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INSTITUT NATIONAL DE RECHERCHE EN INFORMATIQUE ET EN AUTOMATIQUE

Lyapunov exponents of controlled SDE’s and stabilizability property : Some examples Fabien Campillo and Abdoulaye Traore´

N˚ 2397 Novembre 1994

PROGRAMME 5 Traitement du signal, automatique et productique

ISSN 0249-6399

apport de recherche

1994

Lyapunov exponents of controlled SDE's and stabilizability property : Some examples Fabien Campillo and Abdoulaye Traore Programme 5 | Traitement du signal, automatique et productique Projet Me sto Rapport de recherche n2397 | Novembre 1994 | 15 pages

Abstract: We consider a stochastic dierential equation with linear feedback control :

dXt = (A + B K ) Xt dt +

r

X

k=1

(Ak + Bk K ) Xt dWk (t)

where K is the feedback gain matrix. For each value of K , let K be the Lyapunov exponent associated with the solution of the SDE. The set of K , as K describe the set of matrices, is a connected interval of IR. We present some examples where ?1 is the lower bound of this set. For these cases, we say that the corresponding EDS is stabilizable. Key-words: Stochastic dierential equation, stabilizability, Lyapunov exponent. (Resume : tsvp) [email protected] [email protected]

Unite´ de recherche INRIA Sophia-Antipolis 2004 route des Lucioles, BP 93, 06902 SOPHIA-ANTIPOLIS Cedex (France) Te´le´phone : (33) 93 65 77 77 – Te´lećopie : (33) 93 65 77 65

Exposants de Lyapunov d'EDS contrôlees et propriete de stabilisabilite : Quelques exemples Resume : On considere une equation dierentielle stochastique avec contrôle lineaire en boucle fermee :

dXt = (A + B K ) Xt dt +

r

X

k=1

(Ak + Bk K ) Xt dWk (t)

ou K est une matrice de gain. Pour chaque valeur de K , soit K l'exposant de Lyapunov associe a la solution de cette EDS. L'ensemble de valeurs de K lorsque K parcours l'espace des matrices est un intervalle connexe de IR. On presente quelques exemples ou ?1 est la borne inferieure de cet ensemble. Pour ces exemples, nous dirons que l'EDS correspondante est stabilizable. Mots-cle : E quation dierentielle stochastique, stabilisabilite, exposant de Lyapunov.

3

Lyapunov exponents of controlled SDE's and stabilizability property

1 Preliminaries We consider the following linear stochastic dierential equation in IRd

dXt = A Xt dt +

r

X

k=1

Ak Xt dWk (t) ; X = x 2 IRd ; x 6= 0 ; 0

0

(1)

0

where A; A ; : : : ; Ak are d d matrices, W ; : : : ; Wr are independent standard Wiener processes. Here \dW " (resp. \dW ") refer to the Stratonovich (resp. Itô) stochastic integral. We de ne the Lyapunov exponent of the solution of (1) starting at x 1 log kX k : (2) (x ) =4 tlim t !1 t 1

1

0

0

Oseledec's multiplicative ergodic theorem states that the limit (2) exists with probability one and that there are d xed numbers : : : d { called the Lyapunov exponents of (1) { such that the random variable (x ) takes on only these values (see [1] for a review). Moreover, (1) is exponentially stable with probability one if and only if < 0. Let S d? = fx 2 IRd ; kxk = 1g denote the unit sphere of IRd . We can de ne the projection of Xt onto the sphere by 1

2

0

1

1

Ut =4 kXt k? Xt : Ut is the solution of the following SDE on S d? 1

1

dUt = h(A; Ut ) dt +

r

X

k=1

h(Ak ; Ut ) dWk (t) ; U = u =4 kx k? x 0

0

1

0

where

h(C; u) =4 C u ? (C u; u) u : Here (x; y) is the scalar product on IRd and kxk = (x; x).

0

(3) (4)

2

Moreover,

kXtk = kx k exp 0

RR n2397

( Z

t 0

q(Us ) ds +

r

X

k=1

t

Z 0

pk (Us ) dWk (s)

)

(5)

4

Fabien Campillo and Abdoulaye Traore

with

q (u) =4 (A u; u) ; pk (u) =4 (Ak u; u) ; k = 1; : : : ; r ; 4 1 r q (u) = 2 [(Ak u; u) + kAk uk ? 2(Ak u; u) ] ; 0

X

1

4

2

2

2

k=1

q(u) = q (u) + q (u) : For each matrix M , h(M; ?u) = ?h(M; u), so that h(M; ) can be viewed as a vector eld on the projective space P d? (obtained from S d? by identifying u and ?u). Therefore (3) can be considered as a stochastic dierential equation on P d? and (5) is still valid with this de nition. We make the following Hypothesis 1.1 For all u 2 P d? dim Lie Algebrafh(A; :); h(Ak ; :); k = 1; : : : ; rg(u) = d ? 1 : 0

1

1

1

1

1

Theorem 1.2 Under Hypothesis 1.1 (i) The diusion process Ut admits a unique invariant probability measure . Moreover has a C 1 density p with respect to the Lebesgue measure on P d?1 which solves the Fokker{Planck equation Lp = 0, where L is the in nitesimal generator associated with equation (3). (ii) The number

4

=

Z

q(u) (du) is equal to the top Lyapunov exponent . (iii) For all x 2 IRd , x 6= 0, (x ) = with probability one. When < 0, P d?1

1

0

0

0

the system (1) is exponentially stable with probability one.

2 Main result We consider the controlled SDE

dXt = (A Xt + B ut ) dt +

r

X

k=1

(Ak Xt + Bk ut ) dWk (t) INRIA

5


where A, Ak , k = 1; : : : ; r are d d matrices, B , Bk , k = 1; : : : ; r are d p matrices. We suppose that these matrices are given. We restrict ourselves to feedback controls, i.e. ut = K Xt, where K is a p d matrix. The resulting SDE is r

dXt = (A + B K ) Xt dt + (Ak + Bk K ) Xt dWk (t) ; X = x 2 IRd ; x 6= 0 : X

0

k=1

0

0

(6) The problem is to choose the feedback gain matrix K so as to stabilize the system (6). The projection of Xt onto P d? , which now depends on K , satis es 1

dUt = h(A + B K; Ut ) dt + 4 with U = u = kx k? x , and 0

0

0

1

k=1

h(Ak + Bk K; Ut ) dWk (t) ;

(7)

0

kXtk = kx k exp

(

Z

0

with

r

X

t

0

r

q(K; Us ) ds +

X

t

Z

k=1

0

pk (K; Us ) dWk (s)

q (K; u) =4 ((A + B K ) u; u) ; 4 1 r q (K; u) = 2 ((Ak + Bk K ) u; u) + j(Ak + Bk K ) uj

)

(8)

0

Xn

1

k=1

2

2

?2((Ak + Bk K ) u; u) ; 2

o

q(K; u) =4 q (K; u) + q (K; u) ; pk (K; u) =4 ((Ak + Bk K ) u; u) ; k = 1; : : : ; r : 0

2.1 First case :

Bk

1

=0

; k

=1

;:::;r

Here we suppose that only the drift coecient is controlled, i.e. Bk = 0; k = 1; : : : ; r. The equation for Ut reduce to Ut is solution of the following equation

dUt = h(A + B K; Ut ) dt + We make the following RR n2397

r

X

k=1

h(Ak ; Ut ) dWk (t) :

(9)

6


Hypothesis 2.1 For all u 2 P d? and K 2 M (p d) dim Lie Algebrafh(A + BK; :); h(Ak ; :); k = 1; : : : ; rg(u) = d ? 1 ; where M (p d) is the set of p d matrices. Under Hypothesis 2.1, the Theorem 1.2 states that, for all K 2 M (p d) the 1

diusion process Ut admits a unique invariant probability measure K . Let K denote the Lyapunov exponent associated with Equation (6) 4

K =

Z

P d?1

q(K; u) K (du) :

Proposition 2.2 Under Hypothesis 2.1, K 7! K de ne a continuous function de ned on M (p d). Since M (p d) is a connected set, we have the following Corollary 2.3 Under Hypothesis 2.1,

D =4 fK ; K 2 M (p d)g is a connected interval of IR.

In order to prove Proposition 2.2 we need the following

Lemma 2.4 For all t 0, there exist Ct < 1, such that for all K ; K 2 M (p d) supd? E Ut ;u ? Ut ;u Ct kK ? K k ; 1

u2S

1

1

2

2

1

2

2

where Uti;u denote the solution of (9) with control matrix Ki and starting at point u.

INRIA


7

Proof Let Utu denote the solution of Equation (9) with control matrix K and

starting at point u. Utu is solution of the following Itô equation r r dUtu = h(A+B K; Utu ) dt+ 12 h0 (Ak ; Utu ) h(Ak ; Utu ) dt+ h(Ak ; Utu ) dWk (t) ; k k X

X

=1

=1

with h0 (C; u)v = C v ? (C u; u) v ? (C u; v) u ? (C v; u) u. We get

dU = b(K; U dt + u t)

u t

where

b(M; u) =4 h(A + B M; u) + 21

r

X

k=1

r

X

k=1

k (Utu ) dWk (t) ; U u = u ; 0

h0 (Ak ; u) h(Ak ; u) ; k (u) =4 h(Ak ; u) :

The drift coecient b(M; ) and diusion coecients k () are polynomial functions of u, so they are locally Lipschitz. But they are also globally Lipschitz because S d? is a compact set. Hence, there exist L > 0 such that for all u; v 2 S d? and k = 1; : : : ; r 1

1

kb(M; u) ? b(M; v)k + kk (u) ? k (v)k L ku ? vk : Also, for all K; K 0 2 M (p d) and u 2 S d? kb(K; u) ? b(K 0; u)k kBk kK ? K 0k : 1

Now we go back to the proof of the lemma :

Ut ;u ? Ut ;u 1

2

2

Z

0

t

[b(K ; Us ;u ) ? b(K ; Us ;u )] ds

2t

t

t

k

[k (Us

;u )

1

1

1

+2 r RR n2397

2

X Z

0 =1

r

2

2

? k (Us

;u )]

2

dWk (s)

2

2

b(K ; Us ;u ) ? b(K ; Us ;u) ds

0

1

1

+2 Z

2

r

Z X

0 =1

k

2

t

2

2

[k (Us ;u ) ? k (Us ;u )] dWk (s) : 1

2

8


So

E Ut ;u ? Ut ;u 1

2

2

4t

t

Z

+4 t

0 Z

+2 r

2

E b(K ; Us ;u ) ? b(K ; Us ;u ) ds t

1

0

r X k=1

1

1

2

2

E b(K ; Us ;u ) ? b(K ; Us ;u ) ds t

Z 0

1

2

2

2

2

E k (Us ;u ) ? k (Us ;u ) ds : 1

2

So we get

E Ut ;u ? Ut ;u 1

2

2

t

Z

2

(4 t + 2 r) L E Us ;u ? Us ;u ds +4 t kB k kK ? K k : 2

0 2

2

1

2

1

2

2

From this last inequality and Gronwall's inequality we prove the lemma.

2

Proof of Proposition 2.2 Let Kn !n;uK as n ! 1. We want to prove that Kn ! K as n ! 1. Let Utu (resp. Ut ) denote the solution of Equation (9) with control matrix K (resp. Kn ) and starting at point u. From Lemma 2.4, lim sup E kUtn;u ? Utu k = 0 :

(10)

2

n!1 u2S d?1

4 Now we show that the sequence n = Kn admits a weak limit and that = K . First, it is clear that the sequence fn g is tight because S d? is a compact set. So there exist a sub{sequence, denoted fn0 g, and a probability measure de ned on S d? such that n0 ) . Now we prove that = K . Let f be Lipschitz on S d? , then there exists > 0 such that jf (u) ? f (v)j ku ? vk for all u; v 2 S d? , and from (10), we have 1

1

1

1

0

0

supd?1 jEf (Utn ;u ) ? Ef (Utu )j supd?1 E jf (Utn ;u ) ? f (Utu )j

u2S

2

u2S

2

supd? E Utn0;u ? Utu ! 0 : (11) 2

u2S

1

2

INRIA

9


From the fact that n0 ) and that n0 is an invariant probability measure n0 ;u for Ut , we have Z

n0 ;u Ef ( U ) n0 (du) = t P d?1

Z

f (u)n0 (du) ! P d?1

Furthermore Z

P d?Z1

Ef (U

n0 ;u t ) n0 (

du) ?

Z

P d?1 ( tu )

Ef (U (du) u t)

Z

P d?1

f (u) (du) :

jEf (Utn0 ;u) ? Ef U j n0 (du) + d? Ef (Utu ) [n0 (du) ? (du)] P supd? jEf (Utn0;u) ? Ef (Utu )j + P d? Ef (Utu ) [n0 (du) ? (du)] u2S !0:

P d?1

Z

1

Z

1

1

Indeed, the rst term tends to 0 because of (11) and the second term tend to 0 because the function u 7! Ef (Utu ) is continuous and n0 ) . At last, we get Z

P d?1

Ef (U (du) = u t)

Z

P d?1

f (u) (du) ; 8t 0

that is = K . The invariant measure K is unique, so the whole sequence fng converge to K . Finally Z

Z

jKn ? K j P d? q(Kn; u) n(du) ? P d? q(K; u) (du) supd? jq(Kn ; u) ? q(K; u)j + P d? q(K; u) [n (du) ? (du)] u2S

1

1

1

Z

1

which tends to 0.

2.2 General case

We go back to the general set up (6){(8). We make the following RR n2397

2

10


Hypothesis 2.5 For all u 2 P d? and K 2 M (p d) dim Lie Algebra fh(A + B K; :); h(Ak + Bk K; :) ; k = 1; : : : ; rg (u) = d ? 1 : 1

Under this hypothesis, Ut admit a unique invariant measure K and the Lyapunov exponent is given by

K =

Z

P d?1

q(K; u) K (du) :

Proposition 2.6 Under Hypothesis 2.5, K 7! K de nes a continuous function de ned on M (p d). Since M (p d) is a connected set, we have the following Corollary 2.7 Under Hypothesis 2.5, D =4 fK ; K 2 M (p d)g is a connected interval of IR.

Proof of Proposition 2.6 The proof is equivalent to Proposition 2.2. 2

3 Examples

In these examples, we consider feedback gain matrices K () parametrized by a one dimensional parameter 2 IR. We suppose that fK (); 2 IRg is a connected subset of M (p d). Let (resp. ) denote the Lyapunov exponent (resp. the invariant measure) associated with K () and ? = inf 2 IR

We provide in this section two examples where ? = ?1 and one where ? = 1. In all the examples, d = 2 and r = 1. For any 2 2 matrix C , h(C; ) de ned in (4) is a vector eld on P , and h(C; ) = (?c + c ) cos sin ? c sin + c cos for all 2 P . 1

11

22

12

2

21

2

1

INRIA

11


3.1 Example 1

We identify P to [?=2; =2]. Let 1

0 1 1 0

A=

!

; B = 10 01

!

; A = 10 ?01

!

1

; B =0: 1

We consider feedback gain matrices K of the form

K () = ?1 ?21

!

parametrized by 2 IR. The Hypothesis 2.1 is satis ed, since

h(A + BK (); ) = sin cos ; h(A ; ) = 1 ; 8 2 [? 2 ; 2 ] : 1

We have

q (K (); ) = + sin ; q () = 0 ; 8 2 [? 2 ; 2 ] : 2

0

Finally we get

= +

Z

1

sin () (d) :

2

? 2

2

So the system is stabilizable because

= ?1 :

lim !?1

3.2 Example 2

We identify P and [0; ], and we take 1

A = ?01 ?01 ; B = 00 ?01 ; A = 0 ; B = 01 10 !

!

1

and

RR n2397

K () = 0 ?0

!

1

:

!

;

12


So

h(A + BK (); ) = sin cos ; h(A + B K (); ) = ; 8 2 [0; ] ; 1

1

and

q (K (); ) = ?1 + sin ; q (K (); ) = 0 ; 8 2 [0; ] : 2

0

1

The Lyapunov exponent is

Z

= ?1 +

0

sin () (d) : 2

(12)

We want to compute the limit of as ! ?1. We can check that the projected process Ut is solution of dUt = h(A; Ut ) dt + h(A0 ; Ut )dt + h(A00 ; Ut ) dW (t) with A0 = 00 01 ; A00 = 10 ?01 : !

!

4 Ut=2 . U~t is solution of We make the following time scale transformation U~t = t t t U~t = U + 1 h(A; U~s ) ds + 1 h(A0 ; U~s ) ds + h(A00 ; U~s ) dW~ s (13) Z

Z

Z

4 where W~ t = Wt=2 is a standard Wiener process. When ! ?1, we get the following limit equation 0

2

0

0

0

U~t = U + 0

t

Z 0

h(A00 ; U~s ) dW~ s

where W~ t is a standard Wiener process. Let U~t = (cos ~t ; sin ~t ). Because h(A00 ; ) = 1, we get ~t = + W~ t : (14)

Proposition 3.1

0

) U [0; ] as ! ?1 where U [0; ] is the uniform law on [0; ]. INRIA


13

Proof It is clear that ) ?1, where ?1 satis es the following Fokker{

Planck equation L ?1 = 0 and L is the in nitesimal generator associated with equation (14). Moreover, ?1 is the uniform law on [0; ]. 2 Then using this proposition and (12) we have

Corollary 3.2

! ?1 as ! ?1 :

3.3 Example 3

Now we present an example which is not stabilizable. We consider the same coecients as in the previous section except for matrices A, B : 1

0 1 0 0

A=I ; B = 1

!

:

Let Xt = (Xt ; Xt ), we get the following system 1

2

dXt = Xt dt ? Xt dWt ; X = x ; dXt = (1 + ) Xt dt ; X = x ; x 6= 0 ; whose solution is Xt = e t x , and (1+ )

2

2 0

2 0

2

2

t

Z

1 0

1

0

= et x ? x = et (x ? Yt )

et?s Xs dWs 2

Z

2 0

1 0

t 0

et

s

+

dWs

1 0

4

Yt = x

Z 2 0

If x 6= 0, we deduce from

t 0

es dWs :

2 0

kXtk jXt j = e 2

RR n2397

0

2 0

Xt = et x ?

with

1 0

1 0

2

1

1

t

(1+ )

jx j 2 0

14


that 1 + . If x = 0 then x 6= 0 (because, x 6= 0) and = 1. So, for 0, 1. Let us consider the case < 0. By the theorem of convergence of martin4 gales, Yt ! Y1 = X 1 es dWs as t ! 1 and this convergence holds a.s. and in L . We deduce from jjXt jj = e t e tjx j + jx ? Ytj 2 0

1 0

2 0

2

0

R

0

2

2

n

2

2 2 0

1 0

2

o

and from (2) that

= 1 + t!lim1 21t log e tjx j + jx ? Ytj = 1 a:s: h

+

2

2 2 0

1 0

2

i

(this limit is valid whether x = 0 or not). Which proves that the system is not stabilizable. 2 0

References [1] L. ARNOLD. Stabilization by noise revisited. ZAMM, Z. angew. Math. Mech., 70(7):235{246, 1990. [2] L. ARNOLD, H. CRAUEL, and V. WIHSTUTZ. Stabilization of linear systems by noise. SIAM Journal on Control and Optimization, 21(3):451{ 461, 1983. [3] L. ARNOLD, E. OELJEKLAUS, and E. PARDOUX. Almost sure and moment stability for linear itô equations. In L. Arnold and V. Wihstutz, editors, Lyapunov Exponents, Bremen{1984, volume 1186 of Lecture Notes in Mathematics, pages 129{159, Berlin, 1986. Springer Verlag. [4] P.H. BAXENDALE. Moment stability and large deviations for linear stochastic dierential equations. In N.Ikeda, editor, Probabilistic Methods in Mathematical Physics, Katata and Kyoto 1985, pages 31{54, Tokyo, 1987. Kinokuniya. [5] N. IKEDA and S. WATANABE. Stochastic Dierential Equations and Diusion Processes. North{Holland/Kodansha, Amsterdam, 1981. INRIA


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[6] E. PARDOUX and V. WIHSTUTZ. Lyapunov exponent of linear stochastic systems with large diusion term. Stochastic Processes and their Applications, 40(2):289{308, 1992.

RR n2397

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