On explicit integration of two non-holonomic problems

Transcription

1 On explicit integration of two non-holonomic problems Alexey V. Borisov 1 1 Institute of Computer Sciences, Izhevsk, Russia

2 Generalized Chaplygin systems Equations of motion of the generalized Chaplygin system: ( ) d L L ( ) = q 2 S, d L L = q 1 S, dt q 1 q 1 dt q 2 q 2 S = a 1 (q) q 1 + a 2 (q) q 2 + b(q), (1) where Lagrangian L is a function of generalized coordinates q = (q 1, q 2 ) and velocities q = ( q 1, q 2 ). Equations of motion (1) are invariant to the change of time N (q) dt = dτ, if N does not depend on velocities. Indeed, in new time we get: ( ) ( ) d L dτ q 1 L = q d L q 2 S, 1 dτ q 2 L = q q 1 S, 2 ( ) S = N S + 1 N L N q 2 q 1 N L q 1 q 2, L(q, q ) = L(q, N q ). (2)

3 Generalized Chaplygin systems Theorem Let det 2 L 0 and system (1) allow an invariant measure with density q i q j depending only on coordinates. Then there is a change of time N (q) dt = dτ, such that 1 function S defined by (2) depends only on coordinates: S = S(q), 2 in new time equations of motion are Hamiltonian: where dq i dτ = {q i, H}, 2X dp i dτ = {p i, H}, p i = L q i, H = p k q k L, q k=1 i p i and the Poisson bracket is defined by relations {q i, p j } = δ ij, {p 1, p 2 } = S(q), {q 1, q 2 } = 0. (3)

4 Chaplygin ball on a sphere Equations of motion Equations of motion of a ball moving without slipping on a sphere Ṁ = M ω, ṅ = kn ω, k = a a + b, (4) where ω angular velocity of the body, n the normal in the point of contact a radius of the basic sphere, b radius of the ball (fig. 1). The moment with respect to the point of contact M is related with angular velocity ω by a linear equation M = Iω + dn(n ω), d = mb 2, where m mass of the body, I = diag(i 1, I 2, I 3 ) a central tensor of inertia. A coefficient k may be either positive or negative depending on possible cases (fig. 1).

5 Chaplygin ball on a sphere. Rolling of a body with a spherical part on the sphere. Fixed surface is marked by shading.

6 Chaplygin ball on a sphere Integrals of motion and integrable cases Under any value of k, system (4) has three integrals of motion F 0 = (n, n) = 1, H = 1 2 (M, ω), F 1 = (M, M) (5) and invariant measure ρ dω dn with density Integrable cases 1 k = 1 (a ). Rolling on a plane. 2 k = 1 ( b a = 1 2 ). Additional integral of motion ρ 2 = (n, n) d(n, (I + d) 1 n). F 2 = (AM, n), (6) where A = diag( 1 2 ( I 1 + I 2 + I 3 ), 1 2 (I 1 I 2 + I 3 ), 1 2 (I 1 + I 2 I 3 ))

7 Chaplygin ball on a sphere Equations of motion of Chaplygin ball at k = 1 and F 2 = 0 as Chaplygin system We can write the equation of motion (4) as a Chaplygin system d T dt u T u = uφ, d T dt v T v = vφ (7) where T = 1 2 (b uu u 2 + b uv u v + b vv v 2 ), and Φ = (a u u + a v v) a linear homogeneous in velocity function, and u и v sphero-conical coordinates on the sphere n 2 = 1 n 2 i = (J i u)(j i v) (J i J j )(J i J k ), i j k i, J i = I i + d. (8)

8 Chaplygin ball on a sphere Equations of motion of Chaplygin s ball for k = 1 and F 2 = 0 are a conformal Hamiltonian After change of variables N (u, v) dt = dτ system (7) has a Lagrangian form d T dτ u T u = 0, d T dτ v T v = 0, u = du dτ, v = dv dτ. The reducing multiplier: N = 2uv + (u + v)(2d + α 1) + α 2 dα 1 det(i + d dn n) (4α 3 + 2α 1 α 2 α 3 1 dα (α 2 1 2α 2 + 4dα 1 )(u + v) 4d(u + v) 2 ) 1 where α 1 = J i, α 2 = J 2 i, α 3 = J 1 J 2 J 3.

9 Chaplygin ball on a sphere Equations of motion of Chaplygin s ball for k = 1 and F 2 = 0 are a conformal Hamiltonian After change of variables we get a Hamiltonian system on two-dimensional sphere S 2, which can be presented as equations on a special (null) orbit of co-algebra e(3) H = δ det J 8(γ, Bγ) 2 3X i=1 c i = ρ2 δ J i c i m 2 i, F 2 = ρ 2 4(γ, Bγ) 2 (δ2 m 2 4 δ = (γ, JĀγ) d(γ, Āγ) 2, 4 Y k i (J i J k )γ 2 i (ρ2 dδ 4J i det J ), 3X i=1 d i m 2 i ), d i = Y k i(j i J k )γ 2 i (δ(j i + d) (γ, J(J + d)ā 2 γ) + 2d(γ, JĀγ)(γ, Āγ)). (9) where Ā = 2A. Poisson brackets are given by the following expressions {m i, m j } = ε ijk m k, {m i, γ j } = ε ijk γ k, {γ i, γ j } = 0, and the orbit is determined by integrals γ 2 = 1, (m, γ) = 0.

10 Chaplygin ball on a sphere Case F 2 0 Remark We can show that if k = 1 and F 2 0 then equations of motion (7) have the form of generalized Chaplygin system (1). Therefore, using above Theorem, we can write these equations as a Hamiltonian system with gyroscopic forces. But since the equations are rather complicated and general methods of integration in quadratures are not known for such systems, we do not present them here.

11 Chaplygin ball on a sphere Separation of variables We use a canonical presentation of algebra e(3) (Darboux coordinates) m 1 = p 1 (x 2 1) + p 2 (y 2 1), m 2 = ip 1 (x 2 + 1) + ip 2 (y 2 + 1), m 3 = 2p 1 x + 2p 2 y, γ 1 = xy 1 x y, γ 2 = i xy 1 x y, Using (10) we can write the pair (H, F 2 ) in canonical form γ 3 = x + y x y (10) H = a(x, y)p b(x, y)p 1 p 2 + c(x, y)p 2 2, F 2 = A(x, y)p B(x, y)p 1 p 2 + C(x, y)p 2 2 (11) We can show that separating variables are the roots of the equation (B bs) 2 = (A as)(c cs) (12)

12 Chaplygin ball on a sphere Separation of variables In new coordinates functions H and F have the Liouville form where and H = S 1(s 1 ) s 1 s 2 P 2 1 S 2(s 2 ) s 1 s 2 P 2 2, F = s 2S 1 (s 1 ) s 1 s 2 P 2 1 s 1S 2 (s 2 ) s 1 s 2 P 2 2, (13) S(x) = 2(8x 3 + 8(d ɛ)x 2 + (2ɛ 2 β 4dɛ)x 4γ dβ + ) γ(2x ɛ + 2d) 2 (14) α = (J 2 + J 1 J 3 )( J 2 + J 2 J 3 )( J 2 + J 2 + J 3) β = J J J 2 3 2J 1 J 2 2J 2 J 3 2J 3 J 1 γ = J 1 J 2 J 3, ɛ = J 1 + J 2 + J 3, = x 2 (β 2 + 8αd) + 2x(4βγ + dβ 2 2dαɛ + 4αd 2 ) + (4γ + dβ) 2

13 Rubber ball moving on a sphere Equations of motion Let us assume an additional constraint, which does not let the ball twist Equations of motion can be given in the form (ω, n) = 0. (15) J ω = Jω ω + λn + M Q, ṅ = kn ω, J = I + mb 2 E, E = δ ij, (16) where λ = (Jω ω, J 1 n) + (M Q, Jn). (n, J 1 n) and M Q moment of external forces.

14 Rubber ball on a sphere Conservation laws and integrable cases Equations (16) have integral of energy and geometric integral and have invariant measure Integrable cases H = 1 (Jω, ω), (n, n) = 1, 2 1 (n, J 1 n) 2k dω dn. (17) 1 k = 1 (a = ) rolling of a ball on a horizontal plane. Additional integral F = (Jω n, Jω n). The obtained system is equivalent to Veselova s system and can be integrated with the help of sphero-conical coordinates. 2 k = 1 (b = 2a) rolling of dynamically asymmetric sphere by its inner surface on a fixed ball. Additional integral: F = (Jω, Jω)n 2 + det J(ω, Jω)(J 1 n, J 1 n). (18) (n, J 1 n)

15 Rubber ball on a sphere Hamiltonian structure and algebraization Equations of motion (16) can be represented in the form of Veselova s system ( ) d T T dt ξ ξ = ηs, d T T dt η η = ξs, ( ) S = 2k 1 ξ (ξ η) 8k 3 A(ξ) + η (19). A(η) Here (ξ, η) sphero-conical coordinates on a sphere n = 1 n 2 1 = (J 1 ξ)(j 1 η) (J 1 J 2 )(J 1 J 3 ), n2 2 = (J 2 ξ)(j 2 η) (J 2 J 1 )(J 2 J 3 ), n2 3 = (J 3 ξ)(j 3 η) (J 3 J 1 )(J 3 J 2 ), and kinetic energy T = 1 (ω, Jω) in these variables is 2 ( ) T = ξ η 8k 2 η ξ 2 A(ξ) + ξ η2 A(η) (20). (21)

16 Rubber ball on a sphere Hamiltonian structure and algebraization By the above theorem, after the change of time N (u, v) dt = dτ with reducing multiplier ( ) 1 1+ ξη 2k N = (22) det J equations of motion become canonical ξ = T p ξ, η = T p η, p ξ = T ξ, p η = T η. (23) Therefore, system (16) for k = ±1 is conform-hamiltonian. Define isomorphism of systems (16), (19) with the problem of motion of point on a sphere. Introduce three-dimensional vectors M = Ñ J1/2 ω, γ = 1 ρ J 1/2 n, (24) where det J Ñ = N det J k ρ = (n, J 1 n) 2 k 1 2k.

17 Rubber ball on a sphere Hamiltonian structure and algebraization It is evident that (γ, γ) = 1, (M, γ) = Ñρ (ω, n) = 0. (25) Poisson brackets between M and γ: {M i, M j } = ε ijk M k, {M i, γ j } = ε ijk γ k, {γ i, γ j } = 0. (26) Hamiltonian k 2 H = T = 1 2 M 2 = 1 2Ñ 2 det J (γ, Jγ)1/k M 2 = ( ) (2k 1)/k ( = 2k2 ξη A(ξ) ξ η det J η p2 ξ A(η) ) pη 2. (27) ξ

18 Rubber ball on a sphere Hamiltonian structure and algebraization Hamiltonian (27) is the product of two functions, depending on M and γ respectively: H = G(γ)F (M ). Equations of motion can be represented as: Ṁ = G ( M F M ) 1 FGγ G, γ = Gγ F γ M. (28) Make a change of time G(γ)dt = ds and fix the level of integral FG = h. On this level we get a system: dm ds = M H M + γ H γ, H = F (M ) h G(γ). dγ ds = γ H s, (29)

19 Rubber ball on a sphere Trajectory isomorphysim For Hamiltonian (27) we have H = 1 2 M 2 h(γ, Jγ) 1/k, h = const. (30) Therefore, on a fixed level of the energy integral H = h the system (28) is trajectory isomorphic to system (30) for H = 0. Hamiltonian (30) describes the motion of a particle on a surface (on condition that (M, γ) = 0) in potential field of forces with potential V = h(γ, Jγ) 1/k. Integrable potentials correspond to cases 1 k = 1: the Braden system, 2 k = 1: the Neumann system.

20 Rubber ball on a sphere Separation of variables for k = 1 Define sphero-conical coordinates u and v n 2 1 = ρ2 J 1(J 1 u)(j 1 v) (J 1 J 2 )(J 1 J 3 ), n2 2 = ρ 2 J 2(J 2 u)(j 2 v) (J 2 J 1 )(J 2 J 3 ), n2 2 = ρ2 J 3(J 3 u)(j 3 v) (J 3 J 1 )(J 3 J 2 ), where ρ 2 = (n, J 1 n) = ( i T = det Jρ4 (u v) 8 J i u v) 1. Kinetic energy (21): u 2 A(u) v 2 A(v) where A(x) = (J i x), and reducing multiplier is i N = ( J i u v ) 3/2. i Thus, we receive canonical equations with Hamiltonian H = 2 det J ( J i u v ) ( A(u)p 2 u A(v)p 2 ) v. (u v) i!, Therefore, u and v are separating variables.