## The PDP Feedback Control and Stability Analysis of a Diffusive Wave Equation

 基金资助: 国家自然科学基金青年基金.  61603351山西省自然科学基金面上项目.  201801D121027国家自然科学基金青年基金.  201701-D221121中北大学第十七届研究生科技立项.  20201747

 Fund supported: the NSFC.  61603351the NSF of Shanxi Province.  201801D121027the NSF of Shanxi Province.  201701-D221121the 17th Postgraduate Science and Technology Project of The North University of China.  20201747 Abstract

In this paper, a position and delayed position (PDP) feedback controller is established for the Hayami diffusive wave equation. The well-posedness of the closed system is studied firstly, and then, by the method of Lyapunov function, the exponential stability is obtained. Finally, compared with the previous results, the range of the control parameters is expanded in this paper, which shows the efficiency and feasibility of PDP feedback.

Keywords： Diffusive wave equation ; PDP feedback ; Lyapunov function ; Exponential stability

Fan Dongxia, Zhao Dongxia, Shi Na, Wang Tingting. The PDP Feedback Control and Stability Analysis of a Diffusive Wave Equation. Acta Mathematica Scientia[J], 2021, 41(4): 1088-1096 doi:

## 1 引言

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

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

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

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

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

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

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

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

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

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

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

## 3 系统(2.10) 的适定性

结合(2.8) 和(2.9) 式以及分部积分公式可得

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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