Stochastic Control by Functional Analysis Methods

Stochastic Control by Functional Analysis Methods

STUDIES IN MATHEMATICS AND ITS APPLICATIONS VOLUME 11

Editors: J. L. LIONS, Paris G. PAPANICOLAOU,New York R. T. ROCKAFELLAR, Seattle

N O R T H - H O L L A N D PUBLISHING COMPANY - AMSTERDAM

a

NEW YORK

a

OXFORD

STOCHASTIC CONTROL BY FUNCTIONAL ANALYSIS METHODS

ALAIN BENSOUSSAN Universite Paris Dauphine arid I N R I A

19x2

NORTH-HOLLAND PUBLISHING COMPANY - AMSTERDAM

NEW YORK

OXFORE

L

North-Holland Publishing Company, I982

All rights reserved. N o part of this publication may be reproduced, storedin a retrievalsystem,

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ISBN: 0444 86323 x

Publishers: N O R T H - H O L L A N D PUBLISHING COMPANY AMSTERDAM 0 NEW Y O R K 0 O X F O R D Sole distributors f o r the U.S.A. and Canada: ELSEVIER SCIENCE PUBLISHING COMPANY, INC. 5 2 VANDERBILT A V E N U E , NEW Y O R K , N.Y. 10017

Library of Congress Cataloging in Publiration Data

Bensoussan, A l a i n . S t o c h a s t i c c o n t r o l by f u n c t i o n a l a n a l y s i s methods. ( S t u d i e s i n mathematics and i t s a p p l i c a t i o n s ;

v . 11) B ib li o g r a p h y : p . . 1. C o n t r o l t h e o r y .

2. Stochastic analysis. I. T i t l e . 11. S e r i e s . QA402.3. B433 6 2 9 . 8 ' 3 12 81- 19900 ISBN 0-444-86329-X AACFC?

PRINTED IN T H E NETHERLANDS

V

INTRODUCTION

Our objective in this work is to give a presentation of some basic results of stochastic control. It is thus a text intended to advanced students and researchers willing to learn the theory. Stochastic control covers a broad area of disciplines and problems. It is also a field in full development, and some important aspects remain to be cleaned up. That is why in presenting stochastic control a choice is necessary. We have emphasized this choice in the title. The theory of partial differential equations, the semi-group theory, variational and quasi variational inequalities play a very important role in solving problems of stochastic control. We have tried to use them as much as possible, since they bring tools and results which are very important, specially for computational purposes, and which cannot be obtained in an other way, namely regularity results, and weak solution concepts. The books by W. Fleming R. Rishel 1 1 1 , A. Friedman C 1 1 , N. Krylov [ l ] , A . Bensoussan - J.L. Lions [ l ] , [ 2 ] show already the importance of the techniques of Functional Analysis. w. Fleming Rishel, Friedman besides covering many other topics, rely mostly on the classical theory of P.D.E. We try to emphasize here the importance of variational methods. Naturally, the present text has a lot in common with the books of J.L. Lions and the A. But here we have tried to simplify as much as possible the presentation, in particular leaving aside the most technical problems, which are treated there.

-

-

Also the books by J.L. Lions and the A. are devoted to variational and quasi variational inequalities. In the book of Krylov, one will find the study of the general Bellman equation, i.e., when the control enters into the drift as well as into the diffusion term. We do not treat this general case here, although it is certainly one of the nicest accomplishments of P.D.E. techniques in stochastic control. Quite fundamental results have been obtained for the general Bellman equation by P.L. Lions [ l ] , [ 2 ] , and more specialized ones by L.C. Evans A. Friedman [ l ] , H. Brezis - L.C. Evans C l ] . More recently R. Jensen - P.L. Lions [ l ] have introduced new important ideas of approximation. To report on that work would have gone beyond the objectives of the present text, and requested too much material.

There are many important other topics that we have not considered here. We have not reported on the developments of the so called "probabilistic approach" initiated by C. Striebel [ l ] , 1 2 1 and R . Rishel C 1 1 and developed extensively by M. Davis - P. Varayia C 1 3 , M. Davis C 1 1 and many other authors. A good report can be found in N. El Karoui [ l ] , (see also Lepeltier - Marchal [ l ] ) .

vi

INTRODUCTION

This approach is of course fundamental for very general processes, which are not Markov processes. It is certainly the most general one and very satisfactory from the probabilistic point of view. But for the applications, where the processes are mostly Markov, it seems less convenient than the analytic approach, especially for computational purposes. Also it requires technical developments which again would have gone beyond the scope of this text. The interested reader should consult besides the litterature, which has been briefly mentionned the recent book by 1.Gikhman - A. Skorokhod [11. Another very important area, which is in full development is the theory of non linear filtering and control under partial observation. Important results have been obtained recently by several authors in non linear filtering, T. Allinger - S.K. Mitter Cl], E. Pardoux [ l ] , and exploited for the control under partial observation by W. Fleming - E. Pardoux [ l ] , W. Fleming [ I ] . Stochastic P.D.E. play a very important role in this direction, and probably the field will progress fast (see E. Pardoux [2], M. Viot [ l ] , W. Fleming - M. Viot [ l ] ) . For the control of stochastic distributed parameter systems see A.V. Balakrishnan [ l ] , A. BensoussanC11, [ 2 ] , A. Bensoussan - M. Viot [ l ] , R. Curtain A.J. Pritchard [ I ] , S. Tzapestas C 1 3 . We consider in this work some stochastic control problems in discrete time, but mostly as an approximation to continuous time stochastic control. We refer to the books by D. Bertsekas [ I ] , D. Bertsekas S . Shreve [ I ] , E. Dynkin - A. Yushkevitch [ I ] for many more details. In a related direction, we have not discussed the numerical techniques which are used to solve stochastic control problems. We refer to J.P. Quadrat [ l ] , P.L. Lions - B. Mercier [ 1 1 and to the book of H.J. Kushner [ I ] . Let u s also mention the theory of large stochastic systems, with several players, the problems of identification, adaptive control, stochastic realization, stochastic stability etc., as interesting and important areas of research. In Chapter I we present the elements of Stochastic Calculus and Stochastic Differential Equations, in Chapter I1 the theory of partial differential equations, and in Chapter 111 the Martingale problem. This permits to deal with the various formulations of diffusion processes and to interpret the solution of elliptic and parabolic equations as functionals on the trajectory of the diffusion process (in a way similar to the well known method of characteristics This allows u s also to show the Markov for 1 s t order linear P.D.E.). semi-group property of diffusions. In Chapter IV we present the theory of Stochastic Control with complete information (when the control affects only the drift term). We study the Hamilton-Jacobi-Bellman equation, interpret its solution as a value function and solve the stochastic control problem in the stationary as well as non stationary case. We also present a semi group approach to stochastic control for general Markov processes. In Chapter V, we present the theory of filtering and prediction for linear stochastic differential equations, which leads to the Kalman

INTRODUCTION

Vii

filter. We show that the problem reduces to quadratic optimization problems, for which a decoupling argument yields the filter and the Riccati equation. In Chapter VI, we present the variational approach to stochastic control, in two situations, one with complete observation and one with uncomplete observation. We discuss also the separation principle. Chapter VII is devoted to optimal stopping problems which are solved by the theory of variational inequalities. We also develop a semi group approach in the case of general Markov processes. In Chapter VIII we present the theory of impulsive control and its solution by the method of quasi variational inequalities. Also a semi group formulation is given. We have tried to be self contained as such as possible, and have avoided too technical topics. Some basics on probability and Functional Analysis are the only requirements in order to read this book. Nevertheless, we recall the results that we need. We have restricted ourselves to stationary diffusions stopped at the exit of a domain, since one can explain the ideas in the simplest form in that context. But of course the methodology carries over to many other processes, like diffusions with reflexion, diffusions with jumps, random evolutions etc. see A. Bensoussan - J.L. Lions [Z], A. Bensoussan - P.L. Lions 111, A. Bensoussan - J.L. Menaldi I l l ) . When presenting the semi group approach we have kept a certain degree of generality, although we have not tried to describe all the examples which are covered in this approach (cf. M. Robin [ l ] for many examples like semi Markov processes, jump processes, ).

...

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ix

ACKNOWLEDGMENTS

Many of the ideas presented here owe a lot to discussions and joint research with colleagues in France or abroad. I would like to thank particularly J. Frehse, A . Friedman, M. Goursat, J.L. Joly, J.L. Lions, P.L. Lions, J.L. Menaldi, U. Mosco, G. Papanicolaou, E. Pardoux, J.P. Quadrat, M. Robin, M. Viot with whom I had certainly the most fruitful exchanges and from whom I learnt a lot. The material covered here has been presented first as a course at the University of Illinois, during the fall 1980. I would like to thank particularly P. Kokotovic who originated the idea of a course and suggested the writing of a text. I would like to thank also the members of the Control Group, at Coordinated Science Laboratory, University of Illinois, in particular Prof. Cruz, Perkins for their help. Prof. B. Hajek at the University of Illinois read carefully the manuscript and made very useful comments. I would like to thank him very much for that.

I would like to thank very much Professors Lions, Papanicolaou and Rockafellar for accepting to publishing this book in their series with North Holland as well as the publisher for his agreement. Mrs Kurinckx realized a very nice typing of the manuscript. I would like to thank her very much for this excellent job.

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xi

CONTENTS

CHAPTER I .

STOCHASTIC CALCULUS AND STOCHASTIC DIFFERENTIAL EQUATIONS

1

INTRODUCTION

1

1. PRELIMINARIES

2

1.1. 1.2. 1.3. 1.4.

Random v a r i a b l e s Conditional expectation Expectation S t o c h a s t i c processes Martingales

-

2. STOCHASTIC INTEGRALS 2.1. 2.2. 2.3. 2.4. 3.

Wiener process Construction o f the stochastic integral S t o c h a s t i c process d e f i n e d by a s t o c h a s t i c i n t e g r a l Extension o f t h e s t o c h a s t i c i n t e g r a l

ITO'S FORMULA

4. STOCHASTIC DIFFERENTIAL EQUATIONS 4.1. 4.2.

S e t t i n g o f t h e problem L i p s c h i t z case

5. GIRSANOV TRANSFORMATION

5.1. Fundamental lemma 5.2. G i r s a n o v ' s Theorem 5.3. A p p l i c a t i o n t o t h e concept o f weak s o l u t i o n o f a s t o c h a s t i c d i f f e r e n t i a1 e q u a t i o n CHAPTER 11.

PARTIAL DIFFERENTIAL EQUATIONS

2 5 7 9

10 10 12 16 18 20 32 32 33 37 37 42 46 51

INTRODUCTION

51

1. FUNCTIONAL SPACES

52

1.1. Sobolev spaces 1.2. Concept o f t r a c e 1.3. G r e e n ' s f o r m u l a

52 54

57

xii

CONTENTS

2. THE DIRICHLET PROBLEM FOR ELLIPTIC EQUATIONS 2.1. The b a s i c e x i s t e n c e and uniqueness r e s u l t s 2.2. V a r i a t i o n a l t e c h n i q u e s

59 59 64 75

3. PARABOLIC EQUATIONS 3.1. F u n c t i o n a l spaces 3.2. V a r i a t i o n a l f o r m u l a t i o n 3.3. R e g u l a r i t y

75 78 82

3.3.1. R e g u l a r i t y w i t h r e s p e c t t o t i m e 3.3.2. R e g u l a r i t y w i t h r e s p e c t t o space v a r i a b l e s 3.3.3. O t h e r r e g u l a r i t y r e s u l t s 3.4. The Cauchy problem

82 84 87 94

CHAPTER 111.

101

MARTINGALE PROBLEM

INTRODUCTION

101

1. PROPERTIES OF CONTINUOUS MARTINGALES

101

1.1. Square i n t e g r a b l e c o n t i n u o u s m a r t i n g a l e s 1.2. S t o c h a s t i c i n t e g r a l s 1.3. A r e p r e s e n t a t i o n theorem f o r c o n t i n u o u s m a r t i n g a l e s

101 104 105

2. DEFINITION OF THE MARTINGALE PROBLEM

107

2.1. S e t t i n g o f t h e problem 2.2. P r o p e r t i e s o f t h e m a r t i n g a l e 2.3. Some a p r i o r i e s t i m a t e s

107 108 112

problem

3 . EXISTENCE AND UNIQUENESS OF THE SOLUTION OF THE MARTINGALE PROBLEM 3.1. Uniqueness 3.2. E x i s t e n c e 4 . INTERPRETATION OF THE SOLUTION OF P.D.E. 4.1. 4.2.

E l l i p t i c equations Parabolic equations

5. SEMI GROUPS 5.1. Semi group o f d i f f u s i o n s 5.2. Stopped d i f f u s i o n s

117 117 122 125 125 127 129 129 133

CONTENTS

CHAPTER I V .

STOCHASTIC CONTROL WITH COMPLETE INFORMATION

Xiii

139

INTRODUCTION

139

1. SETTING OF THE PROBLEM

139

1.1. N o t a t i o n . Assumptions 1.2. C o n t r o l l e d m a r t i n g a l e 2. THE EQUATION OF DYNAMIC PROGRAMMING Orientation 2.1. N o t a t i o n 2.2. Study o f t h e H.J.B.

equation

139 140 142 142 142 143

3. SOLUTION OF THE STOCHASTIC CONTROL PROBLEM

150

4. EVOLUTION PROBLEMS

156

4.1. P a r a b o l i c e q u a t i o n s 4.2. S t o c h a s t i c c o n t r o l problem 4.3. M a r t i n g a l e p r o p e r t y 5. SEMI GROUP FORMULATION 5.1. 5.2. 5.3. 5.4. CHAPTER V .

156 158 160 16 1

A property o f t h e equation u The problem o f semi group envelope I n t e r p r e t a t i o n o f t h e d i s c r e t i z e d problem A regularity result

161 164 170 178

FILTERING AND PREDICTION FOR LINEAR S.D.E.

191

INTRODUCTION

191

1. SETTING OF THE PROBLEM

191

1.1. S t a t e e q u a t i o n 1.2. The o b s e r v a t i o n process 1.3. Statement o f t h e problem 2. CHARACTERIZATION OF THE BEST ESTIMATE 2.1. Main r e s u l t 2.2. Q u a d r a t i c o p t i m i z a t i o n problems

199 199 203

KALMAN FILTER

208

A g e n e r a l system Recursive formulas Kalman f i l t e r The i n n o v a t i o n process

208 210 213 216

3. RECURSIVITY 3.1. 3.2. 3.3. 3.4.

-

191 195 197

4. PREDICTION

219

CONTENTS

XiV

CHAPTER V I .

VARIATIONAL METHODS I N STOCHASTIC CONTROL

221

INTRODUCTION

221

1. MODEL WITH ADDITIVE N O I S E

22 1

1.1. S e t t i n g o f t h e model 1.2. A d e n s i t y r e s u l t 1.3. Necessary c o n d i t i o n s

221 224 227

2. THE CASE WITH INCOMPLETE OBSERVATION 2.1. 2.2. 2.3. 2.4.

Statement o f t h e problem Prel i m inary r e s u l t s Necessary c o n d i t i o n s The q u a d r a t i c case. S e p a r a t i o n p r i n c i p l e

3. SEPARATION PRINCIPLE Orientation 3.1. The Kalman f i l t e r 3.2. A dynamic programming e q u a t i o n 3.3. S o l u t i o n o f t h e s t o c h a s t i c c o n t r o l problem w i t h incomplete information CHAPTER V I I .

PROBLEFIS OF OPTIMAL STOPPING

234 234 237 240 243 253 253 254 255 26 9 279

INTRODUCTION

279

1. SETTING OF THE PROBLEM

279

1.1. Assumptions. 1.2. Remarks

Notation

2. UNILATERAL PROBLEMS 2.1. P e n a l i z e d problem 2.2. L i m i t problem i n t h e r e g u l a r case 3. VARIATIONAL INEQUALITIES

279 281 281 281 282 287

Orientation 3.1. L i m i t o f t h e p e n a l i z e d problem 3.2. Weakening t h e c o e r c i t i v i t y assumption

287 288 290

4. SOLUTION OF THE OPTIVAL STOPPING TIME PROBLEM

301

4.1. The r e g u l a r case 4.2. The non r e g u l a r case 4.3. I n t e r p r e t a t i o n o f t h e p e n a l i z e d problem 5. S E M I GROUP APPROACH 5.1. S o l u t i o n o f t h e V . I . as a maximum element 5.2. The case o f g e n e r a l semi group

30 1 303 305 307 307 315

CONTENTS 5.3. D i s c r e t i z a t i o n 5.4. Case w i t h o u t c o n t i n u i t y 6. INTERPRETATION AS A PROBLEM OF OPTIMAL STOPPING 6.1. Markov process r e l a t e d t o t h e semi group 6.2. Problem o f o p t i m a l s t o p p i n g 6.3. I n t e r p r e t a t i o n o f t h e d i s c r e t i z e d problem CHAPTER V I I I .

IMPULSIVE CONTROL

xv 326 336 339 339 34 1 345 355

INTRODUCTION

355

1. SETTING OF THE PROBLEM

356

1.1. Assumptions. 1.2. The model 1.3. Some remarks

Notation

356 357 359

2. QUASI VARIATIONAL INEQUALITIES

360

Orientation 2.1. S e t t i n g o f t h e problem 2.2. S o l u t i o n o f t h e Q . V . I .

360 360 36 1

3. SOLUTION OF THE IMPULSIVE CONTROL PROBLEM 3.1. 3.2.

The main r e s u l t I n t e r p r e t a t i o n o f t h e d e c r e a s i n g scheme

4. SEMI GROUP APPROACH 4.1. S o l u t i o n o f t h e Q . V . I . as a maximum element 4.2. Case o f g e n e r a l semi groups 4.3. D i s c r e t i z a t i o n REFERENCES

368 368 387 39 1 391 392 395 399

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1

CHAPTER I

STOCHASTIC CALCULUS AND STOCHASTIC DIFFERENTIAL EQUATIONS

INTRODUCTION This chapter is devoted to the presentation of the stochastic dynamic systems which will be used throughout this work, namely those whose evolution is described by stochastic differential equations. This requires a stochastic calculus and the concept of stochastic integral, originated by K. Ito. The model looks like dy

=

g(y,t)dt

+

a(y,t)dw(t)

and g is called the drift term, 5 the diffusion term. This model generalizes the model of ordinary differential equations

and expresses the fact that the velocity is perturbed by a random term of mean 0. In the standard set up (strong solution) one assumes lipschitz properties of g, u with respect to the space variable. It is important for the applications to control to weaken the concept of solution in order to assume only measurability and boundedness of the drift term. This is achieved through Girsanov transformation. We have kept the presentation to what is essential within the scope of this text. But, aside basic preliminaries in Probability theory, we give complete proofs. We refer to the comments for indications on the natural extensions. Basic references for this chapter are the books by J. NEVEU C l l , I. GIKHMAN-ASKOROKHOD ~ 2 1 ,A. FRIEDMAN r i i , D. STROOCK - S.R.S. VARADHAN E l ] , E.B. DINKIN 1 1 3 .

2

CHAPTER I

R be a set. A 0-algebra

Let

R is a set of subsets

ff on

of

R

such that V

Ai

nAi i if

a

The elements of

, i

I , Ai

E

E

a,

(I countable)

then

, uAied i

Acff

,

@ A c a

are called e v e n t s .

The set

(@,a) is

a o-algebra.

It is contained in all o-algebras on R. It is called the trivial

a-algebra. A probability on i.e., a map

A

+

(Qg)is P(A)

from

a positive measure on 0

a

into

[O,l]

, with

total mass 1,

such that

P(Q) = 1

P(u

n When

P(A)

= 1,

An) = Z P(An) n

one says that A

if the An

are disjoint.

is aZmost certain

(Q is the certain

event). The triple If B of Q

c

(R,Q,P) is called a probability space.

a,and

is also a 0-algebra we say that i3

is a sub o-algebra

*

On

R

(the set of real numbers), the open intervals generate a o-algebra

on

R, which is called the Bore1 0-azgebra on R.

3

STOCHASTIC CALCULUS

On a product space X

1

x X2

, if X I ,X

x1 x X 2

respectively, the product o-algebra by the events of the form A'

x

Hence the Borel u-algebra on

Rn

' A

are u-algebras on

where

X

I

,

X

2

is the u-algebra generated

A'

x1 , A2

E

E

X2 .

is the a-algebra generated by open

cubes. The concept carries over to an infinite set of spaces X1, i

E

I

product u-algebra B X1 , is generated by the events of the form . I , Ai = X1 except for a finite number of i. Ai E

xi

A random variable is a measurable map

R

Borel u-algebra on If

fi

, i

I

E

E

a,ti

B

E

,

R

/3

i.e., if B

I

denotes the

.

It is called the o-algebra generated

fi are measurable.

by the family fi @

f-'(B)

The

TI 'A ,

are random variables, there is a smallest o-algebra for

which all maps u-algebra

,

R -tf

.

,

and denoted by

u(fi,i

E

I).

It is also the product

.

u(fi)

I Note that if is a R.V.

fk

are random variables and

fk(w)

-t

f(w)

br w, then

f

Also a positive random variable is the increasing limit of

a sequence of piecewise constant positive R.V. namely

We will need some results concerning extensions of probabilities,for which we refer to J. NEVEU [ I ] . Let

R be a set. We call algebra a set of subsets of R, satisfying

properties (l.l), ( 1 . 2 ) ,

( 1 . 3 ) except that in (1.1)

the set

I is not

countable, but only finite. We say that a class C5 {C,,

n

in C

t 1)

such that

n

n
Cn

of subsets of

such that =

0

.

n Cn =

n

R is compact, if for any sequence 0 , there exists an integer N

4

CHAPTER I

Theorem 1 . 1 .

Let Q0 be an algebra on

0 and

a d d i t i v e f u n c t i o n on d o w i t h v alue s i n

LO,

such t h a t

Assume t h a t t h e r e e x i s t s a compact c'lass C5 0

P (A)

0 = limp (Cn)

, with

Po be a f i n i t e Z y

11, such t h a t P0 (a)=

Cn

E

?c

n C?

B A 0

n+-

and

Cn c A .

Then

ff generated by

Po

a0, as

E

1

.

ao

,

can be extended i n a unique way t o t h e o-a'lgebra a p r o b a b i l i t y on

W, ff.

0

One can obtain from Theorem 1 . 1 ,

Theorem 1 . 2 .

(Wigi) where

Let us consider an a r b i t r a r y family

o f PoZish spaces (I), provided w i t h t h e i r Bore2 a-algebras. I 1 f i n i t e , the re e x i s t s a p r o b a b i l i t y

for

on

P"

1

and t h a t t h e f a m i l y i s compatible i n t h e following sense, i f are f i n i t e t hen t h e r e s t r i c t i o n o f P (

I1

.

TI

ie I

P

I2

to

ai

TT

ni, @Gi) which e x te nds t h e family I

P

I1

@

I.1

Qi)

= I2

P on

0

.

Let us also state the following useful result (cf. Dynkin [ I J ) .

Let

E , 8 be a topological space provided with its Bore1 o-algebra. A set

51 of functions f(x), x

E

E which are non negative is called a

A-system if the following conditions are satisfied (1.4)

(1.5)

1 €s1

if

fl,f2 E k4

c l f l + c2f2 (1.6)

if

fn

E

51

2

and

0

and

,

c1,c2 are numbers such that then c l f l + c2f2

fn 1. f, then

I,

coincides w i t h

i ' I 1 Then t h e r e e x i s t s one and only one p r o b a b i l i t y I1

E

Assume t h a t

Ri ,

Q1,

i

f

E

E

k4

51.

Then we have (1) A Polish space is a complete denumerable metric space

5

STOCHASTIC CALCULUS

Theorem 1 . 3 .

I f a A-system c ontains a l l bounded non negative continuous

f u n c t i o n s , t hen it a l s o c ontains a l l non ne gative Bore2 f u n c t i o n s .

Let

f be a R.V., f

2 0.

Ef = We say that

fl

=

0

One can define f(w)dP(w)

(possibly

+m) ,

P(olfl#f2) = 0

f2 a.s., if

.

In general, we do not distinguish between R.V. which are a.s. equal. This relation is an equivalence relation among random variables. One defines the spaces LP(n,ff,P)

1

< p <

m

of equivalence classes of R.V.

such that

and

Lm

is the space of R.V. which are a.s. bounded. fk

+

f

a.s.

if

P(w:fk

fk

3

f

(in probability) if

f) = 1

+

P(w1 Ifk(w)-f(w) as

fk

+

f

in LP

(usual meaning)

Let u s consider L2($l,a,P); it is a Hilbe rt space. algebra of

f f , then the subset of LL(fi,Q,P)

We recall that

k+m,

/>E)

+

V e > O

. If 6' is a sub-a-

made of R.V. which are

6' measurable i s a closed sub vector space of L2 (R,ff,P), hence a sub Hilbert space denoted by If

f

E

LL(Q,6',P).

2

L ($l,ff,P), one defines Enf

=

projection of

f

on

It is called the conditiona2 e q e c t a t i o n of

0

L 2 (Q,B,P)

.

f with respect to B

.

6

CHAPTER I

By definitionf ' E

is the unique solution of the equation

for any I-

E

=

E f q

2

L (Q,B,P), andf ' E

the definition of

I?

L2(Q,',P).

E

to R.V.

@f

equation (1.5) with E?f

rl

fE' E

(1.7)

f

It is possible to extend 1

L (n,',P)

belonging to

any element of

Lm(R,',P)

and

uniquely defines

.

Then F?

is a linear operator from

1

L (~,O,P)+ L'(~,B,P)

which is increasing and a contraction. Moreover k l = l .

If

The Holder inequality holds. p,q

E

(1,~)

E

Lp

f

E

,

g

E

-1 +

Lq

'P

1 = 1 q

then one has

The Jensen's inequality holds. from R

f

If

L1

and

is a convez map

@

into R, then

One should also note the following important properties if

(1.10)

if R 1 c R 2

(1.11)

if

f

Ffg

E

f

E

=

is R measurable, then @f

L1

(1.9)

L1

,

f

, 'g

g FPf

E

B L 1 then E If

.

are independent if

(1.12)

E f E g

=

"1

E

measurable such that

We say that B 1 and B 2 E f g

=

V f

g B2 measurable.

,

g

2

0

,

f

=

f

"2

(E f) fg

E

1

L , then

B1 measurable,

STOCHASTIC CALCULUS

Two R.V.

f, g

are independent if

, u(g)

u(f)

are independent. One

can prove ( s e e comments) (1.13)

We note the following result, Let

respect to the product a-algebra of 0,and

B(Rn)

a R.Y. which is B

g(w)

f(x,o) x

: Rn x

s! + R measurable with

0. Let B be a sub-u-algebra

measurable; assume that

1

(I . 1 4 )

f(g(w),w) L f(x,w) E L' T x

fixed ,

Set

then

A map

f(t,w) : R+

X

R

-C

R

,

such that

i s called a stochastic process.

v'

t,

f(t,oi)

is a real R,V.

The trajectory is the family (as w

varies) of maps

Tde interpret

t

as time,

The process is continuous (right, left continuous) if a.s. its trajectories have these properties.

8

CHAPTER I

Let Zt be an increasinE family of sub-o-algebra of d

3'

c Jt

if

s

We say that

2 t.

Y t , f(t)

(1.16)

A positive random variable

It is important to think of at time

t

f

is

$

i.e.

measurable.

is an

T(w)

,

is an adapted process if

Zt stopping t i m e if

Zt as measuring the i n f c m a t i o n available Zt are observable at time

(i.e,, all events belonging to

t, in the sense that we know if they are realized or not at time

t).

The most useful and standard exemple of a stopping time is the following. Let p

be an open or a closed subset of

Rn, assume that

Zt is right continuous (i.e,, 3t = nt3') Tt is complete (i.e,, contains all subsets of R

(1.18)

contained in a set in Q' with probability 0) Let

y(t)

be a stochastic process with values in Rn, which is right

continuous and

Zt adapted then

(1.19)

T = infItly(t)

E

03

= entrance time of

is a

y(t)

in 8

Z~ stopping time.

To a stopping time

T, denoted

T , is associated the o-algebra of e v e n t s p r i o r t o

T 3 , i.e.

Let us list some very useful properties of S . T . (1.21)

Let S.T.

S,T

be two S.T, then

S

A

T

,S

v T

,

S + T

are

9

STOCHASTIC CALCULUS

Let

( I .22)

S

T be a S . T . and

T

S 2

, ST

is a S . T . is gT measurable

(1.23)

Let

T be a S . T . , then T

(1.24)

Let

S,T

be two S.T. and

Let

S,T

be two S.T. such that

(1.25)

measurable, then

A

E

,

3’

S

then

< T

,

then

3s c iT

(1.26)

Let

{ S > T}

Let

be two S . T . ,

then

belong to 3’

and

S,T

{S

<

T), { S = T}

,

ZT .

5 be @ measurable and n ZT measurable, then

(1 .27)

Let If

Tn be a sequence of S.T., then Tn + T , then T is a S . T .

Let Zt be an increasing family of o-algebras, and

sup Tn n

f(t)

is a S.T.

a scalar

stochastic process, which is right continuous and adapted. We assume

Me say that ( I .28)

f

is a martingale if E(f(t) 13’)

a sub martingale if

=

f(s)

Y t

2

s

10

CHAPTER I

super murtingale if

a

( 1 .30)

E(f(t)lZs)

5

V t t s ,

f(s)

Let u s give some very useful estimates on martingales.

f

If

is a

s u b martingale then

(1.31)

05sst If

f

is a martingale then P{ sup

(1 -32)

/f(s) 1 2

A3

5

Elfo/a,

a > l ,

xa

O<s
if the right hand side is finite. Moreover in case

a > 1 , one also has

(1.33)

Assume that

Then if

S

5

T

S,T

is a s u b martingale and

f(t)

are stopping times, f(S),

f(T)

are in L'

and if

a.s. one has

( I .35)

E(f(T)/aS)

2

f(S)

a.s.

This is Doob's optional sampling theorem [ I ] .

2. STOCHASTIC INTEGRALES

Let

(Qp,P)

be a probability space. A scalar stochastic process

w(t)

,

is a standard Wiener process if the following properties

t t 0

hold

(2.1)

... ,tn

V tl,

with mean

0

, w(tl) ,...,w(tn)

is a Gaussian vector

11

STOCHASTIC CALCULUS

(2.2)

E w(t)w(s)

=

.

min(t,s)

We will admit the existence of Wiener processes. Let us check a few properties w(0)

(2.4)

w(t2)

0

=

,

E w(t)2

, w(t,) ,

w(t4)

t l 5 t2 5 t3 5 t4

t

=

.

- w(t3)

are independent if

Indeed E(w(t2)-w(tl))(w(t4)-w(t3))

thus the R.V. w(t2)-w(tl) , w(t4)-w(t3) non correlated hence independent. Let

vt

be the u-algebra generated by

(2.5)

E[w(t)

Indeed w(t) - w ( s )

hs]=

-

= t2

t, - t2 + t l = 0

are Gaussian R.V. which are

w(s)

,

w(h)

,

0

.

s 5 t

5

One has

w(s)

is independent of

B

X

5

s

,

hence ( 2 . 5 1 ,

and ( 2 . 6 ) . A very important result is the following (due to P. Levy, cf. Doob [I:)

Theorem 2 . 1 .

Let

Zt be a f m i l y of sub-U-aZgebras of

0 , verifying

(1.28) and w(t) a s c a l a r process, wh7:ch is c n n t i n u m s , , sush +hnt w(0)

Then

=

0, and w h i c h is an

w(t)

is

Q

zt martingale

satisfginp

Viener process.

We will come back to the proof of Theorem 2.1, after we have established stochastic differential calculus.

r

12

CHAPTER I

In the sequel we will consider an values in Rn, such that

w(0)

=

zt

family, a process

w(t), with

0 , which is a continuous 3

t

martingale satisfying

I is the identity matrix.

where

According t o Levy's Theorem, it is a Wiener process. We call it But we will not use Levy's

the standard n-dimensiona2 Wiener process.

theorem, and its proof will be given as a consequence of Ito's formula (cf.

§

3).

L2 (0,T) the space of stochastic processes @(t:u) , 5

We denote by t

E

[O,T]

such that (@(t)I2dt <

E and a.e. t

@(t;u)

m

is in L2(Q,3;t,P).

We remark that <(O,T)

is a sub Hilbert space of

L 2 ( n x (0,T) : dP B dt).

We say that

c$

is a step function, if there exists a decomposition to = 0 < t l

... < tN

=

T

such that $(t:u) and

@n

is

3;

tn

=

+n

, n

in Ctn,tn+,)

measurable .and E 16n12 <

m

,

=

o

,.. N-I

13

STOCHASTIC CALCULUS

L ema 2.1.

2 L (0,T).

The s e t of step functions is a dense subspace of

5

Proof. Let

N

+.

m,

and set k

=

$

, Let

Then @k is a step function and hence in L3(0,T). 2

Let

@

t$

bk

E

L2 3

,

+.

@

2 2 in L (0,T:L (n,Q,P)),

define

rl

be a step function. We define

Ve have E I = O

and the second sum vanishes.

Indeed assume m > n

,

then m

2

n+l,

and E

C E(hnmm

tm (W(tn+,)-W(t,))(w(tm+l)-W(tm))13

3

=

14

CHAPTER I

Next we have

Hence (2.10)

The operator I is a linear operator defined on a dense subspace of 2 2 L5 with values in L ( f l , a , P ) and is an isom e t ry (by virtue of (2.10)).

Hence it can be extended to

L2

3

and we call it the stochastic integral of

Theorem 2 . 2 . 2 L3(0,T) and Remark 2 . 1 .

and is an isometry.

I$.

We note

We thus have proved the

The s t o c h a s t i c integraZ d e f i n e s a Zinear i so m e t ry betueen 2

L (R,Q,P). We have

which is a consequence of the isometry between Hilbert spaces.

Remark 2 . 2 .

Assume

@(t)

is a matrix

Qij(t)

and

standard n dimensional Wiener process (cf. (2.8)).

iT @(t)dw(t)

w(t)

is the

Then

is the vector whose first component is given by

C j

T

Qij(t)dw.(t) J

We have, as easily checked

.

E

STOCHASTIC CALCULUS

This last relation is also true with

6

instead of

5'

(by linearity),

hence we have proved that

u B

These relations carry over to

stopping times.

Indeed let

be two stopping times such that T ~ , T ~

then by definition

where

x,W

Note that Indeed

if

10

if

belongs to

@(t)x,(t)

{T 5 t}

1

=

and

{T

> t}

ti^

t

<(O,T)

belong

to

t

~

if

@

belongs to

Zt for any

t.

L3(0,T). 2

Then one

16

CHAPTER I

easily checks the relations

From Remark 2.3, we may define

formula which defines a stochastic process.

=

We have

I(s)

hence (2.17)

I(t)

is a martingale.

-

Assume

to b e a step function then

i$

since w

is continuous. Therefore

'is a continuous martingale.

We thus have the estimate (cf. (1.32) with (2.18)

E

lit $(s)dw(s)12

sup

is a continuous process,

I(t)

I(t)

5

a 4 E

=

2)

{T

I@(t)/'dt

.

05 tsT

Take now $

bn

+

4

in

E

There exists a sequence of step functions

G(0,T).

G(o,T).

We may assume

ll@n-@ll

2

1 5 -

2n

hence

mncl - $11

2 5

4 2"

.

17

STOCHASTIC CALCULUS

Now

4 c - 1,6 n 2"

From Bore1 Cantelli's Theorem (l), we can assert that there exists

N(w)

such that for n t N(w)

then

Therefore a.s. the series

converges uniformly in

t, hence its limit is continuous. Therefore

it @,(s)dw(s)

a.s.

+

a continuous function in

t = J(t)

However, possibly extracting a subsequence

(1) Let events

An

P(An) <

such that

m,

n

hence there exists m t N(u)

.

N(U)

such that@ Am

then P(r L A ) m n mtn is realized for

=

0

,

18

CHAPTER I

Hence

Jt

v t

@(s)dw(s)

=

J(t)

a.s

We thus have proved

Let

be an adapted process such that

$(t;w)

Then it is possible

o

define the stochastic integral

{'

is a modification of

ti t

Indeed define

then

We set

where

Then

(1) We say that

g(t)

=

f(t)

g(t)

a.s.

f(t)

if

@(t)dw(t).

19

STOCHASTIC CALCULUS

hence

One then set

Let

o

E

R

then for

N

2

N

SN(cy) =

on

iT

5,

.

(w)

Hence

6, converges

0

NO II RN , hence also a.s. on N O(t)dw(t).

towards a R.V. which is by definition

We can then state the following estimate (2.19)

P{ s u p

! {t

?(s)dw(s)l

> Cl 5 P { i

T 1@(t)/ 2dt

OSt
2

N N} + 7 C

Indeed we have, by the martingale property (1.32)

and

iT $(t)dw(t) hence (2.19).

(2.20)

=

;T ON(t)dw(t)

if

JT

l$(t)12dt < N

It then follows that

,T J

lQn(t)-@(t)1

sup 02 t
2 P dt + 0

Qn(s)dw(s)

implies

- it Q(s)dw(s)l

0

.

20

CHAPTER I

3. ITO's FORMULA Let

a(t)

,

b(t) be processes which are adaptedtost

(3.1)

la(t)I

,

/b(t)I

5,

+

S

C

,

and such that

deterministic constant.

We define (3.2)

where Let

<(t)

6,

Lemma 3. I .

is 3'

=

it a(s)ds

+

,

it b(s)dw(s)

measurable. We say that

c(t)

is a Ito process.

we have

Consider a splitting of

(0,T)

in O,k,

Then

N- I

...,Nk

= T

.

21

STOCHASTIC CALCULUS = I + I I .

Now I

=

N- 1 I: $'(c(nk)) n=O

/n(kn+l)k

a(s)ds

+

Define fk(t) Clearly since

c(t)

=

$'(c(nk))b(t)

for

t c [nk,(n+l)k)

is a continuous process

hence, according to (2.20)

JT $'(S(t)b(t)dt Clearly

Hence (3.6)

Next I1 = I1 1 + 112 + 111

where

,

.

22

CHAPTER I

and

(3.7)

I11

=

.I

N-1

-

C $"(t(nk)+ n=O

S i n c e by v i r t u e o f ( 3 . 1 )

N- 1

El C

5

ti1

f(n+l)k b ( s ) d w ( s )

a(s)ds

n=O

k C2 T

E

J

T

'n k

2

b (t)dt

We can a s s e r t t h a t

111

P

+

0

P 112 + 0

We w r i t e

111

with

=

1111 + HI2

+

0

1'

5

23

STOCHASTIC CALCULUS

Since sup/$"(t(nk)) n

it is easy to check that

1112

-

+ e,(t((n+l)k)-t(nk)))

0

.

Noting that ass.

the proof will be completed if we show that

-P t o . Let xn

then We have

=

1;

if

sup 1 c ( s ) / O<s
M

iCi"(S(nk))

1

0 a.s

+

CHAPTER I

24

N- 1 Z =

=

C

-

+ll(c(nk))[(df+l)k b(s)dw(s))2

n=O

-

k (n+')k k b2(t)dt]

.

Therefore we can assert that

But (3.9)

Next

The second sum tends clearly towards 0 as

N

-+ m .

to showing that (3.10)

which is a consequence of the following Lemma

Therefore a l l amounts

25

STOCHASTIC CALCULUS Lema 3 . 2 . (3.11)

Let

zt

be a

f(t)

continuous martingale such t h a t

,

L 1 1 5 C(t2-tl)

E[(f(t2)-f(tl))213

t l 5 t2

then one has

We will just prove (3.12) for t2 = T , t l = 0 and we assume for simplicity f(0) = 0. Hence we want to prove (3.13)

E f(T) 4

5

We discretize the interval fn

=

.

C T2

, O,k,. ..,Nk

(0,T)

-

f((n+l)k)

=

T

.

Set

.

f(nk)

We have (3.14)

f(T)4

=

Z f n4 + 4

n + 6

C fnfm 3 + 3 C fnfm 2 2 + m#n m#n

Z n,m,P different

f2f f +

c

n,m,p,q different

fnfmfpfq

and 4 fn

(3.15)

n

+

3

4

fnfm

=

3

4(Z fn)

m#n

n

s Now

c n

;(Cn f

fm

-

3

c

4

fn

5

n

2

y + 2a(C fm) m

.

26

CHAPTER I 3

2 fn

/ C fn/ n

n

/fni

Then

By extracting a subsequence, we may assume a.s. convergence, hence by Fatou’s Lemma (3.16)

Ef(T)4

5

%E[2a(C

fm)’ + 3

m

C :f:f

+

m#n

different

different

Now 2

E ( C fm)2 = E C fm s C T

m 2 2 fnfm

E

5

C

2

N(N-1)k

2

< C2T 2

m#n E

2 C f f f n,m,p different

= E

C

2 f f f

=

S E

P

n,m,p different nmvp

C f2ff m,p
2 2 2 E C f ( C fm) s Ck C E ( C fm) n m
= C k C C Ef:= n m
- c2k2 N(N-1)

C2T2 2

5 -

STOCHASTIC CALCULUS

and

Therefore we conclude

and since

c1

is arbitrary, the desired result is proved

We now extend formula ( 3 . 4 ) to its most general form.

Lema 3 . 3 .

Assume t h a t i n s t e a d o f 13.11, one has

( 3 . 1 7 ) a.s.

{T

la(s)/ds <

a.s.

{T

Ib(s))*ds <

m

Then Tormula ( 3 . 4 ) still holds.

Froo:^ Set

a = N

a

N

-,$,

b

bN =

b

N and a.s.

S,(t)

corresponds to

if

/a1 5 N

if

la1 t N

if

Ib/ 5 N

if

lbj

aN,bN.

2

N

Clearly 1

aN(t)

-t

a(t)

in

L (0,T)

bN(t)

+

b(t)

in

L’(O,T)

hence using ( 2 . 2 0 ) ,

we have

21

28

CHAPTER I

By extracting a subsequence, we can assert that ( 3 . 1 8 ) a.s.

aN

-t

a

in L~(o,T)

bN

+

b

in L2(0,T)

sup JSN(t)--S(t)/ OStST Writing ( 3 . 4 ) with (

aN,bN,EN

,

+ .

0

.

using properties of

$

sup /c(t) 1 bounded a.s.) 0s tST and obtain the desired result.

is continuous (hence

limit in N

We can extract formula ( 3 . 4 ) to the case when and

((t)

is in Rn.

$

Considering the process

extend the formula to functions $(x,t).

and the fact that

one can pass to the

0

is a function on Rn (c(t),t)

one can

Here a approximation argument is

9 C2*l is sufficient (and not C2’2)

needed to check that

.

We can then state the most general formula, which is referred as Ito’s formula

Theorem 3.1.

Let

be an n dimensional standard Wiener process,

w(t)

with respect t o a f a m i l y 3 t

.

Let

a.(t),

adapted processes, such t h a t

{T

a.s.

Let

5,

,

lai(t)/dt <

be a R.V. i n

i,j=l,

1J

4T Ibij(t)/

Rn which i s 3’

...,n,

b. .(t),

2

dt <

m

.

measurable, and

c(t)

process (3.19)

Si(t) =

5

+

it ai(s)ds

Oi

Let

$(x,t)

be

C2”(Rn

x

[O,T1

.

+ C J

t

j o

Then one has

b..(s)dw.(s) 1J

be

J

be t h e I t o

29

STOCHASTIC CALCULUS

(3.20)

+

2

9

1 tr

+

bb*(s))(c(s),s)ds

ax

Jt

+

2

Y t

(c(s),s).b(s)dw(s)

E

[O,T] ,a.s

0 Let u s give some applications of Ito's formula. We start by giving a proof of Theorem 2 . 1 (Levy's Theorem), which follows Kunita-Watanabe [ l ] . Proof of Theorem 2.1

IJ, is complex (by linearity).

We first remark that (3.20) holds when We apply it to $(x,t)

=

exp i x u

,

c(t)

= w(t)

hence $(w(t))

-

= 1

l t 2 J u $(w(s))ds

+

Jt i

u $(w(s))dw(s)

Therefore (3.21)

Let

q

lzsl

EC$(w(t))-IJ,(w(s))

be 3'

measurable and bounded. E $(w(t))q

Set next for

= -

=

E $(w(s))~

$ u2

$(w(A)dA1zS1

It follows from (3.21) that

- 71

2

t

Js E $(w(A))ri

t 2 s

f(t) = E $(w(t))n

then we have the following functional equation for 1 u2

f(t) = f(s) - 7 hence

t Js f(A)dA

f

dX

*

.

30

CHAPTER I

f(t)

=

f(s) exp

- -21 u 2(t-s)

.

Therefore we have proved that Erexp i u w(t)

/ssl=

exp i u w(s)

exp

-

1 2 - u (t-s) 2

or

ECexp i u (w(t)-w(s)) This proves that

w(t)

-

13‘1 = exp

5

1 2 - u (t-s) 2

is independant of 3’

w(s)

Gauss probability law with mean 0 and variance t l 2 t2

-

.

and follows a

t-s. Moreover if

t3, one can compute E exp i(ul w(t

and check that Theorem 3 . 2 .

w(tl), w(t2),

w t3)

2 < p <

then

Let

m,

is a Gaussian vector.

c

(3.22) where

prooic

Set $(X,t)

=

1x1‘

and

c(t)

=

it $(s)dw(s)

,

and assume that the right

hand side of (3.22) is finite. By Ito’s formula we have

We want to take the expectation of (3.23).

However, there is a

difficulty due to the integrand of the stochastic integral, which is not

31

STOCHASTIC CALCULUS

a priori in L

2

.

To overcome the difficulty define

This definition implies in particular that if

We can take

t =

-cN

in (3.23) (I).

We can then take the expectation. We

obtain

and by Halder inequality

We use now the fact that

c(tA.rN)

is a martingale (Doob's Theorem), hence

(3.25) Take in particular

t = T

it follows from (3.24),

,

and notice that

T~

s T, hence

(3.25) that

(1) We can apply Ito's formula with stopping times, since we have a.s.

equality of both sides of (3.20) on

[O,T].

32

CHAPTER I

therefore

hence letting N

Remark 3.2.

+. m

and noting that

T~ +

T a.s. we deduce ( 3 . 2 2 ) .

0 0

The estimate ( 3 . 2 2 ) is due to Burkholder [ I ] .

4. STOCHASTIC DIFFERENTIAL EQUATIONS

Let

Zt

be an increasing family of sub-a-algebras of

(G&,P)

Let

w(t)

be a standard m dimensional Wiener process with values in Rm,

which is an Let

T

zt Wiener.

be an

measurable.

zt

stopping time and

be a R.V. which is

zT

We set

We define the following problem. in Rn

a,where

is a probability space.

To find

y(t)

a process with values

such that

-

(4.1)

(1

(4* 2)

y(t)

x,(t))y(t)

is adapted to 3 t

is continuous

33

STOCHASTIC CALCULUS

Clearly 7.

y(t)

=

5 for

0

t 5 T , hence the process starts really after

5

We use the following notation

(4.4)

dy

=

Y (T) = 5 . and we say that

y(t)

is solution of a stochastic differential

equation in the sense of Ito.

We are going to prove the following

Theorem 4 . 1 .

We assume t h a t

Then t her e e x i s t s a process

which i s a s o l u t i o n o f (4.1) ( 4 . 2 ) ( 4 . 3 1 ,

y(t)

and s a t i sf i e s

where

C depends o n l y on

KO and

T.

There i s uniqueness i n t h e follow ing sense. s o l u t i o n s of (4.1), 1 4 . 2 ) , ( 4 . 9 ) a.s.

1

y (t)

=

if y

14.31 and 14.81, then

2 y (t)

ff

t

E

C0,TI

1

, y 2 are two

CHAPTER I

By Ito’s formula

+

it ( I - x T ( s ) ) IO(Y

1

(s),s)-o(Y

2

(s),s)

By virtue of the estimate ( 4 . 8 ) , we can take the expectation, hence

~

2 ds.

35

STOCHASTIC CALCULUS @(t)

= 0

v

t

2

1

is continuous, we also have

i.e.

1

Since

(l-xT(t)) / y (t)-y (t)

hence (4.9).

Ezistence Let

51 be the space of n dimensional processes

z(.)

satisfying

(4.1), (4.2) and

E

(4.10)

sup /z(t)l O
2

=

1121

l2

<

.

is a Banach space with the norm defined by ( 4 . 1 0 ) .

Then

For

z E

we define

Then

I(.;z)

E

since sup 11(t;z,)-I(t;z2)12 OstsT

hence (4.11) Moreover

s 2 T K2 JT /z,(t)-z,(t)) 2dt

+

CHAPTER I

36

We define the iterating scheme (4.13)

Zn+l = l(Z")

z 0 = 5 .

Set

.

s u p I z n+ 1 (s)-zn(s)12

v ( t )= E

O<s
v

n+ 1

(4.11) shows that

(t) < C J t vn(s)ds

,

vo (ct)n n!

Vo

C

constant

hence v (t) n

2

where

=

vo(T)

.

Now

hence

1 lzk+n-zkl 1

as

k+m- 1

< &=k C

vk(T)R(R-I)

k+m-1 c k=k

1

!?.(a-1) <

k + a .

Therefore zk from (4.11)

is a Cauchy sequence in 3$

I is Lipschitz in H

and thus converges. Since

i t follows that the limit

y

is a

31

STOCHASTIC CALCULUS fixed point of

I. Hence the existence.

Defining also

@(t)

=

E

SUP

ly(s)/

2

O5s5t it follows from equation y

=

I(y),

that

0

and from Gronwall's inequality we obtain ( 4 . 8 ) . Remark 4 . 1 .

The horizon

T

does not play a significant role, since we

can start from arbitrary initial conditions. Clearly if we have solved the S.D.E. on data at time

[O,Tl and extend it on

CT,ZTI starting from the

T, then the extension constructed on

with the solution on

[0,2T1 coincides a.s.,

[0,2Tl. Therefore we can define the solution of the

S.D.E. at any finite time, provided of course the assumptions ( 4 . 5 ) , ( 4 . 6 ) hold for any

t E

0

CO,m[.

5. GIRSANOV TRANSFORMATION

Let us consider a system probability space, and

w(t)

zt

(n,O,P,st,w(t))

,

where (n,G,P) is a

an increasing sequence of sub-o-algebras of

an 3t Wiener, standard and n dimensional. In the sequel we

will speak of an

"usu~Z" system to refer to this system. Let now a(t)

be an adapted process with values in Rn, such that

We define

Using Ito's formula, it is easily seen that

38 (5.2)

CHAPTER I f(t)

=

1

+ Jt

f(s)a(s).dw(s)

.

Hence (5.2) is a stochastic differential equation, which differs only from the fact that

is random. However let us assume that (5.3)

/a(t)1

deterministic constant.

_< C

It is not difficult to check that the proof of Theorem 4.1 carries over to equation (5.2).

Hence there is existence and uniqueness of the solution

of (5.2) in the space (5.4)

E

of continuous adapted processes such that

sup lf(t)12 O_
<

m

,

Clearly there is uniqueness among processes of which satisfy a weaker solution than (5.4), namely

Let us show that

f(t)

defined by (5.1) satisfies condition (5.5) when

(5.3) holds. Indeed define as in Theorem 3.2

then from (5.2) taking t

= T ~ we ,

get

and we can take the expectation of both sides, hence E f(TN) = 1

.

39

STOCHASTIC CALCULUS

From which we deduce letting N

+ m

,

and

using

Fatou's Lemma

Now

/f(t)I2

=

exp 2 J t a(s).dw(s)

- it

=

exp 2 it a(s).dw(s)

-

la(s)I2ds

2 J t l a ( s ) i 2ds

.

.

2

exp J

t

/ a ( s ) l ds

Using (5.3) now we obtain If(t) 1 2

2

exp C2 T exp 2

-

2

Jt

hence by (5.6) (which is true with

la(s)

t

Jt

a(s).dw(s)

-

I2ds instead of

T

and

a

replaced by

Za), we get

which proves (5.5). Therefore when (5.3) holds, f(t)

defined by (5.1) is the unique

solution of (5.2) in the class of continuous adapted processes satisfying (5.4).

It is then possible to take the expectation of both

sides of (5.2), hence

E f(t) = I

.

More precisely we can state the Lema 5.1.

Assuming 1 5 . 3 1 , t h e n we have f o r

0 5 t l 5 t2 5

T

CHAPTER I

40

We can replace

t l , t2

by stopping times

T], T~

such that

o S T 1 5 T 2 5 T .

There has been in the litterature a lot of work to study the most general conditions on

a in order that (5.7) to hold.

Although we will not

really use it, let us state the following sufficient condition (which i s still not the best known).

Lemma 5.2. Assume that

then one has (5.9)

E f(T)

=

1

Define

and write fN(T) = exp HN HN

=

- 7I 1T

exp JT a,(s).dw(s)

then we have (5.10)

E exp HN

=

E fN(T)

=

1

Moreover a.s. hence

aN + a

in L’(o,T;R~)

/a,(t)I

2 dt

STOCHASTIC CALCULUS

41

and

fN(T)

f(T).

By extracting a subsequence, we can assert that (5.11)

fN(T)

+

a.s.

f(T)

Now we have

E(fN(T))I+€

=

E exp(l+E)HN

=

E{exp[(l+c)

x

exp($(l+E)

J 2

-

T

~2( I + E )

%dw-

I+E

T

2

T

2

J0 iaN(t)/ dt]

2

T ) ; luNl dtj = E

q1q2

We u s e HElder's inequality and notice that

E

$7

= 1

, hence

E fN(T))l+E

where

1 1 + P q

=

1

5

{E exp(=(l+E) 2

2- q( I CTE )) l T luN(t)12dt}1'q

.

But

provided that we take

q

=

1 +6 - and

6

E

such that

2

E q

1 + ~ ( l + q )- q

2

0

.

x

42

CHAPTER I

This implies that

E P(A)

3

0

,

xA

fN(T)

E fN(T)

hence

+

+

0 uniformly with respect to = 1

Ef(T)

N

as

.

G

5.2.

Assuming (5.8) (or ( 5 . 3 ) ) , then we have (5.9). a probability

PT

d PT

(5.12)

The family

.

f(T) dP

=

PT is compatible, since when PTISS

=

T 2 S

PS ,

which follows from the martingale property of

f.

Provided that we

make some assumptions on the initial probability space

(n,a,P) we

w

assert that there is one and only one extension

P

on C7

= u(u

T

3 )

can

,

such

T

that

?/ZT= PT

.

Assume that there exists a compact class C

with

T,

We may define for any

R, g T by the formula

on

Cn

E

c, n

Then consider

5'

a0 =

and

Cn

c

A

of subsets of

R

such that

.

u @ T which is an algebra and define a finitely T

additive function on

a0

by setting

43

STOCHASTIC CALCULUS

as in ( 5 . 1 3 ) ,

Then consider the sequence Cn

and

Y

Therefore from Theorem 1 . 1 , U(d

T

P

can be extended in a unique way to

aT).

Let u s now define the process

Then we have

Let

0

E

2 L (O,T;Rn) (non random). I(p) =

We define

iT$(s).dw(s)

- JT $(s).a(s)ds

.

We are going to show that (5.15) where

%[exp

t

i 1($)]3 1

=

exp i I ( q x t) exp -

1 T 7 Jt

2 , Q ( s ) ~d s

,

44

and

CHAPTER I

-E

,

1

if

s < t

0

if

s > t , h

denotes the mathematical expectation with respect to

P

.

Let

6 be Zt measurable and bounded. We have

=

E{S exp i I($xt)exp(J ta(s).dw(s)

X

X

Denote by

E[exp(Jt

T

(a+i$).dw(s)

-

T - i Jt $.ads

-

the conditional expectation entering in the last equality.

We can write

But result ( 5 . 7 ) a+i$

with

$

applies (as easily checked) when

a is replaced by

deterministic. Hence 2ds

5 exp i I(@xt)

which proves (5.15). Let

...,e

el,

be an orthonormal basis of

x

I(e k t ) hence if

A

=

(A,, ...,An)

=

wk (t)

- Jt

ak(s)ds

is a vector of

Rn

Rn.

=

,

We have

-wk(t)

45

STOCHASTIC CALCULUS

Therefore, from (5.15) we get

0

which proves the desired result.

Corollary 5 . 1 .

Let

be adapted processes,

a. (t), b.. (t) 13

which are a . s . square i n t e g r a b k on process (5.16)

y(t)

Then on

(n,a,F), z3e

=

y(0)

+

Jt

Let

[O,T].

a(s)ds

+

ft b(s)dw(s)

bk

i,j

= 1,.

,V

t

E

[O,T] P.a.s.

[O,T]

to be a step function which approximates b

-P.a.s.

(cf. Lemma 2.1 . )

For step functions one easily checks

i t bk(s)dw(s)

=

{t

bk(s)dG(s)

+ J

NOW for a subsequence

(1) bk

+

b

2

. .,n

t o be t h e I t o

can m i t e

ti t

Let

y(t)

in L (O,T;&(R";Rn))

a.s.

P or 7

.

t

bk(s)a(s)ds

.

.

46

CHAPTER I

Jt

bk(s)dw“(s)

+

i t bk(s)z(s)ds Since

P,

P”

it b(s)dc(s)

+

Jt

P”

a.s.

b(s)a(s)ds

a.s.

P,

P”

.

are absolutely continuous with respect to each other on

z

3T , the limits are equal a.s. P or P”.

Let go back to 5 4 . 1 . , and consider Bore1 function g(x,t) , o(x,t). We shall weaken the concept of S.D.E. introduced in ( 4 . 1 ) , ( 4 . 2 ) , ( 4 . 3 ) . The main weakening is the following. “usual“ system unknown. Let

t

( R Q , P , 3 ,w(t))

,

Instead of considering an a priori

we will consider it as a part of the

More precisely, we assume given

(T,S)

as in 5 4.1. We search for

-P

(5.18)

is a probability on

(1 - )c,(t))y(t)

(5.19)

and

(nP,at). (F,y(t),G(t))

where

(n&)

f ; ( t ) are continuous and adapted

processes

-w(t)

(5.20)

is an

for

F

y(t)

=

N

(5.21) a.s. P,

zt t

n dimensional, standard Wiener process

5

+

+

it (l-XTis))a(y(s),s)dw(s)

J

(l-XT(s))g(y(s),s)ds

fi t

+

[O,TI.

The main advantage of this weaker concept, still called S.D.E. is that we will need very little regularity for when a

is invertible.

g

Xamely we have

to solve the S.D.E., at least

47

STOCHASTIC CALCULUS

Theorem 5 . 2 .

We assume

m

n,

=

-1

(5.22)

o(x,t) = u

(5.23)

/o(x,t)

(5.24)

g(x,t) measurable and bounded.

-

(x,t) measurable bounded

~(x',t)/ S K(x-x')

L e t us consider a "usual" system

asswrrptions o f Theorem 5 . 1 and 5'

satisfying the

(O,Q,P,Tt,w(t)) T

a B t stopping time,t; a R.V. which is

measurable and

Then t h e r e e x i s t s

. ., ( 5 . 2 0 )

such t h a t (5.181,.

(F,y(t) ,w(t))

Using Theorem 4.1, we define in a unique way the process

y(t)

holds.

such

that (5.26)

ti t

E

C0,TI

This is possible by virtue of ( 5 . 2 2 ) .

a(t)

(5.27)

,

a.s.

We then de ine

= a-](y(t),t)g(y(t),t)(I-X,

t))

which is bounded by a deterministic constant. Clearly

a(t)

(5.28)

-P

satisfies the conditions of applications of Theorem 5 . 1 ,

hence Corollary 5 . 1 holds.

-w(t)

Therefore we define

= w(t)

- It

by ( 5 . 1 2 ) and

a(s)ds

0

and we obtain that desired result.

y(t)

satisfies ( 5 . 2 1 ) , applying ( 5 . 1 7 ) .

Hence the

0

48

CHAPTER I

Remark 5 . 1 .

The solution of a S.D.E. (strong or weak formulation)

will be referred as a diffusion process. a

drift term, and

g

We say that

is the

0

the diffusion term.

COMMENTS ON CHAPTERI 1 . Relation (1.13)

holds with for

can be deduced from Theorem 1.3.

Assume that (1.13)

@,$ continuous bounded, then we must prove that it holds

q ~ , + measurable bounded.

consider the set 24 satisfy (1.13).

I$ t 0

Let us first fix

of functions @ : R

+

R

,

and

which are positive and

This set is a A-system which contains all continuous

bounded functions, therefore it contains all measurable bounded functions.

Hence (1.13)

continuous bounded.

that (1.13) holds for

0 Borel

holds with

Fixing now j,

6

bounded and

bO

and reasoning on

Borel bounded

20

as well.

w,

$

we obtain

We can then

go from any Borel positive bounded function to any Borel positive

function by an approximation argument. 2 . Let

y(t)

be a stochastic process with values in t

Rs

continuous with left links.

Let

open bounded subset of

We write

R".

=

u(y(X),s
This is called the moment of first contact with is a continuous process,

9. Then ?

is a

;

Rn, which is right Let0

be an

Rn - @.. Clearly if

coincides with the first exit time from

stopping time (cf. Dynkin

[

11, p. 1 0 6 ) .

3 . We will slightly develop the theory of martingales in Chapter 111. A

basic reference is C. Dellacherie - P.A. Meyer [ l j . 4 . The existence of a Wiener process can be found in H. ?lac Kean [lj.

can also be studied and constructed using the theory of Markov processes (cf. Dynkin C11, cf. also Chapter 111, section 3 ) . Nevertheless some restrictions on ' SZ,C,P

are necessary.

It

y

49

STOCHASTIC CALCULUS 2 5. I n the proof of Lemma 2.1, we use the fact that C(0,T;L ( Q , f f , P ) ) 2 2 to @ for dense in L (0,T;L (Q,Q,P)). The convergence of r$ 2 elements of C(0,T;L (Q,Q,P)) is easy.

6. Let Let

JT

r$(t)dw(t) be a stochastic integral with T 2 = { W ~ J I@(t,u)! dt = 0 1 , then one has

xA {'

@(t).dw(t)

=

in which we can go to the limit as w(t)

is a

E

L 2 (0,T)

0 a.s.

lndeed consider the approximation Q k ,

7 . If

9

is

k

then

+

m.

Zt standard Wiener process, it is also a Zt+O

standard Wiener process.

This is easily checked from Levy's

Theorem.

8. One can generalize the existence and uniqueness result of S.D.E. to the case when the functions g,c

are only locally Lipschitz.

The

solution does not satisfy ( 4 . 8 ) and assumption ( 4 . 7 ) can be removed. We will not need this result explicitely. A. Friedman [ I ] ,

See for proof, for instance

or A. Bensoussan - J.L. Lions [ l ] .

9. One can replace the growth assumptions ( 4 . 6 ) by the following

We may also assume g(x,t)

continuous in

(g(X,t) - g(x',t)).(X-X')

x 5 0

ti X,x'

This will insure existence and uniqueness of the solution (see A. Bensoussan - J.L. Lions [ I ] ) .

50

CHAPTER I

10. Relation ( 5 . 6 ) holds without assumptions ( 5 . 3 ) .

Indeed consider the

scalar product case to simplify. Set

Clearly a.s.

{T

2

laN(t)-a(t)I

subsequence fN(T)

f(T)

+

dt

+

a.s.

0 as

N

+ m.

At least for a

and since E f (T) N

5

1 , Fatou's

Lemma implies again ( 5 . 6 ) . 1 1 . Lemma 5.1 expresses the fact that

f(t)

is a martingale

12. Assumption (5.13) will be satisfied for the canonical process.

Define Go

qt =

=

Bore1

,

C([O,m);Rn)

,

o(x(s),O<s
x(t;o)

and 4

=

VP".

C(C0,TI;R").

also a probability on

0

= w(t)

Then 7:

coincides with the

Since

defined on RO,R

,7flT , and since

P

C([O,TI;Rn)

is

is a Banach

denumerable space, there exists (cf. Neveu [ l l , p. 61) for any A

E

pT a sequence of compact subsets of C([O,TI;Rn) , Cn such

that

Cn

Clearly

c

A

Cn

and

is

qT

measurable, hence (5.13) is satisfied.

13. We have not discussed at all the problem

of uniqueness, when

dealing with weak solutions. We will come back partially to that question in Chapter 111, with the martingale questions of uniqueness, see T. Yamada

-

problem.

For related

S. Watanabe [ l l .

1 4 . In some cases it is possible to interpret a S.D.E. equation as an

ordinary equation, where w enters as a parameter (cf. H. Sussman

I1 I)

51

CHAPTER I1 PARTIAL DIFFERENTIAL EQUATIONS

INTRODUCTION Partial differential equations play a very important role in the study of diffusion processes, particularly in connection with functionals defined on the trajectory (also in connection with their Markov properties). Actually one can already realize this for ordinary differential equations. Consider the O.D.E.

Assume g Lipschitz and bounded (to simplify). solution of (i) to emphasize the dependance in Consider the functional

Write

yx(t)

for the

X.

Then suppose f bounded, f Lipschitz and n large enough. It is easy is a Lipschitz function. >loreover, we have from (ii), to check that u(x)

From this, one easily deduces,by a limit argument as

(iii)

au - g.Du

=

f

6

+

0 ,

a.e.

which is a first order hyperbolic partial differential equation (P.D.E.). We have emphasized the point of view of computing a functional on the trajectory of the solution of an O.D.E. One can also have the reverse point of view, which consists in giving eqZCo7:t formulas for the solutions of P.D.E. In the litterature on P.D.E., this is known as the method of c h n r a c m r i s s k s . The O.D.E. ( 1 ) indexed by the parameter x, are called the characteristics of (iii). A s we will see, for elliptic and parabolic second order P.D.E., one can introduce characteristics curves, which are diffusion processes. In our presentation, except two difficJit regularity results on P.D.E. we have given full proofs of the results, but we have restricted ourselves to those which will be effectively used. The basic references are A . Friedman [21, O.A. Ladyzhenskaya and N . h . Ural'tseva [ I ; , O.A. Ladyzhenskaya - V . A . Solonnikov, N.J. Ural'ceva [ I ] , J.L. Lions [ I ] , S. Agmon - Douglis - Sirenberg [ I ; , C. Miranda [ I ] , C.B. Morrey [ I ] , G. Stampacchia [I], J.L. Lions -

52

CHAPTER I1

E. Magenes [ I ] .

For Sobolev spaces see for instance F. Trhves [ I ] .

1. FUNCTIONAL SPACES 1.1. &blgd-spaws

Let

We will use constantly the SoboZev

8 be an open subset of R”.

spaces

, WzYp(S) ,

W”p(8)

W’%)

=

2

ff

5

p

E

LP(Cs)

< m,

integer. We recall that

1%

€

LPOS)}

1

,

W”pp)

For

p

=

W2”(0)

are Banach spaces for the norms

2 , we use the notation

HZ

= w192

= w292

They are HiZbert spaces. In the case when 8 follows (1.1)

(p,q

...

if n>p ,

€

is smooth, we have important i n c l u s i o n theorems, as

[I,=)) 2

W’’p(o) c L (8) with continuous injection

Clearly this result implies that if W2*p(c) c W’”((S)

n > p with continuous injection

53

PARTIAL DIFFERENTIAL EQUATIONS n > 2p

and if

(1.2) if p>n with

,

-1

a = 1

P

then

c

~ ~ ( (')6 )with

continuous injections

c

$65)

8 = 2 - 1

w"P(o)

.

Consequently

if n > p E 2

W2'Pf3)

(1.3)

Let

f

E

P

, and f

W2"f&)

# integer

20

for any

E

[ + , I 1 , and

c

I If/

I If1 For example taking

a

lwl,r <

a

=

interesting only when

1

- we get

p < q

.

IflL4

l;2,p

1

If

=

1

E

1 -a

T(;1 +

a = 1,

Lq(0)

1

-),

then, if

1

-

I! # P

*

This choice is

then - = r p

-

n

,

which is

result ( 1 . 1 ) .

(1) space of functions

v

x,y

E

'3

f

I I I I c"= f

C1+'

=

such that SUP XZY

'

If(x)-f(y)l

< Cix-yl"

(x)-f (y) i + sup I f (x) 1 Ix-yl" X

space of functions which are differentiable with derivatives in

c a , o < a < 1 ,

I I f 1 ICl+"

=

If/

C

0

+

IDfI C"

54

CHAPTER I1 When 8 is bounded the injection of

(1.4)

L P ~ ) is compact

, p

t

1

W’”(0)

into

.

We next consider the space

which is a sub Banach space of Similarly we write )B(,H

p >

Assume if

f

for

, then W;”(O)

,

w;’P(o)

E

1

fin

WlYp(@

=

o

c

Co(6)

with continuous injection, hence

.

p 5 2 , then f is no more continuous. However it is still possible to introduce the c m c e p t : 0 t m c e on the boundary r = a 0 (recall that

When

8 is regular). We will just use some formal aspects of the t r a c e c c n c e p ; , that we briefly explain.

Think of 0

introduce local maps), then

as

We need to introduce the space partial derivatives in Let

(1)

f

r

E

{x, > 0 )

? E Rn-l

LL(r)).

H1’2(r)

(otherwise we needs to

. (function L2(r)

,

with “half”

To do this, one uses Fourier transform.

L2(Rn) , the Fourier transform

?(i) is defined by

is provided with a measure, which here is just

dxl,...,dxn-l

(2) J ( 0 ) = space of infinitely differentiable functions with compact

support in

8.

.

55

PARTIAL DIFFERENTIAL EQUATIONS Then

3 is an isometry from

L2 (Rn)

into itself (I)

.

Remark that af

3(-) axk

,

i 5,3(f)

=

and by the isometry property o f 3 af -I axk hence

Therefore an alternative definition of

H ~ ( R ~ =) {f

I n H (R )

is the following

L ~ ( R ~1 )(

This justifies the following definition H1/2 (Rn) = {f

(1.7)

In general

H'l2(r)

E

L2(Rn)

is a HiZbert space of functions defined on

r(2)

,

such that in particular

(1.8)

H1I2(r)

Now the main ideas Consider H'(0.)

.

c

L2(r)

with continuous injection

.

introduce the trace c o n c q t are the following.

to

It contains a dense subspace of smooth f u n c t i o n s ( 3 )

(1) We consider complex

valued functions.

(2) Whose precise definition requires the use of local maps, hence makes

sense only when

(3)

r

is regular.

B ( g ) is dense in H 1 (0).

.

56 When

CHAPTER I1 f

is regular, we can define Yf

=

trace of

f

T

on

We thus have defined a linear operator from a subspace of

H'(e) (made of regular functions) into a space of functions defined on

When

f

is smooth, i t is clear that

yf

considered as a map from a subspace of (therefore also into H1/2( r ) ) .

T.

y can be H'/'(r)

is smooth. Hence

H 1 (8)into a subspace of

The important thing is the following

estimate (1.9)

c

llYfilH1/2

(r)

IIfI I

1 W f H (8)

This estimate permits to extend the definition of

smooth

y

to

.

H 1 (@)

.

The

1

trace operator is a linear continuous mapping from H (8) into H~/~(F). When

f

E

H1@)

,

f 1;

we will write

instead of

yf.

This definition is consistant with the usual meaning of the trace when

f

is smooth.

Hi@)

Naturally, by the definition of

(1.10)

fir

=

o

if

f

E

wa have

.

1 H~

It turns out that the reverse is also true. the operator y

coincides with

HA

In other words the kernel of

.

Let us mention the following important property of inequality)

It follows from PoincarG's inequality that

1

Ho

(PoincarB's

57

PARTIAL DIFFERENTIAL EQUATIONS

is on Ho1 a norm which is equivalent to the norm induced by Remark 1 . 1 . HI(@

c C"(0)

Assume 8 , a = -1

=

.

(0,l)

,

i.e.

The trace yf

check in this case property ( 1 . 9 ) ,

1

,

H1

.

then by ( 1 . 2 )

n

=

=

(f(O),f(l)).

It is easy to

(1.10) and its reverse, (1.11).

This

c3

is left as an exercise.

Let

v

r,

denote the outward unit normal on

and

vi , i=l,...,n, the

v. We have the following integration by parts

components of formula

Formula (1.12) can be proved for regular functions and extended by 1 continuity to functions u , v E H (8); since u,vIr belong to H"2(r) hence to

L2(r)

,

the integral on

r

is well defined.

In the sequel the following Green's formula will play a very important role.

Let

aij(x) , i,j=l,...,n

(1.13)

aij

E

such that

W'~~(R")

We define (1.14)

(1.15)

I

belongs to

2 L (8) when

u

E

.

2 H (8)

58

CHAPTER I1

We write

-avA au -

(1.16) When

u

H2 (C)

E

-

'

, %Ir

aU v . F 1 j

H1'2(r)

E

Define also for u,v

r.

on

.

1 H (8)

E

a(u,v)

(1.17)

aij

au av dx Z J8 a.IJ. ( x ) ax. ax

=

i,j u

Then if

2

H (8)

E

,

v

JS A

(1.18)

E

J

i

1

H (8) , the following Green's formula holds

u v dx

=

a(u,v)

It i s an easy consequence of (1.12).

au - 1r v aVA

dT

Next if

E

u

I

,v

H2(@)

2

H (0)

E

then

In particular if

E

1

HO(8)

J8 A u v dx

(1.20) and if

v

u,v

Jo

A u v dx =

Assume that in (1.12)

tr E

v dx

v

+.

- 10 u

A = A

*

.

=

av

j q dx

is defined as A

u A* v dx

.

1 Ho, then we have

Joe

(1) A*

a(u,v)

1 Ho(B)

E

(1.21)

But the map

=

with

- Jsu

av

dx

i is linear continuous on

.j a., replacing a. 31

.

H

If

1

.

Therefore

aij

=

a. Ji

then

59

PARTIAL DIFFERENTIAL EQUATIONS

Hence

JS where Lu

E

6

,

H-l@)

v dx

=



1 Ho

the dua7, of

,

which is defined on HI

and the map

satisfies

therefore it can be extended as a continuous map from L2@)

.

H-l(O)

Since L

Therefore Au

a

ij

.

&(L~(O);H-'(O))

axi

is defined for u

are just Lm

H'@)

E

and

functions.

2. THE DIRICHLET PROBLEM FOR ELLIPTIC EQUATIONS 2.1.

The_baslc-exlstence-and_regularlty_yesult~

Let 0 be a bounded smooth domain of (2.1)

aij(x) a.. 1J

=

1 aij We write

into

say on smooth functions, this means that

axi

a

(1.22)

when

=

a -

,

R",

...,n

i,j=l,

a.. JI

EiSj

2

B 151

2

.

E

and

a C (0)

,

0 <

a < 1

60

CHAPTER I1

A

-

=

O

C

i,j

a2 aij(x) ax; axj

The first basic result is the Schaiider a prior; estimate, which is the following (2.3) V

where

K

depends only on

r

boundary

=

a

8(1)

r,

B, the C"

norms of the

a.

and the

ij

C2+" (@) -

We have denoted by vanish on

.

the space of functions in C2+"(5)

which

and

I lVIlc2+r

=

lvl

+

i

CO

-

For the proof see Ladyzhenskaya

1-1

C

2

av 0

c

+

x , -I i,j

a v

I c"

Ural'tseva [ l ] .

The second basic result concerns the Dirichlet problem for the Lap lacian

-

(2.4)

Then if

f

E

AU + Xu = f

C"G),

u l r = o

, A 2 0 ,

(2.4) has a unique solution in

C2+"

(c?-)

*

From (2.3), (2.4) one can deduce

We assume ( 2 . 1 ) .

Theorem 2 . 1 .

Let

e x i s t s one and o n l y one s o l u t i o n A

(2.5)

(1) In fact

r

C"

0

u+Au=f

f

u

,

is sufficient.

E

C

Ca(s) 2+a -

E

ulr=O

and

(8) of

h > 0.

Then there

61

PARTIAL DIFFERENTIAL EQUATIONS

Let u s prove uniqueness.

It will follow from the estimate

IlflI ll4

(2.6)

0

A

5-

C

0

C

*

u

Indeed let xo be the point where 0 u(x ) > 0 , then xo E 8, hence

2 Indeed -a(x u 2

o)

reaches its maximum.

0 for any direction

2

y.

Assume

aij(x 0)

Since the matrix

is

aY

positive definite, we can write A

where

Ak(x

0)

U(X

0

0) =

-

o

A

C

k

(X )

are the eigen values of

corresponding eigen vectors,

-a2u (x

o)

a$

k

0 a..(x ) , and

0 yk(x )

the

1J Hence ( 2 . 7 ) . From which and (2.5) it

follows that

If

0

u(x )

E

r

proves that if now minimum, then

0 u(x ) s 0

or if u(x )

xo _>

(2.8) is clear. A similar argument

denotes the point where

.

-

u

reaches its

~

~

Hence ( 2 . 6 ) . (T E [ O , l l )

Consider now the family of operators

and the Dirichlet problem (2.10) Let

u-

A

T

u + A u = f

ujr

be a solution of (2.10).

tion of the matrix

aij

into

=

o ,

9

~

6

)

Since (2.9) corresponds to a modifica-

62

CHAPTER I1

aij(I-'?) + TI

(2.11)

t

Min(l,R) I

we still have as for ( 2 . 8 )

(2.12)


I

1IUT1

0 C

and from the SchaGder estimate

Let

TI

< T , and consider the equation

A T l ur, + Aur,

=

f

which we may write as

or

.

AT u T 1 + AuT, + ( T - T ' ) ( A ~ + A ) u ~ ~= f

(2.14) We start with

,

T = 1

Al

=

-A , and consider

z

E

, and 5 defined

C2'a

as the unique solution of

A 1 < + A.5 + (1-T') (A0+A)z

T'

E

[ T ~ , I I , where

Indeed since

S1

M

1 IA+A,l I

which is a contraction in

T~ < 1

<

E

2

T = 1

1

.

c2+'

c2+a, for

is given below.

is affine, it is enough to consider

applying ( 2 . 1 3 ) with

and if

,

S1

This defines a map

,

= f

and

, we take

f

=

f

=

0. But then

(~-T')(A+A~)z we get

1

if

63

PARTIAL DIFFERENTIAL EQUATIONS

M

1 lA+AOi 1

<

1

we take

T1

= o .

This proves that there exists a solution of (2.10) for -cl

=

0

,

we can take

use (2.14) with T1

M

I lA+Aol 1

? =

2 1

T =

T~

,

, and

T

E

[T~,I]. If

0 and the theorem is proved. Otherwise, we 1 T ’ E [ T ~ , T ~ IT~ , > T~ if

-MTln+Aoll

T~ = 0

if

T~ M

I lA+Aol 1

< 1.

We define S 2 by

solving AT 5 + Xg + (T~-?’)(A~+A)z = f

, 5

E

C2’a

1

and

S2

is a contraction for

T’

E

.

[T~,T~I

Hence there exists a solution of (2.10) for

T E [T2,1]

.

After a finite number of steps, we see that there exists a solution of

(2.10) for

T

E

0

[O,l], hence the desired result.

We now give another important estimate with c2+a

W2”

spacesinstead of

We assume now that

(2.15)

1

aij

=

aji

We have the following estimate W2” n W;”

(2.16) V

v

where K

depends only on

E

with

1 < p

then

< a ,

p,n, the modulus of continuity of the a.

1j

64 and

CHAPTER I1

8

We have t h e n t h e e x i s t e n c e and u n i q u e n e s s r e s u l t (Agmon

-

Douglis

-

N i r e n b e r g [ 11).

Theorem 2.2.

X

ksswne 12.151 and

t

f E Lp(8), p > 1,

0. Let

p <

a.

Then t h e r e exists one and on13 one s o l u t i o n of (2.17)

Remark 2.1.

,

A 0 u + Au = f

The assumption

u

W2"

E

aij = aji

r

n WAY'

d

i s not r e a l l y r e s t r i c t i v e for

problem ( 2 . 5 ) s i n c e we c a n always w r i t e

Let u s assume h e r e

(2.17)

a.

(2.18)

a.

L e m a 2.1.

ij

=

a.

j i

,

a..

aij

L~(R")

E

ij

L ~ ( R ~ ,)

a.

L

o

cisj

2

2

161 , 6 > o

.

Under assumptions 12.171, 12.181 t h e r e exists f o r

one and o n l y one solution

Consider t h e f u n c t i o n a l on

J(v) =

u

E

1

Ho(0)

f

E

L2@)

such that

H;)(O)

C '03ij i,j

a v dx axj 5

From P o i n c a r 6 ' s i n e q u a l i t y i t follows t h a t

+

a.

v2 dx - 2

f v dx

.

65

PARTIAL DIFFERENTIAL EQUATIONS

1

It i s a strictly convex continuous functionalonHo(B) admits a unique minimum.

.

Therefore it

Since clearly (2.19) is the Euler equation

0

for the minimum, it has also one and only one solution.

,?emark 2.2.

The result of Lemma 2.1 holds even when the matrix

not symmetric (wnich is here a restrictive assumption). Lax-Milgram Theorem (cf. for instance K. Yoshida [ I ] ,

a . . is li

This is the

J.L. Lions [ I ] ) .

The proof is different, since (2.8) cannot be interpreted as an Euler

D

equation. Remark 2.3. .The assumption

f

assume

E

f

E

L

2

(8)

is n o t really necessary.

(2.20)

W”~(R~)

a. lj

.

Consider also (2.21)

a

i

E

.

L~(R~)

We write (2.22)

We will set (2.23)

and

bi

b.(x)

E

3

H-l(O) with the same result.

Let us assume now that besides (2.17) the

Lffi(Rn)

.

=

aa. - ai(x) + C ax j j

We can

aij

verify

66

CHAPTER I1

We define next the bilinear form on

H1(0)

(2.24) +

Now if

v

E

1 Ho

1

(2.25)

Theorem 2 . 3 . f

E

,

Lp(0) , 2

fs.a

we deduce from Green’s formula (1.18) and (1.23) that a(u,v)

u

E

=

H1@)

+

,

Au

fsa. E

u v dx

fi v

E

p <

m.

1 Ho

,

.

Let

.

H-’@)

We assume ( 2 . 1 7 1 , (2.181, (2.201, ( 2 . 2 1 ) (l) 5

.

u v dx

Then t h e r e e x i s t s one and o n l y one s o l u t i o n

u

of

Let us first show that without l o s s of generality, we may assume that (2.27)

y

a o t y > O

some positive number

.

Indeed we make the change of unknown function

where yo

to be chosen and

xo

is fixed outside

6.

Clearly (2.26)

is equivalent to

Let us make explicit the operator w A(wz).

We have

(1) Because of (2.20) the symmetry assumption on

a.. is not restrictive 1J

67

PARTIAL DIFFERENTIAL EQUATIONS

= -

C

i,j

a

w - (a..(-

axi

L

Let u s s e t

We have f o r

hence

x

E

0

IJ

ao

ax

+

z

j

I

OJ

aZ

-))

ax. J

t

68

CHAPTER I1

aij(x.-xoi) (x.-x ) J

c

+ yo

i,j

Ix-xol

2

Oj

-

C a.

ii

therefore

exp - ylx-xo/

outside 5, i t is possible to choose yo > 0 such that

Having fixed xo when

.

runs into 8 one has

x

, problem (2.26) is equivalent to the same problem with 2 aij changed into , ai changed into a.w2 , a. changed into aij w(aow + X) 2 y > 0 , and f changed into fw. Therefore we may assume Since

1 5 w 5 2

(2.27)

.

Let u s next prove that for

h

large enough, the problem

has one and only one solution. Let u s first show uniqueness.

f

=

0 , from Green's theorem, it follows that 0 0 a(u , u ) + A luo12

(1) We cannot take

y

=

o

arbitrary large.

Assume

69

PARTIAL DIFFERENTIAL EQUATIONS

A

and we can assume t h a t

hence

L e t u s prove e x i s t e n c e .

uo = 0 .

Define f o r

z

i s l a r g e enough s o t h a t

1

Ho

E

(2.32)

,

5

1

E

'0 a i j =

t o be t h e s o l u t i o n of

Ho

av dx a 1~a. ~ . 1

aZ Is ( f - Z a i %)v

+ J

dx

( a + A ) < v dx 0

8

=

,

i V v r H

Then

5

1 0 '

e x i s t s and i s d e f i n e d i n a unique way, by v i r t u e of Lemma 2 . 1 1 Ho i n t o i t s e l f S ( z ) = Consider

:.

We t h u s have d e f i n e d a map from

cl

z1,z2 and

Take

and i f

v =

tl-c2 ,

A

Let

, c2

=

S(z2).

We have

we deduce t h e e s t i m a t e

i s s u f f i c i e n t l y large it follows t h a t

which proves t h a t point.

S(zl)

=

uo

S

i s a c o n t r a c t i o n mapping i n

be t h e f i x e d p o i n t .

1

Ho, hence i t has a f i x e d

From ( 2 . 3 2 ) we have

70

CHAPTER I1

We next use Theorem 2 . 1 .

l-

(2.34)

Then

zo

i,j

There exists one and only one solution of

XTET

aij

1

+ Azo

=

f +

;

aU 0

bi

j

- a.

u0

1

satisfies -0 au

(2.35)

' 0 aij

~i

Js zo v

+ X

2i $i -0

JS

+ C

av dx K a~ dx

=

aa.

v dx +

auO a u O )v dx (f+Zbjaxj- o

Js

J

y v c H 01 .

But from ( 2 . 3 3 )

satisfies

uo

auO av auO aaij v d x + A J u 0 8 axi d x + C /8 -ax j ax.i -U aU a u 0 )v dx . =Jo(f+;bjaxjc Jo

(2.36)

aij

vdx=

By difference between (2.35) and ( 2 . 3 6 ) we obtain

Z JS aij

v

and since

Go

=

A

- uo

,

ax j

dx

=

(ea")'-' -0

auO av ax j axi

dx + C JS

- -)-

(zo - u 0)v

+ A J8 Taking

a;'

(-

j

0

0 aa..

-

ax j a..~i

v dx +

.

we see that

is sufficiently large,

(1) The existence and uniqueness of consequence of Lax

-

Lo =

uo

Milgram theorem

uo

.

solution of ( 2 . 3 3 ) i s also a

71

PARTIAL DIFFERENTIAL EQUATIONS 1 uo E H2 n Ho

Hence

and

is solution of

uo

AUO + (ao+X)uo

=

f

a.e. in 0 ,

or

If

u

0

0

, (assuming n

H2

E

>

PAq 1 (by ( l . l ) ) ,

hence

g

E

L

and

p,

=

pAq, > 2

uo E’W~’~~(;)

using Theorem 2.1 again we get

L q l , with

axi

2 ) , we have

.

1 = r - 1 q1 2 n

p > 2).

(if

Therefore

By a bootstrap

argument, in a finite number of steps, we obtain uo

E

,

W2”(@)

hence

(2.30) is proved. The next step is to define the following sequence

which is well defined, by virtue of what we have done to study (2.30). We have (2.38)

The important fact to notice is that there is an improvement of regularity at each step. ul-uo

E

u2-u1

obtains for assert that

n

2

E

W2”(0)

n WA”(0)

hence

.

L q l with

Therefore

ul-u’

Indeed

E

-= -- n (if n > 2p) q1 P 2,q1 l’ql W n W, Again using a bootstrap argument, one

no , un - u’-’

.

E

Lm.

Using Lemma 2.2 below, we can then

12

CHAPTER I1

Hence at least for n

2

no, un - un0

is a Cauchy sequence in Lm.

Passing to the limit in (2.37) we conclude to the existence o f

u

solution

of (2.26).

Let us prove the uniqueness of the solution of (2.26). u

Let

f = 0

,

and

to be solution of A u + a u = O 0

then

and by the regularity argument used for (2.37), u

m

E

L

.

From Lemma 2.2

it follows that

hence

I lull

=

0

.

0

Lrn

Lema 2 . 2 .

Consider t h e equation

(f

E

which has one and onZy one s o l u t i o n f o r f

m

E

L , one has

u

m

E

L

Lp(8))

X large enough.

Then if

and

Before proving Lemma 2.2, we will recall an important Lattice property o f the space

1

H (8).We say that

73

PARTIAL DIFFERENTIAL EQUATIONS

Define

then u1

V

,

u2

u1

A

u2

E

1

H (0),

namely the following formula holds (2.41)

where

x

7

=

1

if

u1

u2

2

and

if

0

u 1 < u-

.

Setting in

u zu particular

then u = u

+

- u

-

.

+

, u

Remark that (2.42)

uiHA+u

This is clear when (2.43)

n u

u

B(G). Then one has to notice that

E

1 E H

implies This is clear since

1

6 H O .

, (un)+

in H

u n + 0 +

o

in

H'

1

.

14

CHAPTER I1

ahn)+

-=

ax

aU" axi X n

u LO

hence

Let us also note the relation (2.44)

a(v+,v-)

=

o

Y v

E

I

.

H (8)

W e can then give the

Proof of L e m a 2.2 Let

K

=

llLLl X+Y

then

(u-K)

+

E

I

Ho, hence from (2.39) a(u, (U-K) +) +

or

x (u,(U-K) +)

= (f

, (U-K) +)

PARTIAL DIFFERENTIAL EQUATIONS By a similar argument one obtains u t -K , hence (2.40).

Lemma 2 . 3 .

Assume in Zzenrem 2.3 t h a z

f

t

0. Then

u

0 2

0 (u

scltrtion

of ( 2 . 2 6 1 1 .

This is done by induction on the sequence Assume

un 2 0

,

then multiplying by

un

(un+I)-

defined by (2.37) we get

hence a((un+')-,(un+')-) which implies

n+l (u )

=

0

+

> 1 (un+l)-12

s

o

.

A similar argument holds for

uo, hence the desired result.

3 . PARABOLIC EQUATIONS

3.1. &nc;jmrm-~p~ces

We will use

dz The meaning of d t has to be explained. We note the inclusions

each space being dense in the following, with a continuous injection. Thus we can consider

z

also as an element of

L2(0,T;H-'(O)).

Its

derivative makes sense as a distribution and is defined as follows

76

CHAPTER I1

.

-1

the second integral being with values in H

aZ -6 8’ ( (0,T ) ; H - (0) ~ ). at

@

+

- JT

@ ’dt

z

E&(L~(O,T);H-’(@))

aZliE L2 (O,T;H-’@)) then we say that ; important property of

C

and that

z

E

W(0,T).

0 C ([0,Tl;L2(o))

with continuous injection.

Property (3.1) is true at least after a modification of

0

of measure If

on a set

z

.

z1,z2 E W(O,T),

the following integration by parts formula holds

dz

JT

=

We write

An

is the following

W(0,T)

W(O,T)

(3.1)

Hence a priori

If

< 1x z, 2 > dt + IT < z 1

, dt dz2 >

dt

.

Q = 8 X (0,T) and consider spaces of the form 2 $ 2 $ P ( ~ )=

(3* 3)

12

L~(Q) ,

6

ax; , ax;ax. , aLz

aZ

aZ

6

L~(Q)I

J

with the natural norm of Banach spaces.

...

Similarly we define bl””(Q) b Z y 1= b 2 ” ”

(3.4)

,

= {z E

1 5 p <

m.

2

2

p = 2, we write

When a Z

L (0,T;H (O)),at

A l s o we will denote

1303p(~)=

b O

(3.5)

where

C =

r

X

(O,T),

{ Z

E

bl””(Q)

,

zlz =

01

E

2

2

L (0,T;L

0)))

PARTIAL DIFFERENTIAL EQUATIONS Let u s n o t e t h e f o l l o w i n g

(3.6)

if

z

then

E

lb2’o’p(Q)

z

E

and

bl’o’r(Q)

Lq(Q) p,q z I 1 = 1~ 1( +p-)1 q

with

r

This p r o p e r t y can be e a s i l y deduced from ( 1 . 3 ) .

,

.

Indeed we have

theref ore

and by Holder i n e q u a l i t y

hence

(3.7)

which i m p l i e s ( 3 . 6 ) . We w i l l a l s o u s e Soboleo spaces w i t h Geighss. Define

(3.8)

Lp”

=

{z(x)I

1 Rn

where and

p > O , p > 1 , p < m

/ z ( x ) I p exp - p p ( l + l ~ ~ dx ~ )< ~m /} ~

78

CHAPTER I1

We will use also the following embedding theorems (cf. Ladyzhenskaya Solonnikov- Ural’stseva If

z E

L ~ ’ l ’ p, p >

If

z E

h2””

Let C?

+ I

p > n + 2

, then z

We define

1

E

Co(q)

q

, then z , a z

be an open subset of

(3.9)

( 3 . 10)

,

2

111).

aij , ai ,ao

Rn

and m

E

L (Q)

E

C

o (Q) -

.

Q = 8x (0,T).

We assume

79

PARTIAL DIFFERENTIAL EQUATIONS

(3.11)

a(t;u,v)

au av Jsaij(x,t) ax. - dx ax. 1 1

Z,

=

+ C J

i

Theorem 3 .

+

~

i,J

8

* ax

a.(x,t) 1

v dx + J

8

a (x,t)u v dx

0

I. Assume 13.91 and

ii

(3.13)

.

L20)

E

Then t h e r e erists one and onZy one s o l a t i o n du - <-,v dt

(3.14)

> + a(t;u(t),v)

= <

u

E

W(0,T)

of

f(t),v > a.e. t

in

(0,T) 1

V V E V = H 0 u(T) =

Remark 3 . 1 .

u

Formulation (3.14) is the evolution analogue of a(u,v)

(f,v)

=

, u

E

Here we will not need to split the form

1

Ho

,

V v

a(t;u,v)

E

1 H0 '

in two parts as we

have done in the elliptic case (cf. Lemma 2.1 and Theorem 2.1). we will

not need to assume the symmetry of

gemark 3.2.

a. ij

Problem (3.14) is backward in time.

Also

*

We have considered a

backward evolution problem instead of a forward evolution problem for convenience, since we will consider them more often. the change

t = T - s

,

s

Note that making

[O,T], one can formulate (3.14) as a

E

forward evolution problem.

Remark 3.3. (3.15)

Since

I

A(t)

-;if u

E

d(V;V')

+ A(t)u

'd u(T)

E

=

=

6

W(0,T)

.

,

f(t)

we can write (3.14) as follows

.

80

CHAPTER I1

The differential equality has to be viewed as an equality between

.

2

elements of

L (0,T;V')

3ernark 3 . 4 .

We have not assumed a. 2 0.

In fact we can without loss of

generality assume

(3.15)

a o > y > O

, y arbitrary large

Indeed, make the change of unknown function y

exp - X(T-t)u

=

then problem (3.14) is equivalent to

hence we have the same problem, with a .

changed into

+

a .

I

'

a

changed into

f e-''(T-t)

* *

. 'Wn'

The space

V

being separable, there exists an

* * *

We look for an approximate solution as follows m

(3.16)

U,(t)

=

c

eim(t)wi

i= 1 where the scalar functions gim(t)

are solutions of the following

linear system of ordinary differential equations (3.17)

1

-

and

0

'5

We use Galerkin's method orthonormal basis w

f

du (2 ,w.) J

+ a(t;um,wj) = J

, j

=

I

,...,m

81

PARTIAL DIFFERENTIAL EQUATIONS

where

m

E

.

um u

Cwl,. .,w 1 and m

Multiplying (3.17) by

g

jm

as

m

+ m

.

and adding u p , we obtain

- Id td lUm(t) 1'

(3.18)

in L2(S)

-t

+ a(t;um(t),um(t))

=

.



However by (3.15) we can assume that

hence the estimate

which implies (3.20) Let

$

is bounded in Lm(O,T;H)

u

E

1

C (C0,TI)

such that

@(O)

=

0

.

Set

We can extract a subsequence, still denoted (3.21)

u

Multiply (3.17) by

and letting (3.22)

m

+

u

in L 2 (0,T;V) weakly

u

and integrate.

Q(t)

tend to

and

+a:

$.(t) = $(t)wj J

such that

m ,

We obtain

, yields

1' (u,g')dt j

L 2 (0,T;V)

(U,Oj(T))

+ /' a(t;u,@.)dt = 3 J

.

.

82

CHAPTER I1

is arbitrary it follows from ( 3 . 2 2 ) that

Since j

V V € V

4

E

&(O,T))

hence V v

E

V

Taking

du E dt

,

2 L (0,T;V')

we obtain in the sense of distributions

d (u(t),v)

,

hence

u

L2(0,T)

E

implies (see 5 3.1),

.

W(0,T)

E

, which

Using next integration by parts in W(0,T)

and ( 3 . 2 3 ) yields u ( T )

Hence existence is proved. For uniqueness take

f =

0

,

u

=

0

.

Take v = u ( t )

in ( 3 . 1 4 ) .

We obtain

.

hence

u = 0

3.3.1.

Regularity with respect t o time

Theorem 3 . 2 .

We make t h e assumptions of Theorem 3 . 1 , and

(3.24)

a,. = a 1J ji

(3.25)

a t aij

(3.26)

f E L'(Q)

a

E

L-(Q)

, z

H~I

.

=

G.

83

PARTIAL DIFFERENTIAL EQUATIONS

Then we have

(3.27)

We define (3.28)

ao(t;u,v)

=

aU av dx Z J8 a.1 J.(x,t) ax.1 ax. 1

and

hence a(t;u,v)

=

ao(t,u,v) + (A1u,v)

.

We consider the Galerkin approximation (3.17) which is rewritten as f01lows (3.29)

+ -(dt,w.) J

ao(t;um,wj)

=

(f

-

A1um,wj) ,

j=l,

and we may assume that (3.30)

i +U

Multiplying (3.29) by

in

-

HI:

g! (t) Jm

. and adding up in

j, we obtain

du (3.31)

I t 1

+ ao (t;u, ,-u’) m

= (f

- A u -ul) 1 m, m

.

But if we set

it follows from

(3.21) using the sdr’r%ri’d, a (t;u,v) = ao(t;v,u) 0

...,m

84

CHAPTER I1

hence

T 2 Jt 1- dt

1

d s + - a (t;u (t),u (t)) 2 0 m

+

Jtl

=

( f - A u -u’)ds 1 m, m

and from the assumptions we deduce

c

from which we easily deduce ( 3 . 2 7 ) .

3 . 3 . 2 . Regularity with respect t o space variables

Theorem -

3.3.

We make t h e asswrrptions o f Theorem 3 . 1 , and ( 3 . 2 4 ) , ( 3 . 2 6 )

and (3.32)

a.. 11

Then t h e solution u

E

w””(Q)

.

of (3.141 s a t i s f i e s

The idea is t o use Theorem 3 . 2 and the results on elliptic equations (cf. Theorem 2 . 1 ) . (3.34)

If we fix

Let us write ( 3 . 1 4 ) as

ao(t;u(t),v) t

= (f

+

du dt

A

1

U,V)

(outside a set of Lebesgue measure

( 3 . 3 4 ) as an elliptic equation, where

t

0), we can consider

is a parameter.

Since

85

PARTIAL DIFFERENTIAL EQUATIONS du 2 ;i-i - A1u E L ( Q ) can assert that f

and assumptions of Theorem 2.1 are satisfied, we

+

However there is a slight technical difficulty since this does not prove that t

-t

u(t)

is measurable from (0,T) into H2(8)

.

To overcome this difficulty, we use a time discretization procedure.

du -A u dt ln define a sequence u Set

g = f + -

(3.35)

L

E

2

(9) , We discretize O,k,...,Nk

=

T

by

a0(nk;un,v)

=

(gn,v)

ti v

E

HoI , un

E

1 Ho

where (3.36) Then un

E

2 H (8) and

(3.37) where

C

does not depend on n,k gk(t)

=

gn

for

. t

Define next

E

[(n-l)k,nk[

remains in a bounded subset of It follows from (3.32) that uk 2 2 L (0,T;H 6)) hence, by extracting a subsequence (3.38) Let v

uk

E

L2 (0,T;V)

-t

.

w

in L2(0,T;H2@))

weakly

We deduce from (3.35) that

.

and

86

CHAPTER I1

JT

(3.39)

0

a (t;uk(t),v(t))dt 0

=

JT

Cao(t;uk(t) ,v(t))

nt = integer part of

Therefore letting k

+

0

-

ao(ntk;uk(t),v(t)

(gk(t),v(t))dt

+

where

=

-.kt

But

in ( 3 . 3 9 ) we obtain

hence a.e. t

a0(t;w(t),v)

=

du (f(t) + d t - A1u,v)

.

This and ( 3 . 3 4 ) implies u(t)

=

w(t)

a.e.

which completes the proof of the desired result. R m a r k 3.5.

u

E

#*"(Q)

Under the assumptions of Theorem 3 . 3 we can write and

(3.40)

I

u(x,T)

=

i(x)

Idt +

87

PARTIAL DIFFERENTIAL EQUATIONS

3 . 3 . 3 . Other regularity results We now state without proofs (which are difficult) the analogue of the results stated in

§

2 . 1 for elliptic equations.

spaces of Hglder functions in x,t We denote for

0 < a < 1

,

We need to consider

.

by

+ sup X

1 z (x,t)-2

(x,t )

L

I t-t' 1 a12

t,t'

We then define

c1+a,a/2

and

and

- = (4)

{z(Z E C

0

-

aZ

(Q) , axi

Ca,a/2(Q) - 1

CHAPTER I1

88

with

Let us assume a

(3.41)

ij

(x,t)

Ca , q ? j )

E

,

0 < a<1

aij = aji

ticj

za.

2

R 151

V E E R "

ij

f

Then i f

E

,

C"yc1'2(Q)

z

'a

(3.42)

1

there exists one and only one solution of a. .(x,t) 1J

axiax

f

, ulZ

=

0

, u(x,T)

=

j

2+a,l+a/2 U

€

(Q)

C

Now consider instead of 3.41 (3.43)

Let

f

E

LP(Q)

,

p > 1

,

p <

m,

then there exists one and only one

solution of aU

(3.44)

3t u

E

z

a. .(x,t)

' 1J i,J

axiax

IJ*'''~(Q) n ll~b"'~(Q)

For p r o o f see Ladyzhenskaya - Solonnikov

f

, ulZ

=

0

j

-

. Ural'tseva [ l ] .

, u(x,T)

=

0

0

.

89

PARTIAL DIFFERENTIAL EQUATIONS

From this and Theorem 3 . 3 , we deduce

Theorem 3 . 4 .

We assume 13.91 and

(3.45)

Then t h e r e e x i s t s one and o n l y one s o l u t i o n of

*

-

(3.47)

at

u

E

a; ---

i

,

u, and using ( 3 . 4 4 )

u

=

f

,

u(x,T) = 0

b2””(Q)

Consider the solution of ( 3 . 1 4 ) (3.48)

uIC = 0

+ A(t)u + aou

at

a2G = C a. .(x,t) axiax j

1J

i,j

f

-

-

solution of

a u + C O

Operating as in the proof of Theorem 2 . 3 , we check that

i

au b.(x,t) l ax

-

u = u

We first deduce a priori estimates assuming that the solution u ( 3 . 4 9 ) belongs to (3.49)

b2”’p(Q).

of

We show a priori that

.

L~(o,T;L~(c+))

u

We multiply ( 3 . 4 7 ) by

lulp-2u and integrate over x , using Green’s

formula. We obtain

+ C

i where

lSai

au u

i

lulp-2 dx + /dao /uIpdx=

is

f u lu/p-2dx

90

Since . a

CHAPTER I1

t y

arbitrarily large, it follows that

hence integrating between

t

and

T , we get

and from Gronwall’s inequality it follows (3.50)

Now from the regularity result ( 3 . 4 4 ) we have (3.51)

and from the inclusion result ( 3 . 7 ) (used

q = p)

which with ( 3 . 5 1 ) yields

hence the result. The above argument gives a priori estimates assuming that the solution is b 2 y 1 ’ p . To prove that argument as follows.

11 E

1112 y ” one can

use

a fixed point

91

PARTIAL DIFFERENTIAL EQUATIUNS Consider for

z

E

This defines a map z1,z2 and

c1 =

,<

b2””

5

S(z) =

, c2

S(zl)

to be the solution of

from b2’ y p into itself. Consider = S(z2).

A computation similar to that o f

( 3 . 5 0 ) but more accurate shows that

for

A

2

ho

,

C

does not depend on

X

From the regularity result ( 3 . 4 4 ) one has a l s o

hence

if

A

is large enough.

Therefore at least for X

large enough

has one and o n l y one fixed point.

S

is a contraction, hence it

92

CHAPTER I1 Since we

can modify

. a

into

h

ao+X with

arbitrarily large,

0

the desired result follows.

Corollary 3 . 1 .

Under t h e assumptions of Theorem 3 . 4 and

t h e n t h e r e e x i s t s one and only one s o l u t i o n of

;:

I

(3.53)

+ A(t)u

--

uIc = 0

-

Consider v = u-u

,

-

a

into

f

A(t)E

-

,

+ a0u u(x,T)

=

f

=

;(x)

, u

E

b2””(Q)

it is solution of problem ( 3 . 4 7 ) with

u 0

,

which belongs to

Lp

,

f

changed

by virtue o f (3.52),

17

hence the result

Theorem 3 . 5 . cons t ant s (3.54)

We make t h e assumption of CoroZlary 3. I.

6,,6, such t h a t

Consider

93

PARTIAL DIFFERENTIAL EQUATIONS

prooj We set

then

-

-

au at

-

-+

Hence u

E

5

-

-

+ aou = f

I lf'Ll1

exp-(a2-BI)(T-t)

+

0 .

5

Furthermore uIc

+ A;

0

W(0,T).

;(x,T)

S

0

.

It follows that

-d dt

-+ -+ + a(u ,u

1;+l2

) 5

o

hence

lu+(t)

therefore ?(t)

=

1,

+

1;

-+ -+

a(u ,u )ds

0 , hence (3.55).

5

0

A similar argument leads to (3.56).

3

94

CHAPTER I1

Corollary 3 . 2 .

If f , G

2

then

0

u 2 0

.

x

R"

proos Immediate from (3.56).

The Cauchy problem is the following

-

(3.57)

at

+

AU

=

f

a.e.

6

u(x,T) = U(x) There are of course no boundary conditions. We will assume

We first solve (3.57) in the S o b o l e v spaces with weights

Theorem 3 . 6 . Me assume ( 3 . 5 1 , 1 3 . 2 4 1 , ( 3 . 2 5 1 and 12.581, 13.591. Then there exists one and only one solution oS 13.57) sueh that L~(O,T;V") ,

(3.60)

du

2 L (o,T;H~)

Identical to those of Theorem 3.1 and 3.2. Let us now prove local regularity results with respect to the space variable.

Let

@

E

B(Rn).

Set

z = @I

.

c

95

PARTIAL DIFFERENTIAL EQUATIONS

We have

and

z

*

-

(3.61)

at

+ Az

= $,f

-

2 a

-?axi !$+-ua A(t)$ .

ij axj

vanishes outside a bounded domain. We can state

Theorem 3 . 7 .

We assume 13.91, 1 3 . 4 5 / ,

13.581,

13.591,

E: :;W

.

Then t h e s o l u t i o n of 13.571 s a t i s f i e s b e s i d e s 1 3 . 6 0 1 .

We first prove that (3.63)

u

Lm(Rn

E

x

More precisely we have

(0,T)).

i

i /u+ii

u(x,t) s /jf+i (T-t) +

Indeed define

then

We multiply by

*at

+

(?)exp

A;

2

- 2p(I

dt ~u -+ - d_

,

0 t

( t ) j 2s

P

;(x,T)

5

0

.

IX~*)~'~,and integrate other x.

x

l;+(t)

l2

for some constant A

lJ

hence

-+

By Gronwall's inequality it follows u (t) Similarly we get (3.64).

=

0

,

hence (3.63).

We get

96

CHAPTER I1

Consider now (3.61).

We have

z(x,T)

@G

=

1 W2yp(OQ) n Wo’p(OQ)

E

where OQ is a regular bounded domain containing the support of and

z vanishes on

z

Since

@

E

em.Hence

@

using Corollary 3.1, we see that

QQ

#2’1(Q4)

=

06

(O,T)

x

is arbitrary, we can assert that

u

*

2 L (O,T;H:oc(Rn)).

E

Using the inclusion result (3.7) it follows that $ , 0 9 4 (8x (0,T)) for any bounded domain 8 the right hand side belongs to L4 ( Q $ ) . Hence

implies u

E

LPA4(0,T;W:~~n4(Rn)).

result again we see that exponant by

u E

If

,

p > 4

. z E

p.

2’1yph4(QI).

then using the inclusion

~ 1 ’ 0 ’ 8 x( ~(0,T)).

2 , provided it is less than

Going back to (3.61), L! This We can multiply the

After a finite number

13

of steps we obtain (3.62).

We now state without proofs (see Ladyzhenskaya

-

Solonnikov - Ural’tseva

[ l l ) a result on the Cauchy problem, when the operator A

not in divergence form as follows (3.65)

with the assumptions

a..

=

bi

Bore1 bounded.

13

(3.67)

a

ji

We consider the Cauchy problem

is writen

91

PARTIAL DIFFERENTIAL EQUATIONS with data (3.69) Then we have

Theorem 3 . 8 .

hie assume ( 3 . 6 6 1 ,

13.671,

13-69).

Then t h e r e exists m e

and o n l y one soZution o f 1 3 . 6 8 ) such t h a t

The solution u

of (3.68), (3.70) satisfies the following estimate

(Maximum principle)

which follows from what we have seen in Theorem 3.7.

Indeed for more

regular coefficients (3.71) will hold, and by a limit argument, it holds also under the assumptions of Theorem 3.8. COMMENTS ON CHAPTER I 1

I. By smooth boundary, we mean C2 boundary.

,

2. W l Y p

with

p

3. For Poincar6

=

n

,

W1”

c

Lq

V q c m

.

inequality to hold it is sufficient that

f

vanishes on

some part of the boundary, with positive capacity. 4 . Since

1 Ho c L2

with continuous injection and is dense then we have by

duality

-1

is dense in H with continuous injection. I Now since Ho is a Hilbert space, there is an isomorphism from

and

1 Ho

L2

into

H-l

.

Let us denote by

3

this isomorphism, then for

98

CHAPTER 11

u

E

Ho1 and v

E

1 Ho

we have

<Ju,v>

=

((u,v))

= J

aU av dx --

u v d x + I J

axi

ax;

hence

J

The fact that

is an isomorphism means that for given

L

in H-I,

the Dirichlet problem

-

nu + u

= L

, UEH;)

has one and only one solution. 5 . In the statement of problem ( 2 . 4 ) , it is sufficient t o assume

existence and uniqueness of the solution for

C"(6) .

f

in a dense subset of

Indeed from the Schadder estimate

and therefore one can extend the map linear continuous map from

6. In the case when A 0 estimation (2.16)

=

C"

-A , Theorem

f

+

u

by continuity, as a

c'+'.

in

2.2

and of Theorem 2 . 1 .

is an immediate consequence of Indeed, one has the a priori

estimate

7. In the proof of Theorem 2 . 3 , the idea to use the improvement of n+l n regularity for the sequence u -u was introduced by P.L. Lions [ l ] .

99

PARTIAL DIFFERENTIAL EQUATIONS

8. Property ( 2 . 4 0 ) clearly generalizes property (2.6) that we have seen

in the proof of Theorem 2.1.

This property is known as the Maximum

Principle.

9. If there is more regularity on the coefficients, a . . the data 1J

’

f

and the boundary, one can derive additional regularity properties on the solution. 10.The spaces

ip

,

...

are Frechet spaces.

This Page Intentionally Left Blank

101

CHAPTER I 1 1

MARTINGALE PROBLEM

INTRODUCTION The martingale approach to diffusion processes is due to D. Stroock S . R . S . Varadhan C11. The objective is to define diffusion process with minimal assumptions on the drift and the diffusion term. We have already seen in Chapter I, with the concept of weak solution that a measurable bounded drift was sufficient, but the diffusion term was still Lipschitz. S . V . have solved the problem assuming only continuity of the diffusion term. We will consider the same problem assuming a H6lder condition on the diffusion term. This will allow us to use as much as possible results on P.D.E., in obtaining a priori estimates.

1. PROPERTIES OF CONTINUOUS MARTINGALES

(R,O,P,3t) .

We consider Let

p(t)

be continuous

ztmartingale.

We say that it is square integra-

ble if we have

(1.1)

SUP

E /v(t)12 <

t>O We write

p

E

is that for

M2 E p

E

t

M2(Z ; P ) .

Mg,

We will assume

p ( 0 ) = 0. An important result

then

(1.2)

u(t)

-f

p(m)

and

u(t)

=

E [p(~)15~1

The space M2

.

,

in L '

as t

+ m

.

can be provided with a structure of Hilbert space by setting

102

CHAPTER 111

A second very important result follows from the Doob theorem (see C. Dellacherie - P.A. Meyer [ l ] ) .

If

-

p

Meyer decomposition E

M2,

then we can

write in a unique way

where

v(t)

is a continunus martingale, and

d a p t e d i ncr eas i ng prccess.

One says that

process associated with the martingale

(t) < p , u > (t)

is a continuous is the increasing

p(t).

It can be proved that

in the sense of convergence in

L'

One easily checks the following property

P[

(1.6)

Sup ()l(t)) > tc C O , TI

E l 5

N - + P[(T) 2

t

N1

I

The concept of square integrable martingale being too restrictive, one introduces the concept of Zocali,u square i n t e g r a b l e m a r t i n g a k . that

u

We say

is a locally square integrable martingale, if there exists an

increasing sequence of stopping times

- a.s.

We denote by

kM2

T~

+

+

T~

such that

m

the space of locally square integrable martingales.

The decomposition property ( . 4 ) extends to locally square integrable martingales.

In other words if

p

E

RM2,

there exists one and only one

KARTINGALE PROBLEM

increasing continuous adapted process

2 (tAT) -

!J

ti T

such that

103

such that

< p , u > (t)

(tAT)

p(tA.r)

is a 5 tAT martingale,

E

M2

.

Example Consider the stochastic integral

v(t)

(1.8)

w

where

is

zt

=

it @(s).dw(s)

n dimensional standard Wiener process.

Assume that

E p

Then

E

RM2.

iT ($(t)(’dt

Indeed set

un(t)

=

V T

c

T~ =

.

n and consider

p(tAn)

then

E 1pn(t)

I*

=

E

itAn

I$(s)l

2

ds

9

E

in /@(s)l2ds

.

We define

(1.9)

9 , U >

(t) =

it/$(s)I2ds .

By Ito’s formula

2

lJ (t)

=

2

it l l ( s ) @ ( s ) . d w ( s ) + It

/ @ ( s ) 2ds

hence (1.9) defines the increasing process associated w th the martingale P(t)

*

104

CHAPTER 111

Remark 1.1.

One can justify the notation

(t)

as follows.

Let

1 ~ ~ ~ E1 -M2, 1 ~ then

1

PlU2 = 7 (lJ1+LI2)

(1.10)

=

2

< p , p > (t) 1

-

(t)

2

+

1 2

Pl(t)

- 71

2 U,(t)

martingale

where

Note that

is a difference of two increasing processes.

(t)

property and decomposition (1.10) uniquely defines

(t)

This

.

We also note that

p1,p2 -t (t)

is a bilinear form on

M2

'

and

are orthogonai if pl(t)p 2 (t) is a martingale, This implies that they are also orthonormal 0 in the sense of the Hilbert space M2'

We say that i.e., if

Let now Assume

where

pl,p2

E

(t)

M2 =

.

p ( t ) = (pl(t), . . . , p

pk(t)

E

RM2 , and

(t))

be a vector continuous

Zt martingale.

10s

MARTINGALE PROBLEM

(1.13)

ajk(s)

is an adapted process and

/a.,(s) J

1

S C

.

We can define easily stochastic integrals with respect to integrands

4 be a step function. We define

Let

Then

and as for the case

a

=

I , we can extend

I(@)

to elements of

LL

3

and denote it

Let

b(s)

be a matrix

processes and vector (1.16)

.EM2

JT

5 2 (Rn;Rm) , whose components are adapted <

tr b(s)b*(s)ds

rna:tingale.

<8.

Let

Ct b(s)dp(s) =

( I . 17)

a(s)

(1.18)

b(t)

6

Jt

E

Jt

b(s)dp(s)

Rm , then

,

8.Jt b(s)dp(s)>

=

b(s)a( s)b* ( s )6.€Ids

is i n v e r t i b l e

=

a.s., then

m

It a"2(s)dw(s)

-1

a

(s)

kmciea

V t , a.s

.

is a

106

CHAPTER I11

props Define (1.19) which is an

w(t) =

It a- 1 1 2

n dimensional

(s)dp(s)

martingale. Let

RM2

E Cl'd.(w(t2)-w(tI))l

2 15t l

@

E

Rn

,

then

I= t

=

E Liltt2 a-112(s)€I.dp(s) 121s 'I= 1

= (t2-tl)

181

2

.

From Levy's theorem i t follows that let (assuming all dimensions

1

w(t)

i s a Wiener process.

to simplify)

then

L(T)

=

lim bk(t)

T

1

bk(s)dw(s) =

bn

in

in L2

,

[nk,(n+l)k)

N- 1 =

lim

1 bn (w((n+l)k)-w(nk)) n= 1

where

in L2

Moreover

107

MARTINGALE PROBLEM

Since

bk

and

a

-112

+

all2

in

is bounded, it follows that

i;(~) = U(T)

-

Since

L3(0,T) 2

p(t)

,?ernark 1.2.

,

p(t)

a.s.

are continuous processes we have

3

1.18).

When assumption (1.17) is not satisfied, then there exists a *t

n dimensional Wiener process with respect to a bigger family 5

(i.e.

Zt c 5*t ti

t)

and a matrix process

$(t)

which is adapted t o

5*t , such that 3

2. DEFINITION OF THE MARTINGALE PROBLEM

Let

R0

=

C(IO,m);Rn)

which is provided with the topology of uniform

convergence on compact sets. w E R0

Then

is thus a function o E u(t)

slo

is a metric space. An element

.

The canonical process is defined by

Let

Ocsst , we define

proved that

Let

x:

=

clx(u),s
,

and ?7j

T~ coincides with the Borel o-algebra

aij(x,t) , gi(x,t)

be functions on

a . . = aji

(2.2)

I

aij,bi

, a

Rn

x

.

=T: on

It can b e

"0 .

9

[0,-) such that

E a , . is a non negative matrix

1J

Borel and bounded

.

108

CHAPTER I11

W e write

A(t)

(2.3)

=

-

"f

C

i,j A probability

PXt on Ro ,

d-

a ij

vt A(t)

(2.4)

= 1

is a

m:

is a

pXt,

is a

pXt, 78:

pxt

, p:

E bi a

~* .

if one has

martingale

.

martingale,

wartingaLe, if $

is uniformly p s s i t i v e .

To obtain (2.6) it is clear that we may assume $

then by definition

pxt Also for any

d

is s o l u t i o n of t h e rnnrtingale problem

w i t h r e s p e c t t o t h e operator PXt (x(t)=x)

axiax j

martingale.

A E [t,m)

E

J(Rn

x(t,m)).

But

109

MARTINGALE PROBLEM

Therefore also

We integrate (2.10) with respect to

A

between

t

and

s

, we

get

Property ( 2 . 7 ) follows from the following fact, whose proof is left to the reader (cf, Stroock

-

Varadhan [ 11, or Bensoussan - Lions i 11).

v ( s ) , ~ ( s )scalar processes which are adapted to a family bounded.

Ss

Let

~ ( s ) ,

and

Assume that P(s) +

1; v(X)dA

is a Q s

martingale

.

Then

is also a Q s ~ ( s )exp

JtS

martingale.

r;(A)dA

is a

We apply this result with :(s)

=

9 and P(S)

In particular if

Ss

u

=

u<,

then

martingale.

~ ( s )=

we obtain ( 2 . 7 ) .

$(x(s),s)

, v(s)

=

a+ + A(s)$(x(s)), - as 0

110

CHAPTER 111

L e m a 2.2.

8

Let

(2.1 I )

Rn, then

E

- Itsb(x(A)

expCi8. (x(s)-x

,A)dX) + :I.

8.a(x().) ,A)Bdi.]

rnartizgale.

is a pxt,7.i:

?rcop We apply ( 2 . 7 ) with

$(z) =

exp i€I.(z-x)

expi8. (x(s)-x)

exp[

JS

t

and get

- iB.b(x(A),X)dX

+

+ Is a(x(X),X)S.SdAl t is a

pxt

martingale

u

Hence ( 2 . 1 1 ) .

2 J: a(x(X),X)4.9dX

Replace

B

BB

8 by

and let

B

=

with respect to

0

in ( 2 . 1 1 ) , 3

.

I

scalar.

Differentiate with respect to

We get the first part of ( 2 . 1 2 ) . and let

8

=

0

Differentiate twice

, we obtain the increasing process. J

(1) (2.6) and (2.9) extend to complex valued functions provided 13; 2 B > O .

111

MARTINGALE PROBLEM

then t h e r e e x i s t s a (with w(t)

=

n dimensional Wiener process

w(s)

0) such t h a t

,

tr s

a.s.

This is an immediate consequence of Lemma 2 . 3 and the representation

0

Theorem 1 . 1 .

Remark 2.1.

Matrix

u

can be computed by the following formula

(cf. A . Kato [ I ] ) (2.15)

U(x,t)

This shows that

u

Remark 2 . 2 . then Pxt

=

1

- J X n o

-1/2

(2a(x,t)+h)-l

has the same regularity as

a

a(x,t)dh

.

0

Under the assumptions of Lemma 2 . 4 , given

,

x(s),

w(s)

(QO ? ,,m is a weak solution of the S.D.E. with initial

This is clear from the definition of the concept of

conditions (x,t).

weak solution (cf. (5.18), (5.19), (5.20), (5.21) Chapter I).

The

inverse is also true, since by Ito's formula

$(x(s))

=

$(x) +

it"

+

it" A(A)$(x(X))dX

2

+

(x(X)) .o(x(X) ,A)dw(A)

Therefore if we have a weak solution of the S.D.E. on we have a solution of the martingale problem. Let u s define

Fxt

by

(Ro, pt,p~:)

then

0

112

CHAPTER 111

then we have Lerona 2.5.

(2.17) then

-w(s)

(Qo,A , , F t , 711;)

Consider

is a

w(s)

+ :/

and t h e p r o c e s s

u-’ b(x(X),X)dX

G(s)

=

nl:

iiiener p r o c e s s and

It is an immediate consequence of Theorem 5.1 and Corollary 5.1.

-1

(x,t)

bowaced

(2.19)

a

(2.20)

a , . E C a’a’2(Rn

x

[O,T])

,

0 <

CL

< 1

1J

Let 0 be a b o m d e d srcojth domaiv. of

Rn.

process, we define

the first exit time of

x(s)

from 8

.

If x(s)

is the canonical

5

113

MARTINGALE PROBLEM Theorem 2 . 1 .

Ye assume (2.2j, ( 2 . 1 3 1 , (2.20).

Let

Pxt be a soZz*tion

of t h e m a r t i n g a l e prGbZem and L e t

Then we have zhe estimate TAT t /EXt Jt f(x(s),s)dsi

(2.23)

5

C If1 LP

Ghere

C

does n o t depend on

f.

f

Clearly it is enough to prove ( 2 . 2 3 ) with

smooth.

Assume

f

smooth.

We consider the parabolic problem (2.24)

uIx

=

0

, u(x,T)

=

0

.

According to 5 3.3.3 Chapter I, by virtue of (2.20), we can assert that there exists one and o n l y one solution of ( 2 . 2 4 ) such that

2+z,l+u/2 U € C (Q)

(2.25)

. C2’’(Rn

Consider next a function v which is v = u on Q. Apply Ito’s formula to

Take hence

s =

TAT^

v(x(X),h)

, and note that for :u(x(h),h)

t

v

5

x

[O,T])

and such that

and the process ( 2 . 1 8 ) .

A 5

TAT^,

then

(x(X),X)

We get

E

6,

as well its derivatives, hence using ( 2 . 2 4 )

114

CHAPTER I11

Therefore (2.26)

u(x,t)

rct

E

=

.

itTA-rtf(x(s),s)ds

Now consider problem (2.24) with

f

E

Lr(Q)

,

Then from ( 3 . 4 4 ) Chapter I1 we can assert that

r

2

2

u

E

,

r <

a.

b2y1'r(Q)

.

Now f o r

r >$+ 1 , (2.27)

2,1,r(Q)

c

C0 (Q)

with continuous injection.

This implies that

But when

for

f

f

is smooth, (2.26) applies, hence

smooth. Relation (2.29) can then be extended to all functions of

Lr.

NOW

we can express pXt as

(2.30)

Indeed let

5,

be 5'

measurable and bounded.

To prove (2.30) we have

to prove the formula (2.31)

Ext

5,

=

-x t E

- - 71 fts

5 , expC Its u- 1 b.dw

By (2.16) the right hand side of (2.31) is equal to

10.

- 1 2 bl dX1

.

115

MARTINGALE PROBLEM

(2.32)

Ext SsCexp :J expt-

CS-]

b.d;

-

1 0 -1 b /2d h l

:J

its o- 1 b.dw - 71 Jts

-1

lo

b 'dA1

.

But from (2.17) we have (2.33)

Jts

-1

0

-

b.dw

=

Jts

o

-1

b.dw + Js t

10

2 bl d

-1

w

where the left integral is the stochastic integral with respect to w when

Qo

,m

is provided with

stochastic integral when

F t , and the first right integral is the

Q,,m

is provided with

by approximation and use the fact that

PXt

continuous with respect to each other. expression (2.32) coincides with Next we have, assuming

f

Ext

lEXt

it

XT

exp

PXt

Fxt

.

Indeed proceed

are absolutely

Using (2.33) in (2.32) we see that

5 , , hence (2.30)

.

smooth and bounded

TAT (2.34)

and

f(x(s),s)ds/

=

TAT t

iExt (Jt

f(x(s),s)ds)XTl

where =

tJtT

o-lb.dw -

3I itT

10 -

1 2 bl dsl

hence IExt But

TAT it t f(x(s),s)ds/

TATt

s

[Ext (it

f ds)211/2

(E

X;)"*

116

CHAPTER I11

Next TAT

[Ft( 4

TAT

t f ds)2]1/2

< [T

Pt4

f2 d S P 2

.

Gathering results we see that

and from (2.29) 5

c(/f2 I

1 1 f 2 since p > n+2 PI2

1

This completes the proof of (2.23).

Corollary 2 . 1 . such t h a t

Then if

Q

Let

0 be continuous and bounded on

LP(O,T;W~~~(Rn)),

E

Rn

x

[O,T1

,

, w i t h p > n+2.

LP(O,T;LToc(Rn))

Pxt is a s o l u t i o n o f t h e martingale probleir:, ue kave Ext @(x(TAT~),TAT~) = Q(x,t) +

(2.35)

where

E

Li

T~

is def ine d b y (2.21).

proos We can find a sequence Qn

an Since p > n+2 , Q~

E

C"(6

C0,TI) , such that

x

(Q) " ~

+

Q

in

1

-t

Q

in

c0(Q).

~

1

~

3

Q = 8 X [O,Tj

.

We can write Ito's formula for Q n and the process defined by (2.14)

But

117

MARTINGALE PROBLEM

0

Using Theorem 2.1, we obtain (2.35).

3. EXISTENCE A N D UNIQUENESS OF THE SOLUTION OF THE MARTINGALE PROBLEM

We assume here (3.1) a. ij

c

=

aij

b.

a,. 31

tjtj

2

B /t12

v 5

6

R” , B > o

Bore1 bounded.

Our objective is to prove the following

Theorem 3.1.

Under the assumptions ( 3 . 1 1 , ( 3 . 2 ) the s o k t i c n

the

0

martingale p r ~ b Z e mis unique. We will use several lemmas.

Lemma 3 . 1 .

It is sufficient to prcve uniqueness uhen bi

=

0

Considering the change of probability defined by (2.16), then solution of the martingale problem with unambiguously for

Fxt

0. Since

Fxt

is

is defined -xt by formula ( 2 . 3 0 ) , then the uniqueness of P

implies the uniqueness of We assume that

b

.

PXt

.

=

Pxt

0

118

CHAPTER 111

A(t)

(3.3)

=

-

C

a .(x,t)

iJ

i,j Let

4

E

,

C2+a(Rn)

a2 axiax j

consider the Cauchy problem

which has one and only one solution by Theorem 3.8, Chapter 11.

Lema 3 . 2 .

Let

pxt be a s o l u t i o n o f t h e martingale problem correspcnding

t o t h e operator

A(t)

given by ( 3 . 3 : , then we have

props Since

Property (3.5) is an immediate consequence of Ito's formula applied to the function u

and to the process

regularity of

Lemma 3 . 3 . (3.6)

x(s), which is possible by virtue of the

0

u.

~ e ~t O

YB

Borel subset of

P(x,t;T,B)

=

PXt(x(T)

R"

E

B)

,

t 5 T

then t h e f u n c t i o n P(x,t;T,B) does not depend on t h e p a r t i c u l a r s c l u t i o n of t h e martingale probZem.

119

MARTINGALE PROBLEM

Proof Since on a metric space provided with its Bore1 o-algebra, a probability is uniquely defined by the values of

(see for instance J. Neveu [ I ] , V @

for

p. 60).

,

Co(Rn)

E

E@

@

continuous bounded

It is enough to check that

Ext @(x(T))

does not depend

on the particular solution Pxt. From Lemma 3 . 2 , the result is proved for @

E

C2"(Rn).

Since V I$ ll@kll 0 C

E

Co(Rn)

' ll@l

I

0

,3

a family

@k of smooth functions such that

and

C

Gk

+

I$ uniformly on compact sets,

we clearly have

If now

Ft is

another solution of the martingale problem, from the

equality

and from ( 3 . 7 ) it follows that EXt @(x(T))

=

Ext @ ( x ( T ) )

hence the desired result.

i e m a 3.4.

The function

i.e. it satisfies

P(x,t;T,B)

is

il

Markov transition function

CHAPTER 111

120

x

+

P(x,t;T,B)

is BoreZ

P(x,t;T,B)

7:s

a probability on Rn

(3.9)

B

(3.10)

P(r,T;T,B)

=

xB(x)

(3.11)

P(x,t;T,B)

=

J P(x,t;s,dy) P(y,s;T,B) R"

,

t S s 5 T

(3.9) and (3.10) are obvious from the representation formula (3.6). if

I$ E Co(Rn)

,

x

-Y

E

Xt

@(x(T))

is Bore1 for

approximation result (3.7) and since c2+a,1+a/2 function. Therefore (3.8) is verified if

B

Ext I$~(x(T))

=

uk(x,t)

is a

is a bounded ball, hence also for

compact and from Neveu C11, p. 61, for

B

Now

t < T , since we have the

B

borel.

It remains to prove (3.11), or which is sufficient

0

for any

E

.

O n C (R )

A s for Lemma 3.3, it is sufficient to check (3.12) for function which are @

E

c2+a( R ) .

Bu then considering the function u(x,t)

(3.4) and the function v(x,t,s)

i:

t

5

,

s

solution of

defined by

2

--

c '

i,J

a. .(x,t) a v - 0 ax. ax. 1J

t 5 S

1 - 3

v(x,s;s)

=

u(x,s)

v(x,t;s)

=

u(x,t)

then we have

which is obvious by the uniqueness of the solution of ( 3 . 4 ) .

121

MARTINGALE PROBLEM

Lema 3.5. Is Pxt is avy solution of t h e martingale probLem, t h e n

(3.13)

Pxt [x(T)

E

Blms]

=

P(x(s),s;T,B)

a.s.

V t S s < T

It is sufficient to verify that for

I$

O n C (R )

E

and

cs,n

measurable

and bounded then we have

It is also enough to prove (3.14) with

@

C2'a.

E

In that case (3.14)

reads

But from Ito's formula and (3.4)

0

which easily implies (3.15).

Proof of Theorem 3 . 1 From formula (3.13), recalling that the functions P(x,t;T,B)

does not

depend on the particular solution of the martingale problem, we deduce easily that (3.16)

PxL [x(t,)

E

B ] , . . . , x(tn)

E

BnI =

Fxt [x(tl)EB1 ,..., x(t,)EBn!

V t l,...,t

2

t

,

,...,B

B1

Bore1 sets and where

PXt, pxt are two

solutions.n From Theorem 1n2, Chapter I, it follows that

Pxt

=

pxt . 0

122

CHAPTER 111

Remark 3 . 1 .

We have proved in addition that

x(s)

,m: , PXt

is a

0

Markov process.

We assume here that (3.17)

a

measurable bounded

(3.18)

bi measurable bounded.

ij

,

-1

aij = aji ; a

bounded

We recall that A(t)

Theorem 3 . 2 .

=

-

C

i,j

a. .(x,t) 'J

Under asswnptions 13.171,

a2 axiaxj

Ibi

a

13.181 there e x i s t s a s o l u t i o n of

t h e martingale problem.

Proof Assume first that (3.19)

I

a=-Do

*

2

with

Then according to Theorem 5.2, Chapter I, there exists a weak solution of the S.D.E.

at

Hence on some adequate 0,O ,5

n dimensional Wiener, and

y(t)

t

,

such that

there exists

, w(t)

123

MARTINGALE PROBLEM

F : R

Consider the map w

(3.22)

Ro

-+

y(.;w)

-f

Pxt the measure on 711

and

image of

t

?

F

by

.

Then PXt is a

solution of the martingale problem. Indeed let

...,xk

xl,

@

E

,

J%R")

5,

and

to be a measurable function of

; write

5,

(3.23)

=

t

Ss(Y(S1), ...,y (s,))

s

... s

s1 5

Sk

5 s

.

From Ito's formula we have

hence taking the image measure

and since

5,

generate 711;

,

it follows that

Pxt

is a solution of the

martingale problem. Assume now that we have (3.17), then consider a sequence an (3.26) and

sup lan(x,t)-a(x,t)l X,t

is Lipschitz in

a

(3.17).

x

+

o

as

n

-c

such that

-

uniformly with respect to

t, and satisfies

Such a sequence exists. There exists a solution Pn of the

martingale problem, corresponding to the pair

Therefore

(O,an) and

124

CHAPTER 111

This implies that the sequence Pn

remains in a relatively compact provided with the Ro,mt Let us extract a subsequence, still

subset of the set of probability measures on weak topology (see Parthasaratly [ I ] ) . denoted by

Pn

such that

(3.28)

Pn

-+

P weakly

...,

Consider a continuous function s(xl, xk) which is bounded; writing t 5 s1 5 s sk 5 s then we have, for 5 , = 6(x(sl), x(sk))

...

...,

0

E

B(Rn)

Using (3.26) and (3.28), we can let n

tend to

+m

in (3.29), and

obtain that

P

pair (0,a).

Use Girsanov transformation to obtain a solution of the

is a solution of the martingale problem relative to the

martingale problem relative to the pair

Remark 3.2.

7

(b,a).

We can get rid of the assumption

a

-1

exists and is bounded,

to obtain an existence result, provided that we replace (3.18) by the strongest assumption (3.30) Indeed replace a

SUP

1 x-y 1 s6 by

/bi(x,t)-b.(y,t:l/

p(6)

+

0 as

6

+

0

.

a+EI, then according to Theorem 3.2, there exists

a solution PE relative to the pair

and again

5

b,a+EI.

We have

MARTINGALE PROBLEM

Therefore PE

12s

remains in a relatively weakly compact subset of the set

of measures. Now we have, analogously to (3.29)

By virtue of (3.30) we can let

E +

0 in (3.31) and obtain the desired

0

result.

4. INTERPRETATION OF THE SOLUTION OF P.D.E.

Consider here

aij(x)

a

such that

i Borel bounded on Rn

,

Define (4.3)

Let also (4.4)

. a

Borel bounded

. a t 0 ,

According to Theorems 3.1 and 3.2, there exists one and only one solution of the martingale problem corresponding to the initial condition x

0. More precisely let Q o probability measure

Px

on

, m 0 , n to , there exists one and (Rono) such

that

only one

at

126

CHAPTER I11

(4.5)

PX[x(e)

= XI = 1

(4.6)

@(x(t))

-

@(x) +

Jt A$(x(s))ds &Rn)

.

is a smooth bounded domain of

Rn

,

(Px,?$)

Y @

E

is a martingale

Let

where 0

and consider the solution of

By Sobolev's inclusion theorems, we have (4.9)

u

E

C06)

.

Let

(4.10)

7 =

inf{x(t)

do}

tro the exit time of the process

Theorem 4 . 1 .

x(t)

from 8. We have

Asswne ( 4 . 1 1 , (4.2), 1 4 . 7 1 .

(4.11)

u(x) = EX

JT

f(x(t))(exp

Then we have

- Jt

ao(x(s))ds)dt

Proof 2

,

We can easily extend formula ( 2 . 3 5 ) to the function u(x)exp -

z,

Assume first p <

a.

f

and the process

E

B @ ) and f

2

0 , then u

E

W2yp(8)

for any

p

2

127

MARTINGALE PROBLEM

We obtain (4.12)

=

U(X)

+ EX We let T

+ m

u(x(TAT))

=

0

-

EX u(x(TA?))exp

CAT

ao(x(s))ds

-

ITATf(x(s))(exp

+

isao(x(X))dX)ds

For T > T we have since X(T)

in (4.12).

, hence since

E

a,

is bounded

u

-

EX U(X(TAT) exp

J

TAT ao(x(s))ds

-t

0

.

Formula (4.12) shows that, by the monotone convergence theorem, (4.11) holds. Moreover

Thus formula (4.11) extends to f

E

Lp(8)

,

p >

4.

Let us now give the interpretation of parabolic equations. Let us assume aaij

(4.13)

(x,t)

a

E

C"'"'2(Rn

x

[O,T])

ij

I

aij xai

(4.14)

a.

(4.15)

. a

E

,

-E "k

=

aji

Sitj

2

R

Y 5

1512

L~(R"

x (0,~))

L-(R"

x

(0,~))

.

E

Rn , 8 > 0

Loo

128

CHAPTER I11

Let (4.16)

f

E

Lp(@

(0,T))

x

p >

$+

1

From Theorem 3.4, Chapter 11, there exists one and only one solution of

1

(4.17)

aU - at +

,

0

1 1 1 ~=

+ aou = f

A(t)u

u(x,T)

;(x)

2

where (4.19)

A(t)

-

=

C

a ~

a

axi

a

a.. - + C a 1J ax. i 2,. J i

According to Theorems 3.1 and 3.2, there exists one and only one solution Pxt of the martingale problem relatively to the operator A(t) Let (4.20)

401

S~ = inf{x(s)

S>t and note that since p >

$+

1

,

u

E

Assume 1 4 . 1 3 ) , 1 4 . 1 4 1 ,

Theorem 4 . 2 .

C

0 -

(Q)

,

Q = 8 xl0,TC.

(4.151, (4.161. Then we have

TAT (4.21)

Assume first 2

5

get

p < =.

f

E

u(x,t)

=

B(Q)

, ii

Ext[/

E

f(x(s),s)(exp

t

bo)

.

Then

u

E

- :1 ao(x(X),A)dA)ds

+

b 2 $ ’ * P ( ~ ), for any

Apply formula (2.35) extended asexplained in Theorem 4 . 1 , we

129

MARTINGALE PROBLEM

u(x,t)

=

TAT a ds Ext u(x(TAT~),TAT~) exp - it 0

+

TAT

+ Ext it

f(x(s),s)(exp

- it" aod))ds

from which we easily deduce (4.21).

U = 0 , we first extend (4.21) to functions f We observe the estimate

Taking first p >

+ 1.

Then assume

U

W z Y p n WAY'

(hence in

u

E

W2"

1 n Wo"(0)

COG))

solution of (4.17) with data

and consider a sequence

U +U

,

En, and letting

n

+ 03,

the desired result.

Remark 4.1.

We see that we have the es imate /Ext Jt

f(x(s),s)ds

which improvesthe result of Theorem 2.1

Remark 4.2. We also have the estimate

Assume ( 4 . 1 ) , ( 4 . 2 ) and consider the forward Cauchy problem (5.1)

in

we obtain

0

TAT^ (4.22)

Lp(Q) ,

Un E B O ) . Applying ;L.21) with

and f

E

130

CHAPTER I11

with

4

(5.Ibis)

Borel bounded.

Since setting v(x,t) = u(x,T-t)

-

*

+ Av

at

=

0

then v

, v(x,T)

satisfies

=

@(x)

we can assert that there exists one and only one solution of (5.1)

such

that

Since v(x,t) and noting

(5.3)

PX'O

=

= EXIt $(x(T))

Px, we can assert that

u(x,t)

=

EX $(x(t))

Y t

2

0

.

We write

which defines a family O(t) bounded functions on that

O(t)

of operators on

B , space of Borel

Rn, provided with the sup norm; it is easy to check

satisfies the following properties

131

MARTINGALE PROBLEM

(5.5) O(0)

O(t)@ Hence

I

=

t

0 if

@

z 0

.

i s a semi group of contractions on

@(t)

B , which preserves

positivity.

C

Denote by

the space of uniformly continuous bounded functions on

B.

which is a subspace of in x,t

for

(5.6)

,

t > 0

if :

O(t)

c

Then from 5 . 2 , we know

B.

@

E

+

c ,

that

u

Rn,

is continuous

Let u s check that

if we assume

In that case, we can consider the S.D.E. (5.8)

dy

=

Y(0)

+ a(y)dw

b(y)dt

=

x

(R,Q,P,st ,w(t)) . Since b,a are Lipschitz, according to Theorem 4 . 1 of Chapter I, there is one and only one solution of (5.8). on a system

Denote the solution

yx(t)

.

Then clearly we have

and is an increasing function of

6. We have

132

CHAPTER 111

Therefore

IE O(Yx(t))-EO(y,~

(t))

~

5

E P( IY,(t)-YxI

(t)

5

p(6) +

C(t)

1) /x-x'i2

h2 for any

6 > 0. From this one easily deduces that

uniformly continuous in x, f o r

fixed, t t 0

t

.

@ ( t ) @ ( x ) is

Let us also check

that

(5.10)

in

O(t)b+b

C

as

t + O , ' d ~ c c .

Indeed

from which we obtain (5.10). The infinitesirnaL generator of (5.11)

~7 @

=

lim O(t)@-$ ti0 ~

O t)

is defined by

133

MARTINGALE PROBLEM

The domain of

B , such that (5.11) in the

is the set of functions

0

E

B.

E

C;"

sense of the convergence in EX @(x(t))

=

Assume

$(x)

@

, then from Ito's formula

.

- EX it A @(x(s))ds 0

But when (4.1), (5.7) are satisfied and

I$

E

c2+a b

then

A$

E

CEya

,

hence by the above reasoning we have

Therefore

Thus we have proved that

c ~ +D(c~J ~ ,

(5.13)

Let

b

and Q $

=

-

A@

8 be a smooth bounded domain of Rn. Let

where

T

Assuming

is the exit time from 8

q?

E

x(t)

E

B(6)

we define

.

S(6) , we may consider the non homogeneous Dirichlet

pr ob 1em (5.15)

of

@

1 2+Au=o

CHAPTER 111

134

Since u

-

=

v

satisfies

vIc

=

0

V(X,O)

=

0

then we see that there is one and o n l y one solution of (5.15) such that

Moreover we have

(5.17)

u(x,t) = EX @(x(tAT))

hence by definition of

If

@

E

H

1

@(t)

0 ) , we can give a meaning to (5.15) as follows

(5.19)

1

u - $

E

Hob)

Problem (5.19) has one and only one solution such that

(5.20)

u

Uniqueness i s easy. I

E

2

1

L (0,T;H )

du

, ;ir E L

2

(0,T;H

-1

To prove existence we can consider an approximation

.

+ $ in H (s) , @n E B ( ~ ) Then we can define the solution u (5.15) with @ changed into $n. It is easy to check that we have

@n

of

135

MARTINGALE PROBLEM

'd dt (un(t),v)

a(u (t),v)

t

=

V v

0

E

1 Ho

.

Hence

I2 -

/un(t)-um(t) +

Jt

(un(t)-um(t),$n-($m)

-

a (un(s)-um(s) ,un(s)-um(s))ds

- It a

+

.

(u,(s)-~~(s),m~-~~)ds = 0

Using the fact that

($n- @m + 0 in H 1 as 1 2 is a Cauchy sequence in L (0,T;H ) and

n,m

u

2 C(0,T;L )

If

$

n satisfies (5.19), (5.20).

E

H 1 (8)n Co(g) , then Qn

+

g?

+ m

,

we get that

.

The limit

in Co(g) ,

Since

we obtain (5.21)

u

E

.

CO(<)

Formula (5.18) extends to of (5.19),

If

I$ E

E

1

C O G ) n H (8), u

being the solution

(5.20), (5.21).

0 -

C (8)

un , where show that

g?

,

Qn u

we define u

+

($

0 -

as the limit in C (Q), of the Cauchy sequence

in C O G ) , and

is regular inside Q

$n

,

E

,96)

.

It is then possible to

and satisfies the differential

equation (5.15)

in the sense of distributions i.n Q .

considering us

where

6

E

&@)

This can be done by

and showing that it satisfies an

136

CHAPTER 111

homogeneous Dirichlet problem.

0

At any rate ( 5 . 1 7 ) extends to

E

0 C (8).

Since ( 5 . 1 7 ) expresses the

6,

equality of the values taken by measures on formula ( 5 . 1 7 ) extends to B ( 0 ) , provided u(x,t) is then defined by the integral of @ with respect to a measure. take test functions in

.

COG)

Clearly the first relation ( 5 . 5 ) holds when we

B(B) , and thus also with test functions in

This is enough to assert that it holds with test functions in

B(6) . The remainder of ( 5 . 5 ) is clear. Here

C

I$

,

E

C

coincides with the set of continuous functions on 8. When we have ( 5 . 2 1 ) , which implies that ( 5 . 6 ) still holds.

Note that

( 5 . 2 1 ) also implies ( 5 . 1 0 ) .

We can also check that generator of

O(t).

&O)

belongs to the domain of the infinitesimal

Moreover

Indeed by Ito's formula we have

hence (5.23)

137

MARTIKGALE PROBLEM

Since

@

is with compact support in 0

A$(x) where 6

=

{x

E

=

0 if

B/d(x,T) > 6)

x

we have

O6

.

If x E Q ~, then EX x,
=

PX(.r
Therefore

which completes the proof (5.22).

COMMENTS ON CHAPTER I11 1 . The representation Theorem 1 . 1 is due to Kunita - Watanabe

C11.

See also R. Priouret ill. 2 . The change of probability measures on (2.16) defines a probability on Ro §

since

Ro

, Pxt satisfies the conditions meantionned in Chapter I,

5.2 (see Comments Chapter I, 12).

3. In (2.20) the coefficients may of course be more regular. c 1 f co,l,l

Note that

138

CHAPTER I11

4 . -rt

defined in ( 2 . 2 1 ) is a

since

x(t)

mh

stopping time (cf Chapter I, Comments 2)

is a continuous process.

5 . The proof of uniqueness that we give in Theorem 3 . 1 , uses the Markov

property.

Stroock -Varadhan prove first that

Pxt is unique,

from which the Markov property is easily deduced.

6. To prove the Krylov

[Zl

Lp

estimate in Theorem 2.1, we have relied on P.D.E.

has proved

Lp

estimates for general Ito processes

(not necessarily diffusion processes, and therefore P.D.E. methods cannot be applied).

From this and the parabolic problem ( 3 . 4 4 ) Chapter

11, it is possible to prove the uniqueness of the martingale problem for

aij

E

0

C (Rn x [O,TI), which i s the case considered by S.V.

This idea has been indicated to us by P.L. Lions. 7 . Any Markov process generates a semi group by the formula

Conversely, if

@(t)

is a semi group on

B

satisfying ( 5 . 5 ) and

then

P(x,t,B)

=

@(t)xB(x)

defines a Markov transition function. We refer to E. Dynkin

[l],

[2]

for the theory of Markov processes.

139

CHAPTER I V

STOCHASTIC CONTROL WITH COMPLETE INFORMATION

INTRODUCTION Complete information means that we can observe the evolution of the state of the system as time evolves. Otherwise we say that we have partial information. In this chapter we develop the Dynamic Programming approach. We consider mostly the problem of controlled diffusions. But we study also control problems for general semi groups. In fact, the same methodology carries over to many other types of Markov processes, such as reflected diffusions, diffusions with jumps, and semi Markov processes The control of (see A . Bensoussan - J.L. Lions [ 2 1 , M. Robin [ I ] ) . general non Markov processes can not be treated in that way. One has then to use the "purely probabilistic" approach mentionned in the general introduction. We refer to the corresponding litterature. 1. S E T T I N G OF THE PROBLEM

Let 2 (1.1)

We write

be a subset of a metric space, called the space owz conti.s:s, and f(x,v) : Rn

x

'lf+ R

Bore1 bounded

140

CHAPTER IV

Let next

We note

Px

the solution of the martingale problem relative to

with initial condition x

C a - -a L ij axiax i,j j

A = -

(1.8)

(0,a)

0. Setting

at

f

a

bi

K

then PX(x(0)=x)

(1.9)

1

=

(1.10)

Then also there exists a ;71

t

standard n dimensional Wiener process w ( t )

such that

(1.11)

x(t)

=

x + J

t

o(x(s))dw(s)

+

It b(x(s))ds

,

0

Let

v(s)

be an adapted (to

m t)

controL process, or a control.

process with values in V .

a.s.

Px

.

We call it a

STOCHASTIC CONTROL

141

(1.13)

where we have set

(1.14)

For the probability (1.15)

g(x(s),v(s))

=

PV(S)

,

P,"

is a Wiener process and

wv(t)

x(t)

= x

bi t

a.s.

+

*

{t

(b+gv(s))ds

P;

.

+

Jt

u(x(s),s)dwv(s)

From I t o ' s formula it follows that (1.16)

Therefore P;

@(x)

,

P;

is a A s for ( 2 . 3 0 )

-

O(x(t))

formula ( 1 . 1 3 )

mt

it 2 (x(s)).gv(s)ds

Q

E

nt

t

for ?lo

.

J

t

A @(x(s))ds

can be inverted as follows

J(Rn)

We then set the functional

+

martingale.

is the unique probability on

(1.16) holds for any

(1) We write

-

.

no , 77!, ,

t

such that

142

(1.18)

CHAPTER IV

Jx(v(.))

=

- Jt c(x(s),v(s))ds)dt

EX JT f(x(t),v(t))(exp V Q

where (1.19)

T

=

first exit time of the process

bounded domain of

Rn

from 8,open

x(t)

.

We are interested in characterizing the function (1.20)

u(x)

=

Inf .J~("(.)) V(.)

and find optimal controls.

2. THE EQUATION OF DYNAMIC PROGRAMMING

When there is no control, i.e., then the function u(x)

f(x,v)

=

h(x)

,

a]

=

Au + a 0u

=

h

ulr = 0

We will show in this section that in the control case solution of a P.D.E. of elliptic type, but non linear.

Consider here

= -

where (1.6) and ( 1 . 7 )

0

,

g(x,v)

=

b(x),

defined by (1.20) is the unique solution of

2 d C a. ~C b . ij axiax.' i ax. i,J J hold and

u

E

u(x)

W2%) is also

.

143

STOCHASTIC CONTROL

(2.2) J

We next define a function H(x,X,p) : Rn (2.3)

H(x,A,p)

We call H

=

J

x

R

x

Rn

-f

Rn

as follows

infCf(x,v) + nal(x,v) + p.g(x,v)l

VE 7f

the HamiZtonian function.

Let us give some properties of

H.

is measurable in x,X,p (1)

(2.4)

H

(2.7)

hcLPb)

,

p 2 2 , p i m

Our first objective is to study the equation Au + a u

(2.8)

0

u

E

-

H(x,u,Du) = h

,

uIr = 0

W24(3)

which we call the Hamilton-Jacobi-Bellman equation (H.J.B.).

Our objective is to prove

Theorem 2 . 1 . (1.7).

We make t h e assumptions (1.1), ( 1 , 2 ) , ( i . 3 ) , (1.4), (1.6),

Then t h e r e e x i s t s one and o n l y one s o k i t i o n oc (2.8).

(1) With respect to the Lebesgue measure in

function.

x.

It is not a priori a Bore1

144

CHAPTER I V

We proceed in a way similar to Theorem 2.3 of Chapter 11.

Considering the

change of functions

where

w

i s defined in (2.17) of Chapter 11, we are led to the same

problem with the following changes

Note that

Hence we may assume without any loss of generality

(2.9)

a.

-

a l 1 y. > 0

Let next consider the problem

.

145

STOCHASTIC CONTROL

Auo

(2.10)

with

A

(a +x)uo - H(x,u0,Du 0) 0

+

,

h

=

u0

w29p n w1bP

large enough.

Since

c

/H(x,~,,s) - aOCO1 we can find

X

(1 +

IS,/

+

/El)

large enough so that

2

E(l
+

151

2

)

,

6 > 0 . Let us show uniqueness.

-

(2.12)

A(U-U)

Let

-

u,u

be two solutions of ( 2 . 1 0 ) ,

-

+ (a,+X)(u-u)

H(X,U,DU) + H(~,;,D;)

then

=

o

and

-

Multiplying (2.12) by

u-u

and integrating over

0, by Green’s

theorem we obtain Y

(2.13)

Y

a(u-u,u-u)

2

6

~

t

- -

~

0, hence

Lp(0) , hence there exists

5

E

- .

u=u

Existence is proved by a fixed point argument. E

t

- 2 p-u(

and the left hand side of (2.13) is

H(x,z,Dz)

-

~ 1 u - G ;- (H(. ,u,Du)-H(. ,u,Du),u-u)

W2”

Let ; I

W:”

z

E

W2”(Q)

, then

solution of

CHAPTER IV

146

and (A-ho)

where

C

1
does not depend on

A,

I Iz1

-2

2+

( A > Ao).

Therefore

hence if

X

is large enough

which proves that

S

is a contraction.

This completes the proof of the existence of

u0

The next step i s to define the following sequence

solution of (2.10).

STOCHASTIC CONTROL

147

Aun+’ + (a +X)u n+ 1 - H(x,u n+ 1 ,Dun+’) 0

(2.14)

=

h + Au“

n+ 1

Then by difference

+ H(x,un,Dun)

= h(Un-Un-’)

*

Again there is an improvement of regularity at each step. ul-uo E W2’p(o)

and

u2-u1

E

Indeed

and from

, follows from that D(u 2-u1 )

W2”O)

E

WlYp

hence also

Therefore

2,q1 This implies that

u2-u1

E

W

(@)

.

After a finite number of steps, we obtain in particular Therefore using Lemma 2.1 below we have

Hence at least for

n

2

+

u

no

, un-u

.

i s a Cauchy sequence in

(2.14) we also have

Hence un

in W2”

u n-un-1

weakly.

Lm.

E

Lm .

From

148

CHAPTER IV

Therefore also, since the injection of un

u

+-

W2”

into Wl”

is compact

in W’,P strongly.

This implies +

~ ( x , u , ~ u in )

L‘

strongly

We can pass to the limit in (2.14) which proves that

u

I

is a solution

of (2.8). Let us prove uniqueness.

-

u, u

Let

From Lemma 2 . 1 below it follows that

with

I$

E

Lm

, and A

large enough.

be two solutions, we have

-

u-u = 0

Then

.

u-U

G

E

Lw

and

proos Let

K

be the constant to the right hand side of (2.17).

(2.16) by

-

(u-u-K)

+

E

1

Ho

-

. -

We get

+

-

a ( u - u , ( u - u - ~ ) ) + ~ ( u - u (, ~ - L - K ) + ) =

or

We multiply

149

STOCHASTIC CONTROL

(2.18)

(u-U”-K)+) + A I (u-U”-K)+/

a((u-u-K)+, +

“ + dx I’ K(ao+A)(u-u-K)

=

/(H(x,u,Du)-H(x,;,D~)

+

=

+ @ )(u-u“-K)+dx

- -

+ g.Du + ual + g.D(u-u-K)

= /[inf(f

=

+

V

+ a 1 (u-u-K) + Kal + @ ) - inf(f + g.Du” + u”a,)l(u-u-K)+dx N

V

hence

=

I’

Cinf ( f + g.DC

La, + g.D(u-U”-K) + a (u-L-K) + 1

+

V

+ K(al-a,,-A)

+ @ ) - inf(f + g.Du + U”a,)](u-u-K)+dx

.

V

But recalling that K(a -a - A ) I

0

+

6

S

-

a .

a i r y , it follows from the choice of

0

, hence

S

J [inf(f + g.Du + ual + g.D(u-u-K)

N

“

+ a (u-G-K)

1

V

-

inf(f + g.Dl + ; a V

and for

A

Hence

-

u-u

5 C

J

(ID(u-;-K)

= C

J

(1

D(u-U-K)+l

1

1

)](u-K-K)+dx

K

.

+

-

5

I (u-u-K)+I ) (u-;-K)+dx

A similar reasoning shows that

completes the proof of (2.17).

that

+ Iu-G-KI) (u-L-K)+dx

large enough, this inequality implies 5

K

.-.+ = 0 (u-u-K)

-

u-u

2

.

- K , which

0

150

CHAPTER IV

3. SOLUTION OF THE STOCHASTIC CONTROL PROBLEM Our objective is to prove the

Theorem 2. I.

We make t h e a s s - q t i o n s'0 Theorem 2 . 1 rizd

(3.1)

f(x,v)

(3.3)

a-a 0

1

, g(x,v) , al(x,v) are Z . S . C .

in x,v

> y > O , h = O .

Moreover t h e r e es4sts s i z cptirnal c m t r s l . We start with a Lemma.

Let

P:

0

be the probability defined by (1.13).

Then we have

It is enough to prove (3.5) assuming Y

where

=

E X JrAT $ ( x ( s ) ) d s

4 =

E

.

B(Rn) ,

'42

EX

$(x(s))ds)

0

We have

XV(T)

15 1

STOCHASTIC CONTROL

XV(T)

=

exp CJT

0-l

- 71

gv(s)dw(s)

J

T

IU

-1

gv(s)

1

2

dsl

hence

We easily check that EX XV(T) 2

2

CT

independant of

v

.

Next $(x(s))d~)~]"~ [Ex (ITAT

using (4.22) of Chapter 11.

Hence (3.4)

2

C' T ( E x oJTAr $2(x(s))ds)1'2

2

c;\

IMLP since

p > n+2

.

, 3

l y o o f of Theorem 3 . 1 Since h 2

5

p .<

= m.

,

W2"(O) ,V p From Lemma 3.1, it follows that we can apply Ito's formula,

0

the solution of (2.8) belongs to

under integrated form, to

u

Hence

From equation (2.8) we deduce

and the process

x(t)

verifying (1.15).

152

CHAPTER IV

But H(x,u,Du)

(3.8) Taking

x

=

x(s)

,

v

u(x)

(3.9)

2

v(s)

=

we deduce from (3.7) and (3.8) that

- JTAT

EX u(x(TAr))exp

5

c(x s),v(s))ds

-

+ E~ /TAT f(x(s),v(s))(exp v o Since (3.3) is satisfied we can let (3.10)

u(x)

T

-t

+m

+

Jt c (x(X) ,v(X))dh)ds

in (3.8).

We obtain

.

Jx(v(.))

5

.

f(x,v) + u(x)a 1 (x,v) + Du(x).g(x,v)

Moreover by virtue of ( 3 . 1 ) , (3.2) the function

L(x,v,X,p)

=

f(x,v)

+

Xal(x,v)

is 1.s.c. in all variables and bounded below if Since

+

p.g(x,v) Ihl

2

,

M

R~

x

1 1 ~ 1s

M;

x {Ip( 5

MI

such that (3.11)

L(x,V,X,p)

=

inf L(x,v,i,p) V

u

(3.12)

E

5

M

.

is compact, there exists a Bore1 function with values in 7 ~ ,

?J

v(x,~,p) on

Now

jpl

C1(6)

.

Define ?(x)

=

V(x,u(x) ,Du(x))

=

H(x,i,p)

.

153

STOCHASTIC CONTROL

-

which is a Bore1 function on 8 , with values in "J. 0 ( s ) = C(x(s))

(3.13) then 0 ( s )

Define

is an admissible control.

0 , we

Moreover by the choice of

have

which with ( 3 . 7 ) , taking v

Letting next

T

+

m

,

u(x)

=

0

,

yields

we obtain

=

.Jx(C(.))

which, with (3.10) completes the proof of the desired result

Remark 3 . 1 .

.

The optimal control is defined by a feedback 0(x).

n This

means that to know the control which should be excited at any time the information on the states previous to We also remark that to

b(x) + g(x,?(x))

Pt

and

t,

is unrelevant.

is solution of the martingale problem relative a(x)

Remark 3.2. We can take h # 0

, hence is Markov process.

,

h

E

Lp6)

solution of (2.8) is given by ( 3 . 4 ) with f(x,v) + h(x).

t

,p

f(x,v)

Details are left to the reader.

> n+2.

0

The function u

changed into

0

154

CHAPTER IV

Let us indicate now an other approach to study equation ( 2 . 8 ) ,

called the

method of policy iteration. 0

,...,un...

Let

u

W2"

n W;"

Knowing

, un

,

be a sequence of functions belonging to

p > n

,

defined as follows. We take

define vn(x)

uo

t o be a Bore1 function such that

+ Du"(x).g(x,v"(x)) Define next

un+l

Y x

.

as the solution of the linear equation AU~+'+

(4.15)

arbitrary.

a un+l

=

f(x,vn(x))

+ un+ 1 al(x,vn(x))

+

0

Theorem 3 . 2 .

We make t h e assumptions of Theorem 3.1 and a . 2 0, Lp , p > n , then un c u and in W2'p u e a k l y , where u i s t h e solution of 1 2 . 8 1 .

h

E

Clearly the sequence un al(x,vn(x))

,

g(x,v"(x))

is well defined. Moreover since are bounded,

(3.16)

Next Aun + a un = f(x,vn-l) + unal(x,vn-1)

+

0

+ Dun.g(x,vn-') t

f(x,vn) + unal(x,vn)

+

+

h

2

Dun.g(x,vn)

+

h

f(x,v"(x))

,

155

STOCHASTIC CONTROL

hence ~ ( u ~ + l - u+~ a ) (un+l-un) 0

u which implies u Hence

un

i

u

n -u 1,

n+ 1 +n

5

0

=

0

-

D(un+l -u n ).g(x,vn)

o

(un+l -un)al(x,vn) 5

n+l

-

(recall ( 3 . 3 ) ) .

pointwise and in W2"

weakly.

Therefore also, by compactness (3.17)

un

-t

u

in W'VP

Let us identify the limit.

We have for v

E

V

arbitrary

Aun + aOun

-

f(x,v)

+ aOun

-

f(x,vn)

~u~ + aOun

-

f(x,vn) + (un+l-un)a 1 (x,vn)

S Au"

=

strongly.

-

+ (Du n + l -Dun ).p(x,v")

unal(x,v)

-

unal(x,vn)

n+l)

n

+ (Dun+l-Dun).g(x,vn)

-+

Du".g(x,v)

-

+ (u

n+l

0 in

-

h

Uun.g(x,vn)

Dun+ 1 .g(x,vn)

-

= A(U~-U~+') + a0(u -u

-

-

-

h

=

.

Therefore Au + aou V V € V which means

-

f(x,v)

-

ua ( x , v ) 1

-

Du.g(x,v)

-

h

=

un+'al(x,vn) +

n -u )al(x,vn) +

Lp weakly

-

5

h 5 0

156

CHAPTER IV

(3.18)

AU + a u

-

H(x,u,Du)

S

h

Au + a u

-

H(x,u,Du)

-

h

0

a.e.

Also we have

0

2

Au + a u 0

=

+ m,

-

f(x,vn)

h + Aun + aOun

+

I

Au + a u 0

-

H(x,u,Du)

-

Dun.g(x,vn)

(DU"-D~) .g(x,v")

the right hand side tends to

which with (3.18)

ua (x,vn)

-

-

f(x,vn) - h =

-

h

Z

0 in L p

-

unal(x,vn)

weakly, hence

0

concludes to the desired result.

Let us give the analogue of the situation studied in section 2 and 3 .

4 . 1 . perabgljc-equetjgns

We consider here functions f(x,v,t)

: R" x ?/ x

g(x,v,t)

: R"

c(x,v,t)

: R" x ? / x C0,TI

x T

Bore1 and bounded and set

x

=

+ (un-u)al (x,vn)

4. EVOLUTION PROBLEMS

(4.1)

-

0

= ~(u-u") + a (u-u") 0

n

Du.g(x,vn)

~(u-u") + a (u-u") + (~u"-~u).g(x,v") + (un-u)a,(x,vn)

-

and as

-

[O,Tl

+

R

C0,Tl

+

Rn

+

R

0

157

STOCHASTIC CONTROL

(4.2)

H(x,t,X,p)

=

infCf(x,v,t)

-

Ac(x,v,t) + p.g(x,v,t)]

.

VEV

Next we assume aij

(4.3)

aaij __

CO,a,a/2

E

E

axk

3

Lrn

(4.4)

Let (4.5)

where 8

h

E

Q =8

Lp(Q)

x

, U

(0,T)

is a smooth bounded domain of

E

1

W2” n W0”

Rn.

Then we have

Theorem 4 . 1 .

We assume ( 4 . 1 ) , ( 4 . 3 1 , (4.4), (4.51. Then t h e r e e x i s t s

one and o n l y one s o h t i o n of (4.6)

b2y”p(Q)

u

E

-

au at

uIC

=

+

A(t)u

0

-

H(x,t,u,Du) = 0

, u(x,T)

=

u(x)

.

Similar to Theorem 3.4 of Chapter I1 and Theorem 2 . 1 . We can next give the interpretation o f the function u. (4.7)

f(x,v,t),g(x,v,t),c(x,v,t)

and measurable

We assume

are continuous in v

with respect to x,t,

V v

.

,

a.e.x.t

158

CHAPTER IV

(4.8)

Let

?J is a compact subset of

u

=

f(x,v,t)

which is Lebesgue measurable in a.e.

.

belong to I J J ~ ’ ~ ’ ~ (,Q )and define LU(X,V,t)

x,t.

OU(x,t)

Rd

x,t

-

u(x,t)c(x,v,t)

for any

It is a Caratheodory function.

v

+

and continuous in v

Hence there exists a function

which is Lebesgue measurable and such that

We can take

GU(x,t)

to be a Borel representative.

In the sequel, the

results will not depend on the choice of the Borel representative.

Let

Pxt

operator

be the solution of the martingale problem relatively to the A(t)

, with initial conditions (x,t).

Then if

x(s)

is the

canonical process as usual, we have

A control is an adapted process with values in V .

v(s)

(with respect to the family

We define the measure

:)

Pzt such that

(4.11)

and

Ptt

is the unique solution of the problem of controlled martingales

159

STOCHASTIC CONTROL

(4.12)

And

(4.13)

where

We can state the

Theorem 4.2.

We make the a s s q t i o m of Theorem 4 . 1 and

7de

Then t h e s o l u t i o n of 14.61 is given explicitely by

h = 0.

(4.15)

u(x,t) = inf

J~~(V(.))

p > n+2.

.

V(.) ~ v ~ G P e G V e rzhere ,

(4.16)

exists an cptirnal control O ( s ) = Ou(x(s),s)

.

O(s)

defined by

CHAPTER IV

160

Similar to that of Theorem 3.1.

Theorem 4.3.

We make the assumptions on Theorem 4.2.

control v(.)

, the process SAT

u ( x ( s ~ ~ ~ ) , s ~ ? ~ ) -e xitp

for

t 5 s 2 T,

is a sub martingale

(PVXt

Then for any

t c(x(X),v(X))dX

A;)

-

u(x,t)

+

.

For v = 0 , it is a martingale.

Let

5,

be

measurable and bounded.

Using equation ( 4 . 6 ) we obtain

We have for

t 5 s 5 @ 5

T ,

161

STOCHASTIC CONTROL

This proves the first part of the theorem. Taking to

v

=

0 , we have equality

0

0, hence the second part of the theorem.

Remark 4.1. verifies

1.1~

It is easy to check that if =

0

,

u(x,T)

=

0

then ( 4 . 1 5 ) holds and

u

is Bore1 bounded and

and the properties of Theorem 4 . 3 ,

;(x)

3

is optimal.

5 . S E M I GROUP FORMULATION

5 * 1.

4-eroee r tr -2f - the -E9uatjon- - u

Let us go back t o equation ( 2 . 8 ) , and we will assume here (5.1)

For

a]

v

E

=

, .a

0

= c1

> 0 a constant

2 , a parameter we consider the operator

(5.3)

We note that

Av = A

u

-

g(x,v).D

.

satisfies

(5.4)

1 Moreover let

w

A u + au s f

a.e.

in 8

.

satisfy ( 5 . 4 ) , then we have Aw + uw S f(x,v) + Dw.g(x,v) Au + au = inf[f(x,v) V

hence

fi v

+ Du.g(x,v)l

162

CHAPTER I V

A(w-u) +

U(W-U)

2

f(x,v) + DU g(x,v)

-

-

+ Du.g(x,v)l

inf[f(x,v

+

V

+ (Dw - Du).g(x,v)

< f(x,v) + Du.g(x,v) - inf[f(x

v ) + Du.g(x,v)l

+

V

+ ID(w-u) hence taking the inf in v

,

C

we obtain

(5.5)

Condition (5.5) imp1ies w - U S O .

(5.6)

Relation (5.6) is clear when a to prove the following result.

is large enough. Otherwise we have Let

h

E

0

.

Lp

given and

z

to be the

solution of the H.J.B. equation

I

then (5.8)

Indeed

h z

S

0

implies

z 2

can be obtained as the limit of the following iteration

163

STOCHASTIC CONTROL

(5.9)

zo

starting with

0 , and

=

zn Since

h

0

5

+

z

in

WzYp weakly

, one checks inductively that zn

5

0

,

hence

z

s 0

.

This

proves ( 5 . 8 ) , hence (5.6). We thus have proved the following ble make t h e assumptions o f Theorem 2.1,

Tkeorem 5 . 2 .

Then t h e soZution

u

and (5.1), ( 5 . 2 ) .

~f (2.8) i s t h e maximum element of ;he s e t o f

0

f u n c t i o n s s a t i s f y i n g (5.4). Remark 5 . 1 .

Assumption

(5.1) can be weakened, but it will be sufficient

0

for the remainder of the chapter. We note now

the solution of the martingale problem corresponding to

P:

the operator

Av

,

starting in x

at time

0

controlled martingale problem, with a control S,L.

Let

u

.

It corresponds to a

v(s)

=

s

, independant of

be a function satisfying ( 5 . 4 ) , then from Ito's formula

we have

= o hence u(x) Recalling that

ui,.

=

5

Ez[{tmfv(x(s))e-"sds

0

, we also have

+ E:u(x(tAT))e

-CltAT 1

.

164

CHAPTER IV

.

0

x (x(s~r))e-"~ds] + E:u(x(th~))e-'~l v o

which we have considered in

§

5.2 of Chapter I11 and noting that

(5.12)

+

Bo

u(x)

(5.10)

5

:E

[Jt f

Using the semi group

where

OV(t) : Bo

Bo

is the set of Borel bounded functions on

r,

we see that

(5.13)

u

on

u

6,

which vanish

satisfies the relation

s Jt OV(s)fv

w - " s d s + OV(t)u

This motivates the problem which will be studied in the next paragraph.

We make here the following assumptions.

Let

E

be a polish space

(1)

provided with the Borel a-algebra 8 . We note

B

the space of Borel bounded functions on

E.

uniformly continuous function on

E, C

the space of

We assume given a family

v c 1J, where (5.14)

'V

(5.15)

OV(t) : B

finite set

OV(t)OV(s) OV(t)Q

t

=

o

,

B

-f

Ov(0) = I

.

OV(t+s) if

Q

2

o

(1) This will be needed for the probabilistic interpretation.

OV(t)

,

165

STOCHASTIC CONTROL

We will also assume that (5.16)

aV(t)

(5.17)

t

-t

c

+

c

oV(t)G(x)

E

u

5

To find

u

maximum solution of

+ OV(t)ue-at

u e B

= fV(X)X

(4=

8

R ,

a > 0

1t 0v (s)Lve-"'ds

Lv(x)

-t

C

,

L (x) :L(x,v)

We consider the following problem.

(5.19)

(0,m)

is continuous from

x fixed, Y I$

Y (5.18)

:

f(x,v)x

(x)

0

,

166

CHAPTER IV

t > 0

For

z(x,t)

is a regular function of

hence (5.18) is

x,t

satisfied. We will study ( 5 . 1 9 ) by a discretization procedure. Uh

h > 0

Let

define

by

(5.22)

uh

=

Min!Jh v

e-US QV(s)Lvds

+

QV(h)uhl

-"here mists one and onZg one s c l u t i o r cf (2.22).

Define for

z

E

z =

MinCJh e-"' v

T z h C

.

B

E

C

since 7J'

OV(s)Lvds

+

OV(h)zl

'

is a finite set.

Note also that

T z h

Th

E

C , when

is a contraction, hence has one and only one fixed

point , uh. L e m o 5.2. L e t

0 z c B

such t h a t

z S

T z h

proof Th is increasing we have

2

T z S T z h h hence

t

Moreover

which proves that

Since

uh

B

T and

,

o

S e m a 5.1.

z E

,

z 5

T2z h

and by induction z S

Tnz h

+

u

h a s n - t m '

then

z

< uh '

167

STOCHASTIC CONTROL

Lemma 5 . 3 .

Ue have

(5.23)

Uh 5 UZh

.

procs We have f o r any

v

Uh 5

{

u

LZh e-"'

OV(s)Lvds

+ evah OV(h)uh

hence

which implies <

OV(s)Lvds + e -2ah @ v (2h)uh

h hence

which with Lemma 5 . 2 implies ( 5 . 2 3 ) . We can then state the

Theorem 5 . 2 . u1/2q

+

'

We assume 1 5 . 2 4 1 , 15.15), (5.161, 15.171, 15.18). maxim s o l u t i o n of (5.191, as q + + m .

Let us check that

(5.24)

Assume

z t

-

K

,

then

Then

168

CHAPTER IV

OV(s)Lvds +

Jh e-" t

- e -ah K + Jh e-"'

E

- e -ah K - Max/IL;II

Qv(h)z t QV(s)L;ds t

Jh e-asds

E

V

2 - K

hence

T z > - K h

which i m p l i e s ( 5 . 2 4 ) .

Let u s s e t

q f +

then a s

u

m

q

.

C u

u

Note t h a t

is

U.S.C.

Furthermore we s e e

that uh s

Take

I h = 24

,

imh e-"

OV(s)Lvds + e - a h 0v (rnh)uh

m = j 2q-R

and

R

with

5

R (5.25)

u

q

5

R 5 q

5

hence

q

,

j

integer.

/j/'

e-"

o

,

j

Let

,

q

QV(s)Lvds

+

ti rn

we g e t

9. e-a"2

~ ~ ( j / 2 " . u, ~

integer

.

q

According t o Lemma 5 . 4 below,

-f

a.

integer

.

169

STOCHASTIC CONTROL

(5.26)

ti q.

j

Take next

= Ct

2

9"

I+

1

and let

R

tend to

+a

, we deduce from

( 5 . 2 6 ) using assumption ( 5 . 1 7 )

in which we may again let (5.19).

q

tend to

+m.

This proves that

,.

It is the maximum solution, since assuming u

u

satisfies

to satisfy

( 5 . 1 9 ) then clearly

which implies

-u

5

uh , hence the desired result.

G

Let us now state the result which has been used in the proof of Theorem 5 . 2 .

We refer to Dynkin 121. Let us briefly mention the main elements of the proof.

Let

Banach space.

be the space of bounded measures on We write

Then we have the following result

( E , 6 ) , which is a

170

CHAPTER IV

m

Define next an operator on

One checks that

by the formula

is a contraction semi group o n n , and that the

U(t)

following relation holds,

0

From (5.29) and (5.27) the desired result follows. Remark 5.2.

In example (5.20), we have

uh

Co =

E

T, and of course u

functions which vanish on

E

subspace of

C

of

.

Bo

L

Let u s define

R0

=

E

I

,

x(t;w)

is the canonical process,

Let u s assume for simplicity that

7 To

i

E

b'

=

...,m} .

{I,?,

, we associate a probability PTt

(5.30)

E2t @(x(s))

We will denote by values in V .

W

=

Oi(s-t)6(x)

Ro

, nt

fi s 2 t

such that

.

the class of step processes adapted to

V

E

... 5

7

More precisely, if To =

on

0

s

T1 5

not

with

W , then there exists a sequence 5

...

which is deterministic increasing and convergent to

+x

and

171

STOCHASTIC CONTROL

v

(5.31)

=

, v(t;o)

v(.)

=

vn(w)

t

€

IT,,?

n+ 1 )

T

where

is ?!on

v

measurable.

9"

We next define a family the pair

w,t

(w

E

07; , Oss
Go C E

,

of probabilities on

w,t t 2 0)

(C20,77!o)

vt , then

Let us next define a sequence

n Pz

of probabilities on

(fro$:,) as

follows 0 p;

(5.33)

where

vo

=

vo(x)

=

pX90 "0

, vo being a Bore1 function from E - V .

and by induction

We then note that

Indeed n

and since

7

E

mi"''

indexed by

such that for

we can take

172

CHAPTER IV

r = r l n r2 where

1-0 r l E P ‘n+ / ~ , r2

then for such

u(x(T~+]))

r

Moreover n+ 1 P ;(r)

n = pC(r)

n Then the family

(Qo,

on

Pt(r)

.

mo‘n+ I )

forms a system of compatible

probabilities and thus from Theorem 1.2 Chapter I there exists a unique probability

P;

(C20,7co)

on

~ ; =f

(5.37)

Lema 5 . 5 . Let

$(x)

E

n

P;

B

on

such that

mo‘n+ I .

and

T~ 5 t < T

n+l.

T

(5.38)

E$@(x(t))

1mon1 =

Then w e have

V

@

n(t-?n)@(x(Tn))

.

T

Let

5

be

non

measurable and bounded.

Since

t < T

n+1

,

we have

173

STOCHASTIC CONTROL

But

and

3

from which we deduce (5.38).

We next define for

(5.39)

V

W

E

Jx(V) = El:

Jrn

.

L(x(t),v(t))dt

We have m

Jx(V)

=

C

Et

L(x(t),v(t))dt

n=O

But from Lemma 5.5 we deduce

(5.40)

Ex JTn+l

n '

-Cl(t-Tn)

L(x(t) ,v,)dt

=

CHAPTER IV

174

Define

(5.41)

Lh(x,v)

=

Jh

OV(t)Lv(x)dt

then from (5.40)

and defining

we get the formula -3T

(5.42)

c

. J ~ ( v )= E;

L

e

n=O

(X,(O)

,vn(Ld)

*

‘n+ 1 -Tn

Let

(5.43)

wh

(5.44)

uh(x)

For

V

E

Wh

=

iv

E

WIT,

=

=

:
=

E;

nhl

I

we have m

(5.45)

J~(v)

-anh e ~ ~ ( x ~ , v ~ )

z n=O

and

v

is

nh

0

measurable.

A control V

sequence vo

=

VO(X),

...,Vn,”.

can be considered as a

175

STOCHASTIC CONTROL

where

v

is

nib

measurable. Note that

vo

is not random.

Now from ( 5 . 2 2 ) we have

(5.46) Take

V

u,(x)

to be an arbitrary control in

Replace in (5.46) x

by

uh(xn)

(5.47)

+

e-as OV(s)Lv(x)ds

5

=

(v,(x)

, and v by vn(w).

xn(o) 5

Vh , V

Ov(h)uh(x)

Jh ePas :n(s)Lv

,...,vn(w) ... ) . We get

(xn)ds + edah ;"(h)uh(xn) n

But from Lemma 5.5, we can assert that

hence we deduce from (5.47), (5.48), (5.49) that

Adding up for

Letting N

n = 0

+ m

(5.50) Define now F(x)

,...,

N-l

we get

and recalling that u,(x)

5

Jx(V)

uh

is bounded, we obtain

.

such that

+

Y x .

.-as

O G ( x ) (h)uh(x)

.

.

176

We can find

CHAPTER IV

G(x)

Borel.

G

=

To

0

we associate V

E

Wh

as follows

. . . ,Cn(LI), . . . )

(Go(x),

Go(x) = G(x)

+,(a)

G(x(nh;w))

=

.

... A calculation similar to the one above shows that

u,(x)

Jx(?)

=

which completes the proof of (5.44). We are now in a position to interpret the maximum solution of (5.19) Let us define (5.52)

W

4

= w 1/74

Then (5.53)

w

q+ 1

=w

4

hence (5.54)

and we recover the fact that

*

c

1I 7

STOCHASTIC CONTROL

Theoreq 5 . 4 .

Ye make t h e assumptions of Theorem i . 2 .

Then t h e mmirnum

element o f (5.19) e m be i n t e r p r e t e d a s

VEU

q

Let

u(x)

W g

be the right hand side if (5.55).

Since

we have

-

u > u .

But on the other hand we know that u

and since

u

5

u

4

Jx(V)

S

V V

W

E

4

4 u 5

Jx(V)

V V

E

V

E

W

q

hence u 5 Jx(V)

which implies .?er*Ork 5.3.

u S

-u

'4

u W

q

,

hence the result.

The function u

the solution of (2.8).

q

3

defined by (5.55) is a priori larger than

0

178

CHAPTER IV

In this paragraph we give some regularity results on the maximum solution of (5.19).

We will assume

and

then

Theorem 5 . 5 .

Then if

Let

c1

Ye make the a s s u - v p t i m s of Theorem 5 . 2 ar,d ( 5 . 5 6 ; ) (5.573.

> A

,

.

u c cO"(E)

z c c'"((E).

Let u s fix xo

in

E , then there exists vO

(depending on xo) such that ThZ(XO) * j h e-as CJv~ ( s ) L

(xo)ds +

Ovo(h)a(xO)

vO Let x

arbitrary, we have Thz(x)

5

Jh

(x)ds +

Pvo(s)Lv

0 hence by difference

and from the assumptions (5.56), (5.57) it follows that

Ovo(h)z(x)

179

STOCHASTIC CONTROL

5

Lh e-asei's

Klx-xo

1 'ds

+

hence

and since

xo,x

are arbitrary, this implies

and iterating we see that

and letting

k

tend to

+a

,

it follows that

K

I I ~ h I 5l ~

Taking now h

=

1 and letting -

q

-f

m,

we deduce

24

(5.59)

IlU

Il&q

K

which implies the desired result. Let u s now give an other regularity result. We assume (5.60)

I

I Z I

86 16 Ix-xgl

180

CHAPTER IV The rnuximwn so2ution of 15.191 is a l s o the maximwn s o l u t i o n of

L e m a 5.6. (5.61)

U E B ,

u

5

Jt e-”

tr v

+ e-Bt OV(t)u

Qv(s)(Lv+(G-a)u)ds

Y t

E l J ,

2 0

Proof We first show that (5.61) has a maximum element, which will be denoted by N

u.

Indeed, define for Bhz

=

z

E

B + e-Bh O v ( h ) z l

Min[Jh e-Bs aV(s)(Lv+(B-a)z)ds v o

It is well defined, by virtue if ( 5 . 6 0 ) . This is a contraction, since

’

I / o ~- ~~ ~~

z 5~ / jI z 1j-2 2 ’

= 1l2,-22/

Moreover when

z

,

C

E

Ohz

when

uh

.

C

-

..r

Let

E

be the fixed point, uh

z 2 0

6

.

C

and

Y

One checks as for Theorem 5.2 that

- -

Setting u

q

=

u

1/29

, we get

Y

u

.

J- u

, since

Ohz 2 0

N

uh U

q

-

uh 2 0

5

uZh

.

Then

9, S 4

/j’*

e-Bs OV(s)(Lv+(C-a)u

0

4

)ds

t

L

and as for Theorem 5 . 2 , one checks that

,u.

is a solution of ( 5 . 6 1 ) , and

181

STOCHASTIC CONTROL

that it is the maximum solution, since any other solution satisfies v

5

0hv

,

-

v < uh '

hence

w

Let us show that u = u , where

u

is the maximum element of (5.19).

We will use Lemma 5.7 below to assert that & < I t e-"'

-

hence u

5

u.

w

i e m a 5.7.

,

Let

where (5.63)

E

B.

0

and the desired result is proved. O(t)

be a semi group on Let

w

w < Jt g

+ e-Bt OV(t)u

Jt e-Bs OV(s)(Lv+(B-a)u)ds

( 5 . 1 5 ) , and is.&Ol.

(5.62)

OV(t)u"

However, still using Lemma 5.7 we have u

hence u < u

+

QV(s)Lvds

@(s)g

5

ds +

@(t)w

B > 0 , one has

Then f o r any w

satisfying properties

be such t h a t

B

E

B

it e-as @(s)(g+(B-a)w)ds

+ edBt O(t)w

We set

we have H(0)

=

0

,

H(t)

5

0

In fact, we have the additional property

Y t

,Y

t t 0

182

(5.64)

CHAPTER IV

H(t) s H(s)

for

.

t 2 s

Indeed ( 5 . 6 4 ) amounts t o proving t h a t

(5.65)

e

-us

O(s)w

O ( t ) w + Jst e-"O(A)g

5

,

di,

.

s 5 t

But f r o m ( 5 . 6 2 )

w s /

t-S

e

-a>

dA + e

O(A)g

-a ( t - s )

O(t-s)w

0

0 and

and i n t e g r a t i n g between

we deduce

T

[ l-e-(B-a)Tjw =

JT (B-a)ewbt O ( t ) w d t + 0

+ /T ( 3 - i ) e

+

=

H(t)dt

+

iT ( 3 - a ) e- ( 3 - a ) t ( i t e-" O ( s ) g d s ) d t iT (6-a)e - 6 t O ( t ) w d t + iT ( 0 - a ) e - ( p a ) t =

- e

-(B-a)T /T e - 2t 3 ( t ) g d t +

iT e-r't

H(t)dt

O(t)g d t

hence (5.66)

w =

/T .-3t

O(t)(g+(?-?)w)dt

+ e

-6T

O(T)w

+

0

- (6-3)T H ( T ) + J .T ( 3 - 5 ) e - (5-3) t H ( t ) d t

+ e

a

.

-

183

STOCHASTIC CONTROL

If B 2 u since H(t) < 0 , we clearly have ( 5 . 6 3 ) with 6 < a then using ( 5 . 6 4 ) we have

If

t = T.

hence H(T) +

e-(B-a)T

iT(@-a)e-(B-a)t

H(t)dt

H(T) + (@-a)H(T) JT e-(B-a)t

< e

therefore ( 5 . 6 3 ) holds in all cases for

t = T

.

5

dt

Since T

H(T)

=

5

is

C

arbitrary the desired result is proved.

Theorem 5 . 6 .

Let

z

We make t h e assumptions of Theorem 5.2,

15.561,

Then the m a x i m s o l u t i o n o f (5.19) belongs t o

(5.60).

E

C

5

and

5

1 5 . 5 7 1 and

C.

be the maximum element of

5

it e-Bs Ov(s)(Lv

+ (8-a)z)ds + e-Bt OV(t)<

,

tiv,tit. This set has indeed a maximum element according to Theorem 5 . 2 . defines a map

S : C

we can assert that

.

According to Theorem 5 . 5 , provided S : C o y & + CoP6. +

B

We also know that S(z) =

s1 1im+-(z)

24

as

0

q f

,

ti z

E

C

This

B >

184

CHAPTER IV

Ch

where

=

Sh(z)

is defined by

ch

(5.67)

Min[Jh e-@'

=

v

OV(s)(Lv+(@-a)z)ds

+ e-ah QV(h)ch]

o

< h e C * Sh : B

Note that

B

-t

and

C

C

+

.

One easily checks the estimate

I Ish(zI)-sh(z2) 1 1

(5.68)

5

71

1z1-z21 j

from which one deduces

I IS(zl)-S(z2) 1 I

(5.69)

5

a-a 112 - 2 1 I , p 1 2'

when

We also note the relation, which follows from Lemma 5.7, (5.70)

u

5

sh (u)

.

Define now

,

un = S"(0)

Since

s

maps

coy6

u;

=

.

s;(o)

into itself, u n c

c0j6.

From (5.69) we have n+l IIU

n

' I1

B-a n 1 (T)1Iu 1

and thus un+w We will show that

(5.71)

u = w .

in

C .

1

z 1 ,z2

E

C

185

STOCHASTIC CONTROL

which w i l l prove t h e d e s i r e d r e s u l t .

We f i r s t remark t h a t from ( 5 . 6 8 ) belongs t o

C

,

has a fixed point i n

and

wh

s ~ ( o )+ wh

(5.72)

From ( 5 . 6 9 ) ,

denoted by

Sh

c

in

.

( 5 . 6 8 ) we have

(5.73)

(5.74)

From ( 5 . 7 0 ) we c a n a s s e r t t h a t (5.75)

u s w

h

We check by i n d u c t i o n on

n

that

un h

-

By i n d u c t i o n on

n

we check t h a t

(5.76)

un+un q

un 2h

hence

From ( 5 . 7 3 ) ,

as

q f m ,

=

~

( 5 . 7 4 ) , (5.76) follows t h a t

wq(x) 4- w(x) which w i t h ( 5 . 7 5 ) shows t h a t

(5.77)

Gn

u < w .

v

x

un.

Hence

n

.

B

,

which

186

CHAPTER IV

But a l s o

hence

+ e-pph OV(ph)wh

w < Jph edBS OV(s)(LV+(B-a)wh)ds h - 0 hence also

for

q 2 R.

Using a reasoning as in Theorem 5 . 2 , we obtain easily that w

5

+ e-Rt @"(t)w

J t e-Bs OV(s)(Lv+(p-a)w)ds

hence also, using Lemma 5.7 w

it

5

@"(s)Lvds

+ e-"

QV(t)w

0

which implies w

5 u

,

and from (5.77) we see that ( 5 . 7 1 ) holds.

ci

completes the proof. Let us give an example where ( 5 . 5 7 ) is satisfied, with Consider the S.D.E, dy with (5.78)

This

=

g(y)dt

+

o(y)dw

y(0)

=

x

6

=

0

.

187

STOCHASTIC CONTROL

hence

which proves (5.77).

Remark 5 . 4 .

For other details (cf. M. Nisi0 [l], Bensoussan-Robin C11,

0

Bensoussan-Lions 121). COMMENTS ON CHAPTER I V

I . The method of improvement of regularity used in Theorem 2.1 is due to

P.L. Lions. 2 . Assumption (3.1) can be replaced by Lebesgue measurable in x

continuous in v

as mentionned in the evolution case

§

4.1.

, and In fact

we need a selection theorem. There are two types of such theorems that we may use. Consider Assume

F(x,v)

,

x

E

Rn

,

v

E

3'(compact subset of a metric space).

188

CHAPTER IV

F

1.s.c.

in x,v, F

bounded below.

Then there exists a Borel function C(x) F(x,C(x))

= inf F(x,v)

+V,

: Rn

such that

Y x

V

(see for instance D. Berksekas, S . E . Shreve [ I ] ) . The other theorem uses more explicitely the Lebesgue measure on We assume that F F

Rn.

is a Caratheodory function, i.e.

, continuous in v, a.e. x.

V v

is Lebesgue measurable in x

Then there exists a Lebesgue measurable function C(x)

: Rn

+ V , such

that F(x,?(x))

=

inf F(x,v) a.e. V

We can take a Borel representation of We write

inf

G(x)

,

but it is not unique.

is a Lebesgue measurable

V

function such that if

?(x)

, which

for ess inf F(x,v)

5

then G(x)

F(x,v) 5

a.e.

Y v

ess infF(x,v)

a.e,

V

Note that

inf F(x,v)

when

F(x,v)

is Borel for any v

is not a

V

Borel function (cf. I. Ekeland - R. Temam [ I ] ) . 3 . The method of policy iteration was introduced by R. Bellman [ l ] ,

in the

general context of Dynamic Programming.

4 . For the study of degenerate Dynamic Programming equations (i.e., the matrix

-1

a

does not necessarily exist) we refer to P.L. Lions

-

J.L. Menaldi C11. 5. J.P. Quadrat has formulated a generalized martingale control problem,

which includes degeneracy (cf. J.P. Quadrat [ I ] ,

[Z]).

189

STOCHASTIC CONTROL

6. For numerical techniques t o solve the H.J.B. equation see J.P. Quadrat [ l l , P.L. Lions

- B.

Mercier [I].

7 . A s we have said in the general introduction, the most complete

treatment of the general Bellman equation is due t o P . L . Lions [ll,

[ZI. 8. The problem of semi group enveloppe was introduced by M. Nisi0 C21

9. Nisi0 has also introduced a problem of non linear semi group connected to stochastic control (cf. M. Nisi0

[ll).

10. In the context of Remark 5 . 3 . Under what conditions can we assert

that the solution u

of (5.55) coincides with that of ( 2 . 8 ) .

This Page Intentionally Left Blank

191

CHAPTER

FILTERING AND PREDICTION

V

FOR

LINEAR S.D.E.

INTRODUCTION We present here the classical theory of linear filtering, due to R.E. Kalman [ l ] , R.E. Kalman - R.S. Bucy [ l l . Xe have chosen a presentation which can be easily carried over to infinite dimensional systems, for which we refer to A. Bensoussan [ I ] , R. Curtain - P.L. Falb [ l ] , R. Curtain - A . J . Pritchard [ l ] . For filtering of jump processes For non linear filtering, cf. R. Bucy - P. Joseph (cf. P. Bremaud [ l ] ) . [ I ] , and the recent developments in E. Pardoux [ I ] , T. Allinger S.K. Mitter [ I ] .

1. SETTING OF THE PROBLEM

We consider a usual system

(S2,a,P,5t,w(t)), and

solution of the linear S.D.E.

where

(1.2)

F

E

L~(O,~$(R";R"))

G

E

L~(O,~;~(R";R"))

f(.)

E

L ~ ( o , ~ ; R ,~ )

Clearly the standard theory applies since

x(t)

to be the

192

CHAPTER V

g(x,t)

=

F(x)x

o(x,t)

=

G(t)

+ f(t)

.

5 is gaussian with mean x and covariance matrix

a

To the O.D.E.

corresponds a fundamental matrix

such that the solution of (1.4)

@(t,T)

can be expressed as x(t)

(1.5)

where

g

E

=

2

L (0,m;R").

@(t,O)x

+

Jt

The family

(1.6)

@(t,S)@(S,T)

(1.7)

@(t,t)

=

I

=

@(t,r)g(r)dT has the group property

@(t.T)

@(t,T)

d '

t,S,T

.

It is easy to check that the solution of (1.1) y(t) = @(t,O)c

(1.8)

+

Jt

can be expressed by

@(t,T)f(?)d?

+

It @(t,?)G(~)dw(r)

where the last integral is a stochastic integral. Formula (1.8) is a representation formula for the process

y(t).

It is also useful to

notice the following. Let (1.9)

h

E

Rn

and

-3 dt

=

F*(t)$

,

$(T)

=

h

FILTERING AND PREDICTION

193

then we have (1.10)

@(0).5 +

y(T).h

=

p(t)

O*(T,t)h

@(t).f(t)dt

+

iT @(t).G(t)dw(t)

.

Since

(1.11)

=

it is easy to deduce (1.8) from (1.10) and ( 1 . 1 1 )

It is clear from (1.8) or (1.10) that expectation y ( T ) (1.12)

i s a Gaussian variable with

y(T)

such that

Q(T,O)x + {T @(T,t)f(t)dt

y(T)

=

-dy- -

F(t)y

;(t)

=

i.e. (1.13)

dt

,

y(0) = x

.

Let

Y(t)

-

y(t)

then from (1.10) (1.14)

-Y(T).h

=

O(O).t

where (1.15)

Define next (1.16)

then

- b-'' dt

=

F* (t)+

+

iT @(t).G(t)dw(t)

CHAPTER V

194

hence from ( 1 . 1 4 ) , (1.18)

(1.17)

we deduce

E F(T).h

= Po @ ( O ) . $ ( O )

y(T).k

+

JT

G*(t)@(t).G*(t)$(t)dt

= II(T)h.k

where

II(T)

denotes the covariance operator of

y(T) (or

y(T)).

Hence we have the f o r m u l a (1.19)

n(T)h.k

(1.20)

n(T)

=

Po @ ( o ) . $ ( O )

=

+

O*(T,O) +

O(T,O)PO

J

T

G(t)G*(t)q(t).+(t)dt

iT O(T,t)G(t)G*(t)O*(T,t)dt

We will set for simplicity (1.21)

G(t)G*(t)

=

Q(t)

We can deduce from (1.20) that

.

II is solution of a differential

equation. We have (1.22)

TI(T)h.k

=

Po @*(T,O)h.O*(T,O)k

t

+ JT Q(t) @*(T,t)h.O*(T,t)k 0

The function s

1

O(s,t)h

E

H (t,T;Rn) , and

O*(s,t)h

E

H 1 (t,T;Rn)

+

hence

and

dt

.

.

195

FILTERING AND PREDICTION

(1.24)

d O*(s,t)h ds

=

O*(s,t)F*(s)h

We can approximate (1.22) with respect to

. T , using ( 1 . 2 4 ) .

We

obtain

dT

h.k

= Po

+

O*(T,O)F"(T)h.@*(T,O)k

+ Po O*(T,O)h.O*(T,O)F*(T)k

+

Q(T)h.k

+

+ JT (Q(t)@*(T,t)F*(T)h.@*(T,t)k

.

+Q( t ) @*(T,t ) h O*(T, t) F*( T) k)d t

and from (1.20) we get

We thus have proved

L e m a 1.1.

The process

y

soZution of (1.1) i s

whose mathematical expectation y(t)

covariance matrix n(t) (1.25)

We next define a process

where (1.27)

G

Gaussian process

is solution of 11.131 and whose i s so2zrtion of the equation

z(t)

by setting

+

196

CHAPTER V

(1.28)

is a

n(t)

Rn

and

(1.29)

R

Y 0 is

e.n(t)

Zt continuous martingale with values in E

Rp , the increasing process of

it R(s)B.Bds

,

where

R

is symmetric invertible and R-1

E

Lm(O,m$(Tn;RP)) bounded.

From the representation theorem of continuous martingales, we have q(t) =

Lt R”2(s)db(s)

,

where

b

is a standard 3

Wiener process. We also assume q(t)

(1.30)

It i s clear that

Z(t) (1.31)

is independant from

z(t)

5

and

w(.)

.

is a Gaussian process, whose expectation

is given by

i ( t ) = {t H(s)y(s)ds

.

Set

-z(t)

=

z(t)

- Z(t)

then (1.32)

-

(1.33)

E Y*(s,)?(s,)

z(t) =

it H(s)y(s)ds =

+ q(t)

.

a(s1,s2)~(s2)

if

s 1 2 s2

t

197

F I L T E R I N G AND P R E D I C T I O N

= @(sl,s2)~(s2) +

Jssl @(sl,s)G(s)dw(s) 2

hence

Let h , k

E

R”.

We have from (1.34)

E y(sl).h y(s2).k

=

E Y(s2).0*(s1,s2)h

=

iI(s2)O*(s1,s2)h.k

y(s2).k

therefore

from which we deduce (1.33).

It is easy to deduce from (1.33) and (1.20)

that

From (1.34) and (1.32) it is easy, although tedious to deduce the covariance matrix of

z(t)

We consider that the process be observed.

and the correlation function.

y(t)

cannot be observed, whereas

z(t)

can

The filtering problem consists in estimating the value of

y(t), knowing the past observations. More precisely, we are interested in

We note the following

198

Lema 1.2.

CHAPTER V

Ye have

(1.36)

a(z(s),O<s
=

o(J

T

$(t).dz(t),

Y $

E

2

L (O,T;Rp))

0

Denote by /-31

the o-algebra to the left of ( 1 . 3 6 ) and B2

o-algebra to the right of ( 1 . 3 6 ) .

it is clear that

B 2 measurable

c

,

Y

B2

.

Since

On the other hand

c2 c B I

hence

the

JT $(t).dz(t)

is

.

Since by definition

{T $(t).dz(t) the assertion that

=

LT @(t).H(t)y(t)dt

+

iT @(t).dq(t)

is E 2 measurable is not completely

JT @(t).dz(t) 0

obvious. However when

By convergence in , ' L Lemma 1 . 3 . (1.37)

We have f o r

E =

@

is a step function, this fact is clear.

we see that this fact is true for general

3. C

2

@,$ E L (O,T;Rp)

iT $(t).dy(t) iT +(t).di(t) CT @(t).R(t)$(t)dt .fT JT ~(s).H(s)A(s,t)H*(t)$(t)ds =

dt

+

0

0

proos Immediate from the definition of

A(s,t)

and formula ( 1 . 3 2 ) .

CI

FILTERING AND PREDICTION

2.

CHARACTERIZATION OF THE BEST ESTIMATE

2.1. kljkl-rg;uA;

We w i l l c h a r a c t e r i z e t h e b e s t e s t i m a t e by t h e f o l l o w i n g

Theorem 2 . 1 .

We assume ( 1 . 2 1 ,

Then t h e r e e x i s t s

11.31,

(1.27),

(1.28), 1 1 . 2 3 / , ( 1 . 3 0 ) .

map

CI

I

1

E

Z(L~(O,T;R~);R")

T

such t h a t V h (2.1)

proos Let u s w r i t e

(2.2)

We remark t h a t

E

R"

9 ( T ) . h = JT

i;

h.dy + y(T).h

.

199

200

CHAPTER V

We note that (2.6)

Let

(Q12h,@(.))

C

=

E y(T1.h

@(t).dz(t)

be any element of Z ( L 2 (0,T;R’);R”)

(2.7)

We have since

-

ECy(T1.h Q,

JTC* h.dzI2

=

.

and consider

dh(C)

.

i s invertible

&,(I)

=

Q 1 1 h.h + CQ 22 C* h.h - 2 C Q 12h.h

=

[Qll

+

(Z-Q;2Q;:)Q22(C*-Q;:Q12)

- Q;2Q;:Q,21h.h Clearly we have

zLh(C)

2

gh(iT)

where

* (2.8)

a’’

‘T

=

-1

Q l 2‘22

.

=

-

20 1

FILTERING AND PREDICTION

and

(2.11)

E

Eh (w)

l Let

ek

T

1 C*h2.dg

=

o

be a unit vector i n

Rp.

Let

C Ql2h1.h2 - Z Q,, M

E

to be chosen in order that

M h 2 = ek which is always possible (take for instance

Mkl=l Next define for

I$

E

Then by construction

and

Mkj=O 2

L (O,T;Rp)

j + l .

g(Rn;Rp) and

i$hl.h2 h,

E

Rn

=

0

202

CHAPTER V

Therefore (2.1 I) shows that

V k

,V

s.

is non correlated with

Since both are

Hence for any fixed -independant. ti s , OSsST. Since

h

Rn, E~

E

Y

z(s)

for any

sl,

...,sk

o ( z ( s ) , OSsST)

gk(s)

Gaussian variables, they are

, we see% l

is independant from

z(sl),

...,Z(sk)

that

ch is independant of

form a gaussian vector

= dT '

But Eh

y(T).h

=

-

(JT f;h.dy

+ y(T).h)

and

iT i;h.dy

+ y(T).h

E

2 L ( Q, B T;p)

and

E

~~c

=

V 5

0

2 L (Q,BT;P)

E

.

-

T This means that { ZTh.dz + y(T).h is the projection of y(T).h on 2 L (Q,dT;P), which is by definition the conditional expectation F(T).h.

0

Hence the result. Let u s set (2.12) then (2.13)

63h.h

= a (1 )

=

h T It is natural to call

E~

inf gh(C)

c

.

the estimation e r r o r .

covariance of the error. Introduce the following problem.

Then 6 is the

203

FILTSRING AND PREDICTION

where

q(.)

Clearly

E

2

L (O,T;Rp) and

Jh(q(.))

h

is a parameter.

reaches its minimum in AX

BT(.)

(2.15)

=

CT h

q

at the point

.

Our next step will be to give more convenient computational procedures to obtain

iT

or

i;

.

We first make explicit the functional Jh(q(.)). useful formulas.

Define

- dt ’ _ @ =

(2.16)

9 dt

=

(2.17)

Q

h

(2.18)

Q

F*(t),$

F(t)p

b(T) = h

- Q(t)@(t)

then we have

12

- H(t)Q(t)

h = 11

U(T)

.

Consider also (2.19)

then

I

We first remark some

- dpz

=

F*(t)u

+ H*(t)q(t)

v(0) =

-

Po@(O)

204

CHAPTER V

Next Q 1 1 h.h

-

2(QI2h,q)

=

- $(T).h

+ 2 h.v(T)

=

Set

hence

B

i s solution of

and

^*

Th as the solution of an optimal co n t r ol problem, namely (2.22), (2.23). Moreover (2.13) we can assert Therefore we can compute that

QT(.)

= C

205

FILTERING AND PREDICTION

(2.24)

ph.h

The notation P(T)

=

P(T)h.h

=

inf J ( q ( . ) ) h

.

means the covariance of

We can now solve the control problem ( 2 . 2 2 ) , ( 2 . 2 3 ) Let us consider the system in

‘T ’ BT

(2.25)

then we have

Lema 2 . 1 . The system ( i n aT,BTI ( 2 . 2 5 1 has one and only one solution. Moreover we have (2.26)

4,

(2.27)

P(T)h

= ich = - R-*(t) H(t)aT(t)

=

- aT(T).

We write the Euler equations for the control problem ( 2 . 2 2 ) , ( 2 . 2 3 ) . Let

9, , BT

the optimal control and state, then we have

+ IT Q(t)BT(t).E(t)dt where

=

0

bi q

2

L (o,T;R~)

206

CHAPTER V

- dt ” _ = F*(t)E - H*(t)q(t) Define

uT by ( 2 . 2 5 ) where

BT

E(T) = 0

.

has the meaning of the optimal state, we

deduce from (2.28)

+

iT (F(t)aT(t)

-

dclT ).g(t)dt dt

=

0

and by integration by parts

and since

q

is arbitrary we see that

Hence (2.25) has a solution and we have ( 2 . 2 6 ) . Consider a solution of ( 2 . 2 5 ) , then

3,

of the control problem ( 2 . 2 2 ) , (2.23). hence also

“T’

Let u s compute

and from (2.25)

is necessari.1~the optimal state Therefore it is uniquely defined,

201

FILTERING AND PREDICTION

hence we have (2.29)

P(T)h.h

=

-

.

aT(T).h

T (T) = aT (T;h) and remark that the function h linear. Then using the symmetry of P(T) we have

Write

ct

P(T)(h+k).(h+k)

=

-

+

aT(T;h+k).(h+k)

P(T)h.h

=

-

aT(T;h).h

P(T)k.k

=

-

aT(T;k).k

P(T)h.k

=

- 7 (aT(T;h).k + aT(T;k).h)

hence 1

It is easy to check that (from (2.25))

hence P(T)h.k

=

from which follows ( 2 . 2 7 ) .

-

aT(T;h).k

.

aT(T;h)

is

208

3.

CHAPTER V

RECURSIVITY. KALMAN FILTER

Consider now by analogy with (2.25) the system (3.1)

We have the following

Theorem 3.1. zd

E

2

We make the assumptions Gf Theorem 2 . 2 w i t h 2 E L (o,T;R”). Then system ( 3 . 2 1 has m e and on’,$ one

L (o,T;R~) , f

soZution

yT, pT

and

By analogy with Lemma 2.1, we see that system (3.1) consists in the Euler conditions of an optimal control problem.

(3.3)

- _d A=

dt

F*(t)>

- H*(t)

(q(t) + R-’(t)zd{t))

X(T) = h

with the payoff

-

2 x.X(O)

This problem i s the following

.

209

FILTERING AND PREDICTION

The optimal control 9 (3.5)

G(t)

is given by

=

-

.

R-'(t)H(t)y,(t)

Define next

Y

hence the pair

-

yT,pT

is solution o f

(3.7)

-

- H*(t)R-'(t)(zd

-YT(0)

=

- Po PT(0)

Now from (3.7) and (2.25) we deduce

and from (2.26) we see that

,

H(t)y)

-pT(T)

= 0

.

2 10

CHAPTER V

Y,(T).h

(3.8)

=

(zd

-

H(.)Y(.),?$)

-

H(.)?(.))

therefore

-yT(T)

(3.9)

=

CT(zd

which with ( 2 . 2 7 ) completes the proof of ( 3 . 2 ) .

0

We write (3.10)

r(T) = r(T;zd) = iTzd + y(T)

-

iT H(.)?(.)

then from ( 3 . 2 ) we see that (3.11)

yT(T)

=

-

r(T)

P(T)h

.

Our next task will be to obtain evolution formulas for the pair r(.) P(.).

3.2.

&arsjye-formulas

We will describe a decoupling argument on system ( 3 . 1 ) .

L e m a 3.1.

Functions

(3.12)

yT(t)

,

pT t )

are r e Z a t e d by t h e r e l a t i c n

yT(t) = r(t) - p t)pT(t)

v

t

E

C0,TI.

proos be convenient to denote by y t) ~ ~ , ~ ( the t ) solution of T,h to emphasize the dependance in h Consider next system ( 3 . 1 ) (3.11, on the interval (0,t) instead of (0,T) with final condition pT(t) It wil

instead of (s)

h.

More precisely consider the functions

with

Yt,PT(t)

(s)

,

s 2 t.

Since Y ~ , ~ ( s ), P ~ , ~ ( s )satisfy this system and since the solution is unique, we can assert that

21 1

FILTERING AND PREDICTION

(3.13)

Pt 'PT (t) But by definition of

(s)

=

pT(s)

for

s 5 t

r,P we have

which with (3.13) proves ( 3 . 1 2 ) .

;emu 3 . 2 .

he fitnetion

P(t) c H'(O,T~(R~;R"))

alzd is the unique solzrtion o f t h e eqzration. o f R i c c a t i type

(3.4)

1

P(0)

P 0

=

?roo _" By Lemma 3.1 applied to system ( 2 . 2 5 ) we may assert that

(3.5)

aT(t)

=

.

-

P(t)BT(t)

=

1 DaT.aTdt + PoBT(0).QT(O)

Now from ( 2 . 2 4 ) we have P(T)h.h

T

+

CT QBT.BTdt

where we have set (3.6)

D(t)

=

H*(t)R-'(t)H(t)

From (3.5) we obtain (dropping index T (3.7)

P(T)h.h

=

. on aT,BT)

T (PDP+Q)B.Bdt + PoB(0).B(O)

j 0

.

212

CHAPTER V

Next from the second equation (2.25) we get

- da = dt

(3.8)

(F*-Dp)a

B(T) = h

.

Define ( 3 9)

OP(t,S

9

F(t)

=

fundamental matrix associated to

-

Then the solution of ( 3 . 8) can be expressed as

which, going back to ( 3 . 7 ) yields

hence (3.10)

By analogy with (1.20), (1.25), formula ( 3 . 1 0 ) proves that differentiable in fact that i f c1

P

t

and that ( 3 . 4 ) holds.

is

Uniqueness follows from the

is a solution of ( 3 . 4 ) then defining

by ( 3 . 5 ) then the pair

in a unique way.

P(t)

6 by (3.8) and

a,@ is solution of (2.25) and thus defined

Applying (3.5) at time

T

shows that

P (T)h = P2(T)h

if

Pl,P2 are two solutions. Since h

uniqueness follows

is arbitrary, as well as

T, the

3

213

FILTERING AND PREDICTION

The f u n c t i o n

L e m a 3. 3.

r

E

1 H (0,T;R") and i s t h e unique s o l u t i o n o f

+ f(t) + P(t)H*(t)R-'(t)zd

dr = (F(t)-P(t)D(t))r(t)

(3.11)

r(0)

=

x

.

proos By (3.12) it is clear that

r

E

1 H (0,T;R").

Also

dt =

F~

-

Q~ + f + (FP+PF*

-

PDP+Q)~+

+ P(-F*~-D~+H * R-1 zd)

=

(F-PD)~+ (F-PD)P

P

*R-1

+ f + PH

zd

i.e. (3.11). Moreover r(0)

=

y(0) + P(O)p(O)

= x

which completes the proof of the desired result.

3.3.

Falr!ian-fjlter

Let us consider the observation process (1.26). z(t) where

b

is a

= Jt

H(s)y(s)ds

+

We have

it R''2(s)db(s)

Zt standard Wiener process with values in Rp. We

consider the S.D.E.

214

CHAPTER V

(3.12)

d9

((F(t)-P(t)D(t))?(t)

=

t

P(t)D(t)y(t)

+ f(t))dt

t

+ P( t)H*( t)R-’/’ (t)db (t) 9(0) = x

.

(3.13)

+

IT Op(T,t)P(t)H*(t)R-’/’(t)db(t)

.

Since dz = H y dt

R 1 ” db

t

we can rewrite (3.13) as (3.14)

?(T)

=

+

Op(T,O)x

+

JT

{T

O,(T,t)P(t)H*(t)R-’(t)dz(t)

Op(T,t)f(t)dt

t

.

Now from (3.11) we also can assert (3.15)

r(T)

=

Op(T,O)x +

T Oh(T,t) (f(t)+P(t)H*(t)R-’

(t)zd(t))dt

and (3.16) Take x

r(T).h =

0

,

f

(3.17)

=

0

= (i;h,zd) + y(T).h

then

7

=

0

-

(i;h,H(.)F(.))

, hence

^*

r(T1.h

=

(ZTh,zd)

r (T) =

iT O p ( T ,

and from (3.15) t) P (t) H* (t) R-’zd (t)dt

.

FILTERING AND PREDICTION

which compared to (3.17) yields (3.18)

23

=

.

R-l(t)H(t)P(t)O~(T,t)h

It follows from formula (3.19) that

Let us show that (3.21) Indeed let @(T)

Op(T,O)x

Op(T,t)f(t)

+

=

y(T)

- iTOp(T,t)P(t)D(t)y(t)dt

to be the right hand side of (3.21).

Differentiating in T

, we get

hence -do- - (F - PD)e + f dt O(0) = x

therefore O(T) which proves (3.21).

=

Qp(T,O)x

+

JT Op(T,t)f(t)

Therefore from (3.21) and (3.20) we see that

Qp(T,O)x + which with (3.14) shows that

T 0

Qp(T,t)f(t)dt

=

y(T) - IT H(.)Y(.)

215

216

CHAPER

v

This formula compared to (2.1) shows that of

Y(T)

is the best estimate

y(T).

W e thus have proved the

Theorem 3 . 2 . Ve make the asswnptims of Thecrem 2 # i . Then zi7.e conditional ezpectation 9(T) of y(T) given z ( s ) , O<s
d9

=

(F(t)-P(t)H*(t)R-l

+ f (t)dt +

(t)H(t))F(t)dt

3

9(0) = x

Let

(3.24)

ds = (F

E(O)

=

E

u(0) = 0

*

-

PD)E dt + Gdw - PH R

*

.

-1/2

db

217

F I L T E R I N G AND P R E D I C T I O N

Lema 3 . 4 .

The process

vaZues in Rp

Since v(t)

-

, where

Z

w(t) t

=

is a

Zt standard Wiener process w i t h

o(z(s),s
,

E(S)

is independant of Zs

w(s)

is independant of Z s , for

The process

u(t)

ti s

,

it is easy to check that

t

2

s

.

is clearly Gaussian. We consider the pair

as solution of a system

(dd) @(

(3.26)

=

td):

+ '4 dg

where

@ =

The covariance matrix

(

F-PH*R-' H R- 1 /2H

II of the vector

(:)

is solution of

We can write

in fact

n,2

= 0

.

Indeed w(t)

is adapted to

Zt since

E (t ),

w(t)

218

CHAPTER V

(3.28)

dv = R-”2H(y-9)dt

=

Since

E(t)

hence

nI2

R-’l2(dz

-

i s independant of

.

H 9 dt) t

,

E(t)

v(t)

and

a r e independant,

.

0

=

+ db =

Developping t h e n ( 3 . 2 7 ) w e o b t a i n

il =

(F

-

+ GG*

n2 Hence

II1

= P

PH*R-’H)~,

n l ( ~ *-

H * R - ~ H P )+

+ PH*R-~HP

,

= I

+

rI1(0)

=

which w e a l r e a d y knew, and

Po

,

II,(O)

= 0

I12(t) = I t .

.

T h i s completes

0

t h e proof o f t h e d e s i r e d r e s u l t .

W e c a l l Innovation process t h e p r o c e s s

hence I ( t ) = Jt R’/2(s)dv(s)

(3.30)

and ( 3 . 2 3 ) can b e w r i t t e n a s t h e S . D . E . dy = F ( t ) F ( t ) d t + f ( t ) d t + P ( t ) H * ( t ) R - ” 2 ( t ) d v ( t )

(3.31)

I Remark 3.1.

9(0)

=

The p r o c e s s

x

. 9(t)

i s c a l l e d t h e Kalman f i l t e r .

K(t) = P(t)H*(t)R-1/2(t)

i s c a l l e d t h e g a i n of t h e f i l t e r .

The m a t r i x

FILTERING AND PREDICTION

Remark 3 . 2 .

219

The system (3.1) is the system of Euler conditions of the

control problem (3.3), (3.4). problem as follows.

We can associate to it an other control

Consider

(3.32)

where the control is the pair

(w,v(.)).

The payoff is defined by

+ 2 h.u(T). Then it is easy to check that

yT

is the optimal state, pT

the adjoint

variable, and that the optimal control is given by

5(t)

=

G = - P

For h

=

-

G*(t)pT(t)

O

p(0). T

0 the interpretation of problem (3.32), (3.33) is clear.

the l e a s t square f u n c t i o n a l .

It is

This gives an other justification to the

role of quadratic optimization problems in filtering. Note however that (3.33) makes sense only when

Po

is invertible, which is not requested

0

for (3.3), (3.4).

4. PREDICTION We have computed so far

T 9(T) = ECy(T)(Z 1.

We are now interested by

220

CHAPTER V

which can be easily deduced from the knowledge of

(1.1)

Indeed, from

we have

+

Since

F(T).

Zt c

at

and

w

$+'@(T+B,t)G(t)dw(t)

is a 3

t

.

Wiener, we have

hence

e > o

(4.1)

Hence the prediction of the future state can be easily deduced from the Kalman filter. COMMENTS ON CHAPTER V 1. For the study of general Riccati equations in Hilbert spaces, cf. J.L. Lions [ Z ] , S.K. Mitter [ I ] ,

R. Temam [l], L. Tartar [ll, A . Bensoussan R. Curtain

-

-

M. Delfour

A.J. Pritchard C11, M. Sorine 111.

2 . They are efficient methods to compute the solution of a Riccati

equation, using Chandrasektar's ideas, cf. (among others)

J. Casti 111.

22 1

CHAPTER VI VARIATIONAL METHODS IN STOCHASTIC CONTROL

INTRODUCTION In this chapter we consider the point of view of optimality necessary conditions, which corresponds to Pontryagin's Maximum principle in the case of deterministic systems. We study here the most simple situations. One of the main difficulties to treat the theory in full generality, is that there may be several forms of the stochastic Maximum Principle. The most satisfactory treatment of a general situation is probably that of J . M . Bismut [ I ] . Here we restrict ourselves to controls entering in the drift term only. We also consider some problems of control with partial information and present the separation principle.

1. MODEL WITH ADDITIVE NOISE

Let

g : Rn

(1.1)

Let

X

Rm

X

g

(Q,Q,P)

C0,TI

+

Rn

be a function g(x,v,t)

such that

is measurable

be a probability space.

We consider b(t)

stochastic process with continuous trajectories, 2 with values in Rn, sup Elb(t)/ < CT 0s t
.

222

CHAPTER VI

5 R.V. with values in Rn

(1.3)

Let

Uad

Rm. We consider stochastic processes v(t)

be a subset of

are adapted to g t with values in Uad

which

and square integrable. More

precisely (I.3)bis

v(.)

For any given v(.)

Lemma 1 . 1 .

v(t)

<(O,T)

E

E

Uad

,

Uad closed convex in Rm ,

we solve the equation

There is one and only one s o l u t i o n o f (1.4) which belongs t o and

L2(Q,Q,P;C(0,T;Rn))

is an adapted process.

x(t)

proos Let

y(t)

=

-

x(t)

,

b(t)

then

(1.5)

Then (1.5)

has a unique solution in L2 (R,G,P;C(O,T;Rn))

Indeed define for

z

E

I(t;z)

11zI

I t2

= E

=

5 - b(0) + J t g(z(s)

-

sup Iz(s)l2 0s s< t

B

.

,

B

lI(t;zl)

if

=

I(t;z2)/

,

then

=

K Jt

Iz,(s)

+ b(s),v(s),s)ds

- z2(s)1ds

E I(z)

223

and

Consider

then i t is easy to check that yn + y being unique. Furthermore for any t

fixed point of

I, the fixed point

Define next (1 *6)

X t = o algebra generated by

It is important to notice that

Xt

x(s)

, 0

5

s 5 t

depends on the control.

.

Therefore

these 0-algebras are not given a priori. Let u s set (1.7)

2(* = {v

Once we fix a control v

E

2 L (O,T)/v(t) 3

in

$,

E

Uad

a.s., a.e. t}

.

this defines the a-algebra

define

The set

is not empty, for instance deterministic controls belong t o v

We pose the following control problem

.

224

CHAPTER V I

(1.9)

where

h

(1.10)

: Rn

+

R

,

h

Lipschitz, bounded.

Our objective is to study problem ( 1 . 9 ) .

1 . 2 . Admlty-r~~~~t The set 2; being indirectly defined, problem (1.9)

seems a priori quite

intricate. However we have the following very important property

Theorem 1.1.

Assume (l.l),

(1.11)

(1.2),

InfEh(x(T)) V

=

( 1 . 3 ) , (1.10). Then ue have

Inf*Eh(x(T)) V €3

€2t

proos We shall need several Lemmas.

Lema 1.2. (1.12)

Let

v

then

E

wt = TJt

v t

Ad

proos Indeed

x'

c

Ext

.

Let us prove the reverse inequality. Since

(1) We can add an integral payoff. I

225

VARIATIONAL METHODS

5 is

xv0

measurable.

Next

- 5 - it g(x(s),v(s),s)ds

b(t) = x(t)

x:

and the right hand side is

Zt c

measurable, therefore Let u s consider v E ? ( * with

x:

and

measurable, hence

,

L e m a 1.3.

vk

Clearly vk(t) xk

i~

=

vo

0

for

Uad

E

for

nk

5

=

5

+

zi

,

(n+l)k

measurable.

Vk

ft g(xk(s),vk(s),s)ds

be the 0-algebra generated by

We have 5

<

.

be the trajectory corresponding to

zk

t

.

xk(t) and

...,Nk , vk (t)

0

For

0

2

+

b(t)

xk(t). t < k

hence (1.13) Then for

5

t

2

k

,

1

5

n

5

N-1

' "ad

vO

-k f (n-1)k nk v(s)ds

Let

.

hence (1.12)

O,k,

t is X v

b(t)

b(t) k

2 t <

2k ,

is

x ,t

measurable for

0

5 t 2

k

I

.

226

CHAPTER VI

=

Jk v(s)ds

and sin e v E Q * , vk( ) is g k measurable for k s t < 2k. However from (1.13) it follows that gk Therefore vk(t) is X k k measurable

=xL.

for

k

S

t < Zk, since

By induction

(1.14)

b(t)

measurable

is

V t

hence ;Ft =zL , from which the desired result follows.

Lemma 1.4. xk*

in L’(RP,P;c(o,T;R~))

x

proop We know that

vk

-f

v

in

2 L3(0,T).

Denote

From the relation

and since

E we get

{

T /v,(s)-v(s)l

2

ds 4 0

.

221

VARIATIONAL METHODS

ci

from which we conclude the proof of the Lemma.

Since ?Ac 2(* we have

We can then proceed with the proof of Theorem 1 . 1 . infEh(x(T))

But for any

v

E

V&

%,*, there exists

corresponding trajectory xk

.

inf*Eh(x(T))

t

v&

+

x

,

vk

E

Q , such that denoting by

xk

the

then 2

in L (Q,G,P;c(o,T;R”))

.

But then

Therefore infEh(x(T))

5

Eh(xk(T)) V v

+ Eh(x(T))

E %(*

hence infEh(x (T)) VRi

5

inf*Eh(x (T)) v 4

0

which concludes the proof of Theorem 1.1.

From Theorem 1 . 1 , problem ( 1 . 9 ) , side of (1.11).

it follows that if

u(.)

is an optimal control for

then it is also optimal for the problem on the right hand

228

CHAPTER VI

We first start with a differentiability Lemma

Lemma 1 . 5 .

Let

u,v

E

2 I+(O,T)

and

8

of (1.4) corresponding t o t h e c ontrol

cont r ol

u+&.

(1.15)

E

Let

C0,ll.

u, and

x8(t)

y

be the s o l u t i o n

corresponding t o the

We assume g

continuously d i f f e r e n t i a b l e w i t h respect tc

and

g.',g{

uniform'oy continuous i n

x,v

Then we have

Xe(.

(1.17)

where (1.18)

We have

hence

and

)-y(. ) A

z

+

z(.)

in

c(o,T;R")

i s t he s o l u t i o n of the iine ar equation

x,v

and bcunded.

a . s . and i n

229

VARIATIONAL METHODS

which proves ( 1 . 1 6 ) . We next notice that

and from ( 1 . 1 5 )

we have, with ( 1 . 2 0 )

+

{

t

g:(y(s),u(s),s).v(s)ds

Setting

it follows from (1.23)

and (1.18) that

+ Jt

ds

.

230

and

CHAPTER VI

gi

, :g

are bounded, then

deduce from ( 1 . 2 4 ) ,

is bounded.

p(6)

Therefore we first

(1.25)

It follows from ( 1 . 2 6 ) that sup /xg(s)-y(s)l OSs5T

+

0 a.s.

6

as

+

0

hence from ( 1 . 2 5 )

IT

e

ds

+

0 a.s.

which with ( 1 . 2 4 ) and Cronwall inequality implies s u p /x,(t)I

--t

o

a.s.

as

0

-t

o ,

OSt5T

which proves (1.17). Set (1.27)

J ( v ( . ) ) = Eh(x(T;v(.))

.

We assume (1.28)

h

is continuously differentiable and :h

continuous and bounded.

i s uniformly

23 I

VARIATIONAL METHODS

We have

CoroZZary 1. I. Assume ( 1 . 2 8 ) . The functional J(v( . ) ) differentiable on L 2 (0,T) and 3

where

z

i s t h e solution of ( 1 . 1 8 ) .

We have

hence

From Lemma 1.5 and assumption ( 1 . 2 8 ) ,

If we introduce the adjoint system

we conclude.

is Gateam-

232

CHAPTER VI

I

then it follows from (1.30) and (1.29), (1.18) that

We can then state the following

Uad is conuez cLosed.

(1.32) L et

u(.)

to be

( 1 -33)

ayL

optimal control; u(.)

J(u(.))

=

inf

J(v(.))

EZ(

and

.

V(*)&

Then the following necessary condition holds (1.34)

I

E

p(t).g,'(y(t),u(t),r)(v-u(t))

t

0 a.e. t

,

a.s. w

According to Theorem 1.1, we also have

Since Uad

is convex, Q*

is convex. Therefore from the differentiability

property ( 1 . 2 9 ) , we can assert that (1.35)

( J ' (u( . ) ) ,v( .)-U( , ) ) 2' 0

4' V( . )

E

u* .

233

VARIATIONAL METHODS

then we have

(1.37)

X(t;b) z 0

a.e.

a.s

Indeed, let us set

v(t;h)

v

=

in

A

u(t;w)

outside A

.

This choice is possible, since is

A(t)

st

measurable

Y t

hence At

=

{hih(t;w) < 01

belongs to

But

v(t)

=

v

in

u(t)

*t

outside

st .

234

CHAPTER VI

which shows that

v

eq:

1A h(t;w)dt dP

,

v(.)

Taking this choice of

we get from (1.36)

0

2

which contradicts the definition of

A

A , unless

has measure

0. We

thus have (1.37). Since

u(.)

Remark 1 . 1 .

st =

e Q , we have

vt ,

vtp(t)

is not simple in practice, which

E

The computation of

0

hence property (1.34).

renders the necessary condition (1.34) difficult to apply.

[?

2 . THE CASE WITH INCOMPLETE OBSERVATION 2.1.

statement_of_the_erob1”[11

We consider the following problem.

The state of the system is governed

by the equation (2.1)

!

dx = F(t) x (t) + B(t;v(t))dt x(0) =

+ G(t)dw(t)

5

with the assumption (2.2)

F matrix

n

x

n

measurable and bounded in

G

m

x

m

I

w

matrix

t

,

Zt Wiener process, standard and m dimensional

5 R.V. with values in Rn , 3’ measurable, Gaussian, with mean B(t;v)

x

: (0,m)

subset of

ilk

and covariance matrix

x

Lad

+

R”,

where

Uad

*

is a closed convex

, B measurable bounded.

235

VARIATIONAL METHODS

The process

is the x x t i w l p r o z e s s .

v(t)

Let as in (1.3) Chapter V

Then we can solve (2.1) and obtain a process which is continuous and adapted to

Zt .

We next define the observation process by

dz

(2.7)

z(0)

=

0

H

matrix

dn

=

p

R

zt

(2.11)

x

I = x

the O.U.E.

=

Wiener, standard, p

8(t),

y(t)

t.

0

R- 1

bounded,

dimensional and independant

defined by

d3

=

F(t)8

dt + G(t)dw(t)

dy

=

H(t)B

dt + dn

y(0) Set

q(0)

symmetric invertible

Consider now the processes

1

which is measurable and bounded in

5 , w(t).

of

(2.10)

x n

R”2(t)db(t)

where b

dt + dn

H(t)x

=

=

0

.

B . From (2.1) and (2.10) we see that x I is the solution of

236

CHAPTER V I

(2.12)

Similarly, setting

we have dz

1 = H dt

(2.13)

The processes

xl, z

xi

~ ~ ( =0 0)

contain the control effect, and

We assume in the sequel that (2.16)

st =

(2.17)

, sst)

.

in 2(* we consider the process

For any control v or ( 2 . 1 5 )

o(y(s)

and denote by

z:

z:

= o(z(s)

z defined by ( 2 . 7 ) ,

the o-algebra

,

sst)

.

We define in a way similar to ( 1 . 8 ) (2.18)

The set

=

{v & [ v(t)

zvt

measurable a.e.

is not empty (take deterministic controls).

controls problems as follows

where

is

t}.

We define the

237

VARIATIONAL METHODS

(2.20)

x,v

+

R(x,v,t)

e(o,o,t) (2.21)

h

(2.22)

B(t;v)

Remark 2 . i .

E

is continuously differentiable

L~(o,T)

is continuously differentiable

is continuously differentiable with respect to v

We need to introduce explicitely an integral criterion in

(2.19), since we cannot reduce it to the final criterion, by the usual trick

This restriction is due to the fact that we request a special structure of the state equation (2.1) (linearity with respect to

Lemma 2.1. For any

Since

v(t)

xl(t)

is

z,(t)

is

is

2

vt

v

t

E

Q, 2,

=

st

ti t

x ).

0

.

Zt and Zt measurable a.e. then from (2.12) follows that

andV

zt

and

Zt measurable. Therefore from

measurable Y t.

From (2.13) it follows that ( 2 . 1 5 ) it follows that

238

CHAPTER VI

t

is 2, measurable and

Y(t) and

z(t)

is

st

t

t

Hence Zv c 5

measurable.

Zt c z:.

c

This implies the desired result.

Q is derse in Q*

Lemo. 2.2.

Let

v(.)

%* and consider (as in Lemma 1.3)

c

(2.23)

v(s)ds

k '(n-1)k Then vk(t) c Uad

,

a.e.

a.s.

For

0 s t < k t

function. Hence from (2.14), Zt --Zv k t

2k, we have from (2.14)

For

k

zk

measurable.

6

5

Therefore

Vk k

It 5

zl(t)

for

for

vk(t) is Zk Vk

,

nk

z,(t)

t < (n+l)k

5

.

is a deterministic

O s t s k . is and

st

measurable, hence

Sk measurable for

2k.

t From (2.15) this implies again that 2, = st (t s 2k). k see that 2: = Zt , ti t Therefore vk belongs to 2( k

.

By induction we

.

But

c

VARIATIONAL METHODS

239

240

CHAPTER VI

A similar argument shows that

ci

This completes the proof of ( 2 . 2 4 ) . We can then state the

Theorem 2 . i . (2.201,

>/e a s s m e ( 2 . 2 1 , ( 2 . 3 ) , (2.4), (2.5), ( 2 . 8 / , / 2 . 9 / ,

(2.211, (2.22).

(2.27)

inf

Then 7;e have

J(v(.))

V(.)&

A s the proof of Theorem 1 . 1 ,

=

.

inf*J(v(.)) vn,

using Lemmas ( 2 . 1 ) ,

We start with differentiability Lemmas as in denote by

y

tion) still by

§

(2.29)

Let

u

E

;3

2 S;(O,T)

and

the corresponding state defined by (2.1), and (to save nota-

z

the corresponding observation, defined by (2.7).

1%

=

F y 1 + B(t;u(t))

dz

(2.30)

( 2 . 2 ) , (2.3).

1.3.

We write

where

(2.161,

_ _1

dt -

1'

Z ] ( O )

= 0

.

24 1

VARIATIONAL METHODS

Let

v

E

control

2 L (O,T), we will denote by

3 u+5v

,

x,(t)

x I a defined by (2.12) with

the state corresponding u

replaced by

co

the

u+&.

We define

-

-dy- - F(t)y dt

(2.31)

Y(0) = 0

L e m a 2.4.

The :unctionaZ

+ Bv(t;u(t))v(t)

. J(v(.))

is Gateauz di;lr"ereqtisble on

and r e have t h e formula

?TOO t^

I

We have (2.33)

Now

From assumption ( 2 . 2 2 ) , it i s easy to check, like in Lemma 1.5 that

<(O,T)

242

CHAPTER VI

Also

1

5

CT (iT1v(s)

I 2 ds) 112 .

We can use Lebesgue's Theorem to pass to the limit in ( 2 . 3 3 ) and to obtain ( 2 . 3 2 ) .

U

We define (2.34)

then we deduce from ( 2 . 3 2 ) the formula

We can then state the

Theorem 2 . 2 . We make the asssumptiomof Theorem 2 . 1 . be c solution o f (2.36)

J(u(.))

=

inf

J(v(.))

Let

u(.)

Q

E

.

V(.)4

Then t he f ol l owing necessary c o n d i t i o n h o l d s t

(2.37)

a m

(E?

Y v

~(y(t),u(t),t)+B:(t,u(t))OZ

E

2

Uad , a.e. t , a.s.

By Theorem 2 . 1 , we know that over %*.

t p(t)).(v-u(t))

u(.)

realizes the infimum of

Hence from ( 2 . 3 5 ) we can assert that

J(v(.))

0

(2.38)

Hence also ))dt t 0

Y v(.) € V * . Reasoning as in the proof of Theorem 1 . 2 , we deduce from ( 2 . 3 9 ) that

V V € U ad

' a'e*

€ 3 ,st = Z t

and since u(.)

a*s*

hence relation ( 2 . 3 7 ) .

We assume here that (2.40)

L(x,v,t)

1

h(x)

=

=

1

5 Q(t)x.x

+

1

7 N(t)v.v

+ g(t).x

1

Yx.x 2

where (2.41)

~

Q(t)

M

symmetric

N(t) g

and

symmetric bounded

symmetric bounded, invertible, N-' (t)

bounded

bounded,

244

(2.42)

CHAPTER V I

B(t;v) = B(t)v + f ( t )

where

(2.43)

(2.44)

B

bounded

f

bounded

.

dy = ( F ( t ) y + B ( t ) u ( t ) + f ( t ) ) d t + G ( t ) d w ( t )

Y(0) =

5

Assuming

(2.46)

Uad

= Rk

then ( 2 . 3 7 ) yields t

(2.47)

N ( t ) u ( t ) + B * ( t ) E?

p(t) = 0

and t h e o b s e r v a t i o n p r o c e s s i s g i v e n by

(2.48)

dz = H ( t ) y ( t ) d t + R ” * ( t ) d b ( t )

Replacing ( 2 . 4 7 ) i n ( 2 . 4 4 ) y i e l d s

Y(0) =

5

.

z(0) = 0

.

245

VARIATIONAL METHODS

(2.50)

P(t) = @*(T,t)My(T)

+

:1

.

@*(s,t)(q(s)y(s)+g(s))ds

Let us set (2.51)

t 5 s

Apply (2.50) with obtain

t = s

and take the expectation with respect to

t 5 s

Now from ( 2 . 4 9 ) we have for

t 5 s

Taking the expectation with respect to Z t c 5t

Zt.

Zt

and recalling that since

,

and

we obtain

From Lemma 2 . 5 below we get, considering F(t)

as given and

s

E

Ct,Tl,

We

246

CHAPTER V I

there exists a unique pair

(?$s),fjt(s))

solution of (2.52), (2.53) and

related by the relation

where

II(s), p ( s )

s = t

, we obtain

are deterministic functions. In particular, taking

Therefore going back to (2.47) we get (2.55)

u(t)

=

- N-'(t)B*(t)

(Z(t)p(t) + p(t))

Let u s now derive an equation for ?(t). (2.56)

dy

=

(F(t)y

y(0) =

5

-

where

B(t)

Zt =

+ G(t)dw(t)

f

yl(t)

is

t

measurable

st

i s the Kalman filter of the free system.

according to the S.D.E.

i(0) = x

with

From (2.49) we have

BN-'B*(n(t)F(t)+p(t))+f(t))dt

We know that from (2.28) and the fact that

and since

.

Samely

act)

evolves

241

VARIATIONAL METHODS

(2.59)

dv

-1/2

=

R

(t)(dy(t)-Hg(t)dt)

=

R-1/2(t)dI(t)

where

is the Innovation process. From (2.57), (2/58) and (2.29) we obtain (2.61)

d9 = F(t)?

-

dt

B(t)N-'(t)B"(t)(n(t)9(t)+p(t))dt

+

+ P (t )H* (t ) R-I /2(t ) du (t )

.

Y(0) = x

We can also rewrite (2.61) as follows

+ P (t) H* (t) R-I (t) (dz(t) -H( t)? (t) dt) 9(0) = x

by virtue of the fact that (2.63)

dI = dy

-

Hidt

What wehave proved is that if

=

dz

-

u(.) E Q

Hgdt

.

is an optimal control, then

necessarily it is given by formula (2.55), where

9(t)

is solution of

(2.61). Conversely, we may solve the equation (2.61) which is an equation in 9(t), with external input v(t), where v(t) is adapted to

Zt ,

g(t)

and

9(t)

is given by (2.59).

are adapted to

Since v(t)

Zt. Next setting

248

CHAPTER VI

(2.64)

Y,(t)

=

p(t)

-

i(t)

it follows from ( 2 . 5 8 ) , (2.61) that

yl

by ( 2 . 5 5 ) .

We define a control u

is solution of

Then

u

is adapted to

Zt . The

corresponding state is given by ( 2 . 5 6 ) . Using, ( 2 . 6 4 ) we see that

(2.66)

Y = Y 1 + B

therefore from ( 2 . 6 4 ) we deduce

9(t)

(2.67)

=

Ely(t)1Zt]

Next considering the observation (2.68)

dz

=

,

z

defined by

H y dt + dr,

z(0)

=

0

we still have by ( 2 . 6 6 ) dz

=

H y l d t + H 3 dt + dq

=

H y 1 dt + dy

= H q d t - H E d t +dy

Replacing

dv

by this value in ( 2 . 6 1 ) , we obtain that

9 is solution of

( 2 . 6 2 ) , which has a unique solution, and the. solution is a process

adapted to u(.)

E

Zt . From that and the form of the control we obtain Zt = Zt , wich with ( 2 . 6 7 ) shows that

?.(,and

249

VARIATIONAL METHODS

The control u ( . )

satisfies the necessary (and sufficient by convexity)

optimality conditions. Therefore we have obtained Theorem 2.3. (2.421,

We make the a s s - m p t i o n s of Theorem 2 . 2 and (2.401,

(2.431,

(2.41),

(2.46). Then t h e r e e x i s t s one and o n l y one s o l u t i o n o f t h e

control problem (2.361, g i v e n by (2.55), where

9(t) is

t h e unique

s o l u t i o n o f ( 2 . 6 1 / , ( 2 . 6 2 1 and corresponds t o t h e b e s t e s t i m a t e (2.69).

In the proof of Theorem 2 . 3 , we have used the following result.

Lema 2.5.

Consider t h e two p o i n t boundary value problem

Then i t has a unique s o l u t i o n

(y,p).

Moreover t h e f o l l o w i n g r e l a t i o n

ho Zds

II is t h e unique s o l u t i o n of t h e R i c c a t i equation

where (2.72)

and

p(t)

(2.73)

i s t h e s o l u t i o n of t h e Zinear equation

*

I ,RL. -

=

(F*(t)-n(t)B(t)N-'(t)B*(t))p

+ g(t) + U(t)f(t)

0

CHAPTER VI

250

PrOoJ It is a consequence if the standard quadratic control theory (cf. also Chapter V 5 3 . 2 ) . The state equation is governed by

1

(2.74)

=

x(0)

+ B(t)v(t)

F(t)x

=

Yo

+ f(t)

.

The payoff function is defined by J(v(.))

(2.75)

=

iTQ(t)x(t).x(t)dt +

+ 2

iT N(t)v(t).v(t)dt

There exists an optimal control denoted by

+

u(.)

JT

g(t) .x(t)dt

.

M x(T).x(T)

.

Let

+

y(.)

be the

corresponding optimal trajectory. The Euler condition of optimality is given by

iT Q(t)y(t).f;(t;v)dt

(2.76)

+ 1T g(t).x(t;v)dt

+

where

Defining

p

by the second differential equation ( 2 . 7 0 ) , where

y

stands

for the optimal trajectory, we get from integration by parts in ( 2 . 7 6 )

J and since v (2.77)

T

(Nu + B*p).v dt

=

0

is arbitrary u(t) =

-

from which we obtain that

N-l(t)B*(t)p(t) (y,p)

is a solution of (2.70).

The solution

25 1

VARIATIONAL METHODS

is unique since the control defined by ( 2 . 7 7 ) will be optimal, hence unique, therefore

y

then

p

are unique.

Considering next a system similar to ( 2 . 7 0 ) on the interval

(s,T)

with

initial condition arbitrary and reasoning like in Lemma 3.1 of Chapter V, we define functions lT(t),

p(t)

and we have the relation

Taking in particular the homogeneous system (2.79)

a(0) = h

we have (2.80)

hence

and to

F

-

BN-lB*Il

we can associate a fundamental matrix

such that (2.8

Next from ( 2 . 7 9 ) we deduce

@,(t,s)

252

CHAPTER VI

hence using (2.80), (2.81)

We a l s o have, in a similar fashion

Now from (1.11) Chapter V, the function 1 H (O,T;Rn), and (2.84)

d O*(T,t)h dt

=

t

+

O*(T,t)h

II

-(F*(t)-II(t)B(t)N-l

belongs to

(t)B*(t))@i(T,t)h

II

from which we deduce that

On(T,t)h

E

H 1 (0,T;R")

and

But from (2.83) we deduce

and differentiating with respect to

t, using (2.85) we obtain

253

VARIATIONAL METHODS

i.e. (2.72).

Knowing the equation of

get the equation for

L and using (2.78) and (2.70) we

p.

Uniqueness of the solution of (2.72) follows from an argument similar to that of Lemma 3.2, Chapter V.

Remark 2 . 2 .

0

Hence the result.

Control problem (2.74), (2.75), corresponds to problem (2.l),

(2.19) when there are no noises and when assumptions (2.40). satisfied.

. . (2.43)

are

Combining (2.77) (2.78) we see that the optimal control for the

deterministic control is given by the formula (2.86)

u(t)

=

-

N-l (t)B*(t) (II(t)y(t)+p(t))

.

Therefore the optimal control can be computed via a feedback r u l e . Formula (2.55) shows that for the corresponding stochastic control problem, the optimal control is given by the same f e e d b a c k , where the state y(t)

must be replaced by its best estimate

observations up to time

t.

f(t)

,

given the

This is a very interesting property, which is

known as the s e p a r a t i o n p r i n c i p l e . First of all the optimal control at time

t

can be computed through a d e t e r m i n i s t i c f u n c t i o n of the

best estimate of the state of the system at time

t. More than that this

0

function is the same as in the deterministic case.

3. SEPARATION P R I N C I P L E Orientation We have seen that, for the linear quadratic stochastic control problem with partial observation, considered in 5 2.4, then the optimal control at time

t

is a deterministic function of the best estimate.

We shall see

in this section that this property holds under more general assumptions.

254

CHAPTER V I

However the fact that the deterministic function coincides with that of the corresponding deterministic problem does not extend beyond the case treated in

§

2.4, M. Wonham [ l ]

was the first to obtain such a

property.

We consider the situation of Theorem 2.1. ?(t)

Let v

E

q.

We consider

to be defined by

dP

(3.1)

=

+ B(t;v(t))dt

F(t):(t)dt

+ P(t)H*(t)R-’(t)

+

(dz-H(t)S(t)dt)

S(0) = x

which defines uniquely S

,

assuming v

given and (2.7), (2.1)-

Equation (3.1) shows that

S(t) is

(3.2)

(3.4)

,t

measurable

L

5

E(O)

=

x(t)

- S(t)

S(t)

=

=

B(t)

-

i(t)

hence

and since Zft

=

t

2,

,

t

ELx(t) 15

for v

E

4,

I

we have

V t

.

255

VARIATIONAL METHODS

(3.5)

a(t) = ECx(t)

l c l

which means that the solution of (3.1) is the Kalman filter.

A_dynnrnic_erosralllrin¶-~9~~~ion

3.2.

We will study here the equation of Dynamic Programming which is connected to our control problem.

It will be introduced a priori and

justified in the next paragraph. We need some notation.

We consider the second order differential

operator

(3.6)

-a(t)

where the matrix

is defined by

Y

(3.7) and

a(t)

F..(t) 11

=

1

,

P(t)H*(t)R-l(t)H(t)P(t)

is the matrix

F(t).

In order to avoid degeneracy we will

assume

-a(t)

(3.8)

t a1

,

a >

0

.

Let us remark that even when we assume ( 3 . 8 ) , the operator

is not

X(t)

completely standard, since the drift term (first order term

X F..(t)x

is not a bounded function).

j

E(t)

is the solution of ( 3 . 3 ) .

variable with mean

Note that

0 and covariance matrix

is invertible we have the explicit formula (3.10)

LJ

j

g(x,v,t) = E E(x + E(t),v,t)

(3.9)

where

We also set

P(t).

E(t)

is a Gaussian

Therefore when

P(t)

256

CHAPTER VI

Similarly we define (3.11)

K(x)

=

E h(x + E(T))

.

We introduce next the Hamiltonian (3.12)

E(x,p,t)

=

[x(x,v,t) + p.B(t,v)]

Inf

.

VEUad We now consider the non linear Cauchy problem (3.13)

-3 at

+ x(t)u

u(x,T)

=

-

E(x,Du,t)

=

0

.

K(x)

We will assume (to simplify some technical points) (3.14)

From ( 2 . 2 2 )

Uad and ( 3 . 1 4 )

compact

.

it follows that

We start with an existence and uniqueness result. Since we are dealing with a parabolic problem in the whole space (Cauchy problem), one expects to use the Sobolev spaces with weights, introduced in Chapter 11, 5 3 . 1 . However, since the coefficients of the operator x(t) are unbounded au because of the terms - Fij Xj (see ( 3 . 6 ) ) , the weights

5

are not suitable. This i s due to the fact that

257

VARIATIONAL METHODS

therefore

4J

is boundzd as

0 as

such a ratio tends to

1x1 1x1

+ m.

We will need weights for which A way to achieve that is to use

+ m.

polynomial weights such as

then (3.19)

hence (3.20)

...

We define Sobolev spaces LG HA

We first notice that if A(t)

A(t)

-

=

For instance

is an operator of the form

a

a

C - a. .(x,t) - + C a i axj i,j axi u i

a ax.

with the usual assumptions on the coefficients (in particular the non degeneracy assumption of the

a

ij

)

then we can solve the equation (in a

unique way)

- ?!.a t

(3.21)

u

with data

f

We begin with

E

2

E

+ A(t)u

=

,

f

L 2 (O,T;Hn) I 2

L (O,T;LR) ,

E

2

Ln

I

u(x,T)

=

258

CHAPTER VI

Lema 3 . 1 . Let

f

E

2 L2 (O,T;Ln)

, ii

E

2 Ln

.

Let

a..(t) 1J

such t h a t

(3.23)

be a matrix

measurable and bounded.

g

Then there e x i s t s one arid only one s o l u t i o n o f (3.24)

u(x,T) u

E

=

u(x)

1 L 2 (O,T;Hn)

.

props Since F(t)x

is not a bounded function we cannot a p p l y the variational

theory as for ( 3 . 2 1 ) . We first assume in addition that (3.25)

f ,u

are bounded.

Let us define (3.26)

Pk (x)

=

x

if

1x1

2

k

T?

if

1x1

2

k

.

259

VAR IATIONAL METHOD S

(3.27)

This is an application of the variational theory, since all coefficients are bounded.

We also have the estimates (cf. Chapter 11,

§

3.4)

1 2 Let u s then obtain an additional estimate in L (O,T;Hn); multiply (3.27)

by

uk7;

(3.29)

and integrate of

Rn.

-1 A- u (t) 1; 2 dtl k -

.f

We obtain +

zJ

aij (t)

Rn

F(t)Pk(x).DukukIl 2dx

au au k

p2 ax. ( ax.k J

- 1 Rn

Rn

1

~ . D u II2dx k k

2 2u II sx.

'

k

+

i

2)dx

1x1

=

2 f u k 7 d x .

=

But from (3.28) we can assert that

Therefore we deduce from (3.29)

(3.30)

i IUk/

2 5 c . L (O,T;HP)

We can then consider a subsequence still denoted by uk

-f

u

uk

,

such that

2 1 in L (O,T;HU) weakly and in Lm(Rn

x

(0,T))

-

260

CHAPTER VI

weak star. Since F(t)Pk(x)

-t

F(t)x k

pass to the limit in (3.27) as

in L L ( O , T ; L f I )

strongly, we can

and obtain a solution of (3.24).

-t m,

Let u s now obtain an a priori estimate for such a solution. Multiplying (3.24) by

uI12 and integrating over

Rn

yields

(3.31)

- / F(t)x.Du Rn

u

Ti2 dx

-

2 g.Du u II dx

.f

=

Rn

=JfuI12dx. But

/

F(t).x.Du

1

2

2

2

u II dx = j F(t)x.- 2 D(u )Il dx =

Rn

Rn =

-

1

tr F(t)u 2 JI2dx + Rn u

+ s /

Rn

2 F(t)x.x ~

2 Il dx

1+/xI2

Here we have been using the special property of the weights (3.20).

From this and (3.31) we deduce

where the constant C Therefore we obtain (3.32)

II, stated in

depends only on the Lm

bound of

g , F , aij.

26 1

VARIATIONAL METHODS

where

C

depends only on the

choice of

Lm

g , F, aij, on

bounds of

ci

and the

s.

From this estimate and the linearity of ( 3 . 2 4 ) we deduce uniqueness, and also the fact that the existence and uniqueness result extends to data as

0

in the statement of the Lemma. Remark 3 . 1 .

We do not have the analogue of Theorem 3 . 6 of Chapter 11.

Indeed if we multiply ( 3 . 2 4 ) by

at

I?

, then we get terms of the form

aU au II2 dx 1 F.IJ.(t)x j ax:L at which cannot be estimated as in the case

L e m a 3.2.

ij

-u

j axi

I?dx

.

0

We assume

(3.33)

Me take d a t a

[ F. .(t)x

aij f

E

E

C

0

LP(O,T;LToc(Rn))

,G

E

W:i’(Rn)

,

2 s p <

a.

Then t h e s o l u t i o n o f 1 3 . 2 4 ) s a t i s f i e s (3.34)

We proceed as in the proof of Theorem 3.7 of Chapter 11. Let @

E

&Rn).

Set

z = @u aZ

(3.35)

then from ( 3 . 2 4 ) it follows that

‘

i,j

I

aij

vj

-- O ( f + F(t)x.Du

? ,

z

satisfies

+ g.Du)

n

-

262

CHAPTER VI

Let 8

Q

be a regular bounded domain containing the support of

and

g

on the right hand side of (3.24).

Since from Lemma 3.1, u z

E

#2'1(QG),

Hence

4 , there is no difference between

Note now that because of the term F(t)x

$.

sob.

vanishes on

z

QG

=

-&

2 2 2 2 L ( 0 , T ; L (a$)), c L (0,T;L (0 ) ) , we have

E

$

8$ X (0,T) , by Corollaryl3.1 of Chapter 11.

We can then argue as in the proof of Theorem 3 . 7 of Chapter 11, to

0

complete the proof of (3.24). We turn now to the study of (3.15).

Theorem 3 . 1 .

We have the following

We assume (2.20/, (2.21), ( 2 . 2 2 ) , ( 3 . 6 / , ( 3 . 1 4 ) and ,-

(3.36)

a ij (t)

(3.37)

h

E

E

Co(CO,TI)

W2;Zc(Rn)

,

.

Then t her e exists one and o n l y one soZutiGn o f (3.171 such t h a t

We will use the policy iteration technique.

is continuous (') in

t

for

x,v,p

with respect to fixed.

(1) in fact differentiable

Let

$

x,v,p E

Let us first note that

and measurable (Lebesgue)

1 L2 (O,T;Hn) , we will write

263

VARIATIONAL METHODS

z9(x,v,t)

(3.39)

??(x,v,t) + [email protected](t,v)

=

I

u

Then the function L measurable in x,t.

@

is continuous with respect to v

and (Lebesgue)

It is thus a Caratheodory function. Since Uad

is

compact then there exists a (Lebesgue) measurable function v (x,t) such $

that N

(3.40)

H(x,D@,t)

(x,v (x,t),t)

=

Q

$

a.e. x,t ,

The proof of this property can be found in Ekeland-Temam [ l ] , Chapter VIII. Let us construct an iterative sequence as follows; uo

is an arbitrary

function satisfying properties stated in (3.38). For un

known, define vn(x,t)

=

U

the linear equation (3.41)

, and let un+ 1 be defined by

v .(x,t)

sun+'

z

at

i,j

a;

F(t)x.Du"+'

=

-

= R(x,v"(x, t), t)

nt 1 u (x,T)

=

h(x)

n+ 1 For the existence and uniqueness of u we can apply Lemmas 3.1 and 1 3.2. Indeed h E Hn for s large enough, and h WZ;:(Rn) for any S

Moreover setting

m

it follows from (3.17) that

does not depend on

n).

also

is bounded (note also that the L

bound

Next setting

f"(x,t) it follows from (2.20)

g

=

T(x,vn(x,t

that

f

E

2

L2(0 r;Ln)

for

s

large enough and

p.

264

CHAPTER VI

/f"(x,t)I <

(3.42)

the constant C for any

C(lXl2

+ 1)

being independant of

n, hence

f"

LP(O,T;LToc(Rn))

E

p. is well defined and

Therefore we can assert that the sequence un satisfies (3.43)

and

Q

x

=

(O,T), where 8 is an arbitrary bounded

regular domain of

R"

.

Let next proceed as in the proof of Theorem 3 . 2 of Chpater IV.

-

n+l

F(t)x.D(u

n -u ) s D(un+'-un).B(t,vn)

(u"+'-u")(x,T) We multiply by

n+l n + 2 (u -u ) TI

I (un'l-un)+(t) hence

n+ 1 (u -un)+

=

=

0

.

and integrate over

;1

< :1

R".

1 (un+'-un)+(s)

0. Therefore the sequence u"

the second estimate ( 3 . 4 3 ) we see that

un

We obtain lids

is decreasing. By Q V 8 , hence

is bounded in

we may assert that (3.44)

un

C

u

2

1

point wise, in L (0,T;H )

d"yp(Q)

We have

II

weakly for any

By compactness we may also assert that

p

2

2

,

weakly and p<m.

V A R I A T I O N A L METHODS

265

(3.45)

Let v

E

Uad

arbitrary, we have

-

Dun.B(t,v)

<-aU”at

S

-

C a. .(t) -- F(t)x.Dun IJ ax.ax. i,j 1 J

- l(x,vn,t) -

- Dun.B(t,vn) and from (3.41)

F(t)x.D(un+l-u“)

+

+

0

in

LP(Q)

+ D(un+l-un) .B(t,vn)

weakly

V Q , V p

by virtue of ( 3 . 4 4 ) , (3.45). Therefore letting n aU

at

and since v

+ m,

we obtain

-

1 a. ~-a 2 u ij axiax. i,j J

is arbitrary

F(t)x.Du

- x(x,v,t)

-

Du.B(t,v)

266

CHAPTER VI

We obtain a reverse inequality arguing as in the proof of Theorem 3.2, Chapter IV. Let us now prove uniqueness. Let

u1,u2 be two solutions and

1 2 v ,v

controls such that

v

i

,

v $x,t)

=

i

= 1,2

.

U

We have

-

1 aU

at

2 1

a u c a.ij - F(t)x.Du axiax

i,j

1

=

j

1 1 1 R(x,v ,t) + DU .B(t,v )

-

=

fi(x,Du2,t) + D(u 1-u 2).B(t,v 1 )

2

hence -a( u - u1 ) +1 at 2 1

2-u1 )(x,T)

From this we deduce 1 2 u = u

Hence

Remark 3.2. x,t

=

0

a2 (u2-u') + F(t)x.D(u 2-u1 ) + ij axiax.

J

i,j

+ D(u -U ).B(t,v

(U

-a

C

1

) 2 0

.

u2-u1 t 0. A reverse argument shows that

.

It follows from (3.38) that

ul-u2 2 0.

0

is a Holder function of with an arbitrary hElder constant <1 , on Q UO 0

Let us assume that (3.46)

Du

.

261

VARIATIONAL METHODS

and lB(tl;v)

(3.47)

-

B(t2;v)I

5

Citl-t216

V V € U ad ’ tI,t2 c i 0 , T l

.

From ( 3 . 9 ) and the assumption (2.20) it follows that

From the definition ( 3 . 1 2 ) we then deduce

where

60

=

Yin(&,I/2), and

C

is a constant independant of the

arguments. From ( 3 . 4 9 ) and Remark 3 . 2 it follows that

(3.50)

-H(x,Du,t)

is H6lder in x

and H6lder in subsets of We can then state

Rn

t x

with an arbitrary constant,

with constant

[O,T].

60, uniformly on compact

268

CHAPTER VI

We make t h e assumptions o f Theorem 3 . 1 , as w e l l a s 13.461

Theorem 3.2. and (3.51)

h

(3.52)

F. .(t)

E

1J

Then t h e s o l u t i o n

u

fio,

C2+'6)

E

C 6 ([O,Tl),zij

E

y > 0 , Y'

1

C'(C0,TI).

of ( 3 . 1 3 1 s a t i s f i e s

8 r e g u l a r bcunded donain of Rn.

Set

z = u$

where

@ E B(Rn).

Then

z

satisfies the equation

(3.54)

From Remark 2 . 2 ,

( 3 . 5 0 ) and ( 3 . 5 1 ) , ( 3 . 5 2 ) it follows that the right hand

side of ( 3 . 5 4 ) is hb'lder in x,t

with adequate constants, and

h@

Rn.

E

C2+'

on compact subsets of

an initial condition which is

We can consider

z-h$

0. We can then assert that

z

to obtain is

twice continuously differentiable in x with second derivatives hb'lder, on Q$ , and once continuously differentiable in t, with hb'lder -

derivative on

.

Q@

This f o l l o w s from the result of Ladyzenskaya et alii, mentionned in Chapter 11, Since

q?

§

3.3.3.

is arbitrary we have in particular the property ( 3 . 5 3 ) .

0

269

VARIATIONAL METHODS

With the assumptions of Theorem 3.2, the function L"(x,v,p,t) continuously differentiable in x,v,p

and holder in

There exists a Borel function <(x,p,t)

such that

9

It is hard to give more properties on

is

t.

because the infimum in v

in (3.12) is not necessarily unique. A reasonable candidate for an optimal feedback is the following

(3.56)

0(x,t)

=

<(x,Du(x,t),t)

but this is just a Borel function with values in

"ad

However we would like to solve the following equation (3.57)

where

dy

=

F(t)9

dt + B(t;O(F,t))dt

4

is a continuous process having a Ito differential, assumed to

z

be given.

We would like also to obtain a unique solution in order to

assert that the solution t

family 7

9

is an adapted process with respect to the

generated by the process

z(t).

This unfortunately cannot be

achieved with so weak properties of the function 0. This difficulty motivates the very restrictive following assumption (3.58)

c(x,p, t)

is continuously differentiable with respect to

x,p, with partial derivatives bounded on sets 1x1 5 M

, 1pI

5

M , t

E

C0,TI.

270

CHAPTER VI

When (3.18) is satisfied a s well as the assumptions of Theorem 3.2, then we have (3.59)

C(x,t)

is continuously in

on sets of the form ixI

5

M

x, with derivative bounded

,

t

[O,T].

E

To simplify a little bit we assume that (3.60)

K(t)

=

P(t)H*(t)R-l

is differentiable with bounded

(t)

derivatives. Then (3.57) can be rewritten as an integral equation and solved for each sample. Namely (3.61)

F(t)

=

x +

It

(F-KH)(s)?(s)ds

+ K(t)z(t)

- bCt

+ Jt B(s,O(F,s))ds 0

+

.

ds z(s)ds

The existence and uniqueness of the solution of (3.61) follows from the assumptions and the theory of integral equations. Clearly the process

Y(t)

We now consider a system in equation for (3.62)

z.

is adapted to

9,i

which consists in adding to (3.61) an

Namely we consider

?(t)

=

x +

it (F-KH)(s)?(s)ds

+ K(t).?(t)

.?(t) = I(t)

+

-

+ Jt B(s,;(?,s))ds 0

irt ds i(s)ds

.I

(1)

it H(s)?(s)ds

This system has a unique solution adapted to

Zt.

(?,?)

which are continuous processes

Zt (the o-algebra generated by y(t)).

Of course we also

have (:)

To have a more standard formulation one could make the change of functions

9 - K.? =

$.

27 1

VARIATIONAL METHODS

9 is adapted to -t

(3.63)

.

Therefore from the second equation ( 3 . 6 2 ) we deduce

$.

(3.64)

it

(3.66)

dy = F y dt + B(t,C(t))dt

=

Y(0) =

where

E

5

is the solution of ( 3 . 3 ) . z(t)

+ G dw

From which it follows

it H(s)E(s)ds

=

{t H(s)?(s)ds

+

=

it H(s)F(s)ds

+ I(t)

+ n(t)

which with the second equation ( 3 . 6 2 ) implies (3.69)

z(t)

Hence the process

2

2(t)

=

coincides with the observation process.

and ( 3 . 6 4 ) follows that (3.70)

G E V

.

I

From that

212

CHAPTER VI

0

We are now in a position to show that payoff

J(v(.)),

is optimal with respect to the

given by (2.19). bJe make t 4 e a s e wn p t i o n s of Theorem 3 . 2 , and (3.52),

Thecrem 3 . 3 .

Then zhe admissible c m t r o l defized (3.65) is an cpt?:maZ contro’o

t h e pa3of.f

Let

u(x,O)

v

=

Rn

and

.

inf J ( v ( . ) )

be an arbitrary control v

domain of

;doreover m have

J ( v ( . ) ) , deplned: b . ~( 2 . 1 9 ) .

(3.71)

E

(3.60). fgr

Q. Let CS. be a bounded regular f(t) from 8. We recall that

the exit time of

T

the Kalman filter (3.1) satisfies the Ito equation df

(3.72)

?(O) where

v

is a

3

t

=

F f dt + B(t,v(t))dt

=

+ PH* R -112 dv

x

Wiener process.

We apply Ito’s formula to the function u solution of (3.13) and to the process 8. This is possible since u t C2 9 1 .

We have

We apply this formula with

t

=

T

A

T

and take the expectation, we obtain

273

VARIATIONAL METHODS

u(x,O)

(3.73)

E u(P(TAT),TAT)

=

+ E

H”

and from the definition of (3.74)

u(x,O)

S

JTAT

Assume for a while that h

and ?R

-t

m

E f

? =

it follows that

J0 TATR(x(s),v(s),s)ds

are bounded functions, then u

fTAT

TAT^)

+

u(?(T) ,T) =

-R(x(s),v(s),s)ds

+

K(P(T))

a.s.

x(x(s),v(s),s)ds

a.s.

0

hence by Lebesgue’s Theorem u(x,O)

(3.75)

5

E K(f(T)) + E

IT x(f(s),v(s),s)ds

.

But recall that f ( s ) = x(s)-e(s)

and

E K(f(T))

=

E h(x(T))

E h(x(T))

=

E h(f(T)+E(T))

=

ECE(h(f(T)+e

=

E K(f(T))

since

since

E(T)

is independant of

. is

where OR i s the ball of radius R,

we have

U(P(TAT~)

+

E(Du,s)(f(s),s)ds

(cf. (3.12)),

R

TAT B(s,v(s)).Du(2(s),s)ds

0

E u(~(TAT),TAT) + E

bounded. Note that if as R

-

(T)) I $) 1

gT and P ( T )

is

gT measurable.

274

CHAPTER VI

is 3'

Similarly since v(s)

measurable

E X(?(s),v(s),s)

=

E R(x(s),v(s),s)

.

Therefore we obtain from ( 3 . 7 5 ) (3.76)

u(x,O) i E h(x(T)) =

Ix,(s)

I

5

C

T

R(x(s),v(s),s)

h,2

are bounded functions.

Consider

Remembering that

x ( s ) = X1(S)

and

E

J(v(.))

at least provided we assume that now the general case.

+

+

B(s)

a deterministic constant, since

(cf. (2.12))

-dxl - - F x 1 + B(t;v(t)) dt X1(O) =

0

and B(s)

=

O(s,O)x + O ( s , O ) g + I s O(s,o)G(o)dw(o)

.

Using assumption (2.20) (3.77)

hence (3.78)

where

c1

depends on

Co

and some fixed constants. Now consider a

sequence Qk of bounded functions such that

215

VARIATIONAL METHODS

Rk (x,v,s)

+

R(x,v,s)

pointwise

and

Such a sequence exists. A s

k

-+ a,

it follows from the estimate (3.78)

that

E

(3.79)

JT

Rk(x(s) ,v(s) ,s)ds

-+

E

JT

R(x(s) ,v(s) ,s)ds

Similarly we may consider a sequence of bounded functions hk hk(x)

Now let uk

-+

h(x)

. such that

pointwise

be the solution of (3.13) for the data

Rk, hk.

From the estimates of Theorem 3.1, we have

Using in particular the second estimate (3.49), it is easy to check then that

k

2

1

uk

+

u

in L (0,T;H )

Uk

-f

u

in u ~ ~ ~ ~ P (v QQ)

r!

In particular u (x,O) + u(x,O)

.

weakly

.

But from (3.76)

276

CHAPTER VI

and passing to the limit, we see that (3.76 holds in general. Consider now the control t to u

defined in (3. 5 ) .

Applying Ito's formula

9 , yields

and to the process

There is a difficulty in passing to the limit in a term like

E U(?(TAT~),TAT~) as R

+

-,

To overcome this, consider the equation aU

k Z

IJ

i,j

at

=

-a. .(t)

2 k

a U axiax

F(t)x.Duk

k - Du .B(t,?(t))

=

j

-k

R (x,O(t),t)

k

-k

u (x,T) = h (x)

.

Note that in the above eauation 0 are approximations of

u.

depends on

Moreover

.Ek, hk

R,h as seen before.

It is easy to see that for instance k u (x,O)

-+

But from the boundedness of

u(x,O)

.

hk, ik, one may establish

k k u (x,o) = E h (y(T)) and the right hand side tends to

which with (3.76) shows that u ( . ) the proof of the desired result.

+

E

J(u(.))

{T as

k R (y(t),?(t),t)dt k

-+ m.

Finally we obtain

is an optimal control, and completes

0

277

VARIATIONAL METHODS

Remark 3.3. There are ways to relax the assumptions ( 3 . 5 8 ) , but it is at the price of enlarging the class of admissible controls (see A. Bensoussan-

0

J.L. Lions C11).

COMMENTS ON CHAPTER VI I

1 . The method of density developed in 5 1 . 2 is due to M. Viot CSl.

See also A. Bensoussan 2 . The adjoint process

EZt p(t) does

=

II(t)

n(t)

-

p(t)

Viot E l ] .

M.

is not an adapted process.

Of course

is adapted. An interesting question is the following:

satisfy a S.D.E.

One can show that

II(t)

satisfies an

equation with an additional term which is a martingale (see A. Bensoussan

-

G. Hurst

- B.

Naslhd C11).

This leads naturally to

another question. Can one obtain a stochastic Maximum principle with an adjoint system whose solution is an adapted process? This problem was solved by T.M. Bismut C11.

Unfortunately the equation for n(t)

involves additional martingale terms.

So it is not certain that this is

preferable to a form like ( 1 . 3 4 ) .

3. The proof of Theorem 3.1 given here simplifies that of A. Bensoussan

J.L. Lions C l l . on

x

The reader will check that we can take a depending ij with appropriate assumptions. But the equation that we want

to solve has coefficients which do not depend on x . 4 . For the separation principle in cases of degeneracy, cf. J.L. Menaldi

[11.

-

This Page Intentionally Left Blank

219

CHAPTER V I I

PROBLEMS

OF OPTIMAL STOPPING

INTRODUCTION Optimal stopping problems were first introduced by H. Chernoff [ I 1 for applications in Statistics and soon related to the Stefan problem (problem of the free boundary of ice melted in water), see P. Van Moorbecke [ I ] . The method of Variational inequalities was introduced by A. Bensoussan - J.L. Lions C31. For a full treatment of the applications of V.I. in Stochastic Control, cf. A . Bensoussan J.L. Lions [l]. For a "more" probabilistic approach and connections with the problem of excessive elements, see A . Shyriaev [ I ] and earlier work of E. Dynkin [3]. See also N . Krylov [ I ] . We have restricted ourselves to stationary diffusion processes and general semi groups. For other processes and models (cf. A . Bensoussan - J.L. Lions [I], M. Robin [ I ] , R. Anderson - A. Friedman [ l l , i21, J.L. Menaldi [21). For numerical techniques to solve V.I. R. Tr6moliSres [ 11).

(see R. Glowinski

1. S E T T I N G OF THE PROBLEM

Let

aij(x), bi(x)

(1.2) Let

bi u(x)

be functions on

E

such that

Lm(Rn)

I

Rn

such that

-

J.L. Lions

-

280

CHAPTER VII

We note

Px

the solution of the martingale problem relative to

with initial condition x

(0,a)

0. We set

at

then as already seen (1.5)

PX(x(0)=x)

(1.6)

@(x(t))

There exits a

n;,t

Let

8 be a

(1.8)

not

1

- @(x)

+

it A@(x(s))ds

is a

x +

=

Jt

+ Jt b(x(s))ds

a(x(s))dw(s)

m0t

martingale.

w(t)

such that

Px,

standard n dimensional Wiener process

x(t)

(1.7)

=

,

a.s. Px

.

stopping time, we define a functional

.JX(e)

=

-

EXCJeA' f(x(t))(exp

+

$(x(e))xeCT

exp

Jt ao(x(s))ds)dt

- '1

i

ao(x(s))dsl

where (1.9)

T =

first exit time of the process

bounded regular domain of

(1.10)

a.

(1.11)

f

(1.12)

+

t 0

E

Rn

bounded

LPb)

,

P > " 2

COG)

,

$ir

t

o , r

=

a

We are interested in characterizing the function (1.13)

u(x)

=

inf Jx(S)

e

and to find optimal stopping times.

0.

x(t)

from 8 , open

28 I

PROBLEMS OF OPTIMAL STOPPING

We will formulate an "equation of Dynamic Programming" for the function u(x).

As it will be seen, it will not be an equation but a variational

inequality. We will also according to the plan followed in Chapter IV, consider a semi group formulation of the stopping time problem.

2 . UNILATERAL PROBLEMS

We write here the operator A

in variational form

with

We consider the Hamilton Jacobi equation

(2.3)

This is indeed a Hamilton Jacobi equation, as we have seen in Chapter IV,

(2.8).

Indeed we define

It clearly satisfies all the assumptions of the Hamiltonian H in Chapter IV, ( 2 . 4 ) , (2.5). H(~,o,o) c L~

,

since

It may not satisfy the property

required

282

CHAPTER VII

(u-$)(Au+a 0u-f) = 0

,

ulr = 0

.

0

283

PROBLEMS OF OPTIMAL STOPPING

Lema 2 . 1 .

Setting

without l o s s of g e n e r a l i t y we may assume

u = wz

are led to the

,

where

same

is defined in ( 2 . 1 7 ) of Chapter 11, then we

z

problem with the following changes

aij

+

2

aij 2

a.

+

a. w

. a

+

a . w2 +

1

w

1

'

f

aw c a.1 ax.

+

a (a. aw c axi ij K)

w

i,j

j

fw

and

U

hence the desired result.

Lema 2 . 2 . SoLUtiGn

Let

X Zarge enough.

Then the re e x i s t s one and only one

G f

(2.8)

Au

+ (a0+X)u

5

f

(u-q)(Au+(ao+A)u-f)

, =

u 5

0

proos We recall that

+ /

uvdx

8. a

ulr

)I

, u

E

W2"(8)

=

0

284

CHAPTER VII

A

Then we consider (2.9)

large enough so that

a(v,v)

A lv12

+

2

6 11v11

V v c H 01 .

Let us prove uniqueness. Let

-

u u

be two solutions then we have

J

(Au + aou

+

Au

-

f ) (u-u)dx =

J

(Au + a , u

+ hu

-

f) (9-u)dx s 0

8

0

Similarly

By addition

(2.10)

AU E +

UElr

-

aOuE

0

t ~u~ + 1 (uE +)+

, uE

E

w2q$

.

= f

.

PROBLEMS OF OPTIMAL STOPPING

We m u l t i p l y by

and t h u s f o r

((u'-'$)+)'-'.

A

285

We n o t e t h a t

l a r g e enough

T h e r e f o r e from ( 2 . 1 0 ) we deduce

hence

But t h e n from e q u a t i o n ( 2 . 1 0 ) and Theorem 2 . 1 of Chapter 11, w e deduce t h a t

(2. Let then

uE

be a subsequence such t h a t

U'

hence a l s o i n

(2.13)

Wiyp

+

u

in

W2"

strongly.

weakly

From ( 2 . 1 1 ) i t follows t h a t

U S $ .

From e q u a t i o n (2.10) w e can deduce t h a t

286

CHAPTER VII

hence letting (2.14)

tend to

E

we obtain

Au + (ao+A)u 5 f

We multiply (2.10)

by

J

(2.15)

But

,

0

(uE-+)-

-t

(uE-+)-.

( A U ~+

8

(u-+)-

=

.

Therefore

(ao+A)u E

I$- u

-

in Lp

f) (u'

-

+)- =

strongly.

o

.

Therefore from ( 2 . 1 5 )

we deduce that

which together with ( 2 . 1 4 ) and ( 2 . 1 3 ) implies (Au + (a +A)u - f ) (u-i) = 0 a.e. 0

Hence

L e ma 2 . 3 .

into

f

0

is a solution of ( 2 . 8 ) .

u

f

Let us consider t h e s o l u t i o n @

where

@ E Lw.

N

u

o f 12.81 w i t h

f changed

Then we have the estimate

(2.16)

N

Consider the penalized problems associated to u

E

,

"E

u

the corresponding solutions.

u, u

and denote by

By difference we have

and by Lemma 2 . 1 of Chapter IV, it follows that

from which and the convergence result of Lemma 2.2,

we deduce ( 2 . 1 6 ) .

0

281

PROBLEMS OF OPTIMAL STOPPING Proof o f Theorem 2 . 2 W

Consider for

z

L

E

the solution 5

of

(2.17)

Such a solution exists and is uniquely defined by Lemma 2.2. in particular

E

p >

4

Here we use 0 -

5 E C (8)I hence Thus by (2.17) we have defined a map

explicitely the assumption that Lm@).

to imply that

5 from Lm into itself. Let us show that it is a contraction. x Indeed take z1,z2 E Lm and c l = Sh(zl) , q 2 = S A ( z 2 ) . Then from S

(z) =

Lemma 2 . 3 we deduce that

which proves that

Sx

is a contraction. Since solution of ( 2 . 6 )

coincide with the fixed points of

Remark 2.2.

I3

S A , the proof is completed.

We say that problem ( 2 . 6 ) is a unilateral problem, since it

is defined by one sided conditions. The terminology comes from Mechanics where such problems are met frequently. Our motivation is completely different.

I7

3. VARIATIONAL INEQUALITIES

Assumptions ( 2 . 5 ) are two restrictive f o r the applicationsthat we have in mind.

We will relax them, but it will then not be possible to formulate

the problem like in ( 2 . 6 ) , since clearly regularity of the solution u requested in such a formulation.

is

288

LHAPTER V I I

We make the assumptions of Theorem 2.1, and the coercivity

Theorem 3.1. assumption

Then there exists one a d only one solution of the variational inequality

pi-ovided t h a t t h e set 1 K = { V E H Iv<$j 0

(3.3)

is not empty.

Proof Let

v

0

E

K

.

We multiply (2.3) by E

(3.4)

E

a(u ,u

uE

-

vo

then we get

- v 0 + 1 ((uE-$) + ,uE-vo

=

0 (f,uE-v ) .

But

hence

a(u

E

E

,u )

0

0

a(uE,v ) + (f,uE-v )

from which and the coercivity assumption (3.1), we deduce

< c .

(3.5)

Let

uE

uE

u

+

be a subsequence such that 'u in L~

strongly.

+

u

1

in Ho

weakly, hence also

289

PROBLEMS OF OPTIMAL STOPPING

We deduce from ( 3 . 4 ) and (3.5) ((uE-q)+,uE-vO)

0

+

hence ((u-qJ)

+ ,u-vo )

0

=

or

((u-q)+,u-qJ) 5 0

which implies Let now v

5

$.

(u-$)+

=

0

,

or

u

5

qJ.

We multiply ( 2 . 3 ) by a(uE,v-uE) +

-1

v-uE.

((uE-+)+,v-uE)

It follows =

(f,v-u')

.

But

hence

From the weak 1.s.c. of

a(v,v)

,

we deduce

a(u,v) 2 a(u,u) + (f,v-u)

u,u

hence

u

is a solution of ( 3 . 2 ) .

This solution is unique.

b e two solutions, then we have

Indeed let

290

CHAPTER V I I

and by addition -

N

a(u-u,u-u) t 0

-.

from which and the coercivity we deduce u = u

0

Problem (3.2) is called a uariational inequaZity (V.1.).

In the following,

we will weaken assumption (3.1) which is too strong for our applications. Clearly such an assumption requires that

a .

t

y > 0

some sufficiently

large constant. However since we do not want to rely on such an assumption, we will prove a result similar to Lemma 2 . 1 .

Lennna 3.1.

Without loss of generality, we may asswne

(3.6)

a o t y > O .

Proof Make the change of unknown function u = w z

.

Then we get the same problem with the following changes a

ij

a. 1

f

We recall that

1

+a

+

+

2

ijw

a.w2 1

wf

2 w S 2 ,

The result follows.

29 1

PROBLEMS OF OPTIMAL STOPPING

Let us now consider the problem a(u,v-u) + A(u,v-u)

(3.6)

B v X

For

1

,

Ho

v

5

$ ,

U E

I Ho

,u

5 $.

large enough, the assumptions of Theorem 3.1 are satisfied Therefore there exists one and o n l y one solution of

(namely (3.1)). (3.6).

E

(f,v-u)

2

We have the analogue of Lemma 2 . 3 .

Lema 3 . 2 .

Let

u

be t h e s o l u t i o n of (.3.6), and

corresponding problem w i t h

f changed i n t o

f + @

be t h e s o l u t i o n o f t h e

,

@ E

Lm.

Then we have

Since the penalized problems are the same as in section 2, we still have the estimate

Y

Hence E +

uE-UE

remains bounded in Lm.

+ u-u

0. Hence u"-';'

in Lm

However uE-EE

-+

u-u

in L2

0

( 3 . 7 ) holds.

Lema 3.3.

kssume

f,$ bounded.

I If1

L

+

L

-uLet $, 5

Then t h e s o l u t i o n of 13.61 s a t i s f i e s

/ / ~ / /m < A + yI

(3.9)

-u

lwLa *

be the solution of (3.6) corresponding to hence

as

weak star, from which it follows that

5

1 IqlI

a

L

.

f

=

0. Note that

292

Let

CHAPTER V I I

K

=

I~I) 1

We s h a l l prove that

m.

L (3.10)

-

u + K t O . Y

-

-

v = u+(u+K) as a test function in ( 3 . 6 ) , which is possible by the choice of K. We obtain

Take

or

K(ao+X,(u+K)-)

L

hence

(L+K)-

=

0

t

0

, which proves (3.10).

Using next Lemma 3 . 2 , we obtain ( 3 . 9 ) .

0

We can then obtain the

Theorem 3 . 2 .

and

f E LP@) the V . I . (3.11)

We assume 11.1), i1.2), 1 1 . 1 0 ) ,

,p

>

f.

v

provided t h a t t h e s e t

@

E

H K

1

~ v ,2 Q , u

0

,v)

E

H; n L~

u 5

q

defined i n 13.31 i s n o t empty.

Consider first the equation

a(u

E

Lm

Then t h e r e e x i s t s one and o n l y one s o l u t i o n of

* or

Let

a(u,v-u) t (f,v-u)

v

(3.12)

12.21.

=

(f,v)

,

293

PROBLEMS OF OPTIMAL STOPPING

AU

We know that

uo

0

+ aOuo

,

W2"@)

E

-

u = u-u

u

when

=

,

f

u

0

lT

hence since

.

o

=

p >

5 , uo

E

Now set

COG).

0

is a solution of (3.11).

;;

Then

is a solution of the

following problem

1 0

Y V E H

u

HA

E

n L~

,

v 2 $-u

,

-u

5

0

0 i-u

which is the same problem for data

f

=

0, and

$ = I$-u

O

E

L

m

.

Moreover

we set

KO

=

{v

since it contains vo-uo

1 0

H 1v

E

5

where v

0

I

$-u 0

5

9

is not empty

,

v

0

E

1

Ho.

Therefore without loss of generality, we may assume z

E

Lm, define

5

=

f = 0. Let next

as the solution of

Sh(z)

(3.13)

For

A

large enough, we may apply Theorem 3.1, to ensure the existence

and uniqueness of

5.

Moreover, from Lemma 3.3, we see that

Hence we have defined a map contraction. Indeed if

S,

z1,z2

from Lm

La

and

5

E

Lm.

into itself. This map is a c1,c2 are the corresponding

sohtions of (3.13), it follows from Lemma 3.2 that

294

But clearly the fixed points of (3.11),

when

Remark 3 . 1 .

S A coincide with the solutions of

0

f=O. Hence the existence and uniqueness.

When we make the assumptions of Theorem 2.2, we have one and

only one solution of (2.6) and also-of ( 3 . 1 1 ) .

These solutions

Indeed let us check that the solution u

coincides.

of ( 2 . 6 ) is a

solution of ( 3 . 1 1 ) . Indeed let v

E

1

Ho

,

v

5

$, we have

J0 (Au+aou-f) (v-u)dx and by Green's formula we see that u

=

J

0

(Auta u-f) (v-$)dx

0

satisfies ( 3 . 1 1 ) .

2

0

This justifies

the introduction of V.I. as a weaker formulation to ( 2 . 6 ) , when

$

Lema 3 . 4 .

Let

$,T

L m J and

E

corresponding t o them.

Let u s consider

' U

(3.15)

Au'ta

with (3.16)

X

is

0

not regular.

large and

Then one has

and

Q

u

u,u t o be the s o l u t i o n o f 13.121

0

E

LE

uE

Lm.

to be solutions of

+ AuE +

1 (u'-$)~

Then we have

=

f

, ~

€

=

10 ~

29 5

PROBLEMS OF OPTIMAL STOPPING Set K = Max(/

l$-@l

1

"

m,

L w

Y+X

J8

w+

.

l w E Ho

,

= uE--uE-K

We multiply (3.15) by (3.17)

lMl)

and (3.15)"

by

-w+

and add up.

CA(uE-UE) + a (uE-uE) + A(uE-GE)lw+dx + 0

-E1 X

+ dx

where

x Indeed assume

zE 2 ",)I w

-

= ((U"$)+

2

0

.

then "

C

(U"$)+,W+)

uE-$-K

,.,

< $-$-K

0

S

,

hence ((UE-qJ)',w+) which proves that

X

2

0

.

=

0

Now from (3.17) we deduce

a(w,w+) + X(w,w+) + J

8

hence

w+

=

0

.

[(a0+A)K+qlw+

Therefore uE-zE 2 K

.

By a reverse argument we conclude that (3.16) holds. Therefore considering the solution of

We obtain

dx < 0

=

296

CHAPTER VII

(3.18)

we can assert tha (3.19) Consider next the iterative process

n+l)

+ (f,v-u

When

. t , a ) I then the contraction argument mentionned in Theorem 3.2, guarantees that un Defining similarly

+

u

in L~

.

zn. It follows from estimate (3.19) that

(3.21) Letting a .

2

y

2

n

-t m

we deduce that (3.14) holds true, at least provided that

0. A s this stage, it is not useful to make the change of

unknown function u

= wz

, which changes

$

into $/w.

Indeed we will

only obtain estimate (3.14) with twice the right hand side, which is not the estimate we want.

One proceeds as follows. Consider the V.I.

291

PROBLEMS OF OPTIMAL STOPPING

6 > 0 , will tend to 0.

where

Let also &:

be the solution of the same problem with

1 4

changed into

$.

6 > 0 , we have

Since

And it is enough to show that

u6

-+

u

6

as

+

0

,

in some sense. For

such a result, we may consider the change of functions u therefore it is sufficient to assume

a .

2

y > 0

.

= uz

, and

Consider next the

iterative process

2

J,(un,v-un) 6 6

+

(f,"-Ut++

then we have

from which it follows that

>

k = h+y

with as

6

-t

.

In particular it follows that

u6

0. From this and the V.I., one deduces

is then enough to obtain

u

6

+

u

in

1

us

Ho weakly and

completes the proof of the desired result. We can then state the following regularity result

is bounded in bounded in Lw

Lm

1

Ho

.

It

weak star, which

c

298

CHPATER VII

Theorem 3 . 3 .

We make t h e a s s u m p t i o n s o f Theorem 3 . 2 and

(3.22) Then t h e s o l u t i o n

u

of (3.11) belongs t o

Co(&

.

Define

En = ll$n-j,l

I Lrn

,

En

-+

0

and

Clearly also

satisfy the assumptions of Theorem 2 . 2 .

The functions $n

Let

un

be the

It is also the solution of

solution of the V.I. corresponding to

@n. the unilateral problem ( 2 . 6 ) , hence in particular

u

E

Co(s).

But from

(3.14) we deduce that

lIun-uI Hence

u

E

0 -

.

C (0)

I L"

5

llQn-vl

I L"

*

0

*

0

299

PROBLEMS OF OPTIMAL STOPPING

Let u s prove to end that section, the useful result that the solution uE of the penalized problem converges to the solution

in

.

Co(s)

u

of the V.I.

This result will be generalized in section 5 for general

semi groups, with some slight changes in the assumptions.

We will

however need it in Chapter VII, section 3 .

Theorem 3 . 4 .

Under t h e assumptions ~f Theorem 3 . 3 , t h e n t h e s o l u t i o n

of 1 2 . 3 ) converges towards t h e s o l u t i o n

u

of t h e 7 . 1 . (S.11) i n

Let u s first remark that it is sufficient to prove this result when is regular.

Indeed let

as in Theorem 2 . 3 , @n

@n

Lemma 3 . 4 , we have (noting u

I /Un-U/I

(3.23)

+

$

in

the V.I. corresponding to

Co.

uE

Co(6)

.

y

From

on)

I lon-il I Lrn

5

L

But the proof of Lemma 3 . 4 , in particular estimate ( 3 . 1 6 ) , together with an iterative procedure for the penalized problem, like in ( 3 . 2 0 ) shows that the same estimate is valid for the penalized problem, namely

From ( 3 . 2 3 ) , ( 3 . 2 4 ) it is clear that if we have

1

lu:-unl

1

+

0

in

then the desired result will follow. a We may of course assume . Lemma 3 . 1 ) (1) Now for

)t

large, replacing

Lemma 2 . 2 , that

uE

2

C

0

, for n fixed We may thus assume

$

regular

y > 0 , without loss of generality (cf

a .

by

aO+X

(cf. ( 2 . 1 0 ) ) , we know fron

remains bounded in W2'p.

Now consider the iterative

sequence

(L) We have however to consider a penalized problem with EIUJ

2

.

E

changed into

300

CHAPTER V I I

hence

< - kn

-

k

where

NOW

=

A x+y .

1 /uE>OlI

I-k

Therefore letting p

+ m,

~< c ~, from , Lemma ~ 2.2; similarly

1

lu E, 1

< c .

Hence we have

IIu~-u~'~I I

(3.25)

2

C kn

.

Lrn We a l s o have

I lu-unt 1

(3.26)

< C

kn

L which follows from (3.25) and continuity and convexity of the norm.

It

also follows directly from (2.16) and an iterative scheme. Now for any fixed

since uE'n

remains in a bounded set of

depending on uE + u

in

n, we have by Lemma 2.2,

n).

COG) .

W2"

as

E

+

0

(a priori

From this and (3.25), (3.26) it follows that

0

30 1

PROBLEMS OF OPTIMAL STOPPING

4 . SOLUTION OF THE OPTIMAL STOPPING TIME PROBLEM

4.1.

Ihe-re9ular-case

We are going t o show the following

Theorem 4 . i .

We assume ( 1 . 1 1 , ( i . 2 ) , i l . i O / ,

(1.111, ( 2 . 2 ) , ( 2 . 5 1 .

the solution u

of 12.61 is given explicitely bg

(4.1)

u(x) = Inf Jx(6)

e

Then

.

Moreover there exists an optimal stopping time, characterized as follows. Define

and

then

6

If h

E

is an optimal stopping time.

Lp(&

,p >

5 , we know from Chapter 11, Theorem 4 . 1 ,

that

From this estimate follows that we can deduce the following Ito's the function u

integrated from formula

to

(4.4)

EX u(x(6A-c)exp

u(x)

=

+ EX

where

6

E

- J 6A.r

W2"

,

p >

ao(x(s))ds

4

+

JoAT (Au+aou) (x(s)) (exp-JS ao(x(X))dX)ds

is any stopping time.

302

CHAPTER V I I

Now using the relations ( 2 . 6 ) , it is easy to deduce from ( 4 . 4 ) that (4.5)

u(x)

5

V

JX(8)

8.

On the other hand we may assert that xc(x)

(Au+a u-f) = 0 p.p. 0

hence EX i8A?

xC(Au+aOu-f)

(x(s))

(exp-/’ ao(x(X))dh)ds

0

But for

s <

6

= 1

xc(x(s)) EX

-

(Au+aou)(x(s))(expf(x(s))

/OAT

and applying ( 4 . 4 ) with u(x)

=

8

set

C.

< T

, 6

=

(exp-

is a (x(h))dX)ds 0

=

0

LS aO(x(A))dX)ds

f ( x ( s ) ) (exp - is ao(x(A))dX)ds

u(x(8))~g,~

is finite hence

exp - /8ao(x(s))dsl

x(6)

+

.

belongs to the boundary of the

Therefore

which implies

when used in

.

8, we obtain

EX I

+

But if

0

, therefore

0

(4.6)

=

0

4.6)

u ( x ) = Jx(8

and this comp etes the proof of the desired result.

c

303

PROBLEMS OF OPTIMAL STOPPING

Theorem 4 . 2 . the s e t

Fie assume 11.11,

( 1 . 2 ) , 1 1 . ? 0 ) , 1 1 . 1 1 ) , i 2 . 2 / , (3.221 and

i s not empty. Then t h e s o l u t i o n u of 1 3 . 1 1 ) which i s a continuous f u n c t i o n on 0, can s t i l l be i n t e r p r e t e d by 1 4 . 1 ) . Moreover

6

K

defined by ( 4 . 3 1 is s t i l l an optimal stoppi n g t i m e .

Let

Qn be such as in the proof of Theorem 3.3, and let un

corresponding solution of the V . I .

be the

From Theorem 4 . 1 , we can assert that

where

J:(e)

=

EX [

But

hence

from which it follows that

f(x(t))(exp

-

Jt ao(x(s))ds)dt

+

304

CHAPTER VII

But

u

n

+ u

in Lm.

(4.7)

u(x)

=

inf Jx(S)

e

Let u s prove that If

Hence

u(x)

=

$(x)

boundary of

,

e

.

is an optimal stopping.

then since Px

C, hence

0 = 0

.

a.s.

=

0

hence

If

x

E

x(0)

=

x

,

we have

x(0)

on the

Therefore

u(x)

is optimal.

r

and

u(x)

=

, then ~ ( x )= 0 , and Jx(i)

$(x)

Therefore we may assume that

u ( x ) < v(x).

=

0

=

u(x)

6 > 0 such that

Let

u(x) < Y(X) - 6 and

Let

N,

such that

n t Nc:

implies

i

110,- $ 1 1 Therefore for

s S

6

e 6A T

<$

,

hence

1 /U,-U~ 1

6

< 8 ' ~~

, we have

Therefore 6 2 0 A T , hence as in the end of the proof of Theorem 4 . l , we may assert that

305

PROBLEMS OF OPTIMAL STOPPING

p

6

- is ao(x(X))dX)ds

AT (Aun+aoun)(x(s))(exp

0

6 =

EX

Ie

" f(x(s))(exp

- is ao(x(A))dh)ds

=

.

Therefore from Ito's formula

(4.8)

un(x)

EX un(x(B

=

6

AT))

6 exp - J8

6 + EX

Now since

u

n

+

u

(4.9)

-

ie " f(x(s))(exp

=

'E

6A T ) ) exp

u(x(e

0

6 I. A.

6

C

0

.

JS

ao(x(A))dX)ds

.

6

-

AT ao(x(s))ds

6

+ EX J e Let now

+

ao(x(s))ds

Lm, we deduce from ( 4 . 8 )

in

u(x)

AT

We have

6 I. and since

0

- is .a

" f(x(s))(exp

Now we have since the process

x

8

6

5

+

(x(h))dX)ds

.

-

0 , it follows that

is continuous

hence going to the limit

which implies that

A

t

6.

Hence

66 I.

6.

By passing to the limit in (4.9) we conclude the proof of the desired

D

result.

A s we have said in 5 2 . 1 ,

the penalized problem (2.3) is a Hamilton Jacobi

equation, with Hamiltonian ( 2 . 4 ) .

Therefore its solution can be inter-

preted as the infimum of a payoff corresponding to a stochastic control problem.

The control does not modify the martingale problem.

Therefore

306

CHAPTER VII

from formula (1.18),

Chapter IV and the definition ( 2 . 4 ) of the

Hamiltonian, we have

where

v(s)

is an adapted process such that

0

5

v(s)

s 1

.

We can assert that

(4.11) and there exists an optimal control, namely

One can greatly use the interpretation of the penalized problem in connection with the study of the optimal stopping time problem (we refer to A . Bensoussan

-

J.L. Lions [11 for details).

We will however use it to a

greater extent in the semi group approach developed in the next section.

Remark 4 . i . problem.

We have studied so far a stationary optimal stopping time

One can naturally consider evolution problems like in the

stochastic control case (cf. Chapter IV, section 4 ) . function $

case and lead to an evolution version of ( 2 . 6 ) . the function

The case when the

is smooth caries over by methods similar to the stationary

w

However, the case when

is not regular (especially with respect to time), then

we get a much more complicated situation. ' A treatment of this case will go beyond the scope of the present work.

We refer to A . Bensoussan

J.L. Lions [11 for a presentation of the results existing in this situation.

-

307

PROBLEMS OF OPTIMAL

Remark 4 . 2 .

In the context of mechanical applications the function

is called the obstacZe.

9

To understand the terminology, take the particular

case (in dimension 1) - u " < O

,

u - $ < O

u"(u-y)

=

0

u(0) = u ( l ) = 0

whose solution is drawn in the following picture.

This is the problem of tending an elastic string which passes by points

0 and 1 , below a given obstacle function $. The V.I. is also called the obstacle problem.

For the reader interested

in the applications of V.I. in Mechanics and Physics, we refer to G. Duvaut - J.L. Lions C11. 5 . SEMI GROUP APPROACH

Let u s consider the function z

satisfying

308

CHAPTER V I I

A function

verifying ( 5 . 1 ) will be called a lower s o l u t i o n of t h e V . 1 .

z

We make the asswrrpticns of Theorem 3 . 2 .

Theorem 5 . 1 .

The s o l u t i o n

u

Gf

t h e V . I . ( 3 . 1 1 1 i s a lower s o l u t i o n and t h e m a x i m element among lower solutions.

The fact that

u

is a lower solution is easy. Take

function in ( 3 . 1 1 ) .

Then u

for any

z.

Let

be the solution of

uo

AUO +

(5.3)

Then uo

a uo 0

=

satisfies ( 5 . 1 ) .

, uo

f

w2$P

E

v

,

u

0

lr

=

o

.

2 z

Let indeed

wo

the solution of

AW

0

,

+ (ao+X)wo = f + ~z

wo

E

1

' H

Ho

then

a(wo,$) hence from ( 5 . 1 )

+ x(W

0

-,$I

=

(f,c$)

it follows that 0

0

a(w - z , $ ) + X(w - z , $ ) Take

u

=

0 $ = (w - 2 ) -

0

2

,

V $

0

, then we obtain

-

0

0

a((w - z ) - , (w - z ) - )

-

x

-

c$ as a test

Now we want to prove that

j (w0-.z)-i

z

o

.

309

PROBLEMS OF OPTIMAL STOPPING

hence, since

X is large enough, (w0-2)-

=

.

0

Next consider the iterative sequence (5.4)

AW~+' +

(ao+A)w n+l - f +

wn

,

AW"

1 H2 n H~

6

By the regularity reasoning made in Chapter 11, Theorem 2.1, we know that n n-l for n 2 n w -w E Lm, and then (5.5) Lrn

provided that

a . t y, assumption which can be made without l o s s of

generality to prove ( 5 . 2 ) .

Hence wn-wnO

is convergent in Lm, from which

and the equation ( 5 . 4 ) , it can be easily deduced that wn

+

uo

solution of ( 5 . 3 ) in

1

Ho

weakly.

But let u s check by induction that n w t z . Indeed assuming the property at step n, then we get from ( 5 . 4 ) and (5.1) a(wn+'-z,$)

n+1

n

+ ~ ( w -w , q ) t

o

Y $ t

or

n+ 1 a(wn+l-z,Q) + ~ ( w - z , $ )

0 from which we deduce as for w , that Hence the result is proved for u

0

.

Consider now the iterative sequence

w

n+ 1

2

o 2 z

Y 4 t

.

o

o

310

CHAPTER VII

a(un+',v-un+*) + ~ ( u ~ + ~ , v - uz ~ + ~ )

(5.6)

2 (f

,v-un+5 + X(un,v-un+l)

1 ffvcHo,vS$,

n+l 1 n+l u c H O , u

We know from the contraction property of the map Theorem 3 . 2 (recall that the sequence un

S $ .

considered in

Sx

belongs to

Lm) , that

in L-. We check by induction that

(5.7)

U"LZ.

This is true at step 0. Assume it is true at step n, then t.ake v

=

n+ 1 u + (un+'-z)-

=

Max(un+',z)

+ a(u n+l ,(u n+l - z ) - ) 2

+ (f,(u

_<

@

n + ~ (n+l u -u ,(un+'-z)-)

n+ 1

and from (5.1)

therefore a(u

Letting

n

-z,(u

n+ 1

-z)-)

x(u"-z,(u"+]-z)-)

L

n+ 1 hence (u -2)-

n+ 1

=

0

tend to

, and thus

+ A(U

n+ I

-z,(u

n+l

-2)

L 0

( 5 . 7 ) holds at step

+-, we deduce (5.2).

n.

-

) 2

2

un

+

u

31 1

PROBLEMS OF OPTIMAL STOPPING

Let u s consider the semi group

then we can state the following Theorem 5 . 1 .

We make the assumptions of Theorem 3 . 3 , and

f

(5.10)

Then t h e s o l u t i o n

E

u

(Bore2 bounded cn

B

of t h e V.I. ( 3 . 1 1 1 s a t i s f i e s 0 -

(5.11)

UEC(0)

(5.12)

u

5

6)

Jt

,+=o f

Q(s)

x8

, U S $

ds + @(t)

e-atu

,

Y t t O . Moreover (5.111,

u

i s t h e m a x i m u m element o f t h e s e t o f f u n c t i o n s s a t i s f y i n g

(5.121.

Property (5.11) sequence un

is included in the statement of Theorem 3.3. Consider the in COG). From Ito’s formula

which approximates u

we have u

(X) =

EX

U

(X(tAT))

eXp

-

CltAT +

+ EX JtAT (Aun+aun) (x(s))e-asds hence Un(x)

5

EX Un(X(tAT))e

+ EX

JtAT f(x(s))e-usds 0

Since un

-+ u

in Co(&

,

we deduce

.

312

CHAPTER VII

Using the fact that

uiT = 0

,

we can write the above relation as

and by definition of the semi group

O(t)

(5.8), this is nothing other

than relation (12). Next let

w

E

0 -

C (0)

,

w 2 J

wir t

=

0 , w

x

O(s) f

2 $

ds + @(t)

8

w

Then we have

We now use the fact that Px

,

mt .

€!,: Ro

+

is a stationary Markov process for

x(t)

Then (cf. Dynkin [ I ] ) ,

Ro

consider the shift operator

such that

Note that

r(etw)

(5.14)

Now if

5

(5.15) Since x(t) (5.16)

=

rt(w)

=

infis>tlx(s)

do1

.

is a random variable, we write

"e

=

c(etw)

.

is a stationary Markov process we have the property EX[€ltc~mtl= EX(t)

6

.

We are going to apply that formula to the R.V.

313

PROBLEMS OF OPTIMAL STOPPING

5

=

w(x(s 2 - s 1 )AT))

e

-a(s - s ) 2 1 +

% SAT)) emas ds

s2 t

Then

8 5

-a(s2-s 1) =

w(x(s A T

s1

)) e

-a(s-s

S

+

i12

xo(x(s~-rs ) ) e 1

f

and property (5.16) reads

But from (5.13) we have

hence we have proved that

-as 2 w(x(s,))

e

1

.

This implies that

-aS Z

+

s1

w(x(slAr)) e

1

,

)

ds

S]

314

Indeed l e t

CHAPTER VII

X

be t h e l e f t hand s i d e of ( 5 . 1 8 ) .

= T

T~

1 and

Similarly for

s 2 s

Theref o r e

and from (5.17)

which p r o v e s ( 5 . 1 8 ) . But t h e n t h e p r o c e s s

1

if

T > sl

We u s e t h e f a c t t h a t

315

PROBLEMS OF OPTIMAL STOPPING

is a sub martingale. By Doob's theorem, we deduce

EX[w(x(6A~))e-ae

+ J6 f 0

for any stopping time.

W(X)

x 8(x(XA?))e-aX

dhl

2

w(x)

But this is identical to

4

Ex[w(x(0))

x6,T

+ JeAT f

x

(x(s))e-as

dsl

8

and since w 4 $

Therefore w(x)

5

u(x).

0

This completes the proof of the desired result

As we have done in Chapter IV, for the problem of semi group envelope

we can now define a general formulation of ( 5 . 1 1 ) ,

(5.12).

in Chapter IV 5 5 . 2 , consider a topological space

(E,E) and spaces

B

and

(5.19)

C.

We consider a semi group satisfying @(t)

We also assume that

: B

+

B

@(O) = I

Namely, as

316

CHAPTER VII

:

(5.23)

@(t)

(5.24)

@(t)f

c

-+

c

- + f in

C

as

t 4 0 , Y f E C .

Let now

(5.26)

L

E

B

,

t

-+

O(t)L

is measurable from

Then we consider the set of functions u

into C

[O,m)

.

satisfying

U€C,U
(5.27)

u

~

r

@(s)L

ds +

,

@(t)u

V t

0

2

0

.

Our objective is t o prove the following Theorem 5 . 3 .

We assume 1 5 . 1 9 / ,

. . .(5.261.

Then t h e s e t of f u n c t i o n s

s a t i s f y i n g 1 5 . 2 7 ) i s n o t empty and has a maximwn element. The proof will consist in several Lemmas. Let g

is a regular f u n c t i o n if there exists = jm

(5.28)

@(t)G

.-at

G

E

g E

C

0

C , we will say that

such that

dt

0

Note that this means that generator of

@(t)

in C. g = Jt 0

hence

G

belongs to the domain of the infinitesimal Indeed, we have @ ( s ) G ds +

@(t)g

317

PROBLEMS OF OPTIMAL STOPPING

the opposite of the infinitesimal generator, then we

If we denote by have

(5.29)

c7g + ag = G

g

6

Dk7)

L e m a 5.1. Let g E C. Then t h e r e ezists a sequence gn of regular is dense in Ci. functions converging to g in C (hence D(Q)

Set

and

I Ign-gl I 4 ; e- s I IO(:)S-glI But from ( 5 . 2 4 )

I

I IO(:)g-gl

-t

0

ff

I Ign-gl 1

But we can also write (cf. Lemma 5 . 7 .

gn -

j m

Gn

ng + (a-n)gn

.-at

'

and is bounded by

s

Lebegue's Theorem, it follows that

ds

+

0

2

I /gl 1.

From

.

of Chapter IV)

O (t) (ng+(a-n)gn)dt

and setting

we see that

gn

=

0

is a regular function.

We now introduce the penalized problem connected to ( 5 . 2 7 ) , (5.30)

u

= €

Jm

0

.-at O ( t ) (L

-

1

(uE-+)+)dt

, uE E

namely C

.

318

CHAPTER VII

By the assumptions the right hand side of (5.30) has a meaning.

Lema 5.2.

There e x i s t s one and o n l y one s o l u t i o n of (5.30).

Proof From Lemma 5 . 7 of Chapter IV, (5.30) is equivalent to

Define then a map

-(a+ -1)t

T

(5.32)

J

z = €

0

The fixed points of

T z

check that

T z E

E

e

@(t)(L

~

-Z T ~ ~

=Z

(”~e

1 - ( a + -)t

which proves the desired result

(5.33)

and

5.3.

One has u

E

s u

E’

.

if

Let us

Indeed

hence

L ema -

E

coincide with the solutiomof ( 5 . 3 0 ) .

is a contraction.

T

1

+ - zA$)dt

E S E ’

@(t)(;

1

1 zlAi+L

- ;z2A$)dt

PROBLEMS OF OPTIMAL STOPPING

Proof One has

and

hence

Since

TE

i s non d e c r e a s i n g

T:UEI

and when

n

-t

m,

5

uE '

we o b t a i n ( 5 . 3 3 ) .

Now

uE

5 {me-ut

@(t)Ldt 5

L e t u s prove a r e v e r s e e s t i m a t e .

M a I

We have 1

-(a+ $ t e

T O = / €

0

@ ( t ) ( L-

-(a+ I ) t 2 - / e

@(t)(L-

0

2 - E 1IL-l

I

1 +

+

I I+-/ I

€a

>

319

320

CHAPTER VII

Let us now assume that

Hence

TnO 2 -k

Lema 5.4.

Let

(5.35)

z 2

-k.

, and in the limit u

$ 2 -k

Since

u

2

-k

.

t o be a s o l u t i o n of 1 5 . 2 7 1 .

u s u

, it follows that

0 Then one has

E'

Proof From Lemma 5.7, Chapter I V we have 1

u < J

m

-(a+ -)t O(t) ( L +

e

0

EI u)dt

I

But from (5.32)

and since

u 6 $, we see that

T u

2

u

By induction Tnu 2 u E

and letting

Lemma 5 . 5 .

n

Let

tend to

+m,

0

we get (5.35).

s a t i s f y t h e same assumptions a s

soZution of 15.301 corresponding t o

5.

Then one has

$,

and

-u

t o be t h e

32 1

PROBLEMS OF OPTIMAL STOPPING

proos Let

z,Z c

c

verify

IIz-Zl

(5.37) Let

FE

I

5

be the mapping

I I@-TiI

*

with

-$

TE

replacing

$.

(5.38)

Indeed

But from (5.37) it follows that

I IW - G L I hence (5.38).

5

I Iili-GI I

2

I I ~ J - GI ~ *

But

which implies

1

lTEO

Applying (5.38) with 2 I lT&O

I

z

=

T 0 ,

-2

- -T 0 , z =

I

5

IIG I

pI

5

I

- T&Ol

and generally

1 ITZ0 Letting

n

+ m,

we deduce (5.36).

I$-51 I '

we obtain

We have

322

CHAPTER VII

We have

Lema 5.6. (5.39)

u

.c u

E

I Iu,-u! I

and

From Lemma 5 . 3 , it follows that

uE

j- u

-+

0

pointwise, and

The difficult part is to prove the convergence in C.

we may, without l o s s of generality restrict ourselves to $

Indeed let Let

$,

E

C

.

regular and

1

1

+

0

u

B

E

.

We first check that $

E

D&).

.

Let

u b e the solution of the penalized problem corresponding to En We have uEn + un , un E B

$.,

.

But from Lemma 5 . 5

and thus as

E

tends to

0

,

But

(5.40)

Since $,

is a regular function,

for fixed

n.

We thus assume

I 1uEn -

unl

I

-+

0

,

as

From this and ( 5 . 4 0 ) follows the fact that

i)

E

D@)

.

Then

E

tends to 0 ,

1 luE -u1 1

+

0

.

323

PROBLEMS OF OPTIMAL STOPPING

where we have s e t

Hence

-

u

i s a s o l u t i o n of t h e e q u a t i o n

(5.41)

-

u

E

= Jrn n

.-at

O(t)

(c - 1 ? ) d t E

E

=

hence

fom which i t f o l l o w s t h a t

(5.42)

But we have s e e n i n Lemma 5 . 3 t h a t

and t h u s from ( 5 . 4 1 )

(5.43)

T u

E E'

- u

2 - E '

E

~ ~ ~ + .l l

One has n e x t 1

(5.44)

T

2~

U

-

~

+

,

=

lrne

-(a+ - ) t

O(t)

(x -

1

-

L! +

1 ; TEuE1Ai/i)dt=

324

CHAPTER VII

But from ( 5 . 4 3 ) u

-

TEuE1

$ t uEl -

E'

I

In1

which is negative from ( 5 . 4 1 ) , hence

Therefore 1

~u2

(5.45)

- q t i m e

-(a+ -)t @(t)

E El

But from equation ( 5 . 4 1 ) follows

therefore

which with ( 5 . 4 5 ) implies

or 2 TEuEI

-

uE, t

-

&'

1 IF+l I

uEt 2

-

E'

I lF+l I

We may iterate to obtain k

TEuE1

and letting

k

-

tend to

m , it follows

*

(z+ A uE, - $ I Ir'I 1)dt

.

325

PROBLEMS OF OPTIMAL STOPPING

(5.46)

u

One then let

E

-

UE'

tend to

iiyii ,

2

-

E'

0

,

fixing

Therefore we have proved that one

One obtains

E'.

I+L

, then the following

D(Q)

E

.

if

estimate holds -4

l / ~ E - ~ l /<

(5.47)

E

IIL

I1

'

This completes the proof of Lemma 5.6. proof of Theorem 5 . 3 From the uniform convergence, we have

(5.30) after multiplying by /m 0

E.

.-at @(t)

u

C.

E

Let

tend to

E

0

in

We obtain

(u-$)+dt

=

0 ,

hence

1 /t e-as 0 ( s ) (u-Ji)+ds t

and letting we deduce

t

0

=

o

tend to

0

, from the continuity of

s

-f

@ ( s ) (u-Ji)',

(u-I+L)+ = 0.

On the other hand from (5.30) it follows that u

and letting

E

E

=

it e-as

$(s)(L

st

@(s)

n

tend to

u s / 0

e-"

-

1

+ e-at O(t)uE

- ( u -9)')ds E

L ds +

E

0

0, we obtain

t .-as

0 ( s ) L ds +

O(t

5

326

CHAPTER VII

Therefore

u

belongs to the set ( 5 . 2 7 ) .

From Lemma 5 . 4 , it is the

0

maximum element.

A s we have done in Chapter IV for the control problem, we are going to

define a discretization procedure. (5.48)

To

simplify we will assume that

L E C .

Otherwise, one should replace in the sequel

So

let

h

L

be a parameter, which will tend to

by

0. The discretized

V.I.

will be the following

Equation ( 5 . 4 9 ) has one and only one solution since the map

is a contraction in

C.

We are going to prove the following

Theorem 5.4. We assume 15.191 uh

of (5.49) tends to u

... ( 5 . 2 5 1

and (5.481. Then the s o l u t i o n

0

the m a x i m solution of ( 5 . 2 7 ) in C.

The proof will require several Lemmas.

We will need a penalized version

of the discretized problem as follows (5.50)

uk =

hL - h (ui-$) + + e-ah @(h)uk

,

:u

E

C

.

We first study problem ( 5 . 5 0 ) in a way which is analogue to what has been done in the continuous time case.

We define a map

327

PROBLEMS OF OPTIMAL STOPPING

m

c

T E ~=

(5.51)

n=O on

C.

( E ) n + l e-nah h+ E

@(nh)[hL + h

%I

One checks t h a t i t is a c o n t r a c t i o n , hence h a s a unique f i x e d

p o i n t which i s s o l u t i o n of (5 5 0 ) .

Indeed

I 1 T E Z 1-$z2 I

Let

uk

be t h e f i x e d p o i n t , t h e n i t i s a s o l u t i o n of

E

u;

(5.52)

Lemma 5 . 7 .

=

h+E

E

IcIAuh -ah (hL + h -) + e

One has

(5.53)

and (5.54)

Proof We have

(5.55)

But from (5.50)

h -<

UE

UE'

h

if

E 2 E'

O(h)uE h+E

,

-

328

or

+

h+E

emah @ ( h ) u E '

and t h u s

uE'

(5.56)

E

30

c

=

-nah

n+l

e

(=)

@(nh)lhL+ h

n=O

hence

TEu;'

5 uE'

h

when

E 5 E'

.

I t e r a t i n g we deduce

hence UE < U E '

h -

Let now

0 uh

h

be t h e s o l u t i o n of 0

uh = hL +

Then u s i n g (5.50) we deduce

0

O(h)uh

,

0

uh

E

C

.

+

PROBLEMS OF OPTIMAL STOPPING

'h0

-

Uh E > - .-ah

uh

0

-

uh

uo

=

0

O(h) (uh-u;)

and i t e r a t i n g E

2

O E e-nah O(nh) (uh-uh)

+

0

.

Therefore

(5.57) But

(5.58)

m

o ( n h ) hL n=O

hence

On t h e o t h e r hand o n e h a s m

TEO =

Z n=O

Assume

z

t -k

,

then

(z-) E n+l

e-nah O(nh) (hL - ; h $-)

329

CHAPTER VII

330

.

-k

2

Therefore we can assert that E Uh 2

(5.60)

U

h

-

0

Therefore (5.54) is satisfie-.

Lemma 5.8.

We have

(5.61)

u;

.

2 Uh

proos We have E

co

(5.62)

TLuh

n+l

Z (h+~)

=

-nuh e Q(nh)[hL

$Auh

+ h

7

1

n=O and u < hL + e-uh O(h)uh h or

EhL + h uh + u < e-ah @(h)uh h - h+E h+E h+E hence m

(5.63)

uh

2

C

E n+l (h+~)

h e nahQ(nh) [hL + ; uhl

.

n=O Noting that

uh

f: $ ,

hence

$ A U ~=

uh , it follows from (5.62) and

33 1

PROBLEMS OF OPTIMAL STOPPING

(5.63) t h a t

hence (5.61). N

Lemma 5.3. L e t replacing

Q.

$

C

E

and

-

uk

Then ue have

a o cf

I

Let

-

z,z

be i n

C

such t h a t

t h e n f r o m ( 5 . 5 1 ) i t follows t h a t

hence ( 5 . 6 4 ) .

We have t h e following

be t h e solutim

05 / 5 . , 7 S I

with

-$

332

CHAPTER VII

Then one has

L e m a 5.10. Assume ( 5 . 6 5 ) . (5.66)

E I Iuh-uh/l "(1

lL1

+

~

IiA/l)

*

Boof

I

Let us set

-Lh

=

-

L

.E=uE-$.

Ah

From (5.50) and (5.65) it follows that

(5.67)

ZE

-

h (u-E = h Lh - ,) +

+

.-ah

O(h);;

hence, as already seen

lE

m

Z

=

n=O

(-)

n+l

E

m

5

C

e

h+E

(-)

n+l

E

e

h+E n=O

-nah

-nnh

O(nh) [ h

@(nh) h

Eh - Eh

E -

(u,)

1

5

Eh

hence -E (U,)

(5.68)

+

-+ 7 '1lLhl1

'

Now (cf. Lemma 5.7)) m

(5.69)

TEuE'

-

uE'

=

-

C

E

n+l

(F;~F)

e

-nah

O(nh)h(:

1

1

and from (5.68) t

-

E'

1 -+1 ~ I ~ 1 ,

for

E

<

E'

E'

- z ) ( u h -I))+

n=O

.

PROBLEMS OF OPTIMAL STOPPING

333

Now from (5.65) we deduce m

(5.70)

E n+l (h+~)

1

$ =

-nah h e [hAh++F1

n=O and thus co

E'

(TE)Uh

-

11,

n+l

E

(E)

C

=

e

-nah

-

Ch Lh

-

h E -$Thuh

&'

-+

E'

-I))-]

n=O Using (5.69)

hence (5.71)

-

Now from expression (5.56) of

u;'

EE'

=

n + l e-nah Ch Lh + h -uh E'

(G)E

C

i l I L ~ / /* l

and from (5.65), we can write

-

m

(5.72)

h

h-u$-h

-E'

)+1

n=O which with (5.71) implies m

E'

E

-h-

n=O -

2

1

I '

u;

Therefore (TE)'

u;'

- uh E'

and by iteration, we obtain

b

-

E'

~1~ ~

n+l

(h+~)

-nah e

+

l / ~ ~ l j '

.

334

CHAPTER VII

, from

u:

Let now u i be the decreasing limit of the sequence

(5.73)

we get

0

Therefore

u i = uh

,

ufi

+

u;

*

2 Uh

in

c'

-

t

Uh

-

-+ 1 lLhJ 1 .

E'

Now from (5.50) one easily checks that

C.

hence finally

i.e., (5.66).

(5.74)

C

in C as h + O .

U;+U'

We have m

UE

- u;

z

=

@(nh)[Jh

e-a'

@ ( s )(L +

L $AuE)e-nh'c

td s

n=O n+l

E

- (E+h) m

=

h (L

e-nah O(nh)

C

1 $AU')] ~h

t

[Ch

e-"'

=

@(s)(L

+ 1 $AuE)e-nh'Eds

n=O

-

n+l

- (&) m

+

X

e

-nah

n=O

=I+II. But

1

h(L + -

$AUE)]

+

.

Q(nh)

E+h

E

($,Au'-$Au')

h

=

335

PROBLEMS OF OPTIMAL STOPPING

=

OE(h)

+

0 as

h

-+

0

,

for fixed

E.

But then we get

0

which implies ( 5 . 7 4 ) . We can then give the proof of Theorem 5 . 4 .

iN E Db).

We add the index

they are related to

N

We approximate

$

by

to the quantities u, uh, uE, uk

QN instead of I).

Then we write

when

336

CHAPTER V I I

this last inequality being deduced from ( 5 . 6 4 ) , by letting We may then let

I lu-\/I

+

h

0 as

go to

h

0

+

0, then

to

E

0

,

and

N

E

tend to

We obtain

+ m.

.

c

We have been considering problem ( 5 . 2 7 ) with the restrictive assumptions ( 5 . 2 3 ) , ( 5 . 2 4 ) on the semi group

O(t)

as well as continuous obstacle.

We can relax these assumptions provided that we do not search for a set of continuous functions in ( 5 . 7 7 ) . Here we consider a semi group

which satisfies ( 5 . 1 9 ) ,

O(t)

..., ( 5 . 2 2 )

and (5.75) (5.76)

x,t

j.

t

O(t)f(x)

+

@(t)f(x)

is measurable

is continuous from

fixed , Y f

Y x

Y f

E

B

B (0,m)

+

R ,

.

Let now (5.77)

L t B , $ c E .

We consider the problem ( 5 . 78)

U E B

,

u < i

u < J t e-cLs O ( s ) L ds + 0

(5.79)

@(t)u

,

V t

E

0

.

0.

331

PROBLEMS OF OPTIMAL STOPPING

then assumption 15.76) can be weakened as (5.80)

t

-t

is e o n t i n u o m from

O(t)f(x)

f i x e d , fi f c C

fi x

(0,m)

+

R

,

.

Under t h e asswnpzions o f Theorem 5 . 3 , t h e n t h e maximum element o f (5.78) co
We c o n s i d e r a d i s c r e t i z e d problem

We s e t f o r

z

E

B

and

Th

solution.

O ( t ) L d t + e-"h

Jh

T z = Min[$, h

0

is a contraction i n

B, hence (5.81) h a s one and o n l y one

A s i n Lemmas 5 . 2 and 5 . 3 of Chapter I V , we check t h a t

(5.82) (5.83)

z E B

and

5 U2h

Uh

z

< Thz

implies

.

Let

K

=

Max[

t h e n we have

(5.84)

Indeed

O(h)zl

uh2-K

Ilq-1 1 ,

u, ry

z

s uh

338

CHAPTER VII

-

hence

K

uh

5

from (5.82).

We now set u

q

=

u1/2q

*

From (5.83) and (5.84) we deduce that (5.85)

We let

u

q

q

tend to

tm,

u 5

(5.86)

+ u

,

U C B

u t - K

then from Lemma 5.4, Chapter IV we obtain

Jml2

R

e-"'

Q(s)L ds + e-"mJ2

R

o(m) u . 2R

Let

t > 0.

+a.

By virtue of assumption (5.76) we can assert that

We apply (5.86) with

u 5 J t e-"'

Therefore Indeed let

u

m

=

[t2'

+I],

O(s)L d s +

5

Jh e-"

O(s)L

ds +

-u

5

uh

tend

the maximum element.

O(h)u

hence

and from (5.82), it follows that

R

.

O(t)u

of (5.78). It i s -u isbeananelement other element, then

-u

and we let

-

, hence u

5

u.

to

339

PROBLEMS OF OPTIMAL STOPPING

Now if we do not assume (5.76), but (5.79) and (5.80), then we cannot let R L! tend m in (5.86) (with m = It2 ] + I ) . However since u 2 u we q'

deduce from (5.86)

But from (5.79), it follows that u 2 Jt e-"'

We let then

q

tend to

+m

@(s)L

u

q

Using then (5.80), we obtain

C.

E

ds +

O(t)uq

fi

q

.

as above and obtain the desired result.

The final statement of the theorem follows from the fact that both maximum elements of (5.78) and (5.27) can be approximated by the same sequence

h'

0

*

6 . I N T E R P R E T A T I O N AS A PROBLEM OF O P T I M A L S T O P P I N G

We assume is a semi compact ('1

E,

(6.1)

and the semi group

defined on

@(t)

(5.21), (5.23), (5.24). (6.2)

@(t)l

B

satisfies properties (5.19),

...

We replace (5.22) by =

1.

This assumption and (5.21) imply (5.22) Now in the case when assumption. (6.3)

E

is

ria5

csmpac;, we will need an additional

Let

i:

= {f

,3 K~ compact ~f(x)l < E , for x 4 KE}

continuous i ti

such that

E

(1) Locally Compact Hausdorff space, with denumerable base.

. Example

Rn,

340

The space

CHAPTER VII

is a closed subspace of

C.

Then we will assume that

We next define

for any Bore1 subset of

E.

We consider the canonical space

Ro

=

,

D([O,-);E)

is continuous to the right and

w(.)

has left limits.

no =

u(x(t),t

0)

2

According to the general Theorem of Markov processes, (cf. Dynkin [ 1 ] , [ 2 ] ) there exists a unique probability

Ft ''tl',?=

Px

on

completed ,

Ro, "mo , such that considering

c0 = n o

completed

then

o0,Go,

pX,

n-t

, x(t)

quasi continuous from t h e l e f t (I), PX(x(0)=x)

=

1

is a right continuous,

strong Markov process, and

. -

(1) quasi continuous from the left means that tr A

of stopping times

...,T~ ..., + ~ ( w )<

T ~ ,

rn(w)

~ ( 7 , )+ X ( T )

-

a.s.

E

mo

and tr the sequence

-R, we have

then if on

G , Px (Dynkin [I],

p. 103).

34 1

PROBLEMS OF OPTIMAL STOPPING

We then define the functional

where

+ e-"

Jx(0) = Ex[Ie e-as L(x(s))ds

(6.6) 0

$

is a

stopping time, and

L,$

$(x(0))]

are defined in (5.25), (5.26).

Our objective is to prove the following

Theorem 6.1.

We assume (5.191, 15.201, 15.211, 15.231, ( 5 . 2 4 1 , 1 5 . 2 5 / ,

15.26), 16.11, 16.21, (6.4). Then the m m i m w n solution of the set 15.271

is given explicitely by (6.7)

u(x)

=

Inf JX(e)

.

0

Idoreover there exists an optimal stopping time

8 defined by

.

Consider the shift operator 8, : Ro + Ro Let 5 1 L By the Markov property we have

be a random variable

.

belonging to

Applying that relation to -a(s

5

=

u(x(s2-sl))e

then

(1) the value

6

= +m

is possible.

-s

2 1

)

+

{

s -s

L(x(s))e

-Cis

ds

342

CHAPTER VII

and t h u s

and from (5.27)

Therefore the process

sub martingale. Using Doob's Theorem we deduce

and since

u 5

9, we obtain

Consider now the penalized problem (5.30).

We can write

Reasoning as above we see that the process

is a

ms

pX

martingale, and thus using again Doob's Theorem we can

assert that

for any stopping time 6.

343

PROBLEMS OF OPTIMAL STOPPING Let us define

The set

{xlu,(x)

2

$(x)}

being closed, and the process

standard process (cf. Dynkin [l], p. 104)),

6',

By definition of

6'

is a

-t

x(t)

being a

stopping time.

we have

6',

Now for the same reason as for

we can assert that

6

is a

R-'

stopping time. We want to prove that (6.10)

u(x)

Jx(6)

=

which with (6.7) will complete the proof of the.theorem. When that

,

u ( x ) = $(x)

clearly

Jx(6)

=

=

0

, Px

,

a.s.

x

is such

hence

$(x)

which proves (6.10). We may thus restrict ourselves to the case

Let

such that u(x) < $(x)

-

do.

Take

6 < 60, then u(x) <

(x)

- 6.

Let

which is also a m

-t

such that for

stopping time. E 5 E

6 '

Since u

E

+

u

in

C

,

there exists

344

CHAPTER VII

Then for

and therefore 8 '

6

2

e6,

t <

and

E 2 E~

.

ti6

Applying (6.8) with

and taking (6.9) into

account, we deduce from (6.8)

-ae u,(x)

and since u

+

u

=

in

EX[uE(x(8

))e

' + ie6

L(x(~))e-~~ds]

C

-ua

(6.12)

u(x) = EX[u(x(e

6

Assume

finite, then

86

L(~(s))e-'~ds]

6 c 0 , and

as

f

6+

))e

86

5

* Let 5.

. e6 +

Now by the right continuity of the process

By the quasi left continuity x(8 6)

-t

x(!J

a.s.

Since u,q

continuous

hence

A

2

6.

Rut also

If

(6.13)

G

=

A +=

6,

5

hence

, u(x(t))

Let u s show that in this case

e6

A

A

=

6.

< P(X(t))

+m

a.s.

Indeed let

We have for w

E

A

, by the right continuity

ti t

,

are

h.

345

PROBLEMS OF OPTIMAL STOPPING

and by the quasi left continuity x(8 6)

u(x(A))

2

66

f

- a.s. 6

x(A) a.s., hence

i(x(A))

which contradicts (6.13), then Therefore

+

Px(A)

=

0'

.

We note that using again the quasi left continuity

property

6

x(e )

+

x(6)

a.s. on

i^e <

hence

By Lebesgue's Theorem, since u (6.12),

which yields

(6.14)

U(X)

But when

8 <

m

,

=

=

Q(t)

satisfies (5.19), ( 5 . 2 0 ) , ( 5 . 2 1 )

We also assume (5.75), (5.76), (5.77).

solution of (5.81).

tion.

.

lJJ(x(i))

We now assume that the semi group and (6.2).

+ J e L(x(~))e-"~ds]

EX[u(x(6))e-"8

by the right continuity of the process

u(x(8))

uh

is bounded we can pass to the limit in

We consider the function

We are going to give its probabilistic interpreta-

We define a transition probability by ( 6 . 5 ) .

CHAPTER VII

346

We consider the canonical space

Qo

=

EcO,w)

,

x(t;w) E W(t)

We will assume that

E

(6.15)

is a metrizable u compact topological space (1)

.

According to the general theory of Markov processes, we can assert that (cf. Dynkin [11,[21) there exists a unique probability

Px

Qo,mo

on

such that

Ro,pO , Px, ?$

,

x(t)

is a Markov process,

and PX(x(0)=x) = 1 Naturally we cannot assert that

x(t)

has the properties stated in 5 6 .1 .

By the Markov property we have

Let

8

(6.17)

be a

mt

stopping time, we define JX(8)

=

Ex[ Je

L(x(t))dt

+

+(x(B))I

.

We are going to consider stopping times of the form 8 = vh

(1) cs compact means that it is the sum of a denumerable number of compact spaces.

341

PROBLEMS OF OPTIMAL STOPPING

where

v

is a random integer, satisfying

(6.18)

{v=n}

Note that

Theorem 6 . 2 .

y n

mt

8 i s indeed a

Our objective is

16.151.

mnh

c

stopping time since

prove the following

to

We assume 15.191,

15.201,

(5.21), (6.2),

15.751,

15.77),

Then t h e s o l u t i o n of t h e d i s c r e t i z e d p r o b l e m 15.811 is g i v e n

e z p l i c i t e l y by

(6.19)

u,(x)

Inf

=

Jx(e)

.

B=vh

Moreover t h e r e e x i s t s an o p t i m a l s t o p p i n g t i m e -

(6.20)

A

Bh

= vhh

vh

=

, where

Min[uh(x(nh)) n

From (5.81) it follows that (6.21)

uh 5

{nh

br n

$(x(nh))]

=

.

integer

-at e Q(t)L

dt + e-nffh O(nh)uh

.

Hence from the Markov property (6.16) Ex[uh(x(mh))e-"mh

+ Jmh e-"'

L(x(s))ds

lmnh] =

nh = O((m-n)h)uh(x(nh))e-amh

and from (6.21)

2

e

-anh

uh(x(nh))

+

Jnth

e-as @ (s-nh)L(x(nh) )e-"sds

348

CHAPTER VII

By Doob's Theorem we can replace m,n by random integers v1 satisfying (6.18),

From this it follows taking (6.22)

5

v2

namely

uh(x)

5

vl

= 0

, v2

5

v

.

Jx(Vh)

Now from (5.81) we have

Now from property (6.15) we have

hence /h .-at O(t)L(x(nh))dt

=

Ex[eunh

0

Therefore (6.23)

/n(hn+l)h

emasL(x(s))ds

reads

-anh e uh(x(nh))

= Mince

EX ( J : { + l ) h

-anh

$(x(nh))

,

e-"'

L(x(s))ds

+ e-a(n+l)h uh(x((n+l)h)) Multiply both sizes by

.

xn,j h

Since for n < v h '

+

lmnh)l

imnhl

349

PROBLEMS OF OPTIMAL STOPPING

,

< $(x(nh))

uh(x(nh))

Since %<jh is

we have

mnh

measurable, we can enter

conditional expectation.

xn,jh

into the

Taking then expectation on both sizes we

deduce

Since this relation h o l d s for any

'h- 1 (6.25)

.-anh

n, we may sum up in n.

uh(x(nh))

We obtain

=

n=O

In fact we have implicitely admitted in writing ( 6 . 2 5 ) that

If <,.=

0

,

we have a.s. from ( 6 . 2 0 )

holds assuming that uh(x) < $(x). uh(x)

=

Ex[ J

'hh

uh(x)

=

uh(x)

=

2

1

-as e L(x(s))ds

-avhh + e

uh(x(vhh)) 1

Jx(jhh)

provided again uh(x) < $(x).

.

It clearly follows from ( 6 . 2 5 ) that

0

(6.26)

vh

Therefore ( 6 . 2 5 )

$(x).

When uh(x)

=

@(x),

and ( 6 . 2 6 ) still holds. This completes the proof of the desired result.

jh = 0 , Px a.s

I

350

CHAPTER VII

We write

Then

From this we recover that the sequence u

q

decreases.

We can then state

Theorem 6.3. We make the assumptions of Theorem 6 . 2 and (5.761, or (5.79) (5.80). Then the m a x i m element of the set ( 5 . 7 8 1 i s ezpLicite1y given by (6.28)

Similar to the proof of Theorem 5 . 4 of Chapter IV.

D i f f u s i o n s (strong formulation). Consider as in Chapter 111, section 5, 5 5, the semi group

where

35 1

PROBLEMS OF OPTIMAL STOPPING

assuming Lipschitz conditions on b,U

as in Chapter I, section 4.

know that the semi group satisfies assumptions ( 5 . 1 9 ) ,

We

(5.20), (5.21),

(5.23), (5.24), (6.2) and (6.1) is satisfied. Let us check (6.4).

It is

enough to check that O(t)f

E

C

,

when f

E

C

.

Indeed we have shown that E IyX(t)-xl2 Let

5

C(t2+t)

1x1 > 2N, then

5-

c ( t2+t) N

2

hence

which tends to

0 as

N

tends to

+=, since I$ t E

,

Therefore if we

assume (5.25), (5.26) on the data, Theorem 6.1 will apply.

Stopped d i f f u s i o n s We take @ ( t ) @ ( x )= E ($(YX(tAT))

*

The problem has already been treated in Theorem 5.2.

352

CHAPTER VII

Remark 6.1. One can give examples of processes which are not continuous (cf. A . Bensoussan - J.L. Lions [ I ] ,

M. Robin [ I ] ) .

COMMENTS ON CHAPTER VII

1 luE-ul 1 ,

1 . For estimates on the penalization error J.L. Lions

cf. A . Bensoussan

-

C11.

2 . If the form

a(u,v)

is symmetric, then (3.2) corresponds to the Euler

condition of optimality for the quadratic optimization problem Min a(v,v)

-

2(f,v)

,

v

E

K

In that context, the penalized problem is very much related to the method of constraint penalization, which is well known in optimization theory. 3. The method of increase of regularity used for equations does not carry over easily for V.I.

A natural question in Theorem 3.2 is the

following. Can we weaken the assumption p > 2 2

.

4 . For degenerate diffusions, cf. J.L. Menaldi [ S ] . 5 . The non stationary stopping time problem leads to parabolic variational

inequalities, which are considerably harder than elliptic V.I., when the obstacle $

is not a regular function of time.

treatment is due to F. Mignot A . Benoussan

-

-

J.P. Puel [l],

The first rigorous

cf. also

J.L. Lions [I].

6 . We can mix stopping time and continuous stochastic control problems. This leads to V . I . where the operator A H

is replaced by

Hamiltonian (cf. A . Bensoussan - J.L. Lions [ I ] ,

A . Friedman [l]).

A-H, with

A. Bensoussan

-

For the case when the Hamiltonian has a quadratic

growth, many technical difficulties arise, cf. J. Freshe - U. Mosco [l], A . Benoussan - J . Frehse

-

U. Mosco [l].

353

PROBLEMS OF OPTIMAL STOPPING

7 . The probabilistic interpretation of the solution of V . I .

permits to

prove many properties of that solution (cf. A . Bensoussan

-

J.L. Lions

[I]).

8. For games, cf. A . Friedman [ll, A . Bensoussan

-

J.L. Lions [11,

0. Nakoulima C11.

9. For non continuous obstacles and relation with capacity concepts, cf. P. Charrier

C11.

10. The main advantage of V.I. for solving free boundary value problems, is that the free boundary which is of course the unknown does not appear explicitely in the formulation of the problem. it as a by product.

One recovers

Of course, all free boundary problems do not

lead to V.I.

1 1 . For the separation principle for the stopping time problem, cf J.L. Menaldi [ll.

This Page Intentionally Left Blank

355

CHAPTER V I I I

IMPULSIVE CONTROL

INTRODUCTION

I m p u l s i v e c o n t r o l i s t o some e x t e n t a g e n e r a l i z a t i o n of o p t i m a l s t o p p i n g , I t a r i s e s n a t u r a l l y i n many economic problems, t h e most p e d a g o g i c a l one

b e i n g InventoryTheory. The t h e o r y h a s been i n i t i a t e d by A Bensoussan

-

J.L. L i o n s [41 and h a s m o t i v a t e d numerous r e s e a r c h e r s , among them B . Hanouzet - J.L. J o l y [ I ] , M. Robin 111, P . C h a r r i e r [ I ] , 0 . Nakoulima [ I ] , J . L . Menaldi [ I ] , A . Friedman - L . C a f f a r e l l i [ I ] , P . Mignot - J.P. Puel [I],

L. Barthelemy [ 1 ] , U .

G. T r o i a n i e l l o [ l ] ,

Mosco [ I ] ,

C . Baiochi

-

L. T a r t a r [ 2 ] , 0 . C h a r r i e r

A . Capelo [ I ] ,

-

...

We r e s t r i c t h e r e t o some a s p e c t s of t h e t h e o r y , namely t h e s t a t i o n a r y impulse c o n t r o l problem a r i s i n g i n I n v e n t o r y t h e o r y , and some c o n s i d e r a t i o n f o r g e n e r a l semi-groups

( w i t h o u t p r o b a b i l i s t i c i n t e r p r e t a t i o n ) . For numeri-

c a l t e c h n i q u e s we r e f e r t o M. Goursat

-

G . Maarek [ I ] ,

J.C. M i e l l o u [ I ] ,

H . Kushner [ I ] .

The " p r o b a b i l i s t i c " approach t o impulse c o n t r o l has been developed r e c e n t l y by J . P . L e p e l t i e r

-

B . Marchal [ I ] .

356

CHAPTER V I I I

1 . SETTING OF THE PROBLEM

a

(1.1)

ij

6

~

'

si

c aij

5 J.

2

a . = - b i + C

j

Let

CI

a~i j = a j i

9

~

1

v~ g

6

1R

~

~ ,

>

o

aaxa i j j

such t h a t :

(1.5)

1. C12 = 2

a

Let a l s o : (1.6)

Where # i s

a

2

0

a

bounded

an open bounded r e g u l a r domain o f R".

k > 0,

c o ( s ) : R"+

+

R ~ c, o n t i n u o u s ,

c o ( 0 ) = 0, non d e c r e a s i n g CO(Sl +

s2)

5

co(5,)

+

C0(L2)

Let a l s o :

357

IMPULSIVE CONTROL

1 . 2 . The model

An i m p u l s i v e c o n t r o l i s d e s c r i b e d by a s e t a s f o l l o w s :

(1.10)

w

= (5'

5

I

... < e n .. . .... cn ..

e2 , 52

<

where 5" i s an i n c r e a s i n g sequence of p t s t o p p i n g t i m e s , and Cn i s a n + sequence of (R ) random v a r i a b l e s such t h a t En i s uRn m e a s u r a b l e , ' I n .

We c o n s i d e r a p r o b a b i l i t y P on

Qo,

p o and a g i v e n Wiener p r o c e s s , s t a n d a r d

n d i m e n s i o n a l w i t h v a l u e s i n Rn, w o ( t ) . We may s o l v e i n t h e s t r o n g s e n s e t h e equation :

s i n c e u i s L i p s c h i t z . We may a l s o s o l v e i n t h e s t r o n g s e n s e t h e c o n t r o l l e d equation :

(1.12)

p r o v i d e d we assume t h a t (1.13)

Equation (1.12)

Bn+

+m

a.s.

(5" =

+m

i s possible) (1)

h a s t o be i n t e r p r e t e d a s f o l l o w s . We d e f i n e a sequence of

d i f f u s i o n s w i t h random i n i t i a l c o n d i t i o n s : (1.14)

dxn = u ( x n ) dwo ."(en)

(1)

Condition B n

<en+'

= xn-l

(5")

+ gn

,

i s meaningful o n l y when

n t I

en

<

m.

CHAPTER VIII

358

dxo = a ( x 0 ) dwo x (0) = x . Then we w r i t e :

T h i s d e f i n e s a p r o c e s s which i s r i g h t c o n t i n u o u s w i t h l e f t l i m i t . I t i s e a s y t o check t h a t : o(x(s))dwo(s) + ( 5 ' +

(1.16)

... +

CVt)

where :

v t = max i n

1 on

5

t j

and t h i s i s t h e i n t e g r a l f o r m u l a t i o n of ( 1 . 1 2 ) .

The u n i q u e n e s s i s c l e a r

s i n c e t h e impulses do n o t depend on x .

I n o r d e r t o add a d r i f t term t o e q u a t i o n ( 1 . 1 6 ) ,

we use G i r s a n o v ' s t r a n s f o r -

mation. By v i r t u e of t h e non degeneracy assumption ( 1 . 1 )

u is invertible.

We d e f i n e PG by :

(1.17)

We a l s o c o n s i d e r :

F o r t h e system

no,

po,

p

t

, P Wx , w ,

t h e p r o c e s s x i s s o l u t i o n of t h e e q u a t i o n :

(1.18)

JO

JO

where v t = max i n We w i l l d e f i n e a

I on

5

payoff as f o l l o w s :

tI.

359 (1.19)

where x ( t ) i s t h e p r o c e s s d e f i n e d by ( 1 . 1 6 ) . We s h o u l d w r i t e T;.

Now

s i n c e t h e p r o c e s s x i s r i g h t c o n t i n u o u s and &open bounded,r i s a s t o p p i n g t

time f o r t h e f a m i l y ii

(cf.Dynkin

[I],

p.

188).

Our o b j e c t i v e i s t o s t u d y t h e f u n c t i o n :

x

(1.20)

u ( x ) = I n f J (W) W

and t o c h a r a c t e r i z e an o p t i m a l impulse c o n t r o l

1 . 3 . some remarks We do n o t claim u n i q u e n e s s f o r PG. N e i t h e r d i d we t r y t o d e f i n e a c o n t r o l l e d m a r t i n g a l e p r o b l e m a s we have done i n c h a p t e r I V . T h i s can be done ( c f . A . Bensoussan

-

J . L . Lions [ 2 ] ,

M. Robin [ I ] ) ,

b u t r e q u i r e s t e c h n i q u e s of

patching together p r o b a b i l i t i e s .

We have p r e f e r e d t h e above p r e s e n t a t i o n which p r e s e r v e s t h e i n t u i t i o n . Nothing e s s e n t i a l i s l o s t when d e a l i n g w i t h d i f f u s i o n p r o c e s s e s . However, when d e a l i n g w i t h g e n e r a l Markov p r o c e s s e s , t h i s i s u n a v o i d a b l e .

Let u s a l s o mention t h a t we have r e s t r i c t e d o u r s e l v e s t o one type of impulse c o n t r o l problem, t h a t i n s p i r e d from I n v e n t o r y t h e o r y . Many o t h e r problems c a n be t r e a t e d by s i m i l a r methods. We r e f e r t o A . Bensoussan

-

J.L.

Lions

[ 2 1 f o r a d i s c u s s i o n on d i f f e r e n t s i t u a t i o n s . The s t r u c t u r e of t h e payoff f u n c t i o n (1.19) i s c l e a r enough. There i s an i n t e g r a l c o s t c o r r e s p o n d i n g t o t h e e v o l u t i o n of t h e p r o c e s s and a c o s t p e r impulse. The c o s t p e r impulse i s l a r g e r t h a n a f i x e d c o n s t a n t k > 0, c a l l e d t h e f i x e d c o s t ; co(S) i s c a l l e d t h e v a r i a b l e c o s t .

360

CHAPTER V I I I

2 . QUASI VARIATIONAL INEQUALITIES

Orientation Dynamic Programming l e a d s t o a n a l y t i c problem which resembles t h e t y p e o f problems d i s c u s s e d i n c h a p t e r V I I . This i s n o t s u r p r i s i n g , s i n c e a s t o p p i n g t i m e problem i s a v e r y p a r t i c u l a r c a s e o f impulse contro1,namely when we

impose

e2

=

e3

=

... =

fm.

T h i s w i l l be r e f l e c t e d i n t h e a n a l y t i c as w e l l

a s i n t h e p r o b a b i l i s t i c t r e a t m e n t . We w i l l e n c o u n t e r q u a s i v a r i a t i o n a l i n e q u a l i t i e s ( Q . V . I . ) i n s t e a d of V . I .

We i n t r o d u c e a non l i n e a r o p e r a t o r a s f o l l o w s :

(2.1)

MHx) = k + i n f [cp(x+S) + c o ( S ) l 620

x +

6 €8

which makes s e n s e f o r f u n c t i o n s (Pwhich a r e bounded below. We c a l l Q . V . I . t h e f o l l o w i n g problem :

where a s u s u a l :

(2.3)

+

a.

u v dx.

Remark 2 . 1 . The c o n d i t i o n u t 0 i m p l i e s t h a t Mu i s w e l l d e f i n e d .

I

Remark 2.2. Problem ( 2 . 2 ) i s a n i m p l i c i t V . 1 , s i n c e t h e o b s t a c l e depends on t h e s o l u t i o n . T h i s i s w h y , i t i s c a l l e d a Q . V . 1 , w i t h t h e V.I.

I t s h o u l d n o t be confused

36 1

IMPULSIVE CONTROL Vv 5 Mv,

u s Mu

I

which i s a t o t a l l y d i f f e r e n t problem.

Remark 2 . 3 The o p e r a t o r does n o t p r e s e r v e t h e r e g u l a r i t y o n d e r i v a t i v e s . I t does n o t map H 1 i n t o i t s e l f ' ] ) . However i t h a s a v e r y i m p o r t a n t p r o p e r t y which w i l l p l a y a fundamental r o l e i n t h e s e q u e l . I t i s monotone i n c r e a s i n g i n the following sense :

'Dl ( x ) <(p2(x) a . e . i m p l i e s

xpl (x) <

?Ip2(x) a . e .

I

Remark 2 . 4 . The f a c t t h a t t h e o b s t a c l e i s n o t a p r i o r i r e g u l a r a l s o e x p l a i n s why we do n o t p r e s e n t a r e g u l a r v e r s i o n of t h e problem, a s an u n i l a t e r a l problem, a s we d i d i n s e c t i o n 2 of c h a p t e r V I I .

2.2.

I

S o l u t i o n o f t h e Q.V.I.

Our o b j e c t i v e i s t o prove t h e f o l l o w i n g :

We w i l l u s e s e v e r a l Lemmas.

S e t t i n g $ = Mu, we can c o n s i d e r ( 2 . 2 ) a s a V.I. We a r e i n t h e c o n d i t i o n s of theorem 3 . 2 . of chap. VII. Then (2.5) i s a consequence of t h e proof of theorem 5 . 1 , c h a p t e r VII. Note t h a t s i n c e u i n (3.3) chapter V I I .

(I)

e x c e p t i n dimension 1

2

0 , 0 belongs t o t h e s e t K d e f i n e d

I

362

CHAPTER V I I I

Lemma 2.2. -

We consider an i t e r a t i v e sequence as f o l l o w s : a(,"+',

(2.6)

u

n+l

Vv

E

v

-

1

u n+ 1 ) 2 ( f , v

Ho, u

E

Ho, 1

n+l

-

n+l)

5 MU n

v 5 Mun,

where uo is def i ne d by (2.4). Then un is a decreasing sequence which converges poinedise t o a s o l u t i o n of 1 2 . 2 / , which is t h e mmimum among a l l possible solutions.

Consider f i r s t u

1

.

Note t h a t uo 2 0 , s i n c e f 2 0 ( c h a p t e r 11, Lemma 2 . 3 ) .

Hence Muo 2 0. The assumptions of theorem 3 . 2 . a r e s a t i s f i e d . Hence u ' i s d e f i n e d i n a unique way. S i n c e f 2 0, 0 i s c l e a r l y a lower s o l u t i o n o f t h e (cf.

V.I.

we have u

(5. I ) , 1

c h a p t e r V I I ) . T h e r e f o r e from theorem 5 . I . ,

2 0, and a l s o u 1 5 u

0

. Assume now

Then t h e same argument t e l l s u s t h a t 0

5

u

chapter V I I ,

that a t step n,

n+ 1

5 uo.

Assume now u

n

5

u

n-1

(which i s t r u e a t s t e p n = I ) . L e t u s show t h a t : (2.7)

u

Indeed u

n+ 1

n+l

5 u

n

satisfies :

and :

SinceM i s monotone i n c r e a s i n g . But then u V.I.

v

= 0

n+1

i s a lower s o l u t i o n f o r t h e

d e f i n i n g u n . By theorem 5 . 1 . , c h a p t e r V I I we deduce ( 2 . 7 ) . We can take a s a t e s t f u n c t i o n i n ( 2 . 6 ) . We deduce :

363

IMPULSIVE CONTROL

hence :

I

1

a ( u n + l , u n + l ) + ~ ln +u l 2 s ( f , u n + ' ) + ~ ln +ul 2 5

since u

c

n 1. s a bounded f u n c t i o n . S i n c e A can be t a k e n l a r g e enough i t f o l -

low t h a t :

T h e r e f o r e we can a s s e r t t h a t : 1

p o i n t w i s e and i n H, weakly.

un t u

(2.9)

Let u s check t h a t :

u 5 Mu

(2.10) We have :

hence :

+ c,(S)

U(X) s k

+ un(x+S)

V t 2 0, x +

5

E

T h e r e f o r e going t o t h e l i m i t i n n :

u ( x ) 5 k + c,(S)

+ u(x+S)

hence ( 2 . 1 0 ) by t a k i n g t h e infinum i n

Let now v

E

I

Ho such t h a t v

S

5.

Mu. We have v 5 Mun, hence from ( 2 . 6 )

a ( u n + l , v ) - ( f , v-u n + 1 ) 2 a ( un + l ,un+l ) ,

By weak 2 . s . c .

we deduce :

a ( u , v-u)

t (f, v-u)

:

364

CHAPTER V I I I

hence u i s a s o l u t i o n of t h e Q . V . I . 1,

( 2 . 2 ) . Let us prove t h a t i t i s t h e maxi., S u o . Assume "u Si u n ,

mum s o l u t i o n . L e t u b e a s o l u t i o n , by Lemma 2.1 t h e n w e have :

and : p u " i

u 5 Mu

which i m p l i e s t h a t "i n+l Therefore u 5 u m

Let z

L

E

,

S

Mun

i s a lower s o l u t i o n f o r t h e V . I .

. Letting

n tend t o +

defining u 21

we o b t a i n u 5 u .

m,

z 2 0 , we d e f i n e Tz = 5 a s t h e s o l u t i o n of

n+ I

. I

:

(2. I I )

From theorem 3 . 2 c h a p t e r V I I , we have 5 m

L

.

So we have d e f i n e d a map from

into i t s e l f .

Lemma 2 . 3 .

Let z

"i

5 z

The map T i s i n c r e a s i n g and concave.

" i p u

and 5 = Tz, 5 = T z . We have 5

pu

hence : Let

m

L

E

e<

and :

1 5 5

+

(1-8)

"i

<,

8

E

( 0 , l ) . We have

5

Mz

S

M,:

and :

365

IMPULSIVE CONTROL

a s e a s i l y checked. T h e r e f o r e :

ec

+ (1-e);

5

T(ez +

(1-e):)

which p r o v e s t h a t T i s concave.

Lema 2.4.

Let z ,

Let

( 0 , I ) such t h a t :

II E

be two p o s i t i v e bounded f u n e t i o n s such t h a t :

(2.13)

z(x) -

y z(x) a.e.

5

Then one has : (2. 1 4 )

Tz(x) - T:(x)

5

y(1-p)

Proof From (2.13) we have : (I-y)z

5

2.

S i n c e T i s concave and i n c r e a s i n g

(2.15)

T;

2

T((1-y)z)

t (I-y)Tz

Let u s check t h a t :

(2.16)

Indeed :

To

2 pu

0

+ y To.

Tz(x)

y

6

CO,II

366

CHAPTER VIII

a(uu

0

,cp)

=

u

a(u

0

,cp)

(uf,y)

s (f,@

VY.

0

and : uu

0

2

k

= M(0)

hence ( 2 . 1 6 ) . Using a l s o t h e f a c t t h a t T z

u

0

,

we have from (2.16)

and from (2.15)

I

hence ( 2 . 1 4 ) .

L e m a 2.4.

Let u,

:be

The bounded s o h t i o n o$ (2.2) i s unique.

two s o l u t i o n s . From t h e p o s i t i v i t y , we have : %

u - u i - u

hence ( 2 . 1 3 ) i s v e r i f i e d w i t h y = 1 . Applying ( 2 . 1 4 ) , and u s i n g t h e f a c t that u,

a r e f i x e d p o i n t s f o r T , we o b t a i n u

-; s (1

-p)

U)

which means t h a t (2.13) i s v e r i f i e d w i t h y = ( 1 - p ) . u

and l e t t i n g k t e n d t o

I f cpc Cyo(@ that :

-

:s +m,

I t e r a t i n g , we check :

(l-p)k u u 5

:.

By s p e t r y u =

we e x t e n d i t b y 0 o u t s i d e

6.Then

:.

f o r such a

I

Q,

we may a s s e r t

367

IMPULSIVE CONTROL

(2.17)

MNx) = k + i n f (co(S) +(p(x+5)).

5 LO Indeed :

x+Sk

cr

x+fs 2 k +

inf

s

co(<) = y(x).

520 X+G

r

T h i s f o l l o w s from t h e f a c t t h a t f o r VC such t h a t x + 5 6 E

[ O , l l such t h a t x + 65

inf

r,

E

4 L?, t h e r e e x i s t s

and k + c o ( 5 ) t k + c 0 ( e g ) 2 k +

c o ( 0 =y(x).

520 X+cEr

But on t h e o t h e r hand :

y(x) = k +

(co(5) + p ( x + S ) )

inf

2

Mcp(x)

520 X+cEr

which i m p l i e s ( 2 . 1 7 ) .

L e m a 2.6.

But M q c l e a r l y b e l o n g s t o C

The sequence un

+

0 -

u i n C (W)

0

.

.

0 -

F i r s t , t h e sequence un b e l o n g s t o C (6).I t i s t r u e f o r uo. Assume i t i s n 0 true for u S i n c e un i n f a c t b e l o n g s t o C + o ( U ) , i t followsfrom Lemma 2 . 5

.

t h a t Mun E Cy(8). From theorem 3 . 3 of c h a p t e r V I I , we deduce t h a t n+ 1 E C y o ( B ) . Now we t h a t : u

hence from lemma 2 . 4 :

i.e. u2 -

UI

5

(I-l,J)u 2

CHAPTER V I I I

368 and i t e r a t i n g we o b t a i n :

u

n+l

-

u

n

5

(I-p)

n

u

n+l

<(I-p)

n

u

o

From t h i s e s t i m a t e i t e a s i l y f o l l o w t h a t un i s Cauchy i n L m . Hence t h e

I

desired result.

The proof of theorem 2 . 1 i s t h e c o m p l e t e .

Remark 2 . 1

I f t h e s o l u t i o n u of (2.2) b e l o n g s t o W2",

then one can w r i t e

problem ( 2 . 2 ) a s f o l l o w s :

(2.18)

I t t u r n s o u t t h a t under a d d i t i o n a l r e g u l a r i t y assumption on f , and t h e c o e f f i c i e n t s of A , a. p <

m.

and c o , one can prove t h a t u

E

f o r any p

W2"

2

2,

However t h i s t h e o r y i s t o o e l a b o r a t e f o r t h e scope of t h e p r e s e n t

work. The main

r e s u l t s have been o b t a i n e d by J . L .

Joly

-

U . Mosco -

111 and A . Friedman - L , C a f f a r e l l i [ I ] . We r e f e r t o A . Bensoussan - J.L. Lions C21 f o r a d e t a i l e d p r e s e n t a t i o n of t h e a v a i l a b l e

G. Trojaniello

results

.

3 . SOLUTION OF THE IMPULSIVE CONTROL PROBLEM

3. I . The main r e s u l t Our o b j e c t i v e i s t o prove t h e f o l l o w i n g :

Theorem 3 . ;

Ye make t h e asswvptions of t h e o r e y 2.1. The?? t h e s o l u t i o n u

of (2.2) i s ~ 3 e v e. q l i c i t i y by :

where the functinn J x ( W )

an op'imal

has been d e f i n e d in (1.19). Nsreover t h e r e e d s c s

impwlse c o n t r o i .

I

369

IMPULSIVE CONTROL

W e s t a r t w i t h a complement on s t o p p i n g t i m e problems, which i s i n t e r e s t i n g

i n i t s e l f . Let u s c o n s i d e r a "usual" system 0,a , be t w o 9

t

s t o p p i n g t r u e s such t h a t 8 5 8' a . s . and

P,Ft, n fp e

w ( t ) . L e t 8, 8' measurable w i t h

values i n R ~ . L e t y ( t ) be a c o n t i n u o u s p r o c e s s such t h a t :

where fi i s a n a d a p t e d p r o c e s s , which i s bounded by a d e t e r m i n i s t i c c o n s t a n t . C l e a r l y we have :

e

y ( t ) = rl f o r t

Let T~

T~

time of t h e p r o c e s s y ( t ) a f t e r time 6 from B: Then

be t h e 1st e x i t

i s a s t o p p i n g t i m e . I n d e e d , l e t xo

E

8*,

and s e t :

t h e n z ( t ) i s a d a p t e d , r i g h t c o n t i n u o u s w i t h l e f t l i m i t and i t s unique p o i n t of d i s c o n t i n u i t y i s 8 . Let us be t h e 1st e x i t t i m e of t h e p r o c e s s z ( t ) a f t e r t i m e s from

(3.3)

6.Note t h a t

oe

= T~

sS

=

:

since z ( t ) = y(t) for t t 6.

Let a l s o :

Because of t h e s p e c i a l

(3.4)

a

inf I t t s

I

z(t-)

or z(t)

4

W}

form o f z , i t i s e a s y t o check t h a t :

=

us.

370

CHAPTER V I I I

But s i n c e b i s a bounded open s e t and z r i g h t c o n t i n u o u s , from Dynkin [ I ] , p . 188. {BS > t}

But s i n c e xo

E

r,

Ue

=

c

o(z(A), s s A 5 t )

=yt.

Uo.

F i n a l l y T~ = ii0 which i s a s t o p p i n g t i m e . We make t h e assumption :

B(t) = b ( y ( t ) ) f o r t

(3.5)

[B,B'-.cgC

E

a.s.

We n e x t c o n s i d e r , under t h e assumptions of theorem 4 . 2 , c h a p t e r V I I , t h e V.I.

a(w, v-u)

( 3 -6)

Vv

E

I

Ho,

2 ( f , v-w)

v 5 $, w

E

1

Ho,

w 5 $.

We know t h a t u i s c o n t i n u o u s . Consider :

1

The s e t C = { x l w ( x ) < $ ( x ) } i s open and

0 i s t h e 1st e x i t a f t e r time 0

from C . We may assume t h a t $ i s extended o u t s i d e @ b y a f u n c t i o n which i s s t r i c t l y p o s i t i v e f a r away, and u i s extended by 0 . T h e r e f o r e t h e complement of C i s compact. Now i n t h e d e f i n i t i o n of z ( 3 . 2 ) w e may assume t h a t XOE

c.

We a r e i n a s i t u a t i o n s i m i l a r t o t h a t of T ~ ,e x c e p t t h a t C i s n o t bounded, b u t i t s complement i s compact. By Dynkin [ I ] ,

p . 188, t h i s s u f f i c e s t o 1

a s s e r t t h e a n a l o g u e of ( 3 . 5 ) . We t h u s a g a i n o b t a i n t h a t 0 i s a s t o p p i n g t i m e .

We t h e n have t h e :

Lema 3.1.

The following r e l a t i o n s hoZd

37 1

IMPULSIVE CONTROL

(3.8)

I f 8' = 8 (hence we asswne that ( 3 . 5 ) hoZds with 8 ' =

.

ei then

:

(3.9)

Proof Consider t h e c a s e of a r e g u l a r o b s t a c l e . Then we know t h a t w

E

W2"(6)

(cf.

c h a p t e r I V , theorem 2 . 2 ) . Let wn be a f a m i l y of smooth f u n c t i o n s such t h a t :

(3.10)

wn + w i n W2'p(8)

From I t o ' s f o r m u l a , we g e t :

Taking t h e c o n d i t i o n a l e x p e c t a t i o n w i t h r e s p e c t t o gral contribution

vanishes!2).

We o b t a i n

the s t o c h a s t i c inte-

:

I ) From Lemma 3 . 2 below i t w i l l f o l l o w t h a t T~ < 2) Cf Remark 3 . 2

m

a . s . on 8 <

m.

372

CHAPTER VIII

(3.11)

Now from (3.10) i t f o l l o w s t h a t : A W ~+

aown

+ AW

+ aow

i n L'(LY)

.

From lemma 3 . 2 below we deduce t h a t :

But t h e c o n d i t i o n a l e x p e c t a t i o n i s a c o n t r a c t i o n i n L

1

,

therefore there

e x i s t s a subsequence such t h a t :

*

E

[IeeTAT

e (Aw+aow) ( y ( s ) ) (exp-

/,'

ao(y)dX)dslf

on Since w i s 0 outside w

d,we + w

e1

e<-

can w i t h o u t l o s s of g e n e r a l i t y assume t h a t : u n i f o r m l y i n t h e whole s p a c e .

We c a n t h u s p a s s t o t h e l i m i t a . s . i n (3.11) ( f o r a subsequence) and obtain :

Using t h e f a c t t h a t Aw + aow 5 f a . e . , w e o b t a i n a s s e r t i o n ( 3 . 8 ) .

We a p p l y n e x t (3.12) w i t h 8' = 8. We u s e t h e f a c t t h a t :

a.s,

373

IMPULSIVE CONTROL

(Aw + aow ( c f . proof o f t h e o r e m

-

a.e

f)XC = 0

4.1, chapter IV). Hence,in p a r t i c u l a r

L

a . s . I n t h i s e x p r e s s i o n we may r e p l a c e

xc

by 1, b y d e f i n i t i o n o f '8. From

t h i s we deduce ( 3 . 9 ) . T h e r e f o r e t h e theorem i s proved f o r r e g u l a r o b s t a c l e s , .

I n t h e g e n e r a l c a s e , we approximate j , by

l i t y , s i n c e t h e y b o t h v a n i s h on

5, we

a s i n t h e p r o o f s o f theorem 3.3

$0

and 4.2, c h a p t e r V I I . We have wn +. w i n C

On-

(a), and w i t h o u t loss of genera-

can assume w

whose s p a c e . We can t h e n p a s s t o t h e l i m i t a . s . ,

+

w uniformly i n the

i n e x p r e s s i o n ( 3 . 8 ) . For A

( 3 . 9 ) we have t o be more c a r e f u l , s i n c e t h e s t o p p i n g t i m e 8 depends on n . We have t o make a random v e r s i o n of t h e argument used i n t h e proof of theo-

rem 4.2, c h a p t e r V I I . I t i s t e c h n i c a l l y more c o n v e n i e n t t o u s e t h e p e n a l i z e d problem as an approx i m a t i o n . Consider:

( 3 . 13)

Define :

*

which i s f o r s i m i l a r r e a s o n s as 8 , a s t o p p i n g time A s we d i d t o p r o v e ( 3 . 9 ) f o r r e g u l a r w , we deduce from (3.13)

(3.15)

374

CHAPTER V I I I

S i n c e w E i s a lower s o l u t i o n o f t h e V . I . , (3.16)

WE

2 w.

WE

-+

we have :

Moreover : (3.17)

~'(8)

w in

This r e s u l t has been proved i n c h a p t e r I V , theorem 3 . 4 .

L e t us t h e n show t h a t : 1

eE

(3.18)

1

+

e

a.s.

Indeed f i r s t of a l l we can a s s e r t t h a t :

e r e-

-E

(3.19)

1

,.

1

T h i s i n e q u a l i t y is c l e a r when B = +-. When

e

( c f . (3;7))

e

<

m,

we have by d e f i n i t i o n of

: 1

w ( Y ( ~ ) =) $ ( y ( e ) ) and t h u s from (3.16) :

which c l e a r l y i m p l i e s ( 3 . 1 9 ) .

I n a p o i n t o such t h a t 8 =

e

1

eE

w e c l e a r l y have from ( 3 . 1 9 ) ,

1

=

8 , i.e.,

1

(3.18) s a t i s f i e d . C o n s i d e r now v a l u e s of o such t h a t 0 < 8 . Let 6o be such 1

t h a t 8 > 8 + 60. 1

For 8 5 s 5 8

-

AO,

we have :

By (3.17) w e can f i n d

E(W)

such t h a t f o r

E

5 E~(o)

315

IMPULSIVE CONTROL

I

Therefore f o r

E

5 E~

and 8 5 s

5

8

-

6

which i m p l i e s :

--

F o r any 6 < 6w, w e can show i n a s i m i l a r way t h e e x i s t e n c e of s0(6,u) f o r which

E

5 EO

-E

implies 8

2

8

6 . From t h i s we deduce ( 3 . 1 8 ) .

Now from (3.18) and ( 3 . 1 7 ) i t i s e a s y t o p a s s t o t h e l i m i t as E

+

0 in

I

( 3 . 1 5 ) . We o b t a i n ( 3 . 9 ) .

Remark 3 . 1 .

I f w e want t o use a proof l i k e i n theorem 4 . 2 ,

we need t o i n t r o d u c e random time

e6,

chapter

VII,

w i t h 6 random. We f a c e t h e d i f f i c u l t y

I

t o prove t h a t t h e y a r e s t o p p i n g times.

The following e s t i m a t e holds

Lemma 3 . 2 .

where C does not depend on f non b , u, a

-1

e, e’,

b u t depends on Wand the bounds on

.

Proof We f i r s t make an i n v e r s e G i r s a n o v t r a n s f o r m a t i o n . We d e f i n e :

(3.22)

376

Considering

CHAPTER V I I I

n,

%T, ?T,

Yt,

then G ( t ) f o r t

5

T i s a standard n dimensional

Wiener p r o c e s s , and t h e p r o c e s s y ( t ) s a t i s f i e s :

We a l s o have t h e i n v e r s i o n formula :

(3.24)

:

L e t f 2 = h . We c o n s i d e r t h e P.D.E.

(3.26)

I We know s i n c e a i j

E

u l z = 0, u(x,T) = 0

W1'-,

h ' i s smooth and & i s r e g u l a r t h a t t h e r e i s one and

o n l y one s o l u t i o n of (3.26) i n t h e f u n c t i o n a l s p a c e :

(3.27)

2+a,l+ 5 . 2 U € C

(8).

By c o n s i d e r i n g a f u n c t i o n which i s C2"(Rn

-

Q , we o b t a i n from I t o ' s formula :

x

C0,TI) and c o i n c i d e s w i t h u i n

311

IMPULSIVE CONTROL

(3.28)

But we a l s o know t h a t f o r t h e s o l u t i o n of (3.26)

< C Ihl

(3.29)

(Q) Therefore, f o r p >

:

+

1,l lu/

1 c.0 <-

L~ ( Q ) C ] h i L p which combined wich (3.28)

shows t h a t :

(3.30) Combining t h i s e s t i m a t e w i t h ( 3 . 2 5 ) , f o r h = f 2 , we deduce :

8 ' -TAT (3.31)

e

'8-T

f ( y ( t ) ) d t s CT

If1

,

p > n +2.

LP

Consider now t h e e l l i p t i c P . D . E .

(3.32)

Au = f

Assuming p i s a s i n ( 3 . 3 1 ) , we can deduce from ( 3 . 3 1 ) , ( 3 . 3 2 ) and ( 3 . 5 ) , that : ~',T,T E u(y(8-T)) = E u(y(8'-T-Te))

+ E

ef ( y ( t ) )

dt.

'0 ,T

But n o t i n g t h e e s t i m a t e :

where t h e c o n s t a n t does n o t depend o n T , we deduce a b e t t e r e s t i m a t e t h a t ( 3 . 3 1 ) namely :

378

CHAPTER VIII

Proof of theorem 3 . l . We may c o n s i d e r t h e Q . V . I .

w i t h o b s t a c l e J1 = Mu. I t i s e a s y

(2.2) as a V . I .

t o check t h a t t h e assumptions o f theorem 4 . 2 ,

Qo,

W be a n i m p u l s i v e c o n t r o l . Consider

po,

p

t

c h a p t e r V I I a r e s a t i s f i e d . Let

, P;,

t h e n t h e p r o c e s s xn d e f i -

ned i n ( 1 . 1 4 ) s a t i s f i e s t h e e q u a t i o n :

We t a k e w i t h t h e n o t a t i o n of t h e b e g i n n i n g o f § 3.1

From ( l , l 7 ) ,

( I .18)

:

i t f o l l o w s t h a t assumption ( 3 . 5 ) i s s a t i s f i e d .

T h e r e f o r e from lemma 3 . 1 . ,

p r o p e r t y ( 3 . 8 ) a p p l i e d w i t h w = u we o b t a i n :

Let u s n e x t remark t h a t :

en

(3.38) Indeed i f 8""

(en,On+')

5

< T implies

r n , then x(s)

and x('J"),

x(e"-)

( 3 . 4 8 ) i s v e r i f i e d . I f 8"" x ( r n ) = x ( T ~ - )a @ ,

E

e n+ 1 - T n

d

for s

c &since

T.

If

en

= T~ T.

Ce n ,

1, x(s)

< T , therefore T 2

> T ~ t, h e n when

thus rn =

r n 2 T, which i s i m p o s s i b l e s i n c e 8" <

(3.38).

en

E

en <

< T~

en+',

<

= s(s-)

en+'

on and t h u s

en+',

t h e n x(T")

k 8', hence

This completes t h e proof of

319

IMPULSIVE CONTROL

where t h e c o n s t a n t C does n o t depend on f , n o r 8 , T , 8 ' . L e t t i n g T t e n d +=,

I

we deduce ( 3 . 2 0 ) , by F a t o u ' s Lemma.

Remark 3 . 2 . (3.33)

I t f o l l o w s from lemma 3 . 2 t h a t :

8'-.rg

Indeed t a k e f = I o n f .

<

m

a.s. 8 <

-

From ( 3 . 2 0 ) ,

e)

<

which i m p l i e s ( 3 . 3 3 ) . We a l s o have (which has been used i n t h e proof of lemma 3 . 1 ) :

(3.34)

g(y(t)).dw(t)

ire]0 =

a.s.

f o r s a y g bounded. Indeed we f i r s t n o t e t h a t t a k i n g T f i x e d

(3.35) Also :

T h e r e f o r e by

F a t o d s lemma :

By t h i s e s t i m a t e and Lebesgue's theorem, we can l e t T yelds ( 3 . 3 4 ) .

-f

m

i n ( 3 . 3 5 ) , which

I

380

CHPATER V I I I

We m u l t i p l y b o t h s i d e s of ( 3 . 3 7 ) by

x

which i s 3 en m e a s u r a b l e . We

en< T

o b t a i n from ( 3 . 3 8 ) :

(3.39)

Now we u s e t h e f a c t t h a t u 2 Mu, hence i n p a r t i c u l a r from lemma 2.5

(2.17))

(see

:

n+1

We a p p l y t h i s i n e q u a l i t y , w i t h x = xn(encl), when 8

<

m,

5 = 5

n+ I

and

we o b t a i n :

n+l (,n+l).

from t h e d e f i n i t i o n ( 1 . 1 4 ) of x Now w e have i f

en

<

m

n n

s i n c e u ( x (.c ))x that i f

(3.43)

en

<

m

:

<e

(hence 0

n+,

= 0

n+l

-T

n

n

when 8

i m,

s e e remark 3 . 2 )

< =. From ( 3 . 4 1 ) and ( 3 . 4 2 ) i t f o l l o w s

38 1

IMPIJLSIVE CONTROL

But :

Indeed when 8 n+ I < for s s

E

en+'

E

Ce",e"+'l,

T ,

we have

but xn(s)

(On,8n+11,thereforexn(s) <

E

en+'

< T and

d

for s

for s

E

< T"(indeed

Cen,enc'),

x(s-),

x(s) c @'

xn(s) = x(s-),

for

CBn,en+ll, which implies

E

P).

Now when 8 n+ I 2

but then from x(en+l)

en

= x(s)

=

T,

with

T < T~

xn+l(en+l)

en

< T , 0""

and x(e

< T

~

n+l-) = xn(e"+')

k4,and

necessarily , from ( 3 . 3 8 ) 0

cd,

n+ 1

= T,

i t follows that

therefore u(xn+](enc1))

= 0.

Using ( 3 . 4 4 ) in ( 3 . 4 3 ) and going back to ( 3 . 3 9 ) we obtain :

We add up these relations, after having taken the mathematical expectation, when n runs from 0 to N-I , remarking that by convention x

E

6,we obiain

(3.46)

:

eo

=

0. Assuming

382

CHAPTER V I I I

Now t h e sequence

x

en<

exp

-

loen

a ( x ( s ) ) d s i s d e c r e a s i n g . Hence i t con-

v e r g e s a . s . t o II. I f II = 0 , a . s . ,

t h e n c l e a r l y from ( 3 . 4 6 ) and 8

N

+

-

a.s.

T h e r e f o r e w e have proved t h a t : u ( x ) 5 Jx(W) f o r a l l W such t h a t II = 0

a.s.

Now i f W i s s u c h t h a t X > 0 on a s e t of p o s i t i v e p r o b a b i l i t y , t h e n i t i s e a s y t o check t h a t Jx(W) =

since :

W ,

(3.48) T h e r e f o r e e have proved t h a t :

(3.49)

provided x

u(x) 5 Jx(W)

E

(f

If x

4 &, t h e n

T = 0, u ( x ) = 0 and ( 3 . 4 9 ) i s a l s o s a t i s f i e d

(as an equality). I

L e t u s now p r o v e t h a t we can f i n d W such t h a t : I

(3.50)

u(x) = Jx(W) I

We f i r s t f i n d a f u n c t i o n 5(x) Bore1

and :

L e t u s consider next :

I

0

dxo = U ( X )dwo

2

0 , such t h a t x + S(x)

E

3,Vx

E

5,

383

IMPULSIVE CONTROL

and :

Then : -

(3.52)

1 = T

8

^I

5

=

0

if T

0

0 < S

,

-

0 -1 <(x (8 ) ) , i f 0 -1

+m

<

s',when

S i n c e Mulr = k , t h e p o i n t x ('2 ) E * ^ n- I cess x and en, Cn, w e d e f i n e xn,

otherwise m,

arbitrary i f

el <

en+', in+'

m.

=

+=J

Having d e f i n e d t h e pro-

as follows :

(3.53)

^

sn

=

1st e x i t time of xn from r a f t e r

T"

=

inf { t

in+'=

(3.54)

2

e"

1

en

u ( x n ( t ) ) = Mu(xn(t))]

T" i f Tn < S n ,

+m

otherwise

i f en+'

<

m,

arbitrary i f

in+'=

m

Let u s check t h a t t h e c o n t r o l

w-

^

=

-

( -e1 , -51 , ..., e n , cn,

...)

i s a d m i s s i b l e . The sequence 8" i s a sequence of s t o p p i n g t i m e s . A l s o assume

en

<

m,

t h e n we have : n n u ( x (8 ) ) = u ( x n -

and s i n c e

:

=

u(xn-

=

u(xn-

384

CHAPTER V I I I

we deduce :

and by t h e s u b l i n e a r i t y of co

f o r any 5

2

0. T h e r e f o r e

(3.55)

u ( x n ( e n ) ) 5 Mu(xn(en))

"n+l " Therefore B > B"

a.s. i f

B" <

-

,.

k < Mu(xn(Bn)).

m.

:

L e t u s next show t h a t L

(3.56)

en

+

a.s.

m

Indeed we f i r s t remark t h a t d e f i n i n g :

P(6)

=

sup

lu(x)

-

U(Y)I

145: X,YE

i t f o l l o w from t h e c o n t i n u i t y of u on

which t e n d s t o 0 a . s 6

hence :

which i m p l i e s : (3.57)

+

&

t h a t p i s an i n c r e a s i n g f u n c t i o n

0. But from ( 3 . 5 1 ) , we have :

385

IMPULSIVE CONTROL

iN<

Now we have i f

m

from (3.53)

:

(3.58) If

sN

-,

<

N ^N w e know t h a t x ( 8 )

€8.But

from ( 3 . 5 7 ) and t h e p o s i t i v i t y of t h e

A

components o f

5, i t f o l l o w t h a t

:

12 ... + 5 N I

(3.59)

+

Let us c o n s i d e r t h e s e t 2,

no

(3.60)

%

no

2

fi

L

such t h a t :

= {lim N

GN

= A <

-1

We d e f i n e t h e p r o c e s s :

t h e n (3.58) becomes :

(3.61)

"v

Now on O

(1)

~

we have :

I f Y, i s bounded a d a p t e d ,

5=

xeCm

a stopping time, then :

loe

xe
e

V (t)dw(t)

1

nN

O < N}

c

=

lim T-

x ecm

E

N

and N 2 No

O

OAT cPdw(t) a . s .

and O = U O

ON+],

loeA; dw. F o r w

lo

XN(w)

u (O=+m). Set

N = XNo,

hence t h e r e i s

NO

convergence on U ON a . s . ; on

(O=+m),

N

towards a R . V . which i s by d e f i n i t i o n

X

N

= 0, hence a . s .

x eCrn

re

/,

Y, ( t ) d w ( t ) *

convergence

386

CHAPTER VIII

But t h i s c o n t r a d i c t s (3.61) s i n c e i t would f o l l o w t h a t 5' "u

+

...

+

SN

remains

which i s i m p o s s i b l e by (3.59). Hence P(; 0 ) = 0.

bounded f o r w f i x e d i n Go,

This shows (3.56). We a r e g o i n g t o prove t h a t (3.50) h o l d s . Consider xn, yn a s i n (3.36), (3.37) w i t h en i n s t e a d of en. From lemma 3.1, -n+ 1 p r o p e r t y (3.9) i t f o l l o w s from t h e d e f i n i t i o n of 8 t h a t we g e t (3.37) w i t h

,.

a n i n e q u a l i t y . Next we have i f

en

<

m

:

U ( X " ( ~ " + ~ - T " ) )= ( k + c o (*n+ S I ) + U(X"+'(~"+')))X,n+ I < T n ' ^n+ I < Bu t when 8

m,

from (3.54),

en+'

< Sn 5 T~

and t h u s :

hence we g e t (3.43) w i t h an e q u a l i t y i n s t e a d of on i n e q u a l i t y . T h e r e f o r e

w e o b t a i n (3.45) w i t h a n e q u a l i t y , hence adding up :

-.,

and from t h e p o s i t i v i t y of u :

and l e t t i n g N t e n d t o

+-.

u ( x ) 2 Jx(W)

-

which i m p l i e s ( 3 . 5 0 ) s i n c e W i s a d m i s s i b l e . The proof o f theorem 3. I i s t h e n complete.

387

IMPULSIVE CONTROL

In t h e s t u d y of Q . V . I .

( 2 . 2 ) , w e have i n t r o d u c e d an i t e r a t i v e scheme un

d e f i n e d by ( 2 . 6 ) . We have shown t h a t un d e c r e a s e s and converges i n C o ( a ) towards t h e s o l u t i o n u of t h e Q.V.I.

lemmas 2.2.and

(cf.

2 . 6 ) . One can

g i v e a n i n t e r e s t i n g i n t e r p r e t a t i o n of t h e sequence u n , which e x p l a i n s i n a d i f f e r e n t way t h e convergence of un towards

We c o n s i d e r t h e g e n e r a l s e t up of § 1 . 1 .

U.

We c o n s i d e r t h e r e s t r i c t e d c l a s s of

impulse c o n t r o l s s a t i s f y i n g : ep =

(3.63)

+m

,

f o r p 2 n.

We d e n o t e by :

wn =

(3.64)

c l a s s of impulse c o n t r o l s s a t i s f y i n g ( 3 . 6 3 ) .

Then we have :

We make t h e asswrrptions o f theorem 2 . 1 . Then t h e sequence un

Theorem 3 . 2 .

defined b y (2. 61 s a t i s f i e s (3.65)

u

n-1

(x) =

inf WE

wn

J ~ ( w ), n

E

I

Moreover, t h e r e e x i s t s an optimal c o n t r o l .

For n = l , 0

1

=

*,

0

then x ( t ) = x ( t ) , JX(W) does n o t depend on W f o r W

E

ICC1 ,

and (3.65) i s n o t h i n g o t h e r t h a n t h e i n t e r p r e t a t i o n of t h e s o l u t i o n uo of e q u a t i o n ( 2 . 4 ) . Let u s t a k e n > I . We a p p l y t h e same method a s i n theorem n+ I 3.1, using the f a c t t h a t u i s a s o l u t i o n of a V.I. w i t h o b s t a c l e Mu". LetW

E

'Ikn+and l

take 0

2

j

2

n-l

We f i r s t have a s f o r ( 3 . 3 9 ) :

.

388

CHPATER V I I I

(3.66)

Operating a s f o r ( 3 . 4 5 ) , we o b t a i n :

(3.68)

We t a k e t h e mathematical e x p e c t a t i o n and add up t h e s e r e l a t i o n s , when j r u n s from 0 t o n-I. We o b t a i n :

n

389

IMPULSIVE CONTROL

n+ 1 =

Using t h e f a c t t h a t

(3.69)

un(x)

5

a . s . We o b t a i n :

+m

J ~ ( w ) ,f o r

w

c

V+'. 1

Let u s prove t h e e x i s t e n c e o f a n o p t i m a l c o n t r o l . Definet"(x)

Bore1 2 0

function such t h a t :

(3.70)

+ u n - l ( x + s"(x))

= k + co(;"(x))

Mun-l(x)

Vx

E

8, n

2 1

Consider xo a s

dx

0

=

0

U ( X )dwo

x (0) = x

T:

1

It

= inf

u n ( x o ( t ) ) = Mun-I(xo(t))}

then :

(3.71)

-1

'n -1

5,

=

0

-gn

We have a g a i n x 0 (en) -1 c @ when p r o c e s s x j-1 and -8:,'

if T:

Tn

5: f o r j

0 -1 ( x (en)) i f

;A

<

= 1

,...,n-2,

< So,

;A

+m

otherwise

< =, a r b i t r a r y o t h e r w i s e

-, s i n c e

Mun-]ly = k . Having d e f i n e d t h e

.

we d e f i n e x J ,

I.

€i;+',

^$ +.I n

follows :

(3.72)

Then :

(3.73)

as

CHAPTER VIII

390

and of c o u r s e m t a k e (3.74)

1

arbitrary.

Let u s p r o v e t h a t :

But :

and by t h e s u b l i n e a r i t y of co :

un-j(xj(6d)) i.e.

,"-j(xj(i:)

+ 5) +

c,(<)

+ k

-k

:

un-j(xj(ei)) and s i n c e un-'

(3.76)

u

n-j -1

,

< ~p-j(xj(8:))

-

k

we have :

un-'(xj(ii))

^'+I which by t h e d e f i n i t i o n of 8;

Mun-j-1 ( xj ( ^ e nj ) )

-

k

( c f . ( 3 . 7 3 ) ) proves (3.75)

We have t h u s c o n s t r u c t e d a n a d m i s s i b l e c o n t r o l Wn c W n + ] . We check e a s i l y that :

which c o m p l e t e s t h e proof of t h e d e s i r e d r e s u l t .

39 1

IMPULSIVE CONTROL

Remark 3 . 4 .

The s e t s

hence c l e a r l y u

ur"

satisfy

w"c tun+] n+l 2 u 2 ... 2

u.

We c o n s i d e r lower s o l u t i o n s of t h e Q . V . I . 1 z E Ho

(4.1)

Theorem 4 . 1 . Q.V.I.

12.2)

n Lm,

z 5

as follows :

Mz

We make t h e asswnptions o f theorem 2 . 3 . The s o l u t i o n u of t h e

is a Lower s o l u t i o n . I t is t h e

maxicimwn

element among lower

solutions.

proos The f a c t t h a t u i s a lower s c i t i o n f o l )w from ( 2 . 2 ) , by t a k i n g v = u

-

as a . t e s t f u n c t i o n . To prove t h a t i t i s t h e maximum lower s o l u t i o n , w e prove by i n d u c t i o n t h a t i f z i s a lower s o l u t i o n then :

(4.2)

z 5 un

where t h e sequence un h a s been d e f i n e d i n ( 2 . 6 ) . We f i r s t check t h a t z 5 u

0

.

This h a s been proved i n theorem 5. I . of c h a p t e r V I I . Assume ( 4 . 2 ) i s t r u e

a t s t e p n , t h e n Mz 5 Mun and t h u s z i s a lower s o l u t i o n f o r t h e problem n+ 1 nt1 From theorem 5.1 ., c h a p t e r VII, i t f o l l o w s t h a t z 5 u

defining u

.

Hence s i n c e un

j.

.

u , the desired r e s u l t follows.

We c o n s i d e r t h e n , a s i n theorem 5 . 2 . c h a p t e r V I I , t h e semi group $ ( t ) d e f i ned by :

I

392

CHAPTER VIII

(4.3)

$ ( t ) $ D ( x ) = EXq(x(t..T))

t h e n we can s t a t e t h e f o l l o w i n g .

Theorem 4 . 2 .

We make t h e assumptions o f theorem 2.1,

(4.4)

and :

a o = a > O

(4.5)

f

E

B (Bore2 bounded on

Then t h e s o l u t i o n u o f Q . V . I .

8)

(2.21 s a t i s f i e s

= 0, u 2 M u -a s

x6e

d s + $ ( t ) e-cYt u , V t 2 0.

of t h e s e t o f f u n c t i o n s s a t i s f y i n g ( 4 . 6 1 .

Proof u i s t h e s o l u t i o n of t h e V . I .

5.2, chapter V I I ,

w i t h o b s t a c l e $ = MU. T h e r e f o r e b y theorem

( r e l a t i o n s (5.11),

(5.12))

u s a t i s f i e s (4.6).

L e t us p r o v e t h a t i t i s t h e maximum element among t h e f u n c t i o n s s a t i s f y i n g %

( 4 . 6 ) . T h i s i s proved u s i n g t h e d e c r e a s i n g scheme. L e t u be an element o f ( 4 . 6 ) . We have

%

Assuming t h a t u

%

u

5

uo s i n c e :

%

un, t h e n M u 5 Mun, Therefore?; i s an element of t h e s e t

( 5 . 1 1 ) , (5.12) c h a p t e r V I I , w i t h $ = Mu". % n+l % then implies u < u T h e r e f o r e u S U.

But theorem 5 . 2 , c h a p t e r V I I ,

.

I

We c o n s i d e r now t h e a n a l o g u e of problem (5.27) c h a p t e r VII, w i t h an i m p l i c i t o b s t a c l e . Namely l e t $ ( t ) be a semi group s a t i s f y i n g ( 5 . 1 9 ) , c h a p t e r V I I . We a l s o c o n s i d e r :

..., ( 5 . 2 4 ) ,

393

IMPULSIVE CONTROL

L

(4.7)

E

B , t + $ ( t ) L i s measurable

from [ O , m l

into C

L t O

L e t now M be an o p e r a t o r such t h a t :

M :C

(4.8)

+

C i s L e p s c h i t z , concave, and monotone i n c r e a -

MYl

sing (i.e.

(4.9)

;

M(O)tk>O

5 M(D2 i f

Y1

5

(D2)

a > O

We c o n s i d e r t h e set of f u n c t i o n s : u

(4.10)

E

MU

C, u S e

Theorem 4.3.

We assume ( 5 . 1 9 ) ,

$(s)

Lds + e

..., (5.24)

-at

$(t) u

chapter VII and 1 4 . 7 1 ,

(4.81,

( 4 . 9 ) . Then the s e t of s o l u t i o n of ( 4 . 1 0 ) i s not empty and has a rnadmwn

element, which i s a p o s i t i v e f u n c t i o n .

Let z

E

C and c o n s i d e r 5 t o be t h e maximum s o l u t i o n o f :

(4.11)

S i n c e Mz

E

C, 5 =

Tz i s w e l l d e f i n e d , a c c o r d i n g t o theorem 5 . 3 , c h a p t e r V I I .

I t w i l l be c o n v e n i e n t t o w r i t e :

(4.12) where u(9)

T

=

u o

M

: C -+ C i s t h e maximum element of t h e s e t ( 5 . 2 7 ) , c h a p t e r V I I .

O n e e a s i l y checks t h a t u i s monotone i n c r e a s i n g and concave. From t h e assumpt i o n ( 4 . 8 ) on M i t f o l l o w s t h a t : (4.13)

T i s i n c r e a s i n g and concave.

394

CHAPTER VIII

L e t next

1,

m

uo =

(4.14)

e -at $ ( t ) Ldt

and uo E C , uo 2 0. L e t 1 > p > 0, s u c h t h a t :

I

(4.15)

! ~ I l u ~ l5 k

Then one h a s : T(0) 2 u u

(4.16)

0

Indeed by ( 4 . 1 5 ) and assumption ( 4 . 9 ) .

!J u0 9 M ( 0 )

and from ( 4 . 1 4 )

:

v

uo =

lot lo

e-as $ ( s ) vL d s + e

t

<

ea

S

$ ( s ) Lds + e

-at

-at

a

$(t)v u

$ ( t ) v uo

s i n c e L 2 0 , hence ( 4 . 1 6 ) . We have now a r e s u l t i d e n t i c a l t o lemma 2 . 4 . %

(4.17)

Let z , z

E

C and y

Vx, t h e n Tz(x)

-

E

T;(x)

C0,ll such t h a t z ( x ) S y(1-p) Tz(x)V x .

-

;(x)

< yz(x)

Recalling that :

(4.18)

Tz

A s i n lemma 2 . 4 . , i n lemma 2 . 6 .

S uo

Vz.

one checks t h a t t h e map T has a t most one f i x e d p o i n t . As

one checks t h a t t h e i t e r a t i v e scheme :

395

IMPULSIVE CONTROL

u

n+ 1

= T un, uo g i v e n by ( 4 . 1 4 )

i s d e c r e a s i n g and c o n v e r g e s i n C towards t h e maximum element of ( 4 . 1 1 ) . Moreover Tn uo 5 0 hence u 2 0. This completes t h e proof o f t h e d e s i r e d

I

result.

4.3. gjscretization The d i s c r e t i z e d v e r s i o n of ( 4 . 1 0 ) i s t h e f o l l o w i n g :

We w i l l assume i n s t e a d of ( 4 . 7 )

(4.19)

L E C , L > O

(1)

We can s t a t e t h e f o l l o w i n g :

Theorem 4 . 4 .

Assume (5.191,

. . ,,(5.24) chapter V I I and

14.S), (4.9), ( 4 . 1 9 ) .

Then problem 1 4 . 1 8 ) has one and o n l y one s o l u t i o n . Moreover h

-+

y,

-+

u in C, as

0, where C is the m d m u m element o f (4.10).

Let u s s e t oh($) t o be t h e s o l u t i o n o f ( 5 . 4 9 ) c h a p t e r V I I ( d i s c r e t i z e d V . 1 ) . Then uh : C

(4.20)

+

C , and :

ah i s i n c r e a s i n g and concave.

T h i s w i l l f o l l o w from t h e f o l l o w i n g f a c c :

(4.21)

if i

5

S

E

C v e r i f i e s 5 2 $ and

hL + e

-Uh

$(h)<

,

then

i 5 Oh($)

I ) a s i n § 5 . 3 c h a p t e r V I I , t h i s i s r a t h e r a m a t t e r of convenience.

396

CHAPTER V I I I

Indeed l e t :

s

(2)

= Min

C J I , hL + e -ah $ J ( ~ ) z I

h which i s c o n t r a c t i o n on C , and Ch = ah($) i s t h e f i x e d p o i n t of S h . But t h e assumption ( 4 . 2 1 ) on 5 i m p l i e s :

hence i t e r a t i n g ( s i n c e Sh i s i n c r e a s i n g ) :

therefore :

which p r o v e s t h e d e s i r e d p r o p e r t y ( 4 . 2 1 ) . Then ( 4 . 2 0 )

follows e a s i l y .

Defining next :

(4.22)

Th = uh 0 M

t h e n t h e s o l u t i o n s of ( 4 . 1 8 ) a r e t h e f i x e d p o i n t s of Th. Now we have s e e n i n (5.59) c h a p t e r V I I , t h a t :

hence t h e r e e x i s t s

u

E

( 0 , l ) such t h a t

u

1 I u0h / 1

9

k . Then :

From t h i s and t h e f a c t t h a t Th i s i n c r e a s i n g concave, one deduce t h a t Th h a s one and o n l y one f i x e d p o i n t and i f :

then :

IMPULSIVE CONTROL

397

(4.24) N o w by theorem 5 . 4 , c h a p t e r , V I I , we have :

-

I Since also l/un - ulj uh

+

u i n C.

+

unl

1

+

0 as h + 0 f o r f i x e d n.

0 , u s i n g t h e uniform e s t i m a t e ( 4 . 2 4 ) one o b t a i n s

1

398

CHAPTER V I I I

COMMENTS ON CHAPTER V I I I

I ) The p r o p e r t y g i v e n i n lemma 2 . 4 , which i s v e r y u s e f u l t o s t u d y Q . V . I .

i s due t o B . Hanouzet

-

J . L . Joly [ I ] .

U n f o r t u n a t e l y i t seems t o be

l i m i t e d t o bounded f u n c t i o n s . F o r unbounded f u n c t i o m ( c f . A . BensoussanJ.L.

Lions

[?.I).

2 ) I n t h e proof o f lemma 3 . 2 , we do n o t need t o know t h a t t h e f a m i l y ?T a r e

?.

t h e r e s t r i c t i o n s of a p r o b a b i l i t y

3) We have i n t r o d u c e d a d e c r e a s i n g scheme t o approximate t h e s o l u t i o n of the Q . V . I .

I t i s p o s s i b l e t o d e f i n e an i n c r e a s i n g scheme s t a r t i n g from 0 .

The method of Hanouzet-Joly w i l l show t h a t i t c o n v e r g e s . I t h a s a l s o a probabilistic interpretation.

4 ) For degenerate Q.V.I.

cf. J.L.

Menaldi [ I ] .

5 ) For t h e p r o b a b i l i s t i c i n t e r p r e t a t i o n of t h e Q . V . I . r a l Markov p r o c e s s e s , c f . M. Robin

[]],

c o r r e s p o n d i n g t o gene-

A . Bensoussan- J . L .

Lions [ 2 1 .

6 ) A " p r o b a b i l i s t i c approach" t o impulse c o n t r o l h a s been o b t a i n e d by J.L.

Lepeltier

-

B . Marchal [ I ] .

7 ) The non s t a t i o n a r y problem l e a d t o p a r a b o l i c Q . V . I . , more c o m p l i c a t e d ( c f . F . Mignot J.L.

J.P.

Puel [ I ] ,

A . Bensoussan

Lions [ 2 1 C . Barocchi

t e c h n i q u e s i n Mechanics c f . A . Bensoussan

-

Lions [ Z l ) .

-

A . Capelo [ I ] .

9 ) One can i n t r o d u c e a p e n a l i z e d problem t o t h e Q . V . I . J.L.

-

Lions [ 2 1 ) .

8) For a p p l i c a t i o n s of Q . V . I . J.L.

-

which a r e much

( c f . A . Bensoussan

-

399 R E F E R E N C E S

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