# Perturbed Keplerian Hamiltonian Systems

Authors: Riadh Chteoui
Journal: International Journal of Differential Equations
 (2023)

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## Abstract

This paper deals with a class of perturbation planar Keplerian Hamiltonian systems, by exploiting the nondegeneracy properties of the circular solutions of the planar Keplerian Hamiltonian systems, and by applying the implicit function theorem, we show that noncollision periodic solutions of such perturbed system bifurcate from the manifold of circular solutions for the Keplerian Hamiltonian system.


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## Body

## 1. Introduction

In this paper, we study a class of Hamiltonian systems obtained as perturbation of Keplerian HamiltonianKp,q=1/2p2−q−1. More precisely, we consider the Hamiltonians of the following form:(1)Hp,q,ε=12p2−q−1−ε2Aq,p+ε2Aq24+Vε,q,where p,q∈ℝ2,ε>0 is a perturbation parameter, A is a skew-symmetric matrix A⋆=−A, and V is even in q. The corresponding Hamiltonian system is the following:(2)q¨+qq3+εAq+Vq′ε,q=0.Forε=0, equation (2) becomes(3)q¨+qq3=0.Our aim is to seek noncollision periodic solutions of (2), and we wish to connect them to the circular solutions of the unperturbed system (3). These types of perturbation systems have been the focus of interest by a number of authors and the references therein [1–5]. We mention in particular the works of Poincaré, regarding the three-body problem (these orbits were called “first-view sort solutions”) [6], and the studies by Ambrosetti et al. [7] and Celletti et al. [8] which showed the existence of a skew-T/2 periodic solution of the following problem:(4)q¨+qq3+Vq′ε,q=0.The essential difficulty in studying this problem is the free action of theS1-group acting on equation (2) (if q is a solution of (2), then qs=q.+s is also a solution of (2), for all s∈S1). To overcome this difficulty, we seek solutions of equation (2) near the circular orbits of the Keplerian system (3). These circular orbits are the more stable solutions and by exploiting their nondegeneracy property, we neutralize the free action of the S1-group ([9, 10]). The degenerate solutions of the Keplerian problem are the least stable solutions (KAM theory [11]); we cannot dominate the invariance of the problem under the action of the group S1 in the neighborhood of these solutions.The proofs rely on the implicit function theorem and the nondegeneracy of the circular orbits for (3) in the space of the skew-T/2 periodic functions. Let Ω=ℝ2/0 and V:ℝ×Ω⟶ℝ satisfy the following:(5)V1V∈C2ℝ×Ω,ℝ,V2Vε,−q=Vε,q.For all,q∈Ω,ε∈ℝ.We consider the following perturbed system of ordinary differential equations:(6)q¨+qq3+εt+ε,q=0,q0=qT,q˙0=q˙T,where A is a skew-symmetric matrix A⋆=−A, ε∈ℝ, and T>0 is a fixed period.The unperturbed system corresponding to (2) is the following:(7)q¨+qq3=0,q0=qT,q˙0=q˙T.By a noncollision orbit of (6), we mean a solution of (6) such that qt=0 for all t. We will say that q∈H1ℝ/Tℤ,ℝ2 is skew-T/2 periodic if qt+T/2=−qt for every t. The following result holds.Theorem 1 (see [7, 8, 12]).
There existsε0>0 such that ∀ε<ε0, the perturbed system (6), has at least one noncollision symmetric (skew-T/2 periodic) orbit near the circular orbit of (6).


### 1.1. Bifurcation in the Nondegenerate Case

Actually, we wish to relate the skew-T/2 periodic solutions of (6) to the T-periodic circular solutions of (7). Let(8)S1=ℝℤ,H11=S1,ℝ2,and(9)E1=u∈H11such thatut+12=−ut,∀t∈S1.We consider the open subset ofE1 defined by(10)Λ1=u∈E1such thatut≠0,∀t∈S1.OnΛ×ℝ×ℝ+⋆, we define a functional Φ by setting(11)Φu,ε,T=∫01ut22T+1T2ut−ε2<Au˙t,ut>−εVε,TutTdt.Lemma 1.
The functionsΦbelongs toC2Λ1×ℝ×ℝ+⋆,ℝand for allh,k∈E1, we have(12)aϕu′u,ε,Th=−∫01u¨tT+utT2ut3+εAu˙t+Vu′ε,Tut,htdt,bϕT′u,ε,T=−∫01u˙t22T2+2T3ut+εVu′ε,TuT,ut−Vε,TuT2dt,cϕε′u,ε,T=−∫0112Au˙,u+Vε,TuT+εVε′ε,Tudt,dϕuu″u,ε,T=−∫01<h¨tT+utT2ut3+ht−3utut,htut2=+εAt˙+TVuu″ε,Tuh,kdt,eϕTu′T,ε,uh=∫01u¨tT2+2utT3ut3−εVuu″ε,Tuu,htdt,fϕεu′T,ε,uh=−∫01V′εu′ε,Tu+Au˙t,htdt.Proof.
The proof is left to the reader.
We now show the following lemma.Lemma 2.
The following statements are equivalent:(i)
Φu′u,ε,T=0(ii)
qt=TutT−1is a noncollision skew−T/2periodic solution of(6)Proof.
We prove the lemma in two steps:Step 1.
q∈C2ST,ℝ2. The equation, ϕu′u,ε,T=0, means for every h∈E1,(13)∫01u¨tT+utT2ut3+εAu˙t+Vu′ε,Tut,htdt.
Therefore,(14)∫01u¨tT,htdt=−∫01utT2ut3+εAu˙t+Vu′ε,Tut,htdt.
Let(15)h=h1+h2,where(16)h1=ht+ht+1/22,h1is12periodic,and(17)h2=ht−ht+1/22,h2is skew12periodic.
Hence,(18)∫01u¨tT,htdt=∫01u¨tT,h1tdt+∫01u¨tT,h2tdt.
It is clear that(19)∫01u¨tT,h1tdt=0.
Therefore,(20)∫01u¨tT,htdt=∫01u¨tT,h2tdt,=−∫01utT2ut3+εAu˙t+Vu′ε,Tut,htdt,<+∞.
This impliesq∈H2sT,ℝ2 and q∈C2ST,ℝ2.Step 2.
q is a noncollision skew −1/2 periodic solution. Since u∈Λ1, it is skew-T/2 periodic.
Denote byE1⊥ the orthogonal subspace in H11 to E1 and(21)φt=u¨tT+utT2ut3+εAu˙t+Vu′ε,Tut.
From (13), it follows that(22)φ∈E1⊥.
UsingV2, one finds that(23)φt+12=−φt.
Hence,(24)φ∈E1∩E1⊥,and thus, φ=0,(25)qt=TutT−1.
Then,(26)q˙=u˙tT−1,q¨t=u¨tT−1T.
By substituting inϕ=0, we obtain(27)q¨+qq3+εAq˙+Vq′ε,q=0.
Moreover,(28)qT2=−q0,q˙T2=−q˙0.
This completes the proof.


### 1.2. Finding Critical Points forϕ·,ε,T

LetT0>0, and(29)Y=u∈Λ1,ut=rT0ξe2iπt+ξ¯e−2iπt,ξ∈ℂ2,ξ,ξ=12andξ,ξ¯=0,where(30)1r3=ω02ω0=2πT0,is a manifold of critical points for ϕ·,0,T0 (that is, ϕu′u,0,T0=0, for every u∈Y0). Using Lemma 2, we find that(31)Z0=q∈ΛT0:qt=rξeiε0t,ξ∈ℂ2ξ,ξ=12andξ,ξ¯=0,is a manifold of circular solutions for the unperturbed system,(32)q¨+qq3=0,q0=qT0,q˙0=q˙T.We wish to investigate the situation aroundq0,0,T0,q0∈ℤ0 by applying the inverse function theorem. For this, we need to know more about the derivative of ϕu′ and ϕuu″.Lemma 3.
ϕuu″u,0,T0is the Fredholm operator of index zero,∀u∈Y0.Proof.
Lettingu∈Y0, for all h and k in E1, we have(33)ϕuu″u,0,T0h,k=1T0∫01ht,ktdt+3r5T02∫01ut,ktut,ktdt.
Then,(34)ϕuu″u,0,T0h,k,h,k=h,kH11−1+1r3T02∫0Tht,ktdt+3r5T05∫01ut,htut,ktdt.
Hence,(35)ϕuu″u,0,T0=IEr−ℑ,where ℑ is a linear operator from E1 into E1 and defined by(36)ℑh,k=1+1r3T02∫0Tht,ktdt−3r5T05∫01ut,htut,ktdt.
It is easy to verify thatℑ is a compact operator. S°ϕuu″u,0,T0 is the Fredholm operator of index zero.
Thus, the proof is complete.
The preceding lemma implies(37)E1=Kerϕuu″u,0,T0⊕Imgϕuu″u,0,T0,∀u∈Y0.
We deduce that,ϕuu″u,0,T0 cannot be an onto function. The ultimate reason for this lies in the fact that the function ϕ is invariant by the S1− action which sends ut into ut+θ∈S1 and this induces degeneracy in the derivatives. We can estimate Kerϕuu″u,0,T0 by relating ϕuu″ to the linearized equation (32) around u. This is done as in the following lemma.Lemma 4.
Letu0∈Y0. The following two conditions are equivalent:(a)
ϕuu″u0,0,T0k=0.(b)
ht=T0ktT−1is a skew−T0/2periodic solution of(38)h¨+1q03h−3q0q0,hq02=0,where(39)q0t=T0u0tT0−1.Proof.
Letk∈E1, such that ϕuu″u0,0,T0k=0. By reasoning as in the proof of Lemma 2, we obtain(40)k∈C2S1,ℝ2,and(41)k¨T0+1T02u03k−3u0q0,ku02=0.
Set(42)ht=T0utT0−1,q0t=T0u0tT0−1.
By substituting (42) in (41), we obtain,(43)h¨+1q03q−3q0q0,hq02=0.
This completes the proof.
More precisely, we have the following lemma.Lemma 5.
(44)Kerϕuu″u0,0,T0=Tu0Y0=u˙,∀u0∈Y0.Proof.
According to the above lemma, the dimension ofKerϕuu″u0,0,T0 is equal to that of the set of the solutions for equation (38). Let h be the solution of (38), and by using the fact that q0=r, equation (38) reduces to(45)−h¨=ω02h−3q0q0,hr2.
Set(46)ht=∑k∈ℤckeikωt,ck∈ℂ2,since k∈E1, then c2k=0,∀k∈ℤ. By substituting, we find(47)∑k∈ℤk2−1ckeikωt=−3∑k∈ℤck,ξξ+ck+2,ξξ¯+ck−2,ξ¯ξ+ck,ξ¯ξ¯eikωt,from which we deduce the following equation for ck, where k∈ℤ,(48)afor allk≠±1,bc1,ξ>+<c−1,ξ¯=0,cc1,ξ¯=0.
It is easy to see that conditionsa−c define the tangent manifold to Y0. According to the definition of a nondegenerate critical manifold, this means that ℤ0 is a nondegenerate critical:(49)Kerϕuu″u0,0,T0=Tq0Y0,u˙0∈Kerϕuu″u0,0,T0and dimTu0Y0=1.
Therefore,(50)Kerϕuu″u0,0,T0=u˙0.
This completes the proof.
We denote(51)Nu0=Kerϕuu″u0,0,T0,u0∈Y0,Ru0=Imϕuu″u0,0,T0,u0∈Y0.
This is a fact thatNu0=u˙0 gives us a considerable simplification.Lemma 6.
Foru,ε,Tclose tou0,0,T0, the following statements are equivalent:(52)iϕu′u,ε,T=0.iiϕu′u,ε,T∈Nu0.Proof.
Kerϕuu″u0,0,T0 is spanned by u˙0. The equation Kerϕu′u,0,T∈Kerϕuu″u0,0,T0,(53)u¨tT+utT2ut3+εAu˙t+Vu′ε,Tut=αu˙0.
Multiplying both sides byu˙ and integrating, we get(54)∫01u¨tT,u˙t+ut,u˙tT2ut3+εAu˙t,u˙t+Vu′ε,Tut,u˙tdt=α∫01u˙0,u˙tdt.
Au˙,u˙=0 and if we integrate the first and second terms on the left by parts, we get zero. In the last term, we recognize the time derivative of Vε,Tu, which integrates away to zero. Finally, we get,(55)α∫01u˙0,u˙tdt=0.
Ifu is close to u0, in E1, the integral is strictly positive, and, hence, α must be zero.
We now state our main result.Theorem 2.
Letq0∈ℤ0. If(56)∫01q0teiω0tdt≠0inℂ2,then there are positive numbers r0,ε, a neighborhood V of the path q0 in ℝ2, and a ℂ2 map,(57)v:S1×−ε0,ε0×T0−r0,T0+r0⟶ℝ2,such that(58)q0t=T0vtT−1,0,T0,and for any ε∈−ε0,ε0 and T such T−T0<r0 the curve,(59)qt=TvtT−1,ε,T,is a skew-T/2 periodic solution of equation (6). Conversely, whenever q is a skew −T/2 periodic solution of (6), with ε∈−ε,ε, T−T0<r0, and qt remaining in V for all t, then some θ∈ℝ can be found such that(60)qt=T0vtT−1+θ,ε,T.Proof.
Setu0·=tT0−1q0·T0. It is known that ϕuuu0,0,T0 is a Fredholm map of index zero. Split E1 into Nu0⊕Ru0; then, ϕ′uu′ is an isomorphism of Ru0 onto itself. By the implicit function theorem, the equation(61)ϕu′u,ε,T∈Nu0,determines u∈Ru0 in terms of the remaining variables. These are ε,T and the component α of u in Nu0. By this, we mean(62)u1=u0+αu˙0,∀u1∈u0+αu˙0.
By Lemma6, this means that the equation ϕu′u,ε,T=0 can be solved in E1 as follows in neighborhood of u0,0,T0,(63)u=v1α,ε,T.
We now replaceα by more convenient variables.
For anys∈S1, set ust=ut+s. If ϕu′u,ε,T=0, we also have ϕu′us,ε,T=0. We, thus, have an S1-action which leaves our equations invariants, and we wish to find a coordinate system adapted to this group-invariance. For u near u0∈E1, the complex number,(64)∫01u0te2iπt1∫01u¯0te−2iπt1,has a well-defined argument θu, called the phase of u with respect to u0. In the subsequent paragraphs, we will verify that(65)θus=θu+2πs,∀s∈S1,See7.
I now claim that we can useθ,ε,T as a local coordinate system for u0+NuO×ℝ×ℝT near uo,0,To. Computing the Jacobian at this point gives(66)Dθ,ε,TDα,ε,T=DθDα=θ′u−0,u˙0,See6.
Since(67)θ′u0,u˙0=ddsθu+2πs=2π,so the Jacobian does not vanish. The equation now becomes(68)u=v2α,ε,T.
Using Lemma4 to translate in terms of q and q0, we get the desired result. v is at least a C2 map from ℝ×ℝ+⋆ into the space C2S1,ℝ2; it will then have a Taylor expansion.


## 1.1. Bifurcation in the Nondegenerate Case

Actually, we wish to relate the skew-T/2 periodic solutions of (6) to the T-periodic circular solutions of (7). Let(8)S1=ℝℤ,H11=S1,ℝ2,and(9)E1=u∈H11such thatut+12=−ut,∀t∈S1.We consider the open subset ofE1 defined by(10)Λ1=u∈E1such thatut≠0,∀t∈S1.OnΛ×ℝ×ℝ+⋆, we define a functional Φ by setting(11)Φu,ε,T=∫01ut22T+1T2ut−ε2<Au˙t,ut>−εVε,TutTdt.Lemma 1.
The functionsΦbelongs toC2Λ1×ℝ×ℝ+⋆,ℝand for allh,k∈E1, we have(12)aϕu′u,ε,Th=−∫01u¨tT+utT2ut3+εAu˙t+Vu′ε,Tut,htdt,bϕT′u,ε,T=−∫01u˙t22T2+2T3ut+εVu′ε,TuT,ut−Vε,TuT2dt,cϕε′u,ε,T=−∫0112Au˙,u+Vε,TuT+εVε′ε,Tudt,dϕuu″u,ε,T=−∫01<h¨tT+utT2ut3+ht−3utut,htut2=+εAt˙+TVuu″ε,Tuh,kdt,eϕTu′T,ε,uh=∫01u¨tT2+2utT3ut3−εVuu″ε,Tuu,htdt,fϕεu′T,ε,uh=−∫01V′εu′ε,Tu+Au˙t,htdt.Proof.
The proof is left to the reader.
We now show the following lemma.Lemma 2.
The following statements are equivalent:(i)
Φu′u,ε,T=0(ii)
qt=TutT−1is a noncollision skew−T/2periodic solution of(6)Proof.
We prove the lemma in two steps:Step 1.
q∈C2ST,ℝ2. The equation, ϕu′u,ε,T=0, means for every h∈E1,(13)∫01u¨tT+utT2ut3+εAu˙t+Vu′ε,Tut,htdt.
Therefore,(14)∫01u¨tT,htdt=−∫01utT2ut3+εAu˙t+Vu′ε,Tut,htdt.
Let(15)h=h1+h2,where(16)h1=ht+ht+1/22,h1is12periodic,and(17)h2=ht−ht+1/22,h2is skew12periodic.
Hence,(18)∫01u¨tT,htdt=∫01u¨tT,h1tdt+∫01u¨tT,h2tdt.
It is clear that(19)∫01u¨tT,h1tdt=0.
Therefore,(20)∫01u¨tT,htdt=∫01u¨tT,h2tdt,=−∫01utT2ut3+εAu˙t+Vu′ε,Tut,htdt,<+∞.
This impliesq∈H2sT,ℝ2 and q∈C2ST,ℝ2.Step 2.
q is a noncollision skew −1/2 periodic solution. Since u∈Λ1, it is skew-T/2 periodic.
Denote byE1⊥ the orthogonal subspace in H11 to E1 and(21)φt=u¨tT+utT2ut3+εAu˙t+Vu′ε,Tut.
From (13), it follows that(22)φ∈E1⊥.
UsingV2, one finds that(23)φt+12=−φt.
Hence,(24)φ∈E1∩E1⊥,and thus, φ=0,(25)qt=TutT−1.
Then,(26)q˙=u˙tT−1,q¨t=u¨tT−1T.
By substituting inϕ=0, we obtain(27)q¨+qq3+εAq˙+Vq′ε,q=0.
Moreover,(28)qT2=−q0,q˙T2=−q˙0.
This completes the proof.


## 1.2. Finding Critical Points forϕ·,ε,T

LetT0>0, and(29)Y=u∈Λ1,ut=rT0ξe2iπt+ξ¯e−2iπt,ξ∈ℂ2,ξ,ξ=12andξ,ξ¯=0,where(30)1r3=ω02ω0=2πT0,is a manifold of critical points for ϕ·,0,T0 (that is, ϕu′u,0,T0=0, for every u∈Y0). Using Lemma 2, we find that(31)Z0=q∈ΛT0:qt=rξeiε0t,ξ∈ℂ2ξ,ξ=12andξ,ξ¯=0,is a manifold of circular solutions for the unperturbed system,(32)q¨+qq3=0,q0=qT0,q˙0=q˙T.We wish to investigate the situation aroundq0,0,T0,q0∈ℤ0 by applying the inverse function theorem. For this, we need to know more about the derivative of ϕu′ and ϕuu″.Lemma 3.
ϕuu″u,0,T0is the Fredholm operator of index zero,∀u∈Y0.Proof.
Lettingu∈Y0, for all h and k in E1, we have(33)ϕuu″u,0,T0h,k=1T0∫01ht,ktdt+3r5T02∫01ut,ktut,ktdt.
Then,(34)ϕuu″u,0,T0h,k,h,k=h,kH11−1+1r3T02∫0Tht,ktdt+3r5T05∫01ut,htut,ktdt.
Hence,(35)ϕuu″u,0,T0=IEr−ℑ,where ℑ is a linear operator from E1 into E1 and defined by(36)ℑh,k=1+1r3T02∫0Tht,ktdt−3r5T05∫01ut,htut,ktdt.
It is easy to verify thatℑ is a compact operator. S°ϕuu″u,0,T0 is the Fredholm operator of index zero.
Thus, the proof is complete.
The preceding lemma implies(37)E1=Kerϕuu″u,0,T0⊕Imgϕuu″u,0,T0,∀u∈Y0.
We deduce that,ϕuu″u,0,T0 cannot be an onto function. The ultimate reason for this lies in the fact that the function ϕ is invariant by the S1− action which sends ut into ut+θ∈S1 and this induces degeneracy in the derivatives. We can estimate Kerϕuu″u,0,T0 by relating ϕuu″ to the linearized equation (32) around u. This is done as in the following lemma.Lemma 4.
Letu0∈Y0. The following two conditions are equivalent:(a)
ϕuu″u0,0,T0k=0.(b)
ht=T0ktT−1is a skew−T0/2periodic solution of(38)h¨+1q03h−3q0q0,hq02=0,where(39)q0t=T0u0tT0−1.Proof.
Letk∈E1, such that ϕuu″u0,0,T0k=0. By reasoning as in the proof of Lemma 2, we obtain(40)k∈C2S1,ℝ2,and(41)k¨T0+1T02u03k−3u0q0,ku02=0.
Set(42)ht=T0utT0−1,q0t=T0u0tT0−1.
By substituting (42) in (41), we obtain,(43)h¨+1q03q−3q0q0,hq02=0.
This completes the proof.
More precisely, we have the following lemma.Lemma 5.
(44)Kerϕuu″u0,0,T0=Tu0Y0=u˙,∀u0∈Y0.Proof.
According to the above lemma, the dimension ofKerϕuu″u0,0,T0 is equal to that of the set of the solutions for equation (38). Let h be the solution of (38), and by using the fact that q0=r, equation (38) reduces to(45)−h¨=ω02h−3q0q0,hr2.
Set(46)ht=∑k∈ℤckeikωt,ck∈ℂ2,since k∈E1, then c2k=0,∀k∈ℤ. By substituting, we find(47)∑k∈ℤk2−1ckeikωt=−3∑k∈ℤck,ξξ+ck+2,ξξ¯+ck−2,ξ¯ξ+ck,ξ¯ξ¯eikωt,from which we deduce the following equation for ck, where k∈ℤ,(48)afor allk≠±1,bc1,ξ>+<c−1,ξ¯=0,cc1,ξ¯=0.
It is easy to see that conditionsa−c define the tangent manifold to Y0. According to the definition of a nondegenerate critical manifold, this means that ℤ0 is a nondegenerate critical:(49)Kerϕuu″u0,0,T0=Tq0Y0,u˙0∈Kerϕuu″u0,0,T0and dimTu0Y0=1.
Therefore,(50)Kerϕuu″u0,0,T0=u˙0.
This completes the proof.
We denote(51)Nu0=Kerϕuu″u0,0,T0,u0∈Y0,Ru0=Imϕuu″u0,0,T0,u0∈Y0.
This is a fact thatNu0=u˙0 gives us a considerable simplification.Lemma 6.
Foru,ε,Tclose tou0,0,T0, the following statements are equivalent:(52)iϕu′u,ε,T=0.iiϕu′u,ε,T∈Nu0.Proof.
Kerϕuu″u0,0,T0 is spanned by u˙0. The equation Kerϕu′u,0,T∈Kerϕuu″u0,0,T0,(53)u¨tT+utT2ut3+εAu˙t+Vu′ε,Tut=αu˙0.
Multiplying both sides byu˙ and integrating, we get(54)∫01u¨tT,u˙t+ut,u˙tT2ut3+εAu˙t,u˙t+Vu′ε,Tut,u˙tdt=α∫01u˙0,u˙tdt.
Au˙,u˙=0 and if we integrate the first and second terms on the left by parts, we get zero. In the last term, we recognize the time derivative of Vε,Tu, which integrates away to zero. Finally, we get,(55)α∫01u˙0,u˙tdt=0.
Ifu is close to u0, in E1, the integral is strictly positive, and, hence, α must be zero.
We now state our main result.Theorem 2.
Letq0∈ℤ0. If(56)∫01q0teiω0tdt≠0inℂ2,then there are positive numbers r0,ε, a neighborhood V of the path q0 in ℝ2, and a ℂ2 map,(57)v:S1×−ε0,ε0×T0−r0,T0+r0⟶ℝ2,such that(58)q0t=T0vtT−1,0,T0,and for any ε∈−ε0,ε0 and T such T−T0<r0 the curve,(59)qt=TvtT−1,ε,T,is a skew-T/2 periodic solution of equation (6). Conversely, whenever q is a skew −T/2 periodic solution of (6), with ε∈−ε,ε, T−T0<r0, and qt remaining in V for all t, then some θ∈ℝ can be found such that(60)qt=T0vtT−1+θ,ε,T.Proof.
Setu0·=tT0−1q0·T0. It is known that ϕuuu0,0,T0 is a Fredholm map of index zero. Split E1 into Nu0⊕Ru0; then, ϕ′uu′ is an isomorphism of Ru0 onto itself. By the implicit function theorem, the equation(61)ϕu′u,ε,T∈Nu0,determines u∈Ru0 in terms of the remaining variables. These are ε,T and the component α of u in Nu0. By this, we mean(62)u1=u0+αu˙0,∀u1∈u0+αu˙0.
By Lemma6, this means that the equation ϕu′u,ε,T=0 can be solved in E1 as follows in neighborhood of u0,0,T0,(63)u=v1α,ε,T.
We now replaceα by more convenient variables.
For anys∈S1, set ust=ut+s. If ϕu′u,ε,T=0, we also have ϕu′us,ε,T=0. We, thus, have an S1-action which leaves our equations invariants, and we wish to find a coordinate system adapted to this group-invariance. For u near u0∈E1, the complex number,(64)∫01u0te2iπt1∫01u¯0te−2iπt1,has a well-defined argument θu, called the phase of u with respect to u0. In the subsequent paragraphs, we will verify that(65)θus=θu+2πs,∀s∈S1,See7.
I now claim that we can useθ,ε,T as a local coordinate system for u0+NuO×ℝ×ℝT near uo,0,To. Computing the Jacobian at this point gives(66)Dθ,ε,TDα,ε,T=DθDα=θ′u−0,u˙0,See6.
Since(67)θ′u0,u˙0=ddsθu+2πs=2π,so the Jacobian does not vanish. The equation now becomes(68)u=v2α,ε,T.
Using Lemma4 to translate in terms of q and q0, we get the desired result. v is at least a C2 map from ℝ×ℝ+⋆ into the space C2S1,ℝ2; it will then have a Taylor expansion.


## 2. Conclusion

We can conclude that the class of planar perturbations are Keplerian Hamiltonian systems, as we have shown that the noncollision periodic solutions of this perturbed system radiate from the complex of circular solutions of the Keplerian Hamiltonian system. We have studied a class of Hamiltonian systems obtained as perturbation in the Keplerian Hamiltonian(69)Kp,q=12p2−q−1.Our goal was to search for noncollision periodic solutions of (2), and we wish to relate them to circular solutions for the nonperturbed system (3). These kinds of systems were restless focus of a number of authors and the references therein [1–5]. We mention in particular, the work of Poincaré, on the three-body problem (these orbitals are called “first-view sort solutions”).In an effort to organize another piece of work into a paper, we determine the coefficients of the Taylor expansionup to the second order of the noncollision periodic solutions for the perturbed planar Keplerian Hamiltonian system, which is connected to Kepler Hamiltonian systems by a perturbation parameter. This Taylor expansion is made with respect to a perturbation term ε and the period T of the solution.



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