The restricted three body problem on surfaces of constant curvature

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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 1, Col. Progreso

Tizapán, Ciudad de México, 01080, 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 23 November 2016; revised 25 April 2018 Available online 7 June 2018

Abstract We consider a symmetric restricted three-body problem on surfaces M2κ of constant Gaussian curvature κ = 0, which can be reduced to the cases κ = ±1. 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 M2κ , 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 = ±1 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 2-tori related to these orbits. © 2018 Elsevier Inc. All rights reserved.

MSC: primary 70F07; secondary 70G60, 37D40

* Corresponding author.

E-mail addresses: [email protected] (J. Andrade), [email protected] (E. Pérez-Chavela), [email protected] (C. Vidal). https://doi.org/10.1016/j.jde.2018.06.007 0022-0396/© 2018 Elsevier Inc. All rights reserved.

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Keywords: The restricted three body problem; Surfaces of constant curvature; Hamiltonian formulation; Periodic solutions; KAM tori; Averaging theory

1. Introduction A first approach of the N -body problem on non Euclidean spaces was raised simultaneously by N. Lovachevsky [20], and by J. Bolyai [3] in the 1830’s. In 1860 P.J. Serret extended the gravitational force to the sphere S2 and solved the corresponding Kepler problem [28]. In 1885, W. Killing adapted the Bolyai–Lobachevsky gravitational law to S3 and introduced the cotangent potential as a good extension of the classical potential [16]. One of the fundamental results on the area is due to H. Liebmann, who in 1902 proved that the orbits of the curved Kepler problem are indeed conics in S3 and H3 and a bit later, he proved an important result corresponding to an S2 - and H2 -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 [18,19]). Relevant and more recent works correspond to those obtained by V. Kozlov [17] 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 [1]. The curved N -body problem for N ≥ 2 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 [11], [12], where a clear and detailed deduction of the equations of motion for any number N ≥ 2 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 2-dimensional spaces of constant curvature κ are characterized by the sign of the curvature as the surfaces M2κ = {(x, y, z) ∈ R3 : x 2 + y 2 + σ z2 = κ −1 }, with σ = sign(κ), which induces to define the inner product (ax , ay , az )  (bx , by , bz ) := ax bx + ay by + σ az bz . √ If κ > 0, then the surface is the 2 dimensional sphere of radius R = 1/ κ denoted by S2κ embedded in the Euclidean space R3 . If κ = 0, we recover the Euclidean space R2 . If κ < 0, then the surface is the upper sheet of the hyperboloid x 2 + y 2 − z2 = κ −1 , with z > 0, denoted by H2κ , which corresponds to a surface embedded in the 3 dimensional Minkowski space R2,1 (R3 endowed with the Lorentz inner product). According to [11] the equations of motion for N ≥ 2 bodies on a surface of curvature κ are given by

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q¨ i =

N  j =1,j =i

  mj |κ|3/2 qj − σ (κqi  qj )qi − σ (q˙ i  q˙ i )qi ,  3/2 σ − σ (κqi  qj )2

(1)

with the constraints qi  qi = κ −1 ,

qi  q˙ i = 0,

i = 1, · · · , N,

σ = sgn(κ).

We observe that for κ = 0, equation (1) can be written independently of the curvature by 3/4 and the change of coordinates q → |κ|−1/2 q , so we will means of a rescaling dτ i i dt = |κ| consider along this work that κ = σ = ±1. The restricted three body problem was introduced by L. Euler [14]. In his work, he considered three mutually interacting masses m1 , m2 , m3 , where m3 is negligible so that it does not have influence in the motion of m1 and m2 , but its dynamics is entirely determined by the motion of them. This problem can be extended naturally to the restricted (N + 1)-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 = ±1, we start by considering an elliptic relative equilibrium of the curved 2-body problem, that is, a solution where the primaries are rotating uniformly with constant angular velocity ω on fixed parallels of S21 or H2−1 (this type of solutions will be described in detail in Section 2). As usual we normalized the masses as m1 = μ, m2 = 1 − μ and assume that the primaries are moving on circles of radius r1 and r2 , respectively. We call this problem the restricted three body problem on surfaces of constant curvature (R3BP on S2 or H2 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 r1 = r2 = a. We observe that the above problem depends on two parameters (μ, r), where r is either r1 or r2 . When r1 = r2 and consequently μ = 1/2 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 [22], 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 r1 = r2 = a, where a ∈ (0, 1) if κ = 1; and a ∈ (0, ∞) if κ = −1. We refer it as the symmetric R3BP on S2 or H2 . 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 [26] and [10], we work with intrinsic coordinates on the curved plane, i.e., R2 (in the case of S2 ) or the Poincaré Disk (in the case of H2 ), 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=

2 2 (p + pv2 ) + ω(vpu − upv ) − U (u, v), 8 u

(2)

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,  = 1 + σ (u2 + v 2 ) and U (u, v) is the potential written in intrinsic coordinates. A preliminary study of this model was given by R. Martínez and C. Simó in [22]. In their work, they obtained the relative equilibria depending on the parameters κ and the mass ratio

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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 = 1/2). 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 S2 and the vertex of H2 for certain values of the parameter a (in order to get these families we use the Lyapunov Center Theorem [23]). In the case of S2 we also prove the existence of KAM 2-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 [27], [24], [23]), 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 [2,24]). More precisely, we have used Han, Li and Yi’s Theorem (see details in [15]) that applies in the case of Hamiltonian systems with high-order proper degeneracy. The paper is organized as follows. In Section 2 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 = r1 = r2 , 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 S2 and the vertex of H2 , 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 2-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 7.1. 2. Relative equilibria in the 2-body problem on S2 and H 2 We consider the motion of two particles of normalized masses m1 = μ and m2 = 1 − μ, with 0 < μ ≤ 12 and positions qi = (xi , yi , zi ) ∈ M2σ . We shall assume σ = 1 or σ = −1, where M21 = S2 and M2−1 = H2 . According to (1) for N = 2 and κ = ±1, the equations of motion are given by

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J. Andrade et al. / J. Differential Equations 265 (2018) 4486–4529 1 , q¨ 1 = −σ (q˙ 1  q˙ 1 )q1 + m2  q2 −σ (q1 q22)q3/2

σ −σ (q1 q2 )

q¨ 2 =

2 −σ (q˙ 2  q˙ 2 )q2 + m1  q1 −σ (q1 q22)q3/2 , σ −σ (q1 q2 )

(3)

with the restrictions σ (qi  qi ) = 1,

qi  q˙ i = 0,

i = 1, 2.

(4)

We observe that the above system has singularities at collision (q1 = q2 ). Additionally, on S2 we also have antipodal singularities, that is when (q1 = −q2 ). A complete analysis of the singularities of system (3) can be found in [12]. We start properly this paper with the definition of an elliptic relative equilibrium (see [11]). Definition 2.1. An elliptic relative equilibrium is a solution of (3) of the form qj = (rj cos(ωt + αj ), rj sin(ωt + αj ), zj ),

(5)

where zj is a real  number, with zj ∈ (−1, 1) if σ = 1 or zj ∈ (1, ∞) if σ = −1, αj ∈ [0, 2π ], ω = 0 and rj = σ − σ zj2 . 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 qj = rj ei(ωt+αj ) + zj e3 , with e3 = (0, 0, 1).

(6)

Therefore, in the way to search elliptic relative equilibria we must find relations between the parameters z1 , z2 , m1 , m2 , ω, α1 and α2 , in order that (5) and (3) hold. The elliptic relative equilibria of the curved 2-body problem has been studied in [11] and [22], however, we find an explicit parametrization of the initial conditions generating this kind of solutions. Proposition 2.1. Given 0 < μ ≤ 1/2. The elliptic relative equilibria must have initial conditions such that α1 = α2 + π . For σ = −1, there are exactly two elliptic relative equilibria, both with the same configuration, but with angular velocity ±ω(z2 , μ), respectively. For σ = 1, all elliptic relative equilibria are located on the same hemisphere, i.e., z1 z2 > 0 and there are values 0 < z2− (μ) ≤ z2+ (μ) such that for 0 < μ < 1/2 • • • •

If z2 ∈ (0, z2− ) ∪ (z2+ , 1), then there are 4 elliptic relative equilibria. If z2 ∈ (z2− , z2+ ), then there is not elliptic relative equilibrium. If z2 ∈ {z2− , z2+ }, then there are 2 elliptic relative equilibria. If μ = 1/2, then there are 4 elliptic relative equilibria.

Proof. From Definition 2.1, we have q˙ j = iωrj ei(ωt+αj ) , q˙ j  q˙ j = ω2 rj2 ,

q¨ j = −ω2 rj ei(ωt+αj ) ,

q1  q2 = r1 r2 cos(α1 − α2 ) + σ z1 z2 := α.

(7)

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Let be r = (1 − α 2 )3/2 , then replacing relations (7) into the first equation of (3) we get    −ω2 r1 ei(ωt+α1 ) = −σ ω2 r13 eiα1 + mr2 r2 eiα2 − σ αr1 eiα1 eiωt   − σ ω2 r12 z1 + mr2 (σ αz1 − z2 ) e3 .

(8)

We observe that in order to get a relative equilibrium we must have that the coefficient of e3 vanish, then comparing the two first components we obtain −ω2 r1 = −σ ω2 r13 +

m2 m2 sin(α2 − α1 ), (r2 cos(α2 − α1 ) − σ α1 ) + i r r

(9)

from here we get that α1 − α2 = kπ , k = 0, 1. So, it seems that two types of solutions do exist. With this condition on the angles α1 and α2 , we obtain that α = (−1)k r1 r2 + σ z1 z2 . 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 ω2 r12 z1 = ω2 r22 z2 =

m2 r (σ z2 m1 r (σ z1

− αz1 ), − αz2 ).

(10)

By straightforward computations using (10), the normalized masses and the value of α, we obtain that z1 , z2 and μ are linked through the following equation  r1 z1 k+1 1 = (−1) −1 . r2 z2 μ

(11)

Note that k = 0 implies that z1 z2 < 0, which means that in H2 we must have necessarily k = 1. The same happens in S2 , indeed, taking k = 0 and σ = 1 in the first equation of (10) we obtain  m2 z2 ω2 r2 = r1 − r2 < 0. r z1 Therefore, for both surfaces, we must have k = 1, i.e., the initial conditions of the particles has the form P1 = (r1 , 0, z1 ) and P2 = (−r2 , 0, z1 ). Even more, for S2 , the particles must be located on the same hemisphere, thus, without loss of generality, we can consider z1, z2 ∈ (0, 1). The polynomial form of (11) is p(z1 ) = z14 − z12 + β 2 z22 (1 − z22 ) = 0, with β =

1 − 1, μ

(12)

whose discriminant is given by  = 1 − 4β 2 z22 (1 − z22 ), then, for each solution of (12), we have two relative equilibria with angular velocities ±ω, respectively. For the case σ = −1, we observe p(1) = βz22 (1 − z22 ) < 0,

p (z1 ) = 2z12 (2z12 − 1) > 0 and  > 0,

∀z1 , z2 > 1,

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which implies that there is a unique solution of (12). For σ = 1, there are two values of z2 , 0 < z2− ≤ z2+ < 1, given by

z2±

=

√ 1 − μ ± 1 − 2μ , 2(1 − μ)

such that: • If z2 ∈ (0, z2− ) ∪ (z2+ , 1), then (11) has two solutions z1− , z1+ ∈ (0, 1) given by

z1±

=

√ 1±  . 2

• If z2 ∈ (z2− , z2+ ), then (11) has no solutions. • If z2 ∈ {z2− , z2+ }, then (11) has a unique solution z1 =

√1 . 2

For the special case of equal masses, we have two types of relative equilibria, namely, z1 = z2  (valid for σ = ±1) and z1 = 1 − z22 (valid only for σ = 1). In H2 , the relative equilibria are determined by z1+ , which is defined for all z2 > 1.

2

Remark 2.1. For the case of S2 , we obtain that z1 = 0, if and only if, z2 = 0, while for H2 , z1 = 1, if and only if, z2 = 1. Both cases represent a singularity of equations (3). 3. 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 M2σ 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 = (x1 , x2 , x3 ) are given by q2 (t) − σ (q  q2 (t))q q1 (t) − σ (q  q1 (t))q ˙ + μ q¨ = σ (q˙  q)q 3/2 + (1 − μ)  3/2 , 2 σ − σ (q  q1 (t)) σ − σ (q  q2 (t))2

(13)

with the corresponding restrictions σ (q  q) = 1,

q  q˙ = 0.

(14)

In order to avoid the above restrictions (14), we introduce intrinsic coordinates by means of the stereographic projection (see [10] or [26] for more details)

 : M2σ −→ R2 , Q → q

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with Q = (x1 , x2 , x3 ) and q = (x, y) such that x=

x1 , 1 − x3

y=

x2 . 1 − x3

(15)

The inverse function is given by x1 =

2x , 

x2 =

2y , 

x3 = 

−2 , 

(16)

with  = 1 +σ (x 2 +y 2 ). For σ = −1, we take  = −1; whereas for σ = 1,  = 1 if the projection is taken from the north pole and  = −1 if the projection is taken from the south pole. The metric of M2σ becomes ds 2 = λ(x, y)(dx 2 + dy 2 ), where λ(x, y) = 42 is the conformal factor. We denote by R2sph the Riemann space (R2 , g), where g is the metric  g=

λ 0 0 λ

,

and we call it the curved plane. On R2sph , the kinetic energy takes the form 1 2 ˙ q ˙ g = 2 (x˙ 2 + y˙ 2 ), T = q, 2  while the potential in the coordinates (x, y) becomes U (x, y, t) = 

(1 − μ)α2 + , σ (2 − α12 ) σ (2 − α22 ) μα1

where α1 = 2σ r1 (x, y) · (cos ωt, sin ωt) + z1 ( − 2), α2 = −2σ r2 (x, y) · (cos ωt, sin ωt) + z2 ( − 2). The Euler’s equation associated to the Lagrangian L = T + U gives the equations of motion 4x˙ (x x˙ + y y) ˙ −  4y˙ (x x˙ + y y) ˙ − y¨ = 

x¨ =

2x 2 (x˙ + y˙ 2 ) +  2y 2 (x˙ + y˙ 2 ) + 

2 Ux , 4 2 Uy . 4

In addition, we can consider that the dynamics of the infinitesimal mass particle is governed by a Hamiltonian system (H, R2sph , ), with Hamiltonian H = T − U . The symplectic two form is given by

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= dx ∧ dpx + dy ∧ dpy = d(xdpx + ydpy ), with px =

4 x, ˙ 2

py =

4 y. ˙ 2

(17)

Thus, in terms of the conjugate variables (px , py ), the Hamiltonian function takes the form H (x, y, px , py , t) =

2 2 (p + py2 ) − U (x, y, t). 8 x

(18)

In order to eliminate the dependence of the time in the potential, we introduce rotating coordinates. If q = (x, y), p = (px , py ), Q = (u, v) and P = (pu , pv ), then the symplectic change of coordinates (q, p) = (Rωt Q, Rωt P ), where  cos ωt − sin ωt , Rωt = sin ωt cos ωt transforms (18) into an autonomous Hamiltonian H given by H=

2 2 (p + pv2 ) + ω(vpu − upv ) − U (u, v), 8 u

(19)

where μν1 (1 − μ)ν2 U (u, v) =  + , σ (2 − ν12 ) σ (2 − ν22 ) and ν1 = 2σ r1 u + z1 ( − 2),

ν2 = −2σ r2 u + z2 ( − 2),

 = 1 + σ (u2 + v 2 ).

The corresponding Hamilton’s equations are ∂H u˙ = ∂p = u

v˙ = p˙ u = p˙ v =

2 4 pu + ωv, ∂H 2 ∂pv = 4 pv − ωu, H 2 2 − ∂∂u = − σ u 2 (pu + pv ) + ωpv H 2 2 − ∂∂v = − σ v 2 (pu + pv ) − ωpu

+ Uu ,

(20)

+ Uv ,

with Uu =

∂U ∂u

Uv =

∂U ∂v

 −uν1 +(r1 +z1 u) −uν2 +(−r2 +z2 u) = 2 μ + (1 − μ) , 3/2 3/2 l1 l2  1 2 = 2v μ z1 −ν + (1 − μ) z2 −ν , 3/2 3/2

and lk = 2 − νk2 , k = 1, 2.

l1

l2

(21)

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Remark 3.1. We observe that through the projection −1 , the primaries are located at Pi = (u∗i , 0), with u∗1 =

σ (1 − z1 ) , r1

u∗2 = −

σ (1 − z2 ) . r2

(22)

For the case of S2 the antipodal singularities are at P˜i = (u˜ ∗i , 0), with u˜ ∗1 = −

(1 + z1 ) , r1

u˜ ∗2 =

(1 + z2 ) . r2

(23)

Through the projection + , the values of u∗ and u∗ ˜ 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 r1 = r2 = a. It is easy to check that the angular velocity of the primaries is given by ω2 = on the u-axis. (8a 3 (1 − σ a 2 )3/2 )−1 and they are located symmetrically

4ωv 4ωu All equilibria of system (20) have the form u, v, − 2 , 2 . They are obtained by solving the system F1 (u, v) = F2 (u, v) =

4ω2 u(2−) 3 4ω2 v(2−) 3

+ Uu (u, v) = 0,

(24)

+ Uv (u, v) = 0.

The next result follows easily using (24). Proposition 4.1. For σ = 1, the north and south poles are equilibrium solutions for any 0 < a < 1, whereas for σ = −1, the bottom point (by short the pole) is an equilibrium for any a > 0. Lemma 4.1. A necessary condition to have an equilibrium solution of (20) when r1 = r2 , is that u · v = 0. Proof. Suppose that u · v = 0, then from (24) we obtain that vF1 = uF2 or equivalently  0 = vF1 − uF2 = av

2

1 3/2

l1

−

1



3/2

,

l2

since v = 0, then  0 = l2 − l1 = 8au 1 − σ a 2 (2 − ), and since u = 0, we must necessarily have  = 2; replacing this value into (24) we obtain √  1 − σ a2u F1 = , (σ (1 − a 2 u2 ))3/2

√  1 − σ a2v F2 = , (σ (1 − a 2 u2 ))3/2

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Table 1 Number of isosceles equilibria on S2 . Isos. Eq.

(0, a2 )

a2

(a2 , a3 )

a3

(a3 , a4 )

[a4 , 1)

4

2

0

2

4

2

which contradicts the fact that (u, v) satisfies (24). Therefore u · v = 0 and the proof is complete. 2 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   √ ω2 1 − σ v 2 1 − σ a2 fis (v; a) =  , 4 − l 3/2 1 + σ v2 where l = σ the zeroes of



1 + σ v2

2

(25)

 2   − 1 − σ a 2 1 − σ v 2 , whereas collinear equilibria are given by

  4ω2 u 1 − σ u2 ρ1 ρ2 fcol (u; a) =  4 − 3/2 − 3/2 , l1 l2 1 + σ u2

(26)

where √   ρj = 2u 1 − σ a 2 + (−1)j a 1 − σ u2 , √ 2  2  l1 = σ 1 + σ u2 − 2aσ u + 1 − σ a 2 1 − σ u2 ,  √ 2  2  l2 = σ 1 + σ u2 − −2aσ u + 1 − σ a 2 1 − σ u2 .

(27)

Remark 4.1. We observe that in the case of σ = 1, that is, on S2 , 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 > 0 (of course here the poles have been excluded). Proposition 4.2. For σ = 1, all isosceles equilibria are symmetrically located on the upper hemisphere H + . There are three bifurcation values given by  a2 ≈ 0.396255849301291...,

a3 ≈ 0.797325392579823...,

a4 =

4− 2

√ 2

.

Depending on the value of the parameter a ∈ (0, 1), the number of isosceles equilibria are given in Table 1.

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Proof. From equation (24), the value v must satisfy the equation ω2 (1 − v 2 ) = 4

√ 1 − a2 > 0, r 3/2

(28)

with  = 1 + v 2 , which implies that |v| < 1, i.e., the equilibrium always remains on the upper hemisphere H + (recall the change of coordinates (15)). A solution of (28) is a curve (a, v(a)) 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 v(a) = 0, and by replacing it into (28) we obtain the equation 8a 4 − 16a 2 + 7 = 0, whose unique solution in the interval (0, 1) is given by  √ 1 a= 4 − 2. 2 Now, the solutions of (28) are contained in the set of solutions of the polynomial p(v) + + + + + + + +

14 12 10 − 256a 8 + 63a 6 =   64a14 − 256a 12+ 384a 10 − 2048a 8 + 520a 6 − 12a 4 v 2  512a 14− 2048a 12+ 3072a 10 8 6 4 2 4 1792a 14 − 7168a 12+ 10752a 10− 7168a +8 1764a +6 72a − 448a v 2  6 3584a 14 − 14336a 12 + 21504a 10 − 14336a 8 + 3640a 6 − 180a 4 + 192a 2 − 64 v 8 4480a 14 − 17920a 12 + 26880a 10 − 17920a 8 + 4410a 6 + 240a 4 − 288a 2 + 128 v10 + 3640a − 180a + 192a − 64 v  3584a 14 − 14336a12 + 21504a10 − 14336a 8 + 1764a 6 + 72a 4 − 48a 2 v 12 1792a − 7168a + 10752a − 7168a   14 12 + 3072a 10 − 2048a 8 + 520a 6 − 12a 4 v 14  512a14 − 2048a 64a − 256a 12 + 384a 10 − 256a 8 + 63a 6 v 16 . (29)

The number of roots of p(v) changes when it reaches a double zero, which occurs for the values of the parameter a such that q(a) = Resultant(p(v), p (v)) = 0, where  3  3 q(a) = α(1 − a 2 )78 a 78 8a 4 − 16a 2 + 7 8a 4 − 16a 2 + 9  4 × 16384a 16 − 81920a 14 + 163840a 12 − 163840a 10 + 81920a 8 − 16384a 6 + 27 , (30) which has three zeroes in the interval (0, 1) given by a2 ≈ 0.3962558493012911,

a3 ≈ 0.797325392579823,

a4 =

 √ 1 4 − 2. 2

After taking a test value of the parameter on each interval determined by a2, a3 , a4 , we obtain: • If a ∈ (0, a2 ), then there exist two pairs of symmetric isosceles equilibria, namely, ±Lv1 , ±Lv2 ; • If a = a2 , then there is one pair of symmetric isosceles equilibria, namely, ±Lv12 ; • If a ∈ (a2 , a3 ), then there are no isosceles equilibria; • If a = a3 , then there is one pair of symmetric isosceles equilibria, namely, ±Lv34 ;

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Fig. 1. Parameter bifurcation digram for the existence of equilibria in the symmetric R3BP on S2 : (a) Isosceles equilibria with a ∈ (0, a2 ]; (b) Isosceles equilibria with a ∈ [a3 , 1); (c) Collinear equilibria with a ∈ (0, a1 ]; (d) Collinear equilibria with a ∈ [a5 , 1).

• If a ∈ (a3 , a4 ), then there are two pairs of isosceles symmetric equilibria, namely, ±Lv3 , ±Lv4 ; • If a ∈ [a4 , 1), then there is one pair of symmetric isosceles equilibria, namely, ±Lv5 . Thus, our analysis agrees with Table 1 and the relative position can be viewed in Fig. 1 (a) and (b). 2 Proposition 4.3. For σ = −1 and any value of the parameter a > 0, there are exactly two isosceles equilibria symmetrically located. Proof. From equation (25), the isosceles equilibria for σ = −1 are obtained by solving g1 (v) = g2 (v) where ω(v ˜ 2 + 1) g1 (v) =  4 , 1 − v2

g2 (v) =

1 l 3/2

,

 2 with ω˜ = (8a 3 a 2 + 1 )−1 . The functions g1 and g2 have the following properties: g1 (v) =

    2ωv ˜ 3v 2 + 5 6v a 2 v 2 + a 2 + 2

< 0,  5 > 0, and g2 (v) = − l 5/2 1 − v2

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Table 2 Number of collinear relative equilibria on S2 for different values of the parameter a.  (0, a1 ) a1 (a1 , a5 (a5 , 1) In H + 4 2 0 0 0 0 0 2 In H −

g1 (0) = ω˜ <

1 = g2 (0), a3

g1 (v) → +∞, when v → 1,

which assure that g1 and g2 intersect at a single point in the interval 0 < v < 1 for any value of a > 0. 2 Proposition 4.4. For σ = 1, the collinear relative equilibria are symmetrically located. There are two bifurcation values given by a1 ≈ 0.337849954103973...,

√ 3 a5 = ≈ 0.8660254... 2

The number of relative equilibria in the intervals determined by the bifurcation values are given in Table 2. Proof. We define   4ω2 u 1 − u2 α=  4 , u2 + 1 √ √     a 1 − u2 − 2 1 − a 2 u a 1 − u2 + 2 1 − a 2 u β= − , 3/2 3/2 l1 l2 with l1 and l2 as defined in (27) with σ = 1. Then, solving fcol (u; a) = 0 is equivalent to solve the equation α + β = 0.

(31)

In order to eliminate the roots in the denominators of β, we need to solve the equation α 2 − β 2 = 0.

(32)

In developing β 2 , the term (l1 · l2 )3/2 appears in the denominator and it is given by  (l1 · l2 )3/2 =

(a − 2u + u2 )3 (a + 2u + u2 )3 if u ∈ (0, u∗1 ) ∪ (u˜ ∗2 , +∞), −(a − 2u + u2 )3 (a + 2u + u2 )3 if u ∈ (u∗1 , u˜ ∗2 ).

For u ∈ (0, u∗1 ) ∪ (u˜ ∗2 , +∞), equation (32) becomes   64a 2 1 − a 2 8 −  4  4 = 0, u2 + 1 au2 + a − 2u au2 + a + 2u

1 3 

 4a 6 1 − a 2

(33)

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or equivalently, in its polynomial form p1 (u)p2 (u)p3 (u) = 0, with       p1 (u) = a 2 4a 2 − 5 u4 + 2 4a 4 − 5a 2 + 2 u2 + a 2 4a 2 − 5 , 4 2 2 2 2 p2 (u) = a 2 4a 2 − 3 u4 + 2 4a  8− 3a 2− 2 u6 + a 44a − 32 ,  6 4 4 2 p3 (u) = a 16a − 32a + 17 u + 4a 16a − 32a + 17a − 2 u  8 − 96a 6 + 51a 4 − 8a 2 + 8 u4 + 4a 2 16a 6 − 32a 4 + 17a 2 − 2 u2 +2 48a   +a 4 16a 4 − 32a 2 + 17 .

(34)

We state that the polynomial p3 does not provide solutions of (32), in fact, we obtain that the resultant between p3 (u) and p3 (u) is given by 2 3

12 Resultant(p3 (u), p3 (u)) = αa 28 1 − a 2 16a 4 + 1 16a 4 − 32a 2 + 17 = 0, for all a ∈ (0, 1) and α a nonzero constant, which implies that the number of zeroes of p3 (u) does for any value of a,  8not vary  thus, we can put a = 1/2 and p3 takes the form p3 (u) = 1 6 4 2 8 5u + 4u + 126u + 4u + 5 , which does not have zeroes which proves the statement. The zeroes of p1 and p2 can be easily computed and it is verified that the zeroes of p1(u) do not provide zeroes of (32). Nevertheless, the polynomial p2 gives one positive solution of (32)

1 u3 = a

√ −4a 4 + 3a 2 + 2 + 2 −4a 4 + 3a 2 + 1 , 4a 2 − 3

(35)

√

which is defined for a ∈ (a5 , 1), with a5 = 23 . It is easy to verify that (35) satisfies u > 1, and therefore, the corresponding collinear equilibrium is always located at the lower hemisphere. If u ∈ (u∗1 , u˜ ∗2 ), then equation (32) assumes the form   2 2 4 a 2 u4 − 6a 2 u2 + a 2 + 4u2 u2 1 − u2  3  8 −  4  4 = 0, au2 + a − 2u au2 + a + 2u 4a 6 1 − a 2 u2 + 1 whose polynomial form is given by   p4 (u) =  16a 16 − 48a 14 + 48a 12 − 16a 10 u24  + −64a 16 + 320a 14 − 576a 12 + 448a 10 − 127a 8 u22  + −480a 16 + 1696a 14 − 1952a 12 + 480a 10 + 518a 8 − 272a 6 u20  + 192a 16 − 3008a 14 + 9920a 12 − 13632a 10 + 8589a 8 − 2112a 6 + 96a 4 u18 +a16 u16 + a14 u14 + a10 u10 + a8 u8  6 + 96a 4 u6 + 192a 16 − 3008a 14 + 9920a 12 − 13632a 10 + 8589a 8 − 2112a  8 6 4 + −480a 16 + 1696a 14 − 1952a 12 + 480a 10 + 518a  2 − 272a u 16 14 12 10 8 + −64a + 320a − 576a + 448a − 127a u +16a 16 − 48a 14 + 48a 12 − 16a 10 , with

(36)

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a16 = 5872a 16 − 30928a 14 + 64720a 12 − 67312a 10 + 34824a 8 − 7232a 6 +192a 4 − 256a 2 , a14 = 16256a 16 − 79232a 14 + 154496a 12 − 150656a 10 + 73458a 8 −14272a 6 − 96a 4 + 256, a12 = 21952a 16 − 105280a 14 + 202048a 12 − 193984a 10 + 93156a 8 − 17760a 6 −384a 4 + 512a 2 − 512, a10 = 16256a 16 − 79232a 14 + 154496a 12 − 150656a 10 + 73458a 8 − 14272a 6 −96a 4 + 256, a8 = 5872a 16 − 30928a 14 + 64720a 12 − 67312a 10 + 34824a 8 − 7232a 6 + 192a 4 −256a 2 . The resultant between p4 (u) and p4 (u) is given by Resultant(p4 (u), p4 (u)) = α p(a), ˜

(37)

where α is a nonzero constant and  p(a) ˜ = 16777216a 24 − 100663296a 22 + 264241152a 20 − 398458880a 18 +380436480a 16 − 238288896a 14 + 98435072a 12 − 27525120a 10 4 +6138624a 8 − 1136128a 6 + 54336a 4 − 10560a 2 + 1511 . The zeroes of p(a) ˜ are a1 ≈ 0.33784995410397306 and a˜ ≈ 0.941199983272388. By considering a test value of the parameter on each interval determined by a1 , a5 and a, ˜ it verifies that If a ∈ (0, a1 ), there exist two pairs of symmetric collinear equilibria, namely, ±Lu1 , ±Lu2 ; If a = a1 , then there is one pair of symmetric collinear equilibria, namely, ±Lu12 ; If a ∈ (a1 , a5 ], then there are no collinear equilibria; ˜ then there is one pair of symmetric equilibria, which is the one determined by If a ∈ (a5 , a), (35), namely, ±Lu3 ; ˜ 1), then there is one pair of symmetric equilibria, which is the one determined by • If a ∈ [a, (35), namely, ±Lu3 .

• • • •

From the above, we conclude that a = a˜ 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 = a1 and a = a5 (see Fig. 1(c) and (d)), which completes the proof. 2 Proposition 4.5. For σ = −1 and any value of the parameter a > 0, there are exactly two collinear equilibria symmetrically located. Proof. From equation (26) with σ = −1, we have that the relative equilibria are given by the intersection of two functions f1 and f2 , with

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J. Andrade et al. / J. Differential Equations 265 (2018) 4486–4529

f1 (u) =

4ω2 u(1 + u2 ) , (1 − u2 )4

f2 (u) =

ρ1 3/2 l1

+

ρ2 3/2

.

l2

The function f1 is increasing, positive and f1 (u) → +∞ when u → 1. We separate our analysis in two intervals, (0, u∗1 ) and (u∗ , 1). For the first one, it is easy to verify that ρ1 < 0 and 3/2 3/2 ρ2 > 0, which implies that f¯2 := −ρ1 / l1 + ρ2 / l1 > 0. Since sgn(f2 ) = sgn(f2 · f¯2 ), where √ 16a 1 + a 2 u(1 + u2 )(4u2 + a 2 (1 + 6u2 + u4 )) ¯ f2 · f2 = − < 0, (−4u2 + a 2 (1 − u2 )2 )4 it follows that f2 (u) < 0 for all u ∈ (0, u∗1 ) and therefore, there are no equilibria in this interval. For u ∈ (u∗ , 1), we have ρ1 , ρ2 > 0, then f2 (u) > 0 and f2 (u) → +∞ when u → u∗1 , which implies that the graphics f1 and f2 must intersect at least at one point. Now, we compute f2 (u) =

β1 − β2 5/2 l1

−

β1 + β2 5/2

,

l2

where  2 4   2  2 2+4 , β1 = 4au √ a u + 10a + 8 u + 5a   β2 = 4 a 2 + 1 5a 2 u4 + 10a 2 + 4 u2 + a 2 . It is clear that β1 + β2 > 0 and sgn(β1 − β2 ) = sgn(β12 − β22 ). Since 

2 2 −au2 + a + 2u a 2 + 1 − a 2 u2 < 0, β12 − β22 = −16 −au2 + a − 2u it follows that f2 (u) < 0. Therefore, the graphics of f1 and f2 intersect at a single point located in the interval (u∗1 , 1), which completes the proof. 2 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 S2 . 

4ωu , with uv = 0, has the The linearization matrix at an equilibrium solution u, v, − 4ωv , 2 2   form 

A1 A3

A2 A4

,

where  A1 =

ω−

0 4u2 σ ω 

−ω

4v 2 σ ω 

0

 ,

A2 =

2 I2 , 4

J. Andrade et al. / J. Differential Equations 265 (2018) 4486–4529

  8ω2 u2 + v 2 σ  + 2u2 A3 = H ess(U ) − 0 4

0 σ  + 2v 2

4503

A4 = −AT1 .

,

The characteristic polynomial is given by the bi-quadratic polynomial 4 det B = β4 λ4 + β2 λ2 + β0 , 16

p(λ) =

(38)

where 4 (AT + λI2 )(A1 − λI2 ) = 2 1

B = A3 +



b11 b21

b12 b22

(39)

.

It is easy to verify that Uuv = Uvu = 0 for uv = 0, so the entries bij are given by b11 = b1 −

4λ2 , 2

b12 =

8ω(2 − )λ , 3

b21 = −b12 ,

b22 = b2 −

4λ2 , 2

with b1 = b2 =

4ω2 4 4ω2 4

 2    δ − 4σ u2 − 2 u2 + v 2 δσ + 2u2 − Uuu ,

 2    δ − 4σ v 2 − 2 u2 + v 2 δσ + 2v 2 − Uvv .

(40)

In this way, the characteristic polynomial takes the form p(λ) = λ4 + β2 λ2 + β0 , where β2 =

4ω2 (2−)2 2

β0 =

4 16 b1 b2

=

  2ω2 1+u4 +v 4 − 14 2 (Uuu + Uvv ) , 2       2 4 4 2 4ω2 3v 4 −u4 −8σ v 2 +1 4 4ω 3u −v −8σ u +1 + U uu 4 4 16  

−

2 4 (b1

+ b2 ) =

+ Uvv .

(41)

5.1. Linear stability for the isosceles relative equilibria For u = 0, β2 takes the form 2 √      1 − σ a2 1 − σ v2 1 + σ v2 2ω2 v 4 + 1 β2 = +  2 , l 3/2 σ v2 + 1 and using the condition fis (v; a) = 0, it follows that β2 = ω2 and the characteristic polynomial becomes p(λ) = λ4 + ω2 λ2 + β0 .

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Similarly, we obtain   12a 2 v 2 ω4 1 − σ v 2 q(v; a) β0 = , 2 l 2     with q(v; a) = 3 − 4σ a 2 v 4 + 8a 2 − 10σ v 2 + 3 − 4σ a 2 . The discriminant of p(λ) is given by  δ=ω

4

   48a 2 v 2 1 − σ v 2 q(v; a) 1− . 2 l 2

Theorem 5.1. For σ = 1, there exists a0 ≈ 0.36806586863987895... such that the stability of the isosceles relative equilibria is characterized as follows: • If a ∈ (0, a0 ), then ±Lv1 are unstable (saddle–saddle) and ±Lv2 are unstable (centersaddle); • If a = a0 , then Lv12 is spectrally stable; • If a ∈ (a0 , a2 ), then ±Lv1 are linearly unstable (saddle–saddle) and ±Lv2 are unstable (center-saddle); • If a = a3 , then Lv34 is spectrally stable; • If a ∈ (a3 , a4 ), then ±Lv3 are linearly stable (center–center) and ±Lv4 are unstable (centersaddle); • If a ∈ [a4 , 1), then ±Lv5 are unstable (center-saddle). Proof. Letting x = λ2 , the characteristic polynomial can be written as p(x) ˜ = x 2 + ω2 x + β0 .

(42)

If β0 < 0, then the discriminant δ = ω4 − 4β0 > 0, which implies that both roots of (42), x1 and x2 are real and also, by Descartes’ rule of signs, it verifies that x1 < 0 < x2 . So in this case, it follows that all isosceles relative equilibria with β0 < 0, have two real and two pure imaginary eigenvalues and, therefore, the corresponding relative equilibria are unstable. If β0 = 0, then equation (42) has roots x1 = 0 and x2 = −ω2 , which implies that all isosceles relative equilibria with β0 = 0 have 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 fis (v; a) = 0 and β0 (v; a) = 0 can be solved. In fact, its zeroes are given by solving the polynomial equation q(v; a) = 0, which has a unique solution such that v0 (a) < 1 and it is given by v0 (a) =

√ 4a 2 − 5 + 4 1 − a 2 , 4a 2 − 3

defined for a ∈ (0, a5 ). Replacing (43) into (28) yields the polynomial equation 16384a 16 − 81920a 14 + 163840a 12 − 163840a 10 + 81920a 8 − 16384a 6 + 27 = 0, which has two solutions into the interval (0, 1) given by

(43)

J. Andrade et al. / J. Differential Equations 265 (2018) 4486–4529

a ≈ 0.3962558493012911... = a2 ,

4505

a ≈ 0.797325392579823... = a3 ,

where a2 and a3 are the bifurcation values given in Proposition 4.2. For β0 > 0, the solutions of (42) are either both complex (if δ < 0) or both negative real numbers (if δ ≥ 0). The equation δ = 0 is equivalent to the equation δ˜ = 0, where     ˜ δ(v) = −a 4 v 12 + 136a 2 − 190a 4 v 10 + 769a 4 − 768a 2 − 16 v 8 +     −1156a 4 + 1264a 2 − 32 v 6 + 769a 4 − 768a 2 − 16 v 4 +   136a 2 − 190a 4 v 2 − a 4 ,

(44)

while, solving fis (v; a) = 0 is equivalent to solve (29). The equation ˜ Resultant(δ(v), p(v)) = p2 (a) = 0,

(45)

where  p2 (a) = αa 48 (1 − a 2 )12 21743271936a 40 − 282662535168a 38 + 1695975211008a 36 −6218575773696a 34 + 15546070486976a 32 − 27980269380608a 30 +37298163546624a 28 − 37280430091264a 26 + 27937095457024a 24 −15500373869568a 22 + 6189409662464a 20 − 1686738185216a 18 +284917230288a 16 − 26400323264a 14 + 2957123840a 12 − 1081023168a 10 4 +213652768a 8 − 13340224a 6 − 974976a 4 − 188480a 2 + 41975 , where α is a nonzero constant. The polynomial p2 (a) gives the values of a where the curve γ = (a, v(a)), which satisfies fis (v(a); a) = 0, crosses the boundary of the region of points (a, v) such that δ(a, 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 a0 ≈ 0.36806586863987895.... For a solution (a, v(a)) satisfying (28), we denote by γj = (a, vj (a)), j = 1, 2, 3, 4, 5, the curves of equilibria in the plane (a, v), where Lvj = (0, vj (a)), is defined in its corresponding domain given by Proposition 4.2. For a ∈ (0, a2 ), we consider the two equilibria Lv1 and Lv2 , with v1 (a) < v(a2 ) < v2 (a). It is verified that γ2 is always contained in the region β0 < 0 (region I in Fig. 2) and by the analysis done above, the equilibrium Lv2 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 = a2 (see Proposition 4.2) we have that along the curve fis (v; a) = 0, we have β0 = 0, therefore, the equilibrium Lv12 has one pair of pure imaginary and one double zero eigenvalues, which implies the spectral stability of Lv12 (spectral stability means that all eigenvalues λ satisfy λ2 ≤ 0). For the equilibrium Lv1 , the curve γ1 is always contained in the region β0 > 0 (region II ∪ III in Fig. 2). If a ∈ (0, a0 ), then γ1 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., Lv1 is an unstable equilibrium (saddle–saddle). If a ∈ (a0 , a2 ), the curve γ1 is contained in the region δ > 0, which implies that all its eigenvalues are pure imaginary and different from each other, i.e., Lv1 is a linearly stable center–center. If a = a0 , then along the curve γ1 , we have δ = 0, which implies that Lv1 has a double pure imaginary eigenvalue (i.e., 1 : 1 resonance), therefore, it is spectrally stable.

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Fig. 2. Stability of isosceles relative equilibria in the RC3BP on S2 . Region I: saddle-center. Region II: complex saddle. Region III: center–center.

For a = a3 , along the curve γ4 we have β0 = 0, therefore the equilibrium Lv34 has one pair of pure imaginary and double zero eigenvalues, which gives the spectral stability of Lv34 . For a ∈ (a3 , a4 ), there are two equilibria Lv3 = (0, v3 (a)) and Lv3 = (0, v4 (a)), with v3 (a) ≤ v(a3 ) ≤ v4 (a). The curve γ3 is always contained in the region β0 > 0, δ > 0 (corresponding to the region III in Fig. 2 (c)), this is because the only value of a, for which the curve fis (v; a) = 0 crosses the curve δ = 0 is a = a0 , which, by virtue of the above analysis, implies that the eigenvalues are all pure imaginary and different from each other, i.e., Lv3 is a linearly stable center–center. The curve γ4 is contained in the region β0 < 0 (i.e., region I in Fig. 2 (c)), which implies that the equilibrium Lv4 has one pair of opposite pure imaginary eigenvalues and two

J. Andrade et al. / J. Differential Equations 265 (2018) 4486–4529

4507

opposite nonzero real eigenvalues, so it is an unstable center-saddle. The same occurs with the curve γ5 when a ∈ [a4 , 1) (see Fig. 2), thus, Lv5 is an unstable center-saddle. This finishes the proof of Theorem 5.1. 2 For the case of negative curvature, we have the following result. Theorem 5.2. For σ = −1, all isosceles equilibria are complex saddle, i.e., all of their eigenvalues are complex with nonzero real part. Proof. Since q(v; a) > 0 for σ = −1, 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 fis (v; a) to their polynomial form, namely, p1 (v) and p2 (v), respectively, and computing ω4 the resultant between them, we obtain a polynomial p(a) = Resultant(p1 (v), p2 (v)), which has a unique zero a ≈ 0.360458658... and it generates one intersection between p1 and p2 , but it is outside the domain 0 < v < 1, 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 5.2. 2 5.2. Linear stability for the collinear relative equilibria For v = 0, β2 takes the form 2 β2 = g0 + 2 where g0 =

  2ω2 1+u4 , 2



g1 5/2

l1

g2

+

5/2

 ,

l2

g1 = g11 + g12 , g2 = g11 − g12 ,

   1 − a 2 σ 1 − σ u2 a 2 u4 + 4 − 14a 2 σ u2 + a 2 ,

   g12 = 2au 3σ a 2 − 2 + 8σ − 10a 2 u2 + 3a 2 σ − 2 u4 , g11 =



and

where kj =

b1 =

4ω2 (1 − 8σ u2 4

b2 =

4ω2 (1−σ u2 ) 3

 + 3u4 ) +

− 2



k1

3/2 l1

+

h1 h2 5/2 + 5/2 l1 l2

k2

3/2 l2

, (46)

,

√ 1 − a 2 σ − (−1)j aσ u and hj = h11 − (−1)j h12 , with √     h11 = 4 1 − a 2 σ 2a 2 σ u6 + 5a 2 u4 + 4 − 12a 2 σ u2 + a 2 ,       h12 = 2au a 2 u6 + 4 − 3a 2 σ u4 + 20σ − 25a 2 u2 + 11σ a 2 − 8 .

Lemma 5.1. Let (u, 0), with u > 0 a collinear equilibrium. For σ = 1, b2 < 0 if 0 < u < 1 and b2 > 0 if u > 1. For σ = −1, b2 < 0.

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J. Andrade et al. / J. Differential Equations 265 (2018) 4486–4529

Proof. From condition fcol (u; a) = 0, we obtain that   4ω2 u(1 − σ u2 ) k1 k2 ub2 = − 2u 3/2 + 3/2 3 l1 l2    ρ1 ρ1 k1 k2 =  3/3 + 3/3 − 2u 3/2 + 3/2 l1 l1 l1 l2   ρ1 − 2uk1 ρ2 − 2uk2 = + 3/2 3/2 l1 l2   1 1 = a 3/2 − 3/2 , l2 l1 and the result follows from the fact that l1 < l2 for 0 < u < 1 and l1 > l2 if u > 1 and σ = 1.

2

Theorem 5.3. For σ = 1, the stability of the collinear equilibria is characterized as follows: • If a ∈ (0, a1 ), then Lu1 is unstable (center-saddle) and Lu2 is linearly stable. • If a = a1 , then Lu12 is spectrally stable. • If a ∈ (a5 , 1), then Lu3 is linearly stable. Proof. First, we analyze the sign of β2 , which is equivalent to study the sign of  4β2 / = g˜ 0 − 2

g1 5/2

l1

+

g2



5/2

,

l2

with g˜ 0 = 8ω2 (1 + u4 )/4 . Now we compute  g˜ 02 −

g1 5/2

l1

+

g2 5/2

l2



2 = g˜ 02 −

g12 l25 + g22 l15 2g1 g2 ± 55 l˜110 l˜210 l˜1 l˜2

 ,

(47)

where l˜1 = au2 − 2u + a, l˜2 = au2 + 2u + a, the sign + is for u ∈ (0, u∗1 ) ∪ (u˜ ∗2 , +∞) and the sign − for u ∈ (u∗1 , u˜ ∗2 ). In the first case, (47) can be reduced to its polynomial form β2+ (u) = r(u)(r(u) + s(u)), with

  r(u) = 4a 10 − 8a 8 + 3a 6 u16 + 48a 8 − 102a 6 + 60a 4 u14

 + −80a 10 + 320a 8 − 416a 6 + 208a 4 − 48a 2 u12

 + −256a 10 + 720a 8 − 698a 6 + 292a 4 − 96a 2 + 64 u10

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 + −360a 10 + 912a 8 − 774a 6 + 288a 4 − 96a 2 u8

 + −256a 10 + 720a 8 − 698a 6 + 292a 4 − 96a 2 + 64 u6

 + −80a 10 + 320a 8 − 416a 6 + 208a 4 − 48a 2 u4

 + 48a 8 − 102a 6 + 60a 4 u2 + 4a 10 − 8a 8 + 3a 6 , 3 3 

au2 + 2u + a . s(u) = 2 1 + u4 au2 − 2u + a Now we see that the equation 4

4 Resultant(β2+ (u), p2 (u)) = α(1 − a 2 )24 a 48 1 + 2a 2 8a 4 − 4a 2 − 3 = 0, 

√

with α = 0, has a unique solution a∗ = 1+2 7 and its corresponding equilibrium is given by replacing this value into (35), in this way we obtain the position of the equilibrium

 √ √  1 u3 = 11 + 4 7 + 2 2 19 + 11 7 . 3 Replacing these two values into β2 , we obtain β2 ≈ 15.3865..., thus β2 > 0 along Lu3 . To study the sign of β2 along Lu1 and Lu2 , we consider the second case u ∈ (u∗1 , 1) (both equilibria belong the upper hemisphere). Thus, reducing (47) to its polynomial form β2− (u) and using the fact that the equilibrium points Lu1 and Lu2 are contained in the set of roots of the polynomial p4 (u) defined in (36), we proceed as in the previous case to obtain 0 = Resultant(β2− (u), p4 (u)) = αa 240 (1 − a 2 )120 f1 (a)4 f2 (a)4 , where α is a nonzero constant and f1 (a) = 2048a 16 − 2048a 14 − 1280a 12 + 1408a 10 + 480a 8 − 192a 6 − 8a 4 + 60a 2 + 9, f2 (a) = 524288a 24 − 3145728a 22 + 8257536a 20 − 12451840a 18 + 11907072a 16 − (48) 7520256a 14 + 3168256a 12 − 871680a 10 + 126816a 8 + 21952a 6 − 20928a 4 + 5592a 2 − 1073. By using the Sturm criterion, we obtain that f1 (a) does not have real roots, while f2 (a) has exactly one real root in the interval (0, 1), which is approximately a ≈ 0.980199..., however, the equilibria Lu1 and Lu2 exist only for a ∈ (0, a1 ), thus, the corresponding curves of equilibria do not cross the curve β2 = 0, and even more, β2 > 0 along both equilibria. Finally, we shall determine the sign of β0 . By virtue of Lemma 5.1, it is sufficient to study the sign of b1 . From condition fcol (u; a) = 0, we obtain 4ω2 (1 − 8u2 + 3u4 ) (1 − 8u2 + 3u4 ) = 4 u(1 − u2 )



ρ 1 l1 5/2

l1

+

ρ 2 l2 5/2

l2

 .

(49)

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Replacing condition (49) into b1 , we get 

b1 = =

(1−8u2 +3u4 )ρ1 l1 +u(1−u2 )h1 1 5/2 u(1−u2 ) l1 1 (t + t2 ), u(1−u2 ) 1 5/2

+

(1−8u2 +3u4 )ρ1 l2 +u(1−u2 )h2 5/2 l2



5/2

where t1 = (−t11 + t12 )/ l1 , t2 = (t11 + t12 )/ l2 , with          t11 = a 1 − u2 a 2 u8 + 28 − 44a 2 u6 + 166a 2 − 136 u4 + 28 − 44a 2 u2 + a 2 , √     2 8     t12 = 2 1 − a 2 u 5a u + 12 − 60a 2 u6 + 126a 2 − 40 u4 + 12 − 60a 2 u2 + 5a 2 . Thus, the equation b1 = 0 can be solved through the equation t12 − t22 = 0, which yields up the polynomial equation b˜1 (u) = 0, where       b˜1 (u) = a 4 u12 + 6a 4 − 24a 2 u10 + 15a 4 + 96a 2 − 48 u8 + 20a 4 − 272a 2 + 160 u6 +     15a 4 + 96a 2 − 48 u4 + 6a 4 − 24a 2 u2 + a 4 . Now, we compute Resultant(b˜1 (u), p4 (u)) for a ∈ (0, a1 ], which is proportional to the polynomial  a 80 (1 − a 2 )40 16777216a 24 − 100663296a 22 + 264241152a 20 − 398458880a 18 + 380436480a 16 − 238288896a 14 + 98435072a 12 − (50) 4 10 8 6 4 2 27525120a + 6138624a − 1136128a + 54336a − 10560a + 1511 . Again, applying Sturm criterion, we obtain that (50) has a unique zero in (0, a1 ] and it is given by a = a1 . It is verified that b1 (a1 , u(a1 )) = 0 and defining δj = (a, uj (a)), where Luj = (uj (a), 0), for j = 1, 2, 3, we have that δ1 is contained inside the region b1 (a, u) > 0, for all a ∈ (0, a1 ), which implies β0 < 0, while δ2 is contained inside the region b1 (a, u) < 0, for all a ∈ (0, a1 ), which implies β0 > 0. For the curve δ3 , we must solve the equation Resultant(b˜1 (u), p2 (u)) = 0 for a ∈ (a5 , 1), where this resultant is proportional to the polynomial 4

a 16 (1 − a 2 )8 64a 6 − 48a 4 − 4a 2 + 1 , which has just one zero in the interval (a5 , 1) given approximately by a∗ ≈ 0.8964262021183806... By evaluating b1 at (a∗ , u3 (a∗ )), where u3 is given in (35), we obtain b1 (a∗ , u3 (a∗ )) ≈ −0.0146411.... Therefore, a∗ does not provide simultaneous solutions of (31) and b1 = 0, i.e., the curve δ3 does not cross the boundary of the region b1 < 0. In fact, δ3 is contained inside the region b1 < 0, so, by Lemma 5.1, β0 < 0 along Lu3 . The last step consists in the determination of the sign of the discriminant δ = β22 − 4β0 along the equilibrium Lu2 . For this equilibrium we already know that b1 , b2 < 0 and β0 , β2 > 0 for all 2 a ∈ (0, a1 ). Since b1 + b2 < 0, then from definition of β2 given in (41), we have β2 > − 4 (b1 + b2 ) > 0, then

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Table 3 Type of eigenvalues and stability for the collinear relative equilibria of the R3BP on S2 . Lu1 Lu2 Lu12 Lu3

β2

β0

δ

x

>0 >0 >0 >0

<0 >0 =0 <0

>0 >0 >0 >0

x1 < 0 < x2 x1 < x2 < 0 −β2 , 0 x1 < 0 < x2

Table 4 Stability of the north pole.   √  √ 2 4− 2 a 0, 4− 2 2 Unstable



Linearly unstable

λ √ √ ±i |x1 |, ± x2 √ √ ±i |x1 |, ±i |x2 | √ 0, 0, ±i β2 √ √ ±i |x1 |, ± x2

√  4− 2 , 23 2

Linearly stable

Table 5 Stability of the south pole. √

√  3 a 0, 23 2 Unstable

δ>



Linearly unstable

Type of stability Unstable Linearly stable Spectrally stable Unstable

2 3

Linearly unstable

√



2,1 3

Unstable



3 2 ,1

Linearly stable

4 4 2 4 4 (b1 + b2 )2 − 4β0 = (b1 + b22 ) − b 1 b2 = (b1 − b2 )2 ≥ 0. 16 16 8 16

In order to summarize our analysis, putting x = λ2 , the characteristic polynomial takes the form p(x) ˜ = x 2 + β2 x + β0 . Let x1 and x2 be their roots. We describe the above analysis in Table 3. We observe that when a = a1 , then β0 = 0, therefore the equilibrium Lu12 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. 2 For the case of negative curvature, we have the following result whose proof is easier. Theorem 5.4. For σ = −1, the collinear equilibria are unstable, even more, they are centersaddle. Proof. For σ = −1, it is easy to verify that h1 , h2 > 0, then b1 > 0 and by Lemma 5.1, we have that β0 < 0, and the result follows. 2 We finish this section by studying the linear stability of the poles. Proposition 5.1. For σ = 1, 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 = −1, the pole is unstable for all a > 0.

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Table 6 Eigenvalues and linear stability of the north pole.

λ ρ

I1

 √ 4− 2 2

I2

∈ R+

0

∈ iR+

∈ iR+

2i

Unstable

Linearly unstable

 4

√ 50+44 2 73/4



I3

∈ iR+

2 3 √ i 15 7/4 2√ i 15 27/4

Linearly stable

Linearly unstable

Unstable

∈ R+ + iR ∈ R+ + iR

Table 7 Eigenvalues and linear stability of the south pole. √

3 2

J1

J2

λ

∈ R+

ρ

∈ iR+

√ 2i 5 33/4

∈ iR+

Unstable

Linearly unstable

0

∈ iR+ Linearly stable

Proof. The linearization matrix at (0, 0, 0, 0) is given by ⎛ Aσ

0 −ω 

⎜ ⎜ =⎜ ⎜ − 8 ⎝

1−a 2 σ a3

0

ω 0 

0

4 1−a 2 σ a3

1 4

0

⎞

⎟ ⎟ ⎟ 0 ω⎟ ⎠ −ω 0 0

1 4

(51)

with eigenvalues ±λ, ±ρ, where  λ = ω fσ (a),  fσ = ασ + 4(1 − σ a 2 ) βσ , ασ = −1 − 4(1 − σ a 2 )2 ,

 ρ = ω gσ (a),

 gσ = ασ − 4(1 − σ a 2 ) βεσ ,

(52)

βσ =  + 9(1 − σ a 2 )2 .

For S2 , the north that   polecorresponds to the origin with√ = −1 and σ = 1. It is verified + = 2 − 3a 2 4 − 3a 2 > 0, if and only if, a ∈ (0, 2/3) and in this interval g− (a) < 0,  √ while f−+ (a) > 0, if and only if, a ∈ I1 = (0, 4 − 2/2), then defining the intervals I2 =  √ √ √ ( 4 − 2/2, 2/3) and I3 = ( 2/3, 1), we have that the eigenvalues associated to the north √ √ pole of S2 are characterized in Table 6. For the cases a = 4−2 2 and a = 2/3, it is verified that the linearization matrices are not diagonalizable, thus, in both cases, the north pole is linearly unstable. For S2 , the south pole corresponds to the origin with  = 1 and σ = 1. It is verified that 2  + + β+ = 9 1 − a 2 + 1 > 0 and g+ (a) < 0 for all a ∈ (0, 1), while f++ (a) > 0, if and only if, a ∈ √ √ eigenvalues associated J1 = (0, 3/2). Defining the interval J2 = ( 3/2, 1), we have that the √ 3 2 to the south pole of S are characterized in Table 7. For the case a = 2 , it is verified that the linearization matrix is not diagonalizable, therefore, the south pole is linearly unstable. + β−

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For H2 , studying the pole corresponds to the origin with  = −1 and σ = −1. It is verified that

 

    3a 2 + 2 3a 2 + 4 > 0, f−− (a) = 2a 2 + 1 2a 2 + 3 + 4 a 2 + 1 for all a > 0 and

  − f−− (a) g− (a) = − 8a 4 + 16a 2 + 7 16a 4 + 32a 2 + 17 < 0, − thus, g− (a) < 0 for all a > 0, then λ ∈ R+ and ρ ∈ iR+ for all a > 0, which completes the proof. 2

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. 6.1. 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 [25]. 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 λ1 = iα, and the quotients λj /λ1 for j = 2, 3, · · · , l are not integer numbers, then there exist a family of periodic orbits emanating from the equilibrium whose minimal period tends to 2π/|α|. The proof of this theorem can be found in [23]. 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 

  √  ±(m4 + 6m2 + 1) + m2 + 1 9m4 − 2m2 + 9 1  ± am = 4− , m∈Z 2 2m4 + 5m2 + 2 be a sequence of values of the radial parameter a. Theorem 6.1. For the parameter a, assume that all eigenvalues corresponding to the poles are purely imaginaries, then for the case of positive curvature (on S2 ), the following statements hold: i) If a ∈ I1 ∪ I2 , then there is a family of τ (υ)-periodic orbits x(t, υ), emanating from the north + , for all integer pole, such that limυ→0 τ (υ) =  2π+ . Even more, for a ∈ I2 and a = am ω −g− (a)

m > 1, there is another family of τ (υ)-periodic orbits, with limυ→0 τ (υ) =

 2π . ω −f−+ (a)

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ii) If a ∈ J1 , then there is a family of τ (υ)-periodic orbits x(t, υ), emanating from the south . pole, such that, limυ→0 τ (υ) =  2π ω −g− (a)

iii) If a ∈ J2 , then there exists a family of τ (υ)-periodic orbits x(t, υ), emanating from the south − , for all integer pole, such that, limυ→0 τ (υ) =  2π+ . Even more, for a ∈ J2 and a = am ω −g+ (a)

m > 1, there is another family of τ (υ)-periodic orbits, with limυ→0 τ (υ) =

 2π . ω −f++ (a)

For the case of negative curvature, that is on H2 , there is a family of τ (υ)-periodic orbits x(t, υ), emanating from the pole, such that limυ→0 τ (υ) =  2π− . ω −g− (a)

Proof. To prove part i), we observe in  Table 6, that the intervals the Lyapunov Center  where √ √ √ Theorem might be applied are I1 = (0, 4 − 2/2)  and I2 = ( 4 − 2/2), 2/3). So, when

− a ∈ I1 , 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 ∈ I1 . + For a ∈ I2 , we know that λ, ρ ∈ iR+ , with g− (a) < f−+ (a) < 0, then

 2 |f + (a)| λ 0< = −+ < 1, ρ |g− (a)|

(53)

and it follows that λ/ρ ∈ / Z for all a ∈ I2 . 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 > 1, 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 / Z. Let m ∈ Z with m > 1, by solving the equation (ρ/λ)2 = m2 , we obtain the condition ρ/λ ∈ + solution a = am . By straightforward computations it is easy verify that am ∈ I2 for all integer m√ > 1. The ends of the interval I2 are reached when m = 1 and when m → +∞, that is a1+ = 2/3 and  √ + = 4 − 2/2, here we have resonances. Thus, for a ∈ I , ρ/λ ∈ limm→∞ am / Z, if and only if, 2 + , for all m ∈ Z, m > 1, which assures the existence of the other family of periodic orbits a = am 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 / Z is immediately verified any a ∈ (0, 1) such that a = 3/2. For any a ∈ J1 , the condition λ/ρ ∈ and part ii) follows. For the interval J2 , both eigenvalues λ and ρ are pure imaginary and also it is satisfied + (a) < f++ (a) < 0, which implies immediately that λ/ρ ∈ / Z, which gives the existence of g+ the family of periodic orbits. Since ρ/λ > 1, it is necessary to discard the values of the parameter a such that ρ/λ ∈ Z. Let m > 1 an integer, then the equation (ρ/λ)2 = m2 gives the − and it is verified that a − ∈ J for all integer m > 1, lim − solution a = am 2 m→1 am = 1 and m √ ρ − − limm→∞ am = 3/2. Thus, for a ∈ J2 , λ ∈ / Z, if and only if, a = am , for all m ∈ Z, m > 1, 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 6.1. 2

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6.2. Periodic orbits close to an integrable Hamiltonian We can also obtain periodic orbits by using perturbation theory, i.e., we can continue periodic orbits from an integrable Hamiltonian system through the appropriate introduction of a perturbating parameter and the application of Reeb’s Theorem [27]. Let us introduce the perturbation parameter ε in the symmetric problem by means of scaling the parameter a as a → ε 2 a and considering the time-scaling dt = ε 3 dτ . So, for small values of the parameter ε, the primaries rotate near the north pole (in the case of S2 ) or near the vertex (in the case of H2 ). In other words, this means that we are considering a perturbation of the curved Kepler problem (see [1], [4]). So, after applying the previous two transformations, we expand into power series over ε = 0 and introducing polar coordinates (ρ, θ, R, ) given by the symplectic transformation u = ρ cos θ,

v = ρ sin θ,

pu = R cos θ −

 sin θ, ρ

pv = R sin θ +

 cos θ, ρ

we obtain Hε = H0 () + ε3 H3 (ρ, R, ) + O(ε 4 ),

(54)

where H0 = α1 , and α1 = −

√1 . 2 2a 3/2

H3 =

 2 2    2 1 + σρ 2 + ρ R 2 ρ 1 + σρ 2 + 4σ ρ 2 − σ ,

8ρ 2

We call the Hamiltonian (54) as the limit problem.

Theorem 6.2. Let h be a fixed constant and fix H0 = h. For the Hamiltonian system associated to the Hamiltonian function (54) and for any admissible √ value of the parameter a and for ε small 4π 2a 3/2 ε 3 enough, there is a one-parameter family of T (ε) ≈ + O(ε 6 )-periodic solutions which h are strongly stable. Proof. Fixing h = H0 , the flow associated to the unperturbed Hamiltonian H0 is given by ϕ0t (ρ, θ, R, ) = (ρ, α1 t + , R, ), with  =

h . α1

In order to apply Reeb’s theorem (see [13], [24], [27]), first we compute the average of H3 over the flow ϕ0t , to obtain 1 H3 = T

T H3 (ϕ0s ) ds 0

1 = 2π

2π H3 dθ = H3 . 0

Note that, averaging the first perturbation is equivalent to normalize up to the first perturbation. Now we compute the critic points of H3 by solving ∇(ρ,R) H3 = (0, 0), and we get that there is a unique critical point, given by

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√ σ ( 1 + σ 4 − 1) ρ0 = , R0 = 0, 2

(55)

with  = 0 for σ = 1 and 0 < || < 1 for σ = −1. We verify that the matrix ⎛ ⎜ A = D 2 H3 (ρ0 , R0 ) = ⎜ ⎝

0

2  1+σ 4 − 1+σ 4

  2 1+σ 4 −  2 1− 1+σ 4

8

0

⎞ ⎟ ⎟ ⎠

 2 1+σ 4

has determinant = 0, then, by applying Reeb’s Theorem (see [27]), it follows that for 6 each ε small enough, there is a T (ε)-periodic solution of (54) ϕ(t, ε) = (ρ(t, ε), θ (t, ε), R(t, ε), (t, ε)), such that lim ϕ(t, ε) = (ρ0 , θ0 , 0, h/α1 ).

θ→0

Since, the eigenvalues of A are given by λ± = ±δi, with 

√ 1 + σ 4 δ1 − δ2 δ= 2 ,

√ 3 1 + σ 4 − 1

δ1 = 8 + 8σ 4 + 8,

  δ2 = 4 1 + σ 4 2 + σ 4 ,

it is easy to verify that δ > 0 for all θ = 0 in the case σ = 1 and 0 < || < 1 in the case σ = −1, which shows that, in fact, λ± ∈ iR and then, in virtue of [27], we obtain that the characteristic multipliers are given by 1, 1, 1 ±

2πα1 iδ + O(ε 4 ), h

which implies that the periodic orbit is strongly stable (see [29]). Finally, we approximate the period T¯ (ε) as 2π

T¯ (ε) = α1 + ε 3

2   1+σρ 2 4ρ 2

+ O(ε 4 )

  4 4 2π 3 2π α1 + σ h −ε + O(ε 4 ), = α1 α13 h3

and considering the initial time-scaling t = ε3 τ , we obtain that the period is approximated by T (ε) = ε−3 T¯ (ε), which completes the proof. 2 7. Existence of KAM tori for the symmetric R3BP Using the results related to periodic orbits of the previous section, we shall prove the existence of invariant KAM tori by using a classic Arnold’s result for isoenergetically non-degeneracy condition (see [2]) as well as a more current result proved by Han, Li and Yi in [15].

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Table 8 Values of the parameter a such that resonances up to order 4 appear.

(k1 , k2 ) (j )

aN

(j ) aS

j =1

j =2

j =3

j =4

(1, 0)

(1, −1) 

(2, −1)

  √ 1 36 − 3 41 + 5 145 2 6

(3, −1)





1 6



√

4− 2 2

√

3 2

2 3



  √ 36 − 32 5 145 − 41



√  5

1 418 17 + 15

1− 

1−

√



1 418 15 5 − 17

7.1. Invariant KAM tori near the poles in the symmetric R3BP on S2 We consider the symmetric R3BP on S2 for values of the parameter a where the eigenvalues of the poles are pure imaginary and there are no resonances up to order 4. If k = (k1 , k2 ) ∈ Z2 is the vector of resonance of order j , then, from (52), we must exclude the values of a such that (j ) (j ) k12 f1 (a) = k22 g1 (a), with |k1 | + |k2 | = j , j = 1, 2, 3, 4. If we denote by aN and aS such values of a generating a resonance of order j for the north and south poles, respectively, then they are given by the Table 8. Now, we unified the set where we are going to consider the value of the parameter a for both poles. ⎧  √  (4) ⎪ 4− 2 ⎨ , 23 \ {aN } 2 J = √  ⎪ (4) 3 ⎩ 2 , 1 \ {aS }

if  = −1, if  = 1.

Theorem 7.1. For every a ∈ J , the Hamiltonian system associated to the symmetric problem in a neighbourhood of each pole has invariant tori close to the linearized system, these tori form a set whose relative measure in the polydisk |I | <  tends to 1 as  → 0 (in fact, this measure is at least 1 − O( 1/4 )). In both cases such tori occupy a larger part of each energy level passing near the poles. Proof. We are going to prove that our Hamiltonian is close to an integrable system which is isoenergetically non-degenerate given in [2]. For this purpose, we need to normalize our Hamiltonian up to order 4, such that the truncated Hamiltonian up to order 4 is integrable, i.e., it only depends on the actions. This could be true if we exclude the values of a where a resonance of order ≤ 4 appears. Since for the symmetric problem the Hamiltonian (19) has the symmetries (u, v, pu , pv ) → (−u, v, pu , −pv ) and (u, v, pu , pv ) → (u, −v, −pu , pv ), the terms of odd order in the Taylor expansion around the origin do not appear. Thus, resonances of order 3 do not affect the terms of the normal form up to 4. So, it will be enough to take a ∈ J . If (X, Y, P1 , P2 ) are the coordinates that normalize the quadratic part of the Taylor expansion, then by using the Markeev’s algorithm (see [21]), we obtain that the Hamiltonian function can be written as H = H0 + H2 + . . . ,

(56)

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where H0 = −

ω1 2 ω2 2 (X + P12 ) + (Y + P22 ), 2 2

with   1 ω1 = ω −f (a), ω2 = ω −g1 (a), and the quartic part H2 takes the form H2 =



k

dk1 k2 k3 k4 X k1 Y k2 P1 3 P2k4 ,

(57)

k1 +k2 +k3 +k4 =4

with d2110 = d2101 = d1210 = d1201 = d0121 = d1012 = d0112 = d0121 = 0, d3010 = d3001 = d0310 = d1003 = d0301 = d0130 = 0,

(58)

and the remaining coefficients are given in Appendix A for the north pole and in Appendix B for the south pole. Now, we follow the Lie method described in [5] and [23] to obtain that the normalized quartic part is given by the homogeneous fourth degree polynomial H2 =



k

Dk1 k2 k3 k4 X k1 Y k2 P1 3 P2k4 ,

(59)

k1 +k2 +k3 +k4 =4

where, in virtue of the Lie–Deprit equation {H0 , Hj } = 0,

j = 1, 2, . . .

(60)

we obtain that the coefficients of the normalized quartic part satisfy D0013 = D0031 = D0103 = D0112 = D0121 = D0130 = D0211 = D0301 = D0310 = 0, D1003 = D1012 = D1021 = D1030 = D1102 = D1111 = D1120 = D1201 = D1210 = 0, D1300 = D2011 = D2101 = D2110 = D3001 = D3010 = D3100 = 0.

(61)

Thus, H2 has the form H2 = D4000 X 4 + D0400 Y 4 + D0040 P14 + D0004 P24 + D2200 X 2 Y 2 + D2020 X 2 P12 + D2002 X 2 P22 + D0220 Y 2 P12 + D0202 Y 2 P22 + D0022 P12 P22 ,

(62)

while, in action-angle variables, H2 is given by H2 = ω11 I12 + ω12 I1 I2 + ω22 I22 , with

(63)

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Fig. 3. Isoenergetically non-degeneracy condition.

ω11 = 3D0040 + D2020 + 3D4000 , ω12 = 2D0022 + D0220 + D2002 + D2200 , ω22 = 3D0004 + D0202 + 3D0400 . By the Lie–Deprit normal form algorithm we obtain the remaining terms of the normalized quartic part D4000 = D0040 = D2200 = D2002 = D0202 =

1 4 (3d0040 + d2020 + 3d4000 ), 1 4 (3d0040 + d2020 + 3d4000 ), 1 2 (d0022 + d0220 + d2002 + d2200 ), 1 2 (d0022 + d0220 + d2002 + d2200 ), 1 2 (3d0004 + d0202 + 3d0400 ),

D0400 = D0004 = D2020 = D0220 = D0022 =

1 4 (3d0004 + d0202 + 3d0400 ), 1 4 (3d0004 + d0202 + 3d0400 ), 1 2 (3d0040 + d2020 + 3d4000 ), 1 2 (d0022 + d0220 + d2002 + d2200 ), 1 2 (d0022 + d0220 + d2002 + d2200 ).

(64)

These calculations were made using Mathematica 9®. Now, the Hamiltonian of the symmetric problem can be written as H = H0 + · · · ,

H0 = −ω1 I1 + ω2 I2 + ω11 I12 + ω12 I1 I2 + ω22 I22 .

Here the dots denote terms of order higher than four with respect to the distance from the origin. The isoenergetically non-degeneracy condition is given by 

(a) = det

∂ 2 H0 ∂I 2 ∂ H0 ∂I

∂ H0 ∂I

0



⎛

2ω11 = det ⎝ ω12 −ω1

ω12 2ω22 ω2

⎞ −ω1 ω2 ⎠ = −2H2 (ω2 , ω1 ) = 0. 0

(65)

We observe in Fig. 3 that the condition (65) is in fact verified and the result follows from Theorem 6.23 in [2]. 2

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J. Andrade et al. / J. Differential Equations 265 (2018) 4486–4529

7.2. Invariant KAM close to an integrable Hamiltonian Let us add the term of order 4 of the Taylor expansion in (54) to obtain the Hamiltonian Hε = H0 + ε 3 H3 + ε 4 H4 + O(ε 7 ), where H4 = α2 , α2 = −σ

(66)

√ 3 a √ . 8 2

Theorem 7.2. Around the periodic solutions (given by Theorem 6.2) there are families of invariant KAM 2-tori for the symmetric R3BP on S2 and H2 , of the full system associated to the Hamiltonian (19). The complement of the set in which the KAM tori persists has measure of order O(ε 8 ). Even more, the near circular periodic solutions are orbitally stable and enclosed by invariant KAM 2-tori for ε small enough. Proof. Let us consider the ε −2 -symplectic transformation Q = ε −1 (ρ − ρ0 ),

P = ε −1 R,

where ρ0 is the value given in (55), together with the time-scaling t → εt and expanding in power series of ε we obtain H3 = H30 + ε 2 H32 + O(ε3 ), where H30 = −

1 + σ 4 , 22

H32 = AQ2 + BP 2 ,

with A=



  2 1+σ 4 2 ,

 2 1− 1+σ 4

B=

1+σ 4



2  1− 1+σ 4 28

.

Now, we introduce the action-angle coordinates (I2 , ψ) through  Q=

4

B 2I2 cos ψ, A

 P=

4

A 2I2 sin ψ, B

which satisfies dQ ∧ dP = dI2 ∧ dψ and defining I1 ≡ , we obtain that (66) becomes 1 + σ I14 H = α1 I1 − ε 3 2I12

 + ε α2 I1 + ε 4

5

 1 + σ I1 I2 + O(ε 6 ). I13

(67)

Now, we apply the main Theorem of [15] (see also Theorem 2.4 in [24]) with a = 3, m1 = 3, m2 = 4, m3 = 5, n0 = n1 = n2 = 1, n3 = 2, I n0 = I n1 = I n2 = (I1 ), I n3 = (I1 , I2 ), I¯n0 = I¯n1 = I¯n2 = (I1 ) and I¯n3 = (I2 ). Then we obtain the vector

J. Andrade et al. / J. Differential Equations 265 (2018) 4486–4529



(I ) = α1 , σ I1 +

1 I13

, α2 , σ I1 +

1 I13

4521

T ,

and the matrix whose columns are (I ), ∂I1 (I ) and ∂I2 (I ) (here we choose the integer s = 1 of the main Theorem of [15]) is given by ⎛

α1 ⎜ σ I1 + ⎜ M =⎜ α2 ⎝ σ I1 +

1 I13 1 I13

0 σ− 0 σ−

3 I14 3 I14

⎞ 0 0⎟ ⎟ ⎟, 0⎠ 0

$ which clearly has rank 2. Additionally, since s = 1 and b = ai=1 = mi (ni − ni−1 ) = 8, according to [15], the excluding measure for the existence of quasi-periodic invariant tori is of order O(εb ), which in our case is O(ε 8 ). This completes the proof of Theorem 7.2. 2 Note that due to the high degeneracy of the Hamiltonian (67), we cannot apply Theorem 6.23 of [2] as in the proof of Theorem 7.1, for this reason, we need to make use of Han, Li and Yi’s Theorem. 8. Conclusions In this work we have studied a particular restricted three body problem on surfaces of constant curvature. Using the stereographic projection onto the equatorial plane we work on the curved plane and on the hyperbolic space. We consider an elliptic relative equilibrium of the curved two body problem which in rotating coordinates plays the role of the primaries. A complete study related to bifurcations on the number of equilibria was carried out for the particular case when the primaries move on the same parallel. In this case, the curved problem depends on one parameter given by the radius a (the radius of the circle on the same parallel), while the analogous planar case does not depend on any parameter. For negative curvature the number of relative equilibria do not present bifurcations, its number is the same as in the planar case, i.e., there are three collinear and two triangular relative equilibria (that we have called them as isosceles equilibria by short). In this case we have proved that for μ = 1/2, all of them are unstable, as well as in the planar case. The case of positive curvature differs significantly from the planar case. In fact, we found five bifurcation values with respect to the parameter a ∈ (0, 1); nevertheless in none of the intervals determined by such values of the parameter the number of equilibria coincides with the ones for the planar case. In addition, we proved the existence of periodic orbits surrounding the poles (or vertex) through the application of the Lyapunov Center Theorem for certain intervals of the parameter and also the existence of KAM tori related to these periodic orbits for the case of S2 , where we were able to write the Hamiltonian function as a perturbation of an integrable Hamiltonian system satisfying the isoenergetically non-degeneracy condition. On the other hand, we have applied Averaging Theory to prove the existence of periodic orbits after to introduce an appropriate small perturbation parameter, which becomes the Hamiltonian system close to the curved Kepler problem. Finally, these periodic orbits allowed us to prove the existence of KAM tori surrounding them through the use of Han, Li and Yi’s Theorem.

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Acknowledgments We thank to the anonymous referees, their suggestions and remarks help us to improve the first version of this paper. Jaime Andrade was supported by a CONICYT fellowship (Chile). Ernesto Pérez-Chavela was partially supported by the Asociación Mexicana de Cultura A.C. Claudio Vidal was partially supported by FONDECYT 1180288. This paper is part of Jaime Andrade Ph.D. thesis in the Program Doctorado en Matemática Aplicada, Universidad del Bío-Bío (Chile). Appendix A. Coefficient for H2 at the north pole Defining the quantities √ 8 1 − a2 m2 = , a3

m1 = ω,

(68)

α1 = 9m2 − 32m21 ,  α2 = −9m42 + 81m82 + 768m62 ,

  α3 = α4 = α5 =

16 3 23 √ 3α 2

2

2 3

2/3 √ 3α 2

1− + , m22       √ √   18 22/3 3 3√3 α2 m42 −2 22/3 35/6 √3 α2 m62 27m22 +256 +m22 16 3 232/3 α22/3 −3 81m82 +768m62 +27m62   9m62 −m22 81m82 +768m62  √ √ √ 3 3 3 √ 8 3 3 α2 m22 +9 2m42 − 2 81m82 +768m62 2/3

α2 m22

,

the coefficients are given by     4 m2 − 4ω22 2 m31 − m1 ω22 2     − ω12 14 8ω22 − 2m2 2 − 16m41 + 4m2 m21 2

  √ √ √ 2/3 √  4096 232/3 α3 22/3 3 3α2 + 3 3 α2 − 8 3 232/3 m22 m31 − m1 ω22 4  √3 , 2  2 √  2/3 2/3 √ 2 8 6m√2 −2 α2 5/2 3 α m4 ω 2 8m − 3m α m + α m 2 2 1 2 1 2 1 2 2 3 1

d4000 =

α2 m2

    4m21 −4m21 + m2 + 2ω22 2 −4m21 + m2 + 4ω22 2     d0400 = − √ ω22 α1 m2 − 8m21 − 3m2 α1 m2 2

  √ √ √ 2/3 √  16 232/3 α3 m41 22/3 3 3α2 + 3 3 α2 − 8 3 232/3 m22 −4m21 + m2 + 4ω22 4  √3 , 2  √   2/3 2/3 √ 2 8 6m√2 −2 α2 5/2 3 α m4 ω 2 α m − 8m2 − 3m α m 2 2 2 1 2 2 1 2 2 3 1   4m2 ω2 4m2 + m2 − 4ω22 2 d0040 =  1  2 1 1  1  − 2 − 16m4 + 4m m2 2 2 1 1 4 8ω2 − 2m2

α2 m2

,

J. Andrade et al. / J. Differential Equations 265 (2018) 4486–4529

4523

  √ √ √ 2/3 √  2 232/3 α3 ω12 4 22/3 3 3α2 + 9 3 α2 − 32 3 232/3 m22 4m21 + m2 − 4ω22 4  √3 , 2  2 √  2/3 2/3 √ 2 8 6m√2 −2 α2 5/2 3 α m4 8m − 3m α m + α m 2 2 2 1 2 1 2 2 3 1 α2 m2

  64m21 ω22 −2m21 + m2 + 2ω22 2 √  −  d0004 =  α1 m2 − 8m21 − 3m2 α1 m2 2

  √ √ √ 2/3 √  32 232/3 α3 ω22 4 22/3 3 3α2 + 9 3 α2 − 32 3 232/3 m22 −2m21 + m2 + 2ω22 4  √3 , 2/3  √   √ 8 6m22 −22/3 α2 2 5/2 3 α m4 α m − 8m2 − 3m α1 m2 √ 2 2 1 2 2 2 3 1 α2 m2



 

√ √ √ √ 1 3 6 3 2/3 d2200 = √ (768 22/3 3α4 m21 22/3 3α2 + 3 3 α2 − 8 232/3 m22 · 5/2 3 α α 2 5

2 

  m21 − ω22 −4m21 + m2 + 4ω22 2 ) − 64m42 m21 − ω22 2 −4m21 + m2 + 2ω22 2 −

 

 m42 m2 − 4ω22 2 −4m21 + m2 + 4ω22 2 + 32m42 ω22 − m21 ·

−4m21 + m2 + 2ω22



−4m21 + m2 + 4ω22

 % m2 − 4ω22 ·

1   4 128m1 − 68m2 m21 + 9m22 ω1 ω2

  √ √ √ 2/3 √ 128 232/3 α3 7 22/3 3 3α2 + 18 3 α2 − 56 3 232/3 m22  √3 · d2020 = 2/3  2 √  √ 8 6m22 −22/3 α2 2 5/2 3 α m4 α1 m2 + α1 m2 √ 2 2 8m1 − 3m2 2 3 16m52

α2 m2

 

4m21 + m2 − 4ω22 2 m31 − m1 ω22 2 +



 

−512m61 ω22 + 16m41 −8m2 ω22 + m22 + 32ω24 + 8m21 m2 − 4ω22 3 + m2 − 4ω22 4 + 256m81  

, 4 m2 − 4ω22 2 − 16m41 + 4m2 m21 2

d2002 = −

1   √ 3 α m5 128m4 − 68m m2 + 9m2 ω 2 2 2 1 1 1 2



· √ 2/3 8 3 6m22 −22/3 α2 5/2 √ 3 α m2 2 2

  

√ √ √ √ 3 3 2/3 (32 232/3 α3 ω2 7 22/3 3α2 + 18 3 α2 − 56 232/3 m22 m21 − ω22 2 ·

 −2m21 + m2 + 2ω22 2 ) −

1   · 4 128m1 − 68m2 m21 + 9m22 ω1



  (ω2 −128m61 ω22 + 4m41 −8m2 ω22 + m22 + 32ω24 − 4m21 m2 − 4ω22 2 m2 + 2ω22 +

−2m2 ω22 + m22 − 8ω24



2

4m21 m2

 + 64m81 ,

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J. Andrade et al. / J. Differential Equations 265 (2018) 4486–4529

         ω1 −4m21 + m2 + 2ω22 2 − 4m21 + m2 − 4ω22 2 − 4m41 −4m21 + m2 + 4ω22 2   d0220 = − 4m21 m2 128m41 − 68m2 m21 + 9m22 ω2

  √ √ 2/3 √  32/3 α3 ω1 7 22/3 3 3α2 + 18 3 α2 − 56 3 232/3 m22 4m21 + m2 − 4ω22 2  √3 · 2/3 √ √   8 6m22 −22/3 α2 5 4 2 2 5/2 3 2 α2 m2 128m1 − 68m2 m1 + 9m2 ω2 √ 2 3 α2 m2



−4m21 + m2 + 4ω22 2 ,

  √ √ √ 2/3 √  32 232/3 α3 m21 7 22/3 3 3α2 + 18 3 α2 − 56 3 232/3 m22 −2m21 + m2 + 2ω22 2  √3 · d0202 = 2/3  √   √ 8 6m22 −22/3 α2 2 5/2 3 α m4 α m − 8m2 − 3m α m √ 2 2 1 2 2 1 2 2 3 1 α2 m2



1    · √ √ m2 α1 + 3 α1 m2 − 8m21 α1 m2 2







(16 −8m61 13m2 + 28ω22 + m41 216m2 ω22 + 53m22 + 224ω24 − 12m21 m2 + 2ω22 3 +

  m2 + 2ω22 4 + 80m81 ), &



  d0022 = −4m42 −2m21 + m2 + 2ω22 2 − 8m42 4m21 + m2 − 4ω22 −2m21 + m2 + 2ω22 +

  √ √ √ 2/3 √  3 6 3α4 4 22/3 3 3α2 + 9 3 α2 − 32 3 232/3 m22 4m21 + m2 − 4ω22 2 · √ 5/2 3 √ 2 3 α2 α5 m21 

 %

ω 1 ω2 −2m21 + m2 + 2ω22 2 − m42 4m21 + m2 − 4ω22 2   5 4 m2 128m1 − 68m2 m21 + 9m22 −4m21 + m2 + 4ω22

2

+

Appendix B. Coefficient for H2 at the south pole We define the following quantities β1 = β2 =

 

β3 =

     m2 32m21 + 9m2 8m21 + 3m2 + m2 32m21 + 9m2 , 81m82 + 768m62 − 9m42 ,  16 3 23 − √3 β2

+

2/3 √ 3β 2 23 2



m22

+ 1,

   β4 = 8m21 + 3m2 m2 32m21 + 9m2 − m2 32m21 + 9m2 ,  β5 = 9m62 − m22 81m82 + 768m62 , √ √   √   β62 = β15 18 22/3 3 3 3 β2 m42 − 2 22/3 35/6 3 β2 m62 27m22 + 256 +   √  2/3 m22 16 3 232/3 β2 − 3 81m82 + 768m62 + 27m62 



with m1 , m2 defined in (68) and β6 > 0, then the coefficients in this case are given by

J. Andrade et al. / J. Differential Equations 265 (2018) 4486–4529

4525

  √ √ 2/3 √  √ 32768 32/3 β2 β3 m2 22/3 3 3β2 + 3 3 β2 − 8 3 232/3 m22 m31 − m1 ω22 4

√  d4000 = + 2/3 5/2 β12 ω12 16 3 6m22 − 2 22/3 β2     16 m2 + 4ω22 2 m31 − m1 ω22 2 β12 ω12

  √ √ 2/3 √  √ 128 32/3 β2 β3 m41 m2 22/3 3 3β2 + 3 3 β2 − 8 3 232/3 m22 4m21 + m2 − 4ω22 4

√  d0400 = + 2/3 5/2 β42 ω22 16 3 6m22 − 2 22/3 β2     4m21 4m21 + m2 − 2ω22 2 4m21 + m2 − 4ω22 2 β42 ω22

  √ √ 2/3 √  √ 16 32/3 β2 β3 m2 ω12 4 22/3 3 3β2 + 9 3 β2 − 32 3 232/3 m22 −4m21 + m2 + 4ω22 4

√  d0040 = + 2/3 5/2 β12 16 3 6m22 − 2 22/3 β2   16m21 ω12 −4m21 + m2 + 4ω22 2 β12

  √ √ 2/3 √  √ 256 32/3 β2 β3 m2 ω22 4 22/3 3 3β2 + 9 3 β2 − 32 3 232/3 m22 2m21 + m2 − 2ω22 4

√  d0004 = 2/3 5/2 β42 16 3 6m22 − 2 22/3 β2   64m21 ω22 2m21 + m2 − 2ω22 2 + β42 &    8m22 m21 m21 − ω22 m2 + 4ω22 2 4m21 + m2 − 4ω22 − d3100 = 3/2 √ √ ω1 ω2 (β1 m2 ) 3/2 β4 m2

   8 4m21 + m2 − 2ω22 m2 + 4ω22 m31 − m1 ω22 2 −

 √ √ √ √ √  √   ⎤ 256 3 2 6 3 β2 β6 m2 m41

2/3

8 62/3 − 3 3 2 3 β2 m22 − 2 3 3β2

√ √ 2/3  5/2 8 3 3m22 − 3 2β2

4m21 + m2 − 4ω22 m21 − ω22

    4m21 m22 4m21 + m2 − 4ω22 & 2 2 2 2 2 d1300 = √ 8 m 4m − ω + m − 2ω − 2 1 2 1 2 3/2 √ ω1 ω2 β1 m2 [β4 m2 ) 3/2

   4m21 + m2 − 4ω22 m2 + 4ω22 4m21 + m2 − 2ω22

3

⎦

√ √ √ √ √  √ 2/3   2   ⎤ 4m1 + m2 − 4ω22 2 m21 − ω22 32 3 2 6 3 β2 β6 m21 m2 8 62/3 − 3 3 2 3 β2 m22 − 2 3 3β2 ⎦

√ − √ 2/3  5/2 8 3 3m22 − 3 2β2

   ω2 −4m21 + m2 + 4ω22 & 2 2 2 d0031 = 4m −4m + m + 4ω √ 2 1 1 2 (β1 m2 ) 3/2 β4 m2

 + 8m21 2m21 + m2 − 2ω22 3/2 √

8m22 ω1

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J. Andrade et al. / J. Differential Equations 265 (2018) 4486–4529

√ √ √  √ √ √ 2/3   2   ⎤ 3 2m1 + m2 − 2ω22 −4m21 + m2 + 4ω22 2 2 6 3 β2 β6 m2 32 62/3 − 9 3 2 3 β2 m22 − 8 3 3β2 ⎦

√ − √ 2/3  5/2 8 3 3m22 − 3 2β2

 √ 3/2   32m22 ω1 ω2 2m21 + m2 − 2ω22 & 2 2 2 d0013 = 2m −4m + m + 4ω √ 2 1 1 2 β1 m2 (β4 m2 ) 3/2

 + 4m21 2m21 + m2 − 2ω22

√ √ √  √ √ √ 2/3   2   ⎤ 3 2m1 + m2 − 2ω22 2 −4m21 + m2 + 4ω22 2 6 3 β2 β6 m2 32 62/3 − 9 3 2 3 β2 m22 − 8 3 3β2 ⎦

√ − √ 2/3  5/2 8 3 3m22 − 3 2β2

1  d2200 =  16 128m41 + 68m2 m21 + 9m22 ω1 ω2

    m2 + 4ω22 2 4m21 + m2 − 4ω22 2 m2

  √ √ √ √ 2/3 √    768 22/3 6 3 β2 β6 m21 22/3 3 3β2 + 3 3 β2 − 8 3 232/3 m22 m21 − ω22 2 4m21 + m2 − 4ω22 2

√ + √ 2/3  5/2 8 3 3m22 − 3 2β2

−

     32 m21 − ω22 128m41 ω22 + 2m21 16m2 ω22 + 3m22 − 68ω24 − 24m2 ω24 − 32m61 + m32 + 40ω26 m2

       1 1024m21 m31 − m1 ω22 2 4 m2 + 4ω22 2 −4m21 + m2 + 4ω22 2 d2020 = + 4 β12 β12 −

  √ √ 2/3 √     √ 4096 32/3 β2 β3 m2 7 22/3 3 3β2 + 18 3 β2 − 56 3 232/3 m22 m31 − m1 ω22 2 −4m21 + m2 + 4ω22 2

√  2/3 5/2 β12 16 3 6m22 − 2 22/3 β2

ω2  d2002 =  4 4 128m1 + 68m2 m21 + 9m22 ω1   64 m31 − m1 ω22 2 + m2

    m2 + 4ω22 2 2m21 + m2 − 2ω22 2 m21 m2

  √ √ √ √ 2/3 √    ⎤ 64 22/3 6 3 β2 β6 7 22/3 3 3β2 + 18 3 β2 − 56 3 232/3 m22 m21 − ω22 2 2m21 + m2 − 2ω22 2 ⎦

√ − √ 2/3  5/2 8 3 3m22 − 3 2β2

ω1  d0220 =  4 4 128m1 + 68m2 m21 + 9m22 ω2   4m21 4m21 + m2 − 4ω22 2 + m2 −

    −4m21 + m2 + 4ω22 2 4m21 + m2 − 2ω22 2 m21 m2

  √ √ √ √ 2/3 √     22/3 6 3 β2 β6 7 22/3 3 3β2 + 18 3 β2 − 56 3 232/3 m22 4m21 + m2 − 4ω22 2 −4m21 + m2 + 4ω22 2

√ √ 2/3  5/2 8 3 3m22 − 3 2β2

J. Andrade et al. / J. Differential Equations 265 (2018) 4486–4529

d0202 =

      64m41 4m21 + m2 − 4ω22 2 + 64 2m21 + m2 − 2ω22 2 4m21 + m2 − 2ω22 2 4β42

4527

−

  √ √ 2/3 √    √ 1024 32/3 β2 β3 m21 m2 7 22/3 3 3β2 + 18 3 β2 − 56 3 232/3 m22 4m21 + m2 − 4ω22 2 2m21 + m2 − 2ω22 2

√  2/3 5/2 4β42 16 3 6m22 − 2 22/3 β2

   ω1 ω2 2 64m21 ω22 + 8m2 ω22 − 32m41 − 8m2 m21 + 13m22 − 32ω24  d0022 =  2 128m41 + 68m2 m21 + 9m22 m2

 

√ √ √ √ √ 2/3 3 22/3 6 3 β2 β6 4 22/3 3 3β2 + 9 3 β2 − 32 3 232/3 m22

√ + · √ 2/3  5/2 m21 8 3 3m22 − 3 2β2   %

−4m21 + m2 + 4ω22 2 2m21 + m2 − 2ω22 2  4m22 ωω21

  

4 2 2 2 2 2 2 2 128m m −4m · − ω + m + 4ω + m + 4ω d2011 = √ 2 2 1 1 2 2 1 2 (β1 m2 ) 3/2 β4 m2

 2m21 + m2 − 2ω22 +

√ √ √ √ √  √ 2/3    128 3 2 6 3 β2 β6 m2 28 62/3 − 9 3 2 3 β2 m22 − 7 3 3β2 −4m21 + m2 + 4ω22

√ · √ 2/3  5/2 8 3 3m22 − 3 2β2   

2m21 + m2 − 2ω22 m31 − m1 ω22 2  16m22 ωω21

 

4 2 2 2 2 2 2m 4m d0211 = √ + m − 4ω + −4m + m + 4ω 2 2 1 1 2 1 2 · β1 m2 (β4 m2 ) 3/2

  2m21 + m2 − 2ω22 4m21 + m2 − 2ω22 2 +

√ √ √ √ √  √ 2/3    2 3 2 6 3 β2 β6 m2 m21 28 62/3 − 9 3 2 3 β2 m22 − 7 3 3β2 −4m21 + m2 + 4ω22

√ · √ 2/3  5/2 8 3 3m22 − 3 2β2   

2m21 + m2 − 2ω22 4m21 + m2 − 4ω22 2  4m22 ωω21

   4 2 2 2 2 2 2 32m m 4m − 4m · d1120 = − ω + m − 4ω + m − 2ω √ 2 2 1 1 2 1 2 1 2 (β1 m2 ) 3/2 β4 m2

  m2 + 4ω22 −4m21 + m2 + 4ω22 2 +

√ √ √ √ √  √ 2/3   2 ⎞ 16 3 2 6 3 β2 β6 m2 m21 28 62/3 − 9 3 2 3 β2 m22 − 7 3 3β2 m1 − ω22 ⎠·

√ √ 2/3  5/2 8 3 3m22 − 3 2β2  

−4m21 + m2 + 4ω22 2 4m21 + m2 − 4ω22

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J. Andrade et al. / J. Differential Equations 265 (2018) 4486–4529



ω2 ω1

   4 2 2 2 2 2 2 2 d1102 = √ − ω + m − 4ω + m − 2ω · 8m m 4m − 2m 2 2 1 1 2 1 2 1 2 β1 m2 (β4 m2 ) 3/2  

4m21 + m2 − 2ω22 m2 + 4ω22 +

√ √ √ √ √  √ 2/3  16 3 2 6 3 β2 β6 m2 m21 28 62/3 − 9 3 2 3 β2 m22 − 7 3 3β2

√ · √ 2/3  5/2 8 3 3m22 − 3 2β2

   m21 − ω22 2m21 + m2 − 2ω22 2 4m21 + m2 − 4ω22     16m21 m21 − ω22 4m21 + m2 − 4ω22 1 d1111 = m2 128m41 + 68m2 m21 + 9m22  2       2m1 + m2 − 2ω22 4m21 + m2 − 2ω22 m2 + 4ω22 −4m21 + m2 + 4ω22 − − m21 m2

  √ √ √ √ 2/3 √  8 22/3 6 3 β2 β6 7 22/3 3 3β2 + 18 3 β2 − 56 3 232/3 m22 4m21 + m2 − 4ω22

√ · √ 2/3  5/2 8 3 3m22 − 3 2β2

  % 2m21 + m2 − 2ω22 m21 − ω22 −4m21 + m2 + 4ω22 . 16m22

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