(x,u,v)-依赖噪声的离散时间平均场随机H2/H∞控制 Discrete-Time Mean-Field Stochastic H2/H∞ Control with (x,u,v)-Dependent Noise

doi:10.12677/AAM.2023.129380

Advances in Applied Mathematics
Vol. 12 No. 09 ( 2023 ), Article ID: 72030 , 12 pages
10.12677/AAM.2023.129380

$(x, u, v)$ -依赖噪声的离散时间平均场随机H₂/H_∞控制

贾亚琪，张春梅，杨路，陈柏江

●How to Cite this Article

重庆理工大学理学院，重庆

收稿日期：2023年8月9日；录用日期：2023年9月3日；发布日期：2023年9月8日

摘要

本文研究了一类 $(x, u, v)$ (状态，输入，扰动)-依赖噪声的离散时间平均场随机系统的有限时域混合H₂/H_∞控制问题。首先，给出了已有的平均场随机有界实引理(SBRL)；其次，呈现了离散时间平均场随机线性二次(LQ)最优控制问题可解的充分条件。最后，基于SBRL和LQ最优控制的结果，通过耦合矩阵差分方程的可解性得到了离散时间平均场随机H₂/H_∞控制存在的充分必要条件，延伸了之前的结果。

关键词

离散时间系统，H₂/H_∞控制，平均场

Discrete-Time Mean-Field Stochastic H₂/H_∞ Control with $(x, u, v)$ -Dependent Noise

Yaqi Jia, Chunmei Zhang, Lu Yang, Bojiang Chen

College of Science, Chongqing University of Technology, Chongqing

Received: Aug. 9^th, 2023; accepted: Sep. 3^rd, 2023; published: Sep. 8^th, 2023

ABSTRACT

This paper is concerned with the finite horizon mixed H₂/H_∞ control problem for a class of discrete-time mean-field stochastic systems with $(x, u, v)$ -dependent noise. Firstly, the existing mean-field stochastic bounded real lemma (SBRL) is given. Then, we presented a sufficient condition for the solvability of discrete-time mean-field stochastic linear quadratic (LQ) optimal control problem for the considered system. Finally, based on the results of SBRL and LQ optimal control, a necessary and sufficient condition is obtained for the existence of discrete-time mean-field stochastic H₂/H_∞ control via the coupled matrix difference equations, which extends the previous results.

Keywords:Discrete-Time Systems, H₂/H_∞ Control, Mean-Field

This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).

http://creativecommons.org/licenses/by/4.0/

1. 引言

在近二十年中，混合H₂/H_∞控制是最受欢迎的鲁棒控制设计方法之一，已经吸引了许多学者的广泛关注。例如，关于确定性系统混合H₂/H_∞控制的讨论，见 [1] [2] [3] 。与单纯的H₂控制或鲁棒H_∞控制相比较，混合H₂/H_∞控制不仅能有效地消除外部干扰的影响，而且还能在最坏的外部干扰发生时，使得代价函数达到最小。因此，混合H₂/H_∞控制器能更好地保证系统良好运行，在工程实践中H₂/H_∞控制比单独的H_∞控制使用地更为广泛。

H₂/H_∞控制是将H₂性能设计与H_∞性能设计相结合，使整个系统既可以获得优良的调节性能，又可以保持鲁棒稳定性。利用Nash对策理论，文献 [4] 中研究了具有乘性噪声的连续时间线性随机系统的混合H₂/H_∞控制问题，并且给出了耦合代数Riccati方程的解存在的充分必要条件。更多关于随机系统的混合H₂/H_∞控制的研究请参考文献 [5] [6] [7] [8] [9] 以及其中的参考文献。

虽然有很多关于随机系统混合H₂/H_∞控制问题的研究 [1] - [9] ，但关于平均场随机系统的混合H₂/H_∞控制的结果还很少。在 [10] [11] 中分别对 $(x, v)$ -依赖噪声的连续时间平均场系统和离散时间平均场系统的H₂/H_∞控制问题进行了研究。本文在文献 [11] 的基础上将对应的结果推广到 $(x, u, v)$ -(状态、输入和扰动)均依赖噪声的离散时间平均场系统，它是随机混合H₂/H_∞控制和平均场理论的结合。与经典的随机系统相比，新的特点是系统不仅涉及状态和输入和扰动，而且还涉及它们的期望。平均场理论是为了研究各种物理和社会学动态系统中个体的相互作用所产生的集体行为而发展起来的。在过去的几年中，平均场理论已经被广泛地应用于各个领域，如工程、金融、经济和博弈论等，数学界和控制界对平均场控制理论的兴趣越来越大。特别地，研究了平均场随机系统的有限时域和无限时域线性二次(LQ)最优控制，请参考 [12] [13] [14] [15] 和其中的参考文献。在 [16] 中研究了带有Markov跳跃参数的平均场随机系统的LQ问题。文献 [17] 研究了连续时间平均场随机微分方程的H_∞控制，相应的离散时间H_∞控制在 [18] 中被讨论。

本文的其余组织如下：第2节构造了系统模型并且给出了有用的引理。第3节提供了离散时间平均场随机LQ最优控制问题可解的一个充分条件。第4节呈现了平均场混合H₂/H_∞存在的充分必要条件。最后，在第5节中做出一个简短的结论。

2. 基础知识与模型描述

2.1. 符号说明

$M^{'}$ 是一个矩阵M或者向量M的转置。 $M > 0 (\geq 0)$ 意味着M是一个正定(半正定)的对称矩阵。 $ℝ^{m}$ 是m维的实向量空间赋予通常的内积。 $ℝ^{m \times n}$ 代表所有的 $m \times n$ 维的实矩阵空间。 $N_{T} = {0, 1, \dots, T}$ 。 $N = {0, 1, \dots}$ 。 $S_{n} (ℝ)$ 是所有实对称矩阵 $ℝ^{n \times n}$ 的集合。

2.2. 模型介绍

考虑如下的离散时间平均场随机系统：

${\begin{cases} x (t + 1) = A (t) x (t) + \bar{A} (t) E x (t) + B (t) v (t) + \bar{B} (t) E v (t) \\ + [C (t) x (t) + \bar{C} (t) E x (t) + D (t) v (t) + \bar{D} (t) E v (t)] ω (t) \\ z (t) = ϕ (t) x (t) \\ x (0) = x_{0} \in ℝ^{n}, t \in N_{T} \end{cases}$ (1)

其中， $x (t) \in ℝ^{n}, v (t) \in ℝ^{l}, z (t) \in ℝ^{m}$ 分别是系统的状态、扰动和控制输出。 $A (t), \bar{A} (t), C (t), \bar{C} (t) \in ℝ^{n \times n}$ ， $B (t), \bar{B} (t), D (t), \bar{D} (t) \in ℝ^{n \times l}, ϕ (t) \in ℝ^{m \times n}$ 是给定的矩阵值函数。E是期望算子。 ${ω (t)}_{t \in N_{T}}$ 是一维的定义在完备概率空间 $(Ω, F, μ)$ 上的白噪声过程，有 $E [ω (t)] = 0, E [ω (t) ω (s)] = σ_{t s}, s, t \in N_{T}$ ，其中 $σ_{t s}$ 是克罗内克函数。假设 $ω (t)$ 和 $v (t)$ 不相关。定义由 $ω (\cdot)$ 生成的 $σ$ 代数 $F_{t} = σ {ω (t), t = 0, 1, \dots, T}$ 。令 $L^{2} (Ω, ℝ^{p})$ 是 $ℝ^{p}$ 值随机向量 $ξ$ 且 $E {‖ ξ ‖}^{2} < \infty$ 的空间。 $l_{ω}^{2} (N_{T}; ℝ^{p})$ 表示所有序列 ${y (t)}_{t \in N_{T}}$ 使得 $y (t) \in L^{2} (Ω, ℝ^{m})$ 是 $F_{t - 1}$ 可测的且 $l^{2}$ -范数满足 ${‖ y (t) ‖}_{l_{ω}^{2} (N_{T}; ℝ^{p})} = {(\sum_{t = 0}^{T} E {‖ y (t) ‖}^{2})}^{\frac{1}{2}} < \infty$ 的空间。为了简便，假定 $x_{0}$ 是确定的。对任何 $T \in N$ 和

$(x_{0}, v (t)) \in ℝ^{n} \times l_{ω}^{2} (N_{T}; ℝ^{l})$ ，(1)的唯一解表示为 $x (t) = x (t; x_{0}, v)$ ，初值 $x (0) = x_{0}$ 。

定义1 系统(1)的扰动算子定义为

$\begin{array}{l} L_{T} : l_{ω}^{2} (N_{T}; ℝ^{l}) \to l_{ω}^{2} (N_{T}; ℝ^{m}) \\ L_{T} (v (t)) : = ϕ (t) x (t; 0, v), \forall v (t) \in l_{ω}^{2} (N_{T}; ℝ^{l}) \end{array}$

它的范数由下定义：

$‖ L_{T} ‖ = \sup_{v \in l_{ω}^{2} (N_{T}; ℝ^{l}), v \neq 0, x_{0} \equiv 0} \frac{{‖ z (t) ‖}_{l_{ω}^{2} (N_{T}; ℝ^{m})}}{{‖ v (t) ‖}_{l_{ω}^{2} (N_{T}; ℝ^{l})}} = \sup_{v \in l_{ω}^{2} (N_{T}; ℝ^{l}), v \neq 0, x_{0} \equiv 0} \frac{{(\sum_{t = 0}^{T} E {‖ ϕ (t) x (t) ‖}^{2})}^{\frac{1}{2}}}{{(\sum_{t = 0}^{T} E {‖ v (t) ‖}^{2})}^{\frac{1}{2}}}$

在系统(1)中取期望得到

$E x (t + 1) = A (t) E x (t) + B (t) E v ( t )$

$\begin{matrix} x (t + 1) - E x (t + 1) = A (t) (x (t) - E x (t)) + B (t) (v (t) - E v (t)) + [C (t) (x (t) - E x (t)) \\ + ℂ (t) E x (t) + D (t) (v (t) - E v (t)) + D (t) E v (t)] ω ( t ) \end{matrix}$

其中

$\begin{array}{l} A (t) = A (t) + \bar{A} (t) B (t) = B (t) + \bar{B} (t) \\ ℂ (t) = C (t) + \bar{C} (t) D (t) = D (t) + \bar{D} ( t ) \end{array}$

引理2 [11] 在系统(1)中，假定 $T \in N$ 给定， $P (0), P (1), \dots, P (T + 1); Q (0), Q (1), \dots, Q (T + 1)$ 是 $S_{n} (ℝ)$ 中的任意矩阵族，则对任何 $x_{0} \in ℝ^{n}$ ， $v (t) \in l_{ω}^{2} (N_{T}; ℝ^{l})$ 有

$\begin{matrix} J^{T} (x_{0}, v) : = \sum_{t = 0}^{T} E [γ^{2} {‖ v (t) ‖}^{2} - {‖ z (t) ‖}^{2}] \\ = \sum_{t = 0}^{T} E {{(\begin{array}{l} x (t) - E x (t) \\ v (t) - E v (t) \end{array})}^{'} M (P) (\begin{array}{l} x (t) - E x (t) \\ v (t) - E v (t) \end{array})} + \sum_{t = 0}^{T} {(\begin{array}{l} E x (t) \\ E v (t) \end{array})}^{'} S (P, Q) (\begin{array}{l} E x (t) \\ E x (t) \end{array}) \\ - E {{[x (T + 1) - E x (T + 1)]}^{'} P (T + 1) [x (T + 1) - E x (T + 1)]} \\ - {(E x (T + 1))}^{'} Q (T + 1) E x (T + 1) + {x^{'}}_{0} Q (0) x_{0} \end{matrix}$ (2)

其中

$\begin{array}{l} M (P) = (\begin{matrix} A {(t)}^{'} P (t + 1) A (t) + C {(t)}^{'} P (t + 1) C (t) - ϕ {(t)}^{'} ϕ (t) - P (t) & A {(t)}^{'} P (t + 1) B (t) + C {(t)}^{'} P (t + 1) D (t) \\ B {(t)}^{'} P (t + 1) A (t) + D {(t)}^{'} P (t + 1) C (t) & γ^{2} I + B {(t)}^{'} P (t + 1) B (t) + D {(t)}^{'} P (t + 1) D (t) \end{matrix}) \\ S (P, Q) = (\begin{matrix} A {(t)}^{'} Q (t + 1) A (t) + ℂ {(t)}^{'} P (t + 1) ℂ (t) - ϕ {(t)}^{'} ϕ (t) - Q (t) & A {(t)}^{'} Q (t + 1) B (t) + ℂ {(t)}^{'} P (t + 1) D (t) \\ B {(t)}^{'} Q (t + 1) A (t) + D {(t)}^{'} P (t + 1) ℂ (t) & γ^{2} I + B {(t)}^{'} Q (t + 1) B (t) + D {(t)}^{'} P (t + 1) D ( t ) \end{matrix}) \end{array}$

为了方便，引入下面记号：

${\begin{cases} L (P (t + 1)) = A {(t)}^{'} P (t + 1) A (t) + C {(t)}^{'} P (t + 1) C (t) - ϕ {(t)}^{'} ϕ (t) \\ G (P (t + 1)) = A {(t)}^{'} P (t + 1) B (t) + C {(t)}^{'} P (t + 1) D (t) \\ H (P (t + 1)) = γ^{2} I + B {(t)}^{'} P (t + 1) B (t) + D {(t)}^{'} P (t + 1) D (t) \\ \hat{L} (P (t + 1), Q (t + 1)) = A {(t)}^{'} Q (t + 1) A (t) + ℂ {(t)}^{'} P (t + 1) ℂ (t) - ϕ {(t)}^{'} ϕ (t) \\ \hat{G} (P (t + 1), Q (t + 1)) = A {(t)}^{'} Q (t + 1) B (t) + ℂ {(t)}^{'} P (t + 1) D (t) \\ \hat{H} (P (t + 1), Q (t + 1)) = γ^{2} I + B {(t)}^{'} Q (t + 1) B (t) + D {(t)}^{'} P (t + 1) D ( t ) \end{cases}$

$K (t) = K (t) + \bar{K} (t)$ ， $W (t) = W (t) + \bar{W} ( t )$

引理3 [19] (SBRL)对平均场随机系统(1)， $‖ L_{T} ‖ < γ$ 对某个 $γ > 0$ 当且仅当下列的反向差分方程：

${\begin{cases} P (t) = L (P (t + 1)) - G (P (t + 1)) H {(P (t + 1))}^{- 1} G {(P (t + 1))}^{'} \\ Q (t) = \hat{L} (P (t + 1), Q (t + 1)) - \hat{G} (P (t + 1), Q (t + 1)) \hat{H} {(P (t + 1), Q (t + 1))}^{- 1} \hat{G} {(P (t + 1), Q (t + 1))}^{'} \\ P (T + 1) = Q (T + 1) = 0 \\ H (P (t + 1)) > 0, \hat{H} (P (t + 1), Q (t + 1)) > 0 \end{cases}$ (3)

有唯一解 $(P_{1} (t), Q_{1} (t))$ ，且 $Q_{1} (t) \leq 0$ 。

3. 平均场随机LQ控制

考虑下列的离散时间平均场随机离散系统

${\begin{cases} x (t + 1) = A (t) x (t) + \bar{A} (t) E x (t) + F (t) u (t) + \bar{F} (t) E u (t) \\ + [C (t) x (t) + \bar{C} (t) E x (t) + G (t) u (t) + \bar{G} (t) E u (t)] ω (t) \\ z (t) = (\begin{matrix} ϕ (t) x (t) \\ ψ (t) u (t) \end{matrix}) \\ x (0) = x_{0} \in ℝ^{n}, ψ {(t)}^{'} ψ (t) = I, t \in N_{T} \end{cases}$ (4)

其中 $u (t) \in l_{ω}^{2} (N_{T}; ℝ^{q})$ 是控制输入，有关的代价函数是 $J^{T} (x_{0}, u) = \sum_{t = 0}^{T} E [{‖ z (t) ‖}^{2}]$ 。

定理4 (LQ控制)对于系统(4)，存在 $u^{*} \in l_{ω}^{2} (N_{T}; ℝ^{q})$ 使得 $\min_{u \in l_{ω}^{2} (N_{T}; ℝ^{q})} J^{T} (x_{0}, u) = J^{T} (x_{0}, u^{*}) = {x^{'}}_{0} {\bar{Q}}_{1} (0) x_{0} \geq 0$ 且 ${\bar{Q}}_{1} (t) \geq 0$ ，若下列的反向差分方程：

${\begin{cases} {\bar{P}}_{1} (t) = L_{1} ({\bar{P}}_{1} (t + 1)) - G_{1} ({\bar{P}}_{1} (t + 1)) H_{1} {({\bar{P}}_{1} (t + 1))}^{- 1} G_{1} {({\bar{P}}_{1} (t + 1))}^{'} \\ {\bar{Q}}_{1} (t) = {\hat{L}}_{1} ({\bar{P}}_{1} (t + 1), {\bar{Q}}_{1} (t + 1)) - {\hat{G}}_{1} ({\bar{P}}_{1} (t + 1), {\bar{Q}}_{1} (t + 1)) {\hat{H}}_{1} {({\bar{P}}_{1} (t + 1), {\bar{Q}}_{1} (t + 1))}^{- 1} {\hat{G}}_{1} {({\bar{P}}_{1} (t + 1), Q_{1} (t + 1))}^{'} \\ {\bar{P}}_{1} (T + 1) = {\bar{Q}}_{1} (T + 1) = 0 \\ H_{1} ({\bar{P}}_{1} (t + 1)) > 0, {\hat{H}}_{1} ({\bar{P}}_{1} (t + 1), {\bar{Q}}_{1} (t + 1)) > 0 \end{cases}$ (5)

有唯一解 $({\bar{P}}_{1} (t), {\bar{Q}}_{1} (t)), t \in N_{T}$ ，其中

$\begin{array}{l} u^{*} (t) = - H_{1} {({\bar{P}}_{1} (t + 1))}^{- 1} G_{1} {({\bar{P}}_{1} (t + 1))}^{'} x (t) + [H_{1} {({\bar{P}}_{1} (t + 1))}^{- 1} G_{1} {({\bar{P}}_{1} (t + 1))}^{'} \\ - {\hat{H}}_{1} {({\bar{P}}_{1} (t + 1), {\bar{Q}}_{1} (t + 1))}^{- 1} {\hat{G}}_{1} {({\bar{P}}_{1} (t + 1), Q_{1} (t + 1))}^{'}] E x ( t ) \end{array}$

${\begin{cases} L_{1} ({\bar{P}}_{1} (t + 1)) = A {(t)}^{'} {\bar{P}}_{1} (t + 1) A (t) + C {(t)}^{'} {\bar{P}}_{1} (t + 1) C (t) + ϕ {(t)}^{'} ϕ (t) \\ G_{1} ({\bar{P}}_{1} (t + 1)) = A {(t)}^{'} {\bar{P}}_{1} (t + 1) F (t) + C {(t)}^{'} {\bar{P}}_{1} (t + 1) G (t) \\ H_{1} ({\bar{P}}_{1} (t + 1)) = I + F {(t)}^{'} {\bar{P}}_{1} (t + 1) F (t) + G {(t)}^{'} {\bar{P}}_{1} (t + 1) G (t) \\ {\hat{L}}_{1} ({\bar{P}}_{1} (t + 1), {\bar{Q}}_{1} (t + 1)) = A {(t)}^{'} {\bar{Q}}_{1} (t + 1) A (t) + ℂ {(t)}^{'} {\bar{P}}_{1} (t + 1) ℂ (t) + ϕ {(t)}^{'} ϕ (t) \\ {\hat{G}}_{1} ({\bar{P}}_{1} (t + 1), {\bar{Q}}_{1} (t + 1)) = A {(t)}^{'} {\bar{Q}}_{1} (t + 1) F (t) + ℂ {(t)}^{'} {\bar{P}}_{1} (t + 1) G (t) \\ {\hat{H}}_{1} ({\bar{P}}_{1} (t + 1), {\bar{Q}}_{1} (t + 1)) = I + F {(t)}^{'} {\bar{Q}}_{1} (t + 1) F (t) + G {(t)}^{'} {\bar{P}}_{1} (t + 1) G ( t ) \end{cases}$

$F (t) = F (t) + \bar{F} (t) G (t) = G (t) + \bar{G} ( t )$

证明：计算代价函数

$\begin{array}{l} J^{T} (x_{0}, u) = \sum_{t = 0}^{T} E [{‖ z (t) ‖}^{2}] \\ = \sum_{t = 0}^{T} E {{(\begin{array}{l} x (t) - E x (t) \\ u (t) - E u (t) \end{array})}^{'} (\begin{matrix} - {\bar{P}}_{1} (t) + L_{1} ({\bar{P}}_{1} (t + 1)) & G_{1} ({\bar{P}}_{1} (t + 1)) \\ G_{1} {({\bar{P}}_{1} (t + 1))}^{'} & H_{1} ({\bar{P}}_{1} (t + 1)) \end{matrix}) (\begin{array}{l} x (t) - E x (t) \\ u (t) - E u (t) \end{array})} \\ + \sum_{t = 0}^{T} {(\begin{array}{l} E x (t) \\ E u (t) \end{array})}^{'} (\begin{matrix} - {\bar{Q}}_{1} (t) + {\hat{L}}_{1} ({\bar{P}}_{1} (t + 1), {\bar{Q}}_{1} (t + 1)) & {\hat{G}}_{1} ({\bar{P}}_{1} (t + 1), {\bar{Q}}_{1} (t + 1)) \\ {\hat{G}}_{1} {({\bar{P}}_{1} (t + 1), {\bar{Q}}_{1} (t + 1))}^{'} & {\hat{H}}_{1} ({\bar{P}}_{1} (t + 1), {\bar{Q}}_{1} (t + 1)) \end{matrix}) (\begin{array}{l} E x (t) \\ E u (t) \end{array}) + {x^{'}}_{0} {\bar{Q}}_{1} (0) x_{0} \\ = \sum_{t = 0}^{T} E {{[(u (t) - E u (t)) - (u^{*} (t) - E u^{*} (t))]}^{'} H_{1} ({\bar{P}}_{1} (t + 1)) [(u (t) - E u (t)) - (u^{*} (t) - E u^{*} (t))]} \\ + \sum_{t = 0}^{T} {(E u (t) - E u^{*} (t))}^{'} {\hat{H}}_{1} ({\bar{P}}_{1} (t + 1), {\bar{Q}}_{1} (t + 1)) (E u (t) - E u^{*} (t)) + {x^{'}}_{0} {\bar{Q}}_{1} (0) x_{0} \end{array}$

其中

$\begin{array}{l} u^{*} (t) - E u^{*} (t) = - H_{1} {({\bar{P}}_{1} (t + 1))}^{- 1} G_{1} {({\bar{P}}_{1} (t + 1))}^{'} (x (t) - E x (t)) \\ E u^{*} (t) = - {\hat{H}}_{1} {({\bar{P}}_{1} (t + 1), {\bar{Q}}_{1} (t + 1))}^{- 1} {\hat{G}}_{1} {({\bar{P}}_{1} (t + 1), {\bar{Q}}_{1} (t + 1))}^{'} E x ( t ) \end{array}$

因此 $\min_{u \in l_{ω}^{2} (N_{T}; ℝ^{q})} J^{T} (x_{0}, u) = J^{T} (x_{0}, u^{*}) = {x^{'}}_{0} {\bar{Q}}_{1} (0) x_{0} \geq 0$ 。

4. 主要结果

考虑 $(x, u, v)$ -依赖噪声的离散时间平均场随机系统：

${\begin{cases} x (t + 1) = A (t) x (t) + \bar{A} (t) E x (t) + B (t) v (t) + \bar{B} (t) E v (t) + F (t) u (t) + \bar{F} (t) E u (t) \\ + [C (t) x (t) + \bar{C} (t) E x (t) + D (t) v (t) + \bar{D} (t) E v (t) + G (t) u (t) + \bar{G} (t) E u (t)] ω (t) \\ z (t) = (\begin{matrix} ϕ (t) x (t) \\ ψ (t) u (t) \end{matrix}) \\ ψ {(t)}^{'} ψ (t) = I \\ x (0) = x_{0} \in ℝ^{n}, t \in N_{T} \end{cases}$ (6)

定义平均场有限时域H₂/H_∞控制问题如下：

定义5 考虑平均场随机控制系统(6)， $t \in N_{T}$ ，其中 $u (t) \in l_{ω}^{2} (N_{T}; ℝ^{q})$ 是控制输入。给定 $0 < T < \infty$ 和扰动衰减水平 $γ > 0$ ，若存在一个状态反馈控制

$u^{*} (t) = K (t) x (t) + \bar{K} (t) E x (t) = K (t) (x (t) - E x (t)) + K (t) E x ( t )$

使得：

1) 闭环系统

${\begin{cases} x (t + 1) = (A (t) + F (t) K (t)) x (t) + [\bar{A} (t) + F (t) \bar{K} (t) + \bar{F} (t) K (t) + \bar{F} (t) \bar{K} (t)] E x (t) \\ + B (t) v (t) + \bar{B} (t) E v (t) + {(C (t) + G (t) K (t)) x (t) + [\bar{C} (t) + G (t) \bar{K} (t) \\ + \bar{G} (t) K (t) + \bar{G} (t) \bar{K} (t)] E x (t) + D (t) v (t) + \bar{D} (t) E v (t)} ω (t) \\ z (t) = (\begin{matrix} ϕ (t) x (t) \\ ψ (t) K (t) x (t) + ψ (t) \bar{K} (t) E x (t) \end{matrix}) \\ ψ {(t)}^{'} ψ (t) = I \\ x (0) = x_{0} \in ℝ^{n}, t \in N_{T} \end{cases}$ (7)

满足 $J_{1}^{T} (u^{*}, v) = γ^{2} {‖ v (t) ‖}_{l_{ω}^{2} (N_{T}; ℝ^{l})}^{2} - {‖ z (t) ‖}_{l_{ω}^{2} (N_{T}; ℝ^{m})}^{2} > 0$ ，即：

$‖ L_{T} ‖ = \sup_{v \in l_{ω}^{2} (N_{T}; ℝ^{l}), v \neq 0, x_{0} \equiv 0} \frac{{‖ z (t) ‖}_{l_{ω}^{2} (N_{T}; ℝ^{m})}}{{‖ v (t) ‖}_{l_{ω}^{2} (N_{T}; ℝ^{l})}} < γ$

2) 当最坏情况扰动 $v^{*} (t) = W (t) x (t) + \bar{W} (t) E x (t) = W (t) (x (t) - E x (t)) + W (t) E x (t)$ 存在且施加在(6)上， $u^{*} (t)$ 能够使输出指标 $J_{2}^{T} (u, v^{*}) = {‖ z (t) ‖}_{l_{ω}^{2} (N_{T}; ℝ^{m})}^{2}$ 最小。

若 $(u^{*}, v^{*})$ 存在，则称上述有限时域平均场型的H₂/H_∞控制问题是可解的。有限时域H₂/H_∞控制问题就等价于寻找满足下式的Nash均衡解 $(u^{*}, v^{*})$

$J_{1}^{T} (u^{*}, v^{*}) \leq J_{1}^{T} (u^{*}, v), J_{2}^{T} (u^{*}, v^{*}) \leq J_{2}^{T} (u, v^{*})$

呈现主要结果之前，先引入四个耦合矩阵值方程

${\begin{cases} P_{1} (t) = {(A (t) + F (t) K (t))}^{'} P_{1} (t + 1) (A (t) + F (t) K (t)) \\ + {(C (t) + G (t) K (t))}^{'} P_{1} (t + 1) (C (t) + G (t) K (t)) - ϕ^{'} (t) ϕ (t) - K^{'} (t) K (t) \\ - G_{u} (P_{1} (t + 1)) H {(P_{1} (t + 1))}^{- 1} G_{u} {(P_{1} (t + 1))}^{'} \\ Q_{1} (t) = {(A (t) + F (t) K (t))}^{'} Q_{1} (t + 1) (A (t) + F (t) K (t)) \\ + {(ℂ (t) + G (t) K (t))}^{'} P_{1} (t + 1) (ℂ (t) + G (t) K (t)) - ϕ^{'} (t) ϕ (t) - K^{'} (t) K (t) \\ - {\hat{G}}_{u} (P_{1} (t + 1), Q_{1} (t + 1)) \hat{H} {(P_{1} (t + 1), Q_{1} (t + 1))}^{- 1} {\hat{G}}_{u} {(P_{1} (t + 1), Q_{1} (t + 1))}^{'} \\ P_{1} (T + 1) = Q_{1} (T + 1) = 0, H (P_{1} (t + 1)) > 0, \hat{H} (P_{1} (t + 1), Q_{1} (t + 1)) > 0 \end{cases}$ (8)

${\begin{cases} W (t) = - H {(P_{1} (t + 1))}^{- 1} G_{u} {(P_{1} (t + 1))}^{'} \\ W (t) = W (t) + \bar{W} (t) = \hat{H} {(P_{1} (t + 1), Q_{1} (t + 1))}^{- 1} {\hat{G}}_{u} {(P_{1} (t + 1), Q_{1} (t + 1))}^{'} \end{cases}$ (9)

${\begin{cases} {\bar{P}}_{1} (t) = {(A (t) + B (t) W (t))}^{'} {\bar{P}}_{1} (t + 1) (A (t) + B (t) W (t)) \\ + {(C (t) + D (t) W (t))}^{'} {\bar{P}}_{1} (t + 1) (C (t) + D (t) W (t)) + ϕ^{'} (t) ϕ (t) \\ - G_{v} ({\bar{P}}_{1} (t + 1)) H_{1} {({\bar{P}}_{1} (t + 1))}^{- 1} G_{v} {({\bar{P}}_{1} (t + 1))}^{'} \\ {\bar{Q}}_{1} (t) = {(A (t) + B (t) W (t))}^{'} {\bar{Q}}_{1} (t + 1) (A (t) + B (t) W (t)) \\ + {(ℂ (t) + D (t) W (t))}^{'} {\bar{P}}_{1} (t + 1) (ℂ (t) + D (t) W (t)) + ϕ^{'} (t) ϕ (t) \\ - {\hat{G}}_{v} ({\bar{P}}_{1} (t + 1), {\bar{Q}}_{1} (t + 1)) {\hat{H}}_{1} {({\bar{P}}_{1} (t + 1), {\bar{Q}}_{1} (t + 1))}^{- 1} {\hat{G}}_{v} {({\bar{P}}_{1} (t + 1), {\bar{Q}}_{1} (t + 1))}^{'} \\ {\bar{P}}_{1} (T + 1) = {\bar{Q}}_{1} (T + 1) = 0 \\ H_{1} ({\bar{P}}_{1} (t + 1)) > 0, {\hat{H}}_{1} ({\bar{P}}_{1} (t + 1), {\bar{Q}}_{1} (t + 1)) > 0 \end{cases}$ (10)

${\begin{cases} K (t) = - H_{1} {({\bar{P}}_{1} (t + 1))}^{- 1} G_{v} {({\bar{P}}_{1} (t + 1))}^{'} \\ K (t) = K (t) + \bar{K} (t) = {\hat{H}}_{1} {({\bar{P}}_{1} (t + 1), {\bar{Q}}_{1} (t + 1))}^{- 1} {\hat{G}}_{v} {({\bar{P}}_{1} (t + 1), {\bar{Q}}_{1} (t + 1))}^{'} \end{cases}$ (11)

其中

$\begin{array}{l} G_{u} ({\bar{P}}_{1} (t + 1)) = {(A (t) + F (t) K (t))}^{'} P_{1} (t + 1) B (t) + {(C (t) + G (t) K (t))}^{'} P_{1} (t + 1) D (t) \\ G_{v} ({\bar{P}}_{1} (t + 1)) = {(A (t) + B (t) W (t))}^{'} {\bar{P}}_{1} (t + 1) F (t) + {(C (t) + D (t) W (t))}^{'} {\bar{P}}_{1} (t + 1) G (t) \\ {\hat{G}}_{u} ({\bar{P}}_{1} (t + 1), {\bar{Q}}_{1} (t + 1)) = {(A (t) + F (t) K (t))}^{'} Q_{1} (t + 1) B (t) + {(ℂ (t) + G (t) K (t))}^{'} P_{1} (t + 1) D (t) \\ {\hat{G}}_{v} ({\bar{P}}_{1} (t + 1), {\bar{Q}}_{1} (t + 1)) = {(A (t) + B (t) W (t))}^{'} {\bar{Q}}_{1} (t + 1) F (t) + {(ℂ (t) + D (t) W (t))}^{'} {\bar{P}}_{1} (t + 1) G ( t ) \end{array}$

定理6 对于给定的扰动衰减水平 $γ > 0$ ，平均场随机系统(6)的有限时域H₂/H_∞控制问题有解 $(u^{*} (t), v^{*} (t))$ 为

$u^{*} (t) = K (t) x (t) + \bar{K} (t) E x (t), v^{*} (t) = W (t) x (t) + \bar{W} (t) E x ( t )$

其中 $K (t), \bar{K} (t) \in ℝ^{q \times n}$ 和 $W (t), \bar{W} (t) \in ℝ^{l \times n}$ 是矩阵值函数的充分必要条件是矩阵值方程(8)~(11)有解 $(P_{1} (t), Q_{1} (t); {\bar{P}}_{1} (t), {\bar{Q}}_{1} (t); K (t), \bar{K} (t); W (t), \bar{W} (t))$ ，其中 $Q_{1} (t) \leq 0$ ， ${\bar{Q}}_{1} (t) \geq 0$ ， $t \in N_{T}$ 。

证明：(充分性)

构造 $u^{*} (t) = K (t) x (t) + \bar{K} (t) E x (t)$ ，将 $u^{*} (t)$ 代入系统(6)得到(7)。由引理3和(8)得到 $‖ L_{T} ‖ < γ$ 。通过(8)和引理2，利用完全平方得

$\begin{array}{l} J_{1}^{T} (u^{*}, v) = \sum_{t = 0}^{T} E [γ^{2} {‖ v (t) ‖}^{2} - {‖ z (t) ‖}^{2}] \\ = \sum_{t = 0}^{T} E {{(\begin{matrix} x (t) - E x (t) \\ v (t) - E v (t) \end{matrix})}^{'} (\begin{matrix} L_{0} (P_{1} (t + 1)) & G_{u} (P_{1} (t + 1)) \\ G_{u} {(P_{1} (t + 1))}^{'} & H (P_{1} (t + 1)) \end{matrix}) (\begin{matrix} x (t) - E x (t) \\ v (t) - E v (t) \end{matrix})} \\ + \sum_{t = 0}^{T} {(\begin{matrix} E x (t) \\ E v (t) \end{matrix})}^{'} (\begin{matrix} {\hat{L}}_{0} (P_{1} (t + 1), Q_{1} (t + 1)) & {\hat{G}}_{u} (P_{1} (t + 1), Q_{1} (t + 1)) \\ {\hat{G}}_{u} (P_{1} (t + 1), Q_{1} (t + 1)) & \hat{H} (P_{1} (t + 1), Q_{1} (t + 1)) \end{matrix}) (\begin{matrix} E x (t) \\ E v (t) \end{matrix}) + {x^{'}}_{0} Q_{1} (0) x_{0} \\ = \sum_{t = 0}^{T} E {{[(v (t) - E v (t)) - (v^{*} (t) - E v^{*} (t))]}^{'} H (P_{1} (t + 1)) [(v (t) - E v (t)) - (v^{*} (t) - E v^{*} (t))]} \\ + \sum_{t = 0}^{T} {(E v (t) - E v^{*} (t))}^{'} \hat{H} (P_{1} (t + 1), Q_{1} (t + 1)) (E v (t) - E v^{*} (t)) + {x^{'}}_{0} Q_{1} (0) x_{0} \\ \geq J_{1}^{T} (u^{*}, v^{*}) = {x^{'}}_{0} Q_{1} (0) x_{0} \end{array}$

所以，看出 $v^{*} (t) = W (t) x (t) + \bar{W} (t) E x (t)$ ， $W (t)$ 和 $\bar{W} (t)$ 由(9)给出是最坏情况的扰动。

其中

$\begin{array}{l} L_{0} (P_{1} (t + 1)) = - P_{1} (t) + {(A (t) + F (t) K (t))}^{'} P_{1} (t + 1) (A (t) + F (t) K (t)) \\ + {(C (t) + G (t) K (t))}^{'} P_{1} (t + 1) (C (t) + G (t) K (t)) - ϕ^{'} (t) ϕ (t) - K^{'} (t) K (t) \\ {\hat{L}}_{0} (P_{1} (t + 1), Q_{1} (t + 1)) = - Q_{1} (t) + {(A (t) + F (t) K (t))}^{'} Q_{1} (t + 1) (A (t) + F (t) K (t)) \\ + {(ℂ (t) + G (t) K (t))}^{'} P_{1} (t + 1) (ℂ (t) + G (t) K (t)) - ϕ^{'} (t) ϕ (t) - K^{'} (t) K ( t ) \end{array}$

类似地，由(10)和定理4，利用完全平方得

$\begin{array}{l} J_{2}^{T} (u, v^{*}) = \sum_{t = 0}^{T} E [{‖ z (t) ‖}^{2}] \\ = \sum_{t = 0}^{T} E {{(\begin{matrix} x (t) - E x (t) \\ u (t) - E u (t) \end{matrix})}^{'} (\begin{matrix} L_{2} ({\bar{P}}_{1} (t + 1)) & G_{v} ({\bar{P}}_{1} (t + 1)) \\ G_{v} {({\bar{P}}_{1} (t + 1))}^{'} & H_{1} ({\bar{P}}_{1} (t + 1)) \end{matrix}) (\begin{matrix} x (t) - E x (t) \\ u (t) - E u (t) \end{matrix})} \\ + \sum_{t = 0}^{T} {(\begin{matrix} E x (t) \\ E u (t) \end{matrix})}^{'} (\begin{matrix} {\hat{L}}_{2} ({\bar{P}}_{1} (t + 1), {\bar{Q}}_{1} (t + 1)) & {\hat{G}}_{v} ({\bar{P}}_{1} (t + 1), {\bar{Q}}_{1} (t + 1)) \\ {\hat{G}}_{v} {({\bar{P}}_{1} (t + 1), {\bar{Q}}_{1} (t + 1))}^{'} & {\hat{H}}_{1} ({\bar{P}}_{1} (t + 1), {\bar{Q}}_{1} (t + 1)) \end{matrix}) (\begin{matrix} E x (t) \\ E u (t) \end{matrix}) + {x^{'}}_{0} {\bar{Q}}_{1} (0) x_{0} \\ = \sum_{t = 0}^{T} E {{[(u (t) - E u (t)) - (u^{*} (t) - E u^{*} (t))]}^{'} H_{1} ({\bar{P}}_{1} (t + 1)) [(u (t) - E u (t)) - (u^{*} (t) - E u^{*} (t))]} \\ + \sum_{t = 0}^{T} {(E u (t) - E u^{*} (t))}^{'} {\hat{H}}_{1} ({\bar{P}}_{1} (t + 1), {\bar{Q}}_{1} (t + 1)) (E u (t) - E u^{*} (t)) + {x^{'}}_{0} {\bar{Q}}_{1} (0) x_{0} \\ \geq J_{2}^{T} (u^{*}, v^{*}) = {x^{'}}_{0} {\bar{Q}}_{1} (0) x_{0} \end{array}$

其中

$\begin{array}{l} L_{2} ({\bar{P}}_{1} (t + 1)) = - {\bar{P}}_{1} (t) + {(A (t) + B (t) W (t))}^{'} {\bar{P}}_{1} (t + 1) (A (t) + B (t) W (t)) \\ + {(C (t) + D (t) W (t))}^{'} {\bar{P}}_{1} (t + 1) (C (t) + D (t) W (t)) + ϕ {(t)}^{'} ϕ (t) \\ {\hat{L}}_{2} ({\bar{P}}_{1} (t + 1), {\bar{Q}}_{1} (t + 1)) = - {\bar{Q}}_{1} (t) + {(A (t) + B (t) W (t))}^{'} {\bar{Q}}_{1} (t + 1) (A (t) + B (t) W (t)) \\ + {(ℂ (t) + D (t) W (t))}^{'} {\bar{P}}_{1} (t + 1) (ℂ (t) + D (t) W (t)) + ϕ {(t)}^{'} ϕ ( t ) \end{array}$

所以 $J_{1}^{T} (u^{*}, v^{*}) \leq J_{1}^{T} (u^{*}, v)$ ， $J_{2}^{T} (u^{*}, v^{*}) \leq J_{2}^{T} (u, v^{*})$ 。因此 $(u^{*}, v^{*})$ 是系统(6)的平均场有限时域混合H₂/H_∞控制问题的最优解。

(必要性)

将 $u^{*} (t) = K (t) x (t) + \bar{K} (t) E x (t)$ 代入系统(6)得到(7)，利用引理3得到方程(8)有解 $(P_{1} (t), Q_{1} (t))$ 且 $Q_{1} (t) \leq 0$ 。因为(7)与(1)形式类似，因此参考引理3充分性的证明，可以得出最坏情况扰动为

$\begin{matrix} v^{*} (t) = W (t) x (t) + \bar{W} (t) E x (t) \\ = - H {(P_{1} (t + 1))}^{- 1} G_{u} {(P_{1} (t + 1))}^{'} x (t) + [\hat{H} {(P_{1} (t + 1), Q_{1} (t + 1))}^{- 1} {\hat{G}}_{u} {(P_{1} (t + 1), Q_{1} (t + 1))}^{'} \\ + H {(P_{1} (t + 1))}^{- 1} G_{u} {(P_{1} (t + 1))}^{'}] E x ( t ) \end{matrix}$

将 $v^{*} (t) = W (t) x (t) + \bar{W} (t) E x (t)$ 代入系统(6)，有

${\begin{cases} x (t + 1) = (A (t) + B (t) W (t)) x (t) + [\bar{A} (t) + B (t) \bar{W} (t) + \bar{B} (t) W (t) + \bar{B} (t) \bar{W} (t)] E x (t) \\ + F (t) u (t) + \bar{F} (t) E u (t) + {(C (t) + D (t) W (t)) x (t) + [\bar{C} (t) + D (t) \bar{W} (t) \\ + \bar{D} (t) W (t) + \bar{D} (t) \bar{W} (t)] E x (t) + G (t) v (t) + \bar{G} (t) E v (t)} ω (t) \\ z (t) = (\begin{matrix} ϕ (t) x (t) \\ ψ (t) u (t) \end{matrix}) \\ ψ {(t)}^{'} ψ (t) = I \\ x (0) = x_{0} \in ℝ^{n}, t \in N_{T} \end{cases}$ (12)

此时，问题

$\min_{u \in l_{ω}^{2} (N_{T}; ℝ^{q})} J_{2}^{T} (u, v^{*})$ (13)

变为一个离散时间LQ最优控制问题，其中(13)满足(12)。

根据文献 [20] 的定理3.3知(10)在 $N_{T}$ 上有唯一解 $({\bar{P}}_{1} (t), {\bar{Q}}_{1} (t))$ 且 ${\bar{Q}}_{1} (t) \geq 0$ 。

$\min_{u \in l_{ω}^{2} (N_{T}; ℝ^{q})} J_{2}^{T} (u, v^{*}) = J_{2}^{T} (u^{*}, v^{*}) = {x^{'}}_{0} {\bar{Q}}_{1} (0) x_{0}$

其中

$\begin{array}{l} u^{*} (t) = K (t) x (t) + \bar{K} (t) E x (t) \\ = - H_{1} {({\bar{P}}_{1} (t + 1))}^{- 1} G_{v} {({\bar{P}}_{1} (t + 1))}^{'} x (t) + [{\hat{H}}_{1} {({\bar{P}}_{1} (t + 1), {\bar{Q}}_{1} (t + 1))}^{- 1} {\hat{G}}_{v} {({\bar{P}}_{1} (t + 1), {\bar{Q}}_{1} (t + 1))}^{'} \\ + H_{1} {({\bar{P}}_{1} (t + 1))}^{- 1} G_{v} {({\bar{P}}_{1} (t + 1))}^{'}] E x ( t ) \end{array}$

5. 总结

我们讨论了 $(x, u, v)$ -依赖噪声的离散时间平均场随机系统的混合H₂/H_∞控制问题，通过耦合矩阵差分方程的可解性得到了平均场随机混合H₂/H_∞控制存在的充分必要条件，将经典随机线性系统的结果推广到了平均场随机系统。

文章引用

贾亚琪,张春梅,杨路,陈柏江. (x,u,v)-依赖噪声的离散时间平均场随机H₂/H_∞控制
Discrete-Time Mean-Field Stochastic H₂/H_∞ Control with (x,u,v)-Dependent Noise[J]. 应用数学进展, 2023, 12(09): 3860-3871. https://doi.org/10.12677/AAM.2023.129380

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