Semi-analytical computation of Normal Forms, Centre Manifolds and - PowerPoint PPT Presentation

Centre Manifold of L 1 , 2 Results Efficiency Extensions References Semi-analytical computation of Normal Forms, Centre Manifolds and First Integrals of Hamiltonian systems (II) ` Angel Jorba angel@maia.ub.es University of Barcelona Advanced School on Specific Algebraic Manipulators 1 / 52

Centre Manifold of L 1 , 2 Results Efficiency Extensions References Outline 1 Centre Manifold of L 1 , 2 Introduction Expansion of the Hamiltonian The Lie series method 2 Results Poincar´ e sections 3 Efficiency Transforming the Hamiltonian Efficiency considerations Tests 4 Extensions Intervalar arithmetic 5 References 2 / 52

✖ ☎ ✖ ✕ ✔ ✓ ✒ ✖ ✁ ✁ ✖ ✂ ✝ ✁ ✖ ✝ � ✁ ✁ ✁ � ✂ � ✝ Centre Manifold of L 1 , 2 Results Efficiency Extensions References Centre Manifold of L 1 , 2 Let us consider the dynamics near the points L 1 , 2 of the RTBP. We recall that the linearization of the vectorfield at these points is of the type centre × centre × saddle. ✠✌✑ ✂✞☎✟✝ ✠✌☞ ✠☛✡ ✠✌✍ �✄✂✆☎ ✠✏✎ �✄✂✆☎ ✂✞☎✟✝ ☎✟✝ 3 / 52

Centre Manifold of L 1 , 2 Results Efficiency Extensions References To give an accurate description of the dynamics close to L 1 , 2 one can perform the so-called reduction to the centre manifold. The idea is the following: assume that the diagonal form of H 2 is √ √ H 2 = λ q 1 p 1 + − 1 ω 2 q 2 p 2 + − 1 ω 3 q 3 p 3 , λ, ω 2 , ω 3 ∈ R . Hence, the hyperbolic direction is given (at first order) by the variables ( q 1 , p 1 ). 4 / 52

Centre Manifold of L 1 , 2 Results Efficiency Extensions References Introduction Let us perform canonical transformations on the Hamiltonian, cancelling monomials such that the exponent of q 1 is different from the exponent of p 1 . After a finite number of transformations, H takes the form H = H (0) ( q 1 p 1 , q 2 , p 2 , q 3 , p 3 ) + R ( q 1 , p 1 , q 2 , p 2 , q 3 , p 3 ) , where H (0) is the part that we have arranged and R is the remainder. As H (0) depends on the product q 1 p 1 we can perform the change I 1 = q 1 p 1 to produce H = H (0) ( I 1 , q 2 , p 2 , q 3 , p 3 ) + R ( I 1 , ϕ 1 , q 2 , p 2 , q 3 , p 3 ) , where ϕ is the conjugate variable of I 1 . If we drop R then I 1 is a first integral of the system and putting I 1 = 0 we are skipping the hyperbolic part of the Hamiltonian H (0) . 5 / 52

Centre Manifold of L 1 , 2 Results Efficiency Extensions References Introduction The resulting two degrees of freedom Hamiltonian represents the flow inside the (approximation to the) centre manifold. So, near the origin, the phase space of the original Hamiltonian must be the phase space of H (0) (0 , q 2 , p 2 , q 3 , p 3 ) times an hyperbolic direction. To visualize the phase space of H (0) one can fix the value of the Hamiltonian and then use a Poincar´ e section. Varying the value of the Hamiltonian we will obtain a collection of 2-D plots representing the dynamics in the phase space. 6 / 52

Centre Manifold of L 1 , 2 Results Efficiency Extensions References Expansion of the Hamiltonian Let us start by translating the origin of coordinates to the selected point L 1 , 2 . It is well known that the distance from L j to the closest primary, γ j , is given by the only positive solution of the Euler quintic equation, γ 5 j ∓ (3 − µ ) γ 4 j + (3 − 2 µ ) γ 3 j − µγ 2 j ± 2 µγ j − µ = 0 , j = 1 , 2 , where the upper sign in the first equation is for L 1 and the lower one for L 2 . These equations can be solved numerically by the Newton method, using as starting point ( µ/ 3) 1 / 3 . 7 / 52

Centre Manifold of L 1 , 2 Results Efficiency Extensions References Expansion of the Hamiltonian To have good numerical properties for the coefficients of the Taylor expansion it is very convenient to introduce some scaling. The translation to the equilibrium point plus the scaling is given by X = ∓ γ j x + µ + a , Y = ∓ γ j y , Z = γ j z , where the upper sign corresponds to L 1 , 2 , the lower one to L 3 , a = − 1 + γ 1 for L 1 , a = − 1 − γ 2 for L 2 and a = γ for L 3 . Note that this change redefines the unit of distance as the distance from the equilibrium point to the closest primary. As scalings are not canonical transformations, they have to be applied on the equations of motion. 8 / 52

Centre Manifold of L 1 , 2 Results Efficiency Extensions References Expansion of the Hamiltonian To expand the nonlinear terms, we will use that 1 ( x − A ) 2 + ( y − B ) 2 + ( z − C ) 2 = � � ρ ∞ = 1 � n � Ax + By + Cz � � , P n D D D ρ n =0 where D 2 = A 2 + B 2 + C 2 , ρ 2 = x 2 + y 2 + z 2 and P n is the polynomial of Legendre of degree n . 9 / 52

Centre Manifold of L 1 , 2 Results Efficiency Extensions References Expansion of the Hamiltonian After some calculations, one obtains that the Hamiltonian can be expressed as H = 1 � x � p 2 x + p 2 y + p 2 � c n ( µ ) ρ n P n � � + yp x − xp y − , z 2 ρ n ≥ 2 where ρ 2 = x 2 + y 2 + z 2 and the coefficients c n ( µ ) are given by � ( ± 1) n µ + ( − 1) n (1 − µ ) γ n +1 � c n ( µ ) = 1 j , for L j , j = 1 , 2 γ 3 (1 ∓ γ j ) n +1 j As usual, the upper sign is for L 1 and the lower one for L 2 . P n denotes the Legendre polynomial of degree n . 10 / 52

Centre Manifold of L 1 , 2 Results Efficiency Extensions References Expansion of the Hamiltonian For instance, if we define � x � T n ( x , y , z ) = ρ n P n , ρ then, it is not difficult to check that T n is a homogeneous polynomial of degree n that satisfies the recurrence T n = 2 n − 1 xT n − 1 − n − 1 ( x 2 + y 2 + z 2 ) T n − 2 , n n starting with T 0 = 1 and T 1 = x . 11 / 52

Centre Manifold of L 1 , 2 Results Efficiency Extensions References Expansion of the Hamiltonian The linearization around the equilibrium point is given by the second order terms (linear terms must vanish) of the Hamiltonian that, after some rearranging, takes the form, H 2 = 1 + yp x − xp y − c 2 x 2 + c 2 2 y 2 + 1 z + c 2 p 2 x + p 2 2 p 2 2 z 2 . � � y 2 As c 2 > 0 (for the three collinear points), the vertical direction is an harmonic oscillator with frequency ω 2 = √ c 2 . As the vertical direction is already uncoupled from the planar ones, in what follows we will focus on the planar directions, i.e., H 2 = 1 + yp x − xp y − c 2 x 2 + c 2 p 2 x + p 2 2 y 2 , � � y 2 where, for simplicity, we keep the name H 2 for the Hamiltonian. 12 / 52

Centre Manifold of L 1 , 2 Results Efficiency Extensions References Expansion of the Hamiltonian Next step will be to compute a symplectic change of variable such that Hamiltonian takes a simpler (diagonal) form. This change is given by the symplectic matrix 0 2 λ 1 − 2 λ 1 2 ω 1 1 0 0 0 s 1 s 1 s 2 λ 2 − ω 2 λ 2 1 − 2 c 2 − 1 1 − 2 c 2 − 1 1 − 2 c 2 − 1 B C 0 0 0 B C s 1 s 2 s 1 B 1 C 0 0 0 0 0 B C √ ω 2 B C , λ 2 − ω 2 λ 2 1 +2 c 2 +1 1 +2 c 2 +1 1 +2 c 2 +1 B C 0 0 0 B C s 1 s 2 s 1 B C λ 3 − λ 3 − ω 3 1 +(1 − 2 c 2 ) λ 1 1 − (1 − 2 c 2 ) λ 1 1 +(1 − 2 c 2 ) ω 1 B C 0 0 0 @ s 1 s 1 s 2 A √ ω 2 0 0 0 0 0 and casts the second order Hamiltonian into its real normal form, H 2 = λ 1 xp x + ω 1 y ) + ω 2 2 ( y 2 + p 2 2 ( z 2 + p 2 z ) , where, for simplicity, we have kept the same name for the variables. 13 / 52

Centre Manifold of L 1 , 2 Results Efficiency Extensions References Expansion of the Hamiltonian To simplify the computations, we have used a complex normal form for H 2 because this allows to solve very easily the homological equations that determine the generating functions used during the computations of the center manifold. This complexification is given by q 2 + √− 1 p 2 q 3 + √− 1 p 3 x = q 1 , y = √ , z = √ , 2 2 √− 1 q 2 + p 2 √− 1 q 3 + p 3 p x = p 1 , p y = √ , p z = √ , 2 2 that puts the 2nd order Hamiltonian into its complex normal form, √ √ H 2 = λ 1 q 1 p 1 + − 1 ω 1 q 2 p 2 + − 1 ω 2 q 3 p 3 , being λ 1 , ω 1 and ω 2 real (and positive) numbers. 14 / 52

Centre Manifold of L 1 , 2 Results Efficiency Extensions References Expansion of the Hamiltonian Summarizing: We have a real Hamiltonian H = 1 Z ) + YP X − XP Y − 1 − µ − µ 2( P 2 X + P 2 Y + P 2 , r 1 r 2 with an equilibrium point. 15 / 52

Centre Manifold of L 1 , 2 Results Efficiency Extensions References Expansion of the Hamiltonian Summarizing: We have a real Hamiltonian H = 1 Z ) + YP X − XP Y − 1 − µ − µ 2( P 2 X + P 2 Y + P 2 , r 1 r 2 with an equilibrium point. We want to expand it around that point, composing the expansion with a linear change, U = CV + d , 15 / 52

Centre Manifold of L 1 , 2 Results Efficiency Extensions References The Lie series method Now, the Hamiltonian takes the form � H ( q , p ) = H 2 ( q , p ) + H n ( q , p ) , n ≥ 3 where H 2 = λ 1 q 1 p 1 + √− 1 ω 1 q 2 p 2 + √− 1 ω 2 q 3 p 3 and H n denotes an homogeneous polynomial of degree n . 16 / 52