An Example of an Optimal Control
Problem Whose Extremals Possess
a Continual Set of Discontinuities
of the Control Function
ALEXEI A. MILYUTIN
Institute of Chemical Physics of the Russian Academy
of Sciences, 2 ul. Kosygina, Moscow, Russia
Received March 9, 1992
We present an example of a classical optimal control problem having solutions that are
nonsingular extremals and whose control functions have a continual discontinuity set.
Moreover, for any closed, nowhere-dense set, there exists a solution such that the set of
discontinuities of the control coincides with this set.
W e consider the control system
2
z - (Spy, u),
x = |y| ,
y = u,
\u\^\
(1)
and the optimization problem
V
=
— ~ ~ *
x(T) — x(0)
m
a
x
under the constraint
y(0) = y(T).
(2)
Here Syy = (£iyy,£2yy) is a quadratic vector field on the two-dimensional y-plane.
This problem belongs to a class of problems which is important for higher order condition
theory. However, the problem itself is o f interest since it extends our comprehension o f
singularities in control systems.
W e assume that the field Syy is not a gradient one, since otherwise the functional J
vanishes on any admissible trajectory. Setting
dSiyy
l ( y )
dS yy
2
=
we readily see that l(y) is a linear function o f y. Therefore l(y) = (/, y), where / is some vector
from the plane M . Our assumption means that / ^ 0. W e shall find all solutions o f the problem (1),
(2). Let us write out the most important conditions.
T h e Pontryagin function o f the problem has the form: H = tp (£yy, u) + ^x\y\ + V'y *Since H does not depend explicitly on z, x, and t, it follows from the adjoint equations that
2
2
1
z
ip = const,
z
1061-9208/93/010397--6
© 1994 by John Wiley & Sons, Inc.
ipx — const,
H = const
RUSSIAN JOURNAL OF MATHEMATICAL PHYSICS VOL. 1 NO. 3
397
ALEXEIA. MILYUTIN
on any extremal o f the system ( 1 ) . T h e transversahty conditions give:
a
*'
=
,
*(T)-*(0)'
z(T)-z(0)
=
^
-°
T)-x(0))»'
W
iMO) = ^ ( T ) = / ? ,
(3)
H = 0.
y
A c c o r d i n g to the nontriviality condition, we have a + |/?| > 0. Let us write out the Hamiltonian:
2
H = m a x / f = l ^ y y + V'J + V'xLvl 1
w
1
It follows from condition (3) that the solution of the problem (more precisely, the corre­
sponding extremal) lies on the zero level of the Hamiltonian. Now we can easily prove that
a > 0. In fact, otherwise we would have tp — xp = 0. Then from 7i — 0 it follows that
ip = 0, hence 0 — 0, and the nontriviality condition is violated. Thus, a > 0. Taking into
account the obvious inequalities
z
x
y
z(T)-z(0)>0,
x(T)-x{0)>0,
we obtain \j) > 0, ip < 0.
In order to solve problem ( 1 ) , ( 2 ) , we use the Hamilton-Jacobi equation which has the
form
dS.
dS
,.
dS „
3S
z
x
l 2
oz
x
oy
On account o f the transversahty condition, it is natural to suppose that S — z — jx + s(y).
T h e n the Hamilton-Jacobi equation takes the form:
|£yy + « ; i - 7 l i / l
2
= o.
(5)
Since s'y is related, by virtue of this equation, to the quadratic field Syy, one may guess
that s' is also a quadratic field, i.e., the function s(y) is a cubic form. Put Syy = £yy + s' .
y
y
T h e n it follows from ( 5 ) that
i
ic
SZivy
d^iyy
i i i
1^1 = 7,
-
|y| = i,
v
n
- (/,y).
(6)
Let us find all solutions o f ( 6 ) . Let
£\yy
= aiyj + 2&iyiy + c i y f .
2
2
£syy = «2y + 2 6 y i y + c ^ 2
2
Put yi = cos (f, y = sin <p, then we obtain:
2
£i y(<p)y{(p) = ri (p) = j4i cos 2y? + 5 i sin 2<p + d,
where A\ = (cti — ci)/2, B\ — &i, C i = (ai + c i ) / 2 . Similarly,
C , where
r(<£>) = A
2
2
cos
2
<^+J5 sin 2<p+
2
2
A
= ^
2
,
B
2
= 6 ,
C = ^
2
2
.
(7)
If we set r((p) — {r\{<p), r (<p)), then (7) implies that the condition \r(<p)\ — const is equiv­
alent to one of the following two conditions:
2
a)Ai = Bi = A
2
6)i4? + B
398
2
1
=
= B
^l
2
= 0,
+ B|,
^li5i+^ 5
RUSSIAN JOURNAL OF MATHEMATICAL PHYSICS VOL. 1 NO. 3
2
2
= 0,
Ci = C = 0.
2
(
8
)
AN EXAMPLE OF THE OPTIMAL CONTROL PROBLEM
Put / = (l\,l ),
2
then, in view o f ( 7 ) , we have
k = 2(B
1
In case a), we have C\ = h/2,
- A
C
2
- C ),
l = 2(-B
2
2
= —
2
- A
2
(9)
Therefore,
(f(v,v),-|(y,y)),
Syy=
+ Ci).
t
T = f -
(10)
In case b ) , we have in view o f ( 9 ) :
Y = *i-4»,
I = -£2-^1.
However, it follows from (7) that
(B B )
U
=
2
±(-A ,A ).
2
1
It is clear that only the first possibility is to be retained, since in the second case l\ = l = 0.
Thus, B\ — —A ,B
— A\. Now we can express A], A \B\, B by means o f l\ and l . W e
obtain
2
2
2
2
*=-7-
B
' =7
2
^ = -7-
B 2
2
= -7
4
4
4
Therefore, in the case b ) the following equalities hold:
c
h / 2
2\ , '1
£ i y y = -j(vi
-y2) + -^ym,
^ ( 2
c
4
2\
^2
Kl
hvy =—^ivi -ifi) --^vm,
T = ^ -
(H)
Equalities (10) and (11) contain the complete solution o f equations ( 6 ) . W e consider the
cases (10) and (11) simultaneously. Put
/
v
u{y) =
£yy
\£yy\
T h e function u(y) is well-defined everywhere excluding the point y — 0, since in both cases
(10) and (11) we have 7 > 0. It is obvious that u(y) is smooth for y ^ 0 and
2
Nv)l = l
Vy€M -{0}.
Let z(t), x(t), y(t) be an arbitrary solution defined in some interval [/o,*i] and contained
in the domain y ^ 0 o f the system
2
z = (Eyy,u(y)),
x - |t/| ,
y = «(y).
(12)
Let 5 = 2: — 7X + « ( y ) be a solution of equation ( 5 ) . Then it is easy to see that
S = const
(13)
along the above trajectory. In fact,
^ = (emmMyit)))
= (iy(t)y(t),
u(y(t)))
-
7|y(*)l +
2
2
^(vM)
2
- | < / ( * ) | = I W M O I " 7 l < / ( 0 | = 0.
7
RUSSIAN JOURNAL OF MATHEMATICAL PHYSICS VOL. 1 NO. 3
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ALEXEI A. MILYUTIN
Therefore our assertion is true. On the other hand, if z(t), x(t), y(i), u(t) is an arbitrary
trajectory o f the system (1) such that y(t) ^ 0 for all t in some interval [2o,<i] and the
inequality u(t) ^ u(y(t)) holds on a subset of positive Lebesgue measure, then
S(z(h),
z(h),
y(h))
- S(z(t ),
x(t ),
0
y(t ))
0
< 0.
0
(14)
T h e proof o f this assertion is similar to the proof o f assertion (13). In fact, for such a
trajectory
^
= (£y(t)y(t),u(t))
- |y(*)|
2
T
\£y(t)M\ ~
^
2
T | y ( 0 | = 0,
and the inequality is strict if u(t) ^ u(y(t)). Now, (14) follows easily.
Now we again consider the cases (10) and (11) separately. In case ( 1 0 ) ,
V
l
t \
u{y) = j
T/
0
(
V = [ _
y
1
o
l
and, consequently, there are no cycles among solutions of the equation y = u(y). Therefore,
there are no admissible trajectories of the problem (1),(2) among solutions of system ( 1 2 ) .
Thus, case (10) has nothing to d o with the solution of problem ( 1 ) , ( 2 ) . In case ( 1 1 ) , we
represent u(y) in the form:
u(y) = («i(y),«2(y)),
Wl(y) =
W 2 ( y )
Put C = yi + iy ,e
2
= -h/\l\
n(v)
=
W
y22)+
li("l "
i i ( ~ l
- ih/\l\.
(
y
i
2
" "
2
2
2yiy2
Ff )'
)
" l
2
y
i
y
T
Then
2
= j^WC ),
My)
=
2
^(OC ).
Therefore, the last equation of system (12) can be rewritten in the form
<=
(»,
2
Put dr = dt/\£\ . Then equation (15) takes the form d^/dr
general solution of this equation is given by
< = \ TTlc'
2
= 9( .
Up to a shift in r, the
( 1 6 )
where C is an arbitrary real number. For C = 0 we do not get a cyclic (periodic) solution
of equation ( 1 5 ) . However, for any C ^ 0 the corresponding solution of (15) will be cyclic.
We put
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AN EXAMPLE OF THE OPTIMAL CONTROL PROBLEM
Then the corresponding solution y ( / , C ) of equation (12) takes the form:
er(t)
+ iCJ
\9r(t)
+
iClJ
It is obvious that y(t,C) is a solution of (12) in the closed interval [ 0 , T ( C ) ] where T(C) =
7r/|C|. The solution y(t,C) is continuous in the closed interval [ 0 , T ( C ' ) ] , does not vanish in
the open interval ( 0 , T ( C ) ) , and vanishes at its endpoints. Therefore, the function y{t,C)
is an admissible trajectory for any C ^ 0.
By using (13) and ( 1 4 ) , it is not hard to show that y(t, C) is a solution of problem ( 1 ) , (2)
for any C ^ 0. In fact, let z(t), x(t), y(t, C) be a solution of system ( 1 2 ) . Then, according
to ( 1 3 ) ,
z(T(C))-z(Q)-^(x(T(C))-x(0))
= 0.
Hence,
z(T(C))
- z(0) _ |/|
x(T{C))
- x(0)
(17)
4
Now let z(t), x(t), y(t), ti(t) be any admissible trajectory defined in a closed interval [0, T].
Then, in view o f ( 1 4 ) ,
z(T) - z(0) - | ( x ( T ) - * ( 0 ) )
0.
If y(t) / 0, i.e., if x{T) - x(0) > 0, then it follows that
z(T) - z ( 0 )
l/[
*(T)-x(0) ^
4'
1
;
T h e comparison o f (17) and (18) gives the desired conclusion.
T h e existance of a multiplicity of solutions is related to the invariance o f problem ( 1 ) ,
(2) under a similarity group.
Namely, if y(t) is an admissible function, then the function \y(t/fi), | A / / i | = 1, is also
admissible and gives the same value o f the functional J as the initial one. By using the
solutions y(t,C),C
/ 0 and the fact that problem is invariant with respect to shifts of the
variable
we shall construct the general solution of the problem. Let [ 0 , T ] , T > 0, be
an arbitrary interval, and T C [0, T] be a closed set which contains the endpoints o f the
interval but does not coincide with it. W e define a function y?{t). First we set
yr(t)
= 0
VteF.
Let A C [ 0 , T ] be the complementary interval o f T.
Put
yHOU = y ( < - « L ( A ) , C ( A ) ) ,
where ti(A)
is the left end o f A , | C ( A ) | = 7r|A|. N o w the function y?{t) is completely
defined. It is easy to see that y?(t) satisfies the Lipschitz condition with constant 1 in
[0,T] and that y ^ ( 0 ) = y j - ( T ) = 0. Thus, y?(t) is an admissible function. Since T does not
coincide with the entire interval [0, T ] , this function is not equal to zero on a set of positive
Lebesgue measure.
Put uj:(i) — ^yj{i)It is obvious that
Jo (£yryr,ur)
dt _ E a J a
u
{£yryr> r)dt
RUSSIAN JOURNAL OF MATHEMATICAL PHYSICS VOL. 1 NO. 3
401
ALEXEI A. MILYDTIN
But
Ja {£yryF,u )dt
_ \i\
T
2
LM dt
4
Consequently, y?{t) is a solution of problem ( 1 ) , ( 2 ) .
T h e family yp{t) contains essentially all the solutions of this problem. Let y(t) be a
solution o f the problem in a certain interval [ 0 , T ] . It is not hard to show that there exists
a point t in the interval such that y(t) — 0. Otherwise, according to (14), the equation
y — u(y) must hold on the entire segment. However, as we have already seen, it does not
possess a cyclic solution in the domain y ^ 0. Let t G [ 0 , T ] , y(to) = 0. By a change o f
variables we reduce the situation to the case to = 0. T o this end we put:
0
y(to+t),
y(t
to+t^T,
+ t-T),
0
t +
t>T.
0
Then it is obvious that yi(t) is an admissible trajectory in [ 0 , T ] and y i ( 0 ) = yi(T) = 0. It
is quite clear that J [ y i ( ) ] = J[?/(•)], and thus yi(t) is also a solution.
Denote by T the set o f zeros of the function y\(t) in the interval [0, T ] . It is clear that
T satisfies all the necessary conditions. W e shall prove that
(t)
yi
= y {t).
(19)
T
Let A be the complementary interval of T. Then y\(t) is a cyclic solution of the equation
y — u(y) in A . Hence, there exist C, \C\ = 7r/|A|, such that
yi{t)\
=
A
y(t-t (A),C).
L
Eq. (19) is now proved.
Hence, an arbitrary solution o f the problem differs from the corresponding solution o f
the family { y j - ( ^ ) } only by the choice of the initial point of the cycle. Thus, the problem
( 1 ) , ( 2 ) is completely solved
W e note the following fact. Each solution of problem ( 1 ) , (2) is a trajectory c o m p o n e n t o f
an extremal. T h e adjoint variables of any solution are described, according to the HamiltonJacobi theory, as follows:
xp = 1,
z
where s' (y) = Syy — Syy,
y
4> = - j ,
x
=
s' (y(t)),
y
y(t) is a component of the solution. T h e extremal is singular
on the set of points t where Syy = 0 if the measure of this set is positive. It follows from
(6) that this set coincides with the zero set o f y(t). On the complementary set y(t) ^ 0
(or, what is the same, Syy ^ 0) the extremal is nonsingular since the control is expressed,
according to the m a x i m u m principle by means of y and ip .
Let T be a perfect set and mesJf = 0. Then, since T is just the zero set o f y?{t),
the corresponding extremal is nonsingular. On the other hand, since there are infinitely
complementary intervals in any neighborhood of any point o f JT, any point o f T is a point
of discontinuity of the control uj:{i). Therefore, the system (1) is an example of a control
system having nonsingular extremals such that the control has a continuum o f discontinu­
ities.
In the theory o f higher order conditions for singular extremals due to A . V . Dmitruk,
a somewhat more general problem than ( 1 ) , (2) is important. Namely, instead o f the
constraint |w| ^ 1, one must consider the restriction u £ U, where U is a convex c o m p a c t
set such that 0 G intf/. Using the above ideas, A . V . Dmitruk shows that if the boundary
of U is a shifted ellipse, then solutions of this new problem have a structure similar to that
of problem ( 1 ) , ( 2 ) . This result is unpublished. It seems that this result holds in dimension
2 for any U, while the proof requires much more complicated considerations.
y
Translated by A. I. OVSEEVICH
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RUSSIAN JOURNAL OF MATHEMATICAL PHYSICS VOL. 1 NO. 3