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数学物理学报, 2019, 39(1): 143-155 doi:

论文

平均场正倒向随机控制系统的最大值原理

李瑞敬,

A General Maximum Principle for Forward-Backward Stochastic Control Systems of Mean-Field Type

Li Ruijing,

收稿日期: 2017-12-5  

基金资助: 国家自然科学基金.  11626063

Received: 2017-12-5  

Fund supported: the NSFC.  11626063

作者简介 About authors

李瑞敬,E-mail:li209981@sina.com , E-mail:li209981@sina.com

摘要

该文研究具有时间不连续效用函数的平均场随机系统最优控制问题.其中,扩散项系数包含控制变量且控制区域非凸.借助于延拓的Ekeland变分原理及递归方法,建立平均场理论框架下一般形式的随机最大值原理.最后,求解一个线性二次问题以论证结果的可行性.

关键词: 平均场随机微分方程 ; 最大值原理 ; 伴随方程 ; 延拓的Ekeland变分原理

Abstract

The present paper concerns with optimal control problems allowing for time inconsistent utility functions for instance of mean-field stochastic systems. Moreover, the control variable enters the diffusion coefficient and the control domain is non-convex. Via extended Ekeland's variational principle as well as the reduction method, a general stochastic maximum principle is established in the framework of mean-field theory. Finally, a linear-quadratic example is worked out to illustrate the application of the results.

Keywords: Mean-field SDE ; Maximum principle ; Adjoint equation ; Extended Ekeland's variational principle

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本文引用格式

李瑞敬. 平均场正倒向随机控制系统的最大值原理. 数学物理学报[J], 2019, 39(1): 143-155 doi:

Li Ruijing. A General Maximum Principle for Forward-Backward Stochastic Control Systems of Mean-Field Type. Acta Mathematica Scientia[J], 2019, 39(1): 143-155 doi:

1 引言

近年来,平均场方法被广泛应用于金融学、统计力学、博弈论等多学科领域.这些领域所涉及的系统大多是平均场类型,即:性能指标、漂移项系数、扩散性系数不仅依赖于状态过程,还依赖于状态过程的概率分布.这种情形下,平均场效益函数具有时间不连续性,以致Bellman动态规划原则不再成立.因此,涉及到这类效用函数的最优控制问题不能用经典的HJB方程来解决.故本文通过Pontryagin最大值原理研究平均场背景意义下的委托代理问题.

2009年, Buckdahn等人在文献[4]首次研究了非线性平均场倒向随机微分方程.随后,文献[5]利用经典倒向随机微分方程(BSDEs)方法给出相应于平均场BSDEs的非局部偏微分方程的概率解释.此后,平均场正倒向随机微分方程理论得到广泛发展,且针对McKean-Vlasov类型的系统[8]已取得丰富成果.如:文献[6]利用罚方法构造逼近解证明了最优转换问题中一阶平均场策略方程解的存在性;文献[10]通过将拆分法转化为BSDEs方法给出分布方程的适定性,进一步,通过对系数的正则性假设,证明了其值函数为相应平均场非局部拟线性积分偏微分方程唯一经典解的结论.另一方面,随着平均场正倒向随机微分方程理论的完善,有关平均场框架下随机控制问题的研究得到快速发展.如:文献[3]通过尖变分技术建立了彭-类型的随机最大值原理.当此受控系统包含马尔可夫跳参数时,文献[14]给出了平均场随机线性二次最优控制问题的最大值原理.文献[15]则利用随机最大值原理得到有限时区平均场最优控制问题显式可解的充分必要条件.更多结果可参考文献[9, 11, 13, 16-17].

可以看到,当控制系统具有终端约束且控制区域非凸时,相应的随机最大值原理都是建立在罚泛函的下半连续性要求基础之上的.因此,如何降低这一要求目前还是一个新问题.本文尝试利用延拓的Ekeland变分原理解决这一问题,并在一般控制域情形下借助于递归方法建立平均场正倒向随机微分方程的最大值原理.需要指出的是,由于延拓的Ekeland变分原理的应用,这里所使用的递归方法是不同于文献[18].

2 预备知识

(Ω,F,{Ft}t0,P)是一个满足通常条件的流概率空间.在其上定义一维的标准布朗运动(Bt)t0,且F={Fs,0sT}是由(Bt)t0生成,所有P -零集扩充的自然流,即Fs=σ{Br,rs}Np,s[0,T],其中Np是所有P-零子集组成的集合.记S2F(0,T;R)F -适应和连续且满足E[sup的函数全体所成空间; {{\cal H}}_{{\cal F}}^2(0, T;R){\cal F} -适应且满足E[\int_0^T|\psi_t|^2{\rm d}t] <\infty的函数全体所成空间.

(\bar{\Omega}, \bar{{{\cal F}}}, \bar{P})=(\Omega\times\Omega, {{\cal F}}\bigotimes{{\cal F}} , P\bigotimes P)(\Omega, {{\cal F}}, P)和它自身的乘积空间,具有流\bar{{\bf{F}}}=\{\bar{{{\cal F}}}_t={{\cal F}}\bigotimes{{\cal F}}_t, 0\leq t\leq T\}.则原定义在\Omega上的随机变量\xi\in L^0(\Omega, {{\cal F}}, P)可标准延拓到\bar{\Omega}: \xi'(w', w)=\xi(w'), (w', w)\in\bar{\Omega}=\Omega\times\Omega.对任意的\theta\in L^1(\bar{\Omega}, \bar{{{\cal F}}}, \bar{P}),变量\theta(\cdot, w): \Omega\rightarrow \mathbb{R}属于空间L^1(\Omega, {{\cal F}}, P), P({\rm d}w)-a.s.;记其期望为E'[\theta(\cdot, w)]=\int_{\Omega}\theta(w', w)P({\rm d}w').注意E'[\theta]=E'[\theta(\cdot, w)] ,且

\bar{E}[\theta]=\int_{\bar{\Omega}}\theta {\rm d}\bar{P}=\int_{\Omega}E'[\theta(\cdot, w)]P({\rm d}w)=E[E'[\theta]].

考虑如下受控的平均场正倒向随机微分方程

\begin{equation}\label{eq:a1} \left\{\begin{array}{ll} {\rm d}X_t=E'[b(t, X'_t, X_t, u_t)]{\rm d}t+E'[\sigma(t, X'_t, X_t, u_t)]{\rm d}B_t, \\ {\rm d}Y_t=E'[f(t, X'_t, Y'_t, Z'_t, X_t, Y_t, Z_t, u_t)]{\rm d}t+Z_t{\rm d}B_t, \\ X(0)=a, \ \ Y_T=\Phi(X_T), \end{array}\right.\end{equation}
(2.1)

其中b, \sigma:[0, T]\times\mathbb{R}^2\times U\rightarrow \mathbb{R}; f:[0, T]\times \mathbb{R}^6 \times U\rightarrow \mathbb{R}; \Phi:\mathbb{R}\rightarrow \mathbb{R}. U是实数集\mathbb{R}的非空闭子集和非凸子集.

注2.1  方程(2.1)的扩散项系数可理解为

\begin{eqnarray*}&&E'[f(s, X'_s, Y'_s, Z'_s, X_s, Y_s, Z_s, u_s)](w)\\&=&\int_{\Omega}f(w', w, s, X_s(w'), Y_s(w'), Z_s(w'), X_s(w), Y_s(w), Z_s(w), u_s(w, w'))P({\rm d}w').\end{eqnarray*}

相应于系统(2.1)的性能指标为

J(u)=E\{\int_{0}^{T}{{{E}'}}[l(t,{{{X}'}_{t}},{{{Y}'}_{t}},{{{Z}'}_{t}},{{X}_{t}},{{Y}_{t}},{{Z}_{t}},{{u}_{t}})]\text{d}t\}+E[h({{X}_{T}})+\gamma (Y(0))],
(2.2)

其中l:[0, T]\times \mathbb{R}^6 \times U\rightarrow \mathbb{R}; h, \gamma:\mathbb{R}\rightarrow \mathbb{R}.

定义容许控制集为{{\cal U}}=\{u_t\in L^2_{\bar{\bf{F}}}(0, T, U)| u_t(w', w): [0, T]\times \Omega\times\Omega\rightarrow U\}.本文考虑的最优控制问题为

问题A  找到一个\bar{u}\in {{\cal U}}使得

J(\bar{u})=\inf\limits_{u\in{{\cal U}}}J(u).

\bar{u}取到J的下确界,则称它为最优的,相应于\bar{u}的解(\bar{X}, \bar{Y}, \bar{Z})则称为最优轨道.下面我们给出假设条件

(H_1) b, \sigma, f, l, \Phi, h\gamma分别关于(\tilde{x}, \tilde{y}, x, y)二次连续可微,且它们的一阶和二阶导函数关于(\tilde{x}, \tilde{y}, x, y, u)连续有界;

(H_2) b, \sigma, f, l, \Phi, h\gamma分别关于(\tilde{x}, \tilde{y}, \tilde{z}, x, y, z, u)线性增长,关于(t, u)连续.

注2.2   在条件(H_1)-(H_2)下,由文献[2,定理2]知,平均场系统(2.1)存在唯一适应解(X_t, Y_t, Z_t)_{t\in[0, T]}\in {\mathcal{S}}_{\mathcal{F}}^2(0, T;\mathbb{R})\times {\mathcal{S}}_{\mathcal{F}}^2(0, T;\mathbb{R}) \times {\mathcal{H}}_{\mathcal{F}}^2(0, T;\mathbb{R}).

在系统(2.1)中,倒向方程系数包含变量Zu,而过程Z的正则性(作为状态过程的一部分)不足以得到二阶伴随方程.因此经典方法不再适用.接下来,我们将采用一种递归方法[7]来研究此问题.首先我们对问题A做一变形.

问题B  在(Y_0, Z, u)\in{\cal R}:=\mathbb{R}\times{{\cal H}}_{{\cal F}}^2(0, T;\mathbb{R})\times{\cal U}上最小化性能指标

J(Y_0, Z, u)=E \biggl\{\int_0^TE'[l(t, X'_t, Y'_t, Z'_t, X_t, Y_t, Z_t, u_t)]{\rm d}t\biggr\}+E[h(X_T)+\gamma(Y_0) ],

其状态方程为

\left\{ \begin{array}{*{35}{l}} \text{d}{{X}_{t}}={E}'[b(t,{{{{X}'}}_{t}},{{X}_{t}},{{u}_{t}})]\text{d}t+{E}'[\sigma (t,{{{{X}'}}_{t}},{{X}_{t}},{{u}_{t}})]\text{d}{{B}_{t}}, \\ \text{d}{{Y}_{t}}={E}'[f(t,{{{{X}'}}_{t}},{{{{Y}'}}_{t}},{{{{Z}'}}_{t}},{{X}_{t}},{{Y}_{t}},{{Z}_{t}},{{u}_{t}})]\text{d}t+{{Z}_{t}}\text{d}{{B}_{t}}, \\ X(0)=a,\ \ Y(0)={{Y}_{0}}, \\\end{array} \right.
(2.3)

且具有最优状态约束

E|Y_T-\Phi(X_T)|^2=0.

注2.3  由文献[2,定理2]知,在条件(H_1)-(H_2)下,平均场随机微分方程(2.3)有唯一适应解(X_t, Y_t, Z_t)_{t\in[0, T]}\in {\mathcal{S}}_{\mathcal{F}}^2(0, T;\mathbb{R})\times{\mathcal{S}}_{\mathcal{F}}^2(0, T;\mathbb{R})\times{\mathcal{H}}_{\mathcal{F}}^2(0, T;\mathbb{R}).

对于问题A,系统(2.1)在条件(H_1)-(H_2)下有唯一解,这说明Y(0)是唯一确定的.然而对于问题B来说, Y(0)是任意的且可看做一个控制变量,只需在时刻T满足最优状态约束即可.故问题A嵌入问题B.因此,若(\bar{Y_0}, \bar{Z}, \bar{u})是B的最优控制的话,则\bar{u}必是A的最优控制.下面我们将使用经典的二阶变分技术来求解问题B.在给出主要结果前先介绍一下延拓的Ekeland变分原理.

定理2.1[1]  设K是完备度量空间(S, d)中的一个非空闭子集, F:K\times K\rightarrow \mathbb{R}为一双重函数.假设\varepsilon>0且如下条件成立:

(ⅰ)对所有x\in K, L:=\{y\in K: F(x, y)+\varepsilon d(x, y)\leq0\}是闭的;

(ⅱ)对所有x\in K, F(x, x)=0;

(ⅲ)对所有x, y, z\in K, F(x, y)\leq F(x, z)+F(z, y).

若对某个x_0\in K, \mathop {\inf }\limits_{y \in K} F(x_0, y)>-\infty,则存在\bar{x}\in K使得

\left\{\begin{array}{ll}F(x_0, \bar{x})+\varepsilon d(x_0, \bar{x})\leq0, \\F(\bar{x}, x)+\varepsilon d(\bar{x}, x)>0, \ \ \mbox{对所有的} \ \ x\in K, x\neq\bar{x}.\end{array}\right.

3 最大值原理

现在我们来求解问题B.首先,假定(\bar{Y}_0, \bar{Z}, \bar{u})是B的最优控制,相应的最优状态过程为(\bar{X}, \bar{Y}).对任意的\delta>0(Y_0, Z, u)\in {\cal R},定义罚泛函

J^{\delta}(Y_0, Z, u)=\Bigl\{[J(Y_0, Z, u)-J(\bar{Y}_0, \bar{Z}, \bar{u})+\delta]^2+[E|Y_T-\Phi(X_T)|^2]^2\Bigr\}^ {\frac{1}{2}}.
(3.1)

空间{\cal U}{\mathcal{H}}_{\mathcal{F}}^2(0, T;\mathbb{R})具有度量d:

\begin{eqnarray*}d(v, \tilde{v})=\bar{E}[|\{t\in[0, T]|v_t\neq\tilde{v}_t\}|], \ \ \forall v, \tilde{v}\in{\cal U} \ \ \mbox{或}\ \ {\mathcal{H}}_{\mathcal{F}}^2(0, T;\mathbb{R}), \end{eqnarray*}

其中|\cdot|表示Lebesgue测度.容易验证{\cal R}在如下度量\tilde{d}下是一个完备的度量空间[19]:

\begin{eqnarray*}\tilde{d}=\Bigl[|Y_0-\tilde{Y}_0|^2+d(Z, \tilde{Z})^2+d(u, \tilde{u})^2\Bigr]^{\frac{1}{2}}, \ \ \forall (Y_0, Z, u), (\tilde{Y}_0, \tilde{Z}, \tilde{u})\in {\cal R}.\end{eqnarray*}

显然, (3.1)(2.3)式构成一个无状态约束的正向控制问题.然而, {\cal R}的无界性不能保证J^{\delta}(Y_0, Z, u)({\cal R}, \tilde{d})中的下半连续性.因此,经典的Ekeland变分原理不再适用.本文将利用定理2.1来克服这一困难.首先,我们考虑Y_0, Z分别取值于{{\cal M}}\subset \mathbb{R}{{\cal N}}\subset \mathbb{R}的情形,其中{{\cal M}}是凸闭集, {{\cal N}}是闭集.

F(x, y)=J^\delta(Y_0, Z, u)-J^\delta(\tilde{Y}_0, \tilde{Z}, \tilde{u}), y=(Y_0, Z, u)\in {{\cal R}}_1={{\cal M}}\times {{\cal H}}_{{{\cal F}}}^2 (0, T;{{\cal N}})\times{{\cal U}}, x=(\tilde{Y}_0, \tilde{Z}, \tilde{u}) \in {{\cal R}}_1.则对任意的(\tilde{Y}_0\tilde{Z}, \tilde{u})\in {{\cal R}}_1,

L:=\{(Y_0, Z, u)\in {{\cal R}}_1: J^\delta(Y_0, Z, u)-J^\delta(\tilde{Y}_0, \tilde{Z}, \tilde{u})+\sqrt{\delta} \tilde{d}((Y_0, Z, u), (\tilde{Y}_0, \tilde{Z}, \tilde{u}))\leq0\}

是闭集.事实上,令\{(Y_{0n}, Z_n, u_n)\}L中一个序列满足\tilde{d}((Y_{0n}, Z_n, u_n), (\check{Y}_0, \check{Z}, \check{u}))\rightarrow0 (n\rightarrow0).

J^\delta(Y_{0n}, Z_n, u_n)-J^\delta(\tilde{Y}_0, \tilde{Z}, \tilde{u})+\sqrt{\delta} \tilde{d}((Y_{0n}, Z_n, u_n), (\tilde{Y}_0, \tilde{Z}, \tilde{u}))\leq0.

注意到(3.1)式,估计如下差分

\begin{eqnarray*} &&|J^\delta(Y_{0n}, Z_n, u_n)-J^\delta(\check{Y}_0, \check{Z}, \check{u})|\\ &=& \frac{|{J^\delta(Y_{0n}, Z_n, u_n)}^2-{J^\delta(\check{Y}_0, \check{Z}, \check{u})}^2|}{|J^\delta(Y_{0n}, Z_n, u_n)+J^\delta(\check{Y}_0, \check{Z}, \check{u})|}\\ &=& \frac{|J(Y_{0n}, Z_n, u_n)-J(\check{Y}_0, \check{Z}, \check{u})||J(Y_{0n}, Z_n, u_n)+J(\check{Y}_0, \check{Z}, \check{u})-2J(\bar{Y}_0, \bar{Z}, \bar{u})+2\delta|}{|J^\delta(Y_{0n}, Z_n, u_n)+J^\delta(\check{Y}_0, \check{Z}, \check{u})|}\\ &&+\frac{\Bigl|[E|Y_{Tn}-\Phi(X_{Tn})|^2]^2-[E|\check{Y}_{T}-\Phi(\check{X}_{T})|^2]^2\Bigr|}{|J^\delta(Y_{0n}, Z_n, u_n)+J^\delta(\check{Y}_0, \check{Z}, \check{u})|}. \end{eqnarray*}

其中

\begin{eqnarray*} &&|J(Y_{0n}, Z_n, u_n)-J(\check{Y}_0, \check{Z}, \check{u})|\\ &=&\Bigl|E\int_0^TE'[l(t, X'_{tn}, Y'_{tn}, Z'_{tn}, X_{tn}, Y_{tn}, Z_{tn}, u_{tn})-l(t, \check{X}'_{t}, \check{Y}'_{t}, \check{Z}'_{t}, \check{X}_{t}, \check{Y}_{t}, \check{Z}_{t}, \check{u}_{t})]{\rm d}t\\ &&+ E[h(X_{Tn})-h(\check{X}_T)+\gamma(Y_{0n})-\gamma(\check{Y}_0)]\Bigr|\\ &\leq& CE\int_0^T\Bigl[|X_{tn}-\check{X}_t|+|Y_{tn}-\check{Y}_t|\Bigr]{\rm d}t+Cd(Z_{n}, \check{Z})+Cd(u_{n}, \check{u})\\ &&+CE|X_{Tn}-\check{X}_T|+CE|Y_{0n}-\check{Y}_0|\rightarrow0. \end{eqnarray*}

同理有

\lim\limits_{n\rightarrow\infty}\Bigl|[E|Y_{Tn}-\Phi(X_{Tn})|^2]^2-[E|\check{Y}_{T}-\Phi(\check{X}_{T})|^2]^2\Bigr|=0.

由(H_1)-(H_2)及注2.2知, JJ^\delta是有界的.故当n\rightarrow\infty时, J^\delta(Y_{0n}, Z_n, u_n)\rightarrow J^\delta(\check{Y}_0, \check{Z}, \check{u}),即

J^\delta(\check{Y}_0, \check{Z}, \check{u})-J^\delta(\tilde{Y}_0, \tilde{Z}, \tilde{u})+\sqrt{\delta} \tilde{d}((\check{Y}_0, \check{Z}, \check{u}), (\tilde{Y}_0, \tilde{Z}, \tilde{u}))\leq0.

(\check{Y}_0, \check{Z}, \check{u})\in L. L的闭性得证.

进一步,不难验证F(x, y)满足条件(ⅱ)和(ⅲ).由J^{\delta}(\bar{Y}_0, \bar{Z}, \bar{u})=\delta, J^{\delta}(Y_0, Z, u)>0,可得

\inf\limits_{y\in{{\cal R}_1}}F(x_0, y)=\inf\limits_{(Y_0, Z, u)\in{{\cal R}_1}}\Bigl(J^\delta(Y_0, Z, u)-J^\delta(\bar{Y}_0, \bar{Z}, \bar{u})\Bigr)>-\infty.

从而由定理2.1知,存在x^\delta=(Y^\delta_0, Z^\delta, u^\delta)\in{{\cal R}_1}使得

\begin{eqnarray*} \left\{\begin{array}{ll} J^{\delta}(Y^{\delta}_0, Z^{\delta}, u^{\delta})-J^{\delta}(\bar{Y}_0, \bar{Z}, \bar{u})+\sqrt{\delta} \tilde{d}((Y^{\delta}_0, Z^{\delta}, u^{\delta}), (\bar{Y}_0, \bar{Z}, \bar{u}))\leq 0, \\ J^{\delta}(Y_0, Z, u)- J^{\delta}(Y^{\delta}_0, Z^{\delta}, u^{\delta})+\sqrt{\delta}\tilde{d}((Y^{\delta}_0, Z^{\delta}, u^{\delta}), (Y_0, Z, u))\geq0. \end{array}\right. \end{eqnarray*}

进一步有

\left\{\begin{array}{ll} J^{\delta}(Y^{\delta}_0, Z^{\delta}, u^{\delta})\leq J^{\delta}(\bar{Y}_0, \bar{Z}, \bar{u})=\delta, \\ \tilde{d}((Y^{\delta}_0, Z^{\delta}, u^{\delta}), (\bar{Y}_0, \bar{Z}, \bar{u}))\leq\sqrt{\delta}, \\ J^{\delta}(Y^{\delta}_0, Z^{\delta}, u^{\delta})\leq J^{\delta}(Y_0, Z, u)+\sqrt{\delta}\tilde{d}((Y^{\delta}_0, Z^{\delta}, u^{\delta}), (Y_0, Z, u)). \end{array}\right.
(3.2)

这一结论意味着控制过程(Y^{\delta}_0, Z^{\delta}, u^{\delta})为罚泛函J^{\delta}(Y_0, Z, u)+\sqrt{\delta}\tilde{d}((Y^{\delta}_0, Z^{\delta}, u^{\delta}), (Y_0, Z, u))的全局最小值点.另一方面,因Zu是动态的,故对Z^\delta, u^\delta做针状变分.对充分小的\varepsilon>0,定义

\mu^{\delta, \varepsilon}_t=\left\{\begin{array}{ll}\mu^{\delta}_t, & t \in [0, T]\backslash S_{\varepsilon}, \\ \mu_t, &t \in S_{\varepsilon} , \end{array}\right.

其中S_{\varepsilon}\subseteq[0, T]满足|S_{\varepsilon}|=\varepsilon T, |\cdot|为Lebesgue测度,且\mu=Z, u.由于 Y_0和时间无关且取值于凸控制域,因此对Y^{\delta}_0 做一凸扰动.令控制变量Y_0\in {{\cal M}}满足Y^{\delta}_0+Y_0\in {{\cal M}},则对任意的\varepsilon\in[0, 1], Y^{\delta, \varepsilon}_0=Y^{\delta}_0+\varepsilon Y_0为一取值于{{\cal M}}的控制变量.

为简便记

\phi=\phi(t, (X^{\delta}_t)', X^{\delta}_t, u^{\delta}_t), ~~\phi^\varepsilon=\phi(t, (X^{\delta}_t)', X^{\delta}_t, u^{\delta, \varepsilon}_t),

\Delta\phi=\phi(t, (X^{\delta}_t)', X^{\delta}_t, u_t) -\phi(t, (X^{\delta}_t)', X^{\delta}_t, u^{\delta}_t),

\Delta \psi=\psi(t, (X^{\delta}_t)', (Y^{\delta}_t)', (Z_t)', X^{\delta}_t, Y^{\delta}_t, Z_t, u_t) -\psi(t, (X^{\delta}_t)', (Y^{\delta}_t)', (Z^{\delta}_t)', X^{\delta}_t, Y^{\delta}_t, Z^{\delta}_t, u^{\delta}_t),

\phi=b, \sigma, \Phi, h, \gamma;~~ \psi=f, l.

这些记号对于它们相应导数仍然适用.假设(X^{\delta, \varepsilon}, Y^{\delta, \varepsilon})(X^{\delta}, Y^{\delta})是方程(2.3)相应于(Y^{\delta, \varepsilon}_0, Z^{\delta, \varepsilon}, u^{\delta, \varepsilon})(Y^{\delta}_0, Z^{\delta}, u^{\delta})的解.引入如下一阶和二阶变分方程

\left\{\begin{array}{ll}{\rm d}X^{\delta}_1(t)=E'[b_{\tilde{x}}(X^{\delta}_1(t))'+b_{x}X^{\delta}_1(t)+\Delta bI_{S_{\varepsilon}}(t)]{\rm d}t\\\ \ \ \ \ \ \ \ \ \ \ \ \ +E'[\sigma_{\tilde{x}}(X^{\delta}_1(t))'+\sigma_{x}X^{\delta}_1(t)+\Delta \sigma I_{S_{\varepsilon}}(t)]{\rm d}B_t, \\{\rm d}Y^{\delta}_1(t)=E'[f_{\tilde{x}}(X^{\delta}_1(t))'+f_{x}X^{\delta}_1(t)+f_{\tilde{y}}(Y^{\delta}_1(t))'+f_{y}Y^{\delta}_1(t)+\Delta fI_{S_{\varepsilon}}(t)]{\rm d}t\\\ \ \ \ \ \ \ \ \ \ \ \ \ +E'[(Z^{\delta, \varepsilon}_t-Z^{\delta}_t)'+(Z^{\delta, \varepsilon}_t-Z^{\delta}_t)]{\rm d}B_t, \\ X^{\delta}_1(0)=0, Y^{\delta}_1(0)=\varepsilon Y_0. \end{array}\right.
(3.3)

\left\{\begin{array}{ll}{\rm d}X^{\delta}_2(t)=E'\Bigl\{b_{\tilde{x}}(X^{\delta}_2(t))'+b_{x}X^{\delta}_2(t) +[\Delta b_{\tilde{x}} (X^{\delta}_1(t))'+\Delta b_x X^{\delta}_1(t)]I_{S_{\varepsilon}}(t)\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\frac{1}{2}b_{\tilde{x}\tilde{x}}[(X^{\delta}_1(t))']^2+\frac{1}{2}b_{xx}X^{\delta}_1(t)^2\Bigr\}{\rm d}t\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ +E'\Bigl\{\sigma_{\tilde{x}}(X^{\delta}_2(t))'+\sigma_{x}X^{\delta}_2(t)+[\Delta \sigma_{\tilde{x}} (X^{\delta}_1(t))'+\Delta \sigma_x X^{\delta}_1(t)]I_{S_{\varepsilon}}(t)\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\frac{1}{2}\sigma_{\tilde{x}\tilde{x}}[(X^{\delta}_1(t))']^2+ \frac{1}{2}\sigma_{xx}X^{\delta}_1(t)^2\Bigr\}{\rm d}B_t, \\ {\rm d}Y^{\delta}_2(t)=E'\Bigl\{f_{\tilde{x}}(X^{\delta}_2(t))'+f_{x}X^{\delta}_2(t)+f_{\tilde{y}}(Y^{\delta}_2(t))'+f_{y}Y^{\delta}_2(t)\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +[\Delta f_{\tilde{x}} (X^{\delta}_1(t))'+\Delta f_x X^{\delta}_1(t)+\Delta f_{\tilde{y}} (Y^{\delta}_1(t))'+\Delta f_y Y^{\delta}_1(t)]I_{S_{\varepsilon}}(t)\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\frac{1}{2}f_{\tilde{x}\tilde{x}}[(X^{\delta}_1(t))']^2+\frac{1}{2}f_{xx}X^{\delta}_1(t)^2 +\frac{1}{2}f_{\tilde{y}\tilde{y}}[(Y^{\delta}_1(t))']^2+\frac{1}{2}f_{yy}Y^{\delta}_1(t)^2\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +f_{\tilde{x}\tilde{y}}(X^{\delta}_1(t))'(Y^{\delta}_1(t))'+f_{xy}X^{\delta}_1(t)Y^{\delta}_1(t)\Bigr\}{\rm d}t\\ X^{\delta}_2(0)=0, Y^{\delta}_2(0)=0, \end{array}\right.
(3.4)

其中I_{S_\varepsilon}是集合S_\varepsilon的示性函数.

容易验证方程组(3.3)(3.4)在空间{{\cal H}}_{{{\cal F}}}^2(0, T;\mathbb{R}^2)中存在唯一适应解.为获得变分不等式,我们先给出一些先验估计.

引理3.1  假设(H_1)-(H_2)成立,则

\sup\limits_{t\in[0, T]}E|X^{\delta}_1(t)|^{4}\leq C\varepsilon^2, \ \ \sup\limits_{t\in[0, T]}E|X^{\delta}_1(t)|^{8}\leq C\varepsilon^4,

\sup\limits_{t\in[0, T]}E|X^{\delta}_2(t)|^{2}\leq C\varepsilon^2, \ \ \sup\limits_{t\in[0, T]}E|X^{\delta}_2(t)|^{4}\leq C\varepsilon^4,

\sup\limits_{t\in[0, T]}E\Bigl[|X^{\delta, \varepsilon}_t-X^{\delta}_t-X^{\delta}_1(t)- X^{\delta}_2(t)|^{2}\Bigr] \leq C_\varepsilon\varepsilon^{2},

其中CC_\varepsilon是常数且\lim_{\varepsilon\rightarrow0}C_\varepsilon=0.

  详情可参阅文献[12,引理2.1].

引理3.2  假设(H_1)-(H_2)成立,则

\sup\limits_{t\in[0, T]}E|Y^{\delta}_1(t)|^{4}\leq C\varepsilon^2, \ \ \sup\limits_{t\in[0, T]}E|Y^{\delta}_1(t)|^{8}\leq C\varepsilon^4,

\sup\limits_{t\in[0, T]}E|Y^{\delta}_2(t)|^{2}\leq C\varepsilon^2, \ \ \sup\limits_{t\in[0, T]}E|Y^{\delta}_2(t)|^{4}\leq C\varepsilon^4,

\sup\limits_{t\in[0, T]}E\Bigl[|Y^{\delta, \varepsilon}_t-Y^{\delta}_t-Y^{\delta}_1(t)- Y^{\delta}_2(t)|^{2}\Bigr] \leq C_\varepsilon\varepsilon^{2},

其中CC_\varepsilon为常数且\lim_{\varepsilon\rightarrow0}C_\varepsilon=0.

  由引理3.1、BDG不等式及Gronwall不等式可直接推出结论.

为了建立问题{\cal B}的最大值原理,定义Hamiltonian函数H

\begin{eqnarray*} H(t, \tilde{x}, \tilde{y}, \tilde{z}, x, y, z, u, p, q, k, r, \theta)&=& b(t, \tilde{x}, x, u)p+\sigma(t, \tilde{x}, x, u)q+f(t, \tilde{x}, \tilde{y}, \tilde{z}, x, y, z, u)k\\ &&+ (\tilde{z}+z)r+\theta l(t, \tilde{x}, \tilde{y}, \tilde{z}, x, y, z, u). \end{eqnarray*}

一阶,二阶伴随方程分别为

\left\{\begin{array}{ll} -{\rm d}p_t=E'[b_{\tilde{x}}(p_t)'+b_xp_t+\sigma_{\tilde{x}}(q_t)'+\sigma_xq_t +f_{\tilde{x}}(k_t)'+f_xk_t +\theta l_{\tilde{x}}+\theta l_x]{\rm d}t-q_t{\rm d}B_t, \\ -{\rm d}k_t=E'[f_{\tilde{y}}(k_t)'+f_yk_t+\theta l_{\tilde{y}}+\theta l_y]{\rm d}t-r_t{\rm d}B_t, \\ p_T=\theta h_x(\bar{X}_T), k_T=0. \end{array}\right.
(3.5)

\left\{\begin{array}{ll}-{\rm d}p_2(t)=E'[2f_{\tilde{y}}(p_2(t))'+2f_yp_2(t)+H_{yy}(t)+H_{\tilde{y}\tilde{y}}(t)]{\rm d}t -q_2(t){\rm d}B_t, \\ p_2(T)=2\theta_T. \end{array}\right.
(3.6)

\left\{\begin{array}{ll}-{\rm d}p_3(t)=E'[f_{\tilde{y}}(p_3(t))'+f_yp_3(t)+b_{\tilde{x}}(p_3(t))'+b_xp_3(t)+2f_{\tilde{x}}(p_2(t))'\\\ \ \ \ \ \ \ \ \ \ \ \ \ +2f_xp_2(t)+\sigma_{\tilde{x}}(q_3(t))'+\sigma_xq_3(t)+2H_{xy}(t)+2H_{\tilde{x}\tilde{y}}(t)]{\rm d}t-q_3(t){\rm d}B_t, \\ p_3(T)=0. \end{array}\right.
(3.7)

\left\{\begin{array}{ll}-{\rm d}p_1(t)=E'[2b_{\tilde{x}}(p_1(t))'+2b_xp_1(t)+\sigma_{\tilde{x}}^2(p_1(t))'+\sigma_{x}^2p_1(t) +2\sigma_{\tilde{x}}(q_1(t))'\\\ \ \ \ \ \ \ \ \ \ \ \ \+2\sigma_xq_1(t)+f_{\tilde{x}}(p_3(t))'+f_xp_3(t) +H_{xx}(t)+H_{\tilde{x}\tilde{x}}(t)]{\rm d}t -q_1(t){\rm d}B_t, \\ p_1(T)=\theta h_{xx}(\bar{X}_T)+2\theta_T\Phi^2_{x}(\bar{X}_T). \end{array}\right.
(3.8)

定理3.1  假设(H_1)-(H_2)成立, (\bar{Y}_0, \bar{Z}, \bar{u})是问题B的最优控制,相应的最优轨道为(\bar{X}, \bar{Y}).则存在两个非负参数\theta\theta_T满足\theta^2+\theta^2_T=1

\begin{eqnarray} &&E'\Bigl\{H(t, (\bar{X}_t)', (\bar{Y}_t)', (Z_t)', \bar{X}_t, \bar{Y}_t, Z_t, u_t, p_t, q_t, k_t, r_t, \theta_t)\\ &&- H(t, (\bar{X}_t)', (\bar{Y}_t)', (\bar{Z}_t)', \bar{X}_t, \bar{Y}_t, \bar{Z}_t, \bar{u}_t, p_t, q_t, k_t, r_t, \theta_t)\\ &&+\frac{1}{2}p_3(t)\Delta \sigma[(Z_t-\bar{Z}_t)' +Z_t-\bar{Z}_t]+\frac{1}{2}p_2(t)[(Z_t-\bar{Z}_t)' +Z_t-\bar{Z}_t]^2 \\ &&+\frac{1}{2}p_1(t)(\Delta \sigma)^2\Bigr\}+\theta\gamma_y(\bar{Y}_0)Y_0+k_0Y_0\geq0.\ \ \ \forall Y_0\in {{\cal M}}, Z\in {{\cal N}}, u\in U, \ {\rm a.e., a.s.}.\end{eqnarray}
(3.9)

  证明分为两步.

第一步  令Y_0, Z分别取值于{{\cal M}}\subset \mathbb{R}{{\cal N}}\subset \mathbb{R},其中{{\cal M}}为凸闭集, {\cal N}为闭集.

(3.2)式及\tilde{d}的定义有

\begin{eqnarray}&&J^{\delta}(Y^{\delta, \varepsilon}_0, Z^{\delta, \varepsilon}, u^{\delta, \varepsilon})-J^{\delta}(Y^{\delta}_0, Z^{\delta}, u^{\delta}) +\sqrt{\delta}\tilde{d}((Y^{\delta}_0, Z^{\delta}, u^{\delta}), (Y^{\delta, \varepsilon}_0, Z^{\delta, \varepsilon}, u^{\delta, \varepsilon}))\\&=&\frac{J^{\delta}(Y^{\delta, \varepsilon}_0, Z^{\delta, \varepsilon}, u^{\delta, \varepsilon})^2-J^{\delta}(Y^{\delta}_0, Z^{\delta}, u^{\delta}) ^2}{J^{\delta}(Y^{\delta, \varepsilon}_0, Z^{\delta, \varepsilon}, u^{\delta, \varepsilon})+J^{\delta}(Y^{\delta}_0, Z^{\delta}, u^{\delta}) }+\varepsilon\sqrt{\delta}\sqrt{2+|Y_0|^2}\geq 0.\end{eqnarray}
(3.10)

应用引理3.1、引理3.2及Taylor展式得

\begin{eqnarray*}&&J^{\delta}(Y^{\delta, \varepsilon}_0, Z^{\delta, \varepsilon}, u^{\delta, \varepsilon})^2-J^{\delta}(Y^{\delta}_0, Z^{\delta}, u^{\delta}) ^2\\&=&\big[J(Y^{\delta, \varepsilon}_0, Z^{\delta, \varepsilon}, u^{\delta, \varepsilon})-J(\bar{Y}_0, \bar{Z}, \bar{u}) +\delta\big]^2-[J(Y^{\delta}_0, Z^{\delta}, u^{\delta})-J(\bar{Y}_0, \bar{Z}, \bar{u}) +\delta]^2\\&&+\Big[E|Y^{\delta, \varepsilon}_T-\Phi(X^{\delta, \varepsilon}_T)|^2\Big]^2-\Big[E|Y^{\delta}_T-\Phi(X^{\delta}_T)|^2\Big]^2\\&=&2[J(Y^{\delta}_0, Z^{\delta}, u^{\delta})-J(\bar{Y}_0, \bar{Z}, \bar{u}) +\delta]\Bigl\{E\int_0^TE'\Big[l_{\tilde{x}}(X^{\delta}_1(t)+X^{\delta}_2(t))'\\&&+l_x(X^{\delta}_1(t)+X^{\delta}_2(t))+l_{\tilde{y}}(Y^{\delta}_1(t)+Y^{\delta}_2(t))'+l_y(Y^{\delta}_1(t)+Y^{\delta}_2(t))\\&& +\frac{1}{2}l_{\tilde{x}\tilde{x}}[(X^{\delta}_1(t))']^2+l_{\tilde{x}\tilde{y}}(X^{\delta}_1(t))'(Y^{\delta}_1(t))'+\frac{1}{2}l_{\tilde{y}\tilde{y}}[(Y^{\delta}_1(t))']^2\\&&+\frac{1}{2}l_{xx}[X^{\delta}_1(t)]^2+l_{xy}X^{\delta}_1(t)Y^{\delta}_1(t)+\frac{1}{2}l_{yy}[Y^{\delta}_1(t)]^2+\Delta l^{\delta}I_{S_\varepsilon}(t)\Big]{\rm d}t\\&&+E[h_x(X^{\delta}_T)(X^{\delta}_1(T)+X^{\delta}_2(T))]+\frac{1}{2}E[h_{xx}(X^{\delta}_T)(X^{\delta}_1(T))^2]+E[\gamma_y(Y^{\delta}_0)\varepsilon Y_0]\Bigr\}\\&&+2E|Y^{\delta}_T-\Phi(X^{\delta}_T)|^2\Bigl\{2E[(Y^{\delta}_T-\Phi(X^{\delta}_T))(Y^{\delta}_1(T)+Y^{\delta}_2(T))]+E[Y^{\delta}_1(T)]^2\\&&-2E[(Y^{\delta}_T-\Phi(X^{\delta}_T))\Phi_x(X^{\delta}_T)(X^{\delta}_1(T)+X^{\delta}_2(T))]\\&&-E[(Y^{\delta}_T-\Phi(X^{\delta}_T))\Phi_{xx}(X^{\delta}_T)(X^{\delta}_1(T))^2]+E[\Phi^2_x(X^{\delta}_T)(X^{\delta}_1(T))^2]\Bigr\}+o(\varepsilon).\end{eqnarray*}

\theta^{\delta, \varepsilon}=\frac{2[J(Y^{\delta}_0, Z^{\delta}, u^{\delta})-J(\bar{Y}_0, \bar{Z}, \bar{u}) +\delta]}{J^{\delta}(Y^{\delta, \varepsilon}_0, Z^{\delta, \varepsilon}, u^{\delta, \varepsilon})+J^{\delta}(Y^{\delta}_0, Z^{\delta}, u^{\delta}) }, ~~ \theta^{\delta, \varepsilon}_T=\frac{2E|Y^{\delta}_T-\Phi(X^{\delta}_T)|^2}{J^{\delta}(Y^{\delta, \varepsilon}_0, Z^{\delta, \varepsilon}, u^{\delta, \varepsilon})+J^{\delta}(Y^{\delta}_0, Z^{\delta}, u^{\delta})},

\theta^{\delta, \varepsilon}\geq0, \theta^{\delta, \varepsilon}_T\geq0.进而,由(3.10)式可得变分不等式

\begin{eqnarray*}0&\leq&\theta^{\delta, \varepsilon}\Bigl\{E\int_0^TE'[l_{\tilde{x}}(X^{\delta}_1(t)+X^{\delta}_2(t))'+l_x(X^{\delta}_1(t)+X^{\delta}_2(t))+l_{\tilde{y}}(Y^{\delta}_1(t)+Y^{\delta}_2(t))'\\&&+l_y(Y^{\delta}_1(t)+Y^{\delta}_2(t)) +\frac{1}{2}l_{\tilde{x}\tilde{x}}[(X^{\delta}_1(t))']^2+l_{\tilde{x}\tilde{y}}(X^{\delta}_1(t))'(Y^{\delta}_1(t))'+\frac{1}{2}l_{\tilde{y}\tilde{y}}[(Y^{\delta}_1(t))']^2\\&&+\frac{1}{2}l_{xx}[X^{\delta}_1(t)]^2+l_{xy}X^{\delta}_1(t)Y^{\delta}_1(t)+\frac{1}{2}l_{yy}[Y^{\delta}_1(t)]^2+\Delta l^{\delta}I_{S_\varepsilon}(t)]{\rm d}t\\&&+E[h_x(X^{\delta}_T)(X^{\delta}_1(T)+X^{\delta}_2(T))+\frac{1}{2}h_{xx}(X^{\delta}_T)(X^{\delta}_1(T))^2+\gamma_y(Y^{\delta}_0)\varepsilon Y_0]\Bigr\}\\&&+\theta^{\delta, \varepsilon}_T\Bigl\{2E[(Y^{\delta}_T-\Phi(X^{\delta}_T))(Y^{\delta}_1(T)+Y^{\delta}_2(T))]+E[Y^{\delta}_1(T)]^2\\&&-2E[(Y^{\delta}_T-\Phi(X^{\delta}_T))\Phi_x(X^{\delta}_T)(X^{\delta}_1(T)+X^{\delta}_2(T))]\\&&-E[(Y^{\delta}_T-\Phi(X^{\delta}_T))\Phi_{xx}(X^{\delta}_T)(X^{\delta}_1(T))^2-\Phi^2_x(X^{\delta}_T)(X^{\delta}_1(T))^2]\Bigr\}\\&&+o(\varepsilon)+\varepsilon\sqrt{\delta}\sqrt{2+|Y_0|^2}.\end{eqnarray*}

为了去掉上述不等式中的X^{\delta}_2(\cdot)Y^{\delta}_2(\cdot),对p^{\delta, \varepsilon}_t(X^{\delta}_1(t)+X^{\delta}_2(t))+k^{\delta, \varepsilon}_t(Y^{\delta}_1(t)+Y^{\delta}_2(t))使用Itô公式,其中p^{\delta, \varepsilon}k^{\delta, \varepsilon}为方程组(3.5)相应于u^{\delta, \varepsilon}的解.利用引理3.1及引理3.2,上述变分不等式可化为

\begin{eqnarray}0&\leq&E\int_0^TE'\Bigl\{H(t, (X^{\delta}_t)', (Y^{\delta}_t)', (Z^{\delta, \varepsilon}_t)', X^{\delta}_t, Y^{\delta}_t, Z^{\delta, \varepsilon}_t, u^{\delta, \varepsilon}_t, p^{\delta, \varepsilon}_t, q^{\delta, \varepsilon}_t, k^{\delta, \varepsilon}_t, r^{\delta, \varepsilon}_t, \theta^{\delta, \varepsilon}_t)\\&&-H(t, (X^{\delta}_t)', (Y^{\delta}_t)', (Z^{\delta}_t)', X^{\delta}_t, Y^{\delta}_t, Z^{\delta}_t, u^{\delta}_t, p^{\delta, \varepsilon}_t, q^{\delta, \varepsilon}_t, k^{\delta, \varepsilon}_t, r^{\delta, \varepsilon}_t, \theta^{\delta, \varepsilon}_t)\\&&+\frac{1}{2}H_{\tilde{x}\tilde{x}}[(X^{\delta}_1(t))']^2+\frac{1}{2}H_{xx}[X^{\delta}_1(t)]^2+\frac{1}{2}H_{\tilde{y}\tilde{y}}[(Y^{\delta}_1(t))']^2\\&&+\frac{1}{2}H_{yy}[Y^{\delta}_1(t)]^2+H_{\tilde{x}\tilde{y}}(X^{\delta}_1(t))'(Y^{\delta}_1(t))'+H_{xy}X^{\delta}_1(t)Y^{\delta}_1(t)\Bigr\}{\rm d}t\\&&+\frac{1}{2}E\Bigl\{\theta^{\delta, \varepsilon}h_{xx}(X^{\delta}_T)[X^{\delta}_1(T)]^2\Bigr\}+E\Big[\theta^{\delta, \varepsilon}\gamma_y(Y^{\delta}_0)\varepsilon Y_0+k^{\delta, \varepsilon}_0\varepsilon Y_0\Big]\\&&-E\theta^{\delta, \varepsilon}_T\Bigl\{(Y^{\delta}_T-\Phi(X^{\delta}_T))\Phi_{xx}(X^{\delta}_T)[X^{\delta}_1(T)]^2+\Phi^2_{x}(X^{\delta}_T)[X^{\delta}_1(T)]^2\Bigr\}\\&&+E\theta^{\delta, \varepsilon}_T[Y^{\delta}_1(T)]^2+o(\varepsilon)+\varepsilon\sqrt{\delta}\sqrt{2+|Y_0|^2}, \end{eqnarray}
(3.11)

其中H_j=H_j(t, (X^{\delta}_t)', (Y^{\delta}_t)', (Z^{\delta}_t)', X^{\delta}_t, Y^{\delta}_t, Z^{\delta}_t, u^{\delta}_t, p^{\delta, \varepsilon}_t, q^{\delta, \varepsilon}_t, k^{\delta, \varepsilon}_t, r^{\delta, \varepsilon}_t, \theta^{\delta, \varepsilon}_t).进一步,为了消去(3.11)式中的X^{\delta}_1(\cdot)X^{\delta}_1(\cdot), Y^{\delta}_1(\cdot)Y^{\delta}_1(\cdot)X^{\delta}_1(\cdot)Y^{\delta}_1(\cdot),引入相应于u^{\delta, \varepsilon}的二阶伴随方程(3.6)-(3.8)及随机过程\Theta^{\delta}=X^{\delta}_1(\cdot)X^{\delta}_1(\cdot)\Psi^{\delta}=Y^{\delta}_1(\cdot)Y^{\delta}_1(\cdot).这里\Theta^{\delta}, \Psi^{\delta}分别满足方程

\left\{\begin{array}{ll}{\rm d}\Theta^{\delta}_t=E'\{2b_{\tilde{x}}X_1^{\delta}(t)(X_1^{\delta}(t))'+2b_x\Theta^{\delta}_t+\sigma_{\tilde{x}}^2(\Theta^{\delta}_t)'+\sigma_x^2\Theta^{\delta}_t\\ \ \ \ \ \ \ \ \ \ \ +[2X^{\delta}_1(t)\Delta b +2\sigma_{\tilde{x}}(X^{\delta}_1(t))'\Delta \sigma +2\sigma_{x}X^{\delta}_1(t)\Delta \sigma+(\Delta \sigma)^2] I_{S_{\varepsilon}}(t)\\ \ \ \ \ \ \ \ \ \ \ +2\sigma_{\tilde{x}}\sigma_x(X^{\delta}_1(t))'X^{\delta}_1(t)\}{\rm d}t+E' [2\sigma_{\tilde{x}}X_1^{\delta}(t)(X_1^{\delta}(t))' \\ \ \ \ \ \ \ \ \ \ \+2\sigma_x\Theta^{\delta}_t+2X^{\delta}_1(t)\Delta \sigma I_{S_{\varepsilon}}(t)]{\rm d}B_t, \\ \Theta^{\delta}_0=0, \end{array}\right.

\left\{\begin{array}{ll}{\rm d}\Psi^{\delta}_t=E'\Bigl\{2f_{\tilde{y}}Y_1^{\delta}(t)(Y_1^{\delta}(t))'+2f_y\Psi^{\delta}_t+2f_{\tilde{x}}(X^{\delta}_1(t))'Y^{\delta}_1(t)+2f_{x}X^{\delta}_1(t)Y^{\delta}_1(t)\\\ \ \ \ \ \ \ \ \ \ +2Y^{\delta}_1(t)\Delta fI_{S_{\varepsilon}}(t)+[(Z^{\delta, \varepsilon}_t-Z^{\delta}_t)'+Z^{\delta, \varepsilon}_t-Z^{\delta}_t]^2\Bigr\}{\rm d}t \\ \ \ \ \ \ \ \ \ \ \ + E'\Bigl\{2Y^{\delta}_1(t)[(Z^{\delta, \varepsilon}_t-Z^{\delta}_t)' +Z^{\delta, \varepsilon}_t-Z^{\delta}_t]\Bigr\}{\rm d}B_t, \\ \Psi^{\delta}_0=\varepsilon^2Y^2_0. \end{array}\right.

p^{\delta, \varepsilon}_1(t)\Theta^\delta_t+p^{\delta, \varepsilon}_2(t)\Psi^\delta_t+p^{\delta, \varepsilon}_3(t)X^{\delta}_1(t)Y^{\delta}_1(t)使用Itô公式,则(3.11)式可化简为

\begin{eqnarray}&&E\int_0^TE'\Bigl\{H(t, (X^{\delta}_t)', (Y^{\delta}_t)', (Z^{\delta, \varepsilon}_t)', X^{\delta}_t, Y^{\delta}_t, Z^{\delta, \varepsilon}_t, u^{\delta, \varepsilon}_t, p^{\delta, \varepsilon}_t, q^{\delta, \varepsilon}_t, k^{\delta, \varepsilon}_t, r^{\delta, \varepsilon}_t, \theta^{\delta, \varepsilon}_t)\\&&-H(t, (X^{\delta}_t)', (Y^{\delta}_t)', (Z^{\delta}_t)', X^{\delta}_t, Y^{\delta}_t, Z^{\delta}_t, u^{\delta}_t, p^{\delta, \varepsilon}_t, q^{\delta, \varepsilon}_t, k^{\delta, \varepsilon}_t, r^{\delta, \varepsilon}_t, \theta^{\delta, \varepsilon}_t)\\&&+\frac{1}{2}p^{\delta, \varepsilon}_3(t)\Delta \sigma I_{S_{\varepsilon}}(t)[(Z^{\delta, \varepsilon}_t-Z^{\delta}_t)' +Z^{\delta, \varepsilon}_t-Z^{\delta}_t]+\frac{1}{2}p^{\delta, \varepsilon}_2(t)[(Z^{\delta, \varepsilon}_t-Z^{\delta}_t)' +Z^{\delta, \varepsilon}_t-Z^{\delta}_t]^2 \\ &&+\frac{1}{2}p^{\delta, \varepsilon}_1(t)(\Delta \sigma)^2I_{S_{\varepsilon}}(t)\Bigr\}{\rm d}t+E[\theta^{\delta, \varepsilon}\gamma_y(Y^{\delta}_0)\varepsilon Y_0+k^{\delta, \varepsilon}_0\varepsilon Y_0]\\&&+\frac{1}{2}E[p^{\delta, \varepsilon}_2(0)\varepsilon^2Y^2_0]+o(\varepsilon)+\varepsilon\sqrt{\delta}\sqrt{2+|Y_0|^2}\geq0.\end{eqnarray}
(3.12)

注意到\mathop {\lim }\limits_{\varepsilon \to 0} |\theta^{\delta, \varepsilon}|^2+|\theta^{\delta, \varepsilon}_T|^2=1,故存在(\theta^{\delta, \varepsilon}, \theta^{\delta, \varepsilon}_T)的子列仍记为(\theta^{\delta, \varepsilon}, \theta^{\delta, \varepsilon}_T)使得

\lim\limits_{\varepsilon\rightarrow0}(\theta^{\delta, \varepsilon}, \theta^{\delta, \varepsilon}_T)=(\theta^{\delta}, \theta^{\delta}_T), \ \ \mbox{且} \ \ |\theta^{\delta}|^2+|\theta^{\delta}_T|^2=1.

因此,由平均场倒向随机微分方程解对参数的连续依赖性知,当\varepsilon\rightarrow0(p^{\delta, \varepsilon}, q^{\delta, \varepsilon})\rightarrow(p^{\delta}, q^{\delta}), (k^{\delta, \varepsilon}, r^{\delta, \varepsilon})\rightarrow(k^{\delta}, r^{\delta}), (p^{\delta, \varepsilon}_2, q^{\delta, \varepsilon}_2)\rightarrow(p^{\delta}_2, q^{\delta}_2), (p^{\delta, \varepsilon}_3, q^{\delta, \varepsilon}_3)\rightarrow(p^{\delta}_3, q^{\delta}_3), (p^{\delta, \varepsilon}_1, q^{\delta, \varepsilon}_1)\rightarrow(p^{\delta}_1, q^{\delta}_1), 这里(p^{\delta}, q^{\delta}), (k^{\delta}, r^{\delta}), (p^{\delta}_2, q^{\delta}_2), (p^{\delta}_3, q^{\delta}_3)(p^{\delta}_1, q^{\delta}_1)是(3.5)-(3.8)式相应于(\theta^{\delta}, \theta^{\delta}_T)的解, H_j=H_j(t, (X^{\delta}_t)', (Y^{\delta}_t)', (Z^{\delta}_t)', X^{\delta}_t, Y^{\delta}_t, Z^{\delta}_t, u^{\delta}_t, p^{\delta}_t, q^{\delta}_t, k^{\delta}_t, r^{\delta}_t, \theta^{\delta}_t).然后,令(3.12)\times \frac{1}{\varepsilon}\varepsilon\rightarrow0可得

\begin{eqnarray}&&\bar{E}\Bigl\{H(t, (X^{\delta}_t)', (Y^{\delta}_t)', (Z_t)', X^{\delta}_t, Y^{\delta}_t, Z_t, u_t, p^{\delta}_t, q^{\delta}_t, k^{\delta}_t, r^{\delta}_t, \theta^{\delta}_t)\\&&-H(t, (X^{\delta}_t)', (Y^{\delta}_t)', (Z^{\delta}_t)', X^{\delta}_t, Y^{\delta}_t, Z^{\delta}_t, u^{\delta}_t, p^{\delta}_t, q^{\delta}_t, k^{\delta}_t, r^{\delta}_t, \theta^{\delta}_t)\\&&+\frac{1}{2}p^{\delta}_3(t)\Delta \sigma[(Z_t-Z^{\delta}_t)' +Z_t-Z^{\delta}_t]+\frac{1}{2}p^{\delta}_2(t)[(Z_t-Z^{\delta}_t)' +Z_t-Z^{\delta}_t]^2 \\ &&+\frac{1}{2}p^{\delta}_1(t)(\Delta \sigma)^2\Bigr\}+E[\theta^{\delta}\gamma_y(Y^{\delta}_0)Y_0+k^{\delta}_0Y_0]+\sqrt{\delta}\sqrt{2+|Y_0|^2}\geq0. \end{eqnarray}
(3.13)

类似地,存在(\theta^{\delta}, \theta^{\delta}_T)的子列收敛于(\theta, \theta_T)且满足|\theta|^2+|\theta_T|^2=1.(3.2)式知,当\delta\rightarrow0时, (Y^{\delta}_0, Z^\delta, u^\delta)\rightarrow(\bar{Y}_0, \bar{Z}, \bar{u}).因此(X^{\delta}, Y^\delta)\rightarrow(\bar{X}, \bar{Y}), \bar{Y}_T=\Phi(\bar{X}_T)(p^{\delta}, q^{\delta})\rightarrow (p, q), (k^{\delta}, r^{\delta})\rightarrow (k, r), (p^{\delta}_2, q^{\delta}_2)\rightarrow (p_2, q_2), (p^{\delta}_3, q^{\delta}_3)\rightarrow (p_3, q_3), (p^{\delta}_1, q^{\delta}_1)\rightarrow (p_1, q_1), 这里(p, q), (k, r), (p_2, q_2), (p_3, q_3)(p_1, q_1)分别满足方程组(3.5)-(3.8).最后,在(3.13)式中令\delta\rightarrow0,即可得最优变分不等式(3.9).

第二步  一般控制域情形.

首先,假设

\begin{eqnarray*} \left\{\begin{array}{ll} {{\cal M}}^n=\{Y_0\in R; |Y_0|\leq |\bar{Y}_0|+n\}, \ \ {{\cal N}}^n=\{Z_t\in R; |Z_t|\leq |\bar{Z}_t|+n\}, \\ {{\cal H}}^2_{{\cal F}}(0, T, {{\cal N}}^n)=\{Z_t\in {{\cal H}}^2_{{\cal F}}(0, T, R); Z_t\in {{\cal N}}^n\}, \\ {{\cal U}}^n=\{u\in{{\cal U}}| \sup\limits_{t\in[0, T]}\bar{E}|u_t|\leq\sup\limits_{t\in[0, T]}\bar{E}|\bar{u}_t|+n\}, \ \ n=1, 2, 3, \cdots. \end{array}\right. \end{eqnarray*}

易见{{\cal M}}^n是凸的且\bar{Y}_0\in{{\cal M}}^n\subseteq{{\cal M}}^{n+1}, R=U^{\infty}_{n=1}{{\cal M}}^n, \bar{u}\in{{\cal U}}^n\subseteq{{\cal U}}^{n+1}, {{\cal U}}=U^{\infty}_{n=1}{{\cal U}}^n, \bar{Z}_t\in {{\cal H}}^2_{{\cal F}}(0, T, {{\cal N}}^n)\subseteq {{\cal H}}^2_{{\cal F}}(0, T, {{\cal N}}^{n+1}), {{\cal H}}^2_{{\cal F}}(0, T, R)=U^{\infty}_{n=1}{{\cal H}}^2_{{\cal F}}(0, T, {{\cal N}}^n). {{\cal M}}^n, {{\cal H}}^2_{{\cal F}}(0, T, {{\cal N}}^n), {{\cal U}}^n, n=1, 2, 3, \cdots 都是闭集且控制数组(\bar{Y}_0, \bar{Z}, \bar{u}){{\cal M}}^n\times{{\cal H}}^2_{{\cal F}}(0, T, {{\cal N}}^n)\times{{\cal U}}^n中仍是最优的.则由第一步可得,存在(\theta^n, \theta^n_T, p^n, q^n, k^n, r^n, p^n_2, q^n_2, p^n_3, q^n_3, p^n_1, q^n_1)的一个子列满足|\theta^n|^2+|\theta^n_T|^2=1,方程组(3.5)-(3.8)和最大值条件

\begin{eqnarray} &&E'\Bigl\{H(t, (\bar{X}_t)', (\bar{Y}_t)', (Z_t)', \bar{X}_t, \bar{Y}_t, Z_t, u_t, p^n_t, q^n_t, k^n_t, r^n_t, \theta^n_t)\\ &&- H(t, (\bar{X}_t)', (\bar{Y}_t)', (\bar{Z}_t)', \bar{X}_t, \bar{Y}_t, \bar{Z}_t, \bar{u}_t, p^n_t, q^n_t, k^n_t, r^n_t, \theta^n_t)\\ &&+\frac{1}{2}p^n_3(t)\Delta \sigma[(Z_t-\bar{Z}_t)' +Z_t-\bar{Z}_t]+\frac{1}{2}p^n_2(t)[(Z_t-\bar{Z}_t)' +Z_t-\bar{Z}_t]^2 \\ &&+\frac{1}{2}p^n_1(t)(\Delta \sigma)^2\Bigr\}+\theta\gamma_y(\bar{Y}_0)Y_0+k^n_0Y_0 \geq0. \forall (Y_0, Z, u )\in {{\cal M}}^n\times {{\cal N}}^n\times U, \ {\rm a.e., a.s.}\end{eqnarray}

同样地,存在(\theta^n, \theta^n_T)的子列仍记为它自身使得(\theta^n, \theta^n_T)\rightarrow (\theta, \theta_T)|\theta|^2+|\theta_T|^2=1.因此,在{\mathcal{H}}_{{\cal F}}^2(0, T;\mathbb{R}^2)中有(p^n, q^n, k^n, r^n, p^n_2, q^n_2, p^n_3, q^n_3, p^n_1, q^n_1)\rightarrow(p, q, k, r, p_2, q_2, p_3, q_3, p_1, q_1), 其中(p, q, k, r, p_2, q_2, p_3, q_3, p_1, q_1)满足方程组(3.5)-(3.8).最后在(3.14)式中,令n\rightarrow\infty即可得最优变分不等式(3.9).

(Y_0, Z, u)的任意性知,一般控制域情形下,问题A的随机最大值原理可归结如下.

定理3.2  假设(H_1)-(H_2)成立, (\bar{Y}_0, \bar{Z}, \bar{u})是问题A的最优控制,相应的最优轨道为(\bar{X}, \bar{Y}).则存在两个非负参数\theta\theta_T满足\theta^2+\theta^2_T=1,使得对任意的Y_0\in \mathbb{R}, Z\in {{\cal H}}^2_{{\cal F}}(0, T, \mathbb{R})u\in{\cal U},最大值条件(3.9)几乎必然成立,其中H为哈密顿函数, (p, q, k, r, p_2, q_2, p_3, q_3, p_1, q_1)是方程(3.5)-(3.8)相应于\bar{u}的解.

注3.1  若我们只考虑(2.1)式中的正向控制系统,即: \Phi=0, f=0, Z=0, \gamma=0.\theta=1, \theta_T=0,则k=r=p_2(\cdot)=q_2(\cdot)=p_3(\cdot)=q_3(\cdot)=0.此时定理3.2就是非凸控制域下平均场正向随机控制系统的最大值原理.

注3.2  当系数b, \sigma, f不依赖于w'时,系统(2.1)就退化为经典的正倒向随机微分方程,相应的最大值原理由文献[18]给出.所以我们的结果是文献[18]的一个广泛延拓.

4 实际应用

考虑最优控制问题.

问题C  在{\cal U}上最小化性能指标

J(u)=\frac{1}{2}E\bigg\{\int_0^T(L_1Z^2_t+L_2u^2_t){\rm d}t+H_1X^2_T+\gamma_1Y^2_0\bigg\},

其状态方程为

\begin{equation} \left\{\begin{array}{ll} {\rm d}X_t=\{B_1X_t+B_2EX_t+B_3u_t\}{\rm d}t+\{\sigma_1X_t+\sigma_2EX_t+\sigma_3u_t\}{\rm d}B_t, \\ {\rm d}Y_t=\{F_1X_t+F_2Y_t+F_3Z_t+F_4EX_t+F_5EY_t+F_6u_t\}{\rm d}t+Z_t{\rm d}B_t, \\ X(0)=a, \ \ Y_T=\Phi_1X_T, \end{array}\right. \end{equation}
(4.1)

其中B_i, \sigma_i, F_i, i\in\{1, \cdots , 6\}\Phi_1是实常数,且L_1>0, L_2>0, H_1\geq0\gamma_1>0.在实际应用中,由于某些原因控制变量的取值范围可限定为U=(-\infty, -1]\cup[1, +\infty).假设\bar{u}为问题C的最优控制,则相应的伴随方程为

\begin{equation} \left\{\begin{array}{ll} -{\rm d}p_t=\{B_1p_t+\sigma_1q_t+F_1k_t+B_2Ep_t+\sigma_2Eq_t +F_4Ek_t\}{\rm d}t-q_t{\rm d}B_t, \\ -{\rm d}k_t=[F_2k_t+F_5Ek_t]{\rm d}t-r_t{\rm d}B_t, \\ p_T=\theta H_1\bar{X}_T, k_T=0. \end{array}\right.\end{equation}
(4.2)

\begin{equation}\left\{\begin{array}{ll}-{\rm d}p_2(t)=\{2F_2p_2(t)+2F_5E[p_2(t)]\}{\rm d}t -q_2(t){\rm d}B_t, \\ p_2(T)=2\theta_T.\end{array}\right.\end{equation}
(4.3)

\begin{equation}\left\{\begin{array}{ll}-{\rm d}p_3(t)=\{(B_1+F_2)p_3(t)+(F_5+B_2)E[p_3(t)]+2F_1p_2(t)+\sigma_1q_3(t)\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ +2F_4E[p_2(t)]+\sigma_2E[q_3(t)]\}{\rm d}t-q_3(t){\rm d}B_t, \\ p_3(T)=0.\end{array}\right.\end{equation}
(4.4)

\begin{equation}\left\{\begin{array}{ll}-{\rm d}p_1(t)=\{(2B_1+\sigma_1^2)p_1(t)+(2B_2+\sigma_2^2)E[p_1(t)]+F_1p_3(t)+2\sigma_1q_1(t)\\\ \ \ \ \ \ \ \ \ \ \ \ \ \+F_4E[p_3(t)]+2\sigma_2E[q_1(t)]\}{\rm d}t -q_1(t){\rm d}B_t, \\ p_1(T)=\theta H_1+2\theta_T\Phi^2_{1}. \end{array}\right.\end{equation}
(4.5)

显然,方程组(4.2)-(4.5)分别存在唯一解.因\theta=0的情形是平凡的,本文只考虑非平凡情形,即\theta>0.在最大值条件(3.9)式中,令Z=\bar{Z}u=\bar{u},则有\theta\gamma_1\bar{Y}_0+k_0=0.

\begin{equation} \bar{Y_0}=-\frac{k_0}{\theta\gamma_1}. \end{equation}
(4.6)

其次,在(3.9)式中,令u=\bar{u}, Y_0=\bar{Y}_0,有

\begin{equation} \bar{Z}=-\frac{F_3k+r}{\theta L_1}. \end{equation}
(4.7)

最后,在(3.9)式中,令Z=\bar{Z}, Y_0=\bar{Y}_0,可知候选的最优控制\bar{u}需满足

\begin{eqnarray*} && \frac{1}{2}\theta L_2u^2_t+(B_3p_t+\sigma_3q_t+F_6k_t)u_t+\frac{1}{2}\sigma_3^2p_1(t)u^2_t-\sigma_3^2p_1(t)\bar{u}_tu_t\\& \geq&\frac{1}{2}\theta L_2\bar{u}^2_t+(B_3p_t+\sigma_3q_t+F_6k_t)\bar{u}_t-\frac{1}{2}\sigma_3^2p_1(t)\bar{u}^2_t, \ \ \ \forall u\in U. \end{eqnarray*}

进而有

\begin{equation} \bar{u}_t=\left\{\begin{array}{ll} \mu_t, &\mu_t\in (-\infty, -1]\cup[1, +\infty), \\ 1, & \mu_t\in [0, 1), \\ -1, ~~&\mu_t\in (-1, 0), \end{array}\right. \end{equation}
(4.8)

其中\mu_t=-\frac{B_3p_t+\sigma_3q_t+F_6k_t}{\theta L_2}.

命题4.1  由(4.8)式定义的控制\bar{u}是最优的,由(4.6)(4.7)式给出的(\bar{Y}_0, \bar{Z})及相应的轨道(\bar{X}, \bar{Y})是问题C的最优解.

  假设(X, Y, Z)是系统(4.1)相应于u\in {\cal U}的轨道,由凸函数的性质知

\begin{eqnarray*} \frac{1}{2}H_1X^2_T-\frac{1}{2}H_1\bar{X}^2_T\geq H_1\bar{X}_T(X_T-\bar{X}_T), \ \ \ \frac{1}{2}\gamma_1Y^2_0-\frac{1}{2}\gamma_1\bar{Y}^2_0\geq \gamma_1\bar{Y}_0(Y_0-\bar{Y}_0). \end{eqnarray*}

p_t(X_t-\bar{X}_t)+k_t(Y_t-\bar{Y}_t)应用Itô公式有

\begin{eqnarray*} &&E[\theta H_1\bar{X}_T(X_T-\bar{X}_T)+\theta \gamma_1\bar{Y}_0(Y_0-\bar{Y}_0)] \\ &=&E\int_0^T\Bigl\{(B_3p_t+\sigma_3q_t+F_6k_t)(u_t-\bar{u}_t)+(F_3k_t+r_t)(Z_t-\bar{Z}_t)\Bigr\}{\rm d}t. \end{eqnarray*}

利用(4.6)(4.7)式,可得

\begin{eqnarray*} J(u)-J(\bar{u})&\geq&\frac{1}{2}E\int_0^T\Big\{L_1(Z_t^2-\bar{Z}^2_t)+L_2(u_t^2-\bar{u}^2_t)\Big\}{\rm d}t\\ &&+E\Big[H_1\bar{X}_T(X_T-\bar{X}_T)+\gamma_1\bar{Y}_0(Y_0-\bar{Y}_0)\Big]\\ &=& \frac{1}{2}E\int_0^T\Big\{L_1(Z_t^2-\bar{Z}^2_t)+L_2(u_t^2-\bar{u}^2_t)\Big\}{\rm d}t\\ &&+E\int_0^T\Big\{\frac{B_3p_t+\sigma_3q_t+F_6k_t}{\theta}(u_t-\bar{u}_t)+\frac{F_3k_t+r_t}{\theta}(Z_t-\bar{Z}_t)\Big\}{\rm d}t\\ &=&\frac{1}{2}E\int_0^T\Big\{L_1(Z_t-\bar{Z}_t)^2+L_2(u_t^2-\bar{u}^2_t)\Big\}{\rm d}t\\ &&+E\int_0^T\frac{B_3p_t+\sigma_3q_t+F_6k_t}{\theta}(u_t-\bar{u}_t){\rm d}t. \end{eqnarray*}

另外,由(4.8)式知

\frac{1}{2}L_2u^2_t+\frac{B_3p_t+\sigma_3q_t+F_6k_t}{\theta}u_t\geq\frac{1}{2}L_2\bar{u}^2_t+\frac{B_3p_t+\sigma_3q_t+F_6k_t}{\theta}\bar{u}_t, \ \ \ \forall u\in {\cal U}

这意味着J(u)-J(\bar{u})\geq0, \forall u\in {\cal U}.这即说明了\bar{u}的最优性.

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