Periodic Orbits of the Planar Anisotropic Manev Problem and of the Perturbed Hydrogen Atom Problem

In this paper we use the averaging theory for studying the periodic solutions of the planar anisotropic Manev problem and of two perturbations of the hydrogen atom problem. When a convenient parameter is sufficiently small we prove that for every value e∈(0,1)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$e\in (0,1)$$\end{document} a unique elliptic periodic solution with eccentricity e of the Kepler problem can be continued to the mentioned three problems.


Introduction and Statement of the Results
After the works of Poincaré [32] and Schwarzschild [37] on the periodic solutions of the Hamiltonian systems, the study of the periodic solutions of these systems has been the research of many authors. These last 50 years many works on these periodic solutions has been made numerically thanks to the computers. The analytical study of these periodic orbits is in general more difficult.
In the present paper we study analytically, using the averaging theory, the periodic solutions of three Hamiltonian systems which are perturbation of the Kepler problem. The family of periodic orbits that we find in the three Hamiltonian systems here analyzed as far as we know are new, we cannot find in the literature any result on the periodic solutions here analyzed.
Abouelmagd et al. [1] study the periodic solutions for the planar anisotropic Kepler problem using averaging theory. Motivated by this work, we study the anisotropic Manev problem with a and b arbitrary using the averaging theory.
The equations of the anisotropic Manev problem are where In this paper |ε| > 0 denotes a small parameter, which can be positive or negative, and the prime denotes the derivative with respect to the time t. The Hamiltonian system (1) is defined for all (q 1 , q 2 , p 1 , p 2 ) ∈ R 4 except in the plane q 1 = q 2 = 0. In this paper we study analytically the periodic solutions of the planar anisotropic Manev problem using the averaging theory, and we get the following result.
In [31] the authors studied the hydrogen atom problem in a rotation frame with the following Hamiltonian We consider the same potential but in fixed coordinates, and instead of the perturbation εq 1 with two more general perturbations ε aq 1 + bq 2 1 and ε aq 1 + bq 2 2 , more precisely we shall study the periodic solutions of the Hamiltonian systems associated to the Hamiltonians and We have the following two results.

Proof of Theorem 1
We shall work with the Hamiltonian system of the anisotropic Manev problem defined by the Hamiltonian (2) using the McGehee coordinates (r , θ, v, u) given by for more details on these coordinates see [10,11,24]. The reason for working in the McGehee coordinates is that in these coordinates we can write the solutions of the Kepler problem in an easy way, as we shall see later on.
So in the McGehee coordinates the Hamiltonian equations (1) with Hamiltonian (2) write where and the energy level H = h becomes Note that the differential equations (5) are defined for all (r , θ, v, u) ∈ (0, +∞) × S 1 × R 2 with a collision singularity at r = 0. We can remove this singularity with the change t → τ of the independent variable given by dt/dτ = r 5/2 . The differential system (5) in the new time τ becomė here the dot denotes derivative with respect to τ . Now the differential system (7) is defined for all (r , θ, v, u) ∈ [0, +∞) × S 1 × R 2 .
Since if ε = 0 the Hamiltonian (2) of the anisotropic Manev problem becomes the Hamiltonian of the Kepler problem, we want to detect the periodic solutions of the Kepler problem which can be continued to a fix negative energy level H = h of the anisotropic Manev problem.
Developing in power series of the small parameter ε the function r = r (θ, v, u, h) obtained isolating the variable r from the equality (6) we obtain The solutions of the differential system (7) with ε = 0 (i.e. of the Kepler problem) on the energy level H = h in function of the time τ are complicated. For this reason we take the variable θ as the new independent variable of the differential system, because the expression of the solutions of the Kepler problem in function of the variable θ are easy. Thus the differential equations (7) restricted to the energy level H = h < 0 taking as the new independent variable θ become Indeed, the first equation of the differential system (7) disappears because on the energy level H = h we have obtained the variable r in function of the variables (θ, v, u, h). The second equation of the differential system (7) also disappears because we have taken θ as the new independent variable. The last two equations of the differential system (7) produce the differential system (8) once we develop the derivatives dv/dθ and du/dθ in power series of the small parameter ε.
We claim that computing periodic solutions of system (8), we are obtaining periodic solutions of the Hamiltonian system (1) with Hamiltonian (2) in the energy level H = h, i.e. of the anisotropic Manev problem in the energy level H = h. Now we prove the claim. Roughly speaking the claim follows from the fact that the differential system (8) has been obtained from the differential system (1) through changes of variables and the restriction to the energy level H = h. Assume that (v(θ ), u(θ )) is a periodic solution of the differential system (8). Then this periodic solution provides the periodic solution (r (t), θ (t), v(t), u(t)) of the differential system (7), where θ = θ(t) is obtained using the second equation of (7), i.e. it is obtained from Also from this last equality and the expressions of v(θ) and u(θ ) we can get v = v(t) and u = u(t). Then we obtain of the anisotropic Manev problem in the energy level H = h. Hence the claim is proved.
which has the general solution This is the solution of the planar Kepler problem with eccentricity e and argument of the pericenter θ 0 , i.e. θ 0 is the angle which provides the direction of the pericenter. It is well known that the solutions of the Kepler problem with eccentricity e = 0 are circular periodic solutions, and the solutions with eccentricity e ∈ (0, 1) are elliptic periodic solutions, for more details see [35]. We are interested in knowing what are the periodic solutions of the Kepler problem which can be continued to periodic solutions of the anisotropic Manev problem, i.e. what solutions (10) with eccentricity e ∈ [0, 1) can be extended.
In order to apply the averaging theory to system (8) we identify our variables with those of averaging theory given in [5], and we obtain The first variational equation of the unperturbed system (9) along the periodic solution (10) with e ∈ [0, 1) is where , and consequently The Fundamental matrix M z (θ ) of system (11) such that M z (0) is the identity matrix is Now we compute the functions (G 1 , G 2 ) = M −1 z (θ )F 1 (θ, x(t; z, 0)) which appear in the next integral and we obtain Computing the integrals (12), we obtain +e 2 e 2 + 3 1 − e 2 − 5 (4 cos 3θ 0 + e cos 4θ 0 ), Now we have to study the solutions (e, θ 0 ) of system g 1 (e, θ 0 ) = 0, g 2 (e, θ 0 ) = 0 with e ∈ (0, 1). Note that if e = 0 then g 1 (e, θ 0 ) ≡ 0 and g 2 (e, θ 0 ) ≡ 0, and that in this case the averaging theory does not provide information on the periodic solutions of system (8). We first solve the equation g 2 (e, θ 0 ) with respect to θ 0 , and we obtain ten solutions as follows: Of course, g 1 (θ 0 , e) = 0 if and only if p 3 (θ 0 , e) = 0.
Working as in the previous case we obtain that the periodic solution (v(θ ; e, π), u(θ ; e, π)) which can be continued from the Kepler problem to the anisotropic Manev problem is the same that in Case 1, because (v(0; e, π), u(0; e, π)) = 0, √ 1 + e , i.e. both periodic solutions have the same initial conditions. In conclusion the averaging theory provides a unique periodic solution of the Kepler problem for each value of the eccentricity e ∈ (0, 1) which can be continued to the anisotropic Kepler problem (1). This completes the proof of Theorem 1.

Proof of Theorem 2
In the McGehee coordinates the Hamiltonian system with Hamiltonian (3) become and the energy level H = h writes Then doing the change of time t → τ given by dt/dτ = r 3/2 , the differential system (19) in the new time τ becomeṡ Note that this differential system is defined for all (r , θ, v, u) ∈ [0, +∞) × S 1 × R 2 . Computing r from the energy level H = h < 0, where H is given in (20) we have Then the differential equations (21) in the energy level H = h < 0 become Since hydrogen atom problem (3) for ε = 0 is the Kepler problem (9), as before we take h negative because we want to continue the periodic solutions of the Kepler problem to periodic solutions of the hydrogen atom problem (3).
As in the previous problem we shall use the averaging theory for studying the periodic solutions of system (9) which can be continued to system (22).
We solveḡ 1 (e) = 0 with respect to e, we have The solution e = 1 should be eliminated since we study which periodic solutions of the Kepler problem can be continued to periodic solutions of the perturbed hydrogen atom problem, and fore = 1 the solutions of the Kepler problem are parabolas.
from the averaging theory we obtain that if the parameters a and b of the perturbation satisfy 0 < −3ah/4b < 1, m(−3ah/4b, 0) = 0 and ε = 0 is sufficiently small, then the periodic solution of the Kepler problem can be continued to the energy level h < 0 of the perturbed hydrogen atom problem (3). .
Since m 3ah 4b , π = 3ab 512h 5 4b + 3ah b , as before from the averaging theory we obtain that if the parameters a and b of the perturbation satisfy 3ah/4b ∈ (0, 1), m(3ah/(4b), π ) = 0 and for ε = 0 is sufficiently small, then the periodic solution of the Kepler problem can be continued to the energy level h < 0 of the perturbed hydrogen atom problem (3). This completes the proof of Theorem 2.
uv − ε ar 2 sin θ − 2br 3 cos θ sin θ . (23) and the energy level H = h becomes From this energy relation H = h < 0 we can compute Then the differential system (23) restricted to the energy level H = h < 0 is The unperturbed system when ε = 0 is again the Kepler problem given in (9) having the general solution provided in (10).