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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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R. KAMOCKI
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;
Copyright 2013 John Wiley & Sons, Ltd. Math. Meth. Appl. Sci.2013
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R. KAMOCKI
(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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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
Dkfx.,xn./, u.// fx., Qx./, u.//kLpr0
p.khkLq /
p
Dkfx.,xn./, u.// fx., Qx./, u.//kLpr0
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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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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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
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.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
Copyright 2013 John Wiley & Sons, Ltd. Math. Meth. Appl. Sci.2013
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R. KAMOCKI
for a.e.t2 a, band allx2 Rn,u 2 M, where
C3:D
1 1
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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<
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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
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