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0. (s)ds. ; for a > 0 and t > 0 which are small enough. (iii) If (t) = tln 1 t ln ln 1 t. , for t 2 (0; t0], t0 > 0 is small enough, then from (10), we obtain: u(t) exp ? fln. 1 a ge.
On the existence and uniqueness of solutions to stochastic equations in in nite dimension with integral-Lipschitz coecients Ying Hu, Nicolas Lerner Institut de Recherche Mathematique de Rennes Universite de Rennes 1 Campus de Beaulieu 35042 Rennes Cedex, France E-mail: [email protected], [email protected] ABSTRACT In this paper, we study the existence and uniqueness of solutions to stochastic equations in in nite dimension with an integral-Lipschitz condition for the coecients. Mathematics Subject Classi cation: 60H10

1 Introduction In this paper, we study the solvability of the following stochastic equation on a separable Hilbert space H : 



dX (t) = AX (t) + b(t; X (t)) dt + (t; X (t))dB(t); X (0) = x; or, more precisely,

X (t) = eAtx +

Z

Z t    t A(t?s)  e b s; X (s) ds + eA(t?s)  s; X (s) dB(s); 0

0

(1)

where B () is a cylindrical Wiener process valued in a separable Hilbert space K de ned on a ltered probability space ( ; F ; P ; Ft), A is the in nitesimal generator of a C0-semigroup eAt on H , and x 2 H . 1

It is well known that under a Lipschitz condition on the coecients

b;  (see, e.g., [2] [3]), eq.(1) has a unique mild solution. Furthermore, Da Prato and Zabczyk have studied this equation under the hypothesis that A is almost m-dissipative, and b is continuous and monotone. But they have supposed that  is Lipschitz.

The aim of this paper is to establish the existence and uniqueness of the solution to eq.(1) under the following so-called integral - Lipschitz condition: jb(t; x1) ? b(t; x2)j2 + j(t; x1) ? (t; x2)j2  (jx1 ? x2j2); (2) where  : (0; +1) ! (0; +1) is a continuous, increasing, concave function satisfying Z 1 (0+) = 0; dr (r) = +1: 0

A typical example of (2) is:

jb(t; x ) ? b(t; x )j + j(t; x ) ? (t; x )j  jx ? x j ln jx ?1 x j : In particular, we do not need the Lipschitz condition for  . 1



1

2

1

2

1

2

2

1

2

Furthermore, we prove the existence and uniqueness of the solution to eq.(1) under a \weaker" condition for b, that is: jb(t; x1) ? b(t; x2)j  (jx1 ? x2j): A typical example for this condition is: jb(t; x ) ? b(t; x )j  jx ? x j ln 1 : 1

2

1

2

jx ? x j 1

2

Nevertheless, we suppose that A is m-dissipative; and as a by-product, we obtain a uniqueness result for the one-dimensional stochastic di erential equation which is slightly better than the well-known Yamada-Watanabe one. Finally, we apply our method to the study of backward stochastic di erential equations. The paper is organized as follows: in Sect. 2, we give some preliminaries. In Sect. 3, we prove the existence and uniqueness result to eq. (1) under the integral-Lipschitz condition on the coecients. In Sect. 4, we prove the existence and uniqueness of solution to the dissipative stochastic equations under some \weaker" condition for b, and the corresponding result for the one-dimensional stochastic di erential equation. And in the last section, we turn to the study of backward stochastic di erential equations. 2

2 Preliminaries

2.1 Stochastic equations in in nite dimensions

Let ( ; F ; P ; Ft) be a ltered probability space. Let H; K be two separable Hilbert spaces. We denote their norms by j  j and their scalar products by h; i. A stochastic linear function on K is a linear mapping from K to L0 ( ; F ; P ) (see [2] and [3]).

De nition 2.1 We say that B(t); t 2 R is a cylindrical Wiener process +

in a Hilbert space K , if: (1) 8t 2 R+ , B (t) is a stochastic linear function on K . (2) 8n 2 N ; 8h1; h2;    ; hn 2 K , f(B (t)h1 ; B (t)h2 ;    ; B (t)hn ); t 2 R+ g is a Wiener process (not necessarily standard) with values in Rn . (3) 8h1 ; h2 2 K; 8t 2 R+ ,

E (B(t)h1)(B(t)h2 ) = thh1 ; h2i: We assume that Ft is the natural ltration of B . For any Hilbert space H1 , we denote by L2F (0; T ; H1) the set of all the Ft-progressively measurable processes x(), with values in H1, such that 

jx()j = E [

Z 0

T

jx(t)j dt] 2

=

1 2

< +1:

Obviously L2F (0; T ; H1) is a Hilbert space. We can de ne a stochastic integral for x() 2 L2F (0; T ; H1): t

Z 0

x(s)dB1 (s); 8t 2 [0; T ];

where fB1 (s); s 2 [0; T ]g is a real Wiener process on ( ; F ; P ; Ft). Next, we consider the space L2 (K ; H ) which is the set of Hilbert-Schmidt operators from K into H , i.e.,

L (K ; H ) = f 2 L(K ; H )j 2

1

X

n=1

j enj < +1g; 2

2 where fen g1 n=1 is an orthonormal basis of K . L (K ; H ) is a Hilbert space and its norm is still denoted by j  j.

3

R

T

0

For any () 2 L2F (0; T ; L2(K ; H )) we can de ne the stochastic integral (t)dB (t) : L2F (0; T ; L2(K ; H )) ! L2( ; FT ; P ; H ) as follows: Z

T

0

(t)dB (t) =

1

T

Z

X

n=1

0

(t)end(B (t)en ):

The right hand of the above equality is well de ned since

E [j

1

T

Z

X

n=1

(t)end(B (t)en )j2] = E [

0

= E[

1

X

n=1 Z T 0

Z

T

0

j (t)enj dt] 2

j (t)j dt] < +1: 2

Let b(; x),  (; x) be stochastic processes depending on x 2 H , such that:

8x 2 H; b(; x) 2 LF (0; T ; H ); (; x) 2 LF (0; T ; L (K ; H )); 2

2

2

(3)

8x 2 H; jb(t; x)j + j(t; x)j  (t) + (t)jxj ; (4) where 2 LF (0; T ; R ); 2 L ([0; T ]; R ): For a given C -semigroup eAt with the in nitesimal generator A, we have 2

2

1

+

2

2

2

2 1

2 2

2

+

0

the following well-known result (see, e.g. [2, 3]): Proposition 2.1 We suppose (3), (4), and the following Lipschitz condition: 8x1 ; x2 2 H;

jb(t; x ) ? b(t; x )j + j(t; x ) ? (t; x )j  c jx ? x j ; (5) where c > 0 is a constant; then there exists a unique process X () 2 CF ([0; T ]; L ( ; F ; P ; H )) satisfying the following stochastic equation: for t 2 [0; T ], 1

2

2

1

2

2

2

1

2

2

2

X (t) = eAt x +

Z 0

Z t t A(t?s) e b(s; X (s))ds + eA(t?s) (s; X (s))dB(s): 0

(6)

Such a solution is called a mild solution of (6). The aim of this paper is to study the solvability of eq.(6) under some weaker condition than the Lipschitz condition (5). 4

2.2 Dissipative in nitesimal generator

This subsection is adapted from [3]. The in nitesimal generator A : D(A)  H ! H is said to be dissipative if and only if that for any x 2 D(A),

hAx; xi  0: A is called m-dissipative if the range of I ? A is the whole space H for some  > 0 ( and then for any  > 0 ). Finally A is called almost m-dissipative if A ? I is m-dissipative for some 2 R. The Yosida approximation A ; > 0, of an m-dissipative in nitesimal generator A is de ned by A = AJ = 1 (J ? I ); (7)





where



J = (I ? A)?1 : We list some useful properties of J and A .

(8)

Lemma 2.1 Let A : D(A) ! H be an m-dissipative in nitesimal generator, and let J and A be de ned by (7) and (8) respectively. (i) For any > 0, we have

jJ j  1: (ii) For any > 0, A is dissipative, and bounded: 2 and (iii) We have

jA j  ;

jA xj  jAxj; 8x 2 D(A): lim J x = x; 8x 2 H: !0

5

2.3 Technical lemmas

The following lemma is the starting point of this paper. The rst part of this lemma can be found in [1].

Lemma 2.2 Let  : (0; +1) ! (0; +1) be a continuous, increasing function satisfying

(0+) = 0;

Z 1 0

dr = +1; (r)

and let u be a measurable, non-negative function de ned on (0; +1) satisfying Z t (9) u(t)  a + (s)(u(s))ds; t 2 (0; +1); 0

where a 2 R+ , and 2 L1loc ([0; 1); R+). We have: (1) If a = 0, then u(t) = 0, forR t 2 [0; +1); a:e: (2) If a > 0, we de ne  (t) = tt ds(s) ; t 2 R+ , where t0 2 (0; +1), then, 0

Z t   ? 1 u(t)    (a) + (s)ds : 0

(10)

Proof: We de ne the function on R : +

R(t) = a +

Z

t

0

(s)(u(s))ds;

then as  is non-decreasing, Hence,

dR (t) = (t)(u(t))  (t)(R(t)): dt d  (R(t)) = 1 (t)(u(t))  (t); dt (R(t))

from which we deduce:

 (R(t))   (a) +

Z 0

t

(s)ds:

(11)

On the other hand, from the de nition,  is obviously a strictly increasing function, and  (0+) = ?1. Then  admits a strictly increasing inverse function, and  ?1 (?1) = 0. 6

(1) If a = 0, then, and

 (R(t)) = ?1; R(t) = 0;

from which we deduce the desired result. (2) If a > 0, then from (11), 

R(t)   ?1  (a) +

Z

t

0



(s)ds ;

and we obtain (10).

Examples:

(i) The standard Lipschitz case: (t) = t; t > 0, then from (10), we get:

u(t)  ae

Rt 0

(s)ds:

(ii) The so-called Log-Lipschitz case: (t) = t ln 1t , for t 2 (0; t0], t0 > 0 is small enough, then from (10), we obtain: ? u(t)  ae

Rt 0

(s)ds

;

for a > 0 and t > 0 which are small enough. (iii) If (t) = t ln 1t ln ln 1t , for t 2 (0; t0], t0 > 0 is small enough, then from (10), we obtain: ? u(t)  exp ? fln 1 ge 

Rt

a

0

(s)ds 

;

for a > 0 and t > 0 which are small enough. The second lemma is trivial, we include it for completeness.

Lemma 2.3 We put, for a given  > 0, F (x) = (jxj + ) ; x 2 H; then, F 2 C (H ; R); and 2

1 2

2

F0 (x) =

x

(jxj2 + ) 7

1 2

;

F00 (x) = In particular,

1 (I ? jxxj2 +x ): (jxj + ) 1 2

2

jF0(x)j  1; 2 : jF00(x)j  (jxj + ) 1 2

2

3 Existence and uniqueness of the solution to stochastic equations in in nite dimension Let ( ; F ; P ) be a probability space carrying a cylindrical Wiener process B = fBt; t  0g with values in K (see Sect. 2 for a brief de nition), and let fFtg be the  - eld generated by B (that is, Ft =  (Bs ; 0  s  t)). We make the standard P -augmentation to each Ft such that Ft contains all the P -null sets of F . Then fFtg is right continuous and fFtg satis es the usual hypothesis. Let T > 0 be an arbitrarily xed number. Let b(; x),  (; x) be stochastic processes depending on x 2 H satisfying (3) and (4). The aim of this section is to establish the solvability of (6) under the following so-called integral-Lipschitz condition: for any x1 ; x2 2 H ,

jb(t; x ) ? b(t; x )j + j(t; x ) ? (t; x )j  (t)(jx ? x j ); (12) where 2 L ([0; T ]; R ), and  : (0; +1) ! (0; +1) is a continuous, 1

2

2

2

1

2

2

2

1

2

2

+

non-decreasing, concave function satisfying:

(0+) = 0;

Z 1 0

dr = +1: (r)

(13)

Theorem 3.1 We suppose (3), (4) and the integral - Lipschitz condition (12). Then there exists a unique process X () 2 CF ([0; T ]; L ( ; F ; P ; H ))(= 2

CF ) satisfying the stochastic equation (6). Furthermore, if we de ne Z t  (t) = ds ; t 2 R+ ; t (s) where t0 2 (0; +1), M = maxt2[0;T ]jeAt j; 0

8

and we denote by X (; x) be a mild solution of (6) starting from x at time t = 0, then for any t  0,

E [jX (t; x1) ? X (t; x2)j2] Z t   ? 1 2 2 2    (3M jx1 ? x2j ) + 3M (1 + T ) 2(s)ds : 0

Proof: We begin with the proof of uniqueness. As X (; x) is a mild solution of (6), obviously we have X (t; x ) ? X (t; x ) = eAt(x ? x ) t + eA t?s b(s; X (s; x )) ? b(s; X (s; x )) ds 1

2

1

Z

2

(

+

Z

)

0

t





1

2





eA(t?s) (s; X (s; x1)) ? (s; X (s; x2)) dBs:

0

and

jX (t; x ) ? X (t; x )j  3 jeAt(x ? x )j 1

(

2

1

+ +

Z 0 Z 0

2

2

2

t A(t?s) h  e b s; X (s; x





i

!2

) ? b s; X (s; x2) ds

1

t A(t?s) h  e  s; X (s; x





i

) ?  s; X (s; x2) dBs

!2 )

1

:

Now let us put:

u(t) = sup E [jX (r; x1) ? X (r; x2)j2]; rt

0

then,

u(t)  3M jx1 ? x2j +3M (1+ T ) 2

2

Z

2

0

t

(



(s)E  jX (s; x1) ? X (s; x2)j 2

2

As  is concave and increasing, we deduce:

u(t)  3M jx1 ? x2 j + 3M (1 + T ) 2

2

Z

2

9

0

t

2 (s)(u(s))ds:

)



ds:

From (10), we obtain: Z t   ? 1 2 2 2 u(t)    (3M jx1 ? x2j ) + 3M (1 + T ) 2 (s)ds : 0

In particular, if x1 = x2 , we obtain the uniqueness of the solution to (6). Now we turn to the proof of existence to (6). We de ne the Picard sequence of processes fX n(); n  0g  CF as follows: X 0(t) = x; t 2 [0; T ]; and

X n+1(t)

eAt x +

=

+

Z 0

t

Z

eA(t?s) b(s; X n(s))ds

0

t A(t?s) e (s; X n(s))dB

s;

for t 2 [0; T ]:

(14)

Because of the assumptions (3) and (4), the sequence fX n(); n  0g  CF is well de ned. n h i Furthermore, we establish the a priori estimates for E jX n (t)j2 ; o n1 . From (14), we deduce that

E [jX n+1(t)j2]  3M 2jxj2 + 3M 2(1 + T )E

(

t

Z 0

(s) + 2 1

2 2

(s)jX n(s)j2



)

ds :

Hence,

E [jX n+1(t)j2]

 3M jxj + 3M (1 + T )E [ 2

2

2

+3M 2 (1 + T ) Set n

p(t) = 3M 2jxj2 +3M 2 (1+T )E [

T

Z

0

then p is the solution of

p(t) = 3M jxj + 3M (1 + T )E [ 2

2

2

Z

T

0

Z

t

0

o

Z

T

0

12 (s)ds]

22(s)E [jX n(s)j2]ds: n

12 (s)ds] exp 3M 2 (1+T ) (s)ds] + 3M (1 + T ) 2 1

10

2

Z

t

0

Z

t

0

o

22 (s)ds ;

(15)

22 (s)p(s)ds:

(16)

By recurrence, we prove easily that for any n  0, E [jX n(t)2]  p(t): Set uk+1;n (t) = sup E [jX k+1+n(r) ? X k+1 (r)j2]: rt

0

From the de nition of the sequence fX n(); n  0g,

X k+1+n (t) ? X k+1 (t) =

Z

0

+ Hence, Set then,

t

eA(t?s) (b(s; X k+n(s)) ? b(s; X k(s)))ds

Z 0

t

eA(t?s) ((s; X k+n(s)) ? (s; X k(s)))dBs :

uk+1;n (t)  2M 2(1 + T )

Z 0

t

2 (s)(uk;n(s))ds:

vk (t) = sup uk;n (t); 0  t  T; n Z

t

0  vk+1 (t)  2M 2 (1 + T ) 2 (s)(vk (s))ds: 0 Finally, we de ne: w(t) = lim supk!+1 vk (t); t  0: As  is continuous and vk (t)  2p(t); 0  w(t)  2M 2 (1 + T ) Hence,

Z

t

0

2(s)(w(s))ds; 0  t  T:

w(t) = 0; t 2 [0; T ]:

That is, fX n (); n  0g is a Cauchy sequence in CF . We denote the limit in CF of this sequence by X (). Then there exists a subsequence X nl () and a process X 0() such that lim X nl (t; ! ) = X (t; ! ); for (t; ! ) dt dP ? a:e:; l!+1 and

jX nl (t; !)j  jX 0(t; !)j; for (t; !) dt dP ? a:e: Let l ! +1 in (14), we prove easily that X is a mild solution of (6).

The proof of the existence of the solution to (6) is now complete. 11

4 Existence and Uniqueness of the solution to dissipative stochastic equations in in nite dimension In this section, we give some existence and uniqueness result to (6) under some weaker condition than the condition (12).

4.1 Dissipative stochastic equations in in nite dimension

In this subsection, we assume always that A is m-dissipative. Then the Yosida approximations A ; > 0 are bounded, m-dissipative, and lim eA tx = eAt x; 8x 2 H:

!0

Theorem 4.1 We assume the following one-sided integral-Lipschitz condition for b and  : for any x1 ; x2 2 H , 2hb(t; x1) ? b(t; x2); x1 ? x2 i + j (t; x1) ?  (t; x2 )j2  2 (t)(jx1 ? x2 j2 ): (17) Then there exists at most one mild solution in CF to (6).

Proof: Let us suppose that there exist two mild solutions X (), X () 2 1

CF to (6). And set

t

Z

Z

2

t

X 1 (t) = eA t x+ eA (t?s) b(s; X 1(s))ds+ eA (t?s) (s; X 1(s))dBs; (18) 0

0

and

X 2

(t) = eA t x+

Z

t

0

eA (t?s) b(s; X 2(s))ds+

Z 0

t

eA (t?s) (s; X 2(s))dBs: (19)

As A is bounded, we can rewrite (18) and (19) in their di erential forms, and we deduce: 







dX 1 (t) = A X 1 (t) + b(t; X 1(t)) dt + (t; X 1(t))dBt ; dX 2 (t) = A X 2 (t) + b(t; X 2(t)) dt + (t; X 2(t))dBt : In particular, for any > 0, E [ sup (jX 1 (t)j2 + jX 2 (t)j2)] < +1: tT

0

12

Applying the It^o formula to jX 1 (t) ? X 2 (t)j2, we obtain:

d(jX 1(t) ? X 2 (t)j2) = 2hA (X 1 (t) ? X 2 (t)); X 1(t) ? X 2 (t)idt +2hX 1 (t) ? X 2 (t); b(t; X 1(t)) ? b(t; X 2(t))idt +2hX 1 (t) ? X 2 (t);  (t; X 1(t)) ?  (t; X 2(t))idBt +j (t; X 1(t)) ?  (t; X 2(t))j2dt: As

E

T D

h Z



0

E 2

h

 E ( sup jX (t) ? X (t)j 1

tT

2

T

Z

2

j(t; X (t)) ? (t; X (t))j dt 1

0

0

2

T

2

1 i 2

j(t; X (t)) ? (t; X (t))j dt  21 E [ sup (jX (t) ? X (t)j )]+ 21 E tT < +1; t hX (s) ? X (s);  (s; X (s)) ?  (s; X (s))idB (s); t  0 is a martingale, and 1

2

hZ

2

0

1

2

0

0

R

1

X 1 (t) ? X 2 (t); (t; X 1(t)) ? (t; X 2(t)) dt ]

1

2

1

2

as a consequence,

E

hZ

tD

0

E

i

X 1 (t) ? X 2 (t); (t; X 1(t)) ? (t; X 2(t)) dB(s) = 0;

from which we deduce:

E [jX 1(t) ? X 2 (t)j2] E i hZ t D X 1 (s) ? X 2 (s); b(s; X 1(s)) ? b(s; X 2(s)) ds  2E +E

0 hZ

t

0

i

j(s; X (s)) ? (s; X (s))j ds : 1

2

2

When goes to 0, we get, using (17) and the concavity of , Z

t

E [jX (t) ? X (t)j ]  E [ 2 (s)(jX 1(s) ? X 2(s)j2)ds] 1

2

2



Z 0

t

0

2 (s)(E [jX 1(s) ? X 2(s)j2])ds:

Finally, Lemma 2.2 gives the uniqueness result. 13

2

i

Remark 4.1 The condition (17) is weaker than the condition (12), because the concavity of  implies that

(r)  r(1); r 2 (0; 1]: As for existence, we need some stronger conditions.

Theorem 4.2 We suppose (3) and the following condition: jb(t; x)j + j(t; x)j  (t) + (t)jxj ; 2 LpF (0; T ; H ); 2 Lp([0; T ]; H ); 2

2

1

2

2

1

2

for some p > 2. We assume also

jb(t; x ) ? b(t; x )j  (t) (jx ? x j); j(t; x ) ? (t; x )j  (t) (jx ? x j ); where 2 L ([0; T ]; R );  ;  : (0; +1) ! (0; +1) are continuous, con1

2

1

1

+

2

2

1

1

1

2

2

1

2

2

2

cave and increasing, and both of them satisfy (13). Furthermore, we assume that 2 3(r) = 2 (r ) ; r 2 (0; +1)

r

is also continuous, concave and increasing, and Z 1

3(0+) = 0;

0

dr

1 (r) + 3(r) = +1:

Then there exists a unique solution to the equation (6).

Before the proof of this theorem, let us give one simple example to explain why the condition for b here is \weaker" than that of Theorem 3.1. Example: If 1(r) = r ln 1r ;  (r) = r ln 1 ;

r

2

then the conditions for Theorem 4.2 are satis ed but not for Theorem 3.1. Proof: We de ne a sequence of processes fX n(); n  0g  CF as follows: X 0(t) = x; t 2 [0; T ]; 14

and

X n+1 (t)

=

eAtx + +

Z

Z

t A(t?s) e b(s; X n(s))ds

0

t A(t?s) e (s; X n+1(s))dB

0

s;

for t 2 [0; T ]:

(20)

Because of the assumptions of this theorem and Theorem 3.1, the sequence fX n (); n  0g  CF is well de ned. It is important to note that (14) in Theorem 3.1 is a usual Picard approximation, while given X n , (20) is a stochastic equation and Theorem 3.1 is applied to nd X n+1 in a unique way. Furthermore,

E [jX n+1(t)j2] Z T 2 2 2  3M jxj + 3M (1 + T )E [ 12(s)ds] +3M T

Z

2

t

0

0



2 2

Z t 2 n 2 2 (s)E [jX n+1(s)j2]ds: (s)E [jX (s)j ]ds + 3M 0

Obviously,

2

E [jX 0(t)j2]  p(t):

Let us suppose that

E [jX n(t)j2]  p(t);

where p is de ned by (15) which is the solution of (16). Then

E [jX n+1(t)j2] Z T 2 2 2  3M jxj + 3M (1 + T )E [ 12(s)ds] +3M T

Z

2

0

t

0

(s)p(s)ds + 3M 2 2

Z 2

t

0

22 (s)E [jX n+1(s)j2]ds;

from which we deduce that

E [jX n+1(t)j2]  p(t): The above recurrence prove that:

E [jX n(t)j2]  p(t): 15

Set

uk+1;n (t) = sup E [jX k+1+n(r) ? X k+1 (r)j]: rt

0

From the de nition of the sequence fX n(); n  0g,

X k+1+n (t) ? X k+1 (t) Z t eA(t?s) (b(s; X k+n(s)) ? b(s; X k (s)))ds = 0

+

t

Z 0

eA(t?s) ((s; X k+n+1(s)) ? (s; X k+1(s)))dBs:

We de ne

X n+1 (t)

e A t x +

=

t

Z

+

0

t

Z 0

eA (t?s) b(s; X n(s))ds

eA (t?s) (s; X n+1(s))dBs; for t 2 [0; T ]:

Note that as jxj is not C 2, one has to approximate jxj by F 2 C 2 which is studied in Lemma 2.3. Applying the It^o formula to F X k+1+n ? X k+1(t) , and taking the expectation, we get h



hZ

t

i

E F X k+1+n ? X k+1(t)

i

jb(s; X k n(s)) ? b(s; X k(s))jds t j (s; X k n(s)) ?  (s; X k (s))j +E ds : (jX k n ? X k j + ) Sending ! 0, and then letting  ! 0, we deduce that  E

+

0 " Z

+1+

+1+

0

uk+1;n (t) 

Z

t

0

Set

+1

(s)1(uk;n (s))ds +

+1 2

Z

t

0

2

#

1 2

(s)3 (uk+1;n (s)) ds:

vk (t) = sup uk;n (t); 0  t  T; n

then, 0  vk+1 (t) 

Z

t 0

(s)1(vk (s))ds + 16

Z 0

t

(s)3(vk+1 (s))ds:

Finally, we de ne:

w(t) = lim supk!+1 vk (t); t  0; then

0  w(t) 

t

Z

Hence,

(s)(1 + 3)(w(s))ds; 0  t  T:

0

w(t) = 0; t 2 [0; T ]:

Hence, fX n (); n  0g is a Cauchy sequence in CF ([0; T ];L1( ; F ; P ;H )). We denote the limit in CF ([0; T ]; L1( ; F ; P ; H )) of this sequence by X (). Then there exists a subsequence X nl (), such that:

X nl (t; !) ! X (t; !); (t; !) dt dP ? a:e: Taking into consideration that 1 2 LpF (0; T ; H ); 2 2 Lp (0; T ; H ), we

can prove in a similar way that h

i

sup sup E jX n(t)jp < +1; where p > 2: n0 0tT

Hence, X n (t) is uniformly integrable in L2 ( ; Ft; P ; H ), and thus h

i

lim E jX nl (t; ! ) ? X (t; ! )j2 = 0: l!+1 Let n ! +1 in (20), we prove easily that X is a mild solution of (6). The proof of the existence of the solution to (6) is now complete.

4.2 One-dimensional stochastic di erential equation

In this subsection, we consider the following one-dimensional stochastic differential equation:

X (t) = x +

Z 0

t

b(s; X (s))ds +

Z

t

0

(s; X (s))dBs;

(21)

where b : R+  R ! R,  : R+  R ! R are two measurable functions. We suppose that: (b(t; x1) ? b(t; x2))(x1 ? x2 )  (t)(jx1 ? x2j2 );

(22)

j(t; x ) ? (t; x )j  (t)~(jx ? x j);

(23)

1

2

2

17

1

2

where 2 L1loc (R+ ; R+ );  : (0; +1) ! (0; +1) is a continuous, increasing, concave function satisfying (13); and ~ is a Borel locally bounded function from (0; +1) into itself such that there exists a r0 > 0, r0

Z 0

dr = +1: (r)

Theorem 4.3 Let us assume (22) and (23). Then the pathwise uniqueness holds for the one-dimensional stochastic di erential equation (21).

Proof: Let X ; X be two solutions of (21), then X (t) ? X (t) t t (b(s; X (s)) ? b(s; X (s)))ds + ( (s; X (s)) ?  (s; X (s)))dBs: = 1

1

2

2

Z

Z

1

1

2

2

0

0

As for any xed  > 0, t

Z

10s 1

=



0 Z

t

2

10