Arimoto等人1984年系统地提出了迭代学习控制算法^[1].针对有限时间段上的需要跟踪的理想输出,迭代学习控制设计的基本思想是:构建合适的迭代学习律,并将其作用于有限时间段的重复受控系统上,通过反复迭代,不断地修正输出跟踪误差,并最终使得系统的输出能够完全跟踪给定的理想输出.在设计过程中,常用的是D型学习律^[1-4]和P型学习律^[5-7].由于其构建的简便性和控制的有效性,迭代学习控制算法已成为控制领域研究的热点之一,参见文献[8].

由偏微分方程描述的系统称为分布参数系统,分布参数系统常被应用于描述波运动、热传导等物理现象^[9-12].近年来,描述梁和薄板振动的四阶波动方程受到了广泛关注^[13-19].然而,相关的研究工作至今大多集中于方程的适定性和数值解上^[15-19].最近,文献[20]提出和研究了四阶分布参数系统的迭代学习控制问题,文中的系统是由一维四阶抛物方程或一维四阶波动方程构建而成.当构建得到的P型迭代学习律作用于系统时,跟踪误差于空间有界,进一步,当系统的初始偏差为零时,跟踪误差于$ {L^2} $空间中能够随着迭代次数的增加收敛于零.

对于生物群体行为(如迁徙鸟类的编队飞行、鱼群的聚集行为、蚁群的协同工作等)的研究,使多智能体系统的控制设计成为国际控制领域中的研究热点^[21].近年来,随着计算机通信技术的发展,多智能体系统的协同控制已被应用于无线传感器网络、移动机器人编队、集群航天器深空探测等众多领域中^[22-25].一致性问题作为多智能体系统协调控制中最基本的问题,更是受到了众多研究人员的广泛关注^[26-29].一致性问题考虑的是如何设计合适的分布式控制协议,使得所有智能体的状态(或输出)随着时间趋于无穷而收敛到同一个常数.最近,文献[30]研究了分布参数多智能体系统基于一致性的迭代学习控制问题,考虑的系统中的每个智能体都是由一维二阶抛物型方程或一维二阶双曲型方程构建而成.基于有限时间区间的一致性目的,构建得到P型迭代学习控制协议.文献[31]研究了具有时滞的分布参数模型多智能体系统的一致控制问题,针对多智能体系统中的时滞问题,提出了一种分布式P型迭代学习控制协议,基于所提出的迭代学习控制律,经过足够的迭代次数后,一致性误差于$ {L^2} $范数意义下可以收敛到零.注意到这方面相关的研究工作,尚未涉及到四阶的分布参数系统.

在文献[20, 30]中的工作基础上,本文进一步研究分布参数多智能体系统的迭代学习控制,该类多智能体系统中的每个智能体是由文献[15]中的四阶梁方程构建而成.基于文献[30]中的虚拟领导者方法,构建得到P型学习律.当该学习律作用于系统时,任两个智能体之间于$ {L^2} $空间中的一致性误差有界;进一步,当初始偏差为零时,且迭代次数趋于无穷,则任两个智能体之间的一致性误差于$ {L^2} $空间中能够收敛于零.

本文符号约定如下: $ {I_b} $为$ b \times b $维单位矩阵, $ \otimes $为克罗内克积, $ {{\bf 1}_b} $为每个元素均为$ 1 $的$ {b} $阶列向量.对于给定的向量或矩阵$ P $, $ {P^{\rm{T}}} $表示其转置, $ \left\| P \right\| $表示其2 -范数, $ {P^{ - 1}} $表示其逆(若$ P $可逆).对函数$ W(x, t):[0, l] \times [0, T] \to {{{{\Bbb R}} }^n} $,定义$ {\left\| {W( \cdot , t)} \right\|_{{L^2}}} = \sqrt {\int_0^l {{{\left\| {W(x, t)} \right\|}^2}{\rm{d}}x} } $,进一步记$ {\left\| W \right\|_{{L^2}, s}} = {\sup\limits_{t \in [0, T]}}\left\| {W( \cdot , t)} \right\|_{{L^2}}^2 $.

智能体之间的信息交流常可用图来刻画,由此,一些常用的图论知识给出如下:记$ G = (V, E, A) $为一加权的有向图,其中$ V = \{ 1, 2, \cdots , N\} $为结点集合, $ E \subseteq V \times V $为边集合,而$ A = ({a_{ij}}) \in {R^{N \times N}} $表示图$ G $带权的邻接矩阵,如果$ (j, i) \in E $,则$ {a_{ij}} > 0 $,否则$ {a_{ij}} = 0 $,进一步地,约定$ {a_{ii}} = 0 $.定义有向图$ G $的拉普拉斯阵为$ {L_G} = D - A $,这里$ D = {\rm diag}({\deg _{in}}(1), \cdots , {\deg _{in}}(N)) $,而$ {\deg _{in}}(i) = \sum\limits_{j = 1}^N {{a_{ij}}} $为结点$ i $的入度.在有向图$ G $中,我们用第$ i $个结点表示第$ i $个智能体,用结点$ i $到结点$ j $的有向边$ (i, j) \in E $表示第$ j $个智能体能直接接受到来自第$ i $个智能体的信息.

记

$ \beta = {({a_{12}}, {a_{13}}, \cdots , {a_{1N}})^{\rm{T}}}, $

$ {L_{22}} = \left[ {\begin{array}{*{20}{c}} {{{\deg }_{in}}(2)}&{ - {a_{23}}}& \cdots &{ - {a_{2N}}}\\ { - {a_{32}}}&{{{\deg }_{in}}(3)}& \cdots &{ - {a_{3N}}}\\ \vdots & \vdots & \ddots & \vdots \\ { - {a_{N2}}}&{ - {a_{N3}}}& \cdots &{{{\deg }_{in}}(N)} \end{array}} \right]. $

引理2.1^[30]  若有向图$ G $存在生成树,则矩阵$ {L_{22}} + {{\bf{1}}_{N - 1}} \cdot {\beta ^{\rm{T}}} $的每个特征值均有正实部.

考虑文献[15]中的变系数四阶梁方程

(2.1) $ \begin{equation} \mu (x)\frac{{{\partial ^2}w(x, t)}}{{\partial {t^2}}} + \frac{{{\partial ^2}}}{{\partial {x^2}}}\Big[ {EI(x)\frac{{{\partial ^2}w(x, t)}}{{\partial {x^2}}}} \Big] = q(x, t), x \in (0, l), t > 0, \end{equation} $

其中$ \mu (x) $, $ w(x, t) $, $ EI(x) $和$ q(x, t) $分别表示单位长度质量,侧梁位移,梁抗弯刚度和单位长度荷载.

方程(2.1)的边值条件^[15]如下

$ - \frac{\partial }{{\partial x}}\Big[ {EI(x)\frac{{{\partial ^2}w(x, t)}}{{\partial {x^2}}}} \Big] = 0, \; \; \; \; x = 0, l, $

或者

$ w(x, t) = 0, \; \; \; \; x = 0, l $

和

$ EI(x)\frac{{{\partial ^2}w(x, t)}}{{\partial {x^2}}} = 0, \; \; \; \; x = 0, l $

或者

$ \frac{{\partial w(x, t)}}{{\partial x}} = 0, \; \; \; \; x = 0, l. $

文献[15]利用傅里叶级数展开式解决了方程(2.1)解的存在性问题,并给出了方程(2.1)的解的形式,详见文献[15]中公式(46).

注2.1  文献[15]给出了方程(2.1)解存在的条件: $ \frac{{q(x, t)}}{{\mu (x)}} $在定义区间内能够展开为傅里叶级数.

记

(2.2) $ \begin{equation} {\mu _1} = \mathop {\min }\limits_{0 \le x \le l} \mu (x), \; \; {\mu _2} = \mathop {\max }\limits_{0 \le x \le l} \mu (x). \end{equation} $

(2.3) $ \begin{equation} E{I_1} = \mathop {\min }\limits_{0 \le x \le l} EI(x), \; \; E{I_2} = \mathop {\max }\limits_{0 \le x \le l} EI(x). \end{equation} $

由这些量的物理意义可知, $ {\mu _1} $, $ {\mu _2} $, $ E{I_1} $, $ E{I_2} $均为正常数.

为了控制设计的需要,我们用控制变量$ u(x, t) $代替(2.1)式中的$ q(x, t) $$ \rm(单位长度荷载) $,并添加通常形式的输出项.由此,考虑如下由分布参数系统构建而成的$ N $个智能体

(2.4) $ \begin{equation} \left\{ \begin{array}{l} { } \mu (x)\frac{{{\partial ^2}{w_{i, k}}(x, t)}}{{\partial {t^2}}} + \frac{{{\partial ^2}}}{{\partial {x^2}}}\Big[ {EI(x)\frac{{{\partial ^2}{w_{i, k}}(x, t)}}{{\partial {x^2}}}} \Big] = {u_{i, k}}(x, t), \\ [2mm] {y_{i, k}}(x, t) = F{w_{i, k}}(x, t) + B{u_{i, k}}(x, t), \end{array}\right. \; \; i = 1, 2, \cdots , N. \end{equation} $

其中$ (x, t) \in (0, l) \times [0, T] $, $ ({w_{i, k}}(x, t), \frac{{\partial {w_{i, k}}(x, t)}}{{\partial t}}) $, $ {u_{i, k}}(x, t) $, $ {y_{i, k}}(x, t) $分别表示第$ i $个智能体在$ k $次迭代时的状态(参见文献[10, 11]),控制输入和输出,而$ B \ne 0 $.对于系统(2.4),假设$ k $次迭代时的网络拓扑为已知的,用拉普拉斯阵$ {L_G} $来表示.

注2.2  本文用$ {u_{i, k}}(x, t) $代替$ q(x, t) $,结合傅里叶级数展开需求,我们要求$ {u_{i, k}}(x, t) $与$ \mu (x) $为连续函数,则$ \frac{{{u_{i, k}}(x, t)}}{{\mu (x)}} $为连续函数,满足傅里叶级数展开需求.

假设2.1  图$ G $存在生成树.

假设2.2^[30]  对于所有的$ k $,设重复性初始偏差被设置在容许偏差范围内,即

$ \begin{eqnarray*} &&{\left\| {{{\left. {{w_{i, k + 1}}( \cdot , t)} \right|}_{t = 0}} - {{\left. {{w_{i, k}}( \cdot , t)} \right|}_{t = 0}}} \right\|_{{H^2}}}\\ & = &{\left\| {{{\left. {{w_{i, k + 1}}( \cdot , t)} \right|}_{t = 0}} - {{\left. {{w_{i, k}}( \cdot , t)} \right|}_{t = 0}}} \right\|_{{L^2}}} + {\left\| {{{\left. {\frac{{\partial {w_{i, k + 1}}( \cdot , t)}}{{\partial x}}} \right|}_{t = 0}} - {{\left. {\frac{{\partial {w_{i, k}}( \cdot , t)}}{{\partial x}}} \right|}_{t = 0}}} \right\|_{{L^2}}}\\ &&+ {\left\| {{{\left. {\frac{{{\partial ^2}{w_{i, k + 1}}( \cdot , t)}}{{\partial {x^2}}}} \right|}_{t = 0}} - {{\left. {\frac{{{\partial ^2}{w_{i, k}}( \cdot , t)}}{{\partial {x^2}}}} \right|}_{t = 0}}} \right\|_{{L^2}}} \le {\varepsilon _1}, \end{eqnarray*} $

$ {\left\| {{{\left. {\frac{{\partial {w_{i, k + 1}}( \cdot , t)}}{{\partial t}}} \right|}_{t = 0}} - {{\left. {\frac{{\partial {w_{i, k}}( \cdot , t)}}{{\partial t}}} \right|}_{t = 0}}} \right\|_{{L^2}}} \le {\varepsilon _2}, {\quad} i = 1, 2, \cdots , N, \; k = 0, 1, 2, \cdots , $

这里$ {\varepsilon _1}, {\varepsilon _2} $是非负常数.

相应于方程(2.1)中的边值条件,给出系统(2.4)的如下边值条件假设.

假设2.3  当时$ x = 0, l $,有

$ - \frac{\partial }{{\partial x}}\Big[ {EI(x)\frac{{{\partial ^2}{w_{i, k}}(x, t)}}{{\partial {x^2}}}} \Big] = 0, $

或者$ {w_{i, k}}(x, t) = 0 $和$ EI(x)\frac{{{\partial ^2}{w_{i, k}}(x, t)}}{{\partial {x^2}}} = 0, $或者

$ \frac{{\partial {w_{i, k}}(x, t)}}{{\partial x}} = 0, {\quad} t \in [0, T], \; i = 1, 2, \cdots , N, \; k = 0, 1, 2, \cdots . $

引理2.2^[30]  $ {\rm{\{ }}{\varpi _k}{\rm{\} }} $为实数序列,如果有

$ \left| {{\varpi _{k + 1}}} \right| \le \mathord{\buildrel{\lower3pt\hbox{$ \frown$}} \over \rho } \left| {{\varpi _k}} \right| + \theta, \; \; 0 \le \mathord{\buildrel{\lower3pt\hbox{$ \frown$}} \over \rho } < 1, \; \; \theta > 0, \; \; k = 0, 1, 2, \cdots , $

则

$ \mathop {\lim }\limits_{k \to \infty } \sup \left| {{\varpi _k}} \right| \le \frac{\theta }{{1 - \mathord{\buildrel{\lower3pt\hbox{$ \frown$}} \over \rho } }}. $

构建如下的P型迭代学习律

(3.1) $ \begin{equation} {u_{i, k + 1}}(x, t) = {u_{i, k}}(x, t) + g\sum\limits_{j = 1}^N {{a_{ij}}({y_{j, k}}(x, t) - {y_{i, k}}(x, t))} , \end{equation} $

其中$ g $为学习增益.取$ g $满足

(3.2) $ \begin{equation} \rho \Big( {{I_{N - 1}} - gB\Big( {{L_{22}} + {{\bf{1}}_{N - 1}} \cdot {\beta ^{\rm{T}}}} \Big)} \Big) < 1, \end{equation} $

其中$ \rho $为矩阵的谱半径.

注3.1^[30]  如果$ {L_{22}} + {{\bf{1}}_{N - 1}} \cdot {\beta ^{\rm{T}}} $的每个特征值都有正实部,则满足$ \rm(3.2) $式的学习增益是存在的.

注3.2  取$ {u_{i, 0}}(x, t) $(相应于$ q(x, t) $)为连续函数,由系统(2.4)中第一式可解得$ {w_{i, 0}}(x, t) $为连续函数,再由系统(2.4)中第二式可知$ {y_{i, 0}}(x, t) $也为连续函数,再由学习律(3.1),可知$ {u_{i, 1}}(x, t) $也为连续函数.取$ k = 1, 2, \cdots $,循环往复上述过程,再结合注2.1和注2.2,可知在每步迭代过程中,系统(2.4)的解均存在且连续.

记$ {\delta _{i, k}}(x, t) = {y_{i, k}}(x, t) - {y_{1, k}}(x, t) $, $ i = 1, 2, \cdots , N $,这里$ {y_{1, k}}(x, t) $为虚拟领导者.由系统(2.4)和学习律(3.1)式可知

(3.3) $ \begin{eqnarray} {\delta _{i, k + 1}}(x, t) & = & {y_{i, k + 1}}(x, t) - {y_{1, k + 1}}(x, t){}\\ & = &{y_{i, k}}(x, t) - {y_{1, k}}(x, t) + {y_{i, k + 1}}(x, t) - {y_{i, k}}(x, t) - ({y_{1, k + 1}}(x, t) - {y_{1, k}}(x, t)){}\\ & = & {\delta _{i, k}}(x, t) + F({w_{i, k + 1}}(x, t) - {w_{i, k}}(x, t)) + B({u_{i, k + 1}}(x, t) - {u_{i, k}}(x, t)){}\\ &&- F({w_{1, k + 1}}(x, t) - {w_{1, k}}(x, t)) - B({u_{1, k + 1}}(x, t) - {u_{1, k}}(x, t)){}\\ & = & {\delta _{i, k}}(x, t) + F({w_{i, k + 1}}(x, t) - {w_{i, k}}(x, t)) + gB\sum\limits_{j = 1}^N {{a_{ij}}({y_{j, k}}(x, t) - {y_{i, k}}(x, t))} {}\\ &&- F({w_{1, k + 1}}(x, t) - {w_{1, k}}(x, t)) - gB\sum\limits_{j = 1}^N {{a_{1j}}({y_{j, k}}(x, t) - {y_{1, k}}(x, t))} {}\\ & = &{\delta _{i, k}}(x, t) + F\Delta {w_{i, k}}(x, t) + gB\sum\limits_{j = 1}^N {{a_{ij}}({\delta _{j, k}}(x, t) - {\delta _{i, k}}(x, t))} {}\\ &&- F\Delta {w_{1, k}}(x, t) - gB\sum\limits_{j = 1}^N {{a_{1j}}{\delta _{j, k}}(x, t)} , \; \; i = 2, 3, \cdots , N. \end{eqnarray} $

其中$ \Delta {w_{i, k}}(x, t) = {w_{i, k + 1}}(x, t) - {w_{i, k}}(x, t), $$ i = 1, 2, \cdots , N. $记

$ {\delta _k}(x, t) = \left[ {\begin{array}{*{20}{c}} {{\delta _{2, k}}(x, t)}\\ {{\delta _{3, k}}(x, t)}\\ \vdots \\ {{\delta _{N, k}}(x, t)} \end{array}} \right], \; \; \; \Delta {w_k}(x, t) = \left[ {\begin{array}{*{20}{c}} {\Delta {w_{2, k}}(x, t)}\\ {\Delta {w_{3, k}}(x, t)}\\ \vdots \\ {\Delta {w_{N, k}}(x, t)} \end{array}} \right], $

将$ \rm(3.3) $式写成如下的紧凑形式

$ {\delta _{k + 1}}(x, t) = \Big[ {{I_{N - 1}} - gB\Big( {{L_{22}} + {{\bf{1}}_{N - 1}} \cdot {\beta ^{\rm{T}}}} \Big)} \Big]{\delta _k}(x, t)+ F\Big[ {\Delta {w_k}(x, t) - {{\bf{1}}_{N - 1}} \otimes \Delta {w_{1, k}}(x, t)} \Big]. $

记$ \tilde \Delta {w_k}(x, t) = \Delta {w_k}(x, t) - {{\bf{1}}_{N - 1}} \otimes \Delta {w_{1, k}}(x, t) $,则进一步有

(3.4) $ \begin{equation} {\delta _{k + 1}}(x, t) = \Big[ {{I_{N - 1}} - gB\Big( {{L_{22}} + {{\bf{1}}_{N - 1}} \cdot {\beta ^{\rm{T}}}} \Big)} \Big]{\delta _k}(x, t) + F\tilde \Delta {w_k}(x, t). \end{equation} $

由系统(2.4)和学习律(3.1)式可知

$ \begin{eqnarray*} & &\mu (x)\frac{{{\partial ^2}\Delta {w_{i, k}}(x, t)}}{{\partial {t^2}}} + \frac{{{\partial ^2}}}{{\partial {x^2}}}\Big[ {EI(x)\frac{{{\partial ^2}\Delta {w_{i, k}}(x, t)}}{{\partial {x^2}}}} \Big]\\ & = & {u_{i, k{\rm{ + }}1}}(x, t) - {u_{i, k}}(x, t) = g\sum\limits_{j = 1}^N {{a_{ij}}\Big( {{\delta _{j, k}}(x, t) - {\delta _{i, k}}(x, t)} \Big)}, \; \; \; i = 1, 2, \cdots , N. \end{eqnarray*} $

进一步有

$ \begin{eqnarray*} &&\mu (x)\frac{{{\partial ^2}\Big( {\Delta {w_{i, k}}(x, t) - \Delta {w_{1, k}}(x, t)} \Big)}}{{\partial {t^2}}} + \frac{{{\partial ^2}}}{{\partial {x^2}}}\Big[ {EI(x)\frac{{{\partial ^2}\Big( {\Delta {w_{i, k}}(x, t) - \Delta {w_{1, k}}(x, t)} \Big)}}{{\partial {x^2}}}} \Big]\\ & = & g\sum\limits_{j = 1}^N {{a_{ij}}\Big( {{\delta _{j, k}}(x, t) - {\delta _{i, k}}(x, t)} \Big)} - g\sum\limits_{j = 1}^N {{a_{1j}}} {\delta _{j, k}}(x, t), \; \; \; i = 2, 3, \cdots N. \end{eqnarray*} $

上式写成紧凑形式,可得

$ \begin{eqnarray*} &&\Big( {{I_{N - 1}} \otimes \mu (x)} \Big)\frac{{{\partial ^2}\Big( {\Delta {w_k}(x, t) - {{\bf{1}}_{N - 1}} \otimes \Delta {w_{1, k}}(x, t)} \Big)}}{{\partial {t^2}}}\\ &&+ \frac{{{\partial ^2}}}{{\partial {x^2}}}\Big[ {\Big( {{I_{N - 1}} \otimes EI(x)} \Big)\frac{{{\partial ^2}\Big( {\Delta {w_k}(x, t) - {{\bf{1}}_{N - 1}} \otimes \Delta {w_{1, k}}(x, t)} \Big)}}{{\partial {x^2}}}} \Big]\\ & = & - g\Big( {{L_{22}} + {{\bf{1}}_{N - 1}} \cdot {\beta ^{\rm{T}}}} \Big){\delta _k}(x, t). \end{eqnarray*} $

进一步有

(3.5) $ \begin{eqnarray} &&\Big( {{I_{N - 1}} \otimes \mu (x)} \Big)\frac{{{\partial ^2}\tilde \Delta {w_k}(x, t)}}{{\partial {t^2}}}{\rm{ + }}\frac{{{\partial ^2}}}{{\partial {x^2}}}\Big[ {\Big( {{I_{N - 1}} \otimes EI(x)} \Big)\frac{{{\partial ^2}\tilde \Delta {w_k}(x, t)}}{{\partial {x^2}}}} \Big]{}\\ & = & - g\Big( {{L_{22}} + {{\bf{1}}_{N - 1}} \cdot {\beta ^{\rm{T}}}} \Big){\delta _k}(x, t). \end{eqnarray} $

引理3.1^[32]  对于任意的矩阵$ X \in {{\mathbb R}^{(N - 1) \times (N - 1)}} $,如果谱半径$ \rho (X) < 1 $,则至少存在一种矩阵范数$ {\left\| \cdot \right\|_\kappa } $,使得$ {\left\| X \right\|_\kappa } < 1 $.

引理3.2^[33]  对于任意矩阵范数$ {\left\| \cdot \right\|_\kappa } $,则至少存在与之相容的一种向量范数$ {\left\| \cdot \right\|_\xi } $,使得对任意$ P \in {{\mathbb R}^{(N - 1) \times (N - 1)}} $和$ x \in {{\mathbb R}^{N - 1}} $,有$ {\left\| {Px} \right\|_\xi } \le {\left\| P \right\|_\kappa }{\left\| x \right\|_\xi } $.

后文中为了方便,引理3.1、3.2中的向量范数和与之相容的矩阵范数都用$ {\left\| \cdot \right\|_{\rm{*}}} $表示.

根据$ \rm(3.2) $式和引理3.1,有

$ \bar \rho = {\left\| {{I_{N - 1}} - gB\Big( {{L_{22}} + {{\bf{1}}_{N - 1}} \cdot {\beta ^{\rm{T}}}} \Big)} \right\|_{\rm{*}}} < 1. $

由此,存在$ \varepsilon > 0 $使得

(3.6) $ \begin{equation} \bar \rho = {\left\| {{I_{N - 1}} - gB\Big( {{L_{22}} + {{\bf{1}}_{N - 1}} \cdot {\beta ^{\rm{T}}}} \Big)} \right\|_{\rm{*}}} < \frac{1}{{\sqrt {1 + \varepsilon } }}. \end{equation} $

根据$ \rm(3.4) $式和$ \rm(3.6) $式,有

$ {\left\| {{\delta _{k + 1}}(x, t)} \right\|_{\rm{*}}} \le \bar \rho {\left\| {{\delta _k}(x, t)} \right\|_*} + \left| F \right|{\left\| {\tilde \Delta {w_k}(x, t)} \right\|_*}. $

使用基本不等式,可得

$ \left\| {{\delta _{k + 1}}(x, t)} \right\|_{\rm{*}}^2 \le (1 + \varepsilon ){\bar \rho ^2}\left\| {{\delta _k}(x, t)} \right\|_{\rm{*}}^2 + \Big( {1 + \frac{1}{\varepsilon }} \Big){F^2}\left\| {\tilde \Delta {w_k}(x, t)} \right\|_{\rm{*}}^2. $

进一步有

(3.7) $ \begin{eqnarray} \mathop {\sup }\limits_{t \in [0, T]} \Big\{ {\int_0^l {{{\rm e}^{ - \lambda t}} \left\| {{\delta _{k + 1}}(x, t)} \right\|_{\rm{*}}^2{\rm{d}}x} } \Big\} &\le & (1 + \varepsilon ){\bar \rho ^2}\mathop {\sup }\limits_{t \in [0, T]} \Big\{ {\int_0^l {{{\rm e}^{ - \lambda t}}\left\| {{\delta _k}(x, t)} \right\|_{\rm{*}}^2{\rm{d}}x} } \Big\} {}\\ && + \Big( {1 + \frac{1}{\varepsilon }} \Big){F^2}\mathop {\sup }\limits_{t \in [0, T]} \Big\{ {\int_0^l {{{\rm e}^{ - \lambda t}}\left\| {\tilde \Delta {w_k}(x, t)} \right\|_{\rm{*}}^2{\rm{d}}x} } \Big\}, {\qquad} \end{eqnarray} $

这里$ \lambda $是个正实数.

定理3.1  如果关于系统$ \rm(2.4) $的假设$ \rm2.1–2.3 $都成立,则在学习律$ \rm(3.1) $的作用下,任意两个智能体之间于$ {L^2} $空间中的一致性误差有界;进一步,当系统没有初始偏差时,即

$ {\left\| {{{\left. {{w_{i, k + 1}}( \cdot , t)} \right|}_{t = 0}} - {{\left. {{w_{i, k}}( \cdot , t)} \right|}_{t = 0}}} \right\|_{{H^2}}} = 0, \; \; {\left\| {{{\left. {\frac{{\partial {w_{i, k + 1}}( \cdot , t)}}{{\partial t}}} \right|}_{t = 0}} - {{\left. {\frac{{\partial {w_{i, k}}( \cdot , t)}}{{\partial t}}} \right|}_{t = 0}}} \right\|_{{L^2}}} = 0, $

则系统的一致性误差随着迭代学习次数的增加而趋于零,即, $ \mathop {\lim }\limits_{k \to \infty } {\left\| {{y_{i, k}} - {y_{j, k}}} \right\|_{{L^2}, s}} = 0 $, $ \forall i, $$ j \in \{ 1, 2, \cdots , N\} $.

证  $ \rm(3.5) $式两端左乘$ {\Big( {\frac{{\partial \tilde \Delta {w_k}(x, t)}}{{\partial t}}} \Big)^{\rm{T}}} $,且两端同时对$ x $从0到$ l $积分,有

(3.8) $ \begin{eqnarray} &&\int_0^l {{{\Big( {\frac{{\partial \tilde \Delta {w_k}(x, t)}}{{\partial t}}} \Big)}^{\rm{T}}}} \Big( {{I_{N - 1}} \otimes \mu (x)} \Big)\frac{{{\partial ^2}\tilde \Delta {w_k}(x, t)}}{{\partial {t^2}}}{\rm{d}}x{}\\ &&+ \int_0^l {{{\Big( {\frac{{\partial \tilde \Delta {w_k}(x, t)}}{{\partial t}}} \Big)}^{\rm{T}}}\frac{{{\partial ^2}}}{{\partial {x^2}}}\Big[ {\Big( {{I_{N - 1}} \otimes EI(x)} \Big)\frac{{{\partial ^2}\tilde \Delta {w_k}(x, t)}}{{\partial {x^2}}}} \Big]{\rm{d}}x} {}\\ & = & - g\int_0^l {{{\Big( {\frac{{\partial \tilde \Delta {w_k}(x, t)}}{{\partial t}}} \Big)}^{\rm{T}}}\Big( {{L_{22}} + {{\bf{1}}_{N - 1}} \cdot {\beta ^{\rm{T}}}} \Big){\delta _k}(x, t)} {\rm{d}}x, \end{eqnarray} $

其中

(3.9) $ \begin{eqnarray} &&\int_0^l {{{\Big( {\frac{{\partial \tilde \Delta {w_k}(x, t)}}{{\partial t}}} \Big)}^{\rm{T}}}} \Big( {{I_{N - 1}} \otimes \mu (x)} \Big)\frac{{{\partial ^2}\tilde \Delta {w_k}(x, t)}}{{\partial {t^2}}}{\rm{d}}x {}\\ & = &\frac{1}{2}\frac{{\rm{d}}}{{{\rm{d}}t}}\int_0^l {{{\Big( {\frac{{\partial \tilde \Delta {w_k}(x, t)}}{{\partial t}}} \Big)}^{\rm{T}}}} \Big( {{I_{N - 1}} \otimes \mu (x)} \Big)\frac{{\partial \tilde \Delta {w_k}(x, t)}}{{\partial t}}{\rm{d}}x, \end{eqnarray} $

(3.10) $ \begin{eqnarray} && - g\int_0^l {{{\Big( {\frac{{\partial \tilde \Delta {w_k}(x, t)}}{{\partial t}}} \Big)}^{\rm{T}}}\Big( {{L_{22}} + {{\bf{1}}_{N - 1}} \cdot {\beta ^{\rm{T}}}} \Big){\delta _k}(x, t)} {\rm{d}}x {}\\ & \le &\frac{{{c_1}}}{2}\Big\{ {{{\int_0^l {\left\| {\frac{{\partial \tilde \Delta {w_k}(x, t)}}{{\partial t}}} \right\|} }^2}{\rm{d}}x + \int_0^l {{{\left\| {{\delta _k}(x, t)} \right\|}^2}} {\rm{d}}x} \Big\}, \end{eqnarray} $

这里$ {c_1} = \left| g \right|\left\| {{L_{22}} + {{\bf{1}}_{N - 1}} \cdot {\beta ^{\rm{T}}}} \right\| $.利用分部积分并结合假设$ \rm2.3 $,有

(3.11) $ \begin{eqnarray} &&{\int_0^l {\Big( {\frac{{\partial \tilde \Delta {w_k}(x, t)}}{{\partial t}}} \Big)} ^{\rm{T}}}\frac{{{\partial ^2}}}{{\partial {x^2}}}\Big[ {({I_{N - 1}} \otimes EI(x))\frac{{{\partial ^2}\tilde \Delta {w_k}(x, t)}}{{\partial {x^2}}}} \Big]{\rm{d}}x{}\\ & = &\Big( {{{\Big( {\frac{{\partial \tilde \Delta {w_k}(x, t)}}{{\partial t}}} \Big)}^{\rm{T}}}\frac{\partial }{{\partial x}}\Big[ {({I_{N - 1}} \otimes EI(x))\frac{{{\partial ^2}\tilde \Delta {w_k}(x, t)}}{{\partial {x^2}}}} \Big]} \Big) \Big|_0^l{}\\ && - \int_0^l {{{\Big( {\frac{{{\partial ^2}\tilde \Delta {w_k}(x, t)}}{{\partial t\partial x}}} \Big)}^{\rm{T}}}\frac{\partial }{{\partial x}}\Big[ {\Big( {{I_{N - 1}} \otimes EI(x)} \Big)\frac{{{\partial ^2}\tilde \Delta {w_k}(x, t)}}{{\partial {x^2}}}} \Big]{\rm{d}}x} {}\\ & = & - \int_0^l {{{\Big( {\frac{{{\partial ^2}\tilde \Delta {w_k}(x, t)}}{{\partial t\partial x}}} \Big)}^{\rm{T}}}\frac{\partial }{{\partial x}}\Big[ {\Big( {{I_{N - 1}} \otimes EI(x)} \Big)\frac{{{\partial ^2}\tilde \Delta {w_k}(x, t)}}{{\partial {x^2}}}} \Big]{\rm{d}}x} {}\\ & = & { - \Big( {{{\Big( {\frac{{\partial \tilde \Delta {w_k}(x, t)}}{{\partial t\partial x}}} \Big)}^{\rm{T}}}\Big( {{I_{N - 1}} \otimes EI(x)} \Big)\frac{{{\partial ^2}\tilde \Delta {w_k}(x, t)}}{{\partial {x^2}}}} \Big)} \Big|_0^l\\ &&{\rm{ + }}\int_0^l {{{\Big( {\frac{{{\partial ^3}\tilde \Delta {w_k}(x, t)}}{{\partial t\partial {x^2}}}} \Big)}^{\rm{T}}}\Big( {{I_{N - 1}} \otimes EI(x)} \Big)\frac{{{\partial ^2}\tilde \Delta {w_k}(x, t)}}{{\partial {x^2}}}{\rm{d}}x} \\ & = & \frac{1}{2}\frac{{\rm{d}}}{{{\rm{d}}t}}\int_0^l {{{\Big( {\frac{{{\partial ^2}\tilde \Delta {w_k}(x, t)}}{{\partial {x^2}}}} \Big)}^{\rm{T}}}\Big( {{I_{N - 1}} \otimes EI(x)} \Big)\frac{{{\partial ^2}\tilde \Delta {w_k}(x, t)}}{{\partial {x^2}}}{\rm{d}}x}. \end{eqnarray} $

将$ \rm(3.9)–(3.11) $式代入$ \rm(3.8) $式,并结合$ \rm(2.2) $式,有

$ \begin{eqnarray*} &&\frac{{\rm{d}}}{{{\rm{d}}t}}\Big\{ {\int_0^l {{{\Big( {\frac{{\partial \tilde \Delta {w_k}(x, t)}}{{\partial t}}} \Big)}^{\rm{T}}}} ({I_{N - 1}} \otimes \mu (x))\frac{{\partial \tilde \Delta {w_k}(x, t)}}{{\partial t}}{\rm{d}}x}\\ && { + \int_0^l {{{\Big( {\frac{{{\partial ^2}\tilde \Delta {w_k}(x, t)}}{{\partial {x^2}}}} \Big)}^{\rm{T}}}} ({I_{N - 1}} \otimes EI(x))\frac{{{\partial ^2}\tilde \Delta {w_k}(x, t)}}{{\partial {x^2}}}{\rm{d}}x} \Big\}\\ &\le & {c_1}\Big\{ {{{\int_0^l {\left\| {\frac{{\partial \tilde \Delta {w_k}(x, t)}}{{\partial t}}} \right\|} }^2}{\rm{d}}x + \int_0^l {{{\left\| {{\delta _k}(x, t)} \right\|}^2}} {\rm{d}}x} \Big\}\\ &\le& {c_1}\Big\{ {\frac{1}{{{\mu _1}}}\int_0^l {{{\Big( {\frac{{\partial \tilde \Delta {w_k}(x, t)}}{{\partial t}}} \Big)}^{\rm{T}}}} ({I_{N - 1}} \otimes \mu (x))\frac{{\partial \tilde \Delta {w_k}(x, t)}}{{\partial t}}{\rm{d}}x + \int_0^l {{{\left\| {{\delta _k}(x, t)} \right\|}^2}} {\rm{d}}x} \Big\}\\ & \le &{c_2}\Big\{ {\int_0^l {{{\Big( {\frac{{\partial \tilde \Delta {w_k}(x, t)}}{{\partial t}}} \Big)}^{\rm{T}}}} ({I_{N - 1}} \otimes \mu (x))\frac{{\partial \tilde \Delta {w_k}(x, t)}}{{\partial t}}{\rm{d}}x}\\ &&{ + \int_0^l {{{\Big( {\frac{{{\partial ^2}\tilde \Delta {w_k}(x, t)}}{{\partial {x^2}}}} \Big)}^{\rm{T}}}({I_{N - 1}} \otimes EI(x))\frac{{{\partial ^2}\tilde \Delta {w_k}(x, t)}}{{\partial {x^2}}}} {\rm{d}}x} \Big\} + {c_1}\int_0^l {{{\left\| {{\delta _k}(x, t)} \right\|}^2}} {\rm{d}}x, \end{eqnarray*} $

这里$ {c_2} = \frac{{{c_1}}}{{{\mu _1}}} $.利用Gronwall引理并结合假设$ \rm2.2 $, $ \rm(2.2) $式和$ \rm(2.3) $式,有

$ \begin{eqnarray*} &&\int_0^l {{{\Big( {\frac{{\partial \tilde \Delta {w_k}(x, t)}}{{\partial t}}} \Big)}^{\rm{T}}}} ({I_{N - 1}} \otimes \mu (x))\frac{{\partial \tilde \Delta {w_k}(x, t)}}{{\partial t}}{\rm{d}}x\\ && + \int_0^l {{{\Big( {\frac{{{\partial ^2}\tilde \Delta {w_k}(x, t)}}{{\partial {x^2}}}} \Big)}^{\rm{T}}}} ({I_{N - 1}} \otimes EI(x))\frac{{{\partial ^2}\tilde \Delta {w_k}(x, t)}}{{\partial {x^2}}}{\rm{d}}x\\ & \le& {{\rm{e}}^{{c_2}t}} {\int_0^l {{{\left. {\Big( {{{\Big( {\frac{{\partial \tilde \Delta {w_k}(x, t)}}{{\partial t}}} \Big)}^{\rm{T}}}({I_{N - 1}} \otimes \mu (x))\frac{{\partial \tilde \Delta {w_k}(x, t)}}{{\partial t}}} \Big)} \right|} _{t = 0}}{\rm{d}}x} } \\ && {{\rm{ + }}{{\rm{e}}^{{c_2}t}}{{\int_0^l {\left. {\Big( {{{\Big( {\frac{{{\partial ^2}\tilde \Delta {w_k}(x, t)}}{ {\partial {x^2}}}} \Big)}^{\rm{T}}}({I_{N - 1}} \otimes EI(x))\frac{{{\partial ^2}\tilde \Delta {w_k}(x, t)}}{ {\partial {x^2}}}} \Big)} \right|} }_{t = 0}}{\rm{d}}x} \\ &&{\rm{ + }}{c_1}\int_0^t {{{\rm{e}}^{{c_2}(t - \eta )}}} (\int_0^l {{\left\| {{\delta _k}(x, \eta )} \right\|}^2}{\rm{d}}x) {\rm{d}}\eta \\ & \le& {{\rm{e}}^{{c_2}T}}\Big( {{\mu _2} \int_0^l {{{\left\| {{{\left. {\frac{{\partial \tilde \Delta {w_k}(x, t)}}{{\partial t}}} \right|}_{t = 0}}} \right\|}^2}} {\rm{d}}x + E{I_2}\int_0^l {{{\left\| {{{\left. {\frac{{{\partial ^2}\tilde \Delta {w_k}(x, t)}}{{\partial {x^2}}}} \right|}_ {t = 0}}} \right\|}^2}{\rm{d}}x} } \Big)\\ &&+ {c_1}{{\rm{e}}^{{c_2}T}}\int_0^t {{{\rm{e}}^{\lambda \eta }}{{\rm{e}}^{ - \lambda \eta }}\Big\{ {\int_0^l {{{\left\| {{\delta _k}(x, \eta )} \right\|}^2}{\rm{d}}x} } \Big\}{\rm{d}}\eta } \\ &\le& {{\rm{e}}^{{c_2}T}}{\mu _2}\int_0^l {{{\left\| {{{\left. {\Big( {\frac{{\partial \Delta {w_k}(x, t)}}{{\partial t}} - {{\bf{1}}_{N - 1}} \otimes \frac{{\partial \Delta {w_{1, k}}(x, t)}}{{\partial t}}} \Big)} \right|}_{t = 0}}} \right\|}^2}} {\rm{d}}x\\ && + {{\rm{e}}^{{c_2}T}}E{I_2}\int_0^l {{{\left\| {{{\left. {\Big( {\frac{{{\partial ^2}\Delta {w_k}(x, t)}} {{\partial {x^2}}} - {{\bf{1}}_{N - 1}} \otimes \frac{{{\partial ^2}\Delta {w_{1, k}}(x, t)}}{{\partial {x^2}}}} \Big)} \right|}_{t = 0}}} \right\|}^2}{\rm{d}}x} \\ && + {c_1}{{\rm{e}}^{{c_2}T}}\int_0^t {{{\rm{e}}^{\lambda \eta }}{\rm{d}}\eta } \mathop {\sup }\limits_{t \in [0, T]} \Big\{ {{{\rm{e}}^{ - \lambda t}}\int_0^l {{{\left\| {{\delta _k}(x, t)} \right\|}^2}{\rm{d}}x} } \Big\}\\ &\le& 2{{\rm{e}}^{{c_2}T}}{\mu _2}\int_0^l {\Big( {\sum\limits_{j = 2}^N {{{\Big( {{{\left. {\frac{{\partial \Delta {w_{j, k}}(x, t)}}{{\partial t}}} \right|}_{t = 0}}} \Big)}^2}} + (N - 1){{\Big( {{{\left. {\frac{{\partial \Delta {w_{1, k}}(x, t)}}{{\partial t}}} \right|}_{t = 0}}} \Big)}^2}} \Big)} {\rm{d}}x\\ && + 2{{\rm{e}}^{{c_2}T}}E{I_2}\int_0^l {\Big( {\sum\limits_{j = 2}^N {{{\Big( {{{\left. {\frac{{{\partial ^2} \Delta {w_{j, k}}(x, t)}}{{\partial {x^2}}}} \right|}_{t = 0}}} \Big)}^2}} + (N - 1){{\Big( {{{\left. {\frac{{{ \partial ^2}\Delta {w_{1, k}}(x, t)}}{{\partial {x^2}}}} \right|}_{t = 0}}} \Big)}^2}} \Big)} {\rm{d}}x\\ & &+ {c_1}{{\rm{e}}^{{c_2}T}}\frac{{{{\rm{e}}^{\lambda t}} - 1}}{\lambda }\mathop {\sup }\limits_{t \in [0, T]} \Big\{ {{{\rm{e}}^{ - \lambda t}}\int_0^l {{{\left\| {{\delta _k}(x, t)} \right\|}^2}{\rm{d}}x} } \Big\}\\ & \le& {c_1}{{\rm{e}}^{{c_2}T}}\frac{{{{\rm{e}}^{\lambda t}} - 1}}{\lambda }\mathop {\sup }\limits_{t \in [0, T]} \Big\{ {{{\rm{e}}^{ - \lambda t}}\int_0^l {{{\left\| {{\delta _k}(x, t)} \right\|}^2}{\rm{d}}x} } \Big\}{\rm{ + 4}}l{{\rm{e}}^{{c_2}T}}(N - 1)({\mu _2}\varepsilon _2^2 + E{I_2}\varepsilon _1^2). \end{eqnarray*} $

因此

(3.12) $ \begin{eqnarray} &&\mathop {\sup }\limits_{t \in [0, T]} \Big\{ {{{\rm{e}}^{ - \lambda t}}\int_0^l {{{\Big( {\frac{{\partial \tilde \Delta {w_k}(x, t)}}{{\partial t}}} \Big)}^{\rm{T}}}} ({I_{N - 1}} \otimes \mu (x))\frac{{\partial \tilde \Delta {w_k}(x, t)}}{{\partial t}}{\rm{d}}x} \Big\}\\ &\le &\mathop {\sup }\limits_{t \in [0, T]} \{ {{{\rm{e}}^{ - \lambda t}}\int_0^l {{{\Big( {\frac{{\partial \tilde \Delta {w_k}(x, t)}}{{\partial t}}} \Big)}^{\rm{T}}}} ({I_{N - 1}} \otimes \mu (x))\frac{{\partial \tilde \Delta {w_k}(x, t)}}{{\partial t}}{\rm{d}}x}\\ & &{ + {{\rm{e}}^{ - \lambda t}}\int_0^l {{{\Big( {\frac{{{\partial ^2}\tilde \Delta {w_k}(x, t)}}{{\partial {x^2}}}} \Big)}^{\rm{T}}}} ({I_{N - 1}} \otimes EI(x))\frac{{{\partial ^2}\tilde \Delta {w_k}(x, t)}}{{\partial {x^2}}}{\rm{d}}x} \Big\}\\ & \le& {c_1}{{\rm{e}}^{{c_2}T}}\mathop {\sup }\limits_{t \in [0, T]} \Big\{ {\frac{{1 - {{\rm{e}}^{ - \lambda t}}}}{\lambda }} \Big\}\mathop {\sup }\limits_{t \in [0, T]} \Big\{ {{{\rm{e}}^{ - \lambda t}}\int_0^l {{{\left\| {{\delta _k}(x, t)} \right\|}^2}{\rm{d}}x} } \Big\}\\ &&{\rm{ + }}\mathop {\sup }\limits_{t \in [0, T]} \Big\{ {{{\rm{e}}^{ - \lambda t}}} \Big\}{\rm{4}}l{{\rm{e}}^{{c_2}T}}(N - 1)({\mu _2}\varepsilon _2^2 + E{I_2}\varepsilon _1^2)\\ &\le& {c_1}{{\rm{e}}^{{c_2}T}}\frac{{1 - {{\rm{e}}^{ - \lambda T}}}}{\lambda }\mathop {\sup }\limits_{t \in [0, T]} \Big\{ {{{\rm{e}}^{ - \lambda t}}\int_0^l {{{\left\| {{\delta _k}(x, t)} \right\|}^2}{\rm{d}}x} } \Big\} {\rm{ + }}4l{{\rm{e}}^{{c_2}T}}(N - 1)({\mu _2}\varepsilon _2^2 + E{I_2}\varepsilon _1^2). {\qquad} \end{eqnarray} $

另一方面,应用基本不等式,有

$ \begin{eqnarray*} \frac{{\rm{d}}}{{{\rm{d}}t}}\int_0^l {{{\left\| {\tilde \Delta {w_k}(x, t)} \right\|}^2}{\rm{d}}x} & = & 2\int_0^l {{{(\tilde \Delta {w_k}(x, t))}^{\rm{T}}}\frac{{\partial \tilde \Delta {w_k}(x, t)}}{{\partial t}}{\rm{d}}x} \\ &\le& \int_0^l {{{\left\| {\tilde \Delta {w_k}(x, t)} \right\|}^2}{\rm{d}}x} + \int_0^l {{{\left\| {\frac{{\partial \tilde \Delta {w_k}(x, t)}}{{\partial t}}} \right\|}^2}{\rm{d}}x} . \end{eqnarray*} $

使用Gronwall引理并结合假设$ \rm2.2 $,可得

$ \begin{eqnarray*} \int_0^l {{{\left\| {\tilde \Delta {w_k}(x, t)} \right\|}^2}{\rm{d}}x} & \le& \int_0^t {\Big\{ {{{\rm e}^{t - \eta }}\int_0^l {{{\left\| {\frac{{\partial \tilde \Delta {w_k}(x, \eta )}}{{\partial \eta }}} \right\|}^2}{\rm{d}}x} } \Big\}{\rm{d}}\eta } + {{\rm e}^t}\int_0^l {{{\left\| {\tilde \Delta {w_k}(x, 0)} \right\|}^2}{\rm{d}}x} \\ &\le &{{\rm e}^t}\int_0^t {\Big\{ {{{\rm e}^{(\lambda - 1)\eta }}{{\rm e}^{ - \lambda \eta }}\int_0^l {{{\left\| {\frac{{\partial \tilde \Delta {w_k}(x, \eta )}}{{\partial \eta }}} \right\|}^2}{\rm{d}}x} } \Big\}{\rm{d}}\eta } \\ &&+ {{\rm e}^T}\int_0^l {{{\left\| {\Delta {w_k}(x, 0) - {1_{N - 1}} \otimes \Delta {w_{1, k}}(x, 0)} \right\|}^2}{\rm{d}}x} \\ &\le & {{\rm e}^t}\int_0^t {{{\rm e}^{(\lambda - 1)\eta }}{\rm{d}}\eta \mathop {\sup }\limits_{t \in [0, T]} \Big\{ {{{\rm e}^{ - \lambda t}}\int_0^l {{{\left\| {\frac{{\partial \tilde \Delta {w_k}(x, t)}}{{\partial t}}} \right\|}^2}{\rm{d}}x} } \Big\}} \\ & &+ 2{{\rm e}^T}\int_0^l {\Big\{ {\sum\limits_{j = 2}^N {{{(\Delta {w_{j, k}}(x, 0))}^2}} + (N - 1){{(\Delta {w_{1, k}}(x, 0))}^2}} \Big\}{\rm{d}}x} \\ & = & \frac{{{{\rm e}^{\lambda t}} - {{\rm e}^t}}}{{\lambda - 1}}\mathop {\sup }\limits_{t \in [0, T]} \Big\{ {{{\rm e}^{ - \lambda t}}\int_0^l {{{\left\| {\frac{{\partial \tilde \Delta {w_k}(x, t)}}{{\partial t}}} \right\|}^2}{\rm{d}}x} } \Big\} + 4l{{\rm e}^T}(N - 1)\varepsilon _1^2. \end{eqnarray*} $

取$ k > 1 $,结合$ \rm(2.2) $式,有

(3.13) $ \begin{eqnarray} &&\mathop {\sup }\limits_{t \in [0, T]} \Big\{ {{{\rm e}^{ - \lambda t}}\int_0^l {{{\left\| {\tilde \Delta {w_k}(x, t)} \right\|}^2}{\rm{d}}x} } \Big\}\\ & \le& \mathop {\sup }\limits_{t \in [0, T]} \Big\{ {\frac{{1 - {{\rm{e}}^{ - (\lambda - 1)t}}}}{{\lambda - 1}}} \Big\}\mathop {\sup }\limits_{t \in [0, T]} \Big\{ {{{\rm{e}}^{ - \lambda t}}\int_0^l {{{\left\| {\frac{{\partial \tilde \Delta {w_k}(x, t)}}{{\partial t}}} \right\|}^2}{\rm{d}}x} } \Big\} + \mathop {\sup }\limits_{t \in [0, T]} \Big\{ {{{\rm{e}}^{ - \lambda t}}} \Big\}4l{{\rm{e}}^T}(N - 1)\varepsilon _1^2\\ & = & \frac{{1 - {{\rm{e}}^{ - (\lambda - 1)T}}}}{{\lambda - 1}}\mathop {\sup }\limits_{t \in [0, T]} \Big\{ {{{\rm{e}}^{ - \lambda t}}\int_0^l {{{\left\| {\frac{{\partial \tilde \Delta {w_k}(x, t)}}{{\partial t}}} \right\|}^2}{\rm{d}}x} } \Big\} + 4l{{\rm{e}}^T}(N - 1)\varepsilon _1^2\\ &\le& \frac{{1 - {{\rm{e}}^{ - (\lambda - 1)T}}}}{{\lambda - 1}}\frac{1}{{{\mu _1}}}\mathop {\sup }\limits_{t \in [0, T]} \Big\{ {{{\rm{e}}^{ - \lambda t}}\int_0^l {{{\Big( {\frac{{\partial \tilde \Delta {w_k}(x, t)}}{{\partial t}}} \Big)}^{\rm{T}}}\Big( {{I_{N - 1}} \otimes \mu (x)} \Big)\frac{{\partial \tilde \Delta {w_k}(x, t)}}{{\partial t}}{\rm{d}}x} } \Big\}\\ && + 4l{{\rm e}^T}(N - 1)\varepsilon _1^2. \end{eqnarray} $

将$ \rm(3.12) $式代入$ \rm(3.13) $式,可得

(3.14) $ \begin{eqnarray} &&\mathop {\sup }\limits_{t \in [0, T]} \Big\{ {{{\rm{e}}^{ - \lambda t}}\int_0^l {{{\left\| {\tilde \Delta {w_k}(x, t)} \right\|}^2}{\rm{d}}x} } \Big\}\\ &\le& \frac{{1 - {{\rm{e}}^{ - (\lambda - 1)T}}}}{{\lambda - 1}}\frac{1}{{{\mu _1}}}\Big\{ {{c_1}{{\rm{e}}^{{c_2}T}}\frac{{1 - {{\rm{e}}^{ - \lambda T}}}}{\lambda }\mathop {\sup }\limits_{t \in [0, T]} \Big\{ {{{\rm{e}}^{ - \lambda t}}\int_0^1 {{{\left\| {{\delta _k}(x, t)} \right\|}^2}{\rm{d}}x} } \Big\}} \Big\}\\ &&+ \frac{{1 - {{\rm{e}}^{ - (\lambda - 1)T}}}}{{\lambda - 1}}\frac{1}{{{\mu _1}}}4l{{\rm{e}}^{{c_2}T}}(N - 1)({\mu _2}\varepsilon _2^2 + E{I_2}\varepsilon _1^2) + 4l{{\rm{e}}^T}(N - 1)\varepsilon _1^2. \end{eqnarray} $

由于$ {{\mathbb R}^{N - 1}} $中所有的范数都是等价的,所以存在两个正数$ \tau $和$ \sigma $使得

(3.15) $ \begin{equation} \left\| {\tilde \Delta {w_k}(x, t)} \right\| \ge \tau {\left\| {\tilde \Delta {w_k}(x, t)} \right\|_ * }, \; \; \left\| {{\delta _k}(x, t)} \right\| \le \sigma {\left\| {{\delta _k}(x, t)} \right\|_ * }. \end{equation} $

因此,结合$ \rm(3.14) $式和$ \rm(3.15) $式,有

(3.16) $ \begin{eqnarray} &&\mathop {\sup }\limits_{t \in [0, T]} \Big\{ {{{\rm{e}}^{ - \lambda t}}\int_0^l {\left\| {\tilde \Delta {w_k}(x, t)} \right\|_{\rm{*}}^2{\rm{d}}x} } \Big\} {}\\ & \le& \mathop {\frac{1}{{{\tau ^2}}}\sup }\limits_{t \in [0, T]} \Big\{ {{{\rm{e}}^{ - \lambda t}}\int_0^l {{{\left\| {\tilde \Delta {w_k}(x, t)} \right\|}^2}{\rm{d}}x} } \Big\} {}\\ &\le &\frac{{{\sigma ^2}}}{{{\tau ^2}}}\frac{{1 - {{\rm{e}}^{ - (\lambda - 1)T}}}}{{\lambda - 1}}\frac{1}{{{\mu _1}}}\Big\{ {{c_1}{{\rm{e}}^{{c_2}T}}\frac{{1 - {{\rm{e}}^{ - \lambda T}}}}{\lambda }\mathop {\sup }\limits_{t \in [0, T]} \Big\{ {{{\rm{e}}^{ - \lambda t}}\int_0^l {\left\| {{\delta _k}(x, t)} \right\|_*^2{\rm{d}}x} } \Big\}} \Big\}\\ &&+ \frac{1}{{{\tau ^2}}}\frac{{1 - {{\rm{e}}^{ - (\lambda - 1)T}}}}{{\lambda - 1}}\frac{1}{{{\mu _1}}}4l{{\rm{e}}^{{c_2}T}}(N - 1)({\mu _2}\varepsilon _2^2 + E{I_2}\varepsilon _1^2) + \frac{1}{{{\tau ^2}}}4l{{\rm{e}}^T}(N - 1)\varepsilon _1^2. \end{eqnarray} $

将$ \rm(3.16) $式代入$ \rm(3.7) $式,有

$ \begin{eqnarray*} &&\mathop {\sup }\limits_{t \in [0, T]} \Big\{ {\int_0^l {{\rm e^{ - \lambda t}}\left\| {{\delta _{k + 1}}(x, t)} \right\|_ * ^2{\rm{d}}x} } \Big\}\\ &\le& \Big\{ {(1 + \varepsilon ){{\bar \rho }^2} + (1 + \frac{1}{\varepsilon }){F^2}\frac{{{\sigma ^2}}}{{{\tau ^2}}}\frac{{1 - {{\rm{e}}^{ - (\lambda - 1)T}}}}{{\lambda - 1}}\frac{1}{{{\mu _1}}}\frac{{1 - {{\rm{e}}^{ - \lambda T}}}}{\lambda }{c_1}{{\rm{e}}^{{c_2}T}}} \Big\}\mathop {\mathop {{\rm{sup}}}\limits_{t \in [0, T]} \Big\{ {\int_0^l {{{\rm{e}}^{ - \lambda t}}\left\| {{\delta _k}(x, t)} \right\|_{\rm{*}}^2{\rm{d}}x} } \Big\}}\limits_{} \\ & &+ (1 + \frac{1}{\varepsilon }){F^2}\frac{1}{{{\tau ^2}}}4l(N - 1)\Big( {\frac{{1 - {{\rm{e}}^{ - (\lambda - 1)T}}}}{{\lambda - 1}}\frac{1}{{{\mu _1}}}{{\rm{e}}^{{c_2}T}}({\mu _2}\varepsilon _2^2 + E{I_2}\varepsilon _1^2) + {{\rm{e}}^T}\varepsilon _1^2} \Big)\\ & = & \mathord{\buildrel{\lower3pt\hbox{$ \frown$}} \over \rho } \mathop {\sup }\limits_{t \in [0, T]} \Big\{ {\int_0^l {{{\rm{e}}^{ - \lambda t}}\left\| {{\delta _k}(x, t)} \right\|_{\rm{*}}^2{\rm{d}}x} } \Big\}\\ &&+ (1 + \frac{1}{\varepsilon }){F^2}\frac{1}{{{\tau ^2}}}4l(N - 1)\Big( {\frac{{1 - {{\rm e}^{ - (\lambda - 1)T}}}}{{\lambda - 1}}\frac{1}{{{\mu _1}}}{{\rm e}^{{c_2}T}}({\mu _2}\varepsilon _2^2 + E{I_2}\varepsilon _1^2) + {{\rm e}^T}\varepsilon _1^2} \Big), \end{eqnarray*} $

这里

$ \mathord{\buildrel{\lower3pt\hbox{$ \frown$}} \over \rho } {\rm{ = }}(1 + \varepsilon ){\bar \rho ^2} + (1 + \frac{1}{\varepsilon }){F^2}\frac{{{\sigma ^2}}}{{{\tau ^2}}}\frac{{1 - {{\rm{e}}^{ - (\lambda - 1)T}}}}{{\lambda - 1}}\frac{1}{{{\mu _2}}}\frac{{1 - {{\rm{e}}^{ - \lambda T}}}}{\lambda }{c_1}{{\rm{e}}^{{c_2}T}}. $

由$ \rm(3.6) $式可知$ 0 \le (1 + \varepsilon ){\bar \rho ^2} < 1 $,所以能选取足够大的$ \lambda $使得$ \mathord{\buildrel{\lower3pt\hbox{$ \frown$}} \over \rho } {\rm{ <1 }} $.根据引理$ \rm2.2 $,即得

$ \begin{eqnarray*} &&\mathop {\lim \sup }\limits_{k \to \infty } \Big\{ {\mathop {\sup }\limits_{t \in [0, T]} \Big\{ {\int_0^l {{{\rm{e}}^{ - \lambda t}}\left\| {{\delta _k}(x, t)} \right\|_{\rm{*}}^2{\rm{d}}x} } \Big\}} \Big\}\\ &\le& \frac{1}{{1 - \mathord{\buildrel{\lower3pt\hbox{$ \frown$}} \over \rho } }}\Big( {1 + \frac{1}{\varepsilon }} \Big){F^2}\frac{1}{{{\tau ^2}}}4l(N - 1)\Big( {\frac{{1 - {{\rm{e}}^{ - (\lambda - 1)T}}}}{{\lambda - 1}}\frac{1}{{{\mu _1}}}{{\rm{e}}^{{c_2}T}}({\mu _2}\varepsilon _2^2 + E{I_2}\varepsilon _1^2) + {{\rm{e}}^T}\varepsilon _1^2} \Big), \end{eqnarray*} $

进一步有

$ \begin{eqnarray*} &&\mathop {\lim \sup }\limits_{k \to \infty } \Big\{ {\mathop {\sup }\limits_{t \in [0, T]} \Big\{ {\int_0^l {\left\| {{\delta _k}(x, t)} \right\|_ * ^2{\rm{d}}x} } \Big\}} \Big\}\\ &\le& \mathop {\lim \sup }\limits_{k \to \infty } \Big\{ {\mathop {\sup }\limits_{t \in [0, T]} \Big\{ {\int_0^l {{{\rm{e}}^{\lambda T}}{{\rm{e}}^{ - \lambda t}}\left\| {{\delta _k}(x, t)} \right\|_ * ^2{\rm d}x} } \Big\}} \Big\}\\ &\le& {{\rm{e}}^{\lambda T}}\frac{1}{{1 - \mathord{\buildrel{\lower3pt\hbox{$ \frown$}} \over \rho } }}\Big( {1 + \frac{1}{\varepsilon }} \Big){F^2}\frac{1}{{{\tau ^2}}}4l(N - 1)\Big( {\frac{{1 - {{\rm{e}}^{ - (\lambda - 1)T}}}}{{\lambda - 1}}\frac{1}{{{\mu _1}}}{{\rm{e}}^{{c_2}T}}({\mu _2}\varepsilon _2^2 + E{I_2}\varepsilon _1^2) + {{\rm{e}}^T}\varepsilon _1^2} \Big). \end{eqnarray*} $

再由$ \rm(3.15) $式可得

$ \begin{eqnarray*} \mathop {\lim \sup }\limits_{k \to \infty } {\left\| {{\delta _k}} \right\|_{{L^2}, s}} & = & \mathop {\lim \sup }\limits_{k \to \infty } \Big\{ {\mathop {\sup }\limits_{t \in [0, T]} \Big\{ {\int_0^l {\left\| {{\delta _k}(x, t)} \right\|_2^2{\rm{d}}x} } \Big\}} \Big\}\\ &\le& {\sigma ^2}\mathop {\lim \sup }\limits_{k \to \infty } \Big\{ {\mathop {\sup }\limits_{t \in [0, T]} \Big\{ {\int_0^l {\left\| {{\delta _k}(x, t)} \right\|_ * ^2{\rm{d}}x} } \Big\}} \Big\}\\ &\le& {\sigma ^2}{{\rm{e}}^{\lambda T}}\frac{1}{{1 - \mathord{\buildrel{\lower3pt\hbox{$ \frown$}} \over \rho } }}\Big( {1 + \frac{1}{\varepsilon }} \Big){F^2}\frac{1}{{{\tau ^2}}}4l(N - 1) {\nonumber}\\ &&\times\Big( {\frac{{1 - {{\rm{e}}^{ - (\lambda - 1)T}}}}{{\lambda - 1}}\frac{1}{{{\mu _1}}}{{\rm{e}}^{{c_2}T}}({\mu _2}\varepsilon _2^2 + E{I_2}\varepsilon _1^2) + {{\rm{e}}^T}\varepsilon _1^2} \Big). \end{eqnarray*} $

于是

$ \begin{eqnarray*} \mathop {\lim \sup }\limits_{k \to \infty } {\left\| {{y_{i, k}} - {y_{j, k}}} \right\|_{{L^2}, s}} & = & \mathop {\lim \sup }\limits_{k \to \infty } {\left\| {{y_{i, k}} - {y_{1, k}} - ({y_{j, k}} - {y_{1, k}})} \right\|_{{L^2}, s}}\\ & = & \mathop {\lim \sup }\limits_{k \to \infty } {\left\| {{\delta _{i, k}} - {\delta _{j, k}}} \right\|_{{L^2}, s}} \\ &\le &2\mathop {\lim \sup }\limits_{k \to \infty } \Big\{ {{{\left\| {{\delta _{i, k}}} \right\|}_{{L^2}, s}} + {{\left\| {{\delta _{j, k}}} \right\|}_{{L^2}, s}}} \Big\} \\ &\le &4\mathop {\lim \sup }\limits_{k \to \infty } {\left\| {{\delta _k}} \right\|_{{L^2}, s}}\\ &\le& 4{\sigma ^2}{{\rm{e}}^{\lambda T}}\frac{1}{{1 - \mathord{\buildrel{\lower3pt\hbox{$ \frown$}} \over \rho } }}\Big( {1 + \frac{1}{\varepsilon }} \Big){F^2}\frac{1}{{{\tau ^2}}}4l(N - 1)\\ &&\times \Big( {\frac{{1 - {{\rm{e}}^{ - (\lambda - 1)T}}}}{{\lambda - 1}}\frac{1}{{{\mu _1}}}{{\rm{e}} ^{{c_2}T}}({\mu _2}\varepsilon _2^2 + E{I_2}\varepsilon _1^2) + {{\rm{e}}^T}\varepsilon _1^2} \Big), \\ &&{\qquad} \forall i, j \in \{ 1, 2, \cdots , N\} . \end{eqnarray*} $

由此可知,任两个智能体之间于$ {L^2} $空间中的一致性误差有界;进一步,若

$ {\left\| {{{\left. {{w_{i, k + 1}}( \cdot , t)} \right|}_{t = 0}} - {{\left. {{w_{i, k}}( \cdot , t)} \right|}_{t = 0}}} \right\|_{{H^2}}} = 0, \; \; {\left\| {{{\left. {\frac{{\partial {w_{i, k + 1}}( \cdot , t)}}{{\partial t}}} \right|}_{t = 0}} - {{\left. {\frac{{\partial {w_{i, k}}( \cdot , t)}}{{\partial t}}} \right|}_{t = 0}}} \right\|_{{L^2}}}{\rm{ = }}0, $

$ x \in [0, l], \; \; \; k = 0, 1, 2, \cdots, $

即$ {\varepsilon _i}{\rm{ = }}0 $, $ i = 1, 2 $,我们可以得

$ \mathop {\lim }\limits_{k \to \infty } {\left\| {{y_{i, k}} - {y_{j, k}}} \right\|_{{L^2}, s}} = 0, \; \; \; \forall i, j \in \Big\{ {1, 2 \cdots , N} \Big\}. $

定理3.1证毕.

考虑由四阶梁方程构建而成的如下四个智能体($ \mu (x) = EI(x) \equiv 1 $, $ F = B = 1 $)

$ \left\{ \begin{array}{l} { } \frac{{{\partial ^2}{w_{i, k}}(x, t)}}{{\partial {t^2}}} + \frac{{{\partial ^4}{w_{i, k}}(x, t)}}{{\partial {x^4}}} = {u_{i, k}}(x, t), \\ {y_{i, k}}(x, t) = {w_{i, k}}(x, t) + {u_{i, k}}(x, t), \end{array}\right. i = 1, 2, 3, 4, $

$ (x, t) \in (0, 1) \times [0, 0.5] $,其网络拓扑结构如图 1,则邻接矩阵为

$ A = \left[ {\begin{array}{*{20}{c}} 0&{\quad} 0{\quad} &0&{\quad} 1\\ 1&0&0&{\quad} 0\\ 0&1&0&{\quad} 0\\ 0&0&1&{\quad} 0 \end{array}} \right], $

显然对应的图$ G $存在生成树.当$ 0 < g < 1 $,能推得(3.2)式成立.由此,取$ g = 0.5 $,设置$ k $次迭代时的初、边值条件为: $ {w_{i, k}}(x, 0) = (1 + {\varepsilon _{1, k}}) \times 0.0001 \times i \times {x^2}{(1 - x)^2} $, $ \frac{{\partial {w_{i, k}}(x, 0)}}{{\partial t}} = {\varepsilon _{2, k}} $, $ x \in [0, 1] $; $ {w_{i, k}}(0, t) = {w_{i, k}}(1, t) = {\left. {\frac{{\partial {w_{i, k}}(x, t)}}{{\partial x}}} \right|_{x = 0}} = {\left. {\frac{{\partial {w_{i, k}}(x, t)}}{{\partial t}}} \right|_{t = 0}} \equiv 0 $, $ t \in [0, T] $; $ i = 1, 2, 3, 4 $.设置初始控制为: $ {u_{1, 0}}(x, t) = {u_{2, 0}}(x, t) $$ = {u_{3, 0}}(x, t) = {u_{4, 0}}(x, t) \equiv 0 $.在学习律(3.1)的作用下,有:当$ {\varepsilon _{1, k}} $, $ {\varepsilon _{2, k}} $于$ \Big[ { - 5 \times {{10}^{ - 9}}, 5 \times {{10}^{ - 9}}} \Big] $上随机取值时, $ {\left\| {{\delta _{i, k}}} \right\|_{{L^2}, s}} $则的仿真结果如图 2,从图 2可看出,系统的一致性误差于$ {L^2} $空间中有界;当$ {\varepsilon _{1, k}} = {\varepsilon _{2, k}} = 0 $时,则$ {\left\| {{\delta _{i, k}}} \right\|_{{L^2}, s}} $的仿真结果如图 3,从图 3可看出,系统的一致性误差随着迭代学习次数的增加而趋于零.

本文研究了一类四阶分布参数多智能体系统的迭代学习控制算法,构建了基于一致性的P型学习律.当P型学习律作用于系统时,任两个智能体之间于$ {L^2} $空间中的一致性误差有界,进一步,当无初始偏差时,任两个智能体之间的一致性误差能够沿着迭代轴方向收敛于零.仿真结果验证了理论分析.关于分布参数系统的控制设计,还有一类是边界控制[34],今后将进一步研究高阶偏微分多智能体系统的边界迭代学习控制问题.