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

doi:10.21656/1000-0887.430170

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.

关键词:

Abstract: The random source inverse method and properties for a class of stochastic differential equations driven by the fractional Brownian motion with Hurst index H∈(0, 1). This problem can be obtained from the transform of many stochastic models and is a widely followed problem. For the direct problem, the mild solution to the equation was obtained by means of constant variation, and according to the statistical properties of the mild solution, the well-posedness of the direct problem was discussed. For the inverse problem, some statistics of the random source term were determined from the random data at the final moment, to prove the uniqueness of the inverse problem, and the stability of the inverse problem with a(x) in different ranges was discussed.

Key words:

0. 引言

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), B^H(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u_n(t), φ_n, ψ_n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{e_n(x)}下展开的，λ_n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)=λ_n≥λ₁>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u_t+$u_{x x x x}=\varphi(x) h(t)+\psi(x) \dot{B}^H(t)$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.

1. 理论基础

定义1 ^{[15, 17]}设0 < H < 1, 若存在一个连续的Gauss过程{B^H(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B^H(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B^H(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 α_H=H(2H-1).

2. 正问题

不难得到问题(1)的温和解如下.

定义2 一个随机过程u∈L²(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有

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

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)

由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“a≲b”表示a≤Cb，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‖M‖_L^∞(D)也有界；2) 若a(x)≥0，则‖M‖_L^∞(D)亦有界；3) 若a(x)→-∞，则‖M‖_L^∞(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.

3. 反问题

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.

3.1 唯一性

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).

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

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

当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K_H(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

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

当a(x) < 0时，有

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

(18)

情形3 当H∈(1/2, 1)时，有

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

当a(x)≥0时，有

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

注意到式(13)，有

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

当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‖ψ‖_L²(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.

3.2 稳定性分析

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)

有

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

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K_H(T, τ)大于0，并注意到积分∫_τ^T(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.

4. 结论

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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