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    Geometric Control theoryBavo Langerock

    Department of Mathematical Physics and AstronomyGhent University

    21st INTERNATIONAL WORKSHOP onDIFFERENTIAL GEOMETRIC METHODS IN THEORETICAL MECHANICS

    Madrid, 2006

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    GEOMETRIC CONTROL 1

    Plan of the talk

    Examples of control systems

    Regularity assumptions

    What are interesting problems ?

    Families of vector fields: accessible sets: topology, optimality, smoothness,controllability

    Lie determined systems

    The maximum principle

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    GEOMETRIC CONTROL 2

    Control Theory: examples of open systems

    Open systems: influenceable x = f(x, u)

    future is not determined by current state

    dynamics determined by steering devices

    the system

    state space M

    u1

    u2

    u3

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    GEOMETRIC CONTROL 3

    A (simplified) car

    state space is (x,y,) SE(2)

    suppose that we can(A) steer the car such that we can move forward and backward with velocityv IR

    (B) rotate the car with angular velocity IR...

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    GEOMETRIC CONTROL 3

    A (simplified) car

    state space is (x,y,) SE(2)

    suppose that we can(A) steer the car such that we can move forward and backward with velocityv IR

    (B) rotate the car with angular velocity IR...

    x = v cos y = v sin

    =

    Question: Can we reach any point in the plane ?

    How to concatenate motions to reach these points ?

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    GEOMETRIC CONTROL 4

    A hovercraft

    state space is (x,y,) SE(2)

    A fan exerts a steering force F IR+, S1 on the boat

    x = Fcos( + )

    y = F sin( + )I = F h sin

    F

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    GEOMETRIC CONTROL 5

    The upward pendulum

    state space is (x, ) IR S1

    the system is such that any force can be exerted at point x on the pendulum

    L = 12(m

    2 + I)2 + m sin x + 12mx2 mg sin

    Q = F dx

    x F

    Can we invent F(x,, ) such that the solutions tends to = /2 ?

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    GEOMETRIC CONTROL 6

    Damping a vibrating spring

    state space x IR

    an exterior force is acting on the mass m:

    mx = mg k(x ) + F

    where the force is restricted to |F| 1.

    m

    F

    mg

    Find the force law such that it damps the oscillation in the least amount of time.

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    GEOMETRIC CONTROL 7

    The input-state model

    The state space IRn q = (q1, . . . , q n)

    The inputs u = (u1, . . . , uk) U IRk

    U is called the control domain, is arbitrary, often with boundary !

    Dynamical steering law q = f(q, u) , with f : IRn IRk Rn.

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    GEOMETRIC CONTROL 7

    The input-state model

    The state space IRn q = (q1, . . . , q n)

    The inputs u = (u1, . . . , uk) U IRk

    U is called the control domain, is arbitrary, often with boundary !

    Dynamical steering law q = f(q, u) , with f : IRn IRk Rn.

    Car q = (x,y,), U IR2, u = (v, ) f =

    v cos v sin

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    GEOMETRIC CONTROL 7

    The input-state model

    The state space IRn q = (q1, . . . , q n)

    The inputs u = (u1, . . . , uk) U IRk

    U is called the control domain, is arbitrary, often with boundary !

    Dynamical steering law q = f(q, u) , with f : IRn IRk Rn.

    Car q = (x,y,), U IR2, u = (v, ) f =

    v cos v sin

    Damping q = (x, v), U = [1, 1], u = damping force F,

    f =

    0 1

    k 0

    xv

    + (mg + kl + u)

    01

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    GEOMETRIC CONTROL 8

    Control systems are encountered in population dynamics, economics,electric circuits, quantum mechanics and

    mechanical systems (control forces book by Bullo and Lewis)

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    GEOMETRIC CONTROL 8

    Control systems are encountered in population dynamics, economics,electric circuits, quantum mechanics and

    mechanical systems (control forces book by Bullo and Lewis)

    Can be used to model non-holonomic constraints

    Sub-Riemannian geometry

    Lie algebroids

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    GEOMETRIC CONTROL 9

    What with regularity assumptions ?

    controls are smooth with a finite number of discontinuous jumpsReflects the control nature of u.

    t

    U

    u1

    u2

    u3

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    GEOMETRIC CONTROL 9

    What with regularity assumptions ?

    controls are smooth with a finite number of discontinuous jumpsReflects the control nature of u.

    q = f(q, u(t)) still has continuous unique solutions, q(t) are piecewise

    smooth curves

    u1 u2 u3

    q0

    q1

    M = IRn

    t

    U

    u1

    u2

    u3

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    GEOMETRIC CONTROL 10

    let q(t) be the solution to the time dependent system

    q = f(q, u(t)) through a fixed initial state q0.

    The solution is called a controlled curve.

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    GEOMETRIC CONTROL 11

    More general point of view:

    controls are measurable and bounded curves

    guarantees that q = f(q, u) has absolutely continuous solutions.

    Many results depend on considering tangent maps to f(q, u): how to define

    it within this general framework. . .

    Regularity assumptions are important: results depend on it !

    We do not wish to consider this generality: it is already difficult enough tostudy families of smooth vector fields.

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    GEOMETRIC CONTROL 12

    Control Theory: main problems from a geometers point of

    view

    q0

    q(t)

    What can I say about the set points that I can reach following a controlled q(t) ?

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    GEOMETRIC CONTROL 12

    Control Theory: main problems from a geometers point of

    view

    q0

    q(t)

    What can I say about the set points that I can reach following a controlled q(t) ?

    Topology, Boundary, interior, . . .

    Smoothness:manifold with boundary, corners, stratified space . . .

    sufficient conditions for the accessible sets to be the entireconfiguration space preferable from an engineering pointof view !

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    GEOMETRIC CONTROL 13

    Control Theory: main problems from a geometers point of

    view

    What can I say about the set points that I can reach following a controlled q(t) ?

    What can I say about controls that optimise a given cost ?

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    GEOMETRIC CONTROL 13

    Control Theory: main problems from a geometers point of

    view

    What can I say about the set points that I can reach following a controlled q(t) ?

    What can I say about controls that optimise a given cost ?

    Cost is a certain function that one wants to minimise.

    Typical question: What control minimises time needed to take me from q0 toq1

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    GEOMETRIC CONTROL 14

    A Differential Geometric Framework: families of vector fields

    What are the essential ingredients:

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    GEOMETRIC CONTROL 14

    A Differential Geometric Framework: families of vector fields

    What are the essential ingredients:

    every control u(t) determines a time dependent vector field on the state space

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    GEOMETRIC CONTROL 14

    A Differential Geometric Framework: families of vector fields

    What are the essential ingredients:

    every control u(t) determines a time dependent vector field on the state space

    arbitrary family F of vector fields on configuration manifold M, (forget thestructure given by f(q, u))

    controlled curves are concatenations of integral curves

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    GEOMETRIC CONTROL 14

    A Differential Geometric Framework: families of vector fields

    What are the essential ingredients:

    every control u(t) determines a time dependent vector field on the state space

    arbitrary family F of vector fields on configuration manifold M, (forget thestructure given by f(q, u))

    controlled curves are concatenations of integral curves

    Other possibility:

    control domain translates to bundle over configuration manifoldU M, typical fibre of U is control domain: manifold with boundary,manifold with corners,. . .

    dynamical law translates to bundle map U T M, fibred over identity f is linear in control variables (distribution approach).

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    GEOMETRIC CONTROL 15

    Accessible sets: general definition and observations

    Often family of vector fields picture used.

    M configuration manifold, equipped with an arbitrary family of smooth vectorfields F

    Accq(F) = {t

    1t1(q) | ti IR+, {i} flow of Xi F}

    Acc(q,T)(F){t

    1t1(q) | ti IR+,

    i=1 ti T , {i} flow of Xi F}

    Study the structure of the set Accq(F)

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    GEOMETRIC CONTROL 16

    Accq(F) and Accq,T(F) are subsets of the leaf Lq(F) through q of thesmallest integrable distribution containing FDefinition (Sussmann):

    Lq(F) = {t

    1t1(q) | ti IR , {i} flow of Xi F}

    What is the structure of Accq(F) as a subset of Lq(F) ?

    dimension of accessible set is lower then leaf dimension

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    GEOMETRIC CONTROL 16

    Accq(F) and Accq,T(F) are subsets of the leaf Lq(F) through q of thesmallest integrable distribution containing FDefinition (Sussmann):

    Lq(F) = {t

    1t1(q) | ti IR , {i} flow of Xi F}

    What is the structure of Accq(F) as a subset of Lq(F) ?

    dimension of accessible set is lower then leaf dimension non-smooth boundary

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    GEOMETRIC CONTROL 16

    Accq(F) and Accq,T(F) are subsets of the leaf Lq(F) through q of thesmallest integrable distribution containing FDefinition (Sussmann):

    Lq(F) = {t

    1t1(q) | ti IR , {i} flow of Xi F}

    What is the structure of Accq(F) as a subset of Lq(F) ?

    dimension of accessible set is lower then leaf dimension non-smooth boundary dimension of accessible set is not constant

    (x, y)

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    GEOMETRIC CONTROL 17

    Accessible sets: some results on topology

    Definition: The topology induced by F is the finest topology for which allmaps IR

    + M given by t

    1t1(q) are continuous, with ti 0 and{i} flow of a member of F.

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    GEOMETRIC CONTROL 17

    Accessible sets: some results on topology

    Definition: The topology induced by F is the finest topology for which allmaps IR

    + M given by t

    1t1(q) are continuous, with ti 0 and{i} flow of a member of F.

    This topology on Accq(F) is in general finer than the subset topology w.r.tthe topology of Lq(F).

    X0 = x, defined on IR2;

    X1 = y, defined on ]0, [IR;X2 = x on IR]0, [X3 = yy on IR

    2

    xx

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    GEOMETRIC CONTROL 18

    Full-rank systems or Lie-determined systems

    A diff. geom. condition guaranteeing that locally F will determine the

    structure of Accq(F) (cf. books of Jurdjevic, Agrachev).

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    GEOMETRIC CONTROL 18

    Full-rank systems or Lie-determined systems

    A diff. geom. condition guaranteeing that locally F will determine the

    structure of Accq(F) (cf. books of Jurdjevic, Agrachev).

    Definition: Lie determined or Full-rank systems are families F for whichT(Lq(F)) = Lie(F) with Lie(F) the distribution generated by all iterated Liebrackets of elements of F.

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    GEOMETRIC CONTROL 18

    Full-rank systems or Lie-determined systems

    A diff. geom. condition guaranteeing that locally F will determine the

    structure of Accq(F) (cf. books of Jurdjevic, Agrachev).

    Definition: Lie determined or Full-rank systems are families F for whichT(Lq(F)) = Lie(F) with Lie(F) the distribution generated by all iterated Liebrackets of elements of F.

    Acc(q,T)(F) has non-empty interior interior points are normally accessible (full-rank paths) If Full-rank then F-generated topology and subset topology coincide (at

    least on interior points) int(cl(Accq(F))) = int(Accq(F)) (no separation of interior by boundary) Acc(q,T)(F) cl(int(Acc(q,T)(F))) (no isolated boundary points)

    No dense accessible sets in an orbit occur

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    GEOMETRIC CONTROL 19

    Families with coincident accessible sets (up to boundary)

    Closure of accessible sets should be regarded as the analogue of leaf of

    foliation . . .

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    GEOMETRIC CONTROL 19

    Families with coincident accessible sets (up to boundary)

    Closure of accessible sets should be regarded as the analogue of leaf of

    foliation . . .

    Acc(q,T)(convex(F)) cl(Acc(q,T)(F))

    Acc(q,T)(cl(F)) cl(Acc(q,T)(F))

    Accq(cone(F)) cl(Accq(F))

    A map Diff(M) is a normalizer if

    (Acc1(q)(F)) cl(Accq(F)).

    Then, if N(F) is the family of the pull-back vector fields of F undernormalizers, Accq(N(F)) cl(Accq(F)).

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    GEOMETRIC CONTROL 20

    If cl(Accq(F)) = M then Accq(F) = M. Important for controllability. Closureis essential !

    The Lie saturate S(F) is the largest subset of Lie(F) such thatcl(Accq(S(F))) = cl(Accq(F)).

    Lie saturate is invariant under normalizer

    If Lie saturate = T M then any point can be accessed from an arbitrarypoint Accq(F) = M (:= controllability)

    Not very applicable.

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    GEOMETRIC CONTROL 21

    ControllabilitySufficient conditions for Accq(F) = M or controllability

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    GEOMETRIC CONTROL 21

    ControllabilitySufficient conditions for Accq(F) = M or controllability

    Trick: Extend family F to cl(F), cone(F) and N(F). Results say thatAccq(N(F)) cl(Accq(F)).

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    GEOMETRIC CONTROL 21

    ControllabilitySufficient conditions for Accq(F) = M or controllability

    Trick: Extend family F to cl(F), cone(F) and N(F). Results say thatAccq(N(F)) cl(Accq(F)).

    If N(F) = X(M),

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    GEOMETRIC CONTROL 21

    ControllabilitySufficient conditions for Accq(F) = M or controllability

    Trick: Extend family F to cl(F), cone(F) and N(F). Results say thatAccq(N(F)) cl(Accq(F)).

    If N(F) = X(M),

    then Accq(N(F)) = M,

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    GEOMETRIC CONTROL 21

    ControllabilitySufficient conditions for Accq(F) = M or controllability

    Trick: Extend family F to cl(F), cone(F) and N(F). Results say thatAccq(N(F)) cl(Accq(F)).

    If N(F) = X(M),

    then Accq(N(F)) = M,

    then cl(Accq(F)) = M,

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    GEOMETRIC CONTROL 21

    ControllabilitySufficient conditions for Accq(F) = M or controllability

    Trick: Extend family F to cl(F), cone(F) and N(F). Results say thatAccq(N(F)) cl(Accq(F)).

    If N(F) = X(M),

    then Accq(N(F)) = M,

    then cl(Accq(F)) = M,

    then Accq(F) = M.

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    GEOMETRIC CONTROL 21

    ControllabilitySufficient conditions for Accq(F) = M or controllability

    Example: M = IRn, U = IR, f(x, u) = Ax + ub with A IRnn, b IRn.F = {Ax + ub | u IR}

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    GEOMETRIC CONTROL 21

    ControllabilitySufficient conditions for Accq(F) = M or controllability

    Example: M = IRn, U = IR, f(x, u) = Ax + ub with A IRnn, b IRn.F = {Ax + ub | u IR}

    Full rank ? b,Ax sp(F), [Ax + ub,Ax + ub] = (u u)Ab,

    [Ab, Ax + u

    b] = u

    A2

    b, . . .

    if sp{b,Ab,A2b , . . . , An1b} = IRn then Full rank.

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    GEOMETRIC CONTROL 21

    ControllabilitySufficient conditions for Accq(F) = M or controllability

    Example: M = IRn, U = IR, f(x, u) = Ax + ub with A IRnn, b IRn.F = {Ax + ub | u IR}

    Full rank ? b,Ax sp(F), [Ax + ub,Ax + ub] = (u u)Ab,

    [Ab, Ax + u

    b] = u

    A2

    b, . . .

    sp{b} cl(cone(F)), thus F1 := F sp{b}.

    b

    0

    Ax

    Ax+ b

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    GEOMETRIC CONTROL 22

    the map u(x) = x + ub is a normalizer for F1. (u is flow of (x b) F1)uF

    1 N(F1).

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    GEOMETRIC CONTROL 22

    the map u(x) = x + ub is a normalizer for F1. (u is flow of (x b) F1)uF

    1 N(F1).

    Ab cl(cone(N(F1))) from u(x Ax) = x Ax + uAb, thus extendF2 := F1 sp{Ab} and so on

    if sp{b,Ab,A2b , . . . , An1b} = IRn then Accx(F) = IRn.

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    GEOMETRIC CONTROL 23

    Accessible sets: smooth structure

    Open problem : provide sufficient conditions to show that Accq(F) is a

    smooth manifold with singularities.

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    GEOMETRIC CONTROL 23

    Accessible sets: smooth structure

    Open problem : provide sufficient conditions to show that Accq(F) is a

    smooth manifold with singularities.

    Manifold with boundary ?

    Manifold with corners

    charts are homeomorphisms from open subsets of Accq(F) to opensubsets of IR

    n

    +, such that compatibility holds.

    Stratified SpacesPartitioned topological spaces such that a.o. strata are smooth manifolds....

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    GEOMETRIC CONTROL 24

    Accessible sets: smooth structure

    Example: Given a family of n independent globally defined vector fields on an

    n-dimensional manifold M, then if

    [Xi, Xj] C(Xi, Xj) and

    any point is accessible by a maximal path, then

    Accx(F) is a manifold with corners.

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    GEOMETRIC CONTROL 25

    Example: Consider the family of vector fields{X1 = /x,X2 = /x + ex/y,X3 = /x + ex/z}. It is easily seen that

    [X1, X2] = ex/y C(X1, X2),

    [X1, X3] = ex/z C(X1, X3),

    [X2, X3] = ex(/z /y) C(X2, X3).

    x

    y

    z

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    GEOMETRIC CONTROL 26

    Part II: Optimisation

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    GEOMETRIC CONTROL 27

    Optimisation: variational problems in control theory

    Problem: Given a control system q = f(q, u); an initial state q0, initial time t0,

    a final state q1 and a final time t1. Among all controlled curves(q, u) : [t0, t1] M U, which one minimises a given cost function L(q, u).

    Maximum principle gives necessary conditions for a controlled curve (q, u) tobe optimal:

    Define H(q,u,p,) = pifi

    (q, u) + L(q, u);

    Define a multiplier for (q(t), u(t)) to be a pair (p(t), ) satisfying:

    qi(t) = fi(q(t), u(t)) =H

    pii = 1, . . . , n

    pi(t) = Hqi

    = pj(t)fj

    qi L

    qii = 1, . . . , n

    Fix t, then u H(q(t), u , p(t), ) is maximal for u = u(t) .

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    GEOMETRIC CONTROL 28

    The Maximum Principle If a controlled curve (q, u) is optimal then there is anonzero multiplier (p, ), with either 0, 1.

    Sufficient condition originates from following sequence of arguments:

    Add a coordinate to the configuration space: (q, J) M IR.

    Consider an extended control system: ( q, J) = (f(q, u), L(q, u)).

    A optimal controlled curve (q, u) has to belong to the boundary of theaccessible set of the extended controlled system !

    Study of the boundary of accessible sets is important ! A more detailedinvestigation of the structure of the boundary may lead to stronger MP !

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    GEOMETRIC CONTROL 29

    The Maximum Principle for time optimality

    Find (q, u) such that it minimises time among all controlled curves starting at

    q0 and ending at qf (where the time interval may vary). L = 1

    The Maximum Principle If a controlled curve (q, u) is optimal then there is anonzero multiplier (p, ), with either 0, 1 and such that H(q,u,p,) = 0.

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    GEOMETRIC CONTROL 30

    Maximum Principle: some examplesBang Bang: the rocket car.

    Control Law:

    ddt

    xv

    =

    0 10 0

    xv

    +

    0u

    .

    Hamiltonian: H = pxv +pvu + .

    Equations: px = 0 and pv = px, or px = c0 and pv = c0t + c1.

    Maximality condition: c0v (c0t c1)u attains maximal value ifu = sgn((c0tf + c1)) = 1.

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    GEOMETRIC CONTROL 31

    sgn((c0tf + c1)) changes sign at most once at most one discontinuousjump in u.

    x(t) = t2

    2 + v0t + x0

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    GEOMETRIC CONTROL 32

    Optimal Damping of an oscillation.

    Dynamics: x = x + u, with |u| 1. Control law

    d

    dt x

    v

    = 0 1

    1 0 x

    v

    + 0

    u

    .

    Hamiltonian: H = pxv +pv(x + u) + .

    Equations: px = pv and pv = px,

    px = A cos(t + )pv = A sin(t + )

    .

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    GEOMETRIC CONTROL 33

    Maximality condition: pxv +pv(x + u) attains maximal value ifu = sgn(sin(t )) = 1.

    sgn(sin(t )) changes sign with period .

    x(t) = B sin(t + ) 1 circles in phase space with centre in (1, 0) or(1, 0).

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    GEOMETRIC CONTROL 34

    Lagrangian Equations

    Trivial control system: q = f(q, u) = u (controls are velocities, control domain

    = IRn)

    Hamiltonian: H = piui + L.

    Equations: p = Lqi

    Maximality condition: Hui

    = pi + Lqi

    = 0 ( = 0 gives contradiction, so only

    = 1)

    Reduces to d/dt(L/qi) L/qi = 0.

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    GEOMETRIC CONTROL 35

    References

    [1] A. Agrachev and Y. Sachkov. Control Theory from the GeometricViewpoint, volume 87 of Encyclopaedia of Mathematical Sciences.Springer-Verlag, 2004.

    [2] V. Jurdjevic. Geometric Control Theory, volume 51 of Cambridge Studiesin Advanced Mathematics. Cambridge University Press, 1997.

    [3] L.S. Pontryagin, V.G. Boltyanskii, R.V. Gamklelidze, and E.F. Mishchenko.The Mathematical Theory of Optimal Processes. Wiley, Interscience,1962.

    [4] L.C. Evans. An Introduction to Mathematical Optimal Control Theory.Prepint.