PERIODIC ORBITS OF THE SPATIAL ANISOTROPIC MANEV PROBLEM

We study the periodic orbits of the spatial anisotropic Manev problem which depend on three parameters.


Introduction and statement of the main results
The objective of this paper is to study the periodic orbits of the spatial anisotropic Manev problem given by the Hamiltonian (1) where µ is near 1, β ̸ = 0 is a parameter and ε is small.The dynamics of the planar anisotropic Manev problem was studied in [4] and [6]- [8], including information on its periodic orbits, but as far as we know there are no works on the spatial periodic orbits of the Manev problem.
If µ = 1 and β = 0, we have the spatial Kepler problem, see for instance the book [3].
Note that the Hamiltonian (1) is symmetric with respect to the z-axis.Then it is easy to check that the third component K = xp y − yp x of the angular momentum is a first integral of the Hamiltonian system with Hamiltonian (1).We shall use this integral K to simplify the analysis of the given axially symmetric perturbed system.Since µ is near 1 and ε is small we take µ = 1 − ε, and doing Taylor series in ε at ε = 0 of the Hamiltonian (1), we obtain (2) + O(ε 2 ).
In the following we shall use the Delaunay variables for studying easily the periodic orbits of the Hamiltonian system associated to the Hamiltonian (2), see [3,16] or section 2 for more details on the Delaunay variables.Thus in Delaunay variables the Hamiltonian (2) has the form where l is the mean anomaly, g is the argument of the perigee of the unperturbed elliptic orbit measured in the invariant plane, k is the longitude of the node, L is the square root of the semi-major axis of the unperturbed elliptic orbit, G is the modulus of the total angular momentum, and K is the third component of the angular momentum.
Our main result is the following one.
We note, as we shall see in the proof of Theorem 1, that the averaging method that we shall apply for finding the periodic solutions of Theorem 1 only find periodic solutions when the first integral K = 0. See the appendix for a summary of the averaging method used here.

Delauney variables
The transformation of Delauney is given by x = r(cos(f + g) cos k − c sin(f + g) sin k), y = r(cos(f + g) sin k + c sin(f + g) cos k), z = rs sin(f + g), with c = K/G and s 2 = 1 − K 2 /G 2 .The true anomaly f and the eccentric anomaly E are auxiliary quantities defined by the relations where e is the eccentricity of the unperturbed elliptic orbit.Note that the angular variable k is a cyclic variable for the Hamiltonian (3), and consequently K is a first integral of the Hamiltonian system as we already knew.

Proof of Theorem 1
We shall write the Hamiltonian system on the energy level H = h < 0. From the equation H = h we isolate L = 1/ √ −2h + O(ε).So the Hamiltonian system with Hamiltonian (3) eliminating L becomes Now we take l as new independent variable, and the previous equations of motion restricted to K = K write (5) Since in the previous differential system the right hand side functions do not depend on the angular variable l, its averaged equations with respect to l are the same but without O(ε 2 ).Therefore, from the appendix we have that the equilibrium points of the averaged equations which have non-zero Jacobian will provide periodic solutions of the Hamiltonian system with Hamiltonian (3) in the energy level H = h < 0 and in the level K = K.Now we shall compute these equilibria.
From the first equation of system (5) we have that sin(2g) = 0.So we separate the computation of the equilibria in three cases.
Then, from the appendix, statement (a) of Theorem 1 is proved.
Case 2: G = ±K.First we consider G = K.The last two equations from the averaged system becomes From the last equation we deduce K = cos(2g) − 1 2 √ −2h cos 2 g .We then obtain −2hβ cot 4 g = 0.This equation has the two solutions g = ±π/2.But then K is not defined at these two values of g.So, no equilibria when G = K.
In a similar way we can see that the case G = −K does not provide equilibria of the averaged system.
Subcase A: g = π/2.Then the last two equations from the averaged system becomes