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© 2008 Birkh¨auser Verlag, Basel

1424-3199/08/010001-29,published onlineJanuary 18, 2008 DOI 10.1007/s00028-007-0314-y

Polynomial mixing for the complex Ginzburg–Landau equation perturbed by a random force at random times

Vahagn Nersesyan

Abstract. In this paper we study the problem of ergodicity for the complex Ginzburg–Landau (CGL) equation perturbed by an unbounded random kick-force. Randomness is introduced both through the kicks and through the times between the kicks. We show that the Markov process associated with the equation in question possesses a unique stationary distribution and satisfies a property of polynomial mixing.

1. Introduction

We consider the CGL equation perturbed by a random kick-force on a domainD⊂Rn, n≤4 with∂DC2:

˙

uνu+|u|2u=η(t, x), xD, (1.1)

u|∂D=0, (1.2)

u(0, x)=u0(x), (1.3)

whereu=u(t, x)andν, β >0.We assume thatη(t, x)is a random process of the form η(t, x)=

k=1

ηk(x)δ(tτk), (1.4)

whereδ(t)is the Dirac measure,ηkare independent identically distributed (i.i.d.) random variables with range in the spaceH := H01(D), and the waiting times tk = τkτk1, k≥2 andt1=τ1are i.i.d. random variables exponentially distributed with parameterλ.

Moreover, we assume that the sequencesηk,tkare independent.

Suppose that{gk}k=1 is an orthonormal basis in H. The main result of the present paper is Theorem 4.2, which states that, if the law ofηk is non-degenerate on the space spanned by{gk}Nk=1for sufficiently largeN, then there is a unique stationary measure for the continuous time Markov process associated with (1.1), (1.2), (1.4). Moreover, any solution of the problem polynomially converges to the stationary measure in the dual Lipschitz norm.

Mathematics Subject Classifications(2000): 35K55, 60J25 Key words: Complex Ginzburg-Landau equation, polynomial mixing

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Many authors have studied similar problems for various PDE’s with different random perturbations (e.g., see [16, 2, 17, 18, 21, 15, 14, 26] for discrete forcing and [6, 8, 3, 22, 19, 7, 9, 25, 27, 4, 10] for white noise). Several ideas of this article are taken from [18, 15, 26].

The problem of ergodicity for randomly forced Ginzburg–Landau equation was studied in the following articles. In [9], Hairer considered a real Ginzburg–Landau equation on multidimensional torus. Odasso [25] studied a class of CGL equations with strong nonlinear dissipation. In both of these works the property of exponential mixing is established. In [27], Shirikyan used a sufficient condition for ergodicity of Markov processes to show uniqueness and mixing for a class of CGL equations with linear dispersion. Finally, in [4], Debussche and Odasso proved the polynomial mixing property for a damped 1D Schr¨odinger equation.

The main novelty of the present paper is the condition over the waiting times. Note that the restriction of the solution at timesτk looks like the random dynamical systems considered by Kuksin, Shirikyan [14], [16], [26] and Masmoudi, Young [21] :

uτk =Stk(uτk−1)+ηk, (1.5) but there are some essential differences. As the waiting times can be arbitrarily small, during any time interval the system can receive any number of kicks. This changes the dynamics of the associated process, for example:

• The distance between two trajectories having close initial data can be arbitrary large at any timet >0.

• The phase space of the problem is not bounded even in the case of bounded kicks.

Let us give in a few words the ideas of the proof of Theorem 4.2. An important tool for the proof of the result is the Foias¸–Prodi type estimate. This kind of estimates are often used to prove ergodic properties of PDE’s. Suppose that there are two sequences of kicks ζk andζk, having equal high Fourier modes for kl, such that the solutions of corresponding problems have equal low Fourier modes at kicking timesτk,kl(see Lemma 2.1 for the exact formulation). LetNt be the number of kicks before timet, i.e.

Nt =max{k:τkt}. Then, by Foias¸–Prodi Lemma, we have the following estimate for the distance between solutions at timet, iftτl:

utut1eC(Ntl)

Nt

i=l+1

ti

12

eϕuτluτ

l1, (1.6)

where·1stands for the norm inH,utandutare solutions corresponding to the sequences ζkandζk respectively,ϕis a polynomial function of{uτi1}Ni=tland{uτ

i1}Ni=tlandC >0 is a large constant. Following the ideas from [26], we construct two sequencesζk andζk of i.i.d. random variables inH distributed asη1 such that the conditions of Foias¸–Prodi Lemma are satisfied for a random integer≥1. Moreover, using the law of large numbers

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and some martingale inequalities, we show thatcan be chosen in a such way that the following properties also hold:

(i) Ni=t+1ti

1

2eϕe(Nt), ifNt+1, (ii) uτ1+ uτ

1≤1,

(iii) EpCp.

As we show in Section 4, properties (i)–(iii) and (1.6) imply the polynomial mixing property.

The random variablesζk, ζkandare constructed in Proposition 4.3. In Section 4, we show how Theorem 4.2 is derived from Proposition 4.3. The proof of Proposition 4.3 is carried out in Sections 5 and 6.

Note that an exponential estimate for the random variablein (iii) would immediately imply the exponential mixing property for the system. Finally, using (i)–(iii), one can show that the embedded Markov chainuτk also satisfies a property of polynomial mixing. The stationary measure of the original process and that of embedded chain are connected with the Khasminskii relation (see Section 4).

Notation

LetD⊂Rnbe a bounded domain with smooth boundary and let{gj}jNbe an orthonor- mal basis inH. LetHNbe the vector span of{g1, ..., gN}andHNbe its orthogonal com- plement inH. We denote byPNandQNthe orthogonal projections ontoHN andHNin H. Denote by{ej}jNthe set of normalized eigenfunctions of the Dirichlet Laplacian with eigenvalues{αj}jNand denote byQN the orthogonal projection onto the closure of the vector span of{eN, eN+1, ...}inL2(D).

LetHs(D),s≥0 be the Sobolev space of orders. We denote byu1= ∇u,u2= uthe norms in the spacesH01(D)andH01(D)H2(D)respectively, where · stands for the norm inL2(D).For a Banach spaceX, we shall use the following notation.

B(X)is theσ-algebra of Borel subsets ofX.

C(X)is the space of real-valued continuous functions onX.

Cb(X)is the space of bounded functionsfC(X).

L(X)is the space of functionsfCb(X)such that fL:= f+sup

u=v

|f(u)f(v)|

uv <+∞.

P(X)is the set of probability measures on(X,B(X)). IfµP(X)andfCb(X), we set

(f, µ)=

X

f(u)µ(du).

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Ifµ1, µ2P(X), we set

µ1µ2L=sup{|(f, µ1)(f, µ2)|:fL(X),fL≤1}, µ1µ2var=sup{|µ1()µ2()|:B(X)}. For any1, 2B(X), withP(2)=0, denote

P(1|2)= P(12) P(2) .

The distribution of a random variableξis denoted byD(ξ). We denote byC, Ckunessential positive constants.

2. Preliminaries

It is well known that problem (1.1)–(1.3) withη≡0 andu0Hhas a unique solution in the spaceC(R+, H)L2loc(R+, H2(D)). LetSt :HHbe the resolving semi-group for that problem. Letτ0≡0 and defineutby the relation

ut=

Stτk(uτk), ift∈[τk, τk+1), k≥0, Stk+1(uτk)+ηk+1, ift=τk+1.

Thenutis the unique solution of problem (1.1)–(1.4). Clearly,ut exists for allt >0 with probability 1, asP{

tk = ∞} =1. Let us define a continuous functional onH:

H(u)=

D

α|∇u(x)|2+β

4|u(x)|4

dx, (2.1)

whereαis a positive constant. Ifαis sufficiently small, we have the estimate

H(St(u))eatH(u), t≥0, (2.2) whereais a positive constant, and there is a constantCsuch that

St(u)St(v)1Cexp(C(u61+ v61))uv1, t≥0, (2.3) St(u)St(v)2Ct21exp(C(u61+ v61))uv1, t >0, (2.4) whereu, vH. The proof (2.2), (2.3) and (2.4) is carried out by standard methods and is given in the Appendix.

For any sequenceak,mkn,we set aknm= 1

nm+1 n k=m

ak. (2.5)

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Supposeuk, ukHandtk>0 are arbitrary sequences. Defineζkandζk by the relations uk=Stk(uk1)+ζk, uk =Stk(uk1)+ζk. (2.6) LEMMA 2.1. Suppose that

PNuk =PNuk, QNζk=QNζk, l+1≤kn, (2.7)

ejHN, j=1, ..., N (2.8)

for someN≥1andN≥1. Then

ukuk1(Cα

1 2

N+1)kl

k

i=l+1

ti

12

×exp

C(kl)(ui61kl1+ ui61kl1) ulul1, (2.9) forlkn,whereCis a positive constant not depending onuk, uk, n, l, NandN.

Proof. Using (2.4), (2.7) and (2.8), we see that

ukuk1 = QN(ukuk)1= QN(Stk(uk1)Stk(uk1))1

QN(Stk(uk1)Stk(uk1))1

α

1 2

N+1Stk(uk1)Stk(uk1)2

1 2

N+1t

1 2

k exp

C(uk161+ uk161) uk1uk11.

Iteration of this inequality results in (2.9).

3. Markov chains associated with CGL equation and existence of stationary measures

Letu0be anH-valued random variable, independent of{ηk}and{tk}, and letut be the solution of problem (1.1)–(1.4). Denote byFt,t ≥ 0 theσ-algebra generated byu0and {ζ(s),0≤st}, where

ζ(s)= k=1

I{τks}ηk. (3.1)

LEMMA 3.1. Under the above conditions,ut is a homogeneous Markov process with respect toFt.

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The proof of this lemma is given in the Appendix. For anyuHandB(H), we set Pt(u, )=P{ut(u)}.The Markov operators corresponding to the processuthave the form

Ptf(u)=

H

Pt(u,dv)f(v), Ptµ()=

H

Pt(u, )µ(du), wherefCb(H)andµP(H).

The strong Markov property implies that uτk is a homogeneous Markov chain with respect toσ-algebraGkgenerated by{ηn, tn,1≤nk}. In what follows, we shall write ukinstead ofuτk; this will not lead to confusion.

LEMMA 3.2. (i) For anyε >0there is a constantCε>0such that H(uk)(1+ε)kekH(u0)+Cε

k l=1

ea(τkτl)(1+ε)klH(ηl). (3.2)

(ii) LetEHk)p<for somep≥1. Then

EH(uk)pγkEH(u0)p+ Cp

1−γEHk)p, (3.3) where0< γ <1andCp>0are some constants not depending onk.

Proof. Using (2.2), we obtain

H(uk)(1+ε)etkaH(uk1)+CεH(ηk).

Iteration of this inequality results in (3.2).

To prove (3.3), note that for anyε >0 there is a constantCp,εsuch that

H(uk)p(1+ε)etkapH(uk1)p+Cp,εH(ηk)p. (3.4) Taking the expectation and using the independence oftkanduk1, we obtain

EH(uk)p(1+ε) λ

λ+apEH(uk1)p+Cp,εEHk)p.

Choosingε >0 so small thatγ:=(1+ε)λ+λap <1 and iterating the resulting inequality,

we arrive at (3.3).

LEMMA 3.3. LetEηkp1 <for allp≥1and letu0H. Then

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(i) There is a constantM >0not depending onu0and a random integerT =T(u0)≥1 such that

uk61n0M fornT, (3.5)

ETp<for allp≥1, (3.6)

where·n0is defined by (2.5).

(ii) For anyδ(0,1)andd >0, there is a constantR=R(δ, d) >0such that P{uk61n0R,n≥0} ≥δ, (3.7) for anyu0Bd, whereBd = {uH:u1d}.

Proof. Let us fixε >0.Using (3.4) withp=3, we obtain H(uk)3(1+ε)e3atkH(uk1)3+CεH(ηk)3

=(1+ε)

e3atkλ λ+3a

H(uk1)3 +(1+ε) λ

λ+3aH(uk1)3+CεH(ηk)3. (3.8) Choosingε >0 so small thatq:=(1+ε)λ+λ3a<1 and summing up inequalities (3.8) for 1≤kn,we arrive at

n k=1

H(uk)3(1+ε) n k=1

e3atkλ λ+3a

H(uk1)3+q n k=1

H(uk1)3

+Cε n k=1

H(ηk)3, whence it follows that

α3uk61n0≤ H(uk)3n0≤ 1+ε 1−q

1 n+1

n k=1

e3atkλ λ+3a

H(uk1)3

+ 1 1−q

H(u0)3 n+1 + Cε

1−qm+ Cε 1−q

1 n+1

n k=1

(H(ηk)3−m), (3.9) wherem=EHk)3.To complete the proof, we need the following lemma, whose proof is given in the Appendix.

LEMMA 3.4. Suppose that Mk is a sequence of random variables that satisfies the inequality

E|Mk|2pCpkp for allp≥1. (3.10) Then the following assertions take place.

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(i) There is a random integerT ≥1such that 1

k|Mk| ≤1 forkT, (3.11)

ETp<for allp≥1. (3.12)

(ii) For anyδ(0,1), there is a constantR >0such that P

|Mk|

kR,k≥1

δ. (3.13)

Let us set

Mk = k

i=1

e3atiλ λ+3a

H(ui−1)3,

Mk= k i=1

(H(ηi)3−m), M0 =M0=0.

Clearly Mk and Mk are martingales. For Mk it is easy to verify that (3.10) holds, as MiMi1=H(ηi)3−m,i≥1 are centered i.i.d. random variables. To prove (3.10) for Mk, we need Burkholder’s inequality for martingales ([11], Section 2.4):

C1E k i=1

X2ip≤E|Mk|2pC2E k i=1

X2ip, (3.14)

whereXi =MiMi1,i≥1,p≥1 andC1, C2are positive constants depending only onp. Using (3.14) and (3.3), we obtain

E|Mk|2pC2E k i=1

e3atiλ λ+3a

2

H(ui1)6p

C k i=1

EH(ui1)6pkp−1Ckp,

whereCdepends onu01. Applying Lemma 3.4, letTandTbe the random variables corresponding to martingalesMk andMk. Setting

T =T1T2

H(u0) 1 1−q

, M=1+ε

1−q +Cε(m+1) 1−q +1

α3, it is easy to verify that we have (3.5) and (3.6) forT andM.

To prove (3.7), we apply (3.13) to the sequence Mk = 1+ε

1−qMk + Cε 1−qMk,

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and using (3.9), we see that (3.7) holds with R1=R+Cd 1

1−q+ Cε 1−qm,

whereCd =supuBdH(u)3.

LetτRbe the first hitting time of the ballBR:

τR=min{k≥0 :uk1R}.

LEMMA 3.5. LetEH1) < +∞. Then there are positive constantsδ,C andRnot depending onusuch that

EueδτRC(1+H(u)).

Proof. It suffices to show thatuk possesses a Lyapunov function (see [24]), i.e. there is a continuous functionalF onHsuch that

(i) F(u)≥1 and limu1→∞F(u)= +∞.

(ii) There are positive constantsn, R, Canda <1 that

EuF(un)aF(u) foru1R, (3.15) EuF(uk)C foru1< R,k≥0. (3.16) Let

F(u)=

H(u), ifH(u)A, A, ifH(u) < A, whereA≥1. Then (i) is satisfied. Letu1R.Note that

EuF(un)=EuF(un)I{H(un)<A}+EuF(un)I{H(un)A}

A+EuH(un)γnH(u)+A+CEH1), (3.17) where we used (3.3). ChoosingnandRso large that 2γn<1 andA+CEH1)γnR2α, whereαis the constant in (2.1), we arrive at (3.15) witha=2γn.It remains to note that

(3.16) follows from (3.3).

DEFINITION 3.6. A measure µP(H)is said to be stationary for problem (1.1), (1.2), (1.4), ifPtµ=µfor anyt ≥0.

Using the classical Bogolyubov-Krylov argument and Fatou’s lemma, one can prove the following theorem. Its proof is outlined in the Appendix.

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THEOREM 3.7. LetEHk) <,then problem (1.1), (1.2), (1.4) has at least one stationary measure. Moreover, ifEHk)p <for somep≥1,then for any stationary measureµwe have:

Hp(µ):=

H

H(u)pµ(du) <+∞.

We denote byP1(H)the set of measuresµP(H)such thatH(µ):=H1(µ) <+∞. 4. Main result

To show the uniqueness of stationary measure for (1.1), (1.2), (1.4), we shall need the following condition satisfied forηk:

Condition 4.1.The random variablesηkare i.i.d. and have the form ηk=

j=1

bjξjkgj(x),

where{gi}iNis an orthonormal basis inH,bj≥0are some constants with B:=

j=1

b2j <,

andξjk are independent scalar random variables. Moreover, the distribution ofξjk pos- sesses a densitypj(r)(with respect to the Lebesgue measure), which is a function of bounded variation such that

ε

ε

pj(r)dr >0,

+∞

−∞

|r|ppj(r)drCp<, (4.1)

for allε >0,p≥1,j≥1and for some constantsCp>0.

Clearly, if Condition 4.1 is satisfied, then

Eηkp1 <∞ for allk≥1,p≥1. (4.2) THEOREM 4.2. Suppose that Condition 4.1 is satisfied. For anyB > 0 there is an integerN≥1such that, if

ejHN, j=1, ..., N (4.3)

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for someN≥1, and

bj =0, j=1, ..., N, (4.4)

then there is a unique stationary measureµP(H).Moreover, for any initial measure µP1(H)we have

PtµµLCp(1+H(µ))tp, t >0, (4.5) whereCpis a constant not depending onµ.

Proof. Step 1.It suffices to show that for anyu, uHwe have

|Ptf(u)−Ptf(u)| ≤CpfL(1+H(u)+H(u))tp, (4.6) for anyp≥1, t >0 and some constantCp>0 not depending on(u, u)andt.

Indeed, suppose that (4.6) is already proved. Then for any two initial measuresµ, µP1(H)we derive from (4.6):

Ptµ−PtµLCp(1+H(µ)+H(µ))tp. (4.7) This inequality shows the uniqueness of stationary measure inP1(H). It follows from Theorem 3.7 that any stationary measureµis inP1(H). Takingµ=µin (4.7), we arrive at (4.5).

Step 2.Inequality (4.6) is a direct consequence of the following proposition.

PROPOSITION 4.3. Under the conditions of Theorem 4.2, for anyB > 0there is an integerN ≥ 1 such that, if (4.3) and (4.4) hold for some integer N ≥ 1, then there is a probability space(,F,P)and a sequence of i.i.d. random variables{tk}that are exponentially distributed with parameterλsuch that for anyu, uHone can construct random sequencesuk, ukdefined onwith the following properties:

(i) The initial value of the trajectory(uk, uk)is(u, u):

u0=u, u0=u.

Furthermore, the random variablesζkandζkdefined by (2.6) are i.i.d., and their distribution coincides with that ofηk:

D(ζk)=D(ζk)=D(ηk).

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(ii) There is a random integer=(u, u)and a constantMdepending only onBsuch that

PNuk=PNuk fork+1, (4.8)

QNζk=QNζk fork≥1, (4.9)

ui61+ ui61kM fork+1, (4.10) 1

2 1 k k

i=+1

logtiM fork+1. (4.11)

(iii) There is a positive constantCpnot depending on(u, u)such that

EpCp(1+H(u)+H(u)) for allp≥1, (4.12)

u1u1≤1. (4.13)

To prove (4.6), letuk anduk be the random sequences constructed in Proposition 4.3 and corresponding to the initial value(u, u).Letτk=k

n=1tn, n≥1 andτ0=0.Define ut=

Stτk(uk), ift∈[τk, τk+1), k≥0, Stk+1(uk)+ζk+1, ift=τk+1,

andut is defined in a similar way. Clearly,ut andut have the same distributions as the solutions of (1.1)–(1.4) corresponding touandu, respectively. Thus

|Ptf(u)−Ptf(u)| = |E(f(ut)f(ut))|. (4.14) Let

Nt=max{k≥0 :τkt}, (4.15)

thenNtis a Poisson random variable with parameterλt(e.g., see [13]). DefineGt = {ω: 2+1≤ Nt} = {ω: τ2+1t}. Asτkis a Gamma random variable with parametersλ andk(e.g., see [5]), we have

Eτ2q+1

n=1

E[τ2nq+1I{=n}]≤

n=1

(Eτ2n2q+1)12P{=n}12

C(1+H(u)+H(u)) n=1

1

n2 <, (4.16)

for anyq≥1, where we used the Cauchy–Schwarz inequality and (4.12) withp=2(2+q).

It follows that

P(Gct)Cp(1+H(u)+H(u))tp for anyp≥1. (4.17)

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Using (2.3), we see that

utut1Cexp(C(uτNt61+ uτNt61))uτNtuτNt1, whence, using (4.8)–(4.11), (4.13) and Lemma 2.1, we obtain

E[IGtutut1]≤E[2(CαN21+1)Nte2CM(Nt)].

ChoosingNso large that logαN+1≥2(2CM+logC+2), we arrive at

E[IGtutut1]≤Ee−Nt =ect, (4.18) wherec=λλe. LetfL(H).Then, using (4.14), (4.17) and (4.18), we derive

|Ptf(u)−Ptf(u)| ≤E|f(ut)f(ut)|

fLE[IGtutut1]+EIGc

t2f

fLect+2fLCp(1+H(u)+H(u))tp

CpfL(1+H(u)+H(u))tp. (4.19)

This completes the proof of (4.6).

REMARK 4.4. 1. The embedded Markov chainuτk also satisfies a property of polyno- mial mixing. This follows from Proposition 4.3 and is proved using the same arguments as in the proof of Theorem 4.2. The stationary measures of the original process and that of embedded chain are connected with the Khasminskii relation:

(f, µ)= 1 Eντ1Eν

τ1

0

f(ut)dt,

whereνandµare the stationary measures ofuτk andutrespectively.

2. The integerN in Theorem 4.2 depends on the viscosityν. In the case of the 2D Navier–Stokes equations on the torus, controllability and exponential mixing property hold under condition (4.4) withN not depending onν(see [1, 23, 10]). It would be interesting to establish similar results in the setting of the present paper.

3. The proof of Theorem 4.2. can be generalized to the case of external forcing of the form

η(t, x)=h(x)+

k=1

ηk(x)δ(tτk), (4.20)

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wherehHis a deterministic function. In this case, one needs to replace the first inequality in (4.1) by the following one:

a+ε

aε

pj(r)dr >0 for alla∈R,ε >0 andj≥1. (4.21)

The proofs of all the results, except that of Lemma 5.4, remain literally the same. The proof of Lemma 5.4 in the general case is given at the end of Section 5.

5. Coupling operators

Letηkbe a sequence of random variables with range inHand suppose that Condition 4.1 is satisfied forηk.Clearly, ifbj =0,j=1, ..., N,then the distribution of the random variablePN1)is absolutely continuous with respect to the Lebesgue measure, and its density has the form

p(x):=

N j=1

qj(xj), qj(xj)=bj1pj(xjbj1), x=(x1, ..., xN)HN.

Now we have the following lemma, which is a version of Lemma 3.2 in [18]:

LEMMA 5.1. Suppose that Condition 4.1 is satisfied andbj = 0 for j = 1, ..., N, whereN ≥ 1is an integer. Then there is a probability space(,F,P)such that for any u, uH there areH-valued random variablesζ = ζ(u, u, ω),ζ =ζ(u, u, ω)and a real-valued random variablet=t(ω)with the following properties:

(i) The random variablesζ, ζandη1have the same distributions, andtis exponentially distributed with parameterλ.

(ii) The random variables(PNζ, PNζ)and(QNζ, QNζ)are independent, andζ and ζare independent oft.

(iii) The random variablesQNζandQNζare equal for allωand do not depend on(u, u).

(iv) The random variablesζandζare measurable functions of(u, u, ω)H×H×. Proof. Suppose thatt1=t11)is a random variable that is exponentially distributed with parameterλand is defined on the space(1,F1,P1). Let(v, v)be a maximal coupling for u,ω1, νu1), whereνu,ω1 is a measure onHNgiven by the densityp(xPNSt11)(u)) (see [20], Section I, 5). By Theorem 4.2 in [18], we can assume that the random variables vandvare defined on the same probability space(2,F2,P2)for allu, uH,ω11 and are measurable functions of(u, u, ω1, ω2)H ×H ×1 ×2. Suppose that

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η1is defined on the space (3,F3,P3). We denote by(,F,P)the direct product of (i,Fi,Pi),i=1,2,3,and defineζ,ζandtby the relations:

t(ω)=t11),

PNζ(ω)=v(u, u, ω1, ω2)PNSt(ω1)(u), PNζ(ω)=v(u, u, ω1, ω2)PNSt(ω1)(u), QNζ(ω)=QNζ(ω)=QNη13),

whereω=1, ω2, ω3). Using the definition ofζand Fubini’s theorem, we see that P{PNζ} =EI{PNζ}=E1P2{PNζ(ω1)}

=E1P2{vPNSt1(u)} =

p(x)dx=P{PNη1}, (5.1)

for anyB(HN), whereE1is the expectation corresponding to the measure P1. All assertions of lemma follow from the construction and relation (5.1).

REMARK 5.2. Using inequality (3.8) in Lemma 3.2, [18] for the variational distance betweenνu,ω1 andνu1, we obtain the inequality:

νu,ω1νu1varCNSt(u)St(u)1, which holdsP1-a.s.. Then the definition of maximal coupling gives

P2{v=v} ≤CNSt(u)St(u)1. (5.2) REMARK 5.3. Let(,F,P)be the direct product of(i,Fi,Pi), i=2,3.For any ω11, let Eω1 = {ω : v(u, u, ω1, ω) = v(u, u, ω1, ω)}. As (v, v)is a maximal coupling foru,ω1, νu1), we have

P{v(u, u, ω1,·), v(u, u, ω1,·)|Eω1}

=P{v(u, u, ω1,·)|Eω1}P{v(u, u, ω1,·)|Eω1}, ifP{Eω1}>0 and, B(H).Now it is easy to notice that

P{vω1, vω

1, Eω1} ≥P{vω1, Eω1}P{vω

1, Eω1}. (5.3) Let us define coupling operators by the formulas

R(u, u, ω)=St(ω)(u)+ζ(u, u, ω), R(u, u, ω)=St(ω)(u)+ζ(u, u, ω), whereu, uHandω.

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