# Effect of Perturbations in Coriolis and Centrifugal  Forces on the Nonlinear Stability of Equilibrium Point in  Robe's Restricted Circular Three-Body Problem

Authors: P. P. Hallan; Khundrakpam Binod Mangang
Journal: Advances in Astronomy
 (2008)

---

## Abstract

The effect of perturbations in Coriolis and cetrifugal forces on the nonlinear
stability of the equilibrium point of the Robe's (1977) restricted circular
three-body problem has been studied when the density parameterK is zero. By applying Kolmogorov-Arnold-Moser (KAM) theory, it has been found
that the equilibrium point is stable for all mass ratios μ in the range of
linear stability 8/9+(2/3)((43/25)ϵ1−(10/3)ϵ)<μ<1, where ϵ and ϵ1 are, respectively, the perturbations in Coriolis and centrifugal forces, except for five mass ratios
μ1=0.93711086−1.12983217ϵ+1.50202694ϵ1, μ2=0.9672922−0.5542091ϵ+1.2443968ϵ1, μ3=0.9459503−0.70458206ϵ+1.28436549ϵ1, μ4=0.9660792−0.30152273ϵ + 1.11684064ϵ1, μ5=0.893981−2.37971679ϵ + 1.22385421ϵ1, where the theory is not applicable.


---

## Body

## 1. Introduction

Robe [1] has considered a new kind of restricted
three-body problem in which one of the primaries is a rigid spherical shell m1 filled with a homogeneous   incompressible fluid
of density ρ1.
The second primary is a mass point m2 outside the shell and the third body m3 is a small solid sphere of density ρ3,
inside the shell, with the assumption that the mass and radius of m3 are infinitesimal. He has shown the existence
of an equilibrium point with m3 at the center of the shell, while m2 describes a Keplerian orbit around it.
Further, he has discussed the linear stability of the equilibrium point. Hallan
and Rana [2] considered the effect of perturbations ϵ,
ϵ1 in Coriolis and centrifugal forces,
respectively, on the location and linear stability of the equilibrium points in
Robe's circular three-body problem when the density parameter K is zero. They
have found that  (−μ+(μϵ1/(1+2μ)),0,0) is the only equilibrium point and in the
linear sense it is stable for μc<μ<1 and unstable for 0<μ≤μc,
where μc=8/9+(2/3)((43/25)ϵ1−(10/3)ϵ).
Shrivastava and Garain [3], A. R. Plastino and
A. Plastino [4], Giordano et
al. [5] have also discussed Robe's problem. But all of them
have discussed the linear stability of the equilibrium points. Hallan and
Mangang [6] discussed the nonlinear stability of equilibrium point of Robe's
restricted three-body problem when K=0 in the linear stability range 8/9<μ<1 and they found that the equilibrium point is
stable in nonlinear sense for all mass ratios except for the five mass ratios μ1=0.93711086…, μ2=0.9672922…, μ3=0.9459503…, μ4=0.9660792…, μ5=0.893981…, where the KAM theory is not applicable. Many
authors discussed nonlinear stability of equilibrium points. Recently, Elipe
and López-Moratalla [7] discussed on the Lyapunov
stability of stationary points around a central body. Elipe et
al. [8] studied stability of equilibria in two degrees of
freedom Hamiltonian system. Elipe et al. [9]
discussed nonlinear stability in resonant cases. In the present study, we wish
to discuss the effects of perturbations in Coriolis and centrifugal forces on
the nonlinear stability of equilibrium point (−μ+(μϵ1/(1+2μ)),0,0) found by Hallan and Rana [2] in Robe's
restricted circular three-body problem by taking the density parameter K as zero by applying Moser's version of the
Arnold theorem (KAM theory) and following the procedure as that adopted by
Hallan and Mangang [6].Moser's version [10] of Arnold theorem [11] states
the following.If(1)H=ω1I1+ω2I2+ω3I3+12(aI12+bI22+cI32+2fI2I3+2gI3I1+2hI1I2) is the
normalized Hamiltonian with I1, I2, I3 as the action momenta coordinates and ω1, ω2, ω3  are the basic frequencies for the linear dynamical
system, then on each energy manifold H=ℏ in the neighborhood of an equilibrium point,
there exist invariant tori of quasiperiodic motions which divide the manifold
and consequently the equilibrium point is stable provided that (i)
k1ω1+k2ω2+k3ω3≠0, for all triplets (k1,k2,k3) of rational integers such that(2)|k1|+|k2|+|k3|≤4,(ii)
determinantD≠0,(3)D=det⁡⋅(bij)(i,j=1,2,3,4),bij=(∂2H∂Ii,∂Ij)Ii=Ij=0(i,j=1,2,3),bi4=b4i=(∂H∂Ii)Ii=Ij=0(i=1,2,3),b44=0.Applying  Arnold's theorem, Leontovich [12] proved
that the triangular equilibrium points in the restricted three-body problem are
stable for all permissible mass ratios except for a set of measure zero. Deprit and Deprit-Bartholome [13] discussed nonlinear stability of the
triangular equilibrium points of the classical restricted three-body problem by
applying Moser's theorem. Bhatnagar and Hallan [14] also discussed the
nonlinear stability of the triangular equilibrium points in the same problem after
considering perturbations in Coriolis and centrifugal forces. In another paper,
Bhatnagar and Hallan [15] discussed the nonlinear stability of a cluster of
stars sharing galactic rotation.By applying the Lyapunov theorem  [16] to the linear
stability result obtained by Hallan and Rana [2] in Robe's restricted three-body problem, we can say that the equilibrium point, (−μ+(μϵ1/(1+2μ)),0,0),
is unstable in the nonlinear sense also for 0<μ≤μc.
Therefore, we will study the nonlinear stability of the equilibrium point for μc<μ<1.


## 2. First-order Normalization

Using nondimensional variables and a synodic system of
coordinates(x,y,z) and considering perturbations ϵ, ϵ1,  respectively, in Coriolis and centrifugal
forces, the equations of motion of Robe's restricted problem, when density
parameter K=0 and eccentricity e=0,
are [2](4)x¨−2αy˙−βx=μ(1−μ−x)[(1−μ−x)2+y2+z2]3/2,y¨+2αx˙−βy=−μy[(1−μ−x)2+y2+z2]3/2,z¨=−μz[(1−μ−x)2+y2+z2]3/2,where  α=1+ϵ, β=1+ϵ1, |ϵ|≪1, |ϵ1|≪1,    μ=m2/(m1*+m2) (0<μ<1),  m2 = mass of the second primary, m1* = mass of the first primary along with the mass
of the fluid inside it.LagrangianL of the problem is(5)L=12(x˙2+y˙2+z˙2)+α(xy˙−yx˙)+β2(x2+y2)+μ[(1−μ−x)2+y2+z2]1/2.There is only one equilibrium
point (−μ+p,0,0),
where p=μϵ1/(1+2μ)  [2]. Shifting the origin
to (−μ+p,0,0) and expanding in Taylor series expansion and
neglecting second and higher degree terms in ϵ, ϵ1,
the Lagrangian can be written as(6)L=L0+L1+L2+L3+L4+⋯,where(7)L0=μμ+2β2,L1=py˙−αμy˙,L2=12(x˙2+y˙2+z˙2)+α(xy˙−yx˙)+β2(x2+y2)+μ(x2−y22−z22+3x2p−32y2p−32z2p),L3=μ(x3−32y2x−32z2x+4x3p−6xy2p−6xz2p),L4=μ(x4+38y4+38z4−3x2y2−3x2z2+34y2z2).To the first order, Lagrange's
equations of motion are(8)x¨−2αy˙−x(β+2μ+6μp)=0,y¨+2αx˙−y(β−μ−3μp)=0,z¨=−zμ(1+3p).The characteristic equation of
the first two equations is(9)λ4−λ2[μ−2−ϵ+(2+3μp′)ϵ1]+(1+2μ)(1−μ)+[(1−μ)(1+6μp′)+(1+2μ)(1−3μp′)]ϵ1=0,where p′=μ/(1+2μ).The characteristic equation of the third equation
is(10)λ2+μ(1+3p′ϵ1)=0.Equation (9) has pure imaginary
roots if(11)89+23(4325ϵ1−103ϵ)<μ<1[2] and it is obvious that (10) has
pure imaginary roots. The four characteristic roots of (9) are ±iω1′, ±iω2′ and the two characteristic roots of (10) are ±iω3′,
where ω1′, ω2′, ω3′ represent the perturbed basic frequencies of
the linear dynamical system. We can write(12)ω1′=ω1(1+p1ϵ+q1ϵ1),ω2′=ω2(1+p2ϵ+q2ϵ1),ω3′=ω3(1+p3ϵ+q3ϵ1),where ω1, ω2, ω3 represent the unperturbed basic frequencies of
the linear dynamical system such that(13)ω12+ω22=2−μ,ω12ω22=(1+2μ)(1−μ),ω32=μ,(14)p1=−p2=4ω12−ω22,p3=0,q1=−12(ω12−ω22)[(2+3μp′)(1+ω22)+μ(1−12μp′)ω22],q2=12(ω12−ω22)[(2+3μp′)(1+ω12)+μ(1−12μp′)ω12],q3=3p′2.From (13), we see that 1>ω3>ω2>ω1>0,
therefore, we have 1>ω3′>ω2′>ω1′>0.Following the method given by Whittaker [17], we use
the canonical transformation from the phase space (x,y,z,px,py,pz) into the phase space product (φ1,φ2,φ3,I1,I2,I3) of the angle coordinates φ1, φ2, φ3 and action momenta I1, I2, I3 given by(15)X=AT,where(16)X=(xyzpxpypz),A=(aij′)1≤i,j≤6,T=(Q1Q2Q3P1P2P3),Qi=(2Iiωi′)1/2sin⁡φi,Pi=(2Iiωi′)1/2cos⁡φi(i=1,2,3),aij′=aij(1+αijϵ+aij′ϵ1),a1j=a5j=0(j=1,2,3,6),a2j=a4j=0(j=3,4,5,6),a3j=a6j=0(j=1,2,4,5),a33=a66=0,a13+i=2hili,a2i=−4ωi2hi,a36=2ω3h3,a4i=−2hiωi2mi,a53+i=−2hini,a63=−2μω3h3,α13+i=piωi2limi+4piωi2ni−2pilini−4piωi2li2+4(1−μ)lili(limi+2ni),α13+i′=qiωi2limi+4qiωi2ni−6μp′ni+2nili(limi+2ni)−qiωi2li−qili2mi−2qilinili(limi+2ni),α2i=limi+2ni+pilimi+2pini−piωi2li−piωi2mi+4(1−μ)(limi+2ni),α2i′=qilimi+2qini−qiωi2li−qiωi2mi+(3μp′−1)mi(limi+2ni),α36=0,α36′=−32p′,α4i=pilimi2−4limi+piωi2limi+2pimini−8nimi(limi+2ni)+4piniωi2−piωi2mi2+4mi(1−μ)mi(limi+2ni),α4i′=qilimi(mi+ωi2)+(3μp′−1)(mi−li)mimi(limi+2ni)+2ni(1+2qiωi2+qimi−3μp′)−qiωi2mi2mi(limi+2ni),α53+i=limi(ni+2piωi2−pini)+2ni2(1−pi)ni(limi+2ni)+piniωi2(4−li−mi)+4ni(1−μ)ni(limi+2ni),α53+i′=qiωi2[2limi+ni(4−li−mi)]ni(limi+2ni)+(3μp′−1)[mi(ni+li)+2ni]−niqi(2ni+limi)ni(limi+2ni),α63=0,α63′=32p′,hi2=14ωi2(limi+2ni),h32=14μω32,li=ωi2−μ+1,mi=ωi2−μ−1(i=1,2),ni=ωi2+μ−1(i=1,2).The transformation  changes the
second-order part of the Hamiltonian into the normal form(17)H2=ω1′I1+ω2′I2+ω3′I3.The general solution of the
corresponding equations of motion are(18)Ii=const.(i=1,2,3),φi=ωi′t+const.(i=1,2,3).


## 3. Second-order Normalization

We wish to perform Birkhoff's normalization for which
the coordinates(x,y,z) are to be expanded in double D'Alembert's
series:(19)x=∑n≥1Bn1,0,0,y=∑n≥1Bn0,1,0,z=∑n≥1Bn0,0,1,where the homogeneous components Bn1,0,0, Bn0,1,0, Bn0,0,1 of degree n are of the form(20)∑0<ℓ,m<nI1(1/2)(n−ℓ−m)I2(1/2)ℓI3(1/2)m×∑i,j,k[Cn−ℓ−m,ℓ,m,i,j,kcos⁡(iφ1+jφ2+kφ3)+Sn−ℓ−m,ℓ,m,i,j,ksin⁡(iφ1+jφ2+kφ3)].The double summation over the
indices i, j,  and k is such that (a) i runs over those integers in the interval 0≤i≤n−ℓ−m that have the same parity as n−ℓ−m,
(b) j runs over those integers in the intervals −ℓ≤j≤ℓ that have the same parity as ℓ,
(c) k runs over those integers in the interval −m≤k≤m that have the same parity as m. I1, I2, and I3 are to be regarded as constants of integration
and φ1, φ2, and φ3 are to be determined as linear functions of
time such that(21)φ˙1=ω1′+∑n≥1f2n(I1,I2,I3),φ˙2=ω2′+∑n≥1g2n(I1,I2,I3),φ˙3=ω3′+∑n≥1h2n(I1,I2,I3),where f2n, g2n, h2n are of the form(22)f2n=∑0≤ℓ,m<nf2n−2ℓ−2m,2ℓ,2mI1n−ℓ−mI2ℓI3m,g2n=∑0≤ℓ,m≤ng2n−2ℓ−2m,2ℓ,2mI1n−ℓ−mI2ℓI3m,h2n=∑0≤ℓ,m≤nh2n−2ℓ−2m,2ℓ,2mI1n−ℓ−mI2ℓI3m.As shown by Hallan and
Mangang [6], the first-order components B11,0,0, B10,1,0,
and B10,0,1 are the values of x,y, and z given by (15). B21,0,0, B20,1,0, and B20,0,1 are the solutions of the partial differential
equations(23)Δ1Δ2B21,0,0=Φ2,Δ1Δ2B20,1,0=Ψ2,Δ3B20,0,1=Z2,where(24)Δi=D2+ωi′2(i=1,2,3),Φ2=(D2−β+μ+3μp)X2+2αDY2,Ψ2=(D2−β−2μ−6μp)Y2−2αDX2,D=(ω1′∂∂φ1+ω2′∂∂φ2+ω3′∂∂φ3)and X2, Y2, Z2 are obtained from ∂L3/∂x, ∂L3/∂y, ∂L3/∂z,
respectively, by substituting the first-order components for x, y, z.Equation (23)
can be solved for B21,0,0, B20,1,0, B20,0,1 by using the formulae(25)1Δ1Δ2[cos⁡(ℓφ1+mφ2+nφ3)orsin⁡(ℓφ1+mφ2+nφ3)]=1Δℓ,m,n[cos⁡(ℓφ1+mφ2+nφ3)orsin⁡(ℓφ1+mφ2+nφ3)],1Δ3[cos⁡(ℓφ1+mφ2+nφ3)orsin⁡(ℓφ1+mφ2+nφ3)]=1ω32−(ℓω1+mω2+nω3)2×[cos⁡(ℓφ1+mφ2+nφ3)orsin⁡(ℓφ1+mφ2+nφ3)],where(26)Δℓ,m,n=[ω12−(ℓω1+mω2+nω3)2]×[ω22−(ℓω1+mω2+nω3)2].The second-order components B21,0,0, B20,1,0,
and B20,0,1 are as follows:(27)B21,0,0=r1′I1+r2′I2+r3′I3+r4′I1cos⁡2φ1+r5′I2cos⁡2φ2+r6′I3cos⁡2φ3+r7′(I1I2)1/2cos⁡(φ1+φ2)+r8′(I1I2)1/2cos⁡(φ1−φ2),B20,1,0=s1′I1sin⁡2φ1+s2′I2sin⁡2φ2+s3′I3sin⁡2φ3+s4′(I1I2)1/2sin⁡(φ1+φ2)+s5′(I1I2)1/2sin⁡(φ1−φ2),B20,0,1=t1′(I1I3)1/2cos⁡(φ1+φ3)+t2′(I1I3)1/2cos⁡(φ1−φ3)+t3′(I2I3)1/2cos⁡(φ2+φ3)+t4′(I2I3)1/2cos⁡(φ2−φ3),where(28)rj′=rj+rj1ϵ+rj2ϵ1(j=1,2,3,4,5,6,7,8),sj′=sj+sj1ϵ+sj2ϵ1(j=1,2,3,4,5),tj′=tj+tj1ϵ+tj2ϵ1(j=1,2,3,4),ri=12ω32(w32−1)(li2−2ωi2)hi2ωiω12ω22(i=1,2),r3=−6ω35(w32−1)h32ω12ω22,r3+i=4ω32hi2ωiri′′[3ωi2li2+li3+6ωi4−6ωi2li](i=1,2),r1′′=ω22−4ω12,r2′′=ω12−4ω22,r6=−6ω32(ω32−1)h32(ω12−4ω32)(ω22−4ω32),r7=24ω32h1h2(ω1ω2)1/2(2ω1+ω2)(2ω2+ω1)×[(l1l2+2ω1ω2){(ω12−l1)−(ω1+ω2)2}+2(ω1+ω2)(ω1l2+ω2l1)],r8=24ω32h1h2(ω1ω2)1/2(2ω1−ω2)(2ω2−ω1)×[(l1l2−2ω1ω2){(ω12−l1)−(ω1−ω2)2}+2(ω1−ω2)(ω1l2−ω2l1)],si=−8ω32hi2ri′[−4ωi2li+2li2+4ωi2−(1+2μ)li](i=1,2),s3=−24ω36h32(ω12−4ω32)(ω22−4ω32),s4=24ω32h1h2(ω1ω2)1/2(2ω1+ω2)(2ω2+ω1)×[−(ω1l2+ω2l1){(ω1+ω2)2+1+2μ}+2(ω1+ω2)(l1l2+2ω1ω2)],s5=24ω32h1h2(ω1ω2)1/2(2ω1−ω2)(2ω2−ω1)×[−(ω1l2−ω2l1){(ω1−ω2)2+1+2μ}+2(ω1−ω2)(l1l2−2ω1ω2)],t1=12h1h3l1w37/2(ω1+2ω3)ω11/2,t2=12h1h3l1w37/2(ω1−2ω3)ω11/2,t3=12h2h3l2w37/2(ω2+2ω3)ω21/2,t4=12h2h3l2w37/2(ω2−2ω3)ω21/2and rj1, rj2, sj1, sj2, tj1, tj2 are given in the appendix. We  have checked
that x=B11,0,0+B21,0,0, y=B10,1,0+B20,1,0,
and z=B10,0,1+B20,0,1 transform H3,
the third-order part of the Hamiltonian, to zero.


## 4. Second-order Coefficient in The Frequencies

Proceeding as in the work of Hallan and Mangang
[6], the third-order components B31,0,0, B30,1,0,
and B30,0,1 in the coordinates x, y, z and the second-order polynomials f2, g2,  and h2 in the frequencies φ˙1, φ˙2, and φ˙3 satisfy the partial differential
equations(29)Δ1Δ2B31,0,0=(D2−β+μ+3μp)X3′+2αDY3′,Δ1Δ2B30,1,0=(D2−β−2μ−6μp)Y3′−2αDX3′,Δ3B30,0,1=Z3′,where(30)X3′=X3−2ω1′f2∂2B11,0,0∂φ12−2ω2′g2∂2B11,0,0∂φ22+2f2∂B10,1,0∂φ1+2g2∂B10,1,0∂φ2,Y3′=Y3−2ω2′g2∂2B10,1,0∂φ22−2ω1′f2∂2B10,1,0∂φ12−2g2∂B11,0,0∂φ2−2f2∂B11,0,0∂φ1,Z3′=Z3−2ω3′h2∂2B10,0,1∂φ32and X3,Y3,Z3 are the
homogeneous components of order 3
obtained, respectively, from (∂/∂x)(L3+L4), (∂/∂y)(L3+L4), (∂/∂z)(L3+L4) by substituting(31)x=B11,0,0+B21,0,0,y=B10,1,0+B20,1,0,z=B10,0,1+B20,0,1.The components B31,0,0, B30,1,0, and B30,0,1 are not required to be found out. We find the
coefficients of cos⁡φi, sin⁡φi  (i=1,2,3) on the right-hand side of (29). They are the
critical terms as Δ1,0,0=Δ0,1,0=Δ0,0,1=0.We eliminate these terms by choosing properly the
coefficients in the polynomials(32)f2=f2,0,0I1+f0,2,0I2+f0,0,2I3,g2=g2,0,0I1+g0,2,0I2+g0,0,2I3,h2=h2,0,0I1+h0,2,0I2+h0,0,2I3.We  find that(33)f2,0,0=−(a1+b1)/2[ω1′3α1′+ω1′2(1−2α)β1′+ω1′(β−μ−3μp′−2α)α1′+(β−μ−3μp′)β1′],g0,2,0=−(a2+b2)/2[ω2′3α2′+ω2′2(1−2α)β2′+ω2′(β−μ−3μp′−2α)α2′+(β−μ−3μp′)β2′],f0,2,0=−(a3+b3)/2[ω1′3α1′+ω1′2(1−2α)β1′+ω1′(β−μ−3μp′−2α)α1′+(β−μ−3μp′)β1′],h2,0,0=f0,0,2=−c12ω3′γ′,h0,2,0=g0,0,2=−c22ω3′γ′,h0,0,2=−c32ω3′γ′,where(34)a1=3μ(β−μ−3μp′+ω1′2)×[(1+4p′)(−2r1′α1′−r4′α1′+β1′s1′2)−α1′3+α1′β1′22],a2=3μ(β−μ−3μp′+ω2′2)×[(1+4p′)(−2r2′α2′−r5′α2′+β2′s2′2)−α2′3+α2′β2′22],a3=3μ(β−μ−3μp′+ω1′2)×[(1+4p′)[−2r2′α1′−α2′(r7′+r8′)+β2′(s4′+s5′2)]−α1′(2α2′2−β2′22)],b1=3ω1′αμ[(1+4p′)(−α1′s1′−2β1′r1′+β1′r4′)+3β1′34−α1′2β1],b2=3ω2′αμ[(1+4p′)(−α2′s2′−2β2′r2′+β2′r5′)+3β2′34−α2′2β2′],b3=3ω1′αμ[(1+4p′)(−α2′s4′−α2′s5′−2β1′r2′+r7′β2′−r8′β2′)+3β1′β2′22−4β1′α2′2],c1=−3μ[(1+4p′)(r1′γ′+α1′t1′2+α1′t2′2)−β1′2γ′4+α1′2γ′],c2=−3μ[(1+4p′)(r2′γ′+α2′t3′2+α2′t4′2)−β2′2γ′4+α2′2γ′],c3=−3μ[(1+4p′)(r3′γ′+r6′γ′2)−3γ′38],α1′=α1[1+(α14+p12)ϵ+(α14′+q12)ϵ1],α2′=α2[1+(α15+p22)ϵ+(α15′+q22)ϵ1],β1′=β1[1+(α21−p12)ϵ+(α21′−q12)ϵ1],β2′=β2[1+(α22−p22)ϵ+(α22′−q22)ϵ1],γ′=γ[1−3p′ϵ14],α1=(2ω1)1/2a14,α2=(2ω2)1/2a15,β1=(2ω1)1/2a21,β2=(2ω2)1/2a22,γ=(2ω3)1/2a36.ÒIf the normalized Hamiltonian is written
as(35)H=ω1′I1+ω2′I2+ω3′I3+12×(aI12+bI22+cI32+2fI2I3+2gI3I1+2hI1I2),then, from Hamilton's equations
of motion(36)φi=∂H∂Ii(i=1,2,3)and (21), we find that(37)a=f2,0,0;b=g0,2,0;c=h0,0,2;f=g0,0,2=h0,2,0;g=f0,0,2=h2,0,0;h=g2,0,0=f0,2,0.


## 5. Stability

Now we apply Moser's modified form of Arnold's theorem
[11] to discuss the nonlinear stability. We have(38)1>ω3′>ω2′>ω1′>0.The condition (i) of the theorem
is satisfied provided the basic frequencies do not satisfy the equations (I)
ω2′=2ω1′,(II)
ω2′=3ω1′,(III)
ω3′=2ω2′,(IV)
ω3′=3ω2′,(V)
ω3′=2ω1′,(VI)
ω3′=3ω1′,(VII)
−ω1′+2ω2′−ω3′=0,(VIII)
ω1′+2ω2′−ω3′=0,(IX)
−ω3′+ω2′+2ω1′=0,(X)
ω3′−ω1′−ω2′=0. Out of these
ten equations (I)–(X) in ω1′, ω2′, ω3′,
(IX) and (X) along with (12) and (13) do not give the values of μ in the interval μc<μ<1.
The remaining eight from (I) to (VIII) are the resonance cases.
Taking any of the equations from (I) to (VIII) and eliminating ω1′, ω2′, ω3′ from that equation as well
as (12) and (13), the eliminant is an equation in μ.
Solving those equations, we get only five roots in the range μc<μ<1.
They are(39)μ1=0.93711086…−1.12983217…ϵ+1.50202694…ϵ1,μ2=0.9672922…−0.5542091…ϵ+1.2443968…ϵ1,μ3=0.9459503…−0.70458206…ϵ+1.28436549…ϵ1,μ4=0.9660792…−0.30152273…ϵ+1.11684064…ϵ1,μ5=0.893981…−2.37971679…ϵ+1.22385421…ϵ1.For these values of μ,
the condition (i) of the theorem does not hold.The determinantD occurring in the condition (ii) of the theorem
is(40)D=−[A1ω1′2+B1ω2′2+C1ω3′2+2F1ω2′ω3′+2G1ω3′ω1′+2H1ω1′ω2′],where A1, B1, C1, F1, G1, H1 are the cofactors of a, b, c, f, g, h, respectively, in the
determinant(41)|ahghbfgfc|.D≠0 if the value of μ,
in the range μc<μ<1,
does not satisfy the equation obtained by eliminating ω1′, ω2′, ω3′ from the equation(42)A1ω1′2+B1ω2′2+C1ω3′2+2F1ω2′ω3′+2G1ω3′ω1′+2H1ω1′ω2′=0and (12) and (13).Using Mathematica 5.1, the eliminant isF(μ)/E(μ),
where (43)F(μ)=9[1166400−4195800μ−120985128μ2+795095931μ3+1392485938μ4−24307151051μ5+44309723920μ6+197172775589μ7−851956684990μ8+387526140287μ9+3651296279676μ10−7972159671396μ11+2690851941504μ12+12969939666960μ13−23537135400768μ14+18735235848000μ15−7546662494208μ16+1253826625536μ17]−9ϵ[−2332800+35915400μ+321195312μ2−7793527551μ3+30709728342μ4+198150784114μ5−1924757627794μ6+2843702230896μ7+24498691597546μ8−110305966382026μ9+33587405088162μ10+800597584579759μ11−1881092505237568μ12−277273410185040μ13+7617265518396384μ14−11775715112691360μ15+815590824157440μ16+19637242384981248μ17−28895247571129344μ18+20274247679816448μ19−7392979726368768μ20+1128443962982400μ21]/(2(−1+u)3u2(1+2u)(1−7u+18u2))+7ϵ1[−2332800+35915400μ+321195312μ2−7793527551μ3+30709728342μ4+198150784114μ5−1924757627794μ6+2843702230896μ7+24498691597546μ8−110305966382026μ9+33587405088162μ10+800597584579759μ11−1881092505237568μ12−277273410185040μ13+7617265518396384μ14−11775715112691360μ15+815590824157440μ16+19637242384981248μ17−28895247571129344μ18+20274247679816448μ19−7392979726368768μ20+1128443962982400μ21]/((−1+u)3u2(1+2u)(1−7u+18u2)),E(μ)=64(−1+μ)(1+2μ)3(−8+9μ)2×(1−7μ+18μ2)2(−9−41μ+54μ2)2+ϵ[−801765−38408219μ+293041877μ2+1080522466μ3−12378949640μ4+11333130123μ5+136993003352μ6−269966516193μ7−431307890203μ8+1383753363508μ9−253101310518μ10−2003772335421μ11+2056877130018μ12−619768477598μ13]+ϵ1[1065886+51060841μ−389577148μ2−1436473396μ3+16456883028μ4−15066544594μ5−182121900273μ6+358900190055μ7+573391418865μ8−1839596080857μ9+336479166857μ10+2663864697552μ11−2734463530113μ12+823935603334μ13].So condition (ii) of the theorem is not satisfied for
those values ofμ which satisfy the equation(44)F(μ)=0and also for the value μ1=0.93711086…−1.12983217…ϵ+1.50202694…ϵ1,
where E(μ)=0, F(μ)≠0, and consequently D is not defined. The roots of the equation F(μ)=0 when ϵ,ϵ1=0 are seventeen in number,
[6], out of which nine are real and they are(45)μ={−0.522377,−0.393899,−0.296358,−0.221747,−0.0991954,0.153075,0.350508,0.540579,0.857062}.When ϵ,ϵ1≠0,
let the roots be μi+xiϵ+yiϵ1 (i=1,2,…,9).
Putting these roots in (44) and solving for xi, yi after neglecting higher-order terms in ϵ, ϵ1,
we have(46)μ={−0.522377+1.044754ϵ−1.56713ϵ1,−0.393899+0.787798ϵ−1.181697ϵ1,−0.296358+0.592716ϵ−0.889074ϵ1,−0.221747+0.443494ϵ−0.665241ϵ1,−0.0991954+0.198391ϵ−0.297586ϵ1,0.153075−0.30615ϵ+0.459225ϵ1,0.350508−0.701016ϵ+1.05152ϵ1,0.540579−1.08116ϵ+1.62174ϵ1,0.857062−1.71412ϵ+2.57119ϵ1}.None of these roots lie in the
range μc<μ<1.
Hence, the equilibrium point −μ+μϵ1/(1+2μ),0,0) is stable in the nonlinear sense in the range
of linear stability μc<μ<1 for all values of μ except μ1, μ2, μ3, μ4, μ5,
where the KAM theory is not applicable and consequently no conclusion about
stability can be drawn for the five mass ratios. The result is in agreement
with that result found out by Hallan and Mangang [6] when there is no
perturbations in Coriolis and centrifugal forces (ϵ,ϵ1=0).



---

*Source: 425412-2008-08-13.xml*