## Exponential Tracking Control for a Star-Shaped Network of Euler-Bernoulli Beams with Unknown Internal Disturbance

Zhang Yaxuan,1, Xu Genqi2, Guo Yanni1

 基金资助: 国家自然科学基金.  61503385国家自然科学基金.  11705279中央高校基本科研业务费.  3122018L004天津市教委科研计划项目.  2018KJ253

 Fund supported: the NSFC.  61503385the NSFC.  11705279the Fundamental Research Funds for the Central Universities.  3122018L004the Scientific Research Project of Tianjin Municipal Education Commission.  2018KJ253

Abstract

In this paper, the exponential tracking control for a star-shaped network of Eulerbernoulli beams with unknown internal disturbance is studied. The problem is transformed into the stabilization of the error system between the objective network and the active network. The idea of sliding-mode control is used to design a nonlinear feedback control law. The solvability of the error system is obtained via monotone operator theory under an appropriately chosen space norm. The error system is proved to be exponentially stabilized at any decay rate by a suitable Lyapunov functional. So the objective network can track the active network exponentially at any designated rate.

Keywords： Tracking control ; Unknown internal disturbance ; Star-shaped network of Eulerbernoulli beams ; Sliding-mode control ; Exponential stabilization

Zhang Yaxuan, Xu Genqi, Guo Yanni. Exponential Tracking Control for a Star-Shaped Network of Euler-Bernoulli Beams with Unknown Internal Disturbance. Acta Mathematica Scientia[J], 2019, 39(3): 596-610 doi:

## 1 引言

$$$\label{1} \left\{ \begin{array}{l} w_{j, tt}(x, t)+a_j^2w_{j, xxxx}(x, t)=d_j(x, t), x\in(0, 1), t>0, \\ w_1(0, t)=w_2(0, t)=w_3(0, t), \\ a_1^2w_{1, xxx}(0, t)+a_2^2w_{2, xxx}(0, t)+a_3^2w_{3, xxx}(0, t)=0, \\ w_{j, xx}(0, t)=0, \\ w_{j, xx}(1, t)=0, \\ w_{j, xxx}(1, t)=0, \\ w_j(x, 0)=w_j^0(x), w_{j, t}(x, 0)=w_j^1(x), j=1, 2, 3, \end{array} \right.$$$

$$$\label{2}\left\{\begin{array}{l}\widetilde w_{j, tt}(x, t)+a_j^2\widetilde w_{j, xxxx}(x, t)=0, x\in(0, 1), t>0, \\\widetilde w_1(0, t)=\widetilde w_2(0, t)=\widetilde w_3(0, t), \\a_1^2\widetilde w_{1, xxx}(0, t)+a_2^2\widetilde w_{2, xxx}(0, t)+a_3^2\widetilde w_{3, xxx}(0, t)=0, \\\widetilde w_{j, xx}(0, t)=0, \\\widetilde w_{j}(1, t)=0, \\\widetilde w_{j, x}(1, t)=0, \\\widetilde w_j(x, 0)=\widetilde w_j^0(x), \widetildew_{j, t}(x, 0)=\widetilde w_j^1(x), j=1, 2, 3.\end{array}\right.$$$

$$$\label{3} \left\{\begin{array}{l} a_j^2w_{j, xx}(1, t)=-\alpha_jw_{j, xt}(1, t)+a_j^2\widetilde w_{j, xx}(1, t), \\ a_j^2w_{j, xxx}(1, t)=-\beta_jw_{j, t}(1, t)+a_j^2\widetilde w_{j, xxx}(1, t), j=1, 2, 3, \end{array}\right.$$$

$$$\label{4} \left\{ \begin{array}{l} e_{j, tt}(x, t)+a_j^2e_{j, xxxx}(x, t)=0, x\in(0, 1), t>0, \\ e_1(0, t)=e_2(0, t)=e_3(0, t), \\ a_1^2e_{1, xxx}(0, t)+a_2^2e_{2, xxx}(0, t)+a_3^2e_{3, xxx}(0, t)=0, \\ e_{j, xx}(0, t)=0, j=1, 2, 3, \\ a_j^2e_{j, xx}(1, t)=-\alpha_je_{j, xt}(1, t), \\ a_j^2e_{j, xxx}(1, t)=-\beta_je_{j, t}(1, t), \\ e_j(x, 0)=w_j^0(x)-\widetilde w_j^0(x):=e_j^0(x), \\ e_{j, t}(x, 0)=w_j^1(x)-\widetilde w_j^1(x):=e_j^1(x), j=1, 2, 3. \end{array} \right.$$$

$$$\label{5}u_j(x, t)=-2\rho_j e_{j, t}(x, t)-(\rho_j^2+\rho_j)e_j(x, t)-M_j\frac{e_{j, t}(x, t)+\rho_j e_j(x, t)}{\| e_{j, t}(t)+\rho_je_j(t)\| _{L^2}},$$$

$$$\label{6}\left\{ \begin{array}{l}w_{j, xx}(1, t)=\widetilde w_{j, xx}(1, t), \\w_{j, xxx}(1, t)=\widetilde w_{j, xxx}(1, t), j=1, 2, 3.\end{array}\right.$$$

$$$\label{7}\left\{\begin{array}{l} e_{j, tt}(x, t)+a_j^2e_{j, xxxx}(x, t)=-2\rho_je_{j, t}(x, t)-(\rho_j^2+\rho_j) e_j(x, t)-M_j\frac{e_{j, t}(x, t)+\rho_je_j(x, t)}{\| e_{j, t}(t)+\rho_je_j(t)\| _{L^2}}\\\;\ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad+d_j(x, t), x\in(0, 1), t>0, \\e_1(0, t)=e_2(0, t)=e_3(0, t), \\a_1^2e_{1, xxx}(0, t)+a_2^2e_{2, xxx}(0, t)+a_3^2e_{3, xxx}(0, t)=0, \\e_{j, xx}(0, t)=0, \\e_{j, xx}(1, t)=e_{j, xxx}(1, t)=0, \\e_j(x, 0)=e_j^0(x), e_{j, t}(x, 0)=e_j^1(x), j=1, 2, 3.\end{array}\right.$$$

$$$\label{9}E(t)=\frac{1}{2}\sum\limits_{j=1}^3\int_0^1[a_j^2|e_{j, xx}(x, t)|^2+|e_{j, t}(x, t)|^2]{\rm d}x,$$$

## 2 规范化与有用的引理

${\cal H}$中定义算子${\cal A}$

$$$\label{3-1} D({\cal A})=\left\{(Y, Z)\in{\cal H}\left| \begin{array}{l} Y\in\prod\limits_{j=1}^3H^4[0, 1], Z\in H_e^2[0, 1]\\ a_1^2y'''_1(0)+a_2^2y'''_2(0)+a_3^2y'''_3(0)=0\\ y''_j(0)=y''_j(1)=y'''_j(1)=0, j=1, 2, 3 \end{array} \right.\right\},$$$

$$$\label{3-2-1} {\cal A}(Y, Z)=\left(\{z_j-\rho_jy_j\}_{j=1}^3, \{-a_j^2y^{(4)}_j-\rho_jy_j-\rho_jz_j-M_j\frac{z_j}{\| z_j\| _{L^2}}\}_{j=1}^3\right).$$$

${\mathbbY}(t)=(\{e_j(x, t)\}_{j=1}^3, \{e_{j, t}(x, t)+\rho_je_j(x, t)\}_{j=1}^3)\in{\mathcal H}$,则系统(1.7)可写成如下等价的抽象发展方程

$$$\label{3-2} \left\{\begin{array}{l} \frac{\rm d}{{\rm d}t}{\Bbb Y}(t)={\cal A}{\Bbb Y}(t)+{\cal D}(t), t>0, \\ {\Bbb Y}(0)={\Bbb Y}_0, \end{array}\right.$$$

$$$\label{E5.1} \left\{\begin{array}{l} \frac{\rm d}{{\rm d}t}y(t)+Ay(t)\ni f(t), t\in(0, T), \\ y(0)=y_0, \end{array}\right.$$$

## 3 可解性

对任意的$(Y, Z)=(\{y_j\}_{j=1}^3, \{z_j\}_{j=1}^3), (F, G)=(\{f_j\}_{j=1}^3, \{g_j\}_{j=1}^3)\in D({\cal A})$,成立

对任意的$(F, G)\in{\cal H}$,考虑如下方程

$$$\label{E3.1} (I-{\cal A})(Y, Z)=(F, G),$$$

## 4 指数稳定性

$\begin{eqnarray}V(t)&=&\frac{1}{2}\sum\limits_{j=1}^3\int^1_0[a_j^2|e_{j, xx}(x, t)|^2+|e_{j, t}(x, t)+\rho_je_j(x, t)|^2+\rho_j|e_j(x, t)|^2]{\rm d}x\cr&=&\frac{1}{2}\sum\limits_{j=1}^3\left[a_j^2\| e_{j, xx}(t)\| _{L^2}^2+\| e_{j, t}(t)+\rho_je_j(t)\| _{L^2}^2+\rho_j\| e_j(t)\| _{L^2}^2\right].\end{eqnarray}$

## 5 小结

1)众所周知,边界控制仅能够对不含扰动的网络做到指数镇定.由于边界控制比内部(分布)控制更切合实际,所以能否(如何)将本文控制方案中的前两项替换成边界控制,同时保证指数跟踪的效果,是有意义的研究.这是作者下一步工作的内容之一.

2)从具体证明可以看出,本文的SMC控制设计方案和指数跟踪结论对由$n$个方程构成的一般星形网络也成立.即对如下目标网络

$$$\left\{ \begin{array}{l} w_{j, tt}(x, t)+a_j^2w_{j, xxxx}(x, t)=d_j(x, t), x\in(0, 1), t>0, \\ w_1(0, t)=w_2(0, t)=\cdots=w_n(0, t), \\ \sum\limits_{j=1}^na_j^2w_{j, xxx}(0, t)=0, \\ w_{j, xx}(0, t)=0, \\ w_{j, xx}(1, t)=0, \\ w_{j, xxx}(1, t)=0, \\ w_j(x, 0)=w_j^0(x), w_{j, t}(x, 0)=w_j^1(x), j=1, 2, \cdots, n, \end{array} \right.$$$

$$$\left\{ \begin{array}{l} \widetilde w_{j, tt}(x, t)+a_j^2\widetilde w_{j, xxxx}(x, t)=0, x\in(0, 1), t>0, \\ \widetilde w_1(0, t)=\widetilde w_2(0, t)=\cdots=\widetilde w_n(0, t), \\ \sum\limits_{j=1}^na_j^2\widetilde w_{j, xxx}(0, t)=0, \\ \widetilde w_{j, xx}(0, t)=0, \\ \widetilde w_{j}(1, t)=0, \\ \widetilde w_{j, x}(1, t)=0, \\ \widetilde w_j(x, 0)=\widetilde w_j^0(x), \widetilde w_{j, t}(x, 0)=\widetilde w_j^1(x), j=1, 2, \cdots, n. \end{array} \right.$$$

$e_j(x, t)=w_j(x, t)-\widetilde w_j(x, t), j=1, 2, \cdots, n$表示误差,取内部控制为

$$$u_j(x, t)=-2\rho_j e_{j, t}(x, t)-(\rho_j^2+\rho_j) e_j(x, t)-M_j\frac{e_{j, t}(x, t)+\rho_j e_j(x, t)}{\| e_{j, t}(t)+\rho_j e_j(t)\| _{L^2}}, j=1, 2, \cdots, n,$$$

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