This is a preprint of: “Periodic orbits of the planar anisotropic Manev problem and of the perturbed hydrogen atom problem”, Jaume Llibre, Pengfei Yuan, Qual. Theory Dyn. Syst., vol. 18, 969–986, 2019. DOI: [10.1007/s12346-019-00322-y] PERIODIC ORBITS OF THE PLANAR ANISOTROPIC MANEV PROBLEM AND OF THE PERTURBED HYDROGEN ATOM PROBLEM JAUME LLIBRE1 AND PENGFEI YUAN2 Abstract. 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) a unique elliptic periodic solution with eccentricity e of the Kepler problem can be continued to the mentioned three problems. 1. Introduction and statement of the results The Manev problem is a two-body problem with a potential of the following form a b V (r) = + 2 , r r where r is the distance between the two particles and a, b are arbitrary constants. For a, b > 0 the problem was first considered by Newton. Manev systems were recently reconsidered in a series of studies having as their starting point Diacu’s research [12], see also [2],[6],[17],[20]-[23], [25]- [29],[35]-[38]. 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 motion are ∂H ∂H (1) qi′ = , p′i = − , for i = 1, 2, ∂pi ∂qi Key words and phrases. periodic solutions, averaging theory, anisotropic Manev problem, hydrogen atom problem. 1 2 J. LLIBRE AND P. YUAN 1 1 ε (2) H(q1 , q2 , p1 , p2 ) = (p21 + p22 ) − √ − 2 2 . 2 2 2 (1 + ε)q1 + q2 (1 + ε)q 1 + q2 Here |ε| > 0 denotes a small parameter and the prime denotes the derivative with respect to the time t. The Hamiltonian system (1) is defined for all (q1 , q2 , p1 , p2 ) ∈ R4 except in the plane q1 = q2 = 0. In this paper we study analytically the periodic solutions of the pla- nar anisotropic Manev problem using the averaging theory, and we get the following result. Theorem 1. For ε ̸= 0 sufficiently small a unique elliptic periodic solution of the Kepler problem for each value of the eccentricity can be continued to the energy level h(e) of the planar anisotropic Manev problem (1), where √ (1 − e2 )(1 − 1 − e2 )2 h(e) = − . 4e4 The proof of Theorem 1 is given in Section 2. The hydrogen atom interaction with a circular polarized microwave field is a quiet well known problem in classical mechanics. There are extensive literatures on this problem, for more details, see [3, 4], [7, 8, 9], [13]-[16], [18, 19], [30]-[33] , [39]. In [31] the authors studied the hydrogen atom problem in a rotation frame with the following Hamiltonian 1 1 H = (p2x + p2y ) − xpy + ypx − √ + εx. 2 x + y2 2 We consider the same potential but in fixed coordinates, and instead of the perturbation εx with two more general perturbations ε(ax + bx2 ) and ε(ax + by 2 ). For the perturbation ε(ax + bx2 ) the corresponding Hamiltonian sys- tem is x′ = p x , y ′ = py , (3) x p′x = − − ε(a + 2bx), r3 y p′y = − 3 , r with Hamiltonian 1 1 H = (p2x + p2y ) − √ + ε(ax + bx2 ). 2 x + y2 2 3 For the perturbation ε(ax + by 2 ) the following Hamiltonian system becomes x′ = p x , y ′ = py , (4) x p′x = − − εa, r3 y p′y = − 3 − ε2by, r with Hamiltonian 1 1 H = (p2x + p2y ) − √ + ε(ax + by 2 ). 2 2 x +y 2 We have the following two results. 3ah Theorem 2. For h < 0, 0 < < 1, 4b + 3ah ̸= 0 and ε ̸= 0 4b sufficiently small a unique elliptic periodic solution with eccentricity e = |3ah/(4b)| of the Kepler problem can be continued to the energy level H = h of the hydrogen atom problem (3). 3ah Theorem 3. For h < 0, 0 < < 1, b − 3ah ̸= 0 and ε ̸= 0 b sufficiently small a unique elliptic periodic solution with eccentricity e = |3ah/b| of the Kepler problem can be continued to the energy level H = h of the hydrogen atom problem (4). Theorems 2 and 3 are proved in sections 3 and 4, respectively. 2. Proof of Theorem 1 We change the equations of motion (1) to the McGehee coordinates (r, θ, v, u) as follows (q1 , q2 ) = r(cos θ, sin θ), r−1/2 v = (p1 , p2 ) · (cos θ, sin θ), r−1/2 u = (p1 , p2 ) · (− sin θ, cos θ), 4 J. LLIBRE AND P. YUAN for more details on these coordinates see [10, 11, 24]. So in the McGehee coordinates the equations of motion (1) write r′ = r−1/2 v, θ′ = r−3/2 u, ( ) (5) ′ −3/2 1 2 v =r u + v + V1 (θ) + 2r−5/2 V2 (θ), 2 2 ( ) ′ −3/2 1 dV1 (θ) dV2 (θ) u =r − uv − − r−5/2 , 2 dθ dθ where 1 ε V1 (θ) = − √ , V2 (θ) = − , 1 + ε cos2 θ 1 + ε cos2 θ and the energy level H = h becomes r 2 (6) (u + v 2 ) + rV1 (θ) + V2 (θ) = r2 h. 2 Note that the equations of motion (5) are defined for all (r, θ, v, u) ∈ (0, +∞) × S1 × R2 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 = r5/2 . The equations of motion (5) in the new time τ become ṙ = r2 v, θ̇ = ru, 2 1 2 (7) v̇ = r(u ( + 2 v + V1 (θ)))+ 2V2 (θ), 1 dV1 (θ) dV2 (θ) u̇ = r − uv − − , 2 dθ dθ here the dot denotes derivative with respect to τ . Now the equations of motion (7) are defined for all (r, θ, v, u) ∈ [0, +∞) × S1 × R2 . 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 < 0 of the anisotropic Manev problem. Using (6) from the energy level H = h < 0 we obtain r = r(θ, v, u, h) = f0 + εf1 + O(ε2 ), where u2 + v 2 − 2 f0 = , 2 2h 2 (u + v − 2)(1 + cos 2θ) − 8h f1 = . 4h(u2 + v 2 − 2) 5 Now taking as the new independent variable the variable θ, the equa- tions of motion (7) restricted to the energy level H = h < 0 become dv 2(u2 + v 2 − 2) (u2 + v 2 − 2) cos2 θ − 8h = +ε 2 + v 2 − 2) + O(ε2 ), (8) dθ 2u 2u(u du v cos θ sin θ = − +ε + O(ε2 ). dθ 2 u It is clear that computing periodic solutions of system (8), we are ob- taining periodic solutions of system (1) in the energy level H = h < 0. The unperturbed system when ε = 0 is dv 2u2 + v 2 − 2 du v (9) = , =− , dθ 2u dθ 2 which has the general solution (10) ( ) e sin (θ − θ0 ) √ (v(θ; e, θ0 ), u(θ; e, θ0 )) = √ , 1 + e cos (θ − θ0 ) . 1 + e cos (θ − θ0 ) This is the solution of the planar Kepler problem with eccentricity e and argument of the pericenter θ0 , i.e. θ0 is the angle which provides the direction of the pericenter. Of course, for e = 0 this solution is circular, and for e ∈ (0, 1) it is elliptic, for more details see [34]. 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 ( ) ( ) ( ) e t θ z= , = , θ0 T 2π ( ) ( ) F01 F11 F0 = , F1 = , F02 F12 ( ) v(θ; e, θ0 ) x(t; z, 0) = , u(θ; e, θ0 ) where 2u2 + v 2 − 2 −16h + (−4 + 2u2 + 2v 2 ) cos2 θ F01 = , F11 = , 2u 4u(−2 + u2 + v 2 ) v cos θ sin θ F02 =− , F12 = . 2 u 6 J. LLIBRE AND P. YUAN The first variational equation of the unperturbed system (9) along the periodic solution (10) with e ∈ [0, 1) is ( ′ ) ( ) y1 y2′ y1 y2 (11) =J , y3′ y4′ y3 y4 where   ∂F01 ∂F01  ∂v ∂u    J = Dx F0 (t, x(t; z, 0)) =   ,  ∂F02 ∂F02  ∂u ∂v v=v(θ;e,θ0 ),u=u(θ;e,θ0 ) and consequently   e sin (θ − θ0 ) 3 1 − e2  1 + e cos (θ − θ0 ) +  2 2(1 + e cos (θ − θ0 ))2   J = .  1  − 0 2 The Fundamental matrix Mz (θ) of system (11) such that Mz (0) is the identity matrix is   y1 (θ; e, θ0 ) y2 (θ; e, θ0 ) Mz (θ) =  , y3 (θ; e, θ0 ) y4 (θ; e, θ0 ) where √ 1 + e cos θ0 [4 cos θ + e[cos (2θ − θ0 ) + 3 cos θ0 ] ] y1 (θ; e, θ0 ) = , 4[1 + e cos (θ − θ0 )]3/2 q(θ) y2 (θ; e, θ0 ) = √ , 4[1 + e cos (θ − θ0 )]3/2 1 + e cos θ0 √ sin θ 1 + e cos θ0 y3 (θ; e, θ0 ) = − √ , 2 1 + e cos (θ − θ0 ) e sin θ sin θ0 + 2 cos θ[1 + e cos θ0 ] y4 (θ; e, θ0 ) = √ , 2 (1 + e cos θ0 )(1 + e cos (θ − θ0 )) where θ − θ0 θ q(θ) = 8 sin θ+8e(2+cos θ) cos sin +e2 (cos θ+3 cos (θ − 2θ0 ) sin θ). 2 2 7 Now we compute the functions (G1 , G2 ) = Mz−1 (θ)F1 (θ, x(t; z, 0)) which appear in the next integral ∫ 1 T −1 (12) (g1 , g2 ) = F (z) = Mz (t)F1 (t, x(t; z, 0))dt, T 0 and we obtain p1 (θ) G1 = , 8(1 + e cos θ0 )3/2 (e2 − 1)(1 + e cos (θ − θ0 )) p2 (θ) G2 = 2 √ , 16(e − 1)(1 + e cos (θ − θ0 )) 1 + e cos θ0 where p1 (θ) =[e2 − 1 − 16h + (e2 − 1) cos 2θ − 16eh cos (θ − θ0 )]· [2 cos θ(1 + e cos θ0 ) + e sin θ sin θ0 ] · (1 + e cos (θ − θ0 ))− cos θ sin θ(e2 − 1)[16(1 + e cos θ0 ) sin θ+ e((e + 4 cos θ0 + 3e cos 2θ0 ) sin 2θ+ θ 16(2 + cos θ) sin2 sin θ0 ) + 6e2 sin2 θ sin 2θ0 ], 2 p2 (θ) = sin θ{10(e − 1) − 16h(e2 + 2) + 10(e2 − 1) cos 2θ+ 2 e[2(5e2 − 32h − 5) cos (θ − θ0 ) − 16eh cos (2θ − 2θ0 )+ (e2 − 1)(3 cos (3θ − θ0 ) + 7 cos (θ + θ0 ))]. Computing the integrals (12), we obtain p3 (e, θ0 ) g1 (e, θ0 ) = − , 16πe3 (1 + e cos θ0 )3/2 (1 − e2 )3/2 p4 (e, θ0 ) g2 (e, θ0 ) = − 3 √ , 8e (1 − e2 ) 1 + e cos θ0 where √ p3 = − 8e4 1 − e2 hπ[4 cos θ0 + e(cos 2θ0 + 3)]+ √ √ π(e2 − 1)[e3 ( 1 − e2 + e2 − 1) + 4e2 ( 1 − e2 + e2 − 1) cos θ0 − √ √ √ 6e(2 − 2 1 − e2 + e2 ( 1 − e2 − 2)) cos 2θ0 − (4 − 4 1 − e2 + √ e2 (e2 + 3 1 − e2 − 5))(4 cos 3θ0 + e cos 4θ0 )], √ √ p4 =e2 (1 − 1 − e2 + e2 ( 1 − e2 − 8h − 1)) sin θ0 + √ √ (e2 − 1)(4 − 4 1 − e2 + e2 ( 1 − e2 − 3)) sin 3θ0 . 8 J. LLIBRE AND P. YUAN Let   ∂g1 (e, θ0 ) ∂g1 (e, θ0 )  ∂e ∂θ0 . m(e, θ0 ) = det  ∂g2 (e, θ0 ) ∂g2 (e, θ0 )  ∂e ∂θ0 Now we have to study the solutions (e, θ0 ) of system g1 (e, θ0 ) = 0, g2 (e, θ0 ) = 0 with e ∈ (0, 1). Note that if e = 0 then g1 (e, θ0 ) ≡ 0 and g2 (e, θ0 ) ≡ 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 g2 (e, θ0 ) with respect to θ0 , and we obtain ten solutions as follows: θ01 = 0, θ02 = π, θ0 = − arccos (−F (e)), θ04 = − arccos (−F (e)) + π, 3 θ05 = arccos (−F (e)), θ06 = arccos (−F (e)) + π, 7 θ0 = − arccos F (e), θ08 = − arccos F (e) + π, 9 θ0 = arccos F (e), θ010 = arccos F (e) + π, where √ √ √ √ 2[(1 + 1 − e2 )(1 − 16h) − e2 (1 + 1 − e2 − 4h(3 + 1 − e2 ))] F (e) = √ . 2e 1 − e2 Of course, g1 (θ0 , e) = 0 if and only if p3 (θ0 , e) = 0. Case 1: θ0 = θ01 = 0. Substituting θ01 into p3 (e, θ0 ), we obtain √ p̄3 (e) = 16(e − 1)2 (e + 1)3 − 8(1 + e) 1 − e2 (2 − 3e2 + e4 + 4e4 h). We solve p̄3 (e) = 0 with respect to h, we have √ (1 − e2 )(1 − 1 − e2 )2 (13) h(e) = − . 4e4 By simple computation, we have √ √ ′ 4 − 4 1 − e2 + e2 ( 1 − e2 − 3) (14) h (e) = . 2e5 √ √ Let f (e) = 4 − 4 1 − e2 + e2 ( 1 − e2 − 3), we have √ ′ 3e(1 − 1 − e2 )2 (15) f (e) = √ . 1 − e2 Note that f (0) = 0, then by (15), we deduce that f (e) > 0 for all 0 < e < 1. Then from (14), we have h′ (e) > 0 for all 0 < e < 1, so h(e) is a strictly monotone function on (0, 1). Since h(1) = 0, h(e) < 0 in the interval (0, 1). 9 Next we can compute m(e, θ01 ) and we obtain √ √ √ 4e2 (5 − 3 1 − e2 ) + e4 ( 1 − e2 − 5) + 16( 1 − e2 − 1) (16) − √ . 2e7 (1 + e)( 1 − e2 ) √ √ √ Let g(e) = 4e2 (5 − 3 1 − e2 ) + e4 ( 1 − e2 − 5) + 16( 1 − e2 − 1), we have √ √ ′ 5e(8 − 8e2 + e4 − 8 1 − e2 + 4e2 1 − e2 ) (17) g (e) = − √ . 1 − e2 √ √ Let y(e) = 8 − 8e2 + e4 − 8 1 − e2 + 4e2 1 − e2 , we have √ ′ 4e(1 − 1 − e)3 (18) y (e) = √ > 0, 1 − e2 for e ∈ (0, 1). Note that y(0) = 0, and from (17) and (18) we can deduce that y(e) > 0 and g ′ (e) < 0 in the interval (0, 1). Then, since g(0) = 0 we have g(e) < 0 in the interval (0, 1). So, from (16), we have m(e, 0) > 0 for e ∈ (0, 1). Therefore, from the averaging theory (see [5]) for every value of the eccentricity e ∈ (0, 1) and for ε sufficiently small the periodic solution (v(θ; e, 0), u(θ; e, 0)) (given√in (10)) of the Kepler problem such that (v(0; e, 0), u(0; e, 0)) = (0, 1 + e), can be continued to the energy level h = h(e) < 0 of the anisotropic Manev problem (1). Case 2: θ0 = θ02 = π. 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. Case 3: θ0 = θ03 . Now solving p̄3 (e) = p3 (e, θ03 ) = 0 with respect to h, we obtain the two solutions 1 − e2 h± (e) = s± (e) , 4e6 where s± (e) is given by √ √ − e2 [4 + e4 − 4 1 − e2 + e2 (3 1 − e2 − 5)]± √ √ √ √ −[e6 + e2 (48 − 32 1 − e2 ) + 6e4 ( 1 − e2 − 3) + 32( 1 − e2 − 1)]. Since substituting θ0 = θ03 and h = h± (e) into m(e, θ0 ), we found m(e, θ03 ) = 0 for all e ∈ (0, 1), the averaging theory does not provide information if the corresponding periodic solution of the Kepler prob- lem (v(θ; e, θ03 ), u(θ; e, θ03 )) can be continued or not to the anisotropic Manev problem (1). 10 J. LLIBRE AND P. YUAN For the cases θ0j for j = 4, . . . , 10, we also found that m(e, θ0j ) = 0 for all e ∈ (0, 1), so again the averaging theory does not provide information about the possible continuation of this periodic solution to the anisotropic Manev problem (1). In conclusion the averaging theory provides a unique periodic solu- tion 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. 3. Proof of Theorem 2 In the McGehee coordinates the equations of motion (3) become r′ = r−1/2 v, θ′ = r−3/2 u, ( ) (19) ′ −3/2 1 2 1 v =r u + v − 1 + εr 2 (a cos θ + 2br cos2 θ), 2 2 1 1 u′ = r−3/2 (− uv) + εr 2 (a sin θ + 2br cos θ sin θ), 2 and the energy level H = h writes 1 2 (20) (u + v 2 ) − 1 + ε(ar2 cos θ + br3 cos2 θ) = rh. 2 Then doing the change of time t → τ given by dt/dτ = r3/2 , the equations of motion (19) in the new time τ are ṙ = rv, θ̇ = u, (21) v̇ = u2 + 21 v 2 − 1 + ε(ar2 cos θ + 2br3 cos2 θ), 1 u̇ = − uv − ε(ar2 sin θ + 2br3 cos θ sin θ). 2 Note that these equations of motion are defined for all (r, θ, v, u) ∈ [0, +∞) × S1 × R2 . Computing r from the energy level H = h < 0, where H is given in (20) we have r = r(θ, v, u, h) = f0 + εf1 + O(ε2 ), where u2 + v 2 − 2 f0 = , 2h (u2 + v 2 − 2)2 cos θ[2ah + b(u2 + v 2 − 2) cos θ] f1 = . 8h4 11 Then the equations of motion (21) in the energy level H = h < 0 become (22) dv 2u2 + v 2 − 2 (u2 + v 2 − 2)2 [ah + b(u2 + v 2 − 2) cos θ] cos θ = −ε 3u + O(ε2 ), dθ 2u 2 4h du v (u + v 2 − 2)2 [ah + b(u2 + v 2 − 2) cos θ] sin θ = − +ε + O(ε2 ). dθ 2 4h3 u 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 use the notation of [5] for applying the averaging theory, thus we have from system (22) that 2u2 + v 2 − 2 −16h + (−4 + 2u2 + 2v 2 ) cos2 θ F01 = , F11 = , 2u 4u(−2 + u2 + v 2 ) v cos θ sin θ F02 =− , F12 = . 2 u Now we compute the functions (G1 , G2 ) = Mz−1 (θ)F1 (θ, x(t; z, 0)) which appear in the integral (12) and we obtain p1 (θ) G1 = 3 , 8h3 (1 + e cos (θ − θ0 ))4 (1 + e cos θ0 ) 2 p2 (θ) G2 = 3 √ , 8h (1 + e cos (θ − θ0 ))4 1 + e cos θ0 where p1 (θ) =(e2 − 1)2 (b(e2 − 1) cos θ + ah(1 + e cos (θ − θ0 ))· (cos 2θ − 3 − 2e cos θ0 (2 + cos θ(1 − cos θ + e cos θ0 ))+ e(cos θ − 4 − 3e cos θ0 ) sin θ sin θ0 ), p2 (θ) =(e2 − 1)2 (b(e2 − 1) cos θ + ah(1 + e cos (θ − θ0 ))(cos θ + e cos θ0 ) sin θ, 12 J. LLIBRE AND P. YUAN Computing the integrals (12) for the hydrogen atom problem (22) we get √ 1 − e2 (7be2 + 12ah + 4e(4b + 3ah) cos θ0 + 9be2 cos 2θ0 ) g1 (e, θ0 ) = 3 , 32h3 (1 + e cos θ0 ) 2 √ be sin θ0 1 − e2 g2 (e, θ0 ) = √ . 16h3 1 + e cos θ0 Now we have to study the solutions (e, θ0 ) of the system g1 (e, θ0 ) = 0, g2 (e, θ0 ) = 0 with e ∈ (0, 1), as before e cannot be zero. First we solve the equation g2 (e, θ0 ) with respect to θ0 , and we obtain two solutions θ01 = 0, θ02 = π. Case 1: θ0 = θ01 = 0. Substituting θ01 into g1 (e, θ0 ), we have √ 1 − e2 [16be2 + 12ah + 4e(4b + 3ah)] ḡ1 (e) = 3 . 32h3 (1 + e) 2 We solve ḡ1 (e) = 0 with respect to e, we have 3ah e = 1, e=− . 4b 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. ( ) 3ah 3ab 4b + 3ah By direct computation we have m − ,0 = − . 4b 512h5 b Therefore, 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 (v(θ; −3ab/(512h5 ), 0), u(θ; −3ab/(512h5 ), 0)) of the Kepler problem can be continued to the energy level h < 0 of the perturbed hydrogen atom problem (3). Case 2: θ0 = θ02 = π. Substituting θ02 into g1 (e, θ0 ), we obtain √ 1 − e2 [16be2 + 12ah − 4e(4b + 3ah)] ḡ1 (e) = 3 . 32h3 (1 − e) 2 We solve ḡ1 (e) = 0 with respect to e and we get 3ah e = −1, e= . 4b 13 ( ) 3ah 3ab 4b + 3ah Since m ,π = , as before from the averaging 4b 512h5 b 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 (v(θ; 3ab/(4b), π), u(θ; 3ab/(4b), π)) can be continued to the energy level h < 0 of the perturbed hydrogen atom problem (3). This completes the proof of Theorem 2. 4. Proof of Theorem 3 The proof of Theorem 3 is similar to the proof of Theorem 2. We also take McGehee coordinates for the equations of motion (4) and doing the change of time t → τ given by dt/dτ = r3/2 , the equations of motion (4) in the new time τ become ṙ = rv, θ̇ = u, (23) v̇ = u2 + 21 v 2 − 1 + ε(ar2 cos θ + 2br3 sin2 θ), 1 u̇ = − uv − ε(ar2 sin θ − 2br3 cos θ sin θ). 2 and the energy level H = h becomes 1 2 (24) (u + v 2 ) − 1 + ε(ar2 cos θ + br3 sin2 θ) = rh. 2 From this energy relation H = h < 0 we can compute r = r(θ, v, u, h) = f0 + εf1 + O(ε2 ), where u2 + v 2 − 2 f0 = , 2h (u2 + v 2 − 2)2 [2ah cos θ + b(u2 + v 2 − 2) sin2 θ] f1 = . 8h4 Then the equations of motion (23) restricted to the energy level H = h < 0 are dv 2u2 + v 2 − 2 (u2 + v 2 − 2)2 [ah cos θ + b(u2 + v 2 − 2) sin2 θ] = −ε + O(ε2 ), dθ 2u 4h3 u du v (u2 + v 2 − 2)2 [ah − b(u2 + v 2 − 2) cos θ] sin θ = − +ε + O(ε2 ). dθ 2 4h3 u 14 J. LLIBRE AND P. YUAN The unperturbed system when ε = 0 is again the Kepler problem given in (9) having the general solution provided in (10). Using the notation of [5] we have for system (23) that 2u2 + v 2 − 2 v F01 = , F11 = − , 2u 2 (u2 + v 2 − 2)2 [ah cos θ + b(u2 + v 2 − 2) sin2 θ] F02 = − , 4h3 u (u2 + v 2 − 2)2 [ah − b(u2 + v 2 − 2) cos θ] sin θ F12 = . 4h3 u Now we compute the functions (G1 , G2 ) = Mz−1 (θ)F1 (θ, x(t; z, 0)) which appear in the integral (12) and we obtain p1 (θ) G1 = 3 , 32h3 (1 + e cos (θ − θ0 ))5 (1 + e cos θ0 ) 2 p2 (θ) G2 = √ , 16h (1 + e cos (θ − θ0 ))4 1 + e cos θ0 3 where p1 (θ) =(1 − e2 )2 (−4(ah cos θ(1 + e cos (θ − θ0 ))2 + b(e2 − 1) sin2 θ)· (2 cos θ(1 + e cos θ0 ) + e sin θ sin θ0 ) − ((b(1 − e2 ) cos θ+ ah(1 + e cos (θ − θ0 ))) sin θ(16(1 + e cos θ0 ) sin θ+ θ e((e + 4 cos θ0 + 3e cos 2θ0 ) sin 2θ + 16(2 + cos θ) sin2 sin θ0 )+ 2 6e2 sin2 θ sin 2θ0 ))), p2 (θ) =(e2 − 1)2 (3b − 3be2 + a(e2 + 2)h cos θ − b(e2 − 1) cos 2θ+ ae2 h cos (θ − 2θ0 ) + 3be cos (θ − θ0 ) − 3be3 cos (θ − θ0 )+ aeh cos (2θ − θ0 ) + 3aeh cos θ0 + be cos (θ + θ0 )− be3 cos (θ + θ0 )) sin θ, Computing the integrals (12) for our system (23) we get √ 1 − e2 [2e(b − 3ah) cos θ0 + 3be2 cos 2θ0 − be2 − 6ah] g1 = , 16h3 (1 + e cos θ0 )3/2 √ be sin θ0 1 − e2 g2 = − 3 √ . 4h 1 + e cos θ0 Now we have to study the solutions (e, θ0 ) of the system g1 (e, θ0 ) = 0, g2 (e, θ0 ) = 0 with e ∈ (0, 1). We first solve the equation g2 (e, θ0 ) 15 with respect to θ0 , and we obtain the two solutions θ01 = 0, θ02 = π. Case 1: θ0 = θ01 = 0. Substituting θ01 into g1 (e, θ0 ), we get √ 1 − e2 (2be2 + 2e(b − 3ah) − 6ah) ḡ1 (e) = 3 . 16h3 (1 + e) 2 We solve ḡ1 (e) = 0 with respect to e, we have 3ah e = 1, e= . b ( ) 3ah 3ab b − 3ah By direct computations we have m ,0 = − . b 32h5 b Therefore for every e = 3ah/b ∈ (0, 1), m(3ah/b, 0) ̸= 0 and for ε ̸= 0 sufficiently small the periodic solution of the Kepler problem (v(θ; 3ab/b, 0), u(θ; 3ab/b, 0)) can be continued to the energy level h < 0 of the perturbed hydrogen atom problem (4). Case 2: θ0 = θ02 = π. Substituting θ02 into g1 (e, θ0 ), we obtain √ 1 − e2 (2be2 − 2e(b − 3ah) − 6ah) ḡ1 (e) = 3 . 16h3 (1 − e) 2 We solve ḡ1 (e) = 0 with respect to e, we have 3ah e = −1, e=− . b ( ) 3ah 3ab b − 3ah Since m − ,π = , for every value of the eccentric- b 32h5 b ity e = −3ah/b ∈ (0, 1), m(−3ah/b, π) ̸= 0 and for ε ̸= 0 sufficiently small the periodic solution of the Kepler problem (v(θ; 3ab/b, π), u(θ; 3ab/b, π)) can be continued to the energy level h < 0 of the perturbed hydrogen atom problem (4). This completes the proof of Theorem 3. Acknowledgements The first author is partially supported by the Ministerio de Economı́a, Industria y Competitividad, Agencia Estatal de Investigación grants MTM2016-77278-P (FEDER) and MDM-2014-0445, the Agència de 16 J. LLIBRE AND P. YUAN Gestió d’Ajuts Universitaris i de Recerca grant 2017SGR1617, and the H2020 European Research Council grant MSCA-RISE-2017-777911. 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