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].

设$(\Omega, {{\cal F}}, \{{{\cal F}}_t\}_{t\geq0}, P)$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$(B_t)_{t\geq0}$,且${{\cal F}}=\{{{\cal F}}_s, 0\leq s\leq T\}$是由$(B_t)_{t\geq0}$生成,所有$P$ -零集扩充的自然流,即${{\cal F}}_s=\sigma\{B_r, r\leq s\}\vee {{\cal N}}_p, $$s\in[0, T], $其中${{\cal N}}_p$是所有$P$-零子集组成的集合.记${{\cal S}}_{{\cal F}}^2(0, T;R)$为${\cal F}$ -适应和连续且满足$ E[\sup\limits_{t\in[0, T]}|\psi_t|^2] <\infty$的函数全体所成空间; ${{\cal H}}_{{\cal F}}^2(0, T;R)$为${\cal F}$ -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.

设$(\bar{\Omega}, \bar{{{\cal F}}}, \bar{P})=(\Omega\times\Omega, {{\cal F}}\bigotimes{{\cal F}} , P\bigotimes P)$是$(\Omega, {{\cal F}}, P)$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

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

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

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

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

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

若$\bar{u}$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$_1$) $b$, $\sigma$, $f$, $l$, $\Phi$, $h$和$\gamma$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$_2$) $b$, $\sigma$, $f$, $l$, $\Phi$, $h$和$\gamma$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$Z$和$u$,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

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

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

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

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$\delta>0$和$(Y_0, Z, u)\in {\cal R}$,定义罚泛函

(3.1) $ 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}}. $

空间${\cal U}$和${\mathcal{H}}_{\mathcal{F}}^2(0, T;\mathbb{R})$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, $(3.1)$和$(2.3)$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$J^{\delta}(Y_0, Z, u)$在$({\cal R}, \tilde{d})$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${{\cal M}}\subset \mathbb{R}$和${{\cal N}}\subset \mathbb{R}$的情形,其中${{\cal M}}$是凸闭集, ${{\cal N}}$是闭集.

令$F(x, y)=J^\delta(Y_0, Z, u)-J^\delta(\tilde{Y}_0, \tilde{Z}, \tilde{u})$, $y=(Y_0, Z, u)\in {{\cal R}}_1={{\cal M}}\times {{\cal H}}_{{{\cal F}}}^2 (0, T;{{\cal N}})\times{{\cal U}}$, $x=(\tilde{Y}_0, \tilde{Z}, \tilde{u}) \in {{\cal R}}_1$.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$\{(Y_{0n}, Z_n, u_n)\}$是$L$中一个序列满足$\tilde{d}((Y_{0n}, Z_n, u_n), (\check{Y}_0, \check{Z}, \check{u}))\rightarrow0$ ($n\rightarrow0$).则

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

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, $J$和$J^\delta$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. $

故$(\check{Y}_0, \check{Z}, \check{u})\in L$. $L$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

(3.2) $\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. $

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$Z$和$u$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$(X^{\delta, \varepsilon}, Y^{\delta, \varepsilon})$和$(X^{\delta}, Y^{\delta})$是方程$(2.3)$相应于$(Y^{\delta, \varepsilon}_0, Z^{\delta, \varepsilon}, u^{\delta, \varepsilon})$和$(Y^{\delta}_0, Z^{\delta}, u^{\delta})$的解.引入如下一阶和二阶变分方程

(3.3) $ \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. $

(3.4) $ \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. $

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

容易验证方程组$(3.3)$和$(3.4)$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$C$和$C_\varepsilon$是常数且$\lim_{\varepsilon\rightarrow0}C_\varepsilon=0$.

证  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$C$和$C_\varepsilon$为常数且$\lim_{\varepsilon\rightarrow0}C_\varepsilon=0$.

证  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$H$为

$ \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

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

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

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

(3.8) $\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. $

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$\theta$和$\theta_T$满足$\theta^2+\theta^2_T=1$且

(3.9) $ \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}$

证  证明分为两步.

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

由$(3.2)$式及$\tilde{d}$的定义有

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

应用引理3.1、引理3.2及Taylor展式得

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

令

$\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})}, $

则$\theta^{\delta, \varepsilon}\geq0$, $\theta^{\delta, \varepsilon}_T\geq0$.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$X^{\delta}_2(\cdot)$和$Y^{\delta}_2(\cdot)$,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

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

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$X^{\delta}_1(\cdot)X^{\delta}_1(\cdot)$, $Y^{\delta}_1(\cdot)Y^{\delta}_1(\cdot)$及$X^{\delta}_1(\cdot)Y^{\delta}_1(\cdot)$,引入相应于$u^{\delta, \varepsilon}$的二阶伴随方程(3.6)-(3.8)及随机过程$\Theta^{\delta}=X^{\delta}_1(\cdot)X^{\delta}_1(\cdot)$和$\Psi^{\delta}=Y^{\delta}_1(\cdot)Y^{\delta}_1(\cdot)$.这里$\Theta^{\delta}$, $\Psi^{\delta}$分别满足方程

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

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

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

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

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$\varepsilon\rightarrow0$时$(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})$, $(k^{\delta}, r^{\delta})$, $(p^{\delta}_2, q^{\delta}_2)$, $(p^{\delta}_3, q^{\delta}_3)$和$(p^{\delta}_1, q^{\delta}_1)$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$(3.12)\times \frac{1}{\varepsilon}$且$\varepsilon\rightarrow0$可得

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

类似地,存在$(\theta^{\delta}, \theta^{\delta}_T)$的子列收敛于$(\theta, \theta_T)$且满足$|\theta|^2+|\theta_T|^2=1$.由$(3.2)$式知,当$\delta\rightarrow0$时, $(Y^{\delta}_0, Z^\delta, u^\delta)\rightarrow(\bar{Y}_0, \bar{Z}, \bar{u})$.因此$(X^{\delta}, Y^\delta)\rightarrow(\bar{X}, \bar{Y})$, $\bar{Y}_T=\Phi(\bar{X}_T)$且$(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)$, $(k, r)$, $(p_2, q_2)$, $(p_3, q_3)$和$(p_1, q_1)$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})$在${{\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$(\theta^n, \theta^n_T)\rightarrow (\theta, \theta_T)$且$|\theta|^2+|\theta_T|^2=1.$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)$.

由$(Y_0, Z, u)$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$\theta$和$\theta_T$满足$\theta^2+\theta^2_T=1$,使得对任意的$Y_0\in \mathbb{R}$, $Z\in {{\cal H}}^2_{{\cal F}}(0, T, \mathbb{R})$及$u\in{\cal U}$,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: $\Phi=0$, $f=0$, $Z=0$, $\gamma=0$.令$\theta=1$, $\theta_T=0$,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.

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

(4.1) $\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} $

其中$B_i$, $\sigma_i$, $F_i$, $i\in\{1, \cdots , 6\}$和$\Phi_1$是实常数,且$L_1>0$, $L_2>0$, $H_1\geq0$和$\gamma_1>0$.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

(4.2) $ \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}$

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

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

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

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$Z=\bar{Z}$和$u=\bar{u}$,则有$\theta\gamma_1\bar{Y}_0+k_0=0.$即

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

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

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

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

(4.8) $ \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} $

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$(4.6)$和$(4.7)$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.

证  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*}$

对$p_t(X_t-\bar{X}_t)+k_t(Y_t-\bar{Y}_t)$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$(4.6)$和$(4.7)$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.