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\yihao\color{blue}\centering \li Main Result \\
\centering\xiaosihao\textcolor[rgb]{0.00,0.00,0.00}{The objective is
to design constant gain $C\in R^{m \times (n-m)}$ and a reaching
motion control law u(t),such that}\hspace*{\fill}\\
\vspace{-10mm}
\ding{192}\centering\xiaosihao\textcolor[rgb]{0.00,0.00,0.00}{The
sliding motion is quadratically stable;and}\hspace*{\fill}
\vspace{-10mm}
\ding{193}\centering\xiaosihao\textcolor[rgb]{0.00,0.00,0.00}{The
trajectory of the closed-loop system is convergent into a residual
set of the origin with the control law u(t)}\hspace*{\fill}
\vspace{-10mm}
\begin{itemize}
 \item [\begin{footnotesize}\HandPencilLeft \end{footnotesize}]\raisebox{-1ex}{\xiaoerhao\textcolor[rgb]{0.00,0.00,1.00}{~~Theorem 1}}\\[-10mm]
\xiaosihao \color{black} \centering The reduced order system is
quadratically stable if there exist symmetric positive-definite
matrix \\[-10mm]
$P\in R^{(n-m)\times (n-m)}$,symmetric semi-positive matrix $Z \in R^{(n-m)\times
(n-m)}$,general matrix \\[-10mm]
$V \in R^{m \times (n-m)}$ $Y,M, N,Q \in
R^{(n-m)\times (n-m)}$, $X \in S_D$, and $X_d\in
S_{D_d}$ \\[-13mm]
\sihao such that $\label{theo1.1} \quad \quad \Theta<0  $\\[-10mm]
and$\label{theo1.2} \quad \quad \quad\Omega\ge 0$
\end{itemize}
\end{slide}
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\begin{itemize}
\item[\begin{footnotesize}\HandRight  \end{footnotesize}]\li\sihao\textcolor[rgb]{0.00,0.00,1.00}{Entries of the LMIS}
\end{itemize}
\bea
\Theta_{11} & = & \ol A_{11}\textcolor[rgb]{0.00,0.00,1.00}{P}-\ol A_{12}\textcolor[rgb]{0.00,0.00,1.00}{V}+\textcolor[rgb]{0.00,0.00,1.00}{P}\ol A^T_{11}-\textcolor[rgb]{0.00,0.00,1.00}{V}^T\ol A^T_{12}+\textcolor[rgb]{0.00,0.00,1.00}{\tilde M}+\textcolor[rgb]{0.00,0.00,1.00}{\tilde M^T}+\textcolor[rgb]{0.00,0.00,1.00}{\tilde Q}\nm\\
&~&+\ol \tau \textcolor[rgb]{0.00,0.00,1.00}{\tilde X}+U^T_2G\textcolor[rgb]{0.00,0.00,1.00}{X}G^TU_2+U^T_2G_d\textcolor[rgb]{0.00,0.00,1.00}{X_d}G^T_dU_2 \nonumber \\
\Theta_{12} & = &\Theta^T_{21}=\ol A_{d11}\textcolor[rgb]{0.00,0.00,1.00}{\tilde P}-\ol A^T_{d12}\textcolor[rgb]{0.00,0.00,1.00}{V}-\textcolor[rgb]{0.00,0.00,1.00}{\tilde M}+\textcolor[rgb]{0.00,0.00,1.00}{\tilde N^T}+\ol \tau \textcolor[rgb]{0.00,0.00,1.00}{\tilde Y} \nonumber \\
\Theta_{13}&=&\Theta^T_{31}=\ol\tau\lambda (\textcolor[rgb]{0.00,0.00,1.00}{\tilde P}\ol A^T_{11}-\textcolor[rgb]{0.00,0.00,1.00}{V^T}\ol A^T_{12}+U^T_2G\textcolor[rgb]{0.00,0.00,1.00}{X}G^TU_2\nm\\
&~&+U^T_2G_d\textcolor[rgb]{0.00,0.00,1.00}{X_d}G^T_dU_2)\nm\\
\Theta_{14}&=&\Theta^T_{41}=\textcolor[rgb]{0.00,0.00,1.00}{P}U^T_2H^T-\textcolor[rgb]{0.00,0.00,1.00}{V^T}U^T_1H^T\nm\\
 \Theta_{22} & = &-\textcolor[rgb]{0.00,0.00,1.00}{\tilde N}-\textcolor[rgb]{0.00,0.00,1.00}{\tilde N^T}-(1-d)\textcolor[rgb]{0.00,0.00,1.00}{\tilde
Q}+\ol\tau
\textcolor[rgb]{0.00,0.00,1.00}{\tilde Z} \nonumber \\
\Theta_{23}&=&\Theta^T_{32}=\ol\tau\lambda (\textcolor[rgb]{0.00,0.00,1.00}{\tilde P}\ol A^T_{d11}-\textcolor[rgb]{0.00,0.00,1.00}{V^T}\ol A^T_{d12})\nm\\
\Theta_{25}&=&\Theta^T_{52}=\textcolor[rgb]{0.00,0.00,1.00}{P}U^T_2H^T_d-\textcolor[rgb]{0.00,0.00,1.00}{V^T}U^T_1H^T_d\nm\\
\Theta_{33}&=&-\ol\tau\lambda \textcolor[rgb]{0.00,0.00,1.00}{\tilde
P}+\ol\tau^2\lambda^2(U^T_2G\textcolor[rgb]{0.00,0.00,1.00}{X}G^TU_2+U^T_2G_d\textcolor[rgb]{0.00,0.00,1.00}{X_d}G^T_dU_2)\nm\\
\Theta_{44} & = & -\textcolor[rgb]{0.00,0.00,1.00}{X} \nonumber \\
\Theta_{55} & = & -\textcolor[rgb]{0.00,0.00,1.00}{X_d} \nonumber \\
\Omega&=&\lt[\ba{ccc}X&Y&M\\Y^T&Z&N\\M^T&N^T&\lambda P\ea\rt]\nm\eea

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\begin{slide}
\begin{itemize}
 \item[\begin{footnotesize}\HandRight\end{footnotesize}]\li\erhao\textcolor[rgb]{0.00,0.00,1.00}{Proof of Theorem 1}
    \vspace{-5mm}
    \item[\begin{footnotesize}\ding {226}\end{footnotesize}]\li\sihao\textcolor[rgb]{0.66 ,0.14,0.44}{Shur complementarity}\\%[-5mm]
    if $P^{-1}$ exist,$-Q+MR^{-1}M^T+A^TP^{-1}A<0$ $\Leftrightarrow$ $\lt[\ba{cc}-Q+MR^{-1}M^T&~A^T\\A&-P\ea\rt] < 0\nm$
    $\lt[\ba{ccc}-Q~&A^T~&M\\A&-P&0\\M^T&0&-R\ea\rt] < 0\nm$
    $\Leftrightarrow$ $\lt[\ba{cc}-Q+MR^{-1}M^T&~A^T\\A&-P\ea\rt] < 0\nm$
    \item[\begin{footnotesize}\ding {226}\end{footnotesize}]\li\sihao\textcolor[rgb]{0.66 ,0.14,0.44}{chose the Lyapunov functional candidate}
    \vspace{-5mm}\xiaosihao
    \be\ba{rl}
V(t)&=x^T(t)Px(t)+\int^t_{t-\tau(t)}x^T(s)Qx(s)ds\\[-5mm]
&~~+\lambda\int^0_{-\ol\tau}\int^{t}_{t+\theta}\dot x^T(s)P\dot
x(s)dsd\theta\ea\ee
\end{itemize}
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\begin{itemize}
    \item[\begin{footnotesize}\ding {226}\end{footnotesize}]\li\sihao\textcolor[rgb]{0.66
    ,0.14,0.44}{Free term: 1}\\[-5mm]
    \xiaosihao\color{black}
    It is easy to see that\\
    \vspace{-15mm}\be x(t-\tau(t))=x(t)-\int^t_{t-\tau(t)}\dot x(s)ds\ee then, for any matrices $M$ and $N$ with appropriate dimensions, the following equation is derived:\\
    \vspace{-12mm}\be\label{eq.hewu}2[x^T(t)M+x^T(t-\tau(t))N)][x(t)-x(t-\tau(t))-\int^t_{t-\tau(t)}\dot x(s)ds]=0.\ee\\
    it also can be in the following form \textcolor[rgb]{1.00,0.00,0.00}{\hspace*{\fill}\wuhao \bea\lt[\ba{c}x(t)\\x(t-\tau(t))\ea\rt]^T\lt[\ba{c}M\\N\ea\rt][x(t)-x(t-
    \tau(t))-\int^t_{t-\tau(t)}\dot
    x(s)ds]=0\eea\hspace*{\fill}}\\
    \end{itemize}
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\begin{slide}
\begin{itemize}
    \item[\begin{footnotesize}\ding {226}\end{footnotesize}]\li\sihao\textcolor[rgb]{0.66
    ,0.14,0.44}{Free Term: 2}\\[-5mm]
    \xiaosihao\color{black}
    For any semi-positive definite matrices $X$, $Z$ and
general matrix  $Y$ such that  $W=\lt[\ba{cc}X&Y\\Y^T&Z\ea\rt]\geq
0$, the following result is obvious \be\label{eq.free2}
\lt[\ba{c}x(t)\\x(t-\tau(t))\ea\rt]^TW\lt[\ba{c}x(t)\\x(t-\tau(t))\ea\rt]\nm\geq
0\ee then, \bea&&\ol\tau
\lt[\ba{c}x(t)\\x(t-\tau(t))\ea\rt]^TW\lt[\ba{c}x(t)\\x(t-\tau(t))\ea\rt]\nm-\int^t_{t-\tau(t)}\lt[\ba{c}x(t)\\x(t-\tau(t))\ea\rt]^TW\lt[\ba{c}x(t)\\x(t-\tau(t))\ea\rt]ds\nm\\
&&=(\ol\tau-\tau(t))\lt[\ba{c}x(t)\\x(t-\tau(t))\ea\rt]^TW\lt[\ba{c}x(t)\\x(t-\tau(t))\ea\rt]\geq
0 \eea
    \end{itemize}
\end{slide}
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%                                        The Lyapunov derivative $\dot{V}$(t)                         %
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\begin{slide}
\begin{itemize}
\vspace{-10mm}\item[\begin{footnotesize}\ding{226}\end{footnotesize}]\li\sihao\textcolor[rgb]{0.66,0.14,0.44}{The Lyapunov derivative $\dot{V}$(t)}\\
\vspace{-5mm}\wuhao\be\small\label{eq.lya}\ba{rl}\dot
V(t)=&x^T(t)[P\tilde A+\tilde
A^TP]x(t)+\textcolor[rgb]{0.00,0.00,1.00}{2x^T(t)P\tilde A_dx(t-\tau(t))}\\
&+x^T(t)Qx(t)-(1-\textcolor[rgb]{1.00,0.00,0.00}{\dot\tau(t)})x^T(t-\tau(t))Qx(t-\tau(t))\\
&+\lambda\ol\tau\dot x^T(t)P\dot x(t)-\lambda\int^t_{t-\ol\tau}\dot
x^T(s)P\dot
x(s)ds\\
\le& x^T(t)[P\tilde A+\tilde A^TP]x(t)+2x^T(t)P\tilde
A_dx(t-\tau(t))\\
&+x^T(t)Qx(t)\\
   &-(1-\textcolor[rgb]{1.00,0.00,0.00}{d}\footnote[1]{\wuhao\textcolor[rgb]{1.00,0.00,0.00}{Assumption 3}})x^T(t-\tau(t))Qx(t-\tau(t))+\lambda\ol\tau\dot x(t)P\dot x(t)\\
   &-\lambda\int^t_{t-\tau(t)}\dot x^T(s)P\dot
x(s)ds \ea\ee
\vspace{-10mm}\item[\begin{footnotesize}\HandPencilLeft\end{footnotesize}]\xiaosihao
\quad Let\textcolor[rgb]{0.00,0.00,1.00}{ \bea\alpha(t)
=\lt[\ba{c}x(t)\\x(t-\tau(t))\ea\rt], \beta(t,s
)=\lt[\ba{c}x(t)\\x(t-\tau(t))\\\dot x(s)\ea\rt].\eea}\\
\vspace{10mm}Adding nonnegative terms into right side of the
Lyapunov the derivative inquality,it results in
\end{itemize}
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\begin{itemize}
\vspace{-13mm}\item[\begin{footnotesize}\ding{226}\end{footnotesize}]\li\sihao\textcolor[rgb]{0.66,0.14,0.44}{Adding
nonnegative terms (\ref{eq.hewu}) and (\ref{eq.free2})into the Lyapunov derivative}\\
\vspace{-10mm}\be\small\ba{rl} \dot V(t)&\le
x^T(t)\textcolor[rgb]{1.00,0.40,0.00}{[P\tilde A+\tilde
A^TP]}x(t)+2x^T(t)\textcolor[rgb]{0.67,0.00,1.00}{P\tilde
A_d}x(t-\tau(t))+\\
&x^T(t)\textcolor[rgb]{1.00,0.40,0.00}{Q}x(t)-(1-d)x^T(t-\tau(t))\textcolor[rgb]{0.0,0.50,0.00}{Q}x(t-\tau(t))+\\
&\lambda\ol\tau\dot x(t)P\dot x(t)-\lambda\int^t_{t-\tau(t)}\dot
x^T(s)P\dot
x(s)ds+2[x^T(t)\textcolor[rgb]{1.00,0.40,0.00}{M}\\
 &+x^T(t-\tau(t))\textcolor[rgb]{0.67,0.00,1.00}{N})][x(t)-x(t-\tau(t))-\int^t_{t-\tau(t)}\dot
x(s)ds]\\
&+\ol\tau\alpha^T(t)W\alpha(t)-\int^t_{t-\tau(t)}\alpha^T(t)W\alpha(t)ds\\
=&\alpha^T(t)\Phi\alpha(t)-\int^t_{t-\tau(t)}\beta^T(t,s)\Omega\beta(t,s)ds\\
&\wuhao\textcolor[rgb]{1.00,0.00,0.00}{\kai it~~can~~be~~rewritten:}\\
=&\xiaowuhao\textcolor[rgb]{0.00,0.00,1.00}{\lt[\ba{c}x(t)\\x(t-\tau(t))\ea\rt]^T\Phi\lt[\ba{c}x(t)\\x(t-\tau(t))\ea\rt]-\int^t_{t-\tau(t)}\lt[\ba{c}x(t)\\x(t-\tau(t))\\\dot
x(s)\ea\rt]^T\lt[\ba{ccc}X&Y&M\\Y^T&Z&N\\M^T&N^T&\lambda
P\ea\rt]\lt[\ba{c}x(t)\\x(t-\tau(t))\\\dot x(s)\ea\rt]ds}\ea\nm\ee
where\small \bea
&&\Phi=\lt[\ba{cc}\Phi_{11}&\Phi_{12}\\\Phi^T_{12}&\Phi_{22}\ea\rt]\nm\\
&&\Phi_{11}=\textcolor[rgb]{1.00,0.40,0.00}{P\tilde A+\tilde A^TP+\ol\tau\lambda\tilde A^TP\tilde A+M+M^T+Q+\ol\tau X\nm}\\
&&\Phi_{12}=\Phi_{21}^T=\textcolor[rgb]{0.67,0.00,1.00}{P\tilde A_d+\ol\tau \lambda A^TP\tilde A_d-M+N^T+\ol\tau Y}\\
&&\Phi_{22}=\textcolor[rgb]{0.0,0.50,0.00}{\ol\tau \lambda\tilde
A^T_dP\tilde A_d-N-N^T-(1-d)Q+\ol\tau Z\nm}\\
&&\label{eq.omega}\Omega=\lt[\ba{ccc}X&Y&M\\Y^T&Z&N\\M^T&N^T&\lambda P\ea\rt]\eea\\

\end{itemize}
\end{slide}

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if $\Phi<0$ and $\Omega\geq0$, $\dot V(t)<0$ for any $\alpha(t)\neq
0$. By Schur complement, $\Phi<0$ is equivalent to the following
inequality: \be\small\label{eq.omegap}
\lt[\ba{ccc}\Phi_{11}-\textcolor[rgb]{0.00,0.00,1.00}{\ol\tau\lambda\tilde
A^TP\tilde A}&\Phi_{12}-\textcolor[rgb]{0.00,0.00,1.00}{\ol\tau
\lambda A^TP\tilde
A_d}&\textcolor[rgb]{1.00,0.00,1.00}{\ol\tau\lambda\tilde
A^TP}\\(\Phi_{12}-\textcolor[rgb]{0.00,0.00,1.00}{\ol\tau \lambda
A^TP\tilde
A_d})^T&\Phi_{22}-\textcolor[rgb]{0.00,0.00,1.00}{\ol\tau \lambda\tilde A^T_dP\tilde A_d}&\textcolor[rgb]{1.00,0.00,1.00}{\ol\tau\lambda \tilde A^T_dP} \\
\ol\tau\lambda P\tilde A&\ol\tau\lambda P\tilde
A_d&\textcolor[rgb]{1.00,0.00,0.00}{-\ol\tau\lambda P}\ea\rt]<0\ee\\
\begin{itemize}
\vspace{-10mm}\item[\begin{footnotesize}\ding{226}\end{footnotesize}]\li\sihao\textcolor[rgb]{0.66,0.14,0.44}{Proof of inequality (\ref{eq.omegap})}\\
\vspace{5mm}\sihao For P is a positive matrix and $P = P^T,\Phi < 0 $ is equivalent to the following inequality:\\
\small\be \Phi-\bar{\tau}\lambda\lt[\ba{c}\tilde{A}^T\\
\tilde{A}_d^T\ea\rt] P \lt[\ba{cc}\tilde{A}\quad
\tilde{A}_d\ea\rt]+\ol\tau\lambda\lt[\ba{c}\tilde{A}^T\\
\tilde{A}_d^T\ea\rt]P\lt[\ba{cc}\tilde{A}\quad
\tilde{A}_d\ea\rt]<0\nm\ee
it is easy to rewrite in the following form  %%(24)
\be \lt[\ba{cc}\Phi_{11}-\ol\tau\lambda\tilde A^TP\tilde
A&\Phi_{12}-\ol\tau \lambda A^TP\tilde A_d\\(\Phi_{12}-\ol\tau
\lambda A^TP\tilde A_d)^T&\Phi_{22}-\ol\tau \lambda\tilde
A^T_dP\tilde A_d\ea\rt]+\lt[\ba{c}\ol\tau\lambda\tilde{A}^T P\\
\ol\tau\lambda\tilde{A}_d^T P\ea\rt] (\ol\tau\lambda
P)^{-1}\lt[\ba{cc}\ol\tau\lambda P\tilde{A}\quad \ol\tau\lambda
P\tilde{A}_d\ea\rt]<0\nm\ee \sihao and then, using
\textcolor[rgb]{0.00,0.00,1.00}{Schur Complement} ,yields the
inequality (\ref{eq.omegap})
\end{itemize}
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Multiplying both sides of inequalities (\ref{eq.omega}) and
(\ref{eq.omegap}) with $diag[P^{-1}~P^{-1}~P^{-1}]$, and let $\tilde
P=P^{-1}$, $\tilde M=P^{-1}MP^{-1}$, $\tilde
N=P^{-1}NP^{-1}$,$\tilde X=P^{-1}XP^{-1}$, $\tilde Q=P^{-1}QP^{-1}$,
$\tilde Y=P^{-1}YP^{-1}$, yields \be \lt[\ba{ccc}\tilde X&\tilde
Y&\tilde M\\\tilde Y^T&\tilde Z&\tilde N\\\tilde M^T&\tilde
N^T&\lambda \tilde P\ea\rt]\geq 0\ee\xiaosihao where $\tilde
\Phi_{11}=\tilde A\tilde P+\tilde P\tilde A^T+\tilde M+\tilde
M^T+\tilde Q+\ol\tau \tilde X $, $\tilde \Phi_{12}=\tilde A_d\tilde
P-\tilde M+\tilde N^T+\ol \tau \tilde Y$, $\tilde
\Phi_{13}=\ol\tau\lambda\tilde P\tilde A^T$, $\tilde
\Phi_{22}=-\tilde N-\tilde N^T-(1-d)\tilde Q+\ol\tau \tilde Z
$,$\tilde \Phi_{23}=\ol\tau\lambda \tilde P\tilde A^T_d$, $\tilde
\Phi_{33}=-\ol\tau\lambda \tilde P$. \\
\sihao\color{blue}and\color{blue} \be\label{eq.uncer}
\lt[\ba{ccc}\tilde\Phi_{11}&\tilde\Phi_{12}&\tilde\Phi_{13}\\
 \tilde\Phi^T_{12}&\tilde\Phi_{22}&\tilde\Phi^T_{23}\\
\tilde\Phi^T_{13}&\tilde\Phi^T_{23}&\tilde\Phi_{33}\ea\rt]<0\ee
\color{red}now we prove the inequality(\ref{eq.uncer})
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