On the comparison of short processes Jair Koiller Institute of - - PowerPoint PPT Presentation
Programme on Infinite-Dimensional Riemannian Geometry Riemannian geometry in shape analysis and computational anatomy ESI, Vienna, February 16-20 2015 On the comparison of short processes Jair Koiller Institute of Metrology, Quality and
Programme on Infinite-Dimensional Riemannian Geometry Riemannian geometry in shape analysis and computational anatomy ESI, Vienna, February 16-20 2015
Jair Koiller Institute of Metrology, Quality and Technology Paula Balseiro Universidade Federal Fluminense Alejandro Cabrera Universidade Federal do Rio de Janeiro Teresa Stuchi Universidade Federal do Rio de Janeiro
Supported by CAPES/PVE11-2012 with Darryl Holm/Imperial College
Martins Bruveris (Brunel, UK) Alex Castro (PUC-RJ, Brazil) Velimir Jurdjevic (Toronto) Maria Soledad Aronna (IMPA) Pierre Martinon (INRIA) Kurt Ehlers (TMCC, USA)
Richard Murray (Caltech) Fatima Leite (Coimbra) Ligia Abrunheiro (Aveiro)
Given (v1, q1) , (v2, q2) ∈ TQ Cubic: fix a time T, minimize
T
|∇ ˙
q ˙
q|2 dt L∞: fix a bound for |∇ ˙
q ˙
q|, minimize transition time. Heuristics: cubic splines can attain large normal accelerations (“physiologically” unattainable?). We give a simple 1d example. In order: comparisons in 2d manifolds (we will use INRIA/BOCOP optimization program)
for classification/statistical purposes.
anticipate long run effects?
From Xavier Pennec’s minicourse: two images (2 years apart ) from early stages of Alzheimer’s were extrapolated ±7years.
Sasaki or other natural metrics on TQ must be used if d(qo, q1) is short compared with |vo|, |v1| (thanks to F-X Vialard for this observation!).
anticipate long run effects? One hopes so specially at early stages of neurodegenerative illnesses (M. Miller)
Also for stochastic processes (S. Sommer)
Pennec, Arnaudon
Bookstein, Marsland
Miller, Pottmann, Schulz, Risser, Vialard, Allassonni` ere, Durrleman, Sundaramoorthi
(Ir` end, Barbara, Nina, Nikky, Nardi, Demoures) !! A good surprise (among many) for me: splines on TSn are not just toy models!!
T(TQ) and T ∗TQ using connections are known since 1995 (Andrew Lewis, thesis, Caltech)
to a split bundle when Q = G a Lie group or a par- allelizable manifold has been found more recently (Iyer, 2006, Abrunheiro et al., 2013) Here our aim is just to present a Hamiltonian formulation convenient for Computational Anatomy. For higher order splines: Gay-Balmaz, Holm, Meier, Ratiu, Vialard, Invariant higher-order variational problems (I,II)
Vector bundle with connection (A → Q, ∇) ∗ We pullback the symplectic structure of T ∗A to a split bundle S∇ (a Whitney sum over A). ∗ The symplectic form contains R∇ (curvature) ∗ Reduction of Lie group symmetries Comparing/matching short processes (A ⊂ TQ) ∗ Choice of an adequate cost functional ∗ Cubic Riemannian splines versus L∞ Example: T ∗(TS2−zero section) ≡ T ∗SO(3)×(ℜ+×ℜ). Integrability of the reduced systems?
Part 0: what the 1-dimensional problem can tell us?
¨ x = u , C = 1 2
T
u2du , T = fixed H = −u2 2 + pxv + pvu , Hu = −u + pv ⇒ u∗ = pv H∗ = p2
v
2 + pxv ⇒ ˙ x = v, ˙ v = pv , ˙ px = 0 , ˙ pv = −px
Cubic splines with final time T = 1. Example : x(0) = 0 , ˙ x(0) = −1 , x(1) = 1 , ˙ x(1) = 0 Optimal solution x = −t + 5t2 − 3t3 ⇒ ¨ x = 10 − 18t Range : ¨ x ∈ [−8, 10] Mean|acceleration| =
1
0 |10 − 18t|dt = 41/9 = 4.555555...
Cost : C = 1 2
1
0 (10 − 18t)2dt = 14
Comparing with L∞ acceleration bounded by 1 x =
−t + t2/2 , 0 ≤ t ≤ t∗ 1 − (t − T)2/2 , t∗ ≤ t ≤ T Forcing the splines to match, we get t∗ = 1 +
T = 1 + √ 6 ∼ 3.449 This is the classic bang bang soft landing problem two concatenated parabolas as splines
Cbd ctl = 1 2(1 + √ 6) ∼ 1.73 , Ccubic = 14
that 3.5 times longer than the cubic spline
but this does not not happen in dim ≥ 2.)
than the Riemannian cubic (using its criterion).
However, if we fix the time T = 1 also for the L∞, then its acceleration will be equal to (1 + √ 6)2 ∼ 11.9 (constant). This is more than twice than the mean acceleration of the cubic spline (=4.56). The cost of the bound control solution (using the crite- rion of the cubic spline) will be ˜ Cbd ctl = (1/2)(1 + √ 6)4 ∼ 70.8 This is 5 much more expensive than the minimum cost, achieved by the cubic spline.
Question: are high accelerations harmful? (here I mean mechanically/physiologically,
ation must be bounded (say by 1) The minimum transition time could the serve as a measure of “distance” between tangent vectors.
and the sphere. We will use the INRIA software
Thanks to Pierre Martinon and Soledad Aronna!
to a Whitney sum over A (in the third summand) S∇ = (T ∗Q ⊕ A∗) ⊕ A.
(A = TQ or T ∗Q, a subbundle of TQ, or even a Lie algebroid.)
Λvq ∈ T ∗
vqA
→ (pq , αq) (depending on vq) making easier to apply Pontryagin’s principle on control problems having A as state space.
suited for landmark cometrics (“Mario’s formulae”)
The splitting maps at vx ∈ A for TvxA and T ∗
vxA
φ∇ : TQ ⊕ A ⊕ A → TA , ψ∇ : T ∗Q ⊕ A∗ ⊕ A → T ∗A Z = φ∇(Xx, ux, vx) ∈ TvxA , Λ = ψ∇(px, αx, vx) ∈ T ∗
vxA
Z = hor∇(X)|v +
TvA
uvert|v Λ, Z|vx = px, Xx + αx, ux p ∈ T ∗
xQ,
α ∈ A∗|x separatedly dual to X ∈ TxQ, u ∈ Ax
(px, αx, vx) ∈ S∇ = T ∗Q ⊕ A∗ ⊕ A. For Z1, Z2 ∈ T(p,α,v)S∇ ΩS∇ (Z1, Z2) = ωT ∗Q(π1
∗(Z1), π1 ∗(Z2)) − α, R∇(πo ∗(Z1), πo ∗(Z2))(v)
+[ P˜
∇(π2
∗(Z2)) , P∇(π3 ∗(Z1)) − P˜
∇(π2
∗(Z1)) , P∇(π3 ∗(Z2)) ]
Curvature tensor of ∇: R∇ ∈ Ω2(Q; End(A)) Dual connection: ˜ ∇X(ℓ) = [ s → X(ℓ(s)) − ℓ(∇X(s)) ] P∇, P˜ ∇ vertical projections of ∇ and ˜ ∇ , respectively
For a curve t → (x, v)(t) ∈ A, we denote ∇ ˙
xv = (∇ ˙ xv)aea,
(∇ ˙
xv)a = ˙
va + Γa
ib ˙
xivb. Given a function H = H(xi, pi, va, αa) ∈ C∞(SA,∇) ˙ xi = ∂piH ˙ pi = −∂xiH + Γb
ia(va∂vbH − αb∂αaH) + Rb ija uaαb ˙
xj (∇ ˙
xv)a = ∂αaH ,
(˜ ∇ ˙
xα)a = −∂vaH
˙ xi = ∂piH ˙ pi = −∂xiH + Γb
ia(va∂vbH − αb∂αaH) + Rb ija uaαb ˙
xj (∇ ˙
xv)a = ∂αaH ,
(˜ ∇ ˙
xα)a = −∂vaH
The Poisson bracket on SA,∇ associated to ωSA,∇ is de- fined by the relation {f, g} = ωSA,∇(Xf, Xg) for f, g ∈ C∞(SA,∇). In coordinates, the non vanishing brackets are {xi, pj} = −δij {pi, pj} = Rb
ijavaαb
{va, αb} = −δab {pi, va} = −Γa
ibvb
{pi, αa} = Γc
iaαc
→ 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)/Gµ = T ∗(Q/G) ⊕ A/G ⊕ A∗/G ×A/G ˜
Oµ ⊂ SA,∇/G. (for all the details contact Paula Balseiro)
In favorable cases the action of G is free and proper on Ao = A − zero section.
short review on
Sectional Curvature in Terms of the Cometric with Applications to the Riemannian Manifolds of Landmarks, SIAM J. Imaging Sciences, 5:1, 394433, 2012
Correlations appear in G−1 just soup up the simple problem of 2 particles in 1-dimension 2H = p2
1 + p2 2 + 2κ(q1 − q2) p1p2
κ = exp(−|qi − qj|)
κ =
1 1+(|q1−qj|)2
κ = exp(−|qi − qj|2/σ2)
It is an integrable system!!!
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
In computing Christoffel symbols and curvature components: i) derivatives are taken only on upper indices (gµν’s, cometric) ii) lower indices gµν’s (metric) are obtained by inverting numerically the cometric matrix
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)
Next in order: T ∗(TS3) Note: connectors in homogenous spaces are intrinsically three dimensional!
A “short process” is an element of A = TQ. A connector is a smooth parametrized curve with prescribed tangent vectors in the ends.
T
q ˙
˙ q
q ˙
Riemannian cubics min
T
|∇ ˙
q ˙
q|2 dt (T fixed) Minimal time connector, bounded acceleration ∇ ˙
q ˙
q = u, |u| ≤ b (MTC/BA) [ Equivalently : Linf control (L. Noakes, 2014): Fix T, and then minimize {max |u|, 0 ≤ t ≤ T} ]
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)
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)
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 vo/|vo| to v1/v1 and the second joining v1/v1 to vo/|vo|. This is the “beginners race track”: a closed loop of arbitrary small geodesic curvature.
brake in the tangential direction so that you reach q1 with the desired speed. Note that you may need to do the loop several times!
(zero velocities) with a prescribes limit directions.
Assume that the control set (forces) contains a small ball around the origin at every TqQ.
start from q1 in the direction of −v1; do the bang bang switch at the right moment, so that when coming back you reach q1 with velocity v1.
Let S∇ the splitting of T ∗A. uq is the acceleration (the control) vq is the state to be controlled H = −C(vq, uq, t) + Λ, Z = −C(vq, uq, t) + pq · vq + αq · uq H∗(q, vq, pq, αq) = −C(vq, u∗
q, t) + pq · vq + αq · u∗
u∗ = u∗(vq, αq, t) = arg max [−C(vq, uq, t) + αq · uq ]
more conveniently given in terms of momenta
form on S∇ ≡ T ∗A.
appears naturally. Using S∇ the curvatures appear explicitly in the equations. Playground for “Mario’s formulae” !
¯ g ≡ g−1 co-metric induced in T ∗Q. H = −1 2|ux|2 + px · vx + αx · ux then : u∗
x = ¯
g♯(α) H∗(x, vx, px, αx) = px · vx + 1 2 ¯ g(α, α) ˙ x = v , ∇ ˙
xv = ¯
g♯(α) ˜ ∇ ˙
xp = −i ˙ xα, R∇(v) ,
˜ ∇ ˙
xα = −p
,
These Hamiltonian equations for (x, v, p, α)(t) ∈ SA,∇ are equivalent to the following second-order equa- tions for (x, α)(t) ∈ T ∗Q: ∇ ˙
x ˙
x = ¯ g♯(α) , ˜ ∇ ˙
x(˜
∇ ˙
xα) = i ˙ xα, R∇( ˙
x) ,
H∗ = −1 + p, v + b
g(α, α) ˙ x = v , ∇ ˙
xv =
b
g(α, α) ¯ g♯(α) ˜ ∇ ˙
xp = −i ˙ xα, R∇(v) ,
˜ ∇ ˙
xα = −p
,
These Hamiltonian equations for (x, v, p, α)(t) ∈ SA,∇ are equivalent to the following second-order equa- tions for (x, α)(t) ∈ T ∗Q: ∇ ˙
x ˙
x = b
g(α, α) ¯ g♯(α) , ˜ ∇ ˙
x(˜
∇ ˙
xα) = i ˙ xα, R∇( ˙
x) ,
Are they Arnold-Liouville integrable?
T ∗TS2 = 8 , with SO(3) symmetry Removing the zero section, and reducing we get S2 × (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).
As integrable as it could be...
SE(2) reduction. We can solve directly. An expression for the extra integral is unknown.
the problem where the terminal heading is prescribed but the terminal position is free. They found integrals of motion for cost functions C = |v|ndt.
State variables R ∈ SO(3), v ∈ ℜ+ Control variables u1, u2 State equations (4 dimensional) ˙ v = u1 ˙ R = RX , X =
−u2/v v/r u2/v −v/r
Let ei , i = 1, 2, 3 the columns of R. The last one gives the normals to a sphere of radius r: e3 = q/r. First column: direction of tangent vector, ˙ q = w = v e1 The second column (e2 = e3 × e1) is the unit normal to the path in the sphere, normal inside surface = q r × w v
u1 = scalar acceleration, i.e, ˙ v = u1 . The last column of X gives ˙ q/r = (v/r) e1 or ˙ q = w . Differentiating ˙ q = w = v e1 we get ¨ q = u1 e1 + v ˙ e1 = u1 e1 + u2 e2 − (v2/r) e3 The last term is due to the constraining normal force, comes from differentiating ( ˙ q , q ) ≡ 0. Thus ˙ e1 = (u2/v) e2 − (v/r) e3 .
Why X23 = −X32 = 0 ? Reason: spherical curves have τg ≡ 0 ! There is a general formula: τg = (κ1−κ2) 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.
Key: Gauss : q ↔ N(q) ∈ S2 is a bijection. The third column of R is again a vector N ∈ S2. So the state space remains the same, SO(3) × ℜ+ The equations of motion live in T ∗(SO(3) × ℜ+) = (SO(3) × (ℜ3)∗) × (ℜ+ × ℜ) Everything will be coupled.
˙ R = R X X =
−u2/v v B( t, t) u2/v v τg( t) −v B( t, t) −v τg( t)
Here B = the second fundamental form. Since X depends on t, the SO(3) symmetry is destroyed. S1 symmetry still remains for surfaces of revolution.
ΩM = da ∧ dv + ωT∗SO(3) (a = costate of v)
Recall: symplectic form in T ∗G ≡ G × G∗ at a point (g, µ) For tangent vectors (Xi , wi) ∈ TgG × G∗, i = 1, 2 ωT∗G
(g,µ)((X1, w1) , (X2, w2)) = w1 · Lg−1|∗ X2 − w2 · Lg−1|∗X1−
− µ Lg−1|∗[X1, X2]
Family of left invariant Hamiltonians in T ∗SO(3) × T ∗ℜ+ ≡ (SO(3) × ℜ3) × (ℜ+ × ℜ) . Lie algebra elements: Ω ∈ ℜ3, duals M ∈ ℜ3 Ω = (0 , v/r , u2/v) M = (M1, M2, M3) The Hamiltonian family writes as H = −Cost + a u1 + M2 (v/r) + M3 u2/v .
C = 1 2 ( u2
1 + u2 2)
⇒ u∗
1 = a , u∗ 2 = M3/v
Reduced Hamiltonian H∗ = 1 2 a2 + 1 2 (M3/v)2 + M2 v/r Ω = da ∧ dv + ωcorbit ωcorbit(M)(v1, v2) = det(M; v1, v2)
˙ v = a , ˙ a = −M2/r + M2
3/v3
˙ M = M × (0, v/r, M3/v2) , Casimir : M2
1 + M2 2 + M2 3 = µ2
,
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 ) ωc.o. = M1 dM2 ∧ dM3 + M2 dM3 ∧ dM1 + M3 dM1 ∧ dM2 = µ3 cos φ dφ ∧ dθ Ω = da ∧ dv + µ3 cos φ dφ ∧ dθ H = 1 2 a2 + µ2 2 (cos φ sin θ)2/v2 + µ sin φ (v/r)
It is convenient to introduce z = sin φ , −1 < z < 1 Ω = da ∧ dv + µ3 dz ∧ dθ H = 1 2 a2 + µ2 2 (1 − z2) (sin θ)2/v2 + 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?
˙ v = a , ˙ a = −µz/r + µ2(1 − z2) (sin θ)2/v3 ˙ θ = v µ2 r − z (sin θ)2 µ v2 ˙ z = − sin θ cos θ 1 − z2 µ v2
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.
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?
The equation for ˙ v = 0 gives a = 0. The equation for ˙ z gives the following possibilities z = ±1 , θ = 0, π/2, π, 3π/2 . The equation for ˙ a discards z = ±1. Case 1. sin θ = 0 (θ = 0, π). We show it is impossible. z = 0 and v = 0 from the equations for ˙ a, ˙ θ. But then the last equation (for ˙ z) has v in the denom- inator.
Case 2. cos θ = 0 (θ = π/2, 3π/2) ⇒ (sin θ)2 = 1 z/r = µ(1 − z2)/v3 v µ r = z v2 After some simple manipulations we get two equilibria a = 0 , v6 = r2 µ2 2 , θ = π/2, 3π/2 z = √ 2/2 (ie. φ = π/4) ⇒ M = µ √ 2 2 (0, 1, ±1 )
The Jacobians at the two equilibrium are the same. In the ordering v, a, θ, z we have J =
1
−3 µ2 2v4 − √ 2 µ2 v3
− µ
r √ 2 µ v3 + 1 r µ2
−
1 µ v2 1 2v2µ
To simplify the calculations we take r2µ2 = 2. This makes v = 1. p(λ) = λ4 +
2 + 1 2µ2
2λ2 = − (....) +
where ... = 3 µ2 2 + 1 2 µ2
It is seen immediately that if the two λ2 are real, then they are negative. This occurs for ∆ = (3 µ2
2
+
1 2 µ2)2 − 12 > 0.
In this case the 4 eigenvalues λ are in the imaginary
When ∆ < 0 then we have a quadruple λ = ±α ± βi The equilibrium is of saddle-focus type.
The two equilibria have the same eingenvalues. To facilitate, we took rµ = √ 2 ⇒ v = 1. a = 0, v = 1, θ = π/2, 3π/2, z = √ 2/2 (φ = π/4) The equilibria are:
µ∗ =
√ 3 3 ∼ 1.46789 . At µ∗ the eigenvalues are ± 4 √ 3 i (double each).
We have u2 = ±µ √ 2/2 , v = 1 , r = √ 2/µ. Hence ˙ R = RX with X =
−u2/v v/r u2/v −v/r
=
∓µ √ 2/2 µ √ 2/2 ±µ √ 2/2 −µ √ 2/2
i.e., steady rotations of angular velocity µ about (0 , √ 2 2 , ± √ 2 2 )
Let (ux, uy, uz) an unit vector and θ the rotation angle In our case, θ = µt and ux = 0 , uy = √ 2/2 , uz = ± √ 2/2 The last column of R (we want to multiply it by r) is e3 = ( √ 2 2 sin θ, ±1 2(1 − cos θ) , 1 2(1 + cos θ) ) .
x(t) = 1 µ ( sin(µ t), ± √ 2 2 (1 − cos(µ t)) , √ 2 2 (1 + cos(µ t)) ) ˙ x = ( cos(µ t), ± √ 2 2 sin(µ t) , − √ 2 2 ( sin(µ t) ) ¨ x = µ( − sin(µ t), ± √ 2 2 cos(µ t) , − √ 2 2 ( cos(µ t) ) ˙ x × ¨ x = µ ( 0, √ 2 2 , ± √ 2 2 ) ⇒ perpendicular equators
x × ¨ x .
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.
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 =
(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
0.64 0.66 0.68 0.7 0.72 0.74 0.76 0.78 0.8 0.82 0.84
0.5 1 1.5 Proj.(a,z), mu=2,r=(2)^(1/2)/2, v=1.4,a=0.5,theta=pi/2,z=2^(1/2)/2 "fort.9" u 2:4
Projection in (v, θ) for the same initial conditions
1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 Proj.(v,theta), mu=2,r=(2)^(1/2)/2, v=1.4,a=0.5,theta=pi/2,z=2^(1/2)/2 "fort.9" u 1:3
Projection in (a, z) µ = 1.5, r = √ 2, v = 1.001, a = 0.1, θ = 3π/2, z = √ 2/2
0.6 0.65 0.7 0.75 0.8 0.85 0.9
0.2 0.4 0.6 0.8 Projecao (a,z), mu=1.5,r=sqrt(2), c.i. v=1.001,a=0.1,theta=3pi/2,z=sqrt(2)/2 "fort.9"u 2:4
0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1 1.2 1.4 1.6 1.8 2 2.2 0.6 0.65 0.7 0.75 0.8 0.85 0.9 (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
so(3)∗ ≡ so(3) ≡ ℜ3 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 ∈ ℜ3, duals by M ∈ ℜ3 Ω = (0 , v/r , u2/v) M = (M1, M2, M3) The Hamiltonian family writes as H = −Cost + a u1 + M2 (v/r) + M3 u2/v .
C = 1 ⇒ H = −1 + M2 (v/r) + a u1 + M3 u2/v u∗
1, u∗ 2 = arg max ( a u1 + M3 u2/v ) , s.t. u2 1 + u2 2 ≤ U2
u∗
1 = U a/
2 = U M3/( v
Reduced Hamiltonian H∗ = −1 + M2 v/r + U
Ω = da∧dv+ ωcorbit , ωcorbit(M)(v1, v2) = det(M; v1, v2)
˙ v = U a √... , ˙ a = −M2 r + U M2
3
v3 √... ˙ M = M × ∂H∗ ∂M ∂H∗ ∂M = ( 0 , v/r , U M3 v2 √... )
˙ v = U a √... , ˙ a = −M2 r + U M2
3
v3 √... ˙ M1 = UM2M3 v2√... − M3v r ˙ M2 = −UM1M3 v2√... ˙ M3 = M1v r Casimir : M2
1 + M2 2 + M2 3 = µ2
√... =
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 ) ωc.o. = M1 dM2 ∧ dM3 + M2 dM3 ∧ dM1 + M3 dM1 ∧ dM2 = µ3 cos φ dθ ∧ dφ Ω = da ∧ dv + µ3 cos φ dθ ∧ dφ H = −1 + µ r v cos φ sin θ + U
v2
The equation for ˙ v gives a = 0. The equation for ˙ a produces: M2/r = U M2
3
|M3|/v ⇒ M2 = r U v |M3| . Imposing M
∂H∗ ∂M
[= ( 0 , v/r ,
U M3 v2 √... ) ] and using the Casimirity
(M2
1 + M2 2 + M2 3 = µ2) we get, after a short calculation,
a = 0 , v2 = rU M1 = 0 , M2 = µ √ 2 2 , M3 = ± √ 2 2 . Translated into the a, v, θ, φ coordinates, a = 0 , v = √ rU , θ = π/2 , φ = ±π/4 .
˙ v = U a
˙ a = −µ r cos φ sin θ + Uµ2 (sin φ)2 v3
˙ φ = v cos θ µ2 r ˙ θ = 1 r µ2 v tan φ sin θ − U µ sin φ v2
A =
√ 2 U
3 2 √r
µ
−
√ 2 µ √ U r
3 2
±2 µ
r
−
√ U µ2 √r
± 2
µ2 r 2 √ U µ2 √r
The eigenvalues for the two equilibria are the same: λ2 = U r [−(1 + 1 µ4) ±
µ4)2 − 4(1 + √ 2) µ4 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.
completely integrable
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).
ilarities with the Delauney-Dubins problem.
The “country cousin” of cubic and Linf In ℜn, |u|1 = |ui| ⇒ get concatenated parabolas Actually L1 makes good sense in robotics: the to- tal allowed effort is split among actuators. How to define L1 In infinite dimensions or curved manifolds? A basis for decomposition seems to be needed.
correlation cometrics on Q = (ℜd)n Playground for “Marios formulae”
Ω∇ = the pullback of ΩT ∗(TQ) to S∇ make sense for infinite dimensions?
References for connection and curvatures in spaces of diffeomorphims.
Martin Bauer, Martins Bruveris, Peter W. Michor Overview of the Geometries of Shape Spaces and Diffeomorphism Groups Journal of Mathematical Imaging and Vision, 50: 1/2, 60–97, 2014
Sobolev metrics on diffeomorphism groups and the derived geometry of spaces of submanifolds Izvestiya: Mathematics 77:3 541–570, 2013
Curvatures of Sobolev Metrics on Diffeomorphism Groups Pure and Applied Mathematics Quarterly 9:2, 291–332, 2013 Geometry of Diffeomorphism Groups, Complete integrability and Geometric statistics Geometric and Functional Analysis, 23:1, 334–366, 2013