迭代学习控制由 Arimoto 等人 ^[1]首次提出完整的控制算法后, 已成为近年来控制理论研究的热点问题, 并引起人们的广泛关注. 在迭代学习控制设计中, 采用较多的是 D 型学习律 ^[1-4] 和 P 型学习律 ^[5-9], 根据系统所满足的性质, 在重复受控时间内, 用 D 型学习律或 P型学习律进行控制设计. 事实上, 在不同的受控时间段内, 系统所满足的性质可能是不同的,也就是说, 可能在一个时间段内, 系统更适合用 P 型学习律, 在另一个时间段内, 系统更适合用 D 型学习律. 如何对这类系统进行迭代学习控制设计, 并进行相应的分段控制设计, 据笔者所知, 尚无相关的研究论文.

文[8]针对一类具有最一般形式的非线性系统, 采用P型与Newton型迭代学习控制律相结合的方法, 对此类系统进行了控制设计; 文[9]基于收敛性质分析, 对文[8]中的P型迭代学习控制律, 进一步作了鲁棒优化设计, 并得到最优的学习增益取值. 对具有最一般形式的非线性系统, 如何进行控制设计, 始终是非线性系统控制理论研究深感兴趣的内容.

本文提出分段迭代学习控制的概念, 并对文[9]中的系统进行分段迭代学习控制设计. 根据系统在不同时间段内所满足的不同假设条件, 采用不同的学习控制律, 最终组合成整个重复受控时间内的学习控制律, 当该学习控制律作用于系统时, 系统的输出跟踪误差沿迭代轴方向收敛.

本文给出如下符号约定: 对矩阵$A\in {{R}^{n\times n}}$, 记$\left\| A \right\|$为矩阵$A$的2范数, 即$\left\| A \right\|=\sqrt{\rho ({{A}^{\mathrm{T}}}A)}$, 其中$\rho ({{A}^{\mathrm{T}}}A)$为矩阵${{A}^{\mathrm{T}}}A$的谱半径; 对向量${x}(t)\in {{R}^{n}}$, $t\in [\tilde{T}, T]$, $0\le \tilde{T}<T$, 定义${x}(t)$的上确界范数${{\left\| {x}(t) \right\|}_{s}}={\sup\limits_{t\in [\tilde{T}, T]}}\left\| {x}(t) \right\|$, 其中$\left\| {x}(t) \right\|$ 为${x}(t)$ 的2范数; 对给定的$\lambda >0$, 定义${x}(t)$的$\lambda $ -范数${{\left\| {x}(t) \right\|}_{\lambda }}={\sup\limits_{t\in [\tilde{T},T]}}{{\mathrm{e}}^{-\lambda (t-\tilde{T})}}\left\| {x}(t) \right\|$. 由文[8]可知, 范数${{\left\| {x}(t) \right\|}_{s}}$与${{\left\| {x}(t) \right\|}_{\lambda }}$是等价的, 即可用其中任一种范数来证明收敛性结果.

考虑如下形式的单输入 - 单输出非线性系统 ^[9]:

$\left\{ _{y(t)=g(x(t),u(t),t),}^{\overset{.}{\mathop{x}}\,(t)=f(x(t),u(t),t),} \right.$

(2.1)

这里$t\in \left[ 0,T \right],x(t)\in {{R}^{n}},u(t)\in R,y(t)\in R$分别是系统的状态, 控制输入和输出.

设系统(2.1)的输出$y(t)$在不同的时间段内有不同的表达形式:

$y(t)=\left\{ \begin{aligned} & g({x}(t),u(t),t),&0\le t\le \hat{T}, \\ & g({x}(t),u(\hat{T}),t)+(t-\hat{T}){{\left. \frac{\partial g}{\partial u} \right|}_{t={{{\hat{T}}}^{-}}}}\dot{u}({{{\hat{T}}}^{-}}),&\hat{T}&<t\le T, \\ \end{aligned} \right.$

这里$0<\hat{T}<T$, 而${{\left. \frac{\partial g}{\partial u} \right|}_{t={{{\hat{T}}}^{-}}}}, \dot{u}({{\hat{T}}^{-}})$ 表示函数在$t=\hat{T}$处的左导数.

注 1 在$\hat{T}<t\le T$内, $y(t)$的形式保证了$y(t)$在$[0,T]$内的连续性和可导性.

由此系统(2.1)化为

$\begin{equation} \label{eq:2} \left\{ \begin{aligned} \dot{{x}}(t)&={f}({x}(t),u(t),t),&0\le t\le T, \\ y(t)&=g({x}(t),u(t),t),&0\le t\le \hat{T}, \\ y(t)&=g({x}(t),u(\hat{T}),t)+(t-\hat{T}){{\left. \frac{\partial g}{\partial u} \right|}_{t={{{\hat{T}}}^{-}}}}\dot{u}({{{\hat{T}}}^{-}}),&\hat{T}&<t\le T. \\ \end{aligned} \right. \end{equation}$

(2.2)

参照文[8], 记${{\Omega }_{1}}\triangleq D\times U\times [0,\hat{T}]$, ${{\Omega }_{2}}\triangleq D\times U\times (\hat{T},T]$, 其中$D,U$分别为$R^{n}$和$R$中的紧集. 对系统(2.2), 给出如下假设条件:

假设 1 对于给定的初值${x}(0)$及控制输入$u(t)$, 系统(2.2)的解${x}(t),y(t)$存在 唯一, 而${f}({x}(t),u(t),t)$, $g({x}(t),u(t),t)$是足够光滑函数^[10].

假设 2 对于给定的理想轨迹${{y}_{r}}(t)\in {{C}^{1}}[0,T]$, 存在唯一的控制输入${{u}_{r}}(t)\in C[0,T]\cap {{C}^{1}}[0,\hat{T}]$, 使得

$\left\{ \begin{aligned} {{\dot{x}}_{r}}(t)&={f}({{x}_{r}}(t),{{u}_{r}}(t),t),&0\le t\le T, \\ {{y}_{r}}(t)&=g({{{x}}_{r}}(t),{{u}_{r}}(t),t),&0\le t\le \hat{T}, \\ {{y}_{r}}(t)&=g({{x}_{r}}(t),{{u}_{r}}(\hat{T}),t)+(t-\hat{T}){{\left. \frac{\partial g}{\partial {{u}_{r}}} \right|}_{t={{{\hat{T}}}^{-}}}}{{{\dot{u}}}_{r}}({{{\hat{T}}}^{-}}),&\hat{T}&<t\le T \\ \end{aligned} \right.$

成立.

假设 3 系统的初始定位条件为 ${{{x}}_{k}}(0)={{{x}}_{r}}(0)$, $k=0,1,2,\cdots $.

设动态系统(2.2)在有限区间$t\in [0,T]$内是可重复的, 在迭代学习过程中, 重写系统(2.2)为

$\begin{equation} \label{eq:3} \left\{ \begin{aligned} {{{\dot{{x}}}}_{k}}(t)&={f}({{{x}}_{k}}(t),{{u}_{k}}(t),t),&0\le t\le T, \\ {{y}_{k}}(t)&=g({{{x}}_{k}}(t),{{u}_{k}}(t),t),&0\le t\le \hat{T}, \\ {{y}_{k}}(t)&=g({{{x}}_{k}}(t),{{u}_{k}}(\hat{T}),t)+(t-\hat{T}){{\left. \frac{\partial g}{\partial {{u}_{k}}} \right|}_{t={{{\hat{T}}}^{-}}}}{{{\dot{u}}}_{k}}({{{\hat{T}}}^{-}}),&\hat{T}&<t\le T. \\ \end{aligned} \right. \end{equation}$

(2.3)

学习控制的目的是寻找适当的学习律, 使得迭代学习序列${{y}_{k}}(t)$一致收敛于理想的输出${{y}_{r}}(t)$, 即

$\mathop {\lim }\limits_{k \to \infty } {\mkern 1mu} {\left\| {{e_k}(t)} \right\|_s} = 0,$

其中${{e}_{k}}(t)={{y}_{r}}(t)-{{y}_{k}}(t)$.

设系统(2.2)在不同时间段内满足不同的假设条件, 在$t\in [0,\hat{T}]$内, 有

假设 4 非线性函数${f}({x},u,t)$在${{\Omega }_{1}}$内对变量${x},u$是全局Lipschitz连续的, 即

$\left\| {f}({{{x}}_{1}},{{u}_{1}},t)-{f}({{{x}}_{2}},{{u}_{2}},t) \right\|\le {{L}_{f}}(\left\| {{{x}}_{1}}-{{{x}}_{2}} \right\|+\left| {{u}_{1}}-{{u}_{2}} \right|),$

其中${{L}_{f}}$是未知的Lipschitz常数.

假设 5 $\forall ({x},u,t)\in {{\Omega }_{1}}$, 有$0<{{\alpha }_{1}}\le \frac{\partial g}{\partial u}\le {{\alpha }_{2}},\left\| {{\left( \frac{\partial g}{\partial {x}} \right)}^{\mathrm{T}}} \right\|\le {{\beta }_{1}},\left\| {{\left( \frac{{{\partial }^{2}}g}{\partial u\partial {x}} \right)}^{\mathrm{T}}} \right\|\le {{\beta }_{2}}, \left| \frac{{{\partial }^{2}}g}{\partial {{u}^{2}}} \right|\le {{\beta }_{3}}$成立, 其中${{\alpha }_{1}},{{\alpha }_{2}}$是已知的常数, ${{\beta }_{1}},{{\beta }_{2}},{{\beta }_{3}}$是未知的常数. 记$D=\left\{ \left. a \right|{{\alpha }_{1}}\le a\le {{\alpha }_{2}} \right\}$.

注 2 假设4, 5使得系统(2.2)在$t\in [0,\hat{T}]$内适合用P型学习律.

在$t\in (\hat{T},T]$内, 对系统(2.2)提出如下的假设条件:

假设 6 $\forall ({x},u,t)\in {{\Omega }_{2}},i=1,2,\cdots ,n$, 有$\left\| {f} \right\|\le {{\beta }_{4}}, \left\| \frac{\partial {f}}{\partial {x}} \right\|\le {{\beta }_{5}}, 0<{{\alpha }_{i1}}\le {{\left( \frac{\partial {f}}{\partial u} \right)}_{i}}\le {{\alpha }_{i2}}$成立, 其中${{\alpha }_{i1}},{{\alpha }_{i2}}$ 是已知的常数, ${{\beta }_{4}},{{\beta }_{5}}$是未知的常数, 而${{\left( \frac{\partial {f}}{\partial u} \right)}_{i}}$表示向量$\frac{\partial {f}}{\partial u}$的第$i$个分量.记$\bar{D}=\left\{ \left. ({{a}_{1}},{{a}_{2}},\cdots ,{{a}_{n}}) \right|{{\alpha }_{i1}}\le {{a}_{i}}\le {{\alpha }_{i2}},i=1,2,\cdots ,n \right\}$.

假设 7 $\forall ({x},u(\hat{T}),t)\in {{\Omega }_{2}},j=1,2,\cdots ,n$, 有$\left| \frac{{{\partial }^{2}}g}{\partial t\partial u(\hat{T})} \right|\le {{\beta }_{6}},\left\| \frac{{{\partial }^{2}}g}{\partial {{{x}}^{2}}} \right\|\le {{\beta }_{7}},\left\| {{\left( \frac{{{\partial }^{2}}g}{\partial t\partial {x}} \right)}^{\mathrm{T}}} \right\|\le {{\beta }_{8}},\left\| {{\left( \frac{{{\partial }^{2}}g}{\partial {x}\partial u(\hat{T})} \right)}^{\mathrm{T}}} \right\|\le {{\beta }_{9}},0<{{\gamma }_{j1}}\le \left( \frac{\partial g}{\partial {x}} \right)_{j}^{\mathrm{T}}\le {{\gamma }_{j2}}$ 成立, 其中${{\gamma }_{j1}},{{\gamma }_{j2}}$是已知的常数, ${{\beta }_{6}}, {{\beta }_{7}}, {{\beta }_{8}}, {{\beta }_{9}}$是未知的常数, 而$\left( \frac{\partial g}{\partial {x}} \right)_{j}^{\mathrm{T}}$表示向量${{\left( \frac{\partial g}{\partial {x}} \right)}^{\mathrm{T}}}$的第$j$个分量. 记

$\hat{D}={\left\{ \left. ({{{\hat{a}}}_{1}},{{{\hat{a}}}_{2}},\cdots,{{{\hat{a}}}_{n}}) \right|{{\gamma }_{j1}}\le {{{\hat{a}}}_{j}}\le {{\gamma }_{j2}},j=1,2,\cdots ,n \right\}}.$

注 3 假设6, 7使得系统(2.2)在$t\in [\hat{T},T]$内适合用D型学习律.

引理 1^[11]设$\left\{ {{a}_{k}} \right\}$, $\left\{ {{b}_{k}} \right\}$是满足

${{a}_{k+1}}\le \rho {{a}_{k}}+{{b}_{k}}, 0\le \rho <1$

的非负实数列, 如有$\mathop {\lim }\limits_{k \to \infty } {\mkern 1mu} {b_k} = 0$, 则有$\mathop {\lim }\limits_{k \to \infty } {\mkern 1mu} {a_k} = 0$.

对系统(2.2)构建分段学习律

$\begin{equation} \label{eq:4} \left\{ \begin{aligned} {{u}_{k+1}}(t)&={{u}_{k}}(t)+q{{e}_{k}}(t),&t\in [0,\hat{T}], \\ {{u}_{k+1}}(t)&={{u}_{k}}(t)+p{{{\dot{e}}}_{k}}(t)+q{{e}_{k}}(\hat{T})-p{{{\dot{e}}}_{k}}(\hat{T}),&t\in (\hat{T},T], \end{aligned} \right. \end{equation}$

(3.1)

其中$q,p>0$为学习增益.

注 4 由(3.1)式可知选取${{u}_{0}}(t)\in C[0,T]$, 则${{u}_{k}}(t)\in C[0,T]$, $k=1,2,3,\cdots $, 由此保证了系统(2.3)解的存在性.

注 5 由(2.3), (3.1)式及假设1中${f}({x}(t),u(t),t),g({x}(t),u(t),t)$的足够光滑性可知 选取${{u}_{0}}(t)\in C[0,T]\cap {{C}^{1}}[0,\hat{T}]$, 则${{y}_{k}}(t)\in {{C}^{1}}[0,T]$, 再由假设2, 有${{e}_{k}}(t)\in {{C}^{1}}[0,T]$.

由此给出如下定理:

定理 1 假设1-7成立, 如果有

$\begin{equation} \label{eq:5} \rho =\max \left\{\max_{\frac{\partial g}{\partial u}\in D \atop t\in [0,\hat{T}]}\left|1-q\frac{\partial g}{\partial u} \right|,\max_{{\frac{\partial {f}}{\partial u}\in \bar{D}\atop {\left(\frac{\partial g}{\partial {x}} \right)}^{\mathrm{T}}\in \hat{D}}\atop t\in (\hat{T},T]}\left| 1-p\frac{\partial g}{\partial {x}}\frac{\partial {f}}{\partial u} \right| \right\}<1, \end{equation}$

(3.2)

则系统(2.2)在分段学习律(3.1)作用下是收敛的, 即$\mathop {\lim }\limits_{k \to \infty } {\mkern 1mu} {\left\| {{e_k}(t)} \right\|_s} = 0$.

证 因证明过程需多次用到泰勒展开公式, 为简便起见, 用$*$表示公式中各个不同的中值, 记$\delta {{{x}}_{k}}={{{x}}_{k+1}}(t)-{{{x}}_{k}}(t), \delta {{u}_{k}}={{u}_{k+1}}(t)-{{u}_{k}}(t)$. 由于系统的分段性质, 证明分两部分进行.

(1) 对$t\in [0,\hat{T}]$. 应用泰勒展开公式, 由(2.3), (3.1)式有

$\begin{aligned} {{y}_{k+1}}(t)&=g({{{x}}_{k+1}},{{u}_{k+1}},t)=g({{{x}}_{k}}+\delta {{{x}}_{k}},{{u}_{k}}+\delta {{u}_{k}},t)\\ &=g({{{x}}_{k}},{{u}_{k}},t)+{{\left. \frac{\partial g}{\partial u} \right|}_{*}}\delta {{u}_{k}}+{{\left. \frac{\partial g}{\partial {x}} \right|}_{*}}\delta {{{x}}_{k}}\\ &={{y}_{k}}(t)+q{{\left. \frac{\partial g}{\partial u} \right|}_{*}}{{e}_{k}}(t)+{{\left. \frac{\partial g}{\partial {x}} \right|}_{*}}\delta {{{x}}_{k}}, \end{aligned}$

由此有

$\begin{aligned} {{e}_{k+1}}(t)&={{e}_{k}}(t)+{{y}_{k}}(t)-{{y}_{k+1}}(t)\\ &={{e}_{k}}(t)-q{{\left. \frac{\partial g}{\partial u} \right|}_{*}}{{e}_{k}}(t)-{{\left. \frac{\partial g}{\partial {x}} \right|}_{*}}\delta {{{x}}_{k}}\\ &=\left[ 1-q{{\left. \frac{\partial g}{\partial u} \right|}_{*}} \right]{{e}_{k}}(t)-{{\left. \frac{\partial g}{\partial {x}} \right|}_{*}}\delta {{{x}}_{k}}. \end{aligned}$

上式两端取范数, 由假设5及(3.2)式有

$\left| {{e}_{k+1}}(t) \right|\le \rho \left| {{e}_{k}}(t) \right|+{{\beta }_{1}}\left\| {{{x}}_{k+1}}(t)-{{{x}}_{k}}(t) \right\|.$

取$\lambda$ -范数, 有

$\begin{eqnarray} \label{eq:6} {{\left\| {{e}_{k+1}}(t) \right\|}_{\lambda }}&=&\underset{t\in [0,\hat{T}]}{\mathop{\max }}\,{{\mathrm{e}}^{-\lambda t}}\left| {{e}_{k+1}}(t) \right|\nonumber\\ &\le& \underset{t\in [0,\hat{T}]}{\mathop{\max }}\,{{\mathrm{e}}^{-\lambda t}}\left\{ \rho \left| {{e}_{k}}(t) \right|+{{\beta }_{1}}\left\| {{{x}}_{k+1}}(t)-{{{x}}_{k}}(t) \right\| \right\}\nonumber\\ &\le& \rho \underset{t\in [0,\hat{T}]}{\mathop{\max }}\,{{\mathrm{e}}^{-\lambda t}}\left| {{e}_{k}}(t) \right|+{{\beta }_{1}}\underset{t\in [0,\hat{T}]}{\mathop{\max }}\,{{\mathrm{e}}^{-\lambda t}}\left\| {{{x}}_{k+1}}(t)-{{{x}}_{k}}(t) \right\|\nonumber\\ &=&\rho {{\left\| {{e}_{k}}(t) \right\|}_{\lambda }}+{{\beta }_{1}}{{\left\| {{{x}}_{k+1}}(t)-{{{x}}_{k}}(t) \right\|}_{\lambda }}. \end{eqnarray}$

(3.3)

由系统(2.3)及假设3, 4可知

$\begin{aligned} \left\| {{{x}}_{k+1}}(t)-{{{x}}_{k}}(t) \right\|&=\left\| \int_{0}^{t}{\left( {f}({{{x}}_{k+1}},{{u}_{k+1}},\tau )-{f}({{{x}}_{k}},{{u}_{k}},\tau ) \right)d\tau } \right\|\\ &\le {{L}_{f}}\int_{0}^{t}{\left( \left\| {{{x}}_{k+1}}-{{{x}}_{k}} \right\|+\left| {{u}_{k+1}}-{{u}_{k}} \right| \right)}d\tau. \end{aligned}$

由(3.1)式可得

$\left\| {{{x}}_{k+1}}(t)-{{{x}}_{k}}(t) \right\|\le {{L}_{f}}\int_{0}^{t}{(\left\| {{{x}}_{k+1}}-{{{x}}_{k}} \right\|+q\left| {{e}_{k}} \right|)}d\tau. $

应用Gronwall引理, 有

$\begin{aligned} \left\| {{{x}}_{k+1}}(t)-{{{x}}_{k}}(t) \right\|&\le {{L}_{f}}q\int_{0}^{t}{({{\mathrm{e}}^{{{L}_{f}}(t-\tau )}}}\left| {{e}_{k}} \right|)d\tau\\ &\le {{L}_{f}}q{{\mathrm{e}}^{{{L}_{f}}\hat{T}}}\int_{0}^{t}{\left| {{e}_{k}} \right|}d\tau ={{L}_{f}}q{{\mathrm{e}}^{{{L}_{f}}\hat{T}}}\int_{0}^{t}{{{\mathrm{e}}^{\lambda \tau }}{{\mathrm{e}}^{-\lambda \tau }}\left| {{e}_{k}} \right|}d\tau\\ &\le {{L}_{f}}q{{\mathrm{e}}^{{{L}_{f}}\hat{T}}}\int_{0}^{t}{{{\mathrm{e}}^{\lambda \tau }}}d\tau {{\left\| {{e}_{k}} \right\|}_{\lambda }}={{L}_{f}}q{{\mathrm{e}}^{{{L}_{f}}\hat{T}}}\frac{{{\mathrm{e}}^{\lambda t}}-1}{\lambda }{{\left\| {{e}_{k}} \right\|}_{\lambda }}, \end{aligned}$

由此有

$\begin{aligned} {{\left\| {{{x}}_{k+1}}(t)-{{{x}}_{k}}(t) \right\|}_{\lambda }}&=\underset{t\in [0,\hat{T}]}{\mathop{\max }}\,{{\mathrm{e}}^{-\lambda t}}\left\| {{{x}}_{k+1}}(t)-{{{x}}_{k}}(t) \right\|\\ &\le \underset{t\in [0,\hat{T}]}{\mathop{\max }}\,{{L}_{f}}q{{\mathrm{e}}^{{{L}_{f}}\hat{T}}}\frac{1-{{\mathrm{e}}^{-\lambda t}}}{\lambda }{{\left\| {{e}_{k}} \right\|}_{\lambda }}\\ &={{L}_{f}}q{{\mathrm{e}}^{{{L}_{f}}\hat{T}}}\frac{1-{{\mathrm{e}}^{-\lambda \hat{T}}}}{\lambda }{{\left\| {{e}_{k}} \right\|}_{\lambda }}. \end{aligned}$

记${{O}_{1}}({{\lambda }^{-1}})={{L}_{f}}q{{\mathrm{e}}^{{{L}_{f}}\hat{T}}}\frac{1-{{\mathrm{e}}^{-\lambda \hat{T}}}}{\lambda }$, 则有

$\begin{equation} \label{eq:7} {{\left\| {{{x}}_{k+1}}(t)-{{{x}}_{k}}(t) \right\|}_{\lambda }}\le {{O}_{1}}({{\lambda }^{-1}}){{\left\| {{e}_{k}} \right\|}_{\lambda }}. \end{equation}$

(3.4)

显然, 当$\lambda $足够大时, 能使${{O}_{1}}({{\lambda }^{-1}})$任意小.

将(3.4)式代入(3.3)式, 有

${{\left\| {{e}_{k+1}}(t) \right\|}_{\lambda }}\le (\rho +{{\beta }_{1}}{{O}_{1}}({{\lambda }^{-1}})){{\left\| {{e}_{k}}(t) \right\|}_{\lambda }}.$

因为$\rho <1$, 所以取足够大的$\lambda $, 能使得$\rho +{{\beta }_{1}}{{O}_{1}}({{\lambda }^{-1}})<1$成立, 由压缩映射原理可知

$\mathop {\lim }\limits_{k \to \infty } {\mkern 1mu} {\left\| {{e_k}(t)} \right\|_\lambda } = 0,$

再由范数的等价性, 有

$\mathop {\lim }\limits_{k \to \infty } {\mkern 1mu} {\left\| {{e_k}(t)} \right\|_s} = 0,t \in [0,\hat T],$

(3.5)

由此有

$\mathop {\lim }\limits_{k \to \infty } {\mkern 1mu} {\left\| {{{\dot e}_k}(t)} \right\|_s} = 0,t \in [0,\hat T],$

特别地

$\mathop {\lim }\limits_{k \to \infty } {\mkern 1mu} \left| {{e_k}(\hat T)} \right| = 0,\mathop {\lim }\limits_{k \to \infty } {\mkern 1mu} \left| {{{\dot e}_k}(\hat T)} \right| = \mathop {\lim }\limits_{k \to \infty } {\mkern 1mu} \left| {{{\dot e}_k}({{\hat T}^ - })} \right| = 0.$

(3.6)

由假设1, 2, 有

$\mathop {\lim }\limits_{k \to \infty } {\mkern 1mu} \left| {{y_r}(\hat T) - {y_k}(\hat T)} \right| = 0 \\ \Rightarrow \mathop {\lim }\limits_{k \to \infty } {\mkern 1mu} \left\| {{x_r}(\hat T) - {x_k}(\hat T)} \right\| = 0 \\ \Rightarrow \mathop {\lim }\limits_{k \to \infty } {\mkern 1mu} \left\| {{x_k}(\hat T) - {x_{k + 1}}(\hat T)} \right\| = 0,$

(3.7)

$\mathop {\lim }\limits_{k \to \infty } {\mkern 1mu} {\left\| {{e_k}(t)} \right\|_s} = 0(t \in [0,\hat T]) \\ \Rightarrow \mathop {\lim }\limits_{k \to \infty } {\mkern 1mu} {\left\| {{u_r}(t) - {u_k}(t)} \right\|_s} = 0(t \in [0,\hat T]) \\ \Rightarrow \mathop {\lim }\limits_{k \to \infty } {\mkern 1mu} {\left\| {{{\dot u}_r}(t) - {{\dot u}_k}(t)} \right\|_s} = 0(t \in [0,\hat T]) \Rightarrow \mathop {\lim }\limits_{k \to \infty } {\mkern 1mu} \left| {{{\dot u}_r}({{\hat T}^ - }) - {{\dot u}_k}({{\hat T}^ - })} \right| = 0{\rm{ }} \\ \Rightarrow {\dot u_k}({\hat T^ - }){\rm{有界}}{\mkern 1mu} \Leftrightarrow \exists {\beta _{10}} > 0,\left| {{{\dot u}_k}({{\hat T}^ - })} \right| \le {\beta _{10}}(k = 0,1,2, \cdots ).$

(3.8)

(2) 对$t\in (\hat{T},T]$. 由(2.3)式有

$\begin{aligned} &\quad{{\dot{e}}_{k+1}}(t)={{\dot{e}}_{k}}(t)+{{\dot{y}}_{k}}(t)-{{\dot{y}}_{k+1}}(t)\\ &={{\dot{e}}_{k}}(t)+\frac{\partial g({{{x}}_{k}},{{u}_{k}}(\hat{T}),t)}{\partial {{{x}}_{k}}}{{\dot{{x}}}_{k}}(t)-\frac{\partial g({{{x}}_{k+1}},{{u}_{k+1}}(\hat{T}),t)}{\partial {{{x}}_{k+1}}}{{\dot{{x}}}_{k+1}}(t)\\ &\quad+\frac{\partial g({{{x}}_{k}},{{u}_{k}}(\hat{T}),t)}{\partial t}-\frac{\partial g({{{x}}_{k+1}},{{u}_{k+1}}(\hat{T}),t)}{\partial t}+{{\left. \frac{\partial g}{\partial {{u}_{k}}} \right|}_{t={{{\hat{T}}}^{-}}}}{{\dot{u}}_{k}}({{\hat{T}}^{-}})-{{\left. \frac{\partial g}{\partial {{u}_{k+1}}} \right|}_{t={{{\hat{T}}}^{-}}}}{{\dot{u}}_{k+1}}({{\hat{T}}^{-}})\\ &={{\dot{e}}_{k}}(t)+\frac{\partial g({{{x}}_{k}},{{u}_{k}}(\hat{T}),t)}{\partial {{{x}}_{k}}}{f}({{{x}}_{k}},{{u}_{k}},t)-\frac{\partial g({{{x}}_{k+1}},{{u}_{k+1}}(\hat{T}),t)}{\partial {{{x}}_{k+1}}}{f}({{{x}}_{k}},{{u}_{k}},t)\\ &\quad+\frac{\partial g({{{x}}_{k+1}},{{u}_{k+1}}(\hat{T}),t)}{\partial {{{x}}_{k+1}}}{f}({{{x}}_{k}},{{u}_{k}},t)-\frac{\partial g({{{x}}_{k+1}},{{u}_{k+1}}(\hat{T}),t)}{\partial {{{x}}_{k+1}}}{f}({{{x}}_{k+1}},{{u}_{k+1}},t)\\ &\quad+\frac{\partial g({{{x}}_{k}},{{u}_{k}}(\hat{T}),t)}{\partial t}-\frac{\partial g({{{x}}_{k+1}},{{u}_{k+1}}(\hat{T}),t)}{\partial t}+{{\left. \frac{\partial g}{\partial {{u}_{k+1}}} \right|}_{t={{{\hat{T}}}^{-}}}}({{\dot{u}}_{k}}({{\hat{T}}^{-}})-{{\dot{u}}_{k+1}}({{\hat{T}}^{-}}))\\ &\quad+\left( {{\left. \frac{\partial g}{\partial {{u}_{k}}} \right|}_{t={{{\hat{T}}}^{-}}}}-{{\left. \frac{\partial g}{\partial {{u}_{k+1}}} \right|}_{t={{{\hat{T}}}^{-}}}} \right){{\dot{u}}_{k}}({{\hat{T}}^{-}})\\ &={{\dot{e}}_{k}}(t)+{{({{{x}}_{k}}-{{{x}}_{k+1}})}^{\mathrm{T}}}{{\left. {{\left( \frac{{{\partial }^{2}}g}{\partial {{{x}}^{2}}} \right)}^{\mathrm{T}}} \right|}_{*}}{f}({{{x}}_{k}},{{u}_{k}},t){{\left. +({{u}_{k}}(\hat{T})-{{u}_{k+1}}(\hat{T}))\frac{{{\partial }^{2}}g}{\partial {{{x}}_{k}}\partial u(\hat{T})} \right|}_{*}}{f}({{{x}}_{k}},{{u}_{k}},t)\\ &\quad+\frac{\partial g({{{x}}_{k+1}},{{u}_{k+1}}(\hat{T}),t)}{\partial {{{x}}_{k+1}}}\left( {{\left. \frac{\partial {f}}{\partial {x}} \right|}_{*}}({{{x}}_{k}}-{{{x}}_{k+1}}) +{{\left. \frac{\partial {f}}{\partial u} \right|}_{*}}({{u}_{k}}-{{u}_{k+1}}) \right){{\left. +\frac{{{\partial }^{2}}g}{\partial t\partial {x}} \right|}_{*}}({{{x}}_{k}}-{{{x}}_{k+1}})\\ &\quad{{\left. +\frac{{{\partial }^{2}}g}{\partial t\partial u(\hat{T})} \right|}_{*}}({{u}_{k}}(\hat{T})-{{u}_{k+1}}(\hat{T}))+{{\left. \frac{\partial g}{\partial {{u}_{k+1}}} \right|}_{t={{{\hat{T}}}^{-}}}}({{\dot{u}}_{k}}({{\hat{T}}^{-}})-{{\dot{u}}_{k+1}}({{\hat{T}}^{-}}))\\ &\quad+\left({{\left. \frac{{{\partial }^{2}}g}{\partial u\partial {x}} \right|}_{*}}({{{x}}_{k}}(\hat{T})-{{{x}}_{k+1}}(\hat{T}))+{{\left. \frac{{{\partial }^{2}}g}{\partial {{u}^{2}}} \right|}_{*}}({{u}_{k}}(\hat{T})-{{u}_{k+1}}(\hat{T}))\right){{\dot{u}}_{k}}({{\hat{T}}^{-}}). \end{aligned}$

将(3.1)式代入上式, 得

$\begin{aligned} {{\dot{e}}_{k+1}}(t)=&\left( 1-p\frac{\partial g({{{x}}_{k+1}},{{u}_{k+1}}(\hat{T}),t)}{\partial {{{x}}_{k+1}}}{{\left. \frac{\partial {f}}{\partial u} \right|}_{*}} \right){{\dot{e}}_{k}}(t)\\ &-\frac{\partial g({{{x}}_{k+1}},{{u}_{k+1}}(\hat{T}),t)}{\partial {{{x}}_{k+1}}}{{\left. \frac{\partial {f}}{\partial u} \right|}_{*}}(q{{e}_{k}}(\hat{T})-p{{\dot{e}}_{k}}(\hat{T}))+{{({{{x}}_{k}}-{{{x}}_{k+1}})}^{\mathrm{T}}}{{\left. {{\left( \frac{{{\partial }^{2}}g}{\partial {{{x}}^{2}}} \right)}^{\mathrm{T}}} \right|}_{*}}{f}({{{x}}_{k}},{{u}_{k}},t)\\ &{{\left. -q{{e}_{k}}(\hat{T})\frac{{{\partial }^{2}}g}{\partial {{{x}}_{k}}\partial u(\hat{T})} \right|}_{*}}{f}({{{x}}_{k}},{{u}_{k}},t)+\frac{\partial g({{{x}}_{k+1}},{{u}_{k+1}}(\hat{T}),t)}{\partial {{{x}}_{k+1}}}{{\left. \frac{\partial {f}}{\partial {x}} \right|}_{*}}({{{x}}_{k}}-{{{x}}_{k+1}})\\ &{{\left. +\frac{{{\partial }^{2}}g}{\partial t\partial {x}} \right|}_{*}}({{{x}}_{k}}-{{{x}}_{k+1}})-q{{e}_{k}}(\hat{T}){{\left. \frac{{{\partial }^{2}}g}{\partial t\partial u(\hat{T})} \right|}_{*}}{{\left. -q{{{\dot{e}}}_{k}}(\hat{T})\frac{\partial g}{\partial {{u}_{k+1}}} \right|}_{t={{{\hat{T}}}^{-}}}}\\ &+\left( {{\left. \frac{{{\partial }^{2}}g}{\partial u\partial {x}} \right|}_{*}}({{{x}}_{k}}(\hat{T})-{{{x}}_{k+1}}(\hat{T}))-q{{e}_{k}}(\hat{T}){{\left. \frac{{{\partial }^{2}}g}{\partial {{u}^{2}}} \right|}_{*}} \right){{\dot{u}}_{k}}({{\hat{T}}^{-}}), \end{aligned}$

上式取范数, 由(3.2), (3.8)式及假设5, 6, 7, 可得

$\begin{eqnarray} \label{eq:12} \left| {{{\dot{e}}}_{k+1}}(t) \right|&\le& \rho \left| {{{\dot{e}}}_{k}}(t) \right|+\left( {{\beta }_{4}}{{\beta }_{7}}+{{\beta }_{5}}\sqrt{\sum\limits_{j=1}^{n}{\gamma _{j2}^{2}}}+{{\beta }_{8}} \right)\left\| {{{x}}_{k}}-{{{x}}_{k+1}} \right\|\nonumber\\ &\quad&+{{\beta }_{2}}{{\beta }_{10}}\left\| {{{x}}_{k}}(\hat{T})-{{{x}}_{k+1}}(\hat{T}) \right\|\nonumber\\ &\quad&+q\left( \sqrt{\sum\limits_{i=1}^{n}{\alpha _{i2}^{2}}}\sqrt{\sum\limits_{j=1}^{n}{\gamma _{j2}^{2}}}+{{\beta }_{4}}{{\beta }_{9}}+{{\beta }_{6}}+{{\beta }_{3}}{{\beta }_{10}} \right)\left| {{e}_{k}}(\hat{T}) \right|\nonumber\\ &\quad&+\left( p\sqrt{\sum\limits_{i=1}^{n}{\alpha _{i2}^{2}}}\sqrt{\sum\limits_{j=1}^{n}{\gamma _{j2}^{2}}}+q{{\alpha }_{2}} \right)\left| {{{\dot{e}}}_{k}}(\hat{T}) \right|\nonumber\\ &=&\rho \left| {{{\dot{e}}}_{k}}(t) \right|\!+\!b\left\| {{{x}}_{k}}-{{{x}}_{k+1}} \right\|+{{F}_{k}}(\hat{T}),t\in (\hat{T},T], \end{eqnarray}$

(3.9)

其中$b={{\beta }_{4}}{{\beta }_{7}}+{{\beta }_{5}}\sqrt{\sum\limits_{j=1}^{n}{\gamma _{j2}^{2}}}+{{\beta }_{8}},$

$\begin{aligned} {{F}_{k}}(\hat{T})=&{{\beta }_{2}}{{\beta }_{10}}\left\| {{{x}}_{k}}(\hat{T})-{{{x}}_{k+1}}(\hat{T}) \right\|+q\left( \sqrt{\sum\limits_{i=1}^{n}{\alpha _{i2}^{2}}}\sqrt{\sum\limits_{j=1}^{n}{\gamma _{j2}^{2}}}+{{\beta }_{4}}{{\beta }_{9}}+{{\beta }_{6}}+{{\beta }_{3}}{{\beta }_{10}} \right)\left| {{e}_{k}}(\hat{T}) \right|\\ &+\left( p\sqrt{\sum\limits_{i=1}^{n}{\alpha _{i2}^{2}}}\sqrt{\sum\limits_{j=1}^{n}{\gamma _{j2}^{2}}}+q{{\alpha }_{2}} \right)\left| {{{\dot{e}}}_{k}}(\hat{T}) \right|\ge 0, \end{aligned}$

由(3.6), (3.7)式, 有

$\mathop {\lim }\limits_{k \to \infty } {\mkern 1mu} {F_k}(\hat T) = 0,$

(3.10)

由(2.3)式及注5, 可知${{\dot{e}}_{k}}(t),{{\dot{e}}_{k+1}}(t),{{{x}}_{k}}(t)-{{{x}}_{k+1}}(t)$ 在$t\in [0,T]$上连续, 由此(3.9)式在$t\in [\hat{T},T]$成立, 即

$\begin{equation} \label{eq:14} \left| {{{\dot{e}}}_{k+1}}(t) \right|\le \rho \left| {{{\dot{e}}}_{k}}(t) \right|+b\left\| {{{x}}_{k}}-{{{x}}_{k+1}} \right\|+{{F}_{k}}(\hat{T}),t\in [\hat{T},T]. \end{equation}$

(3.11)

对(3.11)式取$\lambda $ -范数, 则有

$\begin{gathered} {\left\| {{{\dot e}_{k + 1}}(t)} \right\|_\lambda } = \mathop {\max }\limits_{t \in [\hat T,T]} {\mkern 1mu} {{\text{e}}^{ - \lambda (t - \hat T)}}\left| {{{\dot e}_{k + 1}}(t)} \right|\rho \mathop {\max }\limits_{t \in [\hat T,T]} {\mkern 1mu} {{\text{e}}^{ - \lambda (t - \hat T)}}\left| {{{\dot e}_k}(t)} \right| + b\mathop {\max }\limits_{t \in [\hat T,T]} {\mkern 1mu} {{\text{e}}^{ - \lambda (t - \hat T)}} \hfill \\ \left\| {{x_k} - {x_{k + 1}}} \right\| + \mathop {\max }\limits_{t \in [\hat T,T]} {\mkern 1mu} {{\text{e}}^{ - \lambda (t - \hat T)}}{F_k}(\hat T) = \rho {\left\| {{{\dot e}_k}(t)} \right\|_\lambda } + b{\left\| {{x_k} - {x_{k + 1}}} \right\|_\lambda } + {F_k}(\hat T). \hfill \\ \end{gathered} $

(3.12)

同样由连续性, 对$t\in [\hat{T},T]$, 由方程(2.3)有

${{{x}}_{k+1}}(t)-{{{x}}_{k}}(t)={{{x}}_{k+1}}(\hat{T})-{{{x}}_{k}}(\hat{T})+\int_{{\hat{T}}}^{t}{\left( {f}({{{x}}_{k+1}},{{u}_{k+1}},\tau )-{f}({{{x}}_{k}},{{u}_{k}},\tau ) \right)d\tau }.$

取范数, 由假设6及(3.1)式, 有

$\begin{eqnarray*} &&\left\| {{{x}}_{k+1}}(t)-{{{x}}_{k}}(t) \right\|\\ &\le& \left\| {{{x}}_{k+1}}(\hat{T})-{{{x}}_{k}}(\hat{T}) \right\|+\left\| \int_{{\hat{T}}}^{t}{\left( {{\left. \frac{\partial {f}}{\partial {x}} \right|}_{*}}({{{x}}_{k+1}}-{{{x}}_{k}})+{{\left. \frac{\partial {f}}{\partial u} \right|}_{*}}({{u}_{k+1}}-{{u}_{k}}) \right)}d\tau \right\|\\ &\le& \left\| {{{x}}_{k+1}}(\hat{T})-{{{x}}_{k}}(\hat{T}) \right\|+\int_{{\hat{T}}}^{t}{\left({{\beta }_{5}}\left\| {{{x}}_{k+1}}-{{{x}}_{k}} \right\|+\sqrt{\sum\limits_{i=1}^{n}{\alpha _{i2}^{2}}}(p\left| {{{\dot{e}}}_{k}} \right|+q\left| {{e}_{k}}(\hat{T}) \right|+p\left| {{{\dot{e}}}_{k}}(\hat{T}) \right|)\right)}d\tau \\ &\le& \left\| {{{x}}_{k+1}}(\hat{T})-{{{x}}_{k}}(\hat{T}) \right\|+(T-\hat{T})\left(q\left| {{e}_{k}}(\hat{T}) \right|+p\left| {{{\dot{e}}}_{k}}(\hat{T}) \right|\right)\sqrt{\sum\limits_{i=1}^{n}{\alpha _{i2}^{2}}}\\ &&+\int_{{\hat{T}}}^{t}{\left( {{\beta }_{5}}\left\| {{{x}}_{k+1}}-{{{x}}_{k}} \right\|+p\sqrt{\sum\limits_{i=1}^{n}{\alpha _{i2}^{2}}}\left| {{{\dot{e}}}_{k}} \right| \right)}d\tau. \end{eqnarray*}$

再次应用Gronwall引理, 有

$\begin{aligned} \left\| {{{x}}_{k+1}}(t)-{{{x}}_{k}}(t) \right\|&\le {{G}_{k}}(\hat{T})+p\sqrt{\sum\limits_{i=1}^{n}{\alpha _{i2}^{2}}}\int_{{\hat{T}}}^{t}{({{\mathrm{e}}^{{{\beta }_{5}}(t-\tau )}}}\left| {{{\dot{e}}}_{k}} \right|)d\tau\\ &\le {{G}_{k}}(\hat{T})+p{{\mathrm{e}}^{{{\beta }_{5}}(T-\hat{T})}}\sqrt{\sum\limits_{i=1}^{n}{\alpha _{i2}^{2}}}\int_{{\hat{T}}}^{t}{\left| {{{\dot{e}}}_{k}} \right|}d\tau\\ &={{G}_{k}}(\hat{T})+p{{\mathrm{e}}^{{{\beta }_{5}}(T-\hat{T})}}\sqrt{\sum\limits_{i=1}^{n}{\alpha _{i2}^{2}}}\int_{{\hat{T}}}^{t}{{{\mathrm{e}}^{\lambda (\tau -\hat{T})}}{{\mathrm{e}}^{-\lambda (\tau -\hat{T})}}\left| {{{\dot{e}}}_{k}} \right|}d\tau\\ &\le {{G}_{k}}(\hat{T})+p{{\mathrm{e}}^{{{\beta }_{5}}(T-\hat{T})}}\sqrt{\sum\limits_{i=1}^{n}{\alpha _{i2}^{2}}}\int_{{\hat{T}}}^{t}{{{\mathrm{e}}^{\lambda (\tau -\hat{T})}}}d\tau {{\left\| {{{\dot{e}}}_{k}} \right\|}_{\lambda }}\\ &={{G}_{k}}(\hat{T})+p{{\mathrm{e}}^{{{\beta }_{5}}(T-\hat{T})}}\sqrt{\sum\limits_{i=1}^{n}{\alpha _{i2}^{2}}}\frac{{{\mathrm{e}}^{\lambda (t-\hat{T})}}-1}{\lambda }{{\left\| {{{\dot{e}}}_{k}} \right\|}_{\lambda }}, \end{aligned}$

其中

${{G}_{k}}(\hat{T})={{\mathrm{e}}^{{{\beta }_{5}}(T-\hat{T})}}\left( \left\| {{{x}}_{k+1}}(\hat{T})-{{{x}}_{k}}(\hat{T}) \right\|+(T-\hat{T})(q\left| {{e}_{k}}(\hat{T}) \right|+p\left| {{{\dot{e}}}_{k}}(\hat{T}) \right|)\sqrt{\sum\limits_{i=1}^{n}{\alpha _{i2}^{2}}} \right)\ge 0,$

并有

$\mathop {\lim }\limits_{k \to \infty } {\mkern 1mu} {G_k}(\hat T) = 0,$

(3.13)

由此有

$\begin{aligned} {{\left\| {{{x}}_{k+1}}(t)-{{{x}}_{k}}(t) \right\|}_{\lambda }}&=\underset{t\in [\hat{T},T]}{\mathop{\max }}\,{{\mathrm{e}}^{-\lambda (t-\hat{T})}}\left\| {{{x}}_{k+1}}(t)-{{{x}}_{k}}(t) \right\|\\ &\le \underset{t\in [\hat{T},T]}{\mathop{\max }}\,{{\mathrm{e}}^{-\lambda (t-\hat{T})}}{{G}_{k}}(\hat{T})+\underset{t\in [\hat{T},T]}{\mathop{\max }}\,p{{\mathrm{e}}^{{{\beta }_{5}}(T-\hat{T})}}\sqrt{\sum\limits_{i=1}^{n}{\alpha _{i2}^{2}}}\frac{1-{{\mathrm{e}}^{-\lambda (t-\hat{T})}}}{\lambda }{{\left\| {{{\dot{e}}}_{k}} \right\|}_{\lambda }}\\ &={{G}_{k}}(\hat{T})+p{{\mathrm{e}}^{{{\beta }_{5}}(T-\hat{T})}}\sqrt{\sum\limits_{i=1}^{n}{\alpha _{i2}^{2}}}\frac{1-{{\mathrm{e}}^{-\lambda (T-\hat{T})}}}{\lambda }{{\left\| {{{\dot{e}}}_{k}} \right\|}_{\lambda }}. \end{aligned}$

记${{O}_{2}}({{\lambda }^{-1}})=p{{\mathrm{e}}^{{{\beta }_{5}}(T-\hat{T})}}\sqrt{\sum\limits_{i=1}^{n}{\alpha _{i2}^{2}}}\frac{1-{{\mathrm{e}}^{-\lambda (T-\hat{T})}}}{\lambda }$, 则有

$\begin{equation} \label{eq:17} {{\left\| {{{x}}_{k+1}}(t)-{{{x}}_{k}}(t) \right\|}_{\lambda }}\le {{G}_{k}}(\hat{T})+{{O}_{2}}({{\lambda }^{-1}}){{\left\| {{{\dot{e}}}_{k}} \right\|}_{\lambda }}, \end{equation}$

(3.14)

将(3.14)式代入(3.12)式, 有

$\begin{equation} \label{eq:18} {{\left\| {{{\dot{e}}}_{k+1}}(t) \right\|}_{\lambda }}\le (\rho +b{{O}_{2}}({{\lambda }^{-1}})){{\left\| {{{\dot{e}}}_{k}}(t) \right\|}_{\lambda }}+b{{G}_{k}}(\hat{T})+{{F}_{k}}(\hat{T}). \end{equation}$

(3.15)

因为$\rho <1$, 所以取足够大的$\lambda $, 能使得$\rho +b{{O}_{2}}({{\lambda }^{-1}})<1$成立, 而由(3.10), (3.13) 式有$\mathop {\lim }\limits_{k \to \infty } {\mkern 1mu} (b{G_k}(\hat T) + {F_k}(\hat T)) = 0$, 对(3.15)式利用引理1, 得 $\mathop {\lim }\limits_{k \to \infty } {\mkern 1mu} {\left\| {{{\dot e}_k}(t)} \right\|_\lambda } = 0.$ 再由范数的等价性, 有

$\mathop {\lim }\limits_{k \to \infty } {\mkern 1mu} {\left\| {{{\dot e}_k}(t)} \right\|_s} = 0,t \in [\hat T,T],$

由(3.6)式可得

$\mathop {\lim }\limits_{k \to \infty } {\mkern 1mu} {\left\| {{e_k}(t)} \right\|_s}= \mathop {\lim }\limits_{k \to \infty } {\mkern 1mu} {\left\| {{e_k}(\hat T) + \int_{\hat T}^t {{{\dot e}_k}(\tau )d\tau } } \right\|_s} \le \mathop {\lim }\limits_{k \to \infty } {\mkern 1mu} \left| {{e_k}(\hat T)} \right| + \mathop {\lim }\limits_{k \to \infty } {\mkern 1mu} {\left\| {\int_{\hat T}^t {{{\dot e}_k}(\tau )d\tau } } \right\|_s} \\ = \mathop {\lim }\limits_{k \to \infty } {\mkern 1mu} {\left\| {\int_{\hat T}^t {{{\dot e}_k}(\tau )d\tau } } \right\|_s} \le (T - \hat T)\mathop {\lim }\limits_{k \to \infty } {\mkern 1mu} {\left\| {{{\dot e}_k}(t)} \right\|_s} = 0,$

由此可知

$\mathop {\lim }\limits_{k \to \infty } {\mkern 1mu} {\left\| {{e_k}(t)} \right\|_s} = 0,t \in [\hat T,T].$

(3.16)

综上, 由(3.5), (3.5)式得 $\mathop {\lim }\limits_{k \to \infty } {\mkern 1mu} {\left\| {{e_k}(t)} \right\|_s} = 0,t \in [0,T].$

注 6 由假设5, 6, 7, 选取学习增益$q$, 使其满足$0<q<\frac{2}{{{\alpha }_{2}}}$; 选取学习增 益$p$, 使其满足$0<p<\frac{2}{\sum\limits_{i=1}^{n}{({{\alpha }_{i2}}{{\gamma }_{i2}})}}$, 则收敛性条件(3.2)式就能成立.

构建如下非线性系统

$\begin{equation} \label{eq:20} \left\{ \begin{aligned} \dot{x}(t)&=u(t)-\frac{1}{2}\arctan (\frac{1}{2}u(t))+\sin (x(t)),&0\le t\le 2, \\ y(t)&=2x(t)+\cos (x(t))+u(t),&0\le t\le 1, \\ y(t)&=2x(t)+\cos (x(t))+u(1)+(t-1)\dot{u}(1),&1<t\le 2. \\ \end{aligned} \right. \end{equation}$

(4.1)

理想轨迹为${{y}_{r}}(t)={{t}^{2}}, t\in [0,2]$. 通过验证可知系统(4.1)满足假设条件$1\sim7$ (见附录). 由系统(4.1), 有$\frac{\partial g}{\partial u}=1, \frac{\partial f}{\partial u}\in [0.75,1], \frac{\partial g}{\partial x}\in , \forall u,x\in R$, 即${{\alpha }_{1}}=1, {{\alpha }_{2}}={{a}_{12}}={{\gamma }_{11}}=1, {{a}_{11}}=0.75, {{\gamma }_{12}}=3$.

于是由注6可知$q<\frac{2}{{{\alpha }_{2}}}=2,p<\frac{2}{{{\alpha }_{12}}{{\gamma }_{12}}}=\frac{2}{3}$, 选取学习增益为$q=0.5,p=0.3$, 此时$\rho =\max \left\{ \left| 1-0.5\cdot 1 \right|,\left| 1-0.3\cdot 0.75 \right| \right\}=0.775<1$, 满足定理1的条件. 将$p,q$的值代入(3.1)式, 可得系统(4.1)的分段学习律为

$\left\{ \begin{aligned} {{u}_{k+1}}(t)&={{u}_{k}}(t)+0.5{{e}_{k}}(t),&t\in [0,1], \\ {{u}_{k+1}}(t)&={{u}_{k}}(t)+0.3{{{\dot{e}}}_{k}}(t)+0.5{{e}_{k}}(1)-0.3{{{\dot{e}}}_{k}}(1),&t\in (1,2]. \\ \end{aligned} \right.$

取${{u}_{0}}(t)=0,{{x}_{k}}(0)=0,k=0,1,2,\cdots $, 运用Matlab的Simulink模块进行仿真可得仿真结果如图 1 ～ 图 2所示. 从图中可以看出, 随着迭代次数的增加,跟踪误差在分段学习律下呈较好的收敛趋势.

本文提出了分段迭代学习控制的概念, 并对一类满足通常假设条件的非线性系统进行分段迭代学习控制设计. 根据系统在不同时间段内所满足的不同假设条件, 采用不同的学习控制律. 在前段时间段内, 采用 P 型学习控制律, 在后段时间段内, 采用 D 型学习控制律, 通过两个时间段的合理衔接, 最终组合成整个重复受控时间内的学习控制律, 当该学习控制律作用于系统时, 系统在整个重复受控时间内的输出跟踪误差沿迭代轴方向收敛. 仿真算例验证了算法的有效性. 文 [9] 及本文研究的系统均为单输入单输出, 如何将本文的结论推广到多输入多输出系统, 有待作进一步的研究.