# Hausdorff Dimension of a Random Attractor for Stochastic Boussinesq Equations with Double Multiplicative White Noises

Authors: Yin Li; Ruiying Wei; Donghong Cai
Journal: Journal of Function Spaces
 (2016)

---

## Abstract

This paper investigates the existence of random attractor for stochastic Boussinesq equations driven by multiplicative white noises in both the velocity and temperature equations and estimates the Hausdorff dimension of the random attractor.


---

## Body

## 1. Introduction

In this paper, we consider the following 2D stochastic Boussinesq equation onD
=
(
0,1
)
×
(
0,1
)
⊂
R
2:(1)
d
v
+
v
·
∇
v
+
∇
p
-
μ
Δ
v
d
t
=
T
e
2
d
t
+
b
1
v
∘
d
W
1
t
,
d
T
+
v
·
∇
T
-
k
Δ
T
d
t
=
b
2
T
∘
d
W
2
t
,
d
i
v
v
=
0
,
v
t
=
t
0
=
v
0
,
T
t
=
t
0
=
T
0
,where v
=
(
v
1
(
x
,
t
)
,
v
2
(
x
,
t
)
) denotes the velocity of the fluid; T
(
x
,
t
) is the temperature; and function p
(
x
,
t
) represents the pressure. We set (
x
,
t
)
∈
D
×
R, and {
e
1
,
e
2
} is the canonical basis of R
2. Let us consider μ
,
k
>
0 the Grashof and Prandtl numbers. W
(
t
)
=
(
W
1
(
t
)
,
W
2
(
t
)
)
T is two-sided real-valued Wiener process on a probability space (
Ω
,
F
,
{
F
t
}
t
∈
R
,
P
), here (2)
Ω
=
ω
∈
C
R
,
R
2
:
ω
0
=
0
. The Borel σ-algebra F on Ω is generated by the compact open topology, and P is the corresponding Wiener measure; ∘
d
W
i
(
t
) denotes the Stratonovich form white noise.Pu and Guo in [1] studied the global well-posedness of stochastic 2D Boussinesq equations with partial viscosity; Li and Guo [2] consider the stochastic Boussinesq equations, which are influenced by multiplicative white noise in the velocity equations. Recently, Zhao and Li [3] established the existence of a random attractor for stochastic Boussinesq equations with double additive white noises.In many cases, the Hausdorff dimension of a random attractor is finite. Crauel and Flandoli [4] developed a technique to estimate the Hausdorff dimension of attractors for certain dynamical systems. But they required the noise to be bounded. Debussche [5] applied a random squeezing property to estimate the Hausdorff dimension without the assumption of bounded noise. A number of authors [6–13] used this method to estimate the Hausdorff dimension of attractors for many stochastic equations.As far as we know, there are no results on stochastic 2D Boussinesq equations with double multiplicative white noises on bounded domains. In this paper, we prove the existence of random attractor for the corresponding RDS associated with problem (1). And we give an estimate of Hausdorff dimension for the random attractor.The paper is organized as follows. In the first section, we recall some latest results and what we want to do for the 2D stochastic Boussinesq equations. We study the RDS determined by (1) in Section 2. In Section 3, we prove the existence of a random attractor. Finally, we estimate the Hausdorff dimension of the random attractor in Section 4.


## 2. Random Dynamical System

The random dynamical system generated by stochastic Boussinesq equations with double multiplicative noise will be studied in this section. Thus, we introduce a process, which enables us to transform the stochastic equations into a deterministic equation with a random parameter.Let(3)
α
i
t
=
e
-
b
i
W
i
t
,
i
=
1,2
,
ξ
=
α
1
t
v
η
=
α
2
t
T
. From (1), we get the stochastic Boussinesq equations without white noises:(4)
d
ξ
d
t
+
α
1
-
1
t
ξ
·
∇
ξ
-
μ
Δ
ξ
+
α
1
t
∇
p
=
α
1
t
α
2
t
e
2
η
,
(5)
d
η
d
t
+
α
1
-
1
t
ξ
·
∇
η
-
k
Δ
η
=
0
,
(6)
d
i
v
ξ
=
0
,
(7)
ξ
t
=
t
0
=
ξ
0
,
η
t
=
t
0
=
η
0
.In the present paper, we consider the Hilbert space: H
=
H
1
×
H
2, with the scalar product (
·
,
·
) and norm ·, where(8)
H
1
=
ξ
∈
H
2
2
:
d
i
v
ξ
=
0
,
H
2
=
L
2
D
.We also consider the subspace: V
=
V
1
×
V
2. Consider(9)
V
1
=
ξ
∈
V
2
2
:
d
i
v
ξ
=
0
,
V
2
=
c
l
o
s
u
r
e
o
f
H
0
1
D
i
n
L
2
D and its scalar product and norms are (
(
·
,
·
)
) and ·
1, respectively, for any η
1
,
η
2
∈
V
2:(10)
η
1
,
η
2
=
∫
D
g
r
a
d
η
1
g
r
a
d
η
2
d
x
,
η
1
=
η
,
η
1
/
2
=
∇
η
.Now we give some definitions of common operators.a is the bilinear on V:(11)
a
u
1
,
u
2
=
-
μ
Δ
ξ
1
,
ξ
2
-
k
Δ
η
1
,
η
2
,
∀
u
i
=
ξ
i
,
η
i
T
∈
V
,
i
=
1,2
.
A is isomorphism from D
(
A
) into H and from V into the dual space V
′, defined by (12)
A
u
1
,
u
2
=
a
u
1
,
u
2
,
∀
u
i
=
ξ
i
,
η
i
T
∈
V
,
i
=
1,2
, where D
(
A
)
=
D
(
A
1
)
×
D
(
A
2
), being(13)
D
A
1
=
ξ
:
ξ
∈
V
1
∩
H
2
D
2
,
D
A
2
=
η
:
η
∈
V
2
∩
H
2
D
. We consider trilinear forms b on V defined by (14)
b
u
1
,
u
2
,
u
3
=
-
ξ
1
·
∇
ξ
2
,
ξ
3
-
ξ
1
·
∇
η
2
,
η
3
,
∀
u
i
=
ξ
i
,
η
i
T
∈
V
,
i
=
1,2
.
b is continuous on V and (
H
′
(
D
)
)
2
×
H
′
(
D
).We also define the bilinear continuous operatorB, which maps V
×
V into V
′ and D
(
A
)
×
D
(
A
) into H, by (15)
B
u
1
,
u
2
,
u
3
=
b
u
1
,
u
2
,
u
3
,
∀
u
i
=
ξ
i
,
η
i
T
∈
V
,
i
=
1,2
,
3
.Finally, we consider a family of linear continuous operators R
(
t
)
=
R
(
t
,
ω
) on H:(16)
R
t
:
u
=
ξ
,
η
T
⟶
R
u
=
α
1
t
α
2
t
e
2
η
,
0
T
.Assume that u
=
(
ξ
,
η
)
T is a solution of (4)–(7) and y
=
(
f
,
g
)
T is a test function in V. Multiplying (4) by f and (5) by g, integrating over D, and adding the resulting equation, we obtain(17)
d
ξ
d
t
,
f
+
α
1
-
1
t
ξ
·
∇
ξ
,
f
-
μ
Δ
ξ
,
f
+
α
1
t
∇
p
,
f
+
d
η
d
t
,
g
+
α
1
-
1
t
ξ
·
∇
η
,
g
-
k
Δ
η
,
g
=
α
1
t
α
2
t
e
2
η
,
f
.Note that (18)
d
ξ
d
t
,
f
+
d
η
d
t
,
g
=
d
u
d
t
,
y
ξ
·
∇
ξ
,
f
+
ξ
·
∇
η
,
g
=
-
b
u
,
u
,
y
-
μ
Δ
ξ
,
f
-
k
Δ
η
,
g
=
a
u
,
y
. On the other hand, (19)
d
i
v
f
=
0
,
f
=
f
1
,
f
2
T
. Then, (20)
∇
p
,
f
=
0
. Therefore, (9) can also be written as(21)
d
u
d
t
=
-
A
u
+
α
1
-
1
t
B
u
,
u
+
R
t
u
.By the Galerkin method (see [14, Theorem 3.1]), we can prove that (21) has a unique solution u
(
x
,
t
), which satisfies the following:(i) For allt
0
∈
R, if u
0
∈
H, then (22)
u
∈
C
t
0
,
∞
;
H
∩
L
L
o
c
2
t
0
,
∞
;
V
.(ii) Denoting such solution byu
(
t
,
ω
;
t
0
,
u
0
), for any t
≥
t
0, the mapping (23)
u
0
⟶
u
t
,
ω
;
t
0
,
u
0 is continuous. Then, we define a stochastic flow φ
(
t
,
ω
) by (24)
φ
t
,
ω
z
0
=
ξ
,
η
T
=
α
1
t
Q
H
1
z
t
,
ω
;
t
0
,
z
0
,
α
2
t
Q
H
2
z
t
,
ω
;
t
0
,
z
0
T
, where z
0
=
z
(
t
0
)
=
(
v
0
,
T
0
)
T
∈
H and (25)
Q
H
i
:
H
=
H
1
×
H
2
⟶
H
i
,
i
=
1,2are projection operators.We can prove thatφ
(
t
,
ω
) is a continuous RDS determined by (1) easily.


## 3. Random Attractor

In this section, we will establish the existence of nonempty compact tempered random attractor for the RDSφ
(
t
,
ω
) determined by (1). First we prove that φ
(
t
,
ω
) possesses absorbing set in H.Lemma 1.
Letu
=
(
ξ
,
η
)
T be a solution of (4)–(7), for any t
0
≤
t, and one has(26)
ξ
2
≤
ξ
t
0
2
e
-
μ
/
c
0
2
t
-
t
0
+
c
0
2
μ
∫
t
0
t
α
1
2
s
α
2
2
s
η
s
2
e
-
μ
/
c
0
2
t
-
s
d
s
,
η
2
≤
η
t
0
2
e
-
2
k
/
c
1
2
t
-
t
0
.Proof.
Taking the scalar product of (4) with ξ on H
1
(27)
d
ξ
d
t
,
ξ
+
α
1
-
1
t
ξ
·
∇
ξ
,
ξ
-
μ
Δ
ξ
,
ξ
+
α
1
t
∇
p
,
ξ
=
α
1
t
α
2
t
e
2
η
,
ξ
. By the incompressibility condition (6), we have (28)
ξ
·
∇
ξ
,
ξ
=
0
,
∇
p
,
ξ
=
0
.From (27), we obtain(29)
1
2
d
d
t
ξ
2
+
μ
∇
ξ
2
=
α
1
t
α
2
t
η
,
ξ
2
.Using Poincaré inequality, ξ
≤
c
0
∇
ξ (c
0 is constant), we have (30)
1
2
d
d
t
ξ
2
+
μ
c
0
2
ξ
2
≤
α
1
t
α
2
t
η
ξ
2
≤
α
1
t
α
2
t
η
ξ
≤
c
0
2
α
1
2
t
2
μ
α
2
2
t
η
2
+
μ
2
c
0
2
ξ
2
.Then, (31)
d
d
t
ξ
2
+
μ
c
0
2
ξ
2
≤
c
0
2
α
1
2
t
μ
α
2
2
t
η
2
.By the Gronwall inequality, for any t
0
≤
t, we get that (32)
ξ
2
≤
ξ
t
0
2
e
-
μ
/
c
0
2
t
-
t
0
+
c
0
2
μ
∫
t
0
t
α
1
2
s
α
2
2
s
η
s
2
e
-
μ
/
c
0
2
t
-
s
d
s
.Taking the scalar product of (5) with η on H
2 as well (33)
d
η
d
t
,
η
+
1
α
1
t
ξ
·
∇
η
,
η
-
k
Δ
η
,
η
=
0
.Since (
(
ξ
·
∇
)
η
,
η
)
=
0, then(34)
d
d
t
η
2
+
2
k
∇
η
2
=
0
.Using Poincaré inequality η
≤
c
1
∇
η, we have (35)
d
d
t
η
2
+
2
k
c
1
2
η
2
≤
0
.By Gronwall inequality, for any t
0
≤
t, we obtain that (36)
η
2
≤
η
t
0
2
e
-
2
k
/
c
1
2
t
-
t
0
.Lemma 2.
There exist two positive random variablesr
(
ω
) and C
(
ω
), such that, for all ρ
>
0
,
t
0
≤
t
(
ω
,
ρ
)
,
t
∈
[
-
1,0
], and z
0
=
(
v
0
,
T
0
)
T
∈
H with z
0
<
ρ, the solution u
=
(
ξ
,
η
)
T of equations (4)–(7) satisfies(37)
η
t
,
ω
;
t
0
,
α
2
t
0
T
0
2
≤
C
ω
,
ξ
t
,
ω
;
t
0
,
α
1
t
0
v
0
2
≤
r
ω
.Proof.
For-
1
≤
t
≤
0, t
0
≤
-
1, from (26), we have (38)
η
2
≤
α
2
2
t
T
t
0
2
e
2
k
/
c
1
2
1
+
t
0
,taking t
1
(
ω
,
ρ
)
≤
-
1, such that ρ
α
2
2
(
t
)
e
2
k
/
c
1
2
(
1
+
t
1
)
≤
C
(
ω
).
Then, for anyt
0
≤
t
1
(
ω
,
ρ
), (39)
η
t
,
ω
;
t
0
,
α
2
t
0
T
0
2
≤
C
ω
. From (26), we find (40)
ξ
2
≤
ξ
t
0
2
e
μ
/
c
0
2
1
+
t
0
+
c
0
2
μ
∫
t
0
t
α
1
2
s
α
2
2
s
η
s
2
e
μ
/
c
0
2
1
+
s
d
s
,
l
i
m
s
→
-
∞
W
1
s
s
=
0
,
P
-
a
.
s
. Then, α
1
2
(
t
0
)
e
μ
t
0
/
c
0
2
→
0, as t
0
→
-
∞.
Thus, there existst
2
(
ω
,
ρ
)
≤
-
1, for t
0
≤
t
2: (41)
v
0
α
1
2
t
0
e
μ
t
0
/
c
0
2
≤
ρ
2
α
1
2
t
0
e
μ
t
0
/
c
0
2
≤
1
.By (37), we infer that ∀
t
0
≤
t
1, t
∈
[
-
1,0
], P
-
a
.
s is (42)
∫
t
0
t
α
1
2
s
α
2
2
s
η
s
2
e
μ
/
c
0
2
1
+
s
d
s
≤
C
2
ω
∫
-
∞
0
α
1
2
s
α
2
2
s
e
μ
/
c
0
2
1
+
s
d
s
. Since (43)
α
1
2
s
α
2
2
s
=
e
2
b
2
W
2
s
-
2
b
1
W
1
s
=
e
b
3
W
s
,then α
1
2
(
s
)
/
α
2
2
(
s
)
e
μ
s
/
c
0
2
→
0
,
s
→
-
∞. Therefor, (44)
∫
t
0
t
α
1
2
s
α
2
2
s
η
s
2
e
μ
/
c
0
2
1
+
s
d
s
<
∞
. Put (45)
r
ω
=
e
μ
/
c
0
2
1
+
c
0
2
μ
∫
t
0
t
α
1
2
s
α
2
2
s
η
s
2
e
μ
s
/
c
0
2
d
s
.Then, r
(
ω
) exists and it is tempered.
Gettingt
(
ω
,
ρ
)
=
m
i
n
{
t
1
,
t
2
}, for any t
≥
t
0,(46)
η
t
,
ω
;
t
0
,
α
2
t
0
T
0
2
≤
C
ω
,
ξ
t
,
ω
;
t
0
,
α
1
t
0
v
0
2
≤
r
ω
.Now we are going to prove that φ
(
t
,
ω
) is asymptotically compact.Lemma 3.
There exist two positive random variablesC
1
(
ω
) and r
1
(
ω
), such that, for all ρ
>
0
,
t
0
≤
t
(
ω
,
ρ
)
,
t
∈
[
-
1,0
], and z
0
=
(
v
0
,
T
0
)
T
∈
H with z
0
<
ρ, the solution u
=
(
ξ
,
η
)
T of equations (4)–(7) satisfies(47)
∫
-
1
0
∇
η
s
,
ω
;
t
0
,
α
2
s
T
0
2
d
s
≤
C
1
ω
,
∫
-
1
0
∇
ξ
s
,
ω
;
t
0
,
α
1
s
v
0
2
d
s
≤
r
1
ω
.Proof.
Integrating (34) on [
-
1,0
], we have (48)
η
0
2
-
η
-
1
2
+
2
k
∫
-
1
0
∇
η
2
d
s
=
0
.From Lemma 1, we know that (49)
∫
-
1
0
∇
η
2
d
s
≤
1
2
k
η
-
1
2
-
η
0
2
≤
C
ω
k
=
C
1
ω
.We get from (29),(50)
1
2
d
d
t
ξ
2
+
μ
∇
ξ
2
≤
c
0
2
α
1
2
t
2
μ
α
2
2
t
η
2
+
μ
2
∇
ξ
2then (51)
d
d
t
ξ
2
+
μ
∇
ξ
2
≤
c
0
2
α
1
2
t
μ
α
2
2
t
η
2
.Integrating the above formula over [
-
1,0
], (52)
ξ
0
2
-
ξ
-
1
2
+
μ
∫
-
1
0
∇
ξ
s
2
d
s
≤
∫
-
1
0
c
0
2
α
1
2
s
μ
α
2
2
s
η
s
2
d
sthen (53)
∫
-
1
0
∇
ξ
s
d
s
≤
1
μ
ξ
-
1
2
-
ξ
0
2
+
∫
-
1
0
c
0
2
α
1
2
s
μ
α
2
2
s
η
s
2
d
s
.Let (54)
r
1
ω
=
2
r
ω
μ
+
c
0
2
μ
2
∫
-
1
0
α
1
2
s
α
2
2
s
η
s
2
d
sthen (55)
∫
-
1
0
∇
ξ
s
,
ω
;
t
0
,
α
1
s
v
0
2
d
s
≤
r
1
ω
.Lemma 4.
There exist two positive random variablesC
2
(
ω
) and r
2
(
ω
), such that, for all ρ
>
0
,
t
0
≤
t
(
ω
,
ρ
)
,
t
∈
[
-
1,0
], and z
0
=
(
v
0
,
T
0
)
T
∈
H with z
0
<
ρ, the solution u
=
(
ξ
,
η
)
T of (4)–(7) satisfies(56)
∇
η
0
,
ω
;
t
0
,
α
2
s
T
0
2
≤
C
2
ω
,
∇
ξ
0
,
ω
;
t
0
,
α
1
s
v
0
2
≤
r
2
ω
.Proof.
Taking the scalar product of (5) with −Δη on H
2, then (57)
1
2
d
d
t
∇
η
2
+
k
∇
η
2
≤
α
1
-
1
t
ξ
·
∇
η
,
-
Δ
η
≤
α
1
-
1
t
ξ
·
∇
η
Δ
η
. Since (58)
ξ
·
∇
η
≤
ξ
L
4
∇
η
L
4
≤
c
2
∇
ξ
1
/
2
ξ
1
/
2
∇
η
1
/
2
Δ
η
1
/
2
,then we have (59)
1
2
d
d
t
∇
η
2
+
k
Δ
η
2
≤
α
1
-
1
t
c
2
∇
ξ
1
/
2
ξ
1
/
2
∇
η
1
/
2
Δ
η
3
/
2
≤
27
c
2
4
256
k
3
α
1
4
t
ξ
2
∇
ξ
2
∇
η
2
+
k
Δ
η
2
.Hence, (60)
d
d
t
∇
η
2
≤
27
c
2
4
128
k
3
α
1
4
t
ξ
2
∇
ξ
2
∇
η
2
.Using the Gronwall inequality over an arbitrary interval [
s
,
0
]
⊂
[
-
1,0
], we obtain (61)
∇
η
0
2
≤
∇
η
s
2
e
∫
s
0
h
τ
d
τ
.Here, h
(
τ
)
=
27
c
2
4
/
128
k
3
α
1
4
(
τ
)
ξ
(
τ
)
2
∇
ξ
(
τ
)
2.
Integrating the above formula over the interval[
-
1,0
] for s, we obtain (62)
∇
η
0
2
≤
e
∫
-
1
0
h
τ
d
τ
∫
-
1
0
∇
η
s
2
d
s
.Let (63)
C
3
ω
=
e
∫
-
1
0
h
τ
d
τ
C
1
ω
.Thus, C
2
(
ω
) exists and it is tempered and we have (64)
∇
η
0
,
ω
;
t
0
,
α
2
t
0
T
0
2
≤
C
2
ω
.Taking the scalar product of (4) with -
Δ
ξ on H
1
(65)
d
ξ
d
t
,
-
Δ
ξ
+
α
1
t
-
1
ξ
·
∇
ξ
,
-
Δ
ξ
-
μ
Δ
ξ
,
-
Δ
ξ
+
α
1
t
∇
p
,
-
Δ
ξ
=
α
1
t
α
2
t
e
2
η
,
-
Δ
ξ
.Since (
∇
p
,
-
Δ
ξ
)
=
0, then (66)
1
2
d
d
t
∇
ξ
2
+
μ
Δ
ξ
2
≤
α
1
-
1
t
ξ
·
∇
ξ
,
Δ
ξ
+
α
1
t
α
2
t
e
2
η
,
Δ
ξ
≤
c
3
α
1
t
ξ
1
/
2
∇
ξ
Δ
ξ
3
/
2
+
α
1
t
α
2
t
η
Δ
ξ
≤
27
c
3
4
32
α
1
4
t
μ
3
ξ
2
∇
ξ
4
+
α
1
2
t
2
μ
α
2
2
t
η
2
+
μ
Δ
ξ
2
.Namely,(67)
d
d
t
∇
ξ
2
≤
27
c
3
4
16
α
1
4
t
μ
3
ξ
2
∇
ξ
4
+
α
1
2
t
μ
α
2
2
t
η
2
≤
G
1
t
∇
ξ
2
+
G
2
t
,where G
1
t
=
27
c
3
4
/
16
α
1
4
t
μ
3
ξ
2
∇
ξ
2 and G
2
(
t
)
=
α
1
2
(
t
)
/
μ
α
2
2
(
t
)
η
2.
By the Gronwall inequality over an arbitrary interval[
s
,
0
]
⊂
[
-
1,0
], we obtain (68)
∇
ξ
0
2
≤
e
∫
s
0
G
1
σ
d
σ
∇
ξ
s
2
+
∫
s
0
G
2
τ
e
∫
τ
0
G
1
σ
d
σ
d
τ
.Integrating the above formula over the interval [
-
1,0
] for s, we have (69)
∇
ξ
0
2
≤
e
∫
-
1
0
G
1
σ
d
σ
∫
-
1
0
∇
ξ
s
2
d
s
+
∫
-
1
0
G
2
τ
e
∫
-
1
0
G
1
σ
d
σ
d
τ
.Since (70)
∫
-
1
0
G
1
σ
d
σ
≤
27
c
3
4
16
α
1
4
t
μ
3
r
ω
∫
-
1
0
∇
ξ
σ
2
d
σ
≤
27
c
3
4
16
α
1
4
t
μ
3
r
ω
r
1
ω
,let (71)
r
2
ω
=
e
27
c
3
4
/
16
α
1
4
t
μ
3
r
ω
r
1
ω
r
1
ω
+
∫
-
1
0
G
2
τ
d
τ
.Thus, r
2
(
ω
) exists and it is tempered. Then, (72)
∇
ξ
0
,
ω
;
t
0
,
α
1
t
0
u
0
2
≤
r
2
ω
.
LetB
(
ω
) be a random closed ball centered at 0 with radius C
2
(
ω
)
+
r
2
(
ω
).
Then,B
(
ω
) is a random absorbing set for φ
(
t
,
ω
) in V. Since V embedding H is compact, then RDS φ
(
t
,
ω
) is asymptotically compact in H. Thus, the RDS determined by (1) possesses a random attractor.


## 4. Hausdorff Dimensions of Random Attractors

In this section, we will estimate the Hausdorff dimensions ofA
(
ω
). Now, we assume that u
0
,
u
1
(
t
)
,
u
2
(
t
)
,
χ are F
t measurable and prove that the RDS φ
(
t
,
ω
) is differentiable on A
(
ω
).Theorem 5.
The RDSφ
(
t
,
ω
) determined by (21) is almost surely uniformly differentiable on A
(
ω
), ω
∈
Ω, and its derivative is D
φ
(
t
,
t
0
,
ω
;
u
0
)
χ
=
U
(
t
), t
0
≤
t, which is the solution of the following variational equation: (73)
d
U
d
t
=
-
A
U
+
F
t
,
u
U
,
U
t
0
=
χ
,where F
(
t
,
u
)
U
=
α
1
-
1
(
t
)
[
B
(
u
,
U
)
+
B
(
U
,
u
)
]
+
R
(
t
)
U, u
=
φ
(
t
,
t
0
;
ω
)
u
0, t
0
≤
t.Proof.
Letu
1
(
t
)
=
φ
(
t
,
t
0
;
ω
)
u
0
(
1
), u
2
(
t
)
=
φ
(
t
,
t
0
;
ω
)
u
0
(
2
), and δ
(
t
)
=
u
1
(
t
)
-
u
2
(
t
), and we have (74)
d
d
t
δ
t
=
-
A
δ
t
+
α
1
-
1
t
B
u
1
,
δ
t
+
B
δ
t
,
u
2
+
R
t
δ
t
,
d
U
d
t
=
-
A
U
+
α
1
-
1
t
B
u
1
,
U
+
B
U
,
u
2
+
R
t
U
.We also set l
(
t
)
=
δ
(
t
)
-
U
(
t
), and then (75)
d
l
t
d
t
=
-
A
l
t
+
α
1
-
1
t
B
u
2
,
l
t
+
B
l
t
,
u
2
+
B
δ
t
,
δ
t
+
R
t
l
t
.Taking the scalar product of (75) with l
(
t
) on H,(76)
1
2
d
d
t
l
t
2
≤
-
∇
l
t
2
+
α
1
-
1
t
B
l
t
,
u
2
,
l
t
+
B
δ
t
,
δ
t
,
l
t
+
R
t
l
t
2
≤
c
4
α
1
t
∇
u
2
l
t
2
+
δ
t
∇
δ
t
l
t
+
R
t
l
t
2
≤
K
t
l
t
2
+
c
5
δ
t
2
∇
δ
t
2
.Here, K
(
t
) is the function about t. By the Gronwall inequality, we obtain (77)
l
t
2
≤
∫
t
0
t
e
∫
s
t
2
K
s
τ
d
τ
2
c
5
δ
s
2
∇
δ
s
2
d
s
≤
K
1
t
∫
t
0
t
δ
s
2
∇
δ
s
2
d
s
.Taking the scalar product of (74) with δ
(
t
) on H,(78)
1
2
d
d
t
δ
t
2
≤
-
∇
δ
t
2
+
α
1
-
1
t
B
δ
t
,
u
2
,
δ
t
+
R
t
δ
t
2
≤
-
∇
δ
t
2
+
c
6
α
1
t
δ
t
∇
u
2
∇
δ
t
+
R
t
δ
t
2
≤
-
∇
δ
t
2
+
1
2
∇
δ
t
2
+
c
6
2
α
1
2
t
δ
t
2
∇
u
2
2
+
R
t
δ
t
2
≤
-
1
2
∇
δ
t
2
+
K
2
t
δ
t
2
.Then, (79)
d
d
t
δ
t
2
≤
2
K
2
t
δ
t
2
.By the Gronwall inequality, we get (80)
δ
t
2
≤
e
∫
t
0
t
2
K
2
s
d
s
u
0
1
-
u
0
2
2
≤
K
3
t
u
0
1
-
u
0
2
2
.We get from (80) that (81)
∫
t
0
t
∇
δ
t
2
d
s
≤
u
0
1
-
u
0
2
2
+
∫
t
0
t
2
K
2
s
δ
s
2
d
s
≤
u
0
1
-
u
0
2
2
+
u
0
1
-
u
0
2
2
∫
t
0
t
2
K
2
s
K
3
s
d
s
≤
K
4
t
u
0
1
-
u
0
2
2
.From (80) to (81), we have (82)
l
t
2
≤
K
5
t
u
0
1
-
u
0
2
4
.The proof is completed.Theorem 6.
Set(83)
d
=
m
i
n
d
:
-
∑
j
=
1
d
λ
j
+
c
9
d
<
0
,
c
9
∈
R
+
.Then the Hausdorff dimension of random attractor A
(
t
,
ω
) satisfies (84)
d
i
m
H
A
t
,
ω
≤
d
,
∀
t
∈
R
.Proof.
Taking the scalar product of (74) with U
(
t
) on H,(85)
1
2
d
d
t
U
2
=
-
∇
U
2
+
α
1
-
1
t
B
U
,
u
,
U
+
R
t
U
2
≤
α
1
-
1
t
B
U
,
u
,
U
+
R
t
U
2
≤
c
7
α
1
t
U
2
∇
u
+
R
t
U
2
.Then, (86)
d
d
t
U
2
≤
2
K
6
t
U
2
.By the Gronwall inequality, we get (87)
U
t
2
≤
e
∫
t
0
t
2
K
6
s
d
s
U
t
0
2which implies that (88)
D
φ
t
,
t
0
,
u
0
;
ω
≤
e
∫
t
0
t
2
K
6
s
d
s
.Since (89)
a
1
D
φ
t
,
t
0
,
u
0
;
ω
=
D
φ
t
,
t
0
,
u
0
;
ω
,then (90)
a
1
D
φ
t
,
t
0
,
u
0
;
ω
≤
e
∫
t
0
t
2
K
6
s
d
s
.Taking a
′
(
ω
)
=
m
a
x
{
1
,
e
∫
t
0
t
‍
2
K
6
(
s
)
d
s
}, then we have (91)
a
1
ω
D
φ
t
,
0
,
u
0
;
ω
≤
a
1
′
ω
,
E
l
n
a
1
′
ω
<
∞
.Assume that U
j
(
t
)
=
D
φ
(
t
,
t
0
,
u
0
;
ω
)
χ
j, here χ
j
∈
H, and i
=
1,2
,
…
,
d. Q
d
(
t
) is the orthogonal projection in H onto the space spanned by U
1
(
t
)
,
U
2
(
t
)
,
…
,
U
d
(
t
). Let {
Ψ
j
}, j
=
1,2
,
…
,
d, be an orthonormal basis in Q
d
(
t
)
H. Then, (92)
T
r
-
A
+
F
t
,
u
∘
Q
d
t
=
∑
j
=
1
d
-
A
+
F
t
,
u
Ψ
j
,
Ψ
j
≤
∑
j
=
1
d
-
A
Ψ
j
,
Ψ
j
+
α
1
-
1
t
B
Ψ
j
,
u
t
,
Ψ
j
+
R
t
Ψ
j
,
Ψ
j
≤
-
∑
j
=
1
d
λ
j
Ψ
j
2
+
∑
j
=
1
d
c
8
e
b
1
W
1
t
Ψ
j
2
+
e
b
3
W
3
t
Ψ
j
2
≤
-
∑
j
=
1
d
λ
j
+
∑
j
=
1
d
c
8
e
b
1
W
1
t
+
e
b
3
W
3
t
.Set (93)
ϖ
d
D
φ
t
,
0
,
u
0
;
ω
≤
e
x
p
-
∑
j
=
1
d
λ
j
+
∑
j
=
1
d
c
8
e
b
1
W
1
t
+
e
b
3
W
3
t
.Then, (94)
E
l
n
s
u
p
u
0
∈
A
ω
ϖ
D
φ
t
,
t
0
,
u
0
;
ω
≤
-
∑
j
=
1
d
λ
j
+
∫
0
1
∑
j
=
1
d
c
8
E
e
b
1
W
1
t
+
E
e
b
3
W
3
t
d
t
≤
-
∑
j
=
1
d
λ
j
+
c
9
d
.We can get that (95)
E
l
n
s
u
p
u
0
∈
A
ω
ϖ
D
φ
t
,
t
0
,
u
0
;
ω
<
0when (96)
-
∑
j
=
1
d
λ
j
+
c
9
d
<
0
,
c
9
∈
R
+
.In this case, the Hausdorff dimension of random attractor A
(
t
,
ω
) satisfies (97)
d
i
m
H
A
t
,
ω
≤
d
,
∀
t
∈
R
.



---

*Source: 1832840-2016-06-09.xml*