在本节中,我们考虑由fBm驱动的一类FSDE,如下所示:
$$\开始{对齐}X(t)=X_{0}+\int^{t}(t)_{0}页\bigl(s,X(s)\bigr)\,ds+\biggl\langle\int^{t}_{0}克\bigl(s,X(s)\bigr)\,dB_{H}\biggr\rangle,\quad X_{0}=X(0),\end{aligned}$$
(3.1)
哪里\(B_{H}\)是刘维尔形式的fBm\(H \ in(\压裂{1}{2},1)\),\(X_{0}:\Omega\rightarrow{\mathcal{F}}({\mathbb{R}})\)是FRV,\(f:I\times\Omega\times{mathcal{f}}({\mathbb{R}})\rightarrow{\mathcal{f}}、和\(g:I\times\Omega\times{\mathcal{F}}({\mathbb{R}})\rightarrow{\mathbb{R{}}\)相应的近似方程(3.1)是
$$\开始{对齐}X^{\epsilon}(t)=X_{0}+\int^{t}(t)_{0}f\bigl(s,X^{\epsilon}(s)\bigr)\,ds+\biggl\langle\int^{t}(t)_{0}g\bigl(s,X^{\epsilon}(s)\bigr)\,dB^{\ε}_{H}(s-)\biggr\rangle。\结束{对齐}$$
(3.2)
假设3.1
考虑以下关于方程系数的假设:
-
A1)
映射\(f:I\times\Omega\times{mathcal{f}}({\mathbb{R}})\rightarrow{\mathcal{f}}和\(g:I\times\Omega\times{\mathcal{F}}({\mathbb{R}})\rightarrow{\mathbb{R{}}\)是\({\mathcal{N}}\otimes{\mathcal{B}}_{d_{s}}|{\matchcal{B}{{d_}s}}\)-可测量和\({\mathcal{N}}\otimes{\mathcal{B}}_{d_{s}}|{\matchcal{B}({\mathbb{R}})\)-分别是可测量的。
-
A2)
对于每个\(u,v\在{\mathcal{F}}({\mathbb{R}})中\)每\(I中的t),存在一个常量\(L>0\)这样的话
$$\max\bigl\{d_{infty}\bigl(f(t,\omega,u),f(t、\omega、v)\biger),\bigl\ vert g(t,\ omega,u)-g(t、\ omega、v)\bigr\vert\bigr\}\leq Ld_{infty}(u,v)$$
-
A3)
对于每个\(u,v\在{\mathcal{F}}({\mathbb{R}})中\)以及每个\(I中的t),存在一个常量\(C>0\)这样的话
$$\max\bigl\{d_{\infty}\bigl(f(t,\omega,u),\langle 0\rangle\bigr),\bigl\ vert g(t,\ omega,u)\bigr\vert\bigr\}\leq C\bigle(1+d_{\ infty{\bigl$$
提议3.1
([13])
考虑 \(L^{2}中的X,Y\(I\次\Omega,{\mathcal{N}};{\mathbb{R}})\),然后
$$\begin{aligned}{\mathbb{E}}\sup_{u\in[0,t]}d^{2}_{\infty}\biggl(\biggl\langle\int^{u}_{0}X(s)\,dW(s)\biggr\rangle,\biggl\langle\int^{u}_{0}Y(s)\,dW(s)\biggr\rangle\biggr)\leq 4{\mathbb{E}}\int^{t}_{0}天^{2}_{\infty}\bigl(\bigl\langle X(s)\bigr\rangle,\bigl\ langle Y(s)\ bigr\langle\bigr)\,ds,\end{aligned}$$
(3.3)
对于每个 \(t\在I\中).
定理3.2
假设 \(f:I\times\Omega\times{mathcal{f}}({\mathbb{R}})\rightarrow{\mathcal{f}} 和 \(g:I\times\Omega\times{\mathcal{F}}({\mathbb{R}})\rightarrow{\mathbb{R{}}\) 因为映射满足假设(一个1)–(一个3)和 \(X_{0}\在{\mathcal{L}}^{2}中(\Omega,{\mathcal{A}}_{0{,P;{\matchal{F}}({\mathbb{R}})).然后等式. (3.2)具有强大的独特解决方案.
证明
考虑SDE(3.2),
$$\开始{对齐}X^{\epsilon}(t)=X_{0}+\int^{t}(t)_{0}f\bigl(s,X^{\epsilon}(s)\bigr)\,ds+\biggl\langle\int^{t}(t)_{0}g\bigl(s,X^{\epsilon}(s)\bigr)\,dB^{\ε}_{H}(s-)\biggr\rangle。\结束{对齐}$$
(3.4)
通过等式(2.3),我们可以写
$$\开始{对齐}X^{\epsilon}(t)={}&X_{0}+\int^{t}(t)_{0}f\bigl(s,X^{\epsilon}(s)\bigr)\,ds\\&{}+\biggl\langle\int^{t}(t)_{0}\alpha\varphi^{\epsilon}(s)g\bigl(s,X^{\epsilon}(s)\bigr)\,ds+\int^{t}(t)_{0}\epsilon^{\alpha}g\bigl(s,X^{\epsilen}\bigr)\,dW(s)\biggr\rangle。\结束{对齐}$$
(3.5)
让我们考虑Picard迭代
$$\begin{aligned}X^{\epsilon}_{n}(t)={}&X_{0}+\int^{t}(t)_{0}页\bigl(s,X^{\epsilon}_{n-1}(s)\bigr)\,ds\\&{}+\biggl\langle\int^{t}(t)_{0}\alpha\varphi^{\epsilon}g\bigl(s,X^{\ebsilon}_{n-1}(s)\bigr)\,ds+\int^{t}(t)_{0}\epsilon^{\alpha}g\bigl(s,X^{\epsilon}_{n-1}\bigr)\,dW(s)\biggr\rangle,\quad\text{a.e.},\end{aligned}$$
(3.6)
对于\(n=1,2,\ldots\) , 并且对于每个\(I中的t)、和\(X_{0}(t)=X_{0}\)。对于\(t\在I\中)和\(在{\mathbb{n}}\中)我们表示
$$j_{n}(t)={\mathbb{E}}\sup_{u\in[0,t]}d^{2}_{\infty}\bigl(X^{\epsilon}_{n}(u),X^{\ epsilon{{n-1}(u)\bigr)$$
然后,通过命题2.1,3.1,假设A3,可以写成
$$\begin{aligned}j_{1}(t)={}&{\mathbb{E}}\sup_{u\in[0,t]}d^{2}_{\infty}\biggl(\int^{u}_{0}页\bigl(s,X^{\epsilon}_{0}\bigr)\,ds\\&{}+\biggl\langle\int^{u}_{0}\alpha\varphi^{\epsilon}g\bigl(s,X^{\ebsilon}_{0}(s)\bigr)\,ds+\int^{u}_{0}\epsilon^{\alpha}g\bigl(s,X^{\epsilon}_{0}(s)\bigr,dW(s)\ biggr\rangle,\langle 0\rangle\biggr)\\leq{}&3{\mathbb{E}}\sup_{u\in[0,t]}d^{2}_{\infty}\biggl(\int^{u}_{0}f\bigl(s,X^{\epsilon}_{0}\bigr)\,ds,\langle 0\rangle\biggr)\\&{}+3{\mathbb{E}}\sup_{u\in[0,t]}d^{2}_{\infty}\biggl(\biggl\langle\int^{u}_{0}\alpha\varphi^{\epsilon}g\bigl(s,X^{\ebsilon}_{0}(s)\bigr,ds\biggr\rangle,langle 0\rangle\biggr)\\&{}+3{\mathbb{E}}\sup_{u\in[0,t]}d^{2}_{\infty}\biggl(\biggl\langle\int^{u}_{0}\epsilon^{\alpha}g\bigl(s,X^{\epsilon}_{0}(s)\bigr)\,dW(s)\ biggr\rangle,\langle 0\rangle\biggr)\\leq{}&3t{\mathbb{E}}\int^{t}(t)_{0}天^{2}_{\infty}\bigl(f\bigle(s,X^{\epsilon}_{0}\bigr),\langle 0\rangle\biger),ds+3\alpha^{2}{\mathbb{E}}\sup_{u\in[0,t]}d^{2}_{\infty}\biggl(\biggl\langle\int^{u}_{0}\varphi^{\epsilon}(s)g\bigl(s,X^{\epsilon}_{0}(s)\bigr)\,ds\biggr\rangle,\langle0\rangle\biggr)\ \&{}+12\epsilon ^{2\alpha}{\mathbb{E}}}\ int^{t}(t)_{0}天^{2}_{\infty}\bigl(\bigl\langle g\bigle(s,X^{\epsilon}_{0}\bigr)\bigr\rangle,\langle 0\rangle\biger)\,ds\\leq{}&6C^{2}\biggl(T+4\epsilon^{2\alpha}\biger n}\bigr]^{0}\bigrar\vert\!\bigr\vert\}\sup{u\in[0,t]}d^{2}_{\infty}\biggl(\biggl\langle\int^{u}_{0}\varphi^{\epsilon}g\bigl(s,X^{\ebsilon}_{0}(s)\bigr)\,ds\biggr\rangle,langle 0\rangle\biggr),\end{aligned}$$
对于\(α=H-\压裂{1}{2}>0\).因此
$$\开始{对齐}j_{1}(t)\leq{}&6C^{2}\bigl(t+4\epsilon^{2\alpha}\bigr)\bigle(1+{\mathbb{E}}\bigle\vert\!\bigl\vert\bigl[X^{\epsilen}\biger]^{0}\bigrar\vert\[0,t]}中的alpha^{2}{\mathbb{E}}\sup_{u\^{2}_{\infty}\biggl(\biggl\langle\int^{u}_{0}\varphi^{\epsilon}g\bigl(s,X^{\ε}_{0}(s)\bigr)\,ds\biggr\rangle,\langle 0\rangle\biggr)。\结束{对齐}$$
(3.7)
我们有
$$\开始{aligned}&{mathbb{E}}\sup_{u\在[0,t]}d中^{2}_{\infty}\biggl(\biggl\langle\int^{u}_{0}\varphi^{\epsilon}g\bigl(s,X^{\ε}_{0}(s)\bigr,ds\biggr\rangle,langle 0\rangle\biggr)\\&\quad\leq{\mathbb{E}}\sup_{u\in[0,t]}d^{2}_{H} \biggl(\biggl\{\int^{u}_{0}\varphi^{\epsilon}g\bigl(s,X^{\ε}_{0}(s)\bigr)\,ds\biggr\},\{0\}\biggr)\\&\quad\leq{\mathbb{E}}\sup_{u\in[0,t]}\bigbl(\int^{u}_{0}\varphi^{\epsilon}g\bigl(s,X^{\ebsilon}_{0}(s)\bigr)\,ds\biggr)^{2}。\结束{对齐}$$
(3.8)
通过应用(2.4)至(3.8)以及我们得到的Hölder不等式
$$\开始{aligned}&{mathbb{E}}\sup_{u\in[0,t]}\biggl(\int^{u}_{0}\varphi^{\epsilon}g\bigl(s,X^{\ebsilon}_{0}(s)\bigr)^{2}\\&\quad={\mathbb{E}}\sup_{u\in[0,t]}\biggl(\int^{u}_{0}\biggl(\int^{s}_{0}(s-r+\epsilon)^{\alpha-1}\,dW(r)\biggr)g\bigl(s,X^{\epsilon}_{0}\bigr)\,ds\bigger)^{2}\\&\quad={mathbb{E}}\sup_{u\in[0,t]}\biggl(\int^{u}_{0}\int^{u}_{s} g\bigl(r,X^{\epsilon}_{0}(r)\bigr)(r-s+\epsilon)^{\alpha-1}\,dr\,dW(s)\biggr)^{2}\&&quad\leq 4{\mathbb{E}}\ int^{t}(t)_{0}\biggl(\int^{t}(t)_{s} g\bigl(r,X^{\epsilon}_{0}(r)\bigr)(r-s+\epsi隆)^{\alpha-1}\,dr\biggr)^{2}\,ds\\&\quad\leq 4{\mathbb{E}}\int^{t}(t)_{0}\biggl(\int^{t}(t)_{s} g^{2}\bigl(r,X^{\epsilon}_{0}(r)\bigr)(r-s+\epsilen)^{\alpha-1}\,dr\biggr)\biggl(\int^{t}(t)_{s} (r-s+\epsilon)^{\alpha-1}\,dr\biggr^{t}(t)_{0}g^{2}\bigl(r,X^{\epsilon}_{0}(r)\bigr^{t}(t)_{0}g^{2}\bigl \bigr\vert\!\bigr\fort\!\biger\vert^{2}\bigr)T.\end{aligned}$$
(3.9)
因此,从(3.7)和(3.9)我们获得
$$开始{对齐}j_{1}(t)\leq 6C^{2}\bigl{2}\较大)t,\结束{对齐}$$
(3.10)
对于每个\(I中的t)然后,类似地,
$$开始{对齐}j_{n+1}(t)&\leq 3\bigl(t+4\epsilon^{2\alpha}+4(t+\epsilen)^{2\\alpha}\bigr)L^{2}{mathbb{E}}\int^{t}(t)_{0}d^{2}_{\infty}\bigl(X^{\epsilon}_{n}(u),X^{\ epsilon{{n-1}(u)\bigr)^{t}(t)_{0}{\mathbb{E}}\sup_{u\in[0,s]}d^{2}_{\infty}\bigl(X^{\epsilon}_{n}(u),X^{\ epsilon{{n-1}(u)\bigr)^{t}(t)_{0}个_{n} (s)\,ds.\结束{对齐}$$
因此
$$j_{n}(t)\leq 2C^{2}3^{n} \bigl(t+4\epsilon^{2\alpha}+4(t+\epsilen)^{2\\alpha}\biger)^{n}\bigle(1+{\mathbb{E}}\bigl\vert\!\bigl\ vert\ t^{n}}{n!},I中的四元t,{mathbb{n}}中的n$$
应用切比雪夫不等式,如下所示
$$P\biggl(在I}d中为\sup_{u\^{2}_{\infty}\bigl(X^{\epsilon}_{n}(u),X^{\ epsilon{{n-1}(u)\bigr)>\frac{1}{2^{n}}\biggr)\leq 2^{n} j_{n} (T)$$
系列\(总和^{infty}_{n=1}2^{n} j个_{n} (T)\)是收敛的。从Borel–Cantelli引理,我们导出
$$P\biggl(在I}d中为\sup_{u\^{2}_{\infty}\bigl(X^{\epsilon}_{n}(u),X^{\ epsilon{{n-1}(u)\bigr)>\frac{1}{(\sqrt{2})^{n}}\text{无限频繁}\biggr)=0$$
几乎所有人\(欧米茄),存在\(n_{0}(ω)\)这样的话
I}d中的$$\sup_{u\^{2}_{\infty}\bigl(X^{\epsilon}_{n}(u),X^{\ epsilon{{n-1}(u)\bigr)\leq\frac{1}{(\sqrt{2})^{n}},\quad\text{if}n\geqn{0}$$
序列\(\{X^{\epsilon}_{n}(\cdot,\omega)\}\)一致收敛于\(d)^{2}_{\infty}\)-连续模糊过程\(\widetilde{X}^{\epsilon}(\cdot,\omega)\)对于每个\(\omega\in\omega_{c}\),其中\(\Omega_{c}\在{\mathcal{A}}\中)和\(P(Omega_{c})=1)。我们可以定义映射\(X^{\epsilon}:I\times\Omega\rightarrow{\mathcal{F}}({\mathbb{R}})\),作为\(X^{\epsilon}(\cdot,\omega)=宽波浪号{X}^{\epsilon}(\cdot,\omega)\)如果\(欧米茄)和\(X^{\epsilon}(\cdot,\omega)\)作为自由选择的模糊函数,当\(欧米茄) ∖ \(\欧米茄{c}\)。对于每个\(\alpha\在[0,1]\中)以及每个\(I中的t)有了a.e.,我们就有了
$$d_{H}\bigl(\bigl[X^{\epsilon}_{n}(t)\bigr]^{\alpha},\bigl[X^{\ epsilon}(t)\birr]^{\ alpha}\bigr)\rightarrow0\quad\text{as}n\rightarror\infty$$
因此,\(X^{\epsilon}\)将是一个连续的模糊随机过程。然后,通过\(X^{\epsilon}_{n}\在{\mathcal{L}}^{2}中(I\times\Omega,{\mathcal{n}};{\matchcal{F}}({\mathbb{R}})),我们得到\(X^{\epsilon}\在{\mathcal{L}}^{2}中(I\times\Omega,{\mathcal{N}};{\matchal{F}}({\mathbb{R}}))因此,作为n个到无穷大,我们可以验证
$${\mathbb{E}}\sup_{t\inI}\biggl[d^{2}_{\infty}\bigl(X^{\epsilon}_{n}(t),X^{\ epsilonneneneep(t)\bigr)+d^{2}_{\infty}\biggl(X^{\epsilon}_{n}(t),X^{\ epsilon{{0}+\int^{t}(t)_{0}页\bigl(s,X^{\epsilon}\bigr)\,ds+\biggl\langle\int^{t}(t)_{0}克\bigl(s,X^{\epsilon}\bigr)\,dB^{\ebsilon}_{H}\biggr\rangle\biggr)\biggr]^{2}$$
趋于零。然后
I}d中的$${\mathbb{E}}\sup_{t\^{2}_{\infty}\biggl[\biggl(X^{\epsilon}(t),X^{\spilon}_{0}+\int^{t}(t)_{0}页\bigl(s,X^{\epsilon}\bigr)\,ds+\biggl\langle\int^{t}(t)_{0}克\bigl(s,X^{\epsilon}\bigr)\,dB^{\ebsilon}_{H}\biggr\rangle\biggr)\biggr]=0$$
因此
I}d中的$$\sup_{t\^{2}_{\infty}\biggl[\biggl(X^{\epsilon}(t),X^{\spilon}_{0}+\int^{t}(t)_{0}页\bigl(s,X^{\epsilon}(s)\bigr)\,ds+\biggl\langle\int^{t}(t)_{0}克\bigl(s,X^{\epsilon}\bigr)\,dB^{\ebsilon}_{H}\biggr\rangle\biggr)\biggr]=0$$
这表明了强解的存在性。
现在,\(X^{\epsilon},Y^{\ebsilon}:I\times\Omega\rightarrow{\mathcal{F}}({\mathbb{R}})\)假设是强解。考虑
$$j(t)={\mathbb{E}}\sup_{u\in[0,t]}d^{2}_{\infty}\bigl(X^{\epsilon}(u),Y^{\ε}(u)\bigr)$$
然后,通过与存在情况类似的计算,我们得到
$$j(t)\leq 3\bigl(t+4\epsilon^{2\alpha}+4(t+\epsilen)^{2\\alpha}\bigr)L^{2}{mathbb{E}}\int^{t}(t)_{0}天^{2}_{\infty}\bigl(X^{\epsilon}(s),Y^{\ε}\bigr^{t}(t)_{0}个(s) \,ds$$
Gronwall不等式的实现导致\(j(t)=0)对于\(t\在I\中).然后
I}d中的$$\sup_{t\^{2}_{\infty}\bigl(X^{\epsilon}(t),Y^{\ε}(t)\bigr)=0,\quad\text{a.e.}$$
由此完成唯一性的证明。 □
引理3.1
对于每个
\(epsilon>0\)
和
\(0<\alpha<\frac{1}{2}\)
我们有
$$\开始{aligned}\int^{t}(t)_{s} \bigl((r-s+\epsilon)^{\alpha-1-(r-s)^{\ alpha-1}\bigr)\,dr\leq\frac{\alalpha+1}{\alfa}{\epsilen}^{\alpha}。\结束{对齐}$$
(3.11)
证明
我们将有限增量公式应用于函数\(f(x)=x^{α-1})以获得
$$开始{对齐}(x+\epsilon)^{\alpha-1}-x^{\alpha-1}=(\alpha-1)$$
(3.12)
然后
$$开始{aligned}\bigl\vert(r-s+\epsilon)^{\alpha-1}-(r-s)^{\salpha-1{\bigr\vert\leq\vert\alpha-1\vert\vert-r-s\vert^{\阿尔法-2}\epsiron。\结束{对齐}$$
(3.13)
因此
$$\开始{aligned}\int^{t}_{s} \bigl\vert(r-s+\epsilon)^{\alpha-1}-(r-s)^{\ alpha-1{\bigr\vert^{t}(t)_{s+\epsilon}\bigl\vert^{t}(t)_{s+\epsilon}\vert r-s\vert^{\alpha-2}\,dr.\end{aligned}$$
(3.14)
因此
$$\开始{aligned}\int^{t}(t)_{s} \bigl\vert(r-s+\epsilon)^{\alpha-1}-(r-s)^{\alpha-1-}\bigr\vert^{t}(t)_{s+\epsilon}(r-s)^{\alpha-2},dr\\&\leq\frac{1}{\alalpha}{\epsilon}^{\alpha}+\vert\alpha-1\vert\epsilon\biggl。\结束{对齐}$$
(3.15)
□
提议3.2
解决方案 \(X^{\epsilon}(t)\) 等式的. (3.2)收敛到解 \(X(t)\) 等式的. (3.1)在里面 \({\mathcal{L}}^{2}(I\次\Omega)\) 作为 \(\epsilon\rightarrow 0\) 一致地关于 \(t\在[0,t]\中).
证明
考虑等式(3.1)方程的相应近似值如下:
$$\开始{aligned}&X(t)=X_{0}+\int^{t}_{0}f\bigl(s,X(s)\bigr)\,ds+\biggl\langle\int^{t}(t)_{0}g\bigl(s,X(s)\bigr)\,dB_{H}(s)\ biggr\rangle,\end{aligned}$$
(3.16)
$$\开始{对齐}&X^{\epsilon}(t)=X_{0}+\int^{t}(t)_{0}f\bigl(s,X^{\epsilon}(s)\bigr)\,ds+\biggl\langle\int^{t}(t)_{0}g\bigl(s,X^{\epsilon}(s)\bigr)\,dB^{\ε}_{H}(s-)\biggr\rangle。\结束{对齐}$$
(3.17)
我们可以写
$$\开始{aligned}&{mathbb{E}}\sup_{u\在[0,t]}d中^{2}_{\infty}\bigl(X(u),X^{\epsilon}(u)\bigr)\\&\quad\leq 2{\mathbb{E}}\sup_{u\in[0,t]}\int^{u}_{0}天^{2}_{\infty}\bigl(f\bigl[s,X(s)\bigr),f\bigle(s,X^{\epsilon}(s)\figr)\,ds\\&\qquad{}+2{\mathbb{E}}\sup_{u\in[0,t]}d^{2}_{\infty}\biggl(\biggl\langle\int^{u}_{0}g\bigl(s,X(s)\bigr)\,dB_{H}(s)\ biggr\rangle,\biggl\langle\int^{u}_{0}g\bigl(s,X^{\epsilon}\bigr)\,dB^{\ε}_{H}(s)\biggr\rangle\biggr)。\结束{对齐}$$
然后
$$\开始{aligned}&{mathbb{E}}\sup_{u\在[0,t]}d中^{2}_{\infty}\bigl(X(u),X^{\epsilon}(u)\bigr)\\&\quad\leq 2{\mathbb{E}}\sup_{u\in[0,t]}\int^{u}_{0}天^{2}_{\infty}\bigl(f\bigl[s,X(s)\bigr),f\bigle(s,X^{\epsilon}(s)\figr)\,ds\\&\qquad{}+4{\mathbb{E}}\sup_{u\in[0,t]}d^{2}_{\infty}\biggl(\biggl\langle\int^{u}_{0}g\bigl(s,X(s)\bigr)\,dB_{H}(s)\ biggr\rangle,\biggl\langle\int^{u}_{0}g\bigl(s,X(s)\bigr)\,dB^{\epsilon}_{H}\biggr\rangle\biggr)\\&\qquad{}+4{\mathbb{E}}\sup_{u\in[0,t]}d^{2}_{\infty}\biggl(\biggl\langle\int^{u}_{0}g\bigl(s,X(s)\bigr)\,dB^{\epsilon}_{H}(s)\ biggr\rangle,\biggl\langle\int^{u}_{0}g\bigl(s,X^{\epsilon}\bigr)\,dB^{\ε}_{H}(s)\biggr\rangle\biggr)\\&\quad=2{\mathbb{E}}\int^{t}(t)_{0}天^{2}_{\infty}\bigl(f\bigl(s,X(s)\bigr),f\bigle(s,X^{\epsilon}(s)\figr)\,ds\\&\qquad{}+4{\mathbb{E}}\sup_{u\in[0,t]}\biggl\vert\int^{u}_{0}g\bigl(s,X^{\epsilon}(s)\bigr)\ bigl^{u}_{0}(g\bigl(s,X(s)\bigr)-g\bigle。\结束{对齐}$$
应用公式(2.3)得到
$$\开始{aligned}&{mathbb{E}}\sup_{u\在[0,t]}d中^{2}_{\infty}\bigl(X(u),X^{\epsilon}(u)\bigr)\\&\quad\leq 2{\mathbb{E}}\int^{t}(t)_{0}天^{2}_{\infty}\bigl(f\bigl(s,X(s)\bigr),f\bigle(s,X^{\epsilon}(s)\figr)\,ds\\&\qquad{}+8{\epsilon}^{2\alpha}{\mathbb{E}}\sup_{u\in[0,t]}\biggl\vert\int^{u}_{0}g\bigl(s,X^{\epsilon}\bigr)\,dW(s)\biggr\vert^{2}\\&\qquad{}+8{\alpha}^{2{\mathbb{E}}\sup_{u\in[0,t]}\biggl\vert\int^{u}_{0}g\bigl(s,X^{\epsilon}(s)\bigr)\biggl(\int^{s}_{0}\bigl((s-r+\epsilon)^{\alpha-1}-(s-r)^{\alpha-1-}\bigr)\,dW(r)\biggr)\,ds\biggr\vert^{2}\\&\qquad{}+8{\alfa}^{2{\mathbb{E}}\sup_{u\in[0,t]}\biggl\vert\int^{u}_{0}(g\bigl(s,X(s)\bigr)-g\bigle(s^{s}_{0}(s-r+\epsilon)^{\alpha-1}\,dW(r)\biggr)\,ds\biggr\vert^{2}\\&\qquad{}+8{\epsilen}^{2\alpha}{\mathbb{E}}\sup_{u\in[0,t]}\biggl\vert\int^{u}_{0}(g\bigl(s),X(s)\bigr)-g\bigle(s,X^{\epsilon}(s)\biger)\,dW(s)\ biggr\vert^{2}。\结束{对齐}$$
(3.18)
然后
$$\开始{aligned}&{mathbb{E}}\sup_{u\在[0,t]}d中^{2}_{\infty}\bigl(X(u),X^{\epsilon}(u)\bigr)\\&\quad\leq 2{\mathbb{E}}\int^{t}_{0}天^{2}_{\infty}\bigl(f\bigl(s,X(s)\bigr),f\bigle(s,X^{\epsilon}(s)\figr)\,ds\\&\qquad{}+8{\epsilon}^{2\alpha}{\mathbb{E}}\sup_{u\in[0,t]}\biggl\vert\int^{u}_{0}g\bigl(s,X^{\epsilon}\bigr)\,dW(s)\biggr\vert^{2}\\&\qquad{}+8{\alpha}^{2{\mathbb{E}}\sup_{u\in[0,t]}\biggl\vert\int^{u}_{0}\int^{u}_{s} g\bigl(r,X^{\epsilon}(r)\bigr)\bigle((r-s+\epsilen)^{\alpha-1}-(r-s)^{\ alpha-1{\biger)\,dr\,dW(s)\biggr\vert^{2}\\&\qquad{}+8{\alfa}^{2{{\mathbb{E}\sup_{u\in[0,t]}\biggl\vert\int^{u}_{0}\int^{u}_{s} (g\bigl(r,X(r)\bigr)-g\bigle(s,X^{\epsilon}(s)\biger)(r-s+\epsilen)^{\alpha-1}\,dr\,dW(s)\ biggr\vert^{2}\\&\qquad{}+8{\epsilon}^{2\alpha}{\mathbb{E}}\sup_{u\in[0,t]}\biggl\vert\int^{u}_{0}(g\bigl(s),X(s)\bigr)-g\bigle(s,X^{\epsilon}(s)\biger)\,dW(s)\ biggr\vert^{2}。\结束{对齐}$$
(3.19)
应用Doob不等式、Hölder不等式和Itóisometry性质得到
$$\begin{aligned}&{\mathbb{E}}\sup_{u\in[0,t]}d^{2}_{\infty}\bigl(X(u),X^{\epsilon}(u)\bigr)\\&\quad\leq 2{\mathbb{E}}\int^{t}(t)_{0}天^{2}_{\infty}\bigl(f\bigl[s,X(s)\bigr),f\bigle(s,X^{\epsilon}(s)\figr)\,ds+32{\epsilon}^{2\alpha}{\mathbb{E}}\int^{t}(t)_{0}g^{2}\bigl(s,X^{\epsilon}\bigr)\,ds\\&\qquad{}+32{\alpha}^{2{\mathbb{E}}\int^{t}(t)_{0}\biggl[\int^{t}(t)_{s} g^{2}\bigl(r,X^{\epsilon}(r)\bigr)\ bigl^{t}(t)_{s} \bigl((r-s+\epsilon)^{\alpha-1-(r-s)^{\ alpha-1-}\bigr)\,dr\biggr]\,ds\\&\qquad{}+32{\alfa}^{2}{\mathbb{E}}\int^{t}(t)_{0}(\int^{t}(t)_{s} \bigl(g\bigl-(r,X(r)\bigr)-g\bigle(s,X^{\epsilon}(s)\biger)^{2}(r-s+\epsilen)^{\alpha-1}\,dr\bigr)\\&\qquad{}\times\biggl(\int^{t}(t)_{s} (r-s+\epsilon)^{\alpha-1}\,dr\biggr)\,ds\\&\qquad{}+32{\epsilon}^{2\alpha}{\mathbb{E}}\int^{t}_{0}(g\bigl(s),X(s)\bigr)-g\bigle(s,X^{\epsilon}\biger)^{2}\,ds.\end{aligned}$$
(3.20)
通过类似的论据(3.9),来自(3.11)和假设\((A1)\)–\((A3)\),我们推断
$$\开始{aligned}&{mathbb{E}}\sup_{u\在[0,t]}d中^{2}_{\infty}\bigl(X(u),X^{\epsilon}(u)\bigr)\\&\quad\leq 2L^{2}{\mathbb{E}}\int^{t}(t)_{0}天^{2}_{\infty}\bigl(X(s),X^{\epsilon}\bigr)\,ds+64{\epsilon}^{2\alpha}C^{2}\int^{t}(t)_{0}\bigl(1+{\mathbb{E}}\bigle\vert\!\bigr\vert\!\bigr\vert^{2}\bigr)\,ds\\&\qquad{}+64C^{2{(T+\epsilon)^{2\alpha}\frac{\alpha+1}{\ alpha}{\epsilen}^{\alfa}\int^{t}(t)_{0}\bigl(1+{\mathbb{E}}\bigle\vert\!\较大\垂直\!\bigr\vert^{2}\bigr)\,ds\\&\qquad{}+32L^{2{(T+\epsilon)^{2\alpha}{\mathbb{E}}\int^{t}(t)_{0}天^{2}_{\infty}\bigl(X(s),X^{\epsilon}\biger)\,ds\\&\qquad{}+32L^{2}{\epsilon}^{2\alpha}{\mathbb{E}}\int^{t}(t)_{0}天^{2}_{\infty}\bigl(X(s),X^{\epsilon}\bigr)\,ds.\end{aligned}$$
(3.21)
因此
$$\开始{aligned}&{mathbb{E}}\sup_{u\在[0,t]}d中^{2}_{\infty}\bigl(X(u),X^{\epsilon}(u)\biger^{t}(t)_[0,s]}d中的{0}{\mathbb{E}}\sup_{u\^{2}_{\infty}\bigl(X(u),X^{\epsilon}(u)\biger)\,ds\\&\qquad{}+\biggl(64{\epsilon}^{2\alpha}C^{2}+64C^{2](T+\epsilen)^{t}_{0}\bigl(1+{\mathbb{E}}\bigle\vert\!\较大\垂直\!\bigr\vert^{2}\bigr)\,ds.\end{aligned}$$
(3.22)
根据Gronwall引理,\([0,t]}d中的{\mathbb{E}}\sup_{u\^{2}_{\infty}(X(u),X^{\epsilon}(u))\rightarrow 0\)作为\(\epsilon\rightarrow 0\),这就完成了证明。 □
3.1金融学示例
以下清晰的SFDE通常用于财务建模:
$$\开始{对齐}X(t)=X_{0}+\int^{t}(t)_{0}\mu X(s)\,ds+\int^{t}(t)_{0}\sigma X(s)\,dB_{H}(s),\quad X_{0}=X(0),\end{aligned}$$
(3.23)
其中基础随机过程是fBm。fBm的长程相关性和自相似性使得该过程适合于描述金融量。另一方面,我们可以通过包含不确定性的方程来模拟价格动态。这导致在方程中使用模糊过程进行建模。在线性系数的情况下,我们得到了方程(3.1)。因此,考虑满足定理假设的分数FSDE3.2如下:
$$\开始{对齐}X(t)=X_{0}+\int^{t}(t)_{0}\mu X(s)\,ds+\biggl\langle\int^{t}(t)_{0}\frac{\sigma}{2}\bigl(X^{1}_{l} (s)+X^{1}_{u} (s)\较大)\,dB_{H}(s)\biggr\rangle,\end{aligned}$$
(3.24)
哪里\(X:{\mathbb{R}}_{+}\times\Omega\rightarrow{\mathcal{F}}({\mat血红蛋白{R})\),\(B_{H}\)是fBm,\(X)^{1}_{l} ,X^{1}_{u} :{\mathbb{R}}_{+}\times\Omega\rightarrow{\mathbb{R{}}\)这样的话\([X(t)]^{1}=[X^{1}_{l} (t),X^{1}_{u} (t)]\),\(X_{0}\在{\mathcal{L}}^{2}中(\Omega,{\mathcal{A}}_{0{,P;{\matchal{F}}({\mathbb{R}}))、和\({\mathbb{R}}中的\mu,\sigma\)。为了找到解决方案的封闭显式形式(3.24),用于\(\mu\geq 0\),我们得到以下方程组:
$$\开始{aligned}\textstyle\begin{cases}X^{1}_{l} (t)=X^{1}_{l} (0)+\int^{t}(t)_{0}\mu X^{1}_{l} (s)\,ds+\int^{t}(t)_{0}\frac{\sigma}{2}(X^{1}_{l} (s)+X^{1}_{u} (s),dB_{H},\\X^{1}_{u} (t)=X^{1}_{u} (0)+\int^{t}(t)_{0}\mu X^{1}_{u} (s)\,ds+\int^{t}(t)_{0}\frac{\sigma}{2}(X^{1}_{l} (s)+X^{1}_{u} (s)\,dB_{H},\结束{cases}\显示样式\结束{aligned}$$
然后
$$\开始{对齐}X^{1}_{l} (t)+X^{1}_{u} (t)={}&X^{1}_{l} (0)+X^{1}_{u} (0)+\int^{t}(t)_{0}\mu\bigl(X^{1}_{l} (s)+X^{1}_{u} (s)\bigr)\,ds\\&{}+\int^{t}(t)_{0}\sigma\bigl(X^{1}_{l} (s)+X^{1}_{u} (s)\较大)\,dB_{H}(s)。\结束{对齐}$$
(3.25)
通过方程式(2.3)式的近似形式(3.25)是
$$开始{对齐}X^{\epsilon 1}_{l}(t)+X^{\ssilon 1\{u}^{t}(t)_{0}\bigl(\mu+\sigma\alpha\varphi^{\epsilon}(s)\bigr)\bigl(X^{\epsilon 1}_{l}(s)+X^{\epsilon 1}_{u}(s)\bigr)\,ds\\&{}+\int^{t}(t)_{0}\sigma\epsilon^{\alpha}\bigl(X^{\epsiron1}_{l}(s)+X^{\ epsilon1}__{u}(s\bigr)\,dW(s)。\结束{对齐}$$
(3.26)
因此,从清晰线性SDE的显式解来看,有一个唯一的解,
$$开始{对齐}X^{\epsilon 1}_{l}(t)+X^{\spilon 1{{u}^{t}(t)_{0}\varphi^{\epsilon}(s)\,ds-\frac{1}{2}\sigma^{2}\ epsilon^{2\alpha}t+\sigma\epsilen^{\alpha}W(t)\biggr)\\&=\bigl{H}(t)-\压裂{1}{2}\西格玛^{2}\epsilon^{2\alpha}t\biggr)。\结束{对齐}$$
(3.27)
现在,对于每一个\([0,1]\中的alpha\),我们采用类似的程序获得以下系统:
$$\开始{aligned}\textstyle\begin{cases}X^{\epsilon\alpha}_{l}(t)=X^{\ epsilon\ alpha}_{l}_(0)+\int^{t}(t)_{0}\mu X^{\epsilon\alpha}_{l}\,ds+\int^{t}_{0}\frac{\sigma}{2}(X^{\epsilon1}_{l}(s)+X^{\ epsilon1\{u}^{t}(t)_{0}\mu X^{\epsilon\alpha}_{u}\,ds+\int^{t}(t)_{0}\frac{\sigma}{2}(X^{\epsilon1}_{l}(s)+X^{\ epsilon1\{u}(s))。\结束{cases}\displaystyle\end{aligned}$$
对于\(\mu\geq 0\),我们应用该解决方案(3.27)获取以下系统:
$$开始{对齐}X^{\epsilon\alpha}_{l}(t)={}&X^{\ epsilon\ alpha}_{l}(0)+\int^{t}(t)_{0}\mu X^{\epsilon\alpha}_{l}\,ds\\&{}+\int^{t}(t)_{0}\frac{\sigma}{2}\bigl(X^{\epsilon1}_{l}(0)+X^{\ epsilon1}_{u}),\\X^{\epsilon\alpha}_{u}(t)={}&X^{\ epsilon\ alpha}_{u}_(0)+\int^{t}(t)_{0}\mu X^{\epsilon\alpha}_{u}\,ds\\&{}+\int^{t}(t)_{0}\frac{\sigma}{2}\bigl(X^{\epsilon1}_{l}(0)+X^{\ epsilon1}_{u}),\结束{对齐}$$
或者根据维纳过程W公司,我们有
$$开始{对齐}X^{\epsilon\alpha}_{l}(t)={}&X^{\ epsilon\ alpha}_{l}(0)+\int^{t}_{0}\mu X^{\epsilon\alpha}_{l}(s)\,ds\\&{}+\alpha\frac{\sigma}{2}\bigl(X^{\\epsilon1}_{1}(0)+X^{\ epsilon 1}__{u}(O)\bigr)\\&{{}\times\int^{t}(t)_{0}\varphi^{\epsilon}(s)\exp\biggl(\mu s+\sigma\alpha\int^{s}_{0}\varphi^{\epsilon}(u)\,du-\frac{1}{2}\sigma^{2}\epsilon ^{2 \alpha}s+\ sigma\epsilon ^{\alpha}W(s)\biggr)\,ds\\&{}+\ epsilon ^{\alpha}\frac{\sigma}{2}\bigl(X^{\epsilon 1}_{l}(0)+X^{\epsilon 1}_{u}(0)\bigr)\\&{}\times\int^{t}(t)_{0}\exp\biggl(\mu s+\sigma\alpha\int^{s}_{0}\varphi^{\epsilon}(u)\,du-\frac{1}{2}\sigma^{2}\ epsilon^{2\alpha}s+\sigma\epsilon^{\alpha}W(s)\biggr)\,dW(s^{t}(t)_{0}\mu X^{\epsilon\alpha}_{u}(s)\,ds\\&{}+\alpha\frac{\sigma}{2}\bigl(X^{\\epsilon1}_{l}(0)+X^{\ epsilon 1}_}_{u(0)\bigr)\\&{{}\times\int^{t}(t)_{0}\varphi^{\epsilon}(s)\exp\biggl(\mu s+\sigma\alpha\int^{s}_{0}\varphi^{\epsilon}(u)\,du-\frac{1}{2}\sigma^{2}\ epsilon^{2\alpha}s+\sigma\epsillon^{\alpha}W(s)\biggr)\,ds\\&{}+\epsilon^{\ alpha}\frac}{2\sigma}{2{\bigl)\bigr)\\&{}\times\int^{t}(t)_{0}\exp\biggl(\mu s+\sigma\alpha\int^{s}_{0}\varphi^{\epsilon}(u)\,du-\frac{1}{2}\sigma^{2}\ epsilon^{2\alpha}s+\sigma\epsilon^{\alpha}W(s)\biggr)\,dW。\结束{对齐}$$
(3.28)
将定理8.5.2应用于[1]获得唯一的解决方案(3.28)格式为:
$$开始{对齐}X^{\epsilon\alpha}_{l}(t)={}&e^{\mut}X^{\ epsilon\ alpha}_{l}(0)\\&{}+e^{\ mut}\alpha\frac{\sigma}{2}\bigl^{t}(t)_{0}\varphi^{\epsilon}(s)\exp\biggl(\sigma\alpha\int^{s}_{0}\varphi^{\epsilon}(u)\,du-\frac{1}{2}\sigma^{2}\ epsilon^{2\alpha}s+\sigma\epsilon^{\alpha}W(s)\biggr)\,ds\\&{}+e^{\mut}\epsillon^{\ alpha}\ frac{\sigma}{2{\bigl{u}(0)\bigr)\\&{}\times\int^{t}(t)_{0}\exp\biggl(\sigma\alpha\int^{s}_{0}\varphi^{\epsilon}(u)\,du-\frac{1}{2}\sigma^{2}\ epsilon^{2\alpha}s+\sigma\epsilon^{\alpha}W(s)\biggr)\,dW(s t}\alpha\frac{\sigma}{2}\bigl(X^{\epsilon1}_{l}(0)+X^{\ epsilon1}_{u}(O)\bigr)\\&{}\times\int^{t}(t)_{0}\varphi^{\epsilon}(s)\exp\biggl(\sigma\alpha\int^{s}_{0}\varphi^{\epsilon}(u)\,du-\frac{1}{2}\sigma^{2}\ epsilon^{2\alpha}s+\sigma\epsilon^{\alpha}W(s)\biggr)\,ds\\&{}+e^{\mut}\epsillon^{\ alpha}\ frac{\sigma}{2{\bigl{u}(0)\bigr)\\&{}\times\int^{t}(t)_{0}\exp\biggl(\sigma\alpha\int^{s}_{0}\varphi^{\epsilon}(u)\,du-\frac{1}{2}\sigma^{2}\ epsilon^{2\alpha}s+\sigma\epsilon^{\alpha}W(s)\biggr)\,dW。\结束{对齐}$$
然后
$$开始{对齐}X^{\epsilon\alpha}_{l}(t)={}&e^{\mut}\biggl[X^{\ epsilon\ alpha}_{l}(0)+\frac{\sigma}{2}\bigl(X^{epsilon 1}_{1}(0)+X^{\\epsilon1}_{u}(O)\bigr)\\&{}\times\int^{t}(t)_{0}\exp\biggl(\sigma B^{\epsilon}_{H}(s)-\frac{1}{2}\sigma^{2}\ epsilon^{2\alpha}s\biggr)\,dB^{\ epsilen}_{H}(s])\biggr],\\X^{\εalpha}_{u}(t)={}&e^{\mut}\biggl[X^{\alpha}_u}(0)+\frac{\sigma}{2}\bigl(X^{\epsilon1}_{l}(0)+X^{\ epsilon1}_{u}(O)\bigr)\\&{}\times\int^{t}(t)_{0}\exp\biggl(\sigma B^{\epsilon}_{H}(s)-\frac{1}{2}\sigma ^{2}\epsilon ^{2}\alpha}s \biggr)\,dB^{\epsilon}_{H}(s)\biggr]。\结束{对齐}$$
因此\(\mu\geq 0\)是
$$开始{对齐}X^{\epsilon}(t)={}&e^{\mut}X(0)\\&{}+\biggl\langle\frac{\sigma}{2}\bigl^{t}_{0}\exp\biggl(\sigma B^{\epsilon}_{H}(s)-\frac{1}{2}\sigma^{2}\ epsilon^{2\alpha}s\biggr)\,dB^{\ epsilen}_{H}(s-)\biggr范围。\结束{对齐}$$
(3.29)
对于\(\mu<0\),我们可以证明
$$\begin{aligned}\begin{aligned}X^{\epsilon\alpha}_{l}(t)={}&X^{\ epsilon\ alpha}_{l}_(0)+\int^{t}(t)_{0}\mu X^{\epsilon\alpha}_{u}(s)\,ds\\&{}+\alpha\frac{\sigma}{2}\bigl(X^{\\epsilon1}_{l}(0)+X^{\ epsilon 1}_}_{u(0)\bigr)\\&{{}\times\int^{t}(t)_{0}\varphi^{\epsilon}(s)\exp\bigl(\mu s+\sigma\alpha\int^{s}_{0}\varphi^{\epsilon}(u)\,du-\frac{1}{2}\sigma^{2}\ epsilon^{2\alpha}s+\sigma\epsillon^{\alpha}W(s)\biggr)\,ds\\&{}+\epsilon^{\ alpha}\frac}{2\sigma}{2{\bigl)\bigr)\\&{}\times\int^{t}(t)_{0}\exp\biggl(\mu s+\sigma\alpha\int^{s}_{0}\varphi^{\epsilon}(u)\,du-\frac{1}{2}\sigma^{2}\ epsilon^{2\alpha}s+\sigma\epsilon^{\alpha}W(s)\biggr)\,dW(s^{t}_{0}\mu X^{\epsilon\alpha}_{l}(s)\,ds\\&{}+\alpha\frac{\sigma}{2}\bigl(X^{\\epsilon1}_{1}(0)+X^{\ epsilon 1}__{u}(O)\bigr)\\&{{}\times\int^{t}(t)_{0}\varphi^{\epsilon}(s)\exp\biggl(\mu s+\sigma\alpha\int^{s}_{0}\varphi^{\epsilon}(u)\,du-\frac{1}{2}\sigma^{2}\ epsilon^{2\alpha}s+\sigma\epsillon^{\alpha}W(s)\biggr)\,ds\\&{}+\epsilon^{\ alpha}\frac}{2\sigma}{2{\bigl)\bigr)\\&{}\times\int^{t}(t)_{0}\exp\biggl(\mu s+\sigma\alpha\int^{s}_{0}\varphi^{\epsilon}(u)\,du-\frac{1}{2}\sigma^{2}\ epsilon^{2\alpha}s+\sigma\epsilon^{\alpha}W(s)\biggr)\,dW。\end{aligned}\end{alinged}$$
(3.30)
独特的解决方案(3.30)具有以下矩阵形式:
$$开始{对齐}X^{\epsilon\alpha}_{l}(t)={}&X^{\spilon\alha}_{1}(0)\cosh(\mut)+X^{\ epsilon\ alpha}_{u}(O)\sinh(\mu t)\\&{}+e^{\mut}\alpha\frac{\sigma}{2}\bigl{u}(0)\bigr)\\&{}\times\int^{t}(t)_{0}\varphi^{\epsilon}(s)\exp\biggl(\sigma\alpha\int^{s}_{0}\varphi^{\epsilon}(u)\,du-\frac{1}{2}\sigma^{2}\ epsilon^{2\alpha}s+\sigma\epsilon^{\alpha}W(s)\biggr)\,ds\\&{}+e^{\mut}\epsillon^{\ alpha}\ frac{\sigma}{2{\bigl{u}(0)\bigr)\\&{}\times\int^{t}(t)_{0}\exp\biggl(\sigma\alpha\int^{s}_{0}\varphi^{\epsilon}(u)\,du-\frac{1}{2}\sigma^{2}\ epsilon^{2\alpha}s+\sigma\epsilon^{\alpha}W(s)\biggr)\,dW(s{u}(0)\cosh(\mut)\\&{}+e^{\mut}\alpha\frac{\sigma}{2}\bigl\bigr)\\&{}\times\int^{t}(t)_{0}\varphi^{\epsilon}(s)\exp\biggl(\sigma\alpha\int^{s}_{0}\varphi^{\epsilon}(u)\,du-\frac{1}{2}\sigma^{2}\sigma^{2}\sigma^{\alpha}W(s)\biggr)\,ds\\&{}+e^{\mu t}\epsilon^{\alpha}\frac{\sigma}{2}\bigl(X^{\epsilon 1}_{l}(0)+X^{\epsilon 1}_{u}(0)\bigr)\\&{}\times\int^{t}(t)_{0}\exp\biggl(\sigma\alpha\int^{s}_{0}\varphi^{\epsilon}(u)\,du-\frac{1}{2}\sigma^{2}\ epsilon^{2\alpha}s+\sigma\epsilon^{\alpha}W(s)\biggr)\,dW。\结束{对齐}$$
然后
$$\begin{aligned}X^{\epsilon\alpha}_{l}(t)={}&X^{\epsilon\alpha}_{l}(0)\cosh(\mu t)+X^{\epsilon\alpha}_{u}(0)\sinh(\mu t)\&{}+e^{\mu t}\frac{\sigma}{2}\bigl(X^{\epsilon 1}_{l}(0)+X^{\epsilon 1}_{u}(0)\bigr)[int^{t}(t)_{0}\exp\biggl(\sigma B^{\epsilon}_{H}(s)-\frac{1}{2}\sigma^{2}\ epsilon^{2\alpha}s\biggr)\,dB^{\ epsilen}_{H}n\alpha}{u}(0)\cosh(\mut)\\&{}+e^{\mut}\frac{\sigma}{2}\bigl^{t}(t)_{0}\exp\biggl(\sigma B^{\epsilon}_{H}(s)-\frac{1}{2}\sigma^{2}\ epsilon^{2\alpha}s\biggr)\,dB^{\ epsilen}_{H}。\结束{对齐}$$
因此,近似模糊解\(\mu<0\)是
$$开始{对齐}X^{\epsilon}(t)={}&X(0)\cosh(\mut)+X(0^{t}(t)_{0}\exp\biggl(\sigma B^{\epsilon}_{H}(s)-\frac{1}{2}\sigma^{2}\ epsilon^{2\alpha}s\biggr)\,dB^{\ epsilen}_{H}(s-)\biggr范围。\结束{对齐}$$
(3.31)