1/2
•记皿1-a®([t,T];
Rd)为使得Vol.41A恂品…皆时:=|g(t)l
+
0<;勢(dy)
<的连续函数g
:
[t,
T]
T
Rk构成的空间.显然C1-a+e([t,T];Rd)
c
WW
1-a,8([t,T];Rd)
c
C
1-a([t,T];Rd),
Ve
>
0.记Aa(g;[t,T])
:=
~)
sup
|(DS-ags-)(r)1,r
(1
一
a)
0=
Joxa-1e-xdx是Euler函数以及(D1-ags-)(r)于是ein(1-a)(g(r)
—
g(s)
r(a)
s
—
r1-as+
(1
—
a)
/J
rg(y)
—
g(r)
dy)弘)(r).(y
—
r)2-aAa(g;[t,
T])
<
^(1
一
a)r(a)
|g|^i—a®([o,T];Rd)・很明显Aa(g;
[t,
T])
<
Aa(g;
[0,
T])(:=
Aa(g)).假设Wa,1([t,T];Rd)是使得f
(s)
s
—
t)atf(s)-
f(y)l a+1dyi
ds
<
xs
—
y
」的[t,T]上的可测函数
f
构成的空间•显然
Wa,-([t,T];Rd)
c
Wa,1([t,T];Rd)及
||f
||a,1;[t,T]
<
(T
+
)|f
lla,^;[0,T]
•记(。”=宀(毎
+
a]
(2-)定义
2.1
假设
0
<
a
<
1/2.如果
/
G
Wa,1([t,T];Rdxk)和
g
G
IV
1-«,~([t,T];Rk),那
么下面定义的积分[几r)dg(r)
:=
(―1)a
/
(D0+f)(r)(DS-ags-)(r)dr
0
Jo(2.3)对所有的s
G
[t,T]存在,而且/(r)dg(r)|
<0|(DS-ags-)(r)|(D0+f
)(s)|ds(2.4)<
Aa(g;[t,T])f
l|a,1;[0,T].接下来,我们考虑如下的假设.No.1徐丽平等:混合型随机微分方程的传输不等式231(H1)存在正数Ci和一些0
G
(1
-
H,
1]使得对所有的t,
s
G
[0,T]及G便有|a(t,x)|
<
Ci(1
+
|x|),
|a(t,x)
一
a(t,y)|
<
Ci|x
一
y及|a(t,
x)
—
a(s,
x)|
<
Ci|t
—
s|©.(H2)对每一个k
=
1,
•••
和一些0
G
(1
-
H,
1],存在正数C2使得对所有的t,
s
G
[0,
T]
及x,y
G
有|bk(t,x)|
<
C2(1
+
|x|),
|bk(t,x)
—
bk(t,y)|
<
C2〔x
—
y及|bk(t,x)
—
bk
(s,
x)
|
<
C2
|t
一
屮-(H3)对每一个j
=
1,
•••
,r,
i
=
1,
•••
,d和一些0
G
(1
-
H,
1],存在正数C3使得对所有
的
s,t
G
[0,T]及
x,y
G
Rd
有|cj(t,x)|
<
c3(1
+
|x|),|cj(t,x)
-
cj(t,y)|
<
C3|x
-
y|,
和|cj(t,x)
-
cj(s,x)|
<
C3|t
-
s|0,
|dxicj(t,x)
-
dxicj(s,x)|
<
C3|t
-
s|(t,x)
-
dXicj(t,y)|
<
C3W
-
y下面的两个结果来自Mishura和Shevchenko[12].定理2.1在假设(H1)-(H3)下,方程(2.1)存在唯一解,而且llXllg;[0,T〕<
g考虑下面的方程序列Xn(t)
=
Xo
+
[
a(s,Xn(s))ds
+
[
b(s,Xn(s))dW(s)丿0
丿o+
f
c(s,Xn(s))dBH(s),
t
G
[0,T],
(2.5)丿0这里{BH,
n
>
1}
一列Hurst参数为H的分数布朗运动.定理2.2如果||Bh
-
BH||0,〜[0,t]依概率收敛到0,那么Xn(t)对t
一致的依概率收
敛到X(t).3主要结果在这一节,我们将使用逼近方法讨论方程(2.1)解的传输不等式.为此,对x
G
Rd,
n
>
1
记
kn(x)
=
|X|(|x|
A
n)及BH(t)
=
n「
k”(BH(s))ds.
(3.1)V(t-1/n)V0由文献[12,引理2.1]知对a
G
(1
-
H,
1/2)有||BH
-
kn(BH)||0g[0,T]
<
CKh(kn(BH))n1-H-a(3.2)232数学物理学报Vol.41A这里Kh(g)=
sup
气匕閒1是g的HOlder连续常数.显然0
\'
丿IIBH
—
BH
H0,8;[O,T]
<
|BH
—
kn(BH
)|〔0,8;[O,T]
+
IIBH
—
k”(BH
)||0,8;[O,T].注意到BH是连续的,于是它是有界的.因此||BH
—
kn(BH
)|〔0,8;[O,T]
t
0,
n
T结合(3.2)式和不等式Kh(kn(BH))
<
Kh(Bh)
<
8知
IIBf
—
BH
H0,8;[O,T]
T
0,
n
T
8.定理3.1假设条件(H1)-(H3)成立,是方程(2.1)解过程X(•,
Xo)的概率测度.那么概率测度在度量空间C([0,T];
Rd)上满足如下的T2(C)(a)
如果仏(九丁2):=
sup
|yi
-
Y2|,
71,72
G
C([0,T];Rd),OC
=
3TC4e12C2+3T(6+C3)2.(b)
如果d2(Yi,Y2)=
/
f
T
1/2|?i(t)
—
72(t)|2dd
,
71,72
G
C
([0,
T];
Rd),那么
C
=
3T2C4e12C2+3T(6+C3匚这里
C4
=
(M
+
C2)(1
+
T)
+
|b(0,
Xo)|,
M
是
X
的
HOlder
连续常数.
证定理的证明分为三步.第一步
构造序列X\"逼近X.对任意的n
>
1,假设Bf如(3.1)式定义.考虑下面的
随机微分方程Xn(t)
=
Xo
+
[
a(s,Xn(s))ds
+
[
b(s,Xn(s))dW(s)7o
7o+
fc(s,Xn(s))dBH(s),
t
G
[0,T].
Jo(3.3)由定理2.1和定理2.2知对任意的n
>
1,方程(3.3)存在唯一的解Xn(t),且Xn(t)对t
G
[0,
T]
依概率一致收敛到X(t),当n
t
8.进一步,Xn(t)是0
-阶Holder连续的.利用(3.1)式,我们知道(3.3)式能变为下面的ItO随机微分方程Xn(t)
=
Xo
+
/
an(s,Xn(s))ds
+
/
b(s,Xn(s))dW(s),
t
G
[0,T],Jo
Jo这里ran(t,
x)
=
a(t,
x)
+Cj(t,
x)BHn,j(t)j=1r=a(t,x)
+
\"
Cj(t,x)n(kn,j(BH(t))
—
kn,j(BH((t
—
1/n)
V
0))),
j=1(3.4)No.1徐丽平等:混合型随机微分方程的传输不等式233knj(X)是kn(x)的第j个分量”
第二步
对每一个n
>
1,假设是方程(3.3)解过程Xn(.,X°)的概率测度,Q是Rd上使得Q《的任意的概率测度.对每一个n
>
1,定义〜
Qn:=dQ忒(X(•,^))P
是(忆F)上的概率测度•回忆爛的定义、使用测度变换方法利用(3.5)式知H(Q„|P)^Jn
(3.5)d◎”
=
〃n
(联(X(小)))带(X(^,X0))dp=/
l屛黑)器dPX
厶
dPX0
Xo=H(Q|PXo).由文献[3]知,存在可料过程h„(t)0G
R满足fT
||Ms)『ds
<
g
P-a.s.及H(Qn|P)
=
H(Q|PXo
)
=
2Eq”
/
|h„(t)|2dt.〜
1
rT由Girsanov定理知对每一个n
>
1,过程(W”(t))©0,T]Wn(t)
=
W(t)
-
/„(s)ds丿0是概率空间(Q,
F,
Qn)上关于{Ft}t>0的布朗运动•于是对每一个n
>
1,在概率测度Qn下
{Xn(t,X0)}©0,T]满足Xn(t)
=
X0
+
f
an(s,Xn(s))ds
+
f
b(s,Xn(s))h”(s)ds
+
/^(s’Xn(s))dW”(s).
(3.6)
丿0
丿0
丿0现在对每一个n
>
1,考虑下面的方程在Qn下的解YnYn(t)
=
X0
+
f
an(s,Yn(s))ds
+
f
b(s,Yn(s))dW”(s).
丿0
丿0(3.7)因为k„(BH)有界,所以an对x是^pschitz连续和线性增长的.于是,对每二个n
>
1,
(3.7)
式程Qn下有唯一的解Yn,而且在Qn下Yn(.)的分布就是PX0
.因此,在Qn下(Xn,Yn)
是(Qn,
PX0
)的一对耦合,于是我们知<
Eq”
(|d2(X
n,Yn)|2)
=
Eq”
(/T
|X
n(t)
-
Yn(t)|2dt[W加(Q,
PXo)]2
<
Eq”(|d*(Xn,Yn)|2)
=
Eq”
gup」Xn(t)
-
Yn(t)|2接下来我们估计X\"和Yn关于d2和dg的距离.
由(3.6)和(3.7)式知Xn(t)
-
Yn(t)
=
/[an(s,Xn(s))
-
an(s,Yn(s))]ds
+
f
b(s,Xn(s))h”(s)dsJ0
丿0+
f[b(s,Xn(s))
-
b(s,Yn(s))]dWn(s)J0:=厶(t)
+
^2(t)
+
Za(t)
-
(3・8)234数学物理学报Vol.41A注意到an的Lipschitz常数与\"无关•实际上,利用一些简单的计算我们知an的Lipschitz
常数为C1
+
C3•从而,使用Holder\'s不等式得到岭”
o
=
EQ”
0冥0[[an(u,Xn(u))
—
a\"(u,Yn(u))]du『
[sup
|Xn(u)
—
Yn(u)|2ds.
O
O
t(C1
+
C)2E”3q(3.9)根据假设(H2)和X\"的Holder连续性,对任意的t
G
[0,
T]有|b(t,Xn(t))|
<
|b(t,Xn(t))
—
b(0,Xo)|
+
|b(0,Xo)<
|b(t,
Xn(t))
—
b(t,
Xo)|
+
|b(t,
Xo)
—
b(0,
Xo)|
+
|b(0,
Xo)<
C2|Xn(t))
—
X(0)|
+
C2护
+
|b(0,
Xo)<
(C2M
+
C2)护
+
|b(0,Xo)<
(M
+
C2)(1
+
T)
+
|b(0,Xo)|
:=
C4,这里M是Xn的Holder常数,并且与C1,
C2,
C3,
T,
X°,H及0无关”
因此,对每一个n
>
1知Eqn
sup
|/2(s)|2
<
C4TEq
/
|hn(t)|2dt.
\"O
fT\"
Jo(3.10)根据假设(H2)和Burkhold-Davis-Gundy\'s不等式知Eq
sup
|l3(s)|2
<
4Eq
f|b(s,Xn(s))
—
b(s,Yn(s))|2ds
O
%
JO<
4C2Eq
/
sup
|Xn(u)
—
Yn(u)|2ds.n
JO
O
sup
|Xn(s)
—
Yn(s)|2
<
3Eq
sO
O
sup
|I1(s)|2
+
3Eq
O
|/2(s)|2
+
3EqO
I12C2
+
3T(C1
+
C3)2[Eq”
supo
O
n(u)
—
Yn(u)|2ds(3•⑵+3TC:/
Eqn|hn(s)|2ds.于是,Gronwall\'s引理意味着对任意的t
>
0,有Eq”|Xn(t)
—
Yn(t)|2
<
3TC4e12C+3T(6+C3)2
/
Eq”|h”(s)|2ds.2
ftJ
o因此,我们得到广Td;(Xn,Yn)
<
3TC4e12C2
+3T(C1+C3)2
/
EQ」hn(s)|2ds,f
Td2(Xn,Yn)
<
3T2C4e12C2
+3T(6+°3)2
/
EQ」hn(s)|2ds No.1徐丽平等:混合型随机微分方程的传输不等式235及[Wfg(Q,
PJ0)]2
<
3TC4e12C2+3T(Cl+C3)2H(Q|PJ0),
[Wf2
(Q,
PJ0
)]2
<
3T2C4e12C2+3T(Ci+
C3)2H(Q|PJ0).
(3.13)(3.14)第三步
因为Xn(t)对t
G
[0,
T]
一致依概率收敛到X(t),
n
t
8,于是PX0
t
Px0当
n
t
8.由文献[15,定理3]和Wfg的定义知W茫(Q,
PX0)
T
W茫(Q,
PX0),
as
n
t
8及W『(Q,
PJ0)
t W『(Q,
PX0),当
n
t
8.另一方面,由H(Q|P)的定义知H(Q|P^0)
t
H(Q|Px0),当
n
t8.因此,在(3.13)式两边极限得到[Wfg(Q,PX0)]2
<
3TC4e12C2+3T(Cl+C3)2H(Q|Px0)及[W(2
(Q,
PX0
)]2
<
3T2C4e12C2+3T©
+C3)2H(Q|PX0).证毕.
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for
Mixed
Stochastic
Differential
Equa
tionsXu
Liping
Li
Zhi(School of
Information
and
Mathematics,
Yangtze
University,
Hubei
Jingzhou
434023)Abstract:
In
this
paper,
we
discuss
a
class
of
stochastic
differential
equations
containing
both
Wiener
process
and
fractional
Brownian
motion
with
Hurst
parameter
1/2
<
H
<
1.
By
using
some
transformation
technique
and
approximation
argument,
we
establish
the
quadratic
transportation
inequalities
for
the
law
of
the
solution
of
the
equations
under
investigation
under
the
d2
metric
and
the
uniform
metric
words:
Transportation
inequalities;
Mixed
stochastic
differential
equations;
Fractional
Brownian
motion;
Pathwise
(2010)
Subject
Classification:
60H15;
60H05
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