由于在经济和金融中的应用,平均场类型的随机最优控制问题近年来被广泛研究.与经典的随机最优控制问题不同,平均场类型的随机最优控制问题的性能指标函数的漂移和扩散系数不仅取决于状态和控制,而且还取决于状态控制部分的概率分布.平均场的存在使得最优控制问题变得时间不一致,此时动态规划原理(DPP)不再适用,这就推动了人们利用随机最大原理(SMP)对其进行求解.对于平均场类型的SMP更多的研究成果,可以参见文献[1-3, 6, 9, 12, 16-17, 22-23, 25-28]等.2013年,仅由布朗运动驱动的平均场随机微分方程(简称MF-SDE)的线性二次(简称LP)最优控制问题被Yong^[29]系统地研究,并通过引入两个唯一可解的黎卡提方程,得到了最优控制的反馈表示.在文献[29]以后,在相关论题及其应用方面取得了许多成果(例如文献[8, 10, 13-15, 18-21, 26]). 2016年, Tang和Meng^[24]将文献[29]中的结果推广到了跳跃扩散系统,获得相应的理论结果.直观地说,由一个MF-SDE所驱动的受控状态方程的对偶方程是一个平均场倒向随机微分方程(简称MF-BSDE).因此,从文献[4]和[5]建立了关于MF-BSDE的存在唯一性理论以来,平均场类型的随机最优控制问题的最大值原理已经成为了一个研究热点.由于一个MF-BSDE是定义良好的动态系统,在随机控制理论和数学金融学中具有重要应用,因此考虑MF-BSDE的最优控制问题是必要且自然的.2016年, Li, Sun和Xiong^[11]首次研究了仅由布朗运动驱动的MF-BSDE系统的LQ问题.在该研究中,他们通过一种变分法获得了最优化系统,并通过引入两个耦合的黎卡提方程和一个MF-BSDE来解耦最优化系统.最后,他们得到了最优控制的状态反馈表示.本文的目的是将文献[11]的LQ问题推广到平均场类型的倒向跳跃扩散系统,并建立相应的理论结果.更准确地说,第2节介绍本文要使用的基本符号,提出所要讨论的LQ问题.第3节,通过在标准假设下的经典凸变分原理,得到LQ问题最优控制的存在唯一性.第4节,首先建立了相应的随机哈密顿系统最优控制的对偶刻画.这里随机哈密顿系统是一带跳平均场类型的完全耦合的正向随机微分方程,其可解性非常困难.然后,引入两个黎卡提方程和一个带跳的MF-BSDE来解耦,并解出随机哈密尔顿系统.最后,对状态反馈控制的最优性给出了相应的验证.

设$T$是一个固定的严格正实数. $(\Omega, {\cal F}, \{{\cal F}_t\}_{0\leq t\leq T}, P)$是一个完备的概率空间,其上定义一个一维标准布朗运动$\{W(t), 0\leq t\leq T\}$.记${\cal P}$为$[0, T]\times \Omega$上的${\cal F}_t$ -可料$\sigma$ -代数, ${\cal B}(\Lambda)$为任何拓扑空间$\Lambda$的Borel $\sigma$ -代数.设$(E, {\cal B} (E), \nu)$是满足$\nu(E)<\infty $的可测空间, $\eta: \Omega\times D_\eta \longrightarrow E$是具有特征测度$\nu$的${\cal F}_t$ -适应的平稳Poisson点过程,其中$D_\eta$是$(0, \infty)$的可数子集.那么,由$\eta$诱导的计数测度为

$\mu((0, t]\times A):=\#\{s\in D_\eta; s\leq t, \eta(s)\in A\}, ~\mbox{对} t>0, A\in {\cal B} (E).$

$\tilde{\mu}({\rm d}e, {\rm d}t):=\mu({\rm d}e, {\rm d}t)-\nu({\rm d}e){\rm d}t$是独立于布朗运动$\{W(t), 0\leq t\leq T\}$的补偿Poisson随机鞅测度.假设$\{{\cal F}_t\}_{0\leq t\leq T}$是由$\{W(t), 0\leq t\leq T\}$和$\{\iint_{A\times (0, t] }\tilde{\mu}({\rm d}e, {\rm d}s), 0\leq t\leq T, A\in {\cal B} (E) \}$共同生成的$P$ -完备自然流.接下来,给出文章中需要使用的基本记号.

$\bullet$$s$: $s\in[t, T).$

$\bullet$$H$:范数为$\|\cdot\|_H$的Hilbert空间.

$\bullet$$\langle\alpha, \beta\rangle:$$\mathbb{R} ^n$空间上的内积. $ \forall \alpha, \beta\in\mathbb{R} ^n.$

$\bullet$$|\alpha|=\sqrt{\langle\alpha, \alpha\rangle}:$$\mathbb{R} ^n$空间上的范数, $\forall \alpha\in\mathbb{R} ^n.$

$\bullet$$\langle A, B\rangle={\rm tr}(AB^\top):$$\mathbb{R} ^{n\times m}$空间上的内积, $\forall A, B\in \mathbb{R} ^{n\times m}.$这里$B^\top$表示$B$的转置.

$\bullet$$|A|=\sqrt{{\rm tr}(AA^\top)}:$矩阵$A$的范数.

$\bullet$$ S^n:$$n\times n$对称矩阵的全体.

$\bullet$$ S^n_+:$$S^n$中非负定矩阵的全体.

$\bullet$$S_{{\cal F}}^2(t, T;H):$全体$H$ -值的右连左极${{\cal F}}_s$ -适应过程$f=\{f(s, \omega), \ (s, \omega)\in[t, T]\times\Omega\}$组成的空间,且满足

$\|f\|_{S_{{\cal F}}^2(t, T;H)}^2\triangleq{{\Bbb E}\bigg[\displaystyle\sup\limits_{t\leq s \leq T}\|f(s)\|_H^2}\bigg]<+\infty.$

$\bullet$$ M_{{\cal F}}^2(t, T;H):$全体$H$ -值的${{\cal F}}_s$ -适应过程$f=\{f(s, \omega), \(s, \omega)\in[0, T]\times\Omega\}$组成的空间,且满足

$$$\|f\|_{M_{{\cal F}}^2(t, T;H)}^2\triangleq {{\Bbb E}\displaystyle\bigg[\int_t^T\|f(s)\|_H^2{\rm d}t}\bigg]<\infty.$$$

$\bullet$${M}^{\nu, 2}( E; H):$定义在可测空间$(E, {\cal B}(E); \nu)$上的全体$H$ -值可测函数$r=\{r(e), e \in E\}$构成的空间,且满足

$\|r\|_{{ M}^{\nu, 2}( E; H)}^2\triangleq{\displaystyle\int_E\|r(e)\|_H^2{\nu}}({\rm d}e)<~\infty.$

$\bullet$${M}_{{\cal F}}^{\nu, 2}{([t, T]\times E; H)}:$全体${M}^{\nu, 2}( E; H)$ -值的${{\cal F}}_s$ -可料过程$r=\{r(s, \omega, e), \(s, \omega, e)\in[t, T]\times\Omega\times E\}$构成的空间,且满足

$\|r\|_{M_{{\cal F}}^{\nu, 2}([t, T]\times E; H)}^2\triangleq {{\Bbb E}\bigg[\int_t^T\displaystyle\|r(s, \cdot)\|^2_{{M}^{\nu, 2}( E; H)}{\rm d}s}\bigg]<~\infty.$

$\bullet$$L^2(\Omega, {{\cal F}}, P;H):$定义在概率空间$(\Omega, {{\cal F}}, P)$上的全体$H$ -值随机变量$\xi$构成的空间,且满足

$\|\xi\|_{L^2(\Omega, {{\cal F}}, P;H)}\triangleq {\Bbb E}[\|\xi\|_H^2]<\infty.$

考虑如下由布朗运动$\{W(s)\}_{0\leq s\leq T}$和泊松随机鞅测度$\{\tilde{\mu}({\rm d}e, {\rm d}s)\}_{0\leq s\leq T}$共同驱动的线性平均场倒向随机微分方程(MF-BSDE)

(2.1) $\begin{equation}\label{eq:1.1}\left\{\begin {array}{rl} {\rm d}Y(s)=&\bigg\{A(s)Y(s)+\bar A(s){\Bbb E} [Y(s)] +B(s)u(s)+\bar B(s){\Bbb E} [u(s)] \\& +C(s)Z(s)+\bar C(s){\Bbb E} [Z(s)]+\displaystyle\int_{E} D(s, e)R(s, e)\nu ({\rm d}e) \\[4mm] & +\displaystyle\int_{E}\bar D(s, e) {\Bbb E} [R(s, e)]\nu ({\rm d}e)\bigg\}{\rm d}s+ Z(s){\rm d}W(s) \\[4mm]& +\displaystyle \int_{E} R(s, e)\tilde{\mu}({\rm d}e, {\rm d}s), s\in [t, T], \\Y(T)=&\xi\end {array}\right.\end{equation}$

和二次性能指标

(2.2) $\begin{eqnarray}\label{eq:1.2} J(t, \xi; u(\cdot))&=&{\Bbb E}\bigg[\int_t^T\bigg(\langle Q(s)Y(s), Y(s)\rangle+ \langle \bar {Q}(s){\Bbb E}[Y(s)], {\Bbb E}[Y(s)]\rangle+\langle N_1(s)Z(s), Z(s)\rangle\\&&+\langle\bar N_1(s){\Bbb E}[Z(s)], {\Bbb E}[Z(s)]\rangle+ \int _{E}\langle N_2(s, e )R(s, e), R(s, e)\rangle\nu({\rm d}e)\\&&+ \int _{E}\langle \bar N_2(s, e ){\Bbb E}[R(s, e)], {\Bbb E}[R(s, e)]\rangle\nu({\rm d}e)+\langle N_3(s)u(s), u(s)\rangle\\&&+\langle\bar{N}_3(s){\Bbb E}[u(s)], {\Bbb E}[u(s)]\rangle\bigg){\rm d}s+\langle GY(t), Y(t)\rangle +\langle \bar{ G}{\Bbb E}[Y(t)], {\mathbbE}[Y(t)]\rangle \bigg], \end{eqnarray}$

其中$A(\cdot), \bar A(\cdot), $$ B(\cdot), $$\bar B(\cdot), $$C(\cdot), $$ \bar C(\cdot), $$D(\cdot, \cdot), $$\bar D(\cdot, \cdot), $$Q(\cdot), \bar Q(\cdot), N_1(\cdot), $$\bar N_1(\cdot), $$ N_2(\cdot, \cdot), $$\bar N_2(\cdot, \cdot), $$ N_3(\cdot), $$ \bar N_3(\cdot) $是给定的矩阵值函数; $G$和$\bar G$是给定的固定矩阵; $\xi$是一个${\cal F}_T$ -可测随机变量.在上述方程中, $u(\cdot)$是一可允许控制过程.一可料随机过程$u(\cdot)$被称为可允许控制过程,如果$ u(\cdot)\inM_{{\cal F}}^2(t, T;\mathbb{R} ^m)$.记可允许控制过程全体为${\cal A}[t, T]$.显然${\cal A}[t, T]$是一自反的Banach空间,其上的范数$|| \cdot |{|_{{\cal A }[t,T]}}$定义如下:

$||u(\cdot)||_{{\cal A}[t, T]}= \bigg\{{{\Bbb E}\displaystyle\bigg[\int_t^T|u(s)|^2{\rm d}s}\bigg]\bigg\}^{\frac{1}{2}}, \; \forall u(\cdot)\in {\cal A}[t, T].$

对任一可允许控制过程$u(\cdot)$,方程(2.1)相应的强解记作$(Y^{(t, \xi; u)}(\cdot), Z^{(t, \xi; u)}(\cdot), R^{(t, \xi; u)}(\cdot, \cdot))$或$(Y(\cdot), Z(\cdot), R(\cdot, \cdot))$ (如果它对可允许控制$u(\cdot)$的依赖可以从上下文看出),它被称为可允许控制过程$u(\cdot)$所对应的状态过程,且$(u(\cdot); Y(\cdot), Z(\cdot), R(\cdot))$被称为可允许四元组.

下面给出本文所要研究的带跳的平均场倒向随机线性二次最优控制问题.

问题 2.1  对于任意给定的$\xi\in L^2(\Omega, {{\cal F}_T}, P;\mathbb{R} ^n), $找到一可允许控制过程${u}^*(\cdot)\in {\cal A}[t, T]$,使得

(2.3) $\begin{equation}\label{eq:b7}J(t, \xi;{u}^*(\cdot))=\displaystyle\inf\limits_{u(\cdot)\in {{\cal A}[t, T]}}J(t, \xi; u(\cdot))\end{equation}$

成立.

任意满足上述等式的${u}^*(\cdot)\in {{\cal A}[t, T]}, $称为问题2.1的一最优控制过程,同时对应的状态过程$( Y^*(\cdot), Z^*(\cdot), R^*(\cdot, \cdot))$称为一最优状态过程.相应地, $(u^*(\cdot);Y^*(\cdot), Z^*(\cdot), R^*(\cdot, \cdot))$被称为问题2.1的一最优四元组.

接下来,给出本文的基本假设.

假设 2.1  矩阵值函数$A, \bar A, C, \bar C, Q, \bar Q, N_1, \bar N_1:[0, T]\rightarrow \mathbb{R} ^{n\times n};$$B, \bar B:[0, T]\rightarrow \mathbb{R} ^{n\times m}; $$D, \bar D:[0, T]\rightarrow M^{\nu , 2}(E; \mathbb{R} ^{ n\times n}), $$ N_2, \bar N_2:[0, T]\rightarrow M^{\nu, 2}(E; \mathbb{R} ^{n\times n}); N_3, \bar N_3:[0, T]\rightarrow \mathbb{R} ^{m\times m}$是一致有界的可测函数.

假设 2.2  $Q, Q+\bar Q, N_1, N_1+\bar N_1, N_2, N_2+\bar N_2, N_3, N_3+\bar N_3 $是几乎处处的非负定矩阵值函数,且$G, G+\bar G$是非负定矩阵.此外, $N_3$和$N_3+\bar N_3 $是几乎处处一致正定的,即对于$\forall u\in \mathbb{R} ^m$和几乎处处的$s\in [t, T]$, $ \langle N_3(s)u, u \rangle \geq \delta \langle u, u\rangle$和$ \langle (N_3(s)+\bar N_3(s))u, u \rangle \geq \delta \langle u, u\rangle$成立,其中$\delta$是某个正常数.

下面的结果给出了状态方程(2.1)的适定性,以及一些实用的估计.

引理 2.2  设假定2.1成立.则对于任意$(\xi, u(\cdot))\in L^2(\Omega, {{\cal F}_T}, P;\mathbb{R} ^n)\times {\cal A} [t, T]$,状态方程(2.1)存在唯一解$ \Lambda(\cdot):=( Y(\cdot), Z(\cdot), R(\cdot, \cdot))\in M^2[t, T]:=S_{{\cal F}}^2 ( t, T; \mathbb{R} ^n) \times M_{{\cal F}}^2 ( t, T; \mathbb{R} ^n) \times {M}_{{\cal F}}^{\nu, 2}{([t, T]\times E; \mathbb{R} ^n)} .$此外,以下估计式成立：

(2.4) $\begin{equation} \label{eq:1.4} ||\Lambda(\cdot)||^2_{M^2[t, T]}\leq K \bigg\{||u(\cdot)||_{{\cal A}[t, T]}^2+{\Bbb E}\big[|\xi|^2\big]\bigg\}\end{equation}$

和

(2.5) $\begin{equation} \label{eq:1.5} |J(t, \xi; u(\cdot))|< \infty.\end{equation}$

假设$ \bar \Lambda(\cdot):=(\bar Y(\cdot), \bar Z(\cdot), \bar R(\cdot, \cdot))$是对应于另一$(\bar\xi, \bar u(\cdot))\in L^2(\Omega, {{\cal F}}_T, P;\mathbb{R} ^n)\times {\cal A} [t, T]$的状态方程(2.1)的解,则有如下估计

(2.6) $\begin{equation} \label{eq:1.6} ||\Lambda(\cdot)-\bar\Lambda (\cdot)||^2_{M^2[t, T]}\leq K \bigg \{ ||u(\cdot)-\bar u(\cdot)||^2_{{\cal A}[t, T]}+{\Bbb E}\big[ |\xi-\bar \xi|^2\big]\bigg\}, \end{equation} $

这里定义

$||\Lambda(\cdot)||^2_{M^2[t, T]}=:{\Bbb E} \bigg[\sup\limits_{t\leq s\leq T}| Y(s)|^2\bigg] +{\Bbb E} \bigg[\int_t^ T| Z(s)|^2{\rm d}s\bigg] +{\Bbb E} \bigg[\int_t^ T\int_E| R(s, e)|^2 \nu({\rm d}e){\rm d}s\bigg], $

另外, $K$代表一个正常数.

证    方程解的存在唯一性可以由经典的方法即压缩映象原理直接得证.对于估计(2.4)和(2.6),可以通过对$|Y(\cdot)|^2$和$|Y(\cdot)-\bar Y(\cdot)|^2 $应用Itô公式,再结合Gronwall不等式和B-D-G不等式得证.再由假设2.1和估计(2.4),有

(2.7) $\begin{eqnarray}\label{eq:1.7} |J(t, \xi; u(\cdot))|\leq K \bigg\{ ||\Lambda(\cdot)||_{M^2[t, T]}^2 +||u(\cdot)||_{{\cal A}[t, T]}^2\bigg\} \leq K \bigg\{||u(\cdot)||_{{\cal A}[t, T]}^2+ {\Bbb E}[|\xi|^2]\bigg\} < \infty, \end{eqnarray}$

其中,利用了如下的基本不等式:对于任意$\Phi \in {L^2(\Omega, {{\cal F}}, P;H)}, $

(2.8) $\begin{eqnarray} ||{\Bbb E} [\Phi]||^2_H\leq {\Bbb E}[||\Phi||_H^2]. \end{eqnarray}$

证明完毕.

因此,由引理2.2,可以断言问题2.1是定义良好的.

在这一节中,将研究问题2.1最优控制的存在唯一性.为此,首先给出性能指标(2.2)的一些基本性质.

引理 3.1  设假设2.1和2.2成立.则对于任意$(t, \xi)\in [0, T)\times L^2(\Omega, {{\cal F}}_T, P;\mathbb{R} ^n), $性能指标$J(t, \xi;u(\cdot))$在${\cal A}[t, T]$上是连续的.

证    设$(u (\cdot); \Lambda(\cdot))=(u (\cdot); Y(\cdot), Z(\cdot), R(\cdot, \cdot))$和$( \bar u (\cdot); \bar \Lambda(\cdot))=(\bar u (\cdot); \bar Y(\cdot), \bar Z(\cdot), \bar R(\cdot, \cdot))$是任意给定的两个可允许四元组.在假设2.1和2.2下,由性能指标函数$J(t, \xi; u(\cdot))$ (见(2.2)式)的定义,有

(3.1) $\begin{eqnarray}\label{eq:5.10}&&| J (t, \xi; u (\cdot)) - J (t, \xi; \bar u(\cdot) ) |^2 \nonumber\\&\leq& K \Big \{ ||\Lambda(\cdot)-\bar\Lambda (\cdot)||^2_{M^2[t, T]}+||u(\cdot)-\bar u(\cdot)||^2_{{\cal A}[t, T]}\Big \} \nonumber \\&& \times \Big\{ ||\Lambda(\cdot)||^2_{M^2[t, T]}+||u(\cdot)||^2_{{\cal A}[t, T]}+||\bar\Lambda (\cdot)||^2_{M^2[t, T]}+||\bar u(\cdot)||^2_{{\cal A}[t, T]}\Big \} .\end{eqnarray}$

利用估计(2.4)和(2.6)可得

(3.2) $\begin{eqnarray}| J (t, \xi; u (\cdot)) - J (t, \xi; \bar u (\cdot)) |^2 \nonumber&\leq& K \Big \{ ||u(\cdot)-\bar u(\cdot)||^2_{{\cal A}[t, T]}\Big \} \\ && \times \Big\{||u(\cdot)||^2_{{\cal A}[t, T]}+||\bar u(\cdot)||^2_{{\cal A}[t, T]}+{\Bbb E}[|\xi|^2]\Big \} , \end{eqnarray}$

则

(3.3) $ \begin{equation}J (t, \xi; u (\cdot)) - J (t, \xi; \bar u (\cdot)) \rightarrow 0 , \quad {\rm as} \quad u (\cdot) \rightarrow \bar u (\cdot) \quad {\rm in} \quad {{\cal A}[t, T]} . \end{equation} $

因此,性能指标$J(t, \xi;u(\cdot))$在${\cal A}[t, T]$上是连续的.证明完毕.

引理 3.2  设假设2.1和2.2成立.则对于任意给定的$(t, \xi)\in [0, T)\ \times L^2(\Omega, {{\cal F}}_T, P;\mathbb{R} ^n), $性能指标$J(t, \xi; u(\cdot))$在${\cal A}[0, T]$上是严格凸的.进一步,性能指标函数$J(t, \xi; u(\cdot))$在${\cal A}[0, T] $是强制的,即

$\displaystyle\lim\limits_ {\|u(\cdot)\|_{{\cal A}[t, T]}{\rightarrow\infty}}J(t, \xi; u(\cdot))=\infty.$

证    由于加权矩阵是非随机的,故容易验证

(3.4) $\begin{eqnarray}\label{eq:2.4} J(t, \xi; u(\cdot))&=& \displaystyle {\Bbb E}\bigg[\int_t^T\bigg(\langle Q(s)(Y(s)-{\Bbb E}[Y(s)]), Y(s)-{\Bbb E}[Y(s)]\rangle \\ &&+ \langle (Q+\bar {Q})(s){\Bbb E}[Y(s)], {\Bbb E}[Y(s)]\rangle+\langle N_1(s)(Z(s)-{\Bbb E}[Z(s)]), Z(s)-{\Bbb E} [Z(s)]\rangle\\ && +\langle (N_1(s)+\bar{N}_1(s)){\Bbb E}[Z(s)], {\Bbb E}[Z(s)]\rangle\\ &&+ \int_{E}\langle N_2(s, e )(R(s, e)-{\Bbb E}[R(s, e)]), R(s, e)-{\Bbb E} [R(s, e)]\rangle \nu({\rm d}e)\\ &&+ \int_{E}\langle (N_2(s, e)+\bar{N}_2(s, e)){\Bbb E}[R(s, e)], {\Bbb E}[R(s, e)]\rangle \nu({\rm d}e)\\ &&+\langle N_3(s)(u(s)-{\Bbb E}[u(s)]), u(s)-{\Bbb E} [u(s)]\rangle\\ && +\langle (N_3(s)+\bar{N}_3(s)){\Bbb E}[u(s)], {\Bbb E}[u(s)]\rangle\bigg){\rm d}s\bigg]\\&&+{\Bbb E}\Big[\langleG(Y(t)-{\Bbb E} [Y(t)]), Y(t)-{\Bbb E} [Y(t)]\rangle +\langle (G+\bar{ G}){\Bbb E}[Y(t)], { \mathbbE}[Y(t)]\rangle \Big].\end{eqnarray}$

因此由$N_1, N_1+\bar N_1, N_2, N_2+N_2, N_3, N_3+\bar N_3, Q, Q+\bar Q, G, G+\bar G $的非负定性,性能指标$J(t, \xi; u(\cdot))$在${\cal A}[t, T]$上是凸的.再由$N_3$和$N_3+\bar N_3$一致正定性, $J(t, \xi; u(\cdot))$在${\cal A}[t, T]$上是严格凸的.另一方面,由假设2.2和(3.4)式,可得

(3.5) $\begin{eqnarray}\label{eq:2.5} J(t, \xi; u(\cdot)) &\geq& {\Bbb E}\bigg[\int_t^T \bigg (\langle N_3(s)(u(s)-{\Bbb E}[u(s)]), u(s)-{\Bbb E}[u(s)]\rangle\\ && +\langle (N_3(s)+{\bar N}_3(s)){\Bbb E}[u(s)], {\Bbb E}[u(s)]\rangle \bigg ){\rm d}s\bigg]\\& \geq &\delta {\Bbb E}\bigg[\int_t^T\langle u(s)-{\Bbb E}[u(s)], u(s)-{\Bbb E}[u(s)]\rangle {\rm d}s\bigg]+ \delta{\Bbb E}\bigg[\int_t^T \langle {\Bbb E}[u(s)], {\Bbb E}[u(s)]\rangle {\rm d}s\bigg] \\ &=& \delta{\Bbb E} \bigg[\int_t^T |u(s)|^2{\rm d}s\bigg]=\delta ||u(\cdot)||^2_{{\cal A}[t, T]}, \end{eqnarray}$

这意味着$\displaystyle\lim_ {\|u(\cdot)\|_{{\cal A}[t, T]}{\rightarrow \infty}}J(t, \xi; u(\cdot))=\infty.$证明完毕.

引理 3.3  设假设2.1和2.2成立.则对于任意给定的$(t, \xi)\in [0, T)\ \times L^2(\Omega, {{\cal F}}_T, P;\mathbb{R} ^n), $性能指标$J(t, \xi; u(\cdot))$在${\cal A}[t, T]$上是Frèchet可微的,且对应的Frèchet导数$J'(t, \xi; u(\cdot))$由以下给出:

(3.6) $\begin{eqnarray}\label{eq:2.6}&& \langle J'(t, \xi; u(\cdot)), v(\cdot) \rangle\\ &=&2{\Bbb E}\bigg[\int_t^T\bigg(\langle Q(s)Y^{(t, \xi;u)}(s), Y^{(t, 0;v)}(s)\rangle +\langle \bar Q(s){\Bbb E} [Y^{(x, u)}(s)], {\Bbb E}[Y^{(t, 0;v)}(s)]\rangle\\&&+\langle N_1(s)Z^{(t, \xi;u)}(s), Z^{(t, 0;v)}(s)\rangle +\langle \bar N_1(s){\Bbb E} [Z^{(t, \xi;u)}(s)], {\Bbb E}[Z^{(t, 0;v)}(s)]\rangle \\ &&+\int_{E}\langle N_2(s, e)R^{(t, \xi;u)}(s, e), R^{(t, 0;v)}(s, e)\rangle \nu({\rm d}e)\\&&+\int_{E}\langle \bar N_2(s){\Bbb E} [R^{(t, \xi;u)}(s, e)], {\Bbb E}[R^{(t, 0;\xi)}(s)]\rangle \nu({\rm d}e)\\&&+\langle N_3(s)u(s), v(s)\rangle+\langle \bar N_3(s){\Bbb E}[u(s)], {\Bbb E} [v(s)]\rangle \bigg){\rm d}s\bigg]\\&&+2{\Bbb E}\Big[\langle G Y^{(t, \xi;u)}(t), Y^{(t, 0;v)}(t)\rangle+\langle\bar G{\Bbb E}[Y^{(t, \xi;u)}(t)], {\Bbb E}[Y^{(t, 0;v)}(t)]\rangle\Big] , \ \forall u(\cdot), v(\cdot)\in{\cal A}[t, T], \; \end{eqnarray}$

其中$(Y^{(t, 0;v)}(\cdot), Z^{(t, 0;v)}(\cdot), R^{(t, 0;v)}(\cdot, \cdot)) $是如下MF-BSDE的解：

(3.7) $\begin{equation}\label{eq:2.7}\left\{\begin {array}{rl} {\rm d}Y(s)=&\bigg\{A(s)Y(s)+\bar A(s){\Bbb E} [Y(s)] +B(s)v(s)+\bar B(s){\Bbb E} [v(s)]\\ &+C(s)Z(s)+\bar C(s){\Bbb E} [Z(s)]+\displaystyle\int_{E} D(s, e)R(s, e)\nu ({\rm d}e)\\[4mm]& +\displaystyle\int_{E}\bar D(s, e) {\Bbb E} [R(s, e)]\nu ({\rm d}e)\bigg\}{\rm d}s+ Z(s){\rm d}W(s) \\[4mm] &+\displaystyle \int_{E} R(s, e)\tilde{\mu}({\rm d}e, {\rm d}s), s\in [t, T], \\Y(T)=& 0.\end {array}\right.\end{equation}$

证    设$u(\cdot)$和$v(\cdot)$是两个任意给定的可允许控制过程.简单起见,记(3.6)式的左边为$\Delta^{u, v}.$由于状态方程(2.1)是线性的,通过解的唯一性,易得

(3.8) $\begin{eqnarray}\label{eq:2.8}&& Y^{(t, \xi; u+v)}(s)=Y^{(t, \xi; u)}(s)+Y^{(t, 0; v)}(s), \\ &&Z^{(t, \xi; u+v)}(s)=Z^{(t, \xi; u)}(s)+Z^{(t, 0; v)}(s), \\ &&R^{(t, \xi; u+v)}(s)=R^{(t, \xi; u)}(s)+R^{(t, 0; v)}(s), t\leq s\leq T.\end{eqnarray}$

因此,根据(3.8)式和性能指标函数$J(t, \xi; u(\cdot))$ (见(2.2)式)的定义,容易验证

(3.9) $J(t,\xi ;u( \cdot ) + v( \cdot )) - J(t,\xi ;u( \cdot )) = J(t,0;v( \cdot )) + {\Delta ^{u,v}}.$

另一方面,由估计(2.4)可得

(3.10) $\begin{equation} |J(t, 0;v(\cdot))| \leq K||v(\cdot) ||^2_{{\cal A}[t, T]}.\end{equation}$

因此

(3.11) $\begin{equation} \displaystyle\lim\limits_ {\|v(\cdot) \|_{{\cal A}[t, T]} {\rightarrow0}}\frac{|J(t, \xi; u(\cdot) +v(\cdot) )-J(t, \xi; u(\cdot) )-\Delta^{u, v}|}{||v(\cdot) ||_{{\cal A}[t, T]}}=\displaystyle\lim\limits_ {\|v(\cdot) \|_{{\cal A}[t, T]} {\rightarrow0}}\frac{|J(t, 0; v(\cdot)|}{||v(\cdot) ||_{{\cal A}[t, T]}}=0.\end{equation}$

这意味着$J(t, \xi; u(\cdot))$有Fréchet导数$\Delta^{u, v}.$证明完毕.

注 3.1  由于性能指标函数$J(t, \xi; u(\cdot))$是Fréchet可微的,于是其同样是Gâteaux可微的.此外, Gâteaux导数即为Fréchet导数$\langle J'(t, \xi; u(\cdot)), v(\cdot) \rangle.$事实上,由(3.9)式可得

(3.12) $\begin{eqnarray} \label{eq:2.10}\lim\limits_ {\varepsilon {\rightarrow0}}\frac{J(t, \xi; u(\cdot) +\varepsilon v(\cdot) )-J(t, \xi; u(\cdot) )}{\varepsilon}&=&\displaystyle\lim\limits_ {\varepsilon {\rightarrow0}}\frac{J(t, 0; \varepsilon v(\cdot))+\Delta^{u, \varepsilon v}}{\varepsilon }\\&=&\displaystyle\lim\limits_ {\varepsilon {\rightarrow0}}\frac{\varepsilon^2J(t, 0; v(\cdot))+\varepsilon\Delta^{u, v}}{\varepsilon}\\&=&\Delta^{u, v}\\&=&\langle J'(t, \xi; u(\cdot)), v(\cdot) \rangle.\end{eqnarray}$

下面给出并证明问题2.1最优控制的存在唯一性.

定理 3.4  设假设2.1和2.2成立.则问题2.1存在唯一的最优控制.

证    既然可允许控制集${\cal A}[t, T]= M^2_{{\cal F}}(t, T;\mathbb{R} ^m)$是一个自反的Banach空间,根据引理3.1-3.3,问题2.1的最优控制的存在唯一性可以由文献[7]的命题2.12直接获得.(即定义在自反的Banach空间上的强制的,严格凸且下半连续的泛函存在一个唯一的最小值点).证明完毕.

定理 3.5  设假设2.1和2.2成立.则一可允许控制$u(\cdot)\in {\cal A}[t, T]$是问题2.1最优控制的充要条件是对于任意可允许控制$v(\cdot) \in {\cal A}[t, T], $

(3.13) $\begin{equation}\label{eq:b16} \langle J'(t, \xi; u(\cdot) ), v(\cdot) \rangle = 0, \end{equation}$

即

(3.14) $ \begin{eqnarray}\label{eq:2.13} 0&=&2{\Bbb E}\bigg[\int_t^T\bigg(\langle Q(s)Y^{(t, \xi;u)}(s), Y^{(t, 0;v)}(s)\rangle +\langle \bar Q(s){\Bbb E} [Y^{(t, \xi;u)}(s)], {\Bbb E}[Y^{(t, 0;v)}(s)]\rangle\\ &&+\langle N_1(s)Z^{(t, \xi;u)}(s), Z^{(t, 0;v)}(s)\rangle +\langle \bar N_1(s){\Bbb E} [Z^{(t, \xi;u)}(s)], {\Bbb E}[Z^{(t, 0;v)}(s)]\rangle \\ &&+\int_{E}\langle N_2(s, e)R^{(t, \xi;u)}(s, e), R^{(t, 0;v)}(s, e)\rangle \nu({\rm d}e)\\&&+\int_{E}\langle \bar N_2(s){\Bbb E} [R^{(t, \xi;u)}(s, e)], {\Bbb E}[R^{(t, 0;\xi)}(s)]\rangle \nu({\rm d}e)\\&&+\langle N_3(s)u(s), v(s)\rangle+\langle \bar N_3(s){\Bbb E}[u(s)], {\Bbb E} [v(s)]\rangle \bigg){\rm d}s\bigg]\\&&+2{\Bbb E}\Big[\langle G Y^{(t, \xi;u)}(t), Y^{(t, 0;v)}(t)\rangle+\langle\bar G{\Bbb E}[Y^{(t, \xi;u)}(t)], {\Bbb E}[Y^{(t, 0;v)}(t)]\rangle\Big] , \ \forallu(\cdot), v(\cdot)\in{\cal A}[t, T].\\ &&\end{eqnarray} $

证    对于必要性,假设$u(\cdot)$是一个最优控制.由(3.12)式,对于任意可允许控制$v(\cdot)$和$0< \varepsilon \leq 1, $有

(3.15) $\begin{equation} \langle J'(t, \xi; u(\cdot)), v(\cdot) \rangle= \displaystyle\lim\limits_ {\varepsilon {\rightarrow0^{+}}}\frac{J(t, \xi; u(\cdot) +\varepsilon v(\cdot) )-J(t, \xi; u(\cdot) )}{\varepsilon}\geq 0\end{equation}$

和

(3.16) $\begin{equation} -\langle J'(t, \xi; u(\cdot)), v(\cdot)\rangle=\langle J'(t, \xi; u(\cdot)), -v(\cdot) \rangle= \displaystyle\lim\limits_ {\varepsilon {\rightarrow0^{+}}}\frac{J(t, \xi; u(\cdot) +\varepsilon (-v(\cdot)) )-J(t, \xi; u(\cdot) )}{\varepsilon}\geq 0.\end{equation}$

这意味着

(3.17) $ \langle J'(t, \xi; u(\cdot)), v(\cdot) \rangle = 0.$

对于充分性,设$u(\cdot)$是一个给定的可允许控制使得对于任意可允许控制$v(\cdot), $$ \langle J'(t, \xi; u(\cdot) ), $$ v(\cdot) \rangle = 0.$因为性能指标函数$J$在${\cal A}[t, T]$上是凸的,故

(3.18) $ J(t, \xi; v(\cdot))-J(t, \xi; u(\cdot)) \geq \langle J'(t, \xi; u(\cdot) ), v(\cdot) \rangle = 0, $

这意味着$u(\cdot)$是一个最优控制.证明完毕.

本节将利用随机哈密尔顿系统对问题2.1的最优控制进行对偶刻画.为简化的记号,在下文中,在不致混淆的情况下,我们将省略时间变量s.

定理 4.1  设假设2.1和2.2成立.则一可允许四元组$(u(\cdot); Y(\cdot), Z(\cdot), R(\cdot, \cdot))$是问题2.1的最优四元组的充分必要条件是

(4.1) $\begin{eqnarray} \label{eq:3.1000}N_3(s)u(s)+\bar N_3(s){\Bbb E}[u(s)]+B(s)^\top k({s-})+\bar B(s)^\top {\Bbb E} [k({s-})]=0, \quad {\rm a.s.}, \end{eqnarray}$

此处$k(\cdot) $是以下MF-SDE的唯一解

(4.2) $\begin {equation}\label{eq:3.2}\left\{\begin{array}{rl}{\rm d}k(s)=&-\Big[A(s)^\top k(s)+\bar A(s)^\top{\Bbb E} [k(s)]+Q(s)Y(s)+\bar Q(s){\Bbb E}[Y(s)]\Big]{\rm d}s\\&-\Big[C^\top(s)k(s)+\bar C(s)^\top{\Bbb E} [k(s)]+N_1(s)Z(s)+\bar N_1(s){\Bbb E}[Z(s)]\Big]{\rm d}W(s)\\&-\displaystyle\int_ E \Big[D(s, e)^\top k(s-)+\bar D(s, e)^\top{\Bbb E} [k(s-)]\\&+N_2(s, e)R(s, e)+\bar N_2(s, e){\Bbb E}[R(s, e)]\Big]\tilde{\mu}({\rm d}e, {\rm d}s), \\ k(t)=&-GY(t)-\bar G{\Bbb E}[Y(t)], s\in [t, T].\end{array} \right. \end {equation}$

证    设$u(\cdot)\in {\cal A}[t, T]$是一可允许控制过程,其相应的状态过程为$(Y(\cdot), Z(\cdot), R(\cdot, \cdot))$.因此对于任意可允许控制过程$v(\cdot) \in {{\cal A}[t, T]}, $由引理3.3,有

(4.3) $\begin{eqnarray}\label{eq:3.3}& &\langle J'(t, \xi; u(\cdot)), v(\cdot) \rangle \\ &=&2{\mathbbE}\bigg[\int_t^T\bigg(\langle QY^{(t, \xi;u)}, Y^{(t, 0;v)}\rangle +\langle \bar Q{\Bbb E} [Y^{(t, \xi;u)}], {\Bbb E}[Y^{(t, 0;v)}]\rangle +\langle N_1Z^{(t, \xi;u)}, Z^{(t, 0;v)}\rangle\\&&+\langle \bar N_1{\Bbb E} [Z^{(t, \xi;u)}], {\Bbb E}[Z^{(t, 0;v)}]\rangle +\int_{E}\langle N_2R^{(t, \xi;u)}, R^{(t, 0;v)}\rangle \nu({\rm d}e)\\&&+\int_{E}\langle \bar N_2{\Bbb E} [R^{(t, \xi;u)}], {\Bbb E}[R^{(t, 0;\xi)}]\rangle \nu({\rm d}e)+\langle N_3u, v\rangle+\langle \bar N_3{\Bbb E}[u], {\Bbb E} [v]\rangle \bigg){\rm d}s\bigg]\\&&+2{\Bbb E}\Big[\langle G Y^{(t, \xi;u)}(t), Y^{(t, 0;v)}(t)\rangle+\langle\bar G{\Bbb E}[Y^{(t, \xi;u)}(t)], {\Bbb E}[Y^{(t, 0;v)}(t)]\rangle\Big].\end{eqnarray}$

另一方面,由文献[23]可知方程(4.2)存在唯一的适应解$k(\cdot)\in S_{{\cal F}}^2 ( t, T; \mathbb{R} ^n).$对$\langle Y^{t, 0; v}(s), k(s)\rangle$应用Itô's公式并求期望可得

(4.4) $\begin{eqnarray}\label{eq:3.4}&&{\mathbbE}\bigg[\int_t^T\bigg(\langle QY^{(t, \xi;u)}, Y^{(t, 0;v)}\rangle +\langle \bar Q{\Bbb E} [Y^{(t, \xi; u)}], {\Bbb E}[Y^{(t, 0;v)}]\rangle \\&&+\langle N_1Z^{(t, \xi;u)}, Z^{(t, 0;v)}\rangle+\langle \bar N_1{\Bbb E} [Z^{(t, \xi;u)}], {\Bbb E}[Z^{(t, 0;v)}]\rangle \\ &&+\int_{E}\langle N_2R^{(t, \xi;u)}, R^{(t, 0;v)}\rangle \nu({\rm d}e)+\int_{E}\langle \bar N_2{\Bbb E} [R^{(t, \xi;u)}], {\Bbb E}[R^{(t, 0;\xi)}]\rangle \nu({\rm d}e) \bigg){\rm d}s\bigg] \\&&+{\Bbb E}\Big[\langle G Y^{(t, \xi;u)}(t), Y^{(t, 0;v)}(t)\rangle+\langle\bar G{\Bbb E}[Y^{(t, \xi;u)}(t)], {\Bbb E}[Y^{(t, 0;v)}(t)]\rangle\Big]\\& =&{\mathbbE}\bigg [\int_t^T\bigg\langle k, Bv+\bar B{\Bbb E} [v]\bigg\rangle {\rm d}s\bigg ]={ \mathbbE}\bigg[\int_t^T \bigg \langle B^\top k+\bar B^\top {\Bbb E} [k], v\bigg\rangle {\rm d}s\bigg].\end{eqnarray}$

将(4.4)式代入(4.3)式,

(4.5) $\begin{eqnarray}\label{eq:4.5} &&\langle J'(t, \xi;u(\cdot)), v(\cdot) \rangle\\&=&2{\Bbb E}\bigg[\int_t^T \bigg \langle N_3(s)u(s)+\bar N_3(s){\Bbb E}[u(s)] +B(s)^\top k({s-})+\bar B(s)^\top {\Bbb E} [k({s-})], v(s)\bigg\rangle {\rm d}s\bigg].\end{eqnarray}$

对于必要性,设$(u(\cdot); Y(\cdot), Z(\cdot), R(\cdot, \cdot))$是最优四元组.由定理3.5,有$ \langle J'( t, \xi; u(\cdot)), v(\cdot) \rangle=0.$这意味着

(4.6) $\begin{eqnarray} \label{eq:3.60} &N_3(s)u(s)+\bar N_3(s){\Bbb E}[u(s)]+B(s)^\top k({s-})+\bar B(s)^\top {\Bbb E} [k({s-})]=0, \quad {\rm {\rm a.e.}}. \end{eqnarray}$

对于充分性,设$(u(\cdot); Y(\cdot), Z(\cdot), R(\cdot, \cdot))$为一可允许四元组且满足(4.1)式.将(4.1)式代入(4.5)式可得$ \langle J'(t, \xi; u(\cdot)), v(\cdot) \rangle=0.$再由定理3.5可知$(u(\cdot); Y(\cdot), Z(\cdot), R(\cdot, \cdot))$是最优控制四元组.证明完毕.

最后引入所谓的随机哈密顿系统,它由状态方程(2.1),对偶方程(4.2)和对偶刻画(4.1)构成如下

(4.7) $\left\{ {\begin{array}{*{20}{c}}\begin{array}{l}{\rm{d}}Y(s) = \{ A(s)Y(s) + \bar A(s)\Bbb E[Y(s)] + B(s)u(s) + \bar B(s)\Bbb E[u(s)] + C(s)Z(s)\\ + \bar C(s)\Bbb E[Z(s)] - \int_E {D(s,e)R(s,e)\nu ({\rm{d}}e)} + \int_E {\bar D(s,e)\Bbb E[R(s,e)]\nu ({\rm{d}}e)} \} {\rm{d}}s\\ + Z(s){\rm{d}}W(s) + \int_E {R(t,e)\tilde \mu ({\rm{d}}e,{\rm{d}}s)} ,\end{array}\\\begin{array}{l}{\rm{d}}k(s) = - [A{(s)^{\top}}k(s) + \bar A{(s)^{\top}}\Bbb E[k(s)] + Q(s)Y(s) + \bar Q(s)\Bbb E[Y(s)]]{\rm{d}}s\\ - [C{(s)^{\top}}k(s) + \bar C{(s)^{\top}}\Bbb E[k(s)] + {N_1}(s)Z(s) + {{\bar N}_1}(s)\Bbb E[Z(s)]]{\rm{d}}W(s)\\ - \int_E {R\left[ {D{{(s,e)}^{\top}}k(s - ) + \bar D{{(s,e)}^{\top}}\Bbb E[k(s - )] + {N_2}(s,e)R(s,e)} \right.} \\ + {{\bar N}_2}(s,e)\Bbb E[R(s,e)]]\tilde \mu ({\rm{d}}e,{\rm{d}}s),\end{array}\end{array}} \right.$

$\left\{ {\begin{array}{*{20}{c}}{Y(T) = \xi ,k(t) = - GY(t) - \bar G\Bbb E[Y(t)],\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\\{{N_3}(s)u(s) + {{\bar N}_3}(s)\Bbb E[u(s)] + B{{(s)}^{\top}}k(t - ) + \bar B{{(s)}^{\top}}\Bbb E[k(t - )] = 0,\quad s \in [t,T].}\end{array}} \right.$

这是一个带跳跃完全耦合的正倒向随机微分方程(简称MF-FBSDE),它的解由$( u(\cdot), Y(\cdot), Z(\cdot), $$R(\cdot, \cdot), k(\cdot))$组成.

定理 4.2  设假设2.1和2.2成立.则随机哈密顿系统(4.8)存在唯一解$(u(\cdot), Y(\cdot), Z(\cdot), $$ R(\cdot, \cdot), k(\cdot))\in M_{{\cal F}}^2(t, T;\mathbb{R} ^m)\times S_{{\cal F}}^2(t, T;\mathbb{R} ^n)\times M_{{\cal F}}^2(t, T;\mathbb{R} ^n)\times{M}_{{\cal F}}^{\nu, 2}{([0, T]\times {E;}\mathbb{R} ^n)} \times S_{{\cal F}}^2(t, T;\mathbb{R} ^n).$而且, $u(\cdot)$是问题2.1唯一的最优控制, $(Y(\cdot), Z(\cdot), $$R(\cdot, \cdot))$是对应的最优状态过程.

证    由定理3.4可知,问题2.1存在唯一最优的四元组$(u(\cdot), Y(\cdot), Z(\cdot), $$ R(\cdot, \cdot)).$假设$k(\cdot)$是与最优{四元组}相对应的对偶方程(4.2)的唯一解.根据定理4.1的必要性,最优控制$u(\cdot)$有对偶刻画(4.1).因此, $(u(\cdot), Y(\cdot), Z(\cdot), R(\cdot, \cdot), k(\cdot))$组成随机哈密尔顿系统(4.8)的一个适应解.下面证明随机哈密尔顿系统(4.8)解的唯一性.如果随机哈密尔顿系统(4.8)有另一个适应解$(\bar u(\cdot), \bar Y(\cdot), \bar Z(\cdot), \bar R(\cdot, \cdot), \bar k(\cdot)).$根据定理4.1的充分性, $(\bar u(\cdot), \bar Y(\cdot), \bar Z(\cdot), \bar R(\cdot, \cdot))$必定是问题2.1的最优四元组.因此,通过最优控制的唯一性,必有$ u(\cdot)= \baru(\cdot).$此外,根据MF-SDE和MF-BSDE解的唯一性,必有$(\bar Y(\cdot), \bar Z(\cdot), \bar R(\cdot, \cdot), \bar k(\cdot))=(Y(\cdot), Z(\cdot), R(\cdot, \cdot), k(\cdot))$.因此,随机哈密顿系统(4.8)存在唯一适应解.证明完毕.

总之,随机哈密尔顿系统(4.8)完全刻画了问题2.1的最优控制.因此,求解问题2.1等价于求解随机哈密尔顿系统(4.8),并且唯一的最优控制可以通过(4.1)式给出.对(4.1)式的两边取期望可得

(4.8) $\begin{equation} \label{eq:3.8} N_3(s){\Bbb E}[u(s)]+\bar N_3(s){\Bbb E}[u(s)]+B(s)^\top{\Bbb E}[k({s-})]+\bar B(s)^\top {\Bbb E} [k({s-})]=0, \quad {\rm a.e.}, \end{equation}$

这意味着

(4.9) $\begin{equation} \label{eq:3.9} {\Bbb E}[u(s)] =-(N_3(s)+\bar N_3(s))^{-1}(B (s)+\bar B (s))^\top {\Bbb E} [k({s-})], \quad {\rm a.e.}. \end{equation}$

再由(4.1)式可以得到

(4.10) $\begin{equation}\label{eq:3.10}N_3(s)u(s)=-\bar N_3(s){\Bbb E}[u(s)]-B(s)^\top k({s-})-\bar B(s)^\top {\Bbb E} [k({s-})], \quad {\rm a.e.}. \end{equation}$

再将(4.9)式代入(4.10)式,有

(4.11) $\begin{eqnarray} \label{eq:3.6} u(s)&=&-N_3^{-1}(s)\Big [B (s)^\top k({s-})+\bar B (s)^\top {\Bbb E} [k({s-})]\\&&-\bar N_3(s)(N_3(s)+\bar N_3(s))^{-1}(B (s)+\bar B (s))^\top {\Bbb E} [k({s-})]\Big], \quad {\rm {\rm a.e.}}. \end{eqnarray}$

由定理4.2可知假设2.1和2.2下,随机哈密尔顿系统(4.8)存在唯一解$(u(\cdot), Y(\cdot), Z(\cdot), $$ R(\cdot, \cdot), k(\cdot)), $且$(u(\cdot); Y(\cdot), Z(\cdot), R(\cdot, \cdot))$是相应的最优四元组.注意到此时$u(\cdot)$可由$(Y(\cdot), Z(\cdot), $$ R(\cdot, \cdot))$和$k(\cdot)$对偶表示出来.现在想仅用$ k(\cdot)$对$u(\cdot)$进行刻画.为此,如文献[11],利用对FBSDEs的通常的解耦技术,这将推导出两个Riccati方程.接下给出详细的推导过程.设$(u(\cdot), Y(\cdot), Z(\cdot), R(\cdot, \cdot), k(\cdot))$是随机哈密顿系统(4.8)的唯一解.对随机哈密顿系统(4.8)两边求期望,可得$({\Bbb E}[u(\cdot)], {\Bbb E}[Y(\cdot)], {\Bbb E}[Z(\cdot)], {\Bbb E}[R(\cdot, \cdot)], {\Bbb E}[k(\cdot)])$满足如下的正倒向常微分方程

(4.12) $\begin{equation}\label{eq:4.1}\left\{\begin {array}{rl}{\rm d}E[k(s)]=&-\Big[(A^\top(s)+\bar A(s)^\top){\Bbb E} [k(s)]+(Q(s)+\bar Q(s)){\Bbb E}[Y(s)]\Big], \\{\rm d}{\Bbb E}[Y(s)]=&\Big[(A(s)+\bar A(s)){\Bbb E} [Y(s)] +(B(s)+\bar B(s)){\Bbb E} [u(s)]+(C(s)+\bar C(s)){\Bbb E} [Z(s)] \\ &+\displaystyle\int_{E} (D(t, e) +\bar D(s, e)) {\Bbb E} [R(s, e)]\nu ({\rm d}e)\Big]{\rm d}s, \\{\Bbb E}[Y(T)]=& {\Bbb E}[\xi], \; {\Bbb E}[k(t)]=-(G+\bar G){\Bbb E}[Y(t)], \\ 0=&(N_3(s)+\bar N_3(s)){\Bbb E}[u(s)]+(B(s)^\top+\bar B(s)^\top) {\Bbb E} [k({s-})].\end {array}\right.\end{equation}$

此外,容易检验$(u(\cdot)-{\Bbb E} [u(\cdot)], Y(\cdot)-{\Bbb E} [Y(\cdot)], Z(\cdot)-{\Bbb E} [Z(\cdot)], R(\cdot, \cdot)-{\Bbb E} [R(\cdot, \cdot)], k(\cdot)-{\Bbb E} [k(\cdot)])$满足如下正倒向随机微分方程

(4.13) $\begin{equation}\label{eq:3.13}\left\{\begin {array}{rl} {\rm d}Y-{\Bbb E}[Y]=&\bigg\{A(Y-{\Bbb E} [Y]) +B(u-{\Bbb E} [u])+C(Z-{\Bbb E} [Z]) \\ & +\displaystyle\int_{E} D(R-{\Bbb E} [R])\nu ({\rm d}e)\bigg\}{\rm d}s+ Z{\rm d}W+\displaystyle \int_{E} R\tilde{\mu}({\rm d}e, {\rm d}s), \\{\rm d}k-{\Bbb E}[k]=&-\Big[A^\top(k-{\Bbb E} [k])+Q(Y-{\Bbb E}[Y])\Big]{\rm d}s+\Big[C^\top(k-{\Bbb E} [k])\\&+(C^\top+\bar C^\top){\Bbb E} [k]+N_1(Z-{\Bbb E}[Z])+(N_1+\bar N_1){\Bbb E}[Z]\Big]{\rm d}W\\&-\displaystyle\int_ E \Big[D^\top(k-{\Bbb E} [k])+(D^\top+\bar D^\top){\Bbb E} [k]+N_2(R-{\Bbb E}[R])\\&+(N_2+\bar N_2){\Bbb E}[R]\Big]\tilde{\mu}({\rm d}e, {\rm d}s), \\k(t)-{\Bbb E}[k(t)]=&-G(Y(t)-{\Bbb E}[Y(t)]), \; Y(T)-{\Bbb E}[Y(T)]=\xi-{\Bbb E}[\xi], \\ 0=&N_3(u-{\Bbb E}[u])+B^\top(k- {\Bbb E} [k]). \end {array}\right.\end{equation}$

现在假设状态过程$Y(\cdot)$和对偶过程$k(\cdot)$有下列关系:

(4.14) $\begin{equation} Y(s)=P(s)(k(s)-{\Bbb E}[k(s)]) +\Pi(s){\mathbb E}[k(s)]+\varphi(s), \end{equation}$

其中$P(\cdot), \Pi(\cdot):[0, T]\longrightarrow \mathbb{R} ^{n\times n}$是完全连续的确定性函数,且$\varphi(\cdot)$满足如下带跳的BSDE

(4.15) $\begin{equation} {\rm d}\varphi(s)=\alpha(s){\rm d}s+\beta(s){\rm d}W(s) +\Phi(s, e){\rm d}\tilde \mu({\rm d}e, {\rm d}s), \varphi(T)=\xi. \end{equation}$

这里$\alpha$, $\beta$和$\Phi$是某些$\mathbb F$ -循序可测过程,因此,进一步得到下列关系

(4.16) $\begin{equation} \label{eq:4.50} {\Bbb E} [Y(s)]=\Pi(s){\mathbb E}[k(s)] +{\Bbb E}[\varphi(s)] \end{equation}$

和

(4.17) $ \begin{equation} \label{eq:4.6} Y(s)-{\Bbb E} [Y(s)]=P(s)(k(s)-{\Bbb E}[k(s)])+(\varphi(s)-{\Bbb E}[\varphi(s)]). \end{equation} $

定义

(4.18) $\begin{equation}{\rm d}\eta(s)=\gamma {\rm d}s+\beta(s){\rm d}W(s) +\Phi(s, e){\rm d}\tilde \mu({\rm d}e, {\rm d}s), \eta(T)=\xi-{\Bbb E}[\xi], \end{equation}$

其中

$\eta(s)= \varphi (s)-{\Bbb E}[\varphi(s)], \; \gamma(s)= \alpha(s)-{\Bbb E}[\alpha(s)].$

下面将形式地推导$P(\cdot)$和$\Pi(\cdot)$应当满足的黎卡提方程.由关系式(4.17)和(4.13),对{$P(s)(k(s)-{\Bbb E}[k(s)])$}应用Itô公式可得

(4.19) $\begin{eqnarray} \label{eq:4.7}& &\bigg\{A(Y-{\Bbb E} [Y]) +B(u-{\Bbb E} [u])+C(Z-{\Bbb E} [Z]) +\displaystyle\int_{E} D(R-{\Bbb E} [R])\nu ({\rm d}e)\bigg\}{\rm d}s\\ &&+ Z{\rm d}W+\displaystyle \int_{E} R\tilde{\mu}({{\rm d}e}, {\rm d}s)\\&=&{\rm d}\big(Y-{\Bbb E}[Y])\\&=&{\rm d}P(k-{\Bbb E}[k])+{\rm d}\eta\\&=& \Big[\dot{P}(k-{\Bbb E}[k])-PA^\top(k-{\Bbb E} [k])-PQ(Y-{\Bbb E}[Y])\Big]{\rm d}s\\&&-P\Big[C^\top(k-{\Bbb E} [k])+(C^\top+\bar C^\top){\Bbb E} [k]+N_1(Z-{\Bbb E}[Z])+(N_1+\bar N_1){\Bbb E}[Z]\Big]{\rm d}W\\&&-\displaystyle\int_ E P \Big[D^\top(k-{\Bbb E} [k])+(D^\top+\bar D^\top){\Bbb E} [k]\\&&+N_2(R-{\Bbb E}[R])+(N_2+\bar N_2){\Bbb E}[R]\Big]\tilde{\mu}({\rm d}e, {\rm d}s)+\gamma{\rm d}s+\beta {\rm d}W +\Phi{\rm d}\tilde \mu({\rm d}e, {\rm d}s). \end{eqnarray}$

比较上述等式两边的扩散项和漂移项,有

(4.20) $\begin{equation} \label{eq:4.8} Z =-P\Big[C^\top k +\bar C ^\top{\Bbb E} [k ]+N_1 Z +\bar N_1 {\Bbb E}[Z ]\Big]+\beta , \end{equation}$

(4.21) $\begin{equation}\label{eq:4.9} R=-P \Big[D^\top k +\bar D^\top{\Bbb E} [k ]+N_2R+\bar N_2{\Bbb E}[R]\Big] +\Phi \end{equation}$

和

(4.22) $\begin{array}{l}AP(k - \Bbb E[k]) + A\eta - BN_3^{ - 1}{B^{\top}}(k - \Bbb E[k]) + C(Z - \Bbb E[Z]) + \int_E D (R - \Bbb E[R])\nu ({\rm{d}}e)\\ = \dot P(k - \Bbb E[k]) - P{A^{\top}}(k - \Bbb E[k]) - PQP(k - \Bbb E[k]) - PQ\eta + \gamma ,\end{array}$

这意味着

(4.23) $\begin{array}{l}(\dot P - P{A^{\top}} - AP - PQP + BN_3^{ - 1}{B^{\top}})(k - \Bbb E[k])\\ - C(Z - \Bbb E[Z]) - \int_E D (R - \Bbb E[R])\nu ({\rm{d}}e) - (A + PQ)\eta (t) + \gamma = 0.\end{array}$

接着对(4.20)和(4.21)式两边求期望,有下列关系:

(4.24) $0=-(PN_1 +P\bar N_1 +I){\Bbb E}[Z ]-P(C^\top +\bar C ^\top){\Bbb E} [k ]+{\Bbb E}[\beta ], $

(4.25) $0=-(PN_2+\bar PN_2+I){\Bbb E}[R-P(D^\top+\bar D^\top){\Bbb E} [k ]+{\Bbb E}[\Phi], $

(4.26) $0=-(PN_1(t)+I) (Z-{\Bbb E}[Z])-PC^\top (k-{\Bbb E} [k])+(\beta-{\Bbb E}[\beta]), $

(4.27) $0=-(PN_2+I) ( R-E[R])-PD^\top(k-{\Bbb E} [k])+ \Phi-{\Bbb E}[\Phi].$

假设$ PN_1+P\bar N_1+I, PN_2+\bar PN_2+I, PN_2+I$和$ PN_1+I$是可逆的,有

(4.28) ${\Bbb E}[Z]=-(PN_1+P\bar N_1+I)^{-1} \Big\{P(C^\top+\bar C^\top){\Bbb E} [k]-{\Bbb E}[\beta]\Big\}, $

(4.29) ${\Bbb E}[R]=-(PN_2+\bar PN_2 +I)^{-1}\Big\{P(D^\top +\bar D ^\top){\Bbb E} [k]-{\Bbb E}[\Phi ]\Big\}, $

(4.30) $(Z -{\Bbb E}[Z ])=-(PN_1 +I)^{-1}\Big\{PC^\top ) (k -{\Bbb E} [k ])-(\beta -{\Bbb E}[\beta ])\Big\}, $

(4.31) $( R - E[R ])=-(PN_2 +I)^{-1} \Big\{PD^\top (k -{\Bbb E} [k ]-(\Phi -{\Bbb E}[\Phi ])\Big\}.$

将(4.30)和(4.31)式代入(4.23)式可得

(4.32) $\begin{eqnarray} \label{eq:3.32}& &\bigg(\dot{P}-PA^\top-AP -PQP+BN_3^{-1}B^\top+ C(PN_1 +I)^{-1}PC^\top\\&&+\int_ED(PN_2 +I)^{-1}PD^\top\nu({\rm d}e)\bigg)(k-{\Bbb E}[k])-C(PN_1 +I)^{-1}(\beta -{\Bbb E}[\beta ])\\&& -\displaystyle\int_{E} D(PN_2 +I)^{-1} (\Phi -{\Bbb E}[\Phi ])\nu ({\rm d}e) -(A+PQ)\eta(t)+\gamma=0, \end{eqnarray}$

因此形式上有

(4.33) $\begin {equation}\label{eq:4.18}\left\{\begin{array}{ll} &\dot{P}-PA^\top-AP -PQP+BN_3^{-1}B^\top+ C(PN_1 +I)^{-1}PC^\top\\& +\int_ED(PN_2 +I)^{-1}PD^\top\nu({\rm d}e)=0, \\[4mm]& \gamma-C(PN_1 +I)^{-1}(\beta -{\Bbb E}[\beta ]) \\ & -\displaystyle\int_{E} D(PN_2 +I)^{-1} (\Phi -{\Bbb E}[\Phi ])\nu ({\rm d}e) -(A+PQ)\eta =0. \end{array} \right. \end {equation}$

进一步由(4.16)和(4.12)式可得

(4.34) $\begin{eqnarray}&&\bigg\{(A +\bar A ){\Bbb E} [Y ] -(B +\bar B (N +\bar N )^{-1}(B +\bar B )^\top {\Bbb E} [k]+(C +\bar C ){\Bbb E} [Z ]\\&& +\displaystyle\int_{E} (D +\bar D) {\Bbb E} [R]\nu ({\rm d}e)\bigg\}{\rm d}s \\ &=&{\rm d}{\Bbb E} [Y ] \\ &=&{\rm d}\Pi {\mathbb E}[k ] +{\rm d}{\Bbb E}[\varphi ] \\ &=&\Big[\dot{\Pi}{\Bbb E}[k ]-\Pi(A^\top +\bar A ^\top){\Bbb E} [k ]-\Pi(Q +\bar Q ){\Bbb E}[Y ]+{\Bbb E}[\alpha ]\Big]{\rm d}s. \end{eqnarray}$

将(4.16), (4.28)和(4.29)式代入等式左边并比较上述等式两端可得

(4.35) $\begin{eqnarray} \label{eq:4.23} 0&=&\bigg\{\dot{\Pi} -\Pi(A^\top +\bar A ^\top)-(A +\bar A )\Pi -\Pi(Q +\bar Q )\Pi+(B +\bar B )(N_3 +\bar N_3 )^{-1}(B +\bar B )^\top \\ &&+(C +\bar C )(PN_1+P\bar N_1+I)^{-1} P(C^\top +\bar C ^\top) \\ &&+\displaystyle\int_{E} (D+\bar D ) (PN_2+\bar PN_2 +I)^{-1}P(D^\top +\bar D ^\top)\bigg\}k \\ &&+\Big [-\Pi(Q +\bar Q )-(A +\bar A )\Big ]{\Bbb E}[\varphi ]+{\Bbb E}[\alpha ]-(C +\bar C )(PN_1 +P\bar N_1 +I)^{-1} {\Bbb E}[\beta ] \\ &&-\displaystyle\int_{E} (D +\bar D) (PN_2+P\bar N_2+I)^{-1}{\Bbb E}[\Phi]\nu ({\rm d}e). \end{eqnarray}$

因此,形式上有

(4.36) $\begin {equation}\label{eq:3.36}\left\{\begin{array}{rl} &\dot{\Pi} -\Pi(A^\top +\bar A ^\top)-(A +\bar A )\Pi -\Pi(Q +\bar Q )\Pi+(B +\bar B )(N_3 +\bar N_3 )^{-1}(B +\bar B )^\top \\&+(C +\bar C )(PN_1+P\bar N_1+I)^{-1} P(C^\top +\bar C ^\top)\\ &+\displaystyle\int_{E} (D+\bar D ) (PN_2+\bar PN_2 +I)^{-1}P(D^\top +\bar D ^\top)=0, \\& {\Bbb E}[\alpha ]-\big [\Pi(Q +\bar Q )+(A +\bar A )\big]{\Bbb E}[\varphi ]-(C +\bar C )(PN_1 +P\bar N_1 +I)^{-1} {\Bbb E}[\beta ]\\ & -\displaystyle\int_{E} (D +\bar D) (PN_2+P\bar N_2+I)^{-1}{\Bbb E}[\Phi]\nu ({\rm d}e)=0.\end{array} \right. \end {equation}$

此外,比较(4.16)和(4.17)两方程两端的终端值应有

$\Pi(T)=0, \; P(T)=0.$

所以,通过(4.33)和(4.36)式可知$P(\cdot)$和$\Pi(\cdot)$应当分别满足如下黎卡提方程

(4.37) $\begin {equation}\label{eq:4.37}\left\{\begin{array}{lll} &&\dot{P}-PA^\top-AP -PQP+BN_3^{-1}B^\top+ C(PN_1 +I)^{-1}PC^\top\\&&+\int_ED(PN_2 +I)^{-1}PD^\top\nu({\rm d}e)=0, \\&& P(T)=0, \end{array} \right. \end {equation}$

(4.38) $\begin {equation}\label{eq:4.38}\left\{\begin{array}{lll} &&\dot{\Pi} -\Pi(A^\top +\bar A ^\top)-(A +\bar A )\Pi -\Pi(Q +\bar Q )\Pi+(B +\bar B )(N_3 +\bar N_3 )^{-1}(B +\bar B )^\top \\&&+(C +\bar C )(PN_1+P\bar N_1+I)^{-1} P(C^\top +\bar C ^\top) \\ &&+\displaystyle\int_{E} (D+\bar D ) (PN_2+\bar PN_2 +I)^{-1}P(D^\top +\bar D ^\top)=0, \\ &&\Pi(T)=0, \end{array} \right. \end {equation}$

且$\varphi(t)$应当满足如下MF-BSDE:

(4.39) $\begin {equation}\label{eq:4.39}\left\{\begin{array}{rl}{\rm d}\varphi=&\bigg\{(PQ+A)\varphi+C(PN_1+I)^{-1}\beta\\&+\displaystyle\int_{E} D(PN_2+I)^{-1} \Phi\nu ({\rm d}e)+(\bar A+\Pi(Q+\bar Q)-PQ){\Bbb E}[\varphi]\\&+\big[(C+\bar C)(PN_1+P\bar N_1+I)^{-1}-C(PN_1+I)^{-1}\big]{\Bbb E}[\beta]\\&+\displaystyle\int_{E} \big[ (D +\bar D) (PN_2+\bar PN_2+I)^{-1}-D(PN_2+I)^{-1}\big]{\Bbb E}[\Phi]\nu ({\rm d}e)\bigg\}{\rm d}s\\&+\beta {\rm d}W +\Phi {\rm d}\tilde \mu({\rm d}e, {\rm d}s), \\\varphi(T)=& \xi.\end{array} \right. \end {equation}$

和文献[11]的第四部分一样,在假设2.1和2.2下,可以证明黎卡提方程(4.37)和(4.38)分别有唯一解.

本节将通过对黎卡提方程(4.37), (4.38)和MF-BSDE (4.39)的解,给出最优控制的解析公式.主要结果表述如下:

定理 4.3  设假设2.1和2.2成立.设$P(\cdot)$和$\Pi(\cdot)$分别是黎卡提方程(4.37)和(4.38)的唯一解,且$(\varphi(\cdot), \beta(\cdot), \Phi(\cdot))$是方程(4.39)的唯一适应解.则以下的MF-SDE存在一个唯一解$k(\cdot)$:

(4.40) $\begin {equation}\label{eq:3.40}\left\{\begin{array}{rl}{\rm d}k =&-\Big[(A^\top +Q P)k +(\bar A ^\top-Q P+(Q +\bar Q )\Pi ){\Bbb E} [k ]+Q \varphi+\bar Q {\Bbb E}[\varphi]\Big]{\rm d}s\\&- \Big[\big[I-N_1 (PN_1+I)^{-1}P\big]C^\top k+\big[\bar C ^\top+N_1 (PN_1+I)^{-1}PC^\top\\&-(N_1 +\bar N_1 )(PN_1+P\bar N_1+I)^{-1} P(C^\top+\bar C^\top)\big]{\Bbb E} [k ]\\&-N_1 (PN_1+I)^{-1} (\beta-{\Bbb E}[\beta])-(N_1 +\bar N_1 )(PN_1+P\bar N_1+I)^{-1} {\Bbb E}[\beta]\bigg ]{\rm d}W\\ &-\displaystyle\int_ E \Big[\big[I-N_2(PN_2+I)^{-1}P\big]D^\top k+\big[\bar D ^\top+2N_2 (PN_2+I)^{-1}PD^\top\\&-(N_2 +\bar N_2 )(PN_2+P\bar N_2+I)^{-1} P(D^\top+\bar D^\top)\big]{\Bbb E} [k ]\\&-N_2(PN_2+I)^{-1} (\Phi-{\Bbb E}[\Phi])\\ &-(N_2+\bar N_2)(PN_2+P\bar N_2+I)^{-1} {\Bbb E}[\Phi]\bigg\}\tilde{\mu}({\rm d}e, {\rm d}s), \; s\in [t, T], \\ k(t)=&-(I+GP)^{-1}G(\varphi(t)-{\Bbb E}[\varphi(t)])-(I+(G+\bar G)\Pi)^{-1}(G+\bar G){\Bbb E}[\varphi(t)], \end{array} \right. \end {equation}$

且对于终端状态$\xi$,问题2.1的最优控制具有如下显式表达式：

(4.41) $\begin{eqnarray} \label{eq:4.41} u =-N^{-1}_3 B^\top \big[k+ {\Bbb E} [k]\big] +N_3^{-1}\bar N_3(N_3 +\bar N_3 )^{-1}(B+\bar B )^\top {\Bbb E} [k]. \end{eqnarray}$

证    设$P(\cdot)$和$\Pi(\cdot)$分别是黎卡提方程(4.37)和(4.38)的唯一解,且$(\varphi(\cdot), \beta(\cdot), \Phi(\cdot))$是MF-BSDE (4.39)的唯一适应解.显然方程(4.40)有一个唯一解$k(\cdot)$.因此,只需证明由(4.41)式定义的$u(\cdot)$是问题2.1的唯一最优控制.为此,定义

(4.42) $Y = :P(k - \Bbb E[k]) + \Pi \Bbb E[k] + \varphi ,$

(4.43) $\begin{array}{l}Z = : - {(P{N_1} + I)^{ - 1}}[P{C^{\top}})(k - \Bbb E[k]) - (\beta - \Bbb E[\beta ])]\\ - {(P{N_1} + P{{\bar N}_1} + I)^{ - 1}}[P({C^{\top}} + {{\bar C}^{\top}})\Bbb E[k] - \Bbb E[\beta ]],\end{array} $

(4.44) $\begin{array}{l}R&=:&-(PN_2+I)^{-1} \bigg [PD^\top(k-{\Bbb E} [k])- (\Phi-{\Bbb E}[\Phi])\Big] \\ &&-(PN_2+ P\bar N_2+I)^{-1}\Big[P(D^\top+\bar D^\top){\Bbb E} [k]-{\Bbb E}[\Phi]\Big]. \end{array} $

对$Y(s)=P(s)(k(s)-{\Bbb E}[k(s)])+\Pi(s){{\mathbb E}}[k(s)]+\varphi(s)$应用Itô公式, $( Y(\cdot), Z(\cdot), R(\cdot, \cdot))$满足以下带跳的MF-BSDE

(4.45) $\begin{equation}\label{eq:4.45}\left\{\begin {array}{ll} {\rm d}Y(s)=&\bigg\{A(s)Y(s)+\bar A(s){\Bbb E} [Y(s)] +B(s)u(s)+\bar B(s){\Bbb E} [u(s)]\\ &+C(s)Z(s)+\bar C(s){\Bbb E} [Z(s)]+\displaystyle\int_{E} D(s, e)R(s, e)\nu ({\rm d}e)\\& +\displaystyle\int_{E}\bar D(s, e) {\Bbb E} [R(s, e)]\nu ({\rm d}e)\bigg\}{\rm d}t+ Z(s){\rm d}W(s) \\ &+\displaystyle \int_{E} R(s, e)\tilde{\mu}({\rm d}e, {\rm d}s), \; s\in [t, T], \\Y(T)=&\xi.\end {array}\right.\end{equation}$

此外,将(4.41)-(4.44)式代入(4.40)式并根据前面小节中Riccati等式的推导, $ k(\cdot)$满足以下带跳跃MF-SDE:

(4.46) $\begin {equation}\label{eq:4.46}\left\{\begin{array}{rl}{\rm d}k(s)=&-\Big[A(s)^\top k(s)+\bar A(s)^\top{\Bbb E} [k(s)]+Q(s)Y(s)+\bar Q(s){\Bbb E}[Y(s)]\Big]{\rm d}s\\&-\Big[C^\top(s)k(s)+\bar C(s)^\top{\Bbb E} [k(s)]+N_1(s)Z(s)+\bar N_1(s){\Bbb E}[Z(s)]\Big]{\rm d}W(s)\\&-\displaystyle\int_ E \Big[D(s, e)^\top k(s-)+\bar D(s, e)^\top{\Bbb E} [k(s-)]\\&+N_2(s, e)R(s, e)+\bar N_2(s, e){\Bbb E}[R(s, e)]\Big]\tilde{\mu}({\rm d}e, {\rm d}s), \\ k(t)=&-GY(t)-\bar G{\Bbb E}[Y(t)].\end{array} \right. \end {equation}$

由(4.41)式可得

(4.47) $\begin{eqnarray} \label{eq:4.47}N_3(s)u(s)+\bar N_3(s){\Bbb E}[u(s)]+B(s)^\top k({s-})+\bar B(s)^\top {\Bbb E} [k({s-})]=0, \quad {\rm {\rm a.s.}}. \end{eqnarray}$

因此,混合(4.45), (4.46)和(4.47)式,可得$( u(\cdot), Y(\cdot), Z(\cdot), R(\cdot, \cdot), k(\cdot))$满足以下随机哈密尔顿系统

(4.48) $\begin{equation}\left\{\begin {array}{ll}\label{eq:4.48} {\rm d}Y(s)=\bigg\{A(s)Y(s)+\bar A(s){\Bbb E} [Y(s)] +B(s)u(s)+\bar B(s){\Bbb E} [u(s)]\\\; \; +C(s)Z(s)+\bar C(s){\Bbb E} [Z(s)] -\displaystyle\int_{E} D(s, e)R(s, e)\nu ({\rm d}e)\\\; \; +\displaystyle\int_{E}\bar D(s, e) {\Bbb E} [R(s, e)]\nu ({\rm d}e)\bigg\}{\rm d}s + Z(s){\rm d}W(s) +\displaystyle \int_{E} R(t, e)\tilde{\mu}({\rm d}e, {\rm d}s), \nonumber\\{\rm d}k(s)=-\Big[A(s)^\top k(s)+\bar A(s)^\top{\Bbb E} [k(s)]+Q(s)Y(s)+\bar Q(s){\Bbb E}[Y(s)]\Big]{\rm d}s\nonumber\\\quad\quad\quad\quad-\Big[C(s)^\top k(s)+\bar C(s)^\top{\Bbb E} [k(s)]+N_1(s)Z(s)+\bar N_1(s){\Bbb E}[Z(s)]\Big]{\rm d}W(s)\nonumber\\\quad\quad\quad\quad-\displaystyle\int_ E \Big[D(s, e)^\top k(s)+\bar D(s, e)^\top{\Bbb E} [k(s)]\\\; \; +N_2(s, e)R(s, e)+\bar N_2(s, e){\Bbb E}[R(s, e)]\Big]\tilde{\mu}({\rm d}e, {\rm d}s), \\ Y(T)=\xi, k(t)=-GY(t)-\bar G{\Bbb E}[Y(t)], \nonumber \\ N_3(s)u(s)+\bar N_3(s){\Bbb E}[u(s)]+B(s)^\top k({s-})+\bar B(s)^\top {\Bbb E} [k({s-})]=0, \quad s\in [t, T]. \end {array}\right.\end{equation}$

因此,根据定理4.2得出由(4.41)式定义的$ u(\cdot)$是问题2.1的唯一最优控制.证明完毕.

本文发展了具有跳跃和确定性系数的平均场倒向随机微分方程的线性二次最优控制理论.其中最优控制过程可由对偶过程和两个黎卡提方程的解进行反馈表示.在后续的研究中,将对无限时区的这类线性二次问题进行探讨和研究.