众所周知, Euler-Bernoulli梁、Timoshenko梁、Rayleigh梁等系统在航空飞行器及机器人手臂等柔性结构中有着广泛的应用, 引起许多数学与力学爱好者的广泛关注.诸如, 文献[1-9]研究的是Euler-Bernoulli梁、Timoshenko梁, Rayleigh梁的边界及内部控制的稳定性与能控性.然而, 关于Timoshenko梁的最优控制问题工作并不多, 就我们所知只有文献[10-12]所关注的问题.受到文献[7]的启发, 本文研究Timoshenko梁的最优减振问题, 更精确地说, 研究如下最优控制问题：寻找最小的作用(控制)力 $(f_{1}(t), f_{2}(t))$ 在给定的时间段内最大限度地使振动的梁趋于稳定, 并且使代价最小.也就是在容许控制集中, 即在 $f_{i}(t)\in U_{ad}=\left\{f_{i}|f_{i}\in L^{2}(0, T), \ i=1, 2\right\}$ 中, 寻求最优控制 $f_{i}^{\ast}(t)\in U_{ad}, i=1, 2$ , 使代价函数(性能指标)最小

$ \begin{equation} {{\rm minimize}} \quad J(f_{1}(t), f_{2}(t))=\sum\limits_{i=1}^{3} \mu_{i}J_{i}(f_{1}(t), f_{2}(t)), \label{1.1} \end{equation} $

(1.1)

其中, $\mu_{i}>0, i=1, 2, 3, \mu_{1}+\mu_{2}+\mu_{3}=1$ ,

$ \ \ \ J_{1}(f_{1}(t), f_{2}(t))=\int_{0}^{\pi}[w_{t}^{2}(x, T)+\varphi_{t}^{2}(x, T)]{\rm d}x, \\ J_{2}(f_{1}(t), f_{2}(t))=\int_{0}^{\pi}[pw^{2}(x, T)+q\varphi^{2}(x, T)]{\rm d}x, \\ \ \ \ \ \ \ J_{3}(f_{1}(t), f_{2}(t))=\int_{0}^{T}[f_{1}^{2}(t)+f_{2}^{2}(t)]{\rm d}t, $

且满足

$ \begin{equation} \left\{\begin{array}{lll} w_{tt}-pw_{xx}+ p \varphi_{x} =0, \quad 0<x<\pi, \quad 0<t<T, \\ \varphi_{tt}-q\varphi_{xx}-p w_{x}+p\varphi =0, \quad 0<x<\pi, \quad 0<t<T, \\ w(0, t)= \varphi_{x}(0, t)=0, \quad w(\pi, t)=f_{1}(t), \quad\varphi_{x}(\pi, t)=f_{2}(t), \quad 0<t<T, \\ w(x, 0)=w_{0}(x), w_{t}(x, 0)=z_{0}(x), \varphi(x, 0)=\varphi_{0}(x), \varphi_{t}(x, 0)=\psi_{0}(x), \quad 0<x<\pi , \end{array}\right. \label{1.2} \end{equation} $

(1.2)

这里 $w(x, t), \varphi(x, t)$ 分别表示梁在时刻 $t$ 及 $x$ 处的横向位移与全转角, 正常数 $p, q$ 表示波传播速度, 且 $p\neq q$ .为了方便起见, 假定梁的长度为 $\pi$ , Timoshenko梁的力学边界条件是简支的.

本文首先运用线性算子半群理论得到状态约束系统的适定性, 同时提出最优控制问题；再利用相应的伴随系统得到最大值原理；然后对Timoshenko梁系统进行谱的渐近展开, 借助状态变量和伴随变量在终端时刻的关系得到最优控制问题解的显式渐近表示；最后给出具体的计算步骤.

定义系统(1.2)在时刻 $t$ 的有限能量为

$ E(t)=\frac{1}{2}\int_{0}^{\pi}(p| w_{x}-\varphi|^{2} +q|\varphi_{x}|^{2}+|w_{t}|^{2}+|\varphi_{t}|^{2}){\rm d}x. $

引入函数空间

$ W=\{u|u\in H^{1}(0, \pi), u(0)=0\}, $

其中 $H^{k}(0, \pi)$ 是 $k$ 阶Sobolev空间(参见文献[13]),

$ V=W\times W, \quad H=L^{2}(0, \pi)\times L^{2}(0, \pi), $

赋予范数

$ \|(u_{1}, u_{2})\|_{V}^{2}=\int_{0}^{\pi}[p|(u_{1})_{x}-u_{2}|^{2}+q|(u_{2})_{x}|^{2}]{\rm d}x , \quad \|(v_{1}, v_{2})\|_{H}^{2}=\int_{0}^{\pi}(|v_{1}|^{2}+|v_{2}|^{2}){\rm d}x. $

于是, $V, H$ 均为实(复) Hilbert空间, 且具有范数 $\|(\cdot, \cdot)\|_{V}, \|(\cdot, \cdot)\|_{H}$ .

记 ${{\cal H}}=V\times H$ , 赋予范数

$ \|Z\|_{{\cal H}}^{2}=\|(u_{1}, u_{2}, v_{1}, v_{2})\|_{{\cal H}}^{2} =\|(u_{1}, u_{2})\|_{V}^{2}+\|(v_{1}, v_{2})\|_{H}^{2}, $

那么 ${\cal H}$ 也是实(复)Hilbert空间.

设 $Z=(w, \varphi, z, \psi)$ , 在 ${\cal H}$ 中定义线性算子 ${\cal A}$ 如下

$ D({\cal A})=\left\{\begin{array}{lll} Z\in{{\cal H}}|(w, \varphi)\in H^{2}(0, \pi)\times H^{2}(0, \pi), (z, \psi)\in V , \\ w(0)=\varphi_{x}(0)=0, \quad w(\pi)=f_{1}(t), \varphi_{x}(\pi)=f_{2}(t). \end{array}\right\} \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {\cal A}Z=(z, \psi, pw_{xx}-p\varphi_{x}, q\varphi_{xx}+pw_{x}-p\varphi). $

于是, 可以将系统(1.2)改写成 ${\cal H}$ 上的抽象Cauchy问题

$ \left\{\begin{array}{lll} \frac{{\rm d}Z(t)}{{\rm d}t}={\cal A}Z(t), \\ Z(0)=Z_{0}, \end{array}\right. $

其中, $Z_{0}=(w_{0}, \varphi_{0}, z_{0}, \psi_{0})$ .

当 $f_{1}(t)=f_{2}(t)=0$ 时, 线性算子 ${\cal A}$ 变成为线性算子 ${\cal A}_{0}$ , 于是

$ D({\cal A}_{0})=\left\{\begin{array}{lll} Z\in{{\cal H}}|(w, \varphi)\in H^{2}(0, \pi)\times H^{2}(0, \pi), (z, \psi)\in V , \\ w(0)=\varphi_{x}(0)=0, \quad w(\pi)=\varphi_{x}(\pi)=0.\end{array}\right\} \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {\cal A}_{0}Z=(z, \psi, pw_{xx}-p\varphi_{x}, q\varphi_{xx}+pw_{x}-p\varphi). $

容易验证 ${\cal A}_{0}=-{\cal A}_{0}^{\ast}$ (斜伴随算子), 由文献[14, 定理1.10.8] (Stone定理)知, ${\cal A}_{0}$ 生成一个 ${\cal H}$ 上的等距 $C_{0}$ -半群 ${\rm e}^{{\cal A}_{0}t}$ .进一步由文献[14], 我们可以得到下面的经典的适定性结论:

定理1  对 $Z_{0}\in {\cal H}$ 及 $(f_{1}(t), f_{2}(t))\in L^{2}(0, T)\times L^{2}(0, T)$ , 则控制系统(1.2)存在唯一的弱解

$ Z(t)\in C(0, T;V)\cap C^{1}(0, T;H). $

根据文献[15]中的定理1可知, 我们所研究的Timoshenko梁系统的最优边界控制问题是可行的.

因此, Timoshenko梁系统的最优边界控制问题(1.1)-(1.2)可以表述为

$ \begin{equation} J(f_{1}^{\ast}(t), f_{2}^{\ast}(t))=\min\limits_{f_{1}, f_{2}\in U_{ad}} J(f_{1}(t), f_{2}(t)), \label{2.1} \end{equation} $

(2.1)

其中 $(f_{1}^{\ast}(t), f_{2}^{\ast}(t))$ 是最优控制.

为了得到最优控制问题(2.1)的最大值原理, 我们需要引入如下的伴随变量 $(u(x, t), \theta(x, t))$ , 满足方程

$ \begin{equation} \left\{\begin{array}{lll} u_{tt}-pu_{xx}+ p\theta_{x} =0, \quad&0<x<\pi, \quad 0<t<T, \\ \theta_{tt}-q\theta_{xx}-pu_{x}+p\theta =0, \quad &0<x<\pi, \quad 0<t<T, \end{array}\right. \label{3.1} \end{equation} $

(3.1)

边界条件

$ \begin{equation} u(0, t)= \theta_{x}(0, t)=u(\pi, t)=\theta_{x}(\pi, t)=0, \quad 0<t<T, \label{3.2} \end{equation} $

(3.2)

终端条件

$ \begin{equation} \left\{\begin{array}{lll} u(x, T)=2\mu_{1}w_{t}(x, T), \quad u_{t}(x, T)=-2p\mu_{2}w(x, T), \quad &0<x<\pi, \\ \theta(x, T)=2\mu_{1}\varphi_{t}(x, T), \quad \theta_{t}(x, T)=-2q\mu_{2}\varphi(x, T), \quad &0<x<\pi. \end{array}\right. \label{3.3} \end{equation} $

(3.3)

对于 $f_{i}(t)\in U_{ad}, i=1, 2$ , 令

$ w(x, t)=w(x, t, f_{1}, f_{2}), \quad \varphi(x, t)=w(x, t, f_{1}, f_{2}) $

满足方程(1.2), 且记

$ w^{\ast}(x, t)=w(x, t, f_{1}^{\ast}, f_{2}^{\ast}), \quad \varphi^{\ast}(x, t)=\varphi(x, t, f_{1}^{\ast}, f_{2}^{\ast}) $

为最优轨线以及相应的最优伴随变量为

$ u^{\ast}(x, t)=u(x, t, f_{1}^{\ast}, f_{2}^{\ast}), \quad \theta^{\ast}(x, t)=\theta(x, t, f_{1}^{\ast}, f_{2}^{\ast}), $

同时, 它们分别满足方程(3.1)及其边界条件(3.2)和终端条件(3.3).

接下来, 我们给出本文的主要结果(最大值原理):

定理2(最大值原理)  如果最优控制 $f_{i}^{\ast}(t)\in U_{ad}, i=1, 2$ 满足最大值问题

$ \begin{equation} \max\limits_{f_{1}, f_{2}\in U_{ad}}H(T, u^{\ast}, \theta^{\ast}, f_{1}, f_{2})= H(T, u^{\ast}, \theta^{\ast}, f_{1}^{\ast}, f_{2}^{\ast}), \label{3.4} \end{equation} $

(3.4)

其中

$ \begin{equation} H(T, u, \theta, f_{1}, f_{2})= R_{1}(t)f_{1}(t)+R_{2}(t)f_{2}(t)+\mu_{3}f_{1}^{2}(t)+\mu_{3}f_{2}^{2}(t) \label{3.5} \end{equation} $

(3.5)

为Hamilton函数且 $R_{1}(t)=p(\theta(\pi, t)-u_{x}(\pi, t)), \quad R_{2}(t)=q\theta(\pi, t)$ .

那么

$ \begin{equation} J(f_{1}^{\ast}, f_{2}^{\ast})\leq J(f_{1}, f_{2}), \quad f_{i}(t)\in U_{ad}, i=1, 2.\label{3.6} \end{equation} $

(3.6)

证  为了方便起见, 首先我们引入算子

$ \begin{equation} \Psi(w, \varphi)=w_{tt}-pw_{xx}+p\varphi_{x}, \quad \Phi(w, \varphi)=\varphi_{tt}-q\varphi_{xx}-pw_{x}+p\varphi, \label{3.7} \end{equation} $

(3.7)

并且记

$ \Delta w(x, t)=w(x, t)-w^{\ast}(x, t), \quad \Delta \varphi(x, t)=\varphi(x, t)-\varphi^{\ast}(x, t), \\ \ \ \ \ \ \ \ \ \ \ \ \ \Delta f_{1}(t)=f_{1}(t)-f_{1}^{\ast}(t), \quad \Delta f_{2}(t)=f_{2}(t)-f_{2}^{\ast}(t).\label{3.8} $

(3.8)

于是

$ \begin{equation}(\Psi(\Delta w, \Delta\varphi), \Phi(\Delta w, \Delta\varphi))-(\Delta f_{1}(t), \Delta f_{2}(t))=0, \label{3.9} \end{equation} $

(3.9)

于此同时, 边界条件转化为

$ \begin{equation}\Delta w(0, t)=\Delta\varphi(0, t)=0, \quad\Delta w(\pi, t)=\Delta f_{1}(t), \quad\Delta\varphi_{x}(\pi, t)=\Delta f_{2}(t), \label{3.10} \end{equation} $

(3.10)

初始条件转化为

$ \begin{equation}\Delta w(x, 0)=\Delta w_{t}(x, 0)=0, \quad\Delta\varphi(x, 0)=\Delta\varphi_{x}(x, 0)=0, \label{3.11} \end{equation} $

(3.11)

为了寻求最优解, 我们由(1.2), (3.1)和(3.7)-(3.9)式运用乘子、分部积分、边界条件和初值条件得到如下形式

$ \begin{eqnarray}\label{3.12} 0&=&\int_{0}^{\pi}\int_{0}^{T}[\Delta w\Psi(u, \theta)+\Delta\varphi\Phi(u, \theta)-u\Psi(\Delta w, \Delta\varphi)-\theta\Phi(\Delta w, \Delta\varphi)]{\rm d}t{\rm d}x\\ &=&\int_{0}^{\pi}\int_{0}^{T}[\Delta w(u_{tt}-pu_{xx}+p\theta_{x})+\Delta\varphi(\theta_{tt}-q\theta_{xx}-pu_{x}+p\theta)\\ &&-u(\Delta w_{tt}-p\Delta w_{xx}+p\Delta\varphi_{x})-\theta(\Delta\varphi_{tt}-q\Delta\varphi_{xx}-p\Delta w_{x}+p\Delta\varphi)]{\rm d}t{\rm d}x\\ &=&\int_{0}^{\pi}\int_{0}^{T}(\Delta wu_{tt}-u\Delta w_{tt}){\rm d}t{\rm d}x\\ &&+\int_{0}^{\pi}\int_{0}^{T}(\Delta\varphi\theta_{tt}-\theta\Delta \varphi_{tt}){\rm d}t{\rm d}x+q\int_{0}^{\pi}\int_{0}^{T}(\theta\Delta\varphi_{xx}-\Delta\varphi\theta_{xx}){\rm d}t{\rm d}x\\ &&+p\int_{0}^{\pi}\int_{0}^{T}[\Delta w(\theta-u_{x})_{x}-u(\Delta\varphi-\Delta w_{x})_{x}+\Delta\varphi(\theta-u_{x})-\theta(\Delta\varphi-\Delta w_{x})]{\rm d}t{\rm d}x\\ &=&\int_{0}^{\pi}[\Delta w(x, T)u_{t}(x, T)-u(x, T)\Delta w_{t}(x, T)]{\rm d}x\\ &&+\int_{0}^{\pi}[\Delta\varphi(x, T)\theta_{t}(x, T)-\theta(x, T)\Delta\varphi_{t}(x, T)]{\rm d}x\\ &&+q\int_{0}^{T}\theta(\pi, t)\Delta\varphi_{x}(\pi, t){\rm d}t+p\int_{0}^{T}\Delta w(\pi, t)(\theta(\pi, t)-u_{x}(\pi, t)){\rm d}t. \end{eqnarray} $

(3.12)

于是有

$ \begin{eqnarray}\label{3.13} &&\int_{0}^{\pi}[u(x, T)\Delta w_{t}(x, T)-\Delta w(x, T)u_{t}(x, T)]{\rm d}x\\ &&+\int_{0}^{\pi}[\theta(x, T)\Delta\varphi_{t}(x, T)-\Delta\varphi(x, T)\theta_{t}(x, T)]{\rm d}x\\ &=&\int_{0}^{T}[p\Delta w(\pi, t)(\theta(\pi, t)-u_{x}(\pi, t))+q\theta(\pi, t)\Delta\varphi_{x}(\pi, t)]{\rm d}t. \end{eqnarray} $

(3.13)

由将伴随系统终止条件(3.3)代入等式(3.13)的左边得

$ \begin{eqnarray}\label{3.14} &&\int_{0}^{\pi}2\mu_{1}[w_{t}(x, T)\Delta w_{t}(x, T)+\varphi_{t}(x, T)\Delta\varphi_{t}(x, T)]{\rm d}x\\ &&+\int_{0}^{\pi}2\mu_{2}[pw(x, T)\Delta w(x, T)+q\varphi(x, T)\Delta\varphi(x, T)]{\rm d}x\\ &=&\int_{0}^{T}[p\Delta w(\pi, t)(\theta(\pi, t)-u_{x}(\pi, t))+q\theta(\pi, t)\Delta\varphi_{x}(\pi, t)]{\rm d}t. \end{eqnarray} $

(3.14)

现在我们考虑代价函数(性能指标)的增量

$ \begin{eqnarray}\label{3.15} \Delta J(f_{1}, f_{2}) &=&J(f_{1}, f_{2})-J(f_{1}^{\ast}, f_{2}^{\ast})\\ &=&\int_{0}^{\pi}\{\mu_{1}[(w_{t}^{2}(x, T)-w_{t}^{\ast 2}(x, T))+(\varphi_{t}^{2}(x, T)-\varphi_{t}^{\ast 2}(x, T))]\\ &&+\mu_{2}[p(w^{2}(x, T)-w^{\ast 2}(x, T))+q(\varphi^{2}(x, T)-\varphi^{\ast 2}(x, T))]\}{\rm d}x\\ &\quad&+\int_{0}^{T}\mu_{3}[(f_{1}^{2}(t)-f_{1}^{\ast2}(t))+(f_{2}^{2}(t)-f_{2}^{\ast2}(t))]{\rm d}t. \end{eqnarray} $

(3.15)

将(3.15)式中的 $w_{t}^{2}(x, T), $ $\varphi_{t}^{2}(x, T), $ $w^{2}(x, T), $ $\varphi^{2}(x, T)$ 分别关于 $w_{t}^{\ast}(x, T), $ $\varphi_{t}^{\ast}(x, T), $ $w^{\ast}(x, T), $ $\varphi^{\ast}(x, T)$ 按Taylor级数展开得

$ \begin{equation} \begin{array}{ll}\label{3.16} w_{t}^{2}(x, T)-w_{t}^{\ast 2}(x, T)=2w_{t}^{\ast}(x, T)\Delta w_{t}(x, T)+r_{1}, \\ \varphi_{t}^{2}(x, T)-\varphi_{t}^{\ast 2}(x, T)=2\varphi_{t}^{\ast}(x, T)\Delta\varphi_{t}(x, T)+r_{2}, \\ w^{2}(x, T)-w^{\ast 2}(x, T)=2w^{\ast}(x, T)\Delta w(x, T)+r_{3}, \\ \varphi^{2}(x, T)-\varphi^{\ast 2}(x, T)=2\varphi^{\ast}(x, T)\Delta\varphi(x, T)+r_{4}, \end{array}\end{equation} $

(3.16)

其中 $r_{1}=(\Delta w_{t}(x, T))^{2}\geq0, r_{2}=(\Delta\varphi_{t}(x, T))^{2}\geq0, r_{3}=(\Delta w(x, T))^{2}\geq0, r_{4}=(\Delta\varphi(x, T))^{2}\geq0, $ 将(3.16)式代入(3.15)式得

$ \begin{eqnarray}\label{3.17} \Delta J(f_{1}, f_{2}) &=&\int_{0}^{\pi}\{\mu_{1}[(2w_{t}^{\ast}(x, T)\Delta w_{t}(x, T)+r_{1})+(2\varphi_{t}^{\ast}(x, T)\Delta\varphi_{t}(x, T)+r_{2})]\\ &&+\mu_{2}[p(2w^{\ast}(x, T)\Delta w(x, T)+r_{3})+q(2\varphi^{\ast}(x, T)\Delta\varphi(x, T)+r_{4})]\}{\rm d}x\\ &&+\int_{0}^{T}\mu_{3}[(f_{1}^{2}(t)-f_{1}^{\ast2}(t))+(f_{2}^{2}(t)-f_{2}^{\ast2}(t))]{\rm d}t. \end{eqnarray} $

(3.17)

由于 $\mu_{1}(r_{1}+r_{2})+\mu_{2}(pr_{3}+qr_{4})\geq0$ , 因此, 由(3.13), (3.14)和(3.17)式得

$ \begin{eqnarray}\label{3.18} \Delta J(f_{1}, f_{2}) &\geq&\int_{0}^{T}[p(\theta(\pi, t)-u_{x}(\pi, t))\Delta f_{1}(t)+q\theta(\pi, t)\Delta f_{2}(t)]{\rm d}t\\ &&+\int_{0}^{T}\mu_{3}[(f_{1}^{2}(t)-f_{1}^{\ast2}(t))+(f_{2}^{2}(t)-f_{2}^{\ast2}(t))]{\rm d}t\\ &=&\int_{0}^{T}\{[R_{1}f_{1}(t)+\mu_{3}f_{1}^{2}(t)+R_{2}f_{2}(t)+\mu_{3}f_{2}^{2}(t)]\\ &&-[R_{1}f_{1}^{\ast}(t)+\mu_{3}f_{1}^{\ast2}(t)+R_{2}f_{2}^{\ast}(t)+\mu_{3}f_{2}^{\ast2}(t)]\}{\rm d}t, \end{eqnarray} $

(3.18)

这里, $R_{1}=p(\theta(\pi, t)-u_{x}(\pi, t)), R_{2}=q\theta(\pi, t)$ .

由(3.4)-(3.5)式得

$ \begin{equation} R_{1}f_{1}(t)+\mu_{3}f_{1}^{2}(t)+R_{2}f_{2}(t)+\mu_{3}f_{2}^{2}(t) \leq R_{1}f_{1}^{\ast}(t)+\mu_{3}f_{1}^{\ast2}(t)+R_{2}f_{2}^{\ast}(t)+\mu_{3}f_{2}^{\ast2}(t). \label{3.19}\end{equation} $

(3.19)

考察不等式(3.18)-(3.19)有

$ \Delta J(f_{1}, f_{2})\geq0, $

即

$ J(f_{1}^{\ast}, f_{2}^{\ast})\leq J(f_{1}, f_{2}). $

因此, $(f_{1}^{\ast}, f_{2}^{\ast})$ 即为所求的最优控制.

并且, 所求的最优控制可以表示为

$ \begin{equation} (f_{1}^{\ast}(t), f_{2}^{\ast}(t))=-\frac{1}{2\mu_{3}}(p[\theta(\pi, t)-u_{x}(\pi, t)], q\theta(\pi, t)).\label{3.20}\end{equation} $

(3.20)

本节我们给出边界最优控制问题(2.1)的计算步骤.我们分以下几个步骤来处理.

首先计算系统(1.2)对应齐次系统的特征值与特征函数, 我们考虑如下的特征问题

$ \begin{equation}\left\{\begin{array}{lll} \lambda^{2}w-pw_{xx}+p\varphi_{x}=0, \\ \lambda^{2}\varphi-q\varphi_{xx}-pw_{x}+p\varphi=0, \\ w(0)=w(\pi)=\varphi_{x}(0)=\varphi_{x}(\pi)=0, \end{array}\right. \label{4.1}\end{equation} $

(4.1)

分别消去 $\varphi$ 或 $w$ 得到特征值

$ \begin{equation}\left\{\begin{array}{lll} \lambda_{n, 1}^{2}=\frac{1}{2}\left\{-[(p+q)n^{2}+p]+\sqrt{[(p-q)n^{2}+p]^{2}+4pqn^{2}}\right\}, \\ \lambda_{n, 2}^{2}=\frac{1}{2}\left\{-[(p+q)n^{2}+p]-\sqrt{[(p-q)n^{2}+p]^{2}+4pqn^{2}}\right\}, \end{array}\right. \label{4.2}\end{equation} $

(4.2)

和相应的特征函数

$ \begin{equation} w_{n}=\sin nx, \quad \varphi_{0}=1, \quad\varphi_{n}=\cos nx\quad n=1, 2, \cdots. \label{4.3}\end{equation} $

(4.3)

通过求解伴随系统(3.1)-(3.3)可以得到最优控制序列.由伴随系统(3.1) $-$ (3.3)得到如下的最优轨线序列

$ \begin{equation} u_{N}^{\ast}(x, t)=\sum\limits_{n=1}^{N}S_{n}(t)\sin nx, \quad N\in {\Bbb N}, \label{4.4}\end{equation} $

(4.4)

$ \begin{equation} \theta_{N}^{\ast}(x, t)=Q_{0}(t)+\sum\limits_{n=1}^{N}Q_{n}(t)\cos nx, \quad N\in {\Bbb N}, \label{4.5}\end{equation} $

(4.5)

其中, $Q_{0}(t)=b_{0, 1}\cos \sqrt{p}t+b_{0, 2}\sin \sqrt{p}t$ , $S_{n}(t), Q_{n}(t), n=1, 2, \cdots, N$ 有下列四种显式表示

$ \begin{equation} \left\{\begin{array}{lll} ({\rm i})\quad S_{n}(t)=a_{n, 1}\cos\lambda_{n, 1}t+a_{n, 2}\sin\lambda_{n, 1}t, \quad & Q_{n}(t)=a_{n, 3}\cos\lambda_{n, 1}t+a_{n, 4}\sin\lambda_{n, 1}t, \\ ({\rm ii})\quad S_{n}(t)=b_{n, 1}\cos\lambda_{n, 1}t+b_{n, 2}\sin\lambda_{n, 1}t, \quad & Q_{n}(t)=b_{n, 3}\cos\lambda_{n, 2}t+b_{n, 4}\sin\lambda_{n, 2}t, \\ ({\rm iii})\quad S_{n}(t)=c_{n, 1}\cos\lambda_{n, 2}t+c_{n, 2}\sin\lambda_{n, 2}t, \ &Q_{n}(t)=c_{n, 3}\cos\lambda_{n, 1}t+c_{n, 4}\sin\lambda_{n, 1}t, \\ ({\rm iv})\quad S_{n}(t)=d_{n, 1}\cos\lambda_{n, 2}t+d_{n, 2}\sin\lambda_{n, 2}t, \ &Q_{n}(t)=d_{n, 3}\cos\lambda_{n, 2}t+d_{n, 4}\sin\lambda_{n, 2}t, \end{array}\right.\label{4.6}\end{equation} $

(4.6)

这里 $a_{n, j}, b_{n, j}, c_{n, j}, d_{n, j}, j=1, 2, 3, 4$ 均是后面会确定的正常数.

于是

$ \begin{equation} (u_{N}^{\ast})_{x}(x, t)=\sum\limits_{n=1}^{N}nS_{n}(t)\cos nx. \label{4.7}\end{equation} $

(4.7)

将(4.5), (4.6)和(4.7)式代入(3.20)式得最优控制序列

$ \begin{equation} (f_{N, 1}^{\ast}(t), f_{N, 2}^{\ast}(t))=-\frac{1}{2\mu_{3}}(p[Q_{0}(t)+\sum\limits_{n=1}^{N}(-1)^{n}(Q_{n}(t)-nS_{n}(t))], q\sum\limits_{n=1}^{N}(-1)^{n}Q_{n}(t)).\label{4.8}\end{equation} $

(4.8)

接下来, 先齐次化系统(1.2)的边界, 于是引入变量

$ \begin{equation} \left\{\begin{array}{lll} \overline{w}(x, t)=w(x, t)-\frac{f_{1}(t)}{\pi}x, \\ \overline{\varphi}(x, t)=\varphi(x, t)-\frac{f_{2}(t)}{2\pi}x^{2}, \end{array}\right.\label{4.9}\end{equation} $

(4.9)

那么受控系统(1.2)改写为

$ \begin{equation} \left\{\begin{array}{lll} \overline{w}_{tt}(x, t)-p\overline{w}_{xx}(x, t)+p\overline{\varphi}_{x}(x, t)=-\frac{1}{\pi}f''_{1}(t)x-\frac{p}{\pi}f_{2}(t)x, \\ \overline{\varphi}_{tt}(x, t)-q\overline{\varphi}_{xx}(x, t) -p\overline{w}_{x}(x, t)+p\overline{\varphi}(x, t)\\ \ \ = -\frac{1}{2\pi}f''_{2}(t)x^{2}+\frac{p}{\pi}f_{1}(t)+\frac{q}{\pi}f_{2}(t)-\frac{p}{2\pi}f_{2}(t)x^{2}, \\ \end{array}\right.\label{4.10}\end{equation} $

(4.10)

这里"'"表示对时间 $t$ 的导数.

相应的边界条件为

$ \begin{equation} \left\{\begin{array}{lll} \overline{w}(0, t)=\overline{w}(\pi, t)=0, \\ \overline{\varphi}_{x}(0, t)=\overline{\varphi}_{x}(\pi, t)=0, \end{array}\right.\label{4.11}\end{equation} $

(4.11)

初始条件为

$ \begin{equation} \left\{\begin{array}{lll} \overline{w}(x, 0)=w_{0}(x)-\frac{x}{\pi}f_{1}(0), \quad \overline{w}_{t}(x, 0)=z_{0}(x)-\frac{x}{\pi}f'_{1}(0), \\ \overline{\varphi}_{x}(x, 0)=\varphi_{0}(x)-\frac{x^{2}}{2\pi}f_{2}(0), \quad \overline{\varphi}_{t}(x, 0)=\psi_{0}(x)-\frac{x^{2}}{2\pi}f'_{2}(0). \end{array}\right.\label{4.12}\end{equation} $

(4.12)

类似地, (4.10)-(4.11)式对应的齐次系统的近似最优轨线为

$ \begin{equation} \left\{\begin{array}{lll} \overline{w}_{N}(x, t)=\sum\limits_{n=1}^{N}Y_{n}(t)\sin nx, \\ \overline{\varphi}_{N}(x, t)=Z_{0}(t)+\sum\limits_{n=1}^{N}Z_{n}(t)\cos nx. \end{array}\right.\label{4.13}\end{equation} $

(4.13)

将(4.13)式代入(4.10)式得

$ \begin{equation} \left\{\begin{array}{lll} \sum\limits_{n=1}^{N}Y''_{n}(t)\sin nx+p\sum\limits_{n=1}^{N}n^{2}Y_{n}(t)\sin nx -p\sum\limits_{n=1}^{N}nZ_{n}(t)\sin nx\\ \ \ \ =-\frac{1}{\pi}f''_{1}(t)x-\frac{p}{\pi}f_{2}(t)x, \\ Z''_{0}(t)+pZ_{0}(t)+\sum\limits_{n=1}^{N}Z''_{n}(t)\cos nx +q\sum\limits_{n=1}^{N}n^{2}Z_{n}(t)\cos nx\\ -p\sum\limits_{n=1}^{N}nY_{n}(t)\cos nx +p\sum\limits_{n=1}^{N}Z_{n}(t)\cos nx\\ \ \ \ =-\frac{1}{2\pi}f''_{2}(t)x^{2}+\frac{p}{\pi}f_{1}(t)+\frac{q}{\pi}f_{2}(t)-\frac{p}{2\pi}f_{2}(t)x^{2}, \end{array}\right.\label{4.14}\end{equation} $

(4.14)

将(4.14)式的 $x$ 展开成 $[0, \pi]$ 上的正弦级数； $x^{2}$ 展开成 $[0, \pi]$ 上的余弦级数, 并利用正(余)弦级数的正交性得

$ \begin{equation} \left\{\begin{array}{lll} Y''_{n}(t)+pn^{2}Y_{n}(t)-pnZ_{n}(t)=\frac{2}{n\pi}(-1)^{n}f''_{1}(t)x+\frac{2p}{n\pi}(-1)^{n}f_{2}(t), \\ Z''_{n}(t)+(p+qn^{2})Z_{n}(t)-pnY_{n}(t)=\frac{2}{n^{2}\pi}(-1)^{n+1}f''_{2}(t)+\frac{2p}{n^{2}\pi}(-1)^{n+1}f_{2}(t), \\ Z''_{0}(t)+pZ_{0}(t)=\frac{p}{\pi}f_{1}(t)+\frac{q}{\pi}f_{2}(t)-\frac{\pi}{6}f''_{2}(t)-\frac{p\pi}{6}f_{2}(t). \end{array}\right.\label{4.15}\end{equation} $

(4.15)

由(4.15)式得

$ \begin{eqnarray} &&Y_{n}^{(4)}(t)+[(p+q)n^{2}+p]Y''_{n}(t)+pqn^{4}Y_{n}(t)\\ &=&\frac{2}{n\pi}(-1)^{n}f_{1}^{(4)}(t)+\frac{2(p+qn^{2})}{n\pi}(-1)^{n}f''_{1}(t)+\frac{2pqn}{\pi}(-1)^{n}f_{2}(t). \label{4.16}\end{eqnarray} $

(4.16)

(4.16)式对应的齐次方程的通解为

$ \begin{equation} Y_{n}(t)=h_{n, 1}\cos\lambda_{n, 1}t+h_{n, 2}\sin\lambda_{n, 1}t+h_{n, 3}\cos\lambda_{n, 2}t+h_{n, 4}\sin\lambda_{n, 2}t. \label{4.17}\end{equation} $

(4.17)

其中 $\lambda_{n, 1}, \lambda_{n, 2}$ 由(4.2)式给出.

利用常数变易法得方程(4.16)的解为

$ \begin{eqnarray}\label{4.18} Y_{n}(t) =\frac{\cos\lambda_{n, 1}t}{\lambda_{n, 1}(\lambda_{n, 1}^{2}-\lambda_{n, 2}^{2})}\int_{0}^{t}\sin\lambda_{n, 1}\tau\rho_{n}(\tau){\rm d}\tau -\frac{\sin\lambda_{n, 1}t}{\lambda_{n, 1}(\lambda_{n, 1}^{2}-\lambda_{n, 2}^{2})}\int_{0}^{t}\cos\lambda_{n, 1}\tau\rho_{n}(\tau){\rm d}\tau\\ \ \ \ \ \ \ \ \ \ \ \ -\frac{\cos\lambda_{n, 2}t}{\lambda_{n, 2}(\lambda_{n, 1}^{2}-\lambda_{n, 2}^{2})}\int_{0}^{t}\sin\lambda_{n, 2}\tau\rho_{n}(\tau){\rm d}\tau +\frac{\sin\lambda_{n, 2}t}{\lambda_{n, 2}(\lambda_{n, 1}^{2}-\lambda_{n, 2}^{2})}\int_{0}^{t}\cos\lambda_{n, 2}\tau\rho_{n}(\tau){\rm d}\tau, \\ \end{eqnarray} $

(4.18)

这里, $\rho_{n}(t)=\frac{2}{n\pi}(-1)^{n}f_{1}^{(4)}(t) +\frac{2(p+qn^{2})}{n\pi}(-1)^{n}f''_{1}(t)+\frac{2pqn}{\pi}(-1)^{n}f_{2}(t)$ .

将(4.18)式代入(4.15)式的第一式即可解得 $Z_{n}(t), n=1, 2, \cdots$ , 同时, $Z_{0}(t)$ 可以由(4.15)式的最后一式解得.

因此, 由(4.9)和(4.13)式可得

$ \begin{equation} (w_{N})_{t}(x, t)=\sum\limits_{n=1}^{N}Y'_{n}(t)\sin nx+\frac{f'_{1}(t)}{\pi}x, \label{4.19}\end{equation} $

(4.19)

$ \begin{equation} (w_{N})_{x}(x, t)=\sum\limits_{n=1}^{N}nY_{n}(t)\sin nx+\frac{f_{1}(t)}{\pi}, \label{4.20}\end{equation} $

(4.20)

$ \begin{equation} (\varphi_{N})_{t}(x, t)=Z'_{0}(t)+\sum\limits_{n=1}^{N}Z'_{n}(t)\cos nx+\frac{f'_{2}(t)}{2\pi}x^{2}, \label{4.21}\end{equation} $

(4.21)

$ \begin{equation} (\varphi_{N})_{x}(x, t)=-\sum\limits_{n=1}^{N}nZ_{n}(t)\sin nx+\frac{f_{2}(t)}{\pi}x. \label{4.22}\end{equation} $

(4.22)

下面来确定(4.6)式的系数.

1 $^{{\rm o}}$ .对于(4.8)式中的(ⅰ), 根据终端条件(3.3)以及(4.4)和(4.19)-(4.20)式得

$ \left\{\begin{array}{lll} \sum\limits_{n=1}^{N}(a_{n, 1}\cos\lambda_{n, 1}T+a_{n, 2}\sin\lambda_{n, 1}T)\sin nx =2\mu_{1}\bigg[\sum\limits_{n=1}^{N}Y'_{n}(T)\sin nx+\frac{f'_{1}(T)}{\pi}x\bigg], \\ \sum\limits_{n=1}^{N}(-a_{n, 1}\lambda_{n, 1}\sin\lambda_{n, 1}T+a_{n, 2}\lambda_{n, 1} \cos\lambda_{n, 1}T)\sin nx=2p\mu_{2} \bigg[\sum\limits_{n=1}^{N}nY_{n}(T)\cos nx+\frac{f_{1}(T)}{\pi}\bigg], \end{array}\right. $

于是

$ \left\{\begin{array}{lll} a_{n, 1}\cos\lambda_{n, 1}T+a_{n, 2}\sin\lambda_{n, 1}T=2\mu_{1}Y'_{n}(T)+\frac{4\mu_{1}(-1)^{n}f'_{1}(T)}{n\pi}, \\ -a_{n, 1}\lambda_{n, 1}\sin\lambda_{n, 1}T+a_{n, 2}\lambda_{n, 1}\cos\lambda_{n, 1}T=\frac{4p\mu_{2}[(-1)^{n+1}+1]f_{1}(T)}{n\pi^{2}}, \end{array}\right. $

解之得

$ \left\{\begin{array}{rl} a_{n, 1}=&2\mu_{1}Y'_{n}(T)\cos\lambda_{n, 1}T+\frac{4\mu_{1}(-1)^{n}f'_{1}(T)\cos\lambda_{n, 1}T}{n\pi} \\ & -\frac{4p\mu_{2}[(-1)^{n+1}+1]f_{1}(T)\sin\lambda_{n, 1}T}{n\lambda_{n, 1}\pi^{2}}, \\ a_{n, 2}=&2\mu_{1}Y'_{n}(T)\sin\lambda_{n, 1}T+\frac{4\mu_{1}(-1)^{n}f'_{1}(T)\sin\lambda_{n, 1}T}{n\pi} \\ & +\frac{4p\mu_{2}[(-1)^{n+1}+1]f_{1}(T)\cos\lambda_{n, 1}T}{n\lambda_{n, 1}\pi^{2}}. \end{array}\right. $

再根据终端条件(3.3)以及(4.5)和(4.21)-(4.22)式得

$ \left\{\begin{array}{rl} &\sum\limits_{n=1}^{N}(a_{n, 3}\cos\lambda_{n, 1}T+a_{n, 4}\sin\lambda_{n, 1}T)\cos nx \\ =& 2\mu_{1}\bigg[\sum\limits_{n=1}^{N}Z'_{n}(T)\cos nx+\frac{f'_{2}(T)}{2\pi}x^{2}\bigg], \\ &\sum\limits_{n=1}^{N}(-a_{n, 3}\lambda_{n, 1}\sin\lambda_{n, 1}T+a_{n, 4}\lambda_{n, 1} \cos\lambda_{n, 1}T)\cos nx\\ =& 2q\mu_{2} \bigg[-\sum\limits_{n=1}^{N}nZ_{n}(T)\sin nx+\frac{f_{2}(T)}{\pi}x\bigg], \end{array}\right. $

于是

$ \left\{\begin{array}{lll} a_{n, 3}\cos\lambda_{n, 1}T+a_{n, 4}\sin\lambda_{n, 1}T=2\mu_{1}Z'_{n}(T)+\frac{4\mu_{1}(-1)^{n}f'_{2}(T)}{n^{2}\pi}, \\ -a_{n, 3}\lambda_{n, 1}\sin\lambda_{n, 1}T+a_{n, 4}\lambda_{n, 1}\cos\lambda_{n, 1}T=\frac{4q\mu_{2}[(-1)^{n}-1]f_{2}(T)}{n^{2}\pi^{2}}, \end{array}\right. $

解之得

$ \left\{\begin{array}{rl} a_{n, 3}=& 2\mu_{1}Z'_{n}(T)\cos\lambda_{n, 1}T+\frac{4\mu_{1}(-1)^{n}f'_{2}(T)\cos\lambda_{n, 1}T}{n^{2}\pi} \\ & +\frac{4q\mu_{2}[(-1)^{n+1}+1]f_{2}(T)\sin\lambda_{n, 1}T}{n^{2}\lambda_{n, 1}\pi^{2}}, \\ a_{n, 4}=& 2\mu_{1}Z'_{n}(T)\sin\lambda_{n, 1}T+\frac{4\mu_{1}(-1)^{n}f'_{2}(T)\sin\lambda_{n, 1}T}{n^{2}\pi} \\ & -\frac{4q\mu_{2}[(-1)^{n+1}+1]f_{2}(T)\cos\lambda_{n, 1}T}{n^{2}\lambda_{n, 1}\pi^{2}}. \end{array}\right. $

2 $^{{\rm o}}$ .对于(4.6)式中的(ⅱ), 根据终端条件(3.3)以及(4.4)和(4.19)-(4.20)式得

$ \left\{\begin{array}{lll} b_{n, 1}=a_{n, 1}, \\ b_{n, 2}=a_{n, 2}. \end{array}\right. $

再根据终端条件(3.3)以及(4.5)和(4.21)-(4.22)式得

$ \left\{\begin{array}{lll} b_{0, 1}\cos \sqrt{p}T+b_{0, 2}\sin \sqrt{p}T+\sum\limits_{n=1}^{N}(b_{n, 3}\cos\lambda_{n, 2}T+b_{n, 4}\sin\lambda_{n, 2}T)\cos nx\\ \ \ \ =2\mu_{1}\bigg\{Z'_{0}(T)+ \bigg[\sum\limits_{n=1}^{N}Z'_{n}(T)\cos nx+\frac{f'_{2}(T)}{2\pi}x^{2}\bigg]\bigg\}, \\ \sqrt{p}[-b_{0, 1}\sin \sqrt{p}T+b_{0, 2}\cos \sqrt{p}T]+\sum\limits_{n=1}^{N}(-b_{n, 3}\lambda_{n, 2}\sin\lambda_{n, 2}T+b_{n, 4}\lambda_{n, 2}\cos\lambda_{n, 2}T)\cos nx\\ \ \ \ =2q\mu_{2}\bigg[-\sum\limits_{n=1}^{N}nZ_{n}(T)\sin nx+\frac{f_{2}(T)}{\pi}x\bigg], \end{array}\right. $

解之得

$ \left\{\begin{array}{rl} b_{n, 3}=& 2\mu_{1}Z'_{n}(T)\cos\lambda_{n, 2}T+\frac{4\mu_{1}(-1)^{n} f'_{2}(T)\cos\lambda_{n, 2}T}{n^{2}\pi} \\ & +\frac{4q\mu_{2}[(-1)^{n+1}+1]f_{2}(T)\sin\lambda_{n, 2}T}{n^{2}\lambda_{n, 2} \pi^{2}}, \\ b_{n, 4}=& 2\mu_{1}Z'_{n}(T)\sin\lambda_{n, 2}T+\frac{4\mu_{1}(-1)^{n} f'_{2}(T)\sin\lambda_{n, 2}T}{n^{2}\pi} \\ & -\frac{4q\mu_{2}[(-1)^{n+1}+1]f_{2}(T)\cos\lambda_{n, 2}T}{n^{2} \lambda_{n, 2}\pi^{2}}, \\ b_{0, 1}=& 2\mu_{1}Z'_{0}(T)\cos \sqrt{p}T+\frac{\pi}{6}f'_{2}(T)\cos \sqrt{p}T\\ & -\frac{2q\mu_{2}}{\sqrt{p}\pi}\sum\limits_{n=1}^{N}[(-1)^{n}-1]Z_{n}(T) \sin \sqrt{p}T -\frac{f_{2}(T)}{2\sqrt{p}}\sin \sqrt{p}T, \\ b_{0, 2}=& 2\mu_{1}Z'_{0}(T)\sin \sqrt{p}T+\frac{\pi}{6}f'_{2}(T)\sin \sqrt{p}T\\ &+\frac{2q\mu_{2}}{\sqrt{p}\pi}\sum\limits_{n=1}^{N}[(-1)^{n}-1]Z_{n}(T) \cos \sqrt{p}T-\frac{f_{2}(T)}{2\sqrt{p}}\cos \sqrt{p}T. \end{array}\right. $

这里

$ Z_{0}(t)=-\frac{1}{\sqrt{p}}\int_{0}^{t}\eta(\tau)\sin \sqrt{p}\tau {\rm d}\tau\cos\sqrt{p}t+\frac{1}{\sqrt{p}}\int_{0}^{t} \eta(\tau)\cos \sqrt{p}\tau {\rm d}\tau\sin\sqrt{p}t, $

其中, $\eta(t)=\frac{p}{\pi}f_{1}(t)+\frac{q}{\pi}f_{2}(t) -\frac{\pi}{6}f''_{2}(t)-\frac{p\pi}{6}f_{2}(t)$ .

3 $^{{\rm o}}$ .对于(4.6)式中的(ⅲ), 根据终端条件(3.3)以及(4.4)和(4.19)-(4.20)式得

$ \left\{\begin{array}{rl} &\ \ \ \sum\limits_{n=1}^{N}(c_{n, 1}\cos\lambda_{n, 2}T+c_{n, 2}\sin\lambda_{n, 2}T)\sin nx\\ &= 2\mu_{1}\bigg[\sum\limits_{n=1}^{N}Y'_{n}(T)\sin nx+\frac{f'_{1}(T)}{\pi}x\bigg], \\ &\ \ \ \sum\limits_{n=1}^{N}(-c_{n, 1}\lambda_{n, 2}\sin\lambda_{n, 2}T+c_{n, 2}\lambda_{n, 2}\cos\lambda_{n, 2}T)\sin nx \\ &= 2p\mu_{2} \bigg[\sum\limits_{n=1}^{N}nY_{n}(T)\cos nx+\frac{f_{1}(T)}{\pi}\bigg], \end{array}\right. $

于是, 解之得

$ \left\{\begin{array}{rl} c_{n, 1}=& 2\mu_{1}Y'_{n}(T)\cos\lambda_{n, 2}T+\frac{4\mu_{1}(-1)^{n}f'_{1}(T)\cos\lambda_{n, 2}T}{n\pi} \\ & +\frac{4p\mu_{2}[(-1)^{n+1}+1]f_{1}(T)\sin\lambda_{n, 2}T}{n\lambda_{n, 2}\pi^{2}}, \\ c_{n, 2}=& 2\mu_{1}Y'_{n}(T)\sin\lambda_{n, 2}T+\frac{4\mu_{1}(-1)^{n}f'_{1}(T)\sin\lambda_{n, 2}T}{n\pi} \\ & -\frac{4p\mu_{2}[(-1)^{n+1}+1]f_{1}(T)\cos\lambda_{n, 2}T}{n\lambda_{n, 2}\pi^{2}}. \end{array}\right. $

再根据终端条件(3.3)以及(4.7)和(4.21)-(4.22)式得

$ \left\{\begin{array}{lll} c_{n, 3}=a_{n, 3}, \\ c_{n, 4}=a_{n, 4}. \end{array}\right. $

4 $^{{\rm o}}$ .对于(4.6)式中的(ⅳ), 同理可得

$ \left\{\begin{array}{lll} d_{n, 1}=c_{n, 1}, \quad d_{n, 3}=b_{n, 3}, \\ d_{n, 2}=c_{n, 2}, \quad d_{n, 4}=b_{n, 4}. \end{array}\right. $

对于上述1 $^{{\rm o}}$ -4 $^{{\rm o}}$ 的各种情形所确定的 $b_{0, 1}, b_{0, 2}; a_{n, j}, b_{n, j}, c_{n, j}, d_{n, j}, j=1, 2, 3, 4$ .分别代入(4.6)式相应得到 $S_{n}(t)$ 和 $Q_{n}(t)$ 再代入(4.4)和(4.5)式得到对偶系统的最优轨线序列 $\left(u_{N}^{\ast}(x, t), \theta_{N}^{\ast}(x, t)\right)$ , 进而得到最优控制序列 $(f_{N, 1}^{\ast}(t), f_{N, 2}^{\ast}(t))$ .因此, 最优性能指标也可以近似地表示为

$ \begin{eqnarray}\label{4.23} J(f_{N, 1}^{\ast}(t), f_{N, 2}^{\ast}(t)) &=&\int_{0}^{\pi}\mu_{1}[(w_{N}^{\ast})_{t}^{2}(x, T)+(\varphi_{N}^{\ast})_{t}^{2}(x, T)]{\rm d}x\\ &&+\int_{0}^{\pi}\mu_{2}[p((w_{N}^{\ast})_{x}(x, T)-(\varphi_{N}^{\ast})(x, T))^{2}+q(\varphi_{N}^{\ast})_{x}^{2}(x, T)]{\rm d}x\\ &&+\int_{0}^{T}\mu_{3}[f_{N, 1}^{\ast2}(t)+f_{N, 2}^{\ast2}(t)]{\rm d}t. \end{eqnarray} $

(4.23)