THESE DE DOCTORAT DE L’UNIVERSITE DE TOURS

Sp´ ecialit´ e : MATHEMATIQUES

Sujet de la th` ese : Etude des ´ equations de Hamilton-Jacobi avec des conditions de Dirichlet discontinues et applications aux probl` emes de contrˆ ole optimal avec temps de sortie et aux probl` emes de Grandes D´ eviations.

Pr´ esent´ ee par : Alain-Philippe BLANC

Table des Mati` eres

INTRODUCTION 1 Probl`emes de contrˆole optimal avec temps de sortie 1.1 Pr´esentation g´en´erale . . . . . . . . . . . . . . 1.2 Probl`emes de temps de sortie . . . . . . . . . 1.3 Probl´ematique . . . . . . . . . . . . . . . . . . 1.4 Etudes directes des trajectoires . . . . . . . . 1.5 L’´equation de Hamilton-Jacobi-Bellman . . . . 1.5.1 Les solutions de viscosit´e . . . . . . . . 1.5.2 Conditions de Dirichlet discontinues . .

1 . . . . . . . .

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5 5 7 10 14 15 16 18

2 Probl`emes de Grandes D´eviations . . . . . . . . . . . . . . . . . . . . . . 2.1 Pr´esentation g´en´erale . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 M´ethodes de r´esolution . . . . . . . . . . . . . . . . . . . . . . . . . .

21 21 24

I PROBLEMES DE TEMPS DE SORTIE AVEC COUTS DE SORTIE DISCONTINUS 27 3 Deterministic exit time control problems with discontinuous exit cost1 3.1 The exit time control problems with discontinuous exit costs . . 3.1.1 The value functions u− and u+ . . . . . . . . . . . . . . 3.1.2 The properties of the value function u[ϕ] . . . . . . . . . 3.1.3 Partial controllability on the boundary . . . . . . . . . . 3.2 The uniqueness results . . . . . . . . . . . . . . . . . . . . . . . 3.3 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

II

PROBLEMES DE GRANDES DEVIATIONS

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29 33 35 42 46 56 69

73

4 Large deviations estimates for the exit probabilities of a diffusion process through some vanishing parts of the boundary1 . . . . . . . . . . . . . . . . . 75 4.1 The probability estimates in the case of a “regular” limiting subset . 79 4.2 Probability estimates of exiting through vanishing parts of the boundary 87 4.2.1 The case when −ε2 ln(meas(Γε )) → 0 . . . . . . . . . . . . . 88 4.2.2 The case when −ε2 ln(meas(Γε )) → +∞ . . . . . . . . . . . . 94 4.2.3 The case when −ε2 ln(meas(Γε )) → m, (0 < m < +∞) . . . . 104 4.3 The weak comparison result . . . . . . . . . . . . . . . . . . . . . . . 106 4.3.1 The concept of Barron-Jensen solutions . . . . . . . . . . . . . 107 4.3.2 Proof of Theorem 4.1.2 . . . . . . . . . . . . . . . . . . . . . . 112 4.4 Appendix : the gradient estimate . . . . . . . . . . . . . . . . . . . . 123 R´ef´erences bibliographiques . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127

INTRODUCTION

3

1. Probl` emes de contrˆ ole optimal avec temps de sortie

1.1 Pr´ esentation g´ en´ erale La th´eorie du contrˆole optimal d´eterministe permet d’´etudier les contrˆoles α(.) a` appliquer `a un syst`eme dont l’´evolution est r´egie par une ´equation diff´erentielle ordinaire de la forme :  dyx   (t) = b(yx (t), α(t)), dt (1.1)   yx (0) = x ∈ Ω, dans le but d’optimiser une certaine fonctionnelle : J(x, α(.)) (les hypoth`eses pr´ecises sont donn´ees dans la partie suivante). A titre d’illustration, nous allons consid´erer le cas concret de l’alunissage d’un v´ehicule spatial. En supposant qu’il n’y a pas de mouvement lat´eral, le comportement du v´ehicule spatial est caract´eris´e, `a chaque instant t, par la donn´ee de son altitude et de sa vitesse : (z(t), v(t)). Le d´ebut de la manœuvre d’alunissage d´etermine l’origine des temps et `a cet instant d’origine, le v´ehicule spatial se trouve `a une certaine altitude z(0) = z0 et `a une certaine vitesse v(0) = v0 (< 0). z 6

z0 s v0 ? z=0

6

Probl`emes de contrˆole optimal avec temps de sortie

L’engin est soumis `a l’attraction g de la lune que nous supposerons ind´ependante de l’altitude et `a la force de pouss´ee α (positive) du moteur d’alunissage qui constitue le contrˆole. En supposant que la masse M du v´ehicule spatial est ind´ependante du temps, l’´equation de la dynamique s’´ecrit : M

dv = −M g + α(t) dt

et elle conduit au syst`eme diff´erentiel qui r´egit l’´evolution du syst`eme :  dz    (t) = v(t), 

dt

 dv 1    (t) = −g + α(t).

dt

M

Le but est de poser le v´ehicule sur le sol lunaire avec une vitesse faible. Pour cela, il y a une infinit´e de mani`eres de proc´eder mais nous chercherons de plus `a minimiser la consommation de carburant que nous supposerons proportionnelle `a la pouss´ee. Le crit`ere `a minimiser est de la forme : J(α(.)) =

Z

τ

α(t)dt,

0

o` u τ est l’instant d’alunissage et o` u α(.) est un contrˆole qui assure z(τ ) = 0 et v(τ ) ∈ [−ε, 0]. On trouvera dans J.S. Meditch [29] une ´etude plus approfondie sur les probl`emes d’alunissage. Historiquement, ce sont des probl`emes provenant du domaine de l’a´erospatiale qui ont entraˆın´e un important d´eveloppement du contrˆole optimal dans les ann´ees 1950-1960. Mais le contrˆole optimal peut ´egalement ˆetre consid´er´e comme une g´en´eralisation d’une branche plus ancienne des math´ematiques : le calcul des variations qui consiste `a trouver les trajectoires minimisant une fonctionnelle de la forme : Z

t1

L(y(t), y(t), ˙ t)dt.

t0

Le contrˆole optimal ne se limite pas seulement `a d´eterminer des trajectoires optimales pour des avions ou des v´ehicules spatiaux ; il permet de traiter toutes sortes de probl`emes apparaissant dans des domaines aussi vari´es que l’´economie, la m´ecanique ou l’automatique. On trouvera dans les livres de W.H. Fleming et R.W. Rishel [20] et de E.D. Sontag [33], entre autres, des pr´esentations plus d´etaill´ees pour les probl`emes de contrˆole optimal.

1.2 Probl`emes de temps de sortie

7

1.2 Probl` emes de temps de sortie Nous allons maintenant d´ecrire de fa¸con plus pr´ecise les probl`emes de temps de sortie qui forment une classe particuli`ere de probl`emes de contrˆole optimal et qui constituent le cadre principal de notre travail. Nous indiquons maintenant les hypoth`eses faites sur l’´equation diff´erentielle ordinaire (1.1) qui caract´erise le comportement du syst`eme consid´er´e. La fonction b est habituellement born´ee et lipschitzienne pour assurer l’existence d’une solution de (1.1) pour tout temps. La fonction α(.) est une application de IR+ dans l’espace des contrˆoles A ; A est, en g´en´eral, un espace m´etrique compact. La caract´eristique essentielle des probl`emes de temps de sortie est que l’on se place dans un ouvert Ω de IRN . Nous supposerons toujours dans la suite que l’ouvert Ω est born´e. Ainsi, puisque l’on se restreint au domaine Ω, il faut “arrˆeter” les trajectoires quand elles atteignent le bord du domaine Ω ou au moins quand elles sortent de Ω. Pour ce faire, nous consid´erons d’abord l’instant τ qui correspond au premier temps pour lequel la trajectoire yx sort de l’ouvert Ω i.e. τ := inf{t ≥ 0, yx (t) 6∈ Ω}. Nous introduirons d’autres possibilit´es dans la partie suivante. Si la trajectoire ne touche jamais le bord ∂Ω de l’ouvert, nous consid´ererons que τ := ∞. $

'

Ω s

yx (τ )

s

x &

%

Enfin, la fonction coˆ ut J associ´ee au temps τ , est d´efinie de la fa¸con suivante : J(x, α(.)) :=

Z 0

τ

f (yx (t), α(t))e−λt dt + ϕ(yx (τ ))e−λτ

pour x ∈ Ω.

La fonction f est le coˆ ut instantan´e ; nous prendrons une fonction born´ee et N lipschitzienne de IR × A dans IR. Le facteur d’actualisation λ est une constante

8

Probl`emes de contrˆole optimal avec temps de sortie

positive qui permet, entre autres, `a l’int´egrale d’ˆetre toujours d´efinie mˆeme si τ est infini. L’originalit´e du pr´esent travail consiste `a supposer seulement que les fonctions coˆ ut de sortie ϕ sont d´efinies en tout point du bord : elles peuvent donc ˆetre discontinues. Finalement, l’infimum de la fonction J sur l’ensemble des contrˆoles possibles α(.) ∈ L∞ (IR+ , A), d´efinit une fonction-valeur pour tout point x ∈ Ω : u(x) :=

inf

α∈L∞ (IR+ ,A)

J(x, α(.)).

Exemple 1 : Les probl`emes de cible. Les probl`emes de cible forment un cas particulier des probl`emes de temps de sortie pour lesquels, comme leurs noms l’indiquent, le but est d’atteindre une partie donn´ee Γ du bord ∂Ω du domaine. La principale cons´equence est que les fonctions coˆ ut de sortie sont de la forme :   0

ϕ(x) :=  C

si x ∈ Γ si x 6∈ Γ

pour x ∈ ∂Ω,

avec C une constante positive grande ; ce qui se note ´egalement ϕ = C11∂Ω\Γ o` u 11∂Ω\Γ est la fonction indicatrice du compl´ementaire de Γ dans ∂Ω. Ainsi les trajectoires qui sortent `a travers la cible Γ minimisent le coˆ ut de sortie ϕ. Et ils minimisent ´egalement la fonction coˆ ut J si le coˆ ut instantan´e f est de plus strictement positif, ce qui p´enalise les trajectoires restant dans Ω pour tout temps (i.e. pour lesquelles τ = ∞). Exemple 2 : Retour sur le probl`eme d’alunissage. Le probl`eme d’alunissage de la partie pr´ec´edente peut ˆetre d´ecrit comme un probl`eme de cible. En effet, il est naturel de consid´erer que la vitesse du v´ehicule spatial est born´ee (v ∈ [−V, V ]) de mˆeme que son altitude (z ∈ [0, H]). Le domaine Ω est ainsi ´egal `a : Ω := [0, H] × [−V, V ]. Enfin, puisque l’on veut alunir avec une vitesse raisonnable, la cible est la partie du bord du domaine correspondante `a l’altitude nulle et `a la vitesse faible i.e. Γ := {0} × [−ε, 0].

1.2 Probl`emes de temps de sortie

9

z 6

H

Ω −ε 0

−V

+V

-

v

La fonction-valeur associ´ee `a ce probl`eme est donc de la forme : u(x) :=

nZ

inf +

α∈L∞ (IR ,A)

0

τ

o

α(t)dt + C11∂Ω\Γ (z(τ ), v(τ ))

avec A := [0, A] o` u A est la pouss´ee maximale du moteur d’alunissage. Il faut noter que, si le v´ehicule spatial est trop pr`es du sol avec une vitesse de chute trop ´elev´ee, il n’y a pas d’alunissage “en douceur” possible ; la valeur de la fonction u est alors ´egale `a C pour ces points ; sinon u donne la consommation de carburant n´ecessaire `a un alunissage `a partir des autres points de Ω. Nous allons maintenant d´ecrire un probl`eme de cible particuli`erement simple, permettant le calcul explicite de la fonction-valeur. Dans IR2 , nous consid´erons un domaine d´efini par Ω :=]0, 1[2 . La fonction b est ´egale `a b(x, α) := α ; ainsi la dynamique refl`ete exactement ce que permet le contrˆole. Pour A := {(0, −γ), γ ∈ [0, 1]}, toutes les trajectoires issues du point (x1 , x2 ) sont de la forme : Z t (x1 , x2 + α2 (s)ds) pour t ∈ [0, τ ]. 0

Elles peuvent soit atteindre le bord [0, 1] × {0} soit rester dans Ω pour tout temps. x2 6

1

Ω

0

sx

-

1 x1

10

Probl`emes de contrˆole optimal avec temps de sortie

La cible est constitu´ee par le bord x2 = 0 i.e. Γ := [0, 1] × {0}. D’o` u la fonction coˆ ut est :   0 si x2 = 0, pour (x1 , x2 ) ∈ ∂Ω. ϕ(x1 , x2 ) :=  1 sinon, Enfin, le coˆ ut instantan´e f est constant et ´egal `a 1 et la constante λ est fix´ee `a 1 pour simplifier. Alors la strat´egie optimale est d’atteindre le plus vite possible le bord o` u la fonction coˆ ut de sortie ϕ est nulle i.e. [0, 1] × {0}. En effet, la fonction coˆ ut associ´ee `a un contrˆole α ≡ (0, −1) est alors : J((x1 , x2 ), (0, −1)) =

Z

x2

1e−t dt + 0

0

= 1 − e−x2 , alors que si la trajectoire reste dans le domaine Ω pour tout temps, le coˆ ut associ´e est : Z ∞ 1e−t dt = 1. 0

Pour les points du bord, le temps de sortie τ est par d´efinition nul ce qui entraˆıne que u = ϕ sur ∂Ω. D’o` u: u(x1 , x2 ) =

  1 − e−x2

si (x1 , x2 ) ∈]0, 1[2 ∪Γ,

 1

si (x1 , x2 ) ∈ ∂Ω\Γ.

La discontinuit´e de la fonction-valeur u sur le bord ∂Ω\Γ est une cons´equence de sa d´efinition car `a partir de tout point de ∂Ω\Γ, il existe des trajectoires qui atteignent la cible Γ, tout en restant dans Ω.

1.3 Probl´ ematique Comme nous l’avons indiqu´e en d´efinissant la fonction-valeur u, le premier probl`eme provient du fait qu’il est facile de d´efinir d’autres types de fonctions-valeur tout en restant dans un cadre tr`es proche. Nous allons commencer par expliciter les divers choix qui s’offrent `a nous et ´etudier leurs cons´equences pour chercher s’il existe une d´efinition “naturelle” pour la fonction-valeur. La premi`ere possibilit´e de changement dans la d´efinition de u provient du choix du moment o` u l’on d´ecide d’arrˆeter la trajectoire. La fonction-valeur u utilise τ , le premier temps de sortie de l’ouvert Ω. Mais il est aussi simple de prendre le premier temps de sortie du ferm´e Ω c’est-`a-dire : τ¯ := inf{t ≥ 0, yx (t) 6∈ Ω}.

1.3 Probl´ematique

11

Notons, en particulier, que dans l’exemple pr´ec´edent, si l’on prend τ¯ au lieu de τ dans la d´efinition de la fonction-valeur, cette derni`ere devient continue dans tout Ω. Mais la situation n’est pas tout `a fait aussi simple car il peut exister des trajectoires qui touchent plusieurs fois le bord avant de sortir de Ω (si τ 6= τ¯). '

$ syx (τ )

Ω

s

x

&

sy % (¯ τ) x

L’exemple suivant montre que le choix du point o` u l’on consid`ere que la trajectoire “s’arrˆete” sur le bord peut avoir beaucoup d’importance. Dans IR2 , Ω est d´efinit par {(x1 , x2 ) ∈] − 1, 1[2 tel que x1 < 0 ou x2 > 0}. La dynamique est d´etermin´ee par b(x, α) := α ∈ A avec : A := B(0, 1) ∩ (IR+ × IR+ ) = {(x1 , x2 ) ∈ IR2 tel que x21 + x22 ≤ 1 et x1 ≥ 0 et x2 ≥ 0}. Pour la fonction coˆ ut, on choisit f ≡ 1 et λ := 1. Enfin, la fonction coˆ ut de sortie ϕ vaut 1 sur ∂Ω sauf sur [0, 1] × {0} o` u ϕ(x1 , 0) := |1 − 2x1 |. x2

6 1

Ω −1

ϕ≡1

0 ϕ = |1 − 2x1 | 1 x1

-

ϕ≡1

−1

12

Probl`emes de contrˆole optimal avec temps de sortie

Comme dans l’exemple pr´ec´edent, il est facile de voir que la strat´egie optimale est d’atteindre le point ( 12 , 0) o` u la fonction coˆ ut de sortie ϕ est minimale. Mais la dynamique ne permet que de “monter” ou/et d’aller “sur la droite”, si bien que seules les trajectoires issues de points de [−1, 0]2 peuvent atteindre le point ( 12 , 0) et encore faut-il que leurs trajectoires longent une partie du bord et “s’arrˆetent” en ce point. Comme ces trajectoires ne sont pas admises pour la fonction-valeur u, nous obtenons :   1 si x ∈ Ω u(x) =  ϕ(x) si x ∈ ∂Ω. Par contre, si nous d´efinissons une fonction-valeur u− avec le meilleur temps d’arrˆet sur le bord, c’est-`a-dire : u− (x) :=

inf

α(.)∈L∞ (IR+ ,A)

nZ 0

θ

f (yx (t), α(t))e−λt dt + ϕ(yx (θ))e−λθ , τ ≤ θ ≤ τ¯ o

et yx (θ) ∈ ∂Ω , alors : u− (x) =

  1

si x ∈ [−1, 1]×]0, 1] −d(x,0)− 21

 1−e

si x ∈ [−1, 0]2 ,

o` u d(x, 0) correspond `a la distance entre le point x et l’origine du rep`ere. Sur le bord 1 [0, 1] × {0}, on obtient : u(x1 , 0) = 1 − e−( 2 −x1 ) si x1 ≤ 12 et u = ϕ si x1 > 12 . Cet exemple rappelle ´egalement que la fonction-valeur d’un probl`eme de temps de sortie peut ˆetre discontinue mˆeme si la fonction coˆ ut de sortie ϕ est r´eguli`ere (ainsi que les autres donn´ees du probl`eme). En effet, la fonction-valeur d´epend `a la fois de la dynamique et de la “topographie” du domaine consid´er´e. Une autre possibilit´e de choix pour la d´efinition de la fonction-valeur u provient de la discontinuit´e des fonctions coˆ ut sur le bord. Pour les probl`emes de cible ´evoqu´es pr´ec´edemment, cela revient `a se demander si le bord de la cible (dans ∂Ω) doit ˆetre consid´er´e comme faisant partie int´egrante de la cible ou pas. En modifiant un peu l’exemple pr´ec´edent, nous allons montrer que cela peut avoir de l’importance pour la fonction-valeur. La fonction coˆ ut ϕ est maintenant ´egale `a 1 partout sauf sur [0, 1] × {0} o` u elle est nulle. Si la valeur de ϕ au point (0, 0) (le bord de la cible avec le point (1, 0)) vaut 0, alors : u ≡ u−

dans Ω,

car les trajectoires tangentes au bord ne sont plus n´ecessaires. Par contre, si ϕ(0, 0) = 1, alors : u ≡ 1 > u− dans [−1, 0]2 ,

1.3 Probl´ematique

13

puisque l’on retrouve la situation pr´ec´edente. De fa¸con plus g´en´erale, au lieu de prendre simplement la fonction coˆ ut ϕ, nous pouvons consid´erer ses enveloppes semi-continues. On rappelle que l’enveloppe semicontinue sup´erieurement (s.c.s. en abr´eg´e) d’une fonction localement born´ee ζ se note ζ ∗ et est d´efinie par : ζ ∗ (x) := lim sup ζ(y). y→x

De mˆeme, la fonction ζ∗ est l’enveloppe semi-continue inf´erieurement (s.c.i. en abr´eg´e) de ζ d´efinie par : ζ∗ (x) := lim inf ζ(y). y→x

Pour en revenir aux probl`emes de cible, la fonction coˆ ut s’´ecrivant C11∂Ω\Γ , prendre son enveloppe s.c.i. ou s.c.s. revient `a consid´erer ou non le bord de la cible car : (11∂Ω\Γ )∗ = 11

◦

∂Ω\Γ

et

(11∂Ω\Γ )∗ = 11∂Ω\Γ .

Enfin, un aspect plus technique vient du fait que les contrˆoles classiques i.e. L∞ (IR+ , A) n’ont pas de propri´et´e de compacit´e et qu’il faut les plonger dans l’ensemble des contrˆoles dits relax´es c’est-`a-dire appartement `a L∞ (IR+ , P (A)) o` u P (A) est l’espace des mesures de probabilit´es sur A qui ont, eux, une propri´et´e de compacit´e faible. Pour un contrˆole relax´e µ, la trajectoire associ´ee est d´efinie par :  Z dˆ yx   b(ˆ yx (t), α)dµt (α), (t) =

dt A yˆx (0) = x ∈ Ω.

 

Si, dans IRN , les trajectoires associ´ees `a des contrˆoles relax´es peuvent ˆetre approch´ees par des trajectoires classiques, ce n’est plus le cas dans un ferm´e quelconque de IRN comme le montre un exemple dans le premier article de ce travail. Pour synth´etiser toutes les consid´erations vues jusqu’ici, deux fonctions-valeur apparaissent naturelles. La premi`ere est construite en utilisant les contrˆoles relax´es, le coˆ ut de sortie ϕ∗ et le meilleur temps d’arrˆet sur le bord : u− (x) :=

inf+

nZ

µ∈L∞ (IR ,P (A))

θ

Z

0

V

f (ˆ yx (t), α)e−λt dµt (α)dt + ϕ∗ (ˆ yx (θ))e−λθ , τ ≤ θ ≤ τ¯ o

et yˆx (θ) ∈ ∂Ω . La seconde est d´efinie en faisant les choix “inverses”, c’est-`a-dire : u+ (x) :=

inf

α(.)∈L∞ (IR+ ,A)

n

Z

sup{

0

θ

f (yx (t), α(t))e−λt dt + ϕ∗ (yx (θ))e−λθ , τ ≤ θ ≤ τ¯ o

et yx (θ) ∈ ∂Ω} .

14

Probl`emes de contrˆole optimal avec temps de sortie

Il d´ecoule imm´ediatement des d´efinitions des fonctions-valeur u, u− et u+ , la relation suivante : u− ≤ u ≤ u+ dans Ω, ces in´egalit´es pouvant ˆetre strictes.

1.4 Etudes directes des trajectoires Dans cette partie, nous voulons r´esoudre les difficult´es soulev´ees dans la section pr´ec´edente en cherchant des hypoth`eses qui permettent de montrer que ces fonctionsvaleur sont essentiellement les mˆemes. Plus pr´ecis´ement, nous voulons obtenir : (u− )∗ = (u+ )∗

dans Ω.

(1.2)

Il est naturel de se limiter `a l’´egalit´e pour les enveloppes s.c.i. parce que les fonctionsvaleur n’ont pas de raison d’ˆetre continues, les discontinuit´es des coˆ uts de sortie pouvant se prolonger `a l’int´erieur de l’ouvert. Ce point apparaˆıt clairement en reprenant l’exemple du probl`eme de la cible dans ]0, 1[2 avec un coˆ ut de sortie ´egal `a 1 partout 1 u il est nul. sauf sur [0, 2 ] × {0} o` En fait, nous montrons dans le premier article de la th`ese que la fonction-valeur u− est s.c.i. et d’o` u (u− )∗ = u− (alors que u+ est s.c.s.). La principale difficult´e pour obtenir l’´egalit´e (1.2), provient des trajectoires qui touchent plusieurs fois le bord. Pour la r´esoudre, nous faisons une hypoth`ese sur la dynamique pr`es du bord. Nous supposons que l’ouvert est r´egulier et qu’il existe en tout point du bord, un contrˆole α tel que b(x, α) soit dirig´e vers l’ext´erieur du domaine ; en d’autres termes : ∀x ∈ ∂Ω,

∃α ∈ A

tel que

b(x, α) · n(x) > 0

o` u n(x) est la normale au bord en x ∈ ∂Ω dirig´ee vers l’ext´erieur du domaine Ω. (Signalons que dans les exemples pr´ec´edents, les domaines Ω n’´etaient pas r´eguliers pour simplifier leur d´efinitions mais que les mˆemes conclusions resteraient vraies s’ils ´etaient r´eguliers.) Alors, grˆace `a ces contrˆoles “sortants”, une trajectoire relax´ee yˆ qui touche le bord en un point yˆ(θ) ∈ ∂Ω pour θ quelconque dans [τ, τ¯], peut ˆetre approch´ee par une trajectoire classique dont le premier point de sortie de l’ouvert Ω est yˆ(θ). Pour donner une id´ee de la d´emonstration, on consid`ere les trajectoires dans le sens r´etrograde, c’est-`a-dire que l’on inverse le sens de parcours. Alors `a partir du point yˆ(θ), on utilise le contrˆole qui sort du domaine (donc qui entre pour les trajectoires r´etrogrades) pour “pousser” la trajectoire y `a l’int´erieur de Ω pendant un court laps de temps puis on suit la trajectoire relax´ee. Et d`es qu’elle retouche le bord, on utilise de nouveau un contrˆole sortant pour la “repousser” `a l’int´erieur.

1.5 L’´equation de Hamilton-Jacobi-Bellman

15

Ω yˆ

y

s

yˆ(θ)

Une difficult´e suppl´ementaire pour avoir l’´egalit´e (1.2), provient du fait que l’on prend ϕ∗ dans la d´efinition de u− et ϕ∗ dans celle de u+ . Pour r´esoudre ce probl`eme, une premi`ere solution consiste `a se limiter `a des coˆ uts de sortie r´eguliers c’est-`a-dire v´erifiant : (ϕ∗ )∗ = ϕ∗ sur ∂Ω. Mais cette derni`ere condition est en fait assez restrictive et elle sera lev´ee dans la partie suivante. Car si elle est v´erifi´ee par les fonctions s.c.s., elle interdit les coˆ uts de sortie s.c.i. du type 11∂Ω\{x0 } , par exemple associ´es `a une cible ponctuelle {x0 }. En effet, leur enveloppe s.c.s. est : (11∂Ω\{x0 } )∗ ≡ 1. Le point int´eressant x0 puisque le coˆ ut de sortie ϕ y est nul, n’est plus “vu” lorsque l’on prend l’enveloppe s.c.s.

1.5 L’´ equation de Hamilton-Jacobi-Bellman La m´ethode classique pour ´etudier les probl`emes de contrˆole optimal est d’obtenir des ´equations associ´ees au probl`eme : soit par le principe de Pontryagin [31] si on se place du point de vue des trajectoires en cherchant la trajectoire optimale, soit par le principe de la programmation dynamique qui conduit `a l’´equation de HamiltonJacobi-Bellman si on s’int´eresse `a la fonction-valeur (cf. [20]). Dans ce travail, c’est cette derni`ere approche que nous utilisons. Le principe de la programmation dynamique d´ecoule du principe d’optimalit´e de Bellman [12] et il s’´ecrit dans notre cas : Th´ eor` eme 1.1 Pour tout T > 0 et pour tout x ∈ Ω, on a : u(x) =

inf

α(.)∈L∞ (IR+ ,A)

nZ 0

τ ∧T

f (yx (t), α(t))e−λt dt o

+11{τ ≤T } ϕ(yx (τ ))e−λτ + 11{τ >T } u(yx (T ))e−λT .

2

16

Probl`emes de contrˆole optimal avec temps de sortie

Bien que ce principe ne soit pas habituellement ´ecrit pour un coˆ ut de sortie discontinu, sa d´emonstration d´ecoule des d´emonstrations classiques (cf. P.L. Lions [28]). Donnons rapidement une esquisse de la fa¸con d’en d´eduire l’´equation que satisfait la fonction-valeur lorsqu’elle est r´eguli`ere. Pour un point x de l’ouvert Ω, les temps de sortie τ sont minor´es par une constante strictement positive, puisque la fonction b est born´ee. Ainsi pour des temps T assez petits, on en d´eduit : u(x) =

inf

nZ

α(.)∈L∞ (IR+ ,A)

T

0

o

f (yx (t), α(t))e−λt dt + u(yx (T ))e−λT .

On soustrait le terme en inf de chaque cˆot´e et en divisant par T , on obtient : n u(x) − u(y (T ))e−λT x

sup α(.)∈L∞ (IR+ ,A)

T

o 1ZT − f (yx (t), α(t))e−λt dt = 0. T 0

Pour finir, on fait tendre T vers 0 et on obtient ainsi formellement que la fonction u satisfait `a l’´equation : H(x, u(x), Du(x)) = 0 (1.3) avec : H(x, t, p) = sup{−b(x, α) · p + λt − f (x, α)}.

(1.4)

α∈A

La fonction H est habituellement appel´ee l’hamiltonien. Mais la fonction-valeur comme celle des exemples pr´ec´edents, n’est en g´en´erale pas assez r´eguli`ere pour satisfaire l’´equation (1.3) de fa¸con classique, dans C 1 par exemple. Nous sommes donc amen´e `a consid´erer l’´equation dans le sens plus faible et bien adapt´e `a ces probl`emes que constitue la notion de solutions de viscosit´e dont une pr´esentation rapide est l’objet du paragraphe suivant. 1.5.1 Les solutions de viscosit´ e La notion de solutions de viscosit´e a ´et´e introduite en 1981 par M.G. Crandall et P.L. Lions [16] pour les ´equations de Hamilton-Jacobi (et simplifi´ee ensuite par M.G. Crandall, L.C. Evans et P.L. Lions [14]). L’id´ee de base est de remplacer les “d´eriv´ees” de la solution dans l’´equation par les d´eriv´ees au sens classique de fonctions-test ´evalu´ees aux points de minimum et maximum de la diff´erence entre la solution et la fonction-test. Pour expliquer et rattacher ce concept fondamental `a un r´esultat bien connu, nous allons nous placer momentan´ement dans le cadre des ´equations elliptiques du type : H(x, u, Du, D2 u) = 0 dans Ω. (1.5)

1.5 L’´equation de Hamilton-Jacobi-Bellman

17

On rappelle que l’´equation (1.5) est dite elliptique si la condition suivante est satisfaite : H(x, u, p, M1 ) ≤ H(x, u, p, M2 ) si M2 ≤ M1 pour tout x ∈ Ω, p ∈ IRN et M1 , M2 ∈ S N o` u S N est l’espace vectoriel des matrices N × N sym´etriques muni de l’ordre partiel ≤ d´efini par : M2 ≤ M1

⇔

(M2 p, p) ≤ (M1 p, p) ∀p ∈ IRN .

Alors pour l’´equation (1.5), il est bien connu que l’on a le r´esultat suivant, bas´e sur le principe du maximum : Th´ eor` eme 1.2 Une fonction u ∈ C 2 (Ω) est une solution classique de l’´equation (1.5) si et seulement si : ∀φ ∈ C 2 (Ω), si x0 ∈ Ω est un point de maximum local de u − φ, on a : H(x0 , u(x0 ), Dφ(x0 ), D2 φ(x0 )) ≤ 0,

(1.6)

et : ∀φ ∈ C 2 (Ω), si x0 ∈ Ω est un point de minimum local de u − φ, on a : H(x0 , u(x0 ), Dφ(x0 ), D2 φ(x0 )) ≥ 0.

(1.7) t u

Les conditions (1.6) et (1.7) constituent donc une d´efinition ´equivalente de la notion de solution classique. Et la remarque essentielle `a faire est que la r´egularit´e de u n’intervient pas dans (1.6) et (1.7) : il suffit de pouvoir d´efinir un point de maximum ou de minimum (donc d’avoir des fonctions s.c.s. ou s.c.i.). Pour les ´equations du premier ordre, la notion de solution de viscosit´e est ainsi donn´ee par : D´ efinition 1.1 On dit qu’une fonction born´ee u est une sous-solution de viscosit´e de l’´equation (1.3) dans Ω si et seulement si : ∀φ ∈ C 1 (Ω), si x0 ∈ Ω est un point de maximum local de u∗ − φ, on a : H(x0 , u∗ (x0 ), Dφ(x0 )) ≤ 0.

(1.8)

De mˆeme, u est une sursolution de viscosit´e de l’´equation (1.3) dans Ω si et seulement si : ∀φ ∈ C 1 (Ω), si x0 ∈ Ω est un point de minimum local de u∗ − φ, on a : H(x0 , u∗ (x0 ), Dφ(x0 )) ≥ 0.

(1.9)

18

Probl`emes de contrˆole optimal avec temps de sortie

Finalement, on dit que u est une solution de viscosit´e de (1.3) si et seulement si u est ` a la fois une sous- et une sursolution de viscosit´e de (1.3). Dans toute la suite, nous ne consid´ererons plus que des solutions de viscosit´e sans forc´ement le repr´eciser. Ce concept de solution a montr´e depuis une quinzaine d’ann´ees toute sa pertinence pour les ´equations aux d´eriv´ees partielles non-lin´eaire du premier et du second ordre aussi bien pour des questions d’existence que d’unicit´e ou encore de stabilit´e. On trouvera une introduction plus d´etaill´ee dans [4] pour le premier ordre et dans [15] pour le second ordre. La th´eorie des solutions de viscosit´e s’est r´ev´el´ee ´egalement tr`es utile dans le cadre des probl`emes de contrˆole optimal et de Jeux Diff´erentiels, indiquons par exemple les livres de W.H. Fleming et H.M. Soner [21] et celui de M. Bardi et I. Capuzzo Dolcetta [2]. Il reste enfin `a pr´eciser les conditions que l’on doit mettre sur le bord du domaine : le paragraphe suivant est consacr´e `a ce d´elicat probl`eme. 1.5.2 Conditions de Dirichlet discontinues La notion classique de conditions de Dirichlet qui consiste `a supposer qu’en tout point du bord, la solution est ´egale `a la donn´ee au bord i.e. u(x) = ϕ(x)

sur ∂Ω

ne se r´ev`ele pas pertinente. Nous avons vu pr´ec´edemment que les fonctions-valeurs ne v´erifient pas naturellement cette condition. Pour r´esoudre ce probl`eme, la th´eorie des solutions de viscosit´e procure un cadre plus adapt´e. En effet, on dira qu’une fonction u satisfait aux conditions de Dirichlet sur le bord si elle v´erifie les ´equations suivantes : min{H(x, u, Du), u − ϕ∗ } ≤ 0

sur ∂Ω

max{H(x, u, Du), u − ϕ∗ } ≥ 0

sur ∂Ω

(1.10) qui sont `a consid´erer au sens de viscosit´e comme dans la d´efinition 1.1 en autorisant les points de maximum et de minimum `a ˆetre sur le bord. Ce type de condition a ´et´e introduit par H. Ishii [25] pour des probl`emes de temps de sortie mais avec des fonctions coˆ ut de sortie ϕ continues (voir aussi G. Barles et B. Perthame [8]). P. Soravia [34] a r´ecemment propos´e d’autres conditions mais, semble-t-il, inadaptables `a notre probl`eme car toujours pour des fonctions ϕ continues. Nous prouvons que les fonctions-valeur u− , u+ et u d´efinies pr´ec´edemment sont toutes solutions des ´equations (1.3)-(1.10). Le probl`eme de Dirichlet (1.3)-(1.10) ainsi

1.5 L’´equation de Hamilton-Jacobi-Bellman

19

pos´e, apparaˆıt bien adapt´e pour caract´eriser les fonctions-valeur de probl`emes de temps de sortie avec coˆ ut de sortie discontinu. Mais cela entraˆıne aussi que ce probl`eme de Dirichlet a plusieurs solutions, a priori. Il convient donc de d´eterminer dans quel cas nous avons des r´esultats d’unicit´e pour (1.3)-(1.10). Un premier r´esultat d’unicit´e provient de l’´etude que nous avons effectu´ee sur les trajectoires dans une partie pr´ec´edente. En effet, nous pouvons montrer que u+ est la sous-solution maximale de (1.3)-(1.10) lorsque H est d´efinit par (1.4), et qu’inversement u− est la sursolution minimale de (1.3)-(1.10). Ainsi, toute solution w de (1.3)-(1.10) v´erifie l’in´egalit´e suivante : u− ≤ w∗ ≤ w∗ ≤ u+

dans Ω.

(1.11)

Il ne reste plus alors qu’`a utiliser la relation que nous avons obtenue entre u− et u+ ; on rappelle que sous les hypoth`eses : (ϕ∗ )∗ = ϕ∗

sur ∂Ω,

(1.12)

et : ∀x ∈ ∂Ω,

∃α ∈ A

tel que

b(x, α) · n(x) > 0,

(1.13)

on a u− = (u+ )∗ dans Ω. Grˆace `a (1.11), l’unicit´e des enveloppes s.c.i. des solutions de (1.3)-(1.10) s’en d´eduit imm´ediatement : u− = w∗ = (w∗ )∗

dans Ω.

(1.14)

Nous obtenons ainsi un r´esultat d’unicit´e dans un cadre r´eellement discontinu. Les r´esultats classiques de principe de maximum permettent de comparer une sursolution et une sous-solution. Ils entraˆınent, en particulier, pour une solution w l’in´egalit´e suivante : w∗ ≤ w∗ dans Ω, mais comme par d´efinition w∗ ≤ w∗ , cela implique la continuit´e de la solution w. Par exemple, G. Barles et B. Perthame [9] ont obtenu l’unicit´e et la continuit´e de la fonction-valeur, en particulier, s’il existe un champs rentrant i.e. (1.12) et sortant en tout point du bord. Les r´esultats d’unicit´e sont nombreux dans la th´eorie des solutions de viscosit´e. Pour les probl`emes de Dirichlet, il faut retenir le travail de H.M. Soner [32] qui a permit une nouvelle approche de ces probl`emes mˆeme s’il traitait, au d´epart, des probl`emes de contraintes d’´etat ; mentionnons ´egalement l’article [9] que nous venons de d´ecrire et des travaux plus r´ecents comme ceux de M. Bardi et P. Soravia [3]. La condition (1.12) impos´ee `a la fonction ϕ par ce premier r´esultat, peut se r´ev´eler trop restrictive dans certains cas. Mais, par des m´ethodes E.D.P., nous pouvons

20

Probl`emes de contrˆole optimal avec temps de sortie

l’´eviter et obtenir, avec l’hypoth`ese (1.13), un r´esultat d’unicit´e de l’enveloppe s.c.i. des solutions w de (1.3)-(1.10) qui v´erifient la condition suppl´ementaire suivante : lim inf w(y) ≤ ϕ∗ (x)

y→x,y∈Ω

sur ∂Ω.

(1.15)

Ce r´esultat repose sur un principe de comparaison faible qui implique : u∗ ≤ v

dans Ω

si u est une sous-solution s.c.s. de (1.3)-(1.10) satisfaisant (1.15) et si v est une sursolution s.c.i. Notons que l’´egalit´e u− = (u+ )∗ dans Ω peut se d´eduire de ce principe en l’appliquant `a u− et `a u+ car, sous les hypoth`eses (1.12) et (1.13), la fonction u+ v´erifie (1.15). La condition (1.15) assure que la donn´ee au bord ϕ est bien prise en compte, en particulier pour ´eviter les probl`emes soulev´es dans un exemple pr´ec´edent o` u, lorsque l’on prend l’enveloppe s.c.s. de 11∂Ω\{x0 } , on “perd” la valeur int´eressante en x0 car (11∂Ω\{x0 } )∗ ≡ 1. Cette condition apparaˆıt donc naturelle pour les probl`emes de contrˆole optimal mˆeme si elle semble d´elicate `a mettre en œuvre. N´eanmoins dans le second article de la th`ese, nous la v´erifierons explicitement. Pour d´emontrer ce r´esultat, nous sommes amen´es `a utiliser une notion de solution de viscosit´e diff´erente de celle pr´esent´ee pr´ec´edemment qui a ´et´e introduite r´ecemment par E.N. Barron et R. Jensen [10] pour les hamiltoniens convexes.

2. Probl` emes de Grandes D´ eviations

2.1 Pr´ esentation g´ en´ erale Le but de la th´eorie des Grandes D´eviations est de donner des estimations d’´ev´enements “rares”, c’est-`a-dire intervenant avec une probabilit´e faible. Le cadre classique o` u apparaissent de tels ´ev´enements, est celui de la th´eorie des perturbations al´eatoires des syst`emes dynamiques que nous allons donc d´ecrire bri`evement. Consid´erons d’abord une ´equation diff´erentielle ordinaire dans IRN :   dXt 

= b(Xt )dt,

X0 = x ∈ IRN ,

o` u b est un champ lipschitzien sur IRN . Nous la perturbons alors avec un “bruit blanc” c’est-`a-dire avec un mouvement brownien, de la fa¸con suivante :  ε  dXt 

= b(Xtε )dt + εσ(Xtε )dBt ,

Xtε = x ∈ IRN .

(2.1)

Nous obtenons ainsi une ´equation diff´erentielle stochastique o` u σ est une matrice N × p qui est une fonction lipschitzienne et o` u (Bt )t est un processus brownien standard de IRp . Il est bien connu que les trajectoires (Xtε )t convergent en probabilit´e vers la trajectoire non perturb´ee (Xt )t . N´eanmoins elles peuvent avoir un comportement qui diff`ere de celui de (Xt )t mais avec une probabilit´e faible que la th´eorie des Grandes D´eviations permet de d´eterminer de fa¸con pr´ecise. On trouvera une introduction plus compl`ete dans le livre M.I. Freidlin et A.D. Wentzell [37], entre autres. Illustrons notre propos par un exemple : nous consid´erons la couronne Ω := {x ∈ IRN tel que 1 < |x| < 2} avec le champ b(x) := x.

22

Probl`emes de Grandes D´eviations

Le comportement des trajectoires (Xt )t est donc simple : elles sortent par le bord ext´erieur de la couronne i.e. {|x| = 2}. Ainsi, si on appelle Γ le bord int´erieur de la couronne i.e. Γ := {|x| = 1}, la probabilit´e de sortie `a travers Γ pour une trajectoire issue de Ω est nulle : IP x (Xτ ∈ Γ) = 0, o` u IP x est la probabilit´e conditionnelle par rapport `a l’´ev´enement {X0 = x} et o` uτ est le premier temps de sortie de la trajectoire (Xt )t hors de Ω i.e. τ := inf{t ≥ 0, Xt 6∈ Ω}. Perturbons maintenant l’´equation en prenant pour simplifier p = N et la matrice σ ´egale `a la matrice identit´e de N × N . Alors certaines trajectoires perturb´ees (Xtε )t sortent par Γ. La probabilit´e IP x (Xτε ∈ Γ) n’est pas nulle mˆeme si elle tend vers 0. Puisque Xτε ∈ Γ est un ´ev´enement “rare”, la th´eorie des Grandes D´eviations nous permet d’obtenir une estimation plus pr´ecise de la vitesse de cette convergence vers 0 sous la forme de : −ε2 ln(IP x (Xτε ∈ Γ)) → I(x)

quand ε → 0, pour x ∈ Ω,

(2.2)

o` u la fonction I est appel´ee la fonctionnelle d’action. L’´egalit´e pr´ec´edente peut aussi s’´ecrire :  I(x) + o(1)  IP x (Xτε ∈ Γ) = exp − ε2 montrant que la convergence est exponentiellement rapide vers 0 si I > 0. La fonction I est donn´ee dans le probl`eme de la couronne par : I(x) := inf

n1 Z

2

0

τ

1 |y˙ x (s) − b(yx (s))|2 ds ; yx (.) ∈ Hloc (IR+ ; IRN ),

o

yx (0) = x, yx (τ ) ∈ Γ, yx (t) ∈ Ω pour t < τ . Elle peut ˆetre vue comme la fonction-valeur d’un probl`eme de temps de sortie `a travers Γ. La th´eorie g´en´erale des Grandes D´eviations a ´et´e introduite et d´evelopp´ee par A.D. Wentzell et M.I. Freidlin [36] et M.D. Donsker et S.R.S. Varadhan [17]. On trouvera dans le livre de M.I. Freidlin [22] une pr´esentation r´ecente sur l’´etat de l’art en la mati`ere. Dans le pr´esent travail, nous g´en´eralisons le probl`eme de la couronne aux probl`emes de sortie `a travers une partie ouverte quelconque du bord qui peut, de plus, varier avec le param`etre ε ; nous la noterons dans ce cas Γε . Ce qui revient `a l’´etude du comportement asymptotique de la fonction : uε (x) := IP x (Xτε ∈ Γε )

pour x ∈ Ω,

2.1 Pr´esentation g´en´erale

23

o` u (Xtε )t est la solution de (2.1). L’´etude de uε d´epend du comportement des parties (Γε )ε . Si les parties Γε “convergent” essentiellement vers un ouvert de ∂Ω, alors nous montrons que uε a le comportement attendu c’est-`a-dire (2.2) mˆeme si la matrice σ est d´eg´en´er´ee (mais alors le r´esultat s’´ecrit de fa¸con un peu diff´erente). Nous ´etendons ainsi un r´esultat de L.C. Evans et H. Ishii [18] aux matrices σ d´eg´en´er´ees et ´egalement un r´esultat de G. Barles et B. Perthame [9] en consid´erant des parties quelconques du bord. Si l’ensemble Γε d´ecroˆıt jusqu’`a n’ˆetre plus qu’un point du bord par exemple, nous parlerons de parties ´evanescentes et le probl`eme se complique encore. En effet, le comportement de uε est alors soumis `a l’influence de deux effets : la diminution de “l’intensit´e” de la perturbation qui est la cons´equence normale de la d´efinition des probl`emes de perturbation mais aussi la r´eduction de la taille des Γε qui constitue en partie l’originalit´e du pr´esent travail. Pour essayer d’expliquer comment se combinent les deux effets, nous allons d´ecrire un probl`eme semblable mais dans un cadre plus simple. Nous consid´erons, en effet, l’´equation bien connue de la chaleur en dimension N : ∂uε ε2 − ∆uε = 0 dans IRN × (0, +∞), ∂t 2 avec une donn´ee initiale qui est la fonction indicatrice d’une boule centr´ee `a l’origine et de rayon ρε i.e. u0 (x) = 11B(0,ρε ) (x) dans IRN , o` u (ρε )ε est une suite positive qui tend vers 0 ; d’o` u Γε := B(0, ρε ). L’avantage de cet exemple vient du fait que la solution uε se calcule explicitement : Z  |y − x|2  1 1 1 (y) exp − dy. uε (x, t) = B(0,ρε ) (2πtε2 )n/2 IRN 2tε2 Ainsi nous pouvons ´etudier pr´ecis´ement le comportement de uε en ´ecrivant de fa¸con formelle :  |x|2  1 uε (x, t) ' meas (B(0, ρ )) exp − 2 , ε (2πtε2 )n/2 2tε puisque ε tend vers 0. La transformation logarithmique −ε2 ln effectu´ee sur cette relation pour faire apparaˆıtre la forme classique (2.2), montre l’importance de la vitesse de d´ecroissance du domaine cible Γε c’est-`a-dire de la limite du terme : −ε2 ln(meas (B(0, ρε ))) ( ' −N ε2 ln(ρε ) ), 1. Dans le cas o` u −ε2 ln(ρε ) → 0, le comportement classique a lieu : −ε2 ln(uε ) →

|x|2 2t

dans IRN × (0, +∞).

24

Probl`emes de Grandes D´eviations

2. Dans le cas o` u −ε2 ln(ρε ) → +∞, nous obtenons : −ε2 ln(uε ) → +∞

dans IRN × (0, +∞).

3. Enfin, le cas limite o` u −ε2 ln(ρε ) → ρ > 0 entraˆıne : −ε2 ln(uε ) → N ρ +

|x|2 2t

dans IRN × (0, +∞).

Pour le probl`eme stationnaire, nous obtenons ´egalement des r´esultats diff´erents selon la vitesse de d´ecroissance de la taille de la cible Γε .

2.2 M´ ethodes de r´ esolution Historiquement, la th´eorie des Grandes D´eviations s’est d’abord d´evelopp´ee dans un cadre probabiliste. W.H. Fleming [19] a ´et´e le premier `a introduire une approche E.D.P. pour traiter ces questions, en consid´erant, en particulier, les ´equations que satisfont les probabilit´es li´ees aux ´ev´enements rares consid´er´es. Il est, en effet, bien connu qu’il existe un lien ´etroit entre les processus de diffusion et les ´equations elliptiques du second ordre grˆace notamment aux formules FeynmanKac. Le probl`eme qui nous int´eresse est ainsi reli´e `a l’´equation :  X  1 2X ∂uε ∂ 2 uε   − ε bi (x) ai,j (x) −    2 i,j ∂xi ∂xj ∂xi  i       

= 0

dans Ω,

uε = 1

sur Γε ,

uε = 0

sur ∂Ω\Γε ,

o` u la matrice (ai,j ) est d´efinie par σσ T . Notons qu’`a cause des conditions discontinues sur le bord et de la possible d´eg´en´erescence de σ, la fonction : uε (x) := IP x (Xτε ∈ Γε ) n’est solution de cette ´equation dans tout Ω qu’au sens de viscosit´e. Puisque nous voulons une estimation logarithmique et en suivant la m´ethode introduite par W.H. Fleming, nous proc´edons au changement de variable : I ε := −ε2 ln(uε ). Puisque nous nous contentons de donner, ici, une description rapide et formelle de la d´emonstration, nous n’entrons pas dans le d´etail des probl`emes li´es au fait que la fonction uε pourrait ˆetre nulle `a certain endroit.

2.2 M´ethodes de r´esolution

25

La fonction I ε est ainsi solution du probl`eme :  1X ∂2I ε ∂I ε ∂I ε X ∂I ε 1 2X    ε + − a (x) a (x) b (x) − i,j i,j i   2 i,j ∂xi ∂xj 2 i,j ∂xi ∂xj ∂xi   i

= 0

Iε = 0

      

dans Ω, sur Γε ,

I ε = +∞ sur ∂Ω\Γε .

L’approche de W.H. Fleming n´ecessite alors des estimations de la fonction I ε et de ses d´eriv´ees ind´ependantes de ε, pour obtenir que la limite d’une sous-suite de I ε est solution d’un probl`eme du premier ordre. Mais la th´eorie des solutions de viscosit´e permet d’´eviter les estimations de gradient, grˆace `a la m´ethode des “semilimites relax´ees” de [7] qui n´ecessite seulement une borne uniforme sur la fonction I ε. En supposant I ε born´e, la m´ethode des “semi-limites relax´ees” introduit deux fonctions : ¯ I(x) := lim sup I ε (y) et I(x) := lim inf I ε (y) y→x ε→0

y→x ε→0

qui sont respectivement sous- et sursolution de l’´equation :  X 1X ∂I ∂I ∂I   a (x) − bi (x)  i,j   ∂xi ∂xj ∂xi  i  2 i,j       

= 0

dans Ω,

I = 0

sur Γ,

I = +∞

sur ∂Ω\Γ.

(2.3)

Pour obtenir la convergence de I ε , il faut alors classiquement un r´esultat de comparaison pour le probl`eme (2.3) qui donne : I¯ ≤ I dans Ω. Car, comme par ¯ cela implique : I¯ = I dans Ω et la convergence uniforme de I ε vers d´efinition I ≤ I, la fonction continue I := I = I¯ sur tout compact de Ω. Mais, les solutions de (2.3) ´etant discontinues, nous n’aurons pas ce type de r´esultat dans ce cas. Avant de traiter le probl`eme du r´esultat de comparaison entre I¯ et I qui est la principale difficult´e de la preuve, nous allons expliquer comme finir la d´emonstration. Il reste en effet `a obtenir l’expression de la fonctionnelle d’action I. Pour cela, nous identifions le probl`eme (2.3) comme l’´equation de la programmation dynamique d’un probl`eme de contrˆole optimal. La fonctionnelle I est ainsi une fonction-valeur d’un probl`eme de cible comme ceux ´etudi´es dans le premier chapitre. Revenons maintenant sur le r´esultat de comparaison pour I¯ et I. Les ´equations (2.3) correspondent `a un probl`eme de Dirichlet discontinu ; nous retrouvons le cadre d´ecrit dans le premier chapitre de ce travail. Nous devons n´eanmoins adapter les

26

Probl`emes de Grandes D´eviations

r´esultats d’unicit´e obtenus pr´ec´edemment car le probl`eme (2.3) pr´esente des difficult´es suppl´ementaires : l’hamiltonien correspondant ´etant H(x, r, p) :=

X 1X ai,j (x)pi pj − bi (x)pi , 2 i,j i

il ne d´epend, en fait, pas de la variable r et il a par contre un terme `a croissance quadratique. Si la limite de (Γε )ε est, en particulier, un ouvert de Ω, la condition de Dirichlet ϕ sur le bord est une fonction r´eguli`ere selon la terminologie du premier chapitre c’est-`a-dire qu’elle v´erifie (1.12). Nous pouvons alors appliquer le r´esultat d’unicit´e qui entraˆıne : ¯ ∗ ≤ I dans Ω, (I) ¯ ∗ = I dans Ω. Nous n’obtenons pas ce qui permet de conclure la preuve car alors (I) exactement la formule (2.2) parce que nous sommes dans le cas o` u la matrice σ peut ˆetre d´eg´en´er´ee. Dans le cas d’une partie ´evanescente, la fonction ϕ n’´etant pas r´eguli`ere, il faut montrer que I¯ satisfait `a la condition (1.15) i.e. ¯ ≤ 0 lim inf I(y)

y→x,y∈Ω

pour x ∈ Γ.

C’est une des principales difficult´es de la preuve. La d´emonstration de ce r´esultat utilise un argument de changement d’´echelle qui fait apparaˆıtre une condition sur la vitesse de d´ecroissance des parties Γε . Et malheureusement, elle n´ecessite une matrice σ non-d´eg´en´er´ee. Alors, si ε2 ln(meas(Γε )) → 0, l’estimation (2.2) a lieu. Par contre, dans le cas −ε2 ln(meas(Γε )) → ∞, nous obtenons : −ε2 ln(uε ) → +∞

dans Ω.

Ce r´esultat se d´emontre par des m´ethodes diff´erentes qui consistent `a introduire le probl`eme adjoint (comme dans B. Perthame [30]). Enfin, dans le cas limite o` u −ε2 ln(ρε ) → ρ > 0, nous n’avons qu’un r´esultat partiel par rapport `a ce que nous pouvions extrapoler de l’exemple de l’´equation de la chaleur.

Partie I PROBLEMES DE TEMPS DE SORTIE AVEC COUTS DE SORTIE DISCONTINUS

27

3. Deterministic exit time control problems with discontinuous exit cost1

Introduction This paper is concerned with deterministic exit time control problems with discontinuous exit costs. We want to characterize the value functions of this type of control problems as the unique viscosity solutions of the corresponding Hamilton-JacobiBellman problem. We obtain, under some suitable conditions, the uniqueness of the lower semi-continuous envelope of the value function. In order to be more specific, we now describe the optimal control problem. Let Ω be a smooth bounded domain of IRN . We consider a system which state is given by the solution of the ordinary differential equation   dyx (t) = b(yx (t), α(t))dt, 

yx (0) = x ∈ Ω,

where b is a Lipschitz continuous function from Ω×A into IRN and α(.) ∈ L∞ (IR+ , A) is the control, A the control space being a compact metric space. We denote by τ the first exit time of the trajectory yx from Ω i.e. τ = inf{t ≥ 0, yx (t) 6∈ Ω}. In this framework, it is well-know that the value function can be defined in several way : the first value function we want to consider is given by u(x) =

inf

α(.)∈L∞ (IR+ ,A)

nZ 0

τ

o

f (yx (t), α(t))e−λt dt + ϕ(yx (τ ))e−λτ ,

where f is a given continuous real-valued function and λ is a positive constant, the discount factor. Precise assumptions on f and b are detailed in the first part. 1

`a paraˆıtre au SIAM Journal on Control and Optimization.

30

Probl`emes de temps de sortie

The main point is that we assume only that ϕ, the exit cost, is a bounded function on ∂Ω defined pointwise; in particular, it may present discontinuities. It is well-known that, when the value function is continuous, it is a viscosity solution of the corresponding Hamilton-Jacobi-Bellman equation H(x, u, Du) = 0

in Ω,

where H(x, t, p) = sup{−b(x, α) · p + λt − f (x, α)}.

(3.1)

α∈A

This is a consequence of the so-called Dynamic Programming Principle (cf. P.L. Lions [28]). We recall that the notion of viscosity solutions was introduced by M.G. Crandall and P.L. Lions [16] (see also M.G. Crandall, L.C. Evans and P.L. Lions [14]). We refer the reader to W.H. Fleming and H.M. Soner [21] and P.L. Lions [28] where the applications of viscosity solutions to deterministic and stochastic control problems are described. But the value function u may be discontinuous even when the exit cost ϕ is continuous. Therefore we have to use the notion of discontinuous viscosity solution which was introduced by H. Ishii [23, 24] and which requires the concepts of lower semi-continuous (l.s.c. in short) and upper semi-continuous (u.s.c. in short) envelopes of functions. In all the following, ξ∗ (resp. ξ ∗ ) will denote the l.s.c. (resp. u.s.c.) envelope of the locally bounded function ξ which is defined by ξ∗ (x) = lim inf ξ(y) y→x

(resp.

ξ ∗ (x) = lim sup ξ(y) ). y→x

G. Barles and B. Perthame [7] first considered the connections of such discontinuous solutions with optimal control problems. Moreover the concept of viscosity solutions is used to identify the appropriate boundary conditions satisfied by the value function. In the case of a continuous exit cost, H. Ishii [25] and G. Barles and B. Perthame [8] show the connections between the above control problem and the following Hamilton-Jacobi type problem           

H(x, u, Du) = 0

in Ω,

min{H(x, u, Du), u − ϕ} ≤ 0

on ∂Ω,

max{H(x, u, Du), u − ϕ} ≥ 0

on ∂Ω.

(3.2)

When the function ϕ is discontinuous, the definition of viscosity solutions deals with the l.s.c. and the u.s.c. envelope of ϕ. For illustration, we say that the bounded function u is a viscosity subsolution of the equation (3.2) on the boundary ∂Ω if, for every function φ ∈ C 1 (Ω), if x ∈ ∂Ω is a maximum point of u∗ − φ, we have H(x, u∗ (x), Dφ(x)) ≤ 0

or

u∗ (x) ≤ ϕ∗ (x).

Introduction

31

Our first aim is to prove that the value function u is a viscosity solution of (3.2). Then we have to look at “uniqueness” (or characterization) properties for the viscosity solution u. In general, the discontinuous viscosity solution is not unique. One reason is that the Hamilton-Jacobi-Bellman equation is the same for the control problem and for the relaxed control problem. But the value function can be different for these two problems. Moreover, it is quite natural to consider different stopping times on the boundary : for example, the first exit time from the closed set Ω i.e. τ¯ = inf{t ≥ 0, yx (t) 6∈ Ω}. In this paper, since the exit cost ϕ is discontinuous, an additional choice is to use ϕ∗ or ϕ∗ , instead of taking ϕ, in the definition of u. We first prove the existence of a minimum and of a maximum solution which are respectively given by u− (x) = inf

nZ

θ

Z

0

A

f (ˆ yx (t), α)e−λt dµt (α)dt + ϕ∗ (ˆ yx (θ))e−λθ , τˆ ≤ θ ≤ τˆ¯, o

yˆx (θ) ∈ ∂Ω and µ ∈ L∞ (IR+ , P (A)) , and +

u (x) =

inf

α(.)∈L∞ (IR+ ,A)

n

sup{

θ

Z 0

f (yx (t), α(t))e−λt dt + ϕ∗ (yx (θ))e−λθ , τ ≤ θ ≤ τ¯ o

and yx (θ) ∈ ∂Ω} . It is clear on these definitions that one has u− ≤ u ≤ u+

in Ω.

But these functions may be very different, in particular, because of the trajectories which remain on Ω but which touch the boundary several times (τ 6= τ¯). Nevertheless, in the case when the function ϕ is continuous, G. Barles and B. Perthame [9] showed that, essentially, if there exist outer and inner fields at each point of ∂Ω, the value function u is continuous and is the unique solution of (3.2) : hence, we get u− = u = u+ in Ω. (3.3) The underlying idea is that, with these outer and inner fields, one can control a trajectory close to the boundary to make it stay in Ω or to leave Ω. An other aim of this article is to give a similar uniqueness result when ϕ is discontinuous. But, since the value functions may be discontinuous, we first have to explain how to interpret uniqueness of the solution in this case. We want the equalities (3.3) to hold again in a weaker sense, namely u− = u∗ = (u+ )∗

in Ω,

32

Probl`emes de temps de sortie

since u− is l.s.c. on Ω. Therefore uniqueness means in this context that all the solutions have the same l.s.c. envelope. In the case when the exit cost ϕ satisfies (ϕ∗ )∗ = ϕ∗

on ∂Ω,

(3.4)

and when there is an outer field on the boundary, we show that u− = (u+ )∗

in Ω,

by working directly on the control formulae. And therefore, all the solutions of (3.2) have the same l.s.c. envelope since u− and u+ are respectively the minimum and the maximum solution of (3.2). But we want also an uniqueness result without the condition (3.4) on ϕ since, for control problems, it is quite natural to consider exit costs ϕ which are only l.s.c. In fact, one of our motivations for this work is the exit problem from a part Γ of the boundary ∂Ω. If, for instance, Γ is reduced to a point {x0 }, then ϕ is equal to 0 at x0 and 1 elsewhere; therefore it does not satisfy (3.4). To solve this difficulty, we adapt PDE arguments introduced by E.N. Barron and R. Jensen [10, 11] for convex hamiltonians. Under a non-degeneracy condition on the hamiltonian which we interpret as the existence of an outer field on the boundary for control problems, we prove that there exists a unique l.s.c. viscosity solution in the sense of Barron-Jensen in Ω which is a classical viscosity supersolution on the boundary and which satisfies, for every x ∈ ∂Ω, lim inf u(y) ≤ ϕ∗ (x). (3.5) y→x,y∈Ω

We emphasize that this uniqueness result is obtained by PDE arguments. The application of this result to the control problem is the following : since P. Soravia [34] showed that all the value functions are viscosity solutions in the sense of Barron-Jensen, we have the uniqueness of the l.s.c. envelope of the value functions which satisfy the condition (3.5). Besides, when the condition (3.4) holds, we recover the uniqueness result obtained by working directly on the representation formulae of the value functions. We also refer the reader interested in the discontinuous viscosity solution approach to Dirichlet problems to the work of A.I. Subbotin [35], M. Bardi and P. Soravia [3], P. Soravia [34] and G. Barles [4]. These authors consider mainly continuous Dirichlet function on the boundary. In the case of discontinuous data, the pioneering paper of Barron-Jensen [10] was concerned with the Cauchy problem in IRN . G. Barles [5] extended their ideas to stationary optimal stopping time in IRN . To the best of our knowledge, the exit time problems with discontinuous exit costs have not been considered in the literature yet with such a generality.

3.1 The exit time control problems with discontinuous exit costs

33

This paper is organized as follows : the first section is devoted to the study of the exit time problems and its connections with the equation (3.2). We also introduce the condition on the behavior of the controlled vector field at the boundary to obtain an uniqueness result. In the second section, we describe the new idea for discontinuous viscosity solutions for convex hamiltonians and we prove uniqueness results in the case when ϕ is discontinuous by PDE arguments. The third section is devoted to applications of the uniqueness results. The two first parts are nearly independent.

3.1 The exit time control problems with discontinuous exit costs We recall that Ω a smooth bounded domain of IRN and that the state of the system is described by the solution of   dyx (t) = b(yx (t), α(t))dt, 

yx (0) = x ∈ Ω,

or, in the case of relaxed controls, by Z    dˆ b(ˆ yx (t), α)dµt (α)dt, yx (t) = A

 

yˆx (0) = x ∈ Ω,

where α(.) ∈ L∞ (IR+ , A) and µ. ∈ L∞ (IR+ , P (A)); A is a compact metric space and P (A) is the set of probability measures on A. The function b is continuous from Ω × A into IRN and it satisfies   |b(x, α)| ≤ C, 

∀x ∈ Ω, ∀α ∈ A; (3.6)

|b(x, α) − b(y, α)| ≤ C|x − y|,

∀x, y ∈ Ω, ∀α ∈ A.

These assumptions imply the existence and the uniqueness of the solution yx for all t > 0. In the same way as for τ and τ¯, we define τˆ and τˆ¯ for a relaxed trajectory yˆx respectively by τˆ = inf{t ≥ 0, yˆx (t) 6∈ Ω}

and

τˆ¯ = inf{t ≥ 0, yˆx (t) 6∈ Ω}.

The associated cost functions are J(x, α, θ, ϕ) =

Z 0

θ

f (yx (t), α(t))e−λt dt + ϕ(yx (θ))e−λθ ,

34

Probl`emes de temps de sortie

and ˆ µ, θ, ϕ) = J(x,

θ

Z

Z

0

A

f (ˆ yx (t), α)e−λt dµt (α)dt + ϕ(ˆ yx (θ))e−λθ ,

where λ is some positive constant and f is a continuous function from Ω × A into IR satisfying   |f (x, α)| ≤ C, 

∀x ∈ Ω, ∀α ∈ A; (3.7)

|f (x, α) − f (y, α)| ≤ C|x − y|,

∀x, y ∈ Ω, ∀α ∈ A.

We use particularly the three following value functions u− (x) =

inf

µ∈L∞ (IR+ ,P (A))

ˆ µ, θ, ϕ∗ ), τˆ ≤ θ ≤ τˆ¯ and yˆx (θ) ∈ ∂Ω}, {J(x,

which is l.s.c. on Ω, u+ (x) =

inf

α(.)∈L∞ (IR+ ,A)

{sup{J(x, α, θ, ϕ∗ ), τ ≤ θ ≤ τ¯ and yx (θ) ∈ ∂Ω}},

which is u.s.c. on Ω and the value function already introduced in the introduction u[ϕ](x) =

inf

α(.)∈L∞ (IR+ ,A)

{J(x, α, τ, ϕ)}.

These three types of value functions have already been studied in [8] and [23] when the function ϕ is continuous. In the case of a discontinuous exit cost, we introduce different exit cost functions, namely ϕ∗ and ϕ∗ . Our main results are the following. First, we prove that the function u− is the minimum supersolution of (3.2) with the hamiltonian given by (3.1) and the function u+ is the maximum subsolution. We show that the value functions u[ϕ∗ ], u[ϕ] and u[ϕ∗ ] are viscosity solutions of (3.2) and an example shows that all these value functions may be very different. In order to obtain the uniqueness of the l.s.c. envelope of the value functions, we are going to prove that (u+ )∗ = u− on Ω. Therefore we have to show the following series of equalities (u+ )∗ = (u[ϕ∗ ])∗ = (u[ϕ∗ ])∗ = u−

on Ω.

The first equality is proved without additional assumptions; it already holds in the continuous case (see [8]). For the second, since we minimize, “interesting values” may be lost by taking the exit cost ϕ∗ instead of ϕ∗ : for instance, if ϕ is equal to 0 at one point and 1 elsewhere, then ϕ∗ is equal to 1 everywhere. To avoid this difficulty in this section, we consider only “regular” exit costs i.e. satisfying the property (3.4).

3.1 The exit time control problems with discontinuous exit costs

35

Finally, the difficulty with the trajectories which touch the boundary several times, is solved by assuming that there exists an outer field at every point on the boundary or that the boundary is such that there are only inner fields on ∂Ω. This section is divided in three subsections. In the first, we study the value functions u− and u+ . The second is devoted to the properties of the function u[ϕ]. And, in the last one, we prove the uniqueness result. 3.1.1 The value functions u− and u+ In this subsection, we characterize u− and u+ as the minimal and maximal solutions of (3.2). We also prove the connection between u+ and u[ϕ∗ ]. Theorem 3.1.1 We assume that ϕ is a bounded function defined pointwise, that the constant λ is positive and that the assumptions (3.6) and (3.7) hold. 1. The function u− is l.s.c. and the function u+ is u.s.c. on Ω. 2. The functions u+ and u− are viscosity solutions of (3.2). 3. The functions u+ and u− are respectively the maximal subsolution and the minimal supersolution of (3.2). t u Proof. We first consider the case of u+ . We introduce a nonincreasing sequence (ϕn )n of continuous functions such that inf ϕn = ϕ∗ . n

We note u+ [ϕn ] the value function defined by the same formula as u+ except that the exit cost function ϕ∗ is replaced by ϕn (in particular, u+ [ϕ∗ ] is equal to u+ ). Then we need the following lemma. Lemma 3.1.1 We have inf u+ [ϕn ] = u+ [ϕ∗ ] n

on Ω. t u

Proof. It is clear enough that u+ [ϕn ] ≥ u+ [ϕn+1 ] ≥ u+ [ϕ∗ ]

on Ω,

(3.8)

36

Probl`emes de temps de sortie

and thus inf u+ [ϕn ] ≥ u+ [ϕ∗ ] on Ω. n To prove the opposite inequality, we use the definition of u+ [ϕn ]. For any control α(.) and any point x ∈ Ω, we have u+ [ϕn ](x) ≤ sup{J(x, α, θ, ϕn ), τ ≤ θ ≤ τ¯ and yx (θ) ∈ ∂Ω}. Now, for a fixed control α(.), we pick a sequence (θn )n such that 1 Z θn u [ϕ ](x) ≤ + f (yx (t), α(t))e−λt dt + ϕn (yx (θn ))e−λθn . n 0 +

n

(3.9)

First case. If the sequence (θn )n is bounded, considering a subsequence if necessary, ¯ Taking the limsup as n → ∞ in the inequality (3.9), we may assume that θn → θ. we get θ¯

Z

lim sup u+ [ϕn ](x) ≤ n

0

¯

f (yx (t), α(t))e−λt dt + lim sup ϕn (yx (θn ))e−λθ .

(3.10)

n

¯ since the function ϕn is continuous and since inf ϕn = ϕ∗ , But, since yx (θn ) → yx (θ), n we deduce easily ¯ lim sup ϕn (yx (θn )) ≤ ϕ∗ (yx (θ)). n

Moreover, using (3.8), we obtain lim sup u+ [ϕn ](x) = inf u+ [ϕn ](x). n

n

Combining these two results with (3.10), we get +

n

inf u [ϕ ](x) ≤ n

Z 0

θ¯

¯ −λθ¯ = J(x, α, θ, ¯ ϕ∗ ). f (yx (t), α(t))e−λt dt + ϕ∗ (yx (θ))e

And taking the supremum in θ¯ in the right-hand side, we have inf u+ [ϕn ](x) ≤ sup{J(x, α, θ, ϕ∗ ), τ ≤ θ ≤ τ¯ and yx (θ) ∈ ∂Ω}. n

(3.11)

Second case. If the sequence (θn )n is not bounded, then there exists a subsequence, still denoted (θn )n , such that θn → +∞. The inequality (3.9) implies u+ [ϕn ](x) ≤

1 Z θn + f (yx (t), α(t))e−λt dt + ϕ∗ (yx (θn ))e−λθn n 0 +ϕn (yx (θn ))e−λθn − ϕ∗ (yx (θn ))e−λθn

≤

  1 + J(x, α, θn , ϕ∗ ) + ϕn (yx (θn )) − ϕ∗ (yx (θn )) e−λθn . n

3.1 The exit time control problems with discontinuous exit costs

37

Since ϕn and ϕ∗ are bounded, the last term tends to zero and we conclude as before. Hence, finally, since the inequality (3.11) holds for any control α(.) and for any point x ∈ Ω, we conclude inf u+ [ϕn ](x) ≤ u+ [ϕ∗ ](x). n u t Now we deduce the properties of u from this lemma. By using the results of [8], the function u+ [ϕn ] is u.s.c. then the function u+ is u.s.c. too since u+ is equal to the limit of the nonincreasing sequence of function u+ [ϕn ]. Moreover, again by the results of [8], u+ [ϕn ] is the maximal subsolution (and solution) of +

          

H(x, u, Du) = 0

in Ω,

min{H(x, u, Du), u − ϕn } ≤ 0

on ∂Ω,

max{H(x, u, Du), u − ϕn } ≥ 0

on ∂Ω.

(3.12)

Then, using standard stability results (cf. [7]), u+ [ϕ∗ ] is a subsolution of (3.2) because, for every x ∈ Ω, we have lim sup u+ [ϕn ](y) = inf u+ [ϕn ](x) = u+ [ϕ∗ ](x) n→∞,y→x

n

and lim sup ϕn (y) = inf ϕn (x) = ϕ∗ (x). n

n→∞,y→x

And it is also a supersolution of (3.2) since we have, for every x ∈ Ω, lim inf u+ [ϕn ](y) = lim inf u+ [ϕ∗ ](y) = (u+ [ϕ∗ ])∗ (x) y→x

n→∞,y→x

and since lim inf ϕn (y) = (inf ϕn )∗ (x) = (ϕ∗ )∗ (x) n

n→∞,y→x

≥ ϕ∗ (x). Moreover, if w is a subsolution of (3.2), w is also a subsolution of (3.12) since ϕ ≥ ϕ∗ on ∂Ω. This implies, for every n, that n

w ≤ u+ [ϕn ]

on Ω,

because u+ [ϕn ] is the maximal subsolution of (3.12). And therefore, by Lemma 3.1.1, taking the infimum over n yields w ≤ u+ [ϕ∗ ]

on Ω,

38

Probl`emes de temps de sortie

which implies that the function u+ [ϕ∗ ] is a maximal subsolution of (3.2). For the function u− , we proceed exactly as for the value function u+ . Let (ϕn )n be a nondecreasing sequence of continuous functions such that sup ϕn = ϕ∗ . n

Again we introduce the notation u− [ϕn ] to denote the value function defined as u− but with the exit cost ϕn instead of ϕ∗ . Then we consider the following lemma which corresponds to Lemma 3.1.1 for u+ . Lemma 3.1.2 We have sup u+ [ϕn ] = u− [ϕ∗ ]

on Ω.

n

t u With this lemma, we conclude the proof by deducing the properties of u− [ϕ∗ ] from the properties of u− [ϕn ] as before. t u Proof of Lemma 3.1.2. It is clear enough that sup u− [ϕn ] ≤ u− [ϕ∗ ]

on Ω.

n

It remains to prove the opposite inequality. For every x ∈ Ω, we consider a minimizing sequence (µn , θn )n for u− [ϕn ] such that u− [ϕn ](x) +

Z θn Z 1 f (ˆ yxn (t), α)e−λt dµn (α)dt + ϕn (ˆ yxn (θn ))e−λθn . ≥ n 0 A

(3.13)

First case. If the sequence (θn )n is bounded, considering if necessary a subsequence, we may assume that there exists a stopping time θ¯ such that θn → θ¯ as n → ∞. Moreover, using classical arguments relying on the compactness of relaxed controls, we may also assume that µn → µ ¯ weakly in L∞ (IR+ , P (A)) for some relaxed control µ ¯. n Let yˆx be the relaxed trajectory associated to µ ¯. Since yˆx converges locally uniformly ¯ ∈ ∂Ω and yˆn (θn ) → yˆx (θ). ¯ to yˆx , we can check that yˆx (θ) x Then, using again the local uniform convergence of yˆxn to yˆx together with the fact that f is Lipschitz continuous, we replace the trajectory yˆxn by yˆx in the integral of (3.13) i.e. u− [ϕn ](x) +

Z θ¯ Z 1 ¯ ≥ f (ˆ yx (t), α)e−λt d¯ µ(α)dt + ϕn (ˆ yxn (θn ))e−λθ + εn , n 0 A

3.1 The exit time control problems with discontinuous exit costs

39

with a sequence of numbers εn such that εn → 0 when n → ∞. From the above inequality, we deduce u− [ϕn ](x) +

  1 ¯ e−λθ¯ + εn . ≥ u− [ϕ∗ ](x) + ϕn (ˆ yxn (θn )) − ϕ∗ (ˆ yx (θ)) n

Then, taking the liminf, we get o

n

¯

¯ e−λθ . yx (θ)) lim inf u− [ϕn ](x) ≥ u− [ϕ∗ ](x) + lim inf ϕn (ˆ yxn (θn )) − ϕ∗ (ˆ n

n

(3.14)

¯ since the function ϕn is continuous and since sup ϕn = ϕ∗ , But, since yˆxn (θn ) → yˆx (θ), n

we deduce easily ¯ lim inf ϕn (ˆ yxn (θn )) ≥ ϕ∗ (ˆ yx (θ)), n

thus, combining this with (3.14), we obtain limninf u− [ϕn ](x) = sup u− [ϕn ](x) n

≥ u− [ϕ∗ ](x). Second case. If the sequence (θn )n is not bounded, we may assume without loss of generality that θn → ∞ and then the inequality (3.13) implies u− [ϕn ](x) +

  1 ≥ u− [ϕ∗ ](x) + ϕn (ˆ yxn (θn )) − ϕ∗ (ˆ yxn (θn )) e−λθn . n

Since ϕ∗ is bounded and since (ϕn )n is uniformly bounded, letting n → ∞, we have the desired result. t u The following theorem shows that u+ and u[ϕ∗ ] are not very different. Theorem 3.1.2 Under the same assumptions of Theorem 3.1.1, we have (u+ )∗ = (u[ϕ∗ ])∗

in Ω. t u

Proof. We first remark that the proof is inspired from the corresponding one in [8]. It is clear that (u+ )∗ ≥ (u[ϕ∗ ])∗ in Ω since u+ ≥ u[ϕ∗ ] on Ω. It remains to prove the opposite inequality. For any point x of Ω, there exists a sequence (xn )n of points of Ω such that xn → x and lim u[ϕ∗ ](xn ) = (u[ϕ∗ ])∗ (x). n

40

Probl`emes de temps de sortie

For each xn , we consider a control αn (.) such that u[ϕ∗ ](xn ) +

1 ≥ J(xn , αn , τn , ϕ∗ ). n

(3.15)

Then we need the following lemma. Lemma 3.1.3 For every x ∈ Ω and α ∈ L∞ (IR+ , A), there exists a sequence (xp )p of points of Ω such that limpinf u+ (xp ) ≤ J(x, α, τ, ϕ∗ ) where τ is the first exit time of the trajectory yx associated to the control α.

t u

For every n, we apply this lemma to the point xn and to the control αn . Thus there exists a sequence (xpn )p such that xpn → xn as p → ∞ and 1 J(xn , αn , τn , ϕ∗ ) ≥ u+ (xpn ) − . p Combining this with (3.15), we pass to the limit and, by a diagonal procedure, we get (u[ϕ∗ ])∗ (x) = lim u[ϕ∗ ](xn ) n

≥ (u+ )∗ (x), t u

which is the desired result.

Proof of Lemma 3.1.3. If τ = ∞, then u+ (x) ≤ J(x, α, τ, ϕ∗ ) and therefore it suffices to take xp := x for all p. If τ 6= ∞, we consider the map Yτ : z 7→ yz (τ ) which is an homeomorphism from a neighborhood of x onto some neighborhood of yx (τ ). And we introduce the domain c D defined by D := Y −1 (B(yx (τ ), ε) ∩ Ω ), where B(z, r) is the open ball centered at c z and of radius r and Ω is the complementary set of Ω in IRN . For ε small enough, the domain D is a non-empty open subset of IRN and the point x is in its closure D. Moreover, since the function b is Lipschitz continuous, classical ODE estimates yield |yz (s) − yx (s)| ≤ |z − x|eCs for every s ∈ [0, τ ]. And, for every p, since we have n

inf dist(yx (s), ∂Ω), 0 ≤ s ≤ τ −

1o = δ > 0, p

3.1 The exit time control problems with discontinuous exit costs

41

we consider the constant η equal to n1

η := max

p

o

, δ e−Cτ

and therefore we have, for |z − x| < η, |yz (s) − yx (s)| < and yz (s) ∈ Ω,

1 , p

for every s ∈ [0, τ ],

(3.16)

1 for every s ∈ [0, τ − ]. p

Then we choose a point xp ∈ B(x, η) ∩ D. We first remark that the first exit time τp of the trajectory yxp from Ω satisfies τ − p1 ≤ τp since xp ∈ B(x, η). And the exit time τ¯p satisfies τ¯p ≤ τ since xp ∈ D. We may write u+ (xp ) ≤ sup{J(xp , α, θ, ϕ∗ ), θ ∈ [τp , τ¯p ] and yxp (θ) ∈ ∂Ω}. Using the Lipschitz continuity of f and the inequality (3.16), we compute, for every θ ∈ [τp , τ¯p ], J(xp , α, θ, ϕ∗ ) ≤ J(x, α, τ, ϕ∗ ) − ϕ∗ (yx (τ ))e−λτ + ϕ∗ (yxp (θ))e−λτ +kϕk∞ |e−λθ − e−λτ | + kf k∞ |τ − θ| +

Cτ p

 1 ≤ J(x, α, τ, ϕ∗ ) + ρϕ∗ (|yx (τ ) − yxp (θ)|) + C˜ |θ − τ | + (3.17) p

where ρϕ∗ is a continuous function such that ρϕ∗ (t) → 0 as t → 0+ and, for every y ∈ ∂Ω, ϕ∗ (y) − ϕ∗ (yx (τ )) ≤ ρϕ∗ (|yx (τ ) − y|). We pass to the liminf in (3.17) and using again (3.16) and the fact that τ−

1 ≤ τp ≤ θ ≤ τ¯p ≤ τ, p

we get J(x, α, τ, ϕ∗ ) ≥ lim inf J(xp , α, θ, ϕ∗ ) p

≥ limpinf u+ (xp ). And the proof is complete.

t u

42

Probl`emes de temps de sortie

3.1.2 The properties of the value function u[ϕ] In this section, we prove that the value functions u[ϕ∗ ], u[ϕ] and u[ϕ∗ ] are viscosity solutions. Then we give an example which shows that these functions may be very different. Let us begin by a result concerning the values of the solutions of (3.2) on the boundary. Proposition 3.1.1 Let u be a bounded function from Ω into IR. We define   u

uˇ = 

ϕ

  u

in Ω, and on ∂Ω,

u˜ = 

ξ

in Ω, on ∂Ω,

where ξ is a real-valued function such that ϕ∗ ≤ ξ ≤ ϕ∗

on ∂Ω.

Then uˇ is a subsolution (resp. supersolution) of (3.2) if and only if u˜ is a subsolution (resp. supersolution) of (3.2). t u Remark 3.1.1 This proposition shows that the values of the solutions of (3.2) are not prescribed by the viscosity condition on the boundary. So the uniqueness results cannot be extended up to the boundary. Proof. We only prove the proposition in the subsolutions case. The other case may be obtained by the same method. Let us, first, remark that, for every x ∈ ∂Ω, we have u˜∗ (x) = sup{u∗ (x), ξ ∗ (x)} ≤ uˇ∗ (x) = sup{u∗ (x), ϕ∗ (x)}.

(3.18)

We first assume that uˇ is a subsolution and therefore we have to prove that u˜ is a subsolution too. Let φ ∈ C 1 (Ω) be a test-function and let x0 ∈ ∂Ω be a maximum point of u˜∗ − φ. Changing φ in φ + (˜ u∗ (x0 ) − φ(x0 )), we may assume without loss of ∗ generality that u˜ (x0 ) = φ(x0 ) and thus u˜∗ ≤ φ. • If u˜∗ (x0 ) ≤ ϕ∗ (x0 ), there is nothing to prove. • If u˜∗ (x0 ) > ϕ∗ (x0 ) then, by definition of u˜ and since we have ξ ≤ ϕ∗ on ∂Ω, this implies u∗ (x0 ) > ϕ∗ (x0 ). Hence, using the inequality (3.18), we deduce that uˇ∗ (x0 ) = u˜∗ (x0 ).

3.1 The exit time control problems with discontinuous exit costs

43

We introduce the positive constant δ := u˜∗ (x0 ) − ϕ∗ (x0 ). Then, since u˜∗ (x0 ) = φ(x0 ), we get δ δ (3.19) ϕ∗ (x0 ) + = φ(x0 ) − . 2 2 Since φ is continuous, there exists some constant ε1 > 0 such that, for all y ∈ Ω and |x0 − y| < ε1 , we have δ |φ(x0 ) − φ(y)| < . 2 And since ϕ∗ is u.s.c., there exists some constant ε2 > 0 such that, for all y ∈ ∂Ω and |x0 − y| < ε2 , we have δ ϕ∗ (y) − ϕ∗ (x0 ) < . 2 Hence, combining these two preceding estimates with the equality (3.19), we deduce ϕ∗ (y) < ϕ∗ (x0 ) +

δ δ = φ(x0 ) − < φ(y) 2 2

for y ∈ ∂Ω ∩ B(x0 , ε1 ∧ ε2 ),

where a ∧ b := inf{a, b} for a, b ∈ IR. Combining this with the fact that uˇ∗ ≤ φ

for y ∈ Ω ∩ B(x0 , ε1 ∧ ε2 ),

since u∗ ≤ φ in Ω, we deduce that x0 is also a maximum point of uˇ∗ − φ. And thus u˜∗ is a subsolution. Conversely, if u˜ is a subsolution, let x0 ∈ ∂Ω be a maximum point of uˇ∗ − φ with φ ∈ C 1 (Ω). As before, we choose uˇ∗ (x0 ) = φ(x0 ). • If uˇ∗ (x0 ) ≤ ϕ∗ (x0 ), uˇ∗ is a subsolution. • If u˜∗ (x0 ) > ϕ∗ (x0 ) then, using the inequality (3.18) and the fact that ξ ∗ ≤ ϕ∗ on ∂Ω, we get uˇ∗ (x0 ) = u∗ (x0 ) = u˜∗ (x0 ). Since uˇ∗ ≥ u˜∗ on Ω, it is easy to show that uˇ∗ is a subsolution.

t u

Now, we recall a general result in control theory which is the Dynamic Programming Principle. Theorem 3.1.3 Under the same assumptions of Theorem 3.1.1, for any T > 0 and for any x ∈ Ω, we have u[ϕ](x) =

inf

α(.)∈L∞ (IR+ ,A)

nZ 0

τ ∧T

f (yx (t), α(t))e−λt dt o

+11{τ ≤T } ϕ(yx (τ ))e−λτ + 11{τ >T } u[ϕ](yx (T ))e−λT . t u

44

Probl`emes de temps de sortie

Proof. To prove the theorem, we just remark that the continuity of the exit cost is no used in the classical proof (see [28] for instance). t u Theorem 3.1.4 Under the assumptions of Theorem 3.1.1, the value functions u[ϕ∗ ], u[ϕ] and u[ϕ∗ ] are viscosity solutions of the equation (3.2). t u Proof. Since the proof is inspired from the corresponding one in the continuous case, we only point out the modifications we need. Let us denote by ϕ the functions : ϕ∗ , ϕ or ϕ∗ . We only prove that u[ϕ] is a supersolution of (3.2) on ∂Ω. The other properties may be obtained by the same method. We consider some point x0 ∈ ∂Ω. • If (u[ϕ])∗ (x0 ) ≥ ϕ∗ (x0 ), we have nothing to prove. • If (u[ϕ])∗ (x0 ) < ϕ∗ (x0 ), we have to show that the function u satisfies the equation H(x, u, Du) = 0 in viscosity sense. There exists a sequence (xn )n of points of Ω such that xn → x0 and lim u[ϕ](xn ) = (u[ϕ])∗ (x0 ). n

We may assume without loss of generality that xn ∈ Ω for all n. Indeed, if there exists a subsequence (xp )p of (xn )n such that xp ∈ ∂Ω for all p, then, by definition of the value function u[ϕ], u[ϕ](xp ) = ϕ(xp ). And since ϕ is equal to ϕ∗ , ϕ or ϕ∗ , passing to the limit, we obtain lim inf ϕ[xp ] ≥ ϕ∗ (x0 ). p

Hence we get a contradiction. Now we choose a constant T such that T kf k∞ + ρϕ∗ (kbk∞ T ) + kϕk∞ |1 − e−λT | ≤

ϕ∗ (x0 ) − (u[ϕ])∗ (x0 ) , 2

(3.20)

where ρϕ∗ is a nondecreasing continuous function satisfying ρϕ∗ (t) → 0 as t → 0+ and, for every y ∈ ∂Ω, ϕ∗ (x0 ) − ϕ∗ (y) ≤ ρϕ∗ (|x0 − y|), (such a function ρϕ∗ exists because ϕ∗ is l.s.c.).

3.1 The exit time control problems with discontinuous exit costs

45

For every xn , we introduce a control αn (.) such that Z τn ∧T 1 ≥ f (yxn (t), αn (t))e−λt dt + 11{τn ≤T } ϕ(yxn (τn ))e−λτn u[ϕ](xn ) + n 0

+11{τn >T } u[ϕ](yxn (T ))e−λT .

(3.21)

If, for n large enough, we have τn > T , then the inequality (3.21) implies u[ϕ](xn ) +

Z T 1 ≥ f (yxn (t), αn (t))e−λt dt + u[ϕ](yxn (T ))e−λT . n 0

And, since this inequality does not deal with the discontinuous exit cost ϕ, we may apply classical arguments (see [4] for instance). Otherwise, we may assume that, for all n, there exists pn such that τpn ≤ T . Then, the inequality (3.21) implies, for the index pn , that u[ϕ](xpn ) +

Z τp n 1 ≥ f (yxpn (t), αpn (t))e−λt dt + ϕ(yxpn (τpn ))e−λτpn . pn 0

Then we compute ϕ∗ (x0 ) − u[ϕ](xpn ) − ≤

1 pn Z

τpn

0

≤

Z 0

T

kf k∞ dt + ϕ∗ (x0 ) − ϕ(yxpn (τpn )) + kϕk∞ |1 − e−λτpn |

kf k∞ dt + ρϕ∗ (x0 ) (|x0 − yxpn (τpn )|) + kϕk∞ |1 − e−λT |

≤ T kf k∞ + ρϕ∗ (x0 ) (|x0 − xpn | + kbk∞ T ) + kϕk∞ |1 − e−λT |. We pass to the limit and we get a contradiction with the choice of T in (3.20).

t u

Example 3.1.1 We are going to show in this example that the value functions u[ϕ∗ ], u[ϕ] and u[ϕ∗ ] may be not equal. We take in IR2 Ω := {(x, y) ∈ [−1, 1]2 , y > 0 or x < 0}, f ≡ 0, λ = 0, (λ is null for the sake of simplicity but this is not relevant here), and b(x, α) := α ∈ A, with A := B(0, 1) ∩ (IR+ × IR− ) i.e. {(α1 , α2 ) ∈ IR2 , α12 + α22 ≤ 1, α1 ≥ 0, α2 ≤ 0}.

46

Probl`emes de temps de sortie

The exit cost ϕ is equal to 0 except on [0, 1] × {0} where ϕ(0, 0) = −1,

1 ϕ(x, 0) = −2 for x ∈]0, [, 2

1 ϕ(x, 0) = −3 for x ∈ [ , 1]. 2

6y 1

Ω

−1

ϕ≡0

ϕ = −1PP q rϕ ≡ −2 ϕ ≡ −3 0

1

-

x

ϕ≡0

−1

In the region ]−1, 0[×]0, 1[, it is easy to show that the best strategy is to reach the point (0, 0). To do so, we may, for example, use the control (0, −1) until we touch the line y = 0 and then we take the control (1, 0). Hence we compute that u[ϕ∗ ] ≡ −2, u[ϕ] ≡ −1, u[ϕ∗ ] ≡ 0 and u− ≡ −3 in ] − 1, 0[×]0, 1[. The value functions u[ϕ∗ ], u[ϕ] and u[ϕ∗ ] are very different because at the point (0, 0), the exit costs ϕ∗ , ϕ and ϕ∗ take different values. Moreover, the functions u− and u[ϕ∗ ] are not equal because the trajectories which reach [ 12 , 1] × {0} must be tangent to the boundary. The open set Ω is not regular for the sake of simplicity. But we can easily change it. With ϕ continuous, a similar example was used to show that u− and u[ϕ] are not necessarily equal (cf. [8]). This example shows that the discontinuous viscosity solutions may be very different even if we consider only their l.s.c. or u.s.c. envelopes. 3.1.3 Partial controllability on the boundary We introduce now new assumptions which allow us to prove that all the value functions have the same l.s.c. envelope namely u− .

3.1 The exit time control problems with discontinuous exit costs

47

We denote by d(.) the distance function to the boundary ∂Ω. We are given a smooth bounded domain; more precisely, we assume d is a C 1,1 function in the neighborhood V of ∂Ω.

(3.22)

Then we set n(x) := −Dd(x) for x ∈ V. We assume that ∂Ω = ∂Ω1 ∪ ∂Ω2 where ∂Ω1 and ∂Ω2 are unions of connected components of ∂Ω, and, at every point of ∂Ω1 , there exists an outer field i.e. ∀x ∈ ∂Ω1 , ∃α ∈ A, b(x, α) · n(x) ≥ β > 0,

(3.23)

and, on ∂Ω2 , there are only inner fields i.e. ∀x ∈ ∂Ω2 , ∀α ∈ A, b(x, α) · n(x) ≤ −β < 0.

(3.24)

Notice that in (3.23) and (3.24), β can be chosen independent of x since ∂Ω1 and ∂Ω2 are compact subsets of IRN and since the functions b and n are continuous on ∂Ω. Theorem 3.1.5 We assume that ϕ is a bounded function defined pointwise satisfying (3.4), that the constant λ is positive and that the assumptions (3.6), (3.7) and (3.22)(3.24) hold. Then we have u− = (u+ )∗ in Ω ∪ ∂Ω1 . t u Using the representation formula, one easily build special examples where the equality may be wrong on ∂Ω2 . For example, if the exit cost ϕ is equal to 0 on ∂Ω2 and to 1 on ∂Ω1 , the value function u− is null on the boundary ∂Ω2 but not the function u+ since the trajectories cannot exit through ∂Ω2 i.e. τ¯ > τ . Since u+ and u− are the maximum and minimum solutions of (3.2), this result characterizes the discontinuous solutions of (3.2) in Ω∪∂Ω1 . Indeed, if w is a solution of (3.2) then u− ≤ w∗ ≤ w∗ ≤ u+ in Ω ∪ ∂Ω1 , and therefore, taking the l.s.c. envelope of these inequalities and using Theorem 3.1.5 yield u− = w∗ = (w∗ )∗ in Ω ∪ ∂Ω1 . When the exit cost is continuous (cf. [8]), to get uniqueness result, it is enough to know that the first exit time τ is equal to the “best exit time” i.e. which gives the minimal value for the value function or that u− = ϕ on ∂Ω. But this is not the case with discontinuous exit cost since we deal with ϕ∗ and ϕ∗ .

48

Probl`emes de temps de sortie

Example 3.1.2 We show that the assumption (3.23) is necessary to prove Theorem 3.1.5. In fact, we take up the above example 3.1.1 but we change ϕ to satisfy (3.4). Precisely,   −2 on [0, 1] × {0}, ϕ(x) =  0 otherwise. Then we can easily compute : u[ϕ∗ ] ≡ −2 and u[ϕ∗ ] ≡ 0 in the region ] − 1, 0[×]0, 1[. Remark 3.1.2 The assumption (3.24) was introduced and used by H.M. Soner [32] for control problems with state space constraints to prove the continuity and the uniqueness of the value function. Proof of Theorem 3.1.5. It is easy to see that the trajectories cannot exit through the boundary ∂Ω2 because of (3.24). And, since the result holds only in Ω ∪ ∂Ω1 , we do not need to take care of the value of the exit cost on the boundary ∂Ω2 . First, we prove that (u+ )∗ = (u[ϕ∗ ])∗ = (u[ϕ∗ ])∗ = u−

in Ω.

First equality : (u+ )∗ = (u[ϕ∗ ])∗ . This is nothing but Theorem 3.1.2. Second equality : (u[ϕ∗ ])∗ = (u[ϕ∗ ])∗ . It is clear that (u[ϕ∗ ])∗ ≤ (u[ϕ∗ ])∗ in the domain Ω, since u[ϕ∗ ] ≤ u[ϕ∗ ] on Ω. To prove the opposite inequality, let x be a point of Ω. Then there exists a sequence (xn )n of points of Ω such that xn → x and lim u[ϕ∗ ](xn ) = (u[ϕ∗ ])∗ (x). n For each xn , we consider a control αn (.) such that u[ϕ∗ ](xn ) +

1 ≥ J(xn , αn , τn , ϕ∗ ). n

(3.25)

We denote by yxnn the trajectory associated to αn and by zn its first exit point from Ω i.e. zn := yxnn (τn ) ∈ ∂Ω1 . By using the assumption (3.4), we take a sequence (znp )p of points of the boundary such that znp → zn and lim ϕ∗ (znp ) = ϕ∗ (zn ). (3.26) p

Now, we need the following lemma.

3.1 The exit time control problems with discontinuous exit costs

49

Lemma 3.1.4 Let yx be a trajectory such that its first exit time τ from Ω is positive and bounded. Then, for ε small enough and for z˜ ∈ ∂Ω1 such that |yx (τ ) − z˜| < ε, there exists a trajectory y˜x˜ such that z˜ = y˜x˜ (˜ τ ) and |yx (t) − y˜x˜ (t)| < Dε

for t ∈ [0, τ ] t u

with some constant D independent of ε.

For each the trajectory yxnn and for p large enough, using (3.26), we apply Lemma 3.1.4 to the point znp . Then, using again (3.26) and the Lipschitz continuity of the function f , we obtain after tedious but straighforward computations J(xn , αn , τn , ϕ∗ ) ≥ J(xpn , αnp , τnp , ϕ∗ ) − εpn , with a sequence (εpn )p of numbers such that εpn → 0 as p → ∞. Combining this with (3.25), we get u[ϕ∗ ](xn ) +

1 ≥ J(xpn , αnp , τnp , ϕ∗ ) − εpn n ≥ u[ϕ∗ ](xpn ) − εpn .

Hence, passing to the limit, by a diagonal procedure, we obtain (u[ϕ∗ ])∗ (x) = lim u[ϕ∗ ](xn ) n

≥ (u[ϕ∗ ])∗ (x). Since it is true for every x ∈ Ω, the result is complete. Third equality : (u[ϕ∗ ])∗ = u− . By their definitions, u− ≤ u[ϕ∗ ] in Ω. And, since the value function u− is l.s.c., we deduce u− ≤ (u[ϕ∗ ])∗ in Ω. It remains to prove the opposite inequality. Let x0 be a point of Ω. By the compactness of the set of relaxed controls, using classical arguments, there exist µ ¯ ∈ L∞ (IR+ , P (A)) and θ¯ ∈ [0, ∞] such that ¯ ϕ∗ ). ˆ 0, µ u− (x0 ) = J(x ¯, θ, Then we need the following lemma. Lemma 3.1.5 Let yˆx be a trajectory associated to a relaxed control µ and to a point x ∈ Ω. Then, for any bounded exit time θ of yˆx , there exists a classical trajectory y˜x˜ arbitrarily close to yˆx and such that y˜x˜ (˜ τ ) = yˆx (θ) where τ˜ is the first exit time of y˜x˜ from Ω.

t u

50

Probl`emes de temps de sortie

If θ¯ is bounded, using Lemma 3.1.5, there exists a sequence of points xn and of controls αn (.) such that xn → x0 and ¯ ϕ∗ ) = J(xn , αn , τn , ϕ∗ ) − εn ˆ 0, µ J(x ¯, θ, where εn → 0 when n → ∞ using the Lipschitz continuity of f . Then this implies u− (x0 ) ≥ u[ϕ∗ ](xn ) − εn . Passing to the limit, we get u− (x0 ) ≥ (u[ϕ∗ ])∗ (x0 ). It remains the cases when θ¯ is not bounded. We have several cases to consider. The easier is when the trajectory yˆx0 stays far to the boundary i.e. inf d(ˆ yx0 (t)) = δ > 0,

t≥0

then classical arguments imply that yˆx0 can be approximated by classical trajectories. A connected case is when it is true but only after some finite time T i.e. inf d(ˆ yx0 (t)) = δ > 0,

t≥T

then we mix the preceding arguments with ones of Lemma 3.1.4. The last case is when, for any T > 0, we get inf d(ˆ yx0 (t)) = 0.

t≥T

Then we may assume that there exists a sequence of trajectories yn such that their exit times τn tend to ∞ as n → ∞, by using Lemma 3.1.4 for exit times of yˆx0 large enough, or after modifying yˆx0 when yˆx0 is near the boundary, in order that it touches the boundary by using (3.23). Therefore, using the term e−λt which tends to zero as t → ∞ and the boundedness of f and ϕ, there exists a sequence (εn )n of numbers such that εn → 0 as n → ∞ and ¯ ϕ∗ ) = ˆ 0, µ J(x ¯, θ,

Z 0

∞

Z A

f (ˆ yx0 (t), α)e−λt d¯ µ(α)dt

≥ J(xn , αn , τn , ϕ∗ ) − εn . And then we conclude as before. It remains to prove that u− = (u[ϕ∗ ])∗ on the boundary ∂Ω1 . Then we need the following lemma.

3.1 The exit time control problems with discontinuous exit costs

51

Lemma 3.1.6 The function u− satisfies u− (x) = lim inf u− (y), y→x,y∈Ω

for every x ∈ ∂Ω1 . t u

We postpone the proof of this lemma. Moreover, since the function u+ is u.s.c., for every x ∈ ∂Ω1 , we have (u+ )∗ (x) = lim inf u+ (y). y→x,y∈Ω

Therefore, since u− = (u+ )∗ in Ω, we deduce that u− = (u+ )∗ on ∂Ω.

t u

Now we turn to the proof of the lemmas. Proof of Lemma 3.1.4. In order to simplify the notation, we denote by z := yx (τ ), by α(.) the control associated to the trajectory yx and by α ˜ (.) the control associated to y˜x˜ which is the unknown. We first remark that, since the map Yτ : x 7→ yx (τ ) is locally an homeomorphism, we may always find a point x˜ such that yx˜ (τ ) = z˜. Then a possible candidate for α ˜ may directly be α (and thus the constant D is equal to eCτ by standard ODE estimates). But the difficulty is that the trajectory yx˜ may cross the complementary set of Ω to reach z˜. In order to avoid this difficulty, we use the assumption (3.23) of a partial controllability on the boundary. Briefly speaking, we consider a backward trajectory which comes from the point z˜ and which goes in the interior of Ω, during a short time. And after, the trajectory is again associated to α in order to reach a neighborhood of x. First, using the assumption (3.23), we consider the control α ¯ ∈ A such that b(z, α ¯ ) · n(z) > 0. Thus we define the control α ˜ by   α(t)

α ˜ (t) = 

α ¯

for t ∈ [0, τ [, for t ∈ [τ, τ + ε].

Then the trajectory y˜x˜ is the solution of   

d˜ yx˜ (s) = b(˜ yx˜ (s), α ˜ (s))ds,

for s ∈ [0, τ + ε[

y˜x˜ (τ + ε) = z˜.

Using the assumptions (3.22) on n and (3.6) on b, there exists a constant δ > 0 such that β for ξ ∈ B(z, δ) ∩ Ω. b(ξ, α ¯ ) · n(ξ) > 2

52

Probl`emes de temps de sortie

Moreover, using the boundedness of b, standard ODE estimates yield, for t ∈ [τ, τ +ε], |z − y˜x˜ (t)| ≤ |z − z˜| + |˜ z − y˜x˜ (t)| ≤ ε + Cε. Therefore, for ε small enough, the trajectory y˜x˜ stays in B(z, δ) and then we compute, for t ∈ [τ, τ + ε], Z

d(˜ yx˜ (t)) = d(˜ z) + ≥

τ +ε

n(˜ yx˜ (s)) · b(˜ yx˜ (s), α ¯ )ds

t

β (τ + ε − t). 2

(3.27)

Next, we prove that the trajectory y˜x˜ does not touch the boundary during [τ −η, τ ] where η is a positive constant to determine. In order to simplify the notation, we set zε := y˜x˜ (τ ) and y˜zε (.) := y˜x˜ (.). To prove the claim, it suffices to show that d(˜ yzε (t)) is positive for t ∈ [τ − η, τ ]. Since the functions b and n are Lipschitz continuous, at least, in the neighborhood V of the boundary, for η small enough, we compute d(˜ yzε (t)) − d(yx (t)) = d(zε ) − d(z) +

Z τ t



n(˜ yzε (s)) · b(˜ yzε (s), α(s)) − n(yx (s)) · b(yx (s), α(s)) ds

≥ d(zε ) − CK

τ

Z

|˜ yzε (s) − yx (s)|ds.

t

Using standard ODE estimates, we obtain d(˜ yzε (t)) − d(yx (t)) ≥ d(zε ) − CK

Z t

τ

|zε − z|eC(τ −s) ds

≥ d(zε ) − K|zε − z|(eC(τ −t) − 1) ≥ d(zε ) − K(|zε − z˜| + |˜ z − z|)(eCη − 1).

(3.28)

We need an estimate for |zε − z˜|. But, since b is bounded and since zε = y˜zε (τ ), we have Z

|zε − z˜| =

τ +ε

τ



b(˜ yzε (s), α ¯ )ds

≤ Cε. Combining this and (3.27) with (3.28), we get d(˜ yzε (t)) − d(yx (t)) ≥

βε − K(Cε + |˜ z − z|)(eCη − 1). 2

(3.29)

3.1 The exit time control problems with discontinuous exit costs

53

But, since z˜ is such that |˜ z − z| < ε, this implies d(˜ yzε (t)) − d(yx (t)) ≥ ε

β

2



− K(1 + C)(eCη − 1) .

(3.30)

Then we take η such that the right hand side of this inequality is null i.e. η :=

 1  β ln 1 + . C 2K(1 + C)

Hence we conclude that d(˜ yzε (t)) ≥ 0

for every t ∈ [τ − η, τ ],

and the inequality is strict because of the term d(yx (t)). We may remark that the constant η depends only on the functions b and n. Finally, since τ is the first exit time of yx , we have inf{d(yx (t)), t ∈ [0, τ − η]} = ρ > 0. We compute, for t ∈ [0, τ − η], yzε − yx )(t)| d(˜ yzε (t)) ≥ d(yx (t)) − |(˜ ≥ ρ − |zε − z|(eCτ − 1) ≥ ρ − (C + 1)(eCτ − 1)ε. Therefore, for ε small enough, we get the result with a constant D equal to (C + 1)(eCτ − 1). u t Proof of Lemma 3.1.5. First we approximate yˆx by a relaxed trajectory which the first exit point is z˜ := yˆx (θ). We use backward trajectories. We recall that a backward trajectory yˇz is a solution of the ODE  yz (t) = −b(ˇ yz (t), α(t))dt,  dˇ 

yˇz (0) = z.

We begin to choose a control α ¯ ∈ A such that b(˜ z, α ¯ )·n(˜ z ) > 0. Then we consider the backward trajectory yˇ1 with yˇ1 (0) = z˜ and associated to the relaxed control µ1 defined by  ¯ if t ∈ [0, ε1 [,  α µ1 (t) =  µ(t) elsewhere,

54

Probl`emes de temps de sortie

with a parameter ε1 > 0 to be chosen later. Using the estimate (3.27) which can be extended to backward relaxed trajectories, we know that the new trajectory yˇ1 does not touch the boundary during ]0, ε1 [, for ε1 small enough. Now, if the new trajectory yˇ1 touches the boundary at the time τ1 , we set z1 := yˇ1 (τ1 ) and we choose α1 ∈ A such that b(z1 , α1 ) · n(z1 ) > 0. Then we change the control µ1 for µ2 given by µ2 (t) =

  α1 

µ1 (t)

if t ∈ [τ1 − ε2 , τ1 ], elsewhere.

Therefore the trajectory associated to µ2 does not touch ∂Ω, at least, during [τ1 − ε2 , τ1 ]. In fact, by using the arguments of the proof of Lemma 3.1.4, we know that the trajectory lies in the open set Ω during the fixed time η. Next, we do the same thing as many times as the trajectory touches the boundary. Finally, since after each modification, we know that the trajectory does not touch ∂Ω during η, we change the control only a finite number of time. In order to be close to the trajectory yˆx , it suffices to choose the parameters εi small enough since the relaxed control of the new trajectory is equal to µ except during the times εi where µ is replaced by αi . Finally, it remains to approximate the new relaxed trajectory by a classical trajectory. But it is standard since the relaxed trajectory lies in Ω during [ε1 , θ]. t u Proof of Lemma 3.1.6. We consider a point z of ∂Ω1 . Using the assumption (3.23), there exists a control α ¯ ∈ A such that b(z, α ¯ ) · n(z) > 0. As in Lemma 3.1.4, we consider the solution y˜z of for s ∈ [0, 1[,

 yz (s) = b(˜ yz (s), α ¯ )ds,  d˜ 

y˜z (1) = z.

Then we consider the points xn := y˜z (1 −

1 ) n

and the trajectories 1 ). n It is clear enough that, for n large enough, xn ∈ Ω and xn → z as n → ∞. Notice also that the first exit time of y˜xn is equal to n1 . Thus the cost function associated to the trajectory y˜xn is y˜xn (.) := y˜z (. + 1 −

Z 1 1 1 n J(xn , α ¯ , , ϕ∗ ) = f (˜ yxn (t), α ¯ )e−λt dt + ϕ∗ (z)e−λ n . n 0

3.1 The exit time control problems with discontinuous exit costs

55

Therefore this implies 1 u− (xn ) ≤ J(xn , α ¯ , , ϕ∗ ). n Then we pass to the limit 1 limninf u− (xn ) ≤ limninf J(yn , α1 , , ϕ∗ ) n ≤ ϕ∗ (x0 ). Finally, by definition of the liminf, we deduce lim inf u− (y) ≤ limninf u− (xn )

y→z,y∈Ω

≤ ϕ∗ (x0 ) t u

And the proof is complete.

Example 3.1.3 Our aim is to show that the assumption (3.23) is necessary to prove Lemma 3.1.5. More precisely, we give a relaxed trajectory which cannot be approximated by classical trajectories in a closed set. To this end, we consider in IR3 , the field b given, for x = (x1 , x2 , x3 ) ∈ IR3 , by 



1   α b(x, α) =   1 + |x | 2 2 1+α with α ∈ A := [−1, 1]. (This field is already considered in [7]). Around the point O = (0, 0, 0), the domain Ω is reduced by two cylinders C1 and C2 C1 = {(x1 , x2 , x3 ) ∈ IR3 , (x1 − 5)2 + (x3 )2 ≤ 5} , C2 = {(x1 , x2 , x3 ) ∈ IR3 , (x1 − 9)2 + (x3 − 7)2 ≤ 5} . We can define a relaxed trajectory yˆO satisfying (ˆ yO )3 (t) = 2t which is tangent to the two cylinders. We can easily prove that classical trajectories cannot approximate the trajectory yˆO when t ≥ 10 because, since the classical trajectories are such that (y(t)) ˙ 3 >

1 2

for t > 0,

(see [7]), at the time t = 5, they have to be above the cylinder C1 and then they cannot pass below C2 for t = 10.

56

Probl`emes de temps de sortie

3.2 The uniqueness results In this part, we prove uniqueness results for Hamilton-Jacobi equations in the case when the Dirichlet boundary data ϕ is only assumed to be defined pointwise, by using PDE methods. We recall that uniqueness means in this context that all the solutions have the same l.s.c. envelope. For simplicity in what follows, we will henceforth assume that H(x, t, p) = H(x, p) + t and that (x, p) → H(x, p) is a continuous function from Ω × IRN into IR which satisfies (3.31) H(x, p) is convex in p, for every x ∈ Ω, and ∂H ≤ C,

∂p

∂H ≤ C(1 + |p|)

∂x

on Ω × IRN , for some constant C > 0. (3.32)

To explain the problems coming from the discontinuity of the Dirichlet condition, we consider the function ϕ being equal to 0 at some point x0 ∈ ∂Ω and 1 elsewhere, as in the introduction. Then the viscosity subsolution condition on the boundary is satisfied by the function u if u∗ ≤ ϕ∗ ≡ 1

on ∂Ω.

It is clear that this condition is not restrictive enough because the fact that ϕ(x0 ) = 0 is not seen by this boundary condition. Indeed, for illustration, we consider the exit time problem from the point x0 . To this end, we even assume that the controllability is complete i.e. the field b is given by b(x, α) := α with α in the unit ball B(0, 1) of IRN . Then, if we assume that the domain Ω is convex, any point x ∈ Ω may reach the point x0 by following the straight line [x, x0 ]. Finally, we take f ≡ 1 and λ = 1. With the notation of the introduction, the value function u[ϕ] is defined, for every x ∈ Ω, by u[ϕ](x) = =

inf+

α(.)∈L∞ (IR

inf

Z

{ ,B(0,1))

α(.)∈L∞ (IR+ ,B(0,1))

0

τ

1e−t dt + ϕ(yx (τ ))e−τ }

{1 − e−τ + 11∂Ω\{x0 } (yx (τ ))e−τ }.

It is easy to show that the best strategy is to reach the point x0 . Then, the value function is equal to u[ϕ](x) = 1 − e−|x−x0 | . But, by the result of the first part, all the functions uζ (x) := 1 − ζe−|x−x0 | with ζ ∈ [0, 1] are also solutions of (3.2) since ϕ∗ ≤ ζ11∂Ω\{x0 } ≤ ϕ∗ . In particular, the

3.2 The uniqueness results

57

maximal subsolution u+ which is equal to u[ϕ∗ ] ≡ 1, appears to be a pathological solution for this exit time problem. We consider two ways to avoid this difficulty : the first one is to consider only “regular” boundary data ϕ i.e. functions satisfying the condition (3.4) as we did it in the first section. The second one is to impose additional conditions on the solution u : since the viscosity subsolution condition on the boundary turns out to be not restrictive enough, we replace it by the condition (3.5) i.e. for any x ∈ ∂Ω, lim inf u(y) ≤ ϕ∗ (x),

y→x,y∈Ω

and we need, moreover, that the function u is a viscosity solution in the sense of Barron-Jensen inside the domain Ω. For the sake of completeness, we recall the Definition 3.2.1 Let u be a bounded function. We say that u is a Barron-Jensen solution of H(x, Du) + u = 0 in Ω, (3.33) if it satisfies   ∀ φ ∈ C 1 (Ω), at each minimum point x0 ∈ Ω of u∗ − φ, we have 

(3.34)

H(x0 , Dφ(x0 )) + u∗ (x0 ) = 0. t u

In the case of continuous solutions, we have an equivalence between (3.34) and the definition of viscosity solutions of (3.33) : if the function u satisfies (3.34), then u is a viscosity solution of (3.33) and vice versa. For discontinuous solutions, the connections are far less simple (cf. [5]). • If u is a bounded viscosity solution of (3.33) satisfying (u∗ )∗ = u∗ in Ω then the property (3.34) holds for u∗ . (In particular, it holds if u is u.s.c.) • If u is a l.s.c. bounded function satisfying the property (3.34) then u is a viscosity solution of (3.33). To prove the uniqueness result, we need to add non-degeneracy conditions on the hamiltonian on the boundary. To do so, we recall that we assume that the distance function d satisfies (3.22). We assume that ∂Ω = ∂Ω1 ∪ ∂Ω2 where ∂Ω1 and ∂Ω2 are unions of connected components of ∂Ω and   ∀x ∈ ∂Ω1 , ∀R > 0, ∃C R > 0 such that     

1

+

N

if |y − x| ≤ R , y ∈ Ω and λ ≥ C R (1 + |p|), λ ∈ IR , p ∈ IR ,   C     then

H(y, p − λn(y)) ≥ R,

(3.35)

58

Probl`emes de temps de sortie

and   ∀x ∈ ∂Ω2 , ∀R > 0, ∃C R > 0 such that     

1

+

N

if |y − x| ≤ R , y ∈ Ω and λ ≥ C R (1 + |p|), λ ∈ IR , p ∈ IR ,   C     then

(3.36)

H(y, p − λn(y)) ≤ −R,

We recall that n(x) := −Dd(x) for every x in the neighborhood V of ∂Ω. And we also assume that   in a neighborhood of ∂Ω2 , for every p ∈ IRN ,  the map λ → H(x, p + λn(x)) is a nondecreasing function.

(3.37)

In the case of control problems when the hamiltonian is given by (3.1), the assumption (3.35) is equivalent to (3.23) i.e. the existence of an outer field on ∂Ω1 and the assumptions (3.36) and (3.37) are equivalent to (3.24) which implies that there are only inner fields on ∂Ω2 . The assumption (3.36) is used in the comparison result of [9]. Our result is Theorem 3.2.1 We assume that ϕ is a bounded function defined pointwise and that the assumptions (3.22), (3.31), (3.32) and (3.35)-(3.37) hold. 1. If the condition (3.4) holds on ∂Ω1 and if the functions u and v are viscosity solutions of the equation (3.2), then u∗ = v ∗

in Ω.

2. If the functions u and v are Barron-Jensen solutions of (3.33) which are supersolutions of the equation (3.2) on the boundary and which satisfy the condition (3.5) on ∂Ω1 , then u∗ = v∗ in Ω. t u In the first point, we recover the uniqueness result of the first section in the case of control problems. We will detail in the part on applications the adaptation of the second point. In particular, we will show that all the value functions are BarronJensen solutions of the associated Bellman equation. Finally, let us come back to the example of the beginning of the section. The function u+ which is equal to 1, and the functions uζ with ζ ∈]0, 1[ are BarronJensen solutions of the associated Bellman equation in Ω but they do not satisfy the condition (3.5) on ∂Ω. Only the function u[ϕ] satisfies both properties.

3.2 The uniqueness results

59

A viscosity solution u is a Barron-Jensen solution in particular, if u satisfies the property (u∗ )∗ = u∗ in Ω which is a criteria for uniqueness in [8] or if the function u is continuous. We recall that the solution u is continuous if H(x, p) → ∞ when |p| → ∞. This result is similar to that obtained in the case of optimal stopping time problems with discontinuous stopping costs (cf. [5]). The following comparison principle is the keystone of the proof of the uniqueness results. Theorem 3.2.2 Under the assumptions of Theorem 3.2.1, if the function u is a Barron-Jensen solution of (3.33) satisfying (3.5) on ∂Ω1 and if the function v is a bounded supersolution of (3.2), then, u∗ ≤ v ∗

in Ω. t u

Proof of Theorem 3.2.1. 1. First, we remark that all the subsolutions of the equation (3.2) satisfy the condition (3.5) : indeed, using a classical result when the assumption (3.35) holds, we have u∗ ≤ ϕ∗

on ∂Ω1 .

Taking the liminf, we find that (u∗ )∗ ≤ (ϕ∗ )∗ = ϕ∗

on ∂Ω1 ,

(3.38)

since the function ϕ satisfies the assumption (3.4). Finally, since we have, for every x ∈ ∂Ω, lim inf u∗ (y) = (u∗ )∗ (x), y→x,y∈Ω

we deduce from (3.38) that the solution u∗ satisfies, for every x ∈ ∂Ω1 , lim inf u∗ (y) ≤ ϕ∗ (x).

y→x,y∈Ω

Now since the function u∗ is not necessarily a Barron-Jensen solution of (3.33), we are not able to use directly Theorem 3.2.2 with u. Then we introduce the maximal subsolution u+ of the equation (3.2). The function u+ may be obtained by considering the following standard approximate Dirichlet problem  1   − |Dun |2 + H(x, Dun ) + un

n

 

= 0

un = ϕ n

in Ω, on ∂Ω,

(3.39)

60

Probl`emes de temps de sortie

where (ϕn )n is a nonincreasing sequence of continuous functions such that inf ϕn = ϕ∗ . n The existence of solutions of the problem (3.39) is a classical consequence of the Perron’s method (cf. [24]) since the constants Mn := max{kH(x, 0)k∞ , kϕn k∞ } and −Mn are respectively super- and subsolution of this problem. Moreover, the new hamiltonian satisfies the following property 1 − |p|2 + H(x, p) → −∞ n

as

|p| → ∞,

which corresponds to the opposite of the classical coercivity property. Nevertheless, we recover the results of [8] but for supersolutions instead of subsolutions : the supersolutions of (3.39) are uniformly Lipschitz continuous in Ω and we have a comparison result for this equation. Thus we get that the function un is continuous and unique. Thanks to the formulation of the approximate problem, a subsolution w of the equation (3.2) is still subsolution of the new equation (3.39). Then this implies that w ≤ un in Ω. Hence, the maximal subsolution u+ is defined, for x ∈ Ω, by u+ (x) := lim sup un (y) = inf un (x). n n→∞,y→x

Therefore the function u+ is u.s.c.; hence it satisfies the property (3.34) and the condition (3.5) by the first remark. Using the comparison principle of Theorem 3.2.2, we get (u+ )∗ ≤ v∗ in Ω. But, since u+ is the maximal subsolution, we have u+ ≥ u∗ in Ω. Hence v∗ ≥ (u+ )∗ ≥ (u∗ )∗ ≥ u∗

in Ω.

Finally, we exchange the roles of the functions u and v and thus the proof is complete. 2. The second point is a direct consequence of Theorem 3.2.2. t u Now we turn to the proof of the comparison principle. Proof of Theorem 3.2.2. We first recall that the distance function to the boundary d is assumed to be C 1,1 in the neighborhood V of ∂Ω by the assumption (3.22). We still denote by d a nondecreasing C 1,1 function on Ω which is equal to the distance function to ∂Ω in a

3.2 The uniqueness results

61

neighborhood V 0 of ∂Ω included in V. And we set n(x) := −Dd(x) for x ∈ Ω, even if, for x 6∈ V 0 , n(x) is not necessarily unitary. For every α > 0, we introduce the function uα defined, for every (x, t) ∈ Ω × IR+ , by n o ¯ uα (x, t) = inf u∗ (y) + e−Kt φα (x, y) + C(d(x) − d(y)) , y∈Ω

with φα (x, y) :=

|x − y|3 |d(x) − d(y)|4 |x − y|4 +L (d(x) − d(y)) + M , α α α

¯ K, L and M are some positive constants to be chosen later. where C, The properties of uα are described in the ¯ K, L and M such that, for every T > 0 Lemma 3.2.1 There exist constants C, and for α small enough, the following properties hold 1. the function uα is Lipschitz continuous in Ω × [0, T ]. (Thus we may extend it by continuity to the boundary ∂Ω × [0, T ].) 2. The function uα is a viscosity subsolution of √ ∂w + H(x, Dw) + w − B 4 α = 0 ∂t

in (Ω ∪ ∂Ω2 )×]0, T ],

(3.40)

for some constant B > 0, independent of α. 3. We have uα ≤ ϕα on ∂Ω1 × [0, T ], where the function ϕα is defined, for every x ∈ ∂Ω1 and t ≥ 0, by α

ϕ (x, t) =

inf

y∈∂Ω1

n

−Kt |x

ϕ∗ (y) + e

− y|4 o . α t u

We postpone the proof of the lemma. Remark 3.2.1 The inf-convolution procedure which usually leads a supersolution in the viscosity solution theory, allows us to obtain a subsolution following new ideas introduced by Barron-Jensen [10] for convex hamiltonians. The time-dependent formulation of uα is a technical point which permits to treat discontinuous solutions for stationary problems (cf. [5]). The terms with the distance function is a trick to obtain the classical property of the inf-convolution procedure despite of the presence of the boundary, thanks the assumptions (3.35) and (3.36). The distance function to ∂Ω plays here essentially the same role as the time variable in the classical Barron-Jensen approach for Cauchy problem.

62

Probl`emes de temps de sortie

To conclude the proof of Theorem 3.2.2, we compare the functions uα and v. Since ϕ∗ ≥ ϕα on Ω and since the function v∗ does not depend on t, v∗ is a supersolution of  ∂w   + H(x, Dw) + w = 0 in Ω×]0, T ], ∂t (3.41)   w = ϕα on ∂Ω×]0, T ]. Because of the Lipschitz continuity of uα and ϕα , an easy adaptation of the comparison result of [8] for the Cauchy problem (3.41) (see also [28] and [32]) yields √ uα (x, T ) − v∗ (x) ≤ e−T k(uα (., 0) − v∗ (.))+ k∞ + B 4 α. We first let α → 0 in this inequality and then T → ∞. Since the functions u and v are bounded, we obtain u∗ ≤ v∗ in Ω, t u

which is the inequality we wanted to prove. Now we turn to the proofs of the lemmas. Proof of Lemma 3.2.1. 1. Regularity and elementary properties of uα . For every x, z ∈ Ω and t ≥ 0, we have uα (x, t) − uα (z, t) ≤ sup

n



¯ e−Kt φα (x, y) + C(d(x) − d(y))

y∈Ω



o

¯ − e−Kt φα (z, y) + C(d(z) − d(y))

¯ ≤ e−Kt sup{φα (x, y) − φα (z, y)} + C(d(x) − d(z)). y∈Ω

By the compactness of the domain Ω and by Lipschitz regularity of the distance function, we obtain C uα (x, t) − uα (z, t) ≤ |x − z|. α Therefore the function uα is Lipschitz continuous in space variable. It is easy to check the same property in time variable. Then, by Rademacher’s Theorem, uα is differentiable almost everywhere and then, by classical result in optimization theory, ¯ if yα is a point such that uα (x, t) = u∗ (yα ) + e−Kt φα (x, yα ) + C(d(x) − d(yα )), we have ¯ Duα (x, t) = e−Kt Dx φα (x, yα ) − Cn(x) and

∂uα (x, t) = −Ke−Kt φα (x, yα ). ∂t

3.2 The uniqueness results

63

Let us check some elementary properties of the function uα . First, using the fact that the functions u and v are bounded, we fix the constant C¯ to the largest of the constants C R which appear in the assumptions (3.35) and (3.36) for R = max{kuk∞ , kvk∞ }. Lemma 3.2.2 For L large enough and M  L and for every (x, t) ∈ Ω × [0, T ], the following properties hold 1. uα (x, t) ≤ u∗ (x). ¯ 2. If yα is a point such that uα (x, t) = u∗ (yα ) + e−Kt φα (x, yα ) + C(d(x) − d(yα )), there exists a constant E independent of α such that √ |x − yα | ≤ E 4 α. 3. For x in a neighborhood of the boundary, if we write Duα (x, t) = pα − λα n(x) and if we have d(x) ≥ d(yα ), then ¯ + |pα |). λα ≥ C(1 ¯ independent of α and of K such that 4. There exists a constant K ¯ |Duα (x, t)||x − yα | ≤ K(|x − yα | + e−Kt φα (x, yα )). t u We postpone the proof of this lemma. 2. Equation satisfied by uα . Since the hamiltonian is convex, in order to prove that the function uα is a viscosity subsolution of (3.40) in Ω×]0, T [, it is enough to show that uα satisfies the equation (3.40) in the almost everywhere sense (cf. [28]). Let (x0 , t0 ) be a point of Ω×]0, T [ where uα is differentiable. We consider a point yα ∈ Ω such that ¯ uα (x0 , t0 ) = u∗ (yα ) + e−Kt0 φα (x0 , yα ) + C(d(x 0 ) − d(yα )). And by the definition of uα , the point yα is a minimum point of u∗ − ψ with ¯ ψ(y) := −e−Kt0 φα (x0 , y) − C(d(x 0 ) − d(y)). In order to use the equation on u, we have to show that yα ∈ Ω. We first remark that the assumptions (3.22), (3.35) and (3.36) are satisfied in the neighborhood of ∂Ω included in V 0 Ωδ := {x ∈ Ω such that d(x) < δ}, for some positive constant δ small enough.

64

Probl`emes de temps de sortie

• If x0 6∈ Ωδ , then d(x0 ) ≥ δ and, by using the second point of Lemma 3.2.2, we obtain d(yα ) ≥ d(x0 ) − |x0 − yα | √ ≥ δ − E 4 α. Hence, for α small enough, we find that d(yα ) > 0. • If x0 ∈ Ωδ , we need the following lemma Lemma 3.2.3 If x0 ∈ Ωδ , then we have d(x0 ) ≤ d(yα ). t u Using Lemma 3.2.3, we deduce that d(yα ) > 0 since x0 ∈ Ω. Hence, we get H(yα , Dψ(yα )) + u∗ (yα ) = 0,

(3.42)

where ¯ Dψ(yα ) = −e−Kt Dy φα (x, yα ) − Cn(y α ). We want to replace yα by x0 and u∗ (yα ) by uα (x0 , t0 ) in the equality (3.42). By using the assumption (3.32) and the definition of uα , we get H(x0 , Duα (x0 , t0 )) + uα (x0 , t0 ) ≤ C(1 + |Duα (x0 , t0 )|)|x0 − yα | +C|Duα (x0 , t0 ) − Dψ(yα )|

(3.43)

¯ +e−Kt0 φα (x0 , yα ) + C(d(x 0 ) − d(yα )). To estimate the right side of this inequality, we use Lemma 3.2.2 and we compute |Duα (x0 , t0 ) − Dψ(yα )| ≤ |λα ||n(x0 ) − n(yα )| ≤ kDnk∞ |Duα (x0 , t0 )||x0 − yα |. Therefore we deduce from the inequality (3.43) that √ H(x0 , Duα (x0 , t0 )) + uα (x0 , t0 ) ≤ B 4 α + Ke−Kt0 φα (x0 , yα ), ¯ + KkDnk ¯ ¯ ¯ with B := (C(1 + K ∞ ) + C)E and K := C K(1 + kDnk∞ ) + 1. α It remains to introduce the time derivative of u i.e. ∂uα (x0 , t0 ) = −Ke−Kt0 φα (x0 , yα ). ∂t

3.2 The uniqueness results

65

Then, we obtain √ ∂uα (x0 , t0 ) + H(x0 , Duα (x0 , t0 )) + uα (x0 , t0 ) ≤ B 4 α. ∂t Hence, the function uα satisfies the equation (3.40) in the almost everywhere sense. Finally, since uα is a viscosity subsolution of (3.40) in Ω×]0, T [, uα is also a subsolution of (3.40) on the boundary Ω × {T } by classical properties of Cauchy problems (cf. [9]). Moreover, on the boundary ∂Ω2 ×]0, T ], the assumption (3.37) implies Lemma 3.2.4 The function uα is a viscosity subsolution of the equation (3.40) on ∂Ω2 ×]0, T ]. u t We postpone again the proof of this lemma. 3. Boundary properties of uα . By the definition of ϕα , for any x0 ∈ ∂Ω1 and any t ≥ 0, there exists a point y0 of ∂Ω1 such that |x0 − y0 |4 . ϕα (x0 , t) = ϕ∗ (y0 ) + e−Kt α We take a sequence (yn )n of points of Ω such that yn → y0 and lim inf u(y) = lim u(yn ). n

y→y0 ,y∈Ω

By the property (3.5), we know that lim u(yn ) ≤ ϕ∗ (y0 ). n

Now, we introduce the sequence (xn )n defined by xn := x0 − d(yn )n(x0 ) for every n. Since we have d(yn ) → 0 as n → ∞, the sequence (xn )n converges to x0 . Notice also that d(xn ) = d(yn ) for n large enough. By the definition of uα , we may write ¯ uα (xn , t) ≤ u∗ (yn ) + e−Kt φα (xn , yn ) + C(d(x n ) − d(yn )) ≤ u∗ (yn ) + e−Kt +M

h |x − y |4 n n

α

+L

|xn − yn |3 (d(xn ) − d(yn )) α

|d(xn ) − d(yn )|4 i ¯ + C(d(xn ) − d(yn )) α

≤ u∗ (yn ) + e−Kt

|xn − yn |4 α

for n large enough.

66

Probl`emes de temps de sortie

Then, letting n → ∞, we obtain uα (x0 , t) = lim uα (xn , t) ≤ ϕ∗ (y0 ) + e−Kt n

|x0 − y0 |4 = ϕα (x0 , t). α

Since it is true for every x0 ∈ ∂Ω1 and t ≥ 0, the result is proved.

t u

Proof of Lemma 3.2.2. 1. Since φα (x, x) = 0 for any x ∈ Ω, we have u∗ ≥ uα in Ω × [0, T ]. 2. Since the function u∗ is l.s.c., for any α, there exists a point yα ∈ Ω such that ¯ uα (x, t) = u∗ (yα ) + e−Kt φα (x, yα ) + C(d(x) − d(yα )), and since uα ≤ u∗ in Ω × [0, T ], we have ¯ u∗ (yα ) + e−Kt φα (x, yα ) + C(d(x) − d(yα )) ≤ u∗ (x) and then, we deduce ¯ e−Kt φα (x, yα ) ≤ 2kuk∞ + 2Ckdk ∞.

(3.44)

By using the Young inequality L

|x − y|3 3 |x − y|4 1 4 |d(x) − d(y)|4 (d(x) − d(y)) ≤ + L , α 4 α 4 α

we get φα (x, yα ) ≥

1 |x − y|4 1 |d(y) − d(x)|4 + (M − L4 ) 4 α 4 α

1 |x − y|4 ≥ , 4 α

(3.45) (3.46)

for M ≥ 21 L4 . Combining with (3.44), this implies |x − yα | ≤ e

q

Kt 4 4

¯ 2(kuk∞ + Ckdk ∞ )α.

3. If the function uα is differentiable at (x, t), we may write Duα (x, t) = pα − λα n(x) with h

pα := e−Kt 4(x − yα ) and

i |x − yα | |x − yα |2 + 3L(x − yα ) (d(x) − d(yα )) , α α

h |x − y |3 α

λα := e−Kt L

α

+ 4M (d(x) − d(yα ))

|d(x) − d(yα )|2 i ¯ + C. α

3.2 The uniqueness results

67

Again, we use the Young inequality to get 3L

|d(x) − d(yα )|3 |x − yα |3 |x − yα |2 |d(x) − d(yα )| ≤ 2 + L3 . α α α

If we have d(x) ≥ d(yα ), then we deduce 3

3i

¯ + |pα |) − λα ≤ e−Kt (6C¯ − L) |x − yα | + (CL ¯ 3 − 4M ) |d(x) − d(yα )| . C(1 α α And the right hand side of this inequality is negative for L ≥ 6C¯ since we already know that M ≥ 12 L4 . h

4. We compute h |x − y |4 α

|Duα (x, t)||x − yα | ≤ e−Kt 4

α

+4M

+ 3L

|x − yα |3 |x − yα |4 |d(x) − d(yα )| + L α α

i |d(x) − d(yα )|3 ¯ − yα |. |x − yα | + C|x α

(3.47)

Then, using again Young inequalities, we obtain after straightforward computations h

|Duα (x, t)||x − yα | ≤ e−Kt (4 +

|x − yα |4 3 |d(x) − d(yα )|4 i 13 L + M) + ( L + 3M ) 4 α 4 α

¯ − yα |. +C|x ¯ Finally, we use the inequality (3.45) to fix the constant K.

t u

Proof of Lemma 3.2.3. We argue by contradiction assuming that d(x0 ) > d(yα ).

(3.48)

We first consider the case when the point yα is in the domain Ω. Hence, we use the equation on u and we get H(yα , Dψ(yα )) + u∗ (yα ) = 0,

(3.49)

where Dψ(yα ) = pα − λα n(yα ) with the notation of Lemma 3.2.2. But, since d(x0 ) < δ, we get that yα ∈ Ωδ using the inequality (3.48). Then, we may apply the assumptions (3.35) or (3.36) to the equality (3.49) and thus we get a contradiction since, by Lemma 3.2.2, we have ¯ + |pα |). λα ≥ C(1

68

Probl`emes de temps de sortie

It remains the case when the minimum point yα is on the boundary ∂Ω. We set ε , χε (y) := u∗ (y) − ψ(y) + |y − yα |2 + d(y) for y ∈ Ω and ε > 0. We add the term |. − yα |2 to be sure that the point yα is a strict local minimum point of u∗ (.) − ψ(.) + |. − yα |2 . Then, there exists a sequence (yε )ε of local minimum points of χε such that lim yε = yα

ε→0

lim χε (yε ) = u∗ (yα ) − ψ(yα ).

and

ε→0

Since yε ∈ Ω, we may use the equation on u 

H yε , Dψ(yα ) − 2(yε − yα ) −

 ε n(y ) + u∗ (yε ) = 0. ε d(yε )2

(3.50)

But, since d(yε ) → 0 as ε → 0, we have d(yε ) < d(x0 ) for ε small enough. Then, using the same arguments as in preceding case, we get a contradiction with the equation (3.50) since the term |yε − yα | tends to 0 and since the additional term with the normal is negative. Hence, we have d(x0 ) ≤ d(yα ). t u Proof of Lemma 3.2.4. Let φ be a C 1 function on Ω × [0, T ] and let (x0 , t0 ) ∈ ∂Ω2 ×]0, T ] be a maximum point of uα − φ. We may assume (x0 , t0 ) to be a strict local maximum point of uα − φ by changing, if it is needed, φ in φ + |. − x0 |2 + |. − t0 |2 . Following an idea of [6], for ε > 0 and for every (x, t) ∈ Ω×]0, T ], we set ε . χε (x, t) = uα (x, t) − φ(x, t) − d(x) As in Lemma 3.2.3, since the point (x0 , t0 ) is a strict local maximum point of uα − φ, there exists a sequence (xε , tε )ε of local maximum points of χε such that (xε , tε ) → (x0 , t0 )

and

uα (xε , tε ) → uα (x0 , t0 )

as ε → 0.

Since xε ∈ Ω, the equation (3.40) holds for uα and we get   √ ε ∂φ (xε , tε ) + H xε , Dφ(xε , tε ) + n(x ) + uα (xε , tε ) − B 4 α ≤ 0. ε 2 ∂t d(xε )

(3.51)

Using the assumption (3.37), we get 

H(xε , Dφ(xε , tε )) ≤ H xε , Dφ(xε , tε ) +

 ε n(x ) . ε d(xε )2

Combining this with (3.51), we obtain √ ∂φ (xε , tε ) + H(xε , Dφ(xε , tε )) + uα (xε , tε ) − B 4 α ≤ 0. ∂t And we conclude letting ε go to zero.

t u

3.3 Applications

69

3.3 Applications In this section, we apply the uniqueness results of Theorem 3.2.1 to exit time control problems. We use the notations of the first section; in particular, the hamiltonian is given by (3.1). Theorem 3.3.1 We assume that ϕ is a bounded function defined pointwise, that the constant λ is positive and that the assumptions (3.6), (3.7) and (3.22)-(3.24) hold. Then 1. If the property (3.4) holds for ϕ on ∂Ω1 , the l.s.c. envelope of the solutions of (3.2) is equal to the value function u− . 2. All the Barron-Jensen solutions of the equation (3.33) in Ω which are supersolutions of (3.2) on the boundary and which satisfy the condition (3.5) on ∂Ω1 have the same l.s.c. envelope : u− . t u In the second point, the only value functions satisfying the condition (3.5) are, in general, u− and u[ϕ∗ ]. It is clear enough in the example of the introduction where ϕ := 11∂Ω\{x0 } . Besides we can directly show that (u[ϕ∗ ])∗ = u− by Lemma 3.1.5. P. Soravia [34] showed that all the value functions are Barron-Jensen solutions of (3.33) in Ω, using their explicit representation formulas. We have Theorem 3.3.2 Assume (3.6) and (3.7). The value functions u[ϕ∗ ], u[ϕ], u[ϕ∗ ], u− and u+ are Barron-Jensen solutions of (3.33) in Ω. t u It is an open problem to know with which assumptions the other solutions of (3.2) are Barron-Jensen solutions. Now we turn to the proofs. Proof of Theorem 3.3.1. This result is an easy adaptation of Theorem 3.2.1. We just have to verify the assumptions of this theorem. The hamiltonian satisfies (3.32) because of the conditions (3.6) and (3.7) on b and f . Also, the assumptions (3.23) and (3.24) lead (3.35), (3.36) and (3.37). We only prove the claim for (3.35) since the other cases use the same kind of arguments. For any x ∈ ∂Ω1 , using (3.23), we choose a control α ¯ such that b(x, α ¯) · n(x) ≥ β. Then, by definition of the hamiltonian, for R > 0, λ ∈ IR+ and p ∈ IRN , and for y ∈ V, we have H(y, p − λn(y)) − R = sup{−b(y, α) · (p − λn(y)) − f (y, α)} − R α∈A

≥ −b(y, α ¯ ) · (p − λn(y)) − f (y, α ¯) − R

70

Probl`emes de temps de sortie 



H(y, p − λn(y)) − R ≥ λβ − λ b(x, α ¯ ) · n(x) − b(y, α ¯ ) · n(y) −kbk∞ |p| − kf k∞ − R. Using the Lipschitz continuity of the functions b and n, we get

H(y, p − λn(y)) − R ≥ λ(β − K|x − y|) − max{kbk∞ , kf k∞ + R}(1 + |p|). Then, if we take 2 max{K, kbk∞ , kf k∞ + R}, β the right side of the preceding inequality is negative and therefore the assumption (3.35) is satisfied. Now, by Theorem 3.3.2, we already know that u− satisfies (3.34). And Lemma 3.1.6 implies that u− satisfies (3.5). Then the proof is complete. t u C R :=

Proof of Theorem 3.3.2. We already prove that all these value functions are supersolutions of (3.2) in the first part. In order to prove the opposite viscosity inequality, we need the following lemma introduced by P. Soravia [34] as the backwards dynamic programming principle. Lemma 3.3.1 Let x be a point of Ω and α(.) be a control i.e. α ∈ L∞ (IR+ , A). We consider the backward trajectory yˇx solution of the dynamical system  yx (t) = −b(ˇ yx (t), α(t))dt,  dˇ 

yˇx (0) = x ∈ Ω,

We denote by τx the first exit time of the backward trajectory yˇx from Ω. Then, for all T such that 0 < T < τx , we have u(x) ≥ −

Z

T

0

f (ˇ yx (t), α(t))eλt dt + u(ˇ yx (T ))eλT . t u

To prove Theorem 3.3.2, we just have to use the classical arguments as in Theorem 3.1.4 but using the inequality of Lemma 3.3.1 instead of the usual dynamic programming principle. t u Now we explain how use the uniqueness result to pass to the limit. We consider an exit time problem satisfying the assumption (3.23) on ∂Ω with a exit cost satisfying (3.5). Let (ϕn )n be a nondecreasing sequence of continuous functions such that sup ϕn = ϕ∗ = (ϕ∗ )∗ . n

71

We note un a viscosity solution of problem (3.2) where ϕ is replaced by ϕn . By the second part, ((un )∗ )∗ satisfies the property (3.34). And using (3.23) and since ϕn is continuous, we get ((un )∗ )∗ ≤ ϕn

( ≤ ϕn+1 )

on ∂Ω.

Therefore ((un )∗ )∗ satisfies also the property (3.5). Then we apply Theorem 3.3.1 to get (un+1 )∗ ≥ ((un )∗ )∗ ≥ (un )∗

in Ω.

Thus, for x ∈ Ω, we have lim inf (un )∗ (y) = sup (un )∗ (x).

n→∞,y→x

n

Finally, by classical stability results, the function u(x) := n→∞,y→x lim inf (un )∗ (y) = sup (un )∗ (x) n

is a supersolution of (3.2). Moreover it easy to check that u satisfies the properties (3.5) and (3.34). Hence, by using again the preceding uniqueness result, we conclude lim (un )∗ = sup (un )∗ = u = u− n n

in Ω.

72

Partie II PROBLEMES DE GRANDES DEVIATIONS

73

4. Large deviations estimates for the exit probabilities of a diffusion process through some vanishing parts of the boundary1

Introduction In the study of random perturbations of dynamical systems, the aim of the theory of Large Deviations is to give estimates on the events of “small probabilities”; we refer the reader to the works of Wentzell-Freidlin[36], Donsker-Varadhan[17] and Azencott[1] for an introduction and for the basic results of this theory. In general, these events of “small probabilities” are the ones for which the behavior of the perturbed trajectories does not look like the behavior of the unperturbed trajectories. We are going in this article to investigate another possible source of “small probabilities” by estimating the exit probabilities of perturbed trajectories of a deterministic dynamical systems through “small” parts of the boundary of a domain in IRN which typically collapse to a point when the perturbation goes to zero; this is what we call “vanishing parts” of the boundary. In order to be more specific, we consider Ω a smooth bounded domain of IRN and we denote by (Xtε )t the solution of the stochastic differential equation  ε  dXt 

= b(Xtε )dt + εσ(Xtε )dWt ,

X0ε = x ∈ Ω ,

(4.1)

where ε > 0 and where b and σ are Lipschitz continuous functions with values respectively in IRN and in the space of N × p matrices. Finally, (Wt )t denotes a standard p-dimensional Wiener process. Then we consider (Γε )ε a sequence of open subsets of ∂Ω : the aim is to estimate the probability for a sample path of the solution of (4.1) to exit Ω through Γε . To do so, we introduce the real-valued function uε defined on Ω by uε (x) = IP x (Xτεxε ∈ Γε ) = IE x (11Γε (Xτεxε )) , 1

en collaboration avec G. Barles.

76

Probl`emes de Grandes D´eviations

where we denote respectively by IP x and IE x the conditional probability and the conditional expectation with respect to the event {X0ε = x} and where τxε stands for the first exit time of (Xtε )t from Ω i.e. τxε = inf{t ≥ 0, Xtε 6∈ Ω} , where the infimum of the empty set is +∞. The classical results concerning exit probabilities in the theory of Large Deviations read −ε2 ln(uε (x)) → I(x) as ε → 0, for x ∈ Ω , (4.2) where the function I is the so-called Action Functional; this implies that I(x) + o(1) u (x) = exp − ε2

!

ε

in Ω ,

and, since in general I > 0 in Ω, this shows that the exit probabilities are exponentially small when ε → 0. We are looking here for such an asymptotic behavior of uε where the function I has to be determined. This type of exponential estimates was first obtained in different contexts by a probabilistic approach; we refer again the reader to Wentzell-Freidlin[36], DonskerVaradhan[17] and Azencott[1] for the classical results in this direction. The partial differential equation (PDE) approach of these problems was first introduced by Fleming[19]. The PDE method to obtain such kind of results is based on the fact that uε is a solution of a linear elliptic PDE and it can be described as follows 1. To perform the logarithmic change of variable I ε (x) := −ε2 ln(uε (x))

in Ω .

The function I ε solves a non-linear elliptic PDE. 2. To pass to the limit ε → 0, in this non-linear PDE : the aim is to show that I ε converges (in some sense) to a function I which is a solution a certain firstorder PDE (or in certain cases a second-order PDE). It is worth noticing that this is a passage to the limit in a singular perturbation problem. 3. To interpret this limiting first-order PDE by using control theory in order to obtain the formula of representation for I. The limiting step of the method was for a long time the second step : indeed the passage to the limit in the non-linear singular perturbation problem by using classical PDE theory requires the strong convergence of DI ε , the gradient of I ε and

Introduction

77

therefore, in order to pass to the limit, one needs rather strong estimates on the function I ε and these estimates are not true in general. This problem was solved by the introduction of the notion of viscosity solutions by Crandall-Lions[16] which allows passages to the limit in non-linear singular perturbations problems without the strong convergence of the gradient but with only the local uniform convergence of the solutions. For a presentation of this notion of solution, we also refer the reader to Crandall-Evans-Lions[14], Lions[28] and the “Users’ guide” of Crandall-Ishii-Lions[15]. In all the following, we will use this notion of solutions and, to simplify, we will write solutions for viscosity solutions. Using viscosity solutions’ theory, Evans-Ishii[18] gave a new justification of the PDE method for Large Deviations problems and, in particular, of the steps 2 and 3. In the Evans-Ishii[18] work, the local uniform convergence of the solutions was obtained by proving some gradient estimates : these estimates are tedious to obtain and they can be false in problems where the diffusion is degenerate. Then a new method – the so-called “half-relaxed limits” procedure – introduced by Perthame and the first author[7] allows to avoid these gradient estimates; this new method both simplifies the PDE approach but also permits to treat Large Deviations problems for degenerated diffusions (cf. Barles-Perthame[9]). The key ingredient in this method is a so-called strong comparison result for the limiting PDE – i.e. a Maximum Principle type result for discontinuous (viscosity) solutions – which implies a posteriori the uniform convergence of I ε to I. In this article, we follow the method of Barles-Perthame[9] but it is worth mentioning that no strong comparison result will be available in our context since we meet possibly discontinuous action functional or at least we will have to handle discontinuous Dirichlet boundary data. Instead we will use “weak” comparison results inspired from Blanc[13]. Coming back to the problem of estimating exit probabilities, we consider here two types of behavior for the Γε : the first one is when Γε converges in some sense to a “regular” subset Γ of ∂Ω and the second one is typically the case when Γε collapses to a point of ∂Ω. Precise formulations will be given later. The first case may be simplified by considering only fixed subsets i.e. Γε ≡ Γ ⊂ ∂Ω. In this simpler case, Evans-Ishii[18] obtained the asymptotic behavior of uε for non-degenerated diffusions. Then Barles-Perthame[9] extended this result to the case when the diffusion is degenerated but for a subset which is a connected component of ∂Ω. In this article, we generalize these results by allowing both the diffusion to be degenerated and (Γε )ε to converge to any open subset of ∂Ω. To the best of our knowledge, the second case – when typically the Γε collapse to a point – have not been considered in the literature yet. We assume in this case that the diffusion is not degenerated and more precisely that σ is a N × N invertible matrix. To give a flavour of our results, let us consider the case when Γε := B(x0 , ρε ) ∩ ∂Ω

78

Probl`emes de Grandes D´eviations

where x0 ∈ ∂Ω, ρε > 0, ρε → 0 when ε → 0 and B(x0 , ρε ) is the ball of center x0 and of radius ρε . Then the behavior of uε depends strongly on the connections between ε and ρε . More precisely, if lim ε2 ln ρε = 0 ,

ε→0

the function uε has the expected behavior i.e. the function I is given by I(x) = inf

n1 Z

2

τx

0

1 |σ −T (yx (t))[y˙x (t) − b(yx (t))]|2 dt; yx (.) ∈ Hloc (IR+ ; IRN ) ,

o

yx (0) = x , yx (τx ) = x0 , yx (t) ∈ Ω for t < τx , (4.3) T −T T where σ is the transpose matrix of σ and σ is the inverse of σ . The main new feature of the proof of this result – in addition to the “weak” comparison result we need – is the fact that the Dirichlet boundary condition satisfied by I does not come automatically from the passage to the limit : it requires a specific treatment which is one of the main difficulties in the proof. In the case when lim −ε2 ln ρε = +∞ ,

ε→0

the convergence to zero of uε is faster and we show that lim −ε2 ln uε (x) = +∞

ε→0

in Ω .

This result is obtained by considering the adjoint problem to the Dirichlet problem satisfied by uε . In the limiting case i.e. when lim −ε2 ln ρε = ρ > 0 ,

ε→0

we can only prove that lim inf −ε2 ln uε (x) ≥ sup{(N − 1)ρ, I(x)} ε→0

in Ω .

The problem to know if this inequality is optimal is open. This paper is organized as follows. The first section is devoted to the case when the limiting subset is “regular”. In the second, we consider the case of a sequence of “vanishing parts” of the boundary. The third section contains the proof of the “weak” comparison result. Finally, in the appendix, we give a result on a gradient estimate used in the second part.

4.1 The probability estimates in the case of a “regular” limiting subset

79

4.1 The probability estimates in the case of a “regular” limiting subset In this section, we consider the case when the sequence (Γε )ε converges to a “regular” subset of ∂Ω. We first explain what we mean by a “regular” subset of ∂Ω and what kind of convergence we use for the sequence (Γε )ε . Definition 4.1.1 : Γ ⊂ ∂Ω is said to be a “regular” subset of the boundary iff Γ = int(Γ) ,

(4.4) 2

where we denote by int(Γ) the interior of Γ.

Notice that the property (4.4) is always satisfies if Γ is an open subset of ∂Ω. Another way to express (4.4) is in terms of the characteristic function of the set Γ that we denote by 11Γ . We recall that 11Γ (x) = 1 if x ∈ Γ and 0 otherwise. The property (4.4) is equivalent to the fact that the function z := 11Γ satisfies z ∗ = (z∗ )∗

on ∂Ω ,

(4.5)

where z ∗ and z∗ stand respectively for the upper and lower semi-continuous envelope of the function z. These notations will be used throughout the paper for different functions. In order to define the convergence of the sequence (Γε )ε , we first recall the definition of the so-called “half-relaxed limits” in the theory of viscosity solutions. If (zε )ε is a sequence of uniformly bounded functions, we set lim sup∗ zε (x) = lim sup zε (y) ε

and

y→x ε→0

lim inf∗ zε (x) = lim inf zε (y) . y→x ε→0

ε

Definition 4.1.2 : The sequence (Γε )ε is said to converge to Γ ⊂ ∂Ω iff (11Γ )∗ = lim sup∗ 11Γε ε

and

(11Γ )∗ = lim inf∗ 11Γε . ε

(4.6) t u

The convergence property can also be expressed respectively by the following two equalities i \ [ [h \ Γ = ( Γε ) and int(Γ) = int( Γε ) . ε0 ε < ε0

Our main result is the following.

ε0

ε < ε0

80

Probl`emes de Grandes D´eviations

Theorem 4.1.1 : We assume that the boundary ∂Ω is of class C 1,1 , that b and σ are Lipschitz continuous functions on Ω and that the sequence (Γε )ε converges to a regular subset of ∂Ω. If, in addition, the two following properties hold ∃ζ > 0, ∀x ∈ ∂Ω, either |σ T (x)n(x)| ≥ ζ > 0 or b(x) · n(x) ≥ ζ > 0, (4.7) where n(x) denotes the outward unit normal to ∂Ω at x ∈ ∂Ω and where σ T denotes the transpose of the matrix σ, and   ∃T < ∞, ∀ y(.) solution of y(t) ˙ = b(y(t)) with y(0) ∈ Ω,

∃s ≤ T such that y(s) 6∈ Ω,



(4.8)

then we have lim inf∗ (−ε2 ln uε ) ∧ A = (lim sup∗ (−ε2 ln uε ))∗ ∧ A

∀A > 0 ,

ε

ε

= I ∧A

in Ω ,

(4.9)

where α ∧ β = min{α, β} for α, β ∈ IR and where I is the value function of the control problem dyx (t) = 2a(yx (t))α(t) + b(yx (t)), dt I(x) = inf

n1 Z

2

0

θ

yx (0) = x ∈ Ω ,

a(yx (t))α(t) · α(t)dt ; α(.) ∈ L∞ (IR+ , IRN ) , o

(4.10)

yx (θ) ∈ Γ , yx (t) ∈ Ω for t < θ , where a = σσ T .

t u

We first remark that, if the action functional I given by (4.10) is bounded, the property (4.9) reduces to lim inf∗ (−ε2 ln uε ) = (lim sup∗ (−ε2 ln uε ))∗ = I ε

in Ω .

ε

Indeed this equality is obtained by taking A large enough in (4.9). Under the assumptions of Theorem 4.1.1, in particular because of the possible degeneracy of σ, the function I can be discontinuous and therefore one cannot expect the local uniform convergence of −ε2 ln uε to I in Ω. This equality is a way to express, in some weak sense, that this convergence holds. But, in general, the function I can be equal to +∞ at certain points of Ω since nothing ensures that, from any point x ∈ Ω, there is a trajectory yx which exits Ω through Γ. In the same way, the function I can be equal to 0 at certain point of

4.1 The probability estimates in the case of a “regular” limiting subset

81

Ω : so Theorem 4.1.1 is not really a Large Deviation type result since a priori at the points x where I(x) = 0, the probability given by uε does not tend to zero. Of course, if σ is a N × N positive definite matrix and if b(x).n(x) < 0 on Γ, then 0 < I(x) < +∞ in Ω , and so we have a classical Large Deviations type result. The assumptions (4.7) and (4.8) are classical in this context : (4.7) is some “nondegeneracy” type condition like the ones introduced in [9] to prove strong comparison results for the generalized Dirichlet problem (in the viscosity sense); such type of assumption is necessary to get such strong comparison results. We remark that (4.7) is less restrictive than the corresponding assumption in [9] because of the alternative case with b. The condition (4.8) is a classical assumption to obtain the strict subsolution one needs in order to have a comparison result for the limiting PDE (cf. [9], [18], [26]). The importance of such assumptions to get a comparison result will be explained in the section 3 (see also Lemma 4.2.5 in section 2). Proof of Theorem 4.1.1 : We prove the theorem in several steps which follow the basic plan explained in the introduction. The first one is devoted to show that uε is a viscosity solution of a second-order linear equation with a “generalized” Dirichlet boundary condition in the viscosity sense : indeed, since we consider degenerated diffusion, there is no reasons for uε to be a classical solution of this Dirichlet problem and, moreover, losses of boundary data may occur on the boundary because again of this possible degeneracy. In the second step, we perform a logarithmic type change of variable and we let ε go to zero by using the “half-relaxed limits” method. Finally, we conclude by proving a weak comparison result which allows us to identify the action functional. Step 1. It consists in proving the Lemma 4.1.1 : The function uε defined on Ω by uε (x) = IE x (ϕε (Xτεxε )) , where ϕε := 11Γε , is a viscosity solution of the “generalized” Dirichlet problem  1   Luε := − ε2 Tr(a(x)D 2 uε ) − b(x) · Duε

2

 

where Tr denotes the trace operator.

u

ε

= 0 = ϕ

in Ω, ε

(4.11)

on ∂Ω . t u

82

Probl`emes de Grandes D´eviations

We recall that, in (4.11), the Dirichlet boundary condition has to be understood in the viscosity sense i.e. min{Luε , uε − ϕ∗ } ≤ 0

on ∂Ω ,

max{Luε , uε − ϕ∗ } ≥ 0

on ∂Ω .

and We refer to Crandall-Ishii-Lions[15] and Barles-Burdeau[6] for details on these types of “generalized” boundary conditions. Proof of Lemma 4.1.1 : We extend a result of Barles-Burdeau[6] where a similar result was proved in the case when the function on the boundary ϕ is continuous. The ideas are essentially the same as in [6] but we give the proof for the convenience of the reader. We just show how to obtain the supersolution property on the boundary since the other properties are easier to derive. Let φ be a C 2 -function on Ω and let x0 ∈ ∂Ω be a global minimum point of ε u∗ − φ on Ω. We may assume that uε∗ (x0 ) = φ(x0 ) by changing if necessary φ. Then uε∗ (x) ≥ φ(x) on Ω. If uε∗ (x0 ) ≥ ϕε∗ (x0 ), we are done. Otherwise uε∗ (x0 ) < ϕε∗ (x0 ). We may change φ outside a neighborhood of x0 to have φ ≤ ϕε∗ on ∂Ω if this is not the case. We consider a sequence (xp )p of points of Ω such that xp → x0 and lim uε (xp ) = uε∗ (x0 ) . p If there exists a subsequence of (xp )p denoted by (xp0 )p0 such that xp0 ∈ ∂Ω, then, by definition of uε , we get uε (xp0 ) = ϕε (xp0 ). And, since lim 0inf ϕε (xp0 ) ≥ ϕε∗ (x0 ) , p

letting p0 tend to +∞, we have uε∗ (x0 ) ≥ ϕε∗ (x0 ). But since uε∗ (x0 ) < ϕε∗ (x0 ), this cannot happen and therefore xp ∈ Ω for p large enough. We define h2p := |uε (xp ) − φ(xp )| then hp → 0 as p → ∞ and φ(xp ) = uε (xp )+o(hp ). Using the Dynamic Programming Principle, we get h

i

uε (xp ) = IE xp 11{hp < τp } uε (Xhεp ) + 11{hp ≥τp } ϕε (Xτεp ) , where (Xtε )t is a solution of (4.1) with X0ε = xp and τp is the first exit time of Ω for Xtε .

4.1 The probability estimates in the case of a “regular” limiting subset

83

By the definition of hp and since ϕε ≥ ϕε∗ on ∂Ω, uε ≥ uε∗ and uε∗ ≥ φ on Ω, we have i

h

φ(xp ) − o(hp ) = uε (xp ) ≥ IE xp 11{hp < τp } φ(Xhεp ) + 11{hp ≥τp } ϕε∗ (Xτεp ) .

(4.12)

Moreover, since ϕε∗ ≥ φ on ∂Ω, we get φ(xp ) ≥ IE xp [φ(Xhεp ∧τp )] + o(hp ) .

(4.13)

Using the Itˆo’s Lemma, we obtain φ(Xhεp ∧τp ) = φ(xp ) +

hp ∧τp

0

hp ∧τp  ε2

Z

+

Z

0

2

εσ(Xtε )Dφ(Xtε )dWt 

Tr(a(Xtε )D2 φ(Xtε )) + b(Xtε ) · Dφ(Xtε ) ds .

Combining this with (4.13), we have IE xp

hp ∧τp 

hZ 0

 i ε2 − Tr(aD2 φ) − b · Dφ (Xsε )ds ≥ o(hp ) . 2

Because of the regularity of the functions φ, a and b, we get 

−

 ε2 Tr(aD2 φ) − b · Dφ (xp )IE xp [hp ∧ τp ] ≥ o(hp ) . 2

It remains to show that IE xp

hh ∧ τ i p p

hp

→ 1.

To do so, we prove that IP [τp ≤ hp ] → 0. We subtract φ(xp ) in (4.12) and we compute h

i

uε (xp ) − φ(xp ) ≥ IE xp 11{hp