数学物理学报  2016, Vol. 36 Issue (2): 267-286   PDF (438 KB)    
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傅勤
分布参数系统于W1,2空间中的迭代学习控制
傅勤    
苏州科技学院 数理学院 江苏 苏州 215009
收稿日期: 2015-09-21; 修订日期: 2016-01-27
基金项目: 国家自然科学基金(11371013)资助
作者简介: 傅勤,fuqin925@sina.com
摘要  : 研究一类分布参数系统于W1,2空间中的迭代学习控制问题,该类分布参数系统由抛物型偏微分方程或由双曲型偏微分方程构成.针对系统所满足的性质,基于P型学习律构建得到迭代学习控制律,并进一步证明这种学习律能使得系统的输出跟踪误差于W1,2空间内沿迭代轴方向收敛.仿真算例说明了该结论的可行性和有效性.
关键词: P型学习律     迭代学习控制     分布参数系统     W1,2空间    
Iterative Learning Control for Distributed Parameter Systems in Space W1,2
Fu Qin    
School of Mathematics and Physics, Suzhou University of Science and Technology, Jiangsu Suzhou 215009
Foundation Item: Supported by the NSFC (11371013)
Abstract  : In space W1,2, the problem of iterative learning control algorithm for a class of distributed parameter systems is considered. Here, the considered distributed parameter systems are composed of parabolic partial differential equations or hyperbolic partial differential equations. According to system properties, iterative learning control laws are proposed for such distributed parameter systems based on P-type learning scheme. Further, it is shown that the scheme can guarantee the output tracking errors on W1,2 space converge along the iteration axis. Simulation examples show the feasibility and effectiveness of the conclusion.
Key words: P-type learning scheme     Iterative learning control     Distributed parameter systems     W1,2 space    
1 引言

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, 2, 3, 4]和P型学习律[5, 6, 7],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.

由于很多实际问题通常要用偏微分方程所代表的分布参数系统模型描述,近年来,分布参数系统的应用已经渗透到各个领域,相应的研究工作也取得了不少成果[8, 9, 10, 11]. 然而,由于分布参数系统变量涉及无穷维函数空间,所以到目前为止将迭代学习方法应用于分布参数系统的研究成果并不多,且多集中于抛物型分布参数系统上,文献[]研究了抛物型分布参数系统的迭代学习控制问题,在输入、输出有直输通道的情况下,用P型学习律设计出了迭代学习控制器.文献[16]则将迭代学习控制应用到了二阶双曲型分布参数系统上,得到了输出跟踪误差于$L^2$空间内沿迭代轴方向收敛的结论.

研究偏微分方程时,常需研讨其弱解问题,而Sobolev空间是讨论偏微分方程弱解比较合适的空间,所以,于Sobolev空间$W^{1,2}$中研究分布参数系统的迭代学习控制问题是有意义的,然相关的研究成果尚甚少. 文献[12]针对一类抛物型分布参数系统,在证明了输出跟踪误差于$L^2$空间内沿迭代轴方向收敛的定理后,给出了输出跟踪误差于$W^{1,2}$空间内收敛的结论,但限于篇幅,文献[12]略去了其证明过程.

本文研究一类分布参数系统于$W^{1,2}$空间中的迭代学习控制问题,该系统由抛物型偏微分方程或由双曲型偏微分方程构成.当输入,输出有直输通道时,用P型学习律设计出迭代学习控制器,证明这种学习律能使得系统的输出跟踪误差于$W^{1,2}$空间内沿迭代轴方向收敛,并进一步研究输出跟踪误差于$W^{2,2}$空间内的收敛性问题.

本文给出如下符号约定: 用$\left\| \cdot\right\|$表示向量或矩阵的2范数,对向量$Q(x,t)\in {\Bbb R}^n\cap L^2[0, 1]$,$x\in [0, 1]$,$t\in [0,T]$,记

$$\left\| {Q(x,t)} \right\|_{L^2[0, 1]}=\sqrt {\int_0^1 {\left\| {Q(x,t)} \right\|^2} {\rm d} x} , $$

定义$\left\| {Q(x,t)} \right\|_{L^2[0, 1],s} =\sup\limits_{t\in [0,T]} \left\| {Q(x,t)} \right\|_{L^2[0, 1]}^2 $,对给定的$\lambda>0$,定义$\left\| {Q(x,t)} \right\|_{L^2[0, 1],\lambda } = \sup\limits_{t\in [0,T]} {\rm e}^{-\lambda t}\left\| {Q(x,t)}\right\|_{L^2[0, 1]}^2 $. 由文献[7]可知,可用$\left\| {Q(x,t)}\right\|_{L^2[0, 1],s} $与$\left\| {Q(x,t)}\right\|_{L^2[0, 1],\lambda } $其中任一个来证明收敛性结果. 记$\Omega =[0, 1]\times [0,T],$

$$W^{l,2}[0, 1]=\bigg\{ { {\varphi (x,t)}\big|\varphi (x,t)\in L^2[0, 1],} {\frac{\partial\varphi (x,t)}{\partial x}\in L^2[0, 1],\cdots ,\frac{\partial^l\varphi (x,t)}{\partial x^l}\in L^2[0, 1]} \bigg\} . $$
2 问题描述

考虑如下形式的分布参数系统

$\left\{ \begin{gathered} \frac{{{\partial ^\alpha }Q\left( {x,t} \right)}}{{\partial {t^\alpha }}} = D\frac{{{\partial ^2}Q\left( {x,t} \right)}}{{\partial {x^2}}} = + A\left( t \right)Q\left( {x,t} \right) + B\left( t \right)u\left( {x,t} \right), \hfill \\ \left[ {2mm} \right]y\left( {x,t} \right) = C\left( t \right)Q(x,t) + E\left( t \right)u\left( {x,t} \right),\;\;(x,t) \in \left( {0,1} \right) \times \left[ {0,T} \right], \hfill \\ \end{gathered} \right.$ (2.1)
这里$Q(x,t)\in {\Bbb R}^n$,$u(x,t),y(x,t)\in {\Bbb R}$分别是系统的状态,控制输入和输出,其中$D={\rm diag}(d_1 ,d_2 ,\cdots,d_n )$,$0对系统(2.1),给出如下假设条件:

假设2.1   对于给定的初值$Q(x,0)(\alpha =1,2)$,$\left.{\frac{\partial Q(x,t)}{\partial t}} \right|_{t=0} (\alpha =2)$,边值$Q(0,t)$($\alpha =1)$,或$\left. {\frac{\partial Q(x,t)}{\partial x}} \right|_{x=0}(\alpha =1,2)$; $Q(1,t)(\alpha =1)$,或${\left. {\frac{{\partial Q(x,t)}}{{\partial x}}} \right|_{x = 1}}(\alpha = 1,2)$,控制输入$u(x,t)$,系统(2.1)的解$Q(x,t)$在$(0,1)\times [0,T]$内存在唯一.

假设2.2   对$\forall t\in [0,T]$,有$0 < {\alpha _1} \leqslant E(t) \leqslant {\alpha _2}$,即输入、输出有直输通道,而 $ \left\| {A(t)}\right\|\le A,$ $ \left\| {B(t)} \right\|\le B,$ $\left\|{C(t)} \right\|\le C. $

假设2.3   对于给定的理想轨迹$y_r (x,t)$,存在唯一的控制输入$u_r (x,t)$,使得

$$\left\{ \begin{gathered} \frac{{{\partial ^\alpha }{Q_r}(x,t)}}{{\partial {t^\alpha }}} = D\frac{{{\partial ^2}{Q_r}(x,t)}}{{\partial {x^2}}} + A(t){Q_r}(x,t) + B(t){u_r}(x,t), \hfill \\ {y_r}(x,t) = C(t){Q_r}(x,t) + E(t){u_r}(x,t),\;\;\left( {x,t} \right) \in (0,1) \times [0,T]. \hfill \\ \end{gathered} \right.$$

设动态系统(2.1)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(2.1)为

$$\left\{ {\begin{array}{l} \frac{\partial ^\alpha Q_k (x,t)}{\partial t^\alpha }=D\frac{\partial ^2Q_k(x,t)}{\partial x^2}+A(t)Q_k (x,t)+B(t)u_k (x,t) ,\\[2mm] y_k (x,t)=C(t)Q_k (x,t)+E(t)u_k (x,t) ,~~ (x,t)\in (0,1)\times [0,T],\end{array}} \right. $$ (2.2)
其中,$k=0,1,2,\cdots $.

假设2.4   系统的初值定位条件为: $Q_k (x,0)=Q_r(x,0)$ ($\alpha =1,2$),${\left. {\frac{{\partial {Q_k}(x,t)}}{{\partial t}}} \right|_{t = 0}} = {\left. {\frac{{\partial {Q_r}(x,t)}}{{\partial t}}} \right|_{t = 0}}$ $(\alpha =2)$; 边值定位条件为: $Q_k (0,t)=Q_r (0,t)(\alpha=1)$,或$\left. {\frac{\partial Q_k (x,t)}{\partial x}}\right|_{x=0} =\left. {\frac{\partial Q_r (x,t)}{\partial x}}\right|_{x=0}$ $(\alpha =1,2)$; $Q_k (1,t)=Q_r (1,t)(\alpha=1)$,或$\left. {\frac{\partial Q_k (x,t)}{\partial x}} \right|_{x=1}=\left. {\frac{\partial Q_r (x,t)}{\partial x}} \right|_{x=1}(\alpha =1,2)$. $k=0,1,2,\cdots $.

学习控制的目的是寻找适当的学习律,使得迭代学习序列$y_k(x,t)$于$W^{1,2}[0, 1]$空间中一致收敛于理想的输出$y_r (x,t)$,即

$$ \lim \limits_{k\to \infty } \left\| {e_k (x,t)}\right\|_{L^2[0, 1],s} =0 ,\lim \limits_{k\to \infty } \left\| {\frac{\partial e_k(x,t)}{\partial x}} \right\|_{L^2[0, 1],s} =0, $$ 其中$e_k (x,t)=y_r (x,t)-y_k (x,t)$.

引理2.1  [17] 设$\left\{ {a_k } \right\}$,$\left\{ {b_k } \right\}$是满足

$$a_{k+1} \le \rho a_k +b_k ,0\le \rho <1 $$ 的非负实数数列,如有$\mathop {\lim }\limits_{k\to \infty } b_k =0$,则有$\lim \limits_{k\to \infty } a_k =0$.
3 主要结果

对假设2.2中的$\alpha _1 ,\alpha _2 $,取正数$\varepsilon $,使其满足

\begin{equation}\frac{\alpha _2 }{\alpha _1 }<\frac{\sqrt {1+\varepsilon } +1}{\sqrt{1+\varepsilon } -1}.\end{equation} (3.1)

对系统(2.2)构建P型学习律

\begin{equation}u_{k+1} (x,t)=u_k (x,t)+qe_k (x,t),\end{equation} (3.2)
其中$q>0$为学习增益.

选取学习增益$q$,使得

\begin{equation}\gamma =\mathop {\max }\limits_{t\in [0,T]} \left| {1-qE(t)}\right|<\frac{1}{\sqrt {1+\varepsilon } } \end{equation} (3.3)
成立. 记$\delta Q_k (x,t)=Q_{k+1} (x,t)-Q_k (x,t)$,由(2.2),(3.2)式有
\begin{eqnarray}e_{k+1} (x,t)&=&e_k (x,t)+y_k (x,t)-y_{k+1} (x,t)\\&=& e_k (x,t)-C(t)\delta Q_k (x,t)-qE(t)e_k (x,t)\\&=& (1-qE(t))e_k (x,t)-C(t)\delta Q_k (x,t).\end{eqnarray} (3.4)
(3.4)式两端取范数,结合(3.3)式有 $$\left| {e_{k+1} (x,t)} \right|\le \gamma \left| {e_k (x,t)} \right|+\left\|{C(t)} \right\|\left\| {\delta Q_k (x,t)} \right\|. $$ 由基本不等式及假设2.2,得
\begin{equation}\left| {e_{k+1} (x,t)} \right|^2\le (1+\varepsilon )\gamma ^2\left| {e_k(x,t)} \right|^2+(1+\frac{1}{\varepsilon })C^2\left\| {\delta Q_k (x,t)}\right\|^2.\end{equation} (3.5)
记$\rho \mbox{=}(1+\varepsilon )\gamma ^2$,(3.5)式两端对变量$x$从$0$到$1$积分,有
\begin{equation}\left\| {e_{k+1} (x,t)} \right\|_{L^2[0, 1]}^2 \le \rho \left\| {e_k (x,t)}\right\|_{L^2[0, 1]}^2 +(1+\frac{1}{\varepsilon })C^2\left\| {\delta Q_k(x,t)} \right\|_{L^2[0, 1]}^2 .\end{equation} (3.6)
所以
\begin{eqnarray}\left\| {e_{k+1} (x,t)} \right\|_{L^2[0, 1],\lambda } &=&\mathop {\max}\limits_{t\in [0,T]} {\rm e}^{-\lambda t}\left\| {e_{k+1} (x,t)}\right\|_{L^2[0, 1]}^2\\& \le&\rho \mathop {\max }\limits_{t\in [0,T]} {\rm e}^{-\lambda t}\left\| {e_k (x,t)}\right\|_{L^2[0, 1]}^2 +(1+\frac{1}{\varepsilon })C^2\mathop {\max}\limits_{t\in [0,T]} {\rm e}^{-\lambda t}\left\| {\delta Q_k (x,t)}\right\|_{L^2[0, 1]}^2 \\&=&\rho \left\| {e_k (x,t)} \right\|_{L^2[0, 1],\lambda }+(1+\frac{1}{\varepsilon })C^2\left\| {\delta Q_k (x,t)}\right\|_{L^2[0, 1],\lambda } .\end{eqnarray} (3.7)
(3.4)式两端对变量$x$求偏导,可得
\begin{equation}\frac{\partial e_{k+1} (x,t)}{\partial x}=(1-qE(t))\frac{\partial e_k(x,t)}{\partial x}-C(t)\frac{\partial (\delta Q_k (x,t))}{\partial x}.\end{equation} (3.8)
对(3.8)式重复从(3.4)式到(3.7)式的处理过程,相应有
\begin{equation}\left\| {\frac{\partial e_{k+1} (x,t)}{\partial x}}\right\|_{L^2[0, 1],\lambda } \le \rho \left\| {\frac{\partial e_k(x,t)}{\partial x}} \right\|_{L^2[0, 1],\lambda } +(1+\frac{1}{\varepsilon})C^2\left\| {\frac{\partial (\delta Q_k (x,t))}{\partial x}}\right\|_{L^2[0, 1],\lambda } .\end{equation} (3.9)
(3.8)式两端再对变量$x$求偏导,可得
$\frac{{{\partial ^2}{e_{k + 1}}(x,t)}}{{\partial {x^2}}} = (1 - qE(t))\frac{{{\partial ^2}{e_k}(x,t)}}{{\partial {x^2}}} - C(t)\frac{{{\partial ^2}(\delta {Q_k}(x,t))}}{{\partial {x^2}}}.$ (3.10)
同前面的过程,可推得
\begin{equation}\left\| {\frac{\partial ^2e_{k+1} (x,t)}{\partial x^2}}\right\|_{L^2[0, 1],\lambda } \le \rho \left\| {\frac{\partial ^2e_k(x,t)}{\partial x^2}} \right\|_{L^2[0, 1],\lambda } +(1+\frac{1}{\varepsilon})C^2\left\| {\frac{\partial ^2(\delta Q_k (x,t))}{\partial x^2}}\right\|_{L^2[0, 1],\lambda } .\end{equation} (3.11)
对(3.10)式重复从(3.4)式到(3.5)式的处理过程,并在$\Omega $上积分,相应有
\begin{equation}\left\| {\frac{\partial ^2e_{k+1} (x,t)}{\partial x^2}} \right\|_{L^2(\Omega)}^2 \le \rho \left\| {\frac{\partial ^2e_k (x,t)}{\partial x^2}}\right\|_{L^2(\Omega )}^2 +(1+\frac{1}{\varepsilon })C^2\left\|{\frac{\partial ^2(\delta Q_k (x,t))}{\partial x^2}} \right\|_{L^2(\Omega)}^2 .\end{equation} (3.12)
由系统(2.2)及(3.2)式有
\begin{equation}\frac{\partial ^\alpha (\delta Q_k (x,t))}{\partial t^\alpha}=D\frac{\partial ^2(\delta Q_k (x,t))}{\partial x^2}+A(t)\delta Q_k(x,t)+qB(t)e_k (x,t).\end{equation} (3.13)
下面分别就抛物型和双曲型系统给出本文的两个主要结论:

定理3.1   若假设2.1-2.4及(3.3)式成立,则抛物型($\alpha=1)$系统(2.2)在P型学习律(3.2)作用下于$W^{1,2}[0, 1]$空间内是收敛的,即

\[\mathop {\lim }\limits_{k\to \infty } \left\| {e_k (x,t)}\right\|_{L^2[0, 1],s} =0,~~\mathop {\lim }\limits_{k\to \infty }\left\| {\frac{\partial e_k (x,t)}{\partial x}}\right\|_{L^2[0, 1],s} =0.\] 并有 \[\mathop {\lim }\limits_{k\to \infty } \left\| {\frac{\partial ^2e_k(x,t)}{\partial x^2}} \right\|_{L^2(\Omega )} =\mathop {\lim}\limits_{k\to \infty } \sqrt {\int_0^T {\int_0^1 {\left({\frac{\partial ^2e_k (x,t)}{\partial x^2}} \right)^2{\rm d} x {\rm d} t} } } =0.\]

证   (1) 证$\lim \limits_{k\to \infty }\left\| {e_k (x,t)} \right\|_{L^2[0, 1],s} =0$.

对(3.13)式两端用向量$\delta Q_k (x,t)$作内积,得

\begin{eqnarray}\left( {\delta Q_k (x,t)} \right)^{\rm T}\frac{\partial (\delta Q_k(x,t))}{\partial t}&=&\left( {\delta Q_k (x,t)} \right)^{\rm T}D\frac{\partial^2(\delta Q_k (x,t))}{\partial x^2}+\left( {\delta Q_k (x,t)} \right)^{\rm T}A(t)\delta Q_k (x,t)\\&&+q\left( {\delta Q_k(x,t)} \right)^{\rm T}B(t)e_k (x,t). \end{eqnarray} (3.14)
而 \[{\left( {\delta {Q_k}(x,t)} \right)^{\text{T}}}\frac{{\partial (\delta {Q_k}(x,t))}}{{\partial t}} = \frac{1}{2}\frac{\partial }{{\partial t}}{\left\| {\delta {Q_k}(x,t)} \right\|^2}.\] 所以由(3.14)式及假设2.2有 \[\begin{gathered} \frac{\partial }{{\partial t}}{\left\| {\delta {Q_k}(x,t)} \right\|^2} \leqslant 2{\left( {\delta {Q_k}(x,t)} \right)^{\text{T}}}D\frac{{{\partial ^2}(\delta {Q_k}(x,t))}}{{\partial {x^2}}} \\ \;\; + 2\left\| {A(t)} \right\|{\left\| {\delta {Q_k}(x,t)} \right\|^2} + 2q\left\| {B(t)} \right\|\left\| {\delta {Q_k}(x,t)} \right\|\left| {{e_k}(x,t)} \right| \\ \leqslant 2{\left( {\delta {Q_k}(x,t)} \right)^{\text{T}}}D\frac{{{\partial ^2}(\delta {Q_k}(x,t))}}{{\partial {x^2}}} + (2A + qB){\left\| {\delta {Q_k}(x,t)} \right\|^2} + qB{\left| {{e_k}(x,t)} \right|^2}. \\ \end{gathered} \] 上式两端对变量$x$从$0$到$1$积分,有
\[\begin{gathered} \frac{{\text{d}}}{{{\text{d}}t}}\left\| {\delta {Q_k}(x,t)} \right\|_{{L^2}[0,1]}^2 \leqslant 2\int_0^1 {\left\{ {{{\left( {\delta {Q_k}(x,t)} \right)}^{\text{T}}}D\frac{{{\partial ^2}(\delta {Q_k}(x,t))}}{{\partial {x^2}}}} \right\}{\text{d}}x} \\ \;\; + (2A + qB)\left\| {\delta {Q_k}(x,t)} \right\|_{{L^2}[0,1]}^2 + qB\left\| {{e_k}(x,t)} \right\|_{{L^2}[0,1]}^2. \\ \end{gathered} \] (3.15)
由分部积分公式,有 \begin{eqnarray*}&&\int_0^1 {\left\{ {\left( {\delta Q_k (x,t)} \right)^{\rm T}D\frac{\partial^2(\delta Q_k (x,t))}{\partial x^2}} \right\}{\rm d} x} \\&=&\left. {\left\{ {\left({\delta Q_k (x,t)} \right)^{\rm T}D\frac{\partial (\delta Q_k (x,t))}{\partial x}}\right\}} \right|_0^1 -\int_0^1 {\left\{ {\left( {\frac{\partial (\delta Q_k (x,t))}{\partial x}}\right)^{\rm T}D\frac{\partial (\delta Q_k (x,t))}{\partial x}} \right\}{\rm d} x} .\end{eqnarray*} 由假设2.4中的边值定位条件,可知 \[\left. {\left\{ {{{\left( {\delta {Q_k}(x,t)} \right)}^{\text{T}}}D\frac{{\partial (\delta {Q_k}(x,t))}}{{\partial x}}} \right\}} \right|_0^1 = 0.\] 因$D$为正定阵,有 \[\int_0^1 {\left\{ {\left( {\frac{\partial (\delta Q_k (x,t))}{\partial x}}\right)^{\rm T}D\frac{\partial (\delta Q_k (x,t))}{\partial x}} \right\}{\rm d} x} \ge0,\] 所以 \begin{eqnarray*}&&\int_0^1 {\left\{ {\left( {\delta Q_k (x,t)} \right)^{\rm T}D\frac{\partial^2(\delta Q_k (x,t))}{\partial x^2}} \right\}{\rm d} x}\\& =&-\int_0^1 {\left\{ {\left( {\frac{\partial (\delta Q_k (x,t))}{\partial x}}\right)^{\rm T}D\frac{\partial (\delta Q_k (x,t))}{\partial x}} \right\}{\rm d} x} \le0.\end{eqnarray*} (3.15)式结合上式,有
\[\frac{{\text{d}}}{{{\text{d}}t}}\left\| {\delta {Q_k}(x,t)} \right\|_{{L^2}[0,1]}^2 \leqslant (2A + qB)\left\| {\delta {Q_k}(x,t)} \right\|_{{L^2}[0,1]}^2 + qB\left\| {{e_k}(x,t)} \right\|_{{L^2}[0,1]}^2.\] (3.16)
利用假设2.4中的初值定位条件,应用Gronwall引理,有 \begin{eqnarray*}\left\| {\delta Q_k (x,t)} \right\|_{L^2[0, 1]}^2 &\le& qB\int_0^t{{\rm e}^{(2A+qB)(t-\tau )}\left\| {e_k (x,\tau )} \right\|_{L^2[0, 1]}^2 {\rm d} \tau }\\&\le&qB{\rm e}^{(2A+qB)T}\int_0^t {\left\| {e_k (x,\tau )} \right\|_{L^2[0, 1]}^2 {\rm d} \tau}\\& =&qB{\rm e}^{(2A+qB)T}\int_0^t {{\rm e}^{\lambda \tau }{\rm e}^{-\lambda \tau }\left\| {e_k(x,\tau )} \right\|_{L^2[0, 1]}^2 {\rm d} \tau } \\&\le&qB{\rm e}^{(2A+qB)T}\int_0^t {{\rm e}^{\lambda \tau }{\rm d} \tau } \left\| {e_k (x,t)}\right\|_{L^2[0, 1],\lambda } \\&=&qB{\rm e}^{(2A+qB)T}\frac{{\rm e}^{\lambda t}-1}{\lambda }\left\| {e_k (x,t)}\right\|_{L^2[0, 1],\lambda } .\end{eqnarray*} 由此有 \begin{eqnarray*}\left\| {\delta Q_k (x,t)} \right\|_{L^2[0, 1],\lambda } &=&\mathop {\max}\limits_{t\in [0,T]} {\rm e}^{-\lambda t}\left\| {\delta Q_k (x,t)}\right\|_{L^2[0, 1]}^2 \\&\le&qB{\rm e}^{(2A+qB)T}\left\| {e_k (x,t)} \right\|_{L^2[0, 1],\lambda } \mathop {\max}\limits_{t\in [0,T]} \frac{1-{\rm e}^{-\lambda t}}{\lambda }\\&=&qB{\rm e}^{(2A+qB)T}\frac{1-{\rm e}^{-\lambda T}}{\lambda }\left\| {e_k (x,t)}\right\|_{L^2[0, 1],\lambda } .\end{eqnarray*} 记$O(\lambda ^{-1})=qB{\rm e}^{(2A+qB)T}\frac{1-{e}^{-\lambda T}}{\lambda }$,则有
\begin{equation}\left\| {\delta Q_k (x,t)} \right\|_{L^2[0, 1],\lambda } \le O(\lambda^{-1})\left\| {e_k (x,t)} \right\|_{L^2[0, 1],\lambda } .\end{equation} (3.17)
显然,当$\lambda $足够大时,能使$O(\lambda ^{-1})$任意小.

将(3.17)式代入(3.7)式,有

\[\left\| {e_{k+1} (x,t)} \right\|_{L^2[0, 1],\lambda } \le \left\{ {\rho+(1+\frac{1}{\varepsilon })C^2O(\lambda ^{-1})} \right\}\left\| {e_k (x,t)}\right\|_{L^2[0, 1],\lambda } .\] 由(3.3)式有 $\rho \mbox{=}(1+\varepsilon )\gamma ^2<1$,所以取足够大的$\lambda $,能使得$\rho +(1+\frac{1}{\varepsilon})C^2O(\lambda ^{-1})<1$成立,由此可知 \[\mathop {\lim }\limits_{k\to \infty } \left\| {e_k (x,t)}\right\|_{L^2[0, 1],\lambda } =0.\] 即
\begin{equation}\mathop {\lim }\limits_{k\to \infty } \left\| {e_k (x,t)}\right\|_{L^2[0, 1],s} =0.\end{equation} (3.18)

(2) 证$\mathop {\lim }\limits_{k\to \infty } \left\|{\frac{\partial e_k (x,t)}{\partial x}} \right\|_{L^2[0, 1],s} =0$.

对(3.13)式两端用向量$\frac{{{\partial ^2}(\delta {Q_k}(x,t))}}{{\partial {x^2}}}$作内积,得

\begin{eqnarray}&&\left( {\frac{\partial ^2(\delta Q_k (x,t))}{\partial x^2}}\right)^{\rm T}\frac{\partial (\delta Q_k (x,t))}{\partial t}\\&=& \left({\frac{\partial ^2(\delta Q_k (x,t))}{\partial x^2}}\right)^{\rm T}D\frac{\partial ^2(\delta Q_k (x,t))}{\partial x^2}\\&&+\left( {\frac{\partial ^2(\delta Q_k (x,t))}{\partial x^2}}\right)^{\rm T}A(t)\delta Q_k (x,t)+q\left( {\frac{\partial ^2(\delta Q_k(x,t))}{\partial x^2}} \right)^{\rm T}B(t)e_k (x,t).\end{eqnarray} (3.19)
利用分部积分公式,结合假设2.4中的边值定位条件,有
\[\begin{gathered} \int_0^1 {\left\{ {{{\left( {\frac{{{\partial ^2}(\delta {Q_k}(x,t))}}{{\partial {x^2}}}} \right)}^{\text{T}}}\frac{{\partial (\delta {Q_k}(x,t))}}{{\partial t}}} \right\}{\text{d}}x} \hfill \\ = \left. {\left\{ {{{\left( {\frac{{\partial (\delta {Q_k}(x,t))}}{{\partial x}}} \right)}^{\text{T}}}\frac{{\partial (\delta {Q_k}(x,t))}}{{\partial t}}} \right\}} \right|_0^1 \hfill \\ - \int_0^1 {\left\{ {{{\left( {\frac{{\partial (\delta {Q_k}(x,t))}}{{\partial x}}} \right)}^{\text{T}}}\frac{{{\partial ^2}(\delta {Q_k}(x,t))}}{{\partial t\partial x}}} \right\}{\text{d}}x} \hfill \\ = - \int_0^1 {\left\{ {{{\left( {\frac{{\partial (\delta {Q_k}(x,t))}}{{\partial x}}} \right)}^{\text{T}}}\frac{\partial }{{\partial t}}\left( {\frac{{\partial (\delta {Q_k}(x,t))}}{{\partial x}}} \right)} \right\}{\text{d}}x} \hfill \\ = - \frac{1}{2}\frac{{\text{d}}}{{{\text{d}}t}}\int_0^1 {\left\{ {{{\left( {\frac{{\partial (\delta {Q_k}(x,t))}}{{\partial x}}} \right)}^{\text{T}}}\frac{{\partial (\delta {Q_k}(x,t))}}{{\partial x}}} \right\}{\text{d}}x} \hfill \\ = - \frac{1}{2}\frac{{\text{d}}}{{{\text{d}}t}}\left\| {\frac{{\partial (\delta {Q_k}(x,t))}}{{\partial x}}} \right\|_{{L^2}[0,1]}^2. \hfill \\ \end{gathered} \] (3.20)
\begin{eqnarray}&&\int_0^1 {\left\{ {\left( {\frac{\partial ^2(\delta Q_k (x,t))}{\partial {x^2}}} \right)^{\rm T}A(t)\delta Q_k (x,t)} \right\}{\rm d} x} \\&=&\left. {\left\{ {\left( {\frac{\partial (\delta Q_k (x,t))}{\partial x}}\right)^{\rm T}A(t)\delta Q_k (x,t)} \right\}} \right|_0^1 \\&&-\int_0^1 {\left\{ {\left( {\frac{\partial (\delta Q_k (x,t))}{\partial x}}\right)^{\rm T}A(t)\frac{\partial (\delta Q_k (x,t))}{\partial x}} \right\}{\rm d} x} \\&=&-\int_0^1 {\left\{ {\left( {\frac{\partial (\delta Q_k (x,t))}{\partial x}}\right)^{\rm T}A(t)\frac{\partial (\delta Q_k (x,t))}{\partial x}} \right\}{\rm d} x} .\end{eqnarray} (3.21)
(3.19)式两端对变量$x$从$0$到$1$积分,结合(3.20),(3.21)式可得
\[\begin{gathered} \frac{{\text{d}}}{{{\text{d}}t}}\left\| {\frac{{\partial (\delta {Q_k}(x,t))}}{{\partial x}}} \right\|_{{L^2}[0,1]}^2 + 2\int_0^1 {\left\{ {{{\left( {\frac{{{\partial ^2}(\delta {Q_k}(x,t))}}{{\partial {x^2}}}} \right)}^{\text{T}}}D\frac{{{\partial ^2}(\delta {Q_k}(x,t))}}{{\partial {x^2}}}} \right\}{\text{d}}x} \\ = 2\int_0^1 {\left\{ {{{\left( {\frac{{\partial (\delta {Q_k}(x,t))}}{{\partial x}}} \right)}^{\text{T}}}A(t)\frac{{\partial (\delta {Q_k}(x,t))}}{{\partial x}}} \right\}{\text{d}}x} \\ - 2q\int_0^1 {\left\{ {{{\left( {\frac{{{\partial ^2}(\delta {Q_k}(x,t))}}{{\partial {x^2}}}} \right)}^{\text{T}}}B(t){e_k}(x,t)} \right\}{\text{d}}x} .{\text{ }} \\ \end{gathered} \] (3.22)
因$D={\rm diag}(d_1 ,d_2 ,\cdots d_n )$,记$\min\limits_{1\le i\le n} \left\{ {d_i } \right\}=d$,则有
\[\int_0^1 {\left\{ {{{\left( {\frac{{{\partial ^2}(\delta {Q_k}(x,t))}}{{\partial {x^2}}}} \right)}^{\text{T}}}D\frac{{{\partial ^2}(\delta {Q_k}(x,t))}}{{\partial {x^2}}}} \right\}{\text{d}}x} \geqslant d\left\| {\frac{{{\partial ^2}(\delta {Q_k}(x,t))}}{{\partial {x^2}}}} \right\|_{{L^2}[0,1]}^2.\] (3.23)
利用基本不等式,由假设2.2可得
\begin{eqnarray}&&-2q\int_0^1 {\left\{ {\left( {\frac{\partial ^2(\delta Q_k (x,t))}{\partial {x^2}}} \right)^{\rm T}B(t)e_k (x,t)} \right\}{\rm d} x}\\& \le&2q\left| {\int_0^1 {\left\{ {\left( {\frac{\partial ^2(\delta Q_k(x,t))}{\partial x^2}} \right)^{\rm T}B(t)e_k (x,t)} \right\}{\rm d} x} } \right|\\&\le&2q\int_0^1 {\left\{ {\left\| {\frac{\partial ^2(\delta Q_k (x,t))}{\partial {x^2}}} \right\|\left\| {B(t)e_k (x,t)} \right\|} \right\}{\rm d} x}\\& \le&2q\int_0^1 {\left\{ {\frac{d}{2q}\left\| {\frac{\partial ^2(\delta Q_k(x,t))}{\partial x^2}} \right\|^2+\frac{q}{2d}B^2\left | {e_k (x,t)}\right |^2} \right\}{\rm d} x} \\&=&d\left\| {\frac{\partial ^2(\delta Q_k (x,t))}{\partial x^2}}\right\|_{L^2[0, 1]}^2 +\frac{q^2B^2}{d}\left\| {e_k (x,t)}\right\|_{L^2[0, 1]}^2 .\end{eqnarray} (3.24)
而
\[\int_0^1 {\left\{ {{{\left( {\frac{{\partial (\delta {Q_k}(x,t))}}{{\partial x}}} \right)}^{\text{T}}}A(t)\frac{{\partial (\delta {Q_k}(x,t))}}{{\partial x}}} \right\}{\text{d}}x} {\text{ }} \leqslant \left\| {\frac{{\partial (\delta {Q_k}(x,t))}}{{\partial x}}} \right\|_{{L^2}[0,1]}^2.\] (3.25)
将(3.23)-(3.25)式代入(3.22)式,可推得
\begin{eqnarray}&&\frac{\rm d}{{\text{d}}t}\left\| {\frac{\partial (\delta Q_k (x,t))}{\partial x}}\right\|_{L^2[0, 1]}^2 +d\left\| {\frac{\partial ^2(\delta Q_k(x,t))}{\partial x^2}} \right\|_{L^2[0, 1]}^2 \\&\le&2A\left\| {\frac{\partial (\delta Q_k (x,t))}{\partial x}}\right\|_{L^2[0, 1]}^2 +\frac{q^2B^2}{d}\left\| {e_k (x,t)}\right\|_{L^2[0, 1]}^2 .\end{eqnarray} (3.26)
进一步有 \[\frac{\rm d}{{\text{d}}t}\left\| {\frac{\partial (\delta Q_k (x,t))}{\partial x}}\right\|_{L^2[0, 1]}^2 \le 2A\left\| {\frac{\partial (\delta Q_k(x,t))}{\partial x}} \right\|_{L^2[0, 1]}^2 +\frac{q^2B^2}{d}\left\| {e_k(x,t)} \right\|_{L^2[0, 1]}^2 .\] 利用假设2.4中的初值定位条件,应用Gronwall引理,可得 \begin{eqnarray*}\left\| {\frac{\partial (\delta Q_k (x,t))}{\partial x}}\right\|_{L^2[0, 1]}^2 &\le & \frac{q^2B^2}{d}\int_0^t {{\rm e}^{2A(t-\tau )}\left\|{e_k (x,\tau )} \right\|_{L^2[0, 1]}^2 {\rm d} \tau } \\&\le& \frac{q^2B^2}{d}{\rm e}^{2AT}T\left\| {e_k (x,t)} \right\|_{L^2[0, 1],s} .\end{eqnarray*} 进一步有
\begin{eqnarray}\left\| {\frac{\partial (\delta Q_k (x,t))}{\partial x}}\right\|_{L^2[0, 1],s} &=&\mathop {\max }\limits_{t\in [0,T]} \left\|{\frac{\partial (\delta Q_k (x,t))}{\partial x}} \right\|_{L^2[0, 1]}^2 \\&\le& \frac{q^2B^2}{d}{\rm e}^{2AT}T\left\| {e_k (x,t)} \right\|_{L^2[0, 1],s} .\end{eqnarray} (3.27)
对(3.8)式重复从(3.4)式到(3.6)式的处理过程,有 \[\left\| {\frac{\partial e_{k+1} (x,t)}{\partial x}} \right\|_{L^2[0, 1]}^2\le \rho \left\| {\frac{\partial e_k (x,t)}{\partial x}}\right\|_{L^2[0, 1]}^2 +(1+\frac{1}{\varepsilon })C^2\left\| {\frac{\partial(\delta Q_k (x,t))}{\partial x}} \right\|_{L^2[0, 1]}^2 .\] 由此有 \begin{eqnarray*}\left\| {\frac{\partial e_{k+1} (x,t)}{\partial x}} \right\|_{L^2[0, 1],s}&=&\mathop {\max }\limits_{t\in [0,T]} \left\| {\frac{\partial e_{k+1}(x,t)}{\partial x}} \right\|_{L^2[0, 1]}^2 \\&\le&\rho \mathop {\max }\limits_{t\in [0,T]} \left\| {\frac{\partial e_k(x,t)}{\partial x}} \right\|_{L^2[0, 1]}^2 +(1+\frac{1}{\varepsilon})C^2\mathop {\max }\limits_{t\in [0,T]} \left\| {\frac{\partial (\delta Q_k(x,t))}{\partial x}} \right\|_{L^2[0, 1]}^2\\& =&\rho \left\| {\frac{\partial e_k (x,t)}{\partial x}} \right\|_{L^2[0, 1],s}+(1+\frac{1}{\varepsilon })C^2\left\| {\frac{\partial (\delta Q_k(x,t))}{\partial x}} \right\|_{L^2[0, 1],s} .\end{eqnarray*} 将(3.27)式代入上式,有 \[\left\| {\frac{\partial e_{k+1} (x,t)}{\partial x}} \right\|_{L^2[0, 1],s}\le \rho \left\| {\frac{\partial e_k (x,t)}{\partial x}}\right\|_{L^2[0, 1],s} +(1+\frac{1}{\varepsilon})C^2\frac{q^2B^2}{d}{\rm e}^{2AT}T\left\| {e_k (x,t)} \right\|_{L^2[0, 1],s} .\] 对上式应用引理2.1,结合(3.18)式,可知
\begin{equation}\mathop {\lim }\limits_{k\to \infty } \left\| {\frac{\partial e_k(x,t)}{\partial x}} \right\|_{L^2[0, 1],s} =0.\end{equation} (3.28)

(3) 证$\mathop {\lim }\limits_{k\to \infty } \left\|{\frac{\partial ^2e_k (x,t)}{\partial x^2}} \right\|_{L^2(\Omega )}=0$.

(3.26)式两端对变量$t$从$0$到$T$积分,结合假设2.4中的初值定位条件,有

\begin{eqnarray*}&&\left\| {\frac{\partial (\delta Q_k (x,T))}{\partial x}}\right\|_{L^2[0, 1]}^2 +d\int_0^T {\left\| {\frac{\partial ^2(\delta Q_k(x,t))}{\partial x^2}} \right\|_{L^2[0, 1]}^2 {\rm d} t} \\&\le& 2A\int_0^T {\left\| {\frac{\partial (\delta Q_k (x,t))}{\partial x}}\right\|_{L^2[0, 1]}^2 {\rm d} t} +\frac{q^2B^2}{d}\int_0^T {\left\| {e_k (x,t)}\right\|_{L^2[0, 1]}^2 {\rm d} t} .\end{eqnarray*} 由此可得
\begin{eqnarray}\left\| {\frac{\partial ^2(\delta Q_k (x,t))}{\partial x^2}}\right\|_{L^2(\Omega )}^2 &=&\int_0^T {\int_0^1 {\left( {\frac{\partial^2(\delta Q_k (x,t))}{\partial x^2}} \right)^2{\rm d} x {\rm d} t} } \\&=&\int_0^T {\left\|{\frac{\partial ^2(\delta Q_k (x,t))}{\partial x^2}} \right\|_{L^2[0, 1]}^2{\rm d} t} \\&\le&\frac{2A}{d}\int_0^T {\left\| {\frac{\partial (\delta Q_k (x,t))}{\partial x}} \right\|_{L^2[0, 1]}^2 {\rm d} t}\\&& +\frac{q^2B^2}{d^2}\int_0^T {\left\| {e_k(x,t)} \right\|_{L^2[0, 1]}^2 {\rm d} t} .\end{eqnarray} (3.29)
由(3.18),(3.27)式可得 \[\mathop {\lim }\limits_{k\to \infty } \int_0^T {\left\| {e_k (x,t)}\right\|_{L^2[0, 1]}^2 {\rm d} t=\mathop {\lim }\limits_{k\to \infty } \int_0^T{\left\| {\frac{\partial (\delta Q_k (x,t))}{\partial x}}\right\|_{L^2[0, 1]}^2 {\rm d} t} =0} .\] (3.29)式结合上式,可知 \[\mathop {\lim }\limits_{k\to \infty } \left\| {\frac{\partial ^2(\delta Q_k(x,t))}{\partial x^2}} \right\|_{L^2(\Omega )}^2 =0.\] 对(3.12)式应用引理2.1,结合上式,即得
\begin{equation}\mathop {\lim }\limits_{k\to \infty } \left\| {\frac{\partial ^2e_k(x,t)}{\partial x^2}} \right\|_{L^2(\Omega )} =0.\end{equation} (3.30)
综合(3.18),(3.28),(3.30)式,可知定理的结论成立. 证毕.

注3.1   针对抛物型系统,文献[12]使用的是高阶学习律[18],通过先证控制变量$u_k$收敛($L^2$意义下)于理想控制$u_r $,然后再证得输出变量$y_k$收敛($L^2$意义下)于理想控制$y_r $. 相对于高阶学习律,本文使用的是一阶学习律[18],采用的是直接证明输出变量$y_k$收敛($L^2$意义下)于理想控制$y_r $. 由此,本文中的控制器设计和定理3.1中的第1部分证明过程与文献[12]是不同的.

定理3.2   若假设2.1-2.4及(3.3)式成立,则双曲型($\alpha=2)$系统(2.2)在P型学习律(3.2)作用下于$W^{2,2}[0, 1]$空间内是收敛的,即\[\mathop {\lim }\limits_{k\to \infty } \left\| {e_k (x,t)}\right\|_{L^2[0, 1],s} =0,\mathop {\lim }\limits_{k\to \infty }\left\| {\frac{\partial e_k (x,t)}{\partial x}}\right\|_{L^2[0, 1],s} =0,\mathop {\lim }\limits_{k\to \infty }\left\| {\frac{\partial ^2e_k (x,t)}{\partial x^2}}\right\|_{L^2[0, 1],s} =0.\]

证   (1) 由于双曲型($\alpha=2)$系统(2.1)包含于文献[16]研究的系统类型中,由文献[16]的结论有

\begin{equation}\mathop {\lim }\limits_{k\to \infty } \left\| {e_k (x,t)}\right\|_{L^2[0, 1],s} =0.\end{equation} (3.31)

(2) 证$\mathop {\lim }\limits_{k\to \infty } \left\|{\frac{\partial e_k (x,t)}{\partial x}} \right\|_{L^2[0, 1],s} =0$.

对(3.13)式两端用向量$\frac{\partial }{\partial t}\left({\frac{\partial ^2(\delta Q_k (x,t))}{\partial x^2}} \right)$作内积,得

\begin{eqnarray}&&\left( {\frac{\partial }{\partial t}\left( {\frac{\partial ^2(\delta {Q_k}(x,t))}{\partial x^2}} \right)} \right)^{\rm T}\frac{\partial ^2(\delta {Q_k}(x,t))}{\partial t^2}\\&=&\left( {\frac{\partial }{\partial t}\left({\frac{\partial ^2(\delta {Q_k} (x,t))}{\partial x^2}} \right)}\right)^{\rm T}D\frac{\partial ^2(\delta Q_k (x,t))}{\partial x^2}+ \left( {\frac{\partial }{\partial t}\left( {\frac{\partial ^2(\delta {Q_k}(x,t))}{\partial x^2}} \right)} \right)^{\rm T}A(t)\delta Q_k (x,t)\\&&+q\left({\frac{\partial }{\partial t}\left( {\frac{\partial ^2(\delta Q_k(x,t))}{\partial x^2}} \right)} \right)^{\rm T}B(t)e_k (x,t).\end{eqnarray} (3.32)
利用分部积分公式,结合假设2.4中的边值定位条件,有
\[\begin{gathered} \int_0^1 {\left\{ {{{\left( {\frac{\partial }{{\partial t}}\left( {\frac{{{\partial ^2}(\delta {Q_k}(x,t))}}{{\partial {x^2}}}} \right)} \right)}^{\text{T}}}\frac{{{\partial ^2}(\delta {Q_k}(x,t))}}{{\partial {t^2}}}} \right\}{\text{d}}x} \hfill \\ \left. { = \left\{ {{{\left( {\frac{\partial }{{\partial t}}\left( {\frac{{\partial (\delta {Q_k}(x,t))}}{{\partial x}}} \right)} \right)}^{\text{T}}}\frac{{{\partial ^2}(\delta {Q_k}(x,t))}}{{\partial {t^2}}}} \right\}} \right|_0^1 \hfill \\ - \int_0^1 {\left\{ {{{\left( {\frac{\partial }{{\partial t}}\left( {\frac{{\partial (\delta {Q_k}(x,t))}}{{\partial x}}} \right)} \right)}^{\text{T}}}\frac{\partial }{{\partial x}}\left( {\frac{{{\partial ^2}(\delta {Q_k}(x,t))}}{{\partial {t^2}}}} \right)} \right\}{\text{d}}x} \hfill \\ = - \int_0^1 {\left\{ {{{\left( {\frac{\partial }{{\partial t}}\left( {\frac{{\partial (\delta {Q_k}(x,t))}}{{\partial x}}} \right)} \right)}^{\text{T}}}\frac{\partial }{{\partial t}}\left( {\frac{\partial }{{\partial t}}\left( {\frac{{\partial (\delta {Q_k}(x,t))}}{{\partial x}}} \right)} \right)} \right\}{\text{d}}x} \hfill \\ = - \frac{1}{2}\frac{{\text{d}}}{{{\text{d}}t}}\int_0^1 {\left\{ {{{\left( {\frac{\partial }{{\partial t}}\left( {\frac{{\partial (\delta {Q_k}(x,t))}}{{\partial x}}} \right)} \right)}^{\text{T}}}\left( {\frac{\partial }{{\partial t}}\left( {\frac{{\partial (\delta {Q_k}(x,t))}}{{\partial x}}} \right)} \right)} \right\}{\text{d}}x} \hfill \\ = - \frac{1}{2}\frac{{\text{d}}}{{{\text{d}}t}}\left\| {\frac{\partial }{{\partial t}}\left( {\frac{{\partial (\delta {Q_k}(x,t))}}{{\partial x}}} \right)} \right\|_{{L^2}[0,1]}^2. \hfill \\ \end{gathered} \] (3.33)
\begin{eqnarray}&&\int_0^1 {\left\{ {\left( {\frac{\partial }{\partial t}\left({\frac{\partial ^2(\delta Q_k (x,t))}{\partial x^2}} \right)}\right)^{\rm T}A(t)\delta Q_k (x,t)} \right\}{\rm d} x}\\& =&\left. {\left\{ {\left( {\frac{\partial }{\partial t}\left( {\frac{\partial(\delta Q_k (x,t))}{\partial x}} \right)} \right)^{\rm T}A(t)\delta Q_k (x,t)}\right\}} \right|_0^1 \\&&-\int_0^1 {\left\{ {\left( {\frac{\partial }{\partial t}\left({\frac{\partial (\delta Q_k (x,t))}{\partial x}} \right)}\right)^{\rm T}A(t)\frac{\partial (\delta Q_k (x,t))}{\partial x}} \right\}{\rm d} x}\\&=&-\int_0^1 {\left\{ {\left( {\frac{\partial }{\partial t}\left({\frac{\partial (\delta Q_k (x,t))}{\partial x}} \right)}\right)^{\rm T}A(t)\frac{\partial (\delta Q_k (x,t))}{\partial x}} \right\}{\rm d} x} .\end{eqnarray} (3.34)
\begin{eqnarray}&&\int_0^1 {\left\{ {\left( {\frac{\partial }{\partial t}\left({\frac{\partial ^2(\delta Q_k (x,t))}{\partial x^2}} \right)}\right)^{\rm T}B(t)e_k (x,t)} \right\}{\rm d} x} \\&=&\left. {\left\{ {\left( {\frac{\partial }{\partial t}\left( {\frac{\partial(\delta Q_k (x,t))}{\partial x}} \right)} \right)^{\rm T}B(t)e_k (x,t)} \right\}}\right|_0^1 \\&&- \int_0^1 {\left\{ {\left( {\frac{\partial }{\partial t}\left({\frac{\partial (\delta Q_k (x,t))}{\partial x}} \right)}\right)^{\rm T}B(t)\frac{\partial e_k (x,t)}{\partial x}} \right\}{\rm d} x} \\&=&-\int_0^1 {\left\{ {\left( {\frac{\partial }{\partial t}\left({\frac{\partial (\delta Q_k (x,t))}{\partial x}} \right)}\right)^{\rm T}B(t)\frac{\partial e_k (x,t)}{\partial x}} \right\}{\rm d} x} .\end{eqnarray} (3.35)
\[\begin{gathered} \int_0^1 {\left\{ {{{\left( {\frac{\partial }{{\partial t}}\left( {\frac{{{\partial ^2}(\delta {Q_k}(x,t))}}{{\partial {x^2}}}} \right)} \right)}^{\text{T}}}D\frac{{{\partial ^2}(\delta {Q_k}(x,t))}}{{\partial {x^2}}}} \right\}{\text{d}}x} \hfill \\ = \frac{1}{2}\frac{{\text{d}}}{{{\text{d}}t}}\int_0^1 {\left\{ {{{\left( {\frac{{{\partial ^2}(\delta {Q_k}(x,t))}}{{\partial {x^2}}}} \right)}^{\text{T}}}D\frac{{{\partial ^2}(\delta {Q_k}(x,t))}}{{\partial {x^2}}}} \right\}{\text{d}}x} . \hfill \\ \end{gathered} \] (3.36)
(3.32)式两端对变量$x$从$0$到$1$积分,结合(3.33)-(3.36)式及假设2.2,可得 \begin{eqnarray*}&&\frac{{\text{d}}}{{{\text{d}}t}}\left\| {\frac{\partial }{\partial t}\left( {\frac{\partial(\delta Q_k (x,t))}{\partial x}} \right)} \right\|_{L^2[0, 1]}^2+\frac{{\text{d}}}{{{\text{d}}t}}\int_0^1 {\left\{ {\left( {\frac{\partial ^2(\delta Q_k(x,t))}{\partial x^2}} \right)^{\rm T}D\frac{\partial ^2(\delta Q_k(x,t))}{\partial x^2}} \right\}{\rm d} x} \\&=&2\int_0^1 {\left\{ {\left( {\frac{\partial }{\partial t}\left({\frac{\partial (\delta Q_k (x,t))}{\partial x}} \right)}\right)^{\rm T}A(t)\frac{\partial (\delta Q_k (x,t))}{\partial x}} \right\}{\rm d} x} \\&&+2q\int_0^1 {\left\{ {\left( {\frac{\partial }{\partial t}\left({\frac{\partial (\delta Q_k (x,t))}{\partial x}} \right)}\right)^{\rm T}B(t)\frac{\partial e_k (x,t)}{\partial x}} \right\}{\rm d} x}\\& \le&2\left\| {A(t)} \right\|\int_0^1 {\left\{ {\left\| {\frac{\partial}{\partial t}\left( {\frac{\partial (\delta Q_k (x,t))}{\partial x}}\right)} \right\|\left\| {\frac{\partial (\delta Q_k (x,t))}{\partial x}}\right\|} \right\}{\rm d} x} \\&&+2q\left\| {B(t)} \right\|\int_0^1 {\left\{ {\left\| {\frac{\partial}{\partial t}\left( {\frac{\partial (\delta Q_k (x,t))}{\partial x}}\right)} \right\|\left| {\frac{\partial e_k (x,t)}{\partial x}} \right|}\right\}{\rm d} x}\\& \le&A\left\{ {\left\| {\frac{\partial }{\partial t}\left( {\frac{\partial(\delta Q_k (x,t))}{\partial x}} \right)} \right\|_{L^2[0, 1]}^2 +\left\|{\frac{\partial (\delta Q_k (x,t))}{\partial x}} \right\|_{L^2[0, 1]}^2 }\right\}\\&&+qB\left\{ {\left\| {\frac{\partial }{\partial t}\left( {\frac{\partial(\delta Q_k (x,t))}{\partial x}} \right)} \right\|_{L^2[0, 1]}^2 +\left\|{\frac{\partial e_k (x,t)}{\partial x}} \right\|_{L^2[0, 1]}^2 } \right\}.\end{eqnarray*} 上式两端对变量$t$从$0$到$t$积分,结合假设2.4中的初值定位条件,有 \begin{eqnarray*}&&\left\| {\frac{\partial }{\partial t}\left( {\frac{\partial (\delta Q_k(x,t))}{\partial x}} \right)} \right\|_{L^2[0, 1]}^2 +\int_0^1 {\left\{{\left( {\frac{\partial ^2(\delta Q_k (x,t))}{\partial x^2}}\right)^{\rm T}D\frac{\partial ^2(\delta Q_k (x,t))}{\partial x^2}} \right\}{\rm d} x}\\&\le&(A+qB)\int_0^t {\left\{ {\left\| {\frac{\partial }{\partial \xi}\left( {\frac{\partial (\delta Q_k (x,\xi ))}{\partial x}} \right)}\right\|_{L^2[0, 1]}^2 } \right\}{\rm d} \xi }+A\int_0^t {\left\{ {\left\|{\frac{\partial (\delta Q_k (x,\xi ))}{\partial x}} \right\|_{L^2[0, 1]}^2 }\right\}{\rm d} \xi }\\&&+ qB\int_0^t {\left\{ {\left\| {\frac{\partial e_k (x,\xi )}{\partial x}}\right\|_{L^2[0, 1]}^2 } \right\}{\rm d} \xi } .\end{eqnarray*} 因$D$为正定阵,所以有
\begin{eqnarray}&&\left\| {\frac{\partial }{\partial t}\left( {\frac{\partial (\delta Q_k(x,t))}{\partial x}} \right)} \right\|_{L^2[0, 1]}^2 \\&\le& \left\|{\frac{\partial }{\partial t}\left( {\frac{\partial (\delta Q_k(x,t))}{\partial x}} \right)} \right\|_{L^2[0, 1]}^2 +\int_0^1 {\left\{ {\left( {\frac{\partial ^2(\delta Q_k (x,t))}{\partial {x^2}}} \right)^{\rm T}D\frac{\partial ^2(\delta Q_k (x,t))}{\partial {x^2}}}\right\}{\rm d} x}\\& \le&(A+qB)\int_0^t {\left\{ {\left\| {\frac{\partial }{\partial \xi}\left( {\frac{\partial (\delta Q_k (x,\xi ))}{\partial x}} \right)}\right\|_{L^2[0, 1]}^2 } \right\}{\rm d} \xi }\\&&+A\int_0^t {\left\{ {\left\| {\frac{\partial (\delta Q_k (x,\xi ))}{\partial x}} \right\|_{L^2[0, 1]}^2 } \right\}{\rm d} \xi +} qB\int_0^t {\left\{ {\left\|{\frac{\partial e_k (x,\xi )}{\partial x}} \right\|_{L^2[0, 1]}^2 }\right\}{\rm d} \xi } \\&\le&(A+qB)\frac{{\rm e}^{\lambda t}-1}{\lambda }\left\| {\frac{\partial }{\partial t}\left({\frac{\partial (\delta Q_k (x,t))}{\partial x}} \right)}\right\|_{L^2[0, 1],\lambda } \\&&+ A\frac{{\rm e}^{\lambda t}-1}{\lambda}\left\| {\frac{\partial (\delta Q_k (x,t))}{\partial x}}\right\|_{L^2[0, 1],\lambda }+ qB\frac{{\rm e}^{\lambda t}-1}{\lambda }\left\| {\frac{\partial e_k(x,t)}{\partial x}} \right\|_{L^2[0, 1],\lambda } .\end{eqnarray} (3.37)
由此可得 \begin{eqnarray*}&&\left\| {\frac{\partial }{\partial t}\left( {\frac{\partial (\delta Q_k(x,t))}{\partial x}} \right)} \right\|_{L^2[0, 1],\lambda }\\& =&\max\limits_{t\in [0,T]} {\rm e}^{-\lambda t}\left\| {\frac{\partial }{\partial t}\left( {\frac{\partial (\delta Q_k (x,t))}{\partial x}} \right)}\right\|_{L^2[0, 1]}^2 \\&\le&(A+qB)\frac{1-{\rm e}^{\lambda T}}{\lambda }\left\| {\frac{\partial }{\partial t}\left({\frac{\partial (\delta Q_k (x,t))}{\partial x}} \right)}\right\|_{L^2[0, 1],\lambda } \\&&+ A\frac{1-{\rm e}^{\lambda T}}{\lambda}\left\| {\frac{\partial (\delta Q_k (x,t))}{\partial x}}\right\|_{L^2[0, 1],\lambda } + qB\frac{1-{\rm e}^{\lambda T}}{\lambda }\left\| {\frac{\partial e_k(x,t)}{\partial x}} \right\|_{L^2[0, 1],\lambda } .\end{eqnarray*} 取较大的$\lambda $,使得 \[(A+qB)\frac{1-{\rm e}^{\lambda T}}{\lambda }<1\] 成立,则有
\begin{eqnarray}\left\| {\frac{\partial }{\partial t}\left( {\frac{\partial (\delta Q_k(x,t))}{\partial x}} \right)} \right\|_{L^2[0, 1],\lambda }& \le&\frac{A\frac{1-{\rm e}^{-\lambda T}}{\lambda }}{1-(A+qB)\frac{1-{\rm e}^{-\lambda T}}{\lambda}}\left\| {\frac{\partial (\delta Q_k (x,t))}{\partial x}}\right\|_{L^2[0, 1],\lambda } \\&&+\frac{qB\frac{1-{\rm e}^{-\lambda T}}{\lambda }}{1-(A+qB)\frac{1-{\rm e}^{-\lambda T}}{\lambda}}\left\| {\frac{\partial e_k (x,t)}{\partial x}} \right\|_{L^2[0, 1],\lambda} .\end{eqnarray} (3.38)
由假设2.4中的初值定位条件,有 \begin{eqnarray*}\left\| {\frac{\partial (\delta Q_k (x,t))}{\partial x}}\right\|_{L^2[0, 1]}^2 &=&\int_0^t {\frac{\rm d}{\rm {d} \eta }} \left\| {\frac{\partial(\delta Q_k (x,\eta ))}{\partial x}} \right\|_{L^2[0, 1]}^2 {\rm d} \eta \\&=&\int_0^t {\frac{\rm d}{\rm {d} \eta }} \left\{ {\int_0^1 {\left\{ {\left({\frac{\partial (\delta Q_k (x,\eta ))}{\partial x}} \right)^{\rm T}\frac{\partial(\delta Q_k (x,\eta ))}{\partial x}} \right\}{\rm d} x} } \right\}{\rm d} \eta\\& =&\int_0^t {\int_0^1 {\frac{\partial }{\partial \eta }\left\{ {\left({\frac{\partial (\delta Q_k (x,\eta ))}{\partial x}} \right)^{\rm T}\frac{\partial(\delta Q_k (x,\eta ))}{\partial x}} \right\}{\rm d} x} } {\rm d} \eta\\& =&2\int_0^t {\int_0^1 {\left\{ {\left( {\frac{\partial (\delta Q_k (x,\eta))}{\partial x}} \right)^{\rm T}\frac{\partial }{\partial \eta }\left({\frac{\partial (\delta Q_k (x,\eta ))}{\partial x}} \right)} \right\}{\rm d} x} }{\rm d} \eta .\end{eqnarray*} 由Cauchy-Schwarz不等式,有 \begin{eqnarray*}&&\left\| {\frac{\partial (\delta Q_k (x,t))}{\partial x}}\right\|_{L^2[0, 1]}^2 \\&\le& 2\int_0^t {\left\{ {\sqrt {\int_0^1 {\left\|{\frac{\partial (\delta Q_k (x,\eta ))}{\partial x}} \right\|^2} {\rm d} x} \sqrt{\int_0^1 {\left\| {\frac{\partial }{\partial \eta }\left( {\frac{\partial(\delta Q_k (x,\eta ))}{\partial x}} \right)} \right\|^2} {\rm d} x} } \right\}}{\rm d} \eta \\&=&2\int_0^t {\left\{ {\left\| {\frac{\partial (\delta Q_k (x,\eta ))}{\partial x}} \right\|_{L^2[0, 1]} \left\| {\frac{\partial }{\partial \eta }\left({\frac{\partial (\delta Q_k (x,\eta ))}{\partial x}} \right)}\right\|_{L^2[0, 1]} } \right\}} {\rm d} \eta \\&\le&\int_0^t {\left\{ {\left\| {\frac{\partial (\delta Q_k (x,\eta ))}{\partial x}} \right\|_{L^2[0, 1]}^2 +\left\| {\frac{\partial }{\partial \eta }\left({\frac{\partial (\delta Q_k (x,\eta ))}{\partial x}} \right)}\right\|_{L^2[0, 1]}^2 } \right\}} {\rm d} \eta .\end{eqnarray*} 应用Gronwall引理,得 \begin{eqnarray*}\left\| {\frac{\partial (\delta Q_k (x,t))}{\partial x}}\right\|_{L^2[0, 1]}^2 &\le& \int_0^t {\left\{ {{\rm e}^{t-\eta }\left\|{\frac{\partial }{\partial \eta }\left( {\frac{\partial (\delta Q_k (x,\eta))}{\partial x}} \right)} \right\|_{L^2[0, 1]}^2 } \right\}} {\rm d} \eta \\&=&{\rm e}^t\int_0^t {\left\{ {{\rm e}^{(\lambda -1)\eta }{\rm e}^{-\lambda \eta }\left\|{\frac{\partial }{\partial \eta }\left( {\frac{\partial (\delta Q_k (x,\eta))}{\partial x}} \right)} \right\|_{L^2[0, 1]}^2 } \right\}} {\rm d} \eta \\&\le&{\rm e}^t\int_0^t {{\rm e}^{(\lambda -1)\eta }} {\rm d} \eta \left\| {\frac{\partial }{\partial t}\left( {\frac{\partial (\delta Q_k (x,t))}{\partial x}} \right)}\right\|_{L^2[0, 1],\lambda } \\&=&\frac{{\rm e}^{\lambda t}-{\rm e}^t}{\lambda -1}\left\| {\frac{\partial }{\partial t}\left( {\frac{\partial (\delta Q_k (x,t))}{\partial x}} \right)}\right\|_{L^2[0, 1],\lambda } .\end{eqnarray*} 取$\lambda >1$,则
\begin{eqnarray}\left\| {\frac{\partial (\delta Q_k (x,t))}{\partial x}}\right\|_{L^2[0, 1],\lambda } &=&\max \limits_{t\in [0,T]}{\rm e}^{-\lambda t}\left\| {\frac{\partial (\delta Q_k (x,t))}{\partial x}}\right\|_{L^2[0, 1]}^2\\& \le&\mathop {\max }\limits_{t\in [0,T]} \frac{1-{\rm e}^{-(\lambda -1)t}}{\lambda-1}\left\| {\frac{\partial }{\partial t}\left( {\frac{\partial (\delta Q_k(x,t))}{\partial x}} \right)} \right\|_{L^2[0, 1],\lambda } \\&=&\frac{1-{\rm e}^{-(\lambda -1)T}}{\lambda -1}\left\| {\frac{\partial }{\partial t}\left( {\frac{\partial (\delta Q_k (x,t))}{\partial x}} \right)}\right\|_{L^2[0, 1],\lambda } .\end{eqnarray} (3.39)
将(3.39)式代入(3.38)式,得 \begin{eqnarray*}&&\left\| {\frac{\partial }{\partial t}\left( {\frac{\partial (\delta Q_k(x,t))}{\partial x}} \right)} \right\|_{L^2[0, 1],\lambda }\\& \le&\frac{A\frac{1-{\rm e}^{-\lambda T}}{\lambda }}{1-(A+qB)\frac{1-{\rm e}^{-\lambda T}}{\lambda}}\frac{1-{\rm e}^{-(\lambda -1)T}}{\lambda -1}\left\| {\frac{\partial }{\partial t}\left( {\frac{\partial (\delta Q_k (x,t))}{\partial x}} \right)}\right\|_{L^2[0, 1],\lambda } \\&&+\frac{qB\frac{1-{\rm e}^{-\lambda T}}{\lambda }}{1-(A+qB)\frac{1-{\rm e}^{-\lambda T}}{\lambda}}\left\| {\frac{\partial e_k (x,t)}{\partial x}} \right\|_{L^2[0, 1],\lambda} .\end{eqnarray*} 取较大的$\lambda $,使得 \[\frac{A\frac{1-{\rm e}^{-\lambda T}}{\lambda }}{1-(A+qB)\frac{1-{\rm e}^{-\lambda T}}{\lambda}}\frac{1-{\rm e}^{-(\lambda -1)T}}{\lambda -1}<1.\] 则有 \[\left\| {\frac{\partial }{\partial t}\left( {\frac{\partial (\delta Q_k(x,t))}{\partial x}} \right)} \right\|_{L^2[0, 1],\lambda } \le\frac{\frac{qB\frac{1-{\rm e}^{-\lambda T}}{\lambda }}{1-(A+qB)\frac{1-{\rm e}^{-\lambda T}}{\lambda }}}{1-\frac{A\frac{1-{\rm e}^{-\lambda T}}{\lambda}}{1-(A+qB)\frac{1-{\rm e}^{-\lambda T}}{\lambda }}\frac{1-{\rm e}^{-(\lambda -1)T}}{\lambda-1}}\left\| {\frac{\partial e_k (x,t)}{\partial x}}\right\|_{L^2[0, 1],\lambda } .\] 再将上式代入(3.39)式,有 \[\left\| {\frac{\partial (\delta Q_k (x,t))}{\partial x}}\right\|_{L^2[0, 1],\lambda } \le \frac{1-{\rm e}^{-(\lambda -1)T}}{\lambda-1}\frac{\frac{qB\frac{1-{\rm e}^{-\lambda T}}{\lambda }}{1-(A+qB)\frac{1-{\rm e}^{-\lambda T}}{\lambda }}}{1-\frac{A\frac{1-{\rm e}^{-\lambda T}}{\lambda}}{1-(A+qB)\frac{1-{\rm e}^{-\lambda T}}{\lambda }}\frac{1-{\rm e}^{-(\lambda -1)T}}{\lambda-1}}\left\| {\frac{\partial e_k (x,t)}{\partial x}}\right\|_{L^2[0, 1],\lambda } .\] 记 \[O(\lambda ^{-2})=\frac{1-{\rm e}^{-(\lambda -1)T}}{\lambda-1}\frac{\frac{qB\frac{1-{\rm e}^{-\lambda T}}{\lambda }}{1-(A+qB)\frac{1-{\rm e}^{-\lambda T}}{\lambda }}}{1-\frac{A\frac{1-{\rm e}^{-\lambda T}}{\lambda}}{1-(A+qB)\frac{1-{\rm e}^{-\lambda T}}{\lambda }}\frac{1-{\rm e}^{-(\lambda -1)T}}{\lambda-1}},\] 则有
\begin{equation}\left\| {\frac{\partial (\delta Q_k (x,t))}{\partial x}}\right\|_{L^2[0, 1],\lambda } \le O(\lambda ^{-2})\left\| {\frac{\partial e_k(x,t)}{\partial x}} \right\|_{L^2[0, 1],\lambda } .\end{equation} (3.40)
显然,当$\lambda $足够大时,能使$O(\lambda ^{-2})$任意小.

将(3.40)式代入(3.9)式,有

\[\left\| {\frac{\partial e_{k+1} (x,t)}{\partial x}}\right\|_{L^2[0, 1],\lambda } \le \left\{ {\rho +(1+\frac{1}{\varepsilon})C^2O(\lambda ^{-2})} \right\}\left\| {\frac{\partial e_k (x,t)}{\partial x}} \right\|_{L^2[0, 1],\lambda } .\] 由此可得
\begin{equation}\mathop {\lim }\limits_{k\to \infty } \left\| {\frac{\partial e_k(x,t)}{\partial x}} \right\|_{L^2[0, 1],\lambda } =0.\end{equation} (3.41)
即
\begin{equation}\mathop {\lim }\limits_{k\to \infty } \left\| {\frac{\partial e_k(x,t)}{\partial x}} \right\|_{L^2[0, 1],s} =0.\end{equation} (3.42)

(3) 证$\mathop {\lim }\limits_{k\to \infty } \left\|{\frac{\partial ^2e_k (x,t)}{\partial x^2}} \right\|_{L^2[0, 1],s}=0$.

由(3.37)式有

\begin{eqnarray*}&&\int_0^1 {\left\{ {\left( {\frac{\partial ^2(\delta Q_k (x,t))}{\partial x^2}} \right)^{\rm T}D\frac{\partial ^2(\delta Q_k (x,t))}{\partial x^2}}\right\}{\rm d} x}\\& \le&(A+qB)\frac{{\rm e}^{\lambda t}-1}{\lambda }\left\| {\frac{\partial }{\partial t}\left({\frac{\partial (\delta Q_k (x,t))}{\partial x}} \right)}\right\|_{L^2[0, 1],\lambda }\\&&+A\frac{{\rm e}^{\lambda t}-1}{\lambda }\left\| {\frac{\partial (\delta Q_k(x,t))}{\partial x}} \right\|_{L^2[0, 1],\lambda } +qB\frac{{\rm e}^{\lambda t}-1}{\lambda }\left\| {\frac{\partial e_k (x,t)}{\partial x}}\right\|_{L^2[0, 1],\lambda } .\end{eqnarray*} 上式结合(3.23)式,有 \begin{eqnarray*}\left\| {\frac{\partial ^2(\delta Q_k (x,t))}{\partial x^2}}\right\|_{L^2[0, 1]}^2 &\le& (A+qB)\frac{{\rm e}^{\lambda t}-1}{\lambda d}\left\|{\frac{\partial }{\partial t}\left( {\frac{\partial (\delta Q_k(x,t))}{\partial x}} \right)} \right\|_{L^2[0, 1],\lambda } \\&&+A\frac{{\rm e}^{\lambda t}-1}{\lambda d}\left\| {\frac{\partial (\delta Q_k (x,t))}{\partial x}} \right\|_{L^2[0, 1],\lambda }\\&&+qB\frac{{\rm e}^{\lambda t}-1}{\lambda d}\left\| {\frac{\partial e_k(x,t)}{\partial x}} \right\|_{L^2[0, 1],\lambda } .\end{eqnarray*} 进一步有
\begin{eqnarray}\left\| {\frac{\partial ^2(\delta Q_k (x,t))}{\partial x^2}}\right\|_{L^2[0, 1],\lambda } &\le& (A+qB)\frac{1-{\rm e}^{-\lambda T}}{\lambda d}\left\|{\frac{\partial }{\partial t}\left( {\frac{\partial (\delta Q_k(x,t))}{\partial x}} \right)} \right\|_{L^2[0, 1],\lambda } \\&&+A\frac{1-{\rm e}^{-\lambda T}}{\lambda d}\left\| {\frac{\partial (\delta Q_k(x,t))}{\partial x}} \right\|_{L^2[0, 1],\lambda }\\&& +qB\frac{1-{\rm e}^{-\lambda T}}{\lambda d}\left\| {\frac{\partial e_k (x,t)}{\partial x}}\right\|_{L^2[0, 1],\lambda } .\end{eqnarray} (3.43)
由(3.38),(3.40),(3.41)式可得 \begin{eqnarray*}\mathop {\lim }\limits_{k\to \infty } \left\| {\frac{\partial e_k(x,t)}{\partial x}} \right\|_{L^2[0, 1],\lambda }& =&\mathop {\lim}\limits_{k\to \infty } \left\| {\frac{\partial (\delta Q_k (x,t))}{\partial x}} \right\|_{L^2[0, 1],\lambda }\\&=&\mathop {\lim }\limits_{k\to \infty } \left\| {\frac{\partial }{\partial t}\left( {\frac{\partial (\delta Q_k (x,t))}{\partial x}} \right)}\right\|_{L^2[0, 1],\lambda } =0.\end{eqnarray*} (3.43)式结合上式,可知 \[\mathop {\lim }\limits_{k\to \infty } \left\| {\frac{\partial ^2(\delta Q_k(x,t))}{\partial x^2}} \right\|_{L^2[0, 1],\lambda } =0.\] 对(3.11)式应用引理2.1,结合上式,可得 \[\mathop {\lim }\limits_{k\to \infty } \left\| {\frac{\partial ^2e_k(x,t)}{\partial x^2}} \right\|_{L^2[0, 1],\lambda } =0.\] 即
\begin{equation}\mathop {\lim }\limits_{k\to \infty } \left\| {\frac{\partial ^2e_k(x,t)}{\partial x^2}} \right\|_{L^2[0, 1],s} =0.\end{equation} (3.44)
综合(3.31),(3.42),(3.44)式,可知定理的结论成立,证毕.

注3.2   (3.35)式的成立,必须用到第二边值的边值定位条件,所以对于双曲型系统,本文得到的结论仅适用于第二边值的情形.

注3.3  [16] 由假设2.2,选取学习增益$q$,使其满足$\frac{{\sqrt {1 + \varepsilon } - 1}}{{{\alpha _1}\sqrt {1 + \varepsilon } }} < q < \frac{{\sqrt {1 + \varepsilon } + 1}}{{{\alpha _2}\sqrt {1 + \varepsilon } }}$,则收敛性条件(3.3)式就能成立,再由(3.1)式可知,满足条件的学习增益$q$总是存在的.

注3.4   本文中偏微分方程的解,可以是弱解,也可以是强解.有关弱解,强解的定义及保证其存在的充分性条件等可见文献[19].

4 仿真算例

下分别就抛物型和双曲型系统给出算例,并利用数学软件Mathematica进行仿真分析.

(1) 构建如下形式的抛物型系统

\[\left\{ {\begin{array}{l} \frac{\partial Q(x,t)}{\partial t}=\frac{\partial ^2Q(x,t)}{\partial x^2}+Q(x,t)+u(x,t),\\[2mm] y(x,t)=Q(x,t)+\left( {3.5\cos t+4.5} \right)u(x,t) . \end{array}} \right.\] 考虑区域: $(x,t)\in (0,1)\times [0, 1]$. 给出理想输出:$y_{r} (x,t)=\sin (x+t)+\left( {3.5\cos t+4.5} \right)$ $\cos(x+t),$ 相应有,$Q_r (x,t)=\sin (x+t)$,$u_r (x,t)=\cos (x+t)$.

构建迭代

\[\left\{ {\begin{array}{l} \frac{\partial Q_k (x,t)}{\partial t}=\frac{\partial ^2Q_k (x,t)}{\partial x^2}+Q_k (x,t)+u_k (x,t) ,\\[2mm] y_k (x,t)=Q_k (x,t)+\left( {3.5\cos t+4.5} \right)u_k (x,t) . \end{array}} \right.\] 结合理想状态$Q_r (x,t)=\sin (x+t)$,取初、边值(第一边值): $Q_k (x,0)=Q_r (x,0)=\sin (x)$,$Q_k (0,t)=Q_r(0,t)=\sin (t)$,$Q_k (1,t)=Q_r (1,t)=\sin (1+t)$.

取初始控制$u_0 (x,t)=1$,构建迭代控制

\[u_{k+1} (x,t)=u_k (x,t)+qe_k (x,t),\] $k=0,1,2,\cdots .$ 因$E(t)=3.5\cos t+4.5$,结合假设2.2,取$\alpha _1 =1$,$\alpha _2 =8$,由此可取$\varepsilon=\frac{9}{16}$,则有 $$8=\frac{\alpha _2 }{\alpha _1 }<\frac{\sqrt {1+\varepsilon } +1}{\sqrt{1+\varepsilon } -1}=9.$$ 即(3.1)式成立,由注3.3可知,当 $$\frac{1}{5}=\frac{\sqrt {1+\varepsilon } -1}{\alpha _1 \sqrt {1+\varepsilon} }<q<\frac{\sqrt {1+\varepsilon } +1}{\alpha _2 \sqrt {1+\varepsilon }}=\frac{9}{40}$$ 成立时,迭代收敛,由此取$q=\frac{17}{80}$,则当$k\to\infty $时,有 $$\mathop {\lim }\limits_{k\to \infty } \left\| {e_k(x,t)} \right\|_{L^2[0, 1],s} =0,\mathop {\lim }\limits_{k\to\infty } \left\| {\frac{\partial e_k (x,t)}{\partial x}}\right\|_{L^2[0, 1],s} =0,\mathop {\lim }\limits_{k\to \infty }\left\| {\frac{\partial ^2e_k (x,t)}{\partial x^2}}\right\|_{L^2(\Omega )} =0, $$ 见图 1-3.

(2) 构建如下形式的双曲型系统

\[\left\{ {\begin{array}{l} \frac{\partial ^2Q(x,t)}{\partial t^2}=\frac{\partial ^2Q(x,t)}{\partial x^2}+Q(x,t)+u(x,t),\\ y(x,t)=Q(x,t)+\left( {3.5\sin t+4.5} \right)u(x,t) . \end{array}} \right.\] 考虑区域: $(x,t)\in (0,1)\times [0, 1]$,给出理想输出:$y_{r} (x,t)=\cos (x+t)-\left( {3.5\sin t+4.5}\right)\cos (x+t),$ 相应有,$Q_r (x,t)=\cos (x+t)$,$u_r (x,t)=-\cos(x+t)$.

构建迭代

\[\left\{ {\begin{array}{l} \frac{\partial ^2Q_k (x,t)}{\partial t^2}=\frac{\partial ^2Q_k(x,t)}{\partial x^2}+Q_k (x,t)+u_k (x,t),\\ y_k (x,t)=Q_k (x,t)+\left( {3.5\sin t+4.5} \right)u_k (x,t) . \end{array}} \right..\] 结合理想状态$Q_r (x,t)=\cos (x+t)$,取初,边值(第二边值): $Q_k (x,0)=Q_r (x,0)=\cos (x)$,$\left.{\frac{\partial Q_k (x,t)}{\partial t}} \right|_{t=0} =\left.{\frac{\partial Q_r (x,t)}{\partial t}} \right|_{t=0} =-\sin (x)$,$\left. {\frac{\partial Q_k (x,t)}{\partial x}} \right|_{x=0}=\left. {\frac{\partial Q_r (x,t)}{\partial x}} \right|_{x=0}=-\sin(t)$,$\left. {\frac{\partial Q_k (x,t)}{\partial x}} \right|_{x=1}=\left. {\frac{\partial Q_r (x,t)}{\partial x}} \right|_{x=1} =-\sin(1+t)$.

取初始控制$u_0 (x,t)\mbox{=}1$,构建迭代控制

\[u_{k+1} (x,t)=u_k (x,t)+qe_k (x,t),\] $k=0,1,2,\cdots .$ 同样取$q=\frac{17}{80}$,则当$k\to\infty $时,有 $$\mathop {\lim }\limits_{k\to \infty } \left\| {e_k(x,t)} \right\|_{L^2[0, 1],s} =0,\mathop {\lim }\limits_{k\to\infty } \left\| {\frac{\partial e_k (x,t)}{\partial x}}\right\|_{L^2[0, 1],s} =0,\mathop {\lim }\limits_{k\to \infty }\left\| {\frac{\partial ^2e_k (x,t)}{\partial x^2}}\right\|_{L^2[0, 1],s} =0, $$ 见图 4 -6.
5 结论

本文研究了一类分布参数系统于$W^{1,2}$空间中的迭代学习控制问题,该系统由抛物型偏微分方程或由双曲型偏微分方程构成,并具有适定的初值、边值(抛物型: 第一边值或第二边值; 双曲型:第二边值)定解条件. 当输入,输出有直输通道时,用P型学习律设计出迭代学习控制器,证明了这种学习律能使得系统的输出跟踪误差于$W^{1,2}$空间内沿迭代轴方向收敛,并进一步给出了输出跟踪误差于$W^{2,2}$空间内的收敛性结论,得到了很好的结果.仿真算例也说明如此.

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