stability of periodic motions in satellite dynamics ... · stability of nonautonomous hamiltonian...

Stability of Nonautonomous Hamiltonian SystemsConstructive algorithm for the normalization of a periodic Hamiltonian

Stability of Periodic Motions in SatelliteDynamics

Stability Theory for Hamiltonian Systems

Boris S. Bardin

Moscow Aviation Institute (Technical University)Faculty of Applied Mathematics and Physics

Department of Theoretical Mechanics

Course of Computer Algebra and Differential Equations

Boris Bardin


Contents

1 Stability of Nonautonomous Hamiltonian SystemsLinear Problem of StabilityNonlinear Problem of Stability

2 Constructive algorithm for the normalization of a periodicHamiltonian

Construction of the mapNormalization of the area preserving map

Boris Bardin


Linear Problem of StabilityNonlinear Problem of Stability

Contents




Boris Bardin



Floquet theorem

Consider a system of differential equation

dxdt

= A(t)x, A(t + 2π) = A(t) (1)

Theorem.Let X(t) be a fundamental matrix of solution ofsystem (1) satisfying the following initial conditions X(0) = E.Then, for all t ∈ R

X(t) = Y(t)eBt

where B is a constant matrix and Y(t) is a 2π−periodic in tmatrix.

Boris Bardin



Lyapunov-Floquet theorem

Theorem.There exists a linear 2π−periodic change of variablesthat transforms the 2π−periodic linear system to anautonomous linear system.Let us perform linear transformation

x = X(t)e−Bty

where X(t) is the fundamental matrix of solution of system (1)defined above.

In the new variables we have

dydt

= By. (2)

The stability problems for systems (1) and (2) are equivalent.

Boris Bardin



characteristic multipliers and exponent

The eigenvalues ρi of matrix X(2π) are called the characteristicmultipliers of the system.

dxdt

= A(t)x

They are also the eigenvalues of the (linear) Poincare maps.The eigenvalues λi of matrix B are called a characteristicexponent (sometimes called a Floquet exponent).The relation

ρi = e2πλi

and takes place

Boris Bardin



Linear Hamiltonian system

Poincare – Lyapunov Theorem. The characteristic equation

det(X (2π)− ρ) = 0

for linear Hamiltonian system is a reciprocal equation.If linear Hamiltonian system has characteristic multiplier ρithen it also has the multiplier ρ−1

i .Linear periodic Hamiltonian system is stable if and only ifall characteristic multipliers lie on unit circle and Jordannormal form of the matrix X (2π) is diagonal.

Boris Bardin



Contents




Boris Bardin



Let us consider nonlinear 2π−periodic Hamiltonian system withn degrees of freedom. By using 2π−periodic transformation itsHamiltonisn can be brought into the following normal form

H = ω1r1 + · · ·+ ω1rn +n∑

i,j=1

aij ri rj + H(5)(r , ϕ, t)

Let us introduce ’additional’ pair of canonical variables rn+1 andϕn+1 = t and consider autonomous Hamiltonian system withn + 1 degrees of freedom. The Hamiltonian reads

H∗ = ω1r1 + · · ·+ ω1rn + rn+1 +n∑

i,j=1

aij ri rj + H(5)(r , ϕ, ϕn+1)

Boris Bardin



Stability for most of initial conditions

If the Hamiltonian does not satisfy the condition ofnondegeneracy

Dk+1 = det

(∂2H∗

0∂r2

∗

)≡ 0 (3)

Let us notice that the condition of isoenergetic nondegeneracynever fulfilled, i.e

Dk+2 = det

(∂2H∗

0∂r2

∗

) (∂H∗

0∂r∗

)(∂H∗

0∂r∗

)0

6= 0,⇒ Dk = det(∂2H0

∂r2

)6= 0

Boris Bardin



Arnold diffusion in resonant casesMarkeev’s Example

Let us consider the system with Hamiltonian

H = ω1r1 + ω2r2 − 24r21 + 2r1r2 + r2

2 + H(1)(r,ϕ, t),

whereH(1)(r,ϕ, t) = r2

2√

r1 sin(ϕ1 + 4ϕ2 − Nt).

and resonance ω1 + 4ω2 = N takes placeThe conditions of Arnold’s theorem are fulfilled

D3 = a212 − a11a22 = 25 6= 0,

but the Hamiltonian system has the following solution

ϕ1 + 4ϕ2 − Nt = π

r1(t) =14

r2(t) = r1(0)[1− 24r32

3 (0)t ]2/3.

Boris Bardin



Contents




Boris Bardin



Formulation of the problem

Consider 2π−periodic Hamiltonian system with two degrees offreedom with Hamiltonian

H(q1,q2,p1,p2, t) = H2 + H3 + H4 . . .

where Hk is a form of degree k with 2π−periodic coefficients.Notations:

1 q(0)i and p(0)

i are initial values of variables qi and pi

2 q(1)i and p(1)

i are values of variables qi and pi at t = 2π

If q(0)i and p(0)

i are small enough, then q(1)i and p(1)

i are analyticfunctions of q(0)

i and p(0)i . These functions define a map T of

the neighbourhood of equilibrium onto itself.

Boris Bardin



Linear system

Let X(t) be the fundamental matrix of the linear canonicalsystem with Hamiltonian H2. Its elements satisfy the equations

dxjs

dt=

∂H2

∂xj+2,s,

dxj+2,s

dt= −∂H2

∂xjs,

H2 = H2(x1s, x2s, x2s, x2s, t) (j = 1,2; s = 1,2,3,4)

(4)

and the initial conditions

X(0) = E

Boris Bardin



Linear transformstion

Let us introduce new canonical variables∥∥∥∥∥∥∥∥q1q2p1p2

∥∥∥∥∥∥∥∥ = X (t)

∥∥∥∥∥∥∥∥u1u2v1v2

∥∥∥∥∥∥∥∥ (5)

New Hamiltonian G(u1,u2, v1, v2, t) does not contain quadraticterms in u1,u2, v1, v2:

G = G3 + G4 . . .

Boris Bardin



Generating function

Note thatq(0)

1 = u(0)1 , . . . ,p(0)

2 = υ(0)2 .

Thus we have to look for the map

q(0)i ,p(0)

i → u(1)i , υ

(1)i .

q(0)i =

∂S

∂p(0)i

; υ(1)i =

∂S

∂u(1)i

where

S = u(1)1 p(0)

1 + u(1)2 p(0)

2 + S3

(u(1)

1 ,u(1)2 ,p(0)

1 ,p(0)2

)+ . . .

Boris Bardin



Construction of the generating function

In fact

S(

u(1)1 ,u(1)

2 ,p(0)1 ,p(0)

2

)= Φ

(u(1)

1 ,u(1)2 ,p(0)

1 ,p(0)2 ,2π

)where Φ

(u(1)

1 ,u(1)2 ,p(0)

1 ,p(0)2 , t

)satisfies the Hamilton-Jacobi

equation

∂Φ

∂t+ G

(u(1)

1 ,u(1)2 ,

∂Φ

∂u(1)1

,∂Φ

∂u(1)2

, t

)= 0

Boris Bardin




We expend Φ in power series in u(1)i ,p(1)

i (i = 1,2)

Φ = Φ3 + Φ4 + . . .

From the Hamilton-Jacobi equation we have

∂Φ3

∂t= −G3,

∂Φ4

∂t= −G4 −

2∑i=1

∂G3

∂p(0)i

· ∂Φ3

∂u(1)i

, . . . (6)

From equation (6) we calculate coefficients of forms Φk asfunction of t .

Boris Bardin




For example, coefficients of

Φ3 =∑

i1+i2+j1+j2=3

ϕu(1)

1 ,u(1)2 ,p(0)

1 ,p(0)2

i1i2j1j2

satisfy the following differential equation

dϕi1i2j1j2dt

= −gi1i2j1j2(t)

are function of xij . Initial conditions

ϕi1i2j1j2(0) = 0

Coefficients gi1i2j1j2 depend on elements xij(t) of fundamentalmatrix of linear system.

Boris Bardin



Thus, in order to calculate coefficients

si1i2j1j2 = ϕi1i2j1j2(2π)

of the form S3.We have integrate the following system

dxjs

dt=

∂H2

∂xj+2,s,

dxj+2,s

dt= −∂H2

∂xjs(j = 1,2; s = 1,2,3,4),

dϕi1i2j1j2dt

= −gi1i2j1j2(t) (i1 + i2 + j1 + j2 = 3)

with initial conditions

xij(0) = 0 (i 6= j) xij(0) = 1 (i = j) ϕi1i2j1j2(0) = 0.

for interval from t = 0 to t = 2π.

Boris Bardin



In order to calculate the map up to terms:of the second order we solve 16 + 20 = 36 equationsof the third order we solve 16 + 20 + 35 = 71 equations.

Boris Bardin



Form of the area-preserving map

The area-preserving map reads∥∥∥∥∥∥∥∥∥q(1)

1q(1)

2p(1)

1p(1)

2

∥∥∥∥∥∥∥∥∥ = X(2π)

∥∥∥∥∥∥∥∥q̃1q̃2p̃1p̃2

∥∥∥∥∥∥∥∥Where

q̃j = q(0)j − ∂S3

∂p(0)j

+2∑

k=1

∂2S3

∂p(0)j ∂q(0)

k

· ∂S3

∂p(0)k

− ∂S4

∂pj+ O4

p̃j = p(0)j +

∂S3

∂q(0)j

−2∑

k=1

∂2S3

∂q(0)j ∂q(0)

k

· ∂S3

∂p(0)k

+∂S4

∂qj+ O4

Boris Bardin



Contents




Boris Bardin



Linear normalization of the map

∥∥∥∥∥∥∥∥q1q2p1p2

∥∥∥∥∥∥∥∥ = N

∥∥∥∥∥∥∥∥Q1Q2P1P2

∥∥∥∥∥∥∥∥ (7)

In new variables the map reads

Q(1)j = ρj

Q(0)j − ∂W3

∂p(0)j

+ . . .

P(1)j = ρj+2

P(0)j +

∂W3

∂Q(0)j

+ . . .

ρj = ei2πσj ρj+2 = e−i2πσj

Boris Bardin



Nonresonant case

Nonlinear close to identity transformation

Q1,Q2,P1,P2 → ξ1, ξ2, η1, η2

In new variables:

ξ(1)j = ρj

ξ(0)j − ∂Z4

∂η(0)j

+ . . .

η(1)j = ρj+2

η(0)j +∂Z4

∂η(0)j

+ . . .

Z4 = W2020ξ

(0)1

2η(0)1

2+ W1111ξ

(0)1 ξ

(0)2 η

(0)1 η

(0)2 + W0202ξ

(0)2

2η(0)2

2

Boris Bardin



Normal form of Hamiltonian

H = iσ1ξ1η1 + iσ2ξ2η2 − 12π (w2020ξ

21η

21+

+w111ξ1ξ2η1η2 + w0202ξ2η2) + O5

In canonical polar coordinates

H = σ1r1 + σ2r2 + c20r21 + c11r1r2 + c02r2

2 + O5

wherec20 =

12π

w2020, c11 =1

2πw1111,

c02 =1

2πw0202

Boris Bardin

stability of periodic motions in satellite dynamics ... · stability of nonautonomous hamiltonian...

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