The restricted three body problem on surfaces of constant curvature

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1 Available online at ScienceDirect J. Differential Equations The restricted three body problem on surfaces of constant curvature Jaime Andrade a Ernesto Pérez-Chavela b Claudio Vidal c a Departamento de Matemática Facultad de Ciencias Universidad de Bío-Bío Casilla 5-C Concepción VIII-región Chile b Departamento de Matemáticas Instituto Tecnológico Autónomo de México ITAM Río Hondo Col. Progreso Tizapán Ciudad de México 0080 Mexico c Grupo de Investigación en Sistemas Dinámicos y Aplicaciones-GISDA Departamento de Matemática Facultad de Ciencias Universidad de Bío-Bío Casilla 5-C Concepción VIII-región Chile Received 3 November 06; revised 5 April 08 Available online 7 June 08 Abstract We consider a symmetric restricted three-body problem on surfaces M κ of constant Gaussian curvature κ 0 which can be reduced to the cases κ =±. This problem consists in the analysis of the dynamics of an infinitesimal mass particle attracted by two primaries of identical masses describing elliptic relative equilibria of the two body problem on M κ i.e. the primaries move on opposite sides of the same parallel of radius a. The Hamiltonian formulation of this problem is pointed out in intrinsic coordinates. The goal of this paper is to describe analytically important aspects of the global dynamics in both cases κ =±and determine the main differences with the classical Newtonian circular restricted three-body problem. In this sense we describe the number of equilibria and its linear stability depending on its bifurcation parameter corresponding to the radial parameter a. After that we prove the existence of families of periodic orbits and KAM -tori related to these orbits. 08 Elsevier Inc. All rights reserved. MSC: primary 70F07; secondary 70G60 37D40 * Corresponding author. addresses: jandrade@ubiobio.cl J. Andrade ernesto.perez@itam.mx E. Pérez-Chavela clvidal@ubiobio.cl C. Vidal / 08 Elsevier Inc. All rights reserved.

2 J. Andrade et al. / J. Differential Equations Keywords: The restricted three body problem; Surfaces of constant curvature; Hamiltonian formulation; Periodic solutions; KAM tori; Averaging theory. Introduction A first approach of the N-body problem on non Euclidean spaces was raised simultaneously by N. Lovachevsky [0] and by J. Bolyai [3] in the 830 s. In 860 P.J. Serret extended the gravitational force to the sphere S and solved the corresponding Kepler problem [8]. In 885 W. Killing adapted the Bolyai Lobachevsky gravitational law to S 3 and introduced the cotangent potential as a good extension of the classical potential [6]. One of the fundamental results on the area is due to H. Liebmann who in 90 proved that the orbits of the curved Kepler problem are indeed conics in S 3 and H 3 and a bit later he proved an important result corresponding to an S - and H -analogues Bertrand s theorem which states that there exist only two analytic potentials in the Euclidean space such that all bounded orbits are closed see [89]. Relevant and more recent works correspond to those obtained by V. Kozlov [7] and particularly contributions to the two-dimensional case of the Kepler problem belong to J. Cariñena M. Rañada and M. Santander [4] who provided a unified formulation of the Kepler problem for an arbitrary constant curvature κ getting conics orbits depending on the curvature parameter showing that the well known conics orbits in Euclidean spaces can be extended to spaces of constant curvature. A similar approach of the two dimensional curved Kepler problem but with emphasis on the dynamics was treated in []. The curved N-body problem for N was introduced by F. Diacu et al. in a couple of papers The N-body problem in spaces of constant curvature. Parts I and II [] [] where a clear and detailed deduction of the equations of motion for any number N of bodies is presented using variational methods of the theory of constrained Lagrangian dynamics getting the equations of motion depending on the curvature κ. Some other interesting results in this direction can be found in [8] [9] [6] a complete historical background is given in [7]. We know that all -dimensional spaces of constant curvature κ are characterized by the sign of the curvature as the surfaces M κ ={xyz R3 : x y σz = κ } with σ = signκ which induces to define the inner product a x a y a z b x b y b z := a x b x a y b y σa z b z. If κ >0 then the surface is the dimensional sphere of radius R = / κ denoted by S κ embedded in the Euclidean space R 3. If κ = 0 we recover the Euclidean space R. If κ <0 then the surface is the upper sheet of the hyperboloid x y z = κ with z >0 denoted by H κ which corresponds to a surface embedded in the 3 dimensional Minkowski space R R 3 endowed with the Lorentz inner product. According to [] the equations of motion for N bodies on a surface of curvature κ are given by

3 4488 J. Andrade et al. / J. Differential Equations q i = with the constraints N j=j i m j κ 3/ [ q j σκq i q j q i ] [ σ σκqi q j ] 3/ σ q i q i q i q i q i = κ q i q i = 0 i = N σ = sgnκ. We observe that for κ 0 equation can be written independently of the curvature by means of a rescaling dτ dt = κ 3/4 and the change of coordinates q i κ / q i so we will consider along this work that κ = σ =±. The restricted three body problem was introduced by L. Euler [4]. In his work he considered three mutually interacting masses m m m 3 where m 3 is negligible so that it does not have influence in the motion of m and m but its dynamics is entirely determined by the motion of them. This problem can be extended naturally to the restricted N -body problem with N positive masses and the last one of negligible mass. In order to define the restricted three-body problem on a surface of constant curvature κ =± we start by considering an elliptic relative equilibrium of the curved -body problem that is a solution where the primaries are rotating uniformly with constant angular velocity ω on fixed parallels of S or H this type of solutions will be described in detail in Section. As usual we normalized the masses as m = μ m = μ and assume that the primaries are moving on circles of radius r and r respectively. We call this problem the restricted three body problem on surfaces of constant curvature R3BP on S or H by short. In this article we study different aspects of the dynamics of the curved R3BP for the particular case in which the primaries move on the same circle that is when r = r = a. We observe that the above problem depends on two parameters μ r where r is either r or r. When r r and consequently μ / the analysis results really cumbersome in fact the characterization of equilibria becomes hard to carry out by means of analytical methods as well as their stability. This is clearly evidenced in [] where the analysis of this problem had to be complemented with numerical simulations. Due to this difficulty we will consider essentially the symmetric problem that is when r = r = a where a 0 if κ = ; and a 0 if κ =. We refer it as the symmetric R3BP on S or H. Most part of our results concern to this case. In order to avoid the natural constraints of this problem by using the stereographic projection through the equatorial plane as in [6] and [0] we work with intrinsic coordinates on the curved plane i.e. R in the case of S or the Poincaré Disk in the case of H both endowed with the metric induced by the stereographic projection. In both cases we use synodic coordinates in the corresponding surface to write the equations of motion as a Hamiltonian system where the Hamiltonian function is given by H = 8 p u p v ωvp u up v Uuv here u v are the coordinates of the position of the infinitesimal mass particle ω is the angular velocity of the primaries in the synodic frame = σu v and Uu v is the potential written in intrinsic coordinates. A preliminary study of this model was given by R. Martínez and C. Simó in []. In their work they obtained the relative equilibria depending on the parameters κ and the mass ratio

4 J. Andrade et al. / J. Differential Equations μ. Some of the results concerning relative equilibria and their stability properties are proved analytically and the study is completed numerically by using continuation from the κ = 0 case and from other limit cases. In particular they show that the collinear equilibrium points can not be saddle saddle while for κ <0 the triangular equilibria can not be neither saddle saddle nor center-saddle. We have compared their results with the corresponding results present in this paper which of course totally agree even that we have used different coordinates and we have normalized in a different way. Additionally we must point out that when κ >0 Martínez Simó restrict their study to the case when the three bodies remain on the upper hemisphere whereas ours is in both hemispheres. However the big difference in these articles is that they do an exhaustive analysis for all values of the parameter μ whereas we go deeper into the symmetric case and find new families of periodic orbits for the restricted three body problem on spaces of constant curvature. As we have mentioned previously we focus on the symmetric case μ = /. First we look for the relative equilibrium points its number and its linear stability which of course depend on their bifurcations points in the parameter space. We do an analytical and detailed analysis of the type of stability for each equilibrium point in each region determined by the parameters. With this information and using the Hamiltonian structure of our problem by one hand we prove the existence of families of periodic orbits surrounding the poles of S and the vertex of H for certain values of the parameter a in order to get these families we use the Lyapunov Center Theorem [3]. In the case of S we also prove the existence of KAM -tori related to the above family of periodic orbits by using normal forms and Arnold s Theorem for the isoenergetically non-degeneracy condition. On the other hand using Averaging Theory in the Hamiltonian setting see [7] [4] [3] we prove the existence of a family of periodic orbits for sufficiently small values of the radial parameter a. Additionally we study the existence of KAM two-dimensional tori related to these periodic orbits see [4]. More precisely we have used Han Li and Yi s Theorem see details in [5] that applies in the case of Hamiltonian systems with high-order proper degeneracy. The paper is organized as follows. In Section we define and characterize the elliptic relative equilibria. Section 3 is devoted to introduce the equations of motion and the Hamiltonian formulation in intrinsic coordinates given by the stereographic projection through the equatorial plane z = 0. In Section 4 the number of relative equilibria for the symmetric problem is characterized in terms of the parameter a = r = r then we prove the existence of bifurcation values. In Section 5 we analyze the linear stability of the equilibrium points characterized in Section 4. In section 6 we prove the existence of a family of periodic orbits surrounding the poles of S and the vertex of H as well as the existence of periodic orbits for the case when the radial parameter is small enough. Section 7 is devoted to the study of KAM -tori related to the periodic orbits studied in Section 6. Finally we have added in the Appendix the coefficients of the normal form used to prove Theorem 7... Relative equilibria in the -body problem on S and H We consider the motion of two particles of normalized masses m = μ and m = μ with 0 <μ and positions q i = x i y i z i M σ. We shall assume σ = or σ = where M = S and M = H. According to for N = and κ =± the equations of motion are given by

5 4490 J. Andrade et al. / J. Differential Equations q = σ q q q m q σq q q [ σ σq q ] 3/ q = σ q q q m q σq q q [ σ σq q ] 3/ 3 with the restrictions σq i q i = q i q i = 0 i =. 4 We observe that the above system has singularities at collision q = q. Additionally on S we also have antipodal singularities that is when q = q. A complete analysis of the singularities of system 3 can be found in []. We start properly this paper with the definition of an elliptic relative equilibrium see []. Definition.. An elliptic relative equilibrium is a solution of 3of the form q j = r j cosωt α j r j sinωt α j z j 5 where z j is a real number with z j if σ = or z j if σ = α j [0 π] ω 0 and r j = σ σzj. We observe that each particle moves on a circle parallel to the x y-plane not necessarily of the same radius. This induces to introduce the useful notation q j = r j e iωtα j z j e 3 with e 3 = Therefore in the way to search elliptic relative equilibria we must find relations between the parameters z z m m ω α and α in order that 5 and 3 hold. The elliptic relative equilibria of the curved -body problem has been studied in [] and [] however we find an explicit parametrization of the initial conditions generating this kind of solutions. Proposition.. Given 0 <μ /. The elliptic relative equilibria must have initial conditions such that α = α π. For σ = there are exactly two elliptic relative equilibria both with the same configuration but with angular velocity ±ωz μ respectively. For σ = all elliptic relative equilibria are located on the same hemisphere i.e. z z > 0 and there are values 0 < z μ z μ such that for 0 <μ < / If z 0 z z then there are 4 elliptic relative equilibria. If z z z then there is not elliptic relative equilibrium. If z {z z } then there are elliptic relative equilibria. If μ = / then there are 4 elliptic relative equilibria. Proof. From Definition. we have q j = iωr j e iωtα j q j = ω r j e iωtα j 7 q j q j = ω r j q q = r r cosα α σz z := α.

6 J. Andrade et al. / J. Differential Equations Let be r = α 3/ then replacing relations 7into the first equation of 3we get ω r e iωtα = [ σω r 3eiα m r r e iα σαr e iα ] e iωt [ σω r z m r σαz z ] e 3. 8 We observe that in order to get a relative equilibrium we must have that the coefficient of e 3 vanish then comparing the two first components we obtain ω r = σω r 3 m r r cosα α σα i m r sinα α 9 from here we get that α α = kπ k = 0. So it seems that two types of solutions do exist. With this condition on the angles α and α we obtain that α = k r r σz z. Thus from the real part in 9 and repeating this process with the second equation of 3 we obtain that the relative equilibria are given by the solutions of the following system of algebraic equations ω r z = m r σ z αz ω r z = m r σ z αz. 0 By straightforward computations using 0 the normalized masses and the value of α we obtain that z z and μ are linked through the following equation r z = k r z μ. Note that k = 0 implies that z z < 0 which means that in H we must have necessarily k =. The same happens in S indeed taking k = 0 and σ = in the first equation of 0 we obtain ω r = m r z r r < 0. z Therefore for both surfaces we must have k = i.e. the initial conditions of the particles has the form P = r 0 z and P = r 0 z. Even more for S the particles must be located on the same hemisphere thus without loss of generality we can consider z z 0. The polynomial form of is whose discriminant is given by pz = z 4 z β z z = 0 with β = μ = 4β z z then for each solution of we have two relative equilibria with angular velocities ±ω respectively. For the case σ = we observe p = βz z <0 p z = z z >0 and >0 z z >

7 449 J. Andrade et al. / J. Differential Equations which implies that there is a unique solution of. For σ = there are two values of z 0 <z z < given by such that: z ± = μ ± μ μ If z 0 z z then has two solutions z z z ± = ±. 0 given by If z z z then has no solutions. If z {z z } then has a unique solution z =. For the special case of equal masses we have two types of relative equilibria namely z = z valid for σ =± and z = z valid only for σ =. In H the relative equilibria are determined by z which is defined for all z >. Remark.. For the case of S we obtain that z = 0 if and only if z = 0 while for H z = if and only if z =. Both cases represent a singularity of equations The restricted three body problem on curved spaces In this section we introduce a very particular kind of three body problem defined on spaces of constant curvature the restricted three body problem which consists in the study of the dynamics of a particle of infinitesimal mass moving on M σ under the influence of the attraction of two positive masses called primaries whose dynamics is known in our case they move on an elliptic relative equilibrium described in the previous section. The equations of motion for the infinitesimal mass with position q = x x x 3 are given by q = σ q qq μ q t σq q tq [ σ σq q t ] 3/ μ q t σq q tq [ σ σq q t ] 3/ 3 with the corresponding restrictions σq q = q q = 0. 4 In order to avoid the above restrictions 4 we introduce intrinsic coordinates by means of the stereographic projection see [0]or [6]for more details ɛ : M σ R Q q

8 J. Andrade et al. / J. Differential Equations with Q = x x x 3 and q = x y such that The inverse function is given by x = x ɛx 3 y = x ɛx 3. 5 x = x x = y x 3 = ɛ 6 with = σx y. For σ = we take ɛ = ; whereas for σ = ɛ = if the projection is taken from the north pole and ɛ = if the projection is taken from the south pole. The metric of M σ becomes ds = λxydx dy where λx y = 4 is the conformal factor. We denote by R sph the Riemann space R g where g is the metric g = λ 0 0 λ and we call it the curved plane. On R sph the kinetic energy takes the form T = q q g = ẋ ẏ while the potential in the coordinates x y becomes where μα Uxyt = μα σ α σ α α = σr x y cos ωtsin ωt ɛz α = σr x y cos ωtsin ωt ɛz. The Euler s equation associated to the Lagrangian L = T U gives the equations of motion ẍ = 4ẋ xẋ yẏ x ẋ ẏ 4 U x ÿ = 4ẏ xẋ yẏ y ẋ ẏ 4 U y. In addition we can consider that the dynamics of the infinitesimal mass particle is governed by a Hamiltonian system H R sph with Hamiltonian H = T U. The symplectic two form is given by

9 4494 J. Andrade et al. / J. Differential Equations with = dx dp x dy dp y = dxdp x ydp y p x = 4 ẋ p y = 4 ẏ. 7 Thus in terms of the conjugate variables p x p y the Hamiltonian function takes the form Hxyp x p y t= 8 p x p y Uxyt. 8 In order to eliminate the dependence of the time in the potential we introduce rotating coordinates. If q = x y p = p x p y Q = u v and P = p u p v then the symplectic change of coordinates q p = R ωt Q R ωt P where cos ωt sin ωt R ωt = sin ωt cos ωt transforms 8into an autonomous Hamiltonian H given by where H = 8 p u p v ωvp u up v Uuv 9 μν Uuv= μν σ ν σ ν and ν = σr u ɛz ν = σr u ɛz = σu v. The corresponding Hamilton s equations are with u = H p u = 4 p u ωv v = H p v = 4 p v ωu ṗ u = H u = σ u p u p v ωp v U u ṗ v = H v = σ v p u p v ωp u U v U u = U u = [ μ uν r ɛz u l 3/ U v = U v = v [ μ ɛz ν l 3/ μ uν r ɛz u ] l 3/ μ ɛz ν l 3/ ] 0 and l k = νk k =.

10 J. Andrade et al. / J. Differential Equations Remark 3.. We observe that through the projection the primaries are located at P i = u i 0 with u = σ z r u = σ z r. For the case of S the antipodal singularities are at P i = ũ i 0 with ũ = z r ũ = z r. 3 Through the projection the values of u and ũ are exchanged. 4. Relative equilibria for the symmetric R3BP In this section we consider the study of relative equilibria just for the symmetric case that is the primaries are rotating uniformly on opposite sides of the same fixed parallel of radius r = r = a. It is easy to check that the angular velocity of the primaries is given by ω = 8a 3 σa 3/ and they are located symmetrically on the u-axis. All equilibria of system 0 have the form u v 4ωv 4ωu. They are obtained by solving the system F u v = 4ω u 3 U u u v = 0 F u v = 4ω v 3 U v u v = 0. 4 The next result follows easily using 4. Proposition 4.. For σ = the north and south poles are equilibrium solutions for any 0 <a< whereas for σ = the bottom point by short the pole is an equilibrium for any a>0. Lemma 4.. A necessary condition to have an equilibrium solution of 0 when r = r is that u v = 0. Proof. Suppose that u v 0 then from 4we obtain that vf = uf or equivalently since v 0 then 0 = vf uf = av l 3/ l 3/ 0 = l l = 8ɛau σa and since u 0 we must necessarily have = ; replacing this value into 4we obtain F = ɛ σa u σ a u 3/ F = ɛ σa v σ a u 3/

11 4496 J. Andrade et al. / J. Differential Equations Table Number of isosceles equilibria on S. 0a a a a 3 a 3 a 3 a 4 [a 4 Isos. Eq which contradicts the fact that u v satisfies 4. Therefore u v = 0 and the proof is complete. For the equilibrium points that are not located at the poles we define the isosceles equilibria as the equilibria located on the v-axis and the collinear equilibria as the equilibria located on the u-axis. Isosceles equilibria are given by the zeroes of f is v; a = ω σv σv 4 σa l 3/ 5 where l = σ σv σa σv whereas collinear equilibria are given by the zeroes of where f col u; a = 4ω u σu σu 4 ρ l 3/ ρ 6 l 3/ ρ j = u σa j a σu l = σ σu aσu σa σu l = σ σu aσu σa σu. 7 Remark 4.. We observe that in the case of σ = that is on S the equilibria cannot be in the equator in this case we have antipodal singularities. We also observe that the equilibria are given by pairs ±u 0 and 0 ±v so from here on we will consider u v >0of course here the poles have been excluded. Proposition 4.. For σ = all isosceles equilibria are symmetrically located on the upper hemisphere H. There are three bifurcation values given by a a a 4 = 4. Depending on the value of the parameter a 0 the number of isosceles equilibria are given in Table.

12 6384a 6 890a a 63840a 0 890a a J. Andrade et al. / J. Differential Equations Proof. From equation 4 the value v must satisfy the equation ω v a 4 = r 3/ > 0 8 with = v which implies that v < i.e. the equilibrium always remains on the upper hemisphere H recall the change of coordinates 5. A solution of 8is a curve a va in the av-plane. A first possible value of bifurcation for the parameter a appears when the equilibrium collides with the north pole i.e. when va = 0 and by replacing it into 8we obtain the equation 8a 4 6a 7 = 0 whose unique solution in the interval 0 is given by a = 4. Now the solutions of 8are contained in the set of solutions of the polynomial pv = 64a 4 56a 384a 0 56a 8 63a 6 5a 4 048a 307a 0 048a 8 50a 6 a 4 v 79a 4 768a 075a 0 768a 8 764a 6 7a 4 48a v a a 504a a a 6 80a 4 9a 64 v a 4 790a 6880a 0 790a 8 440a 6 40a 4 88a 8 v a a 504a a a 6 80a 4 9a 64 v 0 79a 4 768a 075a 0 768a 8 764a 6 7a 4 48a v 5a 4 048a 307a 0 048a 8 50a 6 a 4 v 4 64a 4 56a 384a 0 56a 8 63a 6 v 6. 9 The number of roots of pv changes when it reaches a double zero which occurs for the values of the parameter a such that qa = Resultantpv p v = 0 where qa = α a 78 a 78 8a 4 6a 7 3 8a 4 6a 9 3 which has three zeroes in the interval 0 given by a a a 4 = 4. After taking a test value of the parameter on each interval determined by a a 3 a 4 we obtain: If a 0 a then there exist two pairs of symmetric isosceles equilibria namely ±L v ±L v ; If a = a then there is one pair of symmetric isosceles equilibria namely ±L v ; If a a a 3 then there are no isosceles equilibria; If a = a 3 then there is one pair of symmetric isosceles equilibria namely ±L v34 ;

13 4498 J. Andrade et al. / J. Differential Equations Fig.. Parameter bifurcation digram for the existence of equilibria in the symmetric R3BP on S : a Isosceles equilibria with a 0 a ]; b Isosceles equilibria with a [a 3 ; c Collinear equilibria with a 0 a ]; d Collinear equilibria with a [a 5. If a a 3 a 4 then there are two pairs of isosceles symmetric equilibria namely ±L v3 ±L v4 ; If a [a 4 then there is one pair of symmetric isosceles equilibria namely ±L v5. Thus our analysis agrees with Table and the relative position can be viewed in Fig. a and b. Proposition 4.3. For σ = and any value of the parameter a>0 there are exactly two isosceles equilibria symmetrically located. Proof. From equation 5 the isosceles equilibria for σ = are obtained by solving g v = g v where g v = ωv v 4 g v = l 3/ with ω = 8a 3 a. The functions g and g have the following properties: g v = ωv 3v 5 v 5 > 0 and g v = 6v a v a l 5/ < 0

14 J. Andrade et al. / J. Differential Equations Table Number of collinear relative equilibria on S for different values of the parameter a. 0a a ] a a 5 a 5 In H In H g 0 = ω< a 3 = g 0 g v when v which assure that g and g intersect at a single point in the interval 0 <v<for any value of a>0. Proposition 4.4. For σ = the collinear relative equilibria are symmetrically located. There are two bifurcation values given by 3 a a 5 = The number of relative equilibria in the intervals determined by the bifurcation values are given in Table. Proof. We define α = 4ω u u u 4 β = a u a u l 3/ a u a u l 3/ with l and l as defined in 7 with σ =. Then solving f col u; a = 0is equivalent to solve the equation α β = 0. 3 In order to eliminate the roots in the denominators of β we need to solve the equation α β = 0. 3 In developing β the term l l 3/ appears in the denominator and it is given by { l l 3/ a u u = 3 a u u 3 if u 0u ũ a u u 3 a u u 3 if u u ũ. 33 For u 0 u ũ equation 3 becomes 4a 6 a 3 u 8 64a a au a u 4 au a u 4 = 0

15 4500 J. Andrade et al. / J. Differential Equations or equivalently in its polynomial form p up up 3 u = 0 with p u = a 4a 5 u 4 4a 4 5a u a 4a 5 p u = a 4a 3 u 4 4a 4 3a u a 4a 3 p 3 u = a 4 6a 4 3a 7 u 8 4a 6a 6 3a 4 7a u 6 48a 8 96a 6 5a 4 8a 8 u 4 4a 6a 6 3a 4 7a u a 4 6a 4 3a We state that the polynomial p 3 does not provide solutions of 3 in fact we obtain that the resultant between p 3 u and p 3 u is given by Resultantp 3 u p 3 u = αa8 a 3 6a 4 6a 4 3a 7 0 for all a 0 and α a nonzero constant which implies that the number of zeroes of p 3 u does not vary for any value of a thus we can put a = / and p 3 takes the form p 3 u = 8 5u 8 4u 6 6u 4 4u 5 which does not have zeroes which proves the statement. The zeroes of p and p can be easily computed and it is verified that the zeroes of p u do not provide zeroes of 3. Nevertheless the polynomial p gives one positive solution of 3 u 3 = a 4a 4 3a 4a 4 3a 4a which is defined for a a 5 with a 5 =. It is easy to verify that 35 satisfies u > and therefore the corresponding collinear equilibrium is always located at the lower hemisphere. If u u ũ then equation 3 assumes the form u u 4a 6 a 3 u 8 4 a u 4 6a u a 4u au a u 4 au a u 4 = 0 whose polynomial form is given by p 4 u = 6a 6 48a 4 48a 6a 0 u 4 64a 6 30a 4 576a 448a 0 7a 8 u 480a 6 696a 4 95a 480a 0 58a 8 7a 6 u 0 9a a 4 990a 363a a 8 a 6 96a 4 u 8 a 6 u 6 a 4 u 4 a 0 u 0 a 8 u 8 9a a 4 990a 363a a 8 a 6 96a 4 u 6 480a 6 696a 4 95a 480a 0 58a 8 7a 6 u 4 64a 6 30a 4 576a 448a 0 7a 8 u 6a 6 48a 4 48a 6a 0 36 with

16 J. Andrade et al. / J. Differential Equations a 6 = 587a a a 673a a 8 73a 6 9a 4 56a a 4 = 656a 6 793a a 50656a a 8 47a 6 96a 4 56 a = 95a a a 93984a a a 6 384a 4 5a 5 a 0 = 656a 6 793a a 50656a a 8 47a 6 96a 4 56 a 8 = 587a a a 673a a 8 73a 6 9a 4 56a. The resultant between p 4 u and p 4 u is given by where α is a nonzero constant and Resultantp 4 u p 4 u = α pa 37 pa = 67776a a 6445a a a a a 7550a 0 The zeroes of pa are 63864a 8 368a a a 5 4. a and ã By considering a test value of the parameter on each interval determined by a a 5 and ã it verifies that If a 0 a there exist two pairs of symmetric collinear equilibria namely ±L u ±L u ; If a = a then there is one pair of symmetric collinear equilibria namely ±L u ; If a a a 5 ] then there are no collinear equilibria; If a a 5 ã then there is one pair of symmetric equilibria which is the one determined by 35 namely ±L u3 ; If a [ã then there is one pair of symmetric equilibria which is the one determined by 35 namely ±L u3. From the above we conclude that a =ã does not provide a bifurcation value with respect to the number of relative equilibria and therefore the number of collinear relative equilibria changes at a = a and a = a 5 see Fig. c and d which completes the proof. Proposition 4.5. For σ = and any value of the parameter a>0 there are exactly two collinear equilibria symmetrically located. Proof. From equation 6 with σ = we have that the relative equilibria are given by the intersection of two functions f and f with

17 450 J. Andrade et al. / J. Differential Equations f u = 4ω u u u 4 f u = ρ l 3/ ρ. The function f is increasing positive and f u when u. We separate our analysis in two intervals 0 u and u. For the first one it is easy to verify that ρ < 0 and ρ > 0 which implies that f := ρ /l 3/ ρ /l 3/ > 0. Since sgnf = sgnf f where l 3/ f f = 6a a u u 4u a 6u u 4 4u a u 4 < 0 it follows that f u < 0for all u 0 u and therefore there are no equilibria in this interval. For u u we have ρ ρ > 0 then f u > 0 and f u when u u which implies that the graphics f and f must intersect at least at one point. Now we compute where f u = β β l 5/ β β l 5/ β = 4au a u 4 0a 8 u 5a 4 β = 4 a 5a u 4 0a 4 u a. It is clear that β β > 0 and sgnβ β = sgnβ β. Since β β au = 6 a u au a u a a u < 0 it follows that f u < 0. Therefore the graphics of f and f intersect at a single point located in the interval u which completes the proof. 5. Linear stability of the relative equilibria for the symmetric R3BP Since we have only one parameter a it is possible to carry out a complete study of the linear stability of each equilibrium point on both surfaces. Moreover a new bifurcation value appears for a certain equilibrium on S. The linearization matrix at an equilibrium solution u v 4ωv 4ωu with uv = 0 has the form A A A 3 A 4 where 0 ω 4v σω A = 4u σω ω 0 A = 4 I

18 J. Andrade et al. / J. Differential Equations A 3 = HessU 8ω u v 4 σ u 0 0 σ v A 4 = A T. The characteristic polynomial is given by the bi-quadratic polynomial pλ = 4 6 det B = β 4λ 4 β λ β 0 38 where B = A 3 4 AT λi b b A λi =. 39 b b It is easy to verify that U uv = U vu = 0for uv = 0 so the entries b ij are given by b = b 4λ b = 8ω λ 3 b = b b = b 4λ with b = 4ω 4 δ 4σu u v δσ u U uu b = 4ω 4 δ 4σv u v δσ v U vv. 40 In this way the characteristic polynomial takes the form where pλ = λ 4 β λ β 0 β = 4ω β 0 = 4 6 b b = b b = ω u 4 v 4 4 U uu U vv 4ω 3u 4 v 4 8σu 4ω U 3v 4 u 4 8σv 4 uu U 4 vv Linear stability for the isosceles relative equilibria For u = 0 β takes the form β = σv σa σv l 3/ ω v 4 σv and using the condition f is v; a = 0 it follows that β = ω and the characteristic polynomial becomes pλ = λ 4 ω λ β 0.

19 4504 J. Andrade et al. / J. Differential Equations Similarly we obtain β 0 = a v ω 4 σv qv; a l with qv; a = 3 4σa v 4 8a 0σ v 3 4σa. The discriminant of pλ is given by δ = ω 4 48a v σv qv; a l. Theorem 5.. For σ = there exists a such that the stability of the isosceles relative equilibria is characterized as follows: If a 0 a 0 then ±L v are unstable saddle saddle and ±L v are unstable centersaddle; If a = a 0 then L v is spectrally stable; If a a 0 a then ±L v are linearly unstable saddle saddle and ±L v are unstable center-saddle; If a = a 3 then L v34 is spectrally stable; If a a 3 a 4 then ±L v3 are linearly stable center center and ±L v4 are unstable centersaddle; If a [a 4 then ±L v5 are unstable center-saddle. Proof. Letting x = λ the characteristic polynomial can be written as px = x ω x β 0. 4 If β 0 < 0 then the discriminant δ = ω 4 4β 0 > 0 which implies that both roots of 4 x and x are real and also by Descartes rule of signs it verifies that x < 0 <x. So in this case it follows that all isosceles relative equilibria with β 0 < 0 have two real and two pure imaginary eigenvalues and therefore the corresponding relative equilibria are unstable. If β 0 = 0 then equation 4 has roots x = 0 and x = ω which implies that all isosceles relative equilibria with β 0 = 0have a double zero eigenvalue and two pure imaginary eigenvalues λ =±iω therefore in this case the relative equilibria are spectrally stable. Moreover the system of algebraic equations f is v; a = 0 and β 0 v; a = 0 can be solved. In fact its zeroes are given by solving the polynomial equation qv; a = 0 which has a unique solution such that v 0 a < and it is given by v 0 a = 4a 5 4 a 4a 43 3 defined for a 0 a 5. Replacing 43into 8 yields the polynomial equation 6384a 6 890a a 63840a 0 890a a 6 7 = 0 which has two solutions into the interval 0 given by

20 J. Andrade et al. / J. Differential Equations a = a a = a 3 where a and a 3 are the bifurcation values given in Proposition 4.. For β 0 > 0 the solutions of 4 are either both complex if δ <0 or both negative real numbers if δ 0. The equation δ = 0is equivalent to the equation δ = 0 where δv = a 4 v 36a 90a 4 v 0 769a 4 768a 6 v 8 56a 4 64a 3 v 6 769a 4 768a 6 v 4 36a 90a 4 v a 4 44 while solving f is v; a = 0is equivalent to solve 9. The equation where Resultant δv pv = p a = 0 45 p a = αa 48 a a a a a a a a a a a a a a a a a a a a a where α is a nonzero constant. The polynomial p a gives the values of a where the curve γ = a va which satisfies f is va; a = 0 crosses the boundary of the region of points a v such that δa v < 0. It is not difficult to verify that equation 45 provides only one solution for the parameter a in the corresponding interval which is given by a For a solution a va satisfying 8 we denote by γ j = a v j a j = the curves of equilibria in the plane a v where L vj = 0 v j a is defined in its corresponding domain given by Proposition 4.. For a 0 a we consider the two equilibria L v and L v with v a < va < v a. It is verified that γ is always contained in the region β 0 < 0 region I in Fig. and by the analysis done above the equilibrium L v is always unstable even more it has one pair of opposite pure imaginary eigenvalues and two opposite nonzero real eigenvalues so it is an unstable center-saddle. Nevertheless when a = a see Proposition 4. we have that along the curve f is v; a = 0 we have β 0 = 0 therefore the equilibrium L v has one pair of pure imaginary and one double zero eigenvalues which implies the spectral stability of L v spectral stability means that all eigenvalues λ satisfy λ 0. For the equilibrium L v the curve γ is always contained in the region β 0 > 0 region II III in Fig.. If a 0 a 0 then γ is contained in the region such that the discriminant δ <0 which implies that all its eigenvalues are complex with nonzero real part i.e. L v is an unstable equilibrium saddle saddle. If a a 0 a the curve γ is contained in the region δ>0 which implies that all its eigenvalues are pure imaginary and different from each other i.e. L v is a linearly stable center center. If a = a 0 then along the curve γ we have δ = 0 which implies that L v has a double pure imaginary eigenvalue i.e. : resonance therefore it is spectrally stable.

21 4506 J. Andrade et al. / J. Differential Equations Fig.. Stability of isosceles relative equilibria in the RC3BP on S. Region I: saddle-center. Region II: complex saddle. Region III: center center. For a = a 3 along the curve γ 4 we have β 0 = 0 therefore the equilibrium L v34 has one pair of pure imaginary and double zero eigenvalues which gives the spectral stability of L v34. For a a 3 a 4 there are two equilibria L v3 = 0 v 3 a and L v3 = 0 v 4 a with v 3 a va 3 v 4 a. The curve γ 3 is always contained in the region β 0 > 0 δ >0 corresponding to the region III in Fig. c this is because the only value of a for which the curve f is v; a = 0 crosses the curve δ = 0is a = a 0 which by virtue of the above analysis implies that the eigenvalues are all pure imaginary and different from each other i.e. L v3 is a linearly stable center center. The curve γ 4 is contained in the region β 0 < 0i.e. region I in Fig. c which implies that the equilibrium L v4 has one pair of opposite pure imaginary eigenvalues and two

22 J. Andrade et al. / J. Differential Equations opposite nonzero real eigenvalues so it is an unstable center-saddle. The same occurs with the curve γ 5 when a [a 4 see Fig. thus L v5 is an unstable center-saddle. This finishes the proof of Theorem 5.. For the case of negative curvature we have the following result. Theorem 5.. For σ = all isosceles equilibria are complex saddle i.e. all of their eigenvalues are complex with nonzero real part. Proof. Since qv; a > 0for σ = it follows that the eigenvalues are either complex with nonzero real part or pure imaginary. Using the same method as in the previous cases we reduce δ and f ω 4 is v; a to their polynomial form namely p v and p v respectively and computing the resultant between them we obtain a polynomial pa = Resultantp v p v which has a unique zero a and it generates one intersection between p and p but it is outside the domain 0 < v< therefore the curve of equilibria does not cross the curve δ = 0 and even more it is contained within the region δ<0 thus any isosceles equilibrium is a complex saddle. This finishes the proof of Theorem Linear stability for the collinear relative equilibria For v = 0 β takes the form β = g 0 g l 5/ g l 5/ where g 0 = ω u 4 g = g g g = g g g = a σ g = au σu a u 4 4 4a σ u a 8σ 0a u 3a σ u 4 3σa and b = 4ω 8σu 3u 4 4 b = 4ω σu 3 k l 3/ k l 3/ h h l 5/ l 5/ 46 where k j = a σ j aσu and h j = h j h with h = 4 a σ a σu 6 5a u 4 4 a σ u a h = au a u 6 4 3a σ u 4 0σ 5a u σa 8. Lemma 5.. Let u 0 with u > 0 a collinear equilibrium. For σ = b < 0 if 0 <u < and b > 0 if u >. For σ = b < 0.

23 4508 J. Andrade et al. / J. Differential Equations Proof. From condition f col u; a = 0 we obtain that ub = 4ω u σu u = = [ = a 3 ρ l 3/3 [ ρ uk [ l 3/ ρ l 3/3 u l 3/ l 3/ k l 3/ k l 3/ ] ρ uk ] l 3/ k k l 3/ l 3/ ] and the result follows from the fact that l <l for 0 <u < and l >l if u > and σ =. Theorem 5.3. For σ = the stability of the collinear equilibria is characterized as follows: If a 0 a then L u is unstable center-saddle and L u is linearly stable. If a = a then L u is spectrally stable. If a a 5 then L u3 is linearly stable. Proof. First we analyze the sign of β which is equivalent to study the sign of 4β / = g 0 with g 0 = 8ω u 4 / 4. Now we compute g l 5/ g l 5/ g 0 g l 5/ g l 5/ = g 0 g l 5 g l5 l 0 l 0 ± g g l 5 l 5 47 where l = au u a l = au u a the sign is for u 0 u ũ and the sign for u u ũ. In the first case 47 can be reduced to its polynomial form β u = ruru su with ru = 4a 0 8a 8 3a 6 u 6 48a 8 0a 6 60a 4 u 4 80a 0 30a 8 46a 6 08a 4 48a u 56a 0 70a 8 698a 6 9a 4 96a 64 u 0

24 Now we see that the equation J. Andrade et al. / J. Differential Equations a 0 9a 8 774a 6 88a 4 96a u 8 56a 0 70a 8 698a 6 9a 4 96a 64 80a 0 30a 8 46a 6 08a 4 48a u 4 48a 8 0a 6 60a 4 u 4a 0 8a 8 3a 6 su = u au u a au u a. u 6 Resultantβ u p u = α a 4 a 48 a 4 8a 4 4a 3 4 = 0 7 with α 0 has a unique solution a = and its corresponding equilibrium is given by replacing this value into 35 in this way we obtain the position of the equilibrium u 3 = Replacing these two values into β we obtain β thus β > 0 along L u3. To study the sign of β along L u and L u we consider the second case u u both equilibria belong the upper hemisphere. Thus reducing 47to its polynomial form β u and using the fact that the equilibrium points L u and L u are contained in the set of roots of the polynomial p 4 u defined in 36 we proceed as in the previous case to obtain 0 = Resultantβ u p 4u = αa 40 a 0 f a 4 f a 4 where α is a nonzero constant and f a = 048a 6 048a 4 80a 408a 0 480a 8 9a 6 8a 4 60a 9 f a = 5488a a a a a a a 87680a 0 686a 8 95a 6 098a 4 559a By using the Sturm criterion we obtain that f a does not have real roots while f a has exactly one real root in the interval 0 which is approximately a however the equilibria L u and L u exist only for a 0 a thus the corresponding curves of equilibria do not cross the curve β = 0 and even more β > 0 along both equilibria. Finally we shall determine the sign of β 0. By virtue of Lemma 5. it is sufficient to study the sign of b. From condition f col u; a = 0 we obtain 4ω 8u 3u 4 4 = 8u 3u 4 ρ l u u l 5/ ρ l l 5/. 49

25 450 J. Andrade et al. / J. Differential Equations Replacing condition 49into b we get b = 8u 3u 4 ρ l u u h u u l 5/ = u u t t where t = t t /l 5/ t = t t /l 5/ with 8u 3u 4 ρ l u u h l 5/ t = a u a u a u 6 66a 36 u a u a t = a u 5a u 8 60a u 6 6a 40 u 4 60a u 5a. Thus the equation b = 0 can be solved through the equation t t = 0 which yields up the polynomial equation b u = 0 where b u = a 4 u 6a 4 4a u 0 5a 4 96a 48 u 8 0a 4 7a 60 u 6 5a 4 96a 48 u 4 6a 4 4a u a 4. Now we compute Resultant b u p 4 u for a 0 a ] which is proportional to the polynomial a 80 a a a 6445a a a a a 7550a a 8 368a a a Again applying Sturm criterion we obtain that 50 has a unique zero in 0 a ] and it is given by a = a. It is verified that b a ua = 0 and defining δ j = a u j a where L uj = u j a 0 for j = 3 we have that δ is contained inside the region b a u > 0 for all a 0 a which implies β 0 < 0 while δ is contained inside the region b a u < 0 for all a 0 a which implies β 0 > 0. For the curve δ 3 we must solve the equation Resultant b u p u = 0 for a a 5 where this resultant is proportional to the polynomial a 6 a 8 64a 6 48a 4 4a 4 which has just one zero in the interval a 5 given approximately by a By evaluating b at a u 3 a where u 3 is given in 35 we obtain b a u 3 a Therefore a does not provide simultaneous solutions of 3 and b = 0 i.e. the curve δ 3 does not cross the boundary of the region b < 0. In fact δ 3 is contained inside the region b < 0 so by Lemma 5. β 0 < 0 along L u3. The last step consists in the determination of the sign of the discriminant δ = β 4β 0 along the equilibrium L u. For this equilibrium we already know that b b < 0 and β 0 β > 0for all a 0 a. Since b b < 0 then from definition of β given in 4 we have β > 4 b b > 0 then

26 J. Andrade et al. / J. Differential Equations Table 3 Type of eigenvalues and stability for the collinear relative equilibria of the R3BP on S. β β 0 δ x λ Type of stability L u > 0 < 0 > 0 x < 0 <x ±i x ± x Unstable L u > 0 > 0 > 0 x <x < 0 ±i x ±i x Linearly stable L u > 0 = 0 > 0 β ±i β Spectrally stable L u3 > 0 < 0 > 0 x < 0 <x ±i x ± x Unstable Table 4 Stability of the north pole. 4 4 a Unstable Linearly unstable Linearly stable Linearly unstable Unstable Table 5 Stability of the south pole. a Unstable Linearly unstable Linearly stable δ> 4 6 b b 4β 0 = 4 6 b b 4 8 b b = 4 6 b b 0. In order to summarize our analysis putting x = λ the characteristic polynomial takes the form px = x β x β 0. Let x and x be their roots. We describe the above analysis in Table 3. We observe that when a = a then β 0 = 0 therefore the equilibrium L u has a double zero eigenvalue and a pair of opposite pure imaginary eigenvalues which implies spectral stability. This finishes the proof of Theorem 5.3. For the case of negative curvature we have the following result whose proof is easier. Theorem 5.4. For σ = the collinear equilibria are unstable even more they are centersaddle. Proof. For σ = it is easy to verify that h h > 0 then b > 0 and by Lemma 5. we have that β 0 < 0 and the result follows. We finish this section by studying the linear stability of the poles. Proposition 5.. For σ = the linear stability of the north and south poles depends on the parameter a and it is given by Tables 4 and 5 respectively. For σ = the pole is unstable for all a >0.

27 45 J. Andrade et al. / J. Differential Equations Table 6 Eigenvalues and linear stability of the north pole. 4 I I 3 I 3 λ R 0 ir i 5 7/4 R ir i ρ ir /4 ir i 5 7/4 R ir Unstable Linearly unstable Linearly stable Linearly unstable Unstable Table 7 Eigenvalues and linear stability of the south pole. J 3 J λ R 0 ir ρ ir i 5 3 3/4 ir Unstable Linearly unstable Linearly stable Proof. The linearization matrix at is given by A σ ɛ = 0 ω 4 0 ω ω a 3 4ɛ a 0 σ ω 0 a 3 8ɛ a σ 5 with eigenvalues ±λ ±ρ where λ = ω fɛ σ a ρ = ω gɛ σ a 5 f σ ɛ = ασ ɛ 4 σa β σ ɛ gσ ɛ = ασ ɛ 4 σa β σ ε α σ ɛ = 4ɛ σa β σ ɛ = ɛ 9 σa. For S the north pole corresponds to the origin with ɛ = and σ =. It is verified that β = 3a 4 3a > 0 if and only if a 0 /3 and in this interval g a < 0 while f a > 0 if and only if a I = 0 4 / then defining the intervals I = 4 / /3 and I 3 = /3 we have that the eigenvalues associated to the north pole of S 4 are characterized in Table 6. For the cases a = and a = /3 it is verified that the linearization matrices are not diagonalizable thus in both cases the north pole is linearly unstable. For S the south pole corresponds to the origin with ɛ = and σ =. It is verified that β = 9 a > 0 and g a < 0for all a 0 while f a > 0 if and only if a J = 0 3/. Defining the interval J = 3/ we have that the eigenvalues associated to the south pole of S are characterized in Table 7. For the case a = 3 it is verified that the linearization matrix is not diagonalizable therefore the south pole is linearly unstable.

28 J. Andrade et al. / J. Differential Equations For H studying the pole corresponds to the origin with ɛ = and σ =. It is verified that 3a f a a = a 3 4 a 3a 4 > 0 for all a >0 and f a g 8a a = 4 6a 7 6a 4 3a 7 < 0 thus g a < 0for all a >0 then λ R and ρ ir for all a >0 which completes the proof. 6. Periodic solutions for the symmetric R3BP In this section we shall prove the existence of periodic orbits for the symmetric R3BP on surfaces of constant curvature in two different ways. We start by looking periodic orbits near the poles of the respective surface as a way of exploiting the nature of the eigenvalues associated to the poles which as we have already proved are equilibrium points of the symmetric problem for any value of the parameter a. On the other hand we will make use of averaging theory for perturbed Hamiltonian systems to obtain periodic orbits close to an integrable Hamiltonian system. 6.. Periodic orbits near the poles One of the most famous theorems concerning the existence of periodic solutions of ordinary differential equations is the Lyapunov Center Theorem [5]. In the Hamiltonian case assumes that the origin is an equilibrium point of elliptic type. The Lyapunov Center Theorem states that if one eigenvalue is purely imaginary λ = iα and the quotients λ j /λ for j = 3 l are not integer numbers then there exist a family of periodic orbits emanating from the equilibrium whose minimal period tends to π/ α. The proof of this theorem can be found in [3]. In this section we will exploit the fact that for certain intervals of the parameter a the eigenvalues corresponding to the poles are purely imaginary this fact give us chance to find periodic orbits by applying the Lyapunov Center Theorem. Let a m ± = ±m 4 6m m 9m 4 m 9 4 m 4 5m m Z be a sequence of values of the radial parameter a. Theorem 6.. For the parameter a assume that all eigenvalues corresponding to the poles are purely imaginaries then for the case of positive curvature on S the following statements hold: i If a I I then there is a family of τυ-periodic orbits xt υ emanating from the north π pole such that lim υ 0 τυ =. Even more for a I and a a g a m for all integer ω m > there is another family of τυ-periodic orbits with lim υ 0 τυ = π. ω f a

29 454 J. Andrade et al. / J. Differential Equations ii If a J then there is a family of τυ-periodic orbits xt υ emanating from the south pole such that lim υ 0 τυ = π ω g a. iii If a J then there exists a family of τυ-periodic orbits xt υ emanating from the south π pole such that lim υ 0 τυ =. Even more for a J and a a g a m for all integer ω m > there is another family of τυ-periodic orbits with lim υ 0 τυ = π. ω f a For the case of negative curvature that is on H there is a family of τυ-periodic orbits π xt υ emanating from the pole such that lim υ 0 τυ =. g a ω Proof. To prove part i we observe in Table 6 that the intervals where the Lyapunov Center Theorem might be applied are I = 0 4 / and I = 4 / /3. So when a I the pure imaginary eigenvalue is ρ = iω g a and due to the fact that λ R the condition λ/ρ / Z holds for all a I. For a I we know that λ ρ ir with g a <f a < 0 then 0 < λ = f a ρ g < 53 a and it follows that λ/ρ / Z for all a I. Now in order to find a new family of periodic orbits we must analyze the reciprocal of the previous ratio. We obtain from 53 that ρ/λ > which means that resonances might occur for certain values of a i.e. it is necessary to determine which values of the parameter a satisfy the condition ρ/λ / Z. Let m Z with m > by solving the equation ρ/λ = m we obtain the solution a = a m. By straightforward computations it is easy verify that a m I for all integer m >. The ends of the interval I are reached when m = and when m that is a = /3 and lim m a m 4 = / here we have resonances. Thus for a I ρ/λ / Z if and only if a a m for all m Z m > which assures the existence of the other family of periodic orbits and the proof of part i is completed. To prove part ii we see from Table 7 that Lyapunov Center Theorem might be applied for any a 0 such that a 3/. For any a J the condition λ/ρ / Z is immediately verified and part ii follows. For the interval J both eigenvalues λ and ρ are pure imaginary and also it is satisfied g a < f a < 0 which implies immediately that λ/ρ / Z which gives the existence of the family of periodic orbits. Since ρ/λ > it is necessary to discard the values of the parameter a such that ρ/λ Z. Let m > an integer then the equation ρ/λ = m gives the solution a = am and it is verified that a m J for all integer m > lim m am = and lim m am = 3/. Thus for a J ρ λ / Z if and only if a a m for all m Z m> where it follows the existence of the other family of periodic orbits which proves part iii. The proof for the case of negative curvature follows similarly to part i by using the fact that λ R and ρ ir which completes the proof of Theorem 6..