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Optimal Control
Prof. Daniela Iacoviello
Lecture 7
19/11/2019 Pagina 2
THESE SLIDES ARE NOT SUFFICIENT
FOR THE EXAM:YOU MUST STUDY ON THE BOOKS
Part of the slides has been taken from the References indicated below
Pagina 219/11/2019Prof. D.Iacoviello - Optimal Control
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
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 3
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
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 4
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
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 5
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
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 6
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
−== )(
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 7
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.
Moreover the associated Hamiltonian is a continuous function
with respect to t and it results:
( )*** ,, ftux
0*
0
fi ttC ,1*
pjRBuB
A
j
pTT
T
,...,2,1,1:***
**
=
−=
**u
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 8
0*
* =ft
H
Proof: The Hamiltonian associated to the problem is:
Applying the minimum principle the theorem is proved.
( ) BuAxtuxH TT ++= 00 ,,,,
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 9
STRONG CONTROLLABILITY
Strong controllability corresponds to the
controllability in
any instant ti,
in any time interval and
by any component of the control vector
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 10
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)
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 11
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 =−
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 12
Characterization of the optimal solution
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 13
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.
Every component of the optimal control is piecewise constant assuming
only the extreme values
and the number of discontinuity instants is limited.
1
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 14
Proof:
the non singularity of the solution is proved by contradiction
arguments.
If the solution were singular, from the necessary condition of the previous
theorem:
there would exist a value
and a subinterval
such that:
pjR
BuB
j
p
TT
,...,2,1,1:
***
=
pj ,...,2,1
],[]'','[ ofi tttt
]'','[,0)()( ttttbt j
oT =
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 15
Pagina 1619/11/2019Prof. D.Iacoviello - Optimal Conrol
Deriving successively this expression and using
we obtain:
On the other hand must be different from zero in
every instant of the interval , otherwise, from the
necessary condition
it should be null in the whole interval and in particular
]'','[,0)()( ttttGt j
oT =
)(to
** TA−=
0)( =fo t
** TA−=
( )1,...,2,1],'','[,0)()(
)()( )1(−===
+nittttbt
dt
tbtd ij
oT
i
joTi
Pagina 1719/11/2019Prof. D.Iacoviello - Optimal Conrol
From the necessary condition:
it would result that also which is absurd.
Therefore, from the condition:
it results that
which contradicts the hypothesis of strong controllability.
the solution is non singular.
0** =ft
H
00 =o
]'','[,0)()( ttttGt j
oT =
]'','[0)(det ttttG j =
From the non singularity of the optimal solution it results that the
quantity
can be null only on isolated points .
Therefore the necessary condition
implies
)()( tbt j
oT
pjRBuB j
pTT ,...,2,1,1:*** =
],[,...,2,1
,)()()(
ofi
joTo
j
tttpj
tbtsigntu
=
−=
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 18
To demonstrate that the number of discontinuity instants is
finite, we can proceed by contradiction argument that
would easily imply
which contradicts the hypothesis of strong controllability.
In fact, let us assume that at least for one component of the
control it is not true.
Then an accumulation point of instants exists and,
for continuity argument
]'','[0)(det ttttG j =
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 19
)(tuoj
ofi tt ,
0)()( = joT b
)( jk
t
For the continuity of between and
there exists an instant, say , in which
Therefore, for continuity, also
Analogously:
That is in contraddiction with the strong controllability hypothesis.
]'','[0)(det ttttG j =
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 20
0)()( =tbt joT
)( jk
t)(1
jk
t+
)( jk
t ( )0
)()(=
dt
tbtd joT
( )0
)()(=
dt
tbtd joT
( ))1(,...,1,0,0
)()(−== ni
dt
tbtd
i
joTi
A control function that assumes only the limit values
is called bang-bang control
and
the instants of discontinuity are called
commutation instants
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 21
UNIQUENESS RESULT
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 22
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.
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 23
Definition
Let be the space of piecewise constant function defined as
follows:
Let be a function such that there exists a sequence
such that , with the exception of isolated points, one has:
is a measurable function.
The space of measurable function is linear.
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 24
( , )i fS t t
1
, 1,2,...,
( )0
j j
m
j
j
c t I j m
s tt I
=
=
=
, ,j i f j kI t t I I =
( )z t ( ) ( , )ki fs S t t
,i ft t t
( )lim ( ) ( )k
ks t z t
→=
( )z t
( , )i fM t t
Remark
These results (characterization of the solution and uniqueness)
hold also to the case in which the control is in
the space of measurable function defined on the space of
real number R.
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 25
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.
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 26
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.
Proof: let us consider the non trivial case in which the number
of admissible solution is infinite and let us indicate with the
minimum extremum of the instants corresponding to the
admissible solutions
The problem is: is there a minimum?
oft
ft
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 27
infof ft t=
It is possible to built a sequence of admissible solutions
such that
Let us indicate with the transition matrix of the linear
system. It results:
( ) of
kf
kf
kk tttux )()()()( ,,,
of
kf
ktt =
→
)(lim
( ) ,t
( ) ( )
( ) ( ) ( ) 0)()(,lim)()(,,lim
,,lim)()(lim
)()()()(
)()()()(
)(
=+−+
−=−
→→
→→
dttutBttdttutBtttt
xtttttxtx
k
t
t
kf
k
k
t
t
of
kf
k
ii
ofi
kf
k
of
kkf
k
kk
f
o
f
o
f
i
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 28
( ) ( )
( ) ( ) dttutBttxtttx
dttutBttxtttx
k
t
t
of
ii
of
of
k
k
t
t
kf
ii
kf
kf
k
of
i
kf
i
)()(,,)(
)()(,,)(
)()(
)()()()()(
)(
+=
+=
0
It is possible to built a sequence of admissible solutions
such that
Let us indicate with the transition matrix of the linear
system. It results:
( ) of
kf
kf
kk tttux )()()()( ,,,
of
kf
ktt =
→
)(lim
( ) ,t
( ) ( )
( ) ( ) ( ) 0)()(,lim)()(,,lim
,,lim)()(lim
)()()()(
)()()()(
)(
=+−+
−=−
→→
→→
dttutBttdttutBtttt
xtttttxtx
k
t
t
kf
k
k
t
t
of
kf
k
ii
ofi
kf
k
of
kkf
k
kk
f
o
f
o
f
i
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 29
( ) ( )
( ) ( ) dttutBttxtttx
dttutBttxtttx
k
t
t
of
ii
of
of
k
k
t
t
kf
ii
kf
kf
k
of
i
kf
i
)()(,,)(
)()(,,)(
)()(
)()()()()(
)(
+=
+=
0
Since it results:
Let us indicate by the function truncated on the interval
This function is limited and L2 , since
The sequence of functions
belongs to the space M of measurable function, L2 and such that
Therefore this sequence admits a subsequence, indicated still with ,
weakly converging to a function
,....2,1,0)()()( == ktx
kf
k0)(lim )( =
→
of
k
ktx
)(ku ofi tt ,
)(ku
ofij tttpjtu ,,,...,2,1,1)( =
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 30
)(ku
Muo
.,...,2,1,1)()(
pjtukj =
)(ku
This implies that for any measurable function in L2 it results:
Weak convergence
If we indicate by the evolution of the state corresponding
to the input from the initial condition we have:
Weak convergence
h
=→
o
f
i
o
f
i
t
t
oT
t
t
kT
kdttuthdttuth )()()()(lim )(
oxou
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 31
( ) ( )
( ) ( ) )()()(,,
)()(,lim,)(lim )()(
of
o
t
t
oof
ii
of
t
t
kof
k
ii
of
of
k
k
txdttutBttxtt
dttutBttxtttx
o
f
i
o
f
i
=+=
+=
→→
Since the elements of
are measurable and L2
( ) )(, tBttof
Therefore, taking into account the expression
we deduce :
An admissible measurable control L2 exists able to transfer
the initial state to the origin.
The optimal solution is
0)(lim )( =→
of
k
ktx
0)( =of
o tx
( )of
oo tux ,,
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 32
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 ,
19/11/2019 Pagina 33
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.
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 34
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
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 35
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
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 36
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
−== )(
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 37
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
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 38
Proof:
We will show that it is possible to determine a neighbor
Ω of the origin such that for any there exists an admissible
solution.
On the basis of the previous results this implies the existence of the optimal
solution.
Let us fix any tf>0. The existence of a control able to transfer the initial
state to the origin at the instant tf corresponds to the possibility
of solving :
ix
ix
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 39
with respect to the control function u.
Let assume for the control the expression:
and substitute it in the preceeding equation:
( ) ( )0)( =+
−−f
i
fif
t
t
tAittAdBuexe
=−
−−−
f
i
Ti
t
t
ATAiAtdeBBexe Gramian
matrix
pfi
AT RtteBuT
= − ,,)(
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 40
From the controllability hypothesis, the Gramian matrix is not singular,
therefore:
a measurable admissible control exists if the initial state is sufficiently
near the origin.
fi
iAt
t
t
ATAAT
tt
xedeBBeeBu i
f
i
TT
−=
−
−
−−−
,)(
1
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 41
=−
−−−
f
i
Ti
t
t
ATAiAtdeBBexe
pfi
AT RtteBuT
= − ,,)(
Remark
From results obtained in the non steady state case it results that
an optimal non singular solution exists
and the control is bang-bang type.
Prof.Daniela Iacoviello- Optimal Control
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 42
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.
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 43
Proof:
Whatever the initial state is chosen, an admissible solution can be defined
as follows:
• Let Ω be a neighbor of the origin constituted by the original state
for which an admissible solution exists;
• Let Ω’ be a closed subset in Ω
• From the asymptotic stability of the system it is possible to reach Ω’ in a
finite time with null input, from any initial state
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 44
Ω
Proof:
Whatever the initial state is chosen, an admissible solution can be defined
as follows:
• Let Ω be a neighbor of the origin constituted by the original state for
which an admissible solution exists;
• Let Ω’ be a closed subset in Ω
• From the asymptotic stability of the system it is possible to reach Ω’ in a
finite time with null input, from any initial state
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 45
Ω
Ω’
Proof:
Whatever the initial state is chosen, an admissible solution can be defined
as follows:
• Let Ω be a neighbor of the origin constituted by the original state for
which an admissible solution exists;
• Let Ω’ be a closed subset in Ω
• From the asymptotic stability of the system it is possible to reach Ω’ in
a finite time with null input, from any initial state
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 46
Ω
Ω’x’
Remark
From the results obtained in the non steady state case
it results that an optimal non singular solution exists and the control is
bang-bang type.
Question: how many commutation instants are possible in
the optimal time control problem?
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 47
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.
Proof: from the previous results, we have:
where the costate in the steady state case is:
ofij
oToj tttpjbtsigntu ,,,...,2,1,)()( =−=
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 48
0
i
ttA iT
et )(0 )(−−
=
Denoting by and the eigenvalues of A and their molteplicity,
the r-th component of can be expressed as follows:
where prs is a polynomial function of degree less than ms.
Substituting this expression in the bang bang control we obtain:
k km0
nretpt
k
s
tt
rsris ,...,2,1,)()(
1
)(0 ===
−−
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 49
ofi
k
s
ttjs
k
s
ttn
r
rsjroj
tttpj
etPsign
etpbsigntu
is
is
,;,...,2,1
,)(
)()(
1
)(
1
)(
1
=
−=
−=
=
−−
=
−−
=
Polynomial function
of degree less than ms
The argument of the sign function has at most
m1 + m2+ …. + mk-1 =n-1
real solutions.
Therefore the control has at most n-1 roots.o
ju
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 50
Remarks
1. If the eigenvalues of A are not all real the number of
switching instants is bounded but not uniformly.
1. In general it is not possible to establish an esplicit relation
between the control and the state
19/11/2019Prof. D.Iacoviello - Optimal Control Pagina 51