Conjugate loci of Riemannian metrics on 2D manifolds related to the - - PowerPoint PPT Presentation

Conjugate loci of Riemannian metrics on 2D manifolds related to the Euler top problem

Nataliya Shcherbakova

Laboratoire de Génie Chimique, ENSIACET - INP Toulouse L2S seminar, April 4th, 2013

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 1 / 45

Collaborators

Bernard Bonnard, Institut de Mathématiques de Bourgogne; Olivier Cots, INRIA Sophia-Antipolis, McTao; Jean-Baptiste Pomet, INRIA Sophia-Antipolis, McTao

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 2 / 45

Related Publications

• B. Bonnard, O. Cots, J.-B. Pomet, N. Shcherbakova. Riemannian metrics on 2D manifolds

related to the Euler-Poinsot rigid body problem (2012, preprint);

• B. Bonnard, O. Cots, N. Shcherbakova. The Serret-Andoer Riemannian metric and

Euler-Poinsot rigid body motion (2012, to appear in Math. Control and Relat. Fields);

• H. Yuan, R. Zeier, N. Khaneja, S. Lloyd. Elliptic functions and efficient control of Ising

spin chains with unequal couplings. Ph. Rev. A 77, 032340, 2008;

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 3 / 45

Plan

Euler’s top problem: classical formulation, optimal control model and Euler’s equations. Integrability via the Serret-Andoyer trasformation. The Seret-Andoyer metric on a 2D surface : geodesics, normal form and conjugate locus. Optimal control of a linear chain of three spin- 1

2 particles as a sub-Riemannian version of

the Euler top problem. First results on the conjugate locus of the tree spins problem. Perspectives.

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 4 / 45

The Euler top problem

Classical Mechanics viewpoint: a rigid body, modeled by its inertia ellipsoid with principal momenta of inertia I1, I2 and I3, freely rotates about its center of mass, which is fixed. Optimal control formulation: the Euler’s top rotations are the extremals of the optimal control problem on SO(3): (P) ˙ R(t) = R(t)   −u3 u2 u3 −u1 −u2 u1   =

3

• i=1

uiAi(R(t)), 1 2

T

• 3
• i=1

u2

i Ii → min u(·) ,

T − fixed, where R(t) ∈ SO(3) and Ai ∈ so(3) are the elements of the orthonormal basis A1 =   −1 1   , A2 =   1 −1   , A3 =   −1 1   verifying [A1, A2] = A3, [A2, A3] = A1, [A3, A1] = A2.

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 5 / 45

Euler’s equations and integrability

Assume R(0) and R(T) are fixed. Denote Hi = p(Ai(R)) and Hu =

3

• i=1

uiHi − ν

3

• i=1

Iiu2

i ,

ν ∈ {0, 1}. Pontryagin’s maximal principle = ⇒ ν = 1, ui = HiI−1

i

and optimal trajectories are projections of the extremals, i.e., solutions of the Hamiltonian system associated to Hn = 1 2 H2

1

I1 + H2

2

I2 + H2

3

I3

• .

Since ˙ Hi = {Hn, Hi}, we get Euler’s equations: dH1 dt = H2H3 1 I3 − 1 I2

• ,

dH2 dt = H1H3 1 I1 − 1 I3

• ,

dH3 dt = H1H2 1 I2 − 1 I1

• .

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 6 / 45

Euler’s equations

Observations: In Classical Mechanics H = (H1, H2, H3) describes the vector of the angular momentum

• f the body written in the moving frame related to the principal axes of inertia of the body,

while u = (u1, u2, u3) is the angular velocity vector. Euler’s equations are integrable by quadratures using two first integrals: the Hamiltonian Hn = h and H2 = H2

1 + H2 2 + H2

• 3. Solutions to Euler’s equations can be seen as the

curves on the energy ellipsoid formed by its intersection with the sphere of the constant angular momentum called polhodes in Classical Mechanics. If I3 > I2 > I1, the physically realizable motions take place iff h ∈

• H2

2I1 , H2 2I3

• .

The full system on SO(3) is integrable by quadratures.

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 7 / 45

Polhodes on the energy ellipsoid

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 8 / 45

Full system

Observations: Remaining equations of motion can be obtained using some suitable coordinates on SO(3), for instance, the Euler angles Φi, i = 1, 2, 3 so that R = exp(Φ1A3) ◦ exp(Φ2A2) ◦ exp(Φ3A3). A complete picture of motion can be seen as the rotation of the energy ellipsoid on a fixed plane (Poinsot model), where the point of contact on the ellipsoid moves along a polhode, while its trace on the plane is called a herpolhode:

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 9 / 45

Serret-Andoer’s transformation

The Serret-Andoyer coordinates is a set of symplectic coordinates x, y, z, px, py, py, defined by H1 =

• p2

x − p2 y sin y,

H2 =

• p2

x − p2 y cos y,

H3 = py, Then Hn = Ha = 1 2 sin2 y I1 + cos2 y I2

• (p2

x − p2 y) + p2 y

2I3 . x and z are cyclic variables, px and pz are first integrals with px = H, and moreover, z = const. The dynamics on the (x, y) plane is described by solutions to the system dx dt = px(A sin2 y + B cos2 y), dpx dt = 0, dy dt = py(C − A sin2 y − B cos2 y), dpy dt = (B − A)(p2

x − p2 y) sin y cos y.

where A = 1 I1 , B = 1 I2 , C = 1 I3 .

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 10 / 45

Serret-Andoer’s transformation

Observations : The Serret-Andoer transformation is not fiber-preserving. The Serret-Andoyer variables are not the action-angle variables. A further transformation using the Hamilton - Jacobi method leads to the standard action-angle representation (Yu. Sadov 1970, H. Konoshita 1972). If A < B < C, then Ha defines a Riemannian metric on a 2D surface: ga = 2 w(y)dx2 + 2 2C − w(y)dy2, where w(y) = 2(A sin2 y + B cos2 y).

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 11 / 45

Extremals of Serret-Andoer metric

(∗) dx dt = px(A sin2 y + B cos2 y), dpx dt = 0, dy dt = py(C − A sin2 y − B cos2 y), dpy dt = (B − A)(p2

x − p2 y) sin y cos y.

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 12 / 45

Extremals of the Serret-Andoer metric

(∗) dx dt = px(A sin2 y + B cos2 y), dpx dt = 0, dy dt = py(C − A sin2 y − B cos2 y), dpy dt = (B − A)(p2

x − p2 y) sin y cos y.

Periodicity: Ha(y, py) = Ha(y + π, py); Symmetry: Ha(y, py) = Ha(y, −py), Ha(y, py) = Ha(−y, py); The phase portrait on the (y, py)-plane is of pendulum type with stable equilibrium at y = π

2 (2h = Ap2 x) and unstable ones at y = 0 mod π (2h = Bp2 x). Thus there are two types

• f periodic trajectories: oscillations (Ap2

x < 2h < Bp2 x)) and rotations (Bp2 x < 2h ≤ Cp2 x)),

and separatrices (2h = Bp2

x).

y(t) describes the trajectory on the 0 energy level set of the natural mechanical system 1 2 ˙ y2 +

• C − w(y)

2 p2

xw(y)

2 − 2h

• = 0.

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 13 / 45

Extremals of the Serret-Andoer metric

(∗) dx dt = px(A sin2 y + B cos2 y), dpx dt = 0, dy dt = py(C − A sin2 y − B cos2 y), dpy dt = (B − A)(p2

x − p2 y) sin y cos y.

Periodicity: Ha(y, py) = Ha(y + π, py); Symmetry: Ha(y, py) = Ha(y, −py), Ha(y, py) = Ha(−y, py); The phase portrait on the (y, py)-plane is of pendulum type with stable equilibrium at y = π

2 (2h = Ap2 x) and unstable ones at y = 0 mod π (2h = Bp2 x). Thus there are two types

• f periodic trajectories: oscillations (Ap2

x < 2h < Bp2 x)) and rotations (Bp2 x < 2h ≤ Cp2 x)),

and separatrices (2h = Bp2

x).

y(t) describes the trajectory on the 0 energy level set of the natural mechanical system 1 2 ˙ y2 +

• C − w(y)

2 p2

xw(y)

2 − 2h

• = 0.

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 13 / 45

Extremals of the Serret-Andoer metric

• Proposition. Set ξ1 =

2h−Ap2

x

p2

x(B−A) , ξ3 = C−A

B−A . Then trajectories starting at the point (x0, y0) can be

parametrized as follows: i). oscillating trajectories: cos2 y(t) = ξ1cn2(Mt + ψ0|m) 1 − ξ1sn2(Mt + ψ0|m) , x(t) − x0 = px

• Bt − (B − A)(1 − ξ1)

M Π(ξ1|am(Mτ + ψ0|m)|m)

• τ=t

τ=0

• ,

m = ξ1(ξ3 − 1) ξ3 − ξ1 , M = (B − A)px

• ξ3 − ξ1,

cn(ψ0|m) = (1 − ξ1) cos2 y0 ξ1 sin2 y0 ; ii). rotating trajectories: cos2 y(t) = ξ1cn2(Mt + ψ0|m) ξ1 − sn2(Mt + ψ0|m) , x(t) − x0 = px

• (A + (B − A)ξ1)t − (B − A)(ξ1 − 1)

M Π( 1 ξ1 |am(Mτ + ψ0|m)|m)

• τ=t

τ=0

• ,

m = ξ3 − ξ1 ξ1(ξ3 − 1) , M = (B − A)px

• ξ1(ξ3 − 1),

cn(ψ0|m) = (ξ1 − 1) cos2 y0 ξ1 − cos2 y0 .

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 14 / 45

Extremals of the Serret-Andoer metric

Solutions of (∗) are quasi-periodic trajectories on the plane (x, y) for all h ∈ [

Ap2

x

2 , Cp2

x

2 ] \ { Bp2

x

2 }, the y-variable being 4K(m) M

• periodic along oscillating

trajectories and 2K(m)

M

• periodic along rotations;

Two oscillating trajectories of (∗) issued from the same initial point (x0, y0) with initial conditions (px, py(0)) and (px, −py(0)) intersect after a half period 2K(m)

M

at y( 2K(m)

M

) = π − y0.

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 15 / 45

Extremals of the Serret-Andoer metric

• Definition. The time t∗ is called conjugate to t0 = 0 if the differential of the end-point mapping

expt

(x0,y0) : (px(0), py(0)) → (x(t), y(t)),

degenerates at t = t∗. The point (x(t∗), y(t∗)) is said conjugate to (x0, y0). We denote t1

∗ = min |t∗| the first conjugate (to 0) time.

Theorem. Conjugate times along oscillating trajectories of (∗) are solutions to Bξ1Mt − (Bξ1 + A(1 − ξ1))

Mt+ψ0

• ψ0

du sn2(u|m)) = 0. Moreover, 2K(m)

M

≤ t1

∗ < 3K(m) M

, and t1

∗ = 2K(m) M

along trajectories starting with py(0) = 0. Rotating trajectories contain no conjugate points; min

px∈ √ 2h

A ,√ 2h B

t1 ∗ = π √ A

√

2h(B−A)(C−A),

lim

px→√ 2h

B

t1

∗ = +∞.

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 16 / 45

The polar form of Serret-Andoyer meric

Let ϕ : dy

• C − (A sin2 y + B cos2 y)

= dϕ. N.B. Since all phase trajectories pass trough the point y = π

2 , one can set y(0) = π 2 , which

leads to sin y = cn(αϕ|k), where k = (B − A) (C − A), α = √ C − A. Proposition. ga admits the Darboux normal form dϕ2 + m(ϕ)dθ2, where θ ≡ x and m(ϕ) = 2 w(ϕ) = (Acn2(αϕ|k) + Bsn2(αϕ|k))−1 ∈ [ √ A−1, √ B−1]. The Gauss curvature of ga along arc-length parametrized geodesics (h = 1

2):

G(ϕ) = (A + B + C)(w(ϕ) − w−)(w(ϕ) − w+) w(ϕ)2 , w± = 2(AB + AC + BC ±

• (AB + AC + BC)2 − 3ABC(A + B + C))

A + B + C .

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 17 / 45

The polar form of Serret-Andoyer meric

1 1

K k Α K k Α

• BA CA

A

Gmin G

Gmax = (B − A)(C − A) A , Gmin = −(A + B + C)(w+w−)2 4w−w+ .

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 18 / 45

The polar form of Serret-Andoyer meric

Π 2Π Θ

• K k

Α K k Α

• Nataliya Shcherbakova (LGC-ENSIACET-INPT)

April 4, 2013 19 / 45

Jacobi fields of Darboux-type metrics

• Definition. Let γ(·) ∈ T∗M, be an integral curve of the Hamiltonian vector field
• H ∈ Vec(T∗M) issued from γ0, where M is a smooth manifold, and et

H is the Hamiltonian flow

generated in T∗M by

• H. Then

J(·) ∈ Tγ(·)(T∗M) : J(t) = et

H ∗ J(0),

J(t) = 0 is called the Jacobi field along γ. 2D Darboux-type metrics. Consider the metric g = dϕ2 + m(ϕ)dθ2, m(ϕ) > 0. Assume that (A1) ϕ = 0 is a parallel solution (i.e. m′(0) = 0); (A2) m(ϕ) = m(−ϕ), and m′′(0) > 0. On the level set h = 1

2 in the neighborhood of ϕ = 0 there is a one-parametric family of

periodic solutions to the equation ˙ ϕ2 + p2

θµ(ϕ) = 1, which describes the evolution of ϕ

along arc-length parametrized geodesics. The Hamiltonian H = 1

2

• p2

ϕ + p2 θµ(ϕ)

• , µ(ϕ) = 1/m(ϕ), and any extremal is a solution

to (∗∗) ˙ ϕ = pϕ, ˙ pϕ = −1 2p2

θµ′(ϕ),

˙ θ = pθµ(ϕ), ˙ pθ = 0.

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 20 / 45

Jacobi fields of Darboux-type metrics

Let J(t) = (δ1(t), δ2(t), δ3(t), δ4(t)), with δ1 = δpϕ, δ2 = δpθ, δ3 = δϕ, δ4 = δθ. Then ˙ δ1 = −pθµ′(ϕ)δ2 − p2

θµ′′(ϕ)

2 δ3, ˙ δ2 = 0, ˙ δ3 = δ1, ˙ δ4 = µ(ϕ)δ2 + pθµ′(ϕ)δ3.

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 21 / 45

Jacobi fields of Darboux-type metrics

Let J(t) = (δ1(t), δ2(t), δ3(t), δ4(t)), with δ1 = δpϕ, δ2 = δpθ, δ3 = δϕ, δ4 = δθ. Then ˙ δ1 = −pθµ′(ϕ)δ2 − p2

θµ′′(ϕ)

2 δ3, ˙ δ2 = 0, ˙ δ3 = δ1, ˙ δ4 = µ(ϕ)δ2 + pθµ′(ϕ)δ3.

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 22 / 45

Jacobi fields of Darboux-type metrics

Let J(t) = (δ1(t), δ2(t), δ3(t), δ4(t)), with δ1 = δpϕ, δ2 = δpθ, δ3 = δϕ, δ4 = δθ. Then ˙ δ1 = −pθµ′(ϕ)δ2 − p2

θµ′′(ϕ)

2 δ3, ˙ δ2 = 0, ˙ δ3 = δ1, ˙ δ4 = µ(ϕ)δ2 + pθµ′(ϕ)δ3. ⇓ ¨ δ3(t) + p2

θµ′′(ϕ(t))

2 δ3(t) = −pθµ′(ϕ(t))δ2.

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 23 / 45

Jacobi fields of Darboux-type metrics

• Proposition. The vector fields J1 =

H, J2 = ∂θ, J3 = pϕ(t)∂ϕ + pθ∂θ + t H, and J4 = (δ1(t), δ2(t), δ3(t), δ4(t)), where w1(t) = p3

θµ′(ϕ(t))

2 Λ(ϕ(t)) − pθµ(ϕ(t)) pϕ(t) , w3(t) = −pθpϕ(t)Λ(ϕ(t)), w4 = p2

ϕ(t)Λ(ϕ(t)),

with Λ(ϕ) =

ϕ

• ϕ0

µ( ¯ ϕ) p3

ϕ( ¯

ϕ)d ¯ ϕ, pϕ(ϕ) =

• 2h − p2

θµ(ϕ),

form a well-defined basis of Jacobi fields along any given solution of (∗∗). N.B. The vectors J3(0), J4(0) are vertical, i.e., Ji(0) ∈ Tγ0(T∗

π(γ0)M), where π : T∗M → M,

while J1(0), J2(0) are horizontal.

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 24 / 45

Jacobi fields of Darboux-type metrics

• Proposition. The vector fields J1 =

H, J2 = ∂θ, J3 = pϕ(t)∂ϕ + pθ∂θ + t H, and J4 = (δ1(t), δ2(t), δ3(t), δ4(t)), where w1(t) = p3

θµ′(ϕ(t))

2 Λ(ϕ(t)) − pθµ(ϕ(t)) pϕ(t) , w3(t) = −pθpϕ(t)Λ(ϕ(t)), w4 = p2

ϕ(t)Λ(ϕ(t)),

with Λ(ϕ) =

ϕ

• ϕ0

µ( ¯ ϕ) p3

ϕ( ¯

ϕ)d ¯ ϕ, pϕ(ϕ) =

• 2h − p2

θµ(ϕ),

form a well-defined basis of Jacobi fields along any given solution of (∗∗). N.B. The vectors J3(0), J4(0) are vertical, i.e., Ji(0) ∈ Tγ0(T∗

π(γ0)M), where π : T∗M → M,

while J1(0), J2(0) are horizontal.

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 24 / 45

Jacobi fields of Darboux-type metrics

• Corollary. For a Darboux type metric g the conjugate (to 0) times along any periodic

trajectory issued from the point (θ0, ϕ0) with pϕ(0) = 0 are solutions to the equation Λ(ϕ(t)) = 0. The points (θ, ±¯ φ), where ±¯ φ are the extremities of the variation of ϕ, are conjugate to each other.

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 25 / 45

Jacobi fields of Darboux-type metrics

• Corollary. For a Darboux type metric g the conjugate (to 0) times along any periodic

trajectory issued from the point (θ0, ϕ0) with pϕ(0) = 0 are solutions to the equation ∂θ(ϕ, pθ) ∂pθ = 0. The points (θ, ±¯ φ), where ±¯ φ are the extremities of the variation of ϕ, are conjugate to each other.

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 26 / 45

Conjugate locus of the Serret-Andoyer metric

• Theorem. The first conjugate locus to a point (0, ϕ0) of the Serret-Andoyer metric ga

is formed by the conjugate points along oscillating trajectories only. It consists of two components, symmetric with respect to the vertical line θ = 0. Each component of the locus is formed by two smooth branches, which asymptotically tend to the horizontal lines ϕ = ± K(k)

α , and form a unique horizontal cusp on the line ϕ = −ϕ0. Moreover,

along arc-length parametrized geodesics (h = 1/2) min

ϕ0 θ∗(ϕ0) = θ∗(0) =

πA √ A

• (B − A)(C − A)

.

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 27 / 45

Conjugate locus of the Serret-Andoyer metric

The right-hand side (pθ > 0) components of the conjugate locus on the (x, y) plane:

2pi 3pi 4pi 5pi y0 pi/2 pi−y0 pi x y

A = 1.5, B = 2, C = 2.8 x0 = 0, y0 = arccos √ 0.1

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 28 / 45

Conjugate locus of the Serret-Andoyer metric

The right-hand side (pθ > 0) components of the conjugate locus on the (x, y) plane:

2pi 3pi 4pi 5pi pi/2 pi x y

A = 1.5, B = 2, C = 2.8 x0 = 0, y0 = π

2 (ϕ0 = 0)

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 29 / 45

Another 2D structure on SO(3)

Let ˙ R(t) = R(t)   −u3 u3 −u1 −0 u1   , R(t) ∈ SO(3). Denote ri = R1i, i = 1, 2, 3, then (P0) ˙ r1 = u3r2, ˙ r2 = −u3r1 + u1r3, ˙ r3 = −u1r2. Consider the problem of optimal transfer (P1) r(0) =   1   , r(T) =   1   , (P2)

T

• (I1u2

1(t) + I3u2 3(t))dt → min,

T − fixed. N.B. Since r = 1, optimal trajectories of (P0) − (P2) are geodesics of an almost-Riemannian metric on S2: gk = dr2

1 + k2dr2 3

r2

2

, k2 = I1 I3

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 30 / 45

Motivating example: fastest transfer in a linear chain of tree weakly coupled spins

Physical model: Consider a linear chain of three spin- 1

2 particles, coupled with unequal Ising

• couplings. The Hamiltonian has the form

H = 2J12I1zI2z + 2J23I2zI3z + Hu, where I1α = Iα ⊗ I0 ⊗ I0, I2α = I0 ⊗ Iα ⊗ I0, I3α = I0 ⊗ I0 ⊗ Iα, Ix = 1 2 1 1

• ,

Iy = 1 2 −i i

• ,

Iz = 1 2 1 −1

• ,

I0 = 1 1

• ,

and Jij are scalar couplings between i-th and j-th particle. Problem: the fastest transfer I1x → 4I1zI2zI3z by the following sequence (P1) I1x → 2I1yI2z → 2I1yI2x → 4I1yI2yI3z.

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 31 / 45

Motivating example: fastest transfer in a linear chain of tree weakly coupled spins

Optimal Control Problem: Denote O = Tr(Oρ), and set x1 = I1x, x2 = 2I1yI2z, x3 = 2I1yI2x, x4 = 4I1yI2yI3z. Then (P1) = ⇒ (P2) : dX dt =     −1 1 −u u −1/k 1/k     X, k = J12 J23 , X(0) = (1, 0, 0, 0)T, X(T) = (0, 0, 0, 1)T, T → min .

• H. Yuan, R.Zeier, N.Khaneja, 2008: geodesics of the metric gk describe solutions to

(P2) for r1 = x1, r2 =

• x2

2 + x2 3,

r3 = x4.

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 32 / 45

Analysis of extremals of (P1) − (P3)

In spherical coordinates such that r2 = cos ϕ, r1 = sin ϕ cos θ, r3 = sin ϕ sin θ, the Hamiltonian associated to gk takes the form Hk = 1 4k2 sin2 ϕ

• p2

ϕ sin2 ϕ(sin2 θ + k2 cos2 θ) + p2 θ cos2 ϕ(cos2 θ + k2 sin2 θ) −

2 (k2 − 1)pϕpθ sin ϕ cos ϕ sin θ cos θ

• .

If k = 1 (∼ equal Ising couplings) we get a Grushin-type metric with H = 1

4

• p2

ϕ + p2 θ cot2 ϕ

• .

The family of metrics gk have a fixed singularity on the equator ϕ = π

2 and a discrete

symmetry group defined by Hk(ϕ, pϕ) = Hk(π − ϕ, −pϕ), Hk(θ, pθ) = Hk(−θ, −pθ).

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 33 / 45

Analysis of extremals of (P1) − (P3)

Consider M0 = {R ∈ SO(3) ; Rt = (r(0), ·, ·), r(0) = (1, 0, 0)}, M1 = {R ∈ SO(3) ; Rt = (r(T), ·, ·), r(T) = (0, 0, 1)}.

• Proposition. The extremals of the Riemannian metric gk on S2 with boundary con-

ditions r(0), r(T) are extremals of the sub-Riemannian problem on SO(3) with k2 = I1/I3, satisfying the boundary conditions (R(0), λ(0)) ∈ M⊥

0 , (R(T), λ(T)) ∈ M⊥ 1 ,

where λ denotes the adjoint vector. N.B. Coming back to SO(3), we get the Hamiltonian Hk = 1 4 H2

1

I1 + H2

3

I3

• .

It can be seen as a limit case of Euler’s top Hamiltonian Hn for I2 → +∞. Setting cos ϑ =

H1 2√I1 , sin ϑ = H3 2√I3 , Euler’s equations imply the generalized pendulum equation

d2ϑ dt2 = sin 2ϑ(k2 − 1) 2I1 , k2 = I1 I3 .

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 34 / 45

Conjugate locus structure, case H = 1, ϕ(0) = π

2 , θ0 = 0: the first insight

Dependence of the first conjugate time upon variations of k ≥ 1 (pθ = 10−4):

1 k1 k2 k3 pi/2 pi

red curve: θ(t1(pθ, k)), blue curve: ϕ(t1(pθ, k)), k1 ≈ 1.061, k2 ≈ 1.250, k2 ≈ 1.429

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 35 / 45

Conjugate locus structure, case Hk = 1, ϕ(0) = π

2 , θ0 = 0: the first insight

Evolution of the conjugate locus: 1 ≤ k < k1 :

pi/2 pi pi/2 pi θ φ

k = 1, k = 1.05

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 36 / 45

Conjugate locus structure, case Hk = 1, ϕ(0) = π

2 , θ0 = 0: the first insight

Evolution of the conjugate locus: k1 ≤ k < k2 :

pi/2 pi pi/2 pi θ φ

k = 1.1, k = 1.2 N.B. A closed loop occurs at ¯ k ∈ [k1, k2]: ϕ(t1(¯ k)) = π

2 ,

¯ k ≈ 1.155.

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 37 / 45

Conjugate locus structure, case Hk = 1, ϕ(0) = π

2 , θ0 = 0: the first insight

Evolution of the conjugate locus: k2 ≤ k < k3 :

pi/2 pi pi/2 pi θ φ

k = 1.3, k = 1.4

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 38 / 45

Conjugate locus structure, case Hk = 1, ϕ(0) = π

2 , θ0 = 0: the first insight

Examples of different types of trajectories for 1 ≤ k < k3 :

−pi −pi/2 pi/2 pi pi/2 pi θ φ −pi −pi/2 pi/2 pi pi/2 pi θ φ

k ∈ [1, k3] k ∈ [k1, k3]

−pi −pi/2 pi/2 pi pi/2 pi θ φ −pi −pi/2 pi/2 pi pi/2 pi θ φ

k ∈ [¯ k, k3] k ∈ [k2, k3]

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 39 / 45

Conjugate locus structure, case Hk = 1, ϕ(0) = π

2 , θ0 = 0: the first insight

Evolution of the conjugate locus, case k ≤ 1 :

pi/2 pi pi/2 pi θ φ

k = 0.8

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 40 / 45

Conjugate locus structure, case Hk = 1, ϕ(0) = π

2 , θ0 = 0: the first insight

Evolution of the conjugate locus, case k ≤ 1 :

pi/2 pi pi/2 pi θ φ

k = 0.5

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 41 / 45

Conjugate locus structure, case Hk = 1, ϕ(0) = π

2 , θ0 = 0: the first insight

Evolution of the conjugate locus, case k ≤ 1 :

pi/2 pi pi/2 pi θ φ

k = 0.2

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 42 / 45

Conjugate locus structure, case Hk = 1, ϕ(0) = π

2 , θ0 = 0: the first insight

Evolution of the conjugate locus, case k ≤ 1 :

pi/2 pi pi/2 pi θ φ

k = 0.1

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 43 / 45

Conjugate locus structure, case Hk = 1, ϕ(0) = π

2 , θ0 = 0: the first insight

Evolution of the conjugate locus on the sphere: magenta curve: k = 1, red curves: k = 0.8, k = 1.15 < ¯ k

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 44 / 45

Conjugate locus structure, case Hk = 1, ϕ(0) = π

2 , θ0 = 0: the first insight

Evolution of the conjugate locus on the sphere: magenta curve: k = 1, red curves: k = 1.2, k = 1.25

Nataliya Shcherbakova (LGC-ENSIACET-INPT) April 4, 2013 45 / 45