Fracrional Pontryagin - Copy

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Research Article

Received 24 August 2012 Published online in Wiley Online Library

(wileyonlinelibrary.com) DOI: 10.1002/mma.2928MOS subject classification: 26A33; 49K15; 49J15

Pontryagin maximum principle for fractionalordinary optimal control problems

Rafal Kamocki*

Communicated by M. A. Lachowicz

In the paper, fractional systems with RiemannLiouville derivatives are studied. A theorem on the existence and unique-ness of a solution of a fractional ordinary Cauchy problem is given. Next, the Pontryagin maximum principle for nonlinearfractional control systems with a nonlinear integral performance index is proved. Copyright 2013 John Wiley& Sons, Ltd.

Keywords: fractional derivatives and integrals; fractional differential equations; fractional ordinary Cauchy problem; maximum

principle

1. Introduction

In the last years, fractional calculus plays an important role in mathematics, physics, electronics, mechanics, and so on [15]. Recent

investigations have shown that the dynamic of many physical processes can be modeled accurately by using fractional differential

equations. If fractional differential equations contain a control variable and a performance index is given, we obtain a fractional opti-

mal control problem (FOCP). It is well known that the classical optimal control is a generalization of the calculus of variations. This is

exactly the same in the fractional context considered in this paper, where the literature concerning the fractional calculus of variations

is very rich and already well established (see [6] and references therein). In 2004 Agrawal, [7], using the Lagrange multipliers technique,

formulated necessary conditions for optimality for the following standard FOCP:D0Cx

.t/ D G.t,x.t/, u.t//, t2 0, 1, (1)

x.0/ Dx0, (2)

J.u/ D

1Z0

F.t,x.t/, u.t//dt! min, (3)

wherex.t/2 Rn is the state variable,u.t/2 Rm is the control variable,

D0Cx

.t/denotes the left RiemannLiouville derivative order

2 .0, 1/andF, Gare two given functions possessing certain regularity properties with respect to xandu.

In the next years, the optimality conditions for the different modifications of the earlier problem were investigated (see, e.g., [8, 9]).

In our article, we consider the following fractional control problem:DaCx

.t/ D f.t,x.t/, u.t//, t2 a, b a.e. (4)

I1aC x

.a/ Dx0, (5)

u.t/ 2M Rm, t2 a, b, (6)

Faculty of Mathematics andComputerScience, Chair of Differential Equations andComputerScience, Universityof Lodz, Banacha 22, 90-238 Lodz, Poland

*Correspondence to: Rafal Kamocki, Faculty of Mathematics and Computer Science, Chair of Differential Equations and Computer Science, University of Lodz, Banacha

22, 90-238 Lodz,Poland.E-mail: [email protected]

Copyright 2013 John Wiley & Sons, Ltd. Math. Meth. Appl. Sci.2013


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R. KAMOCKI

J.x, u/ D

bZa

f0.t,x.t/, u.t//dt! min, (7)

wheref: a, bRn M !Rn,f0: a, bRn M !R,x02 R

n and 0< 0 andf./ 2 L1.a, b, Rn/. The functions

IaCf

.t/:D

1

./

tZa

f./

.t /1d,

Ibf

.t/:D

1

./

bZt

f./

. t/1d

defined for almost everyt2 .a, b/ are called the left-sided RiemannLiouville integraland the right-sided RiemannLiouville integralof the

function f./of order, respectively.

Now, let 2 .0, 1/ and f./ 2 L1.a, b, Rn/. Wesay that the function f./ possesses the left-sided RiemannLiouville derivative

DaCf

./

of order(the right-sided RiemannLiouville derivative

Dbf

./of order), if

I1aC f

./

I1b f

./

has an absolutely continuous rep-

resentant ona, b(i.e., there exists an absolutely continuous function ona, bthat is equal a.e. on a, bto I1aC f ./ I1b f ./). Insuch a case, we write I1aC f ./ 2AC.a, b, Rn/ I1b f ./ 2AC.a, b, Rn/ and putDaCf

.t/:D

d

dt

I1aC f

.t/, t2 a, b a.e.

Dbf

.t/ :D d

dt

I1b f

.t/, t2 a, b a.e.

Letp 1. ByIaC.L

p/andIb.Lp/we denote the sets

IaC.Lp/:D

f./:a, b !Rn; 9g./2Lp.a,b,Rn/; fD I

aCg a.e. on a, b

,

Ib.Lp/:D

f./:a, b !Rn; 9g./2Lp.a,b,Rn/; fD I

bg a.e. on a, b

respectively. We identify functions fromIaC.L

p/ Ib

.Lp/ equal almost everywhere ona, b.Copyright 2013 John Wiley & Sons, Ltd. Math. Meth. Appl. Sci.2013


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We have the following

Proposition 1 ([3, lemmas 2.4, 2.5 (a) and 2.6 (a)])

Let 0< 0, p 1, q 1 and 1p C 1q 1 C

if 1p C

1q D 1C then p 1 and q 1

. Iff./2 Lp.a, b, Rn/andg./2 Lq.a, b, Rn/,

then

bZa

f.t/

IaCg

.t/dtD

bZa

g.t/

Ibf

.t/dt.

Theorem 2

Let2 .0, 1/,p 1,q 1 and 1

p

C 1

q

1C . if 1p C 1q D 1C , then p 1 and q 1. Iff./ 2 Ib.Lp/andg./ 2 IaC.Lq/, thenbZ

a

f.t/

DaCg

.t/dtD

bZa

g.t/.Dbf/.t/dt.

In conclusion of this section, we shall formulate so called smoothconvex optimal control problem and recall the extremumprinciple

for such problem [18]. These facts will be used in the proof of the main result of this paper (Theorem 7).

LetXandYbe Banach spaces, letUbe an arbitrary set, letf0, : : : , fnbe functions onX U, and letF:X U! Ybe a mapping of the

productX UintoY. We consider the problem

f0.x, u/ ! inf; (9)

F.x, u/ D 0, (10)



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R. KAMOCKI

f1.x, u/ 0, : : : , fn.x, u/ 0, (11)

u 2 U. (12)

If the functionsf0, : : : , fnand the mappingFsatisfy certain conditions of smoothness inxand convexity inu(cf. Assumptions (a) and

(b) in Theorem 3), then the earlier problem is called the smoothconvex problem.

The Lagrange function for the earlier problem is given by formula

L.x, u, 0, : : : , n,y/ D

nXiD0

ifi.x, u/C hyF.x, u/i,

where0, : : : , nare real numbers andy 2 Y (dual space toY).

We shall say that a pair.x, u/that satisfies the constraints (10)(12) is a local minimum point of problem (9)(12) if there exists a

neighborhoodVofxsuch that

f0.x, u/ f.x, u/.

for any pair.x, u/ 2X Usatisfying constraints (10)(12).

Theorem 3 (Smoothconvex extremum principle)

Let the point.x, u/satisfy conditions (10)(12). Further, assume that there exists a neighborhoodV Xofxsuch that

(a) for everyu 2 U, the mappingx! F.x, u/and the functionsx! fi.x, u/,iD 0, : : : , n, belong to the classC1 at the pointx;

(b) for everyx2 V, the mappingu ! F.x, u/and the functionsu ! fi.x, u/,iD 0, : : : , n, satisfy the following convexity condition: for

everyu1, u22 Uand2 0, 1, there exists au 2 Usuch that

F.x, u/ D F.x, u1/C .1/F.x, u2/, (13)

fi.x, u/ fi.x, u1/C .1/fi.x, u2/, iD 0, : : : , n; (14)

(c) the rangeImFx.x, u/of the linear operator Fx.x, u/ : X! Y is closed and has finite codimension in Y (i.e., complementary

subspace toImFx.x, u/has finite dimension inY).

Then, if.x, u/is a local minimum point of problem (9)(12), there exist Lagrange multipliers 0 0, : : : , n 0,y

2 Y

, not allzero, such that

Lx.x, u, 0, : : : , n,y/ D

nXiD0

ifix.x, u/C F

x.x, u/y D 0,

L.x, u, 0, : : : , n,y/ D min

u2UL.x, u, 0, : : : , n,y

/,

ifi.x, u/ D 0 for i D 1, : : : , n.

If, in addition to the conditions formulated,

(d) the image of the setX Uunder the mapping

.x, u/ ! Fx

.x, u/xC F.x, u/

contains a neighborhood of the origin ofY, and if there exists a point.x, u/such that

Fx.x, u/xC F.x, u/ D 0,

hfix.x, u/,xi C fi.x, u/


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nw:a, b !Rn; w.t/ D

Qd.ta/1

fort2 .a, b/ a.e. with some Qd2Rno

, satisfying the equation connected with this problem a.e. on

a, band initial condition.

We have the following technical result.

Lemma 1

Ifg./ 2 Lp.a, b, Rn/, 1 p< 1, > 0, then

j IaCg .t/j

p cIaCjgj

p

.t/, t2 a, b a.e., (15)withcD

.ba/

.C1/

p1.

Consequently,

IaCg

./ 2 Lp.a, b, Rn/.

Proof

Let 1


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3. the functiona, b 3 t! h.t, 0/ 2Rn belongs toLp.a, b, Rn/,

then problem (17) possesses a unique solutionx./ 2 IaC.Lp/.

Proof

Let 0< 0 is a fixed constant. Using once again Lemma 1, Assumptions 13, and assumptions of Theorem 1, we obtain

jjS.'/ S. /jjpkD

bZa

ekpth

t,

IaC'

.t/ h

t,

IaC

.t/p

dt NpbZ

a

ekpt

IaC.' /.t/p

dt

cNpbZ

a

ekpt

IaCj' jp

.t/dtD cNpbZ

a

j'.t/ .t/jp

Ibekp

.t/dt

D cNpbZ

a

j'.t/ .t/jp

0@ 1./

bZt

ekp

. t/1d

1Adt.Note that

bZt

ekp

. t/1dD

btZ0

ekp.tCs/

s1 ds D ekpt

btZ0

ekpss1ds D ekpt

kp.bt/Z0

err1 1

.kp/11

kpdr

D ekpt

.kp/

kp.bt/Z0

err1dr ekpt

.kp/

1Z0

err1drD ./ekpt.kp/ .

So

jjS.'/ S./jjpk cNp

bZa

j'.t/ .t/jp

1

././ekpt.kp/

dt

D cNp.kp/bZ

a

ekptj'.t/ .t/jpdtD cNp.kp/ jj' jjpk

,

wherecD .ba/.C1/p1. Let us notice thatN.c.kp// 1p 2 .0, 1/for sufficiently largekand consequently the operatorShas a uniquefixed point. This means that problem (17) possesses a unique solutionx./ 2 IaC.L

p/.

Now, we consider the following Cauchy problem:(DaCy

.t/ DQh.t,y.t//, t2 a, b a.e.

I1aC y

.a/ D d,(18)

whered2Rnnf0g andQh:a, bRn !Rn.

It easy to show that ifx./ 2 IaC.Lp/is a solution to problem (17) with the function hof the form

h.t,x, u/ DQh

t,xC

d

./

1

.t a/1

, (19)



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then the function

y./ Dx./C d

./

1

.t a/1 (20)

is a solution to problem (18). Conversely, ify./ 2 IaC.Lp/C

n Qd

.ta/1; Qd2Rn

ois a solution to problem (18) with thefunction Qh of the

form

Qh.t,y, u/ D h

t,yQd

.t a/1!

,

then QdD d./

and

x./ Dy./ d

./

1

.t a/1

is a solution to problem (17).

So, in an analogous way as in [17, theorem 3.2], we can prove the following

Theorem 5

Let2 .0, 1/and 1 p< 11 . If

1. the functionQh.,y/is measurable for anyy2Rn;

2. there exists a constantN>0 such that

j Qh.t,y1/Qh.t,y2/j Njy1 y2j

fort2 a, ba.e. and ally1,y22 Rn ;

3. the functiona, b 3 t!Qh.t, 0/ 2Rn belongs toLp.a, b, Rn/,

then problem (18) possesses a unique solutiony./ 2 IaC.Lp/C

n Qd

.ta/1; Qd2Rn

o.

4. The Pontryagin maximum principle

Let us consider the following FOCP: D

aCx

.t/ D f.t,x.t/, u.t//, t2 a, b a.e. (21)

I1aC

x

.a/ D 0, (22)

u.t/ 2M Rm, t2 a, b, (23)

J.x, u/ D

b

Za f0.t,x.t/, u.t//dt! min, (24)wheref: a, bRn M !Rn,f0: a, bR

n M !R, 0< 0 such that

jf.t,x1, u/ f.t,x2, u/j Njx1 x2j

fort2 a, ba.e. and allx1,x22 Rn,u 2M;



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(c) there existr./ 2 Lp.a, b, R/and 0 such that

jf.t, 0, u/j r.t/C juj

fort2 a, ba.e. and allu 2 M,

then problems (21)(22) possess a unique solution x./ 2 IaC.Lp/corresponding to any controlu./ 2 Lp.a, b, M/.

Let

UMD fu./ 2 L1.a, b, Rm/; u.t/ 2 M, t2 a, bg.

We say, that a pair.x./, u.//2 IaC.Lp/ UMis a locally optimal solution to problem (21)(24), ifx./is the solution of (21)-(22),

corresponding to the controlu./and there exists a neighborhoodVof pointx./inIaC.Lp/such that

J.x./, u.//J.x./, u.//

for every pair.x./, u.// 2 V UMsatisfying (21)(22).

In the proof of the maximum principle, we use the following three lemmas.

Lemma 2

Let u./ 2 UMand ': a, bM !R be such that '., u/ is measurable on a, b for any u 2 M and '.t, / is continuous on M for t2 a, b

a.e. If

10,p 1. The operatorIaCis

1. Bounded inLp.a, b, Rn/, that is,

jjIaCfjjLp KjjfjjLp

forf2 Lp.a, b, Rn/, whereKD .ba/

.C1/.



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2. If2 .0, 1/and 1


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R. KAMOCKI

Let us notice that Lemma 3, applied to the function h D .f, f0/, guarantees fulfillment of the following condition: for any u1./, u2./ 2

UM,x./ 2 IaC.L

p/and2 0, 1there existseu./ 2 UMsuch thatF0.x./,eu.//D F0.x./, u1.//C .1/F0.x./, u2.//,

F.x./,eu.//D F.x./, u1.//C .1/F.x./, u2.//.Of course, this condition implies conditions (13) and (14). So, Assumption (b) of Theorem 3 is satisfied.

Moreover, the mappingF is continuously differentiable in the Gteaux sense (and consequently continuously Frchet differentiable)

with respect tox./ 2 IaC.Lp/. Indeed, note that

F.x./, u.// D F1.x./, u.// F2.x./, u.//,

where

F1.x./, u.//D

DaCx

./,

F2.x./, u.//D f.,x./, u.//.

It is easy to check that the operatorF1 is linear and continuous with respect to x./, so it is continuously differentiable in the Frchet

sense with respect tox./, and its differential at a point.x./, u.//is given by

F1x.x./, u.//h./ D DaCh ./, h./ 2 I

aC.L

p/.

The mapping

F2G.x./, u.//:IaC.L

p/ ! Lp.a, b, Rn/

given by

F2G.x./, u.//h./ D fx.,x./, u.//h./

is Gteaux differential ofF2.x./, u.//with respect tox./for any fixedu./2 UM. Indeed, it is clear that it is linear. From Lemma 4 p. 1,

Proposition 1 p. 2, and the fact thatfxis bounded (by the constant nN, whereNis a constant from Assumption (b)), it follows that

jjF2G.x./, u.//h./jjpLp D

b

Zajfx.t,x.t/, u.t//j

pjh.t/jpdt .nN/pb

Zajh.t/jpdtD .nN/p

b

ZajIaC

DaCh.t/jpdt

D .nN/pjjIaC

DaCh./jj

pLp .nNK/

pjj

DaCh

./jjpLp D .nNK/

pjjh./jjp

IaC

.Lp/.

This implies the continuity ofF2G.x./, u.//.

Using the dominated convergence theorem, it is easy to show that F2.x./C nh./, u.// F2.x./, u.//n F2G.x./, u.//h./

Lp! 0,

asn !n!1

0. Consequently,F2G.x./, u.//is the Gteaux differential ofF2 with respect tox./.

Now, we shall show that the mapping

IaC.Lp/ 3x./ 7! F2G.x./, u.// 2 L I

aC.L

p/, Lpis continuous (here L

IaC.L

p/, Lp

is the space of linear and continuous mappings F : IaC.Lp/ ! Lp considered with the operator

topology). Indeed, letxn./ !n!1

Qx./inIaC.Lp/. Then, using Lemma 4 p. 1, we obtain

jjxn./Qx./jjLp D jjIaC

DaC.xn.t/Qx.t//

jjLp KjjD

aC.xn./Qx.//jjLp D Kjjxn./Qx./jjIaC.L

p/ !n!10.

It means thatxn./ !n!1

Qx./wLp. Moreover,

bZa

jfx.t,x.t/, u.t//jpdt .b a/.nN/p.

Let us consider three cases:p> 1 , 1


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Letp> 1 . From the earlier inequality follows that, the Nemytskii operator

N :Lp.a, b, Rn/ ! Lp.a, b, Rnn/

given by

N.x.//.t/ D fx.t,x.t/, u.t//, t2 a, b a.e.

forx./ 2 Lp

.a, b, Rn

/is well defined, so it is continuous. Thus

"nD

bZa

jfx.t,xn.t/, u.t// fx.t, Qx.t/, u.t//jpdt !

n!10. (30)

Besides, ifh./ 2 IaC.Lp/, the Hlder inequality (withpandp0 D

pp1 ) implies

jh.t/j D jIaC.

DaC

h/.t/j 1

./

tZa

j

DaCh

.t/j

.t /1 d k

DaCh

./kLp

1

./

0@ tZa

.t /.1/p0d

1A1

p0

(31)

D 1

./.. 1/p0C 1/1

p0

.t a/

1 1p0

jjh./jjI

aC.Lp/

1

./.. 1/p0C 1/1

p0

.b a/

1

p

jjh./jjI

aC.Lp/D cjjh./jjI

aC.Lp/

fort2 a, ba.e., wherecD 1

./..1/p0C1/1

p0.b a/

1p

.

Thus, from (30), for anyh./ 2 IaC.Lp/, we have

bZa

jfx.t,xn.t/, u.t// fx.t, Qx.t/, u.t//jpjh.t/jpdt "nc

pjjh./jjp

IaC

.Lp/,

so

F2G.xn./, u.// F2G.Qx./, u.//

L

IaC

.Lp/,Lp c."n/ 1p !

n!10.

Now, let 1 p is bounded. Thus and because of the fact that h D IaCD

aCh and D

aCh 2 L

p.a, b, Rn/, we assert that

h 2 Lq.a, b, Rn/. Consequently, the function jhjp 2 Lr.a, b, Rn/, whererD qp > 1. Moreover, from the fact that fsatisfies the Lipschitz

condition with respect tox, it follows that

fx.,x./, u.// 2 L1.a, b, Rnn/, (32)

so

jfx.,x./, u.//jp 2 L1.a, b, R/ Lr

0.a, b, R/,

where 1rC 1r0 D 1. Consequently,

bZa

jfx.t,xn.t/, u.t// fx.t, Qx.t/, u.t//jp jh.t/jpdt

0@ bZa

jfx.t,xn.t/, u.t// fx.t, Qx.t/, u.t//j

pr0

dt

1A1r00@ bZ

a

jh.t/jp

rdt

1A1r

Dkfx.,xn./, u.// fx., Qx./, u.//kLpr0

p.khkLpr/

p


p.khkLq /

p


p IaCDaChLqpkfx.,xn./, u.// fx., Qx./, u.//kLpr0

pcppq

DaChLppD

kfx.,xn./, u.// fx., Qx./, u.//kLpr0

pcppq

khkI

aC.Lp/

p

,



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R. KAMOCKI

wherecpqis such that IaC'Lq cpqk'kLp , '2 Lp.a, b, Rn/(existence of the constantcpqfollows from Lemma 4 p. 2). Using (32), we assert that the Nemytskii operator

Lp.a, b, Rn/ 3x./ ! fx.,x./, u.// 2 Lpr0.a, b, Rn/

is well defined, so it is continuous. Then,F2G.xn./, u.// F2G.Qx./, u.//L

IaC

.Lp/,Lp cpq kfx.,xn./, u.// fx., Qx./, u.//kLpr0 !n!1 0.

LetpD 1 . Ifz2 Lp.a, b, Rn/, thenz2 LQp.a, b, Rn/for any 1 < Qp< pand consequently, using Lemma 4 p. 2, IaCz2 L

Qq.a, b, Rn/,

where QqD Qp1Qp

. Let note that ifQp2 .1,p/then Qq2

11 ,1

. It means thatIaCz2 L

Qq.a, b, Rn/for any 11 < Qq< 1. Moreover, if

z2 Lp.a, b, Rn/ LQp.a, b, Rn/, then, using Lemma 4 p. 2 once again, we obtainIaCzLQq cQpQq kzkLQp cpQqkzkLp ,wherec

pQq

D cQpQq

.b a/pQp

pQp Hence, ifp D 1

, then for any 1

1

< Qq< 1 the operatorIaC

: Lp.a, b, Rn/ ! LQq.a, b, Rn/is well defined

and bounded. So let fix any Qq > max p, 11 in considered case p D 1 and repeat argument as in the previous case 1


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Because ImFx.x./, u.// D Lp.a, b, Rn/, the Assumption (d) of Theorem 3 is satisfied and consequently 0 0. Without loss of

generality, we can assume that0D 1. Accounting this fact, we can transform the formula (33) in the following way:

bZa

V.t/h.t/dtC

bZa

T.t/

DaCh

.t/dtD 0, (35)

forh./ 2 IaC.Lp/, whereV.t/ D .f0/x.t,x.t/, u.t//

T.t/fx.t,x.t/, u.t//fort2 a, ba.e.

Note that becausefxis bounded and.f0/x2 Lp0.a, b, Rn/(it follows from the condition (26)), thereforeV./ 2 Lp0 .a, b, Rn/. Putting

w.t/ D .IbV/.t/,

we assert thatw./ 2 Ib.Lp0/. Moreover, from Theorem 2 and Proposition 1 p. 1, we obtain

bZa

V.t/h.t/dtD

bZa

Dbw

.t/h.t/dtD

bZa

w.t/

DaCh

.t/dt,

for anyh./ 2 IaC.Lp/. Thus, condition (35) implies

b

ZaT.t/Cw.t/ DaCh .t/dtD 0,

for anyh./ 2 IaC.Lp/. Consequently,

bZa

T.t/Cw.t/

l.t/dtD 0,

for anyl./ 2 Lp.a, b, Rn/. Thus

T.t/ Dw.t/ D

Ib.V/

.t/, t2 a, b a.e.

So,./ 2 Ib.Lp0/and we obtain the condition (27). Proposition 3 implies the condition (28).

To prove the condition (29), let us observe that the condition (34) can be written as

bZa

f0.t,x.t/, u.t//

T.t/f.t,x.t/, u.t//

dtD minu./2UM

8


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R. KAMOCKI

From Theorem 5, we immediately obtain

Theorem 8

Let2 .0, 1/and 1 p< 11 . If

(a) g.,y, u/is measurable ona, bfor ally2 Rn,u 2 M,g.t,y, /is continuous onMfort2 a, ba.e. and ally2 Rn;

(b) there existsN>0 such that

jg.t,y1, u/ g.t,y2, u/j Njy1 y2j

fort2 a, ba.e. and ally1,y22 Rn,u 2M;

(c) there existr./ 2 Lp.a, b, R/and 0 such that

jg.t, 0, u/j r.t/C juj

fort2 a, ba.e. and allu 2 M,

then problem (36)(37) possesses a unique solution x./ 2 IaC.Lp/ C

n Qd

.ta/1; Qd2Rn

ocorresponding to any control u./ 2

Lp.a, b, M/.

We say, that a pair .y./, u.// 2

IaC.Lp/C

n Qd

.ta/1; Qd2Rn

o UM is a locally optimal solution to problem (36)(39), if

y./is the solution of (36)(37), corresponding to the control u./and there exists a neighborhood QVof point y./in IaC.Lp/Cn Qd

.ta/1; Qd2Rno such that

H.y./, u.// H.y./, u.//

for every pair.y./, u.// 2QV UMsatisfying (36)(37).

It easy to show that if a pair.x./, u.//2IaC

.Lp/UMis a locally optimal solution to problem (21)(24) with functions fand f0of

the form

f.t,x, u/ D g

t,xC

y0

./

1

.t a/1, u

, (40)

f0.t,x, u/ D g0

t,xC

y0

./

1

.t a/1, u

, (41)

then the pair

.y./, u.// D .x./C y0

./

1

.t a/1, u.// 2

IaC

.Lp/C

( Qd

.t a/1; Qd2Rn

)!UM (42)

is a locally optimal solution to problem (36)(39). Conversely, if a pair .y./, u.//2

IaC

.Lp/Cn

Qd.ta/1

; Qd2RnoUMis a locally

solution to problem (36)(39) with functionsgandg0of the form

g.t,y, u/ D f

t,y

Qd

.t a/1, u

!,

g0.t,x, u/ D f0 t,y Qd.t a/1

, u! ,then QdD

y0./

and the pair

.x./, u.// D

y./

y0

./

1

.t a/1, u./

2 I

aC.Lp/UM

is a locally solution to problem (21)(24).

Now, using Theorem 7, we shall derive necessary optimality conditions for a problems (36)(39).

Theorem 9

Let2 .0, 1/, 1


15/19

.2g/ g0.,y, u/is measurable ona, bfor ally2Rn,u 2 Mandg0.t,y, /is continuous onMfor a.e.t2 a, band ally2 R

n,

.3g/ g02 C1 with respect toy2Rn and

jg0.t,y, u/j a1.t/C C1jyjp,

j.g0/x.t,y, u/j a2.t/C C2jyjp1

for a.e.t2 a, band ally2 Rn,u 2 M, wherea22 Lp0

a, b, RC0

,

1

p C 1p0D 1

,a12 L

1

a, b, RC0

,C1, C2 0,

.4g/ gy.,y, u/,.g0/y.,y, u/are measurable ona, bfor ally2 Rn

,u 2 M,.5g/ gy.t,y, /,.g0/y.t,y, /are continuous onMfor a.e.t2 a, band ally2R

n,

.6g/ for a.e.t2 a, band ally2 Rn the set

QZ:Dn

.g0.t,y, u/, g.t,y, u// 2RnC1, u 2 M

ois convex.

If the pair

.y./, u.// 2

IaC

.Lp/C

( Qd

.t a/1; Qd2Rn

)!UM

is a locally optimal solution of problem.36/.39/, then there exists a function 2 Ib.Lp0/, such that

Db

.t/ D gTy.t,y.t/, u.t//.t/ .g0/y.t,y.t/, u.t// (43)

for a.e.t2 a, band I1b

.b/ D 0. (44)

Moreover,

g0.t,y.t/, u.t// .t/g.t,y.t/, u.t// D minu2M

fg0.t,y.t/, u/ .t/g.t,y.t/, u/g (45)

for a.e.t2 a, b.

Proof

To prove the earlier theorem, it suffices to show that if functionsg and g0 satisfy the Assumptions .1g/.6g/, then functionsfandf0given by (40) and (41) satisfy conditions .1f/.6f/. Indeed, condition.6f/ follows directly from condition.6g/. Moreover, functionsf,

f0 2C1 with respect toxas a superposition of functions belonging to C1. The second part of condition.1f/follows from the second

part Assumption.1g/in an analogous way as in [17, proof of theorem 3.2]. The fact that fxand .f0/xare measurable intfollows from

Assumptions .2g/, .4g/ and from measurability of the functionxC y0./

1.ta/1

. Moreover, condition .5g/ implies .5f/ and continuous

of functiong0 with respect tou 2 M implies continuous off0 onM. Now, we shall show that the Assumption .3f/ follows from the

Assumption.3g/. Indeed, we have

jf0.t,x, u/j D

g0

t,xC

y0

./

1

.t a/1, u

a1.t/C C1

xC

y0

./

1

.t a/1

p a1.t/C C12

p1

jy0j

./

1

.t a/1

pC C12

p1jxjp D a1.t/C C1jxjp,

for a.e.t2 a, band allx2 Rn,u 2 M, where

a1.t/:D a1.t/C C12p1

jy0j

./

1

.t a/1

pfor a.e.t2 a, band

C1D C12p1.

Moreover,

j.f0/x.t,x, u/j D

.g0/x

t,xC

y0

./

1

.t a/1, u

a2.t/C C2

xC

y0

./

1

.t a/1

p1 a2.t/C C2C3

jy0j

./

p1 1.t a/.1/.p1/

C C2C3jxjp1 D a2.t/C C2jxj

p1



16/19

R. KAMOCKI

for a.e.t2 a, band allx2 Rn,u 2 M, where

C3:D

1 1


17/19

H.y, u/ D

2Z0

.y1.t/y2.t/C u.t//dt! min, (52)

wherey2R2,A D

0 1

0 0

,B D

1

1

,y0D

0

1

,D 12 ,p D

32 .

In this case

g.t,y, u/ DAyC Bu,g0.t,y, u/ D h.1,1/,yi C u.

Of course,

AkD .AT/kD 0, k 2,

gy.t,y, u/ DA, .g0/y.t,y, u/ D 1,1.

It is easy to check that all assumptions of Theorem 9 are satisfied. Consequently, if.y./, u.//is a locally optimal solution to problem

(49)(52), then there exists./ 2 I12

2.L3/such that

D122

.t/ DAT.t/C

1

1

, t2 0, 2 a.e. (53)

I

12

2

.2/ D 0. (54)

Moreover,

u.t/ .t/Bu.t/ D minu20,1

fu .t/Bug (55)

fort2 0, 2a.e.

Now, we formulate a theorem, which will be used later on.

Theorem 10

Let 0<


18/19

R. KAMOCKI

From Theorem 11, it follows that a solution of problem (53)(54) is given by

1.t/

2.t/

D

264 .2t/12

. 32 /

.2t/12

. 32 /

.2t/.2/

375 , t2 0, 2.Consequently, condition (55) is equivalent to the following one

u.t/ .2 t/u.t/ D minu20,1

fu .2 t/ug D t 1 t2 0, 1

0 t2 1, 2

fort2 0, 2a.e.

Thus

u.t/ D

1 t2 0, 1 a.e.

0 t2 1, 2 a.e.

From Theorem 10, it follows that a solution of system (49)(50), corresponding tou./is given by

y.t/ D .t/y0 C

t

Z0.t s/Bu.s/ds D .t/

0

1 C

8


19/19

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