Download On Stability of Equilibrium Solutions in the Restricted Many

"Dynamics, Topology and Computations"
June 3, 2009, Bedlewo, Poland
On Stability of Equilibrium Solutions
in the Restricted Many-Body Problems
L. Gadomski
The College of Finance and Management, Siedlce, Poland
E-mail: [email protected]
A.N. Prokopenya
The College of Finance and Management, Siedlce, Poland
Brest State Technical University, Belarus
E-mail: [email protected]
Content
Ÿ Introduction
Ÿ Model and its linear analysis
Ÿ Theoretical basis for studying stability of nonlinear
Hamiltonian systems
Ÿ Normalization of the Hamiltonian and its
implementation with Mathematica
Ÿ Results
Introduction
Classical Newtonian many-body problem is a well-known dynamical model
d2 xj
â
mk
â
mk
k=1 H¹jL
dt2
d2 zj
rjk 3
yj - yk
n
k=1 H¹jL
dt2
mk
,
rjk 3
Hj, k = 1, 2, ..., nL
zj - zk
k=1 H¹jL
dt2
â
= -G
,
n
= -G
d2 yj
xj - xk
n
= -G
,
rjk 3
where
Ixj - xk M + Iyj - yk M + Izj - zk M
2
rjk =
2
2
.
It is well-known also that it is not integrable, in general.
To simplify the three-body problem, Leonard Euler introduced the restricted three-body problem.
The essence: the third body has so small mass that it doesn’t influence on the two primaries whose motion is determined by the
corresponding solution of the two-body problem. This idea turned out to be very productive, it stimulated development of
qualitative theory of differential equations and resulted in the KAM-theory. So it was quite natural that different generalizations
of the restricted three-body problem were proposed.
Model
Restricted many-body problem is a generalization of the famous restricted three-body problem.
To formulate a restricted many-body problem it is sufficient to find an exact particular solution of the equations of motion. Then
we can add one more body of infinitesimal mass to the system and the problem is to describe its motion in the gravitational field
of primary bodies.
Here we consider a system of Hn + 1L bodies P0 , P1 , ..., Pn . The body P0 rests at the origin while the others having equal
masses move uniformly about their common center of mass on the same circular orbit and form a regular polygon at any instant
of time.
P2
P1
P3
P6
P0
P4
P5
In the simplest case of the restricted four-body problem Hn = 2L all formulars and calculations are not too bulky and so we
consider this case as an example. The polygon degenerates into a line P1 P2 and the corresponding solution of the three-body
problem is known as Euler collinear solution. We are interested in the motion of the body P3 having negligible mass in the
gravitational field of P0 , P1 , P2 .
y
P3
P1
®
r1
®
r2
®
r
x
P0
P2
A general solution of the equations of motion is not known but some particular solutions can be found. And the problem is to
investigate their stability.
Hamiltonian of the system
It is convenient to analyze motion of the system in the rotating coordinate frame where the primaries P0 , P1 , P2 rest. The
Hamiltonian of the system is defined in a standard way and is given by (we use cylindrical coordinates Ρ, j, z and polar angle Ν as
independent variable, Μ = m  m0 )
H = Ρ ’@ΝD PΡ@ΝD + j ’@ΝD Pj@ΝD + z ’@ΝD Pz@ΝD - Lpol . rul2 
Collect@ð, 8PΡ@ΝD, Pj@ΝD, Pz@ΝD<, SimplifyD &
Pz@ΝD2
PΡ@ΝD2
Pj@ΝD2
+
1
- Pj@ΝD +
2
1
2 Ρ@ΝD2
2
+
4
4+Μ
z@ΝD2
+
Ρ@ΝD2
1
1
Μ
+
1 + z@ΝD2 - 2 Cos@j@ΝDD Ρ@ΝD + Ρ@ΝD2
1 + z@ΝD2 + 2 Cos@j@ΝDD Ρ@ΝD + Ρ@ΝD2
One can easily obtain equations of motion and show that there are six equilibrium positions of the body P3 .
The corresponding positions of the bodies P0 , P1 , P2 , P3 are shown below.
y
S1
N4
N3
N1
P1
N2
P0
x
P2
S2
The points N1 , N2 , N3 , N4 correspond to the radial equilibrium solutions, while the points S1 , S2 represent the bisector
equilibrium solutions.
In the case of n = 3 some of the corresponding equilibrium positions are shown below.
y
S1
P2
S2
N1
S3
N2
P0
P1
x
P3
Hamiltonian in the neighbourhood of the bisector
equilibrium solution
In order to obtain equations of the perturbed motion we add small perturbations q1 , q2 , q3 , p1 , p2 , p3 to the bisector equilibrium
solution according to the following rule.
rul4 =
:Ρ ® HR + ∆ q1 &L, PΡ ® H∆ p1 &L, j ®
Π
& , Pj ® IR 2 + ∆ R p2 &M, z ® H∆ q3 &L, Pz ® H∆ p3 &L>;
q2
+∆
2
R
Here a multiplier ∆ is introduced to simplify further calculations.
In order to get an expansion of the Hamiltonian in powers of perturbations q1 , q2 , q3 , p1 , p2 , p3 in the neighbourhood of the
equilibrium solution it is sufficient to find its expansion in powers of ∆ about the point ∆ = 0. The corresponding expansion up to
the fourth order is given by
H = H2 + H3 + H4 + ...,
where a quadratic term H2 is given by
H2 = Coefficient@Hexp, ∆, 2D  Collect@ð, 8p1, p2, p3, q1, q2, q3<, Simplify@ð, R > 1D &D &
p12
p22
p32
+
+
2
q12
I1 + R2 M
- 2 p2 q1 -
2
3
H4 + ΜL
R3
2
-
I1 +
52
R2 M
R3
+
+
2Μ
I1+R2 M32
+
4+Μ
12 R2 Μ
4
-
H4 + ΜL
52
2
1
2 q32
12 q22 Μ
4Μ
H4 + ΜL
+
I1 +
32
R2 M
H4 + ΜL
The terms of the third and fourth order are obtained in a similar way.
Characteristic exponents
If mass parameter Μ belongs to the interval 0 < Μ < Μmax = 0.0853217 (or equilibrium distance R Î H1, Rmax L) characteristic exponents of the system are purely imaginary numbers IΛ1,2 = ± i Σ1 , Λ3,4 = ± i Σ2 , Λ5,6 = ± i Σ3 M, where Σ1 , Σ2 and Σ3
are given by
eigenval = 8Σ1 ® HΣ . solPolP6TL, Σ2 ® HΣ . solPolP4TL, Σ3 ® HΣ . solPolP2TL<
:Σ1 ®
1
2
1 - 12 b + 4 b2
1-
1
1 - 12 b + 4 b2 , Σ2 ®
+
, Σ3 ® 1>
2
2
Their dependences on the parameter R are shown below.
3Σ2
2.0
2Σ1 -Σ2
1.5
2Σ2
Σ3
1.0
Σ1
0.5
Σ2
1.002
1.004
1.006
1.008
1.010
R
1.012
This picture shows that there are five values of the parameter R in the interval 1 < R < Rmax = 1.01294 for which the
resonance conditions of the third HΣ1 = 2 Σ2 , Σ3 = 2 Σ2 L and the fourth order HΣ1 = 3 Σ2 , Σ3 = 3 Σ2 , 2 Σ1 - Σ2 = Σ3 L
are fulfilled.
Note that stability of the equilibrium solutions in the cases of resonance should be analyzed separately.
Normalization of H2
In order to normalize the quadratic part of the Hamiltonian, we construct the following canonical transformation
rul5 = :q1 ® 2 c@1D p1 + 2 c@2D p2,
q2 ® -
c@1D q1 Σ1 I1 + b - Σ12 M
c@2D q2 Σ2 I1 + b - Σ22 M
+
b
p1 ® -2 c@1D q1 Σ1 + 2 c@2D q2 Σ2 ,
,
b
p2 ® c@1D p1 I1 + b - Σ12 M + c@2D p2 I1 + b - Σ22 M>;
norm = :c@1D ®
Σ1
IΣ12 - Σ22 M I3 - b + Σ12 M
Σ2
IΣ12 - Σ22 M I3 - b + Σ22 M
, c@2D ®
>;
H2norm = HH2a . rul5 . norm . eigenval 
Collect@ð, 8q1, q2, q3, p1, p2, p3<, FullSimplifyD &L;
Then quadratic part of the Hamiltonian H2a takes a form
H2norm =
1
2
Σ1 I p12 + q12 M -
1
2
Σ2 I p22 + q22 M +
1
2
Σ3 I p32 + q32 M;
We see that H2norm is not a positively definite quadratic form and, hence, stability of the bisector equilibrium solutions depends
on higher order terms in the expansion of the Hamiltonian. Thus, the stability problem can be solved only in a strict nonlinear
formulation.
And we’ll invetigate stability of the bisector equilibrium solutions for distances R Î H1, Rmax L considering R as parameter instead
of Μ.
Arnold’s Theorem (for Hamiltonian system with two
degrees of freedom)
Let the Hamiltonian of the system
H = H2 + H3 + H4 + ...,
be represented in the form
H = Σ1 r1 - Σ2 r2 + c20 r1 2 + c11 r1 r2 + c02 r2 2 + O IHr1 + r2 L52 M,
1
2
where Σ j ³ 0 and rj =
Iqj 2 + pj 2 M,
Hj = 1, 2L,
and the following conditions are satisfied:
a) characteristic exponents of the linearized system Λ1,2 = ± i Σ1 , Λ3,4 = ± i Σ2 are purely imaginary numbers;
b) there are no resonances in the system up to the fourth order or
n1 Σ1 + n2 Σ2 ¹ 0,
where n1 , n2 are integers satisfying the condition 0 < n1 + n2
c)
£ 4;
c20 Ω2 2 + c11 Ω1 Ω2 + c02 Ω1 2 ¹ 0,
Then equilibrium solution of the Hamiltonian system is stable in the Liapunov sense.
Arnold’s Theorem (for Hamiltonian system with three
degrees of freedom)
Let the Hamiltonian of the system be represented in the form
H = H0 + O IHr1 + r2 + r3 L52 M,
where
H0 = Σ1 r1 - Σ2 r2 + Σ3 r3 + c11 r1 2 + c12 r1 r2 + c13 r1 r3 + c23 r2 r3 + c22 r2 2 + c33 r3 2 ,
1
2
parameters Σ j ³ 0, rj =
Iqj 2 + pj 2 M,
Hj = 1, 2, 3L
and the following conditions are satisfied:
a) characteristic exponents of the linearized system Λ1,2 = ± i Σ1 , Λ3,4 = ± i Σ2 , Λ5,6 = ± i Σ3 are purely imaginary numbers;
b) there are no resonances in the system up to the fourth order or
n1 Σ1 + n2 Σ2 + n3 Σ3 ¹ 0,
where n1 , n2 , n3 are integers satisfying the condition 0 < n1 + n2 + n3 £ 4;
c) for r1 = r2 = r3 = 0 at least one of the conditions
¶2 H0
D3 = det
¹ or D4 = det
¶ rj ¶ rk
¶2 H0
¶rj ¶rk
¶ H* 0
¶rj
¶ H* 0
¶rk
0
¹ 0.
Then equilibrium solution of the Hamiltonian system is stable for the majority of the initial conditions.
Markeev’s Theorems (Resonance cases)
The cases of third- and fourth-order resonances must be analysed separately on the basis of Markeev’s theorems.
Ÿ Third-order resonance
Let the Hamiltonian of the system
H = H2 + H3 + ...,
be represented in the form
Ž
H = Σ1 r1 - Σ2 r2 + Σ3 r3 + B cos@n1 j1 + n2 j2 + n3 j3 D r1 n1 2 r2 n2 2 r3 n3 2 + H Irj , jj M,
Ž
where n1 Σ1 + n2 Σ2 + n3 Σ3 = 0 H n1 + n1 + n1 = 3L, B is some constant and H Irj , jj M is 2Π-periodic function with
Ž
respect to variables j j and H Irj , jj M = O IHr1 + r2 + r3 L2 M.
Then equilibrium solution of the Hamiltonian system is unstable if B ¹ 0.
Ÿ Fourth-order resonance
Let the Hamiltonian (16) be represented in the form
H = Σ1 r1 - Σ2 r2 + Σ3 r3 + c11 r1 2 + c22 r2 2 + c33 r3 2 + c12 r1 r2 + c13 r1 r3 +
c23 r2 r3 + B r1 n1 2 r2 n2 2 r3 n3 2 cos@n1 j1 + n2 j2 + n3 j3 D + O IHr1 + r2 L52 M,
where n1 Σ1 + n2 Σ2 + n3 Σ3 = 0 H n1 + n1 + n1 = 4L and B is some constant.
Then equilibrium solution of the Hamiltonian system is unstable if
c11 n1 2 + c22 n2 2 + c33 n3 2 + c12 n1 n2 + c13 n1 n3 + c23 n2 n3
B n1 n1 2 n2 n2 2 n3 n3 2
<
Normalization of third-order term in the Hamiltonian
To apply Arnold-Markeev theorems we have to normalize successively the terms H3 , H4 in the Hamiltonian.
The third order term may be represented in the following general form
â
H3 =
i
k l m n
hH3L
ijklmn q1 q2 q3 p1 p2 p3
j
i+j+k+l+m+n=3
where coefficients hH3L
ijklmn are quite cumbersome functions of parameter Μ.
The generating function for canonical transformation normalizing this term is sought in the form
â
Ž
Ž
Ž
S3 = q1 p1 + q2 p2 + q3 p3 +
i j k Žl Žm Žn
sH3L
ijklmn q1 q2 q3 p1 p2 p3
i+j+k+l+m+n=3
H3L
where 56 coefficients sH3L
ijklmn are to be found from the condition that coefficients hijklmn become zeros (this can be done if
there is no third-order resonances in the system).
On substituting all the coefficients found we observe that the third-order term in the Hamiltonian disappears.
H2new + H3new . ∆ ® 1 . sol13a . sol13b . sol13c . sol13d . sol13e . sol13f . sol13g 
Simplify
1
2
IIp12 + q12 M Σ1 - Ip22 + q22 M Σ2 + Ip32 + q32 M Σ3 M
Third-Order resonance
There are two third-order resonances in the system considered, namely, Σ1 = 2 Σ2 and 2 Σ2 = Σ3 = 1. In the second case
equations determining the corresponding coefficients s3 Hi, j, k, l, m, nL do not contain terms of the form h3 Hi, j, k, l, m, nL.
Hence, there exists trivial solution of this system and the corresponding terms h3 Hi, j, k, l, m, nL in the transformed hamiltonian
will disappear.
In the case of Σ1 = 2 Σ2 coefficients s3 Hi, j, k, l, m, nL in the expansion of generating function are chosen in such a way that the
Hamiltonian takes a form
H = 2 r1 Σ2 - r2 Σ2 + r3 Σ3 + B
r1 r2 cos Hj1 + 2 j2 L + O IHr1 + r2 + r3 L2 M,
where
B=
H3L
H3L
IhH3L
000 120 - h020 100 - h110 010 M.
1
2
According to Markeev’s theorem, if paraameter B ¹ 0 then equilibrium solution is unstable. Calculation of parameter B gives the
following result:
B . sol13h . rulH3 . norm . Σ1 ® 2 Σ2 . eigenval . rulb . solΜ . R ® R3res1  N@ð, 10D &
- 0.3658222119
We see that parameter B ¹ 0. On the basis of the Markeev theorem we can conclude that the bisector equilibrium solutions are
unstable for Μ = Μ1 = 0.0529423 when the third order resonance takes place.
Normalization of the fourth-order term
Normalization of the fourth order term can be done similarly but calculations are much more cumbersome.
The fourth-order term is represented in the following general form
â
H4 =
i
k l m n
hH4L
ijklmn q1 q2 q3 p1 p2 p3 .
j
i+j+k+l+m+n=4
with coefficients hH4L
ijklmn depending on the parameter Μ.
The generating function for canonical transformation normalizing this term is sought in the form
â
Ž
Ž
Ž
S4 = q1 p1 + q2 p2 + q3 p3 +
i j k Žl Žm Žn
sH4L
ijklmn q1 q2 q3 p1 p2 p3
i+j+k+l+m+n=4
where 126 coefficients sH4L
ijklmn are to be found from the condition that the fourth-order term H4 takes the simplest form (note
that this term can not be cancelled).
In the case of absence of resonances up to the fourth order the Hamiltonian is reduced to the form
H0 = Σ1 r1 - Σ2 r2 + Σ3 r3 + c11 r1 2 + c12 r1 r2 + c13 r1 r3 + c23 r2 r3 + c22 r2 2 + c33 r3 2 ,
Now we can calculate the determinant
¶2 H0
D3 = det
¶ rj ¶ rk
D3
1000
Μ1
500
Μ2
Μ3
Μ
0.02
0.04
0.06
0.08
-500
-1000
-1500
Here Μ1 corresponds to the case of fourth-order resonance Σ1 = 3 Σ2 , Μ2 corresponds to the case of third-order resonance
Σ1 = 2 Σ2 , and D3 = 0 for Μ = Μ3 .
Similar calculations of the determinant
D4 = det
¶2 H0
¶rj ¶rk
¶ H* 0
¶rj
¶ H* 0
¶rk
0
.
give the following result
D4
150
Μ2
Μ3
100
Μ1
50
Μ
0.02
0.04
0.06
0.08
We see that in the case of Μ = Μ3 the condition D4 ¹ 0 is fulfilled.
Conclusion: In the absence of the resonances up to the fourth order inclusively the equilibrium positions S1 , S2 are stable
for the majority of the initial conditions.
Fourth-Order Resonance Case
Note that only in the case of the fourth-order resonance Σ1 = 3 Σ2 the corresponding terms in the Hamiltonian expansion can not
be eliminated and it takes a form
H = Σ1 r1 - Σ2 r2 + c11 r1 2 + c22 r2 2 + c33 r3 2 + c12 r1 r2 + c13 r1 r3 +
c23 r2 r3 + B r1 n1 2 r2 n2 2 r3 n3 2 cos@n1 j1 + n2 j2 + n3 j3 D + O IHr1 + r2 L52 M,
Now we can easily check the condition
c11 n1 2 + c22 n2 2 + c33 n3 2 + c12 n1 n2 + c13 n1 n3 + c23 n2 n3
<
B n1 n1 2 n2 n2 2 n3 n3 2
The calculations give the following result:
c11 + 9 c22 + 3 c12 = 21.4801 > 3
3 B = 8.99408
Therefore, the fourth-order resonance Σ1 = 3 Σ2 doesn’t influence on stability of the equilibrium positions S1 , S2 .
Conclusion
1. Analyzing stability of the equilibrium positions S1 , S2 in linear approximation, we have shown that they are stable only if
parameter Μ is sufficiently small, namely, 0 < Μ < Μmax = 0.0853217.
2. In the planar case Hq3 = p3 = 0L the equilibrium positions S1 , S2 are stable in Liapunov sense for all values of Μ from the
interval of their linear stability, except for the value Μ2 = 0.0529423 for which the third-order resonance takes place.
3. In the spatial case the equilibrium positions S1 , S2 are stable for the majority of initial conditions for all values of Μ from the
interval of their linear stability, except for the value Μ2 = 0.0529423.
Thank you very much for your attention!!!