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Optimal Control

Prof. Daniela Iacoviello

Lecture

Grading

Project + oral exam

Example of project:

Read a paper on an optimal control problem

1) Study: background, motivations, model, optimal control, solution,

results

2) Simulations

You must give me, al least ten days before the date of the exam:

-A .doc document

-A power point presentation

- Matlab simulation files

Prof. D.Iacoviello- Optimal Control 03/12/2018 Pagina 2

Example of project:

-Read a paper on an optimal control problem (any field will do)

You can choose a topic proposed in the papers I sent you, or from the

examples proposed in the books suggested (i.e. Evans) , or some

specific problem of your interest.


1) Study: background, motivations, model, optimal control, solution,

results

You have to indicate clearly:

-the background of the problem you are studying by a literature

research, motivations (why the problem is important?),

-the model assumed (indicate variables, parameters, their meaning,

etc),

-the cost index and the optimal control introduced (the meaning, what

they represent, different possible choices, etc).

2) Simulations: You have to implement the model and the control by

matlab

Describe the program, propose the results and discuss them.


YOU MUST STUDY ON THE BOOKS

THE SLIDES ARE NOT SUFFICIENT


Schedule

Lecture 1:

Introduction to optimal control: motivations, examples.

Definitions

Unconstrained optimization (first order necessary conditions,

second order necessary conditions)

Weierstrass theorem

Constrained optimization, introduction of Lagrangian function

First order necessary conditions, Second order sufficient

conditions.

Function spaces: First order necessary condition, second

order necessary condition, necessary and sufficient condition


Definitions

Consider a function

And

denotes the Eucledian norm

A point is a local minimum of over

If

RRf n →::

nRD

•

Dx *f

nRD

)()(

0

*

*

xfxf

xxsatisfyingDxallforthatsuch

−


Definitions

Consider a function

And


A point is a strict local minimum of over

If

RRf n →::

nRD

•

Dx * fnRD

**

*

),()(

0

xxxfxf

xxsatisfyingDxallforthatsuch

−


Definitions

Consider a function

And


A point is a global minimum of over

If

RRf n →::

nRD

•

Dx *f

nRD

)()(

0

* xfxf

Dxallforthatsuch


Unconstrained optimization - first order necessary conditions

All points sufficiently near in are in

Assume and its local minimum. Let

an arbitrary vector.

Being in the unconstrained case:

Let’s consider:

0 is a minimum of g

First order necessary condition for optimality

x

nRd

DnR

1Cf *x

)(:)( * dxfg +=

0* toenoughcloseRDdx +

*x

0)( * = xf


Unconstrained optimization- second order conditions

Assume and its local minimum. Let

an arbitrary vector.

Second order Taylor expansion of g around

Since

nRd2Cf *x

0=

0)(

lim),()0(''2

1)0(')0()(

2

2

0

2 =+++=→

oogggg

0)0(' =g

0)0('' g

0)( *2 xf Second order necessary condition for

optimality


Unconstrained optimization- second order conditions

Let and

is a strict local minimum of f

0)( * = xf2Cf 0)( *2 xf

*x


Definitions- Remarks

A vector is a feasible direction at if

If D in not the entire then D is the constraint set over

which f is being minimized

*xnRd

0* + enoughsmallforDdx

nR


Global minimum

Weierstrass Theorem

Let f be a continuous function and D a compact set

there exist a global minimum of f over D


Constrained optimization

Let

Equality constraints

Inequality constraints

Regularity condition:

Lagrangian function

If the stationary point is called normal and we

can assume .

1, CfRD n 1,:,0)( ChRRhxh p →=

1,:,0)( CgRRgxg q →

( )

a

a

a

x

a

qg

ofconstrainactivethearegwhere

qpx

ghrank

dimensionwith

,

*

+=

( ) )()()(,,, 00 xgxhxfxL TT ++=

*x00

10 =Prof. D.Iacoviello- Optimal Control 03/12/2018 Pagina 15

From now on and therefore the Lagrangian is

If there are only equality constraints the are called

Lagrange multipliers

If there are both equality and inequality constrains we have

Kuhn – Tucker multipliers


i

10 =

( ) )()()(,, xgxhxfxL TT ++=



First order necessary conditions for constrained

optimality:

Let and

The necessary conditions for to be a constrained local

minimum are

If the functions f and g are convex and the functions h

are linear these conditions are necessary and

sufficient!!!

i

ixg

x

L

i

ii

T

T

x

=

=

0

,0)(

0

*

*

Dx *1,, Cghf

*x



Second order sufficient conditions for constrained

optimality:

Let and and assume the conditions

is a strict constrained local minimum if

ixgx

Liii

T

T

x

==

0,0)(0 *

*

Dx * 2,, Cghf

*x

piddx

xdhthatsuchdd

x

Ld

x

i

x

T ,...,1,0)(

0**

2

2

==


Function spaces

Functional

Vector space V,

is a local minimum of J over A if there exists an

RVJ →:

VA

Az *

)()(

0

*

*

zJzJ

zzsatisfyingAzallforthatsuch

−


Function spaces

Consider function in V of the form

The first variation of J at z is the linear function

such that

First order necessary condition for optimality:

For all admissible perturbation we must have:

RVz + ,,

RVJz

→:

and

)()()()( oJzJzJ z ++=+

0)(* =z

J


Function spaces

A quadratic form is the second variation

of J at z if

we have:

second order necessary condition for optimality:

If is a local minimum of J over

for all admissible perturbation we must have:

RVJz

→:2

and

)()()()()( 222 oJJzJzJ zz +++=+

0)(*

2 z

J

Az * VA


Function spaces

The Weierstrass Theorem is still valid

If J is a convex functional and is a convex set

A local minimum is automatically a global one and the first

order condition are necessary and sufficient condition for

a minimum

VA


Schedule

Lecture 2:

Calculus of variations- Motivations

The Lagrange problem: Euler equation, Weierstrass

Erdmann condition, transversality conditions


The Lagrange problem

Problem 1

Let us consider the linear space and define the

admissible set:

Introduce the norm:

and consider the cost index:

with L function of C2 class.

Find the global minimum (optimum) for J over D, :

RRRC )(1

( ) ( ) ( ) 111 ),(,),(:)(,, ++ = vf

viiii RDTTzRDttzRRRCTtzD

=T

t

i

i

dtttztzLTtzJ ),(),(),,(

( ) TttztzTtz itt

i +++= )(sup)(sup,,

( )ooi

o Ttz ,,

( ) ( ) DTtzTtzJTtzJ iioo

io ),,(,,,,


03/12/2018 Pagina 25Prof.Daniela Iacoviello- Optimal Control Pagina 2503/12/2018

SCHEME of the theorems

Theorem (Lagrange). If is a local minimum then

1)

2) In any discontinuity point of

Weierstrass-Erdmann condition

3) Transversality conditions different cases

depending on the nature

of the boundary conditions

( ) Dttz fi *** ,,

fiT ttt

z

L

dt

d

z

L,0

**

=

−

t *z

Euler equation

Theorem (Lagrange). If is a local minimum then

In any discontinuity point of the following conditions are

verified:

( ) DTtz i *** ,,

Tttz

L

dt

d

z

Li

T ,0

**

=

−

Euler equation

t *z

****

+−+−

−=

−

=

tttt

zz

LLz

z

LL

z

L

z

L

Weierstrass-

Erdmann

condition


Moreover, transversality conditions are satisfied:

• If are open subset we have:

• If are closed subsets defined respectively by

such that

for

fi DD

0000**

**

**

**

===

=

Tt

T

T

T

t

LLz

L

z

L

i

i

fi DD

( ) ( ) 0),(0),( == ffii ttzttz

( ) ( ) fi

ii TTzrg

ttzrg

=

=

**

),(),(

fi RR

**

****

**

**

,

)(,

)(

Tz

z

LL

tz

z

LL

Tzz

L

tzz

L

T

Ti

T

t

T

Ti

T

t

i

i

=

−

=

−

=

=


• If the sets are defined by the function

affine with respect to such thatfi DandD ))(),(( Tztzw i

)(),( Tztz i

( )=

*

)(),( Tztz

wrg

i

R

Tz

w

z

L

tz

w

z

L T

Ti

T

ti

−=

=

****

)(,

)( **




Problem 2

Consider Problem 1 with

✓ fixed

✓ If are closed sets in defined by the C1

functions

fi tandt

fi DD 1+vR

( )( ) 1dimension of,0),(

1dimension of,0),(

f

i

+=

+=

vTTz

vttz ii

with affine functions and and

( ) ( ) f

o

fi

o

i tzrg

tzrg

=

=

)()(


✓ If the sets are defined by the function

with σ components of C1 class affine with respect to

such that

fi DD ))(),(( fi tztzw

)(),( fi tztz

( )=

o

fi tztz

wrg

)(),(

✓ The function L must be convex with respect to

Find the global minimum (optimum) for J over D:

( ))(),( tztz

oz

( ) ( ) DzzJzJ o

Theorem 2. is the optimum if and only if

In any discontinuity point of the following conditions

are verified:

Moreover, transversality conditions are satisfied

…..

Dzo

fiT

oo

tttz

L

dt

d

z

L,0 =

−

Euler equation

t *z

o

t

o

t

o

t

o

t

zz

LLz

z

LL

z

L

z

L

+−+−

−=

−

=

Weierstrass-

Erdmann

condition

Pagina 31Prof.Daniela Iacoviello- Optimal Control 03/12/2018


Problem 1_bis

Let us consider the admissible set:

consider the cost index:

with L function of C2 class.

Find the global minimum (optimum) for J over D

ffiifi ztzztzttCzD === )(,)(:],[1

=T

t

i

i


Prof.Daniela Iacoviello- Optimal Control



Theorem (Legendre)

Necessary condition for to be local minimum is that Dz *

],[0

*

2

2

fi tttz

L




Problem 3

Let us consider the linear space and define

the admissible set

of dimension

Consider the cost index:

RRRC )(1

( ) ( )

( ) ( ) ( )

==

=

+

+

kdtttztzhttztzgRDTTz

RDttzRRRCTtzD

T

t

v

f

v

iiii

i

),(),(0),(),(,),(

,),(:)(,,

1

11

g v

=T

t

i

i



Define the augmented lagrangian:

( ) ( )

( ) ( )ttztzhtttztzgt

ttztzLtttztz

TT ),(),()(),(),()(

),(),(),(,,),(),( 00

++

=


Theorem 3 (Lagrange). Let be such that

If is a local minimum for J over D, then there exist

not simultaneously null in

such that:

❑

❑

with TRANSVERSALITY

CONDITIONS

( ) Dttz fi *** ,,

**

*

,)(

fi ttttz

grank =

****

,0 fiT ttt

zdt

d

z=

−

( )*** ,, fi ttz

RttCR fi **0**0 ],,[, ],[ *

fi tt

03/12/2018 Pagina 36Prof.Daniela Iacoviello- Optimal Control

****

+−+−

−=

−

=

kkkk tttt

zz

zzzz

where are cuspid points forkt

*z


Problem 4 Let us consider the linear space

and define the admissible set

❑ of dimension

❑ ti tf fixed

❑ g and h affine functions in

❑ L C2 function convex with respect to

Consider the cost index:

RRRC )(1

( ) ( )

=== ],[,),(),(,0),(),(,)(,)(,,1fi

t

t

fiiifi tttkdtttztzhttztzgDtzDtzttCzD

f

i

g v


],[),(),( fi ttttztz

],[),(),( fi ttttztz

=f

i

t

t

fi dtttztzLttzJ ),(),(),,(

Define the augmented lagrangian:

( ) ( )

( ) ( )ttztzhtttztzgt

ttztzLtttztz

TT ),(),()(),(),()(

),(),(),(,,),(),( 0

++

=


Theorem 4 (Lagrange). Let such that

is an optimal normal solution

if and only if

➢

➢ in the instants for which

are open we have:

Dzo

fi

o

ttttz

grank ,

)(=

Dzo

fiT ttt

zdt

d

z,0

**

=

−


fitt and/or fi

DD and/or

To

z0=

Schedule

Lecture 3:

Calculus of variations and optimal control:

The Hamiltonian function- Necessary conditions- Necessary

and sufficient conditions


Problem 1

Let us consider the dynamical sistem described by:

f function of C2 class.

),,( tuxfx =

x(t) state vector in Rn

u(t) control vector in Rp

ii xtx =)(known


1+ nf

( ) 0),( =ff ttx

Vectorial function of C1 class

of dimension

0),,( tuxq


of dimension

Define the cost index

with L function of C2 class

( )=

f

i

t

t

f dttuxLtuxJ ,,),,(

Assume the norm:

( ) ft

t

ttttf ttudutxtxtux

i

++++= )(sup)(sup)(sup)(sup,,


AIM: Find (if it exists)

❑ the instant

❑ the control

❑ the state

that satisfy the previous equations and minimize the cost

index

oft

)(0 RCuo

)(1 RCxo


DEFINE the scalar function

the Hamiltonian function

),,()(),,(),,,,( 00 tuxfttuxLtuxH T +=


Theorem 1

Let be an admissible solution such that )*,*,( *ftux

*

**

,),()),((

fiaactive

fff

ttttu

qrk

ttxrk =

=


IF is a local minimum

there exist

not simultaneously null in such that:

)*,*,( *ftux

],[],,[,0 *0**1**0 fifi ttCttC

],[ *fi tt

dimension of

***

*

TT

x

q

x

H

−

−=

03/12/2018 Pagina 46

***

0

TT

u

q

u

H

+

=

,...,2,1,0)(,0)*,*,()( ** == jttuxqt jjj

❑

❑

❑

f

f

Rtx

t

T

tf

f

−= ,

))(()(*

*

*

*

❑

T

tf

t

f

f tH

*

*

*

*

=❑

The discontinuity of may occurr only in the points

where has a discontinuity and

** and t

*u**+− = tt

HH


Problem 3

Let us consider the dynamical system described by:

f function of C2 class

),,( tuxfx =

x(t) state vector in Rn

u(t) control vector in Rp

ii xtx =)(known

Calculus of variation and optimal control


1+ nf

( ) 0),( =ff ttx


of dimension

( ) =

f

i

t

t

Kdtttutxh ),(),(


of dimension


with L function of C2 class

( )=

f

i

t

t

f dttuxLtuxJ ,,),,(

Assume the norm:

( ) ft

t

ttttf ttudutxtxtux

i

++++= )(sup)(sup)(sup)(sup,,


AIM: Find

❑ the instant

❑ the control

❑ the state

that satisfy the previous equations and minimize the cost

index

oft

)(0 RCuo

)(1 RCxo


DEFINE the scalar function

the Hamiltonian function

( )ttutxhtuxfttuxLtuxH TT ),(),(),,()(),,(),,,,( 00 ++=


Theorem 3

Let be an admissible solution such that )*,*,( *ftux

fff ttx

rk

=

*

)),((


IF is a local minimum

there exist not

simultaneously null in such that:

❑

❑

❑

❑

)*,*,( *ftux

RttCR fi **1**0 ],,[,

],[ *fi tt

T

x

H*

*

−=

f

f

Rtx

t

T

tf

f

−= ,

))(()(*

*

*

*

T

u

H*

0

=

T

tf

t

f

f tH

*

*

*

*

=

The discontinuity of may occurr only in the points

where has a discontinuity

and

* kt*u

**+− =

kk ttHH


Problem 4:

Consider the linear system

Assume:

❑ fixed

❑

❑

utBxtAx )()( +=

Functions of C2 classfi tt ii xtx =)(

nRor pointfixedabeing)( fff DDtx

0),,( tuxq Vectorial function of C2 class of dimension

CONVEX



))((),,(),( f

t

t

txGdttuxLuxJ

f

i

+=

Functions of C3 class- CONVEX

Functions of C2 class CONVEX

Theorem 4

Let be an admissible solution such that

is a normal

optimal solution

),( oo ux

fia

oactive ttttu

qrk ,),( =

),( oo ux

0)(

,...,2,1

,0),,()(

0

=

=

+

=

−

−=

t

j

tuxqt

u

q

u

H

x

q

x

H

o

ooj

oj

ooToT

ooToT

o


nf RD =AND IF

oT

ff

o

tdx

dGt

)()( =

Schedule

Lecture 4:

Pontryagin minimum principle-

Necessary conditions- The convex case- Example


The Pontryagin principle

Problem 1: Consider the dynamical system:

with:

Assume fixed the initial control instant and the initial and final

values :

Define the performance index :

with

( )uxfx ,=

( ) niURCx

ffRUtuRtx n

i

pn ,...,2,1,,)(,)( 0 =

Pagina 5603/12/2018Prof.Daniela Iacoviello- Optimal Control

ff

ii xtxxtx == )()(

( ) ( ) ))(()(),(,, f

t

t

f txGduxLtuxJ

f

i

+=

( ) 20 ,,...,2,1,, CGniURCx

LL n

i

=

Determine:

▪ the value ,

• the control

• the state

that satisfy:

✓ the dynamical system,

✓ the constraint on the control,

✓ the initial and final conditions

✓ and minimize the cost index

( ) ,if tt

)(0 RCuo

)(1 RCxo


Hamiltonian function

( ) ( ) ( )uxftuxLuxH T ,)(,,,, 00 +=



Theorem 1 (necessary condition):

Assume the admissible solution is a minimum

there exist a constant

and a n-dimensional vector

not simultaneously null such that :

( )*** ,, ftux

00

*1* , fi ttC

T

x

H*

*

−=

( ) ( )U

ttutxHttxH

,)(,),(),()(,,),( **0

****0

*


0*=H


Problem 2

Theorem 2

Problem 3

Theorem 3



Problem 4: Consider the dynamical system:

with:

and initial instant and state fixed

( )tuxfx ,,=

( )RURCt

f

x

ffRUtuRtx npn

0,,)(,)(

ii xtx =)(


For the final values assume: where is a

function of dimension of C1 class.

Assume the constraint with


with

( ) 0),( = ff ttx

1+ nf

( ) kduxh

f

i

t

t

= ),(),(

( ) niRURCt

h

tx

hh n ,...,2,1,,

)(, 0 =

( ) ( )=f

i

t

t

f duxLtuxJ ),(),(,,( )RURC

t

L

x

LL n

0,,

Determine

• the value

• the control

• and the state

that satisfy

✓ the dynamical system,

✓ the constraint on the control,

✓ the initial and final conditions

✓ and minimize the cost index .

( ) ,if tt

)(0 RCuo

)(1 RCxo


Hamiltonian function

( ) ( ) ( ) )),(),((,)(,,,, 00 ttutxhuxftuxLuxH TT ++=


Theorem 4 (necessary condition):

Consider an admissible solution such that

IF it is a local minimum

there exist a constant , ,

not simultaneously null such that :

( )*** ,, ftux

00 *1* , fi ttC

( ) f

ff ttxrank =

*

),(


,

**

T

x

H

−=

( ) ( )U

ttutxHttxH

,)(,),(),()(,,),( **

0

****

0

*

Moreover there exists a vectorsuch that:fR

T

ft

T

f tH

txT

f

*

*

*

**

)()(

−=

=

RkkdH

H

ft

t

=

+ ,

*

**

R*

The discontinuities of may occur only in the instants

in which u has a discontinuity and in these instants the

Hamiltonian is continuous

*


Remark

If the set U coincides with Rp the minimum

condition reduces to :

0=

u

H


The Pontryagin principle - convex case

Problem 5: Consider the dynamical linear system:

with A and B of function of C1 class; assume fixed the initial

and final instants and the initial state, and

Assume

where U is a convex set.

utBxtAx )()( +=

nfff Rtxorfixedxtx = )()(

fip tttRUtu ,)(



with

L convex function with respect to x(t), u(t) in

per ogni

G is a scalar function of C2 class and convex with

respect to x(tf).

( ) ( ) ( ))(),(),(, f

t

t

txGduxLuxJ

f

i

+=

( ) nittURCt

L

x

LL fi

n

i

,...,2,1,,,, 0 =

URn

fi ttt ,


Determine:

the control

and the state

that satisfy

the dynamical system,

the constraint on the control,

the initial and final conditions

and minimize the cost index .

fio ttCu ,0

fio ttCx ,1


Theorem 5 (necessary and sufficient condition):

Consider an admissible solution such that

It is a minimum normal (i.e.λ0 =1)

if and only if there exists an n-dimensional vector

such that :

( )oo ux ,

fio ttC ,1

( ) f

o

ff ttxrank =

),(


oTo

x

tuxH

−=

),,,(

( ) ( ) UttutxHttxH ooooo ,)(),(),()(,),(

nf Rtx )(

oT

f

fo

tdx

dGt

)()( =Moreover, if

Schedule

Lecture :

The Hamilton- Jacobi equation- Derivation of the H-J

equation-

Principle of optimality- Example


Principle of optimality

A control policy optimal over the interval [t, tf] is optimal

over all subintervals [t1, tf]

If the shortest path from Rome to Paris passes through

Milan THEN the OBTAINED path between Milan to Paris

is the shortest one between Milan to Paris

Prof. D.Iacoviello - Optimal Control 7203/12/2018

MILAN

Rome

PARIS

Schedule

Lecture 8:

The optimal regulator problem:

The optimal tracking problem

The optimal regulator problem with null final error

The optimal regulator problem with limited control

Examples


The regulator problem

)()()(1 tKtBtR T−

r=0+

-

uo

xi

xo ii

o

oTo

xtx

txtKtBtRtBtAtx

=

−= −

)(

)()()()()()()( 1

03/12/2018Prof. D.Iacoviello - Optimal Control Pagina 74


Let us consider the following linear system:

with fixed, and fixed

are matrix functions of time with

continuous entries; their dimensions are respectively

)()()(

)()()()()(

txtCty

tutBtxtAtx

=

+=

],[ fi tt

pnmnnn ,,

)(),(),( tCtBtA

ii xtx =)(



A linear control law is obtained if we seek to minimize the

quadratic performance index:

where Q(t) and R(t) have continuous entries, symmetric,

nonnegative and positive definite respectively;

F is nonnegative definite matrix simmetric

( )

)()(2

1

)()()()()()(2

1),(

0

ffT

t

t

TT

tFxtx

dttutRtutxtQtxuxJf

+

+=


The Riccati equation

Theorem: The Riccati equation

admits a unique solution, symmetric, semidefinite positive,

in the control interval.


FtK

tQtKtAtAtKtKtBtRtBtKtK

f

TT

=

−−−= −

)(

)()()()()()()()()()()( 1

Solution of the regulator problem

Theorem: The optimal control for the regulator problem is

given by the linear feedback law:

where:

and

)()()()()( 1 txtKtBtRtu oTo −−=

ii

o

oTo

xtx

txtKtBtRtBtAtx

=

−= −

)(

)()()()()()()( 1

( ) i

i

iToo xtKxuxJ )(2

1, =



Theorem: The regulator problem admits a unique solution.

Proof: the uniqueness property is shown by contraddiction

arguments.


Solution of the deterministic linear optimal

regulator problem on

Problem: let us consider the regulator problem over the

infinite time interval .

Assume:

o The matrices A and B are bounded and with elements in

C1 class

o The dynamical system is completely controllable or

exponentially stable

o The matrices Q and R are symmetric, semidefinite and

positive definite respectively, with elements in C1 class

and bounded

),it

),it




Find:

the control

and the state

satisfying the dynamical system , the initial condition and

minimizing the cost index

) ,0i

o tCu

) ,1i

o tCx

( )

+=

it

TT dttutRtutxtQtxuxJ )()()()()()(2

1,

),it




Theorem: The problem admits a unique optimal solution

that can be expressed as follows:

where is the solution of the Riccati equation

with final condition:

and

ii

ooTo

oTo

xtxtxtKtBtRtBtAtx

txtKtBtRtu

=−=

−=

−

−

)(),()()()()()()(

)()()()()(

1

1

)(tK

)()()()()()()()()()()( 1 tQtKtAtAtKtKtBtRtBtKtK TT −−−= −

0)(lim =→

ft

tKf

( ) ii

iToo xtKxuxJ )(2

1, =

),it


The steady state solution of the deterministic

linear optimal regulator problem

Theorem: Let us consider the previous problem with the

additive hypotheses :

o The matrices A, B, Q, R are constant

o The matrix Q is positive definite

Then there exists a unique optimal solution:

where:

i

i

oo

r

To

o

r

To

xtxtxKBBRAtx

xKBRtu

=−=

−=

−

−

)(),()(

)(

1

1


Kr is the constant matrix, unique solution definite positive of

the algebraic Riccati equation:

The minimum value for the cost index is:

01 =−−−− QKAAKKBBRK r

T

rr

T

r

( ) i

r

iToo xKxuxJ2

1, =


Remarks

Consider the system:

It can be defined an optimal regulation problem from the

output y, considering the cost index

)()()(

)()()()()(

txtCty

tutBtxtAtx

=

+=

( )

++

=

f

i

t

t

TT

ffT

dttutRtutytQty

tyFtyuyJ

)()()()()()(2

1

)()(2

1,


The present problem is similar to the previous one, being

satisfied the hypotheses over the matrices involved.

Anyway the state must be accessible, if one want to

realize the feedback action:

ooo BuAxx +=

KBR T1−

r=0 +

-

uo

xi

xo

Cyo



Problem: let us consider the linear system:

where are matrices with elements functions of

time with entries of C1 class.

Consider the reference variable

Determine the optimal control

and the state satisfying the dinamical

constraint and minimizing the cost index:

i

i xtxtutBtxtAtx =+= )(),()()()()(

)(),( tBtA

fi ttCr ,1

fio ttCu ,0

fio ttCx ,1


where:

o Q(t) is a symmetric semi-positive definite matrix

o R(t) is a symmetric positive definite matrix

o The elements of Q(t) and R(t) are functions of C1

class

( ) +−−=

f

i

t

t

TTdttutRtutxtrtQtxtruxJ )()()()()()()()(

2

1,



To the optimal tracking problem it can be associated the

Riccati equation

0)(

)()()()()()()()()()()( 1

=

−−−= −

f

TT

tK

tQtKtAtAtKtKtBtRtBtKtK


Theorem: the optimal tracking problem admits a unique

optimal solution:

i

ioT

oTo

oTo

xtxtgtBtRtB

txtKtBtRtBtAtx

txtKtgtBtRtu

=+

−=

−=

−

−

−

)()()()()(

)()()()()()()(

)()()()()()(

1

1

1


where g is the solution of the differential equation

The minimum value of the cost index is:

where is the solution of the equation:

0)(

)()()()()()()()()( 1

=

−−= −

f

TT

tg

trtQtgtAtBtRtBtKtg

( ) )()()()()()(2

1, iii

oT

i

o

ii

oToo tvtgtxtxtKtxuxJ +−=

v

0)(

)()()(2

1)()()()()(

2

1)( 1

=

−= −

f

TTT

tv

trtQtrtgtBtRtBtgtv


ooo BuAxx +=

K

g +

-

uo

xi

xo

TBR 1−


The optimal regulator problem with null final error

Problem: Let us consider the linear system:

Determine the control

and the state

in order to satisfy the dynamical system, the initial and

final conditions and minimizing the following cost

index:

0)(,)(

),()()()()(

==

+=

fi

i txxtx

tutBtxtAtx

fio ttCu ,0

fio ttCx ,1


( ) +=

ft

TT dttutRtutxtQtxuxJ

0

)()()()()()(2

1,

where:

o Q(t) is symmetric semidefinite positive with elements of C1

class

o R(t) is symmetric definite positive with elements of C1

class


Theorem: Let us introduce the matrix of dimension 2n×2n

and indicate with

its transition matrix partitioned in submatrices of dimension

n×n.

Assume that the submatrix is not singular .

The optimal regulator problem with null final

error admits a unique optimal solution:

−−

−=

−

)()(

)()()()()(

1

tAtQ

tBtRtBtAt

T

T

( )( ) ( )

( ) ( )

=

,,

,,,

2221

1211

tt

ttt

( )if tt ,12


( ) ( ) ( ) ( ) iififii

To xtttttttttBtRtu ,,,,)()()( 111

1222211 −− −−=

( ) ( ) ( ) ( ) iififii

o xtttttttttx ,,,,)( 111

121211 −−=


Schedule

Lecture 9:

The minimum time problem- the minimal time optimal

problem – characterization of the optimal solution-

uniqueness result- Existence- Minimum time problem in case

of steady state system

Switching points


The linear time-optimal control

Let us consider the problem of optimal control of a linear

system

✓ with fixed initial and final state,

✓ constraints on the control variables and

✓ with cost index equal to the length of the time interval


SINGULAR SOLUTIONS

Definition:

Let be an optimal solution of the above problem,

and the corresponding multipliers.

The solution is singular if there exists a subinterval

in which the Hamiltonian

is independent from at least one component of in

( )of

oo tux ,,oo 0

( ) ''','',' tttt ( )tttxH ooo ),(,,),( 0

( )'',' tt


SINGULAR SOLUTIONS

Theorem:

Assume the Hamiltonian of the form

Let be an extremum and the corresponding

multipliers such that is dependent on any

component of in any subinterval of

A necessary condition for to be a singular

extremum is that there exists a subinterval

such that:

( )*** ,, ftux**

0

),,,,(),,,(),,,(),,,,( 002010 tuxNtxHtxHtuxH +=

),,,,( *0

** txN

],[ fi tt

'''],,[]'','[ tttttt fi

]'','[,0),,,( 02 ttttxH =

( )*** ,, ftux


The linear minimum time optimal control

Let us consider the problem of optimal control of a linear

system

✓ with fixed initial and final state,

✓ constraints on the control variables and

✓ with cost index equal to the length of the time interval


Problem: Consider the linear dynamical system:

With and the constraint

Matrices A and B have entries with elements of class

Cn-2 and Cn-1 respectively, at least of C1 class .

The initial instant ti is fixed and also:

)()()()()( tutBtxtAtx +=pn RtuRtx )(,)(

Rtpjtu j = ,,...,2,1,1)(

0)(,)( == fi

i txxtx


The aim is to determine the final instant

the control and the state

that minimize the cost index:

)(RCu oo

Rtof

)(1 RCxo

if

t

t

f ttdttJ

f

i

−== )(


Theorem: Necessary conditions for

to be an optimal solution are that there exist a constant

and an n-dimensional function ,

not simultaneously null and such that:

Possible discontinuities in can appear only in

the points in which has a discontinuity.

( )*** ,, ftux

0*

0 fi ttC ,1*

pjRBuB

A

j

pTT

T

,...,2,1,1:***

**

=

−=

**u


Moreover the associated Hamiltonian is a continuous function

with respect to t and it results:

Proof: The Hamiltonian associated to the problem is:

Applying the minimum principle the theorem is proved.

0*

* =ft

H

( ) BuAxtuxH TT ++= 00 ,,,,


STRONG CONTROLLABILITY

Let us indicate with the j-th column of the matrix B .

The strong controllability is guaranteed by the condition:

where :

)(tb j

( )

i

n

jjjj

ttpj

tbtbtbtG

=

=

,,...,2,1

0)()()(det)(det )()2()1(

nktbtAtbtb

tbtb

k

j

k

j

k

j

jj

,...,3,2)()()()(

)()(

)1()1()(

)1(

=−=

=

−−

Generic column of matrix B(t)


Remark:

Strong controllability corresponds to the

controllability in

any instant ti,

in any time interval and

by any component of the control vector


Remark:

In the NON-steady state case the above condition is

sufficient for strong controllability.

In the steady case it is necessary and sufficient and may

be written in the usual way:

( ) pjbAAbb jn

jj ,...,2,10det 1 =−


Characterization of the optimal solution

Theorem: Let us consider the minimal time optimal problem .

If the strong controllability condition is satisfied and if an

optimal solution exists

it is non singular.

Moreover every component of the optimal control is

piecewise constant assuming only the extreme values

and the number of discontinuity instants is limited.

1


A control function that assumes only the limit values

is called bang-bang control

and

the instants of discontinuity are called

commutation instants


Theorem (UNIQUENESS):

If the hypothesis of strong controllability is satisfied, if an

optimal solution exists it is unique.

Proof: the theorem is proved by contradiction arguments.


Remark

This result holds also to the case in which the control is in

the space of measurable function defined on the space of

real number R.


Existence of the optimal solution

Theorem: if the condition of strong controllability is satisfied

and an admissible solution exists, then there exists a

unique optimal solution nonsingular and the control is

bang-bang.

Proof: the existence theorem is proved on the basis of the

following results.


The following result holds:

Theorem: Let assume that the control functions belong to the

space of measurable functions.

If an admissible solution exists then the optimal solution exists.


The results about the characterization of the control may be

extended to the case in which the control belong to the

space of measurable functions:

Therefore it results in:

If the strong controllability condition is satisfied and

if an optimal control in the space of measurable function

exists then this solution is unique,

non singular and the control is bang bang type

Remark

The existence of the optimal solution is guaranteed only

for the couples for which the admissible solution

exists( )ii xt ,


Existence of the optimal solution

Theorem: if the condition of strong controllability is

satisfied and an admissible solution exists

with the control in the space of measurable functions,

then there exists a unique optimal solution

nonsingular and

the control is bang-bang.


Minimum time problem for steady state system

Let us consider the minimum time problem in case of

steady state system.

In this case it is possible to deduce results about the

number of commutation points


Problem: Consider the linear dynamical system:

With and the constraint

The initial instant ti is fixed and also:

)()()( tButAxtx +=

pn RtuRtx )(,)(

Rtpjtu j = ,,...,2,1,1)(

0)(,)( == fi

i txxtx


The aim is to determine the final instant

the control

and the state

that minimize the cost index:

)(RCu oo

Rtof

)(1 RCxo

if

t

t

f ttdttJ

f

i

−== )(


Theorem:

Let us consider the above time minimum problem and

assume that the control function belong to the space of

measurable function ;

if the system is controllable

there exists a neighbor Ω of the origin such that for any

there exists an optimal solution.ix


Theorem: Let us consider the above time minimum

problem in the steady state case and assume that the

control function belongs to the space of measurable

function M(R).

If the system is controllable and the eigenvalues of

matrix A have negative real part

there exists an optimal solution

whatever the initial state is.


Theorem: If the controllability condition is satisfied and if

all the eigenvalues of the matrix A are real non positive, then

the number of commutation instants for any components

of control is less or equal than n-1, whatever the inititial

state is.


Schedule

Lecture The minimum time problem-

examples: double integrator; harmonic oscillator


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