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Quadratic Transportation Cost Inequality For Scalar

Stochastic Conservation Laws

Rangrang Zhang1, Tusheng Zhang2

1 School of Mathematics and Statistics, Beijing Institute of Technology, Beijing 100081, China2 Department of Mathematics, University of Manchester, Oxford Road, Manchester M13 9PL, United Kingdom

(rrzhang@amss.ac.cn, tusheng.zhang@manchester.ac.uk)

Abstract: In this paper, we established a quadratic transportation cost inequality for scalar stochastic

conservation laws driven by multiplicative noise. The doubling variables method plays an important

role.

AMS Subject Classification: Primary 60F10; Secondary 60H15, 60G40.

Keywords: Quadratic transportation cost inequality; conservation laws; Kinetic solutions.

1 Introduction

Fix T > 0 and let (Ω,F , P, Ftt∈[0,T ], (βk(t)t∈[0,T ])k∈N) be a stochastic basis. Without loss of generality,

here the filtration Ftt∈[0,T ] is assumed to be complete and βk(t)t∈[0,T ](k ∈ N) are one-dimensional i.i.d.

real-valued Ftt∈[0,T ]−Wiener processes. The symbol E denotes the expectation with respect to P. For

fixed N ∈ N, let TN ⊂ RN be the N−dimensional torus with the periodic length to be 1. Consider the

following Cauchy problem for the scalar conservation laws with stochastic forcing

du(t, x) + divA(u(t, x))dt =

∑k≥1 gk(x, u(t, x))dβk(t) in TN × (0, T ],

u(·, 0) = η(·) on TN .(1.1)

where u : (ω, x, t) ∈ Ω × TN × [0, T ] 7→ u(ω, x, t) := u(x, t) ∈ R is a random field, the flux function

A : R → RN and the coefficient gk(·, ·) are measurable and fulfill certain conditions (see Section 2 in

below). Moreover, the initial value η ∈ L∞(TN) is a deterministic function.

The purpose of this paper is to establish the quadratic transportation cost inequality for the solution

of the stochastic conservation laws. Let us recall the relevant concepts. Let (X, d) be a metric space

equipped with the Borel σ-field B. Let µ, ν be two Borel probability measures on the metric space (X, d).

The Lp-Wasserstein distance between µ and ν is defined as

Wp(ν, µ) :=

[inf

"X×X

d(x, y)p π(dx, dy)

] 1p

,

1

where the infimum is taken over all probability measures π on the product space X × X with marginals µ

and ν. Recall that the Kullback information (or relative entropy) of ν with respect to µ is defined by

H(ν|µ) :=

∫

X

log

(dν

dµ

)dν,

if ν is absolutely continuous with respect to µ, and +∞ otherwise.

Definition 1.1. We say that a measure µ satisfies the Lp-transportation cost inequality if there exists a

constant C > 0 such that for all probability measures ν,

Wp(ν, µ) ≤√

2CH(ν|µ). (1.2)

The case p = 2 is referred to as the quadratic transportation cost inequality.

Transportation cost inequalities have close connections with other functional inequalities, e.g.

Poincare inequalities, logarithmic Sobolev inequalities, and they also imply the concentration of measure

phenomenon.

For a measurable subset A ⊂ X and r > 0, we denote by Ar the r-neighborhood of A, namely

Ar = x : d(x, A) < r. We say that µ has normal concentration (or Gaussian tail estimates) on (X, d) if

there are constants C, c > 0 such that for every r > 0 and every Borel subset A with µ(A) ≥ 12 ,

1 − µ(Ar) ≤ Ce−cr2. (1.3)

The fact that the L1-transportation cost inequality implies normal concentration was obtained in [M1,

M2] by Marton and in [T1, T2, T3] by Talagrand. An elegant, simple proof of this fact is also contained

in the book [Le]. The connection of the quadratic transportation cost inequality with other functional

inequalities was studied in [OV] by Otto and Villani (see also [Le]). For other related interesting works,

we refer the reader to [GRS], [LW], [PS], [PS1].

In the past decades, many people established quadratic transportation cost inequalities for various

kinds of interesting measures. Let us mention several papers which are relevant to our work. The

transportation cost inequalities for stochastic differential equations were obtained by H. Djellout, A.

Guillin and L. Wu in [DGW]. The measure concentration for multidimensional diffusion processes

with reflecting boundary conditions was considered by S. Pal in [P]. The quadratic transportation cost

inequalities for stochastic partial differential equations (SPDEs) driven by Gaussian noise which is white

in time and colored in space were obtained by A. S. Ustunel in [U]. In the article [BH], the authors

obtained the quadratic transportation cost inequality under the L2-distance for stochastic heat equations.

In [KS], the authors established the quadratic transportation cost inequality for more general stochastic

partial differential equations (SPDEs) under the L2-distance and under the uniform distance for the case

of additive noise. In [SZ], the authors obtained the quadratic transportation cost inequality for stochastic

heat equations equations driven by multiplicative space-time white noise under the uniform distance.

2

On the other hand, both deterministic (gk = 0) and stochastic conservation laws have been studied

extensively by many people. Conservation law is fundamental to our understanding of the space-time

evolution laws of interesting physical quantities. For more background on this model, we refer the readers

to the monograph [Da], the work of Ammar, Wittbold and Carrillo [AWC] and references therein. As we

know, the Cauchy problem for the deterministic first-order PDE (1.1) does not admit any (global) smooth

solutions, but there exist infinitely many weak solutions to the deterministic Cauchy problem. To solve

the problem of non-uniqueness, an additional entropy condition was added to identify the physical weak

solution. Under this condition, the notion of entropy solutions for the deterministic first-order scalar

conservation laws was introduced by Kružkov [Kr-1, Kr-2]. The kinetic formulation of weak entropy

solution of the Cauchy problem for a general multi-dimensional scalar conservation laws (also called the

kinetic system), was derived by Lions, Perthame and Tadmor in [LPT].

In recent years, the stochastic conservation law has been developed rapidly. We refer the reader

to the references [K], [VW],[FN], [DWZZ] etc. We particularly mention the paper [DV-1] in which

the authors proved the existence and uniqueness of kinetic solution to the Cauchy problem for (1.1) in

any dimension. In addition, the long-time behavior of the first-order scalar conservation laws has been

studied in the paper [DV-2]. Recently, combining techniques used in the context of kinetic solutions as

well as new results on large deviations, Dong et al. [DWZZ] established Freidlin-Wentzell’s type large

deviation principles (LDP) for the kinetic solution to the scalar stochastic conservative laws.

The purpose of this paper is to establish the quadratic transportation cost inequality for the kinetic

solution of the scalar stochastic conservation laws, which in particular implies the concentration phe-

nomenon of the law of the solution. To our knowledge, the present paper is the first work towards

proving the transportation cost inequality directly for the kinetic solutions to the scalar stochastic conser-

vation laws. Due to the lack of viscous term, the kinetic solutions of (1.1) are living in a rather irregular

space L1([0, T ], L1(TN)), we will use the doubling variables method as in the work [DV-1]. Differ from

[DV-1], we need to deal with the martingale term carefully to derive a proper bound, which can ensure

the application of Gronwall inequality to get an appropriate norm estimation (see (3.45)). As an impor-

tant part of the proof, we also need to make some higher order estimates of the error term than [DV-1],

which is nontrivial and completely new.

This paper is organized as follows. In Section 2, we lay out the precise setup for the stochastic con-

servation law and recall some of the known results. Section 3 is devoted to the proof of the transportation

cost inequality.

2 Framework

In this section, we will lay out the precise setup for the stochastic conservation law and recall some

results which will be used later.

3

2.1 Kinetic solution

We will follow closely the framework of [DV-1]. Let ‖ · ‖Lp denote the norm of usual Lebesgue space

Lp(TN) for p ∈ [1,∞]. In particular, set H = L2(TN) with the corresponding norm ‖ · ‖H . Cb represents

the space of bounded, continuous functions and C1b

stands for the space of bounded, continuously differ-

entiable functions having bounded first order derivative. Define the function f (x, t, ξ) := Iu(x,t)>ξ , which

is the characteristic function of the subgraph of u. We write f := Iu>ξ for short. Moreover, denote by

the brackets 〈·, ·〉 the duality between C∞c (TN × R) and the space of distributions over TN × R. In what

follows, with a slight abuse of the notation 〈·, ·〉, we denote the following integral by

〈F,G〉 :=

∫

TN

∫

R

F(x, ξ)G(x, ξ)dxdξ, F ∈ Lp(TN × R),G ∈ Lq(TN × R),

where 1 ≤ p ≤ +∞, q := p

p−1 is the conjugate exponent of p. In particular, when p = 1, we set q = ∞

by convention. For a measure m on the Borel measurable space TN × [0, T ] × R, the shorthand m(φ) is

defined by

m(φ) := 〈m, φ〉([0, T ]) :=

∫

TN×[0,T ]×Rφ(x, t, ξ)dm(x, t, ξ), φ ∈ Cb(TN × [0, T ] × R).

In the sequel, the notation a . b for a, b ∈ R means that a ≤ Db for some constant D > 0 independent

of any parameters.

2.2 Hypotheses

For the flux function A and the coefficients of (1.1), we assume that

Hypothesis H The flux function A belongs to C2(R;RN) and its derivative a := A′ is of polynomial

growth with degree q0 > 1. That is, there exists a constant C(q0) ≥ 0 such that

|a(ξ)| ≤ C(q0)(1 + |ξ|q0), |a(ξ) − a(ζ)| ≤ Υ(ξ, ζ)|ξ − ζ |, (2.1)

where Υ(ξ, ζ) := C(q0)(1 + |ξ|q0−1+ |ζ |q0−1).

Moreover, we assume that gk ∈ C(TN × R) satisfies the following bounds

|gk(x, u)| ≤ C0k ,

∑

k≥1

|C0k |

2 ≤ D0, (2.2)

|gk(x, u) − gk(y, v)| ≤ C1k (|x − y| + |u − v|),

∑

k≥1

|C1k |

2 ≤D1

2, (2.3)

for x, y ∈ TN, u, v ∈ R, where C0k,C1

k,D0,D1 are positive constants.

The hypothesis H implies that

G2(x, u) :=∑

k≥1

|gk(x, u)|2 ≤ D0, (2.4)

∑

k≥1

|gk(x, u) − gk(y, v)|2 ≤ D1

(|x − y|2 + |u − v|2

). (2.5)

4

2.3 Kinetic solution

Let us recall the notion of a kinetic solution to equation (1.1) from [DV-1].

Definition 2.1. (Kinetic measure) A map m from Ω to the set of non-negative, finite measures over

TN × [0, T ] × R is said to be a kinetic measure, if

1. m is measurable, that is, for each φ ∈ Cb(TN × [0, T ] × R), 〈m, φ〉 : Ω→ R is measurable,

2. m vanishes for large ξ, i.e.,

limR→+∞

E[m(TN × [0, T ] × BcR)] = 0, (2.6)

where BcR

:= ξ ∈ R, |ξ| ≥ R,

3. for every φ ∈ Cb(TN × R), the process

(ω, t) ∈ Ω × [0, T ] 7→ 〈m, φ〉([0, t]) :=

∫

TN×[0,t]×Rφ(x, ξ)dm(x, s, ξ) ∈ R

is predictable.

Remark 1. For any φ ∈ Cb(TN × R) and kinetic measure m, define At := 〈m, φ〉([0, t]), then a.s., t 7→ At

is a right continuous function of finite variation. Moreover, the function A has left limits at any point

t ∈ (0, T ]. We write At− = lims↑t As and set A0− = 0. As a result, At− = 〈m, φ〉([0, t)), which is càglàd (left

continuous with right limits).

Definition 2.2. (Kinetic solution) Let η ∈ L∞(TN). A measurable function u : TN × [0, T ] × Ω → R is

called a kinetic solution to (1.1) with initial datum η, if

1. (u(t))t∈[0,T ] is predictable,

2. for any p ≥ 1, there exists Cp ≥ 0 such that

E

(ess sup

0≤t≤T

‖u(t)‖pLp(TN )

)≤ Cp, (2.7)

3. there exists a kinetic measure m such that f := Iu>ξ satisfies: for all ϕ ∈ C1c (TN × [0, T ) × R),

∫ T

0〈 f (t), ∂tϕ(t)〉dt + 〈 f0, ϕ(0)〉 +

∫ T

0〈 f (t), a(ξ) · ∇ϕ(t)〉dt (2.8)

= −∑

k≥1

∫ T

0

∫

TN

gk(x, u(t, x))ϕ(x, t, u(x, t))dxdβk (t)

−1

2

∫ T

0

∫

TN

∂ξϕ(x, t, u(x, t))G2(x, u(t, x))dxdt + m(∂ξϕ), a.s.,

where f0 = Iη>ξ , u(t) = u(·, t, ·) and G2=

∑∞k=1 |gk |

2.

5

Let (X, λ) be a finite measure space. For some measurable function u : X → R, define f : X × R →

[0, 1] by f (z, ξ) = Iu(z)>ξ a.e. we use f := 1 − f to denote its conjugate function. Define Λ f (z, ξ) :=

f (z, ξ) − I0>ξ , which can be viewed as a correction to f . Note that Λ f is integrable on X × R if u is.

It is shown in [DV-1] that almost surely, for each kinetic solution u, the function f = Iu(x,t)>ξ admits

left and right weak limits at any point t ∈ [0, T ], and the weak form (2.8) satisfied by a kinetic solution

can be strengthened to be weak only respect to x and ξ. More precisely, the following results are obtained.

Proposition 2.1. ([DV-1], Left and right weak limits) Let u be a kinetic solution to (1.1) with initial value

η. Then f = Iu(x,t)>ξ admits, almost surely, left and right limits respectively at every point t ∈ [0, T ]. More

precisely, for any t ∈ [0, T ], there exist kinetic functions f t± on Ω × TN × R such that P−a.s.

〈 f (t − r), ϕ〉 → 〈 f t−, ϕ〉 (2.9)

and

〈 f (t + r), ϕ〉 → 〈 f t+, ϕ〉 (2.10)

as r → 0 for all ϕ ∈ C1c (TN × R). Moreover, almost surely,

〈 f t+ − f t−, ϕ〉 = −

∫

TN×[0,T ]×R∂ξϕ(x, ξ)It(s)dm(x, s, ξ).

In particular, almost surely, the set of t ∈ [0, T ] fulfilling f t+, f t− is countable.

For the function f = Iu(x,t)>ξ in Proposition 2.1, define f ± by f ±(t) = f t±, t ∈ [0, T ]. Since we are

dealing with the filtration associated to Brownian motion, both f + and f − are clearly predictable as well.

Also f = f + = f − almost everywhere in time and we can take any of them in an integral with respect to

the Lebesgue measure or in a stochastic integral. However, if the integral is with respect to a measure,

typically a kinetic measure in this article, the integral is not well-defined for f and may differ if one

chooses f + or f −.

The following result was proved in [DV-1].

Lemma 2.1. The weak form (2.8) satisfied by f = Iu>ξ can be strengthened to be weak only respect to x

and ξ. Concretely, for all t ∈ [0, T ) and ϕ ∈ C1c (TN × R), f = Iu>ξ satisfies

〈 f +(t), ϕ〉 = 〈 f0, ϕ〉 +

∫ t

0〈 f (s), a(ξ) · ∇ϕ〉ds

+

∑

k≥1

∫ t

0

∫

TN

∫

R

gk(x, ξ)ϕ(x, ξ)dνx,s(ξ)dxdβk(s)

+1

2

∫ t

0

∫

TN

∫

R

∂ξϕ(x, ξ)G2(x, ξ)dνx,s(ξ)dxds − 〈m, ∂ξϕ〉([0, t]), a.s., (2.11)

and we set f +(T ) = f (T ).

6

Where νx,s(ξ) = −∂ξ f (x, s, ξ) = δu(x,s)=ξ .

Remark 2. By making modification of the proof of Lemma 2.1, we have for all t ∈ (0, T ] and ϕ ∈

C1c (TN × R), f = Iu>ξ satisfies

〈 f −(t), ϕ〉 = 〈 f0, ϕ〉 +

∫ t

0〈 f (s), a(ξ) · ∇ϕ〉ds

+

∑

k≥1

∫ t

0

∫

TN

∫

R

gk(x, ξ)ϕ(x, ξ)dνx,s(ξ)dxdβk(s)

+1

2

∫ t

0

∫

TN

∫

R

∂ξϕ(x, ξ)G2(x, ξ)dνx,s(ξ)dxds − 〈m, ∂ξϕ〉([0, t)), a.s., (2.12)

and we set f −(0) = f0.

The following well-posedness of (1.1) was established in [DV-1].

Theorem 2.2. ([DV-1], Existence and Uniqueness) Let η ∈ L∞(TN). Assume Hypothesis H holds, then

there is a unique kinetic solution u to equation (1.1) with initial datum η.

3 Transportation cost inequality

Let µ be the law of the random field solution u(·, ·) of SPDE (1.1), viewed as a probability measure on

L1([0, T ], L1(TN)). First we state a lemma which is essentially proved in [KS] describing the probability

measures ν that are absolutely continuous with respect to µ.

Let ν ≪ µ on L1([0, T ], L1(TN)). Define a new probability measure Q on the filtered probability

space (Ω,F , Ft0≤t≤T , P) by

dQ :=dν

dµ(u) dP. (3.13)

Denote the Radon-Nikodym derivative restricted on Ft by

Mt :=dQ

dP

∣∣∣∣∣Ft

, t ∈ [0, T ].

Then Mt, t ∈ [0, T ] forms a P-martingale. A variant of the following result was proved in [KS].

Lemma 3.1. There exists an adapted stochastic process h = h(s) = (h1(s), h2(s), · · ·, ) ∈ l2, s ≥ 0 such

that Q − a.s. for all t ∈ [0, T ],∫ t

0|h|2

l2(s) ds < ∞

and βk : [0, T ]→ R defined by

βk(t) := βk(t) −

∫ t

0hk(s) ds, (3.14)

7

are independent Brownian motions under the measure Q. Moreover,

Mt = exp

∞∑

k=1

∫ t

0hk(s) dβk(s) −

1

2

∫ t

0|h|2

l2(s) ds

, Q − a.s., (3.15)

and

H(ν|µ) =1

2EQ

[∫ T

0|h|2

l2(s) ds

], (3.16)

where EQ stands for the expectation under the measure Q.

Theorem 3.1. Let η ∈ L∞(TN). Assume Hypothesis H holds. Then the law µ of the solution of the

stochastic conservation law (1.1) satisfies the quadratic transportation cost inequality on the space

L1([0, T ], L1(TN)).

Proof. Take ν ≪ µ on L1([0, T ], L1(TN)). Let Q be the probability measure defined as in(3.13). Let

h(t) be the corresponding stochastic process appeared in Lemma 3.1. Then, by the Girsanov theorem

the solution u(t) of equation (1.1) satisfies the following stochastic partial differential equation (SPDE)

under the measure Q,

duh(t, x) + divA(uh(t, x))dt =∑

k≥1 gk(x, uh(t, x))dβk(t) +∑

k≥1 gk(x, uh(t, x))hk(t)dt in TN × (0, T ],

uh(·, 0) = η(·) on TN .(3.17)

Similar to Lemma 2.1, we can show that the kinetic solution uh satisfies that for any p ≥ 1, there exists

Cp ≥ 0 such that

EQ(ess sup

0≤t≤T

‖uh(t)‖pLp(TN )

)≤ Cp, (3.18)

and there exists a kinetic measure mh such that for all t ∈ [0, T ) and ϕ ∈ C1c (TN × R), f := Iuh>ξ satisfies

〈 f +(t), ϕ〉 = 〈 f0, ϕ〉 +

∫ t

0〈 f (s), a(ξ) · ∇ϕ〉ds

+

∑

k≥1

∫ t

0

∫

TN

∫

R

gk(x, ξ)ϕ(x, ξ)dνhx,s(ξ)dxdβk(s)

+

∑

k≥1

∫ t

0

∫

TN

∫

R

gk(x, ξ)ϕ(x, ξ)hk(s)dνhx,s(ξ)dxds

+1

2

∫ t

0

∫

TN

∫

R

∂ξϕ(x, ξ)G2(x, ξ)dνhx,s(ξ)dxds − 〈mh, ∂ξϕ〉([0, t]), a.s., (3.19)

where νhx,s(ξ) = −∂ξ f (x, s, ξ) = δuh(x,s)=ξ and we set f +(T ) = f (T ).

Consider the solution of the following SPDE:

du(t, x) + divA(u(t, x))dt =∑

k≥1 gk(x, u(t, x))dβk(t) in TN × (0, T ],

u(·, 0) = η(·) on TN .(3.20)

8

By Lemma 3.1, it follows that under the measure Q, the law of (u, uh) forms a coupling of (µ, ν). There-

fore by the definition of the Wasserstein distance,

W2(ν, µ)2 ≤ EQ

∣∣∣∣∣∣

∫ T

0

∫

TN

|u(t, x) − uh(t, x)|dtdx

∣∣∣∣∣∣

2 .

In view of (3.16), to prove the quadratic transportation cost inequality

W2(ν, µ) ≤√

2CH(ν|µ), (3.21)

it is sufficient to show that

EQ

∣∣∣∣∣∣

∫ T

0

∫

TN

|u(t, x) − uh(t, x)|dtdx

∣∣∣∣∣∣

2 ≤ CEQ[∫ T

0|h|2

l2(s) ds

](3.22)

when the right hand side of (3.22) is finite.

For simplicity, in the sequel we still denote EQ by the symbol E and denote βk by βk. The proof of

(3.22) is technical and lengthy. It is divided into the following two propositions.

Following the idea of the proof Proposition 13 in [DV-1] and using the doubling variables method,

we have the following result relating the two kinetic solution u and uh. As the proof is very similar to

that of Proposition 13 in [DV-1], we omit the proof and refer the reader to [DV-1].

Proposition 3.2. Assume Hypothesis H is in place. Let u and uh be the kinetic solution of (3.20) and

(3.17) , respectively. Then, for all 0 < t < T, and non-negative test functions ρ ∈ C∞(TN), ψ ∈ C∞c (R),

the corresponding functions f1(x, t, ξ) := Iuh(x,t)>ξ and f2(y, t, ζ) := Iu(y,t)>ζ satisfy the following

∫

(TN )2

∫

R2ρ(x − y)ψ(ξ − ζ)( f ±1 (x, t, ξ) f ±2 (y, t, ζ) + f ±1 (x, t, ξ) f ±2 (y, t, ζ))dξdζdxdy

≤

∫

(TN )2

∫

R2ρ(x − y)ψ(ξ − ζ)( f1,0(x, ξ) f2,0(y, ζ) + f1,0(x, ξ) f2,0(y, ζ))dξdζdxdy

+I(t) + J(t) + K(t) + H(t), a.s., (3.23)

where

I(t) =

∫ t

0

∫

(TN )2

∫

R2( f1 f2 + f1 f2)(a(ξ) − a(ζ)) · ∇xαdξdζdxdyds,

J(t) =

∫ t

0

∫

(TN )2

∫

R2α∑

k≥1

|gk(x, ξ) − gk(y, ζ)|2dν1x,s ⊗ ν

2y,s(ξ, ζ)dxdyds,

K(t) = 2∑

k≥1

∫ t

0

∫

(TN )2

∫

R2(gk(x, ξ) − gk(y, ζ))ρ(x − y)χ(ξ, ζ)dν1

x,s ⊗ ν2y,s(ξ, ζ)dxdydβk(s),

H(t) = 2∑

k≥1

∫ t

0

∫

(TN )2

∫

R2gk(x, ξ)hk(s)ρ(x − y)χ(ξ, ζ)dν1

x,s ⊗ ν2y,s(ξ, ζ)dxdyds,

with f1,0(x, ξ) = Iη(x)>ξ , f2,0(y, ζ) = Iη(y)>ζ , α = ρ(x − y)ψ(ξ − ζ), ν1x,s = −∂ξ f1(s, x, ξ) = δuh(x,s)=ξ , ν

2y,s =

∂ζ f2(s, y, ζ) = δu(y,s)=ζ and χ(ξ, ζ) =∫ ξ

−∞ψ(ξ′ − ζ)dξ′ =

∫ ξ−ζ

−∞ψ(y)dy.

9

The statement (3.22) is contained in the next proposition.

Proposition 3.3. For T > 0, it holds that

E

∣∣∣∣∫ T

0‖uh(t) − u(t)‖L1(TN )dt

∣∣∣∣2≤ CE

[ ∫ T

0|h|2

l2(t)dt

], (3.24)

where C = C(T,D0,D1).

Proof. Let ργ, ψδ be approximations to the identity on TN and R, respectively. That is, let ρ ∈ C∞(TN),

ψ ∈ C∞c (R) be symmetric non-negative functions such as∫TN ρ = 1,

∫Rψ = 1 and suppψ ⊂ (−1, 1). We

define

ργ(x) =1

γNρ( x

γ

), ψδ(ξ) =

1

δψ(ξδ

).

Letting ρ := ργ(x − y) and ψ := ψδ(ξ − ζ) in Proposition 3.2, we get from (3.23) that

∫

(TN )2

∫

R2ργ(x − y)ψδ(ξ − ζ)( f ±1 (x, t, ξ) f ±2 (y, t, ζ) + f ±1 (x, t, ξ) f ±2 (y, t, ζ))dξdζdxdy

≤

∫

(TN )2

∫

R2ργ(x − y)ψδ(ξ − ζ)( f1,0(x, ξ) f2,0(y, ζ) + f1,0(x, ξ) f2,0(y, ζ))dξdζdxdy

+I(t) + J(t) + K(t) + H(t), a.s., (3.25)

where I, J, K, H are the corresponding terms I, J,K,H in the statement of Proposition 3.2 with ρ, ψ

replaced by ργ, ψδ, respectively. For simplicity, we still use the notation:

χ(ξ, ζ) =

∫ ξ−ζ

−∞

ψδ(y)dy.

With an eye on the following identity,∫

R

Iuh,±>ξIu±>ξdξ = (uh,± − u±)+,

∫

R

Iuh,±>ξIu±>ξdξ = (uh,± − u±)−, (3.26)

we will start with the estimate (3.25) and eventually let γ, δ appropriately tend to zero to prove the

proposition. For any t ∈ [0, T ], define the error term

Et(γ, δ)

:=

∫

(TN )2

∫

R2( f ±1 (x, t, ξ) f ±2 (y, t, ζ) + f ±1 (x, t, ξ) f ±2 (y, t, ζ))ργ(x − y)ψδ(ξ − ζ)dxdydξdζ

−

∫

TN

∫

R

( f ±1 (x, t, ξ) f ±2 (x, t, ξ) + f ±1 (x, t, ξ) f ±2 (x, t, ξ))dξdx. (3.27)

10

Using∫Rψδ(ξ − ζ)dζ = 1,

∫ ξ

ξ−δψδ(ξ − ζ)dζ = 1

2 and∫

(TN )2 ργ(x − y)dxdy ≤ 1, we find that

∣∣∣∣∫

(TN )2

∫

R

ργ(x − y) f ±1 (x, t, ξ) f ±2 (y, t, ξ)dξdxdy

−

∫

(TN )2

∫

R2f ±1 (x, t, ξ) f ±2 (y, t, ζ)ργ(x − y)ψδ(ξ − ζ)dxdydξdζ

∣∣∣∣

=

∣∣∣∣∫

(TN )2ργ(x − y)

∫

R

Iuh,±(x,t)>ξ

∫

R

ψδ(ξ − ζ)(Iu±(y,t)≤ξ − Iu±(y,t)≤ζ)dζdξdxdy

∣∣∣∣

≤

∫

(TN )2

∫

R

ργ(x − y)Iuh,±(x,t)>ξ

∫ ξ

ξ−δ

ψδ(ξ − ζ)Iζ<u±(y,t)≤ξdζdξdxdy

+

∫

(TN )2

∫

R

ργ(x − y)Iuh,±(x,t)>ξ

∫ ξ+δ

ξ

ψδ(ξ − ζ)Iξ<u±(y,t)≤ζdζdξdxdy

≤1

2

∫

(TN )2ργ(x − y)Iuh,±(x,t)>u±(y,t)

∫ minuh,±(x,t),u±(y,t)+δ

u±(y,t)dξdxdy

+1

2

∫

(TN )2ργ(x − y)Iu±(y,t)−δ<uh,±(x,t)

∫ minuh,±(x,t),u±(y,t)

u±(y,t)−δdξdxdy

=δ

2

∫

(TN )2ργ(x − y)Iuh,±(x,t)>u±(y,t)+δdxdy

+1

2

∫

(TN )2ργ(x − y)Iu±(y,t)<uh,±(x,t)≤u±(y,t)+δ(u

h,±(x, t) − u±(y, t)dxdy

+δ

2

∫

(TN )2ργ(x − y)Iu±(y,t)<uh,±(x,t)dxdy

+1

2

∫

(TN )2ργ(x − y)Iu±(y,t)−δ<uh,±(x,t)≤u±(y,t)(u

h,±(x, t) − u±(y, t) + δ)dxdy

≤ 2δ, a.s.. (3.28)

Similarly, we have∣∣∣∣∫

(TN )2

∫

R

ργ(x − y) f ±1 (x, t, ξ) f ±2 (y, t, ξ)dξdxdy

−

∫

(TN )2

∫

R2f ±1 (x, t, ξ) f ±2 (y, t, ζ)ργ(x − y)ψδ(ξ − ζ)dxdydξdζ

∣∣∣∣ ≤ 2δ, a.s.. (3.29)

Moreover, when γ is small enough, it follows that∣∣∣∣∫

(TN )2

∫

R

ργ(x − y) f ±1 (x, t, ξ) f ±2 (y, t, ξ)dξdydx −

∫

TN

∫

R

f ±1 (x, t, ξ) f ±2 (x, t, ξ)dξdx∣∣∣∣

=

∣∣∣∣∫

(TN )2

∫

R

ργ(x − y) f ±1 (x, t, ξ) f ±2 (y, t, ξ)dξdydx −

∫

TN

∫

|z|<γ

∫

R

ργ(z) f ±1 (x, t, ξ) f ±2 (x, t, ξ)dξdzdx∣∣∣∣

=

∣∣∣∣∫

(TN )2

∫

R

ργ(x − y) f ±1 (x, t, ξ)( f ±2 (y, t, ξ) − f ±2 (x, t, ξ))dξdydx∣∣∣∣

≤ sup|z|<γ

∫

TN

∫

R

f ±1 (x, t, ξ)| f ±2 (x − z, t, ξ) − f ±2 (x, t, ξ)|dξdx

≤ sup|z|<γ

∫

TN

∫

R

|Λ f ±2(x − z, t, ξ) − Λ f ±2

(x, t, ξ)|dξdx. (3.30)

11

As Λ f2 is integrable, we have for a countable sequence γn ↓ 0, (3.30) holds a.s. for all n, hence, passing

to the limit n→∞, we get

limn→∞

∣∣∣∣∫

(TN )2

∫

R

ργn(x − y) f ±1 (x, t, ξ) f ±2 (y, t, ξ)dξdxdy −

∫

TN

∫

R

f ±1 (x, t, ξ) f ±2 (x, t, ξ)dξdx∣∣∣∣ = 0, a.s..(3.31)

Similarly, it holds that

limn→∞

∣∣∣∣∫

(TN )2

∫

R

ργn(x − y) f ±1 (x, t, ξ) f ±2 (y, t, ξ)dξdxdy −

∫

TN

∫

R

f ±1 (x, t, ξ) f ±2 (x, t, ξ)dξdx∣∣∣∣ = 0, a.s..(3.32)

By a similar argument, passing to the limit δ→ 0, it follows from (3.28)-(3.32) that

limn→∞Et(γn, δn) = 0, a.s..

Without confusion, from now on, we write

limγ,δ→0

Et(γ, δ) = 0, a.s.. (3.33)

In particular, when t = 0, it holds that

limγ,δ→0

E0(γ, δ) = 0. (3.34)

Now, we will make some estimates for I(t), J(t), K(t) and H(t). We start with I(t). Set

Γ(ξ, ζ) =

∫ ∞

ζ

∫ ξ

−∞

Υ(ξ′, ζ′)|ξ′ − ζ′|ψδ(ξ′ − ζ′)dξ′dζ′,

where Υ(ξ, ζ) is the function appeared in Hypothesis H. Integration by parts yields that∫ t

0

∫

(TN )2

∫

R2f1 f2(a(ξ) − a(ζ)) · ∇xαdξdζdxdyds

≤

∫ t

0

∫

(TN )2

∫

R2f1 f2Υ(ξ, ζ)|ξ − ζ ||∇xργ(x − y)|ψδ(ξ − ζ)dξdζdxdyds

= −

∫ t

0

∫

(TN )2

∫

R2f1 f2

∂2Γ(ξ, ζ)

∂ξ∂ζ|∇xργ(x − y)|dξdζdxdyds

=

∫ t

0

∫

(TN )2

∫

R2Γ(ξ, ζ)dν1

x,s ⊗ ν2y,s(ξ, ζ)|∇xργ(x − y)|dxdyds

≤ C(q0)δ

∫ t

0

∫

(TN )2

∫

R2(1 + |ξ|q0 + |ζ |q0 )dν1

x,s ⊗ ν2y,s(ξ, ζ)|∇xργ(x − y)|dxdyds

≤ C(q0)δγ−1∫ t

0

∫

(TN )2

∫

R2(1 + |ξ|q0 + |ζ |q0)dν1

x,s ⊗ ν2y,s(ξ, ζ)dxdyds,

where we have used the fact that a(·) is of polynomial growth with degree q0 and (30) in [DV-1]. Namely,

we have obtained that∣∣∣∣∫ t

0

∫

(TN )2

∫

R2f1 f2(a(ξ) − a(ζ)) · ∇xαdξdζdxdyds

∣∣∣∣

≤ C(q0)δγ−1t +C(q0)δγ−1t(ess sup

0≤s≤t

‖uh(s)‖q0

Lq0 (TN )+ ess sup

0≤s≤t

‖u(s)‖q0

Lq0 (TN )

). a.s..

12

Similar calculations lead to∣∣∣∣∫ t

0

∫

(TN )2

∫

R2f1 f2(a(ξ) − a(ζ)) · ∇xαdξdζdxdyds

∣∣∣∣

≤ C(q0)δγ−1t +C(q0)δγ−1t(ess sup

0≤s≤t

‖uh(s)‖q0

Lq0 (TN )+ ess sup

0≤s≤t

‖u(s)‖q0

Lq0 (TN )

). a.s..

Combining the above inequalities, we get

|I(t)| ≤ C(q0)δγ−1t +C(q0)δγ−1t(ess sup

0≤s≤t

‖uh(s)‖q0

Lq0 (TN )+ ess sup

0≤s≤t

‖u(s)‖q0

Lq0 (TN )

). a.s..

By (2.5) in Hypothesis H, we see that

J(t) =

∫ t

0

∫

(TN )2

∫

R2α∑

k≥1

|gk(x, ξ) − gk(y, ζ)|2dν1x,s ⊗ ν

2y,s(ξ, ζ)dxdyds

≤ D1

∫ t

0

∫

(TN )2ργ(x − y)|x − y|2

∫

R2ψδ(ξ − ζ)dν1

x,s ⊗ ν2y,s(ξ, ζ)dxdyds

+D1

∫ t

0

∫

(TN )2

∫

R2ργ(x − y)ψδ(ξ − ζ)|ξ − ζ |2dν1

x,s ⊗ ν2y,s(ξ, ζ)dxdyds

=: J1(t) + J2(t).

Noting that∫

R2ψδ(ξ, ζ)dν1

x,s ⊗ ν2y,s(ξ, ζ) ≤ δ−1, a.s.,

∫

(TN )2ργ(x − y)|x − y|2dxdy ≤ γ2,

we have

J1(t) ≤ D1δ−1γ2t. a.s.. (3.35)

For the term J2, we have

J2 ≤ δD1

∫ t

0

∫

(TN )2

∫

|ξ−ζ |≤δ

ργ(x − y)ψδ(ξ − ζ)|ξ − ζ |dν1x,s ⊗ ν

2y,s(ξ, ζ)dxdyds

≤ δD1Cψt, a.s., (3.36)

where Cψ := supξ∈R ‖ψ(ξ)‖. (3.35) and (3.36) together yield

J(t) ≤ D1δ−1γ2t + D1Cψδt, a.s..

By Hölder inequality and (2.4), we get

H(t) ≤ 2

∫ t

0|h(s)|l2

∫

(TN )2

∫

R2

(∑

k≥1

|gk(x, ξ)|2) 1

2ργ(x − y)χ(ξ, ζ)dν1

x,s ⊗ ν2y,s(ξ, ζ)dxdyds

≤ 2D120

∫ t

0|h(s)|l2

∫

(TN )2ργ(x − y)dxdyds

≤ 2D120

∫ t

0|h(s)|l2 ds, a.s.

13

where we have used the fact that χ(ξ, ζ) ≤ 1.

Combining all the above estimates, we deduce that∫

(TN )2

∫

R2ργ(x − y)ψδ(ξ − ζ)( f ±1 (x, t, ξ) f ±2 (y, t, ζ) + f ±1 (x, t, ξ) f ±2 (y, t, ζ))dξdζdxdy

≤

∫

(TN )2

∫

R2ργ(x − y)ψδ(ξ − ζ)( f1,0(x, ξ) f2,0(y, ζ) + f1,0(x, ξ) f2,0(y, ζ))dξdζdxdy

+D1δ−1γ2t + D1Cψδt +C(q0)δγ−1t + 2D

120

∫ t

0|h(s)|l2 ds

+C(q0)δγ−1t(ess sup

0≤s≤t

‖uh(s)‖q0

Lq0 (TN )+ ess sup

0≤s≤t

‖u(s)‖q0

Lq0 (TN )

)+ K(t), a.s.. (3.37)

For s ∈ (0, T ), set

R(s) :=

∫

(TN )2

∫

R2ργ(x − y)ψδ(ξ − ζ)( f ±1 (x, s, ξ) f ±2 (y, s, ζ) + f ±1 (x, s, ξ) f ±2 (y, s, ζ))dξdζdxdy.

Then, we deduce from (3.37) that

ess sup0≤s≤t

R(s) ≤

∫

TN

∫

R

( f1,0 f2,0 + f1,0 f2,0)dξdx + E0(γ, δ)

+D1δ−1γ2t + D1Cψδt +C(q0)δγ−1t + 2D

120

∫ t

0|h(s)|l2 ds

+C(q0)δγ−1t(ess sup

0≤s≤t

‖uh(s)‖q0

Lq0 (TN )+ ess sup

0≤s≤t

‖u(s)‖q0

Lq0 (TN )

)

+ sup0≤s≤t

|K|(s), a.s.,

where limγ,δ→0 E0(γ, δ) = 0.

Taking the L2(Ω)-norm on both sides and using Hölder inequality, we get that

(E

∣∣∣∣ess sup0≤s≤t

R(s)∣∣∣∣2) 1

2

.

∫

TN

∫

R

( f1,0 f2,0 + f1,0 f2,0)dξdx + E0(γ, δ)

+D1δ−1γ2t + D1Cψδt +C(q0)δγ−1t + 2t

12 D

120

(E

∫ t

0|h(s)|2

l2ds

) 12

+C(q0)δγ−1tR +

E∣∣∣∣ sup

s∈[0,t]|K(s)|

∣∣∣∣2

12

, (3.38)

where

R :=

(Eess sup

0≤s≤T

‖uh(s)‖2q0

L2q0 (TN )

) 12+

(Eess sup

0≤s≤T

‖u(s)‖2q0

L2q0 (TN )

) 12

.

In view of (2.7) and (3.18), we have

R < +∞. (3.39)

14

To estimate the stochastic integral term, we use the Burkholder inequality to get

E

∣∣∣∣ sups∈[0,t]

|K|(s)∣∣∣∣2

(3.40)

= E

∣∣∣∣ sups∈[0,t]

∑

k≥1

∫ s

0

∫

(TN )2

∫

R2χ(ξ, ζ)ργ(x − y)(gk(x, ξ) − gk(y, ζ))dν1

x,r ⊗ ν2y,r(ξ, ζ)dxdydβk(r)

∣∣∣∣2

. E[ ∫ t

0

∑

k≥1

∣∣∣∣∫

(TN )2

∫

R2|gk(x, ξ) − gk(y, ζ)|ργ(x − y)χ(ξ, ζ)dν1

x,r ⊗ ν2y,r(ξ, ζ)dxdy

∣∣∣∣2dr

].

Recalling (2.3) in Hypothesis H

|gk(x, ξ) − gk(y, ζ)| ≤ C1k (|x − y| + |ξ − ζ |),

∑

k≥1

|C1k |

2 ≤D1

2,

it follows from (3.40) that

E

∣∣∣∣ sups∈[0,t]

|K|(s)∣∣∣∣2

. D1E[ ∫ t

0

∣∣∣∣∫

(TN )2

∫

R2(|x − y| + |ξ − ζ |)ργ(x − y)χ(ξ, ζ)dν1

x,r ⊗ ν2y,r(ξ, ζ)dxdy

∣∣∣∣2dr

].

Since∫

(TN )2

∫

R2|x − y|ργ(x − y)χ(ξ, ζ)dν1

x,r ⊗ ν2y,r(ξ, ζ)dxdy ≤ γ, a.s..

it follows that

E

∣∣∣∣ sups∈[0,t]

|K|(s)∣∣∣∣2. D1E

[ ∫ t

0

∣∣∣∣γ +∫

(TN )2|uh,± − u±|ργ(x − y)dxdy

∣∣∣∣2dr

]. (3.41)

With the help of (3.26), (3.28) and (3.29), we deduce that∫

(TN )2|uh,±(x, r) − u±(y, r)|ργ(x − y)dxdy

=

∫

(TN )2

((uh,±(x, r) − u±(y, r))+ + (uh,±(x, r) − u±(y, r))−

)ργ(x − y)dxdy

=

∫

(TN )2

∫

R

( f ±1 (x, r, ξ) f ±2 (y, r, ξ) + f ±1 (x, r, ξ) f ±2 (y, r, ξ))ργ(x − y)dξdxdy

≤ 4δ +

∫

(TN )2

∫

R2( f ±1 (x, r, ξ) f ±2 (y, r, ζ) + f ±1 (x, r, ξ) f ±2 (y, r, ζ))ργ(x − y)ψδ(ξ − ζ)dξdζdxdy

= 4δ + R(r). a.s.. (3.42)

Combining (3.41) and (3.42), we obtain that

E

∣∣∣∣ sups∈[0,t]

|K(s)|∣∣∣∣2. D1E

[ ∫ t

0

∣∣∣∣γ + 4δ + R(r)∣∣∣∣2dr

]

≤ 2D1E[ ∫ t

0

∣∣∣∣γ + 4δ∣∣∣∣2dr +

∫ t

0|R(r)|2dr

]

≤ 2D1t|γ + 4δ|2 + 2D1E( ∫ t

0R2(r)dr

). (3.43)

15

Substitute (3.43) back into (3.38) to get

(E

∣∣∣∣ess sup0≤s≤t

R(s)∣∣∣∣2) 1

2

.

∫

TN

∫

R

( f1,0 f2,0 + f1,0 f2,0)dξdx + E0(γ, δ) + D1δ−1γ2t + D1Cψδt +C(q0)δγ−1t

+2t12 D

120

(E

∫ t

0|h(s)|2

l2ds

) 12

+C(q0)δγ−1Rt + 212 D

121 t

12 |γ + 4δ| + 2

12 D

121

(E( ∫ t

0R2(r)dr

)) 12.

Squaring the above inequality, we get

E

∣∣∣∣ess sup0≤s≤t

R(s)∣∣∣∣2

.

[ ∫

TN

∫

R

( f1,0 f2,0 + f1,0 f2,0)dξdx + E0(γ, δ) + D1δ−1γ2t + D1Cψδt +C(q0)δγ−1t

+2t12 D

120

(E

∫ t

0|h(s)|2

l2ds

) 12

+C(q0)δγ−1Rt + 212 D

121 t

12 |γ + 4δ|

]2

+2D1

∫ t

0

(E

∣∣∣∣ess sup0≤s≤r

R(s)∣∣∣∣2)

dr. (3.44)

Applying Gronwall inequality to (3.44), we get

(E

∣∣∣∣ess sup0≤s≤T

R(s)∣∣∣∣2) 1

2

. eD1T[ ∫

TN

∫

R

( f1,0 f2,0 + f1,0 f2,0)dξdx + E0(γ, δ) + D1δ−1γ2T + D1CψδT +C(q0)δγ−1T

+2T12 D

120

(E

∫ T

0|h(s)|2

l2ds

) 12+C(q0)δγ−1RT + 2

12 D

121 T

12 |γ + 4δ|

]. (3.45)

Let

Q(s) :=

∫

(TN )2

∫

(R)2ργ(x − y)ψγ(ξ − ζ)( f ±2 (s, x, ξ) f ±2 (s, y, ζ) + f ±2 (s, x, ξ) f ±2 (s, y, ζ))dξdζdxdy.

Applying the same arguments to f ±2 and f ±2 (in this case, h = 0 ), we can show that

(E

∣∣∣∣ess sup0≤s≤T

Q(s)∣∣∣∣2) 1

2

. eD1T[E0(γ, δ) + D1δ

−1γ2T + D1CψδT +C(q0)δγ−1T

+C(q0)δγ−1RT + 212 D

121 T

12 |γ + 4δ|

]. (3.46)

On the other hand, from (3.27), it follows that

(E

∣∣∣∣ess sup0≤s≤T

∫

TN

∫

R

( f ±1 (s, x, ξ) f ±2 (s, x, ξ) + f ±1 (s, x, ξ) f ±2 (s, x, ξ))dξdx

∣∣∣∣2) 1

2

.

(E

∣∣∣∣ess sup0≤s≤T

|Es(γ, δ)|∣∣∣∣2) 1

2

+

(E

∣∣∣∣ess sup0≤s≤T

R(s)∣∣∣∣2) 1

2

. (3.47)

16

Now we will provide estimates for

(E

∣∣∣∣ess sup0≤s≤T

|Es(γ, δ)|∣∣∣∣2) 1

2

. For any s ∈ (0, T ), we write

Es(γ, δ) =

∫

(TN )2

∫

R2( f ±1 (x, s, ξ) f ±2 (y, s, ζ) + f ±1 (x, s, ξ) f ±2 (y, s, ζ))ργ(x − y)ψδ(ξ − ζ)dxdydξdζ

−

∫

TN

∫

R

( f ±1 (x, s, ξ) f ±2 (x, s, ξ) + f ±1 (x, s, ξ) f ±2 (x, s, ξ))dξdx

=

[ ∫

(TN )2

∫

R

ργ(x − y)( f ±1 (x, s, ξ) f ±2 (y, s, ξ) + f ±1 (x, s, ξ) f ±2 (y, s, ξ))dξdxdy

−

∫

TN

∫

R

( f ±1 (x, s, ξ) f ±2 (x, s, ξ) + f ±1 (x, s, ξ) f ±2 (x, s, ξ))dξdx]

+

[ ∫

(TN )2

∫

R2( f ±1 (x, s, ξ) f ±2 (y, s, ζ) + f ±1 (x, s, ξ) f ±2 (y, s, ζ))ργ(x − y)ψδ(ξ − ζ)dxdydξdζ

−

∫

(TN )2

∫

R

ργ(x − y)( f ±1 (x, s, ξ) f ±2 (y, s, ξ) + f ±1 (x, s, ξ) f ±2 (y, s, ξ))dξdxdy]

=: H1 + H2.

By (3.28) and (3.29), we have

|H2| ≤ 4δ, a.s.. (3.48)

On the other hand,

|H1| ≤

∣∣∣∣∫

(TN )2ργ(x − y)

∫

R

Iuh,±(x,s)>ξ(Iu±(x,s)≤ξ − Iu±(y,s)≤ξ)dξdxdy

∣∣∣∣

+

∣∣∣∣∫

(TN )2ργ(x − y)

∫

R

Iuh,±(x,s)≤ξ(Iu±(x,s)>ξ − Iu±(y,s)>ξ)dξdxdy∣∣∣∣

≤ 2

∫

(TN )2ργ(x − y)|u±(x, s) − u±(y, s)|dxdy, a.s..

Using (3.28) and (3.29) again, it follows that∫

(TN )2ργ(x − y)|u±(x, s) − u±(y, s)|dxdy

=

∫

(TN )2

∫

R

ργ(x − y)( f ±2 (x, s, ξ) f ±2 (y, s, ξ) + f ±2 (x, s, ξ) f ±2 (y, s, ξ))dξdxdy

≤

∫

(TN )2

∫

R2ργ(x − y)ψδ(ξ − ζ)( f ±2 (x, s, ξ) f ±2 (y, s, ζ) + f ±2 (x, s, ξ) f ±2 (y, s, ζ))dξdζdxdy + 4δ

= Q(s) + 4δ, a.s..

Hence,

|H1| ≤ 2Q(s) + 8δ, a.s.. (3.49)

Collecting (3.48) and (3.49) yields

|Es(γ, δ)| ≤ 2Q(s) + 12δ, a.s.,

17

By (3.46), we deduce that

(E∣∣∣ess sup

0≤s≤T

|Es(γ, δ)|∣∣∣2) 1

2

.

(E|ess sup

0≤s≤T

Q(s)|2) 1

2+ δ

. eD1T[E0(γ, δ) + D1δ

−1γ2T + D1CψδT +C(q0)δγ−1T

+C(q0)δγ−1RT + 212 D

121 T

12 |γ + 4δ|

]+ δ. (3.50)

Combining (3.45) and (3.50), we deduce from (3.47) that

(E

∣∣∣∣ess sup0≤s≤T

∫

TN

∫

R

( f ±1 (s, x, ξ) f ±2 (s, x, ξ) + f ±1 (s, x, ξ) f ±2 (s, x, ξ))dξdx

∣∣∣∣2) 1

2

. eD1T[ ∫

TN

∫

R

( f1,0 f2,0 + f1,0 f2,0)dξdx + 2E0(γ, δ) + 2D1δ−1γ2T + 2D1CψδT + 2C(q0)δγ−1T

+2T12 D

120

(E

∫ T

0|h(s)|2

l2ds

) 12+ 2C(q0)δγ−1RT + 3D

121 T

12 |γ + 4δ|

]+ δ.

Note that we have f ±1 (x, s, ξ) = Iuh,±(s,x)>ξ and f ±2 (x, s, ξ) = Iu±(s,x)>ξ with initial data f1,0 = Iη>ξ and

f2,0 = Iη>ξ , respectively. In view of (3.26), we can rewrite the above inequality as

(E

∣∣∣∣ess sup0≤s≤T

‖uh,±(s) − u±(s)‖L1(TN )

∣∣∣∣2) 1

2

. r(γ, δ), (3.51)

where

r(γ, δ)

:= eD1T[2E0(γ, δ) + 2D1δ

−1γ2T + 2D1CψδT + 2C(q0)δγ−1T

+2T12 D

120

(E

∫ T

0|h(s)|2

l2ds

) 12+ 2C(q0)δγ−1RT + 3D

121 T

12 |γ + 4δ|

]+ δ. (3.52)

Taking

δ = γ43 ,

we have

r(γ, δ)

:= eD1T[2E0(γ, δ) + 2D1γ

23 T + 2D1Cψγ

43 T + 2C(q0)γ

13 T

+2T12 D

120

(E

∫ T

0|h(s)|2

l2ds

) 12+ 2C(q0)γ

13RT + 3D

121 T

12 |γ + 4γ

43 |]+ γ

43 .

Let γ→ 0 to get

limγ→0

r(γ, δ) ≤ 2T12 D

120 eD1T

(E

∫ T

0|h(s)|2

l2ds

) 12.

18

Therefore, we deduce from (3.51) that

(E

∣∣∣∣ess sup0≤s≤T

‖uh,±(s) − u±(s)‖L1(TN )

∣∣∣∣2) 1

2

≤ 2DT12 D

120 eD1T

(E

∫ T

0|h(s)|2

l2ds

) 12,

which implies

E

∣∣∣∣∫ T

0‖uh(t) − u(t)‖L1(TN )dt

∣∣∣∣2≤ 4D2D0T 3e2D1TE

∫ T

0|h(s)|2

l2ds. (3.53)

We complete the proof.

Acknowledgements This work is partly supported by National Natural Science Foundation of China

(No. 11671372, 11971456, 11721101, 11801032, 11971227), Key Laboratory of Random Complex

Structures and Data Science, Academy of Mathematics and Systems Science, Chinese Academy of Sci-

ences (No. 2008DP173182), Beijing Institute of Technology Research Fund Program for Young Schol-

ars.

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