近三十年来, 对于有限维系统在滚动时域控制下的稳定性与最优性, 已受到许多学者的广泛研究, 并得到了许多深刻的结果. 然而对于无穷维控制系统, 近十年才引起学者们的关注. 从目前的研究结果来看, 对于连续时间的无穷维动态系统的最优性, 还没有统一的理论结果. 但是, 对于Timoshenko梁系统的稳定性的研究已有许多很好的结果. 就我们的知识而言, 关于Timoshenko梁系统的最优性结果仍较少, 参见文献[1-4]; 与此同时, 关于板(或Mindlin-Timoshenko板)系统的稳定性研究的结果已有丰富的结果, 参见文献[5-17]. 然而, 具有滚动时域控制的系统的稳定性与最优性结果就比较少, Azmi和Kunisch在文献[18-19]中研究了具有滚动时域控制的Burgers方程和波方程的稳定性与最优性. 基于此, 本文研究的是Mindlin-Timoshenko板系统的最优性与稳定性.

设$ \Omega $是$ {{\Bbb R}} ^{2} $中的有界开集, $ \partial\Omega = \Gamma = \Gamma_{0}\cup\Gamma_{1}(\Gamma_{0}\cap\Gamma_{1} = \emptyset) $满足Lipschitz边界条件, 且$ \Gamma_{1} $是具有非空内部的闭集, $ \Gamma_{0}\neq\emptyset $是相对开的.

我们考虑下面具有边界控制的无限时域的二维Mindlin-Timosahenko系统

(1.1) $ \begin{eqnarray} \left\{\begin{array}{lll} { } \rho_{1}\psi_{tt}-D(\psi_{xx}+\frac{1-\mu}{2}\psi_{yy} +\frac{1+\mu}{2}\phi_{xy}) +K(\psi+\omega_{x}) = 0, & (x, y, t)\in\Omega\times {{\Bbb R}} ^{+}, \\ { } \rho_{1}\phi_{tt}-D(\phi_{yy}+\frac{1-\mu}{2}\phi_{xx} +\frac{1+\mu}{2}\psi_{xy}) +K(\phi+\omega_{y}) = 0, &(x, y, t)\in\Omega\times {{\Bbb R}} ^{+}, \\ { } \rho_{2}\omega_{tt}-K[(\psi+\omega_{x})_{x}+(\phi+\omega_{y})_{y}] = 0, & (x, y, t)\in\Omega\times {{\Bbb R}} ^{+}, \end{array}\right. \end{eqnarray} $

其中, $ \rho_{1} = \frac{\rho h^{3}}{2}, \rho_{2} = \rho h $, $ \rho $是密度, $ h $是板的厚度, $ \mu\in(0, \frac{1}{2}) $是Poisson比, $ D = \frac{Eh^{3}}{12(1-\mu^{2})} $ ($ E $杨氏模量)表示弹性模量, $ K = \frac{kEh}{2(1+\mu)} $ ($ k $剪切校正, $ E $杨氏模量)表示剪切模量, 函数$ \psi, \phi $和$ \omega $依赖于$ (x, y, t)\in\Omega\times {{\Bbb R}} ^{+} $表示板的全转角和板的横向位移.

边界条件和初始条件

(1.2) $ \begin{eqnarray} \left\{\begin{array}{lll} { } \psi = \phi = \omega = 0, & (x, y, t)\in\Gamma_{0}\times {{\Bbb R}} ^{+}, \\ { } D[\nu_{1}\psi_{x}+\mu\nu_{1}\phi_{y}+\frac{1-\mu}{2}(\psi_{y}+\phi_{x})\nu_{2}] = u_{1}, &(x, y, t)\in\Gamma_{1}\times {{\Bbb R}} ^{+}, \\ { } D[\nu_{2}\phi_{y}+\mu\nu_{2}\psi_{x}+\frac{1-\mu}{2}(\psi_{y}+\phi_{x})\nu_{1}] = u_{2}, & (x, y, t)\in\Gamma_{1}\times {{\Bbb R}} ^{+}, \\ { } K(\frac{\partial\omega}{\partial{\bf \nu}}+\nu_{1}\psi+\nu_{2}\phi) = u_{3}, & (x, y, t)\in\Gamma_{1}\times {{\Bbb R}} ^{+}, \\ { } (\psi(x, y, 0), \phi(x, y, 0), \omega(x, y, 0)) = (\psi_{01}, \phi_{01}, \omega_{01}), & (x, y)\in\Omega, \\ { } (\psi_{t}(x, y, 0), \phi_{t}(x, y, 0), \omega_{t}(x, y, 0)) = (\psi_{02}, \phi_{02}, \omega_{02}), & (x, y)\in\Omega, \end{array}\right. \end{eqnarray} $

其中, $ {\bf \nu} = (\nu_{1}, \nu_{2}) $是$ \Gamma_{1} $的单位外法向量, $ U(t) = (u_{1}, u_{2}, u_{3}) $是控制变量. 与系统(1.1)–(1.2)相关的具体技术细节请参阅文献[8].

定义系统(1.1)在$ t $时刻的能量为

$ \begin{eqnarray*} E(t)& = &\frac{1}{2}\int_{\Omega}[D(|\psi_{x}|^{2}+|\phi_{y}|^{2}+2\mu\psi_{x}\phi_{y} +\frac{1-\mu}{2}|\psi_{y}+\phi_{x}|^{2})+K(|\psi+\omega_{x}|^{2}\\ &&+|\phi+\omega_{y}|^{2})]{\rm d}x{\rm d}y +\frac{1}{2}\int_{\Omega}[\rho_{1}(|\psi_{t}|^{2}+|\varphi_{t}|^{2})+\rho_{2}|\omega_{t}|^{2}]{\rm d}x{\rm d}y. \end{eqnarray*} $

为了叙述方便起见, 对足够光滑的$ \Phi = (\psi, \phi, \omega), \widehat{\Phi} = (\widehat{\psi}, \widehat{\phi}, \widehat{\omega}) $, 定义如下线性算子及双线性泛函

$ L_{1}\{\psi, \phi, \omega\} = D(\psi_{xx}+\frac{1-\mu}{2}\psi_{yy} +\frac{1+\mu}{2}\phi_{xy})-K(\psi+\omega_{x}), $

$ L_{2}\{\psi, \phi, \omega\} = D(\phi_{yy}+\frac{1-\mu}{2}\phi_{xx} +\frac{1+\mu}{2}\psi_{xy})-K(\phi+\omega_{y}), $

$ L_{3}\{\psi, \phi, \omega\} = K[(\psi+\omega_{x})_{x}+(\phi+\omega_{y})_{y}], $

$ F_{1}\{\psi, \phi\} = D[\nu_{1}\psi_{x}+\mu\nu_{1}\phi_{y}+\frac{1-\mu}{2}(\psi_{y}+\phi_{x})\nu_{2}], $

$ F_{2}\{\psi, \phi\} = D[\nu_{2}\phi_{y}+\mu\nu_{2}\psi_{x}+\frac{1-\mu}{2}(\psi_{y}+\phi_{x})\nu_{1}], $

$ F_{3}\{\psi, \phi, \omega\} = K(\frac{\partial\omega}{\partial{\bf \nu}}+\nu_{1}\psi+\nu_{2}\phi), $

$ \begin{eqnarray*} a({\psi, \phi, \omega};{\widehat{\psi}, \widehat{\phi}, \widehat{\omega}}) & = &D\int_{\Omega}[\psi_{x}\widehat{\psi}_{x}+\phi_{y}\widehat{\phi}_{y} +\mu(\widehat{\psi}_{x}\phi_{y}+\psi_{x}\widehat{\phi}_{y}) +\frac{1-\mu}{2}(\psi_{y}+\psi_{x})(\widehat{\psi}_{y}+\widehat{\psi}_{x})]{\rm d}x{\rm d}y\\ &&+K\int_{\Omega}[(\psi+\omega_{x})(\widehat{\psi}+\widehat{\omega}_{x}) +(\phi+\omega_{y})(\widehat{\phi}+\widehat{\omega}_{y})]{\rm d}x{\rm d}y. \end{eqnarray*} $

引入函数空间$ W = \{(\psi, \phi, \omega)\in[H^{1}(\Omega)]^{3}|\psi = \phi = \omega = 0, (x, y)\in\Gamma_{0}\} $, 赋予范数$ \left\|(\psi, \phi, \omega)\right\|_{W}^{2} = a(\psi, \phi, \omega; \psi, \phi, \omega) $, 其中$ H^{1}(\Omega) $是一阶Sobolev空间^[20].

空间$ H = (L_{\rho_{1}}^{2}(\Omega))^{2}\times L_{\rho_{2}}^{2}(\Omega) $, 赋予范数$ \|(p, q, r)\|_{H}^{2} = \int_{\Omega}[\rho_{1}(|p|^{2}+|q|^{2})+\rho_{2}|r|^{2}]{\rm d}x{\rm d}y $, 那么$ W, H $都是(复)Hilbert空间.

定义空间$ {\cal H} = W\times H $, 赋予范数$ \|(\psi, \phi, \omega; p, q, r)\|_{{\cal H}}^{2} = \|(\psi, \phi, \omega)\|_{W}^{2}+\|(p, q, r)\|_{H}^{2} $, 因此$ {\cal H} $也是一个Hilbert空间.

定义$ {\cal U} = (L^{2}(\Gamma_{1}))^{3} $, 赋予范数$ \|U\|_{{\cal U}}^{2} = \int_{\Gamma_{1}}(|u_{1}|^{2}+|u_{2}|^{2}+|u_{3}|^{2}){\rm d}\Gamma_{1} $, 那么$ {\cal U} $也是一个Hilbert空间.

在状态空间$ {\cal H} $, 控制空间$ {\cal U} $中, 记$ \Phi = (\psi, \phi, \omega), {\cal Y} = (\Phi, \Phi_{t}), {\cal Y}_{0} = (\Phi_{01}, \Phi_{02}), {\cal Y}, {\cal Y}_{0}\in {\cal H} $和$ U(t) = (u_{1}, u_{2}, u_{3})\in L^{2}((0, \infty);{\cal U}) $. 考虑具有下面的性能指标(代价函数)

(1.3) $ \begin{equation} J_{\infty}({\cal Y}_{0}, U) = \int_{0}^{\infty}\ell({\cal Y}(t), U(t)){\rm d}t, \end{equation} $

(1.4) $ \begin{equation} \ell({\cal Y}(t), U(t)) = \frac{1}{2}\|{\cal Y}(t)\|_{{\cal H}}^{2} +\frac{\beta}{2}\|U(t)\|_{{\cal U}}^{2}, \end{equation} $

这里, $ \beta $是一个正常数.

本文采用文献[17]的滚动时域控制方法, 即对任意给定采样时间$ \delta>0 $和一个合适的预测时间$ T>\delta $, 定义采样时间序列$ t_{k} = k\delta, k = 0, 1, \cdots $, 对于每一个时刻$ t_{k} $, 在有限预测区间$ [t_{k}, t_{k}+T] $内, 求解开环最优控制问题. 由此我们将无限时域的最优控制问题转化为有限时间$ [t_{k}, t_{k}+T] $内最优控制问题. $ {\cal Y}(t), U(t) $分别表示滚动时域的状态和控制变量, $ {\cal Y}_{T}^{\ast}(\cdot, {\cal Y}_{0}, t_{0}), U_{T}^{\ast}(\cdot, {\cal Y}_{0}, t_{0}) $分别表示在有限时域$ [0, T] $内的最优控制问题的最优状态和最优控制. 不难看出, 通过不断改变预测区间, 我们将获得新的最优控制以及最优状态轨线. 具体做法如下:

给定采样时间$ \delta>0 $以及预测时间$ T>\delta $, 当$ k = 0 $时, $ t_{0} = 0 $, $ {\cal Y}(t_{0}) = {\cal Y}_{0} $, 在时间$ [t_{k}, t_{k}, +T] $内求解以下开环最优问题

(1.5) $ \begin{eqnarray} \min\limits_{U\in L^{2}((t_{k}, t_{k}+T);{\cal U})} J_{T}({\cal Y}(t_{k}), U) = \min\limits_{U\in L^{2}((t_{k}, t_{k}+T);{\cal U})} \int_{t_{k}}^{t_{k}+T}\ell({\cal Y}(t), U(t)){\rm d}t, \end{eqnarray} $

满足

(1.6) $ \begin{eqnarray} \left\{\begin{array}{lll} { } \rho_{1}\psi_{tt}-D(\psi_{xx}+\frac{1-\mu}{2}\psi_{yy} +\frac{1+\mu}{2}\phi_{xy}) +K(\psi+\omega_{x}) = 0, (x, y, t)\in\Omega\times (t_{k}, t_{k}+T), \\ { } \rho_{1}\phi_{tt}-D(\phi_{yy}+\frac{1-\mu}{2}\phi_{xx} +\frac{1+\mu}{2}\psi_{xy}) +K(\phi+\omega_{y}) = 0, (x, y, t)\in\Omega\times (t_{k}, t_{k}+T), \\ { } \rho_{2}\omega_{tt}-K[(\psi+\omega_{x})_{x}+(\phi+\omega_{y})_{y}] = 0, {\quad} (x, y, t)\in\Omega\times (t_{k}, t_{k}+T), \\ { } \psi = \phi = \omega = 0, {\quad} (x, y, t)\in\Gamma_{0}\times (t_{k}, t_{k}+T), \\ { } D[\nu_{1}\psi_{x}+\mu\nu_{1}\phi_{y}+\frac{1-\mu}{2}(\psi_{y}+\phi_{x})\nu_{2}] = u_{1}, {\quad} (x, y, t)\in\Gamma_{1}\times (t_{k}, t_{k}+T), \\ { } D[\nu_{2}\phi_{y}+\mu\nu_{2}\psi_{x}+\frac{1-\mu}{2}(\psi_{y}+\phi_{x})\nu_{1}] = u_{2}, {\quad} (x, y, t)\in\Gamma_{1}\times (t_{k}, t_{k}+T), \\ { } K(\frac{\partial\omega}{\partial{\bf \nu}}+\nu_{1}\psi+\nu_{2}\phi) = u_{3}, {\quad} (x, y, t)\in\Gamma_{1}\times (t_{k}, t_{k}+T), \\ { } (\psi(x, y, t_{k}), \phi(x, y, t_{k}), \omega(x, y, t_{k})) = (\psi_{t_{k}1}, \phi_{t_{k}1}, \omega_{t_{k}1}), {\quad}(x, y)\in\Omega, \\ { } (\psi_{t}(x, y, t_{k}), \phi_{t}(x, y, t_{k}), \omega_{t}(x, y, t_{k})) = (\psi_{t_{k}2}, \phi_{t_{k}2}, \omega_{t_{k}2}), {\quad}(x, y)\in\Omega. \end{array}\right. \end{eqnarray} $

本文将无限时域最优性问题转化为有限时域的最优性问题.借助乘子技巧, 对每一有限时域系统的解做先验估计, 证明了系统能量是一致指数衰减的.进一步, 借助对偶方法、变分原理和Bellman最优性原理, 得到系统的次优性条件, 由此证明了最优轨线是指数衰减的.

首先, 我们考虑任意初始值$ {\cal Y}_{0}\in {\cal H} $在有限时域内的最优控制问题(OCP)

(2.1) $ \begin{eqnarray} \min\limits_{U\in L^{2}((0, T);{\cal U})} J_{T}({\cal Y}_{0}, U) = \min\limits_{U\in L^{2}((0, T);{\cal U})}\int_{0}^{T}\ell({\cal Y}(t), U(t)){\rm d}t, \end{eqnarray} $

满足

(2.2) $ \begin{eqnarray} \left\{\begin{array}{lll} { } \rho_{1}\psi_{tt}-D(\psi_{xx}+\frac{1-\mu}{2}\psi_{yy} +\frac{1+\mu}{2}\phi_{xy}) +K(\psi+\omega_{x}) = 0, (x, y, t)\in\Omega\times (0, T), \\ { } \rho_{1}\phi_{tt}-D(\phi_{yy}+\frac{1-\mu}{2}\phi_{xx} +\frac{1+\mu}{2}\psi_{xy}) +K(\phi+\omega_{y}) = 0, (x, y, t)\in\Omega\times (0, T), \\ { } \rho_{2}\omega_{tt}-K[(\psi+\omega_{x})_{x}+(\phi+\omega_{y})_{y}] = 0, {\quad} (x, y, t)\in\Omega\times (0, T), \\ { } \psi = \phi = \omega = 0, {\quad} (x, y, t)\in\Gamma_{0}\times (0, T), \\ { } D[\nu_{1}\psi_{x}+\mu\nu_{1}\phi_{y}+\frac{1-\mu}{2}(\psi_{y}+\phi_{x})\nu_{2}] = u_{1}, {\quad} (x, y, t)\in\Gamma_{1}\times (0, T), \\ { } D[\nu_{2}\phi_{y}+\mu\nu_{2}\psi_{x}+\frac{1-\mu}{2}(\psi_{y}+\phi_{x})\nu_{1}] = u_{2}, {\quad}(x, y, t)\in\Gamma_{1}\times (0, T), \\ { } K(\frac{\partial\omega}{\partial{\bf \nu}}+\nu_{1}\psi+\nu_{2}\phi) = u_{3}, {\quad}(x, y, t)\in\Gamma_{1}\times (0, T), \\ { } (\psi(x, y, 0), \phi(x, y, 0), \omega(x, y, 0)) = (\psi_{01}, \phi_{01}, \omega_{01}), {\quad}(x, y)\in\Omega, \\ { } (\psi_{t}(x, y, 0), \phi_{t}(x, y, 0), \omega_{t}(x, y, 0)) = (\psi_{02}, \phi_{02}, \omega_{02}), {\quad} (x, y)\in\Omega. \end{array}\right. \end{eqnarray} $

定义2.1 (值函数)   对每一初始状态$ {\cal Y}_{0}\in {\cal H} $, 定义无限时域值函数: $ V_{\infty}: {\cal H}\rightarrow {{\Bbb R}} ^{+} $,

$ V_{\infty}({\cal Y}_{0}) = \inf\limits_{U\in L^{2}((0, +\infty);{\cal U})}\{J_{\infty}({\cal Y}_{0}, U)\}. $

其中$ ({\cal Y}, U) $满足系统(1.1)和(1.2)；类似地, 定义有限时域的值函数: $ V_{T}: {\cal H}\rightarrow {{\Bbb R}} ^{+} $,

$ V_{T}({\cal Y}_{0}) = \min\limits_{U\in L^{2}((0, T);{\cal U})}\{J_{T}({\cal Y}_{0}, U)\}, $

且$ ({\cal Y}, U) $满足系统(2.2).

定义2.2 (弱解)   给定$ T>0, {\cal Y}_{0}\in {\cal H}, \quad U\in {\cal U} $, 那么我们称$ \Phi = (\psi, \phi, \omega) $为系统(2.2)的弱解, 如果对于任意$ \widehat{\Phi} = (\widehat{\psi}, \widehat{\phi}, \widehat{\omega})\in C^{1}([0, T]\times\Omega) $, 在$ \Gamma_{0} $上, $ \widehat{\Phi} = 0 $, 以及对于$ \forall t\in[0, T], \tau\in[0, t] $使得

(2.3) $ \begin{eqnarray} & &\int_{\Omega}\{\rho_{1}[\psi_{t}(t, x, y)\widehat{\psi}(t, x, y)+\phi_{t}(t, x, y)\widehat{\phi}(t, x, y)] +\rho_{2}\omega_{t}(t, x, y)\widehat{\omega}(t, x, y)\}{\rm d}x{\rm d}y\\ &&-\int_{\Omega}\{\rho_{1}[\psi_{02}\widehat{\psi}(0, x, y)+\phi_{02}\widehat{\phi}(0, x, y)] +\rho_{2}\omega_{02}\widehat{\omega}(0, x, y)\}{\rm d}x{\rm d}y\\ &&-\int_{0}^{t}\int_{\Omega}\{\rho_{1}[\psi_{t}(\tau, x, y)\widehat{\psi}_{t}(\tau, x, y)+\phi_{t}(\tau, x, y)\widehat{\phi}_{t}(\tau, x, y)] +\rho_{2}\omega_{t}(\tau, x, y)\widehat{\omega}_{t}(\tau, x, y)\}{\rm d}x{\rm d}y{\rm d}\tau\\ &&+\int_{0}^{t}a(\psi, \phi, \omega; \widehat{\psi}, \widehat{\phi}, \widehat{\omega}){\rm d}\tau -\int_{0}^{t}\int_{\Gamma_{1}}(\widehat{\psi}u_{1}+\widehat{\phi}u_{2}+\widehat{\omega}u_{3}){\rm d}\Gamma {\rm d}\tau = 0. \end{eqnarray} $

命题2.1   给定$ T>0, {\cal Y}_{0}\in {\cal H}, \quad U\in L^{2}([0, T];{\cal U}) $, 则系统(2.2)存在唯一的弱解$ \Phi = (\psi, \phi, \omega)\in C^{0}([0, T];W)\cap C^{1}([0, T], H) $满足

(2.4) $ \begin{eqnarray} &&\|\Phi\|_{C^{0}([0, T];W)} +\|\Phi_{t}\|_{C^{0}([0, T];H)}+\|\Phi_{tt}\|_{L^{2}([0, T];W^{\ast})}\\ &\leq &C(\|\Phi_{01}\|_{W} +\|\Phi_{02}\|_{H}+\|U\|_{L^{2}([0, T];{\cal U})}), \end{eqnarray} $

其中$ C $(常数$ C $在不同地方表示不一样的常数)是与$ \Phi_{01}, \Phi_{02}, U $无关的正常数.

证  记$ I(t) = a(\psi, \phi, \omega; \psi, \phi, \omega)+\int_{\Omega}[\rho_{1}(|\psi_{t}|^{2}+|\phi_{t}|^{2}) +\rho_{2}|\omega_{t}|^{2}]{\rm d}x{\rm d}y $, 则有

(2.5) $ \begin{eqnarray} \frac{{\rm d}I(t)}{{\rm d}t} & = &\frac{\partial a(\psi, \phi, \omega; \psi, \phi, \omega)}{\partial t} + 2\int_{\Omega}[\rho_{1}(\psi_{t}\psi_{tt}+\phi_{t}\phi_{tt}) +\rho_{2}\omega_{t}\omega_{tt}]{\rm d}x{\rm d}y\\ & = &2a(\psi, \phi, \omega; \psi_{t}, \phi_{t}, \omega_{t})+2\int_{\Omega}[\psi_{t}L_{1}\{\psi, \phi, \omega\} +\phi_{t}L_{2}\{\psi, \phi, \omega\}+\omega_{t}L_{3}\{\psi, \phi, \omega\}]{\rm d}x{\rm d}y\\ & = &2\int_{\Gamma_{1}}[\psi_{t}F_{1}\{\psi, \phi\} +\phi_{t}F_{2}\{\psi, \phi\}+\omega_{t}F_{3}\{\psi, \phi, \omega\}]{\rm d}\Gamma_{1}\\ &\leq& 2\varepsilon\int_{\Gamma_{1}}[|\psi_{t}|^{2}+|\phi_{t}|^{2}+|\omega_{t}|^{2}]{\rm d}\Gamma_{1} +\frac{1}{2\varepsilon}\int_{\Gamma_{1}}[|u_{1}|^{2}+|u_{2}|^{2}+|u_{3}|^{2}]{\rm d}\Gamma_{1}. \end{eqnarray} $

设$ f = (f_{1}, f_{2}): \overline{\Omega}\rightarrow {{\Bbb R}} ^{2} $是$ C^{1} $的向量值函数, 且在$ \overline{\Gamma}_{0} $上有$ f = 0 $, 则存在$ \sigma>0 $, 在$ \Gamma_{1} $上几乎处处满足$ f\cdot \nu\geq\sigma $.

记$ 2M_{0} = \min\{\rho_{1}\rho_{1}\}, R_{0} = \max\limits_{(x, y)\in\overline{\Omega}}\{|f|, |f_{ix}|, |f_{iy}|\}, i = 1, 2 $, 其中, $ |f|^{2} = f_{1}^{2}+f_{2}^{2} $. 对$ \forall t\in[0, T], \tau\in[0, t] $, 将$ f\cdot\nabla\psi, f\cdot\nabla\phi, f\cdot\nabla\omega $分别乘以系统(2.2)的前三式, 在$ [0, t]\times\Omega $上积分并相加得

(2.6) $ \begin{eqnarray} &&\int_{0}^{t}\int_{\Omega}[\rho_{1}(\psi_{tt}f\cdot\nabla\psi +\phi_{tt}f\cdot\nabla\phi)+\rho_{2}\omega_{tt}f\cdot\nabla\omega]{\rm d}x{\rm d}y{\rm d}\tau\\ &&-\int_{0}^{t}\int_{\Omega}[L_{1}\{\psi, \phi, \omega\}f\cdot\nabla\psi +L_{2}\{\psi, \phi, \omega\}f\cdot\nabla\phi+L_{3}\{\psi, \phi, \omega\}f\cdot\nabla\omega]{\rm d}x{\rm d}y{\rm d}\tau = 0. \end{eqnarray} $

分部积分得

$ \begin{eqnarray*} &&\int_{\Omega}[\rho_{1}(\psi_{t}f\cdot\nabla\psi +\phi_{t}f\cdot\nabla\phi)+\rho_{2}\omega_{t}f\cdot\nabla\omega]_{0}^{t}{\rm d}x{\rm d}y\\ &&-\frac{1}{2}\int_{0}^{t}\int_{\Omega}[\rho_{1}(f\cdot\nabla|\psi_{t}|^{2} +f\cdot\nabla|\phi_{t}|^{2})+\rho_{2}f\cdot\nabla|\omega_{t}|^{2}]{\rm d}x{\rm d}y{\rm d}\tau\\ &&+\int_{0}^{t}a(\psi, \phi, \omega;f\cdot\nabla\psi, f\cdot\nabla\phi, f\cdot\nabla\omega){\rm d}\tau\\ &&-\int_{0}^{t}\int_{\Gamma_{1}}[F_{1}\{\psi, \phi\}f\cdot\nabla\psi +F_{2}\{\psi, \phi\}f\cdot\nabla\phi+F_{3}\{\psi, \phi, \omega\}f\cdot\nabla\omega]{\rm d}\Gamma_{1}{\rm d}\tau = 0, \end{eqnarray*} $

即

(2.7) $ \begin{eqnarray} &&\int_{\Omega}[\rho_{1}(\psi_{t}f\cdot\nabla\psi +\phi_{t}f\cdot\nabla\phi)+\rho_{2}\omega_{t}f\cdot\nabla\omega]_{0}^{t}{\rm d}x{\rm d}y\\ &&-\frac{1}{2}\int_{0}^{t}\int_{\Gamma_{1}}(f\cdot\nu)[\rho_{1}(|\psi_{t}|^{2} +|\phi_{t}|^{2})+\rho_{2}|\omega_{t}|^{2}]{\rm d}\Gamma_{1}{\rm d}\tau\\ &&+\frac{1}{2}\int_{0}^{t}\int_{\Omega}{\rm div}f[\rho_{1}(|\psi_{t}|^{2} +|\phi_{t}|^{2})+\rho_{2}|\omega_{t}|^{2}]{\rm d}x{\rm d}y{\rm d}\tau\\ &&+\int_{0}^{t}a(\psi, \phi, \omega;f\cdot\nabla\psi, f\cdot\nabla\phi, f\cdot\nabla\omega){\rm d}\tau\\ &&-\int_{0}^{t}\int_{\Gamma_{1}}[F_{1}\{\psi, \phi\}f\cdot\nabla\psi +F_{2}\{\psi, \phi\}f\cdot\nabla\phi+F_{3}\{\psi, \phi, \omega\}f\cdot\nabla\omega]{\rm d}\Gamma_{1}{\rm d}\tau = 0. \end{eqnarray} $

又由于

$ \begin{eqnarray*} &&\int_{0}^{t}a(\psi, \phi, \omega;f\cdot\nabla\psi, f\cdot\nabla\phi, f\cdot\nabla\omega){\rm d}\tau\\ & = &D\int_{0}^{t}\int_{\Omega}[\psi_{x}(f_{1x}\psi_{x}+f_{2x}\psi_{y})+\phi_{y}(f_{1y}\phi_{x}+f_{2y}\phi_{y}) +\mu\psi_{x}(f_{1y}\phi_{x}+f_{2y}\phi_{y})\\ &&+\mu\phi_{y}(f_{1x}\psi_{x}+f_{2x}\psi_{y}) +\frac{1-\mu}{2}(\psi_{y}+\phi_{x})(f_{1y}\psi_{x}+f_{2y}\psi_{y}+f_{1x}\phi_{x}+f_{2x}\phi_{y})]{\rm d}x{\rm d}y{\rm d}\tau\\ &&+K\int_{0}^{t}\int_{\Omega}[(\psi+\omega_{x})(f_{1x}\omega_{x}+f_{2x}\omega_{y}) +(\phi+\omega_{y})(f_{1y}\omega_{x}+f_{2y}\omega_{y})]{\rm d}x{\rm d}y{\rm d}\tau\\ &&+\frac{D}{2}\int_{0}^{t}\int_{\Omega}[f\cdot\nabla|\psi_{x}|^{2}+f\cdot\nabla|\phi_{y}|^{2} +\mu f\cdot\nabla(\psi_{x}\phi_{y})+\frac{1-\mu}{2}f\cdot\nabla|\psi_{y}+\phi_{x}|^{2}]{\rm d}x{\rm d}y{\rm d}\tau\\ &&+\frac{K}{2}\int_{0}^{t}\int_{\Omega}[f\cdot\nabla|\psi+\omega_{x}|^{2}+f\cdot\nabla|\phi+\omega_{y}|^{2}]{\rm d}x{\rm d}y{\rm d}\tau\\ & = &D\int_{0}^{t}\int_{\Omega}\nabla\psi\cdot\left(\begin{array}{cc} f_{1x}& f_{1y}\\ f_{2x}&f_{2y} \end{array}\right)\left(\begin{array}{cc} \psi_{x}+\mu\phi_{y}\\ { } \frac{1-\mu}{2}(\psi_{x}+\phi_{x}) \end{array}\right){\rm d}x{\rm d}y{\rm d}\tau\\ &&+D\int_{0}^{t}\int_{\Omega}\nabla\phi\cdot\left(\begin{array}{cc} f_{1x}& f_{1y}\\ f_{2x}&f_{2y} \end{array}\right)\left(\begin{array}{cc} { } \frac{1-\mu}{2}(\psi_{x}+\phi_{x})\\ \phi_{y}+\mu\psi_{x} \end{array}\right){\rm d}x{\rm d}y{\rm d}\tau \\ &&+K\int_{0}^{t}\int_{\Omega}\nabla\omega\cdot\left(\begin{array}{cc} f_{1x}& f_{1y}\\ f_{2x}&f_{2y} \end{array}\right)\left(\begin{array}{cc} \psi+\omega_{x})\\ \phi+\omega_{y} \end{array}\right){\rm d}x{\rm d}y{\rm d}\tau\\ &&+\frac{D}{2}\int_{0}^{t}\int_{\Gamma_{1}}(f\cdot\nu)[|\psi_{x}|^{2}+|\phi_{y}|^{2} +2\mu\psi_{x}\phi_{y}+\frac{1-\mu}{2}|\psi_{y}+\phi_{x}|^{2}]{\rm d}\Gamma_{1}{\rm d}\tau\\ &&+\frac{K}{2}\int_{0}^{t}\int_{\Omega}(f\cdot\nu)[|\psi+\omega_{x}|^{2}+|\phi+\omega_{y}|^{2}]{\rm d}x{\rm d}y{\rm d}\tau\\ &&-\frac{D}{2}\int_{0}^{t}\int_{\Omega}{\rm div}f[|\psi_{x}|^{2}+|\psi_{y}|^{2} +2\mu\psi_{x}\phi_{y}+\frac{1-\mu}{2}|\psi_{y}+\phi_{x}|^{2}]{\rm d}x{\rm d}y{\rm d}\tau\\ &&-\frac{K}{2}\int_{0}^{t}\int_{\Omega}{\rm div}f[|\psi+\omega_{x}|^{2}+|\phi+\omega_{y}|^{2}]{\rm d}x{\rm d}y{\rm d}\tau. \end{eqnarray*} $

因此

$ \begin{eqnarray*} &&\frac{1}{2}\int_{0}^{t}\int_{\Gamma_{1}}(f\cdot\nu)[\rho_{1}(|\psi_{t}|^{2} +|\phi_{t}|^{2})+\rho_{2}|\omega_{t}|^{2}]{\rm d}\Gamma_{1}{\rm d}\tau\\ & = &\int_{\Omega}[\rho_{1}(\psi_{t}f\cdot\nabla\psi +\phi_{t}f\cdot\nabla\phi)+\rho_{2}\omega_{t}f\cdot\nabla\omega]_{0}^{t}{\rm d}x{\rm d}y\\ &&+\frac{1}{2}\int_{0}^{t}\int_{\Omega}{\rm div}f[\rho_{1}(|\psi_{t}|^{2} +|\phi_{t}|^{2})+\rho_{2}|\omega_{t}|^{2}]{\rm d}x{\rm d}y{\rm d}\tau\\ &&+\int_{0}^{t}a(\psi, \phi, \omega;f\cdot\nabla\psi, f\cdot\nabla\phi, f\cdot\nabla\omega){\rm d}\tau\\ &&-\int_{0}^{t}\int_{\Gamma_{1}}[F_{1}\{\psi, \phi\}f\cdot\nabla\psi +F_{2}\{\psi, \phi\}f\cdot\nabla\phi+F_{3}\{\psi, \phi, \omega\}f\cdot\nabla\omega]{\rm d}\Gamma_{1}{\rm d}\tau. \end{eqnarray*} $

为了方便起见, 记

$ \begin{eqnarray*} I_{1}& = &\int_{\Omega}[\rho_{1}(\psi_{t}f\cdot\nabla\psi +\phi_{t}f\cdot\nabla\phi)+\rho_{2}\omega_{t}f\cdot\nabla\omega]_{0}^{t}{\rm d}x{\rm d}y, \\ I_{2} & = &\frac{1}{2}\int_{0}^{t}\int_{\Omega}{\rm div}f[\rho_{1}(|\psi_{t}|^{2} +|\phi_{t}|^{2})+\rho_{2}|\omega_{t}|^{2}]{\rm d}x{\rm d}y{\rm d}\tau\\ &&+D\int_{0}^{t}\int_{\Omega}\nabla\psi\cdot\left(\begin{array}{cc} f_{1x}& f_{1y}\\ f_{2x}&f_{2y} \end{array}\right)\left(\begin{array}{cc} \psi_{x}+\mu\phi_{y}\\ { } \frac{1-\mu}{2}(\psi_{x}+\phi_{x}) \end{array}\right){\rm d}x{\rm d}y{\rm d}\tau\\ &&+D\int_{0}^{t}\int_{\Omega}\nabla\phi\cdot\left(\begin{array}{cc} f_{1x}& f_{1y}\\ f_{2x}&f_{2y} \end{array}\right)\left(\begin{array}{cc} { } \frac{1-\mu}{2}(\psi_{x}+\phi_{x})\\ \phi_{y}+\mu\psi_{x} \end{array}\right){\rm d}x{\rm d}y{\rm d}\tau \\ &&+K\int_{0}^{t}\int_{\Omega}\nabla\omega\cdot\left(\begin{array}{cc} f_{1x}& f_{1y}\\ f_{2x}&f_{2y} \end{array}\right)\left(\begin{array}{cc} \psi+\omega_{x})\\ \phi+\omega_{y} \end{array}\right){\rm d}x{\rm d}y{\rm d}\tau\\ &&-\frac{D}{2}\int_{0}^{t}\int_{\Omega}{\rm div}f[|\psi_{x}|^{2}+|\psi_{y}|^{2} +2\mu\psi_{x}\phi_{y}+\frac{1-\mu}{2}|\psi_{y}+\phi_{x}|^{2}]{\rm d}x{\rm d}y{\rm d}\tau\\ &&-\frac{K}{2}\int_{0}^{t}\int_{\Omega}{\rm div}f[|\psi+\omega_{x}|^{2}+|\phi+\omega_{y}|^{2}]{\rm d}x{\rm d}y{\rm d}\tau, \\ I_{3}& = &\frac{D}{2}\int_{0}^{t}\int_{\Gamma_{1}}(f\cdot\nu)[|\psi_{x}|^{2}+|\phi_{y}|^{2} +2\mu\psi_{x}\phi_{y}+\frac{1-\mu}{2}|\psi_{y}+\phi_{x}|^{2}]{\rm d}\Gamma_{1}{\rm d}\tau\\ &&+\frac{K}{2}\int_{0}^{t}\int_{\Gamma_{1}}(f\cdot\nu)[|\psi+\omega_{x}|^{2}+|\phi+\omega_{y}|^{2}]{\rm d}\Gamma_{1}{\rm d}\tau\\ &&-\int_{0}^{t}\int_{\Gamma_{1}}[F_{1}\{\psi, \phi\}f\cdot\nabla\psi +F_{2}\{\psi, \phi\}f\cdot\nabla\phi+F_{3}\{\psi, \phi, \omega\}f\cdot\nabla\omega]{\rm d}\Gamma_{1}{\rm d}\tau. \end{eqnarray*} $

下面对$ I_{1}, I_{2}, I_{3} $做估计.

由Poincaré和Cauchy-Schwarz不等式知, 存在与$ \psi, \phi, \omega $无关的正常数$ C_{1}, C_{2} $使得

$ \begin{eqnarray*} |I_{1}|&\leq& C_{1}\left|\left[a(\psi, \phi, \omega;\psi, \phi, \omega) +\int_{\Omega}[\rho_{1}(|\psi_{t}|^{2}+|\phi_{t}|^{2})+\rho_{2}|\omega_{t}|^{2}]{\rm d}x{\rm d}y\right]_{0}^{t}\right|\\ &\leq& C_{1}(I(t)+I(0)), \\ |I_{2}|&\leq& \frac{R_{0}}{2}\int_{0}^{t}\int_{\Omega}[\rho_{1}(|\psi_{t}|^{2}+|\phi_{t}|^{2})+\rho_{2}|\omega_{t}|^{2}]{\rm d}x{\rm d}y{\rm d}\tau +\frac{3R_{0}}{2}\int_{0}^{t}a(\psi, \phi, \omega;\psi, \phi, \omega){\rm d}\tau\\ &\leq&2R_{0}\bigg[\int_{0}^{t}\int_{\Omega}[\rho_{1}(|\psi_{t}|^{2}+|\phi_{t}|^{2})+\rho_{2}|\omega_{t}|^{2}]{\rm d}x{\rm d}y{\rm d}\tau +\int_{0}^{t}a(\psi, \phi, \omega;\psi, \phi, \omega){\rm d}\tau\bigg]\\ & = &2R_{0}\int_{0}^{t}I(\tau){\rm d}\tau, \\ |I_{3}|&\leq &\frac{R_{0}C_{2}D^{2}}{2}\int_{0}^{t}\int_{\Gamma_{1}}[|\psi_{x}+\mu\phi_{y}|^{2} +(\frac{1-\mu}{2})^{2}|\psi_{y}+\phi_{x}|^{2}]{\rm d}\Gamma_{1}{\rm d}\tau\\ &&+\frac{R_{0}C_{2}K^{2}}{2}\int_{0}^{t}\int_{\Gamma_{1}}[|\psi+\omega_{x}|^{2} +|\phi+\omega_{y}|^{2}]{\rm d}\Gamma_{1}{\rm d}\tau\\ &&+C_{2}R_{0}\int_{0}^{t}\int_{\Gamma_{1}}[|F_{1}\{\psi, \phi\}|^{2} +|F_{2}\{\psi, \phi\}|^{2}+|F_{3}\{\psi, \phi, \omega\}|^{2}]{\rm d}\Gamma_{1}{\rm d}\tau\\ & = &\frac{C_{2}R_{0}}{2}\int_{0}^{t}\int_{\Gamma_{1}}[|F_{1}\{\psi, \phi\}|^{2} +|F_{2}\{\psi, \phi\}|^{2}+|F_{3}\{\psi, \phi, \omega\}|^{2}]{\rm d}\Gamma_{1}{\rm d}\tau\\ &&+C_{2}R_{0}\int_{0}^{t}\int_{\Gamma_{1}}[|F_{1}\{\psi, \phi\}|^{2} +|F_{2}\{\psi, \phi\}|^{2}+|F_{3}\{\psi, \phi, \omega\}|^{2}]{\rm d}\Gamma_{1}{\rm d}\tau\\ & = &\frac{3C_{2}R_{0}}{2}\int_{0}^{t}\int_{\Gamma_{1}}[|u_{1}|^{2} +|u_{2}|^{2}+|u_{3}|^{2}]{\rm d}\Gamma_{1}{\rm d}\tau. \end{eqnarray*} $

因此, 我们有

(2.8) $ \begin{eqnarray} &&\int_{0}^{t}\int_{\Gamma_{1}}(|\psi_{t}|^{2} +|\phi_{t}|^{2}+|\omega_{t}|^{2}){\rm d}\Gamma_{1}{\rm d}\tau\\ &\leq&\frac{1}{\sigma M_{0}} \bigg[C_{1}(I(t)+I(0))+2R_{0}\int_{0}^{t}I(\tau){\rm d}\tau+\frac{3C_{2}R_{0}}{2}\int_{0}^{t}\int_{\Gamma_{1}}[|u_{1}|^{2} +|u_{2}|^{2}+|u_{3}|^{2}]{\rm d}\Gamma_{1}{\rm d}\tau\bigg]\\ &\leq& C_{3}\bigg[I(t)+I(0)+\int_{0}^{t}I(\tau){\rm d}\tau+\int_{0}^{t}\int_{\Gamma_{1}}[|u_{1}|^{2} +|u_{2}|^{2}+|u_{3}|^{2}]{\rm d}\Gamma_{1}{\rm d}\tau\bigg], \end{eqnarray} $

这里, $ C_{3} $是某个正常数.

由(2.5)和(2.8)式即得

(2.9) $ \begin{eqnarray} I(t)\leq C_{4}\bigg[I(0)+\int_{0}^{t}I(\tau){\rm d}\tau+\int_{0}^{t}\int_{\Gamma_{1}}[|u_{1}|^{2} +|u_{2}|^{2}+|u_{3}|^{2}]{\rm d}\Gamma_{1}{\rm d}\tau\bigg], \end{eqnarray} $

这里, $ C_{4} $是某个正常数.

由Gronwall不等式得

(2.10) $ \begin{eqnarray} I(t)\leq C_{5}\bigg[I(0)+\int_{0}^{t}\int_{\Gamma_{1}}[|u_{1}|^{2} +|u_{2}|^{2}+|u_{3}|^{2}]{\rm d}\Gamma_{1}{\rm d}\tau\bigg], \end{eqnarray} $

这里, $ C_{5} $是某个正常数. 即

$ \|\Phi\|_{C^{0}([0, T];W)} +\|\Phi_{t}\|_{C^{0}([0, T];H)} \leq C_{6}(\|\Phi_{01}\|_{W} +\|\Phi_{02}\|_{H}+\|U\|_{L^{2}([0, T];{\cal U})}), $

这里, $ C_{6} $是某个正常数.

对于任意$ Z = (z_{1}, z_{2}, z_{3})\in W $, 且$ \|Z\|_{W} = 1 $, 将$ Z = (z_{1}, z_{2}, z_{3}) $与系统(2.2)前三式在$ H $中作内积, 并分部积分得

$ \begin{eqnarray*} &&\left|\langle(\psi_{tt}, \phi_{tt}, \omega_{tt}), (z_{1}, z_{2}, z_{3})\rangle_{W^{\ast}\times W}\right| = \left|((\psi_{tt}, \phi_{tt}, \omega_{tt}), (z_{1}, z_{2}, z_{3}))_{H}\right|\\ & = &\left|a(\psi, \phi, \omega;z_{1}, z_{2}, z_{3})-\int_{\Gamma_{1}}(u_{1}z_{1}+u_{2}z_{2}+u_{3}z_{3}){\rm d}\Gamma_{1}\right|\\ &\leq& C_{7}(\|(\psi, \phi, \omega)\|_{W}+\|(u_{1}, u_{2}, u_{3})\|_{{\cal U}})\|(z_{1}, z_{2}, z_{3})\|_{W}, \end{eqnarray*} $

这里$ \langle\cdot, \cdot\rangle_{W^{\ast}\times W} $是对偶积, $ (\cdot, \cdot)_{H} $是內积, $ C_{7} $是某个正常数. 因此, 由(2.10)式得

$ \begin{eqnarray*} \int_{0}^{T}\|(\psi_{tt}, \phi_{tt}, \omega_{tt})\|_{W^{\ast}}{\rm d}t&\leq& C\int_{0}^{T}(\|(\psi, \phi, \omega)\|_{W}+\|(u_{1}, u_{2}, u_{3})\|_{{\cal U}}){\rm d}t\\ &\leq &C(\|\Phi_{01}\|_{W} +\|\Phi_{02}\|_{H}+\|U\|_{L^{2}([0, T];{\cal U})}). \end{eqnarray*} $

于是有

$ \|\Phi\|_{C^{0}([0, T];W)} +\|\Phi_{t}\|_{C^{0}([0, T];H)}+\|\Phi_{tt}\|_{L^{2}([0, T];W^{\ast})} \leq C(\|\Phi_{01}\|_{W} +\|\Phi_{02}\|_{H}+\|U\|_{L^{2}([0, T];{\cal U})}). $

并且有

(2.11) $ \begin{equation} \int_{0}^{t}\int_{\Gamma_{1}}(|\psi_{t}|^{2}+|\phi_{t}|^{2}+|\omega_{t}|^{2}){\rm d}\Gamma_{1}{\rm d}\tau \leq C(I(0)+ \int_{0}^{t}\int_{\Gamma_{1}}(|u_{1}|^{2}+|u_{2}|^{2}+|u{3}|^{2}){\rm d}\Gamma_{1}{\rm d}\tau. \end{equation} $

命题2.1证毕.

定理3.1   对于任意$ T>0 $以及$ {\cal Y}_{0}\in {\cal H} $, 最优控制问题(OCP)存在唯一解.

证  我们利用标准的变分原理进行证明. 由目标函数$ J_{T}({\cal Y}_{0}, U) $的定义知, $ J_{T}({\cal Y}_{0}, U) $是强制的且有下界, 故必有下确界. 记

(3.1) $ \begin{eqnarray} \inf\limits_{U\in L^{2}(0, T;{\cal U})}J_{T}({\cal Y}_{0}, U) = \inf\limits_{U\in L^{2}(0, T;{\cal U})}\int_{0}^{T}\ell({\cal Y}(t), U(t)){\rm d}t = \theta<\infty. \end{eqnarray} $

因此, 存在有界极小化序列$ \{U_{n}\}\subset L^{2}([0, T];{\cal U}) $, 使得

(3.2) $ \begin{eqnarray} J_{T}({\cal Y}_{0}, U_{n})\longrightarrow\inf\limits_{U\in L^{2}(0, T;{\cal U})}J_{T}({\cal Y}_{0}, U) = \theta. \end{eqnarray} $

进一步可知, $ {U_{n}} $存在弱收敛的子列, 不妨仍记为$ {U_{n}} $, 即

(3.3) $ \begin{eqnarray} U_{n}\stackrel{w}{\longrightarrow} U^{\ast} = (u_{1}^{\ast}, u_{2}^{\ast}, u_{3}^{\ast})\in L^{2}(0, T;{\cal U}). \end{eqnarray} $

由(2.4)式知, 对应于$ \{U_{n}\} $, 以及有界序列$ \{\Phi_{n}\}\subset L^{2}(0, T; W) $, $ \{\Phi_{nt}\}\subset L^{2}(0, T; H) $, $ \{\Phi_{ntt}\}\subset L^{2}(0, T:W^{\ast}) $, 存在子列, 仍记为$ \{\Phi_{n}\} $, 使得

(3.4) $ \begin{eqnarray} \begin{array}{lll} \Phi_{n}\stackrel{w^{\ast}}{\longrightarrow}\Phi^{\ast}, & {\rm in}\quad L^{\infty}(0, T; W), \\ \Phi_{nt}\stackrel{w^{\ast}}{\longrightarrow}\Phi_{t}^{\ast}, & {\rm in}\quad L^{\infty}(0, T; H), \\ \Phi_{ntt}\stackrel{w^{\ast}}{\longrightarrow}\Phi_{tt}^{\ast}, {\quad}& {\rm in}\quad L^{2}(0, T; W^{\ast}).\end{array} \end{eqnarray} $

下面, 我们将证明：当$ U = U^{\ast} $时, $ \Phi^{\ast} $是系统(2.2)的弱解.

对于任意$ t\in[0, T], \tau\in[0, t] $, 以及$ \forall\widehat{\Phi} = (\widehat{\psi}, \widehat{\phi}, \widehat{\omega})\in C^{1}([0, T]\times\Omega) $, 且在$ \Gamma_{0} $上, 有$ \widehat{\Phi} = 0 $, 于是, 由(3.3)与(3.4)式有

(3.5) $ \begin{eqnarray} &&\int_{0}^{t}[a(\psi_{n}, \phi_{n}, \omega_{n};\widehat{\psi}, \widehat{\phi}, \widehat{\omega}) -a(\psi^{\ast}, \phi^{\ast}, \omega^{\ast};\widehat{\psi}, \widehat{\phi}, \widehat{\omega})]{\rm d}\tau\longrightarrow 0, \\ &&\int_{0}^{t}\int_{\Omega}\rho_{1}\{[\psi_{nt}(x, y, \tau)-\psi_{t}^{\ast}(x, y, \tau)]\widehat{\psi}(x, y, \tau) +[\phi_{nt}(x, y, \tau)-\phi_{t}^{\ast}(x, y, \tau)]\widehat{\phi}(x, y, \tau)\}{\rm d}x{\rm d}y{\rm d}\tau\\ &&+\int_{0}^{t}\int_{\Omega}\rho_{2}[\omega_{nt}(x, y, \tau)-\omega_{t}^{\ast}(x, y, \tau)]\widehat{\omega}(x, y, \tau){\rm d}x{\rm d}y{\rm d}\tau\longrightarrow 0, \\ &&\int_{0}^{t}\int_{\Omega}\rho_{1}\{[\psi_{nt}(x, y, \tau)-\psi_{t}^{\ast}(x, y, \tau)]\widehat{\psi}_{t}(x, y, \tau) +[\phi_{nt}(x, y, \tau)-\phi_{t}^{\ast}(x, y, \tau)]\widehat{\phi}_{t}(x, y, \tau)\}{\rm d}x{\rm d}y{\rm d}\tau\\ &&+\int_{0}^{t}\int_{\Omega}\rho_{2}[\omega_{nt}(x, y, \tau)-\omega_{t}^{\ast}(x, y, \tau)]\widehat{\omega}_{t}(x, y, \tau){\rm d}x{\rm d}y{\rm d}\tau\longrightarrow 0, \\ &&\int_{0}^{t}\int_{\Gamma_{1}}[(u_{1n}-u_{1}^{\ast})\widehat{\psi}+(u_{2n}-u_{2}^{\ast})\widehat{\phi} +(u_{3n}-u_{3}^{\ast})\widehat{\omega}]{\rm d}\Gamma_{1}{\rm d}\tau\longrightarrow 0. \end{eqnarray} $

又由(2.4)式得, 对于任意$ t\in[0, T] $, $ \{\Phi_{nt}\} $是$ H $中的有界序列, 因此, 必存在弱收敛的子列, 仍记为$ \{\Phi_{nt}\} $, 使得: $ \Phi_{nt}\stackrel{w}{\longrightarrow}\Phi_{t}\in H $.对于任意$ t\in[0, T] $, 定义如下算子$ \Lambda: H^{1}(0, T; W^{\ast})\rightarrow W^{\ast} $, 其中, $ W^{\ast} $为$ W $的对偶空间. 那么算子$ \Lambda $是连续的, 且对于任意$ X\in W $, 有

(3.6) $ \begin{eqnarray} (\Phi_{t}, X)_{H}& = &\lim\limits_{n\rightarrow \infty}\langle\Lambda\Phi_{nt}, X\rangle_{W^{\ast}\times W} \\ & = &\lim\limits_{n\rightarrow \infty}\langle\Phi_{nt}, \Lambda^{\ast}X\rangle_{H^{1}(0, T;W^{\ast})\times (H^{1}(0, T;W^{\ast}))^{\ast}}\\ & = &\langle\Phi_{t}^{\ast}, \Lambda^{\ast}X\rangle_{H^{1}(0, T;W^{\ast})\times (H^{1}(0, T;W^{\ast}))^{\ast}} \\ & = &\langle\Lambda\Phi_{t}^{\ast}, X\rangle_{W^{\ast}\times W}. \end{eqnarray} $

这里, $ \Lambda^{\ast} $表示$ \Lambda $的伴随算子, 且$ \Lambda^{\ast}: W^{\ast}\longrightarrow(H^{1}(0, T;W^{\ast}))^{\ast} $. 因此, 对于任意$ t\in[0, T] $, 以及$ \forall\widehat{\Phi} = (\widehat{\psi}, \widehat{\phi}, \widehat{\omega})\in C^{1}([0, T]\times\Omega) $, 且在$ \Gamma_{0} $上有$ \widehat{\Phi} = 0 $, 我们有

(3.7) $ \begin{eqnarray} & &\int_{\Omega}\{\rho_{1}[\psi_{nt}(x, y, \tau)\widehat{\psi}(x, y, \tau) +\phi_{nt}(x, y, \tau)\widehat{\phi}(x, y, \tau)] +\rho_{2}\omega_{nt}(x, y, \tau)\widehat{\omega}(x, y, \tau)\}{\rm d}x{\rm d}y \\ &&\longrightarrow\int_{\Omega}\{\rho_{1}[\psi_{t}^{\ast}(x, y, \tau)\widehat{\psi}(x, y, \tau) +\phi_{t}^{\ast}(x, y, \tau)\widehat{\phi}(x, y, \tau)] +\rho_{2}\omega_{t}^{\ast}(x, y, \tau)\widehat{\omega}(x, y, \tau)\}{\rm d}x{\rm d}y.{\qquad} \end{eqnarray} $

因此, 当$ U = U^{\ast} $时, 对(3.5)和(3.7)式, 令$ n\rightarrow \infty $取极限得, $ \Phi^{\ast} $是系统(2.2)的弱解. 从而, 对任意控制$ U = (u_{1}, u_{2}, u_{3})\in L^{2}(0, T; {\cal U}) $到$ (\Phi, \Phi_{t})\in L^{\infty}([0, T];{\cal H}) $的映射是连续的. 又由目标函数的定义可知, 目标函数$ J_{T}({\cal Y}_{0}, U) $是弱下半连续的, 这蕴含着

$ 0\leq J_{T}({\cal Y}_{0}, U^{\ast})\leq\lim\limits_{n\rightarrow \infty}\inf\limits_{U\in L^{2}(0, T;{\cal U})}J_{T}({\cal Y}_{0}, U_{n}) = \theta<\infty. $

所以, $ ({\cal Y}^{\ast}, U^{\ast}) $是最优问题(2.1)满足系统(2.2)的最优解.由于$ J_{T}({\cal Y}_{0}, U) $是严格凸的, 于是最优解是唯一的. 证毕.

定理4.1  考虑如下线性系统

(4.1) $ \begin{eqnarray} \left\{\begin{array}{lll} { } \rho_{1}\psi_{tt}-D(\psi_{xx}+\frac{1-\mu}{2}\psi_{yy} +\frac{1+\mu}{2}\phi_{xy}) +K(\psi+\omega_{x}) = 0, (x, y, t)\in\Omega\times (0, T), \\ { } \rho_{1}\phi_{tt}-D(\phi_{yy}+\frac{1-\mu}{2}\phi_{xx} +\frac{1+\mu}{2}\psi_{xy}) +K(\phi+\omega_{y}) = 0, (x, y, t)\in\Omega\times (0, T), \\ { } \rho_{2}\omega_{tt}-K[(\psi+\omega_{x})_{x}+(\phi+\omega_{y})_{y}] = 0, \quad (x, y, t)\in\Omega\times (0, T), \\ { } \psi = \phi = \omega = 0, \quad (x, y, t)\in\Gamma_{0}\times (0, T), \\ { } D[\nu_{1}\psi_{x}+\mu\nu_{1}\phi_{y}+\frac{1-\mu}{2}(\psi_{y}+\phi_{x})\nu_{2}] = u_{1}, {\quad} (x, y, t)\in\Gamma_{1}\times (0, T), \\ { } D[\nu_{2}\phi_{y}+\mu\nu_{2}\psi_{x}+\frac{1-\mu}{2}(\psi_{y}+\phi_{x})\nu_{1}] = u_{2}, {\quad} (x, y, t)\in\Gamma_{1}\times (0, T), \\ { } K(\frac{\partial\omega}{\partial{\bf \nu}}+\nu_{1}\psi+\nu_{2}\phi) = u_{3}, \quad (x, y, t)\in\Gamma_{1}\times (0, T), \\ (\psi(x, y, 0), \phi(x, y, 0), \omega(x, y, 0)) = (0, 0, 0), {\quad} (x, y)\in\Omega, \\ (\psi_{t}(x, y, 0), \phi_{t}(x, y, 0), \omega_{t}(x, y, 0)) = (0, 0, 0), {\quad} (x, y)\in\Omega \end{array}\right. \end{eqnarray} $

与对偶系统

(4.2) $ \begin{eqnarray} \left\{\begin{array}{lll} { }\rho_{1}p_{tt}-D(p_{xx}+\frac{1-\mu}{2}p_{yy} +\frac{1+\mu}{2}q_{xy}) +K(p+r_{x}) = g_{1}, (x, y, t)\in\Omega\times (0, T), \\ { } \rho_{1}q_{tt}-D(q_{yy}+\frac{1-\mu}{2}q_{xx} +\frac{1+\mu}{2}p_{xy}) +K(q+r_{y}) = g_{2}, (x, y, t)\in\Omega\times (0, T), \\ { } \rho_{2}r_{tt}-K[(p+r_{x})_{x}+(q+r_{y})_{y}] = g_{3}, \quad (x, y, t)\in\Omega\times (0, T), \\ { } \psi = \phi = \omega = 0, \quad (x, y, t)\in\Gamma_{0}\times (0, T), \\ { } D[\nu_{1}p_{x}+\mu\nu_{1}q_{y}+\frac{1-\mu}{2}(p_{y}+q_{x})\nu_{2}] = 0, \quad (x, y, t)\in\Gamma_{1}\times (0, T), \\ { } D[\nu_{2}q_{y}+\mu\nu_{2}p_{x}+\frac{1-\mu}{2}(p_{y}+q_{x})\nu_{1}] = 0, \quad (x, y, t)\in\Gamma_{1}\times (0, T), \\ { } K(\frac{\partial r}{\partial{\bf \nu}}+\nu_{1}p+\nu_{2}q) = 0, \quad (x, y, t)\in\Gamma_{1}\times (0, T), \\ { } (p(x, y, T), q(x, y, T), r(x, y, T)) = (p_{0}(T), q_{0}(T), r_{0}(T)), \quad (x, y)\in\Omega, \\ { } (p_{t}(x, y, T), q_{t}(x, y, T), r_{t}(x, y, T)) = (p_{1}(T), q_{1}(T), r_{1}(T)), {\quad}(x, y)\in\Omega, \end{array}\right. \end{eqnarray} $

其中$ (g_{1}, g_{2}, g_{3})\in L^{2}([0, T];W^{\ast}) $, $ ((p_{0}(T), q_{0}(T), r_{0}(T));(p_{1}(T), q_{1}(T), r_{1}(T)))\in H^{\ast}\times W^{\ast} $, $ H^{\ast}, W^{\ast} $分别表示$ H, W $的对偶空间. 且成立着如下的等式

(4.3) $ \begin{eqnarray} & &\int_{0}^{T}\int_{\Gamma_{1}}(pu_{1}+qu_{2}+ru_{3}){\rm d}\Gamma_{1}{\rm d}t\\ & = &\int_{0}^{T}\langle(\psi, \phi, \omega), (g_{1}, g_{2}, g_{3})\rangle_{W\times W^{\ast}}{\rm d}t +\left((p_{0}(T), q_{0}(T), r_{0}(T)), (\psi_{t}(T), \phi_{t}(T), \omega_{t}(T))\right)_{H}\\ &&-\langle(\rho_{1}\psi(T), \rho_{1}\phi(T), \rho_{2}\omega(T)), (p_{1}(T), q_{1}(T), r_{1}(T))\rangle_{W\times W^{\ast}}. \end{eqnarray} $

证  对于$ (p, q, r)\in C^{0}([0, T];W)\cap C^{1}([0, T];H) $, 并用$ (p, q, r) $分别乘以系统(4.1)前三式, 在$ [0, T]\times\Omega $上积分并相加得

$ \begin{eqnarray*} &&\int_{0}^{T}\int_{\Omega}[\rho_{1}(\psi_{tt}p+\phi_{tt}q)+\rho_{2}\omega_{tt}r]{\rm d}x{\rm d}y{\rm d}t\\ & = &\int_{0}^{T}\int_{\Omega}[L_{1}\{\psi, \phi, \omega\}p+L_{2}\{\psi, \phi, \omega\}q+L_{3}\{\psi, \phi, \omega\}r]{\rm d}x{\rm d}y{\rm d}t. \end{eqnarray*} $

分部积分得

$ \begin{eqnarray*} &&\int_{\Omega}\left[\rho_{1}((\psi_{t}p-\psi p_{t})+(\phi_{t}q-\phi q_{t})) +\rho_{2}(\omega_{t}r-\omega r_{t})\right]_{0}^{T}{\rm d}x{\rm d}y\\ &&+\int_{0}^{T}\int_{\Omega}[\rho_{1}(\psi p_{tt}+\phi q_{tt})+\rho_{2}\omega r_{tt}]{\rm d}x{\rm d}y{\rm d}t +\int_{0}^{T}a(\psi, \phi, \omega; p, q, r){\rm d}t\\ & = &\int_{0}^{T}\int_{\Omega}[L_{1}\{\psi, \phi, \omega\}p+L_{2}\{\psi, \phi, \omega\}q+L_{3}\{\psi, \phi, \omega\}r]{\rm d}x{\rm d}y{\rm d}t, \end{eqnarray*} $

即

$ \begin{eqnarray*} &&\int_{\Omega}[\rho_{1}((\psi_{t}(T)p_{0}(T)-\psi(T) p_{1}(T))+(\phi_{t}(T)q_{0}(T)-\phi(T) q_{1}(T))\\ &&+\rho_{2}(\omega_{t}(T)r_{0}(T)-\omega(T) r_{1}(T))]{\rm d}x{\rm d}y\\ &&+\int_{0}^{T}\int_{\Omega}[\rho_{1}(\psi p_{tt}+\phi q_{tt})+\rho_{2}\omega r_{tt}]{\rm d}x{\rm d}y{\rm d}t +\int_{0}^{T}a(\psi, \phi, \omega; p, q, r){\rm d}t\\ & = &\int_{0}^{T}\int_{\Gamma_{1}}[u_{1}p+u_{2}q+u_{3}r]{\rm d}\Gamma_{1}{\rm d}t. \end{eqnarray*} $

进一步有

$ \begin{eqnarray*} &&\int_{\Omega}[\rho_{1}((\psi_{t}(T)p_{0}(T)-\psi(T) p_{1}(T))+(\phi_{t}(T)q_{0}(T)-\phi(T) q_{1}(T))\\ &&+\rho_{2}(\omega_{t}(T)r_{0}(T)-\omega(T) r_{1}(T))]{\rm d}x{\rm d}y+\int_{0}^{T}\int_{\Omega}[\rho_{1}(\psi p_{tt}+\phi q_{tt})+\rho_{2}\omega r_{tt}]{\rm d}x{\rm d}y{\rm d}t\\ &&+\int_{0}^{T}\int_{\Gamma_{1}}(\psi F_{1}\{p, q\}+\phi F_{2}\{p, q\}+\omega F_{3}\{p, q, r\}){\rm d}\Gamma_{1}{\rm d}t\\ & = &\int_{0}^{T}\bigg\{\int_{\Omega}[\psi L_{1}\{p, q, r\}+\phi L_{2}\{p, q, r\}+\omega L_{3}\{p, q, r\}]{\rm d}x{\rm d}y+\int_{\Gamma_{1}}[u_{1}p+u_{2}q+u_{3}r]{\rm d}\Gamma_{1}\bigg\}{\rm d}t. \end{eqnarray*} $

因此有

$ \begin{eqnarray*} &&\int_{0}^{T}\int_{\Gamma_{1}}(pu_{1}+qu_{2}+ru_{3}){\rm d}\Gamma_{1}{\rm d}t\\ & = &\int_{0}^{T}\langle(\psi, \phi, \omega), (g_{1}, g_{2}, g_{3})\rangle_{W\times W^{\ast}}{\rm d}t +\left((p_{0}(T), q_{0}(T), r_{0}(T)), (\psi_{t}(T), \phi_{t}(T), \omega_{t}(T))\right)_{H}\\ &&-\langle(\rho_{1}\psi(T), \rho_{1}\phi(T), \rho_{2}\omega(T)), (p_{1}(T), q_{1}(T), r_{1}(T))\rangle_{W\times W^{\ast}}. \end{eqnarray*} $

由能量估计式及线性双曲方程关于时间的可逆性知系统(4.2)的解$ (p, q, r)\in C^{0}([0, T];H^{\ast})\cap C^{1}([0, T];W^{\ast}) $, 而$ W, H, L^{2}([0, T];H) $分别在$ H^{\ast}, W^{\ast}, L^{2}([0, T];W^{\ast}) $中稠密, 由稠密性理论得

$ (p, q, r)\in C^{0}([0, T];H)\cap C^{1}([0, T];W). $

证毕.

定理4.2   若$ (\overline{\Phi};\overline{U}) = ((\overline{\psi}, \overline{\phi}, \overline{\omega});(\overline{u}_{1}, \overline{u}_{2}, \overline{u}_{3})) $是最优控制问题(OCP)的最优解, 则满足以下最优性条件

(4.4) $ \begin{eqnarray} \left\{\begin{array}{lll} { } \rho_{1}\overline{\psi}_{tt}-D(\overline{\psi}_{xx}+\frac{1-\mu}{2}\overline{\psi}_{yy} +\frac{1+\mu}{2}\overline{\phi}_{xy}) +K(\overline{\psi}+\overline{\omega}_{x}) = 0, (x, y, t)\in\Omega\times (0, T), \\ { } \rho_{1}\overline{\phi}_{tt}-D(\overline{\phi}_{yy}+\frac{1-\mu}{2}\overline{\phi}_{xx} +\frac{1+\mu}{2}\overline{\psi}_{xy}) +K(\overline{\phi}+\overline{\omega}_{y}) = 0, (x, y, t)\in\Omega\times (0, T), \\ { } \rho_{2}\overline{\omega}_{tt}-K[(\overline{\psi}+\overline{\omega}_{x})_{x}+(\overline{\phi}+\overline{\omega}_{y})_{y}] = 0, {\quad} (x, y, t)\in\Omega\times (0, T), \\ { } \overline{\psi} = \overline{\phi} = \overline{\omega} = 0, \quad (x, y, t)\in\Gamma_{0}\times (0, T), \\ { } D[\nu_{1}\overline{\psi}_{x}+\mu\nu_{1}\overline{\phi}_{y}+\frac{1-\mu}{2}(\overline{\psi}_{y}+\overline{\phi}_{x})\nu_{2}] = \overline{u}_{1}, {\quad} (x, y, t)\in\Gamma_{1}\times (0, T), \\ { } D[\nu_{2}\overline{\phi}_{y}+\mu\nu_{2}\overline{\psi}_{x}+\frac{1-\mu}{2}(\overline{\psi}_{y}+\overline{\phi}_{x})\nu_{1}] = \overline{u}_{2}, {\quad}(x, y, t)\in\Gamma_{1}\times (0, T), \\ { } K(\frac{\partial\overline{\omega}}{\partial{\bf \nu}}+\nu_{1}\overline{\psi}+\nu_{2}\overline{\phi}) = \overline{u}_{3}, {\quad}(x, y, t)\in\Gamma_{1}\times (0, T), \\ { } (\overline{\psi}(x, y, 0), \overline{\phi}(x, y, 0), \overline{\omega}(x, y, 0)) = (\psi_{01}, \phi_{01}, \omega_{01}), {\quad} (x, y)\in\Omega, \\ (\overline{\psi}_{t}(x, y, 0), \overline{\phi}_{t}(x, y, 0), \overline{\omega}_{t}(x, y, 0)) = (\psi_{02}, \phi_{02}, \omega_{02}), {\quad}(x, y)\in\Omega \end{array}\right. \end{eqnarray} $

和

(4.5) $ \begin{eqnarray} \left\{\begin{array}{lll} { } \rho_{1}\overline{p}_{tt}-D(\overline{p}_{xx}+\frac{1-\mu}{2}\overline{p}_{yy} +\frac{1+\mu}{2}\overline{q}_{xy}) +K(\overline{p}+\overline{r}_{x})\\ { } = -[\rho_{1}\overline{\psi}_{tt}+D(\overline{\psi}_{xx}+\frac{1-\mu}{2}\overline{\psi}_{yy} +\frac{1+\mu}{2}\overline{\phi}_{xy}) -K(\overline{\psi}+\overline{\omega}_{x})], (x, y, t)\in\Omega\times (0, T), \\ { } \rho_{1}\overline{q}_{tt}-D(\overline{q}_{yy}+\frac{1-\mu}{2}\overline{q}_{xx} +\frac{1+\mu}{2}\overline{p}_{xy}) +K(\overline{q}+\overline{r}_{y})\\ { } = -[\rho_{1}\overline{\phi}_{tt}+D(\overline{\phi}_{yy}+\frac{1-\mu}{2}\overline{\phi}_{xx} +\frac{1+\mu}{2}\overline{\psi}_{xy}) -K(\overline{\phi}+\overline{\omega}_{y})], (x, y, t)\in\Omega\times (0, T), \\ { } \rho_{2}\overline{r}_{tt}-K[(\overline{p}+\overline{r}_{x})_{x}+(\overline{q}+\overline{r}_{y})_{y}]\\ { } = -(\rho_{2}\overline{\omega}_{tt}+K[(\overline{\psi}+\overline{\omega}_{x})_{x} +(\overline{\phi}+\overline{\omega}_{y})_{y}]), {\quad} (x, y, t)\in\Omega\times (0, T), \\ { } \overline{p} = \overline{q} = \overline{r} = 0, \quad (x, y, t)\in\Gamma_{0}\times (0, T), \\ { } D[\nu_{1}\overline{p}_{x}+\mu\nu_{1}\overline{q}_{y}+\frac{1-\mu}{2}(\overline{p}_{y}+\overline{q}_{x})\nu_{2}] = 0, \quad (x, y, t)\in\Gamma_{1}\times (0, T), \\ { } D[\nu_{2}\overline{q}_{y}+\mu\nu_{2}\overline{p}_{x}+\frac{1-\mu}{2}(\overline{p}_{y}+\overline{q}_{x})\nu_{1}] = 0, \quad (x, y, t)\in\Gamma_{1}\times (0, T), \\ { } K(\frac{\partial\overline{r}}{\partial{\bf \nu}}+\nu_{1}\overline{p}+\nu_{2}\overline{q}) = 0, \quad (x, y, t)\in\Gamma_{1}\times (0, T), \\ { } (\overline{p}(x, y, T), \overline{q}(x, y, T), \overline{\omega}(x, y, T)) = (0, 0, 0), \quad (x, y)\in\Omega, \\ { } (\overline{p}_{t}(x, y, T), \overline{q}_{t}(x, y, T), \overline{r}_{t}(x, y, T)) = -(\overline{\psi}_{t}(T), \overline{\phi}_{t}(T), \overline{\omega}_{t}(T)), {\quad} (x, y)\in\Omega. \end{array}\right. \end{eqnarray} $

其中$ (\overline{p}, \overline{q}, \overline{r})\in C^{0}([0, T];H^{\ast})\cap C^{1}([0, T];W^{\ast}) $是对偶系统(4.2)的解, 且在$ [0, T]\times\Gamma_{1} $上有

(4.6) $ \begin{equation} (\overline{p}, \overline{q}, \overline{r}) = -\beta(\overline{u}_{1}, \overline{u}_{2}, \overline{u}_{3}). \end{equation} $

证  由目标函数的表达式得

(4.7) $ \begin{eqnarray} J_{T}({\cal Y}_{0}, U) & = &\int_{0}^{T}(\frac{1}{2}\|{\cal Y}(t)\|_{{\cal H}}^{2}+\frac{\beta}{2}\|U(t)\|_{{\cal U}}^{2}){\rm d}t\\ & = &\frac{1}{2}\int_{0}^{T}a(\psi, \phi, \omega;\psi, \phi, \omega){\rm d}t +\frac{1}{2}\int_{0}^{T}\int_{\Omega}[\rho_{1}(|\psi_{t}|^{2}+|\phi_{t}|^{2})+\rho_{2}|\omega_{t}|^{2}]{\rm d}x{\rm d}y{\rm d}t\\ &&+\frac{\beta}{2}\int_{0}^{T}\int_{\Gamma_{1}}(|u_{1}|^{2}+|u_{2}|^{2}+|u_{3}|^{2}){\rm d}\Gamma_{1}{\rm d}t. \end{eqnarray} $

对于任意的$ (\delta u_{1}, \delta u_{2}, \delta u_{3})\in L^{2}([0, T];{\cal U}) $, 对目标函数作变分得

$ \begin{eqnarray*} &&\frac{\partial J_{T}({\cal Y}_{0}, U)}{\partial u_{1}}\delta u_{1} +\frac{\partial J_{T}({\cal Y}_{0}, U)}{\partial u_{2}}\delta u_{2}+\frac{\partial J_{T}({\cal Y}_{0}, U)}{\partial u_{3}}\delta u_{3}\\ & = &\int_{0}^{T}a(\psi, \phi, \omega;\delta\psi, \delta\phi, \delta\omega){\rm d}t +\int_{0}^{T}\int_{\Omega}[\rho_{1}(\psi_{t}\delta\psi_{t}+\phi_{t}\delta\phi_{t})+\rho_{2}\omega_{t}\delta\omega_{t}]{\rm d}x{\rm d}y{\rm d}t\\ &&+\beta\int_{0}^{T}\int_{\Gamma_{1}}(u_{1}\delta u_{1}+u_{2}\delta u_{2}+u_{3}\delta u_{3}){\rm d}\Gamma_{1}{\rm d}t. \end{eqnarray*} $

分部积分得

$ \begin{eqnarray*} & &\frac{\partial J_{T}({\cal Y}_{0}, U)}{\partial u_{1}}\delta u_{1} +\frac{\partial J_{T}({\cal Y}_{0}, U)}{\partial u_{2}}\delta u_{2}+\frac{\partial J_{T}({\cal Y}_{0}, U)}{\partial u_{3}}\delta u_{3}\\ & = &-\int_{0}^{T}\int_{\Omega}[\delta\psi(\rho_{1}\psi_{tt}+L_{1}\{\psi, \phi, \omega\})+\delta\phi(\rho_{1}\phi_{tt}+L_{2}\{\psi, \phi, \omega\}) +\delta\omega(\rho_{2}\omega_{tt}+L_{3}\{\psi, \phi, \omega\})]{\rm d}t\\ &&+\int_{\Omega}[\rho_{1}(\delta\psi(T)\psi_{t}(T)+\delta\phi(T)\phi_{t}(T))+\rho_{2}\delta\omega(T)\omega_{t}(T)]{\rm d}x{\rm d}y\\ &&+\beta\int_{0}^{T}\int_{\Gamma_{1}}(u_{1}\delta u_{1}+u_{2}\delta u_{2}+u_{3}\delta u_{3}){\rm d}\Gamma_{1}{\rm d}t, \end{eqnarray*} $

其中$ (\delta\psi, \delta\phi, \delta\omega)\in C^{0}([0, T];W)\cap C^{1}([0, T];H) $是以下系统的弱解

(4.8) $ \begin{eqnarray} \left\{\begin{array}{ll} { } \rho_{1}\delta\psi_{tt}-D((\delta\psi)_{xx}+\frac{1-\mu}{2}(\delta\psi)_{yy} +\frac{1+\mu}{2}(\delta\phi)_{xy}) +K(\delta\psi+(\delta\omega)_{x}) = 0, \\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; (x, y, t)\in\Omega\times (0, T), \\ { } \rho_{1}\delta\phi_{tt}-D((\delta\phi)_{yy}+\frac{1-\mu}{2}(\delta\phi)_{xx} +\frac{1+\mu}{2}(\delta\psi)_{xy}) +K(\delta\phi+(\delta\omega)_{y}) = 0, \\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; (x, y, t)\in\Omega\times (0, T), \\ { } \rho_{2}\delta\omega_{tt}-K[(\delta\psi+(\delta\omega)_{x})_{x}+(\delta\phi+(\delta\omega)_{y})_{y}] = 0, \quad(x, y, t)\in\Omega\times (0, T), \\ { } \delta\psi = \delta\phi = \delta\omega = 0, \quad (x, y, t)\in\Gamma_{0}\times (0, T), \\ { } D[\nu_{1}(\delta\psi)_{x}+\mu\nu_{1}(\delta\phi)_{y}+\frac{1-\mu}{2}((\delta\psi)_{y}+(\delta\phi)_{x})\nu_{2}] = \delta u_{1}, (x, y, t)\in\Gamma_{1}\times (0, T), \\ { } D[\nu_{2}(\delta\phi)_{y}+\mu\nu_{2}(\delta\psi)_{x}+\frac{1-\mu}{2}((\delta\psi)_{y}+(\delta\phi)_{x})\nu_{1}] = \delta u_{2}, (x, y, t)\in\Gamma_{1}\times (0, T), \\ { } K(\frac{\partial(\delta\omega)}{\partial{\bf \nu}}+\nu_{1}\delta\psi+\nu_{2}\delta\phi) = \delta u_{3}, \quad (x, y, t)\in\Gamma_{1}\times (0, T), \\ { } (\delta\psi(x, y, 0), \delta\phi(x, y, 0), \delta\omega(x, y, 0)) = (0, 0, 0), \quad (x, y)\in\Omega, \\ { } (\delta\psi_{t}(x, y, 0), \delta\phi_{t}(x, y, 0), \delta\omega_{t}(x, y, 0)) = (0, 0, 0), \quad (x, y)\in\Omega. \end{array}\right. \end{eqnarray} $

则一阶最优性条件等价于

(4.9) $ \begin{eqnarray} & &-\int_{0}^{T}\int_{\Omega}[\delta\psi(\rho_{1}\psi_{tt}+L_{1}\{\psi, \phi, \omega\})+\delta\phi(\rho_{1}\phi_{tt}+L_{2}\{\psi, \phi, \omega\}) +\delta\omega(\rho_{2}\omega_{tt}+L_{3}\{\psi, \phi, \omega\})]{\rm d}t\\ &&+\int_{\Omega}[\rho_{1}(\delta\psi(T)\psi_{t}(T)+\delta\phi(T)\phi_{t}(T))+\rho_{2}\delta\omega(T)\omega_{t}(T)]{\rm d}x{\rm d}y\\ &&+\beta\int_{0}^{T}\int_{\Gamma_{1}}(u_{1}\delta u_{1}+u_{2}\delta u_{2}+u_{3}\delta u_{3}){\rm d}\Gamma_{1}{\rm d}t = 0. \end{eqnarray} $

另一方面, 由定理4.1得

(4.10) $ \begin{eqnarray} & &\int_{0}^{T}\int_{\Gamma_{1}}(p\delta u_{1}+q\delta u_{2}+r\delta u_{3}){\rm d}\Gamma_{1}{\rm d}t\\ & = &\int_{0}^{T}\langle(\delta\psi, \delta\phi, \delta\omega), (g_{1}, g_{2}, g_{3})\rangle_{W\times W^{\ast}}{\rm d}t +\left((p_{0}(T), q_{0}(T), r_{0}(T)), (\delta\psi_{t}(T), \delta\phi_{t}(T), \delta\omega_{t}(T))\right)_{H}\\ &&-\langle(\rho_{1}\delta\psi(T), \rho_{1}\delta\phi(T), \rho_{2}\delta\omega(T)), (p_{1}(T), q_{1}(T), r_{1}(T))\rangle_{W\times W^{\ast}}. \end{eqnarray} $

比较(4.9)式和(4.10)式, 并由$ (\delta u_{1}, \delta u_{2}, \delta u_{3})\in L^{2}([0, T];{\cal U}) $的任意性, 我们可以得到

$ g_{1} = -[\rho_{1}\psi_{tt}+D(\psi_{xx}+\frac{1-\mu}{2}\psi_{yy} +\frac{1+\mu}{2}\phi_{xy})-K(\psi+\omega_{x})], $

$ g_{2} = -[\rho_{1}\phi_{tt}+D(\phi_{yy}+\frac{1-\mu}{2}\phi_{xx} +\frac{1+\mu}{2}\psi_{xy})-K(\phi+\omega_{y})], $

$ g_{3} = -[\rho_{2}\omega_{tt}+K((\psi+\omega_{x})_{x}+(\phi+\omega_{y})_{y})] $

$ p_{0}(T) = q_{0}(T) = r_{0}(T) = 0, $

$ p_{1}(T) = -\psi_{t}(T), q_{1}(T) = -\phi_{t}(T), r_{1}(T) = -\omega_{t}(T), $

$ (p, q, r) = -\beta(u_{1}, u_{2}, u_{3}), (x, y, t)\in[0, T]\times\Gamma_{1}. $

即由变分原理所得到的最优解满足(4.4)–(4.6)式. 证毕.

本节我们主要讨论Mindlin-Timoshenko板系统的能观性与系统能量指数衰减之间的关系. 首先我们采用文献[21]的方法证明如下的结果.

引理5.1  假定存在$ (x_{0}, y_{0})\in {{\Bbb R}} ^{2} $, 满足$ \widetilde{\Gamma} = \{(x, y)\in \Gamma_{1}|(x-x_{0}, y-y_{0})\cdot\nu>0\} $非空, 则存在$ T_{1}>0 $, 使得对任意$ T\geq T_{1} $有

(5.1) $ \begin{eqnarray} C_{1}\|{\cal Y}_{0}\|_{{\cal H}}^{2}\leq\int_{0}^{T}\int_{\widetilde{\Gamma}} (|\widetilde{\psi}_{t}|^{2}+|\widetilde{\phi}_{t}|^{2}+|\widetilde{\omega}_{t}|^{2}){\rm d}\widetilde{\Gamma}{\rm d}t, \end{eqnarray} $

存在$ T_{2}>0 $, 使得对任意$ T\geq T_{2} $有

(5.2) $ \begin{eqnarray} C_{2}\|{\cal Y}_{0}\|_{{\cal H}}^{2}\leq\int_{0}^{T}\int_{\Omega} (|\widetilde{\psi}_{t}|^{2}+|\widetilde{\phi}_{t}|^{2}+|\widetilde{\omega}_{t}|^{2}){\rm d}x{\rm d}y{\rm d}t, \end{eqnarray} $

其中, $ C_{1}, C_{2} $均为仅与$ T $有关的正常数, $ \widetilde{\Phi} = (\widetilde{\psi}, \widetilde{\phi}, \widetilde{\omega}) $是如下齐次系统的弱解, 且$ (\widetilde{\Phi}, \widetilde{\Phi}_{t})\in C^{0}([0, T];{\cal H}) $, 有

(5.3) $ \begin{eqnarray} \left\{\begin{array}{lll} { } \rho_{1}\widetilde{\psi}_{tt}-D(\widetilde{\psi}_{xx}+\frac{1-\mu}{2}\widetilde{\psi}_{yy} +\frac{1+\mu}{2}\widetilde{\phi}_{xy}) +K(\widetilde{\psi}+\widetilde{\omega}_{x}) = 0, \quad (x, y, t)\in\Omega\times (0, T), \\ { } \rho_{1}\widetilde{\phi}_{tt}-D(\widetilde{\phi}_{yy}+\frac{1-\mu}{2}\widetilde{\phi}_{xx} +\frac{1+\mu}{2}\widetilde{\psi}_{xy}) +K(\widetilde{\phi}+\widetilde{\omega}_{y}) = 0, \quad (x, y, t)\in\Omega\times (0, T), \\ { } \rho_{2}\widetilde{\omega}_{tt}-K[(\widetilde{\psi}+\widetilde{\omega}_{x})_{x}+(\widetilde{\phi}+\widetilde{\omega}_{y})_{y}] = 0, \quad (x, y, t)\in\Omega\times (0, T), \\ { } \widetilde{\psi} = \widetilde{\phi} = \widetilde{\omega} = 0, \quad (x, y, t)\in\Gamma_{0}\times (0, T), \\ { } D[\nu_{1}\widetilde{\psi}_{x}+\mu\nu_{1}\widetilde{\phi}_{y}+\frac{1-\mu}{2}(\widetilde{\psi}_{y}+\widetilde{\phi}_{x})\nu_{2}] = 0, \quad (x, y, t)\in\Gamma_{1}\times (0, T), \\ { } D[\nu_{2}\widetilde{\phi}_{y}+\mu\nu_{2}\widetilde{\psi}_{x}+\frac{1-\mu}{2}(\widetilde{\psi}_{y}+\widetilde{\phi}_{x})\nu_{1}] = 0, \quad (x, y, t)\in\Gamma_{1}\times (0, T), \\ { } K(\frac{\partial\widetilde{\omega}}{\partial{\bf \nu}}+\nu_{1}\widetilde{\psi}+\nu_{2}\widetilde{\phi}) = 0, \quad (x, y, t)\in\Gamma_{1}\times (0, T), \\ (\widetilde{\psi}(x, y, 0), \widetilde{\phi}(x, y, 0), \widetilde{\omega}(x, y, 0)) = (\psi_{01}, \phi_{01}, \omega_{01}), \quad (x, y)\in\Omega, \\ (\widetilde{\psi}_{t}(x, y, 0), \widetilde{\phi}_{t}(x, y, 0), \widetilde{\omega}_{t}(x, y, 0)) = (\psi_{02}, \phi_{02}, \omega_{02}), \quad (x, y)\in\Omega. \end{array}\right. \end{eqnarray} $

证  记$ F(x, y) = \eta(x, y)((x-x_{0}), (y-y_{0})) $, 其中, 乘子$ \eta(\cdot)\in C^{1}(\overline{\Omega}) $且满足在$ \overline{\Gamma}_{0} $上, $ \eta = 0 $; 在$ \Gamma_{1} $上, $ \eta = 1 $. 对边界$ \Gamma_{1} $作如下分割

$ \widetilde{\Gamma}_{1} = \{(x, y)\in \Gamma_{1}|F(x, y)\cdot\nu>0\}, $

$ \widehat{\Gamma}_{1} = \Gamma_{1}\backslash \widetilde{\Gamma}_{1} = \{(x, y)\in \Gamma_{1}|F(x, y)\cdot\nu\leq0\}. $

记$ M^{2} = \max\limits_{(x, y)\in\overline{\Omega}}|F(x, y)|^{2} = \max\limits_{(x, y)\in\overline{\Omega}}\{|x-x_{0}|^{2}+|y-y_{0}|^{2}\}, \quad 2M_{1} = \max\{\rho_{1}, \rho_{2}\}. $将$ F\cdot\nabla\widetilde{\psi}, F\cdot\nabla\widetilde{\phi}, F\cdot\nabla\widetilde{\omega} $分别乘以系统(5.3)前三式, 并在$ [0, T]\times\Omega $上积分, 则有

$ \begin{eqnarray*} &&\int_{0}^{T}\int_{\Omega}[\rho_{1}(\widetilde{\psi}_{tt}F\cdot\nabla\widetilde{\psi} +\widetilde{\phi}_{tt}F\cdot\nabla\widetilde{\phi}) +\rho_{2}\widetilde{\omega}_{tt}F\cdot\nabla\widetilde{\omega}]{\rm d}x{\rm d}y{\rm d}t\\ & = &\int_{0}^{T}\int_{\Omega}[L_{1}\{\widetilde{\psi}, \widetilde{\phi}, \widetilde{\omega}\}F\cdot\nabla\widetilde{\psi}+ L_{2}\{\widetilde{\psi}, \widetilde{\phi}, \widetilde{\omega}\}F\cdot\nabla\widetilde{\phi}) +L_{3}\{\widetilde{\psi}, \widetilde{\phi}, \widetilde{\omega}\}F\cdot\nabla\widetilde{\omega}]{\rm d}x{\rm d}y{\rm d}t. \end{eqnarray*} $

分部积分得

$ \begin{eqnarray*} &&\int_{0}^{T}\int_{\Gamma_{1}}(F\cdot\nu)[\rho_{1}(|\widetilde{\psi}_{t}|^{2}+|\widetilde{\phi}_{t}|^{2}) +\rho_{2}|\widetilde{\omega}_{t}|^{2}]{\rm d}\Gamma_{1}{\rm d}t\\ & = &2\int_{\Omega}[\rho_{1}(\widetilde{\psi}_{t}F\cdot\nabla\widetilde{\psi}+\widetilde{\phi}_{t}F\cdot\nabla\widetilde{\phi})+ \rho_{2}\widetilde{\omega}_{t}F\cdot\nabla\widetilde{\omega}]_{0}^{T}{\rm d}x{\rm d}y\\ &&+\int_{0}^{T}\int_{\Omega}{\rm div}F[\rho_{1}(|\widetilde{\psi}_{t}|^{2} +|\widetilde{\phi}_{t}|^{2})+\rho_{2}|\widetilde{\omega}_{t}|^{2}]{\rm d}x{\rm d}y{\rm d}t+2\int_{0}^{T}a(\widetilde{\psi}, \widetilde{\phi}, \widetilde{\omega};\widetilde{\psi}, \widetilde{\phi}, \widetilde{\omega}){\rm d}t\\ &&+D\int_{0}^{T}\int_{\Gamma_{1}}(F\cdot\nu)[|\widetilde{\psi}_{x}|^{2}+|\widetilde{\phi}_{y}|^{2} +2\mu\widetilde{\psi}_{x}\widetilde{\phi}_{y}+\frac{1-\mu}{2}|\widetilde{\psi}_{y}+\widetilde{\phi}_{x}|^{2}]{\rm d}\Gamma_{1}{\rm d}t\\ &&+K\int_{0}^{T}\int_{\Gamma_{1}}(F\cdot\nu)[|\widetilde{\psi}+\widetilde{\omega}_{x}|^{2}+|\widetilde{\phi}+\widetilde{\omega}_{y}|^{2}]{\rm d}\Gamma_{1}{\rm d}t\\ &&-D\int_{0}^{T}\int_{\Omega}{\rm div}F[|\widetilde{\psi}_{x}|^{2}+|\widetilde{\phi}_{y}|^{2} +2\mu\widetilde{\psi}_{x}\widetilde{\phi}_{y}+\frac{1-\mu}{2}|\widetilde{\psi}_{y}+\widetilde{\phi}_{x}|^{2}]{\rm d}x{\rm d}y{\rm d}t\\ &&-K\int_{0}^{T}\int_{\Omega}{\rm div}F[|\widetilde{\psi}+\widetilde{\omega}_{x}|^{2}+|\widetilde{\phi}+\widetilde{\omega}_{y}|^{2}]{\rm d}x{\rm d}y{\rm d}t, \end{eqnarray*} $

即

(5.4) $ \begin{eqnarray} &&\int_{0}^{T}\int_{\Gamma_{1}}(F\cdot\nu)[\rho_{1}(|\widetilde{\psi}_{t}|^{2}+|\widetilde{\phi}_{t}|^{2}) +\rho_{2}|\widetilde{\omega}_{t}|^{2}]{\rm d}\Gamma_{1}{\rm d}t\\ & = &2\int_{\Omega}[\rho_{1}(\widetilde{\psi}_{t}F\cdot\nabla\widetilde{\psi}+\widetilde{\phi}_{t}F\cdot\nabla\widetilde{\phi})+ \rho_{2}\widetilde{\omega}_{t}F\cdot\nabla\widetilde{\omega}]_{0}^{T}{\rm d}x{\rm d}y\\ &&+2\int_{0}^{T}\int_{\Omega}[\rho_{1}(|\widetilde{\psi}_{t}|^{2} +|\widetilde{\phi}_{t}|^{2})+\rho_{2}|\widetilde{\omega}_{t}|^{2}]{\rm d}x{\rm d}y{\rm d}t +2\int_{0}^{T}a(\widetilde{\psi}, \widetilde{\phi}, \widetilde{\omega};\widetilde{\psi}, \widetilde{\phi}, \widetilde{\omega}){\rm d}t\\ &&-2D\int_{0}^{T}\int_{\Omega}[|\widetilde{\psi}_{x}|^{2}+|\widetilde{\phi}_{y}|^{2} +2\mu\widetilde{\psi}_{x}\widetilde{\phi}_{y}+\frac{1-\mu}{2}|\widetilde{\psi}_{y}+\widetilde{\phi}_{x}|^{2}]{\rm d}x{\rm d}y{\rm d}t\\ &&-2K\int_{0}^{T}\int_{\Omega}[|\widetilde{\psi}+\widetilde{\omega}_{x}|^{2}+|\widetilde{\phi}+\widetilde{\omega}_{y}|^{2}]{\rm d}x{\rm d}y{\rm d}t. \end{eqnarray} $

由Poincaré不等式得

(5.5) $ \begin{eqnarray} &&\left|\int_{\Omega}[\rho_{1}(\widetilde{\psi}_{t}F\cdot\nabla\widetilde{\psi}+\widetilde{\phi}_{t}F\cdot\nabla\widetilde{\phi})+ \rho_{2}\widetilde{\omega}_{t}F\cdot\nabla\widetilde{\omega}]_{0}^{T}{\rm d}x{\rm d}y\right|\\ &\leq &c_{1}\left\{\int_{\Omega}[\rho_{1}(|\widetilde{\psi}_{t}|^{2} +|\widetilde{\phi}_{t}|^{2})+\rho_{2}|\widetilde{\omega}_{t}|^{2}]_{0}^{T}{\rm d}x{\rm d}y +a(\widetilde{\psi}, \widetilde{\phi}, \widetilde{\omega};\widetilde{\psi}, \widetilde{\phi}, \widetilde{\omega})|_{0}^{T}\right\}\\ &\leq &2c_{1}I(0). \end{eqnarray} $

将$ \widetilde{\psi}, \widetilde{\phi}, \widetilde{\omega} $分别乘以系统(4.3)前三式, 并在$ [0, T]\times\Omega $上积分并相加, 得

$ \begin{eqnarray*} &&\int_{0}^{T}\int_{\Omega}[\rho_{1}(\widetilde{\psi}_{tt}\widetilde{\psi}+\widetilde{\phi}_{tt}\widetilde{\phi}) +\rho_{2}\widetilde{\omega}_{tt}\widetilde{\omega}]{\rm d}x{\rm d}y{\rm d}t\\ & = &\int_{0}^{T}\int_{\Omega}[L_{1}\{\widetilde{\psi}, \widetilde{\phi}, \widetilde{\omega}\}\widetilde{\psi}+ L_{2}\{\widetilde{\psi}, \widetilde{\phi}, \widetilde{\omega}\}\widetilde{\phi}+L_{3}\{\widetilde{\psi}, \widetilde{\phi}, \widetilde{\omega}\}\widetilde{\omega}]{\rm d}x{\rm d}y{\rm d}t, \end{eqnarray*} $

分部积分得

(5.6) $ \begin{eqnarray} &&\int_{0}^{T}\int_{\Omega}[\rho_{1}(|\widetilde{\psi}_{t}|^{2} +|\widetilde{\phi}_{t}|^{2})+\rho_{2}|\widetilde{\omega}_{t}|^{2}]{\rm d}x{\rm d}y{\rm d}t -\int_{0}^{T}a(\widetilde{\psi}, \widetilde{\phi}, \widetilde{\omega};\widetilde{\psi}, \widetilde{\phi}, \widetilde{\omega}){\rm d}t\\ & = &\int_{\Omega}[\rho_{1}(\widetilde{\psi}_{t}\widetilde{\psi}+\widetilde{\phi}_{t}\widetilde{\phi}) +\rho_{2}(\widetilde{\omega}_{t}\widetilde{\omega}]_{0}^{T}{\rm d}x{\rm d}y. \end{eqnarray} $

再由Poincaré不等式得

(5.7) $ \begin{eqnarray} \left|\int_{0}^{T}\int_{\Omega}[\rho_{1}(|\widetilde{\psi}_{t}|^{2} +|\widetilde{\phi}_{t}|^{2})+\rho_{2}|\widetilde{\omega}_{t}|^{2}]{\rm d}x{\rm d}y{\rm d}t -\int_{0}^{T}a(\widetilde{\psi}, \widetilde{\phi}, \widetilde{\omega};\widetilde{\psi}, \widetilde{\phi}, \widetilde{\omega}){\rm d}t\right|\leq c_{0}I(0). \end{eqnarray} $

由(5.4), (5.5)和(5.7)式得

(5.8) $ \begin{eqnarray} MM_{1}\int_{0}^{T}\int_{\widetilde{\Gamma}}[|\widetilde{\psi}_{t}|^{2} +|\widetilde{\phi}_{t}|^{2}+|\widetilde{\omega}_{t}|^{2}]{\rm d}\widetilde{\Gamma}{\rm d}t &\geq&\frac{1}{2}\int_{0}^{T}\int_{\Gamma_{1}}(F\cdot\nu)[\rho_{1}(|\widetilde{\psi}_{t}|^{2} +|\widetilde{\phi}_{t}|^{2})+\rho_{2}|\widetilde{\omega}_{t}|^{2}]{\rm d}\Gamma_{1}{\rm d}t\\ &\geq&\int_{0}^{T}a(\widetilde{\psi}, \widetilde{\phi}, \widetilde{\omega}; \widetilde{\psi}, \widetilde{\phi}, \widetilde{\omega}){\rm d}t-2c_{1}I(0)-c_{0}I(0). \end{eqnarray} $

又由(5.8)式得

(5.9) $ \begin{eqnarray} \int_{0}^{T}a(\widetilde{\psi}, \widetilde{\phi}, \widetilde{\omega};\widetilde{\psi}, \widetilde{\phi}, \widetilde{\omega}){\rm d}t\geq \int_{0}^{T}\int_{\Omega}[\rho_{1}(|\widetilde{\psi}_{t}|^{2} +|\widetilde{\phi}_{t}|^{2})+\rho_{2}|\widetilde{\omega}_{t}|^{2}]{\rm d}x{\rm d}y{\rm d}t - c_{0}I(0). \end{eqnarray} $

因此, 由(5.8)–(5.9)式得

(5.10) $ \begin{eqnarray} &&MM_{1}\int_{0}^{T}\int_{\widetilde{\Gamma}}[|\widetilde{\psi}_{t}|^{2} +|\widetilde{\phi}_{t}|^{2}+|\widetilde{\omega}_{t}|^{2}]{\rm d}\widetilde{\Gamma}{\rm d}t\\ &\geq&\int_{0}^{T}\int_{\Omega}[\rho_{1}(|\widetilde{\psi}_{t}|^{2} +|\widetilde{\phi}_{t}|^{2})+\rho_{2}|\widetilde{\omega}_{t}|^{2}]{\rm d}x{\rm d}y{\rm d}t-2c_{1}I(0)-2c_{0}I(0). \end{eqnarray} $

由(5.8)和(5.10)式可得

$ \begin{eqnarray*} && MM_{1}\int_{0}^{T}\int_{\widetilde{\Gamma}}[|\widetilde{\psi}_{t}|^{2} +|\widetilde{\phi}_{t}|^{2}+|\widetilde{\omega}_{t}|^{2}]{\rm d}\widetilde{\Gamma}{\rm d}t\\ &\geq&\int_{0}^{T}\int_{\Omega}[\rho_{1}(|\widetilde{\psi}_{t}|^{2} +|\widetilde{\phi}_{t}|^{2})+\rho_{2}|\widetilde{\omega}_{t}|^{2}]{\rm d}x{\rm d}y{\rm d}t+\int_{0}^{T}a(\widetilde{\psi}, \widetilde{\phi}, \widetilde{\omega}; \widetilde{\psi}, \widetilde{\phi}, \widetilde{\omega}){\rm d}t-4c_{1}I(0)-3c_{0}I(0)\\ &\geq& TI(0)-4c_{1}I(0)-3c_{0}I(0). \end{eqnarray*} $

因此, 记$ T_{1} = 4c_{1}+3c_{0}>0 $, 当$ T\geq T_{1} $时, 记$ C_{1} = \frac{1}{2MM_{0}}(T-T_{1})\geq0 $, 则有

$ C_{1}\|{\cal Y}_{0}\|_{{\cal H}}^{2} = C_{1}I(0)\leq\int_{0}^{T}\int_{\widetilde{\Gamma}} (|\widetilde{\psi}_{t}|^{2}+|\widetilde{\phi}_{t}|^{2}+|\widetilde{\omega}_{t}|^{2}){\rm d}\widetilde{\Gamma}{\rm d}t. $

故(5.1)式得证.

又由(5.7)式得

$ \int_{0}^{T}\int_{\Omega}[\rho_{1}(|\widetilde{\psi}_{t}|^{2} +|\widetilde{\phi}_{t}|^{2})+\rho_{2}|\widetilde{\omega}_{t}|^{2}]{\rm d}x{\rm d}y{\rm d}t \geq\int_{0}^{T}a(\widetilde{\psi}, \widetilde{\phi}, \widetilde{\omega};\widetilde{\psi}, \widetilde{\phi}, \widetilde{\omega}){\rm d}t-c_{0}I(0). $

故

$ \begin{eqnarray*} &&2\int_{0}^{T}\int_{\Omega}[\rho_{1}(|\widetilde{\psi}_{t}|^{2} +|\widetilde{\phi}_{t}|^{2})+\rho_{2}|\widetilde{\omega}_{t}|^{2}]{\rm d}x{\rm d}y{\rm d}t\\ &\geq&\int_{0}^{T}\int_{\Omega}[\rho_{1}(|\widetilde{\psi}_{t}|^{2} +|\widetilde{\phi}_{t}|^{2})+\rho_{2}|\widetilde{\omega}_{t}|^{2}]{\rm d}x{\rm d}y{\rm d}t +\int_{0}^{T}a(\widetilde{\psi}, \widetilde{\phi}, \widetilde{\omega};\widetilde{\psi}, \widetilde{\phi}, \widetilde{\omega}){\rm d}t-c_{0}I(0), \end{eqnarray*} $

即

(5.11) $ \begin{equation} 2\int_{0}^{T}\int_{\Omega}[\rho_{1}(|\widetilde{\psi}_{t}|^{2} +|\widetilde{\phi}_{t}|^{2})+\rho_{2}|\widetilde{\omega}_{t}|^{2}]{\rm d}x{\rm d}y{\rm d}t \geq TI(0)-c_{0}I(0). \end{equation} $

因此, 记$ T_{2} = c_{0}>0 $当$ T\geq T_{2} $时, 记$ C_{2} = \frac{1}{2M_{1}}(T-T_{2})\geq0 $, 则由(5.11)式得

$ C_{2}\|{\cal Y}_{0}\|_{{\cal H}}^{2} = C_{2}I(0)\leq\int_{0}^{T}\int_{\Omega} (|\widetilde{\psi}_{t}|^{2}+|\widetilde{\phi}_{t}|^{2}+|\widetilde{\omega}_{t}|^{2}){\rm d}x{\rm d}y{\rm d}t. $

从而, (5.2)式得证. 引理5.1证毕.

定理5.1   假定$ {\cal Y}_{0}\in {\cal H} $, 系统(1.1)–(1.2)能量依$ {\cal H} $范数一致指数收敛到0, 即存在与$ {\cal Y}_{0} $无关, 而与$ T_{1} $有关的正常数$ M, \alpha $使得

$ \left\|{\cal Y}\right\|_{{\cal H}}^{2}\leq M{\rm e}^{-\alpha t}\left\|{\cal Y}_{0}\right\|_{{\cal H}}^{2} $

当且仅当能观性条件(5.1)成立.

证  (Ⅰ) 充分性

首先闭环系统(1.1)和(1.2)是适定的, 且其唯一弱解为

$ \Phi = (\psi, \phi, \omega)\in C^{0}([0, \infty);W)\cap C^{1}([0, \infty);H), $

对于任意$ T>0 $, 考虑如下控制系统

(5.12) $ \begin{eqnarray} \left\{\begin{array}{lll} { } \rho_{1}\psi_{tt}-D(\psi_{xx}+\frac{1-\mu}{2}\psi_{yy} +\frac{1+\mu}{2}\phi_{xy}) +K(\psi+\omega_{x}) = 0, \quad (x, y, t)\in\Omega\times (0, T), \\ { } \rho_{1}\phi_{tt}-D(\phi_{yy}+\frac{1-\mu}{2}\phi_{xx} +\frac{1+\mu}{2}\psi_{xy}) +K(\phi+\omega_{y}) = 0, \quad (x, y, t)\in\Omega\times (0, T), \\ { } \rho_{2}\omega_{tt}-K[(\psi+\omega_{x})_{x}+(\phi+\omega_{y})_{y}] = 0, \quad (x, y, t)\in\Omega\times (0, T), \\ { } \psi = \phi = \omega = 0, \quad (x, y, t)\in\Gamma_{0}\times (0, T), \\ { } D[\nu_{1}\psi_{x}+\mu\nu_{1}\phi_{y}+\frac{1-\mu}{2}(\psi_{y}+\phi_{x})\nu_{2}] = -\psi_{t}, \quad (x, y, t)\in\Gamma_{1}\times (0, T), \\ { } D[\nu_{2}\phi_{y}+\mu\nu_{2}\psi_{x}+\frac{1-\mu}{2}(\psi_{y}+\phi_{x})\nu_{1}] = -\phi_{t}, \quad (x, y, t)\in\Gamma_{1}\times (0, T), \\ { } K(\frac{\partial\omega}{\partial{\bf \nu}}+\nu_{1}\psi+\nu_{2}\phi) = -\omega_{t}, \quad (x, y, t)\in\Gamma_{1}\times (0, T), \\ (\psi(x, y, 0), \phi(x, y, 0), \omega(x, y, 0)) = (\psi_{01}, \phi_{01}, \omega_{01}), \quad (x, y)\in\Omega, \\ (\psi_{t}(x, y, 0), \phi_{t}(x, y, 0), \omega_{t}(x, y, 0)) = (\psi_{02}, \phi_{02}, \omega_{02}), \quad (x, y)\in\Omega. \end{array}\right. \end{eqnarray} $

用$ \psi_{t}, \phi_{t}, \omega_{t} $分别乘以系统(5.12)前三式, 并在$ [0, T_{1}]\times\Omega $上积分并相加, 利用边界条件得到

$ \frac{1}{2}\int_{0}^{T_{1}}\frac{\rm d}{{\rm d}t}I(t){\rm d}t+\int_{0}^{T_{1}}\int_{\Gamma_{1}}[|\psi_{t}|^{2}+|\phi_{t}|^{2}+|\omega_{t}|^{2}]{\rm d}\Gamma_{1}{\rm d}t = 0. $

因此

(5.13) $ \begin{equation} 2\int_{0}^{T_{1}}\int_{\Gamma_{1}}[|\psi_{t}|^{2}+|\phi_{t}|^{2}+|\omega_{t}|^{2}]{\rm d}\Gamma_{1}{\rm d}t = I(0)-I(T_{1}). \end{equation} $

此外系统(5.12)的弱解分解为$ \Phi = \widetilde{\Phi}+\widehat{\Phi} $, 其中, $ \widetilde{\Phi} $是系统(5.3)的弱解, $ \widehat{\Phi} = (\widehat{\psi}, \widehat{\phi}, \widehat{\omega})\in C^{0}([0, T];W)\cap C^{1}([0, T];H) $是以下系统弱解

(5.14) $ \begin{eqnarray} \left\{\begin{array}{lll} { } \rho_{1}\widehat{\psi}_{tt}-D(\widehat{\psi}_{xx}+\frac{1-\mu}{2}\widehat{\psi}_{yy} +\frac{1+\mu}{2}\widehat{\phi}_{xy}) +K(\widehat{\psi}+\widehat{\omega}_{x}) = 0, \quad (x, y, t)\in\Omega\times (0, T), \\ { } \rho_{1}\widehat{\phi}_{tt}-D(\widehat{\phi}_{yy}+\frac{1-\mu}{2}\widehat{\phi}_{xx} +\frac{1+\mu}{2}\widehat{\psi}_{xy}) +K(\widehat{\phi}+\widehat{\omega}_{y}) = 0, \quad (x, y, t)\in\Omega\times (0, T), \\ { } \rho_{2}\widehat{\omega}_{tt}-K[(\widehat{\psi}+\widehat{\omega}_{x})_{x}+(\widehat{\phi}+\widehat{\omega}_{y})_{y}] = 0, \quad (x, y, t)\in\Omega\times (0, T), \\ { } \widehat{\psi} = \widehat{\phi} = \widehat{\omega} = 0, \quad (x, y, t)\in\Gamma_{0}\times (0, T), \\ { } D[\nu_{1}\widehat{\psi}_{x}+\mu\nu_{1}\widehat{\phi}_{y}+\frac{1-\mu}{2}(\widehat{\psi}_{y}+\widehat{\phi}_{x})\nu_{2}] = -\psi_{t}, \quad (x, y, t)\in\Gamma_{1}\times (0, T), \\ { } D[\nu_{2}\widehat{\phi}_{y}+\mu\nu_{2}\widehat{\psi}_{x}+\frac{1-\mu}{2}(\widehat{\psi}_{y}+\widehat{\phi}_{x})\nu_{1}] = -\phi_{t}, \quad (x, y, t)\in\Gamma_{1}\times (0, T), \\ { } K(\frac{\partial\widehat{\omega}}{\partial{\bf \nu}}+\nu_{1}\widehat{\psi}+\nu_{2}\widehat{\phi}) = -\omega_{t}, \quad (x, y, t)\in\Gamma_{1}\times (0, T), \\ { } (\widehat{\psi}(x, y, 0), \widehat{\phi}(x, y, 0), \widehat{\omega}(x, y, 0)) = (0, 0, 0), \quad (x, y)\in\Omega, \\ { } (\widehat{\psi}_{t}(x, y, 0), \widehat{\phi}_{t}(x, y, 0), \widehat{\omega}_{t}(x, y, 0)) = (0, 0, 0), \quad (x, y)\in\Omega. \end{array}\right. \end{eqnarray} $

由能观性条件及命题2.1中(2.4)式知

(5.15) $ \begin{eqnarray} I(0)& = &\|(\Phi_{01}\Phi_{02})\|_{{\cal H}}^{2}\leq\frac{1}{C}\int_{0}^{T_{1}}\int_{\Gamma_{1}}(|\widetilde{\psi}_{t}|^{2}+|\widetilde{\phi}_{t}|^{2}+|\widetilde{\omega}_{t}|^{2}){\rm d}\Gamma_{1}{\rm d}t\\ &\leq&\frac{1}{C}\bigg[\int_{0}^{T_{1}}\int_{\Gamma_{1}}(|\psi_{t}|^{2}+|\phi_{t}|^{2}+|\omega_{t}|^{2}){\rm d}\Gamma_{1}{\rm d}t +\int_{0}^{T_{1}}\int_{\Gamma_{1}}(|\widehat{\psi}_{t}|^{2}+|\widehat{\phi}_{t}|^{2}+|\widehat{\omega}_{t}|^{2}){\rm d}\Gamma_{1}{\rm d}t\bigg]\\ &\leq& C' \int_{0}^{T_{1}}\int_{\Gamma_{1}}(|\psi_{t}|^{2}+|\phi_{t}|^{2}+|\omega_{t}|^{2}){\rm d}\Gamma_{1}{\rm d}t. \end{eqnarray} $

由(5.13)和(5.15)式得

$ I(T_{1})-I(0) = -2\int_{0}^{T_{1}}\int_{\Gamma_{1}}(|\psi_{t}|^{2}+|\phi_{t}|^{2}+|\omega_{t}|^{2}){\rm d}\Gamma_{1}{\rm d}t \leq-\frac{2}{C'}I(0)\leq-\frac{2}{C'}I(T_{1}), $

取$ \alpha = \frac{\ln(1+\frac{1}{C'})}{T_{1}} $, 则有

(5.16) $ \begin{equation} I(T_{1})\leq {\rm e}^{-\alpha T_{1}}I(0). \end{equation} $

因此, 对于$ \forall k\in {\Bbb N}^{+} $, 有$ I(kT_{1})\leq {\rm e}^{-\alpha T_{1}}I((k-1)T_{1}) $. 于是对于$ \forall t\in [0, \infty) $, 存在$ k\in {\Bbb N}^{+} $, 使得$ t\in [kT_{1}, (k+1)T_{1}) $有

$ I(t)\leq I(kT_{1})\leq {\rm e}^{-\alpha kT_{1}}I(0) = (1+\frac{2}{C'}){\rm e}^{-\alpha (k+1)T_{1}}I(0)\leq(1+\frac{2}{C'}){\rm e}^{-\alpha t}I(0). $

记$ M = 1+\frac{2}{C'} $, 则$ I(t)\leq M{\rm e}^{-\alpha t}I(0) $, 即$ \|(\Phi, \Phi_{t})\|_{{\cal H}}^{2}\leq M{\rm e}^{-\alpha t}\|(\Phi_{01}, \Phi_{02})\|_{{\cal H}}^{2} $.

(Ⅱ) 必要性

对于$ \forall t>0 $, 将$ \psi_{t}, \phi_{t}, \omega_{t} $分别乘以系统(5.14) 前三式, 在$ [0, t]\times\Omega $上积分并相加, 分部积分得

$ 2\int_{0}^{t}\int_{\Gamma_{1}}(|\psi_{t}|^{2}+|\phi_{t}|^{2}+|\omega_{t}|^{2}){\rm d}\Gamma_{1}{\rm d}t = I(0)-I(t). $

又由于系统是指数稳定的, 故存在足够大的$ T'>0 $, 使得

(5.17) $ \begin{eqnarray} \int_{0}^{T'}\int_{\Gamma_{1}}(|\psi_{t}|^{2}+|\phi_{t}|^{2}+|\omega_{t}|^{2}){\rm d}\Gamma_{1}{\rm d}t\geq \frac{1}{4}I(0). \end{eqnarray} $

再用$ \widehat{\psi}_{t}, \widehat{\phi}_{t}, \widehat{\omega}_{t} $分别乘以系统(5.14) 前三个等式, 在$ [0, T']\times\Omega $上积分并相加得

$ \begin{eqnarray*} &&\int_{0}^{T'}\int_{\Omega}[\rho_{1}(\widehat{\psi}_{tt}\widehat{\psi}_{t}+\widehat{\phi}_{tt}\widehat{\phi}_{t})+ \rho_{2}\widehat{\omega}_{tt}\widehat{\omega}_{t}]{\rm d}x{\rm d}y{\rm d}t\\ & = &\int_{0}^{T'}\int_{\Omega}[L_{1}\{\widehat{\psi}, \widehat{\phi}, \widehat{\omega}\}\widehat{\psi}_{t} +L_{2}\{\widehat{\psi}, \widehat{\phi}, \widehat{\omega}\}\widehat{\phi}_{t} +L_{3}\{\widehat{\psi}, \widehat{\phi}, \widehat{\omega}\}\widehat{\omega}_{t}]{\rm d}x{\rm d}y{\rm d}t, \end{eqnarray*} $

分部积分得

$ \begin{eqnarray*} &&\int_{\Omega}[\rho_{1}(|\widehat{\psi}_{t}(T')|^{2}+|\widehat{\phi}_{t}(T')|^{2})+ \rho_{2}|\widehat{\omega}_{t}(T')|^{2}]{\rm d}x{\rm d}y\\ &&+a(\widehat{\psi}(T'), \widehat{\phi}(T'), \widehat{\omega}(T');\widehat{\psi}(T'), \widehat{\phi}(T'), \widehat{\omega}(T')) +2\int_{0}^{T'}\int_{\Gamma_{1}}(\widehat{\psi}_{t}\psi_{t}+\widehat{\phi}_{t}\phi_{t}+\widehat{\omega}_{t}\omega_{t}){\rm d}\Gamma_{1}{\rm d}t = 0, \end{eqnarray*} $

亦即

(5.18) $ \begin{equation} 0\leq\frac{1}{2}\|(\widehat{\Phi}(T'), \widehat{\Phi}_{t}(T'))\|_{{\cal H}}^{2} = -\int_{0}^{T'}\int_{\Gamma_{1}}(\widehat{\psi}_{t}(\widehat{\psi}_{t}+\widetilde{\psi}_{t}) +\widehat{\phi}_{t}(\widehat{\phi}_{t}+\widetilde{\phi}_{t})+\widehat{\omega}_{t}(\widehat{\omega}_{t}+\widetilde{\omega}_{t})){\rm d}\Gamma_{1}{\rm d}t. \end{equation} $

因此, 由Cauchy-Schwarz以及Young不等式, 我们有

(5.19) $ \begin{eqnarray} \int_{0}^{T'}\int_{\Gamma_{1}}(|\widehat{\psi}_{t}|^{2}+|\widehat{\phi}_{t}|^{2}+|\widehat{\omega}_{t}|^{2}){\rm d}\Gamma_{1}{\rm d}t \leq C''\int_{0}^{T'}\int_{\Gamma_{1}}(|\widetilde{\psi}_{t}|^{2}+|\widetilde{\phi}_{t}|^{2}+|\widetilde{\omega}_{t}|^{2}){\rm d}\Gamma_{1}{\rm d}t. \end{eqnarray} $

注意到: $ \psi = \widehat{\psi}+\widetilde{\psi}, \phi = \widehat{\phi}+\widetilde{\phi}, \omega = \widehat{\omega}+\widetilde{\omega} $. 于是, 我们有

(5.20) $ \begin{eqnarray} & &\int_{0}^{T'}\int_{\Gamma_{1}}(|\widehat{\psi}_{t}|^{2}+|\widehat{\phi}_{t}|^{2}+|\widehat{\omega}_{t}|^{2}){\rm d}\Gamma_{1}{\rm d}t +\int_{0}^{T'}\int_{\Gamma_{1}}(|\widetilde{\psi}_{t}|^{2}+|\widetilde{\phi}_{t}|^{2}+|\widetilde{\omega}_{t}|^{2}){\rm d}\Gamma_{1}{\rm d}t\\ &\geq&\frac{1}{2}\int_{0}^{T'}\int_{\Gamma_{1}}(|\psi_{t}|^{2}+|\phi_{t}|^{2}+|\omega_{t}|^{2}){\rm d}\Gamma_{1}{\rm d}t. \end{eqnarray} $

上式结合(5.17)和(5.19)式得

$ \begin{eqnarray*} \|(\Phi_{01}, \Phi_{02})\|_{{\cal H}}^{2}& = &I(0) \leq 4\int_{0}^{T'}\int_{\Gamma_{1}}(|\psi_{t}|^{2}+|\phi_{t}|^{2}+|\omega_{t}|^{2}){\rm d}\Gamma_{1}{\rm d}t\\ &\leq&8\bigg[\int_{0}^{T'}\int_{\Gamma_{1}}(|\widehat{\psi}_{t}|^{2}+|\widehat{\phi}_{t}|^{2}+|\widehat{\omega}_{t}|^{2}){\rm d}\Gamma_{1}{\rm d}t +\int_{0}^{T'}\int_{\Gamma_{1}}(|\widetilde{\psi}_{t}|^{2}+|\widetilde{\phi}_{t}|^{2}+|\widetilde{\omega}_{t}|^{2}){\rm d}\Gamma_{1}{\rm d}t\bigg]\\ &\leq&8(1+C'')\int_{0}^{T'}\int_{\Gamma_{1}}(|\widetilde{\psi}_{t}|^{2}+|\widetilde{\phi}_{t}|^{2}+|\widetilde{\omega}_{t}|^{2}){\rm d}\Gamma_{1}{\rm d}t. \end{eqnarray*} $

故能观性条件(5.1)得证. 证毕.

接下来我们研究系统(2.2)的次最优性与最优轨线的指数稳定性.

命题6.1   对于任意初始对$ {\cal Y}_{0} = (\Phi_{01}, \Phi_{02})\in {\cal H} $及任意$ T>0 $, 系统(2.2)存在控制$ \widehat{U} = (\widehat{u}_{1}, \widehat{u}_{2}, \widehat{u}_{3})\in L^{2}([0, T];{\cal U}) $使得

(6.1) $ \begin{eqnarray} V_{T}({\cal Y}_{0})\leq J_{T}({\cal Y}_{0};\widehat{U})\leq \gamma_{1}(T)\|{\cal Y}_{0}\|_{{\cal H}}^{2} , \end{eqnarray} $

其中$ \gamma_{1}(\cdot) $是一个连续、不减的有界函数. 进一步, 存在只与$ T $有关的正常数$ \gamma_{2}(T) $, 使得

(6.2) $ \begin{eqnarray} V_{T}({\cal Y}_{0})\geq \gamma_{2}(T)\|{\cal Y}_{0}\|_{{\cal H}}^{2} . \end{eqnarray} $

证  取$ u_{1} = -\psi_{t}, u_{2} = -\phi_{t}, u_{3} = -\omega_{t} $, 那么

(6.3) $ \begin{eqnarray} I(t)\leq M{\rm e}^{-\alpha t}I(0), \quad \forall t\in[0, T]. \end{eqnarray} $

对(6.3)式两边在$ [0, T] $上积分得

$ \int_{0}^{T}I(t){\rm d}t\leq \frac{M}{\alpha}(1-{\rm e}^{-\alpha T})I(0), $

再由(5.13)式得

$ \int_{0}^{T}\int_{\Gamma_{1}}[|\psi_{t}|^{2}+|\phi_{t}|^{2}+|\omega_{t}|^{2}]{\rm d}\Gamma_{1}{\rm d}t\leq\frac{1}{2} I(0). $

取$ \gamma_{1}(T) = \frac{M}{2\alpha}(1-{\rm e}^{-\alpha T})+\frac{\beta}{4} $, 由值函数的定义, 我们有

$ \begin{eqnarray*} V_{T}({\cal Y}_{0})&\leq&\frac{1}{2}\int_{0}^{T}I(t){\rm d}t+ \frac{\beta}{2}\int_{0}^{T}\int_{\Gamma_{1}}[|\psi_{t}|^{2}+|\phi_{t}|^{2}+|\omega_{t}|^{2}]{\rm d}\Gamma_{1}{\rm d}t\\ &\leq&[\frac{M}{2\alpha}(1-{\rm e}^{-\alpha T})+\frac{\beta}{4}]I(0) = \gamma_{1}(T)\|{\cal Y}_{0}\|_{{\cal H}}^{2}, \end{eqnarray*} $

即(6.1)式成立.

为了证明(6.2)式, 对于任意$ U = (u_{1}, u_{2}, u_{3})\in L^{2}([0, T];{\cal U}) $, 我们应用叠加原理, 将系统(2.2)的弱解表示为$ \Phi = \widetilde{\Phi}+\widehat{\Phi} $, 其中$ \widetilde{\Phi} = (\widetilde{\psi}, \widetilde{\phi}, \widetilde{\omega}) $是系统(5.3)的弱解, $ \widehat{\Phi} = (\widehat{\psi}, \widehat{\phi}, \widehat{\omega}) $是下面系统的弱解

(6.4) $ \begin{eqnarray} \left\{\begin{array}{lll} { } \rho_{1}\widehat{\psi}_{tt}-D(\widehat{\psi}_{xx}+\frac{1-\mu}{2}\widehat{\psi}_{yy} +\frac{1+\mu}{2}\widehat{\phi}_{xy}) +K(\widehat{\psi}+\widehat{\omega}_{x}) = 0, \quad (x, y, t)\in\Omega\times (0, T), \\ { } \rho_{1}\widehat{\phi}_{tt}-D(\widehat{\phi}_{yy}+\frac{1-\mu}{2}\widehat{\phi}_{xx} +\frac{1+\mu}{2}\widehat{\psi}_{xy}) +K(\widehat{\phi}+\widehat{\omega}_{y}) = 0, \quad (x, y, t)\in\Omega\times (0, T), \\ { } \rho_{2}\widehat{\omega}_{tt}-K[(\widehat{\psi}+\widehat{\omega}_{x})_{x}+(\widehat{\phi}+\widehat{\omega}_{y})_{y}] = 0, \quad (x, y, t)\in\Omega\times (0, T), \\ { } \widehat{\psi} = \widehat{\phi} = \widehat{\omega} = 0, \quad (x, y, t)\in\Gamma_{0}\times (0, T), \\ { } D[\nu_{1}\widehat{\psi}_{x}+\mu\nu_{1}\widehat{\phi}_{y}+\frac{1-\mu}{2}(\widehat{\psi}_{y}+\widehat{\phi}_{x})\nu_{2}] = u_{1}, \quad (x, y, t)\in\Gamma_{1}\times (0, T), \\ { } D[\nu_{2}\widehat{\phi}_{y}+\mu\nu_{2}\widehat{\psi}_{x}+\frac{1-\mu}{2}(\widehat{\psi}_{y}+\widehat{\phi}_{x})\nu_{1}] = u_{2}, \quad (x, y, t)\in\Gamma_{1}\times (0, T), \\ { } K(\frac{\partial\widehat{\omega}}{\partial{\bf \nu}}+\nu_{1}\widehat{\psi}+\nu_{2}\widehat{\phi}) = u_{3}, \quad (x, y, t)\in\Gamma_{1}\times (0, T), \\ { } (\widehat{\psi}(x, y, 0), \widehat{\phi}(x, y, 0), \widehat{\omega}(x, y, 0)) = (0, 0, 0), \quad (x, y)\in\Omega, \\ { } (\widehat{\psi}_{t}(x, y, 0), \widehat{\phi}_{t}(x, y, 0), \widehat{\omega}_{t}(x, y, 0)) = (0, 0, 0), \quad (x, y)\in\Omega. \end{array}\right. \end{eqnarray} $

由能观性条件(5.2)和(2.4)得

$ \begin{eqnarray*} \|(\Phi_{01}, \Phi_{02})\|_{{\cal H}}^{2} &\leq&\frac{1}{C_{2}}\int_{0}^{T}\int_{\Omega}(|\widetilde{\psi}_{t}|^{2}+|\widetilde{\phi}_{t}|^{2}+|\widetilde{\omega}_{t}|^{2}){\rm d}x{\rm d}y{\rm d}t\\ &\leq&\frac{1}{C_{2}}\int_{0}^{T} \bigg[\int_{\Omega}(|\psi_{t}|^{2}+|\phi_{t}|^{2}+|\omega_{t}|^{2}){\rm d}x{\rm d}y +\int_{\Omega}(|\widehat{\psi}_{t}|^{2}+|\widehat{\phi}_{t}|^{2}+|\widehat{\omega}_{t}|^{2}){\rm d}x{\rm d}y\bigg]{\rm d}t\\ &\leq& C_{3}\bigg\{\int_{0}^{T}\int_{\Omega}[\rho_{1}(|\psi_{t}|^{2}+|\phi_{t}|^{2})+\rho_{2}|\omega_{t}|^{2}]{\rm d}x{\rm d}y{\rm d}t\\ &&+ \int_{0}^{T}\int_{\Omega}[\rho_{1}(|\widehat{\psi}_{t}|^{2}+|\widehat{\phi}_{t}|^{2}) +\rho_{2}|\widehat{\omega}_{t}|^{2}]{\rm d}x{\rm d}y{\rm d}t\bigg\}\\ &\leq& C_{3}\bigg\{\int_{0}^{T}\int_{\Omega}[\rho_{1}(|\psi_{t}|^{2}+|\phi_{t}|^{2})+\rho_{2}|\omega_{t}|^{2}]{\rm d}x{\rm d}y{\rm d}t \\ &&+ TC^{2}\int_{0}^{T}\int_{\Gamma_{1}}(|u_{1}|^{2}+|u_{2}|^{2})+|u_{3}|^{2}]{\rm d}\Gamma_{1}{\rm d}t\bigg\} +C_{3}\int_{0}^{T}a(\psi, \phi, \omega; \psi, \phi, \omega){\rm d}t\\ &\leq& C''(T)\int_{0}^{T}\bigg[\frac{1}{2}\|{\cal Y}\|_{{\cal H}}^{2} +\frac{\beta}{2}\|U\|_{{\cal U}}^{2}\bigg]{\rm d}t\\ & = &C''(T)\int_{0}^{T}\ell({\cal Y}(t), U(t)){\rm d}t. \end{eqnarray*} $

由控制变量$ U = (u_{1}, u_{2}, u_{3})\in L^{2}([0, T];{\cal U}) $的任意性, 我们得到只与$ T $有关的正常数$ \gamma_{2}(T) = \frac{1}{C''(T)} $使得(6.2)式成立. 证毕.

引理6.1  对于任意初始值$ {\cal Y}_{0}\in {\cal H} $以及$ \delta>0 $, 当$ T>\delta $时, 则, 对所有的$ \widehat{t}\in[\delta, T] $, 有

(6.5) $ \begin{eqnarray} V_{T}({\cal Y}_{T}^{\ast}(\delta, {\cal Y}_{0}, 0)) \leq\int_{\delta}^{\widehat{t}}\ell({\cal Y}_{T}^{\ast}(t, {\cal Y}_{0}, 0), {\cal U}_{T}^{\ast}(t, {\cal Y}_{0}, 0)){\rm d}t +\gamma_{1}(T+\delta-\widehat{t})\left\|{\cal Y}_{T}^{\ast}(\widehat{t}, {\cal Y}_{0}, 0)\right\|_{{\cal H}}^{2} \end{eqnarray} $

和对所有的$ \widetilde{t}\in[0, T] $, 有

(6.6) $ \begin{eqnarray} \int_{\widetilde{t}}^{T}\ell({\cal Y}_{T}^{\ast}(t, {\cal Y}_{0}, 0), {\cal U}_{T}^{\ast}(t, {\cal Y}_{0}, 0)){\rm d}t \leq\gamma_{1}(T-\widetilde{t})\left\|{\cal Y}_{T}^{\ast}(\widetilde{t}, {\cal Y}_{0}, 0)\right\|_{{\cal H}}^{2}. \end{eqnarray} $

证  取$ \xi = \frac{1}{2}\min\{1, \beta\} $, 对于任意$ {\cal Y}_{0}\in{\cal H} $及$ \overline{t}\in[0, T] $, 由(1.4)式得

(6.7) $ \begin{eqnarray} \ell({\cal Y}_{T}^{\ast}(t, {\cal Y}_{0}, 0), U_{T}^{\ast}(t, {\cal Y}_{0}, 0))\geq\xi [\|{\cal Y}_{T}^{\ast}(t, {\cal Y}_{0}, 0)\|_{{\cal H}}^{2}+\|U_{T}^{\ast}(t, {\cal Y}_{0}, 0)\|_{{\cal U}}^{2}]. \end{eqnarray} $

因此, 我们有

$ \begin{eqnarray*} \xi\int_{0}^{\overline{t}}[\|{\cal Y}_{T}^{\ast}(t, {\cal Y}_{0}, 0)\|_{{\cal H}}^{2}+\|U_{T}^{\ast}(t, {\cal Y}_{0}, 0)\|_{{\cal U}}^{2}]{\rm d}t &\leq&\int_{0}^{\overline{t}}\ell({\cal Y}_{T}^{\ast}(t, {\cal Y}_{0}, 0), U_{T}^{\ast}(t, {\cal Y}_{0}, 0)){\rm d}t\\ & = &V_{T}({\cal Y}_{0})-V_{T-\overline{t}}({\cal Y}_{T}^{\ast}(\overline{t}, {\cal Y}_{0}, 0)). \end{eqnarray*} $

又由(2.4)式得

$ \begin{eqnarray*} \|{\cal Y}_{T}^{\ast}(\overline{t}, {\cal Y}_{0}, 0)\|_{{\cal H}}^{2} &\leq& C'[\|{\cal Y}_{0}\|_{{\cal H}}^{2}+ \int_{0}^{\overline{t}}[\|{\cal Y}_{T}^{\ast}(t, {\cal Y}_{0}, 0)\|_{{\cal H}}^{2}+\|U_{T}^{\ast}(t, {\cal Y}_{0}, 0)\|_{{\cal U}}^{2}]{\rm d}t]\\ & \leq &C'[\|{\cal Y}_{0}\|_{{\cal H}}^{2} +\frac{1}{\xi}(V_{T}({\cal Y}_{0})-V_{T-\overline{t}}({\cal Y}_{T}^{\ast}(\overline{t}, {\cal Y}_{0}, 0)))]\\ & \leq& C'[\|{\cal Y}_{0}\|_{{\cal H}}^{2} +\frac{1}{\xi}V_{T}({\cal Y}_{0})]\leq C'(1+\frac{\gamma_{1}(T)}{\xi})\|{\cal Y}_{0}\|_{{\cal H}}^{2}. \end{eqnarray*} $

因此, 对于任意$ \overline{t}\in[0, T] $, 都有$ {\cal Y}_{T}^{\ast}(\overline{t}, {\cal Y}_{0}, 0)\in{\cal H} $.

对于任意$ \widehat{t}\in[\delta, T] $, 我们有

$ \begin{eqnarray*} V_{T}({\cal Y}_{T}^{\ast}(\delta, {\cal Y}_{0}, 0)) & = &V_{T}({\cal Y}_{T}^{\ast}(\delta, {\cal Y}^{\ast}(\delta), \delta))\\ & = &\int_{\delta}^{T+\delta}\ell({\cal Y}_{T}^{\ast}(t, {\cal Y}^{\ast}(\delta), \delta), U_{T}^{\ast}(t, {\cal Y}^{\ast}(\delta), \delta)){\rm d}t\\ & = &\int_{\delta}^{\widehat{t}}\ell({\cal Y}_{T}^{\ast}(t, {\cal Y}^{\ast}(\delta), \delta), U_{T}^{\ast}(t, {\cal Y}^{\ast}(\delta), \delta)){\rm d}t +V_{T+\delta-\widehat{t}}({\cal Y}_{T}^{\ast}(\widehat{t}, {\cal Y}^{\ast}(\delta), \delta)). \end{eqnarray*} $

又$ {\cal Y}_{T}^{\ast}(\cdot, {\cal Y}^{\ast}(\delta), \delta) $是系统在区间$ [\delta, T+\delta] $的最优解, 根据贝尔曼最优性原理, 我们有

$ \begin{eqnarray*} V_{T}({\cal Y}_{T}^{\ast}(\delta, {\cal Y}_{0}, 0)) &\leq&\int_{\delta}^{\widehat{t}}\ell({\cal Y}_{T}^{\ast}(t, {\cal Y}_{0}, 0), U_{T}^{\ast}(t, {\cal Y}_{0}, 0)){\rm d}t +V_{T+\delta-\widehat{t}}({\cal Y}_{T}^{\ast}(\widehat{t}, {\cal Y}_{0}, 0))\\ & \leq&\int_{\delta}^{\widehat{t}}\ell({\cal Y}_{T}^{\ast}(t, {\cal Y}_{0}, 0), U_{T}^{\ast}(t, {\cal Y}_{0}, 0)){\rm d}t +\gamma_{1}(T+\delta-\widehat{t})\|{\cal Y}_{T}^{\ast}(\widehat{t}, {\cal Y}_{0}, 0)\|_{{\cal H}}^{2}. \end{eqnarray*} $

因此, (6.5)式成立.

接下来证明(6.6)式. 对于任意$ \widetilde{t}\in[0, T] $, 我们有

$ \begin{eqnarray*} V_{T}({\cal Y}_{0}) & = &\int_{0}^{\widetilde{t}}\ell({\cal Y}_{T}^{\ast}(t, {\cal Y}_{0}, 0), U_{T}^{\ast}(t, {\cal Y}_{0}, 0)){\rm d}t +\int_{\widetilde{t}}^{T}\ell({\cal Y}_{T}^{\ast}(t, {\cal Y}_{0}, 0), U_{T}^{\ast}(t, {\cal Y}_{0}, 0)){\rm d}t\\ & = &\int_{0}^{\widetilde{t}}\ell({\cal Y}_{T}^{\ast}(t, {\cal Y}_{0}, 0), U_{T}^{\ast}(t, {\cal Y}_{0}, 0)){\rm d}t +V_{T-\widetilde{t}}({\cal Y}_{T}^{\ast}(\widetilde{t}, {\cal Y}_{0}, 0))\\ & \leq&\int_{0}^{\widetilde{t}}\ell({\cal Y}_{T}^{\ast}(t, {\cal Y}_{0}, 0), U_{T}^{\ast}(t, {\cal Y}_{0}, 0)){\rm d}t +\gamma_{1}(T-\widetilde{t})\|{\cal Y}_{T}^{\ast}(\widetilde{t}, {\cal Y}_{0}, 0)\|_{{\cal H}}^{2}. \end{eqnarray*} $

因此, 对所有$ \widetilde{t}\in[0, T] $都有

$ \int_{\widetilde{t}}^{T}\ell({\cal Y}_{T}^{\ast}(t, {\cal Y}_{0}, 0), {\cal U}_{T}^{\ast}(t, {\cal Y}_{0}, 0)){\rm d}t \leq\gamma_{1}(T-\widetilde{t})\left\|{\cal Y}_{T}^{\ast}(\widetilde{t}, {\cal Y}_{0}, 0)\right\|_{{\cal H}}^{2}. $

证毕.

定理6.1  对于任意初始值$ {\cal Y}_{0}\in{\cal H} $, 以及样本时间$ \delta>0 $和预测时间$ T>\delta $, 存在只与$ T $有关的函数

$ \sigma_{1}(T) = 1+\frac{\gamma_{1}(T)}{\xi(T-\delta)}, \quad \sigma_{2} = \frac{\gamma_{1}(T)}{\xi\delta}, $

使得以下估计式

(6.8) $ \begin{equation} V_{T}({\cal Y}_{T}^{\ast}(\delta, {\cal Y}_{0}, 0)) \leq\sigma_{1}(T)\int_{\delta}^{T}\ell({\cal Y}_{T}^{\ast}(t, {\cal Y}_{0}, 0), U_{T}^{\ast}(t, {\cal Y}_{0}, 0)){\rm d}t \end{equation} $

和

(6.9) $ \begin{equation} \int_{\delta}^{T}\ell({\cal Y}_{T}^{\ast}(t, {\cal Y}_{0}, 0), U_{T}^{\ast}(t, {\cal Y}_{0}, 0)){\rm d}t \leq\sigma_{2}(T)\int_{0}^{\delta}\ell({\cal Y}_{T}^{\ast}(t, {\cal Y}_{0}, 0), U_{T}^{\ast}(t, {\cal Y}_{0}, 0)){\rm d}t \end{equation} $

成立.

证  由于$ {\cal Y}_{T}^{\ast}(\cdot, {\cal Y}_{0}, 0)\in C([0, T];{\cal H}) $, 故一定存在$ t_{1}\in[\delta, T], t_{2}\in[0, \delta] $, 使得

$ t_{1} = \arg\min\limits_{t\in[\delta, T]}\left\|{\cal Y}_{T}^{\ast}(t, {\cal Y}_{0}, 0)\right\|_{{\cal H}}^{2}, \quad t_{2} = \arg\min\limits_{t\in[0, \delta]}\left\|{\cal Y}_{T}^{\ast}(t, {\cal Y}_{0}, 0)\right\|_{{\cal H}}^{2}. $

由(6.5)和(6.7)式可得

$ \begin{eqnarray*} V_{T}({\cal Y}_{T}^{\ast}(\delta, {\cal Y}_{0}, 0)) &\leq&\int_{\delta}^{t_{1}}\ell({\cal Y}_{T}^{\ast}(t, {\cal Y}_{0}, 0), U_{T}^{\ast}(t, {\cal Y}_{0}, 0)){\rm d}t +\gamma_{1}(T+\delta-t_{1})\|{\cal Y}_{T}^{\ast}(t_{1}, {\cal Y}_{0}, 0)\|_{{\cal H}}^{2}\\ & \leq&\int_{\delta}^{T}\ell({\cal Y}_{T}^{\ast}(t, {\cal Y}_{0}, 0), U_{T}^{\ast}(t, {\cal Y}_{0}, 0)){\rm d}t +\gamma_{1}(T)\|{\cal Y}_{T}^{\ast}(t_{1}, {\cal Y}_{0}, 0)\|_{{\cal H}}^{2}\\ & \leq&\int_{\delta}^{T}\ell({\cal Y}_{T}^{\ast}(t, {\cal Y}_{0}, 0), U_{T}^{\ast}(t, {\cal Y}_{0}, 0)){\rm d}t +\frac{\gamma_{1}(T)}{T-\delta}\int_{\delta}^{T}\|{\cal Y}_{T}^{\ast}(t, {\cal Y}_{0}, 0)\|_{{\cal H}}^{2}{\rm d}t\\ & \leq&(1+\frac{\gamma_{1}(T)}{\xi(T-\delta)})\int_{\delta}^{T}\ell({\cal Y}_{T}^{\ast}(t, {\cal Y}_{0}, 0), U_{T}^{\ast}(t, {\cal Y}_{0}, 0)){\rm d}t. \end{eqnarray*} $

即(6.8)成立.

同理, 由(6.6)和(6.7)式得

$ \begin{eqnarray*} \int_{\delta}^{T}\ell({\cal Y}_{T}^{\ast}(t, {\cal Y}_{0}, 0), U_{T}^{\ast}(t, {\cal Y}_{0}, 0)){\rm d}t &\leq&\int_{t_{2}}^{T}\ell({\cal Y}_{T}^{\ast}(t, {\cal Y}_{0}, 0), U_{T}^{\ast}(t, {\cal Y}_{0}, 0)){\rm d}t\\ & \leq&\gamma_{1}(T-t_{2})\|{\cal Y}_{T}^{\ast}(t_{2}, {\cal Y}_{0}, 0)\|_{{\cal H}}^{2}\\ & \leq&\frac{\gamma_{1}(T)}{\delta}\int_{0}^{\delta}\|{\cal Y}_{T}^{\ast}(t, {\cal Y}_{0}, 0)\|_{{\cal H}}^{2}{\rm d}t\\ & \leq&\frac{\gamma_{1}(T)}{\xi\delta}\int_{0}^{\delta}\ell({\cal Y}_{T}^{\ast}(t, {\cal Y}_{0}, 0), U_{T}^{\ast}(t, {\cal Y}_{0}, 0)){\rm d}t. \end{eqnarray*} $

因此(6.9)式成立. 证毕.

定理6.2  对于任意初始值$ {\cal Y}_{0}\in{\cal H} $以及样本时间$ \delta>0 $, 存在$ \widetilde{T}>\delta $和$ \kappa\in(0, 1) $, 当$ T\geq\widetilde{T} $时, 使得

(6.10) $ \begin{equation} V_{T}({\cal Y}_{T}^{\ast}(\delta, {\cal Y}_{0}, 0))\leq V_{T}({\cal Y}_{0}) -\kappa\int_{0}^{\delta}\ell({\cal Y}_{T}^{\ast}(t, {\cal Y}_{0}, 0), U_{T}^{\ast}(t, {\cal Y}_{0}, 0)){\rm d}t. \end{equation} $

成立.

证  由定理6.1得

$ \begin{eqnarray*} && V_{T}({\cal Y}_{T}^{\ast}(\delta, {\cal Y}_{0}, 0))- V_{T}({\cal Y}_0)\\ & = &V_{T}({\cal Y}_{T}^{\ast}(\delta, {\cal Y}_{0}, 0))-\int_{0}^{T}\ell({\cal Y}_{T}^{\ast}(t, {\cal Y}_{0}, 0), U_{T}^{\ast}(t, {\cal Y}_{0}, 0)){\rm d}t\\ &\leq&(\sigma_{1}(T)-1)\int_{\delta}^{T}\ell({\cal Y}_{T}^{\ast}(t, {\cal Y}_{0}, 0), U_{T}^{\ast}(t, {\cal Y}_{0}, 0)){\rm d}t -\int_{0}^{\delta}\ell({\cal Y}_{T}^{\ast}(t, {\cal Y}_{0}, 0), U_{T}^{\ast}(t, {\cal Y}_{0}, 0)){\rm d}t\\ &\leq&[\sigma_{2}(T)(\sigma_{1}(T)-1)-1]\int_{0}^{\delta}\ell({\cal Y}_{T}^{\ast}(t, {\cal Y}_{0}, 0), U_{T}^{\ast}(t, {\cal Y}_{0}, 0)){\rm d}t. \end{eqnarray*} $

于是, 当$ T\rightarrow \infty $时, $ 1-\sigma_{2}(T)(\sigma_{1}(T)-1) = 1-\frac{\gamma_{1}^{2}(T)}{\xi^{2}\delta(T-\delta)}\rightarrow1 $. 因此, 存在$ \widetilde{T}>\delta $, 使得当$ T\geq\widetilde{T} $时, 存在$ \kappa\in(0, 1) $, 使得$ 1-\sigma_{2}(T)(\sigma_{1}(T)-1)>\kappa $. 因此(6.10)式成立. 证毕.

定理6.3  对于给定的样本时间$ \delta>0 $, 则存在$ \widetilde{T}>\delta $和$ \kappa\in(0, 1) $, 以及对每一预测时间$ T\geq\widetilde{T} $以及初始值$ {\cal Y}_{0}\in{\cal H} $, 滚动时域控制$ U_{T}^{\ast}(\cdot) $使得

(6.11) $ \begin{equation} \kappa V_{\infty}({\cal Y}_{0})\leq \kappa J_{\infty}({\cal Y}_{0}, U_{T}^{\ast}(\cdot) ) \leq V_{T}({\cal Y}_{0})\leq V_{\infty}({\cal Y}_{0}) \end{equation} $

成立. 进一步, 我们可以得到最优轨线是指数稳定的, 即存在只与$ \delta, \kappa $有关的正数$ M, \eta $使得

(6.12) $ \begin{equation} \|{\cal Y}_{T}^{\ast}(t, {\cal Y}_{0}, 0)\|\leq M{\rm e}^{-\eta t}. \end{equation} $

成立.

证  由于$ \kappa V_{\infty}({\cal Y}_{0})\leq \kappa J_{\infty}({\cal Y}_{0}, U_{T}^{\ast}(\cdot) ) $和$ V_{T}({\cal Y}_{0})\leq V_{\infty}({\cal Y}_{0}) $显然成立. 因此, 我们只需证明$ \kappa J_{\infty}({\cal Y}_{0}, U_{T}^{\ast}(\cdot) ) \leq V_{T}({\cal Y}_{0}) $即可.

对于给定$ \delta>0 $, 取$ t_{k} = k\delta $, 由定理6.2可得, 存在$ \widetilde{T}>\delta $和$ \kappa\in(0, 1) $, 当$ T\geq\widetilde{T} $时, 使得

(6.13) $ \begin{eqnarray} && V_{T}({\cal Y}_{T}^{\ast}(t_{k+1}, {\cal Y}(t_{k}), t_{k}))\\ &\leq& V_{T}({\cal Y}_{T}^{\ast}(t_{k}, {\cal Y}(t_{k}), t_{k})) -\kappa\int_{t_{k}}^{t_{k+1}}\ell({\cal Y}_{T}^{\ast}(t, {\cal Y}(t_{k}), t_{k}), U_{T}^{\ast}(t, {\cal Y}(t_{k}), t_{k})){\rm d}t. \end{eqnarray} $

因此, 由递推得

(6.14) $ \begin{equation} V_{T}({\cal Y}_{T}^{\ast}(t_{k+1}, {\cal Y}(t_{k}), t_{k})) \leq V_{T}({\cal Y}_{0}) -\kappa\int_{0}^{t_{k+1}}\ell({\cal Y}_{T}^{\ast}(t, {\cal Y}_{0}, 0), U_{T}^{\ast}(t, {\cal Y}_{0}, 0)){\rm d}t, k = 0, 1, 2, \cdots. \end{equation} $

上式蕴含着

(6.15) $ \begin{equation} \kappa\int_{0}^{t_{k+1}}\ell({\cal Y}_{T}^{\ast}(t, {\cal Y}_{0}, 0), U_{T}^{\ast}(t, {\cal Y}_{0}, 0)){\rm d}t \leq V_{T}({\cal Y}_{0}). \end{equation} $

令$ k\rightarrow \infty $, 则有

$ \kappa J_{\infty}({\cal Y}_{0}, U_{T}^{\ast}(\cdot)) = \kappa\int_{0}^{\infty}\ell({\cal Y}_{T}^{\ast}(t, {\cal Y}_{0}, 0), U_{T}^{\ast}(t, {\cal Y}_{0}, 0)){\rm d}t \leq V_{T}({\cal Y}_{0}), $

从而(6.11)式成立.

对于任意$ k\in {\Bbb N}^{+} $, 由(6.6)式得

(6.16) $ \begin{eqnarray} && V_{T}({\cal Y}_{T}^{\ast}(t_{k}, {\cal Y}(t_{k-1}), t_{k-1}))-V_{T}({\cal Y}_{T}^{\ast}(t_{k-1}, {\cal Y}(t_{k-1}), t_{k-1}))\\ &\leq&-\kappa\int_{t_{k-1}}^{t_{k}}\ell({\cal Y}_{T}^{\ast}(t, {\cal Y}(t_{k-1}), t_{k-1}), U_{T}^{\ast}(t, {\cal Y}(t_{k-1}), t_{k-1})){\rm d}t\\ &\leq&-\kappa V_{\delta}({\cal Y}_{T}^{\ast}(t_{k-1}, {\cal Y}(t_{k-1}), t_{k-1})). \end{eqnarray} $

又由(6.1)和(6.2)式知, 对于任意$ {\cal Y}_{0}\in{\cal H} $, 我们有

$ V_{\delta}({\cal Y}_{0})\geq\gamma_{2}(\delta)\|{\cal Y}_{0}\|_{{\cal H}}^{2} \geq\frac{\gamma_{2}(\delta)}{\gamma_{1}(T)}V_{T}({\cal Y}_{0}). $

因此

(6.17) $ \begin{equation} V_{\delta}({\cal Y}_{T}^{\ast}(t_{k-1}, {\cal Y}(t_{k-1}), t_{k-1})) \geq\frac{\gamma_{2}(\delta)}{\gamma_{1}(T)} V_{T}({\cal Y}_{T}^{\ast}(t_{k-1}, {\cal Y}(t_{k-1}), t_{k-1})). \end{equation} $

再由(6.16)和(6.17)式得

$ \begin{eqnarray*} &&V_{T}({\cal Y}_{T}^{\ast}(t_{k}, {\cal Y}(t_{k-1}), t_{k-1}))-V_{T}({\cal Y}_{T}^{\ast}(t_{k-1}, {\cal Y}(t_{k-1}), t_{k-1})) \\ &\leq &-\frac{\kappa\gamma_{2}(\delta)}{\gamma_{1}(T)} V_{T}({\cal Y}_{T}^{\ast}(t_{k-1}, {\cal Y}(t_{k-1}), t_{k-1})), \end{eqnarray*} $

即

$ V_{T}({\cal Y}_{T}^{\ast}(t_{k}, {\cal Y}(t_{k-1}), t_{k-1})) \leq(1-\frac{\kappa\gamma_{2}(\delta)}{\gamma_{1}(T)}) V_{T}({\cal Y}_{T}^{\ast}(t_{k-1}, {\cal Y}(t_{k-1}), t_{k-1})). $

记$ \mu = 1-\frac{\kappa\gamma_{2}(\delta)}{\gamma_{1}(T)} $, 注意到$ 0<\gamma_{2}(\delta)\leq\gamma_{1}(\delta)\leq\gamma_{1}(T) $且$ \kappa\in(0, 1) $, 所以, $ \mu\in(0, 1) $, 且对$ k\in{\Bbb N}^{+} $有

(6.18) $ \begin{equation} V_{T}({\cal Y}_{T}^{\ast}(t_{k}, {\cal Y}(t_{k-1}), t_{k-1})) \leq\mu V_{T}({\cal Y}_{T}^{\ast}(t_{k-1}, {\cal Y}(t_{k-1}), t_{k-1})). \end{equation} $

若再记$ \eta = \frac{\ln|\mu|}{\delta} $, 则对$ k\in{\Bbb N}^{+} $有

$ \begin{eqnarray*} \gamma_{2}(T)\|{\cal Y}_{T}^{\ast}(t_{k}, {\cal Y}(t_{k-1}), t_{k-1})\|_{{\cal H}}^{2} &\leq &V_{T}({\cal Y}_{T}^{\ast}(t_{k}, {\cal Y}(t_{k-1}), t_{k-1})) \\ &\leq&\mu V_{T}({\cal Y}_{T}^{\ast}(t_{k-1}, {\cal Y}(t_{k-1}), t_{k-1}))\\ &\leq&\mu^{2} V_{T}({\cal Y}_{T}^{\ast}(t_{k-2}, {\cal Y}(t_{k-2}), t_{k-2}))\leq\cdots \\ &\leq&\mu^{k} V_{T}({\cal Y}_{0}) = {\rm e}^{-\eta t_{k}}V_{T}({\cal Y}_{0})\leq {\rm e}^{-\eta t_{k}}\gamma_{1}(T)\|{\cal Y}_{0}\|_{{\cal H}}^{2}. \end{eqnarray*} $

记$ \gamma = \frac{\gamma_{1}(T)}{\gamma_{2}(T)} $, 则对$ k\in{\Bbb N}^{+} $有

(6.19) $ \begin{eqnarray} \|{\cal Y}_{T}^{\ast}(t_{k}, {\cal Y}(t_{k-1}), t_{k-1})\|_{{\cal H}}^{2} \leq \gamma {\rm e}^{-\eta t_{k}}\|{\cal Y}_{0}\|_{{\cal H}}^{2}. \end{eqnarray} $

故对任意$ t>0 $, 存在$ k\in{\Bbb N}^{+} $, 使得$ t\in[k\delta, (k+1)\delta] = [t_{k}, t_{k+1}] $, 且由(2.4), (6.1)和(6.19)式得

$ \begin{eqnarray*} \|{\cal Y}_{T}^{\ast}(t, {\cal Y}_{0}, 0)\|_{{\cal H}}^{2} &\leq& C(\|{\cal Y}_{T}^{\ast}(t_{k}, {\cal Y}(t_{k}), t_{k})\|_{{\cal H}}^{2} +\int_{t_{k}}^{t_{k}+T}\|U_{T}^{\ast}(t, {\cal Y}(t_{k}), t_{k})\|_{{\cal U}}^{2}{\rm d}t)\\ & \leq& C(\|{\cal Y}_{T}^{\ast}(t_{k}, {\cal Y}(t_{k}), t_{k})\|_{{\cal H}}^{2} +\frac{2}{\beta}V_{T}({\cal Y}_{T}^{\ast}(t_{k}, {\cal Y}(t_{k}), t_{k})))\\ & \leq& C(\|{\cal Y}_{T}^{\ast}(t_{k}, {\cal Y}(t_{k}), t_{k})\|_{{\cal H}}^{2} +\frac{2\gamma_{1}(T)}{\beta}\|{\cal Y}_{T}^{\ast}(t_{k}, {\cal Y}(t_{k}), t_{k})\|_{{\cal H}}^{2})\\ & \leq& C(1+\frac{2\gamma_{1}(T)}{\beta})\gamma {\rm e}^{-\eta t_{k}}\|{\cal Y}_{0}\|_{{\cal H}}^{2}\\ & = &\gamma C(1+\frac{2\gamma_{1}(T)}{\beta})\frac{1}{1-\frac{\kappa\gamma_{2}(\delta)}{\gamma_{1}(T)}} {\rm e}^{-\eta t_{k+1}}\|{\cal Y}_{0}\|_{{\cal H}}^{2}\\ & \leq&\gamma C(1+\frac{2\gamma_{1}(T)}{\beta})\frac{1}{1-\frac{\kappa\gamma_{2}(\delta)}{\gamma_{1}(T)}} {\rm e}^{-\eta t}\|{\cal Y}_{0}\|_{{\cal H}}^{2}. \end{eqnarray*} $

取$ M = \gamma C(1+\frac{2\gamma_{1}(T)}{\beta})\frac{1}{1-\frac{\kappa\gamma_{2}(\delta)}{\gamma_{1}(T)}}\|{\cal Y}_{0}\|_{{\cal H}}^{2} $, 故有$ \|{\cal Y}_{T}^{\ast}(t, {\cal Y}_{0}, 0)\|_{{\cal H}}^{2} \leq M {\rm e}^{-\eta t}. $证毕.