.36) we have

Xn,mj = (1 + β2)−n−1 · A. (2.37)

Now equation (2.35) can be equivalently rewritten as

Φ = ΘΘ1, (2.38)

where

Θ(y, n,m, ν) = (1− βy)n−m+1eνβy, (2.39)

Θ1(y, n,m, ν) =

(

1− β

y

)n+m+1

e−νβy = Θ

(1

y, n,−m,−ν

)

, (2.40)

and ν is defined in (2.24). Notice that, if we find the expansion of Θ, we have auto-matically also the expansion of Θ1. By definition of the exponential function we knowthat

eνβy =

∞∑

k=0

(νβy)k

k!= 1 +

ν

1βy +

ν2

2!(βy)2 + . . . .

Since

d(k)

dy(k)(1− βy)r

∣∣∣∣∣y=0

= (1− βy)r−k(−β)kr(r − 1) . . . (r − k + 1)∣∣∣y=0

= (−β)kr(r − 1) . . . (r − k + 1)

holds, we get by the Taylor Theorem

(1− βy)r =∞∑

k=0

(−1)k(rk

)

(βy)k.

Choosing r = n−m+ 1 and multiplying the two series together we find

Θ(2.39)= eνβy(1− βy)n−m+1 =

∞∑

k=0

(νβy)k

k!

k∑

l=0

(−1)l(rl

)

(βy)l.


The Cauchy product of two series tells us that

Θ =

∞∑

k=0

(−1)k−l νl

l!

(r

k − l

)

(βy)k =

∞∑

k=0

(−1)kQn,mk (βy)k

holds by (2.23). The same for Θ1, i.e. defining r = n+m+ 1 we have that

Θ1(2.40)= Θ

(1

y, n,−m,−ν

)

=∞∑

k=0

(−1)kk∑

l=0

νl

l!

(r

k − l

)(β

y

)k

=:

∞∑

k=0

(−1)kPn,mk

(β

y

)k

holds by (2.22). Since A in (2.37) is the coefficient of yj−m of Φ, and Φ is given by

Φ(2.38)= ΘΘ1

(2.22)&(2.23)=

[ ∞∑

k=0

(−1)kQn,mk (βy)k

][ ∞∑

l=0

(−1)lPn,ml

(β

y

)l]

, (2.41)

we have to take k + l = j −m in order to get A. We distinguish two cases:

(i) If j −m > 0 then we have

A = (−1)j−mβj−m∞∑

k=0

Qn,mj−m+kP

n,mk β2k. (2.42)

(ii) If j −m < 0 then we have

A = (−β)m−j∞∑

k=0

Pn,mm−j+kQ

n,mk β2k. (2.43)

Equation (2.26) follows directly by (2.37), (2.42) and (2.43). This terminates the proofof Lemma 2.3.

Proof. (Proof of Theorem 2.1) By (2.1), (2.2), (2.3) and Remark 2.1 we have

− 1

2ρe(t)3ei2fe(t) =

∑

j 6=0

αj(e)eijt

and, therefore, by (2.25) with n = −3, m = 2 we get

αj = −X−3,2

j

2, for all 0 6= j ∈ Z. (2.44)

In order to find αj in (2.17), we need to find the leading coefficient in the expansion

of −X−3,2j

2 with respect to e, i.e. we must compute X−3,2j,0 in

X−3,2j = X−3,2

j,0 e|j−2| +O(e|j−2|+1).


By (2.44) it follows

αj = −X−3,2

j,0

2, for all 0 6= j ∈ Z. (2.45)

Let us recall that we have to prove (2.17) and (2.18) only for j 6= 0. Since

(1 + β2)−n−1 (2.21)= 1 +O(e), ∀n ∈ Z,

β|j−2| (2.21)=

e|j−2|

2|j−2| +O(e|j−2|+2),

ν(2.24)= j +O(e2),

we get from (2.26) for n = −3, m = 2 that

X−3,2j,0 =

(−12

)j−2Q0

j−2P00 , if j > 2,

(−12

)2−jP 02−jQ

00, if 0 6= j ≤ 2,

(2.46)

holds, where Q0j−2 := Q−3,2

j−2 (j), P02−j := P−3,2

2−j (j). Namely P 02−j and Q

0j−2 are the zero

coefficients in the e−expansion of P−3,22−j , Q

−3,2j−2 defined in (2.22) and (2.23), respec-

tively.Let us consider first the case 0 6= j ≤ 2. Since n = n +m + 1 = −3 + 2 + 1 = 0 by(2.22) we get

P 02−j = P−3,2

2−j (j) =j2−j

(2− j)!.

Since Q00 = 1 (recall that Q0 ≡ 1) it follows

αj(2.45)= −

X−3,2j,0

2

(2.46)=

(−1

2

)3−j

P 02−j =

(−1

2

)3−j j2−j

(2− j)!

(2.19)= P2−j(j) ,

which proves Theorem 2.1 for 0 6= j ≤ 2.In the case j > 2, we have n = n−m+ 1 = −3− 2 + 1 = −4 and from (2.23) we get

Q0j−2 = Q−3,2

j−2 (j) = (−1)jj−2∑

l=0

jl

l!

(j − l + 1)!

3!(j − l − 2)!= (−1)j

j−2∑

l=0

jl

l!

(j − l + 1

3

)

.

(2.47)

Since P 00 = 1 (recall that P0 ≡ 1) we have

αj(2.45)= −

X−3,2j,0

2

(2.46)=

(−1

2

)j−1

Q0j−2

(2.47)= − 1

2j−1

j−2∑

l=0

(j + 1− l

3

)jl

l!

(2.20)= Qj−2(j) .

This concludes the proof of Theorem 2.1 also in the case j > 2.

Remark 2.3. In [29] Kaula uses coefficients Gl,p,q with l, p, q ∈ Z to describe theNewtonian potential. The relation between Gl,p,q and αj in (2.1) is

αj(e) =−G2,0,j−2(e)

2,


i.e. we have to choose l = 2, p = 0 and j = q + 2 to get the spin-orbit case. Here wewrite (without proof) the formula for the coefficients Gl,p,q.If l − 2p + q = 0 define

Glp(2p−l)(e) =1

(1− e2)l−1/2

p′−1∑

d=0

(l − 1

2d+ l − 2p′

)(2d+ l − 2p′

d

)(e

2

)2d+l−2p′

,

where p′ =

p, for p ≤ l/2,

l − p, for p ≥ l/2.If l − 2p + q 6= 0 define

Glpq(e) = (−1)|q|(1 + β2)lβ|q|∞∑

k=0

PlpqkQlpqkβ2k,

where

β =e

1 +√1− e2

,

Plpqk =

h∑

r=0

(2p′ − 2lh− r

)(−1)r

r!

((l − 2p′ + q′)e

2β

)r

,

h =

k + q′, for q′ > 0,k, for q′ < 0,

Qlpqk =

h∑

r=0

(−2p′

h− r

)1

r!

((l − 2p′ + q′)e

2β

)r

,

h =

k, for q′ > 0,

k − q′, for q′ < 0,

and

(p′, q′) =

(p, q), for p ≤ l/2,

(l − p,−q), for p > l/2.

2.2.2 Alternative Proof of Theorem 2.1

Before starting with an alternative proof of Theorem 2.1, which was found indepen-dently of Hansen (see [43]), we need some preparatory lemmata.

Lemma 2.4. Let (Qk)k≥0 be defined as in (2.20). Then

2(k + 1)Qk+1(x) = (x+ k + 4)Qk(x)−x

2Qk−1(x)

holds for all k ≥ 1.

Proof. With some straightforward computations we get

(x+ k + 4)Qk(x)−x

2Qk−1(x) =

(2.20)= −(x+ k + 4)

2k+1

k∑

l=0

1

l!

(k − l + 3

3

)

xl +x

2k+1

k−1∑

l=0

1

l!

(k − l + 2

3

)

xl

= − 1

2k+1k!xk+1 − k + 4

2k+1

(k + 33

)

− 1

2k+1

k∑

l=1

[1

(l − 1)!

(k − l + 4

3

)


+1

l!

(k − l + 3

3

)

(k + 4)− 1

(l − 1)!

(k − l + 3

3

)]

xl

= −(k + 1)

2k+1

k+1∑

l=0

1

l!

(k − l + 4

3

)

xl = 2(k + 1)Qk+1(x).

Lemma 2.5. Let fe = fe(u) be as in Lemma 2.1. In order to simplify the notation we

write D := ∂∂e , ι := tan

(u2

), be :=

√1+e

1−e. Then for all k ≥ 1

Dk fe =

k∑

l=1

Akl (e)gl, (2.48)

holds, where

gl :=ι2l−1

(1 + (beι)2)l

and the coefficients Akl (e) are recursively defined as follows:

A11(e) = 2b′e,

Ak+1l (e) =

DAk1(e), for l = 1,

DAkl (e)− 2(l − 1)beb

′eA

kl−1(e), for 2 ≤ l ≤ k,

−2kbeb′eA

kk(e), for l = k + 1.

(2.49)

Remark 2.4. We observe that

Dgl = −2lbeb′egl+1

holds for all l ≥ 1.

Proof. (Induction with respect to k)

k = 1: By definition of fe in Lemma 2.1 we have Dfe = ∂∂e fe(u) = 2b′

eι

1+(beι)2. On the

other hand by (2.48) and (2.49) we get Dfe = A11(e)g1 = 2b′e

ι1+(beι)2

, which proves the

assertion for k = 1.k → k + 1: Induction step:

Dk+1fe = D[

Dk fe

]

(2.48)= D

[k∑

l=1

Akl (e)gl

]

=

k∑

l=1

[

(DAkl (e))gl +Ak

l (e)Dgl

]

Rmk. 2.4=

k∑

l=1

[

(DAkl (e))gl − 2lbeb

′eA

kl (e)gl+1

]

= (DAk1(e))g1 +

k∑

l=2

[

(DAkl (e))− 2(l − 1)beb

′eA

kl−1(e)

]

gl

−2kbeb′eA

kk(e)gk+1


(2.49)=

k+1∑

l=1

Ak+1l (e)gl.

This concludes the proof of Lemma 2.5.

Remark 2.5. fe defined in Lemma 2.1 is an analytic function of e. Therefore it canbe expanded abound e = 0, i.e.

fe(u) =: u+ e g(u; e),

where

g(u; e) =

∞∑

k=1

gk(u)ek−1

holds for some odd functions gk(u), which will be analysed in details in the next lemma.

Lemma 2.6. Let g(u; e) be as in Remark 2.5. Then for all k ≥ 1

gk(u) =k∑

l=1

gk,l sin(lu), (2.50)

holds with gk,l ∈ R and gk,k = 1k2k−1 .

Proof. From the Taylor expansion of fe around e = 0 we know that

gk(u) =Dk fek!

∣∣∣∣∣e=0

holds for all k ≥ 1. Using the same notation as in Lemma 2.5, we define

Ak,l := Akl (0).

From (2.48) and (2.49) it follows

gk(u) =1

k!

k∑

l=1

Akl (0)(gl)|e=0 =

1

k!

k∑

l=1

Ak,lι2l−1

(1 + ι2)l.

Recall that ι = tan(u2

). So for all l ≥ 1 we have

ι2l−1

(1 + ι2)l=

sin2l−1(u2

)cos2l

(u2

)

cos2l−1(u2

) = cos(u

2

)

sin2l−1(u

2

)

= cos(u

2

)

sin(u

2

) [

sin2(u

2

)]l−1=

1

2sin(u)

[1− cos(u)

2

]l−1

=1

2lsin(u) [1− cos(u)]l−1 .

Using the binomial formula we obtain

ι2l−1

(1 + ι2)l=

1

2lsin(u)

l−1∑

i=0

(−1)i(l − 1i

)

cosi(u), for all l ≥ 1,

and therefore

gk(u) =1

k!

k∑

l=1

Ak,l1

2lsin(u)

l−1∑

i=0

(−1)i(l − 1i

)

cosi(u)


=

k−1∑

i=0

k∑

l=i+1

Ak,l(−1)i(l − 1i

)

k! 2l

sin(u) cosi(u).

Using the trigonometric identity cos(β1) sin(β2) = 12 [sin(β1 + β2)− sin(β1 − β2)] we

get (2.50). Furthermore using equation (2.49) recursively, it is easy to verify that

gk,k =(−1)k−1

k! · 22k−1Ak,k =

(−1)k−1

k! · 22k−1(−2(k − 1))Ak−1,k−1

= . . .

=(−1)k−1

k! · 22k−1(−2)k−1(k − 1)!A1,1 =

1

k2k−1,

since A1,1 = 2.

Remark 2.6. Let g(u; e) as in Remark 2.5. Then we define

g(u; e, j) := 2g(u; e) + j sin(u) =∞∑

k=1

gjk(u)ek−1, (2.51)

where gk(u) are given by

gjk(u) :=

2g1(u) + j sin(u), for k = 1,

2gk(u), for k ≥ 2.(2.52)

From Lemma 2.6 we know that

gjk(u) =

k∑

l=1

gjk,l sin(lu), (2.53)

holds with

gjk,k :=

2 + j, for k = 1,

1k2k−2 , for k ≥ 2.

(2.54)

Lemma 2.7. Let g(u; e, j) be defined as in equation (2.51). Then we have

cos(eg(u;e,j))(1−e cos(u))2 =

∑

k≥0 cjk(u)e

k,

sin(eg(u;e,j))(1−e cos(u))2 =

∑

k≥0 sjk(u)e

k,

(2.55)

where cjk(u), sjk(u) have the Fourier expansions

cjk(u) :=

k∑

l=0

cjk,l cos(lu), (2.56)

sjk(u) :=k∑

l=1

sjk,l sin(lu), (2.57)

for some real coefficients cjk,l, sjk,l.


Proof. Let us define

C(u; e, j) := cos

∑

k≥1

gjk(u)ek

and S(u; e, j) := sin

∑

k≥1

gjk(u)ek

,

where gjk(u) are defined in (2.52). Since g and therefore also g are odd with respectto u, it is easy to verify that C(u; e, j) and S(u; e, j) are even and odd in the variableu, respectively. Therefore C will have only cosinus-terms in is expansion and S onlysinus-terms. Furthermore C(u; e, j) and S(u; e, j) are analytic in e. We make thefollowing formal Ansatz for their power series

C(u; e, j) =∑

k≥0

Cjk(u)e

k, S(u; e, j) =∑

k≥0

Sjk(u)e

k.

In [45] we find recursive formula to compute the coefficients Cjk(u), S

jk(u):

Cj0(u) = 1, Cj

k(u) = − 1k

∑kl=1 l g

jl (u)S

jk−l(u),

Sj0(u) = 0, Sj

k(u) =1k

∑kl=1 l g

jl (u)C

jk−l(u).

(2.58)

Using induction with respect to k and the fact that gjk(u) are trigonometric polynomi-als of degree k (due to (2.53)), from (2.58) it is easy to prove that for some constantCjk,l, S

jk,l ∈ R the functions Cj

k(u), Sjk(u) have the form

Cjk(u) :=

k∑

l=0

Cjk,l cos(lu), (2.59)

Sjk(u) :=

k∑

l=1

Sjk,l sin(lu), (2.60)

Furthermore using the geometric series expansion of the function 1(1−e cos(u))2

withrespect to e, i.e.

1

(1− e cos(u))2=

∞∑

k=1

k cosk−1(u)ek−1,

and the trigonometric identities cos(β1) cos(β2) =12 (cos(β1 + β2) + cos(β1 − β2)) and

cos(β1) sin(β2) =12 (sin(β1 + β2)− sin(β1 − β2)), we get (2.55), (2.56) and (2.57).

Remark 2.7. Recall that: Cjk(u) and Sj

k(u) are even and odd trigonometric polyno-mials of degree k, respectively (see (2.59) and (2.60)); the trigonometric identities

sin(β1) sin(β2) =1

2[− cos(β1 + β2) + cos(β1 − β2)] ,

sin(β1) cos(β2) =1

2[sin(β1 + β2)− sin(β1 − β2)]

hold; gjk(u) has the form (2.53). Then for the coefficients Cjk,k and Sj

k,k in (2.59) and(2.60) it follows that

Cj0,0 = 1, Cj

k,k = 1k

∑k−1l=1

lgjl,lSjk−l,k−l

2 ,

Sj0,0 = 0, Sj

k,k = 1k

∑k−1l=1

lgjl,lCj

k−l,k−l

2 + gjk,k,(2.61)

hold, where gjk,k is given in (2.54).


Remark 2.8. Using (2.61) we compute Cj1,1 = 0 and Sj

1,1 = 2 + j.

Remark 2.9. From the proof of Lemma 2.7 we know that

∑

k≥0

cjk(u)ek =

∑

k≥0

Cjk(u)e

k

( ∞∑

k=0

(k + 1) cosk(u)ek

)

=∞∑

k=0

[k∑

l=0

(k + 1− l)Cjl (u) cos

k−l(u)

]

ek

holds. Therefore we get

cjk(u) =k∑

l=0

(k + 1− l)Cjl (u) cos

k−l(u), for k ≥ 0. (2.62)

Using (2.56), (2.59) and the trigonometric identitiescos(β1) cos(β2) =

12 [cos(β1 + β2) + cos(β1 − β2)] and cosk−l(u) = 1

2k−l−1 cos((k−l)u)+TPk−l−1(u), where TPk−l−1(u) is a trigonometric polynomial in u of degree k− l− 1,from (2.62) it follows:

cjk,k(u) = (k+1)2k−1 +

∑kl=1

Cjl,l(k+1−l)

2k−l , for k ≥ 1,

cj0,0(u) = 1.(2.63)

Analogously we can prove that

sjk(u) =

k∑

l=0

(k + 1− l)Sjl (u) cos

k−l(u), for k ≥ 0. (2.64)

Using (2.57), (2.60) and the trigonometric identitiessin(β1) cos(β2) =

12 [sin(β1 + β2) + sin(β1 − β2)] and cosk−l(u) = 1

2k−l−1 cos((k− l)u)+TPk−l−1(u), where TPk−l−1(u) is a trigonometric polynomial in u of degree k− l− 1,from (2.64) it follows:

sjk,k(u) =∑k

l=1

Sjl,l(k+1−l)

2k−l , for k ≥ 1,

sj0,0(u) = 0.(2.65)

Definition 2.1. Let Cjk,k and Sj

k,k be as in (2.61). Then for k ≥ 0 we define

∆j,±k+1 := Cj

k+1,k+1 ± Sjk+1,k+1,

Ωj,±k+1 := 1

2k+∑k+1

l=1

Cjl,l±Sj

l,l

2k+1−l .

Lemma 2.8. Let ∆j,+k+1 and Ωj,+

k+1 be as in Definition 2.1. Let (Qk)k≥0 be defined asin (2.20) and furthermore define Q−1(x) := 0. Then for k ≥ 0 we have

∆j,+k+1 = −4Qk+1(j) + 4Qk(j) −Qk−1(j),

Ωj,+k+1 = 2Qk(j) − 4Qk+1(j),

(2.66)

and

∆j,−k+1 = (−1)k+1jk−1

k! 2k

[(j+k+1)k+j(j+k+2)

k+1

]

,

Ωj,−k+1 = (−1)k+1jk

k! 2k

(j

k+1 + 1)

.(2.67)


Proof. (Induction with respect to k)k = 0: By (2.20), (2.61) in Remark 2.7, Remark 2.8 and Definition 2.1 we have

∆j,+1 = Cj

1,1 + Sj1,1 = 2 + j = 4 + j − 2 = −4Q1(j) + 4Q0(j)−Q−1(j),

Ωj,+1 = Cj

0,0 + Sj0,0 + Cj

1,1 + Sj1,1 = 3 + j = −1 + 4 + j = 2Q0(j) − 4Q1(j),

∆j,−1 = Cj

1,1 − Sj1,1 = −2− j,

Ωj,−1 = Cj

0,0 − Sj0,0 + Cj

1,1 − Sj1,1 = 1− (2 + j) = −1− j.

This proves the assertion for k = 0.k + 1 → k + 2: By Definition 2.1 we observe that

Ωj,±k+2 =

1

2k+1+

k+2∑

l=1

Cjl,l ± Sj

l,l

2k+2−l=

1

2Ωj,±k+1 +∆j,±

k+2.

On the other side

∆j,±k+2 = Cj

k+2,k+2 ± Sjk+2,k+2

(2.61)=

1

k + 2

k+1∑

l=1

lgjl,l

(

Sjk+2−l,k+2−l ± Cj

k+2−l,k+2−l

)

2± gjk+2,k+2

(2.54)= ± 1

k + 2

(j

2+ 1

)

∆j,±k+1 +

1

k + 2

k+1∑

l=2

Sjk+2−l,k+2−l ± Cj

k+2−l,k+2−l

2l−1

± 2

(k + 2)2k+1

= ± 1

k + 2

(j

2+ 1

)

∆j,±k+1 ±

1

k + 2

[k∑

l=1

Cjl,l ± Sj

l,l

2k+1−l± 1

2k

]

Def.2.1= ± j

2(k + 2)∆j,±

k+1 ±1

k + 2Ωj,±k+1.

As a consequence we get

Ωj,±k+2 =

(12 ± 1

k+2

)

Ωj,±k+1 ± j

2(k+2)∆j,±k+1,

∆j,±k+2 = ± j

2(k+2)∆j,±k+1 ± 1

k+2Ωj,±k+1.

Using the induction step assumption in the previous equation we have

Ωj,−k+2

(2.67)=

k

2(k + 2)

(−1)k+1

2kjk

k!

(j

k + 1+ 1

)

− j

2(k + 2)

(−1)k+1

2kjk−1

k!

[(j + k + 1)k + j(j + k + 2)

k + 1

]

=(−1)k+1jk

(k + 2)! 2k+1[k(j + k + 1)− (j + k + 1)k − j(j + k + 2)]

=(−1)k+2jk+1

(k + 2)! 2k+1[j + k + 2]

=(−1)k+2jk+1

(k + 1)! 2k+1

[j

k + 2+ 1

]

,

∆j,−k+2

(2.67)= − j

2(k + 2)

(−1)k+1

2kjk−1

k!

[(j + k + 1)k + j(j + k + 2)

k + 1

]


− 1

k + 2

(−1)k+1

2kjk

k!

(j

k + 1+ 1

)

=(−1)k+2jk

(k + 2)! 2k+1[(j + k + 1)k + j(j + k + 2) + 2(j + k + 1)]

=(−1)k+2

2k+1

jk

(k + 2)![(j + k + 1)(k + 2) + j(j + k + 2)]

=(−1)k+2jk

(k + 2)! 2k+1[(j + k + 2)(k + 1) + j(j + k + 3)] ,

Ωj,+k+2

(2.66)=

k + 4

2(k + 2)(2Qk(j)− 4Qk+1(j))

+j

2(k + 2)(−4Qk+1(j) + 4Qk(j)−Qk−1(j))

=−1

k + 2

(

2(k + 4 + j)Qk+1(j) − (k + 4 + 2j)Qk(j) +j

2Qk−1(j)

)

Lemma 2.4=

−1

k + 2(2(3 + j)Qk+1(j)− jQk(j))

Lemma 2.4= 2Qk+1(j) − 4Qk+2(j),

and finally

∆j,+k+2

(2.66)=

j

2(k + 2)(−4Qk+1(j) + 4Qk(j)−Qk−1(j))

+1

(k + 2)(2Qk(j)− 4Qk+1(j))

=1

(k + 2)

(

−2(2 + j)Qk+1(j) + (2 + 2j)Qk(j)−j

2Qk−1(j)

)

Lemma 2.4=

1

(k + 2)(−2(1− k + j)Qk+1(j) + (j − k − 2)Qk(j))

Lemma 2.4= −4Qk+2(j) + 4Qk+1(j) −Qk(j).

Remark 2.10. For k ≥ 1 using (2.63), (2.65) and Definition 2.1 it follows

cjk+1,k+1 ± sjk+1,k+1 =k + 2

2k+

k+1∑

l=1

(Cjl,l ± Sj

l,l)(k + 2− l)

2k+1−l

=k + 1

2k+

k+1∑

l=1

(Cjl,l ± Sj

l,l)(k + 1− l)

2k+1−l

+1

2k+

k+1∑

l=1

Cjl,l ± Sj

l,l

2k+1−l

=1

2

(

cjk,k ± sjk,k

)

+Ωj,±k+1, (2.68)

and furthermorecj1,1 ± sj1,1 = cj0,0 ± sj0,0 +Ωj,±

1 (2.69)

holds.


Now we are able to prove Theorem 2.1.

Proof. (Proof of Theorem 2.1) From equation (2.3) and Remark 2.5 we have

αj(e) = − 1

4π

∫ 2π

0

cos(2u− ju+ e(2g(u) + j sin(u)))

(1− e cos(u))2du

(2.51)= − 1

4π

∫ 2π

0

cos((j − 2)u− eg(u; e, j))

(1− e cos(u))2du.

Using the trigonometric identity cos(β1 − β2) = cos(β1) cos(β2) + sin(β1) sin(β2) weget

αj(e) = − 1

4π

∫ 2π

0

cos((j − 2)u) cos(eg(u; e)) + sin((j − 2)u) sin(eg(u; e))

(1− e cos(u))2du.

By Lemma 2.7 we haveαj(e) = I + II,

where

I := − 1

4π

∫ 2π

0cos((j − 2)u)

∑

k≥0

ekcjk(u)du,

and

II := − 1

4π

∫ 2π

0sin((j − 2)u)

∑

k≥0

eksjk(u)du.

We evaluate I and II separately.Using equation (2.56) and the obvious fact that for n,m ∈ Z

1

2π

∫ 2π

0cos(nx) cos(mx)dx =

1, for n = m = 0,12 , for n = m 6= 0,0, otherwise,

we obtain:

I(2.56)= − 1

4π

∑

k≥0

ek∫ 2π

0cos((j − 2)u)

⌊k/2⌋∑

l=0

cjk,k−2l cos((k − 2l)u)du

= − 1

4π

∑

k≥|j−2|ek∫ 2π

0cos(|j − 2|u)

⌊k/2⌋∑

l=0

cjk,k−2l cos((k − 2l)u)du

= − 1

4πe|j−2|

∫ 2π

0cos(|j − 2|u)

⌊|j−2|/2⌋∑

l=0

cj|j−2|,|j−2|−2l cos((|j − 2| − 2l)u)du

+O(e|j−2|+1)

(2.56)= −1

2e|j−2|cj|j−2|,|j−2|

1

2π

∫ 2π

0cos2(|j − 2|u)du+O(e|j−2|+1)


=

−12c

j0,0 +O(e), for j = 2,

−14c

j|j−2|,|j−2|e

|j−2| +O(e|j−2|+1), for j 6= 2,

where the cjk,k are defined in equation (2.63).Analogously using equation (2.57) and the obvious fact that for n,m ∈ Z

1

2π

∫ 2π

0sin(nx) sin(mx)dx =

1, for n = m = 0,12 , for n = m 6= 0,0, otherwise,

we obtain:

II = − 1

4π

∑

k≥0

ek∫ 2π

0sin((j − 2)u)sjk(u)du

(2.57)=

O(e), for j = 2,

−14sign(j − 2)sj|j−2|,|j−2|e

|j−2| +O(e|j−2|+1), for j 6= 2,

where the sjk,k are defined in equation (2.65).Till now we have proved:

αj =

−12 +O(e), for j = 2,

αje|j−2| +O(e|j−2|+1), for j 6= 2,

(2.70)

with

αj := −1

4

(

cj|j−2|,|j−2| + sign(j − 2)sj|j−2|,|j−2|

)

, for all Z ∋ j 6= 2. (2.71)

It remains to prove the formula (2.18). For j = 2 by (2.70) it is trivially true, becauseby definition P0(j) = Q0(j) = −1

2 (see (2.19) and (2.20)). For j 6= 2, we distinguishthe cases j < 2 and j > 2.

• Case 1: j < 2. It is obvious that in this case we have sign(j−2) = −1, therefore(2.71) takes the form

αj := −1

4

(

cj|j−2|,|j−2| − sj|j−2|,|j−2|

)

, for all Z ∋ j < 2.

From equations (2.68) and (2.69) we get

cjk,k − sjk,k =1

2

(

cjk−1,k−1 − sjk−1,k−1

)

+Ωj,−k

=1

4

(

cjk−2,k−2 − sjk−2,k−2

)

+1

2Ωjk−1 +Ωj

k

= . . .

=1

2k−1

(

cj0,0 − sj0,0

)

+

k∑

i=1

1

2k−iΩj,−i .

Using Lemma 2.8 and the fact that cj0,0(u)− sj0,0(u) = 1 we get

cjk,k − sjk,k =1

2k−1+

k∑

i=1

1

2k−i

(−1)i

2i−1

ji−1

(i− 1)!

(j

i+ 1

)


=1

2k−1+

1

2k−1

k∑

i=1

(−1)iji−1

(i− 1)!

(j

i+ 1

)

=1

2k−1+

1

2k−1

k∑

i=1

(−1)iji

i!+

1

2k−1

k∑

i=1

(−1)iji−1

(i− 1)!

=1

2k−1+

1

2k−1

k−1∑

i=1

(−1)iji

i!+

(−1)kjk

2k−1k!

− 1

2k−1+

1

2k−1

k∑

i=2

(−1)iji−1

(i− 1)!

=(−1)kjk

2k−1k!.

Thus we get −14

(

cjk,k − sjk,k

)

= Pk(j), for k ≥ 1. Choosing k = 2 − j proves

formula (2.18) for the case j < 2.

• Case 2: j > 2. It is obvious that in this case we have sign(j − 2) = 1, therefore(2.71) has the form

αj := −1

4

(

cj|j−2|,|j−2| + sj|j−2|,|j−2|

)

, for all Z ∋ j < 2.

From equations (2.68) and (2.69) we get

cjk,k + sjk,k =1

2

(

cjk−1,k−1 + sjk−1,k−1

)

+Ωj,+k

=1

4

(

cjk−2,k−2 − sjk−2,k−2

)

+1

2Ωjk−1 +Ωj

k

= . . .

=1

2k−1

(

cj0,0 + sj0,0

)

+

k∑

i=1

1

2k−iΩj,+i .

Using Lemma 2.8 and the fact that cj0,0(u)− sj0,0(u) = 1 we get

cjk,k + sjk,k =1

2k−1+

k∑

i=1

1

2k−i(2Qi−1(j)− 4Qi(j))

(2.20)=

1

2k−1+

k∑

i=1

1

2k−i

1

2i−1

(

−i−1∑

l=0

1

l!

((i− 1)− l + 3

3

)

jl

+i∑

l=0

1

l!

(i− l + 3

3

)

jl

)

=1

2k−1−

k−1∑

i=0

1

2k−1

i∑

l=0

1

l!

(i− l + 3

3

)

jl

+

k∑

i=1

1

2k−1

i∑

l=0

1

l!

(i− l + 3

3

)

jl

=1

2k−1− 1

2k−1+

1

2k−1

k∑

l=0

1

l!

(k − l + 3

3

)

jl


(2.20)= −4Qk(j).

Thus we get −14

(

cjk,k + sjk,k

)

= Qk(j), for k ≥ 1. Choosing k = 2 − j proves

formula (2.18) for the case j > 2.

This concludes the proof of Theorem 2.1.

3 Generalized Newtonian Potential

In this chapter we study a more general problem than the dissipative spin-orbit prob-lem presented in Subsection 1.3.1. In the equation of motion (1.24) we let the potentialfunction

f ∈ C∞(R2,R)

be any 2π−periodic function in both variables x, t and having zero average. Moreprecisely, we consider the differential equation

0 = x+ η(x− ν) + εfx(x, t), (3.1)

where we assume that f is a real analytic function with Fourier expansion

f(x, t) =∑

Z2∋(m,n)6=(0,0)

fm,neimxeint. (3.2)

We assume fm,n = f−m,−n for all Z2 ∋ (m,n) 6= (0, 0), since we want f to be real forreal arguments. We call this problem the general dissipative spin-orbit problem.

Remark 3.1. Of course the dissipative spin-orbit problem (see equation (1.24)) is aspecial case of the general dissipative spin-orbit problem. Indeed rewriting the Newto-nian potential defined in (1.29) we have

f(x, t)(2.1)=

∑

n 6=0,n∈Zαn cos(2x− nt) =

∑

n 6=0,n∈Zαn

[ei2xe−int + e−i2xeint

2

]

=∑

n 6=0,n∈Z

αn

2eintei(−2)x +

∑

n 6=0,n∈Z

αn

2e−intei2x

=∑

n 6=0,n∈Z

αn

2eintei(−2)x +

∑

n 6=0,n∈Z

α−n

2eintei2x

= f−2(t)ei(−2)x + f2(t)e

i2x.

Therefore the dissipative spin-orbit problem is a special case of the generalized dissipa-tive spin-orbit problem, where m only takes the values ±2. For the Fourier-coefficientsof the functions f2(t) and f−2(t) we get the relations:

f2,n =α−n

2and f−2,n =

αn

2. (3.3)

The aim of this chapter is to find conditions for the existence of a (p, q)−periodic orbitfor the general dissipative spin-orbit problem, which are analogously defined as forthe classical spin-orbit case (cf. Definition 1.1. in Chapter 1). Hence:

Definition 3.1. A (p, q)−periodic orbit (with p and q co–prime non–vanishing inte-gers) for the general dissipative spin-orbit problem is a solution xpq(t) : R → R of(3.1) satisfying

x(t+ 2πq) = x(t) + 2πp , ∀t, (3.4)

which corresponds to the spin-orbit resonance of the system.

45

46 Chapter 3. Generalized Newtonian Potential

3.1 Introduction

In this section we rewrite some important results, which can be found in Sections2.1− 2.4 of [8].

Since the potential f in (3.1) is 2π−periodic in x and t, a (p, q)−periodic solution canbe written as

xpq(t) = ξ +p

qt+ u

(t

q

)

, 〈u〉 := 1

2π

∫ 2π

0u(s)ds = 0, (3.5)

where ξ ∈ R and u is 2π−periodic. Without loss of generality let p and q be positivecoprime integers, then xpq(t) in (3.5) is a solution of (3.1) if and only if u satisfies thefunctional equation

Lηu = Φξ(u), (3.6)

where

Lηu := u′′ + ηu′,[Φξ(u)] (t) := [Φξ(u; ην, ε, p, q)] (t) = ην − εfx(ξ + pt+ u(t), qt),

andη := qη, ν := qν − p, ε := q2ε.

Note that for convenience we again denote the new time t/q by t and u′, u′′ denotethe derivatives with respect to the new time.

Remark 3.2. The nonlinear operator Φξ is 2π−periodic in ξ. Thus we can assumeξ ∈ [0, 2π].

Definition 3.2. Let Ckper be the Banach space of 2π–periodic Ck(R) functions endowed

with the Ck–norm1; let Ckper,0 be the closed subspace of Ck

per formed by functions with

vanishing average over [0, 2π]; finally, denote by B := C0per,0 the Banach space of

2π–periodic continuous functions with zero average (endowed with the sup–norm).

Lemma 3.1. For all 0 < η the linear operator Lη : C2per,0 → B in (3.6) maps injectively

C2per,0 onto B. Furthermore, there exist constants η0 > 0 and κ0 = κ0(η0) > 0 such

that for all η ∈ (0, η0] the inverse operator (the “Green operator”) Gη = L−1η is a

bounded linear isomorphism given by

Gη [g(t)] = Gη

∑

n 6=0

gneint

:=∑

n 6=0

gniηn − n2

eint (3.7)

with‖Gη‖ := ‖Gη‖L(B,C2

per,0)≤ κ0.

Proof. The operator Lη is injective since for η > 0 its kernel ker(Lη) is given by

ker(Lη) =g ∈ C2

per,0 | g′′ + ηg′ = 0= g = 0 .

Equation (3.7) follows since the expression in Fourier series of the operator L is

Lη [g(t)] = Lη

∑

n 6=0

gneint

= g′′ + ηg′ =∑

n 6=0

(iηn − n2)gneint

1‖v‖Ck := sup0≤j≤k

supt∈R

|Djv(t)|.

3.1. Introduction 47

for every function g ∈ C2per,0.

The existence of κ0(η0) > 0 such that ‖Gη‖L(B,C2per,0)

≤ κ(η0) follows easily from the

definition of the operator Gη. For 0 < η ≤ η0 and g ∈ B we have

‖Gη‖L(B,C2per,0)

≤ ‖Gη‖L(B,B) = sup‖g‖

C0=1‖Gη(g)‖

(3.7)= sup

‖g‖C0=1

∣∣∣∣∣∣

∑

n 6=0

gniηn − n2

eint

∣∣∣∣∣∣

≤ 1

1 + η20sup

‖g‖C0=1‖g‖C0 =

1

1 + η20=: κ0(η0).

This proves the boundedness2 of the operator Gη.

3.1.1 Lyapunov-Schmidt decomposition

Performing a Lyapunov-Schmidt decomposition, equation (3.6) can be splitted in abifurcation equation

〈Φξ(u)〉 = 0 ⇔ φu(ξ) :=1

2π

∫ 2π

0fx(ξ + pt+ u(t), qt)dt =

ην

ε, (3.8)

and a range equation

u = Gη (Φξ(u)− 〈Φξ(u)〉) = −εGη (fx(ξ + pt+ u(t), qt)− φu(ξ)) . (3.9)

The unknowns in (3.8), (3.9) are u ∈ B and ξ ∈ [0, 2π]. The idea to find a solutionof the system is: first solve the range equation for arbitrary ξ to get a function u =u(·; ξ, ε); then put this function u into the bifurcation equation in order to determineξ.

3.1.2 The range equation

The range equation (3.9) is solved using a contraction argument.

Lemma 3.2. Let 0 < κ1 ≤ 1 and fix ξ ∈ [0, 2π]. For all η ∈ (0, η0] let κ0 = κ(η0) beas in Lemma 3.1. Let κ2 := 2κ0‖f‖C2 and Bκ1 := v ∈ B| ‖v‖C0 ≤ κ1. For Φξ(v; ε)as in (3.6) define

Φξ(v; ε) :=Φξ(v; ε) − 〈Φξ(v; ε)〉

ε= −fx(ξ + pt+ v(t), qt) + φv(ξ),

where φv(ξ) is as in (3.8). Then for |ε| ≤ ε0 :=κ12κ2

the function

ϕ : v ∈ Bκ1 → ϕ(v) := εGη

(

Φξ(v; ε))

is a contraction with Lipshitz constant |ε|κ2 ≤ ε0κ2 ≤ 12 .

Moreover, for every ξ ∈ [0, 2π] there exists a unique fixed point u := u(·; ξ) ∈ Bκ1 ofthe function ϕ and defining ϕk(0) = ϕ ϕ . . . ϕ

︸︷︷︸

k times

(0) the following holds:

u =

∞∑

k=1

(

ϕk(0)− ϕk−1(0))

, (3.10)

with

‖u‖C0 ≤ 2|ε|κ0‖f‖C2

1− 2|ε|κ0‖f‖C2

≤ 4|ε|κ0‖f‖C2 . (3.11)

2The constant κ0(η0) is not optimal. In Chapter 6 for a particular value of η0 we will find sharpestimates for this bound.


Proof. Notice that Φξ(·; ε) ∈ C1(B,B) and for every u ∈ B and ξ ∈ [0, 2π] we have

‖Φξ(v; ε)‖C0 ≤ 2 supT2

|fx| ≤ 2‖f‖C2 ,

‖DvΦξ(v; ε)‖L(B,B) ≤ 2 supT2

|fxx| ≤ 2‖f‖C2 .

The function ϕ maps Bκ1 into itself, since for every v ∈ B with ‖v‖C0 < κ1 we get

‖ϕ(v)‖C0 = |ε|‖Gη

(

Φξ(v; ε))

‖C0 ≤ 2ε0κ0‖f‖C2 =κ12< κ1.

Furthermore the Lipshitz-continuity follows from the linearity of the operator Gη andfrom the mean value theorem, hence for every v1, v2 ∈ Bκ1

‖ϕ(v1)− ϕ(v2)‖C0 ≤ 2|ε|κ0‖f‖C2‖v1 − v2‖C0

holds. This proves that ϕ is a contraction with Lipshitz constant |ε|κ2 ≤ ε0κ2 =κ12 ≤

12 .The existence and the uniqueness of u follow directly from the Banach fixed-pointtheorem. The sequence (ϕk(0))k∈N converges uniformly to u, since

‖ϕk(0)− ϕk−1(0)‖C0 ≤ (2|ε|κ0‖f‖C2)k → 0, for k → ∞, (3.12)

holds. This proves equation (3.10). Using (3.10), (3.12) and a geometric series ar-gument we find the upper bound for ‖u‖C0 in (3.11). This terminates the proof ofLemma 3.2.

Remark 3.3. (i) Since f in (1.29) is of class C∞ in the variables t, ε, ξ and η, alsothe fixed point u is of class C∞ in all these variables.(ii) Since f in (3.2) is real-analytic in ε the fixed point u(t; ξ, ε) is also real-analyticin ε. From Lemma 3.2 and from the Cauchy estimates we get

u(t; ξ, ε) =

∞∑

k=1

uk(t; ξ)εk, (3.13)

and ‖uk‖ ≤ 4|ε|κ0‖f‖C2ε−k0 on suitable complex domains. To simplify the notation

we will often write uk instead of uk(t; ξ).(iii) From (3.13) we conclude that also the function φu(ξ) in (3.8) can be expressed ina ε power series, i.e.

φu(ξ) =

∞∑

k=0

φ(k)(ξ)εk. (3.14)

3.1.3 The bifurcation equation

For later use we formulate Propositon 2.7 of [8]:

Theorem 3.1. (Propositon 2.7 of [8]) For |ε| ≤ ε0 let u be a solution of (3.9) asin Lemma 3.2. Assume φ in (3.8) can be written as

φu(ξ; ε) = φ(0)(ξ) + εφ(1)(ξ; ε)

= φ(0)(ξ) + εφ(1)(ξ) + ε2φ(2)(ξ; ε)

withsup

|ε|≤ε0, ξ∈[0,2π]

∣∣∣φ(i)(ξ, ε)

∣∣∣ ≤Mi

for suitable Mi > 0 and i = 1, 2.

3.1. Introduction 49

(i) Assume

φ(0)− := min

ξ∈[0,2π]φ(0)(ξ) < max

ξ∈[0,2π]φ(0)(ξ) =: φ

(0)+ .

Let

|ε| ≤ ε1 := min

(

ε0,δ

M1

)

,

where

0 < δ <φ(0)+ − φ

(0)−

2.

Then there exists ξ ∈ [0, 2π] solving the bifurcation equation (3.8) provided

ην

ε∈[

φ(0)− + δ, φ

(0)+ − δ

]

holds.

(ii) Assume ξ → φ(0)(ξ) is identically zero and

φ(1)− := min

ξ∈[0,2π]φ(1)(ξ) < max

ξ∈[0,2π]φ(1)(ξ) =: φ

(1)+ .

Let

|ε| ≤ ε2 := min

(

ε0,δ

M2

)

,

where

0 < δ <φ(1)+ − φ

(1)−

2.


ην

ε∈[

ε(

φ(0)− + δ

)

, ε(

φ(0)+ − δ

)]

holds.

In either case, for the above ε, ξ, η and ν, the couple (u, ξ) solves (3.8)-(3.9), so thatxpq(t) defined in (3.5) solves (3.1).

This procedure can also be extended to higher order of non-degeneration. Hence wehave:

1. Non-degenerate case (at zeroth degree), i.e. ∃ξ such that ∂ξφ(0)(ξ) 6≡ 0.

In this case, we can guarantee the existence of a solution for the bifurcationequation (3.8) under suitable conditions on ε (depending on the maximum andminimum of the function φ(0)(ξ)). From Subsection 3.1.1 and Lemma 3.2 thereexists a solution u(t; ξ, ε) of the functional equation (3.6), which by (3.5) impliesthe existence of a (p, q)−periodic solution xpq(t) of (3.1).

2. Degenerate case (at zeroth degree), i.e. ∂ξφ(0)(ξ) ≡ 0. If that happens,

we analyse the next term of φ, i.e. φ(1)(ξ). Again we need to distinguish twodifferent cases:

2.1 Non-degenerate case (at first degree), i.e. ∃ξ such that ∂ξφ(1)(ξ) 6≡ 0.

In this case, we can guarantee the existence of a solution for the bifurcationequation (3.8) under suitable conditions on ε (depending on the maximumand minimum of the function φ(1)(ξ)). From Subsection 3.1.1 and Lemma3.2 there exists a solution u(t; ξ, ε) of the functional equation (3.6), whichby (3.5) implies the existence of a (p, q)−periodic solution xpq(t) of (3.1).


2.2 Degenerate case (of first degree) if ∂ξφ(1)(ξ) ≡ 0. In this case, we

analyse the next term of φ, i.e. φ(2) and so on.

To study the non-degeneration at higher degree we need the following definition.

Definition 3.3. We say that a (p, q)−periodic orbit xpq is

- degenerate at kth-degree, if φ(l)(ξ) : ξ ∈ [0, 2π] → R are constant functions for

every 0 ≤ l ≤ k.

- non-degenerate at kth-degree, if it is degenerate at (k − 1)th-degree and φ(k)(ξ)is non-constant.

The following theorem is the generalisation of Theorem 3.1 to an arbitrary degree.

Theorem 3.2. For |ε| ≤ ε0 let u be as in Lemma 3.2. Assume xpq is non-degenerateat kth-degree. Furthermore, assume that φ is (3.8) can be expanded with respect to εup to any order, i.e.

φu(ξ; ε) =

k∑

l=1

φ(l)(ξ)εl−1 + εkφ(k+1)(ξ; ε)

with

sup|ε|≤ε0, ξ∈[0,2π]

∣∣∣φ(k+1)(ξ; ε)

∣∣∣ ≤Mk+1

for some suitable Mk+1 > 0.Assume

φ(k)− := min

ξ∈[0,2π]φ(k)(ξ) < max

ξ∈[0,2π]φ(k)(ξ) =: φ

(k)+ .

Let

|ε| ≤ εk := min

(

ε0,δ

Mk+1

)

,

where

0 < δ <φ(k)+ − φ

(k)−

2.


ην

ε∈[

εk(

φ(k)− + δ

)

, εk(

φ(k)+ − δ

)]

holds. For the above ε, ξ, η and ν, the pair (u, ξ) solves (3.8)-(3.9), so that xpq(t)defined in (3.5) solves (3.1).

We need to study the functions φ(k)(ξ) for k ≥ 0 in order to find conditions for theexistence of periodic orbits depending only on the general Newtonian potential f in(3.1). In Section 3.2 we develop a method to compute recursively the functions φ(k)(ξ)for k ≥ 0. Then in Section 3.3 we study the relation between q and k.

3.2. Computing φ(k)(ξ) for k ≥ 0 51

3.2 Computing φ(k)(ξ) for k ≥ 0

First method

We expand fx(ξ + pt+ u, qt) with respect to ε and get

fx(ξ + pt+ u, qt)(3.13)= fx(ξ + pt+

∞∑

k=1

ukεk, qt)

= fx(ξ + pt, qt) + εfxx(ξ + pt, qt)u1

+ε2[

fxx(ξ + pt, qt)u2 + fxxx(ξ + pt, qt)u212

]

+

+ε3[

fxx(ξ + pt, qt)u3 + fxxx(ξ + pt, qt)2u1u2

2+ fxxxx(ξ + pt, qt)

u316

]

+ . . .

=:

∞∑

k=0

f (k)εk. (3.15)

Taking the average with respect to t and comparing coefficients of the same order inε with (3.14) leads to

φ(0) =⟨

f (0)⟩

=1

2π

∫ 2π

0fx(ξ + pt, qt)dt, (3.16)

φ(1) =⟨

f (1)⟩

=1

2π

∫ 2π

0fxx(ξ + pt, qt)u1dt, (3.17)

φ(2) =⟨

f (2)⟩

=1

2π

∫ 2π

0

[

fxx(ξ + pt, qt)u2 + fxxx(ξ + pt, qt)u212

]

dt, (3.18)

φ(3) =⟨

f (3)⟩

=1

2π

∫ 2π

0

[

fxx(ξ + pt, qt)u3 + fxxx(ξ + pt, qt)2u1u2

2+ fxxxx(ξ + pt, qt)

u316

]

dt,

and so on. The problem is that there is no simple formula for φ(k) with k ∈ N.

Second method

A similar method, as the one developed in this subsection, was already used in [14].

This approach allows us to compute uk and φ(k)u recursively for k ∈ N.

Definition 3.4. Define βm,nk (t; ξ) for m,n ∈ Z and k ∈ N to be the coefficients of the

formal expansion of the function ei[(mp+nq)t+mu(t;ε)] with respect to ε, i.e.

ei[(mp+nq)t+mu(t;ξ,ε)] =:

∞∑

k=0

βm,nk (t; ξ)εk, (3.19)

withβm,n0 (t; ξ) := ei(mp+nq)t. (3.20)

To simplify the notation we will often write βm,nk instead of βm,n

k (t; ξ).

Lemma 3.3. For k ≥ 0 let f (k) be as in (3.15) and βm,nk be as in (3.19)-(3.20). Then

the relation between f (k) and βm,nk is

f (k) = i∑

(m,n)6=(0,0)

mfm,neimξβm,n

k , ∀k ≥ 0. (3.21)


Moreover, for uk in (3.13) we get

uk = −Gη(f(k−1) − φ(k−1)), ∀k ≥ 1, (3.22)

where φ(k−1) are given by

φ(k−1)(ξ) =⟨

f (k−1)⟩

=

⟨

i∑

(m,n)6=(0,0)

mfm,neimξβm,n

k−1

⟩

(3.23)

for k ≥ 1.

Proof. Using (3.13) every function ∂lf∂xl (ξ + pt+ u, t) for l ∈ N can be expressed in a

ε power series. In particular, for l = 1 we have from (3.2) and (3.19):

fx(ξ + pt+ u(t; ξ, ε), qt)(3.2)= i

∑

(m,n)6=(0,0)

mfm,neimξei[(mp+nq)t+mu(t;ξ,ε)]

(3.19)= i

∑

(m,n)6=(0,0)

mfm,neimξ

[ ∞∑

k=0

βm,nk εk

]

= i

∞∑

k=0

∑

(m,n)6=(0,0)

mfm,neimξβm,n

k εk.

Equation (3.21) follows directly by comparing the coefficients in the ε−expansion.Putting (3.14) and (3.15) into the range equation (3.9) and using (3.13) and thelinearity of Gη we get

∞∑

k=1

ukεk = −εGη

( ∞∑

k=0

[

f (k) − φ(k)]

εk

)

= −∞∑

k=0

Gη(f(k) − φ(k))εk+1.

Again by comparing coefficients we obtain (3.22). To find φ(k−1)(ξ) in (3.23) we haveto compute the mean over [0, 2π] of the (k − 1)th coefficient of the ε expansion offx(ξ + pt +

∑∞k=1 ukε

k), i.e. by (3.15) we have φ(k−1)(ξ) =⟨f (k−1)

⟩. From formula

(3.21) we have

φ(k−1)(ξ) =

⟨

i∑

(m,n)6=(0,0)

mfm,neimξβm,n

k−1

⟩

for all k ≥ 1. This proves equation (3.23).

Remark 3.4. Furthermore, notice that the derivatives of the Newtonian potential in(3.2) along the unperturbed periodic orbit t→ (ξ + pt, qt) are given by

∂lf

∂xl(ξ + pt, qt) =

∑

(m,n)6=(0,0)

(im)lfm,neinqteim(ξ+pt)

=∑

(m,n)6=(0,0)

(im)lfm,neimξei(mp+nq)t. (3.24)

3.2. Computing φ(k)(ξ) for k ≥ 0 53

Lemma 3.4. The coefficients uk in (3.22) and the coefficients βm,nk in equations

(3.19)-(3.20) satisfy the following recursive relations:

uk = −Gη

i∑

(m,n)6=(0,0)

mfm,neimξβm,n

k−1 −⟨

i∑

(m,n)6=(0,0)

mfm,neimξβm,n

k−1

⟩

, (3.25)

βm,nk =

im

k

k∑

l=1

lulβm,nk−l , for k ≥ 1. (3.26)

Notice that: uk only depends on βm,n0 , . . . , βm,n

k−1; βm,nk only depends on βm,n

0 , . . . , βm,nk−1

and on u1, . . . , uk.

Proof. Taking the derivative of equation (3.19) with respect to ε and using (3.13) weobtain

(im)ei[(mp+nq)t+mu(t;ξ,ε)]

[ ∞∑

k=1

kukεk−1

]

=

∞∑

k=0

kβm,nk εk−1,

⇔ (im)ε−1

[ ∞∑

k=0

βm,nk εk

]

·[ ∞∑

k=0

kukεk

]

=∞∑

k=1

kβm,nk εk−1.

Using the Cauchy-formula for the product of two series we get the following equivalentstatements

⇔ (im)ε−1∑∞

k=0

[∑k

l=0 lulβm,nk−l

]

εk =∞∑

k=1

kβm,nk εk−1,

⇔ (im)∑∞

k=1

[∑k

l=1 lulβm,nk−l

]

εk−1 =

∞∑

k=1

kβm,nk εk−1.

Comparing coefficients in the ε expansions, we find the recursion formula (3.26) forthe functions βm,n

k . Finally equation (3.25) follows from (3.21), (3.22) and (3.23).This terminates the proof of Lemma 3.4.

Lemma 3.5. The function βm,nk (t; ξ) can also be written in the following form:

βm,nk (t; ξ) =

∑

mi,ni1≤i≤k+1

Ξ(ξ;m1, . . . ,mk+1, n1, . . . , nk+1)e[(m1+...+mk+1)p+(n1+...+nk+1)q]t,

for some function Ξ.

Proof. (Induction on k)k = 0: the hypothesis follows directly by the definition of βm,n

0 (t; ξ) in (3.20).k → k + 1: Follows from (3.25), (3.26) and from the inductive step, since for Gη as in(3.7) for all l ≥ 0 we have

Gη

(

e[(m1+...+ml)p+(n1+...+nl)q]t)

= K(m1, . . . ,ml, n1, . . . , nl)e[(m1+...+ml)p+(n1+...+nl)q]t,

where K(m1, . . . ,ml, n1, . . . , nl) ∈ C.


Remark 3.5. (i) Starting from (3.20) and using repeatedly (3.25) and (3.26) we cancompute u1, β

m,n1 , u2, β

m,n2 , . . .. The solution u(t; ξ, ε) of the range equation (3.9) and

the function φu(ξ) in the bifurcation equation (3.8) can be computed recursively witharbitrary accuracy. Their numerical computation for the dissipative spin-orbit problem(using Matlab) is presented in the next chapter and is based on this idea.(ii) In order to compare the first and the second method, we explicitly compute φ(0), φ(1)

and φ(2) using equation (3.23).

φ(0)(ξ)(3.23)=

⟨∑

(m,n)6=(0,0)

imfm,neimξβm,n

0

⟩

(3.20)=

⟨∑

(m,n)6=(0,0)

imfm,neimξei(mp+nq)t

⟩

(3.24)= 〈fx(ξ + pt, qt)〉 .

(3.27)

Using equation (3.26) we have βm,n1 = imu1β

m,n0 and this implies

φ(1)(ξ)(3.23)=

⟨∑

(m,n)6=(0,0)

imfm,neimξβm,n

1

⟩

(3.20)=

⟨

u1∑

(m,n)6=(0,0)

(im)2fm,neimξβm,n

0

⟩

(3.24)= 〈fxx(ξ + pt, qt)u1〉 . (3.28)

Using two times equation (3.26) we get

βm,n2 =

im

2(u1β

m,n1 + 2u2β

m,n0 ) =

im

2

(imu21β

m,n0 + 2u2β

m,n0

)(3.29)

and this implies

φ(2)(ξ)(3.23)=

⟨∑

(m,n)6=(0,0)

imfm,neimξβm,n

2

⟩

(3.29)=

⟨∑

(m,n)6=(0,0)

(u212(im)3fm,ne

imξβm,n0 + u2(im)2fm,ne

imξβm,n0

)⟩

(3.24)=

⟨

fxxx(ξ + pt, qt)u212

+ fxx(ξ + pt, qt)u2

⟩

. (3.30)

Formulas (3.27), (3.28) and (3.30) (computed with the second method) are equal toformulas (3.16), (3.17) and (3.18) (computed with the first method), respectively. How-ever, the second method is easier to handle because of the simple recursion formula.

3.3 Condition for non-degeneration of φu(ξ) at any degree

3.3.1 Condition for non-degeneration of φu(ξ) at zeroth degree

From (3.27) and Definition 3.3 the condition for the non-degeneration at zeroth degreeof φu(ξ) is

φ(0)(ξ) = φ(0)(ξ; p, q) :=1

2π

∫ 2π

0fx(ξ + pt, qt)dt 6= const,

for ξ ∈ [0, 2π]. By (3.24) we have

φ(0)(ξ; p, q) =∑

(m,n)6=(0,0)

imfm,neimξ · 1

2π

∫ 2π

0ei(mp+nq)tdt.

Notice that:

3.3. Condition for non-degeneration of φu(ξ) at any degree 55

(i) 12π

∫ 2π0 ei(mp+nq)tdt =

1, if mp+ nq = 0,0, otherwise.

(ii) Since p and q are coprime mp + nq = 0 is equivalent to m = lq and n = −lpwith l ∈ Z.

So, for φ(0)(ξ) it follows:

φ(0)(ξ; p, q) =∑

l 6=0

ilqflq,−lp eilqξ

=1

2

∑

l 6=0

ilqflq,−lp eilqξ +

∑

l 6=0

ilqflq,−lp eilqξ

=1

2

∑

l 6=0

ilqflq,−lp eilqξ +

∑

l 6=0

(−ilq)f−lq,lp e−ilqξ

=1

2

∑

l 6=0

(ilq)[

flq,−lp eilqξ − f−lq,lp e

−ilqξ]

.

Since by definition of f (see equation (3.2)) flq,−lp = f−lq,lp holds, we have

φ(0)(ξ; p, q) =1

2

∑

l 6=0

(ilq)2i Im(

flq,−lp eilqξ)

= −∑

l 6=0

lq Im(

flq,−lp eilqξ)

= −∑

l>0

[

lq Im(

flq,−lp eilqξ)

− lq Im(

f−lq,lp e−ilqξ

)]

= −2q∑

l>0

l Im(

flq,−lp eilqξ)

. (3.31)

Remark 3.6. If φ(0)(ξ) = const for ξ ∈ [0, 2π] then φ(0)(ξ) ≡ 0.

Remark 3.7. By Remark 3.1 (see in particular (3.3)) equation (3.31) for the specialcase of the dissipative spin-orbit problem is equivalent to

φ(0)(ξ) =

−2α2p sin(2ξ), for q = 1,−2αp sin(2ξ), for q = 2,0, otherwise.

(3.32)

which was already found in [8] (see formula (56)). This is true, since for m = ±2

l ∈ Z | l > 0 and l =m

q

=

2 , for q = 1,1 , for q = 2,∅, otherwise.

holds.

Lemma 3.6. Let p, q be positive coprime integers and assume that there exists l ∈Z \ 0 such that

flq,−lp 6= 0,

where the coefficients fm,n are given in (3.2). Then the general spin-orbit problem in(3.1) is non-degenerate at degree zero, i.e. φ(0)(ξ) 6= const according to Definition 3.3.


Proof. For ξ ∈ [0, 2π] we define flq,−lp =: Al + iBl, with Al, Bl ∈ R. Then we have

flq,−lp eilqξ = Al cos(lqξ)−Bl sin(lqξ) + i (Bl cos(lqξ) +Al sin(lqξ)) .

The degenerate case φ(0) ≡ 0 happens if and only if the imaginary part of the righthand side in the last equation disappears, i.e.

Bl cos(lqξ) +Al sin(lqξ) = 0, for all ξ ∈ [0, 2π] .

Choosing ξ = 0 we get Bl = 0. This means flq,−lp ∈ R for all l 6= 0. So it follows:

φ(0)(ξ; p, q)(3.31)= −2q

∑

l>0

l Im(

flq,−lpeilqξ)

= −2q∑

l>0

lflq,−lp sin(lqξ).

The functions sin(lqξ), sin(lqξ) are linearly independent for l 6= l and l, l > 0. Thisfact implies that

φ(0)(ξ; p, q) = const ⇔ flq,−lp = 0, ∀l 6= 0.

This terminates the proof of Lemma 3.6.

3.3.2 Condition for the non-degeneration of φu(ξ) at first degree

We assume that φ(0)(ξ) ≡ const, which by Remark 3.6 implies φ(0)(ξ) ≡ 0. From (3.28)and Definition 3.3 the condition for the non-degeneration of φu(ξ) at first degree is

φ(1)(ξ) =1

2π

∫ 2π

0fxx(ξ + pt, qt)u1(t)dt 6= const,

for ξ ∈ [0, 2π]. Using (3.25) with k = 1 it follows

u1 = −Gη

(

i∑

(m,n) 6=(0,0)

mfm,neimξβm,n

0 −⟨

i∑

(m,n) 6=(0,0)

mfm,neimξβm,n

0

⟩)

(3.20)= −Gη

(∑

(m,n) 6=(0,0)

(im)fm,neimξei(mp+nq)t −

⟨∑

(m,n) 6=(0,0)

(im)fm,neimξei(mp+nq)t

⟩)

(3.24)= −Gη (fx(ξ + pt, qt)− 〈fx(ξ + pt, qt)〉) . (3.33)

By (3.7), (3.28), (3.33) we get

φ(1)(ξ) = −⟨

fxx(ξ + pt, qt)Gη

fx(ξ + pt, qt)− 1

2π

∫ 2π

0fx(ξ + pt, qt)dt

︸︷︷︸

φ(0)(ξ;p,q)≡0

⟩

.

From the Fourier-expansion of the Newtonian potential f(x, t) in (3.2) and from equa-tions (3.7) and (3.24) we have

φ(1)(ξ) = −〈fxx(ξ + pt, qt)Gη (fx(ξ + pt, qt))〉

=

⟨

∑

(m1,n1)6=(0,0)

(im1)2fm1,n1e

im1ξei(m1p+n1q)t

3.3. Condition for non-degeneration of φu(ξ) at any degree 57

·

∑

(m2,n2)6=(0,0)

(im2)fm2,n2eim2ξeiη(m2p+n2q)t

(m2p+ n2q)2 − iη(m2p+ n2q)

⟩

.

We notice that m1p+ n1q+m2p+ n2q = 0 if and only if n1 =−(m1+m2)p

q − n2. Since

for n ∈ Z we have 12π

∫ 2π0 eintdt =

1, if n = 0,0, otherwise.

, it follows:

φ(1)(ξ) = −im21m2

∑

(m1,m2)6=(0,0)

n1 /∈

0,− (m1+m2)pq

fm1,n1eim1ξf

m2,−(m1+m2)p

q−n1

eim2ξ

(−m1p− n1q)2 − iη(−m1p− n1q)

= −im21m2

∑

(m1,m2)6=(0,0)

n1 /∈

0,− (m1+m2)pq

fm1,n1fm2,−(m1+m2)p

q−n1

ei(m1+m2)ξ

(m1p+ n1q)2 + iη(m1p+ n1q). (3.34)

Furthermore notice that, if m1 +m2 = 0 then φ(1)(ξ) is constant. So we have provedthat:

Lemma 3.7. The existence of exponents mip+ niq for i = 1, 2 in the t−expansion off(x, t) with fm1,n1 , fm2,n2 6= 0 for some n1, n2 ∈ Z\0 such that q divides m1+m2 6= 0is a necessary condition for the non-degeneration of φu(ξ) at first degree, i.e.

φ(1)(ξ) 6= const ⇒ ∃m1,m2 with m1 +m2 6= 0 and

fm1,n1 , fm2,n2 6= 0 for some n1, n2 ∈ Z \ 0such that q|(m1 +m2).

Remark 3.8. As we know from Remark 3.1 in the dissipative spin-orbit problem m =±2. Assuming φ(0)(ξ) ≡ 0 (i.e. q 6= 1, 2 cf. (3.32)) and noticing that ±2 ± 2 =−4, 0, 4 we have

q| ± 4 ⇒ q = 4.

This implies that the only non-degenerate case at first degree for the dissipative spin-orbit problem is q = 4. From the proof of Lemma 3.7 and in particular from (3.34)(with m1 = m2 = 2 or m1 = m2 = −2) the function φ(1)(ξ) is given by

φ(1)(ξ) = const +∑

n 6=0,−p

−i8f2,nf2,−p−nei4ξ

(2p+ nq)2 + iη(2p + 4n)−∑

n 6=0,p

−i8f−2,nf−2,p−ne−i4ξ

(−2p+ 4n)2 + iη(−2p + 4n)

= const +8

i

∑

n 6=0,p

f2,−nf2,−p+nei4ξ

(2p − 4n)2 + iη(2p − 4n)−∑

n 6=0,p

f−2,nf−2,p−ne−i4ξ

(2p − 4n)2 − iη(2p − 4n)

.

Moreover, by (3.2) and (3.3) we have

f2,−n = f−2,n = f−2,n and f2,−p+n = f−2,p−n = f−2,p−n,

since for the dissipative spin orbit problem the coefficients f±2,n of the Newtonianpotential are real. So it follows

φ(1)(ξ)− const = 16 Im

∑

n 6=0,p

f2,−nf2,−p+nei4ξ

(2p − 4n)2 + iη(2p − 4n)


= 16 Im

∑

n 6=0,p

((2p − 4n)2 − iη(2p − 4n))f2,−nf2,−p+nei4ξ

(2p − 4n)4 + η2(2p − 4n)2

= 16∑

n 6=0,p

[

sin(4ξ)(2p − 4n)2f2,−nf2,−p+n

(2p − 4n)4 + η2(2p − 4n)2− cos(4ξ)s(n)f2,−p+n

]

= 16 sin(4ξ)∑

n 6=0,p

f2,−nf2,−p+n

(2p − 4n)2 + η2,

where we used that the function s(n) := 1(2p−4n)3+η2(2p−4n)

has the property s(p−n) =−s(n). By (3.3) we know that f2,−n = αn

2 and f2,−p+n =αp−n

2 hold, so we have

φ(1)(ξ) = const− 4 sin(4ξ)∑

j∈Z,j 6=0,p

αp−jαj

4(p− 2j)2 + η2(3.35)

with const ∈ R, which was already found in [8] (see formula (58)).

3.3.3 Condition for the non-degeneration of φu(ξ) at kth degree

In the following theorem a necessary condition for the non-degeneration of the generalspin-orbit problem in (3.2) at any degree is given. This allows us to predict, at whichlevel of degeneration some (p, q)−orbit may exists.

Theorem 3.3. Let p, q be positive coprime integers. The existence of exponents mip+niq for i = 1, 2, . . . , k+1 in the t−expansion of f(x, t) with fm1,n1 , fm2,n2 , . . . , fmk+1,nk+1

6=0 for some n1, n2, . . . , nk+1 ∈ Z\0 such that q divides m1+m2+ . . .+mk+1 6= 0 is anecessary condition for the non-degeneration at kth−degree of the general spin-orbitproblem, i.e.

φ(k)(ξ) 6= const ⇒ ∃m1,m2, . . . ,mk+1, n1, n2, . . . , nk+1 ∈ Z \ 0with m1 +m2 + . . . +mk 6= 0,

fm1,n1 , fm2,n2 , . . . , fmk+1,nk+16= 0

and q|(m1 +m2 + . . . +mk+1).

Proof. As we know from (3.23) the term φ(k)(ξ) has the form

φ(k)(ξ) =

⟨

i∑

(m,n)6=(0,0)

mfm,neimξβm,n

k

⟩

.

The hypothesis follows from (3.25), (3.26), Lemma 3.5 and the fact that for all Z ∋k 6= 0 we have

∫ 2π0 ektdt = 0.

3.4 Application to the classical spin-orbit problem

The condition of Theorem 3.3 allows us to formulate a conjecture for the degenerationof (p, q)−periodic orbits for the dissipative spin-orbit problem. Since in this specialcase m takes only the values ±2, we have that q divides

±2± 2± . . . ± 2︸︷︷︸

k+1 times

in order to get, that φu(ξ) is non-degenerate at kth degree.

3.4. Application to the classical spin-orbit problem 59

Theorem 3.4. If the spin-orbit problem is non-degenerate at kth−degree, then q|2(k+1) and if k 6= 0 also q ∤ 2l for 0 < l ≤ k hold.

Proof. The proof follows directly from Theorem 3.3.

Remark 3.9. (i) For example if q = 7 the system is non-degenerate at 6th−degree,since 7 | 14 = 2 · (6 + 1) and 7 ∤ 2, 4, 6, 8, 10, 12.

(ii) For q even the system is non-degenerate at( q2 − 1

)

th−degree. For q odd the

system is non-degenerate at (q − 1)th−degree.

(iii) Theorem 3.4 gives a necessary condition for the non-degeneration at kth−degree,for k ≥ 0. The question arises: Is this condition also sufficient?In Chapter 4 we answer this question for the cases q = 1, 2 and 4. Hence inthese special cases we give the proof of “⇐” of Theorem 3.4.

Conjecture 3.5Let k > 0. If q|2(k+1) and q ∤ 2l for 0 < l ≤ k then the dissipative spin-orbitproblem is non-degenerate at kth−degree.

In Chapter 5 we will do a numerical study for 4 6= q ≥ 3, in order to givenumerical evidence for Conjecture 3.5.

4 Quantitative conditions for the existenceof low-order spin–orbit resonances

This chapter will appear in Note di Matematica, (2014), (see [1]).

We consider (p, q)−periodic orbits of a dissipative spin-orbit model. L. Biasco and L.Chierchia in [8] give a sufficient condition for the existence of such orbits for q = 1, 2and 4, imposing that the functions given in (3.32) and in (3.35) are not constant. Inthis chapter we prove that this condition is satisfied provided that the eccentricityand, in the case q = 4, also the dissipation, is smaller than an explicitly given bound.

4.1 Results

Recall that we define αj = αj(e) in (2.1) to be the Fourier coefficients of the Newtonianpotential f given in equation (1.29). Since by Lemma 2.1 we know that α0(e) = 0 fore ∈ [0, 1) this implies

f(x, t; e) =∑

j 6=0,j∈Zαj(e) cos(2x− jt).

From Theorem 1.2 and Proposition 2.10 of [8] L. Biasco and L. Chierchia we canformulate the following theorem:

Theorem 4.1. Let p and q be positive coprime integers1, with q = 1, 2 or 4. Letαj = αj(e) be as in (2.1). There exist ε0 > 0 and η0 > 0 such that if 0 < ε ≤ ε0,0 < η ≤ η0 and

∣∣∣∣ν − p

q

∣∣∣∣<

εη |α2p|, if q = 1,εη |αp|, if q = 2,

32 ε2

η

∣∣∣∑

j∈Z,j 6=0,pαp−jαj

4(p−2j)2+q2η2

∣∣∣ , if q = 4,

hold, then the dissipative spin-orbit problem modelled by equations (1.24)-(1.32) admitsa (p, q)-periodic solution xpq.

Remark 4.1. Theorem 4.1 is applicable if we can prove that it exists ∃e⋆ > 0 suchthat

0 6=

α2p(e), for q = 1,αp(e), for q = 2,∑

j∈Z,j 6=0,pαp−j(e)αj (e)4(p−2j)2+q2η2 , for q = 4,

(4.1)

for all 0 < e ≤ e⋆.

1Equivalently (p, q) = 1.

61

62 Chapter 4. Quantitative conditions for the existence of of low-order spin-orbit resonances

For 0 < b < 1 let

r(b) :=b

cosh band M(b) :=

2

(1− b)5

(

(1 + r(b))(1 + cosh b) + 1− b)2. (4.2)

Moreover define

cp,1 :=|2p||2p−2|

2|2p−1|+1|2p − 2|! , (4.3)

cp,2 :=|p||p−2|

2|p−1|+1|p− 2|! , (4.4)

cp,4 :=

(2−p)4−p

211−p(4−p)2(4−p)! , for p ≤ 1,

− 116 , for p = 3,

(p−2)p−2

2p−1(p−4)2(p−2)!, for p ≥ 5.

(4.5)

Note that

cp,1 ≈1√p

(e

2

)|2p|→ ∞, cp,2 ≈

1√p

(e

2

)|p|→ ∞ and cp,4 ≈

1

p2√p

(e

2

)|p|→ ∞(4.6)

as p→ ±∞.The main result of this chapter is stated in the following theorem.

Theorem 4.2. Besides the assumptions of Theorem 4.1 on p, q, ε, η, ν, for 0 < b < 1let r(b), M(b), cp,q be as in equations (4.2)-(4.5).

- If q = 1, 2, assume

0 < e ≤ cp,q1

M(b)

(r(b)

2

)| 2pq−2|+1

. (4.7)

- If q = 4, assume

0 < e ≤ r(b)π2M2(b)

6|cp,q|r(b)|p−4| + |p− 4|, (4.8)

η <3√

5|cp,q|2πM(b)

e|p−4|/2. (4.9)

Then there exists a (p, q)−periodic orbit of the dissipative spin-orbit problem modelledby equations (1.24)-(1.32).

Remark 4.2. (i) Condition (4.1) depends only on e when q = 1, 2 while it dependsalso on η when q = 4. This is why, for q = 4, we have to assume also the condition(4.9).(ii) Conditions on the existence of (p, q)−periodic orbits are obtained combining The-orem 4.1 and Theorem 4.2. For the case p/q = 1/1 equation (4.7) is equivalent to0 < e ≤ 0.000412055, choosing b = 0.148642. This condition on the eccentricity is sat-isfied by Tethys (a satellite of Saturn), which has an eccentricity approximately equal to0.0001 and is known (from astronomical observation) to be locked in a (1, 1)−periodicspin-orbit.

4.2. Proof of Theorem 4.2 63

For the case p/q = 3/2 equation (4.7) is equivalent to 0 < e ≤ 0.0000609886, choosingb = 0.253122. Mercury (seen as satellite of the Sun), which represents the only ob-served example of the (3, 2)-periodic spin-orbit in our Solar system, doesn’t fulfil thiscondition, since its eccentricity is approximately equal to 0.2056.In Chapter 6 one focus on the cases p/q = 1/1, 3/2 improving the estimate for theexistence of (p, q)−periodic orbits in order to cover the values of the eccentricity of allsatellites in our Solar system, which are observed to be in a (1, 1)− or a (3, 2)−periodicorbit.(iii) We also note that the proof of the theorem above is quite simple in the casesq = 1, 2: one has to prove that some analytic function of e (namely α2p and αp, re-spectively) are not identically zero. Since the behavior of such functions is well known(see Lemma 2.1), this goal is easily obtained. On the other hand the case q = 4 is moredifficult (already in the case η = 0), since one has to prove that a series of functionswith changing signs is not identically zero and compensations could occur.

To prove Theorem 4.2 we have to find a lower bound for the absolute value of theright hand side of (4.1). Since this quantity is a function of e, we evaluate the leadingterm of the e expansion and we prove that, if e satisfies the conditions (4.7)-(4.9), thedifference between the leading term and the rest do not change sign.

4.2 Proof of Theorem 4.2

In order to prepare and to understand the proof of Theorem 4.2 we need some inter-mediate results. The following lemma gives an upper bound on the Fourier coefficientsαj .

Lemma 4.1. Let 0 < b < 1 and let r(b) and M(b) be as in (4.2). The solution ue(t)of the Kepler equation (1.32) is a holomorphic function in the ball

|e| < r(b) =b

cosh b

satisfying

supt∈R

|ue(t)− t| ≤ b . (4.10)

The functions ρe(t) in (1.30) and Ge(t) in (2.2) satisfy

|ρe(t)| ≥ 1− b , ∀ t ∈ R , |e| < r(b), (4.11)

and

|Ge(t)| ≤2

(1− b)5

(

|1− e|(1 + cosh b) + 1− b)2, ∀ t ∈ R , |e| < r(b) , (4.12)

respectively.

Proof. Using that

sup| Im z|≤b

| sin z| = sup| Im z|≤b

| cos z| = cosh b , (4.13)

it is simple to verify that for |e| < r(b) the map

v 7→ χ(v; e)


with (

χ(v; e))

(t) := e sin(v(t) + t)

is a contraction in the closed ball of radius b in

C(R,C)b :=

v(t) ∈ C(R,C) : ‖v(t)‖∞ = supt∈R

|v(t)| ≤ b

,

the space of continuous functions endowed with the sup-norm. In fact, by |e| < r(b) =b

cosh b and (4.13) the function χ maps C(R,C)b into itself, indeed

‖χ(v; e)‖∞ ≤ r(b) cosh(b) = b.

Furthermore for u, v ∈ C(R,C)b we have

‖χ(v; e)− χ(u; e)‖∞ = |e| supt∈R

| sin(v(t) + t)− sin(u(t) + t)|

= |e| supt∈R

∣∣∣∣

∫ 1

0(v(t) − u(t)) cos(u(t) + t+ s(v(t)− u(t))) ds

∣∣∣∣

(4.13)

≤ |e| cosh(b)‖v − u‖∞,

since

| Im (u(t) + t+ s(v(t)− u(t)))| = (1− s)| Im (u(t)) |+ s| Im (v(t)) | ≤ b

holds for s ∈ [0, 1]. By the Banach fixed point theorem there exists a unique ve(t) ∈C(R,C)b such that

ve(t) = e sin(ve(t) + t).

Notice that ue(t) := ve(t) + t satisfies the Kepler equation (1.32), indeed we have

ve(t) + t− e sin(ve(t) + t) = ve(t) + t− ve(t) = t.

Moreover, since χ(v; e) is holomorphic in e, the same holds for the fixed point ve(t) ofχ. Equation (4.10) follows by the fact that ve(t) ∈ C(R,C)b. Since by (4.10) we get

∣∣ Im

(ue(t)

)∣∣ ≤ b , ∀ t ∈ R , |e| < r , (4.14)

estimate (4.11) follows by

|ρe(t)| ≥ 1− |e|| cos(ue(t))|(4.13)

≥ 1− r cosh b = 1− b .

As in Lemma 2.2 define we = w(u; e) :=√

1+e

1−etan u

2 and we(t) := w(ue(t), e). Then

by (2.15) we get

|e2ife(t)| =|we(t)− i|4|we(t)2 + 1|2 =

( |we(t)− i|2|we(t)2 + 1|

)2

(⋆)

≤(

4

|we(t)2 + 1| + 2

)2(⋆⋆)= 4

( |1− e||1 + cos ue(t)||1− e cos ue(t)|

+ 1

)2

. (4.15)

To prove the inequality (⋆) we notice that for x = |w − i| we have

4 + 2|w − i||w + i| − |w − i|2 ≥ 4 + 2x(x− 2)− x2 = (x− 2)2 ≥ 0,


since |w + i| = |w − i + 2i| ≥ |w − i| − 2. To prove (⋆⋆) we used that tan2(β/2) =(1− cos β)/(1 + cos β) holds for every angle β. Then (4.12) follows by (4.11), (4.15),(4.13) and (4.14), i.e.

|Ge(t)| =|e2ife(t)|2|ρe(t)|3

≤ 4

( |1− e||1 + cos ue(t)||1− e cos ue(t)|

+ 1

)2 1

2(1− b)3

≤ 2

(1− b)5

(

|1− e|(1 + cosh b) + 1− b)2.


Lemma 4.2. Let 0 < b < 1 and let r(b) and M(b) be as in (4.2). Then

|αj(e)| ≤2

(1− b)5

(

|1−e|(1+cosh b)+1−b)2

≤M(b) , ∀j ∈ Z ∀ |e| < r(b), (4.16)

holds.

Proof. From the proof of Lemma 2.1 we know that αj = 12π

∫ 2π0 Ge(t)e

−ijtdt holdsfor all j ∈ Z, with Ge(t) defined in (2.2). This proves equation (4.16).

In the following lemma we determine the sign of the leading term in the e expansionof αj(e).

Lemma 4.3. Let αj as in (2.17). Then αj < 0 for all j ∈ Z \ 0, 1, α0 = 0 andα1 =

14 > 0.

Proof. α1 = 14 > 0 follows directly from (2.18) and (2.19). Moreover, Lemma 2.1

implies α0 = 0. If j ≥ 2 we see from (2.20) that all coefficients (qi)0≤i≤k of Qk(x) =q0 + q1x + . . . + qkx

k are strictly negative for every k ≥ 0. By (2.18) we have αj =Qj−2(j) < 0. Otherwise if j ≤ −1 we have from (2.18), (2.19) that

αj =

(−1

2

)3−j j2−j

(2− j)!=

(−1)3−j(−1)2−j

23−j

|j|2−j

(2− j)!=

−1

23−j

|j|2−j

(2− j)!< 0.

In the next lemma we estimate the absolute value of the difference between αj(e) andits leading term in the e expansion.

Lemma 4.4. Let r(b), M(b) be as in (4.2) and αj(e) be as in (2.1) and let

rj(b) :=|j||j−2|

2|j−1|+1|j − 2|!1

M(b)

(r(b)

2

)|j−2|+1

, for all j ∈ Z. (4.17)

Then

|αj(e)− αje|j−2|| ≤ 1

2|αj ||e||j−2| (4.18)

holds for all |e| ≤ rj(b). Moreover, it follows

sgn(αj(e)) = sgn(αj)

for 0 ≤ e ≤ rj(b).


Remark 4.3. Note that rj(b) ≈(e r(b)4

)j→ 0 as j → ∞.

Proof. For j = 0 this lemma is trivial, since by Lemma 2.1 we know that α0(e) =α0 = 0 holds. Fix b and r(b) as in (4.2) and set ρj(b) := r(b) − rj(b). Then by theCauchy estimate and (4.16) we obtain

∣∣∣∣

dn

denαj(e)

∣∣∣∣

≤ n!M(b)

ρj(b)n, ∀ |e| ≤ rj(b), j ∈ Z, n ∈ N . (4.19)

By (2.17) and the integral form for the remainder in the Taylor series we have

αj(e) = αje|j−2| +R|j−2|(e), (4.20)

R|j−2|(e) :=e|j−2|+1

|j − 2|!

∫ 1

0

d|j−2|+1αj(es)

de|j−2|+1(1− s)|j−2| ds. (4.21)

Using (4.19) and (4.21) with n = |j − 2|+ 1 we have

|R|j−2|(e)| ≤|e||j−2|+1

|j − 2|!M(b)(|j − 2|+ 1)!

ρj(b)|j−2|+1

∫ 1

0(1− s)|j−2| ds =

M(b)|e||j−2|+1

ρj(b)|j−2|+1. (4.22)

Notice that by (2.19) and (2.20) we have:

|P2−j(j)| =|j||j−2|

2|j−1||j − 2|! , for j ≤ 2; (4.23)

|Qj−2(j)| =1

2j−1

j−2∑

l=0

jl

l!

(j − l + 1

3

)

≥ |j||j−2|

2|j−1||j − 2|! , for j > 2. (4.24)

From (4.23) and (4.24) we get

|αj | ≥|j||j−2|

2|j−1||j − 2|! , for all j ∈ Z. (4.25)

By (4.20), (4.22) and (4.25) we have that (4.18) follows if we prove that

|Rj−2(e)| ≤1

2|αj ||e||j−2|

or equivalentlyM(b)rj(b)

ρj(b)|j−2|+1≤ |j||j−2|

2|j−1|+1|j − 2|! . (4.26)

Since ρj(b) ≥ r(b)/2 (with r(b) defined in (4.2)), (4.26) follows by the definition ofrj(b) in (4.17).

Remark 4.4. The estimate given in (4.18) is not optimal2 for rj(b).

Lemma 4.5. Let p ∈ Z. It follows

|p− 4| ≤ |p− j − 2|+ |j − 2|, ∀j ∈ Z

and equality holds if one of the following two cases occurs:

• Case 1: p ≥ 4 and 2 ≤ j ≤ p− 2;

2Better estimates in the cases of astronomical interest, as for example Earth-Moon or Mercury-Sunsystem, can be found in Chapter 6.


• Case 2: p ≤ 4 and p− 2 ≤ j ≤ 2.

Proof. Let us define

v(p, j) := |p− j − 2|+ |j − 2| = |j − (p− 2)| + |j − 2|.

In the first case, when p ≥ 4, the function v(p, j) takes is minimum for 2 ≤ j ≤ p− 2and for these values of j we have |p− 4| = |j − (p − 2)| + |j − 2| (see Figure 4.1).

j

y = |j − (p− 2)|y = |j − 2|

2 p− 2

Figure 4.1: Sketch of the function v(p, j) in the case p ≥ 4 (the case p ≤ 4 is analogous).

Analogously, when p ≤ 4, the function v(p, j) takes is minimum for p − 2 ≤ j ≤ 2and, as in the first case, for these values of j we have |p−4| = |j−(p−2)|+ |j−2|.

Lemma 4.6. Let p ∈ Z be odd. Then we have

∑2j=p−2

αp−j αj

4(p−2j)2 >(2−p)4−p

211−p(4−p)2(4−p)! > 0, for p ≤ 1,

∑2j=p−2

αp−j αj

4(p−2j)2= −1

16 < 0, for p = 3,

∑p−2j=2

αp−jαj

4(p−2j)2 >(p−2)p−2

2p−1(p−4)2(p−2)! > 0, for p ≥ 5.

(4.27)

Proof. If p ≤ 1 using Lemma 4.3 we have

2∑

j=p−2

αp−jαj

4(p − 2j)2= 2

2∑

j= p+12

αp−jαj

4(p − 2j)2>

1

2

(αp−1α1

(p− 2)2+αp−2α2

(p− 4)2

)

>1

32

αp−2α2

(p− 4)2.

The last estimate is obtained as follows. Since α2 = −12 , α1 =

14 it is equivalent to

(1

2− 1

32

)αp−2α2

(p− 4)2> −1

2

αp−1α1

(p − 2)2

−15

64αp−2(p− 2)2 > −1

8αp−1(p − 4)2

15 αp−2(2− p)2 < 8 αp−1(4− p)2. (4.28)

Using (2.18) and (2.19) we know that

αp =

(−1

2

)3−p

· p2−p

(2− p)!, for all p ≤ 1 (4.29)

holds. Since p ≤ 1 and p is odd, from (4.29) the equation (4.28) is equivalent to

15(p − 2)4−p

25−p(4− p)!(2− p)2 < − 8(p− 1)3−p

24−p(3− p)!(4− p)2


−15(2 − p)4−p(2− p)2 < −16(1 − p)3−p(4− p)3

15(2 − p)6−p > 16(1 − p)3−p(4− p)3.

With Mathematica it is easy to verify that the last inequality is true for all p ≤ 1 , podd. So we proved that

2∑

j=p−2

αp−jαj

4(p − 2j)2>

1

32

αp−2α2

(p − 4)2(4.29)=

(2− p)4−p

211−p(4− p)2(4− p)!, for p ≤ 1, p odd.

By Theorem 2.1 we have α2 = −12 and α1 =

14 . Therefore the case p = 3 reduces to a

direct computation, i.e.

2∑

j=p−2

αp−jαj

4(p − 2j)2=α1α2

2=

−1

16.

By (2.18) and (2.20) we know that

αp =−1

2p−1

p∑

l=0

1

l!

(p− l + 1

3

)

pl, for p ≥ 3. (4.30)

Furthermore α2 = −12 holds. So in the case p ≥ 5 and p odd, by (4.30) we get

p−2∑

j=2

αp−jαj

4(p − 2j)2≥ αp−2α2

2(p− 4)2>

(p − 2)p−2

2p−1(p− 4)2(p − 2)!.


Now we are able to prove Theorem 4.2.

Proof. (Theorem 4.2) If q = 1, 2 condition (4.1), which guarantees the existenceof a (p, q)−periodic orbit of the dissipative spin-orbit problem modelled by equations(1.24)-(1.32), follows from Lemma 4.3 and 4.4 taking |e| ≤ r 2p

q

(b). By (4.17) this is

equivalent to

|e| ≤ r2p = cp,11

M(b)

(r(b)

2

)|2p−2|+1

, for q = 1,

|e| ≤ rp = cp,21

M(b)

(r(b)

2

)|p−2|+1

, for q = 2,

where cp,1 and cp,2 are defined in (4.3) and (4.4), respectively. This implies (4.7) andso for q = 1, 2 the proof of Theorem 4.2 is finished.


It remains to prove the case q = 4. Since (p, q) = 1 it follows that p is odd, thereforealso p− 2 is odd. Let us define

s(e, ζ) :=∑

j∈Z,j 6=0,p

αp−j(e)αj(e)

4(p − 2j)2 + ζ. (4.31)

Recalling Theorem 4.1 and condition (4.1), we have to prove that s(e, 16η2) 6= 0 for eand η satisfying (4.8) and (4.9), respectively.Let us start with the case η = 0. For 0 < b < 1 and |e| < r(b) = b/ cosh b by (4.16)we get

|s(e, 0)| =

∣∣∣∣∣∣

∑

j∈Z,j 6=0,p

αp−j(e)αj(e)

4(p − 2j)2

∣∣∣∣∣∣

≤ M2(b)

4

∑

k 6=0,k∈Z

1

k2=M2(b)π2

12. (4.32)

Furthermore by (2.17)

αp−j(e)αj(e) =[

αp−je|p−j−2| +O(e|p−j−2|+1)

] [

αje|j−2| +O(e|j−2|+1)

]

= αp−jαje|p−j−2|+|j−2| +O(e|p−j−2|+|j−2|+1). (4.33)

Using Lemma 4.5 and (4.33) we have

s(e, 0) =

∑2j=p−2

αp−j αj

4(p−2j)2 e|p−4| +Rp(e), for p ≤ 3,

∑p−2j=2

αp−j αj

4(p−2j)2e|p−4| +Rp(e), for p > 4 ,

(4.34)

where Rp(e) = O(e|p−4|+1) is the Taylor remainder. Take

0 < r′ < r(b) .

By the Cauchy estimate and (4.32) we get

|Rp(e)| ≤π2M2(b)

12(r − r′)|p−4|+1|e||p−4|+1, ∀ |e| ≤ r′(b) . (4.35)

Let cp,4 be as in (4.5). Choosing r′ = r′p(b) as the unique solution in (0, r(b)) of

r =6|cp,4|π2M2(b)

(r(b)− r)|p−4|+1 =: h(r) (4.36)

(see Figure 4.2), we get by (4.35)

|Rp(e)| ≤ |cp,4|2h(r′p(b))

|e||p−4|+1 (4.36)=

|cp,4|2r′p(b)

|e||p−4|+1

≤ |cp,4|2

|e||p−4| ∀ |e| ≤ r′p(b) . (4.37)

Then for p odd we have

|s(e, 0)| (4.34)=

∣∣∣∑2

j=p−2αp−j αj

4(p−2j)2e|p−4| +Rp(e)

∣∣∣ , if p ≤ 3,

∣∣∣∑p−2

j=2αp−j αj

4(p−2j)2e|p−4| +Rp(e)

∣∣∣ , if p > 4,


(4.27)

≥

∣∣∣

(2−p)4−p

211−p(4−p)2(4−p)

∣∣∣ |e||p−4| − |Rp(e)| , if p ≤ 1,

∣∣−116

∣∣ |e||p−4| − |Rp(e)| , if p = 3,

∣∣∣

(p−2)p−2

2p−1(p−4)2(p−2)!

∣∣∣ |e||p−4| − |Rp(e)| , if p ≥ 5,

(4.5)&(4.37)

≥ |cp,4||e||p−4| − |cp,4|2

|e||p−4|

≥ |cp,4|2

|e||p−4|, ∀ |e| ≤ r′p(b) . (4.38)

r

y

1

y = r

y = g(r)

y = h(r)

r(b)r′r′′

Figure 4.2: Geometric explanation of equation (4.41).

Let us now consider the case η > 0. For |e| < r(b) and ζ > 0 we have by (4.16) and(4.31)

|∂ζs(e, ζ)| ≤∑

j∈Z,j 6=0,p

|αp−j(e)αj(e)|(4(p − 2j)2 + ζ)2

≤ M2(b)

16

∑

06=k∈Z

1

k4≤ π2M2(b)

720. (4.39)

Then by (4.38) and (4.39), for 0 < |e| ≤ r′p(b), we obtain

|s(e, 16η2)| ≥ |s(e, 0)|− π2M2(b)

72016η2 ≥ |cp,4|

2|e||p−4|− π2M2(b)

45η2 ≥ |cp,4|

4|e||p−4| > 0

if we assume that

η ≤ 3√

5|cp,4|2πM(b)

|e||p−4|/2,

which corresponds to (4.9). Notice that (4.8) is equivalent to

0 < e ≤ r′′ :=r(b)

π2M2(b)

6|cp,4|r(b)|p−4| + |p− 4|. (4.40)

Condition (4.1), which guarantees the existence of a (p, 4)−periodic orbit of the dis-sipative spin-orbit problem modelled by equations (1.24)-(1.32), follows if we provethat

r′′ ≤ r′p(b). (4.41)


From (4.36)

h(r) ≥ h(0) + h′(0)r

=6|cp,4|π2M2(b)

(r(b))|p−4|+1 − 6|p − 4||cp,4|π2M2(b)

(r(b))|p−4|r =: g(r)

holds for 0 ≤ r ≤ r(b). Since r′′ (defined in(4.40)) is the unique solution of the equationr = g(r) and r′ is the unique solution of the equation r = h(r), the inequality (4.41)is true (see Figure 4.2). This terminates the proof of Theorem 4.2.

5 Numerical analysis of (p, q)-periodicorbits of the DSOP

The aim of this chapter, as already anticipated in Remark 3.5, is to develop a Matlabprogram in order to compute recursively the functions φ(k) in (3.23) and uk in (3.25).Chapter 5 is organized as follow: In Section 5.1 we explain the basic notation andfunctions, which we will use; In Section 5.2 we develop the programs to find numer-ically the periodic solutions of the dissipative spin-orbit problem (DSOP); Finally,in Section 5.3 we will use this program to verify numerically the Conjecture 3.5 fordiverse values of p and q.

5.1 Basic notation and functions

5.1.1 Functions of t in matrix form

For the programs in Matlab we will use the following notation. Every function u(t) :R ⊇ [0, 2π] → C with N−truncated Fourier-expansion equal to

u(t) =

N∑

n=−N

uneint,

with N ∈ N≥1 and un ∈ C, will be expressed with Matlab using the 2 × (N + 1)complex matrix

U =

u0 u0u−1 u1...

...

u−N uN

∈ C2×(N+1). (5.1)

In the rest of this subsection we implement basic programs (sum, product, ...) to workwith functions as above.

Sum

Given U ∈ C2×N and V ∈ C2×M representing the truncated Fourier expansion of thefunctions u(t) and v(t), respectively (as in (5.1)), we compute W ∈ C2×max(N,M) thematrix representing truncated Fourier expansion of the the sum u(t) + v(t).Idea: If N =M the matrices U and V can be summed up together, henceW = U+V .If (without loss of generality) N < M , then

wi,j = ui,j + vi,j, for i = 1, 2 and 0 ≤ j ≤ N,wi,j = vi,j, for i = 1, 2 and N + 1 ≤ j ≤M,

hold, where ui,j, vi,j and wi,j denote the element in the ith-row and in the jth-columnof the matrices U, V and W , respectively.

73

74 Chapter 5. Numerical analysis of (p, q)-periodic orbits of the DSOP

SUM of u(t) and v(t)

%--------------------------------------------------------------------------

% INPUT: u is a 2xN matrix, v is a 2xM matrix

% OUTPUT: w is a 2xP matrix, where P = maxN,M

% DESCRIPTION: this program sums two truncated Fourier-expansions

%--------------------------------------------------------------------------

function w = sumfourier(u,v)

[N,due_u] = size(u);

[M,due_v] = size(v);

if (M > N)

w = v;

w(1:N,:) = w(1:N,:) + u;

elseif (N > M)

w = u;

w(1:M,:) = w(1:M,:) + v;

else % M = N

w = u + v;

end

end

Product

Given U ∈ C2×N and V ∈ C2×M representing the truncated Fourier expansion of thefunctions u(t) and v(t), respectively (as in (5.1)), we compute W ∈ C2×N+M−1 thematrix representing the truncated Fourier expansion of the the product u(t) · v(t).Idea: Since by (5.1) the functions u(t), v(t) are given by

u(t) =N∑

j=−N

ujeijt and v(t) =

M∑

k=−M

vkeikt,

for the product u(t) · v(t) we have

u(t) · v(t) =

N∑

j=−N

ujeijt

·(

M∑

k=−M

vkeikt

)

=

N+M∑

l=−N−M

cl eilt.

The coefficients cl ∈ C, for −N−M ≤ l ≤ N+M , can be computed using the Cauchyformula for the product of two series, i.e. in the case N ≤M

cl =∑

j+k=l

ujvk =N∑

j=−N

ujvl−j

holds (the case M > N is similar).

PRODFOURIER of u(t) and v(t)

%--------------------------------------------------------------------------

% INPUT: u is a 2xN matrix, v is a 2xM matrix

% OUTPUT: w is a 2xP matrix, where P = N + M - 1

% DESCRIPTION: this program multiplies two truncated Fourier-expansions

%--------------------------------------------------------------------------

function w = prodfourier(u,v)

5.1. Basic notation and functions 75

[N,due_u] = size(u);

[M,due_v] = size(v);

w = zeros(N+M-1,2); % create the bi-vector that will be the output

% convert bi-vectors in vectors

u_vector = [u(end:-1:2,1) ; u(:,2)];

v_vector = [v(end:-1:2,1) ; v(:,2)];

n = length(u_vector);

m = length(v_vector);

w_vector = zeros(2*(N+M-2)+1,1);

for k = 1:n

w_vector(k:(m-1+k)) = w_vector(k:(m-1+k)) + u_vector(k) * v_vector;

end

% convert the solution in a bi-vector

w(:,1) = w_vector((N+M-1):-1:1);

w(:,2) = w_vector((N+M-1):end);

end

Green operator

Given V ∈ C2×N as in (5.1) representing the truncated Fourier expansion of thefunctions v(t), we compute W ∈ C2×N the matrix representing truncated Fourierexpansion of −Gη (v(t)), where the Green operator Gη (see equation (3.7))

Gη

∑

n 6=0

vneint

:=∑

n 6=0

vninη − n2

eint,

is diagonal in Fourier space, being just the multiplication of the n−th Fourier coeffi-cient by 1

inη−n2 . Therefore1 we have

W =

v0 v0v−1

iη+12v1

−iη+12

......

v−N

i(N−1)η+(N−1)2vN

−i(N−1)η+(N−1)2

∈ C2×N .

MINUS GREEN OPERATOR of v(t)

%--------------------------------------------------------------------------

% INPUT: v is a 2xN matrix

% OUTPUT: w is a 2xN matrix

% DESCRIPTION: the Green operator G_eta modify the k-th coefficient of v

% by multiplying it with (i*k*eta - k^2)^(-1)

% Here is implemented -G_eta

%--------------------------------------------------------------------------

function w = geta_op(v,eta)

[N,due] = size(v);

w = v;

w(2:end,1) = w(2:end,1) ./ ( 1i * eta * (1:N-1) + (1:N-1).^2)’;

w(2:end,2) = w(2:end,2) ./ (-1i * eta * (1:N-1) + (1:N-1).^2)’;

end

1The Matlab code works also if v0 6= 0.


Average

Given V ∈ C2×N as in (5.1) representing the truncated Fourier expansion of thefunction v(t), we compute W ∈ C2×N the matrix representing truncated Fourierexpansion of v(t)− 〈v(t)〉, where

〈v(t)〉 = 1

2π

∫ 2π

0v(t)dt =

1

2π

∫ 2π

0

∑

n∈Zvne

intdt = v0.

Therefore it follows

W =

0 0

v−1 v1...

...

v−N vN

∈ C2×(N+1).

MINUS MEDIA: compute < v(t) > and v(t)− < v(t) >

%-------------------------------------------------------------------------------

% INPUT: v is a 2xN matrix

% OUTPUTs: w is a 2xN matrix, phi is a 1x1 matrix

% DESCRIPTION: phi=<v> can be read in the zero coefficient of v,

w=v-<v> can be computed subtracting phi to v.

%-------------------------------------------------------------------------------

function [w,phi] = minus_media(v)

w = v;

phi = v(1,1);

w(1,:) = [0 0];

end

Signum function

Define

signum(x) :=

1, if x ≤ 0,2, if x > 0.

This function is helpful to move a pointer inside a matrix as in (5.1).

SIGNUM: help function

%-------------------------------------------------------------------------------

% INPUT: x is a real number

% OUTPUT: r is a real number

% DESCRIPTION: if x>0 then r=2, otherwise r=1.

%-------------------------------------------------------------------------------

function [r]=signum(x)

if x>0

r=2;

else

r=1;

end

end


5.1.2 Function of ξ and t in cell array form

For the programs in Matlab we will use the following notation. Every function u(ξ, t) :[0, 2π] × [0, 2π] → C with (M, N)−truncated Fourier-expansion equal to

u(ξ, t) =∑

m∈M

(

eimξN∑

n=−N

um,neint

)

,

with N ∈ N≥1, um,n ∈ C and set of ξ−exponents

M =m1,m2, . . . ,m|M|

⊂ Z

with |M| <∞, will be expressed with Matlab using the 2|M|−cell array2

Um1 m1 Um2 m2 . . . Um|M|−1m|M|−1 Um|M|

m|M| ∈ Cell2|M| ,(5.2)

where for m ∈ M we have

Um =

um,0 um,0

um,−1 um,1...

...

um,−N um,N

∈ C2×(N+1).

Sum of functions of ξ and t in cell array form

Let N ∈ N≥1. Let U ∈ Cell2|M1| and V ∈ Cell2|M2| as in (5.2) be (Mi, N)−truncatedFourier expansion for i = 1, 2 of the functions u(ξ, t) and v(ξ, t), respectively. Wecompute W ∈ Cell2|M1∪M2| the cell array representing the (M1 ∪M2, N)−truncatedFourier expansion of the sum u(ξ, t) + v(ξ, t).Idea: For i = 1, 2 let

Mi =

mi1,m

i2, . . . ,m

i|Mi|

⊂ Z

with |Mi| <∞. Define

M1 ∪M2 = (M1 \M2) ⊎ (M1 ∩M2) ⊎ (M2 \M1)

and put l = |M1 ∪M2| ≥ 0. Without loss of generality we assume

m(1)k = m

(2)k , for 1 ≤ k ≤ l.

Then W ∈ Cell2|M| as the following form

W = WM1\M2WM1∩M2 WM2\M1

,

where

WM1\M2= U

m(1)|M1|−l+1

m(1)|M1|−l+1 . . . U

m(1)|M1|

m(1)|M1| ,

WM1∩M2 = Um

(1)1

+ Vm

(2)1

m(1)1 . . . U

m(1)l

+ Vm

(2)l

m(1)l ,

WM2\M1= V

m(2)|M2|−l+1

m(2)|M2|−l+1 . . . V

m(2)|M2|

m(2)|M2| .

Notice that to compute the sums in WM1∩M2 we have to use the program SUM-FOURIER of the previous subsection.

2A cell array is an array that can contain data of varying types and sizes.


SUMCELLS: computes the sum of two functions given as cell arrays

%--------------------------------------------------------------------------

% INPUT: u function u(t,xi) given in cell form (see description)

% v function v(t,xi) given in cell form

% OUTPUT: w function w(t,xi) = u(t,xi) + v(t,xi) given in cell form

%

%

% DESCRIPTION: this program computes the sum of two functions given in this

% form

% __________ _____ _____ __________ _____

% | | | | | |

% | u_1(t) | e_1 | ... | u_n(t) | e_n |

% |__________|_____|_____|__________|_____|

%

%--------------------------------------------------------------------------

function w = sumcells(u,v)

[uno_u,length_u] = size(u);

[uno_v,length_v] = size(v);

%-------------to eliminate potential empty cells---------------------

index_u = length_u;

index_v = length_v;

while ( (index_u > 0) && (isempty(u1,index_u)) )

index_u = index_u - 2;

end

while ( (index_v > 0) && (isempty(v1,index_v)) )

index_v = index_v - 2;

end

if (index_u == 0)

%u = ;

w = v;

return;

else

u = u(1,1:index_u);

length_u = index_u;

end

if (index_v == 0)

%v = ;

w = u;

return;

else

v = v(1,1:index_v);

length_v = index_v;

end

%----------------------------------------------------------------------

index_u = 2;

index_v = 2;

w = cell(1,length_u + length_v);

index_w = 2;

while (index_u ~= length_u + 2) && (index_v ~= length_v + 2)

if (u1,index_u > v1,index_v)

w1 , index_w-1 = u1 , index_u-1;

w1 , index_w = u1 , index_u;

index_u = index_u + 2;

elseif (u1,index_u < v1,index_v)

w1 , index_w-1 = v1 , index_v-1;

w1 , index_w = v1 , index_v;

index_v = index_v + 2;

else % u1,index_u = v1,index_v i.e. the exponent is the same

w1 , index_w-1 = sumfourier(u1 , index_u-1,v1 , index_v-1);

w1 , index_w = v1 , index_v; % = u1,index_u

index_u = index_u + 2;

index_v = index_v + 2;


end

index_w = index_w + 2;

end


w = w(1,1:index_w-2);

if (index_u ~= length_u + 2)

w = [w u(1,index_u-1:end)];

%index_w = index_w + length_u - index_u + 2;

elseif (index_v ~= length_v + 2)

w = [w v(1,index_v-1:end)];

%index_w = index_w + length_v - index_v + 2;

end

end

Product of functions of ξ and t in cell array form

Let N ∈ N≥1. Let U ∈ Cell2|M1| and V ∈ Cell2|M2| as in (5.2) be (Mi, N)−truncatedFourier expansion for i = 1, 2 of the functions u(ξ, t) and v(ξ, t), respectively. Wecompute W ∈ Cell2|M1∪M2| the cell array representing the (M1 ∪M2, N)−truncatedFourier expansion of the product u(ξ, t) · v(ξ, t).Idea: For i = 1, 2 let

Mi =

mi1,m

i2, . . . ,m

i|Mi|

⊂ Z

with |Mi| <∞. By (5.2) it follows

U = Um

(1)1

m(1)1 U

m(1)2

m(1)2 . . . U

m(1)|M1|

m(1)|M1| ,

V = Vm

(2)1

m(2)1 V

m(2)2

m(2)2 . . . V

m(2)|M2|

m(2)|M2| ,

We define

Um

(1)l

:= Um

(1)l

m(1)l , for 1 ≤ l ≤ |M1|,

Vm

(2)k

:= Vm

(2)k

m(2)k , for 1 ≤ k ≤ |M2|.

Using the distributive property we get

W = U ·W =∑

1≤l≤|M1|

∑

1≤k≤|M2|Um

(1)l

Vm

(2)k

.

Notice that in the last formula the sums may be performed using the program SUM-CELLS, since we are adding cell arrays together. Therefore it is sufficient to knowhow to compute U

m(1)l

Vm

(2)k

. By definition of Um

(1)l

and Vm

(2)k

we know that

Um

(1)l

= eim(1)l

ξN∑

n=−N

um

(1)l ,n

eint,

Vm

(2)l

= eim(2)k ξ

N∑

n=−N

vm

(2)k ,n

eint,


for 1 ≤ l ≤ |M1| and for 1 ≤ k ≤ |M2|. Hence

Um

(1)l

Vm

(2)l

= ei(m(1)l

+m(2)k

)ξ

(N∑

n=−N

um

(1)l

,neint

)

·(

N∑

n=−N

vm

(2)k

,neint

)

.

In Matlab the right hand side is represented by the cell array

Um

(1)l

· Vm

(2)k

m(1)l +m

(2)k .

To compute the term Um

(1)l

· Vm

(2)k

the program PRODFOURIER of the previous

section may be used.

PRODCELLS: computes the product of two functions given as cell arrays

%--------------------------------------------------------------------------

% INPUT: u function u(t,xi) given in cell form (see description)

% v function v(t,xi) given in cell form

% OUTPUT: w function w(t,xi) = u(t,xi) + v(t,xi) given in cell form

%

%

% DESCRIPTION: this program computes the sum of two functions given in the

% form

% __________ _____ _____ __________ _____

% | | | | | |

% | u_1(t) | e_1 | ... | u_n(t) | e_n |

% |__________|_____|_____|__________|_____|

%

%--------------------------------------------------------------------------

function w = prodcells(u,v)

[uno_u,length_u] = size(u);

[uno_v,length_v] = size(v);


index_u = length_u;

index_v = length_v;

while ( (isempty(u1,index_u)) )

index_u = index_u - 2;

end

while ( (isempty(v1,index_v)) )

index_v = index_v - 2;

end

if (index_u == 0)

u = ;

else

u = u(1,1:index_u);

length_u = index_u;

end

if (index_v == 0)

v = ;

else

v = v(1,1:index_v);

length_v = index_v;

end

%----------------------------------------------------------------------

w = ;

for index_u = 1:2:length_u-1

for index_v = 1:2:length_v-1

w = sumcells( w , ...

prodfourier(u1,index_u, v1,index_v) , ...

u1,index_u+1 + v1,index_v+1 ...

);

end

end

end

5.2. The DSOP and its representation with Matlab 81

Plot functions given in cell array form

Let u = u(ξ, t) be a function with zero average with respect to t. The followingprogram plots the input function u as in (5.2) on the interval ξ ∈ [0, 2π]

PLOTTARE: plots a cell array function with zero mean

function [dati]=plottare(f,color)

[a,b]=size(f);

dati=0;

for j=2:2:b

ff=fa,j-1;

dati=ff*exp(1i.*fa,j*[0:2 *pi/100:2*pi])+dati;

end

plot([0:2 *pi/100:2*pi],dati,color)

5.2 The DSOP and its representation with Matlab

Recall from (2.1), Lemma 2.1 and Remark 3.1 that the Newtonian potential f(x, t; e)for the dissipative spin-orbit problem is given by

f(x, t; e) =∑

j∈Z,j 6=0

αj cos(2x− jt)

= f−2(t) ei(−2)x + f2(t) e

i2x,

withf±2(t) =

∑

j∈Z

α∓j

2eijt.

Hence the dissipative spin-orbit problem (1.24) is a special case of the generalizedspin-orbit problem in (3.1), where m only admits the values 2 and −2.By Lemma 2.1 the coefficients αj (for j ∈ Z) depend on the eccentricity e ∈ [0, 1) asfollows:

αj(e) = − 1

4π

∫ 2π

0

cos(2fe − ju+ ej sin(u))

ρ2edu,

with

fe = fe(u) := 2 arctan

(√

1 + e

1− etan

(u

2

))

,

ρe = ρe(u) := 1− e cos(u).

Given N ∈ N≥1 the following Matlab program computes an approximation of thecoefficients αj of the Newtonian potential f(x, t; e) in (2.1) for −N ≤ j ≤ N .

COEFF-ALPHA: compute the coefficients αj

%--------------------------------------------------------------------------

% INPUT: N number of Fourier-coefficients we consider

% e the eccentricity

% OUTPUT: alpha vector of the coefficients of f(x,t;e)

%

%--------------------------------------------------------------------------

function [alpha]=coeff_alpha(e,N)


alpha = zeros(N,2);

for j=1:N

g1 = @(x) cos(4.*atan(sqrt((1+e)/(1-e)).*tan(x/2))-j.*x+e.*j.*sin(x))./(1-e.*cos(x)).^2;

g2 = @(x) cos(4.*atan(sqrt((1+e)/(1-e)).*tan(x/2))+j.*x-e.*j.*sin(x))./(1-e.*cos(x)).^2;

alpha(j,2) = -1 / (4*pi) * quad(g1,0,2*pi);

alpha(j,1) = -1 / (4*pi) * quad(g2,0,2*pi);

end

end

Recall that by (3.13), (3.14) and (3.15) we have

u(t; ξ, ε) =∞∑

k=1

ukεk,

φu(u) := 〈fx(ξ + pt+ u(t), qt)〉 =∞∑

k=0

φ(k)εk,

fx(ξ + pt+ u, qt) =

∞∑

k=0

f (k)εk.

By Lemma 3.4 we know that we can compute uk and φ(k) recursively, using equations(3.20), (3.21),(3.23), (3.25) and (3.26). More precisely for the dissipative spin-orbitproblem we have that:

• Equation (3.26) is equivalent to

β2,nl (t) =2i

l

l∑

k=1

kukβ2,nl−k, (5.3)

β−2,nl (t) =

−2i

l

l∑

k=1

kukβ−2,nl−k , (5.4)

for 1 ≤ k ≤ l, where β±2,n0 in (3.20) are given by:

β2,n0 = ei(2p−nq)t,

β−2,n0 = ei(−2p+nq)t.

• Equations (3.21) and (3.23) say

f (k) = 2ie2iξ∑

n 6=0

f2,nβ2,nk − 2ie−2iξ

∑

n 6=0

f−2,nβ−2,nk , (5.5)

φ(k) =⟨

f (k)⟩

, (5.6)

for all k ≥ 0.

• Equation (3.25) implies

uk = uk(t; ξ) = −Gη

(

f (k−1) − φ(k−1))

(5.7)

for all k ≥ 1.

5.2. The DSOP and its representation with Matlab 83

The basic idea of the following program is to implement this recursion, i.e. startingwith β2,n0 , β−2,n

0 (which are explicitly given) we first compute f (0), φ(0) and u1. Then

we are able to compute β2,n1 , β−2,n1 and automatically f (1), φ(1) and u2. And so on.

Idea: Let N ∈ N≥1 be the truncation number. Since from Lemma 2.1 we know thatα0 ≡ 0, the function f (0) in (5.5), can be approximated as follows

f (0) ≈ ei2ξ · 12

N∑

j=−N

[

(2i)αjei(2p−jq)t

]

+ e−i2ξ · 12

N∑

j=−N

[

(−2i)αje−i(2p−jq)t

]

=: ei2ξ · f (0)2 (t;N) + e−i2ξ · f (0)−2 (t;N), (5.8)

where

f(0)±2 (t;N) =

2p+Nq∑

k=−2p−Nq

akeikt (5.9)

with ak ∈ C defined by the relation (5.8). In Matlab we describe the approximatingfunction (5.8) using cell arrays as in (5.2), i.e. for the right hand side of (5.8) we usethe cell array

F(0)2 2 F

(0)−2 −2 ,

where F(0)2 and F

(0)−2 are matrices in C2×(2p+Nq+1) of the form

F(0)±2 =

a±2,0 a±2,0

a±2,−1 a±2,1...

...

a±2,−2p−Nq a±2,2p+Nq

,

representing the truncated Fourier expansion of the functions f(0)2 (t;N) and f

(0)−2 (t;N)

in (5.9), respectively. Then by (5.6) and (5.8) φ(0) is given by

φ(0) =⟨

f (0)⟩

= iei2ξα 2pq

− ie−i2ξα 2pq

and we save it in the cell array

iα 2pq

2 −iα 2pq

−2 .

From the definition of the Green operator (see program MINUS GREEN OPERATORin Section 5.1) and from equation (5.7) the function

u1 = −Gη

(

f (0) − φ(0))

,

is saved in Matlab using the following cell array:

U(1)2 2 U

(1)−2 −2 ,

where

U(1)±2 =

0 0a±2,−1

iη+12a±2,1

−iη+12

......

a±2,−2p−Nq

i(2p−Nq)η+(2p+Nq)2a±2,2p+Nq

−i(2p+Nq)η+(2p+Nq)2

.


We introduce help functions in cell array form

b(0)±2 = B

(0)±2 0 ,

whereB

(0)±2 = 1 1 .

In (5.3) and (5.4) replace βl by b(l). So we have

b(l)±2 =

±2i

l

l∑

k=1

kukb(l−k)±2 , (5.10)

for 1 ≤ k ≤ l. (5.10) can be performed with the programs SUMCELLS and PROD-CELLS of the previous subsection. Then we save the function f (k) in (5.5) with thecell array

b(k)2 · F

(0)2 2 + b

(k)2 · F

(0)−2 −2 .

“ + ” , “ · ” are here the SUMCELLS and PRODCELLS function in Matlab developedabove. Finally φ(k) and uk are computed with (5.6) and (5.7), respectively. The idea

behind this is that the recursion with the b(l)±2 is easier than the recursion with the

β±2,nl and by replacing at the end b

(0)±2 by β±2,n

0 we get the same result.

SPIN-ORBIT: given k, p, q compute the function φ(k−1) in cell format

%--------------------------------------------------------------------------


% k degree


% eta dissipative constant

% p,q constants of the periodic solution

% OUTPUT: phi=(phi_l)_l, where phi_l are cell arrays for 1 <= l <= k.

%--------------------------------------------------------------------------

function [phi]=spin_orbit(k,p,q,e,eta,N)

alpha=coeff_alpha(e,N);

U = cell(k,1);

beta1 = cell(1,2);

beta_1 = cell(1,2);

phi = cell(k,1);

%--------------------------------------------------------------------------

% Create beta0:

N_max = 2*p - (-N)*q;

% the minimal exponent is N_min=-2*p-N*q, obiuvsly N_min = -N_max

beta10 = zeros(N_max+1,2); % this part will be the bi-vector for e^(i2xi)

beta_10 = zeros(N_max+1,2); % this part will be the bi-vector for e^(-i2xi)

j = [(-N:-1) (1:N)];

exponent = 2*p - j*q;

beta10 ( abs(exponent) + 1 + (exponent >= 0)*(N_max+1) ) = alpha(abs(j) + (j>0)*N) * (1i);

beta_10( abs(exponent) + 1 + (exponent < 0)*(N_max+1) ) = alpha(abs(j) + (j>0)*N) * (-1i);

% here the values of the zero-elements are copied on the other

% component of F_2 and F_-2

beta10 (1,1) = beta10 (1,2);

beta_10(1,2) = beta_10(1,1);

% output in the cell form

beta11,1= beta10; beta11,2=2;

beta_11,1=beta_10; beta_11,2=-2;

5.3. Control and results 85

%--------------------------------------------------------------------------

% Compute u1:

[w2,phi02]=minus_media(beta11,1);

[w_2,phi0_2]=minus_media(beta_11,1);

U1,1 = geta_op(w2,eta) , 2 , geta_op(w_2,eta) , -2 ;

phi1,1 = phi02 , 2 , phi0_2 , -2 ;

%--------------------------------------------------------------------------

% Recursive step:

bb1=cell(k,1);

bb11,1=[1 1], 0;

bb_1=cell(k,1);

bb_11,1=[1 1], 0;

for tilde_l =1 :1: k-1

for l=1:1:tilde_l

somma1=;

somma_1=;

[uno,lenght_u]=size(Ul,1);

help_u1=Ul,1;

help_u_1=Ul,1;

for index_u = 1:2:lenght_u-1

help_u11,index_u=help_u11,index_u*l*2*1i*1;

help_u_11,index_u=help_u_11,index_u*l*2*1i*(-1);

end

somma1=sumcells(somma1,prodcells(help_u1,bb1tilde_l-l+1));

somma_1=sumcells(somma_1,prodcells(help_u_1,bb_1tilde_l-l+1));

bb1l+1,1=somma1;

bb_1l+1,1=somma_1;

end

A= prodcells(bb1tilde_l+1,1,beta1);

B= prodcells(bb_1tilde_l+1,1,beta_1);

C=sumcells(A,B);

[uno,lenght_C]=size(C);

for index_C = 1:2:lenght_C-1

[w,phiC]=minus_media(C1,index_C);

w = geta_op(w,eta);

phitilde_l+1,11,index_C = phiC;

phitilde_l+1,11,index_C+1 = C1,index_C +1;

Utilde_l+1,11,index_C= w;

Utilde_l+1,11,index_C+1=C1,index_C +1;

end

end

5.3 Control and results

In the following program we compute φ(0) and φ(1) using equations (3.32) and (3.35),because we want to check if the program SPIN-ORBIT works correctly. We will dothe control only at the zeroth degree and at the first degree of non-degeneration.

CONTROL: given k = 0, 1 compute φ(k) with the formulas (3.32) and (3.35)

%--------------------------------------------------------------------------


% k degree




% OUTPUT: phi=(phi_l)_l, where phi_l are cell arrays for 1 <= l <= k.


%--------------------------------------------------------------------------

function [phik_test,alpha]=control(k,p,q,e,eta,N)

[F,alpha]=f_coeff(N,1,e,p,q);

if k==1

% ZERO DEGREE:

if q==1

PHI_test1,1=-alpha(2*abs(p),signum(2*p))/i; % 1+(p>0)*1

PHI_test1,2=2;

PHI_test1,3=alpha(abs(2*p),signum(2*p))/i;

PHI_test1,4=-2;

elseif q==2

PHI_test1,1=-alpha(abs(p),signum(p))/i;

PHI_test1,2=2;

PHI_test1,3=alpha(abs(p),signum(p))/i;

PHI_test1,4=-2;

else

PHI_test=0,0;

end

phik_test=PHI_test(1,:);

elseif k==2

% FIRST DEGREE:

if q==4

alpha(N+1:1:p+N,:)=0;

uno=-2/p*alpha(abs(p),signum(p))^2/(4*p^2+eta^2);

if p>0

dueA=0;

for j=-N:1:-1

dueA=(alpha(abs(j),signum(j))^2-alpha(abs(p-j),signum(p-j))^2)...

/((p-2*j)*(4*(p-2*j)^2+eta^2))+dueA;

end

dueB=0;

for j=1:1:p-1

dueB=(alpha(abs(j),signum(j))^2-alpha(abs(p-j),signum(p-j))^2)...

/((p-2*j)*(4*(p-2*j)^2+eta^2))+dueB;

end

dueC=0;

for j=p+1:1:N

dueC=(alpha(abs(j),signum(j))^2-alpha(abs(p-j),signum(p-j))^2)...

/((p-2*j)*(4*(p-2*j)^2+eta^2))+dueC;

end

due=dueA+dueB+dueC;

else

dueA=0;

for j=-N:1:p-1

dueA=(alpha(abs(j),signum(j))^2-alpha(abs(p-j),signum(p-j))^2)...

/((p-2*j)*(4*(p-2*j)^2+eta^2))+dueA;

end


dueB=0;

for j=p+1:1:-1

dueB=(alpha(abs(j),signum(j))^2-alpha(abs(p-j),signum(p-j))^2)...

/((p-2*j)*(4*(p-2*j)^2+eta^2))+dueB;

end

dueC=0;

for j=1:1:N

dueC=(alpha(abs(j),signum(j))^2-alpha(abs(p-j),signum(p-j))^2)...

/((p-2*j)*(4*(p-2*j)^2+eta^2))+dueC;

end

due=dueA+dueB+dueC;

end

PARTE1= eta*(uno + due);

quattroA=4/i*sum(alpha(N:-1:p+1,2).*alpha(N-p:-1:1,1)./((4.*(p-2.*(p-N:1:-1)).^2+eta^2))’);

quattroB=4/i*sum(alpha(p-1:-1:1,2) .* alpha(1:p-1,2)./((4.*(p-2.*(1:p-1)).^2+eta^2))’);

quattroC=4/i*sum(alpha(1:N-p,1) .* alpha(p+1:1:N,2)./((4.*(p-2.*(p+1:N)).^2+eta^2))’);

quattro= quattroA + quattroB + quattroC;

PHI_test2,1=quattro/2;

PHI_test2,2=4;

PHI_test2,3=-PARTE1;

PHI_test2,4=0;

PHI_test2,5=-quattro/2;

PHI_test2,6=-4;

phik_test = PHI_test(2,:);

else

uno=0;

for j=-N:1:-1

uno=(alpha(abs(j),signum(j))^2)/((2*p-j*q)^3+eta^2*(2*p-j*q))+uno;

end

for j=1:1:N

uno=(alpha(abs(j),signum(j))^2)/((2*p-j*q)^3+eta^2*(2*p-j*q))+uno;

end

PHI_test2,1=-4*eta*uno;

PHI_test2,2=0;

phik_test = PHI_test(2,:);

end

else

phik_test = ’no informations’;

The next Matlab program compares3 the programs SPIN-ORBIT and CONTROL.The result is plotted with the function PLOTTARE.

LANCIA: SPIN-ORBIT vs CONTROL.

%--------------------------------------------------------------------------


% k degree




% OUTPUT: phi=(phi_l)_l, where phi_l are cell arrays for 1 <= l <= k computed with SPIN-ORBIT

phi_test=(phi_l)_l, where phi_l are cell arrays for 1 <= l <= k computed with CONTROL

%--------------------------------------------------------------------------

3The output φ(k) can only be compared for k = 0, 1. If k ≥ 2 we have only the value obtained withthe program SPIN-ORBIT.


function [phi,phi_test]=lancia(k,p,q)

[phi]=spin_orbit(k,p,q,e,eta,N);

[phi_test]=controllo(k,p,q,e,eta,N);

hold on

plottare(phik,1,’r’)

hold on

plottare(phi_test,’blue’)

hold off

Finally with the following Matlab program we check the hypothesis of Conjecture 3.5for all p, q ∈ N, where (p, q) = 1, 1 ≤ p ≤ pmax and 1 ≤ q ≤ qmax. pmax, qmax ∈ N aregiven as inputs.Idea: If q|2(k + 1) we check that φ(l) = const for all 0 ≤ l ≤ k − 1 and φ(k) 6= const.The output Rpmax,qmax is a 4× |Ipmax,qmax | matrix, where

Ipmax,qmax := (p, q)|p, q ∈ N coprime, 1 ≤ p ≤ pmax, 1 ≤ q ≤ qmax .

For example for pmax = 3 and qmax = 4 we obtain

R3,4 =

1 1 0 12 1 0 13 1 0 11 2 0 13 2 0 11 3 2 12 3 2 11 4 1 13 4 1 1

.

Every row indicates a test. In the example the sixth row of R3,4 says that for p = 1,

q = 3 we have k = 2 and the hypothesis of Conjecture 3.5 is true4, i.e. φ(0), φ(1) areconstant and φ(2) 6= const.

MULTI-SPIN: check the hypothesis.

%--------------------------------------------------------------------------


% k degree




% OUTPUT: prova =

%--------------------------------------------------------------------------

function [R]=multi_spin(p_max,q_max,e,eta,N)

R=[];

for q=1:1:q_max

k=1;

q_test=[-2*k,2*k];

while prod(mod(q_test,q)) ~= 0

k=k+1;

q_test=[-2*k,2*k];

end

4the number in the fourth column can only takes the values 1, when the hypothesis is satisfied, and0 when it is not satisfied.


for p=1:1:p_max

if gcd(p,q)==1

[phi]=spin_orbit(k,p,q,e,eta,N);

result_tot=1;

for j1=1:1:k-1

[uno,b]=size(phij1,1);

j2=1;

while j2<b && ( phij1,11,j2==0 || phij1,11,j2+1 ==0 )

j2=j2+2;

end

if j2==b+1

result=1;

else

result=0;

end

result_tot=result_tot*result;

% if the hypothesis is satisfied result_tot = 1, otherwise zero

end

[uno,b]=size(phik,1);

j2=2;

while j2<b && ( phik,11,j2==0 || phik,11,j2+1 ==0 )

j2=j2+2;

end

if j2<=b

result=1;

else

result=0;

end

result_tot=result_tot*result;

prova=[prova;[p,q,k-1,result_tot]];

[k,p,q,result_tot]

else

% nothing should be done in this case

end

end

end

end


Remark 5.1. (i) The truncation parameter N depends on the choice of p and q. Forexample from equations (3.35), (4.31), (4.33) and (4.34) we see that the leading termin the e−expansion of φ(1) depends on the coefficients αj with

2 ≤ j ≤ p− 2, if p ≤ 3,p− 2 ≤ j ≤ 2, if p > 4,

.

Only a finite number of coefficients αj play a decisive role to determine φ(1), in par-ticular we have N(p) ≥ p− 2.(ii) Till today we have verified the hypothesis of Conjecture 3.5 for all pairs (p, q)of coprime positive integers with 1 ≤ p ≤ 50 and 1 ≤ q ≤ 8. To do that we havechosen N = 100. Notice that N is larger than p − 2 ≤ 48 in order to get a betterapproximation of φ(k), not only considering the leading term in the e−expansion butalso some further terms (although from Theorem 2.1 we know that the coefficients inthe e−expansion of αj(e) decay exponentially fast).(iii) Numerical experiments (see Figures 5.1 and 5.2) suggest that the following con-jecture holds:

Conjecture 5.1Fix k ≥ 0 and q ≥ 1. Then φ(k) corresponding to (q+1, q)−periodic orbit hasthe largest amplitude among all φ(k) corresponding to (p, q)−periodic orbit forany other p.


0 1 2 3 4 5 6−2

−1.5

−1

−0.5

0

0.5

1

1.5

2q=1, k=0

(1,1)(2,1)(3,1)

0 1 2 3 4 5 6−0.4

−0.3

−0.2

−0.1

0

0.1

0.2

0.3

0.4q=2, k=0

(1,2)(3,2)(5,2)(7,2)

0 1 2 3 4 5 6−0.4

−0.3

−0.2

−0.1

0

0.1

0.2

0.3

0.4q=4, k=1

(1,4)(3,4)(5,4)(7,4)

0 1 2 3 4 5 6−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1q=3, k=2

(1,3)(2,3)(4,3)(5,3)

0 1 2 3 4 5 6

−0.2

−0.1

0

0.1

0.2

0.3q=6, k=2

(1,6)(5,6)(7,6)(11,6)

0 1 2 3 4 5 6−4

−3

−2

−1

0

1

2

3

4q=5, k=4

(4,5)(6,5)(7,5)(8,5)

Figure 5.1: Let e = 0.0549 be the moon eccentricity, η = 10−4 and N = 100. For1 ≤ q ≤ 6 we plot the φ(k)(ξ) on the interval [0, 2π], corresponding to the(p, q)−periodic orbit, with the largest amplitude among all p.


0 1 2 3 4 5 6−2

−1.5

−1

−0.5

0

0.5

1

1.5

2q=1, k=0

(1,1)(2,1)(3,1)(4,1)

0 1 2 3 4 5 6−1.5

−1

−0.5

0

0.5

1

1.5q=2, k=0

(1,2)(3,2)(5,2)(7,2)

0 1 2 3 4 5 6−1.5

−1

−0.5

0

0.5

1

1.5q=4, k=1

(1,4)(3,4)(5,4)(7,4)

0 1 2 3 4 5 6−10

−8

−6

−4

−2

0

2

4

6

8

10q=3, k=2

(1,3)(2,3)(4,3)(5,3)

0 1 2 3 4 5 6−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8q=6, k=2

(1,6)(5,6)(7,6)(11,6)

0 1 2 3 4 5 6−40

−30

−20

−10

0

10

20

30

40q=5, k=4

(3,5)(4,5)(6,5)(7,5)(8,5)

Figure 5.2: Let e = 0.2056 be the mercury eccentricity, η = 10−4 and N = 100. For1 ≤ q ≤ 6 we plot the φ(k)(ξ) on the interval [0, 2π], corresponding to the(p, q)−periodic orbit, for some values of p.

6 The spin-orbit resonances of the SolarSystem: a simple mathematical theory

Sections 6.1-6.4 of this chapter are joint work with L. Biasco and L. Chierchia andwill appear in Journal of Nonlinear Science, (2014), (see [3]).

Consider the dissipative spin-orbit problem introduced in Subsection 1.2.1. Recall thatin this framework a p:q spin–orbit resonance (with p and q co–prime non–vanishingintegers) is, by (1.51), a solution t ∈ R → x(t) ∈ R of (1.24) such that

x(t+ 2πq) = x(t) + 2πp ; (6.1)

indeed, for such orbits, after q revolutions of the orbital radius, x has made p completerotations.In Sections 6.1-6.4 of this Chapter we discuss a mathematical theory, which is con-sistent with the existence of all spin–orbit resonances of the Solar System; in otherwords, we prove a theorem, establishing the existence of periodic orbits for parametervalues corresponding to all the satellites (or Mercury) in our Solar system observed inspin–orbit resonance. Furthermore in Section 6.5 we discuss the existence of spin-orbitresonances considering the orbital eccentricity e to be variable.

6.1 Spin–orbit resonances in the Solar System

As already mentioned in the introduction of this thesis there are eighteen moons of theSolar System which are observed in a (1, 1)−periodic orbit. The list is the following:Moon (Earth); Io, Europa, Ganymede, Callisto (Jupiter); Mimas, Enceladus, Tethys,Dione, Rhea, Titan, Iapetus, (Saturn); Ariel, Umbriel, Titania, Oberon, Miranda(Uranus); Charon (Pluto); minor bodies with mean radius smaller than 100 Km arenot considered (see, however, Section 6.4).

93

94 Chapter 6. The spin-orbit resonances of the Solar System: a simple mathematical theory

The known relevant physical parameter values of these 18 moons are reported in Table6.1.

Principal Satellite Eccentricity a b Oblateness ν

body e (km) (km) ε = 3

2

a2−b2

a2+b2

Earth Moon [37] 0.0549 1740.19 1737.31 0.00248454179 1.018088056Jupiter Io [40] 0.0041 1829.7 1819.2 0.00863266715 1.00010086

Europa 0.0094 1561.3 1560.3 0.00096104552 1.000530163Ganymede 0.0011 2632.9 2629.5 0.0019382783 1.00000726Callisto 0.0074 2411.8 2408.8 0.00186698679 1.000328561

Saturn Mimas [21] 0.0193 208.3 196.2 0.08966019091 1.002234993Enceladus [21] 0.0047 257.2 251.2 0.03540026218 1.00013254Tethys [21] 0.0001 538.7 527.0 0.03293212897 1.00000006Dione [21] 0.0022 564.0 560.8 0.00853478156 1.00002904Rhea [21] 0.001 766.8 762.8 0.0098127957 1.000006Titan [27] 0.0288 2575.239 2574.932 0.00017882901 1.00497691Iapetus [21] 0.0283 748.9 743.1 0.011662022156 1.004805592

Uranus Ariel [41] 0.0012 582.0 577.3 0.012162311957 1.00000864Umbriel [41] 0.0039 587.5 581.9 0.01436601227 1.00009126Titania [41] 0.0011 790.7 787.1 0.00684493838 1.00000726Oberon [41] 0.0014 764.0 758.8 0.01024416739 1.00001176Miranda [41] 0.0013 241.0 233.3 0.04869051956 1.00001014

Pluto Charon [38] 0.0022 605.0 602.2 0.00695821306 1.00002904

Table 6.1: Physical data of the moons in 1:1 spin–orbit resonance(see also http://ssd.jpl.nasa.gov/?sat_phys_par and http://ssd.jpl.nasa.gov/?sat_elem )

Mercury is apparently trapped in a 3:2 spin–orbit resonance around the Sun. Itscorresponding data are given in Table 6.2.



2

a2−b2

a2+b2

Sun Mercury 0.2056 2440.7 2439.7 0.00061470369 1.255835458

Table 6.2: Physical data of Mercury (from http://nssdc.gsfc.nasa.gov/planetary/factsheet/

mercuryfact.html and http://solarsystem.nasa.gov/planets/charchart.cfm )

No further pairs of planet-satellite (where the mean radius of the minor body is biggerthan 100 Km) are observed in other spin-orbit resonances in the Solar System.

6.2. Theorem for Solar System spin–orbit resonances 95

6.2 Theorem for Solar System spin–orbit resonances

Our main result can now be stated as follows:

Theorem 6.1. [Moons]The differential equation (1.24) admits spin–orbit resonances

(6.1) with p = q = 1 provided e, ν and ε are as in Table 6.1 and 0 ≤ η ≤ 0.008.[Mercury] The differential equation (1.24) admits spin–orbit resonances (6.1) with

p = 3 and q = 2 provided e, ν and ε are as in Table 6.2 and 0 ≤ η ≤ 0.002.

The proof of this theorem follows closely the proof of (the more general) Theorem 1.2in [8], where, however, no explicit computations of constants (size of admissible ε, sizeof admissible η, ...) have been carried out. Thus, the proof presented here, althoughcomplete and self–contained, is somewhat complementary to that presented in [8].The main idea (as in Chapter 3) is to reduce the existence of spin–orbit resonancesto a fixed point problem containing parameters: then such a problem is solved by aLyapunov–Schmidt or “range–bifurcation equation” decomposition (see Chapter 3).The “range equation” is solved by standard contraction mapping method and the“bifurcation equation” is solved by a topological argument.

Clearly, the main point for the application to the Solar System is to compute ev-erything explicitly with attention to optimal estimates so as to include all cases ofphysical interest.


Step 1. Reformulation of the problem of finding spin–orbit resonances

Let x(t) be a p:q spin–orbit resonance and let u(t) := x(qt) − pt − ξ. Then, by (6.1)and choosing ξ suitably one sees immediately that u is 2π–periodic and satisfies thedifferential equation

u′′(t) + η(u′(t)− ν

)+ εfx

(ξ + pt+ u(t), qt

)= 0, 〈u〉 = 0, (6.2)

where 〈·〉 denotes the average over the period1 and

η := qη, ν := qν − p, ε := q2ε. (6.3)

Separating the linear part from the non–linear one, we can rewrite (6.2) as follows:let

Lηu := u′′ + ηu′,

[

Φξ(u)]

(t) := ην − εfx(ξ + pt+ u(t), qt

),

(6.4)

then, the differential equation in (6.2) is equivalent to the functional equation

Lηu = Φξ(u), (6.5)

(see also (3.6)).

1The parameter ξ represents the average of the periodic part of x(t) and will be our “bifurcationparameter”.


Step 2. The Green operator Gη = L−1η

As in Definition 3.2 let B := C0per,0 be the Banach space of 2π–periodic continuous

functions with zero average (endowed with the sup–norm). The following lemma givessharp estimates, which will be used later.

Lemma 6.1.

v ∈ B ∩ C1 =⇒ ‖v‖C0 ≤ π

2‖v′‖C0 (6.6)

v ∈ B ∩ C2 =⇒ ‖v‖C0 ≤ π2

8‖v′′‖C0 (6.7)

Proof. We first prove (6.6). By a rescaling it is sufficient to prove (6.6) with ‖v′‖C0 =1. Assume by contradiction that

‖v‖C0 =: c > π/2.

Note that it is obvious that c ≤ π since v has zero average and, therefore, it must vanishat some point. Up to a translation we can assume that |v| attains its maximum in −c.Furthermore we can also assume that v attains its minimum in −c (if not multiply vby −1), namely

‖v‖C0 = c = −v(−c).

t

v(t)

−2c −c2π − 2c

π/2

−π/2−c

Figure 6.1: The integral over [−2c, 2π − 2c] of a function with values in the filled areais less than 0 (see [44]).

Since ‖v′‖C0 = 1 we get

v(t) ≤ −c+ |t+ c|, ∀t ∈ [−2c, 0],

and, therefore (see Figure 6.1) we have

v(0) ≤ 0, v(−2c) ≤ 0,

∫ 0

−2cv ≤ −c2. (6.8)

Analogously for t ≥ 0, since ‖v′‖C0 = 1 we get

v(t) ≤ π − c− |t− π + c|, ∀t ∈ [0, 2π − 2c].


Then (see again Figure 6.1) we have

∫ 2π−2c

0v ≤ (π − c)2. (6.9)

From (6.8) and (6.9) we get

∫ 2π−2c

−2cv ≤ (π − c)2 − c2 = π(π − 2c) < 0,

which contradicts the fact that v has zero average, proving (6.6).We now prove (6.7). Up to a rescaling we can prove (6.7) assuming ‖v′′‖C0 = 1.Assume by contradiction that

‖v‖C0 =: c > π2/8. (6.10)

Up to a translation we can assume that |v| attains its maximum at 0. Furthermorewe can also assume that v attains its minimum in 0 (if not multiply v by −1), namely

‖v‖C0 = c = −v(0).

t

v(t)

−√2c

√2c 2π −

√2c

π2/8

−π2/8

−c

Figure 6.2: The integral over[−√2c, 2π −

√2c]of a function with values in the filled

area is less than 0.

Since ‖v′′‖C0 = 1 we get

v(t) ≤ −c+ t2/2 ∀t ∈[

−√2c,

√2c]

.

Since v has zero average there must exist (see Figure 6.2)

t1 ≤ −√2c, t2 ≥

√2c, t2− t1 < 2π, s.t. v(t1) = v(t2) = 0, v(t) < 0 ∀t ∈ (t1, t2).

Moreover∫ t2

t1

v ≤∫ √

2c

−√2c−c+ 1

2t2 = −2

3(2c)3/2. (6.11)

Analogously for t ≥√2c, since ‖v′′‖C0 = 1 we get

v(t) ≤ (t−√2c)(2π −

√2c− t)

2, ∀t ∈ [

√2c, 2π −

√2c].


Then (see again Figure 6.2) we have∫ 2π−

√2c

√2c

v ≤ 2(3√2cπ2 + 2c

√2c− 6cπ − π3)

3. (6.12)

From (6.11) and (6.12) we get∫ 2π−

√2c

−√2c

v ≤ −2

3(2c)3/2 +

2(3√2cπ2 + 2c

√2c− 6cπ − π3)

3

= −c3/2 8√2

3+ 4πc− 2

√2π2

√c+

2

3π3

< −(π2

8

)3/28√2

3+ 4π

π2

8− 2

√2π2√

π2

8+

2

3π3

= 0,

which contradicts the fact that v has zero average, proving (6.7).

Recall that the linear operator Lη : C2per,0 → B defined in (6.4) is the same as in (3.6).

From Definition 3.2 and Lemma 3.1 we know that the inverse operator Gη = L−1η is

a bounded linear isomorphism. The following elementary Lemma finds the optimalbound for the norm of the operator Gη. In its proof we will use Lemma 6.1.

Lemma 6.2. Let η < 2/π. Then it follows:

‖Gη‖L(B,B) = supg:‖g‖

C0=1‖Gη(g)‖C0 ≤

(

1 + ηπ

2

(1− η

π

2

)−1)π2

8. (6.13)

In particular, assuming

η ≤ π

5

(10

π2− 1

)

, (6.14)

which by (6.3) is equivalent to

η ≤

π5

(10π2 − 1

)= 0.0083 · · · , if (p, q) = (1, 1),

π10

(10π2 − 1

)= 0.0041 · · · , if (p, q) = (3, 2),

(6.15)

one gets

‖Gη‖L(B,B) ≤5

4. (6.16)

Proof. Given g ∈ B with ‖g‖C0 = 1 we have to prove that if u ∈ B is the uniquesolution of u′′ + ηu′ = g with 〈u〉 = 0, then (6.13) is equivalent to

‖u‖C0 = ‖Gη(g)‖L(B,B) ≤(

1 + ηπ

2

(1− η

π

2

)−1)π2

8. (6.17)

We note that, setting v := u′, we have that v ∈ B and v′ = −ηv + g. Then we get

‖v‖C0

(6.6)

≤ π

2‖ − ηv + g‖C0 ≤ π

2(η‖v‖C0 + 1),

which implies

‖u′‖C0 = ‖v‖C0 ≤(

1− π

2η)−1π

2. (6.18)

Since u′′ = −ηu′ + g, we have

‖u‖C0

(6.7)

≤ π2

8‖ − ηu′ + g‖C0 ≤ π2

8(1 + η‖u′‖C0)

and (6.17) follows by (6.18). Furthermore (6.16) follows directly by (6.13) and (6.14).


Step 3. Lyapunov–Schmidt decomposition

Here we use the main ideas already developed in Section 3.1 for the general spin-orbitproblem. Solutions of (6.5) are recognized as fixed points of the operator Gη Φξ:

u = Gη Φξ(u), (6.19)

where ξ appears as a parameter. To solve equation (6.19), we shall perform a Lyapunov–Schmidt decomposition. Let us denote by Φξ : C0

per → B = C0per,0 (as in Lemma 3.2)

the operator

Φξ(u) :=1

ε

[Φξ(u)− 〈Φξ(u)〉

]

= −fx(ξ + pt+ u(t; ξ), qt

)+ φu(ξ),

where

φu(ξ) :=1

2π

∫ 2π

0fx(ξ + pt+ u(t; ξ), qt

)dt. (6.20)

Then, equation (6.19) can be splitted into a “range equation” (recall (3.9))

u = εGη Φξ(u), (6.21)

(where u = u(·; ξ)) and a “bifurcation (or kernel) equation” (recall (3.8))

φu(ξ) =ην

ε⇐⇒ 〈Φξ

(u(·; ξ

)〉 = 0. (6.22)

Remark 6.1. (i) If (u, ξ) ∈ B× [0, 2π] solves (6.21) and (6.22), then x(t) = ξ + pq t+

u(t/q; ξ) solves (1.24).

(ii) Recall from the proof of Lemma 3.2 that Φξ ∈ C1(B,B) holds ∀ξ ∈ [0, 2π] and∀(u, ξ) ∈ B× [0, 2π] we have

‖Φξ(u)‖C0 ≤ 2 supT2

|fx|, ‖DuΦξ‖L(B,B) ≤ 2 supT2

|fxx|. (6.23)

Analogously as in Subsection 3.1.1, the usual way to proceed to solve (6.21) and (6.22)is the following:

1. for any ξ ∈ [0, 2π], find u = u(·; ξ) solving (6.21);

2. insert u = u(·; ξ) into the kernel equation (6.22) and determine ξ ∈ [0, 2π] sothat (6.22) holds.

Step 4. Solving the range equation (contracting map method)

For ε small the range equation is easily solved by a standard contraction argument.

Let

R :=5

2ε sup

T2

|fx| (6.24)

and let

BR :=v ∈ B : ‖v‖C0 ≤ R

,

ϕ : v ∈ BR → ϕ(v) := εGη Φξ(v).

(6.25)


Lemma 6.3. Assume that η satisfies (6.14) and that

5

2ε sup

T2

|fxx| < 1. (6.26)

Then, for every ξ ∈ [0, 2π], there exists a unique u := u(·; ξ) ∈ BR such that ϕ(u) = u.

Proof. By (6.14) the map ϕ in (6.25) maps BR into itself, since

‖ϕ(v)‖C0(6.25)= ε‖Gη(Φξ(v))‖C0 ≤ ε‖Gη‖L(B,B)‖Φξ(v)‖C0

(6.16)&(6.23)

≤ ε5

42 sup

T2

|fx|(6.24)= R. (6.27)

From the mean-value theorem we have

‖ϕ(u)− ϕ(v)‖C0 ≤ ε‖Gη‖C0‖Φξ(u)− Φξ(v)‖C0

≤ ε5

4‖DuΦξ‖L(B,B)‖u− v‖C0

(6.23)

≤ ε5

2supT2

|fxx|‖u− v‖C0 (6.28)

and so we proved that ϕ is a contraction with Lipschitz constant smaller than 1 by(6.26). The the existence and uniqueness of u := u(·; ξ) ∈ BR follows by the standardfixed point theorem. This terminates the proof of Lemma 6.3.

Remark 6.2. By (1.29), (1.30) we notice that

supT2

|fx| ≤ 1

|1− e|3 , (6.29)

supT2

|fxx| ≤ 2

|1− e|3 , (6.30)

hold. By (6.29) and (6.30) the “range condition” (6.26) follows from

ε <

(1−e)3

5 , if (p, q) = (1, 1),

(1−e)3

20 , if (p, q) = (3, 2).

(6.31)

Step 5. Solving the bifurcation equation (6.22)

From Taylor’s theorem the function φu(ξ) in (6.20) can be written as

φu(ξ) = φ(0)(ξ) + εφ(1)u (ξ; ε), (6.32)

with

φ(0)(ξ) =1

2π

∫ 2π

0fx(ξ + pt, qt)dt,

φ(1)u (ξ; ε) =1

2πε

∫ 2π

0u(t; ξ)

(∫ 1

0fxx(ξ + pt+ su(t; ξ), qt)ds

)

dt.

Since for ε satisfying (6.26) by Lemma 6.3 we have u(t; ξ) = ϕ(u(t; ξ)) ∈ BR, then

supξ∈[0,2π]

|φ(1)u | ≤ supT2

|fxx|R

ε≤ 5

2

(

supT2

|fx|)(

supT2

|fxx|)

(6.33)


holds. From Remark 6.2 the condition (6.26) follows from (6.31). Hence, by (6.29),(6.30) and (6.33) for ε satisfying (6.31) one finds immediately

supξ∈[0,2π]

|φ(1)u | ≤ 5

(1− e)6=:M1. (6.34)

From (3.32) we know that φ(0)(ξ) can also be written as follows:

φ(0)(ξ) =

−2α2(e) sin(2ξ) , if (p, q) = (1, 1),

−2α3(e) sin(2ξ) , if (p, q) = (3, 2),(6.35)

where αj(e) for j ∈ Z are defined in (2.1). Define

apq :=

2|α2(e)| − εM1 , if (p, q) = (1, 1),

2|α3(e)| − εM1 , if (p, q) = (3, 2).(6.36)

Then, from (6.32), (6.34), (6.35) and (6.36), it follows that φu([0, 2π]) contains theinterval [−apq, apq], which is non empty provided (recall (6.3) and (6.34))

ε <

2(1−e)6

5 |α2(e)| , if (p, q) = (1, 1),

(1−e)6

10 |α3(e)| , if (p, q) = (3, 2).

(6.37)

Therefore, we can conclude that the bifurcation equation (6.22) admits a solution ifone assumes |ηνε | ≤ apq, i.e. (recall again (6.3), (6.34) and (6.36)), if

η <

ε|ν−1|

(

2|α2(e)| − 5ε(1−e)6

)

, if (p, q) = (1, 1),

2ε|2ν−3|

(

2|α3(e)| − 20ε(1−e)6

)

, if (p, q) = (3, 2).

(6.38)

We have proven the following:

Proposition 6.1. Let (p, q) = (1, 1) or (p, q) = (3, 2) and assume (6.15), (6.31), (6.37)and (6.38). Then (1.24) admits a p:q spin-orbit resonance.

Step 6. Lower bounds on |α2(e)| and |α3(e)|In order to complete the proof of the Theorem 6.1, checking the conditions of Propo-sition 6.1 for the resonant satellites of the Solar System, we need to give lower boundson the absolute values of the Fourier coefficients α2(e) and α3(e). To do this we willsimply use the Taylor formula to develop αj(e) in powers of e up to a suitably largeorder2

αj(e) =

h∑

k=0

α(k)j ek +R

(h)j (e) (6.39)

and use the analyticity property of the function

Ge(t) := − e2ife(t)

2ρe(t)3=∑

j∈Zαj(e)e

ijt

2We shall choose h = 4 for the 1:1 resonances and h = 21 for the 3:2 case of Mercury.


(already defined in Definition 2.2), to get an upper bound on R(h)j by means of standard

Cauchy estimates for holomorphic functions. To use Cauchy estimates, we need anupper bound of Ge in a complex eccentricity region. The following simple result willbe enough.

Lemma 6.4. Let R(h)j (e) be as in (6.39), 0 < b < 1 and 0 < e < r(b) := b/ cosh b.

Then,

|R(h)j (e)| ≤ R(h)(e; b)

with

R(h)(e; b) :=2

(1− b)5

(

(1 +b

cosh b− e)(1 + cosh b) + 1− b

)2 eh+1

( bcosh b − e)h+1

.

Proof. For e, > 0 we set

[0, e] := z ∈ C|z = z1 + z2, z1 ∈ [0, e], |z2| < .

Re (z)

Im (z)

1e

r

Figure 6.3: The set [0, e] with = r − e (see [44]).

Since 0 < b < 1 and r(b) := bcosh b are as in (4.2), by Lemma 4.1 and standard

(complex) Cauchy estimates imply, for 0 ≤ s ≤ 1,

|Dh+1αj(se)| ≤(h+ 1)!

(r − e)h+1sup

[0,e]r−e

|αj(e)|

and, therefore,

|R(h)j (e)| ≤ eh+1

(r − e)h+1sup[0,e]

|αj(e)| .

By (4.16) we obtain

sup[0,e]

|αj(e)| ≤2

(1− b)5((1 + r − e)(1 + cosh b) + 1− b

)2

from which Lemma 6.4 follows.

Now, in order to check the conditions of Proposition 6.1 we will expand α2(e) inpowers of e up to order h = 4 and α3(e) up to order h = 21 . Using the representationformula (2.14) and Mathematica we find:

α2(e) = −1

2+

5

4e2 − 13

32e4 +R

(4)2 (e) , (6.40)


α3(e) = −7

4e+

123

32e3 − 489

256e5 +

1763

4096e7 − 13527

327680e9 +

180369

13107200e11

+5986093

734003200e13 +

24606987

3355443200e15 +

33790034193

5261334937600e17

+1193558821627

210453397504000e19 +

467145991400853

92599494901760000e21 +R

(21)3 (e) . (6.41)

From (6.40) and (6.41) in view of Lemma 6.4, we choose, respectively, b = 0.462678and3 b = 0.768368 to get lower bounds:

|α2(e)| ≥∣∣∣1

2− 5

4e2 +

13

32e4∣∣∣− |R(4)(e; 0.462678)| (6.42)

|α3(e)| ≥∣∣∣∣∣

21∑

k=0

α(k)3 ek

∣∣∣∣∣− |R(21)(e; 0.768368)| . (6.43)

Step 7. Checking the conditions and conclusion of the proof

We are now ready to check all conditions of Proposition 6.1 with the parameters ofthe satellite in spin–orbit resonance given in Table 6.1 and 6.2.In Table 6.3 we report:

in column 2: the lower bounds on |α 2pq(e)| as obtained in Step 6 using (6.42)

and (6.43) (with the eccentricities listed in Table 6.1 and 6.2);

in column 3: the difference between the right hand side and the left hand sideof the inequality4 (6.31);

in column 4: the difference between the right hand side and the left hand sideof the inequality (6.37);

in column 5: the right hand side and the left hand side of the inequality (6.38),which is an upper bound for the admissible values of the dissipative parameterη.

3The values for b are rather arbitrary (as long as 0 < b < 1); our choice is made for optimizing theestimates.

4Thus, the inequality is satisfied if the numerical value in the column is positive; the same appliesto the fourth and to the fifth column.


Satellite lower bound r.h.s. - l.h.s. r.h.s. - l.h.s. r.h.s. - l.h.s.on |α2p/q| of Eq. (6.31) of Eq. (6.37) of Eq. (6.38)

Moon 0.45475265 0.1663508 0.127144 0.1225335

Io 0.49997893 0.1889174 0.186489 81.800325

Europa 0.49988598 0.1934518 0.187978 1.8031043

Ganymede 0.49999849 0.1974024 0.196745 264.3751

Callisto 0.49993049 0.1937258 0.189389 5.6260606

Mimas 0.49938883 0.0989819 0.088051 19.852395

Enceladus 0.49997228 0.161793 0.159015 218.44519

Tethys 0.49999999 0.1670079 0.166948 458437.46

Dione 0.49999395 0.1901481 0.188837 281.18521

Rhea 0.49999875 0.1895878 0.18899 1554.7362

Titan 0.49776167 0.1830341 0.166905 0.0357326

Iapetus 0.49790449 0.171834 0.155986 2.2484865

Ariel 0.4999982 0.1871186 0.186401 1321.448

Umbriel 0.499998095 0.1833031 0.180992 145.83674

Titania 0.49999849 0.1924958 0.191838 910.34423

Oberon 0.49999755 0.188917 0.188081 826.10305

Miranda 0.49999789 0.1505305 0.149754 3623.6286

Charon 0.49999395 0.1917247 0.190414 231.15781

Mercury 0.27 0.0244515 0.006171 0.0012363

Table 6.3: Check of the hypotheses of Proposition 6.1 for the satellites in spin–orbitresonance

The positive values reported in the fourth and fifth column means that the rangecondition (6.31) and the “topological condition (6.37) are satisfied for all the moonsin 1:1 resonance and for Mercury; the bifurcation condition (6.38) yields an upperbound on the admissible value for η. Thus, η has to be smaller than the minimumbetween the value in the fifth column of Table 6.3 and the value in the right hand sideof Eq. (6.15) (needed to give a bound on the Green operator): such minimum valuesis seen to be 0.008 · · · for the moons in 1:1 resonance and 0.002 · · · for Mercury.

The proof of the Theorem 6.1 is complete.

6.4. Small bodies 105

6.4 Small bodies

In the Solar System besides the eighteen moons listed in Table 6.2 and Mercury thereare other five minor bodies with mean radius smaller than 100 km observed in 1:1spin–orbit resonance around their planets: Phobos and Deimos (Mars), Amalthea(Jupiter), Janus and Epimetheus (Saturn).



2

a2−b2

a2+b2

Mars Phobos [40],[42] 0.0151 13.4 11.2 0.26616393443 1.00136808Deimos [40],[42] 0.0002 7.5 6.1 0.30558527712 1.00000024

Jupiter Amalthea [40] 0.0031 125 73 0.73704304667 1.00005766Saturn Janus [36] 0.0073 97.4 96.9 0.00771996946 1.000319741

Epimetheus [36] 0.0205 58.7 58.0 0.01799421119 1.002521568

Table 6.4: Physical data of minor bodies in 1:1 spin–orbit resonance (see also http:

//solarsystem.nasa.gov/planets/profile.cfm?Object=Mars&Display=Sats).

Besides being small, such bodies have also a quite irregular shape and only Janus andEpimetheus have a good equatorial symmetry 5. Indeed, for these two small moons,Theorem 6.1 holds as shown by the data reported in Table6 6.5:

Satellite lower bound r.h.s. - l.h.s. r.h.s. - l.h.s. r.h.s. - l.h.s.on |α2p/q| of Eq. (6.31) of Eq. (6.37) of Eq. (6.38)

Phobos 0.4996743 −0.075088 −0.08373 −89.23777Deimos 0.49999995 −0.105705 −0.10583 −674530.2

Amalthea 0.49998797 −0.538897 −0.54074 −35210.01Janus 0.4999324 0.1879319 0.183652 23.167321

Epimetheus 0.49927518 0.1699562 0.158377 6.3987689

Table 6.5: Check of the hypotheses of Proposition 6.1 for the small satellites in spin–orbit resonance

5For pictures, see: http://photojournal.jpl.nasa.gov/catalog/PIA10369 (Phobos),http://photojournal.jpl.nasa.gov/catalog/PIA11826 (Deimos),http://photojournal.jpl.nasa.gov/catalog/PIA02532 (Amalthea),http://photojournal.jpl.nasa.gov/catalog/PIA12714 (Janus),http://photojournal.jpl.nasa.gov/catalog/PIA12700 (Epimetheus).

6Positive values in the 3rd, 4th and 5th column means that the hypotheses of the main theorem aremet.


6.5 Eccentricity Intervals

In this section we let the eccentricity e of the orbit of the satellite be a variable. Thisassumption is plausible since we know that the eccentricity e is a function of the time.For example, in [32] it is explained how the eccentricity of Mercury oscillates duringthe last 3 Gyr.Some natural questions now come into mind: for all of the above satellites what are theintervals of eccentricity for which the (1, 1)−periodic orbit exists? And what are theintervals of eccentricity for which the (3, 2)−periodic orbit exists? Are there satellitesfor which both resonances simultaneously exists?Condition (6.31) is weaker than condition (6.37). Therefore, using a truncated versionof α2(e) and of α3(e) at 6-th and 21−th order in the e-expansion, respectively, we canplot in a e − ε plane an approximation of the right hand side of (6.37). These linescorrespond to the conditions for the existence of the (1, 1)− and of the (3, 2)−periodicorbit, respectively.At the same time for every satellite S in Tables 6.1, Table 6.5 and for Mercury wedraw the point (eS , εS) in the e − ε plane corresponding to its values of eccentricityeS and oblateness εS . If the point (eS , εS) stays below the line corresponding to thecondition for the existence of the (1, 1)− or/and the (3, 2)−periodic orbit, then the(1, 1)− or/and the (3, 2)−periodic orbits exist, respectively. Otherwise, if the point(eS , εS) stays above the line corresponding to the condition for the existence of the(1, 1)− or/and the (3, 2)−periodic orbit, then our model says nothing on the existenceof the (1, 1)− or/and the (3, 2)−periodic orbits, respectively. The results are presentedin Figures 6.4 and 6.5.

0.05 0.10 0.15 0.20 0.25 0.30

0.1

0.2

0.3

0.4

0.5

0.6

0.7

Figure 6.4: e− ε plane. Existence of the (1, 1)− and the (3, 2)−periodic orbit.

In the Figure 6.4 one can see that only three points are above the (blue) line, cor-responding to the condition for the existence of a (1, 1)−periodic orbit. These arethe small satellites Phobos, Deimos and Amalthea studied in Section 6.4 (recall Table6.5). For all the other satellites we are considering the (1, 1)−periodic orbit exists.Moreover the Moon, Titan and Mercury have a large eccentricity, which makes themrecognizable in Figure 6.4. They lie clearly below the (red) line, corresponding to thecondition for the existence of (3, 2)−periodic orbits, i.e. for them both resonances arepossible.

6.5. Eccentricity Intervals 107

The region near the origin in Figure 6.4 is densely filled with points and Figure 6.5 isa zoom of this area.

0.05 0.10 0.15 0.20

0.005

0.010

0.015

0.020

Figure 6.5: Zoom in near to the origin of Figure 6.4.

Besides the objects observed in Figure 6.5 one can see: a (yellow) horizontal linerepresenting the maximum of the condition for the existence of (3, 2)−periodic orbits;a (green) horizontal corresponding to ε = εRhea = 0.0098127957; a (blue) horizontalcorresponding to ε = εCallisto = 0.00186698679. Notice that:

• Rhea is the satellite with the lowest value for the oblateness for which the condi-tion for the existence of (3, 2)−periodic orbits is not satisfied. Every satellite Swith εS ≥ εRhea does not fulfil the condition for the existence of (3, 2)−periodicorbits.

• The point of the satellite Europa lies below the condition for the existence of(3, 2)−periodic orbits. So there are four satellites (Europa, Moon, Titan andMercury), for which both (the 1 : 1 and the 3 : 2) periodic resonances exist atthe same time.

• Seven satellites (Io, Ganymede, Callisto, Dione, Janus, Titania and Charon)lie below the yellow horizontal line. Their eccentricity is however too small tosatisfy the condition for the existence of (3, 2)−periodic orbits. Since from [32]we know that the eccentricity varies as a function of time, it is possible that longtime ago some of those seven satellites were in a (3, 2)−periodic resonances. Forexample the point of Callisto is really near to the line of the condition for theexistence of (3, 2)−periodic orbits, suggesting that he is the latest (among thisseven satellites) to have leaved the region below this curve. Unfortunately wecan not verify this assertion, since the astronomical observations do not go thatback in the past.

• To satisfy both conditions a satellite should have a really small value of oblate-ness (i.e. it should be almost spherical) and a relatively large orbital eccentricity.


In Table 6.6, for all the satellites in Tables 6.1, Table 6.5 and for Mercury we give theeccentricity intervals I1 and I2 for the existence of the (1, 1)− and of the (3, 2)−periodicorbits, respectively.

Satellite I1 I2Moon [0, 0.41721847] [0.02, 0.38]

Phobos ∅ ∅Deimos ∅ ∅

Io [0, 0.36] [0.09, 0.19]

Europa [0, 0.41721847]) [0.01, 0.47]

Ganymede [0, 0.41721847] [0.02, 0.41]

Callisto [0, 0.41721847] [0.02, 0.41]

Amalthea ∅ ∅Mimas [0, 0.11] ∅

Enceladus [0, 0.23] ∅Tethys [0, 0.24] ∅Dione [0, 0.36] [0.09, 0.19]

Rhea [0, 0.35] ∅Titan [0, 0.41721847] [0.01, 0.58]

Iapetus [0, 0.34] ∅Janus [0, 0.37] [0.07, 0.22]

Epimetheus [0, 0.3] ∅Ariel [0, 0.33] ∅

Umbriel [0, 0.32] ∅Titania [0, 0.38] [0.06, 0.24]

Oberon [0, 0.35] ∅Miranda [0, 0.19] ∅Charon [0, 0.38] [0.06, 0.24]

Mercury [0, 0.41721847] [0.01, 0.41721847]

Table 6.6: I1 and I2 are the intervals of the eccentricity for the existence of the (1, 1)−and of the (3, 2)−periodic orbits, respectively. Notice that 0.41721847 =

r(0.462678)(4.2)= b

cosh(b) as in (6.42) and (6.43).

7 Estimates on the basin of attraction

This chapter is joint work with L. Biasco and L. Chierchia, Preprint (2014) (see [2]).

In Theorem 1.3 of [8] Biasco and Chierchia give estimates of the basin of attractionof stable (p, 1)− and (p, 2)−periodic orbits for p ∈ Z. In this chapter we present tworesults, which concern this particular theorem, which we write here for completeness:

Theorem 7.1. Let xpq be a (p, q)−periodic orbit with q = 1, 2. Assume that

θ0 :=1

2π

∫ 2π

0fxx(ξ + pt, qt)dt > 0.

Then for ε0 and η0 as in Theorem 4.1 there exist ε⋆ and η⋆, with 0 < ε⋆ ≤ ε0 and0 < η⋆ ≤ η0, and constants 0 < c1 ≤ c2 such that for all ε, η with 0 < ε ≤ ε⋆,0 ≤ η ≤ η⋆ and

η2 < εmin θ0, 1 ,every solution x(t) with initial conditions sufficiently close to the initial conditions ofxpq tends exponentially to xpq(t); more precisely, if x(t) is a solution of (1.24) with

√ε|x(0) − xpq(0)| + |x(0)− xpq(0)| ≤ c1η,

then √ε|x(t)− xpq(t)|+ |x(t)− xpq(t)| ≤ c2ηe

−ηt/2.

The first result of this chapter, Theorem 7.2, is the generalisation of Theorem 7.1to (p, q)−periodic orbits for arbitrary q ≥ 3. The second result, Theorem 7.3 is aquantitative reformulation of Theorem 7.1. This makes concrete applications possible.We will present the case of Titan (as satellite of Saturn).

7.1 Main results and applications

As already mentioned before, for q = 1, 2 estimates on the basin of attraction ofperiodic orbits are given in Theorem 7.1. In the following theorem we give estimateson the basin of attraction of xpq for q ≥ 3.

Theorem 7.2. Let q ≥ 3 and (p, q) = 1. Let xpq as in (3.5) be a (p, q)−periodicsolution of the spin-orbit problem. Assume that xpq is non-degenerate at kth-degreefor some k ≥ 1 (according to Definition 3.3) and

θk := ∂ξφ(k)(ξ) > 0. (7.1)

Then there exist positive ε⋆ and η⋆ and constants 0 < c1 ≤ c2 such that for all ε, ηwith 0 < ε ≤ ε⋆, 0 ≤ η ≤ η⋆ and

η2 < q2ε2 min([q2ε]k−1

θk, 4), (7.2)

109

110 Chapter 7. Estimates on the basin of attraction

every solution x(t) with initial conditions sufficiently close to the initial conditions ofxpq tends exponentially to xpq(t); more precisely, if x(t) is a solution of (1.24) with

|x(0) − xpq(0)| +|x(0)− xpq(0)|

qε≤ c1η,

then

|x(t)− xpq(t)|+|x(t)− xpq(t)|

qε≤ c2ηe

−ηt/2, ∀t ≥ 0.

The following theorem is a quantitative reformulation of Theorem 7.1. Besides theassumptions (7.3) - (7.5), which were already taken in Theorem 7.1, we have an extracondition (equation (7.6)) on ε. Under these assumptions we give explicit formulasfor the constants c1 and c2 in equation (7.7), which give a lower estimate of the basinof attraction of a (p, 1)− or a (p, 2)−periodic orbit.

Theorem 7.3. Let xpq be a (p, q)−periodic orbit with q = 1, 2 as in (3.5). Assumethat

θ0 :=1

2π

∫ 2π

0fxx(ξ + pt, qt)dt > 0. (7.3)

Furthermore define:

K := 2 supt∈[0,2π]

|fxx(x(t), qt)|,

K := 1 +4K

π2θ0,

λ1 :=

√

1

8π2K,

M1 := 8π2Kλ21 + λ3164π3K

(

2π

∣∣∣∣

2iλ11− e−i4πλ1

∣∣∣∣+ 1

)(

1 + πKλ1

)

,

M2 := λ1

[

2πK + 2λ1M1(1 + πKλ1)]

,

C := (1 +M1)

(

1 +λ1 +M2

|λ1 −M2|

)

,

C ′ := (2M2 + λ1(1 +M1))

(

1 +λ1 +M2

|λ1 −M2|

)

,

S :=1 +M1

|λ1 −M2|,

S′ :=2M2 + λ1(1 +M1)

|λ1 −M2|,

c1 := supt∈[0,2π]

supζ∈[x(t),x(t)+w(t)]

|fxxx(ζ, qt)|2

,

c2 := max

(

C,

√2

π√θ0S,

√

576π2θ047

C ′, S′)

.

Then for ε, ε := q2ε and η so small such, that

η2 < εmin θ0, 1 , (7.4)

1

2θ0 ≤ θ := θ(ε) :=

1

2π

∫ 2π

0fxx(xpq(qt), qt)dt ≤ 2θ0 (7.5)


and

ε ≤ min

1

8θ0π2(

1 + 8Kθ0

) [

4π(

1 + 8Kθ0

)

+ 1] ,

2λ21π2θ0

(7.6)

hold, every solution x(t) with initial conditions sufficiently close to the initial condi-tions of xpq tends exponentially to xpq(t); more precisely, if x(t) is a solution of (1.24)with √

ε|x(0) − xpq(0)| + |x(0)− xpq(0)| ≤ c1η,


−ηt/2,

where

c1 =1

64c1c32

and c2 =3

8c1c22

. (7.7)

Remark 7.1. Theorem 7.3 can be applied to Titan (a satellite of Saturn). For Titanwe have eTitan = 0.0288, εTitan = 1.7882901 · 10−4. Numerical approximations showsthat

θ0 = 1.995068948321280 > 0

andθ(εTitan) = 1.997250241599330,

therefore conditions (7.3) and (7.5) are satisfied. By (1.25) we have

ηTitan ≈ 10−3 · ΩeTitan

=

(

1 + 3e2Titan +3

8e4Titan

)1

(1− e2Titan

)9/2= 0.001006238896743.

Condition (7.4) holds since

1.012516717318596 · 10−6 ≈ η2Titan < εTitan min θ0, 1 ≈ 3.567762049100357 · 10−4.

Computing the right hand side of (7.6) we get that

min

1

8θ0π2(

1 + 8Kθ0

) [

4π(

1 + 8Kθ0

)

+ 1] ,

2λ21π2θ0

≈ 4.679380623399785 · 10−4,

which is larger than εTitan. Therefore also (7.6) is satisfied. Then by (7.7) we getc1 = 0.026316630911581 and c2 = 93.314130269171429.


7.2.1 Introduction

We consider orbits x(t) starting near the (p, q)−periodic solution xpq(t), cf. (3.5), i.e.

x(t) = xpq(t) + w(t). (7.8)

Inserting x(t) in equation (1.24) we get:

(xpq(t) + w(t))′′ + η((xpq(t) + w(t))′ − ν) + εfx(xpq(t) + w(t), t) = 0


x′′pq(t) + η((xpq(t)− ν)︸︷︷︸

=−εfx(x,t)

+w′′(t) + ηw(t)′ + εfx(xpq(t) + w(t), t) = 0.

Setting w(t) := w(qt) and replacing t by qt, the condition that x(t) is a solution of(1.24) implies

w′′(t) + ηw′(t) + εfx(x(t) + w(t), qt) − εfx(x(t), qt) = 0,

where

x(t) = xpq(qt) = ξ + pt+ u(t), (7.9)

η = qη, (7.10)

ε = q2ε. (7.11)

Defining

(Q(w)) (t) := fx(x(t) + w(t), qt)− fx(x(t), qt)− fxx(x(t), qt)w(t),γ(t) := fxx(x(t), qt)− θ,α := η

2 ,(7.12)

for θ as in (7.5) we get

w′′(t) + ηw′(t) + εfx(x(t) + w(t), qt) − εfx(x(t), qt) = 0

⇔ w′′(t) + ηw′(t) + ε (Q(w)) (t) + εfxx(x(t), qt)w(t) = 0

⇔ w′′(t) + ηw′(t) + ε(θ + γ(t))w(t) = −ε (Q(w)) (t).

Setting w(t) = z(t)e−αt the last equation is equivalent to

Lz := z′′ +((εθ − α2) + εγ(t)

)z = −εeαt

(Q(e−αtz)

)(t). (7.13)

Remark 7.2. Q is a quadratic operator, i.e. there exists a constant c1 (dependingonly on f) such that

|(Q(w))(t)| ≤ c1|w(t)|2, ∀t ∈ R.

From the Taylor theorem we get

c1 := supt∈[0,2π]

supζ∈[x(t),x(t)+w(t)]

|fxxx(ζ, qt)|2

.

Remark 7.3. From (3.14) also the function u(t; ξ, ε) can be written in the form

u(t; ξ, ε) =

∞∑

k=1

ukεk.

Using Taylor expansion we define the following formal series in ε:

fxx(x(t), qt) =: f(0)2 + εf

(1)2 + ε2f

(2)2 + . . . ,

θ = θ(ε)(7.5)=

1

2π

∫ 2π

0fxx(x(t), qt)dt =: θ0 + εθ1 + ε2θ2 + . . . ,

γ = γ(t)(7.12)= fxx(x(t), qt)− θ =: γ0 + εγ1 + ε2γ2 + . . . .

Remark 7.4. For φu(ξ) in (3.8) notice that

∂ξφu(ξ) = ∂ξ1

2π

∫ 2π

0fx(x(t), qt)dt =

1

2π

∫ 2π

0fxx(x(t), qt)dt

implies ∂ξφ(l)(ξ) = θl for every Z ∋ l ≥ 1.


7.2.2 The homogeneous equation Lz = 0

The substitution u1 = z, u2 = z′ takes the homogeneous equation Lz = 0 to

(u′1u′2

)

=

(u2

−[

(εk+1θ − α2) + εγ(t)]

u1

)

, (7.14)

where θ := θk + εθk+1 + ε2θk+2 + . . . for θl for l ∈ N as in Remark 7.3. Since ε = q2ε,η = qη and α = η

2 hold, equation (7.2) implies

η2

4= α2 <

εk+1θk4

. (7.15)

The fundamental solution of (7.14), i.e. the two-by-two matrix U satisfying

U ′ =

(0 1

−[

(εk+1θ − α2) + εγ(t)]

0

)

U, with U(0) = Id, (7.16)

is given by

U(t) := U(t; ε) :=

(c(t) s(t)c′(t) s′(t)

)

. (7.17)

We rewrite (7.16) as follows:

U ′ = AU, with U(0) = Id, (7.18)

where

A := T + εD0(t) + ε2D1(t) + . . .+ εkDk−1(t) + εk+1E +O(εk+2),

T :=

(0 10 0

)

,

E(t) :=

(

0 0α2

εk+1 − f(k)2 0

)

,

Dl(t) :=

(0 0

−γl(t) 0

)

, for 0 ≤ l ≤ k − 1.

Lemma 7.1. Let

∆ := −2π

∫ 2π

0(Γ0(s))

2 ds +

(∫ 2π

0Γ0(s)ds

)2

, (7.19)

where Γ0 is defined by

Γ0(t) :=

∫ t

0γ0(s)ds. (7.20)

Then we have ∆ < 0.

Proof. The proof follows from the Cauchy-Schwarz inequality

(∫ b

af(s)g(s)ds

)2

≤∫ b

af(s)2ds

∫ b

ag(s)2ds,

taking f(s) = Γ0(s), g(s) ≡ 1, a = 0 and b = 2π.

In Subsection 7.2.3 we show that the trace tr(U(2π)) of U(2π) plays a crucial role forthe determination of the fundamental solutions of the homogeneous equation Lz = 0.In the following lemma we find a formula for tr(U(2π)) depending on the degree ofnon-degeneration k of the periodic orbit (according to Definition 3.3).


Lemma 7.2. Let (θl)l≥0 be as in Remark 7.3. Let U(t; ε) be the solution of (7.18).Assume (7.15), θ0 = 0, θ1 = 0, . . . , θk−1 = 0 and θk > 0 with k ≥ 1. Then

tr(U(2π)) =

2 + ε2(

−4π2(

θ1 − α2

ε2

)

+∆)

+O(ε3), if k = 1,

2 + ε2∆+O(ε3), if k ≥ 2,

holds, where ∆ is defined in (7.19).

Proof. Introducing the formal expansion U(t, ε) := U0(t)+εU1(t)+ε2U2(t)+ . . . and

comparing terms in equation (7.18) of the same order in ε we have

U ′0 = TU0,

U ′1 = TU1 +D0U0,

U ′2 = TU2 +D0U1 +D1U0,

...

U ′k = TUk +D0Uk−1 +D1Uk−2 + . . .+Dk−1U0,

U ′k+1 = TUk+1 +D0Uk +D1Uk−1 + . . .+Dk−1U1 + EU0.

• The system of ordinary differential equations of zero degree in ε can be easelysolved:

U ′0 = TU0, U0(0) = Id =⇒ U0(t) = eTt =

(1 t0 1

)

.

• For the term of order one in ε we have

U ′1 = TU1 +D0U0, with initial condition U1(0) = 0.

Using the variation of constant formula1 we get

U1(t) = eTt

∫ t

0e−TsD0(s)e

Tsds

= eTt

∫ t

0

(1 −s0 1

)(0 0

−γ0(s) 0

)(1 s0 1

)

ds

= eTt

∫ t

0

(1 −s0 1

)(0 0

−γ0(s) −sγ0(s)

)

ds

= eTt

∫ t

0

(sγ0(s) s2γ0(s)−γ0(s) −sγ0(s)

)

ds.

Since by assumption 〈γ0〉 = θ0 = 0, it follows that

U1(2π) =

(1 2π0 1

)∫ 2π

0

(sγ0(s) s2γ0(s)−γ0(s) −sγ0(s)

)

ds

=

(1 2π0 1

)(a b0 −a

)

=

(a b− 2πa0 −a

)

,

where a :=∫ 2π0 sγ0(s)ds and b :=

∫ 2π0 s2γ0(s)ds. Therefore U1(2π) has zero

trace, i.e.tr (U1(2π)) = a− a = 0.

1Let y′(t)+a(t)y(t) = f(t) be a first order inhomogeneous linear ordinary differential equation. Thehomogeneous solution is given by yhom(t) = ce−A(t), where A(t) is the primitive of a(t). By thevariation of constants formula one gets the solution y(t) = e−A(t)

∫ t

0f(s)eA(s)ds.


Now we have to distinguish two cases:Case 1: k = 1, i.e. q = 4.

• For the term of order two in ε (because k = 1), we have

U ′2 = TU2 +D0U1 + EU0, with initial condition U2(0) = 0.

Again using the variation of constant formula we get

U2(t) =eTt

∫ t

0e−Ts [D0(s)U1(s) + E(s)U0(s)] ds

=eTt

∫ t

0

(1 −s0 1

)

·

·[(

0 0−γ0(s) 0

)(1 s0 1

)∫ s

0

(τγ0(τ) τ2γ0(τ)−γ0(τ) −τγ0(τ)

)

dτ

+

(

0 0

−(f(1)2 (s)− α2

ε2) 0

)(1 s0 1

)]

ds

=eTt

∫ t

0

(1 −s0 1

)[(0 0

−γ0(s) −sγ0(s)

)∫ s

0

(τγ0(τ) τ2γ0(τ)−γ0(τ) −τγ0(τ)

)

dτ+

+

(0 0

−c(s) −c(s)s

)]

ds

=eTt

∫ t

0

(sγ0(s) s2γ0(s)−γ0(s) −sγ0(s)

)∫ s

0

(τγ0(τ) τ2γ0(τ)−γ0(τ) −τγ0(τ)

)

dτds+

+ eTt

∫ t

0

(c(s)s c(s)s2

−c(s) −c(s)s

)

ds,

where c(t) := f(1)2 (t)− α2

ε2. We compute now the trace of U2 for t = 2π:

tr(U2(2π)) = tr(I) + tr(II),

where

I :=

(1 2π0 1

)∫ 2π

0

sγ0(s) s2γ0(s)−γ0(s) −sγ0(s)

∫ s

0

τγ0(τ) τ2γ0(τ)−γ0(τ) −τγ0(τ)

dτds,

II :=

(1 2π0 1

)∫ 2π

0

c(s)s c(s)s2

−c(s) −c(s)s

ds.

For the term II we have

II =

(1 2π0 1

)

∫ 2π0 c(s)sds

∫ 2π0 c(s)s2ds

∫ 2π0 −c(s)ds

∫ 2π0 −c(s)sds

=

( ∫ 2π0 c(s)sds − 2π

∫ 2π0 c(s)ds ⋆

⋆ −∫ 2π0 c(s)sds

)

.

Therefore

tr(II) = −2π

∫ 2π

0c(s)ds = −2π

∫ 2π

0

[

f(1)2 (s)− α2

ε2

]

ds


= −2π

(

2πθ1 − 2πα2

ε2

)

= −4π2(

θ1 −α2

ε2

)

,

where we used the fact that∫ 2π

0f(1)2 (t)dt =

∫ 2π

0fxxx(ξ + pt, qt)dt = ∂ξ

∫ 2π

0fxx(ξ + pt, qt)dt = 2πθ1.

The term I is more complicated to compute:

I =

(1 2π0 1

)∫ 2π

0

(sγ0(s) s2γ0(s)−γ0(s) −sγ0(s)

)∫ s

0

(τγ0(τ) τ2γ0(τ)−γ0(τ) −τγ0(τ)

)

dτds

=

∫ 2π

0

∫ s

0γ0(s)γ0(τ)

(τs− 2πτ − s2 + 2πs ⋆

⋆ −τ2 + τs

)

dτds.

Therefore we have

tr(I) =

∫ 2π

0

∫ s

0γ0(s)γ0(τ)

[τs− 2πτ − s2 + 2πs− τ2 + τs

]dτds

=

∫ 2π

0

∫ s

0γ0(s)γ0(τ)

[2π(s − τ)− (s− τ)2

]dτds

= 2π

∫ 2π

0γ0(s)

∫ s

0γ0(τ)(s − τ)dτ

︸︷︷︸

=:III

ds−∫ 2π

0γ0(s)

∫ s

0γ0(τ)(s − τ)2dτ

︸︷︷︸

=:IV

ds.

The integrals III and IV can be computed by partial integration. Defining

Λ0(t) :=

∫ t

0Γ0(s)ds

with Γ0 in (7.20) and recalling the obvious Γ0(2π)− Γ0(0) = Γ0(2π) = 0 we get

III =

∫ s

0γ0(τ)(s − τ)dτ = Γ0(τ)(s − τ)|s0

︸︷︷︸

=0

+

∫ s

0Γ0(τ)dτ =

∫ s

0Γ0(τ)dτ

and

IV =

∫ s

0γ0(τ)(s − τ)2dτ = Γ0(τ)(s − τ)2|s0

︸︷︷︸

=0

+2

∫ s

0Γ0(τ)(s− τ)dτ

= 2

Λ0(τ)(s − τ)|s0︸︷︷︸

=0

+

∫ s

0Λ0(τ)dτ

= 2

∫ s

0Λ0(τ)dτ.

Again by partial integration we have

tr(I) = 2π

∫ 2π

0γ0(s)

∫ s

0Γ0(τ)dτds − 2

∫ 2π

0γ0(s)

∫ s

0Λ0(τ)dτds

= 2π

Γ0(s)

∫ s

0Γ0(τ)dτ

∣∣2π0

︸︷︷︸

=0

−∫ 2π

0Γ0(s)

2ds

−2

Γ0(s)

∫ s

0Λ0(τ)dτ |2π0

︸︷︷︸

=0

−∫ 2π

0Γ0(s)Λ0(s)ds


= −2π

∫ 2π

0Γ0(s)

2ds+Λ20(2π) − Λ2

0(0)

2

= −2π

∫ 2π

0Γ0(s)

2ds+

(∫ 2π

0Γ0(s)ds

)2(7.19)= ∆.

Case 2: k ≥ 2, i.e. q ≥ 3 and q 6= 4.

• For the second degree in ε (because k ≥ 2) we have

U ′2 = TU2 +D0U1 +D1U0, with initial condition U2(0) = 0.

Using the variation of constant formula we get

U2(t) = eTt

∫ t

0e−Ts [D0(s)U1(s) +D1(s)U0(s)] ds

= eTt

∫ t

0e−TsD0(s)U1(s)ds+ eTt

∫ t

0e−TsD1(s)U0(s)ds

= eTt

∫ t

0

(1 −s0 1

)(0 0

−γ0(s) 0

)(1 s0 1

)

·

·∫ s

0

(τγ0(τ) τ2γ0(τ)−γ0(τ) −τγ0(τ)

)

dτds + eTt

∫ t

0

(sγ1(s) s2γ1(s)−γ1(s) −sγ1(s)

)

ds

= eTt

∫ t

0

(sγ0(s) s2γ0(s)−γ0(s) −sγ0(s)

)∫ s

0

(τγ0(τ) τ2γ0(τ)−γ0(τ) −τγ0(τ)

)

dτds +

+eTt

∫ t

0

(sγ1(s) s2γ1(s)−γ1(s) −sγ1(s)

)

ds.

We compute now the trace of U2 for t = 2π. So we have

tr(U2(2π)) = tr(I) + tr(II),

where

I :=

(1 2π0 1

)∫ 2π

0

sγ0(s) s2γ0(s)−γ0(s) −sγ0(s)

∫ s

0

τγ0(τ) τ2γ0(τ)−γ0(τ) −τγ0(τ)

dτds,

II :=

(1 2π0 1

)∫ 2π0

(sγ1(s) s2γ1(s)−γ1(s) −sγ1(s)

)

ds.

Since by assumption 〈γ1〉 = θ1 = 0, it follows that

II =

(1 2π0 1

)∫ 2π

0

(sγ1(s) s2γ1(s)−γ1(s) −sγ1(s)

)

ds

=

(1 2π0 1

)(a b0 −a

)

=

(a b− 2πa0 −a

)

,

where a :=∫ 2π0 sγ1(s)ds and b :=

∫ 2π0 s2γ1(s)ds.

This implies that the trace of II is zero. Notice that I is equal to I, defined inthe previous subsection. It follows that

tr(U2(2π)) = tr(I) = −2π

∫ 2π

0(Γ0(s))

2 ds+

(∫ 2π

0Γ0(s)ds

)2(7.19)= ∆.



7.2.3 Fundamental solutions of the homogeneous equation Lz = 0

We define 2δ := tr(U(2π)).

Remark 7.5. From Lemma 7.1, 7.2 and condition (7.15) we get 0 < δ < 1 for ε smallenough.

We interpret the second order differential equation Lz = 0 as a Hill equation z′′ +A(t)z = 0 with A(t+ 2π) = A(t) := (εθ − α2) + εγ(t) for all t ∈ R. From the FloquetTheory (see [22], [46]) we know that

tr(U(2π)) = c(2π) + s′(2π),

where c(t), s(t) are the fundamental solutions of Lz = 0 with initial data

c(0) = 1 = s′(0) and c′(0) = 0 = s(0).

The characteristic equation has the form

ρ2 −[c(2π) + s′(2π)

]ρ+ 1 = 0 (7.21)

ρ2 − 2δρ + 1 = 0

and we find

ρ± = δ ±√

δ2 − 1.

Since 0 < δ < 1 the two solutions ρ± are complex conjugate. By [46] (see Theorem atp. 4) the equation Lz = 0 has two independent solutions of the form

z±(t) = eiλtP±(t), (7.22)

where P±(t) are 2π−periodic functions and λ satisfies

cos(2πλ) ± i sin(2πλ) = e±i2πλ = ρ± = δ ±√

δ2 − 1.

Remark 7.6. Comparing the imaginary parts it is easy to see that λ ∼ ε for bothcases (k = 1 and k ≥ 2), i.e. there exist constants c, c ∈ R+ such that |ε| ≤ c|λ| and|λ| ≤ c|ε| for all k ≥ 1.

We define

ω2(ε) := εθ − α2 = εk+1θk − α2 +O(εk+2). (7.23)

Remark 7.7. Of course ω2 ∼ εk+1.

We can resume the results of this section in the following lemma:

Lemma 7.3. Let ω(ε) be as in (7.23). For ω(ε) > 0 small enough there exists 0 < δ <1 such that the solutions of the characteristic equation (7.21) associated to Lz = 0 aregiven by ρ± = δ ±

√δ2 − 1 and, hence, are distinct. Thus λ in (7.22) is real, λ ∼ ε

and ω2 ∼ εk+1 and all solutions of Lz = 0 are bounded together with their derivatives.Finally δ and λ smoothly depend on ε.


7.2.4 Fixed point argument

ζ+(t) = eiλtP+(t) is a solution2 of Lz = 0 with Dirichlet boundary conditions

P+(0) = P+(2π) = 1

if and only if P+ satisfies

P+(t) + 2iλP+(t) +[(εθ − α2) + εγ(t)− λ2

]P+(t) = 0.

Defining γ(t) := γ(t)− γ0(t)− εγ1(t)− . . .− εk−1γk−1(t) (see Remark 7.3) we get

0 = P+(t) + 2iλP+(t) +

[

ω2(1 +εk+1

ω2γ(t))− λ2

]

P+(t)+

+ εγ0(t)P+(t) + . . .+ εkγk−1(t)P+(t)

0 = P+(t) + 2iλP+(t) + λ2G(ε; t)P+(t) + εγ0(t)P+(t) + . . .+ εkγk−1(t)P+(t),

(7.24)

where G(ε; t) := ω2

λ2

(

1 + εk+1

ω2 γ(t))

− 1. We notice that G(ε; t) = −1 + O(εk−1)

smoothly depends on ε. Defining ζ−(t) = e−iλtP+(t) it follows that ζ−(t) is a solutionof Lz = 0. The function P−(t) := P+(t) satisfies the equation

0 = P−(t)− 2iλP−(t) + λ2G(ε; t)P−(t) + εγ0(t)P−(t) + . . .+ εkγk−1(t)P−(t). (7.25)

Lemmata 7.4 - 7.8 are devoted to prove that:

• P+(t) exists and can be extended to a 2π−periodic function over R;

• For P±(t) as in (7.22) there are constants c± ∈ C such that P±(t) = c±P±(t)holds;

• P−(t) = P+(t).

These results will be very useful in Lemma 7.10 in order to understand the behaviourof P±(t) = P±(t;λ) in terms of the parameter λ (cf. equation (7.29)).

Lemma 7.4. Let 0 < |λ| < 12 . Then the two solutions ζ±(t) = eiλtP+(t) are linearly

independent.

Proof. By contradiction assume that there exists c ∈ C such that ζ−(t) = c ζ+(t).Then for t ∈ [0, 2π] we have

e−iλtP+(t) = ceiλtP+(t)

P+(t) = ce2iλtP+(t).

Choosing t = 0 we get c = 1. Analogously for t = 2π it follows

1 = 1 = P+(2π) = ei4λπP+(2π) = ei4λπ,

which is a contradiction, since 0 < |λ| < 12 .

Lemma 7.5. Let 0 < |λ| < 12 . P+(t) and P−(t) := P+(t) can be extended on R to two

2π−periodic functions, such that equations (7.24) and (7.25) still hold.

2The existence of P+(t) will be proved in Lemma 7.7.


Proof. Assume that z±(t) = e±iλtP±(t) are independent solutions of Lz = 0 on R.By hypothesis ζ+(t) = eiλtP+(t) is a solution of Lz = 0 on the interval [0, 2π]. FromLemma 7.4 we know that ζ±(t) = e±iλtP±(t) with P−(t) := P+(t) are independent.Thus on t ∈ [0, 2π] there exists c±, d± ∈ C such that

eiλtP+(t) = c+eiλtP+(t) + c−e

−iλtP−(t),

e−iλtP−(t) = d+eiλtP+(t) + d−e

−iλtP−(t).

For t ∈ [0, 2π] we can therefore write

P+(t) = c+P+(t) + c−e−i2λtP−(t),

P−(t) = d+ei2λtP+(t) + d−P−(t).

Setting t = 0, 2π and defining C± := c±P±(0), D± := d±P±(0) we get

C+ + C− = 1, D+ +D− = 1, C+ + C−e−i4πλ = 1, D+e

i4πλ +D− = 1.

Subtracting the first equation from the third and the second from the fourth yieldsC−(e−i4πλ − 1

)= 0,

D+

(ei4πλ − 1

)= 0.

Since by hypothesis 0 < |λ| < 12 it follows that

(e±i4πλ − 1

)6= 0. This implies that

C− = D+ = 0. So we have

c−P−(0) = 0 and d+P+(0) = 0.

Since z± are independent solution, P+(0) and P−(0) cannot be both zero, thereforec−d+ = 0. If c− = 0 we have P+(t) = c+P+(t). This implies that P+(t) andP−(t) := P+(t) can be extended to two 2π−periodic functions defined on R, whichsolves (7.24) and (7.25). The same arguments also hold for d+ = 0.

The following lemma will be used in the proof of Lemma 7.7.

Lemma 7.6. Let f : [0, 2π] → C, t 7→ f(t) be a continuous function. Then thefollowing holds:

∥∥∥∥∥

∫ t

0f(s)ds

∥∥∥∥∥C0([0,2π],C)

≤ 2π‖f‖C0([0,2π],C),

∥∥∥∥∥

∫ t

0

∫ s

0f(τ)dτds

∥∥∥∥∥C0([0,2π],C)

≤ 4π2‖f‖C0([0,2π],C).

Proof. Recall that for any f ∈ C0([0, 2π],C) we define ‖f‖C0([0,2π]) := supt∈[0,2π] |f(t)|.Therefore we have∥∥∥∥∥

∫ t

0f(s)ds

∥∥∥∥∥C0([0,2π],C)

= supt∈[0,2π]

∣∣∣∣

∫ t

0f(s)ds

∣∣∣∣≤ sup

t∈[0,2π]

∫ t

0|f(s)|ds

≤∫ 2π

0|f(s)|ds ≤ 2π sup

t∈[0,2π]|f(t)| = 2π‖f‖C0([0,2π],C)

and∥∥∥∥∥

∫ t

0

∫ s

0f(τ)dsdτ

∥∥∥∥∥C0([0,2π],C)

= supt∈[0,2π]

∣∣∣∣

∫ t

0

∫ s

0f(τ)dsdτ

∣∣∣∣


≤ supt∈[0,2π]

∫ t

0

∫ s

0|f(τ)|dsdτ

≤∫ 2π

0

∫ 2π

0|f(τ)|dsdτ

≤ 4π2 supt∈[0,2π]

|f(t)| = 4π2‖f‖C0([0,2π],C).

Finally we can state the following fixed point lemma.

Lemma 7.7. The function P+(t) : [0, 2π] → C is a solution of the equation P = ψ(P),where ψ : C0([0, 2π],C) → C0([0, 2π],C) is defined by

[ψ(P;λ)] (t) := 1 + λv0t− 2iλ

∫ t

0[P(s)− 1] ds− λ2

∫ t

0

∫ s

0G(ε, ξ)P(ξ)dξds +

−ε∫ t

0

∫ s

0γ0(ξ)P(ξ)dξds − . . .− εk

∫ t

0

∫ s

0γk−1(ξ)P(ξ)dξds,

(7.26)

where

v0(P;λ) :=i

π

∫ 2π

0[P(s)− 1] ds+

λ

2π

∫ 2π

0

∫ s

0G(ε, ξ)P(ξ)dξds +

+ε

2πλ

∫ 2π

0

∫ s

0γ0(ξ)P(ξ)dξds + . . .+

εk

2πλ

∫ 2π

0

∫ s

0γk−1(ξ)P(ξ)dξds.

(7.27)

Proof. Integrating (7.24) and using the fact that P+(0) = 1 we get

P+(t) = −2iλP+(t)− λ2G(ε; t)P+(t) +

−εγ0(t)P+(t)− . . .− εkγk−1(t)P+(t)

P+(s)− P+(0) = −2iλ [P+(s)− P+(0)]− λ2∫ s

0G(ε, ξ)P+(ξ)dξ +

−ε∫ s

0γ0(ξ)P+(ξ)dξ − . . . − εk

∫ s

0γk−1(ξ)P+(ξ)dξ

P+(s) = P+(0) − 2iλ [P+(s)− 1]− λ2∫ s

0G(ε, ξ)P+(ξ)dξ +

−ε∫ s

0γ0(ξ)P+(ξ)dξ − . . . − εk

∫ s

0γk−1(ξ)P+(ξ)dξ.

Integrating again we obtain

P+(t) = 1 + P+(0)t− 2iλ

∫ t

0[P+(s)− 1] ds− λ2

∫ t

0

∫ s

0G(ε, ξ)P+(ξ)dξds +

−ε∫ t

0

∫ s

0γ0(ξ)P+(ξ)dξds − . . .− εk

∫ t

0

∫ s

0γk−1(ξ)P+(ξ)dξds.

Since the function P+(t) is 2π−periodic, we have P+(2π) = P+(0) = 1. This conditionimplies

1 = 1 + P+(0)2π − 2iλ

∫ 2π

0[P+(s)− 1] ds− λ2

∫ 2π

0

∫ s

0G(ε, ξ)P+(ξ)dξds


−ε∫ 2π

0

∫ s

0γ0(ξ)P+(ξ)dξds − . . .− εk

∫ 2π

0

∫ s

0γk−1(ξ)P+(ξ)dξds

2πP+(0) = 2iλ

∫ 2π

0[P+(s)− 1] ds+ λ2

∫ 2π

0

∫ s

0G(ε, ξ)P+(ξ)dξds

+ε

∫ 2π

0

∫ s

0γ0(ξ)P+(ξ)dξds + . . .+ εk

∫ 2π

0

∫ s


P+(0) = λ

[i

π

∫ 2π

0[P+(s)− 1] ds +

λ

2π

∫ 2π

0

∫ s

0G(ε, ξ)P+(ξ)dξds+

+ε

2πλ

∫ 2π

0

∫ s

0γ0(ξ)P+(ξ)dξds +. . .+

εk

2πλ

∫ 2π

0

∫ s


]

=: λv0(P+; λ),

and so equations (7.26) and (7.27) are satisfied.

Lemma 7.8. For λ small enough ψ(P+; λ) defined in Lemma 7.7 is a contraction inthe space C0([0, 2π] ,C).

Proof. Furthermore the function ψ(P, λ) smoothly depends on ε (because λ(ε) ∼ ε).We need to show that ψ is a contraction. For P+, P+ ∈ C0([0, 2π] ,C)

∥∥∥∥[ψ(P+;λ)] (t)−

[

ψ(P+;λ)]

(t)

∥∥∥∥C0([0,2π])

≤

≤ 2λ

∥∥∥∥

∫ t

0(P+ − P+)ds

∥∥∥∥C0([0,2π])

+ λ2∥∥∥∥

∫ t

0

∫ s

0G[

P+ − P+

]

dξds

∥∥∥∥C0([0,2π])

+ε

∥∥∥∥

∫ t

0

∫ s

0γ0

[

P+ − P+

]∥∥∥∥C0([0,2π])

+ . . .+ εk∥∥∥∥

∫ t

0

∫ s

0γk−1

[

P+ − P+

]∥∥∥∥C0([0,2π])

.

Using Lemma 7.6 we have∥∥∥∥[ψ(P+;λ)] (t)−

[

ψ(P+;λ)]

(t)

∥∥∥∥C0([0,2π])

≤

≤ 4πλ‖P+ − P+‖C0([0,2π]) + 4π2λ2∥∥∥∥G[

P+ − P+

]∥∥∥∥C0([0,2π])

+

+4π2ε

∥∥∥∥γ0

[

P+ − P+

]∥∥∥∥C0([0,2π])

+ . . . + 4π2εk∥∥∥∥γk−1

[

P+ − P+

]∥∥∥∥C0([0,2π])

≤ 4πλ

[

1 + λπmax[0,2π]

∣∣∣∣

ω2

λ2

(

1 +4|γ(t)|θk

)

+ 1

∣∣∣∣+

+ε

λπmax

[0,2π]|γ0(t)|+ . . .+ λk−1 ε

k

λkπmax

[0,2π]|γk−1(t)|

]

· ‖P+ − P+‖C0([0,2π]).

For λ small enough there exist D1,D2 ∈ R+ such that

ε

λ≤ D1,

ω2

λ2≤ D2.

We define the constants M0,M1, . . . ,Mk−1, M ∈ R+ as follow

Mi := maxt∈[0,2π]

|γi(t)| for i = 0, 1, . . . , k − 1,

M := max[0,2π]

∣∣∣∣D2(1 +

4|γ(t)|θk

) + 1

∣∣∣∣≥ max

t∈[0,2π]

∣∣∣∣

ω2

λ2(1 +

4|γ(t)|θk

) + 1

∣∣∣∣.


ψ is a contraction if the the following condition holds:

Hk(λ) := 4πλ ·[

1 + λπM +D1πM0 +D21πλM1 + . . .+Dk

1πλk−1Mk−1

]

< 1.

Hk(λ) is a polynomial of degree k in λ. Since λ = 0 is a zero of this polynom (i.e.Hk(0) = 0), there exists an interval [0, λ∗) such that

|Hk(λ)| < 1, for λ ∈ [0, λ∗).

Therefore ψ is a contraction for λ small enough. This terminates the proof of Lemma7.7.

7.2.5 Estimates on the fundamental solutions c(t) and s(t)

Lemma 7.9. The fixed point P+ = ψ(P+) of Lemma 7.7 is in C∞([0, 2π],C).

Proof. (Bootstrap argument) We know that P+ ∈ C0([0, 2π],C). Furthermore P+

satisfies the equationP+(t) = ψ(P+;λ)(t),

where ψ is defined in Lemma 7.7. Every term on the right-hand side is differentiablewith respect to t, because P+ ∈ C0([0, 2π],C). Therefore P+ ∈ C1([0, 2π],C). Thesame argument can be use to prove recursively that

P+ ∈ C2([0, 2π],C), P+ ∈ C3([0, 2π],C), P+ ∈ C4([0, 2π],C),

and so on.

Lemma 7.10. For λ small enough there exists c2 such that the fundamental solutionsc(t) and s(t) of Lz = 0 satisfy

|c(t)|, ε|s(t)|, |c′(t)|ε

, |s′(t)| ≤ c2, ∀t ≥ 0. (7.28)

Proof. Since ψ smoothly depends on λ, the fixed point P+ also depends smoothly onλ. We assume

P+(t;λ) = 1 +

∞∑

l=1

Al(t)λl,

with Al := Al(t) ∈ C and Al(0) = Al(2π) = 0 for all l ≥ 1. Putting this Ansatz in theequations (7.26) and (7.27) of Lemma 7.7 we get:

• v0 =ε

2πλ

∫ 2π0

∫ s0 γ0 +O(λ);

• λv0t− 2iλ∫ t0 (P+ − 1)− λ2

∫ t0

∫ s0 GP+ = ε

2π

∫ 2π0

∫ s0 γ0 +O(λ2);

• −ε∫ t0

∫ s0 γ0P+ = O(λ) = B1λ+O(λ2), where the term B1 is given by

B1 = − ε

λ

∫ t

0

∫ s

0γ0dξds.

Defining B(t) = (B1(t)− 〈B1〉) ∈ R and recalling that P−(t) = P+(t) we have

P+(t;λ) = 1 +B(t)λ+O(λ2),

P+(t;λ) = B(t)λ+O(λ2),and

P−(t;λ) = 1 +B(t)λ+O(λ2),

P−(t;λ) = B(t)λ+O(λ2).

(7.29)


The fundamental solutions of the equation Lz = 0, i.e. the solutions c(t) and s(t)with initial conditions c(0) = s′(0) = 1 and c′(0) = s(0) = 0, can be expressed aslinear combination of the two known solutions ζ±(t) = e±iλtP±(t). So we have

c(t) = c+ζ+(t) + c−ζ−(t),s(t) = s+ζ+(t) + s−ζ−(t),

(7.30)

where

c+ = 1− c− =iλ− P−(0)

2iλ+ P+(0)− P−(0)=

1

2+O(λ), (7.31)

s+ = −s− =1

2iλ+ P+(0) − P−(0)=

1

2iλ+O(1). (7.32)

c±, s± ∈ C. Equation (7.31) and (7.32) can be proved as follows: Using the initialconditions for the function c(t) we obtain the following system of equations:

1 = c(0) = c+ζ+(0) + c−ζ−(0),0 = c′(0) = c+ζ+(0) + c−ζ−(0).

Since ζ±(t) = ±iλe±iλtP±(t) + e±iλtP±(t) we have

1 = c+P+(0) + c−P−(0),0 = iλc+P+(0) + c+P+(0)− iλc−P−(0) + c−P−(0).

From the first equation it follows c− = 1− c+. This implies that

0 = iλc+ + c+P+(0)− iλ [1− c+] + [1− c+] P−(0)

0 = 2iλc+ − iλ+ P−(0) + c+

[

P+(0) − P−(0)]

.

Solving this equation for c+ and observing that B(0) = 0 we get (7.31). The proof for(7.32) is analogous.Then by equation (7.30) we get:

c(t) = cos(λt) +O(λ),

s(t) = sin(λt)λ +O(1).

(7.33)

From Remark 7.6 we know that λ ∼ ε. We define the constant c ∈ R+ such that∣∣ ελ

∣∣ ≤ c. By (7.33) for 0 < |λ| < 1

2 small enough there exist constants M,M, ˜M,˜M

such that

|c(t)| = | cos(λt) +O(λ)| ≤ 1 +M =: c2,1,

|εs(t)| = |εsin(λt)λ

+O(1)| ≤ c+ M =: c2,2.∣∣∣∣

c′(t)ε

∣∣∣∣

=1

ε| − λ sin(λt) +O(λ)| ≤ c+ ˜M =: c2,3,

|s′(t)| = | cos(λt) +O(λ)| ≤ 1 +˜M =: c2,4.

Lemma 7.10 follows defining c2 := maxi=1,2,3,4 c2,i.


7.2.6 Technical estimates

Recall that z(t) is the solution of the inhomogeneous equation (7.13) with initialcondition

a := z(0), b := z′(0). (7.34)

We definezhom(t) := a · c(t) + b · s(t), v(t) := z(t)− zhom(t). (7.35)

Remark 7.8. Since for the fundamental solutions c(t), s(t) of Lz = 0 we have c(0) =s′(0) = 1 and c′(0) = s(0) = 0, then for v

v(0) = z(0) − zhom(0) = a− (a · c(0) + b · s(0)) = a− a = 0,

v′(0) = z′(0) − z′hom(0) = b− (a · c′(0) + b · s′(0)) = b− b = 0,

hold.

Remark 7.9. By linearity of the operator L in (7.13) we obtain Lz = Lzhom︸︷︷︸

=0

+Lv =

Lv.

Define

(G [h]) (t) := s(t)

∫ t

0c(τ)h(τ)dτ − c(t)

∫ t

0s(τ)h(τ)dτ (7.36)

and notice that

L (G [h]) (t) = L(

s(t)

∫ t


∫ t

0s(τ)h(τ)dτ

)

= s′′(t)∫ t

0c(τ)h(τ)dτ + 2s′(t)c(t)h(t) + s(t)

[c′(t)h(t) + c(t)h′(t)

]

−c′′(t)∫ t

0s(τ)h(τ)dτ − 2c′(t)s(t)h(t) − c(t)

[s′(t)h(t) + s(t)h′(t)

]

+[(εθ − α2) + εγ(t)

](

s(t)

∫ t


∫ t

0s(τ)h(τ)dτ

)

=[s′(t)c(t)− c′(t)s(t)

]h(t),

where we have used that c(t), s(t) solve Lz = 0. Notice that

s′(t)c(t) − c′(t)s(t)(7.17)= det(U(t)) = det(U0)e

∫ t0 tr(A(s))ds = 1, ∀t,

where we have used that for A in (7.18) tr(A) = 0 holds. Furthermore notice that forevery function h we have

(G [h]) (0) = 0,

(G [h])′ (0) = s(0)c(0)h(0) − c(0)s(0)h(0) = 0.

Then (7.13) is equivalent to

Lz = Lv = −εeαtQ(e−αtz)

⇔ v = G[−εeαtQ(e−αtzhom(t) + e−αtv(t))

]. (7.37)

The next lemma contains the most important technical estimate of this section. Theidea is that if the initial data of z(t) are small enough, then v(t) and its derivativev′(t) remain small for every t.


Lemma 7.11. Let c1 be as in Remark 7.2 and c2 be as in Lemma 7.10. Definec3 :=

132c1c32

and c4 :=1

16c1c22. If |a|, |b|ε ≤ c3α then

|v(t)| + |v′(t)|ε

< c4α, ∀t ≥ 0.

Proof. By contradiction we assume that there exists t > 0 such that

|v(t)| + |v′(t)|ε

< c4α, for 0 ≤ t < t,

|v(t)|+ |v′(t)|ε

= c4α.

Using the result of Lemma 7.10 we have

|zhom(t)| = |a · c(t) + b · s(t)| ≤ |a| · |c(t)| + |b| · |s(t)|≤ c3αc2 + εc3α

c2ε

= 2c2c3α,

for all t ≥ 0. By Remark 7.2 we easily get that

|Q(e−αtzhom(t) + e−αtv(t))| ≤ c1e−2αt(|zhom(t)|+ |v(t)|)2 ≤ c1e

−2αt (2c2c3α+ c4α)2

= c1α2e−2αt

(2c2

32c1c32+ c4

)2

= c1α2e−2αt

(1

16c1c22

+ c4

)2

= 4c1c24α

2e−2αt. (7.38)

From Lemma 7.10 it follows that for the operator G in (7.36) the following two in-equalities hold for all t ≥ 0 and for any function h:

(⋆) ε|G [h] (t)| ≤ ε

∣∣∣∣s(t)

∫ t


∫ t

0s(τ)h(τ)dτ

∣∣∣∣

≤ 2εc2εc2

∣∣∣∣

∫ t

0h(τ)dτ

∣∣∣∣≤ 2c22

∫ t

0|h(τ)|dτ ; (7.39)

(⋆⋆)

∣∣∣∣

d

dt(G [h] (t))

∣∣∣∣=

∣∣∣∣s′(t)

∫ t

0c(τ)h(τ)dτ + s(t)c(t)h(t)

−c′(t)∫ t

0s(τ)h(τ)dτ − s(t)c(t)h(t)

∣∣∣∣

≤ 2εc2εc2|∫ t

0h(τ)dτ | ≤ 2c22

∫ t

0|h(τ)|dτ. (7.40)

So it follows that

c4α = |v(t)|+ |v′(t)|ε

(7.37)=

∣∣∣−εG

[

eαtQ(

e−αtzhom(t) + e−αtv(t))]∣∣∣

+1

ε

∣∣∣∣

d

dt

(G[eαtQ

(e−αtzhom(t) + e−αtv(t)

)](t))∣∣∣∣

(7.39)&(7.40)

≤ 4c22

∫ t

0eαt|Q

(e−αtzhom(t) + e−αtv(t)

)|dt


(7.38)

≤ 16c1c22c

24α

2

∫ t

0eαte−2αtdt

= c4α2

∫ t

0e−αtdt = c4 α

2 e−αt

(−α)

∣∣∣∣

t

0

= c4α(

1− e−αt)

< c4α,

which is a contradiction. This terminates the proof of Lemma 7.11.

7.2.7 Proof of Theorem 7.2

Proof. The assumptions (7.1) and (7.2) imply (7.15). From Lemma 7.1, 7.2 andcondition (7.15) we get 0 < δ < 1 for ε small enough (see Remark 7.5). From Lemma7.3 we know that λ ∼ ε, i.e. “for ε small enough” is the same as “for λ small enough”.Therefore Lemmata 7.4 - 7.10 hold.Since (7.8), (7.10), (7.11) hold the assumption

|x(0)− xpq(0)|+|x(0) − xpq(0)|

qε≤ c1η

of Theorem 7.2 is equivalent to

|w(0)| + | ˙qw(0)|ε

≤ c1η

q.

Since w(t) := w(qt) we have w(0) = w(0) and q ˙w(0) = w(0). Therefore we get

|w(0)| + |w(0)|ε

≤ c1η

q.

This implies

|w(0)| ≤ c1η

qand

|w(0)|ε

≤ c1η

q. (7.41)

Let c1 be as in Remark 7.2 and c2 be as in Lemma 7.10. Define c3 :=1

32c1c32. Choosing

c1 :=qc34 equation (7.41) is equivalent to

|w(0)| ≤ c3α

2and

|w(0)|ε

≤ c3α

2, (7.42)

where we used η = 2α, cf. (7.15). Recalling the relation between z(t) and w(t), i.e.z(t) := eαtw(t), it is easy to prove that z(t) has the following initial conditions:

z(0) = w(0), (7.43)

z′(0) =[αeαtw(t) + eαtw′(t)

]

t=0= α · w(0) + w′(0). (7.44)

From (7.42), (7.43) and (7.44) we get

|z(0)| = |w(0)| ≤ c3α

2< c3α, (7.45)

|z′(0)|ε

=|α · w(0) + w′(0)|

ε≤ α · |w(0)|

ε+

|w′(0)|ε

≤ α

ε· c3α

2+c3α

2≤ c3α, (7.46)

where we used that by (7.2) αε ≤ 1 holds3. From (7.34), (7.45), (7.46) we see that the

conditions of Lemma 7.11 are satisfied, i.e.

|a| = |z(0)| ≤ c3α, (7.47)

3Recall that by definition α := η

2.


|b|ε

=|z′(0)|ε

≤ c3α. (7.48)

Then from Lemma 7.11 for c4 :=1

16c1c22and v = z − zhom we get

|v(t)|+ |v′(t)|ε

< c4α, for all t ≥ 0. (7.49)

Furthermore, for zhom(t) defined in (7.35), using (7.47), (7.48) and the estimates inLemma 7.10 we have

|zhom(t)|+|z′hom(t)|

ε≤ |a| · |c(t)| + |b| · |s(t)|+ |a|

ε· |c′(t)|+ |b|

ε· |s′(t)|

≤ c3α · c2 + εc3α · c2ε

+c3α

ε· εc2 + c3α · c2

= 4c2c3α =4c2

32c1c32α =

α

8c1c22= 2c4α. (7.50)

For z(t) in (7.13) we have

|z(t)| + |z′(t)|ε

= |zhom(t) + v(t)| + |z′hom(t) + v′(t)|ε

≤ |zhom(t)|+|z′hom(t)|

ε+ |v(t)| + |v′(t)|

ε(7.49)&(7.50)

< 2c4α+ c4α = 3c4α, (7.51)

for all t ≥ 0. From (7.51) and the fact that w(t) = e−αtz(t) we have

|w(t)| + |w′(t)|ε

= |e−αtz(t)|+ |αe−αtz(t) + e−αtz′(t)|ε

≤ e−αt ·(

|z(t)| + α

ε|z(t)| + |z′(t)|

ε

)

≤ e−αt ·(

2|z(t)| + |z′(t)|ε

)

≤ 2e−αt ·(

|z(t)| + |z′(t)|ε

)

≤ 2e−αt · 3c4α = 6c4αe−αt. (7.52)

Since w(t) = w(qt) and w′(t) = q ˙w(qt) from (7.52) we get

|w(qt)|+ q| ˙w(qt)|ε

≤ 6c4αe−αt.

Choosing c2 := 3qc4 and replacing t by t/q we obtain

|x(t)− xpq(t)|+|x(t)− xpq(t)|

qε≤ c2ηe

−ηt/2, ∀t ≥ 0.

This terminates the proof of Theorem 7.2.


The general idea to prove Theorem 7.3 is the same as for Theorem 7.1. The maindifference is that every estimate is given explicitly.


Lemma 7.12. Let K be as in Theorem 7.3. Let x(t) be as in (7.9) and

γ(t) = fxx(x(t), qt)− θ

be as in (7.12) (see equation (64) of [8]). Let ω2 := εθ− η2

4 as in (7.23) (as in equation(74) of [8]) and let g(t) := ε

ω2γ(t) (as in (75) of [8]). Then the following statementshold:

(i) supt∈[0,2π] |γ(t)| ≤ 2 supt∈[0,2π] |fxx(x(t), qt)| = K;

(ii) 14εθ0 ≤ ω2 ≤ 2εθ0;

(iii) supt∈[0,2π] |1 + g(t)| ≤ 1 + 8Kθ0

.

Proof. (i) By definition of γ(t) and using the triangle inequality we get

supt∈[0,2π]

|γ(t)| = supt∈[0,2π]

|fxx(x(t), qt)− 〈fxx(x(t), qt)〉 | ≤ 2 supt∈[0,2π]

|fxx(x(t), qt)|.

(ii) The hypothesis follows directly from (7.4) and (7.5).(iii) By definition of g(t) we get

supt∈[0,2π]

|1 + g(t)| ≤ 1 + supt∈[0,2π]

|g(t)|(ii)

≤ 1 +4

θ0sup

t∈[0,2π]|γ(t)|

(i)

≤ 1 +8

θ0sup

t∈[0,2π]|fxx(x(t), qt)|.

Lemma 7.13. Let U be the solution of (7.16) or (equivalently of (7.18)) with k = 0.For ω2 as in (7.23) this is equivalent to

U ′ = (T + ω2E(t))U, with U(0) = Id,

where

T =

(0 10 0

)

and E(t) =

(0 0

−1− g(t) 0

)

.

Expanding U = U(t;ω2) =∑

l≥0 Ul(t;ω2)ω2l in powers of ω2 we obtain

U ′0 = TU0, U0(0) = Id, (7.53)

U ′l = TUl + E(t)Ul−1, Ul(0) = 0, for l ≥ 1. (7.54)

Then, for K as in Theorem 7.3, the following holds:

|tr(Ul(2π))| ≤[

4π2(

1 +8K

θ0

)]l

, for all l ≥ 1.

Proof. From (7.53) we know that

U0(t) = eTt =

(1 t0 1

)

.

By (7.54) and the of variation of constants formula we have

Ul(t) = eTt

∫ t

0e−TsE(s)Ul−1(s)ds.


Define

Ul−1(s) =:

(

U l−11,1 (s) U l−1

1,2 (s)

U l−12,1 (s) U l−1

2,2 (s)

)

.

Notice that

e−TsE(s)Ul−1(s) =

(1 −s0 1

)(0 0

−1− g(s) 0

)(

U l−11,1 U l−1

1,2

U l−12,1 U l−1

2,2

)

=

(

s(1 + g)U l−11,1 s(1 + g)U l−1

1,2

−(1 + g)U l−11,1 −(1 + g)U l−1

1,2

)

.

Therefore we have

Ul(t) =

(1 t0 1

)( ∫ t0 s(1 + g)U l−1

1,1 ds∫ t0 s(1 + g)U l−1

1,2 ds

−∫ t0 (1 + g)U l−1

1,1 ds −∫ T0 (1 + g)U l−1

1,2 ds

)

(7.55)

and for t = 2π we get

Ul(2π) =

(1 2π0 1

)( ∫ 2π0 s(1 + g)U l−1

1,1 ds∫ 2π0 s(1 + g)U l−1

1,2 ds

−∫ 2π0 (1 + g)U l−1

1,1 ds −∫ 2π0 (1 + g)U l−1

1,2 ds

)

.

So it follows

tr(Ul(2π)) =

∫ 2π

0s(1 + g)U l−1

1,1 − 2π

∫ 2π

0(1 + g)U l−1

1,1 −∫ 2π

0(1 + g)U l−1

1,2 ds

= −∫ 2π

0(1 + g)

[

U l−11,1 (2π − s) + U l−1

1,2

]

. (7.56)

By applying (7.55) recursively we obtain

U l−11,1 (t) =

∫ t

0(s− t)(1 + g)U l−2

1,1 (s)ds

=

∫ t

0(s1 − t)(1 + g(s1))

∫ s1

0(s2 − s1)(1 + g(s2))U

l−31,1 (s2)ds2ds1

=

∫ t

0(s1 − t)(1 + g(s1))

∫ s1

0(s2 − s1)(1 + g(s2)) . . .

. . .

∫ sl−2

0(sl−1 − sl−2)(1 + g(sl−1)) U

01,1︸︷︷︸

=1

dsl−1 . . . ds2ds1.

This implies

|U l−11,1 (t)| ≤ sup

t∈[0,2π]|1 + g(t)|l−1(4π2)l−1, for all t ∈ [0, 2π] .

Analogously for U l−11,2 (t) we get

|U l−11,2 (t)| ≤ sup

t∈[0,2π]|1 + g(t)|l−1(4π2)l−12π, for all t ∈ [0, 2π] ,

since U01,2(t) = t. Therefore using (7.56) we have an upper bound for the absolute

value of tr(Ul(2π)):

|tr(Ul(2π))| ≤∣∣∣∣

∫ 2π

0(1 + g)

[

U l−11,1 (2π − s) + U l−1

1,2

]

ds

∣∣∣∣


≤ supt∈[0,2π]

|1 + g(t)|l(4π2)l

Lemma 7.12.(iii)

≤(

1 +8K

θ0

)l

(4π2)l, for all l ≥ 1.


Lemma 7.14. Assuming

2εθ0 <1

4π2(

1 + 8Kθ0

) [

4π(

1 + 8Kθ0

)

+ 1] (7.57)

we get

πω2 ≤ |δ − 1| :=∣∣∣∣

tr(U(2π)) − 2

2

∣∣∣∣≤ 3π2ω2.

Proof. From Lemma 7.13 we have∣∣∣∣∣∣

∑

l≥2

tr(Ul(2π))ω2l

∣∣∣∣∣∣

≤∑

l≥2

|tr(Ul(2π))|ω2l

≤∑

l≥2

[

4π2ω2

(

1 +8K

θ0

)]l

Geom. Series=

[

4π2ω2(

1 + 8Kθ0

)]2

1− 4π2ω2(

1 + 8Kθ0

) ,

if ω is so small such that

4π2ω2

(

1 +8K

θ0

)

< 1 or, equivalently ω2 <1

4π2(

1 + 8Kθ0

) . (7.58)

Since by point (ii) of Lemma 7.12 we have ω2 ≤ 2εθ0 and since 4π(

1 + 8Kθ0

)

+ 1 > 1

holds, it is clear that condition (7.57) is stronger and implies (7.58). So we get

|δ − 1| =∣∣∣∣

tr(U(2π)) − 2

2

∣∣∣∣=

∣∣∣∣∣∣

−2π2ω2 +∑

l≥2

tr(Ul(2π))ω2l

∣∣∣∣∣∣

implying

|δ − 1| ≤ 2π2ω2 + π2ω2 = 3π2ω2 and |δ − 1| ≥ 2π2ω2 − π2ω2 = π2ω2,

where we used the triangle inequality

||x| − |y|| ≤ |x+ y| ≤ |x|+ |y|.

Lemma 7.15. (i) Let condition (7.57) be satisfied. Then the following statements

hold: (A) ω2

λ2 ≤ 12π2 , (B) λ2

ω2 ≤ 288π2

47 , (C)√ε

|λ| ≤√2

π√θ0

and (D) |λ|√ε≤√

576π2θ047 .


(ii) Define G(t;λ) := ω2

λ2 (1 + g(t))− 1 (as in equation (Eq±) in [8]). Then we have:

supt∈[0,2π]

|G(t;λ)| ≤ 1 +4K

π2θ0= K

for K defined as in Theorem 7.3.

Proof. (i) By [8] we know the relation

cos(λ) = δ =tr(U(2π))

2= 1− 2π2ω2 +

∑

l≥2

tr(Ul(2π))ω2l.

Since δ > 0 using Lemma 7.14 and the expansion of cos(λ) we get:

(A) 1− λ2

2≤ 1− π2ω2 =⇒ ω2

λ2≤ 1

2π2;

(B) 1− 3π2ω2 ≤ 1− λ2

2+λ4

24

|λ|<1/2=⇒ λ2

ω2≤ 3π2

12 − (1/2)2

24

=288π2

47.

(C) and (D) follows directly from (A), (B) and point (ii) of Lemma 7.12.(ii) Follows from Lemma 7.12 (i) and (ii).

Lemma 7.16. Let K, λ1 and M1 be as in Theorem 7.3. Furthermore assume that:(a) condition (7.57) holds;

(b) λ is so small4 such that 0 ≤ |λ| ≤ λ1 =√

18π2K

.

Then we havesup

t∈[0,2π]|P+(t)| ≤ 1 +M1.

Proof. From [8] we know that P+(t) solves the differential equation

P+(t) + 2iλP+(t) + λ2GP+(t) = 0,

with boundary conditions P+(0) = P+(2π) = 1. P+(t) has the form5

P+(t) = 1 +A2(t)λ2 +A3(t)λ

3,

with A2(t) :=∫ t0

∫ s0 G− t

2π

∫ 2π0

∫ s0 G. Notice that from the definition of K in Theorem

7.3 for A2(t) and A2(t) the following statements hold:

supt∈[0,2π]

|A2(t)| ≤ 8π2K; (7.59)

supt∈[0,2π]

|A2(t)| = supt∈[0,2π]

∣∣∣∣

∫ s

0G− 1

2π

∫ 2π

0

∫ s

0G

∣∣∣∣≤ 4πK. (7.60)

The equation describing A3(t) is

A3(t) + 2iλA3(t) + λ2A3(t) = −2iA2(t)− λGA2(t).

Defining v(t) := A3(t) and f(t) := f(t, A3(t)) = −2iA2(t)− λGA2(t)︸︷︷︸

=:h(t)

−λ2A3(t) we get

v(t) + 2iλv(t) = f(t).

4This condition implies 4π2λ2K ≤ 12.

5This is different from (7.29) (where k ≥ 1) since here k = 0 holds.


With the variation of constant formula we get

v(t) = e−2iλt

(

c+

∫ t

0ei2λsf(s)ds

)

, (7.61)

where

c =2iλ

∫ 2π0

∫ s0 e

i2λ(τ−s)f(τ)dτds

1− e−i4πλ(7.62)

is given by the condition A3(2π) = 0. Estimating the norm of A3 (remember that λis real) we have:

supt∈[0,2π]

|A3(t)| = supt∈[0,2π]

∣∣∣∣

∫ t

0v(s)ds

∣∣∣∣

= |c| supt∈[0,2π]

∣∣∣∣

∫ t

0e−i2λsds

∣∣∣∣+ sup

t∈[0,2π]

∣∣∣∣

∫ t

2πe−i2λs

∫ s

0ei2λτf(τ)dτds

∣∣∣∣

(7.61)&(7.62)

≤ 2π

∣∣∣∣

2iλ

1− e−i4πλ

∣∣∣∣4π2 sup

t∈[0,2π]|f(t)|+ 4π2 sup

t∈[0,2π]|f(t)|.

For real values of λ with 0 ≤ λ < 12 the function n(λ) :=

∣∣∣

2iλ1−e−i4πλ

∣∣∣ is strictly increas-

ing. Since λ1 <12 we get

supt∈[0,2π]

|A3(t)| ≤ 4π2 (2πn(λ1) + 1) supt∈[0,2π]

|f(t)|

≤ 4π2 (2πn(λ1) + 1)

(

supt∈[0,2π]

|h(t)| + λ2K supt∈[0,2π]

|A3(t)|)

≤ 4π2 (2πn(λ1) + 1)

(

8πK + 8π2K2λ+ λ2K supt∈[0,2π]

|A3(t)|)

, (7.63)

where we have used the following calculation:

supt∈[0,2π]

|h(t)| = supt∈[0,2π]

∣∣∣−2iA2(t)− λGA2(t)

∣∣∣

≤ 2 supt∈[0,2π]

|A2(t)|+ λK supt∈[0,2π]

|P2(t)|

(7.59)&(7.60)

≤ 8πK + 8π2K2λ.

Equation (7.63) implies

supt∈[0,2π]

|A3(t)| ≤4π2 (2πn(λ1) + 1)

(

8πK + 8π2K2λ)

1− 4π2Kλ2

(ii)

≤ 64π3K (2πn(λ1) + 1)(

1 + πKλ1

)

.

Therefore now we are able to conclude that

supt∈[0,2π]

|P+(t)| = supt∈[0,2π]

∣∣1 + λ2A2(t) + λ3A3(t)

∣∣

≤ 1 + λ2 supt∈[0,2π]

|A2(t)|+ |λ|3 supt∈[0,2π]

|A3(t)|

≤ 1 + 8π2Kλ2 + |λ|364π3K (2πn(λ1) + 1)(

1 + πKλ1

)

.

Since 0 ≤ |λ| ≤ λ1 this terminates the proof of Lemma 7.16.


Lemma 7.17. As in [8] let v0 :=iπ

∫ 2π0 (P+ − 1) + λ

2π

∫ 2π0

∫ s0 GP+ and let K, λ1, M1,

M2 be defined as in Theorem 7.3. Then under the assumptions of Lemma 7.16 thefollowing statements hold:(i) P+(0) = λv0; (iii) |P+(0) − P−(0)| ≤ 2M2;(ii) |P+(0)| ≤M2; (iv) |P+(t)| ≤ 2M2, for all t ∈ [0, 2π].

Proof. (i) By the fixed point equation we know that P+ satisfies

P+ = λv0 − 2iλ(P+ − 1)− λ2∫ t

0GP+. (7.64)

For t = 0 we getP+(0) = λv0

since by definition P+(0) = 1.(ii) Since

|v0| =

∣∣∣∣

i

π

∫ 2π

0(P+ − 1) +

λ

2π

∫ 2π

0

∫ s

0GP+

∣∣∣∣

≤ 2|P+ − 1|+ 2πK|λ||P+|,

we get by Lemma 7.16 that

|v0| ≤ 2 supt∈[0,2π]

∣∣λ2P2(t) + λ3P3(t)

∣∣+ 2πK|λ| sup

t∈[0,2π]

∣∣1 + λ2P2(t) + λ3P3(t)

∣∣

≤ 2M1 + 2πKλ1 [1 +M1] =M2

λ1

holds. Therefore we have

|P+(0)| = |λ||v0| ≤ λ1M2

λ1=M2.

(iii) We notice that

P−(0) = P+(0) = P+(0) = λv0.

By the triangle inequality we have

|P+(0) − P−(0)| ≤ |P+(0)| + |P−(0)| = 2|P+(0)|(ii)

≤ 2M2.

(iv) From (7.64) we get

|P+| ≤ |λ||v0|+ 2|λ||P+ − 1|+ 2πKλ2|P+|≤ M2 + 2λ1M1 + 2πKλ21(1 +M1)

≤ M2 +M2 = 2M2.

Lemma 7.18. Let c(t) = c+ζ+(t)+ c−ζ−(t) and s(t) = s+ζ+(t)+ s−ζ−(t) as in (7.30)(see also equation (91) of [8]). Let λ1, M1, M2 be as in Theorem 7.3. Under theassumptions of Lemma 7.16 we have:

|c+| = |1− c−| ≤λ1 +M2

|2λ1 − 2M2|, (7.65)

|s+| = |s−| ≤1

|2λ1 − 2M2|. (7.66)


Moreover for all t ∈ [0, 2π] we have:

|c(t)| ≤ C := (1 +M1)

(

1 +λ1 +M2

|λ1 −M2|

)

, (7.67)

|c(t)| ≤ C ′ := (2M2 + λ1(1 +M1))

(

1 +λ1 +M2

|λ1 −M2|

)

, (7.68)

|s(t)| ≤ S :=1 +M1

|λ1 −M2|, (7.69)

|s(t)| ≤ S′ :=2M2 + λ1(1 +M1)

|λ1 −M2|. (7.70)

Proof. (7.65) and (7.66) follow from (7.31), (7.32), the triangle inequality and Lemma7.17. (7.67) - (7.70) follow from (7.30) and the facts that

|P−| = |P+|Lemma 7.16

≤ 1 +M1

and

|P−| = |P+|Lemma 7.17

≤ 2M2

hold, since P− = P+.

Remark 7.10. By Lemma 7.18 and the point (i) of Lemma 7.15, choosing

c2 := max

(

C,

√2

π√θ0S,

√

576π2θ047

C ′, S′)

we have:

|c(t)|,√ε|s(t)|, |c′(t)|√ε, |s′(t)| ≤ c2.

Proof. (Theorem 7.3) From (7.3), (7.4) and (7.5) Theorem 7.1 holds. In the proofTheorem 7.1 (see [8]) the constants c1 and c2 have to satisfy the following conditions:

• The constant c1 (depending only on f) satisfies |(Q(w))(t)| ≤ c1|w(t)|2, ∀t ∈ R,where Q is the quadratic operator in (7.12) (see equation (66) of [8]);

• For ε small enough the constant c2 satisfies |c(t)|,√ε|s(t)|, |c′(t)|√ε, |s′(t)| ≤ c2,

where c(t) and s(t) are the fundamental solutions of Lz = 0 with initial condi-tions c(0) = 1 = s′(0), c′(0) = 0 = s(0) (see Lemma 3.4 of [8]).

As in the proof of Theorem 7.1 (see (7.7)) we take

c1 =1

64c1c32and c2 =

3

8c1c22.

So from Theorem 7.1 we have that for ε and η small enough, if x(t) is a solution of(1.24) with √

ε|x(0) − xpq(0)| + |x(0)− xpq(0)| ≤ c1η,


−ηt/2.

In order to finish the proof of Theorem 7.3 we have give quantitative estimates of theconstants c1 and c2 using the extra assumption (7.6). These estimates are done inLemmata 7.12 - 7.18 and the final results are summarised in Remarks 7.2 and 7.10.

8 Numerical estimates on the basin ofattraction

In this last chapter we aim to improve the numerical results obtained by Celletti andChierchia in [15]. Although our purposes are not yet all achieved, we present somepartial results.

8.1 General setting

The problem (same as in [15]):For all coprime integers p, q determine the intersection of the basin of attraction ofthe (p, q)−periodic orbit with some fixed domain Ω = 0 ≤ x ≤ 2π, 0 ≤ x ≤ 5.

Numerical approach:By a Monte-Carlo method choose 1000 initial conditions (x0, v0) ∈ Ω. For every initialcondition integrate the equation of motion (1.24) numerically for a fixed time TK =103/K, where K is given in (1.49). Determine whether the solution has approchedsome (p, q)−periodic orbit or the quasi-periodic orbit with frequency ω. For everyobserved (p, q)−pair determine the measure of its basin of attraction by counting thepercentage of the initial data which have approached them.

In a similar approach Celletti and Chierchia [15] analyse for the eccentricities of theMoon and of Mercury1 quasi-periodic attractors and (p, q)−periodic orbits with p,q ≤ 5 and p

q ≤ 3. Using Yoshidas algorithm [49] they integrate the equation of motion

(1.24) for a long time TK = 103/K, where K was defined in (1.49) and belongs tothe interval

[5 · 10−6, 10−3

]. As results they give tables with measures of the basin of

attraction for every numerically observed frequency ω of the attractor.

Achivements and possible developments2:

(i) Develop an explicit integrator of order two being symplectic for η = 0;

(ii) Give phase space plots for the dissipative spin-orbit problem for Mercury3;

(iii) For fixed ε we give numerical measure of the basin of attraction for Moon andMercury (computed with the method developed in (i)) and we compared theseresults with those found in [15];

(iv) Give numerical measure of the basin of attraction for all the other satellites inspin-orbit resonance (see list of 18 satellites in Chapter 6);

(v) Compare the numerical result in (ii) with Theorem 7.2;

(vi) Develop explicit integrators of higher order being symplectic for η = 0.

In this chapter we report results on the points (i) and (ii).

1e = 0.0549 and e = 0.2056, corresponding to driving frequencies ν = 1.01809 and ν = 1.25584,respectively.

2Currently I am working on these topics with Prof. Corrado Falcolini.3The phase space of other satellites will look similar to those of Mercury.

137

138 Chapter 8. Numerical estimates on the basin of attraction

8.2 An explicit integrator of order two

Consider the Hamiltonian differential equation

x = g(t, x) (8.1)

where g is 2π−periodic with respect to t. For given initial conditions x(0) = x0 andx(0) = v0 we want to find an approximation to x(2π).The Stormer-Verlet-Method is a second order symplectic integrator for (8.1) (see [25]).Choosing uniform steps of length h := 2π

n and denoting with xk the approximation of

x(tk) = x(2πkn ) for 0 ≤ k ≤ n we have:

xk+1 = −xk−1 + 2xk + h2g(tk, xk),

with starting values

x0 := x(0), x−1 := x0 − hv0 +h2

2g(t0, x0).

Notice that with this choice

v0 =x1 − x−1

2h

holds.However, the dissipative spin-orbit problem x+ η(x− ν) + εfx(x, t) = 0 is not Hamil-tonian (since η > 0). The right-hand side g(t, x, x) = −η(x − ν) − εfx(t, x) of thedifferential equation depends not only on t and x but also on x. Thus, we modify themethod in the following way: Choosing uniform steps of length h := 2π

n and denoting

with xk, vk the approximations of x(tk) = x(2πkn ) and of x(tk) = x(2πkn ) for 0 ≤ k ≤ n,respectively, we have:

xk+1 = −xk−1 + 2xk + h2g(tk, xk, vk),

vk =xk+1−xk−1

2h ,(8.2)

with starting values

x0 := x(0), x−1 := x0 − hv0 +h2

2g(t0, x0, v0).

The scheme (8.2) is an implicit method. For the dissipative spin-orbit problem it isequivalent to

xk+1 = −xk−1 + 2xk + h2g(tk, xk, vk)

= −xk−1 + 2xk + h2 [−η(vk − ν)− εfx(tk, xk)]

= −xk−1 + 2xk + h2[

−η(xk+1 − xk−1

2h− ν)− εfx(tk, xk)

]

.

Solving with respect to xk+1 leads to

xk+1 =−xk−1+2xk+h2[η(

xk−12h

+ν)−εfx(tk ,xk)]1+ ηh

2

vk =xk+1−xk−1

2h ,(8.3)

(8.3) is an explicit integrator of order two being symplectic for η = 0. Given ini-tial conditions (x(0), x(0)) we are now able to compute a numerical approximation of(x(2π), x(2π)) with the method (8.3). Iterating this process we construct the approx-imate Poincare section for the dissipative spin-orbit problem.

8.3. Phase portrait of Mercury 139

8.3 Phase portrait of Mercury

Using (8.3) with h = 2π200 we plot the Poincare section of every (of the 1000) ini-

tial condition in the same compact region 0 ≤ x ≤ 2π, 0 ≤ x ≤ 5. We integrate theequation of motion up to a fixed time TK and therefore ((xk, vk))1≤k≤TK

is the se-quence of points corresponding to an initial condition (x0, v0). In order to make theplots more understandable we paint groups on initial conditions tending to the sameattractor with the same color: the dark blue points correspond to the (1, 1)−periodicresonance; the light blue points correspond to the (2, 1)−periodic resonance; the lightgreen points correspond to the (3, 2)−periodic resonance; the fluorescent green pointscorrespond to the (5, 4)−periodic resonance; the red points converge to the yellowquasi periodic attractor.This procedure leads to the plot (a) in Figure 8.1. To understand what is happening inthis plot we have to enlarge the picture in the regions corresponding to the frequencyω of attractors, which we observe numerically. The plot (b) in Figure 8.1 shows theregion, where ω ≈ ν. The yellow object is the unique quasi-periodic attractor of thesystem, whose existence was proved in [16].The plot (c)-(f) in Figure 8.1 show the regions, where ω is near 1, 1.5, 2, 1.25 andthey correspond to the (1, 1)−, (3, 2)−, (2, 1)− and (5, 4)−resonances of the system,respectively.The theory on (p, q)−periodic orbits developed in this thesis (in particular see Chap-ters 3 and Chapter 7) tells us, that (p, q)−periodic orbits exist also for other valuesof q. Thus, similar plots to the plots (c)-(f) in Figure 8.1 should be observable forevery frequency p

q . Since the basin of attraction of a (p, q)−periodic orbit decreasesas the order of non-degeneration increases, we are not able to observe these attractorsnumerically. This explains, why with this numerical experiment we only find a fewperiodic orbits (the most probable).

140 Chapter 8. Numerical estimates on the basin of attraction

(a) (b)

(c) (d)

0 1 2 3 4 5 6

1.98

1.985

1.99

1.995

2

2.005

2.01

2.015

2.02

(e) (f)

Figure 8.1: Phase space of Mercury for the values e = 0.2056, K = 5 · 10−5, TK = 105

and n = 200.

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