Reduction of symmetries • When G ֒ → A → A/G is a principal bundle, restrict- ing to the zero section it is clear that G ֒ → Q → Q/G is also principal bundle. The reduced symplectic structure can be described in J − 1 ∇ ( µ ) /G µ = T ∗ ( Q/G ) ⊕ A/G ⊕ A ∗ /G × A/G ˜ O µ ⊂ S A, ∇ /G. (for all the details contact Paula Balseiro) • What if the action of G on Q is not free? In favorable cases the action of G is free and proper on A o = A − zero section .
Part II: Cometrics in landmark spaces short review on M. Michelli, P. Michor, D. Mumford Sectional Curvature in Terms of the Cometric with Applications to the Riemannian Manifolds of Landmarks, SIAM J. Imaging Sciences, 5:1, 394433, 2012
Cometrics in landmark space Q = ( ℜ d ) n Correlations appear in G − 1 just soup up the simple problem of 2 particles in 1-dimension 2 H = p 2 1 + p 2 2 + 2 κ ( q 1 − q 2 ) p 1 p 2 • Camassa Holm: κ = exp( −| q i − q j | ) 1 • Cauchy: κ = 1+( | q 1 − q j | ) 2 κ = exp( −| q i − q j | 2 /σ 2 ) • Gaussian:
Camassa-Holm (c.1995) • n particles in the line, with the CH cometric: It is an integrable system!!! • Solitons (peakons) for a shallow water PDE
“Mario’s formula” M. Micheli, P. Michor, D. Mumford: Sectional Curvature in Terms of the Cometric with Applications to the Riemannian Manifolds of Landmarks SIAM J. Imaging Sciences, 5:1, 394–433, 2012
where
“Mario’s formula” will be key ingredient for the ODEs in T ∗ ( T ( ℜ d ) n ) = ℜ 4 nd In computing Christoffel symbols and curvature components: i) derivatives are taken only on upper indices ( g µν ’s, cometric) ii) lower indices g µν ’s (metric) are obtained by inverting numerically the cometric matrix
A code, a code! My kingdom for a code!! i) symbolic computation for Christoffel symbols and curvatures ii) but need numerical inversions .... we know G − 1 , compute G = ( G − 1 ) − 1 : d blocks of size n × n ii) Write the ODEs for the state variables iii) Implement a dynamic optimization method (Pontryagin, Bellman, or direct simulation)
T ∗ ( TS 2 ) Part III. T ∗ ( TS 3 ) Next in order: Note: connectors in homogenous spaces are intrinsically three dimensional!
Connectors (splines) for short processes A “short process” is an element of A = TQ . A connector is a smooth parametrized curve with prescribed tangent vectors in the ends. • Riemannian cubics: minimize over all connectors � T | u | 2 dt where ∇ ˙ q ˙ q = u 0 (with T = fixed , prescribed ends) ∇ (3) Leads to : q = R ( ˙ ˙ q ˙ q ) ˙ q, ∇ ˙ q q ˙
Comparing two control problems on TQ Riemannian cubics � T q | 2 dt ( T fixed) min q ˙ |∇ ˙ 0 Minimal time connector, bounded acceleration ∇ ˙ q ˙ q = u, | u | ≤ b (MTC / BA) [ Equivalently : L inf control (L. Noakes, 2014): Fix T, and then minimize { max | u | , 0 ≤ t ≤ T } ]
L inf -acceleration splines C.Y. Kaya, J. L. Noakes, Finding Interpolating Curves Min- imizing L ∞ Acceleration in the Euclidean Space via Optimal Control Theory, SIAM J. Control Optim., 51(1), 442464 (2013) J. L. Noakes, L ∞ accelerations in Riemannian manifolds, Computational Mathematics, 40: 4, 839–863 (2014) Alex Castro, JK, On the dynamic Markov-Dubins problem: From path planning in robotics and biolocomotion to compu- tational anatomy, Regular and Chaotic Dynamics 18:1, 1–20 (2013)
Comments on accessibility Assume that the metric is complete and Q has finite volume. Then any pair of tangent vectors can be connected smoothly by a “near geodesic”. The bounded acceleration problem is accessible.
Joining smoothly two tangent vectors of arbitrary length with an arbitrary small force control. • Pick two arc length RGCE- δ curves, the first joining v o / | v o | to v 1 /v 1 and the second joining v 1 /v 1 to v o / | v o | . This is the “beginners race track”: a closed loop of arbitrary small geodesic curvature. • Take δ < ǫ . Use the ǫ − δ margin to accelerate or brake in the tangential direction so that you reach q 1 with the desired speed. Note that you may need to do the loop several times! • The argument extends to departures or landings (zero velocities) with a prescribes limit directions.
Accessibility moving only along (reparametrized) geodesics Assume that the control set (forces) contains a small ball around the origin at every T q Q . • Brake along the geodesic through v o until full stop. • Rest as much as you wish. • Restart along the geodesic in the direction of q 1 . • Make a full stop at q 1 . • Do the soft landing in reverse: start from q 1 in the direction of − v 1 ; do the bang bang switch at the right moment, so that when coming back you reach q 1 with velocity v 1 .
Control on A = TQ (or A = T ∗ Q ) Let S ∇ the splitting of T ∗ A . u q is the acceleration (the control) v q is the state to be controlled H = − C ( v q , u q , t ) + � Λ , Z � = − C ( v q , u q , t ) + p q · v q + α q · u q H ∗ ( q, v q , p q , α q ) = − C ( v q , u ∗ q , t ) + p q · v q + α q · u ∗ u ∗ = u ∗ ( v q , α q , t ) = arg max [ − C ( v q , u q , t ) + α q · u q ]
Remarks • Most common: A = TQ, ∇ = Levi-Civita • A could be instead T ∗ Q if the state equations are more conveniently given in terms of momenta • We will describe in a moment the symplectic form on S ∇ ≡ T ∗ A . • In Computational Anatomy landmark cometrics appears naturally. Using S ∇ the curvatures appear explicitly in the equations. Playground for “Mario’s formulae” !
Cubic splines g ≡ g − 1 co-metric induced in T ∗ Q . ¯ H = − 1 2 | u x | 2 + p x · v x + α x · u x u ∗ g ♯ ( α ) then : x = ¯ H ∗ ( x, v x , p x , α x ) = p x · v x + 1 2 ¯ g ( α, α ) g ♯ ( α ) x = v , ∇ ˙ ˙ x v = ¯ ˜ ˜ ∇ ˙ x p = − i ˙ x � α, R ∇ ( v ) � , ∇ ˙ x α = − p ,
Equivalent second-order equations These Hamiltonian equations for ( x, v, p, α )( t ) ∈ S A, ∇ are equivalent to the following second-order equa- tions for ( x, α )( t ) ∈ T ∗ Q : g ♯ ( α ) ∇ ˙ x ˙ x = ¯ , ˜ x (˜ ∇ ˙ ∇ ˙ x α ) = i ˙ x � α, R ∇ ( ˙ x ) � ,
Time minimal, bounded acceleration H ∗ = − 1 + � p, v � + b � ¯ g ( α, α ) b g ♯ ( α ) x = v , ∇ ˙ ˙ x v = ¯ � ¯ g ( α, α ) ˜ ˜ ∇ ˙ x p = − i ˙ x � α, R ∇ ( v ) � , ∇ ˙ x α = − p ,
Equivalent second-order equations These Hamiltonian equations for ( x, v, p, α )( t ) ∈ S A, ∇ are equivalent to the following second-order equa- tions for ( x, α )( t ) ∈ T ∗ Q : b g ♯ ( α ) ∇ ˙ x ˙ x = ¯ � g ( α, α ) ¯ , ˜ x (˜ ∇ ˙ ∇ ˙ x α ) = i ˙ x � α, R ∇ ( ˙ x ) � ,
Two control problems on TS 2 • Cubic splines with fixed connecting time • Minimal time problem with bounded acceleration Are they Arnold-Liouville integrable?
Reduction T ∗ TS 2 = 8 , with SO (3) symmetry Removing the zero section , and reducing we get S 2 × ( ℜ + × ℜ )
Working assumption: zero velocities are not at- tained for most initial conditions. (of course there are is a manifold of solutions that attain zero velocities, in particular soft landings). For integrability, besides the reduced Hamiltonian we need an extra integral of motion (not coming from symmetry). Is there one?
What we know from ℜ 2 • Cubic splines = cubic polynomials As integrable as it could be... • Bounded acceleration problem: no need to do SE (2) reduction. We can solve directly. An expression for the extra integral is unknown. A. Venkatraman and S.P. Bhat (2006 to 2009) considered the problem where the terminal heading is prescribed but the terminal position is free. They found integrals of motion for cost functions C = � | v | n dt .
A control system on Q = SO (3) × ℜ + State variables R ∈ SO (3) , v ∈ ℜ + Control variables u 1 , u 2 State equations (4 dimensional) v = u 1 ˙ 0 − u 2 /v v/r ˙ R = RX , X = u 2 /v 0 0 − v/r 0 0
What is the meaning? Let e i , i = 1 , 2 , 3 the columns of R . The last one gives the normals to a sphere of radius r : e 3 = q/r. First column: direction of tangent vector, q = � ˙ w = v e 1 The second column ( e 2 = e 3 × e 1 ) is the unit normal to the path in the sphere, normal inside surface = q r × � w v
Meaning of the state equations u 1 = scalar acceleration, i.e, v = u 1 . ˙ The last column of X gives q/r = ( v/r ) e 1 or ˙ q = � ˙ w . Differentiating ˙ q = � w = v e 1 we get e 1 = u 1 e 1 + u 2 e 2 − ( v 2 /r ) e 3 ¨ q = u 1 e 1 + v ˙ The last term is due to the constraining normal force, comes from differentiating ( ˙ q , q ) ≡ 0. Thus e 1 = ( u 2 /v ) e 2 − ( v/r ) e 3 . ˙
Geodesic torsion for spherical curves Why X 23 = − X 32 = 0 ? Reason: spherical curves have τ g ≡ 0 ! There is a general formula: τ g = ( κ 1 − κ 2 ) sin phi cos φ ( φ is the angle between � t and a principal direction) This means: the geodesic torsion cannot be con- trolled. But we can control the normal acceleration.
Convex surfaces Gauss : q ↔ N ( q ) ∈ S 2 is a bijection. Key: The third column of R is again a vector N ∈ S 2 . So the state space remains the same, SO (3) × ℜ + The equations of motion live in T ∗ ( SO (3) × ℜ + ) = ( SO (3) × ( ℜ 3 ) ∗ ) × ( ℜ + × ℜ ) Everything will be coupled.
State equations ˙ R = R X v B ( � t,� 0 − u 2 /v t ) v τ g ( � X = u 2 /v 0 t ) − v B ( � t,� − v τ g ( � t ) t ) 0 Here B = the second fundamental form. Since X depends on � t , the SO (3) symmetry is destroyed. S 1 symmetry still remains for surfaces of revolution.
Symplectic form: Ω M = da ∧ dv + ω T ∗ SO(3) ( a = costate of v ) Recall: symplectic form in T ∗ G ≡ G × G ∗ at a point ( g, µ ) For tangent vectors ( X i , w i ) ∈ T g G × G ∗ , i = 1 , 2 ω T ∗ G ( g,µ ) (( X 1 , w 1 ) , ( X 2 , w 2 )) = w 1 · L g − 1 | ∗ X 2 − w 2 · L g − 1 | ∗ X 1 − − µ L g − 1 | ∗ [ X 1 , X 2 ]
Family of left invariant Hamiltonians in T ∗ SO (3) × T ∗ ℜ + ≡ ( SO (3) × ℜ 3 ) × ( ℜ + × ℜ ) . Lie algebra elements: Ω ∈ ℜ 3 , duals M ∈ ℜ 3 Ω = (0 , v/r , u 2 /v ) M = ( M 1 , M 2 , M 3 ) The Hamiltonian family writes as H = − Cost + a u 1 + M 2 ( v/r ) + M 3 u 2 /v .
Cubic splines C = 1 2 ( u 2 1 + u 2 u ∗ 1 = a , u ∗ 2 ) ⇒ 2 = M 3 /v Reduced Hamiltonian H ∗ = 1 2 a 2 + 1 2 ( M 3 /v ) 2 + M 2 v/r Ω = da ∧ dv + ω corbit ω corbit ( M )( v 1 , v 2 ) = det( M ; v 1 , v 2 )
Reduced equations for cubic splines (2 degrees of freedom) a = − M 2 /r + M 2 3 /v 3 v = a , ˙ ˙ M = M × (0 , v/r, M 3 /v 2 ) ˙ , Casimir : M 2 1 + M 2 2 + M 2 3 = µ 2 ,
Introduce polar coordinates on the momentum sphere, say, with poles on the second axis M = µ ( cos φ cos θ , sin φ , cos φ sin θ ) ω c . o . = M 1 dM 2 ∧ dM 3 + M 2 dM 3 ∧ dM 1 + M 3 dM 1 ∧ dM 2 µ 3 cos φ dφ ∧ dθ = Ω = da ∧ dv + µ 3 cos φ dφ ∧ dθ 2 a 2 + µ 2 H = 1 2 (cos φ sin θ ) 2 /v 2 + µ sin φ ( v/r )
It is convenient to introduce z = sin φ , − 1 < z < 1 Ω = da ∧ dv + µ 3 dz ∧ dθ 2 a 2 + µ 2 H = 1 2 (1 − z 2 ) (sin θ ) 2 /v 2 + µ z v/r In order: surfaces of section try for instance ( a, z ) plane ( − 1 < z < 1 ; fix H ; perhaps section θ = 0 .) In a few minutes: planar projections of solutions. Chaos or Integrability?
Reduced equations of motion a = − µz/r + µ 2 (1 − z 2 ) (sin θ ) 2 /v 3 v = a , ˙ ˙ µ 2 r − z (sin θ ) 2 v ˙ θ = µ v 2 z = − sin θ cos θ 1 − z 2 ˙ µ v 2
In order: numerical code for reconstruction and matching Initial conditions of reduced system (4) + µ = 5 + 3 initial Euler angles for reconstruction total = 8 In principle we can match any two tangent vectors in the sphere, using a shooting method.
Questions Equilibria? Eigenvalues? Invariant manifolds of unstable equilibria? Here it would be interesting to find if there is a general behavior of v ( t ) along solutions. Monotonous? Diverging? Zero velocity points?
Equilibria of reduced system The equation for ˙ v = 0 gives a = 0 . The equation for ˙ z gives the following possibilities z = ± 1 , θ = 0 , π/ 2 , π, 3 π/ 2 . The equation for ˙ a discards z = ± 1 . Case 1. sin θ = 0 ( θ = 0 , π ) . We show it is impossible. a, ˙ z = 0 and v = 0 from the equations for ˙ θ . But then the last equation (for ˙ z ) has v in the denom- inator.
(sin θ ) 2 = 1 Case 2. cos θ = 0 ( θ = π/ 2 , 3 π/ 2) ⇒ z/r = µ (1 − z 2 ) /v 3 v z µ r = v 2 After some simple manipulations we get two equilibria a = 0 , v 6 = r 2 µ 2 θ = π/ 2 , 3 π/ 2 , 2 √ √ 2 z = 2 / 2 (ie . φ = π/ 4) ⇒ M = µ 2 (0 , 1 , ± 1 )
Stability of equilibria The Jacobians at the two equilibrium are the same. In the ordering v, a, θ, z we have 0 1 0 0 √ − 3 µ 2 2 µ 2 − − µ 0 0 2 v 4 v 3 r J = √ 2 1 1 µ v 3 + 0 0 − r µ 2 µ v 2 1 0 0 0 2 v 2 µ
Characteristic polynomial To simplify the calculations we take r 2 µ 2 = 2. This makes v = 1. 3 µ 2 � � 1 p ( λ ) = λ 4 + λ 2 + 3 . + 2 µ 2 2 � 2 λ 2 = − ( .... ) + ( .... ) 2 − 12 where ... = 3 µ 2 1 + 2 µ 2 2
Stability analysis It is seen immediately that if the two λ 2 are real, then they are negative. This occurs for ∆ = ( 3 µ 2 2 µ 2 ) 2 − 12 > 0. 1 + 2 In this case the 4 eigenvalues λ are in the imaginary axis. The hamiltonian is linearly stable. When ∆ < 0 then we have a quadruple λ = ± α ± βi The equilibrium is of saddle-focus type.
Summary: stability analysis The two equilibria have the same eingenvalues. √ To facilitate, we took rµ = 2 ⇒ v = 1 . √ a = 0 , v = 1 , θ = π/ 2 , 3 π/ 2 , z = 2 / 2 ( φ = π/ 4) The equilibria are: • center-center for µ > µ ∗ • saddle-focus for 0 < µ < µ ∗ . where √ � 3 µ ∗ = 1 + 2 3 ∼ 1 . 46789 . √ At µ ∗ the eigenvalues are ± 4 3 i (double each).
We observe that the reconstructed curves in S 2 are two orthogonal great circles √ √ We have u 2 = ± µ 2 / 2 , v = 1 , r = 2 /µ . Hence ˙ R = RX with √ √ 0 2 / 2 2 / 2 0 − u 2 /v v/r ∓ µ µ √ X = u 2 /v 0 0 = ± µ 2 / 2 0 0 √ − v/r 0 0 − µ 2 / 2 0 0 i.e., steady rotations of angular velocity µ about √ √ 2 2 (0 , 2 ) 2 , ±
Let ( u x , u y , u z ) an unit vector and θ the rotation angle In our case, θ = µt and √ √ u x = 0 , u y = 2 / 2 , u z = ± 2 / 2 The last column of R (we want to multiply it by r ) is √ 2 sin θ, ± 1 2 2(1 − cos θ ) , 1 e 3 = ( 2(1 + cos θ ) ) .
√ √ x ( t ) = 1 2 2 µ ( sin( µ t ) , ± 2 (1 − cos( µ t )) , 2 (1 + cos( µ t )) ) √ √ 2 2 x ˙ = ( cos( µ t ) , ± sin( µ t ) , − 2 ( sin( µ t ) ) 2 √ √ 2 2 x ¨ = µ ( − sin( µ t ) , ± cos( µ t ) , − 2 ( cos( µ t ) ) 2 √ √ 2 2 x × ¨ ˙ x = µ ( 0 , 2 , ± 2 ) ⇒ perpendicular equators • We can apply any rotation to this pair of great circles. • Momentum along the special solutions: M = ˙ x × ¨ x .
When is a curve in the unit sphere a Riemannian cubic? A convenient representation of the cubic spline equations on the unit sphere was given by Crouch, Yan, Silva Leite, Brunnett, “On the construction of spline elements on spheres”, Proceedings of the European Control Conference 1993, Groningen, Netherlands, June 28–July 1, 1993, ed. by J.W. Nieuwenhuis, C. Praagman, H.L. Trentelman, vol. 4, pages 1930–1934. Given x ( t ) a curve on the sphere, compute: y = ˙ x , w = ˙ y + ( y, y ) x , z = ˙ w + ( w, y ) x The crucial test to be a Riemannian cubic on the sphere: z + ( z, y ) x − ( w, y ) y + ( y, y ) w ≡ 0 . ˙ Can be rewritten as a system for ˙ x, ˙ y, ˙ z, ˙ w . √ The great circle with µ = 2 so r = 1 passed.
Pretend you did not know (that curve is an equator) syms t real x = [ ( sqrt(2)/2)*sin(sqrt(2)*t); (1/2)*(1-cos(sqrt(2)*t)); (1/2)*(1+cos(sqrt(2)*t))]; y = diff(x,t); w = diff(y,t) + (y*y)*x; w = simplify(w); z = diff(w,t) + (w*y)*x; z = simplify(z); test = diff(z,t) + (z’*y)*x - (w’*y)*y + (y’*y)*w test = cos(2^(1/2)*t)*((2^(1/2)*sin(2^(1/2)*t)*sin((2^(1/2)*t)/2)^2)/2 - (2^(1/2)*cos((2^(1/2)*t)/2)^2*sin(2^(1/2)*t))/2 + (2^(1/2)*cos(2^(1/2)*t)*sin(2^(1/2)*t))/2) + 2^(1/2)*sin(2^(1/2)*t) - 2^(1/2)*sin(2^(1/2)*t)*(cos(2^(1/2)*t)^2 + sin(2^(1/2)*t)^2)
(cos(2^(1/2)*t)/2 - 1/2)*(cos(2^(1/2)*t)^2 + sin(2^(1/2)*t)^2) - cos(2^(1/2)*t) + cos((2^(1/2)*t)/2)^2*(cos(2^(1/2)*t)^2 + sin(2^(1/2)*t)^2) + (2^(1/2)*sin(2^(1/2)*t)*((2^(1/2)*sin(2^(1/2)*t)*sin((2^(1/2)*t)/2)^2)/2 - (2^(1/2)*cos((2^(1/2)*t)/2)^2*sin(2^(1/2)*t))/2 + (2^(1/2)*cos(2^(1/2)*t)*sin(2^(1/2)*t))/2))/2 cos(2^(1/2)*t) - (cos(2^(1/2)*t)/2 + 1/2)*(cos(2^(1/2)*t)^2 + sin(2^(1/2)*t)^2) + sin((2^(1/2)*t)/2)^2*(cos(2^(1/2)*t)^2 + sin(2^(1/2)*t)^2) - (2^(1/2)*sin(2^(1/2)*t)*((2^(1/2)*sin(2^(1/2)*t)*sin((2^(1/2)*t)/2)^2)/2 - (2^(1/2)*cos((2^(1/2)*t)/2)^2*sin(2^(1/2)*t))/2 + (2^(1/2)*cos(2^(1/2)*t)*sin(2^(1/2)*t))/2))/2 test = simplify(test) test = 0 0 0
GALLERY (some trajectories with µ in the stable regime numerically computed by Teresa Stuchi)
Projection in ( a, z ) √ √ µ = 2 , r = 2 / 2 , v = 1 . 4 , a = 0 . 5 , θ = π/ 2 , z = 2 / 2 Proj.(a,z), mu=2,r=(2)^(1/2)/2, v=1.4,a=0.5,theta=pi/2,z=2^(1/2)/2 0.84 "fort.9" u 2:4 0.82 0.8 0.78 0.76 0.74 0.72 0.7 0.68 0.66 0.64 -1.5 -1 -0.5 0 0.5 1 1.5
Projection in ( v, θ ) for the same initial conditions Proj.(v,theta), mu=2,r=(2)^(1/2)/2, v=1.4,a=0.5,theta=pi/2,z=2^(1/2)/2 2 "fort.9" u 1:3 1.9 1.8 1.7 1.6 1.5 1.4 1.3 1.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8
Projection in ( a, z ) √ √ µ = 1 . 5 , r = 2 , v = 1 . 001 , a = 0 . 1 , θ = 3 π/ 2 , z = 2 / 2 Projecao (a,z), mu=1.5,r=sqrt(2), c.i. v=1.001,a=0.1,theta=3pi/2,z=sqrt(2)/2 0.9 "fort.9"u 2:4 0.85 0.8 0.75 0.7 0.65 0.6 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8
3d Projection in ( v, θ, z ) (v,theta,z), mu=1.5,r=sqrt(2), c.i. v=1,a=0,theta=pi/2,z=sqrt(2)/2 "fort.9"u 1:3:4 0.9 0.85 0.8 0.75 0.7 0.65 0.6 2.2 2 1.8 0.6 1.6 0.7 0.8 0.9 1.4 1 1.1 1.2 1.2 1.3 1.4 1.5 1.6 1
Recall: setting for splines on S 2 so (3) ∗ ≡ so (3) ≡ ℜ 3 via the bi-invariant metric Bi-invariant metric is not even necessary, but since here there is no Legendre transform, it is convenient On an arbitrary Lie group, if we have a left invariant control problem can use a left invariant metric.) Lie algebra elements denoted Ω ∈ ℜ 3 , duals by M ∈ ℜ 3 Ω = (0 , v/r , u 2 /v ) M = ( M 1 , M 2 , M 3 ) The Hamiltonian family writes as H = − Cost + a u 1 + M 2 ( v/r ) + M 3 u 2 /v .
Minimal time with bounded acceleration (these results have not been checked C = 1 ⇒ H = − 1 + M 2 ( v/r ) + a u 1 + M 3 u 2 /v u ∗ 1 , u ∗ 2 = arg max ( a u 1 + M 3 u 2 /v ) , s . t . u 2 1 + u 2 2 ≤ U 2 � � a 2 + ( M 3 /v ) 2 , u ∗ a 2 + ( M 3 /v ) 2 ) u ∗ 1 = U a/ 2 = U M 3 / ( v Reduced Hamiltonian � H ∗ = − 1 + M 2 v/r + U a 2 + ( M 3 /v ) 2 Ω = da ∧ dv + ω corbit , ω corbit ( M )( v 1 , v 2 ) = det( M ; v 1 , v 2 )
Reduced equations for MTBA + U M 2 v = U a a = − M 2 3 ˙ √ ... , ˙ v 3 √ ... r M = M × ∂H ∗ ˙ ∂M ∂H ∗ ∂M = ( 0 , v/r , U M 3 v 2 √ ... )
+ U M 2 v = U a a = − M 2 3 ˙ ˙ √ ... , v 3 √ ... r M 1 = UM 2 M 3 − M 3 v ˙ v 2 √ ... r M 2 = − UM 1 M 3 ˙ v 2 √ ... M 3 = M 1 v ˙ r Casimir : M 2 1 + M 2 2 + M 2 3 = µ 2 √ ... = � a 2 + ( M 3 /v ) 2
We now use the usual spherical coordinates on the mo- mentum sphere with poles on the third axis M = µ ( cos φ cos θ , cos φ sin θ , sin φ ) ω c . o . = M 1 dM 2 ∧ dM 3 + M 2 dM 3 ∧ dM 1 + M 3 dM 1 ∧ dM 2 µ 3 cos φ dθ ∧ dφ = Ω = da ∧ dv + µ 3 cos φ dθ ∧ dφ � a 2 + µ 2 (sin φ ) 2 H = − 1 + µ r v cos φ sin θ + U v 2
(Relative) equilibria for MTBA The equation for ˙ v gives a = 0. The equation for ˙ a produces: M 2 3 M 2 /r = U ⇒ M 2 = r U v | M 3 | . | M 3 | /v ∂H ∗ U M 3 Imposing M � [= ( 0 , v/r , v 2 √ ... ) ] and using the Casimirity ∂M ( M 2 1 + M 2 2 + M 2 3 = µ 2 ) we get, after a short calculation, a = 0 , v 2 = rU √ √ 2 2 M 1 = 0 , M 2 = µ , M 3 = ± . 2 2 Translated into the a, v, θ, φ coordinates, √ a = 0 , v = rU , θ = π/ 2 , φ = ± π/ 4 .
Equations of motion in v, a, φ, θ U a v = ˙ � a 2 + µ 2 (sin φ ) 2 /v 2 (sin φ ) 2 a = − µ r cos φ sin θ + Uµ 2 ˙ � a 2 + µ 2 (sin φ ) 2 /v 2 v 3 φ = v cos θ ˙ µ 2 r r µ 2 v tan φ sin θ − U 1 sin φ ˙ θ = � a 2 + µ 2 (sin φ ) 2 /v 2 µ v 2
Jacobian at the equilibria (order v, a, φ, θ ) √ 3 2 √ r 2 U 0 0 0 µ √ 2 µ ± 2 µ − 0 0 √ 3 r A = U r 2 √ U 0 0 0 − µ 2 √ r √ ± 2 2 U 0 0 µ 2 √ r µ 2 r
Stability analysis The eigenvalues for the two equilibria are the same: √ � � λ 2 = U r [ − (1 + 1 � (1 + 1 µ 4 ) 2 − 4(1 + 2) � µ 4 ) ± µ 4 The conclusions here are similar to the cubic splines. We have either saddle-focus or linear stability. We will check the calculations (we simplified Matlab by hand) and make some simulations.
Conclusions for splines in T ∗ S 2 • We believe that the cubics in the sphere are completely integrable • We believe the minimal time, bounded curvature problem is nonintegrable (chaotic). For one reason, we suspect that there are arbi- trarily small initial perturbations a = ǫ such that solutions still run the circles, attaining zero veloc- ity and turning back (bang bang solutions). V. Jurdjevic (personal communication) seems sim- ilarities with the Delauney-Dubins problem.
Final remarks
And... how about L 1 bounded control? The “country cousin” of cubic and L inf In ℜ n , | u | 1 = � | u i | ⇒ get concatenated parabolas Actually L 1 makes good sense in robotics: the to- tal allowed effort is split among actuators. How to define L 1 In infinite dimensions or curved manifolds? A basis for decomposition seems to be needed.
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