G x ,y = 1 p. 2 d exp ,k x ,yk2. 2. : We consider an interacting particle system associated with (1.2). The configuration of the system at time t is denoted by Zn t = Zn.
A LAW OF LARGE NUMBERS FOR RELAXED SCALAR CONSERVATION LAWS Ren´e FERLAND* and Gaston GIROUX University of Quebec in Montreal and University of Sherbrooke
1. Introduction. Probabilistic methods have been used to obtain results about nonlinear evolution equations (see [BFG, COR, Fer, Rob]). Starting with an evolution equation one builds up a system composed of a large number of randomly interacting particles in such a way that formally the empirical laws converge to a weak solution of the equation. In some sense, this validates the equation. Usually, one has to modify or regularize the evolution equation to obtain such a validation. Here we start with the quasilinear scalar conservation law
X
@tu +
d
A @x u = 0
(1:1)
1
and we consider the Cauchy problem u(x; 0) = u0 (x), assuming that u0 is a probability density. Generally, one considers weak solutions of special nature: irregular functions which are solutions in the sense of distributions. It would be nice to show that if one starts with a probability density
u0 then one has a unique solution u such that u t is a probability density for all t > 0.
Here we are not so precise. First we relax the equation. Written in the weak sense the equation becomes the following
ht; 'i = h0; 'i + where v : R+
Zt 0
hs; v() rx'i ds +
Zt 0
h˜ s; 'i ds;
(1:2)
! Rd is a given function and ˜ s is a nonlinear term of the form h˜ s; 'i = hs; 1(s)'i :
1 and v , that starting with a probability measure 0 , this relaxed equation has one and only one solution such that t is a probability measure for all t.
We then prove for some particular
2. The Setup. We study the relaxed equation when 1 ()' is given by
1 ()'(x; ) = where
Z 1h 0
[G ](x) =
i
'(x; [G ](x)) , '(x; ) d
Z Rd R
G(x , y )(dy; d)
+
* Research supported by a grant from NSERC, Canada. 1
and G is a regular nonegative function which we take here equal to
G(x , y) =
p 1 d exp ( 2 )
kx , yk2 , : 2
We consider an interacting particle system associated with (1.2). The configuration of
t is denoted by Zn (t) = (Z1n (t); : : :; Znn (t)) where Zjn (t) = (xnj(t); jn (t)) is Rd R+ -valued. The process f Zn (t); t 2 [0; T ] g is Markov. We may assume that the , n trajectories are cadlag functions from [0; T ] to Rd R+ . The law Pn of this process is such , that for any f 2 Cb1 Rd R+ ]n , the process the system at time
Zt
f (Z n (t)) , f (Z n (0)) , is a Pn -martingale where, for zn
0
(Ln + Jn)f (Z n (s)) ds
(2:1)
, = (xn1 ; 1n ); : : :; (xnn ; nn ) ,
Lnf (z n ) = Jnf (z n ) =
n X
v (jn ) rxj f (zn )
j =1 n Z X j =1
1
f (z n;j ()) , f (zn ) d
0
and z n;j ( ) is the vector obtained from zn by replacing the component jn by [G n ](xnj ) and
leaving the other ones unchanged. The term Ln corresponds to a free movement for the xnj ’s
while the term Jn indicates that the jn ’s entail jumps at random times driven by an external field which is given here by the empirical measure of zn :
n 1X n = n (xnj ;jn ) : j =1 Let
' 2 Cb1(Rd R+ ) and consider the martingale f Mtn ('); t 2 [0; T ] g obtained from (2.1)
using the function
,
n X
f (xn1 ; 1n ); : : :; (xnn ; nn ) = n1 '(xnj ; jn ) : j =1 Simple computations give that
Mtn (') = hnt ; 'i , h0n ; 'i ,
Zt 0
hns ; ,'i ds ,
Zt 0
hns ; 1 (ns )'i ds
(2:2)
,'(x; ) = v () rx '(x; ) and nt is the empirical measure of Zn (t). Moreover, the quadratic variation hM n (')it is given by
where
hM n (')it = 1
n
Zt 0
2
hns ; 2(ns )'i ds
(2:3)
where
2 ()'(x; ) =
Z1 0
('(x; [G ](x) ) , '(x; ))
2
d :
We shall use this martingale a few times in the sequel.
3. Preliminaries. In this section we study the continuity of some mappings on a particular set of measures. The main result is Proposition 3.2 which we shall use later in the proof of Lemma 4.4.
Let P1 be the set of
2 M1(Rd R+ ) such that m1() :=
Z
RdR
kxk + (dx; d) < 1 :
+
We give P1 the Polish topology induced by the metric
(; ) = inff h; kx , y k + j , ji j is a coupling of and g : Proposition 3.1 below characterizes this topology. Proposition 3.1. If n ! in P1 then hn ; 'i ! h; 'i for any continuous function j'(x; )j C (1 + kxk + ). In particular, n ! weakly.
' such that
Let be a probability measure on Rd . Put
[G ](x) = where
Z
Rd
G(x , y ) (dy )
G(x , y) = p 1 d expf , kx ,2 y k g : ( 2 ) 2
! weakly then G n ! G uniformly on compact subsets of Rd . Proof. Let K be a compact subset of Rd and " > 0. There exists > 0 such that kz , z k < ) jG (z ) , G (z )j < "
Lemma 3.1. If n
1
2
1
2
3
for any probability measure . Indeed, the function G is uniformly continuous on Rd . Hence
there exists
> 0 such that
kz1 , z2k < ) kG(z1) , G(z2)k < 3" :
Consequently we have
kz1 , z2 k < ) kG(z1 , y) , G(z2 , y)k 0; 9 > 0; k(x; ) , (x0; 0 )k < ) j'(x; ) , '(x0; 0 )j < " : In particular, we obtain
8 " > 0; 9 > 0; j , 0 j < ) j'(x; ) , '(x; 0)j < "; 8 x 2 Rd :
(3:1)
Let K be a compact subset of Rd R+ . There exists a compact subset K0 of Rd and a > 0 such
K 0 [0; a]. Write n and for the marginals of n and on Rd . Since G n converge to G uniformly on K 0 , there exists an integer n0 such that that K
n n0 ) jG n (x) , G (x)j < ; 8 x 2 K 0 : Hence, recalling (3.1),
8 n n0; 8 x 2 K 0 ; 8 2 [0; 1]; j'(x; G n (x)) , '(x; G (x))j < " : Therefore, provided n n0 and x 2
j1(n )'(x) , 1 ()'(x)j
K,
Z1 0
j'(x; [G n ](x)) , '(x; [G ](x))j d < " :
! Rd be a Lipschitz function with constant c > 1: kv() , v( )k c j , j. For ' 2 Cb1 (Rd R+ ), recall that ,'(x; ) = v () rx '(x; ). Now let v : R+
Proposition 3.2. Let
mappings on P1 .
' 2 Cb1(Rd R+ ).
Then
7! h; 1 ()'i and 7! h; ,'i are continuous 4
Proof. For the first mapping, it suffices to show that
hn ; 1(n )'i ! h; 1 ()'i if n !
weakly. Let us write
hn ; 1(n )'i , h; 1()'i = hn ; 1(n )' , 1 ()'i + (hn ; 1 ()'i , h; 1 ()'i) : The second term goes to zero because 1 ()' is bounded and continuous. For the first one, fix
" > 0 and choose K compact in Rd R+ and such that n (K ) > 1 , " for all n. This is possible because (n ) is tight (being weakly convergent). Now write
hn ; 1(n )' , 1 ()'i = hn ; (1 (n )' , 1()') 1K i + hn ; (1(n )' , 1()') 1Kc i : On one hand, we have k1 (n )' , 1 ()'k1
2k'k1 so
h ; ( ( )' , ()') 1 c i 2k'k (K c ) 2k'k " : 1 K 1 n 1 n 1 n
On the other hand, ((n )') converges uniformly to 1 ()' on K . Hence, there exists n0 such
that j1 (n )'(x) , 1 ()'(x)j < " for all x 2 K and n n0 . For those n, we have
h ; ( )' , ()'i " (K ) + 2k'k " : 1 n n 1 n
Since " is arbitrary, we obtain the result.
The continuity of the second mapping is immediate because, by the hypothesis made on v ,
,' is a continuous function such that j,'(x; )j C'
kv(0)k + c
.
f nt ; t 2 [0; T ] g where nt is the empirical measure of Zn (t). We look at nt as an element of P1 . From the definition of , 4. The Law of Large Numbers. We are interested by the process one easily deduce the inequality
n 1X n n n (s ; t ) n kxj (t) , xnj (s)k + jjn (t) , jn (s)j j =1 using the coupling 1
=n Therefore, the trajectories t
n X j =1
Zjn (s) Zjn (t) :
7! nt belong to D([0; T ]; P1). The law of f nt; t 2 [0; T ] g is denoted
by P n . Our main result is the following.
Theorem 4.1. Assume the following conditions hold: (a)
v : R+ !2 Rd is Lipschitz with constant 3 c > 1 : kv() , v( )k c j , j.
n 1 X n 2 n (0)2 5 < 1; n 4 (b) sup E 0 ) k + k x ( j n j =1 j n
5
f0n g converges in distribution to 0 2 P1. Then the sequence (P n ) converges weakly to where is the unique solution of (1.2) (c)
The proof of Theorem 4.1 is mainly based on Lemma 4.1 which gives the propagation of the moment condition (b). As usual, the method is to prove first that the sequence
(P n ) is
relatively compact. This is Proposition 4.1. Then we prove in Proposition 4.2 that any limit point of the sequence (P n ) is concentrated on the solutions of (1.2). To end the proof it is then sufficient to show the uniqueness of the solutions of (1.2) (Proposition 4.3). Lemma 4.1. Under conditions (a) and (b) of Theorem 4.1, we have sup
3 2 n X 1 sup En 4 kxnj (t)k2 + jn (t)2 5 < 1 : n j=1
tT n
0
Proof. Consider the function '(x; ) = kxk2 + 2 . First we have
hns ; ,'i = n2
n X
j =1 n 2X
v (jn (s)) xnj (s)
n
kv(jn (s))k kxnj(s)k
j =1 n 1 X,
n
kv(jn (s))k2 + kxnj (s)k2 :
j =1
Because of (a), kv ()k kv (0)k + c, so
hns ; ,'i n1
n h, X j =1
2 kv(0)k and therefore
Second, we have
2 + kxn (s)k2i
kv(0)k + c jn (s)
2
+
1
j
n X
n j=1
2c2 jn (s)2 + kxnj (s)k2
hns ; ,'i 2kv(0)k2 + 2c2hns ; 'i : [G ns (x)]2 (ns )'(x; ) = 3
(4:1)
, 2
n 1 X [G ns (xnj (s))]2 n n + hns ; 2 i : hs ; (s )'i n 3 j =1 But G is bounded and ns is a probability measure. Thus G ns (z ) is bounded by a constant c0 so that
which does not depend on n; s or z . It implies that
, hns ; (ns )'i c0 2 + hns ; 'i : 6
(4:2)
We now use the martingale (2.2), (4.1) and (4.2) to obtain
En hnt ; 'i En h0n ; 'i + t (c0 )2 + 2kv (0)k2 + (2c2 + 1)
Zt En hns ; 'i ds : 0
Gronwall lemma implies that
En
hn ; 'i En hn ; 'i + t (c0 )2 + 2kv(0k2 e 2c +1 t : t
(
0
2
)
2 n 3 X 1 En 4 kxnj (t)k2 + jn (t)25 = En hnt ; 'i ;
Since
n j =1
the result follows from hypothesis (b). Remark. The function
' above is not bounded.
Strictly speaking the previous calculations
would have to be done using truncation. Lemma 4.2. Under conditions (a) and (b) of Theorem 4.1, there exists a constant CT such that
8 n 1; 8 m 1;
P
n
sup
n h1 X
tT n j =1
0
i
o
kxnj (t)k2 + jn (t)2 > CT 2m 21m :
Proof. Again we take '(x; ) = kxk2 + 2 in (2.2) to obtain the inequality: sup
n h1 X
0tT n j =1
i
kxnj (t)k2 + jn (t)2 hn0 ; 'i +
sup
tT
0
+
jMtn (')j
ZTn 0
Put
DT =
sup
3 2 n X 1 sup En 4 kxnj (t)k2 + jn (t)2 5 ; n j=1
tT n
0
(4.1) and (4.2) implies that
En so
P
"Z T n
(Z T n
0
0
o
jhns ; ,'ij + jhns ; (ns )'ij ds : (4:3)
o #
jhns ; ,'ij + jhns; (ns )'ij ds [(c0)2 + 2kv(0)k2 + (2c2 + 1)DT ]T
o
jhns ; ,'ij + jhns ; (ns )'ij ds > a3
)
a3 [(c0)2 + 2kv(0)k2 + (2c2 + 1)DT ]T : 7
(4:4)
Next we have
jMTn (')j h0n ; 'i + hnT ; 'i + Therefore,
En
jM n(')j 2D T
ZTn
o
jhns ; ,'ij + jhns ; (ns )'ij ds :
0
T + [(c0 )2 + 2kv (0)k2 + (2c2 + 1)DT ]T
and Doob’s inequality gives
(
a P sup jMTn (')j > 3 0tT
)
,
a3 En jMTn (') a3 2DT + [(c0)2k + (c + 1)DT ]T :
(4:5)
Finally, P f h0n ; 'i > a=3 g 3DT =a. But inequality (4.3) implies that the probability
n
P
sup
0
n h1 X
tT n j =1
is bounded above by the sum
(
)
a +P D + P sup jMTn (')j > T a 3 0tT 3
If we take
i
kxnj (t)k2 + jn (t)2 > a
o
(Z T n
o
jhns ; ,'ij + jhns ; (ns )'ij
0
)
ds > a3 :
o
CT = 3 2DT + 2 (c0 )2 + 2kv (0)k2 + (2c2 + 1)DT T
the result follows from (4.4) and (4.5). To prove the relative compactness of (P n ), it will be necessary to show that several auxillary processes are tight. This is done in the following lemma. Recall that a sequence of real processes is called C -tight if it is tight and any limit process has continuous paths. Lemma 4.3. Suppose ' 2
d
Cb (R R+ ) and that conditions (a) and (b) of Theorem 4.1 hold. Then the R Rt t n processes f h ; ,'i ds; t 2 [0; T ] g, f hn ; (n )'i ds; t 2 [0; T ] g and f hM n (')i ; t 2 [0; T ] g
are C -tight.
0
1
s
s
0
s
t
Proof. Consider the first sequence of processes. According to [Jac, Prop. 2.11, p. 309], we have to prove that a)
8 " >, 9 a > 0, 9 n0 2 such that
Z t n sup Pf sup hs ; ,'i ds > a g " ; 0 nn 0tT 0
b)
8 " >, 8 > 0, 9 > 0, 9 n0 2 such that
Z
sup Pf w(
nn0
0
hns ; ,'i ds; ) > g " 8
where w is the modulus of continuity:
w(x; ) = Since j,'(x; )j C'
sup
s;t2[0;T ] jt,sj a 2 2kv (0)k T + 2Tc E hs ; i ds : a 0 0 0tT
By Lemma 4.1, the expectation on the right-hand side is uniformly bounded in n and s. Property a) then follows. Next remark that
Z
w(
0
hns ; ,'i ds; ) C'
!
kv(0)k + c sup hnt ; i 0tT
:
Choose small enough to have C' kv (0)k =2. Then,
P
Z
w( hns ; ,'i ds; ) > P 0
(
2 sup hnt ; 2 i > 2 2 2 4c C' 0tT
)
and property b) follows from Lemma 4.2.
R t n n h ; ( ) ' i ds 0 s s and sup0tT jhM n (')itj are bounded by 2T k'k1
For the other processes, we must show similar statements. This is easier. Indeed, the random variables sup0tT
and 4T k'k21 respectively. The analogues of property a) then follow trivially by taking a large enough. Similarly, the modulus of continuity is bounded by 2k'k1 and 4k'k21 respectively. This time property b) is obtained by taking small enough.
Proposition 4.1. Under conditions (a) and (b) of Theorem 4.1, the sequence (Pn ) is relatively compact. Proof. According to [Frn], it suffices to prove that:
1) There exists a sequence (Km ) of compact subsets of P1 such that
8 m; 8 n;
Pn f 9 t 2 [0; T ] j nt
62 Km g 2,m :
2) For any ' 2 Cb1 (Rd R+ ), the real processes f hnt ; 'i;
But for fixed c the set
t 2 [0; T ] g are tight.
L(c) = f 2 P1 j h; kxk2 + 2 i c g 9
is compact in P1 . Hence property 1) follows from Lemma 4.2 by taking Km
CT is the constant appearing in that lemma. For 2), recall that
hnt ; 'i = hn0 ; 'i +
Zt 0
hns ; ,'i ds +
Zt 0
= L(CT 2m ) where
hns ; 1(ns )'i ds + Mtn (') :
We then use [JS, Cor. 3.33, p. 317] which says that the real processes
f hnt ; 'i; t 2 [0; T ] g are
tight provided
n
f h0 ; 'i g is tight, R thn ; ,'i ds; t 2 [0; T ] is C -tight, b) s 0 R t c) hn ; (n )'i ds; t 2 [0; T ] is C -tight, a)
s 1 s n d) f Mt ('); t 2 [0; T ] g is tight. 0
But a) is fulfilled because the random variables are bounded by k'k1 ; b) and c) were proved in Lemma 4.3. For d), we use [JS, Theo. 4.13, p. 322]:
f Mtn('); t 2 [0; T ] g is tight if
f hM n (')it; t 2 [0; T ] g is C -tight. Again, this was proved in Lemma 4.3. Lemma 4.4. Let ' 2 Cb1 (Rd R+ ). The mapping ' : D([0; T ]; P1) !
' (w)(t) = hw(t); 'i , hw(0); 'i ,
Zt 0
hw(s); ,'i ds ,
Zt 0
D([0; T ]; R) defined by
hw(s); 1(w(s))'i ds
is continuous. Proof. Since 0 is a continuity point for any w 2 D([0; T ]; P1), the mapping ('1) (w)(t) = hw(0); 'i is continuous. Next, we use the fact that the mapping
f (x)(t) =
Zt 0
x(s) ds
from D([0; T ]; R) into D([0; T ]; R) is continuous. Therefore, it remains to show that
('2) (w)(t) = hw(t); 'i; ('3) (w)(t) = hw(t); ,'i and ('4) (w)(t) = hw(t); (w(t))'i
D([0; T ]; P1) into D([0; T ]; R). This is fulfilled if ! h; ,'i and ! h; ()'i are continuous mappings from P1 into R. These properties were proved are continuous mappings from in Proposition 3.2. Proposition 4.2. Under conditions (a), (b) and (c) of Theorem 4.1, any limit point of (Pn ) is concentrated on the solutions of (1.2). Proof. Consider the mapping ' of Lemma 4.4. It follows from (2.2) that ' (n )(t) =
Mtn (').
Doob’s inequality and (2.3) give
En
"
#
4 sup jMtn (')j2 = 4En MTn (')2 = n 0tT 10
Zt 0
16T En [ hns ; 2 (ns )')i ] ds k'k21
n
so that lim
n!1
En
#
" sup
tT
0
"
n j '()(t)j2 = nlim j '(n )(t)j2 !1 E 0sup tT " #
= lim En n!1
sup
tT
0
#
jMtn (')j2 = 0 :
(4:6)
P be a limit point of (P n ) and (P nk ) a subsequence converging to P . The mapping w ! sup0tT j ' (w)(t)j is continuous on D([0; T ]; P1) because ' is. Moreover, by the h i n computations above, supk E k sup0tT j ' ()(t)j2 < 1. Therefore, the weak convergence of (P nk ) yields " # " # n (4:7) lim E k sup j ' ()(t)j = E sup j ' ()(t)j : n!1 Let
tT
0
tT
0
h
i
The limits (4.6) and (4.7) imply that E sup0tT j ' ()(t)j = 0. Therefore, there exists a set
' D([0; T ]; P1) such that P ( ' ) = 1 and for any w 2 ' we have
hw(t); 'i = hw(0); 'i +
Zt 0
Zt
hw(s); ,'i ds +
0
hw(s); 1 (w(s))'i ds; 8 t 2 [0; T ] :
S be a countable subset of Cb1(Rd R+ ) which separates the probability Rd R+ and define Let
(4:8)
measures on
0 1 \
0 = @ ' A \ f w 2 D([0; T ]; P1) j w(0) = 0 g : '2S
P ( 0) = 1, and every w 2 0 satisfies (4.8) for all ' 2 S . Since S separates the probability measures, (4.8) is true for all ' 2 Cb (Rd R+ ). Thus the measure P Then by (c) of Theorem 4.1,
is concentrated on the solutions of (1.2). To study the uniqueness of the solutions of (1.2), we use the bounded Lipschitz metric BL
on M1 (Rd R+ ):
(1; 2 ) := BL
where
k'k := max BL
(
sup
'2Cb (Rd R+ ); k'kBL 1
sup j'(x; )j;
x;)
(
For ' 2 Cb (Rd R+ ) we put
J ( )'(x; ) =
sup
x;)6=(y; )
(
Z1 0
jh1 , 2 ; 'ij
'(x; [G ](x)) d
and we define a probability measure K ( ) by hK ( ); 'i = h; J ( )'i. 11
)
j'(x; ) , '(y; )j : kx , yk + j , j
Lemma 4.5. Let '
2 Cb(Rd R+ ) such that k'k 1. Then for any probability measure , 1 and BL
2 we have a) J ( )' 2 Cb (Rd R+ ) and kJ ( )'k kGk + 1, b) kJ (1 )' , J (2 )'k1 kGk (1 ; 2 ), c) (K (1 ); K (2)) (2 kGk + 1) (1 ; 2 ). Proof. First observe that kGk < 1 and jG (x) , G (y )j kGk kx , y k. Clearly, kJ ( )'k1 k'k1 1. If k'k 1 then j'(x; ) , '(y; ))j kx , yk + j , j and consequently BL
BL
BL
BL
BL
BL
BL
BL
BL
BL
we have
j'(x; [G ](x)) , '(y; [G ](y))j kx , yk + jG (x) , G (y)j kx , yk + kGk kx , yk BL
which implies
jJ ( )'(x; ) , J ( )'(y; )j
Z1
j'(x; [G ](x)) , '(y; [G ](y))j d
0
kGk
BL
+1
kx , y k + j , j :
Conclusion a) follows. For b), we have
Z Z jG 1 (x) , G 2 (x)j = d G(x , y)1(dy; d) , d G(x , y)2 (dy; d) R R R R Z Z G(x , y ) (dy; d) , G(x , y ) (dy; d) = kGk 1 2 R d R kGk Rd R kGk h1 , 2; i = kGk (1 ; 2) : k Gk sup +
+
BL
BL
+
BL
Therefore
BL
+
2Cb (Rd R+ );k kBL 1
BL
BL
j'(x; [G 1](x)) , '(x; [G 2 ](x))j kGk (1; 2) BL
which implies
BL
Z1
kJ (1 )' , J (2 )'k1 j'(x; [G 1 ](x)) , '(x; [G 2 ](x))j d 0 kGk (1 ; 2) : BL
BL
Now c) follows from a) and b). Indeed, suppose k'kBL
1, then hK ( ) , K ( ); 'i = h ; J ( )'i , h ; J ( )'i 1 2 1 1 2 2 h1; J (1 )'i , h1 ; J (2)'i + h1 ; J (2)'i , h2; J (2 )'i h1; jJ (1)' , J (2 )'ji + h1 , 2 ; J (2)'i kJ (1 )' , J (2 )'k1 + (kGk + 1) (1 ; 2) 2 kGk + 1 (1 ; 2) : BL
BL
BL
12
BL
Proposition 4.3. If v satisfies hypothesis (a) of Theorem 4.1, the equation (1.2) has at most one solution for a given 0 .
Proof. Fix ' 2 Cb (Rd R+ ) such that k'kBL
1. Put T (t)'(x; ) = '(x + tv(); ). An argument
similar to that of [TW] shows that any solution of (1.2) must satisfy
8 t 2 [0; T ]; ht; 'i = e,t h0; T (t)'i +
Zt
e,(t,s) hK (s ); T (t , s)'i ds :
0
Let (1) and (2) be two solutions of (1.2) with the same initial data 0 . Then we have
8 t 2 [0; T ]; jht1 , t2 ; 'ij ( )
( )
Zt 0
e,(t,s) hK ((s1)) , K ((s2)); T (t , s)'i ds :
Using hypothesis (a), we can easily show that kT (t)'kBL
8 t 2 [0; T ]; jht , t ; 'ij (1)
(2)
Zt 0
tc + 1 so
e,(t,s) (t , s)c + 1 (K ((s1)); K (s(2))) ds :
Therefore, by Lemma 4.6,
8 t 2 [0; T ]; (t ; t ) C (2 kGk (1)
(2)
BL
BL
+ 1)
Zt
BL
((s1) ; s(2)) ds BL
0
and the result follows from Gronwall lemma.
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