Dynamics of multiple pendula without gravity

June 13, 2013 Institute of Physics University of Zielona Góra Chaos 2013 Istambul, 11-15 June Dynamics of multiple pendula without gravity Wojciech Szumiński http://midori-yama.pl For Anna

Finding new integrable systems; Why it is so difficult...?

Finding new integrable systems; Why it is so difficult...? Lack of sufficiently strong theorems; Dependence on many parameters: Masses m i ; Lengths of arms a i ; Young modulus of the springs k i ; and unstretched lengths of the springs.

The Idea! Simple double pendulum 1 The configuration space T 2 = S 1 S 1 Local coordinates (φ 1, φ 2 ) mod 2π; Regular as well as chaotic trajectories; Proof of its non-integrability is still missing. On the other hand, 1 W. Szumiński. Dynamics of multiple pendula. Bachelor thesis, University of Zielona Góra, 2012. accessible at http://midori-yama.pl/?q=publications.

The Idea! Simple double pendulum 1 The configuration space T 2 = S 1 S 1 Local coordinates (φ 1, φ 2 ) mod 2π; Regular as well as chaotic trajectories; Proof of its non-integrability is still missing. On the other hand, in the absence of gravity system has S 1 symmetry. Introducing new variables θ 1 = φ 1, θ 2 = φ 2 φ 1, we note that θ 1 is a cyclic variable and the corresponding momentum is a missing first integral. 1 W. Szumiński. Dynamics of multiple pendula. Bachelor thesis, University of Zielona Góra, 2012. accessible at http://midori-yama.pl/?q=publications.

The idea is simple! Lets look for new integrable systems among planar multiple pendula in the absence of gravity when the S 1 symmetry is present. Methods: numerical: The Poincaré cross sections; analytical:...?

The idea is simple! Lets look for new integrable systems among planar multiple pendula in the absence of gravity when the S 1 symmetry is present. Methods: numerical: The Poincaré cross sections; analytical: Morales-Ramis theory 2 Proofs of all theorems can be found at http://midori-yama.pl 2 J. J. Morales Ruiz. Differential Galois theory and non-integrability of Hamiltonian systems. Progress in Mathematics, volume 179, Birkhäuser Verlag, Basel, 1999.

Double spring pendulum Figure: Geometry of the system and Poincaré section: E = 0.001, m 1 = m 2 = a 1 = a 2 = 1, k = 0.1, c = 0 cross-plain x = 1, p 3 > 0. Theorem: Assume that a 1 m 1 m 2 0, and c = 0. Then the reduced system descended from double spring pendulum is non-integrable in the class of meromorphic functions of coordinates and momenta.

Two rigid spring pendula Figure: Geometry and Poincaré cross-section: E = 0.12, m 1 = m 2 = a 1 = a 2 = 1, k 1 = k 2 = 1/10, c = 1/10, cross-plain x 1 = 0, p 1 > 0. Theorem: If m 1 m 2 k 1 c 0, and k 2 = 0, then the reduced two rigid spring pendula system is non-integrable in the class of meromorphic functions of coordinates and momenta.

Two rigid spring pendula: integrable cases Case I: c = 0, then the reduced Hamilton equations becomes linear equations with constant coefficients. Case II: k 1 = k 2 = 0, then original Hamiltonian simplifies to H = 1 ( p 2 1 + p2 2 p 2 ) 3 + 2 m 1 m 2 m 1 x 2 1 + m 2x 2, (1) 2 with two additional first integrals F 1 = p 3, F 2 = m 2 p 2 x 1 m 2 p 1 x 2.

Triple flail pendulum Figure: Geometry and Poincaré cross-section: E = 0.01, m 1 = 1, m 2 = 3, m 3 = 2, a 1 = 1, a 2 = 2, a 3 = 3, c = 1, cross-plain θ 2 = 0, p 2 > 0. Theorem: Assume that a 1 a 2 a 3 m 2 m 3 0, and m 2 a 2 = m 3 a 3. If either (i) m 1 0, c 0, a 2 a 3, or (ii) a 2 = a 3, and c = 0, then the reduced flail system is not integrable in the class of meromorphic functions of coordinates and momenta. 3 3 M. Przybylska, W. Szumiński. Non-integrability of flail triple pendulum. Chaos, Solitons & Fractals. DOI: 10.1016

Examples of Poincare sections (a) E = 0.011, (b) E = 0.012. Figure: Poincare sections for parameters: m1 = 1, m2 = 3, m3 = 2, a1 = 1, a2 = 2, a3 = 3, c = 1, cross-plain θ2 = 0, p2 > 0.

Figure: Enlargement of the central part from the previous picture.

Examples of Poincaré sections (a) E = 0.0035, (b) E = 0.0363. Figure: Poincaré sections for parameters: m 1 = 1, m 2 = m 3 = 2, a 1 = 2, a 2 = a 3 = 1, c = 1/2, cross-plain θ 2, p 2 > 0.

Figure: Enlargement of the central part from the previous picture.

Triple bar pendulum Figure: Geometry and Poincaré section: E = 0.008, m 1 = m 2 = 1, m 3 = 2, a 1 = 1, a 2 = 2, a 3 = 1, d 1 = d 2 = 1, c = 1 2, cross-plain θ 2, p 2 > 0 Theorem: Assume that a 2 a 3 m 1 m 2 m 3 0, and m 2 a 2 = m 3 a 3, d 1 = d 2. If either (i) c 0, a 2 a 3 or (ii) a 2 = a 3, and c = 0, then the reduced triple bar system is not integrable in the class of meromorphic functions of coordinates and momenta.

Simple triple pendulum Figure: Geometry and Poincaré section: E = 0.011, m 1 = 2, m 2 = 1, m 3 = 1, a 1 = 2, a 2 = a 3 = 1, c = 1, cross-plain θ 2, p 2 > 0. Problem: We are not able to find a non-trivial particular solution. Thus we can not apply the Morales-Ramis theory.

Chain of mass points Figure: geometry and Poincaré section: E = 0.045; m 1 = m 3 = 1, m 2 = 2, m 4 = 3, a 2 = a 3 = 1, a 4 = 3, c = 3 2, cross-plain θ 3, p 3 > 0. n + 1 degrees of freedom: Generalized coordinates: (x 1, y 1 ) of the first mass calculated with respect to the center of the mass of all n bodies; Relative angles θ i between i-th and i 1-th arms.

Vector radii of corresponding masses r 1 = (x 1, y 1 ), r i = r i 1 + a i cos In the center of mass Thus, we can express Then i θ j, sin j=2 i j=2 θ j, i = 2,..., n n m i r i = 0, (2) i=1 (x 1, y 1 ) [x 1 (θ i ), y 1 (θ i )]. V = n m i gy i = 0. i=1 three masses - integrable; four masses - not integrable. When m 3 a 4 = m 3 a 2 a non-trivial particular solution is know: The non-integrability analysis is in progress.

Unfixed triple flail pendulum Figure: Geometry and Poincaré section: E = 0.24, m 1 = m 3 = 1, m 2 = 2, m 4 = 3, a 2 = a 3 = 1, a 4 = 3, c = 3 2, cross-plan θ 3, p 3 > 0. One can also find a non-trivial particular solution when a 3 = a 4. The non-integrability analysis is in progress. http://midori-yama.pl

General procedure to get integrability or non-integrability proofs... to apply the Morales-Ramis theory! Let H : C 2n C be a holomorphic Hamiltonian, and d dt x = ν H(x), ν H (x) = I 2n x H, x C 2n, t C, the associated Hamilton equations. Let t φ(t) C 2n be a non-equilibrium solution of (3). (3)

General procedure to get integrability or non-integrability proofs Variational equations along φ(t) have the form d dt ξ = A(t)ξ, A(t) = ν H (φ(t)). (4) x We can attach to equations (4) the differential Galois group G. Theorem (Morales-Ruiz and Ramis): Assume that a Hamiltonian system is meromorphically integrable in the Liouville sense in a neigbourhood of a phase curve Γ. Then the identity component of the differential Galois group of NVEs associated with Γ is Abelian.

Example: double spring pendulum Hamiltonian of the reduced system for p 1 = c = 0 H = [ m 2 p 2 2x 2 + 2a 1 m 2 p 2 x(p 2 cos θ 2 + p 3 x sin θ 2 ) + a 2 1(m 1 (p 2 2 + x 2 (p 2 3 + km 2 (x a 2 ) 2 )) + m 2 (p 2 cos θ 2 + p 3 x sin θ 2 ) 2] /(2a 2 1m 1 m 2 x 2 ), (5) Hamilton equations have particular solutions given by θ 2 = p 2 = 0, ẋ = p 3 m 2, ṗ 3 = k(a 2 x). (6) We chose a solution on the level H(0, x, 0, p 3 ) = E, and [Θ 2, X, P 2, P 3 ] T be variations of [θ 2, x, p 2, p 3 ] T. Then the variational equations along this particular solution are following

Θ 2 Ẋ P 2 = P 3 p 3 (a 1 +x) a 1 m 1 x 0 a 2 1 m 1+m 2 (a 1 +x) 2 a 2 1 m 1m 2 x 2 0 0 0 0 1 m 2 p2 3 m 1 0 p 3(a 1 +x) a 1 m 1 x 0 0 k 0 0 Θ 2 X P 2, (7) P 3 where x and p 3 satisfy (6). Equations for Θ 2 and P 2 form a subsystem of normal variational equations and can be rewritten as one second-order differential equation Θ + P Θ + QΘ = 0, Θ Θ 2, P = 2a 1p 3 (a 1 (m 1 + m 2 ) + m 2 x) m 2 x(a 2 1 m 1 + m 2 (a 1 + x) 2 ), Q = k(a 1 + x)(x a 2 ) m 1 a 1 x 2a 2 1 p2 3 m 2 a 1 x(a 2 1 m 1 + m 2 (a 1 + x) 2 ). (8)

The following change of independent variable t z = x(t) + a 1, and then a change of dependent variable [ Θ = w exp 1 z ] p(ζ) dζ 2 z 0 transforms this equation into an equation with rational coefficients w = r(z)w, r(z) = q(z)+ 1 2 p (z)+ 1 4 p(z)2, p(z), q(z) C(z). (9)

Kovacic algorithm 4. Lemma A : Let G be the differential Galois group of Eq.(9). Then one of the four cases can occur. Case I. Eq. (9) has an exponential solution w = P exp[ ω], P C[z], ω C(z) and G is a triangular group. Case II. (9) has solution w = exp[ ω], where ω is algebraic function of degree 2 and G is the dihedral group. Case III. G is finite; in this case all solutions of equations (9) are algebraic. Case IV. G = SL(2, C) and equation (9) has no Liouvillian solution. 4 J. J. Kovacic. An algorithm for solving second order linear homogeneous differential equations. J. Symbolic Comput., 2(1):3 43, 1986.

Kovacic algorithm: necessary conditions Let us write r(z) C(z) in the form r(z) = s(z) t(z), s(z), t(z) C[z]. The roots of t(z) are poles of r(z). The order of the pole is the multiplicity of the zero of t; The order of r at is deg(t) deg(s).

Lemma B : The necessary conditions for the respective cases in Lemma A are the following Case I. Every pole of r must have even order or else have order 1. The order of r at must be even or else be grater than 2. Case II. r must have at least on pole that either has odd order greater than 2 or else has order 2. Case III. The order of a pole of r cannot exceed 2 and the order of r at must be at least 2. If the partial fraction expansion of r is r(z) = i a i (z c i ) 2 + j b j z d j, then i = 1 + 4a i Q, for each i, j b j = 0 and if g = i a i + j b j d j, then 1 + 4g Q.

Results Equation (9) can only fall into cases (I) or (IV), its degree of infinity is 1. Moreover, one can show that there is no algebraic function ω of degree 2 such that ω = exp [ ω ] satisfies (9). Thus G = SL(2, C) with non-abelian identity component and the necessary integrability condition is not satisfied!

Results Equation (9) can only fall into cases (I) or (IV), its degree of infinity is 1. Moreover, one can show that there is no algebraic function ω of degree 2 such that ω = exp [ ω ] satisfies (9). Thus G = SL(2, C) with non-abelian identity component and the necessary integrability condition is not satisfied! Thank You for Your attention! :)

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