of Hamiltonian systems Boris S. Bardin Moscow Aviation Institute (Technical University) Faculty of Applied Mathematics and Physics Department of Theoretical Mechanics Hamiltonian Dynamics and Celestial Mechanics To honour Professor Kenneth Meyer in his 75th year 30. May 3. June, 2011
Contents 1 Linear problem of stability 2 3 4
Problem statement Let us consider Hamiltonian system dq dt = H p, dp dt = H q, (1) where q = (q 1,..., q n ) T and p = (p 1,..., p n ) T Suppose that Hamiltonian H is analytic in a small neighborhood of the origin q = p = 0, which is an equilibrium H = H 2 + H 3 + + H m +... (2) Hamiltonian H is autonomous or periodically depends on t
Linear system Linear Hamiltonian system dx dt = Ax (3) where A = JH 2, x = (q 1,..., q n, p 1,..., p n ) T
Linear system Linear Hamiltonian system dx dt = Ax (3) where A = JH 2, x = (q 1,..., q n, p 1,..., p n ) T Autonomous case: A is a constant matrix Periodic case: A(t + 2π) = A(t) Lyapunov-Floquet Theorem.There exists a linear 2π periodic change of variables that transforms the 2π periodic linear system to an autonomous linear system.
Let us perform linear transformation x = X(t)e Bt y where X(t) is the fundamental matrix of solution of the system linear system, X(0) = E. In the new variables we have dy dt = By. (4) The eigenvalues λ i of matrix B are called a characteristic exponent. The stability problems for systems are equivalent for both x and y variables.
Stability in the First Approximation Theorem 1. If at least one of the characteristic exponent of the linear system has a non-zero real part, then irrespective of terms of order higher than one, the equilibrium is unstable. Theorem 2. If the Hamiltonian system autonomous and the quadratic part H 2 of the Hamiltonian is positive (or negative) definite, then the equilibrium is stable in the sense of Liapunov.
When nonlinear stability analysis is necessary? In what follows We suppose that real parts of all characteristic exponents are equal to zero ±iω 1, ±iω 2,..., ±iω n, In autonomous case the quadratic part H 2 of the Hamiltonian is not definite We consider autonomous Hamiltonian systems with two degrees of freedom or periodic systems with one degree of freedom
Nonresonant case Arnold Moser Theorem. If the Hamiltonian satisfied the following conditions: 1 There is no resonance up to fourth order, i.e. m 1 ω 1 + m 2 ω 2 0, 0 < m 1 + m 2 4 where m 1, m 2 are integer numbers 2 The coefficients of the Hamiltonian normal form H = ω 1 r 1 ω 2 r 2 + a 11 r 2 1 + 2a 12 r 1 r 2 + a 22 r 2 2 + O((r 1 + r 2 ) 5 2 ) q i = 2r i sin ϕ i, p i = 2r i cos ϕ i (i = 1, 2) satiesfy the following inequality a 11 ω 2 2 + 2a 12 ω 1 ω 2 + a 22 ω 2 2 0, then the Hamiltonian system is stable in the sense of Lyapunov. If a 11 ω 2 2 + 2a 12 ω 1 ω 2 + a 22 ω 2 2 = 0, then nonlinear analysis up to terms of order higher then four (with respect to q i, p i ) is necessary.
Third order resonance ω 1 = 2ω 2. Let us suppose that the third order resonance takes place, i.e. ω 1 = 2ω 2 then the Hamiltonian normal form reads H = 2ω 2 r 1 ω 2 r 2 + a 11 r 2 1 + 2a 12 r 1 r 2 + a 22 r 2 2 + + ar 2 r1 cos(ϕ 1 + 2ϕ 2 ) + H (ϕ 1, r 1, ϕ 2, r 2 ), (5) Markeev s Theorem. If a 0 of Hamiltonian (5) is not equal to zero, then equilibrium is unstable in the sense of Liapunov. If a = 0 and the second condition of the Arnold - Moser Theorem is fulfilled a 11 ω 2 2 + 2a 12 ω 1 ω 2 + a 22 ω 2 2 0, then equilibrium is stable. If a = 0 and the second condition of the Arnold - Moser Theorem is not fulfilled, then additional study of stability is necessary.
Fourth order resonance ω 1 = 3ω 2. Suppose that the fourth order resonance takes place, i.e. ω 1 = 3ω 2 Hamiltonian takes the form H = 3ω 2 r 1 ω 2 r 2 + a 11 r 2 1 + 2a 12 r 1 r 2 + a 22 r 2 2 + + dr 2 r1 r 2 cos(ϕ 1 + 3ϕ 2 ) + H (ϕ 1, r 1, ϕ 2, r 2 ), (6) Markeev s Theorem. If the coefficients of the Hamiltonian (6) satisfy the following inequality then equilibrium is stable; a 11 + 6a 12 + 9a 22 > 3 3d, If the coefficients of the Hamiltonian satisfy the following inequality then equilibrium is unstable. a 11 + 6a 12 + 9a 22 < 3 3d, In the case a 11 + 6a 12 + 9a 22 = 3 3d an additional study of stability is necessary.
How to go on? In nonresonant case the Arnold - Moser Theorem is still valid; In the case of higher order resonance (the order higher then four) Markeev s theorem 1 or Cabral Meyer theorem 2 can be applied; For strong degeneration in the case of fourth order resonance Markeev formulated stability condition. For other resonances the stability question is in the case of strong degeneration still open. 1 A. P. Markeev. Libration Points in Celestial Mechanics and Space Dynamics. Nauka, Moscow, 1978. (in Russian). 2 H. E. Cabral and K. R. Meyer. Stability of equilibria and fixed points of conservative systems. Nonlinearity, 12(5):1351 1362, 1999.
Low order resonances ω 1 = ω 2, ω 1 = 0, or ω 2 = 0 In general position the Jordan canonical form of the matrix of the linear system is not diagonal. Let us consider resonance ω 1 = 0. The Hamiltonian takes the form H = δ 2 η2 1 + 1 2 ω(ξ2 2 + η 2 2) + [ M m 2 ] a m 2l,2l (ξ2 2 + η2) 2 l ξ m 2L 1 + H, (7) m=3 l=0 Sokolskii s Theorem 3. Let us suppose that a k,0 = 0 for k = 1,..., M 1 and a M,0 0. If M is odd, then the equilibrium is unstable. If M is even, then the equilibrium is unstable for δa M,0 < 0 and stable for δa M,0 > 0. 3 Sokolsky, A. G. On stability of an autonomous Hamiltonian system with two degrees of freedom under first-order resonance J. Appl. Math. Mech., 1977, 41, 20 28
Transcendental cases M = 3: M = 4: H = δ 2 η2 1 + 1 2 ω(ξ2 2 + η 2 2) + a 3,0 ξ 3 1 + a 1,2 ξ 1 (ξ 2 2 + η 2 2) + H, (8) H = δ 2 η2 1 + 1 2 ω(ξ2 2 + η 2 2) + a 4,0 ξ 4 1 + a 2,2 ξ 2 1(ξ 2 2 + η 2 2) + a 0,4 (ξ 2 2 + η 2 2) 2 + H, (9) Definition. If for any integer m the coefficients of the normal form a m,0 = 0 for any integer m, then we shall say that a transcendental case takes place.
Transcendental cases When transcendental cases can appear? Is equilibrium position of system stable or unstable in transcendental case?
When transcendental cases can appear? A. J. Maciejewski 4 has shown that transcendental case takes place if the canonical system with Hamiltonian (1) has invariant manifold M = {(ξ 1, ξ 2, η 1, η 2 ) : ξ 1 = constant, ξ 2 = η 1 = η 2 = 0} of equilibrium positions. The above sufficient condition for the existence of the transcendental case can be generalized. In fact the following lemma takes place. 4 A.J. Maciejewski. Ruch obrotowy bryly sztywnej umieszczonej w punkcie libracyjnym ograniczonego zagadnienia trzech cial. PhD thesis, Torun, 1987
When transcendental cases can appear? Lemma. If the equilibrium position ξ 1 = η 1 = ξ 2 = η 2 = 0 is not isolated then the transcendental case takes place. We shall prove the Lemma by contradiction. Let us suppose that the equilibrium ξ 1 = η 1 = ξ 2 = η 2 = 0 is not isolated, but the exists an integer number M such that the coefficients a k,0 (k = 1,..., M) of the normal form H = δ 2 η2 1 + 1 2 ω(ξ2 2 + η 2 2) + [ M m 2 ] a m 2l,2l (ξ2 2 + η2) 2 l ξ m 2L 1 m=3 l=0 satisfy the conditions a k,0 = 0 for k = 1,..., M 1 and a M,0 0 First we calculate the Jacobian ( H, H, H ) η 1 ξ 2 η 2 δ 0 0 = (η 1, ξ 2, η 2 ) = 0 ω 0 0 0 ω = δω2 where H = H + H η1 =ξ 2 =η 2 =0
Hence 0 then by using implicate function theorem for for small values of ξ 1 the system of equations H = 0 η 1 H = 0 ξ 2 H = 0 η 2 can be solved with respect to η 1, ξ 2, η 2. For ξ 1 small enough the solution is analytic with respect to ξ 1 and the following asymptotic expressions take place η 1 = η 1 (ξ 1 ) = Aξ α 1 + O(ξ α+1 1 ) ξ 2 = ξ 2 (ξ 1 ) = Bξ β 1 + O(ξβ+1 1 ) η 2 = η 2 (ξ 1 ) = Cξ γ 1 + O(ξγ+1 1 ) where A, B, C are constant, α, β, γ N. (10)
Moreover, the truncated system of equations H = 0 η 1 H = 0 ξ 2 H = 0 η 2 where H is the normalized part of Hamiltonian (9), has the solution η 1 = ξ 2 = η 2 = 0. It follows from the above fact that α M, β M, γ M.
The equilibrium is not isolated, i.e. there exists a nontrivial equilibrium position in a small neighbourhood of η 1 = ξ 2 = η 2 = 0. Let us look for such an equilibrium. To this end we substitute in the equation Then we get η 1 = η 1 (ξ 1 ), ξ 2 = ξ 2 (ξ 1 ), η 2 = η 2 (ξ 1 ). H ξ 1 = 0 H ξ 1 (ξ 1, η 1 (ξ 1 ), ξ 2 (ξ 1 ), η 2 (ξ 1 )) = ξ M 1 1 (a M,0 + O(ξ 1 )) = 0 Hence a M,0 0, then for ξ 1 small enough the last equation has a unique root ξ 1 = 0. Thus, at a M,0 0 system has isolated equilibrium ξ 1 = η 1 = ξ 2 = η 2 = 0. This contradicts our assumption.
Suppose that the transcendental case takes place and the Hamiltonian normal form reads H = 1 2 η2 1 + ωr 2 + M [m/2] m=3 l=0 where ξ 2 = 2r 2 sin ϕ 2, η 2 = 2r 2 cos ϕ 2. a m 2l,2l ξ m 2l 1 r l 2 + H (M+1) (ξ 1, η 1, ϕ, r) Isoenergetic reduction. Consider the motion on fixed energy level H = 0. For small values of ξ 1, η 1 r 2 = K (ξ 1, η 1, ϕ 2 ) = 1 2ω η2 1 + K (3) (ξ 1, η 1, ϕ 2 ) (11) K (3) is periodic function of variable ϕ. Whittaker equations dξ 1 = K, dϕ 2 η 1 dη 1 = K (12) dϕ 2 ξ 1
By a proper choice of variables ξ 1, η 1, ξ 2, η 2 we can represent the invariant manifold of equilibrium positions in the form M = {(ξ 1, ξ 2, η 1, η 2 ) : ξ 1 = constant, ξ 2 = η 1 = η 2 = 0} This manifold lies on the energy level H = 0. That is why equations (12) have solution ξ 1 = const, η 1 = 0. It means that ( K (ξ 1, η 1, ϕ 2 ) = η1 2 1 2ω a ) 1,2 2ω ξ 1η1 2 +... = η1 2 G(ξ 1, η 1, ϕ 2 ), (13) where G = 1 2ω a 1,2 2ω ξ 1η 2 1 +...
Let us consider Chetaev function V = ξ 1 η 1 V is positive in the domain ξ 1 > 0, η 1 > 0 and is equal to zero on its border. Let us calculate the derivative of V with respect to equations (12). dv = dξ ( ) ( ) 1 dη 1 η 1 + ξ 1 = η 1 2η 1 G + η 2 G 1 + ξ 1 η 2 G 1 = dϕ 2 dϕ 2 dϕ 2 η 1 ξ 1 ( ) = η1 2 G G 2G + η 1 ξ 1 = η1 2 F (ξ 1, η 1, ϕ 2 ) η 1 ξ 1 The above derivative is positive in the Domain V > 0 because F = 1 ω + > 0 where by dots we denote terms of order one and higher with respect of canonical variables ξ, η.
Accourding to Chetaev theorem the equilibrium ξ 1 = η 1 = 0 of the reduced system is unstable. It follows that the equilibrium ξ 1 = η 1 = ξ 2 = η 2 = 0 of the original system is also unstable. By summarizing the results we can formulate Theorem. If the equilibrium position ξ 1 = η 1 = ξ 2 = η 2 = 0 of the canonical system with Hamiltonian H = 1 2 η2 1 + ω(ξ 2 2 + η 2 2) + H (3) (ξ 1, η 1, ϕ, r) is not isolated, then it is unstable.
Motion of a Satellite with respect to its Center of Mass Main assumptions: Satellite is a rigid body Center of mass moves in circular or elliptic orbit Motion of mass center does not depend on motion with respect to center of mass Reference frames orbital frame (axes OX, OY, OZ directed along the radius vector of the center of mass O, the transversal and normal of the orbit, respectively) principal axes frame Oxyz We use the Euler angles ψ, θ, ϕ to describe the position of the the satellite in orbital system
Euler angles
Hamiltonian of the problem The Hamiltonian of the problem reads H = p2 ψ 2 sin 2 θ + p2 θ 2 p cos θ ψ cot θ cos ψ αβp ψ sin 2 θ p θ sin ψ + αβ cos ψ sin θ + + α2 β 2 2 sin 2 θ + 3 2 (α 1) cos2 θ. (14) Parameters: α = C/A and A, B, C(A = B) are the principal moments of inertia (0 < α 2). β = Ω 0 /ω 0 ; ω 0 and Ω 0 are the orbital angular velocity and the projection of the angular velocity of the satellite onto its symmetry axis respectively. The parameter β can take any real value. The angle ϕ is a cyclic coordinate so in the Hamiltonian we have put p ϕ = αβ.
The equilibria solution (4 3α) cos θ 0 = αβ, ψ 0 = π 2, p θ 0 = 0, p ψ0 = 1 3(α 1) cos 2 θ 0. (15) Solution (15) describes the so-called conical precession of the satellite representing the steady-state rotation in which the satellite symmetry axis is perpendicular to the velocity vector of the center of mass and forms the angle θ 0 with the normal to the orbit plane.
The stability problem of the conical precession has been studied in detail. The stability question of the conical precession is still open 5 only for the special case α = 4/3, β = 0, when the angle θ 0 may take any constant value. In this case relations (15) describe not a unique equilibria solution, but a one parametric family of the following equilibria solutions 0 < θ 0 π 2, ψ 0 = π 2, p θ 0 = 0, p ψ0 = sin 2 θ 0. (16) 5 A. P. Markeev, A. G. Skolskii, and T. N. Chekhovskaia. Stability of the conical precession of a dynamically symmetric rigid body. Soviet Astronomy Letters, 3:178 180, August 1977.
The local canonical variables q i, p i (i = 1, 2) ψ = π 2 + q 1, p ψ = sin 2 θ 0 + p 1, θ = θ 0 + q 2, p θ0 = p 2. (17) The Hamiltonian quadratic part H 2 = p 2 1 2 sin 2 θ 0 + p2 2 2 1 2 sin2 θ 0 q 2 2 2 cot θ 0 p 1 q 2 (18) The CE of the linear system with Hamiltonian H 2 λ 4 + (4 cos 2 θ 0 1)λ 2 = 0 (19) If π/3 < θ 0 π/2, then the CE has two real roots Instability. If 0 < θ 0 < π/3 then roots of the CE are pure imaginary. Matrix of the linear system is not diagonal Nonlinear stability analysis is necessary. The case θ 0 = π/3, when all roots of the CE are zero we do not consider here.
The linear canonical change of variables q 1 = 2 cot θ 0 x 1 + ω 1 ω sin ω y 2, q 2 = 2 cos θ 0 ω x 2 + 1 ω y 1, p 1 = sin θ 0 x 2 + sin 2θ 0 y 1, p 2 = 1 ω ω ω x 1 + 2 cos θ (20) 0 y 2 ω The normal form of the Hamiltonian quadratic part H 2 = 1 2 x 2 1 + 1 2 ω(x 2 2 + y 2 2 ) (21) Conclusion: The conical precession is unstable for the case 0 < θ 0 < π/3
Other problems Three body problem. Stability of L 4 at µ = 0. Stability of regular precession of a satellite at α = 0 (spherically symmetric satellite). Stability of planar periodic pendulum-like motions of a rigid body with a fixed point in the case of Lagrange and in the case of Gorychev-Chaplygin 6. 6 Bardin, B. S. On stability problem of rigid body penduliform motions in Goraychev-Chaplygin s case Akademiia Nauk SSSR, Izvestiia, Mekhanika Tverdogo Tela, 2007, 14-21
Open questions What are conditions for transcendental situation nonresonant case and in cases of other resonances? Is equilibrium can be stable in the transcendental case? What are relation between transcendental situation and integrability of the system? What is dynamical behavior of the system in the neighborhood of equilibria in transcendental case?
Thank you for attention and Happy Birthday, Ken!!!