Singularities in the one control problem. S.I.S.S.A., Trieste August 16, 2007.
The geometry of strokes arises in the control problems of Reeds–Shepp car, Dubins’ car, modeling of vision and some others. The main problem is to characterize the shortest paths and minimal distances on the plane, equipped with the structure of geometry of strokes.
This problem is formulated as an optimal control problem in 3-space with 2 dimensional control and a quadratic integral cost. Here is studied the symmetries of the sub-Riemannian structure, extremals of the optimal control problem, the Maxwell stratum, conjugate points and boundary value problem for the corresponding Hamiltonian system.
Singularities in the one control problem. S.I.S.S.A., Trieste August 16, 2007.
1. 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
2. Abstract
The geometry of strokes arises in the control problems of Reeds – Shepp
car, Dubins’ car, modeling of vision and some others.
The main problem is to characterize the shortest paths and minimal
distances on the plane, equipped with the structure of geometry of strokes.
This problem is formulated as an optimal control problem in 3-space with
2 dimensional control and a quadratic integral cost.
Here is studied the symmetries of the sub-Riemannian structure, extremals
of the optimal control problem, the Maxwell stratum, conjugate points and
boundary value problem for the corresponding Hamiltonian system.
Moiseev Igor (SISSA) Singularities in the one control problem August 16, 2007 2 / 25
3. 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. One
should find the shortest path which connects given points and the curve
must 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 → min
u,v
q = (x, y, θ) ∈ M = R2
x,y × S1
θ , (u, v) ∈ R2
Moiseev Igor (SISSA) Singularities in the one control problem August 16, 2007 3 / 25
4. Continuous symmetries
Figure: The Strokes algebra
The solution of the system (1) can be lifted to the group of motion of
the plane – the Euclidean group E2 and it acts on M as
x1
y1
θ1
·
x2
y2
θ2
=
x1 + x2 cos θ1 − y2 sin θ1
y1 + x2 sin θ1 + y2 cos θ1
θ1 + θ2
(2)
Moiseev Igor (SISSA) Singularities in the one control problem August 16, 2007 4 / 25
5. Theorem (Symmetries of the sub-Riemannian structure)
Symmetries of the sub-Riemannian structure (∆, · , · ) on the group
M = R2
x,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
6. PMP system
System which solves the optimization problem using the concept of
Pontryagin Maximum Principle can be written as
˙x
˙y
= ρ Rβ
cos2(θ − β)
1
2 sin 2(θ − β)
,
˙θ = pθ,
˙ρ = 0,
˙β = 0,
˙pθ = 1
2 ρ2
sin 2(θ − β) .
x(0) = y(0) = 0, θ(0) = θ0,
˙θ2
(0) = p2
θ (0) = 1 − ρ2
cos2
(θ0 − β).
where Rβ = cos β −sin β
sin β cos β is a rotation matrix and the polar change of
variables in the plane (px , py ) had performed: ρ = p2
x + p2
y , px = ρ cos β
and py = ρ sin β.
Moiseev Igor (SISSA) Singularities in the one control problem August 16, 2007 6 / 25
7. 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 the
cylinder (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 sn ψ,
pθ = − cn ψ.
ψ
def
= ρϕ = 0, 4K
cos (θ − β) = s2 sn ϕ,
pθ = −s2 dn ϕ.
ϕ = 0, 2K
cos (θ − β) = s1s2 tanh ϕ,
pθ = −s2 sech ϕ.
ϕ = (−∞, ∞)
Moiseev Igor (SISSA) Singularities in the one control problem August 16, 2007 7 / 25
8. Parameterization of the geodesics
Case 1, ρ > 1. Let ψt = ψ + ρt, where ψ=F arcsin[ρ cos θ(0)] | 1
ρ
and the
phase β can be expressed in terms of elliptic coordinates as
β = θ0 − π
2 + arcsin 1
ρ sn ψ , then
xt = ψt − ψ + E(ψ) − E(ψt),
yt = 1
ρ(cn ψ − cn ψt),
θt = θ0 + arcsin 1
ρ sn ψ − arcsin 1
ρ sn ψt .
Case 2, ρ < 1. Let ϕt = ϕ + t, where ϕ = F π
2 − θ(0) | ρ and
β = θ0 − π
2 + am ϕ, then
xt = 1
ρ ϕt − ϕ + E(ϕ) − E(ϕt ) ,
yt = 1
ρ dn ϕ − dn ϕt ,
θt = θ0 + am ϕ − am ϕt.
Moiseev Igor (SISSA) Singularities in the one control problem August 16, 2007 8 / 25
9. Figure: (a) Two general type geodesics (Cases 1, 2) and one limit are
illustrated; (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
10. 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 = 1
2 ρ2
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
11. Reflection of the extremals
ε1
: δ → δ1
= (θ1
s , p1
s , ρ, β1
) = (θt−s , −pt−s, ρ, β) ,
ε2
: δ → δ2
= (θ2
s , p2
s , ρ, β2
) = (π − θt−s, pt−s , ρ, −β) ,
ε3
: δ → δ3
= (θ3
s , p3
s , ρ, β3
) = (π − θs, −ps, ρ, −β) .
(5)
Proposition
The reflections εi
is acting on the space (θ, pθ, ρ, β) and generates the following
action in the space (x, y)
(x1
s , y1
s ) = (xt − xt−s , yt − yt−s ),
(x2
s , y2
s ) = (xt − xt−s , yt−s − yt), (x3
s , y3
s ) = (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
12. Action of reflections in M
One can see that the reflection ε1 preserves the initial and final points in
the plane (x, y). The others two ε2 and ε3 have to be corrected by the
rotation on the angle 2χ, where χ is defined as
sin χ =
yt
x2
t + y2
t
, cos χ =
xt
x2
t + y2
t
.
Table: The action of discrete symmetries on extremals. The equalities of theta
should be read modulo π
Reflection ε1 Reflection ε2 Reflection ε3
θ1
s = θt−s
x1
s = xt − xt−s
y1
s = yt − yt−s
θ2
s = π − θt−s + 2χ
x2
s
y2
s
= R2χ
xt − xt−s
yt−s − yt
θ3
s = π − θs + 2χ
x3
s
y3
s
= R2χ
xs
−ys
Moiseev Igor (SISSA) Singularities in the one control problem August 16, 2007 12 / 25
13. Moiseev Igor (SISSA) Singularities in the one control problem August 16, 2007 13 / 25
14. Cut Locus
Figure: Cut locus image on the pendulum phase portrait. The green line
indicate the Cut locus formed by reflection ε1
, trajectories (a) and (c); The red
line designate ε2
-reflection, type (b) trajectories on the previous figure; Points
mark conjugate points.
Moiseev Igor (SISSA) Singularities in the one control problem August 16, 2007 14 / 25
15. Proposition (Characterization of the cut time)
The moment of loss of global optimality by extremals in the problem (1) is
completely described by the following statements.
1. The cut time t = t(θ0 | ρ, β) is a positive root of
C2(t, θ0 | ρ, β) = π + 2χ − θt − θ0, when ρ > 1
C1(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 2
ρK 1
ρ , 4
ρK 1
ρ 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
16. 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 when
geodesics coincide and intersect being not tangent to the final vector). Applying the equivalent
transformation 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 root
of the equation sin C2(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 be
represented 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
17. Proof.
xt
yt
= Rβ
¯xt
¯yt
¯xt = ψt − ψ + E(ψ) − E(ψt ),
¯yt = 1
ρ
(cn ψ − cn ψt ),
θt = θ0 + arcsin 1
ρ
sn ψ − arcsin 1
ρ
sn ψt .
β = θ0 − π
2
+ arcsin 1
ρ
sn ψ ,
(9)
where Rβ = cos β −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 β
sin β
= Rθ0− π
2
dn ψ
1
ρ
sn ψ
⇒ Rβ = Rθ0− π
2
Qψ. (10)
Where Qψ =
dnψ − 1
ρ
snψ
1
ρ
snψ dnψ
, note here that det Qψ = 1, therefore Qψ ∈ SO(2). Then in
a similar way
cos(θt + θ0)
sin(θt + θ0)
= R2θ0
dn ψ dnψt + 1
ρ2 sn ψ snψt
1
ρ
sn ψ dn ψt − 1
ρ
dn ψ snψt
= R2θ0
Qψ
dn ψt
− 1
ρ
sn ψt
(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
18. Proof.
R2θ0
Qψ
dn ψt
− 1
ρ
sn ψt
,
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
R2θ0
Qψ
dn ψt
− 1
ρ
sn ψt
, 0 −1
1 0
R2β
¯x2 − ¯y2
2 ¯x ¯y
= 0. (14)
Fortunately, the matrix R2β can be represented in a natural way
R2β = −R2θ0
Q2
ψ. (15)
And finally, noting that
QT
ψ
0 −1
1 0
Qψ = RT
2θ0
0 −1
1 0
R2θ0
= 0 −1
1 0
(16)
we conclude with the scalar product
1
ρ
sn ψt
dn ψt
, Qψ
¯x2 − ¯y2
2 ¯x ¯y
= 0. (17)
The last is valid, since the matrices R2θ0
and Qψ are orthogonal.
Moiseev Igor (SISSA) Singularities in the one control problem August 16, 2007 18 / 25
19. 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
0
dn2
w dw.
The superfluous information is represented by the roots of the equation sn τ = 0, therefore
t = 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 obtain
are absolutely identical. Further, the function
E(p) − p =
p
0
dn2
w − 1 dw = − 1
ρ2
p
0
sn2
w dw
is always negative for any positive p. The rest of functions are always sign-definite, except of the
last 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
22. 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, yt )
r
2
x
y
0
32
x
y
0
(a)
(b)
(c)
1 2
4
3
1
Period
Period1/2
31
Moiseev Igor (SISSA) Singularities in the one control problem August 16, 2007 22 / 25
23. 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
24. A. Agrachev, B. Bonard, M. Chyba and I. Kupka, Sub – Riemanian sphere in
Martinet flat case. J. ESAIM: Control, Optimization and Calculus of
Variations, 1997, v. 2, 377–448.
L. E. Dubins, On curves of minimal length with a constraint on average
curvature 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 contact
structure, J. Physiology – Paris 97 (2003), 265–309.
J. A. Reeds, L. A. Shepp, Optimal paths for a car that goes both forward and
backwards, Pacific J. Math. 145 (1990), 2: 367-378
Yu. L. Sachkov, Symmetries of flat rank two distributions and
sub – Riemanninan structures, Transactions of the American Mathematical
Society, 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
25. Acknowledgment
I’m grateful to Professor A. Agrachev for the valuable discussions and
constant support during my stay at S.I.S.S.A.
I’d like to thank Professor Yu. Sachkov for the kind replies on my
questions related his work.
I’m grateful to Professor A. Davydov for the help during my scientific
carrier.
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