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Singularities in the one control problem Moiseev Igor S.I.S.S.A., Trieste August 16, 2007 Moiseev Igor (SISSA) Singularities in the one control problem August 16, 2007 1 / 25

Singularities in the one control problem. S.I.S.S.A., Trieste August 16, 2007

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Page 1: Singularities in the one control problem. S.I.S.S.A., Trieste August 16, 2007

Singularities in the one control problem

Moiseev Igor

S.I.S.S.A., Trieste

August 16, 2007

Moiseev Igor (SISSA) Singularities in the one control problem August 16, 2007 1 / 25

Page 2: Singularities in the one control problem. S.I.S.S.A., Trieste August 16, 2007

Abstract

The geometry of strokes arises in the control problems of Reeds – Sheppcar, Dubins’ car, modeling of vision and some others.

The main problem is to characterize the shortest paths and minimaldistances on the plane, equipped with the structure of geometry of strokes.This problem is formulated as an optimal control problem in 3-space with2 dimensional control and a quadratic integral cost.

Here is studied the symmetries of the sub-Riemannian structure, extremalsof the optimal control problem, the Maxwell stratum, conjugate points andboundary value problem for the corresponding Hamiltonian system.

Moiseev Igor (SISSA) Singularities in the one control problem August 16, 2007 2 / 25

Page 3: Singularities in the one control problem. S.I.S.S.A., Trieste August 16, 2007

Problem statement

It is given two points on the plane (x0, y0), (x1, y1) ∈ R2 and to each of

them attached a vector, parameterized by the angles θ0 and θ1. Oneshould find the shortest path which connects given points and the curvemust be tangent to the attached vectors.

x = v cos θ, x(0) = x0 x(t1) = x1

y = v sin θ, y(0) = y0 y(t1) = y1

θ = u. θ(0) = θ0 θ(t1) = θ1

(1)

ℓ =

∫ t1

0

u2 + v2 dt → minu,v

q = (x , y , θ) ∈ M = R2x ,y × S1

θ , (u, v) ∈ R2

Moiseev Igor (SISSA) Singularities in the one control problem August 16, 2007 3 / 25

Page 4: Singularities in the one control problem. S.I.S.S.A., Trieste August 16, 2007

Continuous symmetries

Figure: The Strokes algebra

The solution of the system (1) can be lifted to the group of motion ofthe plane – the Euclidean group E2 and it acts on M as

x1

y1

θ1

·

x2

y2

θ2

=

x1 + x2 cos θ1 − y2 sin θ1y1 + x2 sin θ1 + y2 cos θ1

θ1 + θ2

(2)

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Page 5: Singularities in the one control problem. S.I.S.S.A., Trieste August 16, 2007

Theorem (Symmetries of the sub-Riemannian structure)

Symmetries of the sub-Riemannian structure (∆, 〈 · , · 〉) on the groupM = R

2x ,y × S1

θ form the Strokes algebra.

Sym(∆, 〈 · , · 〉) = span{X1,X2,X3}

where ∆ = span{ξ1, ξ2} ⊂ TM and

X1 =∂

∂x, X2 = −y

∂x+ x

∂y+

∂θ, X3 =

∂y.

with the multiplication rules

[X1,X2] = X3, [X1,X3] = 0, [X2,X3] = X1

Moiseev Igor (SISSA) Singularities in the one control problem August 16, 2007 5 / 25

Page 6: Singularities in the one control problem. S.I.S.S.A., Trieste August 16, 2007

PMP system

System which solves the optimization problem using the concept ofPontryagin Maximum Principle can be written as

(

xy

)

= ρRβ

(

cos2(θ − β)12 sin 2(θ − β)

)

,

θ = pθ,

ρ = 0,

β = 0,

pθ = 12 ρ

2 sin[

2(θ − β)]

.

x(0) = y(0) = 0, θ(0) = θ0,

θ2(0) = p2θ(0) = 1 − ρ2 cos2(θ0 − β).

where Rβ =[ cos β −sinβ

sinβ cos β

]

is a rotation matrix and the polar change of

variables in the plane (px , py ) had performed: ρ =√

p2x + p2

y , px = ρ cos βand py = ρ sinβ.

Moiseev Igor (SISSA) Singularities in the one control problem August 16, 2007 6 / 25

Page 7: Singularities in the one control problem. S.I.S.S.A., Trieste August 16, 2007

Elliptic Coordinates

Selecting the pendulum equation in the PMP system in coordinates(θ, pθ, ρ, β), we perform integration subject to the value of ρ. That is thecylinder (θ, pθ) ∈ C = S1

θ × Rpθ splits into three subsets

C1 ={

(θ, pθ) ∈ C | ρ > 1}

, C2 ={

(θ, pθ) ∈ C | ρ < 1}

,

C3 ={

(θ, pθ) ∈ C | ρ = 1}

.

C1, ρ ∈ (1,+∞) C2, ρ ∈ (0, 1) C3, ρ = 1{

ρ cos (θ − β) = s1 snψ,

pθ = − cnψ.

ψdef= ρϕ =

[

0, 4K]

{

cos (θ − β) = s2 snϕ,

pθ = −s2 dnϕ.

ϕ =[

0, 2K]

{

cos (θ − β) = s1s2 tanhϕ,

pθ = −s2 sechϕ.

ϕ = (−∞,∞)

Moiseev Igor (SISSA) Singularities in the one control problem August 16, 2007 7 / 25

Page 8: Singularities in the one control problem. S.I.S.S.A., Trieste August 16, 2007

Parameterization of the geodesics

Case 1, ρ > 1. Let ψt = ψ + ρt, where ψ=F(

arcsin[ρ cos θ(0)] | 1ρ

)

and thephase β can be expressed in terms of elliptic coordinates asβ = θ0 −

π2 + arcsin

(

snψ)

, then

xt = ψt − ψ + E(ψ) − E(ψt),

yt = 1ρ(cnψ − cnψt),

θt = θ0 + arcsin(

snψ)

− arcsin(

snψt

)

.

Case 2, ρ < 1. Let ϕt = ϕ+ t, where ϕ = F(

π2 − θ(0) | ρ

)

andβ = θ0 −

π2 + amϕ, then

xt = 1ρ

(

ϕt − ϕ+ E(ϕ) − E(ϕt))

,

yt = 1ρ

(

dnϕ− dnϕt

)

,

θt = θ0 + amϕ− amϕt .

Moiseev Igor (SISSA) Singularities in the one control problem August 16, 2007 8 / 25

Page 9: Singularities in the one control problem. S.I.S.S.A., Trieste August 16, 2007

Figure: (a) Two general type geodesics (Cases 1, 2) and one limit areillustrated; (b) Elliptic coordinates on the cylinder C and the projection (in color)of the level set of maximized hamiltonian H(p, q) = 1

2 .

Moiseev Igor (SISSA) Singularities in the one control problem August 16, 2007 9 / 25

Page 10: Singularities in the one control problem. S.I.S.S.A., Trieste August 16, 2007

Transformation to the pendulum equation

The following transformation of state variables

µ = π − 2θ, c = −2pθ,

β = −2β, α = ρ2.(3)

drives the original system

θ = pθ, θ ∈ S1 ⊂ M,

pθ = 12 ρ

2 sin[

2(θ − β)]

, pθ ∈ R,

ρ = 0, ρ ≥ 0,

β = 0, β ∈ S1.

(4)

to the system of generalized pendulum equation.

Moiseev Igor (SISSA) Singularities in the one control problem August 16, 2007 10 / 25

Page 11: Singularities in the one control problem. S.I.S.S.A., Trieste August 16, 2007

Reflection of the extremals

ε1 : δ 7→ δ1 ={

(θ1s , p

1s , ρ, β

1)}

={

(θt−s ,−pt−s , ρ, β)}

,

ε2 : δ 7→ δ2 ={

(θ2s , p

2s , ρ, β

2)}

={

(π − θt−s , pt−s , ρ,−β)}

,

ε3 : δ 7→ δ3 ={

(θ3s , p

3s , ρ, β

3)}

={

(π − θs ,−ps , ρ,−β)}

.

(5)

Proposition

The reflections εi is acting on the space (θ, pθ, ρ, β) and generates the followingaction in the space (x , y)

(x1s , y

1s ) = (xt − xt−s , yt − yt−s),

(x2s , y

2s ) = (xt − xt−s , yt−s − yt), (x3

s , y3s ) = (xs , −ys).

Where s ∈ [0, t] and the integer i = 1, 3 corresponds to the symmetry εi .

Moiseev Igor (SISSA) Singularities in the one control problem August 16, 2007 11 / 25

Page 12: Singularities in the one control problem. S.I.S.S.A., Trieste August 16, 2007

Action of reflections in M

One can see that the reflection ε1 preserves the initial and final points inthe plane (x , y). The others two ε2 and ε3 have to be corrected by therotation on the angle 2χ, where χ is defined as

sinχ =yt

x2t + y2

t

, cosχ =xt

x2t + y2

t

.

Table: The action of discrete symmetries on extremals. The equalities of thetashould be read modulo π

Reflection ε1 Reflection ε2 Reflection ε3

θ1s = θt−s

x1s = xt − xt−s

y1s = yt − yt−s

θ2s = π − θt−s + 2χ

(

x2s

y2s

)

= R2χ

(

xt − xt−s

yt−s − yt

)

θ3s = π − θs + 2χ

(

x3s

y3s

)

= R2χ

(

xs

−ys

)

Moiseev Igor (SISSA) Singularities in the one control problem August 16, 2007 12 / 25

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Moiseev Igor (SISSA) Singularities in the one control problem August 16, 2007 13 / 25

Page 14: Singularities in the one control problem. S.I.S.S.A., Trieste August 16, 2007

Cut Locus

Figure: Cut locus image on the pendulum phase portrait. The green lineindicate the Cut locus formed by reflection ε1, trajectories (a) and (c); The redline designate ε2-reflection, type (b) trajectories on the previous figure; Pointsmark conjugate points.

Moiseev Igor (SISSA) Singularities in the one control problem August 16, 2007 14 / 25

Page 15: Singularities in the one control problem. S.I.S.S.A., Trieste August 16, 2007

Proposition (Characterization of the cut time)

The moment of loss of global optimality by extremals in the problem (1) iscompletely described by the following statements.

1. The cut time t = t(θ0 | ρ, β) is a positive root ofC2(t, θ0 | ρ, β) = π + 2χ− θt − θ0, when ρ > 1C1(t, θ0 | ρ, β) = π + θt − θ0, when ρ < 1.

2. The cut time is invariant w.r.t. values of β and θ0, that is t = t(ρ).

3. t(ρ) takes its values from the interval[

2ρK(

)

, 4ρK(

)

)

while

ρ ∈ [1,+∞) and it is a monotonically decreasing function of ρ.

4. t(ρ) is a monotonically increasing function of ρ, such that t = 2K (ρ),when ρ < 1.

Where K (m) is a complete elliptic integral of the first kind.

Moiseev Igor (SISSA) Singularities in the one control problem August 16, 2007 15 / 25

Page 16: Singularities in the one control problem. S.I.S.S.A., Trieste August 16, 2007

Proof.

The formulas derived from geometric interpretation of the cut locus, collected in the Table 1,possess redundant information about self-intersections of geodesics (it includes cases whengeodesics coincide and intersect being not tangent to the final vector). Applying the equivalenttransformation to these formulas we derive the refined version of cut locus condition formulas,which are free of superfluous information.At first we recall the following formulas

sin 2χ =2 x y

x2 + y2, cos 2χ =

x2 − y2

x2 + y2. (6)

Then we say that to find a root of the equation C2(t, θ0 | ρ, β) = 0 it is equivalent to find a rootof the equation sinC2(t, θ0 | ρ, β) = 0. Expanding sinus we obtain

2 x y cos(θt + θ0) −(

x2 − y2)

sin(θt + θ0) = 0, (7)

keeping in mind that x2 + y2 > 0 at any positive moment of time. The equation (7) can berepresented in terms of scalar product, for the matter of the convenience

((

cos(θt + θ0)sin(θt + θ0)

)

,

(

2 x y

−x2 + y2

))

= 0. (8)

Further, the geodesics are represented as

Moiseev Igor (SISSA) Singularities in the one control problem August 16, 2007 16 / 25

Page 17: Singularities in the one control problem. S.I.S.S.A., Trieste August 16, 2007

Proof.

(

xt

yt

)

= Rβ

(

xt

yt

)

xt = ψt − ψ + E(ψ) − E(ψt),

yt = 1ρ(cnψ − cnψt),

θt = θ0 + arcsin(

snψ)

− arcsin(

snψt

)

.

β = θ0 − π2

+ arcsin(

snψ)

,

(9)

where Rβ =[ cosβ −sin β

sin β cos β

]

is a rotation matrix. Later we shall try to represent the scalar product

as a function of variables θ0, ρ and ψ. At first, the formulas (9) give that

(

cos βsinβ

)

= Rθ0−π

2

(

dnψ1ρ

snψ

)

⇒ Rβ = Rθ0−π

2Qψ . (10)

Where Qψ =[

dnψ −

1ρsnψ

1ρsnψ dnψ

]

, note here that det Qψ = 1, therefore Qψ ∈ SO(2). Then in

a similar way

(

cos(θt + θ0)sin(θt + θ0)

)

= R2θ0

(

dnψ dnψt + 1ρ2 snψ snψt

snψ dnψt −1ρ

dnψ snψt

)

= R2θ0Qψ

(

dnψt

− 1ρ

snψt

)

(11)

Thereafter we transform the scalar product (8) to the following form

Moiseev Igor (SISSA) Singularities in the one control problem August 16, 2007 17 / 25

Page 18: Singularities in the one control problem. S.I.S.S.A., Trieste August 16, 2007

Proof.

(

R2θ0Qψ

(

dnψt

− 1ρ

snψt

)

,

(

2 x y

−x2 + y2

))

= 0. (12)

Now we transform the second part of scalar product.

x2 − y2 =(

x2 − y2)

cos 2β − 2 x y sin 2β

2 x y =(

x2 − y2)

sin 2β + 2 x y cos 2β(13)

So, we can make step forward with a transformation(

R2θ0Qψ

(

dnψt

− 1ρ

snψt

)

,(

0 −11 0

)

R2β

(

x2 − y2

2 x y

))

= 0. (14)

Fortunately, the matrix R2β can be represented in a natural way

R2β = −R2θ0Q2ψ . (15)

And finally, noting that

QTψ

(

0 −11 0

)

Qψ = RT2θ0

(

0 −11 0

)

R2θ0=(

0 −11 0

)

(16)

we conclude with the scalar product(( 1

ρsnψt

dnψt

)

, Qψ

(

x2 − y2

2 x y

))

= 0. (17)

The last is valid, since the matrices R2θ0and Qψ are orthogonal.

Moiseev Igor (SISSA) Singularities in the one control problem August 16, 2007 18 / 25

Page 19: Singularities in the one control problem. S.I.S.S.A., Trieste August 16, 2007

Proof.

Now passing from elliptic coordinates representation to the the following coordinates

τ = ψ +ρ t

2, p =

ρ t

2(18)

we conclude transformation of (7) with the following equation

8 sn τ dn τ

ρ(

1−(1−dn2p)sn2τ)

[

E(p) − p][

cn p (E(p) − p) − dn p sn p]

= 0 (19)

where E(u) =∫ u

0dn

2w dw .The superfluous information is represented by the roots of the equation sn τ = 0, thereforet = 2

ρ(2K − ψ). This root corresponds to the stationary point of the map obtained by the

reflection ε2 of trajectories. We exclude it from consideration because the trajectories we obtainare absolutely identical. Further, the function

E(p) − p =

∫ p

0dn

2w − 1 dw = − 1ρ2

∫ p

0sn

2w dw

is always negative for any positive p. The rest of functions are always sign-definite, except of thelast term. So the final, refined expression of cut locus condition can be written as

F(t, ρ) = cn p (E(p) − p) − dn p sn p = 0. (20)

The roots of this equation describes all possible self-intersections of sub-Riemannian sphere and

wavefront in the region of oscillations. The first positive root describes the moment of time

when the extremals lose their global optimality.

Moiseev Igor (SISSA) Singularities in the one control problem August 16, 2007 19 / 25

Page 22: Singularities in the one control problem. S.I.S.S.A., Trieste August 16, 2007

The BVP scheme

Theorem (Symmetric BVP)

The solution of the symmetric boundary value problem stated for the PMP

system is given by the set of symmetric w.r.t. the cuspidal points (cases a and b

on Figure) and the hump (case a) pieces of geodesics.

x

y

0

(xt , y

t )

r

2

x

y

0 32

x

y

0

(a)

(b)

(c)

1 24

3

1

Period

Period1/2

31

Moiseev Igor (SISSA) Singularities in the one control problem August 16, 2007 22 / 25

Page 23: Singularities in the one control problem. S.I.S.S.A., Trieste August 16, 2007

Bifurcation Diagram

Figure: The bifurcation diagram for symmetric cases, described by the Theorem.

Moiseev Igor (SISSA) Singularities in the one control problem August 16, 2007 23 / 25

Page 24: Singularities in the one control problem. S.I.S.S.A., Trieste August 16, 2007

A. Agrachev, B. Bonard, M. Chyba and I. Kupka, Sub – Riemanian sphere inMartinet flat case. J. ESAIM: Control, Optimization and Calculus ofVariations, 1997, v. 2, 377–448.

L. E. Dubins, On curves of minimal length with a constraint on averagecurvature and with prescribed initial and terminal positions and tangents.Am. J. Math., 1957, 79:497–516.

J. Petitot, The neurogeometry of pinwheels as a sub – Riemannian contactstructure, J. Physiology – Paris 97 (2003), 265–309.

J. A. Reeds, L. A. Shepp, Optimal paths for a car that goes both forward andbackwards, Pacific J. Math. 145 (1990), 2: 367-378

Yu. L. Sachkov, Symmetries of flat rank two distributions andsub –Riemanninan structures, Transactions of the American MathematicalSociety, 356 (2004), 2: 457–494.

Yu. L. Sachkov, Discrete symmetries and Maxwell set in generalized Dido’s

problem, SISSA preprint 2005.

Moiseev Igor (SISSA) Singularities in the one control problem August 16, 2007 24 / 25

Page 25: Singularities in the one control problem. S.I.S.S.A., Trieste August 16, 2007

Acknowledgment

I’m grateful to Professor A. Agrachev for the valuable discussions andconstant support during my stay at S.I.S.S.A.

I’d like to thank Professor Yu. Sachkov for the kind replies on myquestions related his work.

I’m grateful to Professor A. Davydov for the help during my scientificcarrier.

I’d like to thank my wife and parents for their encourage during all my life.

Moiseev Igor (SISSA) Singularities in the one control problem August 16, 2007 25 / 25