# Stabilization for a joint string equation with input disturbance

Stabilization for a joint string equation with input disturbance Abstract The stabilization problem for a joint string subject to pointwise input disturbance is concerned in this article. First, a method like sliding mode control is adopted to resist the disturbance. Observation blind point which leads to zero output is found in this non-linear system. Second, the existence and uniqueness of the solution for this non-linear-system is obtained based on the extension of Lions–Lax–Milgram Theorem and the developed Variational Methods. Finally, the system’s asymptotic behaviour is shown by the LaSalle’s Invariance Principle-like. 1. Introduction In the past 30 years, there has been a steady increase in the study of PDE systems, since it can be widely used in engineering. The stability and stabilization by feedback of joint wave equations in bounded domains are important topics which have drawn much attention. A number of literatures can be found in Chen et al. (1987), Liu (1994) and Ammari et al. (2000). Among these works, the controller is designed in an ideal environment, that’s to say, there is not disturbance. However, in practice, the disturbance may occur in both the boundary and the internal of the system. Thus, it is of great importance to design the controller as well as to attenuate the uncertainty. Sliding mode control, which is first used in infinite dimensional systems by Orlov & Utkin (1998), has been applied to some PDEs in Guo & Jin (2013). Active disturbance rejection control is another control strategy that has been successfully used to reject the disturbance in wave, beam and Schrödinger equations (Guo & Liu, 2008; Guo & Jin, 2013; Guo et al., 2014). Some other methods that can be used to deal with uncertainties is the adaptive control, Lyapunov function approach in Guo & Guo (2013) and Ge & Zhang (2011). These methods show it is possible to tackle the problem with disturbance in the indefinite system. But we also notice that it is a challenge to obtain the stability and existence of solution to the closed-loop system because applying slide-mode controller can lead to a non-linear system. In most literatures authors use Riesz methods or approximate approach to obtain the Filippov solution in finite-dimensional system and take limit to get the solvability of the closed-loop system (Dieci & Lopez, 2009), the classical LaSalle invariance set (LaSalle, 1960) and Lyapunov method (Xu & Yung, 2003) seem not to give the asymptotic behaviour of the closed-loop system. Notice that the closed-loop system might have observation blind point if the control has disturbance. The limit functions of disturbance might fall in the set of observation blind point. So such control designs do not guarantee asymptotical stability of the closed-loop system. Another challenge is the redesign of the feedback control law. Applying the feedback controller is to stabilize the system with disturbance asymptotically or exponentially. From the stability analysis we see that the closed-loop system does not satisfy the asymptotical stability. Therefore we need to modify the feedback control law. Since the disturbance is unknown, we cannot determine explicitly its weak limit functions, including limits of state of the closed-loop system. We must redesign controller so that it stabilizes the system asymptotically. We note that some literatures obtain the stability in the sense of uniformly bounded. Obviously, this is not the aim of control. The final challenge comes from non-collocated control with disturbance. The previous challenges are based on the collocated feedback controller. If the collocated information is invalid, the classical method of anti-non-collocated control is to design the Luenberger observer. However, when the control has disturbance, the Luenberger observer does not give state reconstruction of the system, so the full state observer for the system with unknown input is necessary. However, the existing results seem to be not suitable to the system with control disturbance (e.g., see Darouach, 2009; Hassan & Zied, 2010; Demetriou, 2005; Kyung-soo & Eeun-ho, 2013; Cui et al., 2016). Again the designs of the full state observer and feedback control law are new challenges. In this article, we are concerned with the following hyperbolic equation: $$\label{oringinal} \def\arraystretch{1.2}\left\{\begin{array}{@{}ll} \hat{w}_{tt}(x,t)=\hat{w}_{xx}(x,t), & x\in(0,1)\cup (1,2), t >0,\\ \hat{w}(1^{-},t)=\hat{w}(1^{+}, t),& t\ge 0, \\ \hat{w}_x(1^{-},t)-\hat{w}_x(1^{+}, t)=u(t)+r(t),& t\ge 0, \\ \hat{w}(0,t)=0,& t\geq 0,\\ \hat{w}_{x}(2,t)=0,& t\geq 0,\\ y(t)=\hat{w}_{t}(1^{-},t),& t\geq 0, \end{array}\right.$$ (1.1) where $$u(t)$$ is a controller, $$r(t)$$ is an unknown bounded and continuous disturbance and $$y(t)$$ is the output of the system. This model can be conceived as segments of power transmission lines, aerial cable/railway systems or the upper cable part of an idealized suspension bridge. By introducing a new variable $$w(x,t)=\left[w_1(x,t),w_2(x,t)\right]^{T}$$ for $$x\in[0,1]$$ and $$t\geq 0$$, where \begin{equation*} w_1(x, t)=\hat{w}(x,t), \;\; w_2(x,t)=\hat{w}(2-x,t). \end{equation*} Then system (1.1) is transformed into an equivalent system of wave equations: $$\label{target1} \def\arraystretch{1.2}\left\{\begin{array}{@{}ll} w_{tt}(x,t)=w_{xx}(x,t), & 0<x<1, t >0, \\ w_1(1,t)=w_2(1, t),& t\ge 0, \\ w_{1x}(1,t)+w_{2x}(1, t)=u(t)+r(t),& t\ge 0, \\ w_{1}(0,t)=0,& t\geq 0,\\ w_{2x}(0,t)=0,& t\geq 0,\\ w(x,0)=w^{0}(x)=(w_{1}^{0}(x),w_{2}^{0}(x))^{T},w_{t}(x,0)=w^{1}(x)=(w_{1}^{1}(x),w_{2}^{1}(x))^{T},\\ y(t)=w_{1t}(1,t),&t\geq 0. \end{array}\right.$$ (1.2) Our main contribution of this article are: (a) sliding mode control is adopted to resist the disturbance and observation blind point is found in this non-linear system. (b) We redesign the controller and give the well-posedness of the closed-loop systems by developed variational methods. (c) Asymptotic behaviour of the system is shown by the LaSalle’s Invariance Principle-like. The article is organized as follows. In Section 2, we analysed the observation blind point of the system. In Section 3, the controller is designed based on the observation blind point. In Section 4, we studied the well-posedness of the closed-loop system. In Section 5, we proved the asymptotic behaviour of the system. 2. Observation blind point In the past decades, the controller design and the output feedback law of the partial differential equations (PDEs) relies on the exact input and the stability relies on the exact form of the feedback. If there is unknown input disturbance, the output feedback controller may not work. For convenience, we introduce the concept of the observation blind point. Definition 2.1 Suppose that $${\mathcal H}$$ is a Hilbert space. A linear dynamic system in space $${\mathcal H}$$ is described by the following system $$\label{b} \def\arraystretch{1.2}\left\{\begin{array}{@{}c} \dot{x}(t)=Ax(t)+Bu(t),\\ x(0)=x_{0}, \\ y(t)=Cx(t). \end{array}\right.$$ (2.3) For any given initial value $$x_{0}$$, if there exists a controller $$u_{0} (t)\in L_{loc}^{2}(0,\infty)$$ such that $$y(t)\equiv 0$$, then $$(x_{0},u_{0}(t))$$ is an observation blind point of the system. If there exists a constant $$\tau>0$$, $$y(t)\equiv 0$$ when $$t\geq \tau$$, then $$(x_{0},u_{0}(t))$$ is a final observation blind point of the system. For a control-observation system, the observation blind point may exist or not, if there is an unknown input or disturbance, then system (2.3) becomes $$\label{bd} \def\arraystretch{1.2} \left\{\begin{array}{@{}c} \dot{x}(t)=Ax(t)+B[u(t)+r(t)],\\ x(0)=x_{0}, \\ y(t)=Cx(t). \end{array}\right.$$ (2.4) Obviously, if $$(x_{0},r(t))$$ is an observation blind point, then the output feedback controller $$u(t)=Ky(t)$$ will not work. We can also say that the output feedback is not robust. In this part, we are going to consider the observation blind point problem of the following wave system governed by $$\label{target} \def\arraystretch{1.2}\left\{\begin{array}{@{}ll} w_{tt}(x,t)=w_{xx}(x,t), & 0<x<1, t >0, \\ w_1(1,t)=w_2(1, t),& t\ge 0, \\ w_{1x}(1,t)+w_{2x}(1, t)=u(t)+r(t),& t\ge 0, \\ w_{1}(0,t)=0,& t\geq 0,\\ w_{2x}(0,t)=0,& t\geq 0,\\ w(x,0)=w^{0}(x),w_{t}(x,0)=w^{1}(x),\\ y(t)=w_{1t}(1,t),&t\geq 0, \end{array}\right.$$ (2.5) where $$u(t)$$ is a controller, $$r(t)$$ is an unknown uniformly bounded disturbance and $$y(t)$$ is the output of the system. We will firstly deal with the underlying observation blind point, then design the controller based on the form of the observation blind point. Now we will discuss the observation blind point. $$\label{targetblind} \def\arraystretch{1.2}\left\{\begin{array}{@{}ll} w_{tt}(x,t)=w_{xx}(x,t), & 0<x<1, t >0, \\ w_1(1,t)=w_2(1, t),& t\ge 0, \\ w_{1x}(1,t)+w_{2x}(1, t)=u_{0}(t),& t\ge 0, \\ w_{1}(0,t)=0,& t\geq 0,\\ w_{2x}(0,t)=0,& t\geq 0,\\ w(x,0)=w^{0}(x),w_{t}(x,0)=w^{1}(x),\\ y(t)=w_{1t}(1,t),&t\geq 0. \end{array}\right.$$ (2.6) First we will consider the existence of the observation blind point of system (2.6). We consider system (2.5) in an energy Hilbert space $${\mathcal H}$$ $${\mathcal H}=H_{E}^{1}(0,1)\times L^2(0,1)\times H^{1}(0,1)\times L^2(0,1),H_{E}^{1}(0,1)=\left\{f(x)\in H^1(0,1) \mid f(0)=0\right\}$$ with the inner product induced norm: for $$\forall (f_1,g_1,f_2,g_2)\in {\mathcal H}$$, $$\left\|(f_{1},g_{1},f_{2},g_{2})\right\|^2=\int_0^1 f_{1}'(x)^{2}+g_{1}^2(x)+ f_{2}'(x)^{2}+g_{2}^2(x)dx.$$ Theorem (2.1) gives the existence of the observation blind point and the relationship between $$u_{0}(t)$$ and the initial value $$(w_{0},w_{1})$$. Theorem 2.1 For any boundary value $$(w_{1}(0,t),w_{1t}(0,t),w_{2}(0,t),w_{2t}(0,t)) \in {\mathcal H}$$, there exists an input $$u_{0}(t)\in L_{loc}^{2}(R_{+})$$ so that the output of the system is zero. Moreover, $$u_{0}(t)$$ is determined by the boundary value $$(w_{1}(0,t),w_{1t}(0,t),w_{2}(0,t),w_{2t}(0,t))$$. Proof. We will consider the case when $$w_{1t}(1,t)\equiv0$$. That is $$\label{targetw} \def\arraystretch{1.2}\left\{\begin{array}{@{}ll} w_{tt}(x,t)=w_{xx}(x,t), & 0<x<1, t >0, \\ w_1(1,t)=w_2(1, t),& t\ge 0, \\ w_{1}(0,t)=0,& t\geq 0,\\ w_{2x}(0,t)=0,& t\geq 0,\\ w_{1t}(1,t)=0,& t\geq 0,\\ w(x,0)=w^{0}(x)=(w_{1}^{0}(x),w_{2}^{0}(x)),w_{t}(x,0)=w^{1}(x)=(w_{1}^{1}(x),w_{2}^{1}(x)). \end{array}\right.$$ (2.7)$$w_{1t}(1,t)=0$$ implies that $$w_{1}(1,t)=c$$. In this case, $$v(x,t)=w(x,t)-(x w_{1}^{0}(1),w_{2}^{0}(1))$$ satisfies $$\label{targetv} \def\arraystretch{1.2}\left\{\begin{array}{@{}ll} v_{tt}(x,t)=v_{xx}(x,t), & 0<x<1, t >0, \\ v_1(1,t)=v_2(1, t)=0,& t\ge 0, \\ v_{1}(0,t)=0,& t\geq 0,\\ v_{2x}(0,t)=0,& t\geq 0,\\ v(x,0)=v^{0}(x)=(w_{1}^{0}(x)-xw_{1}^{0}(1),w_{2}^{0}(x)-w_{2}^{0}(1)),v_{t}(x,0)=v^{1}(x)=w^{1}(x). \end{array}\right.$$ (2.8) The eigenvalue pairs of (2.8) are $$(\lambda_{1,n},\lambda_{2,n})=\bigg(in\pi,\frac{2n-1}{2}\pi i \bigg),n=1,2,3\ldots$$ and corresponding eigenfunctions are $$(\varphi_{1,n},\varphi_{2,n})=\bigg(\sin n\pi x,\cos\frac{2n-1}{2}\pi x\bigg)$$ and the family $$\{(\varphi_{1,n},\varphi_{2,n}),n=1,2,3\ldots\}$$ forms an orthogonal basis for $$L^{2}[0,1]\times L^{2}[0,1]$$. Hence \begin{eqnarray*} w_{1}^{0}(x)-xw_{1}^{0}(1)&=&\sum_{n=1}^{\infty}a_{1,n}\sin n\pi x, \;\;w_{1}^{1}(x)= \sum_{n=1}^{\infty}b_{1,n}\sin n\pi x,\\ w_{2}^{0}(x)-w_{2}^{0}(1)&=&\sum_{n=1}^{\infty}a_{2,n}\cos(n\pi -\pi/2)x, \;\;w_{2}^{1}(x)= \sum_{n=1}^{\infty}b_{2,n}\cos(n\pi - \pi/2)x, \end{eqnarray*} where \begin{eqnarray*} a_{1,n} &=& 2\int_{0}^{1}(w_{1}^{0}(x)-xw_{1}^{0}(1))\sin n\pi xdx,\;\;b_{1,n}=2\int_{0}^{1} w_{1}^{1}(x)\sin n\pi xdx,\\ a_{2,n} &=& 2\int_{0}^{1}(w_{2}^{0}(x)-w_{2}^{0}(1))\cos(n\pi -\pi/2)xdx,\;\;b_{2,n}=2\int_{0}^{1}w_{2}^{1}(x)\cos(n\pi -\pi/2)xdx. \end{eqnarray*} The solution to (2.8) is \begin{eqnarray*} &&v(x,t)=\left( \begin{array}{c} v_{1}(x,t) \\ v_{2}(x.t) \\ \end{array} \right)\\ &=&\left( \begin{array}{c} \sum_{n=1}^{\infty}\bigg[a_{1,n}\cos n\pi t +b_{1,n}\frac{\sin n\pi t}{n\pi}\bigg]\sin n\pi x \\ \sum_{n=1}^{\infty}\bigg[a_{2,n} \cos(n\pi -\pi/2) t +b_{2,n}\frac{\sin(n\pi -\pi/2)t}{n\pi -\pi/2}\bigg]\cos(n\pi -\pi/2) x \end{array} \right)\!. \end{eqnarray*} Hence, \begin{align*} &w(x,t)=\left( \begin{array}{c} w_{1}(x,t) \\ w_{2}(x.t) \\ \end{array} \right)\\ &\quad =\left( \begin{array}{c} xw_{1}^{0}(1)+\sum_{n=1}^{\infty}\bigg[a_{1,n}\cos n\pi t +b_{1,n}\frac{\sin n\pi t}{n\pi}\bigg]\sin n\pi x \\ w_{2}^{0}(1)+\sum_{n=1}^{\infty}\bigg[a_{2,n} \cos(n\pi -\pi/2) t +b_{2,n}\frac{\sin(n\pi -\pi/2)t}{n\pi -\pi/2}\bigg]\cos(n\pi -\pi/2) x \end{array} \right) \end{align*} is the solution to (2.7). From this expression we get \begin{align*} w_{1x}(1,t)+w_{2x}(1,t)&=w_{1}^{0}(1)+\sum_{n=1}^{\infty}\bigg[a_{1,n} n\pi\cos n\pi t +b_{1,n}\sin n\pi t\bigg]\\ &\quad+\sum_{n=1}^{\infty}\bigg[a_{2,n}(n\pi -\pi/2) \cos(n\pi -\pi/2) t +b_{2,n}\sin(n\pi -\pi/2)t\bigg]. \end{align*} We take $$u_{0}(t)=r(t)=w_{1x}(1,t)+w_{2x}(1,t)$$. Obviously, the pair $$(w^{0},w^{1},r(t))$$ is an observation blind point to (2.6). From the above discussion we see that $$r(t)$$ is uniquely determined by the initial value $$(w^{0}(x),w^{1}(x))$$. Note that \begin{eqnarray*} a_{1,n}(n\pi) &=& 2\int_{0}^{1}(w_{1}^{0'}(x)-w_{1}^{0}(1))\cos n\pi xdx +w_{1}^{0}(0)\\ a_{1,n}(n\pi -\pi/2) &=&(-1)^{n+1}-2\int_{0}^{1} w_{2}^{0'}(x)\sin(n\pi -\pi/2)x\,dx \end{eqnarray*} so $$r(t)$$ depends continuously on $$(w^{0},w^{1})$$. From Theorem 2.1 we can denote \begin{eqnarray*} \hat{r}(t)=r(t,w^{0},w^{1})&=&w_{1}^{0}(1)+\sum_{n=1}^{\infty}\bigg[a_{1,n} n\pi\cos n\pi t +b_{1,n}\sin n\pi t\bigg]\\ &&+\sum_{n=1}^{\infty}\bigg[a_{2,n}(n\pi -\pi/2) \cos(n\pi -\pi/2) t +b_{2,n}\sin(n\pi -\pi/2)t\bigg]. \end{eqnarray*} We define a set composed of all observation blind points of (2.6) $${\mathcal G}=\{\hat{r}(t)=r(t,w^{0},w^{1})|\forall(w^{0},w^{1})\in {\mathcal H}\}.$$ It can be proved that $${\mathcal G}$$ is a closed space in $$L_{loc}^{2}(R^{+})$$. The desired result follows. $$\quad\Box$$ When $$r(t)$$ is in the position of the observation blind point, the used feedback control law $$u(t)=-kw_{1t}(1,t)-M{\rm sign}(w_{1t}(1,t))$$ is invalid. A question arises: how do we remove the observation blind point and what is the feedback controller? In the following we will discuss the design of the feedback controller to the system (2.5). Let’s consider the following system $$\label{stru0} \def\arraystretch{1.2}\left\{\begin{array}{@{}ll} z_{tt}(x,t)=z_{xx}(x,t), & 0<x<1, t >0, \\ z_1(1,t)=z_2(1, t)=0,& t\ge 0, \\ z_{1}(0,t)=0,& t\geq 0,\\ z_{2x}(0,t)=0,& t\geq 0,\\ z(x,0)=z^{0}(x),z_{t}(x,0)=z^{1}(x),& t\geq 0,\\ \hat{r}(t)=z_{1x}(1,t)+z_{2x}(1,t),& t\geq 0. \end{array}\right.$$ (2.9) where $$z^{0}(1)=0$$. Based on Theorem 2.1, $$\hat{r}(t)$$ is uniquely determined by $$(z^{0},z^{1})$$. Noting that $$F(f)=f(1)$$ is a bounded linear functional, $$N(F)$$ is a closed subset in $$H_{E}^{1}(0,1)\times H^{1}(0,1)$$ $$N(F)=\{(f_{1},f_{2})| f_{1}(0)=0,f_{1}(1)=f_{2}(1)=0\}.$$ A direct computation shows that $$N(F)^{\bot}=span\{(x,1)^{T}\}.$$ So for any $$w^{0}(x)\in H_{E}^{2}(0,1)$$, we have $$w^{0}=z^{0}(x)+(xw_{1}^{0}(1),w_{2}^{0}(1))^{T}\in {\mathcal N}(F)\oplus {\mathcal N}(F)^{\perp}$$ where $$z^{0}(x)=[w^{0}(x)-(xw_{1}^{0}(1),w_{2}^{0}(1))^{T}]$$ and $$z^{0}(1)=0$$. Suppose $$z(x,t)$$ is a solution of (2.9), the initial value is $$(z^{0},w^{1})$$ and $$w(x,t)=z(x,t)+(xw_{1}^{0}(1),w_{2}^{0}(1))^{T}.$$ Obviously $$w(x,t)$$ satisfies equation (2.8) and $$w(x,0)=w_{0}(x),w_{t}(x,0)=w_{1}(x)$$. So $$u_{0}(t)=w_{1x}(1,t)+w_{2x}(1,t)=\hat{r}(t)+w_{1}^{0}(1).$$ For any $$w(x)\in H_{E}^{1}[0,1]$$, we have the following inequality $$w(1)=\int_{0}^{1}w_{x}(x)dx\leq \sqrt{\int_{0}^{1}|w_{x}(x)|^{2}dx},$$ thus $$\label{ur} |u_{0}(t)-\hat{r}(t)|=|w_{1}^{0}(1)|\leq \parallel(w^{0},w^{1})\parallel.$$ (2.10) 3. Control design The energy of system (2.5) is $$E(t)=\frac{1}{2}\int_{0}^{1} w_{1x}^{2}(x,t)+w_{1t}^{2}(x,t)+w_{2x}^{2}(x,t)+w_{2t}^{2}(x,t) dx.$$ The derivation of $$E(t)$$ with respect to $$t$$ is \begin{eqnarray*} \dot{E}(t)&=&\int_{0}^{1} w_{1x}(x,t)w_{1xt}(x,t)+w_{1t}(x,t)w_{1tt}(x,t)+w_{2x}(x,t)w_{2xt}(x,t)+w_{2t}(x,t)w_{2tt}(x,t)dx,\\ &=& w_{1x}w_{1t}|_{0}^{1}+w_{2x}w_{2t}|_{0}^{1},\\ &=&(u(t)+r(t))w_{1t}(1,t). \end{eqnarray*} In order to make $$\dot{E}(t)$$ dissipative, let $$u(t)=-k w_{1t}(1,t)+m_{1}(t)+m_{2}(t)$$ where $$k>0$$, $$m_{i},i=1,2$$ are to be determined. Then we have $$\dot{E}(t)=-kw_{1t}^{2}(1,t)+w_{1t}(1,t)(m_{1}(t)+m_{2}(t)+r(t)).$$ Since $$r(t)$$ is an unknown function, we are not sure whether it is an observation blind point of system (2.5). Suppose that the initial value $$(w^{0},w^{1})$$ satisfies $$\parallel(w^{0},w^{1})\parallel_{{\mathcal H}}\leq R$$ and $$R\geq {\rm sup}_{t\geq 0}|r(t)|$$. In this case $$r(t)$$ might be an observation blind point. That is because $$w_{1}^{0}(1)=\int_{0}^{1}w_{1,x}^{0}(x)dx\leq R.$$ In order to reject the disturbance $$r(t)$$, we choose $$m_{2}(t)= 3R$$. So $$m_{2}(t)+r(t)\geq0$$ and $$\mid m_{2}(t)+r(t)-\hat{r}(t)\mid > R.$$ So for any $$(w^{0},w^{1})\in {\mathcal H}$$, we have $$\parallel(w^{0},w^{1})\parallel\leq R$$, for any $$\tau > 0$$, we have $$((w^{0},w^{1}),m_{2}(t+\tau)+r(t+\tau))\notin {\mathcal G}$$ which shows that $$w_{1t}(1,t+\tau)\neq \theta(t)\equiv 0$$. Finally, we choose $$m_{1}(t)=-M {\rm{sign}}(w_{1t}(1,t))$$, where $$M\geq 4R\geq sup_{t\geq 0}\mid m_{2}(t)+r(t)\mid.$$ Then $$w_{1t}(1,t)(m_{1}(t)+m_{2}(t)+r(t))\leq 0$$, and $$\dot{E}(t)=-kw_{1t}^{2}(1,t)+w_{1t}(1,t)(m_{1}(t)+m_{2}(t)+r(t))\leq 0.$$ Thus we choose $$u(t)=-kw_{1t}(1,t)-M {\rm{sign}}(w_{1t}(1,t))+3R,$$ (3.11) where $$M\geq 4R.$$ Then we get the closed-loop system $$\def\arraystretch{1.2}\left\{\begin{array}{@{}ll} w_{tt}(x,t)=w_{xx}(x,t), & 0<x<1, t >0, \\ w_1(1,t)=w_2(1, t),& t\ge 0, \\ w_{1x}(1,t)+w_{2x}(1, t)=-kw_{1t}(1,t)-M {\rm{sign}}(w_{1t}(1,t))+r(t)+3R,& t\ge 0, \\ w_{1}(0,t)=0,& t\geq 0,\\ w_{2x}(0,t)=0,& t\geq 0,\\ w(x,0)=w^{0}(x),w_{t}(x,0)=w^{1}(x),& t\geq 0, \end{array}\right.$$ (3.12) where $$\|(w^{0},w^{1})\|_{{\mathcal H}}\leq R$$ and $$\tilde{r}(t)=r(t)+3R$$. It can be easily seen that $$\mid \tilde{r}(t)-\hat{r}(t)\mid> R$$, which shows that $$((w^{0},w^{1}),\tilde{r}(t))$$ is not an observation blind point of system (3.12). 4. The well-posedness of the closed-loop system In this part, we are going to show the well-posedness of the close-loop system (3.12). As can been seen, (3.12) is a non-linear system, so we adopt the calculus of variations to deal with the existence of the solution. The result is based on the following lemma (Cui et al., 2016), which can be seen as an expansion of linear Lions–Lax–Milgram to the semi-linear case. Lemma 4.1 Suppose $${\mathcal F}$$ is a Hilbert space with the norm $$\|v\|_{{\mathcal F}}$$, $${\it{\Phi}}$$ is a subspace of $${\mathcal F}$$, and $${\it{\Phi}}$$ is a normed linear space with the norm $$\|\phi\|_{{\it{\Phi}}}$$, $$E(v_{1},v_{2},\phi)$$ is a triple linear functional defined on $${\mathcal F} \times {\mathcal F}\times {\it{\Phi}}$$ which satisfies \begin{eqnarray*} E(\alpha v_{1}+\beta u_{1},v_{2},\phi)=\alpha E(v_{1},v_{2},\phi)+\beta E(u_{1},v_{2},\phi),\\ E(v_{1},\alpha v_{2}+\beta u_{2},\phi)=\alpha E(v_{1},v_{2},\phi)+\beta E(v_{1},u_{2},\phi),\\ E(v_{1},v_{2},\alpha \phi+\beta \psi)=\alpha E(v_{1},v_{2},\phi)+\beta E(v_{1},v_{2},\psi), \end{eqnarray*} and $$N(v_{1},v_{2},\phi)$$ is semi-linear in space $${\mathcal F}\times {\mathcal F}\times{\it{\Phi}}$$, and it is linear with respect to $$\phi$$, the multiple range is admissible. Suppose $$E(v_{1},v_{2},\phi)$$ and $$N(v_{1},v_{2},\phi)$$ satisfy the following form: (1) There exists a constant $$C_{0}$$ such that $$\|\phi\|_{{\mathcal F}}\leq C_{0}\|\phi\|_{{\it{\Phi}}},\forall \phi\in{\it{\Phi}}$$; (2) For any $$\phi\in{\it{\Phi}}$$, $$E(v_{1},v_{2},\phi)$$ is continuous with respect to $$v_{1},v_{2}$$; (3) There exists a constant $$C_{2}$$ such that $$E(\phi,\phi,\phi)\geq C_{2}\|\phi\|_{{\it{\Phi}}},\forall \phi\in{\it{\Phi}}$$; (4) $$N(v_{1},v_{2},\phi)$$ is $${\it{\Phi}}$$-monotonous \begin{eqnarray*} &&N(\phi,\phi,\phi)+N(\psi,\psi,\psi)+ N(\phi,\phi,\psi)+N(\psi,\psi,\phi)\\ &&-N(\psi,\phi,\phi)-N(\phi,\psi,\psi)-N(\phi,\psi,\phi)-N(\psi,\phi,\psi)\geq 0,\forall \phi,\psi\in{\it{\Phi}}. \end{eqnarray*} is full range, for $$\lambda>0$$, $$N(v_{1},v_{2},\eta)=[\lim_{\lambda\longrightarrow0}N(v_{1}-\lambda \eta,\eta),\lim_{\lambda\longrightarrow0}N(v_{1}+\lambda \eta,\eta),\lim_{\lambda\longrightarrow0}N(v_{2}-\lambda \eta,\eta),\lim_{\lambda\longrightarrow0}N(v_{2}+\lambda \eta,\eta)],$$$$\forall \eta\in{\it{\Phi}} ;$$ (5) There exists a constant $$C_{3}$$ such that $$N(v_{1},v_{2},\phi)\leq C_{3}\|\phi\|_{{\it{\Phi}}},\forall \phi\in{\it{\Phi}}$$. Then for any $$L(\phi)$$, which is a bounded linear functional on space $${\it{\Phi}}$$, there exists a $$u=(u_{1},u_{2})\in{\mathcal F}\times{\mathcal F}$$ such that $$E(u_{1},u_{2},\phi)+N(u_{1},u_{2},\phi)=L(\phi),\forall \phi\in{\it{\Phi}}.$$ 4.1. The existence of the solution Now we are going to show the existence of system (3.12). First, we standardize the equation and the initial conditions, let $$v(x,t)=w(x,t)-w^{0}(x).$$ Then (3.12) becomes $$\def\arraystretch{1.2}\left\{\begin{array}{@{}ll} v_{tt}(x,t)=v_{xx}(x,t)+w_{xx}^{0}(0), & 0<x<1, t >0, \\ v_1(1,t)=v_2(1, t),& t\ge 0, \\ v_{1x}(1,t)+v_{2x}(1, t)=-kv_{1t}(1,t)-M {\rm{sign}}(v_{t}(1,t))\\ +\tilde{r}(t)-w_{1x}^{0}(1)-w_{2x}^{0}(1),& t\ge 0, \\ v_{1}(0,t)=0,& t\geq 0,\\ v_{2x}(0,t)=0,& t\geq 0,\\ v(x,0)=v^{0}(x),v_{t}(x,0)=v^{1}(x),& t\geq 0,\\ y(t)=v_{t}(1,t). \end{array}\right.$$ (4.13) In order to prove the existence of the solution, we first transform equation (4.13) into a variational equations. Suppose $$\eta >0$$, for any $$T>0$$, integrating the first equation of (4.13) by multiplying $$e^{-\eta t}\phi_{t}(x,t)$$ on domain $$[0,1]\times [0,T]$$, we get \begin{eqnarray*} 0 &=& \int_{0}^{T}\int_{0}^{1}e^{-\eta t}[v_{1tt}(x,t)-v_{1xx}(x,t)-w_{1xx}^{0}(x)+v_{2tt}(x,t)-v_{2xx}(x,t)-w_{2xx}^{0}(x)]\phi_{t}(x,t)dx dt \\ &=& \int_{0}^{1}e^{-\eta T}v_{1t}(x,T)\phi_{t}(x,T)dx-\int_{0}^{1}v_{1t}(x,0)\phi_{t}(x,0)dx-\int_{0}^{T}\int_{0}^{1}v_{1t}(x,t)[e^{-\eta t}\phi_{t}(x,t)]_{t}dxdt\\ &&+\int_{0}^{T}[e^{-\eta t}v_{1x}(0,t)\phi_{t}(0,t)-e^{-\eta t}v_{1x}(1,t)\phi_{t}(1,t)]dt+\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{xt}(x,t)v_{1x}(x,t)dxdt\\ &&+\int_{0}^{T}[e^{-\eta t}\phi_{t}(0,t)w_{1x}^{0}(0)-e^{-\eta t}\phi_{t}(1,t)w_{1x}^{0}(1)]dt+\int_{0}^{1}\int_{0}^{T}[e^{-\eta t}\phi_{tx}(x,t)]w_{1x}^{0}(x)dtdx\\ &&+ \int_{0}^{1}e^{-\eta T}v_{2t}(x,T)\phi_{t}(x,T)dx-\int_{0}^{1}v_{2t}(x,0)\phi_{t}(x,0)dx-\int_{0}^{T}\int_{0}^{1}v_{2t}(x,t)[e^{-\eta t}\phi_{t}(x,t)]_{t}dxdt\\ &&+\int_{0}^{T}[e^{-\eta t}v_{2x}(0,t)\phi_{t}(0,t)-e^{-\eta t}v_{2x}(1,t)\phi_{t}(1,t)]dt+\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{xt}(x,t)v_{2x}(x,t)dxdt\\ &&+\int_{0}^{T}[e^{-\eta t}\phi_{t}(0,t)w_{2x}^{0}(0)-e^{-\eta t}\phi_{t}(1,t)w_{2x}^{0}(1)]dt+\int_{0}^{1}\int_{0}^{T}[e^{-\eta t}\phi_{tx}(x,t)]w_{2x}^{0}(x)dtdx\\ &=& \int_{0}^{1}e^{-\eta T}v_{1t}(x,T)\phi_{t}(x,T)dx-\int_{0}^{1}v_{1t}(x,0)\phi_{t}(x,0)dx-\int_{0}^{T}\int_{0}^{1}v_{1t}(x,t) [e^{-\eta t}\phi_{t}(x,t)]_{t}dxdt\\ &&+\int_{0}^{T}e^{-\eta t}v_{1x}(0,t)\phi_{t}(0,t)-e^{-\eta t}v_{1x}(1,t)\phi_{t}(1,t)dt+\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{xt}(x,t)v_{1x}(x,t)dxdt\\ &&+\int_{0}^{T}[e^{-\eta t}\phi_{t}(0,t)w_{1x}^{0}(0)-e^{-\eta t}\phi_{t}(1,t)w_{1x}^{0}(1)]dt+\eta\int_{0}^{1}\int_{0}^{T}[e^{-\eta t}\phi_{x}(x,t)]w_{1x}^{0}(x)dtdx\\ &&+\int_{0}^{1}e^{-\eta t}\phi_{x}(x,t)w_{1x}^{0}(x)|_{t=0}^{T}dx\\ &&+\int_{0}^{1}e^{-\eta T}v_{2t}(x,T)\phi_{t}(x,T)dx-\int_{0}^{1}v_{2t}(x,0)\phi_{t}(x,0)dx-\int_{0}^{T}\int_{0}^{1}v_{2t} (x,t)[e^{-\eta t}\phi_{t}(x,t)]_{t}dxdt\\ &&+\int_{0}^{T}[e^{-\eta t}v_{2x}(0,t)\phi_{t}(0,t)-e^{-\eta t}v_{2x}(1,t)\phi_{t}(1,t)]dt+\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{xt}(x,t)v_{2x}(x,t)dxdt\\ &&+\int_{0}^{T}[e^{-\eta t}\phi_{t}(0,t)w_{2x}^{0}(0)-e^{-\eta t}\phi_{t}(1,t)w_{2x}^{0}(1)]dt +\eta\int_{0}^{1}\int_{0}^{T}[e^{-\eta t}\phi_{x}(x,t)]w_{2x}^{0}(x)dtdx\\ &&\int_{0}^{1}e^{-\eta t}\phi_{x}(x,t)w_{2x}^{0}(x)|_{t=0}^{T}dx. \end{eqnarray*} Suppose $$\phi(x,0)=0,\phi_{t}(0,t)=0,\phi(x,T)=0,\phi_{t}(x,T)=0$$, we have \begin{eqnarray} &&\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{xt}(x,t)v_{1x}(x,t)dxdt-\int_{0}^{T}\int_{0}^{1}v_{1t}(x,t)[e^{-\eta t}\phi_{t}(x,t)]_{t}dxdt\nonumber\\ &&-\int_{0}^{T}\int_{0}^{1}v_{2t}(x,t)[e^{-\eta t}\phi_{t}(x,t)]_{t}dxdt+\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{xt}(x,t)v_{2x}(x,t)dxdt\nonumber\\ &&+\int_{0}^{T}e^{-\eta t}\phi_{t}(1,t)k v_{1t}(1,t)dt+M\int_{0}^{T}e^{-\eta t}\phi_{t}(1,t){\rm {\rm{sign}}} v_{1t}(1,t)dt\nonumber\\ &=&\int_{0}^{1}w_{1}^{1}(x)\phi_{t}(x,0)dx+\int_{0}^{1}w_{2}^{1}(x)\phi_{t}(x,0)dx+\int_{0}^{T}e^{-\eta t}\phi_{t}(1,t)\tilde{r}(t)dt\nonumber\\ && -\eta\int_{0}^{1}\int_{0}^{T}[e^{-\eta t}\phi_{x}(x,t)]w_{1x}^{0}(x)dtdx-\eta\int_{0}^{1}\int_{0}^{T}[e^{-\eta t}\phi_{x}(x,t)]w_{2x}^{0}(x)dtdx. \end{eqnarray} (4.14) Obviously equation (4.14) includes three parts \begin{gather} \begin{aligned}[b] E(v_{1},v_{2},\phi)&=\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{xt}(x,t)v_{1x}(x,t)dxdt-\int_{0}^{T}\int_{0}^{1}v_{1t}(x,t)[e^{-\eta t}\phi_{t}(x,t)]_{t}dxdt\\ &\quad-\int_{0}^{T}\int_{0}^{1}v_{2t}(x,t)[e^{-\eta t}\phi_{t}(x,t)]_{t}dxdt+\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{xt}(x,t)v_{2x}(x,t)dxdt\\ &\quad+\int_{0}^{T}e^{-\eta t}\phi_{t}(1,t)k v_{1t}(1,t)dt, \end{aligned}\\ \end{gather} (4.15) \begin{gather} N(v_{1},v_{2},\phi)=M\int_{0}^{T}e^{-\eta t}\phi_{t}(1,t){\rm sign} v_{1t}(1,t)dt,\\ \end{gather} (4.16) \begin{gather} \begin{aligned}[b] L(\phi)&=\int_{0}^{1}w_{1}^{1}(x)\phi_{t}(x,0)dx+\int_{0}^{1}w_{2}^{1}(x)\phi_{t}(x,0)dx+\int_{0}^{T}e^{-\eta t}\phi_{t}(1,t)\tilde{r}(t)dt\\ &\quad -\eta\int_{0}^{1}\int_{0}^{T}[e^{-\eta t}\phi_{x}(x,t)]w_{1x}^{0}(x)dtdx-\eta\int_{0}^{1}\int_{0}^{T}[e^{-\eta t}\phi_{x}(x,t)]w_{2x}^{0}(x)dtdx. \end{aligned} \end{gather} (4.17) Thus (4.14) can be written as $$E(v_{1},v_{2},\phi)+N(v_{1},v_{2},\phi)=L(\phi).$$ Suppose \begin{eqnarray} {\mathcal F} &=& \def\arraystretch{1.3}\left \{v(x,t) \left |\begin{array}{c} e^{-\eta t}v(x,t),e^{-\eta t}v_{x}(x,t),e^{-\eta t}v_{t}(x,t)\in L^{2}(0,1),\\ e^{-\eta t}v_{t}(1,t)\in L^{2}(0,T), \\ v(0,t)=0, \\ v(x,0)=0 \end{array} \right. \right\} \end{eqnarray} (4.18) with the norm $$\|v\|_{{\mathcal F}}^{2}=\int_{0}^{T}\int_{0}^{1} e^{-\eta t}[|v(x,t)|_{x}^{2}+|v(x,t)|_{t}^{2}]dxdt+\int_{0}^{T}e^{-\eta t}|v_{t}^{2}(1,t)|dt,$$ and \begin{eqnarray} {\it{\Phi}} &=& \def\arraystretch{1.3}\left \{\phi(x,t) \left |\begin{array}{c} e^{-\eta t}\phi_{t}(x,t),e^{-\eta t}\phi_{xt}(x,t),e^{-\eta t}\phi_{tt}(x,t)\in L^{2}(0,1),\\ e^{-\eta t}\phi_{t}(1,t)\in L^{2}(0,T), \\ \phi(0,t)=0, \\ \phi(x,0)=\phi(x,T)=\phi_{t}(x,T)=0 \end{array} \right. \right\} \end{eqnarray} (4.19) with the norm $$\|\phi\|_{{\it{\Phi}}}^{2}=\int_{0}^{T}\int_{0}^{1} e^{-\eta t}[|\phi(x,t)|_{x}^{2}+|\phi(x,t)|_{t}^{2}]dxdt+\int_{0}^{T}e^{-\eta t}|\phi_{t}(1,t)|^{2}dt+\int_{0}^{1}|\phi_{t}(x,0)|^{2}dx.$$ It can be seen that $${\it{\Phi}}$$ is a linear subspace of $${\mathcal F}$$, $${\mathcal F}$$ is Hilbert space and $${\it{\Phi}}$$ is a normed linear space. In addition, $$\|\phi\|_{{\mathcal F}}^{2}\leq\|\phi\|_{{\it{\Phi}}}^{2},\forall \phi \in{\it{\Phi}}$$. In order to discuss the existence of the solution, we give the definition of weak solution. Definition 4.1 Suppose that $$w(x,t)$$ is differentiable with respect to $$x,t$$, $$w_{xx}(x,t),w_{t}(x,t)\in L^{2}[0,1]$$,$$w_{t}(1,t)\in L^{2}[0,T]$$, and $$\|w_{xx}(\cdot,t)-w_{xx}^{0}\|_{L^{2}[0,1]}\longrightarrow 0, \|w_{t}(\cdot,t)-w^{1}\|_{L^{2}[0,1]}\longrightarrow 0, t\longrightarrow 0.$$ Let $$v(x,t)=w(x,t)-w^{0}\in {\mathcal F}$$, if for any $$\phi(x,t)\in{\it{\Phi}}, v(x,t)$$ satisfies \begin{align} &\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{xt}(x,t)v_{1x}(x,t)dxdt-\int_{0}^{T}\int_{0}^{1}v_{1t}(x,t)[e^{-\eta t}\phi_{t}(x,t)]_{t}dxdt\nonumber\\ &\qquad-\int_{0}^{T}\int_{0}^{1}v_{2t}(x,t)[e^{-\eta t}\phi_{t}(x,t)]_{t}dxdt+\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{xt}(x,t)v_{2x}(x,t)dxdt\nonumber\\ &\qquad+\int_{0}^{T}e^{-\eta t}\phi_{t}(1,t)k v_{1t}(1,t)dt+M\int_{0}^{T}e^{-\eta t}\phi_{t}(1,t){\rm sign} v_{1t}(1,t)dt\nonumber\\ &\quad=\int_{0}^{1}w_{1}^{1}(x)\phi_{t}(x,0)dx+\int_{0}^{1}w_{2}^{1}(x)\phi_{t}(x,0)dx+\int_{0}^{T}e^{-\eta t}\phi_{t}(1,t)\tilde{r}(t)dt\nonumber\\ &\qquad -\eta\int_{0}^{1}\int_{0}^{T}[e^{-\eta t}\phi_{x}(x,t)]w_{1x}^{0}(x)dtdx-\eta\int_{0}^{1}\int_{0}^{T}[e^{-\eta t}\phi_{x}(x,t)]w_{2x}^{0}(x)dtdx, \end{align} (4.20) then we call $$w(x,t)$$ a weak solution of system (3.12). From the definition, we can see $$v(x,t)$$ is the solution of the variational equations. Next, we are going to prove the existence of the weak solution using Lemma (4.1). First we demonstrate the all conditions of Lemma (4.1). Proposition 4.1 Let $$E(v_{1},v_{2},\phi)$$ be defined by (4.15), then $$E(v_{1},v_{2},\phi)$$ is a bilinear form in space $${\mathcal F}\times{\it{\Phi}}$$. It is also continuous for any $$\phi\in{\it{\Phi}}$$ with respect to $$v$$. Moreover, there exists a constant $$C_{2}>0$$ so that $$E(\phi,\phi,\phi)\geq C_{2}\|\phi\|_{{\it{\Phi}}}^{2}$$. Proof. It is easy to see $$E(v_{1},v_{2},\phi)$$ is a bilinear form in space $${\mathcal F}\times{\it{\Phi}}$$. So we only need to demonstrate that it is continuous with respect to $$v=(v_{1},v_{2})^{T}$$ for any $$v_{1},v_{2}\in {\mathcal F},\phi \in{\it{\Phi}}$$ \begin{eqnarray*} |E(v_{1},v_{2},\phi)|^{2} &\leq&\bigg(\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{xt}(x,t)v_{1x}(x,t)dxdt+\eta\int_{0}^{T}\int_{0}^{1}e^{-\eta t}v_{1t}(x,t)\phi_{t}(x,t)dxdt\nonumber\\ &&-\int_{0}^{T}\int_{0}^{1}e^{-\eta t}v_{1t}(x,t)\phi_{tt}(x,t)dxdt-\int_{0}^{T}\int_{0}^{1}e^{-\eta t}v_{2t}(x,t)\phi_{tt}(x,t)dxdt\nonumber\\ &&+\eta \int_{0}^{T}\int_{0}^{1}e^{-\eta t}v_{2t}(x,t)\phi_{t}(x,t)dxdt+\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{xt}(x,t)v_{2x}(x,t)dxdt\nonumber\\ &&+\int_{0}^{T}e^{-\eta t}\phi_{t}(1,t)k v_{1t}(1,t)\bigg)^{2}dt\\ &\leq& \bigg(\int_{0}^{T}\int_{0}^{1}e^{-\eta t}v_{1x}^{2}(x,t)dxdt+\eta\int_{0}^{T}\int_{0}^{1}e^{-\eta t}v_{1t}^{2}(x,t)dxdt\nonumber\\ &&+\int_{0}^{T}\int_{0}^{1}e^{-\eta t}v_{1t}^{2}(x,t)dxdt+\int_{0}^{T}\int_{0}^{1}e^{-\eta t}v_{2t}^{2}(x,t)dxdt\nonumber\\ &&+\eta \int_{0}^{T}\int_{0}^{1}e^{-\eta t}v_{2t}^{2}(x,t)dxdt+\int_{0}^{T}\int_{0}^{1}e^{-\eta t}v_{2x}^{2}(x,t)dxdt\nonumber\\ &&+k\int_{0}^{T}e^{-\eta t} v_{1t}^{2}(1,t)\bigg)^{1/2}dt\\ &&\times \bigg(\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{xt}^{2}(x,t)dxdt+\eta\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{t}^{2}(x,t)dxdt\nonumber\\ &&+\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{tt}^{2}(x,t)dxdt+\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{tt}^{2}(x,t)dxdt\nonumber\\ &&+\eta \int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{t}^{2}(x,t)dxdt+\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{xt}^{2}(x,t)dxdt\nonumber\\ &&+k\int_{0}^{T}e^{-\eta t}\phi_{t}^{2}(1,t)\bigg)^{1/2}dt\\ &\leq&\max\{2(1+\eta)^{2},k^{2}\} \bigg(\int_{0}^{T}\int_{0}^{1}e^{-\eta t}v_{1x}^{2}(x,t)dxdt\nonumber\\ &&+\int_{0}^{T}\int_{0}^{1}e^{-\eta t}v_{1t}^{2}(x,t)dxdt+\int_{0}^{T}\int_{0}^{1}e^{-\eta t}v_{2t}^{2}(x,t)dxdt\nonumber\\ &&+\int_{0}^{T}\int_{0}^{1}e^{-\eta t}v_{2x}^{2}(x,t)dxdt+k\int_{0}^{T}e^{-\eta t} v_{1t}^{2}(1,t)\bigg)^{1/2}dt\\ &&\times \bigg(\int_{0}^{T}\int_{0}^{1}e^{-\eta t}|\phi_{xt}(x,t)|^{2}dxdt+\int_{0}^{T}\int_{0}^{1}e^{-\eta t}|\phi_{t}(x,t)|^{2}dxdt\\ &&+\int_{0}^{T}\int_{0}^{1}e^{-\eta t}|\phi_{tt}(x,t)_{t}|^{2}dxdt+\int_{0}^{T}e^{-\eta t}|\phi_{t}(1,t)|^{2}dt\bigg)^{1/2}. \end{eqnarray*} So $$E(v_1,v_2,\phi)\leq C_{1}(\parallel v_{1}\parallel_{{\cal F}}+\parallel v_{2}\parallel_{{\cal F}})(\parallel \phi\parallel_{{\cal F}}+\parallel \phi_{t}\parallel_{{\cal F}})$$ where $$C_{1}=\max\{2(1+\eta)^{2},k^{2}\}$$. For any $$\phi$$, \begin{eqnarray*} E(\phi,\phi,\phi) &=& \int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{xt}(x,t)\phi_{x}(x,t)dxdt+\eta\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{t}(x,t)\phi_{t}(x,t)dxdt\nonumber\\ &&-\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{t}(x,t)\phi_{tt}(x,t)dxdt-\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{t}(x,t)\phi_{tt}(x,t)dxdt\nonumber\\ &&+\eta \int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{t}(x,t)\phi_{t}(x,t)dxdt+\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{xt}(x,t)\phi_{x}(x,t)dxdt\nonumber\\ &&+\int_{0}^{T}e^{-\eta t}\phi_{t}(1,t)k \phi_{t}(1,t)\nonumber\\ &=&2\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\frac{d}{dt}\left(\frac{1}{2}|\phi_{x}(x,t)|^{2}\right)dxdt-2\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\frac{d}{dt}(\frac{1}{2}\phi_{t}(x,t))dxdt\\ &&+2\eta\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{t}(x,t)^{2}dxdt+k\int_{0}^{T}e^{-\eta t}\phi_{t}(1,t)\phi_{t}(1,t)dt\\ &=&\eta\int_{0}^{T}\int_{0}^{1}e^{-\eta t}|\phi_{x}(x,t)|^{2}dxdt+\eta\int_{0}^{T}\int_{0}^{1}e^{-\eta t}|\phi_{t}(x,t)|^{2}dxdt\\ &&+k\int_{0}^{T}e^{-\eta t}|\phi_{t}(1,t)|^{2}dt+\int_{0}^{1}|\phi_{t}(x,0)|^{2}dx\\ &\geq&\min \{\eta,k,1\}\bigg(\int_{0}^{T}\int_{0}^{1} e^{-\eta t}[|\phi_x(x,t)|^{2}+|\phi_t(x,t)|^{2}]dxdt\\ &&+\int_{0}^{T}e^{-\eta t}|\phi_{t}(1,t)|^{2}dt+\int_{0}^{1}|\phi_{t}(x,0)|^{2}dx\bigg) \end{eqnarray*} So $$E(\phi,\phi,\phi)$$ satisfies that $$E(\phi,\phi,\phi)\geq C_{2}\|\phi\|,\forall \phi\in{\it{\Phi}}$$ where $$C_{2}=\min \{\eta,k,1\}$$. □ Proposition 4.2 Let $$N(v_{1},v_{2},\phi)$$ be defined by (4.16), then for any $$v\in{\cal F}$$, $$N(v_{1},v_{2},\phi)$$ is a bounded and linear, specially, there exists a constant $$C_{3}>0$$ such that $$|N(v_{1},v_{2},\phi)|\leq C_{3}\|\phi\|_{{\it{\Phi}}}$$ for any $$v\in {\cal F}$$. $$N(v_{1},v_{2},\phi)$$ is monotonous with respect to $${\it{\Phi}}$$. \begin{eqnarray*} &&N(\phi,\phi,\phi)+N(\psi,\psi,\psi)+ N(\phi,\phi,\psi)+N(\psi,\psi,\phi)\\ &&-N(\psi,\phi,\phi)-N(\phi,\psi,\psi)-N(\phi,\psi,\phi)-N(\psi,\phi,\psi)\geq 0,\forall \phi,\psi\in{\it{\Phi}}. \end{eqnarray*} Moreover, $$N(v_{1},v_{2},\phi)$$ is full range. Proof. Let $$N(v_{1},v_{2},\phi)$$ be defined by (4.16). Obviously, for any $$\phi\in{\it{\Phi}}$$, $$N(v_{1},v_{2},\phi)$$ might be multi-valued function with respect to $$v_{1}$$. $$N(v_{1},v_{2},\phi)$$ is non-linear with respect to $$v_{1}$$ and it is linear with respect to $$\phi$$, and $$|N(v_{1},v_{2},\phi)|^{2}\leq M^{2}\bigg(\int_{0}^{T}e^{-\eta t}|{\rm sign}(v_{1t}(1,t))\phi_{t}(1,t)|\bigg)^{2}\leq M^{2}\frac{1-e^{-\eta t}}{\eta}\int_{0}^{T}e^{-\eta t}|\phi_{t}(1,t)|^{2}dt.$$ Thus, for all $$v_{1},v_{2},\in {\cal F}$$, $$N(v_{1},v_{2},\phi)$$ satisfies $$|N(v_{1},v_{2},\phi)|\leq C_{3}\|\phi\|_{{\it{\Phi}}},\forall \phi\in{\it{\Phi}}$$ where $$C_{3}=M\sqrt{\frac{1-e^{-\eta t}}{\eta}}$$, so $$N(v_{1},v_{2},\phi)$$ is continuous with respect to $$\phi$$. For any $$\phi,\psi\in{\it{\Phi}}$$, \begin{align*} & N(\phi,\phi,\phi)+N(\psi,\psi,\psi)+ N(\phi,\phi,\psi)+N(\psi,\psi,\phi)\\ &\qquad-N(\psi,\phi,\phi)-N(\phi,\psi,\psi)-N(\phi,\psi,\phi)-N(\psi,\phi,\psi) \\ &\quad= 2M\int_{0}^{T}e^{-\eta t}[|\phi_{t}(1,t)|+|\psi_{t}(1,t)|]dt\\ &\qquad-2M\int_{0}^{T}e^{-\eta t}[{\rm{sign}}(\phi_{t}(1,t))\psi_{t}(1,t)+{\rm{sign}}(\psi_{t}(1,t))\phi_{t}(1,t)]dt\\ &\quad=2M\int_{0}^{T}e^{-\eta t}[1-{\rm{sign}}(\psi_{t}(1,t)){\rm{sign}}(\phi_{t}(1,t))]|\phi_{t}(1,t)|dt\\ &\qquad+2M\int_{0}^{T}e^{-\eta t}[1-{\rm{sign}}(\phi_{t}(1,t)){\rm{sign}}(\psi_{t}(1,t))]|\psi_{t}(1,t)|dt\\ &\quad\geq 0. \end{align*} In order to prove that $$N(v_{1},v_{2},\phi)$$ is full range, we suppose $$\lambda >0$$, for every $$v_{1},v_{2}\in{\cal F}$$, and any $$\eta\in{\it{\Phi}}$$, $${\rm sign}(v_{1t}(1,t)\pm \lambda \eta_{t}(1,t))=\def\arraystretch{1.3}\left\{\begin{array}{@{}c} 1, v_{1t}(1,t)\pm \lambda \eta_{t}(1,t)>0,\\ \beta(t)\in[-1,1],v_{1t}(1,t)\pm \lambda \eta_{t}(1,t)=0,\\ -1,v_{1t}(1,t)\pm \lambda \eta_{t}(1,t)<0. \end{array}\right.$$ (4.21) Thus \begin{eqnarray*} N(v_{1}+\lambda \eta,v_{2},\eta)&=&\int_{E(v_{1t}+\lambda \eta_{t}\neq 0)}e^{-\eta t} sgn(v_{1t}(1,t)\pm \lambda \eta_{t}(1,t))\eta_{t}(1,t)dt\\ && +\int_{E(v_{1t}+\lambda \eta_{t}=0)}e^{-\eta t}\beta(t)\eta_{t}(1,t)dt;\\ N(v_{1}-\lambda \eta,v_{2},\eta)&=&\int_{E(v_{1t}-\lambda \eta_{t}\neq 0)}e^{-\eta t} sgn(v_{1t}(1,t)\pm \lambda \eta_{t}(1,t))\eta_{t}(1,t)dt\\ &&-\int_{E(v_{1t}-\lambda \eta_{t}=0)}e^{-\eta t}\beta(t)\eta_{t}(1,t)dt\\ N(v_{1},v_{2}+\lambda \eta,\eta)&=&\int_{E(v_{2t}+\lambda \eta_{t}\neq 0)}e^{-\eta t} sgn(v_{2t}(1,t)\pm \lambda \eta_{t}(1,t))\eta_{t}(1,t)dt\\ &&+\int_{E(v_{2t}+\lambda \eta_{t}=0)}e^{-\eta t}\beta(t)\eta_{t}(1,t)dt;\\ N(v_{1},v_{2}-\lambda \eta,\eta)&=&\int_{E(v_{2t}-\lambda \eta_{t}\neq 0)}e^{-\eta t} sgn(v_{2t}(1,t)\pm \lambda \eta_{t}(1,t))\eta_{t}(1,t)dt\\ &&-\int_{E(v_{2t}-\lambda \eta_{t}=0)}e^{-\eta t}\beta(t)\eta_{t}(1,t)dt \end{eqnarray*} where $$E(v_{it}+\lambda \eta_{t}=0)=\{t\in[0,T]|v_{it}(1,t),i=1,2\pm \lambda \eta_{t}(1,t)=0\}$$, when $$\lambda\longrightarrow0$$, we have \begin{gather*} \lim_{\lambda\longrightarrow0}N(v_{1}+\lambda \eta,v_{2},\eta)=\int_{E(v_{1}\neq0)}e^{-\eta t}sgn(v_{1t}(1,t))\eta_{t}(1,t)dt+\int_{E(v_{1}=0)}e^{-\eta t}|\eta_{t}(1,t)|,\\ \lim_{\lambda\longrightarrow0}N(v_{1},v_{2}+\lambda \eta,\eta)=\int_{E(v_{2}\neq0)}e^{-\eta t}sgn(v_{1t}(1,t))\eta_{t}(1,t)dt+\int_{E(v_{2}=0)}e^{-\eta t}|\eta_{t}(1,t)| \end{gather*} and \begin{gather*} \lim_{\lambda\longrightarrow0}N(v_{1}-\lambda \eta,v_{2},\eta)=\int_{E(v_{1}\neq0)}e^{-\eta t}sgn(v_{t}(1,t))\eta_{t}(1,t)dt+\int_{E(v_{1}=0)}e^{-\eta t}|\eta_{t}(1,t)|,\\ \lim_{\lambda\longrightarrow0}N(v_{1},v_{2}-\lambda \eta,\eta)=\int_{E(v_{2}\neq0)}e^{-\eta t}sgn(v_{t}(1,t))\eta_{t}(1,t)dt+\int_{E(v_{2}=0)}e^{-\eta t}|\eta_{t}(1,t)| \end{gather*} Since \begin{eqnarray*} N(v_{1},v_{2},\eta) &=& \int_{0}^{T}e^{-\eta t} sign(v_{t}(1,t))\eta_{t}(1,t)dt\\ &=& \int_{E(v\neq 0)}e^{-\eta t} sgn(v_{t}(1,t))\eta_{t}(1,t)dt+\int_{E(v=0)}e^{-\eta t} \beta(t)\eta_{t}(1,t)dt \end{eqnarray*} for all $$\beta(t)\in L^{1}(E(v=0))$$, where $$|\beta(t)|\leq 1$$, so we have $$N(v_{1},v_{2},\eta) =[\lim_{\lambda\longrightarrow0}N(v_{1}-\lambda \eta,\eta),\lim_{\lambda\longrightarrow0}N(v_{1}+\lambda \eta,\eta),\lim_{\lambda\longrightarrow0}N(v_{2}-\lambda \eta,\eta),\lim_{\lambda\longrightarrow0}N(v_{2}+\lambda \eta,\eta)],$$ $$\forall \eta\in{\it{\Phi}}$$ Specially, if $$v_{1t}(1,t)=v_{2t}(1,t)\equiv 0=\theta(t)$$, $$N(\theta,\theta,\phi)=\int_{0}^{T}e^{-\eta t} {\rm{sign}}(\theta(t))\eta_{t}(1,t)dt=\int_{0}^{T}e^{-\eta t}\beta(t)\eta_{t}(1,t)dt$$ the result is proved. □ Proposition 4.3 Let $$L(\phi)$$ be defined by (4.17), then $$L(\phi)$$ is a bounded linear functional in space $${\it{\Phi}}$$. Proof. It is obviously to see that $$L({\it{\Phi}})$$ a linear functional in space $${\it{\Phi}}$$. For any $$\phi\in{\it{\Phi}}$$, \begin{eqnarray*} |L(\phi)|^{2} &\leq&\bigg(\int_{0}^{1}|w_{1}^{1}(x)\phi_{t}(x,0)|dx+\int_{0}^{1}|w_{2}^{1}(x)\phi_{t}(x,0)|dx+\int_{0}^{T}e^{-\eta t}|\phi_{t}(1,t)\tilde{r}(t)|dt\nonumber\\ && +\eta\int_{0}^{1}\int_{0}^{T}e^{-\eta t}|\phi_{x}(x,t)w_{1x}^{0}(x)|dtdx+\eta\int_{0}^{1}\int_{0}^{T}e^{-\eta t}|\phi_{x}(x,t)w_{2x}^{0}(x)|dtdx\bigg)^{2}\nonumber\\ &\leq& \bigg(\int_{0}^{T}e^{-\eta t}|\tilde{r}(t)|^{2}dt+\eta \int_{0}^{1}\int_{0}^{T}e^{-\eta t}(|w_{1x}^{0}(x)|^{2}+|w_{2x}^{0}(x)|^{2})dxdt\nonumber\\ &&+\int_{0}^{1}(|w_{1}^{1}(x)|^{2}dx+|w_{2}^{1}(x)|^{2})dx\nonumber\\ &&\times\bigg(\int_{0}^{T}e^{-\eta t}|\phi_{t}(1,t)|^{2}dt+\eta \int_{0}^{1}\int_{0}^{T}e^{-\eta t}|\phi_{x}(x,t)|^{2}dxdt+\int_{0}^{1}|\phi_{t}(x,0)|^{2}dx\bigg) \end{eqnarray*} So $$|L(\phi)|\leq C_{5}\|\phi\|_{{\it{\Phi}}},$$ where $$C_{5}=C_{4}\bigg(\int_{0}^{T}e^{-\eta t}|\tilde{r}(t)|^{2}dt+\eta \int_{0}^{1}\int_{0}^{T}e^{-\eta t}(|w_{1x}^{0}(x)|^{2}+|w_{2x}^{0}(x)|^{2})dxdt+\int_{0}^{1}(|w_{1}^{1}(x)|^{2}dx+|w_{2}^{1}(x)|^{2})dx\bigg)^{1/2}$$ and $$C_{4}=\max\{1,\eta\}$$. Therefore $$L(\phi)$$ is a bounded linear functional in space $${\it{\Phi}}$$. □ Based on the above propositions and Lemma (2.1), we have the following theorem. Theorem 4.1 There exists a weak solution to the system (4.13) for any $$(w^{0},w^{1})$$ and $$r(t)$$. 4.2. The uniqueness of the solution We will consider the uniqueness of the solution of system (3.12). Suppose that there are two solutions $$z(x,t),v(x,t)$$. Consider the error function $$e(x,t)=v(x,t)-z(x,t)$$, obviously the error function satisfies the following equation $$\label{error} \def\arraystretch{1.2}\left\{\begin{array}{@{}ll} e_{tt}(x,t)=e_{xx}(x,t), & 0<x<1, t >0, \\ e_1(1,t)=e_2(1, t),& t\ge 0, \\ e_{1x}(1,t)+e_{2x}(1, t)=-k e_{1t}(1,t)-M[{\rm{sign}} (v_{1t}(1,t))-{\rm{sign}} (z_{1t}(1,t))],& t\ge 0, \\ e_{1}(0,t)=0,& t\geq 0,\\ e_{2x}(0,t)=0,& t\geq 0,\\ e(x,0)=e^{0}(x),e_{t}(x,0)=e^{1}(x). \end{array}\right.$$ (4.22) Consider the energy function \begin{eqnarray*} \varepsilon(t) &=& \frac{1}{2}\int_{0}^{1} |e_{1x}(x,t)|^{2}+|e_{2x}(x,t)|^{2}+|e_{1t}(x,t)|^{2}+|e_{2t}(x,t)|^{2}dx \end{eqnarray*} Differentiate the energy function with respect to $$t$$, we have \begin{eqnarray*} \dot{\varepsilon}(t) &=& \int_{0}^{1} e_{1t}(x,t)e_{1tt}(x,t)+ e_{2t}(x,t)e_{2tt}(x,t)+e_{1x}(x,t)e_{xt}(x,t)dx+e_{2x}(x,t)e_{2xt}(x,t)dx\\ &=& e_{1t}(x,t)e_{1x}(x,t)|_{0}^{1}+e_{2t}(x,t)e_{2x}(x,t)|_{0}^{1}\\ &=& e_{1t}(1,t)e_{1x}(1,t)+e_{2t}(1,t)e_{2x}(1,t)\\ &=& e_{1t}(1,t)(-k e_{1t}(1,t)-M[{\rm {\rm{sign}}} (v_{1t}(1,t))-{\rm sign} (z_{1t}(1,t))])\\ &=& -ke_{1t}^{2}(1,t)-M[{\rm{sign}} (v_{1t}(1,t))-{\rm{sign}} (z_{1t}(1,t))]e_{1t}(1,t)\\ &=& -ke_{t}^{2}(1,t)-M[{\rm{sign}} (v_{1t}(1,t))-{\rm{sign}} (z_{1t}(1,t))][v_{1t}(1,t)-z_{1t}(1,t))]\\ &\leq& -ke_{1t}^{2}(1,t) \end{eqnarray*} So the energy function satisfies that \begin{eqnarray*} \varepsilon(t) +k\int_{0}^{t}e_{t}^{2}(1,t)dt&\leq& \varepsilon(0)=0 \end{eqnarray*} which shows that $$e(x,t)=0$$. We then immediately have the following theorem. Theorem 4.2 For any $$(w^{0},w^{1})$$ and disturbance $$r(t)$$, the closed-loop system (3.12) has a unique solution. 5. The asymptotical stability of the closed-loop system In this part, we are going to elaborate the asymptotically stability of the closed-loop system. Note that (3.12) is a quasi-autonomous system. The asymptotic behaviour of the solution depends strongly on the disturbance. In the case of autonomous systems, the LaSalles invariant theorem shows that the trajectory of the system approaches a large invariant set $$\omega$$ on which the $$\dot{V} (t) = 0$$. In the case of non-autonomous systems, it may not be clear how to define such a set $$\omega$$. The difficulty is that the positive limit set is not invariant and the zero is not equilibrium. Here we shall find a way to discuss the asymptotic behaviour of the system. We rewrite the closed-loop system as following: $$\label{reclose} \def\arraystretch{1.2}\left\{\begin{array}{@{}ll} w_{tt}(x,t)=w_{xx}(x,t), & 0<x<1, t >0, \\ w_1(1,t)=w_2(1, t),& t\ge 0, \\ w_{1x}(1,t)+w_{2x}(1, t)=-kw_{1t}(1,t)-M {\rm {\rm{sign}}}(w_{1t}(1,t))+\tilde{r}(t),& t\ge 0, \\ w_{1}(0,t)=0,& t\geq 0,\\ w_{2x}(0,t)=0,& t\geq 0,\\ w(x,0)=w^{0}(x),w_{t}(x,0)=w^{1}(x), \end{array}\right.$$ (5.23) where $$\|(w^{0},w^{1})\|\leq R$$, and $$\tilde{r}(t)=r(t)+3R$$. The energy function in defined by \begin{eqnarray*} E(w)(t) &=& \frac{1}{2}\int_{0}^{1}(w_{1x}^{2}(x,t)+w_{2x}^{2}(x,t)+w_{1t}^{2}(x,t)+w_{2t}^{2}(x,t))dx \end{eqnarray*} In the previous part, we have proved that \begin{eqnarray*} \frac{dE(w)(t)}{dt} &=& -kw_{1t}^{2}(1,t)-(M {\rm{sign}}(w_{1t}(1,t)-\tilde{r}(t))w_{1t}(1,t)\leq 0. \end{eqnarray*} For a fixed $$r(t)$$, we assume that the initial data is in the domain $$S(R)=\{(w^{0}(x),w^{1}(x))\in{\mathcal H}|\|(w^{0},w^{1})\|_{{\mathcal H}}\leq R\}.$$ Since $$((w^{0},w^{1}),\tilde{r}(t))$$ is not an observer blind spot of system (3.12), $$E(w(t))$$ is monotonically decreasing. $$S(R)$$ is an invariant set. In light of the energy function, for any $$t,\tau>0$$, $$E(w)(t+\tau)+\int_{0}^{\tau}kw_{1t}^{2}(1,t)+(M {\rm{sign}}(w_{1t}(1,t)-\tilde{r}(t))w_{1t}(1,t)ds=E(w)(t),$$ so $$\lim_{t\longrightarrow\infty}\int_{0}^{\tau}kw_{1t}^{2}(1,t)+(M {\rm{sign}}(w_{1t}(1,t)-\tilde{r}(t))w_{1t}(1,t)ds=0$$ which shows that $$\lim_{t\longrightarrow\infty}w_{1t}(1,t)=0$$. Suppose $$t_{n}>0$$ and $$\lim\limits_{t\longrightarrow +\infty}t_{n}=+\infty$$. For any given $$\tau$$, consider the function sequence $$z_{n}(x,t)=w(x,t+t_{n}),n\in N, \;\; t\in[0,\tau]$$ where $$w(x,t)$$ is the solution of the system (5.23). We assume $$r_{n}(t)=\tilde{r}(t+t_{n})$$ $$\label{reclosez} \def\arraystretch{1.2}\left\{\begin{array}{@{}ll} z_{ntt}(x,t)=z_{nxx}(x,t), & 0<x<1, t >0, \\ z_{1n}(1,t)=z_{2n}(1, t),& t\ge 0, \\ z_{1nx}(1,t)+z_{2nx}(1, t)=-kz_{n}(1,t)-M {\rm sign}(z_{1nt}(1,t))+\tilde{r}_{n}(t),& t\ge 0, \\ z_{1n}(0,t)=0,& t\geq 0,\\ z_{2nx}(0,t)=0,& t\geq 0,\\ z_{n}(x,0)=w(x, t_{n}),z_{nt}(x,0)=w_{t}(x, t_{n}),\\ y_{n}=z_{1n}(1,t),& t\geq 0. \end{array}\right.$$ (5.24) Noticing that $$E(z_{n})(t)=\frac{1}{2}\int_{0}^{1}|z_{1n,x}(x,t)|^{2}+|z_{2n,x}(x,t)|^{2}+|z_{1n,t}(x,t)|^{2}+|z_{2n,t}(x,t)|^{2}dx$$ and $$E(z_{n})(t)+\int_{0}^{t}kz_{1n,t}^{2}(1,t)+(M {\rm{sign}}(z_{1n,t}(1,t)-\tilde{r}_{n}(t))z_{1n,t}(1,t)ds=E(w)(t_{n}).$$ Hence, $$\{(z_{n}(x,t),z_{n,t}(x,t)),n\in N\}$$ is a bounded sequence in Hilbert space $${\mathcal H}$$, and it has a weak convergent subsequence. Without loss of generality, we assume $$(z_{n}(x,t),z_{n,t}(x,t))\longrightarrow (z(x,t),\xi(x,t))\in {\mathcal H}$$. Because $$\{(w(x,t_{n}),w_{t}(x,t_{n})),n\in N\}$$ is also a bounded sequence in Hilbert space $${\mathcal H}$$ and it has a subsequence. Without loss of generality, we assume the subsequence to be itself, and it convergents to $$(w(x),\eta(x))$$. Moreover, $$r(t)$$ is uniformly bounded, so it has a weak limitation in space $$L^{2}(0,1)$$. Thus, $$z_{t}(x,t)=\xi(x,t)$$, so $$z(x,t)$$ satisfies \begin{eqnarray} &&\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{xt}(x,t)v_{1x}(x,t)dxdt-\int_{0}^{T}\int_{0}^{1}v_{1t}(x,t)[e^{-\eta t}\phi_{t}(x,t)]_{t}dxdt\nonumber\\ &&-\int_{0}^{T}\int_{0}^{1}v_{2t}(x,t)[e^{-\eta t}\phi_{t}(x,t)]_{t}dxdt+\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{xt}(x,t)v_{2x}(x,t)dxdt\nonumber\\ &=&\int_{0}^{1}\eta_{1}(x)\phi_{t}(x,0)dx+\int_{0}^{1}\eta_{2}(x)\phi_{t}(x,0)dx\int_{0}^{1}d(t)\phi_{t}(1,t)dx\nonumber\\ && -\eta\int_{0}^{1}\int_{0}^{T}[e^{-\eta t}\phi_{x}(x,t)]w_{1x}^{0}(x)dtdx-\eta\int_{0}^{1}\int_{0}^{T}[e^{-\eta t}\phi_{x}(x,t)]w_{2x}^{0}(x)dtdx \end{eqnarray} (5.25) where $$\phi(x,t)$$ is detection function. So $$z(x,t)$$ satisfies the following equation in the sense of weak solution $$\label{closez} \def\arraystretch{1.2}\left\{\begin{array}{@{}ll} z_{tt}(x,t)=z_{xx}(x,t), & 0<x<1, t >0, \\ z_1(1,t)=z_2(1, t),& t\ge 0, \\ z_{1x}(1,t)+z_{2x}(1, t)=d(t),& t\ge 0, \\ z_{1}(0,t)=0,& t\geq 0,\\ z_{2x}(0,t)=0,& t\geq 0,\\ z(x,0)=w(x),z_{t}(x,0)=\eta(x). \end{array}\right.$$ (5.26) If $$d(t)=0$$, then (5.26) has a solution if and only if $$(w(x),\eta(x))=0$$. In this case, the system (5.26) is asymptotically stable; If $$d(t)\neq0$$, then equation (5.26) has a non-zero solution if and only if $$d(t)$$ is an observation blind point. Suppose $$\lim_{t\longrightarrow\infty}E(w)(t)=c(w^{0},w^{1})$$ and $$c={\rm sup}_{(w^{0},w^{1})\in S(R)} c(w^{0},w^{1})$$. We have $$\|(z(x,t),\xi(x,t))\|_{{\mathcal H}}^{2}\leq \lim_{n\longrightarrow\infty} \|(z_{n}(x,t),z_{n,t}(x,t))\|_{{\mathcal H}}^{2}.$$ If $$\|(z(x,t),\xi(x,t))\|_{{\mathcal H}}^{2}=\lim_{n\longrightarrow\infty} \|(z_{n}(x,t),z_{n,t}(x,t))\|_{{\mathcal H}}^{2}=2c(w^{0},w^{1})\neq 0$$, we have $$\|(z_{n}(x,t),z_{n,t}(x,t))-(z(x,t),\xi(x,t))\|_{{\mathcal H}}^{2}\longrightarrow0,n\longrightarrow\infty.$$ Since $$\|(w,\eta)\|_{H}\leq R,|\tilde{r}(t)-\hat{r}(t)|>R,$$ we have $$(w(x),\eta(x),d(t))$$ is not a observer blind spot. So $$(z(x,t),\xi(x,t))\equiv 0, \lim_{t\longrightarrow\infty}E(w_{0},w_{1})(t)=0$$, so the solution of the system converges to zero. In summary, we have the following main result. Theorem 5.1 Suppose $$r(t)$$satisfies that $$sup_{t}|r(t)|<\infty, R\in {R}_{+},R\geq {\rm sup}_{t}|r(t)|$$ and $$M\geq 4R$$, so for any $$(w^{0},w^{1})\in {\mathcal H}$$, and $$\|(w^{0},w^{1})\|_{{\mathcal H}}\leq R$$, thus the solution of the system converges to zero. 6. Concluding remarks In this article, we study the problem of a joint wave equation with point-wise input disturbance. A novel feedback control strategy $$u(t) = -kw_{1t}(1,t)-M{\rm{sign}}(w_{1t}(1,t)) + c$$ where c is a suitable constant that depends on estimation of the initial value is attempted to stabilize the system. we obtain the existence and unique of the closed-loop system based on expansion of the Lions–Lax–Milgram Theorem and the developed Variational Method, we used the Lasalle’s Invariance Principle-like together with observation blind point to study asymptotical stability of the closed loop system. Under the improved controller, we proved that the solution of the system converges to zero in state space $${\mathcal H}$$ for any initial value in a bounded set. Funding National Natural Science Foundation of China (NSFC-61503276, 61174080, 61503275). References Ammari K. , Tucsnak M. & Henrot A. 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# Stabilization for a joint string equation with input disturbance

, Volume Advance Article – Sep 1, 2017
22 pages

/lp/ou_press/stabilization-for-a-joint-string-equation-with-input-disturbance-UmUaR8bhPb
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Oxford University Press
© The authors 2017. Published by Oxford University Press on behalf of the Institute of Mathematics and its Applications. All rights reserved.
ISSN
0265-0754
eISSN
1471-6887
D.O.I.
10.1093/imamci/dnx029
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### Abstract

Abstract The stabilization problem for a joint string subject to pointwise input disturbance is concerned in this article. First, a method like sliding mode control is adopted to resist the disturbance. Observation blind point which leads to zero output is found in this non-linear system. Second, the existence and uniqueness of the solution for this non-linear-system is obtained based on the extension of Lions–Lax–Milgram Theorem and the developed Variational Methods. Finally, the system’s asymptotic behaviour is shown by the LaSalle’s Invariance Principle-like. 1. Introduction In the past 30 years, there has been a steady increase in the study of PDE systems, since it can be widely used in engineering. The stability and stabilization by feedback of joint wave equations in bounded domains are important topics which have drawn much attention. A number of literatures can be found in Chen et al. (1987), Liu (1994) and Ammari et al. (2000). Among these works, the controller is designed in an ideal environment, that’s to say, there is not disturbance. However, in practice, the disturbance may occur in both the boundary and the internal of the system. Thus, it is of great importance to design the controller as well as to attenuate the uncertainty. Sliding mode control, which is first used in infinite dimensional systems by Orlov & Utkin (1998), has been applied to some PDEs in Guo & Jin (2013). Active disturbance rejection control is another control strategy that has been successfully used to reject the disturbance in wave, beam and Schrödinger equations (Guo & Liu, 2008; Guo & Jin, 2013; Guo et al., 2014). Some other methods that can be used to deal with uncertainties is the adaptive control, Lyapunov function approach in Guo & Guo (2013) and Ge & Zhang (2011). These methods show it is possible to tackle the problem with disturbance in the indefinite system. But we also notice that it is a challenge to obtain the stability and existence of solution to the closed-loop system because applying slide-mode controller can lead to a non-linear system. In most literatures authors use Riesz methods or approximate approach to obtain the Filippov solution in finite-dimensional system and take limit to get the solvability of the closed-loop system (Dieci & Lopez, 2009), the classical LaSalle invariance set (LaSalle, 1960) and Lyapunov method (Xu & Yung, 2003) seem not to give the asymptotic behaviour of the closed-loop system. Notice that the closed-loop system might have observation blind point if the control has disturbance. The limit functions of disturbance might fall in the set of observation blind point. So such control designs do not guarantee asymptotical stability of the closed-loop system. Another challenge is the redesign of the feedback control law. Applying the feedback controller is to stabilize the system with disturbance asymptotically or exponentially. From the stability analysis we see that the closed-loop system does not satisfy the asymptotical stability. Therefore we need to modify the feedback control law. Since the disturbance is unknown, we cannot determine explicitly its weak limit functions, including limits of state of the closed-loop system. We must redesign controller so that it stabilizes the system asymptotically. We note that some literatures obtain the stability in the sense of uniformly bounded. Obviously, this is not the aim of control. The final challenge comes from non-collocated control with disturbance. The previous challenges are based on the collocated feedback controller. If the collocated information is invalid, the classical method of anti-non-collocated control is to design the Luenberger observer. However, when the control has disturbance, the Luenberger observer does not give state reconstruction of the system, so the full state observer for the system with unknown input is necessary. However, the existing results seem to be not suitable to the system with control disturbance (e.g., see Darouach, 2009; Hassan & Zied, 2010; Demetriou, 2005; Kyung-soo & Eeun-ho, 2013; Cui et al., 2016). Again the designs of the full state observer and feedback control law are new challenges. In this article, we are concerned with the following hyperbolic equation: $$\label{oringinal} \def\arraystretch{1.2}\left\{\begin{array}{@{}ll} \hat{w}_{tt}(x,t)=\hat{w}_{xx}(x,t), & x\in(0,1)\cup (1,2), t >0,\\ \hat{w}(1^{-},t)=\hat{w}(1^{+}, t),& t\ge 0, \\ \hat{w}_x(1^{-},t)-\hat{w}_x(1^{+}, t)=u(t)+r(t),& t\ge 0, \\ \hat{w}(0,t)=0,& t\geq 0,\\ \hat{w}_{x}(2,t)=0,& t\geq 0,\\ y(t)=\hat{w}_{t}(1^{-},t),& t\geq 0, \end{array}\right.$$ (1.1) where $$u(t)$$ is a controller, $$r(t)$$ is an unknown bounded and continuous disturbance and $$y(t)$$ is the output of the system. This model can be conceived as segments of power transmission lines, aerial cable/railway systems or the upper cable part of an idealized suspension bridge. By introducing a new variable $$w(x,t)=\left[w_1(x,t),w_2(x,t)\right]^{T}$$ for $$x\in[0,1]$$ and $$t\geq 0$$, where \begin{equation*} w_1(x, t)=\hat{w}(x,t), \;\; w_2(x,t)=\hat{w}(2-x,t). \end{equation*} Then system (1.1) is transformed into an equivalent system of wave equations: $$\label{target1} \def\arraystretch{1.2}\left\{\begin{array}{@{}ll} w_{tt}(x,t)=w_{xx}(x,t), & 0<x<1, t >0, \\ w_1(1,t)=w_2(1, t),& t\ge 0, \\ w_{1x}(1,t)+w_{2x}(1, t)=u(t)+r(t),& t\ge 0, \\ w_{1}(0,t)=0,& t\geq 0,\\ w_{2x}(0,t)=0,& t\geq 0,\\ w(x,0)=w^{0}(x)=(w_{1}^{0}(x),w_{2}^{0}(x))^{T},w_{t}(x,0)=w^{1}(x)=(w_{1}^{1}(x),w_{2}^{1}(x))^{T},\\ y(t)=w_{1t}(1,t),&t\geq 0. \end{array}\right.$$ (1.2) Our main contribution of this article are: (a) sliding mode control is adopted to resist the disturbance and observation blind point is found in this non-linear system. (b) We redesign the controller and give the well-posedness of the closed-loop systems by developed variational methods. (c) Asymptotic behaviour of the system is shown by the LaSalle’s Invariance Principle-like. The article is organized as follows. In Section 2, we analysed the observation blind point of the system. In Section 3, the controller is designed based on the observation blind point. In Section 4, we studied the well-posedness of the closed-loop system. In Section 5, we proved the asymptotic behaviour of the system. 2. Observation blind point In the past decades, the controller design and the output feedback law of the partial differential equations (PDEs) relies on the exact input and the stability relies on the exact form of the feedback. If there is unknown input disturbance, the output feedback controller may not work. For convenience, we introduce the concept of the observation blind point. Definition 2.1 Suppose that $${\mathcal H}$$ is a Hilbert space. A linear dynamic system in space $${\mathcal H}$$ is described by the following system $$\label{b} \def\arraystretch{1.2}\left\{\begin{array}{@{}c} \dot{x}(t)=Ax(t)+Bu(t),\\ x(0)=x_{0}, \\ y(t)=Cx(t). \end{array}\right.$$ (2.3) For any given initial value $$x_{0}$$, if there exists a controller $$u_{0} (t)\in L_{loc}^{2}(0,\infty)$$ such that $$y(t)\equiv 0$$, then $$(x_{0},u_{0}(t))$$ is an observation blind point of the system. If there exists a constant $$\tau>0$$, $$y(t)\equiv 0$$ when $$t\geq \tau$$, then $$(x_{0},u_{0}(t))$$ is a final observation blind point of the system. For a control-observation system, the observation blind point may exist or not, if there is an unknown input or disturbance, then system (2.3) becomes $$\label{bd} \def\arraystretch{1.2} \left\{\begin{array}{@{}c} \dot{x}(t)=Ax(t)+B[u(t)+r(t)],\\ x(0)=x_{0}, \\ y(t)=Cx(t). \end{array}\right.$$ (2.4) Obviously, if $$(x_{0},r(t))$$ is an observation blind point, then the output feedback controller $$u(t)=Ky(t)$$ will not work. We can also say that the output feedback is not robust. In this part, we are going to consider the observation blind point problem of the following wave system governed by $$\label{target} \def\arraystretch{1.2}\left\{\begin{array}{@{}ll} w_{tt}(x,t)=w_{xx}(x,t), & 0<x<1, t >0, \\ w_1(1,t)=w_2(1, t),& t\ge 0, \\ w_{1x}(1,t)+w_{2x}(1, t)=u(t)+r(t),& t\ge 0, \\ w_{1}(0,t)=0,& t\geq 0,\\ w_{2x}(0,t)=0,& t\geq 0,\\ w(x,0)=w^{0}(x),w_{t}(x,0)=w^{1}(x),\\ y(t)=w_{1t}(1,t),&t\geq 0, \end{array}\right.$$ (2.5) where $$u(t)$$ is a controller, $$r(t)$$ is an unknown uniformly bounded disturbance and $$y(t)$$ is the output of the system. We will firstly deal with the underlying observation blind point, then design the controller based on the form of the observation blind point. Now we will discuss the observation blind point. $$\label{targetblind} \def\arraystretch{1.2}\left\{\begin{array}{@{}ll} w_{tt}(x,t)=w_{xx}(x,t), & 0<x<1, t >0, \\ w_1(1,t)=w_2(1, t),& t\ge 0, \\ w_{1x}(1,t)+w_{2x}(1, t)=u_{0}(t),& t\ge 0, \\ w_{1}(0,t)=0,& t\geq 0,\\ w_{2x}(0,t)=0,& t\geq 0,\\ w(x,0)=w^{0}(x),w_{t}(x,0)=w^{1}(x),\\ y(t)=w_{1t}(1,t),&t\geq 0. \end{array}\right.$$ (2.6) First we will consider the existence of the observation blind point of system (2.6). We consider system (2.5) in an energy Hilbert space $${\mathcal H}$$ $${\mathcal H}=H_{E}^{1}(0,1)\times L^2(0,1)\times H^{1}(0,1)\times L^2(0,1),H_{E}^{1}(0,1)=\left\{f(x)\in H^1(0,1) \mid f(0)=0\right\}$$ with the inner product induced norm: for $$\forall (f_1,g_1,f_2,g_2)\in {\mathcal H}$$, $$\left\|(f_{1},g_{1},f_{2},g_{2})\right\|^2=\int_0^1 f_{1}'(x)^{2}+g_{1}^2(x)+ f_{2}'(x)^{2}+g_{2}^2(x)dx.$$ Theorem (2.1) gives the existence of the observation blind point and the relationship between $$u_{0}(t)$$ and the initial value $$(w_{0},w_{1})$$. Theorem 2.1 For any boundary value $$(w_{1}(0,t),w_{1t}(0,t),w_{2}(0,t),w_{2t}(0,t)) \in {\mathcal H}$$, there exists an input $$u_{0}(t)\in L_{loc}^{2}(R_{+})$$ so that the output of the system is zero. Moreover, $$u_{0}(t)$$ is determined by the boundary value $$(w_{1}(0,t),w_{1t}(0,t),w_{2}(0,t),w_{2t}(0,t))$$. Proof. We will consider the case when $$w_{1t}(1,t)\equiv0$$. That is $$\label{targetw} \def\arraystretch{1.2}\left\{\begin{array}{@{}ll} w_{tt}(x,t)=w_{xx}(x,t), & 0<x<1, t >0, \\ w_1(1,t)=w_2(1, t),& t\ge 0, \\ w_{1}(0,t)=0,& t\geq 0,\\ w_{2x}(0,t)=0,& t\geq 0,\\ w_{1t}(1,t)=0,& t\geq 0,\\ w(x,0)=w^{0}(x)=(w_{1}^{0}(x),w_{2}^{0}(x)),w_{t}(x,0)=w^{1}(x)=(w_{1}^{1}(x),w_{2}^{1}(x)). \end{array}\right.$$ (2.7)$$w_{1t}(1,t)=0$$ implies that $$w_{1}(1,t)=c$$. In this case, $$v(x,t)=w(x,t)-(x w_{1}^{0}(1),w_{2}^{0}(1))$$ satisfies $$\label{targetv} \def\arraystretch{1.2}\left\{\begin{array}{@{}ll} v_{tt}(x,t)=v_{xx}(x,t), & 0<x<1, t >0, \\ v_1(1,t)=v_2(1, t)=0,& t\ge 0, \\ v_{1}(0,t)=0,& t\geq 0,\\ v_{2x}(0,t)=0,& t\geq 0,\\ v(x,0)=v^{0}(x)=(w_{1}^{0}(x)-xw_{1}^{0}(1),w_{2}^{0}(x)-w_{2}^{0}(1)),v_{t}(x,0)=v^{1}(x)=w^{1}(x). \end{array}\right.$$ (2.8) The eigenvalue pairs of (2.8) are $$(\lambda_{1,n},\lambda_{2,n})=\bigg(in\pi,\frac{2n-1}{2}\pi i \bigg),n=1,2,3\ldots$$ and corresponding eigenfunctions are $$(\varphi_{1,n},\varphi_{2,n})=\bigg(\sin n\pi x,\cos\frac{2n-1}{2}\pi x\bigg)$$ and the family $$\{(\varphi_{1,n},\varphi_{2,n}),n=1,2,3\ldots\}$$ forms an orthogonal basis for $$L^{2}[0,1]\times L^{2}[0,1]$$. Hence \begin{eqnarray*} w_{1}^{0}(x)-xw_{1}^{0}(1)&=&\sum_{n=1}^{\infty}a_{1,n}\sin n\pi x, \;\;w_{1}^{1}(x)= \sum_{n=1}^{\infty}b_{1,n}\sin n\pi x,\\ w_{2}^{0}(x)-w_{2}^{0}(1)&=&\sum_{n=1}^{\infty}a_{2,n}\cos(n\pi -\pi/2)x, \;\;w_{2}^{1}(x)= \sum_{n=1}^{\infty}b_{2,n}\cos(n\pi - \pi/2)x, \end{eqnarray*} where \begin{eqnarray*} a_{1,n} &=& 2\int_{0}^{1}(w_{1}^{0}(x)-xw_{1}^{0}(1))\sin n\pi xdx,\;\;b_{1,n}=2\int_{0}^{1} w_{1}^{1}(x)\sin n\pi xdx,\\ a_{2,n} &=& 2\int_{0}^{1}(w_{2}^{0}(x)-w_{2}^{0}(1))\cos(n\pi -\pi/2)xdx,\;\;b_{2,n}=2\int_{0}^{1}w_{2}^{1}(x)\cos(n\pi -\pi/2)xdx. \end{eqnarray*} The solution to (2.8) is \begin{eqnarray*} &&v(x,t)=\left( \begin{array}{c} v_{1}(x,t) \\ v_{2}(x.t) \\ \end{array} \right)\\ &=&\left( \begin{array}{c} \sum_{n=1}^{\infty}\bigg[a_{1,n}\cos n\pi t +b_{1,n}\frac{\sin n\pi t}{n\pi}\bigg]\sin n\pi x \\ \sum_{n=1}^{\infty}\bigg[a_{2,n} \cos(n\pi -\pi/2) t +b_{2,n}\frac{\sin(n\pi -\pi/2)t}{n\pi -\pi/2}\bigg]\cos(n\pi -\pi/2) x \end{array} \right)\!. \end{eqnarray*} Hence, \begin{align*} &w(x,t)=\left( \begin{array}{c} w_{1}(x,t) \\ w_{2}(x.t) \\ \end{array} \right)\\ &\quad =\left( \begin{array}{c} xw_{1}^{0}(1)+\sum_{n=1}^{\infty}\bigg[a_{1,n}\cos n\pi t +b_{1,n}\frac{\sin n\pi t}{n\pi}\bigg]\sin n\pi x \\ w_{2}^{0}(1)+\sum_{n=1}^{\infty}\bigg[a_{2,n} \cos(n\pi -\pi/2) t +b_{2,n}\frac{\sin(n\pi -\pi/2)t}{n\pi -\pi/2}\bigg]\cos(n\pi -\pi/2) x \end{array} \right) \end{align*} is the solution to (2.7). From this expression we get \begin{align*} w_{1x}(1,t)+w_{2x}(1,t)&=w_{1}^{0}(1)+\sum_{n=1}^{\infty}\bigg[a_{1,n} n\pi\cos n\pi t +b_{1,n}\sin n\pi t\bigg]\\ &\quad+\sum_{n=1}^{\infty}\bigg[a_{2,n}(n\pi -\pi/2) \cos(n\pi -\pi/2) t +b_{2,n}\sin(n\pi -\pi/2)t\bigg]. \end{align*} We take $$u_{0}(t)=r(t)=w_{1x}(1,t)+w_{2x}(1,t)$$. Obviously, the pair $$(w^{0},w^{1},r(t))$$ is an observation blind point to (2.6). From the above discussion we see that $$r(t)$$ is uniquely determined by the initial value $$(w^{0}(x),w^{1}(x))$$. Note that \begin{eqnarray*} a_{1,n}(n\pi) &=& 2\int_{0}^{1}(w_{1}^{0'}(x)-w_{1}^{0}(1))\cos n\pi xdx +w_{1}^{0}(0)\\ a_{1,n}(n\pi -\pi/2) &=&(-1)^{n+1}-2\int_{0}^{1} w_{2}^{0'}(x)\sin(n\pi -\pi/2)x\,dx \end{eqnarray*} so $$r(t)$$ depends continuously on $$(w^{0},w^{1})$$. From Theorem 2.1 we can denote \begin{eqnarray*} \hat{r}(t)=r(t,w^{0},w^{1})&=&w_{1}^{0}(1)+\sum_{n=1}^{\infty}\bigg[a_{1,n} n\pi\cos n\pi t +b_{1,n}\sin n\pi t\bigg]\\ &&+\sum_{n=1}^{\infty}\bigg[a_{2,n}(n\pi -\pi/2) \cos(n\pi -\pi/2) t +b_{2,n}\sin(n\pi -\pi/2)t\bigg]. \end{eqnarray*} We define a set composed of all observation blind points of (2.6) $${\mathcal G}=\{\hat{r}(t)=r(t,w^{0},w^{1})|\forall(w^{0},w^{1})\in {\mathcal H}\}.$$ It can be proved that $${\mathcal G}$$ is a closed space in $$L_{loc}^{2}(R^{+})$$. The desired result follows. $$\quad\Box$$ When $$r(t)$$ is in the position of the observation blind point, the used feedback control law $$u(t)=-kw_{1t}(1,t)-M{\rm sign}(w_{1t}(1,t))$$ is invalid. A question arises: how do we remove the observation blind point and what is the feedback controller? In the following we will discuss the design of the feedback controller to the system (2.5). Let’s consider the following system $$\label{stru0} \def\arraystretch{1.2}\left\{\begin{array}{@{}ll} z_{tt}(x,t)=z_{xx}(x,t), & 0<x<1, t >0, \\ z_1(1,t)=z_2(1, t)=0,& t\ge 0, \\ z_{1}(0,t)=0,& t\geq 0,\\ z_{2x}(0,t)=0,& t\geq 0,\\ z(x,0)=z^{0}(x),z_{t}(x,0)=z^{1}(x),& t\geq 0,\\ \hat{r}(t)=z_{1x}(1,t)+z_{2x}(1,t),& t\geq 0. \end{array}\right.$$ (2.9) where $$z^{0}(1)=0$$. Based on Theorem 2.1, $$\hat{r}(t)$$ is uniquely determined by $$(z^{0},z^{1})$$. Noting that $$F(f)=f(1)$$ is a bounded linear functional, $$N(F)$$ is a closed subset in $$H_{E}^{1}(0,1)\times H^{1}(0,1)$$ $$N(F)=\{(f_{1},f_{2})| f_{1}(0)=0,f_{1}(1)=f_{2}(1)=0\}.$$ A direct computation shows that $$N(F)^{\bot}=span\{(x,1)^{T}\}.$$ So for any $$w^{0}(x)\in H_{E}^{2}(0,1)$$, we have $$w^{0}=z^{0}(x)+(xw_{1}^{0}(1),w_{2}^{0}(1))^{T}\in {\mathcal N}(F)\oplus {\mathcal N}(F)^{\perp}$$ where $$z^{0}(x)=[w^{0}(x)-(xw_{1}^{0}(1),w_{2}^{0}(1))^{T}]$$ and $$z^{0}(1)=0$$. Suppose $$z(x,t)$$ is a solution of (2.9), the initial value is $$(z^{0},w^{1})$$ and $$w(x,t)=z(x,t)+(xw_{1}^{0}(1),w_{2}^{0}(1))^{T}.$$ Obviously $$w(x,t)$$ satisfies equation (2.8) and $$w(x,0)=w_{0}(x),w_{t}(x,0)=w_{1}(x)$$. So $$u_{0}(t)=w_{1x}(1,t)+w_{2x}(1,t)=\hat{r}(t)+w_{1}^{0}(1).$$ For any $$w(x)\in H_{E}^{1}[0,1]$$, we have the following inequality $$w(1)=\int_{0}^{1}w_{x}(x)dx\leq \sqrt{\int_{0}^{1}|w_{x}(x)|^{2}dx},$$ thus $$\label{ur} |u_{0}(t)-\hat{r}(t)|=|w_{1}^{0}(1)|\leq \parallel(w^{0},w^{1})\parallel.$$ (2.10) 3. Control design The energy of system (2.5) is $$E(t)=\frac{1}{2}\int_{0}^{1} w_{1x}^{2}(x,t)+w_{1t}^{2}(x,t)+w_{2x}^{2}(x,t)+w_{2t}^{2}(x,t) dx.$$ The derivation of $$E(t)$$ with respect to $$t$$ is \begin{eqnarray*} \dot{E}(t)&=&\int_{0}^{1} w_{1x}(x,t)w_{1xt}(x,t)+w_{1t}(x,t)w_{1tt}(x,t)+w_{2x}(x,t)w_{2xt}(x,t)+w_{2t}(x,t)w_{2tt}(x,t)dx,\\ &=& w_{1x}w_{1t}|_{0}^{1}+w_{2x}w_{2t}|_{0}^{1},\\ &=&(u(t)+r(t))w_{1t}(1,t). \end{eqnarray*} In order to make $$\dot{E}(t)$$ dissipative, let $$u(t)=-k w_{1t}(1,t)+m_{1}(t)+m_{2}(t)$$ where $$k>0$$, $$m_{i},i=1,2$$ are to be determined. Then we have $$\dot{E}(t)=-kw_{1t}^{2}(1,t)+w_{1t}(1,t)(m_{1}(t)+m_{2}(t)+r(t)).$$ Since $$r(t)$$ is an unknown function, we are not sure whether it is an observation blind point of system (2.5). Suppose that the initial value $$(w^{0},w^{1})$$ satisfies $$\parallel(w^{0},w^{1})\parallel_{{\mathcal H}}\leq R$$ and $$R\geq {\rm sup}_{t\geq 0}|r(t)|$$. In this case $$r(t)$$ might be an observation blind point. That is because $$w_{1}^{0}(1)=\int_{0}^{1}w_{1,x}^{0}(x)dx\leq R.$$ In order to reject the disturbance $$r(t)$$, we choose $$m_{2}(t)= 3R$$. So $$m_{2}(t)+r(t)\geq0$$ and $$\mid m_{2}(t)+r(t)-\hat{r}(t)\mid > R.$$ So for any $$(w^{0},w^{1})\in {\mathcal H}$$, we have $$\parallel(w^{0},w^{1})\parallel\leq R$$, for any $$\tau > 0$$, we have $$((w^{0},w^{1}),m_{2}(t+\tau)+r(t+\tau))\notin {\mathcal G}$$ which shows that $$w_{1t}(1,t+\tau)\neq \theta(t)\equiv 0$$. Finally, we choose $$m_{1}(t)=-M {\rm{sign}}(w_{1t}(1,t))$$, where $$M\geq 4R\geq sup_{t\geq 0}\mid m_{2}(t)+r(t)\mid.$$ Then $$w_{1t}(1,t)(m_{1}(t)+m_{2}(t)+r(t))\leq 0$$, and $$\dot{E}(t)=-kw_{1t}^{2}(1,t)+w_{1t}(1,t)(m_{1}(t)+m_{2}(t)+r(t))\leq 0.$$ Thus we choose $$u(t)=-kw_{1t}(1,t)-M {\rm{sign}}(w_{1t}(1,t))+3R,$$ (3.11) where $$M\geq 4R.$$ Then we get the closed-loop system $$\def\arraystretch{1.2}\left\{\begin{array}{@{}ll} w_{tt}(x,t)=w_{xx}(x,t), & 0<x<1, t >0, \\ w_1(1,t)=w_2(1, t),& t\ge 0, \\ w_{1x}(1,t)+w_{2x}(1, t)=-kw_{1t}(1,t)-M {\rm{sign}}(w_{1t}(1,t))+r(t)+3R,& t\ge 0, \\ w_{1}(0,t)=0,& t\geq 0,\\ w_{2x}(0,t)=0,& t\geq 0,\\ w(x,0)=w^{0}(x),w_{t}(x,0)=w^{1}(x),& t\geq 0, \end{array}\right.$$ (3.12) where $$\|(w^{0},w^{1})\|_{{\mathcal H}}\leq R$$ and $$\tilde{r}(t)=r(t)+3R$$. It can be easily seen that $$\mid \tilde{r}(t)-\hat{r}(t)\mid> R$$, which shows that $$((w^{0},w^{1}),\tilde{r}(t))$$ is not an observation blind point of system (3.12). 4. The well-posedness of the closed-loop system In this part, we are going to show the well-posedness of the close-loop system (3.12). As can been seen, (3.12) is a non-linear system, so we adopt the calculus of variations to deal with the existence of the solution. The result is based on the following lemma (Cui et al., 2016), which can be seen as an expansion of linear Lions–Lax–Milgram to the semi-linear case. Lemma 4.1 Suppose $${\mathcal F}$$ is a Hilbert space with the norm $$\|v\|_{{\mathcal F}}$$, $${\it{\Phi}}$$ is a subspace of $${\mathcal F}$$, and $${\it{\Phi}}$$ is a normed linear space with the norm $$\|\phi\|_{{\it{\Phi}}}$$, $$E(v_{1},v_{2},\phi)$$ is a triple linear functional defined on $${\mathcal F} \times {\mathcal F}\times {\it{\Phi}}$$ which satisfies \begin{eqnarray*} E(\alpha v_{1}+\beta u_{1},v_{2},\phi)=\alpha E(v_{1},v_{2},\phi)+\beta E(u_{1},v_{2},\phi),\\ E(v_{1},\alpha v_{2}+\beta u_{2},\phi)=\alpha E(v_{1},v_{2},\phi)+\beta E(v_{1},u_{2},\phi),\\ E(v_{1},v_{2},\alpha \phi+\beta \psi)=\alpha E(v_{1},v_{2},\phi)+\beta E(v_{1},v_{2},\psi), \end{eqnarray*} and $$N(v_{1},v_{2},\phi)$$ is semi-linear in space $${\mathcal F}\times {\mathcal F}\times{\it{\Phi}}$$, and it is linear with respect to $$\phi$$, the multiple range is admissible. Suppose $$E(v_{1},v_{2},\phi)$$ and $$N(v_{1},v_{2},\phi)$$ satisfy the following form: (1) There exists a constant $$C_{0}$$ such that $$\|\phi\|_{{\mathcal F}}\leq C_{0}\|\phi\|_{{\it{\Phi}}},\forall \phi\in{\it{\Phi}}$$; (2) For any $$\phi\in{\it{\Phi}}$$, $$E(v_{1},v_{2},\phi)$$ is continuous with respect to $$v_{1},v_{2}$$; (3) There exists a constant $$C_{2}$$ such that $$E(\phi,\phi,\phi)\geq C_{2}\|\phi\|_{{\it{\Phi}}},\forall \phi\in{\it{\Phi}}$$; (4) $$N(v_{1},v_{2},\phi)$$ is $${\it{\Phi}}$$-monotonous \begin{eqnarray*} &&N(\phi,\phi,\phi)+N(\psi,\psi,\psi)+ N(\phi,\phi,\psi)+N(\psi,\psi,\phi)\\ &&-N(\psi,\phi,\phi)-N(\phi,\psi,\psi)-N(\phi,\psi,\phi)-N(\psi,\phi,\psi)\geq 0,\forall \phi,\psi\in{\it{\Phi}}. \end{eqnarray*} is full range, for $$\lambda>0$$, $$N(v_{1},v_{2},\eta)=[\lim_{\lambda\longrightarrow0}N(v_{1}-\lambda \eta,\eta),\lim_{\lambda\longrightarrow0}N(v_{1}+\lambda \eta,\eta),\lim_{\lambda\longrightarrow0}N(v_{2}-\lambda \eta,\eta),\lim_{\lambda\longrightarrow0}N(v_{2}+\lambda \eta,\eta)],$$$$\forall \eta\in{\it{\Phi}} ;$$ (5) There exists a constant $$C_{3}$$ such that $$N(v_{1},v_{2},\phi)\leq C_{3}\|\phi\|_{{\it{\Phi}}},\forall \phi\in{\it{\Phi}}$$. Then for any $$L(\phi)$$, which is a bounded linear functional on space $${\it{\Phi}}$$, there exists a $$u=(u_{1},u_{2})\in{\mathcal F}\times{\mathcal F}$$ such that $$E(u_{1},u_{2},\phi)+N(u_{1},u_{2},\phi)=L(\phi),\forall \phi\in{\it{\Phi}}.$$ 4.1. The existence of the solution Now we are going to show the existence of system (3.12). First, we standardize the equation and the initial conditions, let $$v(x,t)=w(x,t)-w^{0}(x).$$ Then (3.12) becomes $$\def\arraystretch{1.2}\left\{\begin{array}{@{}ll} v_{tt}(x,t)=v_{xx}(x,t)+w_{xx}^{0}(0), & 0<x<1, t >0, \\ v_1(1,t)=v_2(1, t),& t\ge 0, \\ v_{1x}(1,t)+v_{2x}(1, t)=-kv_{1t}(1,t)-M {\rm{sign}}(v_{t}(1,t))\\ +\tilde{r}(t)-w_{1x}^{0}(1)-w_{2x}^{0}(1),& t\ge 0, \\ v_{1}(0,t)=0,& t\geq 0,\\ v_{2x}(0,t)=0,& t\geq 0,\\ v(x,0)=v^{0}(x),v_{t}(x,0)=v^{1}(x),& t\geq 0,\\ y(t)=v_{t}(1,t). \end{array}\right.$$ (4.13) In order to prove the existence of the solution, we first transform equation (4.13) into a variational equations. Suppose $$\eta >0$$, for any $$T>0$$, integrating the first equation of (4.13) by multiplying $$e^{-\eta t}\phi_{t}(x,t)$$ on domain $$[0,1]\times [0,T]$$, we get \begin{eqnarray*} 0 &=& \int_{0}^{T}\int_{0}^{1}e^{-\eta t}[v_{1tt}(x,t)-v_{1xx}(x,t)-w_{1xx}^{0}(x)+v_{2tt}(x,t)-v_{2xx}(x,t)-w_{2xx}^{0}(x)]\phi_{t}(x,t)dx dt \\ &=& \int_{0}^{1}e^{-\eta T}v_{1t}(x,T)\phi_{t}(x,T)dx-\int_{0}^{1}v_{1t}(x,0)\phi_{t}(x,0)dx-\int_{0}^{T}\int_{0}^{1}v_{1t}(x,t)[e^{-\eta t}\phi_{t}(x,t)]_{t}dxdt\\ &&+\int_{0}^{T}[e^{-\eta t}v_{1x}(0,t)\phi_{t}(0,t)-e^{-\eta t}v_{1x}(1,t)\phi_{t}(1,t)]dt+\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{xt}(x,t)v_{1x}(x,t)dxdt\\ &&+\int_{0}^{T}[e^{-\eta t}\phi_{t}(0,t)w_{1x}^{0}(0)-e^{-\eta t}\phi_{t}(1,t)w_{1x}^{0}(1)]dt+\int_{0}^{1}\int_{0}^{T}[e^{-\eta t}\phi_{tx}(x,t)]w_{1x}^{0}(x)dtdx\\ &&+ \int_{0}^{1}e^{-\eta T}v_{2t}(x,T)\phi_{t}(x,T)dx-\int_{0}^{1}v_{2t}(x,0)\phi_{t}(x,0)dx-\int_{0}^{T}\int_{0}^{1}v_{2t}(x,t)[e^{-\eta t}\phi_{t}(x,t)]_{t}dxdt\\ &&+\int_{0}^{T}[e^{-\eta t}v_{2x}(0,t)\phi_{t}(0,t)-e^{-\eta t}v_{2x}(1,t)\phi_{t}(1,t)]dt+\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{xt}(x,t)v_{2x}(x,t)dxdt\\ &&+\int_{0}^{T}[e^{-\eta t}\phi_{t}(0,t)w_{2x}^{0}(0)-e^{-\eta t}\phi_{t}(1,t)w_{2x}^{0}(1)]dt+\int_{0}^{1}\int_{0}^{T}[e^{-\eta t}\phi_{tx}(x,t)]w_{2x}^{0}(x)dtdx\\ &=& \int_{0}^{1}e^{-\eta T}v_{1t}(x,T)\phi_{t}(x,T)dx-\int_{0}^{1}v_{1t}(x,0)\phi_{t}(x,0)dx-\int_{0}^{T}\int_{0}^{1}v_{1t}(x,t) [e^{-\eta t}\phi_{t}(x,t)]_{t}dxdt\\ &&+\int_{0}^{T}e^{-\eta t}v_{1x}(0,t)\phi_{t}(0,t)-e^{-\eta t}v_{1x}(1,t)\phi_{t}(1,t)dt+\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{xt}(x,t)v_{1x}(x,t)dxdt\\ &&+\int_{0}^{T}[e^{-\eta t}\phi_{t}(0,t)w_{1x}^{0}(0)-e^{-\eta t}\phi_{t}(1,t)w_{1x}^{0}(1)]dt+\eta\int_{0}^{1}\int_{0}^{T}[e^{-\eta t}\phi_{x}(x,t)]w_{1x}^{0}(x)dtdx\\ &&+\int_{0}^{1}e^{-\eta t}\phi_{x}(x,t)w_{1x}^{0}(x)|_{t=0}^{T}dx\\ &&+\int_{0}^{1}e^{-\eta T}v_{2t}(x,T)\phi_{t}(x,T)dx-\int_{0}^{1}v_{2t}(x,0)\phi_{t}(x,0)dx-\int_{0}^{T}\int_{0}^{1}v_{2t} (x,t)[e^{-\eta t}\phi_{t}(x,t)]_{t}dxdt\\ &&+\int_{0}^{T}[e^{-\eta t}v_{2x}(0,t)\phi_{t}(0,t)-e^{-\eta t}v_{2x}(1,t)\phi_{t}(1,t)]dt+\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{xt}(x,t)v_{2x}(x,t)dxdt\\ &&+\int_{0}^{T}[e^{-\eta t}\phi_{t}(0,t)w_{2x}^{0}(0)-e^{-\eta t}\phi_{t}(1,t)w_{2x}^{0}(1)]dt +\eta\int_{0}^{1}\int_{0}^{T}[e^{-\eta t}\phi_{x}(x,t)]w_{2x}^{0}(x)dtdx\\ &&\int_{0}^{1}e^{-\eta t}\phi_{x}(x,t)w_{2x}^{0}(x)|_{t=0}^{T}dx. \end{eqnarray*} Suppose $$\phi(x,0)=0,\phi_{t}(0,t)=0,\phi(x,T)=0,\phi_{t}(x,T)=0$$, we have \begin{eqnarray} &&\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{xt}(x,t)v_{1x}(x,t)dxdt-\int_{0}^{T}\int_{0}^{1}v_{1t}(x,t)[e^{-\eta t}\phi_{t}(x,t)]_{t}dxdt\nonumber\\ &&-\int_{0}^{T}\int_{0}^{1}v_{2t}(x,t)[e^{-\eta t}\phi_{t}(x,t)]_{t}dxdt+\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{xt}(x,t)v_{2x}(x,t)dxdt\nonumber\\ &&+\int_{0}^{T}e^{-\eta t}\phi_{t}(1,t)k v_{1t}(1,t)dt+M\int_{0}^{T}e^{-\eta t}\phi_{t}(1,t){\rm {\rm{sign}}} v_{1t}(1,t)dt\nonumber\\ &=&\int_{0}^{1}w_{1}^{1}(x)\phi_{t}(x,0)dx+\int_{0}^{1}w_{2}^{1}(x)\phi_{t}(x,0)dx+\int_{0}^{T}e^{-\eta t}\phi_{t}(1,t)\tilde{r}(t)dt\nonumber\\ && -\eta\int_{0}^{1}\int_{0}^{T}[e^{-\eta t}\phi_{x}(x,t)]w_{1x}^{0}(x)dtdx-\eta\int_{0}^{1}\int_{0}^{T}[e^{-\eta t}\phi_{x}(x,t)]w_{2x}^{0}(x)dtdx. \end{eqnarray} (4.14) Obviously equation (4.14) includes three parts \begin{gather} \begin{aligned}[b] E(v_{1},v_{2},\phi)&=\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{xt}(x,t)v_{1x}(x,t)dxdt-\int_{0}^{T}\int_{0}^{1}v_{1t}(x,t)[e^{-\eta t}\phi_{t}(x,t)]_{t}dxdt\\ &\quad-\int_{0}^{T}\int_{0}^{1}v_{2t}(x,t)[e^{-\eta t}\phi_{t}(x,t)]_{t}dxdt+\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{xt}(x,t)v_{2x}(x,t)dxdt\\ &\quad+\int_{0}^{T}e^{-\eta t}\phi_{t}(1,t)k v_{1t}(1,t)dt, \end{aligned}\\ \end{gather} (4.15) \begin{gather} N(v_{1},v_{2},\phi)=M\int_{0}^{T}e^{-\eta t}\phi_{t}(1,t){\rm sign} v_{1t}(1,t)dt,\\ \end{gather} (4.16) \begin{gather} \begin{aligned}[b] L(\phi)&=\int_{0}^{1}w_{1}^{1}(x)\phi_{t}(x,0)dx+\int_{0}^{1}w_{2}^{1}(x)\phi_{t}(x,0)dx+\int_{0}^{T}e^{-\eta t}\phi_{t}(1,t)\tilde{r}(t)dt\\ &\quad -\eta\int_{0}^{1}\int_{0}^{T}[e^{-\eta t}\phi_{x}(x,t)]w_{1x}^{0}(x)dtdx-\eta\int_{0}^{1}\int_{0}^{T}[e^{-\eta t}\phi_{x}(x,t)]w_{2x}^{0}(x)dtdx. \end{aligned} \end{gather} (4.17) Thus (4.14) can be written as $$E(v_{1},v_{2},\phi)+N(v_{1},v_{2},\phi)=L(\phi).$$ Suppose \begin{eqnarray} {\mathcal F} &=& \def\arraystretch{1.3}\left \{v(x,t) \left |\begin{array}{c} e^{-\eta t}v(x,t),e^{-\eta t}v_{x}(x,t),e^{-\eta t}v_{t}(x,t)\in L^{2}(0,1),\\ e^{-\eta t}v_{t}(1,t)\in L^{2}(0,T), \\ v(0,t)=0, \\ v(x,0)=0 \end{array} \right. \right\} \end{eqnarray} (4.18) with the norm $$\|v\|_{{\mathcal F}}^{2}=\int_{0}^{T}\int_{0}^{1} e^{-\eta t}[|v(x,t)|_{x}^{2}+|v(x,t)|_{t}^{2}]dxdt+\int_{0}^{T}e^{-\eta t}|v_{t}^{2}(1,t)|dt,$$ and \begin{eqnarray} {\it{\Phi}} &=& \def\arraystretch{1.3}\left \{\phi(x,t) \left |\begin{array}{c} e^{-\eta t}\phi_{t}(x,t),e^{-\eta t}\phi_{xt}(x,t),e^{-\eta t}\phi_{tt}(x,t)\in L^{2}(0,1),\\ e^{-\eta t}\phi_{t}(1,t)\in L^{2}(0,T), \\ \phi(0,t)=0, \\ \phi(x,0)=\phi(x,T)=\phi_{t}(x,T)=0 \end{array} \right. \right\} \end{eqnarray} (4.19) with the norm $$\|\phi\|_{{\it{\Phi}}}^{2}=\int_{0}^{T}\int_{0}^{1} e^{-\eta t}[|\phi(x,t)|_{x}^{2}+|\phi(x,t)|_{t}^{2}]dxdt+\int_{0}^{T}e^{-\eta t}|\phi_{t}(1,t)|^{2}dt+\int_{0}^{1}|\phi_{t}(x,0)|^{2}dx.$$ It can be seen that $${\it{\Phi}}$$ is a linear subspace of $${\mathcal F}$$, $${\mathcal F}$$ is Hilbert space and $${\it{\Phi}}$$ is a normed linear space. In addition, $$\|\phi\|_{{\mathcal F}}^{2}\leq\|\phi\|_{{\it{\Phi}}}^{2},\forall \phi \in{\it{\Phi}}$$. In order to discuss the existence of the solution, we give the definition of weak solution. Definition 4.1 Suppose that $$w(x,t)$$ is differentiable with respect to $$x,t$$, $$w_{xx}(x,t),w_{t}(x,t)\in L^{2}[0,1]$$,$$w_{t}(1,t)\in L^{2}[0,T]$$, and $$\|w_{xx}(\cdot,t)-w_{xx}^{0}\|_{L^{2}[0,1]}\longrightarrow 0, \|w_{t}(\cdot,t)-w^{1}\|_{L^{2}[0,1]}\longrightarrow 0, t\longrightarrow 0.$$ Let $$v(x,t)=w(x,t)-w^{0}\in {\mathcal F}$$, if for any $$\phi(x,t)\in{\it{\Phi}}, v(x,t)$$ satisfies \begin{align} &\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{xt}(x,t)v_{1x}(x,t)dxdt-\int_{0}^{T}\int_{0}^{1}v_{1t}(x,t)[e^{-\eta t}\phi_{t}(x,t)]_{t}dxdt\nonumber\\ &\qquad-\int_{0}^{T}\int_{0}^{1}v_{2t}(x,t)[e^{-\eta t}\phi_{t}(x,t)]_{t}dxdt+\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{xt}(x,t)v_{2x}(x,t)dxdt\nonumber\\ &\qquad+\int_{0}^{T}e^{-\eta t}\phi_{t}(1,t)k v_{1t}(1,t)dt+M\int_{0}^{T}e^{-\eta t}\phi_{t}(1,t){\rm sign} v_{1t}(1,t)dt\nonumber\\ &\quad=\int_{0}^{1}w_{1}^{1}(x)\phi_{t}(x,0)dx+\int_{0}^{1}w_{2}^{1}(x)\phi_{t}(x,0)dx+\int_{0}^{T}e^{-\eta t}\phi_{t}(1,t)\tilde{r}(t)dt\nonumber\\ &\qquad -\eta\int_{0}^{1}\int_{0}^{T}[e^{-\eta t}\phi_{x}(x,t)]w_{1x}^{0}(x)dtdx-\eta\int_{0}^{1}\int_{0}^{T}[e^{-\eta t}\phi_{x}(x,t)]w_{2x}^{0}(x)dtdx, \end{align} (4.20) then we call $$w(x,t)$$ a weak solution of system (3.12). From the definition, we can see $$v(x,t)$$ is the solution of the variational equations. Next, we are going to prove the existence of the weak solution using Lemma (4.1). First we demonstrate the all conditions of Lemma (4.1). Proposition 4.1 Let $$E(v_{1},v_{2},\phi)$$ be defined by (4.15), then $$E(v_{1},v_{2},\phi)$$ is a bilinear form in space $${\mathcal F}\times{\it{\Phi}}$$. It is also continuous for any $$\phi\in{\it{\Phi}}$$ with respect to $$v$$. Moreover, there exists a constant $$C_{2}>0$$ so that $$E(\phi,\phi,\phi)\geq C_{2}\|\phi\|_{{\it{\Phi}}}^{2}$$. Proof. It is easy to see $$E(v_{1},v_{2},\phi)$$ is a bilinear form in space $${\mathcal F}\times{\it{\Phi}}$$. So we only need to demonstrate that it is continuous with respect to $$v=(v_{1},v_{2})^{T}$$ for any $$v_{1},v_{2}\in {\mathcal F},\phi \in{\it{\Phi}}$$ \begin{eqnarray*} |E(v_{1},v_{2},\phi)|^{2} &\leq&\bigg(\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{xt}(x,t)v_{1x}(x,t)dxdt+\eta\int_{0}^{T}\int_{0}^{1}e^{-\eta t}v_{1t}(x,t)\phi_{t}(x,t)dxdt\nonumber\\ &&-\int_{0}^{T}\int_{0}^{1}e^{-\eta t}v_{1t}(x,t)\phi_{tt}(x,t)dxdt-\int_{0}^{T}\int_{0}^{1}e^{-\eta t}v_{2t}(x,t)\phi_{tt}(x,t)dxdt\nonumber\\ &&+\eta \int_{0}^{T}\int_{0}^{1}e^{-\eta t}v_{2t}(x,t)\phi_{t}(x,t)dxdt+\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{xt}(x,t)v_{2x}(x,t)dxdt\nonumber\\ &&+\int_{0}^{T}e^{-\eta t}\phi_{t}(1,t)k v_{1t}(1,t)\bigg)^{2}dt\\ &\leq& \bigg(\int_{0}^{T}\int_{0}^{1}e^{-\eta t}v_{1x}^{2}(x,t)dxdt+\eta\int_{0}^{T}\int_{0}^{1}e^{-\eta t}v_{1t}^{2}(x,t)dxdt\nonumber\\ &&+\int_{0}^{T}\int_{0}^{1}e^{-\eta t}v_{1t}^{2}(x,t)dxdt+\int_{0}^{T}\int_{0}^{1}e^{-\eta t}v_{2t}^{2}(x,t)dxdt\nonumber\\ &&+\eta \int_{0}^{T}\int_{0}^{1}e^{-\eta t}v_{2t}^{2}(x,t)dxdt+\int_{0}^{T}\int_{0}^{1}e^{-\eta t}v_{2x}^{2}(x,t)dxdt\nonumber\\ &&+k\int_{0}^{T}e^{-\eta t} v_{1t}^{2}(1,t)\bigg)^{1/2}dt\\ &&\times \bigg(\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{xt}^{2}(x,t)dxdt+\eta\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{t}^{2}(x,t)dxdt\nonumber\\ &&+\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{tt}^{2}(x,t)dxdt+\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{tt}^{2}(x,t)dxdt\nonumber\\ &&+\eta \int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{t}^{2}(x,t)dxdt+\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{xt}^{2}(x,t)dxdt\nonumber\\ &&+k\int_{0}^{T}e^{-\eta t}\phi_{t}^{2}(1,t)\bigg)^{1/2}dt\\ &\leq&\max\{2(1+\eta)^{2},k^{2}\} \bigg(\int_{0}^{T}\int_{0}^{1}e^{-\eta t}v_{1x}^{2}(x,t)dxdt\nonumber\\ &&+\int_{0}^{T}\int_{0}^{1}e^{-\eta t}v_{1t}^{2}(x,t)dxdt+\int_{0}^{T}\int_{0}^{1}e^{-\eta t}v_{2t}^{2}(x,t)dxdt\nonumber\\ &&+\int_{0}^{T}\int_{0}^{1}e^{-\eta t}v_{2x}^{2}(x,t)dxdt+k\int_{0}^{T}e^{-\eta t} v_{1t}^{2}(1,t)\bigg)^{1/2}dt\\ &&\times \bigg(\int_{0}^{T}\int_{0}^{1}e^{-\eta t}|\phi_{xt}(x,t)|^{2}dxdt+\int_{0}^{T}\int_{0}^{1}e^{-\eta t}|\phi_{t}(x,t)|^{2}dxdt\\ &&+\int_{0}^{T}\int_{0}^{1}e^{-\eta t}|\phi_{tt}(x,t)_{t}|^{2}dxdt+\int_{0}^{T}e^{-\eta t}|\phi_{t}(1,t)|^{2}dt\bigg)^{1/2}. \end{eqnarray*} So $$E(v_1,v_2,\phi)\leq C_{1}(\parallel v_{1}\parallel_{{\cal F}}+\parallel v_{2}\parallel_{{\cal F}})(\parallel \phi\parallel_{{\cal F}}+\parallel \phi_{t}\parallel_{{\cal F}})$$ where $$C_{1}=\max\{2(1+\eta)^{2},k^{2}\}$$. For any $$\phi$$, \begin{eqnarray*} E(\phi,\phi,\phi) &=& \int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{xt}(x,t)\phi_{x}(x,t)dxdt+\eta\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{t}(x,t)\phi_{t}(x,t)dxdt\nonumber\\ &&-\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{t}(x,t)\phi_{tt}(x,t)dxdt-\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{t}(x,t)\phi_{tt}(x,t)dxdt\nonumber\\ &&+\eta \int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{t}(x,t)\phi_{t}(x,t)dxdt+\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{xt}(x,t)\phi_{x}(x,t)dxdt\nonumber\\ &&+\int_{0}^{T}e^{-\eta t}\phi_{t}(1,t)k \phi_{t}(1,t)\nonumber\\ &=&2\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\frac{d}{dt}\left(\frac{1}{2}|\phi_{x}(x,t)|^{2}\right)dxdt-2\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\frac{d}{dt}(\frac{1}{2}\phi_{t}(x,t))dxdt\\ &&+2\eta\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{t}(x,t)^{2}dxdt+k\int_{0}^{T}e^{-\eta t}\phi_{t}(1,t)\phi_{t}(1,t)dt\\ &=&\eta\int_{0}^{T}\int_{0}^{1}e^{-\eta t}|\phi_{x}(x,t)|^{2}dxdt+\eta\int_{0}^{T}\int_{0}^{1}e^{-\eta t}|\phi_{t}(x,t)|^{2}dxdt\\ &&+k\int_{0}^{T}e^{-\eta t}|\phi_{t}(1,t)|^{2}dt+\int_{0}^{1}|\phi_{t}(x,0)|^{2}dx\\ &\geq&\min \{\eta,k,1\}\bigg(\int_{0}^{T}\int_{0}^{1} e^{-\eta t}[|\phi_x(x,t)|^{2}+|\phi_t(x,t)|^{2}]dxdt\\ &&+\int_{0}^{T}e^{-\eta t}|\phi_{t}(1,t)|^{2}dt+\int_{0}^{1}|\phi_{t}(x,0)|^{2}dx\bigg) \end{eqnarray*} So $$E(\phi,\phi,\phi)$$ satisfies that $$E(\phi,\phi,\phi)\geq C_{2}\|\phi\|,\forall \phi\in{\it{\Phi}}$$ where $$C_{2}=\min \{\eta,k,1\}$$. □ Proposition 4.2 Let $$N(v_{1},v_{2},\phi)$$ be defined by (4.16), then for any $$v\in{\cal F}$$, $$N(v_{1},v_{2},\phi)$$ is a bounded and linear, specially, there exists a constant $$C_{3}>0$$ such that $$|N(v_{1},v_{2},\phi)|\leq C_{3}\|\phi\|_{{\it{\Phi}}}$$ for any $$v\in {\cal F}$$. $$N(v_{1},v_{2},\phi)$$ is monotonous with respect to $${\it{\Phi}}$$. \begin{eqnarray*} &&N(\phi,\phi,\phi)+N(\psi,\psi,\psi)+ N(\phi,\phi,\psi)+N(\psi,\psi,\phi)\\ &&-N(\psi,\phi,\phi)-N(\phi,\psi,\psi)-N(\phi,\psi,\phi)-N(\psi,\phi,\psi)\geq 0,\forall \phi,\psi\in{\it{\Phi}}. \end{eqnarray*} Moreover, $$N(v_{1},v_{2},\phi)$$ is full range. Proof. Let $$N(v_{1},v_{2},\phi)$$ be defined by (4.16). Obviously, for any $$\phi\in{\it{\Phi}}$$, $$N(v_{1},v_{2},\phi)$$ might be multi-valued function with respect to $$v_{1}$$. $$N(v_{1},v_{2},\phi)$$ is non-linear with respect to $$v_{1}$$ and it is linear with respect to $$\phi$$, and $$|N(v_{1},v_{2},\phi)|^{2}\leq M^{2}\bigg(\int_{0}^{T}e^{-\eta t}|{\rm sign}(v_{1t}(1,t))\phi_{t}(1,t)|\bigg)^{2}\leq M^{2}\frac{1-e^{-\eta t}}{\eta}\int_{0}^{T}e^{-\eta t}|\phi_{t}(1,t)|^{2}dt.$$ Thus, for all $$v_{1},v_{2},\in {\cal F}$$, $$N(v_{1},v_{2},\phi)$$ satisfies $$|N(v_{1},v_{2},\phi)|\leq C_{3}\|\phi\|_{{\it{\Phi}}},\forall \phi\in{\it{\Phi}}$$ where $$C_{3}=M\sqrt{\frac{1-e^{-\eta t}}{\eta}}$$, so $$N(v_{1},v_{2},\phi)$$ is continuous with respect to $$\phi$$. For any $$\phi,\psi\in{\it{\Phi}}$$, \begin{align*} & N(\phi,\phi,\phi)+N(\psi,\psi,\psi)+ N(\phi,\phi,\psi)+N(\psi,\psi,\phi)\\ &\qquad-N(\psi,\phi,\phi)-N(\phi,\psi,\psi)-N(\phi,\psi,\phi)-N(\psi,\phi,\psi) \\ &\quad= 2M\int_{0}^{T}e^{-\eta t}[|\phi_{t}(1,t)|+|\psi_{t}(1,t)|]dt\\ &\qquad-2M\int_{0}^{T}e^{-\eta t}[{\rm{sign}}(\phi_{t}(1,t))\psi_{t}(1,t)+{\rm{sign}}(\psi_{t}(1,t))\phi_{t}(1,t)]dt\\ &\quad=2M\int_{0}^{T}e^{-\eta t}[1-{\rm{sign}}(\psi_{t}(1,t)){\rm{sign}}(\phi_{t}(1,t))]|\phi_{t}(1,t)|dt\\ &\qquad+2M\int_{0}^{T}e^{-\eta t}[1-{\rm{sign}}(\phi_{t}(1,t)){\rm{sign}}(\psi_{t}(1,t))]|\psi_{t}(1,t)|dt\\ &\quad\geq 0. \end{align*} In order to prove that $$N(v_{1},v_{2},\phi)$$ is full range, we suppose $$\lambda >0$$, for every $$v_{1},v_{2}\in{\cal F}$$, and any $$\eta\in{\it{\Phi}}$$, $${\rm sign}(v_{1t}(1,t)\pm \lambda \eta_{t}(1,t))=\def\arraystretch{1.3}\left\{\begin{array}{@{}c} 1, v_{1t}(1,t)\pm \lambda \eta_{t}(1,t)>0,\\ \beta(t)\in[-1,1],v_{1t}(1,t)\pm \lambda \eta_{t}(1,t)=0,\\ -1,v_{1t}(1,t)\pm \lambda \eta_{t}(1,t)<0. \end{array}\right.$$ (4.21) Thus \begin{eqnarray*} N(v_{1}+\lambda \eta,v_{2},\eta)&=&\int_{E(v_{1t}+\lambda \eta_{t}\neq 0)}e^{-\eta t} sgn(v_{1t}(1,t)\pm \lambda \eta_{t}(1,t))\eta_{t}(1,t)dt\\ && +\int_{E(v_{1t}+\lambda \eta_{t}=0)}e^{-\eta t}\beta(t)\eta_{t}(1,t)dt;\\ N(v_{1}-\lambda \eta,v_{2},\eta)&=&\int_{E(v_{1t}-\lambda \eta_{t}\neq 0)}e^{-\eta t} sgn(v_{1t}(1,t)\pm \lambda \eta_{t}(1,t))\eta_{t}(1,t)dt\\ &&-\int_{E(v_{1t}-\lambda \eta_{t}=0)}e^{-\eta t}\beta(t)\eta_{t}(1,t)dt\\ N(v_{1},v_{2}+\lambda \eta,\eta)&=&\int_{E(v_{2t}+\lambda \eta_{t}\neq 0)}e^{-\eta t} sgn(v_{2t}(1,t)\pm \lambda \eta_{t}(1,t))\eta_{t}(1,t)dt\\ &&+\int_{E(v_{2t}+\lambda \eta_{t}=0)}e^{-\eta t}\beta(t)\eta_{t}(1,t)dt;\\ N(v_{1},v_{2}-\lambda \eta,\eta)&=&\int_{E(v_{2t}-\lambda \eta_{t}\neq 0)}e^{-\eta t} sgn(v_{2t}(1,t)\pm \lambda \eta_{t}(1,t))\eta_{t}(1,t)dt\\ &&-\int_{E(v_{2t}-\lambda \eta_{t}=0)}e^{-\eta t}\beta(t)\eta_{t}(1,t)dt \end{eqnarray*} where $$E(v_{it}+\lambda \eta_{t}=0)=\{t\in[0,T]|v_{it}(1,t),i=1,2\pm \lambda \eta_{t}(1,t)=0\}$$, when $$\lambda\longrightarrow0$$, we have \begin{gather*} \lim_{\lambda\longrightarrow0}N(v_{1}+\lambda \eta,v_{2},\eta)=\int_{E(v_{1}\neq0)}e^{-\eta t}sgn(v_{1t}(1,t))\eta_{t}(1,t)dt+\int_{E(v_{1}=0)}e^{-\eta t}|\eta_{t}(1,t)|,\\ \lim_{\lambda\longrightarrow0}N(v_{1},v_{2}+\lambda \eta,\eta)=\int_{E(v_{2}\neq0)}e^{-\eta t}sgn(v_{1t}(1,t))\eta_{t}(1,t)dt+\int_{E(v_{2}=0)}e^{-\eta t}|\eta_{t}(1,t)| \end{gather*} and \begin{gather*} \lim_{\lambda\longrightarrow0}N(v_{1}-\lambda \eta,v_{2},\eta)=\int_{E(v_{1}\neq0)}e^{-\eta t}sgn(v_{t}(1,t))\eta_{t}(1,t)dt+\int_{E(v_{1}=0)}e^{-\eta t}|\eta_{t}(1,t)|,\\ \lim_{\lambda\longrightarrow0}N(v_{1},v_{2}-\lambda \eta,\eta)=\int_{E(v_{2}\neq0)}e^{-\eta t}sgn(v_{t}(1,t))\eta_{t}(1,t)dt+\int_{E(v_{2}=0)}e^{-\eta t}|\eta_{t}(1,t)| \end{gather*} Since \begin{eqnarray*} N(v_{1},v_{2},\eta) &=& \int_{0}^{T}e^{-\eta t} sign(v_{t}(1,t))\eta_{t}(1,t)dt\\ &=& \int_{E(v\neq 0)}e^{-\eta t} sgn(v_{t}(1,t))\eta_{t}(1,t)dt+\int_{E(v=0)}e^{-\eta t} \beta(t)\eta_{t}(1,t)dt \end{eqnarray*} for all $$\beta(t)\in L^{1}(E(v=0))$$, where $$|\beta(t)|\leq 1$$, so we have $$N(v_{1},v_{2},\eta) =[\lim_{\lambda\longrightarrow0}N(v_{1}-\lambda \eta,\eta),\lim_{\lambda\longrightarrow0}N(v_{1}+\lambda \eta,\eta),\lim_{\lambda\longrightarrow0}N(v_{2}-\lambda \eta,\eta),\lim_{\lambda\longrightarrow0}N(v_{2}+\lambda \eta,\eta)],$$ $$\forall \eta\in{\it{\Phi}}$$ Specially, if $$v_{1t}(1,t)=v_{2t}(1,t)\equiv 0=\theta(t)$$, $$N(\theta,\theta,\phi)=\int_{0}^{T}e^{-\eta t} {\rm{sign}}(\theta(t))\eta_{t}(1,t)dt=\int_{0}^{T}e^{-\eta t}\beta(t)\eta_{t}(1,t)dt$$ the result is proved. □ Proposition 4.3 Let $$L(\phi)$$ be defined by (4.17), then $$L(\phi)$$ is a bounded linear functional in space $${\it{\Phi}}$$. Proof. It is obviously to see that $$L({\it{\Phi}})$$ a linear functional in space $${\it{\Phi}}$$. For any $$\phi\in{\it{\Phi}}$$, \begin{eqnarray*} |L(\phi)|^{2} &\leq&\bigg(\int_{0}^{1}|w_{1}^{1}(x)\phi_{t}(x,0)|dx+\int_{0}^{1}|w_{2}^{1}(x)\phi_{t}(x,0)|dx+\int_{0}^{T}e^{-\eta t}|\phi_{t}(1,t)\tilde{r}(t)|dt\nonumber\\ && +\eta\int_{0}^{1}\int_{0}^{T}e^{-\eta t}|\phi_{x}(x,t)w_{1x}^{0}(x)|dtdx+\eta\int_{0}^{1}\int_{0}^{T}e^{-\eta t}|\phi_{x}(x,t)w_{2x}^{0}(x)|dtdx\bigg)^{2}\nonumber\\ &\leq& \bigg(\int_{0}^{T}e^{-\eta t}|\tilde{r}(t)|^{2}dt+\eta \int_{0}^{1}\int_{0}^{T}e^{-\eta t}(|w_{1x}^{0}(x)|^{2}+|w_{2x}^{0}(x)|^{2})dxdt\nonumber\\ &&+\int_{0}^{1}(|w_{1}^{1}(x)|^{2}dx+|w_{2}^{1}(x)|^{2})dx\nonumber\\ &&\times\bigg(\int_{0}^{T}e^{-\eta t}|\phi_{t}(1,t)|^{2}dt+\eta \int_{0}^{1}\int_{0}^{T}e^{-\eta t}|\phi_{x}(x,t)|^{2}dxdt+\int_{0}^{1}|\phi_{t}(x,0)|^{2}dx\bigg) \end{eqnarray*} So $$|L(\phi)|\leq C_{5}\|\phi\|_{{\it{\Phi}}},$$ where $$C_{5}=C_{4}\bigg(\int_{0}^{T}e^{-\eta t}|\tilde{r}(t)|^{2}dt+\eta \int_{0}^{1}\int_{0}^{T}e^{-\eta t}(|w_{1x}^{0}(x)|^{2}+|w_{2x}^{0}(x)|^{2})dxdt+\int_{0}^{1}(|w_{1}^{1}(x)|^{2}dx+|w_{2}^{1}(x)|^{2})dx\bigg)^{1/2}$$ and $$C_{4}=\max\{1,\eta\}$$. Therefore $$L(\phi)$$ is a bounded linear functional in space $${\it{\Phi}}$$. □ Based on the above propositions and Lemma (2.1), we have the following theorem. Theorem 4.1 There exists a weak solution to the system (4.13) for any $$(w^{0},w^{1})$$ and $$r(t)$$. 4.2. The uniqueness of the solution We will consider the uniqueness of the solution of system (3.12). Suppose that there are two solutions $$z(x,t),v(x,t)$$. Consider the error function $$e(x,t)=v(x,t)-z(x,t)$$, obviously the error function satisfies the following equation $$\label{error} \def\arraystretch{1.2}\left\{\begin{array}{@{}ll} e_{tt}(x,t)=e_{xx}(x,t), & 0<x<1, t >0, \\ e_1(1,t)=e_2(1, t),& t\ge 0, \\ e_{1x}(1,t)+e_{2x}(1, t)=-k e_{1t}(1,t)-M[{\rm{sign}} (v_{1t}(1,t))-{\rm{sign}} (z_{1t}(1,t))],& t\ge 0, \\ e_{1}(0,t)=0,& t\geq 0,\\ e_{2x}(0,t)=0,& t\geq 0,\\ e(x,0)=e^{0}(x),e_{t}(x,0)=e^{1}(x). \end{array}\right.$$ (4.22) Consider the energy function \begin{eqnarray*} \varepsilon(t) &=& \frac{1}{2}\int_{0}^{1} |e_{1x}(x,t)|^{2}+|e_{2x}(x,t)|^{2}+|e_{1t}(x,t)|^{2}+|e_{2t}(x,t)|^{2}dx \end{eqnarray*} Differentiate the energy function with respect to $$t$$, we have \begin{eqnarray*} \dot{\varepsilon}(t) &=& \int_{0}^{1} e_{1t}(x,t)e_{1tt}(x,t)+ e_{2t}(x,t)e_{2tt}(x,t)+e_{1x}(x,t)e_{xt}(x,t)dx+e_{2x}(x,t)e_{2xt}(x,t)dx\\ &=& e_{1t}(x,t)e_{1x}(x,t)|_{0}^{1}+e_{2t}(x,t)e_{2x}(x,t)|_{0}^{1}\\ &=& e_{1t}(1,t)e_{1x}(1,t)+e_{2t}(1,t)e_{2x}(1,t)\\ &=& e_{1t}(1,t)(-k e_{1t}(1,t)-M[{\rm {\rm{sign}}} (v_{1t}(1,t))-{\rm sign} (z_{1t}(1,t))])\\ &=& -ke_{1t}^{2}(1,t)-M[{\rm{sign}} (v_{1t}(1,t))-{\rm{sign}} (z_{1t}(1,t))]e_{1t}(1,t)\\ &=& -ke_{t}^{2}(1,t)-M[{\rm{sign}} (v_{1t}(1,t))-{\rm{sign}} (z_{1t}(1,t))][v_{1t}(1,t)-z_{1t}(1,t))]\\ &\leq& -ke_{1t}^{2}(1,t) \end{eqnarray*} So the energy function satisfies that \begin{eqnarray*} \varepsilon(t) +k\int_{0}^{t}e_{t}^{2}(1,t)dt&\leq& \varepsilon(0)=0 \end{eqnarray*} which shows that $$e(x,t)=0$$. We then immediately have the following theorem. Theorem 4.2 For any $$(w^{0},w^{1})$$ and disturbance $$r(t)$$, the closed-loop system (3.12) has a unique solution. 5. The asymptotical stability of the closed-loop system In this part, we are going to elaborate the asymptotically stability of the closed-loop system. Note that (3.12) is a quasi-autonomous system. The asymptotic behaviour of the solution depends strongly on the disturbance. In the case of autonomous systems, the LaSalles invariant theorem shows that the trajectory of the system approaches a large invariant set $$\omega$$ on which the $$\dot{V} (t) = 0$$. In the case of non-autonomous systems, it may not be clear how to define such a set $$\omega$$. The difficulty is that the positive limit set is not invariant and the zero is not equilibrium. Here we shall find a way to discuss the asymptotic behaviour of the system. We rewrite the closed-loop system as following: $$\label{reclose} \def\arraystretch{1.2}\left\{\begin{array}{@{}ll} w_{tt}(x,t)=w_{xx}(x,t), & 0<x<1, t >0, \\ w_1(1,t)=w_2(1, t),& t\ge 0, \\ w_{1x}(1,t)+w_{2x}(1, t)=-kw_{1t}(1,t)-M {\rm {\rm{sign}}}(w_{1t}(1,t))+\tilde{r}(t),& t\ge 0, \\ w_{1}(0,t)=0,& t\geq 0,\\ w_{2x}(0,t)=0,& t\geq 0,\\ w(x,0)=w^{0}(x),w_{t}(x,0)=w^{1}(x), \end{array}\right.$$ (5.23) where $$\|(w^{0},w^{1})\|\leq R$$, and $$\tilde{r}(t)=r(t)+3R$$. The energy function in defined by \begin{eqnarray*} E(w)(t) &=& \frac{1}{2}\int_{0}^{1}(w_{1x}^{2}(x,t)+w_{2x}^{2}(x,t)+w_{1t}^{2}(x,t)+w_{2t}^{2}(x,t))dx \end{eqnarray*} In the previous part, we have proved that \begin{eqnarray*} \frac{dE(w)(t)}{dt} &=& -kw_{1t}^{2}(1,t)-(M {\rm{sign}}(w_{1t}(1,t)-\tilde{r}(t))w_{1t}(1,t)\leq 0. \end{eqnarray*} For a fixed $$r(t)$$, we assume that the initial data is in the domain $$S(R)=\{(w^{0}(x),w^{1}(x))\in{\mathcal H}|\|(w^{0},w^{1})\|_{{\mathcal H}}\leq R\}.$$ Since $$((w^{0},w^{1}),\tilde{r}(t))$$ is not an observer blind spot of system (3.12), $$E(w(t))$$ is monotonically decreasing. $$S(R)$$ is an invariant set. In light of the energy function, for any $$t,\tau>0$$, $$E(w)(t+\tau)+\int_{0}^{\tau}kw_{1t}^{2}(1,t)+(M {\rm{sign}}(w_{1t}(1,t)-\tilde{r}(t))w_{1t}(1,t)ds=E(w)(t),$$ so $$\lim_{t\longrightarrow\infty}\int_{0}^{\tau}kw_{1t}^{2}(1,t)+(M {\rm{sign}}(w_{1t}(1,t)-\tilde{r}(t))w_{1t}(1,t)ds=0$$ which shows that $$\lim_{t\longrightarrow\infty}w_{1t}(1,t)=0$$. Suppose $$t_{n}>0$$ and $$\lim\limits_{t\longrightarrow +\infty}t_{n}=+\infty$$. For any given $$\tau$$, consider the function sequence $$z_{n}(x,t)=w(x,t+t_{n}),n\in N, \;\; t\in[0,\tau]$$ where $$w(x,t)$$ is the solution of the system (5.23). We assume $$r_{n}(t)=\tilde{r}(t+t_{n})$$ $$\label{reclosez} \def\arraystretch{1.2}\left\{\begin{array}{@{}ll} z_{ntt}(x,t)=z_{nxx}(x,t), & 0<x<1, t >0, \\ z_{1n}(1,t)=z_{2n}(1, t),& t\ge 0, \\ z_{1nx}(1,t)+z_{2nx}(1, t)=-kz_{n}(1,t)-M {\rm sign}(z_{1nt}(1,t))+\tilde{r}_{n}(t),& t\ge 0, \\ z_{1n}(0,t)=0,& t\geq 0,\\ z_{2nx}(0,t)=0,& t\geq 0,\\ z_{n}(x,0)=w(x, t_{n}),z_{nt}(x,0)=w_{t}(x, t_{n}),\\ y_{n}=z_{1n}(1,t),& t\geq 0. \end{array}\right.$$ (5.24) Noticing that $$E(z_{n})(t)=\frac{1}{2}\int_{0}^{1}|z_{1n,x}(x,t)|^{2}+|z_{2n,x}(x,t)|^{2}+|z_{1n,t}(x,t)|^{2}+|z_{2n,t}(x,t)|^{2}dx$$ and $$E(z_{n})(t)+\int_{0}^{t}kz_{1n,t}^{2}(1,t)+(M {\rm{sign}}(z_{1n,t}(1,t)-\tilde{r}_{n}(t))z_{1n,t}(1,t)ds=E(w)(t_{n}).$$ Hence, $$\{(z_{n}(x,t),z_{n,t}(x,t)),n\in N\}$$ is a bounded sequence in Hilbert space $${\mathcal H}$$, and it has a weak convergent subsequence. Without loss of generality, we assume $$(z_{n}(x,t),z_{n,t}(x,t))\longrightarrow (z(x,t),\xi(x,t))\in {\mathcal H}$$. Because $$\{(w(x,t_{n}),w_{t}(x,t_{n})),n\in N\}$$ is also a bounded sequence in Hilbert space $${\mathcal H}$$ and it has a subsequence. Without loss of generality, we assume the subsequence to be itself, and it convergents to $$(w(x),\eta(x))$$. Moreover, $$r(t)$$ is uniformly bounded, so it has a weak limitation in space $$L^{2}(0,1)$$. Thus, $$z_{t}(x,t)=\xi(x,t)$$, so $$z(x,t)$$ satisfies \begin{eqnarray} &&\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{xt}(x,t)v_{1x}(x,t)dxdt-\int_{0}^{T}\int_{0}^{1}v_{1t}(x,t)[e^{-\eta t}\phi_{t}(x,t)]_{t}dxdt\nonumber\\ &&-\int_{0}^{T}\int_{0}^{1}v_{2t}(x,t)[e^{-\eta t}\phi_{t}(x,t)]_{t}dxdt+\int_{0}^{T}\int_{0}^{1}e^{-\eta t}\phi_{xt}(x,t)v_{2x}(x,t)dxdt\nonumber\\ &=&\int_{0}^{1}\eta_{1}(x)\phi_{t}(x,0)dx+\int_{0}^{1}\eta_{2}(x)\phi_{t}(x,0)dx\int_{0}^{1}d(t)\phi_{t}(1,t)dx\nonumber\\ && -\eta\int_{0}^{1}\int_{0}^{T}[e^{-\eta t}\phi_{x}(x,t)]w_{1x}^{0}(x)dtdx-\eta\int_{0}^{1}\int_{0}^{T}[e^{-\eta t}\phi_{x}(x,t)]w_{2x}^{0}(x)dtdx \end{eqnarray} (5.25) where $$\phi(x,t)$$ is detection function. So $$z(x,t)$$ satisfies the following equation in the sense of weak solution $$\label{closez} \def\arraystretch{1.2}\left\{\begin{array}{@{}ll} z_{tt}(x,t)=z_{xx}(x,t), & 0<x<1, t >0, \\ z_1(1,t)=z_2(1, t),& t\ge 0, \\ z_{1x}(1,t)+z_{2x}(1, t)=d(t),& t\ge 0, \\ z_{1}(0,t)=0,& t\geq 0,\\ z_{2x}(0,t)=0,& t\geq 0,\\ z(x,0)=w(x),z_{t}(x,0)=\eta(x). \end{array}\right.$$ (5.26) If $$d(t)=0$$, then (5.26) has a solution if and only if $$(w(x),\eta(x))=0$$. In this case, the system (5.26) is asymptotically stable; If $$d(t)\neq0$$, then equation (5.26) has a non-zero solution if and only if $$d(t)$$ is an observation blind point. Suppose $$\lim_{t\longrightarrow\infty}E(w)(t)=c(w^{0},w^{1})$$ and $$c={\rm sup}_{(w^{0},w^{1})\in S(R)} c(w^{0},w^{1})$$. We have $$\|(z(x,t),\xi(x,t))\|_{{\mathcal H}}^{2}\leq \lim_{n\longrightarrow\infty} \|(z_{n}(x,t),z_{n,t}(x,t))\|_{{\mathcal H}}^{2}.$$ If $$\|(z(x,t),\xi(x,t))\|_{{\mathcal H}}^{2}=\lim_{n\longrightarrow\infty} \|(z_{n}(x,t),z_{n,t}(x,t))\|_{{\mathcal H}}^{2}=2c(w^{0},w^{1})\neq 0$$, we have $$\|(z_{n}(x,t),z_{n,t}(x,t))-(z(x,t),\xi(x,t))\|_{{\mathcal H}}^{2}\longrightarrow0,n\longrightarrow\infty.$$ Since $$\|(w,\eta)\|_{H}\leq R,|\tilde{r}(t)-\hat{r}(t)|>R,$$ we have $$(w(x),\eta(x),d(t))$$ is not a observer blind spot. So $$(z(x,t),\xi(x,t))\equiv 0, \lim_{t\longrightarrow\infty}E(w_{0},w_{1})(t)=0$$, so the solution of the system converges to zero. In summary, we have the following main result. Theorem 5.1 Suppose $$r(t)$$satisfies that $$sup_{t}|r(t)|<\infty, R\in {R}_{+},R\geq {\rm sup}_{t}|r(t)|$$ and $$M\geq 4R$$, so for any $$(w^{0},w^{1})\in {\mathcal H}$$, and $$\|(w^{0},w^{1})\|_{{\mathcal H}}\leq R$$, thus the solution of the system converges to zero. 6. Concluding remarks In this article, we study the problem of a joint wave equation with point-wise input disturbance. A novel feedback control strategy $$u(t) = -kw_{1t}(1,t)-M{\rm{sign}}(w_{1t}(1,t)) + c$$ where c is a suitable constant that depends on estimation of the initial value is attempted to stabilize the system. we obtain the existence and unique of the closed-loop system based on expansion of the Lions–Lax–Milgram Theorem and the developed Variational Method, we used the Lasalle’s Invariance Principle-like together with observation blind point to study asymptotical stability of the closed loop system. Under the improved controller, we proved that the solution of the system converges to zero in state space $${\mathcal H}$$ for any initial value in a bounded set. Funding National Natural Science Foundation of China (NSFC-61503276, 61174080, 61503275). References Ammari K. , Tucsnak M. & Henrot A. 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IMA Journal of Mathematical Control and InformationOxford University Press

Published: Sep 1, 2017

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