自从1871年法国科学家圣维南提出了圣维南方程组, 圣维南方程组一直在水力及其他工程领域中有着广泛的应用, 许多学者致力于研究它的稳定性和各类渠道边界控制问题^[1-6]. 但由于圣维南方程组是一组非线性的偏微分方程组, 无法得出解析解, 从而对其控制是非常困难的. 为此, 人们做出一系列的假设对其简化并线性化处理, 从而衍生了著名的Hayami模型

(1.1) $ \begin{equation} {\frac{\partial{q}}{\partial{t}}} = {\alpha}{\frac{\partial^2{q}}{\partial{x^2}}}-{\beta}{\frac{\partial{q}}{\partial{x}}}, \end{equation} $

其中, $ x $表示空间位置(m), $ t $表示时间(s), $ q(x, t) $表示流量($ m^3/s $), 正常数$ {\alpha} $和$ {\beta} $分别表示扩散率和波速. 该模型不必测量太多的渠道物理参数, 只需要两个参数来表征水流. 对于模型(1.1) 的控制和稳定问题已经有很多学者进行了研究^[7-9], 其中文献[8] 和[9]通过如下分布式输入来控制波方程(1.1) 的输出稳定性

(1.2) $ \begin{equation} \left\{\begin{array}{ll} { } {\frac{\partial{q}}{\partial{t}}} = {\alpha}{\frac{\partial^2{q}}{\partial{x^2}}}-{\beta}{\frac{\partial{q}}{\partial{x}}}+b(x)U(t), \;\; x\in (0, l), \\ { } q(0, {t}) = {\frac{\partial{q}}{\partial{x}}}(l, {t}) = 0, \;\;\;\; y(t) = {q(l, {t})}. \end{array}\right. \end{equation} $

其中, $ U(t) $代表输入控制, $ b(x)\in L^2(0, l) $是控制项的空间分布函数, $ y(t) $代表输出观测. 文献[8] 本质上是采用比例控制器$ U(t) = {\kappa{q(l, t)}} $去镇定系统, 其中$ {\kappa}\neq 0 $代表反馈增益, 利用算子半群理论和谱分析的方法讨论了系统的适定性, 谱的构成以及系统的指数稳定性. 进而, 考虑到实际中系统的控制器、传感器之间在通信时不可避免地会产生时滞, 因此文献[9] 讨论了输入控制中带有时滞的情形: $ U(t) = {\kappa{q(l, t-\tau)}} $, 其中, $ {\tau}>0 $表示时滞. 结果表明, 只要控制参数满足某个约束条件, 同样可以建立系统的指数稳定性.

另一方面, 对于不稳定ODE系统, 文献[10-12] 采用PDP(position and delayed position) 控制器, 即瞬时位置和时滞位置的线性组合来设计控制器, 取得了一系列指数稳定性的研究成果. 受上述文献启发, 本文考虑PDP反馈

(1.3) $ \begin{equation} U(t) = {k_1{q(l, t)}}+{k_2{q(l, t-\tau)}}, \;\;\;\;k_1, k_2\in {{\Bbb R}} , \;\;\tau>0 \end{equation} $

对系统(1.2) 稳定性的影响.

将(1.3) 式代入(1.2) 式, 并给定初始值$ q_0 $和$ u_0 $, 从而可得如下时滞PDE闭环系统

(2.1) $ \begin{equation} \left\{\begin{array}{ll} { } \frac{\partial{q}}{\partial{t}} = \alpha{\frac{\partial^2{q}}{\partial{x^2}}}-{\beta}{\frac{\partial{q}}{\partial{x}}} +b(x)\big(k_1q(l, t)+k_2 q(l, t-\tau)\big), \;\; (x, t)\in(0, l)\times(0, \infty), \\ { } q(0, {t}) = {\frac{\partial{q}}{\partial{x}}}(l, {t}) = 0, \;\;t>0, \\ q(x, 0) = q_0(x), \;\;x\in(0, l), \\ q(l, \theta) = u_0(\theta), \;\;\theta\in(-\tau, 0). \end{array}\right. \end{equation} $

考虑到一阶双曲PDE初值问题

(2.2) $ \begin{equation} \left\{\begin{array}{ll} { }\tau{\frac{\partial{u}}{\partial{t}}}(\rho, t)+{\frac{\partial{u}}{\partial{\rho}}}(\rho, t) = 0, \;\;(\rho, t)\in(0, 1)\times(0, \infty), \\ u(\rho, 0) = q(l, -\tau\rho) \end{array}\right. \end{equation} $

的解可以表示为

(2.3) $ \begin{equation} u(\rho, t) = q(l, t-\tau\rho). \end{equation} $

于是可将时滞PDE系统(2.1) 改写为如下PDE-PDE耦合系统的形式

(2.4) $ \begin{equation} \left\{\begin{array}{ll} { }\frac{\partial{q}}{\partial{t}} = \alpha{\frac{\partial^2{q}}{\partial{x^2}}}-{\beta}{\frac{\partial{q}}{\partial{x}}} +b(x)\big(k_1u(0, t)+k_2 u(1, t)\big), \;\; (x, t)\in(0, l)\times(0, \infty), \\ { }\tau{\frac{\partial{u}}{\partial{t}}}(\rho, t)+{\frac{\partial{u}}{\partial{\rho}}}(\rho, t) = 0, \;\;(\rho, t)\in(0, 1)\times(0, \infty), \\ { } q(0, {t}) = {\frac{\partial{q}}{\partial{x}}}(l, {t}) = 0, \;\;t>0, \\ q(x, 0) = q_0(x), \;\;x\in(0, l), \\ u(\rho, 0) = u_0(-\tau\rho), \;\;\rho\in(0, 1). \end{array}\right. \end{equation} $

为后续计算方便, 不妨令$ l = 1 $, 定义

(2.5) $ \begin{equation} \phi(t) = (q(\cdot, t), u(\cdot, t)), \;\;\;\;\phi_0 = (q_0, u_0), \end{equation} $

那么系统(2.4) 转化为

(2.6) $ \begin{equation} \left\{\begin{array}{ll} { } \dot{\phi}(t) = \Big(\alpha q''-\beta q'+b(x)\big(k_1u(0, t)+k_2 u(1, t)\big), -\frac{1}{\tau}u'\Big), \\ { } q(0, {t}) = {\frac{\partial{q}}{\partial{x}}}(1, {t}) = 0, \;\;t>0, \\ \phi(0) = \phi_0, \end{array}\right. \end{equation} $

其中, $ ' $表示对变量$ x $或$ \rho $的一阶导数, $ '' $表示对变量$ x $或$ \rho $的二阶导数.

设Hilbert状态空间为

(2.7) $ \begin{equation} {\cal H} = {L^2(0, 1)}\times{L^2(0, 1)}, \end{equation} $

其中的内积定义为: $ \forall \phi_1 = (q_1, u_1), \;\phi_2 = (q_2, u_2)\in {\cal H} $, 有

(2.8) $ \begin{equation} \langle\phi_1, \phi_2\rangle_{{\cal H}} = \int_0^1 q_1(x)\overline{q_2(x)}{\rm d}x+\tau\xi\int_0^1 u_1(\rho)\overline{u_2(\rho)}{\rm d}\rho, \end{equation} $

其中, $ \xi<\beta $是一个正常数.

定义线性算子$ {\cal A}:{\cal H}\longrightarrow{\cal H} $如下

(2.9) $ \begin{equation} \left\{\begin{array}{ll} { } {\cal A}(q, u) = \Big(\alpha q''-\beta q'+b(x)\big(k_1 u(0)+k_2 u(1)\big), -\frac{1}{\tau}u'\Big), \\ { } D({\cal A}) = \{(q, u)\in{\cal H}\mid q\in H^2(0, 1), \;\;u\in H^1(0, 1), \;\;q(0) = q'(1) = 0\}. \end{array}\right. \end{equation} $

那么系统(2.6) 可以写成Hilbert状态空间$ {\cal H} $上的抽象发展方程的形式

(2.10) $ \begin{equation} \left\{\begin{array}{ll} \dot{\phi}(t) = {\cal A}\phi(t), \; t>0, \\ \phi(0) = \phi_0. \end{array}\right. \end{equation} $

定理3.1  由(2.9) 式定义的算子$ {\cal A} $生成$ {\cal H} $上的一个$ C_0 $半群.

证   结合(2.8) 和(2.9) 式以及分部积分公式可得

(3.1) $ \begin{eqnarray} \langle{\cal A}{\phi}, {\phi}\rangle_{{\cal H}} & = &-\alpha\int_{0}^{1}{q'}^2(x){\rm d}x-\frac{\beta}{2}u^2(0)+(k_1u(0)+k_2u(1))\int_{0}^{1}b(x)q(x){\rm d}x \\ &&-\frac{\xi}{2}[u^2(1)-u^2(0)]\\ & = &-\alpha\int_{0}^{1}{q'}^2(x){\rm d}x+\frac{\xi-\beta}{2}u^2(0)- \frac{\xi}{2}u^2(1)+k_1u(0)\int_{0}^{1}b(x)q(x){\rm d}{x}\\ &&+k_2u(1)\int_{0}^{1}b(x)q(x){\rm d}x\\ &\leq&-\alpha\int_{0}^{1}{q'}^2(x){\rm d}x+\frac{\left|{k_1}\right|A_1}{2}{u^2(0)} +\frac{\left|{k_1}\right|\|b\|^2}{2A_1}\int_{0}^{1}{q^2(x)}{\rm d}x +\frac{\left|{k_2}\right|A_2}{2}{u^2(1)}\\ &&+\frac{\left|{k_2}\right|\|b\|^2}{2A_2}\int_{0}^{1}{q^2(x)}{\rm d}x +\frac{\xi-\beta}{2}u^2(0)-\frac{\xi}{2}u^2(1)\\ & = &-\alpha\int_{0}^{1}{q'}^2(x){\rm d}x+\frac{(\left|{k_1}\right|A_2 +\left|{k_2}\right|A_1)\|b\|^2}{2A_1 A_2}\int_{0}^{1}{q^2(x)}{\rm d}x \\ &&+\frac{\xi-\beta+\left|{k_1}\right|A_1}{2}u^2(0)+\frac{\left|{k_2}\right|A_2-\xi}{2}u^2(1), \end{eqnarray} $

其中, $ A_1 $和$ A_2 $为正常数, 且满足$ A_1 = \frac{\beta-\xi}{\left|{k_1}\right|}, \;\; A_2 = \frac{\xi}{\left|{k_2}\right|} $, 我们可以选择$ \xi<\beta $, 于是可得

(3.2) $ \begin{eqnarray} Re\langle{\cal A}{\phi}, {\phi}\rangle_{{\cal H}}&\leq&-\alpha\int_{0}^{1}{q'}^2(x){\rm d}x+\frac{({\left|{k_{1}}\right|}\cdot\frac{\xi}{\left|k_{2}\right|}+ \left|k_{2}\right|\cdot\frac{\beta-\xi}{\left|k_{1}\right|})\|b\|^2}{2\cdot\frac{\xi}{\left|{k_{2}}\right|}\cdot\frac{\beta-\xi}{\left|{k_{1}}\right|}}\int_{0}^{1}q^2(x){\rm d}x\\ & = &-\alpha\int_{0}^{1}{q'}^2(x){\rm d}x+\frac{(\xi k_1^2+(\beta-\xi)k_2^2){\|b\|^2}}{2\xi\cdot(\beta-\xi)}\int_{0}^{1}q^2(x){\rm d}x\\ &\leq&\frac{(\xi k_1^2+(\beta-\xi)k_2^2){\|b\|^2}}{2\xi\cdot(\beta-\xi)}\int_{0}^{1}q^2(x){\rm d}x. \end{eqnarray} $

显然, $ {\cal A} $是闭稠定的线性算子. 通过计算可得, $ {\cal A} $的伴随算子$ {\cal A}^{\ast} $定义如下

(3.3) $ \begin{equation} \left\{\begin{array}{ll} { } {\cal A}^{\ast}(q, u) = \Big(\alpha q''+\beta q', \frac{1}{\tau}u'\Big), \\ { } D({\cal A}^{\ast}) = \Big\{(q, u)\in{\cal H}\mid q\in H^2(0, 1), \;\;u\in H^1(0, 1), \;\;q(0) = 0, \;\\ { } \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\alpha q'(1)+\beta q(1) = 0, \;\;u(0)+\frac{k_1}{k_2}u(1) = 0, \;\;\xi u(1) = k_{2}\int_{0}^{1}b(x)q(x){\rm d}x\Big\}. \end{array}\right. \end{equation} $

设$ (q, u)\in D({\cal A}^{\ast}) $, 结合(2.8) 和(3.3) 式以及分部积分公式可得

(3.4) $ \begin{eqnarray} \langle{\cal A}^{\ast}(q, u), (q, u)\rangle_{{\cal H}} & = &\alpha\int_{0}^{1}q''q{\rm d}x+\beta\int_{0}^{1}q'q{\rm d}x+\tau\xi\int_{0}^{1}\frac{u'}{\tau}ud{\rho}\\ & = &\alpha q'(1)q(1)-\alpha\int_{0}^{1}{q'}^{2}(x){\rm d}x+\frac{\beta}{2}q^{2}(1)+\frac{\xi}{2}[u^2(1)-u^2(0)]\\ & = &\alpha q'(1)q(1)-\alpha\int_{0}^{1}{q'}^{2}(x){\rm d}x+\frac{\beta}{2}q^{2}(1) +\frac{\xi}{2}\cdot\frac{k_{2}^2-k_1^2}{\xi^{2}}\Big(\int_{0}^{1}b(x)q(x){\rm d}x\Big)^2\\ & = &-\frac{\beta}{2}q^2(1)-\alpha\int_{0}^{1}{q'}^{2}(x){\rm d}x+\frac{k_{2}^2-k_1^2}{2\xi} \Big(\int_{0}^{1}b(x)q(x){\rm d}x\Big)^2\\ &\leq&\frac{(k_{2}^2-k_1^2)||b||^2}{2\xi}\int_{0}^{1}q^2(x){\rm d}x. \end{eqnarray} $

令

(3.5) $ \begin{equation} {\eta}_{1} = \frac{(\xi k_1^2+(\beta-\xi)k_2^2){\|b\|^2}}{2\xi\cdot(\beta-\xi)}, \;\;\;\; {\eta}_{2} = \frac{(k_{2}^2-k_1^2)||b||^2}{2\xi}, \end{equation} $

取$ \eta = \max\{{\eta}_{1}, {\eta}_{2}\} $, 则

$ \begin{eqnarray*} \langle{\cal A}(q, u), (q, u)\rangle_{{\cal H}} &\leq&\eta\int_{0}^{1}q^2(x){\rm d}x, \;\;\forall (q, u)\in D({\cal A}), \\ \langle{\cal A}^{\ast}(q, u), (q, u)\rangle_{{\cal H}} &\leq&\eta\int_{0}^{1}q^2(x){\rm d}x, \;\;\forall (q, u)\in D({\cal A}^{\ast}). \end{eqnarray*} $

由文献[9] 中的定理2得, 算子$ {\cal A} $生成$ {\cal H} $上的一个$ C_{0} $半群$ S(t) $满足$ \|S(t)\|_{{\cal \psi}({\cal H})}\leq {\rm e}^{\eta t} $.

在这一部分, 我们致力于构造合适的Lyapunov函数并寻求调控参数$ k_1 $, $ k_2 $的取值范围, 从而使得闭环系统(2.1) 达到指数稳定.

构造系统(2.1) 的Lyapunov函数为

(4.1) $ \begin{equation} E(t) = {E}_1(t)+{E}_2(t), \end{equation} $

其中

(4.2) $ \begin{equation} {E}_1(t) = \frac{1}{2}\Big(\int_{0}^{1}{q^2(x, t)}{\rm d}x+{\tau}{\xi}\int_{0}^{1}{q^2(1, t-\tau\rho)}{\rm d}{\rho}\Big), \end{equation} $

(4.3) $ \begin{equation} E_2(t) = \nu\tau\int_{0}^{1}{\rm e}^{-2\gamma\tau\rho}q^2(1, t-\tau\rho){\rm d}{\rho}, \end{equation} $

其中, $ \nu $是待定的正常数, $ \gamma $是任意的正常数. 显然, 存在两个正常数$ L_1 $和$ L_2 $, 使得对于任意的$ t>0 $, 有

(4.4) $ \begin{equation} {L_1}{{E}_1(t)}\leq E(t)\leq{L_2}{{E}_1(t)}. \end{equation} $

定理4.1  对任给的正常数$ \gamma $以及系统参数$ \alpha, \beta, \xi, \tau, b(x) $, 如果调控参数$ k_1 $, $ k_2 $满足

(4.5) $ \begin{equation} |k_2|\beta-(|k_1|+|k_2|)\xi>0 \end{equation} $

和

(4.6) $ \begin{equation} \frac{\alpha\pi^2[\xi+(\beta-\xi){\rm e}^{-2\gamma\tau}]}{2||b||^2}>(|k_1|+|k_2|)(|k_2|+|k_1|{\rm e}^{-2\gamma\tau}). \end{equation} $

则闭环系统(2.1) 是指数稳定的, 即$ \exists\;\zeta>0 $使得Lyapunov能量函数满足

$ \begin{eqnarray*} \dot{E}(t)\leq -\zeta E(t), \end{eqnarray*} $

其中, $ \zeta = \frac{\varepsilon}{L_2} $, $ L_2 $和$ \varepsilon $分别由(4.4) 和(4.15) 式给出.

证   结合系统(2.1) 和(2.4), 可计算得$ \dot{E}_1(t) $和$ \dot{E}_2(t) $如下

(4.7) $ \begin{eqnarray} \dot{E}_1(t)& = &\int_{0}^{1}{q(x, t)\frac{\partial{q(x, t)}}{\partial{t}}}{\rm d}x+{\tau}{\xi} \int_{0}^{1}{u(\rho, t)\frac{\partial{u(\rho, t)}}{\partial{t}}}{\rm d}{\rho}\\ & = &-\alpha\int_{0}^{1}{q_x}^2(x, t){\rm d}x-\frac{\beta}{2}q^2(1, t)+k_1q(1, t)\int_{0}^{1}q(x, t)b(x){\rm d}x\\ &&+k_2q(1, t-\tau)\int_{0}^{1}q(x, t)b(x){\rm d}x-\frac{\xi}{2}q^2(1, t-\tau)+\frac{\xi}{2}q^2(1, t), \end{eqnarray} $

(4.8) $ \begin{eqnarray} \dot{E}_2(t)& = &2\nu\tau\int_{0}^{1}{\rm e}^{-2\gamma\tau\rho}q(1, t-\tau\rho)q_t(1, t-\tau\rho){\rm d}{\rho}\\ & = &2\nu\tau\int_{0}^{1}{\rm e}^{-2\gamma\tau\rho}q(1, t-\tau\rho)(-\frac{1}{\tau})q_{\rho}(1, t-\tau\rho){\rm d}{\rho}\\ & = &-2\nu\int_{0}^{1}{\rm e}^{-2\gamma\tau\rho}d\big(\frac{1}{2}q^2(1, t-\tau\rho)\big)\\ & = &-\nu{\rm e}^{-2\gamma\tau}q^2(1, t-\tau)+{\nu}q^2(1, t)-2\nu\gamma\tau\int_{0}^{1}{\rm e}^{-2\gamma\tau\rho}q^2(1, t-\tau\rho){\rm d}{\rho}. \end{eqnarray} $

于是, 利用Cauchy-Schwartz不等式和Young's不等式可得

(4.9) $ \begin{eqnarray} \dot{E}(t)& = &\dot{E}_1(t)+\dot{E}_2(t)\\ &\leq&{-\alpha}\int_{0}^{1}{q_x}^{2}(x, t){\rm d}x-\frac{\beta}{2}q^2(1, t) +\frac{\left|{k_1}\right|A}{2}q^2(1, t)+\frac{\left|{k_1}\right|\|b\|^2}{2A}\int_{0}^{1}q^2(x, t){\rm d}x\\ &&+\frac{\left|{k_2}\right|A}{2}{{q^2(1, t-\tau)}}+\frac{\left|{k_2}\right|\|b\|^2}{2A}\int_{0}^{1}q^2(x, t){\rm d}x-\frac{\xi}{2}q^2(1, t-\tau)+\frac{\xi}{2}q^2(1, t)\\ &&-\nu{\rm e}^{-2\gamma\tau}q^2(1, t-\tau)+{\nu}q^2(1, t)-2\nu\gamma\tau\int_{0}^{1}{\rm e}^{-2\gamma\tau\rho}q^2(1, t-\tau\rho){\rm d}{\rho}. \end{eqnarray} $

进而, 根据Wirtinger's不等式

(4.10) $ \begin{equation} \int_{0}^{1}\varphi^2(x, t){\rm d}x\leq{\frac{4}{\pi^2}}\int_{0}^{1}{\varphi_x}^{2}(x, t){\rm d}x, \;\;\forall{\varphi}\in\{{f}\in{H^1(0, 1)}:{f}(0) = 0\}. \end{equation} $

整理可得

(4.11) $ \begin{eqnarray} \dot{E}(t) &\leq&\big(-\frac{\alpha\pi^2}{4}+\frac{(\left|{k_1}\right|+|k_2|)\|b\|^2}{2A}\big)\int_{0}^{1}q^{2}(x, t){\rm d}x+\big(-\frac{\beta}{2}+\frac{\xi}{2} +\frac{\left|{k_1}\right|A}{2}+\nu\big)q^2(1, t)\\ &&+\big(\frac{\left|{k_2}\right|A}{2}-\frac{\xi}{2}-\nu {\rm e}^{-2\gamma\tau}\big)q^2(1, t-\tau) -2\nu\gamma\tau{\rm e}^{-2\gamma\tau}\int_{0}^{1}q^2(1, t-\tau\rho){\rm d}{\rho}. \end{eqnarray} $

选取参数$ A, \nu $使得

(4.12) $ \begin{equation} -\frac{\beta}{2}+\frac{\xi}{2}+\frac{\left|{k_1}\right|A}{2}+\nu = 0, \;\; \frac{\left|{k_2}\right|A}{2}-\frac{\xi}{2}-\nu {\rm e}^{-2\gamma\tau} = 0, \end{equation} $

即

(4.13) $ \begin{equation} A = \frac{\xi+(\beta-\xi){\rm e}^{-2\gamma\tau}}{|k_2|+|k_1|{\rm e}^{-2\gamma\tau}}, \;\; \nu = \frac{|k_2|\beta-(|k_1|+|k_2|)\xi}{2(|k_2|+|k_1|{\rm e}^{-2\gamma\tau})}, \end{equation} $

于是有

(4.14) $ \begin{eqnarray} \dot{E}(t) &\leq&\big(-\frac{\alpha\pi^2}{4}+\frac{(|k_1|+|k_2|)\|b\|^2}{2A}\big)\int_{0}^{1}q^{2}(x, t){\rm d}x -2\nu\gamma\tau{\rm e}^{-2\gamma\tau}\int_{0}^{1}q^2(1, t-\tau\rho){\rm d}{\rho}\\ & = &\frac{1}{2}(-\varepsilon_1 \int_{0}^{1}q^{2}(x, t){\rm d}x-\varepsilon_2\tau\xi\int_{0}^{1}q^2(1, t-\tau\rho){\rm d}{\rho})\\ &\leq& -\varepsilon E_1(t), \end{eqnarray} $

其中

(4.15) $ \begin{equation} \varepsilon = \min\{\varepsilon_1, \varepsilon_2\}, \;\; \varepsilon_1 = \frac{\alpha\pi^2}{2}-\frac{(|k_1|+|k_2|)\|b\|^2}{A}, \;\; \varepsilon_2 = \frac{4\nu\gamma{\rm e}^{-2\gamma\tau}}{\xi}. \end{equation} $

参数$ A, \nu $由(4.13) 式给出. 结合条件(4.5) 和(4.6) 可知, $ \varepsilon>0. $于是, 结合(4.4)和(4.14) 式可得: 存在正常数$ \zeta = \frac{\varepsilon}{L_2}>0 $使得$ \dot{E}(t)\leq-\zeta{E}(t) $, 因此系统(2.1) 指数稳定.

注4.1  在对(4.9) 式进行放缩时, 可运用不同的方法, 从而得到如下定理.

定理4.2  对任给的常数$ c_1>0 $, $ c_2>1 $, 如果调控参数$ k_1 $, $ k_2 $满足

(4.16) $ \begin{equation} (2c_1+1){k_1}^2+c_2{k_2}^2+(2c_1+c_2+1)|k_1k_2|<\frac{{\alpha}^2{\pi}^2}{4\|b\|^2}, \end{equation} $

则闭环系统(2.1) 是指数稳定的, 即$ \exists\; \hat{\zeta}>0 $使得Lyapunov能量函数满足

$ \dot{E}(t)\leq -\hat{\zeta} E(t), $

其中, $ \hat{\zeta} = \frac{\hat{\varepsilon}}{L_2} $, $ L_2 $和$ \hat{\varepsilon} $分别由(4.4) 和(4.24) 式给出.

证   由$ q(1, t) = \int_{0}^{1}q_x(x, t){\rm d}x $可得

(4.17) $ \begin{equation} q^2(1, t)\leq{\int_{0}^{1}{q_x}^2(x, t){\rm d}x}. \end{equation} $

于是结合(4.9) 式、(4.10) 式和(4.17) 式, 整理可得

(4.18) $ \begin{eqnarray} \dot{E}(t)&\leq&\frac{\pi^2 |k_1|A^2+(\pi^2\xi+2\pi^2\nu -2\pi^2\alpha)A+4||b||^2(|k_1|+|k_2|)}{2\pi^2A}\int_{0}^{1}{q_x}^2(x, t){\rm d}x\\ &&-{\frac{\beta}{2}}q^2(1, t)+(\frac{{\left|{k_2}\right|}A-\xi}{2}-\nu{\rm e}^{-2\gamma\tau})q^2(1, t-\tau) -2\nu\gamma\tau{\rm e}^{-2\gamma\tau}\int_{0}^{1}q^2(1, t-\tau){\rm d}{\rho}.{\qquad} \end{eqnarray} $

取$ \nu = c_1{\left|{k_1}\right|}A $, $ \xi = c_2{\left|{k_2}\right|}A $, 其中, $ c_1>0 $, $ c_2>1 $, 于是(4.17) 式转化为

(4.19) $ \begin{eqnarray} \dot{E}(t)&\leq&\frac{\pi^2\big[(2c_1+1){\left|{k_1}\right|}+c_2{\left|{k_2}\right|}\big]A^2-2\pi^2\alpha{A} +4\|b\|^2({\left|{k_1}\right|}+{\left|{k_2}\right|})}{2\pi^2A}\int_{0}^{1}{q_x}^{2}(x, t){\rm d}x\\ &&-2c_1{\left|{k_1}\right|}A\gamma\tau{\rm e}^{-2\gamma\tau}\int_{0}^{1}q^2(1, t-\tau){\rm d}{\rho}. \end{eqnarray} $

令

(4.20) $ \begin{equation} g(A) = \pi^2\big[(2c_1+1){\left|{k_1}\right|}+c_2{\left|{k_2}\right|}\big]A^2-2\pi^2\alpha{A} +4\|b\|^2({\left|{k_1}\right|}+{\left|{k_2}\right|}), \end{equation} $

这是一个关于$ A $的二次多项式, 根据条件(4.16) 可知, 其判别式

(4.21) $ \begin{equation} \Delta = 4\pi^4\alpha^2-16\pi^2||b||^2\big[(2c_1+1){\left|{k_1}\right|}+c_2{\left|{k_2}\right|}\big](|k_1|+|k_2|)>0. \end{equation} $

那么当正常数$ A $满足

(4.22) $ \begin{equation} A\in\Big(\frac{2\pi^2\alpha-\sqrt{\Delta}}{2\pi^2 \big[(2c_1+1){\left|{k_1}\right|}+c_2{\left|{k_2}\right|}\big]}, \frac{2\pi^2\alpha+\sqrt{\Delta}}{2\pi^2 \big[(2c_1+1){\left|{k_1}\right|}+c_2{\left|{k_2}\right|}\big]}\Big) \end{equation} $

时, 有$ g(A)<0. $于是结合(4.10) 式可将(4.19) 式改写为

(4.23) $ \begin{eqnarray} \dot{E}(t)&\leq&\frac{\pi^2\big[(2c_1+1){\left|{k_1}\right|}+c_2{\left|{k_2}\right|}\big]A^2-2\pi^2\alpha{A} +4\|b\|^2({\left|{k_1}\right|}+{\left|{k_2}\right|})}{8A}\int_{0}^{1}q^2(x, t){\rm d}x\\ &&-2c_1{\left|{k_1}\right|}A\gamma\tau{\rm e}^{-2\gamma\tau}\int_{0}^{1}q^2(1, t-\tau){\rm d}{\rho}. \end{eqnarray} $

取

(4.24) $ \begin{eqnarray} \hat{\varepsilon}_1& = &-\frac{\pi^2\big[(2c_1+1){\left|{k_1}\right|}+c_2{\left|{k_2}\right|}\big]A^2-2\pi^2\alpha{A} +4\|b\|^2({\left|{k_1}\right|}+{\left|{k_2}\right|})}{4A}, \\ \hat{\varepsilon}_2& = &\frac{4c_1{\left|{k_1}\right|}\gamma{A}{\rm e}^{-2\gamma\tau}}{\xi}, \;\;\;\; \hat{\varepsilon} = \min\{\hat{\varepsilon}_1, \hat{\varepsilon}_2\}>0, \end{eqnarray} $

则得

(4.25) $ \begin{equation} \dot{E}(t)\leq -\hat{\varepsilon}{E}_1(t). \end{equation} $

进而结合(4.4) 式可得$ \dot{E}(t)\leq -\frac{\hat{\varepsilon}}{L_2} E(t), $因此系统(2.1) 指数稳定.

注4.2  当$ k_2 = 0 $时, 控制项$ U(t) = {k_1{q(1, t)}}+{k_2{q(1, t-\tau)}} $变为单纯状态反馈$ U(t) = {k_1{q(1, t)}} $, 稳定性条件(4.16) 式退化为$ \left|{k_1}\right|<\frac{\pi\alpha}{2\|b\|\sqrt{2c_1+1}}. $与文献[8]得到的参数结果$ \left|{\kappa}\right|<\frac{\sqrt{2}\alpha}{\|b\|} $相比较, 可得: 当$ 0<c_1<\frac{\pi^2-8}{16} $时

$ \frac{\pi\alpha}{2\|b\|\sqrt{2c_1+1}}>\frac{\sqrt{2}\alpha}{\|b\|}. $

显然, 本文得到的参数范围更广.

注4.3  当$ k_1 = 0 $时, 控制$ U(t) = {k_1{q(1, t)}}+{k_2{q(1, t-\tau)}} $变为纯时滞状态反馈$ U(t) = {k_2{q(1, t-\tau)}} $, 稳定性条件(4.16) 式退化为$ \left|{k_2}\right|<\frac{\pi\alpha}{2\|b\|\sqrt{c_2}}. $与文献[9] 得到的参数结果$ \left|{\kappa}\right|<\frac{\pi\alpha}{2\|b\|\sqrt{2c_1+c_2}} $作比较, 显然

$ \frac{\pi\alpha}{2\|b\|\sqrt{c_2}}>\frac{\pi\alpha}{2\|b\|\sqrt{2c_1+c_2}}, $

即本文得到的参数范围更广.

注4.4  在定理$ 4.1 $中, 调控参数$ k_1 $和$ k_2 $满足的不等式组(4.5)–(4.6)有解. 例如: 令

$ \begin{eqnarray*} \gamma = 0.2, \; \alpha = 1, \;\beta = 2, \; \xi = 1, \; \tau = 0.4, \; b(x) = x/\pi-1, \end{eqnarray*} $

我们可以取到$ k_1 = 0.5, \; k_2 = 0.9 $满足不等式组(4.5)–(4.6). 在定理$ 4.2 $中, 参数$ \alpha, \; b(x) $取法同前, $ c_1 = 0.5, \; c_2 = 1.2 $, 则可取$ k_1 = 0.2, \; k_2 = 0.3 $满足不等式$ (4.16) $.

本文选取了不同于文献[8] 和文献[9] 的PDP反馈控制, 描述了所得时滞PDE闭环系统的适定性, 并对系统进行了稳定性分析, 通过Lyapunov方法得出了系统稳定参数所满足的条件, 通过与文献[8] 和文献[9] 的比较结果可知, 我们得到的参数范围更为广泛.