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一类随机微分方程的随机源反演方法和性质

陈琛 冯晓莉 陈汉章

陈琛, 冯晓莉, 陈汉章. 一类随机微分方程的随机源反演方法和性质[J]. 应用数学和力学, 2023, 44(7): 847-856. doi: 10.21656/1000-0887.430170
引用本文: 陈琛, 冯晓莉, 陈汉章. 一类随机微分方程的随机源反演方法和性质[J]. 应用数学和力学, 2023, 44(7): 847-856. doi: 10.21656/1000-0887.430170
CHEN Chen, FENG Xiaoli, CHEN Hanzhang. The Random Source Inverse Method and Properties for a Class of Stochastic Differential Equations[J]. Applied Mathematics and Mechanics, 2023, 44(7): 847-856. doi: 10.21656/1000-0887.430170
Citation: CHEN Chen, FENG Xiaoli, CHEN Hanzhang. The Random Source Inverse Method and Properties for a Class of Stochastic Differential Equations[J]. Applied Mathematics and Mechanics, 2023, 44(7): 847-856. doi: 10.21656/1000-0887.430170

一类随机微分方程的随机源反演方法和性质

doi: 10.21656/1000-0887.430170
基金项目: 

陕西省自然科学基础研究计划项目 2023-JC-YB-054

中央高校基本科研业务费 XJS220702

详细信息
    作者简介:

    陈琛(1998—),男,硕士(E-mail: c.chen@stu.xidian.edu.cn)

    陈汉章(2000—),男(E-mail: hzchen2000@foxmail.com)

    通讯作者:

    冯晓莉(1981—),女,博士(通讯作者. E-mail: xiaolifeng@xidian.edu.cn)

  • 中图分类号: O175.26

The Random Source Inverse Method and Properties for a Class of Stochastic Differential Equations

  • 摘要: 该文考虑了一类由分式Brown运动驱动的随机微分方程的随机源反演方法及其性质,其中分式Brown运动对应的Hurst参数H∈(0, 1).该问题可由很多随机模型转化而得,是一种比较广泛的随机问题.对于正问题,通过常数变易法得到方程的温和解,根据温和解的统计性质讨论其适定性.对于反问题,根据终止时刻的随机数据的统计量反演随机源项的部分统计量,证明了反演的唯一性,并讨论了当a(x)在不同范围时反问题的稳定性情况.
  • 众所周知,反问题的应用非常广泛,相应的研究成果也越来越多[1-5]. 然而,绝大多数反问题都是不适定的,尤其是稳定性往往不成立,这也是研究反问题的困难和意义所在.不仅如此,实际应用中会有很多随机因素的影响,比如测量数据带有噪声,数学模型的建立含有模型误差等.这些因素对于不适定反问题的研究不容小觑,否则求得的数值结果会出现灾难性的问题.一种自然而又巧妙的处理方法是将这些因素视为随机的,相应的问题就看作随机问题.此外,随机技术的引入可以很好地耦合不同度量尺度之间的干扰[6-9]. 综上可知,关于随机反问题的研究既非常重要又很有实际意义.

    本文关心的是如下一类随机微分方程:

    $$ \begin{cases}\frac{\partial u(x, t)}{\partial t}+a(x) u(x, t)=\varphi(x) h(t)+\psi(x) \dot{B}^H(t), & (x, t) \in D \times(0, T), \\ u(x, t)=0, & (x, t) \in \partial D \times[0, T], \\ u(x, 0)=0, & x \in \bar{D}, \end{cases} $$ (1)

    其中a(x), h(t)已知,且a(x)≢0, h(t)≥h0>0, D=(0, 1), BH(t)为概率空间($\varOmega, \mathcal{F}, \left\{\mathcal{F}_t\right\}_{t \geqslant 0}, \mathbb{P}$) 上的分式Brown运动,H∈(0, 1)为Hurst参数,$\dot{B}^H(t)$为彩噪声.为了简便,后文将u(x, t, ω)简记为u(x, t).本文主要探讨:

    (P1) 正问题  已知φ(x), ψ(x), 求温和解u(x, t);

    (P2) 反问题  给定T时刻的样本数据u(x, T, ω), 反演问题(1)中随机源项的部分统计量φ(x)和ψ2(x).

    对于正问题(P1),由于随机源项$F(x, t):=\varphi(x) h(t)+\psi(x) \dot{B}^H(t)$的低正则性[10],该方程不是几乎处处存在的,所以方程的适定性与经典意义下截然不同,相应的解在什么样的空间存在是值得研究的.而关于反问题(P2),F(x, t)为随机场,如何由u(x, t, ω)的统计量来反演F(x, t)的统计量φ(x)与ψ2(x)是很困难且有意义的,而其不适定性又将使得该反问题的求解难上加难.

    由于以上的众多原由,近年来反随机源问题倍受学者们的重视[11-12],关于分式Brown运动驱动下的随机微分方程的反源问题,例如,随机的时空分数阶方程[13],随机波动方程[14],一个随机的时间分数阶扩散方程[15],一维随机对流扩散方程[16]等已有一些研究结果.这些工作均先是利用Laplace变换或Fourier变换或广义Fourier级数展开得到相应的温和解,然后再进一步讨论.本文所要讨论的问题(1),是一种更为广泛的随机问题,比如随机热方程:

    $$ \begin{cases}\frac{\partial u(x, t)}{\partial t}+\left\llcorner u(x, t)=\varphi(x) h(t)+\psi(x) \dot{B}^H(t), \right. & (x, t) \in D \times(0, T), \\ u(x, t)=0, & (x, t) \in \partial D \times[0, T], \\ u(x, 0)=0, & x \in \bar{D}, \end{cases} $$ (2)

    其中$\mathcal{L}$为一致椭圆算子. 借助于广义Fourier级数展开,问题(2)可转化为

    $$ \left\{\begin{array}{l} \frac{\mathrm{d} u_n(t)}{\mathrm{d} t}+\lambda_n u_n(t)=\varphi_n h(t)+\psi_n \dot{B}^H(t), \quad t \in(0, T), \\ u_n(0)=0, \end{array}\right. $$ (3)

    这里un(t), φn, ψn分别为u(x, t), φ(x), ψ(x)在椭圆算子$\mathcal{L}$的特征函数所形成的正交系{en(x)}下展开的,λn为相应的特征值.通过比较不难发现问题(3)是问题(1)的一种特殊情况,其中a(x)=λnλ1>0. 又如对流方程$u_t+b u_x=\varphi(x) h(t)+\psi(x) \dot{B}^H(t)$,Airy方程$u_t+u_{x x x}=\varphi(x) h(t)+\psi(x) \dot{B}^H(t)$,Beam方程ut+$u_{x x x x}=\varphi(x) h(t)+\psi(x) \dot{B}^H(t)$等均可转化为问题(1)去研究.注意问题(1)中的a(x)≢0,但是正负号不定,是一种非常广泛的情形.

    定义1  [15, 17]设0 < H < 1, 若存在一个连续的Gauss过程{BH(t), t≥0}满足如下条件:

    $$ B^H(0)=0, E\left[B^H(t) B^H(s)\right]=\frac{1}{2}\left\{|t|^{2 H}+|s|^{2 H}-|t-s|^{2 H}\right\}, $$

    则称BH(t)为分式Brown运动,其中H是Hurst参数,E[·]表示期望.特别地,当H=1/2时,即为标准Brown运动B(t).

    对于分式Brown运动BH(t),其中H∈(0, 1),相应的随机积分有如下公式[15]

    1) $E\left[\int_0^t \psi(s) \mathrm{d} B^H(s)\right]=0, \quad H \in(0, 1)$.

    2) 当H∈(0, 1/2)时,

    $$ \begin{aligned} E[ & \left.\int_0^t \psi(s) \mathrm{d} B^H(s) \int_0^t \phi(s) \mathrm{d} B^H(s)\right]= \\ & \int_0^t\left[K_H(t, s) \psi(s)+\int_s^t(\psi(u)-\psi(s)) \frac{\partial K_H(u, s)}{\partial u} \mathrm{d} u\right] \times \\ & {\left[K_H(t, s) \phi(s)+\int_s^t(\phi(u)-\phi(s)) \frac{\partial K_H(u, s)}{\partial u} \mathrm{d} u\right] \mathrm{d} s, } \end{aligned} $$ (4)

    其中

    $$ \begin{aligned} & K_H(t, s)=C_H\left[\left(\frac{t}{s}\right)^{H-1 / 2}(t-s)^{H-1 / 2}-\left(H-\frac{1}{2}\right) s^{1 / 2-H} \int_s^t u^{H-3 / 2}(u-s)^{H-1 / 2} \mathrm{d} u\right], \\ & \frac{\partial K_H(u, s)}{\partial u}=C_H\left(\frac{u}{s}\right)^{H-1 / 2}(u-s)^{H-3 / 2}, C_H=\left(\frac{2 H}{(1-2 H) \beta(1-2 H, H+1 / 2)}\right)^{1 / 2} \cdot \end{aligned} $$

    3) 当H=1/2时,

    $$ \begin{aligned} & E\left[\int_0^t \psi(s) \mathrm{d} B^H(s) \int_0^t \phi(s) \mathrm{d} B^H(s)\right]= \\ & \quad E\left[\int_0^t \psi(s) \mathrm{d} W(s) \int_0^t \phi(s) \mathrm{d} W(s)\right]=\int_0^t \psi(s) \phi(s) \mathrm{d} s . \end{aligned} $$ (5)

    4) 当H∈(1/2, 1)时,

    $$ E\left[\int_0^t \psi(s) \mathrm{d} B^H(s) \int_0^t \phi(s) \mathrm{d} B^H(s)\right]=\alpha_H \int_0^t \int_0^t \psi(r) \phi(u)|r-u|^{2 H-2} \mathrm{d} u \mathrm{d} r, $$ (6)

    其中  αH=H(2H-1).

    不难得到问题(1)的温和解如下.

    定义2   一个随机过程uL2(D)称为问题(1)的温和解,如果

    $$ u(x, t)=\int_0^t \mathrm{e}^{-a(x)(t-\tau)} \varphi(x) h(\tau) \mathrm{d} \tau+\int_0^t \mathrm{e}^{-a(x)(t-\tau)} \psi(x) h(\tau) \mathrm{d} B^H(\tau) $$ (7)

    几乎处处成立.

    下面进一步说明在一定条件下,温和解u(x, t)是适定的.为了推导的方便,先给出如下引理.

    引理1[15]  当H∈(0, 1/2)时,对∀t>τ

    $$ \int_\tau^t u^{H-3 / 2}(u-\tau)^{H-1 / 2} \mathrm{d} u \lesssim \tau^{2 H-1}-t^{2 H-1}. $$

    由式(7)知,

    $$ \begin{aligned} E \| & u(\cdot, t) \|_{L^2(D)}^2= \\ & E\left[\int_0^1\left(\int_0^t \mathrm{e}^{-a(x)(t-\tau)} \varphi(x) h(\tau) \mathrm{d} \tau+\int_0^t \mathrm{e}^{-a(x)(t-\tau)} \psi(x) h(\tau) \mathrm{d} B^H(\tau)\right)^2 \mathrm{d} x\right] \leqslant \\ & 2 \int_0^1\left(\int_0^t \mathrm{e}^{-a(x)(t-\tau)} h(\tau) \mathrm{d} \tau\right)^2 \varphi^2(x) \mathrm{d} x+2 \int_0^1 E\left(\int_0^t \mathrm{e}^{-a(x)(t-\tau)} \mathrm{d} B^H(\tau)\right)^2 \psi^2(x) \mathrm{d} x= \\ & 2 I_1(t)+2 I_2(t) . \end{aligned} $$ (8)

    h(t)≥h0>0及积分中值定理可知,存在某个ξ∈(0, t),有

    $$ I_1(t)=\int_0^1 \varphi^2(x) h^2(\xi)\left(\int_0^t \mathrm{e}^{-a(x)(t-\tau)} \mathrm{d} \tau\right)^2 \mathrm{d} x, $$

    其中

    $$ 0<\int_0^t \mathrm{e}^{-a(x)(t-\tau)} \mathrm{d} \tau \leqslant \begin{cases}t, & a(x) \geqslant 0, \\ \mathrm{e}^{-a(x) T} t, & a(x)<0 .\end{cases} $$

    若令M(x): =max{1, e-a(x)T},则

    $$ I_1(t) \leqslant\|\varphi\|_{L^2(D)}^2\|h\|_{L^{\infty}(0, T)}^2\|M\|_{L^{\infty}(D)}^2 t^2 . $$

    关于I2(t),有

    $$ I_2(t)=\int_0^1 \psi^2(x) \cdot E\left(\int_0^t \mathrm{e}^{-a(x)(t-\tau)} \mathrm{d} B^H(\tau)\right)^2 \mathrm{d} x . $$

    情形1   当H∈(0, 1/2)时,由式(4)可知

    $$ \begin{aligned} & E\left(\int_0^t \mathrm{e}^{-a(x)(t-\tau)} \mathrm{d} B^H(\tau)\right)^2 \leqslant \int_0^t\left[\left(\frac{t}{\tau}\right)^{H-1 / 2}(t-\tau)^{H-1 / 2} \mathrm{e}^{-a(x)(t-\tau)}\right]^2 \mathrm{d} \tau+ \\ & \int_0^t \tau^{1-2 H}\left[\left(\int_\tau^t u^{H-3 / 2}(u-\tau)^{H-1 / 2} \mathrm{d} u\right) \mathrm{e}^{-a(x)(t-\tau)}\right]^2 \mathrm{d} \tau+ \\ & \int_0^t\left[\int_\tau^t\left(\mathrm{e}^{-a(x)(t-u)}-\mathrm{e}^{-a(x)(t-\tau)}\right) \frac{\partial K_H(u, \tau)}{\partial u} \mathrm{d} u\right]^2 \mathrm{d} \tau= \\ & J_1(t)+J_2(t)+J_3(t) \text {, } \\ & \end{aligned} $$ (9)

    这里“ab”表示aCb,其中C为某个正常数.显然有

    $$ J_1(t) \leqslant \int_0^t(t-\tau)^{2 H-1} \mathrm{d} \tau \cdot M^2(x) \leqslant t^{2 H} M^2(x), $$ (10)
    $$ J_2(t) \leqslant \int_0^t \tau^{1-2 H}\left(\int_\tau^t u^{H-3 / 2}(u-\tau)^{H-1 / 2} \mathrm{d} u\right)^2 \mathrm{d} \tau \cdot M^2(x) . $$ (11)

    由引理1可知

    $$ J_2(t) \lesssim \int_0^t \tau^{1-2 H}\left(t^{4 H-2}+\tau^{4 H-2}\right) \mathrm{d} \tau \cdot M^2(x) \lesssim t^{2 H} M^2(x). $$

    关于J3(t),由于

    1) 当|a(x)|≤1时,

    $$ \left|\mathrm{e}^{-a(x)(t-u)}-\mathrm{e}^{-a(x)(t-\tau)}\right| \leqslant|a(x)| M(x)|u-\tau| \leqslant M(x)|u-\tau|; $$

    2) 当|a(x)|>1时,

    $$ \left|\mathrm{e}^{-a(x)(t-u)}-\mathrm{e}^{-a(x)(t-\tau)}\right|=\left|\int_{t-\tau}^{t-u} \frac{\mathrm{e}^{-a(x) s}}{-a(x)} \mathrm{d} s\right| \leqslant \frac{1}{|a(x)|} M(x)|u-\tau| \leqslant M(x)|u-\tau|. $$

    根据$\partial K_H(u, \tau) / \partial u$的表达式,易得

    $$ \begin{aligned} & J_3(t) \lesssim M^2(x) \int_0^t\left(\int_\tau^t u^{H-1 / 2}\left(\frac{u}{\tau}-1\right)^{H-1 / 2} \mathrm{d} u\right)^2 \mathrm{d} \tau \leqslant \\ & M^2(x) \int_0^t \tau^{2 H-1}\left(\int_\tau^t\left(\frac{u}{\tau}-1\right)^{H-1 / 2} \mathrm{d} u\right)^2 \mathrm{d} \tau \leqslant M^2(x) t^{2 H} . \end{aligned} $$ (12)

    将式(10)—(12)代入式(9),可得

    $$ E\left(\int_0^t \mathrm{e}^{-a(x)(t-\tau)} \mathrm{d} B^H(\tau)\right)^2 \leqslant M^2(x) t^{2 H}. $$

    此时有

    $$ I_2(t) \lesssim\|M\|_{L^{\infty}(D)}^2\|\psi\|_{L^2(D)}^2 t^{2 H}. $$

    情形2   当H=1/2时,由式(5)可知

    $$ E\left(\int_0^t \mathrm{e}^{-a(x)(t-\tau)} \mathrm{d} B^H(\tau)\right)^2=\int_0^t \mathrm{e}^{-2 a(x)(t-\tau)} \mathrm{d} \tau \lesssim M^2(x) t. $$

    所以

    $$ I_2(t) \lesssim\|M\|_{L^{\infty}(D)}^2\|\psi\|_{L^2(D)}^2 t. $$

    情形3   当H∈(1/2, 1)时,由式(6)可知

    $$ E\left(\int_0^t \mathrm{e}^{-a(x)(t-\tau)} \mathrm{d} B^H(\tau)\right)^2=\alpha_H \int_0^t \int_0^t \mathrm{e}^{-a(x)(t-\tau)} \mathrm{e}^{-a(x)(t-u)}|\tau-u|^{2 H-2} \mathrm{d} \tau \mathrm{d} u \lesssim \\ M^2(x) \int_0^t \int_0^t|\tau-u|^{2 H-2} \mathrm{d} \tau \mathrm{d} u=2 M^2(x) \int_0^t \int_u^t(\tau-u)^{2 H-2} \mathrm{d} \tau \mathrm{d} u \lesssim M^2(x) t^{2 H}. $$ (13)

    因此

    $$ I_2(t) \leqslant\|M\|_{L^{\infty}(D)}^2\|\psi\|_{L^2(D)}^2 t^{2 H}. $$

    综上可知如下定理成立.

    定理1   如果函数f(x), g(x)∈L2(D), 则有估计式

    $$ \sup\limits_{0 \leqslant L T}\|u(\cdot, t)\|_{L^2(D)} \lesssim\|h\|_{L^{\infty}(0, T)}\|M\|_{L^{\infty}(D)}\|\varphi\|_{L^2(D)}+\|M\|_{L^{\infty}(D)}\|\psi\|_{L^2(D)}. $$

    注1  由定理1可知:1) 若a(x)有界,则‖ML(D)也有界;2) 若a(x)≥0,则‖ML(D)亦有界;3) 若a(x)→-∞,则‖ML(D)无界. 因此第1)、2)两种情况下问题(1)的解是稳定的,但第3)种情况下问题(1)的解不稳定.

    本节我们考虑通过u(x, T, ω)的统计量来反演问题(1)随机源项中的统计量φ(x)和ψ2(x).若h(t)≡1时,φ(x)和ψ2(x)就是随机源项的期望和方差. 由式(7)可知

    $$ E[u(x, T)]=\varphi(x) \int_0^T \mathrm{e}^{-a(x)(T-\tau)} h(\tau) \mathrm{d} \tau, $$ (14)
    $$ \operatorname{var}[u(x, T)]=\psi^2(x) E\left(\int_0^T \mathrm{e}^{-a(x)(T-\tau)} \mathrm{d} B^H(\tau)\right)^2, $$ (15)

    其中var(·)表示方差.

    进一步可得

    $$ \varphi(x)=\frac{E[u(x, T)]}{\int_0^T \mathrm{e}^{-a(x)(T-\tau)} h(\tau) \mathrm{d} \tau}, \psi^2(x)=\frac{\operatorname{var}[u(x, T)]}{E\left(\int_0^T \mathrm{e}^{-a(x)(T-\tau)} \mathrm{d} B^H(\tau)\right)^2}. $$ (16)

    下面分析由式(14)和(15)来反演φ(x)与ψ2(x)的唯一性与稳定性如何.

    从式(16)可知,要反演φ(x)和ψ2(x),则式(16)中两式分母都不能为0,同时注意到E[u(x, T)], $\operatorname{var}[u(x, T)], \int_0^T \mathrm{e}^{-a(x)(T-\tau)} h(\tau) \mathrm{d} \tau, E\left(\int_0^T \mathrm{e}^{-a(x)(T-\tau)} \mathrm{d} B^H(\tau)\right)^2$的值都是唯一的.因此,如果能证得$\int_0^T \mathrm{e}^{-a(x)(T-\tau)} h(\tau) \mathrm{d} \tau>0 \text { 和 } E\left(\int_0^T \mathrm{e}^{-a(x)(T-\tau)} \mathrm{d} B^H(\tau)\right)^2>0$,那么由式(16)可以唯一地确定φ(x)和ψ2(x).

    h(τ)≥h0>0易知,对于固定的x,总有

    $$ \int_0^T \mathrm{e}^{-a(x)(T-\tau)} h(\tau) \mathrm{d} \tau \geqslant \int_0^T \mathrm{e}^{-a(x)(T-\tau)} \mathrm{d} \tau \cdot h_0>0, $$

    所以$\varphi(x)=\frac{E[u(x, T)]}{\int_0^T \mathrm{e}^{-a(x)(T-\tau)} h(\tau) \mathrm{d} \tau}$存在且唯一.

    下面针对不同的H,讨论$E\left(\int_0^T \mathrm{e}^{-a(x)(T-\tau)} \mathrm{d} B^H(\tau)\right)^2$的情况.

    情形1   当H∈(0, 1/2)时,

    $$ \begin{aligned} & E\left(\int_0^T \mathrm{e}^{-a(x)(T-\tau)} \mathrm{d} B^H(\tau)\right)^2= \\ & \quad \int_0^T\left[K_H(T, \tau) \mathrm{e}^{-a(x)(T-\tau)}+\int_\tau^T\left(\mathrm{e}^{-a(x)(T-u)}-\mathrm{e}^{-a(x)(T-\tau)}\right) \frac{\partial K_H(u, \tau)}{\partial u} \mathrm{d} u\right]^2 \mathrm{d} \tau. \end{aligned} $$

    对于固定的x,当a(x)≥0时,e-a(x)(T-s)关于s是递增的,所以

    $$ \mathrm{e}^{-a(x)(T-u)}-\mathrm{e}^{-a(x)(T-\tau)} \geqslant 0. $$

    又由KH(u, τ)的表达式可知KH(T, τ)与$\partial$KH(u, τ)/$\partial$u均大于0且连续,所以一定存在一个c1(x)>0,有

    $$ E\left(\int_0^T \mathrm{e}^{-a(x)(T-\tau)} \mathrm{d} B^H(\tau)\right)^2 \geqslant c_1(x)>0 . $$

    a(x) < 0时,e-a(x)(T-s)关于s是递减的,所以(d/ds)e-a(x)(T-s) < 0,此时

    $$ \begin{aligned} & K_H(T, \tau) \mathrm{e}^{-a(x)(T-\tau)}+\int_\tau^T\left(\mathrm{e}^{-a(x)(T-u)}-\mathrm{e}^{-a(x)(T-\tau)}\right) \frac{\partial K_H(u, \tau)}{\partial u} \mathrm{d} u= \\ & \quad \lim\nolimits_{\varepsilon \rightarrow 0^{+}}\left\{K_H(T, \tau) \mathrm{e}^{-a(x)(T-\tau)}+\int_{\tau+\varepsilon}^T\left(\mathrm{e}^{-a(x)(T-u)}-\mathrm{e}^{-a(x)(T-\tau)}\right) \frac{\partial K_H(u, \tau)}{\partial u} \mathrm{d} u\right\}. \end{aligned} $$

    根据分部积分,可知

    $$ \begin{aligned} & K_H(T, \tau) \mathrm{e}^{-a(x)(T-\tau)}+\int_{\tau+\varepsilon}^T\left(\mathrm{e}^{-a(x)(T-u)}-\mathrm{e}^{-a(x)(T-\tau)}\right) \frac{\partial K_H(u, \tau)}{\partial u} \mathrm{d} u= \\ & \quad\left[\mathrm{e}^{-a(x)(T-\tau)}-\mathrm{e}^{-a(x)(T-(\tau+\varepsilon))}\right] K_H(\tau+\varepsilon, \tau)+K_H(u, T)-\int_{\tau+\varepsilon}^T \frac{\mathrm{d}}{\mathrm{d} u} \mathrm{e}^{-a(x)(T-u)} K_H(u, \tau) \mathrm{d} u. \end{aligned} $$

    由e-a(x)(T-s)关于s是递减的,且KH(u, τ)>0,可知

    $$ E\left(\int_0^T \mathrm{e}^{-a(x)(T-\tau)} \mathrm{d} B^H(\tau)\right)^2 \geqslant \int_0^T K_H^2(u, \tau) \mathrm{d} \tau=c_1>0. $$ (17)

    情形2   当H=1/2时,由Itô等距公式(5)可知:当a(x)≥0时,有

    $$ E\left(\int_0^T \mathrm{e}^{-a(x)(T-\tau)} \mathrm{d} B^H(\tau)\right)^2 \geqslant \mathrm{e}^{-2 a(x) T} T=c_2(x)>0 . $$

    a(x) < 0时,有

    $$ E\left(\int_0^T \mathrm{e}^{-a(x)(T-\tau)} \mathrm{d} B^H(\tau)\right)^2 \geqslant T>0. $$ (18)

    情形3   当H∈(1/2, 1)时,有

    $$ E\left(\int_0^T \mathrm{e}^{-a(x)(T-\tau)} \mathrm{d} B^H(\tau)\right)^2=\alpha_H \int_0^T \int_0^T \mathrm{e}^{-a(x)(T-\tau)} \mathrm{e}^{-a(x)(T-u)}|\tau-u|^{2 H-2} \mathrm{d} \tau \mathrm{d} u. $$

    a(x)≥0时,有

    $$ E\left(\int_0^T \mathrm{e}^{-a(x)(T-\tau)} \mathrm{d} B^H(\tau)\right)^2 \geqslant \alpha_H \mathrm{e}^{-2 a(x) T} \int_0^T \int_0^T|\tau-u|^{2 H-2} \mathrm{d} \tau \mathrm{d} u. $$

    注意到式(13),有

    $$ E\left(\int_0^T \mathrm{e}^{-a(x)(T-\tau)} \mathrm{d} B^H(\tau)\right)^2 \geqslant \alpha_H \frac{T^{2 H}}{H(2 H-1)} \mathrm{e}^{-2 a(x) T}=c_3(x)>0 . $$

    a(x)≤0,类似地有

    $$ E\left(\int_0^T \mathrm{e}^{-a(x)(T-\tau)} \mathrm{d} B^H(\tau)\right)^2 \geqslant \alpha_H \frac{T^{2 H}}{H(2 H-1)}>0. $$ (19)

    综上分析可知,如下定理成立.

    定理2   若φ(x), ψ(x)∈L2(D)且‖ψL2(D)≠0,h(x)≥h0>0,h(x)∈L(0, T),则源项中φ(x)与ψ2(x)能由u(x, T, ω)的期望和方差唯一确定.

    反源问题在医学、地质勘探等方面都有非常广泛的应用,然而反源问题往往是不稳定的.本小节将分析反演φ(x)与ψ2(x)时的稳定性. 先分析a(x)>0时的稳定性.

    一方面,我们有

    $$ \int_0^T \mathrm{e}^{-a(x)(T-\tau)} h(\tau) \mathrm{d} \tau=h(\xi) \int_0^T \mathrm{e}^{-a(x)(T-\tau)} \mathrm{d} \tau=h(\xi) \frac{1-\mathrm{e}^{-a(x) T}}{a(x)} \lesssim \frac{1}{a(x)}. $$

    另一方面,关于$E\left(\int_0^T \mathrm{e}^{-a(x)(T-\tau)} \mathrm{d} B^H(\tau)\right)^2$,我们分两种情形来讨论.

    情形1   若a(x)>1/T2,由式(9)可知

    $$ E\left(\int_0^T \mathrm{e}^{-a(x)(T-\tau)} \mathrm{d} B^H(\tau)\right)^2 \leqslant J_1(T)+J_2(T)+J_3(T). $$

    对于H∈(0, 1/2),借助于不等式

    $$ \mathrm{e}^{-a(x)(T-\tau)} \leqslant \begin{cases}1, & 0<T-\tau<\frac{1}{\sqrt{a(x)}}, \\ \frac{1}{a(x)(T-\tau)}, & T-\tau>\frac{1}{\sqrt{a(x)}}\end{cases} $$ (20)

    $$ J_1(T)=\left(\int_0^{T-1 / \sqrt{a(x)}}+\int_{T-1 / \sqrt{a(x)}}^T\right)\left(\frac{T}{\tau}\right)^{2 H-1}(T-\tau)^{2 H-1} \mathrm{e}^{-2 a(x)(T-\tau)} \mathrm{d} \tau=J_{11}(T)+J_{12}(T). $$

    根据式(20)易得

    $$ \begin{aligned} & J_{11}(T) \leqslant \int_0^{T-1 / \sqrt{a(x)}} T^{2 H-1} \tau^{1-2 H}(T-\tau)^{2 H-1} \frac{1}{2 a(x)(T-\tau)} \mathrm{d} \tau \leqslant \\ & T^{2 H-1}\left(T-\frac{1}{\sqrt{a(x)}}\right)^{1-2 H} \frac{1}{2 a(x)} \int_0^{T-1 / \sqrt{a(x)}}(T-\tau)^{2 H-2} \mathrm{d} \tau \leqslant \frac{1}{a^{H+1 / 2}(x)} . \end{aligned} $$

    并且

    $$ J_{12}(T) \leqslant T^{2 H-1} T^{1-2 H} \int_{T-1 / \sqrt{a(x)}}^T(T-\tau)^{2 H-1} \mathrm{d} \tau \lesssim \frac{1}{a^H(x)}. $$

    因此

    $$ J_1(T) \lesssim \frac{1}{a^{H+1 / 2}(x)}+\frac{1}{a^H(x)} \lesssim \frac{1}{a^H(x)}. $$

    此外,

    $$ \begin{aligned} & J_2(T) \leqslant \int_0^{T-1 / \sqrt{a(x)}} \tau^{1-2 H}\left(\int_\tau^{T-1 / \sqrt{a(x)}} u^{H-3 / 2}(u-\tau)^{H-1 / 2} \mathrm{d} u\right)^2 \mathrm{e}^{-2 a(x)(T-\tau)} \mathrm{d} \tau+ \\ & \int_0^{T-1 / \sqrt{a(x)}} \tau^{1-2 H}\left(\int_{T-1 / \sqrt{a(x)}}^T u^{H-3 / 2}(u-\tau)^{H-1 / 2} \mathrm{d} u\right)^2 \mathrm{e}^{-2 a(x)(T-\tau)} \mathrm{d} \tau+ \\ & \int_{T-1 / \sqrt{a(x)}}^T \tau^{1-2 H}\left(\int_\tau^T u^{H-3 / 2}(u-\tau)^{H-1 / 2} \mathrm{d} u\right)^2 \mathrm{d} \tau= \\ & J_{21}(T)+J_{22}(T)+J_{23}(T) . \end{aligned} $$

    根据引理1,有

    $$ J_{21}(T) \lesssim \int_0^{T-1 / \sqrt{a(x)}} \tau^{1-2 H} \frac{1}{a(x)(T-\tau)}\left[\tau^{2 H-1}-\left(T-\frac{1}{\sqrt{a(x)}}\right)^{2 H-1}\right]^2 \mathrm{d} \tau \leqslant \\ \int_0^{T-1 / \sqrt{a(x)}} \tau^{1-2 H} \tau^{4 H-2} \frac{1}{a(x)(1 / \sqrt{a(x)})} \mathrm{d} \tau \lesssim \frac{T^{2 H}}{a^{1 / 2}(x)}. $$

    利用证明引理1相同的方法,我们得到

    $$ \begin{aligned} & J_{22}(T) \lesssim \int_0^{T-1 / \sqrt{a(x)}} \frac{1}{a(x)(T-\tau)} \tau^{1-2 H}\left(\int_{T-1 / \sqrt{a(x)}}^T u^{2 H-2}\left[\sum\limits_{n=0}^{\infty}\left(\begin{array}{c} H-1 / 2 \\ n \end{array}\right)\left(-\frac{\tau}{u}\right)^n\right] \mathrm{d} u\right)^2 \mathrm{d} \tau \leqslant \\ & \frac{1}{a(x)(1 / \sqrt{a(x)})} \int_0^{T-1 / \sqrt{a(x)}} \tau^{1-2 H}\left(\sum\limits_{n=0}^{\infty}\left(\begin{array}{c} H-1 / 2 \\ n \end{array}\right)(-1)^n \times\right. \end{aligned} \\ \begin{aligned} & \left.\frac{(T-1 / \sqrt{a(x)})^{2 H-1-n}-T^{2 H-1-n}}{n+1-2 H} \tau^n\right)^2 \mathrm{d} \tau \leqslant \\ & \frac{1}{a^{1 / 2}(x)} \int_0^{T-1 / \sqrt{a(x)}} \tau^{1-2 H}\left(\sum\limits_{n=0}^{\infty}\left(\begin{array}{c} H-1 / 2 \\ n \end{array}\right)(-1)^n \frac{(T-1 / \sqrt{a(x)})^{2 H-1}}{n+1-2 H}\right)^2 \mathrm{d} \tau \lesssim \frac{T^{2 H}}{a^{1 / 2}(x)} \cdot \end{aligned} $$

    同样地,利用引理1,有

    $$ J_{23} \lesssim \int_{T-1 / \sqrt{a(x)}}^T \tau^{1-2 H}\left(\tau^{4 H-2}+T^{4 H-2}\right) \mathrm{d} \tau \lesssim \int_{T-1 / \sqrt{a(x)}}^T \tau^{2 H-1} \mathrm{d} \tau \lesssim \frac{T^{2 H}}{a^{1 / 2}(x)}. $$

    所以,对于J2(T),我们有

    $$ J_2(T) \lesssim \frac{T^{2 H}}{a^{1 / 2}(x)}. $$

    此外,

    $$ \begin{gathered} J_3(T)=\int_0^T\left[\int_\tau^T \int_{T-\tau}^{T-u} \frac{\mathrm{e}^{-a(x) s}}{-a(x)} \mathrm{d} s\left(\frac{u}{\tau}\right)^{H-1 / 2}(u-\tau)^{H-3 / 2} \mathrm{d} u\right]^2 \mathrm{d} \tau \leqslant \\ \int_0^T\left[\int_\tau^T \frac{1}{a(x)}\left(\frac{u}{\tau}\right)^{H-1 / 2}(u-\tau)^{H-1 / 2} \mathrm{d} u\right]^2 \mathrm{d} \tau \lesssim \frac{T^{2 H+2}}{a^2(x)} . \end{gathered} $$

    因此

    $$ E\left(\int_0^T \mathrm{e}^{-a(x)(T-\tau)} \mathrm{d} B^H(\tau)\right)^2 \lesssim \frac{1}{a^H(x)}+\frac{1}{a^{1 / 2}(x)}+\frac{1}{a^2(x)} \lesssim \frac{1}{a^H(x)}. $$

    对于H=1/2,利用Itô等距公式(5)并注意到a(x)>1/T2可得

    $$ E\left(\int_0^T \mathrm{e}^{-a(x)(T-\tau)} \mathrm{d} B^H(\tau)\right)^2=\int_0^T \mathrm{e}^{-2 a(x)(T-\tau)} \mathrm{d} \tau \lesssim \frac{1}{a(x)} \lesssim \frac{T}{a^{1 / 2}(x)}. $$

    对于H∈(1/2, 1),利用式(6)并注意到估计式(20),将其化为4个部分,每个部分分别借用估计式(20)进行计算,所以有

    $$ \begin{aligned} & E\left(\int_0^T \mathrm{e}^{-a(x)(T-\tau)} \mathrm{d} B^H(\tau)\right)^2 \lesssim \\ & \quad\left(\int_0^{T-1 / \sqrt{a(x)}} \int_0^{T-1 / \sqrt{a(x)}}+\int_0^{T-1 / \sqrt{a(x)}} \int_{T-1 / \sqrt{a(x)}}^T+\int_{T-1 / \sqrt{a(x)}}^T \int_0^{T-1 / \sqrt{a(x)}}+\right. \\ & \left.\quad \int_{T-1 / \sqrt{a(x)}}^T \int_{T-1 / \sqrt{a(x)}}^T\right) \mathrm{e}^{-a(x)(T-u)} \mathrm{e}^{-a(x)(T-\tau)}|u-\tau|^{2 H-2} \mathrm{d} u \mathrm{d} \tau=N_1+N_2+N_3+N_4, \end{aligned} $$

    其中

    $$ \begin{aligned} N_1 \leqslant & \int_0^{T-1 / \sqrt{a(x)}} \int_0^{T-1 / \sqrt{a(x)}} \frac{1}{a(x)(T-u)} \frac{1}{a(x)(T-\tau)}|u-\tau|^{2 H-2} \mathrm{d} u \mathrm{d} \tau \leqslant \\ & \frac{2}{a(x)} \int_0^{T-1 / \sqrt{a(x)}} \int_\tau^{T-1 / \sqrt{a(x)}}(u-\tau)^{2 H-2} \mathrm{d} u \mathrm{d} \tau \leqslant \frac{T^{2 H}}{a(x)} . \end{aligned} $$

    由于对称性

    $$ N_2=N_3 \leqslant \int_0^{T-1 / \sqrt{a(x)}} \int_{T-1 / \sqrt{a(x)}}^T \frac{1}{a(x)(T-\tau)}(u-\tau)^{2 H-2} \mathrm{d} u \mathrm{d} \tau \leqslant \frac{T^{2 H-1}}{a(x)}. $$

    根据估计式(20)和积分中值定理可知,存在$\tilde{u} \in(T-1 / \sqrt{a(x)}, T)$, 有

    $$ \begin{aligned} N_4 \leqslant & \int_{T-1 / \sqrt{a(x)}}^T \int_{T-1 / \sqrt{a(x)}}^T|u-\tau|^{2 H-2} \mathrm{d} u \mathrm{d} \tau= \\ & \frac{1}{\sqrt{a(x)}} \int_{T-1 / \sqrt{a(x)}}^T|\tilde{u}-\tau|^{2 H-2} \mathrm{d} \tau \lesssim \frac{1}{a^H(x)}. \end{aligned} $$

    因此

    $$ E\left(\int_0^T \mathrm{e}^{-a(x)(T-\tau)} \mathrm{d} B^H(\tau)\right)^2 \lesssim \frac{1}{a(x)}+\frac{1}{a^H(x)} \lesssim \frac{1}{a^H(x)}. $$

    情形2   若0 < a(x)≤1/T2,当H∈(0, 1/2)时,根据式(4)及KH(T, τ)大于0,并注意到积分∫τT(u-τ)H-3/2du的奇异性,计算可得

    $$ E\left(\int_0^T \mathrm{e}^{-a(x)(T-\tau)} \mathrm{d} B^H(\tau)\right)^2= \\ \begin{aligned} & \int_0^T\left[K_H(T, \tau) \mathrm{e}^{-a(x)(T-\tau)}+\int_\tau^T\left(\int_{T-\tau} \frac{\mathrm{e}^{-a(x) s}}{-a(x)} \mathrm{d} s\right) C_H\left(\frac{u}{\tau}\right)^{H-1 / 2}(u-\tau)^{H-3 / 2} \mathrm{d} u\right]^2 \mathrm{d} \tau \geqslant \\ & \int_0^T\left[K_H(T, \tau) \mathrm{e}^{-1 / T}+\int_\tau^T T^2 \mathrm{e}^{-1 / T} C_H\left(\frac{u}{\tau}\right)^{H-1 / 2}(u-\tau)^{H-1 / 2} \mathrm{d} u\right]^2 \mathrm{d} \tau=c_4>0 . \end{aligned} $$

    对于H=1/2,利用Itô等距公式(5)易得

    $$ E\left(\int_0^T \mathrm{e}^{-a(x)(T-\tau)} \mathrm{d} B^H(\tau)\right)^2=\int_0^T \mathrm{e}^{-2 a(x)(T-\tau)} \mathrm{d} \tau \geqslant T \mathrm{e}^{-2 / T}=c_5>0 . $$

    对于H∈(1/2, 1),利用公式(6)并进行简单的计算可得

    $$ \begin{aligned} & E\left(\int_0^T \mathrm{e}^{-a(x)(T-\tau)} \mathrm{d} B^H(\tau)\right)^2 \geqslant \alpha_H \mathrm{e}^{-2 / T} \int_0^T \int_0^T|u-\tau|^{2 H-2} \mathrm{d} u \mathrm{d} \tau= \\ & \quad 2 \alpha_H \mathrm{e}^{-2 / T} \int_0^T \int_u^T(\tau-u)^{2 H-2} \mathrm{d} u \mathrm{d} \tau=\frac{\alpha_H}{H(2 H-1)} \mathrm{e}^{-2 / T} T^{2 H}=c_6>0 . \end{aligned} $$

    综合本小节的推导以及3.1小节中的式(17)—(19),我们得到以下定理.

    定理3   对于任意的a(x),有估计式

    $$ \int_0^T \mathrm{e}^{-a(x)(T-\tau)} h(\tau) \mathrm{d} \tau \lesssim \frac{1}{a(x)}. $$

    对于a(x)≥1/T2,有

    $$ E\left(\int_0^T \mathrm{e}^{-a(x)(T-\tau)} \mathrm{d} B^H(\tau)\right)^2 \lesssim \frac{1}{a^H(x)} ; $$

    对于a(x) < 0或0 < a(x)≤1/T2,存在一个与x无关的正常数C,使得

    $$ E\left(\int_0^T \mathrm{e}^{-a(x)(T-\tau)} \mathrm{d} B^H(\tau)\right)^2 \geqslant C . $$

    注2   由定理3可知,当a(x) < 0或0 < a(x)≤1/T2时,由式(15)反演ψ2(x)是稳定的;当a(x)趋于∞时,由式(14)反演φ(x)和由式(15)反演ψ2(x)都是不稳定的.

    本文讨论了一类随机微分方程的反源问题,根据方程的温和解,讨论了正问题的适定性.利用温和解的统计性质,根据T时刻的数据反演方程的源项,证明了源项反演的唯一性,并分析了关于源项φ(x)和ψ2(x)反演的稳定性情况.后面我们将进一步考虑相应的数值结果.

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  • 收稿日期:  2022-05-19
  • 修回日期:  2022-06-28
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